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The Proof is Trivial!

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Problem 506

This is a very nice one indeed, evaluate:

∫01ln⁡(x+1)x2+1 dx\displaystyle \int_0^1 \frac{\ln (x+1)}{x^2 + 1} \, \mathrm{d}x
Original post by Zacken
It's called Latex, and here's how you use it on TSR. :h:


I thank you, good sir :h:
Original post by Alexion
I thank you, good sir :h:


You're welcome, have a stab at my previous integral, you'll find it a valiant challenge. :colone:
Original post by Zacken
You're welcome, have a stab at my previous integral, you'll find it a valiant challenge. :colone:


Haha, I'll pass :wink: it's late, and I'm not the best at these sorta challenges anyway :lol:
Original post by Alexion
Haha, I'll pass :wink: it's late, and I'm not the best at these sorta challenges anyway :lol:


Fair enough. :rofl:
I'm heading to bed myself!

Anyways, what's your story, A-Levels, degree maths, prospective mathmo? :smile:
Original post by Zacken
Fair enough. :rofl:
I'm heading to bed myself!

Anyways, what's your story, A-Levels, degree maths, prospective mathmo? :smile:


A-level Further Mathematician :h: prospective Electronic Engineer actually...
Doesn't stop me from being that nerd who aced C3 though :tongue:
Original post by Alexion
A-level Further Mathematician :h: prospective Electronic Engineer actually...
Doesn't stop me from being that nerd who aced C3 though :tongue:


Filthy engineer. :colonhash:

Kidding, kidding. Y12 or 13?
Original post by Zacken
Filthy engineer. :colonhash:

Kidding, kidding. Y12 or 13?


Y13... only 31/2 people in our FM class, it's great :h:

All the easier to brag about my higher test and exam scores :lol:
Original post by Alexion
Y13... only 31/2 people in our FM class, it's great :h:


1/2? Do I want to ask? xD
Original post by Zacken
1/2? Do I want to ask? xD


Yeah, we sliced him up and threw the other half out the window :mwuaha:

Nah, it's just that he dropped the subject at A2, so he's only in half of our lessons now (hence the .5)
Original post by Alexion
Yeah, we sliced him up and threw the other half out the window :mwuaha:

Nah, it's just that he dropped the subject at A2, so he's only in half of our lessons now (hence the .5)


What's the M/F ratio in that class? :rofl:
Original post by Zacken
What't the M/F ratio in that class? :rofl:


#DIV/0!

All guys :tongue:
Original post by Zacken
Problem 506

This is a very nice one indeed, evaluate:

∫01ln⁡(x+1)x2+1 dx\displaystyle \int_0^1 \frac{\ln (x+1)}{x^2 + 1} \, \mathrm{d}x
Set x=1+t1−t\displaystyle x = \frac{1+t}{1-t} then I=∫01ln⁡(x+1)x2+1 dx=∫01ln⁡(21+t)t2+1 dt=∫01ln⁡(2)t2+1 dt−∫01ln⁡(1+t)t2+1 dt.\begin{aligned} I = \int_0^1 \frac{\ln (x+1)}{x^2 + 1} \, \mathrm{d}x = \int_0^1 \frac{\ln (\frac{2}{1+t})}{t^2 + 1} \, \mathrm{d}t = \int_0^1 \frac{\ln (2)}{t^2 + 1} \, \mathrm{d}t - \int_0^1 \frac{\ln (1+t)}{t^2 + 1} \, \mathrm{d}t.\end{aligned}

So
Unparseable latex formula:

\displaystyle 2I = \frac{\pi}{4} \log(2)}

therefore I=π8log⁡(2).\displaystyle I = \frac{\pi}{8}\log(2).
Original post by Kummer
Set x=1+t1−t\displaystyle x = \frac{1+t}{1-t} then I=∫01ln⁡(x+1)x2+1 dx=∫01ln⁡(21+t)t2+1 dt=∫01ln⁡(2)t2+1 dt−∫01ln⁡(1+t)t2+1 dt.\begin{aligned} I = \int_0^1 \frac{\ln (x+1)}{x^2 + 1} \, \mathrm{d}x = \int_0^1 \frac{\ln (\frac{2}{1+t})}{t^2 + 1} \, \mathrm{d}t = \int_0^1 \frac{\ln (2)}{t^2 + 1} \, \mathrm{d}t - \int_0^1 \frac{\ln (1+t)}{t^2 + 1} \, \mathrm{d}t.\end{aligned}

So
Unparseable latex formula:

\displaystyle 2I = \frac{\pi}{4} \log(2)}

therefore I=π8log⁡(2).\displaystyle I = \frac{\pi}{8}\log(2).


That's certainly one way to do it, bravo!
Problem 507

Does the series

(ln⁡x)+(ln⁡x)22!+(ln⁡x)33!+...+(ln⁡x)nn! (\ln x) + \frac{(\ln x)^2}{2!} + \frac{(\ln x)^3}{3!} +... + \frac{(\ln x)^n}{n!}

Converge as n⟶+∞ n \longrightarrow + \infty? If so, to what value.
Problem 508

M M is an N x N matrix with N distinct eigenvalues all in the range −1<λ<1 - 1 < \lambda < 1 . Prove that:

I+∑r=1∞Mr I + \displaystyle \sum_{r = 1}^{\infty} M^r

Converges. {NB: I denotes the N x N identity matrix}
Problem 509

Let a,b,c a,b,c be the sides of a triangle. Prove that
ab+c+bc+a+ca+b+3abc(a+b)(b+c)(c+a)<2 \dfrac{a}{b+c}+\dfrac{b}{c+a}+ \dfrac{c}{a+b}+\dfrac{3abc}{(a+b)(b+c)(c+a)}<2
Original post by 16Characters....
Problem 507


Solution 507

You like your convergence questions, don't you? :wink:

By the ratio test, we have

(ln⁡x)n+1(n+1)!⋅n!(ln⁡x)n=lim⁡n→∞ln⁡xn+1=0<1\displaystyle \frac{(\ln x)^{n+1}}{(n+1)!} \cdot \frac{n!}{(\ln x)^n} = \lim_{n \to \infty} \frac{\ln x}{n+1} = 0 < 1.

So our series converges. To determine what is converges to is slightly easier to see if we define y=ln⁡xy = \ln x so that our series becomes

y+y22!+y33!+⋯+ynn!+⋯=∑n=1∞ynn!\displaystyle y + \frac{y^2}{2!} + \frac{y^3}{3!} + \cdots + \frac{y^n}{n!} + \cdots = \sum_{n=1}^{\infty} \frac{y^n}{n!}

This looks suspiciously like a certain well-known McLaurin series, so lo and behold:

∑n=1∞ynn!=ey−e0=x−1\displaystyle \sum_{n=1}^{\infty} \frac{y^n}{n!} = e^{y} - e^0 = x - 1.

This was fun!
(edited 8 years ago)
Original post by Zacken
Solution 507

You like your convergence questions, don't you? :wink:
This was fun!


Haha maybe... :-)
Original post by 16Characters....
Haha maybe... :-)


I don't know much about eigenvalues to attempt your other question, sadly! :tongue:

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