Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Suppose $$F(x)=\sum_{n \geq 0}{f_nx^n}=\frac{x}{1-3x}$$ We are asked to find an explicit formula for $f_n$. My working is , since$$\frac{1}{1-3x}=\sum_{n \geq 0}{3^nx^n} \\ \frac{x}{1-3x}=\sum_{n \geq 0}{3^nx^{n+1}}= \sum_{n \geq 0}{3^{n-1}x^n}$$ So we know that $f_n=3^{n-1}$ . I would like to ask when we do 'shifting' , wouldn't we also need to shift the index ? Like in this case , why we do not shift the index from $n \geq 0$ to $n \geq 1$ ?

share|improve this question
add comment

4 Answers 4

up vote 2 down vote accepted

You need to shift the index to $n\ge 1$. (If you don't do it, comparing the coefficient of $x^0$ - i.e. plugging $x=0$ - gives contradiction). The reasoning is as follows:

In $\sum_{n \geq 0}{3^nx^{n+1}}$ you did the substitution $m=n+1$ (equivalent to $n=m-1$), and so it becomes $\sum_{m-1 \geq 0}{3^{m-1}x^{m}}=\sum_{m \geq 1}{3^{m-1}x^{m}}$.

This is not very different from doing a variable substitution in an integral and then changing the limits of the integrals.

share|improve this answer
add comment

Yes, you do need to shift the index - what you've written is incorrect. You should have $$\frac{x}{1-3x}=x\sum_{n\geq 0}3^n x^n=\sum_{n\geq 0} 3^nx^{n+1}=0x^0+3^0x^1+3^1x^2+\cdots=\sum_{n\geq 0} f_nx^n$$ where $$f_n=\begin{cases} 0 & \text{ if }n=0,\\ 3^{n-1} & \text{ if }n\geq 1.\end{cases}$$

share|improve this answer
add comment

By shifting I think you mean $$\sum_{n \geq 0}{3^nx^{n+1}}= \sum_{n \geq 0}{3^{n-1}x^n}$$ Is this valid? The left hand side is $$ x^{1}+3x^2+9x^3+....$$ while the right hand side is $$ \frac13+x^{1}+3x^2+9x^3+....$$ As you can see they are not equal. What is valid is the following $$\sum_{n \geq 0}{3^nx^{n+1}}= \sum_{n \geq 1}{3^{n-1}x^n}=\sum_{n \geq 0}{3^{n-1}x^n}-\frac13$$ It can be easily seen that $f_0=0$ and $f_n=3^{n-1}$, $n\ge 1$

share|improve this answer
add comment

$$ \sum_{n\geq 0}3^nx^{n+1}=\sum_{n\geq 1}3^{n-1}x^{n}=\sum_{n\geq 0}f_n x^n, $$ with $f_0=0$ and $f_n=3^{n-1}$ for $n\geq 1$.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.