Taylor Polynomials

**isp_of_doom** · February 8th 2010, 09:53 PM

hi all,
I fear this will be long, but i'll try to keep it short.
the main question I am having trouble with is:

Evalutate f(0), f'(0), f''(0), f'''(0) for the function
$f(x) = (1+x)^{1/3}$
and thus find the Taylor Polynomials of degree 3 about the center c = 0 for this function.

I've been having difficulty with the Taylor Polynomials - here is what I have:

$f(x) = (1+x)^{1/3}$

$f'(x) = \frac{1}{3}(1+x)^{-2/3}$

$f''(x) = \frac{-2}{9}(1+x)^{-5/3}$

$f'''(x) = \frac{10}{27}(1+x)^{-8/3}$

$f''''(x) = \frac{-80}{81}(1+x)^{-11/3}$

Therefore in the Taylor Polynomial we get:

$(1+0)^{1/3}+\frac{1}{3}(1+0)^{-2/3}(x-0)+\frac{-2}{9}(1+0)^{-5/3} \frac{(x-0)^2}{2!}+$ $\frac{10}{27}(1+0)^{-8/3}\frac{(x-0)^3}{3!}+\frac{-80}{81}(1+0)^{-11/3}\frac{(x-0)^4}{4!}$

which gives:
$1 + \frac{1}{3}x + (\frac{-2}{9})(\frac{x^2}{2!}) + (\frac{10}{27})(\frac{x^3}{3!}) + (\frac{-80}{81})(\frac{x^4}{4!})$

from here I start trying to find the $\sum$ notation to express this. I can get as far as:

$1 + \sum \frac{(-1)^{n+1}...}{3^n} * \frac{x^n}{n!}$

As you can probably tell from the '...' that is the part I am having difficulty with. Hopefully my conundrum makes sense.

**Prove It** · February 8th 2010, 10:49 PM

Originally Posted by isp_of_doom

hi all,
I fear this will be long, but i'll try to keep it short.
the main question I am having trouble with is:

Evalutate f(0), f'(0), f''(0), f'''(0) for the function
$f(x) = (1+x)^{1/3}$
and thus find the Taylor Polynomials of degree 3 about the center c = 0 for this function.

I've been having difficulty with the Taylor Polynomials - here is what I have:

$f(x) = (1+x)^{1/3}$

$f'(x) = \frac{1}{3}(1+x)^{-2/3}$

$f''(x) = \frac{-2}{9}(1+x)^{-5/3}$

$f'''(x) = \frac{10}{27}(1+x)^{-8/3}$

$f''''(x) = \frac{-80}{81}(1+x)^{-11/3}$

Therefore in the Taylor Polynomial we get:

$(1+0)^{1/3}+\frac{1}{3}(1+0)^{-2/3}(x-0)+\frac{-2}{9}(1+0)^{-5/3} \frac{(x-0)^2}{2!}+$ $\frac{10}{27}(1+0)^{-8/3}\frac{(x-0)^3}{3!}+\frac{-80}{81}(1+0)^{-11/3}\frac{(x-0)^4}{4!}$

which gives:
$1 + \frac{1}{3}x + (\frac{-2}{9})(\frac{x^2}{2!}) + (\frac{10}{27})(\frac{x^3}{3!}) + (\frac{-80}{81})(\frac{x^4}{4!})$

from here I start trying to find the $\sum$ notation to express this. I can get as far as:

$1 + \sum \frac{(-1)^{n+1}...}{3^n} * \frac{x^n}{n!}$

As you can probably tell from the '...' that is the part I am having difficulty with. Hopefully my conundrum makes sense.

I use a different approach when trying to find Taylor Polynomials.

Assume that $(1 + x)^{\frac{1}{3}}$ can be expressed as a polynomial.

Then $(1 + x)^{\frac{1}{3}} = a + bx + cx^2 + dx^3 + \dots$ .

Let $x = 0$ and we find that $a = 1$ .

Take the derivative of both sides...

$\frac{1}{3}(1 + x)^{-\frac{2}{3}} = b + 2cx + 3dx^2 + 4ex^3 + \dots$ .

Let $x = 0$ and we find that $b = \frac{1}{3}$ .

Take the derivative of both sides...

$-\frac{2}{9}(1 + x)^{-\frac{5}{3}} = 2c + 3\cdot 2 dx + 4\cdot 3 ex^2 + 5 \cdot 4 fx^3 + \dots$ .

Let $x = 0$ and we find that $2c = -\frac{2}{9}$ .

Therefore $c = -\frac{2}{18}$ .

Take the derivative of both sides...

$\frac{10}{27}(1 + x)^{-\frac{8}{3}} = 3!d + 4!ex + 5 \cdot 4 \cdot 3fx^2 + 6\cdot 5 \cdot 4 gx^3 + \dots$

Let $x = 0$ and we find that $3!d = \frac{10}{27}$ .

Therefore $d = \frac{10}{27\cdot 3!}$ .

Take the derivative of both sides...

$-\frac{80}{81} = 4!e + 5!fx + \dots$ .

Let $x = 0$ and we find that $4!e = -\frac{80}{81}$

Therefore $e = -\frac{80}{81\cdot 4!}$ .

If we were to continue this way, we would see that

$(1 + x)^{\frac{1}{3}} = 1 + \frac{1}{3}x - \frac{2}{18}x^2 + \frac{10}{27 \cdot 3!}x^3 - \frac{80}{81\cdot 4!}x^4 + \dots$

$= 1 + \frac{1}{3^1 \cdot 1!} - \frac{2}{3^2 \cdot 2!}x^2 + \frac{10}{3^3 \cdot 3!}x^3 - \frac{80}{3^4 \cdot 4!}x^4 + \dots$

I agree with what you have ended up with. I am also having trouble writing this sum using sigma notation...

**lgstarn** · February 8th 2010, 11:41 PM

Originally Posted by isp_of_doom

hi all,
Therefore in the Taylor Polynomial we get:

$(1+0)^{1/3}+\frac{1}{3}(1+0)^{-2/3}(x-0)+\frac{-2}{9}(1+0)^{-5/3} \frac{(x-0)^2}{2!}+$ $\frac{10}{27}(1+0)^{-8/3}\frac{(x-0)^3}{3!}+\frac{-80}{81}(1+0)^{-11/3}\frac{(x-0)^4}{4!}$

which gives:
$1 + \frac{1}{3}x + (\frac{-2}{9})(\frac{x^2}{2!}) + (\frac{10}{27})(\frac{x^3}{3!}) + (\frac{-80}{81})(\frac{x^4}{4!})$

from here I start trying to find the $\sum$ notation to express this. I can get as far as:

$1 + \sum \frac{(-1)^{n+1}...}{3^n} * \frac{x^n}{n!}$

As you can probably tell from the '...' that is the part I am having difficulty with. Hopefully my conundrum makes sense.

Hi,

The problem is how to represent the sequence

$a_0 = 1/3 = 1/3$
$a_1 = -2/9 = (1/3)(-2/3)$
$a_2 = 10/27 = (1/3)(-2/3)(-5/3)$
$a_3 = -80/81 = (1/3)(-2/3)(-5/3)(-8/3)$
$a_n = (1/3)(-2/3)(-5/3)(-8/3)\ldots(-(3n-1)/3)$

The easiest way to do this is using the $\prod$ notation, called "product notation" or "pi notation." This is very similar to the $\sum$ notation (called sum or sigma notation), but instead of a sum, it is a product, i.e. you multiply all the terms together.

Using this notation, we can write $a_0 = 1/3$ and $a_n = \frac{(-1)^n}{3^{n+1}} \prod \limits_{i = 1}^{n} (3i-1)$ for $n > 0$ .

So the fancy way to write your Taylor polynomial is

$1 + \frac{x}{3} + \sum\limits_{n = 1}^{\infty} \frac{(-1)^n}{(n+1)!}(\frac{x}{3})^{n+1} \prod \limits_{i = 1}^{n} (3i-1)$

But to be honest, this is a little bit too much work for you to get all the indexes lined up (hopefully I didn't make a mistake) and for your readers to decode the dense symbols. Readability is often more important than getting things really compact. So it would probably be better for everyone to just write it as you had it originally:

$1 + \frac{1}{3}x - (\frac{2}{9})(\frac{x^2}{2!}) + (\frac{10}{27})(\frac{x^3}{3!}) - (\frac{80}{81})(\frac{x^4}{4!}) + \ldots$

As you can see, the second way takes up about the same amount of space as the first, but the second is actually a lot easier to understand, don't you think? Which would you rather see on a Calculus exam?

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