# Thread: Finding a series ?

1. ## Finding a series ?

Consider the series for 1/1+x = 1+x^2+.....+x^n (1)

now use that to find the series for 1/1+x^2 (which is basically replace x with x^2

which produces

1/1+x^2 = 1+x^2+x^4+....+x^2n (2)

and now find a series with 2x/1+x^2

basically multiply (2) with 2x ....

no my question is how do i find the series for

ln(1+x^2) using those facts above me

i used equation (2)

1+x^2 = 1/(1+x^2+x^4+....+x^2n)
now i log both sides

ln(1+x^2) = ln(1/(1+x^2+x^4+....+x^2n)

which gives me

ln(1+x^2) = -ln(1+x^2+x^4+....+x^2n)

is this correct ??

or am i going no where with this proof ?

Tanks

2. Originally Posted by Khonics89
Consider the series for 1/1+x = 1+x^2+.....+x^n (1)

now use that to find the series for 1/1+x^2 (which is basically replace x with x^2

which produces

1/1+x^2 = 1+x^2+x^4+....+x^2n (2)

and now find a series with 2x/1+x^2

basically multiply (2) with 2x ....

no my question is how do i find the series for

ln(1+x^2) using those facts above me

i used equation (2)

1+x^2 = 1/(1+x^2+x^4+....+x^2n)
now i log both sides

ln(1+x^2) = ln(1/(1+x^2+x^4+....+x^2n)

which gives me

ln(1+x^2) = -ln(1+x^2+x^4+....+x^2n)

is this correct ??

or am i going no where with this proof ?

Tanks
Note that $\displaystyle \int \frac{2x}{1+x^2} \, dx = \ln (1 + x^2) + C$ ....

3. Hmm a little confused... why do i need to integrate ?

4. Originally Posted by mr fantastic
Note that $\displaystyle \int \frac{2x}{1+x^2} \, dx = \ln (1 + x^2) + C$ ....
I had hoped from my first reply that you would realise that you need to integrate the series for $\displaystyle \frac{2x}{1+x^2}$. For reasons already explained the result is equal to $\displaystyle \ln (1 + x^2) + C$. To get the value of C, substitute x = 0 into the result (unsurprisingly you find C = 0).

5. mr fantastic.. you are hopeless at explaining stuff you know that!

6. Originally Posted by Khonics89
mr fantastic.. you are hopeless at explaining stuff you know that!
My replies are not intended to be explanations. They are hints (and blunt hints at that). You're meant to think about what I posted and see how it relates to the question. Read the replies again and then, after you have applied appropriate thought, say what you don't understand if you're still stuck.

7. Originally Posted by Khonics89
Consider the series for 1/1+x = 1+x^2+.....+x^n (1)

now use that to find the series for 1/1+x^2 (which is basically replace x with x^2

which produces

1/1+x^2 = 1+x^2+x^4+....+x^2n (2)

and now find a series with 2x/1+x^2

basically multiply (2) with 2x ....

no my question is how do i find the series for

ln(1+x^2) using those facts above me

i used equation (2)

1+x^2 = 1/(1+x^2+x^4+....+x^2n)
now i log both sides

ln(1+x^2) = ln(1/(1+x^2+x^4+....+x^2n)

which gives me

ln(1+x^2) = -ln(1+x^2+x^4+....+x^2n)

is this correct ??

or am i going no where with this proof ?

Tanks
First the series are infinite series not finite series.

Alternative method (not compliant with some of the restrictions you imply on method):

$\displaystyle \frac{1}{1+x}=\sum_{r=0}^{\infty}x^r,\ \ \ |x|<1$

Now:

$\displaystyle \frac{d}{dt}\ln(1+x)=\frac{1}{1+x}$

so:

$\displaystyle \ln(1+x)=\int_{\zeta=0}^x \frac{1}{1+\zeta}\; d\zeta=\int_{\zeta=0}^x \sum_{r=0}^{\infty}\zeta^r\; d\zeta$

Now interchange the integral and summations, do the integrals to get a series representation for $\displaystyle \ln(1+x)$, ...

CB