How do I find the Taylor Series of ##\frac{q}{\sqrt{1+x}}## around x = 0?

[TheFerruccio]

This is rather embarrassing, because I should have known how to do this for years.

Question:

Compute the Taylor Series of $\frac{q}{\sqrt{1+x}}$ about x = 0.

Attempt at Solution:

Term-wise, I have gotten...

$f(0)+f'(0)+f''(0)+... = 1+1\left(-\frac{1}{2}\right)x+1\left(-\frac{1}{2}\right)\left(-\frac{3}{2}\right)x^2+1\left(-\frac{1}{2}\right)\left(-\frac{3}{2}\right)\left(-\frac{5}{2}\right)x^3+...$

I have gotten this to reduce to...

$\sum\limits_{k=0}^{\infty}x^k\left(-\frac{1}{2}\right)^k\frac{(2k-1)!}{2^k}$

There is definitely a better way to do this. I am not thinking clearly. Additionally, I am not that confident in my answer, given the time of night.

 

[Simon Bridge]

At first glance - shouldn't there be a q in there?

If your termwise relation is supposed to be the series, you seem to have missed some bits.

Apart from that - it all seems pretty standard.

 

[chingel]

This is the binomial series with a=-0.5

http://en.wikipedia.org/wiki/Binomial_series

 

[HallsofIvy]

Writing q/\sqrt{1+ x} as q(1+ x)^{-1/2} we have:
The value at 0 is q. The first derivative is -(1/2)q(1+ x)^{-3/2} so
its value at 0 is -(1/2)q. The second derivtive is (3/4)q(1+ x)^{-5/2} so
its value at 0 is (3/4)q. The third derivative is -(15/8)q(1+ x)^{-7/2} so
its value at 0 is -(15/8)q. The fourth derivative is (105/16)q(1+ x)^{-9/2} so
its value at 0 is (105/16)q...

That should be enough to see the pattern: it alternates sign with positive sign when n, the order of the derivative, is even so that is (-1)^n. The denominator of the fraction is a power of 2: 2^n. There is, obviously, a factor of "q". The numerator of the fraction is the only "tricky" part. It is 1(3)(5)(7)..., a sort of "factorial" except that the even integers are missing. We can write that as 1(3)(5)(7)= 1(2)(3)(4)(5)(6)(7)/[2(4)(6)]= 7!/([2(1)][(2)(2)][(2)(3)])= 7!/[2^3(3!)] (when n= 4). So that is (2n- 1)!/[2^{n-1}(n-1)!].

Putting those together, the nth derivative, at x= 0, is
(-1)^n\frac{(2n-1)!}{2^{n-1}(2^n)(n- 1)!}q= (-1)^n\frac{(2n-1)!}{2^{2n-1}(n-1)!}q
and the coefficient of x^n in the Taylor series expansion about x= 0 is that divided by n!.

 

[mathman]

This is rather embarrassing, because I should have known how to do this for years.

Question:

Compute the Taylor Series of $\frac{q}{\sqrt{1+x}}$ about x = 0.

Attempt at Solution:

Term-wise, I have gotten...

$f(0)+f'(0)+f''(0)+... = 1+1\left(-\frac{1}{2}\right)x+1\left(-\frac{1}{2}\right)\left(-\frac{3}{2}\right)x^2+1\left(-\frac{1}{2}\right)\left(-\frac{3}{2}\right)\left(-\frac{5}{2}\right)x^3+...$

I have gotten this to reduce to...

$\sum\limits_{k=0}^{\infty}x^k\left(-\frac{1}{2}\right)^k\frac{(2k-1)!}{2^k}$

There is definitely a better way to do this. I am not thinking clearly. Additionally, I am not that confident in my answer, given the time of night.

You can get the same result, without calculating the derivatives, by getting the binomial expansion for (1+x)^-1/2.

 

[TheFerruccio]

Sorry guys. I meant to type 1 instead of q. My keyboard broke and the function key messed up. I did not catch that error.

 

[Simon Bridge]

You can get the same result, without calculating the derivatives, by getting the binomial expansion for (1+x)^-1/2.

You have to know already that the binomial series is the Taylor series at x = 0 of the function f given by f(x) = (1 + x) α, where α ∈ ℂ is an arbitrary complex number.

I suppose that answers the original question.

 

[mathman]

You have to know already that the binomial series is the Taylor series at x = 0 of the function f given by f(x) = (1 + x) α, where α ∈ ℂ is an arbitrary complex number.

I suppose that answers the original question.

The binomial theorem series and the Taylor series (for this case) are both power series around 0. Therefore they must be identical for the same function.

 

[Office_Shredder]

If there was a simple explanation for why the binomial series worked for an arbitrary power then it would be a good answer, but as is it's basically just a way of telling someone what the answer is without explanation which isn't great.

 

[mathman]

http://people.sju.edu/~pklingsb/genbinom.pdf

Above has proof.