something i came across while working with summations

[void kelp]

i noticed that the term after the leading term of the summation of x^n always has a coefficient of ±1/2(depending on if you sum to x or x-1), is there any simple reason for this?

also because every function can be written as a taylor series, does this mean summation of any function f(x) can be approximated as F(x)±f(x)/2 + c (where F(x) is the anti derivative of f(x))

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[swift dagger]

What are you actually talking about? "The summation of x^n"?

[timid hill]

Look up Faulhaber's formula.
What you've noticed is B(1) = -1/2, where B(1) is the Bernoulli number with index 1.

Note that your generalization doesn't work.

[swift dagger]

Yeah, it's just plain untrue that "every function can be written as a Taylor series".

[void kelp]

sorry for using this weird terminalogy but im defining the summation of f(x) as summation of f(t) from 1 to x

[swift dagger]

...so you're talking about $\sum_{i = 1}^m i^n$?

[heavy bone]

Techie Literate

[void kelp]

yup

[swift dagger]

Wait, but then what are you talking about coefficients? There aren't any coefficients, the coefficients are all 1.

[void kelp]

wait no

[timid hill]

I'm fairly sure Lokey means the coefficients in the general formula for this sum, which are related to Bernoulli numbers.

Hold on, let me show Faulhaber's formula.

[swift dagger]

@void kelp In case you missed it, since it wasn't a direct reply to you.

[timid hill]

Here you go.
Here B(j) are the Bernoulli numbers. The minus subscript denotes one of the conventions.

For the first two terms we always have a simple representation:
(1/(p + 1))(-1)^0 C(p + 1, 0) B(0) n^(p + 1) = n^(p + 1)/(p + 1)
(1/(p + 1))(-1)^1 C(p + 1, 1) B(1) n^p = n^p/2

After that it's not as easy.

The first two highest terms are always just n^(p + 1)/(p + 1) + n^p/2, though.

By the way, note that B(j) for odd j more than 1 is always 0.

[void kelp]

looks like ill be having a long night ahead of me

y'all tysm for replying

[timid hill]

You're welcome!

[void kelp]

and working with my sloppy notations

[timid hill]

It's quite an interesting topic, I agree.