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Mathematical induction again

  1. Oct 26, 2006 #1
    I'm connected from a phone so it would be pretty hard to write out MY equation for An...here's the equation in words: square root of 3 over 4 + square root of 3 over 12 * ((9/5 -9/5(4/9)^n)...

    using your general expression for A_n and the iterative (what does that mean?) relation between A_n and A_n+1 already established, prove the general expression for A_n by induction?

    if you've read my previous post, I was looking for n when the value for the area (this relates to fractals) reaches a limit to the millionth decimal place, and I found it. at terms 17 (A_n+1) and 16 (A_n). now I'm confused, do not know how to prove my equation true without knowing the sum formula for the geometric sequence .
     
  2. jcsd
  3. Oct 26, 2006 #2

    radou

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    The sum formula for a geometric sequence is giben by [tex]S_{n} = a_{1} \frac{q^n-1}{q-1}[/tex], where the general term of the sequence is [tex]a_{n}=a_{1} q^{n-1}[/tex]. I hope this helps.
     
  4. Oct 27, 2006 #3
    ok thanks radou but my equation isn't set in general term format.
    square root of 3 over 4 + square root of 3 over 12 * (9/5 -9/5(4/9)^n)
    what do I do with the value that's added to the equation? square root of 3 over 4
    this however can be the ratio: (9/5 -9/5(4/9)^n)
    and the first term: square root of 3 over 12
     
  5. Oct 27, 2006 #4

    radou

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    Just a sec, is your expression [tex]\sqrt{\frac{3}{4}}+\sqrt{\frac{3}{12}}\cdot\left(\frac{9}{5}-\frac{9}{5}\cdot\left(\frac{4}{9}\right)^n\right)[/tex]?
     
  6. Oct 27, 2006 #5
    the four and the twelve of the fractions are NOT square root-ed. only the numerators are irrational (hence square root-ed)
     
    Last edited: Oct 27, 2006
  7. Oct 27, 2006 #6

    Integral

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    Your dial up connection has no effect on the ability to post formulas. Please read this thread.
     
  8. Oct 27, 2006 #7
    umm, it's not the dial up connection that's troubling me, but the small keyboard i got to write on...
     
  9. Oct 28, 2006 #8

    HallsofIvy

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    I feel sure that neither your dial up connection nor you small keyboard has anything to do with the fact that that is not an equation!
    Even after we understand that you mean
    [tex]\frac{\sqrt{3}}{4}+\frac{\sqrt{3}}{12}\cdot\left(\frac{9}{5}-\frac{9}{5}\cdot\left(\frac{4}{9}\right)^n\right)[/tex]
    we still have no idea what you are trying to prove!
     
  10. Oct 28, 2006 #9
    I'm as confused as you are, Halls of Ivy...This was the question we were asked....Perhaps you would understand the wording better than I did...
    using your general expression for [tex]\frac{\sqrt{3}}{4}+\frac{\sqrt{3}}{12}\cdot\left(\ frac{9}{5}-\frac{9}{5}\cdot\left(\frac{4}{9}\right)^n\right)[/tex] and the iterative relation between A_nand A_n+1 already established, prove the general expression for A_n by induction?
     
    Last edited: Oct 28, 2006
  11. Oct 28, 2006 #10

    HallsofIvy

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    So presumably at some place earlier in the text or whatever, you were given or asked to derive an "iterative relation between An and An+1. That was "already established". What is it? You are asked to prove that the expression
    [tex]\frac{\sqrt{3}}{4}+\frac{\sqrt{3}}{12}\cdot\left(\frac{9}{5}-\frac{9}{5}\cdot\left(\frac{4}{9}\right)^n\right)[/tex]
    satisfies a given equation. Before we can make any comments on that we need to know what that equation is!
     
  12. Oct 28, 2006 #11
    We were asked earlier in the text to find the value of n to within one millionth of a square unit when An = An+1 ...and at the 17th and 16th term the values are the same....do you see an iterative relation?
     
  13. Oct 29, 2006 #12
    help..........
     
  14. Oct 29, 2006 #13

    HallsofIvy

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    Obviously, then, You we given some iteration: An+1= some function of An and asked to find n such that An+1= An to within one millionth= .000001. In other words where the right side differs from the left by less than .000001.
    You need to read the text more carefully to find where you were given that iteration.
    The fact that it says "of a square unit" is interesting. Is this some kind of applications problem where you are given units?
     
  15. Oct 29, 2006 #14
    im not given any units. I did find the terms where An+1 and An are different by .000001
    16 and 17
     
  16. Oct 29, 2006 #15
    nevermind, found the answer!
    Thanks for the help anyway!
     
  17. Oct 30, 2006 #16

    HallsofIvy

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    Great. Just out of curiosity, what was the iteration equation?
     
  18. Oct 30, 2006 #17
    In question six, we are told that An=An+1, so I had this one equation on area which wasn't the general term, but it was another method of finding the area of the koch snowflake. I susbstituted n+1 for n. Then I added An to An+1 and ended up with equation An, hence An is true whenever An+1 is true. (am i making an sense? lol)
     
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