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7

Observations. Let us consider only when $a > 0$. We have three observations: The function $f(x) = \frac{1-x}{1+x}$ is positive if and only if $|x| < 1$. $f(-x) = 1/f(x)$. So the set of solution of $a^{x} = f(x)$ must be symmetric around the origin. Numerical observation shows that the set of solutions $(a, x)$ of $a^{x} = f(x)$ is Solution. If we ...


7

Yes, using polar coordinates the boundaries are: $$0 \leq r \leq R \\ 0 \leq \theta \leq 2 \pi$$ Since $D(R)$ is the disk of radius $R$ with center at $(0,0)$: $D(R)=\{(x,y): x^2+y^2 \leq R^2\}$ So we have the following: $$x = r\cos \theta, y = r \sin\theta \\ \renewcommand{\intd}{\,\mathrm{d}} \intd x \intd y = r \intd r \intd \theta$$ $$ ...


6

This is an interesting problem. Since you say you're "not very good at mathematics", I'll give the result first and then explain how I came up with it. If $x$ and $y$ are greater than $2.71828\dots$, then $x^y < y^x$ if and only if $x > y$. If $x$ and $y$ are less than $2.71828\ldots$, then $x^y < y^x$ if and only if $x < y$. If $x$ ...


6

If $a\not=0$, let $k=ah$ and note that $h\to0$ iff $k\to0$, which gives $$\lim_{h\to0}{(b^a)^h-1\over h}=a\lim_{h\to0}{b^{ah}-1\over ah}=a\lim_{k\to0}{b^k-1\over k}$$ If $a=0$, then $(b^a)^h-1=0$, so the limit is obviously $0$.


5

Hint: \begin{align*} \frac{f(x + h) - f(x)}{h} &= f(x) \frac{f(h) - 1}{h} \\ &= f(x) \frac{h g(h)}{h} \end{align*} Can you finish from here?


5

You have to apply the chain rule: if $f(x)$ is a differentiable function the the derivative of $e^{f(x)}$ is $f'(x)e^{f(x)}$.


5

this holds because for $-1\lt u \lt 1$, $-u-u^2/2-...=\ln(1-u)$ thus $$e^{-u-u^2/2-...}=e^{\ln(1-u)}=1-u$$


5

Here is yet another one, which is one of my favorite irrationality/transcendence proofs : The confluent hypergeometreic series $$_{0}F_{1}(k; z) = \sum_{n = 0}^\infty \frac1{(k)_n} \frac{z^n}{n!}$$ Satisfies the more-or-less easily verifiable identity $$_0F_1(k-1;z) - {}_0F_1(k; z) = \frac{z}{k(k-1)}{}_0F_1(k+1;z)$$ Iterating this, one ends up with ...


4

As Kaj Hansen said the derivative of $e^x$ is $e^x$ Assume $e^x=\frac{f(x)}{g(x)}$. Then $e^xg(x)=f(x)$. But none of the n'th derivatives of $e^xg(x)$ are zero, on the other hand if $f(x)$ has degree $d$ then the $(d+1)$'th derivative of $f(x)$ is zero. Using a similar reasoning we have the derivative of $\log(x)$ is $\frac{1}{x}$. So if ...


4

Let $A = \begin{bmatrix}0&6\pi&0\\-6\pi&0&0\\0&0&0\end{bmatrix}$ and $B = \begin{bmatrix}0&0&0\\0&0&8\pi\\0&-8\pi&0\end{bmatrix}$. Using the formula I derived here, $e^A = e^B = e^{A+B} = I$. Hence, $e^{A+B} = e^Ae^B$. However, $AB-BA = ...


4

In principle, your proof looks pretty good and I might have to use it when I teach calculus! There are definitely important details that would need to be addressed if it were to be 100% rigorous, however. Here are the obvious things I notice: You should be very precise about what assumptions you are making. How are you defining $e^z$? As the unique ...


4

Try $\begin{cases} x=3\cos^3\theta\\ y=3\sin^3\theta\\ \end{cases}$ or, for rational solutions, $\begin{cases} x=3\left({{2t}\over{t^2+1}}\right)^3\\ y=3\left({{t^2-1}\over{t^2+1}}\right)^3\\ \end{cases}$


4

There is an explicit solution for $x$ using Lambert $W$ function. The solution is given by $$x=\frac{p}{p-1}-\frac{W\left(-\frac{2^{\frac{p}{p-1}} R \log (2)}{p-1}\right)}{\log (2)}$$ In the case where the argument of the Lambert $W$ function is small or large, there are very nice approximations which at least would give you a reasonable estimate of the ...


4

For the range $0 \leq y \leq 1$, you may have a very accurate estimation of the integral expanding first the integrand as a Taylor series built at $x=0$. This gives $$e^{\sqrt{x(1-x)}}=1+\sqrt{x}+\frac{x}{2}-\frac{x^{3/2}}{3}-\frac{11 x^2}{24}-\frac{11 x^{5/2}}{30}-\frac{59 x^3}{720}-\frac{13 x^{7/2}}{630}+\frac{1513 x^4}{40320}-\frac{311 ...


4

The function $f(x)=3^{x-1}+5^{x-1}$ increases monotonically for all $x\in\mathbb{R}$. Hence, if $f(x_0)=34$ for some $x_0\in\mathbb{R}$, then $f(x)<34$ for all $x<x_0$ and $f(x)>34$ for all $x>x_0$.


4

Yes, you did apply well this property of logarithm: $ln(ab)=ln(a) + ln(b)$, and the fact that $ln(x)$ is the inverse of $e^x$


4

No, this is not true. For a fixed $n$ you have $2^n \le \alpha_n$ for some $\alpha_n \gt 0$. This is generally obvious, since $2^n$ is some number, but your "proof by induction" basically proves the same. Note that $\alpha_n$ is depending on $n$. But: $2^n = \mathcal{O}(1)$ would imply that $2^n \le \alpha$ for all $n$ and just one $\alpha$. That‘s not ...


3

A rigorous way to define the sine function is to consider it as the solution to the IVP: $$ \begin{cases} y^{\prime \prime} + y = 0\\ y(0) = 0 \\ y^{\prime}(0) = 1 \end{cases} $$


3

You have to use the chain rule here. Writing $f(x) = e^x$ and $g(x) = -2x$ we have $h(x) := f(g(x)) = e^{-2x}$, hence by the chain rule $$ h'(x) = f'(g(x))g'(x) $$ Now $f'(x) = e^x$, hence $f'(g(x)) = e^{-2x}$, and $g'(x) = -2$, this gives $$ h'(x) = f'(g(x))g'(x) = e^{-2x} \cdot (-2) = -2e^{-2x} $$


3

You want to solve for $X$, but it may be more useful to first solve for $A$. Note that $\dfrac{1-X}{1+X} > 0$ if $-1 < X < 1$, $< 0$ if $X < -1$ or $X > 1$. If $-1 < X < 0$ or $0 < X < 1$, you could have $$ A = \left(\dfrac{1-X}{1+X}\right)^{1/X}$$ with $A$ real. Plotting $A$ as a function of $X$, we get this: You can check ...


3

Hint: Let $x = r\cos \theta$ and $y = r\sin \theta$, then $$I = \int_{0}^{2\pi} \int_{0}^R r e^{-r^2} \,\mathrm{d}r \,\mathrm{d}\theta$$


3

You have a defined region to integrate about, which is $D(R)=\{(x,y) \in \mathbb{R}^{2}| x^{2}+y^{2} \leq R^{2}\}$ as the disc with radius $R$. At first, you should choose an adequate parametrization for making the problem easier. Here you might use the transformation $(r, \theta) \mapsto (x, y)=(r \cos \theta, r \sin \theta)$ with $r \in [0, R]$, $\theta ...


3

I must say that at first I was quite skeptical that such function could exist, but it turned out that one can prove its existence. I have written down the proof but I will give here only sketch it (you should not have any problems computing it for yourself, nice exercise) Let us define $b_1=2\pi \rm i m_1$ and $\displaystyle b_n=2\pi \rm i m_n+ \log ...


3

For $|x|<1,$ using infinite Geometric Series, $$(1-x)^{-1}=1+x+x^2+\cdots$$ Integrate either sides to find $$-\ln(1-x)=\sum_{r=1}^\infty\frac{x^r}r$$ Now $$\ln y=z\implies y=e^z$$ If $z$ is finite real, real $y>0$


3

I would say that "Feynman physics lectures" here: http://www.feynmanlectures.caltech.edu/I_22.html


2

For rational numbers, the number $a^{\frac pq}$ is defined as $$\displaystyle\sqrt[q]{a^p}$$


2

If $P=0\text{ or }1$ then the growth rate is $0$, so the population does not change. If $P$ is between $0$ and $1$ then the growth rate is positive, so the population is getting bigger. So it approaches $P=1$ and then stops growing. So $P=1$ is the carrying capacity i.e. 10,000 buffalo.


2

In light of JJacquelin's hint, you may play with the equation as follows: $$kdt=\frac{dP}{P(1-P)}=\left(\frac{1}P+\frac{1}{1-P}\right)dP$$ And a simple integration from both sides, gives us: $$kt+C=\ln|P|-\ln|1-P|=\ln|\frac{P}{1-P}|$$ So we have: $$P(t)=\frac{\exp(kt+C)}{1+\exp(kt+C)},~~\text{or}~~P(t)=\frac{\exp(kt+C)}{-1+\exp(kt+C)}$$


2

Hint: $$2^{x}=16=2^{4}$$ ${}{}{}{}{}{}{}{}$


2

$$2^{x}=16$$ Take logarithm from both sides. \begin{align} \require{cancel} x\ln 2&=4\ln2\\ x\color{red}{\cancel{\color{black}{\ln 2}}}&=4\color{red}{\cancel{\color{black}{\ln 2}}}\\ x&=4\\ \end{align}



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