Calculate $f^{(25)}(0)$ for $f(x)=x^2 \sin(x)$

Question

Calculate $f^{(25)}(0)$ for $f(x)=x^2 \sin(x)$.

The answer is too short for me to understand, and the answer is

$- 25 \cdot 24 \cdot 8^{23}$

@DonLarynx : That would not be a better idea and i believe OP knows the definition of what it is written and what you have said is nothing more than a definition — , Dec 23 '13 at 12:34
@DavidMitra : I do not quite understand your idea... you need that derivative to write maclaurian series right? I am missing something I guess... — , Dec 23 '13 at 12:38
@PraphullaKoushik You can find the Maclauren series of $f(x)=x^2\sin x$ by multiplying the Maclauren series for $\sin$ by $x^2$. — David Mitra, Dec 23 '13 at 12:40
@DavidMitra : That would be very good if that is true (I do not know if you can get like that)... This is very nice thing to know... I would look into that in more detail... Thank you :) — , Dec 23 '13 at 12:41
MacLaurin series as David Mitra said, or Leibniz rule will also do. — Jean-Claude Arbaut, Dec 23 '13 at 12:41

score 11 · Accepted Answer · edited Dec 23 '13 at 13:00

11

$$f(x) = x^2\sin x$$ $$ = x^2 \left(x - \frac{x^3}{3!} +...- \frac{x^{23}}{23!} +...\right)$$ $$ = -\frac{x^{25}}{23!} +...$$ $$f^{(25)}(x) = -\frac{25!}{23!} + ...$$ $$f^{(25)}(0) = -600$$

WolframAlpha verifies that the $8^{23}$ should not be there.

edited Dec 23 '13 at 13:00

Ahaan S. Rungta

7,706

answered Dec 23 '13 at 12:51

Yiyuan Lee

14,435

Suraj M S · Answer 2 · 2013-12-23T13:02:08.920

5

use the leibnitz rule to find the nth derivative of the function. $$\frac{d^n}{dx^n}(uv) = ^nC_0u \frac{d^n}{dx^n} (v)+^nC_1 \frac{d}{dx} (u) \frac{d^{n-1}}{dx^{n-1}} v+...... ^nC_n \frac{d^n}{dx^n} (u)v$$ here $ u=x^2$ and $v=\sin x$ also note $$ \frac{d^n}{dx^n} (\sin x)=sin(x+n\frac{\pi}{2})$$ on solving you get $$f^{n}(x)=x^2\sin(x+n\frac{\pi}{2})+2nx\sin(x+(n-1)\frac{\pi}{2})+n(n-1)\sin (x+(n-2)\frac{\pi}{2})$$

edited Dec 23 '13 at 13:02

answered Dec 23 '13 at 12:46

Suraj M S

1,891

wow, very fast downvote (not mine!) – Norbert Dec 23 '13 at 12:47
i dont know why some are interested in downvoting without even noticing the idea behind the answer. – Suraj M S Dec 23 '13 at 13:11
I upvote, and I really don't understand the downvote. – Jean-Claude Arbaut Dec 23 '13 at 13:31
2

This seems to be a very good answer as it does not use the Taylor's theorem which is probably at a higher level than Leibniz rule for differentiation. I wonder why it is becoming common to down-vote without giving even a few words of comment. – Paramanand Singh Dec 23 '13 at 14:00

score 3 · Answer 3 · answered Dec 23 '13 at 12:48

3

The answer you have stated is wrong. It should be $-25*24$. Note that in the Taylor expansion for $f(x)$ the only non-zero term that matters is $-x^{25}/23!$. Differentiate this last expression $25$ terms and plug in zero, you will get

$$f^{(25)}(0) = -25!/23! = -600.$$

answered Dec 23 '13 at 12:48

I think it is $\frac{-x^{25}}{25!}$. – lsp Dec 23 '13 at 12:50
Ok get it ! There's another $x^2$ to be multiplied with ! – lsp Dec 23 '13 at 12:52

score 1 · Answer 4 · edited Apr 13 '17 at 12:21

1

Here is an approach. Follow the steps

i) use the identity $\sin x = \frac{e^{ix}-e^{-ix} }{2i} $

ii) use the product rule for differentition.

Note that the function $(x^2)^{(m) }$ vanishes for $m=3$. Chech this related problem.

edited Apr 13 '17 at 12:21

Community

1

answered Dec 23 '13 at 13:10

Mhenni Benghorbal

47,431

Calculate $f^{(25)}(0)$ for $f(x)=x^2 \sin(x)$

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