complete parts a), b), c), d) below (power series question)

Question

a) find a power series expansion for (1/(1+x²<) valid for |x|<1

b) integrate to obtain a power series expansion for tan^-1x valid for |x|<1

c) show that the power series obtained in part (b) converges at the endpoint x=1

d) show that pi/4 = 1-1/3+1/5-1/7

Answer 1

Alyssa:

We know that 1/(1-x) = ∑xⁿ provided |x| <1

1/(1+x²) = 1/(1-(-x²)) in other words we are transforming 1/(1+x²) into a form that resembles 1/(1-x) so to speak.

Then 1/(1+x²) = ∑(-x²)ⁿ = ∑(-1)ⁿ*x²ⁿ This is valid for |x|<1

∑(-1)ⁿ*x²ⁿ = 1 -x²+x⁴-x⁶+....

b) We know that d/dx(tan^-1(x)) = 1/(x²+1)

So tan^-1(x) = ∫ ∑(-1)ⁿ *x²ⁿ dx = (-1)ⁿ* ∫∑x²ⁿ dx = (-1)ⁿ ∑x²ⁿ⁺¹/(2n+1)

c) Plug in x=1 to get ∑(-1)ⁿ*(1)²ⁿ⁺¹/(2n+1) = ∑(-1)ⁿ/(2n+1). This converges by alternating series test; I'll let you show the steps.

d) You know that tan(π/4) =1 so that tan^-1(1) = π/4

So π/4 = tan^-1(1). Use the fact that tan^-1(x) = ∑(-1)ⁿ x²ⁿ⁺¹/(2n+1) and plug in x =1.

Hope this helps!