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Posers and Puzzles

Posers and Puzzles

  1. Subscriber joe shmo On Vacation
    Strange Egg
    09 Jan '16 15:36 / 1 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. Subscriber AThousandYoung
    It's about respect
    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. Subscriber joe shmo On Vacation
    Strange Egg
    09 Jan '16 20:45 / 1 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. Subscriber AThousandYoung
    It's about respect
    10 Jan '16 00:13
    Is n a constant or a variable?
  5. Subscriber joe shmo On Vacation
    Strange Egg
    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. Subscriber AThousandYoung
    It's about respect
    10 Jan '16 06:45 / 1 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. Subscriber joe shmo On Vacation
    Strange Egg
    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. Subscriber joe shmo On Vacation
    Strange Egg
    10 Jan '16 22:11 / 1 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. Subscriber joe shmo On Vacation
    Strange Egg
    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. Subscriber AThousandYoung
    It's about respect
    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 member Soothfast
    0,1,1,2,3,5,8,13,21,
    19 Jan '16 02:01 / 1 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. Subscriber joe shmo On Vacation
    Strange Egg
    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!