# `f(x)=ln(x^2+1), c=0` Use the definition of Taylor series to find the Taylor series, centered at c for the function.

Taylor series is an example of infinite series derived from the expansion of `f(x) ` about a single point. It is represented by infinite sum of `f^n(x)`  centered at `x=c` . The general formula for Taylor series is:

`f(x) = sum_(n=0)^oo (f^n(c))/(n!) (x-c)^n`

or

`f(x) = f(c) + f'(c) (x-c)+ (f'(c))/(2!) (x-c)^2+ (f'(c))/(3!) (x-c)^3+ (f'(c))/(4!) (x-c)^4+...`

To determine the Taylor series for the function `f(x)=ln(x^2+1) ` centered at `c=0` , we may list the `f^n(x)` as:

`f(x)=ln(x^2+1)`

Applying derivative formula for logarithmic function: d`/(dx) ln(u) = 1/u *(du)/(dx)` .

Let `u = x^2+1` then `(du)/(dx)=2x`

`f'(x) = d/(dx)ln(x^2+1)`

`= 1/(x^2+1) *2x`

`=(2x)/(x^2+1)`

Applying Quotient rule for differentiation: `d/(dx) (u/v) = (u' *v - u*v')/v^2` .

Let `u = 2x` then `u'= 2`

`v = x^2+1` then `v'=2x`  and` v^2 = (x^2+1)^2`

`f^2(x) = d/(dx)((2x)/(x^2+1))`

`= ( 2*(x^2+1)-(2x)(2x))/(x^2+1)^2`

`=( 2x^2+2-4x^2)/(x^2+1)^2 `

`= (2-2x^2)/(x^2+1)^2 `

Let `u =2-2x^2` then `u'= -4x`

` v = (x^2+1)^2`

then ` v^2 = ((x^2+1)^2)^2=(x^2+1)^4`

and `v'=2*(x^2+1)^(2-1)*2x=4x(x^2+1)`

`f^3(x) = ((-4x)(x^2+1)^2 -(2-2x^2)*4x(x^2+1))/(x^2+1)^4`

`=(x^2+1)^2((-4x)(x^2+1) -(2-2x^2)*4x)/(x^2+1)^4`

`=((-4x)(x^2+1) -(2-2x^2)*4x)/(x^2+1)^3`

`=((-4x^3-4x) -(8x-8x^3))/(x^2+1)^3`

`=(-4x^3-4x -8x+8x^3)/(x^2+1)^3`

` =(4x^3-12x)/(x^2+1)^3`

Let `u =(4x^3-12x)`  then `u'= 12x^2-12`

`v =(x^2+1)^3`

then `v^2 = ((x^2+1)^3)^2`

`=(x^2+1)^(3*2)`

`=(x^2+1)^6`

then `v'=3*(x^2+1)^(3-1)*2x`

`=6x(x^2+1)^2`

`f^4(x) = ((12x^2-12)*(x^2+1)^3 - (4x^3-12x)*6x(x^2+1)^2)/(x^2+1)^6`

`=(x^2+1)^2((12x^2-12)*(x^2+1) -(4x^3-12x)*6x)/(x^2+1)^6`

`=((12x^4-12)-(24x^4-72x^2))/(x^2+1)^4`

`=(12x^4-12-24x^4+72x^2)/(x^2+1)^4`

`=(-12x^4+72x^2-12)/(x^2+1)^4`

`f^5(x)=(-480x^3+28x^5+240x)/(x+1)^5`

`f^6(x)=(-240x^6+3600x^4-3600x^2+240)/(x+1)^6`

Plug-in `x=0` for each `f^n(x)` , we get:

`f(0)=ln(0^2+1)`

`= ln(1)`

`=0`

`f'(0)=(2*0)/(0^2+1)`

`=0/1`

`=0`

`f^2(0)= (2-2*0^2)/(0^2+1)^2 `

`= 2/1`

`= 2`

`f^3(0) =(4*0^3-12*0)/(0^2+1)^3`

`=0/1`

`=0`

`f^4(0)=(-12*0^4+72*0^2-12)/(0^2+1)^4`

`= -12/1`

`= -12`

`f^5(0)=(-480*0^3+28*0^5+240*0)/(0+1)^5`

`=0/1`

`=0`

`f^6(0)=(-240*0^6+3600*0^4-3600*0^2+240)/(0+1)^6`

`=240/1`

`=240`

Applying the formula for Taylor series, we get:

`ln(x^2+1) =sum_(n=0)^oo (f^n(0))/(n!) (x-0)^n`

` =sum_(n=0)^oo (f^n(0))/(n!) x^n`

` = f(0) + f'(0) x+ (f'(0))/(2!) x^2+(f'(0))/(3!) x^3+ (f'(0))/(4!) x^4+...`

` =0+ 0*x+2/(2!) x^2+ 0/(3!) x^3+ (-12)/(4!) x^4+ 0/(5!) x^5+ (240)/(6!) x^6+...`

` =0+ 0*x+2/2x^2+ 0/6 x^3-12/24 x^4+ 0/120 x^5+ 240/720x^6+...`

` =0+0+ x^2+0-1/2x^4+0+1/3x^6`

`= x^2-1/2x^4+1/3x^6+... `

The Taylor series of the function` f(x)=ln(x^2+1)` centered at `c=0` is:

`ln(x^2+1) =x^2-1/2x^4+1/3x^6+...`

or

`ln(x^2+1)= sum_(n=1)^oo (-1)^(n+1) x^(2n)/n`

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