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18

One indirect approach: Write $$f(a,b) = \int_0^\infty \frac{e^{ax}-e^{bx}}{(1+e^{ax})(1+e^{bx})}dx$$ Then changing variables by $x = ku$, for some positive $k$, $$f(a,b) = k \int_0^\infty \frac{e^{kau}-e^{kbu}}{(1+e^{kau})(1+e^{kbu})}du = k\, f(ka,kb) $$ The only answer which obeys the relation $$f(a,b) = k \,f(ka,kb)$$ is Option E.


11

Ultimately, it suffices to evaluate the limit $$ \lim_{n \to \infty} \frac{g(n)}{f(n)} $$ if it's $\infty$, then $g$ grows faster. If it's $0$, then $f$ grows faster. If it's anything else, then they grow the same (up to some constant), i.e. $f = \Theta(g)$. As far as intuition goes, it helps to rewrite things, or sometimes to divide/multiply both sides ...


10

The integral being considered is, and is evaluated as, the following. \begin{align} I &= \int_{0}^{\infty} \frac{e^{ax}-e^{bx}}{(1+e^{ax})(1+e^{bx})}dx \\ &= \int_{0}^{\infty} \frac{dx}{1 + e^{bx}} - \int_{0}^{\infty} \frac{dx}{1 + e^{ax}} \\ &= \left( \frac{1}{b} - \frac{1}{a} \right) \, \int_{1}^{\infty} \frac{dt}{t(1+t)} \mbox{ where $t = ...


5

This is simply false. $\sin$ is locally square integrable, i.e. $\sin \in L^2([a,b])$ for all $a<b\in\mathbb R$. This can also be written as $\sin \in L^2_{\text{loc}}(\mathbb R)$. To see that estimate $$\int_a^b \sin^2 x \ \mathrm dx \le \int_a^b 1 \ \mathrm dx = (b-a) < \infty$$ Thus $\sin \in L^2([a,b])$ for any $a<b\in\mathbb R$, so we are done. ...


5

I do not believe there is a name for your specific inequality (which I rewrite as following): $$ |x|^p+|y|^p\le \big(|x|+|y|\big)^p, \ p \ge1 \iff \boxed{\ \ \left(|x|^p +|y|^p\right)^{\frac{1}{p}} \le |x|+|y|, \quad p \ge 1 \ \ } $$ However, it can be viewed as a special case of multiple more general statements, such as Jensen, AMGM, Hölder, and probably ...


4

Yes, you can apply the formula of the radius of convergence. Your solution is essentially correct. Here is a more direct way to present it. Given the series $$ S_1(x)= \sum^{+\infty}_{n=0}a_n(x-x_0)^n $$ its radius of convergence $\rho_1$ is $$\frac{1}{\rho_1} = \limsup_{n \to \infty} \sqrt[n]{\vert a_n \vert}$$ Given the series $$ S_2(x)= ...


4

It isn't true that "Peano's axioms are all we need". Landau's beautiful account involves some set theory (the Dedekind cuts are sets or pairs of sets of rational numbers). There are many elegant ways of getting to the real numbers from the integers, but they all need a bit more than Peano's postulates. One of my favourites is this paper by Behrend, which ...


3

Tale $a = 1$ and let $b\to 0^+.$ In the limit you get $$\int_0^\infty\frac{e^x-1}{(1+e^x)2}\,dx.$$ That integral equals $\infty$ because the integrand has a positive limit at $\infty.$ The only answer that fits this phenomenon is E.


3

Consider the sequence: $$1-\frac{1}{2}-\frac{1}{2}+\frac{1}{\sqrt[3]2}-\frac{1}{2\sqrt[3]2}-\frac{1}{2\sqrt[3]2}+\frac{1}{\sqrt[3]3}-\frac{1}{2\sqrt[3]3}-\frac{1}{2\sqrt[3]3}+\frac{1}{\sqrt[3]4}-\frac{1}{2\sqrt[3]4}-\frac{1}{2\sqrt[3]4}+\ldots$$ It clearly converges to $0$, but the sequence of its cubes does not (by combining terms you can see that the ...


3

I think this should be called an special case of Minkowski's Inequality. 1 - For finite sequences, the Minkowski Inequality states that $$ \left(\sum_{k=1}^n |x_k + y_k |^p \right) ^{1/p} \le \left(\sum_{k=1}^n |x_k|^p \right)^{1/p} + \left(\sum_{k=1}^n |y_k|^p\right)^{1/p} $$ Your inequality then follows if $x_1 = x, x_i = 0, \; i \ge 2$, and $y_2 = ...


3

The local minima is at $x=\frac{3\pi}{2}$ This is very obvious from the graph of $f(x)=\sin{x}$ On a second look at the graph below, I believe $x=\frac{\pi}{4}$ is also a local minimum. This is because it is lesser than all other values within its locality. Thus we have two local minima: $x=\frac{\pi}{4}, \frac{3\pi}{2}$


3

Correct for the first two. For the third you need the product rule, which says that $\frac{\partial}{\partial x}(uv)=u_xv+uv_x$ if $u,v$ are functions of $x$. Se we have: $$2\frac{\partial}{\partial x}(uu_x)-2u_x^2$$ $$=2(u_x\cdot u_x+u\cdot u_{xx})-2u_x^2$$ $$=2uu_{xx}$$


2

Let us prove that $2uu_{xx} = 2\frac{\partial }{\partial x}\big ( uu_{x} \big ) -2u_{x}^{2}$. First, we elaborate the first term of the right hand side : $$ 2\,\frac{\partial }{\partial x}\big ( uu_{x} \big ) = 2 \big( u_x \cdot u_x + u \cdot u_{xx}\big) = \boxed{\ 2\left( u_x^2 + u u_{xx}\right) \ } $$ Second, we subtract $2u_{x}^{2}$ from the ...


2

Solution without trigonometric functions. Given $$ \int \frac{dx}{ \sqrt{ a^2 - x^2 }^3 }. $$ Write this as $$ \int \frac{dx}{ \sqrt{ a^2 - x^2 }^3 } = \frac{1}{a^2} \int \frac{a^2 - x^2 + x^2}{ \sqrt{ a^2 - x^2 }^3 } dx $$ or $$ \int \frac{dx}{ \sqrt{ a^2 - x^2 }^3 } = \frac{1}{a^2} \int \frac{dx}{ \sqrt{ a^2 - x^2 } } + \color{blue}{ ...


2

In my opinion, the "calculus side" of functional analysis is... linear algebra. Functional analysis deals with vector spaces endowed with a topology (normed spaces, topological vector spaces), and the most elementary example of such spaces are finite-dimensional vector spaces. Linear operators on finite-dimensional spaces are represented by matrices, ...


2

Lef $f:\mathbb{R}^n\longrightarrow \mathbb{R}^n$ a function such that $df$ exists. Let $J_f$ the jacobin matrix associate to $f$. Let now $\phi:(\mathbb{R}^n, y_1,...,y_n)\longrightarrow (\mathbb{R}^n,x_1...,x_n)$ an orthogonal change of coordinates. So $x_i=\phi_i(x_1,...,x_n)$, and define $J_\phi$ matrix associate to $\phi$. Set $g:\phi\circ f\circ ...


2

Here is what they mean. $f$ is continuous at $a \in \mathbb{R}$ if and only if for any sequence $x_1, x_2, x_3, \ldots,$, if $x_n$ converges to $a$, then $f(x_n)$ converges to $f(a)$. When you say $x$ preserves $y$, that means, "$x$ does not change $y$", or "$x$ keeps $y$ the same". In mathematics if you say that a function $f$ preserves some property $P$, ...


2

I think they simply mean that $f$ is continuous (at $x_0$) iff $$ \lim\limits_{x\to x_0} f(x) = f(\lim\limits_{x\to x_0} x) = f(x_0). $$ Using sequences then for any sequence $(x_n)_{n\in\mathbb{N}}$ where $x_n\to x_0$ as $n\to\infty$, we have that $$ \lim\limits_{n\to\infty} f(x_n) = f(\lim\limits_{n\to\infty} x_n) = f(x_0). $$ By the word 'preserves', ...


2

There are interesting elements in your proof. However, the way you state it can be improved: You mention Cauchy criteria, but what you use is not Cauchy criteria. You suppose that $\lim x_n$ is not equal to $2$. You cannot say $|x_n-2|>\epsilon$ without mentionning of what $n$ your're speaking. You should state that you have a subsequence ...


2

To answer your question: yes, it still can be beneficial to read old textbooks or publications, especially from a history of math point of view it is even necessary to read the original works. You get an idea on how the original ideas have been developed. Very interesting - no question about it. However, you have to take into account that these old books ...


2

$$ \cos\left(A+B\right) = \cos(A)\cos(B) - \sin(A)\sin(B)$$ $$\Downarrow$$ $$ \dfrac{\cos\left(\frac{\pi}{2}+h\right)-\cos\left(\frac{\pi}{2}\right)}{h} \ = \ \dfrac{\cos\left(\frac{\pi}{2}\right)\cos(h)-\sin\left(\frac{\pi}{2}\right)\sin(h)}{h} \ = \ \dfrac{-\sin(h)}{h}$$


2

A cosine wave is just a sine wave shifted to the left by $\frac{\pi}{2}$, so use the following trig identity: $$\cos(\theta + \frac{\pi}{2}) = \sin(\theta + \frac{\pi}{2} + \frac{\pi}{2}) = \sin(\theta + \pi) = -\sin(\theta)$$


2

Your solution seems to be reasonable, however I believe that there is an easier way to prove the statement. Note that the second series is, actually, a derivative of the first. Indeed, denote $F(x) = \sum^{+\infty}_{n=0}a_n(x-x_0)^n$ and $G(x) = \sum^{+\infty}_{n=0}(n+1) a_{n+1}(x-x_0)^n$. Take derivative of $F$: $$ F'(x) = ...


1

Regarding the last question on whether or not one can get a formula, the Cauchy--Hadamard theorem says that the radius of convergence $R$ of $\sum_{n=0}^\infty a_n (x-x_0)^n$ satisfies $$\frac{1}{R} = \limsup_{n \to \infty} \lvert a_n \rvert^{1/n},$$ where the possibilities $0$ and $\infty$ are treated in the natural way.


1

$$x \to 0 , cos x \to 1 \\cos^m(ax) \sim 1-\frac{m*(ax)^2}{2} $$ my proof for this :base on combining tailor expansion of cos x and this $x \to 0 ,so,(1+x)^m \sim 1+mx$ example for that formula $$\lim_{x \to 0}\frac{cos^3(5x)-\sqrt[3]{cos(3x)}}{1-\sqrt{cosx}}=\\\lim_{x \to ...


1

$$\lim_{n\to +\infty}\sum_{r=1}^{n}\frac{r^2}{n^3}=\lim_{n\to +\infty}\frac{1}{n}\sum_{r=1}^{n}\left(\frac{r}{n}\right)^2 = \int_{0}^{1}u^2\,du = \frac{1}{3}$$ by a Riemann sum argument, but neither $\lim_{x\to +\infty}(\sin x)^x$ or $\lim_{x\to +\infty}|\sin x\,|^{x}$ exist, as pointed in the comments.


1

So you have $X^\prime=T.X$. Therefore $$g(X^\prime)=T.f[T^t.X^\prime]$$ You shoudn't forget to multiply $f[T^t.X^\prime]$ on the left by $T$ to move the result of $f$ in the "new" coordinate system. Then you have using chain rule and the fact that the derivative of a linear map is itself $$Dg = T.Df.T^t$$ You can then apply the trace. Knowing that ...


1

Noting $$\lim_{a\to\infty} f(a,b) = \lim_{a\to\infty}\int_0^\infty \frac{e^{ax}-e^{bx}}{(1+e^{ax})(1+e^{bx})}dx=\int_0^\infty \frac{1}{1+e^{bx}}dx=\frac1b\ln 2,$$ we have to choose (E).


1

We know that $\sin \theta=x/a$. From the identity $(\sin \theta)^2+(\cos\theta)^2=1$, we find $\cos\theta=\sqrt{1-(x/a)^2}$ (assuming $\theta$ is in the first quarter). From this you can find $\tan\theta=\sin \theta/\cos \theta$, which agrees with what the book says.


1

Ultimately you want to re-express your answer $\frac{\tan\theta}{a^2}$ in terms of $\sin\theta$ and then convert back using the substitution you already did. Observe that $\tan\theta = \frac{\sin\theta}{\cos\theta}$ and that $\cos\theta = \sqrt{1 - \sin^2\theta}$. Then $$ \tan\theta \;\; =\;\; \frac{\sin\theta}{\cos\theta} \;\; =\;\; ...



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