Some thoughts about the sums

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:

i=1niana+1a+1\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 i=1nia\sum^n_{i=1}i^a is na+1a+1\frac{n^{a+1}}{a+1}"

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

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

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

j=1njk+1=i=1k+2(k+2i)nii=0k((k+2i)j=1nji)k+2\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 nk+2k+2\frac{n^{k+2}}{k+2}, that is obtained when, on the first sum, the i=k+2i=k+2, so (k+2i)ni=(k+2k+2)nk+2=nk+2\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 nk+1k+1\frac{n^{k+1}}{k+1}, whose order is not the same, the term becomes nk+2k+2\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 na2\frac{n^a}{2}

I will have to use the Faulhaber's formula:

i=1nia=1a+1i=0a(a+1i)Bi+na+1i\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 aa degree appears when i=1 on the right sum, so it is:

1a+1(a+11)B1+na=B1+na\frac{1}{a+1}\binom{a+1}{1}B^+_1n^a=B^+_1n^a

Since B1+=12B^+_1=\frac{1}{2} (I am using this definition), the term is:

12na=na2\frac{1}{2}n^a=\frac{n^a}{2}

Note by Matheus Jahnke
2 years, 6 months ago

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These 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 - 2 years, 6 months ago

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