I have been trying to put $0.\lceil{x}$ in wolfram alpha, but it keeps thinking that I am trying to type $0*\lceil{x}$. I've tried experimenting with the idea that when $0<x\le10$, I can write it as $\frac{\lceil{x}}{10}$, and when $10<x\le100$, I can write that as $\frac{\lceil{x}}{100}$. But to do that, I need a function that turns $x$ into its base, such that the base of 97 is 10, the base of 193 is 100, etc. Then I need to write $\frac{\lceil{x}}{base}$, and then that will be in a notation wolfram alpha will understand. Is there a way to get the base of any number $x$ easily? Also keep in mind that currently my highest math class is algebra 2, so if possible could you explain it in less advanced terms?
Rewrite 0.ceil(x) in a way that wolfram alpha will accept it
ceiling-and-floor-functionswolfram alpha
Related Solutions
You might be able use boole, which returns $0$ or $1$ depending on whether it's argument is True or False. Thus, for $n=10$, Euler's phi function $\varphi(n)$ (which counts the number of positive integers less than and relatively prime to $n$) can be expressed as
sum boole(gcd(i,10)=1), i=1..10
Of course, it's hard to say for sure, since you don't mention your specific application. This method of computing $\varphi(n)$, however, happens to be Wikipedia's first example application of Iverson brackets.
Yes, you can simply include those constraints: minimize ax^2+bx+c, a>0, b>0, c>0
Alternatively, you can complete the square and observe the extreme value occurs when $x=-\frac b{2a}$: complete the square ax^2+bx+c
Best Answer
Disregard my answer below.
Type StringJoin["0", "x"] into Mathematica.
$$f(x) = \frac{\lceil x \rceil}{10^{\lfloor \log_{10} x \rfloor + 1}}, 0.1 ≤ x; \ f(x) = 0.1, 0 < x < 0.1$$
works.
As you were saying, $0$ point $x$ is $\lceil x \rceil$ divided by a suitable power of $10$. So a first attempt might be $10^{\lceil \log_{10} x \rceil}$, which shifts the decimal point left (making it smaller) by the number of digits of $x$. For example, $102 > 10^2$, but $102$ has $3$ digits, not $2$, so $\lceil \log_{10} 102 \rceil = 3$, and we divide by $10^3$ to get $0.102$. It works!
But try this out for say, $x = 100$, and $\lceil \log_{10} 100 \rceil = 2$, so we end up with $\frac{100}{100^2} = 1$. To fix this problem, we adjust the denominator $10^{\lfloor \log_{10} x \rfloor + 1}$, which has the same effect for all other numbers.
And there are other problems when $0 < x < 0.1$, as now $\log_{10} x ≤ -2$, so the denominator is less than $1$ and we end up multiplying by powers of $10$. However, the ceiling of these values of $x$ is always $1$, so the result is $0.1$.
There you have it: that's how we got the function above.