# Homework Help: Binomial Expansion Question - fractional powers

1. May 3, 2010

### Asphyxiated

1. The problem statement, all variables and given/known data

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!

2. May 3, 2010

### 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.

3. May 3, 2010

### 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?

4. May 3, 2010

### Asphyxiated

Can someone please explain this to me?

5. May 3, 2010

### rock.freak667

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 nCr

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

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

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

6. May 3, 2010

### Asphyxiated

so does

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

7. May 3, 2010

### Staff: Mentor

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

8. May 3, 2010

### 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 nCr as I've typed above right?

9. May 3, 2010

### Asphyxiated

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

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

is:

$$b^{n-1}a$$

then multiplied by:

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

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?

Last edited: May 3, 2010
10. May 3, 2010

### Staff: Mentor

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.

11. May 3, 2010

### 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,

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

12. May 3, 2010

### 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?

13. May 3, 2010

### VeeEight

14. May 3, 2010

### 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.

15. May 3, 2010

### Staff: Mentor

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.

16. May 3, 2010

### Asphyxiated

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

17. May 3, 2010

### rock.freak667

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.