Solving a function through series (SAS) - function

Question

I need some help in checking my program in SAS that uses loops to solve a function as i am struggling. The function which i need to solve is in terms of a series expansion and it is:
ln(1+x)
I need to evaluate the function at x0 = 0.8 and the summation must stop when the next term is less than 0.05.
This is what I have so far but I'm very far off from where I need to be I believe:
data sumseries;
numerator = x;
denom =1;
x = 0.8;
do while (tot>abs(0.05));
numerator=x*numerator;
denom=denom+1;
sign=sign*-1
term=numerator/denom;
tot=sign*tot+term;
output;
end;
proc print data=sumseries;
run;
Any help is greatly appreciated!
Edit: Apologies if my formatting/presentation of the coding etc isn't up to scratch as I'm not familiar with this site
Edit: Change of names/values but replied answer still stands

Answer 1

I'll first give you pointers with your method, then provide my preferred solution.
Your solution, updated with my fixes:
data q2;
*indent please!;
denom =1;
x = 0.3;
*Move this assignment to after x is assigned;
numerator= x;
*initialize sign;
sign = 1;
*initialize tot;
tot=x;
*need an output to start out with;
output;
*WHILE change to UNTIL (req. switching > to <);
*tot changed to term;
do until (term < abs(0.000001));
numerator=x*numerator;
denom=denom+1;
sign=sign*-1;
term=numerator/denom;
*you want to add (sign*term) to the total, right?;
tot=tot + sign*term;
output;
end;
*always include a RUN after data steps, even though not required;
run;
proc print data=q2;
run;
My preferred solution:
*Use the macro language to define things like starting values;
%let x=0.3;
*let us see how close we get to correct!;
%put %sysfunc(log(1.3));
data want;
*initialize sum;
sum = 0;
*loop until we hit the limit. The until here is technically unneeded;
*but I like to include it for clarity;
do term = 1 by 1 until (abs(new) lt 1e-6);
*make the new value to be added (x to the power of term, divided by term;
*times -1 to the power of term plus one to get the right value for the sign);
new = (((&x)**term)/term) * (-1)**(term+1);
*stop iterating if the new term would be too small;
if abs(new) lt 1e-6 then leave;
*add to the sum now and output;;
sum = sum + new;
output;
end;
stop; *stop the data step loop;
run;
You might want to not output that last line, by the way, but that should be easy to add.

Answer 2

You can use the DIF function to compare the difference between two items that are FIFO stacked.
Example:
Presume the fuzz (1e-5) will be reached within 100 iterations.
data quotient;
x = 0.3;
guard = 100;
numerator = 1;
denom = 0;
sign = -1;
sum = 0;
* Mercator series:
* ln(1+x) = x^1/1 - x^2/2 + x^3/3 - x^4/4 + ... ;
ln_1_3 = log(1+0.3);
do iteration = 1 to guard until (iteration > 1 and abs(dif(sum)) < 1e-5);
numerator = x * numerator;
denom = denom + 1;
sign = sign * -1;
term = sign * numerator / denom;
sum = sum + term;
output;
end;
stop;
run;
proc print data=q2;
run;