1. Subscribersonhouse
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    28 Sep '12 16:23
    So a guy gets a free car worth $30,000 US and the IRS sends a tax bill for 20% or $6,000. So the company giving away the car pays the tax. The IRS sends a bill for 20% of that, now he owes $1200. So the company pays that. The IRS now sends a bill for $240, 20% of the last one, etc.

    How much money does the company have to pay so the guy getting the car pays no tax using that rule?
  2. Standard memberSwissGambit
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    28 Sep '12 21:061 edit
    Originally posted by sonhouse
    So a guy gets a free car worth $30,000 US and the IRS sends a tax bill for 20% or $6,000. So the company giving away the car pays the tax. The IRS sends a bill for 20% of that, now he owes $1200. So the company pays that. The IRS now sends a bill for $240, 20% of the last one, etc.

    How much money does the company have to pay so the guy getting the car pays no tax using that rule?
    In general:

    sn = x + x^2 + x^3 + ... + x^n

    Multiply all terms by x

    x(sn) = x^2 + x^3 + ... + x^(n+1)

    Subtract the second equation from the first:

    (1-x)sn = x - x^(n+1)

    For x != 1, sn = (x - x^(n-1))/(1-x)

    For |x| < 1, the limit as n goes to infinity is:

    sn = x/(1-x)

    In this case, x = 0.2, so

    sn = 0.2/(1 - 0.2) = 0.25

    So the tax the company pays is:

    30000 * 0.25 = $7500

    I must credit http://www.math.hmc.edu/calculus/tutorials/convergence/ for the calculus refresher.
  3. Standard memberGrampy Bobby
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    28 Sep '12 23:20
    Originally posted by SwissGambit
    In general:

    sn = x + x^2 + x^3 + ... + x^n

    Multiply all terms by x

    x(sn) = x^2 + x^3 + ... + x^(n+1)

    Subtract the second equation from the first:

    (1-x)sn = x - x^(n+1)

    For x != 1, sn = (x - x^(n-1))/(1-x)

    For |x| < 1, the limit as n goes to infinity is:

    sn = x/(1-x)

    In this case, x = 0.2, so

    sn = 0.2/(1 - 0.2) = 0.25

    So t ...[text shortened]... must credit http://www.math.hmc.edu/calculus/tutorials/convergence/ for the calculus refresher.
    Wow!
  4. Subscribersonhouse
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    29 Sep '12 15:13
    Originally posted by SwissGambit
    In general:

    sn = x + x^2 + x^3 + ... + x^n

    Multiply all terms by x

    x(sn) = x^2 + x^3 + ... + x^(n+1)

    Subtract the second equation from the first:

    (1-x)sn = x - x^(n+1)

    For x != 1, sn = (x - x^(n-1))/(1-x)

    For |x| < 1, the limit as n goes to infinity is:

    sn = x/(1-x)

    In this case, x = 0.2, so

    sn = 0.2/(1 - 0.2) = 0.25

    So t ...[text shortened]... must credit http://www.math.hmc.edu/calculus/tutorials/convergence/ for the calculus refresher.
    Very good! I just did it the hard way, went down to where there was less than a penny and found it was converging on 7500, clearly the limit.
  5. Subscribersonhouse
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    30 Sep '12 17:51
    Originally posted by SwissGambit
    In general:

    sn = x + x^2 + x^3 + ... + x^n

    Multiply all terms by x

    x(sn) = x^2 + x^3 + ... + x^(n+1)

    Subtract the second equation from the first:

    (1-x)sn = x - x^(n+1)

    For x != 1, sn = (x - x^(n-1))/(1-x)

    For |x| < 1, the limit as n goes to infinity is:

    sn = x/(1-x)

    In this case, x = 0.2, so

    sn = 0.2/(1 - 0.2) = 0.25

    So t ...[text shortened]... must credit http://www.math.hmc.edu/calculus/tutorials/convergence/ for the calculus refresher.
    Where did the X !=1 come from? It seems to be just stuck in there.
  6. Standard memberSwissGambit
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    30 Sep '12 19:09
    Originally posted by sonhouse
    Where did the X !=1 come from? It seems to be just stuck in there.
    "!=" means "not equal to" in C++. It came from the desire to avoid a 0 in the denominator.
  7. Subscribersonhouse
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    01 Oct '12 10:08
    Originally posted by SwissGambit
    "!=" means "not equal to" in C++. It came from the desire to avoid a 0 in the denominator.
    I was wondering, because in math (!)= factorial.
  8. Standard memberforkedknight
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    02 Oct '12 05:54
    Originally posted by sonhouse
    I was wondering, because in math (!)= factorial.
    yeah, my math professors preferred us to use ~= or =/=, but I always liked programming notation better (!=)
  9. Subscribersonhouse
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    03 Oct '12 20:16
    Originally posted by forkedknight
    yeah, my math professors preferred us to use ~= or =/=, but I always liked programming notation better (!=)
    You never confused that with factorial?
  10. Subscribersonhouse
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    03 Oct '12 20:182 edits
    Originally posted by SwissGambit
    In general:

    sn = x + x^2 + x^3 + ... + x^n

    Multiply all terms by x

    x(sn) = x^2 + x^3 + ... + x^(n+1)

    Subtract the second equation from the first:

    (1-x)sn = x - x^(n+1)

    For x != 1, sn = (x - x^(n-1))/(1-x)

    For |x| < 1, the limit as n goes to infinity is:

    sn = x/(1-x)

    In this case, x = 0.2, so

    sn = 0.2/(1 - 0.2) = 0.25

    So t ...[text shortened]... must credit http://www.math.hmc.edu/calculus/tutorials/convergence/ for the calculus refresher.
    The line with the ! in it, isn't that infinity? x/1-1, or X/0?
  11. Subscribertalzamir
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    04 Oct '12 12:01
    != is what some use for non-equal.. and not equal to one is necessary as otherwise the method has you divide by zero.

    With combinations of inequalities in use, like =< for less than or equal to.. would make sense to me to use <> for !=, not equal to meaning less than or more than. After all, =< is not done by writing > and then a slash over it either.
  12. Subscribersonhouse
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    04 Oct '12 13:31
    Originally posted by talzamir
    != is what some use for non-equal.. and not equal to one is necessary as otherwise the method has you divide by zero.

    With combinations of inequalities in use, like =< for less than or equal to.. would make sense to me to use <> for !=, not equal to meaning less than or more than. After all, =< is not done by writing > and then a slash over it either.
    Well, you could use /= just as well. Just can't put them on top of one another like we are used to. Seems more intuitive to me than an exclamation point which I confused with the factorial operation.
  13. Standard memberSwissGambit
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    05 Oct '12 20:27
    Originally posted by sonhouse
    The line with the ! in it, isn't that infinity? x/1-1, or X/0?
    Here's something I couldn't look up:

    The limit, as x goes to 1, of...

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

    ...is n-2, but why?
  14. Standard memberSwissGambit
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    06 Oct '12 21:041 edit
    Originally posted by SwissGambit
    Here's something I couldn't look up:

    The limit, as x goes to 1, of...

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

    ...is n-2, but why?
    OK, here is as spiffy a solution as I could find.

    Convert the equation

    f(x) = (x-x^(n-1))/(1-x)
    = x(x^(n-2)-1))/(x-1)

    Let's take the denominator off for a bit, so it's easier to read.

    x(x^(n-2)-1))

    Now we start adding and subtracting the same term from the numerator over and over to factor out x-1 from every term. This will create new terms, but we'll do the same trick on them until EVERY term has an (x-1) in it.

    x(x^(n-2) - x^(n-3) + x^(n-3) - 1)
    x(x^(n-3)(x-1) + x^(n-3) - x^(n-4) + x^(n-4) - 1)
    x(x^(n-3)(x-1) + x^(n-4)(x-1) + x^(n-4) - 1)

    Keep doing this until you have dropped x's exponent down to one and you're left with an x-1 at the end.

    x(x^(n-3)(x-1) + x^(n-4)(x-1) + ... + x(x-1) + x-1)

    Now add the denominator back in:

    x((x^(n-3)(x-1) + x^(n-4)(x-1) + ... + x(x-1) + x-1))/(x-1)

    Cancel out all the (x-1)'s:

    x(x^(n-3) + x^(n-4) + ... + x + 1)

    Multiply the outer x through:

    f(x) = x^(n-2) + x^(n-3) + ... + x^2 + x

    Now that the denominator is gone, there is no more divide by 0 problem. With x=1 in each term, and n-2 total terms, the limit of f(x) as x approaches 1 is n-2.
  15. Subscribersonhouse
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    07 Oct '12 13:011 edit
    Originally posted by SwissGambit
    OK, here is as spiffy a solution as I could find.

    Convert the equation

    f(x) = (x-x^(n-1))/(1-x)
    = x(x^(n-2)-1))/(x-1)

    Let's take the denominator off for a bit, so it's easier to read.

    x(x^(n-2)-1))

    Now we start adding and subtracting the same term from the numerator over and over to factor out x-1 from every term. This will create new ter x=1 in each term, and n-2 total terms, the limit of f(x) as x approaches 1 is [b]n-2
    .[/b]
    What does that do, if anything, to the original answer of $7500?

    Nice bit of analysis for sure.
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