We cannot find the value of exp(x) directly, except for a very few values of x. However, we can create a table of values using Taylor polynomials as approximations:
for some z in [0,x]. If x is sufficiently small, this gives a decent error bound. For instance,
So, *** Error Below: it should be 6331/3840 instead of 6331/46080 ***
since exp(x) is an increasing function, 0 <= z <= x <= 1/2, and
That tells us that *** Error Below: it should be 6331/3840 instead of 6331/46080 ***
or *** Error Below: it should be 6331/3840 instead of 6331/46080 ***
to at least three decimal places. Taking a larger-degree Taylor Polynomial will make the approximation closer. For instance, the 10th degree polynomial is off by at most (e^z)*x^10/10!, so for sqrt(e), that makes the error less than .5*10^-9, or good to 7decimal places.
Similarly, you can find values of trigonometric functions. Say you wanted to find sin(0.1). Use a Taylor expansion of sin(x) with a close to 0.1 (say, a=0), and find the 5th degree Taylor polynomial.
Since |cos(z)| <= 1, the remainder term can be bounded.
So, for x=0.1,
with an error of at most
or sin(0.1) = 0.09983341666... with an error of at most .139*10^-8, or good to seven decimal places.
Let's try a more complicated example. Suppose you needed to find
Since exp(x^2) doesn't have a nice antiderivative, you can't do the problem directly. Instead, use Taylor polynomials to find a numerical approximation. Let's try a Taylor polynomial of degree 5 with a=0:
, , , , , ,
(where z is between 0 and x)
The error is
(with z between 0 and x)
so the answer .54479 is accurate to within .0006588, or at least to two decimal places. Note that the inequality comes from the fact that f^(6)(x) is increasing, and 0 <= z <= x <= 1/2 for all x in [0,1/2].
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