The Proof is Trivial!

That's the solution I was looking for

The complex analysis route is really quite messy in this case.

Solution:



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I am new to this but I had a quick question: Why is it possible to add the two integrals when the first one has the 'dx' operator and the second has the 'du' operator on it ?

I am new to this but I had a quick question: Why is it possible to add the two integrals when the first one has the 'dx' operator and the second has the 'du' operator on it ?

∫ f(x) dx = ∫ f(u) du = ∫ f(t) dt = ∫ f(k) dk etc
as the limits are the same they represent the same number

Problem 551
A,B are fixed points on a circle, not diametrically opposite each other.P is a variable point of the circle and Q is the point diametrically opposite P. Find the Locus of the point of intersection of AP and BQ.

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Problem 552***

Let $\alpha, \beta \in [0,\pi]$ and $k > 0$

By considering the Maclaurin series of $\ln(1-z)$, or otherwise, show that

$\displaystyle \sum_{k=1}^{\infty} \dfrac{x^k \cos(k\alpha)}{k} = -\dfrac{1}{2} \ln \left(x^2 - 2x \cos(\alpha) + 1\right)$

for $|x| < 1$

Show that

$\displaystyle \int_{0}^{\infty} \ln \left(\dfrac{x^2 + 2kx \cos \beta + k^2}{x^2 + 2kx \cos \alpha + k^2}\right) \ dx = \alpha^2 - \beta^2$

Solution 551

The locus is a circle through A and B centered on the intersection of the tangents to the original circle at A and B.
There are two cases to consider, but I'll write a short argument, the details can be added.
Let the point of intersection of AP and BQ be T.
We know that as Q varies, the angle AQB stays constant.
We also know that the angle QAP, and thus QAT are right angles, as QAP lies in a semicircle.
Thus in the triangle QAT, the remaining angle QTA is constant, thus BTA is constant and the locus is a circle.

Now let the the intersection of the tangents to the original circle through A and B meet at S and let O be the center of the original circle.
ASB = 180 - AOB = 180 - 2AQB = 2(90 - AQB) and let T lie outside the original circle (if it doesn't in your sketch, swap the positions of P and Q), so ATB = 90 - AQB (considering the triangle AQT), this and symmetry implies that S is the center of the locus circle.
(I haven't done this last part in a particularly beautiful way, but notice that it implies that if we used the locus circle as the generating circle, keeping A and B constant, the locus generated will be the original circle, and notice also that the two circles intersect each other at right angles.)

What level of mathematics is required for this?

Problem 552***

Let $\alpha, \beta \in [0,\pi]$ and $k > 0$

By considering the Maclaurin series of $\ln(1-z)$, or otherwise, show that

$\displaystyle \sum_{k=1}^{\infty} \dfrac{x^k \cos(k\alpha)}{k} = -\dfrac{1}{2} \ln \left(x^2 - 2x \cos(\alpha) + 1\right)$

for $|x| < 1$

Show that

$\displaystyle \int_{0}^{\infty} \ln \left(\dfrac{x^2 + 2kx \cos \beta + k^2}{x^2 + 2kx \cos \alpha + k^2}\right) \ dx = \alpha^2 - \beta^2$

I'm confused about the integral in the second part, because in the first part you prove a formula for |x| < 1, but the integral ranges from 0 to infinity. Also, when I substitute x = ku, the result seems to imply that the integral is not independent on the value of k.

Problem 553 (*)

Here's a nice, easy to start with problem

Let $a_{1}, a_{2},...$ be a sequence of positive integers such that every (positive) integer appears in the sequence only once. Suppose further that $n$ divides $S_{n} = a_{1} + ... + a_{n}$. Show that such a sequence exist.

Solution 553

We construct a sequence that works. Start with $a_1=1$ then let $a_2$ be some large number such that we may have $a_3=2$, i.e $a_2 \equiv 5 \pmod{6}$. Now let $a_4$ be some even large number compared to $a_2$ such that we may have $a_5=3$, and such a number exists by CRT. Hence, we keep repeating this, alternating the sequence between large number and the sequence $\{ 1,2,3... \}$, missing out a part of the sequence if that number has already been used as one of our large numbers, and hence, we have constructed a sequence that works.

I have never studied the Chinese remainder theorem so this question may be stupid, but I'm trying to understand your construction.
I understand that $a_{2n-1}=n$ so $a_{1}=1, a_{2}=k>1$ but then $a_{2k-1}=k=a_{2}$ so $k$ appears twice in the sequence.

I have never studied the Chinese remainder theorem so this question may be stupid, but I'm trying to understand your construction.
I understand that $a_{2n-1}=n$ so $a_{1}=1, a_{2}=k>1$ but then $a_{2k-1}=k=a_{2}$ so $k$ appears twice in the sequence.

I've just used CRT to justify the existence of such a large number with the property I want. We then miss it out in our sequence as physicsmaths said.

OK, I'm still struggling with the construction though. If $a_1=1, a_2=6n-1$ then $S_2=6n, S_3=6n+2$ which isn't divisible by $3$.