One Oier elementary school girl asked me a question about sigma I ^k, I <= 1E8,k<= 1E6. I thought Bernoulli series might be able to solve his problem, so I arranged the following article for my students to learn ~~~ I have limited skills, and I can only help here
Enter the body below:
Calculation of ∑ {i= 1,n}i^kThe value of is required to introduce the concept of Bernoulli sequence
I’m going to expand out B minus 1 to the k, and I’m going to write B to the k as the KTH term of the sequence, B of k.
When k is greater than or equal to 2, the expansion of B minus 1 to the k is equal to B of k
B minus 1 to the k is equal to B to the k, and then B to the k is B of k, and then you figure out what the terms are.
Bernoulli numbers.
For example,
Calculation of B (1)
Make (1 B) ^ 2 = B ^ 2
B^2-2B+1=B^2
Let’s write B to the k as the KTH term of the sequence, B of k.
B (2) – 2 B (1) + 1 = B (2)
B (1) = 0.5
Similarly, if you compute B of 2,
Make (1 B) ^ ^ 3 3 = B
There are
B^3-3B^2+3B-1=B^3
Let’s write B to the k as the KTH term of the sequence
There are
B(3)-3B(2)+3B(1)-1=B(3)
B(2)=[3B(1)-1]/3
namely
B(2)=1/6
From this we can calculate any term in the sequence \
Define B (0) = 1
From the above:
(x+B)^(k+1)
=∑{I =0, k+1}C{I,k+ 1}B^ I *x^(k+1*-i*)
= x ^ (k + 1) + C {1, k + 1} Bx ^ k + ∑ {I = 2 and k + 1} {I, k + 1} C * B ^ I * x ^ (k + 1 – I)
= x ^ (k + 1) + 0.5 C {1, k + 1} x ^ k + ∑ {I = 2 and k + 1} {I, k + 1} C * B ^ I * x ^ (k + 1 – I)
again
(x+B-1)^(k+1)
=∑{I =0, k+1}C{I,k+ 1}(B-1)^ I *x^(k+1*-i*)
= x ^ (k + 1) + C {k + 1, 1} (1 B) x ^ k + ∑ C {I = 2 and k + 1} {I, k + 1} * I * x (1 B) ^ ^ (k + 1 – I)
= x ^ (k + 1) – 0.5 – C {1, k + 1} x ^ k + ∑ C {I = 2 and k + 1} {I, k + 1} * I * x (1 B) ^ ^ (k + 1 – I)
Because B of k is a Bernoulli sequence
There are
(B-1)^i=B^i
namely
(x+B-1)^(k+1)
= x ^ (k + 1) – 0.5 – C {1, k + 1} x ^ k + ∑ {I = 2 and k + 1} {I, k + 1} C * B ^ I * x ^ (k + 1 – I)
so
(x+B)^(k+1)-(x+B-1)^(k+1)=(k+1)x^k
Set x= 1,2,3… And I,… N.
There are
(1+B)^(k+1)-B^(k+1)=(k+1)
(2+B)^(k+1)-(1+B)^(k+1)=(k+1)*2^k
(3 +B) ^ (k+ 1) – (2 +B) ^ (k+ 1) = (k3 + 1) * ^k
…
(i+B)^(k+1)-(i-1+B)^(k+1)=(k+1)*i^k
…
(n+B)^(k+1)-(n-1+B)^(k+1)=(k+1)*n^k
Sum from the above formula, get:
(n + B)^(k+1) -B^(k+1)=(k+1)∑{I =1,n} I ^k
namely
∑{I =1,n} I ^k=[(n + B)^(k+1) -B^(k+1)]/(k+1)
\
Note:
This n plus B to the k plus 1 doesn’t mean n plus B to the k plus 1 star
I’m referring to the expansion of alpha as the KTH term of the Bernoulli sequence of **B^k
Just like I said before. If you want a rigorous algorithm, it’s Euler
The algorithm, which involves infinite series, is cumbersome but very rigorous.
Symbols used in this article:
A (1)+a(2)+a(3)+… + a (n) said
∑ {I = 1, n} a (n)
The number of combinations of m from n is
C{m,n}
The following is the text sorted by Word. I don’t know why sometimes the formula is wrong, so I’m afraid I can’t see it clearly. I’ll save the picture and make it convenient for myself to check and learn
The picture below is of mathematician Euler’s more rigorous algorithm for infinite series:
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