A community for students.
Here's the question you clicked on:
 0 viewing
 2 years ago
Find the Taylor series for \[f(x)=sinx\] centered at \[a=\frac{\pi}{2}\]
\[\sum_{n=0}^{\infty}(1)^n\frac{x^{2n+1}}{(2n+1)!}\]
 2 years ago
Find the Taylor series for \[f(x)=sinx\] centered at \[a=\frac{\pi}{2}\] \[\sum_{n=0}^{\infty}(1)^n\frac{x^{2n+1}}{(2n+1)!}\]

This Question is Closed

MathSofiya
 2 years ago
Best ResponseYou've already chosen the best response.1I've figured out what the sum is

gedtajia
 2 years ago
Best ResponseYou've already chosen the best response.0are u from socrates's class?

gedtajia
 2 years ago
Best ResponseYou've already chosen the best response.0i have exam 5 on this tmr, and final on the day after that =]

MathSofiya
 2 years ago
Best ResponseYou've already chosen the best response.1Oh I think what I wrote is a maclaurin series

gedtajia
 2 years ago
Best ResponseYou've already chosen the best response.0i don get this section of the chapter either

TuringTest
 2 years ago
Best ResponseYou've already chosen the best response.3I have to write it from scratch every time...

TuringTest
 2 years ago
Best ResponseYou've already chosen the best response.3oh I messed up, the factorials should be 1 less, I skipped n=1 factorial...\[f(x)=\sum_{n=0}^\infty{f^{(n)}(a)\over n!}(xa)^n\]\[=\frac1{1!}(x\frac\pi2)\frac1{3!}(x\frac\pi2)^3+\frac1{5!}(x\frac\pi2)^5...\]

TuringTest
 2 years ago
Best ResponseYou've already chosen the best response.3so where does the pi go ? (no pun intended :P)

MathSofiya
 2 years ago
Best ResponseYou've already chosen the best response.1give me a second to look it this...I'm kinda new and slow at this.

TuringTest
 2 years ago
Best ResponseYou've already chosen the best response.3oh no I've messed up terribly, this is all wrong... I should probably have just slept :P lol

MathSofiya
 2 years ago
Best ResponseYou've already chosen the best response.1No it's not that. I guess the x and a's and n confuse me.

TuringTest
 2 years ago
Best ResponseYou've already chosen the best response.3The formula for the Taylor expansion around \(x=a\) is\[f(x)=\sum_{n=0}^\infty{f^{(n)}(a)\over n!}(xa)^n\]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

MathSofiya
 2 years ago
Best ResponseYou've already chosen the best response.1Oh 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

MathSofiya
 2 years ago
Best ResponseYou've already chosen the best response.1I get the first line....now on to the second line

TuringTest
 2 years ago
Best ResponseYou've already chosen the best response.3one sec, I have to deal with a potentially problematic user...sorry brb

MathSofiya
 2 years ago
Best ResponseYou've already chosen the best response.1all odd n terms are 0 though

TuringTest
 2 years ago
Best ResponseYou've already chosen the best response.3right I said it backwards :/ sorry I'm pretty tired I guess

TuringTest
 2 years ago
Best ResponseYou've already chosen the best response.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\[\{a_1\}={1\over1!}(0)(x\frac\pi2)^1=0\]

MathSofiya
 2 years ago
Best ResponseYou've already chosen the best response.1yep that makes sense

TuringTest
 2 years ago
Best ResponseYou've already chosen the best response.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\[\{a_2\}={1\over2!}(1)(x\frac\pi2)^1={1\over2!}(x\frac\pi2)\]

TuringTest
 2 years ago
Best ResponseYou've already chosen the best response.3next one will be zero...

TuringTest
 2 years ago
Best ResponseYou've already chosen the best response.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)

TuringTest
 2 years ago
Best ResponseYou've already chosen the best response.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

MathSofiya
 2 years ago
Best ResponseYou've already chosen the best response.1yep I see the pattern now. Thanks @TuringTest

experimentX
 2 years ago
Best ResponseYou've already chosen the best response.0my 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

experimentX
 2 years ago
Best ResponseYou've already chosen the best response.0let x = u + pi/2, u=0 ... it is equivalent to expansion of cos u at u=0, change back u = x  pi/2

MathSofiya
 2 years ago
Best ResponseYou've already chosen the best response.1I won't be able to use wolfram on my final exam though

TuringTest
 2 years ago
Best ResponseYou've already chosen the best response.3yeah 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

experimentX
 2 years ago
Best ResponseYou've already chosen the best response.0yep ... math hacks!!

mahmit2012
 2 years ago
Best ResponseYou've already chosen the best response.0dw:1343925856092:dw
Ask your own question
Sign UpFind more explanations on OpenStudy
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
 Engagement 19 Mad Hatter
 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.