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2

Dalzell's integral is related to the rational approximation $\pi\approx \frac{22}{7}$. $$\pi=\frac{22}{7}-\int_0^1 \frac{x^4(1-x)^4}{1+x^2}dx\approx\frac{22}{7}$$ Similar small integrals are related to simple irrational approximations using $\sqrt{2}$ and $\sqrt{3}$. $$\pi=\frac{20\sqrt{2}}{9}-\frac{2\sqrt{2}}{3} \int_0^1 ...


2

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}} ...


2

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 $\lambda$, $$\int_0^\infty e^{-u^2/2\lambda^2}e^{-u}du\approx\int_0^\infty e^{-u}du.$$


2

Assuming you have access to the IEEE recommended function $\mathrm{log1p}$ or sometimes $\mathrm{logp1}$ or $\mathrm{ln1p}$ defined by $\mathrm{log1p}(p)=\log(1+p),\,$ I can give a numerical stable evaluation of the usual function $$\mathrm{logit}(p)=\log \frac{p}{1-p}$$ The instable/inaccurate region is near $p=1/2$. You can safely use the definition for ...


2

You can get a good approximation by writing that $\cos(x)\approx 1-\frac{x^2}{2}$ and thus $\cos^k(\alpha/k)\approx \left(1-\frac{\alpha^2}{2k^2}\right)^{k}$ and then use that $\left(1-\frac{x}{n}\right)^n \approx e^{-x}$, so you get that, for $k$ reasonably large: $$\cos^k\left(\frac{\pi}{2k}\right)\approx e^{-\pi^2/(8k)}$$ Setting this to $99/100$, we ...


2

There are two reasonable interpretations of this problem. More experienced students would likely say to do $$s=\int ds=\int_0^2\sqrt{1+y^{\prime}(x)}dx\approx\frac h2\sum_{i=0}^8w_i\sqrt{1+(y^{\prime}(x_i))^2}\approx14.09254$$ Where $w_0=w_8=1$ and all other $w_i=2$. But I betcha that the questioner is doing something like ...


2

Your problem reminds me of a similar one developed here, in Chapter 4. It is solved with a graph theory approach, more specifically with a shortest path algorithm that gives you the optimal set of knots. It is pretty well detailed, take a look!


1

You are asked to compute the two appoximations at $x=1$ using $h=1/16$, which you can then compare with the exact value. Using the approximation in (1): $$ f'(1)\approx \frac{f(17/16)-f(1)}{1/16}=\frac{\sqrt{17/16+1}-sqrt(2)}{1/16}\approx 0.35083 $$ which you can compare with the value of $1/(2\sqrt{2})\approx 0.35355$. Now do the same with (2).


1

According to https://en.wikipedia.org/wiki/Fresnel_integral, for large $x$, $S(x) =\sqrt{\dfrac{\pi}{2}}\left(\dfrac{sign(x)}{2} -(1+O(x^{-4}))\left(\dfrac{\cos(x^2)}{x\sqrt{2\pi}} +\dfrac{\sin(x^2)}{x^3\sqrt{8\pi}} \right)\right) $ and $C(x) =\sqrt{\dfrac{\pi}{2}}\left(\dfrac{sign(x)}{2} +(1+O(x^{-4}))\left(\dfrac{\sin(x^2)}{x\sqrt{2\pi}} ...


1

The first one is easy. If you compare the two sides, it's really just asserting that $${n\choose t} \leq\frac{n^t}{t!}$$ But that follows quickly from the fact that $${n\choose t}=\frac{n!}{t!(n-t)!}=\frac{\overbrace{n(n-1)(n-2)\cdots(n-t+1)}^{t{\textrm{ factors}}}}{t!}$$ since each of the factors in the last numerator are no more than $n$.


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What you're seeing is that$\frac{n!}{(n-t)!}$ is best understood as having $n$ terms above and $n-t$ of those are cancelled by the denominator. So this expression is a product of $t$ terms all of which are less than $n$. Hence $\frac{n!}{(n-t)!}\leq n^t$ and so $\binom n t =\frac{n!}{(n-t)!t!}\leq \frac{n^t}{t!}$. The second similarity is by the Sterling ...


1

For all $d \geq e$ and $0 < p < 1$ the equation $$ N^d = pe^N $$ has two positive solutions $N$, call them $N_1$ and $N_2$. Since the maximum of $x \mapsto x^d e^{-x}$ occurs at $x=d$ and $x^d e^{-x} \to 0$ as $x \to \infty$ we know that $N_1 < d < N_2$. We'll first investigate the behavior of $N_1$, then $N_2$ afterwards. Because the maximum ...


1

'my problem is to approximate' - this strongly depends on what exactly you want to do, e.g. approximate around which value. For $t \approx 0 \ \log(1-t) \sim -t +\frac{t^2}{2}$ if $t$ is positive. BTW in the expansion you gave $t=-1$ works too.


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The usual result (I don't know of a particular name for it) is that $$\lim_{n \to \infty} \left( 1 + \dfrac{t}{n}\right)^n = e^t $$ This is sometimes taken to be the definition of the exponential function.


1

Your final inequality seems to be false. For instance, $\dfrac{\Gamma(2;a)}{\Gamma(2)} = (a+1)\mathrm{e}^{-a}$ and $\dfrac{\Gamma(3;a)}{\Gamma(3)} = \dfrac{1}{2}(a^2 + 2a + 2)\mathrm{e}^{-a}$. The bound $\dfrac{\Gamma(x;a)}{\Gamma(x)} \leq O(a^{x-1}\mathrm{e}^{-a})$ seems possible, but I haven't taken the time to find a proof.


1

$$\int_{a}^{+\infty}t^{x-1}e^{-t}\,dt = e^{-a}\int_{0}^{+\infty}(t+a)^{x-1}e^{-t}\,dt\leq e^{-a}a^{x-1}\int_{0}^{+\infty}\exp\left(\frac{t(x-1)}{a}-t\right)\,dt$$ hence a simple upper bound is: $$ \Gamma(x;a)\leq \frac{a^x e^{-a}}{a+1-x}.$$ That approximation can be refined, producing a continued fraction representation for the incomplete $\Gamma$ function: ...


1

Assume X is a rational number for which you would like to approximate its square root with a rational number. Say you have an estimate $x_1$ and would like a better estimate. You know that if $x_1$ is an estimate for $\sqrt{X}$ then $\frac{X}{x_1}$ is an estimate as well, and their geometric mean is your answer. You cannot calculate the geometric mean of ...


1

A series can be matched to your inequality. The number $\sqrt{10}$ is $\sqrt{\frac{4^8}{3^8}+\frac{74}{6561}}$, which is $\sqrt{\frac{4^8}{3^8}+x}$ for $x=\frac{74}{6561}$. The expansion has alternating sign $\sqrt{\frac{4^8}{3^8}+x} = \frac{256}{81}+\frac{81}{512}x-\frac{531441}{134217728}x^2+...$ and its first term is $\frac{256}{81}=\frac{4^4}{3^4}$, ...


1

For $c>0$ and $n$ sufficient large the dominator becomes $W(0)$ : $W(-(\ln n)^{-c})\to W(0)=0$ for $n\to \infty$. It’s $\frac{1}{W(x)}= \frac{1}{x}e^{W(x)}$ by definition. Substitute $x$ by $-(\ln n)^{-c}$. Then your term becomes $(\ln n)^{c+1}$; that’s very different to $\frac{\ln n}{\ln\ln n}$. EDIT: I don’t know exactly what you mean, but I will ...


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A more accurate asymptotic series can be obtained with more terms. but, before, one can observe that $y=\frac{-\ln n}{W(- \ln^{-c}n)}$ is not real if $n<e^{e^{1/c}}$ This means that, for small values of $c$, $n$ has to be very large : In case of $c=0.2$ for example, $y$ is not real if $n<10^{64}$ (roughly) In case of $c=0.5$ for example, $y$ is not ...



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