# Finding roots without a calculator?

• eNathan
In summary: Do you mean any mathematical statement can be written in a single equation? If so, that's not true at all. In fact, a lot of mathematical statements can't even be expressed as equations. For example, the statement "every prime number greater than 2 is odd" can't be expressed as a single equation.
eNathan
hello. I was wanting to know how to raise a number to any power without using a calculator. More specifically, I was wanting to raise numbers to the .5 power, (and all the other roots, 1/2, 1/3, 1/4). How can this be done?

<---------------------->

My fellow 633|<5

Newton's method is one of the more general ways. For an example of getting a square root, see http://planetmath.org/encyclopedia/NewtonRaphsonMethod.html .

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Horner's Method

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hmn, it all seems very complicated. I will try to go over those sites tho :) But isn't there just an equation that can do it? It might be an extreamly large equation, but there should be one right?

Not really.Handy methods would involve logarithm tables and slide rule...

Daniel.

P.S.Do u know the algorithm of extracting sq.root from any real #...?If not,learn it...

Yes, those references do look a bit complicated. Newton's method for finding the square root of N is to use the iterative formula x:=(x+N/x)/2, and for kth roots I think it is x:=((k-1)x+N/(x^(k-1))). There is also an algorithm for finding the square root from the decimal expansion, given at http://www.geocities.com/cnowlen/Cathy/Emat4680/Squareroot.htm (This used to be taught in schools in the days before calculators) There are similar algorithms for higher roots, but they do get a bit difficult.

Yes, those references do look a bit complicated. Newton's method for finding the square root of N is to use the iterative formula x:=(x+N/x)/2, and for kth roots I think it is x:=((k-1)x+N/(x^(k-1))). There is also an algorithm for finding the square root from the decimal expansion, given at http://www.geocities.com/cnowlen/Ca.../Squareroot.htm (This used to be taught in schools in the days before calculators) There are similar algorithms for higher roots, but they do get a bit difficult.

notice in the equation x:=(x+N/x)/2 you yse the variable "x" when the whole purpose of the equation is to get the value of "x". What is the logic in that?

chronon, would you happen to be a Delphi programmer? lol, I just notice how you use the ":=" operator instead of "=". := is the variable assignment operator in Delphi, and is quite unique from any other language (exept Pascal which is the same), so that is why I ask.

And dextercioby, I have a theory that anything can be expressed in a single equation. I find it hard to believe that it is imposible to make an equation to find the square root, no matter how long it is.

$$:= \ \mbox{or} \ =:$$ could also stand for definitions...

Daniel.

P.S.If u have a theory,publish it...

P.S.If u have a theory,publish it...

lol. yes, I have found my theory to be true (and when its not, its a result of my own mind not being able to make the euqations). Did you know you can do boolean operations in an equation? It is equavilent to running "if" statements in computer code. And what makes this posible is the fact that a negative times a negative is a positive, while a positive times a positive is a positive. Do you agree with my theory?

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U know,every mathematical theory needs "acta,non verba",meaning formulas/ae.

Daniel.

Here is a very simple example of my theory.

//in code...
If x < 0 then R = -i
Else If x > 0 then R = i

//Translation: If x is negative, then R equals the negative form of i. If x is
//positive, then R equals the positive form of i.

Or, in an equation...

R=i(x/abs(x))

eNathan said:
chronon, would you happen to be a Delphi programmer? lol, I just notice how you use the ":=" operator instead of "=". := is the variable assignment operator in Delphi, and is quite unique from any other language (exept Pascal which is the same), so that is why I ask.

Yes, I am a Delphi programmer, and I used := deliberately because, to me, it shows that it is an assignment (that is you start with x having some value and repeatedly replace x with (x+N/x)/2 ) rather than an equation.

eNathan said:
notice in the equation x:=(x+N/x)/2 you yse the variable "x" when the whole purpose of the equation is to get the value of "x". What is the logic in that?
It is iterative. In sequential form, one guesses an a0 that is close to the root (this can be made more precise, but such machinery is unnecessary at this level) and uses it to generate a series of ai's that converge to your root. For a square root of some number b, the relation is an = (1/2)(an-1 + b/an-1) . As shown in the link, 3 or 4 iterations will give more digits than a simple scientific calculator.

notice in the equation x:=(x+N/x)/2 you yse the variable "x" when the whole purpose of the equation is to get the value of "x". What is the logic in that?

It's an iterated algorithm. It isn't really

$$x = \frac{x + \frac{N}{x}}{2},$$

but

$$x_{n+1} = \frac{x_n + \frac{N}{x_n}}{2}$$

where $N$ is the number you want the root of. Say, for example, I want $\sqrt{2}$, so $N=2$. Then I make the initial guess $x_1 = 1$, so

$$x_2 = \frac{x_1 + \frac{2}{x_1}}{2} = \frac{1+\frac{2}{1}}{2} = \frac{3}{2} = 1.5$$

$$x_3 = \frac{x_2 + \frac{2}{x_2}}{2} = \frac{\frac{3}{2} + \frac{2}{\frac{3}{2}}}{2} = \frac{17}{12} \approx 1.4167$$

$$x_4 = \frac{x_3 + \frac{2}{x_3}}{2} = \frac{\frac{17}{12} + \frac{2}{\frac{17}{12}}}{2} \approx 1.414216$$

and you can repeat as necessary to get arbitrarily close. The exact root is $1.414213562...$.

In general, if you want to solve for a root of a differentiable real-valued function $f$ of a single variable $x$, the algorithm is

$$x_{n+1} = x_n - \frac{f(x_n)}{f^\prime(x_n)},$$

though this algorithm is by no means guaranteed to converge (it is quite easy to put conditions on its convergence though).

Anyways, what exactly do you mean by

...anything can be expressed in a single equation.

?

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I have tried that method over and over...and it actually gets the root! I think I can implement this method into an asm code using a loop. There is one probem tho. I thougt that I could use this method to raise any number to any power. For instance, if I wanted to do 25^.4 I would run the equation x:=(x+n/x)/2.5 (I use 2.5 because it is the Multiplicative Inverse of .4) and the closest I get is 4.0... Is there a way around this?

Thanks everyone :)

You need to use the general algorithm to find different roots. As I said above, this is

$$x_{n+1} = x_n - \frac{f(x_n)}{f^\prime(x_n)}.$$

In the square root case, you take $f(x) = x^2 - N$, and you're trying to find a root, ie. to solve $f(x)=0$. Thus you use

$$x_{n+1} = x_n - \frac{f(x_n)}{f^\prime(x_n)} = x_n - \frac{x_n^2 - N}{2x_n} = \frac{x_n + \frac{N}{x_n}}{2}$$

If you want to find $N^{0.4}$, then you would be solving $f(x) = x^{2.5} - N = 0$, which gives

$$f^\prime(x) = 2.5x^{1.5}$$

so

$$x_{n+1} = x_n - \frac{f(x_n)}{f^\prime(x_n)} = x_n - \frac{x_n^{2.5}-N}{2.5x_n^{1.5}} = x_n - \frac{2}{5}x_n + \frac{2N}{5x_n^\frac{3}{2}}$$

$$= \frac{3}{5}x_n + \frac{2N}{5x_n{}^{1.5}}$$

so to use the algorithm to find $N^{0.4}$, you're going to need to know how to calculate $1.5$th roots anyways, unfortunately. But that's not too hard, since $1.5$th roots can be found using $f(x) = x^2 - n^3$, for which the algorithm works nicely

Edit: Silly me, of course you can just use $f(x) = x^5 - N^2$ to find $N^{0.4}$! Then

$$x_{n+1} = x_n - \frac{x_n^5 - N^2}{5x_n^4} = \frac{4}{5}x_n + \frac{N^2}{5x_n^4}.$$

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If you memorize the logarithmic table for single digits to about 4 decimal places, you can approximate roots and powers in your head. I actually did this when I was a youth [I didn't have a very glamorous life back then either]. It was fun watching the shocked looks you get upon rattling off fairly accurate cube roots [or other fractional powers] at the drop of a hat.

The "shocked looks" were probably people wondering why in the world you would DO such a thing!

lets say you need a root of 65 so y=f(x) where x is 65
f(x) = f(a)+f'(a)(x-a)
sqrt 65 = sqrt64+1/2Sqrt(64)* (65-64) = ~ 8.0625
where a is the closest known root ie a=64 ... sqrt 64 = 8
but this is only good to like 3 significant digits

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Sorry to bring this from the grave but it was the only suitable thread for me to ask if there is an easier way to find cube roots without a calculator than this?

Also square roots without a calculator. This website helped but is there an easier way?

Cheers.

## 1. How do you find the roots of a quadratic equation without a calculator?

To find the roots of a quadratic equation without a calculator, you can use the quadratic formula: x = (-b ± √(b^2 - 4ac)) / 2a. Here, a, b, and c represent the coefficients of the quadratic equation (ax^2 + bx + c = 0). Simply plug in the values and solve for x to find the roots.

## 2. What is the difference between finding real and complex roots?

Real roots are the values of x that make the quadratic equation equal to 0 when plugged in. Complex roots, on the other hand, are solutions that involve imaginary numbers (such as √-1). Real roots can be found using the quadratic formula, while complex roots require the use of more advanced methods.

## 3. Is it possible to find the roots of a cubic or higher degree equation without a calculator?

Yes, it is possible to find the roots of a cubic or higher degree equation without a calculator. However, it becomes increasingly complex and time-consuming as the degree of the equation increases. In these cases, it is often more efficient to use a calculator or computer program.

## 4. Can you use the graph of a quadratic equation to find its roots?

Yes, the graph of a quadratic equation can be used to estimate the roots. The x-intercepts of the graph represent the roots of the equation. However, this method is not as accurate as using the quadratic formula.

## 5. What are some strategies for finding roots without a calculator?

Aside from using the quadratic formula, some strategies for finding roots without a calculator include factoring, completing the square, and using the rational root theorem. These methods can be used for both quadratic and higher degree equations.

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