# Tag Info

32

One way to convert any decimal fraction to base $16$ is as follows (taking $\pi$ as an example).$$\pi=\color{blue}3.141592...$$ Take the whole number part and convert it to base $16$ as usual. In this case $\color{blue}3$ will remain as $3$. So we have so far got $3.14159..._{10}=\color{red}{3...._{16}}$ This now leaves us with $0.141592...$ - ...

11

Note that: $$10000\pi=31415.92653\dots$$ which means that the decimal version of $\pi$ begins $3.1415\dots$. Similarly: $$16^4\pi=205887.46145\dots$$ Since $205887$ is $3243F$ in hexadecimal, the hexadecimal version begins $3.243F\dots$.

11

For the particular base of $16$, there is this remarkable formula: $$\pi=\sum_{n=0}^\infty \left(\frac{4}{8n+1}-\frac{2}{8n+4}-\frac{1}{8n+5}-\frac{1}{8n+6}\right)\frac{1}{16^n}$$ It allows the computation of any base 16 digit of $\pi$ without the need to compute all the preceding digits. The discovery of this formula by Bailey, Borwein and Plouffe in 1995 ...

5

You could express this as $\log(1/2) > -\dfrac{253}{365}$. The series $$\log(1/2) = \log(1-1/2) = -\sum_{n=1}^\infty \dfrac{1}{n 2^n}$$ converges quickly, and has nice bounds: $$\log(1/2) \ge - \sum_{n=1}^{N-1} \dfrac{1}{n 2^n} - \sum_{n=N}^\infty \dfrac{1}{N 2^n} = - \sum_{n=1}^{N-1} \dfrac{1}{n 2^n} - \dfrac{1}{N 2^{N-1}}$$ EDIT: Another way to ...

5

Sometimes, simpler is better ... So, let's just start with the identity that Chappers provided, namely $$\int_{0}^{\infty} (J_1(x)^2+J_1(x)J_1(x)'')\,dx= \frac{1}{4}\int_0^{\infty} J_1(x) (J_1(x)+J_3(x)) \, dx$$ Then, let's substitute the recurrence relation $$J_2(x)=\frac{x}{4} (J_1(x)+J_3(x))$$ to find that $$\int_{0}^{\infty} ... 4 Since J_1 is a solution of the Bessel differential equation:$$ x^2 f'' + x f' + x^2 f = f \tag{1}$$by exploiting integration by parts we have that:$$ \int_{0}^{+\infty}\left(J_1(x)^2+J_1(x)\,J_1''(x)\right)\,dx = \frac{1}{2}\int_{0}^{+\infty}\left(\frac{J_1(x)}{x}\right)^2\,dx \tag{2} $$so we just need to recall that the Fourier transform of ... 3 Mathematica says it converges to 2/(3\pi). You can probably get this from this identity:$$ \int_{0}^{\infty}\frac{J_{\mu}(at)J_{\nu}(at)}{t^{\lambda}}dt= \frac{(\frac{1}{2}a)^{\lambda-1} \Gamma\left(\frac{1}{2}\mu+\frac{1}{2}\nu-\frac{1}{2}\lambda+\frac{1}{2}\right) \Gamma(\lambda)}{ 2\Gamma\left(\frac{1}{2}\lambda+\frac{1}{2}\nu- ...

3

The reason is that your sequence $$1 - x^2 + \frac{x^4}{2!}- \frac{x^6}{3!} + .... = \sum_{k=0}^\infty \frac{(-1)^k x^{2k}}{k!}$$ is alternating. In this case, if you use the first $n$-th term to approximate the value, then the error will be bounded by the $n+1$- term. (I hope this is proved in your textbook). In your case, the next term will be ...

3

The alternating series theorem says that the error is less than the absolute value of the first neglected term, so you just need to find the first term less than $10^{-4}$. You don't know that this term is the first one for which the error is less than $10^{-4}$, but you know it works.

3

This seems like an improper application of the tangent line approximation. The usual approximation is $$f(x+\varepsilon)\approx f(x)+\varepsilon f'(x)\tag{*}$$ for $\varepsilon$ small. Your choice of $f$ doesn't match up with $(*)$. But using $f(x)=\sqrt x$ gives $$\sqrt{x+\varepsilon}=f(x+\varepsilon)\approx\sqrt x+{\varepsilon\over{2\sqrt x}}$$ and ...

2

A method for computing the curvature is described for example in: Driscoll MK, McCann C, Kopace R, Homan T, Fourkas JT, et al. (2012) "Cell Shape Dynamics: From Waves to Migration." PLoS Comput Biol 8(3): e1002392. doi:10.1371/journal.pcbi.1002392 They have 400 points (they call them boundary points) and: At each boundary point, we calculate the ...

2

Let $\displaystyle S_n = 4\sum_{k=0}^n \frac{(-1)^k}{2k+1}$. Then $$|S_{10001}-S_{10000}| = |a_{10001}| = \left|\frac{-4}{20003}\right| < 0.0002.\tag 1$$ We know that $S_{10000}>\pi>S_{10001}$ but $\pi$ is somewhat closer to the latter term than to the former, and $S_{10002}>\pi>S_{10001}$ and $\pi$ is somewhat closer to the former term ...

2

You can use the binomial expansion, which approximates the sum of reciprocals of the factorials: $$(1+0.0001)^{10000} = 1^{10000}+10000\times0.0001+{10000\times9999\over2}\times0.0001^2+{10000\times9999\times9998\over2\times3}\times0.0001^3+{10000\times9999\times9998\times9997\over2\times3\times4}\times0.0001^4+...\simeq1+1+1/2+1/6+1/24+1/120\simeq 3$$ Also ...

2

Since $\log 2=2\operatorname{arctanh}\frac{1}{3}$, we may use the series: $$\log 2 = \sum_{n\geq 0}\frac{2}{(2n+1)\,3^{2n+1}}$$ and the bound: $$\sum_{n\geq 4}\frac{2}{(2n+1)\,3^{2n+1}}\leq\sum_{n\geq 4}\frac{2}{3^{2n+3}}=\frac{1}{157464}$$ to prove that (we just need the terms of the previous series till $n=4$): $$\log 2\approx \frac{4297606}{6200145} ... 2 Well your series should be$$\sum_{n=0}^{\infty}\frac{(-1)^n(x^{2n+1})}{(2n+1)!} = x-\frac{1}{3!}x^3+\frac{1}{5!}x^5-\dots$$simply plug in x=\frac{6\pi}{7} and x=\frac{20\pi}{7}. Then compare the difference (if the series were perfectly accurate they would be the same, because \frac{20\pi}{7}=\frac{6\pi}{7}+2\pi) The error is told by ... 2 We can use the three identities$$ \sum_{i=1}^n1=n  \sum_{i=1}^ni=\frac{n^2+n}2  \sum_{i=1}^ni^2=\frac{2n^3+3n^2+n}6 $$to evaluate$$ \begin{align} \frac1n\sum_{i=1}^n\left(\frac{2i-1}{2n}\right)^2 &=\frac1{4n^3}\sum_{i=1}^n(4i^2-4i+1)\\ &=\frac1{4n^3}\left(\frac{4n^3+6n^2+2n}3-2n^2-2n+n\right)\\ &=\frac1{4n^3}\frac{4n^3-n}3\\ ...

2

No, in fact $\sup_{|z|<1}|f(z)-u(z)|\ge 1$ for all such $u.$ Proof: Observe that as $t \to 0^+,$ $f(e^{it})$ takes on every value in the unit circle infinitely many times. Using the continuity of $f$ at all boundary points save $1,$ we see there exists sequences $z_n,w_n$ in the open disc, $z_n, w_n \to 1,$ such that $f(z_n) \to 1, f(w_n)\to -1.$ Thus ...

1

If $x_2-x_1 = \frac12$ then $x_2^2-x_1^2 = (x_2-x_1)(x_2+x_1) = \frac12$ is equivalent to $x_2+x_1 = 1$ The book may have wanted you to use Euler Lagrange equations?

1

We know that $$\int_0^1 x^2 \, dx = \frac{1}{3}$$Thus, proof by induction would be most efficient if we just make this replacement in the formula. So, prove $$\frac{1}{n}\sum_{i=1}^n\left(\frac{2i-1}{2n}\right)^2 + \frac{1}{12n^2} = \frac{1}{3}$$for all $n\in\mathbb{N}$. Proof by induction: First, we must show the base case works for $n=1$. So, ...

1

Both expression are obviously cubic. If they both have the same value and the same first and second derivatives at $x=1$, your set of expressions fits the definition of a cubic spline. A pretty simple spline, but there it is.

1

$\log 2 = \log (1 + 1/3) + \log (1 + 1/2)$, and each of those terms can be easily evaluated to the desired predictions using $\log (1+x) = x - x^2/2 + x^3/3 - \cdots$. There are other ways to write $\log 2$ as a sum of logarithms that will let you write it as a sum of more, but faster-converging series; I'm not sure what minimizes the arithmetic. In ...

1

Observe that your number is $f(10^4)$, where $$f(n) := \left( 1 + \frac{1}{n} \right)^n$$ and recall that $e = \lim\limits_{n \to \infty} f(n)$. Further, observe that $f(n)$ is strictly increasing for $n>1$, therefore $f(10^4) < e \approx 2.71828$. If you don't already know that $f(n)$ is increasing, you could prove it by recalling that $\log(1 + ... 1 You have only five digits of accuracy in this computation, so$x$rounds to$x^* = 2.1437$and$y$rounds to$y^* = 2.1436$. Now compute$x^* - y^*.$1 Crank-Nicolson for the PDE$\frac{∂u}{∂t}=L(t,u)$plus boundary conditions first performs a discretization in the space dimension(s) forming a vector of sample point values$\vec u(t)=\{u(t,x_k)\}_{k=0,…,N}$,$x_k=x_0+k·Δx$, and transforming the operator$L$plus the boundary conditions into a function$f(t,\vec u)$of this vector approximating them. Then ... 1 This is the consequence of three elements: First you have$\Vert D(c_0,\dots,c_n) \Vert_\infty \le 2\Vert f\Vert_\infty +1$for points belonging to$S$.$(c_0,\dots,c_n) \mapsto \Vert D(c_0,\dots,c_n) \Vert_\infty$is a norm of the finite dimensional space$\mathbb{R}^{n+1}$. And by point 1.$S$is bounded for that norm. Finally as in finite dimensional ... 1 Abbreviate $$\sum_{i=0}^{\infty}F(i)\frac{\ell(t_0)-i}{\sum_{u\in T}\rho(u)\ell(u)-i}=\sum_{i=0}^{\infty}F_i\frac{a-i}{b-i}$$ We assume that$F_i$is a decreasing sequence. Note that both$a-i$and$b-i$are also decreasing sequences, but that the ratio$\frac{a-i}{b-i}$can be increasing or decreasing depending on whether$a<b$or$b<a$, ... 1 No, your understanding is not correct. All simple functions are measurable, but in most$\sigma$-algebras of interest, far more functions are also measurable. For instance, all continuous functions are measurable for the Lebesgue$\sigma$-algebra on$\mathbb{R}$, but only constant continuous functions are simple. What is true is that all measurable ... 1 The important question to ask is what do you mean by "approximate"? Think of the Riemann integral for example. There the partial sums you use to define a Riemann integral are all really integrals of step functions. Yet the Riemann integral of trigonometric functions works just fine. There are many other ways in which step functions can approximate trig ... 1 You could start with Taylor expansion on each term: $$f(x-h)=f(x)+f'(x)(-h)+\frac{1}{2!}f''(x)(-h)^2+O(h^3)\\ f(x)=f(x)\\ f(b)=f(x)+f'(x)(b-x)+\frac{1}{2!}f''(x)(b-x)^2+O((b-x)^3)$$ Then add three equations with weights$w_0,w_1,w_2$:$\$w_0f(x-h)+w_1f(x)+w_2f(b)=w_0f(x)+w_1f(x)+w_2f(x)+w_0f'(x)(-h)+w_2 f'(x)(b-x)+w_0 \frac{1}{2!}f''(x)(-h)^2 + w_2 ...

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