# Tag Info

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None of those numbers can be represented exactly in double precision floating point. Because of the nature of the representation, 0.4 has exactly the same fractional part as 0.2 - only the binary exponent is one larger. Therefore, 0.4-0.2 is computed to give 0.2. However, 0.6 has a different fractional part which is such that when 0.6-0.4 is computed it ...

3

The adjoint of an operator, $L$, on an inner product space, $V$, written $L^{\ast}$ is the operator such that the following holds, for all $v,w \in V$. $$\langle v, Lw\rangle=\langle L^{\ast}v,w\rangle$$ Let $f,g$ be $C^{\infty}(\Omega)$ with zero boundary conditions (a vector space) . We simply apply the definition and then integration by parts. $$\left ... 3 Here a proof that a root exists: first, note that f(1)<0. Then, note that$$ \lim_{x\to \infty}\frac{x}{(x+k)^{\epsilon}} > 1 $$So, there exists a c>0 with c>(c+k)^{\epsilon}. By the intermediate value theorem, f has a root. 2 Consider the following model stiff problem$$ \dot x = -0.5 x + 20 y\\ \dot y = -20 y\\ x(0) = 0,\quad y(0) = 1 $$The plot of x(t), y(t) has two regions 0 \leq t \leq 0.5. Here y(t) vanishes quickly transforming into x(t). I would refer to this region (maybe incorrectly) as a boundary layer. t > 0.5. Here y(t) is almost zero and does not ... 2 So your question is satisfactorily answered if you are given initial conditions for Newton's method which find each root. Here I will assume that two roots exist, i.e. p^2-4q>0. Given an initial condition not exactly on the vertex, the Newton iteration will stay on that side of the vertex, because of the fact that there is only one turning point. Also, ... 2 Taking a shot at this question... You have four points, so we need a third-degree polynomial, p_3(x). We have p(0)=2, p(2)=4, p(3)=-4 and p(5)=82. Calculating the divided differences...$$p[0,2]=\frac{p(2)-p(0)}{2-0}=\frac{2}{2}=1p[2,3]=\frac{p(3)-p(2)}{3-2}=\frac{-8}{1}=-8p[3,5]=\frac{p(5)-p(3)}{5-3}=\frac{86}{2}=43$$... 2 This is not intend to be a canonical answer, but rather a guide of discusion I still think there are many more roots. The reason is as follows: Let's imagine q is purely imaginary, let q=i(t-\pi/2), t\in \mathcal{R}, z=\sinh q=i\sin(t-\pi/2)=-i\cos t. Then we get the following equation for t:$$ \frac{1}{2}(\frac{t}{a}-\frac{a}{t})=\cot t $$Note ... 2 Well I guess you answered the question yourself, minus the intuitive explanation. If you do accept that the "Reimann integration" error will be like O( {} \frac{1}{n^{1/d}} ) and the MC error like  {} \frac{C}{\sqrt{n}} , then it's obvious that MC will in general give lower errors for  \mathbb d > 2. The only thing that I suppose might still ... 1 As already noted, this equation preserves the quantity e^g(g'^2+ag-b). So it is worth trying to formulate the problem as Hamilton's equations for this Hamiltonian in some system of coordinates. The coordinates cannot be (p,q)=(g',g) as they would often be, so we have to try something else. If we try to keep p=g', we have \frac{\partial H}{\partial ... 1 Following the method from our previous discussion, we use divided differences to find p_4(x)=f[x_0]+f[x_1](x-x_0)+f[x_2](x-x_0)(x-x_1)+f[x_3](x-x_0)(x-x_1)(x-x_2) We know that p(1)=-1=f[x_0], p(2)=-1/3, p(2.5)=3/32, p(3)=4/3, and p(4)=25.$$p[1,2]=\frac{-1/3+1}{1}=\frac{2}{3}=f[x_1]p[2,2.5]=\frac{3/32+1/3}{0.5}=\frac{41}{48}$$... 1 Start with the original Newton's method recurrence:$$x_{n+1}=x_n - \frac{f(x_n)}{f'(x_n)} $$and write x_n=r+e_n, where r is the root f(r)=0 and we assume e_n is small in the sense that we may perform Taylor expansions, viz.$$\begin{align}e_{n+1} &= e_n - \frac{f(r)+e_n f'(r) + \frac12 e_n^2 f''(r)+\cdots}{f'(r)+e_n f''(r)+\frac12 e_n^2 ...

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The condition number is scaling invariant. That is, for each non-singular $A$ and $\alpha > 0$ we have $$\operatorname{cond} (\alpha A) = \alpha\alpha^{-1} \| A\| \|A^{-1}\| = \operatorname{cond}(A).$$ So, by multiplication with a scalar you can make the norm of the inverse arbitrarily large (or small) without changing the condition number. However, if ...

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You could use Newton's Method to solve the equation numerically: Set $f(n) = 1−0.955^n −0.005^n\cdot0.995^{n −1}\cdot n− 0.005^2\cdot0.995^{n −2}\cdot \frac{n(n−1)}{2} -0.5$ Then $n$ is a solution of your equation exactly when $n$ is a root of $f$,that is $f(n) = 0$. Newton's method goes like this: $x_{m+1} = x_m - \frac{f(x_m)}{f'(x_m)}$ where $f'(n)$ ...

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I would plot the function: > n=0.001:0.001:20; > plot(n,1−(0.995.^n) −(0.005.^n).*(0.995.^(n −1)).*n− (0.005.^2).*(0.995.^(n −2)).*n.*(n−1)/2)) Then see where the graph cuts 0.5

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In Matlab and Octave "fsolve" is a nice function for finding roots to functions. First you need a "function handle" for your function, you can do this defining it as a lambda expression and assigning it a name, say f: f = @(n) 1−(0.955^n) −(0.005^n)*(0.995^(n −1))n− (0.005^2)(0.995^(n −2))n((n−1)/2) - 0.5; now typing n_solved = fsolve(@f, ...

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$c_n$ is the mid point of $[a_n,b_n]$. $c_{n+1}$ is the mid point of either $[a_n, c_n]$ or $[c_n,b_n]$. Compute $|c_n-c_{n+1}|$ in terms of $a_n,b_n$. You know that $b_{n+1}-a_{n+1} = {1 \over 2} (b_n-a_n)$, hence

1

We solve the equation $3x=1$ using the Bisection Method, with initial values $a_0=0$ and $b_0=1$. The root is given by $x=1/3$. The root is in $[0,1/2]$, so $a_1=0=a_0$ and $b_1=1/2$. The root is in $[1/4,1/2]$, so $a_2=1/4\gt a_1$ and $b_2=1/2$. Now look at the points $1/4,1/3,1/2$. The distance from $1/4$ to the root is $1/12$ and the distance from ...

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if $ɛ\Vert A_n\Vert =\Vert - ɛA_n\Vert =\Vert I_n-(I_n + ɛA_n)\Vert< 1$ then $I_n + ɛA_n$ is invertible.

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You have $x=\sqrt{p+\text{something}}$, and "something" is equal to $x$ itself. So $$x = \sqrt{p + x}. \tag 1$$ Thus $x$ is a fixed point of the mapping $$w\mapsto \sqrt{p+w}. \tag 2$$ We can ask whether that mapping is a contraction. $$\frac d {dw} \sqrt{p+w} = \frac{ 1 }{2 \sqrt{p+w}}.$$ If the absolute value of the derivative remains less than ...

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This is more a long comment than an answer. If I may make the problem more general, you need to solve for $n$ the equation $$\frac{a^{n+1}}{(n+1)!}=\epsilon$$ You can do this graphically (with some approximations) rewriting $n+1=x$ and make the equation $$x!=\frac{a^{x}}{\epsilon}$$ Taking logarithms gives $$\log(x!)=x\log(a)-\log(\epsilon)$$ Now, use the ...

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Its very hard to test, but this code should work. basically, I'm using a Matlab minimisation function and ODE solver, and finding the parameters that fit the data best. %// Some sample data X=0:0.1:4*pi; YI=2*exp(-X/5).*cos(X)+0.2*rand(size(X)); f=@(x) cos(x); %// Use a Matlab minimisation routine and ode45 to do the integration CC=fminunc(@(C) ...

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The root is on the very rightmost end. For example, solve $1-x^2 = 0$ in $[0, 1]$.

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Consider a function $f$ continuos in a compact interval $[a,b]$ with a single root in it, WLOG take $[0,1]$, and assume that $a_0>a_1>\dots$ and that the root we are seeking is in $(0,1)$ open. Since bisection actually bisects the last interval, the sequence $\{a_i\}_{i\in\mathbb{N}}$ we are looking for must necessary be $a_{i-1}=2^{-i}$ for ...

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In the real world situations I have encountered (and I have encountered several), evaluating the function is by far the most expensive thing you can do. Even massive amounts of side calculation to avoid a single function evaluation is well worth the while. So the faster a method converges, the better choice it can be - provided you can meet its requirements ...

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