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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 $A$


5

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 large $n$, $e^{\sqrt n} - e^{\sqrt{n-1}} \approx \left.\frac{d}{dx}[e^\sqrt x]\right|_{x=n} = \frac{e^\sqrt n}{2\sqrt n}$


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 function $g(x)$ 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

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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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