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How can $S_n=n$ for the series $S_n=\frac{1}{2}+\frac{3}{4}+\frac{7}{8}+\frac{15}{16}+\cdots$ How can $S_n=n$ for the series $$S_n=\frac{1}{2}+\frac{3}{4}+\frac{7}{8}+\frac{15}{16}+\cdots$$
My try:
We can Re-write $S_n$ as
$$S_n=\frac{0+1}{2}+\frac{1+2}{2^2}+\frac{3+2^2}{2^3}+\frac{7+2^3}{2^4}+\cdots \frac{2^{n-1}-1+2^{n-1}}{2^n}$$
So we get
$$S_n=\frac{1}{2}+\frac{1}{2}+\cdots+\frac{1}{2}+\left(\frac{1}{2^2}+\frac{3}{2^3}+\frac{7}{2^4}+\cdots\right)$$
Hence
$$S_n=\frac{n}{2}+\frac{S_n}{2}$$
Hence
$$S_n=n$$
But the actual answer is entirely different. Whats wrong in my solution?
| Given:
$$S_n=\frac{1}{2}+\frac{3}{4}+\frac{7}{8}+\frac{15}{16}+\cdots+a_n,$$
as pointed out by John Omielan, you are missing the first (or last) term:
$$S_n=\frac{\color{red}0+\color{blue}1}{2}+\frac{\color{red}1+\color{blue}2}{2^2}+\frac{\color{red}3+\color{blue}{2^2}}{2^3}+\frac{\color{red}7+\color{blue}{2^3}}{2^4}+\cdots +\underbrace{\frac{\color{red}{2^{n-1}-1}+\color{blue}{2^{n-1}}}{2^n}}_{a_n=\frac{2^n-1}{2^n}}=\\
\left(\frac{\color{red}0}2+\frac{\color{red}1}{2^2}+\frac{\color{red}3}{2^3}+\frac{\color{red}7}{2^4}+\cdots +\frac{\color{red}{2^{n-1}-1}}{2^n}\right)+
\left(\frac{\color{blue}1}2+\frac{\color{blue}2}{2^2}+\frac{\color{blue}{2^2}}{2^3}+\frac{\color{blue}{2^4}}{2^4}+\cdots +\frac{\color{blue}{2^{n-1}}}{2^n}\right)=\\
\frac12
\left(\color{red}0+\frac{\color{red}1}{2}+\frac{\color{red}3}{2^2}+\frac{\color{red}7}{2^3}+\cdots +\frac{\color{red}{2^{n-1}-1}}{2^{n-1}}\right)+
\left(\frac{\color{blue}1}2+\frac{\color{blue}1}{2}+\frac{\color{blue}{1}}{2}+\frac{\color{blue}{1}}{2}+\cdots +\frac{\color{blue}{1}}{2}\right)=\\
\frac12(0+S_{n-1})+\frac12n=\\
\frac12(S_n-a_n)+\frac12n \Rightarrow \\
S_n=n-a_n=n-\frac{2^{n-1}-1+2^{n-1}}{2^n}=n-\frac{2^n-1}{2^n}=n-\left(1-\frac{1}{2^n}\right)=n-1+\frac1{2^n}.$$
You can prove by mathematical induction: $S_{n}=n-1+\frac1{2^n}$. Indeed:
$$S_n=\color{red}{S_{n-1}}+\color{blue}{a_n}=\color{red}{(n-1)-1+\frac1{2^{n-1}}}+\color{blue}{\frac{2^n-1}{2^n}}=\\
n-1+\frac{-2^n+2+2^n-1}{2^n}=n-1+\frac1{2^n}.$$
| {
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"url": "https://math.stackexchange.com/questions/3331620",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Deriving the vector field from a curl If $B= (x/r^3,y/r^3,z/r^3)$ find a vector $A$ such that $curl A = B$.
My attempt:
$∂A_3∂Y−∂A_2∂Z=x/r^3$
$∂A_1∂Z−∂A_2∂y=y/r^3$
$∂A_2∂x−∂A_1∂y=z/r^3$
I assumed $A_3=0$, ending up having to solve:
$A_2=\int xr^{-3}dz$
Here, I got stuck. Is there an easier way of doing this?
| I assume your book asks for any solution even though physically meaningless. Take for instance $A_3 = 0$. Then
\begin{equation}
B_1 = \frac{\partial A_3}{\partial y}-\frac{\partial A_2}{\partial z} = \frac{x}{(\sqrt{x^2+y^2+z^2})^3}
\end{equation}
Thus,
\begin{equation}
A_2 = -\int \frac{x}{(\sqrt{x^2+y^2+z^2})^3} \mathrm{d}z = \frac{-xz}{(x^2+y^2)\sqrt{x^2+y^2+z^2}}
\end{equation}
Similarly,
\begin{equation}
B_2 = \frac{\partial A_1}{\partial z}-\frac{\partial A_3}{\partial x} = \frac{y}{(\sqrt{x^2+y^2+z^2})^3}
\end{equation}
Thus,
\begin{equation}
A_1 = \int \frac{y}{(\sqrt{x^2+y^2+z^2})^3} \mathrm{d}z = \frac{yz}{(x^2+y^2)\sqrt{x^2+y^2+z^2}}
\end{equation}
Solving $\nabla \times \mathbf{A} = \mathbf{B}$ is a coupled partial differential equation, which is not complete unless boundary conditions are given to make solution $\mathbf{A}$ unique. Otherwise, the following answer assumes $\mathbf{A}$ is zero on boundary (decays at infinity).
By Helmholtz decomposition, every vector field $\mathbf{B}$ is decomposed to a curl-free and divergence-free component, by
\begin{equation}
\mathbf{B} = \nabla \times \mathbf{A} - \nabla \phi,
\end{equation}
where $\phi$ is the potential (the curl-free part, since $\nabla \times (\nabla \phi) = 0$), and $\mathbf{A}$ is vector potential (the divergence-free part, since $\nabla \cdot (\nabla \times \mathbf{A}) = 0$). Here
\begin{equation}
\mathbf{B} = \mathbf{0} - \nabla \left(\frac{1}{\sqrt{x^2 + y^2 + z^2}} \right) = -\nabla \left(\frac{1}{r}\right).
\end{equation}
It means $\mathbf{B}$ can be decomposed into a potential field $\phi=r^{-1}$ and no vector potential $\mathbf{A}$.
The potential is unique upto a constant, $\phi = \frac{1}{r}+c$ for any constant $c$. The vector potential is unique up to a gradient field, $\mathbf{A} = \mathbf{0} + \nabla \psi$ for any scalar field $\psi$.
Indeed, $\mathbf{B} = \frac{1}{r^2}\mathbf{e}_r$ is the Coulomb force, which is curl free and represented by potential above.
| {
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"timestamp": "2023-03-29T00:00:00",
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How do I find the real or imaginary part of this complex equation? I am trying to solve the below equation for surface plasmon polariton(SPP) wave
$ \sqrt{\dfrac{1}{\beta^2-(\frac{\omega^2}{c^2})}} + \sqrt{\dfrac{\epsilon_r}{\beta^2-(\frac{\omega^2}{c^2}\epsilon_r)}} = -j \dfrac{\sigma(w)}{\omega \epsilon_0}$
Because i can't solve the problem otherwise I have solved the equation assuming ${\dfrac{1}\epsilon_r} = 1$ present in the second term of the equation $ \frac{\omega^2}{c^2}\epsilon_r $
which looks like these,
$ \dfrac{1}{\beta^2-(\frac{\omega^2}{c^2})} +2\sqrt{\dfrac{1}{\beta^2-(\frac{\omega^2}{c^2})}} \sqrt{\dfrac{\epsilon_r}{\beta^2-(\frac{\omega^2}{c^2})}} + \dfrac{\epsilon_r^2}{\beta^2-(\frac{\omega^2}{c^2})} = - \dfrac{\sigma^2(w)}{\omega^2 \epsilon_0^2}$
Then after some doing some calculation, I have come to this solution
$ \beta = \sqrt{\dfrac{\omega^2}{c^2} + \dfrac{(1 + (\epsilon_r)^2)\omega^2 \epsilon_0) ^2}{{\sigma^2(w)}} }$
But the problem is that it's not been matched with my simulated result, I am thinking it's because of that ${\epsilon_r} $ that I have not taken into account.
Is there any branch of mathematics I should be aware of, Any kind of help will be highly appreciated. Thanks in Advance.
| You replaced $\epsilon_r$ with $1$ in some but by no means all places, but more to the point, you committed a fallacy of the form $a+b=c\implies a^2+b^2=c^2$. Squaring should give $$2\sqrt{\frac{\epsilon_r}{(\beta^2-\omega^2/c^2)(\beta^2-\epsilon_r\omega^2/c^2)}}=-\frac{\sigma^2}{\omega^2\epsilon_0^2}-\frac{1}{\beta^2-\omega^2/c^2}-\frac{\epsilon_r}{\beta^2-\epsilon_r\omega^2/c^2}.$$Squaring again gives$$\frac{2\epsilon_r+\frac{2\sigma^2}{2\epsilon_r\omega^2/c^2-\omega^2\epsilon_0^2}\left((1+\epsilon_r)\beta^2\right)}{(\beta^2-\omega^2/c^2)(\beta^2-\epsilon_r\omega^2/c^2)}=\frac{\sigma^4}{\omega^4\epsilon_0^4}+\frac{1}{(\beta^2-\omega^2/c^2)^2}+\frac{\epsilon_r^2}{(\beta^2-\epsilon_r\omega^2/c^2)^2}.$$Multiplying both sides by $(\beta^2-\omega^2/c^2)^2(\beta^2-\epsilon_r\omega^2/c^2)^2$,$$\left[2\epsilon_r+\frac{2\sigma^2}{2\epsilon_r\omega^2/c^2-\omega^2\epsilon_0^2}\left((1+\epsilon_r)\beta^2\right)\right](\beta^2-\omega^2/c^2)(\beta^2-\epsilon_r\omega^2/c^2)=\frac{\sigma^4}{\omega^4\epsilon_0^4}(\beta^2-\omega^2/c^2)^2(\beta^2-\epsilon_r\omega^2/c^2)^2+(\beta^2-\epsilon_r\omega^2/c^2)^2+\epsilon_r^2(\beta^2-\omega^2/c^2)^2.$$This is a quartic in $\beta^2$, so you can solve it but not easily. Then we have to get rid of spurious roots from the repeated squaring.
| {
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"timestamp": "2023-03-29T00:00:00",
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Find value for $k$ such that $(x^2-k)$ is a factor for $f(x)=2x^4+(3k-4)x^3+(2k^2-5k-5)x^2+(2k^3-2k^2-3k-6)x+6$ Find value for $k$ such that $(x^2-k)$ is a factor for, $$f(x)=2x^4+(3k-4)x^3+(2k^2-5k-5)x^2+(2k^3-2k^2-3k-6)x+6$$
My Try
Since $x^2-k=0$ when we substitute $x=\pm k$ to $f(x)$ it should be equal to $0.$
But this gives a polynomial of k where it has $\sqrt{k}$ terms as well. Is my approach correct or is there a simpler way? Please any hint would be highly appreciated.
| I would begin by replacing $x^2$ with $k$ in the polynomial. Where there is an odd power of $x$ you leave the remaining power of $x$ as is. Thus
$2x^4=2k^2$
$(3k-4)x^3=(3k^2-4k)x$
$(2k^2-5k-5)x^2=2k^3-5k^2-5k$
$(2k^3-2k^2-3k-6)x=(2k^3-2k^2-3k-6)x$
$6=6$
Add the right sides together and collect all $x$ terms to get
$\color{blue}{(2k^3+k^2-7k-6)}x+\color{blue}{(2k^3-3k^2-5k+6)}$
If $x^2-k$ is to be a factor of the original polynomial, then the last expression above must be zero for *both roots** of $x^2=k$. This is possible only if the blue polynomials are both zero for some common root $k$. You are to find this common root. It turns out that both blue polynomials have all their roots rational, so you should be able to find the answer with the usual method of searching for rational roots.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How to factorize $\frac{4x^3+4x^2-7x+2}{4x^4-17x^2+4}$? How to factorize and simplify the following?
$$\frac{4x^3+4x^2-7x+2}{4x^4-17x^2+4}$$
I've tried everything I know. Trying to factorize the numerator first then denominator, but I get no where. Usual identities like $(x+y)^2=x^2+2xy+y^2$ don't work either, and neither does long division. I'm pretty stuck.
The answer from wolfram is
$(2x-1)/((2x+1)(x-2))$.
But I can't get there.
| The polynomial in the denominator can be rewritten as $4t^2 - 17t + 4$ where $t = x^2$
Use the quadratic formula to find that this polynomial has roots $t_1 = 4, t_2 = \frac{1}{4}$
Since $t = x^2$, we can factor the bottom polynomial as $(x-2)(x+2)(x-\frac{1}{2})(x + \frac{1}{2})$. The roots are solutions for $x^2 = t_1$ and $x^2 = t_2$
Find out whether any of these 4 roots is also a root of the polynomial in the numerator and factorize
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3334164",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to equalize correctly? If i have this number:
$2 \sqrt{2-\sqrt{3}}$ and i want to find some $x,y$ nonzero real numbers such that $2\sqrt{2-\sqrt{3}} = \sqrt{x} + \sqrt{y}$
And for that, i do this:
$(2 \sqrt{2-\sqrt{3}})^2 = x + 2\sqrt{xy} + y$
$4(2-\sqrt{3})=(x+y)+2\sqrt{xy}$
$(8)+(-4\sqrt{3})=(x+y)+(2\sqrt{xy})$
Then:
$i) 8 = x+y$,
$ii)-4\sqrt{3} = 2\sqrt{xy} => -2\sqrt{3}=\sqrt{xy}$
$ii) = (-2\sqrt{3})^2= (\sqrt{xy})^2 => 4\cdot 3=xy , x = 12/y$
And solving the equation $y^2-8y+12=0$ gives $y_{1,2} = \{6,2\}$
But $2\sqrt{2-\sqrt{3}} \neq \sqrt{6} + \sqrt{2}$
I know that the correct value must be $\sqrt{6} - \sqrt{2}$ but my result is different. What is wrong with my development?
| You should assume a binomial of the form whose root you desire. In this case, you should have supposed the root is $$\sqrt x -\sqrt y$$ instead.
To be specific, the problem in the above calculation is in your step ii, where you set $$-\sqrt{12}=\sqrt{xy}.$$ But this is impossible if you're dealing only with real numbers. It seems you need to note that the symbol $\sqrt{}$ denotes a function which, by definition, assumes nonnegative values. Thus, you can see that your equation is false, for it says a negative number is equal to a nonnegative one. That's a contradiction.
| {
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"timestamp": "2023-03-29T00:00:00",
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Evaluating $\int_{0}^{1} \int_{0}^{1} \int_{0}^{1} \sqrt{{x^2+y^2+z^2}}\, dx \,dy \,dz$ by converting to spherical coordinates
I would like to know how to evaluate the following triple integral with the help of spherical coordinates
$$\int_{0}^{1} \int_{0}^{1} \int_{0}^{1} \sqrt{{x^2+y^2+z^2}} \,dx \,dy\, dz$$
The relations between Cartesian coordinates and spherical ones are given by
${\displaystyle {\begin{aligned}x&=r\,\sin \theta \,\cos \varphi \\y&=r\,\sin \theta \,\sin \varphi \\z&=r\,\cos \theta \end{aligned}}}$
I know that a function is generally integrated over $\mathbb{R}^3$ by the following triple integral
$$ \ \int \limits _{\varphi =0}^{2\pi }\ \int \limits _{\theta =0}^{\pi }\ \int \limits _{r=0}^{\infty }f(r,\theta ,\varphi )r^{2}\sin \theta \,\mathrm {d} r\,\mathrm {d} \theta \,\mathrm {d} \varphi .$$
I found a numerical solution with Wolfram Alpha (0.960592), I tried to change the bounds of integration from Cartesian to spherical but I got a different numerical value.
Could someone please give a detailed solution showing how to change the limits of integration?
Thanks.
| Substitute:
$$z=\sqrt{x^2+y^2} t$$
$$I=\int_{0}^{1} \int_{0}^{1} \int_{0}^{1/\sqrt{x^2+y^2}} (x^2+y^2) \sqrt{{1+t^2}} dt dy dx$$
Integrating w.r.t. $t$:
$$I=\frac{1}{2} \int_{0}^{1} \int_{0}^{1} (x^2+y^2) \left(\frac{1}{\sqrt{x^2+y^2}}\sqrt{1+\frac{1}{x^2+y^2}}+\sinh^{-1} \frac{1}{\sqrt{x^2+y^2}} \right) dy dx$$
Using the symmetry:
$$I= \int_{0}^{1} \int_{0}^{x} (x^2+y^2) \left(\frac{1}{\sqrt{x^2+y^2}}\sqrt{1+\frac{1}{x^2+y^2}}+\sinh^{-1} \frac{1}{\sqrt{x^2+y^2}} \right) dy dx$$
Now we can use polar coordinates:
$$x=r \cos \phi$$
$$y=r \sin \phi$$
$$0<\sin \phi<\cos \phi, \qquad 0<\phi< \frac{\pi}{4}$$
$$0<r< \frac{1}{\cos \phi}$$
$$I= \int_{0}^{\frac{\pi}{4}} \int_{0}^{\frac{1}{\cos \phi}} r^3 \left(\frac{1}{r}\sqrt{1+\frac{1}{r^2}}+\sinh^{-1} \frac{1}{r} \right) dr d \phi$$
$$I= \int_{0}^{\frac{\pi}{4}} \int_{0}^{\frac{1}{\cos \phi}} \left(r\sqrt{r^2+1}+r^3 \sinh^{-1} \frac{1}{r} \right) dr d \phi$$
We have:
$$\int_{0}^{\frac{1}{\cos \phi}} r\sqrt{r^2+1} dr= \frac{1}{3} \left(\frac{1}{\cos^2 \phi}+1 \right)^{3/2}-\frac{1}{3}$$
$$\int_{0}^{\frac{1}{\cos \phi}} r^3 \sinh^{-1} \frac{1}{r} dr= \frac{1}{4 \cos^4 \phi} \sinh^{-1} \cos \phi+ \frac{1-2 \cos^2 \phi}{12 \cos^3 \phi} \sqrt{1+\cos^2 \phi}+\frac16$$
So we have a complicated expression:
$$I= \frac{\pi}{24}+ \frac13 \int_{0}^{\frac{\pi}{4}} \left(\left(\frac{1}{\cos^2 \phi}+1 \right)^{3/2}-1 \right) d \phi+ \\+ \frac{1}{4} \int_{0}^{\frac{\pi}{4}} \left( \frac{1}{\cos \phi}\sinh^{-1} \cos \phi+ \frac{1}{3} (1-2 \cos^2 \phi)\sqrt{1+\cos^2 \phi} \right) \frac{d \phi}{\cos^3 \phi}$$
These are elliptic kind of integrals, though some of them might be elementary. Substitution $\cos \phi=s$ seems prudent here.
I hope this might be helpful.
Maybe using spherical coordinates from the start is better, but I haven't figured out the correct bounds yet either.
| {
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"timestamp": "2023-03-29T00:00:00",
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For which $a$, $b$, $c$ does this linear system have exactly one solution? $x+ay+a^2z=0$, $x+by+b^2z=0$, $x+cy+c^2z=0$
For which $a$, $b$, $c$ does this linear system have exactly one solution?
$$x + ay + a^2z = 0$$
$$x + by + b^2z = 0$$
$$x + cy + c^2z = 0$$
I started this problem by recognizing that if the RREF of a linear system's augmented matrix has a leading 1 in every column except the last, then the system must have exactly one solution. The system's augmented matrix reduces to
$$
\begin{matrix}
1 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 \\
0 & 0 & 1 & 0 \\
\end{matrix}
$$
which indicates that no matter the values of $a, b, c$, $x = 0, y = 0,$ and $z = 0$. But one can see that if $a = 0$ and $b = 0$, then $x = 0$ and the third equation of the system becomes the equation of a line, indicating an infinite number of solutions, meaning that no two of $a,b,c$ van be equal to $0$.
So my best guess is that the system has exactly one solution when $a,b,c \in \mathbb{R}$ and $a, b \neq 0$. Is this correct? Close at all? How would I go about proving this?
| HINT
According to Cramer's rule, this system admits a unique solution iff the determinant of the coefficient matrix is different from zero. Precisely,
\begin{align*}
\begin{vmatrix}
1 & a & a^{2}\\
1 & b & b^{2}\\
1 & c & c^2\\
\end{vmatrix}\neq 0
\end{align*}
Such matrix is widely known as the Vandermonte's matrix, and its determinant equals
\begin{align*}
\begin{vmatrix}
1 & a & a^{2}\\
1 & b & b^{2}\\
1 & c & c^2\\
\end{vmatrix} = (c-b)(c-a)(b-a)
\end{align*}
Can you take it from here?
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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The units digit of a power tower of consecutive numbers, from 2019 to 1
Is it possible to find the units digit of
$2019^{2018^{2017^{.^{.^{.^{3^{2^{1}}}}}}}}$?
Where the expression contains all natural numbers $[1,2018]$ as powers and $2019$ as the main base.
Any help would be appreciated. THANKS!
| HINT : Assume $a_n=n^{n-1^{n-2^{.^{.^{.^{3^{2^{1}}}}}}}} \mod 10$, then the sequence will be :
$$
1,2,9,4,5,6,9,8,1,0,1,2,9,4,5,6,9,8,1,0,1,2,9,4,5,6,9,8,1,0,1,2,9,4,5,6,9,8,1,0,1,2,9,4,5,6,9,8,1,0,1,2,9,4,5,6,9,8,1,0,1,2,9,4,5,6,9,8,1,0...
$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 6,
"answer_id": 4
} |
Knowing that $m$ and $n$ are positive integers satisfying $mn \mid m^2 + n^2 + m$, prove that $m$ is a square number.
Knowing that $m$ and $n$ are positive integers satisfying $$\large mn \mid m^2 + n^2 + m$$, prove that $m$ is a square number.
We have that $mn \mid m^2 + n^2 + m \implies mn \mid (m^2 + n^2 + m)(n + 1)$
$\implies mn \mid m^2n + n^3 + mn + m^2 + n^2 + m$ $\implies \left\{ \begin{align} &mn \mid n^3\\ &mn \mid n^3 + m^2 + n^2 + n \end{align} \right.$
$\implies \left\{ \begin{align} m &\mid n^2\\ mn &\mid m(m + 1) + n^2(n + 1) \end{align} \right.$ $\implies \left\{ \begin{align} mm' &= n^2\\ mn &\mid m(m + 1) + mm'(n + 1) \end{align} \right. (m \in \mathbb Z^+)$
$\implies \left\{ \begin{align} mm' &= n^2\\ m'n &\mid m'(m + 1) + m'^2(n + 1) \end{align} \right.$ $\implies \left\{ \begin{align} mm' &= n^2\\ m'n &\mid n^2 + m' + m'^2 \end{align} \right.$
That also means that if the $m$ is proven to be a square number, $m'$ is also a perfect square. Moreover, $mm' = n^2 \implies (m, m') = 1$ needs to be proven. But I don't know how to.
| Let $p$ be prime and let $k\ge0 $ be maximal with $p^k\mid m$. Similarly, let $l\ge 0$ be maximal with $p^l\mid n$. So $m=p^kr$, $n=p^ls$ for some positive integers $r,s$.
So $$p^{k+l}rs\mid p^{2k}r^2+p^{2l}s^2+p^kr. $$
Conclude that $2l\ge k$ and
$$p^{l}rs\mid p^{k}r^2+p^{2l-k}s^2+r. $$
of $k>0$ then also $l>0$ and so $p\mid p^{2l-k}s+r$, which leads to $2l-k=0$. Therefore $k$ must be even.
As this holds for all primes $p$, we conclude that $m$ is a perfect square.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3342214",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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If $ a + b + c = 90^{\circ}$, prove $ \tan(a) \cdot \tan(b) + \tan(b) \cdot \tan(c) + \tan(c) \cdot \tan(a) = 1 $ If $ a + b + c = 90^{\circ}$, prove $ \tan(a) \cdot \tan(b) + \tan(b) \cdot \tan(c) + \tan(c) \cdot \tan(a) = 1 $
Attempt:
Notice that $$ \tan(a+b+c) = \frac{\tan(a+b) + \tan(c)}{1 - \tan(a+b)\tan(c)} $$
$$ = \frac{\frac{\tan(a) + \tan(b)}{1 - \tan(a)\tan(b)} + \tan(c)}{1 - \frac{\tan(a) + \tan(b)}{1 - \tan(a)\tan(b)} \tan(c)} $$
$$ = \frac{\tan(a) + \tan(b) + \tan(c) - \tan(a) \tan(b) \tan(c)}{ (1-\tan(a)\tan(b)) - \tan(a) \tan(c) -\tan(b) \tan(c) } $$
Then the denomenator must be $0$, so it is proven?
Another way:
$$ \sin(a+b+c) = 1 \implies \sin(a+b) \cos(c) + \sin(c) \cos(a+b) = 1 $$
$$ \sin(a)\cos(b)\cos(c) + \sin(b)\cos(a)\cos(c) + \sin(c) \cos(a) \cos(b) - \sin(c) \sin(a) \sin(b) = 1$$
$$ \tan(a) \cos(a) \cos(b)\cos(c) + \tan(b) \cos(a) \cos(b) \cos(c) + \tan(c) \cos(a) \cos(b) \cos(c) - \sin(a) \sin(b) \sin(c) = 1$$
$$ \cos(a)\cos(b)\cos(c) (\tan(a) + \tan(b) + \tan(c) - \tan(a)\tan(b)\tan(c)) = 1 $$
| It's wrong for $a=90^{\circ}$ for example.
But on the domain of our tangents we obtain:
$$1-\sum_{cyc}\tan{a}\tan{b}=1-\tan{a}\tan{b}-\tan{c}(\tan{a}+\tan{b})=$$
$$=\left(1-\tan{a}\tan{b}\right)\left(1-\frac{\tan{c}(\tan{a}+\tan{b})}{1-\tan{a}\tan{b}}\right)=$$
$$=\left(1-\tan{a}\tan{b}\right)\left(1-\tan{c}\tan(a+b)\right)=$$
$$=\left(1-\tan{a}\tan{b}\right)\left(1-\tan{c}\cot{c}\right)=0.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3342989",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 2
} |
Show that $\operatorname{Aff}(3)$ is isomorphic to $S_3$, the symmetric group of all permutations of 3 objects. Show that $\operatorname{Aff}(3)$ is isomorphic to $S_3$, the symmetric group of all permutations of 3 objects.
Where
$$\operatorname{Aff}(3):={\{( \begin{array}{cc}
a & b \\
0 & 1
\end{array}): a,b\in\mathbb{Z}_3}, a\neq0\}$$, $\mathbb{Z}_3$ are the integers module 3.
Idea: I know that $ S_3 $ has 6 elements, and the $\operatorname{Aff}(3)$ matrices are: $( \begin{array}{cc}
1 & 0 \\
0 & 1
\end{array}) $, $( \begin{array}{cc}
1 & 1 \\
0 & 1
\end{array})$ $( \begin{array}{cc}
1 & 2 \\
0 & 1
\end{array}) $, $( \begin{array}{cc}
2 & 0 \\
0 & 1
\end{array})$ $( \begin{array}{cc}
2 & 1 \\
0 & 1
\end{array}) $, $( \begin{array}{cc}
2 & 2\\
0 & 1
\end{array})$
But how do I find an isomorphism? Can you help me please
| If $a\mapsto \begin{pmatrix} 1 & 1 \\ 0 & 1\end{pmatrix}$ and $b\mapsto \begin{pmatrix} 2 & 0 \\ 0 & 1\end{pmatrix}$, then $\operatorname{Aff}(3)$ has as a presentation $$\langle a, b\mid a^3, b^2, bab=a^{-1}\rangle,$$ which is, in turn, a presentation for $S_3$. Hence $\operatorname{Aff}(3)\cong S_3.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3349518",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Expression of $x^n+\frac1{x^n}$ by $x+\frac1{x}$ where $n$ is a positive odd number. There was a problem in a book:
Denote that $y=x+\dfrac{1}{x}$, express $x^7+\dfrac{1}{x^7}$ using $y$.
It's not a hard question, but I find a special sequence:
$x+\dfrac{1}{x}=y\\x^3+\dfrac{1}{x^3}=\left(x+\dfrac{1}{x}\right)^3-3\left(x+\dfrac{1}{x}\right)=y^3-3y \\ x^5+\dfrac{1}{x^5}=\left(x+\dfrac{1}{x}\right)^5-5\left(x^3+\dfrac{1}{x^3}\right)-10\left(x+\dfrac{1}{x}\right)=y^5-5\left(y^3-3y\right)-10y=y^5-5y^3+5y\\x^7+\dfrac{1}{x^7}=\left(x+\dfrac{1}{x}\right)^7-7\left(x^5+\dfrac{1}{x^5}\right)-21\left(x^3+\dfrac{1}{x^3}\right)-35\left(x+\dfrac{1}{x}\right)=y^7-7\left(y^5-5y^3+5y\right)-21\left(y^3-3y\right)-35y=y^7-7y^5+14y^3-7y$
I find that the coefficient has some relationship between the Pascal Triangle, such as $y^7-7y^5+14y-7y=y^7-7 \binom{2}{0}y^5+7\binom{2}{1}y^3-7\binom{2}{2}y$. That's strange but funny! However, I can't really prove this, or show that it is false. Hope there is someone who can answer me. Thank you!
| Let $x=e^{iz}$. We want to express $x^n+\frac{1}{x^n}=2\cos(nz)$ in terms of $x+\frac{1}{x}=2\cos(z)$, which can be done through Chebyshev polynomials of the first kind:
$$ x^n+\frac{1}{x^n}=2\cos(nz) = 2\,T_n(\cos z) = 2\, T_n\left(\frac{x+\frac{1}{x}}{2}\right).$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3353877",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 1
} |
Continuity of $f(x,y)=\dfrac{(xy+\sqrt{x^2+y^2})\sin^2 (x+y)}{x^2+y^2}$
$f(x,y)=\dfrac{(xy+\sqrt{x^2+y^2})\sin^2 (x+y)}{x^2+y^2}$ if $xy\neq0$ and $f(x,y)=0$ if $xy=0$, check whether $f(x,y)$ is continous at origin.
For it to be continuous $$\lim_{(x,y)\rightarrow (0,0)} f(x,y)=0$$
using epsilon-delta definition,$$\begin{align}\left|\dfrac{(xy+\sqrt{x^2+y^2})\sin^2 (x+y)}{x^2+y^2}\right|&\\\leq\dfrac{xy+\sqrt{x^2+y^2}}{x^2+y^2}&\\\leq\left|\dfrac{xy}{x^2+y^2}\right|+\left|\dfrac{1}{\sqrt{x^2+y^2}}\right|&\\\leq\dfrac{1}{2}+\left|\dfrac{1}{\sqrt{x^2+y^2}}\right|\end{align}$$
So I'm unable to put this in delta does limit DNE? But if I draw its graph, it looks to me continuous.
Please help.
| $\sin^{2}(x+y)\leq (x+y)^{2} \leq 2(x^{2}+y^{2})$. Hence $|f(x,y)|\leq 2|xy|+2\sqrt {x^{2}+y^{2}} \to 0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3355889",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Calculate the limit without l'Hopital rule I have the following limit:
$$
\lim_{x\to 7}\dfrac{x^2-4x-21}{x-4-\sqrt{x+2}}
$$
I could easily calculate the limit = 12 using the l'Hopital rule.
Could you please suggest any other ways to solve this limit without using the l'Hopital rule?
Thank you
| Just another way to do it.
Consider
$$A=\dfrac{x^2-4x-21}{x-4-\sqrt{x+2}}$$ and let $x=y+7$ to work around $y=0$. So,
$$A=\frac{y (y+10)}{y+3-\sqrt{y+9}}$$ Now, using the binomial theorem or Taylor series
$$\sqrt{y+9}=3+\frac{y}{6}-\frac{y^2}{216}+O\left(y^3\right)$$ making
$$A=\frac{y (y+10)}{\frac{5 y}{6}+\frac{y^2}{216}+O\left(y^3\right)}=\frac{ (y+10)}{\frac{5 }{6}+\frac{y}{216}+O\left(y^2\right)}$$
Now, using long division
$$A=12+\frac{17 y}{15}+O\left(y^2\right)$$ which, for sure, shows the limit but also how it is approached.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3356594",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
If $a$, $b$, $c$ are three positive integers such that $a^3+b^3=c^3$ then one of the integer is divisible by $7$ Let on contrary that none of the $a$, $b$, $c$ is divisible by $7$. Then either $a^3\equiv b^3\pmod{7}$ or $b^3\equiv c^3\pmod{7}$ or $c^3\equiv a^3\pmod{7}$.
Now how to go further?
| For any integer $x$ we have $x^3 \equiv -1 \text{ or } 0 \text{ or }1\pmod{7}$.
$a^3+b^3=c^3$
$a^3+b^3+(-c)^3=0 \equiv 0\pmod{7}$ [Take modulo 7 on the whole equaton]
This sum can be zero iff one of them is $-1 \bmod7$, one is $1 \bmod 7$ and one is $0 \bmod 7$. Or if each one of them is $0 \bmod 7$.
The result follows.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3356927",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
Solutions to $\left( \frac{1+3x}{1+2x} \right)^{\frac{1+x}{x}}=2$? I am looking for all non-negative real solutions to
$$
\left( \frac{1+3x}{1+2x} \right)^{\frac{1+x}{x}}=2.
$$
Numerically it seems that there is a unique solution $x \approx 0.4256$. Any ideas how to prove/find it?
| This is probably nowhere near a full answer, but steps could potentially be useful.
Recognize that $\frac{1+x}{x}=\frac1{x}+1$. Then, the equation can be rewritten as
\begin{align}
\bigg(\frac{1+3x}{1+2x}\bigg)^{\frac{1+x}{x}}=\bigg(\frac{1+3x}{1+2x}\bigg)^{\frac1{x}+1}&=2\\
\bigg(\frac{\frac1{x}+3}{\frac1{x}+2}\bigg)^{\frac1{x}+1}&=2
\end{align}
Let $A=\frac1{x}+2$. Then,
\begin{align}
\bigg(\frac{A+1}{A}\bigg)^{A-1}&=2\\
\bigg(1+\frac1{A}\bigg)^{A-1}&=2\\
\bigg(1+\frac1{A}\bigg)^{A}&=2\bigg(1+\frac1{A}\bigg)
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3360276",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
} |
What is $\alpha^{4} + \beta^{4} + \gamma^{4}$? If
$$ \alpha + \beta + \gamma = 14 $$
$$ \alpha^{2} + \beta^{2} + \gamma^{2} = 84 $$
$$ \alpha^{3} + \beta^{3} + \gamma^{3} = 584 $$
What is $\alpha^{4} + \beta^{4} + \gamma^{4}$?
Attempt:
Notice that
$$ \alpha + \beta + \gamma = 14 \implies \alpha^{2} + \beta^{2} + \gamma^{2} + 2(\alpha \beta + \beta \gamma + \alpha \gamma) = 196$$
$$\alpha^{2} + \beta^{2} + \gamma^{2} + 2(\alpha \beta + \beta \gamma + \alpha \gamma) = 196 \implies 84 + 2(\alpha \beta + \beta \gamma + \alpha \gamma) = 196 $$
$$ \alpha \beta + \beta \gamma + \alpha \gamma = 56 $$
Also
$$ \alpha + \beta + \gamma = 14 \implies \alpha^{3} + \beta^{3} + \gamma^{3} + 3(\alpha^{2} \beta + \alpha^{2} \gamma + \beta^{2} \alpha + \beta^{2} \gamma + \gamma^{2} \alpha + \gamma^{2} \beta) + 3 \alpha \beta \gamma = 2744$$
$$ 584 = 2744 - 3(\alpha + \beta + \gamma)(\alpha \beta + \alpha \gamma + \beta \gamma) + 6 \alpha \beta \gamma$$
$$ \alpha \beta \gamma = 32$$
So that
$$ (\alpha + \beta + \gamma)^{4} = ( \alpha^{2} + \beta^{2} + \gamma^{2} + 2(\alpha \beta + \beta \gamma + \alpha \gamma) )^{2} $$
$$ 38416 = (\alpha^{2} + \beta^{2} + \gamma^{2})^{2} + 4 (\alpha \beta + \alpha \gamma + \beta \gamma)(\alpha^{2} + \beta^{2} + \gamma^{2}) + 4 (\alpha \beta + \alpha \gamma + \beta \gamma)^{2}$$
$$ 38416 = \left[ \alpha^{4} + \beta^{4} + \gamma^{4} + 2 \left( (\alpha \beta)^{2} + (\beta \gamma)^{2} + (\alpha \gamma)^{2} \right) \right] + 4 (56)(84) + 4 (56^{2}) $$
$$ 7056 = \alpha^{4} + \beta^{4} + \gamma^{4} + 2 \left( (\alpha \beta + \alpha \gamma + \beta \gamma)^{2} - 2\alpha \beta \gamma(\alpha + \beta + \gamma ) \right) $$
$$ 7056= \alpha^{4} + \beta^{4} + \gamma^{4} + 2 \left( 56^{2} - 2(32)(14) \right) $$
$$\alpha^{4} + \beta^{4} + \gamma^{4} = 2576$$
| Well, its visible clearly that numbers are 8, 4 and 2.
From first two equations, $$y = 1/2 (-\sqrt{-3 x^2 + 28 x - 28} - x + 14)$$
$$z = 1/2 (\sqrt
{-3x^2 + 28 x - 28} - x + 14)$$
Putting them into third equation, we get a cubic equation
$$3x^3 - 42 x^2 + 168 x + 392=584$$ solving which by cardano or any method or writing the symmetric equation for sum and product of roots etc, we see that roots are 2, 4 and 8.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3360541",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Contradiction when solving a linear system with Gauss-Jordan elimination Consider a linear system with an unknown constant $a$:
$$
\left\{
\begin{aligned}
x+(a-1)y+az&=1 \\
ax+ay+az&=1 \\
a^2x+y+z&=a
\end{aligned}
\right.
$$
This gives us an augmented matrix:
$$A= \left[
\begin{array}{ccc|c}
1&a-1&a&1\\
a&a&a&1\\
a^2&1&1&a
\end{array}
\right] $$
From the augmented matrix, it is clear to me that when $a = 1$, we will have one parameter, since there are three unknowns but only two equations.
Let $a = 1$:
$$ \left[
\begin{array}{ccc|c}
1&0&1&1\\
1&1&1&1\\
1&1&1&1
\end{array}
\right] \xrightarrow{R_3 - R_2}
\left[
\begin{array}{ccc|c}
1&0&1&1\\
1&1&1&1\\
0&0&0&0
\end{array}
\right] \xrightarrow{R_2 - R_1}
\left[
\begin{array}{ccc|c}
1&0&1&1\\
0&1&0&0\\
0&0&0&0
\end{array}
\right]
$$
So we get the solution $z = s, y = 0, x = 1 - s \implies \text{Infinite number of solutions}$.
However, if instead I don't consider $a = 1$ but only $a \ne 0$, then I can proceed with Gauss-Jordan elimination to eventually get the following RREF after these series of elementary row operations:
$$
\frac1aR_2\longrightarrow
R_1 \leftrightarrow R_2 \longrightarrow
R_2-R_1 \longrightarrow
R_1-\frac{1}{a^2}R_3 \longrightarrow
\frac{1}{a^2}R_3 \longrightarrow
R_3+R_1 \longrightarrow
a^2R_1 \longrightarrow
\frac{1}{a^2-1}R_1 \longrightarrow
R_2-aR_1 \longrightarrow
R_2+2R_1 \longrightarrow
R_3\leftrightarrow
R_1 \longrightarrow
R_2\leftrightarrow
R_1 \longrightarrow
R_3-R_2 \longrightarrow
R_2-R_1
$$
$$A_\text{RREF}=
\left[
\begin{array}{ccc|c}
1&0&0&\frac1a\\
0&1&0&\frac{-(a-1)}{a}\\
0&0&1&\frac{a-1}{a}
\end{array}
\right]
$$
I had assumed that there was an error in my Gauss-Jordan elimination because this RREF implies that when $a = 1$, we have one and only one unique solution of $x = 1, y = 0, z = 0$. However, I plugged this matrix into MatLab and got the same RREF.
How can it be that when $a = 1$, we get two different solutions from the two row-equivalent matrices?
| If you want to rref and avoid any division at all you'll get
$$\begin{pmatrix}
a(a-2)(a^2-1)& 0 & 0 & (a-2)(a^2-1)\\
0 & -a(a-2)(a+1) & 0 & (a-2)(a^2-1)\\
0 & 0 & a(a^2-1) & (a+1)(a-1)^2
\end{pmatrix}.$$
From here it's eazy to consider all different cases.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3360866",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Given concurrent cevians $AD$, $BE$, $CF$ of $\triangle ABC$, show $\text{area }\triangle DEF\leq \tfrac14\text{}\triangle ABC$
Let $M$ be a point in $\triangle ABC$. Lines $AM$, $BM$, $CM$ respectively intersect $BC$, $CA$, $AB$ at $D$, $E$, $F$. Prove that
$$S_{DEF}\leq \tfrac{1}{4}S_{ABC}$$
where $S_{XYZ}$ indicates the area of $\triangle XYZ$.
| Let the segment ratios be
$$x= \frac {CD}{DB}, \>\>\> y= \frac {AE}{EC}, \>\>\> z= \frac {BF}{FA}$$
According to the Ceva's theorem,
$$xyx=1\tag{1}$$
Evaluate the area ratio
$$\frac{S_{AEF}}{S_{ABC}} = \frac{ \frac 12 AE\cdot AF\sin A}{\frac 12 AB\cdot AC\sin A}
=\frac{1}{(1+z)(1+\frac 1y)}=\frac{y}{(1+z)(1+y)}$$
Similarly,
$$\frac{S_{BDF}}{S_{ABC}} = \frac{z}{(1+x)(1+z)}, \>\>\>\>\> \frac{S_{CDE}}{S_{ABC}} = \frac{x}{(1+y)(1+x)}$$
Then,
$$\frac{S_{DEF}}{S_{ABC}} = 1 - \frac{S_{AEF} + S_{BDF}+S_{CDE} }{S_{ABC}} $$
$$= 1- \frac{y}{(1+z)(1+y)}-\frac{z}{(1+x)(1+z)}-\frac{x}{(1+y)(1+x)}$$
Simplify the expression,
$$\frac{S_{DEF}}{S_{ABC}} = \frac{2}{(1+x)(1+y)(1+z)} $$
Apply the AM-GM inequality to the denominator,
$$(1+x)(1+y)(1+z)=1+x+y+z+xy+yz+zx+xyz \ge 8[(xyz)^4]^{\frac 18}=8$$
Thus
$$\frac{S_{DEF}}{S_{ABC}} \le \frac 14$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3366208",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Angle between a vector and cross product of two vectors
The angle between the vectors $\overrightarrow {a}$ and $\overrightarrow {b}$ is $\pi/3$, $\overrightarrow {b}$ and $\overrightarrow {c}$ is $\pi/4$, $\overrightarrow {c}$ and $\overrightarrow {a}$ is $\pi/6$.
Find the angle between $\overrightarrow {a}$ and $\overrightarrow {b}\times \overrightarrow {c}$
I tried to calculate the angle by drawing the diagram and geometry but it didn't work out for me.
The brute Force method of fixing two of the vectors as convenient position vectors also got messy.
How can we evaluate this? I would prefer a general approach, but if there isn't any, a brute Force approach would also be fine.
| I just got another solution.
Consider $\overrightarrow{a},\overrightarrow{b},\overrightarrow{c}$ to be unit vectors.
$$\left| \overrightarrow {a}\cdot \left( \overrightarrow {b}\times \overrightarrow {c}\right) \right| ^{2}=\begin{vmatrix} \overrightarrow {a}\cdot \overrightarrow {a} & \overrightarrow {a}\cdot \overrightarrow {b} & \overrightarrow {a}\cdot \overrightarrow {c} \\ \overrightarrow {b}\cdot \overrightarrow {a} & \overrightarrow {b}\cdot \overrightarrow {b} & \overrightarrow {b} \cdot \overrightarrow {c} \\ \overrightarrow {c}\cdot \overrightarrow {a} & \overrightarrow {c}\cdot \overrightarrow {b} & \overrightarrow {c} \cdot \overrightarrow{a} \end{vmatrix}$$ (product of two determinants)
$$=\begin{vmatrix} 1 & \dfrac {1}{2} & \dfrac {\sqrt {3}}{2} \\ \dfrac {1}{2} & 1 & \dfrac {1}{\sqrt {2}} \\ \dfrac {\sqrt {3}}{2} & \dfrac {1}{\sqrt {2}} & 1 \end{vmatrix}$$
$$=\frac{\sqrt3}{2\sqrt{2}}-\frac{1}{2}$$
$\implies \cos^2(\theta) \times \frac{1}{2} = \frac{\sqrt3}{2\sqrt{2}}-\frac{1}{2}$
$\implies \theta = \cos^{-1} \left(\sqrt{\sqrt{\frac{3}{2}}-1} \right)$ is the required angle.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3372171",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 3
} |
Find the sum to n terms $$S=1^2+3^2+6^2+10^2+15^2+.......$$
My attempt is as follows:
$$T_n=\left(\frac{n\cdot\left(n+1\right))}{2}\right)^2$$
$$T_n=\frac{n^4+n^2+2\cdot n^3}{4}$$
$$S=\frac{1}{4}\cdot\sum_{n=1}^{n}\left(n^4+n^2+2\cdot n^3\right)$$
Now to solve this one has to calculate $\sum_{n=1}^{n}n^4$ which will be a very lengthy process, is there any shorter method to solve this question?
By the way I calculated $\sum_{n=1}^{n}n^4$ and it came as $\dfrac{\left(n\right)\cdot\left(n+1\right)\cdot\left(2\cdot n+1\right)\cdot\left(3\cdot n^2+3\cdot n-1\right)}{30}$, then I substituted this value into the original equation.
Then I got final answer as $\dfrac{\left(n\right)\cdot\left(n+1\right)\cdot\left(n+2\right)\cdot\left(3\cdot n^2+6\cdot n+1\right)}{60}$
But it took me a very long time to calculate all of this, is there any shorter way to solve this problem?
| Once you’ve rewritten your sum as $\frac14\sum_{k=1}^n k^4+2k^3+k^2$ it’s not terribly difficult to compute this using generating functions if you use some key tools for manipulating them. To wit, if $g(x)$ is the ordinary generating function for the sequence $\{a_n\}_{n=0}^\infty$, then $g(x)/(1-x)$ is the o.g.f. for the sequence of partial sums $\{\sum_{k=0}^na_k\}_{n=0}^\infty$, and similarly, $x\frac d{dx}g(x)$ is the o.g.f. for $\{na_n\}_{n=0}^\infty$. So, starting with the o.g.f. $(1-x)^{-1}$ for the sequence of all ones, we have $$\left\{\sum_{k=0}^n k^2\right\}_{n=0}^\infty \stackrel{ogf}{\longleftrightarrow} \frac1{1-x}\left(x\frac d{dx}\right)^2\frac1{1-x} = {x+x^2 \over (1-x)^4},$$ therefore $$\sum_{k=0}^n k^2 = [x^n]{x+x^2\over(1-x)^4} = [x^{n-1}]\frac1{(1-x)^4} + [x^{n-2}]\frac1{(1-x)^4},$$ which you can compute using the generalized binomial theorem. Similarly, $$\sum_{k=0}^n k^3 = [x^n]{x+4x^2+x^3\over(1-x)^5} \\ \sum_{k=0}^n k^4 = [x^n]{x+11x^2+11x^3+x^4\over(1-x)^6}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3374300",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
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Differentiation using l´Hopital I need to use L´Hopital's rule with this functions:
$$\lim_{x\rightarrow\frac{\pi}{2}} (1-\sin(x))^{\cos(x)}$$
$$\lim_{x\rightarrow\frac{\pi}{4}} (\tan(x))^{\tan(2x)}$$
I take the exponent down using the properties of logarithms and then make the denominator like: $\lim_{x\rightarrow\frac{\pi}{2}} \frac{\cos(x)}{\frac{1}{\ln(1-\sin(x))}}$ but I still get stuck.
| For the first one we have
$$(1-\sin x)^{\cos x}= (1-\sin x)^{\cos x}\frac{(1+\sin x)^{\cos x}}{(1+\sin x)^{\cos x}}=\frac{(\cos^2 x)^{\cos x}}{(1+\sin x)^{\cos x}}\to \frac 11=1$$
indeed by standard limits
*
*$\lim_{x\rightarrow\frac{\pi}{2}} (\cos^2 x)^{\cos x}=\lim_{t\to 0}(t^2)^t=1$
*$\lim_{x\rightarrow\frac{\pi}{2}} (1+\sin x)^{\cos x}=2^0=1$
For the second one
$$\lim_{x\rightarrow\frac{\pi}{4}} (\tan x)^{\tan(2x)}=\lim_{x\rightarrow\frac{\pi}{4}} \left[(1+(\tan x-1))^\frac{1}{\tan x-1}\right]^{\tan(2x)(\tan x-1)}\to e^{-1}=\frac1e$$
indeed by standard limits
*
*$\lim_{x\rightarrow\frac{\pi}{4}} (1+(\tan x-1))^\frac{1}{\tan x-1}\to e$
*$\lim_{x\rightarrow \frac{\pi}{4}} \tan(2x)(\tan x-1)=\lim_{x\rightarrow \frac{\pi}{4}} \frac{2\tan x}{1-\tan^2 x}(\tan x-1)=\lim_{x\rightarrow \frac{\pi}{4}} \frac{-2\tan x}{1+\tan x}=-1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3374957",
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"source": "stackexchange",
"question_score": "4",
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Find a $k$ such that $3^k \equiv -6 \pmod{43}$ I have been trying to find this $k$, but I am stuck.
The only information I could extract was from the Fermat's Little Theorem:
Since $43$ doesn't divide 3 and it is a prime, it follows $3^{42} \equiv 1 \pmod{43}$
However, I have no idea how to proceed from now.
All help is appreciated!
| Notice $3$ and $43$ are relatively prime. Thos if we hae $3m \equiv 3n \pmod{43}$ we can safely conclude that $m \equiv n\pmod{43}$
So if $3^k \equiv -6 \pmod {43}$ then
$3^{k-1} \equiv -2\equiv -45\pmod {43}$
$3^{k-2} \equiv -15$
$3^{k-3}\equiv -5\equiv -48$
$3^{k-4} \equiv -16\equiv 27$
$3^{k-7} \equiv 1\pmod {43}$.
So we can let $k = 7$.
Another way of looking add it is that as $43 \equiv 1 \pmod 3$ we can find a multiple of $3$ by adding or subtraction $43$.
$-6 \equiv 3(-2)\equiv$
$3(-45)\equiv 3^3(-5)\equiv 3^3(-48)$
$3^4(-16)\equiv 3^4(27)\equiv 3^7$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Roots in equation In the equation
$\sqrt{x-2} - \sqrt{6x-11} + \sqrt{x+3} =0$
I got the roots of $x$ being $6$ and $7\sqrt{3}$.
Considering the graph shows only $6$ as being a valid solution, how should I go as figuring this out in the equation itself?
| Square both sides
$$\sqrt{x-2} + \sqrt{x+3} = \sqrt{6x-11} $$
$$2x+1 + 2\sqrt{(x-2)(x+3)}=6x-11$$
$$\sqrt{(x-2)(x+3)}=2x-6$$
Square again
$$x^2+x-6 = 4x^2 - 24x+36$$
$$ 3x^2 - 25x+42=0$$
$$ (x-6)(3x-7)=0$$
$$x=6,\>\>\>x=\frac 73$$
Plug into the original equation and check validity. Only $x=6$ is the true solution. (Edit: See comment below by @Joe as to how the spurious solution $x=\frac 73$ came about.)
| {
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Prove that $(5^{\frac 12} + 2)^{\frac 13}- (5^{\frac 12} - 2)^{\frac 13}$ is an integer and find its value I had proceed this question by taking
$$x =(5^{\frac 12} + 2)^{\frac 13}- (5^{\frac 12} - 2)^{\frac 13}$$
Then
$$x + (5^{\frac 12} -2)^{\frac 13} =(5^{\frac 12} + 2)^{\frac 13}$$
And then cubing both sides and then solving for $x$ by using Cardano's method but I got the same equation at last that is
$$x =(5^{\frac 12} + 2)^{\frac 13}- (5^{\frac 12} - 2)^{\frac 13}$$
Now I don't know how to solve this
| Let $$x=\sqrt[3]{\sqrt5+2}-\sqrt[3]{\sqrt5-2},$$
raise it to the third power:
$$\begin{align}
x^3&=\sqrt5+2-3\sqrt[3]{\sqrt5+2}^2\sqrt[3]{\sqrt5-2}
+3\sqrt[3]{\sqrt5+2}\sqrt[3]{\sqrt5-2}^2-(\sqrt5-2)\\
&=4-3\cdot\sqrt[3]{\sqrt5+2}\sqrt[3]{\sqrt5-2}\cdot\bigl(\sqrt[3]{\sqrt5+2}-\sqrt[3]{\sqrt5-2}\bigr)\\
&=4-3\cdot\sqrt[3]{5-4}\cdot x\\
&=4-3\cdot1\cdot x.
\end{align}
$$
Now show that $x=1$ is the only solution of $x^3+3x-4=0$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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In an integral domain prove that $a=b$
Let R be an integral domain and $a,b\in R$ are elements satisfying $a^7=b^7$ and $a^{12} = b^{12}$. Prove that $a=b$.
My attempt:-
Since $R$ is an integral domain, R is commutative. That is all I could conclude...could you please help me out?
| If $b=0$ then $a^6=0$ but $R$ is an integral domain, so $a=0=b$
We suppose that $b\neq 0$. Then
$a^{12}-b^{12}=(a^6-b^6)(a^6+b^6)=0$
but $R$ is an integral domain, so
$a^6-b^6=0$ or $a^6+b^6=0$
If $a^6=b^6$ then
$b^7=a^7=aa^6=ab^6$ that implies
$b^6(b-a)=0$
but $b\neq 0$ then $b=a$
Conversely, if $a^6=-b^6$, then
$b^7=a^7=aa^6=-ab^6$ so
$b^6(b+a)=0$ that means $b=-a$
In this case we get $b^7=a^7=(-1)^7b^7$. Thus
$(-1)^7=1$ that means $2=0$. So your ring must be $\mathbb{Z}/2\mathbb{Z}$ and we have $b=1=a$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Find an approximate sum of the series $\sum_{k=0}^{\infty}\frac{1}{2^{2k+1}(2k+1)}$ $$
S=\sum_{k=0}^{\infty}\frac{1}{2^{2k+1}(2k+1)}\ \ \ (*)
$$
I have to find $S_n: |S-S_n|<\epsilon=10^{-4}$
Also, I cannot use expansion series for any functions.
This is what I came up with so far:
$$
a_k=\frac{1}{2^{2k+1}(2k+1)}\leqslant\frac{1}{(2k+1)^2}=b_k\\
\sum_{k=0}^{\infty}b_k \text{ converges. Therefore, $(*)$ converges too.}
$$
Also, I noticed that $a_k$ is a monotonically decreasing function. But I still do not know how to apply all these calculations to the problem
| $S
=\sum_{k=0}^{\infty}\frac{1}{2^{2k+1}(2k+1)}\ \ \ (*)
$
Let
$t_n
=\sum_{k=n}^{\infty}\frac{1}{2^{2k+1}(2k+1)}
$.
Then
$\begin{array}\\
t_n
&\lt\sum_{k=n}^{\infty}\frac{1}{2^{2k+1}(2n+1)}\\
&=\frac1{2n+1}\sum_{k=n}^{\infty}\frac{1}{2^{2k+1}}\\
&=\frac1{2n+1}\sum_{k=0}^{\infty}\frac{1}{2^{2k+2n+1}}\\
&=\frac1{(2n+1)2^{2n+1}}\sum_{k=0}^{\infty}\frac{1}{4^k}\\
&=\frac1{(2n+1)2^{2n+1}(1-1/4)}\\
&=\frac{4}{3(2n+1)2^{2n+1}}\\
&=\frac{1}{3(2n+1)2^{2n-1}}\\
\end{array}
$
Therefore,
you want
$\frac{1}{3(2n+1)2^{2n-1}}
\le 10^{-4}
$
or
$3(2n+1)2^{2n-1}
\ge 10^{4}
$.
Since
$2^{14} = 16384$,
start at $n=7$
and go down.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3382064",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Solve in $\mathbb N^{2}$ the following equation : $5^{2x}-3\cdot2^{2y}+5^{x}2^{y-1}-2^{y-1}-2\cdot5^{x}+1=0$ Question :
Solve for natural number the equation :
$5^{2x}-3\cdot2^{2y}+5^{x}2^{y-1}-2^{y-1}-2\cdot5^{x}+1=0$
My try :
Let : $X=5^{x}$ and $Y=2^{y}$ so above equation
equivalent :
$2X^{2}+(Y-4)X-6Y^{2}-Y+2=0$
We solve this equation for $X$
$\Delta =(7Y)^{2}$ mean : $X_{1}=\frac{3}{2}Y+1$ and $X_{2}=1-2Y$
From here how I can find $X$ and $Y$ , this is
all my effort ?
Thanks!
| Now substitute back: you get that $5^x=3\cdot 2^{y-1}+1$ or $5^x=1-2^{y+1}$. The second one is impossible cause the RHS is $\leq 0$. For the first one to hold, looking mod 3 you see that $x$ needs to be even, so write $x=2x'$. Then $(5^{x'}-1)(5^{x'}+1)=3\cdot 2^{y-1}$. So the only chance is that $5^{x'}-1=2^{y_1}$ and $5^{x'}+1=3\cdot 2^{y_2}$ for some $y_1,y_2\in \mathbb N$, or the other way round. Substract term by term and you'll see immediately that the minimum of $y_1,y_2$ has to be either $0$ or $1$. Now you can check by hand case by case.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Find all $n$ natural numbers such that $10\mid n^{10}+1$ Determine all natural numbers $n$ such that :
$10$ divisor of $n^{10}+1$
My attempt :
Let $n=r(\mod{10})$ so $n^{10}+1=(r^{10}+1)(\mod{10})$
This mean : $r^{10}+1=0(\mod{10})$
Now $r\in {0,1,2,3,4,5,6,7,8,9}$ after try I get $r=3,7$
So : $n=10k+3,10k+7$
Is my work correct ?
Please I need other simple method to computing
| By Fermat's theorem, $n^5 \equiv n \bmod 5$. Therefore, $n^{10} \equiv n^2 \bmod 5$ and so $n^{10} +1 \equiv n^2 +1 \bmod 5$. Thus, if $10\mid n^{10}+1$, then $n^2 \equiv 4 \bmod 5$ and so $n \equiv \pm 2 \equiv 2,3 \bmod 5$. Thus, $n \equiv 2,3,7,8 \bmod 10$. Since $n$ is odd, we're left with $n \equiv 3,7 \bmod 10$.
Alternatively,
By Euler's theorem, $n^4 \equiv 1 \bmod 10$, and so $n^{10} \equiv n^2
\bmod 10$. Thus, if $10\mid n^{10}+1$, then $n^2 \equiv 9 \bmod 10$. Therefore, $n \equiv 3,7 \bmod 10$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Find the explicit form of $ \sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n(n+2)}x^{n-1} $. Find the explicit form of
$$
\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n(n+2)}x^{n-1}.
$$
Let $S(x)=\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n(n+2)}x^{n-1}$. It has radius of convergence $1$.
Let $S_1(x)=xS(x)$. Then $S_1'(x)=\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{(n+2)}x^{n-1}$ for $|x|<1$.
Let $S_2(x)=x^3S_1'(x)$. Then $S_2'(x)=\sum_{n=1}^{\infty}(-x)^{n-1}=\frac{x^2}{1+x}$.
By integration, I obtained $S_1'(x)=\frac{1}{2x}-\frac{1}{x^2}+\frac{\ln (x+1)}{x^3}$. Then how to obtain $S(x)$? Or there is other method to do this problem?
| @ Tao X : Your method is not the simplest. But what you did is correct :
$$S_1'(x)=\frac{1}{2x}-\frac{1}{x^2}+\frac{\ln (x+1)}{x^3}$$
Then integrate :
$$S_1(x)=\int S_1'(x)dx=\frac{1}{2x}-\frac{\ln (x+1)}{2x^2}+\frac12\ln(x+1)+C$$
We know that $S_1(0)=0$. This allows to find $C$. Expand $\ln(x+1)$ to series around $x=0$ :
$\frac{x-\ln (x+1)}{2x^2}+\frac12\ln(x+1)=\frac{x-(x-\frac12 x^2+...)}{2x^2}+\frac12(x-\frac12 x^2+...)=\frac14+\epsilon(x)$ with $\epsilon(0)=0$.
Thus $C=-\frac14$ and $S_1(x)=\frac{1}{2x}-\frac{\ln (x+1)}{2x^2}+\frac12\ln(x+1)-\frac14$ .
$$S(x)=\frac{1}{2x^2}-\frac{\ln (x+1)}{2x^3}+\frac{1}{2x}\ln(x+1)-\frac{1}{4x}$$ .
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3386371",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Divergence theorem, Gauss's theorem $$\mathbf F(x,y,z)=x\mathbf i + y\mathbf j+z\mathbf k$$ and
$$ D=\left\{x,y,z\in \mathbf R^3 : 0\le z \le1-x^2-y^2\right\}.$$
I want to calculate both
$$\iiint_D\nabla\cdot\mathbf F\,dV = \oint_{\delta D}\mathbf F \cdot\mathbf N\,dS.$$
So far:
$$\nabla f=1+1+1=3$$
$$\oint_{\delta D}f\cdot n\,dS=\iiint_D=3\,dV$$
and after that
$$v(D)=\int_0^\pi\int_0^{2\pi}\int_0^rr^2\sin(\phi)\,dr\,d\theta \,d\phi=\frac{4\pi r^3}{3}.$$
But what is the $r$ in there? Is it $r=1$?
$$\frac{4\pi1^3}{3}=\frac{4\pi}{3}?$$
And after that:
$$\oint_D\mathbf F\cdot \mathbf N\,dS=\int_0^\pi\int_0^{2\pi}r(\theta,\phi)\cdot \mathbf n\,dS=\int_0^\pi\int_0^{2\pi}r^3\sin(\phi)\,d\theta\,d\phi=4\pi r^3,$$ and this $4\pi r^3$ should be equal to $\dfrac{4\pi}{3}$ but is it not. There must be some mistake somewhere but where?
| As I said in the comments, $D$ is not a ball. It's bounded below by the unit disk in the $xy$-plane, and above by the paraboloid $z=1-x^2-y^2$. (You may be thinking of $z = \sqrt{1-x^2-y^2}$; that surface is the upper unit hemisphere.) In cylindrical coordinates,
$$
D = \left\{(r,\theta,z) : 0 \leq r \leq 1,\ 0 \leq z \leq 1-r^2 \right\}
$$
So
\begin{align*}
\iiint\limits_{D} \nabla \cdot \mathbf{F}\,dV
&= \int_0^{2\pi} \int_0^1 \int_0^{1-r^2} 3 r\,dz\,dr \,d\theta
\\&= 3\cdot 2\pi \int_0^1 r(1-r^2) \,dr
\\&= 3 \cdot 2\pi \cdot \frac{1}{4} = \frac{3\pi}{2}
\end{align*}
The boundary of $D$ is the union of two surfaces:
\begin{align*}
S_1 &= \left\{(r,\theta,0) : 0 \leq r \leq 1 \right\} \\
S_2 &= \left\{(r,\theta,1-r^2) : 0 \leq r \leq 1 \right\} \\
\end{align*}
As oriented surfaces, $\partial D = S_2 - S_1$. On $S_1$, the normal vector is vertical, and $\mathbf{F}$ has no vertical component. Therefore,
$$
\iint\limits_{S_1} \mathbf{F}\cdot \mathbf{N}\,dS
= \iint\limits_{S_1} (x \mathbf{i} + y \mathbf{j}+z\mathbf{k}) \cdot \mathbf{k}\,dA
= \iint\limits_{S_1} 0\,dA = 0
$$
On top, $S_2$ is the graph of $z=1-x^2-y^2$, so the vector $\left<2x,2y,1\right>$ is normal to $S_2$. Therefore
\begin{align*}
\iint\limits_{S_2} \mathbf{F}\cdot \mathbf{N}\,dS
&= \iint\limits_{S_1}\left<x,y,1-x^2-y^2\right>\cdot\left<2x,2y,1\right>\,dA
\\&= \iint\limits_{S_1}(1+x^2+y^2)\,dA = \iint\limits_{S_1}(1+r^2)\,dA
\\&= \int_0^{2\pi} \int_0^1 (1+r^2)r\,dr\,d\theta
\\&= 2\pi \int_0^1 \left(r + r^3\right)\,dr = 2\pi \cdot\frac{3}{4} = \frac{3\pi}{2}
\end{align*}
Therefore
$$
\iint\limits_{\partial D} \mathbf{F}\cdot \mathbf{N}\,dS = \frac{3\pi}{2}
$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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To prove that $a_n=[x^n]\frac {x-2x^2+x^3}{1-2x+x^2-x^4}$ To prove: $$a_n=[x^n]\frac {x-2x^2+x^3}{1-2x+x^2-x^4}$$
Given that, $a_n$is the number of compositions of $n$ in which the number of parts is odd with the first part being equal to the number of parts.
| For $k\geq 1$ and $n\geq 3$, the number of positive integer solutions of
$$(2k+1)+x_2+x_3+\dots+x_{2k+1}=n$$
that is of
$$x_2+x_3+\dots+x_{2k+1}=n-(2k+1)$$
which is equal to
$$\binom{n-(2k+1)-1}{2k-1}.$$
Hence
$$a_n=\sum_{1\leq k\leq \lfloor (n-1)/4\rfloor}\binom{n-(2k-1)-1}{2k-1}.$$
Now by partial fraction decomposition
$$\frac{x-2x^2+x^3}{1-2x+x^2-x^4}=
\frac{1}{2(1-x+x^2)}-\frac{1-2x}{2(1-x-x^2)}$$
and
$$\begin{align}[x^n]\frac{1}{1-x+x^2}&=[x^n]\sum_{m=0}^{\infty}(x-x^2)^m=[x^n]\sum_{m=0}^{\infty}
\sum_{j=0}^m(-1)^j\binom{m}{j}x^{m+j}=\sum_{0\leq j\leq \lfloor n/2\rfloor}
(-1)^j\binom{n-j}{j},\\
[x^n]\frac{1}{1-x-x^2}&=[x^n]\sum_{m=0}^{\infty}(x+x^2)^m=[x^n]\sum_{m=0}^{\infty}
\sum_{j=0}^m\binom{m}{j}x^{m+j}=\sum_{0\leq j\leq \lfloor n/2\rfloor}\binom{n-j}{j}.
\end{align}$$
Can you take it from here?
| {
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"timestamp": "2023-03-29T00:00:00",
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Maximum value of $a+b+c$ in an inequality Given that $a$, $b$ and $c$ are real positive numbers, find the maximum possible value of $a+b+c$, if
$$a^2+b^2+c^2+ab+ac+bc\le1.$$
From the AM-GM theorem, I have
$$a^2+b^2+c^2+ab+ac+bc\geq 6\sqrt[6]{a^4b^4c^4} = 6\sqrt[3]{a^2b^2c^2} \\
6\sqrt[3]{a^2b^2c^2} \le1 \\
a^2b^2c^2 \le \frac{1}{216} \\
abc \le \frac{\sqrt{6}}{36}$$
However, I don't know where to go from here.
| Notice that
$$ab+bc+ca\leq\frac{(a+b+c)^2}{3}.$$
Hence
\begin{align}
a^2+b^2+c^2+ab+bc+ca&=(a+b+c)^2-(ab+bc+ca)\\&\geq (a+b+c)^2-\frac{(a+b+c)^2}{3}\\&=\frac{2}{3}(a+b+c)^2.
\end{align}
That is
$$\frac{2}{3}(a+b+c)^2\leq 1,$$
which we get
$$a+b+c\leq\sqrt{\frac{3}{2}}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3401308",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Rational with minimal denominator between two rationals My question from an easy problem.
$p,q$ are positive integers such that $$
\frac{5}{9}<\frac{p}{q}<\frac{4}{7}
$$ find $p,q$ such that $q$ is the smallest number that satisfies this inequality.
Draw the line of $ y<\frac{9}{5}x$ and $y>\frac{7}{4}x$ , we can "observe" that $\frac{9}{16}$ is such number.
However, if the question becomes
$a,b,c,d$ are positive integers such that $$\frac{a}{c}<\frac{b}{d}
$$ find $p$,$q$ such that $q$ is the smallest number that satisfies the inequality
$$\frac{a}{c}<\frac{p}{q}<\frac{b}{d}$$
No idea about this.
| What do we get with continued fractions?
$\dfrac{5}{9}=\dfrac{1}{1+\dfrac{1}{1+\dfrac{1}{4}}}$
$\dfrac{4}{7}=\dfrac{1}{1+\dfrac{1}{1+\dfrac{1}{3}}}$
The fractions are identical until we get to the last layer where one has a $4$ and the other has a $3$. Were there an integer between $3$ and $4$ we could replace the last layer with the smallest such integer, following the accepted answer here.
We don't have such an integer between $3$ and $4$ so this does not appear to work. But we can force the issue by rendering
$4=3+\dfrac{1}{1}$
$3=3+\dfrac{1}{M}$
where $M$ is taken as approaching infinity. Then we have
$\dfrac{5}{9}=\dfrac{1}{1+\dfrac{1}{1+\dfrac{1}{3+\dfrac{1}{1}}}}$
$\dfrac{4}{7}=\dfrac{1}{1+\dfrac{1}{1+\dfrac{1}{3+\dfrac{1}{M}}}}$
Now we put $2$ as the smallest integer between $1$ and $M$ to get an intervening fraction with a minimal denominator. Thus
$\dfrac{1}{1+\dfrac{1}{1+\dfrac{1}{3+\dfrac{1}{2}}}}=\dfrac{9}{16}$
| {
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The sum of the series
$$\frac{1}{\log_24}+\frac{1}{\log_44} + \frac{1}{\log_84}.....\frac{1}{\log_{2^n}4}$$
MY SOLUTION
We can write it as
$$\frac{\log2}{\log4} + \frac{\log4}{\log4} + ....\frac{\log 2^n}{\log4}$$
$$=\frac{1}{\log4}\left[\log 2 + \log 4....\log 2^n\right]$$
$$=\frac{1}{\log 4}\left [\log(2.4.8....2^n)\right]$$
$$\frac{1}{\log 4}[log(2^n.2^{\frac{(n)(n+1)}{2}}
)]$$
$$\frac{2n+n^2+n}{4}$$
But the answer is $\frac{n^2+n}{4}$
| $\dfrac{1}{2\log 2}[\log 2(1+2+3+....n)]=$
$\dfrac{n(n+1)}{2\cdot 2}$;
Used:
1)$\log 4= \log 2^2=2\log/2$;
2)$\log 2+\log 2^2+.......\log 2^n=$
$\log 2 +2\log 2 +....+n\log 2=$
$\log 2(1+2+3+.....n)=$
$\log 2\dfrac{n(n+1)}{2}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3402896",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
How do I solve a problem with term $a^{n} + b^{n}$?
Given two non-zero numbers $x$ and $y$ such that $x^{2} + xy + y^{2} = 0$.
Find the value of
$$\left(\frac{x}{x + y}\right)^{2013} + \left(\frac{y}{x + y}\right)^{2013}$$.
I found out that $(x + y)^2 = xy$ and I'm stuck at $\frac{x^{2013} + y^{2013}}{(x + y)^{2013}}$
Does anyone know how to solve this?
| We have
$${y}^{2}=-{x}^{2}-xy\\
{y}^{3}=-{x}^{2}y-xy^2
$$
Then after substition and elimination $y, y^2,y^3$ we get
$$
(x+y)^3={x}^{3}+3\,{x}^{2}y+3\,x{y}^{2}+{y}^{3}=-x^3.
$$
Thus
$$
\left(\dfrac x{x+y}\right)^{2013}=\left(\left(\dfrac x{x+y}\right)^{3}\right)^{671}=\left(\dfrac{ x^3}{(x+y)^3}\right)^{671}=\left(\dfrac{ x^3}{-x^3}\right)^{671}=-1.
$$
Similarly $$\left(\dfrac y{x+y}\right)^{2013}=-1.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3404062",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 7,
"answer_id": 4
} |
Following Fibonacci Prove that $ F_ {n + 2} = \sqrt{\frac{F_n {F_ {n + 1} ^ 2}(3 {F_n} +4 {F_ {n + 1}}) + 1 }{{F_n} ^ 2 + {F_ {n + 1}} ^ 2}} $?
I discovered this property from an attempt to solve the following problem:
Defining $ A_n = \sqrt{{F_n} ^ 2 + {F_{n + 2}} ^ 2}, $ the numbers $ A_n, $ $ A_ {n + 1} $ and $ A_{n + 2} $ are the length measures of the sides of a triangle whose area is $ \frac{1}{2} $ unit.
(I also put this in the Wikipedia article about the Fibonacci Sequence).
| The hypothesis is equivalent to $g(n)=0$ with$$g(n):=F_nF_{n+1}^2(3F_n+4F_{n+1})+1-F_n^2F_{n+2}^2-F_{n+1}^2F_{n+2}^2\\=-F_n^4-2F_n^3F_{n+1}+F_n^2F_{n+1}^2+2F_nF_{n+1}^3-F_{n+1}^4+1.$$Since $F_0=0,\,F_1=1$, we have the base case $g(0)=0$. For the inductive step use the abbreviations $a:=F_n,\,b:=F_{n+1}$, so$$g(n+1)-g(n)=a^4-b^4+2a^3b-2b^3(a+b)-a^2b^2+b^2(a+b)^2-2ab^3+2b(a+b)^3+b^4-(a+b)^4\\=\color{red}{a^4}\color{blue}{-b^4}\color{orange}{+2a^3b}\color{limegreen}{-2ab^3}\color{blue}{-2b^4}\color{hotpink}{-a^2b^2+a^2b^2}\color{limegreen}{+2ab^3}\color{blue}{+b^4}\color{limegreen}{-2ab^3}\color{orange}{+2a^3b}\color{hotpink}{+6a^2b^2}\color{limegreen}{+6ab^3}\color{blue}{+2b^4}\color{blue}{+b^4}\color{red}{-a^4}\color{orange}{-4a^3b}\color{hotpink}{-6a^2b^2}\color{limegreen}{-4ab^3}\color{blue}{-b^4}.$$I've colour-coded the terms progressing through the rainbow, except for yellow being changed to pink as yellow is hard to read, so red is $a^4$, orange is $a^3b$, pink is $a^2b^2$, green is $ab^3$ and blue is $b^4$. And you can verify each colour has zero coefficient overall.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3407394",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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} |
Limiting value of the coefficients of the power series $(1-x)^{-\frac{1}{2}}$ Suppose we have the power series $(1-x)^{-\frac{1}{2}}$. We know that this power series converges iff $|x|<1$. Suppose $a_n$ denote the coefficieis nt of this power series. So $a_0=1, a_1=\frac{1}{2}, a_2= \frac{3}{8}, a_3=\frac{5}{16} \ldots \text{ so on}$. My question is what is $\lim_{n\to \infty} a_n$. It is clear that $a_n>0$ for every $n$ and $a_n$ is a decreasing sequence, so limit does exist. But I couldn't find the exact value of the limit. Since for $x=1$ the power series doesn't converge, we cannot say anything from there.
Thanks in advance for any kind of help.
| Determine a general formula for the coefficients
How do we compute these coefficients? The way to expand this is:
$$(1-x)^{-1/2}=\frac1{0!}(-x)^0+\frac1{1!}(-x)^1\left(-\frac12\right)+\frac1{2!}(-x)^2\left(-\frac12\right)\left(-\frac32\right)+\frac1{3!}(-x)^3\left(-\frac12\right)\left(-\frac32\right)\left(-\frac52\right)+\cdots$$
Notice that as we go from $a_n$ to $a_{n+1}$, we multiply each coefficient by $-\frac1{n+1}\left(-\frac12-n\right)=\frac1{n+1}\left(\frac{2n+1}2\right)$. So We have a recurrance relation $$\begin{align}a_{n+1}&=\frac{2n+1}{2(n+1)}a_n\\&=\frac{(2n+1)(2n-1)\cdots(1)}{2^{n+1}(n+1)!}a_0\\&=\frac{(2n+1)!}{2^{n+1}(n+1)!}\cdot\frac{1}{(2n)(2n-2)\cdots(2)}\\&=\frac{(2n+1)!}{2^{n+1}(n+1)!}\cdot\frac{1}{2^nn!}\\&=\frac{(2n+1)!}{2^{2n+1}n!(n+1)!}\end{align}$$
So we have, for $n>0$ $$a_n=\frac{(2n-1)!}{2^{2n-1}n!(n-1)!}$$
Evaluate coefficients in large $n$ limit
Stirling's approximation tells us that for large $n$, $\,\,n!\sim n^{1/2}n^ne^{-n}$. Using this, for large $n$, $$\begin{align}a_n&\sim\left(\frac{2n-1}{n(n-1)}\right)^{1/2}\frac{2^{2n-1}n^{2n-1}e^{-2n+1}}{2^{2n-1}n^ne^{-n}n^{n-1}e^{-n+1}}\\&\sim\left(\frac{2n-1}{n^2-n}\right)^{1/2}\\&\sim\left(\frac2n\right)^{1/2}\to 0\end{align}$$
So $$\lim_{n\to\infty}a_n=0$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Prove that $\sqrt[n]{n} < (1 + \frac{1}{\sqrt{n}})^2$ for all n in the naturals. I need to prove that $\sqrt[n]{n} < (1 + \frac{1}{\sqrt{n}})^2$ for all n in the naturals.
I started by using Bernoulli's inequality:
$(1+\frac{2}{\sqrt{n}}) < (1 + \frac{1}{\sqrt{n}})^2$
I can say that:
$(1+\frac{2}{\sqrt{n}}) = (1+\frac{2\sqrt{n}}{n})$
I can also subtract the one and divide by 2 on the left side without changing the inequality (because it makes it even smaller): $(\frac{\sqrt{n}}{n}) < (1 + \frac{1}{\sqrt{n}})^2$
But now I am stuck...
| $(1+x)^n \ge (n^2/4) x^2 $, $n \ge 1$, $x \ge 0$.
Proof:
$(1+x)^n=$
$1+nx + (n(n-1)/2!)x^2+...\gt (n(n-1)/2)x^2 \ge (n^2/4)x^2,$
since we have
$(n)((n-1)/2) \ge n( n/4)$, for $n\ge 2.$
Let $x=2/√n$, in
$(1+x)^n > (n^2/4)x^2$;
$1+2/√n \gt \sqrt[n]{n}$, $n \ge 1$.
Finally
$(1+2/√n)^2 >(1+2/√n) \gt \sqrt[n]{n}$, $n \ge 1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3409831",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 4
} |
$\int \frac{1}{(x^2-4)^2}dx$ Calculate:
$$\int \frac{1}{(x^2-4)^2}dx.$$
I tried Partial Fractions method first I write:
$$\frac{1}{(x^2-4)^2}=\frac{A}{X-2}+\frac{Bx+C}{(x-2)^2}+\frac{D}{x+2}+\frac{Ex+F}{(x+2)^2}.$$
We have:
$$A(x-2)(x+2)^2+(Bx+C)(x+2)^2+D(x+2)(x-2)^2+(Ex+F)(x-2)^2=1.$$
$$(A+B+D+E)x^3+(4A-2A+4B+C-4D+2D-4E+F)x^2+(4A-8+4B+4C+4D-8D+4E-4F)x+(-8A+4C+8D+4F)=1$$
So:
$$A+B+C+D+E=0$$
$$2A+4B+C-2D-4E+F=0$$
$$A+B+C-D+E=2$$
$$-8A+4C+8D+4F=1.$$
But how to find $A$, $B$, $C$, $D$, $E$, $F$?
I also tried substitution $$x=2\sec t$,$
but It caused some difficulty.
| It is $$\frac{1}{(x^2-4)^2}=1/16\, \left( x-2 \right) ^{-2}-1/32\, \left( x-2 \right) ^{-1}+1/16\,
\left( x+2 \right) ^{-2}+1/32\, \left( x+2 \right) ^{-1}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3409950",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 3
} |
Find the minimum of :$P=a+b+c-ab-bc-ca$ soluLet $a,b,c$ be positive real numbers and $a+b+c+abc=4$.
We can rewrite the first equation as $a+b+c=4-abc.$ Then,
\begin{align*}
P&=a+b+c-ab-bc-ca\\&=(4-abc)-ab-bc-ca\\&=4-abc-ab-bc-ca-(a+b+c+1)+(a+b+c+1)\\&=4-(abc+ab+bc+ca+a+b+c+1)+a+b+c+1\\&=4-(a+1)(b+1)(c+1)+a+b+c+1\\&=5+a+b+c-(a+1)(b+1)(c+1)\\&=5+(4-abc)-(a+1)(b+1)(c+1)\\&=9-abc-(a+1)(b+1)(c+1).
\end{align*}It seems that both $abc$ and $(a+1)(b+1)(c+1)$ are maximized when $a=b=c,$ which would give $a=b=c=1$ and then $P=9-(1)(1)(1)-(2)(2)(2)=9-(1+8)=0.$ However, I'm not sure how to prove that statement.
| We will show that $a+b+c \geq ab+bc+ca$, from which it follows that the minimum is 0. This is attained when $a=b=c=1$ or permutations of $ (2,2,0)$.
The equality conditions suggest that Shur's inequality is involved, and we likely want to use
$$ a^3 + b^3 + c^3 + 3abc \geq \sum ab(a+b)$$.
Let's rewrite this in terms of $P=a+b+c$, $Q=ab+bc+ca$, and $R=abc$ which are the terms that appear in the given inequality and condition. Schur's inequality can be rewritten as:
$$P^3 -4PQ + 9 R \geq 0.$$
The given condition can be written as $P + R = 4$.
Proof that $P \geq Q$.
Proof by contradiction. Suppose not, then $ P < Q$. Then,
$$ \begin{array} \text{9}R & \geq 4PQ - P^3 & \text{(Schur's)} \\
& > 4P^2 - P^3 & \text{(Contradiction assumption)} \\
& = P^2 (4-P) \\
& = P^2 R. & \text{(Given condition)} \\ \end{array}$$
So $9 > P^2 \Rightarrow 3 > P$.
Then, $R \leq \sqrt[3]{\frac{P}{3}} < 1$,
which contradicts $4 = P+R < 3 + 1 = 4$.
Hence $P \geq R$.
Note: I'm not happy with the structure of this proof. I wish there was a way to write it up more directly.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3410321",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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Calculate the approximate value I am supposed to calculate the approximate value of $$\cos 151^\circ$$ My idea was that I can divide it in the form: $$\cos 90^\circ+ 61^\circ= \cos \frac{\pi}{2}+ \left (\frac{\pi}{3} +\frac{\pi}{180} \right )$$ Then I use the derivation for cosx:
$$-\sin \frac{\pi}{2}\left ( \frac{\pi}{3} +\frac{\pi}{180} \right )=-\frac{61\pi}{180}+61^\circ$$
But I guess, it is not correct.
Can anyone help me?
| \begin{equation}
\cos\left(150^\circ\right)=\cos\left(90^\circ-\left(-60^\circ\right)\right)=\sin\left(-60^\circ\right)=-\sin\left(60^\circ\right)=-\frac{\sqrt3}2\\
\sin\left(150^\circ\right)=\sin\left(90^\circ-\left(-60^\circ\right)\right)=\cos\left(-60^\circ\right)=\cos\left(60^\circ\right)=\frac12
\end{equation}
Using the fact that $\cos\left(a+b\right)=\cos\left(a\right)\cos\left(b\right)-\sin\left(a\right)\sin\left(b\right)$, we get
\begin{equation}
\cos\left(151^\circ\right)=\cos\left(150^\circ+1^\circ\right)=\cos\left(150^\circ\right)\cos\left(1^\circ\right)-\sin\left(150^\circ\right)\sin\left(1^\circ\right)\\
=-\left(\frac{\sqrt3}2\cos\left(1^\circ\right)+\frac12\sin\left(1^\circ\right)\right)
\end{equation}
Since
\begin{equation}
\sin\left(x\right)=x-\frac{x^3}{3!}+\frac{x^5}{5!}+\cdots\\
\cos\left(x\right)=1-\frac{x^2}{2!}+\frac{x^4}{4!}+\cdots\\
\end{equation}
We have
\begin{equation}
\sin\left(1^\circ\right)=\sin\left(\frac\pi{180}\right)\approx\frac\pi{180}-\frac1{3!}\left(\frac\pi{180}\right)^3\\
\cos\left(1^\circ\right)=\cos\left(\frac\pi{180}\right)\approx1-\frac1{2!}\left(\frac\pi{180}\right)^2
\end{equation}
Plugging these approximate values into the above formula for $\cos\left(151^\circ\right)$ we get
\begin{equation}
\cos\left(151^\circ\right)\approx-\left(\frac{\sqrt3}2\left(1-\frac1{2!}\left(\frac\pi{180}\right)^2\right)+\frac12\left(\frac\pi{180}-\frac1{3!}\left(\frac\pi{180}\right)^3\right)\right)\\
\approx-0.8746
\end{equation}
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove the equality (Taylor series).
Prove the equality:
$$
\frac{1}{3}\left(e^x+2e^{-x/2}\cos\frac{x\sqrt{3}}{2}\right)=
\sum_{n=0}^{\infty}\frac{x^{3n}}{(3n)!},\ \ -\infty<x<+\infty
$$
I tried to apply Euler's formula ($e^{ix}=\cos x+i\sin x$) to this problem but it went rather unsuccessful. Here is what I did:
$$
e^{-x/2}=e^{i(ix/2)}=\cos\frac{ix}{2}+i\sin\frac{ix}{2}\Rightarrow\\
\Rightarrow 2e^{-x/2}\cos\frac{x\sqrt{3}}{2}=2\cos\frac{ix}{2}\cos\frac{x\sqrt{3}}{2}+
2i\sin\frac{ix}{2}\cos\frac{x\sqrt{3}}{2}=\\
=\cos\frac{x(i+\sqrt{3})}{2}+\cos\frac{x(i-\sqrt{3})}{2}+
i\sin\frac{x(i+\sqrt{3})}{2}+i\sin\frac{x(i-\sqrt{3})}{2}=\\
=e^{ix(i+\sqrt{3})/2}+e^{ix(i-\sqrt{3})/2}=
e^{x(-1+i\sqrt{3})/2}+e^{x(-1-i\sqrt{3})/2}
$$
Then I tried to use Maclaurin series for $e^{x(-1+i\sqrt{3})/2}$ and $e^{x(-1-i\sqrt{3})/2}$ after which I got completely befuddled because it seemed to me that I had only complicated the initial problem.
So, if anyone could help me, I would appreciate it.
| Hint: (A followup to Lord Shark the Unknown's observation) You're already half-way there. You've established that
\begin{align}
\frac{1}{3}\left(e^x + 2e^\frac{-x}{2}\cos \frac{x\sqrt{3}}{2}\right) &=\frac{1}{3}\left(e^x + e^\frac{x\left(-1+i\sqrt{3}\right)}{2}+ e^\frac{x\left(-1-i\sqrt{3}\right)}{2}\right)\\
&= \frac{1}{3}\left(e^{z_1 x} + e^{z_2x} + e^{z_3}x
\right)\ ,
\end{align}
where $\ z_1=1\ $, $\ z_2=\frac{x\left(-1+i\sqrt{3}\right)}{2}\ $, and $\ z_3=\frac{\left(-1-i\sqrt{3}\right)}{2}\ $ are the three cube roots of unity. If you now use the expansions
\begin{align}
e^{z_ix}&= \sum_{n=0}^\infty\frac{z_i^nx^n}{n!}\\
&=\sum_{n=0}^\infty\left(\frac{z_i^{3n}x^{3n}}{(3n)!}+\frac{z_i^{3n+1}x^{3n+1}}{(3n+1)!}+\frac{z_i^{3n+2}x^{3n+2}}{(3n+2)!}\right)\\
&=\sum_{n=0}^\infty \frac{x^{3n}}{(3n)!}\left(1 +\frac{z_ix}{3n+1}+\frac{z_i^2x^2}{3n+2}\right)\ ,
\end{align}
and the observations that $\ z_1^2=z_2\ $, $\ z_2^2=z_1\ $, and $\ z_1 + z_2 + z_3=0\ $, you should be able to complete the demonstration.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Find eigenvalues of $x^2 \frac{d^2\phi}{dx^2} + x \frac{d\phi}{dx} + \lambda \phi = 0$ with boundary conditions $\phi(1) = \phi(b) = 0$
Since this is an equidimensional equation, determine all positive eigenvalues.
\begin{align*}
x^2 \frac{d^2\phi}{dx^2} + x \frac{d\phi}{dx} + \lambda \phi &= 0 \\
\phi(1) &= 0 \\
\phi(b) &= 0 \\
\end{align*}
The given answer that I need to find the solution for is for $n = 1,2,\ldots$:
\begin{align*}
\lambda_n &= \left( \frac{n \pi}{\ln b} \right)^2 \\
\end{align*}
If we multiply by $1/x$ and rearrange we can get this in Sturm-Liouville form with $p(x) = x, q(x) = 0, \sigma(x) = 1/x$:
\begin{align*}
x \frac{d^2\phi}{dx^2} + \frac{d\phi}{dx} + \frac{1}{x} \lambda \phi &= 0 \\
\frac{d}{dx} \left( x \frac{d\phi}{dx} \right) + \lambda \frac{1}{x} \phi &= 0 \\
\end{align*}
If we look at the Rayleigh quotient we can see there are no negative eigenvalues and we can manually verify that zero is not an eigenvalue.
Assume a trial solution of $\phi(x) = x^m, \phi'(x) = m x^{m-1}, \phi''(x) = m(m-1)x^{m-2}$ such that:
\begin{align*}
x^2 \frac{d^2\phi}{dx^2} + x \frac{d\phi}{dx} + \lambda \phi &= 0 \\
m (m-1) x^m + m x^m + \lambda x^m &= 0 \\
\left( m^2 + \lambda \right) x^m &= 0 \\
\lambda &= -m^2 \\
\end{align*}
Since we know that $\lambda > 0$, then $m^2 < 0$, which means that $m = i r$ where $r \in \mathbb{R}$. However, this can't satisfy the boundary conditions:
\begin{align*}
\phi(x) &= e^{ir} \\
\phi(1) = 0 &\neq 1^{ir} = 1 \\
\end{align*}
I'm stuck on what else to try.
| You have to take the change of variable $y=\ln x$ and the equation becomes
$$
\frac{d^2\phi}{dy^2}+\lambda\phi=0
$$
that is a standard form having as a solution
$$
\phi(y)=A\cos\sqrt{\lambda} y+B\sin\sqrt{\lambda} y.
$$
I think now you can go on from here.
| {
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Square-root equation
Solve square-root equation: $\left (\sqrt{5+2\sqrt{6}} \right )^x+\left (\sqrt{5-2\sqrt{6}} \right )^x=10$
$\left (\sqrt{5+2\sqrt{6}} \right )^x+\left (\sqrt{5-2\sqrt{6}} \right )^x=10\\
\left (\sqrt{\left (\sqrt{3}+\sqrt{2} \right )^2} \right )^x+\left (\sqrt{\left (\sqrt{3}-\sqrt{2} \right )^2} \right )^x=10\\
\left (\sqrt{3}+\sqrt{2} \right )^x+\left (\sqrt{3}-\sqrt{2} \right )^x=10$
at the moment I don't know what to do
| As an alternative,
quadratic with roots $\alpha=\sqrt{3}+ 2$ and $\beta=\sqrt{3}-2$ is $x^2-2\sqrt{3}x+1=0$
Sum of roots is $\alpha+\beta=2\sqrt{3}$
$\alpha^2+\beta^2=(\alpha+\beta)^2 - 2\alpha\beta = 12-2(1) = 10$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3420373",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "2",
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Writing polynomial using powers of (x-a) I have a question related to writing a polynomial using powers of binomial of form $(x-a).$
I found an example: polynomial $P(x) = x^4 + 2x^3-3x^2-4x+1$ can be written as
$ (x+1)^4-2(x+1)^3-3(x+1)^2+4(x+1)+1$ using powers of $(x+1)$ and Horner's Method. How do we obtain this representation of polynomial? How is Horner's Method used for that?
| You want an expression $a_4(x+1)^4+a_3(x+1)^3+a_2(x+1)^2+a_1(x+1)+a_0$. Repeatedly apply the method for $x=-1$. The last number in each row will give you the next coefficient starting from the lowest power.
\begin{array}{|c|c|c|c|c|c|}
\hline
& 1 & 2 & -3 & -4 & 1 \\ \hline
-1 & 1& 1& -4 & 0 & 1 \rightarrow a_0\\ \hline
-1 & 1 & 0& -4 & 4 \rightarrow a_1\\ \hline
-1 & 1 & -1& -3 \rightarrow a_2\\ \hline
\end{array}
and so on.
| {
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Find integers $x$ and $y$ such that $8^x-9^y=431$ Find integers $x$ and $y$ such that $8^x-9^y=431$
My working:
By taking mod 9 and 16, I got $x$ odd and $y$ even.
Also $8^x>431\implies x\ge 3$
For $x=3$ I got $y=2$
| SECOND ANSWER: We can stick with smaller primes in the other direction
We suspect that $512-81$ is the largest solution. Proof by contradiction:
Giving new names to $x,y,$ we say
$$ 512(8^x - 1) = 81 (9^y - 1) $$
We ASSUME both $x \geq 1, y \geq 1.$
First, $9^y \equiv 1 \pmod {512}.$ A calculation shows that $y$ must be divisible by $64$
Next,
$$ 9^{64} - 1 = 3^{128} - 1 = 2^9 \cdot 5 \cdot 17 \cdot 41 \cdot 193 \cdot ... $$
wWe use the 193.
$$ 8^x \equiv 1 \pmod{193} $$
so that $x$ is divisible by $32$
Then $8^x - 1$ is divisible by
$$8^{32} - 1 = 2^{96}-1 = 3^2 \cdot 5 \cdot 7 \cdot 13 \cdot 17 \cdot 97 \cdot 193 \cdot 241 \cdot 257 \cdot 673 \cdot 65537 \cdot 22253377
$$
We notice some Fermat primes here, in particular $257 = 2^8 + 1$ is a factor of $2^{16} - 1,$ in turn this divides $2^{96} - 1.$
That's all we needed. We find $9^y-1$ divisible by $257.$
This tells us that $y$ is divisible by $128.$ But $9^{128} - 1 = 3^{256}-1$ is divisible by 1024.
Therefore $512(8^x - 1) $ is divisible by $1024,$ which is a contradiction of $x \geq 1.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3421712",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
Determine the value of $k$ such that the line determined by the points $(3,2)$ and $(1,-2)$ is tangent to the graph of $y=\frac{k}{x+1}$. I have done the following work but I am stuck on solving for k. Am I doing this right? If so what do I do next?
| You are given a function $$y=\frac{k}{x+1}.$$
The line defined by the two points has a slope of $$m=\frac{\Delta y}{\Delta x}=\frac{2-(-2)}{3-1}=2$$
and therefore has an equation $$y-2=2(x-3)\Rightarrow2x-y-4=0$$
We want the original function and the tangent line to have exactly one point of intersection (by the definition of a tangent line), so we have the set of equations
\begin{cases} y=\dfrac{k}{x+1} \\ 2x-y-4=0
\end{cases}
Solving by substitution we have
\begin{align}
2x-\frac{k}{x+1}-4&=0\\
2x(x+1)-k-4(x+1)&=0\\
2x^2-2x+(-4-k)&=0
\end{align}
This quadratic equation has exactly one solution when the discriminant $b^2-4ac=0$. Thus, we have
\begin{align}
b^2-4ac=0&=(-2)^2-4(2)(-4-k)\\
0&=4+32+8k\\
8k&=-36\\
k&=-\frac92
\end{align}
A graph of the two functions:
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Solving the differential equation $y''=\frac{1}{y}$ I tried to solve the differential equation $\frac{d^2y}{dx^2}=\frac{1}{y}$ by assuming $y(0)=1$ and $y'(0)=0$ and finding a taylor expansion for $y$ at $x=0$. By differentiating $\frac{d^2y}{dx^2}$, I got:
$\frac{d^3y}{dx^3}=-\frac{1}{y^2}\frac{dy}{dx}$
$\frac{d^4y}{dx^4}=
\frac{2}{y^3}\frac{dy}{dx}
-\frac{1}{y^2}\frac{d^2y}{dx^2}$
$\frac{d^5y}{dx^5}=
-\frac{6}{y^4}\frac{dy}{dx}
+\frac{4}{y^3}\frac{d^2y}{dx^2}
-\frac{1}{y^2}\frac{d^3y}{dx^3}$
$\frac{d^6y}{dx^6}=
\frac{24}{y^5}\frac{dy}{dx}
-\frac{18}{y^4}\frac{d^2y}{dx^2}
+\frac{6}{y^3}\frac{d^3y}{dx^3}
-\frac{1}{y^2}\frac{d^4y}{dx^4}$
Evaluating those, I got the following values:
$y^{(3)}(0)=0$
$y^{(4)}(0)=1$
$y^{(5)}(0)=4$
$y^{(6)}(0)=-17$
I then created a Python program to automatically derive more values:
$y^{(7)}(0)=84$
$y^{(8)}(0)=-483$
$y^{(9)}(0)=3192$
$y^{(10)}(0)=-23919$
I used the derivatives to get a Taylor series, and I found that the series does not converge when $x>1$. Did I make a mistake? If not, is there an approach I could use to approximate $y(2)$ or other values where $x>1$?
| Multiply both sides of $y''= 1/y $ by $ y'$ to get $$y''y' = \frac {y'}{y}$$
Integrate to get $$(y')^2 = 2\ln y $$
which gives us $$y'=\sqrt {2\ln y+K}$$
Separate variables to get $$\int \frac {dy}{\sqrt {2\ln y+K}} =x+C$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3427545",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Find the units digit of $572^{42}$ The idea of this exercise is that you use the modulus to get the right answer.
What I did was:
$$572\equiv 2\pmod {10} \\
572^2 \equiv 2^2 \equiv 4\pmod{10} \\
572^3 \equiv 2^3 \equiv 8\pmod{10} \\
572^4 \equiv 2^4 \equiv 6\pmod{10} \\
572^5 \equiv 2^5 \equiv 2\pmod{10} \\
572^6 \equiv 2^6 \equiv 4\pmod{10} \\
(...)$$
I can see that this goes 2,4,8,6 and then repeats. I remember that the gist of the exercise is to find the remainder based on this repetition. How do I do that? I know that $572^{42} \equiv 2^{42}\equiv ? \pmod {10}$. How do I simplify that 42 and answer this using that repetition?
| Notice the pattern in the indices, the indices that get mapped to each of $2,4,8$ or $6.$
Note that the indices $1,5,9,\cdots, 1+4n$ get mapped to $2.$ The indices $2,6,10,\cdots,2+4n$ to $4,$ the indices $3,7,11,\cdots 3+4n$ get mapped to $8,$ and finally all multiples of $4,$ of the form $4n$ get mapped to $6,$ where $n$ is a nonnegative integer.
Then you only need reduce $42$ modulo $4,$ which shows you that it is of the second form. Thus, you have that to the $42$nd power, your number is $4\pmod {10}.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3428727",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 1
} |
Let $f(x)=(x+1)(x+2)(x+3)(x+4)+5$ where $x\in[-6,6]$. If the range of the function is [a,b] where $a,b\in N$, then find the value of (a+b) Let $f(x)=(x+1)(x+2)(x+3)(x+4)+5$ where $x\in[-6,6]$. If the range of the function is [a,b] where $a,b\in N$, then find the value of (a+b)
My attempt is as follows:-
Rewrite $f(x)=g(x)+5$ where $g(x)=(x+1)(x+2)(x+3)(x+4)$
Let's find the maximum value of g(x)
It can be clearly seen that from $x=-6$ to $x=6$, maximum value of $g(x)$ is at $x=6$.
$$g(6)=5040$$
Let's find the minimum value of $g(x)$
From the sign scheme one can see that negative value of $g(x)$ occurs in the interval $(-4,-3)$ and $(-2,-1)$
Hence intuitively it feels that the minimum value of g(x) would be at $x=-\dfrac{3}{2}$ or at $x=-\dfrac{7}{2}$ as in case of parabola also, minimum value is at average of both the roots.
So $g\left(-\dfrac{3}{2}\right)=-\dfrac{1}{2}\cdot\dfrac{1}{2}\cdot\dfrac{3}{2}\cdot\dfrac{5}{2}=-\dfrac{15}{8}$
At $x=-\dfrac{7}{2}$, $g\left(-\dfrac{7}{2}\right)=-\dfrac{5}{2}\cdot\dfrac{3}{2}\cdot\dfrac{1}{2}\cdot\dfrac{1}{2}=-\dfrac{15}{8}$
So range of $g(x)$ would be $[\dfrac{-15}{8},5040]$
Hence range of $f(x)$=$\left[\dfrac{25}{8},5045\right]$
But it is given that the range is $[a,b]$ where $a,b\in N$
I am stuck here. I am also not able to prove mathematically that at $x=-\dfrac{3}{2}$ or $x=-\dfrac{7}{2}$, minimum value of $g(x)$ will occur.
Please help me in this.
| Like you have suggested, consider the function $$g(x)=(x+1)(x+2)(x+3)(x+4).$$
$g'(x)$ potentially has minimum values at its critical numbers, which we can find by setting the derivative equal to $0$.
\begin{align}
g(x)&=(x+1)(x+2)(x+3)(x+4)\\
\log g(x)&=\log\big((x+1)(x+2)(x+3)(x+4)\big)\\
\log g(x)&=\log(x+1)+\log(x+2)+\log(x+3)+\log(x+4)\\
\frac{d}{dx}\big(\log g(x)\big)&=\frac{d}{dx}\big(\log(x+1)+\log(x+2)+\log(x+3)+\log(x+4)\big)\\
\frac{g'(x)}{g(x)}&=\frac1{x+1}+\frac1{x+2}+\frac1{x+3}+\frac1{x+4}\\
g'(x)&=g(x)\bigg(\frac1{x+1}+\frac1{x+2}+\frac1{x+3}+\frac1{x+4}\bigg)\\
g'(x)&=(x+1)(x+2)(x+3)(x+4)\bigg(\frac1{x+1}+\frac1{x+2}+\frac1{x+3}+\frac1{x+4}\bigg)
\end{align}
Solving the equation $$\frac1{x+1}+\frac1{x+2}+\frac1{x+3}+\frac1{x+4}=0$$ gives us $x=-\dfrac52-\dfrac{\sqrt5}2$, $-\dfrac52$, and $-\dfrac52+\dfrac{\sqrt5}2$. To find maximum and minimum values on an interval, we then need to compare the values at these critical numbers and at the bounds of the interval.
The maximum value is $g(6)=5040$ (as you found) and the minimum value is $g(-\frac52+\frac{\sqrt5}2)=-1$.
Since $f(x)=g(x)+5$, the range of $f(x)$ on $[-6,6]$ is therefore $[-1+5,5040+5]$ or $\boxed{[4,5045]}.$ As a result, the value of $a+b$ is $\boxed{5049}$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 0
} |
Prove convergence of $n \cdot \left(\sqrt{1+ \frac{1}{n}} -1\right)$ I am working on some old analysis exams and i got stuck on this exercise :
Using the epsilon definition show that $a_{n} = n \cdot \left(\sqrt{1+ \frac{1}{n}} -1\right)$ converges and determine its limit.
Knowing that the limit is 1/2, I know need to find an $ N \in \mathbb{N}$ so that : $ \forall \epsilon > 0 n > N \implies \left| a_n - \frac{1}{2} \right| < \epsilon $
Next step I simplify $a_n$ : $ a_n = \frac{n \cdot \left(\sqrt{1+ \frac{1}{n}} -1\right) \cdot \sqrt{1+ \frac{1}{n}} + 1 } {\sqrt{1+ \frac{1}{n}} + 1} = \frac{1}{\sqrt{1+ \frac{1}{n}} + 1}$
And then I got stuck,what am I supposed to do with : $ \left|\frac{1}{\sqrt{1+ \frac{1}{n}} + 1} - \frac{1}{2} \right|$
| $$ a_n = \sqrt{n^2+n}-n = \frac{n}{\sqrt{n^2+n}+n} $$
is bounded between $$ \frac{n}{\left(n+\frac{1}{2}\right)+n}\quad\text{and}\quad\frac{n}{n+n} $$
so $\frac{1}{2}-a_n$ is bounded between
$$ 0\quad\text{and}\quad\frac{1}{2(4n+1)}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3432944",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
} |
Finding the minimal polynomial of an $n \times n$ matrix How would one find the minimal polynomial of
$$ A= \begin{pmatrix}0&-1&-2&\cdots&1-n\\1&0&-1&\cdots&2-n\\2&1&0&\cdots&3-n\\\vdots&\vdots&\vdots&\ddots&\vdots\\n-1&n-2&n-3&\cdots&0\end{pmatrix}$$
Where $A$ is an $n \times n$ matrix with $n\ge 3$?
| Let
$$
U=\begin{pmatrix} 0 & -1 \\ 1 & -1 \\ 2 & -1 \\ \vdots & \vdots \\ n-1 & -1
\end{pmatrix}
\;\;\;\text{and}
\;\;\;
V=\begin{pmatrix} 1 & 1 & 1 & \cdots & 1 \\ 0 & 1 & 2 & \cdots & n-1
\end{pmatrix}
$$
Then $A=UV.$ Therefore, $\mathrm{rank}(A)=2,$ and $A$ has only $2$ non-zero eigenvalues. $iA$ is hermitian. This means $A$ is diagonalizable (in $\mathbb{C}$) and the minimal polynomial is square-free. As $iA$ has real eigenvalues ($iA$ is hermitian), the non-zero eigenvalues of $A$ are pure imaginary. Furthermore, the non-zero eigenvalues must appear in conjugate pairs, because $A$ is real.
If we put all this together, we get $\mu_A(x) = x^3+ax$ for a suitable $a.$
In order to find $a$, we take a look at $\mu_A(A)$ :
$$
\mu_A(A) = A^3+aA = UVUVUV+aUV = U\left((VU)^2+aI\right)V = 0
$$
$(VU)^2$ can be computed using Faulhaber's formulas. We find:
$$
(VU)^2 = \begin{pmatrix}
-\frac{n^2(n^2-1)}{12} & 0 \\ 0 & -\frac{n^2(n^2-1)}{12}
\end{pmatrix}
$$
Therefore, $a=\frac{n^2(n^2-1)}{12}$ and we have found our minimal polynomial.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Convergence of $\left( \frac{1}{1} \right)^2+\left( \frac{1}{2}+\frac{1}{3} \right)^2+\cdots$ Does the following series converge? If yes, what is its value in simplest form?
$$\left( \frac{1}{1} \right)^2+\left( \frac{1}{2}+\frac{1}{3} \right)^2+\left( \frac{1}{4}+\frac{1}{5}+\frac{1}{6} \right)^2+\left( \frac{1}{7}+\frac{1}{8}+\frac{1}{9}+\frac{1}{10} \right)^2+\dots$$
I have no idea how to start. Any hint would be really appreciated. THANKS!
| Since for moderately large values of $n$ we have
$$ H_n \approx \log(n)+\gamma+\frac{1}{2n}-\frac{1}{12n^2} $$
we also have
$$ H_{n(n+1)/2}-H_{n(n-1)/2} \approx \frac{2}{n}-\frac{4}{3n^3}$$
and
$$ \sum_{n\geq 1}\left[H_{n(n+1)/2}-H_{n(n-1)/2}\right]^2 \approx 1+\sum_{n\geq 2}\left(\frac{2}{n}-\frac{4}{3n^3}\right)^2=1+\frac{2 \left(-1890+2835 \pi ^2-252 \pi ^4+8 \pi ^6\right)}{8505} $$
is finite and approximately equal to $\color{green}{3.17}151$. I doubt there is a simple closed form.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3436804",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 3,
"answer_id": 0
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Solve for positive real solutions of cyclic equations $x+y^2+2xy=9$, $y+z^2+2yz=47$, $z+x^2+2xz=16$
Solve over positive reals $$x+y^2+2xy=9$$
$$y+z^2+2yz=47$$
$$z+x^2+2xz=16$$
With standard manipulation we get that $x+y+z=8$. Thus $x=8-y-z$ and we have two equations,
$$(8-y-z)^2+y^2+2(8-y-z)y=9$$
$$y+z^2+2yz=47$$
Which becomes
$$-y^2-2yz+15y-z=1$$
$$y+z^2+2yz=47$$
Adding up gets
$$-y^2+16y+z^2-z=1$$
$$\implies z^2=z+1+y^2-16y$$
And this substitutes to
$$y+z+1+y^2-16y+2yz=47$$
$$\implies y+\bigg(\frac{-y^2+15y-1}{2y+1}\bigg)+1+y^2-16y+2y\bigg(\frac{-y^2+15y-1}{2y+1}\bigg)=47$$
However, this results into a nasty quartic which I don't know how to simplify. Thanks!
| Continue by substituting $z=\frac{-y^2+15y-1}{2y+1}$ into $y+z^2+2yz=47$,
$$y+\left(\frac{-y^2+15y-1}{2y+1}\right)^2+2y\left(\frac{-y^2+15y-1}{2y+1}\right)=47$$
Simplify,
$$-3y^4+32y^3+69y^2-219y-46=0$$
and factorize
$$(y-2)(3y^3-26y^2-121y-23)=0$$
which yields the positive real solutions $x=1,\>y=2,\>z=5$. The other three sets of solutions are also real, but containing negative values.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3437528",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
If $\sin(x+28^\circ)=\cos(3x-78^\circ)$, then find x $$\sin(x+28^\circ)=\sin (90^\circ-3x+78^\circ)$$
$$x+28^\circ=168^\circ-3x$$
$$x=35^\circ$$
Pretty straightforward question, but then answer is $35^\circ$ and $8^\circ$. How is $8^\circ$ the answer?
| Hint: Use $$\sin(x)-\cos(y)=-2 \sin \left(-\frac{x}{2}-\frac{y}{2}+\frac{\pi
}{4}\right) \sin
\left(-\frac{x}{2}+\frac{y}{2}+\frac{\pi }{4}\right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3438094",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
How do I solve this exercise with limits? I have some difficulties solving this exercise:
\begin{align}
a_n & = \sum_{k=1}^n \frac 1 {\sqrt {n^2+k}} \\[6pt]
b_n & = n\\[6pt]
\lim_{n→∞} (a_n)^{b_n} & = \text{?}
\end{align}
I believe that I have to find the limit of $a_n$ first but then I have a limit of something at power infinity and I'm pretty stuck. I don't know how to find the an's limit and what to do next.
| For $1 \le k \le n$, we have that $n < \sqrt{n^2+k} < \sqrt{n^2+n+\frac{1}{4}} = n+\frac{1}{2}$. Therefore, $$\frac{n}{n+1/2} = \sum_{k=1}^n \frac{1}{n+1/2} < \sum_{k=1}^n \frac{1}{\sqrt{n^2+k}} < \sum_{k=1}^n \frac{1}{n} = 1.$$
Thus, by the Squeeze Theorem, we have that $\lim_{n\to \infty} a_n = 1$. However, this only shows us that the limit is in the indeterminate form $1^\infty$.
To make more progress, we need more detailed information on the asymptotic behavior of $a_n$ as $n \to \infty$. To this end, let us write:
$$1 - a_n = \sum_{k=1}^n \left( \frac{1}{n} - \frac{1}{\sqrt{n^2+k}} \right) = \sum_{k=1}^n \frac{\sqrt{n^2+k}-n}{n\sqrt{n^2+k}} = \sum_{k=1}^n \frac{k}{n\sqrt{n^2+k}(\sqrt{n^2+k}+n)}.$$
Now, in the final sum, we have
$$\frac{k}{n (n+1/2) (2n+1/2)}< \frac{k}{n\sqrt{n^2+k}(\sqrt{n^2+k}+n)} < \frac{k}{n \cdot n (2n)}.$$
Taking the sum, we get
$$\frac{n(n+1)/2}{n(n+1/2)(2n+1/2)} < 1 - a_n < \frac{n(n+1)/2}{n\cdot n(2n)}.$$
Looking at both ends, they are both asymptotically equivalent to $\frac{1}{4n}$. More formally, by the Squeeze Theorem we can conclude that $\lim_{n\to \infty} n(1 - a_n) = \frac{1}{4}$, or equivalently:
$$a_n = 1 - \frac{1}{4n} + o\left(\frac{1}{n}\right).$$
It now follows that $\log(a_n) = -\frac{1}{4n} + o(1/n)$, so $\log(a_n^{b_n}) = n \log(a_n) = -\frac{1}{4} + o(1)$. In other words, $\log(a_n^{b_n}) \to -\frac{1}{4}$ as $n \to \infty$, from which you should hopefully be able to conclude what the limit of $a_n^{b_n}$ is.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3441104",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Solve the equation $\cos^{-1}\frac{x^2-1}{x^2+1}+\tan^{-1}\frac{2x}{x^2-1}=\frac{2\pi}{3}$ $\cos^{-1}\dfrac{x^2-1}{x^2+1}+\tan^{-1}\dfrac{2x}{x^2-1}=\dfrac{2\pi}{3}$
Let's first find the domain
$$-1<=\dfrac{x^2-1}{x^2+1}<=1$$
$$\dfrac{x^2-1}{x^2+1}>=-1 \text { and } \dfrac{x^2-1}{x^2+1}<=1$$
$$\dfrac{x^2-1+x^2+1}{x^2+1}>=0 \text { and } \dfrac{x^2-1-x^2-1}{x^2+1}<=0$$
$$\dfrac{2x^2}{x^2+1}>=0 \text { and } \dfrac{-2}{1+x^2}<=0$$
$$x\in R$$
$$x^2-1\ne0$$
$$x\ne\pm1$$
$$\cos^{-1}\dfrac{x^2-1}{x^2+1}+\tan^{-1}\dfrac{2x}{x^2-1}=\dfrac{2\pi}{3}$$
$$\pi-\cos^{-1}\dfrac{1-x^2}{1+x^2}-\tan^{-1}\dfrac{2x}{1-x^2}=\dfrac{2\pi}{3}$$
$$\dfrac{\pi}{3}=\cos^{-1}\dfrac{1-x^2}{1+x^2}+\tan^{-1}\dfrac{2x}{1-x^2}$$
Substituting $x$ by $\tan\theta$
$$x=\tan\theta$$
$$\tan^{-1}x=\theta$$
$$\theta\in \left(-\dfrac{\pi}{2},\dfrac{\pi}{2}\right)-\{-\dfrac{\pi}{4},\dfrac{\pi}{4}\}$$
$$\dfrac{\pi}{3}=\cos^{-1}\dfrac{1-\tan^2\theta}{1+\tan^2\theta}+\tan^{-1}\dfrac{2\tan\theta}{1-\tan^2\theta}$$
$$\dfrac{\pi}{3}=\cos^{-1}(\cos2\theta)+\tan^{-1}(\tan2\theta)$$
$$2\theta\in(-\pi,\pi)-\{-\dfrac{\pi}{2},\dfrac{\pi}{2}\}$$
Now we break the range of $2\theta$ into various parts:-
Case $1$: $2\theta\in\left(-\pi,-\dfrac{\pi}{2}\right)$,$\theta\in\left(-\dfrac{\pi}{2},-\dfrac{\pi}{4}\right)$
$$\dfrac{\pi}{3}=2\pi+2\theta+\pi+2\theta$$
$$-\dfrac{8\pi}{3}=4\theta$$
$$-\dfrac{2\pi}{3}=\theta$$
But it is not the range of $\theta$ we assumed
Case $2$: $2\theta\in\left(-\dfrac{\pi}{2},0\right]$,$\theta\in\left(-\dfrac{\pi}{4},0\right]$
$$\dfrac{\pi}{3}=-2\theta+2\theta$$
$$\dfrac{\pi}{3}=0 \text { not possible }$$
Case $3$: $2\theta\in\left(0,\dfrac{\pi}{2}\right)$,$\theta\in\left(0,\dfrac{\pi}{4}\right)$
$$\dfrac{\pi}{3}=2\theta+2\theta$$
$$\dfrac{\pi}{12}=\theta$$
It is coming in the range of $\theta$, so its a valid solution.
$$\tan^{-1}x=\dfrac{\pi}{12}$$
$$x=\tan\dfrac{\pi}{12}$$
$$x=2-\sqrt{3}$$
Case $4$: $2\theta\in\left(\dfrac{\pi}{2},\pi\right)$, $\theta\in\left(\dfrac{\pi}{4},\dfrac{\pi}{2}\right)$
$$\dfrac{\pi}{3}=2\pi-2\theta+2\theta-\pi$$
$$\dfrac{\pi}{3}=\pi \text { not possible }$$
So only solution is $2-\sqrt{3}$, but actual answer is $2-\sqrt{3}, \sqrt{3}$
| Let me propose an alternative pathway.
Suppose there exists $y$ such that
$$
\frac{2x}{x^2-1} = \tan(y).
$$
This seems resonable as $\tan$ range is the whole $\mathbb{R}$.
Now observe
$$
\frac{1}{\cos^2(y)} = \frac{\sin^2(y) + \cos^2(y)}{\cos^2(y)} = 1+\tan^2(y)
= 1 + \frac{4x^2}{(x^2-1)^2} = \frac{x^4 - 2x^2 + 1+4x^2}{(x^2-1)^2}
= \left(\frac{x^2+1}{x^2-1}\right)^2.
$$
Thus we can choose $y$ such that
$$
\cos(y) = \frac{x^2-1}{x^2+1}.
$$
Now the first equation simplifies to
$$
\cos^{-1}(\cos(y)) + \tan^{-1}(\tan(y)) = \frac{2\pi}{3}.
$$
This is the sum of two periodic functions, one with period $2\pi$ and one with period $\pi$.
Thus it suffices to solve the problem for $[0;2\pi]$ interval to get the full solution.
The problem is $\arccos:[-1;1]\rightarrow[0;\pi]$ and $\arctan:\mathbb{R}\rightarrow[-\pi/2;\pi/2]$, thus we should take into account $\arccos \circ \cos: y \in[\pi;2\pi] \mapsto 2\pi-y$, $\arctan \circ \tan: y \in[\pi/2;3\pi/2] \mapsto y - \pi$, and $\arctan \circ \tan: y \in[\pi/2;3\pi/2] \mapsto y - 2\pi$.
Thus
$$
\begin{aligned}
y \in [\pi;3\pi/2] \Rightarrow \cos^{-1}(\cos(y)) + \tan^{-1}(\tan(y)) = 2\pi - y + y - \pi = \pi,\\
y \in [3\pi/2;2\pi] \Rightarrow \cos^{-1}(\cos(y)) + \tan^{-1}(\tan(y)) = 2\pi - y + y - 2\pi = 0.\\
\end{aligned}
$$
Obviously, there are no solutions for $y \in [\pi;2\pi]$, but this result may be handy if you would ever like to plot the function.
The leftover is in essence equivalent to
$$
\begin{aligned}
y \in [0;\pi/2] \Rightarrow y + y = \frac{2\pi}{3} \Rightarrow y &= \frac{\pi}{3},\\
y \in [\pi/2;\pi] \Rightarrow y + y - \pi = \frac{2\pi}{3} \Rightarrow y &= \frac{5\pi}{6}.\\
\end{aligned}
$$
Taking the overall period into account we get
$$
y = \frac{\pi}{3} + 2\pi n,\quad\text{and}\quad y = \frac{5\pi}{6} + 2\pi n.
$$
and you now only need to solve
$$
\left\{
\begin{aligned}
\frac{2x}{x^2-1} = \tan\left(\frac{\pi}{3} + 2\pi n\right) = \sqrt{3}\\
\frac{x^2-1}{x^2+1} = \cos\left(\frac{\pi}{3} + 2\pi n\right) = \frac{1}{2}
\end{aligned}
\right.
\quad\text{and}\quad
\left\{
\begin{aligned}
\frac{2x}{x^2-1} = \tan\left(\frac{5\pi}{6} + 2\pi n\right) = -\frac{1}{\sqrt{3}}\\
\frac{x^2-1}{x^2+1} = \cos\left(\frac{5\pi}{6} + 2\pi n\right) = -\frac{\sqrt{3}}{2}
\end{aligned}
\right.
$$
that are simple quadratic equations.
The result reads
$$
x = \sqrt{3}\quad\text{and}\quad x = 2-\sqrt{3}.
$$
| {
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Prove that $(x + \sqrt[3]{abc})^3 \le (x + a)(x + b)(x + c) \le ( x + \frac{a + b + c}{3})^3$ Let $x,$ $a,$ $b,$ $c$ be nonnegative real numbers. Prove that
$$(x + \sqrt[3]{abc})^3 \le (x + a)(x + b)(x + c) \le \left( x + \frac{a + b + c}{3} \right)^3.$$
I know that this problem is a RHS-AM-GM-HM problem, but I am unsure how to solve it. I think that the leftmost part is AM, but that is as far as I was able to get. I also think that taking the cube root of everything might change it into something more manageable, but I am still unsure about that.
Can anyone help me with this or give me tips about how to solve it? Thank you.
| $$F=(x+a)(x+b)(x+c)=x^3+(a+b+c)x^2+(ab+bc+ca)x+abc~~~~(1)$$ By AM-GM
$$F \ge x^3 + 3(abc)^{1/3} x^2+3(abc)^{2/3} x+ abc= (x+(abc)^{1/3})~~~~(2)$$
Using $$\frac{(a+b+c)^2}{3} \ge (ab+bc+ca)$$ and AM-GM in (1), we get
$$F \le x^3+3 \frac{(a+b+c)}{3} x^2 +3\frac{(a+b+c)^2}{9}x+ \left(\frac{a+b+c}{3}\right)^3 = \left(x+\frac{a+b+c}{3}\right)^3$$
| {
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How do we prove this inequality? Suppose $a,b,c > 0$. Prove that
$$\frac{a^2}{b^2} +\frac{b^2}{c^2} + \frac{c^2}{a^2} \geq \frac ab + \frac bc + \frac ca.$$
I've tried multiplying everything by the denominator and then I tried to use the rearrangement inequality, but it didn't yield the result I was looking for. I couldn't really think of anything else to do.
| Because by AM-GM we obtain: $$\sum_{cyc}\frac{a^2}{b^2}=\frac{1}{6}\sum_{cyc}\left(\frac{4a^2}{b^2}+\frac{b^2}{c^2}+\frac{c^2}{a^2}\right)\geq\sum_{cyc}\sqrt[6]{\left(\frac{a^2}{b^2}\right)^4\cdot\frac{b^2}{c^2}\cdot\frac{c^2}{a^2}}=\sum_{cyc}\frac{a}{b}.$$
Just another way:
Let $x^3=\frac{a}{b}$, $y^3=\frac{b}{c}$ and $z^3=\frac{c}{a}$.
Thus, $xyz=1$ and we need to prove that
$$x^6+y^6+z^6\geq(x^3+y^3+z^3)xyz,$$ which is true by Muirhead because
$$(6,0,0)\succ(4,1,1).$$
| {
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Compute $\int\frac{du}{\sqrt{a^2-u^2}}$ Please, can you explain why isn't the $$\int\frac{du}{\sqrt{a^2-u^2}} = \int\frac{du}{a\sqrt{1-(u/a)^2}}=\frac1a\sin^{-1}\bigg(\frac{u}{a}\bigg)+c$$ if $a > 0$ is a positive constant.
| We have $$\int\frac{du}{\sqrt{a^2-u^2}}.$$
Let $u=a\sin\theta\implies\theta=\arcsin\bigg(\dfrac{u}{a}\bigg)$. Then $du=a\cos\theta\ d\theta$ and we have
$$\int\frac{du}{\sqrt{a^2-u^2}}=\int\frac{a\cos\theta\ d\theta}{\sqrt{a^2-a^2\sin^2\theta}}=\int\frac{a\cos\theta\ d\theta}{a\sqrt{1-\sin^2\theta}}=\int\frac{\cos\theta\ d\theta}{\sqrt{\cos^2\theta}}=\int d\theta$$
$$=\theta=\boxed{\arcsin\bigg(\frac{u}{a}\bigg)+C}.$$
| {
"language": "en",
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$\underset{n}\lim{\Big(\sqrt[3\;]{\frac{\sin n}{n}+n^3-n^2}-\sqrt[3\;]{n^3+n}\Big)}$ $\underset{n}\lim{\Big(\sqrt[3\;]{\frac{\sin n}{n}+n^3-n^2}-\sqrt[3\;]{n^3+n}\Big)}$
I thought of using the squeeze theorem for$\;\frac{\sin n}{n},$ but I think it might incorrect here.
Another attempt was:
$$\underset{n}\lim{\frac{\frac{\sin n}{n}-n^2-n}{\sqrt[3\;]{\Big(\frac{\sin n}{n}+n^3-n^2}\Big)^2+\sqrt[3\;]{{\Big(\frac{\sin n}{n}+n^3-n^2}\Big)\Big(n^3+n}\Big)+\sqrt[3\;]{(n^3+n)^2}}}$$ I got stuck here.
| Starting from Paramanand Singh's first equation
$$A=n\left(\sqrt[3]{1-\frac{1}{n}+\frac{\sin (n)} {n^4}}-\sqrt[3]{1+\frac{1}{n^2}}\right)$$ since $\sin(n)$ is bounded and very small when divided by $n^4$, consider, for the time being
$$B=n\left(\sqrt[3]{1-\frac{1}{n}+\frac{a} {n^4}}-\sqrt[3]{1+\frac{1}{n^2}}\right)\quad \text{where} \quad -1 \leq a \leq 1$$ and use the binomial expansion or Taylor series (which I shall push to high order until $a$ appears.
This would give
$$B=-\frac{1}{3}-\frac{4}{9 n}-\frac{5}{81
n^2}+\frac{81 a+17}{243 n^3}+O\left(\frac{1}{n^4}\right)$$
| {
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} |
Solve for $y'\cos{y}=\sin(x+y)$ Solve for $y'\cos{y}=\sin(x+y)$
My attempt
$$y'\cos{y}=\sin x \cos y + \cos x \sin y$$
Divide both side by $\cos y$
$$\frac{dy}{dx}=\sin x + \cos x \tan y$$
Integrate both sides and I got
$$y = -\cos x + \int \cos x \tan y \,dx$$
As Maximilian Janisch said that $\int\cos() \tan(())d≠\sin() \tan(())$ I didn't know what to do next.
| As said in comments, power series solution seems to be the only way.
Writing
$$y=\sum_{n=0}^\infty \frac{a_n}{n!} x^n$$ the very first terms would be
$$\left(
\begin{array}{cc}
1 & t \\
2 & t^3+t+1 \\
3 & 3 t^5+4 t^3+t^2+1 \\
4 & 15 t^7+27 t^5+7 t^4+9 t^3+8 t^2-3 t \\
5 & 105 t^9+240 t^7+63 t^6+144 t^5+103 t^4+6 t^3+34 t^2-2 t-6 \\
6 & 945 t^{11}+2625 t^9+693 t^8+2250 t^7+1476 t^6+566 t^5+847 t^4+49 t^3+48 t^2+53
t-15
\end{array}
\right)$$ where $t=\tan(a_0)$.
Not very funny, isn't it ?
| {
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Transforming coordinates system to obtain an ellipse in the standard form Let $E$ be the ellipse $x^2+xy +y^2 = 1$. I would like to obtain the area of $E$ with the formula $ab\pi$. I am not able to transform the equation in the form $\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2} = 1$. The textbook I am following uses a transformation to describe $E$ as $$\dfrac{X^2}{\left(\dfrac{1}{\sqrt{3}}\right)^2}+\dfrac{Y^2}{1^2}=1$$
I do not understand what intermediate steps lead to this transformation.
EDIT : As some were curious, this is the original question:
Use Stoke's Theorem to to evaluate $\oint_C\vec{F}\cdot\vec{dr}$. Assume C is oriented counterclockwise as viewed from above.
$$\vec{F} = z\vec{i}+x\vec{j} + y\vec{k}$$
$C$ is the curve of intersection of the plane $x + y + z = 0$ and the sphere $x^2 + y^2 + z^2 = 1$.
Hint : recall that the area of an ellipse $\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2} = 1$ is $ab\pi$
The textbook solution : 3 * Area of R = $3 \cdot \dfrac{\pi}{\sqrt(3)}$=$\sqrt3\pi$
| Let $X=\dfrac x2$ and $Y=\dfrac x2 + y$.
Then $\dfrac{X^2}{\left(\dfrac{1}{\sqrt{3}}\right)^2}+\dfrac{Y^2}{1^2}=3X^2+Y^2=\dfrac34x^2+\dfrac14x^2+xy+y^2=1$.
Addendum:
The above explains the intermediate steps that led to the transformation, as the original post requested, but it should be noted, as alluded to in the comments, that the transformation
$\begin{pmatrix}X\\Y\end{pmatrix}=\begin{pmatrix}\frac12 & 0 \\ \frac12 &1\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix}$ has determinant $\frac12$, so this transformation halves area, and that must be taken into account when addressing the underlying question about the area of the ellipse.
| {
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Prove that $\sum_\text{cyc} \frac{ab}{ab+b^2+ca}\le 1$ Hello ladies and gentlemen, here I have another inequality that I am struggling with:
Let $a,b,c>0$ Then
$$\sum_\text{cyc} \frac{ab}{ab+b^2+ca}\le 1.$$
I try to show $$\frac{ab}{ab+b^2+ca}\le\frac{a}{a+b+c}$$ but this is doesn't work because $$\frac{ab}{ab+b^2+ca}=\frac{a}{a+b+\frac{ca}b}$$ which is in general not less than $\frac{a}{a+b+c}$.
So what can I do? Do I have to use some Hölder?
| Let $a=x^3$, $b=y^3$ and $c=z^3$.
Thus, by Muirhead $$\sum_{cyc}\frac{ab}{ab+b^2+ca}=\sum_{cyc}\frac{x^3y^3}{x^3y^3+\left(y^2\right)^3+(zx)^3}\leq$$
$$\leq\sum_{cyc}\frac{x^3y^3}{x^3y^3+\left(y^2\right)^2\cdot(zx)+(zx)^2\cdot(y^2)}=\sum_{cyc}\frac{x^2y}{x^2y+y^2z+z^2x}=1.$$
| {
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Logical statements: $A+B+C=A+B+C+AB+BC+AC$
Let $A,B, C$ be logical statements.
Then:
$A+B+C=A+B+C+AB+BC+AC$
Prove it without a table.
My attempt:
$$A=A\cdot(1+B+C)$$
$$B=B\cdot(1+A+C)$$
$$C=C\cdot(1+A+B)$$
$$\implies A+B+C= A+AB+AC+B+AB+BC+C+BC+AC$$
$$\implies A+B+C=A+B+C+AB+AB+BC+BC+AC+AC$$
$$\implies A+B+C=A+B+C+AB+BC+AC$$
Is this correct?
| It's correct, if you are using Monotone laws of Boolean algebra or Logical equivalence, we can write$:$
\begin{align}
&\hspace{3ex}A+B+C\\
&=A+AB+B+BC+C+CA\tag*{Absorption law}\\
&=A+B+C+AB+BC+AC\tag*{Commutative law}
\end{align}
| {
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Show that $\frac{(m_a)^4+(m_b)^4+(m_c)^4}{a^4+b^4+c^4} = \frac{9}{16}$ Let $ m_a, m_b $ and $ m_c $ be the medians relative to the $ a, b, c $ sides of a triangle, show that:
$$\frac{(m_a)^4+(m_b)^4+(m_c)^4}{a^4+b^4+c^4} = \frac{9}{16}$$
What i tried:
ust use Stewart’s theorem. We have $m=n=a/2,d=m_a$ so
$$\frac{a^3}{4}+m_a^2a=\frac{ab^2}{2}+\frac{ac^2}{2}\implies m_a^2=\frac{2b^2+2c^2-a^2}{4},$$and likewise for the other two medians.
How to proceed?
| Use that $$m_a=\frac{1}{2}\sqrt{2(b^2+c^2)-a^2}$$ etc.
| {
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Determine the minimum of:$\frac{\sqrt{(a^2+1)(b^2+1)(c^2+1)(d^2+1)}}{a+b+c+d},\text{ if }a,b,c,d>0$ Determine the minimum of:$\frac{\sqrt{(a^2+1)(b^2+1)(c^2+1)(d^2+1)}}{a+b+c+d}$, if $a,b,c,d>0$
I tried partial derivatives of the unction and also its natural log aI came up with system of equation that is not easy to solve. Using AM GM easily you can show the given fraction is greater or equal than 1
| Let $a=\frac{x}{\sqrt3},$ $b=\frac{y}{\sqrt3},$ $c=\frac{z}{\sqrt3}$ and $d=\frac{t}{\sqrt3}$.
Thus, by C-S we obtain:
$$\frac{\sqrt{(a^2+1)(b^2+1)(c^2+1)(d^2+1)}}{a+b+c+d}=\frac{\sqrt{(x^2+3)(y^2+3)(z^2+3)(t^2+3)}}{9(a+b+c+d)}\geq$$
$$\geq\frac{\sqrt{2((x+y)^2+4)\cdot2(4+(z+t)^2)}}{9(a+b+c+d)}\geq\frac{2(2(x+y)+2(y+t))}{9(a+b+c+d)}=\frac{4}{3\sqrt3}.$$
The equality occurs for $a=b=c=d=\frac{1}{\sqrt3},$ which says that we got a minimal value.
I used the following inequality:
$$(x^2+3)(y^2+3)\geq2((x+y)^2+4),$$ which is
$$(xy-1)^2+(x-y)^2\geq0.$$
| {
"language": "en",
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Use DeMoivre's Theorem to prove $ \cos 5x = 16 \cos^5x - 20\cos^3x+5\cos x$ I need to prove the following equalities using DMT:
$ \cos 5x = 16 \cos^5x$$ - 20\cos^3x$$+5\cos x$
and
$ \sin 5x = 16 \sin^5x$$ - 20\sin^3x$$+5\sin x$
Can someone help me with this question?
(Attempt: $\cos5x+i\sin5x=(\cos x+i\sin x)^5$)
| Use the short-hands $c= \cos x$, $s=\sin x$ and continue with
$$\cos5x+i\sin5x=(c+is)^5
=c^5-10s^2c^3+5s^4c+i(s^5-10s^3c^2+5sc^4)$$
$$=c^5-10(1-c^2)c^3+5(1-c^2)^2c+i(s^5-10s^3(1-s^2)+5s(1-s^2)^2)$$
$$=16c^5-20c^3+5c+i(16s^5-20s^3+5s)$$
where $c^2+s^2=1$ is used. Thus,
$$ \cos 5x = 16 \cos^5x - 20\cos^3x+5\cos x$$
$$ \sin 5x = 16 \sin^5x - 20\sin^3x +5\sin x$$
| {
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Convergence of $\int_0^{\infty}\frac{\ln (1 + x^2)}{\sqrt {2x^5 + x^6}}\,dx $
Find out if the following integral diverges or converges:
$$
\int_0^\infty \frac{\ln (1 + x^2)}{\sqrt {2x^5 + x^6}}\,dx
$$
First I split the integral as $\displaystyle \int_0^1 \frac{\ln (1 + x^2)}{\sqrt {2x^5 + x^6}}\,dx + \int_1^\infty \frac{\ln (1 + x^2)}{\sqrt {2x^5 + x^6}}\,dx $.
*
*For $\int_0^{1}\frac{\ln (1 + x^2)}{\sqrt {2x^5 + x^6}}\,dx $ , I prove $\ln(1+x^2)< x^2$, using that I can prove $\int_0^1\frac{\ln (1 + x^2)}{\sqrt {2x^5 + x^6}}\,dx $ converges, but it take too long so is there shorter way to do this problem?
*For $\int_1^{\infty}\frac{\ln (1 + x^2)}{\sqrt {2x^5 + x^6}}\,dx $ , I have no idea how to do this problem.
| For small $x$, $$\frac{\log(1+x^2)}{x^2\sqrt{2x+x^2}}=\frac1{\sqrt{2x}}+o(x^{-1/2}).$$
For large $x$, $$\frac{\log(1+x^2)}{x^2\sqrt{2x+x^2}}=\frac{\log(x)}{x^3}+o\left(\frac{\log(x)}{x^3}\right)=o\left(x^{-2}\right).$$
In both cases, the integrals of the asymptotic expressions do converge.
| {
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combinatorics - arranging consecutive numbers One is arranging the $10$ numbers $0$ to $9$ in a row.
what is the probability of not having a consecutive of $7$ numbers or more?
for example, $2034567891$ is forbidden as it has $7$ consecutive numbers.
Edit :
What I have tried:
for $0-6,$ I treat it as one block along with $7,8,9.$ So the number of permutation for $0-6,7,8,9$ is $ 4!=24.$
For $1-7,$ we already counted $0-7$ for example so we need to consider it, so the number of uncounted permutations of $0,1-7,8,9$ is: $4!-3!=18 .$
Same for $2-8,$ (now we need to remove $1$ or $9$ or both). Number of permutations: $ 4!-3!-3!-2=12.$
For $3-9$ : $ 4!-3!=18.$
In total:
$$ P = \frac{10!-72}{10!} = \frac{50399}{50400} $$
| Might not be the adaptable way to solve and certainly wouldn't be practical for just about any numbers but certainly the easiest is:
There is $1$ way to have all ten consecutive.
There are$2$ ways to have exactly nine ways consecutive: ($0$ to $8$ after a $9$ or $1$ to $9$ before a $0$.
exactly eight. There are $3$ choices for the numbers: $0$ to $7$, $1$ to $8$, or $2$ to $9$. $0-7$ can precede $9,8$; follow $9,8$ or $8,9$; or be between $8\*\*9$. similarly $2-9$ can follow $1,0$; precede $0,1$ or $1,0$; or be between $0\*\*1$. And $1-8$ can precede $0,9$; be between $9**0$; or following $0,9$. That is $11$.
exactly $7$: There $4$ choices for the numbers. $0-6$ can be in $0,1,2,3,4,5,6,\*,\*,\*$ provided the next number isn't $7$; so there are $4$ ways to arrange $7,8,9$ so that it doesn't start with $7$. Likewise for $1,2,3,4,5,6,7,\*,\*,\*$ and $2,3,4,5,6,8,\*,\*,\*$ provided the next numbers aren't $8$ or $9$.
And the same is true for $\*,\*,\*,1,2,...,7$, or $\*,\*,\*,2,....,8$, or $\*,\*,\*,3,....9$.
So that's $24$ more.
$3,4,5,6,7,8,9,\*\*\*$ and $\*,\*,\*,0,1,2,3,4,5,6$ can have $6$ ways each so that is $12$ more.
Now for $\*,a,b,c,d,e,f,g,\*,\*$ and $\*,\*,a,b,c,d,e,f,g,\*$. So long as the it isn't preceded by $a-1$ and followed by $g+1$, this is good. So for $--,0,.....,6,\*--$ or $-\*,3,....,9\*--$ there are $4$ ways (not followed by $7$ or not preceded by $2$ so that is $4*4 = 16$ more ways.
If $a\ne 0$ and $g \ne 9$ (i.e. $a= 1,2$ and $7,8$) then you you can arrange the three remaining numbers, $a-1, g+1, x$ as 1)$g+1$ before $a$ the $a-1$ after the $g$; 2)$g+1$ before the $a$ and $x$ after the $g$ 3)$x$ before the $a$ and $a-1$ after the $g$. So that is $2\cdot 2\cdot 3=12$ more ways.
So there are $1+2 +11+(24 +12+16+ 12) = 78$ ways to have $7$ or more consecutive.
So there are $10! -78$ ways not to.
So the probability of not is $\frac {10!-78}{10!}$
THis really is not a good way to do it and it quickly becomes a nightmare unless you reign it in quickly with some recursion formulas. But they look farirly simply. Can't have $\*\*\*\*\*, a,b,....,h,\*\*\*\*$ preceded by $a-1$ and $h$ followed by $h+1$ and so.....
| {
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Closed expression for sum $\sum_{k = 1}^{\infty} \frac{\left\lfloor \sqrt{k} \right \rfloor}{k^2}$ Generalizing a recent post Closed expression for sum $\sum_{k=1}^{\infty} (-1)^{k+1}\frac{\left\lfloor \sqrt{k}\right\rfloor}{k}$ where convergence was assured by alternating the sign here's a similar problem in which convergence in forced by the increased power in the denominator.
Question: is there a closed form of this sum
$$\begin{align} s_2
&=\sum_{k=1}^{\infty} \frac{\left\lfloor \sqrt{k}\right\rfloor}{k^2}\simeq 2.33198\tag{1}\\
\end{align}\tag{1}$$
The sum is obviously convergent, and obeys the following inequality
$$1.64493\simeq\zeta(2)=\sum_{k=1}^{\infty} \frac{1}{k^2}\lt s_{2} \lt \sum_{k=1}^{\infty} \frac{\sqrt{k}}{k^2}=\zeta(\frac{3}{2})\simeq 2.61238\tag{2}$$
| My solution attempt
I have not found a closed form expression but the following integral representation
$$s_{2} =\sum_{k = 1}^{\infty} \frac{\left\lfloor \sqrt{k} \right \rfloor}{k^2} = \int_0^{\infty } \frac{t \left(\vartheta _3\left(0,e^{-t}\right)-1\right)}{2 \left(1-e^{-t}\right)} \, dt\tag{1}$$
Here
$$\vartheta _3(u,q)=1+2 \sum _{n=1}^{\infty } q^{n^2} \cos (2 n u)$$
is a Jacobi theta function.
Derivation
The drivation starts with a similar method as in $[1]$.
We find that the partial sum from $k=1$ to $k=m^2-1$ ($m \in N$) can be written as
$$\begin{align} p(m)
&= \sum_{k=1}^{m^2-1} \frac{\left\lfloor \sqrt{k}\right\rfloor}{k^2}= f(m) - g(m)
\end{align}\tag{2}$$
where
$$\begin{align}
f(m)
& =m H(m^2-1,2)\tag{3}\\
g(m)
&= \sum_{k=1}^{m} H(k^2-1,2)\tag{4}
\end{align}$$
Here $H(n,2)=H_{n,2}=\sum_{k=1}^n \frac{1}{k^2}$ is the generalized harmonic number of order $2$ of $n$.
Indeed, writing (dropping the second index $2$ in $H$ for simplicity)
$m=2\to k=1..3$ :
$\frac{\left\lfloor \sqrt{1} \right \rfloor}{1^2}+\frac{\left\lfloor \sqrt{2} \right \rfloor}{2^2}+\frac{\left\lfloor \sqrt{3} \right \rfloor}{3^2}= \frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}= H_{3}= H(2^2-1)$
$m=3\to k=4..8$ : $\frac{\left\lfloor \sqrt{4} \right \rfloor}{4^2}+\frac{\left\lfloor \sqrt{5} \right \rfloor}{5^2}+\frac{\left\lfloor \sqrt{6} \right \rfloor}{6^2}+\frac{\left\lfloor \sqrt{7} \right \rfloor}{7^2}+\frac{\left\lfloor \sqrt{8} \right \rfloor}{8^2}=2 \frac{1}{4^2}+2\frac{1}{5^2}+2\frac{1}{6^2}+2\frac{1}{7^2}+2\frac{1}{8^2}= 2(H(8)-H(3))=2 (H(3^2-1)-H(2^2-1))$
Together up to $m=3$
$p(3) = H(3) + 2 (H(8)-H(3))=2 H(8) - H(3) $
and so on gives an the first place
$p(m) = (m-1) H(m^2-1) - \sum_{k=1}^{m-1} H(k^2-1)$
but then shifting $-H(m^2-1)$ from the first term to the second, i.e. including it into the sum we get $(2)$,$(3)$ and $(4)$.
Now we need the limit $m\to\infty$.
This is no problem for $f$ where we have
$$f(m) \sim \zeta(2)-\frac{1}{m}-\frac{1}{2 m^3}+ O(\frac{1}{m^5})\tag{5}$$
Now since
$$ H(k^2-1,2) = \sum_{j=1}^{k^2-1} \frac{1}{j^2} = \sum_{j=1}^{\infty} \frac{1}{j^2}-\sum_{j=k^2}^{\infty} \frac{1}{j^2}=\zeta(2)-\sum_{j=k^2}^{\infty} \frac{1}{j^2} $$
$g$ can be written as
$$g(m) = m \zeta(2) - \sum_{k=1}^m \sum_{j=k^2}^\infty \frac{1}{j^2}\tag{6}$$
Hence we have
$$s_{2}=\lim_{m\to \infty } \, p(m) =\lim_{m\to \infty } \,(\sum_{k=1}^m \sum_{j=k^2}^\infty \frac{1}{j^2})\tag{7} $$
and we have to calculate the asymptotic behaviour of the double sum
$$g_1(m) = \sum_{k=1}^m \sum_{j=k^2}^\infty \frac{1}{j^2}\tag{8}$$
With
$$\frac{1}{j^2}=\int_0^{\infty } t \exp (-j t) \, dt$$
we can do the $j$-sum
$$\sum _{j=k^2}^{\infty } \exp (-j t)=\frac{e^{-k^2 t}}{1-e^{-t}}$$
and subsequently do the $k$-sum extending the limit $m\to\infty$
$$\sum _{k=1}^{\infty } e^{-k^2 t}=\frac{1}{2} \left(\vartheta _3\left(0,e^{-t}\right)-1\right)$$
Putting this back into the the $t$-integral gives $(1)$ QED.
Discussion
In the previous problem $[1]$ other users have provided interesting results with other approaches to "remove" the floor function. I'm sure this can be done here as well.
Maybe also a head-on attack on the double sum $(8)$ can lead to "sum-based" simplifications.
Proximity to the harmonic numbers makes the existence of a closed form for $s_2$ probable.
References
$[1]$ Closed expression for sum $\sum_{k=1}^{\infty} (-1)^{k+1}\frac{\left\lfloor \sqrt{k}\right\rfloor}{k}$
| {
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"question_score": "2",
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"answer_id": 0
} |
Show that following determinant is divisible by $\lambda^2$ and find the other factor. Show that $\begin{vmatrix}
a^2+\lambda &ab &ac \\
ab & b^2+\lambda & bc \\
ac & bc & c^2+\lambda
\end{vmatrix}=0$ is divisible by $\lambda^2$ and find the other factor.
My attempt is as follows:-
$$R_1\rightarrow R_1+R_2+R_3$$
$$\begin{vmatrix}
a(a+b+c)+\lambda &b(a+b+c)+\lambda &c(a+b+c)+\lambda \\
ab & b^2+\lambda & bc \\
ac & bc & c^2+\lambda
\end{vmatrix}=0$$
$$C_1\rightarrow C_1-\dfrac{a}{b}C_2$$
$$C_2\rightarrow C_2-\dfrac{b}{c}C_3$$
$$\begin{vmatrix}
\lambda-\dfrac{a\lambda}{b}&\lambda-\dfrac{b\lambda}{c} &c(a+b+c)+\lambda \\
-\lambda & \lambda & bc \\
0 & -\lambda & c^2+\lambda
\end{vmatrix}=0$$
Taking $\lambda^2$ common
$$\lambda^2\begin{vmatrix}
1-\dfrac{a}{b}&1-\dfrac{b}{c} &c(a+b+c)+\lambda \\
-1 & 1 & bc \\
0 & -1 & c^2+\lambda
\end{vmatrix}=0
$$
$$\dfrac{\lambda^2}{bc}\begin{vmatrix}
b-a&c-b &c(a+b+c)+\lambda \\
-b & c & bc \\
0 & -c & c^2+\lambda
\end{vmatrix}=0
$$
$$R_1\rightarrow R_1-R_3$$
$$\dfrac{\lambda^2}{bc}\begin{vmatrix}
b-a&2c-b &ca+bc \\
-b & c & bc \\
0 & -c & c^2+\lambda
\end{vmatrix}=0$$
$$R_1\rightarrow R_1-R_2$$
$$\dfrac{\lambda^2}{bc}\begin{vmatrix}
2b-a&c-b &ca \\
-b & c & bc \\
0 & -c & c^2+\lambda
\end{vmatrix}=0$$
Now expanding it
$$\dfrac{\lambda^2}{bc}\left(c(2b^2c-abc+abc)+(c^2+\lambda)(2bc-ac+bc-b^2)\right)=0$$
$$\dfrac{\lambda^2}{bc}\left(2b^2c^2+(c^2+\lambda)(3bc-ac-b^2)\right)=0$$
$$\dfrac{\lambda^2}{bc}\left(2b^2c^2+3bc^3-ac^3-b^2c^2+3bc\lambda-\lambda ac-\lambda b^2\right)=0$$
$$\dfrac{\lambda^2}{bc}\left(b^2c^2+3bc^3-ac^3+3bc\lambda-\lambda ac-\lambda b^2\right)=0$$
$$\dfrac{\lambda^2}{bc}\left(c^2(b^2+3bc-ac\right)+\lambda(3bc-ac-b^2)=0$$
So another factor seems to be $\dfrac{1}{bc}\left(c^2(b^2+3bc-ac)+\lambda\left(3bc-ac-b^2\right)\right)$
But actual answer is $a^2+b^2+c^2+\lambda$.
I tried to find my mistake, but everything seems correct. What am I missing here? Please help me in this.
| The determinant in your problem is equal to $p(-\lambda),$ where $p$ is the characteristic polynomial of $A = vv^T,$ where $ v = [a, \ b, \ c]^T.$ The $0$-eigenspace (i.e, the kernel of $A$) is the two dimensional hyperplane given by $v^Tx = 0,$ so $0$ is an eigenvalue with algebraic multiplicity at least two and the final eigenvalue is given by $$\operatorname{tr}(A) = \operatorname{tr}(vv^T) = \operatorname{tr} (v^Tv) = \| v \|^2.$$
| {
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"source": "stackexchange",
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"answer_count": 6,
"answer_id": 5
} |
Two Primitive Stochastic Matrices with Eventually Equal Sequence of Powers Let $A$ and $B$ be two $n\times n$ primitive row-stochastic matrices. That is, all of their entries are non-negative, all the rows sum up to $1$, and there is an integer $p\geq 1$ such that all the entries of $A^p$ and $B^p$ are strictly positive. Assume that $A_{i,j}=0\iff B_{i,j}=0$ for all $i,j$. Here is my question:
Suppose that there exists $k_0\geq 1$ such that $A^k=B^k$ for all
$k\geq k_0$. Does this imply that $A=B$?
Remark: Note that if any of $A$ and $B$ is invertible, then the answer is positive. Say for instance that $B$ is invartible, then
$$AB^{k_0}=AA^{k_0}=A^{k_0+1}=B^{k_0+1}=BB^{k_0}.$$
Multiplying by $B^{-k_0}$ we get that $A=B$.
| The specified conditions are not sufficient to force $A=B$.
Here's an example . . .
Choose $a,b\in (0,\frac{1}{2})$ with $a\ne b$, and let $A,B$ be given by
$\\[5.5pt]$
$$
A =
\pmatrix
{
a &\frac{1}{2}-a&\frac{1}{2}\cr
a &\frac{1}{2}-a&\frac{1}{2}\cr
\frac{1}{2}-a&a&\frac{1}{2}
}
\\
$$
$$
B =
\pmatrix
{
b &\frac{1}{2}-b&\frac{1}{2}\cr
b &\frac{1}{2}-b&\frac{1}{2}\cr
\frac{1}{2}-b&b&\frac{1}{2}
}
$$
Then $A\ne B$, but for all $k\ge 2$, we have
$$
A^k=B^k=
\pmatrix
{
{\large{\frac{1}{4}}}\;&{\large{\frac{1}{4}}}\;&{\large{\frac{1}{2}}}\cr
{\large{\frac{1}{4}}}&{\large{\frac{1}{4}}}&{\large{\frac{1}{2}}}\cr
{\large{\frac{1}{4}}}&{\large{\frac{1}{4}}}&{\large{\frac{1}{2}}}
}
$$
| {
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"answer_id": 0
} |
Evaluate $\int_0^1 \frac{\log ^2(x+1) \log \left(x^2+1\right)}{x^2+1} dx$ How can we evaluate
$$\int_0^1 \frac{\log ^2(x+1) \log \left(x^2+1\right)}{x^2+1} dx$$
Any kind of help is appreciated.
| I found the answer:
$$\small \int_0^1 \frac{\log ^2(x+1) \log \left(x^2+1\right)}{x^2+1} \, dx=\frac{\pi ^2 C}{48}-\frac{15}{4} C \log ^2(2)+20 \Im(\text{Li}_4(1+i))+2 \log (2) \Im(\text{Li}_3(1+i))+\frac{35 \pi \zeta (3)}{64}-\frac{5}{24} \pi \log ^3(2)-\frac{21}{64} \pi ^3 \log (2)-\frac{3}{256} \left(\psi ^{(3)}\left(\frac{1}{4}\right)-\psi ^{(3)}\left(\frac{3}{4}\right)\right)$$
A generalization:
$$\scriptsize \int_0^1 \frac{\log ^3(x+1) \log \left(x^2+1\right)}{x^2+1} \, dx=-\frac{39}{8} C \log ^3(2)+\frac{1}{32} \pi ^2 C \log (2)-27 \beta (4) \log (2)-48 \Im\left(\text{Li}_5\left(\frac{1}{2}+\frac{i}{2}\right)\right)-6 \log ^2(2) \Im\left(\text{Li}_3\left(\frac{1}{2}+\frac{i}{2}\right)\right)+6 \log (2) \Im\left(\text{Li}_4\left(\frac{1}{2}+\frac{i}{2}\right)\right)+\frac{105}{128} \pi \zeta (3) \log (2)+\frac{119 \pi ^5}{1024}+\frac{5}{16} \pi \log ^4(2)+\frac{105}{256} \pi ^3 \log ^2(2)$$
| {
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"answer_id": 0
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How to compute $\int \frac{16 x^3 - 42 x^2+2x}{\sqrt{-16x^8+112x^7-204x^6+28x^5-x^4+1}}\,\mathrm dx.$ I want to compute the following integral:
$$\int \frac{16 x^3 - 42 x^2+2x}{\sqrt{-16x^8+112x^7-204x^6+28x^5-x^4+1}}\,\mathrm dx.$$
First I tried substituting $y=\text{denominator}$ but it gets very messy.
Also, I tried using partial fractions but it doesn’t work because of the square root.
What can I do here?
| $-16x^8+112x^7-204x^6+28x^5-x^4+1=-(4x^4-14x^3+x^2+1)(4x^4-14x^3+x^2-1)$
Set $a=4x^4-14x^3+x^2.$ The integral rewrites
$$\int \frac{\mathrm da}{\sqrt{-a^2+1}}.$$
| {
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If $ab+bc+ca\ge1$, prove that $\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\ge\frac{\sqrt{3}}{abc}$ The following problem is from CHKMO 2018 Problem 1:
If $ab+bc+ca\ge1$, prove that $$\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\ge\frac{\sqrt{3}}{abc}$$
I tried to use Cauchy–Schwarz inequality, by try multiplying different things, such as $1^2+1^2+1^2$, $(abc)^2+(abc)^2+(abc)^2$. But I still can’t solve it. Can someone help me?
| AM-GM leads to
\begin{align*}
\frac{a^2b^2}{c^2}+\frac{b^2c^2}{a^2}&\geqslant 2\sqrt{\frac{a^2b^2}{c^2}\cdot \frac{b^2c^2}{a^2}}=2b^2
\end{align*}
Similarly
\begin{align*}
\frac{a^2b^2}{c^2}+\frac{b^2c^2}{a^2}\geqslant 2b^2\qquad \frac{b^2c^2}{a^2}+\frac{c^2a^2}{b^2}&\geqslant 2c^2\qquad \frac{c^2a^2}{b^2}+\frac{a^2b^2}{c^2}\geqslant 2a^2\\
\\
\therefore \frac{a^2b^2}{c^2}+\frac{b^2c^2}{a^2}+\frac{c^2a^2}{b^2}&\geqslant a^2+b^2+c^2
\end{align*}
Hence
\begin{align*}
\frac{a^2b^2}{c^2}+\frac{b^2c^2}{a^2}+\frac{c^2a^2}{b^2}+2a^2+2b^2+2c^2&\geqslant 3a^2+3b^2+3c^2\\
\iff \left(\frac{ab}{c}+\frac{bc}{a}+\frac{ca}{b}\right)^2&\geqslant 3a^2+3b^2+3c^2\tag{1}
\end{align*}
And since $a^2+b^2+c^2\geqslant ab+bc+ac$, we obtain $$\left(\frac{ab}{c}+\frac{bc}{a}+\frac{ca}{b}\right)^2\geqslant 3\cdot(ab+bc+ac)\geqslant3$$ Can you finish now?
| {
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"answer_count": 4,
"answer_id": 2
} |
Can any number of squares sum to a square? Suppose
$$a^2 = \sum_{i=1}^k b_i^2$$
where $a, b_i \in \mathbb{Z}$, $a>0, b_i > 0$ (and $b_i$ are not necessarily distinct).
Can any positive integer be the value of $k$?
The reason I am interested in this: in a irreptile tiling where the smallest piece has area $A$, we have $a^2A = \sum_{i=1}^k b_i^2A$, where we have $k$ pieces scaled by $b_i$ to tile the big figure, which is scaled by $a$. I am wondering what constraints there are on the number of pieces.
Here is an example tiling that realizes $4^2 = 3^2 + 7 \cdot 1^2$, so $k = 8$.
| Yes.
For $k = 2$:
$3^2 + 4^2 = 5^2$
For $k > 2$:
Start with a solution for $k-1$
Multiply both sides by $5^2$
Replace one $(5a)^2$ on the left with $(3a)^2$ + $(4a)^2$.
$3^2 + 4^2 = 5^2$
$15^2 + 20^2 = 25^2$
$9^2 + 12^2 + 20^2 = 25^2$ (k = 3)
$45^2 + 60^2 + 100^2 = 125^2$
$27^2 + 36^2 + 60^2 + 100^2 = 125^2$ (k = 4)
Repeat until k is as desired.
| {
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"question_score": "46",
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In how many ways could six of the colored Easter eggs been crushed? Problem:
Mother was carrying a basket with 11 red, 8 blue, 5 green and 5 yellow decorated Easter eggs (eggs
of the same color are identical).
When entering the room of her son Thomas, she tripped
over door step and crushed six eggs.
Thomas paused in thought for a moment, and than he said to
his mom the number expressing, all the options of which various colors six eggs could be crushed.
Determine the number given by Thomas.
Here what have I done so far:
$$Omega = ^{29}C_6$$
*
*The first case (2 red, 2 blue, 1 green, 1 yellow)
We have : $$^{11}C_2\cdot ^8C_2\cdot ^5C_1\cdot ^5C_1$$
*
*The second case (2 red, 1 blue, 2 green, 1 yellow)
We have : $$^{11}C_2\cdot ^8C_1\cdot ^5C_2\cdot ^5C_1$$
*
*The third case (2 red, 1 blue, 1 green, 2 yellow)
We have : $$^{11}C_2\cdot ^8C_1\cdot ^5C_1\cdot ^5C_2$$
*
*The fourth case (1 red, 2 blue, 2 green, 1 yellow)
We have : $$^{11}C_1\cdot ^8C_2\cdot ^5C_2\cdot ^5C_1$$
*
*The fifth case (1 red, 2 blue, 1 green, 2 yellow)
We have : $$^{11}C_1\cdot ^8C_2\cdot ^5C_1\cdot ^5C_2$$
*
*The sixth case (1 red, 1 blue, 2 green, 2 yellow)
We have : $$^{11}C_1\cdot ^8C_1\cdot ^5C_2\cdot ^5C_2$$
*
*The seventh case (3 red, 1 blue, 1 green, 1 yellow)
We have : $$^{11}C_3\cdot ^8C_1\cdot ^5C_1\cdot ^5C_1$$
*
*The eighth case (1 red, 3 blue, 1 green, 1 yellow)
We have : $$^{11}C_1\cdot ^8C_3\cdot ^5C_1\cdot ^5C_1$$
*
*The ninth case (1 red, 1 blue, 3 green, 1 yellow)
We have : $$^{11}C_1\cdot ^8C_1\cdot ^5C_3\cdot ^5C_1$$
*
*The tenth case (1 red, 1 blue, 1 green, 3 yellow)
We have : $$^{11}C_1\cdot ^8C_1\cdot ^5C_1\cdot ^5C_3$$
I have read the problem so many time, and have some confusion about "various colors six eggs", and I feel something wrong with my solution.
I do not know if all those case I have done above are enough, if not please show step if you know how to do it.
| It doesn't matter which specific red, blue, green, or yellow were crushed. Eggs the same color are considered equivalent.
Let $R = $ then number of red eggs. $R$ can be $0$ or it can be as high as $6$.
Let $B = $ be the number of blue eggs. $B$ can be as low as $0$ or as high as $6-R$
Let $G = $ the number of green eggs. $G$ can be as low as $0$ or as high as $\min(6-R-B, 5)$. However if $R=B =0$ then as there are only $5$ yellow eggs, the would have to be at least $1$ green egg. So $G$ can be as low as $\max(0, 1-R-B)$..
Let $Y=$ the number of yellow eggs. $Y = 6-R-B-G$. Once $R,B,G$ are determined there is only $1$ value of this.
Number ways this can be is:
$\sum\limits_{R=0}^6 \sum\limits_{B=0}^{6-R}\sum\limits_{G=\max(0, 1-R-B)}^{\min(6-R-B, 5)}1$
THose conditionals make it hard so I think we should divide it into cases
Case 1: $R = 0; B=0$.
Then the number of ways to do it is:
$\sum\limits_{R=0}^0 \sum\limits_{B=0}^{0}\sum\limits_{1}^{5}1=$
$ \sum\limits_{1}^{5}1=5$.
Case 2: There is at least one red or blue eggs.
Case 2a: There are no red eggs but at least one blue egg.
Then the numbers of ways to do this is
$\sum\limits_{R=0}^0 \sum\limits_{B=1}^{6}\sum\limits_{0}^{5}1=$
$\sum\limits_{B=1}^{6}\sum\limits_{0}^{5}1 = $
$\sum\limits_{B=1}^{6}6 = 36$
Case 2b: There is at least one red egg
Then the number of ways to do it are:
$\sum\limits_{R=1}^6 \sum\limits_{B=0}^{6-R}\sum\limits_{G=0}^{6-R-B}1$
Now $\sum\limits_{G=0}^{6-R-B}1= 6-R-B + 1=7-R-B$.
So the number of ways to do the eggs is:
$\sum\limits_{R=0}^6 \sum\limits_{B=0}^{6-R}(7-R-B)$
Now $\sum\limits_{B=0}^{6-R}(7-R-B)=$
$\sum\limits_{B=0}^{6-R}(7-R) - \sum\limits_{B=0}^{6-R}B =$
$(7-R)(6-R+1)- \frac {(6-R)(6-R+1)}2=$
$(7-R)^2 - \frac {(7-R)(6-R)}2 =\frac{2(7-R)^2 - (7-R)(6-R)}2=$
$\frac {(7-R)(2(7-R) - (6-R))}2=\frac {(7-R)(8-R)}2=$
$\frac {R^2 - 15R + 56}2$
So the number of ways to do the eggs if there is at least one egg is:
$\sum\limits_{R=1}^6 \frac {R^2 - 15R + 56}2=$
$\frac 12\sum\limits_{R=1}^6 R^2 -\frac {15}2\sum\limits_{R=1}^6 R + 28\sum\limits_{R=1}^61=$
$\frac 12\frac {6*7*13}6 - \frac {15}2\frac {6*7}2 + 28*6 =$
$\frac {13}2 - \frac{15*21}2 + 168 =$
$6\frac 12 - 157\frac 12 + 168 = 17$.
So there are $17+36+5 =58$ ways to do this.
....
| {
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Does exist $a,b,c \in \Bbb N$ such that $(a+b)(b+c)(c+a)=340$? Does exist $a,b,c \in \Bbb N$ such that $(a+b)(b+c)(c+a)=340$?
$340=2\cdot2\cdot5\cdot17$
I just noticed that $(a+b)+(b+c)+(c+a)=2(a+b+c)$ can it be useful to prove first equation?
Also I tried to construct numbers $a,b,c$ such that $a+b$ would be even, $b+c$ odd, $c+a$ odd and I couldn't get $340$ so I made a prediction that it is impossible to get it. But how should I prove it?
| Your observation that $(a+b)+(b+c)+(c+a)=2(a+b+c)$ i.e. even, could help. Since $a,b,c\in \mathbb{N}$, so the least value of each of the factors in LHS is $2$. So, there are four possibilities $340=2\times 2\times 85=2\times 5\times 34=2\times 10\times 17=4\times 5\times 17$. The first three are not acceptable as their sum is odd. Assume $a\leq b\leq c$. Then $a+b\leq b+c$. Also $a+b\leq c+a$. So, $a+b$ is the smallest. Thus, $a+b=4$. If $b+c=5$ then we get $c=9$ which is not possible. If $c+a=5$ then we get $c=9$ which is not possible.
| {
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"timestamp": "2023-03-29T00:00:00",
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Inequality Proof $(x^2+1)(y^2+1)(z^2+1)\leq...$ I want to show, that for positive numbers $x,y,z$ with $xy,yz,zx\geq1$, $(x^2+1)(y^2+1)(z^2+1)\leq\left(\left(\frac{x+y+z}{3}\right)^2+1\right)^3$.
Using the AM-QM inequality for $(x^2+1),(y^2+1)$ and $(z^2+1)$, I already got $(x^2+1)(y^2+1)(z^2+1)\leq\left(\frac{x^2+y^2+z^2}{3}+1\right)^3$.By equivalent transformations of $\left(\frac{x^2+y^2+z^2}{3}+1\right)^3\leq\left(\left(\frac{x+y+z}{3}\right)^2+1\right)^3$ I get $0\leq\frac{2}{3}\left(xy+yz+zx-x^2-y^2-z^2\right)$, which cannot be true. Where did I miss something? Is the AM-QM inequality too weak?
Thank you!
| By the Carl Schildkraut's hint we have: $$(1+x^2)(1+y^2)\leq\left(1+\left(\frac{x+y}{2}\right)^2\right)^2$$ it's
$$(x-y)^2(x^2+y^2+6xy-8)\geq0,$$ which is true by AM-GM:
$$x^2+y^2+6xy-8\geq2xy+6xy-8\geq0.$$
Thus, it's enough to prove that
$$\prod_{cyc}\left(1+\left(\frac{x+y}{2}\right)^2\right)\leq\left(1+\left(\frac{x+y+z}{3}\right)^2\right)^3$$ or
$$\frac{\sum\limits_{cyc}\ln\left(1+\left(\frac{x+y}{2}\right)^2\right)}{3}\leq\ln\left(1+\left(\frac{\sum\limits_{cyc}\frac{x+y}{2}}{3}\right)^2\right),$$
which is true by Jensen because by AM-GM $$\frac{x+y}{2}\geq\sqrt{xy}\geq1$$ and for any $x\geq1$ we have:
$$\left(\ln(1+x^2)\right)''=\left(\frac{2x}{1+x^2}\right)'=\frac{2(1-x^2)}{(1+x^2)^2}\leq0.$$
| {
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Find $\lim\limits_{x \to \infty} \left( \frac{\sqrt{x^2+2x-3}}{x+2} \right)^{3-2x}$ How can I find this limit?
$$\lim\limits_{x \to \infty} \bigg ( \dfrac{\sqrt{x^2+2x-3}}{x+2} \bigg )^{3-2x}$$
Firstly I thought I can use the limit:
$$\lim\limits_{x \to \infty} \bigg ( 1 + \dfrac{1}{x} \bigg )^x=e$$
by adding $1$ and subtracting $1$ from the original limit. However, since $3-2x$ $\rightarrow - \infty$ and not $+\infty$, I got nowhere. Then I tried finding the logarithm of this limit. It resulted in a $\dfrac{0}{0}$ indeterminate form, I tried L'Hospital, but again, it led me nowhere. Either I made some mistakes in the calculations, or I should use a different approach.
| Let $y =\bigg ( \dfrac{\sqrt{x^2+2x-3}}{x+2} \bigg )^{3-2x}$. Then,
$$\ln y = (3-2x)\ln \sqrt{\dfrac{x^2+2x-3}{(x+2)^2}} $$
$$= \frac{3-2x}2 \ln \left(1-\frac2{x+2}+O(\frac1{(x+2)^2})\right)
= \frac{\ln\left( 1-\frac2{x+2}\right)+O(\frac1{(x+2)^2})}{\frac1{-(x+2)}+O(\frac1{(x+2)^2})}$$
Therefore,
$$\lim\limits_{x \to \infty} \ln y
= \lim\limits_{x \to \infty}\frac{\ln\left( 1-\frac2{x+2}\right)}{\frac1{-(x+2)}} =2$$
which leads to $\lim\limits_{x \to \infty} y = e^2$.
| {
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For real positive $a,b$ given $a^2b^2(a^2b^2+4)=2(a^6+b^6)$, show that at least one of the numbers is irrational I think I am close to the answer, but still not sure how to finish it.
$$a^2b^2(a^2b^2+4)=2(a^6+b^6)$$
I know I can rewrite that equation as
$$(a^2-b\sqrt2)(a^2+b\sqrt2)(a\sqrt2-b^2)(a\sqrt2+b^2)=0$$
Thus $$a^2=\pm b\sqrt2 \space\space\space\space\space\space b^2=\pm a\sqrt2$$
Now, am I allowed to substitute one equation into the other? This would mean that $b=0$ or $b=\pm\frac{\sqrt[3]4}{\sqrt[3]{\sqrt2}}$. But then when I try $b=0$ I get $a=0$ as well.
How can I prove at least one of them is irrational?
| Lets do it the long way. Distribute everything and move it to one side
$$a^4b^4+4a^2b^2-2a^6-2b^6=0$$
Lets substitute
$$x=ab$$
Note x is always positive by the positive definition of $1$ and $b$. Then,
$$a=\frac{x}{b}$$
$$b=\frac{x}{a}$$
Which are also positive definite relations. Now substitute to get
$$x^4 + 4x^2 -2(\frac{x^6}{a^6})-2(\frac{x^6}{b^6})=0$$
Factor out an $x^2$
$$x^2(x^2+4-x^4(\frac{2}{a^6} + \frac{2}{a^6}))=0$$
Note that $(\frac{2}{a^6} + \frac{2}{a^6})$ is also positive definite, so lets replace it with another positive definite substitution $c = \frac{2}{a^6} + \frac{2}{a^6}$. Now we have
$$x^2(-cx^6+x^2+4)=0$$
Lets find what the roots of the inside will be, since $x^2=0$ is trivial.
$$-cx^4+x^2+4=0$$
Complete the square
$$x^4-\frac{1}{c}x^2=\frac{4}{c}$$
$$x^4-\frac{1}{c}x^2 +\frac{1}{4c^2}=\frac{4}{c}+\frac{1}{4c^2}$$
$$(x^2-\frac{1}{2c})^2=\frac{16c+1}{4c^2}$$
$$x^2-\frac{1}{2c}=\sqrt{\frac{16c+1}{4c^2}} = \pm \frac{\sqrt{16c+1}}{2c}$$
Finally,
$$x^2=\frac{1}{2c}\pm \frac{\sqrt{16c+1}}{2c}$$
and since everything is positive definite, note that
$$\frac{1}{2c} - \frac{\sqrt{16c+1}}{2c} < 0$$
so we must have at least 1 complex root.
| {
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Generating Function & Sequence Find the generating functions of the sequences
2, 1, 2, 1, 2, 1, . . .
I get $\frac{1}{1+x} + \frac{1}{1-x} = \frac{2}{1-x^2}$
But the solution ends up with $\frac{2}{1-x^2} + \frac{x}{1-x^2} = \frac{2+x}{1-x^2}$.
The solutions starts with $\sum_{n\ge 0} (2)x^{2n}+\sum_{n\ge 0} (1) x^{2n+1}$
I couldn't come up with anything like that. I feel like I'm confused with something.
| The sequence $2,1,2,1,2,1,...$ alternates between $2$ and $1$, being $2$ for even-numbered terms and $1$ for odd-numbered terms. The generating function is thus $$\sum\limits_{n \ge0}(2)x^{2n}+\sum\limits_{n\ge0}(1)x^{2n+1}=\sum\limits_{k\ge0}(1)x^k+\sum\limits_{k\ge0}(1)x^{2k}=\dfrac{1}{1-x}+\dfrac{1}{1-x^2}=\dfrac{2+x}{1-x^2}$$
| {
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How do you evaluate $\int_{0}^{1} \frac{(3x^3-x^2+2x-4)}{\sqrt{x^2-3x+2}} \, dx$? Saw this problem on a FaceBook meme that said the pin code to his ATM debit card is the solution to the following problem:
$$\int_{0}^{1} \frac{(3x^3-x^2+2x-4)}{\sqrt{x^2-3x+2}} \, dx$$
I was trying to see how we could break this up into an easier integrals but nothing comes to mind at first glance. Perhaps complex integration is possible?
| As an alternative to the substitution described in the comments, the anti-derivative of expressions of the form $P(x)/\sqrt{ax^2+bx+c}$, $(a\ne 0)$, where $P(x)$ is a non-constant polynomial is:
$$\int \frac{P(x)}{\sqrt{ax^2+bx+c}}\mathrm{d}x=Q(x)\sqrt{ax^2+bx+c}+\lambda\int\frac{1}{\sqrt{ax^2+bx+c}}\mathrm{d}x $$
where $Q(x)$ is a polynomial with undetermined coefficients of one degree less than $P(x)$ and $\lambda$ is an unknown number. To find the coefficients, differentiate both sides, get rid of the square root, and equate coefficients for the powers of $x$.
In this case:
$$\int \frac{3x^3-x^2+2x-4}{\sqrt{x^2-3x+2}}\mathrm{d}x=\left(x^2+\frac{13}{4}x+\frac{101}{8}\right)\sqrt{x^2-3x+2}+\frac{135}{16}\int \frac{1}{\sqrt{x^2-3x+2}}\mathrm{d}x$$
and
$$\int \frac{1}{\sqrt{x^2-3x+2}}\mathrm{d}x=\int \frac{1}{\sqrt{\left(x-\frac{3}{2}\right)^2-\frac{1}{4}}}\mathrm{d}x=\ln\left|x-\frac{3}{2}+\sqrt{x^2-3x+2}\right|+C $$
Update: In your case, $P(x)$, the polynomial in the numerator, has degree $3$, so $Q(x)$ has degree $2$: $Q(x)=Ax^2+Bx+C$.
So you have
$$\int \frac{3x^3-x^2+2x-4}{\sqrt{x^2-3x+2}}\mathrm{d}x=\left(Ax^2+Bx+C\right)\sqrt{x^2-3x+2}+\lambda\int \frac{1}{\sqrt{x^2-3x+2}}\mathrm{d}x$$
and after differentiation:
$$\frac{3x^3-x^2+2x-4}{\sqrt{x^2-3x+2}}=(2Ax+B)\sqrt{x^2-3x+2}+(Ax^2+Bx+C)\frac{2x-3}{2\sqrt{x^2-3x+2}}+\frac{\lambda}{\sqrt{x^2-3x+2}} $$
Now, multiply both sides by the square root to remove it, and equate coefficients for the powers of $x$.
| {
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Find area of region bounded by $xy = a^2$, $xy = b^2$, $x = py$, $x = qy$ I need to obtain the area of the region bounded by the curves
$xy = a^2$, $xy = b^2$, $x = py$, $x = qy$, $0 < a < b$, $0 < p < q.$
I guess I should use Jacobian of the transformation. I tried $u = \sqrt{xy}$ and $v = \frac{x}{y}$, but it didn't work out. Can you suggest an appropriate substitution?
| Here is to integrate in polar coordinates. The boundaries are
$a^2\le r^2\sin\theta\cos\theta\le b^2$ and $\frac1q\le \tan\theta \le \frac1p$, or, the corresponding integral limits,
$$\theta_1= \tan^{-1}\frac1q,\>\>\>\>\> \theta_2= \tan^{-1}\frac1p,
\>\>\>\>\>r_1^2(\theta)=\frac{a^2}{\sin\theta\cos\theta},
\>\>\>\>\>r_2^2(\theta)=\frac{b^2}{\sin\theta\cos\theta}$$
Thus, the integral is
$$\int_{\theta_1}^{\theta_2}
\int_{r_1(\theta)}^{r_2(\theta)} rdrd\theta
=\frac{b^2-a^2}2\int_{\theta_1}^{\theta_2} \frac{d\theta}{\sin\theta\cos\theta}
=\frac{b^2-a^2}2\ln(\tan\theta)\bigg|_{\theta_1}^{\theta_2}=\frac{b^2-a^2}2\ln\frac qp$$
| {
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What is the angle between the asymptotes of the hyperbola $5x^2-2\sqrt 7 xy-y^2-2x+1=0$?
What is the angle between the asymptotes of this hyperbola?
$$5x^2-2\sqrt 7 xy-y^2-2x+1=0$$
I used $S+\lambda=0$ and used straight line condition to find combined equation to asymptotes. Then how to find angle between them?
| Solve the quadratic equation for $y$:
$$y=-x\sqrt{7}-\sqrt{12x^2-2x+1},y=-x\sqrt{7}+\sqrt{12x^2-2x+1}$$
Let the assymptotes be $y=a_1x+b_1$ and $y=a_2x+b_2$. Then:
$$\lim_{x\to\infty} \frac{-x\sqrt{7}-\sqrt{12x^2-2x+1}}{a_1x+b_1}\stackrel{LR}{=}\lim_{x\to\infty} \frac{-\sqrt{7}-\frac{12x-1}{\sqrt{12x^2-2x+1}}}{a_1}= \lim_{x\to\infty} \frac{-\sqrt7-2\sqrt3}{a_1}=1\Rightarrow \\
a_1=-\sqrt7-2\sqrt3$$
Similarly, $a_2=-\sqrt7+2\sqrt3$. The angle between the lines with such slopes is:
$$\tan \phi=\left|\frac{a_1-a_2}{1+a_1a_2} \right|=\sqrt3 \Rightarrow \phi =60^\circ.$$
| {
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$ \lim_{n \to \infty} (1+ \frac{3}{n^2+n^4})^n$ I've tried to solve the limit
$$ \lim_{n \to \infty} (1+ \frac{3}{n^2+n^4})^n$$
but I'm not sure.
$ (1+ \frac{3}{n^2+n^4})^n = \sqrt [n^3]{(1+ \frac{3}{n^2+n^4})^{n^4}} \sim \sqrt [n^3]{(1+ \frac{3}{n^4})^{n^4}} \sim \sqrt [n^3]{e^3 } \rightarrow 1$
Is it right?
| Yes, I think it's right. Of course, you might have to justify that your approximations still preserve the limit.
Another way would be to write
$$
\left(1+ \frac{3}{n^2+n^4}\right)^n=\exp\left( n\log\left(1+ \frac{3}{n^2+n^4}\right)\right)\sim\exp\left( \frac{3n}{n^2+n^4}\right)\to e^0=1.
$$
| {
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Simplify $\frac{d}{dx}\frac{x^2}{1+\sqrt{x^2+1}}$ Simplify $$\frac{d}{dx}\frac{x^2}{1+\sqrt{x^2+1}}$$
I have manage to arrive to $$\frac{x^3+2x+2x\sqrt{x^2+1}}{(x^2+2)\sqrt{x^2+1}+2x^2+2}$$
But Wolfram manage to simplify to $$\frac{x}{\sqrt{x^2+1}}$$
| Hint: Expand the fraction with $1-\sqrt{x^2+1}$:
$$\frac{x^2}{1+\sqrt{x^2+1}} =\frac{x^2(1-\sqrt{x^2+1})}{1-(x^2+1)} = \sqrt{x^2+1}-1$$
| {
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How to solve $x^{\log(x)-1 } = \frac{1}{\sqrt[4]{10}}$ Question:
$x^{\log(x)-1 } = \frac{1}{\sqrt[4]{10}}$
My attempts to solve:
$x^{\log(x)-1 } = \frac{1}{\sqrt[4]{10}}$
$x^{\log(x)-\log(10) } = 10^\frac{-1}{4}$
$x^{\log(\frac{x}{10}) } = 10^\frac{-1}{4}$
$-\frac{1}{4}\log{_x 10} = \log(\frac{x}{10})$
$-\frac{1}{4}\log{_x 10} = \frac{\log{_x}x}{\log{_x}10}$
$-\frac{1}{4}\log{_x 10} = \frac{1}{\log{_x}10}$
$-\frac{1}{4}\log{_x 10^2} = 1$
$-\frac{1}{2}\log{_x 10} = 1$
$\log{_x 10} = -2$
$10 = x^{-2}$
$10 = \frac{1}{x^2}$
$x^2 = \frac{1}{10}$
$x = {\sqrt{\frac{1}{10}}}$
But this is still wrong. the correct answer is $x = {\sqrt{10}}$. Can someone please point out which part i have done wrong? Thanks.
| $$ x^{\log_{10} x-1}=10^{-1/4} \implies 10^{\log_{10} (x^{\log_{10} x-1})}=10^{1/4} $$ $$\implies
10^{\log^2_{10} x -\log_{10} x-1/4}=10^0$$ $$\implies (\log_{10} x-1/2)^2=0 \implies \log_{10}x=1/2 \implies x=10^{1/2}=\sqrt{10}.$$
| {
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Finding $\int \sqrt{3-2x-x^2}dx$ $\int \sqrt{3-2x-x^2}\,dx$
First I did:
$$\begin{align}\int \sqrt{3-2x-x^2}\,dx &= \int \sqrt{-((x+1)^2-4)}\,dx \\
&=\int \sqrt{4-(x+1)^2}\,dx \\
\end{align}$$
Then I set $(x+1)=2\sin(t)$, $dx = 2\cos(t)\,dt$
$$\begin{align}
\int \sqrt{4-(2\sin(t))^2}2\cos(t)\,dt &= \int\sqrt{4-4\sin^2(t)}\,\,2\cos(t)\,dt \\
&= \int4\sqrt{1-\sin^2(t)}\cos(t)\,dt \\
&=4\int \sqrt{1-\sin^2(t)}\cos(t)\,dt \\
&=4\int\cos^2(t)\,dt \\
&=4\int\frac{1+\cos(2t)}\,2\,dt \\
&=2 \int 1+\cos(2t) \,dt \\
&=2t + \sin(2t) + C
\end{align}$$
$x+1 = 2\sin(t) \Leftrightarrow \frac{x+1}{2} = \sin(t) \Leftrightarrow t = \arcsin(\frac{x+1}{2})$
Replacing,
$$2\arcsin\left(\frac{x+1}{2}\right) + \sin\left(2\arcsin\left(\frac{x+1}{2}\right)\right) + C$$
But my book's solution is $$\frac{x+1}{2}\sqrt{3-2x-x^2}+2\arcsin\left(\frac{x+1}{2}\right)$$
So... what went wrong?
| HINT:
Note that $\sin(2x)=2\sin(x)\cos(x)$, $\sin(\arcsin(x))=x$, and $\cos(\arcsin(x))=\sqrt{1-x^2}$.
Can you reconcile now?
| {
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Evaluate $\int_0^{2\pi}{dx/(1+\epsilon \cos{x})}$ Please give hints or help to solve this integral, $\epsilon^2<1$
$$\int_0^{2\pi}\frac{dx}{1+\epsilon\cos x}$$
| Let $a^2<1$, then
$$I=\int_{9}^{2\pi} \frac{dx}{1+a \cos x}= 2 \int_{0}^{\pi} \frac{dx}{1+a\cos x}~~~~(1)$$
$$I=2\int_{0}^{\pi}\frac{dx}{1-a \cos x}~~~~(2)$$
Adding (1) and (2), we get
$$2I=4\int_{0}^{\pi}\frac{dx}{1-a^2\cos^2 x} =8 \int_{9}^{\pi/2} \frac{sec^2 xdx}{sec^2 x-a^2}=8\int_{0}^{\pi/2} \frac{sec^2 x dx}{\tan^2x+1-a^2}.$$
$$I=4\int_{0}^{\infty} \frac{ du}{u^2+(1-a^2)}=\frac{4}{\sqrt{1-a^2}}
\left . \tan^{-1} \frac{u}{\sqrt{1-a^2}}\right|_{0}^{\infty}=\frac{2\pi}{\sqrt{1-a^2}} $$
| {
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Determine all pairs $(a, b)$ for which $s (an + b) - s (n)$ assumes finitely many values For every positive integer n we define $s (n)$ as the sum of the digits of $n. $ Determine all pairs $(a, b)$ of
positive integers for which$$s (an + b) - s (n)$$assumes a finite number of values by varying $n$ in positive integers.
Solution: The solution is $a = 10^k, 1 \le b < 10^k$ for $k \ge 1.$ This clearly works: $s(10^k n) = s(n),$ and adding $b$ on only affects the last $k$ digits, which are all zeroes, so $0 \le s(10^k n + b) - s(n) \le 9k.$
If $a$ is not of the form $2^u 5^v,$ let $n = \lceil 10^m / a \rceil.$ Now $10^m + b \le an+b \le 10^m + (a+b)$ means that $an+b$ is $1$ followed by a bunch of zeroes followed the expansion of $k$ for $b \le k \le a+b$ for $m$ large enough. Thus, $s(an+b) \le 1 + s(M)$ where $M = \max\limits_{a \le i \le a+b} s(i).$ However, $s(\lceil 10^m / a \rceil)$ is unbounded* as $m$ increases, contradiction.
Suppose $a = 2^u 5^v.$ If $u \ne v,$ WLOG $u>v$ (other case is similar). Let $c_m = (10^m - 1)/9.$ Take $n = 10^{r-v}c_m$ so that $s(n) = m.$ We want to show $s(2^u 5^v n+b) = s(10^r 2^{u-v} c_m + b) \ge cm$ for some $c>1$ for $m$ large enough (this part is incomplete). Let $t = u-v > 0$ and take $r$ large enough so that $s(10^r 2^{u-v} c_m + b) = s(2^t c_m + b).$ Now we want to show $s(2^t c_m + b) \ge cm$ for some $c>1$ for $m$ large enough.
The expansion of $2^t c_m$ is a bunch of junk digits followed by a long repetition of the same digit, followed by some junk at the end from $b.$ We can take $t$ large enough to ignore the junk, so we just require that the repeating digit is not $1.$ For example, $111,111,111,111 \cdot 128 = 14,222,222,222,208$ and we ignore the $14, 08$ at the start and end. This rules out $t=7.$ But we cannot rule out every $t.$ For example, $111,111,111,111 \cdot 64 = 7,111,111,104.$ Now the junk is $7, 04$ but the main string is all ones. In these cases, we go back to the start and take $c_m = 2(10^m - 1)/9$ instead, resolving the issue. This should be the last fix to the proof.
Now if $a = 10^k,$ let $n = 10^m - 1$ so that $an+b = 10^{m+k} + (b-10^k)$ and $s(n) = m$ is unbounded. For $m$ large enough and $b \ge 10^k,$ $an+b$ is $1$ followed by a bunch of zeroes followed by the expansion of $b-10^k.$ Thus, $s(an+b) = 1 + s(b-10^k),$ which is bounded, contradiction.
Proof of $(*)$: We are essentially doing long division on $1/a$ with a decimal shift. The decimal expansion does not terminate unless $a = 2^u 5^v$ for $u, v \ge 0,$ so the sum of digits after a decimal shift is unbounded as $m$ increases and reveals more decimal places.
Any more elegant solution than this?
| If $a \ne 10^k$, then choose $L$ greater than both $a$ and $b$ and let $n=10^L+10^{2L}+... +10^{mL}$. Then $S(an+b)-S(n)=m(S(a)-1)+S(b)$ takes infinitely many values.
If $a =10^k$ and $b\ge a$, then choose $m$ greater than $k$ and let $n=10^m-1$. Then $S(an+b)-S(n)=1+S(b-10^k)-9m$ takes infinitely many (negative) values.
If $a =10^k$ and $b<a$, then $S(an+b)-S(n)=S(b)$ takes only one value.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3482492",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find $\lim_{x\to0}\frac{1-\cos x\cos2x\cos3x}{x^2}$
Find $$\lim_{x\to0}\dfrac{1-\cos x\cos2x\cos3x}{x^2}$$
My attempt is as follows:-
$$\lim_{x\to0}\dfrac{1-\dfrac{1}{2}\left(2\cos x\cos2x\right)\cos3x}{x^2}$$
$$\lim_{x\to0}\dfrac{1-\dfrac{1}{2}\left(\cos3x+\cos x\right)\cos3x}{x^2}$$
$$\lim_{x\to0}\dfrac{1-\dfrac{1}{2}\left(\cos^23x+\cos x\cos3x\right)}{x^2}$$
$$\lim_{x\to0}\dfrac{1-\dfrac{1}{2}\left(\dfrac{1+\cos6x}{2}+\dfrac{1}{2}\left(\cos4x+\cos2x\right)\right)}{x^2}$$
$$\lim_{x\to0}\dfrac{1-\dfrac{1}{4}\left(1+\cos6x+\cos4x+\cos2x\right)}{x^2}$$
Applying L'Hospital as we have $\dfrac{0}{0}$ form
$$\lim_{x\to0}\dfrac{-\dfrac{1}{4}\left(-6\sin6x-4\sin4x-2\sin2x\right)}{2x}$$
Again applying L'Hospital as we have $\dfrac{0}{0}$ form
$$\lim_{x\to0}\dfrac{-\dfrac{1}{4}\left(-36\cos6x-16\cos4x-4\cos2x\right)}{2}=\dfrac{36+16+4}{8}=\dfrac{56}{8}=7$$
But actual answer is $3$, what am I messing up here?
| Using $\cos x=1-x^2/2+O(x^4)$ gives
$\cos 2x=1-2x^2+O(x^4)$ and
$\cos 3x=1-9x^2/2+O(x^4)$. Therefore
$$\cos x\cos 2x\cos 3x=1-7x^2+O(x^4)$$
and so
$$\lim_{x\to0}\frac{1-\cos x\cos 2x\cos 3x}{x^2}=7.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3483609",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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$\frac{x}{y}-\frac{y}{x}=\frac56$ and $x^2-y^2=5$
Solve the system: $$\begin{array}{|l}
\dfrac{x}{y}-\dfrac{y}{x}=\dfrac{5}{6} \\ x^2-y^2=5 \end{array}$$
First, we have $x,y \ne 0$. Let's write the first equation as:
$$\dfrac{x}{y}-\dfrac{y}{x}=\dfrac{5}{6} \Leftrightarrow \dfrac{x^2-y^2}{xy}=\dfrac{5}{6}$$
We have $x^2-y^2=5$, therefore $xy=6$. What to do next?
| $xy = 6 \implies y= \dfrac6x$ . So ,
$$x^2-y^2 = 5 \implies x^2 - \dfrac{36}{x^2} =5$$
Taking $x^2 = a$ , we get :
$$a^2-36=5a\implies (a-9)(a+4) = 0$$
We have $a=9\implies x=\pm3$ and $y = \pm2$
And we have $a=-4 \implies x = \pm 2i$ and $y=\pm 3i$
So the solutions are $(x,y) = (3,2)$ , $(x,y) = (-3,-2)$ ,
$(2i,-3i)$ and $(x,y) = (-2i,3i)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3484688",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 0
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Justification for Rudin exercise 3.2
Rudin 3.4. Calculate $\lim\limits_{n \to \infty} \left(\sqrt{n^2 + n} - n\right)$.
The solution to this seems obvious: after enough arithmetic, let $n \to \infty$ and get $\frac{1}{2}$. I am not sure how to rigorously justify the final step, though. Here is what I have:
Leveraging multiplication by the conjugate, we get:
\begin{align*}
\sqrt{n^2 + n} - n & = \left(\sqrt{n^2 + n} - n\right) \cdot \frac{\sqrt{n^2 + n} + n}{\sqrt{n^2 + n} + n} \\
& = \frac{n^2 + n + n \sqrt{n^2 + n} - n \sqrt{n^2 + n} - n^2}{\sqrt{n^2 + n} + n} \\
& = \frac{n^2 + n - n^2}{\sqrt{n^2 + n} + n} \\
& = \frac{n}{\sqrt{n^2 + n} + n}
\end{align*}
Multiplying through by $\frac{\frac{1}{n}}{\frac{1}{n}}$ gives:
\begin{align*}
& = \frac{\frac{1}{n} \cdot n}{\frac{1}{n} \left(\sqrt{n^2 + n} + n\right)} \\
& = \frac{1}{\frac{1}{n} \sqrt{n^2 + n} + 1} \\
& = \frac{1}{\sqrt{\frac{1}{n^2} \left(n^2 + n\right)} + 1} \\
& = \frac{1}{\sqrt{1 + \frac{1}{n}} + 1}
\end{align*}
Clearly, a limit of a quotient is the quotient of the limits. The limit of a constant is $1$, so that take scare of the numerator. The limit of a sum is the sum of the limits, so that gets us the limit of $\sqrt{1 + \frac{1}{n}}$ and the limit of $1$, which is a constant.
However, obvious though it is that this limit is $\frac{1}{2}$ because $\frac{1}{n} \to 0$, I cannot find a theorem in Rudin justifying this. Proving it from scratch would not be too much of a problem, but my question is: is there a general, all-purpose justification for a result like this that says, in effect, that we can "plug in" the limiting argument? If the function were continuous, this would be fine, but we don't have continuity for another chapter.
EDIT: This isn't a duplicate because the linked question does not address my specific questions on this exercise.
| Here's how I solved it back when I self-studied my way through Baby Rudin. This has never been seen by anyone other than me before, so apologies if there are any errors in it.
As a lemma, we will show that $\lim_{n\rightarrow\infty}\sqrt{n+\alpha}-\sqrt{n}=0$
for any $\alpha\in\mathbb{P}$. To that end, let $\epsilon\in\mathbb{P}^{\times}$
be given. Then let $x\in\mathbb{N}$
be given and assume that $x>\left(\frac{\alpha-\epsilon^{2}}{2\epsilon}\right)^{2}$. As none of the terms that follow can be negative,
$$\sqrt{x} > \frac{\alpha-\epsilon^{2}}{2\epsilon}\\
2\epsilon\sqrt{x} > \alpha-\epsilon^{2}\\
\epsilon^{2}+2\epsilon\sqrt{x} > \alpha\\
\epsilon^{2}+2\epsilon\sqrt{x}+x > x+\alpha\\
\sqrt{\epsilon^{2}+2\epsilon\sqrt{x}+x} > \sqrt{x+\alpha}\\
\epsilon+\sqrt{x} > \sqrt{x+\alpha}\\
\epsilon > \sqrt{x+\alpha}-\sqrt{x}$$
Since $x$
was an arbitrary “sufficiently large” number, we may conclude that $\lim_{n\rightarrow\infty}\sqrt{n+\alpha}-\sqrt{n}=0$
as that expression is clearly not strictly negative for any value of $n$.
With that done, the limit in question is quite easily calculated. $$\lim_{n\rightarrow\infty}(\sqrt{n^{2}-n}-n) = 0+\lim_{n\rightarrow\infty}(\sqrt{n^{2}-n}-n)\\
= \lim_{n\rightarrow\infty}(\sqrt{n^{2}-n+\frac{1}{4}}-\sqrt{n^{2}-n})+\lim_{n\rightarrow\infty}(\sqrt{n^{2}-n}-n)\\
= \lim_{n\rightarrow\infty}(\sqrt{n^{2}-n+\frac{1}{4}}-\sqrt{n^{2}-n}+\sqrt{n^{2}-n}-n)\\
= \lim_{n\rightarrow\infty}(\sqrt{n^{2}-n+\frac{1}{4}}-n)\\
= \lim_{n\rightarrow\infty}((n+\frac{1}{2})-n)=\lim_{n\rightarrow\infty}\frac{1}{2}=\frac{1}{2}$$
So I think that you're in the same boat I was in, where I did a little epsilon wrangling using the definition in 3.1 to justify the part where I wanted to "plug in infinity".
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3484802",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Why does $x^2 \equiv 2 \pmod{5}$ have no solution? Can someone explain why $x^2 \equiv 2 \pmod{5}$ has no solution other than trial and error?
So I've shown that modulo 5, we have $0^2 \equiv 0, 1^2 \equiv 1, 2^2 \equiv 4, 3^2 \equiv 4, 4^2 \equiv 1$ and I'm seeing that it doesn't look like there is a solution but what is the actual reason? Since I can't go through infinitely many numbers.
| You have already shown what you need. Every integer is of the form $0, 1, 2, 3$, or $4$ modulo $5$, and $0^2$ mod $5 = 0$, $1^2$ mod $5 = 1$, $2^2$ mod $5 = 4$, $3^2$ mod $5 = 4$, and $4^2$ mod $5 = 1$, so we know that there is no integer $x$ for which $x^2 \equiv 2(mod\;5)$ i.e. $x^2(mod\;5) \neq 2$ for all integers $x$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3485558",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Jordan canonical form of a matrix $A$ Can you help me to find solution to this problem?
I have matrix $ A \in M_n(\mathbb{C}): A^2 + A^4 = 0$ and I know, that maximal number of linear independent eigenvectors is $4$. Also known, that minimal polynomial not equal to characteristic polynomial.
What we can say about Jordan canonical form of this matrix $ A$?
| Here is my second answer that I believe is a sufficient guide to a full classification of all Jordan Canonical forms for arbitrary dimension $n$.
So first we need to find the possible minimal polynomials, which are:
\begin{align*}
x, \ x+i, \ x−i, \ x^2, \ x^2+1, \ x(x+i), \ x(x−i), & \\
x(x^2+1), \ x^2(x−i), \ x^2(x+i), x^2(x^2+1). & \quad (1)
\end{align*}
Next, note that for $n<4$, the minimal polynomial must have a degree equal or less than $n$. So for example, $n=0$ is trivial, and for $n=2$, we can only have $x, x+i, x−i, x^2, x^2+1, x(x+i), x(x−i)$.
For the case $n \geq 4$, all the possible minimal polynomials listed in $(1)$ are included. To determine the possibilities based on a given minimal polynomial for a dimension $n$, we include at least one block of each root of the multiplicity of the minimal polynomial, finding the rest of the possibilities is then a combinatorial exercise of finding all the possible combinations of extra Jordan blocks such that the sum of lengths of each Jordan block adds up to $n$.
As in my previous answer, I'll do the cases for $x(x^2+1)$ and $x^2(x-i)$, when $n \geq 3$.
For $x(x^2+1)$, we find that there must be at least one $0$-block, $i$-block, and $−i$-block of length one. The rest of the possible blocks can be any $0$-block $i$-block, $-i$-block of length one. Each of the possibilities are diagonal matrices of the form:
$$\begin{pmatrix}
0 & & & & \\
& \underbrace{\ddots}_{n_1} & & & \\
& & 0 & & \\
& & & i & \\
& & & & \underbrace{\ddots}_{n_2} \\
& & & & & i \\
& & & & & & -i \\
& & & & & & & \underbrace{\ddots}_{n_3} \\
& & & & & & & & -i
\end{pmatrix},$$
with $n_1, n_2, n_3 \geq 1$ $n_1+n_2+n_3=n$.
For $x^2(x-i)$, we find that there must be at least one $0$-block of length $2$, and one $i$-block of length $1$. And the rest can be any arrangements of $0$-blocks up to length $2$ and $i$-blocks up to length $1$. What we get is
$$\begin{pmatrix}
0 & & & & \\
& \underbrace{\ddots}_{n_1} & & & \\
& & 0 & & \\
& & & \begin{pmatrix}
0 & 1 \\
0 & 0
\end{pmatrix} & \\
& & & & \underbrace{\ddots}_{n_2} \\
& & & & & \begin{pmatrix}
0 & 1 \\
0 & 0
\end{pmatrix} \\
& & & & & & i \\
& & & & & & & \underbrace{\ddots}_{n_3} \\
& & & & & & & & i
\end{pmatrix},$$
with $n_1 \geq 0$ and $n_2, n_3 \geq 1$ such that $n_1+2n_2+n_3=n$.
Hope I didn't make any clumsy misreads, and I hope you can see how to replicate the other cases (which are of similar nature to the ones I did).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3485785",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 2
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If $p, q$ and $r$ are distinct roots of $x^3-x^2+x-2=0$, find the value of $p^3+q^3+r^3$. If $p, q$ and $r$ are distinct roots of $x^3-x^2+x-2=0$, find the value of $p^3+q^3+r^3$.
Here's what I have got,
By Vieta's rule
$p+q+r=1\text{. ...........}(1)$
$pq+qr+pr=1\text{. ...........}(2)$
$pqr=2\text{. ...........}(3)$
Squaring $(1)$,
$p^2+q^2+r^2+2(pq+qr+pr)=1\text{. ...........}(4)$
From $(2)$,
$p^2+q^2+r^2=-1\text{. ...........}(5)$
Putting the roots and adding these equations,
$p^3-p^2+p-2=0$
$q^3-q^2+q-2=0$
$r^3-r^2+r-2=0$
We get,
$(p^3+q^3+r^3)-(p^2+q^2+r^2)+(p+q+r)-6=0$
Putting the values,
$(p^3+q^3+r^3)-(-1)+1-6=0$
$(p^3+q^3+r^3)=4$
Am I doing something wrong in my solution?
Because the answer given is -5.
Any help would be appreciated.
| $$p³+q³+r³ = (p+q+r)³-3(p+q)(q+r)(r+p)$$
$$ = (1)³- 3(1-r)(1-p)(1-q)$$
$$ = 1-3(1-1+1-2) =4$$
Here, 1-1+1-2= the sum of coefficients of the given equation
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3486173",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
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