How to obtain Roots from indicial equation!?

Discussion in 'Homework Help' started by mcdara, Jan 2, 2013.

  1. mcdara

    Thread Starter New Member

    Feb 23, 2011
    11
    0
    Hi

    I am stuck on a Frobunius equation. The part is where I have to obtain roots from a indical equation as follows

    4r^2 -4r + 2r = 0

    => 2r(2r-1) = 0

    Roots = r1 = 0 r2 = 1/2

    If anyone could tell me how these roots were obtained it would be great, when I google help the level of sums are in great detail and over my head
    Any step by step method would be great, im sure its simple but I cant get my head around it.

    Thanks in advanced!
     
  2. tshuck

    Well-Known Member

    Oct 18, 2012
    3,531
    675
    Wolfram Alpha

    it's nothing more than a quadratic formula:
    \frac{-b+-\sqrt{b^{2} - 4ac}}{2a}


    4r^2 -4r + 2r = 0 => 4r^2 - 2r = 0

    a = 4
    b = -2
    c = 0

    \frac{-(-2)+-\sqrt{(-2)^{2} - 4(4)(0)}}{2(4)}

    =

    \frac{2+-2}{8}

    r = 0 & 1/2
     
  3. mcdara

    Thread Starter New Member

    Feb 23, 2011
    11
    0
    I had a feeling it was something simple!!

    Thank you!
     
  4. WBahn

    Moderator

    Mar 31, 2012
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    Which part of the process don't you understand?

    I have no idea what a Frobunius equation or indical equation are, but you don't need the quadratic equation to find the roots of the equation you've given.

    Is getting from 4r^2 -4r + 2r = 0 to 2r(2r-1) = 0 the stumbling point?

    If so, the distributive property says you can factor out the same thing from each term in a group of terms. All of your terms are divisible by 2r, so factoring out 2r gives you your second equation.

    4r^2 -4r + 2r = 0
    2r(2r) - 2r(2) + 2r(1) = 0
    2r(2r - 2 + 1) = 0
    2r(2r - 1) = 0

    Note that we could have combined the last two terms earlier since

    -4r + 2r = -2r

    Making the equation

    4r^2 - 2r

    And, after factoring out 2r, we again get

    2r(2r - 1)=0

    Is the problem that you don't understand what a "root" of an equation is?

    If so, a root is nothing more than a value of the variable that results in the equation being equal to zero. Well, if I have xy=0, then the equation overall is equal to zero if either x is zero or y is zero. So I can set each factor equal to zero and solve for the roots that way. In your equation you have two factors (technically three), namely (2r) and (2r-1). I say "technically three" because the three factors are really (2), (r), and (2r-1) and, if you recognize this, then things become even a bit easier. But let's say we didn't recognize this. So our root are:

    The values of r that make
    2r = 0
    or
    2r-1 = 0

    2r = 0 => r=0
    2r-1 = 0 => r = 1/2
     
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  5. WBahn

    Moderator

    Mar 31, 2012
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    I just did a quick look and Frobunius method is a means of finding an infitite power series solution to an second order ordinary differential equation. And its "indicial", not "indical" (easy typo to make).

    Is this the level of math you are supposed ot be working at, namely solving differential equaitons?
     
  6. justtrying

    Active Member

    Mar 9, 2011
    329
    349
    knowing how to solve quadratic equations is essential for survival. The basic approaches are factoring, completing the square, and using the formula. I suggest you practice all of them.
     
    mcdara likes this.
  7. Papabravo

    Expert

    Feb 24, 2006
    10,140
    1,790
    It seems odd to me that you would be playing with ininite series and differential equations, having slept through Algebra I, which is practically dedicated to the quadratic equation, factoring, synthetic division, and Newton's method.
     
  8. WBahn

    Moderator

    Mar 31, 2012
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    Let's not jump too quickly to that conclusion. It's possible that it is a much lower level course and these terms are just being thrown about in some effort to convince the students that what they are studying is relevant and will be useful down the road. But it certainly would be useful to know what level the OP is really at, because the indications thus far paint a contradictory picture, making it hard to choose a suitable way to describe things.
     
  9. Papabravo

    Expert

    Feb 24, 2006
    10,140
    1,790
    I agree that the evidence does not paint a clear picture and that multiple interpretations are possible.
     
  10. mcdara

    Thread Starter New Member

    Feb 23, 2011
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    0
    Thanks!! it was the best class for a nap!
     
  11. mcdara

    Thread Starter New Member

    Feb 23, 2011
    11
    0
    I'm a 4th year electrical engineering student. It was a very long 12 hour day and I was stressed over the sum and I came to the site for help. After a sleep I was able to realize I should have use that simple equation
    Thanks for the reply before!
     
  12. WBahn

    Moderator

    Mar 31, 2012
    17,743
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    That's not helping to clear up the picture any!
     
  13. WBahn

    Moderator

    Mar 31, 2012
    17,743
    4,792
    Okay, now that IS helping clear things up. Lag in the server meant I didn't see this when making my last (off the cuff and not-to-be-taken-too-seriously) remark.
     
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