# Tag Info

## Hot answers tagged numerical-methods

6

If $$z = e^z$$ then $$-ze^{-z} = -1$$ so $$-z = W(-1)$$ and thus $$z = - W(-1),$$ where $W$ is any branch of the Lambert W function.

5

Here is an extremely simple idea of $$\frac{a}{b}\to\frac{a+2b}{a+b}$$ For instance to get the square root of 2 you can start with any 2 counting integers like a = 1 and b = 2. $\frac{1}{2}\to\frac{1+2 (2)}{1+2}=\frac{5}{3}\to \frac{5+2(3)}{5+3}=\frac{11}{8}\to \frac{11+2(8)}{11+8}=\frac{27}{19}\to \frac{27+2(19)}{27+19}=\frac{65}{46}...$ continue as ...

3

It will depend on the application. In most practical problems, you are likely to have some idea of the order of magnitude of the solution you expect to find. You take the initial value to be the best guess you have available. If you're lucky, Newton-Raphson might still work even if this initial guess is quite far from the actual solution. If you're ...

2

What could $\sqrt 2$ be? You know that $1$ is too small because $1^2=1<2$. And you know that $2$ is too big becasue $2^2=4>2$. So the truth must be somewehre inbetween. The average of $1$ and $2$, that is $1.5$ suggests itself as a better guess. In general, if we have two different positive numbers $a,b$ such that $a\cdot b=2$, then one of them is too ...

2

This problem is difficult for numerical rather than computational reasons. Part of the problem is that you really need to be confident that the matrix is full rank, because if it is not, then a single error can make a determinant very large when it should actually be zero. For illustration, suppose we were trying to compute the determinant of $$A ... 1 The answer to "is it always true" is always "no", especially in numerical methods. The issue is in how closely the integrand resembles a polynomial function. If the integrand is analytic in a large neighborhood of the interval of integration, then Gaussian quadrature converges extremely fast. But for integrals like \int_{-1}^1\sqrt{1-x^4}\,dx or worse yet, ... 1 The following paper discusses the three commonly known numerical methods, Bisection, Newton-Raphson and Secant on their rate of convergence and computational efficiency. I hope this helps. http://www.jcbsc.org/journal/Paper/Vol_2_I_1_2011/V2I1_P4.pdf 1 To show that the method cannot achieve certain precision in general, it suffices to exhibit an equation for which it does not do it. The world's favorite equation for this purpose is x'=\lambda x. In practice x'=x is usually enough. Let's start from x_0=1, so the exact solution is x(t)=e^t. The numerical method gives$$ \psi^h x=1+h \gamma_1 + h ...

1

The Pade approximation of order $(m,n)$ to a function $f$ is a local rational approximation of the form $$f(x) \approx \frac{P_m(x)}{Q_n(x)}= \frac{\sum_{j=0}^{m}p_jx^j}{\sum_{j=0}^{n}q_jx^j}.$$ Expand $f(x)$ in a Maclaurin series $$f(x) = \sum_{j=0}^{\infty}a_jx^j$$ and look at the difference f(x) - \frac{P_m(x)}{Q_n(x)}= ...

1

Given the roots of a polynomial, call them $r_1, r_2, r_3, \ldots r_{201}$, then the coefficients are given by the elementary symmetric functions on the roots. If you are not familiar with these, read this article on Wikipedia So, $a_{2k+1} = \dfrac{e_{201-2k -1}(r_1, r_2, \ldots r_{201})}{a_{201}}$ This gives a "general" solution. Unless he/she gave you ...

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