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 June 16th, 2017, 03:37 PM #1 Newbie   Joined: Oct 2015 From: Greece Posts: 24 Thanks: 0 Custom function help! Hello! Say I have a number x. I want to make a function which will take this number and add some value to it in order to make it power of two (the closest value that is a power of two). f(x) --> (power of two) x can be any integer number. For example, if I have x = 1, then I must add to it 1 in order to make it power of two because (1+1 = 2). If x = 9, I must add 7 (9+7 = 16). I tried, but I can't think of any mathematical way of doing it. So my point is, how can I find which number I must add to x in order to create a new number that is gonna be power of two? Thank you! Last edited by skipjack; June 17th, 2017 at 03:18 AM.
 June 16th, 2017, 03:52 PM #2 Senior Member   Joined: Aug 2012 Posts: 1,620 Thanks: 411 How can we find the next largest power of $2$? Given $x$, its exact power of $2$ is $\log_2 x$ and the next integer power of $2$ is $2^{\lceil \log_2 x \rceil}$ where $\lceil \cdot \rceil$ is the ceiling function, by definition the integer greater than or equal to a given real number. Then the difference function is $2^{\lceil \log_2 x \rceil} - x$. For example with $x = 9$ we have: * $\log_2 9 \approx 3.1699 \dots$ [Wolfram Alpha] * $\lceil 3.1699 \dots \rceil = 4$ * $2^4 = 16$ * $16 - 9 = 7$ This method works for any positive real number, not just positive integers. However we have problems if $x$ is negative. For example what's the smallest power of $2$ greater than or equal to $-5$? Is it $\frac{1}{2}$, or $\frac{1}{4}$, or $\frac{1}{8}$? There is no "next" power of $2$ greater than $-5$. I'm too lazy at the moment to sort this out in terms of complex logs but if we stick to real numbers, then $x$ has to be strictly positive. Last edited by Maschke; June 16th, 2017 at 04:13 PM.
 June 16th, 2017, 04:08 PM #3 Math Team     Joined: Jul 2013 From: काठमाडौं, नेपाल Posts: 872 Thanks: 60 Math Focus: सामान्य गणित $\displaystyle f(x)=\left \lceil \sqrt{x} \right \rceil^{2}$ for a positive number 'x' if you take the square root of 'x' and again square it will give the same value. If you square a value less than the square root of 'x' it will give lesser value than 'x' and if you square a value greater than the square root of 'x' it will give greater value than 'x' so, if you square a whole number just greater than the square root of 'x' you will get the desired value. for example: if $\displaystyle x = 98$ $\displaystyle f(x)=\left \lceil \sqrt{98} \right \rceil^{2}$ $\displaystyle f(x)=\left \lceil 9.8994949.......\right \rceil^{2}$ $\displaystyle f(x)=10^{2}$ $\displaystyle \therefore f(x)=100$
 June 16th, 2017, 04:11 PM #4 Newbie   Joined: Oct 2015 From: Greece Posts: 24 Thanks: 0 Thank you guys so much! That was easy but i don't know much about logarithms so this is why i couldn't think about it myself.
 June 16th, 2017, 04:12 PM #5 Math Team     Joined: Jul 2013 From: काठमाडौं, नेपाल Posts: 872 Thanks: 60 Math Focus: सामान्य गणित Seems like I misunderstood the question ....... Last edited by skipjack; June 17th, 2017 at 03:19 AM.

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