How do you find the taylor series for #f(x) = cos x # centered at a=pi?

Answer 1

The general formula for the Taylor Series is as follows:

#sum_(n=0)^(N) f^((n))(a)/(n!)(x-a)^n# with #f^((n))(a)# being the #n#th derivative of #f(x)# at #x->a#.

Thus, we have to take the derivative multiple times. You should also consider the following:

Let us go to #n = 4# (one might also say "truncate the series at #n = 4#").
#f^((0))(x) = color(green)(f(x) = cosx)# #color(green)(f'(x) = -sinx)# #color(green)(f''(x) = -cosx)# #color(green)(f'''(x) = sinx)# #color(green)(f''''(x) = cosx)#

Now, writing it out, we get:

#= (f(a))/(0!)(x-a)^0 + (f'(a))/(1!)(x-a)^1 + (f''(a))/(2!)(x-a)^2 + (f'''(a))/(3!)(x-a)^3 + (f''''(a))/(4!)(x-a)^4 + ...#
#= (cosa)/(0!)(x-a)^0 + (-sina)/(1!)(x-a)^1 + (-cosa)/(2!)(x-a)^2 + (sina)/(3!)(x-a)^3 + (cosa)/(4!)(x-a)^4#
#= cospi + (-sinpi)/(1!)(x-pi) + (-cospi)/(2!)(x-pi)^2 + (sinpi)/(3!)(x-pi)^3 + (cospi)/(4!)(x-pi)^4#
but #sinpi = 0#... so the odd terms go away (we talked about this!):
#= cospi + cancel((-sinpi)(x-pi)) + (-cospi)/(2)(x-pi)^2 + cancel((sinpi)/(6)(x-pi)^3) + (cospi)/(24)(x-pi)^4#
#= color(blue)(cospi - (cospi)/(2)(x-pi)^2 + (cospi)/(24)(x-pi)^4 - ...)#
Sign up to view the whole answer

By signing up, you agree to our Terms of Service and Privacy Policy

Sign up with email
Answer 2

To find the Taylor series for ( f(x) = \cos x ) centered at ( a = \pi ), we can use the formula for the Taylor series expansion:

[ f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(a)}{n!} (x - a)^n ]

First, we find the derivatives of ( f(x) = \cos x ) up to the desired order at ( a = \pi ), which is the center of the Taylor series. Then, we substitute these derivatives into the Taylor series formula.

Here are the derivatives of ( \cos x ):

[ f(x) = \cos x ] [ f'(x) = -\sin x ] [ f''(x) = -\cos x ] [ f'''(x) = \sin x ] [ f^{(4)}(x) = \cos x ]

Now, evaluate these derivatives at ( x = \pi ):

[ f(\pi) = \cos \pi = -1 ] [ f'(\pi) = -\sin \pi = 0 ] [ f''(\pi) = -\cos \pi = -1 ] [ f'''(\pi) = \sin \pi = 0 ] [ f^{(4)}(\pi) = \cos \pi = -1 ]

Substitute these values into the Taylor series formula:

[ \cos x = \sum_{n=0}^{\infty} \frac{(-1)^n}{n!} (x - \pi)^n ]

This is the Taylor series for ( \cos x ) centered at ( a = \pi ).

Sign up to view the whole answer

By signing up, you agree to our Terms of Service and Privacy Policy

Sign up with email
Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

Not the question you need?

Drag image here or click to upload

Or press Ctrl + V to paste
Answer Background
HIX Tutor
Solve ANY homework problem with a smart AI
  • 98% accuracy study help
  • Covers math, physics, chemistry, biology, and more
  • Step-by-step, in-depth guides
  • Readily available 24/7