are we allowed to use in exam? (1 Viewer)

[freaking_out]

i want to know a certain answer, i know it might 've been asked a thousand times bfore, but i wanna know if we are allowed to assume/quote the cauchy's am-gm (whatever its called) rule. u know the inequality with the arithmetic mean and geometric mean...

 

[Hotdog1]

I remember calling the HSC help line and they said it depends on the question.

If they obviously wanted you to use another method but instead, you did it this way in one line, then you will get marks deducted. Also when applying any theorem beyond the scope of the syllabus, you should state exactly what it is and how/why you are using it.

 

[MyLuv]

QUOTE]i want to know a certain answer, i know it might 've been asked a thousand times bfore, but i wanna know if we are allowed to assume/quote the cauchy's am-gm (whatever its called) rule. u know the inequality with the arithmetic mean and geometric mean...[/QUOTE]
<--- I think we cant,anyway if U run out of time and couldnt think of any other way then just assume it ( and qoute Cauchy's Inequality )
Another nice Inequality:
(ax+by)^2 <= (a^2+b^2)(x^2+y^2) (called Bunhiacopxki 's in.)
-->this on1 true for n Number like Cauchy but I cant prove it

 

[Affinity]

MyLuv, the one you've posted is the cauchy-schwarz
in general it's

[Sum {i=1 to n} [a(i)]^2 ][Sum {i=1 to n} [b(i)]^2 ]
>=[Sum {i=1 to n} [a(i)b(i)] ]^2

you can prove it by considering the discriminant of the quadratic

Sum {i=1 to n} [a(i)x + b(i)]^2

 

[freaking_out]

Originally posted by Affinity
MyLuv, the one you've posted is the cauchy-schwarz
in general it's

[Sum {i=1 to n} [a(i)]^2 ][Sum {i=1 to n} [b(i)]^2 ]
>=[Sum {i=1 to n} [a(i)b(i)] ]^2

you can prove it by considering the discriminant of the quadratic

Sum {i=1 to n} [a(i)x + b(i)]^2

is the proof very long??

i was thinking, if its short u can just use the theorem by prooving it, might save u time. :S

 

[MyLuv]

Originally posted by Affinity
MyLuv, the one you've posted is the cauchy-schwarz
in general it's

[Sum {i=1 to n} [a(i)]^2 ][Sum {i=1 to n} [b(i)]^2 ]
>=[Sum {i=1 to n} [a(i)b(i)] ]^2

you can prove it by considering the discriminant of the quadratic

Sum {i=1 to n} [a(i)x + b(i)]^2

Err...I always think the Cauchy is this one:
(a1+a2+...+an)>= n* (a1a2..an)^1/n
Btw Affinity,I've tried Ur way,all I got is a discriminant=0,doesnt help me anything,can u show me the proof plz

 

[redslert]

this may help you work it out

QUESTION
a) f(x) = (ax - c)^2 + (bx - d)^2
find its discriminant. Leave in unsimplify form

b) Hence, show that
(a^2 + b^2)(c^2 + d^2) >= (ac + bd)^2

c) using the quadratic function
f(x) = (ax - d)^2 + (bx - e)^2 + (cx - f)^2
prove that
(a^2 + b^2 + c^2)(d^2 + e^2 + f^2) >= (ad + be +cf)^2

isn't that hard....

ANSWER
a) find the discriminant
= 4(ac + bd)^2 - 4(a^2 + b^2)(c^2 + d^2)

b) since (ax - c)^2 >= 0 ; (bx - d)^2 >= 0
.'. f(x) >= 0
.'. discriminant is <= 0
because there are no roots, the graph of f(x) just touches the x-axis

c) using f(x) given
expand
simplify

take discriminant <=0
and you should have it

 

[redslert]

there's actually only two kind of questions they can ask about this....Bunhiacobski (sp?)

the one above is one of them
the other one is quite similar, except it's not using the discriminant but just expand and sub in appropriate substitutions...

 

[ND]

Originally posted by redslert
because there are no roots, the graph of f(x) just touches the x-axis

If f(x) touches the x-axis, then is does have a root.

 

[redslert]

Originally posted by ND
If f(x) touches the x-axis, then is does have a root.

ar my bad
doh

one root..

 

[Affinity]

incidentally that also specifies the equality condition:

a(i)/b(i) = constant.

 

[redslert]

Originally posted by Affinity
incidentally that also specifies the equality condition:

a(i)/b(i) = constant.

i honestly dont get your notations?!

what exactly does that mean?
it's not complex numbers.......what's the 'i'

 

[Affinity]

i= arbitary integer, used traditionally for the dummy variable in sigma notation, also, usual variable for the 'for' loop in programming languages.

a(i) is just like a_i

 

[redslert]

so it's basically

a(1) = a_1

why make things so complicated!?!

your stressing me out!!!!
lol

 

[Affinity]

a_1^2 is confusing

and the subscript takes too long to type

 

[freaking_out]

Originally posted by Affinity
...and the subscript takes too long to type

 

[Harimau]

How do you type subscripts and supercripts anyway?

 

[redslert]

Originally posted by Harimau
How do you type subscripts and supercripts anyway?

Originally posted by McLack



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