The Road to Reality
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Exercise [04.05]
http://www.roadtoreality.info/viewtopic.php?f=19&t=131
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Author:  ZZR Puig [ 24 May 2008, 17:42 ]
Post subject:  Exercise [04.05]

We have:
{s}_{n}=1+x^2+x^4+x^6+x^8+x^{10}+x^{12}+...=\frac{1}{1-x^2}
{s}_{m}=1-x^2+x^4-x^6+x^8-x^{10}+x^{12}-...=\frac{1}{1+x^2}
Let's define s_k as s_n+s_m so
s_k=s_n+s_m=2+2x^4+2x^8+2x^{12}+...=\frac2{1-x^4}
Then,
s_m=s_k-s_n=\frac{2}{1-x^4}-\frac1{1-x^2}=\frac{2}{(1+x^2)(1-x^2)}-\frac1{1-x^2}=\frac{2-(1+x^2)}{(1+x^2)(1-x^2)}
s_m=\frac{1-x^2}{(1+x^2)(1-x^2)}=\frac1{1+x^2}

P.S.: This is only a way to get the result without the use of complex numbers. The most elemental relation between both expressions, as stated by Sameed on the exercise discussion subforum, is the possibility to substitute x with ix to get the other one.

Author:  nroop [ 14 Nov 2011, 18:45 ]
Post subject:  Re: Exercise [04.05]

what is the point in showing a relationship between the two in this particular manner? what kind of relationship between the two is he looking for?

when i tried to do this problem i just ended up proving, like with the previous one, that moving the (1+x^2) over to the left side equaled 1.

from there i gathered that alternating symbols with a negative coming first tends to allow a + in the equation on the right side, that being (1+x^2)^-1


and could you do the imaginary version of the solution too?

Author:  vasco [ 15 Nov 2011, 08:11 ]
Post subject:  Re: Exercise [04.05]

Have a look at my solution, which I have just posted, and see if it makes more sense.
PS I noticed that you deleted your post about exercise 4.6. Is that because you realised that a solution had already been posted?

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