binomial coefficients series

**buckeye1973** · February 13th 2010, 02:46 PM

Knowing that:
$(1+x)^k = \sum_{n=0}^{\infty}\binom{k}{n}x^n = 1 + kx + \frac{k(k-1)}{2!}x^2 + \frac{k(k-1)(k-2)}{3!}x^3 + ...$

Give the Maclaurin series for the function:
$f(x) = \frac{x^2}{\sqrt{2+x}}$

I figure I can rewrite the equation as $f(x) = x^2(1+(1+x))^{-1/2}$ so that my series becomes:
$x^2(1+(x+1))^{-\frac1{2}} = x^2\sum_{n=0}^{\infty}\binom{(-\frac1{2})}{n}(x+1)^n =$ $x^2 [1 + (-\frac1{2})(x+1) + \frac{(-\frac1{2})((-\frac1{2})-1)}{2!}(x+1)^2 +$ $\frac{(-\frac1{2})((-\frac1{2})-1)((-\frac1{2})-2)}{3!}(x+1)^3 + ... ]$

which is

$= x^2 [1 -\frac{x+1}{2} + \frac{3(x+1)^2}{{2^2}2!} - \frac{15(x+1)^2}{{2^3}3!} +$
$... + \frac{(-1)^n(1\cdot3\cdot5\cdot\cdot\cdot(2n-1))(x+1)^n}{{2^n}n!}]$

I thought this might be okay until I checked my answer on WolframAlpha and got a different answer:
(x^2)*(2+x)^(-1/2) - Wolfram|Alpha
$\sum_{n=2}^{\infty}x^n 2^{3/2-n} \binom{-1/2}{-2+n}$
$\frac{x^2}{\sqrt{2}}-\frac{x^3}{4\sqrt{2}} + \frac{3x^4}{32\sqrt{2}}-\frac{5x^5}{128\sqrt{2}}+ ...$

So are both answers somehow correct? Or did I do something wrong?

Thanks,
Brian

**Prove It** · February 13th 2010, 06:40 PM

Do you have to use the Binomial Series?

I'd find it easier just to find the Taylor series for $\frac{1}{\sqrt{2 + x}}$ and then multiply everything by $x^2$ .

**vince** · February 13th 2010, 07:22 PM

where $\Gamma$ is the usual gamma function. this allows for complex or real arguments.

i think u need to use this because binomial coeffs are usually defined for z>k>0 where z,k are integers. that $z=\frac{-1}{2}$ is gnarly. ill keep thinking on this one.

**vince** · February 13th 2010, 08:00 PM

well, as "luck" would have it:
Binomial coefficient - Wikipedia, the free encyclopedia
The definition of the binomial coefficients can be extended to the case where n is real and k is integer.
In particular, the following identity holds for any non-negative integer k :

This shows up when expanding

into a power series using the Newton binomial serie :

cleary now this is very doable...just express it like this,
$\frac{x^2}{(1+(x+1))^\frac{1}{2}}$
use the result above and youre done.
kinda sleazy, but what can u do. in no mood to attempt deriving the above.

**buckeye1973** · February 13th 2010, 08:16 PM

Originally Posted by Prove It

Do you have to use the Binomial Series?

I'd find it easier just to find the Taylor series for $\frac{1}{\sqrt{2 + x}}$ and then multiply everything by $x^2$ .

Well, the problem says "Use a Maclaurin series from Table 1 to obtain the Maclaurin series for the given function."

Table 1 has the series for:
$\frac1{1-x} = \sum_{n=0}^{\infty} x^n$
$e^x = \sum_{n=0}^{\infty} \frac{x^n}{n!}$
$sin x = \sum (blah)$
$cos x = \sum (blah)$
$tan^{-1}x = \sum (blah)$
$(1+x)^k = \sum_{n=0}^{\infty} \binom{k}{n}x^n = 1 + kx + \frac{k(k-1)x^2}{2!} + ...$

And I figured that the given function best fit the binomial series. I got that the x^2 would be factored out first, and then multiplied back in, and that's what I did, using k = -1/2 and x+1 in place of x.

I'm new to series in general, so I'm just not sure if I'm making correct use of this binomial series. The idea of "just finding the Taylor series" has no meaning to me yet.

**buckeye1973** · February 13th 2010, 08:22 PM

Originally Posted by vince

In particular, the following identity holds for any non-negative integer k
<snip>
cleary now this is very doable...just express it like this,
$\frac{x^2}{(1+(x+1))^\frac{1}{2}}$

That is how I expressed it, but then k=-1/2, so it is neither non-negative nor an integer, right? (Are we mixing Ks and Ns here? I see a k in the bottom of your binomial expression, and no Ns at all? I don't get it.)

Oh, and I have absolutely no idea what the "usual gamma function" is. This being a section where we are just being introduced to Maclaurin, I don't think it applies to this problem.

**awkward** · February 14th 2010, 05:04 AM

Originally Posted by buckeye1973

Knowing that:
$(1+x)^k = \sum_{n=0}^{\infty}\binom{k}{n}x^n = 1 + kx + \frac{k(k-1)}{2!}x^2 + \frac{k(k-1)(k-2)}{3!}x^3 + ...$

Give the Maclaurin series for the function:
$f(x) = \frac{x^2}{\sqrt{2+x}}$

I figure I can rewrite the equation as $f(x) = x^2(1+(1+x))^{-1/2}$ so that my series becomes:
$x^2(1+(x+1))^{-\frac1{2}} = x^2\sum_{n=0}^{\infty}\binom{(-\frac1{2})}{n}(x+1)^n =$ $x^2 [1 + (-\frac1{2})(x+1) + \frac{(-\frac1{2})((-\frac1{2})-1)}{2!}(x+1)^2 +$ $\frac{(-\frac1{2})((-\frac1{2})-1)((-\frac1{2})-2)}{3!}(x+1)^3 + ... ]$

which is

$= x^2 [1 -\frac{x+1}{2} + \frac{3(x+1)^2}{{2^2}2!} - \frac{15(x+1)^2}{{2^3}3!} +$
$... + \frac{(-1)^n(1\cdot3\cdot5\cdot\cdot\cdot(2n-1))(x+1)^n}{{2^n}n!}]$

I thought this might be okay until I checked my answer on WolframAlpha and got a different answer:
(x^2)*(2+x)^(-1/2) - Wolfram|Alpha
$\sum_{n=2}^{\infty}x^n 2^{3/2-n} \binom{-1/2}{-2+n}$
$\frac{x^2}{\sqrt{2}}-\frac{x^3}{4\sqrt{2}} + \frac{3x^4}{32\sqrt{2}}-\frac{5x^5}{128\sqrt{2}}+ ...$

So are both answers somehow correct? Or did I do something wrong?

Thanks,
Brian

Brian,

Your answer, although it may be correct (I haven't checked the details), is not a Maclaurin series, because it is in terms of powers of 1+x. A Maclaurin series is the Taylor series about 0, i.e. it is in terms of powers of x. I suggest you start with

$\frac{1}{\sqrt{2+x}} = 2^{-1/2} (1+ \frac{x}{2})^{-1/2}$ ,

then apply the Binomial Theorem.

**vince** · February 14th 2010, 07:39 AM

The path i suggested would lead to a messy formula that would equal the one provided by wolfram. Still, with Awkward's wise choice of factorization the following steps will lead to wolfram's answer.

$\frac{x^2}{\sqrt{2+x}} = {x^2}2^{-1/2} (1+ \frac{x}{2})^{-1/2}$

Using the binomial theorem, this is then equal to:

$\sum_{k=0}^{\infty}2^\frac{-1}{2}x^2\binom{(-\frac1{2})}{k}(x/2)^k$

$\sum_{k=0}^{\infty}2^\frac{-1}{2}\binom{(-\frac1{2})}{k}(\frac{x^\frac{2}{k}x}{2})^k$

let k=n-2

$\Rightarrow$
$\sum_{n=2}^{\infty}2^\frac{-1}{2}\binom{(-\frac1{2})}{n-2}(\frac{x^\frac{2}{n-2}x}{2})^{n-2}$

=

$\sum_{n=2}^{\infty}2^{\frac{3}{2}-n}\binom{(-\frac1{2})}{n-2}(x^{\frac{2+n-2}{n-2}})^{n-2}$

=

$\sum_{n=2}^{\infty}2^{\frac{3}{2}-n}\binom{(-\frac1{2})}{n-2}(x^{\frac{n}{n-2}})^{n-2}$

=

$\sum_{n=2}^{\infty}x^n 2^{3/2-n}\binom{(-\frac1{2})}{n-2}$

**buckeye1973** · February 14th 2010, 08:02 AM

Originally Posted by awkward

Brian,

Your answer, although it may be correct (I haven't checked the details), is not a Maclaurin series, because it is in terms of powers of 1+x. A Maclaurin series is the Taylor series about 0, i.e. it is in terms of powers of x. I suggest you start with

$\frac{1}{\sqrt{2+x}} = 2^{-1/2} (1+ \frac{x}{2})^{-1/2}$ ,

then apply the Binomial Theorem.

Ah-ha! It's amazing how I can get the big concepts, but it's missing the basic algebra that so often trips me up.

It seems like there may be something wrong with using an offset of x like I did, while multiples and powers of x are okay, as they still center around zero. Not just in terms of technically being called a Maclaurin, but actually not being mathematically correct. I don't know.

Thanks,
Brian

**buckeye1973** · February 14th 2010, 08:14 AM

Originally Posted by vince

The path i suggested would lead to a messy formula that would equal the one provided by wolfram. Still, with Awkward's wise choice of factorization the following steps will lead to wolfram's answer.

$\frac{x^2}{\sqrt{2+x}} = {x^2}2^{-1/2} (1+ \frac{x}{2})^{-1/2}$

Using the binomial theorem, this is then equal to:

$\sum_{k=0}^{\infty}2^\frac{-1}{2}x^2\binom{(-\frac1{2})}{k}(x/2)^k$

$\sum_{k=0}^{\infty}2^\frac{-1}{2}\binom{(-\frac1{2})}{k}(\frac{x^\frac{2}{k}x}{2})^k$

let k=n-2

$\Rightarrow$
$\sum_{n=2}^{\infty}2^\frac{-1}{2}\binom{(-\frac1{2})}{n-2}(\frac{x^\frac{2}{n-2}x}{2})^{n-2}$

<snip>

I haven't learned binomial theorem well enough to understand the way you are using it. I don't understand this having the k in the bottom, only the top, as shown in my original post:

$(1+x)^k = \sum_{n=0}^{\infty}\binom{k}{n}x^n = 1 + kx + \frac{k(k-1)}{2!}x^2 + \frac{k(k-1)(k-2)}{3!}x^3 + ...$

Where k is simply -1/2 and just gets plugged in to the product sequences on the right side above, and x/2 is used in place of x. (And this is about all the explanation I'm given in the text as well, so I don't think I'm expected to understand anything any fancier yet.)

I'd multiply the terms of the resulting series by $\frac{x^2}{\sqrt{2}}$ to get my original function $f(x) = \frac{x^2}{\sqrt{2+x}}$

I'll have to try that, and see if it gets me to the same answer. If it doesn't, I'll do some more research on binomial theorem and look closer at your solution.

Thanks,
Brian

**vince** · February 14th 2010, 08:28 AM

Please note and remember that the letters/symbols ascribed to terms in a function, theorem, whatever, are never absolute, they are only relative. by this i mean any pair of letters $k,n$ have no menaing unless they are related to each other -- otherwise, any choice of letter/symbols will work, such as $x$ and $y$ , etc.

Now look at this:
$(1+ \frac{x}{2})^{-1/2}$

By comparing this relative to the statement of the binomial theorem, is not the $k$ u see in the statement meant to mean ${-1/2}$ ? is not the $x$ equal to $\frac{x}{2}$ ?(Where the $x$ and the $\frac{x}{2}$ here are not the same, they're just referring to the same argument.)

so all in all, you really should not ever get attached to symbols or letters used to relate variables in a mathematical expression. they can be switched, twirled, and discarded if it suits the progression of the operations to make things simpler.

**Renji Rodrigo** · February 14th 2010, 08:55 AM

from the identity
$\frac{1}{\sqrt{1+x}}= \sum^{\infty}_{k=0}{2k \choose k}\frac{(-1)^{k}x^{k}}{4^{k}} $ , we change
$x $ by
$\frac{x}{2} $ , so

$\frac{\sqrt{2}}{\sqrt{2+x}}= \sum^{\infty}_{k=0}{2k \choose k}\frac{(-1)^{k}x^{k}}{8^{k}} $ so

$\frac{x^{2}}{\sqrt{2+x}}= \frac{1}{\sqrt{2}}\sum^{\infty}_{k=0}{2k \choose k}\frac{(-1)^{k}x^{k+2}}{8^{k}}. $

**buckeye1973** · February 14th 2010, 09:04 AM

Originally Posted by vince

Please note and remember that the letters/symbols ascribed to terms in a function, theorem, whatever, are never absolute, they are only relative. by this i mean any pair of letters $k,n$ have no menaing unless they are related to each other -- otherwise, any choice of letter/symbols will work, such as $x$ and $y$ , etc.

Now look at this:
$(1+ \frac{x}{2})^{-1/2}$

By comparing this relative to the statement of the binomial theorem, is not the $k$ u see in the statement meant to mean ${-1/2}$ ? is not the $x$ equal to $\frac{x}{2}$ ?(Where the $x$ and the $\frac{x}{2}$ here are not the same, they're just referring to the same argument.)

so all in all, you really should not ever get attached to symbols or letters used to relate variables in a mathematical expression. they can be switched, twirled, and discarded if it suits the progression of the operations to make things simpler.

Well, yes, I understand this, but you have things like "n-2" in the lower half of your binomial expression, and I do not know how to evaluate that. It has no meaning to me, and it is made more confusing when I use the variable k in one role and you use it in a different role. Like I said, I'm using k=-1/2 and x/2 in place of x, and plugging those into the sequence
$ (1+x)^k = \sum_{n=0}^{\infty}\binom{k}{n}x^n = 1 + kx + \frac{k(k-1)}{2!}x^2 + \frac{k(k-1)(k-2)}{3!}x^3 + ... $

It turns out this ultimately did lead me to
$ \frac{x^2}{\sqrt{2}}-\frac{x^3}{4\sqrt{2}} + \frac{3x^4}{32\sqrt{2}}-\frac{5x^5}{128\sqrt{2}}+ ... $

but my final expression looks different, although I do think it is correct: [EDIT: No, it isn't, see below..]
$ \frac{x^2}{\sqrt{2}} + \sum_{n=1}^{\infty}\frac{(-1)^n x^{n+2}}{n 2^{2n+\frac{1}{2}}} $

I get this because factoring out the $(-1)^n$ allows me to cancel out $\frac{(n-1)!}{n!}$ to get $\frac1{n}$ , thus eliminating the need for the binomial coefficient expression.

Brian

**vince** · February 14th 2010, 09:22 AM

i hear ya man. but what is unclear about the function,
$\binom{(-\frac1{2})}{n-2}$ is the $-\frac1{2}$ not the $n-2$ , at least to me. that is why i suggested the gamma function earlier, and then happened to run into the following formula

which would have done the trick.

The problem with your most recent answer, as Awkward pointed out, is that "A Maclaurin series is the Taylor series about 0, i.e. it is in terms of powers of x."

Your answer includes higher order powers of x, and by that i mean it needs to be cast as powers of x and not the sum of $x^2$ and the sum of powers of $x^{2+n}$ .

Best regards

**Renji Rodrigo** · February 14th 2010, 09:33 AM

Using the definition

${x\choose k}=\frac {\prod\limits^{k-1}_{s=0}(x-s)}{k!}$
and the identity
$\prod^{k-1}_{s=0}(1+2s)=\frac{(2k)!}{2^{k}k!}$
we have

$(1+x)^{-\frac{1}{2}}=\sum^{\infty}_{k=0}{-\frac{1}{2}\choose k}x^{k}=\sum^{\infty}_{k=0}\prod^{k-1}_{s=0}(-\frac{1}{2}-s)\frac{x^{k}}{k!} = $

$=\sum^{\infty}_{k=0}\frac{(-1)^{k}}{2^{k}}\prod^{k-1}_{s=0}(1+2s)\frac{x^{k}}{k!}=\sum^{\infty}_{k=0} \frac{(-1)^{k}}{2^{k}}\frac{(2k)!}{2^{k}k!}\frac{x^{k}}{k! }= \sum^{\infty}_{k=0}\frac{(-1)^{k}(2k)!x^{k}}{(k!^{2})4^{k}} $

Thread: binomial coefficients series

LinkBack

Thread Tools

Display

binomial coefficients series

Similar Math Help Forum Discussions

Series of Binomial Coefficients

Series : binomial coefficients

Proof of a series with binomial coefficients

binomial coefficients

Binomial coefficients

Search Tags