Expansion of series and possible induction

[CAF123]

Homework Statement

Consider the infinite series $$\frac{x}{e^x - 1} = A_o + A_1 x + \frac{A_2}{2!}x^2 + ... + \frac{A_n}{n!}x^n + ...$$ Determine that $1 = A_o,\,\,\,\,\,0 = A_o/2! + A_1,\,\,\,\,\,0 = A_o/3! + A_1/2! + A_2/2!$.
Show that for $n > 1$, one can write the relations as $$(A+1)^n - A^n,$$ where $A^n \rightarrow A_n$.

2. Homework Equations
Taylor expansions and induction

The Attempt at a Solution

First part is fine. I thought about using induction on the second part since $n$ is confined to an integer. I understand $A^n \rightarrow A_n$ as meaning the powers of the series converge to the actual terms in the series, which is why I thought $n$ was an integer, labeling the terms in the series. However, I can't seem to make sense of the base case (n=2). It is $$(A+1)^2 - A^2 = 2A + 1$$ What is $A$?

thanks

 

[CAF123]

Anyone any comments or help? thanks.

 

[pasmith]

Homework Statement

Consider the infinite series $$\frac{x}{e^x - 1} = A_o + A_1 x + \frac{A_2}{2!}x^2 + ... + \frac{A_n}{n!}x^n + ...$$ Determine that $1 = A_o,\,\,\,\,\,0 = A_o/2! + A_1,\,\,\,\,\,0 = A_o/3! + A_1/2! + A_2/2!$.
Show that for $n > 1$, one can write the relations as $$(A+1)^n - A^n,$$ where $A^n \rightarrow A_n$.

2. Homework Equations
Taylor expansions and induction

The Attempt at a Solution

First part is fine. I thought about using induction on the second part since $n$ is confined to an integer. I understand $A^n \rightarrow A_n$ as meaning the powers of the series converge to the actual terms in the series, which is why I thought $n$ was an integer, labeling the terms in the series. However, I can't seem to make sense of the base case (n=2). It is $$(A+1)^2 - A^2 = 2A + 1$$ What is $A$?

thanks

What this is saying (I think) is that you can expand the left hand side of $(A + 1)^n - A^n = 0$, and if you then replace $A^k$ with $A_k$ where $0 \leq k < n$, you can find $A_{n-1}$ in terms of $A_0, \dots, A_{n-2}$.

Thus: $(A + 1)^2 - A^2 = 2A + 1$ and indeed $2A_1 + A_0 = 0$.

 

[CAF123]

What this is saying (I think) is that you can expand the left hand side of $(A + 1)^n - A^n = 0$, and if you then replace $A^k$ with $A_k$ where $0 \leq k < n$, you can find $A_{n-1}$ in terms of $A_0, \dots, A_{n-2}$.

I see, is there a reason why you neglected the case where $k=n$?

Thus: $(A + 1)^2 - A^2 = 2A + 1$ and indeed $2A_1 + A_0 = 0$.

I nearly see what you did there except how did you get that $A=A_1$? Is this just a notation $A = A^1 = A_1$?

Would you say induction is the way to go? I can try further given I now understand what the base case means, thanks.

 

[pasmith]

I see, is there a reason why you neglected the case where $k=n$?

The $A^n$ terms cancel.

I nearly see what you did there except how did you get that $A=A_1$? Is this just a notation $A = A^1 = A_1$?

$x^1 = x$ for every $x$.

Would you say induction is the way to go?

It is the obvious method.

 

[haruspex]

I nearly see what you did there except how did you get that $A=A_1$? Is this just a notation $A = A^1 = A_1$?

It's not that $A^1 = A_1$, but that it represents it in the analogy between the two sets of equations.

 

[CAF123]

I tried the induction argument, but seemed to run into a mistake. Assume $P(n) = (A + 1)^n - A^n = 0$ is true for $n$. Then $$(A+1)^{n+1} - A^{n+1} = (A+1)^n (A+1) - A^n A $$$$= ((A+1)^n - A^n)A + (A+1)^n = A^n \neq 0$$ The $A^n$ are just numbers so I am not sure where the mistake is.

 

[pasmith]

I think you want to be using induction to establish that $$\sum_{k=0}^{n-1} \binom{n}{k} A_k = 0$$ since $$(A+ 1)^n - A^n = \sum_{k=0}^{n-1} \binom{n}{k} A^k.$$