
#1
Feb2112, 01:38 PM

P: 343

Find the Taylor series for f(x) centered at a. Assume power series expansion.
f(x) = sin(x) at a= pi/2 My answer ended up being the summation from n=0 to infinity of ((xpi/2)^n ) / n! And the radius of convergence is R=infinity This is right ? 



#2
Feb2112, 01:42 PM

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No, that is not right. (In fact it looks like a series for [itex]e^{x\pi/2}[/itex]. If you show how you got that, we might be able to point out an error.




#3
Feb2112, 02:12 PM

P: 343

Oh yea I followed the example in the book for e^x that's probly why
So where do I start then? All I know about Taylor series stuff is: F(x)= summation [ (f^n)(a)*(xa)^n ] / n! Ok well when I followed it in the book you have (f^n)(pi/2) = sin(pi/2) and then you put the a value which is pi/2 into the definition of a Taylor series so you get: Summation [ sin(pi/2)(xpi/2)^n ] / n! 



#4
Feb2112, 03:45 PM

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Taylor seriesf(x) = sin(x), f(pi/2) = 1 f'(x) = cos(x), f'(pi/2) = 0 f''(x) = sin(x), f''(pi/2) = 1 etc. 



#5
Feb2112, 07:19 PM

P: 343

Ok so how does it go (I don't have my book with me)
It's like f(x) = f'(x)*(xa) + f''(x)*(xa)^2 + f'''(x)*(xa)^3 + .... With the first part has 1! Under it, the second part has 2! Under it, and the third part had 4! Under it?? So you just plug in that stuff? Idk 



#6
Feb2112, 07:51 PM

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It goes like this:
f(x) = f(a) + f'(a) * (x  a)/1! + f''(a) * (x  a)^{2}/2! + f'''(a) * (x  a)^{3}/3! + ... + f^{(n)}(a) * (x  a)^{n}/n! + ... 



#7
Feb2212, 10:22 AM

P: 343

Oh yea I think that's what I meant!
So when you plug in values you get 1+01(((xpi/2)^3)/2)+0+... right? So it has (1)^n. I don't really know what else the series would be... Umm I get this: Summation n=0 to infinity: (1)^n * ((xpi/2)^(n+2)) / (n+2)! Does that work? 



#8
Feb2212, 12:41 PM

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Make a table with the values for n, f^{(n)}(x), and f^{(n)}([itex]\pi/2[/itex]), similar to the one I did in post #4, but with more entries than I showed. 



#9
Feb2312, 08:10 AM

P: 343

oh, yea i meant this:
1+01(((xpi/2)^2)/2)+0+... i just had a typo. but okay, i did all the way up to f'''''''(x) and when you plug in the a value (pi/2) you get" 1, 0, 1, 0, 1, 0, 1, 0 i ended up getting: f(x) = 1  (xpi/2)^{2} / 2! + (xpi/2)^{4} / 4! + (xpi/2)^{6} / 6! so i dont know what the series would be. all i know is (1)^{n} should be there. 



#10
Feb2312, 11:09 AM

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1) You have a sign error on the last term you show. 2) This is supposed to be an infinite series, so there is no last term. To indicate this, add "+ ..." after the last term you show. With these corrections, that is your Taylor series for f(x) = sin(x). Does the series have to be shown in closed form (i.e., as a summation)? If so, you need to be able to identify the general term of the series. Some things to note: Only even degree terms show up. IOW, the exponents are 0, 2, 4, 6, 8, ... This means that the exponent should be 2n. The terms alternate in sign. You can show this by having a factor of (1)^{n} or possibly (1)^{n + 1}. 



#11
Feb2312, 11:15 AM

P: 343

Yes I realize those corrections. Typo and the second is just me typing slower than im thinking.
But it doesn't say if it had to be in closed form but I think that's how all the answers are so yea. Ok so te answer is just summation (1)^n * ((xpi/2)^(2n))/(2n)! Right? 



#12
Mar412, 10:13 PM

P: 343

Is my series in the last post at the end the correct answer?




#13
Mar412, 10:22 PM

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#14
Mar412, 10:52 PM

P: 343

Cool awesome, that's what I wanted!



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