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# Factorials of Decimals

Can Anyone please explain the process of calculation of factorials of Decimal numbers? Any help would be sincerely appreciated. Thank You.

Note by Swapnil Das
2 years, 9 months ago

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Besides nonnegative integers, the factorial function can also be defined for non-integer values, but this requires more advanced tools from mathematical analysis. One function that "fills in" the values of the factorial (but with a shift of 1 in the argument) is called the Gamma function, denoted $$Γ(z)$$, defined for all complex numbers z except the non-positive integers, and given when the real part of z is positive by $$\Gamma(z)=\int_0^\infty t^{z-1} e^{-t}\, \mathrm{d}t. \!$$

Its relation to the factorials is that for any natural number n $$n!=\Gamma(n+1).\,$$

Euler's original formula for the Gamma function was $$\Gamma(z)=\lim_{n\to\infty}\frac{n^zn!}{\displaystyle\prod_{k=0}^n (z+k)}. \!$$

An alternative notation, originally introduced by Gauss, is sometimes used. The Pi function, denoted Π(z) for real numbers z no less than 0, is defined by $$\Pi(z)=\int_0^\infty t^{z} e^{-t}\, \mathrm{d}t\,.$$

In terms of the Gamma function it is $$\Pi(z) = \Gamma(z+1) \,.$$

It truly extends the factorial in that $$\Pi(n) = n!\text{ for }n \in \mathbf{N}\, .$$

In addition to this, the Pi function satisfies the same recurrence as factorials do, but at every complex value z where it is defined $$\Pi(z) = z\Pi(z-1)\,.$$

In fact, this is no longer a recurrence relation but a functional equation. Expressed in terms of the Gamma function this functional equation takes the form $$\Gamma(n+1)=n\Gamma(n)\,.$$

Since the factorial is extended by the Pi function, for every complex value z where it is defined, we can write: $$z! = \Pi(z)\,$$

The values of these functions at half-integer values is therefore determined by a single one of them; one has $$\Gamma\left (\frac{1}{2}\right )=\left (-\frac{1}{2}\right )!=\Pi\left (-\frac{1}{2}\right ) = \sqrt{\pi},$$ from which it follows that for n ∈ N,)

$$\Gamma\left (\frac{1}{2}+n\right ) = \left (-\frac{1}{2}+n\right )! = \Pi\left (-\frac{1}{2}+n\right ) = \sqrt{\pi} \prod_{k=1}^n {2k - 1 \over 2} = {(2n)! \over 4^n n!} \sqrt{\pi} = {(2n-1)! \over 2^{2n-1}(n-1)!} \sqrt{\pi}.$$

For example, $$\Gamma\left (4.5 \right ) = 3.5! = \Pi\left (3.5\right ) = {1\over 2}\cdot{3\over 2}\cdot{5\over 2}\cdot{7\over 2} \sqrt{\pi} = {8! \over 4^4 4!} \sqrt{\pi} = {7! \over 2^7 3!} \sqrt{\pi} = {105 \over 16} \sqrt{\pi} \approx 11.63.$$

It also follows that for n ∈ N, $$Gamma\left (\frac{1}{2}-n\right ) = \left (-\frac{1}{2}-n\right )! = \Pi\left (-\frac{1}{2}-n\right ) = \sqrt{\pi} \prod_{k=1}^n {2 \over 1 - 2k} = {(-4)^n n! \over (2n)!} \sqrt{\pi}.$$

For example, $$\Gamma\left (-2.5 \right ) = (-3.5)! = \Pi\left (-3.5\right ) = {2\over -1}\cdot{2\over -3}\cdot{2\over -5} \sqrt{\pi} = {(-4)^3 3! \over 6!} \sqrt{\pi} = -{8 \over 15} \sqrt{\pi} \approx -0.9453.$$

The Pi function is certainly not the only way to extend factorials to a function defined at almost all complex values, and not even the only one that is analytic wherever it is defined.

Nonetheless it is usually considered the most natural way to extend the values of the factorials to a complex function. For instance, the Bohr–Mollerup theorem states that the Gamma function is the only function that takes the value 1 at 1, satisfies the functional equation$$Γ(n + 1) = nΓ(n)$$, is meromorphic on the complex numbers, and is log-convex on the positive real axis.

A similar statement holds for the Pi function as well, using the $$Π(n) = nΠ(n − 1)$$ functional equation.

However, there exist complex functions that are probably simpler in the sense of analytic function theory and which interpolate the factorial values.

For example, Hadamard's 'Gamma'-function (Hadamard 1894) which, unlike the Gamma function, is an entire function.

Euler also developed a convergent product approximation for the non-integer factorials, which can be seen to be equivalent to the formula for the Gamma function above:

\begin{align}n! = \Pi(n) &= \prod_{k = 1}^\infty \left(\frac{k+1}{k}\right)^n\!\!\frac{k}{n+k} \\ &= \left[ \left(\frac{2}{1}\right)^n\frac{1}{n+1}\right]\left[ \left(\frac{3}{2}\right)^n\frac{2}{n+2}\right]\left[ \left(\frac{4}{3}\right)^n\frac{3}{n+3}\right]\cdots. \end{align}

However, this formula does not provide a practical means of computing the Pi or Gamma function, as its rate of convergence is slow.

- 2 years, 9 months ago

Several Brilliant members wrote up a great article on the Gamma Function.

Staff - 2 years, 9 months ago

Thank You Sir for providing the link. I read up the whole article and found it quiet interesting. But the weakness of mine is that I don't know trigonometry and calculus as a ninth grader. What should I do now?

- 2 years, 9 months ago

That's the great thing about Brilliant! Once you find that you're interested in a particular area, you can go and learn more about it. For example, look up the Trigonometry Wiki.

Staff - 2 years, 9 months ago

Thank You Sir! Now I have decided to learn trigonometry and advanced mathematics. Hope that I would also make a place in the world among great people like you. ( My aim is to become a Theoretical Physicist and a Master Challenger in Mathematics.)

- 2 years, 9 months ago

All the best! Look to the Brilliant community for assistance if necessary :)

Staff - 2 years, 9 months ago