- Definitions of Euler's Gamma
- General Formula and Pi Function
- Gamma of One
- Gamma and Factorials
- Gamma One Half (Factorial Minus One Half)
- Graph of some factorials

We can define the binomial coefficients as follows:

[1.01] [from Binomial Theorem 6.2]

We read the left-hand-side as r over k when it is a binomial coeffient, and r choose k, when it is a combination.

k is normally a positive integer or zero. If k=0, then the binomial coefficient B(r,0)=1. the upper index, r can be positive, negative (or a complex number).

When B(r,k) is a combination, we can write:

[1.02]

On
the left is a graph of the factorials. I have plotted the positive
integers up to 5. We can now wonder whether the graph is a continuous one, including fractions or whether it should be a discrete graph, perhaps drawn with bars for each t value. We suspect that there are fractional factorials from the Binomial Theorem. However, we can wonder about the function, which we will call gamma which relates the t values on the x-axis with the factorials (on the y axis). We could now seek some function that might fit the curve. Doing this is quite difficult, but Leonhard Euler in the late 1720's discovered a function. Even without such a function, we can learn some things from gamma. Let us define it like this: [0.01] Where t is a positive real number. Because t!=t·(t−1)!, and Γ(t)=(t-1)! by our definition [0.01], then [0.02] We now have a recursive formula for gamma. Consider: [0.03], from the definition [0.01] Where n is a positive integer or zero. The reason we have introduced this n is, that as we will see later, to make gamma big enough to obtain a sufficiently accurate approximation. By continually applying [0.01], we obtain: [0.04] |

Making Γ(t) the subject, we have:

[0.05]

And because Γ(t+n+1)=(t+n)! (By our definition 0.04), we have a formula for gamma in terms of a factorial and some factors:

[0.06]

Of course, none of the factors can be equal to zero (otherwise we would divide by zero). This formula enables us to compute certain gamma values. The next one enables to compute certain factorials.

By noting that t·Γ(t)=Γ(t+1)=t!, and by multiplying [0.06] throughout by t, we obtain:

[0.07]

So if we knew what (t+n)! was, we could compute any gamma value from [0.06], except when any of the factors are equal to zero, as they would be if we tried to compute the gamma values (or factorials) of the negative integers. Similarly, if we knew (t+n)! we could compute any factorial (except the negative integers).

We can compute (t+n)! provided it is greater than 1 using Stirling's Approximation.

With a means of calculating factorials (such as (2.5)!), as with using Stirling's Approximations, we can compute all the gamma values excluding those of the negative integers.

However, we can gain greater understanding (and compute some gamma values exactly) using Euler's Gamma Function.

All of the above equations will be derived again, using Euler's Gamma.

In the 1720's Euler discovered a formula relating gamma to the factorials. One way to express this is:

[2.01]

Where t is a constant, such that t≥ (0). While the gamma function is defined for all real (and complex) numbers, except negative integers, it converges only when the power of x is greater than or equal to zero (Stated here without proof).

Or equivalently, setting t=t−1

[2.02]

Where t is a constant, such that t ≥1

We can integrate [2.02] by parts.

Let v=-e^{-x} so, dv=e^{-x} dx,

And let u=x^{t-1}, so du=(t−1)·x^{t-2}
dx

When we substitute these u and v values in ∫v·du=u·v−∫u·dv, we have, :

[2.03]

If t=1, then the uv part is 1, and the integral on the right is zero. If t≠1, then the 'uv' part above is zero, so [2.03] becomes

[2.04]

Note that the new integral on the right is:

[2.05]

That is, substituting t=t−1 in 2.02 above, we obtain the right-hand side of [2.05], which is Γ(t−1). Hence:

, t>1 [2.06] ■

If t=1, then, as mentioned above, Γ(1)=1, not zero. See: Gamma of One, below.

Had we integrated [2.01], we would have obtained:

By making Γ(t) the subject of [2.07], we obtain:

[2.08]

Which can be used to calculate some negative values of Γ(t).

An equivalent form of the gamma function, is the Pi function:

[2.09]

From [2.07], using x as any real number (but does include complex numbers when the real part is positive) we have for a positive integer n (1, 2, 3...), when n+x>0 :

[2.10]

And from [2.08], using x as a real number, and n+x>0, as before, and none of the factors is zero:

[2.11]

This formula can be used to calculate gamma for negative numbers, except the negative integers, because, for instance, if we have x=−1, then (x+1)=0, and we cannot use the formula because we would be dividing by zero. Similarly, the other negative integers cannot be calculated.

If we substitute t=1 in [2.02], we obtain

[3.01]

This is a simple standard integral, and gives us:

[3.02]

Which, evaluated, gives:

[3.03] ■

We know that Γ(1)=1, which we showed above.

Γ(2)=1·Γ(1)=1, from [2.06]

Similarly, for 3:

Γ(3)=2·Γ(2)=2·1

And for 4:

Γ(4)=3·Γ(3)=3·2·1

We might suppose, therefore, that

Γ(t)=(t−1)! [4.01]

Let us suppose, then (and we proceed by induction), that if [4.01] is true, then so is:

Γ(t+1)=t! [4.01]

Now if [4.01] is true then so is:

t·Γ(t)=t·(t−1)! [4.03], because this is 4.01 multiplied throughout by t

We know that t·Γ(t)=Γ(t+1), from [2.07].

And we know that t·(t−1)!=t!

So, if [4.01] is true, then so is [4.02].

Now, when t=1, then Γ(1)=1=0!, so Γ(2)=1! is also true

And so on. Therefore:

[4.01] ■

Using equation [2.02], repeated below:

[2.02, repeated]

And substituting t=1/2, we have:

[5.01]

Which is, with minor simplification:

[5.02]

If we let x=u^{2} then dx=2·u·du.
Substituting these:

[5.03]

After simplifying, we have the standard integral.

[5.04]

This is sometimes called the Gaussian Integral. And its integral from 0 to ∞ is √(π)/2

Hence, gamma(1/2) is:

■ [5.05]

From:

[4.01, repeated]

We note Γ(1/2)=(-1/2)!=√(π), so factorial of minus one half is the square root of pi!

Also, Γ(3/2)=(1/2)!=(1/2)·Γ(1/2)=√(π)/2

Ken Ward's Mathematics Pages

Ken's book is packed with examples and explanations that enable you to discover more than 150 techniques to speed up your arithmetic and increase your understanding of numbers. Paperback and Kindle: