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33

Assume that $f(x)$ is a function such that $f'(x)=f(x)$ for all $x\in\Bbb{R}$. Consider the quotient $g(x)=f(x)/e^x$. We can differentiate $$g'(x)=\frac{f'(x)e^x-f(x)D e^x}{(e^x)^2}=\frac{f(x)e^x-f(x)e^x}{(e^x)^2}=0.$$ By the mean value theorem it follows that $g(x)$ is a constant. QED.

10

Like square root, log is treated as a "multi-valued function." It's just that when we restrict to the positive real line, we can make it single-valued. (Just as we do with square root in algebra 1.) So the first few things you derived are right: one of the logarithms of $-1$ is $\pi i$. The others all differ from it by multiples of $2\pi i$. So when you ...

8

Consider the equation $y'=y$. Our goal is to solve for the function $y=f(x)$. Roughly speaking $$\frac{dy}{dx}=y \implies \frac{dy}{y}=dx \implies \int\frac{dy}{y}=\int dx \implies\ln(y)=x+C \implies y=e^{x+C}=Ae^x$$ for some constant $A$

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The first identity is true. Pretty neat, right? However, in general for the complex extension of the logarithm $$\log(ab) \ne \log(a) + \log(b)$$ which clears up the other troubling identities you arrive at. The log function is constructed from the exponential function as follows: You want a function $\log(z)=u(x,y) + iv(x,y)$ such that $z = e^{u+iv} = e^... 4 The equation $$\frac{\mathrm{d}}{\mathrm{d}x} f(x) = f(x)$$ is a linear (thus Lipschitz continuous), first-order ordinary differential equation on$\mathbb{R}$. By the Picard-Lindelöf theorem, such an equation has a unique solution for any initial condition of the form $$f(0) = y_0$$ with$y_0 \in \mathbb{R}. In particular, for the condition $$f(... 4 This may not be an answer you are looking for, but its a nice one to consider. Consider y=\cos(ix)-i\sin(ix). You may find that:$$\frac{dy}{dx}=-i\sin(ix)-i^2\cos(ix)=\cos(ix)-i\sin(ix)$$Thus, y'=y is satisfied. Since y(0)=1, y'(0)=1, \dots, then by Taylor's theorem, we have e^x=\cos(ix)-i\sin(ix), or more commonly known as$$e^{ix}=\cos(... 4 We have \begin{align} e^{\sqrt{n}}-e^{\sqrt{n-1}} & = e^{\sqrt{n}}\left(1-e^{\sqrt{n-1}-\sqrt{n}}\right) \approx e^{\sqrt{n}}\left(1-\left(1+\sqrt{n-1}-\sqrt{n}\right)\right)\\ & = e^{\sqrt{n}} \left(\sqrt{n}-\sqrt{n-1}\right) = \dfrac{e^{\sqrt{n}}}{\sqrt{n}+\sqrt{n-1}} \approx \dfrac{e^{\sqrt{n}}}{2\sqrt{n}} \end{align} 4 For largen$,$e^{\sqrt n} - e^{\sqrt{n-1}} \approx \left.\frac{d}{dx}[e^\sqrt x]\right|_{x=n} = \frac{e^\sqrt n}{2\sqrt n}$3 Hint. One may recall that, as$u \to 0$, by the Taylor series expansion one has, $$e^u=1+u+\frac{u^2}2+O(u^3).$$ Can you take it from here? 3 Here is a slightly different variation to OPs example, followed by another one. Suppose$p$is an even function, i.e.$p(x)=p(-x)$and$q(x)q(-x)=1. Then \begin{align*} \int_{-a}^{a}\frac{p(x)}{1+q(x)}\,dx=\int_{0}^{a}p(x)\,dx \end{align*} A proof of the statement together with an application can be found in this answer. Note: This technique ... 2 Any functiong(x)$such that$g(c+a)+g(c-a)=k$for all$a$on the interval$(0,b)$with any function$f(x)$such that$f(c-a)=f(c+a)$for all$a$on the interval$(0,b)\$ will satisfy the equation $$\int_{c-b}^{c+b}{f(x)g(x)dx}=k*\int_{c}^{c+b}{f(x)dx}$$ because, using a trapezoidal Riemann sum after splitting the integrals into $$\int_{c-b}^{c}{f(x)g(x)dx}... 2 The fundamental fact is that, intuitively, sin(X) \approx X when X is small. Here you replace X with \frac{sin(x) + tan(x)}{2} and \frac{sin(x) - tan(x)}{2} and that's how you get rid of the outter sin. Formally, you can say that sin(x) \sim_{0} x meaning that sin(x) = \epsilon(x) x or equivalently x = \epsilon'(x)sin(x) where \epsilon(x)... 2 HINT: As e^y=\sum_{r=0}^\infty\dfrac{y^r}{r!}$$e^{ax}-e^x-x=1-1+x(a-1-1)+x^2\cdot\dfrac{a^2-1}2+\text{terms containing higher powers of }x$$1 If f_n(t)=\sin(nt)/n then f_n\to0 uniformly although f_n'(0)=1 for all n. 1 Applying once L'Hôpital yields$$\lim_{x\to0}\frac{a e^{ax}-e^x-1}{2x},$$which can be finite only if the numerator tends to 0, i.e. a=2. Another application, with this assumption, shows that the limit is finite iff a=2. 1 For small angles( \theta \rightarrow  0), we can say sin \theta=\theta..(i) For  \theta, \rightarrow  0, sin\theta \rightarrow0 tan\theta \rightarrow0 sin\theta + tan\theta \rightarrow0 Using ..(i), sin\frac{(sin\theta + tan\theta)}{2}=\frac{(sin\theta + tan\theta)}{2} 1 In the second problem, it is easier to use memorylessness. The distribution of X-1, given that X\gt 1, is the same as the unconditional distribution of X^3. Thus our conditional expectation is \int_0^\infty x^3e^{-x}\,dx. In the first problem, I would prefer to say the distribution of Y, given that Y\gt 3, is the same as the distribution of 3+Y^... 1 Put A = \left(\sqrt{t+1} + \sqrt{t-1}\right)^{\frac{x}{2}}, B = \left(\sqrt{t+1}-\sqrt{t-1}\right)^{\frac{x}{2}}, then apply the AM-GM inequality: A+B \ge 2\sqrt{AB} and note that AB = 2^{\frac{x}{2}}. Thus the left side \ge  the right side, and we have equality so A = B \implies t = 1 \implies x^2-5x+5 = 1 \implies (x-1)(x-4) = 0 \implies x = 1,4. 1 \newcommand{\angles}[1]{\left\langle\,{#1}\,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{\mathrm{i}} \... 1 The following two steps are not legal operations involving the logarithm function. 1) From \ln\frac{12}{1+e^r} to \frac{\ln12}{\ln{1+e^r}} 2) From \ln{1+e^r} to \ln1+\ln e^r You can only use the logarithm rules you have learnt in class: A) \ln (a\times b)=\ln a+\ln b B) \ln \frac{a}{b} = \ln a - \ln b C) \ln a^n = n\ln a D) \log_x a=\... 1 Logarithms do not "distribute" over multiplication and division like that. The two rules are this:$$\ln(AB) = \ln A + \ln B$$and$$\ln\left(\frac AB\right) = \ln A - \ln B$$It appears that you tried to do this:$$ \ln \frac{12}{1 +e^{-0.6(t-6)}} = \frac{\ln 12}{\ln 1 + \ln e^{-0.6(t-6)}}$$But that is not correct. The best we can do is this:$$ \...

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