28 Sep 12
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?
28 Sep 12
1 edit
Originally posted by sonhouseIn general:
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?
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.
28 Sep 12
Originally posted by SwissGambitWow!
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.
29 Sep 12
Originally posted by SwissGambitVery 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.
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.
30 Sep 12
Originally posted by SwissGambitWhere did the X !=1 come from? It seems to be just stuck in there.
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.
03 Oct 12
2 edits
Originally posted by SwissGambitThe line with the ! in it, isn't that infinity? x/1-1, or X/0?
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.
04 Oct 12
!= 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.
04 Oct 12
Originally posted by talzamirWell, 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.
!= 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.
06 Oct 12
1 edit
Originally posted by SwissGambitOK, here is as spiffy a solution as I could find.
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?
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.
07 Oct 12
1 edit
Originally posted by SwissGambitWhat does that do, if anything, to the original answer of $7500?
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]
Nice bit of analysis for sure.