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