12/30/2022, 11:44 PM

tommy's summability method

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I might not be the first to consider it so the name is perhaps disputable or temporarily anyway ,

This method is for divergeant strictly increasing positive integer sums

consider the geometric sum : 1 + 2 + 4 + 8 + ... + 2^n + ... = x

now we see 2 x = 2 + 4 + 8 + ...

notice x - 1 = 2 + 4 + 8 + ...

so

2 x = x - 1.

thus

x = -1.

lets generalize for real a > 1 :

1 + a + a^2 + ... = x

then

a x = x - 1

so

x = -1/(a-1) = 1/(1-a).

This is all old stuff.

So whats next ?

consider an entire function f(x) given by a taylor series : f_1 x + f_2 x^2 + f_3 x^3 + ... + f_i x^i + ...

such that f(i+1) > f(i) for every integer i > 0.

Notice f(0) = 0.

Now we can sum f(1) + f(a) + f(a^2) + f(a^3) + ... + f(a^n) + ... = x

Here is how

sum f(1) + f(a) + f(a^2) + f(a^3) + ... + f(a^n) + ... = x

sum_n sum_i f_i (a^n)^i = x

so

x = sum_n sum_i f_i (a^n)^i = sum_i sum_n f_i (a^i)^n

= sum_i f_i * - 1/(a^i - 1) = sum_i f_i / (1 - a^i)

As example consider

dexp(x) := exp(x) - 1.

dexp(1) + dexp(2) + dexp(2^2) + dexp(2^3) + dexp(2^4) + ... = x

Then

x = sum_i f_i /( 1 - 2^i ) = sum_i 1/( i! * (1 - 2^i) )

this clearly converges.

(I got an estimate of x = -1.19355 btw)

I call this a linear summability method because it has some linear properties.

regards

tommy1729

**************************

I might not be the first to consider it so the name is perhaps disputable or temporarily anyway ,

This method is for divergeant strictly increasing positive integer sums

consider the geometric sum : 1 + 2 + 4 + 8 + ... + 2^n + ... = x

now we see 2 x = 2 + 4 + 8 + ...

notice x - 1 = 2 + 4 + 8 + ...

so

2 x = x - 1.

thus

x = -1.

lets generalize for real a > 1 :

1 + a + a^2 + ... = x

then

a x = x - 1

so

x = -1/(a-1) = 1/(1-a).

This is all old stuff.

So whats next ?

consider an entire function f(x) given by a taylor series : f_1 x + f_2 x^2 + f_3 x^3 + ... + f_i x^i + ...

such that f(i+1) > f(i) for every integer i > 0.

Notice f(0) = 0.

Now we can sum f(1) + f(a) + f(a^2) + f(a^3) + ... + f(a^n) + ... = x

Here is how

sum f(1) + f(a) + f(a^2) + f(a^3) + ... + f(a^n) + ... = x

sum_n sum_i f_i (a^n)^i = x

so

x = sum_n sum_i f_i (a^n)^i = sum_i sum_n f_i (a^i)^n

= sum_i f_i * - 1/(a^i - 1) = sum_i f_i / (1 - a^i)

As example consider

dexp(x) := exp(x) - 1.

dexp(1) + dexp(2) + dexp(2^2) + dexp(2^3) + dexp(2^4) + ... = x

Then

x = sum_i f_i /( 1 - 2^i ) = sum_i 1/( i! * (1 - 2^i) )

this clearly converges.

(I got an estimate of x = -1.19355 btw)

I call this a linear summability method because it has some linear properties.

regards

tommy1729