# Math Help - Taylor Polynomial

1. ## Taylor Polynomial

hey..just wanted to check if im doing this correctly

Find the Taylor polynomial of degree 4 for the function f(x) = x^(3/2) about the centre c=1

the 1st 5 terms i got is...

1, 3/2, 3/4, -3/8, 9/16

Now im not quite sure if im writing the series correctly..

1 + 3/2(3/2 - 1) + 3/4(3/4 -1) - 3/8(-3/8 -1)^2/2! + 9/16(9/16 - 1)^3/3!

am i correct?? or should i just leave the bracket as (x - 1)^2 / 2! and so on......

Thanks

2. Originally Posted by b00yeah05
hey..just wanted to check if im doing this correctly

Find the Taylor polynomial of degree 4 for the function f(x) = x^(3/2) about the centre c=1

the 1st 5 terms i got is...

1, 3/2, 3/4, -3/8, 9/16

Now im not quite sure if im writing the series correctly..

1 + 3/2(3/2 - 1) + 3/4(3/4 -1) - 3/8(-3/8 -1)^2/2! + 9/16(9/16 - 1)^3/3!

am i correct?? or should i just leave the bracket as (x - 1)^2 / 2! and so on......

Thanks
If you have to find a polynomial approximation for f(x), your polynomial must have x's in it ...... !

3. Leave the bracket as (x-1).

$1+\frac{3}{2}(x-1)+\frac{3}{16}(x-1)^{2}-\frac{1}{16}(x-1)^{3}+\frac{3}{128}(x-1)^{4}-\frac{3}{256}(x-1)^{5}+.....$

4. Originally Posted by b00yeah05
Now im not quite sure if im writing the series correctly..

1 + 3/2(3/2 - 1) + 3/4(3/4 -1) - 3/8(-3/8 -1)^2/2! + 9/16(9/16 - 1)^3/3!

am i correct?? or should i just leave the bracket as (x - 1)^2 / 2! and so on......

Thanks
No, you should leave the factors of $(x - 1)$ alone. The Taylor series expansion of a function $f(x)$ about a point $x = a$ is given by

$f(a) + f'(a)(x - a) + \frac{f''(a)}{2!}(x - a)^2 + \cdots + \frac{f^{(n)}(a)}{n!}(x - a)^n + \cdots$

In your case, we have the following:

$f(x) = x^{3/2}$

$a = 1$

and

$\begin{array}{rclcrcl}
f(x) & = & x^{3/2} & \Rightarrow & f(1) & = & 1\\
f'(x) & = & \frac32x^{1/2} & \Rightarrow & f'(1) & = & \frac32\\
f''(x) & = & \frac34x^{-1/2} & \Rightarrow & f''(1) & = & \frac34\\
f'''(x) & = & -\frac38x^{-3/2} & \Rightarrow & f'''(1) & = & -\frac38\\
f^{(4)}(x) & = & \frac9{16}x^{-5/2} & \Rightarrow & f^{(4)}(1) & = & \frac9{16}
\end{array}$

So, you should get

$f(a) + f'(a)(x - a) + \frac{f''(a)}{2!}(x - a)^2 + \frac{f'''(a)}{3!}(x - a)^3 + \frac{f^{(4)}(a)}{4!}(x - a)^4$

$= 1 + \frac32(x - 1) + \frac38(x - 1)^2 - \frac3{48}(x - 1)^3 + \frac9{384}(x - 1)^4$

$= 1 + \frac32(x - 1) + \frac38(x - 1)^2 - \frac1{16}(x - 1)^3 + \frac3{128}(x - 1)^4$

Note that the series expansion produces a polynomial in $x$ and not a constant.