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12

Either integrate by parts, or write that, when $P(x)$ is a polynomial, $$\int P(x)e^x \mathrm{d}x=Q(x)e^x+C$$ Where $Q$ is a polynomial and $C$ a constant. You have thus, differentiating: $$(Q(x)+Q'(x))e^x=P(x)e^x$$ Or $Q+Q'=P$. Hence, $Q$ has same degree as $P$, and same coefficient of highest degree, so $Q(x)=x^3+ax^2+bx+c$, and $Q'(x)=3x^2+2ax+b$, ...


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 ...


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 ...


10

Repeated integrations by parts show that, for every positive $a$ and $x$, $$\int_0^x\mathrm e^{-t}t^{a-1}\mathrm dt=\Gamma(a)\mathrm e^{-x}\sum_{n\geqslant0}\frac{x^{n+a}}{\Gamma(n+a+1)}.$$ When $x\to\infty$, the LHS converges to $\Gamma(a)$, hence the series in the RHS is equivalent to $\mathrm e^x$. Now, ...


8

Your integral depend of $y$ only, then you can put $e^{x^2}$ out of your integral. $$\int ye^{x^2}\mathrm{d}y=e^{x^2}\int y\mathrm{d}y=e^{x^2}\left(\frac{y^2}{2}+C\right)$$ where $C$ is a constant.


7

Hint : Simplification yields $$ \frac{e^{-4x} + 3e^{-2x}}{e^{-4x}-9}=\frac{e^{-2x}\left(e^{-2x} + 3\right)}{\left(e^{-2x} + 3\right)\left(e^{-2x} - 3\right)}, $$ where the denominator is using $a^2-b^2=(a+b)(a-b)$. Now, set $u=e^{-2x}-3$ to integrate it.


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

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

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

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


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

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


3

If $x = e^c$ with $c \neq 1$, then $\lim_{h \to 0}\dfrac{x^h - 1}{h} = \lim_{h\to 0}\dfrac{e^{ch} - 1}{ch}c = c \neq 1$


3

$\dfrac{5}{80}=(\dfrac{300}{600})^x$. Can you proceed from here?


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 ...


3

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


3

Here 3 Integrations by parts will help, as $e^x$ will always reproduce itself and $x^3$ is constant after 3 derivatives.


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 $C=0$.


3

$x^3 + i = 0 \to x^3 - i^3 = 0 \to (x-i)(x^2 + xi - 1) = 0$. Can you use quadratic formula to complete the answer.


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 ...


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

It follows from the fact $\mathbb{R}$ has no "zero divisors"; i.e. if $x,y \in \mathbb{R}$ and $xy = 0$, then at least one of $x$ and $y$ must be zero.


2

Hint: Use integration by parts twice.


2

Hint: $$\frac{e^{-4x}+3e^{-2x}}{e^{-4x}-9} \equiv \frac{e^{-4x}+3e^{-2x}}{e^{-4x}-9} \cdot \underbrace{\left(\frac{e^{4x}}{e^{4x}}\right)}_{1} \equiv \frac{1+3e^{2x}}{1-9e^{4x}} \equiv \require{cancel}\frac{\cancel{1+3e^{2x}}}{\cancel{(1+3e^{2x}})(1-3e^{2x})} \equiv \frac{1}{1-3e^{2x}}.$$ Now, using the substitution $u=1-3e^{2x},$ have a go at integrating ...


2

The max of these random variables is less than $x$ if and only if all of them are less than $x$, and since they are independent and identically distributed, the probability of that is the $n$th power of the probability that the first one is less than $x$.


2

You can do this "trick" All the roots of $x^3=1$ are $1,\omega,\omega^2$ where $\omega=e^{2i\pi/6}$. So, all roots of $x^3=-i$ would be roots of $1$ times $i$ (since $i^3=-i$). Thus the roots are $i,i\omega,i\omega^2$


2

In the case of equation like $$ x^n-\alpha=0 $$ where $\alpha\in\mathbb C$ the best trick is to write $\alpha$ in polar form: $\alpha=re^{i\theta}$. But $e^{i\theta}=e^{i(\theta+2k\pi)},\;k\in\mathbb Z$. Then you have $$ x^n=re^{i\theta+2k\pi i} $$ from which immediately $$ x=\sqrt[n]re^{i(\frac{\theta+2k\pi }{n})} $$ and as $k=0,1,\dots,n-1$ you get all and ...


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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