# Proving a function is larger than another.

• Sep 10th 2009, 11:50 AM
don.bertone
Proving a function is larger than another.
let 0>i>1

prove that (1+i)^t < 1+i*t for 0>t>1.

I am a little bit clueless where to start. At first I thought about integrating the area in between and if that area was positive then it would be true, if negative then it would be false and that would basically prove it. But it turned out to be a lot more complicated than it first seemed which led me to believe there is probably a much better way. Any help would be appreciated.
• Sep 10th 2009, 12:58 PM
Opalg
Quote:

Originally Posted by don.bertone
let 0<x<1

prove that (1+x)^t < 1+x*t for 0<t<1.

I have tried to make sense of the question by reversing four of the inequality signs. I have also replaced i by x, partly because i tends to mean the square root of –1 and partly because it's easier to prove the result if you think of x as a variable and t as a constant.

The Mean Value theorem (with remainder) for the function $f(x) = (1+x)^t$ tells you that $(1+x)^t = 1+tx+\tfrac12t(t-1)(1+y)^{t-2}$ for some y between 0 and x. Since $t-1<0$, it follows that $(1+x)^t < 1+tx$.

(If you don't know the Mean Value theorem, we'll have to think of another approach.)
• Sep 10th 2009, 02:13 PM
don.bertone
I did make a mistake and your sign reversal is correct. I do know the Mean Value Theorem roughly but I must admit that I have no clue what you did there.

Just as a side note, this is not a calculus class it's my actuarial class, so the question might have a non calculus way of being solved(i doubt that though).

If you could give a little bit more explanation as to how you got:

$

(1+x)^t = 1+tx+\tfrac12t(t-1)(1+y)^{t-2}
$

I would really appreciate that.
• Sep 11th 2009, 06:02 AM
Opalg
Quote:

Originally Posted by don.bertone
If you could give a little bit more explanation as to how you got:

$

(1+x)^t = 1+tx+\tfrac12{\color{red}x^2}t(t-1)(1+y)^{t-2}
$

I would really appreciate that.

The particular case of Taylor's theorem that I was quoting says that if f is a twice differentiable function then $f(x) = f(0) + xf'(0) + R_2(x)$, where $R_2(x)$ ( the remainder after the first 2 terms of the Taylor series) is equal to $\tfrac12x^2f''(y)$ for some y lying between 0 and x. (And I was quoting that result wrongly, because I left out the $x^2$ that I have inserted in red above.)
• Sep 16th 2009, 07:09 PM
Volcanicrain
Is there a simpler way to solve this question?
Using only derivatives? The Mean Value theorem is above me presently.
• Sep 17th 2009, 03:34 AM
Opalg
Quote:

Originally Posted by Volcanicrain
Is there a simpler way to solve this question?
Using only derivatives? The Mean Value theorem is above me presently.

We want to show that $(1+x)^t<1+tx$ when x and t both lie between 0 and 1.

Fix t, with 0 < t < 1, and let $f(x) = 1+tx - (1+x)^t$. Then $f(0)=0$. If we can show that f is an increasing function, it will follow that f(x)>0 whenever x>0, which is what we want to prove.

So differentiate f(x), to get $f'(x) = t - t(1+x)^{t-1}$. Notice that $(1+x)^{t-1} = \frac1{(1+x)^{1-t}}$. But 1+x>1, and 1–t>0. It follows that $(1+x)^{1-t}>1$, so that $(1+x)^{t-1}<1$. That tells us that $f'(x)>0$, so that f is an increasing function, as required.