How can we write the function x/(ex−1) as the sum of a Taylor Series?

Answer 1

First write x/(e^x − 1) as 1/{ x^-1[(∑_{(n = 0 to n = ∞)}(f⁽ⁿ⁾(0)(x − 0)ⁿ)/n!) − 1] } where [∑_{(n = 0 to n = ∞)}(f⁽ⁿ⁾(0)(x − 0)ⁿ)/n!] is the Taylor Series Expansion Formula.

The 2nd form of x/(e^x − 1) will expand to 1/{x^-1[e⁰x⁰/0! + e⁰x¹/1! + e⁰x²/2! + e⁰x³/3! + e⁰x⁴/4! + e⁰x⁵/5! + e⁰x⁶/6! + e⁰x⁷/7! + e⁰x⁸/8! + e⁰x⁹/9! + e⁰x¹⁰/10! + e⁰x¹¹/11! + e⁰x¹²/12! + ... − 1]}.

This expansion of the 2nd form of x/(e^x − 1) can be written as:

1/{e⁰ (or 1) times [∑_{(n = 0 to n = ∞)}x^{(n − 1)}/n!] − x^-1} (Expression A).

With x = 1 as a test value, Expression A for 0 ≤ n ≤ 12 will come to a limiting value of 1/(2.718281828 − 1) or 0.581976707. This is the value of the given function, x/(e^x − 1), when x = 1.

Answer 2

Since f(x)=∑f⁽ⁿ⁾(a)/n! (x-a)ⁿ all you have to do is pick "a" the point you want the series to expand about and calculate the f⁽ⁿ⁾(a)/n! terms for n=0,1,2,3,....for as many terms as needed.

In your case f(x)=x/(e^x-1) and the f⁽ⁿ⁾(a)'s require the use of l'Hospital's rule to evaluate if you pick a=0 as the expansion point, otherwise it's straight forward.

Let me know if you have any questions.

Regards

Jim

Answer 3

Taylor series definition:

f(x) = Σ^∞_n=0 ^{f⁽ⁿ⁾(a)}/_n! = f(a) + ^f'(a)/_1!(x – a) + ^f''(a)/_2!(x – a)² + ^f'''(a)/_3!(x – a)³ + ...

f'(a) = ^{eˣ – 1 – xeˣ}/_{(eˣ – 1)²} = ^{eˣ – 1}/_{(eˣ – 1)²} – ^xeˣ/_{(eˣ – 1)²} = ¹/_{eˣ – 1} – ^xeˣ/_{(eˣ – 1)²}

f''(a) = ^-eˣ/_{(eˣ – 1)²} – ^{(eˣ + xeˣ)(eˣ – 1)² – 2xe²ˣ(eˣ – 1)}/_{(eˣ – 1)³}

etc.

Therefore...

^x/_{eˣ – 1} = ^a/_eᵃ – 1 + (¹/_{eˣ – 1} – ^xeˣ/_{(eˣ – 1)²})(x – a) + ¹/₂(^-eˣ/_{(eˣ – 1)²} – ^{(eˣ + xeˣ)(eˣ – 1)² – 2xe²ˣ(eˣ – 1)}/_{(eˣ – 1)³})(x – a)² + ...

There are ways to simplify that second derivative, but it's certainly getting messy, isn't it?