Hundredth Derivative of Sinc Function

this approach does not use the series expansion....

(NOTE: $y^n$ represents 'n'th derivative of y)

-let y=sin(x)/x

then yx=sin(x)

after differentiating once we have

x ${ y }^{ 1 }$ +y=cos(x)

the next derivative (2nd)

x ${ y }^{ 2 }$ +2 ${ y }^{ 1 }$ = -sin(x)

and so on.......

we get the general form to be

x $y^{n}$ +n $y^{n-1}$ =sin(x) * (where * n=4z and n,z belong to integers)

hence to find the 100th derivative at '0' we need to put n=101 giving us

0 $y^{101}$ + 101 $y^{100}$ =cos(0)

[cos(0) because 100 is of the form '4z' and 101 th is the next derivative]

finally 101 $y^{100}$ =1

and the hundredth derivative is $1/101$

This is how i solved this:

Note that, we can write,

$\frac{\sin(x)}{x}= \int_{0}^{1}\cos(ux)du$ Now, using leibiniz rule repeatedly, gives , $\left( \frac{\sin(x)}{x} \right)^{(100)} = \int_{0}^{1} u^{100}\cos(ux) du$ Putting $x =0$ , $\left(\frac{\sin(x)}{x} \right)^{(100)} = \int_{0}^{1} u^{100} du = \frac{1}{101} \Box$

The series expansion of sinx is

$\displaystyle sinx$ = $\displaystyle \sum_{n = 0}^\infty \frac{(-1)^nx^{2n+1}}{(2n+1)!}$

So $\displaystyle \frac{sinx}{x} = \sum_{n = 0}^\infty \frac{(-1)^nx^{2n}}{(2n+1)!}$

Then the first derivate is $\displaystyle \frac{d}{dx} (\frac{sinx}{x}) = \sum_{n = 1}^\infty \frac{(-1)^nx^{2n-1}}{(2n+1)(2n-1)!}$

So is easy to see that

$\displaystyle \frac{d^{100}}{dx^{100}} (\frac{sinx}{x}) = \sum_{n = 50}^\infty \frac{(-1)^nx^{2n-100}}{(2n+1)(2n-100)!}$

Note that the sum starts with n = 50, this is because in each derivate a term (2n - a) appears and if n = $\frac{a}{2}$ and n is a integer this term is zero. So the sum need to start with n > $\frac{a}{2}$ and since the last a that appears is a = 99, n needs to be iqual to 50.

If x = 0 the only term that is not zero is for $x^0$ , i.e., when n = 50:

$\displaystyle \frac{d^{100}}{dx^{100}}(\frac{sinx}{x}) = \frac{1}{2 \times 50 + 1} = \frac {1}{101}$

Which means that m = 1, n = 101 and m + n = 102.

Use series expansion of $\sin x$ .