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Hint: The Taylor Series is a polynomial in $h$, let $$a+bh+ch^2+dh^3+eh^4+fh^5\cdots$$ Evaluating it for the first multiples of the step gives 0:a\\ h:a+bh+ch^2+dh^3+eh^4+fh^5+\cdots\\ 2h:a+2bh+4ch^2+8dh^3+16eh^4+32fh^5+\cdots\\ 3h:a+3bh+9ch^2+27dh^3+81eh^4+243fh^5+\cdots\\ 4h:a+4bh+16ch^2+64dh^3+256eh^4+1024fh^5+\cdots\\ ... 5 In fact this series admits a very nice closed form. We have $$\sum_{n=1}^{\infty}\frac{(-1)^{\frac{n(n+1)}{2}}}{n!}\int_0^1x(x+1)\cdots(x+n-1)\: dx=\color{blue}{\frac{8 \ln 2 }{\pi ^2+4 \ln^2 2}-1} \tag1$$ with the numerical evaluation \begin{align} \color{blue}{-0.529727623071144825135721674689648876963...} .\tag2 \end{align} I've found the ... 3 Your suspicions are correct, you can convert ann-th$order differential equation into an n-dimensional system of first order equations. Let$x_1 = x, and then we have: \begin{align} x_1' &= x' = x_2 \\ x_2' &= x'' = x_3 \\ x_3' &= x''' = -t^2 x_3 - 4 x_1 \end{align} Our new system is: \begin{align} x_1' &= x_2 \\ x_2' &= ... 2 f(n) = \frac{1}{(2n+1)!(2n+1)} $$we know that$$ f(n+1) < f(n) i.e. strictly decreasing function with n. thus trying a few values of n we find \begin{align} \text{n=4: } \frac{1}{9!9} &=& \frac{1}{3265920}\approx 3\times10^{-7}\\ \text{n=5: } \frac{1}{11!11} &=& \frac{1}{439084800}\approx 2\times10^{-9}\\ \text{n=6: } ... 2 In the bisection method, and any root-finder that brackets the root, you can take the error to be half the distance between your brackets. This is because you can report the center of the interval as the root and you know the true root is no farther away than this. You can stop when the length of the interval is less than 2\cdot 10^{-6}, so make that a ... 2 The Fixed Point Iteration Method takes an equationf(x)=0$$and converts it into the form$$x=g(x)$$You then make an initial guess, say x_0, and recursively compute$$x_{n+1}= g(x_n)$$Continue this process until one of the following criteria is met: A specific number of iterations are done (which you define yourself) The error E= | x_{n+1}-g(x_n)| ... 2 To properly start Newton's method we begin by first localizing the root, finding a compact interval I that contains it, ideally that contains only that root. Then we take the first approximation in that interval. We can say that the initial error e_0 is smaller than the length of the interval. Now we can use the estimation of the n-th error ... 2 Your teacher's example cannot work in general as I present a counter example below. Nonetheless, I think that your teacher's approach is a reasonable way to explain the intuition behind what happens in a typical case, provided that the proper caveats are given. I think a more reasonable stopping condition, for programming purposes, is to iterate until the ... 1 This is a complement to Pp.'s answer. Newton's method converges, and with the rate given in Pp.'s answer, only if certain assumptions are met. In simple numerical examples these assumptions are usually not tested in advance. Instead, one chooses a reasonable starting point x_0 and will then soon find out whether the process converges to the intended root. ... 1 If f is a continuous function and a sequence x_n defined by the recurrence x_{n+1} = f(x_n) happens to have a limit X, then$$ X = \lim_{n \to \infty} x_{n+1} = \lim_{n \to \infty} f(x_n) = f(X)$$so X is a fixed point. In general there is no guarantee that the limit will exist, but in some cases it will. In particular, suppose there is a ... 1 In your particular case, I can't say very much, but notice this looks very much like Bezout's identity for the greatest common divisor.$$ \mathrm{gcd}(a,b) = 1 \longleftrightarrow \exists \; x,y \in \mathbb{Z} : ax + by =1$$We can clear denominators:$$ \frac{234}{24621} - x^{\frac{1}{3456}- \frac{1}{12345}} = \frac{1}{24621 \times ... 1 The problem is that althought$is monotone increasing, you don't know that the sequential$t$values are in equal steps. So the best thing to do is to try to fit a quadric to your data, and then compute the curvature of that quadric. Details: Suppose the points are$x_i, y_i$. You want to take a few points to either side of$i$...say$x_{i-2}, y_{i-2}$... 1 A very general method is to reconstruct (for example using splines) the parametric curve from the discrete data, and then compute the curvature of the reconstructed curve. The end answer can be understood as some finite difference operator on the discrete points. Different choice of interpolation will give different finite difference operators. For example, ... 1 Your equations amount to a system of$8$scalar equations with$8$scalar unknowns. Writing$x$and$y$as separate vectors is not helping clarity here. Stack them into one$8$-dimensional vector$z$, and similarly stack$f$and$g$into one function$h$. Formally speaking, the equation$h(z)=0$could be solved by the Newton-Raphson method as $$z_{n+1} = ... 1 Thanks for your answer, Shaun, that's exactly what I needed to know! I spent countless hours yesterday trying to figure out what I was doing wrong. Should have searched on here first. This is what I used that worked out perfectly! f(x)-g(x)=0 1 For Fixed Point Iteration, we want to get the form:$$x = g(x)$$This can produce several results where some converge and some do not converge. Do you know the rules of when you get convergence or divergence? The first step is to figure out how to get the given problem into this desired form. Case 1:$$\cos x = e^{-x} \implies \cos^{-1}( \cos x) = ... 1 The first method that I applied successfully with function calculator was approximation of circle by$2^k$-polygon with approximating sides with one point on the circle and corners outside the circle. I started with unit circle that was approximated by square and the equation$\tan(2^{-k} \pi/4) \approx 2^{-k} \pi/4$, that gives$\pi \approx \frac{8}{2} = 4$... 1 You can expect the error in$S_n$to be of the order of$a_{n+1}$. The problem is that the sequence $$\frac1{n!}\int_0^1x(x+1)\cdots(x+n-1)\,dx$$ decreases very slowly. In fact $$a_{2^{20}}=0.06865413726465990.$$ I have done calculations that suggest that$a_n$decreases like$C/\log n$for some constant$C$. This implies that to get an error of order ... 1 Since: $$x(x+1)\cdot\ldots\cdot(x+n-1) = \frac{\Gamma(x+n)}{\Gamma(x)}=\frac{1}{\Gamma(x)}\int_{0}^{+\infty}t^{x+n-1}e^{-t}\,dt$$ we have: $$\sum_{\substack{n\geq 1\\n\equiv 1\!\!\pmod{\!\!4}}}\!\!\!\!\!\!\frac{x(x+1)\cdot\ldots\cdot(x+n-1)}{n!}=\frac{1}{\Gamma(x)}\int_{0}^{+\infty}t^x e^{-t}\sum_{k=0}^{+\infty}t^{4k}\,dt$$ so:$$\sum_{n\geq ... 1 Yes, this is correct. For one thing, there are numbers that cannot be represented with a finite length. One of the most notable is$1/10$. In a binary representation, this is the repeating, non-terminating decimal$0.0\overline{0011}$. Also, for a given precision (Let's say$24$, since that is that precision of 32-bit IEEE-$754\$ floating point numbers, like ...