# Tag Info

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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