Tim's Weekly #1

Calculus Level 3

A = ∫ 0 1 f ( x ) d x \large A = \int_{0}^{1} f(x)\ dx

Find A A as defined above, where f ( x ) = { x 2 x ∈ Reals \Transendentals x x ∈ Transendentals 1 x ∈ Algebraics f(x) = \begin{cases} x^{2} & x\in \text{Reals\Transendentals}\\ x & x\in \text{Transendentals} \\ 1 & x\in \text{Algebraics}\\ \end{cases}

Notes

  1. An algebraic number is any complex number (including real numbers) that is a root of a non-zero polynomial.
  2. Reals\Trasendentals mean the set of reals with the set of transcendentals removed from it.


The answer is 0.5.

Timothy Cao by Timothy Cao , 21, Canada

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

1 solution

Jason Carrier
Jan 22, 2019

The set of transendentals is the reals which are not algebraic, so the first and third conditions are the same? It turns out not to matter, since most real numbers are tramsendental. More formally, algebraic numbers are countable, but the real numbers are uncountable, so transendental numbers are also uncountable. We can fully disredard the contribution of the algebraic numbers, in which case our answer is just the integral of x from 0 to 1, or 1/2.

0 Helpful 0 Interesting 0 Brilliant 0 Confused

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...