series representation for tan^(-1)x

Question

amistre64 · Answer

whats the derivative of tan^-1 x ?

idku · Answer

I am going to start from the elementary geometric series.  $\large\color{black}{ \displaystyle  \frac{1}{1-x}=\sum_{n=0}^{\infty}x^n}$

making a substitution of -x instead of x on both sides.
$\large\color{black}{ \displaystyle  \frac{1}{1-(-x)}=\sum_{n=0}^{\infty}(-x)^n}$

$\large\color{black}{ \displaystyle  \frac{1}{x+1}=\sum_{n=0}^{\infty}(-1)^nx^n}$

substitution of x^2 instead of x

$\large\color{black}{ \displaystyle  \frac{1}{x^2+1}=\sum_{n=0}^{\infty}(-1)^nx^{2n}}$

idku · Answer

then I integrate both sides

idku · Answer

$\large\color{black}{ \displaystyle  \int \frac{1}{x^2+1} dx=\int \sum_{n=0}^{\infty}(-1)^nx^{2n}dx}$

idku · Answer

$\large\color{black}{ \displaystyle  \tan^{-1}x=\sum_{n=0}^{\infty}\left[(-1)^n\frac{x^{2n+1}}{2n+1}\right]+C}$

geerky42 · Answer

C shouldn't be included;
$\large\color{black}{ \displaystyle  \int \frac{1}{x^2+1} dx=\int \sum_{n=0}^{\infty}(-1)^nx^{2n}dx}$

$\large\color{black}{ \displaystyle  \tan^{-1}x~+C=\sum_{n=0}^{\infty}\left[(-1)^n\frac{x^{2n+1}}{2n+1}\right]+C}$

$\large\color{black}{ \displaystyle  \tan^{-1}x=\sum_{n=0}^{\infty}\left[(-1)^n\frac{x^{2n+1}}{2n+1}\right]}$

idku · Answer

but those are not necessarily the same C's on v=both sides

geerky42 · Answer

Yeah but C isn't supposed to be there, I just don't know how to explain lol...

amistre64 · Answer

tan^-1 (x)  is not a family of functions ...

amistre64 · Answer

you are trying to detemrine the powere series of a specific function

amistre64 · Answer

tan^-1(0) = power series at zero if you want to "solve for C"

idku · Answer

So no +C ever, in these cases?

amistre64 · Answer

you arent defining a family of functions, so its not pertinent

idku · Answer

so lets say ln(1-x)

$\large\color{black}{ \displaystyle  \ln(1-x)=\sum_{n=0}^{\infty}\left[\frac{x^{n+1}}{n+1}\right]}$

amistre64 · Answer

int [1/(x^2+1)] dx = tan^-1 (x) +C

but C = 0 when we are looking for the value of tan^-1 (x) right?

idku · Answer

ok, as I understand we don't put the +C here....

idku · Answer

yes, right...

amistre64 · Answer

the +C is irrelevant, we can either ignore it, or solve for it .. solving for it will show its equal to 0 if you want to insist on keeping it around :)

idku · Answer

So, if          $\large\color{black}{ \displaystyle  f_1(x)=\sum_{n=0}^{\infty}f_2({x,n})}$

then,         $\large\color{black}{ \displaystyle  F_1(x)=\sum_{n=0}^{\infty}F_2({x,n})}$

(without +C, for any function in general), such that   F`=f

idku · Answer

it does look confusing, I can elaborate on what I'm asking.

idku · Answer

$\large\color{black}{ \displaystyle  f_1(x)=\sum_{n=0}^{\infty}f_2({x,n})}$

where:
$\large\color{black}{ \displaystyle  f_1(x)}$ some function y=f(x)
$\large\color{black}{ \displaystyle  f_2(x,n)}$ equivalent to f(x) power series representation (function of x and n)
$\large\color{black}{ \displaystyle  F_1(x)}$ general antideriv. (not a family of functions) for the f(x)
$\large\color{black}{ \displaystyle  F_2(x,n)}$ is the antideriv. (not a family of functions) for the power series $\large\color{black}{ \displaystyle f_2(x,n)}$ (where both the function and antideriv. function are both functions of x and n)

then if I wanted a power series to $\large\color{black}{ \displaystyle  F_1(x)}$ (thatis the antideriv. of $\large\color{black}{ \displaystyle  f_1(x)}$ ), then I could integrate both sides, without adding +C's, and get

$\large\color{black}{ \displaystyle F_1(x)=\sum_{n=0}^{\infty}F_2({x,n})}$

amistre64 · Answer

$$f(x)=a_0+a_1x+a_2x^2+a_3x^3+a_4x^4+...$$
because integration is a linear thing
$$\int f(x)=\int a_0+\int a_1x+\int a_2x^2+\int a_3x^3+\int a_4x^4+...$$
$$F(x)+c_0=a_0x+c_1+\frac12a_1x^2+c_2+\frac13a_2x^3+c_3+\frac14 a_3x^4+c_4+\frac15a_4x^5+c_5+...$$

$$F(x)+c_0=(a_0x+\frac12a_1x^2+\frac13a_2x^3+\frac14 a_3x^4+\frac15a_4x^5+...)+k_0$$

what is F(x)?  well, F(x) + 0 = F(x) so c_0 = 0

$$F(x)=(a_0x+\frac12a_1x^2+\frac13a_2x^3+\frac14 a_3x^4+\frac15a_4x^5+...)+k_0$$

now what is F(0)?  if F(0) = 3, then k_o = 3 seems fair

but k_o is just the constant term of the power series

idku · Answer

so, no +C :) tnx

idku · Answer

even if I wrote on my teacher wouldn't deduct, but it is still nice to know.... tnx again

amistre64 · Answer

yep ...

amistre64 · Answer

more succinctly, i would say k_o = F(0), which just represents the constant term of the power series.  or k_o = F(a) for some series centered around x=a

idku · Answer

I am starting to make sense from the math I am doing:D
Don't ever or going to be very good at this, but at least I can be decent:)

tnx for the help

amistre64 · Answer

good luck :)