- #26

etotheipi

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- #26

etotheipi

- #27

StoneTemplePython

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The proof without this method can be given in a single line.

It took me 3 lines, courtesy of ##\text{GM} \leq \text{AM}##, with a little clever padding

##\big(\frac{2}{c}\big)^\frac{1}{3} = \big(\frac{1}{r}\frac{1}{r}\frac{2}{h}\big)^\frac{1}{3} \leq \frac{1}{3}\big(\frac{1}{r} + \frac{1}{r} + \frac{2}{h}\big) = \frac{2}{3}\big(\frac{h +r}{rh}\big) \cdot \frac{r}{r}= \frac{2}{3}\cdot \frac{rh +r^2}{r^2 h} = \frac{2}{3}\cdot \frac{rh +r^2}{c}##

Rescale by ##3 c \pi## for standard form. Lower bound equality is met

- #27

fresh_42

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Or a bit more compact:It took me 3 lines, courtesy of ##\text{GM} \leq \text{AM}##, with a little clever padding

##\big(\frac{2}{c}\big)^\frac{1}{3} = \big(\frac{1}{r}\frac{1}{r}\frac{2}{h}\big)^\frac{1}{3} \leq \frac{1}{3}\big(\frac{1}{r} + \frac{1}{r} + \frac{2}{h}\big) = \frac{2}{3}\big(\frac{h +r}{rh}\big) \cdot \frac{r}{r}= \frac{2}{3}\cdot \frac{rh +r^2}{r^2 h} = \frac{2}{3}\cdot \frac{rh +r^2}{c}##

Rescale by ##3 c \pi## for standard form. Lower bound equality is metiff##\frac{1}{r} = \frac{2}{h}\longrightarrow h = 2r##

$$\dfrac{A}{3\pi} = \dfrac{2r^2+rh+rh}{3}\stackrel{AM\geq GM}{\geq}\sqrt[3]{2r^2\cdot rh\cdot rh}= \sqrt[3]{\dfrac{2V^2}{\pi^2}} =: \text{ const. } > 0$$

with equality iff ##2r^2=rh##.

- #28

etotheipi

\begin{align}

\lim_{x\to0} \frac{\tan{(\sin{x})}}{\sin{(\tan{x})}} &= \lim_{x\to0} \frac{\sec^2{(\sin{x})}\cos{x}}{\cos{(\tan{x})}\sec^2{x}}

\\ &= \lim_{x\to0} (\cos^3{x}) \cdot \lim_{x\to0} (\sec^2{(\sin{x})}) \cdot \lim_{x\to0} \frac{1}{\cos{(\tan{x})}}

\\ &= \lim_{x\to0} (\cos^3{x}) \cdot \sec^2{(\lim_{x\to0} \sin{x})} \cdot \frac{1}{\cos{(\lim_{x\to0} \tan{x})}}

\\ &= 1 \cdot \sec^2{0} \cdot \frac{1}{\cos{0}} = 1 \cdot 1 \cdot 1 = 1

\end{align}$$Sorry if I've made a mistake somewhere, I haven't done too much with limits as of yet.

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- #29

etotheipi

$$\Delta f = f_{0}(\frac{c}{c+v} - \frac{c}{c-v}) = -\frac{2cv}{c^{2}-v^{2}}f_{0} = -\frac{17}{144} f_{0}$$

- #30

fresh_42

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No. And the question asks for the acoustic(!) interval, which is a ratio, not a difference. There is no way to deduce an initial condition to calculate the absolute frequencies.

$$\Delta f = f_{0}(\frac{c}{c+v} - \frac{c}{c-v}) = -\frac{2cv}{c^{2}-v^{2}}f_{0} = -\frac{17}{144} f_{0}$$

- #31

etotheipi

No. And the question asks for the acoustic(!) interval, which is a ratio, not a difference. There is no way to deduce an initial condition to calculate the absolute frequencies.

In that case would the factor instead be$$\frac{f_{2}}{f_{1}} = \frac{c-v}{c+v} = \frac{8}{9}$$which corresponds to a decrease by a major second as per the harmonic series?

- #32

fresh_42

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Yes. It is ##\nu'= \dfrac{\nu}{1-\dfrac{v}{s}} ## versus ##\nu''= \dfrac{\nu}{1+\dfrac{v}{s}}##In that case would the factor instead be$$\frac{f_{2}}{f_{1}} = \frac{c-v}{c+v} = \frac{8}{9}$$which corresponds to a decrease by a major second as per the harmonic series?

- #33

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if (1) is wrong, then (2) must also be wrong, so (1) is correct.

(3) & (4)

since (1) told us that ##a## is rational, then ##a## can't be ##\sqrt13## or ##-\sqrt13##, so (4) is correct.

Now, if (6) is correct, ##a## can be any even natural number divisible by ##7##, which is wrong.

if (5) is right otherwise, ##a## can only be ##7## or ##14##, but since (6) is wrong at the same time, ##a## can only be ##7##, which is the answer.

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- #35

fresh_42

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Yes and no. Yes, as this is part of the question, and no, since this procedure has to be done with any second point. We have a fixed single point ##P## and a set ##\mathbb{S}^2##. Then we consider all lines ##\overline{PQ}## from ##P## to ##Q## to another point ##Q## on the sphere. This line has a center ##M_Q##. The problem asks for ##\{\,M_Q\in \overline{PQ}\,|\,Q\in \mathbb{S}^2\,\}\,.##

- #36

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Like this?Yes and no. Yes, as this is part of the question, and no, since this procedure has to be done with any second point. We have a fixed single point ##P## and a set ##\mathbb{S}^2##. Then we consider all lines ##\overline{PQ}## from ##P## to ##Q## to another point ##Q## on the sphere. This line has a center ##M_Q##. The problem asks for ##\{\,M_Q\in \overline{PQ}\,|\,Q\in \mathbb{S}^2\,\}\,.##

- #37

fresh_42

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Yes, but which equation describes those points? What is it what you have written down?Like this?

- #38

lekh2003

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Question 14

Isn't this just a scaling of the original sphere? So this would be a sphere at point P whose slope at P is tangent to the original sphere and has radius half of the original sphere.

Okay so I thought about an equation but for the love of me I cannot seem to translate the sphere into the correct position to be a tangent to P.

Okay so I thought about an equation but for the love of me I cannot seem to translate the sphere into the correct position to be a tangent to P.

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- #39

fresh_42

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Well, it's not a scaling anymore if the center is shifted. Your description is right, as is the set which @archaic defined. What I wanted to see is a simple transformation from one description to the other one. As you both have basically solved it, let me give the solution I had in mind. (Please note the two directions! The centers are on the smaller ball,Question 14

Okay so I thought about an equation but for the love of me I cannot seem to translate the sphere into the correct position to be a tangent to P.

The variable endpoint ##X## of the chord is on the sphere, so for its position vector we have ##\vec{x}^2=r^2##. The position vector of the center of the chord ##\overline{PX}## is thus

$$

\vec{c}=\dfrac{\vec{p}+\vec{x}}{2} \;\Longleftrightarrow \; \vec{x}=2\vec{c}-\vec{p}

$$

hence ##r^2=(2\vec{c}-\vec{p})^2## or ##\left( \vec{c}-\dfrac{\vec{p}}{2} \right)^2=\dfrac{r^2}{4}##.

So the set of points we were looking for are all on a sphere with center ##\overline{OP}/2=\vec{p}/2## and radius ##r/2##. All points of this sphere are on the other hand a center of some chord of the original sphere with endpoint ##P##, since we can go back. The point ##P## itself is the center of the chord ##\overline{PP}##.

- #40

- #41

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4 challenge problems left, who can solve them?

New ones on Friday!

New ones on Friday!

- #42

lekh2003

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I would try, but i don't fancy speed learning topology right now :P4 challenge problems left, who can solve them?

New ones on Friday!

- #43

benorin

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This is The Theorem of Mertens, Knopp, Theory and Applications of Infinite Series, pg 321,5.Let ##A=\sum_{k=0}^\infty a_k\, , \,B=\sum_{k=0}^\infty b_k## be two convergent series one of which absolutely.

Prove: The Cauchy-product ##C=\sum_{k=0}^\infty c_k## with ##c_k=\sum_{j=0}^ka_jb_{k-j}## converges to ##AB##.

Give an example that absolute convergence of at least one factor is necessary.

An example when absolute convergence of at least one factor is necessary is the other challenge problem I just did, the Cauchy product ##\left(\sum_{n=0}^\infty \frac{(-1)^n}{\sqrt{n+1}}\right) ^2##.

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