Maximise the Absolute

Let $m$ and $n$ be distinct positive integers.

Find the maximum value of $|x^m - x^n|$ , where $x$ is a real number in the interval $(0, 1)$ .

#Algebra

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Math	Appears as
Remember to wrap math in `$` ... `$` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$

Comments

Simplest thing I can think of $|1^1-0^1|$ .... It has to be more complicated than this...

Trevor Arashiro - 6 years, 8 months ago

That doesn't work at all. I think you typoed.

Sharky Kesa - 6 years, 8 months ago

He put (0,1), which means 0 & 1 not included, so it must be something else!!

Vivek Bhagat - 6 years, 8 months ago

Asama Zaldy Jr. - 6 years, 8 months ago

I dont think it will have a numeric value as answer as "x" is variable and there are infinite nos. between 0 and 1...

Sparsh Goyal - 6 years, 8 months ago

Note that it asks for a maximum value.

Sharky Kesa - 6 years, 8 months ago

whats the answer

manish bhargao - 6 years, 8 months ago

In terms of m and n?

Kenny Lau - 6 years, 8 months ago

The answer should be "x" !

Sparsh Goyal - 6 years, 8 months ago

Note that $x^r$ is a strictly decreasing function when $x\in(0,1)$ , and the maximum value would be achieved when $m$ is the least and $n$ is the most, in this case, $m=1$ and $n\to\infty$ , where $|x^m-x^n|=|x^1-0|=x$ .

Note: @Sparsh Goyal posted the numerical answer first.

Kenny Lau - 6 years, 6 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$