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4

Actually, the answer to this lies in quite a bit more advanced topic called rough path theory (beware: PDF). A rough path is a way of "enhancing" a $\alpha$-Hölder continuous path with some extra information. A rough path is an ordered pair, $\textbf{X}=(X, \mathbb{X})$ where $X\colon [0,T]\to V$ where $V$ is some Banach space (typically $\Bbb{R}$) and a ...


3

You can find a least squares estimate of $x$ and $y$. You want to "solve" the overdetermined system $Az = c$, where \begin{equation} A = \begin{bmatrix} a_1 & b_1 \\ a_2 & b_2 \\ a_3 & b_3 \\ a_4 & b_4 \end{bmatrix} \end{equation} and \begin{equation} c = \begin{bmatrix} c_1 \\ c_2 \\ c_3 \\ c_4 \end{bmatrix}. \end{equation} In a least ...


3

You can rephrase your question as finding $c$ to minimize $E[|e^X-c|]$ where $X$ is a uniform random variable. In general, The $c$ that minimizes $E[|Y-c|]$ is the median of the distribution of $Y$. Some intuition can be found in this answer. Here, we see the median of $X$ is $1/2$, and since $x \mapsto e^x$ is increasing, the median of $e^X$ is $e^{1/...


3

The question specifically asks to find an equation for the constant $c$ that minimizes this expression. The typical method here is to let this expression be a function of $c$ and find the minimum of this function with respect to $c$. Thus, we have $$ f(c) = \int_0^1 |e^x - c| \ dx$$ We have to be careful here, as we are not sure if the expression inside of ...


1

As always with iterative systems $x_{n+1}=f(x_n)$ you get convergence around the fixed point $x_*$ if $|f'(x_*)|<1$. In this case, $$ f(x)=2-(1+c)x+cx^3\implies f'(x)=-(1+c)+3cx^2,\quad f'(1)=-1+2c $$ Thus for $c\in (0,1)$ you get a locally contractive iteration around $x_*=1$.


1

Since you're wondering whether $x_n\to1$ it may make things more transparent if you set $x_n=y_n+1$ and check whether $y_n\to0$ (just because "small" can be easier to see than "close to $1$"). You get $$y_{n+1}=(c-1)y_n+cy_n^2.$$ Now regardless of $c$, if $y_n$ is small enough then $cy_n^2$ is even smaller. When could you conclude that $(c-1)y_n$ is smaller ...


1

This is not an answer, but expands on a comment I made to the question. First, some background. Let's assume function $f$ is a product of $k$ subfunctions $g$, in $n$ variables, $$f(x_1, x_2, \dots, x_n) = \prod_{i=1}^{k} g_i(x_1,\dots,x_n)$$ where each subfunction $g_i$ is $$g_i(x_1, x_2, \dots, x_n) = H \left ( c_{i,0} + \sum_{j=1}^{n} c_{i,j} x_j \right ...


1

It depends on how good your initial guess is. In your program, you take an initial guess of $x=N$. Consequently there will be an initial period where $x/2$ is way bigger than $N/x$, so that the method is essentially just cutting the number in half over and over. So it will take roughly $\log_2(N/\sqrt{N})=1/2 \log_2(N)$ steps to get to within some fixed ...


1

The average theoretical complexity is $log_2(N)$. However, two notes: $log_2(N)$ is just the highest order term. The actual number of steps to convergence might be: constant term + coefficient*$log_2(N)$ Secondly, there can be some nuance about the exact floating point computations, etc. where convergence might occur a step sooner or later, before x ...


1

Gradient methods (such as the gradient descent method) for minimizing a differentiable function $f:\mathbb R^n \to \mathbb R$ require evaluating the gradient of $f$ at each iteration. Unfortunately, sometimes there is no nice formula for the gradient of $f$. So what do you do? One option is to estimate the gradient of $f$ numerically, using finite ...


1

It's easier to think about in the integration context. Good integration rules proceed by exactly integrating a certain function which is close to the desired integrand. For example, the trapezoidal rule proceeds by exactly integrating a piecewise linear interpolant of the integrand. Its error depends on how far the integrand deviates from the piecewise ...


1

The number of iterations measures how long you choose to run the algorithm. The rate of convergence measures how accurate your answer will be depending on how long you choose to run the algorithm.


1

As in @mweiss 's answer we use repeatedly the approximation $$\sqrt{a^2+b}\approx a+{b\over 2a}\qquad(|b|\ll a^2)\ .$$In this way we obtain on the one hand $$\eqalign{ \sqrt{20}&=\sqrt{25-5}\approx 5-{1\over2},\quad 20+\sqrt{20}\approx25-{1\over2},\cr \sqrt{20+\sqrt{20}}&\approx5-{1\over20},\quad 20+\sqrt{20+\sqrt{20}}\approx 25-{1\over20},\cr \sqrt{...


1

The other answer does not really address the question of how to systematically determine when cancellation error occurs. Here's a reasonably accurate method. We assume that the machine computes at a fixed floating-point precision with relative error $ε$. Let "$[r]$" denote $\def\rr{\mathbb{R}}$"$\{ x : x \in \rr \land |x| \le r \}$" for any $r \in \rr$, for ...


1

Yes, numerical methods can be used to derive results about the continuous problem, but has to be done a bit carefully. In particular, the same questions of regularity, notion of a solution, what is meant by convergence, etc. have to be addressed. This family of methods are known variously as Rothe's method or method of finite differences, among others. For ...



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