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Help to solve and understand Simultaneous equations. Could anyone help me solve this simultaneous equation:
$$6x + 7y = 12.5\tag1$$
$$7x + 5y = 14\tag2$$
I have watched loads of videos on YouTube, and still don't understand how to do it. I am trying to learn it. And practice with different equations. But don't understand how to do it. So any help will be much appreciated.
|
We have \begin{align}6x + 7y &= 12.5\tag{1}\\
7x + 5y &= 14\tag{2}\end{align}
One way to solve this, would be to rearrange equation $(1)$ for $x$:
\begin{align}6x + 7y &= 12.5\\
6x&=12.5-7y\\
x&=\frac{12.5-7y}{6}\end{align}
We can then put this into equation $(2)$ and solve for $y$:
\begin{align}7x + 5y &= 14\\
7\left(\frac{12.5-7y}{6}\right)+5y&=14\\
\frac{87.5-49y}{6}+5y&=14\\
87.5-49y+30y&=84\\
87.5-19y&=84\\
19y&=3.5\\
y&=\frac{3.5}{19}\\
&=\frac{7}{38}\end{align}
Finally we can put this into equation $(1)$ to find that
\begin{align}6x + 7\left(\frac{7}{38}\right) &= 12.5\\
6x+\frac{49}{38}&=12.5\\
6x&=12.5-\frac{49}{38}\\
6x&=\frac{213}{19}\\
x&=\frac{71}{38}\end{align}
We can then check this using equation $(2)$:
\begin{align}7x + 5y &=7\left(\frac{71}{38}\right)+5\left(\frac{7}{38}\right)\\
&=\frac{497}{38}+\frac{35}{38}\\
&=\frac{532}{38}\\
&=14\end{align}
Therefore we can be confident that the answer is $x=\dfrac{71}{38}$ and $y=\dfrac{7}{38}$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Proof for sum of product of four consecutive integers I had to prove that
$(1)(2)(3)(4)+\cdots(n)(n+1)(n+2)(n+3)=\frac{n(n+1)(n+2)(n+3)(n+4)}{5}$
This is how I attempted to do the problem:
First I expanded the $n^{th}$ term:$n(n+1)(n+2)(n+3)=n^4+6n^3+11n^2+6n$.
So the series $(1)(2)(3)(4)+\cdots+(n)(n+1)(n+2)(n+3)$ will be equal to
$(1^4+2^4+\cdots+n^4)+6(1^3+2^3+\cdots+n^3)+11(1^2+2^2+\cdots+n^2)+6(1+\cdots+n)$
$=\frac{n(n+1)(2n+1)(3n^2-3n-1)}{30}+6*\frac{n^2(n+1)^2}{4}+11*\frac{(n+1)(2n+1)(n)}{6}+6*\frac{n(n+1)}{2}$
I tried to factor out $\frac{n(n+1)}{5}$ and tried to manipulate the expressions in ways which further complicated things. How should I proceed? Can anyone please give me some clues? Any help is appreciated.
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If I were you, I would try induction. First establish that it is true for $n=1$:
$$1*2*3*4=24=\frac{1(2)(3)(4)(5)}{5}$$
Then use induction by first assuming that, for some $k$,
$$1*2*3*4+...+k(k+1)(k+2)(k+3)=\frac{k(k+1)(k+2)(k+3)(k+4)}{5}$$
Then add $(k+1)(k+2)(k+3)(k+4)$ to both sides:
$$1*2*3*4+...+(k+1)(k+2)(k+3)(k+4)=\frac{k(k+1)(k+2)(k+3)(k+4)}{5}+(k+1)(k+2)(k+3)(k+4)$$
If you want to try this on your own, stop reading here and try the induction yourself.
Factor out $(k+1)(k+2)(k+3)$ from the right side:
$$1*2*3*4+...+(k+1)(k+2)(k+3)(k+4)=(k+1)(k+2)(k+3)(\frac{k(k+4)}{5}+(k+4))$$
$$1*2*3*4+...+(k+1)(k+2)(k+3)(k+4)=(k+1)(k+2)(k+3)(\frac{(k+5)(k+4)}{5})$$
$$1*2*3*4+...+(k+1)(k+2)(k+3)(k+4)=\frac{(k+1)(k+2)(k+3)(k+4)(k+5)}{5}$$
And there you have it.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Find short and simple methods to solve $24x^4+1=y^2$
Find all this diophantine equation $$24x^4+1=y^2\tag{1} $$ postive integers solution
it is clear $(x,y)=(1,5)$
I know $y^{2}=Dx^{4}+1$, where $D>0$ and is not a perfect square, has at most two solutions in positive integers (cf. L. J. Mordell, Diophantine equations, p. 270.
Does this equation have another proof such Lucas's assertion, with short and simple methods? Like this paper: Anglin, W. S. "The Square Pyramid Puzzle." Amer. Math. Monthly 97, 120-124, 1990. The square pyramid puzzle
In the paper,Following two question have simple methods to solve it.
There are no positive integers $x$ such $2x^4+1$ is a square.
and
There is exactly one positive integer $x$,namely $1$, such that $8x^4+1$ is a square?
But How can I find simple methods to solve $(1)$?
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$\frac{y-1}{2} \cdot \frac{y+1}{2} = 6x^4$
So $\frac{y+1}{2} = p a^4, \frac{y-1}{2} = q b^4$ where $pq = 6$ and we need to solve the equation $p a^4 - q b^4 = 1
Case 1: $p = 6, q = 1$
This is impossible modulo 3.
Case 2: $p = 2, q = 3$
This is impossible modulo 3.
Case 3: $p = 1, q = 6$
We will show that there is no solution using the method of infinite descent. Take the minimal solution in positive integers $a, b$.
Moving sides and factoring we get $\frac{a - 1}{2} \cdot \frac{a+1}{2} \cdot \frac{a^2 + 1}{2} = 12 (\frac{b}{2})^4$, therefore there exist coprime positive integers $\alpha, \beta, \gamma$ and coprime integers $m,n,k$ such that $\frac{a-1}{2} = \alpha m^4, \frac{a+1}{2} = \beta n^4, \frac{a^2 + 1}{2} = \gamma k^4$ such that $\alpha \beta \gamma = 12$.
Notice that $\gamma | \frac{a^2 + 1}{2}$, which means that neither $2$ nor $3$ can divide $\gamma$, so we must have $\gamma = 1$ and so $\alpha \beta = 12$. Therefore, $\frac{a^2 + 1}{2} - 2 \cdot \frac{a+1}{2} \cdot \frac{a-1}{2} = k^4 - 24(mn)^4 = 1$
Now we can repeat the same argument again: $\frac{k-1}{2} \cdot \frac{k+1}{2} \cdot \frac{k^2 + 1}{2} = 3(mn)^4$, so we get $\frac{k^2 + 1}{2} = u^4$, and $\frac{k - 1}{2}, \frac{k + 1}{2}$ are equal to $3v^4, w^4$ in some order. From this we get $u^4 - 6(vw)^4 = 1$, which is a smaller solution to our original equation, a contradiction.
Case 4: $p = 3, q = 2$
In this case we have to solve the equation $3a^4 - 2b^4 = 1$. Unfortunately, I do not know of a proof of this fact which is as simple as Case 3 and entirely elementary (nor am I sure such a proof exists). However there is this paper by R.T. Bumby, which solves the more general equation $3x^4 - 2y^2 = 1$ (which has besides the trivial solution $(1,1)$ also the more surprising $(3, 11)$) using essentially elementary methods, relying only on unique factorization in $\mathbb{Z}[\sqrt-2]$.
I have only skimmed this paper by Paolo Ribenboim, but it claims to give an algorithm to find all solutions to the equation $x^2 - Dy^4 = 1$ with fixed $D$ apparently with an elementary proof.
This article is a great survey by P.G. Walsh on questions like these about Pell equations: it contains good explanations about many results and probably contains all the references you'll run across studying this topic.
Hope this answer is of help.
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|
Characteristic curves : Why do I get different results? With a given $h(x)$ we want to solve $$xu_y-yu_x=u \\ u(x,0)=h(x)$$
I have solved it using two ways.
$$$$
First way:
For $x\neq 0$ we get $u_y-\frac{y}{x}u_x=\frac{u}{x}$.
We have that $$\frac{du}{ds}=\frac{du}{dx}\cdot \frac{dx}{ds}+\frac{du}{dy}\cdot \frac{dy}{ds}$$
Therefore, we have $\frac{dy}{ds}=1$, $\frac{dx}{ds}=-\frac{y}{x}$ and $\frac{du}{ds}=\frac{u}{x}$.
From $\displaystyle{\frac{dy}{ds}=1}$ we get $\displaystyle{y=s}$.
From $\displaystyle{\frac{dx}{ds}=-\frac{y}{x}}$ with $x(0)=x_0$ we get $\displaystyle{\frac{dx}{dy}=-\frac{y}{x} \Rightarrow x dx=-y dy \Rightarrow x^2=-y^2+x_0^2 \Rightarrow x=\pm \sqrt{-y^2+x_0^2}}$.
From $\displaystyle{\frac{du}{ds}=\frac{u}{x}}$ we get \begin{align*}&\frac{du}{dy}=\frac{u}{\pm \sqrt{-y^2+x_0^2}} \Rightarrow \frac{1}{u}du=\frac{1}{\pm \sqrt{x_0^2-y^2}}dy\Rightarrow \frac{1}{u}du=\frac{1}{\pm |x_0|\sqrt{1-\left (\frac{y}{x_0}\right )^2}}dy \\ & \Rightarrow \int \frac{1}{u}du=\int \frac{1}{\pm |x_0|\sqrt{1-\left (\frac{y}{x_0}\right )^2}}dy\end{align*}
We set $\displaystyle{\frac{y}{x_0}=\cos k \Rightarrow \frac{1}{x_0}dy=-\sin k dk \Rightarrow dy=-x_0\sin k \ dk}$.
Therefore we get:
\begin{align*}\int \frac{1}{\pm |x_0|\sqrt{1-\left (\frac{y}{x_0}\right )^2}}dy&=\int \frac{-x_0\sin k}{\pm |x_0|\sqrt{1-\cos^2k}}dk=\int \frac{-x_0\sin k}{\pm |x_0|\sqrt{\sin^2 k}}dk=\int \frac{-x_0\sin k}{\pm |x_0| |\sin k|}dk \\ & =\mp \int \frac{x_0\sin k}{|x_0 \sin k|}dk=\mp \int \pm 1 dk=k+c\end{align*}
So, we get \begin{align*}&\int \frac{1}{u}du=\int \frac{1}{\pm |x_0|\sqrt{1-\left (\frac{y}{x_0}\right )^2}}dy \Rightarrow \ln |u|=k+c \Rightarrow e^{\ln |u|}=e^{k+c} \Rightarrow |u|=e^ce^k \Rightarrow u=\pm e^ce^k \Rightarrow u=Ce^k \\ & \Rightarrow u=Ce^{\arccos \left (\frac{y}{x_0}\right )}\end{align*}
For $y=0$ we have $\displaystyle{u=h(x_0) \Rightarrow Ce^{\arccos \left (\frac{y}{x_0}\right )}=h(x_0) \Rightarrow C=\frac{2h(x_0)}{\pi}}$.
Therefore, the solution is $$u=\frac{2h(x_0)}{\pi}e^{\arccos \left (\frac{y}{x_0}\right )} \Rightarrow u=\frac{2h(\pm \sqrt{x^2+y^2
})}{\pi}e^{\arccos \left (\frac{y}{\pm \sqrt{x^2+y^2
}}\right )}$$
$$$$
Second way:
From $\displaystyle{xu_y-yu_x=u}$ and $\displaystyle{\frac{du}{ds}=\frac{du}{dx}\cdot \frac{dx}{ds}+\frac{du}{dy}\cdot \frac{dy}{ds}}$ we get : $\displaystyle{\frac{dx}{ds}=-y, \ \frac{dy}{ds}=x, \ \text{ and } \ \frac{du}{ds}=u}$.
From $\displaystyle{\frac{dx}{ds}=-y}$ we get that $\displaystyle{ds=-\frac{1}{y}dx}$ and from $\displaystyle{\frac{dy}{ds}=x}$ we get $\displaystyle{ds=\frac{1}{x}dy}$.
Therefore we get (using that $x(0)=x_0$) $\displaystyle{-\frac{1}{y}dx=\frac{1}{x}dy \Rightarrow -xdx=ydy \Rightarrow -\frac{x^2}{2}=\frac{y^2}{2}-\frac{x_0^2}{2} \Rightarrow x^2=-y^2+x_0^2 \Rightarrow x=\pm \sqrt{x_0^2-y^2}}$.
From $\displaystyle{\frac{du}{ds}=u}$ using that $ds=\frac{1}{x}dy$ we get that \begin{align*}\frac{xdu}{dy}&=u \Rightarrow \frac{1}{u}du=\frac{1}{x}dy \Rightarrow \frac{1}{u}du=\frac{1}{\left (\pm \sqrt{x_0^2-y^2}\right )}dy \Rightarrow \ln |u|=\pm \arctan \left (\frac{y}{x_0^2-y^2}\right )+C \\ & \Rightarrow u=\pm e^{\pm C}e^{\arctan \left (\frac{y}{x_0^2-y^2}\right )}\end{align*}
Using the condition for $y=0$ then $\displaystyle{u=h(x_0)}$ we get that $\displaystyle{h(x_0)=\pm e^{\pm C}}$.
Therefore the solution is $$u=h(x_0)e^{\arctan \left (\frac{y}{x_0^2-y^2}\right )} \Rightarrow u=h(\pm \sqrt{x^2+y^2})e^{\arctan \left (\frac{y}{x^2}\right )}$$
Why do I get different results? Have I done something wrong?
|
They are a few typos in your answer.
And :
Also, the arctan and arccos must be considered as multivalued in the final equations. With convenient choice of signs $\quad \cos^{-1}\left(\frac{y}{\pm\sqrt{x^2+y^2}}\right)\pm\frac{\pi}{2}= \tan^{-1}\left(\frac{y}{x}\right)$
So, the result is the same for both ways.
|
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|
Existence/Limit of Improper Integral I got the improper integral:
$$ \int_0^\infty \frac{x^2}{x^4+1} \, dx $$
On one hand one could of course just evaluate the integral and then apply limits.
However, it's not always practical to find an antiderivative.
I got a hint that one can trace back improper integrals to sequences and thus just calculate the limit of a sequence which is way easier, could someone explain to me how this works or link me to some theorems?
Thanks.
|
$$I=\int_{0}^{1}\frac{x^2}{x^4+1}\,dx + \int_{1}^{+\infty}\frac{x^2}{x^4+1}\,dx = \int_{0}^{1}\left(\frac{x^2}{x^4+1}+\frac{1}{x^4+1}\right)\,dx \tag{1}$$
leads to:
$$ I = \int_{0}^{1}\frac{1+x^2-x^4-x^6}{1-x^8}\,dx = \sum_{k\geq 0}\left(\frac{1}{8k+1}+\frac{1}{8k+3}-\frac{1}{8k+5}-\frac{1}{8k+7}\right)\tag{2}$$
but due to the digamma reflection formula we have:
$$ \sum_{k\geq 0}\left(\frac{1}{8k+1}-\frac{1}{8k+7}\right)=\frac{\pi}{8}\cot\frac{\pi}{8}, \quad \sum_{k\geq 0}\left(\frac{1}{8k+3}-\frac{1}{8k+5}\right)=\frac{\pi}{8}\cot\frac{3\pi}{8}\tag{3}$$
hence it follows that:
$$ I = \frac{\pi}{8}\left[\cot\frac{\pi}{8}+\cot\frac{3\pi}{8}\right]=\color{red}{\frac{\pi}{2\sqrt{2}}}.\tag{4}$$
The same can be achieved by partial fraction decomposition, since $x^4+1=\Phi_8(x)=(x^2-x\sqrt{2}+1)(x^2+x\sqrt{2}+1)$.
Still another way is to apply Glasser's master theorem:
$$ \int_{0}^{+\infty}\frac{dx}{x^2+\frac{1}{x^2}} = \int_{0}^{+\infty}\frac{dx}{\left(x-\frac{1}{x}\right)^2+2}=\int_{0}^{+\infty}\frac{dx}{x^2+2}.\tag{5} $$
|
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|
Square root of Matrix $A=\begin{bmatrix} 1 &2 \\ 3&4 \end{bmatrix}$ Find Square root of Matrix $A=\begin{bmatrix} 1 &2 \\ 3&4 \end{bmatrix}$ without using concept of Eigen Values and Eigen Vectors
I assumed its square root as $$B=\begin{bmatrix} a &b \\ c&d \end{bmatrix}$$
hence
$$B^2=A$$ $\implies$
$$\begin{bmatrix} a^2+bc &b(a+d) \\ c(a+d)&d^2+bc \end{bmatrix}=\begin{bmatrix} 1 &2 \\ 3&4 \end{bmatrix}$$
So
$$a^2+bc=1 \tag{1}$$
$$b(a+d)=2 \tag{2}$$
$$c(a+d)=3 \tag{3}$$
$$d^2+bc=4 \tag{4}$$ From $(2)$ and $(3)$ we get
$$\frac{b}{c}=\frac{2}{3}$$
Let $b=2k $ and $c=3k$ Then
$$a^2=1-6k^2$$ and $$d^2=4-6k^2$$ So
$$B=\begin{bmatrix} \sqrt{1-6k^2} &2k\\ 3k&\sqrt{4-6k^2} \end{bmatrix}$$
So
$$B^2=\begin{bmatrix} 1 &2k \left(\sqrt{1-6k^2}+\sqrt{4-6k^2}\right) \\ 3k \left(\sqrt{1-6k^2}+\sqrt{4-6k^2}\right)&4 \end{bmatrix}=\begin{bmatrix} 1 &2 \\ 3&4 \end{bmatrix}$$
Now we have to solve for $k$ using equation
$$ \left(\sqrt{1-6k^2}+\sqrt{4-6k^2}\right)=\frac{1}{k} \tag{5}$$ Also
$$\left(\sqrt{1-6k^2}+\sqrt{4-6k^2}\right) \times \left(\sqrt{4-6k^2}-\sqrt{1-6k^2}\right)=3$$ So from $(5)$
$$\left(\sqrt{4-6k^2}-\sqrt{1-6k^2}\right)=3k \tag{6}$$
Subtracting $(6)$ from $(5)$ we get
$$2\sqrt{1-6k^2}=\frac{1}{k}-3k$$ Squaring Both sides we get
$$33k^4-10k^2+1=0$$ we get
$$k^2=\frac{5 \pm i \sqrt{8}}{33}$$ From this we get $k$. is there any other simpler way to find ?
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Eliminate $bc$ from the first equation (the uppercase letters are the RHS),
$$a^2+\frac{BC}{(a+d)^2}=A,$$ and from the first and the last, $$d^2-a^2=D-A.$$
Let $p:=(d+a)/2,m:=(d-a)/2$, giving
$$(p-m)^2+\frac{BC}{4p^2}=A,\\mp=4(D-A)$$
and multiplying by $4p^2$,
$$4(p^2-mp)^2+BC=4(p^2-4(D-A))^2+BC=4Ap^2.$$
This biquadratic equation will give the values of $p$, then $m, a$ and $d$, followed by $b$ and $c$.
|
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|
Evaluate the closed form of $\int_{0}^{\pi/2}{\arctan(a\tan^2x)\over b\sin^2x+c\cos^2x}\mathrm dx=f(a,b,c)$ Last question of this form
I am very curious to what is the closed form of:
Assume where $a,b,c > 0$
$$\int_{0}^{\pi/2}{\arctan(a\tan^2x)\over b\sin^2x+c\cos^2x}\mathrm dx=f(a,b,c)\tag1$$
$$b\sin^2x+c\cos^2x$$
$$=b(1-\cos^2 x)+c\cos^2 x$$
$$=b-(b-c)\cos^2 x$$
$$\int_{0}^{\pi/2}{\arctan(a\tan^2x)\over b-(b-c)\cos^2x}\mathrm dx\tag2$$
$u=a\tan^2x\implies du=2a\tan x\sec^2 x dx$
${u-a\over a}=\sec^2 x$
$${\sqrt{a}\over 2}\int_{0}^{\infty}{\arctan(u)\over \sqrt{u}(bu+ac-2ab)}\mathrm du\tag3$$
$u=y^2\implies du=2ydy$
$$\sqrt{a}\int_{0}^{\infty}{\arctan(y^2)\over by^2+ac-2ab}\mathrm dy\tag4$$
$b=P$ and $ac-2ab=Q$
Take the form of: $$\int_{0}^{\infty}{\arctan(y^2)\over Py^2+Q}\mathrm dy\tag5$$
Apply $arctan(y)$ series
$$\sum_{n=0}^{\infty}{(-1)^n\over 2n+1}\int_{0}^{\infty}{y^{4n+2}\over Py^2+Q}\mathrm dy\tag6$$
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We will use Fourier transformation and Parseval's identity.
Say that
\begin{equation*}
\hat{g}(\xi) = \int_{-\infty}^{\infty}e^{-i\xi x}g(x)\, dx .
\end{equation*}
Then the Fourier transform of $g(x) = \arctan\dfrac{2}{x^2}$ will be
\begin{equation*}
\hat{g}(\xi) = 2\pi\dfrac{\sin \xi}{\xi}e^{-|\xi|}\tag{1}
\end{equation*}
(integration by parts followed by residue calculus).
We are now prepared for the given integral. We start with the substituition $y=\tan x$.
\begin{equation*}
f(a,b,c) = \int_{0}^{\infty}\dfrac{\arctan(ay^2)}{by^2+c}\, dy =\left[y =\dfrac{\sqrt{2}}{z\sqrt{a}}\right] = \dfrac{1}{c\sqrt{2a}} \int_{-\infty}^{\infty}\dfrac{1}{z^2+d^2}\arctan\dfrac{2}{z^2}\, dz \tag {2}
\end{equation*}
where $d=\sqrt{\dfrac{2b}{ac}}$. But Parseval's identity combined with (1) give
\begin{gather*}
\int_{-\infty}^{\infty}\dfrac{1}{z^2+d^2}\arctan\dfrac{2}{z^2}\, dz =[\text{ Parseval ]} = \dfrac{1}{2\pi}\int_{-\infty}^{\infty}\dfrac{\pi}{d}e^{-d|\xi|}\, 2\pi\dfrac{\sin \xi}{\xi}e^{-|\xi|}\, d\xi = \\[2ex] \dfrac{2\pi}{d}\dfrac{1}{2\pi}\int_{-\infty}^{\infty}\pi e^{-(1+d)|\xi|}\, \dfrac{\sin \xi}{\xi}\, d\xi = [\text{ Parseval, H = Heaviside ]} = \\[2ex]\dfrac{2\pi}{d}\int_{-\infty}^{\infty}\dfrac{1+d}{x^2+(1+d)^2}\dfrac{1}{2}({\rm H}(\xi+1)-{\rm H}(\xi-1))\, d\xi = \dfrac{\pi}{d}\int_{-1}^{1}\dfrac{1+d}{x^2+(1+d)^2} = \\[2ex]\dfrac{2\pi}{d}\arctan\dfrac{1}{1+d}.
\end{gather*}
Finally this and (2) give
\begin{equation*}
f(a,b,c) = \dfrac{\pi}{\sqrt{bc}}\arctan\left(\dfrac{\sqrt{ac}}{\sqrt{ac}+\sqrt{2b}}\right).
\end{equation*}
|
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|
Circular track problem with LCM I am not able to understand this question based on circular tracks:
$A$ and $B$ start together from the same point on a circular track, and walk in the same direction until they both again arrive together at the starting point. $A$ completes one round in $224$ seconds and $B$ in $364$ seconds. How many times $A$ might have passed $B$?
Firstly I took the LCM of $224$ and $364$ in order to find the meeting point:
$2^5*7*13$.
Then I divided this by $224$ to get the number of rounds taken by $A$ as $13$. Similarly I divided the LCM by $364$ to get the number of rounds taken by $B$ as $8$. So I think $A$ and $B$ must meet $8$ times, but the answer is $5$ i.e. difference between $13$ and $8$. Why? How can they meet just $5$ times? Can anyone explain?
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Lets make a table, of laps completed, and the floor of laps ahead (times when they must have passed):
$\begin{array}{|c|c|c|}
\hline
A & B& \lfloor{\text {laps ahead}}\rfloor \\ \hline
1 & \frac{8}{13}& 0\\ \hline
2 & 1\frac{3}{13}&0\\ \hline
3 & 1\frac{11}{13}&1\\ \hline
4 & 2\frac{6}{13}&1\\ \hline
5 & 3\frac{1}{13}&1\\ \hline
6 & 3\frac{9}{13}&2\\ \hline
7 & 4\frac{4}{13}&2\\ \hline
8 & 4\frac{12}{13}&3\\ \hline
9 & 5\frac{7}{13}&3\\ \hline
10 & 6\frac{2}{13}&3\\ \hline
11 & 6\frac{10}{13}&4\\ \hline
12 & 7\frac{5}{13}&4\\ \hline
13 & 8 & 5\\ \hline
\end{array}$
So, we can show, that the the number of laps ahead, hits 5 full laps when A did 13 laps, and B did 8 laps.
|
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A right triangle has leg $n$, hypotenuse $2n+6$, and perimeter $60$. Find the lengths of all sides. This problem comes from a Grade 11 math textbook in a section on solving quadratic functions by factoring. What follows is my attempt at a solution.
A right triangle has one leg, $n$ and a hypotenuse, $2n+6$. The challenge is to determine the lengths of all the sides, given that the perimeter is $60$.
I worked through this problem a few times with a student. Finally I suggested we step away from it. I decided to post it to this forum for everyone's feedback. As I progressed through the problem, a strange thing happened -- a solution emerged. As is often the case, slowing down and meticulously formulating a question is all that is needed to find a solution.Rather than include my attempt in the question body, I've added it as a solution below.
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Let's correct your solution. Following your notation, the other leg will have length $x$. Since the perimeter is $60$,
\begin{align*}
x + n + 2n + 6 & = 60\\
x + 3n + 6 & = 60\\
x & = 54 - 3n
\end{align*}
By the Pythagorean Theorem,
\begin{align*}
x^2 + n^2 & = (2n + 6)^2\\
x^2 + n^2 & = 4(n + 3)^2\\
x^2 + n^2 & = 4(n^2 + 6n + 9)\\
x^2 + n^2 & = 4n^2 + 24n + 36\\
x^2 & = 3n^2 + 24n + 36\\
x^2 & =3(n^2 + 8n + 12)
\end{align*}
On the other hand, we know that
\begin{align*}
x^2 & = (54 - 3n)^2\\
x^2 & = 9(18 - n)^2\\
x^2 & = 9(324 - 36n + n^2)\\
\end{align*}
Hence,
\begin{align*}
3(n^2 + 8n + 12) & = 9(n^2 - 36n + 324)\\
n^2 + 8n + 12 & = 3(n^2 - 36n + 324)\\
n^2 + 8n + 12 & = 3n^2 - 108n + 972\\
0 & = 2n^2 - 116n + 960\\
0 & = n^2 - 58n + 480\\
0 & = (n - 10)(n - 48)
\end{align*}
giving $n = 10$ and $n = 48$ as possible solutions.
If $n = 10$, then $2n + 6 = 26$, and $x = 54 - 3n = 24$. As you can check, $(10, 24, 26)$ is a Pythagorean triple.
If $n = 48$, then $2n + 6 = 102$, and $x = 54 - 3n = -90$. Since a side length cannot be negative, we reject this solution.
You confused the roles of $x$ and $n$, which led you astray.
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geometry question - Prove that $KB = KD$ Let $\Delta ABC$ have $AB = AC$. On the line perpendicular to $AC$ at $C$ take $D$ so that points $B, D$ is different side $AC$. Let $K$ is the intersection of the straight line through $B$ perpendicular to $AB$ and the line through the center of $M$ of the $CD$ perpendicular to $AD$. Prove that $KB = KD$
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Let $AB=AC=a$, $CD=2b$ and $ML\perp AD$, where $L\in AD$.
Hence, $MD=b$ and since $\Delta MLD\sim \Delta ACD$, we obtain:
$$\frac{LD}{CD}=\frac{MD}{AD}$$ or
$$\frac{LD}{2b}=\frac{b}{\sqrt{a^2+4b^2}},$$ which gives
$$LD=\frac{2b^2}{\sqrt{a^2+4b^2}}.$$
Hence, $$AL=AD-LD=\sqrt{a^2+4b^2}-\frac{2b^2}{\sqrt{a^2+4b^2}}=\frac{a^2+2b^2}{\sqrt{a^2+4b^2}}.$$
Id est, $$KD^2=KL^2+LD^2=AK^2-AL^2+LD^2=BK^2+AB^2-AL^2+LD^2=$$
$$=BK^2+a^2-\left(\frac{a^2+2b^2}{\sqrt{a^2+4b^2}}\right)^2+\left(\frac{2b^2}{\sqrt{a^2+4b^2}}\right)^2=BK^2$$
and we are done!
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Radius of converge of Laurent series for $\frac{1}{\sin z}$ I want to show that the radius of convergence of Laurent series for $\frac{1}{\sin z}$ is $\pi$
I showed that:
\begin{align}
\frac 1 {\sin z} & = \frac 1 {z - \frac{z^3}{3!} + \frac{z^5}{5!} + \cdots } = \frac 1 {z\big(1 - (\frac{z^2}{3!} - \frac{z^4}{5!} + \cdots )\big)} \\[10pt]
& = \frac{1}{z} \bigg( 1 + \bigg(\frac{z^2}{3!} - \frac{z^4}{5!} + \cdots \bigg) + \bigg(\frac{z^2}{3!} - \frac{z^4}{5!} + \cdots \bigg)^2 + \cdots \bigg) \\[10pt]
& = \frac{1}{z} +\frac{z}{3!} + \bigg( \frac{1}{3!}- \frac{1}{5!}\bigg)z^3 + \cdots
\end{align}
This equality holds if and only if:
$$\bigg|\frac{z^2}{3!} - \frac{z^4}{5!} + \cdots \bigg| < 1
\iff \bigg|\frac{\sin z}{z} - 1\bigg| < 1$$
How can I show that the last inequality implies $|z| < \pi$ ?
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The radius of convergence of the Laurent series of $1/\sin z$ about $z=0$ is $\pi$. The reason is that its poles are at $n\pi$, $n\in\Bbb Z$
(because that's where the sine has its zeroes). The nearest poles to zero
are at $\pm \pi$, as distance $\pi$ from the origin.
|
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Solve $\sin(5\theta)=1$, $0<\theta<2\pi$. Show that the roots of $16x^4+16x^3-4x^2-4x+1=0$ are $x=\sin{\frac{(4r+1)\pi}{10}}$, $r=0,2,3,4$. This is a very interesting problem that I came across. I know it's got something to do with trigonometry identities, polynomials and complex numbers, but other than that, I'm not too sure how to approach this. Any guidance hints or help would be greatly appreciated. Thanks in advance. Here's the problem:
Solve the equation $\sin(5\theta)=1$ for $0<\theta<2\pi$ and hence show that the roots of the equation $16x^4+16x^3-4x^2-4x+1=0$ are $x=\sin{\frac{(4r+1)\pi}{10}}$, where for $r=0,2,3,4$. Determine the exact value of $\sin\frac\pi{10}\sin\frac{3\pi}{10}$.
It might be useful to find $\sin(5\theta)$ in terms of $sin^n \theta$ and so on.
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Use$$\sin5\theta=\sin{x}\cos4\theta+\cos{x}\sin4\theta=\sin{\theta}(2\cos^22\theta-1)+4\cos^2\theta\cos2x\sin{\theta}=$$
$$=\sin{\theta}(2\cos^22\theta-1+2(1+\cos2\theta)\cos2\theta))=\sin{\theta}(4\cos^22\theta+2\cos2\theta-1)$$
and $16x^4+16x^3-4x^2-4x+1=(4x^2+2x-1)^2$
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How to find the sum of this series: $\sum_{n=1}^{\infty}(-1)^{n+1} \frac{n^2}{n^3+1}$? $$\frac{1^2}{1^3+1}-\frac{2^2}{2^3+1}+\frac{3^2}{3^3+1}-\frac{4^2}{4^3+1}+\cdots$$
in terms of summation i can write it as
$$S_{n}=\sum_{n=1}^{\infty}(-1)^{n+1} \frac{n^2}{n^3+1}$$
How to continue from this point?
used partial fraction:
$$\frac{n^2}{n^3+1}=\frac{1}{3}\cdot\frac{2n-1}{n^2-n+1}+\frac{1}{3}\cdot \frac{1}{n+1}$$
I'm stuck with the first term in the partial fraction , the second term simply yields $1-\log(2)$
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A route. One may recall the standard series representation of the digamma function,
$$
\psi(z+1)+\gamma=\sum_{n=1}^\infty\left(\frac1n-\frac1{n+z}\right),\quad \text{Re}\: z>-1,
$$
giving
$$
2\sum _{n=1}^{\infty } \frac{(-1)^n}{n+z}=\psi\left(\frac{z}{2}+\frac{1}{2}\right)-\psi\left(\frac{z}{2}+1\right),\quad \text{Re}\: z>-1,
$$ then by writting
$$
(-1)^{n+1}\frac{n^2}{n^3+1}=a\cdot\frac{(-1)^{n+1}}{n+1}+b\cdot\frac{(-1)^{n+1}}{n+e^{2i\pi/3}}+\bar{b}\cdot\frac{(-1)^{n+1}}{n+e^{-2i\pi/3}}
$$ one may conclude with special values of the digamma function.
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Base $5$ representation is the last $k$ digits of its base $2$ representation for some $k$
Compute the largest integer between $1$ and $1000$ whose base $5$ representation consists of the last $k$ digits of its base $2$ representation, for some $k$.
I thought about taking cases based on the number of leading $1$s such a number $n$ has in base $5$. Since $1000 = 13000_5$, there are many possibilities to check using this approach. For example, if the number of leading $1$s is $1$, then \begin{align*}n &= 10_5, 100_5,101_5,1000_5,1010_5,1011_5,1001_5,10000_5,10001_5,10010_5,10011_5,10100_5,10101_5,\\&=10110_5,10111_5.\end{align*}
The number of cases here is $2^0+2^1+2^2+2^3 = 15$. Then we would also have to check if each of these satisfies the requirements.
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If a number $n$ has $k$ digits (quits?) in base $5$, then let
$$n = \sum_{i=0}^{k-1} a_i 5^i = a_{k-1} 5^{k-1} + \cdots + a_15 + a_0,$$ where $a_{k-1} > 0$. Since the same digits $a_i$ are used as bits, $a_i \in \{0, 1\}$.
Consider the last $k$ bits in base $2$:
$$\begin{align*}
n &\equiv \sum_{i=0}^{k-1} a_i 2^i \pmod{2^k}\\
\sum_{i=0}^{k-1} a_i 5^i &\equiv \sum_{i=0}^{k-1} a_i 2^i\\
\sum_{i=0}^{k-1} a_i (5^i-2^i)&\equiv 0\\
\end{align*}$$
For $k=5$, which is the largest possible $k$,
$$\begin{align*}
609a_4 + 117a_3 + 21a_2 + 3 a_1 &\equiv 0 \pmod {32}\\
a_4 + 21a_3 + 21 a_2 + 3a_1 &\equiv 0
\end{align*}$$
Which does not look possible given that $a_4$ must be $1$.
For $k = 4$,
$$\begin{align*}
117a_3 + 21a_2 + 3 a_1 &\equiv 0 \pmod {16}\\
5a_3 + 5 a_2 + 3a_1 &\equiv 0
\end{align*}$$
Which does not look possible given that $a_3$ must be $1$.
For $k = 3$,
$$\begin{align*}
21a_2 + 3 a_1 &\equiv 0 \pmod {8}\\
5 a_2 + 3a_1 &\equiv 0
\end{align*}$$
So $a_2 = a_1 = a_0 = 1$ is a solution (and gives a larger $n$ than setting $a_0 = 0$). The result is
$$n = 5^2+5^1+1 = 111_5 = 31 = 11111_2$$
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|
Evaluating $\int_0^1 \sqrt{1 + x ^4 } \, d x $ $$
\int_{0}^{1}\sqrt{\,1 + x^{4}\,}\,\,\mathrm{d}x
$$
I used substitution of tanx=z but it was not fruitful. Then i used $ (x-1/x)= z$ and
$(x)^2-1/(x)^2=z $ but no helpful expression was derived.
I also used property $\int_0^a f(a-x)=\int_0^a f(x) $
Please help me out
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Consider the ${}_{2}F_{1}$ hypergeometric integral form given by
$${}_{2}F_{1}(a, b; c; x) = \frac{\Gamma(c)}{\Gamma(b) \, \Gamma(c-b)} \, \int_{0}^{1} t^{b-1} \, (1-t)^{c-b-1} \, (1-x \, t)^{-a} \, dt$$
leads to, with $a=-1/2$, $b=1/4$, $c=5/4$, $x=-1$,
$${}_{2}F_{1}\left(-\frac{1}{2}, \frac{1}{4}; \frac{5}{4}; -1\right) = \frac{1}{4} \, \int_{0}^{1} t^{-3/4} \, \sqrt{1+ t} \, dt.$$
Now let $t = x^{4}$ to obtain
$$\int_{0}^{1} \sqrt{1 + x^4} \, dx = {}_{2}F_{1}\left(-\frac{1}{2}, \frac{1}{4}; \frac{5}{4}; -1\right) = 1.08943...$$
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|
Prove that the product is never a perfect square
Prove that for nonnegative integers $x_1,\ldots,x_{2011}$ and $y_1,\ldots,y_{2011}$ the product $$(2x_1^2+3y_1^2)(2x_2^2+3y_2^2) \cdots (2x_{2011}^2+3y_{2011}^2)$$ is never a positive perfect square.
I thought about generalizing this question to any odd subscript $n$ instead of $2011$. Thus, $$(2x_1^2+3y_1^2)(2x_2^2+3y_2^2) \cdots (2x_n^2+3y_n^2)$$ is never a perfect square. For $n = 1$ we have $2x^2+3y^2 = z^2$ and I want to show the only solution is $x = y = z = 0$. If $x$ is even, then $y$ must be even by taking modulo $4$. If $x$ is odd, then $y$ must be odd. I didn't see how to continue from here.
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Modulo $3$, the equation for $n=1$ reduces to $2x^2=z^2 \pmod3$.
If $x$ were not divisible by $3$ then we would find $2=u^2 \pmod3$ for some $u$. But $2$ is not a quadratic residue $\pmod 3$ so this is impossible.
But if $x$ were divisible by $3$ this would make $z$ divisible by $3$, and thus $z^2$ divisible by $3^2 = 9$. This would in turn force $y$ to be divisible by $3$ otherwise $x^2+3y^2$ would be divisible by $3$ but not $9$.
Dividing such a solution by $9$ would give another solution $(x/3,y/3,z/3)$ with $z$ divisible by a smaller power of $3$, and eventually we reach the same case as above (infinite descent), or conclude that $x,y,z$ are infinitely divisible by $3$ and are thus equal to $0$.
So the only solution to $2x^2 + 3y^2 = z^2$ over the integers is $(x, y, z) = (0, 0, 0)$.
For the case of arbitrary odd $n$, we proceed by induction on $n$.
Now suppose we had a solution to the equation $(2x_1^2 + 3y_1^2)...(2x_{n}^2 + 3y_{n}^2)(3k + 1) = z^2$ for $n$ odd. Note that we have loosened the equation by a factor of $3k + 1$ for any $k \in \mathbb{Z}$. Also note that the base case $n = 1$ for this equation is proven exactly the same way as above because the factor $3k + 1$ does not affect the residues modulo $3$. So proceed with assuming we proved that there are no nonzero solutions for $n-2$.
As long as we have some $i$ such that both $x_i$ and $y_i$ are divisible by $3$, replace $(x_i, y_i, z)$ with $(x_i/3, y_i/3, z/3)$ to obtain another valid equation (dividing by $9$ on both sides). If the process never terminates this means $z = 0$ and thus one of the $(x_i, y_i)$ must be $(0, 0)$ as well.
Otherwise we have reached a point where every factor is divisible at most by $3$ but not by $9$ (this would require both $x_i$ and $y_i$ divisible by $3$).
We consider two cases for the number $N$ of $i$ for which $2x_i^2 + 3y_i^2$ is divisible by $3$.
1. Case $N = 0$
Suppose we have a solution to $(2x_1^2 + 3y_1^2)...(2x_{n}^2 + 3y_{n}^2)(3k + 1) = z^2$ with none of the terms divisible by $3$. Then the equation reduces to $-1 = (-1)^n = 2^n = (zx_1^{-1}...x_n^{-1})^2 \pmod 3$ which is impossible because $-1$ is not a residue $\pmod 3$.
2. Case $N \geq 1$
We must then have $x_i$ divisible by $3$ for some $i$. Because of the process above, this means $y_i$ is not divisible by $3$. Since $z^2$ is divisible by an even power of $3$, there must be another factor $2x_j^2 + 3y_j^2$ divisible by $3$ (exactly once). Now divide the equation by $9$ on both sides to replace $z$ by $z/3$ and $(2x_i^2 + 3y_i^2)(2x_j^2 + 3y_j^2)$ by $(6(x_i/3)^2 + y_i^2)(6(x_j/3)^2 + y_j^2)$. Both of these factors are of the form $3k + 1$ because $y_i^2, y_j^2 = 1 \pmod 3$.
We have therefore obtained an equation of the form $(2x_1^2 + 3y_1^2)...(2x_{n-2}^2 + 3y_{n-2}^2)(3k + 1) = z^2$. Such an equation has been assumed to have no nonzero solutions, and so we are done by induction.
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simplify :$\frac{1}{f_5}+\frac{1}{f_6}+\frac{1}{f_{12}}+\frac{1}{f_{20}}$
let :
$$f_n=\sqrt[4]{2}+\sqrt[n]{4} \ \ \ n\geq2 \in \mathbb{N}$$
then simplify :
$$\dfrac{1}{f_5}+\dfrac{1}{f_6}+\dfrac{1}{f_{12}}+\dfrac{1}{f_{20}}=?$$
MyTry :
$$f_5=\sqrt[4]{2}+\sqrt[5]{4} \ \ \ , \ \ f_6=\sqrt[4]{2}+\sqrt[6]{4} \ \ , \ \ f_{20}=\sqrt[4]{2}+\sqrt[20]{4}$$
$$\dfrac{1}{\sqrt[4]{2}+\sqrt[5]{4}}+\dfrac{1}{\sqrt[4]{2}+\sqrt[6]{4}}+\dfrac{1}{\sqrt[4]{2}+\sqrt[12]{4}}+\dfrac{1}{{\sqrt[4]{2}+\sqrt[20]{4}}}=?$$
Now :
$$\dfrac{1}{\sqrt[4]{2}+\sqrt[5]{4}}+\dfrac{1}{\sqrt[4]{2}+\sqrt[6]{4}}=\dfrac{\sqrt[4]{2}+\sqrt[6]{4}+\sqrt[4]{2}+\sqrt[5]{4}}{(\sqrt[4]{2}+\sqrt[5]{4})(\sqrt[4]{2}+\sqrt[6]{4})}$$
And :
$$\dfrac{1}{\sqrt[4]{2}+\sqrt[12]{4}}+\dfrac{1}{\sqrt[4]{2}+\sqrt[20]{4}}=\dfrac{\sqrt[4]{2}+\sqrt[20]{4}+\sqrt[4]{2}+\sqrt[12]{4}}{(\sqrt[4]{2}+\sqrt[12]{4})(\sqrt[4]{2}+\sqrt[20]{4})}$$
now what ?
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You get that $$\frac1{\sqrt[20]{32}+\sqrt[20]{256}}+\frac1{\sqrt[12]{8}+\sqrt[12]{16}}+\frac{1}{\sqrt[12]{8}+\sqrt[12]{4}}+\frac{1}{\sqrt[20]{32}+\sqrt[20]{4}}$$
If we group first and fourth and second and third term we get
$$\frac1{\sqrt[20]{32}(1+\sqrt[20]{8})}+\frac{1}{\sqrt[20]{4}(1+\sqrt[20]{8})}+\frac1{\sqrt[12]{8}(1+\sqrt[12]{2})}+\frac1{\sqrt[12]{4}(1+\sqrt[12]{2})}=\\=\frac{\sqrt[20]{8}+1}{\sqrt[20]{32}(\sqrt[20]{8}+1)}+\frac{\sqrt[12]{2}+1}{\sqrt[12]{8}(\sqrt[12]{2}+1)}\\=\frac{1}{\sqrt[20]{32}}+\frac{1}{\sqrt[12]{8}}=\frac{2}{\sqrt[4]{2}}=2^{\frac34}$$
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For $a,b,c>0$ . Prove $ (a+b+c)(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})\geq 9+8\frac{(a-b)^{2}+(b-c)^{2}+(c-a)^{2}}{(a+b+c)^{2}}$ For $a,b,c>0$ . Prove
$$(a+b+c)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\geq 9+8\frac{(a-b)^{2}+(b-c)^{2}+(c-a)^{2}}{(a+b+c)^{2}}$$
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Let $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.
Hence, we need to prove that
$$\frac{(a+b+c)(ab+ac+bc)}{abc}\geq9+\frac{16(a^2+b^2+c^2-ab-ac-bc)}{(a+b+c)^2}$$ or
$$\frac{9uv^2}{w^3}\geq9+\frac{16(u^2-v^2)}{u^2},$$
which is $f(w^3)\geq0$, where $f$ is an decreasing function.
Thus, it's enough to prove our inequality for a maximal value of $w^3$, which happens for equality case of two variables.
Let $b=c=1$.
Hence, we need to prove that $(a-1)^2(a-2)^2\geq0.$
Done!
Another way.
$$(a-b)^2(a-c)^2(b-c)^2=27(3u^2v^4-4v^6-4u^3w^3+6uv^2w^3-w^6)\geq0,$$
which gives
$$3uv^2-2u^3-2\sqrt{(u^2-v^2)^3}\leq w^3\leq3uv^2-2u^3+2\sqrt{(u^2-v^2)^3}.$$
we need to prove that
$$\frac{9uv^2}{w^3}\geq9+\frac{16(u^2-v^2)}{u^2}$$ or
$$\frac{9uv^2}{w^3}\geq\frac{25u^2-16v^2}{u^2}$$ or
$$w^3\leq\frac{9u^3v^2}{25u^2-16v^2},$$ for which it's enough to prove that
$$3uv^2-2u^3+2\sqrt{(u^2-v^2)^3}\leq\frac{9u^3v^2}{25u^2-16v^2}$$ or
$$25u^3-49u^3v^2+24uv^4\geq(25u^2-16v^2)\sqrt{(u^2-v^2)^3}$$ or
$$u(25u^2-24v^2)\geq(25u^2-16v^2)\sqrt{(u^2-v^2)}$$ or
$$u^2(25u^2-24v^2)^2\geq(25u^2-16v^2)^2(u^2-v^2)$$ or
$$(15u^2-16v^2)^2\geq0.$$
Done again!
Also we have the following solution.
$$(a+b+c)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)-9-8\cdot\frac{(a-b)^{2}+(b-c)^{2}+(c-a)^{2}}{(a+b+c)^{2}}=$$
$$=\frac{\sum\limits_{cyc}c(a-b)^2(a+b-3c)^2}{abc(a+b+c)^2}\geq0.$$
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Find complex number Z in $\lvert Z\rvert= Z+3-2i$ $$\lvert Z\rvert = Z+ 3-2i$$
what I did so far is
let $Z = a +bi$
so $$\sqrt{a^2 + b^2} = a+bi+3-2i$$
$$\sqrt{a^2 + b^2} = a+3 + i (b-2)$$
now what I'm thinking is squaring both sides but that doesn't work, any tips?
|
Solve
$$
\lvert z \rvert = z + 3 - 2i
\tag{1}
$$
1
Observation: $\lvert z \rvert$ is real, while the right hand side has the imaginary term $-2i$.
For the equality to hold, the imaginary part of $z$ must be $2i$ to cancel the imaginary component, that is
$$z=x+2i\tag{2}$$
2
Use $(2)$ in $(1)$ to obtain
$$
\lvert z \rvert = \lvert x + 2i \rvert = \sqrt{x^{2}+4} = x + 3
\tag{3}
$$
3
Solve $(3)$ to recover
$$
x = -\frac{5}{6}
$$
4
The value of $z$ which solves $(1)$ is
$$
z = -\frac{5}{6} + 2 i
$$
Confirmation
$$
\begin{align}
\require{cancel}
\lvert z \rvert &= z + 3 - 2i \\[10pt]
\Big\lvert -\frac{5}{6} + 2 i \Big\rvert &= \left(-\frac{5}{6} \cancel{+ 2 i} \right) + 3 \cancel{- 2i} \\[10pt]
\sqrt{\frac{25}{36} + 4} &= -\frac{5}{6} + 3\\[10pt]
\sqrt{\frac{169}{36}} &= \frac{13}{6}\\[10pt]
\frac{13}{6} &= \frac{13}{6}
\end{align}
$$
|
{
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Find Angle Between Two Curves at Point of Intersection Find angle between these two curves at point of intersection :
$$K_1: x^2y^2 + y^4 = 1$$ and
$$K_2 : x^2 + y^2 = 4 $$
Thanks!
|
Problem
Given the equations
$$
\color{blue}{x^{2} + y^{2} = 4}
\tag{1}
$$
and
$$
x^{2}y^{2} + y^{4} = 1
\tag{2}
$$
find the angles between the curves at the points of intersection.
Symmetry analysis
Each term in both equations has a definite parity; this invites investigation.
The $x^{2}$ terms are invariant under the transformation $x\to-x$ because
$$
(-x)^{2} = x^{2}
$$
A similar argument holds for both $y^{2}$ and $y^{4}$.
From this we see invariance in form when both figures are reflected through each of the $x-$axis and the $y-$axis. Therefore, both figures are invariant under reflection through the origin. These ideas are demonstrated in the sequence of images below.
Conclusion: we only need to solve the problem at one intersection point.
Find intersections
Rewrite $(2)$:
$$
y^{2} \left( x^{2} + y^{2} \right) = 1
\tag{2a}
$$
and inserting $\color{blue}{(1)}$ we have
$$
y^{2}\left( 4 \right) = 1
$$
or
$$
y = \frac{1}{2}
\tag{3}
$$
Insert $(3)$ into $(1)$ to find
$$
x = \frac{\sqrt{15}}{2}.
$$
by the symmetry analysis, the locus of intersecting points is
$$
\left\{
\frac{1}{2} \left( \sqrt{15}, 1 \right ),
\frac{1}{2} \left( \sqrt{15}, -1 \right ),
\frac{1}{2} \left( -\sqrt{15}, 1 \right ),
\frac{1}{2} \left( -\sqrt{15}, -1 \right )
\right\}
$$
We will focus on the point $p = \frac{1}{2} \left( \sqrt{15}, 1 \right )$.
Compute angles
Here are the slopes at the intersection point:
The slope calculation is detailed in the answer by @Manuel Guillen. Now to relate the slopes to angles.
Given the line
$$
y = mx + b
$$
the slope is related to the angle by
$$
m = \tan \theta
$$
The figure below shows the intersection angle as an arc between the tangent lines. The arc goes from
$$
-\tan ^{-1}\left(\frac{\sqrt{15}}{17}\right) \le \theta \le
\pi -\tan ^{-1}\left(\sqrt{15}\right)
$$
The difference $\Delta$, is the angle of intersection:
$$
\Delta = \pi +\tan ^{-1}\left(\frac{\sqrt{15}}{17}\right)-\tan ^{-1}\left(\sqrt{15}\right) =
\pi -\cot ^{-1}\left(\frac{2}{\sqrt{15}}\right)
\approx 117^{\circ}
$$
|
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|
Finding the number of ways of choosing 4 distinct integers from first n natural numbers so that no two are consecutive. Question:
Show that the number of ways of choosing 4 distinct integers from the first $n$ numbers so that no two are consecutive is $\binom{n-3}{4}$.
I tried to find it by the principle of inclusion and exclusion.
The total number of ways of choosing 4 integers here is $\binom{n}{4}$, number of ways to choose them such that at least 2 are consecutive is $\binom{n-2}{2}×(n-1)$ and at least three are consecutive is $\binom{n-3}{1}×(n-2)$ and all 4 can be consecutive in $n-3$ ways.
Thus total number of ways such that no two are consecutive by principle of inclusion and exclusion should be $$\binom{n}{4} - \binom{n-2}{2}\times(n-1) + \binom{n-3}{1}\times(n-2) - (n-3) $$
How can I show this is $\binom{n-3}{4}$? Or did I do anything wrong? Please help. Thanks in advance.
|
You must have done something wrong, since for $n=7$ your expression is $35-60+20-4=-9.$
To use inclusion-exclusion, your events should be something like "includes $i$ and $i+1$". Now the number of choices which include $i$ and $i+1$ is $\binom{n-2}{2}$, and there are $n-1$ possible $i$, so you should start $$\binom{n}{4}-(n-1)\binom{n-2}{2}...$$
Then how many ways are there to have $i$ and $i+1$ and $j$ and $j+1$, for $i<j$? If $j=i+1$ there are $n-3$ ways (and there are $n-2$ choices for $i$). If not there is $1$ way, and $\binom{n-1}{2}-(n-2)=\binom{n-1}2$ choices for $i,j$ (the number of pairs minus the number where $i,j$ are consecutive). So it should continue
$$\binom{n}{4}-(n-1)\binom{n-2}{2}+(n-3)(n-2)+\binom{n-2}2...$$
For the final term, the only way three events can happen is if $i+1=j,j+1=k$, for which there are $n-3$ options. So overall you get
$$\binom{n}{4}-(n-1)\binom{n-2}{2}+(n-3)(n-2)+\binom{n-2}2-(n-3).$$
[This is very close to what you had, the only difference is the extra term for having two separate consecutive pairs of $+\binom{n-2}2$.]
To see that the above is the same as $\binom{n-3}{4}$, note that $(n-1)\binom{n-2}2=3\binom{n-1}3$. Also $(n-3)(n-2)=2\binom{n-2}{2}$, so this can be written as
$$\binom{n}4-\binom31\binom{n-1}3+\binom32\binom{n-2}2-\binom33\binom{n-3}1.$$
This is precisely the inclusion-exclusion formula for the number of choices of $4$ numbers from the first $n$ which don't include any of the last $3$, so is $\binom{n-3}4$.
|
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|
A polynomial is divisible with another one For what $m$ and $n$ is the polynomial $2X^{19}+X^{13}+mX^{11}+X^8+2X^6+nX^2+2$ divisible by $X^4+X^3+X^2+X+1$.
I tried to find the real solutions for g but couldn't
|
Note that $x^5-1=(x-1)(x^4+x^3+x^2+x+1)$
\begin{align}
&\;2x^{19}+x^{13}+mx^{11}+x^8+2x^6+nx^2+2\\
=&\;2x^4(x^5-1+1)^3+(x^3+mx)(x^5-1+1)^2+(x^3+2x)(x^5-1+1)+nx^2+2\\
=&\;2x^4[(x^5-1)^3+3(x^5-1)^2+3(x^5-1)+1]\\
&\quad+(x^3+mx)[(x^5-1)^2+(x^5-1)+1]+(x^3+2x)[(x^5-1)+1]+nx^2+2\\
=&\;P(x)(x^5-1)+2x^4+x^3+mx+x^3+2x+nx^2+2
\end{align}
for some polynomial $P(x)$.
$2x^{19}+x^{13}+mx^{11}+x^8+2x^6+nx^2+2$ is divisible by $x^4+x^3+x^2+x+1$ if and only if $2x^4+x^3+mx+x^3+2x+nx^2+2$ is divisible by $x^4+x^3+x^2+x+1$.
\begin{align}
&\;2x^4+x^3+mx+x^3+2x+nx^2+2\\
=&\;2x^4+2x^3+nx^2+(m+2)x+2
\end{align}
It is obvious that $m=0$ and $n=2$.
|
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|
If $f (x) +f'(x) = x^3+5x^2+x+2$ then find $f (x)$
If $f (x) +f'(x) = x^3+5x^2+x+2$ then find $f(x)$.
$f'(x)$ is the first derivative of $f (x)$.
I have no idea about this question, please help me.
|
If $f(x)$ has degree $n $ then $ f'(x) $ has $n-1$ . Here highest degree is $3$ thus $ f (x) $ is a polynomial of degree $3$. Thus let $f (x)=ax^3+bx^2+cx+d $ thus we have $f (x)+f'(x)=ax^3+bx^2+cx+d+3ax^2+2bx+c $ Now comparing we have $ a=1,3a+b=5,c+2b=1,d+c=2$ thus $ f(x)=x^3+2x^2-3x+5.$
|
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|
Find $P$ such that $P^TAP = D$ where $D$ is a diagonal matrix.
Let $$A = \left(\begin{array}{cc} 2&3 \\ 3&4 \end{array}\right) \in
M_n(\mathbb{C})$$
Find $P$ such that $P^TAP = D$ where $D$ is a diagonal matrix.
How can I find $P$? I am doing Gauss but it does not work?$$A = \left(\begin{array}{cc|cc} 2&3&1&0\\ 3&4&0&1 \end{array}\right) \sim \left(\begin{array}{cc|cc} 2&0&-8&6\\ 0&-1/2&-3/2&1 \end{array}\right)$$
What am I doing wrong? Steps would be much appreciated.
|
The characteristic polynomial of $A$ is $x^2-6x-1$, whose roots are $3\pm\sqrt{10}$. An example of an eigenvector with norm $1$ whose eigenvalue is $3+\sqrt{10}$ is $\frac1{\sqrt{20-2\sqrt{10}}}\bigl(\sqrt{10}-1,3\bigr)$ and an example of an eigenvector with norm $1$ whose eigenvalue is $3-\sqrt{10}$ is $\frac1{\sqrt{20-2\sqrt{10}}}\bigl(3,1-\sqrt{10}\bigr)$. So, take$$P=\frac1{\sqrt{20-2\sqrt{10}}}\begin{pmatrix}\sqrt{10}-1&3\\3&1-\sqrt{10}\end{pmatrix}$$and then $P^TAD=\left(\begin{smallmatrix}3+\sqrt{10}&0\\0&3-\sqrt{10}\end{smallmatrix}\right)$.
|
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|
Generalised Stirling number that each partition has more than one component We know that the Stirling number of the second kind is the number of ways to partition a set of $n$ objects into $k$ non-empty subsets and is denoted by $s(n,k)=\frac{1}{k!}\sum_{j=0}^{k}(-1)^{k-j}\begin{pmatrix}
k\\
j
\end{pmatrix}j^n$
This formula is for the case that the partitions have at least one component. I am trying to find the similar formula for the case that each partition has at least $m$ component. Does anyone have any ideas?
|
The first question you should be asking is where does the formula
$$ s(n,k) = \frac1{k!} \sum_{j = 0}^k \binom{k}{j} j^n (-1)^{k - j} $$
come from? Then once you understand that, perhaps you will be able to generalize.
Note that the generating function for nonempty sets is $e^x - 1 = x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots$. The generating function for a set of size $k$ is $x^k/k!$. Thus the generating function for a set of size $k$ whose elements are non-empty sets is
$$ \sum_{n \ge 0} s(n,k)\frac{x^n}{n!} = \frac{(e^x - 1)^k}{k!} $$
and therefore
$$ \sum_{k \ge 0} \sum_{n \ge 0} s(n,k)\frac{x^n}{n!}y^k = \exp(y(e^x - 1)). $$
Therefore
\begin{align}
s(n,k) &= \left[ \frac{x^n}{n!}y^k \right] \exp(y(e^x - 1)) \\
&= \frac{1}{k!} \left[ \frac{x^n}{n!} \right] (e^x - 1)^k \\
&= \frac{1}{k!} \left[ \frac{x^n}{n!} \right] \sum_{j = 0}^k \binom{k}{j} e^{jx}(-1)^{k - j} \\
&= \frac{1}{k!} \sum_{j = 0}^k \binom{k}{j} j^n(-1)^{k - j}.
\end{align}
Let us agree to write $s_m(n,k)$ for the number of ways to partition a set of size $n$ into $k$ subsets of size at least $m$.
Then the same thing happens as before except instead of using $e^x - 1$ for non-empty sets, we use $$ e^x - 1 - x - \frac{x^2}{2!} - \dots - \frac{x^{m - 1}}{(m - 1)!} = \frac{x^m}{m!} + \frac{x^{m + 1}}{(m + 1)!} + \frac{x^{m + 2}}{(m + 2)!} + \cdots $$
for sets with at least $m$ elements. So we have
$$ \sum_{n,k} s_m(n,k)\frac{x^n}{n!}y^k = \exp \left[ y \left( e^x - 1 - x - \frac{x^2}{2!} - \dots - \frac{x^{m - 1}}{(m - 1)!} \right) \right]. $$
And like before,
\begin{align}
s_m(n, k) &= \frac{1}{k!} \left[ \frac{x^n}{n!} \right] \left( e^x - 1 - x - \frac{x^2}{2!} - \dots - \frac{x^{m - 1}}{(m - 1)!} \right)^k \\
&= \frac{1}{k!} \left[ \frac{x^n}{n!} \right] \sum_{i + j_0 + \dots + j_{m - 1} = k} \binom{k}{i,j_0,\dots,j_{m-1}} e^{ix}(-1)^{j_0}\left( -x \right)^{j_1} \cdots \left( -\frac{x^{m - 1}}{(m - 1)!} \right)^{j_{m - 1}}
\end{align}
Now we group together the minus signs: $(-1)^{j_0 + \dots + j_{m - 1}} = (-1)^{k - i}$ giving
\begin{align}
&\frac{1}{k!}\left[ \frac{x^n}{n!} \right] \sum_{i + j_0 + \dots + j_{m - 1} = k} \binom{k}{i,j_0,\dots,j_{m-1}} e^{ix}(-1)^{k - i}\left( x \right)^{j_1} \cdots \left( \frac{x^{m - 1}}{(m - 1)!} \right)^{j_{m - 1}} \\
&= \frac{n!}{k!} \sum_{i + j_0 + \dots + j_{m - 1} = k} \binom{k}{i,j_0,\dots,j_{m-1}} \frac{1}{0!^{j_0}\cdots (m-1)!^{j_{m - 1}}} [x^n] e^{ix}(-1)^{k - i} x^{0j_0 + \dots + (m - 1)j_m} \\
&= \frac{n!}{k!} \sum_{i + j_0 + \dots + j_{m - 1} = k} \binom{k}{i,j_0,\dots,j_{m-1}} \frac{1}{1!^{j_1}\cdots (m-1)!^{j_{m - 1}}} [x^{n - 1j_1 - \dots - (m - 1)j_m}] e^{ix}(-1)^{k - i} \\
&= \frac{n!}{k!} \sum_{i + j_0 + \dots + j_{m - 1} = k} \binom{k}{i,j_0,\dots,j_{m-1}} \frac{1}{1!^{j_1}\cdots (m-1)!^{j_{m - 1}}} i^{n - j_1 - 2j_2 - \dots - (m - 1)j_{m - 1}} (-1)^{k - i}.
\end{align}
|
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|
Proof that $x^3+y^4=z^{31}$ has infinitely many solutions This is a question from RMO 2015.
Show that there are infinitely many triples (x,y,z) of integers such that $x^3+y^4=z^{31}.$
This is how I did my proof:
Suppose $z=0,$ which is possible because $0$ is an integer. Then $x^3+y^4=0 \Rightarrow y^4=-x^3.$ Now, suppose $y$ is a perfect cube such that it is of the form $a^3$ where $a$ is any integer. Then $(a^3)^4=-x^3 \Rightarrow a^{12}=-x^3,$ whereby $x=-(-a)^4.$ Hence there exists infinitely many triples {x,y,z}={$-(-a)^4, a^3, 0$}, which satisfy $x^3+y^4=z^{31}$ for every integer $a$.
However the solution that they have given is quite different from mine. What I want to know is that, is my solution valid, and is this a convincing method to do proofs of such kind?
Thanks in advance!
|
This is the solution they gave...
Choose $x = 2^{4r}$ and $y = 2^{3r}$. Then the left side is $2^{12r+1}$. If we take$ z = 2^k$,
then we get$ 2^{12r+1} = 2^{31k}$. Thus it is sufficient to prove that the equation 12r + 1 = 31k
has infinitely many solutions in integers. Observe that$ (12.18) + 1 = 31.7$. If we choose
$r = 31l + 18$ and $k = 12l + 7$, we get
$12(31l + 18) + 1 = 31(12l + 7)$;
for all$ l$. Choosing $l\in N$, we get infinitely many$ r = 31l + 18$ and$ k = 12l + 7$ such that
$12r + 1 = 31k$. Going back we have infinitely many (x; y; z) of integers satisfying the given
equation.
|
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|
Series expansion of Elliptic integral $F(\sin^{-1}(1-x)|-1)$ where $0I understand the leading term $F(\pi/2\mid-1)$ would be $K(-1)$. I also want the next term contributing from $x$. Actually I could not expand $\sin^{-1}(1-x)$ (when $0<x\ll1$) around 1. Any help would be highly appreciated.
Thanks a lot.
|
We may put $1-y=\sin{\theta}$ in the integral, so $dy = -\cos{\theta} d\theta $, $\cos{\theta}=\sqrt{y(2-y)}$ on the interval in question, to find
$$ F(\arcsin{(1-x)}\mid -1) = \int_x^1 \frac{dy}{\sqrt{y(2-y)(1+(1-y)^2)}} = \int_x^1 \frac{dy}{\sqrt{y(2-y)(2-2y+y^2)}}. $$
Of course,
$$F(\arcsin(1) \mid -1) = F(\pi/2 \mid -1) = \int_0^{\pi/2} \frac{d\theta}{\sqrt{1+\sin^2{\theta}}} = K(-1) = \frac{\Gamma(1/4)^2}{4\sqrt{2\pi}}, $$
so this is the constant term (this particular value is found using the Beta-function integral). The rest comes from
$$ F(\arcsin{(1-x)}\mid -1)-F(\arcsin{(1)}\mid -1) = -\int_0^x \frac{dy}{\sqrt{y(2-y)(2-2y+y^2)}}, $$
and since $x$ is small, we can do by expanding the integrand:
$$ \frac{1}{\sqrt{y(2-y)(2-2y+y^2)}} = \frac{1}{2} y^{-1/2} \left(1 - y \left(\frac{3}{2} + y - \frac{1}{4} y^2 \right) \right)^{-1/2} = \frac{1}{2} y^{-1/2} + O(y^{1/2}) $$
by using the binomial expansion on the bracket. Integrating term-by-term gives
$$ F(\arcsin{(1-x)}\mid -1) = K( -1) - x^{1/2} + O(x^{3/2}) = \frac{\Gamma(1/4)^2}{4\sqrt{2\pi}} - x^{1/2} + O(x^{3/2}), $$
and you can find more of the terms by a taking more terms of the expansion of $\left(1 - y \left(\frac{3}{2} + y - \frac{1}{4} y^2 \right) \right)^{-1/2}$.
As for $\arcsin{(1-x)}$, we can use the same trick on its integral: putting $y=1-u$, $1-u^2 = 1-(1-y)^2 = y(2-y)$
$$ \arcsin{(1-x)} = \int_0^{1-x} \frac{du}{\sqrt{1-u^2}} = \int_x^1 \frac{du}{\sqrt{y(2-y)}}, $$
and then you can expand the square root binomially and get an expansion in odd powers of $x^{1/2}$ in the same way.
|
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|
Area of a circle inscribed in a polygon If a circle is inscribed in a polygon, show that,
$$\dfrac{\text{(Area of inscribed circle)}}{\text{(Perimeter of inscribed circle)}} = \dfrac{\text{(Area of Polygon)}}{\text{(Perimeter of Polygon)}}$$
|
For a regular polygon with $n$ sides with side length $l$. The ends of each side when connected to the centre of the polygon forms a triangle with an angle of $\frac{2\pi}{n}$ at the centre. There will be $n$ such triangles. The altitude of each triangle starting from the centre of the polygon has a length of $\frac{a}{2\tan\frac{\pi}{n}}$ with the opposite side (base of the triangle) of length $a$. This altitude height will also be the radius of the circle inscribed in it.
So,
$\text{Area of incribed circle} = \pi \left(\frac{a}{2\tan\frac{\pi}{n}}\right)^2$
$\text{Perimeter of incribed circle} = 2\pi \left(\frac{a}{2\tan\frac{\pi}{n}}\right)$
$\text{Area of polygon} = n \cdot \frac{1}{2}a\left(\frac{a}{2\tan\frac{\pi}{n}}\right)$
$\text{Perimeter of polygon} = n \cdot a$
|
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|
Determine if $\lim_{(x,y)\to(0,0)} \frac{x^3y^4}{(x^4 + y^2)^2}$ exist What I tried:
Let $$\ f(x,y) = \frac{x^3y^4}{(x^4 + y^2)^2}$$
For points of the form$\ (x,0)$ then $\ f(x,0)=0$, similarly, for$\ (0,y)$ then $\ f(0,y)=0$, so lets suppose that:
$$\lim_{(x,y)\to(0,0)} \frac{x^3y^4}{(x^4 + y^2)^2} =0$$
So, for$\ ε>0$ if$\ δ=ε$ we have to prove that:
$$ |\frac{x^3y^4}{(x^4 + y^2)^2} -0|<ε$$
But I'm having a hard time trying to prove the last part, I tried:
$$ |\frac{x^3y^4}{(x^4 + y^2)^2} -0|=|\frac{x^3y^4}{(x^4 + y^2)^2}| ≤ |\frac{(x^3y^4)(x^4 + y^2)^2}{(x^4 + y^2)^2}|=|(x^3y^4)|$$
|
Let $u=x^4+y^2$. Then $|x|\le u^{1/4}$ and $|y|\le u^{1/2}$. So
$|x^3y^4|\le u^{11/4}$ and, for $(x,y)\ne(0,0)$, $|f(x,y)|\le u^{3/4}$.
|
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|
Combinatorial proof of Recurrence for the number of permutations of odd order Let $a_n$ denote the number of permutations of odd order in $S_n$. This is sequence A000246 in OEIS. Then there is the recurrence relation on OEIS that $a_{2n} = (2n-1)a_{2n-1}$ and $a_{2n+1} = (2n+1)a_{2n}$, or equivalently $a_n = a_{n-1} + (n-1)(n-2)a_{n-2}$.
It seems like there should be a combinatorial proof for this result.
Since there isn't a simple recurrence for the number of partitions with odd parts, it doesn't seem like working on the level of partitions will work.
|
These permutations are precisely the ones that do not contain even
cycles, giving the species
$$\mathfrak{P}(\mathfrak{C}_{=1}(\mathcal{Z})
+ \mathfrak{C}_{=3}(\mathcal{Z})
+ \mathfrak{C}_{=5}(\mathcal{Z})
+ \mathfrak{C}_{=7}(\mathcal{Z})
+ \cdots)$$
This yields the EGF
$$G(z) = \exp\left(\sum_{q\ge 1} \frac{z^q}{q}
- \sum_{q\ge 1} \frac{z^{2q}}{2q}\right)
\\ = \exp\log\frac{1}{1-z}
\exp\left(- \frac{1}{2} \log\frac{1}{1-z^2}\right)
\\ = \frac{\sqrt{1-z^2}}{1-z}.$$
Differentiate to obtain
$$G'(z) = \frac{\sqrt{1-z^2}}{1-z} \frac{1}{1-z}
- \frac{\sqrt{1-z^2}}{1-z} \frac{z}{1-z^2}
= G(z) \frac{1}{1-z^2}.$$
Extracting coefficients we find for $n\ge 2$
$$n! [z^n] G'(z) (1-z^2) = n! [z^n] G(z)
\quad\text{or}\quad
a_{n+1} - n! [z^{n-2}] G'(z) = a_n$$
which finally yields
$$a_{n+1} = a_n + n! \times \frac{a_{n-1}}{(n-2)!}
\quad\text{or}\quad
a_{n+1} = a_n + n (n-1) a_{n-1}$$
as desired.
For the second recurrence we start with the closed form from the EGF
$$n! \sum_{q=0}^{\lfloor n/2\rfloor} [z^{2q}] \sqrt{1-z^2}
= n! \sum_{q=0}^{\lfloor n/2\rfloor} (-1)^q {1/2\choose q}.$$
Since $\lfloor (2n+1)/2\rfloor = \lfloor 2n/2\rfloor$ we have
$$a_{2n+1} = (2n+1) a_{2n}$$ by inspection.
The first recurrence now yields
$$a_{2n+1} = a_{2n} + 2n (2n-1) a_{2n-1}
\quad\text{or}\quad
(2n+1) a_{2n} = a_{2n} + 2n (2n-1) a_{2n-1}$$
and finally
$$2n a_{2n} = 2n (2n-1) a_{2n-1}
\quad\text{or}\quad
a_{2n} = (2n-1) a_{2n-1}.$$
|
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|
Finding angle in triangle I have a triangle $\triangle ABC$ with vertices $A$, $B$ and $C$ with $\angle BAC = 45^\circ$ and $\angle ABC = 30^\circ$. Point $M$ is the center of the $BC$. Find $\angle AMC$.
This is not my homework, it's a little girl's, so she can't use trigonometric equations or anything after special lines in triangles.
Thanks in advance!
|
Let $X$ on $AB$ be the foot of the altitude through $C$.
*
*Since $\angle BAC = 45^\circ$, $\angle ACX = 45^\circ$, $\triangle AXC$ is isoceles and $AX = CX$.
*Since $\angle CBA = 30^\circ$, $\angle XCB = 60^\circ$, $\triangle XBC$ is half of an equilateral triangle and $CX : BC = 1 : 2$.
*Since $M$ is midpoint of $BC$, $CX = CM$ and $\triangle XCM$ is equilateral.
*This implies $XM = CM = BM$, $\triangle XBM$ is isosceles and $\angle BXM = \angle MBX = 30^\circ$.
*Since $AX = CX = XM$, $\triangle AXM$ is also isosceles with $\angle XAM = \angle AMX = \frac12 \angle BXM = 15^\circ$.
From this, we find
$$\begin{align}
\angle CMA &= 180^\circ - \angle AMX - \angle XMB \\
&= 180^\circ - \angle AMX - (180^\circ -\angle BXM - \angle MBX )\\
&= \angle BXM + \angle MBX - \angle AMX\\
&= 30^\circ + 30^\circ - 15^\circ\\
&= 45^\circ\end{align}$$
|
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|
prove that : if $a, b \in \mathbb{R}^+$ : then : $a^2b^2(a^2+b^2-2)\geq (a^2+b^2)(ab-1)$ prove that : if $a, b \in \mathbb{R}^+$ : then :
$$a^2b^2(a^2+b^2-2)\geq (a^2+b^2)(ab-1)$$
$$a^4b^2+b^4a^2-2a^2b^2 \geq a^3b-a^2b+b^2a-a^2$$
$$a^4b^2+b^4a^2-2a^2b^2 \geq a^3b-a^2+b^3a-b^2$$
now what ?
|
it is equivalent to $$(a-b)^2((ab)^2-ab+1)+2ab(ab-1)^2\geq 0$$ which is true.
|
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|
Solve $x^2 + (y-1)^2 + (x-y)^2 - \frac{1}{3} = 0.$ I came across an interesting equation with two variables $x,y\in\mathbb{R}$, $x^2 + (y-1)^2 + (x-y)^2 - \frac{1}{3} = 0.$
This, once expanded, can be simplified to $3x^2 - 3xy + 3y^2 - 3y + 1=0.$
How can one proceed to solve it algebraically? The solution according to Wolfram is $(x,y)=(\frac{1}{3},\frac{2}{3}).$
|
$$3x^2+3(y-1)^2+3(x-y)^2-1=(x+y-1)^2+(2x-y)^2+(x-2y+1)^2\geq0.$$
The equality occurs for
$$x+y-1=2x-y=x-2y+1=0,$$ which gives $x=\frac{1}{3}$ and $y=\frac{2}{3}$.
|
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|
Coefficient in expansion - Discrete Math
Let $\alpha$ be the coefficient of $x^2y^2z^2$ in $ (x^2+3y-2z)^5$. Show that $800\leq|\alpha|<1100$.
By my calculation, $\alpha=\binom {5}{2} \times (3^{2})\binom {3}{2} \times(-2)^2 \binom {1}{2}$
However $\binom {1}{2}=0$, so $\alpha$ should be $0$.
What am I doing wrong?
|
$$
\begin{align}
1^1\cdot3^2\cdot(-2)^2\cdot\overbrace{\binom{5}{2,2}}^{\substack{\text{trinomial}\\\text{coefficient}}}
&=1^1\cdot3^2\cdot(-2)^2\cdot\frac{5!}{2!2!1!}\\
&=1080
\end{align}
$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Range in an Interval I want to find the range of the function, $$f(x) = {{x + 1} \over {{x^2} + 1}}\,\,\;{\rm{when }}\;x \in \left[ { - 1,1} \right].\;\;\;$$
I have isolated $x$ as - $$x = {{1 \pm \sqrt {1 - 4y(y - 1)} } \over {2y}}$$
But I am unable to solve this inequality, $$ - 1 \le {{1 \pm \sqrt {1 - 4y(y - 1)} } \over {2y}} \le 1$$
as it involved $\, \pm \sqrt {} \,$ part. Please tell me how to proceed from here!
|
$$\frac{x+1}{x^2+1}\geq0.$$
The equality occurs for $x=-1$, which says that $0$ is a minimum.
Now, by AM-GM for $x\neq-1$ we obtain:
$$\frac{x+1}{x^2+1}=\frac{x+1}{(x+1)^2-2(x+1)+2}=\frac{1}{x+1+\frac{2}{x+1}-2}\leq\frac{1}{2\sqrt2-2}=\frac{1}{2}(\sqrt2+1).$$
The equality occurs for $x+1=\frac{2}{x+1}$, which says that we got a maximal value because for $x=-1$ we get a value $0$.
Id est, the answer is
$$\left[0,\frac{1+\sqrt2}{2}\right]$$
|
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|
Condition on a matrix so that its determinant is $2$
Is the following statement true or false? There exists $A \in M_{3,3}(\mathbb{Z})$ with determinant $2$ such that $$A\begin{pmatrix}2\\1\\4\end{pmatrix} = \begin{pmatrix}4\\-8\\16\end{pmatrix}$$
I first thought of eigenvalues but it doesn't look like the second vector is a multiple of the first. What are the conditions on $A$ so that it's "true"? Or is it always false?
|
False. Let us say $A=(a_{i,j})$ with $1\le i,j\le 3$. Then for all $i$
$$
4\,\,\text{ divides }\,\, 2a_{i,1}+a_{i,2}+4a_{i,3},
$$
which happens if and only if $a_{i,2}$ is even and $a_{i,1}$ has the same parity of $a_{i,2}/2$.
Then
\begin{align}
\mathrm{det}(A)&=
\begin{bmatrix}a_{1,1}&a_{1,2}&a_{1,3}\\
a_{2,1}&a_{2,2}&a_{2,3} \\
a_{3,1}&a_{3,2}&a_{3,3}
\end{bmatrix}
=2\mathrm{det}
\begin{bmatrix}a_{1,1}&a_{1,2}/2&a_{1,3}\\
a_{2,1}&a_{2,2}/2&a_{2,3} \\
a_{3,1}&a_{3,2}/2&a_{3,3}
\end{bmatrix}\\
&=2\mathrm{det}
\begin{bmatrix}a_{1,1}&a_{1,2}/2-a_{1,1}&a_{1,3}\\
a_{2,1}&a_{2,2}/2-a_{2,1}&a_{2,3} \\
a_{3,1}&a_{3,2}/2-a_{3,1}&a_{3,3}
\end{bmatrix}\\
&=4\mathrm{det}
\begin{bmatrix}a_{1,1}&\frac{1}{2}(a_{1,2}/2-a_{1,1})&a_{1,3}\\
a_{2,1}&\frac{1}{2}(a_{2,2}/2-a_{2,1})&a_{2,3} \\
a_{3,1}&\frac{1}{2}(a_{3,2}/2-a_{3,1})&a_{3,3}
\end{bmatrix}.\\
\end{align}
|
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|
$\sin(x) = \sin(x − \frac\pi3)$, solve for $x$ on interval $[-2\pi, 2\pi]$ According to the answer sheet:
$\sin(x) = \sin(x -\frac\pi3)$ gives:
$x = x-\frac\pi3 + k \cdot 2\pi$ or $x = \pi-(x-\frac\pi3) + k \cdot 2\pi$
^ How did they go from $\sin(x) = \sin(x-\frac\pi3)$ to the equations above?
Thanks in advance!
|
The first comes from the fact that $\sin(x) = \sin(x \pm 2\pi n) \,\, n\in \mathbb{Z}$, because $\sin$ has period $2\pi$ The second comes from $\sin(x) = \sin(\pi - x)$.
To answer the question in your title, however, about the intersections on the interval $[-2\pi, 2\pi]$, we'll reduce this to a formula for $x$.
\begin{align*}
\sin(x) &= -\cos(x + \pi/6) \\
1/2 (\sqrt{3} \cos x + \sin x) &= 0 \\
\tan x &= -\sqrt{3} \\
x &= \pi n - \frac{\pi}{3} \,\, n\in\mathbb{Z}
\end{align*}
which has values on $[-2\pi, 2\pi]$ of $x=-\frac{4\pi}{3}, -\frac{\pi}{3}, \frac{2\pi}{3}, \frac{5\pi}{3}$.
|
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|
Derivative of $(2x-1)(x+3)^{\frac{1}{2}}$ Find the derivative of
$(2x-1)(x+3)^{\frac{1}{2}}$
My try -
$(2x-1)(\frac{1}{2} (x+3)^{\frac{-1}{2}} (x+0) + (x+3)^{\frac{1}{2}} (2)$
$ = (x+3)^{\frac{-1}{2}} (\frac{1}{2}(2x-1) + 2(x+3) $
$= \frac{3x+5.5}{2 (x+3)^{\frac{1}{2}}} $
My numerator is wrong and should be
$6x+11$ . Where did I go wrong ? Thanks !!
|
The derivative of $2x-1$ is $2$.
The derivative of $(x+3)^{1/2}$ is
$$
\frac{1}{2}(x+3)^{-1/2}\cdot \color{red}{1}=\frac{1}{2(x+3)^{1/2}}
$$
Not $(x+0)$ as you wrote, but this disappeared in the second step.
The product rule gives
$$
2(x+3)^{1/2}+(2x-1)\frac{1}{2(x+3)^{1/2}}
=\frac{4x+12+2x-1}{2(x+3)^{1/2}}
=\frac{6x+11}{2(x+3)^{1/2}}
$$
You divided the numerator by $2$ and forgot to cancel it in the denominator.
|
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|
How to properly simplify and trig substitute this integral? $\int\frac{{\sqrt {25-x^2}}}x\,dx$
I see that it is the form $a-x^2$.
Let $x = 5\sin\theta$
Then substitute this $x$ value in and
get $\int\frac {\sqrt {25-(5 \sin\theta)^2}}{5 \sin\theta}\, d\theta $
Take the root, simplified to $\int\frac { {5-(5 \sin\theta)}}{5 \sin\theta}\,d\theta $
Factor out 5 and cancel, simplified to $\int\frac { {5(1-( \sin\theta))}}{5 (\sin\theta)}\, d\theta $
Divide by $\sin\theta$ to get $\int\frac { (1-( \sin\theta))}{ (\sin\theta)}\, d\theta $
Simplified to $\int\frac {1}{ \sin\theta}d\,\theta - 1 $
I am stuck here and I can't figure it out. Supposedly the answer simplifies to $\int-5 \sin x \tan x \,dx$
|
Following your steps, with corrections:
$\int\frac{{\sqrt {25-x^2}}}x\,dx$
I see that it is the form $a-x^2$.
Let $x = 5\sin\theta$
Then $dx=5\cos\theta d\theta$
Then substitute this $x$ and $dx$ value in and
get $\int\frac {\sqrt {25-(5 \sin\theta)^2}}{5 \sin\theta}\, 5\cos\theta d\theta $
Using the trig identity, simplify the root to get $\int\frac {\sqrt {25\cos^2\theta}}{5 \sin\theta}\, 5\cos\theta d\theta $
Factor out 5 and cancel, simplified to get $\int\frac {5|cos\theta|}{ \sin\theta}\, \cos\theta d\theta $
(Note that we're taking the square root of a square here, so we should get the absolute value. Unless you're on a domain where cosine is positive, you can't neglect this.)
This seems to not be where you want to go.
Let's try instead a substitution of $x=5\cos\theta$. Then $dx=-5\sin\theta d\theta$
Thus, we have $-\int\frac{\sqrt{25-(5\cos\theta)^2}}{5\cos\theta}5\sin\theta d\theta$
After simplification, we get $-5\int\frac{|\sin\theta|}{\cos\theta}\sin\theta d\theta$
The simplification you get works if x is such that $\sin\theta$ is always positive, so we can drop the absolute value sign.
|
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|
Find volume bounded by 3 equations using integration The prompt is to find the volume of the solid bounded by the equations $x^2 + y^2 -2y = 0$, $z = x^2 + y^2$ and $z \ge 0$
Plotting the equations, I get something like this
We get a paraboloid and a cylinder.
How to know if I have to use double or triple integration to solve this problem? and if I should use spherical or double polar or cylindrical coordinate system?
Also how to find the limits of integration in this particular problem as an example?
This is what the cross section looks like
It is clearly visible that $\theta$ goes from $0$ to $\pi$
|
First, complete the square to see that
$$
x^2+y^2-2y=x^2 +(y-1)^2-1=0.
$$
So $x^2+(y-1)^2=1$. The cylindrical change-of-coordinate we will be using is:
$$
x=r\cos \theta, \hspace{4mm} y=1+r\sin \theta,\hspace{4mm} z=z, \hspace{4mm} \mbox{ where } 0\leq r\leq 1,\hspace{4mm} 0\leq \theta\leq 2\pi.
$$
Note that the absolute value of the Jacobian is:
$$
\left|\frac{\partial(x,y)}{\partial(r,\theta)}\right|
= \left| \pmatrix{
x_r& x_\theta \\
y_r & y_\theta \\
}
\right|
= r.
$$
Now, we calculate the volume of the solid:
$$
\begin{align*}
\iint_{D} \int_0^{x^2+y^2} dz\: dA
&= \iint_{D} x^2+y^2 \: dA \\
&= \int_{0}^{2\pi}\int_{0}^{1} \left( (r\cos\theta)^2 + (1+r\sin \theta)^2 \right)
r\: dr d\theta \\
&= \int_{0}^{2\pi}\int_{0}^{1} \left( r^2\cos^2\theta+ r^2\sin^2 \theta + 1+2 r\sin \theta \right)
r\: dr d\theta \\
&= \int_{0}^{2\pi}\int_{0}^{1} \left( r^2 + 1+2 r\sin \theta \right)
r\: dr d\theta \\
&= \int_{0}^{2\pi}\int_{0}^{1} \left( r^3 + r + 2 r^2 \sin \theta \right)
dr d\theta \\
&= \int_{0}^{2\pi} \left( \frac{r^4}{4} + \frac{r^2}{2} + 2 \frac{r^3}{3} \sin \theta \right) \Bigg|_{0}^1
d\theta \\
&= \int_{0}^{2\pi} \left( \frac{1}{4} + \frac{1}{2} + \frac{2}{3} \sin \theta \right)
d\theta \\
&= \frac{\pi}{2}+\pi -\frac{2}{3}\cos\theta\Bigg|_{0}^{2\pi} \\
&= \frac{3\pi}{2}.
\end{align*}
$$
|
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|
Evaluate the sum of two complex conjugates to the power of $n$ Please help me out with the following problem for I'm stuck and don't even have any idea on how to proceed.
Find all integers $n$ such that
$$\left(\frac {-1 +i{\sqrt3} } {2}\right)^n+\left(\frac {-1 -i{\sqrt3} } {2}\right)^n=2$$
|
Hint: let $\displaystyle z_{1,2}=\frac {-1 \pm i{\sqrt3} } {2}\,$, then $z_1+z_2=-1$ and $z_1z_2=1\,$, so $z_{1,2}$ are the roots of $z^2+z+1\,$ which implies that they are roots of $z^3-1=(z-1)(z^2+z+1)$ i.e $z_1$ and $z_2 = \overline{z_1}$ are the complex cube roots of unity, therefore $z_1^n = z_1^{n \bmod 3}\,$.
[ EDIT ] expanded hint: let $\displaystyle z_{1}=\frac {-1 + i{\sqrt3} } {2}\,$, then $\displaystyle z_{2}=\frac {-1 - i{\sqrt3} } {2}=\overline{z_1}\,$, and the sum $S_n=z_1^n+z_2^n=z_1^n+\bar z_1^n=z_1^n+\overline{z_1^n}=2 \operatorname{Re}(z_1^n)$. Working out the calculations for $n=1,2,3\,$:
*
*$\displaystyle \;S_1=2 \operatorname{Re}(z_1)=-1$
*$\displaystyle \;z_1^2=\frac{(-1+i\sqrt{3})^2}{4}=\frac{(-1)^2-2 \cdot 1 \cdot i \sqrt{3} + i^2 \left(\sqrt{3}\right)^2}{4}=\frac{1-2 \sqrt{3} i-3}{4}=\frac{-1 - i \sqrt{3}}{2}\,$, therefore $S_2=2 \operatorname{Re}(z_1^2) = -1$
*$\displaystyle \;z_1^3 = z_1 \cdot z_1^2=\frac {(-1 + i{\sqrt3})(-1 - i{\sqrt3}) } {4} = \frac{(-1)^2 - i^2 \left(\sqrt{3}\right)^2}{4}=\frac{1+3}{4}=1\,$, therefore $S_3=2 \operatorname{Re}(z_1^3)=2$
Let $n=3k+m$ where $m = n \bmod 3 \in \{0,1,2\}\,$, then using the just established fact that $z_1^3=1$ it follows that $z_1^n=z_1^{3k+m}=\left(z_1^3\right)^k \cdot z_1^m = 1 \cdot z_1^m=z_1^m\,$. Therefore, the sum $S_n=2 \operatorname{Re}(z_1^n)$ always takes one of the values $-1$ or $2\,$, depending on the remainder $n \bmod 3\,$. In particular, $S_n$ is $2$ iff $n$ is a multiple of $3\,$, and $-1$ otherwise.
|
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|
How to find $x$ of this equation I am a beginner to logarithms.
By using logarithms, in this equation, I want to find the value of $x$:
$$ \ 3^x = 4^{x-1} \ $$
I am using 10 as base of logarithm.
And I want $x$ in terms of $\log2$ and $\log3$, as far as possible.
|
As mentioned logarithm use is desired. Here, for a solution in terms of natural logarithms,
\begin{align}
\ln(3^{x}) &= \ln(4^{x-1}) \\
x \, \ln(3) &= (x-1) \, \ln(4) \\
x \, (\ln(3) - \ln(4)) &= - \ln(2^{2}) = - 2 \, \ln(2) \\
x &= - \frac{2 \, \ln(2)}{\ln\left(\frac{3}{4}\right)}
\end{align}
Converting from natural logarithm to a base 10 logarithm:
Use
$$\log_{b}(x) = \frac{\log_{d}(x)}{\log_{d}(b)}$$
to convert the base of the logarithms. Here, let $b = e$ and $d = 10$ to yield
\begin{align}
x = - 2 \, \cdot \frac{\log_{10}(2)}{\log_{10}(e)} \, \frac{\log_{10}(e)}{\log_{10}\left(\frac{3}{4}\right)} = -2 \, \frac{\log_{10}(2)}{\log_{10}\left(\frac{3}{4}\right)}.
\end{align}
Strictly solving by base 10 logarithm:
\begin{align}
\log_{10}(3^{y}) &= \log_{10}(4^{y-1}) \\
y \, \log_{10}(3) &= y \, \log_{10}(4) - \log_{10}(4) \\
y \, (\log_{10}(3) - \log_{10}(4)) &= - \log_{10}(4) \\
y &= \frac{- \, \log_{10}(4)}{\log_{10}(3) - \log_{10}(4)} = \frac{2 \, \log_{10}(2)}{\log_{10}\left(\frac{4}{3}\right)}.
\end{align}
|
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|
Trig Identity Proof $\frac{1 + \sin\theta}{\cos\theta} + \frac{\cos\theta}{1 - \sin\theta} = 2\tan\left(\frac{\theta}{2} + \frac{\pi}{4}\right)$ I've been working on this for like half an hour now and don't seem to be getting anywhere. I've tried using the double angle identity to write the LHS with $\theta/2$ and have tried expanding the RHS as a sum. The main issue I'm having is working out how to get a $\pi$ on the LHS or removing it from the right to equate the two sides.
$$\frac{1 + \sin\theta}{\cos\theta} + \frac{\cos\theta}{1 - \sin\theta} = 2\tan\left(\frac{\theta}{2} + \frac{\pi}{4}\right)$$
|
LHS:
$$\frac{(1 + \sin\theta)(1-\sin\theta)}{\cos\theta(1-\sin\theta)} + \frac{\cos\theta}{1 - \sin\theta} = \frac{2\cos\theta}{1 - \sin\theta}.$$
RHS:
$$2\tan\left(\frac{\theta}{2} + \frac{\pi}{4}\right)=2\frac{\frac{\sin\frac{\theta}{2}}{\cos\frac{\theta}{2}}+1}{1-\frac{\sin\frac{\theta}{2}}{\cos\frac{\theta}{2}}}=2\frac{\sin\frac{\theta}{2}+\cos\frac{\theta}{2}}{\cos\frac{\theta}{2}-\sin\frac{\theta}{2}}=$$
$$2\frac{\left(\sin\frac{\theta}{2}+\cos\frac{\theta}{2}\right)\left(\cos\frac{\theta}{2}-\sin\frac{\theta}{2}\right)}{\left(\cos\frac{\theta}{2}-\sin\frac{\theta}{2}\right)^2}=\frac{2\cos\theta}{1 - \sin\theta}.$$
|
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|
If $x$ is real, evaluate $k$ in absolute inequality
If $x$ is real and $$y=\frac{(x^2+1)}{x^2+x+1}$$ then it can be shown that $$\left|y-\frac{4}{3}\right|\leq k$$ Evaluate $k$
My attempt,
\begin{align}\left(y-\frac{4}{3}\right)^2&\leq k^2\\
\sqrt{\left(y-\frac{4}{3}\right)^2}&\leq k\end{align}
I don't know how to proceed anymore. Thanks in advance.
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$$\frac{x^2+1}{x^2+x+1}-\frac 43=1-\frac x{x^2+x+1}-\frac 43=-\frac 13-\frac x{x^2+x+1}$$
So you need to find $m$ such that $$\frac{x}{x^2+x+1}\leq m$$
Since both $m,x^2+x+1>0$ we can multiply it out we get $$ mx^2+(m-1)x+m\geq 0$$
Now we must have that $m>0$ and that $\Delta\leq 0$(discriminant), now just find the minimal $m$ that satisfies those requirements and then the minimum will be $$k=\frac 13+m$$
I'll leave it to you to find the $k$.
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|
Baby Rudin: Theorem 3.20 (c) and (d) In theorem 3.20, Rudin offers a proof to the limits of the following sequences:
$$u_n = n^{\frac{1}{n}}$$
$$v_n = \frac{n^{\alpha }}{(1+p)^n}$$ with $\alpha$ a real number and $p > 0$.
In both the proofs, Rudin uses inequalities which I am not familiar with. For $u_n$, the following inequality is used:
$$(1+x)^n \ge \frac{n(n-1)}{2}x^2$$
For $v_n$:
$${n \choose k}p^k > \frac{n^kp^k}{2^kk!}\quad \quad(n>2k) $$
I have painstakingly come up with proofs for both of those inequalities, however, and this is especially true for the second inequality, my proofs are not very elegant nor do they give me any idea of the inequality came to be.
Surely Rudin doesn't pull these inequalities out of thin air. Where can I find a more comprehensive derivation of these inequalities? Where do they come from exactly (which field)?
Update:
Here are my proofs. They both rely on induction.
For the first one:
Suppose $x\ge0$. Then, for $n = 2$:
$$ (1+x)^2 = 1+2x+x^2 $$
$$\frac{2(2-1)}{2} x^2 = x^2$$
Since $x\ge0$, we have $(1+x)^n \ge \frac{n(n-1)}{2}x^2 $ for n = 2.
Suppose our inequality is true for a certain n. Then:
$$ (1+x)^n + nx^2 \ge \frac{n(n+1)}{2}x^2$$
Let $D = (1+x)^{n+1} - (1+x)^n - nx^2 $. Since $\forall n \in \mathbb{N}, \forall m = 0,1, ... n, {n+1 \choose m} \ge {n \choose m} $, and ${n+1 \choose 2} - {n \choose 2} - n = 0$, we have:
$$D \ge 0 \Rightarrow (1+x)^{n+1} \ge (1+x)^n + nx^2$$
So:
$$(1+x)^{n+1} \ge \frac{n(n+1)}{2}x^2$$ The theorem follows by induction.
For the second inequality:
We want to prove that $n(n-1)...(n-k+1) > \frac{n^k}{2^k}$, with $n > 2k$
Let $n\in\mathbb{N}, k$ and integer greater than 0 such that $2n > k$. Then for $k=1$, we have: $$n > \frac{n}{2} $$
Now suppose our inequality holds for a certain $k$. Then:
$$n(n-1)...(n-k+1)(n-k) > \frac{n^k}{2^k}(n-k)$$
Since $n > 2k$, $n-k > \frac{n}{2} > 0$ such that:
$$n(n-1)...(n-k+1)(n-k) > \frac{n^k}{2^k}(n-k) > \frac{n^{k+1}}{2^{k+1}} $$
By induction, our inequality holds for all $k$ such that $n > 2k$
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The first inequality can be derived, for $x\ge0$, from the binomial theorem:
$$
(1+x)^n=\sum_{k=0}^n \binom{n}{k}x^k
$$
If we remove all the (non negative) terms except the one for $k=2$, we get
$$
(1+x)^n\ge \binom{n}{2}x^2
$$
|
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|
BMO2 1997 - Combinatorics Find the number of polynomials of degree 5 with distinct coefficients from the set {1, 2, 3, 4, 5, 6, 7, 8} that are divisible by $x^2-x+1$.
I tried multiplying $x^2-x+1$ by $ax^3+bx^2+cx+d$ to get $ax^5+(b-a)x^4+(a-b+c)x^3+(b-c+d)x^2+(c-d)x+d$, but I am struggling to count from here.
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It's divisible by $x^2-x+1$ if and only if it vanishes at $\zeta=e^{\pi i/3}$. Now $$a\zeta^5+b\zeta^4+c\zeta^3+d\zeta^2+e\zeta+f=(d-a)\zeta^2+(e-b)\zeta+f-c$$ so we must have $d-a=b-e=f-c$. So you just have to count solutions of that system with the variables taking on distinct values from $1,\dots,8$.
|
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|
Solution of differential equation using Laplace transform Can someone please help me identify my mistake as I am not able to find it.
$y''+y'=3x^2$ where $y(0)=0,y'(0)=1$
$L[y'']+L[y']=3L[x^2]$
$L[y'']=p^2L[y]-py(0)-y'(0)=p^2L[y]-1$
$L[y']=pL[y]-y(0)=pL[y]$
$p^2L[y]-1+pL[y]=\frac{6}{p^3}$
$p^2L[y]+pL[y]=\frac{6}{p^3}+1$
$(p^2+p)L[y]=\frac{6}{p^3}+1$
$L[y]=\frac{6}{p^3(p^2+p)}+\frac{1}{p^2+p}$
$y=x^3-e^{-x}+1-e^{-x}$
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Using
\begin{align}
\mathcal{L}\{y'\} &= s \, \overline{y} - y(0) \\
\mathcal{L}\{y''\} &= s^2 \, \overline{y} - s \, y(0) - y'(0)
\end{align}
then
$$y'' + y' = a \, t^2$$
transforms to
\begin{align}
s \, (s+1) \, \overline{y} &= s \, y(0) + y'(0) + y(0) + \frac{2 a}{s^3} \\
\overline{y} &= \frac{y(0) + y'(0)}{s \, (s+1)} + \frac{y(0)}{s+1} + \frac{2 a}{s^4 \, (s+1)} \\
&= \frac{y(0) + y'(0) - 2 a}{s} + \frac{2 a - y'(0)}{s+1} + \frac{2 a}{s^4} - \frac{2 a}{s^3} + \frac{2 a}{s^2} \\
&= \mathcal{L}\{ (2a - y'(0)) \, e^{-t} + \frac{a}{3} \, t^3 - a \, t^2 + 2 a \, t + y(0) + y'(0) - 2 a \}.
\end{align}
This leads to the general solution
$$y(t) = (2a - y'(0)) \, e^{-t} + \frac{a}{3} \, t^3 - a \, t^2 + 2 a \, t + y(0) + y'(0) - 2 a. $$
Now applying the conditions $y(0) = 0$, $y'(0) =1$, $a=3$ then
$$y(t) = 5 \, e^{-t} + t^3 - 3 \, t^2 + 6 \, t - 5. $$
Note that a key component is
$$\frac{1}{s^4 \, (s+1)} = \frac{1}{s+1} + \frac{1}{s^4} - \frac{1}{s^3} + \frac{1}{s^2} - \frac{1}{s}.$$
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|
Prove the inequality using AM-GM only. By considering "Arithmetic mean $\geq $ Geometric mean" prove the trigonometric inequality:
$$\sin A + \sin B + \sin C \leq \frac{3\sqrt{3}}{2}. $$
where $A+B+C=180°$.
My try: By using transformation formulae, I proved that
$$\sin A + \sin B + \sin C = 4\cos\left(\frac{A}{2}\right)\cos\left(\frac{B}{2}\right)\cos\left(\frac{C}{2}\right)=y(let)$$
Next using AM-GM inequality
$$\cos\left(\frac{A}{2}\right)+\cos\left(\frac{B}{2}\right)+\cos\left(\frac{C}{2}\right)\geq 3 \left(\frac{y}{4}\right)^{\frac{1}{3}}.$$
I'm unable to proceed further. Please help me.
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You can use Jensen Inequality on concave function.
Let $f(x)=\sin x$, then $f''(x)=-\sin x<0$, as $x\in (0,\pi)$.
Then $$\frac{f(A)+f(B)+f(C)}{3}\leq f\Big(\frac{A+B+C}{3}\Big)\\\text{i.e.}\space\frac{\sin A+\sin B+\sin C}{3}\leq \sin \Big(\frac{A+B+C}{3}\Big)=\sin \frac{\pi}{3}=\frac{\sqrt{3}}{2}\\\text{i.e.}\space \sin A+\sin B+\sin C\leq \frac{3\sqrt{3}}{2}\space\space\space \blacksquare$$
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|
Solving Telescoping Series $$\ \sum_{n=5}^\infty \frac{8}{n^2-1} $$
I tried the following:
$$\ \sum_{n=5}^\infty \frac{8}{n^2-1} = \sum_{n=5}^\infty \frac{8}{n-1} - \frac{8}{n} =$$ $$\left(2-\frac{8}{5}\right) + \left(\frac{8}{5} - \frac{8}{6}\right) + \left(\frac{8}{6} - \frac{8}{7}\right) + \cdots + \left(-\frac{8}{n}\right)$$
Terms cancelled each other out, therefore we are left with:
$$ \ (2 - \frac{8}{n}) $$
I could think the series converges to 2 since:
$$\lim_{n \to \infty} \left(2-\frac 8 n \right) = 2$$
However, the correct answer is $$ \frac{9}{5} $$ what am I doing wrong?
|
Your partial fraction is wrong:
$$\frac{8}{n^2-1}=\frac{-4}{n+1}+\frac{4}{n-1}$$
Using similar trick, you will see that most terms cancel out and you will left with $1+\frac45.$
Edit:
to obtain the partial fractions,
Since $n^2-1=(n-1)(n+1)$,
$$\frac{8}{(n-1)(n+1)}= \frac{A}{n+1}+\frac{B}{n-1}.$$
We can for instance equate the two and solve for $A$ and $B$ by comparing coefficients.
I use a trick call heaviside cover method.
To determinte $A$, $n+1=0$, $n=-1$. I will cover up the term $n+1$ in the denominator of the left hand side and evalute $\frac{8}{n-1}$ with $n=-1$, hence $A=-4$.
We can do similar stuff for $B$ as well.
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|
what is smallest possible value of range of $7$ values given that their mean is $12$ and their median is $9$
what is smallest possible value of range of $7$ values given that their mean is $12$ and their median is $9$ :-
a) $3$
B) $6$
C) $7$
D) $8$
What is the proper approach to solve this problem , is there any relation or inquality that help ? What if we need the largest possible value?
Thank you for your help
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List the numbers in order: $a_1\le a_2\le a_3\le a_4\le a_5\le a_6\le a_7$.
You know the median is 9, so you now have $a_1,a_2,a_3,9,a_5,a_6,a_7$
The mean is $12$, so you know that $7\times 12=84=a_1+a_2+a_3+9+a_5+a_6+a_7$.
So $a_1+a_2+a_3+a_5+a_6+a_7=75$
You want $a_7-a_1$ to be as small as possible. This means you want $a_1,a_2,a_3,a_5,a_6$ to be assigned with values as large as possible and $a_7$ to be as small as possible.
The maximum for $a_1,a_2,a_3$ is $9$. This leaves $a_5+a_6+a_7=48$. You want $a_7$ as small as possible, and this occurs when $a_5=a_6=a_7$. So set them to $16$.
Now you have as your list $9,9,9,9,16,16,16$.
This gives $7$ as the answer.
In general, let's say you wanted to find the smallest range of the finite set $\{a_k \in \mathbb R, k\in\{1,...,n\}: i<j \implies a_i\le a_j\}$ which has mean $\mu$ and median $m$. If $n = 1$ or $2$, then the answer is $0$ unless $\mu \neq m$, in which case no such set is possible.
$n$ is odd $\implies$ Using the same reasoning as before, you assign $a_k=m$ for $k\lt \frac{n+1}{2}$ (and we know that $a_{\frac{n+1}{2}}=m$).
Using the mean,
$$n\mu=\sum^n_{k=1}{a_k}=\frac{n+1}{2}m+\sum^n_{k=\frac{n+3}{2}}{a_k}$$
$\implies$
$$\sum^n_{k=\frac{n+3}{2}}{a_k}=n\mu-\frac{n+1}{2}m$$
Again, using the same reasoning as before, you let $a_\frac{n+3}{2}=a_\frac{n+5}{2}=...=a_{n}$.
So the range is
$$a_n-a_1=\frac{n\mu-\frac{n+1}{2}m}{\frac{n-1}{2}}-m$$
$$=\frac{n\mu-mn}{\frac{n-1}{2}}$$
$$=2(\mu-m)\frac{n}{n-1}$$
Testing this with the above example gives $2(12-9)\frac{7}{6}=7$.
$n$ is even $\implies$ The average of the middle two numbers is the median. However, you want the smallest one of them as large as possible, as you later want to make all of the numbers smaller than it as large as possible. This happens when the middle two numbers are equal.
So we have $a_{\frac{n}{2}}=a_{\frac{n+2}{2}}=m$ assign $a_k=m$ for $k\lt \frac{n}{2}$.
Using the mean,
$$n\mu=\sum^n_{k=1}{a_k}=\frac{n+2}{2}m+\sum^n_{k=\frac{n+4}{2}}{a_k}$$
$\implies$
$$\sum^n_{k=\frac{n+4}{2}}{a_k}=n\mu-\frac{n+2}{2}m$$
As before, let $a_\frac{n+4}{2}=a_\frac{n+6}{2}=...=a_{n}$.
The range is
$$a_n-a_1=\frac{n\mu-\frac{n+2}{2}m}{\frac{n-2}{2}}-m$$
$$=\frac{n\mu-mn}{\frac{n-2}{2}}$$
$$=2(\mu-m)\frac{n}{n-2}$$
You can write this as $2(\mu-m)\frac{n}{2\lfloor\frac{n-1}{2}\rfloor}$ for all $n\ge 3$.
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|
Explanation of hyperbola drawing Here is an image of a hyperbola:
I'm wondering why $c^2=a^2+b^2$ in this drawing given that $c$ is the distance from the origin to both of the foci and $a$ is half the length of the absolute distance between any given point and the two foci.
Sorry if this seems like a stupid question.
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Take the foci of the hyperbola $F_1=(c,0)$ and $F_2=(-c,0)$ and the two vertices : $V_1=(a,0)$ and $V_2=(-a,0)$.
From the definition we have that a point $P$ of the hyperbola is such that $\overline{PF_1}-\overline{PF_2}=k$, and using as $P$ a vertex we find that $k=2a$.
So, for $P=(x,y)$ the equation is
$$
\sqrt{(x-c)^2+y^2}-\sqrt{(x+c)^2+y^2}=2a
$$
squarin two times and with ambit of algebra you can find the equation:
$$
\frac{x^2}{a^2}-\frac{y^2}{c^2-a^2}=1
$$
that becomes
$$
\frac{x^2}{a^2}-\frac{y^2}{b^2}=1
$$
with $b^2=c^2-a^2$
and this means that $c^2=a^2+b^2$. Then note that the asymptotes of the hyperbola have slope $\pm\frac{b}{a}$, so the asymptotes intrecept the circle with center the origin and radius $c$ at the points of coordinates $(\pm a,\pm b)$ where $a^2+b^2=c^2$
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|
Greatest term in the expansion $(2+3x)^9$ I have got a question which is stated as
Find numerically the greatest term in the expansion of $(2+3x)^9$ where $x=\frac{3}{2}$
I haven't any idea how to proceed but I just plugged in value of $x$ and got $$(2+3x)^9=\bigg(\frac{13}{2}\bigg)^9$$
How can I determine greatest term here?
Please help!!!
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You have
$$(3 x+2)^9=19683 x^9+118098 x^8+314928 x^7+489888 x^6+489888 x^5+326592 x^4+145152 x^3+41472 x^2+6912 x+512$$
Plugging $x=1.5$ the terms, from degree $0$ up are
$$512,10368,93312,489888,1.65337\times 10^6,3.72009\times 10^6,5.58013\times 10^6,5.38084\times 10^6,3.02672\times 10^6,756681$$
the largest terms is for $x^6$ and is
$5.58013\times 10^6$
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|
If $b>a>0$ , how to prove that $\left(\frac {\sqrt a +\sqrt b}{2}\right )^2 < \frac 1e \left(\frac {b^b} {a^a}\right)^{\frac 1{b-a}}$? If $b>a>0$ , then how to prove that $\left(\dfrac {\sqrt a +\sqrt b}{2}\right )^2 < \dfrac 1e \left(\dfrac {b^b} {a^a}\right)^{\dfrac 1{b-a}}$ ?
I tried applying the known standard inequalities but without success . Please help .
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Let $b=ax$, where $x>1$.
Hence, we need to prove that
$$ea\left(\frac{\sqrt{x}+1}{2}\right)^2<\left(\frac{(ax)^{ax}}{a^a}\right)^{\frac{1}{a(x-1)}}$$ or
$$ea\left(\frac{\sqrt{x}+1}{2}\right)^2<\left(\frac{(ax)^{x}}{a}\right)^{\frac{1}{x-1}}$$ or
$$e\left(\frac{\sqrt{x}+1}{2}\right)^2<\left(x^{x}\right)^{\frac{1}{x-1}}$$
or $f(x)>0$, where
$$f(x)=x\ln{x}+1-x+2(x-1)\ln\frac{\sqrt{x}+1}{2},$$
which is obvious after two differentiations.
|
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|
Interval of convergence of $\sum_{n=0}^\infty (\frac{n}{n+1})^{n^2}(2x)^n$ I want to find the interval of convergence of $$\sum_{n=0}^\infty (\frac{n}{n+1})^{n^2}(2x)^n$$
By the root test, $\sqrt[n]{|(\frac{n}{n+1})^{n^2}(2x)^n|}=|(\frac{n}{n+1})^{n}(2x)|$
And $(\frac{n}{n+1})^n=(1-\frac{1}{n+1})^n\to\frac1e$ as $n\to\infty$
So I've got an open interval(of convergence) s.t. $|\frac1e 2x|<1\implies |x|<\frac e2$.
Now I need to figure out whether the series is convergent or not at the end points, $x=\frac e2, -\frac e2$.
So there are two series I need to deal with, namely, $\sum_{n=0}^\infty (\frac{n}{n+1})^{n^2}e^n$ and $\sum_{n=0}^\infty (\frac{n}{n+1})^{n^2}(-1)^ne^n$.
Suppose $(\frac{n}{n+1})^{n^2}$ is increasing. Then $$(\frac{n}{n+1})^{n^2}<(\frac{n+1}{n+2})^{(n+1)^2}\iff 1<(\frac{n+1}{n+2})^{(n+1)^2}(\frac{n+1}{n})^{n^2}$$
$$\iff 1<(1-\frac{1}{n+2})^{2n+1}(1+\frac{1}{n^2+2n})^{n^2}$$
But $(1-\frac{1}{n+2})^{2n+1}\to\frac{1}{e^2}$ and $(1+\frac{1}{n^2+2n})^{n^2}<(1+\frac{1}{n^2})^{n^2}\to e$
So I have $1<\frac1e<1$, which is a contradiction. Hence $(\frac{n}{n+1})^{n^2}$ is not increasing. But I don't know if it's decreasing. (is it true that $(1+\frac{1}{n^2+2n})^{n^2}\to e$?)
If $(1+\frac{1}{n^2+2n})^{n^2}\to e$ is true, then $(\frac{n}{n+1})^{n^2}e^n$ converges to some positive number, so $\sum(\frac{n}{n+1})^{n^2}e^n$ diverges. But I don't know if it's decreasing: can't apply the alternating series convergence test(Leibniz's criterion).
So, can you help me with the two end points?
|
One can use the Taylor expansion of $\log$ to get an equivalent of the sequence :
$$
\begin{align}
\left(\frac{n}{n+1}\right)^{n^2}e^n &= \exp{\left[n+n^2\log\frac{n}{n+1}\right]} \\
&= \exp{\left[n+n^2\left(-\frac{1}{n+1}-\frac{1}{2(n+1)^2}+o\left(\frac{1}{n^2}\right)\right)\right]} \\
&= \exp{\left[\frac{n^2+2n}{2(n+1)^2}+o(1)\right]} \\
&=\sqrt{e}+o(1).
\end{align}
$$
That should give you the information you need to conclude about the convergence of both your series.
|
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Verifying $|F(r)| \geq \frac{1}{1-r}\log(\frac{1}{1-r}) $ and $|F(re^{i \theta})| \geq c_{q/r}\frac{1}{1-r}\log({\log(\frac{1}{1-r})})$ I'm attempting to take a Tauberian route in verifying the proposition in $(1)$ below, which is from Complex Analysis, by Elias M Stein and Rami M. Shakarchi.
Let $F(z)$ be the following series:
$$F(z) = \sum_{n=1}^{\infty}d(n)z^{n} \, \, \text{for} \, |z| < 1$$
$\text{Remark}$
One can also observe the following relationship:
$$\sum_{n=1}^{\infty}d(n)z^{n} = \sum_{n=1}^{\infty} \frac{z^{n}}{1-z^{n}}$$
$(1)$
If $z=r$ with $0 < r < 1$, then as $r \rightarrow 1$, another case that be considered is $\theta=\frac{2\pi p}{q}$, where $p$ and $q$ are positive integers and then:
$(1.2)$
$$|F(r)| \geq \frac{1}{1-r}\log\left(\frac{1}{1-r}\right)$$
$(1.3)$
$$\lvert F(re^{i \theta})\rvert \geq c_{q/r}\frac{1}{1-r}\log\left(\frac{1}{1-r}\right)$$
$\text{Lemma}$
Formally attacking $(1)$, one can make the initial observations for the case seen in $(1.2)$
$$\left\lvert \sum_{n=1}^{\infty} \frac{z^{n}}{1-z^{n}}\right\rvert \leq \frac{1}{1-r}\log\left( \frac{1}{1-r}\right)$$
$$\left\lvert\frac{z^{1}}{1-z^{1}} + \frac{z^{2}}{1-z^{2}} + \frac{z^{3}}{1-z^{3}} + \frac{z^{4}}{1-z^{4}} + \cdot \cdot \cdot + \frac{z^{n}}{1-z^{n}}\right\rvert \leq \frac{1}{1-r}\log\left( \frac{1}{1-r}\right)$$
Recall the archetypal technique of Abel summability as formally developed in $(2)$:
$(2)$
$\text{Definition (0.2)}$
A series $A(r)= \sum_{n=1}^{\infty}a_{n}r^{n}$ is said to Abel summable to $L$ if $f(r)$ is convergent for all $\lvert r\rvert < L$ and if $f(r)$ converges to some limit $L$ as $r \rightarrow 1^{-}$:
$$A(r)= \sum_{n=1}^{\infty}a_{n}r^{n}$$
$\text{Remark}$:
The developments of Abel summability, expressed within a prior definition, can be fully expressed as follows:
$$\lim_{r \rightarrow 1^{}} A(r)= \lim_{r \rightarrow 1^{}} \sum_{n=1}^{\infty}a_{n}r^{n} = L$$
Now recall our previous observation with some added developments:
$$\left|\frac{z^{1}}{1-z^{1}} + \frac{z^{2}}{1-z^{2}} + \frac{z^{3}}{1-z^{3}} + \frac{z^{4}}{1-z^{4}} + \cdot \cdot \cdot + \frac{z^{n}}{1-z^{n}} = d(1)z^{1} + d(2)z^{2} + d(3)z^{3} + \cdot \cdot \cdot + d(4)z^{4} + d(z)z^{n}\right| \leq \frac{1}{1-r}\log\left( \frac{1}{1-r}\right)$$
$$\left|d(1)z^{1} + d(2)z^{2} + d(3)z^{3} + \cdot \cdot \cdot + d(4)z^{4} + d(z)z^{n}\right| \leq \frac{1}{1-r}\log\left( \frac{1}{1-r}\right)$$
$$\lim_{r \rightarrow 1^{}}|d(1)z^{1} + d(2)z^{2} + d(3)z^{3} + \cdot \cdot \cdot + d(4)z^{4} + d(z)z^{n}| \leq \frac{1}{1-r}\log\left( \frac{1}{1-r}\right)$$
Are the recent developments valid so far? I feel like this approach is too "archetypal." If the developments are wrong, may I have an answer on an alternate approach?
|
For the case of (1.2), note that
$$1-r^n = (1-r)(1+r+r^2+\cdots + r^{n-1}) \le (1-r)\cdot n.$$
Therefore
$$ \sum_{n=1}^{\infty}\frac{r^n}{1-r^n}\ge \sum_{n=1}^{\infty}\frac{r^n}{(1-r)n} = \frac{1}{1-r}\sum_{n=1}^{\infty}\frac{r^n}{n}.$$
But the sum on the right is precisely $\ln\left (\dfrac{1}{1-r}\right ).$ This gives your (1.2).
|
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|
If $ a=\frac{\sqrt{x+2} + \sqrt {x-2}}{\sqrt{x+2} -\sqrt{x-2}}$ then what is the value of $a^2-ax$ is equal to
If $$ a=\frac{\sqrt{x+2} + \sqrt {x-2}}{\sqrt{x+2} -\sqrt{x-2}}$$ then
the value of $a^2-ax$ is equal to:
a)2 b)1 c)0 d)-1
Ans. (d)
My attempt:
Rationalizing $a$ we get,
$ x+ \sqrt {x^2-4}$
$a^2=(x+\sqrt{x^2-4)^2}=2x^2-4+2x\sqrt{x^2-4}$
Now,
$a^2-ax=2x^2-4+2x\sqrt{x^2-4}-x^2-x\sqrt{x^2-4}=x^2+x\sqrt{x^2-4}-4=xa-4$
Why am I not getting the intended value?
|
$$a=\frac{\left(\sqrt{x+2}+\sqrt{x-2}\right)^2}{\left(\sqrt{x+2}+\sqrt{x-2}\right)\left(\sqrt{x+2}+\sqrt{x-2}\right)}=\frac{\left(\sqrt{x+2}+\sqrt{x-2}\right)^2}{4},$$ which gives $\sqrt{x+2}+\sqrt{x-2}=2\sqrt{a}$ and from here $\sqrt{x+2}-\sqrt{x-2}=\frac{2}{\sqrt{a}}.$
After summing of last two equalities we obtain $\sqrt{x+2}=\sqrt{a}+\frac{1}{\sqrt{a}}$ or
$x=a+\frac{1}{a}$,
which gives the answer: $-1$ because
$$a^2-ax=a^2-a^2-1=-1$$
|
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|
What can be said if $A^2+B^2+2AB=0$ for some real $2\times2$ matrices $A$ and $B$? Let $A,B\in M_2(\mathbb{R})$ be such that $A^2+B^2+2AB=0$ and $\det A= \det B$. Our goal is to compute $\det(A^2 - B^2)$. According to the chain of comments on Art of Problem Solving, the following statements are true:
*
*$\det(A^2+B^2)+\det(A^2-B^2)=2(\det A^2+\det B^2)$. (Is this well known?)
*(1) $\implies \det(A^2-B^2)=0$.
*If $A,B\in M_2(\mathbb{C})$ satisfy $A^2+B^2+2AB=0$, then $AB=BA$.
*$(A+B)^2=0 \implies \det(A^2-B^2)=0$.
Can someone help me with justifying these statements?
Edit:
Doug M provided an explanation for (1) in the answers. Here is an explanation for (2):
$A^2+B^2+2AB=O_2 \implies A^2+B^2=-2AB$. So $\det(A^2+B^2)=4\det(AB)$. Now using (1), $\det(A^2-B^2)= 2(\det(A^2)-2\det(AB)+\det(B^2)) = 2((\det(A)^2-\det(B)^2) = 0$.
|
Of course, the OP did not understand much about the proposed exercise. In particular, in $M_2(\mathbb{C})$, $A^2 +B^2+2AB=0$ implies $\det(A)=\det(B)$.
In fact, the interesting result is
Proposition. Let $A,B\in M_2(\mathbb{C})$ be such that $(1)$ $A^2+B^2+2AB=0$; then $AB=BA$. Note that the result works only in dimension $2$.
Proof. If $H=A+B$, then $H^2=HA-AH$ and $trace(H^2)=0$. Moreover $H^3=H^2A-HAH=HAH-AH^2$ implies that $trace(H^3)=0$. We deduce that the eigenvalues of $H$ are $0$, $H$ is nilpotent, $H^2=0$, $AH=HA$, $AB=BA$. $\square$
Remark. According to Proposition, $(1)$ implies that $(A+B)^2=0$, $\det(A+B)=0$, $\det(A^2-B^2)=0$; moreover $AB=BA$ implies that $A,B$ are simultaneously triangularizable; since $A+B$ is nilpotent, we may assume that $A=\begin{pmatrix}a&c\\0&b\end{pmatrix},B=\begin{pmatrix}-a&d\\0&-b\end{pmatrix}$ and, consequently, $\det(A)=\det(B)$.
|
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|
How to find $1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{6} + \frac{1}{8} + \frac{1}{9} + \frac{1}{12}+\cdots$? Find the sum of series $$1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{6} + \frac{1}{8} + \frac{1}{9} + \frac{1}{12}+\cdots,$$ where the terms are the reciprocals of positive integers whose only prime factors are two's and three's:
|
$$\sum_{k=0}^{\infty}\frac{1}{2^k}\sum_{k=0}^{\infty}\frac{1}{3^k}=\frac{1}{1-\frac{1}{2}}\cdot\frac{1}{1-\frac{1}{3}}=3$$
|
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|
What is the maximum number of distinct positive integer's square that sums up to $2002$?
What is the maximum number of distinct positive integer's square that sums up to $2002$ ?
My tries:
$$\frac{n(n+1)(2n+1)}{6} = 2002$$
$$\implies n\approx 17 $$ but am clueless as to how to proceed any further.
|
Your estimate gives an upper bound: Even the smallest sum obtainable from $18$ distinct positive squares is $1^2+2^2+\ldots+18^2=2109>2002$, hence we can have at most $17$ squares.
Let's try to find a solution with $17$ squares.
If $k\le 17$ is the first number the sequence $x_1<x_2<\ldots< x_{17}$ omits, then $x_1^2+\ldots +x_{17}^2\ge 2109 -k^2$. We conclude $k\ge 11$ and want to solve
$$ x_{11}^2+\ldots +x_{17}^2=2002-(1^2+2^2+\ldots+10^2)=1617$$
Also, $2109-k^2=2002$ has no integer solution, so that there must be at least a second omitted numbre $l$. Can we find $l>k\ge 11$ such that $1^2+2^2+\ldots+19^2-k^2-l^2=2002$? That is, $k^2+l^2=468$?
First try, $468-11^2$ is not a square. But the second try gives $468-12^2=18^2$.
So
$$2002=1^2+2^2+3^3+4^4+5^2+6^2+7^2+8^2+9^2+10^2+11^2+13^2+14^2+15^2+16^2+17^2+19^2 $$
shows that we can write $2002$ a sum of $17$ distinct positive squares.
|
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|
Second order nonhomgeneous equation euler stuck at nonhomogeneous part there is second order nonhomogeneous differential equation, euler
Im stuck at find nonhomogeneous equation,
$x^2y"-3xy'+4y=log x$ and i have to find the solution
my attempt :
first i found solution for the homogeneous equation, it is a repeated root so
$yp=Ax^2lnx+Bx^2$
here for the nonhomgenous i used variation parameter
$y1= x^2ln x$ and $y2=x^2$
using Wronski and cramer rule:
$yp=u1.y1+u2.y2$
W=$ \left[
\begin{array}{ c c }
x^2lnx & x^2 \\
2xlnx+x & 2x
\end{array} \right] $
$=-x^3$
W1=$ \left[
\begin{array}{ c c }
0 & x^2 \\
log x & 2x
\end{array} \right]$
$=-x^2lnx$
W2=$ \left[
\begin{array}{ c c }
x^2lnx & 0 \\
2xlnx+x & log x
\end{array} \right]$
$=x^2ln^2x$
$u'=\frac{W_1}{W}\frac{-x^2lnx}{-x^3}=\frac{lnx}{x}$
$u'=\frac{W_2}{W}\frac{x^2ln^2x}{-x^3}=\frac{-ln^2x}{x}$
integrate to find u1, u2
$u1=\frac{1}{2}ln^2x$
$u1=-\frac{1}{3}ln^3x$
$yp=u1.y1+u2.y2$
$=\frac{1}{2}ln^2x$$.x^2lnx$ $-\frac{1}{3}ln^3x$ $.x^2$
$=x^2ln^3x\frac{1}{6}$
is this right???
however the book said that the answer is $y=x^2(Alogx+B)+\frac{1}{4}(logx+1)$
why this is not the same??
|
I believe your error was in finding $u_1^{\prime}$ and $u_2^{\prime}$;
you should have
$\displaystyle\frac{\ln x}{x^2}$ instead of logx in the expressions
for $W1$ and $W2$, so that $u_1^{\prime}=-\frac{(\ln x)^2}{x^3}$ and $u_2^{\prime}=\frac{\ln x}{x^3}$.
This will give $u_1=\left(\frac{1}{2}(\ln x)^2+\frac{1}{2}\ln x+\frac{1}{4}\right)(x^{-2})$ and $u_2=-\left(\frac{1}{2}\ln x+\frac{1}{4}\right)(x^{-2})$, so
$y_{p}=u_1y_1+u_2y_2=\frac{1}{2}(\ln x)^2+\frac{1}{2}\ln x+\frac{1}{4}-\frac{1}{2}(\ln x)^2-\frac{1}{4}\ln x=\frac{1}{4}\left(\ln x+1\right)$.
If you use the method of undetermined coefficients instead, with $y_p=A\ln x+B$,
you will find that $A=\frac{1}{4}$ and $B=\frac{1}{4}$.
|
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|
Find minimum of $a+b+c+\frac1a+\frac1b+\frac1c$ given that: $a+b+c\le \frac32$
Find minimum of $a+b+c+\frac1a+\frac1b+\frac1c$ given that: $a+b+c\le \frac32$ ($a,b,c$ are positive real numbers).
There is a solution, which relies on guessing the minimum case happening at $a=b=c=\frac12$ and then applying AM-GM inequality,but what if one CANNOT guess that?!
|
your inequality is equivalent to
$$2\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)+2(a+b+c)\geq 15$$
By $AM-HM$ we get
$$a+b+c\geq \frac{9}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}$$ thus $$2(a+b+c)\geq \frac{18}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}$$ thus we have
$$2\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)+\frac{18}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}\geq 15$$
Setting $$t=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}$$ so we have to prove
$$2t+\frac{18}{t}\geq 15$$ this is equivalent to
$$2t^2-15t+18\geq 0$$ or $$t\le \frac{3}{2}$$ or $$t\geq 6$$
or we can consider the function $$f(t)=2t+\frac{18}{t}$$ and $$f'(t)=2-\frac{18}{t^2}$$ and $$f'(t)=0$$ for $t=3$
|
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|
Why two characterizations of the arcsine distribution are equiv? According to Wikipedia the CDF of arcsine dist. is:
$$F(x)=\frac2{\pi}\arcsin(\sqrt{x})=\frac{\arcsin(2x-1)}{\pi}+\frac12$$
So, why are these two equivalent?
Thanks in advance.
|
One way to prove this trigonometric identity is to $(1)$ first show that the derivatives of the functions on both sides of the equality are the same, and $(2)$ show that the two sides are equal when $x=0.$
\begin{align}
& \frac d {dx}\,\frac 2 \pi \, \arcsin\sqrt x = \frac 2 \pi\cdot\frac 1 {\sqrt{1 - x}} \cdot \frac d {dx} \sqrt x = \frac 2 \pi \cdot \frac 1 {\sqrt{1-x}} \cdot \frac 1 {2\sqrt x} = \frac 1 \pi \cdot \frac 1 {\sqrt{x - x^2}}. \\[12pt]
& \frac d {dx} \, \frac 1 \pi \arcsin(2x-1) = \frac 1 \pi \cdot \frac 1 {\sqrt{1- (2x-1)^2}} \cdot 2 = \frac 2 \pi \cdot \frac 1 {\sqrt{4x-4x^2}} = \frac 1 \pi \cdot \frac 1 {\sqrt{x-x^2}}.
\end{align}
(To make them equal when $x=0,$ one must of course add $\dfrac 1 2$ to the second one; that is omitted above because it's not involved in finding the derivative.)
A more conventional way (without calculus):
\begin{align}
\text{Let } u & =\arcsin\sqrt x. \\[10pt]
\text{Then } \sin u & = \sqrt x \\[10pt]
x & = \sin^2 u \\[10pt]
2x-1 & = 2\sin^2 u - 1 \\
& = -\cos(2u) \text{ (This is the double-angle formula for the cosine.)} \\[10pt]
\arcsin(2x-1) & = \arcsin(-\cos(2u)) = - \arcsin(\cos(2u)) \\[10pt]
& = \arccos(\cos(2u)) - \frac \pi 2 = 2u - \frac \pi 2 = (2\arcsin\sqrt x) - \frac \pi 2.
\end{align}
(That $\arccos(\cos(2u)) = 2u$ relies on the fact that $2u$ is between $0$ and $\pi$.)
|
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|
Rotate the parabola $y=x^2$ clockwise $45^\circ$. I used the rotation matrix to do this and I ended up with the equation: $$x^2+y^2+2xy+\sqrt{2}x-\sqrt{2}y=0$$
I tried to plot this but none of the graphing softwares that I use would allow it.
Is the above the correct equation for a parabola with vertex (0,0) and axis of symmetry $y=x$ ?
$$\left(
\begin{array}{cc}
\cos\theta&-\sin\theta\\
\sin\theta&\cos\theta\\
\end{array}
\right)\left(
\begin{array}{cc}
x\\
y\\
\end{array}
\right)=\left(
\begin{array}{cc}
X\\
Y\\
\end{array}
\right)$$
For a clockwise rotation of $\frac{\pi}{4}$, $\sin{-\frac{\pi}{4}}=\frac{-1}{\sqrt{2}}$ and $\cos{-\frac{\pi}{4}}=\frac{1}{\sqrt{2}}$
$$\left(
\begin{array}{cc}
\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}}\\
-\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}}\\
\end{array}
\right)\left(
\begin{array}{cc}
x\\
y\\
\end{array}
\right)=\left(
\begin{array}{cc}
X\\
Y\\
\end{array}
\right)$$
$$X=\frac{x}{\sqrt{2}}+\frac{y}{\sqrt{2}}$$
$$Y=\frac{-x}{\sqrt{2}}+\frac{y}{\sqrt{2}}$$
$$y=x^2$$
$$\left(\frac{-x}{\sqrt{2}}+\frac{y}{\sqrt{2}}\right)=\left(\frac{x}{\sqrt{2}}+\frac{y}{\sqrt{2}}\right)^2$$
$$\frac{-x}{\sqrt{2}}+\frac{y}{\sqrt{2}}=\frac{x^2}{2}+\frac{2xy}{2}+\frac{y^2}{2}$$
$$-\sqrt{2}x+\sqrt{2}y=x^2+2xy+y^2$$
$$x^2+2xy+y^2+\sqrt{2}x-\sqrt{2}y=0$$
Have I made a mistake somewhere?
|
Your result is corect. This is the plot in geogebra.org
|
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|
Simplifying Complex Numbers in Exponential Form
$(a)$ Suppose $p$ and $q$ are points on the unit circle such that the line through $p$ and $q$ intersects the real axis. Show that if $z$ is the point where this line intersects the real axis, then $z = \dfrac{p+q}{pq+1}$.
$(b)$ Let $P_1 P_2 \dotsb P_{18}$ be a regular $18$-gon. Show that $P_1 P_{10}$, $P_2 P_{13}$, and $P_3 P_{15}$ are concurrent.
I already proved part $(a)$, and used it to prove part $(b)$ by making the diameter of the $18$-gon one unit and defining $P_1 P_{10}$ as the real axis and then plugging both $P_2 P_{13}$ and $P_3 P_{15}$ into $p$ and $q$ to get
$$
z_{1} = \dfrac{\mathrm{e}^{\mathrm{i}\frac{\pi}{9}} + \mathrm{e}^{12\mathrm{i}\frac{\pi}{9}}}{\mathrm{e}^{\mathrm{i}\frac{\pi}{9}} \cdot \mathrm{e}^{12\mathrm{i}\frac{\pi}{9}} + 1}
$$
and
$$
z_{2} = \dfrac{\mathrm{e}^{2\mathrm{i}\frac{\pi}{9}} + \mathrm{e}^{14\mathrm{i}\frac{\pi}{9}}}{\mathrm{e}^{2\mathrm{i}\frac{\pi}{9}} \cdot \mathrm{e}^{14\mathrm{i}\frac{\pi}{9}} + 1} \;.
$$
Then, I set $z_{1}$ and $z_2$ equal and simplified to get
$$
0 = \mathrm{e}^{8\mathrm{i}\frac{\pi}{9}} - \mathrm{e}^{6\mathrm{i}\frac{\pi}{9}} - \mathrm{e}^{5\mathrm{i}\frac{\pi}{9}} + \mathrm{e}^{3\mathrm{i}\frac{\pi}{9}} + \mathrm{e}^{2\mathrm{i}\frac{\pi}{9}} - 1 \;.
$$
However, I can't seem to get past this part to get something like $0=0$ that proves part $(b)$. Any tips?
|
Let $x=e^{i\pi /9}.$ We have $x^9=-1.$ $$\text { We have }\quad x^3-x^6=x^{-3}(x^6-x^9)=x^{-3}(x^6+1)=x^{-3}+x^3.$$ $$\text { So }\quad x^2-x^5=x^{-1}(x^3-x^6)=x^{-1}(x^{-3}+x^3).$$ $$\text { We have } \quad x^8-1=x^{-1}(x^9-x)=x^{-1}(-1-x)=(1+x^{-1})(-1).$$ $$\text { Therefore }\quad x^8-x^6-x^5+x^3+x^2-1=$$ $$=(x^8-1)+(x^3-x^6)+(x^2-x^5)=$$ $$= (1+x^{-1})(-1)+(x^{-3}+ x^3)+x^{-1}(x^{-3}+x^3)=$$ $$=(1+x^{-1})(-1)+(1+x^{-1}) (x^{-3}+x^3)=$$ $$=(1+x^{-1})(-1+x^{-3}+x^3)=$$ $$=(1+x^{-1})\cdot x^{-3}\cdot (1-x^3+x^6)=$$ $$=(1+x^{-1})\cdot x^{-3}\cdot (1+x^9)(1+x^3)^{-1}=0$$ because $1+x^9=0.$
|
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|
Find all possible values of $x+y+z$. Let $x, y, z$ be non-zero integers such that ${x\over y}+{y\over z}+{z\over x}$ is an integer. Find all possible values of $x+y+z$. Please provide a proof with all solutions.
|
If $x = \frac{z}{4}$, and $y= \frac{z}{2}$, then $\frac{x}{y}+\frac{y}{z}+\frac{z}{x} = \frac{1}{2}+\frac{1}{2} + 4 = 5$. Thus, whenever $z = 4t, t \in \mathbb{N}$, the solution set $(x,y,z) = (t,2t,4t)$ satisfies, and thus are infinitely many forms of $x+y+z$ with the given constraints - notably of the form $7t$.
These aren't all the solutions, but it's at least one class.
EDIT: Doing some further work, we can utilize this method to greatly expand the solution class. Say that $x | z$ and that $\frac{x}{y} < 1, \frac{y}{z} < 1$. Thus, in order for the original equation to be an integer, $\frac{x}{y} + \frac{y}{z} = 1$. Then for some $m,n \in \mathbb{N}$ where $m< n$, $\frac{x}{y} = \frac{m}{n}, \frac{y}{z} = \frac{n-m}{n}$.
Thus $nx = my$ and $ny = (n-m)z$, implying that $y = (1-\frac{m}{n})z$ and that $x = (\frac{m}{n} - \frac{m^{2}}{n^{2}})z$. Therefore $x+y+z = (\frac{m}{n} - \frac{m^{2}}{n^{2}})z + (1-\frac{m}{n})z + z = (2-\frac{m^{2}}{n^{2}})z$. Set $z = n^{2}t$ and the sum becomes $(2n^{2}-m^{2})t$, where $m,n,t \in \mathbb{N}$.
Thus the solution set $(mnt - m^{2}t,n^{2}t-mnt,n^{2}t)$ suffices to greatly expand the number of possible sums of $x+y+z$. In short, any multiple of any integer of the form $2n^{2} - m^{2}$ where $m < n$ can be expressed as a sum of $x,y,$ and $z$.
|
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|
Find n so that the following converges $\int_1^{+ \infty} \left( \frac{nx^2}{x^3 + 1} - \frac 1 {13x + 1} \right) dx$ Question
Determine $n$ such that the following improper integral is convergent
$$ \int_1^{+ \infty} \left(
\frac{nx^2}{x^3 + 1}
-\frac{1}{13x + 1}
\right) dx
$$
I'm not sure how to go about this.
Working
This is convergent if
$$
\lim_{b \to + \infty} \int_1^b \left(\frac{nx^2}{x^3 + 1} - \frac{1}{13x + 1} \right) dx
$$
exists.
The indefinite integral is
\begin{equation*}
\begin{aligned}
\int \left(\frac{nx^2}{x^3 + 1} - \frac{1}{13x + 1} \right)dx
& = \int \left(\frac{nx^2}{x^3 + 1}
\right) - \int \left(\frac{1}{13x + 1} \right)dx \\
&= \frac{n}{3} \cdot \ln(x^3 + 1) - \frac{1}{13} \cdot \ln(13x + 1)
\end{aligned}
\end{equation*}
Which gives
\begin{equation*}
\begin{aligned}
&\lim_{b \to + \infty}
\int_1^b \left(\frac{nx^2}{x^3 + 1} - \frac{1}{13x + 1} \right) dx \\
&= \lim_{b \to + \infty} \left[\frac{n}{3} \cdot \ln(x^3 + 1) - \frac{1}{13} \cdot \ln(13x + 1) \right]_1^b \\
&= \lim_{b \to + \infty}
\left(\left[\frac{n}{3} \cdot \ln(b^3 + 1) - \frac{1}{13} \cdot \ln(13b + 1) \right] - \left[\frac{n}{3} \cdot \ln(2) - \frac{1}{13} \cdot \ln(14) \right] \right)
\\ &= \lim_{b \to + \infty}
\left(\frac{n}{3} \cdot \ln(b^3 + 1) - \frac{1}{13} \cdot \ln(13b + 1) - \frac{n}{3} \cdot \ln(2) + \frac{1}{13} \cdot \ln(14) \right)
\\ &= \lim_{b \to + \infty}
\left(
\frac{n}{3}
\left(
\ln(b^3 + 1) - \ln(2)
\right)
- \frac{1}{13}
\left(
\ln(13b + 1)
- \ln(14)
\right)
\right)
\\ &= \lim_{b \to + \infty}
\left(
\frac{n}{3}
\left(
\ln \left(\frac{b^3 + 1}{2}\right)
\right)
- \frac{1}{13}
\left(
\ln \left(\frac{13b + 1}{14}\right)
\right)
\right)
\end{aligned}
\end{equation*}
I've tried to use L'Hopital's from here as I have the form $(+ \infty ) - ( +
\infty)$. But things went pretty south.
So I'm sure there's a better approach.
|
Hint: note $\ln (b^3+1) \sim \ln b^3=3\ln b$ and $\ln (13b+1) \sim \ln 13b=\ln 13+\ln b$ for $b\to+\infty$.
Hence: $$\lim_\limits{b\to +\infty} [\frac{n}{3} \ln (b^3+1)-\frac{1}{13}\ln (13b+1)]=\lim_\limits{b\to+\infty} [\left(n-\frac{1}{13}\right) \ln b - \frac{\ln 13}{13}]=\begin{cases} -\frac{\ln 13}{13}, \ if \ n=\frac{1}{13} \\ -\infty, \ if \ n<\frac{1}{13} \\ +\infty, \ if \ n>\frac{1}{13}\end{cases}.$$
|
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|
Clarification for a simple argument in an inequality Suppose $a,b,c,d \in \mathbb{R}^+$ are such that $$ a \leq b \\ c \leq d \\ ac = bd $$ then is the following true? $$ a= b \\ c = d $$
Attempt:
Suppose not true, i.e. $a < b$ or $c < d$, then we have $ac < bd$ thus contradicting the fact that $ac=bd$. Thus $a= b$ and $c = d$.
So, is my argument correct? Sorry for posting this if this is a trivial fact.
|
Okay, FINAL final answer:
If $0 \le a \le b$ and $0 \le c \le d$ then
$ac \le bc$ with equality holding only if $a = b$ or $c = 0$.
And $bc \le bd$ with equality holding only if $c = d$ or $b= 0$.
So $ac \le bd$ with equality holding only if one of the four occur:
1) $a = b$ and $c = d$
2) $a =b$ and $b = 0$ so $a = b =0$.
3) $c = 0$ and $c= d$ so $c = d = 0$.
4) $c =0$ and $b= 0$ and $0 \le a \le b = 0$ so $a = b = c = 0$
So if $ac = bd$ then either i) $a=b$ and $c=d$ or ii) $a=b=0$ or $c=d=0$.
And if $ac = bd \ne 0$ then $a=b$ and $c=d$.
Otherwise $ac < bd$.
|
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|
Proof by strong induction of $(a + b\sqrt{2})^n = a_n + b_n \sqrt{2}$ I'm not very familiar with proofs by strong induction. I have a sketch for this one but Iot quite sure about it's validaty.
Let $a$, $b$, $a_n$, $b_n$ be integers such that.
$$(a + b\sqrt{2})^n = a_n + b_n \sqrt{2}$$
where $a$ is the integer closest to $b\sqrt{2}$. Prove that $a_n$ is the integer closest to $b_n\sqrt{2}$.
My sketch:
Suppose the statement holds for all $k\le n $ we have to show that it holds for $n+1$
$(a + b\sqrt{2})^n = a_n + b_n \sqrt{2} \Rightarrow$
$ (a + b\sqrt{2})^{n+1} = (a_n + b_n \sqrt{2})(a + b\sqrt{2}) = a a_n + ab_n\sqrt{2} + \sqrt{2} (a_nb + bb_n\sqrt{2}) $
Since $a_n$ is the integer closest to $b_n \sqrt{2}$, $aa_n$ is the integer closest to $ab_n\sqrt{2} $ and $ba_n$ is the integer closest to $b b_n \sqrt{2}$.
My intuition tells that I can prove this with what I wrote, but I can't progress now.
|
Hint.
Observe that, if
$$
(a+b\sqrt{2})^n=a_n+b_n\sqrt{2},
$$
then (it can be shown inductively)
$$
(a-b\sqrt{2})^n=a_n-b_n\sqrt{2},
$$
and if $|a-b\sqrt{2}|<1/2$, then clearly, $(a-b\sqrt{2})^n<1/2$.
|
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|
Find $\tan(π/16)+\tan(5π/16)+\tan(9π/16)+\tan(13π/16)$ I had to calculate $\tan(π/16)+\tan(5π/16)+\tan(9π/16)+\tan(13π/16)$
I tried writing $π/16=x$ and then writing the sum as $\tan x+\tan(9x-4x)+\tan9x+\tan(9x+4x)$ and then simplifying using $\tan(x+y)$ and $\tan(x-y)$ but it didn't simplify to anything, can anybody please give a hint on how should I approach this.
|
$$a=\tan(\frac{\pi}{16})+\tan(\frac{5\pi}{16})+\tan(\frac{9\pi}{16})+\tan(\frac{13\pi}{16})\\=
\tan(\frac{\pi}{16})+\tan(\frac{5\pi}{16})+\tan(\frac{1\pi}{16}+\frac{\pi}{2})+\tan(\frac{5\pi}{16}+\frac{\pi}{2})\\=\\
\tan(\frac{\pi}{16})-\cot(\frac{\pi}{16})+\tan(\frac{5\pi}{16})-\cot(\frac{5\pi}{16})\\=?$$
now let me note that $\color{red} {2\cot(2x)=\cot x- \tan x}\tag{1}$ so
$$a=-(-\tan(\frac{\pi}{16})+\cot(\frac{\pi}{16})-\tan(\frac{5\pi}{16})+\cot(\frac{5\pi}{16}))\\=
-(2\cot(\frac{2\pi}{16})+2\cot(\frac{10\pi}{16}))\\=$$can you go on ?
$$a=-2(\cot(\frac{\pi}{8})+\cot(\frac{5\pi}{8}))\\=-2(\cot(\frac{\pi}{8})+\cot(\frac{\pi}{8}+\frac{\pi}{2}))\\\text{from formula (1)}\\
-2(\cot(\frac{\pi}{8})-\tan(\frac{\pi}{8}))\\=-2(2\cot(2(\frac{\pi}{8}))\\=-4$$
|
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|
If $\cos A+\cos B+\cos C=\frac{3}{2}$, prove $ABC$ is an equilateral triangle If in $\Delta ABC$ $$\cos A+\cos B+\cos C=\frac{3}{2}$$prove that it is an equilateral triangle without using inequalities. I tried using the cosine rule as follows:
$$\frac{b^2+c^2-a^2}{2bc}+\frac{a^2+c^2-b^2}{2ac}+\frac{a^2+b^2-c^2}{2ab}=\frac{3}{2}$$ $\implies$
$$a(b^2+c^2-a^2)+b(a^2+c^2-b^2)+c(a^2+b^2-c^2)=3abc$$ $\implies$
$$(a+b+c)(a^2+b^2+c^2)=2(a^3+b^3+c^3)+3abc$$
Any clues as to how I can take it from here?
|
A different kind of trick : I thought you will like it.
So multiply the above equation by $a,b,c$ and add these up. Next , use the identity that $c = a \cos B + b \cos A$, and likewise for the other sides. After pairing these, you get:
()
$$
\frac{3(a+b+c)}{2} = a \cos A + b \cos B + c \cos C + a + b +c
$$
which then simplifies to:
$$
\frac{a+b+c}{2} = a\cos A + b\cos B + c \cos C \leq \frac{(a+b+c)(\cos A + \cos B + \cos C)}3
$$
by Chebyshev's inequality. But then, equality is attained here : this happens only under certain conditions. I leave you to conclude.
|
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|
Find the area bounded by $ \ x^4+y^4=4(x^2+y^2) \ $ . Find the area bounded by $ \ x^4+y^4=4(x^2+y^2) \ $ .
Answer:
The graph is above :
Since the region is symmetrical ,
$ Area =4 \times \int_{0}^{2} \int_{0}^{\sqrt{2+\sqrt{4x^2-x^4+4}}} dxdy $
Am I right ? Is there any help ?
|
According to your own sketch, the integral you wrote would not cover the area $|y| > 2$, never mind that the order of integration does not match the limits of integration. So the setup is problematic in multiple ways.
The suggestion to use polar coordinates is worth exploring. Let $(x,y) = (r \cos \theta, r \sin \theta)$ so that we have $$x^2 + y^2 = r^2,$$ and $$x^4 + y^4 = (x^2 + y^2)^2 - 2 x^2 y^2 = r^4 - 2 r^4 \sin^2 \theta \cos^2 \theta = r^4 (1 - \tfrac{1}{2} \sin^2 2\theta).$$ Thus we have $$0 = (x^4 + y^4) - 4(x^2 + y^2) = r^4 (1 - \tfrac{1}{2} \sin^2 2\theta) - 4r^2,$$ and eliminating the trivial solution $r = 0$, we get $$r^2 = 4(1 - \tfrac{1}{2} \sin^2 2\theta)^{-1}.$$ In the first quadrant, $0 \le \theta \le \pi/2$ and we can write $$\frac{A}{4} = \int_{\theta=0}^{\pi/2} \frac{1}{2}r^2 \, d\theta = 2 \int_{\theta=0}^{\pi/2} \frac{1}{1-\frac{1}{2}\sin^2 2\theta} \, d\theta = 8\int_{\theta=0}^{\pi/2} \frac{d\theta}{3+\cos 4\theta},$$ and the rest is quite straightforward with the standard trigonometric substitutions.
|
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|
How to simplify an expression involving several square roots without a calculator? $$\frac{5 \sqrt{7}}{4\sqrt{3\sqrt{5}}-4\sqrt{2\sqrt{5}}}- \frac{4 \sqrt{5}}{\sqrt{3\sqrt{5}}-\sqrt{2\sqrt{5}}}$$
This type of questions are common in the university entrance examinations in our country but the calculators are not allowed can someone help me to find the way to simplify the expression.
|
Assuming that your expression contains a typo :)
Actually we have to simplify
$$\frac{5 \sqrt{5}}{4\left(\sqrt{3\sqrt{5}}-\sqrt{2\sqrt{5}}\right)}- \frac{4 \sqrt{5}}{\sqrt{3\sqrt{5}}-\sqrt{2\sqrt{5}}}$$
Notice that both fractions can be simplified multiplying numerator and denominator by
$\left(\sqrt{3\sqrt{5}}+\sqrt{2\sqrt{5}}\right)$
Indeed
$$\left(\sqrt{3\sqrt{5}}-\sqrt{2\sqrt{5}}\right)\left(\sqrt{3\sqrt{5}}+\sqrt{2\sqrt{5}}\right)=3\sqrt{5}-2\sqrt{5}=\sqrt{5}$$
the expression becomes
$$\frac{\left(5 \sqrt{5}\right)\left(\sqrt{3\sqrt{5}}+\sqrt{2\sqrt{5}}\right)}{4\sqrt 5}- \frac{4 \sqrt{5}\left(\sqrt{3\sqrt{5}}-\sqrt{2\sqrt{5}}\right)}{\sqrt 5}=\\=\frac{5 \left(\sqrt{3\sqrt{5}}+\sqrt{2\sqrt{5}}\right)}{4}- 4\left(\sqrt{3\sqrt{5}}-\sqrt{2\sqrt{5}}\right)=\frac{21 \sqrt{2 \sqrt{5}}}{4}-\frac{11 \sqrt{3 \sqrt{5}}}{4}$$
If the $\sqrt{7}$ is not a typo the numerator is a bit more complicated, but the basics are always the same
|
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|
Is $\displaystyle\sum_{k=1}^n \frac{k \sin^2k}{n^2+k \sin^2k}$ convergent? Let $\displaystyle x_n=\sum_{k=1}^n\frac{k \sin^2k}{n^2+k \sin^2k}$ for all $n>0$. How can we prove that $(x_n)$ is convergent?
|
$$
\begin{align}
\sum_{k=1}^n\frac{k\sin^2(k)}{n^2+k\sin^2(k)}
&=n-\sum_{k=1}^n\frac{n^2}{n^2+k\sin^2(k)}\\
&=n-\sum_{k=1}^n\frac1{1+\frac{k\sin^2(k)}{n^2}}\\
&=n-\sum_{k=1}^n\left(1-\frac{k\sin^2(k)}{n^2}+O\!\left(\frac{k^2}{n^4}\right)\right)\\
&=\frac1{n^2}\sum_{k=1}^nk\sin^2(k)+O\!\left(\frac1n\right)\\
&=\frac1{n^2}\sum_{k=1}^nk\,\frac{1-\cos(2k)}2+O\!\left(\frac1n\right)\\
&=\frac14+O\!\left(\frac1n\right)
\end{align}
$$
Since
$$\newcommand{\Re}{\operatorname{Re}}
\begin{align}
\sum_{k=1}^nk\cos(2k)
&=\Re\left(\sum_{k=1}^nke^{2ik}\right)\\
&=\Re\left(\tfrac{e^{2i(n+1)}\left(ne^{2i}-n-1\right)+e^{2i}}{\left(1-e^{2i}\right)^2}\right)\\[6pt]
&=O(n)
\end{align}
$$
|
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|
Show that $ \sum\limits_{k=1}^\infty \frac{3^k-2^k}{k\cdot6^k}=\ln\frac43$
$\sum_{k=1}^\infty \frac{3^k-2^k}{k\cdot6^k}$ i must prove this sum converges to $\ln(4/3)$.
i tried to write expresion :$\frac{3^k-2^k}{k\cdot6^k}=\frac 1 k \left(\frac 1 {2^k-3^k}\right)$ and make two sums but i tink this sums is a dezvoltation in series of a function. $\ln(4/3)=\sum_{k=1}^\infty \frac{(-1/3)^k} k$ => we must prove $\sum_{k=1}^\infty (-1/3)^k$ is equal with $\sum_{k=1}^\infty (1/2^k-3^k)$ and that is not inductive and that is hard to. Have any ideea?
|
Work with two sums separately:
\begin{align}
\sum_{k=1}^\infty \frac{3^k-2^k}{k\cdot6^k} & = \sum_{k=1}^\infty \frac{(3/6)^k} k - \sum_{k=1}^\infty \frac{(2/6)^k} k \\[10pt]
& = \sum_{k=1}^\infty \frac{(1/2)^k} k - \sum_{k=1}^\infty \frac{(1/3)^k} k \\[10pt]
& = \sum_{k=1}^\infty \frac{(-1)^k(-1/2)^k} k - \sum_{k=1}^\infty \frac{(-1)^k(-1/3)^k} k \\[10pt]
& = \sum_{k=1}^\infty \frac{(-1)^k(x-1)^k} k - \sum_{k=1}^\infty \frac {(-1)^k(y-1)^k} k \text{ where } x = \frac 1 2 \text{ and } y = \frac 2 3 \\[10pt]
& = \ln x - \ln y \\[10pt]
& = \ln \frac 1 2 - \ln \frac 2 3 = \ln \frac{1/2}{2/3} = \ln \frac 3 4
\end{align}
|
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|
If $x^2+x+1$ is a factor of $f (x)= ax^3+bx^2+cx+d$ , then the real root of $f (x)=0$ is If $x^2+x+1$ is a factor of $f (x)= ax^3+bx^2+cx+d$ , then the real root of $f (x)$ is
A) $-d/a$
B) $d/a$
C) $a/d $
D) none of these
My try
I take $f (x) = ( x^2+x+1)(x+1) = x^3+2x^2+2x+1$
Real root $x= -d/a =-1$
And also
$f (x) = ( x^2+x+1)(x-2) = x^3-x^2-x-2$
Real root $x = -d/a =2$
But how can I prove it in general?
|
If you divide $f (x)= ax^3+bx^2+cx+d$ to $x^2+x+1$ : you will have
$$f (x)= ax^3+bx^2+cx+d=(x^2+x+1)(ax+(b-a))+x(a+c-b+a)+(a-b+d)$$ when remainder is $0 $ so $$ax+(b-a)=0 \to x=\frac{a-b}{a} \tag{1}\\
\forall x:\begin{cases}a+c-b+a=0 \\a-b+d=0\end{cases}$$ now from 3rd equation we have $d=-(a-b)$ put into (1)
$$x=\frac{a-b}{a}=x=\frac{-d}{a}$$
|
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|
How to turn the ellipse $x^2 - xy + y^2 - 3y - 1 = 0$ to the canonical form using an isometric transformation? There is an exam problem I'm having trouble with, it is as follows:
Turn the equation $x^2 - xy + y^2 - 3y -1 = 0$ into the canonical form using an isometric transformation and write down the transformation.
Comparing to the general equation for conic sections I see that this is a rotated ellipse. I assume that the solution would be to multiply a rotation matrix (to align the ellipse's axes to the $x$ and $y$ axes) and a translation matrix (to bring the ellipse's center to the origin), and then somehow apply the result to the ellipse.
How would I go about doing this? How do I get the coordinates of the ellipse's center and the angle of its rotation to construct the matrices, and then how would I apply the transformation to the ellipse itself?
|
First center the equation by definting the function $f(x,y) = x^2 - xy + y^2 - 3y - 1 = 0$ and solving the following system of equations for $(x,y)$
$$\left. \begin{aligned}
\frac{\partial f(x,y)}{\partial x} & = 2x-y = 0 \\
\frac{\partial f(x,y)}{\partial y} & = -x+2y-3 = 0 \\
\end{aligned} \right\} \pmatrix{x & y} = \pmatrix{1 & 2} $$
So now we reset the coordinates to $\pmatrix{x&y} \rightarrow \pmatrix{x+1&y+2}$
The new equation is $g(x,y) = x^2-x y + y^2 -4 = 0$
Now to re-orient the conic. Use $\pmatrix{x&y} \rightarrow \pmatrix{x \cos \theta - y \sin \theta & x \sin\theta + y \cos \theta}$ and set the coefficient of $x y$ equal to zero.
$$ g = \left( 1- \frac{\sin 2 \theta}{2} \right) x^2 + \left( 1 + \frac{\sin 2 \theta}{2} \right) y^2 +2 \left( -\frac{\cos 2\theta}{2} \right) x y -4 =0 $$
$$ \left. -\frac{1}{2}\cos(2 \theta)=0 \right\} \theta =\pm \frac{\pi}{4} $$
$$\boxed{ d(x,y) = \frac{1}{2} x^2 + \frac{3}{2} y^2 -4 =0 }$$
|
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|
What are the roots of this equation? I have a quadratic equation $ ax^2 +bx+c =0 $, where $ a,b,c $ all are positives and are in Arithmetic Progression.
Also, the roots $\alpha$ and $\beta$ are integers.
I need to find out $ \alpha + \beta + \alpha\beta $.
I have tried taking $ a = a' - d , b = a' , c = a' + d $ because I have supposed $ a' $ is the first term and $ d $ is common difference, I used sum of roots and product of roots rule, but nothing helped..
|
We see that the equation $$ax^2+(a+d)x+a+2d=0$$
has integer roots, which says $1+\frac{d}{a}\in\mathbb Z$.
Let $\frac{d}{a}=k$.
Hence, we have $$x^2+(1+k)x+1+2k=0$$
has integer roots, which gives
$$(1+k)^2-4(1+2k)=n^2,$$ where $n$ is non-negative integer number,
which gives
$$k^2-6k-3=n^2$$ or
$$(k-3)^2=n^2+12$$ or
$$(k-3-n)(k-3+n)=12$$ and the rest is smooth:
Since $n$ is non-negative integer, we obtain two cases only.
*
*$n-k+3=6$ and $n+k-3=-2$, which gives $k=-1$;
*$n-k+3=-2$ and $n+k-3=6$, which gives $k=7$.
|
{
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|
If $\sum_{k=0}^n \frac{A_k}{x+k}=\frac{n!}{x(x+1)(x+2)(x+3)\cdots(x+n)}$ Then find $A_7$ If $$\sum_{k=0}^n \frac{A_k}{x+k}=\frac{n!}{x(x+1)(x+2)(x+3)\cdots(x+n)}$$ Then find $A_7$
My Try:
I have considered a function
$$f(x)=x^{A_0}(x+1)^{A_1}(x+2)^{A_2}\cdots(x+n)^{A_n}$$
taking natural log on both sides and then differentiating we get
$$\frac{f'(x)}{f(x)}=\sum_{k=0}^n \frac{A_k}{x+k}$$
hence
$$\frac{f'(x)}{f(x)}=\frac{n!}{x(x+1)(x+2)(x+3)\cdots(x+n)}$$
any clue here?
|
By using partial fractions, chance upon the solution $A_{0}=1,$ and $A_{1}=-1$. For $n>1$, in particular for all $n=7,$ it seems that
$$A_{7}=(x+7)(\frac{(7)!}{x(x+1)...(x+7)}-\frac{6!}{x(x+1)...(x+6)})
\\=6!\frac{7-(x+7)}{x(x+1)...(x+6)}=6!\frac{-x}{x(x+1)...(x+6)}
\\=-\frac{6!}{(x+1)...(x+6)}.$$
|
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|
Find $\lvert m\rvert$ given $\sqrt[3]{m+9}=3+\sqrt[3]{m-9}$ I've already found the solution, but I was wondering if there is a faster or alternative method. My solution is found below:
$$\sqrt[3]{m+9}=3+\sqrt[3]{m-9}$$
$$(m+9)^\frac{1}{3}-(m-9)^\frac{1}{3}=3$$
$$\bigl((m+9)^\frac{1}{3}-(m-9)^\frac{1}{3}\bigr)^3=27$$
$$(m+9)-3(m+9)^\frac{2}{3}(m-9)^\frac{1}{3}+3(m+9)^\frac{1}{3}(m-9)^\frac{2}{3}-(m-9)=27$$
$$-3(m+9)^\frac{2}{3}(m-9)^\frac{1}{3}+3(m+9)^\frac{1}{3}(m-9)^\frac{2}{3}=9$$
$$-(m+9)^\frac{2}{3}(m-9)^\frac{1}{3}+(m+9)^\frac{1}{3}(m-9)^\frac{2}{3}=3$$
$$-(m+9)^\frac{1}{3}(m-9)^\frac{1}{3}\bigl((m+9)^\frac{1}{3}-(m-9)^\frac{1}{3}\bigr)=3$$
$$-(m+9)(m-9)\bigl((m+9)^\frac{1}{3}-(m-9)^\frac{1}{3}\bigr)^3=27$$
$$-(m+9)(m-9)(27)=27$$
$$m^2-81=-1$$
$$m=\pm4\sqrt{5}$$
$$\lvert m\rvert=4\sqrt{5}$$
In the third line, I cubed both sides, and then in the forth line, I expanded the left side of the equation using the binomial theorem. Is there a faster or alternative way to do this type of question? If in the question there was a higher-index root (ie. instead of the cube roots there is a higher index), I don't think my method would work, because it would take too long to apply the binomial theorem. How would solve a question in this form?
|
Go ahead and cube:
$$m+9=27+27\sqrt[3]{m-9}+9\sqrt[3]{(m-9)^2}+m-9.$$
Denote: $\sqrt[3]{m-9}=t$ to get:
$$t^2+3t+1=0 \Rightarrow t=\frac{-3\pm \sqrt{5}}{2}$$
Now:
$$m-9=t^3$$
will produce the answers.
|
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|
Find the real and imaginary part of z let $z=$ $$ \left( \frac{1 + \sin\theta + i\cos\theta}{1 + \sin\theta - i\cos\theta} \right)^n$$
Rationalizing the denominator:
$$\frac{1 + \sin\theta + i\cos\theta}{1 + \sin\theta - i\cos\theta}\cdot\left( \frac{1 + \sin\theta + i\cos\theta}{1 + \sin\theta + i\cos\theta}\right) = \frac{(1 + \sin\theta + i\cos\theta)^2}{(1 + \sin\theta)^2 + \cos^2\theta}$$
$$=\frac{(1 + \sin\theta)^2 + \cos^2\theta + 2i(1 + \sin\theta)\cos\theta }{(1 + \sin\theta)^2 + \cos^2\theta}$$
thus
$$x = \frac{(1 + \sin\theta)^2 + \cos^2\theta }{(1 + \sin\theta)^2 + \cos^2\theta} $$
$$y= \frac{2i(1 + \sin\theta)\cos\theta }{(1 + \sin\theta)^2 + \cos^2\theta}$$
According to the binomial theorem,
$$(x+y)^n = \sum_{k=0}^n \binom{n}{k} x^{n-k}y^k$$
we get
$$z = \frac{1}{(1 + \sin\theta)^2 + \cos^2\theta}\sum_{k=0}^n \binom{n}{k} ((1 + \sin\theta)^2 + \cos^2\theta)^{n-k}\cdot(2i(1 + \sin\theta)\cos\theta)^k$$
...and that is where I'm stuck. What do you think? Thanks for the attention.
|
Let's get a picture:
$|1 + \sin\theta + i\cos\theta| = |1 + \sin\theta - i\cos\theta|$
$\left|\frac {1 + \sin\theta + i\cos\theta}{1 + \sin\theta - i\cos\theta}\right| = 1$
When we divide complex numbers the argument of the ratio equals the difference between the arguments.
let $\phi = \frac {\pi}{2} - \theta$
$1 + \sin\theta + i\cos\theta = 1 + \cos\phi + i\sin\phi$
$\frac {1 + \cos \phi + i\sin\theta}{1 + \cos\phi - i\sin\phi} = e^{\phi i}$
$\left(\frac {1 + \cos \phi + i\sin\phi}{1 + \cos\phi - i\sin\phi}\right)^n = e^{n\phi i}$
or
$e^{n(\frac \pi 2 - \theta) i}$
|
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|
Finding $n$th power of a $3\times 3$ matrix Find the $A^n$ if $$A=\begin{bmatrix}1 & a & b \\0 & 1 &a\\0 &0 &1\end{bmatrix}$$
I tried inductive method to show $$A^n=\begin{bmatrix}1 & na & nb+\frac{n(n-1)}{2}a^2 \\0 & 1 &na\\0 &0 &1\end{bmatrix}$$
now : My question is : Is there other method (idea ) to find $A^n$ ?
Thanks in advance.
Can the idea apply for $$A=\begin{bmatrix}1 & a & b \\0 & 1 &c\\0 &0 &1\end{bmatrix}$$ when $c \neq a$ ?
|
$$A=I+aJ+bJ^2$$
where
$$J=\pmatrix{0&1&0\\0&0&1\\0&0&0}.$$
Then $J^3=O$.
Now
$$A^n=\sum_{k=0}^n{n\choose k}(aJ+bJ^2)^n
=I+n(aJ+bJ^2)+{n\choose 2}(aJ+bJ^2)^2$$
as $(aJ+bJ^2)^3=O$. But $(aJ+bJ^2)^2=a^2J^2$ so
$$A^n=I+naJ+nbJ^2+\frac{n(n-1)}2a^2J^2
=\pmatrix{1&na&nb+\frac{n(n-1)}{2}a^2\\0&1&na\\0&0&1}.$$
|
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|
$x \equiv 1$ mod $4$ iff $x^3 \equiv 1$ mod $4$. This is my solution; I was wondering if there was a better or a neater solution than this.
Suppose $x \equiv 1$ (mod $4$).
Then $x = 4k + 1$ for some integer $k$.
Hence, $x^3 = 4(integer) + 1 \equiv 1$ (mod $4$).
Now suppose $x^3 \equiv 1$ (mod $4$) $\iff$ $x^3 - 1 \equiv 0$ (mod $4$).
$(x-1)(1+x+x^2) \equiv 0$ (mod $4$).
By the Division Algorithm, $x$ is in the form of $4k + i$ for some $i \in \{0,1,2,3\}$.
Hence,
$(4k+i-1)(1+4k+i+(4k+i)^2)\equiv 0$ (mod $4$).
$\iff (i-1)(1+i+i^2) \equiv 0$ (mod $4$). \quad $(\star)$
If $i = 1$ then this is shown to true, and implies $x\equiv 1$ (mod $4$).
If $i = 0,2,3$, then the left hand side in $(\star)$ equals $3,3,2$ (mod $2$) respectively which contradicts the orignal assumption.
Thanks!
|
$$x^3-1=(x-1)(1+x+x^2)$$
If $4$ divides $x-1$, then from the factorization above we can see that it also divides $x^3-1$.
If $4$ divides $x^3-1$, then $x$ must be odd. This implies that $1+x+x^2$ must also be odd, which means that $x-1$ must be divisible by $4$.
|
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|
Solving $\frac{1}{(x-1)} - \frac{1}{(x-2)} = \frac{1}{(x-3)} - \frac{1}{(x-4)}$. Why is my solution wrong? I'm following all hitherto me known rules for solving equations, but the result is wrong. Please explain why my approach is not correct.
We want to solve:
$$\frac{1}{(x-1)} - \frac{1}{(x-2)} = \frac{1}{(x-3)} - \frac{1}{(x-4)}\tag1$$
Moving the things in RHS to LHS:
$$\frac{1}{(x-1)} - \frac{1}{(x-2)} - \frac{1}{(x-3)} + \frac{1}{(x-4)} = 0\tag2$$
Writing everything above a common denominator:
$$\frac{1}{(x-4)(x-1)(x-2)(x-3)}\bigg[(x-2)(x-3)(x-4) - (x-1)(x-3)(x-4) - (x-2)(x-1)(x-4) + (x-1)(x-2)(x-3)\bigg] = 0\tag3$$
Multiplying both sides with the denominator to cancel the denominator:
$$(x-2)(x-3)(x-4) - (x-1)(x-3)(x-4) - (x-2)(x-1)(x-4) + (x-1)(x-2)(x-3) = 0\tag4$$
Multiplying the first two factors in every term:
$$(x^2-3x-2x+6)(x-4) - (x^2-3x-x+3)(x-4) - (x^2-x-2x+2)(x-4) + (x^2-2x-x+2)(x-3) = 0\tag5$$
Simplifying the first factors in every term:
$$(x^2-5x+6)(x-4) - (x^2-4x+3)(x-4) - (x^2-3x+2)(x-4) + (x^2-3x+2)(x-3) = 0\tag6$$
Multiplying factors again:
$$(x^3-4x^2-5x^2+20x+6x-24) - (x^3-4x^2-4x^2+16x+3x-12) - (x^3-4x^2-3x^2-12x+2x-8) + (x^3-3x^2-3x^2+9x+2x-6) = 0\tag7$$
Removing the parenthesis yields:
$$x^3-4x^2-5x^2+20x+6x-24 - x^3+4x^2+4x^2-16x-3x+12 - x^3+4x^2+3x^2+12x-2x+8 + x^3-3x^2-3x^2+9x+2x-6 = 0\tag8$$
Which results in:
$$28x - 10 = 0 \Rightarrow 28x = 10 \Rightarrow x = \frac{5}{14}\tag9$$
which is not correct. The correct answer is $x = \frac{5}{2}$.
|
The arithmetic error is already pinpointed. The result is a special case of the Theorem below
since we have $\ \overbrace{x\!-\!1\, +\, x\!-\!4}^{\Large a\ \ +\ \ b} \, =\, \overbrace{x\!-\!2\, +\, x\!-\!3}^{\Large c\ \ +\ \ d}\ =\ \overbrace{\color{#c00}{2x\!-\!5}}^{\LARGE\color{#c00}s}\ \ $ and $\ \ \color{#0a0}{\overbrace{{x\!-\!2\neq x\!-\!1,\,x\!-\!4}}^{\Large c\ \neq\ a,\,b\quad\ \ }}\,$
therefore $\quad\ \ \ \dfrac{1}{x\!-\!1}+\dfrac{1}{x\!-\!4} = \dfrac{1}{x\!-\!2}+\dfrac{1}{x\!-\!3}\!\iff\! \color{#c00}{2x\!-\!5} = 0$
Theorem $\ \ $ If $\ a\!+\!b = c\!+\!d =: \color{#c00}s\ $ then $\ \dfrac{1}a+\dfrac{1}b =\dfrac{1}c+\dfrac{1}d \iff \color{#c00}{s = 0}\,$ or $\,\color{#0a0}{c = a}\,$ or $\,\color{#0a0}{c=b}$
$\begin{align}{\bf Proof}\qquad \dfrac{1}a+\dfrac{1}b\, &=\,\dfrac{1}c+\dfrac{1}d\\[.3em]
\iff \dfrac{s}{ab}&=\dfrac{s}{cd}\quad{\rm by} \ \ \ \color{#c00}s = a+b = c+d\\[.3em]
\iff\, \ \ 0\, &=\, s(ab-cd)\\[.3em]
&=\, s(ab-c(a+b-c))\\[.3em]
&=\, \color{#c00}s\color{#0a0}{(a-c)(b-c)}\\
\end{align}$
|
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|
Fast/smart way to write polar curve in cartesian Is there a fast way to write the curve:
$$r=\frac{a}{1-\frac{1}{\sqrt{2}}\cos(\theta)}$$
as a cartesian curve $f(x,y)=0$?
It seems I can take
$$r(1-\frac{1}{\sqrt{2}}\cos(\theta)) = a$$
$$r-\frac{x}{\sqrt{2}}=a$$
$$\sqrt{x^2+y^2}-\frac{x}{\sqrt{2}}=a$$
$$x^2+y^2=(a+\frac{x}{\sqrt{2}})^2$$
and then expand out, and complete the square. But I seem to get an error. Perhaps there is a smart way to do this?
|
HINT: we get
$$\sqrt{x^2+y^2}=\frac{a}{1-\frac{1}{\sqrt{2}}\frac{x}{\sqrt{x^2+y^2}}}$$
and we get by squaring
$$2(x^2+y^2)=2a^2+x^2+2\sqrt{2}ax$$
and then we have $$(x-\sqrt{2}a)^2+2y^2=2a^2$$
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
Find all real triples $(x,y,z)$ such that: $x^2+1=2y,y^2+1=2z,z^2+1=2x$
Find all real triples $(x,y,z)$ such that: $x^2+1=2y,y^2+1=2z,z^2+1=2x$.
Can it be solved avoiding equations of orders higher than 2?
|
Adding all the equations, we get -
$$x^2+1+y^2+1+z^2+1=2x+2y+2z$$
$$x^2-2x+1+y^2-2y+1+z^2-2z+1=0$$
$$(x-1)^2+(y-1)^2+(z-1)^2=0$$
$$\implies x=y=z=1$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Finding the remainder when $p(x^7)$ is divided by $p(x)$, where $p(x)=x^6+x^5+ \cdots + x + 1$? I know there are one or two similar questions to this one but I did not want to look at them since I would like just a hint.
Find the remainder when $p(x^7)$ is divided by $p(x)$, where $p(x)=x^6+x^5+ \cdots + x + 1$?
I wrote $p(x)= \dfrac {x^7-1}{x-1}$, so $p(x^7)=\dfrac {x^{49}-1}{x^7-1}$. We are looking for the unique polynomial $R(x)$ with $deg (R)<6$ such that
$$\dfrac {x^{49}-1}{x^7-1} =Q(x)\left(\dfrac {x^7-1}{x-1}\right)+R(x)$$
Multiplying both sides by $x^7-1$,
$$x^{49}-1=Q(x)p(x)(x^7-1)+R(x)(x^7-1)$$
I know one way to do it would be to find $R(x)$ at $6$ different points and then set up a system of $6$ equations. I want to plug in values which will make the $Q(x)$ term zero because I can't evaluate $Q(x)$. But the values which will make the $Q(x)$ term zero are the $7^{th}$ roots of unity, which will also make the $R(x)$ term zero.
|
Write $p(x^7) = q(x)p(x) +r(x)$ where $deg(r) <7$.
If we say $1,a_1,...a_6$ are all zeros (which are obviously different) of $x^7=1$ then for each $i$ we get:
$$ p(1) = p(a_i^7) = q(a_i)p(a_i) +r(a_i) = r(a_i)$$
Thus $r$ takes the same value for 6 different values and thus $r(x) \equiv p(1) =7$.
|
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|
If $x$, $y$ and $z$ are distinct positive integers and $x+y+z=11$ then what is the maximum value of $(xyz+xy+yz+zx)$?
If $x$, $y$ and $z$ are distinct positive integers and $x+y+z=11$ then what is the maximum value of $(xyz+xy+yz+zx)$?
We know that product is maximum when difference between $x$, $y$ and $z$ is minimum.
So, I assumed $x=3$, $y=4$ and $z=4$.
Now putting this value in $xyz+xy+yz+zx$ I got my answer $88$. But actual answer is $78$. Where am I doing it wrong?
|
$11$ is reasonably small. So we can search exhaustively through the options to find maximum $f=xyz+xy+yz+zx$.
Take $x<y<z$. Then smallest values for $x,y$ are $1,2$ and thus greatest $z=8$. Then we can work down values of $z$ and partition the residue accordingly, avoiding duplicates.
$\begin{array}{|c|c|} \hline
(x,y,z) & f\\ \hline
(1,2,8) & 42 \\
(1,3,7) & 52 \\
(1,4,6) & 58 \\
(2,3,6) & 72 \\
(2,4,5) & 78 \\ \hline
\end{array}$
|
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|
Solve the differential equation $\left(\arctan(xy)+\frac{xy-2xy^{2}}{1+x^{2}y^{2}}\right)dx+\left(\frac{x^{2}-2x^{2}y}{1+x^{2}y^{2}}\right)dy=0$ This is problem 9, exercise 10 from Tannebaum and Pollard's ODE book. I have deduced that the differential equation is exact, but I can't find all the integrable combinations. Any hints that would help me to move forward would be great!
Solve the differential equation $\left(\arctan(xy)+\frac{xy-2xy^{2}}{1+x^{2}y^{2}}\right)dx+\left(\frac{x^{2}-2x^{2}y}{1+x^{2}y^{2}}\right)dy=0$
Solution.
We have,
$\begin{align}
P&=\arctan(xy)+\frac{xy-2xy^{2}}{1+x^{2}y^{2}}\\
\frac{\partial P}{\partial y}&=\frac{x}{1+x^{2}y^{2}}+\left[\frac{(1+x^{2}y^{2})(x-4xy)-(xy-2xy^{2})(2x^{2}y)}{(1+x^{2}y^{2})^{2}}\right]\\
&=\frac{x}{1+x^{2}y^{2}}+\left[\frac{x-4xy+x^{3}y^{2}-4x^{3}y^{3}-2x^{3}y^{2}+4x^{3}y^{3}}{(1+x^{2}y^{2})^{2}}\right]\\
&=\frac{x}{1+x^{2}y^{2}}+\left[\frac{x-4xy-x^{3}y^{2}}{(1+x^{2}y^{2})^{2}}\right]\\
&=\frac{x(1+x^{2}y^{2})+x-4xy-x^{3}y^{2}}{(1+x^{2}y^{2})^{2}}\\
&=\frac{x+x^{3}y^{2}+x-4xy-x^{3}y^{2}}{(1+x^{2}y^{2})^{2}}\\
&=\frac{2x-4xy}{(1+x^{2}y^{2})^{2}}\\
Q&=\frac{x^{2}-2x^{2}y}{1+x^{2}y^{2}}\\
\frac{\partial Q}{\partial x}&=\frac{(1+x^{2}y^{2})(2x-4xy)-(x^{2}-2x^{2}y)(2xy^{2})}{(1+x^{2}y^{2})^{2}}\\
&=\frac{2x-4xy+2x^{3}y^{2}-4x^{3}y^{3}-2x^{3}y^{2}+4x^{3}y^{3}}{(1+x^{2}y^{2})^{2}}\\
&=\frac{2x-4xy}{(1+x^{2}y^{2})^{2}}
\end{align}$
Since $\partial{P}/\partial{y}=\partial{Q}/\partial{x}$, this is an exact differential equation.
We know that-
$\begin{align}
d(x\cdot \arctan(xy))&=\arctan(xy)dx+x\frac{1}{1+x^{2}y^{2}}(xdy+ydx)\\
&=\arctan(xy)dx+\frac{x^{2}dy}{1+x^{2}y^{2}}+\frac{xydx}{1+x^{2}y^{2}}
\end{align}$
I am not able to find an integrable combination for the remaining two terms.
|
Your solution is good, here is another approach:
$$\left(\arctan(xy)+\frac{xy-2xy^{2}}{1+x^{2}y^{2}}\right)dx+\left(\frac{x^{2}-2x^{2}y}{1+x^{2}y^{2}}\right)dy=0$$
$$\arctan(xy)dx +\frac{xy}{1+x^{2}y^{2}}dx +\frac{x^{2}}{1+x^{2}y^{2}}dy -\frac{2xy^{2}}{1+x^{2}y^{2}}dx -\frac{2x^{2}y}{1+x^{2}y^{2}}dy=0$$
$$\arctan(xy)dx +\frac{xy\,dx+x^2\,dy}{1+x^{2}y^{2}} -\frac{2xy^{2}\,dx+2x^{2}y\,dy}{1+x^{2}y^{2}}=0$$
$$\arctan(xy)dx +x\frac{d(xy)}{1+x^{2}y^{2}} -\frac{2xy\,d(xy)}{1+(xy)^{2}}=0$$
$$d(x.\arctan(xy))-\frac{2xy\,d(xy)}{1+(xy)^{2}}=0$$
$$x\arctan(xy)-\ln(1+(xy)^{2})+C=0$$
|
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|
If $450^\circ<\alpha<540^\circ$ and $\cot\alpha=-\frac{7}{24},$ calculate $\cos\frac{\alpha}{2}$. Why is my solution wrong? The problem says:
If $450^\circ<\alpha<540^\circ$ and $\cot\alpha=-\frac{7}{24},$ calculate $\cos\frac{\alpha}{2}$
I solved it in the following way: $$\begin{align} -\frac{7}{24}&=-\sqrt{\frac{1+\cos2\alpha}{1-\cos2\alpha}}\\ \frac{49}{576}&=\frac{1+\cos2\alpha}{1-\cos2\alpha}\\ 625\cos2\alpha&=527\\ 2\cos^2\alpha-1&=\frac{527}{625}\\ \cos\alpha&=-\frac{24}{25}, \end{align}$$ therefore, $$\begin{align} \cos\frac{\alpha}{2}&=\sqrt{\frac{1-\frac{24}{25}}{2}}\\ &=\sqrt{\frac{1}{50}}\\ &=\frac{\sqrt{2}}{10}. \end{align}$$ But there is not such an answer:
A) $0.6$
B) $\frac{4}{5}$
C) $-\frac{4}{5}$
D) $-0.6$
E) $0.96$
I have checked the evaluating process several times. While I believe that my answer is correct and there is a mistake in the choices, I want to hear from you.
|
It's probably easier to use $$\cos^2\alpha = \frac{1}{\sec^2\alpha} = \frac{1}{1+\tan^2\alpha} = \frac{1}{1+\frac{1}{\cot^2\alpha}}$$ to find $\cos\alpha$, this gives $\cos\alpha=\pm\frac7{25}$. The given range for $\alpha$ tells which of these applies.
I'd suggest you check your calculation of $\cos\alpha$.
|
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|
Evaluate $\int_0^{\pi/2}\int_0^{\pi/2} \sin x \sin^{-1}(\sin x \sin y) \ dx \ dy$
Evaluate $$\int_0^{\pi/2}\int_0^{\pi/2} \sin x \sin^{-1}(\sin x \sin y) \ dx \ dy$$
My attempt. Taking $\sin x \sin y=t$, $\iint\sin x \ dx \ dy$.
I am not sure but I think the limits would change to $x=0$ to $\pi/2$ and $t=0$ to $1$.
I do not know how to change the elemental area i.e $dx\, dy$. Can I apply Jacobian where I am only changing one variable (i.e $y$ to $t$)?
Is this the correct way. Is there any easier way to go about it?
|
Perform integration by parts,
$\begin{align}J&=\int_0^{\frac{\pi}{2}}\int_0^{\frac{\pi}{2}} \sin x \arcsin(\sin x \sin y) \ dx \ dy\\
&=\int_0^{\frac{\pi}{2}}\left(\Big[-\cos x\arcsin(\sin x \sin y)\Big]_0^{\frac{\pi}{2}}+\int_0^{\frac{\pi}{2}}\frac{\cos^2x\sin y}{\sqrt{1-\sin^2 x\sin^2 y }}\ dx\right)\ dy\\
&=\int_0^{\frac{\pi}{2}}\int_0^{\frac{\pi}{2}}\frac{\cos^2x\sin y}{\sqrt{1-\sin^2 x\sin^2 y }}\ dx \ dy\\
&=\int_0^{\frac{\pi}{2}}\left(\Big[-\frac{\cos x\text { arcsinh} \left(\tan x\cos y\right)}{\tan x}\Big]_{y=0}^{y=\frac{\pi}{2}}\right)\ dx\\
&=\int_0^{\frac{\pi}{2}} \frac{\cos x\text{ arcsinh}(\tan x)}{\tan x}\ dx
\end{align}$
Perform the change of variable $u=\tan x$,
$\begin{align}J&=\int_0^{\infty}\frac{\text{ arcsinh } x}{x\left(1+x^2\right)^{\frac{3}{2}}}\ dx
\end{align}$
Perform integration by parts,
$\begin{align}J&=\left[\left(\frac{1}{\sqrt{1+x^2}}-\text{arcsinh}\left(\frac{1}{x}\right)\right)\text{arcsinh }x\right]_0^{\infty}-\int_0^{\infty}\left(\frac{1}{\sqrt{1+x^2}}-\text{arcsinh}\left(\frac{1}{x}\right)\right)\frac{1}{\sqrt{1+x^2}}\ dx\\
&=\int_0^{\infty}\frac{\text{arcsinh}\left(\frac{1}{x}\right)}{\sqrt{1+x^2}}\ dx-\int_0^{\infty}\frac{1}{1+x^2}\ dx\\
&=\int_0^{\infty}\frac{\text{arcsinh}\left(\frac{1}{x}\right)}{\sqrt{1+x^2}}\ dx-\frac{\pi}{2}
\end{align}$
In the following integral perform the change of variable $u=\frac{1}{x}$,
$\begin{align}
K&=\int_0^{\infty}\frac{\text{arcsinh}\left(\frac{1}{x}\right)}{\sqrt{1+x^2}}\ dx\\
&=\int_0^{\infty}\frac{\text{arcsinh }x}{x\sqrt{1+x^2}}\ dx\\
\end{align}$
Perform the change of variable $u=\text{arcsinh }x$,
$\begin{align}
K&=\int_0^{\infty}\frac{x}{\sinh x}\ dx
\end{align}$
and according to Contour integral of $\int_{0}^{\infty}\frac{x}{\sinh x}\operatorname{dx}$ ,
$\displaystyle K=\frac{\pi^2}{4}$
Therefore,
$\boxed{\displaystyle J=\frac{\pi^2}{4}-\frac{\pi}{2}}$
Addenda:
1)Another way to compute $K$.
Performing integration by parts,
$\begin{align}K&=\left[x\ln\left(\tanh\left(\frac{x}{2}\right)\right)\right]_0^{\infty}-\int_0^{\infty}\ln\left(\tanh\left(\frac{x}{2}\right)\right)\ dx\\
&=-\int_0^{\infty}\ln\left(\tanh\left(\frac{x}{2}\right)\right)\ dx\\
\end{align}$
Perform the change of variable $y=\tanh\left(\frac{x}{2}\right)$,
$\begin{align}K&=2\int_0^1 \frac{\ln x}{x^2-1}\ dx\\
&=\int_0^1 \frac{\ln x}{x-1}\ dx-\int_0^1 \frac{\ln x}{x+1}\ dx
\end{align}$
$\begin{align}L&=\int_0^1 \frac{\ln x}{x-1}\ dx+\int_0^1 \frac{\ln x}{x+1}\ dx\\
&=\frac{1}{2}\int_0^1 \frac{2x\ln\left(x^2\right)}{x^2-1}\ dx
\end{align}$
Perform the change of variable $u=x^2$,
$\begin{align}L&=\frac{1}{2}\int_0^1 \frac{\ln x}{x-1}\ dx
\end{align}$
Therefore,
$\begin{align}\int_0^1 \frac{\ln x}{x+1}\ dx=-\frac{1}{2}\int_0^1 \frac{\ln x}{x-1}\ dx
\end{align}$
Therefore,
$\begin{align}K&=\frac{3}{2}\int_0^1 \frac{\ln x}{x-1}\ dx
\end{align}$
Knowing that,
$\begin{align}\int_0^1 \frac{\ln x}{x-1}\ dx=\dfrac{\pi^2}{6}\end{align}$
Therefore,
$\boxed{\displaystyle K=\dfrac{\pi^2}{4}}$
2)Playing around with lindep, intnum, functions of GP-PARI, i suspect that the following formulae are true:
$\begin{align}
\int_0^{\frac{\pi}{2}}\int_0^{\frac{\pi}{2}} \sin x \arcsin(\cos x \cos y) \ dx \ dy&=\frac{1}{8}\pi^2-\ln 2\\
\int_0^{\frac{\pi}{2}}\int_0^{\frac{\pi}{2}} \sin x \arccos(\sin x \sin y) \ dx \ dy&=\frac{1}{2}\pi\\
\int_0^{\frac{\pi}{2}}\int_0^{\frac{\pi}{2}} \sin x \arccos(\cos x \cos y) \ dx \ dy&=\frac{1}{8}\pi^2+\ln 2\\
\int_0^{\frac{\pi}{2}}\int_0^{\frac{\pi}{2}} \sin x \arctan(\sin x \sin y) \ dx \ dy&=\frac{1}{8}\pi^2-\frac{1}{4}\pi\ln 2\\
\int_0^{\frac{\pi}{2}}\int_0^{\frac{\pi}{2}} \sin x \arctan^2(\sin x \sin y) \ dx \ dy&=\frac{1}{32}\pi^3+\frac{1}{8}\pi^2\ln 2-\frac{1}{2}\text{G}\pi\\
\end{align}$
$\text{G}$ is the Catalan constant.
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
Proving $x-\frac{x^3}{6} < \sin(x) < x - \frac{x^3}{6} + \frac{x^5}{120} ~~ \forall x \in \Bbb R^+$ using Taylor's expansion I'm trying to prove
$$x-\frac{x^3}{6} < \sin(x) < x - \frac{x^3}{6} + \frac{x^5}{120} ~~ \forall x \in \Bbb R^{+}$$
I wrote $\sin(x)=x-x^3/3!+x^5/5!-x^7/7!+\dots$ and then the expression for $\sin(x)-(x-\frac{x^3}{6})$ i.e. $x^5/5!-x^7/7!+x^9/9!+\dots$.
I don't see why $x^5/5!-x^7/7!+x^9/9!+\dots$ should be positive for all positive real $x$. Any idea?
|
Define
$$f(x) = \sin x - x + \frac{1}{6}x^3$$
and
$$g(x) = x - \frac{1}{6}x^3 + \frac{1}{120}x^5 - \sin x$$
We want to show $f(x) >0$ and $g(x) > 0$ for $x > 0$.
Notice that $f$ and its first four derivatives vanish at $x=0$. If $f$ has a positive zero, say $f(a_0) = 0$, then by Rolle's Theorem $f'$ has a zero $a_1$ with $0 < a_1 < a_0$. Repeating this argument, we have $0 < a_4 < a_3 < a_2 < a_1 < a_0$ with $f^{(n)}(a_n) = 0$. Since $f^{(4)}(x) = \sin x$, we must have $\pi \le a_4$, so $\pi < a_0$. This shows $f(x)$ does not change sign for $0<x<\pi$. Since $-x + (1/6) x^3 > 3/2$ and is increasing for $x > 3$ while $\sin x \ge -1$, we see that $f(x) > 0$ for all $x > 0$.
To see that $g(x) > 0$ for $x>0$, observe that $g(0) = g'(0) = 0$, $g''(x) = f(x)$, and $f(x) > 0$ for $x>0$, as we just showed.
|
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|
Find $x\in \Bbb Z$ such that $x=\sqrt[3]{2+\sqrt{5}}+\sqrt[3]{2-\sqrt{5}}$
Find $x\in \Bbb Z$ such that $x=\sqrt[3]{2+\sqrt{5}}+\sqrt[3]{2-\sqrt{5}}$
Tried (without sucess) two different approaches: (a) finding $x^3$ by raising the right expression to power 3, but was not able to find something useful in the result that simplifies to an integer; (b) tried to find $a$ and $b$ such that $(a+\sqrt{b})^3=2+\sqrt{5}$ without success.
The answer stated for the problem in the original source (a local Math Olympiad Constest) is $x=1$.
|
We can use $a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-ac-bc)$.
Since $a^2+b^2+c^2-ab-ac-bc=\frac{1}{2}((a-b)^2+(a-c)^2+(b-c)^2)$, we see that
$$a^2+b^2+c^2-ab-ac-bc=0\Leftrightarrow a=b=c.$$
Thus, since $\sqrt[3]{2+\sqrt5}\neq\sqrt[3]{2-\sqrt5}$, we obtaion
$$x=\sqrt[3]{2+\sqrt5}+\sqrt[3]{2-\sqrt5}$$ it's
$$x-\sqrt[3]{2+\sqrt5}-\sqrt[3]{2-\sqrt5}=0$$ or
$$x^3-2-\sqrt5-2+\sqrt5-3x\sqrt[3]{(2+\sqrt5)(2-\sqrt5)}=0$$ or
$$x^3+3x-4=0$$ or
$$x^3-x^2+x^2-x+4x-4=0$$ or
$$(x-1)(x^2+x+4)=0$$ or
$$x=1.$$
We can make also the following.
$$\sqrt[3]{2+\sqrt5}+\sqrt[3]{2-\sqrt5}=\frac{1}{2}\left(\sqrt[3]{16+8\sqrt5}+\sqrt[3]{16-8\sqrt5}\right)=$$
$$=\frac{1}{2}\left(\sqrt[3]{1+3\sqrt5+3\cdot5+5\sqrt5}+\sqrt[3]{1-3\sqrt5+3\cdot5-5\sqrt5}\right)=$$
$$=\frac{1}{2}\left(1+\sqrt5+1-\sqrt5\right)=1.$$
|
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
If $x^3+y^3+(x+y)^3+33 xy=2662$, $x,y\in \Bbb R$, find $S=x+y$.
If $x^3+y^3+(x+y)^3+33 xy=2662$ and $\{x,y\}\subset \Bbb R$, find $S=x+y$.
This question from an olympiad contest. Answer stated: $S=x+y=11$
Tried to develop $(x+y)^3$ to find something useful for the situation, but without success.
|
$$ (x+y)^3 + (x+y)(x^2-xy+y^2)+33xy=2662 $$
$$ S^3 + S(x^2+y^2-xy) + 33xy=2662 $$
$$ S^3 + S(S^2 - 3xy) + 33xy=2662 $$
$$ 2S^3 + 3xy(11 - S) = 2662 $$
$$ \frac{3}{2}xy(11-S)=11^3 - S^3$$
if $S \ne 11$:
$$ \frac{3}{2}xy = 121 + 11S + S^2 $$
Now try to write down roots of this equation on $S$, add condition of their existence ($D \ge 0$, cause roots are real) and remember that $S = x + y$. Hope this will lead to contradiction.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Evaluate $(\sqrt{3}-3i)^6$
Evaluate $$(\sqrt{3}-3i)^6.$$
So I assume that we should write the following in polar form
$r=\sqrt{(\sqrt{3})^2+(-3)^2}=\sqrt{3+9}=\sqrt{12}=2\sqrt{3},$
$\theta=\arctan{\frac{-3}{\sqrt{3}}}=-\frac{\pi}{3}.$
So $$(\sqrt{3}-3i)^6=[2\sqrt{3}e^{-i\frac{\pi}{3}+2\pi k}]^6=1728e^{-2\pi i+2\pi m},$$ where $m\in \mathbb{Z}.$
So $$1728e^{-2\pi i+2\pi m}=1728cos(-2\pi)=1728.$$
Is it correct?
|
Yes, it is correct. In fact
$$(\sqrt{3}-3i)^6=3^3(1-\sqrt{3}i)^6=2^63^3(e^{-i\pi/3})^6=1728.$$
P.S. There is no need of the term $+2\pi k$.
|
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|
Find an invertible matrix $S$ and a matrix $J$ in Jordan form, such that $S^{-1}AS = J$ $A = \begin{pmatrix}
0 & -1 & 0 & 0 \\
2 & 0 & -1 & 0 \\
3 & -1 & -2 & -1 \\
-1 & 0 & 1 & 1 \\
\end{pmatrix}$
We need to find an invertible matrix $S$ and a matrix in Jordan form $J$, such that $S^{-1}AS=J$.
I've found that the characteristic polynomial is $P_A(\lambda) = \lambda^3(\lambda+1)$. Therefore, $\lambda = 0,-1$.
The minimal polynomial is $m_A(\lambda) = P_A(\lambda).$
At this stage, we know that $J = \begin{pmatrix}
-1 & 0 & 0 & 0 \\
0 & 0 & 1 & 0 \\
0 & 0 & 0 & 1 \\
0 & 0 & 0 & 0 \\
\end{pmatrix}$ (up to blocks' order).
Next, I've found the Null Spaces:
For $\lambda=0$:
$N(A) = N \begin{pmatrix}
0 & -1 & 0 & 0 \\
2 & 0 & -1 & 0 \\
3 & -1 & -2 & -1 \\
-1 & 0 & 1 & 1 \\
\end{pmatrix} = span \begin{Bmatrix} \begin{pmatrix}
-1 \\ 0 \\ -2 \\ 1
\end{pmatrix} \end{Bmatrix}$
$N(A^2) = N \begin{pmatrix}
-2 & 0 & 1 & 0 \\
-3 & -1 & 2 & 1 \\
-7 & -1 & 4 & 1 \\
2 & 0 & -1 & 0 \\
\end{pmatrix} = span \begin{Bmatrix}
\begin{pmatrix}
1 \\ 1 \\ 2 \\ 0
\end{pmatrix},
\begin{pmatrix}
0 \\ 1 \\ 0 \\ 1
\end{pmatrix}
\end{Bmatrix}$
$N(A^3) = N \begin{pmatrix}
3 & 1 & -2 & -1 \\
3 & 1 & -2 & -1 \\
9 & 3 & -6 & -3 \\
3 & -1 & 2 & 1 \\
\end{pmatrix} = span \begin{Bmatrix}
\begin{pmatrix}
-1 \\ 3 \\ 0 \\ 0
\end{pmatrix},
\begin{pmatrix}
2 \\ 0 \\ 3 \\ 0
\end{pmatrix},
\begin{pmatrix}
1 \\ 0 \\ 0 \\ 3
\end{pmatrix}
\end{Bmatrix}$
For $\lambda=-1$:
$N(A+I) = N \begin{pmatrix}
1 & -1 & 0 & 0 \\
2 & 1 & -1 & 0 \\
3 & -1 & -1 & -1 \\
-1 & 0 & 1 & 2 \\
\end{pmatrix} = span \begin{Bmatrix} \begin{pmatrix}
1 \\ 1 \\ 3 \\ -1
\end{pmatrix} \end{Bmatrix}$
I'm not sure how to continue from here...
|
The eigenvector for $-1$ you already have, call it $v_1$. For $0$ what you want is a vector $v_4$ such that $A^3v_4 = 0$ but $A^2v_4 \ne 0$. I'll let you think of how to do this. (Incidentally all three vectors you have for a basis of $N(A^3)$ work.)
Once you have $v_4$, let $v_3 = Av_4$ and $v_2 = A^2v_4 = Av_3$. Then take
$$ S = (v_1, v_2, v_3, v_4). $$
|
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|
A question of trigonometry on how to find minimum value.
Find The minimum value of $$2^{\sin^2 \alpha} + 2^{\cos^2 \alpha}.$$
I can easily get the maximum value but minimum value is kinda tricky. Please help.
|
$x:= \sin^2(\alpha) $; $ 1-x = \cos^2(\alpha)$ .
Then: $ f(x):= 2^x + \frac{2}{2^x} $, $0 \le x \le1$.
$z := 2^x$ ;
$ g(z): = z + \frac{2}{z} , 1 \le z \le 2$.
AM GM inequality:
$ (1/2) ( z + \frac{2}{z}) \ge (z \frac{2}{z})^{1/2} =$
$ \sqrt{2}$.
$g(z) = z + \frac{2}{z} \ge 2 \sqrt{2}$.
Equality for $z = √2$.
Back substitution: $2^x = 2^{1/2}$.
Minimum at $x= 1/2$, I.e
$x = \sin^2(\alpha) = 1/2 = \cos^2(\alpha)$ ,
hence $\alpha = 45°$.
$\min(f(x)) = f(x =1/2) = 2√2$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
How to differentiate $\left(x^2 + \frac{1}{x^2}\right)^5$ How can I differentiate this expression
$$\left(x^2 + \frac{1}{x^2}\right)^5$$
I will appreciate any help.
|
Start with the Chain Rule:$\frac{d}{dx}f(g(x))=f'(g(x))g'(x)$
where $f(x)=x^5$ and $g(x)=x^2+\frac{1}{x^2}$. $$\\$$
For $f'(g(x))$ we get:
$$f'(x)= 5(x^2+\frac{1}{x^2})^4.$$
For $g'(x)$ we can separate the expression:
$$\frac{d}{dx} (x^2) + \frac{d}{dx} (\frac{1}{x^2})$$
$\frac{d}{dx} (x^2) = 2x$ and $\frac{d}{dx} (\frac{1}{x^2}) = \frac{-2}{x^3}$.
So, For $g'(x)$ we get:
$$2x-\frac{2}{x^3}.$$
Thus, we get $f'(g(x))g'(x)$ as:
$$(5(x^2+\frac{1}{x^2})^4)(2x-\frac{2}{x^3}).$$
|
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|
Functional equation : $ f(1)^3 + f(2)^3 + \ldots + f(n)^3 = (f(1) + f(2) + \ldots + f(n))^2$ Find all function $f:\mathbb{N}\rightarrow\mathbb{N}\cup\{0\}$ satisfying
$$ f(1)^3 + f(2)^3 + \ldots + f(n)^3 = (f(1) + f(2) + \ldots + f(n))^2$$
$\forall n \in \mathbb{N}$.
Thank you, Batominovski and Guy Fabrice..
Is my understanding correct ? Please let me know if there is any mistake.
Substitute $n=1$, $f(1)^3 =f(1)^2$, so $f(1)=0$ or $1$
$f(1)=0$ : substitute $n=2$, $f(2)^3 =f(2)^2$, so $f(2)=0$ or $1$
If $f(2) = 0$, let $l$ be the maximal value such that $f(l) =0$.
so $ f(1)^3 + f(2)^3 + \ldots + f(l+1)^3 = (f(1) + f(2) + \ldots + f(l+1))^2$
then $f(l+1)^3=(f(l+1))^2$ so $f(l+1) = 1$
If $f(2) = 1$, then $0+1+f(3)^3=(f(1)+f(2)+f(3))^2$ , so $f(3) = 0$ or $2$
Since $ f(1)^3 + f(2)^3 + \ldots + f(n)^3 = (f(1) + f(2) + \ldots + f(n))^2$
and $ f(1)^3 + f(2)^3 + \ldots + f(n+1)^3 = (f(1) + f(2) + \ldots + f(n+1))^2$
so $f(n+1)^3 = 2(f(1)+f(2)+\ldots+f(n))f(n+1)+f(n+1))^2$
so $f(n+1)^2-f(n+1)=2(f(1)+f(2)+\ldots+f(n))$
i.e., if $f(n) \not= 0$, then $f(n)=n, \forall n \in \mathbb{N}$
so if $f(x)=0, \forall x \in \mathbb{N}$ then $f(n) = 0, \;\forall n \in \mathbb{N}$
If $f(x)$ be the value such that Max$\{f(1), f(2), \ldots ,f(x)\}=k$ then $f(x+1) = k+1$ or $0$ ---[1]
By induction, let $P(n)$ denotes [1]
Basic step , it's obvious that if $f(1) = 0$, then $f(l_i)=1$, $1\leq l_i\leq x$
so $1+f(l_i+1)^3 = (f(l_i+1) +1)^2$ so $f(l_i+1)=2$ or $0$
Inductive step, suppose that $P(k)$ is true.
$1^3+2^3+\ldots+k^3+f(x+1)^3=(1+2+\ldots+k+f(x+1))^2$,
so $f(x+1)=k+1$ or $0$, so $P(n)$ is true.
The sequence of $f$ is $\underbrace{0\ldots0}_{\text{$k_1$}}1\underbrace{0\ldots0}_{\text{$k_2$}}2\ldots$, where $k_1, k_2, \ldots \in \mathbb{N}$
Answer check :
$ f(1)^3 + f(2)^3 + \ldots + f(n)^3 = 1^3+2^3+\ldots+m^3=\left(\frac{m(m+1)}{2}\right)^2=(f(1) + f(2) + \ldots + f(n))^2$
$\blacksquare$
|
Find all function $f:\mathbb{N}\rightarrow\mathbb{N}\cup\{0\}$ satisfying
$$ f(1)^3 + f(2)^3 + \ldots + f(n)^3 = (f(1) + f(2) + \ldots + f(n))^2$$
This at $n+1$ gives
$$ f(1)^3 + f(2)^3 + \ldots + f(n+1)^3 = (f(1) + f(2) + \ldots + f(n+1))^2$$
i.e
$$ (f(1) + f(2) + \ldots + f(n))^2+ f(n+1)^3 = (f(1) + f(2) + \ldots + f(n+1))^2$$
hence
we have
$$ f(n+1)^3 = (f(1) + f(2) + \ldots + f(n+1))^2-(f(1) + f(2) + \ldots + f(n))^2= 2f(n+1)(f(1) + f(2) + \ldots + f(n)) +f(n+1)^2$$
we then get the relationships
$$ f(n+1)^2 -f(n+1)=2(f(1) + f(2) + \ldots + f(n)) $$
Hence $f(n+1)(f(n+1)-1) $ must be an even number . this way you check the values of $f(n)$
by induction.
|
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|
How prove infinitely many postive integers triples $(x,y,z)$ such $(x+y+z)^2+2(x+y+z)=5(xy+yz+zx)$
show that there exsit infinitely many postive integers triples $(x,y,z)$
such
$$(x+y+z)^2+2(x+y+z)=5(xy+yz+zx)$$
May try it is clear $(x,y,z)=(1,1,1)$ is one solution，and
$$(x+y+z+1)^2=5(xy+yz+xz)+1$$
|
For the equation.
$$(x+y+z)^2+2(x+y+z)=5(xy+xz+yz)$$
It is possible to reduce the parameterization of the solutions to some equivalent to the Pell equation.
It has the form.
$$x=3a^2-(b+c)a+b^2-3bc+c^2$$
$$y=a^2-(b+3c)a+3b^2-bc+c^2$$
$$z=a^2-(3b+c)a+b^2-bc+3c^3$$
These parameters can be recorded through the solution of the equation Pell.
$$p^2-5(2k^2+t^2)s^2=-1$$
$$a=ks$$
$$b=p-(3k-t)s$$
$$c=p-(3k+t)s$$
|
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|
$\sin(40^\circ)<\sqrt{\frac{3}7}$
Prove without using of calculator, that $\sin40^\circ<\sqrt{\frac{3}7}$.
My attempt.
Since $$\sin(40^\circ)=2\sin(20^\circ)\cos(20^\circ)<2\sin(20^\circ)$$
$$=2\sin(60^\circ-40^\circ)=\sqrt{3} \cos(40^\circ)-\sin(40^\circ),$$
$$2\sin(40^\circ)<\sqrt{3} \cos(40^\circ).$$
Hence, $$4\sin^2(40^\circ)<3\cos^2(40^\circ)=3(1-\sin^2(40^\circ))$$
$$7\sin^2(40^\circ)<3$$
$$\sin(40^\circ)<\sqrt{\frac{3}7}$$
Is there another way to prove this inequality?
|
We need to prove that
$$\frac{1-\cos80^{\circ}}{2}<\frac{3}{7}$$ or
$$\sin10^{\circ}>\frac{1}{7}.$$
Let $\sin10^{\circ}=x$.
Thus, $$3x-4x^3=\frac{1}{2}$$ or $f(x)=0$, where
$$f(x)=x^3-\frac{3}{4}x+\frac{1}{8}$$ and since $$f\left(\frac{1}{7}\right)=\frac{1}{343}-\frac{3}{28}+\frac{1}{8}=\frac{57}{2744}>0,$$ we are done!
Indeed, $f'(x)=3x^2-\frac{3}{4}=3\left(x-\frac{1}{2}\right)\left(x+\frac{1}{2}\right)$,
which says that $\sin10^{\circ}$ is an unique root of the equation on $\left(0,\frac{1}{2}\right].$
|
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Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.