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Where is the fallacy?

Please help me to find out the fallacy.

Note by Trishit Chandra
2 years, 10 months ago

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See Trishit, a special form of the gamma function (the one you have mentioned in your problem) is actually an improper integral of the following form: \[\Gamma (t)=\int _{ 0 }^{ \infty }{ { x }^{ t-1 }{ e }^{ -x }dt } \] the domain of this function being the set of all complex numbers t, with Re(t)>=0. The gamma function reduces to the simple factorial function defined by \[n!=\prod _{ i=1 }^{ n }{ i } ,when\quad n\ge 1\] and \[n!=1,when\quad n=0\] which has its domain the set of non-negative integers. Factorial function turns out to be a very special form of the gamma function in the set of non-negative integers. Clearly, the domain of the factorial function does not contain fractions like 2.5 and thus (2.5)! is not defined. Your fallacy is that you are considering the factorial and gamma functions to be the same. Well, you cannot call two functions the same if their domains are different.

Kuldeep Guha Mazumder - 2 years, 5 months ago

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Got it. And thank you very much.

Trishit Chandra - 2 years, 5 months ago

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Welcome. You can learn more about functions like gamma and beta in the topic of improper integrals There you will learn why \[\Gamma \left( \frac { 1 }{ 2 } \right) =\sqrt { \pi } \]. This result can be proved from the Euler's reflection formula. I give a short sketch of the proof here. The reflection formula states that(this can also be proved):\[\Gamma (t)\Gamma (1-t)=\frac { \pi }{ sin\quad \pi t } \\ \Longrightarrow \Gamma \left( \frac { 1 }{ 2 } \right) \Gamma \left( 1-\frac { 1 }{ 2 } \right) =\Gamma \left( \frac { 1 }{ 2 } \right) \Gamma \left( \frac { 1 }{ 2 } \right) ={ \left\{ \Gamma \left( \frac { 1 }{ 2 } \right) \right\} }^{ 2 }=\frac { \pi }{ sin\quad \frac { \pi }{ 2 } } =\pi \]\[\Longrightarrow \Gamma \left( \frac { 1 }{ 2 } \right) =\sqrt { \pi } \]

Kuldeep Guha Mazumder - 2 years, 4 months ago

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@Kuldeep Guha Mazumder Thanks for providing the insight!

Could you look over this Gamma Function wiki page and provide us with feedback about areas that it can be improved?

Calvin Lin Staff - 2 years, 4 months ago

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Definitions can be extended to include more cases than what you initially dealt with.

A typical example that you should be familiar with, is that in (say) 2nd grade you were told that "multiplication is just repeated addition. To find \( 4 \times 5 \), you draw out 4 groups of 5 items, and count them to get 20". However, that "logic" quickly fails when you try and extend it to frations / irrational numbers. Pray tell, how do you draw out \( \sqrt{2} \) groups of \( \pi \) items and count them?

The next example would be exponentiation, which is (similarly) often introduced as "exponentiation is just repeated multiplication" To find \( 4 ^ 5 \), we do \( 4 \times 4 \times 4 \times 4 \times 4 \). Pray tell, how would you write \( \sqrt{2} ^ \pi \)?

Calvin Lin Staff - 2 years, 10 months ago

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So you are saying that the factorial of fraction should be possible but this can not be possible according to the definition of it. And hence the derivation is correct.

Trishit Chandra - 2 years, 10 months ago

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We can take the factorial of fractions, and or irrational numbers. The definition of it is not \( x ! = x \times (x-1) \times (x-2) \times \ldots \) where the last term is between 0 and 1.

We can also extend this to negative numbers, but not to the negative integers.

The above derivation is correct, but it does not explain how it calculated \( \Gamma ( \frac{1}{2} ) \). With that as the working assumption (ie not proven), the rest of the steps look correct.

Calvin Lin Staff - 2 years, 10 months ago

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@Calvin Lin Okay. Thank you Calvin Lin for helping me. Now I've understood. Can you refer me some link or book where I can find this conception of factorial because Wikipedia is saying that factorial means the product of the positive integers less the or equal to the number. And you said in the last line bracket that something is not proven, what is that?

Trishit Chandra - 2 years, 10 months ago

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So, is the process of this derivation wrong ?

Trishit Chandra - 2 years, 10 months ago

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There's another definition for the gamma function. Check it out here: Gamma Function

Marc Vince Casimiro - 2 years, 10 months ago

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I saw that there is another definition of gamma function but there is not another definition of factorial which implies the above derivation. And as the formulas I have written in the first portion are correct then the derivation should be correct but there is a contradiction between the definition of factorial and the formulas.

Trishit Chandra - 2 years, 10 months ago

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