1 + x + x^{2} + ... + x^{n1}
is (x^{n}  1)/(x  1). One of my beliefs (as also shown by that post) is that you should never take a formula at its face value. Here I quickly attempt to show why this particular formula should be correct using a very commonly used proof. Let us assume that S denotes the desired sum. That is:
S = 1 + x + x^{2} + ... + x^{n1}
If we multiply both sides of this equation by x, we get:
If we subtract each side of the first equation from the corresponding side of the second equation we get:
This simplifies to:
which gives:
as we had claimed.
By the way, while attending a conference call last night where it was particularly hard to concentrate on what was being discussed, I began to doodle and stumbled upon a simple result which might have been interesting about a millennium ago and is probably just an idle curiosity now. I am also sure it is contained somewhere in our high school text books for mathematics, but I cannot recall seeing it before. (Despite what might be indicated by these blog posts, I do not have much of an interest in mathematics itself and have never pursued it for its own sake.)
Anyways, looking at the factorisation of (x^{2}  y^{2}) into (x  y)×(x + y) and (x^{3}  y^{3}) into (x  y)×(x^{2} + x×y + y^{2}), I wondered if (x  y) is always a factor of (x^{n}  y^{n}). By playing around a little bit with the terms, I could write:
confirming my hypothesis. If you set y=1, this reduces to:
which can be rewritten as:
which is the same formula that we derived just a while back for calculating the sum of a geometric series!
If we multiply both sides of this equation by x, we get:
x×S = x + x^{2} + x^{3} + ... + x^{n}
If we subtract each side of the first equation from the corresponding side of the second equation we get:
(x  1)×S = x + x^{2} + x^{3} + ... + x^{n}  1  x  x^{2}  ...  x^{n1}
This simplifies to:
(x  1)×S = x^{n}  1
which gives:
S = (x^{n}  1)/(x  1)
as we had claimed.
By the way, while attending a conference call last night where it was particularly hard to concentrate on what was being discussed, I began to doodle and stumbled upon a simple result which might have been interesting about a millennium ago and is probably just an idle curiosity now. I am also sure it is contained somewhere in our high school text books for mathematics, but I cannot recall seeing it before. (Despite what might be indicated by these blog posts, I do not have much of an interest in mathematics itself and have never pursued it for its own sake.)
Anyways, looking at the factorisation of (x^{2}  y^{2}) into (x  y)×(x + y) and (x^{3}  y^{3}) into (x  y)×(x^{2} + x×y + y^{2}), I wondered if (x  y) is always a factor of (x^{n}  y^{n}). By playing around a little bit with the terms, I could write:
x^{n}  y^{n} = (x  y)×(x^{n1} + x^{n2}×y + ... + x×y^{n2} + y^{n1})
confirming my hypothesis. If you set y=1, this reduces to:
x^{n}  1^{} = (x  1)×(x^{n1} + x^{n2} + ... + x + 1^{})
which can be rewritten as:
(x^{n}  1)/(x  1) = x^{n1} + x^{n2} + ... + x + 1
which is the same formula that we derived just a while back for calculating the sum of a geometric series!

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i have used this formula, it works, especially for yearly rests, it does not work with monthly rests. is there something missing?
ReplyDeleteMatthew, Kenya, Nairobi
I believe you're commenting on the other post rather than this one. In any case, the formula holds for both yearly rests and monthly rests  just make sure you put in the rates and the periods correctly. You might want to carefully read the other post once again.
ReplyDeleteWith reference to your earlier post on EMI formula, i hope u agree it will not work if the rate is zero?
ReplyDeleteramkumar: Yes. In that case the EMI should simply be equal to "P / n".
ReplyDeleteDear Ranjit,
ReplyDeleteGreat job done. Thanks to you I am able to calculate the EMI. I am indebted to you for guiding educated fools like me on how to use simple algebra to derive at a formula.
Felt like school days. Keep it up dear and God bless you.