Another try at a difficult proof

[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!

 

[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?

 

[sebasalekhine7]

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

 

[learningphysics]

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

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.

 

[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.

 

[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

 

[dextercioby]

How do you make that simplification...?

Daniel.

 

[learningphysics]

How do you make that simplification...?

Daniel.

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

 

[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.

 

[sebasalekhine7]

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

 

[learningphysics]

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.

well, yours is the faster solution.

 

[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.

 

[Galileo]

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.

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}