yet another induction question (1 Viewer)

[kpq_sniper017]

Q15 from fitzpatrick:

Prove:
(1 + x)^n >= 1 + nx, x>0

i did the question, but i just want to make sure my proof was valid (or if theres a quicker, easier or better way of proving it).

ill skip step 1 (since u can assume it will true for n=1 anyway)


STEP 2:
Assume S(k) is true
.'. (1+x)^k >= 1 + kx, x>0
Consider S(k+1):
To prove: (1+x)^(k+1) >= 1 + (k+1)x

LHS = (1+x)^k.(1+x)
>= (1+kx)(1+x)
= 1 + x + kx + kx^2
= 1 + (k+1)x + kx^2

And then, since kx^2 is always > 0 (x^2 > 0 and k > 0):

(1+x )^k.(1+x) >= 1 + (k+1)x
.'. (1+x)^(k+1) >= 1 + (k+1)X, If S(k) is true

.'. S(k) => S(k+1)

BTW. My teacher told us to use => (implication symbol) - do teacher's recognise this symbol, or do you have to say what it means? i.e. => denotes "implies the truth of".

Anyways, I hope my proof is correct. The only thing I think I might have gone wrong in was the final step in my proof. What do you guys think?

 

[Affinity]

looks alright

 

[CM_Tutor]

This can be done in a much much much easier way without induction. The proof of this result should be two lines long. If they ever say to prove this 'by induction or otherwise', go otherwise.

 

[nike33]

Originally posted by CM_Tutor
This can be done in a much much much easier way without induction. The proof of this result should be two lines long. If they ever say to prove this 'by induction or otherwise', go otherwise.

can you post it please?

 

[CM_Tutor]

Expanding (1 + x)ⁿ using the binomial theorem, we get (1 + x)ⁿ = 1 + ⁿC₁x + ⁿC₂x² + ... + ⁿC_nxⁿ.
Now, all terms ⁿC_kx^k > 0 for k = 1, 2, 3, ..., n, as x > 0, and ⁿC₁ = n, and so (1 + x)ⁿ => 1 + nx

 

[kimmeh]

is that valid in an exam though ?

 

[Grey Council]

i'm quite sure it is. It's the definition of binomial theorem, i don't see why we wouldn't be able to expand it.

i think that we even have to memorise that formula thing.

 

[CM_Tutor]

Originally posted by kimmeh
is that valid in an exam though ?

Yes it is, as binomial theorem is in the Extn 1 syllabus - many people won't have done it yet, but that's another story...

Note, my answer is provided the question allows this solution. If it said 'Prove that, for all x > 0, (1 + x)ⁿ => 1 + nx, for integers n > 0', then mine is a valid (and the best) solution. If it said 'Prove by induction on n, a positive integer, that (1 + x)ⁿ => 1 + nx for x > 0', then it would not be a valid solution, as I haven't used induction, and you'd need to write something along the lines of pcx_demolition017's solution. If it started 'Prove by induction, or otherwise' then my solution is again the easiest way to go.

 

[kpq_sniper017]

so i take it my solution is correct?

i find inequality inductions to be the most difficult.....maybe its just me, but i find i have to kind of twist it around a bit (that's the hardest bit).

do any of u guys have any tips for solving inequality inductions? ive done some inequality proofs before (and i think im right), but it just seems like the proof isnt "valid enough" - probably is, but i find that with ones like the one above, it doesn't seem like enough.......probably just me

 

[CM_Tutor]

Your solution would be fine in an exam.

 

[Grey Council]

yeah, i have a tip.

do heaps of inequality induction questions.