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0

I have shown below the derivation of this model from Newton's law of heating and cooling. If the temperature difference, ΔT, between the two bodies or between an object and the ambient is not too large, the rate of change of this temperature difference is nearly proportional to that temperature difference: $\frac{dΔT}{dt} = - K ΔT$ $\frac{dΔT}{ΔT} = - ...


1

$\int_0^y e^{-\alpha\sqrt{x(1-x)}}~dx$ $=\int_0^y e^{-\alpha\sqrt{-(x^2-x)}}~dx$ $=\int_0^y e^{-\alpha\sqrt{-\left(x^2-x+\frac{1}{4}-\frac{1}{4}\right)}}~dx$ $=\int_0^y e^{-\alpha\sqrt{\frac{1}{4}-\left(x-\frac{1}{2}\right)^2}}~dx$ $=\int_{-\frac{1}{2}}^{y-\frac{1}{2}}e^{-\alpha\sqrt{\frac{1}{4}-x^2}}~dx$ ...


1

The advantage of defining $\ln x$ as an integral is that it gives a simple way to define $e^x$ in elementary calculus (without resorting to infinite series, taking limits of rational powers of $e$, or other methods). Once $\ln x$ has been defined in this way, you can establish that $y=\ln x$ is an increasing function whose range is $\mathbb{R}$, so it has ...


1

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 ...


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 ...


2

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


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 $$\ln be^{x\ln b}$$


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: $$f'(x)=e^u \frac{du}{dx}=e^{x\ln b}\ln b$$


0

Put: $$X = \log\left(\frac{\alpha}{\alpha + \beta}\right) = \log\left(\frac{1}{1 + \frac{\beta}{\alpha}}\right)$$ Then working in terms of X will ensure that the logarithm is always well defined. You then need to define another independent variable Y such that a linarization of Y in terms of small changes in $\alpha$ and $\beta$ does not become almost ...


0

Hint: $\int_0^\infty\dfrac{e^{-xf}}{m+x}\gamma(a,hx)~dx$ $=\int_0^\infty\dfrac{e^{-fx}}{x+m}\sum\limits_{n=0}^\infty\dfrac{(-1)^nh^{n+a}x^{n+a}}{n!(n+a)}dx$ $=\int_0^\infty\sum\limits_{n=0}^\infty\dfrac{(-1)^nh^{n+a}x^{n+a}e^{-fx}}{n!(n+a)(x+m)}dx$ $=\int_m^\infty\sum\limits_{n=0}^\infty\dfrac{(-1)^nh^{n+a}(x-m)^{n+a}e^{-f(x-m)}}{n!(n+a)x}d(x-m)$ ...


0

The linear function in the exponent is an easy example. In general you could have a much more complicated function for $g(x)$. For example if $f(x) = e^{\textrm{sin}(x)}$ then $f'(x) = \textrm{cos}(x)e^{\textrm{sin}(x)}$ In every case the derivative of $f(g(x))$ is just $f(g(x))*g'(x)$.


0

Finding a closed form for this integral seems difficult. Let us try to obtain an analytic expression. We may start to write $$ \begin{align} y & =\int_{0}^{m}\frac{e^{- x f}}{m+x}\gamma\left(a,h x\right) dx +\int_{m}^{+\infty}\frac{e^{- x f}}{m+x}\gamma\left(a,h x\right) dx \\\\ & =\sum_{k=0}^{\infty} ...


1

No, the case $\mathrm{g}(x) = kx+b$, i.e. where $\mathrm{g}$ is a linear function, is really quite special. Even amongst all possible polynomials $a_nx^n + a_{n-1}x^{n-1}+\cdots + a_1x + a_0$, the linear functions are a special example. When you include trigonometric function, logarithms, rational functions, hyperbolic functions and other non-elementary ...


0

Your formula $$100000*0.9691^t$$ is actually an increasing function. Try $$f(t)=100000\times N^{-t}$$ Spoiler:Answer: From the formula get no. of tigers in 2000 and 2010, then find decrease percent.


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 ...


1

The Binomial Expansion is given by $$\begin{eqnarray*} \left(1+\frac{1}{n}\right)^{\!n} &=& 1+n\left(\frac{1}{n}\right)+\frac{n(n-1)}{2!}\left(\frac{1}{n}\right)^{\!2}+\frac{n(n-1)(n-2)}{3!}\left(\frac{1}{n}\right)^{\!3}+\cdots \\ \\ &=& 1+1+\frac{n(n-1)}{n^2}\cdot\frac{1}{2!}+\frac{n(n-1)(n-2)}{n^3}\cdot \frac{1}{3!}+\cdots \\ \\ ...


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


1

$2^x+17=y^2$ leads to the three elliptic curves $y^2=u^3+17$, $y^2=2u^3+17$, and $y^2=4u^3+17$ (depending on $x$ modulo $3$). There are standard techniques for solving these, that is, for finding all the integral solutions. Indeed, you can write each one as a Mordell equation, $Y^2=U^3+k$; for example, from $y^2=4u^3+17$ we get $16y^2=64u^3+272$, so ...


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 ...


0

Note that $$\frac{n!}{(n-k)!n^{k}}=\frac{n(n-1)(n-2)....(n-k+1)}{n*n*n... k \, times}$$ So, as n tends to infinity, it becomes 1. I think you can figure out the rest


1

$x=e^y$ does not imply $y=e^x$, because if it did, then for any finite number $x\in \mathbb{R}$, $$x=e^y = e^{e^x}=e^{e^{e^y}} =e^{e^{e^{e^{x}}}} = \cdots = \infty $$ Which is a contradiction (i.e. non-sense). What you really want is, $$x=\ln (y)$$ implies $$e^x = e^{ln(y)}$$ And then remember that exponential and natural log ''cancel'' i.e. ...


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 ...


0

What we have given is a rotation matrix R. We know there exits a form $e^B$ for every R. Now as per the paper $B=log R \tag 1$ $ \theta = \arccos\left( \frac{\mathrm{trace}(R) - 1}{2} \right) \tag 2$ $ \log R = \{ \begin{matrix} 0 & \mathrm{if} \; \theta = 0 \\ \frac{\theta}{2 \sin(\theta)} (R - R^\top) & \mathrm{if} \; \theta \ne 0 \; ...


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 ...


1

If you prefer a closed form with functions without $i$ in the argument, better use the functions Ci($x$) and Si($x): http://mathworld.wolfram.com/CosineIntegral.html http://mathworld.wolfram.com/SineIntegral.html $$ \int_{-n}^{N-n} \left( \frac{e^{j\pi\alpha\tau}}{\tau} - \frac{e^{j\pi\beta\tau}}{\tau} \right) d\tau=\int_{-n}^{N-n} ...


2

Here is an approach. Expand the exponential functions in terms of their Taylor series and note that constant terms cancel each other so we have $$I= \int_{-n}^{N-n} \left( \frac{e^{j\pi\alpha\tau}}{\tau} - \frac{e^{j\pi\beta\tau}}{\tau} \right) d\tau = \sum_{k=1}^{\infty}\frac{(j\pi\alpha)^k-(j\pi\beta)^k}{k!} \int_{-n}^{N-n}\tau^{k-1}=\dots $$ $$ I= ...


0

No, this is not a polynomial as a polynomial necessarily has non-negative integer exponents. I've heard it referred to as a "hyperpower" function and you can read about it, and similar topics, on this page. Also, you could think of this function as $e^{x\log x}$ which is a composition of polynomial, exponential and logarithmic functions.


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.


0

There really isn't much you can do. But here's something that might help. Let $y=e^x$. Clearly $$y=\log_{0.001}(x)$$ Which is the same as saying that $$x=0.001^y$$ And so your one equation is the same as solving the system of equations $$\begin{cases}y=e^x\\x=0.001^y\end{cases}$$


0

I do not see any way to solve this exactly. Even Wolfram-Alpha gives only a numerical solution (the other algebraic answers are just ways of transforming the question into other forms). Multiple numeric methods would work here--just choose your favorite.


0

Here is the general case, \[ y=a\log_{b} x \] \[ \frac{y}{a}=\log_{b} x \] \[ b^{\frac{y}{a}}=b^{\log_{b} x}=x \] Usually \[ \log x=\log_{10} x \] So if $a=2$ and $b=10$, then \[ x= 10^{\frac{y}{2}}=\sqrt{10^{y}} \]


2

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


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 $10^{y/2} = x$.


0

For $(x,y) \in A=\{(x,y) |x^2+y^2=2\}$ you have: $$f(x,y)=x^2e^{-x^2-y^2}=x^2e^{-(x^2+y^2)}=x^2e^{-2}$$ so $f(x,y)$ is maximal when $x^2$ is maximal, so for $(\sqrt{2},0)$ and $(-\sqrt{2},0)$.


1

Let us consider a general form $$ I = \int_{-\infty}^\infty f(x+a) \, {\rm d}x. $$ With a change of variables $z=x+a$, we get ${\rm d}z = {\rm d}x$, and $$ I= \int_{-\infty+a}^{\infty+a} f(z) \, {\rm d}z = \int_{-\infty}^\infty f(z) \, {\rm d}z. $$ This change of variables is intuitively a translation of the $x$-axis. But because we are integrating the ...


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 ...


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 $e^{-25}$, ...


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

As shown in equation $(6)$ from this answer, $$ \prod_{k=1}^h\frac{c^kx-1}{c^k-1}=\sum_{p=0}^h\frac{\displaystyle c^{p(p+1)/2}}{\displaystyle\prod_{k=1}^p(c^k-1)\prod_{k=1}^{h-p}(1-c^k)}x^p\tag{1} $$ Plugging $x=1$ into $(1)$, we get that $$ \sum_{p=0}^h\frac{\displaystyle c^{p(p+1)/2}}{\displaystyle\prod_{k=1}^p(c^k-1)\prod_{k=1}^{h-p}(1-c^k)}=1\tag{2} $$ ...


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


0

Edited Since $\log(c)<0$ we have $x=1/o(1/\log(n))$ For fixed const $c$ assuming $x$ is integer this sum is $\frac{c^x-c^{n}}{1-c}$. Since $c^n=o(1/n)$ the condition for $x$ remains the same.


0

Hint: Write $20^{50}=e^{50\ln 20}$


1

To define the complex numbers as a field, one only needs the definition of them as $a+bi$ where $i^2 = -1$. The definition of a complex exponential is not necessary to construct the complex numbers. As far as the definition of the complex exponential, this is usually done by extending the definition of the exponential function as the power series $$ e^z = ...


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 ...


1

Isolate $ 1.00531^{\large t}$ on the left side of the equation. Take the $\ln$ of each side of that equation. And use the fact that $\ln a^b = b \ln a$. Solve for $t$ as you would any first degree polynomial.


0

Hint: $$8.513 \cdot 1.00531^t=10 \iff 1.00531^t=\frac{10}{8.513} \iff \ln[1.00531^t]=\ln\left[\frac{10}{8.513}\right] $$ $$\iff t\ln[1.00531]=\ln\left[\frac{10}{8.513}\right] \iff t= \quad?$$


1

Hint: Use the fact that $\log(ab^c)=\log a + c\log b$


0

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1

If you differentiate $f(x)=e^{g(x)}=e^{xln(x)}$ you will see, that $g'(x)$ is not a constant. Thus $f(x)=x^x$ is not an exponential function.



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