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Derivative of a infinite product (Challenge Q)

  1. Jan 24, 2012 #1
    1. The problem statement, all variables and given/known data

    f(x) = (1-x3)(2+x4)(3-x5)(4+x6)(5-x7)(6+x8)...(2n+1-x2n+3)

    Find f ' (1).

    2. Relevant equations

    3. The attempt at a solution

    f (1) = 0 because of the first term. Also the pattern is that terms with odd numbers have a subtraction sign and terms with even numbers have an addition sign.

    I used logarithmic differentiation ("a" & "b" denote the terms of the original function in order to avoid rewriting all that):

    ln y = ln a + ln b ...
    y' * (1/y) = (1/a) + (1/b)...
    y' = [(1/a) + (1/b)] * [y]
    y'(1) = [(1/a) + (1/b)] * [y(1)

    Since we already know y(1) = 0, y'(1) also equals 0.

    Is this correct? Any help is much appreciated.

    Edited: I just noticed that after differentiating, my first term would be [1/(1-x3)] and plugging 1 to x would give a fraction where 0 is the denominator. I guess that makes my solution incorrect?
    Last edited: Jan 24, 2012
  2. jcsd
  3. Jan 24, 2012 #2


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    welcome to pf!

    hi kscplay! welcome to pf! :smile:
    nooo, that's dooomed :redface:

    the LHS is something divided by zero (when x = 1), and so is the first term on the RHS :wink:
  4. Jan 24, 2012 #3
    Yes, i only noticed that after posting. Thanks for replying. Any suggestions on how to solve this problem?
  5. Jan 24, 2012 #4


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    product rule:

    (AB)' = A'B + AB'

    (ABC)' = A'BC + AB'C + ABC'

    etc :wink:
  6. Jan 24, 2012 #5
    First of all I didn't know that the product rule "expanded" to more factors in the same way so thanks for mentioning that. If I did that then all the terms would be zero except for the first one. But then how would I solve for the first term since it's an infinite product. I am sorry, this is my first time dealing with an infinite product. Thanks for your reply.
  7. Jan 25, 2012 #6


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    hi kscplay! :smile:

    (just got up :zzz: …)

    actually, the way you wrote it, it's not an infinite product :wink:
    but if it was, it'd be one of those cases where you have to say "assuming that everything converges …" :smile:

    (oh, and you only need the first one, A'B + AB', where B is everything-except-the-first-bracket)
  8. Jan 25, 2012 #7
    I wrote the problem exactly as in the textbook. I thought that this would be an infinite product since they don't tell us what the last term is. I still don't how to do it with what you told me. Like I previously mentioned, I'm pretty weak in this and I really need to figure out this problem by tomorrow. Can anyone please elaborate (and again this is all that was given so forget about what I told you about it being an "infinite product" if it doesn't seem to be that) Thanks.

    P.S. Does this question require factorials (idk if its the right name but I'm referring to the exclamation mark thing)
  9. Jan 25, 2012 #8


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    Yes they are fishing for a factorial. If n=1 they want f'(1) if f(x)=(1-x^3)(2+x^4)(3-x^5). Warm up by figuring out what that is. DON'T MULTIPLY IT OUT. If n=2 they are asking what is f'(1) if f(x)=(1-x^3)(2+x^4)(3-x^5)(4+x^6)(5-x^7). Now do n=3.
    Last edited: Jan 25, 2012
  10. Jan 25, 2012 #9
    I think I get it now Dick. i was thinking that the answer has to be a numerical value that it converges to. I misunderstood the question. Now I believe they're asking for an expression given any number of terms BUT it has to finish on one with an odd number.

    So here it goes...

    No matter what n is, when we apply the product rule, all the resulting terms will be zero (because they contain 1-x^3) except for the one where it's (1-x^3) '.

    Now if we plug in 1 for x, we'll get (-3) * [b*c*d*e...]
    But b,c,d,e... follow a simple pattern when x is one. They are all the numbers from 2 to (2n+1).

    Finally, we end up with f' (1) = -3[(2n+1)!]

    Is that correct?
  11. Jan 25, 2012 #10


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    Yes, that is correct. Good job!
  12. Jan 25, 2012 #11
    YAY :) Thanks a lot for both of your help. You guys are the best.

    P.S. @tinytim: Now I see what you were trying to get at the whole time. I just completely misunderstood the question until now. Sorry about that.
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