Polynomials and the exponential

conjecturally a_m=1/m!

>>8756790
good observation anon, though I'm not sure how to prove this, as it is difficult to see the connection between [math]p_m(z)[/math] and [math]p_{m+1}(z) [/math], if we could find that an induction would be easy.

>>8756840
p_m-p_{m-1}
=(1/m!)(z)(z-1)...(z-[m-1])

>>8756875
I see that, very nice, lots of desirable results come from this, including the conjecture given, and:

[math]a_1 = (-1)^{m-1}m! H_m [/math] where H_m is the mth harmonic number

generally though

[math]a_k = (-1)^{m-k} m! sum_{p \in C_k} \frac{1}{p}, C_k = \{b_1 b_2 \dots b_k : \forall i. b_i \in \mathbb{Z}_m \} [/math]

and it looks like that sum would not have a closed form

>>8756932
I mean:

[math]\displaystyle a_k = (-1)^{m-k} m! \sum_{p \in C_k} \frac{1}{p}, \ C_k = \{b_1 b_2 \dots b_k | (\forall i) \ b_i \in \mathbb{Z}_m \} [/math]

>>8756875
Also we can make this into a second order recurrence, making it:

[math]p_{m+1}(z) = \frac{z}{m} p_m(z) + \left(\frac{z}{m} - 1 \right)p_{m-1}(z) [/math]

>>8756948
I made a mistake, it should be:

[math]p_{m+1}(z) = \frac{z+1}{m+1}p_m(z) + \frac{z-m}{m+1} p_{m-1}(z) [/math]

but anyway the generating function:
[math]P(x) = \sum_{n=0}^{\infty} p_n(z) x^n [/math]
satisfies the following first order differential equation:
[math]P'(x) = \frac{(z+1) + x(z-1)}{1 + x^2}P(x) [/math]
which should satisfy the equation:
[math]P(x,z) = (1+x^2)^{\frac{z-1}{2}} e^{(z+1)tan^{-1}x} [/math]
I will double check these later

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Building on >>8756875, the first point to emphasize point is probably that in

[math] p_m(i) = 2^i [/math]

the left hand side is such that it doesn't depend on m.
More abstractly, given functions f, g and h with a rule for an evaluation at v given as
f(v) = h(v)
g(v) = h(v)
you have
f(v) - g(v) = 0
i.e. the function
f-g
has a zero at v.
Thus it results that the different [math] p_{m+1} - p_m [/math] is forced to have all zeroes at, in your example, the integers and so
[math] p_{m+1} - p_m = c \prod_{j=0}^{m-1} (z-j) [/math]

That all wouldn't work if it was something like [math] p_m(i) = 2^i + 3m [/math].
And the fact that you evaluate at integers isn't particularly crucial, you'd just have a difference rule that's a product over the whatever evaluation points.

Now I think the necessety of 1/m!, the factor leading to 2^i is explained as follows
[math]\prod_{j=0}^{m-1}(z-j) [/math]
evaluated at z=i, being m+k for some m (so that the smallest factor is k+1) is just (m+k)!/k!. So dividing it by m! leaves you with i over m. The recusion then summs over m and the sum of those binomial coefficients is (1+1)^i = 2^i.

In pic related I've added, for the lulz, some coefficients to the recurrence relation to see what happens if you don't set the to 1.

>>8757050
>being m+k for some m (so that the smallest factor is k+1) is just (m+k)!/k!. So dividing it by m! leaves you with i over m. The recusion then summs over m and the sum of those binomial coefficients is (1+1)^i = 2^i.
I was wondering why it was specific to 2^i, nice explanation anon

>>8757068
I wrote it down more explicitly

So the recurrence is

[math] p_{m+1} = p_m + c(m) \prod_{j=0}^{m-1}(z-j) [/math]

and we wonder what [math] c(m) [/math] has to be so that for any m,

[math] p_m (i) = 2^i[/math].

Since to evaluate p_m we must unpack all p_{m-1}, p_{m-2}, etc. this just amounts to saying that the sum over those strange Vieta products is 2^i

For z=i, which for any of the m's in the sum we may write as i=m+k for some k, the product is a product of integers that starts at k+1 and ends at i. This can be written as i!/k!.

We have
[math] (a+b)^i = \sum_{m=0}^{i} \frac{1}{m!}\frac{i!}{(i-m)!} a^{i-m}b^m [/math].
so
[math] 2^i = \sum_{m=0}^{i} \frac{1}{m!}\frac{i!}{(i-m)!} [/math].

As i-m = k, we see we must insert 1/m!

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Oh, and I realize this makes the generalization to

[math] p_m(i) = a^i[/math]

evident.

Instead of summing

[math] \sum_{m=0}^i \frac{1}{m!} \frac{i}{(i-m)} 1^m 1^{i-k} = (1+1)^i = 2^i [/math]

sum

[math] \sum_{m=0}^i \frac{1}{m!} \frac{i}{(i-m)} (a-1)^m 1^{i-k} = (a-1+1)^i = a^i [/math]

I.e. choose

[math] c(m) = \frac{1}{m!} (a-1)^m [/math]

for any a, not just a=2.

It works, pic related

(forgot some exclamation marks)

>>8757101
>>8757134
nice, take my (You)s

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>>8757408
Thx. I had worked on it more

So the [math] p_m(z) [/math] with [math] p_m(i) = a^i [/math] are

[math] p_m(z) = \sum_k^{m} \frac {1}{k!} (a-1)^k \prod_{j=0}^{k-1} (z-j) [/math]

and Mathematica says this is also

[math] p_m(z) = a^z - \frac {1}{(m+1)!} (a-1)^{m+1} \dfrac{ \Gamma (z+1)}{\Gamma (z-m)}{}_2\tilde{F}_1(1,m-z+1,m+2;1-a) [/math]

with F given by the the hypergeometric series
https://en.wikipedia.org/wiki/Hypergeometric_function#The_hypergeometric_series

So really

[math] p_m(z) = a^z + R(z) [/math]

and R(z) has gammas of -m in the nominator which kills this rest at all integers. And btw. the limit m to infinity.

And the limit m to infinity indeed just gives

[math] p_\infty(z) = \sum_k^{\infty} \frac {1}{k!} (a-1)^k \prod_{j=0}^{k-1} (z-j) = a^z [/math]

see the expansion in my pic.

Btw. I had been playing with polynomial approximations to exponential functions at various ends, although very different ones. My note on them are here

https://axiomsofchoice.org/finite_exponential_power

These notes may also be of interest
https://axiomsofchoice.org/generalized_hypergeometric_function?s[]=hypergeometric&s[]=series