You are welcome.
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Evaluate this integral:
Yes that is almost correct!
]]>Hi;
Evaluate this integral:
Prove the inequality:
1/2[x - log(sinx + cosx)]
]]>Hi;
Another integral this one is also easy:
(e^2x)/2 - (e^-2x)/2
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Another integral this one is also easy:
]]>The inequality was the easy one. Now how about the integral it is even easier. If you start substituting you will go mad.
]]>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.
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.
Ever seen this theorem?
Yes Jane, I have heard of that theorem but I forgot it.
]]>Are we allowed to express the LHS of the inequality as a Taylor series? If so, the problem becomes trivial.
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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[/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.
]]>Thats how I would do it. But in this Putnam problem book he doesn't use that method. Why, I can't say.
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