Is it true $\left|\dfrac{1}{2^a}-\dfrac{1}{2^b}\right|<b$

Question

Let $0 \leq a <b<1$ be two real numbers

Is it true that $$ \left| \dfrac {1} {2^a} - \dfrac {1} {2^b} \right| <b $$

Answer 1

I think it's true.

We need to prove that $$b2^{a+b}+2^a-2^b>0$$ and since $2^a\geq1$, it's enough to prove that $$b2^b+1-2^b>0$$ or $f(b)>0$, where $$f(b)=b-1+2^{-b}.$$ Indeed, $$f'(b)=1-2^{-b}\ln2=\frac{2^b-\ln2}{2^b}>\frac{1-\ln2}{2^b}>0,$$ which says $$f(b)>f(0)=0$$ and we are done!

Answer 2

For fixed $b$, the maximum of the left side is when $a=0$, so you want:

$$1-\frac{1}{2^b}<b$$

for $b\in(0,1)$. (The absolute values are unnecessary.)

This is equivalent to:

$$2^b\leq \frac{1}{1-b} $$

for $b\in(0,1)$.

This can be seen by Taylor series:

$$2^b=e^{b\log 2}=1+\sum_{k=1}^{\infty}\frac{\log^k 2}{k!}b^k$$

and $$\frac{1}{1-b}=1+\sum_{k=1}^{\infty}b^k$$

But $0<\log 2<1$ so $\frac{\log^k 2}{k!}<\frac{1}{k!}\leq 1$ and thus $$2^b<\frac{1}{1-b}$$

Answer 3

Here is another version $f(x)=\frac{1}{2^x}$. Using MVT, $\exists c\in(a.b)$ s.t. $$\left|f(a)-f(b)\right|=\left|f'(c)\right|\left|a-b\right|$$ Or $$\left|\frac{1}{2^a}-\frac{1}{2^b}\right|=\left|-\frac{1}{2^c} \ln{2}\right|\left|a-b\right|<\left|\frac{1}{2^c} \right|\left|b-a\right|=\frac{1}{2^c} (b-a)\leq b-a<b$$

Answer 4

As others have observed, it's enough to show that $1-{1\over2^b}\lt b$ for $b\gt0$. We'll show a stronger inequality.

Since $b\ge0$, we have

$${1\over 2^{bx}}=e^{-(b\ln2)x}\lt1$$

for $0\lt x$, and thus

$$1\gt\int_0^1e^{-(b\ln 2)x}dx={1\over b\ln2}{1\over2^{bx}}\Big|_1^0={1\over b\ln2}\left(1-{1\over2^b}\right)$$

which, since $0\lt\ln2\lt1$, gives us the stronger inequality

$$1-{1\over2^b}\lt b\ln2\quad\text{for }b\gt0$$