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# Why does $\lim_{n\to\infty} z_n=A$ imply $\lim_{n\to\infty}\frac{1}{n}(z_1+\cdots+z_n)=A$?

I'm self-studying a bit of complex analysis, and I'm attempting to figure out the following.

Suppose $\lim_{n\to\infty}z_n=A$. How can I show that $$\lim_{n\to\infty}\frac{1}{n}(z_1+\cdots+z_n)=A.$$

Is there a clever way to write the limit to make it more approachable? Thank you.

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It'll also be helpful, math.columbia.edu/~nironi/stolz-cesaro.pdf – Ehsan M. Kermani Jan 30 '12 at 4:11
Thanks for the link, @ehsanmo. – Dedede Jan 30 '12 at 4:19

Let $\epsilon>0$. Since $\lim_{n\to\infty}z_n=A$, there exists an positive integer $N$ such that $$\tag{1}|z_n-A|<\frac{\epsilon}{2}\mbox{ whenever }n\geq N.$$ For this fixed $N$, we can find another positive integer $N'$ such that $$\tag{2}\sum_{k=1}^{N-1}|z_i-A|\leq \frac{N'\epsilon}{2}.$$

Hence, if $m\geq\max\{N, N'\}$, then $$\left|\frac{1}{m}(z_1+\cdots+z_m)-A\right|=\left|\frac{1}{m}\sum_{k=1}^{m}(z_i-A)\right|$$ $$\leq\left|\frac{1}{m}\sum_{k=1}^{N-1}(z_i-A)\right|+\left|\frac{1}{m}\sum_{k=N}^{m}(z_i-A)\right|$$ $$\leq\frac{1}{m}\sum_{k=1}^{N-1}|z_i-A|+\frac{1}{m}\sum_{k=N}^{m}|z_i-A|$$ $$\leq \frac{\epsilon}{2}+\frac{1}{m}(m-N+1)\frac{\epsilon}{2}\leq \epsilon$$ where we have used $(1)$ and $(2)$ in the last inequality. This proves that $$\lim_{n\to\infty}\frac{1}{n}(z_1+\cdots+z_n)=A.$$

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Thank you Paul, I find this answer very clear and easy to understand. – Dedede Jan 30 '12 at 2:50
@Dedede: You are welcome. – Paul Jan 30 '12 at 3:45

We have that

$$\frac{z_1+\cdots+z_n}{n}-A=\frac{(z_1-A)+\cdots+(z_n-A)}{n},$$

so it suffices to prove the statement when $A=0$. The $\epsilon$-$N$ definition of the limit tells us that for any positive $\epsilon>0$, there is an $N$ such that $n>N\implies |z_n|<\epsilon$, which subsequently implies

$$\left|\frac{z_1+\cdots+z_{k+N}}{k+N}\right|=\left|\frac{z_1+\cdots+z_N}{k+N}\,+\,\frac{z_{N+1}+\cdots+z_{N+k}}{k+N}\right|$$

$$\le \left|\frac{z_1+\cdots+z_N}{k+N}\right|\,+\,\left|\frac{z_{N+1}+\cdots+z_{N+k}}{k+N}\right|\le \left|\frac{z_1+\cdots+z_N}{k+N}\right|+\epsilon\stackrel{k\to\infty}{\longrightarrow}\epsilon.$$

(We also used the triangle inequality above.) Now take $\epsilon\to0^+$ and use the squeeze theorem.

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Thanks again for your help, anon. – Dedede Jan 30 '12 at 2:50


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Courses

# RD Sharma Solutions - Ex - 4.3 & Ex - 4.4, Rational Numbers, Class 7, Math Class 7 Notes | EduRev

## Class 7 : RD Sharma Solutions - Ex - 4.3 & Ex - 4.4, Rational Numbers, Class 7, Math Class 7 Notes | EduRev

The document RD Sharma Solutions - Ex - 4.3 & Ex - 4.4, Rational Numbers, Class 7, Math Class 7 Notes | EduRev is a part of the Class 7 Course RD Sharma Solutions for Class 7 Mathematics.
All you need of Class 7 at this link: Class 7

## Exercise 4.3

#### QUESTION 1:

Determine whether the following rational numbers are in the lowest form or not:

(i) 65/84
(ii) −15/32
(iii) 24/128
(iv) −56/−32

(i) We observe that 65 and 84 have no common factor i..e., their HCF is 1.

Thus, 65/84 is in its lowest form.

(ii) We observe that −-15 and 32 have no common factor i..e., their HCF is 1.

Thus, −15/32 is  in  its lowest form.

(iii) HCF of 24 and 128 is not 1.

Thus, given rational number is not in its simplest form.

(iv) HCF of 56 and 32 is 8.

Thus, given rational number is not in its simplest form.

#### QUESTION 2:

Express each of the following rational numbers to the lowest form:

(i) 4/22
(ii) −36/180
(iii) 132/−428
(iv) −32/−56

Lowest form of:
(i) 4/22 is:
4 =2×2
22 = 2×11
HCF of 4 and 22 is 2.
Dividing the fraction by 2 , we get  2/11.

(ii)  −36/180 is:
36 = 3×3×2×2
180 = 5×3×3×2×2
HCF of  36 and 180 is 36.
Dividing the fraction by 36, we get −1/5

(iii) 132/−428 is:
132 = 2×3×2×11
428 = 2×2×107
HCF of 132 and 428 is 4.
Dividing the fraction by 4 , we get   33/−107

(iv) −32/−56 is:
32 =2×2×2×2×2
56 = 2×2×2×7
HCF of 32 and 56 is 8.
Dividing the fraction by 8, we get 4/7

#### QUESTION 3:

Fill in the blanks:

## Exercise 4.4

#### QUESTION 1:

Write each of the following rational numbers in the standard form:

(i) The denominator is positive and HCF of 2 and 10 is 2.

Dividing the numerator and denominator by 2, we get:

(ii) The denominator is positive and HCF of 8 and 36 is 4.

Dividing the numerator and denominator by 4, we get:

(iii)
The denominator is negative.

HCF of 4 and 16 is 4.
Dividing the numerator and denominator by 4, we get:

(iv) The denominator is negative.

HCF of 15 and 35 is 5.
Dividing the numerator and denominator by 5, we get:

(v) The denominator is negative.

HCF of 299 and 161 is 23.

Dividing the numerator and denominator by 23, we get:

(vi) The denominator is negative.

HCF of 63 and 210 is 21.

Dividing the numerator and denominator by 21, we get:

(vii) The denominator is negative.

HCF of 68 and 119 is 17.

Dividing the numerator and denominator by 17, we get:

(viii) The denominator is positive and HCF of 195 and 275 is 5.

Dividing divide the numerator and denominator by 5, we get:

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# BdMO 2016 National Junior Question 8.

$\triangle ABC$ is isosceles with $AB = AC$ ans $\angle A = 20^{\circ}$ and $BC = 12$. $BP \perp AC$ ans $Q$ is a point on $AB$ such that $QB=6$. Find $\angle CPQ$.

I was not able to solve this problem. Tried it many times. It can be solved very easily if I use trigonometric functions and calculator. But I need to solve it in Olympiad Math's way. The Diagram is something like this -

• What is the way you need to solve this? Dec 30 '16 at 17:52
• @Fimpellizieri Just without using Calculator. Dec 30 '16 at 17:52
• the answer is $$120^{\circ}$$ Dec 30 '16 at 18:15
• I found that same answer using laws of cosines but this is terribly long. Dec 30 '16 at 19:16
• Let $X \in AP$ be such that $\angle CBX=20^{\circ}$. Then $\triangle CBX$ is isosceles and $BX=BC=12$. Now let $Y \in AB$ be such that $\angle BYX=90^{\circ}$. We claim that $Y=Q$. Indeed, $\angle BXA=110^{\circ}$, so $\angle AXY=80^{\circ}$ and hence $\angle BXY = 30^{\circ}$. It follows that $\triangle BXY$ is a $30-60-90$ right triangle; since $BX=12$ then $BY=12\cos(60^{\circ})=6$ and the claim follows. I have not managed to finish yet, but this seemed important enough that I thought to post it while I work on the rest. Dec 30 '16 at 20:13

A short solution from the construction suggested by @Fimpellizieri in the comments.

Let $X\in AP$ such that $\Delta CBX$ is isosceles, and $Y\in AB$ such that $XY\bot AB$. Then $\angle XBY=\angle CBA-\angle CBX=80^\circ-20^\circ=60^\circ$, and, hence, $|BY|=12\cdot\cos 60^\circ=6$, i.e. $Y=Q$.

Now consider the quadrilateral $BQXP$. Since $\angle BQX=\angle BPX=90^\circ$, all four points lay on the same (circumscribed) circle. From the inscribed angles theorem it follows that $\angle BPQ=\angle BXQ$, but the latter is $90^\circ-60^\circ=30^\circ$. Finally, $\angle CPQ=90^\circ+30^\circ=120^\circ$.

• Nice! I was having trouble coming up with a simple solution near the end. Yours is great. +1 Dec 30 '16 at 21:47
• @Fimpellizieri It was only possible after your great construction, so it is more than on half your solution.
– A.Γ.
Dec 30 '16 at 21:53

Let us consider $\triangle APB$:

$$\angle ABP = 180° - (\angle APB + \angle PAB)$$ $$\angle ABP = 70°$$

Let us consider $\triangle BPQ:$

Let $\alpha=\angle BPQ$

$$\dfrac{\sin{\alpha}}{BQ}= \dfrac{\sin{70°}}{QP}$$ $$\sin{\alpha}= \dfrac{\sin{70°}}{QP}BQ$$ $$\sin{\alpha}= 6\dfrac{\sin{70°}}{QP}$$

Now let us find $QP$

Let us consider $\triangle APQ:$

$$QP^2 = AP^2 + AQ^2 - 2\cdot AP\cdot AQ \cdot \cos(\angle PAQ)$$

Now let us find $AP \text{ and } AQ$

Let us consider $\triangle BPC:$

$$\dfrac{\sin{\angle PBC}}{PC}= \dfrac{\sin{\angle PCB}}{BP}= \dfrac{\sin{\angle BPC}}{BC}$$ $$BP = \dfrac{\sin{\angle PCB}}{\sin{\angle BPC}}\cdot BC$$ $$BP = 11.82$$ $$PC = \dfrac{\sin{\angle PBC}}{\sin{\angle PCB}}\cdot PB$$ $$BP = 2.08$$

Let us consider $\triangle ABP\text{ to find } AP$

$$\dfrac{\sin{\angle BAP}}{BP}= \dfrac{\sin{\angle ABP}}{AP}$$ $$AP= \dfrac{\sin{\angle ABP}}{\sin{\angle BAP}}BP$$ $$AP= 32.48$$

$P \in [AC] \Rightarrow AC = AP + PC \Leftrightarrow AC = 34.56$

Since $\triangle ABC \text{ is isosceles, then } AB = AC = 34.56$

$Q \in [AB] \Rightarrow AQ = AB - BQ \Leftrightarrow AQ = 28.56$

$$QP^2 = AP^2 + AQ^2 - 2 AP \cdot AQ \cdot \cos(\angle QAP)$$ $$QP = 11.28$$

Now let us consider $\triangle BPQ$

$$\dfrac{\sin{\angle \alpha}}{BQ}= \dfrac{\sin{\angle QBP}}{QP}$$ $$\sin{\alpha}= \dfrac{\sin{\angle QBP}}{QP}BQ$$ $$\sin{\alpha}= \dfrac{1}{2} \Rightarrow \alpha = 30°$$

$$\angle CPQ = \angle CPB + \angle BPQ$$ $$\angle CPQ = 90° + 30°$$ $$\angle CPQ = 120°$$

• This solution is really huge -_- There should be some easier solution.. Btw. Nice (Y) (+1) Dec 30 '16 at 21:00
• I already got solution from Fimpellizieri's comment :) Dec 30 '16 at 21:01

Since $\angle BPC = 90^{\circ}$, it follows that $P$ lies on the circle whose diameter is $BC$. As such, let $D$ be the midpoint of $BC$, and let $E$ be a point inside $\Delta ABC$ such that $\Delta BDE$ is equilateral. Then $E$ also lies on the circle with diameter $BC$, so quadrilateral $BEPC$ is cyclic. Thus, $\angle BEP$ and $\angle PCB$ are supplementary, and since $\angle PCB = 80^{\circ}$, it follows that $\angle BEP = 100^{\circ}$. Now, $BC = 12$, so $BE = BD = \frac{1}{2}BC = 6$. Thus, $BE = 6 = QB$, so $\triangle QBE$ is isosceles. Since $$\angle QBE = \angle QBC - \angle EBC = 80^{\circ} - 60^{\circ} = 20^{\circ}$$ it follows that $\angle QEB = 80^{\circ}$. Thus, $\angle QEB$ and $\angle BEP$ are supplementary, so $E$ lies on line $QP$. As such, we have $$\angle CPQ = \angle CPE$$ and $\angle CPE = 120^{\circ}$, either by noting that it subtends an arc of $240^{\circ}$, or by noting that it is supplementary to angle $\angle EBC = 60^{\circ}$.


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## CryptoDB

### Richard Lindner

#### Publications

Year
Venue
Title
2010
EPRINT
We propose a new, efficient decoding algorithm for square-free (irreducible or otherwise) Goppa codes over $\F_p$ for any prime $p$. If the code in question has degree $t$ and its average code distance is at least $(4/p)t + 1$, the proposed decoder can uniquely correct up to $(2/p)t$ errors with high probability. The correction capability is higher if the distribution of error magnitudes is not uniform, approaching or reaching $t$ errors when any particular error value occurs much more often than others or exclusively. This makes the method interesting for (semantically secure) cryptosystems based on the decoding problem for permuted and punctured Goppa codes.
2008
EPRINT
Building upon a famous result due to Ajtai, we propose a sequence of lattice bases with growing dimension, which can be expected to be hard instances of the shortest vector problem (SVP) and which can therefore be used to benchmark lattice reduction algorithms. The SVP is the basis of security for potentially post-quantum cryptosystems. We use our sequence of lattice bases to create a challenge, which may be helpful in determining appropriate parameters for these schemes.
2008
EPRINT
Building upon a famous result due to Ajtai, we propose a sequence of lattice bases with growing dimension, which can be expected to be hard instances of the shortest vector problem (SVP) and which can therefore be used to benchmark lattice reduction algorithms. The SVP is the basis of security for potentially post-quantum cryptosystems. We use our sequence of lattice bases to create a challenge, which may be helpful in determining appropriate parameters for these schemes.
2007
EPRINT
The NTRU encryption scheme is an interesting alternative to well-established encryption schemes such as RSA, ElGamal, and ECIES. The security of NTRU relies on the hardness of computing short lattice vectors and thus is a promising candidate for being quantum computer resistant. There has been extensive research on efficient implementation of the NTRU encryption scheme. In this paper, we present a new algorithm for enhancing the performance of NTRU. The proposed method is between $11$\% and $23$\% faster on average than the best previously known method. We also present a highly efficient implementation of NTRU within the Java Cryptography Architecture.
2007
EPRINT
Micciancio defined a generalization of cyclic lattices, called ideal lattices. These lattices can be used in cryptosystems to decrease the number of parameters necessary to describe a lattice by a square root, making them more efficient. He proves that the computational intractability of classic lattice problems for these lattices gives rise to provably secure one-way and collision-resistant hash functions. This provable security relies on the assumption that reducing bases of ideal lattices is similar to reducing bases of random lattices. We give an indication that lattice problems in ideal lattices do not represent the general case by providing a distinguisher, which decides in time $O(n^4)$ whether a given basis of rank $n$ spans an ideal lattice or not. Using this algorithm we perform a statistical analysis for several dimensions and show that randomly generated lattices are practically never ideal.


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A 292 kg motorcycle is accelerating up along a ramp that is inclined 30.0° above the horizontal. The propulsion force pushing the motorcycle up the ra

Falak Kinney 2020-11-05 Answered
A 292 kg motorcycle is accelerating up along a ramp that is inclined 30.0° above the horizontal. The propulsion force pushing the motorcycle up the rampis 3020 N, and air resistance produces a force of 250 N that opposes the motion. Find the magnitude of the motorcycles
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Laaibah Pitt

What are the sum of forces on the motorcycle? Theproblem states the following:
The motorcycle is pushed up the ramp with force 3020 N
Air resists the motorcyles forward motion, therefore it pusheswith a force of 250 N down the ramp.
The only other force is that due to the motorcycles weight(a. force of gravity). Then the force diagram looks like);

Where;
F1=3020 Newtons
F3=250 Newtons
F4 is force due to gravity and F2 is the component of that force parallel to the ramp.
So,
${F}_{2}=\left(292\right)\left(9.8\right)\mathrm{sin}\left(30\right)=1430.8$
now the sum of the forces is equal to the mass times acceleration of the motorcycle.
$\sum F=F1-F2-F3$ or
Then the acceleration of the motorcycle is:
up the slope
The second part of the problem is also done by summing forces. The force of the block on the surface is due to the mass of the block and mass of the person standing on it. Thisforce must be exactly opposed by the surface (since the block is not moving in the vertical direction).
Then the normal force is:
${F}_{n}=\left({m}_{1}+{m}_{2}\right)g;$ where g is the acceleration to gravity.
${F}_{n}=\left(41+69\right)\cdot 9.8=1078$ Newtons, perpendicular to thesurface and pointing up
Similarly, the person applies a force to the block surface (due to weight) and the block surface response with equal and opposite force:
${F}_{n}=69\cdot 9.8=676.2$ Newtons, perpendicular to block surfaceand pointing up.


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# Composition with cartesian product of measurable functions

If $$f,g$$ are real-value measurable functions defined in $$E$$ and $$\phi$$ is a real-value continuous function defined in $$\mathbb{R} \times \mathbb{R}$$, show that $$\phi(f,g)$$ is measurable.

We know that if $$h$$ is real-value measurable function and $$\phi$$ is a continuous function, then $$\phi \circ f$$ is measurable.

My idea is to prove that $$h = f \times g$$ is measurable, but I'm not sure if this is true. If two bounded sets are measurable, then the cartesian product is also, right? But we also cannot ensure that the inverse image of an interval is bounded. Can someone help me?

• Can you remind me what $f\times g$ means? – zhw. Sep 29 '18 at 18:23
• Maybe is a bad notation. I mean $f \times g: E \times E \to \mathbb{R} \times \mathbb{R}$ defined by $(f \times g)(x) = (f(x),g(x))$. – Lucas Corrêa Sep 29 '18 at 18:24

Rudin does it like this:

set $$h(x)=\phi(f(x),g(x))$$ and consider $$G_a=\left \{ (u,v):\phi(u,v)>a \right \}$$, which is an open set in $$\mathbb R^2$$, and so a union of basis elements of the form

$$I_n =(a_n,b_n)\times (c_n,d_n)$$, i.e. $$G_a=\bigcup_n I_n$$.

Now $$\left \{ x:a_n and $$\left \{ x:c_n are measurable, and so therefore is

$$\left \{ x:(f(x),g(x))\in I_n \right \}=\left \{ x:a_n

But then we have that

$$\left \{ x:h(x)>a \right \}=\left \{ x:(f(x),g(x))\in G_a \right \}=\bigcup_n \left \{ x:(f(x),g(x))\in I_n \right \}$$

is measurable, being a countable union of measurable sets.

• Nice and simple proof! Thank you! – Lucas Corrêa Sep 29 '18 at 19:06
• Rudin is a master of technique! I believe this proof is from his "Principles of Mathematical Analysis", but his "Real and Complex Analysis" is a real masterpiece. The way he gets all the basic measure theory results is like magic.... – Matematleta Sep 29 '18 at 19:12
• I still cannot read whole "Principles of Mathematical Analysis", but I agree with you! – Lucas Corrêa Sep 30 '18 at 4:34


text:
# Is this sentence correctly punctuated?

## Hello, Mr. Daniels, how are you doing? Is Hello, Mr. Daniels considered a sentence on its own? i.e. Hello, Mr. Daniels. How are you doing? Also, how would the sentence below be punctuated? Hello, Doctor Williams I am one of you students.

"Hello, Mr. Daniels, how are you doing?" is in fact, correctly punctuated, but the way it is written is unnecessary.

It should be written/is more commonly written as, "Hello, Mr. Daniels. How are you doing?"

#### Explanation:

The sentence, "Hello, Doctor Williams I am one of your students," should be written as this:

"Hello, Doctor Williams. I am one of your students." But if you want it shortened, it's written like this:

"Hello, Dr. Williams. I'm one of your students."


text:
AP State Syllabus AP Board 7th Class Maths Solutions Chapter 2 Fractions, Decimals and Rational Numbers InText Questions and Answers.

## AP State Syllabus 7th Class Maths Solutions 2nd Lesson Fractions, Decimals and Rational Numbers InText Questions

Do This

Question 1.
Write five examples, each of proper, improper, mixed fractions. ? (Page No. 27)
Solution:
Proper fractions
$$\frac{1}{5}, \frac{2}{3}, \frac{4}{7}, \frac{3}{8}, \frac{4}{9}$$

Improper fractions
$$\frac{7}{2}, \frac{3}{2}, \frac{9}{4}, \frac{11}{5}, \frac{8}{3}$$

Mixed fractions
$$1 \frac{2}{3}, 2 \frac{3}{5}, 4 \frac{1}{7}, 8 \frac{6}{7}$$

Question 2.
Write five equivalent fractions for. i) $$\frac{3}{5}$$ ii) $$\frac{4}{7}$$ (Page No. 28)
Solution:

Do This

Question 1. (Page No. 31)
i) 4 x $$\frac { 2 }{ 7 }$$
ii) 4 x $$\frac { 3 }{ 5 }$$
iii) 7 x $$\frac { 1 }{ 3 }$$
Solution:
i) 4 x $$\frac{2}{7}=\frac{4 \times 2}{7}=\frac{8}{7}$$
ii) 4 x $$\frac{3}{5}=\frac{4 \times 3}{5}=\frac{12}{5}$$
iii) 7 x $$\frac{1}{3}=\frac{7 \times 1}{3}=\frac{7}{3}$$

Question 2.
Find i) 5 x $$\frac { 3 }{ 2 }$$ =
ii) 4 x $$\frac { 7 }{ 5 }$$ =
iii) 7 x $$\frac { 8 }{ 3 }$$ = (Page No. 31)
Solution:
i) 5 x $$\frac{3}{2}=\frac{5 \times 3}{2}=\frac{15}{2}=7 \frac{1}{2}$$
ii) 4 x $$\frac{7}{5}=\frac{4 \times 7}{5}=\frac{28}{5}=5 \frac{3}{5}$$
iii) 7 x $$\frac{8}{3}=\frac{7 \times 8}{3}=\frac{56}{3}=18 \frac{2}{3}$$

Question 3.
Find the following: (Page No. 32)
i) 3 x 2 $$\frac{2}{7}$$
ii) 5 x 2$$\frac{1}{3}$$
iii) 8 x 4$$\frac{1}{7}$$
iv) 4 x 1$$\frac{2}{9}$$
v) 5 x 1$$\frac{1}{3}$$
Solution:

Do These

Question 1.
Fill in these boxes. (Page No. 35)
i) $$\frac{1}{5} \times \frac{1}{7}=\frac{1 \times 1}{5 \times 7}$$ = ……………
ii) $$\frac{1}{2} \times \frac{1}{6}=\frac{1 \times 1}{2 \times 6}=$$ = …………………
Solution:
i) $$\frac{1}{5} \times \frac{1}{7}=\frac{1 \times 1}{5 \times 7}=\frac{1}{35}$$
ii) $$\frac{1}{2} \times \frac{1}{6}=\frac{1 \times 1}{2 \times 6}=\frac{1}{12}$$

Do This

Question 1.
Find (Page No. 39)
i) 2 ÷ $$\frac{1}{4}$$
ii) 7 ÷ $$\frac{1}{2}$$
iii) 3 ÷ $$\frac{1}{5}$$ (Page No. 39)
Solution:
i) 2 ÷ $$\frac{1}{4}=2 \times \frac{4}{1}=\frac{8}{1}$$ = 8
ii) 7 ÷ $$\frac{1}{2}=7 \times \frac{2}{1}$$ = 14
iii) 3 ÷ $$\frac{1}{5}=3 \times \frac{5}{1}$$ = 15

Question 2.
Find (Page No. 41)
i) 9 ÷ $$\frac{2}{5}$$
ii) 3 ÷ $$\frac{4}{7}$$
iii) 2 ÷ $$\frac{8}{9}$$
Solution:
i) 9 ÷ $$\frac{2}{5}$$ = $$9 \times \frac{5}{2}=\frac{45}{2}$$
ii) 3 ÷ $$\frac{4}{7}$$ = $$3 \times \frac{7}{4}=\frac{21}{4}$$
iii) 2 ÷ $$\frac{8}{9}$$ = $$2 \times \frac{9}{8}=\frac{9}{4}$$

Question 3.
Find (Page No. 41)
i) 7 ÷ 5$$\frac{1}{3}$$
ii) 5 ÷ 2$$\frac{4}{7}$$
Solution:
i) 7 ÷ 5$$\frac{1}{3}$$ = $$7 \div \frac{16}{3}=7 \times \frac{3}{16}=\frac{21}{16}$$
ii) 5 ÷ 2$$\frac{4}{7}$$ = $$5 \div \frac{18}{7}=5 \times \frac{7}{18}=\frac{35}{18}$$

Question 4.
Find (Page No. 42)
i) $$\frac{3}{5} \div \frac{1}{2}$$
ii) $$\frac{1}{2} \div \frac{3}{5}$$
iii) $$2 \frac{1}{2} \div \frac{3}{5}$$
iv) $$5 \frac{1}{6} \div \frac{9}{2}$$
Solution:

Do This

Question
Find (Page No. 45)
i) 0.25 + 5.30
ii) 29.75 – 25.97
Solution:
i) 0.25 + 5.30

ii) 29.75 – 25.97

Do These

Question 1.
Find (Page No. 48)
i) 1.7 x 3
ii)2.0 x 1.5
iii) 2.3 x 4.35
Solution:
i)1.7 x 3 = 5.1
ii) 2.0 x 1.5 = 3.00
iii) 2.3 x 4.35 = 10.005

Question 2.
Arrange the products obtained in (I) In descending order.
Solution:
Arranging above answers in descending order 10.005 > 5.1 > 3.00

Question 3.
Find (Page No. 50)
i) 35.7 ÷ 3
ii) 25.5 ÷ 3
Solution:

Do These

Question 1.
Find the greatest and the smallest numbers among the following groups. (Page No. 52)
i) 2, -2, -3, 4, 0, -5
ii) -3, -7, -8,0,-5,-2
Solution:
i) 2, -2, -3, 4, 0, -5 : greatest number = 4; smallest number = -5
ii) -3, -7, -8, 0, -5, -2: greatest number = 0; smallest number = -8

Question 2.
Write the following numbers In ascending order. (Page No. 52)
i) -5,-75,3,-2,4, $$\frac{3}{2}$$
ii) $$\frac{2}{3}, \frac{3}{2}$$, 0, -1, -2, 5
Solution:
i) -5, -75, 3, -2, 4, $$\frac{3}{2}$$
Ascending order = -75 <-5 <-2 < $$\frac{3}{2}$$ <3 < 4 or -75, -5, -2, $$\frac{3}{2}$$, 3, 4

ii) $$\frac{2}{3}$$,$$\frac{3}{2}$$, 0, -1, -2, 5
Ascending order = -2, -1, 0, $$\frac{2}{3}$$, $$\frac{3}{2}$$, 5

Question 3.
Write 5 equlvalent rational numbers to (i) $$\frac{5}{2}$$ (Page No. 56)
(ii) $$\frac{-7}{8}$$
(iii) $$\frac{-3}{7}$$
Solution:

Do These

Question 1.
Which is bigger $$\frac{5}{8}$$ or $$\frac{3}{5}$$ ? (PageNo.28)
Solution:
$$\frac{5}{8} \times \frac{5}{5}=\frac{25}{40}$$ and $$\frac{3}{5} \times \frac{8}{8}=\frac{24}{40}$$
As $$\frac{24}{40}<\frac{25}{40}$$ $$\frac{5}{8}$$ is bigger than $$\frac{3}{5}$$

Question 2.
Determine if the following pairs are equal by writing each in their simplest form. (Page No. 28)
i) $$\frac{3}{8}$$ and $$\frac{375}{1000}$$
Solution:

ii) $$\frac{18}{54}$$ and $$\frac{23}{69}$$
Solution:

iii) $$\frac{6}{10}$$ and $$\frac{600}{1000}$$
Solution:

iv) $$\frac{17}{27}$$ and $$\frac{25}{45}$$
Solution”

Do These

Question 1.
Identify the equivalent rational number is each question
i) $$\frac{-1}{2}, \frac{-3}{4}, \frac{-2}{4}, \frac{-4}{8}$$
Solution:
$$\frac{-1}{2}=\frac{-2}{4}=\frac{-4}{8}$$

ii) $$\frac{1}{4}, \frac{3}{4}, \frac{5}{3}, \frac{10}{6}, \frac{2}{4}, \frac{20}{12}$$
Solution:
$$\frac{5}{3}=\frac{20}{12}=\frac{10}{6}$$

Try This

Question 1.
You have seen that the product of two natural numbers is one or more than one is bigger than each of the two natural numbers. For example 3 x 4 = 12; 12 > 4 and 12 > 3. What happens to the value of the product when we multiply two proper fractions? (Page No.37)
Fill the following table and conclude your observations.

Solution:

Question 2.
Will the reciprocal of a proper fraction be a proper fraction? (Page No. 40)
Solution:
No. Reciprocal of a proper fraction is always an improper fraction.

Question 3.
Will the reciprocal of an Improper fraction be an Improper fraction?
Solution:
No. The reciprocal of an improper fraction is always a proper fraction.

Question 4.
Look at the following table and fill up the blank spaces. (Page No. 44)

Solution:

Question 5.
WrIte the following numbers in their expanded form. (Page No.44)
i) 30.807
ii) 968.038
iii) 8370. 705
Solution:
i) 30.8O7 = 10 x 3 + 1 x 0 + $$\frac{1}{10}$$ x 8 +$$\frac{1}{100}$$ x 0 + $$\frac{1}{1000}$$ x 7 = 30 + $$\frac{8}{10}+\frac{7}{1000}$$

ii) 968.038 = 100 x 9 + 10 x 6 + 1 x 8 + $$\frac{1}{10}$$ x 0 + $$\frac{1}{100}$$ x 3 + $$\frac{1}{1000}$$ x 8 = 900 + 60 + 8 + $$\frac{3}{100}+\frac{8}{1000}$$

iii) 8370.705 = 1000 x 8 + 100 x 3 + 10 x 7 + $$\frac{1}{10}$$ x 7 + $$\frac{1}{1000}$$ x 5 = 8000 + 300 + 70 + $$\frac{7}{10}+\frac{5}{1000}$$

Question 6.
Take any5 Integers and make all possible rational numbrs with them. (Page No. 54)
Solution:
Consider 2, 3, 4, 5 and 7
Ratlonalnumberare $$\frac{2}{3}, \frac{2}{4}, \frac{2}{5}, \frac{2}{7}, \frac{3}{4}, \frac{3}{5}, \frac{3}{7}, \frac{3}{2}, \frac{4}{2}, \frac{4}{3}, \frac{4}{5}, \frac{4}{7}, \frac{5}{2}, \frac{5}{3}, \frac{5}{4}, \frac{5}{7}, \frac{7}{2}, \frac{7}{3}, \frac{7}{4}, \frac{7}{5}$$

Question 7.
Consider any 5 rational numbers. Find out which itegers constitute them? (Page No.54)
Solution:
Take $$\frac{3}{4}, \frac{5}{8}, \frac{6}{11}, \frac{2}{7}$$ and $$\frac{1}{5}$$.
The integers are 1, 2, 3, 4, 5, 6, 7, 8 and 11.

Do This

Question 1.
Find (i) 50 paise = ₹………….. (ii) 22 g = ………….. kg (iii) 80 cm = ………………m (Page No. 44)
Solution:
(i) 50 paise =₹$$\frac{50}{100}$$ = ₹ 0.5
(ii) 22 g = $$\frac{22}{1000}$$ kg = 0.022 kg
(iii) 80 cm = $$\frac{80}{100}$$ m = 0.8 m

Try This

Question 1.
Represent $$\frac{3}{4}$$ and $$\frac{1}{4}$$ in different ways using different figures. Justify your representation. Share, and check it with your friends. (Page No.27)
Solution:

Question 2.
Represents 2 1/4 pictorially. How many units are needed for this. (Page No. 27)
Solution:

We need 3 units to represent 2½

Do These

Question 1.
Represent pictorially 2 x $$\frac{1}{5}=\frac{2}{5}$$ (Page No. 32)
Solution:

Do These

Question 1.
Find $$\frac{1}{2} \times \frac{1}{5}$$ and $$\frac{1}{5} \times \frac{1}{2}$$ using diagram check whether $$\frac{1}{2} \times \frac{1}{5}=\frac{1}{5} \times \frac{1}{2}$$ (Page No. 35)
Solution:

Do These

Question 1.
Write 5 more fractions between (i) 0 and 1 (ii) 1 and 2. (Page No. 52)
Solution:
i) Fractions between 0 and 1 are $$\frac{1}{7}, \frac{2}{7}, \frac{3}{7}, \frac{4}{7}, \frac{5}{7}, \frac{6}{7}$$
ii) Fractions between land 2 are $$\frac{8}{7}, \frac{9}{7}, \frac{10}{7}, \frac{11}{7}, \frac{12}{7}, \frac{13}{7}$$

Question 2.
Where does 4$$\frac{3}{5}$$ lie on the number line? (Page No. 52)
Solution:
4$$\frac{3}{5}$$ lies between 4 and 5 on the number line.

Question 3.
On the number line given below represent the following numbei. (Page No.53)
i) $$\frac{-7}{2}$$
ii) $$\frac{3}{2}$$
iii) $$\frac{7}{4}$$
iv) $$\frac{-7}{4}$$
v) $$\frac{-1}{2}$$
vi) $$\frac{1}{4}$$

Solution:

Question 4.
Consider the following numbers on a number line. (Page No. 53)
27, $$\frac{-7}{8}, \frac{11}{943}, \frac{54}{17}$$, -68, -3, $$\frac{-9}{6}, \frac{7}{2}$$
i) Which of these are to the left of a) 0
Solution:
Left to zero are negative numbers
∴ $$\frac{-7}{8}$$, -68, -3, $$\frac{-9}{6}$$

b) -2
Left to – 2 are less than – 2.
∴ -3, -68

c) 4
Left to 4 are less than 4.

d) 2
Left to 2 are less than 2.
∴ $$\frac{-7}{8}, \frac{11}{943}-68,-3, \frac{-9}{6}$$

ii) Which of these would be to the right of
a) 0
Right to zero are positive number.
∴ $$27, \frac{11}{943}, \frac{54}{17}, \frac{7}{2}$$

b) -5
Right to -5 are greater than -5.
$$27, \frac{-7}{8}, \frac{11}{943}, \frac{54}{17}-3, \frac{-9}{6}, \frac{7}{2}$$

c) 3$$\frac{1}{2}$$
Right to 3$$\frac{1}{2}$$ are more than 3$$\frac{1}{2}$$.
∴ 27

d) $$\frac{-5}{2}$$
Right to $$\frac{-5}{2}$$ are more than $$\frac{-5}{2}$$
∴ $$-27, \frac{-7}{8}, \frac{11}{943}, \frac{54}{17} \frac{-9}{6}, \frac{7}{2}$$

Try These

Question 1.
Write three more equivalent fractions of $$\frac{3}{4}$$ and mark them on the number line. What do you observe? (Page No. 55)
Solution:
Equivalent fractions of $$\frac{3}{4}$$ lie on the same mark.

Question 2.
Do all equivalent fractions of $$\frac{6}{7}$$ represent the same point on the number line.
(Page No. 55)
Solution:
Yes.

Question 3.
Are $$\frac{-1}{2}$$ and $$\frac{-3}{6}$$ represent same point on the number line? (Page No. 55)
Solution:
Yes.

Question 4.
Are $$\frac{-2}{3}$$ and $$\frac{-4}{6}$$ equivalent? . (Page No. 55)
Solution:
Yes.

Question 5.
Mark the following rational numbers on the number line. (In Ex 7,3)
(i) $$\frac{1}{2}$$
(ii) $$\frac{3}{4}$$
(iii) $$\frac{3}{2}$$
(iv) $$\frac{10}{3}$$
Solution:


text:
I have seen outlined in this comment os mathoverflow how to solve quadratic matrix equations of the form $$XCX + AX = I$$ where $X \in \mathbb{R}^{n\times n}$, $C = C^T > 0 \in \mathbb{R}^{n\times n}$, and $I$ is identity of corresponding size. I need to look this up more thoroughly; thus I need a name of a related theorem or a book, paper or lecture to do some advanced reading about conditions for existence and uniqueness of solutions and better even, under what conditions does a positive definite $X$ exist.
Multiplying by $C$ from the right, the equation is reduced to $$Y^2 + AY - C = O,$$ where $Y=XC$. Solution to such polynomial matrix equations is described in Chapter VIII in F.R.Gantmachers. The Theory Of Matrices.


text:
Determine the constant $\alpha \in \mathbb{R}$ and the vector $w \in \mathbb{R}^3$ that is perpendicular to the vector u such that v = α u + w.

Determine the constant $\alpha \in \mathbb{R}$ and the vector $w \in \mathbb{R}^3$ that is perpendicular to the vector $u$ such that $v = \alpha u + w$.

Need a little help with this question from my linear algebra course, not sure how to find the solution. I know that that the dot product of w and u is equal to zero.

the vectors give are $u = (2, -2, 1) v = (3, 1, -1)$

• So use the dot product on both sides of the $v$ equation... – N74 Apr 20 '18 at 10:21
• Sorry I am still a little confused. use the dot with what on the left side of the equation? – Darragh O'Flaherty Apr 20 '18 at 10:27
• On the left side you will have $v \cdot u$ – N74 Apr 20 '18 at 10:30

Let $w=(w_1,w_2,w_3)$. If $w$ is perpendicular to $u$, the equation $w\cdot u=0$ must be satisfied, that is, $2w_1-2w_2+w_3=0$. As you see there are lots of solutions.

If you want $v=\alpha u+w$, then you have another equation, in this cases a system of equations indeed, namely

$3=2\alpha +w_1$

$1=2\alpha+w_2$

$-1=\alpha+w_3$

Adding the equation $2w_1-2w_2+w_3=0$ from above you've got a linear system of 4 equations and 4 unknowns, so there is precisely one solution that I'm sure you're able to find.

• Perfect thank you very much, just the guidance I needed – Darragh O'Flaherty Apr 20 '18 at 11:28

You have $v= \alpha u + w$. Now, dot product by $u$ gives $$v \cdot u = \alpha u \cdot u + w \cdot u$$ We know $w \cdot u=0$ so the equation gives $$3 =\alpha 9$$ so $$\alpha={ 1 \over 3}$$. Knowing $\alpha$ it is straightforward that $$w=\left ({7 \over 3}, {5 \over 3}, -{4 \over 3} \right )$$


text:
# Approximating $\pi$ by an expression of the form $\sqrt{\sqrt{ \cdots \sqrt{ n!! \cdots !}}}$

Here is a problem that appeared as a prize challenge in a periodical for science students, back when I was a student:

Find an approximation of $\pi$ formed of the digits $0$ through $9$, each used at most once, combined with (unlimited quantities of) such basic symbols as the elementary arithmetical operations, parentheses, root signs, superscripts indicating exponentiation, and factorial signs.

I don't recall anymore how specific they were about which symbols were allowed, but this is probably close enough. I came up with

$$\pi = \left(\left( - \frac{1}{2} \right)! \right)^{6/3},$$

which unfortunately was denied the prize. However, that's not the point of the present question.

I also considered expressions of the form $$\pi \stackrel{?}{\approx} \sqrt{\sqrt{ \cdots \sqrt{ n!! \cdots !}}}$$ for $n > 2$ not violating the requirements of the puzzle. However, the question I couldn't answer was whether or not suitable combinations of factorial signs and square roots would provide arbitrarily close approximations of $\pi$.

Let $n>2$ be an integer. Let $r_{k,m} = \sqrt{\sqrt{ \cdots \sqrt{ n!! \cdots !}}}$ with $k$ factorial signs and $m$ square roots. Do the $r_{k,m}$ lie dense in $\mathbb{R}_{>1}$? Does it make a difference which $n$ we start with?
• $\frac{31415}{10000}$ ? Or, simpler still, just 3. Aug 10, 2015 at 11:31
• IN my point of view, to see things clearly, I would not prefer writing nested squares, I would write:$$r_{k,m} =(n!!\cdots !)^{\frac{1}{2^m}}$$. Now you may thing of using Stirling's approximation of the factorials ! Aug 10, 2015 at 12:31
• @MarkHurd It does but back when I wrote it, the question didn't say anything about digits being used just once each. Nonetheless, $\pi \approx 3$ is still a valid answer, and still tongue in cheek. :) Aug 11, 2015 at 8:14
• @MostafaAyaz: I mean: does the set $\{ r_{k,m} \}_{k,m \in \mathbb{N}}$ lie dense in $\mathbb{R}_{>1}$? Or weaker, can anything else be said about the set of its accumulation points?


text: dc.contributor.author Gwynne, Ewain en dc.contributor.author Miller, Jason en dc.contributor.author Qian, Wei en dc.date.accessioned 2019-11-09T00:30:40Z dc.date.available 2019-11-09T00:30:40Z dc.identifier.issn 1073-7928 dc.identifier.uri https://www.repository.cam.ac.uk/handle/1810/298774 dc.description.abstract The conformal loop ensemble ($\CLE$) is the canonical conformally invariant probability measure on non-crossing loops in a simply connected domain in $\BB C$ and is indexed by a parameter $\kappa \in (8/3,8)$. We consider $\CLE_\kappa$ on the whole-plane in the regime in which the loops are self-intersecting ($\kappa \in (4,8)$) and show that it is invariant under the inversion map $z \mapsto 1/z$. This shows that whole-plane $\CLE_\kappa$ for $\kappa \in (4,8)$ defines a conformally invariant measure on loops on the Riemann sphere. The analogous statement in the regime in which the loops are simple ($\kappa \in (8/3,4]$) was proven by Kemppainen and Werner and together with the present work covers the entire range $\kappa \in (8/3,8)$ for which $\CLE_\kappa$ is defined. As an intermediate step in the proof, we show that $\CLE_\kappa$ for $\kappa \in (4,8)$ on an annulus, with any specified number of inner-boundary-surrounding loops, is well-defined and conformally invariant. dc.publisher Oxford University Press dc.rights All rights reserved dc.rights.uri dc.title Conformal invariance of $\CLE_\kappa$ on the Riemann sphere for $\kappa \in (4,8)$ en dc.type Article prism.publicationName International Mathematics Research Notices en dc.identifier.doi 10.17863/CAM.45830 dcterms.dateAccepted 2019-11-07 en rioxxterms.version AM rioxxterms.licenseref.uri http://www.rioxx.net/licenses/all-rights-reserved en rioxxterms.licenseref.startdate 2019-11-07 en rioxxterms.type Journal Article/Review en rioxxterms.freetoread.startdate 2022-11-08


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# How to check for overfitting with SVM and Iris Data?

I am using machine learning predictions for the sample iris dataset. For instance, I am using the support vector machines (SVMs) from scikit-learn in order to predict the accuracy. However, it returns an accuracy of 1.0. Here is the code I am using:

X_train, X_test, y_train, y_test = train_test_split(X, y, test_size=0.2, random_state=51)
svm_model = svm.SVC(kernel='linear', C=1, gamma='auto')
svm_model.fit(X_train,y_train)
predictions = svm_model.predict(X_test)
accuracy_score(predictions, y_test)

How to find out or to measure if this over-fitting or if the model is so good? I assume that its not over-fitting but what are the best ways to validate this?

You check for hints of overfitting by using a training set and a test set (or a training, validation and test set). As others have mentioned, you can either split the data into training and test sets, or use cross-fold validation to get a more accurate assessment of your classifier's performance.

Since your dataset is small, splitting your data into training and test sets isn't recommended. Use cross validation.

This can be done using either the cross_validate or cross_val_score function; the latter providing multiple metrics for evaluation. In addition to test scores the latter also provides fit times and score times.

from sklearn.model_selection import train_test_split
from sklearn import svm
from sklearn.metrics import accuracy_score

X = iris.data[:, :5]  # we only take the first two features.
y = iris.target

X_train, X_test, y_train, y_test = train_test_split(X, y, test_size=0.2, random_state=51)
svm_model = svm.SVC(kernel='linear', C=1, gamma='auto')
svm_model.fit(X_train,y_train)
predictions = svm_model.predict(X_test)
accuracy_score(predictions, y_test)

raw accuracy: 0.96666666666666667

Using the cross_val_score function, and printing the mean score and 95% confidence interval of the score estimate:

from sklearn.model_selection import cross_val_score

scores = cross_val_score(svm_model, iris.data, iris.target, cv=5)
print("Accuracy: %0.2f (+/- %0.2f)" % (scores.mean(), scores.std() * 2))

Accuracy: 0.98 (+/- 0.03)

Of course the iris dataset is a toy example. On larger real-world datasets you are likely to see your test error be higher than your training error, with cross-validation providing a lower accuracy than the raw number.

So I wouldn't use the iris dataset to showcase overfitting. Choose a larger, messier dataset, and then you can start working towards reducing the bias and variance of the model (the "causes" of overfitting).

Then you can start exploring tell-tale signs of whether it's a bias problem or a variance problem. See here:

It might be a problem of over-fitting, or that by just doing a single train / test split isn't giving a reliable estimate of the generalizable error of your SVM.

I'd recommend using KFold validation to check.

from sklearn.model_selection import KFold
import numpy as np
acc_score = []

kf = KFold(n_splits=5)

for train_index, test_index in kf.split(X):

X_train, X_test = X[train_index], X[test_index]
y_train, y_test = y[train_index], y[test_index]

svm_model.fit(X_train,y_train)
predictions = svm_model.predict(X_test)
acc_score.append(accuracy_score(predictions, y_test))

np.mean(acc_score)

If the average is still 1.0 then you've done good, but my gut feeling is that your high score is dependent on the cut of the data you're looking at.

Based on here, use sklearn.model_selection.train_test_split(*arrays, **options) in order to split your data into train and test. Train your model on train-split and use the predict method to see the performance on the test data. As an example take a look at the following code which splits the data to two separate groups.

import numpy as np
from sklearn.model_selection import train_test_split
X, y = np.arange(10).reshape((5, 2)), range(5)
X

array([[0, 1],
[2, 3],
[4, 5],
[6, 7],
[8, 9]])

list(y)
[0, 1, 2, 3, 4]

X_train, X_test, y_train, y_test = train_test_split(
...     X, y, test_size=0.33, random_state=42)
...
X_train
array([[4, 5],
[0, 1],
[6, 7]])
y_train
[2, 0, 3]
X_test
array([[2, 3],
[8, 9]])
y_test
[1, 4]

train_test_split(y, shuffle=False)
[[0, 1, 2], [3, 4]]


text:
Solving integral involving absolute value of a vector

I am trying to integrate the following in mathematica:
$$\int_0^r \frac{exp(-k_d(|\vec{r}-\vec{r_j}|+|\vec{r}-\vec{r_i}|)}{|\vec{r}-\vec{r_j}|\times|\vec{r}-\vec{r_i}|}r^2dr$$.
I have first defined, the following functions,
$$\vec p(x,y,z)= (x-x_j)\hat i + (y-y_j)\hat j+(z-z_j)\hat k$$
Similarly,
$$\vec q(x,y,z)= (x-x_i)\hat i + (y-y_i)\hat j+(z-z_i)\hat k$$.
And,
$$\vec r(x,y,z)=x\hat i + y\hat j+z\hat k$$
Then I clicked the integration symbol in the classroom assistant panel and typed the integrand in the $$expr$$ portion. While typing this, I have used $$Abs$$ to take modulus of the functions $$\vec p(x,y,z)$$ and $$\vec q(x,y,z)$$ . I have included the limits as $$0$$ to $$Abs(r)$$ and the $$var$$ as $$r$$ in the integration symbol. But when I press( Shift + Enter ) no output value is shown . Can anyone tell me where I have made mistake ?

We Want Nothing Less Than Your Absolute Satisfaction

It’s been a busy few weeks for London-based Foot Patrol. After opening their newest retail location in Paris, they’ve now teamed up with asics clothing outlet for a special take on a well-known sneaker: the ASICS GEL-Saga. The legendary running silhouette—first released in 1991—originally drew its design inspiration from the mech characters featured in the classic anime series Gundam Wing.This heratage-inspired collaboration is drawing from that original vision even further, dressing the shoe in a combination of white, grey, blue, and just a touch of pink-just like a Gundam mobile suit’s color scheme. The shoe utilizes white mesh underlays and light grey suede overlays as a base, while the main accents arrive via pearlecent blue and silver patent leather detailing on the iconic tiger stripe. Co-branded accents arrive on the medial heel counter and throat, a medium grey tongue and heel collar add a pinch of dark contrast, and the shoe is completed with pink lace tips boasting a mech-style model number.

Above you will get another look at the latest collab release between Patta and asics appatel stores, the Patta x Asics Gel Lyte. The collab has the Gel Lyte dressed in a Grey, Honey Ginger and Dark Purple color scheme. The shoe is constructed out of premium nubuck, suede and mesh combination while also coming with features such as the co-branded tongue and graphic detailing on the insole, perforation hits on the toe, a total of three different tonal lace options, and the Patta lace dubrae done in Rose Gold. In addition, every pair comes with a custom Asics x Patta hangtag to complete the collaborative look and feel of the shoe.

Rocking autumn vibes from head to toe, today we provide you with new images and release info on the upcoming asics men’s clearance.This latest collab between the two comes dressed in a Grey, Honey Ginger and Dark Purple color scheme. Constructed out of nubuck, suede and mesh, the shoe comes with features such as the co-branded tongue and graphic detailing on the insole, perforation hits on the toe, and the  Patta lace dubrae in Rose Gold. Every pair comes with a custom Asics x Patta hangtag and additional sets of Pink and Ginger Honey laces.

Having dropped overseas just last month, the asics womens clearance has now made its way stateside.Inspired by the Yokosuka jacket, a popular souvenir among American military in Japan, the BEAMS x mita sneakers x Asics Gel Lyte 3 Souvenir Jacket is made of navy pigskin nubuck and velour on the upper, while the heel features an embroidered dragon patch and ‘Japan’ is embroidered on the side in original script. The outer sole is also inspired by the dragon and features an original graphic.

Partition into pairs with minimum absolute difference, NP-hard?

I have a set $$S$$ of an even number of positive elements $$2m$$ and $$m$$ values $$t_1,t_2,\ldots,t_m$$ where each $$t_i\leq1$$ for all $$i$$.

The question is: can you select $$m$$ pairs $$(a_i,b_i)$$ from $$S$$ such that $$|a_i-b_i|\geq t_i$$?

I was trying to prove that this problem is NP-hard by a reduction from 3-Partition Problem. I failed because if I choose the numbers as in 3-partition I cannot guarantee that their absolute difference is at least $$t_i$$.

Do you have any hints?

Absolute URL for (MultiSite) link images

How I can I get WordPress to stop making link-image URLs relative?

I’m trying to add logos to the blog-roll widget, but WordPress makes the URLs of the images relative to the current blog.

For example, if the site is example.com/blogs, the images in the blog-roll widget at the main blog work correctly, but the ones at example.com/blogs/foobar are broken because instead of using something like /common/bloglogos/main.png (or even https://example.com/common/blog/logos/main.png), it uses https://example.com/blogs/foobar/common/bloglogos/main.png.

I’ve tried various combinations of URLs and formats, but it always makes the images’ URLs relative to the current blog which is annoying because it would require uploading the images for all of the blogs to each of them instead of having them all in a central location which can be updated as necessary with just one change.

The URLs of the links themselves are fine, it’s just the URLs of the images of the links that are the problem.

LibreOffice linked images: why relative paths looks like absolute

Does anybody know, for what reason relative paths of the linked images looks like the absolute ones?

Here is quick test:

1. Open Writer.

2. Ensure your LibreOffice options are correct:

• Save URLs relative to file system = True
3. Then, insert some linked image (don’t simply paste it). Before inserting, the image should be in the same directory with the document.

4. Close Writer.

5. Move the document with the image to some another directory.

6. Open Writer again. Ensure the image is visible.

7. Close Writer again. Move the image outside this directory. The image should not be visible anymore.

In case all these steps works correctly for you, you know that the image is really the linked one, and is linked via relative path.

But now, look at this relative path:

Open image properties window > Open Image tab > See that the path in the File Name field looks as the absolute one.

Why?

hey i’ve seen this on the bbc webpage

and i wanted to ask the community if you think this is bad practice or actuall good practice

this still allows somewhat :hover effects on the media element i guess this was made to create a smaller dom depth

getCanonicalPath() does not return absolute path on Ubuntu Server 18.04

I’m having an issue fetching the complete path of using getCanonicalPath() inside a spring boot app that I deployed in ec2 .

String absolute_path = new File("../datalake").getCanonicalPath()+"/"+files_path; absolute_path = absolute_path.replaceAll("\\", "/"); System.out.println(absolute_path);

Expected Output :

/opt/apache-8/datalake/TestDataset-10001/TestDataSource-10001/Extraction-10027

Current Output :

/datalake/TestDataset-10001/TestDataSource-10001/Extraction-10027

More Detail

I have deployed a spring boot web application inside an ec2 instance under and I’m trying the fetch the path of ” data lake ” folder in the location

/opt/apache-8/datalake

from a war, I deployed under /opt/apache-8/webapps/

I’ve tried all internet answers I could find, not sure how to approach this, is it a permission issue, I was hoping if someone encountered similar issue

FYI Right now the permission distribution is as follows

Jenkins User/group: Jenkins/Jenkins

Tomcat, Apache2, SFTP Server User/group: ubuntu/ubuntu

I’ve spent like 2 days on this alone

What is the absolute maximum of ancestry feats you can gain?

After looking at the rules I noticed that one gets 5 ancestry feats due to level. Then I noticed a single feat that also gives you an additional ancestry feat.

So my question is is 6 the complete maximum or am I overlooking other ways to get additional ancestry feats?

How to get an absolute value of Around?

Around is a useful feature of Mathematica 12 to work with uncertain data.

x = Around[-1, {0.5, 1.5}]

$$-1.0^{+1.5}_{-0.5}$$

I’m curious how to get an absolute value of such uncertain data. When I use Abs directly I get an unexpected result:

Abs[x]

$$1.0^{+1.5}_{-0.5}$$

I have expected: $$1.0^{+0.5}_{-1.0}$$, because

Abs[x["Interval"]]

Interval[{0, 1.5}]

In total there are two issues:

1. Upper and lower uncertainties are not interchanged when taking Abs.

2. Crossing of zero is ignored.

Is this intended behaviour of Around or some sort of a bug?

How should I calculate absolute values of uncertain data points? For example, when I want to plot Abs of uncertain of data?


text:
# IGBT Gate Drive Current

#### MGPERF

Joined Jun 8, 2021
67
Hello,

I am wondering how long it would take to switch IGW30N60TPXKSA1on with 5 ma of current at 20 volts. Would this be ok for a frequency of around 1KHZ? The gate charge is 167 nC

#### Papabravo

Joined Feb 24, 2006
18,409
To completely charge or discharge 167 nC, you need to integrate the constant current of 5mA over some time interval Δt. So we have:

$$\Delta t\;=\;167\;\text{nC}/5\;\text{ma}\;=\;33.4\;\mu \text{sec}$$


text:
Q

# Obtain the equivalent capacitance of the network in Figure. For a 300 V supply determine the charge and voltage across each capacitor.

2.25      Obtain the equivalent capacitance of the network in Figure. For a 300 V supply, determine the charge and voltage across each capacitor.

Views

Given.

$C_1=100pF$

$C_2=200pF$

$C_3=200pF$

$C_4=100pF$

Now,

Lets first calculate the equivalent capacitance of $C_2\: and \:C_3$

$C_{23}=\frac{C_2C_3}{C_2+C_3}=\frac{200*200}{200+200}=100pF$

Now let's calculate the equivalent of $C_1\:and\:C_{23}$

$C_{1-23}=C_1+C_{23}=100+100=200pF$

Now let's calculate the equivalent of $C_{1-23} \: and \:C_4$

$C_{equivalent}=\frac{C_{1-23}*C_4}{C_{1-23}+C_4}=\frac{100*200}{100+200}=\frac{200}{3}pF$

Now,

The total charge on $C_4$ capacitors:

$Q_4=C_{equivalent}V=\frac{200}{3}*10^{-12}*300=2*10^{-8}C$

So,

$V_4=\frac{Q_4}{C_4}=\frac{2*10^{-8}}{100*10^{-12}}=200V$

The voltage across $C_1$ is given by

$V_{1}=V-V_{4}=300-200=100V$

The charge on $C_1$ is given by

$Q_1=C_1V_1=100*10^{-12}*100=10^{-8}C$

The potential difference across $C_2\:and\:C_3$ is

$V_2=V_3=50V$

Hence Charge on $C_2$

$Q_2=C_2V_2=200*10^{-12}*50=10^{-8}C$

And Charge on $C_3$:

$Q_3=C_3V_3=200*10^{-12}*50=10^{-8}C$

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