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1

Since $f$ is analytic on $\mathbb{C}\setminus\{0\}$, the fact that $\lim_{z\to0}zf(z)=0$ would imply $g(z)=zf(z)=e^{-1/z^2}$ has a removable singularity at $0$. But it is well known that the Taylor series at $0$ for the real function $$ h(x)=\begin{cases} 0 & \text{if $x=0$}\\ e^{-1/x^2} & \text{if $x\ne0$} \end{cases} $$ doesn't converge to $h$ ...


0

Your problem is that the exponential can become greater than $1$, which happens when the argument is greater than $0$. As long as both $x, x_b \gt \log H$ you don't have a problem. The problem is when one of them is less than $\log H$, which shouldn't happen if that is a barrier. Maybe it is roundoff error in assessing the barrier?


1

For a complete answer, it would be helpful to know where the question is taken from. As far as I would say, an "arrival" describes the time until the next person enters the bank. As you pointed out, this is independent of the number of customers in the bank and therefore the expected time until an arrival is $1/\lambda$. For your second question, you have ...


2

In special cases, an exact solution is possible. Suppose the $x$ values are $x$ and $2x$. Then you have $u = a^x + b^x$ and $v = a^{2x}+b^{2x}$. Squaring, $u^2 = a^{2x}+2a^xb^x+b^{2x} = v+2a^xb^x $, so $a^xb^x =(u^2-v)/2 $. Since $a^x + b^x =u $, $a^x$ and $b^x$ are the roots of $0 =(z-a^x)(z-b^x) =z^2-z(a^x+b^x)+a^xb^x =z^2-uz+(u^2-v)/2 $. Setting $d = ...


1

Assuming that you have $n$ data points $(x_i,y_i)$, the fit of the model $$y=a^x+b^x$$ (which is nonlinear) requires reasonable estimates of parameters $a$ and $b$ to start the nonlinear regression. Suppose that the data are not in too much error. You can probably find among the data points two of them $(x_1,y_1)$, $(x_2,y_2)$ such that $x_2\approx 2x_1$. ...


1

So you need to find when its decreasing the fastest. In other words, you need to find when the derivative is negative and has the largest negative value. (Decreasing means its derivative is negative) So first perform the derivative to get a new function: the rate of change. Now we need to find when the rate of change has the largest negative value. ...


6

I don't think this has anything to do with the complex logarithm function. If the series $\sum a_i$ converges, then the left-hand side will exist (because $\exp$ is continuous): $\exp (\sum \limits _{i=1} ^\infty a_i) = \exp (\lim \limits _{N \to \infty} \sum \limits _{i=1} ^N a_i) = \lim \limits _{N \to \infty} \exp (\sum \limits _{i=1} ^N a_i) = \lim ...


5

Another way to look at it (that probably isn't simpler) : Let $u=x^2$, then the integral becomes: $$\frac{1}{2} \int_0^\infty u^{3/2}e^{-\alpha u}du$$ which is the Laplace transform of $u^{3/2}$, times ${1}/{2}$. We know that the Laplace transform of $u^n$ is $n!*\alpha^{-n-1}$ or generally $\Gamma(n+1)*\alpha^{-n-1}$. Here, $n=3/2$ so $$\Gamma(3/2 ...


3

Here, you'll want to integrate by part to find the Gauss integral : $$I = \int_0^{\infty} x^4 e^{-\alpha x^2} dx = \int_0^{\infty} x^3 \times x e^{-\alpha x^2} dx$$ Now, you let $u = x^3$ and $v' = x e^{-\alpha x^2} $, and this gives you $$I = \left [ -x^3 \frac{e^{-\alpha x^2}}{2\alpha} \right]_0^{\infty}+\int_0^{\infty} \frac{3}{2\alpha} x^2 e^{-\alpha ...


9

One way is to use 'Feynman's trick', and differentiate under the integral. Note $$\int_0^\infty x^4e^{-\alpha x^2}=\int_0^\infty \frac{\partial^2}{\partial \alpha^2}e^{-\alpha x^2}=\frac{\partial^2}{\partial \alpha^2}\int_0^\infty e^{-\alpha x^2}$$ Calculate the inner definite integral first, then differentiate it twice with respect to $\alpha$.


1

Get rid of the $x^4$ by integrating by parts twice. Then look for one of the many ways (some probably explained on this site) to evaluate the integral over a Gaussian. My favorite is squaring it and transforming the resulting double integral to polar coordinates – which, by the way, you can also do directly with your integral without first integrating ...


0

if you are looking for any analytical solution it is very difficult and may consume too much of time, 1)you can try solving this by any of the numerical method by taking any approximate initial value. 2) second thing you can draw a graph of the function in some commercial package and try to find out the solution of the problem. hope that is useful


4

Hint: Put $-\alpha x^2=u \Rightarrow -2\alpha x \mathrm dx=\mathrm du$,then $x\mathrm dx=\frac {-1}{2\alpha}\mathrm du$. you will have, $$\displaystyle\int xe^{-\alpha x^2}\,dx=\frac {-1}{2\alpha}\displaystyle\int e^u \mathrm du$$


0

For $x > 0$, consider the mean value theorem on the interval $[0,x]$. Then $$e^x - e^0 \geq \inf_{(0,x)} e^c \cdot (x - 0) = x,$$ implying $e^x \geq 1+x$. For $x < 0$, apply MTV on $[x,0]$: $$e^0 - e^x \leq \sup_{(x,0)} e^c \cdot (0 - x) = -x,$$ giving us $-e^x \leq -x - 1$, or $e^x \geq x + 1$. Then check $x = 0$ and the equality is proven.


1

Are you familiar with integrals of the form $\int f'(x).g(f(x)) \text{d}x$, where $f$ and $g$ are ''well-behaved'' functions? If yes, try to see how you can apply this idea when $g(x)=e^x$ and $f(x)=\alpha x^2$, where $\alpha$ is a constant of course. If not, find out how to use this method because it is an essential tool. Actually, knowing about ...


1

If $\alpha=0$ the integral diverges (it is $\infty$). Now if $\alpha \not =0$, the derivative of $\frac{-1}{2\alpha}e^{-\alpha x^{2}}$ is $xe^{-\alpha x^{2}}$. Now $\int\limits_{0}^{\infty} xe^{-\alpha x^{2}}dx=\frac{1}{2\alpha}$. The answer should be a real number, and not a function of $x$


1

Find the derivative of $x\mapsto e^{-\alpha x^2}$. You'll see that it's easy to find an antiderivative of $x\mapsto xe^{-\alpha x^2}$ and thus to solve this integral, which by the way is not a function of $x$.


1

Hint: I don't think integration by parts is the right strategy (and I don't understand the way you used it). Try performing the substitution $u=x^2$, $du = 2xdx$.


1

This is technique applies the product rule for differentiation in reverse. Essentially you are utilizing the fact that the exponential function is proportional to its derivative to create an expression that is the derivative of a product. This technique is known as the method of integrating factors. The product rule states $$\frac {d (uv)}{dx}=u\frac ...


1

Proceed as follows: Note that, by the distributive law, $e^{\lambda t}(\dfrac{dN}{dt} + \lambda N) = e^{\lambda t} \dfrac{dN}{dt} + \lambda e^{\lambda t} N; \tag{1}$ also note that, by the Leibniz rule for differentiation of a product, $\dfrac{d}{dt}(Ne^{\lambda t}) = e^{\lambda t} \dfrac{dN}{dt} + \lambda e^{\lambda t} N, \tag{2}$ since ...


1

Instead of thinking about going from the first equation to the second, let's go the other way. That means we have to take the derivative in your second equation: $$\frac{d}{dt}\left(Ne^{\lambda t}\right) = \frac{dN}{dt}e^{\lambda t} + N\frac{d}{dt}e^{\lambda t} = e^{\lambda t}\frac{dN}{dt} + \lambda N e^{\lambda t} = e^{\lambda t}\left(\frac{dN}{dt}+\lambda ...


1

A straightforward method (not iterative, no initial guessed values required) is described with a numerical example in the paper : https://fr.scribd.com/doc/14674814/Regressions-et-equations-integrales , in chapter "Double exponential regression - Double power regression", pages 71-73. The method (addapted to the present case) is shown below with in addition ...


0

$$f(x) = a^x + b^x $$ You need two equations to solve for both $a$ and $b$. $$f(x_1)=c=a^{x_1}+b^{x_1}$$ and $$f(x_2)=d=a^{x_2}+b^{x_2}$$ Of course, there are some obvious restrictions on what points can be used to yield unique solutions. $$x_1 \not = x_2 \not =0 $$ Of course, in all likelihood, you'll end up having to solve these equations using ...


0

HINT: it is $$\frac{6^{5/2}\cdot 2^{3/2}}{3^{3/2}}=\frac{2^{5/2}\cdot 3^{5/2}\cdot 2^{3/2}}{3^{3/2}}=2^4\cdot 3$$


0

Notice, $$\left(\frac{a^2}{27}\right)^{1/3}\cdot\left(\frac{64}{a}\right)^{2/3}$$ $$=\left(\frac{(a)^{2/3}}{(27)^{1/3}}\right)\cdot\left(\frac{(64)^{2/3}}{(a)^{2/3}}\right)$$ $$=\left(\frac{1}{3}\right)\cdot\left(4^2\right)=\color{blue}{\frac{16}{3}}$$


0

Assuming that $a\neq 0$; then $$\Big(\frac{64}{a}\Big)^{\frac{2}{3}}=\Big(\frac{4^6}{a^2}\Big)^{\frac{1}{3}};$$ then $$\Big(\frac{a^2}{3^3}\Big)^{\frac{1}{3}}\Big(\frac{16^3}{a^2}\Big)^{\frac{1}{3}}=\frac{16}{3}.$$ So your answer is correct. Clearly if $a=0$ this expression is not define.


0

I suspect there was a typo in the OP (couldn't tell until the formatting got fixed). Surely it should be $A(\phi)=\binom{\ \cos\phi\,\, -\sin\phi}{\sin \phi\,\,\cos\phi}$. Assuming so: Hint: This is a funny case where calculating $A(\phi)^\nu$ is easier than $A(\pi/4)^\nu$, because you can see the pattern easier. In fact the thing to do is this: Calculate ...


0

Note that the differences on the left side of the equals sign doubles between each consecutive pair of equations. It should be fairly clear from this that $b=2$. Indeed by taking the difference of equations 3 and 2 and then dividing by the differences of equations 2 and 1 we get: $${1+1\over-1+2}={ab^1-ab^0\over ab^{0}-ab^{-1}}=b$$ Or in other words, ...


0

First of all, $b^0=1$, whatever $b$ may be, so the second equation gives $a=-1-c$. The third equation gives us that $b=\frac{1-c}{a}$. Now the first equation gives us that $ab^{-1}+c=a\frac{a}{1-c}+c=\frac{a^2}{1-c}+c=\frac{(-1-c)^2}{1-c}+c=-2$. Assuming that $c\neq 1$, we can use this to obtain $c$, and then also $a$ and $b$. To be thorough, we check what ...


0

Indeed, the order is $q$. Write $Q(z):=\sum_{j=0}^qc_jz^j$, where $c_q\neq 0$. We have for $|z| =r$: $$\tag{*} |f(z)|\leqslant \exp\left(\sum_{j=0}^q|c_j|r^j\right),$$ and since $$\lim_{r\to \infty} \frac 1{r^{q+\varepsilon}} \sum_{j=0}^q|c_j|r^j=0,$$ we have $\sum_{j=0}^q|c_j|r^j\lt r^{q+\varepsilon} $ for $r$ large enough, hence by (*), it follows that ...


2

For your first question, you must know by now that $e^{i\pi}=-1$, if you do not, then note that it follows from Euler's identity, since $$ e^{i\pi}=\cos(\pi)+i\sin(\pi) = -1+i0=-1 \tag{1} $$ Hence \begin{align} e^{\pi(1-i)}-e^{-\pi(1-i)} & = e^{\pi}e^{-\pi i}-e^{-\pi}e^{\pi i} \\ & = \frac{e^{\pi}}{e^{\pi i}}-e^{-\pi}e^{\pi i} \\ & ...


1

Recall that, for any $w, z \in \Bbb C$, $e^{w + z} = e^w e^z \tag{1}$ and of course the Euler formula $e^{i\theta} = \cos \theta + i \sin \theta; \tag{2}$ then via (1), $e^{\pi(1-i)}-e^{-\pi(1-i)} = e^{\pi - i\pi} - e^{i\pi - \pi} = e^\pi e^{-i\pi} - e^{-\pi}e^{i\pi} = e^{-\pi} - e^\pi, \tag{3}$ since, by (2), $e^{i\pi} = \cos \pi + i \sin \pi = -1; ...


2

Let $m=\lceil -\mathrm ep\rceil$. Then the ratio of magnitudes of successive terms beyond $m$ is less than $1/\mathrm e$. Now choose $n$ such that $$ \frac{|p|^m}{m!}\frac1{1-1/\mathrm e}\mathrm e^{-(n-m)}\le\frac12\mathrm e^p\;. $$ Since the $m$-th term has magnitude $|p|^m/m!$ and the magnitudes of the remaining terms are bounded by a geometric sequence ...


1

For the most general case, the root of equation $$f(x)=a^x+b^x-c$$ does not show any closed form and numerical methods should be used. The root is bounded by the roots of $2a^x=c$ and $2b^x=c$ which are simple and then we have a range which contains the solution we look for. So, a root finder method, such as Newton method could start,a the mid point, using ...


3

We can show that the only $nice$ solutions to $$f(x^y)=f(x)^{f(y)}\qquad(1)$$ are the identity function, and the constant functions $1$ and $-1$. There are also a few $ugly$ solutions. To avoid problems with $0^0$, let's first find all functions $f$ such that $f:(0,+\infty)\to(-\infty,0)\cup(0,+\infty)$. (Note that because the question is about nonnegative ...


0

Let $f(t)=e^t$, as defined by the power series. Then by differentiating term by term, we find that $f'(t)=t$. Now let $y$ be fixed, and let $$g_y(x)=\frac{f(x+y)}{f(x)}.$$ Differentiate. We get $g_y'(x)=0$, so $g_y$ is a constant function. Now let $x=0$. We find that $g_y(x)=e^y$, and the result follows.


5

Let $F(a)$ be defined as $$F(a)=\int_0^{\pi}e^{i(x+a\cos x)}\sin x\,dx \tag 1$$ Now, it is easy to show that the real part of $F$ is identically $0$. To do this, we write the real part of $(1)$ as $$\text{Re}\lbrace F(a)\rbrace=\int_0^{\pi/2}\cos(x+a\cos x)\sin x\,dx+\int_{\pi/2}^{\pi}\cos(x+a\cos x)\sin x\,dx \tag 2$$ Then, enforcing the substitution ...


1

In the same spirit as Josh Broadhurst's answer, you should arrive to $$\int_{0}^{\pi}e^{i(x+a\cos(x))} \sin(x)\,dx=\frac{\pi a J_1(a)-2 a \cos (a)+ 2 \sin (a)}{a^2}\,i$$ I do not find any way to compute the antiderivative itself.


2

If you leave out the (what I assume is a) constant $a$ in your original problem and type it into WolframAlpha, it tells you that the solution involves the Bessel function of the first kind ($J_1$) and gives the following answer $$ (\pi J_1(1)-2 \cos(1)+2 \sin(1))i \approx 1.9848i$$ If you set $a=2$ as the input to WolframAlpha this gives $$ \frac{1}{2} ...


0

Let $\lambda$ be the interarrival rate and $\mu$ the service rate (so the mean interarrival and service times are $\frac1\lambda$ and $\frac1\mu$, respectively). We can model the $M/M/1$ queue as a continuous-time Markov chain $\{X(t) : t\geqslant 0\}$, where $X(t)$ is the number of entities in the system at time $t$, with state space $\mathbb N\cup\{0\}$ ...


0

The exponential distribution with mean $1$ also has standard deviation $1$. If you collect statistics for $1000$ time units, you should get a sample mean of about $1$ with a standard error of about $0.032$. When you say your values are not so close to the mean, how far off are they? By the way, you should be aware that if both the inter-arrival and the ...


2

Let us consider the antideivative firt $$I=\int\Big[e^{-i\omega (n-1)t}-e^{-i\omega (n+1)t}\Big]dt$$ and integrate each term; then $$I=\frac{i e^{-i (n-1) t \omega }}{(n-1) \omega }-\frac{i e^{-i (n+1) t \omega }}{(n+1) \omega }$$ When you compute the integral, the value at $t=0$ is just $$\frac{i}{(n-1) \omega }-\frac{i}{(n+1) \omega }$$ and the value at ...


0

Hint: $$\int_{0}^{\pi/\omega_0}\left(e^{-j\omega_0(n-1)t}-e^{-j\omega_0(n+1)t}\right)dt\\=-j\left(\frac{1-e^{-j\pi(n-1)}}{n-1}-\frac{1-e^{-j\pi(n+1)}}{n+1}\right)\\=-j\left(1+e^{-jn\pi}\right)\left(\frac{1}{n-1}-\frac{1}{n+1}\right)$$ where I have used the Euler's formula $e^{\pm j \pi}=-1$. I hope this clears things.


0

As a side note, I think it is worth mentioning that by the Lagrange inversion formula we have: $$ \forall x\in\left(-\frac{1}{e},\frac{1}{e}\right)\qquad W(x) = \sum_{n\geq 1}\frac{(-n)^{n-1}}{n!} x^n $$ with $W$ being the Lambert $W$ function, i.e. the inverse function of $x e^x$ in a neighbourhood of the origin. Trivial manipulations then leads to: $$ ...


1

Let $x=1/e$, and let $a_k=\frac{k^k}{k!}(1/e)^k$. We want to show that $\sum a_k$ diverges. Let $\frac{k!}{\sqrt{2\pi k}(k/e)^k}=c_k$. We are told that $\lim_{k\to\infty} c_k=1$. With some algebra, we can see that $$a_k=\frac{k^k}{k!}(1/e)^k=\frac{1}{c_k}\cdot \frac{1}{\sqrt{2\pi k}}.$$ Note that the limit of $\frac{a_k}{1/\sqrt{k}}$ as $k\to\infty$ is a ...


3

Just expanding Daniel Fischer's comment, given that $f(x)=e^x$ is a positive and continuous function for which $f(x+y)=f(x)\,f(y)$, we have: $$f\left(\frac{x+y}{2}\right) = \sqrt{f(x)\cdot f(y)}\color{red}{\leq} \frac{f(x)+f(y)}{2}\tag{1}$$ where $\color{red}{\leq}$ follows from the AM-GM inequality. But $(1)$ just gives the midpoint-convexity of $f(x)$, ...


0

We can use the definition of convexity itself. A space S is convex if for any $u,v \in S$ $$\lambda u + (1- \lambda)v \in S \ \forall \ \lambda \in [0,1] $$ Intuitively this means if two points are in the space, then every point between them is in the space. (I can explain further if requested in comments). From here note that the space we want to ...


1

Let's check condition $$ f\Big(\frac{x+y}{2}\Big) \le \frac12 (f(x) + f(y)) $$ for $f(x)=e^x$. So, $$ e^{(x+y)/2} = e^{x/2} e^{y/2} \le \frac12 (e^x + e^y). $$ Denote $a=e^{x/2}$, $b=e^{y/2}$; then, $$ ab \le \frac12 (a^2 + b^2) \Longrightarrow 2ab \le a^2 + b^2 \Longrightarrow (a-b)^2\ge 0 $$ And... it's true!


1

Hint Here is a first step. Direct definition of convexity between points $x,y$ is $$f(hx + (1-h)y) \le hf(x) + (1-h) f(y)$$ Let $y=0$ and use Taylor expansion $e^x = \sum_{k=0}^\infty x^n/n!$ to note that $$ e^{hx} = 1 + (hx) + h^2 x^2 + h^3 x^3 + \ldots $$ and $$ (1-h) + he^x = 1+h + (h + hx + hx^2 + hx^3 + \ldots) = 1 + hx + hx^2 + hx^3 + ...


3

For first, get rid of the extra parameter by setting $c=\frac{b}{a^2}$: $$I= \int_{\mathbb{R}}e^{ibx^2}\frac{\sin(ax)}{x}\,dx = \int_{\mathbb{R}}e^{icx^2}\frac{\sin x}{x}\,dx=\text{Im PV}\int_{\mathbb{R}}e^{icx^2+ix}\frac{dx}{x} $$ then translate the $x$ variable in order to get: $$ I = ...



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