# Thread: Taylor series question confusion

1. ## Taylor series question confusion

There is this problem that has been appearing a lot lately but I cannot find any examples that help me solve this.

It states:

"Find (at least) the first 3 nonzero terms in the Taylor (power) series for

$f(x) = \frac{x^3}{3+5x}$ centered at $a = 0.$

(This is the same as the Maclaurin series for the function $f$)"

Now I know how to use the Taylor series and get an expansion, but my method involves continuously taking the derivative of the function which can get pretty messy after even the second derivation. And even at that point, I'm still getting zero for my expansion terms.

What is this question asking exactly?

2. Originally Posted by freyrkessenin
There is this problem that has been appearing a lot lately but I cannot find any examples that help me solve this.

It states:

"Find (at least) the first 3 nonzero terms in the Taylor (power) series for

$f(x) = \frac{x^3}{3+5x}$ centered at $a = 0.$

(This is the same as the Maclaurin series for the function $f$)"

Now I know how to use the Taylor series and get an expansion, but my method involves continuously taking the derivative of the function which can get pretty messy after even the second derivation. And even at that point, I'm still getting zero for my expansion terms.

What is this question asking exactly?
I agree that Taylor series can be fairly nasty at time, but sometimes (as in your case you can make life alot easier on yourself)

here you are given x^3/( 3 + 5x) = x^3 * ( 1/(3 + 5x) )

Here we observe that the only component that truely requires a taylor expansion is (1/(3+ 5x)), whose derivatives are far easier to compute than that of the original function.

Once this is obtained you simply multiple the series through by x^3 and viola you have answered the problem with only a 10th of the pain!!

Hope this points you in the right direction,

Let me know if you require any further assistance,

David

ps - make life even easier on yourself by simplifying the function even further, i.e.
1/(3 + 5x) = 1/( 3 (1 + ((5/3)x)) = (1/3)(1/(1 + (5/3)x)) i.e bring the 1/3 out the front
furthermore let t = 5/3x and thus you you need only solve for 1/(1+t) = a0 + a1*t + ....

To then get in back in terms of x, sub t = 5/3x into the the taylor expansion and your done!

3. ## Hi

I tried performing some hand calculations, and got the following result.
$f(x) = \frac{x^3}{3+5x} = x^3 \cdot \frac{1}{3+5x}$

Now I only performing a maclaurin series on the $\frac{1}{3+5x}$
term.
Let $h(x) \, \mbox{be the maclaurin series of }\frac{1}{3+5x}$

Then $f(x) \approx x^3 \cdot h(x)$

If I dont bother with the restterm, then we get something like (see picture).

And the two functions seem to be pretty close around $x=0$

You can see the maclaurin series in the picture...

4. Originally Posted by Twig
I tried performing some hand calculations, and got the following result.
$f(x) = \frac{x^3}{3+5x} = x^3 \cdot \frac{1}{3+5x}$

Now I only performing a maclaurin series on the $\frac{1}{3+5x}$
term.
Let $h(x) \, \mbox{be the maclaurin series of }\frac{1}{3+5x}$

Then $f(x) \approx x^3 \cdot h(x)$

If I dont bother with the restterm, then we get something like (see picture).

And the two functions seem to be pretty close around $x=0$

You can see the maclaurin series in the picture...
Yes, I prefer Twiggy's method.

$\forall{x}\backepsilon|x|<1~\frac{1}{1-x}=\sum_{n=0}^{\infty}x^n$

And

\begin{aligned}\forall{x}\backepsilon|5x|<1~\frac{ 1}{1+5x}&=\frac{1}{1-(-5x)}\\
&=\sum_{n=0}^{\infty}\left(-5x\right)^n\\
&=\sum_{n=0}^{\infty}(-1)^n5^nx^n
\end{aligned}

\begin{aligned}\therefore~\forall{x}\backepsilon|5 x|<1~\frac{x^3}{1+5x}&=x^3\cdot\frac{1}{1+5x}\\
&=x^3\cdot\sum_{n=0}^{\infty}(-1)^n5^nx^n\\
&=\sum_{n=0}^{\infty}(-1)^n5^nx^{n+3}
\end{aligned}

So now truncate to the number of terms you need for your series.