Summation simpilification process

[baby_1]

TL;DR

convert a summation to a specific formula

Hello,
Here is my summation:
sum(1/(25+n*B)),n=0 to (N/2)-1:=A
where A is between .01 to 2, N is between 10 to 2000 and I need to find the B for different values of N. I calculate this summation online (check here)
But the Digamma function makes the output function complex and it is not easy to solve with a calculator or even by hand. I need to calculate the result of this summation with hand easily. Is there any way to convert my summation to simple formula? something like this

 

[BvU]

$$sum(1/(25+n∗B)),n=0to(N/2)−1:=A$$

A few hashes to enclose wolfram input does not a decent $\TeX$ ouput make, as you can see ...

I don't understand what you mean with $:=A$ but I can imagine you mean the equation
$$\sum_{n=0}^{{N\over 2}-1} {1\over Bn+25} = A $$ That looks a lot better, don't you think ?

So am I right in assuming you want to find B(N, A) satisfying$$
{\psi^{(0)}\Bigl ({N\over 2} + {25\over B}\Bigr ) -
\psi^{(0)} \Bigl ( {25\over B}\Bigr )\over B }= A
$$where N = 10 ... 20000 and A = 0.01 ... 2 $\qquad$ ?

not easy to solve with a calculator or even by hand

Quite !
Good thing we have computers nowadays.

I need to calculate the result of this summation with hand easily

Well, then you have a problem ! You could make a table and write it on your hand, but if you don't have very big hands it will be hard to read.

Any chance of a compromise ? Can you compute the table and do a fit to a suitable function ? What accuracy is required ?

Is there any way to convert my summation to simple formula?

Wouldn't Wolfram have shown that if it was so easy as in your examples ?

(Disclaimer: no expert with di gamma. If x is real, is $\psi^{0}(x) \$ real ? Must be isn't it ? )

$\$

 

[baby_1]

I really appreciate your time and explanation.
Yes, I need to find B based on the given variables values. what is the simplest function of digamma to solve with the small computers ( I need to run a simple program on an embedded computer which has a 1Mhz CPU(Yes 1Mhz not 1Gihz), the result should be real not a complex value.

 

[BvU]

You still have a problem. And (for me) an undetermined problem: if you can allow 20% inaccuracy it's a different problem than if you need double precision accuracy.

I would try to evaluate/solve on a real computer and then fit a suitable approximating function that can be ported to the embedded device.

The result is real because all terms in the sums are real, right ?

You don't even need the digamma: just run the summation for N = 10 ... 20000 and do a hit and miss (or something better) on B. Keep the B(N,A) that have A in the right range. Once you have enough points, search for an approximating function with Tablecurve or similar.

An exploring excursion with excel can get you started for the first step:

[edit] checking -- after 'discovering' that $\ A < 0.04 \$ does not solve, no matter which B and N

I didn't try any fitting yet (don't have Tablecurve).

$\$

 

[pasmith]

The Wikipedia article on the digamma function includes a section on "computation and approximation" which might prove useful.

(The digamma function is defined as the derivative of a function which is real valued for positive reals, so will itself be real-valued for positive reals.)

 

[BvU]

Still here ?