# Square root of 144400

1. Jun 6, 2010

### smslca

1. square root of 144400 using Taylor series

2. I used the equation of sqrt(x) for x>2 given in http://www.understandingcalculus.com/chapters/23/23-1.php

3. I have tried it upto the 7th power and I got sqrt(144400)=1/0.209482810=4.773661

I know sqrt(144400)=380; and 1/380=0.00263157894.

So What is the power upto I have to go. Is that very large. Is 7th power not enough.
In case If it is not enough, then we can observe that for sqrt(144400) got different answers for different powers

2. Jun 6, 2010

### estro

I'm not sure if I understood you problem but,
You know that 300*300 = 90000.
Or even 350*350 = 122500.
So I think it will be a good idea to use Taylor series on these points.
The general idea is to pick "known a" as close as you can to the solution.

Last edited: Jun 6, 2010
3. Jun 6, 2010

### smslca

I have studied Taylor series in wikipedia, I did not understand how to use "a", i.e what value I should use. I dont know how to use Taylor series.

I just used the formula given in the url I mentioned above.

So Please clearly explain what u have said. And what does "use taylor series on these points". I mean how to use these points.

4. Jun 6, 2010

### arildno

Write:
$$\sqrt{144400}=\sqrt{14,4*10000}=100*\sqrt{\frac{144}{10}}=\frac{1200}{\sqrt{10}}=\frac{1200}{3\sqrt{1+\frac{1}{9}}}=\frac{400}{\sqrt{1+\frac{1}{9}}}$$

The square root in the denominator can be approximated by:
$$\sqrt{1+\frac{1}{9}}\approx{1}+\frac{1}{2}*\frac{1}{9}=\frac{19}{18}$$

Thus, we get:
$$\sqrt{144400}\approx\frac{7200}{19}\approx{378,9}$$

5. Jun 6, 2010

### estro

I think this can be very helpful to you.
http://tutorial.math.lamar.edu/Classes/CalcII/TaylorSeries.aspx
Basically speaking "a" is a point where you know all the information (numerical) about the function and it's derivatives.

Last edited: Jun 6, 2010
6. Jun 6, 2010

### smslca

how can we choose the value of "a" in f(a).

7. Jun 6, 2010

### arildno

I made a mess in my previous post, using 14,4, rather than 14,44 at one point.

By a similar argument as the one above, we get that we can approximate the square root by $$\frac{7210}{19}$$ instead, which equals about 379.4

8. Jun 6, 2010

### arildno

In whatever way you want!

9. Jun 6, 2010

### estro

R_n (x) is the reminder. (look for Lagrange reminder theorem.)
When you use Taylor series you basically need 3 things.

1. Numerical value of f(a) up to the relevant f^k (a). (the k'th derivative, you choose point a where you know exactly: f(a), f'(a), f"(a)...)
2. You should choose k depending on the desirable accuracy. (for this you must use the reminder)
3. x is the point where you want to approximate the function.

* Your first step should be understanding the theory behind the equation. (use the url)

10. Jun 6, 2010

### smslca

As I am new to Taylor series, and im little confused, and in an urgent to expansion of sqrt
I want Taylor expansion for

sqrt{[(100*c)+21-p]^2-[8400*c]}

where "p" is constant. and "c" is variable.
and c>=1,

I know I should do it by myself. But I need it in urgent.can anyone please expand it for me.

Last edited: Jun 6, 2010
11. Jun 6, 2010

### vela

Staff Emeritus
The Taylor expansion is usually of $\sqrt{1+x}$ about x=0 rather than $\sqrt{x}$ about x=1:

$$\sqrt{1+x} = 1+\frac{1}{2}x-\frac{1}{8}x^2+\frac{1}{16}x^3-\cdots$$

The series converges for |x|<1, so the idea is to manipulate the radicand to get it into the right form. If you want to find the square root of 47, for example, you could write

$$\sqrt{47} = \sqrt{49-2} = \sqrt{49}\sqrt{1-2/49} = 7\sqrt{1-2/49}$$

At this point, you'd expand the square root as a series with x=-2/49. If you wanted a series for $$\sqrt{a^2+b^2}$$ when |a|>|b|, you would write

$$\sqrt{a^2+b^2} = \sqrt{a^2}\sqrt{1+(b/a)^2}$$

and expand with x=(b/a)2 since |x|<1. On the other hand, if |a|<|b|, you'd pull the b2 out instead:

$$\sqrt{a^2+b^2} = \sqrt{b^2}\sqrt{1+(a/b)^2}$$

so that x=(a/b)2 and again |x|<1.

Your problem is a bit of an algebraic mess, so I doubt anyone's going to do it for you.