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Posers and Puzzles

Posers and Puzzles

  1. Subscriber sonhouse
    Fast and Curious
    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 member SwissGambit
    Caninus Interruptus
    28 Sep '12 21:06 / 1 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 member Grampy Bobby
    Boston Lad
    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. Subscriber sonhouse
    Fast and Curious
    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. Subscriber sonhouse
    Fast and Curious
    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 member SwissGambit
    Caninus Interruptus
    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. Subscriber sonhouse
    Fast and Curious
    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 member forkedknight
    Defend the Universe
    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. Subscriber sonhouse
    Fast and Curious
    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. Subscriber sonhouse
    Fast and Curious
    03 Oct '12 20:18 / 2 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. Standard member talzamir
    Art, not a Toil
    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. Subscriber sonhouse
    Fast and Curious
    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 member SwissGambit
    Caninus Interruptus
    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 member SwissGambit
    Caninus Interruptus
    06 Oct '12 21:04 / 1 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. Subscriber sonhouse
    Fast and Curious
    07 Oct '12 13:01 / 1 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.