# Solving the Gamma Function: Using Recursion & Tables

• Erbil
In summary: Gamma(a/b)For s>0 this can be derived by the change of variable u=s t^bIn summary, The homework equation for Gamma is Gamma(n+1)=n!+(1/P)+Gamma(P+1). I found it from a table, and then used recursion and table to find it. I found the same equation using recursion again. The attempt at a solution is to find Gamma(0.7) which
Erbil

## Homework Statement

Questions are in picture.

## Homework Equations

$$\int _{0}^{\infty }x^{n}e^{-x}dx$$ = $$Gamma (n+1) = n!$$ Gamma(P+1) $$=$$Gamma(P) Gamma(P) = (1/P) $$Gamma(P+1)$$

## The Attempt at a Solution

2) I have found it from table.
3) I have used recursion and table to find it.
4) Again With recursion.
5) $$\Gamma(0.7)$$ = 1/p(p+1) with this formula.
8) If $$\ Gamma (p+1)$$ is equal to this integral,I think it can be written as $$\ Gamma (2/3+1)$$ later we can found the value from table.Am I right?
Same logic again for 9,10?
But what next? How can I convert them to Gamma function?

#### Attachments

• gama function.jpg
23.3 KB · Views: 717
11) I found 1/2 but I'm not sure(?)

Edit : No it's not true.I think true solution is on down.

Last edited:
$$\int _{0}^{\infty }x^{2}e^{-x^{2}}dx=2\int _{0}^{\infty }u^{2}e^{-u}du=2\cdot \Gamma 3$$

Erbil said:
$$\int _{0}^{\infty }x^{2}e^{-x^{2}}dx=2\int _{0}^{\infty }u^{2}e^{-u}du=2\cdot \Gamma 3$$
No. How did you get that?
What did you get for (5)?
I don't follow your logic for (8). Where did 2/3 come from? I suspect what you posted is not what you meant.

5. 1/0.7(Gamma1.7) = 1.30 ? Isn't it?
I'm sorry questions mess up.It's solution of question 11.
Here is for (8)

For n= 2/3 Gamma(n+1) = Gamma(1+2/3)=Gamma(1.6) = 0.89

Erbil said:
5. 1/0.7(Gamma1.7) = 1.30 ? Isn't it?
Yes.
I'm sorry questions mess up.It's solution of question 11.
But you haven't done the substitution correctly. What will dx become?
Here is for (8)
For n= 2/3 Gamma(n+1) = Gamma(1+2/3)=Gamma(1.6) = 0.89
OK, but 1.6 is a bit inaccurate. The answer should be > 0.9.

Yes.There's a problem.But I can't figure. if we say u=x^2 --> 2xdx=du --> dx = du/2x Where to go from here? I can't remember how can I escape from x variable?

For the ones of the type

$$\int_0^\infty t^a b \, e^{-t^b} \, \dfrac{\mathrm{dt}}{t}=\Gamma(a/b)$$

this can be derived by the change of variable u=t^b

Erbil said:
Yes.There's a problem.But I can't figure. if we say u=x^2 --> 2xdx=du --> dx = du/2x Where to go from here? I can't remember how can I escape from x variable?
Just replace the x using u=x2

haruspex said:
Just replace the x using u=x2

I understand but what about dx ? Would be? x^2=u 2xdx = du dx = du/2x (I'm asking about this x?)

$$\int _{0}^{\infty }ue^{-u}?$$

dx = du/(2x)
or
dx = du/(2sqrt(u))

lurflurf said:
dx = du/(2x)
or
dx = du/(2sqrt(u))

Thanks!

Can I use same logic for 12?

x^3 = u
3x^2dx= du
dx = du / 3*√u

∫xe^-x^3 = ∫u^1/3*e^-u du/3*√u= 1/3 ∫ u^-1/6 e^-udu = 1/3 * Gamma -5/6 ...?+ No idea for 15?

Last edited:
Erbil said:
Can I use same logic for 12?

x^3 = u
3x^2dx= du
dx = du / 3*√u
Not √u, a little more complicated than that. What is x as a function of u?
+ No idea for 15?
In line with the other substitutions, u=8x looks obvious. Did you try that?

haruspex said:
Not √u, a little more complicated than that. What is x as a function of u?

In line with the other substitutions, u=8x looks obvious. Did you try that?

Oh what I was did.it will be u^1/3 =?
Not yet.But I will.

^Yes above I said

$$\int_0^\infty t^a b \, e^{-t^b} \, \dfrac{\mathrm{dt}}{t}=\Gamma(a/b)$$

this can be derived by the change of variable u=t^b

also

$$\int_0^\infty (st)^a b \, e^{-(s t)^b} \, \dfrac{\mathrm{dt}}{t}=\Gamma(a/b)$$

s>0 this can be derived by the change of variable u=s t

## 1. What is the gamma function and why is it important in mathematics?

The gamma function is a mathematical function that is defined as an extension of the factorial function for non-integer values. It is denoted by the Greek letter gamma (Γ) and is used to calculate the area under the curve of certain mathematical functions. It has applications in various fields of mathematics, including number theory, statistics, and physics.

## 2. What is the recursive approach to solving the gamma function?

The recursive approach to solving the gamma function involves breaking down the calculation of the gamma function into smaller, simpler steps. This involves using the identity Γ(z+1) = zΓ(z) and repeatedly applying it to reduce the value of z until it reaches a known value. This approach is useful for calculating the gamma function for non-integer values.

## 3. How can tables be used to solve the gamma function?

Tables can be used to solve the gamma function by pre-calculating and storing the values of the gamma function for a range of values. This saves time and effort in calculating the gamma function for each individual value. The table can then be used to quickly look up the value of the gamma function for any given input.

## 4. What are the benefits of using recursion and tables to solve the gamma function?

The use of recursion and tables to solve the gamma function offers several benefits. It allows for a more efficient calculation of the gamma function, especially for non-integer values. It also reduces the chances of human error in calculations. Additionally, having pre-calculated values stored in a table allows for faster look-up times and can save time in solving the gamma function for multiple inputs.

## 5. Are there any limitations to using recursion and tables for solving the gamma function?

While using recursion and tables can be beneficial in solving the gamma function, there are also limitations to these methods. The recursive approach may not be suitable for very large values of z, as it can lead to an infinite loop or exceed the memory capacity of a computer. Additionally, the use of tables may not be feasible for values that are not included in the pre-calculated range, requiring an alternative method for calculation.

• Calculus and Beyond Homework Help
Replies
3
Views
2K
• Calculus and Beyond Homework Help
Replies
1
Views
957
• Topology and Analysis
Replies
2
Views
385
• Calculus and Beyond Homework Help
Replies
1
Views
2K
• Calculus and Beyond Homework Help
Replies
3
Views
1K
• Calculus and Beyond Homework Help
Replies
5
Views
1K
• Calculus and Beyond Homework Help
Replies
4
Views
1K
• Calculus and Beyond Homework Help
Replies
5
Views
1K
• Calculus and Beyond Homework Help
Replies
8
Views
1K
• Calculus and Beyond Homework Help
Replies
1
Views
1K