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11

The function $z \mapsto e^{-z}$ doesn't satisfy the conditions. Actually the limit $$\lim_{z\to \infty} e^{-z}$$ doesn't exist. (For exemple let $z$ go to infinity on $i\mathbb{R}$) Hence it is not a counterexample. The proposition is correct, by Liouville's theorem the function must by $1$ but then the limit is not $0$ so there are no such functions.


8

Let's look at a simpler (but otherwise analogous) example: $$ \frac{dy}{dx} = y. $$ The usual method of solution is to separate variables, integrate, and solve for $y$: $$ \frac{dy}{y} = dx,\qquad \ln|y| = x + C,\qquad y = Ae^{x},\quad A = \pm e^{C}. $$ (The choice of signs arises when dropping the absolute value from $y$.) Now, $e^{C}$ cannot be zero, but ...


7

Since $e^x$ is convex function (as can be checked with the second derivative), any tangent line will be less than the function. By taking the tangent at $x=0$, we get $e^x\geq 1+x$. Now, set $x=b^2-1$.


5

Hint: Look at $g(x) = (e^x-1)/x$ instead. Then $g$ equals a power series that converges on all of $\mathbb R,$ hence is $C^\infty.$ Show that $g$ is never $0.$ Therefore …


5

This can be written as $$\int_{-\pi}^{\pi}\frac{x(\cos x+i\sin x)dx}{1+\cos^2 x}$$ $$=\int_{-\pi}^{\pi}\frac{x\cos xdx}{1+\cos^2 x}+i\int_{-\pi}^{\pi}\frac{x\sin xdx}{1+\cos^2 x}$$ The first integral evaluates to $0$ (Odd function) Whereas the second can be written as $$2i\int_{0}^{\pi}\frac{x\sin xdx}{1+\cos^2 x} \space\space\text{(even function)}$$ ...


4

As you suggested, this is through a logarithm. We see that $$\frac{e^{b^2-1}}{b^2} \geq 1 \leftarrow e^{b^2-1} \geq b^2 \leftarrow \ln(e^{b^2-1}) = b^2-1 \geq \ln(b^2)$$ $$\leftarrow b^2-1 \geq 2\ln(b) \leftarrow b^2 \geq 2\ln(b) + 1.$$ The condition $b^2 \geq 2\ln(b) + 1$ holds for all $b > 0.$


2

$$\lim_{x\to \infty}\left(\frac{ax+1}{bx+2}\right)^x=\lim_{x\to \infty}\left(\frac ab\right)^x\left(\frac{1+\frac{1/a}{x}}{1+\frac{2/b}{x}}\right)^x \tag 1$$ Using the limit definition of the exponential function, we see that $$\lim_{x\to \infty}\left(\frac{1+\frac{1/a}{x}}{1+\frac{2/b}{x}}\right)^x=e^{1/a-2/b}$$ However, we have $$\lim_{x\to \infty} ...


2

The basic point, I think is this. A family of solutions of an analytic differential equation, given by an analytic function with a parameter, will be valid on any connected open set of values of the independent variable and the parameter, as long as you stay in the region of analyticity. So even if you assumed $C > 0$ in your derivation of the solution, ...


1

Hints/Ideas: $xe^{ix} = x\cos x + ix\sin x$, and you integrate on an interval symmetric around $0$. The function $x\mapsto \frac{x\cos x}{1+\cos^2 x}$ is odd, and the function $x\mapsto \frac{x\sin x}{1+\cos^2 x}$ is even. $$ \int_{-\pi}^\pi f(x) dx = i\int_{-\pi}^\pi dx\frac{x\sin x}{1+\cos^2 x} = 2i\int_{0}^\pi dx\frac{x\sin x}{1+\cos^2 x} $$ Now, ...


1

Note that $$\frac d{dx}e^{ax}=ae^{ax}$$ We also have that, for $n\in\mathbb N$ $$\frac{d^n}{dx^n}e^{ax}=a^ne^{ax}$$ or, one could assume that this works for $n\in\mathbb R$, and put into notations, $$D^ne^{ax}=a^ne^{ax}$$ This is the result we want to get, probably through the methods you are given. I will say the above result should be correct via ...


1

We need to assume that both $\cos a, \sin a$ are positive in order to evaluate this limit. This means that $a$ is in first quadrant. Assuming that this is so, we can simply assume that $\cos a = A, \sin a = B$ where $A, B$ are fixed positive numbers with $A^{2} + B^{2} = 1$ we can put $x = 2 + h$ so that as $x \to 2$ we have $h \to 0$. Then we have ...


1

Note that $\cos kz = \mathrm{Re}(\cos kz + i \sin kz) = \mathrm{Re}(e^{ikz})$. Thus: $$ \sum_{k \geq 1} e^{-tk} \cos kz = \sum_{k \geq 1} \mathrm{Re}(e^{-tk})\mathrm{Re}(e^{ikz}) = \sum_{k \geq 1} \mathrm{Re}(e^{k(-t + iz)}) = \mathrm{Re}\left(\sum_{k \geq 1} \left(e^{-t + iz}\right)^k\right) $$ which is a geometric series.


1

Equations with messy powers are usually dealt with better by taking logs of everything in sight. So instead of looking for $\lim_{x \to \infty} f(x)$, look for $\lim_{x \to \infty} log(f(x))$ Once you know that, you can easily find the original limit easily. It is easy to start. $$\begin{align}\log(f(x)) &= \log((\frac{ax + 1}{bx + 2})^x) \\ &= x ...


1

We assume that $y$ is always positive. Let $z$ be the logarithm of $y$. Which base? It does not matter. I will use the natural logarithm $\ln$, but others might use the logarithm to the base $10$. Note that if $y=KC^x$ then $z=\ln y=\ln K+(\ln C)x$. Suppose that we are given values $x_1,x_2,\dots, x_n$ of $x$, and the associated values $y_1,y_2,\dots,y_n$ ...



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