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#1
11-10-2006, 12:18 PM
 Mayo Speaks! Guest Join Date: Jan 2006 Location: My Old Kentucky Home Posts: 718
Proof that (1+x)^n >= 1+nx?

Prove that (1+x)^n >= 1+nx for x>-1 and n any natural number.

I just got done with a math test*, and I could not for the life of me figure out the above. I was totally stumped. I rewrote it by defining y:=x+1 to obtain y^n >= 1+n(y-1) for y>0.

I then noticed that for the trivial case where n = 1, the statement is y >= 1+1(y-1) which is obviously true.

Also, when y=1, the statement is 1^n >= 1+n(1-1) which is also true.

Aside from these obvious observations, I made no more headway. I assume that a proof by contradiction or a proof by induction is possible, as those two techniques were what the test was about. If anyone out there could give a proof, or an outline of a proof, or some helpful advice, I'd be grateful.

* Meaning that this is not a homework question, and I'm only asking so that I may gain knowledge.
#2
11-10-2006, 12:26 PM
 ftg Guest Join Date: Feb 2001 Location: Not the PNW :-( Posts: 15,757
Since it is impossible to tell the real reason why someone asks such homework-type questions on "Teh Intranets", I only give hints. (One would think this would be SDMB policy but it isn't.)

You didn't happen to be studying the Binomial Theorem did you?
#3
11-10-2006, 12:37 PM
 Thudlow Boink Charter Member Join Date: May 2000 Location: Lincoln, IL Posts: 24,020
Quote:
 Originally Posted by Mayo Speaks! I assume that a proof by contradiction or a proof by induction is possible, as those two techniques were what the test was about.
Ah, this helps. Whenever you have to prove a statement about "any natural number n," consider using induction on n.

Try starting with the assumption that it's true for n and multiply both sides of the inequality by (1+x) (which is positive if x > -1) to get (1+x)n+1 >= ... something which can be shown to be >= 1 + (n+1)x. I'll let you work out the details.
#4
11-10-2006, 12:45 PM
 Giles Charter Member Join Date: Apr 2004 Location: Newcastle NSW Posts: 12,821
The OP is incomplete, because it has to be about the limit as x approaches something. (And those that have responded seem to be assuming that x is approaching 0, which is probably the right assumption).
#5
11-10-2006, 12:48 PM
 Thudlow Boink Charter Member Join Date: May 2000 Location: Lincoln, IL Posts: 24,020
Quote:
 Originally Posted by Giles The OP is incomplete, because it has to be about the limit as x approaches something. (And those that have responded seem to be assuming that x is approaching 0, which is probably the right assumption).
I was assuming that x was any fixed number > -1 (not approaching anything).
#6
11-10-2006, 12:51 PM
 Giles Charter Member Join Date: Apr 2004 Location: Newcastle NSW Posts: 12,821
Quote:
 Originally Posted by Thudlow Boink I was assuming that x was any fixed number > -1 (not approaching anything).
Sorry -- I've reread more carefully, and I've realised that the equation is an inequality, not a limit. Just ignore my babble, and carry on without me.
#7
11-10-2006, 02:24 PM
 C K Dexter Haven Right Hand of the Master Charter Member Join Date: Feb 1999 Location: Chicago north suburb Posts: 16,078
Following Thudlow: Restrict x > -1, so 1+x > 0.
- For n = 1, obviously (1+x)1 = (1+x) = 1 + nx.

- Suppose true for n.
Then (1+x)n+1 = (1+x)n(1+x)
>= (1+nx)(1+x) since we're supposing the statement true for n and since (1+x) > 0
= 1 + nx + x + nx2 by multiplication
>= 1 + nx + x since x2 >= 0
= 1 + (n+1)x QED.

Note that substituting for a multiplicative quantity in inequalities only works when all elements are positive, so we need (1+x) > 0
#8
11-10-2006, 04:30 PM
 ReuvenB Charter Member Join Date: Sep 2001 Location: Portland, OR Posts: 757
You can also prove it using calculus, though I leave it as an exercise to the reader (snerk).
#9
11-10-2006, 04:40 PM
 Giles Charter Member Join Date: Apr 2004 Location: Newcastle NSW Posts: 12,821
Quote:
 Originally Posted by XWalrus2 You can also prove it using calculus, though I leave it as an exercise to the reader (snerk).
Presumably by showing that the derivative exists and is non-negative in the desired region.
#10
11-10-2006, 05:12 PM
 Mayo Speaks! Guest Join Date: Jan 2006 Location: My Old Kentucky Home Posts: 718
Quote:
 Originally Posted by C K Dexter Haven Following Thudlow: Restrict x > -1, so 1+x > 0. - For n = 1, obviously (1+x)1 = (1+x) = 1 + nx. - Suppose true for n. Then (1+x)n+1 = (1+x)n(1+x) >= (1+nx)(1+x) since we're supposing the statement true for n and since (1+x) > 0 = 1 + nx + x + nx2 by multiplication >= 1 + nx + x since x2 >= 0 = 1 + (n+1)x QED. Note that substituting for a multiplicative quantity in inequalities only works when all elements are positive, so we need (1+x) > 0
Of course. Thank you very much. Somehow, I had forgotten that a^(b+c)=(a^b)*(a^c), which was making this much harder. Don't tell my professor that I'm an idiot, please.
#11
10-22-2013, 01:15 PM
 DA6righthand Guest Join Date: Oct 2013 Posts: 1
Quote:
 Originally Posted by C K Dexter Haven Following Thudlow: Restrict x > -1, so 1+x > 0. - For n = 1, obviously (1+x)1 = (1+x) = 1 + nx. - Suppose true for n. Then (1+x)n+1 = (1+x)n(1+x) >= (1+nx)(1+x) since we're supposing the statement true for n and since (1+x) > 0 = 1 + nx + x + nx2 by multiplication >= 1 + nx + x since x2 >= 0 = 1 + (n+1)x QED. Note that substituting for a multiplicative quantity in inequalities only works when all elements are positive, so we need (1+x) > 0
I'm confused as to where the nx2 went in the second to last line of Dexter's proof. I understand x2>=0 but I don't understand why it's omitted in the following lines of the proof. Any help is much appreciated!
#12
10-22-2013, 02:07 PM
 Thudlow Boink Charter Member Join Date: May 2000 Location: Lincoln, IL Posts: 24,020
Quote:
 Originally Posted by DA6righthand I'm confused as to where the nx2 went in the second to last line of Dexter's proof. I understand x2>=0 but I don't understand why it's omitted in the following lines of the proof. Any help is much appreciated!
Those two next-to-last lines, taken together, say
1 + nx + x + nx2 >= 1 + nx + x

Since n and x2 are both nonnegative, the left side consists of the same thing as what's on the right, but with something nonnegative added to it.
#13
10-22-2013, 05:24 PM
 Half Man Half Wit Guest Join Date: Jun 2007 Posts: 6,203
Funnily enough, I just used that example in the exercises yesterday to discuss proof by induction. Otherwise, everything of substance has already been said in this thread, so all that's left for me is to point out that the inequality in the OP is generally known as Bernoulli's inequality.
#14
10-22-2013, 06:07 PM
 Andy L Member Join Date: Oct 2000 Posts: 4,693
This is a very practical question, actually, since it amounts to answering the question of why compound interest is better than simple interest.
#15
10-22-2013, 07:05 PM
 Jaxon Guest Join Date: Aug 2008 Posts: 9
OP....

You have not included

Z^%√7

Later,
Jaxon..
#16
10-22-2013, 08:57 PM
 Indistinguishable Guest Join Date: Apr 2007 Posts: 10,519
Lest anyone not realize, this is a zombie thread. Not that there's any harm in that.
#17
10-22-2013, 09:35 PM
 ShibbOleth Guest Join Date: Jul 2001 Location: Schlaraffenland Posts: 20,611
I wonder if the homework ever got turned in?

(I love zombie threads in October)
#18
10-22-2013, 10:22 PM
 JBDivmstr Guest Join Date: Jun 1999 Location: Texas... Need I say more? Posts: 2,042
Quote:
 Originally Posted by C K Dexter Haven Following Thudlow: Restrict x > -1, so 1+x > 0. - For n = 1, obviously (1+x)1 = (1+x) = 1 + nx. - Suppose true for n. Then (1+x)n+1 = (1+x)n(1+x) >= (1+nx)(1+x) since we're supposing the statement true for n and since (1+x) > 0 = 1 + nx + x + nx2 by multiplication >= 1 + nx + x since x2 >= 0 = 1 + (n+1)x QED. Note that substituting for a multiplicative quantity in inequalities only works when all elements are positive, so we need (1+x) > 0
Auuuughhhh!!! <brain implodes>

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#19
10-23-2013, 12:57 AM
 Indistinguishable Guest Join Date: Apr 2007 Posts: 10,519
For what it's worth, we can also splay out the argument like this:

Suppose n >= 1 and b >= 0. Then (b - 1)2 * [1 + (1 + b) + (1 + b + b2) + ... + (1 + b + b2 + ... + b(n - 2))], being a square times a sum of terms each >= 0, is >= 0 as well.

Expanding this out and cancelling and collecting terms, we get (n - 1) - nb + bn = bn - (1 + n(b - 1)).

Thus, we've shown that bn >= 1 + n(b - 1) for b >= 0, which is to say, (1 + x)n >= 1 + nx for x >= -1, as desired.

[This is essentially the same calculation as powers the inductive argument, just set down differently. (Incidentally, the calculus argument is also just a slight reframing of the same reasoning as in the inductive argument or this one)]

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