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Here we are providing Class 12 Geography Important Extra Questions and Answers Chapter 3 Population Composition. Geography Class 12 Important Questions are the best resource for students which helps in class 12 board exams.

## Class 12 Geography Chapter 3 Important Extra Questions Population Composition

### Population Composition Important Extra Questions Very Short Answer Type

Question 1.
When is sex ratio unfavourable ?
When number of men is more than number of women.

Question 2.
What is the average world sex ratio ?
990 females per 1000 males.

Question 3.
In which country, is the highest sex ratio found ?
In Estonia—1190 females per 1000 males.

Question 4.
In which country, is the lowest sex ratio found ? (CJ3.S.E. 2011)
In Qatar—303 females per 1000 males.

Question 5.
In how many countries (according to U.N.O.), the sex ratio is favourable ?
139 countries.

Question 6.
In how many countries (according to U.N.O.), the sex ratio is unfavourable ?
Ans.
72 countries.

Question 7.
What is age structure ?
Or
Define the term ‘Age structure’. (Delhi 2019)
Age structure represents the number of people of different age groups.

Question 8.
Which diagram shows age-sex structure ?
A population pyramid.

Question 9.
Which country’s population shows a constant Pyramid ?
Australia.

Question 10.
Which country shows a declining pyramid ?
Japan.

### Population Composition Important Extra Questions Short Answer Type

Question 1.
What is the main characteristic of people of a country ? What are the distinguishing attributes of people ?
People of India are diverse and unique in many respects. People can be distinguished by their age, sex and their place of residence. Other distinguishing attributes of the population are :

• occupation
• education
• life expectancy.

Question 2.
What is sex ratio ? How is it calculated in India ? (C.B.S.E. 2013)
The ratio between the number of women and men in the population is called the sex ratio. It is number of females per 1000 males. In India the sex ratio is worked out using the formula.
sex Ratio = $$\frac{\text { Female Population }}{\text { Male Population }} \times 1000$$

Question 3.
Why Japan has a declining population Pyramid ?
Japan’s pyramid has a narrow base and it tapers topward. It shows low birth and low death rates. The population growth is zero or negative.

Question 4.
What are the characteristics of population in developed countries ?
(i) Share of older people is large.
(ii) Population in higher age group has increased due to increased life expectancy.
(iii) The proportion of children has declined due to reduction in birth rates.

Question 5.
Describe the Ageing Population.
This is the process by which the older population share becomes proportionally larger. This is a new phenomena of 20th century. Population in higher age groups in most of the developed countries has increased because of increased life expectancy.

Question 6.
Describe the distribution of world pattern of sex ratio.

• On an average, the world population reflects a sex ratio of 990 females per 1000 males.
• The highest sex ratio in the world has been recorded in Latvia which is 1187 females per 1000 males.
• The lowest sex ratio occurs in Qatar which is 320 females per 1000 males.
• The sex ratio is favourable for females in 139 countries of the world.
• It is unfavourable sex ratio for females in 72 countries listed by the United Nations.
• In general, Asia has a low sex ratio. Countries like China, India, Saudi Arabia, Pakistan, Afghanistan have a lower sex ratio.
• On the other extreme is greater part of Europe (including Russia) where males are in minority.

Question 7.
What is meant by literacy rate ? Why is there a wide variation in the literacy rate among different countries of the world ?
Literacy rate is the percentage of people above 7 years age who can read and write in their everyday life. Literacy rate is different in different countries due to different levels of economic development, urbanisation and standard of living.

Question 8.
Describe the characteristics of world population as revealed by the age structure.
(i) World population is more youthful with about 36% population in age group below 15 years. In developed countries, it is 23 per cent while in developing countries, it is 40 per cent. This is an unproductive group.
(ii) The adult age group (15-59 years) is higher in number and is a productive group.
(iii) Aged people (above 60 years) increase with increase in population.

Question 9.
Discuss the factors responsible for imbalance in the sex-ratio found in different parts of the world.

• In developing countries, infant mortality is higher among males than females.
• In developed countries, male mortality is higher than females.
• Migration of males or females also affects sex ratio.
• In developing countries, male migration from villages to towns also affects sex-ratio.

Question 10.
Explain the inter-relationship between population and development.
Since Malthus’ projection, the study of population and development has become important. A further growth of population, puts pressure on land for food production. Large population is a negative factor in development. However, it depends upon its quality. Growth of population creates an imbalance in population and resources.

Other factors like technology also affects this balance. Thus, development depends upon a set of social, economic, technological and political conditions. A new concept of Human Development Index (HDI) has been introduced to assess it.

Question 11.
Write a short note on literacy in the world.
Or
‘The proportion of literate population of a country is an indicator of its socioeconomic development.” Evaluate the statement. (Outside Delhi 2019)
Literacy is an index of socio-economic development of an area. It reflects quality of populations. There is a wide variation over the world in the literacy rates which denotes the percentage of people aged 7 and above who can, with understanding, both read and write a short, simple statement in their everyday life.

Major factors affecting this rate are

• levels of economic development
• urbanisation and
• standard of living
• social status of females
• availability of educational facilities and
• the policies of the government.

Level of economic development is both a cause and a consequence of literacy.

The developed and urban economies reflect higher literacy rate and higher standards of education. Low levels of literacy and education indicate rural-farm economies.

### Population Composition Important Extra Questions Long Answer Type

Question 1.
Write a note on Age-Sex structure in the world.
Age Composition. The composition of population according to age or sex is called age and sex structure. It is fundamental to understand demographic characteristics. People are grouped according to age groups. The age is written in round number. The differences in age groups create economic disparities in the world.

Importance. Age composition unfolds many mysteries of population and affects long range future results. It helps to determine the labour force of a region and its efficiency. It estimates the needs and employment of a region. It helps to know the dependent population. It helps to know the future growth of population of a region.

Determinants of Age Structure
The following elements are the main determinants of age structure :

• Fertility
• Mortality
• Migration.

Distribution of age groups

1. Young age group. It includes the population below 15 years. In developing countries, 40% of population consists of young population; whereas this percentage is about 20 per cent in developed countries.

2. Adult group. It includes the population between 15-64 years age. It is the productive population group. In developed countries, this group is 60% and it is 35% in developing countries.

3. The Old age group. It includes the population above 65 years age. In developed countries, this group is 10 to 12%. It is only 4% in developing countries. In India, old age group is 8%. This is dependent population.

Question 2.
What do you mean by age-sex Pyramid ? Describe the different types of age-sex Pyramids.
Age-Sex Pyramid (Imp)
Or
What is age structure? Interpret with example. (C.B.S.E. Delhi 2017)

The age-sex structure of a population refers to the number of females and males in different age groups. A population pyramid is used to show the age-sex structure of the population.

The shape of the population pyramid reflects the characteristics of the population. The left side shows the percentage of males, while the right side shows the percentage of women in each age group.

Different types of age-sex Pyramid

(1) Expanding Populations. This is a triangular shaped pyramid with a wide base and is typical of less developed or under-developed countries. These have a larger population in the lower age groups due to high birth rates.
Examples. The Pyramids for Bangladesh, Mexico and Nigeria.

(2) Constant Population. It is bell shaped structure and taper towards the top. This shows that the birth and death rates are almost equal, leading to a near constant population. Example. Australia’s age-sex pyramid.

(3) Declining Pyramid. It has a narrow base and a tapered top showing low birth and death rates. The population growth in developed countries is usually zero or negative.
Example. Pyramid of Japan.

Question 3.
Describe the different types of Human occupations.
Human activities. Man adopts some ways to make a livelihood. He may be engaged in certain economic activities. These activities are called human occupations. These human occupations are considerably influenced by physical conditions. Lumbering is carried on in Taiga forests. Sheeps are reared in temperate grasslands. Agriculture is the mainstay of people in tropical lowlands.

Cultural conditions help in the choice of occupations. Man adapts his occupations according to his needs and cultural development. Religion, education, social systems, technical knowledge and the standard of living also influences human occupations. The economic activities can be divided into four groups :

(i) Primary Activities. These activities are directly related to physical conditions. Man obtains products directly from nature through these activities. The primary activities include, gathering, hunting, fishing, lumbering, mining, cattle-rearing and agriculture.

(ii) Secondary Activities. In secondary activities, man changes the form of raw materials. This increases the value and utility of products. Manufacturing, dairy farming and commercial fishing are secondary activities.

(iii) Tertiary Activities. Tertiary activities include all services and occupations. These are used for exchange and distribution of products. These include transport and trade.

(iv) Quarternary Activities. Some high services which indirectly influence the human activities are included in Quarternary Activities. Some high services like Education, Research, Medical services, Administrative, Defence, etc. are included in Quarternary Services.

### Population Composition Important Extra Questions HOTS

Question 1.
In which area a deficit of males is found ? Give two reasons.


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## anonymous 3 years ago Find the domain: $h(x) = \sqrt{4-x} + \sqrt{x^2 - 1}$

1. anonymous

i smell a race....

2. ParthKohli

Hint: you can't use positive numbers more than 4.

3. anonymous

^i wish you would have let me try first....

4. anonymous

5. anonymous

^and then you did it again @Jonask

6. anonymous

sorry,like you said its a race

7. anonymous

i hate races.....

8. anonymous

Ok just try to see when each square root is negative

9. anonymous

anyway....

10. anonymous

i suppose $$\sqrt{x^2 - 1}$$ will have no restrictions?

11. anonymous

yes x=0

12. anonymous

hmm oh yeah

13. anonymous

didn't think of that

14. anonymous

then $$\sqrt{4-x}$$ would be 4 above?

15. anonymous

I would say greater than 4

16. anonymous

@lgbasallote was that the question of 100 level???

17. anonymous

greater than 4...4 above...same shiz

18. anonymous

Take Seperately and Connect it by union (U)

19. anonymous

to be the best, one needs to master the basics @sauravshakya

20. anonymous

$f(x)=x^2,g(x)=\sqrt{1-x}$ domain of$gof,fog$ domain

21. anonymous

^?

22. anonymous

just a question that i wont interupt you can try

23. anonymous

Domain is just 1 and 0

24. anonymous

g o f would be sqrt(1 - x^2) it would only be non-negative if 0 <= x <= 1

25. anonymous

so the domain is 0<=x<=1

26. anonymous

for f(g(x))

27. anonymous

f o g would be 1- x so all real numbers

28. anonymous

$Domain=(-\infty,0) U (0,4]$

29. ParthKohli

0 isn't included in the domain since $$\sqrt{0^2 - 1} = \sqrt{-1}$$ and this is a real function.

30. anonymous

Lol...@ParthKohli u see it is a open Bracket.....

31. anonymous

"real" function?

32. ParthKohli

@Yahoo! I wasn't talking to you.

33. ParthKohli

Yeah, which has all ranges as real numbers.

34. anonymous

oh.....Sorry...) @ParthKohli

35. anonymous

now you confuse me...you're telling me they're wrong?

36. ParthKohli

swissgirl Best Response 1 Domain is just 1 and 0 that ^^

37. anonymous

hmm seems one of the answers here is wrong then...wonder which

38. ParthKohli

Okay, let me ask you a question. Why didn't you include all real numbers in the domain? @lgbasallote

39. anonymous

i did

40. anonymous

when didn't i?

41. ParthKohli

lgbasallote Best Response 0 i suppose sqrt(x62−1) will have no restrictions? swissgirl Best Response 1 yes x=0 lgbasallote Best Response 0 hmm oh yeah

42. ParthKohli

0 is included in all real numbers.

43. anonymous

since when was $$\sqrt{-1}$$ real @ParthKohli ?

44. ParthKohli

45. anonymous

....what exactly are you saying?

46. ParthKohli

ParthKohli 0 0 isn't included in the domain since sqrt(0^2 - 1) = sqrt(-1) and this is a real function. <argument happens> ParthKohli 0 Okay, let me ask you a question. Why didn't you include all real numbers in the domain? @lgbasallote lgbasallote Best Response 0 i did lgbasallote Best Response 0 when didn't i? <some more convo> lgbasallote Best Response 0 since when was √−1 real @ParthKohli ?

47. anonymous

48. ParthKohli

You are contradicting yourself. First, you say that you have included all real numbers in the domain. Then, by "since when was √−1 real @ParthKohli ?," you mean that 0 is not in the domain.

49. anonymous

by the way 1) none of us said 0 is included in the domain (we were talking about restrictions) 2) sqrt -1 is not a real function

50. anonymous

$D_g=(-\infty,1),D_f=\mathbb{R}$ $fog=\left| 1-x \right|$$gof=\sqrt{1-x^2}$

51. anonymous

^?

52. ParthKohli

No, real function means that for all domains, the range will be real. We have to get a domain which keeps all ranges real.

53. anonymous

54. ParthKohli

That 0 is not in the domain.

55. anonymous

and isn't that what we said?

56. ParthKohli

You said that you included all real numbers in the domain. -.-

57. anonymous

when???

58. anonymous

59. ParthKohli

ParthKohli 0 Okay, let me ask you a question. Why didn't you include all real numbers in the domain? @lgbasallote lgbasallote i did 10 minutes ago lgbasallote Best Response 0 when didn't i?

60. ParthKohli

I was talking about the domain then.

61. anonymous

$D_{gof}=[-1,1]$ $Dfog=x \in (-\infty,1]$

62. anonymous

you really tend to make things big when you misread, don;t you @ParthKohli

63. anonymous

guys i think we can close the question i did not mean to create a big issue

64. anonymous

it has been closed a long time ago @jonask

65. anonymous

oh i dint notice,cos commentsn are still running

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Open access peer-reviewed chapter

# Role of Bacterial Biofilms in Catheter-Associated Urinary Tract Infections (CAUTI) and Strategies for Their Control

By Mary Anne Roshni Amalaradjou and Kumar Venkitanarayanan

Submitted: April 25th 2012Reviewed: April 16th 2013Published: July 10th 2013

DOI: 10.5772/55200

## 1. Introduction

Urinary tract infections (UTI’s) can be defined as bacteriuria (>105 CFU/mL in adults; >104 CFU/mL in children) of an uropathogen with associated clinical signs that include dysuria and urgency [18]. According to the United States Centers for Disease Control and Prevention (CDC), a symptomatic urinary tract infection must meet at least one of the following criteria:

• Patients had/did not have an indwelling catheter in place at the time of specimen collection or onset of signs or symptoms

• Patient has at least one of the following signs or symptoms with no other recognized cause: fever (>38oC), urgency, frequency, dysuria, suprapubic tenderness or costovetebral angle pain or tenderness

• Patient has a positive urine culture of ≥105 with no more than 2 species of microorganisms [20].

### 5.1. Need for and duration of catheterization

It is estimated that about 21-50% of catheters are placed without justified need and catheters are inappropriately retained for 33-50% of total device days [73, 101]. The most effective ways for the preventing CAUTI are by reducing the duration of catheterization and its early removal [51]. Use of interventions such as nurse prompted removal suggestions and computer based reminders to the patients have resulted in a decline in catheter retention and a concomitant reduction in bacteriuria [164]. Thus, it is important to refrain from using an indwelling catheter without an appropriate indication. A study conducted in an emergency department indicated that use of pre-insertion checklists have led to an improved adherence to indications for placement resulting in the increase in the number of appropriately placed catheters from 37% to 51% [50].

### 5.2. Catheter placement and management

Since the catheter provides a connection between the highly colonized perineum and the sterile bladder, sterility during catheter handling and placement is of greatest importance. In this regard, hand hygiene plays a vital role in the prevention of CAUTI [16]. Insertion of a catheter in the emergency room rather than an operating room has been shown to be associated with higher rates of catheter associated bacteriuria (CAB; 158). Use of an aseptic insertion technique reduces the risk of acquiring resistant organisms in the hospital [63]. A randomized study conducted by Platt and others [1983; 118] demonstrated that hospitalized patients intubated with a catheter without a pre-sealed junction were 2.7 times more likely to develop CAB than patients with pre-connected catheter drainage bags and sealed junctions. Therefore, the use of closed catheter drainage systems universally is recommended [63]. Similarly, any breach in the closed drainage system would also increase the risk for CAB. Any manipulation of the indwelling catheter should be avoided so that breaches in the closed drainage and shear trauma can be minimized [25].

### 5.3. Catheter design

Catheter design has not changed significantly since the inception of the Foley catheter in the 1930s [97]. In addition to the catheter design, biocompatibility of the material is crucial. Catheter material can also impact the rate of biofilm formation. Scanning electron microscopy imaging of latex catheters revealed that presence of more uneven surfaces on it than other silicone counterparts which can promote bacterial adhesion [150]. Additionally latex has been associated with toxic effects in vitro and proinflammatory reactions in vivo leading to polypoid cystitis on chronic exposure [49]. Moreover, silicone catheters are more popular to avoid allergic reactions associated with latex use. Besides being hypoallergenic, silicone catheters have a larger lumen and are minimally prone to encrustation by crystalline biofilms [36]. A newly engineered silicone catheter with a trefoil cross-section was shown to reduce CAB and inflammation when compared to a standard urinary catheter [153]. The trefoil conformation helps to minimize the surface area of contact between the catheter and the urethra, thereby decreasing friction and trauma and increasing drainage of urethral secretions [137].

### 5.4. Hydrogel coated catheters

Cross linked insoluble polymers that are hydrophilic and trap water are known as hydrogels. Use of hydrophilic coating on catheters has been shown to improve patient comfort, reduce bacterial adherence and encrustation. The presence of hydrogels also increases lubrication and decreases bacterial adhesion to the interface of the tissue and the catheter [11]. However, conflicting data exist on the ability of hydrogel coated catheters to reduce CAUTI, which could be attributed to the type of hydrogel incorporated. Tunney and Gorman [2002; 169] used in vitro models to demonstrate that Poly(vinyl pyrollidone)-coated polyurethane catheters had a lower rate of encrustation when compared to uncoated polyurethane and silicone catheters. Another study showed that the use of poly(ethylene oxide)-based multiblock copolymer and segmented polyurethane increased the time to encrustation and catheter blockage from 7.8 h to 20.1 h [116]. These findings collectively suggest that the type of hydrogel coating can affect the rate of encrustation and the resulting catheter blockage.

### 5.5. Antimicrobial coating

Antimicrobial modification of catheters is achieved by coating, matrix loading and immersion in an antimicrobial solution. The primary objective behind the incorporation of antimicrobial on a catheter is to reduce bacterial attachment and biofilm formation. Additionally, release of antimicrobials from the catheters into the milieu is also another potential approach to control planktonic cells of uropathogens [56].

#### 5.5.1. Nanoparticles and iontophoresis

Nanoparticles by virtue of their small size have the ability to penetrate bacterial cells, disrupt cell membranes and bind to the chromosomal DNA. Lelouche and others [2009; 84] demonstrated that glass surfaces coated with magnesium fluoride nanoparticles inhibited biofilm formation by S. aureus and E. coli, whereas magnesium fluoride solutions did not affect biofilm formation. This highlights the size dependent effect of nanoparticles.

The application of low intensity direct current (Ionotophoresis) in vitro has been shown to increase the antimicrobial activity of antibiotics on bacteria embedded in biofilms [27]. Chakravarti and others [2005; 21] used a urinary flow model to test the in vitro antibiofilm efficacy of iontophoretic silver wire containing silicone catheters. These catheters were challenged with P. mirabilis and then exposed to a steady current of 150 µA. It was observed that application of the electric field increased the time to blockage from 22 h to 156 h, and reduced the viable count from 109 CFU/ml to 104 CFU/ml. Similar in vivo study in sheep intubated with catheters containing platinum electrodes showed a decline in pathogen count from 107 CFU/ml to 103 CFU/ml on application of a direct current of 400 µA [33].

#### 5.5.2. Antimicrobials

A variety of antimicrobials applied on urinary catheters have been investigated for their efficacy in controlling UTIs using in vitro and in vivo models.Nitrous oxide is known to exhibit bactericidal activity [123]. Urinary catheters impregnated with gaseous nitrous oxide, a known antimicrobial, and challenged with E. coli resulted in the slow release of nitrous oxide into the urine for over 14 days, and decreased biofilm formation by E. coli. Chlorhexidine is a common antimicrobial used against oral plaques. In vivo studies in rabbits intubated with genidine (combination of chlorhexidine and gentian violet) coated silicone catheters showed a reduction in biofilm formation by E. coli, E. faecium, P. aeruginosa, K. pneumoniae and Candida in comparison to silver coated and uncoated catheters [54]. Catheter associated bacteriuria was noticed in 60% and 71% of the rabbits with uncoated catheters and silver hydrogel coated catheters, respectively, whereas CAB did not occur in any of the rabbits with genidine coated catheters. Similar to chlorhexidine, triclosan is another antibacterial ingredient in toothpastes and cleaners used in health care settings. Triclosan exerts its antibacterial effect by inhibiting bacterial fatty acid synthesis [147]. Incorporation of triclosan in the balloon of catheters resulted in its release and diffusion through latex and silicon catheter balloons. The balloon served as a reservoir and the membrane helped in controlled release of triclosan. This in turn slowed encrustation and maintained the lumen patent for 7 days as compared to 24 h in saline-filled catheters [150]. Another antibacterial shown to possess antibiofilm effect is nitrofurazone, which interferes with bacterial ribosomes, DNA and cell wall. When nitrofurazone coated catheters were compared with standard catheters, it was observed that nitrofurazone significantly reduced CAB [133]. Besides nitrofurazone, norfloxacin coated catheters were also shown to inhibit the growth of E. coli, K. pneumoniae and P. vulgaris for up to 10 days [115]. Similarly, gentamicin coated catheters were also effective in reducing CAB in rabbits [23]. Another study demonstrated that sparfloxacin coated and heparin coated catheters reduced colonization by S. aureus, E. coli and S. epidermidis for greater than 26 days compared to control catheters [79]. However, the use of antibiotics on catheters to control bacterial biofilms could potentially lead to the emergence of antibiotic resistant bacteria [126]. Repeated use of antibiotics for treating UTIs has been linked to the emergence of antibiotic resistant UPEC [41, 126]. Therefore, there is an increasing interest in the use of natural antimicrobials for controlling microbial infections, including UTIs.

#### 5.5.3. Plant molecules

Plants are capable of synthesizing a large number of molecules [47], most of which are produced as a defense mechanism against predation by microorganisms and insects. A variety of plant-derived polyphenols are active components in traditional medicines [178]. A significant body of literature exists on the positive effects of dietary intake of berry fruits on human health, performance and disease [134]. Cranberry products such as its juice and tablets have been used as an alternative medicine to prevent UTIs in humans for decades. Clinical and epidemiological studies support the use of cranberry in maintaining a healthy urinary tract [117]. Although several studies have tested the antimicrobial effect of cranberries against multiple uropathogens, it was found to be most effective against UPEC.

Cranberries exert anti-adhesive effects on certain uropathogens [112] and this effect is specific to certain components of cranberry [110]. Cranberries contain three different flavonoids (flavonols, anthocyanins and PAC), catechins, hydroxycinnamic and other phenolic acids and triterpenoids. The anthocyanins are absorbed in the human circulatory system and transported without any chemical change to the urine [117]. Cranberry products do not inhibit bacterial growth, but reduced bacterial adherence to uroepithelial cells, thereby decreasing the development of UTI. The anti-adhesive effects of p-fimbriated UPEC to uroepithelial cells are related with A-linked PAC as compared with lack of anti-adhesion activities of B-linked PAC from grape, apple juice, green tea and chocolate [67]. The A-type PAC in cranberries enhances the anti-adhesive effects in vitro and in urine. PAC binds to lipopolysaccharide in gram-negative bacteria. When E. coli was grown in the presence of cranberry components, the bacterial morphology changed to a more spherical cell-like form. These changes cause them to be repelled by the human cells [88]. Similar study by Tao and others [2011; 159] have also demonstrated that consumption of cranberry juice cocktail reduced the adhesion of UPEC to a silicon nitride probe.

Cranberry has undergone extensive evaluation in the management of UTIs. However, currently there is no evidence that cranberry can be used to treat UTIs. Hence, the focus has been on its use as a prophylactic agent in the prevention of UTIs [52]. The consumption of cranberry juice can help to prevent the adhesion of UPEC to the uroepithelium and thereby help reduce the incidence of UTIs. With rising concerns of antibiotic resistance among UPEC, cranberry could serve as an effective alternative in controlling UTIs.

Trans-cinnamaldehyde (TC) is a major component of the bark extract of cinnamon [1]. It is a generally recognized as safe (GRAS) molecule approved for use in foods by the Food and Drug Administration (FDA). The U. S. Flavoring Extract Manufacturers’ Association reported that TC has a wide margin of safety between conservative estimates of intake and no observed adverse effect levels, from sub-chronic and chronic studies [1]. The report also indicated no genotoxic or mutagenic effects due to TC. Although, cinnamon or cinnamon oil has been used for ages in the treatment of UTIs, no scientific study was undertaken to investigate its antimicrobial efficacy against uropathogens. Amalaradjou and group [2010; 4] investigated the efficacy of TC for controlling UPEC biofilm formation. They observed that TC as a catheter lock solution or as a coating significantly inactivated UPEC and prevented biofilm formation when compared to untreated catheters. In a follow up study, these researchers reported that TC decreased the attachment and invasion of UPEC in cultured urinary tract epithelial cells by down-regulating several virulence genes in the pathogen [5].

Besides the use of cranberry and TC, other plant derived natural antimicrobials have also been shown to be effective against uropathogens. Sosa and Zunino [2009; 141] demonstrated that Ibicella lutea (Devils claw or Rams horn) extracts had an effect on bacterial growth rate and morphology of P.mirabilis by affecting its swarming differentiation, hemagglutination and biofilm formation on glass and polystyrene. Similarly, the use of Coccinia grandis (Ivy gourd) plant extracts have been reported to inhibit growth of UPEC in vitro [119]. Several other herbs that are used for the treatment of UTIs, but lacking scientific basis include Agrimonia eupatoria (agrimony), Althea officinalis (marshmallow), Apium graveolens (celery seed), Arctium lappa (burdock), Elymus repens (couchgrass), Hydrangea aborescens (hydrangea), Juniperus communis (juniper), Mentha piperita (peppermint), Taraxacum officinalis leaf (dandelion), Ulmus fulva (slippery elm) and Zea mays (corn silk; 3).

#### 5.5.4. Silver coated catheters

Silver is a well-known antimicrobial exerting its bactericidal action by inactivating bacterial enzymes and causing cell wall damage [96]. Silver alloy and silver oxide coatings on catheters were investigated for reducing CAB, where silver alloy coating was found to be more effective [131]. In addition to reducing CAB, other studies also demonstrated the ability of silver alloy to decrease CAUTI compared to silver oxide or latex catheters [143]. However other researchers have observed conflicting results with no difference in antibiofilm effect of silver alloy and silver oxide [122, 143].

### 5.6. Enzyme inhibitors

Urease producing bacteria are known to produce crystalline biofilms and encrustation on catheters. Use of urease inhibitors such as acetohydroxamic acid and fluorofamide have been reported to reduce encrustation and thereby prevent CAB [98]. These urease inhibitors have been also shown to prevent urea break down and pH increase in vitro by P. mirabilis besides decreasing the associated encrustation. Another enzyme target is N-acetyl-D-glucosamine-1-phosphate acetyltransferase, which is essential for peptidoglycan, lipopolysaccharide and adhesion synthesis. Inhibitors of the enzyme belonging to the N-substituted maleimide family have produced antibiofilm activity against P. aeruginosa and S. epidermidis compared to silver hydrogel coated catheters [17].

#### 5.6.1. Bacterial interference

Use of nonpathogenic microorganisms to counteract pathogenic bacteria is known as bacterial interference [137]. Colonization of catheter surfaces with nonpathogenic bacteria can prevent adhesion and colonization by pathogens. The nonpathogenic E. coli 83972 has been extensively investigated both in vitro and in vivo in bacterial interference protocols [68]. Initially, studies with this nonpathogenic strain were done by instilling the bacteria into the bladder of patients. Colonization by E. coli 83972 protected these patients from symptomatic UTI. To reduce the need for instillation of bacteria into the bladder of patients, experiments were later conducted with catheters coated with the nonpathogenic strain [168]. This study also revealed that E. coli 83972 was effective in reducing symptomatic UTI similar to previous experiments with direct infusion of the bacteria.

#### 5.6.2. Bacteriophages

Another potential approach investigated for controlling CAUTI is the use of bacteriophages. Catheters coated with T4 bacteriophage against E. coli and coli-proteus bacteriophage active against Proteus were exposed to E. coli ATCC 11303, P. mirabilis or saline. It was observed that phage treatment of catheters led to approximately 90% reduction in biofilm formation compared to control catheters [19]. It was also observed that the application of phage cocktail on catheters was more effective against bacteria than the use of a single phage [19]. When hydrogel coated catheters were pretreated with a five-phage cocktail, P. aeruginosa biofilm formation was reduced by 99% after 48 h [45].

#### 5.6.3. Liposomes

Liposomes are carrier or delivery vehicles that can carry both hydrophilic and hydrophobic molecules to their target site for delivery. This helps to increase the half life of the drugs besides protecting them from the environment. Liposomes containing ciprofloxacin embedded in a hydrogel coated catheter were evaluated in a rabbit model to investigate its antibiofilm effect against E. coli induced CAUTI [121]. The results from this study revealed that liposomal ciprofloxacin treated group had a delayed onset of positive urine cultures compared to the control group.

#### 5.6.4. Quorum sensing inhibitors

Quorum sensing between bacterial cells in a biofilm have been shown to be essential for biofilm formation and maintenance. Inhibition of quorum sensing could therefore provide a potential route for the control of biofilms. Delisea pulchra, an algal species has been shown to produce furanones that interfere with autoinducer signaling and biofilm formation [92]. In vitro and in vivo sheep experiments using furanone containing catheters have been evaluated against S. epidermidis [35]. Similarly, use of azithromycin has been shown to inhibit the production of quorum sensing signals, swimming, swarming and twitching motilities, and biofilm formation in vitro [9].

#### 5.6.5. Surface vibroacoustic stimulation

Catheters containing peizo elements can generate low energy acoustic waves that can lead to the formation of a vibrating coat along the catheter and prevent bacterial attachment and biofilm formation [60]. Scanning electron microscopy studies demonstrated that application of surface acoustic waves led to reduced biofilm formation by E. coli, E. faecalis, Candida albicans and P. mirabilis. An in vivo study in rabbits demonstrated that peizo element containing catheters with acoustic vibration led to a delayed positive urine culture compared to control animals [60]. The acoustic waves generated resulted in bacterial vibration at the same frequency, thereby preventing bacterial attachment and eventual biofilm formation.

## 6. Conclusion

Catheter associated urinary tract infections are the most common nosocomial infections and a vast majority of them are caused by biofilms formed on catheters. The complications caused by biofilms can undermine the patient’s quality of life and threaten their health. The high incidence of CAUTI and the consequent complications warrants the development and application of effective control strategies. Prevention is predominantly based on enforcing guidelines for appropriate catheter placement and early removal. However, a comprehensive understanding of bacterial biofilm formation, pathogenesis and other key factors essential for development of UTIs would help in the development of novel and effective control strategies.

## More

© 2013 The Author(s). Licensee IntechOpen. This chapter is distributed under the terms of the Creative Commons Attribution 3.0 License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

## How to cite and reference

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Mary Anne Roshni Amalaradjou and Kumar Venkitanarayanan (July 10th 2013). Role of Bacterial Biofilms in Catheter-Associated Urinary Tract Infections (CAUTI) and Strategies for Their Control, Recent Advances in the Field of Urinary Tract Infections, Thomas Nelius, IntechOpen, DOI: 10.5772/55200. Available from:

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text:
# Equivariant handle decompositions

Suppose I have some smooth closed high-dimensional manifold $M$ acted on smoothly by a finite group $G$. By a metric averaging procedure, we can equip $M$ with a smooth Riemannian metric so that $G$ acts by isometries. I can't necessarily pick a $G$-invariant morse function $f:M\to\mathbb R$, but nevertheless, I can certainly pick a smooth function $f:M\to\mathbb R$ which, though perhaps not Morse, still has only isolated "nice" critical points in some precise sense. We therefore conclude:

There is a "handle" decomposition of $M$ (where I haven't said what I mean by "handle") which is preserved by $G$. Thus $G$ just permutes (and/or acts on individually) the handles.

I am interested in knowing to what extent this can be generalized to the case of an "action up to homotopy". More specifically, suppose we have $G\to\operatorname{Homeo}(M)/\text{homotopy}$. To what extent can we "decompose" M into simple pieces in a $G$-invariant way? If it helps, then it is OK to assume that the action of $G$ is "close to the identity" in a vague coarse sense.

(I am essentially just interested on the case of high-dimensional $M$, but of course the question makes sense in any dimension).

• This does not answer your question, but provides a different approach to the Morse theory in the first part. A theorem of Illman says that if $M$ is a manifold and $G$ is a finite group acting on $M$, then $M$ has a $G$-invariant triangulation. You could get an equivariant handle decomposition from this, but it is in some sense more rigid. See MR1770606 (2001j:57032) Illman, Sören(FIN-HELS) Existence and uniqueness of equivariant triangulations of smooth proper G-manifolds with some applications to equivariant Whitehead torsion. J. Reine Angew. Math. 524 (2000), 129–183. – Andy Putman Aug 9 '12 at 3:15
• Can you be more precise about the meaning of "$G\to\operatorname{Homeo}(M)/\text{homotopy}.$" ? – John Klein Aug 9 '12 at 13:40
• Say that two elements of $\operatorname{Homeo}(M)$ are equivalent iff they are homotopic as maps $M\to M$. This is an equivalence relation, and the group operation on $\operatorname{Homeo}(M)$ descends to the quotient. – John Pardon Aug 9 '12 at 15:45
• @JohnPardon do you mean "two maps of $Homeo(M)$ are equivalent iff they are homotopic through homeomorphisms as maps $M \to M$? Because otherwise it's not clear that the inverses of equivalent maps are equivalent. – Tim Campion Apr 12 '18 at 17:09
• When $M=S^n$ is a sphere, then the space of orientation preserving homeomorphism is arc connected. In this case $G \to \textrm{Homeo}^+(M)/\textrm{homotopy} = pt$ seems to be no data at all? Is that really what you mean? – Chris Schommer-Pries Apr 12 '18 at 17:58

• I'm upvoting because if somebody googles "equivariant handle decompositions" (which is, after all, the title of the question), they will be well-served to find this answer. Perhaps it's worth mentioning that an equivariant handle consists of an equivariant disc bundle over a homogenous space $G/H$. – Tim Campion Apr 12 '18 at 17:02


text:
# quotient group inheritance of some properties from original group

Let us say that there is the group with addition modulo 6:

$G = \{0,1,2,3,4,5\}$ and let $N = \{0,3 \}$.

Then the quotient group $G/N$ would be $G/N = \{ \{0,3\}, \{1,4 \}, \{2,5\} \}$.

According to my sources, they say that as $N$ is normal subgroup, $G/N$ must be also group. As the group $G$ is abelian, according to my sources (Wikipedia and textbook), they say that $G/N$ must be abelian group.

The question is, by quotient group $G/N$ being group, is group operation done by assuming that in the example, $0$ and $3$ are treated same, $1$ and $4$ are treated same and $2$ and $5$ are treated same? And abelian-ness can be checked by adding $(3+4) \pmod 6 = (4+3) \pmod 6 = 1 = 4?$

Or is this something else?

-
Yeah, essentially you are right that you have to treat them the same. So, for example, using the fact that $1$ and $4$ are the same we have $1+2=3\text{ mod }6$ while $4+2=0\text{ mod }6$ and this makes sense as $0$ and $3$ are treated the same. You should realise though that what is really going on is that you are working $\text{mod }3$. –  user1729 Mar 28 '13 at 10:14
You may find interesting this question about quotient sets. –  A.P. Mar 28 '13 at 12:48

## 1 Answer

This is something else: in the quotient you have equivalence classes which you will multiply (sum) according to the rules you were surely told. There are no more mere elements as in the mother group, yet it is always possible to pick up some representative from each class andwork with it...

This subject is better understood in calss, with a decent blackboard and some chalk, and doing many examples, but in your case we can say, for example, that

$$\{1,4\}+\{2,5\}=\{0,3\}\iff \bar 1+\bar 2=\bar 0$$

where the bar above means that we'reworking the equiv. class, not merely with the bare elements.

-
The point here is that elements which are in the same equivalence classes are "treated the same". Equivalence classes just formalise this. –  user1729 Mar 28 '13 at 10:17
Yes, but the operation is not more modulo $\,6\,$ , as my example shows, and this is what the OP asked, among other things. –  DonAntonio Mar 28 '13 at 12:47


text:
# Eigenvalues of a non-symmetric matrix

Let $M \in \mathbb{R}^{n \times n}$ be a symmetric, positive definite matrix, such that $M \preccurlyeq I$ (i.e., $I-M$ is positive semi-definite). Prove or disprove that the matrix $$\left[\begin{matrix} I-M & -M \\ I-M & I-M \end{matrix}\right]$$ has all eigenvalues inside the unit circle.

Comments. I can only prove the claim if $n=1$.

## 2 Answers

Call your block matrix $A$. Since the blocks commute, its characteristic polynomial is $p(\lambda) = \det(A - \lambda I) = \det(((1-\lambda) I - M)^2 + M(I-M)) = \det((2\lambda - 1)M + (\lambda - 1)^2)$. This is $0$ when $2 \lambda - 1 \ne 0$ and $(\lambda - 1)^2/(1-2\lambda)$ is an eigenvalue of $M$. Now solving the equation $$\dfrac{(\lambda - 1)^2}{1-2\lambda} = t$$ we get $$\lambda = 1-t \pm \sqrt{t^2-t}$$ For $t \in (0,1)$ (where the eigenvalues of $M$ are), this is in the unit circle because $t^2-t < 0$ and $|\lambda|^2 = (1-t)^2 + t - t^2 = 1 - t < 1$.

• Thanks a lot your the answer. Can you please further clarity what is meant by "Since the blocks commute"? – user693 Jul 10 '15 at 16:11

Since $M$ is symmetric, it is orthogonally similar to a diagonal matrix: $M=QDQ^T$ where $Q$ is orthogonal and $D$ is diagonal. We have $$\pmatrix{Q&0\\0&Q}^T\pmatrix{I-M&-M\\I-M&I-M}\pmatrix{Q&0\\0&Q}=\pmatrix{I-D&-D\\I-D&I-D}=:C.$$ Note that $C$ (having the same eigenvalues as the original matrix) is very special, it is a $2\times 2$ block matrix where the blocks are diagonal matrices. It is useful to know that the eigenvalues of such a matrix are given by the eigenvalues of $2\times 2$ matrices composed of the corresponding diagonal entries in the diagonal blocks (this can be shown simply by choosing "special eigenvectors"). Hence the eigenvalues of $C$ are the eigenvalues of matrices $$\pmatrix{1-d&-d\\1-d&1-d},$$ (and vice versa) where $d$ are diagonal entries of $D$ (the eigenvalues of $M$). Since you can deal with the case $n=1$, you can now deal with any $n>1$.


text:
## My 2018 Mathematics A To Z: Limit

I got an irresistible topic for today’s essay. It’s courtesy Peter Mander, author of Carnot Cycle, “the classical blog about thermodynamics”. It’s bimonthly and it’s one worth waiting for. Some of the essays are historical; some are statistical-mechanics; many are mixtures of them. You could make a fair argument that thermodynamics is the most important field of physics. It’s certainly one that hasn’t gotten the popularization treatment it deserves, for its importance. Mander is doing something to correct that.

# Limit.

It is hard to think of limits without thinking of motion. The language even professional mathematicians use suggests it. We speak of the limit of a function “as x goes to a”, or “as x goes to infinity”. Maybe “as x goes to zero”. But a function is a fixed thing, a relationship between stuff in a domain and stuff in a range. It can’t change any more than January, AD 1988 can change. And ‘x’ here is a dummy variable, part of the scaffolding to let us find what we want to know. I suppose ‘x’ can change, but if we ever see it, something’s gone very wrong. But we want to use it to learn something about a function for a point like ‘a’ or ‘infinity’ or ‘zero’.

The language of motion helps us learn, to a point. We can do little experiments: if $f(x) = \frac{sin(x)}{x}$, then, what should we expect it to be for x near zero? It’s irresistible to try out the calculator. Let x be 0.1. 0.01. 0.001. 0.0001. The numbers say this f(x) gets closer and closer to 1. That’s good, right? We know we can’t just put in an x of zero, because there’s some trouble that makes. But we can imagine creeping up on the zero we really wanted. We might spot some obvious prospects for mischief: what if x is negative? We should try -0.1, -0.01, -0.001 and so on. And maybe we won’t get exactly the right answer. But if all we care about is the first (say) three digits and we try out a bunch of x’s and the corresponding f(x)’s agree to those three digits, that’s good enough, right?

This is good for giving an idea of what to expect a limit to look like. It should be, well, what it really really really looks like a function should be. It takes some thinking to see where it might go wrong. It might go to different numbers based on which side you approach from. But that seems like something you can rationalize. Indeed, we do; we can speak of functions having different limits based on what direction you approach from. Sometimes that’s the best one can say about them.

But it can get worse. It’s possible to make functions that do crazy weird things. Some of these look like you’re just trying to be difficult. Like, set f(x) equal to 1 if x is rational and 0 if x is irrational. If you don’t expect that to be weird you’re not paying attention. Can’t blame someone for deciding that falls outside the realm of stuff you should be able to find limits for. And who would make, say, an f(x) that was 1 if x was 0.1 raised to some power, but 2 if x was 0.2 raised to some power, and 3 otherwise? Besides someone trying to prove a point?

Fine. But you can make a function that looks innocent and yet acts weird if the domain is two-dimensional. Or more. It makes sense to say that the functions I wrote in the above paragraph should be ruled out of consideration. But the limit of $f(x, y) = \frac{x^3 y}{x^6 + y^2}$ at the origin? You get different results approaching in different directions. And the function doesn’t give obvious signs of imminent danger here.

We need a better idea. And we even have one. This took centuries of mathematical wrangling and arguments about what should and shouldn’t be allowed. This should inspire sympathy with Intro Calc students who don’t understand all this by the end of week three. But here’s what we have.

I need a supplementary idea first. That is the neighborhood. A point has a neighborhood if there’s some open set that contains it. We represent this by drawing a little blob around the point we care about. If we’re looking at the neighborhood of a real number, then this is a little interval, that’s all. When we actually get around to calculating, we make these neighborhoods little circles. Maybe balls. But when we’re doing proofs about how limits work, or how we use them to prove things, we make blobs. This “neighborhood” idea looks simple, but we need it, so here we go.

So start with a function, named ‘f’. It has a domain, which I’ll call ‘D’. And a range, which I want to call ‘R’, but I don’t think I need the shorthand. Now pick some point ‘a’. This is the point at which we want to evaluate the limit. This seems like it ought to be called the “limit point” and it’s not. I’m sorry. Mathematicians use “limit point” to talk about something else. And, unfortunately, it makes so much sense in that context that we aren’t going to change away from that.

‘a’ might be in the domain ‘D’. It might not. It might be on the border of ‘D’. All that’s important is that there be a neighborhood inside ‘D’ that contains ‘a’.

I don’t know what f(a) is. There might not even be an f(a), if a is on the boundary of the domain ‘D’. But I do know that everything inside the neighborhood of ‘a’, apart from ‘a’, is in the domain. So we can look at the values of f(x) for all the x’s in this neighborhood. This will create a set, in the range, that’s known as the image of the neighborhood. It might be a continuous chunk in the range. It might be a couple of chunks. It might be a single point. It might be some crazy-quilt set. Depends on ‘f’. And the neighborhood. No matter.

Now I need you to imagine the reverse. Pick a point in the range. And then draw a neighborhood around it. Then pick out what we call the pre-image of it. That’s all the points in the domain that get matched to values inside that neighborhood. Don’t worry about trying to do it; that’s for the homework practice. Would you agree with me that you can imagine it?

I hope so because I’m about to describe the part where Intro Calc students think hard about whether they need this class after all.

All right. Then I want something in the range. I’m going to call it ‘L’. And it’s special. It’s the limit of ‘f’ at ‘a’ if this following bit is true:

Think of every neighborhood you could pick of ‘L’. Can be big, can be small. Just has to be a neighborhood of ‘L’. Now think of the pre-image of that neighborhood. Is there always a neighborhood of ‘a’ inside that pre-image? It’s okay if it’s a tiny neighborhood. Just has to be an open neighborhood. It doesn’t have to contain ‘a’. You can allow a pinpoint hole there.

If you can always do this, however tiny the neighborhood of ‘L’ is, then the limit of ‘f’ at ‘a’ is ‘L’. If you can’t always do this — if there’s even a single exception — then there is no limit of ‘f’ at ‘a’.

I know. I felt like that the first couple times through the subject too. The definition feels backward. Worse, it feels like it begs the question. We suppose there’s an ‘L’ and then test these properties about it and then if it works we say we’re done? I know. It’s a pain when you start calculating this with specific formulas and all that, too. But supposing there is an answer and then learning properties about it, including whether it can exist? That’s a slick trick. We can use it.

Thing is, the pain is worth it. We can calculate with it and not have to out-think tricky functions. It works for domains with as many dimensions as you need. It works for limits that aren’t inside the domain. It works with domains and ranges that aren’t real numbers. It works for functions with weird and complicated domains. We can adapt it if we want to consider limits that are constrained in some way. It won’t be fooled by tricks like I put up above, the f(x) with different rules for the rational and irrational numbers.

So mathematicians shrug, and do enough problems that they get the hang of it, and use this definition. It’s worth it, once you get there.

This and other Fall 2018 Mathematics A-To-Z posts can be read at this link. And I’m still taking nominations for discussion topics, if you’d like to see mathematics terms explained. I know I would.

## Wronski’s Formula For Pi: My Boring Mistake

Previously:

So, I must confess failure. Not about deciphering Józef Maria Hoëne-Wronski’s attempted definition of π. He’d tried this crazy method throwing a lot of infinities and roots of infinities and imaginary numbers together. I believe I translated it into the language of modern mathematics fairly. And my failure is not that I found the formula actually described the number -½π.

Oh, I had an error in there, yes. And I’d found where it was. It was all the way back in the essay which first converted Wronski’s formula into something respectable. It was a small error, first appearing in the last formula of that essay and never corrected from there. This reinforces my suspicion that when normal people see formulas they mostly look at them to confirm there is a formula there. With luck they carry on and read the sentences around them.

My failure is I wanted to write a bit about boring mistakes. The kinds which you make all the time while doing mathematics work, but which you don’t worry about. Dropped signs. Constants which aren’t divided out, or which get multiplied in incorrectly. Stuff like this which you only detect because you know, deep down, that you should have gotten to an attractive simple formula and you haven’t. Mistakes which are tiresome to make, but never make you wonder if you’re in the wrong job.

The trouble is I can’t think of how to make an essay of that. We don’t tend to rate little mistakes like the wrong sign or the wrong multiple or a boring unnecessary added constant as important. This is because they’re not. The interesting stuff in a mathematical formula is usually the stuff representing variations. Change is interesting. The direction of the change? Eh, nice to know. A swapped plus or minus sign alters your understanding of the direction of the change, but that’s all. Multiplying or dividing by a constant wrongly changes your understanding of the size of the change. But that doesn’t alter what the change looks like. Just the scale of the change. Adding or subtracting the wrong constant alters what you think the change is varying from, but not what the shape of the change is. Once more, not a big deal.

But you also know that instinctively, or at least you get it from seeing how it’s worth one or two points on an exam to write -sin where you mean +sin. Or how if you ask the instructor in class about that 2 where a ½ should be, she’ll say, “Oh, yeah, you’re right” and do a hurried bit of erasing before going on.

Thus my failure: I don’t know what to say about boring mistakes that has any insight.

For the record here’s where I got things wrong. I was creating a function, named ‘f’ and using as a variable ‘x’, to represent Wronski’s formula. I’d gotten to this point:

$f(x) = -4 \imath x 2^{\frac{1}{2}\cdot \frac{1}{x}} \left\{ e^{\imath \frac{\pi}{4}\cdot\frac{1}{x}} - e^{- \imath \frac{\pi}{4}\cdot\frac{1}{x}} \right\}$

And then I observed how the stuff in curly braces there is “one of those magic tricks that mathematicians know because they see it all the time”. And I wanted to call in this formula, correctly:

$\sin\left(\phi\right) = \frac{e^{\imath \phi} - e^{-\imath \phi}}{2\imath }$

So here’s where I went wrong. I took the $4\imath$ way off in the front of that first formula and combined it with the stuff in braces to make 2 times a sine of some stuff. I apologize for this. I must have been writing stuff out faster than I was thinking about it. If I had thought, I would have gone through this intermediate step:

$f(x) = -4 \imath x 2^{\frac{1}{2}\cdot \frac{1}{x}} \left\{ e^{\imath \frac{\pi}{4}\cdot\frac{1}{x}} - e^{- \imath \frac{\pi}{4}\cdot\frac{1}{x}} \right\} \cdot \frac{2\imath}{2\imath}$

Because with that form in mind, it’s easy to take the stuff in curled braces and the $2\imath$ in the denominator. From that we get, correctly, $\sin\left(\frac{\pi}{4}\cdot\frac{1}{x}\right)$. And then the $-4\imath$ on the far left of that expression and the $2\imath$ on the right multiply together to produce the number 8.

So the function ought to have been, all along:

$f(x) = 8 x 2^{\frac{1}{2}\cdot \frac{1}{x}} \sin\left(\frac{\pi}{4}\cdot \frac{1}{x}\right)$

Not very different, is it? Ah, but it makes a huge difference. Carry through with all the L’Hôpital’s Rule stuff described in previous essays. All the complicated formula work is the same. There’s a different number hanging off the front, waiting to multiply in. That’s all. And what you find, redoing all the work but using this corrected function, is that Wronski’s original mess —

$\pi = \frac{4\infty}{\sqrt{-1}}\left\{ \left(1 + \sqrt{-1}\right)^{\frac{1}{\infty}} - \left(1 - \sqrt{-1}\right)^{\frac{1}{\infty}} \right\}$

— should indeed equal:

$2\pi$

All right, there’s an extra factor of 2 here. And I don’t think that is my mistake. Or if it is, other people come to the same mistake without my prompting.

Possibly the book I drew this from misquoted Wronski. It’s at least as good to have a formula for 2π as it is to have one for π. Or Wronski had a mistake in his original formula, and had a constant multiplied out front which he didn’t want. It happens to us all.

Fin.

## Wronski’s Formula For Pi: How Close We Came

Previously:

Józef Maria Hoëne-Wronski’s had an idea for a new, universal, culturally-independent definition of π. It was this formula that nobody went along with because they had looked at it:

$\pi = \frac{4\infty}{\sqrt{-1}}\left\{ \left(1 + \sqrt{-1}\right)^{\frac{1}{\infty}} - \left(1 - \sqrt{-1}\right)^{\frac{1}{\infty}} \right\}$

I made some guesses about what he would want this to mean. And how we might put that in terms of modern, conventional mathematics. I describe those in the above links. In terms of limits of functions, I got this:

$\displaystyle \lim_{x \to \infty} f(x) = \lim_{x \to \infty} -2 x 2^{\frac{1}{2}\cdot \frac{1}{x}} \sin\left(\frac{\pi}{4}\cdot \frac{1}{x}\right)$

The trouble is that limit took more work than I wanted to do to evaluate. If you try evaluating that ‘f(x)’ at ∞, you get an expression that looks like zero times ∞. This begs for the use of L’Hôpital’s Rule, which tells you how to find the limit for something that looks like zero divided by zero, or like ∞ divided by ∞. Do a little rewriting — replacing that first ‘x’ with ‘$\frac{1}{1 / x}$ — and this ‘f(x)’ behaves like L’Hôpital’s Rule needs.

The trouble is, that’s a pain to evaluate. L’Hôpital’s Rule works on functions that look like one function divided by another function. It does this by calculating the derivative of the numerator function divided by the derivative of the denominator function. And I decided that was more work than I wanted to do.

Where trouble comes up is all those parts where $\frac{1}{x}$ turns up. The derivatives of functions with a lot of $\frac{1}{x}$ terms in them get more complicated than the original functions were. Is there a way to get rid of some or all of those?

And there is. Do a change of variables. Let me summon the variable ‘y’, whose value is exactly $\frac{1}{x}$. And then I’ll define a new function, ‘g(y)’, whose value is whatever ‘f’ would be at $\frac{1}{y}$. That is, and this is just a little bit of algebra:

$g(y) = -2 \cdot \frac{1}{y} \cdot 2^{\frac{1}{2} y } \cdot \sin\left(\frac{\pi}{4} y\right)$

The limit of ‘f(x)’ for ‘x’ at ∞ should be the same number as the limit of ‘g(y)’ for ‘y’ at … you’d really like it to be zero. If ‘x’ is incredibly huge, then $\frac{1}{x}$ has to be incredibly small. But we can’t just swap the limit of ‘x’ at ∞ for the limit of ‘y’ at 0. The limit of a function at a point reflects the value of the function at a neighborhood around that point. If the point’s 0, this includes positive and negative numbers. But looking for the limit at ∞ gets at only positive numbers. You see the difference?

… For this particular problem it doesn’t matter. But it might. Mathematicians handle this by taking a “one-sided limit”, or a “directional limit”. The normal limit at 0 of ‘g(y)’ is based on what ‘g(y)’ looks like in a neighborhood of 0, positive and negative numbers. In the one-sided limit, we just look at a neighborhood of 0 that’s all values greater than 0, or less than 0. In this case, I want the neighborhood that’s all values greater than 0. And we write that by adding a little + in superscript to the limit. For the other side, the neighborhood less than 0, we add a little – in superscript. So I want to evalute:

$\displaystyle \lim_{y \to 0^+} g(y) = \lim_{y \to 0^+} -2\cdot\frac{2^{\frac{1}{2}y} \cdot \sin\left(\frac{\pi}{4} y\right)}{y}$

Limits and L’Hôpital’s Rule and stuff work for one-sided limits the way they do for regular limits. So there’s that mercy. The first attempt at this limit, seeing what ‘g(y)’ is if ‘y’ happens to be 0, gives $-2 \cdot \frac{1 \cdot 0}{0}$. A zero divided by a zero is promising. That’s not defined, no, but it’s exactly the format that L’Hôpital’s Rule likes. The numerator is:

$-2 \cdot 2^{\frac{1}{2}y} \sin\left(\frac{\pi}{4} y\right)$

And the denominator is:

$y$

The first derivative of the denominator is blessedly easy: the derivative of y, with respect to y, is 1. The derivative of the numerator is a little harder. It demands the use of the Product Rule and the Chain Rule, just as last time. But these chains are easier.

The first derivative of the numerator is going to be:

$-2 \cdot 2^{\frac{1}{2}y} \cdot \log(2) \cdot \frac{1}{2} \cdot \sin\left(\frac{\pi}{4} y\right) + -2 \cdot 2^{\frac{1}{2}y} \cdot \cos\left(\frac{\pi}{4} y\right) \cdot \frac{\pi}{4}$

Yeah, this is the simpler version of the thing I was trying to figure out last time. Because this is what’s left if I write the derivative of the numerator over the derivative of the denominator:

$\displaystyle \lim_{y \to 0^+} \frac{ -2 \cdot 2^{\frac{1}{2}y} \cdot \log(2) \cdot \frac{1}{2} \cdot \sin\left(\frac{\pi}{4} y\right) + -2 \cdot 2^{\frac{1}{2}y} \cdot \cos\left(\frac{\pi}{4} y\right) \cdot \frac{\pi}{4} }{1}$

And now this is easy. Promise. There’s no expressions of ‘y’ divided by other expressions of ‘y’ or anything else tricky like that. There’s just a bunch of ordinary functions, all of them defined for when ‘y’ is zero. If this limit exists, it’s got to be equal to:

$\displaystyle -2 \cdot 2^{\frac{1}{2} 0} \cdot \log(2) \cdot \frac{1}{2} \cdot \sin\left(\frac{\pi}{4} \cdot 0\right) + -2 \cdot 2^{\frac{1}{2} 0 } \cdot \cos\left(\frac{\pi}{4} \cdot 0\right) \cdot \frac{\pi}{4}$

$\frac{\pi}{4} \cdot 0$ is 0. And the sine of 0 is 0. The cosine of 0 is 1. So all this gets to be a lot simpler, really fast.

$\displaystyle -2 \cdot 2^{0} \cdot \log(2) \cdot \frac{1}{2} \cdot 0 + -2 \cdot 2^{ 0 } \cdot 1 \cdot \frac{\pi}{4}$

And 20 is equal to 1. So the part to the left of the + sign there is all zero. What remains is:

$\displaystyle 0 + -2 \cdot \frac{\pi}{4}$

And so, finally, we have it. Wronski’s formula, as best I make it out, is a function whose value is …

$-\frac{\pi}{2}$

… So, what Wronski had been looking for, originally, was π. This is … oh, so very close to right. I mean, there’s π right there, it’s just multiplied by an unwanted $-\frac{1}{2}$. The question is, where’s the mistake? Was Wronski wrong to start with? Did I parse him wrongly? Is it possible that the book I copied Wronski’s formula from made a mistake?

Could be any of them. I’d particularly suspect I parsed him wrongly. I returned the library book I had got the original claim from, and I can’t find it again before this is set to publish. But I should check whether Wronski was thinking to find π, the ratio of the circumference to the diameter of a circle. Or might he have looked to find the ratio of the circumference to the radius of a circle? Either is an interesting number worth finding. We’ve settled on the circumference-over-diameter as valuable, likely for practical reasons. It’s much easier to measure the diameter than the radius of a thing. (Yes, I have read the Tau Manifesto. No, I am not impressed by it.) But if you know 2π, then you know π, or vice-versa.

The next question: yeah, but I turned up -½π. What am I talking about 2π for? And the answer there is, I’m not the first person to try working out Wronski’s stuff. You can try putting the expression, as best you parse it, into a tool like Mathematica and see what makes sense. Or you can read, for example, Quora commenters giving answers with way less exposition than I do. And I’m convinced: somewhere along the line I messed up. Not in an important way, but, essentially, doing something equivalent to divided by -2 when I should have multiplied by that.

I’ve spotted my mistake. I figure to come back around to explaining where it is and how I made it.

## Wronski’s Formula For Pi: Two Weird Tricks For Limits That Mathematicians Keep Using

Previously:

So now a bit more on Józef Maria Hoëne-Wronski’s attempted definition of π. I had got it rewritten to this form:

$\displaystyle \lim_{x \to \infty} f(x) = \lim_{x \to \infty} -2 x 2^{\frac{1}{2}\cdot \frac{1}{x}} \sin\left(\frac{\pi}{4}\cdot \frac{1}{x}\right)$

And I’d tried the first thing mathematicians do when trying to evaluate the limit of a function at a point. That is, take the value of that point and put it in whatever the formula is. If that formula evaluates to something meaningful, then that value is the limit. That attempt gave this:

$-2 \cdot \infty \cdot 1 \cdot 0$

Because the limit of ‘x’, for ‘x’ at ∞, is infinitely large. The limit of ‘$2^{\frac{1}{2}\cdot\frac{1}{x}}$‘ for ‘x’ at ∞ is 1. The limit of ‘$\sin(\frac{\pi}{4}\cdot\frac{1}{x})$ for ‘x’ at ∞ is 0. We can take limits that are 0, or limits that are some finite number, or limits that are infinitely large. But multiplying a zero times an infinity is dangerous. Could be anything.

Mathematicians have a tool. We know it as L’Hôpital’s Rule. It’s named for the French mathematician Guillaume de l’Hôpital, who discovered it in the works of his tutor, Johann Bernoulli. (They had a contract giving l’Hôpital publication rights. If Wikipedia’s right the preface of the book credited Bernoulli, although it doesn’t appear to be specifically for this. The full story is more complicated and ambiguous. The previous sentence may be said about most things.)

So here’s the first trick. Suppose you’re finding the limit of something that you can write as the quotient of one function divided by another. So, something that looks like this:

$\displaystyle \lim_{x \to a} \frac{h(x)}{g(x)}$

(Normally, this gets presented as ‘f(x)’ divided by ‘g(x)’. But I’m already using ‘f(x)’ for another function and I don’t want to muddle what that means.)

Suppose it turns out that at ‘a’, both ‘h(x)’ and ‘g(x)’ are zero, or both ‘h(x)’ and ‘g(x)’ are ∞. Zero divided by zero, or ∞ divided by ∞, looks like danger. It’s not necessarily so, though. If this limit exists, then we can find it by taking the first derivatives of ‘h’ and ‘g’, and evaluating:

$\displaystyle \lim_{x \to a} \frac{h'(x)}{g'(x)}$

That ‘ mark is a common shorthand for “the first derivative of this function, with respect to the only variable we have around here”.

This doesn’t look like it should help matters. Often it does, though. There’s an excellent chance that either ‘h'(x)’ or ‘g'(x)’ — or both — aren’t simultaneously zero, or ∞, at ‘a’. And once that’s so, we’ve got a meaningful limit. This doesn’t always work. Sometimes we have to use this l’Hôpital’s Rule trick a second time, or a third or so on. But it works so very often for the kinds of problems we like to do. Reaches the point that if it doesn’t work, we have to suspect we’re calculating the wrong thing.

But wait, you protest, reasonably. This is fine for problems where the limit looks like 0 divided by 0, or ∞ divided by ∞. What Wronski’s formula got me was 0 times 1 times ∞. And I won’t lie: I’m a little unsettled by having that 1 there. I feel like multiplying by 1 shouldn’t be a problem, but I have doubts.

That zero times ∞ thing, thought? That’s easy. Here’s the second trick. Let me put it this way: isn’t ‘x’ really the same thing as $\frac{1}{ 1 / x }$?

I expect your answer is to slam your hand down on the table and glare at my writing with contempt. So be it. I told you it was a trick.

And it’s a perfectly good one. And it’s perfectly legitimate, too. $\frac{1}{x}$ is a meaningful number if ‘x’ is any finite number other than zero. So is $\frac{1}{ 1 / x }$. Mathematicians accept a definition of limit that doesn’t really depend on the value of your expression at a point. So that $\frac{1}{x}$ wouldn’t be meaningful for ‘x’ at zero doesn’t mean we can’t evaluate its limit for ‘x’ at zero. And just because we might not be sure that $\frac{1}{x}$ would mean for infinitely large ‘x’ doesn’t mean we can’t evaluate its limit for ‘x’ at ∞.

I see you, person who figures you’ve caught me. The first thing I tried was putting in the value of ‘x’ at the ∞, all ready to declare that this was the limit of ‘f(x)’. I know my caveats, though. Plugging in the value you want the limit at into the function whose limit you’re evaluating is a shortcut. If you get something meaningful, then that’s the same answer you would get finding the limit properly. Which is done by looking at the neighborhood around but not at that point. So that’s why this reciprocal-of-the-reciprocal trick works.

So back to my function, which looks like this:

$\displaystyle f(x) = -2 x 2^{\frac{1}{2}\cdot \frac{1}{x}} \sin\left(\frac{\pi}{4}\cdot \frac{1}{x}\right)$

Do I want to replace ‘x’ with $\frac{1}{1 / x}$, or do I want to replace $\sin\left(\frac{\pi}{4}\cdot \frac{1}{x}\right)$ with $\frac{1}{1 / \sin\left(\frac{\pi}{4}\cdot \frac{1}{x}\right)}$? I was going to say something about how many times in my life I’ve been glad to take the reciprocal of the sine of an expression of x. But just writing the symbols out like that makes the case better than being witty would.

So here is a new, L’Hôpital’s Rule-friendly, version of my version of Wronski’s formula:

$\displaystyle f(x) = -2 \frac{2^{\frac{1}{2}\cdot \frac{1}{x}} \sin\left(\frac{\pi}{4}\cdot \frac{1}{x}\right)}{\frac{1}{x}}$

I put that -2 out in front because it’s not really important. The limit of a constant number times some function is the same as that constant number times the limit of that function. We can put that off to the side, work on other stuff, and hope that we remember to bring it back in later. I manage to remember it about four-fifths of the time.

So these are the numerator and denominator functions I was calling ‘h(x)’ and ‘g(x)’ before:

$h(x) = 2^{\frac{1}{2}\cdot \frac{1}{x}} \sin\left(\frac{\pi}{4}\cdot \frac{1}{x}\right)$

$g(x) = \frac{1}{x}$

The limit of both of these at ∞ is 0, just as we might hope. So we take the first derivatives. That for ‘g(x)’ is easy. Anyone who’s reached week three in Intro Calculus can do it. This may only be because she’s gotten bored and leafed through the formulas on the inside front cover of the textbook. But she can do it. It’s:

$g'(x) = -\frac{1}{x^2}$

The derivative for ‘h(x)’ is a little more involved. ‘h(x)’ we can write as the product of two expressions, that $2^{\frac{1}{2}\cdot \frac{1}{x}}$ and that $\sin\left(\frac{\pi}{4}\cdot \frac{1}{x}\right)$. And each of those expressions contains within themselves another expression, that $\frac{1}{x}$. So this is going to require the Product Rule, of two expressions that each require the Chain Rule.

This is as far as I got with that before slamming my hand down on the table and glaring at the problem with disgust:

$h'(x) = 2^{\frac{1}{2}\frac{1}{x}} \cdot \log(2) \cdot \frac{1}{2} \cdot (-1) \cdot \frac{1}{x^2} + 2^{\frac{1}{2}\frac{1}{x}} \cdot \cos( arg ) bleah$

Yeah I’m not finishing that. Too much work. I’m going to reluctantly try thinking instead.

(If you want to do that work — actually, it isn’t much more past there, and if you followed that first half you’re going to be fine. And you’ll see an echo of it in what I do next time.)

## Wronski’s Formula For Pi: A First Limit

Previously:

When I last looked at Józef Maria Hoëne-Wronski’s attempted definition of π I had gotten it to this. Take the function:

$f(x) = -2 x 2^{\frac{1}{2}\cdot \frac{1}{x}} \sin\left(\frac{\pi}{4}\cdot \frac{1}{x}\right)$

And find its limit when ‘x’ is ∞. Formally, you want to do this by proving there’s some number, let’s say ‘L’. And ‘L’ has the property that you can pick any margin-of-error number ε that’s bigger than zero. And whatever that ε is, there’s some number ‘N’ so that whenever ‘x’ is bigger than ‘N’, ‘f(x)’ is larger than ‘L – ε’ and also smaller than ‘L + ε’. This can be a lot of mucking about with expressions to prove.

Fortunately we have shortcuts. There’s work we can do that gets us ‘L’, and we can rely on other proofs that show that this must be the limit of ‘f(x)’ at some value ‘a’. I use ‘a’ because that doesn’t commit me to talking about ∞ or any other particular value. The first approach is to just evaluate ‘f(a)’. If you get something meaningful, great! We’re done. That’s the limit of ‘f(x)’ at ‘a’. This approach is called “substitution” — you’re substituting ‘a’ for ‘x’ in the expression of ‘f(x)’ — and it’s great. Except that if your problem’s interesting then substitution won’t work. Still, maybe Wronski’s formula turns out to be lucky. Fit in ∞ where ‘x’ appears and we get:

$f(\infty) = -2 \infty 2^{\frac{1}{2}\cdot \frac{1}{\infty}} \sin\left(\frac{\pi}{4}\cdot \frac{1}{\infty}\right)$

So … all right. Not quite there yet. But we can get there. For example, $\frac{1}{\infty}$ has to be — well. It’s what you would expect if you were a kid and not worried about rigor: 0. We can make it rigorous if you like. (It goes like this: Pick any ε larger than 0. Then whenever ‘x’ is larger than $\frac{1}{\epsilon}$ then $\frac{1}{x}$ is less than ε. So the limit of $\frac{1}{x}$ at ∞ has to be 0.) So let’s run with this: replace all those $\frac{1}{\infty}$ expressions with 0. Then we’ve got:

$f(\infty) = -2 \infty 2^{0} \sin\left(0\right)$

The sine of 0 is 0. 20 is 1. So substitution tells us limit is -2 times ∞ times 1 times 0. That there’s an ∞ in there isn’t a problem. A limit can be infinitely large. Think of the limit of ‘x2‘ at ∞. An infinitely large thing times an infinitely large thing is fine. The limit of ‘x ex‘ at ∞ is infinitely large. A zero times a zero is fine; that’s zero again. But having an ∞ times a 0? That’s trouble. ∞ times something should be huge; anything times zero should be 0; which term wins?

So we have to fall back on alternate plans. Fortunately there’s a tool we have for limits when we’d otherwise have to face an infinitely large thing times a zero.

I hope to write about this next time. I apologize for not getting through it today but time wouldn’t let me.

## As I Try To Figure Out What Wronski Thought ‘Pi’ Was

A couple weeks ago I shared a fascinating formula for π. I got it from Carl B Boyer’s The History of Calculus and its Conceptual Development. He got it from Józef Maria Hoëne-Wronski, early 19th-century Polish mathematician. His idea was that an absolute, culturally-independent definition of π would come not from thinking about circles and diameters but rather this formula:

$\pi = \frac{4\infty}{\sqrt{-1}}\left\{ \left(1 + \sqrt{-1}\right)^{\frac{1}{\infty}} - \left(1 - \sqrt{-1}\right)^{\frac{1}{\infty}} \right\}$

Now, this formula is beautiful, at least to my eyes. It’s also gibberish. At least it’s ungrammatical. Mathematicians don’t like to write stuff like “four times infinity”, at least not as more than a rough draft on the way to a real thought. What does it mean to multiply four by infinity? Is arithmetic even a thing that can be done on infinitely large quantities? Among Wronski’s problems is that they didn’t have a clear answer to this. We’re a little more advanced in our mathematics now. We’ve had a century and a half of rather sound treatment of infinitely large and infinitely small things. Can we save Wronski’s work?

Start with the easiest thing. I’m offended by those $\sqrt{-1}$ bits. Well, no, I’m more unsettled by them. I would rather have $\imath$ in there. The difference? … More taste than anything sound. I prefer, if I can get away with it, using the square root symbol to mean the positive square root of the thing inside. There is no positive square root of -1, so, pfaugh, away with it. Mere style? All right, well, how do you know whether those $\sqrt{-1}$ terms are meant to be $\imath$ or its additive inverse, $-\imath$? How do you know they’re all meant to be the same one? See? … As with all style preferences, it’s impossible to be perfectly consistent. I’m sure there are times I accept a big square root symbol over a negative or a complex-valued quantity. But I’m not forced to have it here so I’d rather not. First step:

$\pi = \frac{4\infty}{\imath}\left\{ \left(1 + \imath\right)^{\frac{1}{\infty}} - \left(1 - \imath\right)^{\frac{1}{\infty}} \right\}$

Also dividing by $\imath$ is the same as multiplying by $-\imath$ so the second easy step gives me:

$\pi = -4 \imath \infty \left\{ \left(1 + \imath\right)^{\frac{1}{\infty}} - \left(1 - \imath\right)^{\frac{1}{\infty}} \right\}$

Now the hard part. All those infinities. I don’t like multiplying by infinity. I don’t like dividing by infinity. I really, really don’t like raising a quantity to the one-over-infinity power. Most mathematicians don’t. We have a tool for dealing with this sort of thing. It’s called a “limit”.

Mathematicians developed the idea of limits over … well, since they started doing mathematics. In the 19th century limits got sound enough that we still trust the idea. Here’s the rough way it works. Suppose we have a function which I’m going to name ‘f’ because I have better things to do than give functions good names. Its domain is the real numbers. Its range is the real numbers. (We can define functions for other domains and ranges, too. Those definitions look like what they do here.)

I’m going to use ‘x’ for the independent variable. It’s any number in the domain. I’m going to use ‘a’ for some point. We want to know the limit of the function “at a”. ‘a’ might be in the domain. But — and this is genius — it doesn’t have to be. We can talk sensibly about the limit of a function at some point where the function doesn’t exist. We can say “the limit of f at a is the number L”. I hadn’t introduced ‘L’ into evidence before, but … it’s a number. It has some specific set value. Can’t say which one without knowing what ‘f’ is and what its domain is and what ‘a’ is. But I know this about it.

Pick any error margin that you like. Call it ε because mathematicians do. However small this (positive) number is, there’s at least one neighborhood in the domain of ‘f’ that surrounds ‘a’. Check every point in that neighborhood other than ‘a’. The value of ‘f’ at all those points in that neighborhood other than ‘a’ will be larger than L – ε and smaller than L + ε.

Yeah, pause a bit there. It’s a tricky definition. It’s a nice common place to crash hard in freshman calculus. Also again in Intro to Real Analysis. It’s not just you. Perhaps it’ll help to think of it as a kind of mutual challenge game. Try this.

1. You draw whatever error bar, as big or as little as you like, around ‘L’.
2. But I always respond by drawing some strip around ‘a’.
3. You then pick absolutely any ‘x’ inside my strip, other than ‘a’.
4. Is f(x) always within the error bar you drew?

Suppose f(x) is. Suppose that you can pick any error bar however tiny, and I can answer with a strip however tiny, and every single ‘x’ inside my strip has an f(x) within your error bar … then, L is the limit of f at a.

Again, yes, tricky. But mathematicians haven’t found a better definition that doesn’t break something mathematicians need.

To write “the limit of f at a is L” we use the notation:

$\displaystyle \lim_{x \to a} f(x) = L$

The ‘lim’ part probably makes perfect sense. And you can see where ‘f’ and ‘a’ have to enter into it. ‘x’ here is a “dummy variable”. It’s the falsework of the mathematical expression. We need some name for the independent variable. It’s clumsy to do without. But it doesn’t matter what the name is. It’ll never appear in the answer. If it does then the work went wrong somewhere.

What I want to do, then, is turn all those appearances of ‘∞’ in Wronski’s expression into limits of something at infinity. And having just said what a limit is I have to do a patch job. In that talk about the limit at ‘a’ I talked about a neighborhood containing ‘a’. What’s it mean to have a neighborhood “containing ∞”?

The answer is exactly what you’d think if you got this question and were eight years old. The “neighborhood of infinity” is “all the big enough numbers”. To make it rigorous, it’s “all the numbers bigger than some finite number that let’s just call N”. So you give me an error bar around ‘L’. I’ll give you back some number ‘N’. Every ‘x’ that’s bigger than ‘N’ has f(x) inside your error bars. And note that I don’t have to say what ‘f(∞)’ is or even commit to the idea that such a thing can be meaningful. I only ever have to think directly about values of ‘f(x)’ where ‘x’ is some real number.

So! First, let me rewrite Wronski’s formula as a function, defined on the real numbers. Then I can replace each ∞ with the limit of something at infinity and … oh, wait a minute. There’s three ∞ symbols there. Do I need three limits?

Ugh. Yeah. Probably. This can be all right. We can do multiple limits. This can be well-defined. It can also be a right pain. The challenge-and-response game needs a little modifying to work. You still draw error bars. But I have to draw multiple strips. One for each of the variables. And every combination of values inside all those strips has give an ‘f’ that’s inside your error bars. There’s room for great mischief. You can arrange combinations of variables that look likely to break ‘f’ outside the error bars.

So. Three independent variables, all taking a limit at ∞? That’s not guaranteed to be trouble, but I’d expect trouble. At least I’d expect something to keep the limit from existing. That is, we could find there’s no number ‘L’ so that this drawing-neighborhoods thing works for all three variables at once.

Let’s try. One of the ∞ will be a limit of a variable named ‘x’. One of them a variable named ‘y’. One of them a variable named ‘z’. Then:

$f(x, y, z) = -4 \imath x \left\{ \left(1 + \imath\right)^{\frac{1}{y}} - \left(1 - \imath\right)^{\frac{1}{z}} \right\}$

Without doing the work, my hunch is: this is utter madness. I expect it’s probably possible to make this function take on many wildly different values by the judicious choice of ‘x’, ‘y’, and ‘z’. Particularly ‘y’ and ‘z’. You maybe see it already. If you don’t, you maybe see it now that I’ve said you maybe see it. If you don’t, I’ll get there, but not in this essay. But let’s suppose that it’s possible to make f(x, y, z) take on wildly different values like I’m getting at. This implies that there’s not any limit ‘L’, and therefore Wronski’s work is just wrong.

Thing is, Wronski wouldn’t have thought that. Deep down, I am certain, he thought the three appearances of ∞ were the same “value”. And that to translate him fairly we’d use the same name for all three appearances. So I am going to do that. I shall use ‘x’ as my variable name, and replace all three appearances of ∞ with the same variable and a common limit. So this gives me the single function:

$f(x) = -4 \imath x \left\{ \left(1 + \imath\right)^{\frac{1}{x}} - \left(1 - \imath\right)^{\frac{1}{x}} \right\}$

And then I need to take the limit of this at ∞. If Wronski is right, and if I’ve translated him fairly, it’s going to be π. Or something easy to get π from.

I hope to get there next week.

## L’Hopital’s Rule Without End: Is That A Thing?

I was helping a friend learn L’Hôpital’s Rule. This is a Freshman Calculus thing. (A different one from last week, it happens. Folks are going back to school, I suppose.) The friend asked me a point I thought shouldn’t come up. I’m certain it won’t come up in the exam my friend was worried about, but I couldn’t swear it wouldn’t happen at all. So this is mostly a note to myself to think it over and figure out whether the trouble could come up. And also so this won’t be my most accessible post; I’m sorry for that, for folks who aren’t calculus-familiar.

L’Hôpital’s Rule is a way of evaluating the limit of one function divided by another, of f(x) divided by g(x). If the limit of $\frac{f(x)}{g(x)}$ has either the form of $\frac{0}{0}$ or $\frac{\infty}{\infty}$ then you’re not stuck. You can take the first derivative of the numerator and the denominator separately. The limit of $\frac{f'(x)}{g'(x)}$ if it exists will be the same value.

But it’s possible to have to do this several times over. I used the example of finding the limit, as x grows infinitely large, where f(x) = x2 and g(x) = ex. $\frac{x^2}{e^x}$ goes to $\frac{\infty}{\infty}$ as x grows infinitely large. The first derivatives, $\frac{2x}{e^x}$, also go to $\frac{\infty}{\infty}$. You have to repeat the process again, taking the first derivatives of numerator and denominator again. $\frac{2}{e^x}$ finally goes to 0 as x gets infinitely large. You might have to do this a bunch of times. If f(x) were x7 and g(x) again ex you’d properly need to do this seven times over. With experience you figure out you can skip some steps. Of course students don’t have the experience to know they can skip ahead to the punch line there, but that’s what the practice in homework is for.

Anyway, my friend asked whether it’s possible to get a pattern that always ends up with $\frac{0}{0}$ or $\frac{\infty}{\infty}$ and never breaks out of this. And that’s what’s got me stuck. I can think of a few patterns that would. Start out, for example, with f(x) = e3x and g(x) = e2x. Properly speaking, that would never end. You’d get an infinity-over-infinity pattern every derivative you took. Similarly, if you started with $f(x) = \frac{1}{x}$ and $g(x) = e^{-x}$ you’d never come to an end. As x got infinitely large both f(x) and g(x) would go to zero and all their derivatives would be zero over and over and over and over again.

But those are special cases. Anyone looking at what they were doing instead of just calculating would look at, say, $\frac{e^{3x}}{e^{2x}}$ and realize that’s the same as $e^x$ which falls out of the L’Hôpital’s Rule formulas. Or $\frac{\frac{1}{x}}{e^{-x}}$ would be the same as $\frac{e^x}{x}$ which is an infinity-over-infinity form. But it takes only one derivative to break out of the infinity-over-infinity pattern.

So I can construct examples that never break out of a zero-over-zero or an infinity-over-infinity pattern if you calculate without thinking. And calculating without thinking is a common problem students have. Arguably it’s the biggest problem mathematics students have. But what I wonder is, are there ratios that end up in an endless zero-over-zero or infinity-over-infinity pattern even if you do think it out?

And thus this note; I’d like to nag myself into thinking about that.

## Also, Some Mathematics Answers, Whatever Those Are

Solved Problems In Mathematics appears to be a new mathematics blog. It’s a few posts old, but I’m charmed by it. Each post takes some problem and shows how to solve it. It’s been heavy on calculus so far. Many limits, a bunch of evaluate-the-integral problems, and so on. But it may be a good bit of reading if you want to brush up on stuff you once knew, like how to calculate the length of peculiar curves. Or if you have a calculus problem you need to solve and haven’t got any idea what to do. A similar problem may give you a good hint.

Less generally, John Quintanilla’s Mean Green Math has looked at one specific calculus problem, a limit, in several different ways. He’s grouped the essays about that together. Most mathematics problems, once you’re away from the cutting edge of human knowledge, have many different possible solutions. Here’s an example of such.

And, what the heck. Math With Bad Drawings hardly needs my publicity. But a recent post discusses the challenge of saying just what mathematics is. The comments get into some decent discussion about what mathematics is. And I’m not sure what mathematics is. There are many facets which seem barely to have anything to do with one another; but they do all feel related in some way. It’s hard to say just what unites all this.

## Error

This is one of my A to Z words that everyone knows. An error is some mistake, evidence of our human failings, to be minimized at all costs. That’s … well, it’s an attitude that doesn’t let you use error as a tool.

An error is the difference between what we would like to know and what we do know. Usually, what we would like to know is something hard to work out. Sometimes it requires complicated work. Sometimes it requires an infinite amount of work to get exactly right. Who has the time for that?

This is how we use errors. We look for methods that approximate the thing we want, and that estimate how much of an error that method makes. Usually, the method involves doing some basic step some large number of times. And usually, if we did the step more times, the estimate of the error we make will be smaller. My essay “Calculating Pi Less Terribly” shows an example of this. If we add together more terms from that Leibniz formula we get a running total that’s closer to the actual value of π.

## Quick Little Calculus Puzzle

fluffy, one of my friends and regular readers, got to discussing with me a couple of limit problems, particularly, ones that seemed to be solved through L’Hopital’s Rule and then ran across some that don’t call for that tool of Freshman Calculus which you maybe remember. It’s the thing about limits of zero divided by zero, or infinity divided by infinity. (It can also be applied to a couple of other “indeterminate forms”; I remember when I took this level calculus the teacher explaining there were seven such forms. Without looking them up, I think they’re $\frac00, \frac{\infty}{\infty}, 0^0, \infty^{0}, 0^{\infty}, 1^{\infty}, \mbox{ and } \infty - \infty$ but I would not recommend trusting my memory in favor of actually studying for your test.)

Anyway, fluffy put forth two cute little puzzles that I had immediate responses for, and then started getting plagued by doubts about, so I thought I’d put them out here for people who want the recreation. They’re both about taking the limit at zero of fractions, specifically:

$\lim_{x \rightarrow 0} \frac{e^x}{x^e}$

$\lim_{x \rightarrow 0} \frac{x^e}{e^x}$

where e here is the base of the natural logarithm, that is, that number just a little high of 2.71828 that mathematicians find so interesting even though it isn’t pi.

The limit is, if you want to be exact, a subtly and carefully defined idea that took centuries of really bright work to explain. But the first really good feeling that I really got for it is to imagine a function evaluated at the points near but not exactly at the target point — in the limits here, where x equals zero — and to see, if you keep evaluating x very near zero, are the values of your expression very near something? If it does, that thing the expression gets near is probably the limit at that point.

So, yes, you can plug in values of x like 0.1 and 0.01 and 0.0001 and so on into $\frac{e^x}{x^e}$ and $\frac{x^e}{e^x}$ and get a feeling for what the limit probably is. Saying what it definitely is takes a little more work.