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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 104,813

Hi;

Then you must know that the answer is pi^2 / 6.

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**

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**gAr****Member**- Registered: 2011-01-09
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Yes, I was tring to prove that without using series.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 104,813

I get it now!

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**

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

**17)**

*Last edited by gAr (2011-08-02 06:14:13)*

"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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**reconsideryouranswer****Member**- Registered: 2011-05-11
- Posts: 171

gAr & my edit wrote:

16)

The limit needs to approach from the right side of 0,

as amended above. There aren't any real values for

ln(x) from the left side of 0, so it is undefined there.

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**gAr****Member**- Registered: 2011-01-09
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"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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**reconsideryouranswer****Member**- Registered: 2011-05-11
- Posts: 171

gAr wrote:

I am not "forgetting about" ln(1 - x). I was looking right at it and working with it, regardless if I mishandled it. x approaching 0 from either side is not an issue for ln(1 - x), as it is 0. But x approaching 0 from the left side of ln(x) is a problem, just as it is for x approaching from the left of, say, x^x, as those limits do not exist. And where is ln(-x) = ln(x) + ipi coming from? And then, why isn't your alleged expression this

instead of what you typed, because you assumed x --->0-? ---------------------------------------------------------- ------------------------------------------- And ----------------------------------------------------- on using*Last edited by reconsideryouranswer (2011-08-02 17:56:25)*

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**gAr****Member**- Registered: 2011-01-09
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"Data! Data! Data!" he cried impatiently. "I can't make bricks without clay."

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**gAr****Member**- Registered: 2011-01-09
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I see that you edited your post instead of replying.

Anyway,

Have you read about complex numbers? It exists and it is -∞

and this is 1

*Last edited by gAr (2011-08-02 20:14:25)*

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**reconsideryouranswer****Member**- Registered: 2011-05-11
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gAr wrote:

and this is 1

Look at all of those negative x-values approaching 0 from the left where x^x is real.

For example, x =

-1/5, -1/25, -1/125, -1/625. -1/3125, ...

For these, x^x is approaching -1.

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**gAr****Member**- Registered: 2011-01-09
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I would suggest you to go to www.wolframalpha.com

and enter this:

`Limit[x^x,{x->0},Direction->1]`

That would show you a graph along with the answer.

If you observe the graph, you'll see that the complex part approaches 0, and the real part approaches 1.

How did you calculate that it would approach -1? They should have been complex numbers, with positive real part.

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**reconsideryouranswer****Member**- Registered: 2011-05-11
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gAr wrote:

How did you calculate that it would approach -1?

They should have been complex numbers, with positive real part.

Notice in the folowing lines how I will use fractions that

necessarily have odd denominators (for odd indices)

Let me show three of my examples worked out:

-------------------------------------------------------------

-------------------------------------------------------------

-------------------------------------------------------------

I will use a fraction relatively much closer to 0:

-------------------------------------------------------------

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

That is not correct.

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**

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**reconsideryouranswer****Member**- Registered: 2011-05-11
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bobbym wrote:

Hi;

That is not correct.

No, the odd root of a negative integer is some type of negative real number.

It must be.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
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No, the odd root of a negative integer is some type of negative real number.

It must be.

Yes but check this out! You are forgetting there are also complex answers.

This is one answer for (-1)^(1/3)

Are not the roots of

, ,What I am saying is just because (-1)^3 = -1 that does not mean ( -1) ^(1 / 3 ) is -1 solely. There are other answers. 3 of them as a matter of fact.

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**

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**gAr****Member**- Registered: 2011-01-09
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reconsideryouranswer, do you admit that you do not know complex numbers?

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

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**reconsideryouranswer****Member**- Registered: 2011-05-11
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bobbym wrote:

Yes but check this out! You are forgetting there are also complex answers.

This is one answer for (-1)^(1/3)

Are not the roots of

, ,

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**anonimnystefy****Real Member**- From: Harlan's World
- Registered: 2011-05-23
- Posts: 16,015

sqrt(-1) doesnt exist because it is indefined.

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**reconsideryouranswer****Member**- Registered: 2011-05-11
- Posts: 171

anonimnystefy wrote:

sqrt(-1) doesnt exist because it is indefined.

----------------------------------------------------------------------------------------------

*** Edit*** I will stop posting to this subject thread for the foreseeable future.

*Last edited by reconsideryouranswer (2011-08-04 06:12:39)*

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

You are making heavy use of the principal value of a square root being only the positive answer. But does that apply for cube roots and higher?

http://mathworld.wolfram.com/CubeRoot.html

Both Mathematica and Maple do not agree.

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**

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**gAr****Member**- Registered: 2011-01-09
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**18) **

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

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 7,613

hi guys,

Sorry to jump in on your thread but would you mind casting an eye over

http://www.mathisfunforum.com/viewtopic … 03#p184503

Thanks,

Bob

Children are not defined by school ...........The Fonz

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

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

Hi Bob,

I was also confused with that, so couldn't reply!

Anyway, I'll try again.

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

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 7,613

hi gAr

Thanks for the response.

I'm happy with the substitution leading to the final DE. I'm really rusty on second order DEs having last done them about 30 years ago, so I was hoping for a second opinion on my final post.

Thanks

ps. I'm being called into the garden to help my wife so I'll log back in later.

Bob

*Last edited by bob bundy (2011-08-04 22:28:35)*

Children are not defined by school ...........The Fonz

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

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**anonimnystefy****Real Member**- From: Harlan's World
- Registered: 2011-05-23
- Posts: 16,015

hi gAr

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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