Monday, March 06, 2006
Maths Qn
After getting home, dwin asked me a maths qn, not because he doesn't know how to do it, but just to let me try. Sadly, my maths is getting really rusty and it took me a long time and lots of hints from zq to work this one out.
The question was first to find a nice simplified expression for 1/(1*2) + 1/(2*3) + 1/(3*4) +...
Each term is basically 1/[n(n+1)] which is 1/n - 1/(n+1). When it is all summed up, all the intermediate terms conveniently cancel out, leaving 1 - 1/(n+1).
Ok, so that's part 1. Part 2 is to find the sum of 2/(1*2*3) + 2/(2*3*4) + 2/(3*4*5) +...+2/(1997*1998*1999) + 2/(1998*1999*2000).
My first instinct was to write a program to solve it. After all, there's an obvious pattern and the number of terms are finite. In Java, the exact code would be
double sum=0;
double n=1;
for (n=n;n<1998;n++) {
sum += 2/(n*(n+1)*(n+2));
}
System.out.println(sum);
and in a couple of seconds, the program will compile and give you the answer. How wonderful.
Unfortunately, I don't think that's what they were looking for. The examiners would probably want a nice answer in a very precise fraction, and I wouldn't have my java compiler in a maths exam.
After many hints from zq (ok he basically told me the answer), I figured out that I shouldn't split it fully into partial fractions, but instead express it as 1/n(1/(n+1) - 1/(n+2)). The 2 can be multiplied back later.
This gives 2 * sigma[n=1,1998] {1/n - 1/(n+1)} - sigma[n=1,1998] {1/n - 1/(n+2)}. The 2 in the second term cancels out the denominator you would get by splitting 1/[n(n+2)].
If you write out a few terms for each sigma bit, you would realise that the terms cancel out. The first one is an exact duplicate of the 1st part of the question, so you get 2*(1-1/1999).
The 2nd halfway is a little trickier, but nearly all terms still cancel out, leaving (1+1/2-1/2000).
Thus the final expression is 2*(1 - 1/1999) - (1 + 1/2 - 1/2000).
Simplified, it gives 1/2 - 2/1999 + 1/2000.
While this answer is very close to the one I got from my java program, it's not exactly the same. Shall have to ask dwin if the discrepancy is due to rounding errors or maybe my answer is wrong. If anyone finds any errors, please inform me, thanks.
For those who want to know, the answer given by my java prog is 0.49999974962456195.
Calculating a decimal number from the fractional expression in java gives 0.499499499749875.
So the difference is 0.00050024987468695.
The question was first to find a nice simplified expression for 1/(1*2) + 1/(2*3) + 1/(3*4) +...
Each term is basically 1/[n(n+1)] which is 1/n - 1/(n+1). When it is all summed up, all the intermediate terms conveniently cancel out, leaving 1 - 1/(n+1).
Ok, so that's part 1. Part 2 is to find the sum of 2/(1*2*3) + 2/(2*3*4) + 2/(3*4*5) +...+2/(1997*1998*1999) + 2/(1998*1999*2000).
My first instinct was to write a program to solve it. After all, there's an obvious pattern and the number of terms are finite. In Java, the exact code would be
double sum=0;
double n=1;
for (n=n;n<1998;n++) {
sum += 2/(n*(n+1)*(n+2));
}
System.out.println(sum);
and in a couple of seconds, the program will compile and give you the answer. How wonderful.
Unfortunately, I don't think that's what they were looking for. The examiners would probably want a nice answer in a very precise fraction, and I wouldn't have my java compiler in a maths exam.
After many hints from zq (ok he basically told me the answer), I figured out that I shouldn't split it fully into partial fractions, but instead express it as 1/n(1/(n+1) - 1/(n+2)). The 2 can be multiplied back later.
This gives 2 * sigma[n=1,1998] {1/n - 1/(n+1)} - sigma[n=1,1998] {1/n - 1/(n+2)}. The 2 in the second term cancels out the denominator you would get by splitting 1/[n(n+2)].
If you write out a few terms for each sigma bit, you would realise that the terms cancel out. The first one is an exact duplicate of the 1st part of the question, so you get 2*(1-1/1999).
The 2nd halfway is a little trickier, but nearly all terms still cancel out, leaving (1+1/2-1/2000).
Thus the final expression is 2*(1 - 1/1999) - (1 + 1/2 - 1/2000).
Simplified, it gives 1/2 - 2/1999 + 1/2000.
While this answer is very close to the one I got from my java program, it's not exactly the same. Shall have to ask dwin if the discrepancy is due to rounding errors or maybe my answer is wrong. If anyone finds any errors, please inform me, thanks.
For those who want to know, the answer given by my java prog is 0.49999974962456195.
Calculating a decimal number from the fractional expression in java gives 0.499499499749875.
So the difference is 0.00050024987468695.
posted by cosine @ 12:16 PM 
