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**Yusuke00****Member**- Registered: 2013-11-19
- Posts: 43

Hey guys,i'm new here. Hope u can help me with those 3 problems:

1)Let it be f,g:[-1,1]-> R; f,g - continous functions.Show that exist a,b ∈[-1,1],a<b so that f(a)=g(b) and f(b)=g(a) then there is c∈[-1,1],so that f(c)=g(c).

My guess here it's Lagrange but i have no clue on how to apply it.

2)Calculate lim n->inf ( 1/(2ln2)+1/(3ln3)+...+1/(nln(n)) )

3)a,b,c > 0 so that a^x+b^x+c^x>=3, with any x∈R. Show that a*b*c=1

Ty for help xD

*Last edited by Yusuke00 (2013-11-20 02:50:41)*

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 81,616

Hi Yusuke00

For 2)

**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: 14,836

I think he might have meant 1/(2log2)+1/(3log3)+...+1/(nlogn).

For the first problem, did you mean f'(c)=g'(c)?

*Last edited by anonimnystefy (2013-11-19 10:55:11)*

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

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**Yusuke00****Member**- Registered: 2013-11-19
- Posts: 43

Yes i am sorry at 2) was +...+

@anonimnystefy No it is f(c)=g(c)

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 81,616

Hi;

Have you tried some tests on 2) which is really only:

The integral test works well to prove divergence.

**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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**Yusuke00****Member**- Registered: 2013-11-19
- Posts: 43

They don't ask to prove the divergence but a result.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 81,616

If it diverges it means the sum is infinity. That is what I am talking about. Use the integral test and you will prove the sum diverges.

**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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**Yusuke00****Member**- Registered: 2013-11-19
- Posts: 43

Actually the sum is constant,that's a thing i know.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 81,616

Constant what does that mean?

Infinity is not a constant.

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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**Yusuke00****Member**- Registered: 2013-11-19
- Posts: 43

it's 7e/16 i think.

ok the thing is that i did this exercise 2 years ago but i lost the notebook and now i can't figure out how to solve it anymore

i know that the sum is constant and i know it's something with xe/12 or xe/12 but i cannot remember exactly.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 81,616

Why do you think that sum converges besides from your memory?

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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**Yusuke00****Member**- Registered: 2013-11-19
- Posts: 43

I am sorry you were write,i just found the notebook.apologize

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 81,616

Hunches are fine, but in math you will have to back them up with numerical evidence or proof.

Using the integral test:

http://en.wikipedia.org/wiki/Integral_t … onvergence

If that integral converges then so does the sum, if it diverges so does the sum:

Say u = log(x), then du/dx = 1 / u, and the integral becomes

Substituting back u = log(x)

log(log(x)). Taking the limits of integration.

The integral is infinity, so it diverges. The sum also equals infinity, so it diverges.

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: 14,836

The way the question is posed (with the limit instead of infinity as the upper sum limit), I would guess there is a missing term in front of the sum.

Hi bobbym

How fast does the sum converge?

*Last edited by anonimnystefy (2013-11-20 06:38:17)*

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

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**Yusuke00****Member**- Registered: 2013-11-19
- Posts: 43

Yes that's a part i don't understand yet.I'm on last year of high school so i started learning about integrals just now.I still have to learn a bit more about integrals because at the time i solved it i used just Lagrange and derivates.

I also found out how to solve 3) if you are interested but still no clue for 1).

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 81,616

How fast does the sum converge?

The sum as given does not converge.

Lagrange what?

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: 14,836

Not converge, sorry. What's its growth rate?

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

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 81,616

Slower than 1 / n that is for sure.

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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**Yusuke00****Member**- Registered: 2013-11-19
- Posts: 43

Lagrange Theoreme for derivates

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 81,616

Lagrange probably has 2000 theorems named after him. Do you mean the mean value theorem?

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: 14,836

bobbym wrote:

Slower than 1 / n that is for sure.

I'm thinking it's O(log log n). Can you do the same limit, but with an 1/n in front of the sum.

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

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**Yusuke00****Member**- Registered: 2013-11-19
- Posts: 43

Yes the mean value theorem.Sorry they don't call like that here.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 81,616

That is okay.

Are you sure that you have the right question?

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: 14,836

Hi bobbym

anonimnystefy wrote:

bobbym wrote:Slower than 1 / n that is for sure.

I'm thinking it's O(log log n). Can you do the same limit, but with an 1/n in front of the sum.

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

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 81,616

Hi;

Yes, I saw that. Why are we doing that limit?

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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