On this note, I said about some patterns, even though the Faulhaber's formula makes this somewhat useless, I found some patters, first of all, as someone commented:

\(\sum^n_{i=1}i^a\approx\frac{n^{a+1}}{a+1}\)

I will prove the following statement:

"The highest order term of the polynomial that represents \(\sum^n_{i=1}i^a\) is \(\frac{n^{a+1}}{a+1}\)"

I will use a strong induction to prove this, starting with the base case of \(a=0\), we have:

\(\displaystyle \sum^n_{i=1}i^a=n\) So the highest term is \(n=\frac{n^{a+1}}{a+1}\)

We will assume that this will work for \(a=0,1,2,\ldots,k\), then for \(a=k+1\), we have:

\(\displaystyle \sum^n_{j=1}j^{k+1}=\frac{\sum^{k+2}_{i=1}\binom{k+2}{i}n^i-\sum^k_{i=0}\left (\binom{k+2}{i}\sum^n_{j=1}j^i \right )}{k+2}\)

The highest term of this will be \(\frac{n^{k+2}}{k+2}\), that is obtained when, on the first sum, the \(i=k+2\), so \(\binom{k+2}{i}n^i=\binom{k+2}{k+2}n^{k+2}=n^{k+2}\), no other term has this order on the numerator, because the highest one on the other sum is \(\frac{n^{k+1}}{k+1}\), whose order is not the same, the term becomes \(\frac{n^{k+2}}{k+2}\) when I consider the denominator;

The other pattern I saw is that the second term of the sum is \(\frac{n^a}{2}\)

I will have to use the Faulhaber's formula:

\(\sum^n_{i=1}i^a=\frac{1}{a+1}\sum^a_{i=0}\binom{a+1}{i}B^+_in^{a+1-i}\)

The term of \(a\) degree appears when i=1 on the right sum, so it is:

\(\frac{1}{a+1}\binom{a+1}{1}B^+_1n^a=B^+_1n^a\)

Since \(B^+_1=\frac{1}{2}\) (I am using this definition), the term is:

\(\frac{1}{2}n^a=\frac{n^a}{2}\)

## Comments

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TopNewestThese are great thoughts! Mathematicians look for patterns in the data, and then try to prove that the pattern exists, as you have done above. – Calvin Lin Staff · 3 months, 4 weeks ago

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