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5

OP, you are correct, $$\int_1^n\frac{\sin \sqrt{x}}{\sqrt{x}}dx=-2\cos\sqrt{n}+2\cos 1$$ Hence the improper integral is $$\int_1^\infty\frac{\sin \sqrt{x}}{\sqrt{x}}dx=\lim_{n\to\infty}\int_1^n\frac{\sin \sqrt{x}}{\sqrt{x}}dx=\lim_{n\to\infty}-2\cos\sqrt{n}+2\cos 1$$ And the latter limit does not exist. Your computer algebra system (mathematica) is giving ...

1

If you expand $\cot2\theta$, you get $$\cot\theta=\frac{\cot^2\theta-1}{\cot\theta}$$ that's clearly an inconsistent equation. On the other hand, if you rewrite it as $$\tan2\theta=2\tan\theta$$ and expand, you get $$\frac{\tan\theta}{1-\tan^2\theta}=\tan\theta$$ that has the solutions $\tan\theta=0$, but they aren't solutions of the original ...

1

Reciprocate both sides: $$\tan(\theta)=\frac12\tan(2\theta)=\frac{\sin(2\theta)}{2\cos(2\theta)}$$ $$=\frac{2\sin(\theta)\cos(\theta)}{2\cos(2\theta)}=\frac{\sin(\theta)\cos(\theta)}{2\cos^2(\theta)-1}$$ $$\tan(\theta)=\frac{\sin(\theta)}{\cos(\theta)}$$ Which allows us to cancel the sines and move all the cosines to the right. ...

0

HINT: your equation is equivalent to $$\frac{1}{\cot(\theta)}=0$$

0

Any two vectors lying on the plane will do as long as they're linearly independent. A simple way to find such vectors is, as mentioned to choose two variable and solve for the third: let $x=y=1$, then by substitution into the equation of the plane, $z=-1$ This tells you the vector $(1,1,-1)$ lies on the given plane. You can alsocheck your answer by ...

0

For $f^{-1}$ to exist $f$ must be injective. This means that if $x \neq y$ then $f(x) \neq f(y)$. But $f(x) = x^3$ isn't injective since $$f\left(\frac{1}{2} + i\frac{\sqrt{3}}{2}\right) = -1,\ f\left(\frac{1}{2} - i\frac{\sqrt{3}}{2}\right) = -1, \ f(-1) = -1$$ So $f(x) = x^3$ doesn't have an inverse! If you try to define one you realize that ...

2

EXTRA: Alright, Mathematica works with the rows as generating vectors. A vector in their lattice will be $k A,$ where $k$ is a row vector of integers. Try $k = (1,0,0)$ and see what happens. This means that reduction follows the recipe $A = LB,$ so that $L = A B^{-1}.$ ? opa = a %16 = [1 0 1345] [1 0 35] [0 1 154] ? opb = [ 9,-2,7; -2,-8,8; -2,9,6] %17 ...

1

It depends which root of the equation the software considers by default. The point is that for real $y$, the equation $x^3-y=0$ always has $3$ complex roots. If $y>0$, there is no controversy about taking the positive real root, which we normally call $\sqrt[3]{y}$. On the other hand, if $y<0$, then there are no positive real roots; instead the roots ...

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