Anonymous functions not acting as they should?

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Cameron am 9 Apr. 2013
In the following code I have two anonymous functions that work fine when the function is worked out by hand, iteration by iteration, but choke up by the fourth iteration as apre depreciates linearly to zero (not supposed to happen) and macexp levels off at 10.667 (not supposed to happen) before going directly to zero (not supposed to happen). In the version of this code which does not include any anonymous functions apre diminishes exponentially towards zero and macexp increases towards 55 (with x=4, se=.01). The code is to model the taylor series for exp(x).
x = input('Input the value of x to be approximated: ');
se = input('Input the specified error (se): ');
fprintf('\nn\t\tSeries Approx\t\tapre\n');
fmacexp=@(x,n) macexp(n)+((x^n)/factorial(n));
fapre=@(approx,n) (100*((approx(n+1)-approx(n))/approx(n+1)));
for n=1:100;
if apre<se
  2 Kommentare
Sean de Wolski
Sean de Wolski am 9 Apr. 2013
is the largest factorial that can be evenly represented as an integer in double precision:

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Jonathan Epperl
Jonathan Epperl am 9 Apr. 2013
I think the mistake happens here:
fmacexp=@(x,n) macexp(n)+((x^n)/factorial(n));
The value of macexp is now "hardwired" with its current valued into your function famcexp, i.e. when you call macexp in your loop, macexp executes with a value of macexp=zeros(1,100) every time, not with the updated value.
What Yao commented you should also consider though, those factorials are going to get gigantic very quickly.
  3 Kommentare
Teja Muppirala
Teja Muppirala am 9 Apr. 2013
For something this simple, it doesn't really matter, it's just a matter of milliseconds anyways. If efficiency is really an issue, then certainly you don't want to be redefining function handles all the time.
I would have probably written it like this outside the loop just once:
fmacexp=@(x,n) ((x^n)/factorial(n));
And then inside the loop do
macexp(n+1) = macexp(n+1) + fmacexp(x,n)

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