harder 3U Q (1 Viewer)

mojako

Active Member
Joined
Mar 27, 2004
Messages
1,333
Gender
Male
HSC
2004
Rorix said:
You inspect it.

hence, 'by inspection'.
ok, teach me how to inspect stuff ^_^

btw my sixth sense suggests that this will be one of those pointless thread
 

Logix

Member
Joined
May 6, 2004
Messages
177
does any1 know how nike got from
-------nlnn - n + 1 > 1/2(0+ln2) + 1/2(ln2+ln3) + .... + 1/2 (ln(n-1) + lnn)
to
-------1/2lnn + nlnn - n + 1 > lnn! (adding 1/2lnn to each side)
????

im confused
 

mojako

Active Member
Joined
Mar 27, 2004
Messages
1,333
Gender
Male
HSC
2004
oh its by inspection!! (again)
:p :p :p :p :p

the inspection method is:
you see on the RHS of the first line that ln2 will be repeated twice, ln3 will be repeated twice etc (but lnn will not be repeated)
or you can also use the full trapezium rule to work that out (the rule which combines all trapeziums into 1 formula).

so now on RHS you have ln2 + ln3 + ln4 + ... + ln(n-1) + 1/2ln(n)
= ln(2*3*4*...*[n-1]) + 1/2ln(n) by the log rule
 
Last edited:

Logix

Member
Joined
May 6, 2004
Messages
177
thx for introducing me to this "inspection method" :)

is trapezia stuff in the course? coz i've hardly ever seen it b4
 

mojako

Active Member
Joined
Mar 27, 2004
Messages
1,333
Gender
Male
HSC
2004
its in the 2U course..
remember when doing simpson's rule and trapezium / trapezia / trapezoid / trapezoidal rule (im not sure which term is the most correct haha)
 

Logix

Member
Joined
May 6, 2004
Messages
177
oh........but thats a totally diff. thing, isnt it?
coz thats used to approximate the area under the curve, whereas this is 2 do with inequality or something
 

mojako

Active Member
Joined
Mar 27, 2004
Messages
1,333
Gender
Male
HSC
2004
its a totally identical thing ^_^

the idea is that the approximation will be less than the exact area.
why? because the curve of ln(x) is concave down and if you try to draw the trapeziums they'll be all under the curve, so the approx area will be less than actual area.

some other Q's use approximation by upper or lower rectangles and compare it to the exact area.

oh btw you also need to memorise Simpson's rule. I have seen it to be necessary to do 1 question on volumes.
 

Logix

Member
Joined
May 6, 2004
Messages
177
ok

so how many methods r used to solve these inequalities? apart from the trapezia.
 

mojako

Active Member
Joined
Mar 27, 2004
Messages
1,333
Gender
Male
HSC
2004
um what do u mean? this particular question or all inequality questions in general?
 

mojako

Active Member
Joined
Mar 27, 2004
Messages
1,333
Gender
Male
HSC
2004
there are many...
some are:
area under curve (trapezia, rectangle, etc)
squares cant be negative (such as when proving LHS-RHS > 0)

also the results that a^2 + b^2 >= 2ab and (a + b)^2 >= 4ab are sometimes used

and swapping variables

also the ideas that inequalities can be added and multiplied.

sample Q (btw, Im asking how to do it):
show that (b + c - a) (c + a - b) <= c^2.
hence show that (b + c - a) (c + a - b) (a + b - c) <= abc


also the use of calculus
sample question: show that cos(x) >= 1 - 1/2 x^2 for all x.
 

Logix

Member
Joined
May 6, 2004
Messages
177
inequalities can be solved by adding and multiplying?
can u give an example?

i am lost in the whole 3 unit topic coz may teacher 'taught' the topic in 1 lesson
 

ngai

Member
Joined
Mar 24, 2004
Messages
223
Gender
Male
HSC
2004
mojako said:
hence show that (b + c - a) (c + a - b) (a + b - c) <= abc
clearly, by inspection :), when a = b = 0, c = -1, the question states that (-1)(-1)(1) <= 0, which is not true, hence the question is wrong
 

Logix

Member
Joined
May 6, 2004
Messages
177
From inspection, this question is in the cambridge book, and its an example, so there is a worked solution :)
 

Logix

Member
Joined
May 6, 2004
Messages
177
theres yet another question in the cambridge book that i cannot do!

ok here it is:

Prove that (lx+my+nz)^2=<(l^2+m^2+n^2)(x^2+y^2+z^2)
Deduce that (a+b+c)^2=<3(a^2+b^2+c^2)
Deduce that (a^3+b^3+c^3)^2=<(a^2+b^2+c^2)(a^4+b^4+c^4)
 

Constip8edSkunk

Joga Bonito
Joined
Apr 15, 2003
Messages
2,398
Location
Maroubra
Gender
Male
HSC
2003
use the inequality a^2+b^2 >= 2ab several times to prove the expansions of the 1st 1(eg. 2lxmy <= l^2y^2 + m^2x^2)

or 2nd q, add up inequalities a^2+b^2 >= 2ab, a^2+c^2 >= 2ac, b^2+c^2 >= 2bc , then add a^2 +b^2 +c^2 to both sides, factorise

again use a^2+b^2 >= 2ab for q3, multiply both sides by a^2b^2 to get 2a^3b^3 <= a^2b^4 +a^4b^2... repeat for a&c, and b&c.... add up the 3 inequalities, add a^6 +b^6 +c^6, factorize
 

mojako

Active Member
Joined
Mar 27, 2004
Messages
1,333
Gender
Male
HSC
2004
ngai said:
clearly, by inspection :), when a = b = 0, c = -1, the question states that (-1)(-1)(1) <= 0, which is not true, hence the question is wrong
Clearly, by inspection, I forgot to mention that a,b,c must all be positive :)
 
Last edited:

Users Who Are Viewing This Thread (Users: 0, Guests: 1)

Top