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Author Topic: TT's Maths Thread  (Read 133116 times)  Share 

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TrueTears

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Re: TT's Maths Thread
« Reply #150 on: November 24, 2009, 11:15:38 pm »
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Thanks kamil, I was playing around and gets a "better" estimate :P

The LHS can be rewritten into:



But we know

Reciprocate:



But we require sum from 0 to infinity.



However this is false since the inequality holds for

Thus we require:



The RHS is a geometric sum, denote this as .



Now

Since we are after the upper bound of



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TrueTears

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Re: TT's Maths Thread
« Reply #151 on: November 25, 2009, 01:30:11 am »
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Quote
4) Prove that where are non negative real numbers for which

I've proved that however I can't seem to get the RHS inequality. Here's my working so far:

Applying AM-GM on





But





Now if we can prove

Then that implies since

Applying AM-GM on yields:







Now let's try to prove that



But we showed that

Sub in the upper bound into yields

Which is clearly true.

Thus

Hence



Now the trouble is how to prove ...


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kamil9876

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Re: TT's Maths Thread
« Reply #152 on: November 25, 2009, 01:34:11 am »
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4.)

I have developed a method in the past of making these substitutions:

let

with obviously.

Now expand the expression and after some algebra you eventually get:

 (*)

then using AM-GM:




and so

which proves * is less than or equal to
« Last Edit: November 25, 2009, 01:46:09 am by kamil9876 »
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TrueTears

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Re: TT's Maths Thread
« Reply #153 on: November 25, 2009, 01:47:00 am »
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LOL thanks kamil but that's soooooooooooooooooooooooooo needle in the haystack for me :( (the substitution part, I would not have thought of that in a million years =S)
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kamil9876

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Re: TT's Maths Thread
« Reply #154 on: November 25, 2009, 01:58:41 am »
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lol i think the questions are rigged for it:

http://vcenotes.com/forum/index.php/topic,2398.msg140527.html#msg140527

sort of developed the technique here, it ussually happens when the expression is symmetric in the variables and in symmetric curves/surfaces the highest point is commonly the centre, and so what I did here is try to find a nice centre ie: (1/3,1/3,1/3) and show that if you move to another point (1/3 +t,1/3 +s, 1/3 + u) you 'fall off' from the top.
Voltaire: "There is an astonishing imagination even in the science of mathematics ... We repeat, there is far more imagination in the head of Archimedes than in that of Homer."

kamil9876

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Re: TT's Maths Thread
« Reply #155 on: November 25, 2009, 02:18:26 am »
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let

   (triangle inequality)

from which the result immediately follows.


I once did a proof of the triangle inequality without using the Cauchy Schwartz ienquality, and hence used it to prove Cauchy Schwartz. However many textbooks go the other way and so you can try this yourself.
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TrueTears

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Re: TT's Maths Thread
« Reply #156 on: November 25, 2009, 02:39:45 am »
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lol i think the questions are rigged for it:

http://vcenotes.com/forum/index.php/topic,2398.msg140527.html#msg140527

sort of developed the technique here, it ussually happens when the expression is symmetric in the variables and in symmetric curves/surfaces the highest point is commonly the centre, and so what I did here is try to find a nice centre ie: (1/3,1/3,1/3) and show that if you move to another point (1/3 +t,1/3 +s, 1/3 + u) you 'fall off' from the top.
Right, do you think that substitution only works for symmetrical questions such as this one? Ie, you can switch the variables around yet you get the same thing.

So say you had a symmetrical question like this but it had 4 variables, let's say and given that

Would you then do ...?

What about if the question was NOT symmetrical. Yet still, would you do the same substitution?

Sorry there are no examples here to illustrate my point but I'm just speaking in 'general' terms.

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humph

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Re: TT's Maths Thread
« Reply #157 on: November 25, 2009, 03:54:09 am »
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3) Show that for all real values of the variables. Furthermore, give a condition for equality to hold.
This is just the triangle inequality, as kamil said. I'm guessing you're using the Cauchy-Schwarz inequality in the form

If we let and , then this is equivalent to

where is the dot product of and , and is the magnitude of respectively.
Now note that , and hence

as the dot product is linear and commutative. Then by the Cauchy-Schwarz inequality,

and taking square roots yields the triangle inequality. By induction, this implies that

where . It remains to note that

and that


Note that you can prove all this without writing it in the language of vectors, dot products, and magnitudes, but it's in some sense the most "natural" way. Also as kamil said, you don't need the Cauchy-Schwarz inequality to prove the triangle inequality, but it's normal to do so in that order, rather than the other way around.

The Cauchy-Schwarz inequality is a particular feature of a mathematical structure called an inner product space, which is a vector space with an inner product, a function that measures angles between vectors and magnitudes of vectors; the dot product on the vector space is the standard example. Inner products in particular give rise to norms, which is the magnitude of a vector, and norms satisfy the triangle inequality. This stuff is very important mathematically; the theory of Banach spaces (complete normed vector spaces) and Hilbert spaces (complete inner product spaces) are the foundations of an area of mathematics called functional analysis, and have applications to many areas of mathematics and physics.
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Ahmad

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Re: TT's Maths Thread
« Reply #158 on: November 25, 2009, 11:12:37 am »
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  (*)

I get , I could be wrong but it'd be worth checking your working, I would like to see this work out!  :)

Also for the original problem the left hand side inequality isn't too strong. Notice that not all of x, y and z can be greater than 1/2, which means that we can suppose one of them is say z. Then as required.  :)
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kamil9876

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Re: TT's Maths Thread
« Reply #159 on: November 25, 2009, 12:50:38 pm »
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yep ahmad you're right, i missed
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Re: TT's Maths Thread
« Reply #160 on: November 25, 2009, 12:58:38 pm »
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Fucking inequalities. I just spent ages stuck trying to show that
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Re: TT's Maths Thread
« Reply #161 on: November 25, 2009, 01:19:10 pm »
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It's asymptotically

where I used which is true because . Is that how you did it?  :)
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Re: TT's Maths Thread
« Reply #162 on: November 25, 2009, 01:21:06 pm »
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Kamil another thing which might make your approach tricky (or maybe not, haven't invested so much thought into it) is that s and t can be negative so AM-GM isn't applicable.
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kamil9876

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Re: TT's Maths Thread
« Reply #163 on: November 25, 2009, 01:32:22 pm »
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I didn't realise the restriction that x,y,z is positive, so i havn't took that into account. I think that ruins it more.
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Re: TT's Maths Thread
« Reply #164 on: November 25, 2009, 01:37:22 pm »
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Lagrange multipliers? Calculus is boss!

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« Last Edit: November 25, 2009, 01:40:37 pm by /0 »