Question

Find the expansion of cos x about the point x=0

Answer 1

Answer:

Cos x = 1 - $\frac{x^2}{2!}$ + $\frac{x^4}{4!}$ - $\frac{x^6}{1!}$ + ...

Step-by-step explanation:

We use Taylor series expansion to answer this question.

We have to find the expansion of cos x at x = 0

f(x) = cos x, f'(x) = -sin x, f''(x) = -cos x, f'''(x) = sin x, f''''(x) = cos x

Now we evaluate them at x = 0.

f(0) = 1, f'(0) = 0, f''(0) = -1, f'''(0) = 0, f''''(0) = 1

Now, by Taylor series expansion we have

f(x) = f(a) + f'(a)(x-a) + $\frac{f''(a)(x-a)^2}{2!}$ + $\frac{f'''(a)(x-a)^3}{3!}$ + $\frac{f''''(a)(x-a)^4}{4!}$ + ...

Putting a = 0 and all the values from above in the expansion, we get,

Cos x = 1 - $\frac{x^2}{2!}$ + $\frac{x^4}{4!}$ - $\frac{x^6}{1!}$ + ...