Binomial Expansion Question - fractional powers

[Asphyxiated]

Homework Statement

My question is simple is there a formula for the bi/tri-nomial expansion of bi/tri-nomials raised to fractional powers. that is,

(x^{2}+1)^{1/2} or (x^{2}+x+1)^{1/2}

I know pascals triangle for integer exponents but i can't really find anything about fraction exponents. I also know that 1/2 or 1/3 are square or cube roots but in that form i don't see anyway of expanding them.

thanks!

 

[rock.freak667]

If you have a fractional or negative power, you have an infinite number of terms. Normally you'd expand it the usual way. But you work out ⁿC₁ and ⁿC₂ to get results such as:

ⁿC₁ =n

ⁿC₂= n(n-1)/2!


and then you'd just substitute for n.

 

[Asphyxiated]

I don't really understand what you are saying, can you use those rules to show me on one of the examples that I listed?

 

[Asphyxiated]

Can someone please explain this to me?

 

[rock.freak667]

I don't really understand what you are saying, can you use those rules to show me on one of the examples that I listed?

Can someone please explain this to me?

The expansion of (a+b)ⁿ is as follows:

(a+b)^n= b^n +\left( ^n _1 \right)b^{n-1}a +\left( ^n _2 \right)b^{n-2}a^2+...

If you insert in your calculator something like ^1/2C₁, it won't give you a number. But you can simplify ⁿC₁ from the definition of ⁿC_r

\left( ^n _r \right) = \frac{n!}{(n-r)!r!}

\left( ^n _1 \right) = \frac{n!}{(n-1)!1!}

n!=n(n-1)(n-2)...3.2.1 = n(n-1)!

You can similarly simplify ⁿC₂, in the same manner.


(a+b)^n = b^n + nb^{n-1}a + ...
Sp

 

[Asphyxiated]

so does

ⁿC₁ read n choose 1?

I really have no experience with n choose k equations, that's why I am having a hard time understanding you.

 

[Mark44]

Yes, ⁿC₁ is read as "n choose 1." It also means the number of combinations of n things taken 1 at a time.

It is also written as
\left( \begin{array}{c} n \\ 1 \end{array} \right)

Take a look at this Wike article, especially the section on Newton's generalized binomial theorem - http://en.wikipedia.org/wiki/Binomial_theorem

 

[rock.freak667]

Well yes they are essentially the same idea. Except you are not really 'choosing' in a binomial expansion.

But you do understand the definitions of n! and ⁿC_r as I've typed above right?

 

[Asphyxiated]

Ill take a shot at understanding it though, so in:

<br /> (a+b)^n= b^n +\left( ^n _1 \right)b^{n-1}a +\left( ^n _2 \right)b^{n-2}a^2+...<br />

is:

b^{n-1}a

then multiplied by:

<br /> \left( ^n _1 \right) = \frac{n!}{(n-1)!1!}<br />

is that how it goes? I got the pattern, just those n choose r phrases are a mystery to me at the moment

or rather

\left (^n _k \right)

become the binomial coefficients, is that right?

 

[Mark44]

Right. The snag here is that for "n choose k" expressions, both n and k are normally integer values, with n >= r. For your problem, n is going to be 1/2.

 

[Asphyxiated]

ok, so i guess the only other question I have is how would you do factorials with fractions,

what would

\frac{1}{2}! be?

I know integer factorials are easy,

4! = 1*2*3*4...

 

[Asphyxiated]

On my calculator I found a factorial button, at least I think that's what it is, if i put in:

(1/2)! it gives 0.8862269255, does that sound right? It doesn't work out to a fraction but if i solve the equation:

\left(^n _k \right) = \frac{n!}{(n-k)!k!}

for

\left(^n _1 \right)

with

n! = n(n-1)!

you get:

\frac{\frac{1}{2}(-\frac{1}{2})!}{(-\frac{1}{2})!(1)!} = \frac {1}{2}

does it look like i have a hold on this here?

 

[VeeEight]

ok, so i guess the only other question I have is how would you do factorials with fractions

See here: http://en.wikipedia.org/wiki/Gamma_function

 

[Asphyxiated]

ok then... I really don't know if this is what I am suppose to use but would this be appropriate?

\Gamma (n + \frac {1}{2}) = \frac {(2n-1)!}{2^{n}} \sqrt{ \pi } = \left(^{n-\frac{1}{2}} _{\;\;n} \right) n! \sqrt{ \pi }

If that's the right equation would I use that were the normal binomial theorem calls for a factorial of a fraction?

This is much more complicated that I had originally hoped.

Also how do I know when to stop? With the normal theorem using whole integers there should be n+1 terms for a binomial raised to the n powers, but when n = 1/2 n+1 = 3/2 or 1+ 1/2 terms, which does make sense.

 

[Mark44]

As rock.freak667 said in post 2, with fractional or negative exponents, you get an infinite number of terms, unlike what happens when n is a positive integer.

 

[Asphyxiated]

ah so this is hopeless? No way to get a finite number of terms? Basically its an infinite sum?

 

[rock.freak667]

ah so this is hopeless? No way to get a finite number of terms? Basically its an infinite sum?

The only way to get a finite number of terms is if you neglect certain powers and higher. So if you need to approximate a square root for example, to a certain degree of accuracy, higher terms will become negligible.