2
$\begingroup$

We have function $f:\mathbb{R}\rightarrow \mathbb{R}$ with $$f\left(x\right)=\frac{1}{3x+1}\:$$ $$x\in \left(-\frac{1}{3},\infty \right)$$ Write the Maclaurin series for this function.

Alright so from what I learned in class, the Maclaurin series is basically the Taylor series for when we have $x_o=0$ and we write the remainder in the Lagrange form. It has this shape: $f\left(x\right)=\left(T_n;of\right)\left(x\right)+\left(R^Ln;of\right)\left(x\right)=\sum _{k=0}^n\left(\frac{f^{\left(k\right)}\left(0\right)}{k!}x^n+\frac{f^{\left(n+1\right)}\left(c\right)}{\left(n+1\right)!}x^{n+1}\right)$

So when I compute derivatives of my function I can see that the form they take is:$$f^{\left(n\right)}\left(x\right)=\left(-1\right)^n\cdot \frac{3^n\cdot n!}{n!}x^n$$

Does that mean that the Maclaurin series is basically: $$f\left(x\right)=1-3x+3^2x^2-3^3x^3+....+\left(-1\right)^n\cdot 3^n\cdot x^n$$ ?

But what about that remainder in Lagrange form? I don't get that part. We didn't really have examples in class, so I've no idea if what I'm doing is correct. Can someone help me with this a bit?

$\endgroup$
1
$\begingroup$

Hint: Remember that

$$\frac{1}{1 + x} = \sum_{n=0}^{\infty} (-1)^n x^n \,\,\,\, \text{for} \,\,\,\ |x| <1$$

Then $$\frac{1}{1 +3x} = \ldots$$

for $|3x| < 1 \implies |x| < \frac{1}{3}$

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Wouldn't that just be $\frac{1}{1+3x}=\sum _{n=0}^{\infty }\left(-1\right)^n\cdot \:3^nx^n$ ? (What I wrote in my post) $\endgroup$ – MikhaelM Feb 4 '16 at 14:50
  • $\begingroup$ That's right, you took the long way though. Compositions of functions works faster. $\endgroup$ – Aaron Maroja Feb 4 '16 at 14:52
  • $\begingroup$ So is this all I need to do for this exercise? Is that my final MacLaurin series? $\endgroup$ – MikhaelM Feb 4 '16 at 14:53
  • $\begingroup$ Yup, pretty much. $\endgroup$ – Aaron Maroja Feb 4 '16 at 14:53
  • $\begingroup$ Oh, thank you! I thought it was harder than this. $\endgroup$ – MikhaelM Feb 4 '16 at 14:54
0
$\begingroup$

HINT: Use the fact that $\displaystyle \sum_\limits{n=0}^{\infty} x^n=\frac{1}{1-x}$ for $|x|<1$.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.