# Tag Info

5

I would factor out the $2x$ from both terms, and at a crucial point use the Binomial expansion for $(1+a)^{1/2}$: \begin{align} \sqrt{4x^2+x}-2x&=2x\left[\frac{\sqrt{4x^2+x}}{\sqrt{4x^2}}-1\right]\\ &=2x\left[\sqrt{1+\frac1{4x}}-1\right]\\ &=2x\left[-1+1+\frac12\cdot\frac1{4x}-\frac18\left(\frac{1}{4x}\right)^2+\cdots\right]\\ &=2x\left[ ...

5

The method is entirely valid! The problem is that it becomes rather impractical for digits greater than say $4$. Also it may not be very practical for numbers with many digits in them. Let me try to explain the method. Basic principle: parallel lines crossing If $m$ parallel lines intersect $n$ parallel lines they will intersect in $m\cdot n$ points. So ...

4

It's easier than you are making it. To show degree of exactness $r$, it suffices to check exactness on a basis of polynomials of degree $r$, e.g. check $1,x,\ldots,x^r$. This check is also a necessary condition, so it's pretty efficient. Added: Let's try to address the example given in the Question: $$I_3(f) = \frac{1}{4}[f(-1)+f(-1/3)+f(1/3)+f(1)]$$ ...

3

$$e^x-x-1 = \frac{x^2}{2} + \frac{x^3}{6} + \frac{x^4}{24} + \cdots$$ Dividing that by $x^2$, you get $$\frac 1 2 + \frac x 6 + \frac{x^2}{24} + \cdots$$ If you mean limit as $x\to0$, that should now become apparent. (If you mean limit as $x\to\text{something else}$, then you need to clarify that.)

2

Setting $\frac1x=h,$ $$\lim_{x\to\infty^+}\sqrt{4x^2+x}-2x=\lim_{h\to0}\frac{\sqrt{4+h}-2}h\ \ \ \ (1)$$ Method $1:$ Now, $$\frac{\sqrt{4+h}-2}h=\frac{(4+h)-2^2}{h(\sqrt{4+h}+2)}=\frac1{\sqrt{4+h}+2}\text{ if }h\ne0$$ Here as $h\to0,h\ne0$ Can you take it from here? Method $2:$ This has strong resemblance with your method $\displaystyle ... 2 $$e^{\pi\sqrt{163}}\ \simeq\ 640,320^3+744\qquad\iff\qquad\pi\ \simeq\ \frac{\ln(640,320^3+744)}{\sqrt{163}}$$ See Heegner number for more details. The precision is$30$decimals. 2 Start from the definition of a Gauss quadrature: $$\int_{-1}^1 f(x)\ dx \approx \sum_{i=1}^3 g_i f(x_i).$$ We perform a change of variables to change the range of integration, so $$\int_a^b f(x)\ dx = \frac{b-a}{2}\int_{-1}^1 f\left(\frac{b-a}{2}z+\frac{b+a}{2}\right)\ dz.$$ Apply gaussian quadrature to the new integral on the right. ... 1 In MO there was an answer indicating, that there shall be no more information than that of Richard Fischer's site, where he lists, that indeed that pair$(68,113)$is the only pair up to about$p \le 3.6 \cdot 10^6$and where also$b \lt p$which gives a fermat-quotient greater than 2 , so I think I should "close the case" here. For the casual reader ... 1 Some hints. Hint 1: Since$f$is continuous, for any$\epsilon\gt0$, there is a$\delta\gt0$so that $$|x-\bar{x}|\le\delta\implies|f(x)-f(\bar{x})|\le\epsilon$$ Hint 2:$x\in\partial B_a(\bar{x})\implies|x-\bar{x}|=a\$. Hint 3: $$\left|\int_{\partial B_a(\bar{x})}f(x)\,\mathrm{d}x-\int_{\partial ... 1 If \phi: [0,1]\times [a,b]\rightarrow \mathbb R, (x,t)\mapsto \varphi(x,t) is continuous on [0,1]\times [a,b] with its partial derivative \frac{\partial \phi}{\partial x}(x,t), then M=M(t) is differentiable and$$\frac{dM}{dt}(t)=\int_0^1\frac{\partial \phi}{\partial x}(x,t)dx. $$The proof uses the Mean value theorem and the fact that any ... 1 It's likely just for computational efficiency. Some totally unverified ideas: 327,843,840 = 2^{15} \times 3 \times 5 \times 23 \times 29. This concentrates a whole bunch of left-shifts (with the powers of 2). 100,100,025 is a perfect square and also has a whole bunch of zeroes in it, which makes it faster to multiply with an arbitrary-precision ... 1 If you have a value x between 0 and 2\pi, this implies that the segment [0,x] represents \frac{x}{2\pi} of the whole interval [0,2\pi] (as a fraction). Therefore, you want the corresponding point y\in [1,44100] to cover the same fraction. This means that you want y to satisfy$$\frac{y-1}{44100-1}=\frac{x}{2\pi}.The solution is ... 1 We have writing C := f(a) + f(b) \begin{align*} \sum_{k=1}^{n-1} f(x_k) &= \frac 12\left( \sum_{k=1}^{n-1} f(x_k) + \sum_{k=1}^{n-1} f(x_{n-k})\right)\\ &= \frac 12 \sum_{k=1}^{n-1} \bigl(f(x_k) + f(x_{n-k})\bigr)\\ &= \frac{n-1}2 \cdot C \end{align*} Hence \begin{align*} T(n) &= \frac{b-a}{2n}\cdot C + \frac{b-a}n \cdot ... 1 First, you should splity'''+y^2y''-y'=0$$into a system of three first order ordinary differential equations by letting$$y'= u$$and$$ u'=v $$such that you get$$v'+y^2v-u = 0$$After taking another look, I think I might now see what you are trying to do with Simpson's rule. Starting with the definition of Simpson's rule$$\int_{x}^{x+2\Delta ...

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