New answers tagged

0

We want to find the inverse Laplace transform using convolution for: $$\dfrac{1}{\left(\left(s+\dfrac{1}{2}\right)^2+\dfrac{1}{4}\right)*s}$$ Using the linearity of the inverse Laplace transform and the convolution theorem, we have: $$\mathscr{L}^{-1}\left( \dfrac{1}{\left(\left(s+\dfrac{1}{2}\right)^2+\dfrac{1}{4}\right)*s}\right) = ...


0

Convolution in time domain corresponds to multiplication in the s-domain. You can decompose the laplace transform as $$ \frac{1}{((s+1/2)^2+1/4)s} = \frac{1}{((s+1/2)^2+1/4)}\cdot\frac{1}{s} $$ The second term corresponds to the laplace transform of the step function $u(t)$. The first term is $$ \frac{1}{((s+1/2)^2+1/4)} $$ The inverse laplace transform of ...


5

If $a\gt b$ then this implies:$$a^b\gt b^b\tag{1}$$ Also, if $a\gt b$ then this implies:$$a^a\gt a^b\tag{2}$$ Now, making use of result (1) in result (2) yields:$$a^a\gt a^b\gt b^b$$


2

You're asking to prove that the map $f(x):=x^x=\exp(x\log x)$ is strictly increasing. But since $x\mapsto x$ and $x\mapsto\log x$ are both strictly increasing, their product is such. Then the exponential is strictly increasing too, so $f$ is the composition of strictly increasing maps, thus itself is strictly increasing too.


2

Note that the derivative of $x^x$ is $(\ln(x)+1)x^x>0$(see here) if $x>e^{-1}$, implying $x^x$ is a strictly increasing function if $x>e^{-1}$. This implies your claim is true not only when $a,b>1$ but also $a,b>e^{-1}$


2

Try this sequence of inequalities: $$b^b < b^a < a^a $$


1

However, since there exists an element $cx \in \{xn\}$, and $cx>c$, it is a contradiction. This is false, and I cannot figure out how you contrived this claim. $\def\nn{\mathbb{N}}$ Instead, since $c = \sup(\{x_n:n\in\nn\}) \ge x > 1$, let $y \in \{x_n:n\in\nn\}$ such that $y>\frac{c}{x}$ since $\frac{c}{x} < c$. Then $xy > c$, which is ...


1

Here is another approach. Since $x >1$, we have $x=1+t$ for some $t>0$. Then $x^n = (1+t)^n = \sum_k \binom{n}{k} t^k \ge 1 + n t$. Since $1+nt \to \infty$, we have the desired result.


0

If $x \geq 0 \Rightarrow 3^x \geq 2^x \Rightarrow 3^x + \dfrac{1}{2} \geq 2^x + \dfrac{1}{2} > 2^x$, thus its true for $x \geq 0$. If $x < 0$, put $y = - x \Rightarrow y > 0$. So you prove: $3^{-y} + \dfrac{1}{2} > 2^{-y}\iff \dfrac{1}{3^y} + \dfrac{1}{2} > \dfrac{1}{2^y}\iff 2^{y+1} + 6^y > 2\cdot 3^y\iff 2^{y+1} > ...


1

One idea $3^{x} + 0.5 > 2^{x} \Leftrightarrow 3^{x} - 2^{x} > -0.5$. Let $g(x) = 3^{x} - 2^{x}$. What is the min of $g$?


3

To continue with your line of reasoning (which so far is correct), you need to solve $e^{2iz} + 4 k \pi e^{iz} - 1 = 0$. With $x = e^{iz}$, this becomes the quadratic equation $x^2 + 4kx - 1 = 0$. The discriminant is $\Delta = (4k\pi)^2 + 4 = 4 (4k^2\pi^2 + 1)$, which is always positive ($k$ is real). The two solutions are therefore $x = -2k\pi + ...


-1

$$e^{\sin(z)}=1\Longleftrightarrow$$ $$\ln\left(e^{\sin(z)}\right)=\ln(1)\Longleftrightarrow$$ $$\sin(z)\ln\left(e\right)=\ln(1)\Longleftrightarrow$$ $$\sin(z)=2i\pi n_1\Longleftrightarrow$$ $$\arcsin\left(\sin(z)\right)=\arcsin\left(2i\pi n_1\right)\Longleftrightarrow$$ $$z=\begin{cases} \pi-i\text{arcsinh}(2\pi n_1)+2\pi n_2\\ i\text{arcsinh}(2\pi ...


3

$$\lim_{x\to 0} (1+x)$$ is simply equal to $1$, so the equation you wrote is equal to the equation $$\lim_{x\to0} \exp(x) = 1$$ which is most certainly not enough to characterise the exponential function, for example $\cos(x)$. As for what you proved in points (a) and (b), I don't even know what you were intending to prove so I cannot comment on that.


0

When $x=0$ $f(x)=1$ and it decreases from there to zero in the limit. So $0$ and $1$ are bounds.


1

We have $$ |f(x)|=\left|\exp(-\frac{1}{2}x^2)\right|<1, \quad x>0, $$ and $$ \lim_{x \to +\infty}|f(x)|=\lim_{x \to +\infty}\left|\exp(-\frac{1}{2}x^2)\right|=0. $$


0

Your last table is incorrect, and as others have mentioned, tetration would be the next operation. Let's try to figure out what these kinds of operations are and how they work, to begin with. We will be considering $x+2$, $x\cdot2$, and $x^2$. What is $x+2$? In simple terms, it means $x+1+1$ where we have $2$ ones. What is $x\cdot2$? In terms of ...


2

Note that your table can be written as $$ \begin{array}{c|l} \text{Input} & \text{Output} \\ \hline 1 & 2^{2^{1-1}}\\ 2 & 2^{2^{2-1}}\\ 3 & 2^{2^{3-1}}\\ 4 & 2^{2^{4-1}} \end{array} $$ Tetration is the operation you could say that succeeds exponentiation. As an example, notationally, if we want to represent "a to the a to the a to the ...


1

The map satisfies the recurrence relation $x_{n+1}={x_n}^2$ with base case $x_1=2$. By backtracking or induction one can show that $x_n=2^{2^{n-1}}$ for all $n\in\mathbb{N}$.


1

Each of your tables is found by taking 2 to the power of the previous table (possibly with a shift by 1). As the formula for your second table is $2^x$, the formula for your third table is $2^{2^{x-1}}$ (for $x \geq 1$).


0

When we rotate the part from $y=0$ to $y=1$ about the $x$-axis, we get a cylinder. The volume of the cylinder can be found using cylindrical shells, but can be found more simply. As to the part between $y=1$ and $y=e$, the radius of the cylindrical shell at height $y$ is $y$. The "height" of the cylinder is $1-x$. In terms of $y$ it is $1-\ln y$. So the ...


4

$$2^x+5^x=3^x+4^x \iff5^x-4^x=3^x-2^x$$ Lagrange's theorem applies (intermediate value theorem) functions: $$f:[4, 5]\to R,\ g:[2, 3]\to R,\ f(u)= u^x,\ g(v)=v^x$$ Exist $c\in [4, 5]$ and exist $ d\in [2, 3] $ so $5^x-4^x = xc^{x-1}$ and $3^x-2^x=xd^{x-1}$. The equation is written as equivalent: $$xc^{x-1}= xd^{x-1}.$$ It follows that equation solutions ...


0

Put $f(x)=2^x+5^x-(3^x+4^x)$ and $g(n)=\frac{d^n}{dx^n} f(0)=(\ln2)^n+(\ln5)^n-(\ln3)^n-(\ln4)^n$ . Now $g(n)$ is positive for $n>2$ and negative for $n=1,2$ and then because Taylor series of $f$ have only two negative terms , then $f$ have at most $2$ roots . It remains that we show $g(n)$ is positive for $n>2$ and negative for $n=1,2$ : ...


1

In that step when you first took the derivative of $y$ you should get $\frac{dy}{dx}= \frac{y}{(y+1)x}$, right?


1

I don't understand what you did, I (and Wolfram Alpha) would do this: $${d\over dx}\left(ye^y\right) = {d\over dx}\left(x\right)$$ $$\iff {d\over dx}\left(ye^y\right) = 1$$ $$\iff y'e^{y} + y'e^{y}y = 1$$ $$\iff y'\left(e^{y} + e^{y}y\right) = 1$$ $$\iff y'(x) = {1 \over e^{y} + e^{y}y}$$ $$\iff y'(x) = {1 \over e^{y}(1 + y)}$$


0

Note that $$|\cot(\pi z)|^2={\cosh(2\pi y)+\cos(2\pi x)\over \cosh(2\pi y)-\cos(2\pi x)}\ .$$ It follows that $$|\cot(\pi z(t))|^2={1+q(t)\over 1-q(t)}$$ with $$q(t)={\cos\bigl((2n+1)\pi\cos t\bigr)\over \cosh\bigl((2n+1)\pi\sin t\bigr)}\ .$$ We have to determine the maximum of the function $q$. This $q$ starts off in the negative and then has a first peak ...


0

$$E_n = (1-a)^n E_0 +a \sum_{i=1}^n (1-a)^{n-i}S_i$$


0

As Lubos Motl said it is doubtful in the extreme that you will be able to solve this equation analytically. Suppose that the numbers $A_i$ are positive, then continuity implies that there is at least one solution. Simply evaluate the right hand side and the left hand side separately at both $x=0$ and infinity. Moreover, your right hand side is rapidly ...


0

You have just stumbled upon the Exponential integral (Ei) function, which is a non-elementary function. Without getting into too much detail, that means that the function cannot be fully simplified, and the best we can do is express it as an series, as you did. I believe however that the following expansion is more widely used, due to simplicity: $$ \int ...


1

If you actually want to use a substitution to check your answer, then if $y=f^{-1}(x)$ and $f(x)=3e^{2x}+4$ then $$y=\dfrac12\log_e\left(\dfrac{x-4}{3}\right)$$ which will give $$\frac{1}{6e^{2y}} = \frac{1}{6e^{\log_e\left(\frac{x-4}{3}\right)}}= \frac{1}{6\left(\frac{x-4}{3}\right)}= \frac{1}{2x-8}$$ but this is not strictly necessary here


1

$3e^{2y} + 4 = x \implies 3e^{2y} = x - 4 \implies 6e^{2y} = 2x - 8 $ Therefore : $\frac{1}{2x - 8 }=\frac{1}{6e^{2y}}$


1

From the second line, $x= 3e^{2y} + 4$ $\Rightarrow 3e^{2y} = x-4$ $\Rightarrow 6e^{2y} = 2x-8$.


1

As mfl commented, a linear fit $y=a+bx$ looks more appropriate than $y=ae^{bx}$. For the linear model, as already given by mfl, we should get $$y=76.6228+0.0944818 x\qquad SSQ=7363.8\qquad R^2=0.995485$$ Because of the curvature for low values of $x$, a better fit could be obtained using $y=a+bx^c$. For such a case $$y=37.6705+0.57678 x^{0.804092}\qquad ...


0

There are a few problems with addition. $$W(x+a)=?$$ $$\log(x+a)=?$$ The inverse functions of exponential related functions don't like addition on the inside. Addition on the outside is perfectly fine though. $$\log(x)+a=\log(e^ax)$$ Except even then, the Lambert W Function won't simplify if you have addition on the outside... most of the time. And ...


0

Consider the function $$f(x)=e^x-x^e$$ The derivative is $$e(e^{x-1}-x^{e-1})$$ It is clear that $$f'(1)=f'(e)=0$$ and it is verify that $x=1$ goes for a maximun and $x=e$ goes for a minimun. Furthermore $f(1)=e-1>0$ and $f(e)=e^e-e^e=0$. Taking in account that $f(0)=1$ and the domain of $f$ is $x\ge 0$ it follows the conclusion $$e^x\ge x^e$$


3

Note that the ceiling changes when its argument passes through an integer. This happens when the uniform variable passes through $1/n$ for some $n$. The ceiling will be $n$ when the uniform variable is between $1/n$ and $1/(n-1)$. The length of this interval is $\frac{1}{n-1}-\frac{1}{n}=\frac{1}{n(n-1)}$, and the probability of a uniform random variable on ...


0

The graph $\gamma$ of $\log$ is concave, hence $\gamma$ stays below the tangent to $\gamma$ at $(e,\log e)$: $$\log x\leq \log e+{1\over e}(x-e)\qquad(x>0)\ .$$ This simplifies to $\log x\leq{x\over e}$, and exponentiation gives $$x^e\leq e^x\qquad(x>0)\ .$$ The case $x=0$ is obvious.


2

If you allow, you can solve for $x$ through Lagrange Inversion Theorem. $$y=x^{x+1}$$ Invert it... I now have time to work the problem out. $$f(x)=x^{x+1}$$ $$f(1)=1,f'(1)\ne0$$ $$f^{-1}(x)=1+\sum_{n=1}^{\infty}\lim_{w\to1}\frac{(x-1)^n}{n!}\frac{d^{n-1}}{dw^{n-1}}\left(\frac{w-1}{w^{w+1}-1}\right)^n$$ This is not reducible as far as I know, so you ...


0

If $x=0$, then given inequality holds. Suppose that $x > 0$ and define a function $f(x)=x - e\ln x$. The derivative of $f(x)$ is $f'(x)=1-\frac{e}{x}$. Thus $f(x)$ decreases when $x<e$ and increases when $x > e$. $f(e)=e-e\ln e =0$ and $f(e)$ is minimum of $f(x)$, so $f(x)\ge 0$ for all $x > 0$. Thus, $x\ln e \ge e\ln x$ for all $x > 0$ and so ...


2

Hint Since $x \mapsto \log x$ is a strictly increasing function, applying it to both sides of the inequality gives that it is equivalent to $$x \geq e \log x .$$ There are at least two options here: (1) Rearrange the inequality as $\frac{\log x}{x} \leq \frac{1}{e}$, and analyze the expression on the l.h.s. of this equality. In particular, show that its ...


0

Hint: change the lower bound to 1/x and the upper bound to x, and write the integral as a function of x. Plug e^x into this function and differentiate using the chain rule.


0

Reading Courant-Robbins I was surprised to see that proving the irrationality of $$ e = \sum_{n=0}^{\infty} \frac{1}{n!} = 2 + \frac{1}{2} + \frac{1}{2 \cdot 3} + \frac{1}{2 \cdot 3 \cdot 4} + \dots $$ isn't harder than proving the irrationality of $\sqrt{2}$, just a bit more laborious (I fear that with $\pi$ things go in a very different way...). I found ...


1

In general, let $X_1,\dotsc, X_n$ be iid eponential distribution with mean $1/\lambda$. Then the distribution of the minimum $M$ is $$P(M\leq m) = 1-P(M>m) = 1-(e^{-\lambda m})^{n} = 1-e^{-\lambda nm}.$$ Notice that this shows that $M$ follows an exponential distribution with mean $\frac{1}{\lambda n}$. In our case, $\lambda = 1/22.5, n =4$, and so $$E[M] ...


1

Yes your solution is correct $$2^{-k} = \frac 1 n \implies n = 2^k \implies \log_{2} n = k$$


-1

$$k=\frac{\log\left(n\right)+2i\pi c_1}{\log\left(2\right)}$$ for any integer $c_1$, where $\log$ is the natural logarithm and $i$ is the imaginary unit.


6

Note that if $x\to 0^-$ and $t=-1/x$ then $x=-1/t$ and $t\to+\infty$, hence $$\lim_{x\to 0^-}\frac{e^{1/x}}{x^2} = \lim_{t\to\infty}\frac{e^{-t}}{(-1/t)^2} = \lim_{t\to\infty}\frac{t^2}{e^t}.$$ And the last limit should be a proud member of the collection of limits you know.


0

HINT: $$\lim_{x\to0^-}\space\frac{e^{\frac{1}{x}}}{x^2}=\lim_{u\to-\infty}\space\frac{e^u}{x^{\frac{1}{u^2}}}=$$ $$\frac{\lim_{u\to-\infty}e^u}{\lim_{u\to-\infty}x^{\frac{1}{u^2}}}=\frac{\exp(\lim_{u\to-\infty}u)}{x^{\lim_{u\to-\infty}{\frac{1}{u^2}}}}$$


1

You're correct: $$\begin{align} 4\sinh{2\log{2}} - \cosh{\log{2}} &= \cosh{(\log{2})} \left( 8\sinh{(\log{2})}-1 \right) \\ &= \frac{1}{2}(2+1/2)(4 \cdot 2 - 4 \cdot 1/2-1) \\ &= \frac{5}{4}(8-2-1) \\ &= \frac{25}{4} \end{align}$$


2

The answer should be $6.25$. \begin{align} & 4 \sinh (2 \ln 2) - \cosh(\ln2 ) \\ =& 2 \left(e^{2\ln2}-e^{-2\ln2} \right) - \frac{e^{\ln2}+e^{-\ln2}}{2}\\ =& 2(4-0.25)-\frac{2+0.5}{2}\\ =& 7.5-1.25 = 6.25. \end{align}


1

The order in which you do the operations is vital here. Unlike addition and multiplication, exponentiation is not associative: $$a^{(b^{\large c})} \neq \left(a^b\right)^c$$ for "most" values of $a$, $b$, and $c$. It is true that $2^{1/n} = 2^{n^{\large -1}}$, but that's because $2^{n^{\large -1}}$ means $2^{(n^{\large -1})}$. As you saw for yourself, ...


0

The exponents cannot be multiplied, as the first exponent is raised to the second. You are right that $(x^a)^b=x^{ab}$, but you do not have $x^{a^b}=x^{ab}$, as $a^b \neq ab$.



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