### 2 Answers

- JOHNLv 74 weeks ago
Macaulin’s theorem:

f(x) f= f(0) + x f’(0) + (x²/2!)f’’(0) + (x³/3!)f’’’(0) +…..

Her f(x) = cosh2x, f’(x) = 2sinh2x, f’’(x) = 2²cosh2x, etc

f(0) = cosh0 = 1, f’(0) = 2sinh0 =0, f’’(0) = 2²cosh0 = 2², f’’’(0) = 2³sinh0 = 0, f’’’’(0) = 2⁴cosh0, etc

coshx = 1 + 2²x²/2! + 2⁴x⁴/4! +….+ 2²ⁿ⁻²x²ⁿ⁻²/(2n – 2)! +…

- 4 weeks ago
cosh(t) = (1/2) * (e^(t) + e^(-t))

The Maclaurin series for e^(t) is t^0 / 0! + t^1 / 1! + t^2 / 2! + t^3 / 3! + ...

cosh(t) =>

(1/2) * (e^(t) + e^(-t)) =>

(1/2) * (1 + t + t^2/2! + t^3/3! + t^4/4! + ... + 1 - t + t^2/2! - t^3/3! + t^4/4! - t^5/5! + ...) =>

(1/2) * (2 + 0t + 2 * t^2/2! + 0 * t^3/3! + 2 * t^4/4! + ....) =>

(1/2) * (2 + 2 * t^2/2! + 2 * t^4/4! + 2 * t^6/6! + ....) =>

1 + t^2/2! + t^4/4! + t^6/6! + t^8/8! + ....

cosh(t) = 1 + t^2 / 2! + t^4 / 4! + t^6 / 6! + t^8 / 8! + ....

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

sum((2x)^(2k) / (2k)! , k = 0 , k = inf)