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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 ...


6

Since $$\int \frac{e^{-2x-x^2}}{(x+1)^2}\,dx = e\cdot\int \frac{e^{-(x+1)^2}}{(x+1)^2}$$ your question is equivalent to finding a primitive for $\frac{e^{-x^2}}{x^2}$. Integration by parts gives: $$\int \frac{e^{-x^2}}{x^2} = -\frac{e^{-x^2}}{x}-\sqrt{\pi}\cdot\operatorname{Erf}(x).$$


6

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 ...


5

Consider $e^{ik\pi x} = \cos k\pi x + i \sin k\pi x$. The integrand becomes $e^{(ik\pi + 1)x}$, then take the imaginary part after integrating. We get: $$ \int e^{(ik\pi + 1) x} dx=\frac{e^{(ik\pi+1)x}}{ik\pi + 1}+C = \frac{(1-ik\pi)e^{(ik\pi+1)x}}{1+k^2\pi^2}+C=\frac{(1-ik\pi)e^x (\cos k\pi x+ i \sin k\pi x)}{1+k^2\pi^2}+C $$ Take the imaginary parts of ...


5

Assuming you are allowed to use that $e>2$, you have that $$e^2>2^2=4$$ and therefore$$2^{(e^2)}>2^4=16$$ thus $$\ln 2^{(e^2)} > \ln 16 > \ln e =1$$ (however, here I used that $\ln$ is a monotone increasing function). Now since the left hand side of the above inequality can be written as $e^2\cdot \ln 2$ you have that $$e^2 \ln 2> 1$$ which ...


4

$\pi^{-\pi x} = e \iff -\pi x \ln(\pi) = 1 \iff x = \frac{1}{-\pi \ln(\pi)}$


4

$$ \exp(x) = \lim_{n \to \infty} \left( 1 + \frac{x}{n} \right)^n $$


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 ...


4

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


4

We first observe that $$ y=\frac{1}{(n+1)(n+2)}\mathrm{e}^{nx}, $$ is a particular solution of $$ y''+3y'+2y=\mathrm{e}^{nx}. $$ Hence $$ \sum_{n=0}^\infty \frac{1}{(n+1)(n+2)n!}\mathrm{e}^{nx}=\mathrm{e}^{-2x}\left(\exp(\mathrm{e}^x)-1-\mathrm{e}^x\right) $$ is a particular solution of $$ ...


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)


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 ...


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

$$ y''e^{-y}=1 \quad\Longrightarrow\quad y''y'=e^yy'\quad\Longrightarrow\quad\frac{1}{2}(y')^2=e^y+c\quad\Longrightarrow\quad y'=\pm\sqrt{2e^y+c'}. $$ The general solution is obtainable if the indefinite integral $$ \int\frac{dy}{\sqrt{e^y+c}} $$ is computable.


4

$$y''e^{-y}=1\Rightarrow y''y'=e^yy'\Rightarrow \int y''(x)y'(x)dx=\int e^{y(x)}y'(x)dx\Rightarrow \frac{1}{2}(y'(x))^2=e^{y(x)}+c$$ The above result is equivalent to $$|y'(x)|=\sqrt{2e^{y(x)}+c}$$ Without loss of generality lets consider the case $y'(x)=\sqrt{2e^{y(x)}+c}$ then $$\frac{dy}{dx}=\sqrt{2e^y+c}\Rightarrow \int\frac{dy}{\sqrt{2e^y+c}}=\int dx$$ ...


3

If you have $a=0$ then $z'\leq b$. So you just have to integrate the inequality \begin{equation} \int_{0}^{t}z'dt \leq \int_{0}^{t}bdt, \end{equation} and obtain \begin{equation} z(t)\leq z_0 +bt. \end{equation}


3

If $$ z'\le az+b,\quad\text{then}\quad \mathrm{e}^{-at}\big(z'(t)-az(t)\big)\le b\,\mathrm{e}^{-at}, $$ or $$ \big(\mathrm{e}^{-at}z(t)\big)'\le -\frac{b}{a}\big(\mathrm{e}^{-at}\big)' $$ and integrating in $[0,t]$ we obtain $$ \left(\mathrm{e}^{-at}z(t)+\frac{b}{a}\mathrm{e}^{-at}\right)- \left(z(0)+\frac{b}{a}\right)\le 0, $$ or $$ z(t)\le ...


3

Notice that: \begin{align*} 3^{x-5} + 3^{x-7} + 3^{x-9} &= 3^{4 + (x-9)} + 3^{2 + (x-9)} + 3^{x-9} \\ &= (3^4)3^{x-9} + (3^2)3^{x-9} + 3^{x-9} \\ &= (81)3^{x-9} + (9)3^{x-9} + (1)3^{x-9} \\ &= (81 + 9 + 1)3^{x-9} \\ &= (91)3^{x-9} \\ \end{align*}


3

$$I = \int e^x \sin (k \pi x) dx\\=\sin k\pi x\int e^xdx-\int k\pi\cos k\pi x \left(\int e^xdx \right)dx\\=e^x\sin k\pi x-k\pi\int e^x\cos k \pi x\\=e^x\sin k\pi x-k\pi\left(\cos k\pi x\int e^x dx-\int k\pi \sin k\pi x\left(\int e^x dx\right)dx\right) \\ I=e^x(\sin k \pi x -k\pi\cos k\pi x)+k^2\pi^2 I+c$$ $$I=\frac{e^x(\sin k\pi x-k\pi\cos k\pi ...


3

The equation can be written $x\log2=2\log|x|$. Let's consider the function $$ f(x)=x\log2-2\log|x| $$ defined for $x\ne0$. We have easily $$ \lim_{x\to-\infty}f(x)=-\infty, \qquad \lim_{x\to\infty}f(x)=\infty $$ and $$ \lim_{x\to0}f(x)=\infty. $$ Moreover $$ f'(x)=\log2-\frac{2}{x}=\frac{x\log2-2}{x} $$ Set $\alpha=2/\log2$; then $f'(x)$ is positive for ...


3

This is just an application of the chain rule $$ \frac{\partial\mathrm{log}(\mathrm{exp}(w_1 * x_1 + b_1) + \mathrm{exp}(w_2 * x_2 + b_2))}{\partial w_1} = \frac{\partial}{\partial w_1}(log(f(w_1)) = \frac{\frac{\partial f(w_1)}{\partial w_1}}{f(w_1)} $$ So your final answer is: $$ \frac{x_1\mathrm{exp}(w_1x_1+b_1)}{\mathrm{exp}(w_1 * x_1 + b_1) + ...


3

If $$ A=\left(\begin{matrix} \lambda & 1 & 0 & \cdots & 0&0 \\ 0 & \lambda & 1 & \cdots &0& 0 \\ \vdots &\vdots&\vdots&&\vdots&\vdots \\ 0 & 0 &0 &\cdots &1&0 \\ 0 & 0 &0 &\cdots &\lambda &1 \\ 0 & 0 &0 &\cdots &0 &\lambda \end{matrix} ...


3

$$e^{\pi-\ln\frac{x+4}{x}}=e^{\pi}.\dfrac{1}{\dfrac{x+4}{-x}}=e^{\pi}\dfrac{-x}{x+4}$$ Hence $$\lim_{x \to 0^{-}}\frac{e^{\pi-\ln\frac{x+4}{-x}}}{x}=\lim_{x \to 0^{-}}\frac{-e^{\pi}}{x+4}$$


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 ...


2

l'Hopital's rule is for computing the limit of a ratio when the numerator and denominator both approach $\infty$ or $0$ (this is called an indeterminant form). Often after an application of l'Hopital's rule, a simplification of the resulting function will be necessary. Other times, the resulting limit will also be in this indeterminant form. Specifically ...


2

I think the problem with your solution might be here: $2^2x * 3^x => 2^2 * 2*x * 3^x$ $4^{3x+1} => 4^3 * 4^x * 4$ since $a^{bc} = (a^b)^c$ and not $a^b * a^c$. So for instance: $4^{3x+1} => 4^x * 4^x * 4^x * 4$


2

Once you have defined $exp(x)=e^x$ we can work our way trhough the definition of derivative (which is quite intuitive) for demostrating that $exp'(x)=exp(x)$. The formal definition of derivative is: $$f'(x)=\lim_{h\to0}\frac{f(x+h)-f(x)}{h}$$ When $h$ tends to $0$. So, substituting $f(x)$ by $e^x$ we get that: ...


2

$$2^{2x}3^x=4^{3x+1},$$ $$(e^{\log2})^{2x}(e^{\log3})^x=(e^{\log4})^{3x+1},$$ $$e^{2x\log2}e^{x\log3}=e^{(3x+1)\log4},$$ $$e^{2x\log2+x\log3}=e^{(3x+1)\log4}.$$ Take the log, $$2x\log2+x\log3=(3x+1)\log4,$$and $$x=\frac{\log4}{2\log2+\log3-3\log4}=\frac1{\frac{\log3}{\log4}-2}.$$


2

step 1. learn the rules of exponentiation. step 2. learn the rules of logarithms. step 3. apply that knowledge and solve for $x$



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