# Tag Info

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=\log34.185x\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$$2 The minimum or maximum must lie either in the interior region, or on the boundary. For the interior, you find where the derivative of the function is zero, for both x and y. \frac{d}{dx}e^{-x^2-y^2} = -2xe^{-x^2-y^2} = 0  => x=0 and similarily y=0. This means we have a possible min or max in f(0,0) = 1 Now for the boundary. You can ... 2 If we are given log(x), it is generally assumed that this is log base 10. We can re-write your problem as y=2 log_{10}(x), which is equivalent to (1/2)y = log_{10}(x). Then we have 10^{y/2} = x. 2 y=2\mbox{log}x \ \Leftrightarrow \mbox{log}x=\frac{y}{2} \ \Leftrightarrow x=10^{\frac{y}{2}}. 2 Given any b > 0 then f: \mathbb{R} \to (0, \infty) by f(x) = b^x is a bijective, that is one-to-one and onto, function. This means f^{-1}:(0, \infty) \to \mathbb{R} exists. Further more we denote this inverse function by f^{-1}(x) = \log_b(x). That is \log_b is exactly the inverse function to the exponential function with base b. So, why do ... 2 Notice that x-x^2/2\leq \ln(1+x) \leq x for x\geq 0t hen :$$\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)

Only top voted, non community-wiki answers of a minimum length are eligible