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9

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 ...


8

$\textbf{Possible direction}$ $$ \sqrt{x(1-x)} = \sqrt{\frac{1}{4}-\left(x-\frac{1}{2}\right)^2} = \frac{1}{2}\sqrt{1-4\left(x-\frac{1}{2}\right)^2}. $$ thus making a change of variable $$ \cos(t) = 2\left(x-\frac{1}{2}\right),\\ -\sin(t)dt = 2dx. $$ we can re-write the integral as $$ -\frac{1}{2}\int_{\pi}^{0} \mathrm{e}^{\frac{\alpha}{2}\sin(t)}\sin(t)dt ...


8

$$x^x=e^{\ln x^x}=e^{x \ln x}$$ Therefore, it is a composition of an exponential and the product of $x \cdot \ln x$


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: $$x = r\cos \theta, y = r \sin\theta \\ \renewcommand{\intd}{\,\mathrm{d}} \intd x \intd y = r \intd r \intd \theta$$ $$ ...


7

It's neither. A poynomial is a function that is of the form $\sum_i c_ix^i$ where the $c_i$ are constants. An exponential function is one of the form $Ca^x$ for some constant $a$ and nonzero constant $C$ Note that $x$ is not a constant, and so $x^x$ is of neither form.


7

We have with $u=e^{-x}$ so $du=-e^{-x}dx\implies dx=-\frac{du}{u}$ $$\int \frac{dx}{1+e^{-x}}=-\int\frac{du}{u(1+u)}=\int\frac{du}{1+u}-\int\frac{du}{u}=\ln(1+u)-\ln u+C\\=\ln\left(1+e^x\right)+C$$


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 $0$.


6

$e^{x^2+4x-7}(6x^2+12x+3)=0 \Rightarrow e^{x^2+4x-7}=0 \text{ or } \ 6x^2+12x+3=0$ $$\text{It is known that } e^{x^2+4x-7} \text{ is non-zero }$$ therefore,you have to solve : $$6x^2+12x+3=0$$ The solutions are: $$x=-1-\frac{1}{\sqrt{2}} \\ x=-1+\frac{1}{\sqrt{2}}$$


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

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 ...


6

Starting with $e^{i \theta}=e^{i \psi}$, multiply both sides by $e^{-i \psi}$ to get $e^{i \theta}e^{-i \psi}=1$. Using rules of exponents this means $e^{i( \theta - \psi)}=1$. Now applying Euler's identity to the exponent means $cos(\theta - \psi)+isin(\theta - \psi)=1$. Since the RHS has no imaginary component, it follows that $isin(\theta - \psi)=0$, ...


5

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 ...


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

A rotation matrix $R$ is orthogonally diagonalizable with eigenvalues of absolute value one, i.e., $$ R=U^* D U, $$ where $D=\mathrm{diag}(d_1,\ldots,d_n)$, with $\lvert d_j\rvert=1$, for all $j=1,\ldots,n$, and $U^*U=I$. Clearly, as $\lvert d_j\rvert=1$, there exists a $\vartheta_j\in\mathbb R$, such that $$ d_j=\mathrm{e}^{i\vartheta_j}, \quad ...


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

$F(x,y)=e^{-x^2-y^2}=e^{-(x^2+y^2)}$. Note that $e^{-z}$ is strictly decreasing with respect to $z$. So to maximize and minimize $F(x,y)$, just minimise and maximise $x^2+y^2$ respectively. By the domain of definition, $x^2+y^2$ is minimised at $0$ and maximised at $25$, so the maximum and minimum values of $F(x,y)$ are $e^{-0}=1$ and $e^{-25}$, ...


4

It is not taking the limit "outside," really, but pushing it inside. So: $$\lim_{n\to\infty} f(n)^{g(n)} = (\lim f(n))^{\lim g(n)}$$ When both $\lim f(n)$ and $\lim g(n)$ exist. This is true because (for some values $x,y$ at least) the function $(x,y)\to x^y$ is a continuous function. (Specifically, it is continuous at $(x,y)$ when $x>0$.


4

Solve $$2e^{-x} = e^{-2x}$$ by taking the $\ln$ of each side of the equation The logarithm function is an increasing and injective function, so using it is legitimate: the equality remains unchanged when applied to each side. So you can solve $$\begin{align} &\qquad\underbrace{\ln(2e^{-x})}_{\large \ln 2 + \ln(e^{-x})} = \ln (e^{-2x})\tag{1}\\ \\ & ...


4

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


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 ...


4

Considering @cooper's comment and your first comment above: The function $y=\exp(x)$ and the relation $x=\exp(y)$ are defined differently. Just look at their plots: But sometimes we change the alphabet $x$ with $y$ just to read the relation we got easily. For example, when we want to find the inverse of an strictly increasing function $y=f(x)$ we do ...


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

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

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 ...


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


3

Hint: transform the inequality to a quadratic inequality with the substitution $y = 2^x$, then $2^{-x} = \dfrac{1}{y}$


3

Your steps are not quite right. It seems you're skipping a couple of them. From the substitution $u=1+e^{-x}$, you should get: $$du=d(1+e^{-x}) = -e^{-x}dx$$ So that: $$dx = - \frac {du}{e^{-x}}$$ Since $u=1+e^{-x}$ implies $e^{-x}=u-1$, this becomes: $$dx = - \frac {du}{u-1}$$ So your integral should be: $$\int \frac {dx} {1+e^{-x}} =\int \frac 1 u \cdot ...


3

I'm writing this just to show a possibility. If $e^{x^2}$ represents $(e^x)^2$ in the textbook, which should be written as $e^{2x}$, then the answer is $(e^x)^2\cdot x^2\cdot (3+2x)$. (By the way, since $e^{x^2}$ means $e^{(x^2)}$ in general, your calculation has no mistakes.)



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