The Proof is Trivial!

Problem 169*/** (Finally a question that normal people can do without Stokes' theorem or Lagrange multipliers).

This is a short but simple problem for you guys (your mathematical ability is far more superior to mine). Did take me a fair time though to realize what to do, but it was nice.

Prove the following result:

$[br][br][br]\cos \frac{\pi}{7} - \cos \frac{2\pi}{7}+\cos \frac{3\pi}{7} = \frac{1}{2}.$

I love how you've shown all your working

Seriously, if I ever see you, don't take the first 15 minutes of swearing personally.

The first part is IBP using u = x and dv/dx = cotx. The second part is more complicated (A couple of $\frac{\pi }{2} - x$ substitutions in order to obtain the result, has come up in STEP before) but I'll type it up if you want it.

The first part is IBP using u = x and dv/dx = cotx. The second part is more complicated (A couple of $\frac{\pi }{2} - x$ substitutions in order to obtain the result, has come up in STEP before) but I'll type it up if you want it.

LOLOLOLOLOLOL!

I actually don't understand the sentence though. If it's referring to the Gaussian error integral, surely the authors do know how Mathematica (successfully) computes the integral?

Seriously, if I ever see you, don't take the first 15 minutes of swearing personally.

Bro do you lift? $f'(\lambda) =x^{\lambda} \ln x$

You must note that $\displaystyle \lim_{x \to 0} x \ln(\sin x) = \lim_{x \to 0} x \ln(x) = 0$ which must be shown using calculus or some sort of clever approach.

Solution 171:

$cos(\pi - x) = cos \pi cosx + sin \pi sinx$

$cos(\pi - x) = -cosx$

So we have:

$cos\frac{\pi }{7} = - cos\frac{6\pi }{7}$

$cos\frac{2\pi }{7} = -cos\frac{5\pi }{7}$

$cos\frac{3\pi }{7} = -cos\frac{4\pi }{7}$

So we have:

$\cos \frac{\pi}{7} - \cos \frac{2\pi}{7}+\cos \frac{3\pi}{7} = \cos \frac{\pi}{7} + \cos \frac{3\pi}{7}+\cos \frac{5\pi}{7}$

We can then evaluate this using De Moivre's theorem:

$\cos \frac{\pi}{7} + \cos \frac{3\pi}{7}+\cos \frac{5\pi}{7} = \Re ( e^{\frac{i \pi }{7}}(1+e^\frac{i 2 \pi }{7}+e^\frac{i 4\pi }{7} ))$

$= \Re ( e^\frac{i \pi }{7}( \dfrac{1-e^\frac{i 6\pi }{7}}{1-e^\frac{i 2\pi }{7}}))$

$= \Re ( \dfrac{-1-e^\frac{i \pi }{7}}{e^\frac{i \pi }{7}-e^\frac{i -\pi }{7}}))$

$= \Re ( \dfrac{-1-cos\frac{i \pi }{7}-isin\frac{i \pi }{7}}{2isin\frac{\pi }{7}} )$

= $\frac{1}{2}$

They're nothing difficult! The justification/adjusting is often excruciatingly tedious and difficult but with theorem like the Leibniz Integral Rule (aka. Feynman Integration aka. "Differentiating under the integral sign") you need only apply it without much thought (though you must assure the function is continuous in the relevant range - as usual) so it's definitely worth learning! Wikipedia is a good place to start, that's where I found a lot of these problems in fact

In a nut shell: $\frac{\partial}{\partial \alpha} f(\alpha, x)$ is done by assuming all other variables except the subject of your partial differentiation are constant.

But $\displaystyle f'(\lambda)=\int_0^{\infty} \lambda \frac{x^{\lambda-1}\,dx}{(x^3+1)^2 }$ ? But we havent proved that this would be equal to the thing with the log in? Have I missed something?

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On your marks...
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Problem 169*/**/*** (Looks friendly... at first)

Evaluate $\displaystyle\int_0^{\frac{\pi}{2}}x\cot(x) dx$

Problem 170​***

Prove that $\displaystyle \int_0^{\frac{\pi}{2}} (\sin \theta)^{2x-1} (\cos \theta)^{2y-1} d\theta = \frac{x+y}{2xy} \prod_{n=1}^{\infty} \left(1+\frac{xy}{n(x+y+n)} \right)^{-1}$

Disprove my claim by counterexample then

Problem 169 is STEP III 1991 Q7. In the STEP on you are lead through it a bit though i suppose.

I would be comfortable quoting that sort of thing anywhere, except in a first year analysis exam where the point is to make you prove everything rigorously. This kind of limit, where you are essentially using the fact that a logarithm is slower than a polynomial is common knowledge.

By making a leap from one correct statement to another, you're not lacking rigour (although you can argue you're lacking clarity, which is different).

But the difficulty of the question lies in the fact that, if you use integration by parts, you have $\left[x \ln(\sin x) \right]_0^{\frac{\pi}{2}}$ which requires some thought to evaluate.

Alternatively, there is another really nice approach that avoids this altogether!

What about problem 140? Unless the Basel problem's on the A-level spec, though I do take your point, some of these things could - hypothetically be done, but emphasis on hypothetical.

That's where the difficult of the question lies? Please. Also, I'm failing to see how "it is acceptable to quote" that integral but not this limit.

I know what you are referring to. If you're going to be picky and require justification for the limit above, then differentiating under the integral requires a lot more justification. So the only thing you've avoided is the easier path.

And in any case, this thread is not an exam, it's for fun. So unless something is terribly obscure/incorrect, it's hardly worth pointing out.

Edit: voilà, you've just posted exactly what I was referring to.

Some more (*) problems for people who want them:

Problem 171*

$f(x) = 6x^{7} \sin^{2} \left( x^{1000} \right) e^{x^{2}}$

Find $f^{2013} (0)$ (note $f^{n} (x)$ is the nth derivative of $f(x)$)

Hasn't this already been posted by LOTF somewhere?