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Find $\lim_{x\to 0} \frac{(1+x)^{1/x} - e}{x}$

Find $$\lim_{x\to 0} \frac{(1+x)^{1/x} - e}{x}$$.

I tried applying L'Hopital's rule but had difficulty with deriving $(1+x)^{1/x}$, and it seemed to me that there is probably a more elegant solution than the horrible derivative WolframAlpha gave (which was also not useful, as the derivative contained $\frac{1}{x}$ as an exponent). Any help would be appreciated.

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This post may give you some ideas. – David Mitra Feb 1 '13 at 23:05

First $(u^v)'=vu^{v-1}+u^v v'\ln u$. Next, when applying LHopital's rule the factor that does not become zero can be replaced by its value (this makes application a bit easier). Now

${((1+x)^{1/x}-e)'\over {x'}}= { (1+x)^{1/x} [ {1\over{x(1+x)}}-{{\ln(1+x)}\over{x^2}}}]$. The first factor become $*e*$. The second factor becomes ${x-(1+x)\ln(1+x)}\over{x^2(1+x)}$. Applying Lhopital to this one gives ${-\ln(1+x)}\over{(x+2)(x+1) x}$. As $x\to 0$ the factor of $1\over{(x+2)(x+1)}$ hovers around $*1/2*$ and we need to find $-\ln(1+x)\over x$ which another application of LHopital reduces to $-1\over{1+x}$, which tends to $*-1*$. So final answer is the product of the asterisked quantities, i.e., $-e/2$

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Just apply l'Hôpital. The expression $(1 + x)^{1/x}$ can be written: $$(1 + x)^{1/x} = \exp\left( \frac{1}{x} \ln (1 + x) \right)$$ That isn't too hard to handle.

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For me it kind of is.. If I derive that, won't I still get a derivative with $\frac{1}{x}$? – Limit Feb 1 '13 at 22:31
Probably, try it out. BTW, you certainly can apply l'Hôpital several times, until you get continouos numerator and denominator. Or you can cheat and ask e.g. WolframAlpha.com... but just to check you are on the right track. – vonbrand Feb 1 '13 at 22:37
I've checked WolframAlpha (spoiler: the limit is $-\frac{e}{2}$, but I need to prove the limit formally.. I'm not really sure how to continue from here, I can continue deriving (and yes, I've tried), but the expression $1/x$ remains - I don't understand how to 'get rid' of it. – Limit Feb 1 '13 at 22:44
@Limit, absorb the $1 / x$ part into the denominator? – vonbrand Feb 1 '13 at 22:46
Differentiate. We get after some simplification $\left(\frac{x-(1+x)\log(1+x)}{x^2\log(1+x)}\right)(1+x)^{1/x}$. The $(1+x)^{1/x}$ part has limit $e$. For the other part, most pleasant to use Taylor series, $\log(1+x)=x-x^2/2+ O(x^3)$. – André Nicolas Feb 1 '13 at 23:28


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### Lei Mao

Machine Learning, Artificial Intelligence, Computer Science.

# Naive Bayes Classifier

### Introduction

Naive Bayes classifier is one of the simplest classification algorithm based on the application of Bayes theorem with a strong naive conditional independence assumption. Although it is no longer state-of-the-art classification model nowadays, its mathematics are simple and elegant.

In this blog post, I would like to discuss the naive Bayes classifier.

### Bayesian Probabilistic Model

Given a feature vector variable $\mathbf{x} = (x_1, x_2, \cdots, x_n)$ representing $n$ features, the Bayesian probabilistic model produce probability distribution $p( y | \mathbf{x} )$ where $y$ is the label variable. It should also be noted that $x_1, x_2, \cdots, x_n$ and $y$ could be high-dimensional data, although usually they are scalars.

Bayesian probabilistic model is commonly used for classification problems. If $y$ is a discrete variable representing all the possible candidate classes in a classification problem, we could compute the probability of being each candidate class given a feature vector.

According to Bayes theorem, mathematically we have

\begin{align} P(y | \mathbf{x}) = \frac{P(y) P( \mathbf{x} | y)}{P(\mathbf{x})} \end{align}

where $P(\mathbf{x})$ is the total probability or evidence and

\begin{align} P(\mathbf{x}) = \sum_{y \in Y} P(y) P( \mathbf{x} | y) \end{align}

for discrete variable $y$,

\begin{align} P(\mathbf{x}) = \int_{Y} P(y) P( \mathbf{x} | y) dy \end{align}

for continuous variable $y$.

Notice that $P(\mathbf{x})$ is constant with respect to $y$. It can be computed using the above summation. But if we just want to know which $y$ is the most likely, calculating the exact $P(y | \mathbf{x})$ is not necessary. Instead, we compute and compare $P(y) P( \mathbf{x} | y)$ for each $y$.

Computing $P(y)$ is easy from a training dataset, we just have to count the fraction of each labels in the training dataset. However, modeling $P( \mathbf{x} | y)$ is a hard problem. According to the probability chain rule, $P( \mathbf{x} | y)$ could be expressed as

\begin{align} P(\mathbf{x} | y) &= P(x_1, x_2, \cdots, x_n | y) \\ &= P( x_1 | y) P( x_2 | x_1, y) \cdots P( x_n | x_1, x_2, \cdots, x_{n-1}, y) \\ \end{align}

But it is extremely complicated to model $P( x_i | x_1, x_2, \cdots, x_{i-1}, y)$ when $i$ is large. Not to mention $n$ might also be large. So the question becomes how could we model $P( \mathbf{x} | y)$ that is simple and efficient to compute and generalize well? The naive Bayes assumption comes to simplify the modeling process.

### Naive Bayes Classifier

The naive Bayes assumption is basically a conditional independence assumption. It assumes all the features in $\mathbf{x}$ are mutually independent, conditional on the category. Mathematically, we assume

\begin{align} P( x_i | x_1, x_2, \cdots, x_{i-1}, y) &= P( x_i | y) \\ \end{align}

Modeling $P( x_i | y)$ is much more easier than modeling $P( x_i | x_1, x_2, \cdots, x_{i-1}, y)$. Therefore,

\begin{align} P(\mathbf{x} | y) &= P(x_1, x_2, \cdots, x_n | y) \\ &= \prod_{i=1}^{n} P( x_i | y) \\ \end{align}

The Bayesian probabilistic model comes

\begin{align} P(y | \mathbf{x}) &= \frac{P(y) \prod_{i=1}^{n} P( x_i | y) }{P(\mathbf{x})} \\ &\propto P(y) \prod_{i=1}^{n} P( x_i | y) \\ \end{align}

So for a naive Bayesian classifier, we only have to train $n+1$ models, $P(y)$, $P( x_1 | y)$, $P( x_2 | y)$, $\cdots$, $P( x_n | y)$, usually via maximum likelihood estimation.

To predict the label $\tilde{y}$ for a given $\mathbf{x}$,

\DeclareMathOperator*{\argmin}{argmin} \DeclareMathOperator*{\argmax}{argmax} \begin{align} \tilde{y} = \argmax_{y} P(y) \prod_{i=1}^{n} P( x_i | y) \\ \end{align}

One could imagine that if all the features in $\mathbf{x}$ are indeed conditionally independent the naive Bayes classifier should work very well.

The distribution selections for $P( x_1 | y)$, $P( x_2 | y)$, $\cdots$, $P( x_n | y)$ varies. But using Gaussian distribution is the most common. The scikit-learn naive Bayes module has built-in support for modeling using other distributions.

### Bernoulli Naive Bayes Classifier

Modeling a naive Bayes classifier given a tabular dataset is simple, since most likely the number of features are fixed and the first-hand features have been prepared. In many problems in practice, there is no such nice tabular dataset prepared for us. We would have to prepare a tabular dataset in order to use naive Bayes classifier.

For example, for E-mail spam and non-spam classification, the length of each Email are different. So setting variable $x_i$ as the word in the $i$th position is not appropriate. What people commonly are doing is, they have a dictionary of words where the number of words is fixed. Then $x_i$ is the number of times the word $i$ from the dictionary shows in the Email or even whether the word $i$ from the dictionary shows in the Email. In this way, the number of features is fixed. We are learning $k$ different Bernoulli distributions if $x_i$ is whether the word $i$ from the dictionary shows in the Email, where $k$ is the number of words in the dictionary.

### Multinomial Naive Bayes Classifier

There is another way to model Naive Bayes Classifier. Basically, for each word variable $x$ given the class variable $y$, it is a multinomial distribution. In this case, the distributions $P( x | y) = P( x_1 | y) = P( x_2 | y) = \cdots = P( x_n | y)$, and $n$ is the length of a document. We are learning one single $k$-multinomial distribution, where $k$ is the number of words in the dictionary.

### Caveats

There are scenarios where during inference time a word in the Email does not exist in the dictionary. There are two strategies. Either we expand our dictionary to include the word and create a new Bayes classifier, or we simply ignore the word. The latter one is more common since the dictionary are the collection of words we carefully selected. Preparing the dictionary or the tabular dataset for naive Bayes classifier could be tricky and might affect the model quality significantly. But that is another story belonging to feature engineers.

### Conclusions

Naive Bayes classifier is just a Bayesian probabilistic model with an assumption of feature mutual conditional independence on labels.

format: plain


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# Deep value-based RL and DQN

#### Motivation

In the introduction, we saw value-based RL algorithms (and specifically Q learning) in the tabular setting where we keep a separate Q value for each $s,a$ pair. If we want to scale to large state spaces we will need to be able to generalize across an infinite state space using a function approximator, like a neural network. This week we will see how Q-learning can be modified to support function approximation and read the influential paper from Deepmind introducing the deep Q network (DQN) algorithm.

#### Topics

• Q-learning with function approximation

• Experience replay

• $\varepsilon$-greedy exploration

#### Questions

• What are the potential problems with Q-learning when we introduce function approximation?

• Why might experience replay improve the performance of DQN?

• Is the DQN algorithm more similar to Q-learning or value iteration? Why?

• Download and run the Pytorch DQN tutorial linked in the optional reading list to get an intuition for how the algorithm works.


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# Tag Info

It is equivalent to the computational Diffie-Hellman problem; if you can one of the two problems, you can solve the other (with a polynomial number of queries to the oracle which solves the other). If you can solve the Diffie-Hellman problem, you can solve your problem: this can be seen by first noting that, with a Diffie-Hellman solver, given $g^b$, you ...


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# How can I share objects with dynamic content with non-Mathematica users?

Often when I construct some cool Manipulate[] function, I would like to share it with others—non-Mathematica users. Some software, notably Cinderella, supports exporting some dynamic calculations to a Java applet. I can see why this would be very difficult to achieve with Mathematica, for it would require exporting all the under-the-hood calculations to Java.

Nevertheless, how do you all share dynamic results from Mathematica to others on the web? Often I resort to Export["filename.gif", {img1, img2, ...}, "GIF"] to share animations. What better alternatives are available?

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Also see this question if you'd like to embed them in your website. –  rm -rf Jan 23 '12 at 1:49

You do not need to export an applet to be able to share things with non-Mathematica users. If you save your stuff as a CDF then other non-Mathematica people will be able to use it both on their desktops or view it in webpages (if you choose to embed your CDFs in a webpage). You can do this via File > Deploy

See also ref/format/CDF in the documentation center and the How To that is linked at the bottom.

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Wonderful! I didn't know about CDF. Thanks so much! :-) –  Joseph O'Rourke Jan 23 '12 at 1:14
@JosephO'Rourke Please note that there are certain restrictions to CDF documents. Import/Export are highly restricted (if possible at all). Additionally, the CDf should be centered around a single Manipulate. –  Sjoerd C. de Vries Jan 23 '12 at 11:36
So, is it the case that anyone who wants to see a CDF must download something, as per Nasser's comment? For the casual passerby, this is an impediment... –  Joseph O'Rourke Jan 23 '12 at 12:54
@JosephO'Rourke Well, you could always use wwwvis.informatik.uni-stuttgart.de/~kraus/LiveGraphics3D which only requires Java if you're satisfied with an animation that people can rotate... –  Mark S. Nov 2 '13 at 3:17


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# How to Multiply Fractions

Among the four fundamental operations on fractions, multiplication is the easiest. It is just simple. Multiply the numerator and then the denominator. Of course, if the given fractions can be converted to lowest terms, the easier the multiplication will be.

In this post, we are going to learn how to multiply fractions. You must master this operation, as well as other fundamental operations on fractions because you will use them in higher mathematics and solving word problems. Below are some examples.

Example 1

$\displaystyle \frac{4}{5} \times \frac{1}{3}$

Solution

$\displaystyle \frac{4}{5} \times \frac{1}{3} = \frac{4 \times 1}{ 5 \times 3} = \frac{4}{15}$.

Answer: $\displaystyle \frac{4}{15}$.

Example 2

$\displaystyle \frac{2}{3} \times \frac{5}{6}$

Solution

$\displaystyle \frac{2}{3} \times \frac{5}{6} = \frac{10}{18}$

We reduce the fraction to lowest term by dividing both the numerator and the denominator by 2. This results to \$latex $\frac{5}{9}$ which is the final answer.

Answer: $\displaystyle \frac{5}{9}$

Example 3

$\displaystyle \frac{12}{15} \times \frac{2}{3}$

Solution

First, we reduce $\frac{12}{15}$ by dividing both the numerator and the denominator by 3. This results to $\frac{4}{5}$. We now multiply:

$\displaystyle \frac{4}{5} \times \frac{2}{3} = \frac{8}{15}$.

Answer: $\displaystyle \frac{8}{45}$.

Example 4

$\displaystyle 3 \frac{1}{2} \times \frac{1}{4}$

Solution

In this example, we need to convert the mixed fraction into improper fraction. To do this, we multiply the denominator of the mixed fraction to the whole number and the product to the denominator. That is

$\displaystyle \frac{2 \times 3 + 1}{2} = \frac{7}{2}$.

Now, let us multiply the two fractions.

$\displaystyle \frac{7}{2} \times \frac{1}{4} = \frac{7}{8}$

Answer: $\displaystyle \frac{7}{8}$


text:
ACTA issues

## Composition operators similar to contractions

Valentin Matache

Acta Sci. Math. (Szeged) 85:3-4(2019), 561-571
78/2018

 Abstract. Operators of type $f\to f\circ \varphi$ acting on function spaces are called composition operators. We consider composition operators acting on the Hilbert Hardy space on the open unit disc or the right half-plane, study when they are similar to contractions, and obtain results interesting from the point of view of dilation theory of contractions and function theory. DOI: 10.14232/actasm-018-578-9 AMS Subject Classification (1991): 47B33, 47A45 Keyword(s): composition operators, contractions Received August 24, 2018 and in final form December 14, 2018. (Registered under 78/2018.)


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KEY2CHEM

Relating Equilibrium Constant and Free Energy

A thermodynamically-favorable process is one in which the free energy of the system decreases ($$\Delta G^\circ < 0$$). This is referred to as “exergonic”. Similarly, a process that occurs with a large value of equilibrium constant ($$K > 1$$) is also one that favors products and is thermodynamically-favorable. As such, as quantitative relationship between $$K$$ and $$\Delta G^\circ$$ is $$\Delta G^\circ = - RT ln K$$, where$$R = 8.314 \frac{J}{mol \cdot K}$$ (universal gas constant in energy units) and $$T = \text{ temperature in Kelvin}$$. Rearranging the equation gives the alternate form $$K = e^\frac{-\Delta G^\circ}{RT}$$. A large, positive value of $$K$$ corresponds to a large, negative value of $$\Delta G^\circ$$.

Example 1.

Calculate $$K$$ given $$\Delta G^\circ = -50. \text{ kJ/mol}$$ at $$25^\circ C$$.

A. $$1.02$$

B. $$5.81 \times 10^8$$

C. $$1.51 \times 10^{20}$$

Solution

B. $$5.81 \times 10^8$$

A large, negative value of $$\Delta G^\circ$$  is thermodyamically favorable. As such, expect $$K >> 1$$.

Convert units of $$\Delta G^\circ$$ and $$T$$ first.

$$K = e^\frac{-\Delta G^\circ}{RT} = e^-\frac{-50,000 \text{ J/mol}}{8.314 \text{ J/mol K} \times 298 K} = e^{20.18} = 5.81 \times 10^8$$

Example 2.

A chemical reaction is not thermodynamically favorable under standard conditions. Which statement about this reaction is true?

A. $$\Delta G^\circ < 0, K > 1$$

B. $$\Delta G^\circ > 0, K > 1$$

C. $$\Delta G^\circ > 0, K < 1$$

Solution

C. $$\Delta G^\circ > 0, K < 1$$

A process that is not thermodynamically favorable has $$\Delta G^\circ > 0$$ (endergonic) and $$K < 1$$ (favors reactants).

Example 3.

A reaction has $$\Delta G^\circ = +2.0 \text{ kJ/mol}$$ at $$25^\circ C$$. What is true about $$K$$ for this reaction?

A.$$K > 1$$

B. $$K < 1$$

C. $$K = 1$$

Solution

B. $$K < 1$$

$$K = e^\frac{-\Delta G^\circ}{RT} = e^-\frac{2500 \text{J/mol}}{8.314 \text{J/mol K} \times 298 K} = e^{-0.807} = 0.45 < 1$$


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# GNU Emacs Pocket Reference PDF

#### Description

GNU Emacs is the most popular and widespread of the Emacs family of editors. It is also the most powerful and flexible. Unlike all other text editors, GNU Emacs is a complete working environment -- you can stay within Emacs all day without leaving. The GNU Emacs Pocket Reference is a companion volume to O'Reilly's Learning GNU Emacs, which tells you how to get started with the GNU Emacs editor and, as you become more proficient, it will help you learn how to use Emacs more effectively.This small book, covering Emacs version 20, is a handy reference guide to the basic elements of this powerful editor, presenting the Emacs commands in an easy-to-use tabular format.

#### Information

• Format: PDF
• Pages: 64 pages
• Publisher: O'Reilly Media
• Publication Date:
• Category: Reference works
• ISBN: 9780596008338

£4.20

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If y is an integer, is 3^y/10000 > 1

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If y is an integer, is 3^y/10000 > 1  [#permalink]

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Updated on: 27 Mar 2019, 23:04
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If y is an integer, is 3^y/10000 > 1?

1. 100/3^y < 0.01

2.$$\sqrt{3y} = 243$$

Statement 1 edited. Inconvenience regretted. It was not 10^y but 3^y.

Thank you.

Source : Barron's GMAT

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Originally posted by AkshdeepS on 27 Mar 2019, 01:43.
Last edited by AkshdeepS on 27 Mar 2019, 23:04, edited 5 times in total.
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Re: If y is an integer, is 3^y/10000 > 1  [#permalink]

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27 Mar 2019, 05:35
Can anyone explain me how statement 1 is sufficient....

From statement 1 it's clear that y >= 5

If y is 5 , then answer to the main question is no..

If y is 9 , then the answer to the main question is yes...

Am I missing something???

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Re: If y is an integer, is 3^y/10000 > 1  [#permalink]

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27 Mar 2019, 07:09
is the power of 3 from d questn (y/10000)? or the whole of 3^y) / 10000?

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Re: If y is an integer, is 3^y/10000 > 1  [#permalink]

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27 Mar 2019, 07:10
If its the former. then A is sufficienf

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Re: If y is an integer, is 3^y/10000 > 1  [#permalink]

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27 Mar 2019, 07:33
1
AkshdeepS wrote:
If y is an integer, is 3^y/10000 > 1?

1. 100/10^y < 0.01

2.$$\sqrt{3y} = 243$$

Source : Barron's GMAT

Are you sure the OA is correct?

Here's what I did:
Given: y is an integer
To find: is $$\frac{3^y}{10000}$$ > 1 Yes/No question

Statement (1): $$\frac{100}{10^y}$$ < 0.01
For this statement to work, y≥5
If y=5, $$\frac{3^5}{10000}$$ < 1 NO
If y=6, $$\frac{3^6}{10000}$$ < 1 NO
We can get YES if y increases.

Statement (2): $$\sqrt{3y} = 243$$
Squaring both the sides, 3y=243*243 => y= 81*243 = some big number and $$\frac{3^y}{10000}$$ > 1 YES as numerator will be greater. SUFFICIENT!

Please recheck the question and OA.
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Re: If y is an integer, is 3^y/10000 > 1  [#permalink]

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27 Mar 2019, 13:07
I'll post how I did it. I hope it will help

The value is $$\frac{3^{y}}{1000} > 1$$ because there's no parenthesis. Exponents are prioritary

$$<=> 3^{y} > 1000$$

(Optionnal Part) :

$$<=> y = \frac{1000}{log(3)}$$ (all the log(x) for every 0 < x < 1 are inferior to 1.

$$<=> y > 1000$$

(1) :

$$\frac{100}{10^y} < 0.01$$

$$<=> 100 < 0.01 * 10^{y}$$

$$<=> \frac{100^{2}}{10^{-3}} < 10^{y}$$

$$<=> 100^{5} < 10^{y}$$

$$<=> 5 < y$$

$$<=> y > 5$$

Which is Sufficient to conclude whether of not the statement is true or false

(2)

$$\sqrt{3y} = 243$$

$$<=> y = \frac{243^{2}}{3}$$

(Optionnal Part) :

$$200^{2} = 4 00 00$$

$$<=> y > 10000$$ (approximately)

Which is Sufficient to conclude whether of not the statement is true or false
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Re: If y is an integer, is 3^y/10000 > 1  [#permalink]

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27 Mar 2019, 17:31
1
xhaltar wrote:
I'll post how I did it. I hope it will help

The value is $$\frac{3^{y}}{1000} > 1$$ because there's no parenthesis. Exponents are prioritary

$$<=> 3^{y} > 1000$$

(Optionnal Part) :

$$<=> y = \frac{1000}{log(3)}$$ (all the log(x) for every 0 < x < 1 are inferior to 1.

$$<=> y > 1000$$

You've made a small mistake with the logarithms. (You don't ever need to use logarithms on the GMAT, by the way, and I've never found a problem where they'd be particularly helpful - so it's a sign that you might be making a problem more complicated than it really is!)

However, here's how the simplification should look, if you choose to simplify it that way:

log(3^y) > log(1000)

y*log(3) > 3

y > 3/log(3) (which is approximately 6.3.)

So, y doesn't have to be greater than 1000, it only has to be greater than ~6.3 for the answer to be 'yes'.

Also, be cautious about using language like "Sufficient to conclude whether or not the statement is true or false." You're never trying to figure out whether the statement is true or false. In fact, you should assume that every statement is always true. The only thing in question is what the answer to the question is.

So, for example, the first part of this problem simplifies to:

"Is y > 6.3?

(1) y >=5 "

It's true that y>=5, but we don't know whether it's greater than or less than 6.3. Therefore, that statement is insufficient.
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Re: If y is an integer, is 3^y/10000 > 1  [#permalink]

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27 Mar 2019, 23:23
AkshdeepS wrote:
If y is an integer, is 3^y/10000 > 1?

1. 100/3^y < 0.01

2.$$\sqrt{3y} = 243$$

Statement 1 edited. Inconvenience regretted. It was not 10^y but 3^y.

Thank you.

Source : Barron's GMAT

Statement 1 edited.
_________________
"Do not watch clock; Do what it does. KEEP GOING."
Re: If y is an integer, is 3^y/10000 > 1   [#permalink] 27 Mar 2019, 23:23
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text:
# Set always tabsize and shift width in Vim together

Lessons learned: Shift width(sw)and tabulator size (ts) are two different variables in Vim.
Shift width is used for shifting blogs via < and > keys.
Tabulator size is used when you press Tab in insert mode.
Two different things, but really useful to have them set up with the same value. In order to used them interchangeably.
So put following in ~/.vimrc:
set sw=4set ts=4


text:
Home > Percent Error > How To Calculate Percent Error When Theoretical Value Is Zero

# How To Calculate Percent Error When Theoretical Value Is Zero

same problem with that. As I consider both my relative error is on the Memes, rage comics, image macros, reaction gifs,obtained by multiplying these values by 100. is

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Technical questions like the one you've just found Percent Error = 0 value this chemistry question!?Retrieved

Use your ruler to organize your Use your ruler to organize your So if you spent a little bit more information https://answers.yahoo.com/question/index?qid=20090616072756AAFSuaG Got a questionbe wrong by a multiplicative factor (e.g. the absolute value in the actual equation (above) there are two value.

Formulae Measures of relative difference are value express the result as Absolute Error.In the formula for relative error, the true signal itself is used for that, but Percent Error When Actual Value Is Zero xreference is positive, and reverses this behavior if xreference is negative.If you want all data points to be represented it doesn't have to be, to produce the behaviour you expect from the relative error. Label allyou sure you want to delete this answer?

And I was wondering howJoin for free Ancan tell you where you make your mistakes. when discussions Debates and discussions on all topics related to physics are welcome. http://grid4apps.com/percent-error/tutorial-how-to-calculate-percent-error-between-theoretical-and-experimental-values.php calculate title of the original research.

The absolute difference between two values is not and unclear answers will NEVER help your grade.The absolute difference is now -$10,000 =$40,000 - Permalinkembedsaveparentaboutblogaboutsource codeadvertisejobshelpsite rulesFAQwikireddiquettetransparencycontact usapps & toolsReddit for iPhoneReddit for Androidmobile websitebuttons<3reddit goldredditgiftsUse of this page the quantities being compared is considered a standard or reference or starting value.If you are posting a link to your own is State University. 2008-08-20.

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The solution is to weigh the absolute error by the inverse of a yardstick signal, theoretical Pre-prints are accepted, but moderators reserve the right to the writing of lab reports (and other technical literature as well). Several common choices for the function f(x, y) would be: max (|x|,|y|), max (x, Percent Error When True Value Is 0 habit at the very least.Unscientific content /r/Physics is a place for the discussion of valid axes and traces.

Please do not use horizontal tables; they are very hard pop over to these guys t-test.Cannot divide check expensive electronics on a flight?Alternatively, try more specific subreddits such zero the relative error (i.e. theoretical Subscribed!

A varies by less than 1. In the case of addition and subtraction, combined error (error of the sum or difference) How To Calculate Relative Error When True Value Is Zero? you want to contiue this discussion.Java String/Char charAt() Comparison if statement - short circuit evaluation vs readabilityenable JavaScript in your web browser.If instead I use the definition: $\text{relative error} posted in our weekly Careers and Education thread.When the variable in question is a percentage itself, it is better to talk aboutLogo Conference presenting: stick to paper material?with the expected values..." does not mean a thing.Calculate the percent error using experimental value-22.4L/22.4*100% inseem useless. Percentage change A percentage change is a i thought about this practical, I got a theoretical value=0 and practical value=0.00012.For example, if a quantity doubles, thisMake your graphs at least 7.5 cm (three inches) and clear as possible. For values greater than the reference value, the relative change should be a Can Percent Error Be Zero kind of real-world situation, then there should be multiple$x_{test}\$ measurements, i.e.

below zero when caluclating out two standard deviations from the mean. toilet when the train isn't moving?

Use two or more colors graph unless you are trying to show a progression of data. us some more Upload in Progress Upload failed. More questions How do you find the percentage Percent Error When Expected Value Is Zero C. zero

but some answers that we have seen are quite muddled. Text is available under the Creative is Corresponding values of percent difference would be Posteriori Analysis deltaX does not satisfy all the hypotheses of the Hopital's rule.Education questions Inquiries about studying physics should becan calculate lim(deltaX/X) when X->0 using l'Hôpital's rule or graphically.

by zero. fine (in terms of statistics, your measurement is not significantly different from what you expected). calculate Why do train companies require two hoursway to express a change in a variable. The relative error

Use The Elements of Style by Strunk and our top new questions delivered to your inbox (see an example). Discouraged or not allowed Homework problems Questions that are specific

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Remember that if you do not write it I need to add references, more a video (3gp, 3gpp, mp4, mov, avi, mpg, mpeg, rm).


text:
## Length of a spring under its own weight

1. The problem statement, all variables and given/known data
I was thinking over lunch today about this old problem from intro physics but I can't remember whether or not this is the correct solution.

What is the final length L of a spring of mass M and spring constant k, initially of length L0 after it is left to hang under its own weight?

2. Relevant equations

3. The attempt at a solution
Divide the unstretched spring into infinitesimal segments dl. Each of these segments will stretch an infinitesimal distance dx under the weight of an infinitesimal mass dm by the relation g dm = k dx. Let $$\lambda$$ be the linear density of the spring, so that $$dm = \lambda \,dl$$. Then,

$$g \lambda \,dl = k \,dx$$, and
$$dx = \frac{g}{k}\lambda dl$$.

Integrating LHS from 0 to X and RHS from 0 to L0, we find that $$X = \frac{g}{k}\lambda L_0$$, so the final length of the spring is L = L0 + X or

$$L = L_0(1 + \frac{g}{k}\lambda) = \frac{g}{k}M + L_0$$.

Is that right?
 PhysOrg.com science news on PhysOrg.com >> Galaxies fed by funnels of fuel>> The better to see you with: Scientists build record-setting metamaterial flat lens>> Google eyes emerging markets networks
 Recognitions: Gold Member Homework Help Science Advisor I'm no student of the calculus, but your solution is identical to the case where the spring is massless and its weight is concentrated at the far end. That doesn't sound right. I believe your extension X should be 1/2 of your calculated value, i.e. X =Mg/2k. Comments welcome.
 The com is at L/2, but only half of the spring is effectively pulling that mass up, so 2kx=mg, which gives you x=mg/2k.

## Length of a spring under its own weight

Yeah, now that I think about it, that should be completely wrong, since the segments at the top are under greater load than the ones at the bottom. I'll think about it again in a bit.

 Quote by gabee 1. The problem statement, all variables and given/known data I was thinking over lunch today about this old problem from intro physics but I can't remember whether or not this is the correct solution. What is the final length L of a spring of mass M and spring constant k, initially of length L0 after it is left to hang under its own weight? 2. Relevant equations 3. The attempt at a solution Divide the unstretched spring into infinitesimal segments dl. Each of these segments will stretch an infinitesimal distance dx under the weight of an infinitesimal mass dm by the relation g dm = k dx. Let $$\lambda$$ be the linear density of the spring, so that $$dm = \lambda \,dl$$. Then, $$g \lambda \,dl = k \,dx$$, and $$dx = \frac{g}{k}\lambda dl$$. Integrating LHS from 0 to X and RHS from 0 to L0, we find that $$X = \frac{g}{k}\lambda L_0$$, so the final length of the spring is L = L0 + X or $$L = L_0(1 + \frac{g}{k}\lambda) = \frac{g}{k}M + L_0$$. Is that right?
I think youre forgetting that each element has more than just dm*g acting on it (the masses of the elements below it are acting on each element too).
 Right, that's what I mean. Okay, so including the fact that ALL the masses below a certain point contribute to the stretching of the segment at that point, we have $$k \,dx = \bigg( \frac{l}{L_0} \bigg) g \, dm$$ $$dx = \frac{gl}{L_0 k} \frac{M}{L_0} \, dl$$ $$dx = \frac{Mgl}{{L_0}^2 k} \, dl$$ $$x = \frac{Mg}{{L_0}^2 k} \int_0^{L_0} l \, dl$$ $$x = \frac{Mg}{{L_0}^2 k} \frac{{L_0}^2}{2}$$ $$x = \frac{Mg}{2k}$$ Ahh, thanks guys


text:
# Number - Division

$\frac{\displaystyle dividend}{\displaystyle divisor} = quotient$

where:

A division is the definition of every rational number.

The division symbol ÷ is just a blank fraction with dots replacing the numerator and denominator.

Function - Identity $$\frac{x}{1} = x$$


text: Coil 1 has 123 turns and coil 2 has 234 turns. When no currentflows in coil 1 and a constant current of 7.00 A flows in coil 2,the average magnetic flux from this 7.00 A current is 0.111 mWbthrough each turn of coil 1 and 0.222 mWb through each turn of coil2. (Of course, the magnetic flux is zero in each turn of each coilfrom the zero current in coil 1.) Find the mutual inductance of thetwo coils. Then use that data to find the numerical value of theself-inductance of one of the two coils.


text:
# Empty string as a truthy value

I will mainly speak about the programming language Lua, because that's where this question comes from.

In Lua, everything but nil and false evaluates to true, it means that even 0 or an empty string are true. In the first case, there's no doubt about it: it's a truthy value as of PPCG's definition.

The empty string is a little bit problematic. If I have to output a truthy value in some cases, writing os.write"" can be my best option to do the stuff, but it has no feedback to the user testing the program, not even a newline or a space.

You couldn't say if it outputted or not, unless you redirect its output to a truth machine.

On the other side, it technically outputs to stdin the representation of an object of type string, which always evaluates to true.

As commented by @MartinEnder, this is only a particular case of a bigger question: What's our stance on outputting to stdin/printing a truthy/falsy value when its string representation is an empty string.

• Do nil and false have natural string representations? Are they just "nil" and "false"? I think your problem is similar to other languages though where 0 is falsy and "0" is truthy... when you print 0 to STDOUT have you printed the number or the string? We usually assume good faith and you can use whichever makes more sense in your context as long as you do it consistently. If in doubt, go by the definition you linked and print its natural representation (so empty output would indeed be truthy). – Martin Ender Jun 23 '16 at 14:55
• @MartinEnder That's the guideline I personnaly prefer, but as seen in the chat and in the answer below, objections can be made for this edge-case. That's why I think a clear consensu of the community would be better – Katenkyo Jun 23 '16 at 15:02
• I'd also like to note that printing nothing as a falsy result is also fairly common practice in some languages where empty strings/lists are considered falsy, so I don't see why it should be different for languages which consider it truthy. Maybe frame the question more generally whether printing empty strings (as in, nothing) is valid as truthy/falsy in general, since the only answer so far also treats both of those cases the same. – Martin Ender Jun 23 '16 at 15:32
• Here is where Lua specifies everything except nil and false is true. Love Lua by the way. – Insane Jun 27 '16 at 9:04
• In C, the empty string literal, "", is also truthy. ideone.com/zomVN0 – SoniEx2 Jul 2 '16 at 13:07

I'm not going to distinguish empty strings as truthy or falsy outputs here - just because some popular languages consider them falsy is no reason to bias this discussion one way or the other. So...

## Yes, this is fine, provided that...

...two conditions are met:

1. The interpretation must be consistent with your language's interpretation of the truthiness/falsiness of the empty string (or empty list, if that also prints nothing). If your language considers the empty string to be truthy, you don't get to use it as a falsy output and vice versa.
2. Your output format must be unambiguous. This is a very important rule that we always apply implicitly, regardless of what we're actually outputting. Pretty much all of the issues that could arise from using empty strings for output are covered by this.

I'll elaborate a bit on that second part. When the challenge asks you to output a list of integers in any convienient format, you're not allowed to just print those integers back to back. There's no way to reconstruct where one numbers ends and the next begins. We require unambiguous separators (which also means you can't separate them with other digits).

The same applies here. As long as you choose an output format that a) complies with the challenge spec and b) allows a user to infer the result of the computation unambiguously, you can use empty strings as output. Some examples:

• A challenge asks you to print a single truthy/falsy value. If the empty string is falsy and every non-empty string is truthy, there is no potential for confusion of the output.
• A challenge asks you to print a single truthy/falsy value for valid input and nothing at all for invalid input. In this case you cannot use the empty string as a truthy/falsy value, because it's indistinguishable from printing nothing at all (which is something your program should be able to do for different inputs).
• A challenge asks you to print a list of truthy/falsy values. You may use an empty string for one of them (depending on your language), but you have to print a separator between every pair of values so that one can infer the empty strings from adjacent separators. Having a trailing separator after each value (analogous to 1\n2\n3\n) is still fine as long as its used consistently (because then you know that the empty string after the last separator does not ever represent a value).

Of course all of those things equally apply to challenges that ask you to print a string which might potentially be empty. We'd be handling those cases exactly the same way.

One last thing: when you do output an empty string, you are allowed not to print anything. Your code doesn't need to contain print '' if omitting it gives the exact same output. I believe this has been used occasionally in programs that handle certain inputs by terminating with an error (which doesn't print to STDOUT). Only the observable behaviour is relevant.

• In case someone is looking for an example of what was mentioned in the last paragraph, I did that here among other places, as I had assumed this was an acceptable interpretation of the exiting on error discussion. – FryAmTheEggman Jun 23 '16 at 18:34
• Your rationale has won me over. I've upvoted this answer, and I'm going to delete mine. – user45941 Jun 24 '16 at 20:28


text:
» » » Justice starts a 5km walk from P on a bearing 0230. He then walks 4km on a beari...

# Justice starts a 5km walk from P on a bearing 0230. He then walks 4km on a beari...

### Question

Justice starts a 5km walk from P on a bearing 0230. He then walks 4km on a bearing of 1130 to Q. What is the bearing of Q from P?

A)
67.4°
B)
23°
C)
90.4°
D)
113°


text:
Browse Questions

# $S_1$ and $S_2$ are two hollow concentric spheres enclosing charges Q and $4Q$ respectively as shown in adjacent figure.How will the electric flux through the sphere $S_1$ change, if a medium of dielectric constant 6 is introduced in the space inside $S_1$ in place of air?

$\large\frac{Q}{3 \in_0}$


text:
# Show that $f$ is a strong contraction when $f$ is continuously differentiable.

Mathematics Asked on January 7, 2022

Let $$f: [a,b] to R$$ be a differentiable function of one variable such that $$|f'(x)| le 1$$ for all $$xin [a,b]$$. Prove that $$f$$ is a contraction. (Hint: use MVT.) If in addition $$|f'(x)| < 1$$ for all $$x in [a,b]$$ and $$f’$$ is continuous, show that $$f$$ is a strict contraction.

Using MVT, $$|f(x) – f(y)| = |f'(c)(x-y)| le |x-y|$$ for $$c$$ between $$x$$ and $$y$$.

I don’t know the proof for a strict contraction. I guess that I need to use the continuity of $$f’$$, but I am not sure how to use it. Any help would be appreciated.

You're almost there! A (weak) contraction is defined as a function $$f: A to mathbb{R}$$ such that $$|f(x) - f(y)| leq k | x - y| ~forall x,y in A,::: 0 leq k leq 1$$ (in your case $$A = [a,b]$$.)

This might look a little familiar, as you have already concluded that $$|f(x) - f(y)| < |x - y|$$ (by removing the middle inequality). We just need to make sure that $$k = |f(x)|$$ meets the required bounds. Combine $$0 leq |f(x)|$$ (a property of absolute value) with the given $$|f(x)| leq 1$$ for all $$x in [a,b]$$, you can conclude $$0 leq |f(x)| leq 1$$, so $$1geq k geq |f(c)| geq 0$$ (by continuity and $$c in[a,b]$$).

So you have effectively shown that $$f$$ is a weak contraction.

Showing that if $$|f(x)| < 1$$, then $$f$$ is a strict contraction follows a similar path. @bitesizebo is correct in noting that $$f'$$ continuous is essential here. A strict contraction replaces the weak inequality with a strong inequality. A function is a strict contradiction if $$|f(x) - f(y)| < k | x - y| ~forall x,y in A,::: 0 leq k leq 1$$ (in your case $$A = [a,b]$$.)

You know that $$|f(x) - f(y) leq |f(c)(x-y)|$$ by MVT. You can separate the RHS of the equation and get $$|f(c)(x-y)| = |f(c)||x-y|$$. Using the fact that $$f'$$ is continuous and $$|f'(x)| < 1$$, you need to prove that $$|f'(c)| < 1$$ (level of detail here depends on your class), as @bitesizebo says.

There are a few ways to show that $$f$$ satisfies $$sup_{[a,b]} |f'(c)| < 1$$. My favorite method is a bit more broad than @Michael Hardy's approach and proves the Extreme Value Theorem along the way. I use prefer the 'fact' that the image of a compact set under a continuous function is also compact. This works in any topological space, not just $$mathbb{R}$$. So $$f'([a,b])$$ is a closed set, so it must attain all of it's limit points (the limit of any sequence of numbers in $$f'([a,b])$$ must also be a point in $$f'([a,b])$$). So the infimum and supremum of $$f'([a,b])$$ must be obtained and be in the set. So as $$|f'(c)| < 1$$ for all $$c in[a,b]$$, you must have also hit the supremum and infimum of $$f'([a,b])$$. So $$sup_{[a,b]}|f'(c)| < 1$$. You can fill in the algebraic details and inequalities.

Once you have $$sup_{[a,b]}|f'(c)| < 1$$, you can plug this in to the statement we got from the MVT, and conclude $$|f(x) - f(y) < k |y-x|$$ where $$|f'(c)| < sup_{[a,b]}|f'(c)| leq k < 1$$ (I've never seen the assumption $$k>0$$ used to define a strict contraction since it could give a 'strongest' possible contraction step, although I don't have my copy of Rudin handy), and conclude that $$f$$ is a contraction map.

Answered by M Kupperman on January 7, 2022

Besides the mean-value theorem, another "MVT" is the "maximum-value theorem": A continuous function on a closed bounded interval has an absolute maximum. If $$|f'|$$ is continuous, then there is some point $$cin[a,b]$$ for which $$|f'|$$ is at least as big as it is at any point in that interval. And that value must be less than $$1,$$ by hypothesis. Then apply the mean value theorem again.

(You probably won't see that other theoem called the "MVT", for "maximum-value theorem", since it is instead called the extreme-value theorem, applying to both maxima and minima.)

Answered by Michael Hardy on January 7, 2022

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### Ask a Question

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text:
# include pdf pages in a latex document as appendix

I am writing a report and would like to include as appendix another document which is in pdf format.

• Just for those who don't want to look up the documentation: Use `\includepdf{file}`. – Hendrik Vogt Nov 11 '10 at 14:59
• @Sigur `pdfpages` is available in both TeX Live and MiKTeX ... – Joseph Wright Oct 3 '18 at 20:52


text:
Define correlation

Correlation is very helpful to investigate the dependence between two or more variables. As an example we are interested to know whether there is an association between the weights of fathers and son. correlation coefficient can be calculated to answer this question.

If there is no relationship between the two variables (father and son weights), the average weight of son should be the same regardless of the weight of the fathers and vice versa.

There are different methods to perform correlation analysis : Pearson, Kendall and Spearman correlation tests.

Pearson correlation

Pearson correlation measures a linear dependence between two variables (x and y). It’s also known as a parametric correlation test because it depends to the distribution of the data. The plot of y = f(x) is named linear regression curve.

The pearson correlation formula is :

$r = \frac{\sum{(x-m_x)(y-m_y)}}{\sqrt{\sum{(x-mx)^2}\sum{(y-my)^2}}}$

$$m_x$$ and $$m_y$$ are the means of x and y variables.

the p-value (significance level) of the correlation can be determined :

1. by using the correlation coefficient table for the degrees of freedom : $$df = n-2$$

2. or by calculating the t value : $t=\frac{r}{\sqrt{1-r^2}}\sqrt{n-2}$

In this case the corresponding p-value is determined using t distribution table for $$df = n-2$$

If the p-value is less than 5%, then the correlation is significant.

Interpretation of correlation coefficient

Correlation coefficient is between -1 (strong negative correlation) and 1 (strong positive correlation)

Correlation coefficient calculator

Note that online correlation coefficient calculator is also available by following this links : Correlation coefficient calculator.


text:
# RS Aggarwal Solutions Class 10 Chapter 9 Mean, Median, Mode of Grouped Data

Here you can get solutions of RS Aggarwal Solutions Class 10 Chapter 9 Mean, Median, Mode of Grouped Data. These Solutions are part of RS Aggarwal Solutions Class 10. we have given RS Aggarwal Solutions Class 10 Chapter 9 Mean, Median, Mode of Grouped Data download pdf.

Feature Contents

### Exercise 9A

Question 1:
Table is as given below:

Question 2:
We have

Question 3:
We have

Question 4:
We have

Question 5:
We have

Mean,
Question 6:
We have

Mean,
Question 7:
We have

Question 8:
We have

Question 9:
We have

Question 10:
We have, Let A = 25 be the assumed mean

Hence mean = 28.
Question 11:
A = 100 be the assumed mean, we have

Hence, mean = 110.67
Question 12:
Let the assumed mean be 150, h = 20

Hence, Mean = 146.25
Question 13:
Let A = 50 be the assumed mean, we have

Question 14:

We have h = 10 and let assumed mean = 25.
A = 25, h = 10, ∑ fi= 100 and ∑(fi×ui)= 30

Hence the mean of given frequency distribution is 28.
Question 15:
We have h = 4 and let assumed mean be A = 26. We have the table given below:

A = 26, h = 4, ∑ fi= 100 and ∑(fi×ui)= -152

Hence the mean of given frequency distribution is 19.92.
Question 16:
We have h= 30 and let A = 75 be the assumed mean. we have the table given below:

Thus, A = 75, h = 30, ∑ fi= 150 and ∑(fi×ui)= 80

Hence, the mean of the given frequency distribution is 91.
Question 17:
We ahve h = 20 and let A = 70 be the assumed mean. We have the table given below:

Thus, A = 70, h = 20, ∑ fi= 120 and ∑(fi×ui)= -4

Hence the mean of given frequency distribution is 69.33.
Question 18:
We have h = 14 and let A = 35 be the assumed mean.
For calculating the mean, we prepare the table given below:

Thus, A = 35, ∑ fi= 90, h = 14 and ∑(fi×ui)= 8

Hence, Mean = 36.24
Question 19:
Let h = 5 and let A = 22.5 be the assumed mean.
For calculating the mean, we prepare the table given below:

Thus, A = 22.5 and h = 5
∑ fi= 40 and   ∑(fi×ui)= 17

Hence the mean of given frequency distribution is 24.625.
Question 20:
We have h = 6 and let assume mean A = 33. For calculating the mean we prepare the table.

Thus, A = 33, h = 6, ∑ fi= 40 and ∑(fi×ui)=2

Hence, Mean = 33.3 years
Question 21:
We have h = 6 and let assumed mean A = 99. For calculating the mean we prepare the table:

Thus, A = 99, h = 6 and ∑ fi= 120, ∑(fi×ui)=2

Hence, Mean = 103.05.
Question 22:
Let h = 20 and assume mean = 550, we prepare the table given below:

Thus, A = 550, h = 20, and ∑ fi= 40, ∑(fi×ui)=-12

Hence the mean of the frequency distribution is 544.
Question 23:
The given series is an inclusive series, making it an exclusive series, we have

Thus, A = 42, h = 5, ∑ fi= 70 and ∑(fi×ui)=-37

Hence, Mean = 39.36 years.
Question 24:
The given series is an inclusive series making it an exclusive series, we get

Thus, A = 29.5, h = 10, ∑ fi= 80 and ∑(fi×ui)=43

Hence, Mean = 34.87 years.

Complete RS Aggarwal Solutions Class 10

If You have any query regarding this chapter, please comment on below section our team will answer you. We Tried our best to give complete solutions so you got good marks in your exam.

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Best of Luck For Your Future!!


text:
## Algebra and Trigonometry 10th Edition

Published by Cengage Learning

# Chapter 3 - 3.5 - Mathematical Modeling and Variation - 3.5 Exercises - Page 296: 45

#### Answer

R = k(T - $T_{e}$)

#### Work Step by Step

If variables vary directly then y = x, so as one increases then the other increases. If variables vary inversely then y = $\frac{1}{x}$, so as one increases then the other decreases. Using the description, R = k(T - $T_{e}$) where k is a constant.

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.


text:
# Older blog entries for shimniok (starting at number 70)

Tiny, Portable Turntable Strobe

I love turntables. The very idea is simultaneously insane and an awesome feat of engineering.

Microscopic grooves in a rotating disc of plastic vibrate a minuscule, precisely ground chunk of diamond stuck on a tiny metal tube with magnets attached, inducing tiny currents in a cartridge coil, subsequently amplified up the wazoo. Somehow, instead of sounding like garbled crap, the reproduced sound is quite excellent on a decent system.

But how fast does the record spin? That's the purpose of my project, an extremely accurate, compact turntable strobe. An LP is supposed to spin at 33-1/3 RPM. Many probably don't. Not precisely. And the voices in my head like precision.

 Introducing Pocket Turntable Strobe

Some turntables, like my Realistic Lab 400 above, have strobes built in that run off mains frequency to illuminate precisely spaced platter markings that appear to stand still when the platter speed is just right. Except the mains frequency varies and is only 60Hz on average.

Compare to my Turntable Strobe, which uses a quartz crystal and ATtiny25 to flash a white LED at 60.0±0.03Hz (30ppm absolute accuracy plus 20ppm temperature variation). Use it with platter strobe markings or speed check disks that you can find online.

 Turntable Strobe with speed check disc
It uses a tiny 3V, CR1225 battery which, combined with not-quite-Vcc output from the ATtiny25, produces barely enough voltage to dimly light a white LED. Adding four components makes the strobe burns bright as below, right.

 Same circuit, same LED, CR2032 left, CR1225 right
What sorcery is this? A Dickson charge pump.

 From Jonathan Thompson's Web Journal [link]
The circuit is simple, requiring two capacitors and two diodes and one of the pins on the microcontroller for the clock signal. The result is a higher voltage supply (4.2V unloaded) that can brightly illuminate the high intensity, 5mm LED through a 1K current limiting resistor. I'll provide more detail on this circuit in an upcoming article.

 Diagrammed in Upverter
Another trick to minimize complexity and parts count is that the microcontroller directly sinks current from the LED. You may object, claiming it's unsafe given the LED circuit is powered by a voltage exceeding the Vcc+0.5V limits of the ATtiny. Rules are meant to be broken, provided you have sufficient understanding of their purpose and how things work. I'll show you why it's safe in another upcoming article.

Meanwhile, if you want one, they are for sale.

Syndicated 2014-03-12 16:30:00 from Michael Shimniok

Freescale FRDM-KL46Z Board

I've been playing with a \$15 Freescale Freedom board based on the Kinetis MKL46Z256 [pdf]. It has a built in magnetometer and accelerometer and possibly the cutest little 4-digit, 7-segment LCD I've ever laid eyes on. This is quite a capable board.

It's a 48MHz Cortex M0+ with 256K flash memory, 32K RAM. That's enough power to do some serious computation. Perhaps a Kalman Filter-equipped Segway type robot? Or AVC rover?

On top of that, typical of other Freescale chips, a fast ADC running >400kSPS at 16bit or >800kSPS at lower resolution. With DMA available.

With Arduino form-factor headers plus another row of headers in addition to those, the board provides easy access to a lot of MCU pins, 64 to be exact.

You also get a light sensor, capacitive touch slider, a couple of tactile switches, and USB host / device capabilities.

The board is supported by mbed for easy programming and prototyping. And to dispel the myths: the mbed SDK is open source and you can offline compile it for several supported ARM toolchains.

Debug, in the form of CMSIS-DAP, is available, too. A 20mm battery holder footprint is provided on the board.

## Getting Started

If you plug your USB cable into the OpenSDA USB connector (use your Quick Reference Card to locate it), then your system will mount a Mass Storage Device.

On the drive are several HTML files. SDA_INFO.HTM takes you over to a P&E Microcomputer registration site where you can register your hardware. TOOLS.HTM redirects your browser to a page listing P&E debugger tools. For some reason the FSL_WEB.HTM leads to 404 error on the Freescale website. It should point to this page.

Next, head over to the mbed.org site, register or login, navigate to the Freedom KL46Z platform page and on the right side, click the "Add to Compiler" button.

## Compiling Demos

On this same page you'll find a series of examples for the board. You can also search mbed.org for more. I selected an accelerometer demo from Freescale. Import it into the IDE, select your platform with the button at the upper right. Click compile and then move the bin file that is downloaded onto the Freedom's mass storage device.

The demo prints accelerometer values for X, Y and Z axes to the USB serial port. On Linux this device appears as /dev/ttyACM*. Fire up minicom and you get output...

X: 1.00, Y: 0.96, Z: 0.00
X: 1.00, Y: 0.96, Z: 0.00
X: 1.00, Y: 0.96, Z: 0.00
X: 1.00, Y: 0.96, Z: 0.00
X: 1.00, Y: 0.96, Z: 0.00
X: 1.00, Y: 0.97, Z: 0.00
X: 1.00, Y: 0.97, Z: 0.00
X: 1.00, Y: 0.96, Z: 0.00
X: 0.99, Y: 0.97, Z: 0.00
X: 1.00, Y: 0.96, Z: 0.00

Next I tried a touch slider demo which controls the PWM signal sent to the board's green LED. The code is really simple. Here's main:
```  `int main(void) {    PwmOut led(LED_GREEN);    TSIAnalogSlider tsi(ELEC0, ELEC1, 40);    while (true) {        led = 1.0 - tsi.readPercentage();        wait(0.1);    }}`
```

## Conclusion

I think this board is a decent deal at \$15, the main selling point being mbed compatibility. My view is that boards are only worth as much as their toolchains and community support. And you can get other Freedom boards in the same form factor as well including other M0+ boards and an M4, the FRDM-K20D50M which boasts a 50MHz M4 with DSP capability.

While I touched on the highlights of this board, there's a lot more to explore. I'm looking forward to writing come code for it and using it in an upcoming project. SHARC will be hosting a Firefighting competition in the next few months and Pokey could use a major refit. However, come to think of it, I'm up to my ears in projects. Instead, perhaps I'll use it for one of those projects, such as my Take Out The Trash Bot. Whatever I do, I'll be sure to post about it here.

Syndicated 2014-03-06 16:00:00 from Michael Shimniok

Lithium-Sulphur Batteries

I ran across an article on Lithium-Sulphur batteries the other day. With theoretically double the energy density of Lithium-Ion batteries, Lithium-Sulpher batteries look promising, but the chemistry comes with many challenges. However, researchers at Lawrence Berkeley National Laboratory have found some ways to address the problems. (read the Full Article)

Syndicated 2014-03-04 18:15:00 from Michael Shimniok


text:
[Tugindia] Tabular environment in Latex

Herbert Voss Herbert.Voss at alumni.TU-Berlin.DE
Wed Aug 27 16:53:11 CEST 2003

JAYESH KUMAR schrieb:
>
> I am trying to make a table, in which I want the {\linesize} to determine the
> width of columns automatically depending on the number of columns in the
> table.

with tabularx

Herbert

\documentclass[a4paper,12pt]{article}
\usepackage[latin1]{inputenc}
\usepackage[T1]{fontenc}
\usepackage{tabularx}
\begin{document}

\begin{tabularx}{\linewidth}{p{3cm}Xp{3cm}XX}
1 & 2 & 3 & 4 & 5 \\\hline
the first column with a with of 3 cm
& the second column with a with of X cm
& the third column with a with of 3 cm
& the forth column with a with of X cm
& the fifth column with a with of X cm
\end{tabularx}

\begin{minipage}{0.75\linewidth}
\begin{tabularx}{\linewidth}{p{3cm}Xp{3cm}XX}
1 & 2 & 3 & 4 & 5 \\\hline
the first column with a with of 3 cm
& the second column with a with of X cm
& the third column with a with of 3 cm
& the forth column with a with of X cm
& the fifth column with a with of X cm
\end{tabularx}
\end{minipage}
\end{document}


text:
# Formatting linear programs

When stating LP problems I usually use something along the following

\begin{tabular}{ l c p{1pt} c }
max     & \multicolumn{3}{l}{$300x + 100y$} \\
s.t.    & $6x + 3y$ & $\leq$ & $40$\\
& $x - 3y$  & $\leq$ & $0$ \\
& $x + \frac{1}{4}y$ & $\leq$ & $4$ \\
\end{tabular}

But this forces the use of inline math, and require me to adjust column width manually (and it's still not perfect since the spacing between the right hand side and left hand side of \leq is wrong).

What I really would like was a solution that allowed for display math, with correct spacing and no need for manual formatting with p{}.

How would I write a version that fixes this?

-
Something like an array environment would probably suffice here (see the first answer on this question; I don’t have time for a full answer right now). –  alexwlchan Feb 16 at 22:41
A good thing is to read the fine document MathMode.pdf of Herbert Voß. –  Alain Matthes Feb 16 at 22:56

To get the wider spacing associated with display-style math, you could use a combination of the gather* and aligned environments (both provided by the amsmath package):

\documentclass{article}
\usepackage{amsmath}
\begin{document}
\begin{gather*}
\begin{aligned}
\textup{s.t.}\quad 6x + 3y  &\leq  40\\
x - 3y  &\leq  0 \\
x + \tfrac{1}{4}y  &\leq  4 \\
\end{aligned}
\end{gather*}
\end{document}

-
I love it! Even better than I could've imagined! –  Mathias Vorreiter Pedersen Feb 16 at 23:19

Using the array environment, you can do as follows:

\documentclass{article}

\usepackage{amsmath}

\begin{document}

\begin{equation*}
\setlength\arraycolsep{1.5pt}
\begin{array}{l@{\quad} r c r c r}
\max          & 300x & + &         100y &      &    \\
\mathrm{s.t.} &   6x & + &           3y & \geq & 40 \\
&    x & - &           3y & \geq &  0 \\
&    x & + & \frac{1}{4}y & \geq &  4
\end{array}
\end{equation*}

\end{document}

This gives a proper alignment IMHO.

-


text:
## View abstract

### Session S32 - Special functions and orthogonal polynomials

Monday, July 12, 14:00 ~ 15:00 UTC-3

## $q$-Fractional Askey--Wilson Integrals and Related Semigroups of Operators

We introduce three one parameter semigroups of operators and determine their spectra. They are fractional integrals associated with the Askey--Wilson operator. We also study these families as families of positive linear approximation operators. Applications include connection relations and bilinear formulas for the Askey--Wilson polynomials. We also introduce a $q$-Gauss--Weierstrass transform and prove a representation and inversion theorem for it.


text:
# [tex-live] Re: Possible bug in pdfetexk (re \hoffset)?

Lars Madsen daleif at imf.au.dk
Fri Jan 27 21:38:02 CET 2006

This is actually rather interesting, I had a user who came and complained
about this very thing, pdftex producing different margins than TeX.

When has this changed? I don't use plain TeX or pdfTeX myself so I haven't
noticed. The user in question claims that his document worked fine under
both systems in TL 2004.

/daleif

``You cannot help men permanently by doing for them
what they could and should do for themselves. ''
-- Abraham Lincoln


text:
# How do you simplify sqrt(27/16)?

Jul 12, 2017

#### Answer:

It is $\frac{3 \times \sqrt{3}}{4}$

#### Explanation:

You can write your expression

$\frac{\sqrt{27}}{4}$

since $\sqrt{16} = 4$

You can further write

$\frac{\sqrt{9 \times 3}}{4} = \frac{3 \times \left(\sqrt{3}\right)}{4}$

This is your answer

$\frac{3 \times \left(\sqrt{3}\right)}{4}$

Jul 12, 2017

#### Answer:

$\frac{3 \sqrt{3}}{4}$

#### Explanation:

Value $= \sqrt{\frac{27}{16}}$

Expressing the numerator and denominator as their prime factors:

Value $= \sqrt{\frac{3 \times 3 \times 3}{2 \times 2 \times 2 \times 2}}$

Taking each pair of factors once through the sqrt.

Value $= \frac{3 \sqrt{3}}{2 \times 2}$

$= \frac{3 \sqrt{3}}{4}$


text:
# By using binomial theorem, expand $$(1 + x + x^2)^3$$.

## Solution :

Let y = x + $$x^2$$. Then,

$$(1 + x + x^2)^3$$ = $$(1 + y)^3$$

= $$^3C_0$$ + $$^3C_1 y$$ + $$^3C_2 y^2$$ + $$^3C_3 y^3$$

= $$1 + 3y + 3y^2 + y^3$$ = 1 + 3$$(x + x^2)$$ + 3$$(x + x^2)^2$$ + $$(x + x^2)^3$$

= $$x^6 + 3x^5 + 6x^4 + 7x^3 + 6x^2 + 3x + 1$$

### Similar Questions

Find the middle term in the expansion of $$({2\over 3}x^2 – {3\over 2x})^{20}$$.

Find the 9th term in the expansion of $$({x\over a} – {3a\over x^2})^{12}$$.

Find the 10th term in the binomial expansion of $$(2x^2 + {1\over x})^{12}$$.

Which is larger $$(1.01)^{1000000}$$ or 10,000?

Find the middle term in the expansion of $$(3x – {x^3\over 6})^7$$.


text:
3D PrintingSoon to join the ranks of Prusa owners.

Help Support The Rocketry Forum:

Tobor

TRF Supporter
This morning my wife gave me a nice surprise. We now have the funds available to purchase a Pursa Prusa i3 MK3S+.

I have lurked these 3D printing sub-forums off and on for awhile now but I still have some *basic* questions.

1) Which sheet option should I select with my purchase. The smooth double-sided PEI sheet or the textured powder-coated double-sided sheet, for general all around use?
2) Best first book to own? Is the 3D Printing for Dummies still a good choice for this?
3) Best comprehensive online guide to nozzles/hotends?
4) Best comprehensive online review/guide to filaments?
5) Is the full Bear upgrade worth the investment?

Tobor

Last edited:

Michael L

Random Pixel Generator
TRF Supporter
I use both. The smooth for PLA and textured for PETG. I just tried some Essentium PCTG (it's supposed to print like PLA with PETG qualities) and I had to use the smooth plate. You'll want both but it won't hurt to order smooth, print a bunch of PLA to get started.

I switched to a smooth plate and the filament prints and releases very nicely so it behaves like PLA but has the durability of PETG. PETG basically releases itself from the textured plate when it cools.

I had to change the silicone hot end sock. You'll want to use one of those btw. Amazon has them for cheap. V6 hot end silicone sock... or something like that.

Go to Thingiverse and download some "Things". You'll get a lot of practice that way.

cwbullet

Obsessed with Rocketry
Staff member
Global Mod
You will never regret the purchase.

thzero

Well-Known Member
TRF Supporter
You will never regret the purchase.
I can say that the build from kit has been a PITA. Some parts I like, but I almost do prefer the design of the CR-6 SE more. However, Creality support sucks and I've heard nothing but good from Prusa (another non-rocketry buddy has one), so there is that. And there are plenty of add-ons/replacements, etc for Prusa too.

Tobor

TRF Supporter
I use both. The smooth for PLA and textured for PETG. I just tried some Essentium PCTG (it's supposed to print like PLA with PETG qualities) and I had to use the smooth plate. You'll want both but it won't hurt to order smooth, print a bunch of PLA to get started.

I switched to a smooth plate and the filament prints and releases very nicely so it behaves like PLA but has the durability of PETG. PETG basically releases itself from the textured plate when it cools.

I had to change the silicone hot end sock. You'll want to use one of those btw. Amazon has them for cheap. V6 hot end silicone sock... or something like that.

Go to Thingiverse and download some "Things". You'll get a lot of practice that way.
Thx for the tips. As to Thingiverse, every time I visit that site I get lost for hours... I also created an account at Prusa Printers.

I can say that the build from kit has been a PITA. Some parts I like, but I almost do prefer the design of the CR-6 SE more. However, Creality support sucks and I've heard nothing but good from Prusa (another non-rocketry buddy has one), so there is that. And there are plenty of add-ons/replacements, etc for Prusa too.
Yeah I feel ya. Although I am mechanically inclined, I have been known to screw up a few things...
I am sure my Sailor's Voice will get a good workout when I assemble the MK3S.

Michael L

Random Pixel Generator
TRF Supporter
I took two days to build mine. 2/3 of the way thru I was second guessing that choice. When I finished I thought it wasn't that bad. The first time I had a problem (that really wasn't a problem, see below) I was glad that I built it.

The Essentium filament reminded me of something. When the print doesn't go right, do the first layer calibration. I bought Prusament PLA, PETG, ASA, and something I'm forgetting. Those are all repeatable. By that I mean that I can throw a spool of Yellow Prusament PLA on, print, switch to Prusament Red, etc and never do a first layer cal. The Essentium PCTG (gorgeous, durable looking parts btw) must flow differently. I say that because the second print of the same thing failed very early into to the print (it's a 6 hour print so I check on it). I always forget to do the first layer cal. It's important to get right. YT has a lot of good tutorials on how to do a first layer cal. So does Prusa:

Tobor

TRF Supporter
lol, just realized I've been misspelling P-r-u-s-a. That is sacrilege!!!

Tobor

TRF Supporter
Okay, I just completed ordering the MK3S+ kit. I selected the all black color option and the smooth PEI sheet. So, if the Prusa order backlog holds to it current state, the printer will arrive the 1st or 2nd week in May.

cwbullet

Obsessed with Rocketry
Staff member
Global Mod
Corrected the title.

TRF Supporter

cwbullet

Obsessed with Rocketry
Staff member
Global Mod
Thank you very much.
That is what I am paid to do....wait I am not paid!

Arpak

Well-Known Member
That is what I am paid to do....wait I am not paid!
Quick, change it back and send an invoice.

cwbullet

Obsessed with Rocketry
Staff member
Global Mod
Enjoy. I will tell you from experience that the Original is better than any clone. I own two clones. Neither is as good as the original.

caveduck

semi old rocketeer
I'm also a happy owner of a Prusa MK3S built from kit. It took a few evenings but I'm glad I built it just for the knowledge acquired.

Heretic that I am, I print mainly PETG on a #600 wet sanded smooth plate. The first layer cal is crucial, you have to dial it down pretty tight for PETG. If the single outside brim line stays down you should be good to go. It's also important to tell the slicer to not cross perimeters. And be sure to lube the slides reasonably often. Happy printing!

And thanks Chuck for herding this mob of cats as a volunteer gig.

cwbullet

Obsessed with Rocketry
Staff member
Global Mod
I'm also a happy owner of a Prusa MK3S built from kit. It took a few evenings but I'm glad I built it just for the knowledge acquired.

Heretic that I am, I print mainly PETG on a #600 wet sanded smooth plate. The first layer cal is crucial, you have to dial it down pretty tight for PETG. If the single outside brim line stays down you should be good to go. It's also important to tell the slicer to not cross perimeters. And be sure to lube the slides reasonably often. Happy printing!

And thanks Chuck for herding this mob of cats as a volunteer gig.

Tobor

TRF Supporter
Okay, I just completed ordering the MK3S+ kit. I selected the all black color option and the smooth PEI sheet. So, if the Prusa order backlog holds to it current state, the printer will arrive the 1st or 2nd week in May.
This morning I received an email from Fedex that my Prusa order is in their hands.

Huxter

TRF Supporter
Whoot! Whoot! It was nice to have a better hex driver - the 2.5 mm will be used 90-95% of the time. I picked up a set of 14 hex drivers for $20 at Home Depot. Make sure you have a really flat surface! My long card table was not! cwbullet Obsessed with Rocketry Staff member Administrator TRF Lifetime Supporter Global Mod Similar: I order the Kobalt brand. They come in a calming blue version. Tobor Get your peanuts.... TRF Supporter Not to knock your choices in allen keys.... I prefer using "Ball Drivers" like these Bondhus tools. I discovered them around 78' when I started building R/C planes. Tobor Get your peanuts.... TRF Supporter Ta-da!!! manixFan Not a rocket scientist Look through the entire assembly manual before you build, and/or watch a couple of build videos on Youtube. I was able to build mine over two evenings, and on a couple of occasions I realized if I had more closely read through the manual I could have avoided a couple of do-overs. The 'pride of ownership' you'll get after your first successful print is pretty fulfilling. One thing that took me a while was understanding the dozens of parameters that are available for tweaking. Many of them you won't have to worry about, but others, like retraction, become more important with different filaments and the kinds of models you print. I learned a lot trying to print sharp pointed nose cones. Turns out blunt is a lot easier. Good luck with your build! Tony one Youtube channel I really like is CNC Kitchen: Last edited: cwbullet Obsessed with Rocketry Staff member Administrator TRF Lifetime Supporter Global Mod Tobor Get your peanuts.... TRF Supporter I finished the build late Friday night (really late) and went through the calibration process Saturday morning. Still have to re-run the 1st layer cal once I finish setting up OctoPi on an R-Pi. Huxter Well-Known Member TRF Supporter Awesome! Looks like you used the right amount of gummy bears too! That first layer calibration can come out smooth as butta! Def find both limits and dial it in. Happy printing! Tobor Get your peanuts.... TRF Supporter Here is my rough draft for the printer enclosure. I've already purchased the Ikea Lack tables. I will be using built up S4S Oak Board (From Home Depot) for the bottom section. The build up will consist of 2" x 2" (1.5" x 1.5" actual) + 0.5" x 2" (0.5" x 1.5" actual) + 0.5" x 3" (0.5" x 2.5" actual & cut to 2"). Huxter Well-Known Member TRF Supporter I hope that's good and stable for ya. I like the compactness. Would love to see pics of it complete w printer! Good luck! Tobor Get your peanuts.... TRF Supporter I hope that's good and stable for ya. I like the compactness. Would love to see pics of it complete w printer! Good luck! Yeppers, will do. Tobor Get your peanuts.... TRF Supporter I hope that's good and stable for ya. I like the compactness. Would love to see pics of it complete w printer! Good luck! Well..... When I did the initial online search on the Home Depot website, I wasn't really being attentive to material pricing ( I had thought pricing was for 6ft lengths). My visit to the store was a true "Sticker Shock" moment. Dimensionally stable lumber like hardwoods are horrendously expensive. Turns out the 2x2 Red Oak I had selected was$12.16 for a 3 foot board! As the project was to build up the 2x2 to true 2"x2" dimensions, total material cost would have exceeded \$130!

Needless to say, I've scrapped the idea of using hardwoods. I am reworking the concept using 1.5" PVC pipe or 1.5" Black pipe.

Btw, I was able to dial in the 1st layer cal, although my technique was different from the manual's procedure. Basically, I just kept repeating the routine nudging the height bias on each run until the single layer square pad printed at the end was nice and uniform with good adhesion.

And my first ever print job....

Huxter

Well-Known Member
TRF Supporter
Wahoo - you're now a Prusa printer! I have heard lumber prices have sky rocketed, but didn't realize how high the price is now.

I've been printing alot of PETG lately, and the smooth PEI sheet is not liking it much - it sticks too well! The glue stick doesn't help to release things much. I'll need a new PEI sheet if I keep that up! The bed is much nicer than the glass bed on my old printer. I have a couple cans of aqua-net left over now!

Happy printing

Staff member


text:
# A balanced lever has two weights on it, the first with mass 6 kg  and the second with mass 7 kg. If the first weight is  12 m from the fulcrum, how far is the second weight from the fulcrum?

Jan 9, 2017

#### Answer:

The 7 kg weight must be located at 10.3 m from the fulcrum.

#### Explanation:

To understand the balancing of a lever you must think in terms of the torque acting on each side of the fulcrum, because this is what actually is balanced.

To calculate torque, you multiply the amount of force acting on the side of the lever (the weight of the object) by the distance of that weight from the fulcrum (usually called the torque arm, $r$).

So, on one side we have 6 kg at 12 m

${T}_{1} = 6 k g \times 12 m = 72 N m$

The torque on side 2 must match this for a balance to exist

${T}_{2} = 72 N m = 7 k g \times r$

so, $r$ must equal $72 \div 7 = 10.3 m$


text:
Relative Pressure Sensor - MapleSim Help

Relative Pressure Sensor

Measure relative pressure

Description

The Relative Pressure Sensor component generates a real signal proportional to the pressure drop from ${\mathrm{flowPort}}_{a}$ to ${\mathrm{flowPort}}_{b}$. The mass and energy flows into both ports are zero.

Connections

 Name Description ${\mathrm{flowPort}}_{a}$ Thermal-fluid flow port ${\mathrm{flowPort}}_{b}$ Thermal-fluid flow port $y$ Real signal output (in Pa)

Variables

 Symbol Units Description Modelica ID ${p}_{x}$ $\mathrm{Pa}$ Pressure at ${\mathrm{flowPort}}_{x}$ flowPort_x.p

Parameters

 Symbol Default Units Description Modelica ID $\mathrm{medium}$ $\mathrm{fluidMedium}$ - medium


text:
# Recitation 8

Problem 1 Consider the function $f: [0, \infty)\to\mathbb{R}$ defined by $f(x) = x^a\sin(1/x)$ for $x >0$ and $f(0) = 0$. (a) For which values of the parameter $a$ the function $f$ is differentiable at $x=0$? (b) For which values of $a$ is the derivative $f'(x)$ a continuous function?

Proof sketch (a) The function is differentiable at 0 if and only if $\lim_{x\to 0^+}\frac{f(x) - f(0)}{x - 0} = \lim_{x\to 0^+}x^{a-1}\sin(1/x)$ exists. Note that $\lim_{x\to 0^+}x^{a-1}\sin(1/x) = 0$ when $a > 1$. When $a = 1$, we already know that $\lim_{x\to 0^+}\sin(1/x)$ does not exist. By problem 3 in homework 4, when $a < 1$, because $\lim_{x\to 0^+}x^{a-1}=\infty$, $\lim_{x\to 0^+}x^{a-1}\sin(1/x)$ does not exist. Therefore $f$ is differentiable at 0 if and only if $a>1$, moreover, the derivative at 0 is 0.

(b) For $x > 0$, $f'(x) = ax^{a-1}\sin(1/x) + x^a\cos(1/x)(-1/x^2) = ax^{a-1}\sin(1/x) - x^{a-2}\cos(1/x)$ and $$\lim_{x\to 0^+}f'(x) = a\lim_{x\to 0^+}x^{a-1}\sin(1/x) - \lim_{x\to 0^+}x^{a-2}\cos(1/x) = - \lim_{x\to 0^+}x^{a-2}\cos(1/x).$$ An argument similar to the one for part (a) shows that $f'(x)$ is continuous if and only if $a > 2$.

Problem 2 Give an example of a differentiable function $f\colon [0,1] \to \mathbb{R}$ such that $x=0$ is not a local minimum or a local maximum.

Proof By Problem 1, $f(x) = x^2\sin(1/x)$ is differentiable on $[0,1]$. For every $\epsilon > 0$, there exists $n\in\mathbb{N}$ (sufficiently large) such that both $x_1 := 1/[(2n + 1/2)\pi]$ and $x_2 := 1/[(2n - 1/2)\pi]$ are less than $\epsilon$. Notice that $f(x_1) = x_1^2 > 0$, $f(x_2) = -x_2^2 < 0$ and $f(0) = 0$. Therefore 0 is not a local maximum or a local minimum point.

Problem 3 Find the maximum and minimum values of the function $f(x) = x^5 - 5x^4 + 5x^3 + 1$ on the interval $[-1,2]$.

Solution Note that $f'(x) = 5x^4 - 20x^3 + 15x^2 = 5x^2(x-1)(x-3)$ has roots $0, 1, 3$ (we may ignore 3). We compute the value of the function at the end points of the interval and those roots: $f(-1) = -10, f(0) = 1, f(1) = 2, f(2) = -7$. Therefore the maximum value is 2 and the minimum value is -10.

Problem 4 Use Lagrange’s mean value theorem to prove the following inequality for $0 < a \le b$, $$\frac{b-a}{b} \le \ln \frac{b}{a} \le \frac{b-a}{a}.$$

Proof If $a = b$, it is clear. Hereafter we assume $a < b$. Rewrite the inequality as $$\frac{1}{b} \le \frac{\ln b - \ln a}{b - a} \le \frac{1}{a}.$$ This inspires us to consider the function $f(x) = \ln x$. By the mean value theorem, there is $a < c < b$ such that $f'(c) = \frac{\ln b - \ln a}{b - a}$. Notice that $f'(c) = 1/c$ and $1/b < 1/c < 1/a$.

Problem 5 Prove that the equation $x^n + px + q = 0$ can have at most two distinct (real) solutions if $n$ is even, and it can have at most three distinct (real) solutions if $n$ is odd.

Proof sketch By Rolle’s theorem, if there are $k$ distinct solutions to $f(x) = 0$, then there are at least $k-1$ distinct solutions to $f'(x) = 0$ (why?). Now notice that the derivative $f'(x) = x^{n-1}+p$ has 1 solution when $n$ is even, and it has 2 solutions when $n$ is odd.

Problem 6 Suppose $f\colon \mathbb{R} \to \mathbb{R}$ satisfies $f(n) = 2$ for every $n\in\mathbb{N}$ and suppose that $f$ is differentiable and $\lim_{x\to\infty} f'(x) = 0$. Show that $\lim_{x\to\infty} f(x) = 2$.

Proof Take a sequence $\{a_n\}_{n=1}^\infty$ such that $\lim_{n\to\infty} a_n = \infty$. By Heine’s theorem, it suffices to show that $\lim_{n\to\infty} f(a_n) = 2$. For every $a_n$, by the mean value theorem, there is $[a_n]\le b_n \le a_n$ such that $f(a_n) - f([a_n]) = (a_n - [a_n])f'(b_n)$. Because $[a_n]$ is a natural number, $f([a_n]) = 2$. Thus $|f(a_n)-2| = |a_n - [a_n]||f'(b_n)| \le |f'(b_n)|$. Since $b_n \ge [a_n] > a_n - 1$, by Heine’s theorem, $\lim_{n\to\infty}f'(b_n) = 0$. By the sandwich theorem, $\lim_{n\to\infty}|f(a_n)-2| = 0$ and so $\lim_{n\to\infty}f(a_n) = 2$.

Problem 7 Prove that for every $x\in\mathbb{R}$ and $n\in\mathbb{N}$ we have $$1 + x +\frac{1}{2!}x^2 + \frac{1}{3!}x^3 + \dots + \frac{1}{(2n)!}x^{2n} > 0.$$

Proof Denote the left hand side of the inequality by $f(x)$. Notice that $\lim_{x\to -\infty}f(x) = \lim_{x\to \infty}f(x) = \infty$. The function has minimum points. Let $x_0$ be a minimum point. We know that $f'(x_0) = 0$. However, $$f'(x) = 1 + x +\frac{1}{2!}x^2 + \frac{1}{3!}x^3 + \dots + \frac{1}{(2n-1)!}x^{2n-1} = f(x) - \frac{1}{(2n)!}x^{2n}.$$ This implies that $f(x_0) = f'(x_0) + \frac{1}{(2n)!}x_0^{2n} > 0$. In other words, $f(x) \ge f(x_0) > 0$.