1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Square root of 144400

  1. Jun 6, 2010 #1
    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. jcsd
  3. Jun 6, 2010 #2
    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
  4. Jun 6, 2010 #3
    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.
  5. Jun 6, 2010 #4


    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    Dearly Missed


    The square root in the denominator can be approximated by:

    Thus, we get:
  6. Jun 6, 2010 #5
    I think this can be very helpful to you.
    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
  7. Jun 6, 2010 #6
    how can we choose the value of "a" in f(a).
  8. Jun 6, 2010 #7


    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    Dearly Missed

    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 [tex]\frac{7210}{19}[/tex] instead, which equals about 379.4
  9. Jun 6, 2010 #8


    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    Dearly Missed

    In whatever way you want! :smile:
  10. Jun 6, 2010 #9

    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)
  11. Jun 6, 2010 #10
    As I am new to Taylor series, and im little confused, and in an urgent to expansion of sqrt
    I want Taylor expansion for


    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
  12. Jun 6, 2010 #11


    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    The Taylor expansion is usually of [itex]\sqrt{1+x}[/itex] about x=0 rather than [itex]\sqrt{x}[/itex] about x=1:

    [tex]\sqrt{1+x} = 1+\frac{1}{2}x-\frac{1}{8}x^2+\frac{1}{16}x^3-\cdots[/tex]

    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

    [tex]\sqrt{47} = \sqrt{49-2} = \sqrt{49}\sqrt{1-2/49} = 7\sqrt{1-2/49}[/tex]

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

    [tex]\sqrt{a^2+b^2} = \sqrt{a^2}\sqrt{1+(b/a)^2}[/tex]

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

    [tex]\sqrt{a^2+b^2} = \sqrt{b^2}\sqrt{1+(a/b)^2}[/tex]

    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.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook