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What is the value of:

$$1+2x+3x^2+4x^3+5x^4+6x^5+...$$ in function of x,

With $$|x|<1$$ ( x ∈ ]-1;1[ )

difficulty advanced
EinsteinJr  May 12, 2015

Best Answer 

 #1
avatar+26328 
+15

Let f = 1 + 2x + 3x^2 + 4x^5 + ...

 

First consider y = x + x^2 + x^3 + x^4 + x^5 + ...

y = x(1 + x + x^2 + x^3 + x^4 + ...) or y = x(1 + y)  (as long as y converges, which it does if |x| < 1)

so y(1 - x) = x

y = x/(1 - x)

 

Now from the original series expression for y we have dy/dx = 1 + 2x + 3x^2 + 4x^3 + ... or in other words

f = dy/dx.  Since y also equals x/(1 - x) we have dy/dx = 1/(1-x) + x/(1-x)^2 or dy/dx = (1-x+x)/(1-x)^2 or dy/dx = 1/(1-x)^2, so, f = 1/(1-x)^2 or

 

1 + 2x + 3x^2 + 4x^5 + ... = 1/(1 - x)^2 

.

Alan  May 12, 2015
Sort: 

5+0 Answers

 #1
avatar+26328 
+15
Best Answer

Let f = 1 + 2x + 3x^2 + 4x^5 + ...

 

First consider y = x + x^2 + x^3 + x^4 + x^5 + ...

y = x(1 + x + x^2 + x^3 + x^4 + ...) or y = x(1 + y)  (as long as y converges, which it does if |x| < 1)

so y(1 - x) = x

y = x/(1 - x)

 

Now from the original series expression for y we have dy/dx = 1 + 2x + 3x^2 + 4x^3 + ... or in other words

f = dy/dx.  Since y also equals x/(1 - x) we have dy/dx = 1/(1-x) + x/(1-x)^2 or dy/dx = (1-x+x)/(1-x)^2 or dy/dx = 1/(1-x)^2, so, f = 1/(1-x)^2 or

 

1 + 2x + 3x^2 + 4x^5 + ... = 1/(1 - x)^2 

.

Alan  May 12, 2015
 #2
avatar+886 
0

Exactly.

Here's your cookie :

EinsteinJr  May 12, 2015
 #4
avatar+91001 
0

Thanks Alan,  I have not quite got my head around your answer yet, I will try some more:/

Einstein, does 'exacly' mean 'thank you' to you?

OR  did you already know the answer and thought Alan, or someone else, needed some brain food?

 

I will certainly admit that it was a excellent question. :)

Melody  May 13, 2015
 #5
avatar+886 
0

I already knew the answer.

But I rarely (not to say "never") submit questions when I don't already know the answer. It's just, as you said, "brain food"

EinsteinJr  May 13, 2015
 #6
avatar+91001 
0

I have finally looked at it properly 

That is really neat Alan.  Thankyou   

I have added this one to the "Great anwswers to Learn from" Thread

Melody  May 14, 2015

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