Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Expansion of functions

  1. Aug 6, 2011 #1
    In the following attachment file, I have trouble trying to understand a step in sum no.2. In the Step 2 category, 3rd line, I didn't understood how they simplified from the previous step.

    Also, wanted to ask one basic question. What exactly is the purpose of that sum in mathematical world? I mean how can I apply this knowledge of expanding functions in practical life?

    Attached Files:

    • IMG.jpg
      File size:
      29.7 KB
  2. jcsd
  3. Aug 6, 2011 #2
    Help me guys!
  4. Aug 6, 2011 #3
    [tex]\left(x + \frac{x^3}{3} + \cdot\cdot\cdot\right) - \frac{1}{2}x^2 + \frac{1}{3}x^3 + \cdot\cdot\cdot[/tex][tex]= x - \frac{1}{2}x^2 + \frac{x^3}{3} + \frac{x^3}{3} \cdot\cdot\cdot[/tex]
    They're removing the parentheses and rearranging the four terms shown.

    You'll probably see this kind of function expanding again in calculus when finding integrals that can't be expressed with a finite number of elementary functions or in a somewhat closed form in summation notation, and in differential equations for solutions of some equations that again can't be expressed with a finite number of elementary functions or in summation notation. Other than that, I can't tell you how it's practical in real life (someone else may be able to tell you more about their uses).
  5. Aug 7, 2011 #4
    How did the first step that you have shown above come in the first step? That's where my doubt lies.
  6. Aug 7, 2011 #5

    I like Serena

    User Avatar
    Homework Helper

    Hi RoughRoad! :smile:

    Simplifying it, you have something like:
    [tex](x + x^3 + ...)^2 = (x x + x x^3 + x^3 x + x^3 x^3 + ...) = (x^2 + 2x^4 + x^6 + ...)[/tex]
    If we ignore all terms with a power greater than x3 has, we're left with just x2.
  7. Aug 7, 2011 #6
    Here, I have edited the image and ow have cropped down the two steps in which I am facing problems. Now, I understood the first step, but can anyone help me how to simplify the first step into the second one? I am still clueless :-(

    Thanks for bearing with my stupidity..

    Attached Files:

    • lll.jpg
      File size:
      13.3 KB
  8. Aug 7, 2011 #7

    I like Serena

    User Avatar
    Homework Helper

    I thought that I explained that step.
    I took the large parenthesized squared term, stripped all the fractions (to spare myself a lot of LaTeX typing), and expanded it.
    Can you clarify what it is that you do not understand?
  9. Aug 7, 2011 #8

    I like Serena

    User Avatar
    Homework Helper

    Since I have no idea what it is that you do not understand, I'm going to gamble.
    I suspect that you do not know how to take the square of an expression with more than one term.

    So let me give a few examples:

    (a+b)(c+d) = ac + ad + bc + bd

    = (a+b)(a+b)
    = a2 + ab + ba + b2
    = a2 + 2ab + b2

    = (a+b+c)(a+b+c)
    = a2 + ab + ac + ba + b2 + bc + ca + cb + c2
    = a2 + 2ab + 2ac + b2 + 2bc + c2

    Are these equation familar?
    Or should I explain in more detail?
  10. Aug 8, 2011 #9
    [tex]\left(x + \frac{x^3}{3} + \frac{2}{15}x^5 + \cdot\cdot\cdot\right) -\frac{1}{2}\left(x + \frac{x^3}{3} + \frac{2}{15}x^5 + \cdot\cdot\cdot\right)^2 + \frac{1}{3}\left(x + \frac{x^3}{3} + \frac{2}{15}x^5 + \cdot\cdot\cdot\right)^3[/tex]
    So you want to just keep terms up to x3 and ignore any terms with higher powers.
    Looking at the terms in the first set of parentheses, only the first two terms meet this requirement, so that's why they're shown in the parentheses in the second line.

    Now looking at the second set of parentheses, they have an exponent of 2, so (ignoring the -1/2 in front) that really means you have
    [tex]\left(x + \frac{x^3}{3} + \frac{2}{15}x^5 + \cdot\cdot\cdot\right)\left(x + \frac{x^3}{3} + \frac{2}{15}x^5 + \cdot\cdot\cdot\right)[/tex]
    If you were to multiply out just the terms shown, you'd see that multiplying the first two terms gives you x2 which turns out to be the only term where the exponent isn't larger than 3; all other "combinations" of multiplying the other terms become terms with exponents larger than 3 and you ignore those. Now going back to the -1/2 in front, you multiply that with the x2 and that gives you the term after the parentheses in the second line.

    Do the same thing for the last set of parentheses:
    [tex]\left(x + \frac{x^3}{3} + \frac{2}{15}x^5 + \cdot\cdot\cdot\right)\left(x + \frac{x^3}{3} + \frac{2}{15}x^5 + \cdot\cdot\cdot\right)\left(x + \frac{x^3}{3} + \frac{2}{15}x^5 + \cdot\cdot\cdot\right)[/tex]
    Again multiplying together the first term of each set of parentheses gives x3, the only term with the exponent not greater than 3; all the others are larger and you ignore those as before.
    Multiplying x3 by the 1/3 in front finally gives you the last term shown in that second line.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook