# How can this be proven? (maths)

**URL:** <https://boards.straightdope.com/t/how-can-this-be-proven-maths/292505>\
**Category:** Factual Questions\
**Created:** [March 1, 2005, 11:06pm UTC](https://boards.straightdope.com/t/how-can-this-be-proven-maths/292505 "2005-03-01T23:06:42Z")\
**Posts on this page:** 20\
**Page:** 1

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**Author:** ![ReuvenB](https://avatars.discourse-cdn.com/v4/letter/r/9fc348/32.png) [@ReuvenB](https://boards.straightdope.com/u/ReuvenB)\
**Post date:** [March 1, 2005, 11:06pm UTC](https://boards.straightdope.com/t/how-can-this-be-proven-maths/292505/1 "2005-03-01T23:06:42Z")

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I was having a conversation with my maths teacher, and he mentioned in passing a maths problem he had in his higher level class that he wasn’t able to solve. I asked him about it, and he gave it to me. Now, granted, I knew it was pretty high above my level, but I thought maybe I might see something he missed. Unfortunately, I didn’t see any way to do it. So I thought I might turn to the dopers for assistance. Here’s the problem:

Prove  
(1 + x)[sup]k[/sup] =\< 1 + kx

given:  
0\< k =\< 1  
x \> 0

(note: I don’t know how to make the sign “less than or equal to”, so I’ve used =\<. I have no idea if this is proper notation.)

If any of you math dopers could help out, he’d be most appreciative. As for me, I’m lost.

(note #2: I know this sounds suspicious, but I swear upon my honor as a doper that this is NOT a homework problem.)

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**Author:** ![Cunctator](https://avatars.discourse-cdn.com/v4/letter/c/43a26b/32.png) [@Cunctator](https://boards.straightdope.com/u/Cunctator)\
**Post date:** [March 1, 2005, 11:20pm UTC](https://boards.straightdope.com/t/how-can-this-be-proven-maths/292505/2 "2005-03-01T23:20:57Z")

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It looks like the old example of simple interest being better than compound interest for periods of less than a year. Could it not be demonstrated graphically?

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**Author:** ![Omphaloskeptic](https://avatars.discourse-cdn.com/v4/letter/o/bcef8e/32.png) [@Omphaloskeptic](https://boards.straightdope.com/u/Omphaloskeptic)\
**Post date:** [March 1, 2005, 11:36pm UTC](https://boards.straightdope.com/t/how-can-this-be-proven-maths/292505/3 "2005-03-01T23:36:10Z")

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Can you prove that 1+kx =\< (1+x)[sup]k[/sup] for x\>0 and k\>=1? (I don’t know exactly what level of math you’re using so I don’t know how hard this would be.) If so, can you see a way to use this fact?

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**Author:** ![ultrafilter](https://avatars.discourse-cdn.com/v4/letter/u/3d9bf3/32.png) [@ultrafilter](https://boards.straightdope.com/u/ultrafilter)\
**Post date:** [March 1, 2005, 11:40pm UTC](https://boards.straightdope.com/t/how-can-this-be-proven-maths/292505/4 "2005-03-01T23:40:12Z")

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Hmm…the binomial theorem is your friend.

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**Author:** ![ultrafilter](https://avatars.discourse-cdn.com/v4/letter/u/3d9bf3/32.png) [@ultrafilter](https://boards.straightdope.com/u/ultrafilter)\
**Post date:** [March 1, 2005, 11:42pm UTC](https://boards.straightdope.com/t/how-can-this-be-proven-maths/292505/5 "2005-03-01T23:42:59Z")

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> [@Cunctator](#):
>
> It looks like the old example of simple interest being better than compound interest for periods of less than a year. Could it not be demonstrated graphically?

No, because that’s not a proof, unless you graph it for all values of x. Since the range is infinite, you’ll need a lot of paper.

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**Author:** ![Pedro](https://avatars.discourse-cdn.com/v4/letter/p/919ad9/32.png) [@Pedro](https://boards.straightdope.com/u/Pedro)\
**Post date:** [March 1, 2005, 11:56pm UTC](https://boards.straightdope.com/t/how-can-this-be-proven-maths/292505/6 "2005-03-01T23:56:32Z")

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> [@ultrafilter](#):
>
> Hmm…the binomial theorem is your friend.

You mean the [binomial series](http://en.wikipedia.org/wiki/Binomial_series) right?

But then it diverges for |x|\>1.

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**Author:** ![Fat\_Marrow](https://avatars.discourse-cdn.com/v4/letter/f/ebca7d/32.png) [@Fat\_Marrow](https://boards.straightdope.com/u/Fat_Marrow)\
**Post date:** [March 2, 2005, 12:24am UTC](https://boards.straightdope.com/t/how-can-this-be-proven-maths/292505/7 "2005-03-02T00:24:02Z")

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Please bear with me on this one, I’ve not attempted maths like this in 7 years but I think I can explain. I’ve used n instead of k as I couldn’t find how to superscript k!!!

Prove that:

(1+x)ⁿ ≤ 1+nx where x\>0 and 0\<n≤1

For n=1: Anything to the power of 1 is the number being raised. Likewise, any number mupltiplied by 1 is itself.

e.g. for x=3 LHS: (1+3)^1 = 4 RHS: 1+(1\*4) = 5 Q.E.D

For values of n\<1, xⁿ = 1/n√x  
Now I can’t use the correct notation. What the above means is that xⁿ = (the reciprocal of n root) x. i.e. if n=2 then square root x. if n=3 then cube root x etc.  
Therefore any value of 0\<n\<1, (1+x)ⁿ will be less than 1+x.

Finding the way to prove this part is more tricky, I though I had it but I don’t. I think I’ve proved n=1 part ok. Apologies for the anticlimax.

You realise I’m now gonna be awake for the rest of the night trying to solve this.  
Cheers!!

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**Author:** ![Xema](https://avatars.discourse-cdn.com/v4/letter/x/9de053/32.png) [@Xema](https://boards.straightdope.com/u/Xema)\
**Post date:** [March 2, 2005, 12:35am UTC](https://boards.straightdope.com/t/how-can-this-be-proven-maths/292505/8 "2005-03-02T00:35:59Z")

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> [@XWalrus2](#):
>
> (note: I don’t know how to make the sign “less than or equal to”, so I’ve used =\<. I have no idea if this is proper notation.)

It’s often given as “\<=”, but your version is comprehensible.

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**Author:** ![gregongie](https://avatars.discourse-cdn.com/v4/letter/g/919ad9/32.png) [@gregongie](https://boards.straightdope.com/u/gregongie)\
**Post date:** [March 2, 2005, 12:47am UTC](https://boards.straightdope.com/t/how-can-this-be-proven-maths/292505/9 "2005-03-02T00:47:22Z")

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I would take **ultrafilter’s** advice. This website should help:

[http://www.krysstal.com/binomial.html](http://www.krysstal.com/binomial.html)

Scroll down to “The Binomial Theorem”

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**Author:** ![ultrafilter](https://avatars.discourse-cdn.com/v4/letter/u/3d9bf3/32.png) [@ultrafilter](https://boards.straightdope.com/u/ultrafilter)\
**Post date:** [March 2, 2005, 12:52am UTC](https://boards.straightdope.com/t/how-can-this-be-proven-maths/292505/10 "2005-03-02T00:52:43Z")

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> [@Pedro](#):
>
> You mean the [binomial series](http://en.wikipedia.org/wiki/Binomial_series) right?
> 
> But then it diverges for |x|\>1.

Bah, details!

Seriously, that is a problem.

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**Author:** ![silverfish](https://avatars.discourse-cdn.com/v4/letter/s/e480ec/32.png) [@silverfish](https://boards.straightdope.com/u/silverfish)\
**Post date:** [March 2, 2005, 1:02am UTC](https://boards.straightdope.com/t/how-can-this-be-proven-maths/292505/11 "2005-03-02T01:02:25Z")

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I think I have a method. The idea is that we fix k to be one of the allowable ones, then vary x. This assumes you know about differentiation.

We wish to prove (1+x)[sup]k[/sup] \<= 1+ k\*x

First, we can see what happens when x = 0. Then  
LHS = (1+x)[sup]k[/sup] = (1+0)[sup]k[/sup] = 1[sup]k[/sup] = 1  
RHS = 1+ k\*0 = 1

So LHS = RHS. Now we will show that the LHS increases more slowly than the RHS, as x increases, by taking derivatives, with respect to x.

The derivative of the LHS is k(1+x)[sup]k-1[/sup]  
The derivative of the RHS is k

(1+x) \>= 1 for x \>= 0, so as k - 1 \<= 0, (1+x)[sup]k-1[/sup] \<= 1, the derivative of the LHS is less than the derivative of the RHS, for x \>= 0.

To sum up, when x = 0, both sides are equal and for positive x (x \>= 0), the derivative of the LHS is less than that of the RHS, so the RHS increases faster. so the RHS cannot be any lower than the LHS (for x \> 0), as they start at the same point.

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**Author:** ![Thudlow\_Boink](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/thudlow_boink/32/320_2.png) [@Thudlow\_Boink](https://boards.straightdope.com/u/Thudlow_Boink)\
**Post date:** [March 2, 2005, 4:18am UTC](https://boards.straightdope.com/t/how-can-this-be-proven-maths/292505/12 "2005-03-02T04:18:20Z")

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That looks good to me, **silverfish**.

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**Author:** ![moes\_lotion](https://avatars.discourse-cdn.com/v4/letter/m/7ba0ec/32.png) [@moes\_lotion](https://boards.straightdope.com/u/moes_lotion)\
**Post date:** [March 2, 2005, 5:30am UTC](https://boards.straightdope.com/t/how-can-this-be-proven-maths/292505/13 "2005-03-02T05:30:12Z")

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**silverfish** beat me to it, but his/hers is the approach I would have used with the following slight variation.

Consider the function f(x) = 1 + kx - (1 + x)[sup]k[/sup].

WIth this, the original question becomes show that for x \> 0 and 0 \< k \<= 1, f(x) \>= 0

Taking the derivative with respect to x gives f’(x) = k(1 - (1+ x)[sup]k-1[/sup])

Now, for 0 \< k \<=1, k - 1 \<= 0 and so (1 + x)[sup]k-1[/sup] \<= 1, thus f’(x) \> 0

Since f(0) = 0 and for x \> 0, 0 \< k \<=1, f’(x) \> 0 the function is strictly increasing and thus the inequality is proven.

This reminds me of a similar question from my dim, dark past… which is larger, e to the power pi or pi to the power e?

_(as an aside, how does one code to get the greek letter pi?)_

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**Author:** ![moes\_lotion](https://avatars.discourse-cdn.com/v4/letter/m/7ba0ec/32.png) [@moes\_lotion](https://boards.straightdope.com/u/moes_lotion)\
**Post date:** [March 2, 2005, 5:50am UTC](https://boards.straightdope.com/t/how-can-this-be-proven-maths/292505/14 "2005-03-02T05:50:18Z")

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Arrghhhh… forgot to account for the boundary condition, k = 1

For k = 1,  
f(x) = 1 + x - (1 + x)[sup]1[/sup] = 0

Which also satisfies the inequality.

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**Author:** ![Malacandra](https://avatars.discourse-cdn.com/v4/letter/m/45deac/32.png) [@Malacandra](https://boards.straightdope.com/u/Malacandra)\
**Post date:** [March 2, 2005, 8:58am UTC](https://boards.straightdope.com/t/how-can-this-be-proven-maths/292505/15 "2005-03-02T08:58:43Z")

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> [@moes lotion](#):
>
> _(as an aside, how does one code to get the greek letter pi?)_

Easy as p 🙂

Use \<font=symbol\>p\<font\> but with square brackets

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**Author:** ![scm1001](https://avatars.discourse-cdn.com/v4/letter/s/c4cdca/32.png) [@scm1001](https://boards.straightdope.com/u/scm1001)\
**Post date:** [March 2, 2005, 10:01am UTC](https://boards.straightdope.com/t/how-can-this-be-proven-maths/292505/16 "2005-03-02T10:01:17Z")

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> [@Malacandra](#):
>
> Easy as p 🙂
> 
> Use \<font=symbol\>p\<font\> but with square brackets

on my computer I get a funny exagerated serif font, not greek symbols. Is this because I dont have the right fonts installed (I have symbol in word, so it exist somewhere on this computer)

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**Author:** ![scm1001](https://avatars.discourse-cdn.com/v4/letter/s/c4cdca/32.png) [@scm1001](https://boards.straightdope.com/u/scm1001)\
**Post date:** [March 2, 2005, 10:05am UTC](https://boards.straightdope.com/t/how-can-this-be-proven-maths/292505/17 "2005-03-02T10:05:00Z")

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sorry it works on IE, but not on Opera. still a few bugs in Opera I guess

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**Author:** ![Tyrrell\_McAllister](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/tyrrell_mcallister/32/16772_2.png) [@Tyrrell\_McAllister](https://boards.straightdope.com/u/Tyrrell_McAllister)\
**Post date:** [March 2, 2005, 2:46pm UTC](https://boards.straightdope.com/t/how-can-this-be-proven-maths/292505/18 "2005-03-02T14:46:49Z")

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> [@scm1001](#):
>
> on my computer I get a funny exagerated serif font, not greek symbols. Is this because I dont have the right fonts installed (I have symbol in word, so it exist somewhere on this computer)

If you are willing to go to a little trouble, you can use unicode to get Greek characters. [Here](http://www.alanwood.net/demos/ent4_frame.html) is a table of codes for many different characters, including Greek. For example, typing &# followed by 960 followed by ; will get you π.

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**Author:** ![glee](https://avatars.discourse-cdn.com/v4/letter/g/b5a626/32.png) [@glee](https://boards.straightdope.com/u/glee)\
**Post date:** [March 2, 2005, 3:14pm UTC](https://boards.straightdope.com/t/how-can-this-be-proven-maths/292505/19 "2005-03-02T15:14:58Z")

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When come back, bring p :eek:

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**Author:** ![Chronos](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/chronos/32/134_2.png) [@Chronos](https://boards.straightdope.com/u/Chronos)\
**Post date:** [March 2, 2005, 7:40pm UTC](https://boards.straightdope.com/t/how-can-this-be-proven-maths/292505/20 "2005-03-02T19:40:30Z")

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There is no method for producing Greek letters which will work on all browsers. The various Mozillas don’t recognize Symbol font[sup]\*[/sup], and older browsers (which, believe it or not, some folks still use) won’t recognize Unicode or escape sequences for Greek letters.

\*This is actually deliberate, since the Symbol font is not officially listed in the HTML standards. Sometimes, I think that standards compliance can go too far.

Back on topic, it’s clear that **Silverfish** et al.'s method works, but it does seem like it should be possible without differentiation.

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