sinc function
, denoted by
$\def\sinc{\text{sinc }} \sinc x$
, is defined by
$\def\sinc{\text{sinc }}
\sinc x=\frac{\sin x}{x}.$
The term "sinc" is a contraction of the function's full Latin name, the
*
sinus
*
*
cardinalis
*
(cardinal sine). It was introduced by Phillip M. Woodward in his 1952 paper "Information Theory and Inverse Probability in Telecommunication" in which he said the function "occurs so often in Fourier analysis and its applications that it does seem to merit some notation of its own" and his 1953 book "Probability and Information Theory, with Applications to Radar".

Compute $m+n$ , where $m$ and $n$ are relatively prime , where

$\def\sinc{\text{sinc }} \left.\frac{d^{100}}{dx^{100}}(\sinc x)\right|_{x=0}=\frac{m}{n}.$

The answer is 102.

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Truly cool, good job !

Aditya Raut
- 6 years, 10 months ago

Why isn't this at the top? This is so beautiful!

Kenny Lau
- 6 years, 11 months ago

I learned from this. :)

Esrael Santillan
- 6 years, 10 months ago

Brilliant solution. I love it so much! A little note, it would be convenience if you denote the $n$ -derivative of $y$ as $y^{(n)}$ . It's widely accepted & common notation too. ≥Ö‿Ö≤

Anastasiya Romanova
- 6 years, 8 months ago

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@Valentina Moy thank you . i'll follow the notation :)

Abhinav Raichur
- 6 years, 8 months ago

Nice and innovative! +1

Kartik Sharma
- 5 years, 8 months ago

Excelent, I understood as well. Thanks!

Carlos David Nexans
- 6 years, 10 months ago

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I thank everyone for your generous appreciation :) ......... It has certainly boosted my confidence.I will try further to write more simplified and distinctive solutions :)

Abhinav Raichur
- 6 years, 10 months ago

This a wonderfully explained!!!

erica phillips
- 3 years, 2 months ago

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 $x =0$ , $\left(\frac{\sin(x)}{x} \right)^{(100)} = \int_{0}^{1} u^{100} du = \frac{1}{101} \Box$

21 Helpful
2 Interesting
3 Brilliant
0 Confused

@shivang jindal awesome!!

Abhinav Raichur
- 6 years, 7 months ago

Excellent Solution!

A Former Brilliant Member
- 4 years, 10 months ago

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.

17 Helpful
1 Interesting
1 Brilliant
0 Confused

This is the manly way to solve this problem.

A Former Brilliant Member
- 6 years, 9 months ago

Use series expansion of $\sin x$ .

6 Helpful
0 Interesting
0 Brilliant
0 Confused

Short and sweet answer! very nice. (>‿◠)b

Anastasiya Romanova
- 6 years, 12 months ago

please explain.

Ajitesh Mishra
- 6 years, 12 months ago

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

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

y=sin(x)/xthen

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)and n,z belong to integers)* (where *n=4zhence to find the 100th derivative at '0' we need to put n=101giving us0 $y^{101}$ + 101 $y^{100}$ =cos(0)[cos(0) because 100 is of the form '4z' and 101 th is the next derivative]finally

101 $y^{100}$ =1and the hundredth derivative is $1/101$