# Another try at a difficult proof

1. Feb 14, 2005

### sebasalekhine7

I found something very strange working with Riemann sums and expansion series.
Can anyone tell me why
$$\frac{x^n}{\sum_{i=1}^n x^{n-i}}=\frac{(x-1)x^n}{x^n-1}$$

Excuse the profanity, but I thought in the beginning this was a disgusting joke! :rofl: :yuck:

2. Feb 14, 2005

### dextercioby

Define a new variable of summation
$$j=:n-i$$

Transform the sum $\sum_{i=1}^{n} x^{n-i}$ into a sum over "j" and the result will come out immediately.

Daniel.

P.S.What profanity?

3. Feb 14, 2005

### sebasalekhine7

Sorry, but I don't quite understand how to do what you just wrote.

4. Feb 14, 2005

### learningphysics

The way I did, just multiply the numerator and denominator by x-1. So in the denominator you get 2 summations (one minus the other). Simplify the denominator.

5. Feb 14, 2005

### dextercioby

It doesn't make any sense to me.
$$j=:n-i$$ (1)
$$i=1 \Rightarrow j=n-1$$ (2)
$$i=n \Rightarrow j=0$$ (3)

The new sum (over "j") becomes
$$\sum_{j=n-1}^{j=0} x^{j}=\sum_{j=0}^{j=n-1} x^{j}=\frac{x^{n}-1}{x-1}$$ (4)

Now plug it in the initial expression and u'll find the desired result...

Daniel.

6. Feb 14, 2005

### learningphysics

If you multiply the denominator through by x-1 you get:

$$\sum_{i=1}^{n} x^{n-i+1} - \sum_{i=1}^{n}x^{n-i}$$

You can simplify the above (try to make the two sums look the same) and get $$x^n-1$$

7. Feb 14, 2005

### dextercioby

How do you make that simplification...?

Daniel.

8. Feb 14, 2005

### learningphysics

Similar to your method:

$$\sum_{i=1}^{n} x^{n-i+1} - \sum_{i=1}^{n}x^{n-i} =$$

$$\sum_{i=1}^{n} x^{n-(i-1)} - \sum_{i=1}^{n}x^{n-i}=$$

For the first sum, let j=i-1

$$\sum_{j=0}^{n-1} x^{n-j} - \sum_{i=1}^{n}x^{n-i} =$$

$$(x^{n-0} + \sum_{j=1}^{n-1} x^{n-j}) - (\sum_{i=1}^{n-1}x^{n-i} + x^{n-n}) =$$

I take out the first term from the first sum and last term from the second sum.

The two sums that are left cancel each other out.

$$x^n-1$$

9. Feb 14, 2005

### dextercioby

Wow,that's so simple!*admirative*...Gosh,why didn't i think of that...?*envy*

Daniel.

P.S.We should let the OP decide which solution he likes best.

10. Feb 14, 2005

### sebasalekhine7

I think they are both equally creative, well, thanks a lot for your help.

11. Feb 14, 2005

### learningphysics

well, yours is the faster solution.

12. Feb 15, 2005

### Curious3141

Why can't one just use the sum for Geometric series ?

The denominator is $$x^{n-1} + x^{n-2} + ... + x^0 = \frac{x^n - 1}{x - 1}$$

using the well worn sum for GPs, and the answer immediately follows.

13. Feb 15, 2005

### Galileo

Indeed, you can just cancel the $x^n$ on both sides. Invert to get:
$$\frac{x^n-1}{x-1}=\sum_{i=1}^n x^{n-i}}$$
A well known fact.

PS: $\sum_{i=1}^n x^{n-i}=\sum_{i=0}^{n-1} x^{i}$

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