Taylor's Approximation playing up

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Daniel Petkov
Daniel Petkov 2018-5-13
编辑： Jan 2018-5-13
I'm trying to approximate a function (e^x) to a 10th order approximate about x = 0. I have made my code compatible with anonymous functions and it works for the most part. When I approximate e^x to a 8th order, it gives the correct answer, however when I go higher than the 8th order, the answer gets weird.
Here's the code:
Func = @(x) exp(x);
a = 0;
N = 10;
FuncApprox = 0;
for i = 0:N
syms x
f_derrived = matlabFunction( diff(Func(x),i) );
FuncApprox = FuncApprox + ( f_derrived(a)/factorial(i) )*( x-a )^i;
end
disp(FuncApprox)
When I run the code, this is what I get:
(1301357606610903*x^10)/4722366482869645213696 + (1626697008263629*x^9)/590295810358705651712 + x^8/40320 + x^7/5040 + x^6/720 + x^5/120 + x^4/24 + x^3/6 + x^2/2 + x + 1
When I should get:
x^10/3628800 + x^9/362880 + x^8/40320 + x^7/5040 + x^6/720 + x^5/120 + x^4/24 + x^3/6 + x^2/2 + x + 1

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采纳的回答

Jan
Jan 2018-5-13
编辑：Jan 2018-5-13
The result is mathematically correct, but the display could be simplified. 1301357606610903 / 4722366482869645213696 equals 1 / 3628800. Now it matters, what exactly your question is. Do you want to know how to simplify the output or why the values are displayed in this way?
Maybe simplifyFraction helps. ([EDITED] Nope, it does not.)
Move the factorial out of the symbolic part to the numerical part:
FuncApprox = FuncApprox + f_derrived(a) * (x - a)^i / factorial(i)
This works also, when i! exceeds flintmax.
[EDITED] Another solution:
FuncApprox = FuncApprox + (f_derrived(a) / factorial(sym(i))) * (x - a)^i;
  2 个评论
Ameer Hamza
Ameer Hamza 2018-5-13
编辑：Ameer Hamza 2018-5-13
"1301357606610903 / 4722366482869645213696 equals 1 / 3628800"
there is a small difference caused by calculation in floating points
digits(20)
vpa(str2sym('4722366482869645213696/1301357606610903'))
ans =
3628800.000000000313
also, a numerator ending in 3 and denominator ending in 6 can never simply to a fraction like that.
Jan
Jan 2018-5-13
Ameer, you are right. 3628800 cannot be 4722366482869645213696 / 1301357606610903 obviously. I was tired.

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更多回答（1 个）

Ameer Hamza
Ameer Hamza 2018-5-13
You are getting this error because floating point precision issues. The result of (f_derrived(a)/factorial(i)) is initially a double value, which later becomes a sym after multiplying with ( x-a )^i. To prevent this calculation from happening in floating point numbers, convert the coefficient to sym explicitly like this (f_derrived(a)/sym(factorial(i))). This will prevent the floating point calculation and hence loss of precision. The following will give correct answer
Func = @(x) exp(x);
a = 0;
N = 10;
FuncApprox = 0;
for i = 0:N
syms x
f_derrived = matlabFunction( diff(Func(x),i) );
FuncApprox = FuncApprox + ( f_derrived(a)/sym(factorial(i)) )*( x-a )^i;
end
disp(FuncApprox)
Result:
x^10/3628800 + x^9/362880 + x^8/40320 + x^7/5040 + x^6/720 + x^5/120 + x^4/24 + x^3/6 + x^2/2 + x + 1
For more details of why you lose precision, refer to the answer on this question: https://www.mathworks.com/matlabcentral/answers/396433-i-want-to-do-this-2-38987696573583658648235686-5-6387649837698237687649879932-and-i-need-all-digits

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