# Find the Taylor polynomials p3 and p4 centered at `a=0 for f(x)=(3+2)^-2`

Hello!

For a function `f ( x ) , ` `n ` times differentiable in some neighborhood of a point `a , ` its n-th Taylor polynomial is by definition

`sum_( k = 0 )^n (f^( ( k ) ) ( a )) / ( k ! ) ( x...

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Hello!

For a function `f ( x ) , ` `n ` times differentiable in some neighborhood of a point `a , ` its n-th Taylor polynomial is by definition

`sum_( k = 0 )^n (f^( ( k ) ) ( a )) / ( k ! ) ( x - a )^k ,`

where `f^( ( k ) ) ( a ) ` is the k-th derivative of `f ( x ) ` at `x=a .`

These polynomials are not arbitrary, they approximate `f ( x ) ` near `a ` to the degree `n + 1 . ` For `n = 1 ` the polynomial is the tangent to the graph of `f ( x ) ` at `x = a .`

This way the procedure of finding 3-th and 4-th Taylor polynomial is straightforward, we need to find all derivatives up to 4. Here `a=0 ` and I suppose `f(x) = (3+2x)^(-2), ` so

`f' ( x ) = -4 ( 3 + 2 x )^( -3 ), f'' ( x ) = 24 ( 3 + 2 x )^( -4 ), `

Further,

`f( 0 ) = 3^( -2 ), f'( 0 ) = -4*3^( -3 ), f''( 0 ) = 24*3^( -4 ), ` `f'''( x ) = -192 ( 3 + 2 x )^( -5 ), f^( ( 4 ) ) ( x ) = 1920 ( 3 + 2 x )^( -6 ).`

Finally,

`p_3(x) = 3^(-2)-4*3^(-3)x+12*3^(-4)x^2-32*3^(-5)x^3 =`
`= 1/9-4/27x+4/27x^2-32/243x^3`

and `p_4(x)=p_3(x) + 80*3^(-6)x^4 = 1/9-4/27x+4/27x^2-32/243x^3 + 80/729x^4.`

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