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

User Ammad Ali
by
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1 Answer

4 votes

Answer:

Cos x = 1 -
(x^2)/(2!) +
(x^4)/(4!) -
(x^6)/(1!) + ...

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) +
(f''(a)(x-a)^2)/(2!) +
(f'''(a)(x-a)^3)/(3!) +
(f''''(a)(x-a)^4)/(4!) + ...

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

Cos x = 1 -
(x^2)/(2!) +
(x^4)/(4!) -
(x^6)/(1!) + ...

User Nilloc
by
4.6k points