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Iitjee

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Simplicity
I don't know who he is.


Okay, the fourier series bit is a bit of a throw out idea. I was wondering apart from calculus can somebody solve it using a more advanced method.

I mean use induction to show it converges to what it does.

he goes by the alter ego 'question boy'

he's in the year above me and tries to prove everything by induction.

and he hasn't accepted my friend request on fb yet >:[
Reply 21
Totally Tom
i don't quite see your problem. the LHS factorises nicely, on the RHS look at it as 1+2x(1+x+x^2+...) sum the series and then fiddle around with fractions.



Thanks, I see now.
rnd
My way was as follows.

Spoiler


I'm sure I've seen something like this before, otherwise I probably wouldn't have solved it.


I think the method was in a STEP III (possibly II) question from a paper definitely within the last ten years. If I may say so you choose the most elegant solution by far - it is just such an unexpected and clever method :yes:
Reply 23
DeanK22
I think the method was in a STEP III (possibly II) question from a paper definitely within the last ten years. If I may say so you choose the most elegant solution by far - it is just such an unexpected and clever method :yes:


You may say :smile: so but I'm taking no credit. I've been trying to remember and I think that what I saw before was probably 1+3x+5x2+7x3...1+3x+5x^2+7x^3... which meets the other problem halfway so to speak.

Has anyone tried any of the other problems from that paper? There are a few easy ones but also some more interesting ones.
Reply 24
The next problem up in a sequence of sums is 1+5x+14x2+30x3+55x4....1+5x+14x^2+30x^3+55x^4.... which is now no harder than the other.
Here the coefficients a(i) are defined inductively by

a(0)=1
a(i) = a(i-1)+(i+1)^2 for i>0

So then we can express the sum S as

S = a(0) + (a(0) + 2^2)x + (a(1) + 3^2)x^2 + (a(2) + 4^2)x^3 +...
= 1 + xS + 4x + 9x^2 +16x^3 +...
= xS + the old series

So we can solve for S; S= (1+x)/(1-x)^4
Reply 26
I really quite liked this one that went like this:

[br]1+3is a root of3x3+ax2+bx+12=0[br][br]1+ \sqrt{3} \qquad \text{is a root of} \qquad 3x^3 + ax^2 + bx +12=0 [br]

where a and b are integers. Find the values of a and b. Oviously, as they're integers, one of the other roots is ±(13)\pm (1- \sqrt{3})
Reply 27
jbeacom600
I really quite liked this one that went like this:

[br]1+3is a root of3x3+ax2+bx+12=0[br][br]1+ \sqrt{3} \qquad \text{is a root of} \qquad 3x^3 + ax^2 + bx +12=0 [br]

where a and b are integers. Find the values of a and b. Oviously, as they're integers, one of the other roots is ±(13)\pm (1- \sqrt{3})


I did that one too.

±\pm ?
Reply 28
You could conceivably have -1 - sqrt 3...

i think the brackets are misplaced.
Reply 29
My reasoning went like:
[br](1+3)(13)=13=2[br](1+3)(13)=2[br][br](1+\sqrt{3})(1-\sqrt{3}) = 1 -3 =-2[br]-(1+\sqrt{3})(1-\sqrt{3}) =2[br]

it wouldn't be 13 -1 -\sqrt{3} as that is (1+3) -(1+\sqrt{3}) so you would get (1+3)2-(1+\sqrt{3})^2 and this is clearly neither rational nor an element of the integers.

I.e. we're looking at conjugate surds. As for the other question, I think I did it the same way as tommm, but using the limits on the integral. That is of course, assuming that we can actually integrate/differentiate an infinite series term by term.
Reply 30
jbeacom600

As for the other question, I think I did it the same way as tommm, but using the limits on the integral.


What's that about? I don't think I did it that way, I KNOW I did it that :s-smilie: me am tired
rnd
Has anyone tried any of the other problems from that paper? There are a few easy ones but also some more interesting ones.


I did 13 using the beta and gamma functions, but it's a bit high-powered, having to at least justify two inversions. Is there a more elementary (possibly easier) way?
The general term in the sequence is r!/(r+5)! = 1/((r+1)(r+2)(r+3)(r+4)(r+5)).

Maybe you can express this in partial fractions which will then cancel in the whole sum..

Not that I've bothered to do this.
Drederick Tatum
.

Given exam conditions, I might actually slog through that... Maybe there's a trick from this step, though?

Spoiler

DFranklin

Spoiler


That's really nice! :yes:

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