Integer value of x?

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The notation means $[x]$ is the least integer not greater than $x$. Thus $[2.9] = 2$ and $[-2.9] = -3$ and $[3.0] = 3$.

This is a bit old fashioned; modern practice is to call this the "floor" function $\lfloor x \rfloor$.

So $[2^x]$ is $1$ until $x=1$, then $2$ until $x=\log_2 3$, then $3$ until $x=\log_2 4 = 2$ and so forth.

The integral becomes a discrete sum; don't forget that the domains of $x$ in which the integrand takes on each successive integer are not all the same length.

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Gentleman_Narwhal
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Gentleman_Narwhal

I was thoroughly disappointed when I realised I couldn't put maths in this box.

Updated on December 02, 2020

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  • Gentleman_Narwhal
    Gentleman_Narwhal almost 3 years

    I got a problem this evening which I might well be able to solve, but am unable to confidently interpret the information it has given me. It goes as follows:

    If $n\in\Bbb{N}$ and $[x]$ is the integer value of $x$, show that

    $$\int_0^n[2^{x}]dx=n2^n-log_2((2^n)!)$$

    I have emboldened the bit that I don't understand. Does it only want me to take integer values for $x$, i.e: what would the graph of $y=[2^x]$ look like?