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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 107,801

Hi;

Evaluate this integral:

Prove the inequality:

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**** Always satisfy the Prime Directive of getting the right answer above all else.**

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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: 107,801

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.****If it ain't broke, fix it until it is.**** Always satisfy the Prime Directive of getting the right answer above all else.**

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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: 107,801

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.****If it ain't broke, fix it until it is.**** Always satisfy the Prime Directive of getting the right answer above all else.**

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 107,801

Hi;

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

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**** Always satisfy the Prime Directive of getting the right answer above all else.**

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 107,801

Hi;

Another integral this one is also easy:

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**** Always satisfy the Prime Directive of getting the right answer above all else.**

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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: 107,801

Hi Jane;

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

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**** Always satisfy the Prime Directive of getting the right answer above all else.**

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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: 107,801

Hi 123ronnie321;

Yes that is almost correct!

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**** Always satisfy the Prime Directive of getting the right answer above all else.**

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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: 107,801

Hi gAr;

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**** Always satisfy the Prime Directive of getting the right answer above all else.**

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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: 107,801

HI gAr;

You are welcome.

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**** Always satisfy the Prime Directive of getting the right answer above all else.**

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