Taylor Series Expansion

Question

Could you please explain how could I represent the infinite series of cos(x) using Taylor series expansion.

Note that we're advised to use the following for simplification of the Taylor expansion.

d/dx(sin x)=cos x

It is given that the infinite series of sin x sums to sigma n=0-->infinity[((-1)*(x^(2n+1)))/((2n+1)!)], using Taylor series expansion.

Hence we have to show that cos x = sigma n=0-->infinity[((-1)^n)*((2n+1)*(x^(2n)))/((2n+1)!)]

Thank you!

Answer 1

Given that you have the series for sin(x), just take the derivative of the series.

Note for continuous functions, d(∑f(x))/dx = ∑df(x)/dx. In other words, you can swap the ∑ and d/dx. So you can take the derivative of each member of the series.

f(x) = (-1)ⁿx²ⁿ⁺¹/(2n+1)!

f '(x) = (-1)ⁿ(2n+1)x²ⁿ/(2n+1)! = (-1)ⁿ)x²ⁿ/(2n)! --> cos(x) = ∑(n=0-->∞)(-1)ⁿ(2n+1)x²ⁿ/(2n+1)!