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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
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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.

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 15,673

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

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
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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 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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it's x+x^3/3!+x^5/5!+...

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
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That is not correct.

**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
- Posts: 15,673

sry.copied the wrong series.

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

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

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

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

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
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don't now.

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**bobbym****Administrator**- From: Bumpkinland
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Look at the first two terms of the Mclaurin series, what do you see?

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

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**bobbym****Administrator**- From: Bumpkinland
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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?

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

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

next?

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

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

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

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**anonimnystefy****Real Member**- From: The Foundation
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nice!

next?

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**bobbym****Administrator**- From: Bumpkinland
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What next? You are done for now!

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**anonimnystefy****Real Member**- From: The Foundation
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next question?i'm not done.

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**bobbym****Administrator**- From: Bumpkinland
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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.

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**anonimnystefy****Real Member**- From: The Foundation
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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
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I disagree, you can do it all by yourself. I do not know the answer right now either.

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