- 28 Sep '12 16:23So 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 21:06 / 1 edit

In general:*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?

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 23:20

Wow!*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. - 29 Sep '12 15:13

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.*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. - 30 Sep '12 17:51

Where did the X !=1 come from? It seems to be just stuck in there.*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. - 03 Oct '12 20:18 / 2 edits
*Originally posted by SwissGambit*

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 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. - 04 Oct '12 13:31

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.*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. - 06 Oct '12 21:04 / 1 edit

OK, here is as spiffy a solution as I could find.*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?

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 13:01 / 1 edit

What does that do, if anything, to the original answer of $7500?*Originally posted by SwissGambit***OK, here is as spiffy a solution as I could find.**.[/b]

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

Nice bit of analysis for sure.