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Differential Calculus

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13 Mar 2008 10:44:16 IST

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Due to an internet filter, I am unable to post the solution to the nth derivative problem in that thread. So I am posting it here. Please bear with me. $ext{One way to arrive at a higher order derivative of a function is to look at its polynomial series expansion} \ \ ext {i.e. its Taylor expansion} \ \ f(x) = f(x_0) + (x-x_0) rac {f$

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Hari Shankar

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Joined: 28 Feb 2007

Posts: 2173

13 Mar 2008 10:50:51 IST

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By coincidence this qn has been asked in another forum at about the same tim. And of course I referred this link to them. So now a nice loop has been created!

Please see http://www.mathlinks.ro/viewtopic.php?p=1063282#1063282.

@someone who nudged me:

In many cases, nth derivative is easily computed using the Taylor series, because it forms part of the coefficient of xⁿ in that series

You have already encountered this series, as the polynomial expansion for many functions such as sinx, cosx, logx, e^x etc.

e.g. sinx = x-x³/3! + x⁵/5!+.. is nothing but an application of Taylor series.

You can verify each term of the expression easily.

That is why in this problem taylor series is useful.

The variable change x=y+1 is done so that the derivative has to be found for y=0. The left hand-side becomes easy to handle then.

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