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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
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

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MathSofiyaBest ResponseYou've already chosen the best response.1
I've figured out what the sum is
 one year ago

gedtajiaBest ResponseYou've already chosen the best response.0
are u from socrates's class?
 one year ago

gedtajiaBest ResponseYou've already chosen the best response.0
i have exam 5 on this tmr, and final on the day after that =]
 one year ago

MathSofiyaBest ResponseYou've already chosen the best response.1
Oh I think what I wrote is a maclaurin series
 one year ago

gedtajiaBest ResponseYou've already chosen the best response.0
i don get this section of the chapter either
 one year ago

TuringTestBest ResponseYou've already chosen the best response.3
I have to write it from scratch every time...
 one year ago

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

TuringTestBest ResponseYou've already chosen the best response.3
so where does the pi go ? (no pun intended :P)
 one year ago

MathSofiyaBest ResponseYou've already chosen the best response.1
give me a second to look it this...I'm kinda new and slow at this.
 one year ago

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

MathSofiyaBest ResponseYou've already chosen the best response.1
No it's not that. I guess the x and a's and n confuse me.
 one year ago

TuringTestBest ResponseYou've already chosen the best response.3
The 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
 one year ago

MathSofiyaBest ResponseYou've already chosen the best response.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

MathSofiyaBest ResponseYou've already chosen the best response.1
I get the first line....now on to the second line
 one year ago

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

MathSofiyaBest ResponseYou've already chosen the best response.1
all odd n terms are 0 though
 one year ago

TuringTestBest ResponseYou've already chosen the best response.3
right I said it backwards :/ sorry I'm pretty tired I guess
 one year ago

TuringTestBest 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\[\{a_1\}={1\over1!}(0)(x\frac\pi2)^1=0\]
 one year ago

MathSofiyaBest ResponseYou've already chosen the best response.1
yep that makes sense
 one year ago

TuringTestBest 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\[\{a_2\}={1\over2!}(1)(x\frac\pi2)^1={1\over2!}(x\frac\pi2)\]
 one year ago

TuringTestBest ResponseYou've already chosen the best response.3
next one will be zero...
 one year ago

TuringTestBest 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)
 one year ago

TuringTestBest 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
 one year ago

MathSofiyaBest ResponseYou've already chosen the best response.1
yep I see the pattern now. Thanks @TuringTest
 one year ago

experimentXBest ResponseYou've already chosen the best response.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

experimentXBest ResponseYou've already chosen the best response.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

MathSofiyaBest ResponseYou've already chosen the best response.1
I won't be able to use wolfram on my final exam though
 one year ago

TuringTestBest ResponseYou've already chosen the best response.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

experimentXBest ResponseYou've already chosen the best response.0
yep ... math hacks!!
 one year ago

mahmit2012Best ResponseYou've already chosen the best response.0
dw:1343925856092:dw
 one year ago
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