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The Proof is Trivial!

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Reply 3020
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Original post by james22
You haven't actually given such an f that works yet, but you did the hard bit.

ff such that f(x)=1f(x) = 1 when xx is rational and f(x)=0f(x) = 0 when xx is irrational works right?
Original post by 0x2a
ff such that f(x)=1f(x) = 1 when xx is rational and f(x)=0f(x) = 0 when xx is irrational works right?


No, but it does if you replace rational and irrational by the sets you defined.
Reply 3022
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Original post by james22
No, but it does if you replace rational and irrational by the sets you defined.

Oh woops, apparently I don't understand periodicity.

I'll just set f(x)=0f(x) = 0 when xx is algebraic, and create some bijection between T/∼T/\sim and (0,1](0,1], and if a,a′∈[a‾]a,a' \in [\overline{a}], then f(a)=f(a′)f(a) = f(a').

I guess this one doesn't even work, periodicity can't be defined for some xx right?
(edited 9 years ago)
Original post by 0x2a
Oh woops, apparently I don't understand periodicity.

I'll just set f(x)=0f(x) = 0 when xx is algebraic, and create some bijection between T/∼T/\sim and (0,1](0,1], and if a,a′∈[a‾]a,a' \in [\overline{a}], then f(a)=f(a′)f(a) = f(a').

I guess this one doesn't even work, periodicity can't be defined for some xx right?


Looking over your proof again, it isn't obvious how it leads to a function with uncountably many periods. You haven't got any additive properties in there which are essential.
Problem 490**

∫0πxsin⁡x1+cos⁡2x dx \displaystyle \int^{\pi}_{0} \dfrac{x \sin x}{1 + \cos^2 x} \ dx
Original post by ThatPerson
Problem 490**

∫0πxsin⁡x1+cos⁡2x dx \displaystyle \int^{\pi}_{0} \dfrac{x \sin x}{1 + \cos^2 x} \ dx


∫0πxsin⁡x1+cos⁡2x=−∫0πddx(cos⁡x)x1+cos⁡2x\displaystyle \int_{0}^{\pi}\frac{x \sin x}{1 + \cos^2 x} = -\int_{0}^{\pi}\frac{\mathrm{d}}{\mathrm{d}x}(\cos x)\frac{x}{1+\cos^2 x}

By IBP, letting u=x,dv=(cos⁡x)′1+cos⁡2x\displaystyle u = x, dv = \frac{(\cos x)'}{1+\cos^2x}, we arrive at:

[xarctan⁡cos⁡x]0π+∫0πarctan⁡cos⁡x dx\displaystyle \bigg[x \arctan \cos x \bigg]_0^{\pi} + \int_{0}^{\pi}\arctan \cos x \, \mathrm{d}x

But ∫0πarctan⁡cos⁡x dx=0\displaystyle \int_{0}^{\pi}\arctan \cos x \, \mathrm{d}x = 0 since arctan⁡cos⁡x\arctan \cos x is symmetric x=π2x= \frac{\pi}{2}.

Then the integral evaluates to π24\displaystyle \frac{\pi^2}{4}. Which is absolutely beautiful.
(edited 9 years ago)
Original post by Zacken
∫0πxsin⁡x1+cos⁡2x=−∫0πddx(cos⁡x)x1+cos⁡2x\displaystyle \int_{0}^{\pi}\frac{x \sin x}{1 + \cos^2 x} = -\int_{0}^{\pi}\frac{\mathrm{d}}{\mathrm{d}x}(\cos x)\frac{x}{1+\cos^2 x}

By IBP, letting u=x,dv=(cos⁡x)′1+cos⁡2x\displaystyle u = x, dv = \frac{(\cos x)'}{1+\cos^2x}, we arrive at:

[xarctan⁡cos⁡x]0π+∫0πarctan⁡cos⁡x dx\displaystyle \bigg[x \arctan \cos x \bigg]_0^{\pi} + \int_{0}^{\pi}\arctan \cos x \, \mathrm{d}x

But ∫0πarctan⁡cos⁡x dx=0\displaystyle \int_{0}^{\pi}\arctan \cos x \, \mathrm{d}x = 0 since arctan⁡cos⁡x\arctan \cos x is symmetric x=π2x= \frac{\pi}{2}.

Then the integral evaluates to π24\displaystyle \frac{\pi^2}{4}. Which is absolutely beautiful.


Nice method :smile:
Original post by ThatPerson
Nice method :smile:


Thanks! :smile:

I think you can use the substitution x=π−yx = \pi - y as well to evaluate the integral.

Also, we could have used ∫0πxf(sin⁡x)dx=π2∫0πf(sin⁡x)dx\displaystyle \int_0^\pi xf(\sin x )\mathrm{d}x=\frac \pi2\int_0^\pi f(\sin x )\mathrm{d}x to evaluate the integral. We were asked to prove this in a past STEP paper, if I remember correctly. :biggrin:
Original post by Zacken
Thanks! :smile:

I think you can use the substitution x=π−yx = \pi - y as well to evaluate the integral.

Also, we could have used ∫0πxf(sin⁡x)dx=π2∫0πf(sin⁡x)dx\displaystyle \int_0^\pi xf(\sin x )\mathrm{d}x=\frac \pi2\int_0^\pi f(\sin x )\mathrm{d}x to evaluate the integral. We were asked to prove this in a past STEP paper, if I remember correctly. :biggrin:


Yeah that's the approach I took.
Problem 491:**

The enthusiasts amongst you may recognize this:
Let: (xn)(x_n) be a sequence of positive real numbers such that:
∑n=1∞n2xn2\displaystyle\sum_{n=1}^{\infty} n^2x_n^2 converges.
Show that:
∑n=1∞xn\displaystyle\sum_{n=1}^{\infty} x_n is convergent. And the converse?
Reply 3030
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Original post by joostan
Problem 491:**

The enthusiasts amongst you may recognize this:
Let: (xn)(x_n) be a sequence of positive real numbers such that:
∑n=1∞n2xn2\displaystyle\sum_{n=1}^{\infty} n^2x_n^2 converges.
Show that:
∑n=1∞xn\displaystyle\sum_{n=1}^{\infty} x_n is convergent. And the converse?

Can you use the ratio test to show that the ratio must be less than or equal to 1 for the first part?
Original post by 0x2a
Can you use the ratio test to show that the ratio must be less than or equal to 1 for the first part?


What if the ratio test is inconclusive?
Reply 3032
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Original post by james22
What if the ratio test is inconclusive?

Can that be covered by the = 1 part? My logic was that if the ratio was bigger than 1 then it must diverge, so it must be less than or equal to 1?

Ah of course, it can fail to exist right?
(edited 9 years ago)
Original post by 0x2a
Can that be covered by the = 1 part? My logic was that if the ratio was bigger than 1 then it must diverge, so it must be less than or equal to 1?

Ah of course, it can fail to exist right?


It can fail to exist, but also if the limit of (n+1)xn+12n2xn2\frac{(n+1)x_{n+1}^2}{n^2 x_n^2} is 1, we can only say that the limit of x_(n+1)/x_n is 1 (if it exists), and if the ratio test gives an answer of 1 you cannot say anything about convergence or divergence.
Original post by 0x2a
Can you use the ratio test to show that the ratio must be less than or equal to 1 for the first part?


Perhaps, it is however not absolutely necessary.
(edited 9 years ago)
Reply 3035
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Original post by james22
It can fail to exist, but also if the limit of (n+1)xn+12n2xn2\frac{(n+1)x_{n+1}^2}{n^2 x_n^2} is 1, we can only say that the limit of x_(n+1)/x_n is 1 (if it exists), and if the ratio test gives an answer of 1 you cannot say anything about convergence or divergence.

So something like n(xn+1)2n(xn)2<(n+1)(xn+1)2n(xn)2≤1\dfrac{n(x_{n + 1})^2}{n(x_n)^2} < \dfrac{(n + 1)(x_{n + 1})^2}{n(x_n)^2} \leq 1 and so xn+1xn<1\dfrac{x_{n + 1}}{x_n} < 1 doesn't work since the limits of n(xn+1)2n(xn)2\dfrac{n(x_{n + 1})^2}{n(x_n)^2} and (n+1)(xn+1)2n(xn)2 \dfrac{(n + 1)(x_{n + 1})^2}{n(x_n)^2} are the same?
Original post by 0x2a
So something like n(xn+1)2n(xn)2<(n+1)(xn+1)2n(xn)2≤1\dfrac{n(x_{n + 1})^2}{n(x_n)^2} < \dfrac{(n + 1)(x_{n + 1})^2}{n(x_n)^2} \leq 1 and so xn+1xn<1\dfrac{x_{n + 1}}{x_n} < 1 doesn't work since the limits of n(xn+1)2n(xn)2\dfrac{n(x_{n + 1})^2}{n(x_n)^2} and (n+1)(xn+1)2n(xn)2 \dfrac{(n + 1)(x_{n + 1})^2}{n(x_n)^2} are the same?


Limits don't have to respect strong inequalities, so those << can become ≥\geq after taking the limit.
From IMO 2013 Q1.

Problem 492**

Prove that for any pair of positive integers kk and n n, there exist k k positive integers m1,m2,...,mk m_1,m_2,...,m_k (not necessarily different) such that

1+2k−1n=(1+1m1)(1+1m2)...(1+1mk) 1 + \dfrac{2^k - 1}{n} = \left(1 + \dfrac{1}{m_1}\right)\left(1 + \dfrac{1}{m_2}\right)...\left(1 + \dfrac{1}{m_k}\right)
Problem 493***

Let ζ(s)\zeta(s) denote the Riemann-Zeta function in the usual way, for complex s=σ+its=\sigma + it

Prove that, for σ>1\sigma > 1,

∣ζ(σ)∣3∣ζ(σ+it)∣4∣ζ(σ+2it)∣≥1|\zeta(\sigma)|^3 |\zeta(\sigma + it)|^4 |\zeta(\sigma + 2it)| \geq 1
Problem 494 ***

Evaluate:
∫0∞cos(ax)1+x2dx\displaystyle \int_{0}^{\infty }\frac{cos(ax)}{1+x^2}dx for a≥0a\geq 0.

Sorry if it's been asked before.

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