My Math Forum Possible Logarithmic Method to Find Roots

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 May 11th, 2016, 09:11 AM #1 Senior Member   Joined: Nov 2015 From: USA Posts: 107 Thanks: 6 Possible Logarithmic Method to Find Roots This is an interesting thing I discovered but not yet refined. Have fun with it. I may return if I make any developments. I looked at the ratio of numbers to their roots, basically asking, the roots are what percentage of their square? Interestingly, if we plot according to the squares, the percentage of their roots, we find that the percantage halves exactly at each order of magnitude when counting in base 4. I.E. 1 is 100% 4 or 10 base 4 is 50% 16 or 100 base 4 is 25% 64 or 1000 base 4 is 12.5% etc. I'm not up to par on logarithmics yet, but I figured this might be a usable fact to look at a number in base 4 to more easily determine the square root. Also, if such a method is feasible and does work, then that might lead to similar possibilities for higher roots.
 May 11th, 2016, 09:56 AM #2 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,674 Thanks: 2654 Math Focus: Mainly analysis and algebra Consider the squares as the area of a square and the square roots as the side of the square. What happens to your ratio when you double the side of the square? Can you thing of a similar analogy for cube roots? Thanks from manus
 May 12th, 2016, 04:20 AM #3 Senior Member   Joined: Nov 2015 From: USA Posts: 107 Thanks: 6 Cube roots are easy, volume to edge. Probably follows base 8 progression, but haven't fully checked yet. It is my guess that each added dimension has a base twice that of the previous. Basically following the powers of two progression. Which is interesting going in reverse, does 1 dimensional for a length become double, itself, or 1? What about 0 dimension? x/x(2^y) x is the root, y is the dimension For any change x in the root, you will get a change in the result equal to the change x multiplied by 2 to the same power as the result is of the root. I.E. Increasing the edge of a cube by (x) increases it's area on a single face (edge squared) by x*4 (x*2^2). Likewise, the volume of the cube (edge cubed) would increase by x*8 (x*2^3). Thus also, an increase of x in the 0 dimension is an increase of x as x*2^0 = x*1. This would be the edge of the cube. Thus in the single dimension, increasing the edge by x, increases the result by x*2, as x*2^1 is x*2. But where is this on the cube? I'm not sure yet how to put this in more solid terms describing root to square/cube/etc yet, but that might be due to tiredness. Thanks from manus Last edited by MystMage; May 12th, 2016 at 04:22 AM.
 May 12th, 2016, 04:47 AM #4 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,674 Thanks: 2654 Math Focus: Mainly analysis and algebra I don't know that the base is likely to be massively useful although logarithms are probably used by calculators to estimate $n$th roots using $\sqrt[n] x = x^{\frac1n} = \mathrm e^{\log x^{\frac1n}}=\mathrm e^{\frac1n \log x}$. Calculators have efficient algorithms to approximate logs and exponentials. Thanks from manus
 May 12th, 2016, 11:09 AM #5 Senior Member   Joined: Nov 2015 From: USA Posts: 107 Thanks: 6 Three reasons to pursue this, 1, by hand calculations with simpler math. I don't have fancy calculators with really nice scientific functions. If this fixes up solid, then I'll have a reliable method of getting a root by hand. And while it uses a logarithm, you don't actually need to calculate with a logarithm to use it. You simply convert bases the division way, then each order of magnitude brings you closer to the percentage the root is of the number, then you multiply by that percentage. Probably is much slower for computers, but this way is easier by hand and potentially simple enough to forgoe table lookups altogether even by hand. 2, alternative methods are always good to follow up on, as they increase overall understanding, but also, even if an alternative is at first less efficient, it may someday lead to something that is more efficient or accurate, or better in some way, even if only situationally (situational options aren't given enough credit in my opinion, at least out here away from professional circles.). 3, Exploration is fun. Edit, why do the emotes keep turning into stupid smileys. I hate that.
 July 9th, 2016, 11:28 AM #6 Senior Member   Joined: Nov 2015 From: USA Posts: 107 Thanks: 6 x=x y=square root of x z=temp value p=percentage y is of x x=4^z then 2^(-z)= p then p*x=y Example x=8 8=4 ^ 1.5 2 ^ (-1.5) = 0.35355339 0.35355339 * 8 = 2.82842712 2.82842712 ^ 2 = 8 Thus 2.82842712 is the square root of 8. I have found this to be true for a few numbers so far. Feel free to break it. Thanks from manus
 July 9th, 2016, 11:50 AM #7 Senior Member   Joined: Nov 2015 From: USA Posts: 107 Thanks: 6 Check this out! x=x y=square root of x z=temp value p=percentage y is of x x=4^z then 2^(-z)= p then p*x=y Example x=8 8=4 ^ 1.5 2 ^ (-1.5) = 0.35355339 0.35355339 * 8 = 2.82842712 2.82842712 ^ 2 = 8 Thus 2.82842712 is the square root of 8. I have found this to be true for a few numbers so far. Feel free to break it.
 July 10th, 2016, 09:32 AM #8 Math Team   Joined: Apr 2010 Posts: 2,780 Thanks: 361 x = 8 y = sqrt(x) = sqrt(8 ) = sqrt(2^3) = 2^1.5 1/y = 2^-1.5 = 1/sqrt(x) = sqrt(x) / x. 1/sqrt(8 ) * 8 = sqrt(8 ). Thanks from manus
 July 11th, 2016, 12:50 PM #9 Senior Member   Joined: Nov 2015 From: USA Posts: 107 Thanks: 6 What is your point? My method shows that the percentage a root is of the square can be found from the square itself. That is what makes it interesting. More, my method shows that the link between 4 and 2 and roots are not dependent on the square and can therefore be applied even when simple shortcuts won't suffice. What you did implies that 2 was used as a base only because it was a factor of 8, which is not the case in what I did. The key here is that 0.35.... is the percentage the root is of 8. Further, the scale between 4 and 2 are consistent, meaning you can easily use that to directly find the root of any square without guessing or using repetition to refine a estimate. (perhaps even make a slide rule to go straight to the answer.) The only piece remaining is to easily convert a base 4 number into the exponential notation to find the exponent. Once I manage that, it is goodbye to the days of finding roots through guesses and estimates, however well they can be refined.
 July 11th, 2016, 06:09 PM #10 Math Team   Joined: Jan 2015 From: Alabama Posts: 3,264 Thanks: 902 His point is that this is nothing more than basic, secondary school, algebra. You have defined x to be $\displaystyle 4^z$ and p to be $\displaystyle 2^{-z}$. From $\displaystyle x= 4^z$, $\displaystyle \sqrt{x}= 2^z$ so $\displaystyle p= 2^{-z}= \frac{1}{\sqrt{x}}$. So, of course, $\displaystyle px= \frac{x}{\sqrt{x}}= \sqrt{x}$. In your example, to find the square root of 8, you use the -1.5 root of 2. Why would you consider that to be simpler that finding the square root of 8? Thanks from topsquark and manus

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