Use MVT to Prove that $.99^5>= .95$
I realize I should find a function before using MVT. However, the only function I can think about is $f(x)=x^5$, which doesn't work in this case. Any idea how about how to find a proper function?
Asked
Active
Viewed 143 times
1 Answers
2
The trick is to consider $f(x)=a^x$, where $a=0.99$. Then, for the interval $[1,5]$, there exists $y\in(1,5)$ such that $$\frac{a-a^5}{4}=-\frac{f(5)-f(1)}{5-1}=-f'(y)=-a^y\ln a=a^y\ln\frac{1}{a}.$$ Since $a<1$ and $y>1$, we have that $a^y\leq a$. Also, $\ln\frac{1}{a}>0$, therefore $$a-a^5=4a^y\ln\frac{1}{a}\leq 4a\ln\frac{1}{a}=4a\left(\ln100-\ln99\right).$$ For the last term, note that from the mean value theorem for $\ln x$ in $[99,100]$, there exists $z\in(99,100)$ such that $$\ln100-\ln 99=\frac{1}{z}\leq\frac{1}{99}.$$ Therefore $$a-a^5\leq\frac{4a}{99}=\frac{4}{100},$$ which finally shows that $$a^5\geq a-\frac{4}{100}=0.95.$$
detnvvp
- 8,237
-
The step $\ln(1+{1\over99})\leq{1\over 99}$ does not look trivial to me. It is equivalent of showing $(1+{1\over99})^99\leq e$ so it is equivalent of sequence $(1+{1\over n})^n$ is increasing, but that does not really seem trivial to me though. – cr001 Dec 02 '15 at 09:21
-
Ok, I'll add a proof of that. – detnvvp Dec 02 '15 at 09:22
-
Got it. Clean and nice proof. – cr001 Dec 02 '15 at 09:28