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Pace_T

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Prove that ab(a^2 + b^2) >= 2(a^2)(b^2)

thanx
 

FinalFantasy

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Prove that ab(a^2 + b^2) >= 2(a^2)(b^2)
consider (a-b)²>=0
a²-2ab+b²>=0
a²+b²>=2ab
.: ab(a²+b²)>=2a²b²
 

Pace_T

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ta
The message you have entered is too short. Please lengthen your message to at least 10 characters.
 

who_loves_maths

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okay... well, if you really want a question to pass time acmilan, then,

let the second question be this:

If lim(x->infinity) F(x) = lim(x->infinity) S(r=2, x-2)[(xCr)*x^r] , then find F(x).

where 'S' represents the "Sigma" notation, and S(r=2, x-2) means the sum of the terms in the series starting with r=2 to r=(x-2).

and where 'lim(x->infinity)' obviously means the limit as 'x' approaches infinity...

hope that'll occupy a bit of your time :)

P.S. you still haven't answered my last 3u geometry question that you said you were going to take one day to answer yet, lol. but it's hard, so don't worry about it.

Edit: 'xCr' is the combinations notation... {this question's for you too FinalFantasy.}
 

FinalFantasy

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who_loves_maths said:
Edit: 'xCr' is the combinations notation... {this question's for you too FinalFantasy.}
hahaha i don't need any questions to past time!! i have more den i can handle already and at a glance it doesn't look like i can do ur q. kekeke
 

SeDaTeD

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Hmm, is it (e-2)*x^x -x^2 -1? Although F(x) xould be anything just as long as its limit -> infinity is the same as the RHS.
 
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SeDaTeD

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Rewrote RHS as a binomial thing, subtracted the terms not in the sum, and used lim of x-> inf of (1 + 1/x). The x^2 and 1 term seem a bit redundant though. It would've been a bit neater if your q used S(r=0, x-2).

Edit: Hurlstone Ag HS
 

who_loves_maths

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Rewrote RHS as a binomial thing, subtracted the terms not in the sum, and used lim of x-> inf of (1 + 1/x). The x^2 and 1 term seem a bit redundant though. It would've been a bit neater if your q used S(r=0, x-2).
yea but it's not general/common knowledge for high schoolers to know the limit of (1 + 1/x)^x as 'x' approaches infinity.

and as to the x^2 and 1 term, i think it's still neater to leave them out on the LHS, since i'm a preferrer of mathematical symmetry (RHS) over mathematical "neatness".
 

SeDaTeD

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Hmm being a little nitpicky here. In your q, the RHS obviously approaches infinity as x_> inf. Thus F(x) can be any function where its limit as x->inf is also infinity, eg. F(x) = x, or x^2 or lnx.
Maybe you should write Find F(x) if lim(x->infinity) ( S(r=2, x-2)[(xCr)*x^r] - F(x) } = 0
It's also a bit more conventional to use n instead of x if you're dealing with integers rather than real numbers.
 

who_loves_maths

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Originally Posted by Sedated
Hmm being a little nitpicky here. In your q, the RHS obviously approaches infinity as x_> inf. Thus F(x) can be any function where its limit as x->inf is also infinity, eg. F(x) = x, or x^2 or lnx.
Maybe you should write Find F(x) if lim(x->infinity) ( S(r=2, x-2)[(xCr)*x^r] - F(x) } = 0
It's also a bit more conventional to use n instead of x if you're dealing with integers rather than real numbers.
ah, yes, point taken. thanks for that, i'll make amends next time... i knew that flaw at the time too, except i didn't know how to write it in a way so as to overcome that problem... so thanks for the tip.
 

acmilan

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Hahaha damnit i just saw the question now. The fact that SeDaTeD got it in 30 mins is not surprising :) Oh well i have plenty of questions to keep me occupied until the start of 2nd semester
 

Templar

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Hmm, since SeDaTeD seems to feel like doing some maths, I'll throw in a few questions.

Here's something for warm up. Prove that any convex polyhedron will have at least 2 faces with the same number of sides.
 

Jago

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SeDaTeD said:
Rewrote RHS as a binomial thing, subtracted the terms not in the sum, and used lim of x-> inf of (1 + 1/x). The x^2 and 1 term seem a bit redundant though. It would've been a bit neater if your q used S(r=0, x-2).

Edit: Hurlstone Ag HS
hey hey, nice.

and a parra fan as well!
 

who_loves_maths

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Originally Posted by Templar
Hmm, since SeDaTeD seems to feel like doing some maths, I'll throw in a few questions.

Here's something for warm up. Prove that any convex polyhedron will have at least 2 faces with the same number of sides.
is this question specifically for SeDaTed? or is it for all the 4u ppl out there??? ie. can you do this using 4u techniques??? or does it involve too much uni level maths such as matrices and vectors?
 

acmilan

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who_loves_maths said:
^ well acmilan, there's always that 3u triangle question i posted that you said you were going to solve but have not done so yet... :D
maybe that can occupy a bit more of your time? :uhhuh:
Haha I will get to it. Lately ive been staring at these usyd maths competition questions
 

Templar

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who_loves_maths said:
is this question specifically for SeDaTed? or is it for all the 4u ppl out there??? ie. can you do this using 4u techniques??? or does it involve too much uni level maths such as matrices and vectors?
I believe it's doable with 4u techniques, although I don't really know what exactly they are. But SeDaTeD has a head start because some of his subjects can be applied to get the result really easily.

I don't exactly recommend every 4u person to do it though.
 

who_loves_maths

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Originally Posted by Templar
I believe it's doable with 4u techniques, although I don't really know what exactly they are. But SeDaTeD has a head start because some of his subjects can be applied to get the result really easily.

I don't exactly recommend every 4u person to do it though.
and you're certain that we don't need additional theory??? (eg. what exactly is a convex polyhedron?)

and other things like the Euler Formula: V + F - E = 2 ?
 

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