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.)
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?
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?
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!!
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.
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 is a table of codes for many different characters, including Greek. For example, typing &# followed by 960 followed by ; will get you π.
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.
