## Example 8.4.4 (a): Taylor Series

Find the Taylor series centered at

*c = 0*for*f(x) = x*. Then find^{3}+ 2x^{2}+ 3x + 4*T*. Confirm that_{f}(x, 1)*T*for all_{f}(x, 0) = T_{f}(x, 1)*x*.To find a Taylor series we need to find for the given function at the specified center. First, let's find the various derivatives:

c=0c=1f(x) = x^{3}+ 2x^{2}+ 3x + 4f(0) = 4f(1) = 10f '(x) = 3x^{2}+ 4x + 3f '(0) = 3f '(1) = 10f ''(x) = 6x + 4f ''(0) = 4f ''(1) = 10f '''(x) = 6f '''(0) = 6f '''(1) = 6f^{(n)}(x) = 0, n > 3f^{(n)}(0) = 0f^{(n)}(1) = 0

Now we can put together the Taylor series, which in this case will reduce to
finite polynomials since *f ^{(n)}(x) = 0*
for

*n > 3*:

T_{f}(x, 0) = 4 + 3/1! x + 4/2! x^{2}+ 6/6! x^{3}=

= 4 + 3x + 2x^{2}+ x^{3}

In other words, *T _{f}(x, 0)* is the original polynomial.
For our second Taylor series we get

T_{f}(x, 1) = 10 + 10/1! (x-1) + 10/2! (x-1)^{2}+ 6/6! (x-1)^{3}=

= 10 + 10(x-1) + 5(x-1)^{2}+ (x-1)^{3}

But

so that indeedT_{f}(x, 1) = 10 + 10(x-1) + 5(x-1)^{2}+ (x-1)^{3}=

= 10 + 10x - 10 + 5x^{2}-10x + 5 + x^{3}- 3x^{2}+ 3x - 1 =

= 10 - 10 + 5 - 1 + 10x - 10x + 3x + 5x^{2}- 3x^{2}+ x^{3}=

= 4 + 3x + 2x^{2}+ x^{3}

*T*for all

_{f}(x, 0) = T_{f}(x, 1)*x*.