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## #1 2008-02-27 05:23:23

tony123
Member
Registered: 2007-08-03
Posts: 189

### a and r

find a and r

Last edited by tony123 (2008-02-27 05:26:08)

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## #2 2008-02-27 06:12:59

JaneFairfax
Member
Registered: 2007-02-23
Posts: 6,868

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## #3 2008-02-27 08:53:36

tony123
Member
Registered: 2007-08-03
Posts: 189

but how olivia

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## #4 2008-02-27 11:09:09

JaneFairfax
Member
Registered: 2007-02-23
Posts: 6,868

I guessed.

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## #5 2011-07-28 06:27:22

gAr
Member
Registered: 2011-01-09
Posts: 3,479

### Re: a and r

"Believe nothing, no matter where you read it, or who said it, no matter if I have said it, unless it agrees with your own reason and your own common sense"  - Buddha?

"Data! Data! Data!" he cried impatiently. "I can't make bricks without clay."

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## #6 2011-07-28 07:57:36

bobbym
bumpkin
From: Bumpkinland
Registered: 2009-04-12
Posts: 109,606

### Re: a and r

Hi tony123, wonderful Jane, and gAr the great who finally got me to learn how to differentiate under the integral, wunderbar!

To prove that there are no more real roots for r we eliminate the a variable in the equations. We are left with a single 6th degree polynomial.

We could deflate out the two roots they found and solve a 4th degree equation. If this were 1670 that would be cool. Powerful computational techniques make life easier.

We use a Cauchy bound to bound the roots.
They are found to be.

To find out how many roots are in the interval { -3 , 3 } we form a Sturm chain.

Go down the first column and count the number of sign changes and call it A. Go down the second column and count the number of sign changes and call it B. The theorem says the number of real roots in { -3, 3 } is A - B = 2. Multple roots count as 1 root.

So Jane and gAr have found the only 2 real roots of r, the rest are complex.

In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.

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## #7 2011-07-28 15:06:36

gAr
Member
Registered: 2011-01-09
Posts: 3,479

### Re: a and r

Hi bobbym,

Those things are new to me. Thanks!

"Believe nothing, no matter where you read it, or who said it, no matter if I have said it, unless it agrees with your own reason and your own common sense"  - Buddha?

"Data! Data! Data!" he cried impatiently. "I can't make bricks without clay."

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## #8 2011-07-28 18:16:01

bobbym
bumpkin
From: Bumpkinland
Registered: 2009-04-12
Posts: 109,606

### Re: a and r

Hi gAr;

You are welcome!

Hi tony123;
We can continue in the same manner to prove that the roots found for a are the only real roots. By eliminating r we are left with the polynomial.

Using the Cauchy bound we get:

We see that all the real roots are located in the closed interval [-12398,12398]. Not as tight a bounds as the other but adequate. We form the Sturm chain.

A = 4 and B = 2, so A - B = 2. There are two real roots in that interval. So Jane's and gAr's are the only real roots of both r and a. We are done!

In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.

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