Taylor Series/Mathematica

[DivGradCurl]

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.

[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?

[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.

[DivGradCurl]

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.

[Hurkyl]

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

[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.

[DivGradCurl]

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

[Hurkyl]

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)