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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 82,549

Hi anonimnystefy;

That thread got closed by mistake. It is open now for business.

Did you get as far as a catenary? If you find the equation of the catenary and Taylorize it you will be close. See if you can get it from there. Tell me what you think about it.

**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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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 14,859

i found that it's y=cosh(x)

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 82,549

Hi;

But you must get the series for it and then you will see.

**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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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 14,859

it's x+x^3/3!+x^5/5!+...

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 82,549

That is not correct.

**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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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 14,859

sry.copied the wrong series.

1+x^2/2!+x^4/4!+...

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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
- Registered: 2009-04-12
- Posts: 82,549

Hi

what about?i see the pic.it looks parabolish.

Now what can you say about that? Be right back!

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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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 14,859

don't now.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 82,549

Look at the first two terms of the Mclaurin series, what do you see?

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

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
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x^2/2 +1,which is a parabola.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
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That is correct! Since that series is an approximation of cosh(x) anywhere we truncate it, is also an approximation. The further we go the better we expect to do. Now you see why Galilieo was fooled into thinking the catenary was a parabola.

That series is a good approximation at x = 0 and close to that. How about for any x?

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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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 14,859

hi bobbym

it's not so good for very large terms because the difference grows 'cubicly' so as x gets larger,the difference becomes much much larger.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
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Yes, that is true for that particular series.You can get around that and we are just trying to demonstrate that a parabola is a good approximation. We do not need a high precision algorithm.

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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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 14,859

yup.

next?

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
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Hold it! You did not provide any indication that it is so.

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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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 14,859

hi bobbym

|cosh(x)-1-x^2/2|<c*|x^3|

so (cosh(x)-1-x^2/2) is O(x^3).thus,my statement is correct.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
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Yes, but that is only the expansion around zero. It is not the series that we want at 5 for an example.

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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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 14,859

maybe we could take the taylor's series expansion arounf 5 or a in a general case.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
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Exact a Mundo! We can expand around c.

Now since cosh(c) and sinh(c) are constants we can substitute. a = cosh(c) and b = sinh(c).

Truncating at x^2 we get:

which again means for any value of x it can be approximated by a quadratic. This is not exactly true in a rigorous sense but for the purposes of your question we will allow it.

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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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 14,859

nice!

next?

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 82,549

What next? You are done for now!

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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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 14,859

next question?i'm not done.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
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It will take me some time to think of one. I think the other question you have in the other will be a good source of more questions.

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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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 14,859

ok,but i will need help with it.that's of course assuming that you mean the one with the squares and stuff.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 82,549

I disagree, you can do it all by yourself. I do not know the answer right now either.

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.

**Online**