Taylor Expansion of $x\sqrt{x}$ at x=9

Question

How can I go about solving the Taylor expansion of $x\sqrt{x}$ at x=9?

I solved the derivative down to the 5th derivative and then tried subbing in the 9 value for a using this equation

$$\sum_{n=0}^\infty \frac{f^{(n)}(a)}{n!}(x-a)^n.$$

Can someone walk me through the process (in neatly put-together LaTEX/Jax?)

I know I'm asking for a lot! Thanks. :)

Answer 1

When it's possible, it's wise to avoid explicit computation.

Note that $$ x\sqrt{x} = [(x - 9)+9]\sqrt{x-9 + 9}=\\ 9[(x/9 - 1)+1]\sqrt{9[x/9-1 + 1]}=\\ 27[(x/9 - 1)+1]\sqrt{[x/9-1 + 1]} $$ Let's define $z = x/9 - 1 = (x-9)/9$. Find the Taylor expansion (in terms of $z$) for $$ 27[z + 1]\sqrt{z+1} $$ By looking up the expansion for $\sqrt{z+1}$ in a table. Then, make the above substitution, and put the resulting series in the correct form.

Answer 2

Lets write down the first couple of derivatives first:

$f(x)=x\cdot \sqrt x$

$f'(x)=\frac{3\cdot\sqrt x }{2}$

$f''(x)=\frac{3}{4\cdot \sqrt x}$

You mentioned the taylor expansion in your opening post:

$\sum_{n=0}^\infty \frac{f^{(n)}(a)}{n!}(x-a)^n.$

So your first three terms will be:

$\frac{f^{o}(9)}{0!}(x-9)^0=27$

$\frac{f^{1}(9)}{1!}(x-9)^1=\frac{9(x-9)}{2}$

$\frac{f^{2}(9)}{2!}(x-9)^2=\frac{1}{8}(x-9)^2$

(Just so you don't get confused $f^2=f''$)

Now it's your turn. Try to calculate the next couple of terms.