Quantcast

Got Homework?

Connect with other students for help. It's a free community.

  • across
    MIT Grad Student
    Online now
  • laura*
    Helped 1,000 students
    Online now
  • Hero
    College Math Guru
    Online now

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

MathSofiya

Find the Taylor series for \[f(x)=sinx\] centered at \[a=\frac{\pi}{2}\] \[\sum_{n=0}^{\infty}(-1)^n\frac{x^{2n+1}}{(2n+1)!}\]

  • one year ago
  • one year ago

  • This Question is Closed
  1. MathSofiya
    Best Response
    You've already chosen the best response.
    Medals 1

    I've figured out what the sum is

    • one year ago
  2. gedtajia
    Best Response
    You've already chosen the best response.
    Medals 0

    are u from socrates's class?

    • one year ago
  3. MathSofiya
    Best Response
    You've already chosen the best response.
    Medals 1

    who's that?

    • one year ago
  4. gedtajia
    Best Response
    You've already chosen the best response.
    Medals 0

    nvm

    • one year ago
  5. gedtajia
    Best Response
    You've already chosen the best response.
    Medals 0

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

    • one year ago
  6. MathSofiya
    Best Response
    You've already chosen the best response.
    Medals 1

    oh I see.

    • one year ago
  7. MathSofiya
    Best Response
    You've already chosen the best response.
    Medals 1

    Oh I think what I wrote is a maclaurin series

    • one year ago
  8. gedtajia
    Best Response
    You've already chosen the best response.
    Medals 0

    i don get this section of the chapter either

    • one year ago
  9. TuringTest
    Best Response
    You've already chosen the best response.
    Medals 3

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

    • one year ago
  10. TuringTest
    Best Response
    You've already chosen the best response.
    Medals 3

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

    • one year ago
  11. TuringTest
    Best Response
    You've already chosen the best response.
    Medals 3

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

    • one year ago
  12. MathSofiya
    Best Response
    You've already chosen the best response.
    Medals 1

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

    • one year ago
  13. TuringTest
    Best Response
    You've already chosen the best response.
    Medals 3

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

    • one year ago
  14. MathSofiya
    Best Response
    You've already chosen the best response.
    Medals 1

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

    • one year ago
  15. TuringTest
    Best Response
    You've already chosen the best response.
    Medals 3

    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

    • one year ago
  16. MathSofiya
    Best Response
    You've already chosen the best response.
    Medals 1

    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

    • one year ago
  17. MathSofiya
    Best Response
    You've already chosen the best response.
    Medals 1

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

    • one year ago
  18. TuringTest
    Best Response
    You've already chosen the best response.
    Medals 3

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

    • one year ago
  19. MathSofiya
    Best Response
    You've already chosen the best response.
    Medals 1

    all odd n terms are 0 though

    • one year ago
  20. TuringTest
    Best Response
    You've already chosen the best response.
    Medals 3

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

    • one year ago
  21. MathSofiya
    Best Response
    You've already chosen the best response.
    Medals 1

    no worries :P

    • one year ago
  22. TuringTest
    Best Response
    You've already chosen the best response.
    Medals 3

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

    • one year ago
  23. MathSofiya
    Best Response
    You've already chosen the best response.
    Medals 1

    yep that makes sense

    • one year ago
  24. TuringTest
    Best Response
    You've already chosen the best response.
    Medals 3

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

    • one year ago
  25. TuringTest
    Best Response
    You've already chosen the best response.
    Medals 3

    next one will be zero...

    • one year ago
  26. TuringTest
    Best Response
    You've already chosen the best response.
    Medals 3

    *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)

    • one year ago
  27. TuringTest
    Best Response
    You've already chosen the best response.
    Medals 3

    \[\{a_4\}={f^{(4)}(\frac\pi2)\over4!}(x-\frac\pi2)^4\]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

    • one year ago
  28. MathSofiya
    Best Response
    You've already chosen the best response.
    Medals 1

    yep I see the pattern now. Thanks @TuringTest

    • one year ago
  29. TuringTest
    Best Response
    You've already chosen the best response.
    Medals 3

    welcome :)

    • one year ago
  30. experimentX
    Best Response
    You've already chosen the best response.
    Medals 0

    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

    • one year ago
  31. experimentX
    Best Response
    You've already chosen the best response.
    Medals 0

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

    • one year ago
  32. MathSofiya
    Best Response
    You've already chosen the best response.
    Medals 1

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

    • one year ago
  33. TuringTest
    Best Response
    You've already chosen the best response.
    Medals 3

    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

    • one year ago
  34. experimentX
    Best Response
    You've already chosen the best response.
    Medals 0

    yep ... math hacks!!

    • one year ago
  35. mahmit2012
    Best Response
    You've already chosen the best response.
    Medals 0

    |dw:1343925856092:dw|

    • one year ago
  36. experimentX
    Best Response
    You've already chosen the best response.
    Medals 0

    thnx!!

    • one year ago
    • Attachments:

See more questions >>>

Your question is ready. Sign up for free to start getting answers.

spraguer (Moderator)
5 → View Detailed Profile

is replying to Can someone tell me what button the professor is hitting...

23

  • Teamwork 19 Teammate
  • Problem Solving 19 Hero
  • You have blocked this person.
  • ✔ You're a fan Checking fan status...

Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.

This is the testimonial you wrote.
You haven't written a testimonial for Owlfred.