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1 vote

How can I argue that $f(x) = (2-x)^3-x+\frac{3}{2}$ is decreasing with a polynomial?

$$f'(x)=-3x^2+12x-13 = -3(x^2-4x+4)-1 = -3(x-2)^2-1<0 \text{ for all } x\in\mathbb R$$
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Simplifying expression with definite integral

Take $$f(t)=\frac{sin(t)}{1+cos(t)} +1$$ $$=>f(-t)=-\frac{sin(t)}{1+cos(t)}+1$$ $$=>f(t)+f(-t)=2$$ Now the expression becomes $$\frac{d}{dx}\left(\int_{-x^2}^{x^3}f(t)dt+\int_{x^3}^{x^2} (f(t)-2)...
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1 vote

Is it hard to evaluate $\int_{0}^{\infty} \frac{\ln \left(1-x+x^{2}\right)}{1+x^{2}} d x$?

Utilize still $I’(a)$ to integrate \begin{aligned} &\int_{0}^{\infty} \frac{\ln \left(1-x+x^{2}\right)}{1+x^{2}} d x= I\left(-\frac{\pi}{6}\right)\\ &={I(0)}+\int_{0}^{-\frac{\pi}{6}} \cot a\...
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Is there a closed form for $\int_{0}^{\frac{\pi}{2}} \frac{\sin x}{(\sin x+\cos x)^{2 n+1}} d x$?

Something which could be interesting $$I_n=\frac 1 {2^n} \sum_{k=0}^{n-1}\binom{n-1}{k}\frac{1}{2 k+1}=2^{-n} \, \,_2F_1\left(\frac{1}{2},1-n;\frac{3}{2};-1\right)$$ generates the unknown (?) sequence ...
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2 votes

Computing $\int_{0}^{\pi} \ln (\sin x+2) d x$ and $\int_{0}^{\pi} \ln (2-\sin x) d x$

$$I(a)=\int_{0}^{\pi} \log (a\sin (x)+2)\, dx$$ $$I'(a)=\int_{0}^{\pi} \frac{\sin (x)}{a \sin (x)+2}\, dx=\frac \pi a-\frac{2 \pi }{a \sqrt{4-a^2}}+\frac{4 \tan ^{-1}\left(\frac{a}{\sqrt{4-a^2}}\right)...
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3 votes

How to calculate the following expectation?

$$ \lambda\sum_{k=0}^{m}\binom{m}{k}k\lambda ^{k-1}\mu ^{m-k} = \lambda\sum_{k=0}^{m}\binom{m}{k}\frac{\partial}{\partial \lambda}\lambda ^{k}\mu ^{m-k} = \lambda\frac{\partial}{\partial \lambda}\...
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2 votes

How to calculate the following expectation?

$$\sum_{k=1}^{m}\binom{m}{k}k\lambda^k\mu^{m-k}=\sum_{k=1}^{m}m\binom{m-1}{k-1}\lambda^k\mu^{m-k}=m\lambda\sum_{k=0}^{m-1}\binom{m-1}{k}\lambda^k\mu^{(m-1)-k}=m\lambda(\lambda+\mu)^{m-1}.$$
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1 vote
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Question on the natural logarithm laws

Yes. Just like the comment to your question, if we know two objects are equal, then the "reverse case" is also true. Here is a more rigorous proof for fun: "Let $n \in \mathbb{R}$, $a \...
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0 votes

Spivak, Ch. 20, Problem *4(i): Write down a sum which equals $\sin{1}$ with an error of less than $10^{-10^{10}}$.

Since $\sin1=\sum_{k=0}^\infty(-1)^k\!/(2k+1)!$, the error by truncating to $n$ terms is less in magnitude than $1/(2n+1)!$. Therefore, to reduce the error below $10^{-10^{10}}$, it is enough to ...
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5 votes
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Continuity of $f$ doesn't imply differentiability

You are right: from the fact that$$\lim_{x\to a}\frac{f(x)-f(a)}{x-a}(x-a)=0,$$you cannot deduce that$$\lim_{x\to a}\frac{f(x)-f(a)}{x-a}\cdot\lim_{x\to a}(x-a)=0.\tag1\label a$$The equality \eqref{a} ...
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1 vote

Is it hard to evaluate $\int_{0}^{\infty} \frac{\ln \left(1-x+x^{2}\right)}{1+x^{2}} d x$?

Let $$x^2-x+1=(x-a)(x-b)\qquad \text{where} \qquad a=\frac{1+i \sqrt{3}}{2} \quad \text{and} \quad b=\frac{1-i \sqrt{3}}{2}$$ and $$\frac 1 {x^2+1}=\frac 1{(x-i)(x+i)}=\frac i 2\left(\frac{1}{x+i}-\...
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0 votes

Spivak, Ch. 20, Problem *4(i): Write down a sum which equals $\sin{1}$ with an error of less than $10^{-10^{10}}$.

$\sin 1 = \sum_{n=0}^\infty (-1)^{2n+1} {1 \over (2n+1)!}$ which is alternating so $|\sin 1 - \sum_{n=0}^N (-1)^{2n+1} {1 \over (2n+1)!} | \le {1 \over (2N+2)!} $. You want $(2N+2)! > 10^{10^{10}}$ ...
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1 vote

Question on the natural logarithm laws

Alternative approach: Assume that $\log(x) = r \implies e^r = x.$ Then $\displaystyle e^{(3r)} = \left[e^r\right]^3 = x^3.$ Thus, $3r = \log(x^3).$ Thus $3 \times \log(x) = \log(x^3).$
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Is it hard to evaluate $\int_{0}^{\infty} \frac{\ln \left(1-x+x^{2}\right)}{1+x^{2}} d x$?

Define: $$x=\frac{1+t}{1-t}$$ The integral goes to $$\begin{align} I&=\int_{-1}^1 \frac{\ln(1+3t^2)}{1+t^2}dt-2\int_{-1}^1 \frac{\ln(1-t)}{1+t^2}dt\\ \\ &=2\int_{0}^1 \frac{\ln(1+3t^2)}{1+t^2}...
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  • 5,536
0 votes

How To Measure Work Done Against Friction On A Non-Linear Path?

I am obtaining a slightly different result for the work done by the force of friction on the rolling ball. I'm assuming $c(t)=\left(x(t),y(t)\right):a\leq t \leq b$ is the parametric curve which ...
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1 vote

Showing the existence of $\int_0^1\sqrt{x}\ln (x)\, dx$ and calculating it

You could make it simpler using $x=t^2$ $$I=\int_0^1\sqrt{x}\log (x)\, dx=4\int_0^1 t^2\log(t) \,dt$$ and we know that $\lim_{t\to 0} \, t\log(t)=0$. Now, one integration by parts to obtain the result....
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1 vote

Question on the natural logarithm laws

Let $x \in \mathbb{R}^+$. Since $3 \cdot log(x) = log(x^3)$, it follows that $log(x^3) = 3 \cdot log(x)$.
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2 votes

How do we show that $(10^{10})!$ is larger than $10^{10^{10}}$?

Using algebra, you want to find when $$(a^a)! \geq a^{a^a}$$ To make life easier because of the huge numbers, when is $$f(a)=\log\Bigg[\frac{(a^a)!}{a^{a^a}}\Bigg] >0$$ By inspection or plotting, ...
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0 votes

Showing the existence of $\int_0^1\sqrt{x}\ln (x)\, dx$ and calculating it

One way to show just existence is as follows: Note that $\ln(0) = -\infty$ so we need to show that $$\lim_{c \to 0^+} \int_c^1 \sqrt{x}\ln(x)dx\;\;\text{exists}$$ We can use the fact that $\sqrt{x},\...
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6 votes

How do we show that $(10^{10})!$ is larger than $10^{10^{10}}$?

$10^{10^{10}}$ is the product of $10^{10}$ factors of $10$ while the first $\frac 12 \cdot 10^{10}$ factors of $(10^{10})!$ are larger than $10^2$ so have a product larger than $10^{10^{10}}$.
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5 votes

How do we show that $(10^{10})!$ is larger than $10^{10^{10}}$?

$$\left(10^{10}\right)! = 100!\cdot\left(\prod\limits_{i=101}^{10^{10}}i\right) \gt 100!\cdot\left(\prod\limits_{i=101}^{10^{10}}100\right) = 100!\cdot 100^{10^{10}-100} = 100!\cdot 10^{2\cdot10^{10} -...
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2 votes
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On the gradient of the squared Euclidean norm

It looks like he is taking the gradient of equation 1 with respect to $y$ and then taking the minimum: $$ \nabla_y\left(\nabla f(x)^T (y-x) + \frac{\mu}{2} \| y - x \|^2\right)= \nabla f(x) + \mu(y-x)=...
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1 vote

Showing the existence of $\int_0^1\sqrt{x}\ln (x)\, dx$ and calculating it

Then we get $$\int_0^1\sqrt{x}\log (x)\, dx=\int_{-\infty}^0\sqrt{e^z}z\, e^zdz=\int_{-\infty}^0e^{z/3}z\, dz$$ The last step should be $$I=\int_{-\infty}^0e^{\frac{3}2z}z\, dz$$ you can define: $z=-\...
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2 votes

Showing the existence of $\int_0^1\sqrt{x}\ln (x)\, dx$ and calculating it

By using this substitution, you are using the fact that the map $x \mapsto e^x$ is differentiable with a continuous derivative. But no, this is not the most efficient way. You may use integration by ...
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0 votes

find $\int_0^\infty \frac{|\cos (\pi x)|}{4x^2 - 1} dx$

We can also solve the integral via contour integration... $$\int_{0}^{\infty}{{|\cos(\pi x)|}\over{4x^2-1}}dx \rightarrow \oint_{C}{{|\cos(\pi z)|}\over{4z^2-1}}dz=\oint_{C}{{|\cos(\pi z)|}\over{(2z+1)...
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  • 151
0 votes

Proving that the maximum of two convex functions is also convex

Here is another way of seeing this. First notice we can rewrite the maximum function as follows: $$\max(x,y) \,=\, \frac{x+y+|x-y|}{2}.$$ So if $f,g$ are convex functions, using the fact that the ...
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1 vote
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Variational Calculus - Derivation of Lagrangian Equation

About the reason for imposing the condition that the added variation $\eta(x)$ should be continuous and differentiable: The way that Jacob Bernoulli solved the Brachistochrone problem also underlines ...
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1 vote

if $f(x)f(y) = y^h f(x/2) + x^k f(y/2)$ for all positive reals $x,y$, then $f$ must be identically zero

Partial Solution: First, note that $$f(0)^2 = 0^hf(0) + 0^kf(0) = 0.$$ The remainder of this answer assumes that $f(1) = 0$. Now with $x = y = 1$, $$0 = f(1/2)+f(1/2) \implies f(1/2) = 0.$$ Finally, ...
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1 vote
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I need some help with a specific probability question. It's about radioactive decay.

Since the particle can only decay once, the probability of decay in a particular interval is equal to the expected number of decays in that interval. Because of this, you can use linearity of ...
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uniform continuity on $(a, b]$ implies limit at $a^+$ exists and finite

Here's a way without Cauchy sequences. We prove first that $f$ is bounded on $(a,b]$. By uniform continuity, there is some $\eta \in (0,b-a)$ such that $|x-y|<\eta \implies |f(x)-f(y)|<1$. Since ...
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3 votes
Accepted

How can I solve the integral equation $\pi a^2 (f(x))^2 = \int_{0}^{f(x)/a} \sqrt{(f(x))^2 -a^2t^2} \ \mathrm dt $ for the function $f(x)$?

Your integral on the RHS has a closed form, which is relatively easy to find after making the substitution $t\mapsto t |f(x)|/a$ (I'll assume $a>0$). The resulting equation you obtain is $$\pi a^2f(...
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5 votes
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Explaining to a student what we mean by "let the variable $x$ be [some physical quantity]"

This question can definitely be interpreted differently by different people, so I can only answer from my perspective. Additionally, without talking with the student, it is difficult to address their ...
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3 votes

Explaining to a student what we mean by "let the variable $x$ be [some physical quantity]"

When thinking about and explaining these applications I find it useful to imagine the underlying set $S$ of all possible physical states of the system. Then "variables" like $x, v, m, t$ are ...
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0 votes

I need help finding the domain of $\left(\cot(5x + 3)(\cot5 + \cot3x) - (\cot3x)^{1/2} + 1\right)^{1/2}$

Well, notice first of all that the domain of $\sqrt{x}$ is given by: $$\left\{x\in\mathbb{R}\space:\space x\ge0\right\}\tag1$$ So we can see that the domain of $\sqrt{\cot\left(3x\right)}$ is given by:...
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-2 votes
Accepted

Solve the integral, perhaps by parts

This is the famous wifi integral
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7 votes
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Are there any non-constant differentiable functions $f : \mathbb R \to \mathbb R$ where each $t \in \mathbb R$ has $f(t)f(f'(t))=1$?

The property that every $t\in\mathbb{R}$ satisfies \begin{equation}\tag{1}\label{functional-identity} f(t)f(f'(t))=1 \end{equation} is very restrictive. In particular, we can deduce uniqueness of ...
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0 votes

Find $m$ if $f(x)=x^m\sin\frac{1}{x}$ is continuous and is not differentiable

Your approach in correct in all aspects except your considering fractional powers with even denominators as base. The original question does not imply any condition to the domain of the function, it ...
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2 votes

Can anyone explain this process of solving? (Differentiation)

If it gets too messy, you can always use logarithmic differentiation. Take the natural logarithm of both sides to get: $$\ln y = \ln((x+1)^3) - \ln(x^2)$$ $$\ln y = 3 \ln(x+1) - 2 \ln x$$ and so using ...
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1 vote

Can anyone explain this process of solving? (Differentiation)

Here is one way to solve it. Use the fact that the derivative of $(ax+b)^n=n(ax+b)^{n-1}(ax+b)'$. This just follows from the chain rule. Rewrite the equation as $y=(x+1)^3x^{-2}$. Then apply the ...
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0 votes

Can anyone explain this process of solving? (Differentiation)

Before we start, let's recall the first derivative of a quotient: Let: $$y(x)=\frac{u(x)}{v(x)}, v(x)\ne 0$$ then: $$y'(x)=\frac{u'(x)v(x)-u(x)v'(x)}{v^2(x)}$$ Apply for your case: $$y(x)=\frac{(x+1)^...
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-1 votes

Derivative of an integral with respect to a function

I just see this question posted 5 years ago because I got into a similar problem. I struggled for days but I think I got a reasonable answer that convince me. So your integration variable x is ...
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1 vote

How to compute $\lim\limits_{x\to 0} (-1)^{n+1} \frac{\int_0^x \frac{t^{2n+2}}{1+t^2}dt}{x^{2n+1}}$?

Too much advanced but this is just for your curiosity. Sonner or later, you will learns that $$\int \frac{t^{2n+2}}{1+t^2}\,dt=\frac{t^{2 n+1} }{2 n+1}\left(1-\, _2F_1\left(1,n+\frac{1}{2};n+\frac{...
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2 votes
Accepted

$f(x):[0,\infty)\to\mathbb R$ is twice differentiable such that $f(0)=0$ and $f''(x)<0$ $\forall x\in(0,\infty)$ and $f(1)=\alpha,\alpha>0.$

For an unbounded above counterexample, we can take: $$f(x)=e^{-a}+Cx-e^{-x-a},\,0\le x$$ For any constants $C>0$, $a>0$. Motivation: although $f’’(x)=-e^{-x-a}<0$, $f’’$ rapidly decays and $...
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1 vote

How to compute $\lim\limits_{x\to 0} (-1)^{n+1} \frac{\int_0^x \frac{t^{2n+2}}{1+t^2}dt}{x^{2n+1}}$?

You can expand ${1\over 1+t^2}$ as a geometric series as $x \rightarrow 0$. $$\lim_{x \rightarrow 0} {1\over x^{2n+1}}{\int_{0}^{x}{t^{2n+2} \over {1+t^2}}dt}\space =\space\lim_{x \rightarrow 0} {1\...
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1 vote

In Maple, how do we compute $\sin{1}$ with an error of less than $10^{-17}$ using this very manual way?

Just for your curiosity. There is an analytical solution to this problem. Writing $$\sin(x)=\sum_{n=0}^\infty \frac{(-1)^n}{(2 n)!}\left(x-\frac{\pi }{2}\right)^{2 n}=\sum_{n=0}^p \frac{(-1)^n}{(2 n)!}...
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2 votes
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In Maple, how do we compute $\sin{1}$ with an error of less than $10^{-17}$ using this very manual way?

Your wrote, 10^(17)*[1 - 1/2*Pi]^(2*n + 2) which is incorrect syntax for what you're trying to accomplish. In Maple group of terms is accomplished by using round-...
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3 votes

Evaluating $\int_0^\pi x\frac{\sin{\frac{x}{2}} - \cos{\frac{x}{2}}}{\sqrt{\sin{x}}} dx$

Evaluate \begin{align} I=& \int_0^{\pi/2} \frac{x(\sin x- \cos x)}{\sqrt{\sin{2x}}} dx =\int_0^{\pi/2} x\ d\bigg(- \tanh^{-1}\frac{\sqrt{2\tan x}}{1+\tan x}\bigg)\\ \overset{ibp}=&\int_0^{\pi/...
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2 votes

How to solve the following Fourier Transform, knowing the note below?

You could use several properties of the Fourier transform to reduce it to just a few standard transforms. If will assume $$\mathcal F(f)(\xi) = \int_{-\infty}^\infty f(x) e^{-i \xi x}\, dx.$$ ...
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3 votes
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Question about integration by parts for quotiens

The differentiation rule for quotients is $$\left(\frac{F(x)}{g(x)}\right)' = \frac{F'(x)g(x)-F(x)g'(x)}{g^2(x)}.$$ By rearranging the terms (we solve for the first term on the right hand side), this ...
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  • 1,685
0 votes

Asymptotic expansion as $r\to 0^+$ for $ (2re^r -3(e^r-1))^2 $ using small o

Take your initial expansions out a little further: $2re^r = 2r+2r^2+r^3+O(r^4)$ and $3(e^r-1) = 3r+3r^2/2+r^3/2+O(r^4)$; you now get $2re^r-3(e^r-1) = -r+r^2/2-r^3/2+O(r^4)=-r+r^2/2+O(r^3)=-r+O(r^2)$....
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