Solving the Gamma Function: Using Recursion & Tables

Erbil
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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?
 

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11) I found 1/2 but I'm not sure(?)

Edit : No it's not true.I think true solution is on down.
 
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$$ \int _{0}^{\infty }x^{2}e^{-x^{2}}dx=2\int _{0}^{\infty }u^{2}e^{-u}du=2\cdot \Gamma 3 $$
 
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
 
  • #10
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?)

$$ \int _{0}^{\infty }ue^{-u}? $$
 
  • #11
dx = du/(2x)
or
dx = du/(2sqrt(u))
 
  • #12
lurflurf said:
dx = du/(2x)
or
dx = du/(2sqrt(u))

Thanks!
 
  • #13
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?
 
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  • #14
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?
 
  • #15
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.
 
  • #16
^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
 
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