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0

The cross-product $U\times V$ of vectors $U$ and $V$ can be written in coordinates as the sum $$dy\wedge dz(U,V)\textbf{ i }- dx \wedge dz(U,V)\textbf{ j }+ dx \wedge dy(U,V)\textbf{ k }$$ (note the sign). Since the length of the cross product $U\times V$ is the area of the parallelogram spanned by them, the formula you wrote is quite suggestive. Note that ...


0

This is more like the base $b$ notation for a number, but with the base varying per digit. For example: $$ \begin{align}\frac{17}{10} &= 1 + \frac{7}{10} \\&=1 + \frac{1}{2!} + \frac{1}{5} \\ &= 1+\frac{1}{2!} +\frac{1}{3!} + \frac{1}{30}\\ &=1+\frac{1}{2!}+\frac{1}{3!}+\frac{4}{5!} \end{align}$$ If $x$ is a real number, define the ...


0

The answer of D. Fischer is shorter and elegant. Let me nevertheless provide my own elementary answer: First note that $$ \sum_{n=1}^\infty \left(\frac{1}{(2n-1)^a}-\frac{1}{(2n)^a}\right)=\sum_{n=1}^\infty \left(\frac{(2n)^a-(2n-1)^a}{(2n-1)^a(2n)^a}\right)=\sum_{n=1}^\infty \frac{(1+\frac{1}{2n-1})^a-1}{(2n)^a} \\ ...


2

Having looked at your previous question, I see the issue here. The key point is that we are computing a partial derivative here. $\partial$ is not a number that gets canceled out. The derivative of $f(k) = k^{\frac12}$ is \begin{equation*} f'(k) = \frac12 k^{-\frac12}. \end{equation*}.


0

First note that \[ \frac{\partial}{\partial x}[x^{n}]=nx^{n-1} \] So here's how the $2$ appears \[ \frac{\partial}{\partial k}[k^{0.5}]= \frac{\partial}{\partial k}[k^{\frac{1}{2}}]=\frac{1}{2}k^{\frac{1}{2}-1} \] \[ = \frac{1}{2}k^{-\frac{1}{2}}= \frac{1}{2k^{\frac{1}{2}}}=\frac{1}{2\sqrt{k}} \]


2

The series is, for $\operatorname{Re} a > 0$, the Dirichlet $\eta$ function, $$\eta(a) = \left(1-2^{1-a}\right)\zeta(a).$$ Since $\eta$ is an entire function, by continuity, the limit is $$\eta(0) = -\zeta(0) = \frac{1}{2}.$$


0

$\quad\quad\quad\quad\quad\quad\quad\quad$ If we look at it like this, with the condition that the ratios of the lengths are the same, we can get the following equality. $$\frac{x-x_{min}}{x_{max}-x_{min}}=\frac{y-y_{min}}{y_{max}-y_{min}}$$ We can then rearrange for $y$ & substitute the other values.


2

The $2$ shouldn't be there. It's a mistake.


0

The formula you need is $$ valueY = lowerlimitY + (valueX - lowerlimitX)* \frac{upperLimitY - lowerlimitY}{upperLimitX - lowerlimitX} $$ There's a geometric reason for this: the point $(valueX, valueY)$ needs to lie on the straight line between the two points $(lowerLimitX,lowerLimitY)$ and $(upperLimitX,upperLimitY)$. That's what the formula above ...


0

You want linear interpolation. $m=\frac{y_{1}-y_{}}{x_{1}-x_{0}}$ then $y=y_{0}+m x_{0}$


5

Then you need the derivative of $$y=\frac{e^{2x}+e^{-2x}}{e^{2x}-e^{-2x}}$$ Multiply both numerator and denominator by $e^{2x}$; this gives you $$y=\frac{e^{4x}+1}{e^{4x}-1}$$ To make your life simpler, define $u=e^{4x}$ so $$y=\frac{u+1}{u-1}$$ and now apply $$\frac{dy}{dx}=\frac{dy}{du}\times\frac{du}{dx}$$ I am sure that you can take from here.


3

Using $\cos(5x)=16\cos^5(x)-20\cos^3(x)+5\cos(x)$ and assuming that $\cos(18^\circ)\ne0$, we get $$ 16\cos^4(18^\circ)-20\cos^2(18^\circ)+5=0 $$ therefore $$ \cos(18^\circ)=\sqrt{\frac{5+\sqrt5}8} $$ Then, use $\cos(3x)=4\cos^3(x)-3\cos(x)=(4\cos^2(x)-3)\cos(x)$ to get $$ \begin{align} \cos(54^\circ) &=\frac{-1+\sqrt{5}}2\sqrt{\frac{5+\sqrt5}8}\\ ...


2

$\cos27^\circ=\cos(72^\circ-45^\circ)=\cos72^\circ\cos45^\circ+\sin72^\circ\sin45^\circ$ where $\cos72^\circ=\frac{\sqrt5-1}4$ and $\cos45^\circ=\frac{\sqrt2}2$.


3

Using $\cos2A=2\cos^2A-1$ and as $0<27^\circ<90^\circ,$ $$\cos27^\circ=+\sqrt{\frac{1+\cos54^\circ}2}$$ and $\cos54^\circ$ has been calculated here


0

$\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{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} ...


0

As you wrote $$\frac{d}{dx}x^r=rx^{r-1}$$ there is no problem at $x=0$ if $(r-1) \geq 0$ that is to say if $r \geq 1$. But now, consider the case where $r \lt 1$. So the resulting exponent in the rhs becomes negative that is to say that the derivative looks like $\frac{1}{x^q}$ with $q \gt 0$ and you are looking at the slope at $x=0$.


0

Handle it case by case: $$ x\ge0\text{ and }x\ge\frac13\text{ and }4x-1\lt5\implies x\in\left[\frac13,\frac32\right) $$ $$ x\ge0\text{ and }x\lt\frac13\text{ and }1-2x\lt5\implies x\in\left[0,\frac13\right) $$ $$ x\lt0\text{ and }x\ge\frac13\text{ and }2x-1\lt5\implies x\in\varnothing $$ $$ x\lt0\text{ and }x\lt\frac13\text{ and }1-4x\lt5\implies ...


0

Hint As answered by Martín-Blas Pérez Pinilla, you just need to focus on the first six terms. You can apply the standard method computing the values of the function and the first fifth derivatives or replace in your expression $x$ by $(t+2)$ and expand. Doing so, you will arrive to $$t^5+7 t^4+18 t^3+19 t^2+3 t-5$$ that is to say that $$f(x) = -5+3 (x-2)+19 ...


2

Mathematics courses geared towards engineers are (generally) more applied, focusing on applications rather than theory. For example, at UBC, the Linear Algebra course for engineers concerns mostly solving systems of equations, geometry, determinants, eigenvalues/eigenvectors, and applications of the aforementioned topics. Conversely, the course for ...


0

it all depends. At the start of your undergrad, most math majors and engineers will take calculus 1 and 2 (the same one). However sometimes math majors may take the honours versions of those. By multivariable calc it may be a bit different with engineers focusing on vectors. and ofcourse engineers and math majors alike will take courses on ODE's and lots of ...


0

Actually, the succinct Leibniz product rule $$D^m(f(x)g(x))=\sum_{n=0}^\infty \binom{m}{n}(D^{m-n}f(x))(D^ng(x))$$ works quickly and transparently, so $$D^9(x^8\log(x))=\sum_{n=0}^\infty \binom{9}{n}(D^{9-n}x^8)D^nlog(x)$$ $$=\sum_{n=1}^\infty \binom{9}{n}8!\frac{x^{n-1}}{(n-1)!}(-1)^{n+1}\frac{(n-1)!}{x^n}=\frac{8!}{x}\sum_{n=1}^\infty ...


0

Notice $$\require{cancel} \frac{\partial^9}{\partial x^9}\left[ x^\beta\log x\right] = \frac{\partial^{10}}{\partial x^9\partial\beta}\left[x^\beta\right] = \frac{\partial^{10}}{\partial\beta\partial x^9}\left[x^\beta\right] = \frac{\partial}{\partial\beta}\left[\prod_{k=0}^8(\beta-k)x^{\beta-9}\right] = ...


1

For any analytic function $f$, $$\sum_{n=0}^\infty \dfrac{z^n}{n!} \dfrac{d^n}{dx^n} f(x) = f(x+z) $$ Thus $\dfrac{d^n f}{dx^n}$ is $n!$ times the coefficient of $z^n$ in the Maclaurin series of $f(x+z)$. In this case $f(x+z) = (x+z)^8 \ln(x+z)$. Now $$ \eqalign{(x+z)^8 &= \sum_{j=0}^8 {8 \choose j} x^{8-j} z^j \cr \ln(x+z) &= \ln(x) + \ln(1 + ...


0

I am filling in the details of the calculation for the Hessian, as we can derive a couple of relations that will save some work further along. You presumably found the critical points from $$ yz \ = \ 2x \ \ , \ \ xz \ = \ 2y \ \ , \ \ xy \ = \ 2z \ \ \Rightarrow \ \ xyz^2 \ = \ 4xy \ = \ 8z \ \ ; $$ [multiplying the first two equations together, and also ...


9

Let us define $$D_n=\frac{d^n}{dx^n}(x^8\ln x)$$ and follow what illysial suggested. You then have $$D_1=x^7+8 x^7 \log (x)$$ $$D_2=15 x^6+56 x^6 \log (x)$$ $$D_3=146 x^5+336 x^5 \log (x)$$ $$D_4=1066 x^4+1680 x^4 \log (x)$$ and so on. When going to $D_9$, the derivative of the polynomial part is zero and the only thing which remains in $D_8$ is $$8 \times ...


-1

Consider negative values of r. Let r=-1. Then $d \over dx$$x^{-1}$= $-x^{-2}$=$-1 \over x^2$. What happens when x=0?


5

The derivative of $x^n\log x$ is $x^{n-1}(n\log x+1)$ by the product rule. The first derivative is: $x^7+8x^7\log x$. Edit: For further steps note that we can ignore the $x^n$ terms (they will die out after $9$ derivatives) So the second derivative (with the $7x^6$ ignored) is: $8 \cdot x^6(7\log x +1)$ Ignoring the $x^n$ term here again yields $56x^6 ...


1

HINT: Try multiplying by $\frac{3+\sqrt{c+2}}{3+\sqrt{c+2}}$


0

Think of the domain of this function for various values of r, and their corresponding graphs. For instance, r=-1. http://www.sagemath.org/calctut/pix-calctut/onesided05.png


1

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1

Note that for large $x$ $$ \frac{1}{(1+x^2)^\alpha} \sim \frac{1}{x^{2\alpha}} $$ Now check for what values of alpha the last integrand converge.


2

Outline: Use Comparison. Break up into two parts at $0$. By symmetry the two integrals both exist or both fail to exist, and if they exist they have the same value. Since our function is nicely behaved on the interval $[-1,1]$, we only need to worry about $\int_1^\infty \frac{1}{(1+x^2)^\alpha}\,dx$. For $\alpha \gt 1/2$, we have convergence, by ...


3

The integral is evaluated in terms of the Gamma function as follows $${\frac {\sqrt {\pi }\Gamma \left( -1/2+\alpha \right) }{\Gamma \left( \alpha \right) }} $$ Then $\alpha >1/2$ is domain with convergence. When $\alpha =1/2$ the integral does not converge. When $-1/2 + \alpha = -n$ where $n$ is a positive integer, the integral does not converge. ...


1

By simetry, then $$\int_0^{70 \pi} \left|\cos^{2}\!\left(x\right)\sin\!\left(x\right)\right| dx=70\cdot\int_0^{ \pi} \cos^{2}\!\left(x\right)\sin\!\left(x\right) dx$$


0

Hint: $\int_0^{70\pi}=\sum_{k=0}^{34}\int_{2k\pi}^{2k\pi+2\pi}$.And $$\int_{2k\pi}^{2k\pi+2\pi}|\cos^2(x)\sin(x)|dx=\int_{2k\pi}^{2k\pi+\pi}\sin(x)\cos^2(x)dx-\int_{2k\pi+\pi}^{2k\pi+2\pi}\sin(x)\cos^2(x)dx$$ and $$\sin(x)\cos^2(x)=-\dfrac{1}{3}(\cos^3(x))'$$


3

If you take the integral from $0$ to $\pi$ and then multiply by $70$, you've got it, since it's periodic with period $\pi$. On the interval from $0$ to $\pi$ you can drop the absolute value since the function is nonnegative there. And $$ \int_0^{\pi/2} (\cos^2 x) \Big(\sin x\,dx\Big) = \int_1^0 u^2 \Big(-du\Big). $$ And finally, ...


1

As you said in the preamble, $f(\theta,\phi)$ is just a (periodic) function of two variables. It comes from the parametrization of a sphere, but this knowledge is just ours. So we find the stationary points of $f(\theta,\phi)$ as if we were looking for the stationary points of a smooth function $g(x,y)$. If $$f(\theta,\phi) = A\cos\theta\sin\phi + ...


1

Eric: You are correct that one needs $n\ge 0$ for the proof to work. Indeed, as this plot suggests, there are no zeroes on $x<0$.


0

The only case where the formula doesn't work is $n=1$ (because of a $0$ in the denominator). It's perfectly valid to say, e.g., $$\int \tan^{5/2}(x)\; dx = - \int \tan^{1/2}(x)\; dx + \dfrac{\tan^{3/2}(x)}{3/2}+ C$$ (at least on an interval where $\tan(x) > 0$). It might not be too useful unless you can evaluate one of those integrals by other means ...


4

You pretty much have all the right pieces. Note that we're not dealing with a biconditional though; we want to show that: $$ 0 < |x - 3| < \delta \implies |x^2 - 9| < \epsilon $$ Here's a cleaned up version of your proof. Given any $\epsilon > 0$, consider $\delta = \min\{1, \epsilon/7\} > 0$. Then observe that if $0 < |x - 3| < ...


1

Using the method of disks($V=π\int_{\alpha}^{\beta} f^2(x) dx$); The function $y=\sqrt{\frac{2-x^2}{2}}$ has two distinct real roots $(\alpha,\beta)=(-\sqrt{2},\sqrt{2})$ and the solid of revolution of this half-elipsoid will have the same volume with the revolution of the whole elipsoid as the whole elipsoid will just mark the solid of revolution's boundary ...


0

$\begin{align} f(x) &= \frac 4 x \\[1ex] f(x+h) &= \frac 4 {x+h} \\[1ex] f(x+h)- f(x) &= \frac 4 {x+h} - \frac 4 x \\[1ex] &= \frac {4x - 4(x+h)}{(x+h)\times x} \\[1ex] &= \frac{-4h}{x(x+h)} \\[1ex] \frac{f(x+h)-f(x)}{h} &= \frac{-4}{x(x+h)} \\[1ex] \lim_{h\to 0}\frac{f(x+h)-f(x)}{h} &= \frac{-4}{x^2} \\[1ex] \therefore ...


0

Just write it: $$ 4/(x+h) - 4/x = 4(x-x-h)/x(x+h) = -4h/x(x+h) $$and divide by $h$.


1

We will start by defining $$\log(x) \equiv \int_1^x \frac{dt}{t}$$ and then showing that this is indeed the inverse function of the exponential function $e^{x}$. By definition and for $x>0$, $\log(x)$ is continious and monotonely increasing with $\log(1) = 0$. We first start by showing that $\log(x) + \log(y) = \log(xy)$: $$\log(x) + \log(y) \equiv ...


3

They are the same up to arbitrary integration constant. HINT: $$\log a - c = \log a - \log e^c = \log \frac{a}{e^c}.$$


1

As $f^{(k)}=0$ for $k>5$, the Taylor series is a finite sum and $=f$.


2

Hint: For $h\neq0$, we have $~\displaystyle\int_1^xt^{~h-1}~dt=\bigg[\frac{t^h}h\bigg]_1^x=\frac{x^h-1}h.~$ For $h=0$, we have $~\displaystyle\int_1^x\frac{dt}t=$ $=\displaystyle\lim_{h\to0}\frac{x^h-1}h.~$ At the same time, $~\Big(a^x\Big)'=\displaystyle\lim_{h\to0}\dfrac{a^{x+h}-a^x}h=a^x~\lim_{h\to0}\dfrac{a^h-1}h.~$ Do you notice anything ...


1

Hint $\,\ (x\!-\!a)^2\mid P(x)\,\color{#C00}\Rightarrow\, \color{#0a0}{x\!-\!a}\mid P'(x)\,\Rightarrow\, 0 = P'(a) = \frac{a^2}2+a+1,\,$ contra $\,a\in R\ \ $ QED ${\rm\color{#c00}{Indeed}}\,\ P = (x\!-\!a)^2 Q\,\Rightarrow\, P'\! = (\color{#0a0}{x\!-\!a})^2 Q' + 2(\color{#0a0}{x\!-\!a}) Q\,\Rightarrow\,P'(a) = 0$ Remark $\ $ It generalizes as follows. Let ...


0

Cop-out answer Define $\ln(x):=\int_1^x \frac{1}{t}dt$. Then the result follows immediately. Serious answer $$ \int \color{green}{\frac{1}{x}}dx = \int \color{green}{\sum_{k=0}^\infty (-1)^k (x-1)^k} dx \\$$ $$= \sum_{k=0}^\infty (-1)^k \int (x-1)^k dx \tag{pulling out the constant }\\ = \sum_{k=0}^\infty \frac{(-1)^k}{k+1} (x-1)^{k+1} + C = \ln ...


0

Here is the proof I wrote with Ian's guiding. Ian, thank you for pointing me in the right direction for what I was looking for; $ln(x)$ is defined to be the inverse function of $e^x$. The inverse function theorem states that $(f^{-1})'(b)=\frac{1}{(f'(a))}$, $b=f(a)$. Let $f(x)=\ln(x), f^{-1}(f^{-1}(x))=f(x)$, thus, $f^{-1}(x)=e^x$. $(f^{-1})'=(e^x)'=e^x$. ...



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