how do i find square root of 33 (or any other natural number) using differential calculus?

Question

please give step by step instructions

Answer 1

Need to use a Taylor expansion:

The Taylor expansion of a function f(x) at a given point x0 is given by

f(x0)+(x-x0)*f'(x0)+(x-x0)^2*f''(x0)/2!+(x-x0)^3*f'''(x0)/3!+...

For example, the Taylor expansion of the function f(x)=x^3 at x=2 is given by

8+(x-2)*12+(x-2)^2*12/2+(x-2)^3*6/6

(f(2)=8, f'(2)=12, f''(2)=12, f'''(2)=6)

Now, for the square root, you use the square root function f(x)=sqrt(x)

You need to calculate the Taylor expansion of f(x) at the point closest to 33 that is a perfect square. In this case, it is 36.

So, the Taylor expansion at 36 is

6+(x-36)*1/12+
(x-36)^2*(-1/864)/2+...

(note that: f'(x)=1/2/sqrt(x), hence f'(36)=1/12 and, similarly, f''(36)=-1/864)

To get the aproximation of sqrt(33), plug x=33 in the above expansion. The more term you take, the better (note that in my calculations, I took 3 terms but, in most calculus books, they only use 2 terms)

Let's take only two terms

6+(33-36)*1/12=23/4=5.75

Three terms

6+(33-36)*1/12+(33-36)^2*(-1/864)/2=1103/192=5.744791667

You can do more terms. The actual value is 5.744562647

Pretty good for an approximation!

Answer 2

You begin finding the more aproximated root that is less than 33. In this case, 5.

33....l__ 5.7__
.800 l 10. x . =

107 x 7 is less than 800

and 108 x 8 is bigger. Then you put a 7 after the decimal dot and you calculate the remainder.

Then you use 2 more 0 and you double the 57, you have to find the 3rd digit, so that 57. x. is less than your remainder when you have added two 0.

If the number has 3 or more digits instead of 2, like the 33, you divide it in groups from 2 digits.

Ana

Answer 3

y = x^2
dy/dx = lim(x^2 - x0^2)/(x - x0) = 2x
let y = x0^2 and solve for x0:
2x(x - x0) = x^2 - y
x0 = (y - x^2)/(2x) + x
The iterative solution is now set up.
33 is between 25 & 36 so use a 1st estimate x = 5.5
x0 = (33 - 30.25)/11 + 5.5 = 5.75
x1 = (33 - 33.0625)/11.5 + 5.75 = 5.7445652
x3 = (33 - 32.999998)/11.48913 + 5.7445652 = 5.74456265
x4 = (33 - 32.999999999563)/11.48912593 + 5.74456265 = 5.7445626465, which is accurate to 6 decimal places.

Note that the iteratiion converges very slowly if the original estimate is very far from the actual value.

Answer 4

y = x^2
dy/dx = lim(x^2 - x0^2)/(x - x0) = 2x
let y = x0^2 and solve for x0:
2x(x - x0) = x^2 - y
x0 = (y - x^2)/(2x) + x
The iterative solution is now set up.
33 is between 25 & 36 so use a 1st estimate x = 5.5
x0 = (33 - 30.25)/11 + 5.5 = 5.75
x1 = (33 - 33.0625)/11.5 + 5.75 = 5.7445652
x3 = (33 - 32.999998)/11.48913 + 5.7445652 = 5.74456265
x4 = (33 - 32.999999999563)/11.48912593 + 5.74456265 = 5.7445626465, which is accurate to 6 decimal places

Answer 5

you may easily try this one by making use of factorization: x^2 + 4x - 21 = (x + 7)(x -3) = 0 x = -7, 3 yet once you're able to do it making use of the quadratic formulation then a = one million b = 4 c = 21 the discriminant is b^2 - 4ac = 4^2 - 4(one million)(-21) = sixteen + 80 4 = a hundred because of the fact the discriminant is beneficial you recognize you have 2 genuine roots given by making use of x = (one million/2a)[-b ± ?a hundred ] x = (one million/2)[ -4 ± 10 ] = -2 ± 5 x= -7, 3

Answer 6

I think what you're looking for is called Newton's method (also the Newton-Raphson method). You can find it in pretty much any Calculus book, and there is a good explanation at
http://www.ugrad.math.ubc.ca/coursedoc/math100/notes/approx/newton.html

Also some other interesting articles at
http://en.wikipedia.org/wiki/Newton's_method
http://mathworld.wolfram.com/NewtonsMethod.html
and there's a cute applet at
http://www.cs.utah.edu/~zachary/isp/applets/Root/Newton.html


Doug

Answer 7

you fool it's 33*33