## definition – Chapter 2, Exercise 7. Rudin Functional Analysis. Clarification of multiplier property.

Can you clarify what is meant by “multiplier property” in the statement of this exercise?

Let $$C(T)$$ be the set of all continuous complex functions on the unit circle $$T$$. Suppose $$left{ gamma_n right}, (n in mathbb{Z})$$ is a complex sequence that associates to each $$f in C(T)$$ a function $$Lambda f in C(T)$$ whose fourier coefficients are
$$(Lambda f){hat{}}(n) = gamma_n hat{f}(n) ;; (n in mathbb{Z}).$$
Prove that $$left{ gamma_n right}$$ has this multiplier property if there’s a complex Borel measure $$mu$$ on $$T$$ such that
$$gamma_n = int e^{-intheta} dmu(theta) ;; (n in mathbb{Z})$$

Is it meant which $$gamma_n$$ makes the Fourier series exist?

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## Spivak’s Calculus Chapter 1, Question 19a

I found this post as solution to the question. Here’s a quote for easy reference.

Supposing $$y_1$$ and $$y_2$$ are not both $$0$$, and that there is no number $$lambda$$ such that $$x_1=lambda y_1$$ and $$x_2=lambda y_2$$, then $$begin{array}{tcl}0 &<& (lambda y_1-x_1)^2 + (lambda y_2-x_2)^2 \ &=& lambda^2 (y_1^2+y_2^2)-2lambda(x_1y_1+x_2y_2)+(x_1^2+x_2^2),end{array}$$ and the equation $$lambda^2 (y_1^2+y_2^2)-2lambda(x_1y_1+x_2y_2)+(x_1^2+y_1^{2 *})=0 \$$
has no solution $$lambda$$. So by problem 18(a) we must have $$Bigg(frac{2(x_1y_1+x_2y_2)}{({y_1}^2+{y_2}^2)}Bigg)^2-frac{4({x_1}^2+{x_2}^{2 *})}{({y_1}^2+{y_2}^2)} < 0,****** \$$ which yields the Schwarz inequality.

Notice the heavily asterisked line. I don’t understand how we derive this. I recognize that this is “completing the square.” Question 18 emphasized that $$b^2 – 4c < 0$$ means $$x^2 + bx + c > 0$$. Except, in this problem, it’s not clear why he choose the $$b$$ the way he did. Where does the 2 come from? In the sense that, isn’t $$b = frac{-2(x_1y_1+x_2y_2)}{({y_1}^2+{y_2}^2)}$$.

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## Real Analysis – Spivak's Calculus Chapter 7, Task 19 (a): A Logical Question

The problem is as follows:

Suppose that $$f$$ is continuously on $$(0.1)$$ and $$f (0) = f (1)$$. To let $$n$$ be any natural number. Prove that there is a number $$x$$ so that $$f (x) = f (x + frac {1} {n}).$$

I was wondering if my proof is logically sound, especially the last piece. I tried to apply the logic of the intermediate value theorem, but I was curious to see whether the contradiction actually gave the desired result:

To let $$g$$ be a function that $$g (x) = f (x + frac {1} {n}) – f (x)$$. We want to show that there is one $$x in (0.1)$$ so that $$g (x) = 0$$. We prove by contradiction.

1. Accept $$forall x in (0,1), g (x) <0.$$ Then for $$i = 0, 1, 2, …, n$$ we have that $$g ( frac {i} {n}) <0$$. Therefore, $$f (0)> f ( frac {1} {n})> f ( frac {2} {n})> …> f (1),$$ which means, that $$f (0) neq f (1)$$.
2. Following a similar logic follows from this $$g (x)> 0$$ can't keep for everyone $$x in (0.1)$$.

Therefore it must exist $$a, b in (0.1)$$ so that $$a$$ and $$b$$ have a different sign, i.e. $$g (a) leq 0 leq g (b)$$. According to the intermediate value theorem exists $$c$$ so that $$g (c) = 0$$and we're done.

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## Create a different heading for each chapter on pages 8.2.1

Can I change the heading of the current chapter automatically? In picture we can see that the previous chapter (title) in the header (I text in the cell, not automatically) and when start title 2, the header still titleIf I change title With title 2 Modify the entire document.

## Introduction to Modern Geometry shively Chapter 3 Exercise 10

Intersects a circle at the points P, P & #; to the sides BC, CA and AB of the triangle ΔABC; Q, Q & # 39 ;; R, R & # 39 ;, and if AP, BQ, and CR are simultaneous, then AP & quot ;, BQ & quot; YCR & # 39; simultaneously.

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## Is there an Android audio player that will allow you to jump to the next chapter in a file?

OK, here is what I mean.

I ripped a collection of .flac files from a CD.

I used a MKA tool on Windows to merge them into a large MKA file with chapter markers between each file.

With VLC Player I can jump from mark to mark on my desktop in this one big MKA file as if this file was a CD.

The problem is, for some reason, VLC for Android does not have this functionality.

Also, I can not find an audio player for Android that offers the "Next Chapter" functionality I'm looking for.

I understand that all available audio players have the "Next item in the playlist" button.

But that does not interest me.

Does anyone know an Android audio player that allows you to jump from chapter to chapter within an MKA file?

Many thanks.

## I could not find Chapter 1 of the JMU open source textbook on the cs361 computer system fundamentals course

Thank you for sending a reply to Computer Science Stack Exchange!

But avoid

• Make statements based on opinions; Cover them with references or personal experience.

Use MathJax to format equations. Mathjax reference.

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## Python – CTCI Chapter 1: Palindrome Permutation

Below is my code for problem 1.4 in CTCI. I would like a review of my code and if my approach to the problem is correct.

Problem:
Palindrome permutation: Write a function for a given string to see if it is a permutation
a palindrome. A palindrome is a word or phrase that is the same forwards and backwards. ON
Permutation is a rearrangement of letters. The palindrome does not have to be limited to just
Dictionary words.

EXAMPLE

Input: Tact Coa

Output: True (permutations: "taco cat", "atco cta", etc.)

My solution (Python):

def checkPalindromeAndPermutation(inputstr):
lengthOfInputString = len(inputstr)
counterforodd = 0
actualCharactersInInput = 0
inputstr = inputstr.lower()
hashTable = dict()

for i in inputstr:
if i != " ":
hashTable(i) = hashTable.get(i, 0) + 1
actualCharactersInInput = actualCharactersInInput + 1
print(hashTable)

for item in hashTable:
# if input has even length, but each character's frequency is not even, then it's not a plaindrome
if actualCharactersInInput % 2 == 0 and hashTable(item) % 2 != 0:
return False
# if input has odd length, but more than one character's frequency is greater than 1 , then it's not a plaindrome
if actualCharactersInInput % 2 == 1 and hashTable(item) % 2 == 1:
counterforodd = counterforodd + 1
if counterforodd > 1:
return False
return True

print("Answer : " , checkPalindromeAndPermutation("abc bac"))


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## To Atiyah-Macdonald Chapter 3 Exercise 27 ii)

To let $$B = Pi _ { alpha} B _ { alpha}$$ is the direct product of finitely many $$A$$ algebras $$f _ { alpha}: A to B _ { alpha}$$, Then, when $$f: A to B$$ is given by $$f (x) = (f _ { alpha} (x)) _ { alpha}$$ show that
$$f ^ * (Spec (B)) = bigcup _ { alpha} f _ { alpha} ^ * (Spec (B))$$

My proof: When all tensors are over $$A$$ and for rent $$k$$ be the rest field on a certain main ideal $$p in Spec (A)$$ we have that

begin {align *} f ^ {* – 1} (p) & = Spec (B otimes k) \ & = Spec left ( bigoplus _ { alpha} B _ { alpha} right) otimes k right) \ & = Spec left ( bigoplus _ { alpha} B _ { alpha} otimes k right) \ & = bigsqcup _ { alpha} Spec left (B _ { alpha} otimes k right) \ & = bigsqcup _ { alpha} f _ { alpha} ^ {* – 1} (p) quad quad (1) end {align *}

So we have that $$f ^ {* – 1} (p) neq varnothing iff exists alpha , ,, , , f _ { alpha} ^ {* – 1} (p) neq varnothing$$, That gives us that
$$f ^ * (Spec (B)) = bigcup _ { alpha} f _ { alpha} ^ * (Spec (B))$$

My only concern is that there is a disjoint union in Equation (1). I was wondering if this disjointed union is more external than internal.

Many Thanks,
Vatsa

## 3. l1

BlackHatKings: Crypto speculation and investment
Posted by: ThomasLound
Post Time: June 29, 2019 at 17:42.

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