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1

Just as clarification? I think the 'equations' you wrote down are programming instructions, where you modifie the 'value' variable and store it again in 'value'? If you want the 'amplification' as a factor, you could do it like this (pseudocode): for i = 0 upto 25 value[i] = value[i] * 2^(i/25) end This way you multiply the first element by the ...


0

Yes, if we regard $x \mapsto \lceil e^x \rceil$ as a function with codomain its image, it is bijective, and this doesn't depend on whether we include $0$ in $\mathbb{N}$ or not. For $x \in [0, \infty)$, $\frac{d}{dx} e^x = e^x \geq 1$, and so $$e^{x + 1} = e^x + \int_x^{x + 1} e^x \, dt \geq e^x + \int_x^{x + 1} \,dt = e^x + 1.$$ In particular, $$\lceil e^{x ...


-1

While that function is injective, it is not surjective, so it can't has an inverse in the common sense. What it does have is a right inverse, that is, a function $g$ such that $f \circ g = id$, and you can simply define it as $g(x) = y$ such that $f(y) =x$, which is guaranteed existing a unique. Finding a closed form for that is another thing, and in ...


0

$7^x = e^{x \ln 7}$. Now you want $$e^{x \ln 7} = -49 = e^{\Re (x \ln 7)} e^{i \Im (x \ln 7)} \equiv e^R e^{i \theta}$$ where $R \equiv \Re(x) \ln 7$ and $\theta \equiv \Im(x) \ln 7$. Since the final form of the product can be expressed as the product of a magnitude and direction in the complex plane, and we know the magnitude is 49 and direction is left, ...


4

Let $(x, e^{bx})$ be the point of contact of $y = e^{bx}$ and $y = 10x$. $$10x = e^{bx} \qquad (1)$$ Similarly, using your method, we get $$10 = be^{bx} \qquad (2)$$ Now you have two equations and two unknowns. (Hint, divide two equations)


2

Let $a\in G$. We need to show that $h$ is differentiable at $a$. As $G$ is open there exists a disc $D(a,r)\subset G$. Now as $f$ does not vanish in $G$, and hence in $D(a,r)$, we define the function $$ F(z)=\int_{[a,z]}\frac{f'}{f}. $$ Then $F$ is holomorphic in $D(a,r)$ and the product $\exp(-F(z))f(z)$ is constant in $D(a,r)$, as $$ ...


2

Hint: try with branch of logarithmic function.


2

Note that $t\mathrm{e}^t$ is asymptotically larger than the other solutions, and tends to $\infty$ as $t$ tends to $\infty$. Hence $A_2=0$. Similarly $A_1=0$, as $\mathrm{e}^t\to\infty$, while the rest of the terms tend to 0. Hence $$ x(t)=(at+b)\mathrm{e}^{-t}. $$ The constants $a,b$ can be found from the initial data.


5

As $r$ is an eigenvalue of multiplicity 2 of the $2\times 2$ matrix $A$, then $A$ does not have any other eigenvalues, and its characteristic polynomial should be $$ p(x)=(x-r)^2, $$ and by Cayley-Hamilton Theorem we have that $(A-rI)^2=0$. Next observe that $$ \mathrm{e}^{tA}=\mathrm{e}^{rt}\mathrm{e}^{t{(A-rI)}}=\mathrm{e}^{rt}\sum_{n=0}^\infty ...


4

They key observation is to write the relation as $$e^{(A-rI)t}=I+(A-rI)t. $$ This is clearly true once you notice that $(A-rI)^2=0$. You can prove this using the Jordan form as Slade mentioned: a $2\times2$ matrix with both eigenvalues equal to zero has its square equal to zero.


4

This is much simpler if you observe that what you're trying to prove is invariant under conjugation/similarity. That is, it suffices to show this for some matrix similar to $A$, not necessarily $A$ itself. Note that if $A$ has a repeated eigenvalue of $r$, then it must be similar either to $\begin{pmatrix}r & 0 \\ 0 & r\end{pmatrix}$ or to ...


11

It is not currently known whether or not $\pi+e$ is rational. It is also not known whether or not $\pi e$ is rational. However, we can say that $\pi+e$ and $\pi e$ cannot both be rational. For suppose to the contrary that they are both rational. Then $(\pi+e)^2-4\pi e$ is rational. But this is $(\pi-e)^2$, so $\pi-e$ is algebraic. But then adding and ...


3

Since $$ \log(1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+\mathcal{O}(x^5), \quad \lvert x\rvert<1, $$ we have that $$ \log\left(1+\frac{2}{n}\right)=\frac{2}{n}-\frac{2}{n^2}+\frac{8}{3n^3}-\frac{4}{n^4}+{\mathcal O}(n^{-5}). $$ In particular, using the Taylor expansion theorem, we obtain that the ${\mathcal O}(n^{-5})$ term is of the form ...


0

For purely algebraic proof: Suppose $m\ge n.$ Note that should there exist a positive integer $k$ such that $$m=n^k.$$ Then $$m^n=n^{kn}=n^m.$$ That is $$m=kn.$$ $$kn=n^k$$ $$k=n^{k-1}.$$ If $k=1,$ it gives $n=m.$ If $k=2,$ it gives $n=2, m=4$ If $k\ge3$ and $n\ge2,$ then $n^{k-1}>k.$ (By mathematical induction we can show that $2^{k-1}>k$ for ...


1

HINT : $$m^n=n^m\iff n\ln m=m\ln n\iff \frac{\ln m}{m}=\frac{\ln n}{n}.$$ Then, consider the graph of $\frac{\ln x}{x}$.


1

As already answered by Eoin, there is no solution in the real domain. However, this equation has a solution which can be expressed using the Lambert function; this solution is $$x=1-\frac{W\left(-\frac{3 \log (3)}{2}\right)}{\log (3)} \approx 0.968921-1.44932 i$$ In a more general manner, the equation $$a x=b^x+c$$ has a solution which is ...


0

Let a be the difference error between $e^n$ and maclaurins series the value of the expression now becomes $1-\frac a{e^n}$ and now as n tends to $\infty $ the difference error a tends to zero and $e^n$ tends to $\infty$ and therefore the limit tends to 1 which implies that they both grow at same rate.


2

Observe that $3^x$ is always positive the right side of the equality is strictly greater than 2. The left side takes this form when $x>1$. So we need only look for a solution in this area. Now we can approximate the solution for this as follows. Try $x=1$. Then $2(1) =2$ and $3^1+2 =5$. Since the exponential increases faster than the left, linear side, ...


1

Hint: First find the inverse of the function $t/(1+2t)$.


4

Applying the logarithm to both sides gives $$y \log x = x \log y,$$ and rearranging gives $$\frac{\log x}{x} = \frac{\log y}{y}$$, so we get a nontrivial solution (i.e., one for which $x \neq y$) for any value assumed twice by $f(x) := \frac{\log x}{x}$. More explicitly, if $x$ and $y$ are distinct values such that $f(x) = f(y)$, then by construction we have ...


-6

the solution is $\left\{\left\{y\to -\frac{x W\left(-\frac{\log (x)}{x}\right)}{\log (x)}\right\}\right\}$


4

Your conjecture is false, take for example $$ f(x) = \exp(x^2/2) \frac{\sin x^4}{1+x^2}. $$


0

I do not think that, in the real domain at least, any of these equations can be solved analytically (even using complex functions). The second one can be solved numerically considering, as you did, the function $$f(x)=3^{x+2} + 2^x - 5$$ By inspection, there is a root between $x=-1$ and $x=0$ (since $f(-1)=-\frac{3}{2}$ and $f(0)=5$). For the computation ...


0

For the second, Alpha doesn't find a symbolic answer. For the first, Alpha again doesn't. I didn't know if the second log is base $10$ or base $e$, but it doesn't find one either way. For equations of this sort, usually you find a solution by inspection or you don't find one at all.


-1

Given $ 2^x $, $ 3^x $, $ 5^x $,... are integers. Now suppose $ x $ is not an integer. Suppose it is rational of the form $ p/q $ where $ gcd(p,q) = 1 $. So $ 2^{p/q} $, $ 3^{p/q} $, ... can't be integer, since base of exponents are all prime.


0

What are you stuck with exactly? Is the wording confusing you? Sometimes I find it helpful to start breaking the question down into really simple chunks. Start from the beginning, X is an r.v. conditioned on A ... so See where Dilip Sarwate added in the |A notation? Just start breaking it down like that. It's been helping me a LOT lately.


0

$1/3$ and $2/6$ are the same number, so if we need to evaluate $(-8)^{1/3}$ and $(-8)^{2/6}$, we must make sure that the result is the same in each case. So yes, you need to reduce a fractional exponent to its lowest terms before applying it to a negative operand. Having said that, the exponential function $x^y$ doesn't seem to be useful for negative $x$ ...


0

As long as $a\ge 0$, the function $a^x$ is well defined and therefore it doesn't matter how you represent x, either as $\frac 1 3$ or $\frac 2 6$, etc. Once a hits negative numbers, you need to be careful because of the inability to take even powered roots of negative numbers, so for instance $(x^2)^\frac 1 2$ is not equal to x, but instead |x|.


1

By the way: Looking at the involved areas under the curve $y={1\over t}$ one verifies easily that $$\int_1^{x\cdot y}{dt\over t}=\int_1^x{dt\over t}+\int_x^{x\cdot y}{dt\over t}=\int_1^x{dt\over t}+\int_1^y{dt\over t}\ .$$ This says that the function $$\ell(x):=\int_1^x{dt\over t}\qquad(x\geq1)$$ satisfies the functional equation of the logarithm; ...


3

Let $f(x)=\int_1^x t^{-1}dt$, for $x>0$. Then, by FTC, $$f'(x)=x^{-1}.$$ Then: $$\left(f({\rm e}^x)\right)'=f'({\rm e}^x){\rm e}^x={\rm e}^{-x}{\rm e}^x=1.$$ Hence: $f({\rm e}^x)=x+ C.$ As $f(1)=0$, we have $C=0$.


11

Since $f(x)=e^x$ is a solution of the ODE $$ f' = f $$ and it is an increasing positive function, given that $g(x)$ is the inverse function of $f(x)$ we have: $$ g(f(x)) = x,$$ hence by differentiating we get: $$ f'(x)\, g'(f(x)) = 1, $$ or: $$ g'(f(x)) = \frac{1}{f'(x)} = \frac{1}{f(x)}, $$ $$ g'(t) = \frac{1}{t}. $$ Since $f(0)=1$ implies $g(1)=0$, the ...


0

The very last step where you find the inverse is incorrect, but you have the right idea. Let me remind that the identity you want to use here reads $a \cdot \ln (x) = \ln(x^a)$ and not $a \cdot \ln(x) = \ln(x)^a$. Notice the difference between the right hand side of the two identities.


0

While it may not lead to a better formal proof than other approaches, I find the following quite intuitive. It also compares nicely to Euler's original proof using the continued fraction representation of$~e$. It is easy to see that every rational number $\alpha\in(0,1)_\Bbb Q$ can be uniquely expressed as $$\alpha = ...


2

You can approach this problem with a simple inequality: if $a>b>0$, then: $$ n(a-b)\,b^{n-1}<a^n-b^n < n(a-b)\,a^{n-1},\tag{1}$$ so, if we choose $a=2^{1/n}$ and $b=1$, from the $RHS$ we get: $$ 2^{1/n}-1 > \frac{1}{n\cdot 2^{\frac{n-1}{n}}}>\frac{1}{2n}\tag{2}$$ so the series diverges by comparison with the harmonic series.


1

$a=e^{\ln a}, a>0$ $$(e^{\ln a})^{ix}=e^{ix \ln a}=\cos(x \ln a)+i\sin(x\ln a)$$


3

Hint:If $a>0$ then $a=e^{\ln a}$


1

The idea is this: we know that $x \log (1 + x^{-1}) = \log (1+x^{-1})^x$. Now exponentiating both sides gives $$(1+x^{-1})^x = \exp \left(1 - \frac{1}{2x} + \frac{1}{3x^2} - \frac{1}{4x^3} + \cdots \right) = e \cdot e^{-1/2x} \cdot e^{1/3x^2} \cdot e^{-1/4x^3} \cdots.$$ Now consider each factor separately: $$\exp \frac{1}{(k+1)(-x)^k} = \sum_{n=0}^\infty ...


0

By definition, for $k=2$ and $|a|<1$, we have $\displaystyle\sum_{n=-\infty}^\infty a^{n^2}=\theta_3(0,a)$, and $\displaystyle\sum_{n=-\infty}^\infty a^{\big(n+\frac12\big)^2}=\theta_2(0,a)$ where $\theta$ is the Jacobi $\theta$ function.


0

Whenever you see factorials popping up mysteriously in integral evaluations, you should always think of the gamma function, and in particular that fact that for any integer $n$ $$n! = \Gamma (n+1) = \int_0^{\infty}x^{n}e^{-x} dx$$ If you make a variable transform $u = \gamma x$ you should be able to massage your integral into something that looks like ...


0

We could try to compute $$L_1=\lim_{n\to \infty} \frac{a^{b^n}}{b^{a^n}}\text{,}$$ but it is harder than computing the limit $$L_2=\lim_{n\to \infty} \log \left(\frac{a^{b^n}}{b^{a^n}} \right) = \lim_{n\to \infty} \left(\log (a^{b^n}) - \log (b^{a^n})\right)\text{.}$$ Since $\log x^y = y\log x$ for all $x,y>0$, it holds that $$L_2=\lim_{n\to \infty} ...


0

The curve you are trying to find an equation for also seems to pass through the point $(-5,15)$. Given three points, it's always possible to find a parabola that passes through them, that is, an equation of the form $y=ax^2+bx+c$. To find the coefficients $a$, $b$, and $c$, you need to plug in the $x$ and $y$ coordinates of the three points, $(5,17)$, ...


1

a) First note that if we have a sum $$f(a)=\sum_{k=1}^N \lambda_k \exp(iax_k)$$ with the $x_k\in \mathbb{R}$ distincts, then this function is zero on $\mathbb{R}$ if and only if all $\lambda_k$ are zero. To see this, use induction on $N$, and if $f(a)=0$ for all $a$, compute the derivative of $g(a)=f(a)\exp(-iax_N)$. b) Now prove your assertion by ...


1

As I said in a comment, I do not think that you could obtain an explicit solution for $$f(t)= e^{\alpha t}\cos(\omega t + \phi)+\frac{\alpha^2}{\omega^2}=0$$ where $\phi = \tan^{-1}\left(\frac{\alpha}{\omega}\right)$. However, if you look for the first solution, you could expand $f(t)$ as a Taylor series around $t=0$ and obtain $$f(t)=\left(\frac{\alpha ...


1

The factor $Ae^{\alpha t}$ is never zero (unless $A=0$), so you are actually looking for the roots of $\cos (\omega t+\phi)$ But $$\cos (\omega t+\phi)=0 \iff \omega t+\phi=\frac{\pi}2+k\pi, k\in\Bbb Z$$


1

If $\alpha, t, \omega, \phi$ are all real numbers and $A\neq 0$ (if $A=0$, then $i(t)=0$ for all $t$), then $$i(t)=0\iff \cos(\omega t +\phi) = 0,$$ since $e^{\alpha t}$ is never $0$. From here on, it should be simple to find the values of $t$.


0

I am trying a differential equation based approach for solution, though I am not sure about it. So it is better someone correct me if I am wrong. Let's say grow rate is: $$\frac{dx}{dt}=kx$$ Therefore: $$x=Ae^{kt}$$ $t=0\:\Longrightarrow\:x=2$ thus we find $A=2$. $t=5\:\Longrightarrow\:x=4$ thus we find $k=\frac{ln2}{5}$ If $t=25$: ...


0

Every five days it doubles 25 days go This means it doubles $\frac{25}5=5$ times It starts as $2$ on each side Let's double the value $2$, and do that $5$ times $$2\cdot2\cdot2\cdot2\cdot2\cdot2=64$$ The final size is then 64 cm by 64 cm


3

The original picture tells you the curve has the form $y=a+c\log_b(dx)$ for some constants $a$, $b$, $c$, and $d$, and that it passes through the points $(-27,17)$, $(-9,13)$, and $(-3,9)$. Since $27$, $9$, and $3$ are powers of $3$, it makes sense to try $b=3$ for the base of the logarithm (as you did). If you let $d=-1$ (it has to be negative because ...


0

Let $$f(a)=\frac{2-a}{2+a}-e^{-a}$$ Now $f'(a)=e^{-a}-\frac4{(a+2)^2}$ $f'(a)<0\;\forall\; a\ge0$ and also $f(0)=f'(0)=0$ You can do that probably by differentiating further.


2

$$\frac{2-a}{2+a}\lt e^{-a}\tag1$$ We know that $(1)$ holds for $a\lt -2\ \text{or}\ a\ge 2$ trivially. So, let us consider the case when $0\lt a\lt 2$ (Note that $(1)$ does not hold for $a=0$). Since $$(1)\iff f(a)=a+2+(a+2)e^a-4e^a\gt 0,$$ we have $$f'(a)=e^a(a-1)+1,\ \ f''(a)=ae^a\gt 0.$$ Since $f'(a)$ is increasing with $f'(0)=0$, we know that ...



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