Posers and Puzzles

Posers and Puzzles

  1. Subscriberjoe shmo
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    09 Jan '16 15:361 edit
    What is the proper way to find the limit of this function Y(x) as "a" → 1

    Y(x) = [1/( a^n - 1)] * ( 1 - a^x ) + 1

    where "n" is a constant.

    Lets let the limit of Y(x) as "a" → 1 = y(x)

    Graphically,I know that y(x) is linear in "x", but I cant seem to find the correct mathematical approach to get there.

    Thanks
  2. SubscriberAThousandYoung
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    09 Jan '16 20:22
    Normally limits don't work like that. If the function is Y(x) then you would have x approach some limit, not a different variable.
  3. Subscriberjoe shmo
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    09 Jan '16 20:451 edit
    Originally posted by AThousandYoung
    Normally limits don't work like that. If the function is Y(x) then you would have x approach some limit, not a different variable.
    Well, I could call it Y(a,x).

    The limit of Y(a,x) as a→1 = y(a→1,x) = m(b,n)*x+b

    where m(b,n) is the slope of the functional limit, and is itself a function of n, and b.

    I understand its a bit different from a typical limit, but it appears that a functional limit exists, and is of this form. So how does one get there?
  4. SubscriberAThousandYoung
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    10 Jan '16 00:13
    Is n a constant or a variable?
  5. Subscriberjoe shmo
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    10 Jan '16 01:45
    Originally posted by AThousandYoung
    Is n a constant or a variable?
    "n" is a parameter i'm going to vary for the application, but it is to be treated as a constant for this particular question.

    It represents the zero of Y(x) once "a" is fixed. Y(n) = 0 and y(n) = 0

    The linear equation that it transforms into as a→1 is: y(x) = b - b/n*x

    Just trying to figure out the math procedure to get there, or get an idea on how these problems are typically handled.
  6. SubscriberAThousandYoung
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    10 Jan '16 06:451 edit
    Lim Y(a) as a=>1

    Y(a) = (1-a^x)/(a^n-1)+1

    = (1-a^x+a^n-1)/(a^n-1)

    = (a^n-a^x)/(a^n-1)

    Perhaps L'Hopital's Rule can be used on the above. Take the derivative of top and bottom as many times as needed?

    If we take the nth derivative, we get

    (1-a^(x-n))/1

    lim as a=>1 of 1-a^(x-n) is...0?
  7. Subscriberjoe shmo
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    10 Jan '16 16:33
    Originally posted by AThousandYoung
    Lim Y(a) as a=>1

    Y(a) = (1-a^x)/(a^n-1)+1

    = (1-a^x+a^n-1)/(a^n-1)

    = (a^n-a^x)/(a^n-1)

    Perhaps L'Hopital's Rule can be used on the above. Take the derivative of top and bottom as many times as needed?

    If we take the nth derivative, we get

    (1-a^(x-n))/1

    lim as a=>1 of 1-a^(x-n) is...0?
    I know this isn't allowed, but if we take the derivative of 1 - a^(x-n) and let a=1:

    1 - a^(x-n) => -(x-n)*a^(x-n-1) => -x+n => -1/n*x + 1 which is the functional limit I am after. A very nice and unfortunate coincidence I suppose.
  8. Subscriberjoe shmo
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    10 Jan '16 22:111 edit
    Originally posted by AThousandYoung
    Lim Y(a) as a=>1

    Y(a) = (1-a^x)/(a^n-1)+1

    = (1-a^x+a^n-1)/(a^n-1)

    = (a^n-a^x)/(a^n-1)

    Perhaps L'Hopital's Rule can be used on the above. Take the derivative of top and bottom as many times as needed?

    If we take the nth derivative, we get

    (1-a^(x-n))/1

    lim as a=>1 of 1-a^(x-n) is...0?
    How did you get to that final form by taking n derivatives?

    f(x)/g(x) = (a^n - a^x)/(a^n - 1)

    f'(x)/g'(x) = (n*a^(n-1) - x*a^(x-1))/(n*a^(n-1))

    f"(x)/g"(x) = (n*(n-1)*a^(n-2) - x*(x-1)*a^(x-2))/(n*(n-1)*a^(n-2))
    .
    .
    .
    fⁿ(x)/gⁿ(x) = (n! - x*(x-1)* ... *(x-n+1)*a^(x-n))/(n!)

    fⁿ(x)/gⁿ(x) = 1 - (x*(x-1)* ... *(x-n+1)*a^(x-n))/n!

    How does the rest simplify?
  9. Subscriberjoe shmo
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    11 Jan '16 00:33
    Originally posted by joe shmo
    How did you get to that final form by taking n derivatives?

    f(x)/g(x) = (a^n - a^x)/(a^n - 1)

    f'(x)/g'(x) = (n*a^(n-1) - x*a^(x-1))/(n*a^(n-1))

    f"(x)/g"(x) = (n*(n-1)*a^(n-2) - x*(x-1)*a^(x-2))/(n*(n-1)*a^(n-2))
    .
    .
    .
    fⁿ(x)/gⁿ(x) = (n! - x*(x-1)* ... *(x-n+1)*a^(x-n))/(n!)

    fⁿ(x)/gⁿ(x) = 1 - (x*(x-1)* ... *(x-n+1)*a^(x-n))/n!

    How does the rest simplify?
    Sorry for the notation confusion, that should be:

    f(a)/g(a) = (a^n - a^x)/(a^n - 1)

    f'(a)/g'(a) = (n*a^(n-1) - x*a^(x-1))/(n*a^(n-1))

    f"(x)/g"(x) = (n*(n-1)*a^(n-2) - x*(x-1)*a^(x-2))/(n*(n-1)*a^(n-2))
    .
    .
    .
    fⁿ(a)/gⁿ(a) = (n! - x*(x-1)* ... *(x-n+1)*a^(x-n))/(n!)

    fⁿ(a)/gⁿ(a) = 1 - (x*(x-1)* ... *(x-n+1)*a^(x-n))/n!
  10. SubscriberAThousandYoung
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    14 Jan '16 21:09
    Originally posted by joe shmo
    How did you get to that final form by taking n derivatives?

    f(x)/g(x) = (a^n - a^x)/(a^n - 1)

    f'(x)/g'(x) = (n*a^(n-1) - x*a^(x-1))/(n*a^(n-1))

    f"(x)/g"(x) = (n*(n-1)*a^(n-2) - x*(x-1)*a^(x-2))/(n*(n-1)*a^(n-2))
    .
    .
    .
    fⁿ(x)/gⁿ(x) = (n! - x*(x-1)* ... *(x-n+1)*a^(x-n))/(n!)

    fⁿ(x)/gⁿ(x) = 1 - (x*(x-1)* ... *(x-n+1)*a^(x-n))/n!

    How does the rest simplify?
    I might have made some mistakes. I need to think about it.
  11. Standard memberSoothfast
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    19 Jan '16 02:011 edit
    Originally posted by joe shmo
    What is the proper way to find the limit of this function Y(x) as "a" → 1

    Y(x) = [1/( a^n - 1)] * ( 1 - a^x ) + 1

    where "n" is a constant.

    Lets let the limit of Y(x) as "a" → 1 = y(x)

    Graphically,I know that y(x) is linear in "x", but I cant seem to find the correct mathematical approach to get there.

    Thanks
    For n and x fixed (n not equal to 0), and "a" the only variable, you can use L'Hopital's Rule (differentiating with respect to a) just once to evaluate the limit:

    lim_{a → 1} (1 - a^x)/(a^n - 1) = lim_{a → 1} [-x*a^(x-1)]/[n*a^(n-1)] = -x/n.

    So

    lim_{a → 1}Y(a,x) = -x/n+1
  12. Subscriberjoe shmo
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    21 Jan '16 00:17
    Originally posted by Soothfast
    For n and x fixed (n not equal to 0), and "a" the only variable, you can use L'Hopital's Rule (differentiating with respect to a) just once to evaluate the limit:

    lim_{a → 1} (1 - a^x)/(a^n - 1) = lim_{a → 1} [-x*a^(x-1)]/[n*a^(n-1)] = -x/n.

    So

    lim_{a → 1}Y(a,x) = -x/n+1
    Aha...There it is! Haven't thought about finding limits for some time, and it turns out we were on wild goose chase with the multiple derivatives. Thank You Soothfast!
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