## MathSofiya 3 years ago Find the Taylor series for $f(x)=sinx$ centered at $a=\frac{\pi}{2}$ $\sum_{n=0}^{\infty}(-1)^n\frac{x^{2n+1}}{(2n+1)!}$

1. MathSofiya

I've figured out what the sum is

2. gedtajia

are u from socrates's class?

3. MathSofiya

who's that?

4. gedtajia

nvm

5. gedtajia

i have exam 5 on this tmr, and final on the day after that =]

6. MathSofiya

oh I see.

7. MathSofiya

Oh I think what I wrote is a maclaurin series

8. gedtajia

i don get this section of the chapter either

9. TuringTest

I have to write it from scratch every time...

10. TuringTest

oh I messed up, the factorials should be 1 less, I skipped n=1 factorial...$f(x)=\sum_{n=0}^\infty{f^{(n)}(a)\over n!}(x-a)^n$$=\frac1{1!}(x-\frac\pi2)-\frac1{3!}(x-\frac\pi2)^3+\frac1{5!}(x-\frac\pi2)^5-...$

11. TuringTest

so where does the pi go ? (no pun intended :P)

12. MathSofiya

give me a second to look it this...I'm kinda new and slow at this.

13. TuringTest

oh no I've messed up terribly, this is all wrong... I should probably have just slept :P lol

14. MathSofiya

No it's not that. I guess the x and a's and n confuse me.

15. TuringTest

The formula for the Taylor expansion around $$x=a$$ is$f(x)=\sum_{n=0}^\infty{f^{(n)}(a)\over n!}(x-a)^n$in your case $$a=\frac\pi2$$ let's go check each term:$\{a_0\}={f^{(0)}(\frac\pi2)\over0 !}(x-\frac\pi2)^0=\sin\frac\pi2=1$$\{a_1\}=\frac1{1!}\cancel{\cos\frac\pi2}^{\huge0}(x-\frac\pi2)^1=0$$\{a_2\}=-\frac1{2!}\sin\frac\pi2(x-\frac\pi2)^2=-\frac1{2!}(x-\frac\pi2)^2$$\{a_3\}=-\frac1{3!}\cancel{\cos\frac\pi2}^{\huge0}(x-\frac\pi2)^3=0$so the pattern is that all odd n terms stay

16. MathSofiya

Oh darn! I get it now. For a second I forgot what $f^{(0)}$ and $f^{(1)}$ meant. Durrr so the zero derivative is sin(pi/2) and $(x-\frac{\pi}{2})^0 =1$ makes sense

17. MathSofiya

I get the first line....now on to the second line

18. TuringTest

one sec, I have to deal with a potentially problematic user...sorry brb

19. MathSofiya

all odd n terms are 0 though

20. TuringTest

right I said it backwards :/ sorry I'm pretty tired I guess

21. MathSofiya

no worries :P

22. TuringTest

$\{a_1\}={f^{(1)}(\frac\pi2)\over1!}(x-\frac\pi2)^1$and$f(x)=\sin x\implies f'(x)=\cos x\implies f'(\frac\pi2)=0$so$\{a_1\}={1\over1!}(0)(x-\frac\pi2)^1=0$

23. MathSofiya

yep that makes sense

24. TuringTest

$\{a_2\}={f^{(2)}(\frac\pi2)\over2!}(x-\frac\pi2)^2$and$f'(x)=\cos x\implies f''(x)=-\sin x\implies f''(\frac\pi2)=-1$so$\{a_2\}={1\over2!}(-1)(x-\frac\pi2)^1=-{1\over2!}(x-\frac\pi2)$

25. TuringTest

next one will be zero...

26. TuringTest

*another typo, the last one should be$\{a_2\}=-\frac1{2!}(x-\frac\pi2)^2$(I forgot the n=2 in the exponent on the parentheses)

27. TuringTest

$\{a_4\}={f^{(4)}(\frac\pi2)\over4!}(x-\frac\pi2)^4$and$f(x)=\sin x\implies f^{(4)}(x)=\sin x\implies f^{(4)}(\frac\pi2)=1$so$\{a_4\}={1\over4!}(1)(x-\frac\pi2)^4={1\over4!}(x-\frac\pi2)^4$by now a pattern should be emerging

28. MathSofiya

yep I see the pattern now. Thanks @TuringTest

29. TuringTest

welcome :)

30. experimentX

my suggestion ... keep it easy as much as possible http://www.wolframalpha.com/input/?i=expand+sin%28x%29+at+pi%2F2&dataset=&equal=Submit http://www.wolframalpha.com/input/?i=expand+cos+x+at+0

31. experimentX

let x = u + pi/2, u=0 ... it is equivalent to expansion of cos u at u=0, change back u = x - pi/2

32. MathSofiya

I won't be able to use wolfram on my final exam though

33. TuringTest

yeah I realized that @experimentX but I though it good practice to do it manually first time around thanks for reminding me to point that out though, I had almost forgotten

34. experimentX

yep ... math hacks!!

35. mahmit2012

|dw:1343925856092:dw|

36. experimentX

thnx!!