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

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I've figured out what the sum is

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

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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Oh I think what I wrote is a maclaurin series

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

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

TuringTest
 4 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
 4 years ago
Best ResponseYou've already chosen the best response.3so where does the pi go ? (no pun intended :P)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0give me a second to look it this...I'm kinda new and slow at this.

TuringTest
 4 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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0No it's not that. I guess the x and a's and n confuse me.

TuringTest
 4 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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Oh 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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I get the first line....now on to the second line

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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0all odd n terms are 0 though

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

TuringTest
 4 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\[\{a_1\}={1\over1!}(0)(x\frac\pi2)^1=0\]

TuringTest
 4 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
 4 years ago
Best ResponseYou've already chosen the best response.3next one will be zero...

TuringTest
 4 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
 4 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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0yep I see the pattern now. Thanks @TuringTest

experimentX
 4 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
 4 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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I won't be able to use wolfram on my final exam though

TuringTest
 4 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
 4 years ago
Best ResponseYou've already chosen the best response.0yep ... math hacks!!

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1343925856092:dw
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