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**zetafunc.****Guest**

I don't know, it is compulsory. She also had a differential equations lecture today too. All first-years have to do the compulsory timetable as listed at least.

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 81,622

Too bad you could not be there as a heckler.

**In mathematics, you don't understand things. You just get used to them.I have the result, but I do not yet know how to get it.All physicists, and a good many quite respectable mathematicians are contemptuous about proof.**

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**zetafunc.****Guest**

Yes, but I probably wouldn't do it, as that'd make a whole load of people upset...

I used to have this stupid idea that I could get to know F through maths. She could tell me about an idea and I would tell her something she did not know in return.

**zetafunc.****Guest**

Also, I had a thought today, about complex derivatives. As we recall, you can use the gamma function to find the ith derivative of a function (such as a polynomial). It follows that if you take the ith derivative, and then take the (1-i)th derivative of that, that is the same as taking the first derivative (and very easy to prove).

But, supposing you wanted to take the ith derivative of x^k, i times. How does one reason or even compute this? We know that if you take, for example, the 3rd derivative of something 3 times, that is the same as taking the 9th (3^2) derivative. So surely taking the ith derivative i times would yield the (-1)th derivative, also known as an integration.

My problem lies in the first part. How do you do anything i times, and how could you write that or compute it?

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 81,622

Hi;

When you say the ith derivative are you using i as an iterator or as √(-1) ?

**In mathematics, you don't understand things. You just get used to them.I have the result, but I do not yet know how to get it.All physicists, and a good many quite respectable mathematicians are contemptuous about proof.**

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**zetafunc.****Guest**

I am using it as √(-1).

**zetafunc.****Guest**

Taking the ith derivative is itself doable:

But it is doing that i times which is the issue.

**zetafunc.****Guest**

Of course doing this would require computation if i!, which can be written in the form of a + bi, but a and b are irrational and only expressible with infinite series. However, taking the absolute value of it yields a much nicer value for i!:

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 81,622

Hi;

What is the answer if you do it one more time?

**In mathematics, you don't understand things. You just get used to them.I have the result, but I do not yet know how to get it.All physicists, and a good many quite respectable mathematicians are contemptuous about proof.**

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**zetafunc.****Guest**

You mean, taking the ith derivative of x^k again? If we do that n times;

and letting n = i yields

which seems to reduce to

since gamma(k+1)/gamma(k+2) = k!/(k+1)! = 1/k...

But this then seems quite strnage, because the denominator should equate to k+1 if this is true (we are integrating x^k here using differentiation). But I do not see how that infinite product will approach that.

**zetafunc.****Guest**

At the moment, I do not know if I have made any sense at all, or if I have just said a bunch of meaningless mumbo-jumbo...

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 81,622

Hi;

Okay, let me play with that formula a bit.

I have the result, but I do not yet know how to get it.

All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

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**zetafunc.****Guest**

Okay, let me know what you get.

Interestingly I am seeing my maths teacher (the one I talked about, over e-mail) tomorrow, maybe I can bring this up with her. She said she liked complex analysis.

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 81,622

Yes, do that. Might bring an interesting turn of events.

I have the result, but I do not yet know how to get it.

All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

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**zetafunc.****Guest**

Unfortunately we won't be alone, we will be in the maths office. But at least there will not be any Cs and Fs in there.

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 81,622

Sounds interesting. You could all listen to F give her lecture in person.

I have the result, but I do not yet know how to get it.

All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

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**zetafunc.****Guest**

F giving lectures...? Why would F be giving lectures?

Also, I just found out that the ith derivative can't be defined in the way that I defined it, so looks like everything I did was a waste of time... sigh.

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 81,622

LIke the one she is giving at Cambridge.

That saves me from having to investigate that formula.

I have the result, but I do not yet know how to get it.

All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

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**zetafunc.****Guest**

Huh? F is not giving lectures, she is only a student. She is receiving them and will be for at least the next 4 years...

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 81,622

Oh, I am sorry. I thought they had her giving the lecture. Sort of like an exercise.

I have the result, but I do not yet know how to get it.

All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

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**zetafunc.****Guest**

I think you are allowed to do that after your 4th year but I am not sure.

I don't know why I am still talking about her. There is no reason why I should be, yet, she is still in my head. It is silly.

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 81,622

Talking is okay just as long as there is no tears or weeping.

I have the result, but I do not yet know how to get it.

All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

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**zetafunc.****Guest**

I am not weeping, I just still feel disappointed somewhat, since I never got to talk to her properly, not even about maths.

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 81,622

It takes two to make a conversation. Maybe you will find someone else interested in math.

I have the result, but I do not yet know how to get it.

All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

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**zetafunc.****Guest**

Before I asked her out she would have happily talked to me. She even used to smile at me if she saw me.

Honestly, I think that is impossible. Almost everyone in my school hates maths, and the one girl who is applying for maths at uni doesn't even seem to like it (nor does she seem to know much about it...). There is one 'maths club/masterclass' where an old man offers lectures to any students interested, but there is only one person who attends that and he is a guy. This is another reason for my 2013 prediction.