Inequalities (1 Viewer)

[YannY]

can someone explain to me how HSC 1997 Q6a iii) is done and the need of the condition it gives you.

 

[Affinity]

helps if you post the question.

You integrate the result from part 2 with limits 0 to y.

the condition >= 0 is to ensure the integral does not flip signs
andthe condition <= 1 is there to allow you to estimate the last bit.

 

[YannY]

Ah, sorry i didn't post it - cause its quite tedious.

But i dont get why you need the estimate the second bit and not the first part with the >=1

 

[Affinity]

integral [0 to y] x^(4n+2) = 1/(4n+3) y^(4n+3)

assuming y <=1 allows you to conclude that it is less than 1/(4n+3)

 

[FreakFace]

i have to agree this is a tough question.

 

[conics2008]

question

The series 1-x^2+x^4-...........+x^4n has 2n+1 terms..

part i) just use the sum of geo series...

ii) i dont even know where to begin...

iii) integrate between y and 0 the whole equaion and you should come to the required position... ( basic look and give knowledge)

iv) it looks like u have to sub some shit in.. prolly a number on iii...

sorry i just have the questions no solutions...

 

[u-borat]

can anyone post the question?

 

[heybashme]

Quick question is proof by contradiction for some inequalities a legitimate way of proving an inequality is true in an exam?

 

[Affinity]

Quick question is proof by contradiction for some inequalities a legitimate way of proving an inequality is true in an exam?

Done correctly -> yes.

In particular, with inequalities.. they usually want you to prove that

LHS < RHS

where LHS and RHS depends on some parameter(s) say x.

so the statement they want you to prove is
for all x, LHS(x) < RHS(x)


to prove it by contradiction, you would first assume that

there exist a particular x, such that

LHS(x) >= RHS(x)

and derive a contradiction from there.. problem is usually this will be quite hard/ more tedious.