Differential Calculus

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8 Dec 2009 14:10:29 IST
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If f(0) = 1 and f (1) = 2 and f(x) = 1/2{ f(x+1)+f(x+2) } , Then find f(100) =
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If f(0) = 1 and f (1) = 2 and f(x) = 1/2{ f(x+1)+f(x+2) } , Then find f(100) =



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The HellBlaizer's Avatar

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8 Dec 2009 14:14:17 IST
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http://www.goiit.com/posts/list/differential-calculus-hard-question-951552.htm

This question has already been discussed on the forum.Check the link above!!!!!!!!!!

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8 Dec 2009 14:21:56 IST
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I use a short-cut method called annihilators, but here's a good way to understand how to deal with such recursive definitions of sequencesWe have to find a general term for a sequence \{ a_n \} with the property 2a_n = a_{n+1} + a_{n+2} Consider the polynomial f(x) = a_1x + a_2x^2+...+a_k x^k+... i.e. a polynomial that has the terms of the sequence as its coefficients. This is called the generating polynomial or more generally the generating function of the sequence. This will enable us to obtain an explicit function for the sequence. We have: f(x) = a_1x + a_2x^2+a_3x^3...+a_k x^k+... \\ \\x f(x) = a_1 x^2 + a_2x^3+...+a_{k-1} x^k +... \\ \\-2x^2 f(x) = -2a_1x^3-...-2a_{k-2}x^k-..... Adding up the three expressions, we get f(x) (1+x-2x^2) = a_1x +(a_1+ a_2)x^2 or f(x)= \frac{a_1x}{(1+2x)(1-x)} +\frac{(a_1+ a_2)x^2}{(1+2x)(1-x)} \\ \\= \frac{1}{3}[a_1 x +(a_1+a_2)x^2]\left[\frac{2}{1+2x} + \frac{1} {1-x} \right] We can expand 1/(1+2x) and 1/(1-x) and the compare coefficients on both sides to obtain a_n = \frac{4}{3} - \frac{(-2)^n}{3} Substituting n=100 gives the value of a100

New kid on the Block

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8 Dec 2009 14:22:22 IST
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f(100)=20100

shiva  kumar's Avatar

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8 Dec 2009 14:26:20 IST
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4-2^100/3 this que is been done already
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8 Dec 2009 14:36:40 IST
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FORESTS's Avatar

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8 Dec 2009 17:37:36 IST
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please solve by short-cut method called annihilators  and give me full explaination 

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8 Dec 2009 19:13:52 IST
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 THANKS     DEEPAK




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