9

The first order approximation means you use the Taylor series of $f(x) = \sqrt{x}$ up to the term of degree 1 (=first order). In this case it is very convenient to develop it around $x_0=4$, as it is close to $3.9$ and we already know that $\sqrt{4}=2$: $$f(x_0+h)\approx f(x_0) + hf'(x_0)$$ In our case $x_0=4$ and $h=-0.1$ and with $f(x) = \sqrt{x}$ we get ...


5

It's just $$\frac xa\cdot\frac1{1+\frac ba}=\frac xa\sum_{k=0}^\infty \left(-\frac ba\right)^n$$ et cetera.


3

Using PARI/GP you can get a thousand decimal digits using default(realprecision,10^3); c=sumalt(n=3, (-1)^n*(1-n^(1./n))); in under 30 seconds. It takes much longer for more digits. It Seems to be an $O(n^3)$ time algorithm where $n$ is the number of digits. In case you are interested, the documentation for PARI/GP describes the algorithms used for the ...


3

Let $\, S_n := \sum_{i=2}^{n-1} \csc(\pi/i). \,$ Note the Laurent series expansion in powers of $\,1/i\,$: $$ \csc\left(\frac{\pi}i\right) = \frac{i}{\pi} + \frac{\pi}{6i} + \frac{7\pi^3}{360i^3} + O\left(\frac1{i^5}\right). $$ By summing we get $$ S_n \approx (n^2-2)/(2\pi) - C + H_{n-1}\pi/6 \quad \text{ where } \quad C \approx 0.7.$$ We can get closer ...


2

PART 1. Firstly, your matrix is only non-negative; consequently, $\lambda=\rho(A)$ is an eigenvalue of $A$ associated to a non-negative eigenvector $u$; yet, $\rho(A)$ is not necessarily simple and there may be other eigenvalues ​​of A of same modulus. See for example $A=\begin{pmatrix}0&0&1\\1&0&0\\0&1&0\end{pmatrix}$. Anyway, the ...


2

Have you asked the right question? I think it unlikely that either of the calculations you want to do quickly to two decimal place accuracy is likely to come up on exam where data is given in the form of a pie chart the quarterly sales of a company and you can't use a calculator. I think you might get more help here if you posted some sample questions ...


1

As Peter already answered $$\log(a+b)=\log(a)+\log\left(1+\frac{b}{a}\right)$$ Now let $$1+\frac{b}{a}=\frac{1+x}{1-x}\implies x=\frac{b}{2 a+b}$$ and use the very fast convergent $$\log\left(\frac{1+x}{1-x}\right)=2 \left(x+\frac{x^3}3 +\frac{x^5}5+\cdots\right)$$ Let us compute $$\log(123)=\log(100)+\log\left(1+\frac{23}{100}\right)\implies x=\frac{23}{...


1

Your lectures notes are covering what is known as the prime number theorem. Let $\pi(N)$ denote the number of primes $\leq N$. The prime number theorem states that $$\pi(N) \sim \frac{N}{\log N} \ \text{as} \ N \to +\infty$$ So, if you compare $\pi(20)$ with ${20}/{\log 20}$, you will find that $\pi(20)=8$ while ${20}/{\log 20}\approx 6.676$. But, what ...


1

We calculate the equivalent limit $$L=\lim_{x\to \infty}x\Big[\frac{I_0(x)}{I_1(x)}-\frac{I_1(x)}{I_0(x)}\Big]$$ The asymptotic expansions of the two modified Bessel functions are: $$I_0(x)\sim\frac{e^x}{\sqrt{2\pi x}}(1+\frac{1}{8x}+\mathcal{O}(x^{-2}))+i\frac{e^{-x}}{\sqrt{2\pi x}}(1-\frac{1}{8x}+\mathcal{O}(x^{-2}))\\I_1(x)\sim\frac{e^x}{\sqrt{2\pi x}}...


1

I'll give you the steps to get started, as well as the answer at the end (in the form of a spoiler). You want to do Gaussian quadrature for this problem. Define the weighted $L^2$ inner product $$(f,g)=\int\limits_0^1 f(x)g(x)\sqrt{x}\, dx,$$ then perform Gram-Schmidt with respect to this inner product to get up to the linear term in the basis. The root of ...


1

Hints. $\displaystyle \sum\limits_{n=3}^\infty (-1)^{n-1}(\sqrt[n]{n}-1) = \sqrt{2}-1 + \sum\limits_{k=1}^\infty \frac{1}{k!} \sum\limits_{n=1}^\infty (-1)^{n-1}\left(\frac{\ln n}{n}\right)^k $ $\displaystyle \sum\limits_{n=1}^\infty (-1)^{n-1}\left(\frac{\ln n}{n}\right)^k |_{k=1} = \frac{\ln 2}{2}(\ln 2~-2\gamma) $ $\displaystyle \sum\limits_{n=1}^\...


1

If you want an infinitely differentiable function, translate the piecewise function so that the point where the two lines meet is at the origin. Then rotate the two lines clockwise around the origin so that one of them has a slope of $0$ and is on the left side of the y-axis (ie $x<0$). The other line should have a positive slope. Let the new slope of the ...


1

Copson's Asymptotic Expansions considers all three cases (no stationary points, a stationary point on the boundary, a stationary point inside the interval), with the assumption that the functions are complex analytic. For the first case, the idea is just to apply integration by parts to $$\frac 1 {i k} \int_a^b \frac {f(x)} {p'(x)} d(e^{i k p(x)}).$$ In ...


1

$$\dfrac{11789}{234558} \approx \dfrac{12000}{235000}$$ In fact: $$\left| \dfrac{11789}{234558} - \dfrac{12000}{235000}\right| < 0.001$$ It should be easy to round before you divide and still get very close.


1

It's $$(45+0.11)(68-0.11)=3060+23\cdot0.11-0.0121=$$ $$=3062.53-0.0121=3062.5179\approx3062.52$$


1

This is a generalized hypergeometric function, with \begin{align}\sum_{k=0}^\infty\frac{(-1)^{2k+1}}{\Gamma(2k+v+2)(2k+1)!}\left(\frac z2\right)^{4k+v+2}&=-\left(\frac z2\right)^{v+2}\sum_{k=0}^\infty\frac1{\Gamma(2k+v+2)(2k+1)!}\left(\frac z2\right)^{4k}\\&=-\left(\frac z2\right)^{v+2}{}_0F_2\left(;-\frac32,\frac{v+3}2;\frac{z^4}{64}\right)\end{...


1

So you are looking for a "nice" form to express $$ \sum\limits_{j = 0}^n {P(j)} = \sum\limits_{k = 0}^n {c_{\,k} \sum\limits_{l = 0}^n {l^{\,k} } } $$ Now, as far as I know, the different formulations for the sums of powers can be summarized in $$ \eqalign{ & S_m (n + 1) = \sum\limits_{0\, \le \,l\, \le \,n} {l^{\,k} } \quad \left| {\;0 \le {\rm ...


1

Note that this essentially comes down to prove that $CPL\{0,1,2,3\}$ has dimension 4 (because 4 linearly independant vectors in a 4-dimensional space form a basis of this space, and you proved already they were linearly independant) Consider the map $T : CPL\{0,1,2,3\} \to \mathbb{R}^4$ such that $T(f) = (f(0),f(1),f(2),f(3))$. It is clear that this map is ...


1

For $k=3$: $$\sum _{n=1}^k \frac{1}{n^s}+\frac{1}{k^{s-1} (s-1)}=1+2^{-s}+3^{-s}+\frac{1}{3^{s-1} (s-1)}$$ For $s=0$: $$1+2^{-s}+3^{-s}+\frac{1}{3^{s-1} (s-1)}=0$$ Note for $s=0$, $\sum\limits_{n=1}^k \frac{1}{n^s}+\frac{1}{k^{s-1} (s-1)}=0$ for all values of $k$. For $F(s)=1+2^{-s}+3^{-s}+\frac{1}{3^{s-1} (s-1)}$, the contour plots of $\Re(F(s))=0$ (...


1

\begin{align} S&=\sum_{n=1}^\infty\left(H_n^{(2)}-\zeta(2)\right)^2=\sum_{n=1}^\infty\left(-\int_0^1\frac{x^n\ln x}{1-x}\ dx\right)^2\\ &=\int_0^1\int_0^1\frac{xy\ln x\ln y}{(1-x)(1-y)}\sum_{n=1}^\infty(xy)^n\ dx=\int_0^1\int_0^1\frac{xy\ln x\ln y}{(1-x)(1-y)(1-xy)}\ dx\\ &=\int_0^1\frac{x\ln x}{1-x}\left(\int_0^1\frac{y\ln y}{(1-y)(1-xy)}\ dy\...


1

I'll try to obtain a more clear form of the series first, then maybe something about the integral. Let's work with the series: $$e^\pi=\sum_{k=0}^\infty\frac{2^{k-1}\left(e^{\pi}-\left(-1\right)^ke^{-\pi}\right)\Gamma\left(\frac{k}{2}+i\right)\Gamma\left(\frac{k}{2}-i\right)}{\pi k!}$$ First we explicitly separate even and odd terms for clarity: $$e^\pi=\...


1

As far as I have been able to tell, there isn't a rigorous rational for the Padé approximant. The most recent work that I found about these approximants are the two books written by Baker Padé Approximants (Encyclopedia of Mathematics and its Applications) Essentials of Padé Approximants I haven't found any paper in the archive which answers why Padé ...


1

$\approx$ is used mostly for the approximate (read: calculated) value of a mathematical expression like $\pi \approx 3.14$ In LaTeX it is coded as \approx. $\cong$ is used to show a congruency between two mathematical expressions, which could be geometrical, topological, and when using modulo arithmetic you can get different numbers that are congruent, e.g.,...


1

Assume $\dfrac mn\ne\sqrt2;$ otherwise $\dfrac mn$ is $\sqrt2$, not an approximation. Then $d'\ne0$ so we can compute $\dfrac {d''}{d'}=\dfrac{\dfrac{m+2n}{m+n}-\sqrt2}{\dfrac mn-\sqrt2}= \dfrac n{m+n}\dfrac{m+2n-\sqrt2(m+n)}{m-\sqrt2n}$ $=\dfrac n{m+n}\dfrac{m-\sqrt2n-\sqrt2(m-\sqrt2n)}{m-\sqrt2n}=\dfrac {1}{1+\dfrac mn}\left(1-\sqrt2\right).$ We could ...


Only top voted, non community-wiki answers of a minimum length are eligible