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37

As some people on this site might be aware I don't always take downvotes well. So here's my attempt to provide more context to my answer for whoever decided to downvote. Note that I will confine my discussion to functions $f: D\subseteq \Bbb R \to \Bbb R$ and to ideas that should be simple enough for anyone who's taken a course in scalar calculus to ...


17

I'll first give a intuitive answer, then an analytic answer. Intuitively, the tangent goes in the same direction as the function, following it as closely as possible for a line. Any other line immediately starts to diverge from the function. Analytically: Consider the Taylor aproximation at $x$: $f(x+h) =f(x)+hf'(x)+h^2f''(x)/2+... $. This means that, ...


12

There is a sense in which the derivative is the best linear approximation. You just have to define "best" approximation in a proper way, taking into account that the derivative is a very local property. In particular, suppose we are trying to approximate $f$ at $x_0$. Then, we make the following definition: A function $g$ is a at least as good of an ...


6

Think about the derivative in the sense that If you zoom in very close to any differentialiable (smooth curve) you'll get a straight line. The slope of that line is the derivative and is the best linear approximation for the function near that point. If any linear approximations fit better zoomed in that closely then by definition it would be closer to the ...


5

Let's assume that $k \ll N$. By using the Taylor series expansion $\ln(1-x) = -x+O(x^2)$, we have: $$\displaystyle\sum_{i = 1}^{k-1}\ln\left(1-\dfrac{i}{2N}\right) \approx \displaystyle\sum_{i = 1}^{k-1}-\dfrac{i}{2N} = -\dfrac{1}{2N}\cdot\dfrac{(k-1)k}{2} = -\dfrac{\binom{k}{2}}{2N}.$$ Then, by exponentiating both sides, and using the Taylor series ...


5

I think you might be confusing the derivative as a linear operator vs. the derivative evaluated at a point (a linear functional). The derivative itself does not approximate anything, it just gives you a function that tells you the rate of change of the original function for every value of x in the domain. Now, when you evaluate the derivative at a single ...


5

This depends a lot on how we measure error. So we could turn the question around and ask for what definition of error will a first order Taylor approximation give the least error? You already have gotten good explanations from others on this, considering what happens as we take limits close to the point. So I could maybe contribute by adding something new. ...


4

With $t=\dfrac u\lambda+\lambda$, $$\int_\lambda^\infty e^{-t^2/2}dt=\int_0^\infty e^{-(u/\lambda+\lambda)^2/2}\frac{du}\lambda=\frac{e^{-\lambda^2/2}}\lambda\int_0^\infty e^{-u^2/2\lambda^2}e^{-u}dt.$$ For large $\lambda$, the second factor decreases faster and $$\int_0^\infty e^{-u^2/2\lambda^2}e^{-u}du\approx\int_0^\infty e^{-u}du.$$


3

Take the function f (x) = $x^2$. At x = 0, the derivative gives you the function g (x) = 0. On every interval $[-a, +a]$ the function g (x) = $a^2/2$ will give a better approximation with a maximum error of $a^2/4$ instead of $a^2$. However, that approximation will be worse on any interval smaller than $[-a/2, +a/2]$. g (x) = 0 will beat any other ...


3

Here's a simple explanation of what's wrong with your argument. You're not understanding what is meant by "near". The claim isn't that for a given small interval it is the best approximation. But this is what you are arguing --- given a $\delta>0$, it is true that you can find a better (or at least as good) approximation in ...


3

Let's try to find the best fitting line to the parabola $$\text{$y = f(x) = x^2$ at the point $(1,1)$ of $f$.}$$ We require that $$f(1) = L(1).$$ So the line must look like $$L(x) = m(x-1) + 1 = mx - (m-1).$$ The difference between the two curves will be $$E(x) = f(x) - L(x) = x^2 - mx + (m - 1).$$ In order to emphasize that we are interested in the ...


3

The claim is false without additional assumptions on $\frac mn$. For example if $\frac mn=\frac1{10}$ then your next guess is $10.05$ which is further away from $\sqrt 2$. The claim is true, however, if we impose a condition like $\frac mn>1$. Write $x=\frac mn$. You want to show that $\frac x2 +\frac1x$ is closer to $\sqrt 2$ than $x$ is. In other ...


3

$\frac{m}{n}$ approximately equals $\sqrt 2$ Suppose $\epsilon$ is our the error in our estimate. i.e. $\frac{m}{n} + \epsilon = \sqrt 2$ or $(\frac{m}{n} + \epsilon)^2 = 2$ $\frac{m^2}{n^2} + 2\frac{m}{n}\epsilon + \epsilon^2 = 2$ Now we need to solve for $\epsilon.$ If $\epsilon$ is small then then $\epsilon^2$ is very small. Which is a little ...


3

Assume $w>-1$ and $x \in \mathbb{N}$. One may recall the classic identity $$ \Gamma(y+1)=y \:\Gamma(y),\quad y>0. \tag1 $$ from which, setting $\psi (y):= \left(\log \Gamma (y) \right)'$, one deduces $$ \psi(y+1)-\psi(y)=\frac1y,\quad y>0. \tag2 $$ Next, putting $y=w+k$ and summing $(2)$ from $k=1$ to $k=x$, terms telescope, one gets $$ ...


3

Almost from definition $$I=\int e^{-\frac{t^2}{2}}\,dt=\sqrt{\frac{\pi }{2}} \text{erf}\left(\frac{t}{\sqrt{2}}\right)$$ So $$J=\int_\lambda^\infty e^{-\frac{t^2}{2}}\,dt=\sqrt{\frac{\pi }{2}} \text{erfc}\left(\frac{\lambda }{\sqrt{2}}\right)$$ Now, look here for the asymptotic expansion for large $\lambda$ to get $$J=e^{-\frac{\lambda ^2}{2}} ...


3

This is not a solution but approximation. First, note that $\frac{3}{8}=0.3750000000$. The function $f(x)=x^x(1-x)^{2x}$ is convex on $(0,c_1)\cup(c_2,1)$ and concave on $(c_1,c_2)$, where $c_1\approx 0.2718247$ and $c_2\approx 0.5243816$. So that applying the hermite-hadamard inequality \begin{eqnarray} f\left( {\frac{{a + b}}{2}} \right) \le \frac{1}{{b ...


2

A simpler dynamic programming approach is to keep a single vector of counts of zig-zag sequences that ended at a certain value in, say, a down movement, and invert it in each update. Then with $a_{ki}$ denoting the number of zig-zag sequences of length $k$ ending in $i$ on a down movement, we have $$ a_{k+1,N+1-i}=\sum_{j=1}^ia_{kj} $$ (or with upper limit ...


2

This isn't a definitive answer, but hopefully points you in the right direction. The short of it is, I don't know how the coefficients were derived, but it is likely related to some Pade, Remez, interpolating, or least-squares approximant. Read on for details. The function we're talking about is $$ Q(x) = \frac{1}{\sqrt{2\pi}}\int_{x}^\infty e^{-t^2/2}\;dt ...


2

You can also use harmonic numbers since $$S_x=\sum_{i=1}^x \frac{1}{w+i}=H_{w+x}-H_w$$ Now, using the asymptotics you would get $$S_x=\left(\log \left(x\right)+\gamma-H_w \right)+\frac{w+\frac{1}{2}}{x}-\frac{6 w^2+6 w+1}{12 x^2}+O\left(\frac{1}{x^3}\right)$$ For illustration purposes, using $w=123$ and $x=456$, the exact result would be $\approx ...


2

Consider: $$ \begin{align*} e^{-(c\ln\frac{1}{x})^s} &= e^{-(\ln(\frac{1}{x})^c)^s} \\ &= e^{-(\ln(\frac{1}{x^c}))^s} \\ &= \frac{1}{e^{(\ln(\frac{1}{x^c}))^s}} \\ &= \frac{1}{e^{\ln(\frac{1}{x^c})} \cdot e^{\ln(\frac{1}{x^c})} \ldots \text{s times} } \\ &= \frac{1}{\frac{1}{x^c} \cdot \frac{1}{x^c} \ldots \text{s times}} \,\,\,\,\, ...


2

Yes, there is a problem in your approximations. Recall that $$\lim_{n \to \infty} \biggl(1 + \frac{x}{n}\biggr)^n = e^x.$$ Effectively, you have replaced this limit with $1$ four times, and since the various $x$s don't cancel, you got a wrong result. The square root does indeed tend to $1$, so we can ignore that. Assuming that $a_2$ and $p$ are constant ...


2

If $x>0$ is close to zero $$\frac{1}{4+x}=\frac{1}{4}\cdot\frac{1}{1+\frac{x}{4}}=\frac{1}{4}\sum_{n\geq 0}\frac{(-1)^n x^n}{4^n}$$ holds, and since the terms of the last series are decreasing in absolute value, $$ \frac{1}{4}\left(1-\frac{x}{4}\right) \leq \frac{1}{4+x}\leq \frac{1}{4}\left(1-\frac{x}{4}+\frac{x^2}{16}\right) $$ holds, too. Just plug in ...


1

The problem is that the indefinite integral $\int \dfrac{dx}{x\sqrt{1-ax-bx^2}}$ is $\ln(x)-\ln(2\sqrt{1-ax-bx^2}-ax+2)+C$. With the bounds provided in the problem, the integral would not converge as $\lim_{x\rightarrow 0} \ln(x) =-\infty $


1

Taking the logarithms, $$n\log(a)=m\log(b),$$ or $$\frac{\log(a)}{\log(b)}=\frac mn\in\mathbb Q.$$ So the ratio of the logarithms in the desired bases should be well approximated by a rational. In your example $$\frac{\log(2)}{\log(10)}=0.3010299957\cdots\approx\frac3{10}.$$ Any real number can be approximated as closely as you want by rationals using ...


1

Hint 1 : Putting $x=\frac{m}{n}$, try to show that if $x$ is an approximation to $\sqrt{2}$ then $f(x)=\frac{x}{2}+\frac{1}{x}$ is a better approximation to $\sqrt{2}$. Hint 2: $\frac{f(x)-f(\sqrt{2})}{x-\sqrt{2}}=f'(c)$ for some $c$ between $x$ and $\sqrt{2}$ (this is the mean value theorem) If you don't know about derivatives, you can also compute that ...


1

We are given the equation the nonlinear equation $f(x) = 0$ where $f : [0,\infty) \rightarrow \mathbb{R}$ is given by \begin{equation} f(x) = a x^p + bx^q + cx + d, \quad p = 2.56, \quad q=1.78 \end{equation} It is known that $a, c, d$ are (strictly) negative and that $b$ is (strictly) positive. We first consider the question of the existence of solutions. ...


1

$$ \sum_{i=1}^{x}\frac{1}{w+i}=\text{area of rectangles}\leq\int_{0}^{x}\frac{1}{w+t}dt=\ln\left(w+x\right)-\ln\left(w\right) $$


1

If your endgame is to get the posterior density for y, I'm not understanding why you need an approximation when you can get the exact posterior density for any particular set of data. Here's how to do it using Mathematica: (* Data *) n = 10; p = {2/10, 3/10, 222/1000, 15/100, 19/100, 4/10, 44/100, 21/100, 34/100, 1/10}; x = {1, 1, 0, 0, 1, 1, 1, 1, 0, 1}; ...


1

HINT Required rational approximation for $\operatorname{erfc}$ function can be constructed analytically, on the basis of known Maclaurin series $$\sqrt\pi ze^{z^2}\operatorname{erfc}z = v,$$ ...


1

Fix $n \in \mathbb{N}-\left\{0\right\}$. One may use Stolz–Cesàro theorem: $$\lim_{m \to \infty} \frac{\sum \limits_{k=n}^{m} \frac{1}{\sqrt k}}{2\sqrt{ m-n}} = \lim_{m \rightarrow \infty}\frac{\frac{1}{\sqrt m}}{2(\sqrt{ m-n} - \sqrt{ m-n-1})} = \frac{1}{2} \lim_{n \rightarrow \infty} \frac{\sqrt{ m-n} + \sqrt{ m-n-1}}{\sqrt m} = 1,$$ giving, as $m \to ...



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