# Taylor Series/Mathematica

1. Nov 11, 2004

In order to obtain with the aid of Mathematica, say, an 8-th degree Taylor polynomial of $$\sqrt{x}$$ centered at $$4$$, I use the following command:

Normal[Series[Sqrt[x], {x, 4, 8}]]

and I get

$$\sqrt{x} \approx 2 + \frac{1}{2^2} \left( x - 4 \right) - \frac{1}{2^6} \left( x - 4 \right) ^2 + \frac{1}{2^9} \left( x - 4 \right) ^3 - \frac{5}{2^{14}} \left( x - 4 \right) ^4 + \frac{7}{2^{17}} \left( x - 4 \right) ^5 - \frac{3\cdot 7}{2^{21}} \left( x - 4 \right) ^6 + \frac{3\cdot 11}{2^{24}} \left( x - 4 \right) ^7 - \frac{3\cdot 11 \cdot 13}{2^{30}} \left( x - 4 \right) ^8$$

This is ok, but what I really need is to write the series in sigma notation. Unfortunately, its pattern is not obvious, although I tried to find it by factoring the coefficients and also browsed the help of Mathematica. Anyway, does anyone know a command that gives it?

Thank you.

Last edited: Nov 11, 2004
2. Nov 11, 2004

### GrapesOfWrath

are you trying to find a Mathematica command to do that in general, or do you just need that particular function? Couldn't you type in the general definition of the nth Taylor function term?

3. Nov 11, 2004

### krab

Why don't you just use Taylor's theorem? The coefficients will have a product of odd numbers in the numerator and a factorial in the denominator.
Mathematica is fine for doing math, but not as helpful for learning math. Do some of the work yourself.

4. Nov 11, 2004

I see what you mean:

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

and yes, I know how to take the derivatives and get to the same place that Mathematica got me. The problem is finding the pattern for the n-th term so that I can write the series in sigma notation. I tried to obtain it by factoring the coefficients, though. Then, the only alternative I have is to ask about a command that would give me this. The pattern is not obvious, and I spent quite some time working on it myself. Thanks, anyway.

5. Nov 11, 2004

### Hurkyl

Staff Emeritus
Doing it by hand, though, suggests a more intelligent way to organize the product than simply looking at the resulting number and factoring it.

6. Nov 11, 2004

### krab

So all you need now is the general form of the nth derivative of square root. Just start writing them out. You'll see the pattern right away.

7. Nov 11, 2004

I can now see... this is what I've got:

$$\sqrt{x} = 2 + \frac{1}{4} \left( x-4 \right) + \sum _{n=2} ^{\infty} \left( -1 \right)^{n+1} \frac{1\cdot 3 \cdot 5 \cdots \cdot \left( 2n-3 \right)}{n!2^{3n-1}} \left( x-4 \right)^n$$

thanks

8. Nov 12, 2004

### Hurkyl

Staff Emeritus
And as a final trick, notice that:

1 * 3 * 5 * 7 = 1 * 2 * ... * 8 / (2 * 4 * 6 * 8)
= 1 * 2 * ... * 8 / (1 * 2 * 3 * 4 * 2^4)