Find −4 + (64* π^2) /2! − (1024* π^4)/4! + (16384* π^6)/6! −…

Thanks!

Use the Maclaurin series for cosine
cos x = Σ(n = 0 to ∞) (-1)^n x^(2n)/(2n)!.
.........= 1 - x^2/2! + x^4/4! - x^6/6! + ...

So, -4 + (64π^2)/2! - (1024π^4)/4! + (16384π^6)/6! - …
= -4 [1 - (16π^2)/2! + (256π^4)/4! - (4096π^6)/6! + …]
= -4 [1 - (4π)^2/2! + (4π)^4/4! - (4π)^6/6! + …]
= -4 cos(4π), letting x = 4π in the Maclaurin series for cos x
= -4.

I hope this helps!

Answer: -4

Start with the Taylor series for sin(x):

sin(x) = x - x^3/3! + x^5/5! - ...

Integrate both sides from 0 to some constant a:

1 - cos(a) = a^2/2 - a^4/4! + a^6/6! - ...

Put a = 4 pi,

1 - cos(4pi) = 16 pi^2 / 2! - 256 pi^4/4! + 4096 pi^6/6! - ...

Since cos(4pi) = 1, LHS is zero.

0 = 16 pi^2 / 2! - 256 pi^4/4! + 4096 pi^6/6! - ...

Multiply both sides by 4:

0 = 64 pi^2 / 2! - 1024 pi^4/4! + 16384 pi^6/6! - ...

Add -4 to both sides:

-4 = -4 + 64 pi^2 / 2! - 1024 pi^4/4! + 16384 pi^6/6! - ...

The right side is exactly what you wanted to calculate, so the answer is -4.

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