# Tag Info

1

$$e^x=1+\frac x{1!}+\frac {x^2}{2!}+\cdots$$ For $x\to0,$ $$e^x=1+x+O(x^2)$$

1

The large discrepancy between the computed data and the experimental data is due to the supposed relationship $y=\alpha e^{\beta x}$ which is not convenient. You cannot correctly fit this function with your data. This is obvious when drawing $\ln(y)$ as a function of $x$. The curve is far from a straight line $\ln(y)=a+\beta x$ where $a=\ln(\alpha)$ as it is ...

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I don't think going from step (2) to step (3) is obvious. You can do it with a trick. I'm renaming the constants so that they're not confused with yours. For $\alpha > 0$ and $\beta > 0$, define: $$F(\beta)=\int_{0}^{\infty}e^{-\alpha x^{2}}\cos(\beta x)\,dx.$$ Notice that F'(\beta)=-\int_{0}^{\infty}e^{-\alpha x^{2}}x\sin(\beta x)\,dx ... 2 Try taking the indefinite integral of both sides instead: \begin{align*} \frac{ds}{s} &= \mu \, dt \\ \int \frac{1}{s} \, ds &= \int \mu \, dt \\ \ln|s| &= \mu t + C &\text{for some constant C \in \mathbb R}\\ e^{\ln|s|} &= e^{\mu t + C}\\ |s| &= e^{\mu t} \cdot e^C \\ s &= \underbrace{(\pm e^C)}_{A}e^{\mu t} \\ s(t) &= ... 0 Okay, The professor told me that since the general Fourier Transform is: \mathfrak{F}[f(x)] = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}f(t)e^{i\xi x}dx=\sqrt{\frac{2}{\pi}}\int_{0}^{\infty}f(x)cos(\xi x)dx works when it is even. If you look at the form of (step 2). it can be tought of this way: the \xi in the above cosine transform can be \xi+b ... 0 \int_1^x \frac{1}t dt = 1 so e is the unique x. 2 Proposition: Let E=\{A\in M_n(\mathbb{C})|A\text{ has algebraic entries }\}. Then the exponential map is injective on E. Proof: Assume that e^A=e^B. Here A,B have algebraic entries ; according to "Wermuth, 2 remarks on matrix exponentials" (in free access) http://www.sciencedirect.com/science/article/pii/0024379589905545 AB=BA. Thus A,B are ... 0 To get a nicer form, use \left(\frac{3x+2}{3x-2}\right)^{2x} =\left(\frac{\left(1+\frac{2}{3x}\right)^x}{\left(1-\frac{2}{3x}\right)^x}\right)^2 $$to get the exponential limit twice in standard form. 2 The commutativity of A and B is used where the double summation becomes a single summation. Consider as an example the term in that sum with \ell =2: (A+B)^2 = A^2+AB+BA+B^2. 3 The binomial formula$$(A+B)^l=\sum_{k=0}^l{l\choose k}A^kB^{l-k}$$is valid if AB=BA. 6 The step that claims$$\sum_{m=0}^l{\frac{l!}{m!(l-m)!}A^{m}B^{l-m}}=(A+B)^l$$For example, we would usually say that$$(A+B)^2=A^2+AB+BA+B^2$$Only because$$AB=BA$$can this be simplified. 1 Yes. This is the famous Gaussian integral, which, by a simple change in variable, becomes the \Gamma function of argument \dfrac12. But this function also possesses a logarithmic form over the unit interval (0,1). In general, we have$$\int_0^\infty e^{-x^n}dx=\int_0^\infty\sqrt[n]x\cdot ...

3

Sure. Try, for example, with the change of variable $\displaystyle y = \frac{2}{\pi} \arctan(x); \quad x = \tan\left( \frac{\pi y}{2} \right)$. In this example, the $\arctan$ function is continuous, when $x \rightarrow \infty$, $y \rightarrow 1$, and when $x \rightarrow 0$, $y \rightarrow 0$. You can express $\displaystyle dx = \frac{\pi}{2} \sec^2\left( ... 0 Write $$\frac{3x+2}{3x-2}=\frac{3x-2+4}{3x-2}=1+\frac{4}{3x-2}$$ and then make the substitution$t=3x-2$, so$x=(t+2)/3$and your limit becomes $$\lim_{t\to\infty}\left(1+\frac{4}{t}\right)^{2\frac{t+2}{3}}= \left(\lim_{t\to\infty} \left(1+\frac{4}{t}\right)^t\cdot \lim_{t\to\infty}\left(1+\frac{4}{t}\right)^2 \right)^{2/3}$$ 0 We know that for this particular situation , where x->infinity and the indeterminate form is 1 ^Infinity, we cand use Euler's limit : (1+1/u(x))^u(x)=e . Our example needs a little more polishing . Therefore , by adding 1 and substract 1 from the base function , we arrive here (1 + 1/(3x-2)/4)^(2x) . Next up , we continue by building our exponent so that it ... 0 Hint: $$\lim_{x \to +\infty}\exp\left(2x\ln\left(\frac{3x+2}{3x-2}\right)\right)=\exp\left(2\lim_{t \to 0}\dfrac{\ln\left(\dfrac{3+2t}{3-2t}\right)}{t}\right)$$ 0 Since$X=e^{\log X}$, we can write $$\lim_{n \to \infty} \bigg(1+\frac{x}{n}\bigg )^n=e^{\log\lim_{n \to \infty} \bigg(1+\frac{x}{n}\bigg )^n} =e^{\lim_{n \to \infty} n\log \bigg(1+\frac{x}{n}\bigg )}$$ Notice that the exponent on$e$is an indeterminate form of type$\infty\cdot 0$, so rewrite the limit in the exponent as$\frac{\log ...

0

Let's say you define $e^x$ by the derivative $\frac{d}{dx} e^x=e^x$. Then, we try the same to what you have here. $\frac{d}{dx}(1+\frac{x}{n})^n=(1+\frac{x}{n})^{n-1}$. And, as n approaches infinity, n and n-1 essentially boils down to the same thing. Then, $\frac{d}{dx}(1+\frac{x}{n})^n=(1+\frac{x}{n})^n$.

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Months, days, years: too many time units! Let's say the bug has an average of $\alpha$ offspring per day from age $A$ days to age $B$ days. We can't describe the growth just by the number of bugs, we need the age distribution. Let's say $f(t)$ is the rate at which bugs are born at time $t$. The parents of these bugs were born between times $t-B$ and ...

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Let $t = 0$ represent your starting point with $N(0)$. Well, if $N = N_0 e^{kt}$ is the right model for their growth then we would need $N(0) = N_0 e^{0}$, thus $N_0$ is your starting number of bugs. Now, choose a unit for $t$, for instance seconds. Then calculate the number of seconds in a month, $m$. For any number $N_0$ of starting bugs we have ...

1

As @d.k.o.'s comment hints at, your expression is: $$\frac{1}{1 + \exp(b - a) + exp(c - a) + \exp(d - a)}$$ If the resulting $b - a$, $c - a$, $d - a$ are still (very) negative (small exponentials), a few terms of the geometric series should be accurate enough. If some turn out positive, call them $A$, $B$, $C$ in order of decreasing value: $$\frac{1}{1 + ... 3 This is a tautology since \log_b(x) is defined as the exponent m such that b^m = x. 1 What exponent a can you put on b so that b^a=x? \log_b(x) is just the name for an a that makes b^a be equal to x. So by definition b^{\log_b(x)}=x. 2 By definition, if A^x=B, then x=\log_AB. Now, replace the value of x from the latter identity into the former. 0 The inverse of a function is found by substituting y for x and vice versa. So, we alter our equation to look like this: x = -2 * 4^{2(y-3)} - 1 and solve for y. x + 1 = -2 * 4^{2(y-3)} \frac{x + 1}{-2} = 4^{2(y-3)} log_{4}(\frac{x + 1}{-2}) = 2(y-3) \frac{log_{4}(\frac{x + 1}{-2})}{2} = y - 3 y = \frac{log_{4}(\frac{x + 1}{-2})}{2} + 3 ... 1 Hint Try to express x with y and you find$$y=f(x)=-2\cdot4^{2(x-3)}-1\iff x=\frac12\log_4\left(-\frac{y+1}2\right)+3$$0 You have for simple interest at a fixed interest rate per time period R:$$I=\sum_{i=1}^{T}PR=PRT$$Where I is the total interest after T time periods. Therefore your other formula should read:$$P(T)=P(0)(1+RT)$$Where P(T) is the principle after T time periods. 0 (To state my solution more rigorously and clearly, I put my comment into an answer.) First, it is easy to calculate:$$ f'(x) = e^x (x^2+5x+1) - 2x -2 \\ f''(x) = (x+1)(x+6)e^x-2 \\ f'''(x)=e^x(x^2+9x+13) $$Second, we will check how many real roots f''(x)=0 has: Case -\infty<x \leq -1: Calculate f'''(x)=0, we get the stationary ... 0 You can also use the fact that the function is differentiable (and continuous) at all R, to find a contradiction: Suppose that 0 <a <b with a, b, roots of f(x), with 0, a, b, consecutive roots, then there exists c_1\in[0,a] and c_2 \in[a, b] such that f(x) reaches maximum or minimum, as the function is twice differentiable we ... 1 e(n) is an approximation to the Euler's number e=1 + \frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!} + \ldots. You have e(n) = 1 + \frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!} + \ldots + \frac{1}{n!} and e(n-1) = 1 + \frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!} + \ldots + \frac{1}{(n-1)!}, therefore,$$e(n) - e(n-1) = 1 + \frac{1}{1!} + \frac{1}{2!} + ...

1

e(1) = 1 + 1/1! = 1 + 1 = 2 e(2) = 1 + 1/1! + 1/2! = 1 + 1 + 0.5 = 2.5 e(3) = 1 + 1/1! + 1/2! + 1/3! = 1 + 1 + 0.5 + 0.167 = 2.667 e(4) = 1 + 1/1! + 1/2! + 1/3! + 1/4! = 1 + 1 + 0.5 + 0.167 + 0.042 = 2.709 For N = 2, delta = e(2) - e(1), or 0.5 For N = 3, delta = e(3) - e(2), or 0.167 The above have accumulated rounding errors, but the idea is there. ...

1

your e(n) series is the Tylor series for $e^x$ evaluated at $x=1$, so it will converge to $e$. As you keep adding terms, the error between your current sum and $e$ will decrease (since this series converges), and so will the differences between a term and the previous one. $e=2.718281828...$ 'the delta, e(n), and e(n-1) variable must doubles' ok 'if e(n) = ...

0

I bet, you have to find any two positive numbers a and bsuch that f(a)<0 and f(b)>0. Thus, it will prove that there is one positive root that is between a and b. Same logic applies for the negative root.

3

How about $f(x) = 100 + e^{-x}$? It has a limit of $100$ as $x\to\infty$ (by the way, it is better to speak of limits as $x$ approaches infinity, rather than being equal to it.

0

We have that $e>0$. Then for every $n\in \Bbb N$ we have that $e^n>0$.Because $\sqrt[m] .$ is increasing then $\sqrt [m] {e^n}>0$. Thus $$e^q>0$$ for every $q\in \Bbb Q$. Now $\Bbb Q$ is dense in $\Bbb R$ thus $e^x>0$ for every $x\in \Bbb R$.

2

Every definition of $\exp(x)$ leads to the property, $\exp(a+b)= \exp(a)\exp(b)$. Furthermore every definition also tells us that $\exp$ is defined on the entire real line and takes real values when given a real number as an argument. So we will consider the exponential function evaluated at $x$ and notice that$x=x/2+x/2$. $$\exp(x) = \exp(x/2)\exp(x/2) ... 3 How do you define e^x? The most common way to define it is$$e^x = \sum_{n = 0}^{\infty} \frac{x^n}{n!}$$If you take this definition, then if x > 0, e^x is the sum of an infinite number of positive terms. One should show that the series converges, but if you take that for granted, then of course if converges to a positive number. If x < ... 1 The usual elementery proof that e is irrational works also for e^2. Suppose that e^2 is irrational. We know that$$ e^2 = \sum_{k=0}^\infty \frac{2^k}{k!}. Let q\ge2 be a positive integer, and let Q=2^q. If q is sufficiently large then \dfrac{Q!}{2^{Q-q+1}}\cdot e^2 is an integer. (Notice that the exponent of 2 in the prime ... 1 Answer: yes, it is the mean of the sample data. (I leave the single data case below as it is illustrative of the technique.) Let x be a single observed sample, f the likelihood function for an exponentially distributed random variable with parameter 1/\lambda. Maximising the log-likelihood: \begin{align} \ell(\lambda) := \log(\ f(x)) &= - ... 1 Try to write it out in words. The rate of change in population is the population we have minus the loss ratio of that population (of course, we could have other factors, but that is what we are working with here), so we have:\dfrac{dP}{dt} = P - \alpha P = P(1 - \alpha)$$Now how we can we find the loss ratio \alpha of the population per year? The ... 0$$\begin{align*} \rm N=b\cdot g^{\rm\displaystyle t}&\iff g^{\rm\displaystyle t}=\rm\dfrac Nb\\ &\iff\color{grey}{\boxed{\color{black}{\rm\, t=\log_{\rm\displaystyle g}\left(\rm \dfrac Nb\right)}}} \end{align*}$$where \rm\log_{\rm\displaystyle g} is the logarithmic function to the base \rm g which is basically the function that tells us for ... 0 We want to solve for y, after switching x and y. y=-3(3^{-x})+9 becomes x=-3(3^{-y})+9\\ \implies x-9=-3(3^{-y})\\ \implies \dfrac{x-9}{-3}=3^{-y}\\ \implies -\dfrac{1}{3}x+3=3^{-y}\\ \implies \ln\left(-\dfrac{1}{3}x+3\right)=\ln (3^{-y})\\ \implies \ln\left(-\dfrac{1}{3}x+3\right)=-y\cdot \ln 3\\ \implies -\dfrac{ ... 1 Divide both sides by -\ln 3 and you are there 1 Hints for you to justify:$$n\log\left(1+\frac xn\right)=\log\left(1+\frac xn\right)^n\xrightarrow[n\to\infty]{}\log e^x=x$$0 Using the limit definition of e^x$$e^x = \lim_{n\to\infty}\left(1+\frac{x}{n}\right)^n$$12 @Robert Israel was very close. Assume e^2=p/q. Rewrite as qe-pe^{-1}=0, or$$2q+{q-p\over2!}+{q+p\over3!}+{q-p\over4!}+\cdots=0$$This isn't exactly in the form of the expansion in the first paragraph of the question, but the numerators are bounded in absolute value, which is more than enough to make the proof go through. 0 xyz's method has the right idea. If you consider the equation$$a^x \ge x^a$$we can rewrite this as the equivalent$$\ln(a)/a \ge \ln(x)/x$$by using the properities of the logarithm and the positivity of x and a. Thus if you find the x that maximizes \ln(x)/x, that is the a you are looking for. 0 I would do it this way: Let f(x)=\dfrac{x}{\log x} for some x>0. Hence, f^\prime(x)=\dfrac{\log x-1}{\log^2 x}. I can conclude that f(x)\geq e \forall\, x>0. Therefore,$$f(x)=\dfrac{x}{\log x}\geq e,$$so,$$x\geq \log x^e,$$hence,$$e^x\geq x^e.$$Finally, \exists a\,>0 where a=e and a^x\geq x^a 0 How have you find \displaystyle \ln e? Applying logarithm wrt e on \displaystyle a^x=b^{1-x} we get$$ x\ln a=(1-x)\ln b\implies x(\ln a+\ln b)=\ln b

0

$\large \lim_{t\to \infty}\dfrac{t^m}{e^{\alpha t}}$ since the limit is in the inderteminate form $\frac{\infty}{\infty}$ you can use l'hopital \$\lim_{t\to \infty}\dfrac{t^m}{e^{\alpha t}}=\frac{m}{\alpha}\lim_{t\to \infty}\dfrac{t^{m-1}}{e^{\alpha t}}=\frac{m(m-1)}{\alpha^2}\lim_{t\to \infty}\dfrac{t^{m-2}}{e^{\alpha ...

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