Do you have to use the Binomial Series?
I'd find it easier just to find the Taylor series for and then multiply everything by .
Give the Maclaurin series for the function:
I figure I can rewrite the equation as so that my series becomes:
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
So are both answers somehow correct? Or did I do something wrong?
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,
use the result above and youre done.
kinda sleazy, but what can u do. in no mood to attempt deriving the above.
Table 1 has the series for:
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.
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.
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
then apply the Binomial Theorem.
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.
Using the binomial theorem, this is then equal to:
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.
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 to get my original function
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.
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 have no menaing unless they are related to each other -- otherwise, any choice of letter/symbols will work, such as and , etc.
Now look at this:
By comparing this relative to the statement of the binomial theorem, is not the u see in the statement meant to mean ? is not the equal to ?(Where the and the 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.
It turns out this ultimately did lead me to
but my final expression looks different, although I do think it is correct: [EDIT: No, it isn't, see below..]
I get this because factoring out the allows me to cancel out to get , thus eliminating the need for the binomial coefficient expression.
i hear ya man. but what is unclear about the function,
is the not the , 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 and the sum of powers of .