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Differential Calculus

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2 Sep 2008 18:46:16 IST

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Non-zero derivative

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Let f(x) = sin (x³). Find out for what n will fⁿ(x) i.e. the nth derivative of f(x) be non-zero at x = 0?

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Anant Kumar

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Joined: 10 Jul 2008

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3 Sep 2008 01:09:25 IST

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First, we note that the Taylor series of an infinitely often differentiable function $f(x)$ in an open interval

containing the point $x=0$ is

$\sum_{n=0}^\infty \dfrac{f^{(n)}(0)}{n!}\;x^n \qquad (1)$

From there it is easily seen that the coefficient of $x^k$ can be used to find out the derivative of the $n$ -th order

for $x=0$ .

Now consider the series representation of $\sin x$ :

$\sin x = \sum_{k=0}^\infty \dfrac{(-1)^k }{(2k+1)!}\; x^{2k+1} \qquad \textrm{for all }x$

Therefore,

$f(x)=\sin x^3 = \sum_{k=0}^\infty \dfrac{(-1)^k }{(2k+1)!}\; (x^3)^{2k+1} = \sum_{k=0}^\infty \dfrac{(-1)^k }{(2k+1)!}\; x^{6k+3} \qquad (2)$

By comparing (1) and (2), we see that the $n$ -th derivative of $f(x)=\sin x^3$ at $x=0$ is non-zero only for those $n$ which are of the form $6k+3$ for $k=0$ , $1$ , $2$ , $\ldots$ . And when $n$ is of this form, the $n$ -th derivative is $f^n(0)=(-1)^k \dfrac{(6k+3)!}{(2k+1)!}$

For example $45 = 6(7)+3$ , therefore the 45-th derivative at $x=0$ is non-zero and

$f^{(45)}(0) = -\dfrac{45!}{15!}$

Anant Kumar

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Joined: 10 Jul 2008

Posts: 598

13 Sep 2008 11:57:08 IST

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Actually, in the line after the equation (1), it should be read " ..... the coefficient of $x^k$ can be used to find the $k$ -th order derivative at $x=0$ ."..

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