The Student Room Group

The Proof is Trivial!

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Since this one hasn't been answered yet:

Solution 16

Note, applying x+x2+1xx+\sqrt{x^2+1} \to x to the former and xx2+1xx-\sqrt{x^2+1} \to x to the latter:

I(f)(x)dx=ddx[x+x2+1]f(x+x2+1)dx+\displaystyle\int _{-\infty}^{\infty} \displaystyle \mathfrak{I}(f)(x) dx = \displaystyle\int _{-\infty}^{\infty} \dfrac{d}{dx}[x+\sqrt{x^2+1}]f(x+\sqrt{x^2+1}) dx +ddx[xx2+1]f(xx2+1)dx\displaystyle\int _{-\infty}^{\infty} \dfrac{d}{dx}[x-\sqrt{x^2+1}]f(x-\sqrt{x^2+1}) dx

=0f(x)dx+0f(x)dx=\displaystyle\int _{0}^{\infty} f(x)dx + \displaystyle\int _{-\infty}^{0}f(x)dx

=f(x)dx=\displaystyle\int _{-\infty}^{\infty} f(x) dx _{\square}

Observe now that cosxsin(x2+1)x2+1dx=12sin(x+x2+1)sin(xx2+1)x2+1dx\displaystyle\int_{-\infty}^{\infty} \dfrac{\cos x \sin (\sqrt{x^2+1})}{\sqrt{x^2+1}}dx = \dfrac{1}{2} \displaystyle\int_{-\infty}^{\infty}\dfrac{\sin (x+ \sqrt{x^2+1} ) - \sin (x-\sqrt{x^2+1})}{\sqrt{x^2+1}} dx

=12=\dfrac{1}{2} \displaystyle\int_{-\infty}^{\infty} (x+x2+1x2+1)sin(x+x2+1)x+x2+1+(x2+1xx2+1)sin(xx2+1)xx2+1dx\left(\dfrac{x+ \sqrt {x^2+1}}{\sqrt{x^2+1}}\right) \dfrac{\sin (x +\sqrt{x^2+1})}{x+\sqrt{x^2+1}} + \left( \dfrac{\sqrt{x^2+1} - x}{\sqrt{x^2+1}} \right) \dfrac{\sin(x-\sqrt{x^2+1})}{x-\sqrt{x^2+1}} dx

=12=\dfrac{1}{2} \displaystyle\int_{-\infty}^{\infty}(sin(x+x2+1)x+x2+1+sin(xx2+1)xx2+1)\left(\dfrac{\sin (x +\sqrt{x^2+1})}{x+\sqrt{x^2+1}} + \dfrac{\sin (x -\sqrt{x^2+1})}{x-\sqrt{x^2+1}}\right)+xx2+1(sin(x+x2+1)x+x2+1sin(xx2+1)xx2+1)dx + \dfrac{x}{\sqrt{x^2+1}}\left( \dfrac{ \sin (x +\sqrt{x^2+1})}{x+\sqrt{x^2+1}} - \dfrac{\sin (x -\sqrt{x^2+1})}{x-\sqrt{x^2+1}}\right) dx

=12sinxxdx=\dfrac{1}{2} \displaystyle\int_{-\infty}^{\infty} \dfrac{\sin x}{x} dx

=π2=\dfrac{\pi}{2}, by applying the above result in the case f(x)=sinxxf(x) =\dfrac{\sin x}{x}.
(edited 10 years ago)
Problem 24

Evaluate n=11006sin(nπ2013)\displaystyle \prod_{n=1}^{1006} \sin \left(\frac{n\pi}{2013}\right)
Original post by Star-girl
Problem 14**

Prove that 26012^{60}-1 is divisible by 61.


2 to the power of 60 - 1 = 1152921504606846795
1152921504606846795 / 61 = 18900352534538475
Therefore it's divisible by 61. Problem solved :smile:
Original post by Indeterminate
x

Solution 20

Integral



circle

(edited 10 years ago)
Reply 84
Original post by Felix Felicis
x

Damn, I was just about to start writing this one up! :tongue:
Reply 85
Original post by Boy_wonder_95
2 to the power of 60 - 1 = 1152921504606846795
1152921504606846795 / 61 = 18900352534538475
Therefore it's divisible by 61. Problem solved :smile:


The funny thing is... my method of using FLT also only takes 2 lines of work :lol:
Original post by und
Damn, I was just about to start writing this one up! :tongue:

Ninja'd :ninja: :biggrin:
Original post by Felix Felicis
....


Hmm I'm really tempted to ask problems that are well-known theorems under disguise (as I think they serve to educate interesting points) but it's hard to judge the difficulty :frown:
Original post by ukdragon37
Hmm I'm really tempted to ask problems that are well-known theorems under disguise (as I think they serve to educate interesting points) but it's hard to judge the difficulty :frown:

Lay it on me :biggrin: (Not today though, tired, sleepy time :biggrin: )
Original post by Noble.
The funny thing is... my method of using FLT also only takes 2 lines of work :lol:


Not fair, FLT isn't part of the A Level syllabus is it?
Reply 90
Original post by Boy_wonder_95
2 to the power of 60 - 1 = 1152921504606846795
1152921504606846795 / 61 = 18900352534538475
Therefore it's divisible by 61. Problem solved :smile:

We were having trouble proving that 2^700^70^7-1 is divisible by 101. Show us your ways.
Original post by Boy_wonder_95
Not fair, FLT isn't part of the A Level syllabus is it?

Spoiler

Original post by Felix Felicis

Spoiler


In the context Noble uses it, FLT means Fermat's Little Theorem here (which you don't need to know at A-Level :p: ).
Original post by Farhan.Hanif93
In the context Noble uses it, FLT means Fermat's Little Theorem here (which you don't need to know at A-Level :p: ).

Haha, I know, I was just trying to mess with Boy Wonder xD
Original post by Felix Felicis
Haha, I know, I was just trying to mess with Boy Wonder xD

You managed to mess with me in the process.

I can see this thread becoming prime procrastination material for my revision...
Reply 95
Original post by Farhan.Hanif93
You managed to mess with me in the process.

I can see this thread becoming prime procrastination material for my revision...


I know, I was going to say the same. It's sad when you have to class doing 'other' maths as procrastination :lol:
Problem 25**/***

1.

Show that for any (potentially infinite) set AA, there is no surjection f:AP(A)f : A \to \mathcal{P}(A), where P(A)\mathcal{P}(A) is the powerset of AA.

2.

Hence show there is no surjection from the the set AA to set of functions AAA \to A if A>1|A|>1.

3.

Determine the relationship between R\mathbb{R} and NN\mathbb{N} \to \mathbb{N}.

(edited 10 years ago)
Original post by Noble.
I know, I was going to say the same. It's sad when you have to class doing 'other' maths as procrastination :lol:

If only I could understand groups by doing integrals, then I'd be on to something...

Out of interest, are questions on things like groups etc. suitable for this thread?
Reply 98
Original post by Farhan.Hanif93
If only I could understand groups by doing integrals, then I'd be on to something...

Out of interest, are questions on things like groups etc. suitable for this thread?


If we're going to procrastinate, I don't see the harm in combining a bit of revision :lol:
Reply 99
Original post by Farhan.Hanif93
If only I could understand groups by doing integrals, then I'd be on to something...

Out of interest, are questions on things like groups etc. suitable for this thread?

It's fine as long as it's marked accordingly. If possible it would be good if everything could be defined so at a stretch anyone could have a go, but it's not a requirement.

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