ਗਾਮਾ ਫੰਕਸ਼ਨ: ਰੀਵਿਜ਼ਨਾਂ ਵਿਚ ਫ਼ਰਕ

ਸਮੱਗਰੀ ਮਿਟਾਈ ਸਮੱਗਰੀ ਜੋੜੀ
"Gamma function" ਸਫ਼ੇ ਨੂੰ ਅਨੁਵਾਦ ਕਰਕੇ ਬਣਾਇਆ ਗਿਆ
"Gamma function" ਸਫ਼ੇ ਨੂੰ ਅਨੁਵਾਦ ਕਰਕੇ ਬਣਾਇਆ ਗਿਆ
ਲਾਈਨ 21:
 
A plot of the first few factorials makes clear that such a curve can be drawn, but it would be preferable to have a formula that precisely describes the curve, in which the number of operations does not depend on the size of {{ਹਿਸਾਬ|''x''}}. The simple formula for the factorial, {{ਹਿਸਾਬ|''x''! {{=}} 1 × 2 × … × ''x''}}, cannot be used directly for fractional values of {{ਹਿਸਾਬ|''x''}} since it is only valid when {{ਹਿਸਾਬ|''x''}} is a [[ਕੁਦਰਤੀ ਅੰਕ|natural number]] (''i.e.'', a positive integer). There are, relatively speaking, no such simple solutions for factorials; no finite combination of sums, products, powers, exponential functions, or [[ਲਘੂਗਣਕ|logarithms]] will suffice to express {{ਹਿਸਾਬ|''x''!}}; but it is possible to find a general formula for factorials using tools such as integrals and limits from [[ਕੈਲਕੂਲਸ|calculus]]. A good solution to this is the gamma function.
 
== ਪਰਿਭਾਸ਼ਾ ==
 
=== ਮੁੱਖ ਪਰਿਭਾਸ਼ਾ ===
The notation {{ਹਿਸਾਬ|Γ(''z'')}} is due to Legendre. If the real part of the complex number {{ਹਿਸਾਬ|''z''}} is positive ({{ਹਿਸਾਬ|Re(''z'') > 0}}), then the integral
 
: <math> \Gamma(z) = \int_0^\infty x^{z-1} e^{-x}\, dx</math>
 
converges absolutely, and is known as the '''Euler integral of the second kind''' (the Euler integral of the first kind defines the beta function). Using integration by parts, one sees that:
 
: <math>
\begin{align}
\Gamma(z+1) & = \int_0^\infty x^{z} e^{-x} \, dx \\
& = \left[-x^z e^{-x}\right]_0^\infty + \int_0^\infty z x^{z-1} e^{-x}\, dx \\
& = \lim_{x\to \infty}(-x^z e^{-x}) - (0 e^{-0}) + z\int_0^\infty x^{z-1} e^{-x}\, dx
\end{align}
</math>
: Recognizing that as <math>x\to \infty, -x^z e^{-x}\to 0,</math>
: <math>
\begin{align}
\Gamma(z+1) & = z\int_0^\infty x^{z-1} e^{-x}\, dx \\
& = z\Gamma(z)
\end{align}
</math>
 
: <math>
\begin{align}
\Gamma(1) & = \int_0^\infty x^{1-1} e^{-x}\,dx \\
& = \left[-e^{-x}\right]_0^\infty \\
& = \lim_{x\to \infty}(-e^{-x}) - (-e^{-0}) \\
& = 0 - (-1) \\
& = 1
\end{align}
</math>
 
Given that <math>\Gamma(1) = 1</math> and <math>\Gamma(n+1) = n\Gamma(n) </math>
 
: <math>\Gamma(n) = 1 \cdot 2 \cdot 3 \cdots (n-1) = (n-1)!\,</math>
 
for all positive integers <math>n</math>. This can be seen as an example of proof by induction.
 
== Notes ==