Something neat I read, but not quite understand:

Spose you have a sequence and need to find its generating function:

$$a_1,a_2,a_3,a_4,a_5,...$$

find the differences between $a_n$ and $a_{n+1$, such that they are

$$b_1,b_2,b_3,b_4,b_5,...$$

continue finding differences till you end up with a constant difference:

$$k,k,k,k,k....$$

the generating function can be determined by:

$$a_1*n\choose 0 + b_1*n\choose 1 + c_1*n\choose 2 + ... + k* n\choose kth.row$$

maybe ....

Something neat I read, but not quite understand:

Spose you have a sequence and need to find its generating function:

$$a_1,a_2,a_3,a_4,a_5,...$$

find the differences between $a_n$ and $a_{n+1$, such that they are

$$b_1,b_2,b_3,b_4,b_5,...$$

continue finding differences till you end up with a constant difference:

$$k,k,k,k,k....$$

the generating function can be determined by:

$$a_1*n\choose 0 + b_1*n\choose 1 + c_1*n\choose 2 + ... + k* n\choose kth.row$$

maybe ....

@Mathematics

Question

Something neat I read, but not quite understand:

Spose you have a sequence and need to find its generating function:

$$a_1,a_2,a_3,a_4,a_5,...$$

find the differences between $a_n$ and $a_{n+1$, such that they are

$$b_1,b_2,b_3,b_4,b_5,...$$

continue finding differences till you end up with a constant difference:

$$k,k,k,k,k....$$

the generating function can be determined by:

$$a_1*n\choose 0 + b_1*n\choose 1 + c_1*n\choose 2 + ... + k* n\choose kth.row$$

maybe ....

Something neat I read, but not quite understand:

Spose you have a sequence and need to find its generating function:

$$a_1,a_2,a_3,a_4,a_5,...$$

find the differences between $a_n$ and $a_{n+1$, such that they are

$$b_1,b_2,b_3,b_4,b_5,...$$

continue finding differences till you end up with a constant difference:

$$k,k,k,k,k....$$

the generating function can be determined by:

$$a_1*n\choose 0 + b_1*n\choose 1 + c_1*n\choose 2 + ... + k* n\choose kth.row$$

maybe ....

@Mathematics

amistre64 · Answer

ack!! ... latex failed me lol

anonymous · Answer

Yes I think that would work.

amistre64 · Answer

Something neat I read, but not quite understand:

Spose you have a sequence and need to find its generating function:

$$a_1,a_2,a_3,a_4,a_5,...$$


find the differences between $a_n$ and $a_{n+1}$, such that they are

$$b_1,b_2,b_3,b_4,b_5,...$$


continue finding differences till you end up with a constant difference:

$$k,k,k,k,k....$$


the generating function can be determined by:

$$ a_1 {n\choose 0} + b_1 {n\choose 1}+ c_1{n\choose 2} + ... + k{ n\choose kth.row}$$


maybe ....

amistre64 · Answer

for example

the sequence: 1, 4, 10, 23, 47,  ...


r0: 1    4     10      23        47  ...
r1:    3    6      13        24
r2:       3      7       11      
r3:          4       4

$$1{n\choose0}+3{n\choose1}+3{n\choose2}+4{n\choose3}$$

$$1 + 3n + \frac{3}{2}n(n-1)+\frac{4}{6}n(n-1)(n-2)$$

amistre64 · Answer

n=0,1,2,3,...  of course

amistre64 · Answer

its reminiscent of a taylor series to me .... almost

anonymous · Answer

so i have a question. suppose we look at the simplest case 1, 2, 3, 4, 5, ... what do we get?

anonymous · Answer

or can we use it to find summation as in 
$$\sum_{k=1}^nk$$

amistre64 · Answer

1 2  3  4  5
 1  1  1  1

$$1{n\choose0}+1{n\choose1}=1+n$$
for the function that generates terms, n=0,1,2,3,...

anonymous · Answer

r0  1    3    6     10     15     21 
         2    3     4      5       6
             1   1       1      1

anonymous · Answer

I think the way this works is as follows.

Say I have a, a+d, a+2d, a+3d, ... (an arithmetic progression)

The generating function is obviously a+(n-1)d. The difference of each term from the previous one is always d.

Now suppose I have a, a+(a+d), a+(a+d)+(a+2d), a+(a+d)+(a+2d)+(a+3d), ..., where the difference of the terms form an arithmetic progression.

This is simplified to a, 2a+d, 3a+3d, 4a+6d, ...

The difference between a_n and a_{n-1} now is a+(n-1)d. Hence the generating function is a + summation of all these (n-1) a+(n-1)d's, which is

a + [summation of (n-1) a's] + [summation of (n-1) (n-1)d's]

The summation of (n-1) a's is a(n-1). The summation of (n-1) (n-1)d's is d(n-1)n/2.

Hence general term is a + (n-1)a + d(n-1)n/2 = an + dn(n-1)/2.

I think that if this argument is continued, the result you mention will follow inductively.

anonymous · Answer

so let me see if i understand. it should be, for the one i wrote 
$$1\dbinom{n}{0}+2\dbinom{n}{1}+\dbinom{n}{2}$$

amistre64 · Answer

$$1{n\choose0}+2{n\choose1}+1{n\choose2}$$
$$1+2n+\frac{n(n-1)}{2}=\frac{2+4n+n^2-n}{2}$$
$$\frac{2+4n+n^2-n}{2}=\frac{1}{2}(n^2+3n+2)$$

amistre64 · Answer

n=0, a0=1

n=1, a1=3

amistre64 · Answer

good explanation xact

amistre64 · Answer

$$\frac{1}{2}(5^2+3.5+2)=\frac{1}{2}(42)=21$$

anonymous · Answer

yes a good explanation. 
but i seem to have confused myself. why don't i just get 
$$\frac{n(n+1)}{2}$$ instead of 
$$\frac{(n+1)(n+2)}{2}$$or am i just being dense?

amistre64 · Answer

i think its the result of n=0,1,2,3,... as the domain

anonymous · Answer

ooooh starting at n = 0 i see!

anonymous · Answer

how about fibonacci??

amistre64 · Answer

if the fib can get differences down to a constant ...

might have to start at something other than the first term maybe

anonymous · Answer

ok  i see this is "if you can do this" got it. you cannot for fibonacci because successive differences are fibonacci again, all the way down the line

anonymous · Answer

i have read maybe 20 times how to get generating functions for sequences and even gave a brief lesson on it, but for some reason it does not stick.  to get the one for fib you solve a basic quadratic equation, but i forget the procedure

amistre64 · Answer

i cant recall fibs either ;)