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  #1  
Old 11-10-2006, 11:18 AM
Mayo Speaks! Mayo Speaks! is offline
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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.
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  #2  
Old 11-10-2006, 11:26 AM
ftg ftg is offline
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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?
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  #3  
Old 11-10-2006, 11:37 AM
Thudlow Boink Thudlow Boink is offline
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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.
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  #4  
Old 11-10-2006, 11:45 AM
Giles Giles is online now
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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).
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  #5  
Old 11-10-2006, 11:48 AM
Thudlow Boink Thudlow Boink is offline
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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).
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  #6  
Old 11-10-2006, 11:51 AM
Giles Giles is online now
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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.
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  #7  
Old 11-10-2006, 01:24 PM
C K Dexter Haven C K Dexter Haven is offline
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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
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  #8  
Old 11-10-2006, 03:30 PM
ReuvenB ReuvenB is offline
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You can also prove it using calculus, though I leave it as an exercise to the reader (snerk).
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  #9  
Old 11-10-2006, 03:40 PM
Giles Giles is online now
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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.
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  #10  
Old 11-10-2006, 04:12 PM
Mayo Speaks! Mayo Speaks! is offline
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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.
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  #11  
Old 10-22-2013, 12:15 PM
DA6righthand DA6righthand is offline
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Quote:
Originally Posted by C K Dexter Haven View Post
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!
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  #12  
Old 10-22-2013, 01:07 PM
Thudlow Boink Thudlow Boink is offline
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Quote:
Originally Posted by DA6righthand View Post
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.
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  #13  
Old 10-22-2013, 04:24 PM
Half Man Half Wit Half Man Half Wit is online now
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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.
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  #14  
Old 10-22-2013, 05:07 PM
Andy L Andy L is offline
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This is a very practical question, actually, since it amounts to answering the question of why compound interest is better than simple interest.
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  #15  
Old 10-22-2013, 06:05 PM
Jaxon Jaxon is offline
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OP....

You have not included

Z^%√7

Later,
Jaxon..
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  #16  
Old 10-22-2013, 07:57 PM
Indistinguishable Indistinguishable is online now
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Lest anyone not realize, this is a zombie thread. Not that there's any harm in that.
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  #17  
Old 10-22-2013, 08:35 PM
ShibbOleth ShibbOleth is offline
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I wonder if the homework ever got turned in?

(I love zombie threads in October)
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  #18  
Old 10-22-2013, 09:22 PM
JBDivmstr JBDivmstr is offline
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Quote:
Originally Posted by C K Dexter Haven View Post
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>

Note to self... Do.Not.Read.'Math'.Threads.
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  #19  
Old 10-22-2013, 11:57 PM
Indistinguishable Indistinguishable is online now
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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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