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[ )

EinsteinJr
May 12, 2015

#1**+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

#1**+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

#4**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**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