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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

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.


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 $isin(\theta - \psi)=0$, ...


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: $$x=-1-\frac{1}{\sqrt{2}} \\ x=-1+\frac{1}{\sqrt{2}}$$


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

$F(x,y)=e^{-x^2-y^2}=e^{-(x^2+y^2)}$. 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

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

Here I use the Daniel's method: Take the logarithm of $\prod _{k=1}^n (1+\dfrac{kx}{n^2} ),$ then we have $$\ln\left(\prod_{k=1}^n\ \left(1+\dfrac{kx}{n^2 }\right) \right)=\sum_{k=1}^n\ln\left(1+\dfrac{kx}{n^2 }\right).$$ By the Taylor series expansion of logarithms, $$ \ln\left(1+\dfrac{kx}{n^2} \right)=\sum_{m=1}^\infty\dfrac{(-1)^{m-1}}{m}\left(\dfrac{ ...


4

Given an event whose frequencies open the Poisson distribution and occurs an average of $n$ times per trial, the probability that it occurs $k$ times in a given trial is $e^{-n} \frac{n^k}{k!}$. So, the sum in the limit is the probability that the event (which now must have an integer average) occurs no more than the mean number of times. For large $n$, ...


4

This could help here. $$ \int^{y_2}_{y_1}\mathrm{e}^{-\alpha x}x\sqrt{1-x^2}dx = -\frac{d}{d\alpha}\int\mathrm{e}^{-\alpha x}\sqrt{1-x^2}dx $$ using $x = \cos u$ then $$ \int^{y_2}_{y_1}\mathrm{e}^{-\alpha x}x\sqrt{1-x^2}dx = -\frac{d}{d\alpha}\int_{\cos^{-1}y_1}^{\cos^{-1}y_2}\mathrm{e}^{-\alpha \cos u}\sin^2 u du $$ here the last bit was edited due to ...


4

$$\begin{align} \left|\frac{1}{e^{i\omega t} - 1}\right| &= \left|\frac{1}{e^{i\omega t/2}(e^{i\omega t/2} - e^{-i\omega t/2})}\right| \\ &= \frac{1}{|e^{i\omega t/2}|} \left|\frac{1}{2i \sin(\omega t/2)}\right|\\ &= \frac{1}{2|\sin(\omega t/2)|} \end{align}$$


4

As you said: $$f(x)=b^x\\\ln f(x)=x\ln b \\f(x)=e^{x\ln b}$$ Now: $$\frac{d}{dx}e^u=e^u \frac{du}{dx}$$ Hence: $$u=x\ln b \\ \frac{du}{dx}=\ln b$$ So: $$f'(x)=e^u \frac{du}{dx}=e^{x\ln b}\ln b$$


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 ...


3

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


3

The definition for the derivative of a real valued function $f$ is $$\frac{df}{dx}=\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}.$$ Letting $f(x) = e^{x \ln b }$, we have \begin{eqnarray} \frac{df}{dx}&=&\lim_{h\to 0}\frac{e^{(x+h) \ln b }-e^{x \ln b }}{h}\\ &=& \lim_{h\to 0}\frac{e^{x \ln b }(e^{h\ln b}-1)}{h}\\ & = & e^{x \ln b }\lim_{h\to ...


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.)


3

There's a lot of uses for logarithms There are things you can solve without logarithms. Take for example: $$4.9^x=34.185$$ Without logarithms you might try taking the $x$'th root, and really just end up going in circles, with logarithms, we can solve this with logarithms this way. $$\log 4.9^x=\log34.185$$ $$x\log 4.9=\log34.185$$ ...


3

Most of the arguments given in other answers have a curious fallacy. In the binomial expansion of $(1+1/n)^{n}$ the number of terms as well as each term is dependent on $n$ hence taking limits term by term is not justified. A proper proof requires more analysis. I have presented this proof in detailed manner in my blog post. Update: Upon OP's request (see ...


2

I understood nothing of what you have written, but: $$\left(1+\frac{1}{n}\right)^n=\sum_{k=0}^n \frac{n!}{(n-k)!k!}\left(\frac{1}{n}\right)^k=\sum_{k=0}^n \frac{1}{k!}\prod_{i=n-k+1}^n \frac{i}{n}\leq e$$ Because: $\frac{n!}{n-k!}=\prod_{i=n-k+1}^n i$ are exactly $k$ terms, so I can bring in the $\frac{1}{n^k}$. If we take the limit with $n\to\infty$ then ...


2

$\ln[f(x|\theta)] = \eta(\theta)T(X) - \psi(\theta) + \ln h(X)$ $E_\theta[T(X)]= \frac{d}{d(\theta)}\ln[f(x|\theta)] = \eta'(\theta)T(X) - \psi'(\theta)$ = 0 $\eta'(\theta)T(X) = \psi'(\theta)$ $T(X) = \frac{\psi'(\theta)}{\eta'(\theta)}$


2

$$e^{2x} \cdot (2e^{-x}-e^{-2x})=0 \Rightarrow 2e^x-1=0 \Rightarrow e^x=\frac{1}{2} \Rightarrow \ln e^x=\ln \frac{1}{2} \Rightarrow x=\ln 1-\ln 2=-\ln 2$$ Therefore: $$x=-\ln 2$$


2

The minimum or maximum must lie either in the interior region, or on the boundary. For the interior, you find where the derivative of the function is zero, for both $x$ and $y$. $\frac{d}{dx}e^{-x^2-y^2} = -2xe^{-x^2-y^2} = 0 $ => $x=0$ and similarily $y=0$. This means we have a possible min or max in $f(0,0) = 1$ Now for the boundary. You can ...


2

If we are given log(x), it is generally assumed that this is log base 10. We can re-write your problem as y=2 $log_{10}$(x), which is equivalent to $(1/2)y = log_{10}(x)$. Then we have $10^{y/2} = x$.


2

$y=2\mbox{log}x \ \Leftrightarrow \mbox{log}x=\frac{y}{2} \ \Leftrightarrow x=10^{\frac{y}{2}}.$


2

Given any $b > 0$ then $f: \mathbb{R} \to (0, \infty)$ by $f(x) = b^x$ is a bijective, that is one-to-one and onto, function. This means $f^{-1}:(0, \infty) \to \mathbb{R}$ exists. Further more we denote this inverse function by $f^{-1}(x) = \log_b(x)$. That is $\log_b$ is exactly the inverse function to the exponential function with base $b$. So, why do ...


2

Notice that $x-x^2/2\leq \ln(1+x) \leq x$ for $x\geq 0$t hen : $$\frac{xk }{n^2} \geq \ln\left(1+\frac{ xk}{n^2}\right) \geq \frac{x k}{n^2} -\frac{x^2 k^2}{n^4} $$ Sum form $k=1$ to $n$ : $$\frac{(n+1)x}{2n} \geq \sum_{k=1}^n \ln\left(1+\frac{ xk}{n^2}\right) \geq \frac{(n+1)x}{2n} - \frac{(n+1)(2n+1)x}{6n^3} $$ Then the limit of the middle sum is $x/2$ ...


2

There are many ways to define the complex numbers. Each such definition should consist of the following: A "list" of all complex numbers. Two examples are the formal expressions $x+yi$ for real $x, y$ (or, which is the same, pairs $(x, y) $), and polar representation, i.e. $(r,\theta) $ for real $r> 0$ and $\theta$ plus the number $0$. Each complex ...


2

If $f(x)=\log x$ is defined as a primitive of $\frac{1}{x}$ for which $f(1)=0$, then $f(ab)=f(a)+f(b)$ holds for any couple of positive real numbers. This gives that the inverse function $g(x)$ is a $C^1$ function, satisfies $g(0)=1$ and the functional equation: $$ g(a)g(b)=g(a+b)\tag{1} $$ for any couple of real numbers $a,b$. $(1)$ and differentiability ...


2

$(e^u)'=u'e^u$. Then, if you set $u(x)= x\ln b$, we have $u'(x)=\ln b$. If you apply the formula, you arrive at $$\ln be^{x\ln b}$$


2

$$f'(x) = b^xln(b)$$ $$f'(x) = e^{ln(b)x}ln(b)$$



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