Induction - inequalitage (1 Viewer)

[illin]

Induction - inequalities

Examples:

Show that n ! > 2n for n > 4.
Show that 3n>= 1+2n

What is the best approach when doing these types of questions?
like LHS - RHS > 0, that sort of stuff?

when u get to prove for (k+1)th term, how do u use ur assumption properly to prove the statement.

like for 1.

1.Show true for 1st term

LHS=5! RHS=2^5

LHS>RHS
Therefore statement is true for 1st term

2. Assume true for kth term
k!>2^k

3. Show true for (k+1)th term

(k+1)!>2^(k+1)

k!(k+1)>2.2^k

k!(k+1) - 2.2^k>0
LHS= ...
is this how u do it?
what aboutfor >=? (greater than or equal to)?

cheers

 

[acmilan]

It should really work both ways, either if you move everything to one side or keep them on opposite sides, its up to you

 

[Slidey]

Yeah, LHS>RHS is the same as LHS-RHS>0

 

[withoutaface]

Working one side only is more advisable because it's easier to follow and HSC markers generally prefer it. ie LHS=stuff>stuff=...=RHS

 

[dawso]

LHS=stuff>stuff=...=RHS

gotta love that great mathematical jargon we use here....lol

 

[Slidey]

Working one side only is more advisable because it's easier to follow and HSC markers generally prefer it. ie LHS=stuff>stuff=...=RHS

That's true but for questions such as:

Prove:
(a+b)/2 >= sqrt(ab)

It's far easier to do something like:
LHS-RHS
(a+b)/2-sqrt(ab)
(a-2sqrt(ab)+b)/2
(sqrta-sqrtb)^2/2 >=0
Q.E.D.

 

[ngai]

Examples:

Show that n ! > 2n for n > 4.
Show that 3n>= 1+2n

dont we just hate using induction for these kind of questions?
what a waste of ink and paper

 

[illin]

Could someone help me out and do like the first question up there?

 

[KFunk]

Show that 3n>= 1+2n

Call me wrong, but isn't that just trying to show that n ≥ 1 (after subtracting 2n from both sides) given the condition that n is an integer where n is ≥ 1 ? It seems as though it's proven from the outset.

 

[LaCe]

But the thing is that we will get questions like this, and we would have to prove it by induction.
I think you may have to use the condition of k is ≥ 1 when u are trying to prove that
3(k+1)≥1+2(k+1)
But how would u approach it? how do u use ur assumption of
3k≥1+2k??

 

[withoutaface]

But the thing is that we will get questions like this, and we would have to prove it by induction.
I think you may have to use the condition of k is ≥ 1 when u are trying to prove that
3(k+1)≥1+2(k+1)
But how would u approach it? how do u use ur assumption of
3k≥1+2k??

3(k+1)=3k+3>2k+3 (since k>0)

QED

 

[illin]

And that's is it?

but dont u have to use ur assumption that
3k≥1+2k??

 

[Slidey]

You basically are.

If 3k≥1+2k, then 3k+3>=4+2k>=3+2k

 

[withoutaface]

In lace's post the RHS= 3+2k, I don't get what assumption we're making

 

[LaCe]

the assumption that we make
3k≥1+2k

and then we have to prove for (k+1)th term, which is
3(k+1)3≥1+2(k+1)

[ie.(RHS= 3+2k)]
How do we use the assumption?