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## Main Question or Discussion Point

What is the easiest way to solve "the square root of 150" etc.. without using calculator?

- Thread starter Ephratah7
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- #1

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What is the easiest way to solve "the square root of 150" etc.. without using calculator?

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- #3

HallsofIvy

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Choose some "starting" value that is close to the square root. Since 12

If you want more accuracy, just keep going.

Those who are aware of Newton's method should recognize that as Newton's method applied to the equation f(x)= x

- #4

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It sounds like the bisection method to me.

Choose some "starting" value that is close to the square root. Since 12^{2}= 144 is close to 150, let's start with 12. 150/12= 12.5 (and I didn't use a calculator to do that!) Notice that says 12(12.5)= 150. If x^{2}= 150, x must bebetween12 and 12.5. Just because it is easy, let take half way between: 12.25. Now 150/12.25= 12.249 (To 3 decimal places. If you want more accuracy, just keep going- but you are going to wish you could use a calculator!). Again, the square root of 150 must be between 12.25 and 12.49. Halfway between is 12.247. 150/12.247= 12.247 again, to 3 decimal places. Since that is the same as the previous number, the square root of 150, to 3 decimal places, is 12.247.

If you want more accuracy, just keep going.

Those who are aware of Newton's method should recognize that as Newton's method applied to the equation f(x)= x^{2}- 150= 0.

http://en.wikipedia.org/wiki/Bisection_method

- #5

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No, bisection is much slower to converge.It sounds like the bisection method to me.

http://en.wikipedia.org/wiki/Bisection_method

- #6

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Sorry I saw the above poster taking a midpoint and thought he was using a bisection method. To me the above method is Newton–Raphson method, while from what I learned Newtons method does not look for a mid point as an intermediate step.No, bisection is much slower to converge.

Using the mid point helps to ensure convergence but I'm not sure it is [a good idea since. Consider the problem of finding a [tex]\Delta[/tex]:

in

[tex](Y_o + \Delta )^2=X[/tex] (1)

expanding:

[tex]Y_o^2+2Y_o \Delta + \Delta^2=X[/tex] (2)

Now if neglect [tex]\Delta^2[/tex] (3)

and solve for \Delta we get what is equivalent to newtons method.

[tex]\Delta= \frac{X-Y_o^2}{2Y_o}[/tex] (4)

Notice though that the ratio of [tex]\Delta[/tex] to [tex]\Delta^2[/tex] increases as delta gets small. This makes me wonder if newtons method accelerates in convergence (for finding square roots) when [tex]\Delta[/tex] gets small.

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- #7

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Some more interesting thoughts. A better approximation then neglecting [tex]\Delta^2[/tex] would be to write equation (2) as:

[tex]Y_o^2+(2Y_o+\Delta)\Delta=X[/tex] (5)

[tex]\Delta=\frac{X-Y_o^2}{2Y_o+\Delta}[/tex] (6)

Now I wonder if it is worth while to iterate the above expression. If we do so algebraically we could get a higher order root finding method. If we do so numerically, it allows us to reduce the amount of carry operations that we need to do.

For instance. Say the above expression is comparable to newtons method for the rate of convergence. Then we can pick a Y_o to start with. When \Delta is computed to a large number of significant digits. We just add Y_o to delta to get a new Y_o and then search for a new value of Delta and so on.

[tex]Y_o^2+(2Y_o+\Delta)\Delta=X[/tex] (5)

[tex]\Delta=\frac{X-Y_o^2}{2Y_o+\Delta}[/tex] (6)

Now I wonder if it is worth while to iterate the above expression. If we do so algebraically we could get a higher order root finding method. If we do so numerically, it allows us to reduce the amount of carry operations that we need to do.

For instance. Say the above expression is comparable to newtons method for the rate of convergence. Then we can pick a Y_o to start with. When \Delta is computed to a large number of significant digits. We just add Y_o to delta to get a new Y_o and then search for a new value of Delta and so on.

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- #8

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[tex]F(x) = x^2 - A = 0[/tex]

If you have'nt seen it before, the reulting NR iteration process results to a conveinent computation scheme:

[tex]x_{\nu+1} = x_{\nu}-\frac{F(x_{\nu})}{F'(x_{\nu})}[/tex]

with

[tex]F'(x_{\nu}) = 2x_{\nu}[/tex]

the RHS reduces to

[tex]x_{\nu+1} = x_{\nu}-\frac{x_{\nu}^2-A}{2x_{\nu}} = \frac{1}{2}(x_{\nu}+\frac{A}{x_{\nu}})[/tex]

The "trick" here is to reconize the simple, but effective, starting point Xo=12. With this choice, the number of required iterations to reach the stated level of convergence (3 decimals) is quite small when compared to other methods.

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....Thanks, guys.

- #10

HallsofIvy

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Sorry I saw the above poster taking a midpoint and thought he was using a bisection method. To me the above method is Newton–Raphson method, while from what I learned Newtons method does not look for a mid point as an intermediate step.

IHal's "method" is indeed a Newton-Raphson search, for this particular problem of solving

[tex]F(x) = x^2 - A = 0[/tex]

Those who are aware of Newton's method should recognize that as Newton's method applied to the equation f(x)= x^{2}- 150= 0.

- #11

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"Euclid alone has looked on beauty bare"

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we have x^2 = 150

an educated guess would be to choose x = 12

transform the eqn, x^2 = xy = 150, just as the poster did above. then y = 12.5 if x = 12

now, take the partial derivative of xy = z, in respect to x, which is y = (dz/dx), and evaluate at y = 12.5, and multiply this number by x = 12. Now, take the partial derivative in respect to y and evaluate at x = 12. x = (dz/dy) = 12, and multiply this number by y = 12.5. Now sum these two terms and divide by the the sum of (dz/dx) and (dz/dy).

it should look like,

(12(12.5) + 12.5(12))/(12 + 12.5) = 300/(24.5) = 12.24

and now make this number equal to x, and reiterate.

in general, if you have a function f(x) = 0, make an educated guess for x (this is b), and transform the function into g(x,y)= 0, and solve for y (this is c) when x = b (educated guess).

a good approximation operator would be:

{c(dg/dy) + b(dg/dx)}/{(dg/dy) + (dg/dx)}

where (dg/dy) and (dg/dx) are evaluated at x=b, and y=c

follow?

- #13

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But it makes things easier, you could just use prime factorization and reach easier numbers.

(150)^0.5=(5*3*5*2)^0.5=5*(6)^0.5 or 5*(3)^0.5*(2)^0.5

Now all you have to do is remember the square root of 2 and 3.

- #14

Gib Z

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As for mathis314, all you are doing is taking the linear part of the taylor series, which is quite a bit harder than Newtons method. And You can not improve your approximations easily with that method, with Newtons method you can.

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THanks for the information.

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