f(x)=sin x, centered at x=pi/2 Question

linux666

hi, im trying to get the Taylor Series for f(x)=sin x, centered at x=pi/2,
but i am seem to be getting an incorrect taylor series, any help?
Thanks

f'(x) = cos(x)
f(2)(x) = - sin(x)
f(3)(x) = - cos(x)
f(4)(x) = sin(x)

Cyrus

Do you want to offset sin(o) = pi/2?

linux666

Taylor series for f(x) centered about a=pi/2

the a is :

f(x) = sum[ f^(n)*(a)/n! * (x-a)^n ], n=0.....infinity

lurflurf

The formula for taylor series centered about a is easily derived by integration by parts.
[tex]\sum_{i=0}^{\infty} \frac{(x-a)^i}{i!}f^{(i)}(a)[/tex]
a nice way to write the sine derivatives is
[tex]\sin^{(n)}(x)= \sin(x+n\frac{\pi}{2})[/tex]

Benny

You've listed the first four derivatives of sin(x). Evaluated at x = pi/2, the values of the first four derivatives are 0, -1, 0, 1 and the cycle repeats itself. What this tells you is that the 'even numbered' derivatives(ie. 2nd, 4th, 6th etc derivatives) are non-zero whilst the others are. Further, these even numbered derivatives alternate between -1 and 1. So it seems reasonable to suggest that the nth, where n is a natural number, derivative evaluated at x = pi/2 is given by [tex]\left( { - 1} \right)^{2n} [/tex].

Edit: Nevermind the error message before. What I have said above alludes to a possible way to representing the nth derivative evaluated at x = pi/2. I haven't done the question myself, but if you try variations of what I suggested above you'll probably get the answer.

Edit 2: You should probably just ignore what I said, it's a bit misleading now that I think about it.

linux666

Thanks for the suggestions, i was google-ing this question, and the problem is that ALL solutions for this question are answered using : a is centered at 0.

In other words, Maclaurin Series.

When a != 0, it totally throws me off.

bao_ho

replace x with x-pi/2

kant

life might be better if you center yourself at zero.. lol

lurflurf

No this it is no harder. Plus this is nice if one wants to compute
sin(pi/2+.00001) approximately.