- #1

linux666

- 5

- 0

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)

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- Thread starter linux666
- Start date

- #1

linux666

- 5

- 0

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)

- #2

Cyrus

- 3,150

- 16

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

- #3

linux666

- 5

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the a is :

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

- #4

lurflurf

Homework Helper

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The formula for taylor series centered about a is easily derived by integration by parts.linux666 said:

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)

[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]

- #5

Benny

- 584

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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.

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.

Last edited:

- #6

linux666

- 5

- 0

In other words, Maclaurin Series.

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

:grumpy:

- #7

bao_ho

- 42

- 0

replace x with x-pi/2

- #8

kant

- 388

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

- #9

lurflurf

Homework Helper

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No this it is no harder. Plus this is nice if one wants to computekant said:life might be better if you center yourself at zero.. lol

sin(pi/2+.00001) approximately.

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