# Gift taxes paid by benefactor:

sonhouse
Posers and Puzzles 28 Sep '12 16:23
1. 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. SwissGambit
Caninus Interruptus
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. Grampy Bobby
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. 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. 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. 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. 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. 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. 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. sonhouse
Fast and Curious
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. 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. 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. 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. SwissGambit
Caninus Interruptus
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. sonhouse
Fast and Curious
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.