Can someone show me the formula for calculating 1+(1+2)+(1+2+3)+(1+2+3+4)+...
Yes, I presume you have a finite sum.Originally Posted by OReilly
The key in the the fact that,
Therefore, that entire sum is,
Thus, subdivide the summation,
Use the fact above, and the sum of squares formula,
Add, common denominator and add,
Just for refernce, the sum of squares formula is,
I think it is cool that,
That means you can calculate the sum by taking 2 more than number of terms and finding the number of combinations of forming three.
TPHacker is absolutely correct . . . Nice job, T.P. !
If we are desperate, we could find the formula "from scratch".
Can someone show me the formula for calculating:
. . .
Crank out a list of the first few sums:
We have a function . .For
. . the function has consecutive values: .
Take differences of consecutive terms: . .
Take differences again: . . . . . . . . . . . . . .
Take differences again: . . . . . . . . . . . . . . .
We have a series of constants at the third differences.
. . Hence, is a third-degree polynomial ... a cubic.
Hence, the function is of the form: .
. . and we must determine
Use the first four values from our list.
Now we must solve this system of equations . . . but it's easy!
Subtract (1) from (2): .
Subtract (2) from (3): .
Subtract (3) from (4): .
Subtract (5) from (6): .
Subtract (6) from (7): .
Subtract (8) from (9): .
Substitute into (8): .
Substitute into (5): .
Substitute into (1): .
~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~
This "scratch"method can be used
. . when we suspect we have a polynomial function.
It is long and tedious, but it works.
Of course, it is more efficient to learn a few sums-of-powers formula:
Note that the third formula is the square of the first formula,
. . making it easier to memorize.
But what about the implications?
. . This means: .
It looks like a bad joke or a terrible blunder, doesn't it?
Tell me something, is this a famous method? This is the second time I seen it used on this forum. When I was younger I developed many theorems concerning differences of a sequence. One of them you just used. (A sequence is a polynomial of degree n if and only if it takes n steps to reach a constanct sequence).Originally Posted by Soroban
I have a elegant prove with this rule that,
is always a polynomial sequence.
Consider an infinite sequence,
Its diffrence sequence is,
A polynomial sequence of degree therefore subtractions are required. In total we used subtractions on our original sequence. Thus there exists a polynomial sequence of degree . Furthermore, the first coefficient is . Based on the fact the constant sequence follows the factorial.
Tell me something, is this a famous method?
I'm sure it is . . .
I ran across many years ago in some book. .Since then, I've seen more efficient methods,
but I like that very primitive method. .(But I don't use it unless I am forced to.)
In case anyone is interested . . .
I was shown this method in graduate school . . . quite an eye-opener.
To find, for example, , we are expected to know the three "preceding" formulas:
Consider the next-higher power and form: .
We have: /
Now let and "stack" the equations.
. . . . . .
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
. . .
. . .
. . .
Add the stack (most of the left side cancels out):
We have: .
Then, after an enormous amount of algebra, we get:
. . . .
Quick! . . . Add that to your list!
This problem with the generalized exponent sum, I think, is a very elegant problem. The method I used is solving a system of equations which is the most primitive and complicated method. I then tried to find the patterns for these numbers... and failed. Later I learned the problem was solved in the 18th Century using bernouilli numbers.
However, I believe I found an elegant way how to solve that nasty system of equations. I have not yet worked it out but I believe it might work. My idea, is that the matrix that you get is also the solution of the polynomial of best fit which can be calculated with least squares.