IS this coincidence? Integral adn summation

[jd12345]

The sum of 1 + 2 + 3...n = n(n+1) / 2 - highest power term is n^2
sum of 1^2 + 2^2 + 3^2...n^2 - n(n+1)(2n+1) / 6 - highest power term is n^3
sum of 1^3 + 2^3 + 3^3...n^3 - it has highest power term of n^4
similarly 1^k +2^k ...n^k - it has highest power term of n^(k+1)

Is it a coincidence that ∫x^k dx = x^(k+1) / (k+1) - power is k+1 ?

 

[Number Nine]

Do you understand how the Reimann integral is derived? (i.e. its relationship to summation)

 

[chiro]

The sum of 1 + 2 + 3...n = n(n+1) / 2 - highest power term is n^2
sum of 1^2 + 2^2 + 3^2...n^2 - n(n+1)(2n+1) / 6 - highest power term is n^3
sum of 1^3 + 2^3 + 3^3...n^3 - it has highest power term of n^4
similarly 1^k +2^k ...n^k - it has highest power term of n^(k+1)

Is it a coincidence that ∫x^k dx = x^(k+1) / (k+1) - power is k+1 ?

Hey jd12345.

For the summation part there is a formula for a finite n which is known as the Bernoulli polynomials that basically allow you to derive the closed form solution for x^p where p >= 0 and p is a integer and x ranges from 1 to n.

 

[Stephen Tashi]

similarly 1^k +2^k ...n^k - it has highest power term of n^(k+1)

Suppose $F(n) = \sum_{i=1}^n f(n)$

then $F(n+1) - F(n) = f(n+1)$

Denote $F(n+1) - F(n)$ by $\triangle F(n)$.

Then $\triangle F(n)$ resembles a finite version of the derivative of $F(x)$

To sum $f(n)$ you must solve $\triangle F(n) = f(n+1)$ for $F(n)$. i.e. you must find the anti-$\triangle$ of f(n+1). So this resembles integration.

The study of stuff like this is called "The Calculus Of Finite Differences". (George Boole himself wrote an interesting book about it.)

 

[jd12345]

Ok - thanks Stephen

 

[AlephZero]

Incidentally, if you want to see neater analogy between sums of series and integrals, find the formulas for
1 + 2 + 3 + ...
1.2 + 2.3 + 3.4 + ...
1.2.3 + 2.3.4 + 3.4.5 + ...
Or in general
$$\sum_{n=0}^N \frac{(n+k)!}{ n!}$$
for k = 1, 2, 3, ...