[Guest]

Einsein had once said that the strongest power in this world is the power of compounding.

The story (not by Einstein) goes as: When a great mathematician developed the game of chess, he brought it to the court of the king and explained the game. The king as well as the courtiers were amazed and the king promised him any thing on the earth, he wished, as the reward. The mathematician said, I am a poor man and I have no great wishes, but since if you want to reward me, just put one paisa (Indian cent) on the first square, 2 paisa on the second, and so on, go on doubling the amount of money on every square till the 64th one. The king said, I pity on you great man, you could have asked for much more. Anyway, keep this bag of 10,000 rupees, arrange the petty paisa on the squares the way you like and keep the balance money. The mathematician insisted that this operation be done by the king's men. Anyway, the king agreed and ordered his treasurer to do the needful.

After sometime, the treasurer came to the king and said, Your highness, 10,000 rpees are all gone and still squares are left over. The king, a bit surprised, told him to use whatever extra money is required and not to disturb him for these small issues. After some more time, the treasurer again came running, in a nervous state, and said, excuse me Your highness, but our treasury is exhausted and still squares are left over. The king was aghast - we have billions of billions of rupees in our treasury and you mean all of it is exhausted by putting some petty paisa on the squares. Yes my lord ...

Well, the problem is as follows:

If you deposit $100 in a bank on simple interest rate of 10% per anum, the money will be doubled in 10 years. If another bank offers compound interest - compounded anually, it will be doubled in much less time. Better still, if it is coumpounded monthly. Find out the absolute minimum time in which it will double. No approximation please.

[Guest]

6.9315 Years

[superprismatic]

6.9315 Years

10×ln(2)

[Guest]

10*ln(2) years

[Aaryan]

10*In(2) years.

[Guest]

Thanks Simon, superprismatic, magician and Aryan.

If it was too easy try this.

X promised to pay to Y, $1 on the first day, $1/2 on the second day, $1/3 on the third day and so on. If they live forever, how much money will X end up pauing to Y?

[Guest]

ln(2)/(ln(1+10/1200)) months = 83.52376 months = 6.9603 years

[Guest]

ln(2)/(ln(1+10/1200)) months = 83.52376 months = 6.9603 years but if interest is only posted at the end of each month then 84 months is the minimum time ( 7 years)

[Guest]

ln(2)/(ln(1+10/1200)) months = 83.52376 months = 6.9603 years but if interest is only posted at the end of each month then 84 months is the minimum time ( 7 years)

inert, go on reducing the period of compounding, and find the least period.

[Guest]

There is no limit for the the sum of 1/n for integer n from 1 to infinity. Though X may have trouble paying the tiny slivers of a penny as time passes. More problematic is what to do when the payment requires less than a single atom from a penny. But those atoms do add up given enough days. An atom saved is an atom earned. But what happens when the payment is so small that it is significantly affected by quantum mechanics and wave functions?

Thanks Simon, superprismatic, magician and Aryan.

If it was too easy try this.

X promised to pay to Y, $1 on the first day, $1/2 on the second day, $1/3 on the third day and so on. If they live forever, how much money will X end up pauing to Y?

[Guest]

lim(a->inf) (1/(a*log2(1+0.1/a))) ~ 6.93147 years

since harmonic series is a divergent one, X will end up with infinite amount of money

[Guest]

How about combining the first two questions? X agrees to pay Y 1 dollar on day 1, 1/2 dollar on day 2, and so on. And Y agrees that repay X each amount after 2 days, but Y keeps the compound interest earned based on continuous compounding with the same .10 interest rate from question 1? So on days 1 and 2, Y pays X nothing and has 1.50. On day 3, Y repays the 1 dollar to X and is paid 1/3 dollar, for a balance total of .83 and a third plus the interest earned so far. On day 4, Y repays the 1/2 dollar from day 2 and is paid 1/4 dollar. How much money will Y have as forever approaches? Assume all transactions happen at the same time every day.

Edited May 30, 2011 by Nana7

[Guest]

Your answer to the logical part, - "There is no limit for the the sum of 1/n for integer n from 1 to infinity" is correct. The sum of the series is divergent and hence, not withstanding the Quantum Mechanics' limitation, X will end up paying infinite sum to Y.

There is no limit for the the sum of 1/n for integer n from 1 to infinity. Though X may have trouble paying the tiny slivers of a penny as time passes. More problematic is what to do when the payment requires less than a single atom from a penny. But those atoms do add up given enough days. An atom saved is an atom earned. But what happens when the payment is so small that it is significantly affected by quantum mechanics and wave functions?

[Guest]

Interest earned by Y in 2 days will be proportional to the principal amt lent by X, the proportionality constant being exp (1/1825). Being more specific, On day-1, Y receives $1 and returns it on day-3. The interest earned by Y in 2 days =1*exp(1/1825). The interest earned on $1/2 lent by X on the 2nd day and returned by Y on the 4th day is (1/2)*exp(1/1825), ... Therefore, the sum of the series becomes exp(1/1825)*Sum (1, 1/2, 1/3, ...). Remember, Sum (1, 1/2, 1/3, ...) tends to infinity as the no. of terms increase. Therefore, profit of Y tends to becomes infinity with passage of time (hell of a lot of time though)

How about combining the first two questions? X agrees to pay Y 1 dollar on day 1, 1/2 dollar on day 2, and so on. And Y agrees that repay X each amount after 2 days, but Y keeps the compound interest earned based on continuous compounding with the same .10 interest rate from question 1? So on days 1 and 2, Y pays X nothing and has 1.50. On day 3, Y repays the 1 dollar to X and is paid 1/3 dollar, for a balance total of .83 and a third plus the interest earned so far. On day 4, Y repays the 1/2 dollar from day 2 and is paid 1/4 dollar. How much money will Y have as forever approaches? Assume all transactions happen at the same time every day.

[pdqkemp]

lim(a->inf) (1/(a*log2(1+0.1/a))) ~ 6.93147 years

since harmonic series is a divergent one, X will end up with infinite amount of money

I don't believe the second one is right. (I have no idea about the first.)

It would depend on what is possible for a minimum payment. If you have to use conventional methods, and the minimum is 1 of ANY form of real currency, then of course the answer is infinity. If, however, you can go to fractions of the smallest currency, then the number approaches $3.00.

[Guest]

All,

Please dont bother for currency limitations and also quantum physics' limitations of subdividing atoms. X and Y keep their books of accounts, and when amount becomes more than 1 cent, they clear it off.

[Guest]

At any instant of time, if balance is x, the interest accrued in time dt, = dx = x*intt rate*dt.

Therefore, (1/x)*dx = (1/10)*dt

Integrating both sides for time = 0 to t, we get t = 10*ln(x_final/x_initial). As x_final = 2*x_initial,

t = 10*ln2

lim(a->inf) (1/(a*log2(1+0.1/a))) ~ 6.93147 years

since harmonic series is a divergent one, X will end up with infinite amount of money

[pdqkemp]

I don't believe the second one is right. (I have no idea about the first.)

It would depend on what is possible for a minimum payment. If you have to use conventional methods, and the minimum is 1 of ANY form of real currency, then of course the answer is infinity. If, however, you can go to fractions of the smallest currency, then the number approaches $3.00.

OK - forget what I said. I was thinking of something different. I was thinking 1 + 1/2 + 1/4 + 1/8 + 1/16 + 1/32......in which case the amount would approach $2.00, not $3.00.

For this problem, it needs to be 1 + 1/2 + 1/3 + 1/4 + 1/5 + 1/6 + 1/7 + 1/8..... right?

I'm not quite sure how to solve this one, but I'm sure it'll be easy for someone.

[Guest]

Thanks everybody. Summing up:

1. The 1st question was solved by Simon, superprismatic, magician and Aryan. The proof is:

At any instant of time, if balance is x, the interest accrued in time dt, = dx = x*intt rate*dt.

Therefore, (1/x)*dx = (1/10)*dt

Integrating both sides for time = 0 to t, we get t = 10*ln(x_final/x_initial). As x_final = 2*x_initial,

t = 10*ln2 years

2. The second question has been solved by det and Nana7. The proof is as follows:

S = 1 + 1/2 + 1/3 + 1/4 + ...

S = 1 + 1/2 + (1/3+1/4) + (1/5+1/6+1/7+1/8) + (1/9 + ... 1/16) .. Go on doubling the numbers under the parantheses.

Be modest and replace all nos. in the 1st paranthesis by 1/4, all nos. in 2nd paranthesis by 1/8, 3rd paranthesis by 1/16, we get:

S > 1 + 1/2 + (1/4+1/4) + (1/8+1/8+1/8+1/8) + (1/16+ ... +1/16) ...

S > 1 + 1/2 +1/2 + 1/2 + ...

Therefore S can be made greater than any given number by choosing appropriate no. of terms.

Therefore, S tends to infinity.

3. Thanks Nana7 for supplementing the question. The solution is as follows:

Interest earned by Y in 2 days will be proportional to the principal amt lent by X, the proportionality constant being exp (1/1825). Being more specific, On day-1, Y receives $1 and returns it on day-3. The interest earned by Y in 2 days =1*exp(1/1825). The interest earned on $1/2 lent by X on the 2nd day and returned by Y on the 4th day is (1/2)*exp(1/1825), ... Therefore, the sum of the series becomes exp(1/1825)*Sum (1, 1/2, 1/3, ...). Remember, Sum (1, 1/2, 1/3, ...) tends to infinity as the no. of terms increase. Therefore, profit of Y tends to becomes infinity with passage of time (hell of a lot of time though)

P.S. Qu no. 2 shows that sum of a convergent series can be divergent. Can you think of any other example??

Edited May 31, 2011 by Mukul Verma

[Guest]

another example: sum(n=2->inf) 1/(n*ln(n))

to the 1st problem: my method gives the same result as yours, I simply started from finite amounts and then shifted (with lim) to the realm of infinitesimals

problem #4:

Y after collecting the money from X decided to give back every "second group" of money (1/2, 1/4, 1/6 ...) so Y has this series of money: 1, -1/2, 1/3, -1/4, 1/5, ... How much money will Y end up? (don't bother with the interest)

Edited May 31, 2011 by det

[Guest]

Problem: S = 1 - 1/2 + 1/3 - 1/4 + 1/5 - ...infinite terms

Solution:

Let S1 = 1/2 + 1/4 + 1/6 + ...

S2 = 1/3 + 1/5 + 1/7 + ...

Therefore, S = 1 - S1 + S2

Now subtract 1 from the denominator of each term of expansion of S2, the resulting series should be > S2.

S3 = 1/2 + 1/4 + 1/6 + ...

Since S3 > S2

Therefore, S < 1 - S1 + S3

Substituting for S1 and S3, S < 1

Now add 1 to the denominator of each term of expansion of S2, the resulting series should be < S2.

S4 = 1/4 + 1/6 + 1/8 + ...

Since S4 < S2,

S > 1 - S1 + S4

Substituting for S1 and S4, S > 1/2

Therefore, S is between 1/2 and 1 (both exclusive)

another example: sum(n=2->inf) 1/(n*ln(n))

to the 1st problem: my method gives the same result as yours, I simply started from finite amounts and then shifted (with lim) to the realm of infinitesimals

problem #4:

Y after collecting the money from X decided to give back every "second group" of money (1/2, 1/4, 1/6 ...) so Y has this series of money: 1, -1/2, 1/3, -1/4, 1/5, ... How much money will Y end up? (don't bother with the interest)

det,

1. The series sum(n=2->inf) 1/(n*ln(n)) is indeed convergent, but I am not sure whether the sum is divergent (tending to infinity). Can you give a proof?

2. Your answer to the 1st question of finding the shortest time in which the money will be doubled, is numerically correct, but I have 2 questions - (i) How did you get the expression lim(a->inf) (1/(a*log2(1+0.1/a))), and (ii) lim(a->inf) (1/(a*log2(1+0.1/a))) tends to 0 and not to 6.93147 years as given in your answer. Can you please clarify these 2 points?

3. Thanks for your qu no. 4; it is indeed very interesting.

I have tried to solve it. see the spoiler.

Edited June 1, 2011 by Mukul Verma

[Guest]

1.

imagine the sum as the sum of rectangles (width = 1, height = 1/(n*ln(n)) )

take this continuous function: y(x) = 1/(x*ln(x)), now its clear that y(x) is always under rectangles upper lines, so the area (which is actually the sum of the series in case of the rectangles) under it is smaller than the sum area of rects.

if we can show that the improper integral of y(x) is infinite (2->inf) we proved that the original series is divergent; actually the integral is infinite

2. (i)

lets start with yearly compounding interest, to get years we have to solve this equation:

100 * (1,1)^y = 200 (y - years)

y = 1/ log2(1,1)

( log2(1,1) = log₂(1,1) )

do it with months:

100 * (1+0,1/12)^m = 200 (m - months)

m = 1/ log2(1+0,1/12)

in years: y = m/ 12 = 1/ (log2(1+0,1/12) * 12)

if we generalize this:

y = 1/ (log2(1+0,1/a) * a)

and with the time interval tends to zero (a->inf):

y = lim(a->inf) (1/ (log2(1+0,1/a) * a)) ~ 6,93147 years (10 * ln(2))

thats why I said I shifted to the infinitesimals at the end; in your solution you started with them

2. (ii)

cause its hard to read the expression:

this tends to the above mentioned number (try it with a big number)

3. nicely done, but

actually the upper and lower bound value of the problem (stated by you : (S = 1 - 1/2 + 1/3 - 1/4 + 1/5 - ...)) you have found are both solutions of my original problem

to help you (the interesting part of the question is not the numeric calculation), here is the sum of your series: S = 1 - 1/2 + 1/3 - 1/4 + 1/5 - ... = ln(2)

so how about the solution of the original problem #4? (you solved a bit different one, or only a part of it)

try to add the numbers in a different order ie.:

S = 1 + 1/3 - 1/2 + 1/5 + 1/7 - 1/4 + 1/9 + 1/11 - 1/6 + ...

does it matter how you add the same numbers?

[Guest]

1.

imagine the sum as the sum of rectangles (width = 1, height = 1/(n*ln(n)) )

take this continuous function: y(x) = 1/(x*ln(x)), now its clear that y(x) is always under rectangles upper lines, so the area (which is actually the sum of the series in case of the rectangles) under it is smaller than the sum area of rects.

if we can show that the improper integral of y(x) is infinite (2->inf) we proved that the original series is divergent; actually the integral is infinite

2. (i)

lets start with yearly compounding interest, to get years we have to solve this equation:

100 * (1,1)^y = 200 (y - years)

y = 1/ log2(1,1)

( log2(1,1) = log₂(1,1) )

do it with months:

100 * (1+0,1/12)^m = 200 (m - months)

m = 1/ log2(1+0,1/12)

in years: y = m/ 12 = 1/ (log2(1+0,1/12) * 12)

if we generalize this:

y = 1/ (log2(1+0,1/a) * a)

and with the time interval tends to zero (a->inf):

y = lim(a->inf) (1/ (log2(1+0,1/a) * a)) ~ 6,93147 years (10 * ln(2))

thats why I said I shifted to the infinitesimals at the end; in your solution you started with them

2. (ii)

cause its hard to read the expression:

this tends to the above mentioned number (try it with a big number)

3. nicely done, but

actually the upper and lower bound value of the problem (stated by you : (S = 1 - 1/2 + 1/3 - 1/4 + 1/5 - ...)) you have found are both solutions of my original problem

to help you (the interesting part of the question is not the numeric calculation), here is the sum of your series: S = 1 - 1/2 + 1/3 - 1/4 + 1/5 - ... = ln(2)

so how about the solution of the original problem #4? (you solved a bit different one, or only a part of it)

try to add the numbers in a different order ie.:

S = 1 + 1/3 - 1/2 + 1/5 + 1/7 - 1/4 + 1/9 + 1/11 - 1/6 + ...

does it matter how you add the same numbers?

1. I am still trying to figure out item#1. Will revert back.

2. Agreed 100%

3. I see the trick. However, still the answer is between 1/2 and 1 and my approximation holds.

[Guest]

1. I am still trying to figure out item#1. Will revert back.

2. Agreed 100%

3. I see the trick. However, still the answer is between 1/2 and 1 and my approximation holds.

look at the second series I added to consider:

S = 1 + 1/3 - 1/2 + 1/5 + 1/7 - 1/4 + 1/9 + 1/11 - 1/6 + ... = 1,5 * ln(2) > 1

so what do you think now?

[dark_magician_92]

if we take rate=10%, then whats the point of minimum time??

anyways 6 years 351 days i guess

[dark_magician_92]

ok i got it, i m wrong