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The Taylor polynomials for ln(1 + x) only provide accurate approximations in the range −1 < x ≤ 1. For x > 1, Taylor polynomials of higher degree provide worse approximations. The Taylor approximations for ln(1 + x) (black). For x > 1, the approximations diverge. Pictured is an accurate approximation of sin x around the point x = 0. The ...
This function was plotted above to illustrate the fact that some elementary functions cannot be approximated by Taylor polynomials in neighborhoods of the center of expansion which are too large. This kind of behavior is easily understood in the framework of complex analysis.
of which, as shown in a later 1748 work, [2] the right hand side can be obtained by setting x = 1 in the Taylor series expansion log ( 1 1 − x ) = ∑ n = 1 ∞ x n n . {\displaystyle \log \left({\frac {1}{1-x}}\right)=\sum _{n=1}^{\infty }{\frac {x^{n}}{n}}.}
In probability theory, it is possible to approximate the moments of a function f of a random variable X using Taylor expansions, provided that f is sufficiently differentiable and that the moments of X are finite. A simulation-based alternative to this approximation is the application of Monte Carlo simulations.
The natural logarithm of x is generally written as ln x, log e x, or sometimes, if the base e is implicit, simply log x. [2] [3] Parentheses are sometimes added for clarity, giving ln(x), log e (x), or log(x). This is done particularly when the argument to the logarithm is not a single symbol, so as to prevent ambiguity.
Thus to -approximate () = using a polynomial with lowest degree 3, we do so for () with < / by truncating its Taylor expansion. Now iterate this construction by plugging in the lowest-degree-3 approximation into the Taylor expansion of g ( x ) {\displaystyle g(x)} , obtaining an approximation of lowest degree 9, 27, 81...
The Taylor expansion would be: + where / denotes the partial derivative of f k with respect to the i-th variable, evaluated at the mean value of all components of vector x. Or in matrix notation , f ≈ f 0 + J x {\displaystyle \mathrm {f} \approx \mathrm {f} ^{0}+\mathrm {J} \mathrm {x} \,} where J is the Jacobian matrix .
The log-tangent integral on the right-hand side can be evaluated in terms of the Clausen function (of order 2), as is shown below: 2 π log ( G ( 1 − z ) G ( 1 + z ) ) = 2 π z log ( sin π z π ) + Cl 2 ( 2 π z ) {\displaystyle 2\pi \log \left({\frac {G(1-z)}{G(1+z)}}\right)=2\pi z\log \left({\frac {\sin \pi z}{\pi }}\right ...