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13

The accuracy of any Taylor expansion of a given order is given by the next order of the expansion. For this very case, the answer is that it depends on how far out you want to take the expansions. For example, $$\frac1{1-x} = 1+x+x^2+O(x^3)$$ This has twice the error as the expansion $1+x$, whose next term is $x^2/2!$. However, ...


4

This is the same as proving that $$\frac{1}{16}(133-37\sqrt{5})>\pi$$ and it follows from the fact that the continued fraction of the LHS is: $$ [3;7,15,1,660,\ldots] $$ while the continued fraction of $\pi$ is: $$ [3;7,15,1,292,\ldots].$$


4

First, let's split up the integral: $$ \begin{align} \int_0^\infty\frac{x^n}{1+e^{x-t}}\mathrm{d}x &=\int_{-t}^\infty\frac{(x+t)^n}{1+e^x}\mathrm{d}x\\ &=\color{#C00000}{\int_{-t}^0\frac{(x+t)^n}{1+e^x}\mathrm{d}x}+\color{#00A000}{\int_0^\infty\frac{(x+t)^n}{1+e^x}\mathrm{d}x}\tag{1} \end{align} $$ Note that the first integral on the right side of ...


3

We have: $$ F_0(t) = \log(e^t+1) = t+\log(1+e^{-t}) = t+o(1) $$ and since your identity gives: $$ F_n(t) = n\int_{0}^{t} F_{n-1}(u)\,du $$ we have: $$ F_1(t) = \frac{t^2}{2}+o(t), $$ $$ F_2(t) = \frac{t^3}{3}+o(t^2), $$ and so on, so the claim holds by induction for any $n\in\mathbb{N}_{>0}$. Moreover, since: $$ F_n(t) = ...


3

Here's one quick observation. Letting $$f_{1,n}(x)=\sum_{k=0}^n \frac{x^n}{n!}$$ and $$f_{2,n}(x)=1/\left(\sum_{k=0}^n \frac{(-x)^n}{n!}\right),$$ we have $$\lim_{x\rightarrow\pm\infty}f_{1,n}(x) = \pm\infty \: \text{ and } \: \lim_{x\rightarrow\pm\infty}f_{2,n}(x) = 0.$$ Thus, I suppose we'd expect $f_{1,n}(x)$ to be a better approximation for positive ...


3

My favorite method would be continued fractions. If the fraction is exact the process will (should! ;-)) converge : (illustrations using the free pari/gp) contfrac(0.14285) = [0, 7, 2857] with the exact answer : $0+\dfrac 1{7+\dfrac 1{2857}}=\dfrac{2857}{20000}$ else you'll get a large integer after some terms (the difficult part is to decide when ...


3

Say you're trying to approximate a number by rational numbers $p/q$. Usually, the bigger $q$ is, the better your chance of approximating the number closely. On the other hand, the smaller $q$ is, the simpler the approximation. In the case of $\pi$, if you want to have a better approximation than $22/7$, you have to go all the way up to $q = 57$. (See this.) ...


2

3.14 is two decimal places. That's the only justification I can give. When doing hand calculations (or sliderule calculations), carrying out more than three significant digits is cumbersome. Remember this: with $N$ digits in a multiplication, you have $N^2$ digits in the final answer before you can round again. In the world of engineering, often you can't ...


2

I don't agree that "Evaluating decimals are always better...", but $$ x_{n+1} = \frac{1}{2}\left(x_n+\frac{x}{x_n}\right) $$ converges to $\sqrt{x}$ quite quickly.


2

Let's try this using the classical expansion of $\;\ln\Gamma$ (in A&S expression $6.1.41$) : $$\ln\,\Gamma(z)\sim \left(z-\frac 12\right)\ln z-z+\frac 12\ln(2\pi)+\frac 1{12z}+\cdots,\quad \text{for}\;\; z\to\infty,\ |\arg z|<\pi$$ (this is an asymptotic expression and the error doesn't exceed the first term omitted) Let's apply this formula to the ...


2

While the first one approaches from below the other one oscillates near the actual values, it also depends how many terms (say t) you evaluate, for most of the case with large t both are approximately equal but with smaller t the first one is more accurate see graphs: In case of odd t it is above $e^x$ since negative terms dominate and denominator becomes ...


2

In general, we want the approximation to go to (at least) first nontrivial order. This is why it was not necessary for the "minus" approximation since it was finite, as oppose to the "positive" case.


2

You must define what you mean by "better". If you mean closer to the value, then yes $\frac{22}{7} \approx 3.14285714286$ and $3.14$ are closer to $\pi \approx 3.14159265359$ and $\frac{22}{7} > \pi$ is closer than $3.14 < \pi$. However, if you mean a rational number whose decimal representation is, for example, less than $5$ digits (not including ...


1

Hi.Recall $a^{log_a(b)}=b$, so $10^{log_{10}(29^{1312000})}=29^{1312000}$. A better estimate would be $10^{1918666}$.


1

I'm not sure what you mean by "more accuracy." But the second series has alternating addition and subtraction. Subtraction is subject to loss of precision. (Example: $1.234-1.2=0.034$, a loss of two significant digits.) So the first is more accurate in that sense.


1

$$\log\frac{r+l_1}{r-l_2}=\log\frac{1+(l_1/r)}{1-l_2/r}\\=\log(1+l_1/r)-\log(1-l_2/r)\\ \approx l_1/r-(-l_2/r) = (l_1+l_2)/r$$


1

Hint $$\frac{r+l_2}{r-l_1}=\frac{r-l_1+l_1+l_2}{r-l_1}=1+\frac{l_1+l_2}{r-l_1}\approx 1+\frac{l_1+l_2}{r}$$ Now, use, for small $x$, $\log(1+x)\approx x$.


1

Here is a method to calculate the least absolute deviation solution that goes back at least to Gauss (see Least Absolute Deviations). Please note that the least absolute deviation solution is rarely unique (one of the differences to linear least squares). Your over-determined linear equation system is given by $Ax=b$ with: $$A = \left( \begin{array}{ccc} ...


1

for the numerator of the first derivative i have got after simplification i have got $$2\,{F}^{2} \left( \cos \left( x \right) \right) ^{4}-3\,{F}^{2} \left( \cos \left( x \right) \right) ^{3}-{F}^{2} \left( \cos \left( x \right) \right) ^{2}-2\, \left( \cos \left( x \right) \right) ^{4}+3\,{F}^{2}\cos \left( x \right) +3\, \left( \cos \left( x ...


1

Multiplying top and bottom leads to the following $$ (1+\cos x)\sin x\left[F^2 +\cot ^2 x\right]\frac{1}{F} $$ The derivative with respect to x of the above is $$ -\sin^2 x \left[F^2 +\cot ^2 x\right]\frac{1}{F} +(\cos x+\cos^2x)\left[F^2 +\cot ^2 x\right]\frac{1}{F} + (1+\cos x)\sin x\left[2\cot x \left(-\mathrm{csc}^2 x\right)\right]\frac{1}{F} $$ This ...


1

$$ S = k_B \ln{N \choose N_+} \approx k_B N \,H\left(p \right)$$ where $H(p)$ is the binary entropy function (in nats) and $$p=\frac{N_+}{N}=\frac{1}{2}\left(1+\frac{L}{aN}\right)$$ Now, if we can do the (additional) assumption $\frac{L}{aN} \ll 1$, we can do a Taylor expansion of $H(p)$ around $p=1/2$, so that $$H(p)\approx \ln 2 - \frac{(1-2p)^2}{2 }= ...


1

Note that we can reformulate $$\lim_{n\rightarrow \infty}\frac{P(n)}{F(n)}=1,$$ with a little bit of algebra as $$P(n)\underset{n\rightarrow\infty}{\sim} F(n)\Leftrightarrow \lim_{n\rightarrow \infty}\frac{P(n)-F(n)}{F(n)}=0.$$ That is the difference between the functions becomes negligible as $n\rightarrow\infty$. For example consider $F(n)=10^n$ and ...


1

Don you want the following: $$\sin x=x-\frac{x^3}{3!}+\frac{x^5}{5!}+\cdots$$ $$\cos x=1-\frac{x^2}{2!}+\frac{x^4}{4!}+\cdots $$


1

Near to zero it is usual to use the following approximation: $$\sin x \approx x $$ $$\cos x \approx 1-\displaystyle \frac{x^2}{2}$$ $$\tan x \approx x $$ you can convince yourself, for example, using the relationships that @paul gives.


1

There is a general construction that can be used here. Consider any nonnegative measurable function $f$. Now define $$A_{n,m} = \left \{ x : \frac{m-1}{n} \leq f(x) \leq \frac{m}{n} \right \}, \quad m = 1,2,\dots,n^2, \\ A_{n,n^2+1} = \{ x : f(x) \geq n \}.$$ Then the sequence of simple functions $$s_n(x) = \sum_{m=1}^{n^2} \frac{m-1}{n} 1_{A_{n,m}}(x) + ...


1

You can establish quite nice approximations using Pade expansions of the functions. For example $$\sin(x)\approx \frac{x-\frac{7 x^3}{60}}{1+\frac{x^2}{20}}$$ $$\cos(x)\approx\frac{1-\frac{5 x^2}{12}}{1+\frac{x^2}{12}}$$ $$\tan(x)\approx\frac{x-\frac{x^3}{15}}{1-\frac{2 x^2}{5}}$$ are quite good. For sure, if you increase the degrees of numerator and ...


1

I agree with @Hebrik that "decimals are better" is dubious. But: $\sin(x) \approx x$ for small $x$; for not-small $x$, the half-angle formula (together with rules like $\sin(x) = \sin (\pi - x)$ get you to small $x$ fairly quickly. And $\ln(1+x) \approx x$ for small $x$, and you can use $\ln(x/e) = \ln(x) - 1$ to rapidly reduce the size of $x$ until it's ...


1

Let $$E_n(f)=a,\qquad E_n(g)=b$$ There exist two polynomials $p_1$ and $p_2$ such that $$\|f-p_1\|<a+\epsilon,\qquad \|g-p_2\|<b+\epsilon$$ so $$E_n(f+g)\le\|f+g-p_1-p_2\|\le a+b+2\epsilon$$ for all epsilon, so $$E_n(f+g)\le a+b=E_n(f)+E_n(g)$$


1

For the first question, note that $\frac{1+3x}{2+x} = \frac{1}{2+x} + 3x * \frac{1}{2+x}$. Now, try to put $\frac{1}{2+x}$ into the form $\frac{1}{1-y}$, and use the fact that, for $|y|<1$, we have $\frac{1}{1-y} = \sum_{n=0}^{\infty} y^n$. The second question can be answered similarly. You just have to determine the series for $\frac{1}{(1-y)^2}$ using ...


1

Use Urysohn to construct $g$ with $g \equiv 2$ on $U$ (multiply your $g$ by $2$). Then take a sequence $(f_n)$ with$f_n \to g$ uniformly. For $n$ large, you will then have $f_n \geq 1$ on $U$.



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