# Recursive method to get the differential not working.

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Hampus Toft on 14 Mar 2021
Commented: Hampus Toft on 17 Mar 2021
Im trying to make a recursive method to get the n:th-order differential equation.
what i have currently is 2 methods im my .m file first one being the simple 1st order differential.
function func = differential(f) % callculates the n:th-order differential
arguments
f function_handle
end
h = 10^(-5);
func = @(x)((f(x+h)-f(x))./h);
end
then im trying to use this in my recursive method
function output = differentialPower(f,n)
arguments
f function_handle
n
end
if(n==0)
output = f;
return;
else
f = differentialPower(differential(f),n-1);
output = f;
return;
end
end
to get the n:th differential
Problem i have is that my output will allways be either the original function (f)
or the first order differential of:
f = @(x)((f(x+h)-f(x))./h)
gooten from the differential method.
what i want to happen is that each time it gose deeper it will replace f(x) with ((f(x+h)-f(x))./h) and there for going deeper.
is this possible without using syms? or do i have to use the syms methods?

Uday Pradhan on 17 Mar 2021
Hi Hampus,
You will need to use the output function handle to evaluate the nth derivative approximation:
f = @(x) x^5;
df3 = differentialPower(f,3);
%Find approximation of f'''(1)
>> df3(1)
ans =
60.174087934683492
This approximation will get highly inaccurate as you increase the order of the derivative. Instead, a neat way to calculate higher order derivatives is using the diff function.
Hampus Toft on 17 Mar 2021
Yes i noticed that the resault got worse and worse for each derivative when using a function like e^x
where dfn(0) always should be 0 but when using this approximation the answer started to giving answers ≠ 1...
I wanted to use this to calculate a Taylor Polynomial for e^(-x), but in the end i wrote a custom function to get the Taylor Polynomial only for e^(-x).
creating a function that would convert x^k/k! to a Pn(x) function for later use. Code for anyone intreasted:
% Method to give n:th Taylor Polynom String
function output = taylor(taylorFunction,n)
fstring = "";
for loop = 0:n
if(loop==0)
varfix = strrep(func2str(taylorFunction),'@(n,x)',"@(x)");
fstring = strrep(varfix,'n',"(" + num2str(loop) + ")");
else
fstring = fstring + "+" + strrep(strrep(func2str(taylorFunction),'@(n,x)',""),'n',"(" + num2str(loop) + ")");
end
end
output = str2func(fstring);
end
Usage would be:
% Taylor function for e^(-x)
fTaylor = @(n,x) ((-1).^n*(x.^n./factorial(n)));
% gets the Taylor Function of order n using the fTaylor
Pn = taylor(fTaylor,n)
PnFunction = stringFormat(Pn);
PnFunction is a optional poorly written function_handle -> latex text converter
function out = stringFormat(in)
instring = func2str(in);
displayString = strrep(instring,'@(x)',"");
displayString = regexprep(displayString,"\.",""); % finds . and replace with nothing
% removes redundant (-1)^n
displayString = regexprep(displayString,"(\+*$$\(-1$$\^$$(.?)$$\*)","+");
displayString = regexprep(displayString,"(\+*$$\(-1$$\^$$(.?)$$\*)","-");
% formats exponetials to latex format i.e from 1^(15) to 1^{15} etc
displayString = regexprep(displayString,"\^$$(?<exponetial>.*?)$$","\^\{$<exponetial>}"); % formats matlabs factorial(n) function into n! displayString = regexprep(displayString,"factorial$$(?<factorial>.*?)$$","$<factorial>!");
% an attempt to fix fractions in latex, proabably wrong but works in
% this use case...
expression = "$$(?<numerator>.*?)\/\((?<denominator>.*?)$$([\+\-\*\!]?)\)\)";
displayString = regexprep(displayString,expression,"\\frac{$<numerator>}{$<denominator>}");
% dont remeber what this part even fixes :/
out = regexprep(displayString,"^\+","");
end
The reason for trying to get n:th derivative numerical was due tou the assignment not allowing the usage os the syms toolbox, and i wanted a method to get the generic Taylor polynomial.
Anyway i'm going to consider this answard.