# Thread: I just solved this. Can you? (MX2 students should have a go)

1. ## I just solved this. Can you? (MX2 students should have a go)

And I'm feeling quite happy about it because... it was long and arduous. It's not that hard in hindsight, I figure most HSC Extension 2 Maths students would be able to do it - I think it'd a pretty good test of their integration mastery.

I don't think there's anything abstractly difficult in it - it's just knowing what to do and being smart about it all.

So here it is:

$\textup{Show that}\;E(X)=\alpha\beta\textup{, where:}$

$E(X)=\lim_{R \to \infty}\int_{0}^{R} x\frac{e^{-x/\beta}x^{\alpha - 1}}{\Gamma (\alpha) \beta^{\alpha}} dx$

You may also need the results:

$\Gamma(\phi) = (\phi - 1)!, \;\;\phi = 0, 1, 2, 3,...$
$\lim_{k \to \infty} k^{n}e^{-k/m} = 0, \textup{where k, m and n are real numbers}$

Have fun!

EDIT: Also part of why I like this particular integral: "It looks difficult".

2. ## Re: I just solved this. Can you? (MX2 students should have a go)

Haha fun question!

I like how it reduces (reduction formulae) and everything just cancels out...

Indeed, this is do-able by a good 4U student who won't get intimidated by all the 'strange' notation and symbols.

3. ## Re: I just solved this. Can you? (MX2 students should have a go)

yeh this is pretty decent, cambridge has a funner one of these type of cancellation reduction formula ones, a lot more challenging

4. ## Re: I just solved this. Can you? (MX2 students should have a go)

If the Gamma function is used in the question, surely its definition as an integral can be assumed? From there its one substitution and a couple of lines...

5. ## Re: I just solved this. Can you? (MX2 students should have a go)

Originally Posted by seanieg89
If the Gamma function is used in the question, surely its definition as an integral can be assumed? From there its one substitution and a couple of lines...
First Class Honours (Pure Mathematics) at USYD

6. ## Re: I just solved this. Can you? (MX2 students should have a go)

Surprised an enthusiastic 2012er hasnt attempted this yet

7. ## Re: I just solved this. Can you? (MX2 students should have a go)

Originally Posted by seanieg89
If the Gamma function is used in the question, surely its definition as an integral can be assumed? From there its one substitution and a couple of lines...
i did that originally, but then the question is too trivial, its a much funner exercise than a couple of lines doing it this way

8. ## Re: I just solved this. Can you? (MX2 students should have a go)

Originally Posted by seanieg89
If the Gamma function is used in the question, surely its definition as an integral can be assumed? From there its one substitution and a couple of lines...
As a uni student method, sure - if it works, then it works.

Umm, but for here I kinda gauged it towards HSC students - so I simplified it down a bit, because the original integral I got had identity functions and was most notably an improper integral - sort of things that MX2 students wouldn't have covered yet.

So yes, you could do it that way - but for the... intent of this thread, I'd say it's tantamount to cheating =P (because as math man said, apparently it becomes trivial if we do that substitution)

Plus there is a perfectly long and horrible HSC way to do it so... why not? Who knows, Board of Studies people might be watching and think "omg thats an awesome question. brb putting it in this year's question 8"

Originally Posted by mirakon
Surprised an enthusiastic 2012er hasnt attempted this yet
They're probably still gawking about how ugly the integral looks and going 'HOW THE #@%#@5 DOES THAT SIMPLIFY TO THAT"

9. ## Re: I just solved this. Can you? (MX2 students should have a go)

well if they were planning to put it in question 6 as of this year then you just ruined it for them by exposing it

10. ## Re: I just solved this. Can you? (MX2 students should have a go)

I'm guessing that this is a question that Shadowdude got from Uni.

The fact that he is so proud about it implies that he did it the long way without use of the (integration) definition of the Gamma Function (ie: Integration by parts), otherwise the proof would've been fairly straightforward and I doubt he would have been so proud...

11. ## Re: I just solved this. Can you? (MX2 students should have a go)

Originally Posted by math man
well if they were planning to put it in question 6 as of this year then you just ruined it for them by exposing it
Question 6?

But come to think of it, doing it the 'HSC way' might take a bit long. Or they'd have to at least split it into parts.

12. ## Re: I just solved this. Can you? (MX2 students should have a go)

Originally Posted by Carrotsticks
I'm guessing that this is a question that Shadowdude got from Uni.

The fact that he is so proud about it implies that he did it the long way without use of the (integration) definition of the Gamma Function (ie: Integration by parts), otherwise the proof would've been fairly straightforward and I doubt he would have been so proud...
Well, when I first saw the question saw the gamma thing as a constant and chucked it out the front intuitively.

But umm, yes it's from my stats homework.

And yes I did the long and horrible way.

13. ## Re: I just solved this. Can you? (MX2 students should have a go)

But come to think of it, doing it the 'HSC way' might take a bit long. Or they'd have to at least split it into parts.
HSC has multiple choice now, so instead of written response going up to Q8, it goes up to Q6.

And the 'HSC way' wasn't too long. Took me a page and a bit (but I have massive handwriting)

14. ## Re: I just solved this. Can you? (MX2 students should have a go)

Ugh multiple choice in 4u

WHY

15. ## Re: I just solved this. Can you? (MX2 students should have a go)

Originally Posted by Carrotsticks
HSC has multiple choice now, so instead of written response going up to Q8, it goes up to Q6.

And the 'HSC way' wasn't too long. Took me a page and a bit (but I have massive handwriting)
eww multiple choice

And... yeah I suppose you're right. Looking back on my scribbles - didn't take that long.

16. ## Re: I just solved this. Can you? (MX2 students should have a go)

I got down to this but can't see what to do from here:

$I(\alpha) = \frac{I(\alpha -1)}{\beta^\alpha (\alpha -1)!} [\alpha \beta]$

If only I could show that the fraction = 1 then I would have the answer, but I don't know if that equals to 1 lol.

EDIT: waitttttttttttttttttttttttttt

I know what to do now.

17. ## Re: I just solved this. Can you? (MX2 students should have a go)

ok I'm at:

$I(\alpha) = \alpha \beta .I(\alpha -1)$

Am I on the right track?

18. ## Re: I just solved this. Can you? (MX2 students should have a go)

Just ended up with $\alpha \beta (-1)^{\alpha}$ :/

19. ## Re: I just solved this. Can you? (MX2 students should have a go)

Originally Posted by RealiseNothing
ok I'm at:

$I(\alpha) = \alpha \beta .I(\alpha -1)$

Am I on the right track?
Aren't their limits as R approaches infinity equal?

20. ## Re: I just solved this. Can you? (MX2 students should have a go)

Originally Posted by RealiseNothing
ok I'm at:

$I(\alpha) = \alpha \beta .I(\alpha -1)$

Am I on the right track?
Yes, that's on the right track definitely.

Originally Posted by nightweaver066
Just ended up with $\alpha \beta (-1)^{\alpha}$ :/
How? Could you show working? I think you might've taken a limit wrong or something.

Originally Posted by bleakarcher
Aren't their limits as R approaches infinity equal?
Limits of...?

21. ## Re: I just solved this. Can you? (MX2 students should have a go)

Yes, that's on the right track definitely.

How? Could you show working? I think you might've taken a limit wrong or something.

Limits of...?

Working is a bit long, but i combined the x and x^(alpha - 1), took out the constants (the T(alpha) and beta thing), considered the integral seperately, applied tabular integration, deduced what the integral would result in:
$\alpha!(-1)^{\alpha + 2}\beta^{\alpha + 1}$

Subbed that back in to the integral, simplified and got $\alpha \beta (-1)^{\alpha}$

Edit: Nevermind, figured it out. I forgot to include something while integrating and now i have the answer.

22. ## Re: I just solved this. Can you? (MX2 students should have a go)

I got:

lim[R->infinity] I(A)=lim[R->infinity] AB*I(A-1) where A is alpha and B is beta.

?

23. ## Re: I just solved this. Can you? (MX2 students should have a go)

I give up with this problem...

24. ## Re: I just solved this. Can you? (MX2 students should have a go)

Are there any restrictions on $\alpha$ and $\beta$

25. ## Re: I just solved this. Can you? (MX2 students should have a go)

Originally Posted by bleakarcher
I got:

lim[R->infinity] I(A)=lim[R->infinity] AB*I(A-1) where A is alpha and B is beta.

?
You don't take limits, you keep working it out. I(A) and I(A-1) are difficult to work with, so you simplify it down.

Originally Posted by jetblack2007
Are there any restrictions on $\alpha$ and $\beta$
As in...? I don't think so... or at least when I did the problem, I didn't think of that.

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