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**George,Y****Member**- Registered: 2006-03-12
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*Last edited by George,Y (2012-04-30 19:36:56)*

**X'(y-Xβ)=0**

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
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Hi;

I do not think there are closed forms for either of those. But if you could put some bounds on some of the constants an asymptotic form might be possible.

**In mathematics, you don't understand things. You just get used to them.**

**I agree with you regarding the satisfaction and importance of actually computing some numbers. I can't tell you how often I see time and money wasted because someone didn't bother to run the numbers.**

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**George,Y****Member**- Registered: 2006-03-12
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There actually is closed form, but through a different integration.

**X'(y-Xβ)=0**

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**bobbym****Administrator**- From: Bumpkinland
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A different integration?

**In mathematics, you don't understand things. You just get used to them.**

**I agree with you regarding the satisfaction and importance of actually computing some numbers. I can't tell you how often I see time and money wasted because someone didn't bother to run the numbers.**

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**George,Y****Member**- Registered: 2006-03-12
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bobbym wrote:

A different integration?

I have found a way to integrate this directly, but it is very tricky.

**X'(y-Xβ)=0**

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
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I doubt that.

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Taking a new step, uttering a new word, is what people fear most. ― Fyodor Dostoyevsky, Crime and Punishment

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**bobbym****Administrator**- From: Bumpkinland
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I would say it is unlikely to but the whole question could be answered by posting the solution. Then it can be checked.

**In mathematics, you don't understand things. You just get used to them.**

**I agree with you regarding the satisfaction and importance of actually computing some numbers. I can't tell you how often I see time and money wasted because someone didn't bother to run the numbers.**

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
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http://m.wolframalpha.com/input/?i=inte … E2&x=0&y=0

*Last edited by anonimnystefy (2013-12-17 01:29:39)*

Here lies the reader who will never open this book. He is forever dead.

Taking a new step, uttering a new word, is what people fear most. ― Fyodor Dostoyevsky, Crime and Punishment

The knowledge of some things as a function of age is a delta function.

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**bobbym****Administrator**- From: Bumpkinland
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Hi;

That is not his integral.

But even if it were, that is not in closed form.

**In mathematics, you don't understand things. You just get used to them.**

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**anonimnystefy****Real Member**- From: The Foundation
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It is not his integral, but his can be manipulated into that one with substitutions and manipulations. The point is that Alpha says there's no closed form...

Here lies the reader who will never open this book. He is forever dead.

Taking a new step, uttering a new word, is what people fear most. ― Fyodor Dostoyevsky, Crime and Punishment

The knowledge of some things as a function of age is a delta function.

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**bobbym****Administrator**- From: Bumpkinland
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Hi;

Have you looked at the integral you sent to Alpha in post #8?

**In mathematics, you don't understand things. You just get used to them.**

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**anonimnystefy****Real Member**- From: The Foundation
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Yes, I have.

Taking a new step, uttering a new word, is what people fear most. ― Fyodor Dostoyevsky, Crime and Punishment

The knowledge of some things as a function of age is a delta function.

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**bobbym****Administrator**- From: Bumpkinland
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Then you know it is not and can never be his integral.

**In mathematics, you don't understand things. You just get used to them.**

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**anonimnystefy****Real Member**- From: The Foundation
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Fixed.

Taking a new step, uttering a new word, is what people fear most. ― Fyodor Dostoyevsky, Crime and Punishment

The knowledge of some things as a function of age is a delta function.

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**bobbym****Administrator**- From: Bumpkinland
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Brings us back to post #7. The fact that both M's can not do the integral does mean there is a high probability that it is intractable. But they are not infallible, so I think if George would post his answer the whole thing can resolved quickly.

**In mathematics, you don't understand things. You just get used to them.**

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**anonimnystefy****Real Member**- From: The Foundation
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Do you think some progress can be made using DUIT?

Taking a new step, uttering a new word, is what people fear most. ― Fyodor Dostoyevsky, Crime and Punishment

The knowledge of some things as a function of age is a delta function.

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**bobbym****Administrator**- From: Bumpkinland
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I have not tried but it sure does respond well to numerical integration for any T.

**In mathematics, you don't understand things. You just get used to them.**

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**George,Y****Member**- Registered: 2006-03-12
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Sorry guys, I made a mistake, the question should be:

*Last edited by George,Y (2013-12-29 01:12:55)*

**X'(y-Xβ)=0**

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**anonimnystefy****Real Member**- From: The Foundation
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That does not change much.

Taking a new step, uttering a new word, is what people fear most. ― Fyodor Dostoyevsky, Crime and Punishment

The knowledge of some things as a function of age is a delta function.

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**bobbym****Administrator**- From: Bumpkinland
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Hi George,Y;

That does not change much.

Yes, I think it is time to show your solution. I am willing to bet 21% of my bankroll, a whopping $1.16 that the solution is wrong.

**In mathematics, you don't understand things. You just get used to them.**

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**George,Y****Member**- Registered: 2006-03-12
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*Last edited by George,Y (2013-12-30 22:11:21)*

**X'(y-Xβ)=0**

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
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Hi;

What is d?

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**George,Y****Member**- Registered: 2006-03-12
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After this change of variable, I think now the question is easier.

**X'(y-Xβ)=0**

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**George,Y****Member**- Registered: 2006-03-12
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bobbym wrote:

Hi;

What is d?

d is the differential operator

**X'(y-Xβ)=0**

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
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What is the new integral?

**In mathematics, you don't understand things. You just get used to them.**

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