Maclaurin series for functions which are not infinitely differentiable

Question

I was introduced to Maclaurin series through $\sin(x)$, $\cos(x)$ and $e^x$. I have always thought that Maclaurin series works for these functions because they are infinitely differentiable.

My question is; Does this also work for functions which aren't infinitely differentiable? Like for example $$\dfrac{1}{7} x^4 - 12x^2 + 1$$

My intuition tells me that with maclaurin series you can find a function which somewhat resembles these types of functions, but not completely/ well.

What makes you think that function is not infinitely differentiable? — Robert Israel, Jun 03 '13 at 14:59
Try to find the MacLaurin series for this function. I think you will be surprised. — Kris Williams, Jun 03 '13 at 15:01
@RobertIsrael I meant it becomes 0. I am not too familair with mathematical lingo — DeriveMe, Jun 03 '13 at 15:02
@KrisWilliams How would I be able to do that sir? (apart from tedious pen and paper stuff, sorry for being lazy..) — DeriveMe, Jun 03 '13 at 15:03
This is a very nice function. You may find its Maclaurin series a little boring. But it will tell you a fair bit about Maclaurin series. Compute. — André Nicolas, Jun 03 '13 at 15:04
You will need to do the "tedious pen and paper stuff." Mathematics is an adventure. Sometimes the trip is the boring part: think of great explorers sailing for months across the ocean then having a great surprise when they arrive at the destination. Other times, the trip itself is interesting. This is the former. — Kris Williams, Jun 03 '13 at 15:06
A function whose $n$'th derivative is $0$ everywhere is a polynomial of degree $< n$. — Robert Israel, Jun 03 '13 at 15:06

score 1 · Answer 1 · edited Dec 15 '21 at 20:14

1

I'm sorry to answer my own question. But I was so wrong. I found out you can do it using wolfram alpha (chronic laziness, sorry for that):

http://www.wolframalpha.com/input/?i=maclaurin+expansion+of+17x^4%E2%88%9212x^2%2B1

Spot on...

edited Dec 15 '21 at 20:14

Wolgwang

1,563

answered Jun 03 '13 at 15:06

DeriveMe

91

Btw, this answer is just to show I know the answer now. If anyone has a really good answer to this, be my guest. – DeriveMe Jun 03 '13 at 15:07

score 1 · Answer 2 · answered Jun 03 '13 at 15:23

Maclaurin's formula is simply a tool to help you find a power series representation of a function centered at zero, ie: $$f(x) = a_0 + a_1 x + a_2 x^2 + \cdots + a_n x^n + \cdots,$$ where $a_n = f^{(n)}(0)/n!$. Your function is already a power series, though, it's just that only three terms are non-zero:

$$1-12x^2 +\frac{1}{7}x^4 = 1 + 0x - 12 x^2 + 0x^3 + \frac{1}{7}{x^4} + 0x^5 + 0x^6 +\cdots$$

All the remaining terms are zero, so there's not much to do. As you even observed yourself, though, $f^{(n)}(x)=0$ for large enough $n$, so Maclaurin's formula yields the same result.

score 0 · Answer 3 · answered Nov 07 '17 at 15:14

0

The Taylor/Maclaurin series of a polynomial is simply the same polynomial :)

answered Nov 07 '17 at 15:14

Mohamed Ouda

407

Maclaurin series for functions which are not infinitely differentiable

3 Answers3