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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,251

Hi;

Evaluate this integral:

Prove the inequality:

**In mathematics, you don't understand things. You just get used to them.Of course that result can be rigorously obtained, but who cares?Combinatorics is Algebra and Algebra is Combinatorics.**

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**mathsyperson****Moderator**- Registered: 2005-06-22
- Posts: 4,900

Are we allowed to express the LHS of the inequality as a Taylor series? If so, the problem becomes trivial.

Why did the vector cross the road?

It wanted to be normal.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,251

Hi mathsyperson;

Thats how I would do it. But in this Putnam problem book he doesn't use that method. Why, I can't say.

*Last edited by bobbym (2009-09-14 11:42:02)*

**In mathematics, you don't understand things. You just get used to them.Of course that result can be rigorously obtained, but who cares?Combinatorics is Algebra and Algebra is Combinatorics.**

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

Ever seen this theorem?

Let be convex and differentiable. Then

[align=center]

[/align]So you just put

and .(What Im saying is that no point on a convex and differentiable curve lies below the corresponding point on a straight line that is tangent to any part of the curve.)

PS: I just looked up Wikipedia. This theorem is stated there without proof.

*Last edited by JaneFairfax (2009-09-14 12:56:36)*

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

mathsyperson wrote:

Are we allowed to express the LHS of the inequality as a Taylor series? If so, the problem becomes trivial.

bobbym wrote:

Thats how I would do it.

The inequality holds for

as well, in which case the method will not be trivial for .And note that there is an apostrophe in **Thats**.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,251

Hello Jane;

Jane wrote:

The inequality holds for

as well, in which case the method will not be trivial for .

But the problem states x > 0 and mathsy is right using the expansion of e^x makes it trivial.

Jane wrote:

And note that there is an apostrophe in

Thats.

That is true Jane, poor typing and high speeds, a lethal combination. Thanks for the correction.

Jane wrote:

Ever seen this theorem?

Yes Jane, I have heard of that theorem but I forgot it.

*Last edited by bobbym (2009-09-14 18:23:55)*

**In mathematics, you don't understand things. You just get used to them.Of course that result can be rigorously obtained, but who cares?Combinatorics is Algebra and Algebra is Combinatorics.**

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,251

Hi;

The inequality was the easy one. Now how about the integral it is even easier. If you start substituting you will go mad.

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,251

Hi;

Another integral this one is also easy:

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,251

Hi Jane;

*Last edited by bobbym (2009-09-25 11:14:14)*

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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**123ronnie321****Member**- Registered: 2010-09-28
- Posts: 128

bobbym wrote:

Hi;

Another integral this one is also easy:

(e^2x)/2 - (e^-2x)/2

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**123ronnie321****Member**- Registered: 2010-09-28
- Posts: 128

bobbym wrote:

Hi;

Evaluate this integral:

Prove the inequality:

1/2[x - log(sinx + cosx)]

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,251

Hi 123ronnie321;

Yes that is almost correct!

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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

Hi bobbym,

Evaluate this integral:

"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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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,251

Hi gAr;

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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

Thank you!

"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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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,251

HI gAr;

You are welcome.

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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