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`f(x)=1/x^2 , n=4,c=2` Find the n'th Taylor Polynomial centered at c

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


To determine the Taylor polynomial of degree `n=4 ` from the given function `f(x)=1/x^2` centered at `c=2` , we may apply the definition of Taylor series.


For the list of `f^n(x)` up to `n=4` , we may apply Law of Exponent: `1/x^n = x^-n`  and  Power rule for derivative: `d/(dx) x^n= n *x^(n-1)` .


`f(x) = 1/x^2 or x^(-2)`


`f'(x) = d/(dx)x^(-2)`


           `=-2 *x^(-2-1)`


           `=-2x^(-3) or -2/x^3`


`f^2(x)= d/(dx) -2x^(-3)`


           `=-2 *d/(dx) x^(-3)`


           `=-2 *(-3x^(-3-1))`


           `=6x^(-4) or 6/x^4`


`f^3(x)= d/(dx) 6x^(-4)`


           `=6 *d/(dx) x^(-4)`


           `=6 *(-4x^(-4-1))`


           ` =-24x^(-5) or -24/x^5`


`f^4(x)= d/(dx) -24x^(-5)`


           ` =-24 *d/(dx) x^(-5)`


           `=-24 *(-5x^(-5-1))`


           `=120x^(-6) or 120/x^6`


Plug-in `x=2` , we get:


`f(2)=1/2^2 =1/4`


`f'(2)=-2/2^3 = -1/4`


`f^2(2)=6/2^4 =3/8`


`f^3(2)=-24/2^5 = -3/4`


`f^4(2)=120/2^6= 15/8`


Applying the formula for Taylor series, we get:


`sum_(n=0)^4 (f^n(2))/(n!) (x-2)^n`


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


`=1/4+(-1/4)(x-2) +(3/8)/(2!)(x-2)^2 +(-3/4)/(3!)(x-2)^3+(15/8)/(4!)(x-2)^4`


`=1/4-1/4(x-2) +(3/8)/2(x-2)^2 -(3/4)/6(x-2)^3+(15/8)/24(x-2)^4`



`=1/4-1/4(x-2) + 3/16(x-2)^2 -1/8(x-2)^3+5/64(x-2)^4`


The Taylor polynomial of degree `n=4`  for the given function `f(x)=1/x^2 ` centered at `c=2` will be:


`P_4(x)=1/4-1/4(x-2) + 3/16(x-2)^2 -1/8(x-2)^3+5/64(x-2)^4`

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