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Find real parts of the complex roots of this $9^{th}$ order polynomial in explicit form I have a following polynomial. (See WolframAlpha ):
$$x^9-6x^8+14x^7-16x^6+36x^5-56x^4+ 24x^3-320x+\frac{640}{9}=0 \tag{1}$$
Wolfram says that $(1)$ has three real roots and three pairs of complex conjugate roots.
I know one of the real roots of this polynomial is equal to:
$$x_1=\tan \left(\frac{1}{3} \arctan \frac{1}{3} \right)+\tanh \left(\frac{1}{3} \text{arctanh} \frac{1}{3} \right) =0.22267663636945$$
I know this is a root, because I derived the polynomial from this expression.
Edit
By @IvanNeretin's comment we also have two other real roots:
$$x_2=\tan \left(\frac{\pi}{3} +\frac{1}{3} \arctan \frac{1}{3} \right)+\tanh \left(\frac{1}{3} \text{arctanh} \frac{1}{3} \right)$$
$$x_3=\tan \left(\frac{2\pi}{3} +\frac{1}{3} \arctan \frac{1}{3} \right)+\tanh \left(\frac{1}{3} \text{arctanh} \frac{1}{3} \right)$$
A very surprising (for me) discovery: all six complex roots have the same absolute value of the imaginary part:
$$b=Im(x)= \pm \frac{\sqrt[3]{2}}{\sqrt{3}} (1+\sqrt[3]{2})= \pm 1.643902181980216,~~~~\text{for all } x \notin \mathbb{R}$$
I found it using ISC, but hadn't found closed forms for the real parts.
How can we find explicitly the real parts of the complex roots?
My thoughts - if we substitute in $(1)$:
$$x=a \pm i b$$
then we obtain two equations for one real varibale $a$ (for each three pairs of complex roots of course), and by adding and subtracting these equations we can lower their order.
Since there should be three distinct real parts $a$, we could probably bring it down to a cubic equation.
But I need a CAS for that and I'm not sure it will lead to a solution.
|
The other two real roots, quite expectedly, are $\tan \left({\pi\over3}+\frac{1}{3} \arctan \frac{1}{3} \right)+\tanh \left(\frac{1}{3} \text{arctanh} \frac{1}{3} \right)$ and $\tan \left({2\pi\over3}+\frac{1}{3} \arctan \frac{1}{3} \right)+\tanh \left(\frac{1}{3} \text{arctanh} \frac{1}{3} \right)$. The complex roots are produced in a similar manner, remembering that $\tanh$ is a periodic function with imaginary period.
Long story short, all nine roots are
$$\tan \left({\pi\over3}\cdot m+\frac{1}{3} \arctan \frac{1}{3} \right)+\tanh \left({\pi i\over3}\cdot n+\frac{1}{3} \text{arctanh} \frac{1}{3} \right),\;n,m\in\{0,1,2\}$$
A rigorous explanation is beyond my expertise. Informally speaking, inverse trigonometric and inverse hyperbolic functions are multivalued, but the polynomial "knows nothing" of this; to it, they are all the same, so if one value is a root, so must be the others.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Solve in integers the equation $\sqrt{x^3-3xy^2+2y^3}=\sqrt[3]{13x+8}$
Solve in integers the equation
$$\sqrt{x^3-3xy^2+2y^3}=\sqrt[3]{13x+8}$$
My work so far:
I used www.wolframalpha.com. Then $x=9,y=8 -$ solution.
My attempt:
1) Let $\sqrt{x^3-3xy^2+2y^3}=a, \sqrt[3]{13x+8}=b$. Then
$$\begin{cases}
a=b\\
a^2-b^3=x^3-3xy^2+2y^3-13x-8 \in \mathbb Z
\end{cases}$$
2) $$(x^3-3xy^2+2y^3)^3=(13x+8)^2$$
Addition:
We have $$A^6\cdot B^6=(13x+8)^2$$
and $$n^6 \equiv 0, \pm 1 (\bmod 13)$$
|
There are no integer solutions with $x<0$, since $13x+8<0$ for such $x$. When $0\leq x\leq 8$ then $13x+8$ is not an integer cubic. Since the latter is easily seen to be a necessary condition it follows that there are no solutions with $x$ in this range. On the other hand you already have found the solution $(9,8)$. In the following I shall prove that there are no solutions with $x\geq10$.
I'm considering $x\geq10$ as "primary variable" and write $y:=x-t$ with a new integer variable $t$. The equation at stake then reads
$$a:=\sqrt{t^2(3x-2t)}=\root 3\of{13x+8}=:b\ .$$
If $t\leq-1$ then
$$a=|t|\sqrt{3x+2|t|}\geq\sqrt{3x}\ .$$
For $x=10$ we therefore have $a^3\geq\bigl(\sqrt{3x}\bigr)^3=30^{3/2}>164>138=13x+8=b^3$, and things get worse for larger $x$.
If $t=1$ and $x=10$ then $a=\sqrt{3x-2}=\sqrt{28}$.
For $x=10$ we therefore have $a^3\geq28^{3/2}>148>138=b^3$, and things get worse for larger $x$.
If $2\leq t\leq x$ then
$$a\geq2\sqrt{3x-2t}\geq2\sqrt{x}\ .$$
For $x=10$ we therefore have $a^3\geq8\cdot10^{3/2}>252>138=b^3$, and things get worse for larger $x$.
Finally, if $2x\leq 2t\leq 3x-1$ then
$$a=\sqrt{t^2(3x-2t)}=t\sqrt{3x-2t}\geq x\ .$$
For $x=10$ we therefore have $a^3\geq10^3>138$, and things get worse for larger $x$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Value of $\left(\frac{a}{b-c}+\frac{b}{c-a}+\frac{c}{a-b}\right)\cdot \left(\frac{b-c}{a}+\frac{c-a}{b}+\frac{a-b}{c}\right)$
If $a+b+c=0,$ Then value of $\displaystyle \left(\frac{a}{b-c}+\frac{b}{c-a}+\frac{c}{a-b}\right)\cdot \left(\frac{b-c}{a}+\frac{c-a}{b}+\frac{a-b}{c}\right)$
$\bf{My\; Try::}$ Given $$\displaystyle \left(\frac{a}{b-c}+\frac{b}{c-a}+\frac{c}{a-b}\right)\cdot \left(\frac{b-c}{a}+\frac{c-a}{b}+\frac{a-b}{c}\right) = 1+1+1+\frac{b-c}{a}\left(\frac{b}{c-a}+\frac{c}{a-b}\right)+\frac{c-a}{b}\left(\frac{a}{b-c}+\frac{c}{a-b}\right)+\frac{a-b}{c}\left(\frac{a}{b-c}+\frac{b}{c-a}\right)$$
Now How can I Solve after that, Is there is any less complex method plz explain here , Thanks
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Expanding $$A= \left(\frac{a}{b-c}+\frac{b}{c-a}+\frac{c}{a-b}\right)\cdot \left(\frac{b-c}{a}+\frac{c-a}{b}+\frac{a-b}{c}\right)$$ we get $$A=\frac{a^3-a^2 (b+c)-a \left(b^2-3 b c+c^2\right)+(b-c)^2 (b+c)}{a b c}$$ Replace $c$ by $-(a+b)$, expand and simplify.
The result is a whole number.
|
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|
When is $\sqrt{x/y^2}$ equal to $\sqrt{x}/y$? The solution to the quadratics is given by
$r = -\dfrac{b}{2a}\pm\sqrt{\dfrac{b^2-4ac}{4a^2}}$, which is shortened to
$r = -\dfrac{b}{2a}\pm\dfrac{\sqrt{b^2-4ac}}{2a}$, but I'm wondering how if this is justified, given that $4a^2$ can be negative if $a \in \mathbb{C}$, and $\sqrt{\dfrac{x}{y}} \neq \dfrac{\sqrt{x}}{\sqrt{y}}$ if $x$ and $y$ are negative, but given that we have $a^2$, is this justified?
Is $\sqrt{x/y^2} = \sqrt{x}/y$? Is $r = -\dfrac{b}{2a}\pm\dfrac{\sqrt{b^2-4ac}}{2a}$ always true?
|
When $y> 0$, you have
$$\sqrt{\frac x{y^2}}=\frac{\sqrt{x}}{\sqrt{y^2}}=\frac{\sqrt{x}}{y}.$$
When $y<0$, you have
$$\sqrt{\frac x{y^2}}=\frac{\sqrt{x}}{\sqrt{y^2}}=\frac{\sqrt{x}}{-y}.$$
|
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|
Factoring $x^5+B x^4+C x^3+D x^2+E x+F=(x^2+a x+b)(x^3+p x+q)$ over $\mathbb{Q}$ For a quintic polynomial to be reducible to the following form over $\mathbb{Q}$:
$$x^5+B x^4+C x^3+D x^2+E x+F=(x^2+a x+b)(x^3+p x+q)$$
we need to match the coefficients ($a=B$ obviously, so we write the system of equations for the rest):
$$b+p=C \\ Bp+q=D \\ bp+Bq=E \\ bq=F$$
The system contains $4$ equations, so we get the condition on one of the coefficients of the original quintic. For convenience we choose $F$:
$$F=\frac{1}{2} \left(B^5-B^3C+3B^2D-2BE+CD \color{blue}{\pm} (B^3+D) \sqrt{(B^2-C)^2+4(BD-E)} \right)$$
$$b=\frac{1}{2} \left(B^2+C \color{blue}{\pm} \sqrt{(B^2-C)^2+4(BD-E)} \right)$$
$$p=\frac{1}{2} \left(C-B^2 \color{blue}{\mp} \sqrt{(B^2-C)^2+4(BD-E)} \right)$$
$$q=\frac{1}{2} \left(B^3-BC+2D \color{blue}{\pm} B \sqrt{(B^2-C)^2+4(BD-E)} \right)$$
For $b,p,q,F \in \mathbb{Q}$ we need:
$$E=BD+\frac{1}{4}(B^2-C)^2-n^2,~~~~~n \in \mathbb{Q}$$
My question is: Is the following statement correct?
A quintic polynomial in the form:
$$x^5+B x^4+C x^3+D x^2+E x+F,~~~~~B,C,D,E,F \in \mathbb{Q}$$
can be factored into $(x^2+a x+b)(x^3+p x+q)$ over $\mathbb{Q}$ iff:
$$E=BD+\frac{1}{4}(B^2-C)^2-n^2,~~~~~n \in \mathbb{Q}$$
$$F=\frac{1}{2} \left(B^5-B^3C+3B^2D-2BE+CD \color{blue}{\pm} 2(B^3+D)n \right)$$
So we get four free parameters $B,C,D,n$.
Or did I miss something?
Edit
If we allow $b,p,q$ to be irrational and/or complex, we can still keep $F$ rational by setting $D=-B^3$, which expands the set of quintics solvable this way.
|
The statement is true.
You already proved the following :
If a quintic polynomial in the form:
$$x^5+B x^4+C x^3+D x^2+E x+F,~~~~~B,C,D,E,F \in \mathbb{Q}$$
can be factored into $(x^2+a x+b)(x^3+p x+q)$ over $\mathbb{Q}$, then
$$E=BD+\frac{1}{4}(B^2-C)^2-n^2,~~~~~n \in \mathbb{Q}$$
$$F=\frac{1}{2} \left(B^5-B^3C+3B^2D-2BE+CD \pm 2(B^3+D)n \right)$$
In the following, let us prove the following :
If $$E=BD+\frac{1}{4}(B^2-C)^2-n^2$$
$$F=\frac{1}{2} \left(B^5-B^3C+3B^2D-2BE+CD \pm 2(B^3+D)n \right)$$
where $B,C,D,n\in\mathbb Q$, then a quintic polynomial in the form:
$$x^5+B x^4+C x^3+D x^2+E x+F$$
can be factored into $(x^2+a x+b)(x^3+p x+q)$ over $\mathbb{Q}$.
$$\begin{align}&x^5+Bx^4+Cx^3+Dx^2+Ex+F
\\&=x^5+Bx^4+Cx^3+Dx^2+\left(BD+\frac{1}{4}(B^2-C)^2-n^2\right)x
\\&\qquad +\frac{1}{2} \left(B^5-B^3C+3B^2D-2BE+CD\pm 2(B^3+D)n\right)
\\&=x^5+Bx^4+Cx^3+Dx^2+\left(BD+\frac{1}{4}(B^2-C)^2-n^2\right)x
\\&\qquad +\frac{1}{2} \left(B^5-B^3C+3B^2D-2B\left(BD+\frac{1}{4}(B^2-C)^2-n^2\right)+CD\pm 2(B^3+D)n\right)
\\&=x^5+Bx^4+Cx^3+Dx^2+\left(BD+\frac{1}{4}(B^2-C)^2-n^2\right)x
\\&\qquad +\frac{1}{4} \left(B^5+2B^2D-BC^2+4Bn^2+2CD\pm 4(B^3+D)n\right)
\\&=\left(x^2+Bx+\frac{B^2+C\pm 2n}{2}\right)\left(x^3+\frac{C-B^2\mp 2n}{2}x+\frac{B^3-BC+2D\pm 2nB}{2}\right)\end{align}$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Eliminate the parameter given $x = \tan^{2}\theta$ and $y=\sec\theta$ $x = \tan^{2} (\theta)$ and $y = \sec (\theta)$
knowing that $\tan^{2} (\theta) = (\tan (\theta))^2 = \dfrac{\sin^{2}\theta}{\cos^{2}\theta}$
and that $\sec(\theta) = \dfrac{1}{\cos(\theta)}$ $\to$ $y=\dfrac{1}{\cos(\theta)}$
For $y$ we can get an even more simplified expression that will be useful for substituting into $x$ by multiplying both sides by $\cos(\theta)$ and then dividing both sides by $y$ to obtain: $\cos(\theta)=\frac{1}{y}$
With the trig simplified throught identities know we can now use more identities that become obvious and then simplify to a final expression through substitution into the x component:
$x = \dfrac{1-(\cos(\theta))^{2}}{(\cos(\theta))^{2}}$
Now substituting $\cos(\theta)$ for $\frac{1}{y}$ we get:
$x = \dfrac{1-(\frac{1}{y})^2}{(\frac{1}{y})^2}$
which simplifies to:
$x = \dfrac{1-\frac{1}{y^2}}{\frac{1}{y^2}}$
which can be simplified to:
$x = \dfrac{1-(\frac{1}{y^2})}{1} \cdot \dfrac{y^2}{1} \to y^2 - 1$
So removing the parameter we get:
$x= y^2 - 1$
I feel very confident that this answer is correct but I would like verification.
|
You can save some passages:
$$
x=\frac{\sin^2\theta}{\cos^2\theta}=\frac{1-\cos^2\theta}{\cos^2\theta}=
\frac{1}{\cos^2\theta}-\frac{\cos^2\theta}{\cos^2\theta}=y^2-1
$$
|
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|
Baricenter of a region bounded by a closed parametric curve I've always known how to get the center of mass of any region, but now i met a new question with a region bounded by a parametric curve and the question is to get its baricenter! My question is what changes in this case than the normal way of getting the baricenter? Because i dont have any idea now how to solve this question any help please?
The region is bounded by the parametric curve :
$ x(t)=cos(2\pi t)$
$ y(t)=t-t^3 $
$t\in [0,1] $
|
So we are looking for the baricenter (centroid) of the region enclosed by the curve
$$
C:\left\{ \begin{gathered}
0 \leqslant t \leqslant 1 \hfill \\
x(t) = \cos \left( {2\pi t} \right) \hfill \\
y(t) = t - t^{\,3} \hfill \\
\end{gathered} \right.
$$
Among the methods exposed in answering to
Baricenter of a region bounded by a parametric curve
we can in this case profitably choose the last.
Thus the area of the region enclosed by $C$ will be:
$$
\begin{gathered}
A = \frac{1}
{2}\oint\limits_C {\left| {\,\begin{array}{*{20}c}
x & y \\
{dx/dt} & {dy/dt} \\
\end{array} \,} \right|dt} = \frac{1}
{2}\int_{t\, = \,0}^{\;1} {\left| {\,\begin{array}{*{20}c}
{\cos \left( {2\pi t} \right)} & {t - t^{\,3} } \\
{ - 2\pi \sin \left( {2\pi t} \right)} & {1 - 3t^{\,2} } \\
\end{array} \,} \right|dt} = \hfill \\
= \frac{1}
{2}\int_{t\, = \,0}^{\;1} {\left( {\left( {1 - 3t^{\,2} } \right)\cos \left( {2\pi t} \right) + 2\pi \left( {t - t^{\,3} } \right)\sin \left( {2\pi t} \right)} \right)dt} = - \frac{3}
{{2\,\pi ^{\,2} }} \hfill \\
\end{gathered}
$$
which tells us that, for $t$ going from $0$ to $1$, we are tracing the curve clock-wise.
Then the baricenter $\mathbf{b}$ will be:
$$
\begin{gathered}
\mathbf{b} = \frac{1}
{{3A}}\oint\limits_C {\mathbf{r}\left| {\,\begin{array}{*{20}c}
x & y \\
{dx/dt} & {dy/dt} \\
\end{array} \,} \right|dt} = \hfill \\
= - 2\left( {\frac{\pi }
{3}} \right)^{\,2} \int_{t\, = \,0}^{\;1} {\left( \begin{gathered}
\cos \left( {2\pi t} \right) \\
t - t^{\,3} \\
\end{gathered} \right)\left( {\left( {1 - 3t^{\,2} } \right)\cos \left( {2\pi t} \right) + 2\pi \left( {t - t^{\,3} } \right)\sin \left( {2\pi t} \right)} \right)dt} = \hfill \\
= - 2\left( {\frac{\pi }
{3}} \right)^{\,2} \left( \begin{gathered}
- \frac{9}
{{32\,\pi ^{\,2} }} \\
\frac{{9\,\pi ^{\,2} - 135}}
{{4\,\pi ^{\,4} }} \\
\end{gathered} \right) = \left( \begin{gathered}
\frac{1}
{{16}} \\
\frac{{15}}
{{2\,\pi ^{\,2} }} - \frac{1}
{2} \\
\end{gathered} \right) \hfill \\
\end{gathered}
$$
which matches with the answer above.
|
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|
Eliminate $\theta$
Eliminate $\theta$ in
$$\sin \theta + \mbox{cosec} \, \theta = m$$
$$\sec \theta - \cos \theta = n$$
My approach-
I multiplied the first equation by $\sin \theta$ and the second equation by $\cos \theta$ but it doesn't give me the desired answer..
|
Squaring and adding,we get
$1+cosec^2\theta+\sec^2\theta=m^2+n^2.....(1)$
We need to find $cosec\theta$ and $\sec\theta$ from the given equations
First equation becomes $\frac{1}{\csc\theta}+\csc\theta=m$
$cosec^2\theta-m$ $cosec\theta+1=0$
Similarly second equation becomes
$\sec^2\theta-n\sec\theta-1=0$
Solving these using quadratic formula,
$cosec\theta=\frac{m\pm\sqrt{m^2-4}}{2}$
$\sec\theta=\frac{n\pm\sqrt{n^2+4}}{2}$
Put these values in equation $(1)$.
|
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|
Finding the parameters of an ellipsoid given its quadratic form Suppose we have the quadratic form of an ellipsoid of the form
$$ax^2 + by^2+cz^2+dxy+eyz+fxz+gx+hy+iz+j=0$$
I want to find centroid of the arbitrarily oriented ellipsoid, its semi-axes, and the angles of rotation.
For the 2D case I found an answer here. I was wondering if someone can help me do the same for 3D.
|
The equation of an ellipsoid centered at the origin has the form $q_A(\mathbf{v}) = \mathbf{v}^T A \mathbf{v} = C$ where $C > 0$ and $A$ is a symmetric matrix whose eigenvalues are all positive. If the eigenvalues of $A$ are $\lambda_i$ with a corresponding orthonormal basis of eigenvectors $v_i$ (so $Av_i = \lambda_i v_i$) then the semi-principal axes of the ellipsoid are the $v_i$'s with lengths $\lambda_i$.
If the ellipsoid is centered at $\mathbf{p}$ instead of the origin, the equation describing it is
$$ q_A(\mathbf{v} - \mathbf{p}) = (\mathbf{v} - \mathbf{p})^T A (\mathbf{v} - \mathbf{p}) = \mathbf{v}^T A \mathbf{v} - 2 \mathbf{v}^T A \mathbf{p} + \mathbf{p}^T A \mathbf{p} = C.$$
Thus, the first part of the equation is quadratic in $\mathbf{v}$, the second part of the equation is linear in $\mathbf{v}$ and the third part is constant. To rewrite your equation in the form above, we set
$$ \mathbf{v} = \begin{pmatrix} x \\ y \\ z \end{pmatrix}, A = \begin{pmatrix} a & \frac{d}{2} & \frac{f}{2} \\
\frac{d}{2} & b & \frac{e}{2} \\
\frac{f}{2} & \frac{e}{2} & c\end{pmatrix}, \mathbf{p} = A^{-1} \begin{pmatrix} \frac{-g}{2} \\ \frac{-h}{2} \\ \frac{-i}{2} \end{pmatrix}, C = \mathbf{p}^T A \mathbf{p} - j.$$
If the eigenvalues of $A$ are all positive and $C > 0$, then your equation indeed describes an ellipsoid whose center is $\mathbf{p}$ and whose semi-principal axes are given by an orthonormal basis of eigenvectors of $A$ with lengths the corresponding eigenvalues.
|
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Why is my solution wrong? Question: "Find all x such that ${\frac{x-a}{b}}+{\frac{x-b}{a}}={\frac{b}{x-a}}+{\frac{a}{x-b}}$ , where a and b are constants."
My attempt:
${\frac{x-a}{b}}+{\frac{x-b}{a}}={\frac{b}{x-a}}+{\frac{a}{x-b}}$
Let $m=x-a$ ,Let $w=x-b$ . Therefore:
${\frac{m}{b}}+{\frac{w}{a}}={\frac{b}{m}}+{\frac{a}{w}}$
${\frac{m}{b}}-{\frac{b}{m}}={\frac{a}{w}}-{\frac{w}{a}}$
${\frac{m^2-b^2}{bm}}={\frac{a^2-w^2}{aw}}$
${\frac{(m+b)(m-b)}{bm}}={\frac{(a+w)(a-w)}{aw}}$
Substituting back $m=x-a$ , $w=x-b$ :
${\frac{(x-a+b)(x-a-b)}{b(x-a)}}={\frac{(a+x-b)(a-x+b)}{a(x-b)}}$
Factoring out $-1$ from $(a-x+b)$:
${\frac{(x-a+b)(x-a-b)}{b(x-a)}}={\frac{-(a+x-b)(x-a-b)}{a(x-b)}}$
Dividing by $(x-a-b)$:
${\frac{(x-a+b)}{b(x-a)}}={\frac{-(a+x-b)}{a(x-b)}}$
Cross multiplying:
$(x-a+b)(ax-ab)=(-x-a+b)(bx-ab)$
Distributing;
$ax^2-abx-a^2x+a^2b+abx-ab^2=-bx^2+abx-abx+a^2b+b^2x-ab^2$
$ax^2-a^2x+a^2b-ab^2=-bx^2+a^2b+b^2x-ab^2$
$ax^2-a^2x+a^2b-ab^2-(-bx^2+a^2b+b^2x-ab^2)=0$
$ax^2+bx^2-a^2x-b^2x=0$
$(a+b)x^2-(a^2+b^2)x=0$
Factoring out $x$:
$x[(a+b)x-(a^2+b^2)]=0$
Therefore:
$x=0$
or
$(a+b)x-(a^2+b^2)=0$
$(a+b)x=(a^2+b^2)$
$x=\frac{(a^2+b^2)}{(a+b)}$
My two solutions are $x=0$ and $x=\frac{(a^2+b^2)}{(a+b)}$; however, I am missing the solution $x=a+b$. Where did I eliminate this solution? How do I prevent this from happening (eliminating solutions) in the future? Is there a better way to solve this equation?
|
When you divided by $x-a-b$ you threw away the root $x=a+b$. Just like in the equation $t(t-1)=t(2t-5)$, if we cancel the $t$ we are losing the root $t=0$.
|
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|
Solve $\frac{\mathrm{d}y}{\mathrm{d}x} = (x-y)/(x+y)$ Solve
$$\frac { { d }y }{ { d }x } =\frac { x-y }{ x+y } $$
It is homogeneous, thus let $y = vx$. From this, $\frac{\mathrm{d}y}{\mathrm{d}x} = x\frac{\mathrm{d}v}{\mathrm{d}x} + v$
Thus,
$v'x + v = (1-v)/(1+v)$ thus,
$\frac{2}{1-v} + \ln(v - 1) = \ln(x) + C$.
Which by substituion,
$\frac{2x}{x-y} + \ln(y - x) = 2\ln(x) + C$
But I cant get it any further
|
$$\frac { dy }{ dx } =\frac { x-y }{ x+y } \\ \frac { dy }{ dx } =\frac { 1-\frac { y }{ x } }{ 1+\frac { y }{ x } } \\ y=zx\\ z^{ \prime }x+z=\frac { 1-z }{ 1+z } \\ z^{ \prime }x=\frac { 1-z }{ 1+z } -z=\frac { 1-2z-z^{ 2 } }{ 1+z } \\ \int { \frac { 1+z }{ 1-2z-z^{ 2 } } dz } =\int { \frac { dx }{ x } } \\ -\int { \frac { 1+z }{ { z }^{ 2 }+2z-1 } dz } =\int \frac { dx }{ x } \\ -\frac { 1 }{ 2 } \int \frac { d\left( z^{ 2 }+2z-1 \right) }{ z^{ 2 }+2z-1 } =\int { \frac { dx }{ x } } \\ \\ \ln { \left| z^{ 2 }+2z-1 \right| } =\ln { \frac { C }{ x^{ 2 } } } \\ z^{ 2 }+2z-1=\frac { C }{ x^{ 2 } } \\ \frac { y^{ 2 } }{ x^{ 2 } } +2\frac { y }{ x } -1=\frac { C }{ x^{ 2 } } $$
so the answer is :
$$\\ y^2 +2xy-x^2 =C\\ $$
|
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|
Solving the Inequality $\frac{14x}{x+1}<\frac{9x-30}{x-4}$ The question says to find all the integral values of x for which the inequality holds.
the question is
$$\frac{14x}{x+1}<\frac{9x-30}{x-4}$$
My Solution
\begin{align}
& \frac{14x}{x+1} < \frac{9x-30}{x-4} \\[6pt]
& \frac{14x(x-4)-(9x-30)(x+1)}{(x+1)(x-4)}<0 \\[6pt]
& \frac{14x^2-64x-9x^2+21x+30}{(x+1)(x-4)} < 0 \\[6pt]
& \frac{5x^2-43x+30}{(x+1)(x-4)}<0
\end{align}
using quadratic formula, roots of $5x^2-43x+30$ comes $7.83$(approx.) and $0.763$(approx.)
so rewriting the equation as
$$\frac{(x-7.83)(x-0.763)}{(x+1)(x-4)}<0$$
I then did the plotting of zeroes and poles on number line for finding the values for $x$ but I donot get 2 integral values (which is the answer). Can anyone tell where I did wrong?
|
Basic approach. Perhaps easier would be to rewrite the original inequality as
$$
14 - \frac{14}{x+1} < 9 + \frac{6}{x-4}
$$
This leads to
$$
\frac{14}{x+1} + \frac{6}{x-4} > 5
$$
which becomes
$$
\frac{4x-10}{(x+1)(x-4)} > 1
$$
You can now (a) consider the cases $x = 0, 1, 2, 3$ separately, and then otherwise, (b) for $x < -1$ or $x > 4$, we have
$$
(x+1)(x-4) < 4x-10
$$
which becomes
$$
x^2-7x+6 < 0
$$
which you should be able to handle. Keep in mind that this inequality is only valid for the subcase (b) $x < -1$ or $x > 4$.
(There's probably a simpler way, incidentally. See alans' comment, for instance. This is just what I wrote up.)
|
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|
Solving an Itô Integral Can someone please show me how to solve this Itô Integral?
$$\begin{align}\int_{1}^{t}\frac{dB_s}{B_s^2 + B_s^4} && \end{align} $$
|
Set $f(x)=-\frac{1}{x}-\tan^{-1}x$. we have
$$f'(x)=\frac{1}{x^2}-\frac{1}{1+x^2}=\frac{1}{x^2+x^4}$$
and
$$f''(x)=-\frac{2x+4x^3}{(x^2+x^4)^2}$$
By application of Ito's lemma we have
$$f(B_t)=f(B_1)+\int_{1}^{t}f'(B_s)dB_s+\frac{1}{2}\int_{1}^{t}f''(B_s)ds$$
therefore
$$-\left(\frac{1}{B_t}+\tan^{-1}(B_t)\right)=-\left(\frac{1}{B_1}+\tan^{-1}(B_1)\right)+\int_{0}^{t}\frac{1}{B_s^2+B_s^4}dB_s-\int_{0}^{t}\frac{B_s+2B_s^3}{(B_s^2+B_s^4)^2}ds$$
In other words
$$\int_{0}^{t}\frac{1}{B_s^2+B_s^4}dB_s=\int_{0}^{t}\frac{B_s+2B_s^3}{(B_s^2+B_s^4)^2}ds-\left(\frac{1}{B_t}+\tan^{-1}(B_t)\right)+\left(\frac{1}{B_1}+\tan^{-1}(B_1)\right)$$
|
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|
$A_{mn} = \frac{1}{\pi}\int_0^{\pi}d\theta\, \sin(2m\theta)\, \frac{1-\cos^{2n}(\theta)}{\tan(\theta)} = $ ? $m$, $n$ integers > 0 The integral
$$
A_{mn} = \frac{1}{\pi}\int_0^{\pi}d\theta\, \sin(2m\theta)\, \frac{1-\cos^{2n}(\theta)}{\tan(\theta)}
$$
popped up when I was playing around with the integral representation of the harmonic numbers $H_n$.
I find that Mathematica can give a result for specific values of $m$ and $n$ but I can't get a general formula. Some observations that appear to be true, based on Mathematica results for specific cases:
1) $A_{mn} > 0$.
2) When $m = n$, $A_{nn} = \frac{1}{4^n}$.
3) When $m > n$, $A_{mn} = 0$. (I think this should follow from expanding the $\cos^{2n}(\theta)$ part out in a finite sum of terms like $\sin$ or $\cos$ of integers times $\theta$.)
I tried doing this via contour integration by extending the integral (which is symmetric around $\pi$) to $[0,2\pi)$, and making the substitution $z = e^{i\theta}$. Unless I've made a mistake, this leads to:
$$
A_{mn} = \frac{1}{2\pi i}\oint dz\,
\frac{\left(z^{2m} - z^{-2m}\right)\left(z+z^{-1}\right)\left[1 - \frac{1}{2^{2n}}{\left(z + z^{-1}\right)}^{2n}\right]}{z^2 - 1}\, ,
$$
where the integral is over the counter-clockwise unit circle. This doesn't seem to work: The zeroes appear to cancel on the top and bottom at $z = \pm 1$, so the residues are zero.
Have I made a mistake in the contour integration approach? Is there a better way to evaluate this whole thing?
|
Using the identities $$\frac{\sin (2m \theta)}{\sin (\theta)} = 2 \sum_{k=0}^{m-1} \cos[(2k+1)\theta]$$
and
$$\cos^{2n+1}(\theta) = \frac{1}{4^{n}}\sum_{j=0}^{n} \binom{2n+1}{j} \cos [2n+1-2j)\theta], $$ we get
$$ \begin{align} A_{mn} &= \frac{1}{\pi} \int_{0}^{\pi} \frac{\sin (2 m \theta)}{\sin (\theta)}\left(\cos (\theta) - \cos^{2n+1}(\theta) \right) \, d \theta \\ &= \frac{2}{\pi} \sum_{k=0}^{m-1} \int_{0}^{\pi} \cos[(2k+1)\theta] \cos(\theta) \, d \theta \\ &- \frac{2}{\pi} \, \frac{1}{4^{n}}\sum_{k=0}^{m-1} \sum_{j=0}^{n} \binom{2n+1}{j}\int_{0}^{\pi} \cos[(2k+1)\theta ] \cos[(2n+1-2j) \theta] \, d \theta. \end{align} $$
First assume that $m <n$.
Then using the fact that
$$ \int_{0}^{\pi} \cos(mx) \cos(nx) \, dx = \begin{cases}
\frac{\pi}{2} & m = n \\
0 & \text{otherwise}
\end{cases} $$ we get
$$A_{mn} = \frac{2}{\pi} \left(\frac{\pi}{2} \right) - \frac{2}{\pi} \, \frac{1}{4^{n}} \sum_{j=n-m+1}^{n} \binom{2n+1}{j} \frac{\pi}{2} = 1 - \frac{1}{4^{n}} \sum_{j=n-m+1}^{n} \binom{2n+1}{j}. $$
Now if $m> n$, then
$$ A_{mn} = \frac{2}{\pi} \left(\frac{\pi}{2} \right)- \frac{2}{\pi} \, \frac{1}{4^{n}} \sum_{j=0}^{n} \binom{2n+1}{j} \frac{\pi}{2} = 1 - \frac{1}{4^{n}} \left(4^{n}\right) = 0. \tag{1}$$
And if $m=n$, $$A_{mn} = 1- \frac{1}{4^{n}} \sum_{j=1}^{n} \binom{2n+1}{j} = 1 - \frac{1}{4^{n}} \left(4^{n}-1\right) = \frac{1}{4^{n}}. $$
$(1)$ Prove the identity $\binom{2n+1}{0} + \binom{2n+1}{1} + \cdots + \binom{2n+1}{n} = 4^n$
|
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|
Integral value of $n$ that makes $n^2+n+1$ a perfect square. Find all integers $n$ for which $n^2+n+1$ is a perfect square.
By hit and trial we get $n=-1,0$ but could someone suggest any genuine approach as how to approach this problem?
|
As other solutions stated , one of your values can be obtained as like ,
$ n^2+n+1=k^2 \equiv 1 (mod 3) \implies n(n+1)\equiv 0 (mod 3)$
So $3|n$ or $3| (n+1)$ , put $n=3k$ , $f(k)=9k^2+3k+1=(3k+1)^2 - 3k$ which implies $n=k=0$ .
Secondly if $3|(n+1)$ put $n=3k-1$ & we get $ f(k)=9k^2-3k+1=(3k-1)^2+3k$ which forces $k=0$ & thus $n=3.0-1=-1$
We have $n^2<n^2+n+1<(n+1)^2$ and $n^2+n+1$ is perfect square for $n=0,-1$ .
|
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|
Find the coefficient of $ x^{12}$ in $(1-x^2)^{-5}$ Find the coefficient of $x^{12}$ in $(1-x^2)^{-5}$
What can be said for $x^{17}$
Tried $\frac{1}{(1-x^2)^{5}}$=$\sum_{n=0}^\infty \binom{n+5-1}{n}x^n$
not sure that i can do that with $x^2$
|
$$y=\frac{1}{1-x}=\sum_{n=0}^{\infty }x^n$$
the first derivative is
$$\frac{-1}{(1-x)^2}=\sum_{n=0}^{\infty }nx^{n-1}$$
the fourth derivative
$$\frac{24}{(1-x)^5}=\sum_{n=4}^{\infty }n(n-1)(n-2)(n-3)x^{n-4}$$
let $x\rightarrow x^2$
$$\frac{24}{(1-x^2)^5}=\sum_{n=4}^{\infty }n(n-1)(n-2)(n-3)x^{2n-8}$$
to find coefficient of $x^{12}$
let
$$2n-8=12$$
$$n=10$$
so the coefficient is
$$(10)(10-1)(10-2)(10-3)/24=210$$
|
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|
How to find $\tan x $ from $(a+1)\cos x + (a-1)\sin x=2a+1$? How do I find $\tan x$ from this equation?
$$(a+1)\cos x + (a-1)\sin x=2a+1$$
Thanks for any help!!
|
using the so-called Weierstrass substitution we get this here
$${\frac { \left( a+1 \right) \left( 1- \left( \tan \left( x/2 \right)
\right) ^{2} \right) }{1+ \left( \tan \left( x/2 \right) \right) ^{2
}}}+2\,{\frac { \left( a-1 \right) \tan \left( x/2 \right) }{1+
\left( \tan \left( x/2 \right) \right) ^{2}}}=2\,a+1
$$
further use that $$\tan(x+y)={\frac {\tan \left( x \right) +\tan \left( y \right) }{1-\tan \left( x
\right) \tan \left( y \right) }}
$$
|
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|
integrate $\int \frac{x^4-16}{x^3+4x^2+8x}dx$
$$\int \frac{x^4-16}{x^3+4x^2+8x}dx$$
So I first started with be dividing $p(x)$ with $q(x)$ and got:
$$\int x-4+\frac{8x^2+32x-16}{x^3+4x^2+8x}dx=\frac{x^2}{2}-4x+\int \frac{8x^2+32x-16}{x^3+4x^2+8x}dx$$
Using partial sum I have received:
$$\int \frac{8x^2+32x-16}{x^3+4x^2+8x}dx=8\int -\frac{1}{4x}+\frac{5x}{4(x^2+4x+8)}+\frac{5}{x^2+4x+8} dx=-2ln|x|+8(\frac{5}{4}\int\frac{x}{(x^2+4x+8)} +5\int \frac{1}{x^2+4x+8})=-2ln|x| +10\int\frac{x}{(x^2+4x+8)}dx +40\int \frac{1}{x^2+4x+8}dx$$
How do I continue from here?
|
HINT:
As $x^2+4x+8=(x+2)^2+2^2$ and $\dfrac{d(x^2+4x+8)}{dx}=2(x+2)$
for $\dfrac{Ax+B}{x^2+4x+8},$ express it as $$a\cdot\dfrac{2(x+2)}{x^2+4x+8}+b\cdot\dfrac1{(x+2)^2+2^2}$$
Choose $x+2=2\tan y$ for the second term
|
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|
Determine how many roots are real, and finding all roots of a quintic: $-2y^5 +4y^4-2y^3-y=0$ Using a computer we can see that the only real root of $f(y)=-2y^5 +4y^4-2y^3-y=0$ is $0$. Furthermore, we know from algebra that since this polynomial lives in $\Bbb R[y]$ that the roots come in complex conjugate pairs. I.e. we knew that there were either $1,3$ or $5$ real roots of this polynomial. Furthermore, if we could guess factors, we could complete polynomial long division to break up the polynomial. Noting that $y=0$ is a root, we want the $4$ roots of:
$$-2y^4+4y^3-2y^2-1=0$$
*
*How would we deduce that the remaining roots are not real?
*How would we find these by hand.
|
Note that$$-2y^4+4y^3-2y^2-1=-2y^2(y-1)^2-1<0$$
for real $y$, so the remaining roots are not real. We need
$$
y^2(y-1)^2=-\frac{1}{2},
$$
or
$$
y(y-1)=\pm\frac{i\sqrt{2}}{2}.
$$
For each choice of sign, this is a quadratic that you can solve with the usual formula. Specifically,
$$
y^2-y(\pm)_1\frac{i\sqrt{2}}{2}=0\implies y=\frac{+1(\pm)_2\sqrt{1(\mp)_12i\sqrt{2}}}{2}=\frac{1}{2}(\pm)_2\sqrt{\frac{1}{4}(\mp)_1\frac{i\sqrt{2}}{2}}.
$$
|
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|
Evaluate $\int \frac{\sin^4 x}{\sin^4 x +\cos^4 x}{dx}$ $$\int \frac{\sin^4 x}{\sin^4 x +\cos^4 x}{dx}$$
$$\sin^2 x =\frac{1}{2}{(1- \cos2x)}$$
$$\cos^2 x =\frac{1}{2}{(1+\cos2x)}$$
$$\int \frac{(1- \cos2x)^2}{2.(1+\cos^2 2x)}{dx}$$
$$\frac{1}{2} \int \left[1-\frac{2 \cos2x}{1+\cos^22x}\right] dx$$
What should I do next ?
Please also tell me alternative way to do this .
|
$$I=\int \frac{\sin^4 x}{\sin^4 x +\cos^4 x}{dx}=\int \frac{1-2\cos^2(x)+\cos^4(x)}{1-2\cos^2(x)+2\cos^4(x)}dx$$
$$=\int\frac{1}{2}-\frac 12\frac{-1+2\cos^2(x)}{1-2\cos^2(x)+2\cos^4(x)}dx$$
$$=\int\frac{1}{2}-\frac 12\frac{\cos(2x)}{1-2\cos^2(x)\sin^2(x)}dx$$
$\color{blue}{\text{This last step above is a few manipulations away from being the last step in your question.}}$
Let $u=\cos(x)\sin(x)$ then $du=\cos(2x)dx$
$$\boxed{I=\frac x2-\frac 12\int \frac 1{1-2u^2}du=\frac x2 -\frac12\frac{\tanh^{-1}(\sqrt2 u)}{\sqrt 2} +k_1}$$
Replace $u$ and you're good to go...
|
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|
Proving that $x^{16} > 5$ when given a polynomial of degree $15$. I am unable to prove the following
If $x^{15}-x^{13}+x^{11}-x^9+x^7-x^5+x^3-x = 7$ prove that $x^{16} > 15$.
|
$ x^{14} - x^{12} + \ldots + x^2 -1 = \frac{7}{x}, $ I am considering $ x \neq 0 $ multiply both side by $ x^2 $ and add you get $ x^{16} = 1+ 7x + \frac{7}{x}. $ Take minimum of R.H.S.
|
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|
If $n$ is a positive integer, then $(-2^n)^{-2} + (2^{-n})^2 = 2^{-2n+1}$ I'm not sure why $$(-2^n)^{-2} + (2^{-n})^2=2^{-2n+1}$$
I have been going over this equation for a while now, noticing, and have successfully got quite far in the equation, finding that
$$ (-2^n)^{-2} + (2^{-n})^2 \implies \frac{1}{(-2^n)^2} + \frac{1}{(2^n)^2} $$
which I think then becomes $ \dfrac{2}{2^{2n}}$
But then I get stuck.
|
The right hand side of the implies sign above can be studied, viz;
\begin{align}
\frac{1}{(-2^n)^2} + \frac{1}{(2^n)^2} &= \frac{2^{2n}+(-1)^{2n}2^{2n}}{(-1)^{2n}2^{4n}}\\
&= \frac{2.2^{2n}}{2^{4n}} \qquad (-1)^{2n}=1 \quad \forall n\\
&= \frac{2}{2^{2n}}
\end{align}
Thus,
\begin{align}
\frac{1}{(-2^n)^2} + \frac{1}{(2^n)^2} &= \frac{2}{2^{2n}} \\
\implies (-2^n)^{-2} + (2^n)^{-2}&=2.2^{-2n} \\
&= 2^{-2n+1}
\end{align}
Hope that helps.
|
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|
Monotonicity of the sequence $(a_n)$, where $a_n=\left ( 1+\frac{1}{n} \right )^n$ Define $a_n=\left ( 1+\frac{1}{n} \right )^n$ for $n\geq 1$. I want to show that it is increasing. First, we have
$$\frac{a_{n+1}}{a_n}=\left ( \frac{1+\frac{1}{n+1}}{1+\frac{1}{n}} \right )^n\left ( 1+\frac{1}{n+1} \right )=\left ( 1-\frac{1}{(n+1)^2} \right )^n\left ( \frac{n+2}{n+1} \right )$$
Using the Bernoulli inequality, we see that, for all $n\geq 1$,
$$\left ( 1-\frac{1}{(n+1)^2} \right )^n\geq 1-\frac{n}{(n+1)^2}=\frac{n^2+n+1}{n^2+2n+1}.$$
How do you then show that
$$\left ( 1-\frac{n}{(n+1)^2} \right )\left ( \frac{n+2}{n+1} \right )>1?$$
Edit: This is not a duplicate question; the question is not to show about the existence of the definition of Euler number $e$. The question is about showing that it is increasing, the way I have shown that I have been stuck with, not other ways. It seems that the question is too easy that I have been too tired to think at this late.
|
The left hand side is
$$\frac1{(n+1)^3}\cdot((n+1)^2-n)(n+2)=\frac{n^3+3n^2+3n+2}{(n+1)^3} $$
|
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|
Show $\cos\left( \pi n^{2}\ln\left(\frac{n}{n-1} \right) \right)=(-1)^{n+1}\frac{\pi}{3n}+\mathcal{O}\left( \frac{1}{n^2}\right) $
I would like to show :
$$\cos\left( \pi n^{2}\ln\left(\dfrac{n}{n-1} \right) \right)=(-1)^{n+1}\dfrac{\pi}{3n}+\mathcal{O}\left( \dfrac{1}{n^2}\right) $$
by starting from the left side and get the right side
My attempt:
\begin{align*}
\cos\left( \pi n^{2}\ln\left(\dfrac{n}{n-1} \right) \right)&=\cos\left( \pi n^{2}\ln\left(1+\dfrac{1}{n-1} \right) \right)\\
&=\cos\left( \pi n^{2}\left(\dfrac{1}{n-1}+\mathcal{O}\left(\dfrac{1}{(n-1)^{2}} \right) \right) \right)
\end{align*}
|
You may write
$$
\begin{align*}
\cos\left( \pi n^{2}\ln\left(\dfrac{n}{n-1} \right) \right)&=\cos\left( -\pi n^{2}\ln\left(\dfrac{n-1}{n} \right) \right)\\
&=\cos\left( -\pi n^{2}\ln\left(1-\frac1n \right) \right)\\
&=\cos\left( \pi n^{2}\left(\frac{1}{n}+\frac{1}{2 n^2}+\frac{1}{3 n^3}+\mathcal{O}\left(\dfrac{1}{n^4} \right) \right) \right)
\\&=\cos\left( \pi n+\frac{\pi}2+\frac{\pi}{3 n}+\mathcal{O}\left(\dfrac{1}{n^2} \right) \right)
\\&=-\sin\left( \pi n+\frac{\pi}{3 n}+\mathcal{O}\left(\dfrac{1}{n^2} \right) \right)
\\&=(-1)^{n+1}\sin\left(\frac{\pi}{3 n}+\mathcal{O}\left(\dfrac{1}{n^2} \right) \right)
\\&=(-1)^{n+1}\frac{\pi}{3 n}+\mathcal{O}\left(\dfrac{1}{n^2} \right)
\end{align*}
$$ as announced.
|
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Use lagrange multipliers to calculate the maximum and minimum $f(x,y,z)=x^2y^2z^2$ constrained by $x^2+y^2+z^2=1$
$\nabla f_x$ $=$ $2xy^{2}z^{2}$,
$\nabla f_y$ $=$ $2yx^{2}z^{2}$,
$\nabla f_z$ $=$ $2zx^{2}y^{2}$
$\nabla g_x$ $=$ $2x$,
$\nabla g_y$ $=$ $2y$,
$\nabla g_z$ $=$ $2z$
setting the sets equal to each other and multiplying each to get the equality where $\nabla f_{x..z} = 2x^2y^2z^2$ gives me(namely multiplying $x$ on $\nabla f_x = \lambda g_x$...and so on) :
$\lambda2x^2 = \lambda2y^2=\lambda2z^2 \to x=y=z$ after division and taking the roots. So:
$x=y=z=a \to 3a^2=1 \to a= \pm \frac{1}{\sqrt{3}}$
Thus a maximum at $f\Big(\pm\frac{1}{\sqrt{3}},\pm\frac{1}{\sqrt{3}},\pm\frac{1}{\sqrt{3}}\Big) = \frac{1}{27}$.
However looking at wolrfram alpha, it also gives a minimum at $f\Big(-\frac{1}{2},-\frac{\sqrt{3}}{2},0\Big) = 0$. I would like to know how there are more solutions from $\lambda 2x^2=\lambda 2y^2=\lambda 2z^2$ than just x=y=z
Using the corrections from posters lets say I let $x=0$ now I have $y^2+z^2=1$ Then if I wanted one scenario just to show there is a minimum I could say $y^2=z^2$ and then say $a^2+a^2=1 \to 2a^2=1 \to a^2=\frac{1}{2} \to a=\pm\frac{1}{\sqrt{2}}$ therefore have a minimum at $f\Big(0,\pm\frac{1}{\sqrt{2}},\pm\frac{1}{\sqrt{2}}\Big)$ Is this correct. I feel the only correct solution thought is to use set notation and proof to show there are infinitely many cases. But does my above asnwer for one of those cases hold?
|
You have three equations from the first order conditions:
$$2xy^2z^2=2\lambda x \qquad 2x^2yz^2=2\lambda y\qquad 2x^2y^2z=2\lambda z $$
Suppose $x=0$. Then the first equation is satisfied, and the other two equations imply that $\lambda=0$ (since we cannot have $x=y=z=0$ as that does not satisfy the constraint). This gives us infinitely many solutions with $\lambda=x=0$, since we can choose any $y$ and $z$ such that $y^2+z^2=1$. (Geometrically, these solutions are just intersections of the constraint set - a sphere of radius $1$ centred at the origin - and the $x=0$ plane. The intersection is a circle of radius $1$ centred at the origin in the $x=0$ plane.) Mathematically, the set of such minimizers is:
$$\{(0,y,z)\in\mathbb{R}^3\ \ |\ \ y^2+z^2=1\}$$
Similar arguments hold for the cases $y=0$ and $z=0$ (everything is symmetric). These critical points are all minimizers of $f$ on the constraint set. The minimum is $0$. (Notice that $\lambda=0$ here because even if we did not have the equality constraint, these points would minimize $f$.)
You lost all these solutions because you inadvertently assumed all the variables were nonzero when you divided the equations by one another.
Now suppose that $x$, $y$ and $z$ are all nonzero. Then $\lambda\neq0$ (this is implied by any of the three equations). Now since all the variables are nonzero, we can divide the equations by $2x$, $2y$ and $2z$ respectively to get
$$y^2z^2=\lambda \qquad x^2z^2=\lambda \qquad x^2y^2=\lambda. $$
Then dividing each equation by each other equation we find that we require $$x^2=y^2=z^2.$$
Plugging into the constraint then implies that $$x^2=y^2=z^2=\frac{1}{3}$$
and $\lambda =1/9$ (from any of the three equations). This gives us $8$ solutions, which are maximizers of $f$ on the constraint set. The maximum is $1/27$.
|
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How to solve the equation $xy = 1, x^{2x-y} = y^{2(x-y)}$ I have the following equation that I don't know how to solve:
$$ \begin{cases} xy = 1 \\ x^{2x-y} = y^{2(x-y)} \end{cases} $$
Here's what I've tried (but my mathematical instinct tells me that I didn't solve it right):
$$ \begin{cases} xy = 1 \\ x^{2x-y} = y^{2(x-y)} \end{cases} \rightarrow \begin{cases} x = \frac {1}{y} \\ \frac {1}{y}^{2x-y} = y^{2(x-y)} \end{cases} \rightarrow y^{y-2x} = y^{2(x-y)} \rightarrow y-2x = 2x-2y \rightarrow 4x=3y \rightarrow \frac{4}{y} = 3y \rightarrow 3y^2-4 = 0\rightarrow \begin{cases} y_1=\frac{-2\sqrt{3}}{3} \\ y_2= \frac{2\sqrt{3}}{3}\end{cases} \rightarrow \begin{cases} y_1=\frac{-2\sqrt{3}}{3} \\ y_2= \frac{2\sqrt{3}}{3}\end{cases} \rightarrow \begin{cases} x_1=\frac{-\sqrt{3}}{2} \\ x_2= \frac{\sqrt{3}}{2}\end{cases}$$
|
Your attempted solution is correct. But you missed the trivial solution $(x,y)=(1,1)$.
|
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|
Evaluating the sums $\sum\limits_{n=1}^\infty\frac{1}{n \binom{kn}{n}}$ with $k$ a positive integer
How to evaluate the sums $\sum\limits_{n=1}^\infty\frac{1}{n \binom{kn}{n}}$ with $k$ a positive integer?
For $k=1$, the series does not converge.
When $k=2$, I can prove that:
$$\sum_{n=1}^\infty\frac{1}{n \binom{2n}{n}}=\frac{\pi}{3\sqrt{3}}$$
Usually, this can be proven by differentiating $\sum_{n=1}^\infty \frac{x^{2n}}{n^2 \binom{2n}{n}}=2(\arcsin{\frac{x}{2}})^2$, but I have an alternative proof.
Using the result of:
$$\int_0^\infty \frac{x^ndx}{(x+1)^{y+n+1}}=\frac{1}{y \binom{y+n}{n}} \tag1$$
, which can be easily proved.
I can substitute $y=n$ to obtain
$$\int_0^\infty \frac{x^ndx}{(x+1)^{2n+1}}=\frac{1}{n \binom{2n}{n}}$$
$$\sum_{n=1}^\infty\int_0^\infty \frac{x^ndx}{(x+1)^{2n+1}}=\sum_{n=1}^\infty\frac{1}{n \binom{2n}{n}}$$
Therefore,
\begin{align}
\sum_{n=1}^\infty\frac{1}{n \binom{2n}{n}} & = \int_0^\infty \frac{xdx}{(x+1)(x^2+x+1)} \\
& = \lim_{L\to \infty} \frac{1}{2}\ln(x^2+x+1)-\ln(x+1)+\frac{\tan^{-1}(\frac{2x+1}{\sqrt{3}})}{\sqrt{3}}\large{|_0^L} \\
&= \frac{\pi}{3\sqrt{3}}
\end{align}
Now for $k=3$, I tried to substitute $y=2n$ into $(1)$:
$$\int_0^\infty \frac{x^ndx}{(x+1)^{3n+1}}=\frac{1}{2n \binom{3n}{n}}$$
$$\sum_{n=1}^\infty\int_0^\infty \frac{x^ndx}{(x+1)^{3n+1}}=\sum_{n=1}^\infty\frac{1}{2n \binom{3n}{n}}$$
So we can have
$$\sum_{n=1}^\infty\frac{1}{n \binom{3n}{n}}=\int_0^\infty\frac{2xdx}{(x+1)(x^3+3x^2+2x+1)}$$
However, by partial fraction $$\frac{2xdx}{(x+1)(x^3+3x^2+2x+1)}=-\frac{2}{1+x}+\frac{2x^2+4x+2}{x^3+3x^2+2x+1}$$
The left part does not seem to converge.
Feeling frustrated, Wolfram Alpha plays its part. It spits out these results:
$$\sum_{n=1}^\infty\frac{1}{n \binom{3n}{n}}=\frac{1}{3}{}_3F_2\left(\left.\begin{array}{c} 1,1,\frac{3}{2}\\ \frac{4}{3}, \frac{5}{3} \end{array}\right| \frac{4}{27}\right)$$
$$\sum_{n=1}^\infty\frac{1}{n \binom{4n}{n}}=\frac{1}{4}{}_4F_3\left(\left.\begin{array}{c} 1,1,\frac{4}{3},\frac{5}{3}\\ \frac{5}{4}, \frac{6}{4}, \frac{7}{4} \end{array}\right| \frac{27}{256}\right)$$
However, I am not very familiar with hypergeometric function.
The pattern suggests that $$\sum_{n=1}^\infty\frac{1}{n \binom{2n}{n}}=\frac{1}{2}{}_2F_1\left(\left.\begin{array}{c} 1,1\\ \frac{3}{2} \end{array}\right| \frac{1}{4}\right)=\frac{\pi}{3\sqrt{3}}$$
Thus, $${}_2F_1\left(\left.\begin{array}{c} 1,1\\ \frac{3}{2} \end{array}\right| \frac{1}{4}\right)=\frac{2\pi}{3\sqrt{3}}$$
Arriving these results, I have the following questions:
How can ${}_2F_1\left(\left.\begin{array}{c} 1,1\\ \frac{3}{2} \end{array}\right| \frac{1}{4}\right)$ be expressed into this simple elementary form?
How can we arrive to the result for $\sum_{n=1}^\infty\frac{1}{n \binom{3n}{n}}$ given by Wolfram Alpha?
Ultimately, can we evaluate $\sum_{n=1}^\infty\frac{1}{n \binom{kn}{n}}$ for all integers k $\ge$ $2$?
|
I'm late for this party but the appearance of the plastic constant's minpoly, or $x^3-x-1=0$, got me interested.
The binomial sum can be expressed as a concise finite sum of logarithms. For $k>1$,
$$\sum_{n=1}^\infty\frac1{n\binom{kn}n} =\sum_{n=1}^k \frac{\ln(1-x_n)}{k-(k-1)x_n}=\int_1^\infty\frac1{x(x^k-x+1)}$$
and where the $x_n$, naturally enough, are the roots of $x^k-x+1=0$. Example, for $k=3$,
$$A=\sum_{n=1}^\infty\frac1{n\binom{3n}n} = \frac{\ln(1-x_1)}{3-2x_1}+ \frac{\ln(1-x_2)}{3-2x_2}+ \frac{\ln(1-x_3)}{3-2x_3}=0.371216\dots$$
and the $x_n$ are the three roots of $x^3-x+1=0$, one of which is the negated plastic constant $x\approx-1.32472$. It was pointed out that equivalently,
$$3A ={_3F_2}\left(\frac32,1,1;\ \frac43,\frac53;\ \frac4{27}\right)$$
Interestingly, the plastic constant also appears in a similar generalized hypergeometric function,
$$2B ={_3F_2}\left(\frac12,1,1;\ \frac43,\frac53;\ \frac4{27}\right)$$
discussed by Reshetnikov in this post.
|
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Evaluate the expression $\sqrt{6-2\sqrt5} + \sqrt{6+2\sqrt5}$ $$\sqrt{6-2\sqrt5} + \sqrt{6+2\sqrt5}$$
Can anyone tell me the formula to this expression.
I tried to solve in by adding the two expression together and get $\sqrt{12}$ but as I insert each expression separately in calculator the answer is above $\sqrt{12}$.
|
More generally,
if
$t
=\sqrt{a+\sqrt{b}}+\sqrt{a-\sqrt{b}}
$,
$\begin{array}\\
t^2
&=(\sqrt{a+\sqrt{b}}+\sqrt{a-\sqrt{b}})^2\\
&=a+\sqrt{b}+2(\sqrt{a+\sqrt{b}}\sqrt{a-\sqrt{b}})+a-\sqrt{b}\\
&=2a+2\sqrt{(a+\sqrt{b})(a-\sqrt{b})}\\
&=2a+2\sqrt{a^2-b}\\
\text{so}\\
t
&=\sqrt{2a+2\sqrt{a^2-b}}\\
\end{array}
$
In this case,
$a=6$
and
$b=20$
so
$t
=\sqrt{2\cdot 6+2\sqrt{36-20}}
=\sqrt{12+8}
=\sqrt{20}
=2\sqrt{5}
$.
|
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Proof that $P(x)=x-\frac{1}3 x^3+\frac{1}{5}x^5-\frac{1}{7}x^7+\cdots$ has radius of convergence $1$
Proof that
$P(x)=x-\frac{1}{3}x^3+\frac{1}{5}x^5-\frac{1}{7}x^7+\cdots$ has
radius of convergence $1$
First of all, I need to convert this to a series:
$$\sum_{k=1}^\infty \frac{x^{2k-1}(-1)^k}{1-2k}$$
(I hope this is correct so far?)
On this series, you use ratio test, count the limit and then take the reciprocal of the limit which will hopefully equal 1:
$$\lim_{k\rightarrow\infty}\left (\frac{x^{2(k+1)-1}(-1)^{k+1}}{1-2(k+1)}: \frac{x^{2k-1}(-1)^k}{1-2k}\right ) = \lim_{k\rightarrow\infty}\frac{\left(x^{2k+1}(-1)^{k+1}\right)(1-2k)}{(-2k-1) (x^{2k-1}(-1)^k)}$$
$$=\lim_{k\rightarrow\infty}\left(\frac{x^{2k+1}}{x^{2k-1}}\right ) \cdot \left(\frac{(-1)^{k+1}}{(-1)^k}\right)\cdot\left(\frac{1-2k}{-2k-1}\right)$$
$$=\lim_{k\rightarrow\infty}\left (x^{2k+1-2k+1} \right )\cdot\left ((-1)^{k+1-k} \right )\cdot\left (\frac{k(\frac{1}{k}-2)}{k(-2-\frac{1}{k})} \right )$$
$$=\lim_{k\rightarrow\infty}x^2\cdot((-1)^k)\cdot\left (\frac{\frac{1}{k}-2}{-2-\frac{1}{k}} \right )=-x^2\cdot\left(\frac{0-2}{-2-0} \right) = -x^2$$
$$R=-\frac{1}{x^2}$$
After all, I don't get 1 as result... :(
Where is my mistake? I cannot imagine it's as complicated and long as I did?
Or maybe, at the end, I ignore the $x^2$? Then result would be 1!?
|
Taking the absolute value of your algebra gives $x^2$. Since the ratio test looks for when this is $<1$, we ask ourselves when is $|x^2|<1$. The answer is when $-1<x<1$, and if $|x|>1$ it does not converge. This suffices to find the radius; it is $1$.
If you want the interval of convergence as well, then you need to test when $|x|=1$. In particular, just test $x=1$. Then your sum becomes $$1-\frac13+\frac15-\frac17+\ldots$$
which must converge by the alternating series test. Therefore, the interval of convergence is $-1 \leq x \leq 1$.
Note, the radius is the same
|
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|
properties of ratio and rule of cross multiplication If $$\frac{l}{\sqrt a-\sqrt b}+\frac{m}{\sqrt b-\sqrt c}+\frac{n}{\sqrt c-\sqrt a} =0$$
$$\frac{l}{\sqrt a+\sqrt b}+\frac{m}{\sqrt b+\sqrt c}+\frac{n}{\sqrt c+\sqrt a} =0$$
Show that $$ \frac{l}{(a-b)(c-\sqrt ab)}=\frac{m}{(b-c)(a-\sqrt bc)}=\frac{n}{(c-a)(b-\sqrt ac)}$$
I try to solve it with all the properties of ratio and rule of cross muliplication ,which I know.
I get $(a-b)$ term in denominator but I can't able to get $(c-\sqrt ab)$ term. I want to solve it on my own so give me hint to solve it. Thanks in advance.
|
For $a,b,c>0,$ let $a =A^2$ etc.
As $(B+C)(C-A)-(B-C)(C+A)=2(C^2-AB),$
$$\dfrac l{\dfrac n{(B-C)(C+A)}-\dfrac n{(B+C)(C-A)}}=\dfrac m{\cdots}=\dfrac1{\dfrac1{(A-B)(B+C)}-\dfrac1{(A+B)(B-C)}}$$
$$\implies\dfrac {l(B^2-C^2)(C^2-A^2)}{2(C^2-AB)}=\dfrac m{\cdots}=\dfrac{n(A^2-B^2)(B^2-C^2)}{2(B^2-CA)}$$
$$\iff\dfrac {l(C^2-A^2)}{C^2-AB}=\dfrac{n(A^2-B^2)}{B^2-CA}$$
Can you take it from here?
|
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|
Find the real and imaginary parts of an equation
Find the real and imaginary parts of $\frac{1}{3z+2}$
So I have expanded it out to get $\frac{1}{3x+3iy+2}$
Thus giving $Re(\frac{1}{3z+2})=\frac{1}{3x+2}$ and $Im(\frac{1}{3z+2})=\frac{1}{3y}$
However in my answer book it says: $Re(\frac{1}{3z+2})=\frac{3x+2}{(3x+2)^2+9y^2}$ and $Im(\frac{1}{3z+2})=\frac{-3y}{(3x+2)^2+9y^2}$
Is the book incorrect/outdated or if not could someone explain how to gain these answers, thanks
|
Note quite, remember that $\frac{1}{a} + \frac{1}{b} \neq \frac{1}{a+b}$. In general, the real and imaginary part of $z$ is $x$ and $y$ where $z = x+iy$. That is, you must seek to turn your expression for $z$ into the form $x+iy$. You can then multiply $\frac{1}{3x + 3iy + 2}$ with a fraction that consists of its conjugate on the numerator and denominator. In general $$\frac{1}{a+ib} = \frac{a-ib}{(a+ib)(a-ib)} = \frac{a-ib}{a^2 + b^2}.$$
In your case, try $$\frac{1}{(3x+2) +3iy} \times \frac{(3x+2) - 3iy}{(3x+2) - 3iy} = \frac{3x+2 - 3iy}{(3x+2)^2 + 9y^2}$$
|
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Minimum value of $f(x) = \sqrt{x^2 + (1-x)^2} + \sqrt{(1-x)^2 +(1+x)^2}$ If $f(x) = \sqrt{x^2 + (1-x)^2} + \sqrt{(1-x)^2 +(1+x)^2}$
Find the minimum value of the function
I tried using the AMGM inequality and differentiation but didn't know how to solve it any ideas?
This is from a math competition. ( I would like to see the most efficient way as I think differentiation in a math competition is not that efficient)
Using Mogjals comment and using the AM-GM inequality , setting $A = \sqrt{x^2 + (1-x)^2}$ and $B=\sqrt{(1-x)^2 +(1+x)^2}$
then
$A+B \geq 2\sqrt{AB}$
$$ \sqrt{x^2 + (1-x)^2} + \sqrt{(1-x)^2 +(1+x)^2} \geq 2\sqrt{2}\sqrt{x^2+1}\sqrt{2x^2-2x+1}$$
With equality if and only $A=B$ so $x=-\frac{1}{2}$
|
Write $f(x)=g(x)+h(x)$. Now $g(x)$ has minima at $x=1/2$ and $h(x)$ is increasing function with minima at $x=0$.Then $f(x)$ has minima at ?
|
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For which $a$, equation $4^x-a2^x-a+3=0$ has at least one solution.
Find all values of $a$ for which the equation $4^x-a2^x-a+3=0$ has at least one solution.
$\bf{My\; Try::}$ We can write it as $$2^{x}-a-\frac{a}{2^x}+\frac{3}{2^x}=0$$
So $\displaystyle \left(2^x+\frac{3}{2^x}\right)=a\left(1+\frac{1}{2^x}\right).$
Now for the existance of solution $\displaystyle 2^x+\frac{3}{2^x}\geq 2\sqrt{3}$ Using $\bf{A.M\geq G.M}$
So $\displaystyle a\left(1+\frac{1}{2^x}\right)\geq 2\sqrt{3}\Rightarrow a\geq \sqrt{3}\cdot \frac{2^x}{2^x+1}\geq \sqrt{3}$
But answer given as $a\geq 2,$ please explain me whats wrong with that, Thanks
|
You found a necessary condition, but not a sufficient one. You've concluded that
$$a \ge \frac{2^x + \frac{3}{2^x}}{1 + \frac 1 {2^x}}$$
Your application of AM-GM is all about minimizing the numerator, but it doesn't minimize the entire fraction. In fact, AM-GM is sharp when $2^x = \sqrt{3}$, in which case the fraction is $6 / (\sqrt{3} + 1) \approx 2.2$.
To minimize the whole thing: Recognize that it's equivalent to
$$a \ge \frac{4^x + 3}{2^x + 1}$$
As $x \to \infty$, this blows up; as $x \to -\infty$, this tends to $3$. Standard calculus techniques (e.g. root of first derivative) shows that this is minimized when $x = 0$, and the minimum is in fact $2$. By continuity, each value of $a \ge 2$ has a corresponding $x$. Hence $a \ge 2$ is sufficient.
|
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|
$x^4 - 4x^3 + 6x^2 - 4x + 1 = 0$ Determine all the possibilities for rational roots of the polynomial $x^4 - 4x^3 + 6x^2 - 4x + 1 = 0$. Then determine how many of the real roots of the polynomial may be positive and how many may be negative. Factor the polynomial to confirm your results.
The answer is possible rational roots: $+-1$; number of possible real roots - positive: four or two or zero, negative: zero; actual roots: $x = 1, 1, 1, 1$ (a quadruple root).
Using the rational root theorem, you divide the factors of the constant, $1$, by the factors of the lead coefficient, also a 1. That step gives you only two different possibilities for rational roots: $1$ and $-1$.
The signs change four times in the original polynomial, indicating $4$ or $2$ or $0$ positive real roots. Replacing each $x$ with $-x$, you get $x^4 + 4x^3 + 6x^2 + 4x + 1 = 0$. The signs never change. The polynomial is the fourth power of the binomial $(x - 1)$, so it factors into $(x - 1)^4 = 0$, and the roots are $1, 1, 1, 1$. There are four positive roots (all the same number, of course).
Can someone explain, the factorization of the polynomial? I do not understand, how it factors into $(x - 1)^4$.
|
$$
\begin{align}
n^4 - 4n^3 + 6n^2 - 4n + 1 = 0\\
n^4 - n^3 - 3n^3 + 3n^2 + 3n^2 - 3n - n + 1 = 0\\
n^3(n - 1) - 3n^2(n - 1) + 3n(n - 1) - 1(n-1) = 0\\
(n - 1)(n^3 - 3n^2 + 3n - 1) = 0\\
(n - 1)(n^3 - n^2 -2n^2 + 2n + n - 1) = 0\\
(n - 1)[n^2(n - 1) - 2n(n - 1) + 1(n - 1)] = 0\\
(n - 1)(n - 1)(n^2 - 2n + 1) = 0\\
(n - 1)(n - 1)(n - 1)^2 = 0\\
(n - 1)^4 = 0\\
n - 1 = 0\\
\therefore n = 1
\end{align}
$$
Hope you get it.....
.....--))
|
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Two tangents to the hyperbola $4x^2-y^2=36$ intersect at the point $(0,4)$. Find the coordinates for the points on the hyperbola for this to occur. The question is two tangents to the hyperbola $4x^2-y^2=36$ intersect at the point $(0,4)$. Find the coordinates for the points on the hyperbola for this to occur.
My attempt:
Consider the line $y=mx+c$ as it passes through $(0,4)$ then $c=4$ --> Condition 1
Plug in $y=mx+c$
$$ 4x^2-(mx+c)^2=36$$
$$4x^2-36=m^2x^2+2mcx+c^2$$
$$(m^2-4)x^2+(2mc)x^2+c^2+36=0$$
For tangency $b^2-4ac=0$
Hence
$$ 4m^2c^2-4(c^2+36)(m^2-4)=0$$
$$ 16c^2-144m^2+576=0$$
But $c=4$ due to the line condition hence
$$ m = \pm \frac{2\sqrt{13}}{3}$$
So the lines are
$y= \frac{2\sqrt{13}}{3}x + 4$ and $y= -\frac{2\sqrt{13}}{3}x + 4$
How to find the coordinates for the points on the hyperbola for this to occur tho?
|
The hyperbola can be written as $\frac{x^2}{9} - \frac{y^2}{36} = 1$ and let the tangent at $(3\sec\theta, 6\tan \theta)$ pass through $(0,4)$. The equation of the tangent is $\frac{x}{3}\sec\theta - \frac{y}{6}\tan\theta = 1$ and since $(0,4)$ lies on this, we have $\tan\theta = -\frac{3}{2}$. Thus the points are given by
$\left(\frac{3}{2}\sqrt{13}, -9\right), \left(-\frac{3}{2}\sqrt{13}, -9\right)$ according as we take $\cos \theta$ as positive or negative.
|
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|
Simplifying Ramanujan-type Nested Radicals Ramanujan found many awe-inspiring nested radicals, such as...
$$\sqrt{\frac {1+\sqrt[5]{4}}{\sqrt[5]{5}}}=\frac {\sqrt[5]{16}+\sqrt[5]{8}+\sqrt[5]{2}-1}{\sqrt[5]{125}}\tag{1}$$$$\sqrt[4]{\frac {3+2\sqrt[4]{5}}{3-2\sqrt[4]{5}}}=\frac {\sqrt[4]{5}+1}{\sqrt[4]{5}-1}\tag{2}$$$$\sqrt[3]{\sqrt[5]{\frac {32}{5}}-\sqrt[5]{\frac {27}{5}}}=\frac {1+\sqrt[5]{3}+\sqrt[5]{9}}{\sqrt[5]{25}}\tag{3}$$$$\sqrt[3]{(\sqrt{2}+\sqrt{3})(5-\sqrt{6})+3(2\sqrt{3}+3\sqrt{2})}=\sqrt{10-\frac {13-5\sqrt{6}}{5+\sqrt{6}}}\tag{4}$$$$\sqrt[6]{4\sqrt[3]{\frac {2}{3}}-5\sqrt[3]{\frac {1}{3}}}=\sqrt[3]{\sqrt[3]{2}-1}=\frac {1-\sqrt[3]{2}+\sqrt[3]{4}}{\sqrt[3]{9}}\tag{5}$$
And there's more!
Question: Is there a nice algebraic way to denest each radical such as above?
For me, I've only been able to prove such identities by raising both sides to the appropriate exponents and use Algebra to simplify them. But sometimes, that can be very difficult for identities such as $(1)$.
|
Landau's algorithm: http://ieeexplore.ieee.org/xpl/articleDetails.jsp?arnumber=63496 $\qquad\qquad$
https://en.wikipedia.org/wiki/Susan_Landau
|
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|
Problem in the solution of a trigonometric equation $\tan\theta + \tan 2\theta+\tan 3\theta=\tan\theta\tan2\theta\tan3\theta$
I needed to solve the following equation:
$$\tan\theta + \tan 2\theta+\tan 3\theta=\tan\theta\tan2\theta\tan3\theta$$
Now, the steps that I followed were as follows.
Transform the LHS first:
$$\begin{split}
\tan\theta + \tan 2\theta+\tan 3\theta
&= (\tan\theta + \tan 2\theta)
+ \dfrac{\tan\theta + \tan 2\theta}
{1-\tan\theta\tan2\theta} \\
&= \dfrac{(\tan\theta + \tan 2\theta)(2-\tan\theta\tan2\theta)}
{1-\tan\theta\tan2\theta}
\end{split}$$
And, RHS yields
$$\begin{split}
\tan\theta\tan2\theta\tan3\theta
&= (\tan\theta\tan2\theta)\dfrac{\tan\theta + \tan 2\theta}
{1-\tan\theta\tan2\theta}
\end{split}$$
Now, two terms can be cancelled out from LHS and RHS, yielding the equation:
$$
\begin{split}
2-\tan\theta\tan2\theta &= \tan\theta\tan2\theta\\
\tan\theta\tan2\theta &= 1,
\end{split}$$
which can be further reduced as:
$$\tan^2\theta=\frac{1}{3}\implies\tan\theta=\pm\frac{1}{\sqrt3}$$
Now, we can yield the general solution of this equation:
$\theta=n\pi\pm\dfrac{\pi}{6},n\in Z$. But, setting $\theta=\dfrac{\pi}{6}$ in the original equation is giving one term $\tan\dfrac{\pi}{2}$, which is not defined.
What is the problem in this computation?
|
Set, provisionally, $x=\tan\theta$ and $y=\tan2\theta$. Then the equation becomes
$$
x+y+\frac{x+y}{1-xy}=xy\frac{x+y}{1-xy}
$$
Note that the expression only makes sense for $1-xy\ne0$.
After removing the denominator, we get
$$
(x+y)(2-xy)=xy(x+y)
$$
so this becomes
$$
2(x+y)(1-xy)=0
$$
Since $1-xy\ne0$, we remain with $x+y=0$. So $\tan\theta=-\tan2\theta$ and therefore
$$
\tan\theta=\tan(\pi-2\theta)
$$
Hence
$$
\theta=\pi-2\theta+k\pi
$$
that gives
$$
\theta=\frac{\pi}{3}+k\frac{\pi}{3}
$$
and this can be simplified into $\theta=k\pi/3$.
|
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|
What is the $\lim_\limits{x \to 0} \frac{\cos(x)-1+\frac{x^2}{2}}{x^4}$? What is the limit of
$\lim_\limits{x \to 0} \frac{\cos(x)-1+\frac{x^2}{2}}{x^4}$?
I attempted the problem via L^Hopital's Rule so I rewrote it as
$$y=\frac{\cos(x)-1+\frac{x^2}{2}}{x^4}$$
then took the natural of both sides
$$\ln(y)=\ln(\frac{\cos(x)-1+\frac{x^2}{2}}{x^4})$$
then using the properties of logarithms I came to the conclusion that
$$\ln(y)= \ln(\cos(x)-1+\frac{x^2}{2})+\ln(x^4)$$
$$\ln(y)= \ln(\cos(x)-1+\frac{x^2}{2})+4\cdot \ln(x)$$
So then I took the limit as the $\ln(y)$ approaches 0.
$$\lim_\limits {x \to 0} (\ln(\cos(x)-1+\frac{x^2}{2})+4\cdot \ln(x))$$
Here I used L'Hopital's Rule and got
$$\lim_\limits {x \to 0} \frac{-\sin(x)+x}{\cos(x)-1+x^2}+\frac{4}{x}$$
then got a common denominator
$$\lim_\limits {x \to 0} \frac{-\sin(x)+x+4 \cdot (\cos(x)-1+x^2)}{x\cdot (\cos(x)-1+x^2)}$$
I clearly made a mistake somewhere because the denominator is 0. The answer by the way is $\frac{1}{24}$. I have no idea how to arrive at that conclusion.
|
I believe the problem should read $$\lim_{x \to 0} \frac{\cos(x) - 1 + \tfrac {x^2}2}{x^4};$$ i.e., the sign of the $1$ should be flipped. Otherwise the top goes to $2$ and the bottom to $0$ so the limit is $+\infty$. As I have written it, you can simply use l'Hopital's rule several times: \begin{align*}\lim_{x \to 0} \frac{\cos(x) - 1 + \tfrac {x^2}2}{x^4} &= \lim_{x \to 0} \frac{-\sin(x) +x }{4x^3} \\
&= \lim_{x \to 0} \frac{-\cos(x) + 1}{12x^2} \\
&= \lim_{x\to 0} \frac{\sin(x)}{24x}\\ &= \lim_{x\to 0}\frac{\cos(x)}{24} = \frac 1 {24}.\end{align*}
|
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|
How to prove that $\frac{a^ab^b}{a!b!}\le\frac{(a+b)^{(a+b)}}{(a+b)!}$? As stated in the question. Thank you!
$$\frac{a^ab^b}{a!b!}\le\frac{(a+b)^{(a+b)}}{(a+b)!}$$
|
Just using calculus.
If you use Stirling approximation, that is to say $$n! \sim \sqrt{2\pi n}\left(\frac n e\right)^n$$ the expression $$\frac{a^ab^b}{a!b!}\le\frac{(a+b)^{(a+b)}}{(a+b)!}$$ becomes $$\frac{e^{a+b}}{2 \pi \sqrt{a} \sqrt{b}}\le \frac{e^{a+b}}{\sqrt{2 \pi } \sqrt{a+b}}$$ which becomes $$1 \le \frac{\sqrt{2 \pi } \sqrt{a} \sqrt{b}}{\sqrt{a+b}}=\frac{\sqrt{2 \pi a b} }{\sqrt{a+b}}\implies a+b \leq 2\pi a b$$ which looks simpler to analyze.
|
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|
find the number of pairs $(a,b)$ where $a,b,n\in\mathbb N$ and $1\le aGiven a positive integer $n$, find the number of pairs $(a,b)$ where $a,b\in\mathbb N$ such that $(a^2-1)$ is divisible by $b$ and $(b^2-1)$ is divisible by $a$, and where $1 \le a < b \le n$.
Please help me how to solve this problem in constant time complexity. Thank you.
|
Clearly $(1,b)$ works always, so we concentrate on the other solutions.
Suppose we have $a$ and $b$ such that $a|(b+1)(b-1)$ and $b|(a+1)(a-1$).
Lemma $1$: Since $(a+1,a-1)=2$ or $1$ we have $a|2(b+1)$ or $a|2(b-1)$.
Lemma $2$: If $a$ is odd then $a|(b+1)$ or $a|(b-1)$. In particular $a\leq b+1$
Clearly $a$ and $b$ can't both be even.
If $a$ and $b$ are both odd by the lemma we have $a\leq b+1$ and $b\leq a+1$, so the only options are $(n,n+1)$ and $n,n$, of course only $(n,n)$ can have both odd. And clearly $(n,n)$ only works if $n=1$.
We now assume $a$ is odd and $b$ is even.
Case $1: a|(b-1)$, if $a\neq b-1$ then $a\leq (b-1)/3$ (because $(b-1)$ is odd). Notice that we must have $b|2(a+1)$ or $b|2(ba-1)$ which implies $2a+2\geq b$.
We have $a\leq(b-1)/3\implies 3a+1\leq b$ and $2a+2\geq b$, which would imply $a=1$. So the only possible solution is $a=b-1$, or $(a,a+1)$ which clearly always works.
Case $2$: $a|b+1$, if $a\neq b+1$ then $a\leq(b+1)/3$ (because $b+1$ is odd). Notice we have $b|2(a+1)$ or $b|2(a-1)$ which implies $2a+1\geq b$.
We have $a\leq(b+1)/3\implies 3a-1\leq b$ and $2a+2\geq b$, which would imply $a=1,2$ or $3$. When $a=2$ clearly $b$ must divide $2^2-1=3$, and $(2,3)$ works. When $a=3$ clearly $b$ must be $1,2,4$ or $8$. We notice $(3,8)$ works. So the only other option is $a=b+1$.
So the only solutions are: $(1,b),(a,1),(a,a+1),(b+1,b),(3,8)$
|
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|
Solve $x \equiv 32 \pmod{81}$ and $x \equiv 59 \pmod{64}$.
Solve $x \equiv 32 \pmod{81}$ and $x \equiv 59 \pmod{64}$.
$32 + 81k = 59 + 64n \implies 81k - 64n = 27$
$17k \equiv 27 \pmod{64}$.
$64 = 3(17) + 13$
$17 = 1(13) + 4$
$13 = 3(4) + 1$
So $1 = 13 - 3(4) = 13 - 3[17 - 13] = 4(13) - 3(17) = 4(64 - 3*17) - 3*17 =
4(64) - 15(17)$. Thus $k \equiv 1 \pmod{64} \implies k = 1 + 64y$ so we have $n = \frac{54 + 5184y}{64}$
But this is not possible . Help?
EDIT $k\equiv 43 \pmod{64}$ thus,
$k = 43 + 64y$ thus $x = 32 + 81(43 + 64y) = 3515 + 5824y$ so then $x \equiv 3515 \pmod{5824}$.
|
$$x = 59 + 64k$$
$$x = 32 + 81j$$
$\text{lcm}(81, 64) = 5184$
so
$$81x = 4779 + 5184k$$
$$64x = 2048 + 5184j$$
Subtract equations
$$17x = 2731 + 5184(k-j)$$
or
$$17x \equiv 2731 \bmod 5184$$
$\gcd(17, 5184)=1$, so:
$$x \equiv 2731 \cdot 17^{-1} \bmod 5184$$
$$x \equiv 2731 \cdot 305 \equiv 832955 \equiv 3515 \bmod 5184$$
Answer is
$$x \equiv 3515 \bmod 5184$$
|
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|
Range of function involving fractional part and integer part of $x$
The range of function $$f(x) = \left\{\frac{x}{4}\right\}+\cos \left(\frac{1-2\lfloor x \rfloor }{2}\right)+\sin \left(\frac{\pi \lfloor x \rfloor }{2}\right)$$
Where $\{x\} = x-\lfloor x \rfloor$ and $\lfloor x \rfloor $ represent floor function of $x$.
$\bf{My\; Try::}$ If $x$ is a multiple of $4\;,(x=4k\; ,k\in \mathbb{Z})$ Then $$f(x) = 0+\cos \left(\frac{1-8k}{2}\right)$$
But i did not understand If $x\neq 4k\;, k\in \mathbb{Z},$ Help required, Thanks
|
We consider four cases.
i) Let $x=4k+t$ with $k\in \mathbb{Z}$ and $t\in [0,1)$ then
$$f(x) = \frac{t}{4}+\cos \left(\frac{1-8k}{2}\right)=\frac{t}{4}+\cos \left(\frac{8k-1}{2}\right).$$
Hence if $A_0:=\cup_{k\in\mathbb{Z}}[4k,4k+1)$ then
$$f(A_0)=\bigcup_{k\in\mathbb{Z}}\left[\cos \left(\frac{8k-1}{2}\right),\frac{1}{4}+\cos \left(\frac{8k-1}{2}\right)\right)$$
ii) Let $x=(4k+1)+t$ with $k\in \mathbb{Z}$ and $t\in [0,1)$ then
$$f(x) = \frac{1+t}{4}+\cos \left(\frac{1-2(4k+1)}{2}\right)+1
=\frac{5+t}{4}+\cos \left(\frac{8k+1}{2}\right)
.$$
Hence if $A_1:=\cup_{k\in\mathbb{Z}}[4k+1,4k+2)$ then
$$f(A_1)=\bigcup_{k\in\mathbb{Z}}
\left[\frac{5}{4}+\cos \left(\frac{8k+1}{2}\right),\frac{3}{2}+\cos \left(\frac{8k+1}{2}\right)\right)$$
iii) Let $x=(4k+2)+t$ with $k\in \mathbb{Z}$ and $t\in [0,1)$ then
$$f(x) = \frac{2+t}{4}+\cos \left(\frac{1-2(4k+2)}{2}\right)
=\frac{2+t}{4}+\cos \left(\frac{8k+3}{2}\right)
.$$
Hence if $A_2:=\cup_{k\in\mathbb{Z}}[4k+2,4k+3)$ then
$$f(A_2)=\bigcup_{k\in\mathbb{Z}}
\left[\frac{1}{2}+\cos \left(\frac{8k+3}{2}\right),\frac{3}{4}+\cos \left(\frac{8k+3}{2}\right)\right)$$
iv) Let $x=(4k+3)+t$ with $k\in \mathbb{Z}$ and $t\in [0,1)$ then
$$f(x) = \frac{3+t}{4}+\cos \left(\frac{1-2(4k+3)}{2}\right)-1
=\frac{-1+t}{4}+\cos \left(\frac{8k+5}{2}\right)
.$$
Hence if $A_3:=\cup_{k\in\mathbb{Z}}[4k+2,4k+3)$ then
$$f(A_3)=\bigcup_{k\in\mathbb{Z}}
\left[-\frac{1}{4}+\cos \left(\frac{8k+5}{2}\right),\cos \left(\frac{8k+5}{2}\right)\right)$$
Now, $f(\mathbb{R})=f(A_0)\cup f(A_1)\cup f(A_2)\cup f(A_3)$.
In order to conclude we use the fact that the sequences of all these cosine values are in $(-1,1)$ and they are dense in $[-1,1]$. Therefore $f(\mathbb{R})=(-1/4-1,3/2+1)=(-5/4,5/2)$.
Does it make sense?
|
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|
Counting even and odd numbers in the columns of a triangular arrangement of the integers I write the positive numbers starting at $1$ in a triangle:$$\mathbb{N}_\triangle = \begin{matrix}
&&&&&21&\ldots \\
&&&&15&20&\ldots \\
&&&10&14&19&\ldots \\
&&6&9&13&18&\ldots \\
&3&5&8&12&17&\ldots \\
1&2&4&7&11&16&\ldots
\end{matrix}$$
I would like some help proving the following claim
I denote the count of even numbers in column $n$ of
$\mathbb{N}_\triangle$ by $e(n)$. Then for $n>1$ $$e(n) = \Bigg\lfloor{n+2
\above 1.5pt 4}\Bigg\rfloor+\Bigg\lfloor{n+1
\above 1.5pt 4}\Bigg\rfloor$$ Note the $n^{th}-column$ of $\mathbb{N}_\triangle$ has $n$ integers so I can set $f(n)$ to count the number of odd integers in
$\mathbb{N}_\triangle$ and $$f(n) = n-e(n)$$ It appears that
$e(n)\in$ A004524
This is being driven by hunches and computer checks. $e(n)$ can be simplified but I think it is easier to remember it the way it is written.
|
Note that each column forms an interval of integers, and so the numbers in the column alternate between being odd and even.
Hence if $n$ is even, so that there are an even number of integers in the $n$th column, exactly half of them will be even, so $e(2k) = k$ for every $k \in \mathbb{N}$.
If $n$ is odd, there will either be $\frac{n-1}{2}$ or $\frac{n+1}{2}$ even numbers, depending on whether the column starts with an odd or even number respectively. Notice that the difference between the initial numbers of the $2k-1$st and $2k+1$st columns is $4k-1$, since the $2k-1$st column has $2k-1$ numbers, and the $2k$th column in between has a further $2k$ numbers. As $4k-1$ is odd, that means that the initial numbers in the $2k-1$st and $2k+1$st odd columns have different parities.
Hence the first odd column starts with an odd number (namely $1$), the next odd column starts with an even number, the next odd column after that starts with an odd number again, and so on. This means we have the formula
$$\begin{array}{c|cccccccccccccc}
n & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & ... & 4k + 1 & 4k + 2 & 4k + 3 & 4k + 4 & ... \\ \hline
e(n) & \frac{n-1}{2} & \frac{n}{2} & \frac{n+1}{2} & \frac{n}{2} & \frac{n-1}{2} & \frac{n}{2} & \frac{n+1}{2} & \frac{n}{2} & ... & 2k & 2k + 1 & 2k + 2 & 2k + 2 & ... \\
e(n) - \frac{n}{2} & - \frac12 & 0 & \frac12 & 0 & - \frac12 & 0 & \frac12 & 0 & ... & - \frac12 & 0 & \frac12 & 0 & ...
\end{array}. $$
As you can see, $e(n) - \frac{n}{2}$ is periodic with period $4$, as explained by the dependence on the parity of $n$, and then the alternating behaviour in the odd case. Your formula captures this periodicity, and is correct. Other ways to express it, which I find delightfully ridiculous, include
$$ e(n) = \frac12 \operatorname{Re} \left( n + i^{n+1} \right) = \frac12 \left( n - \sin \left(\frac12 n \pi\right)\right). $$
|
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|
Analytic Geometry Question: Computing the equation of a circle given 2 points and center I want to derive the equation of a circle passing through the intersection of the two circles. My method gave me a nonsensical solution. I would love if someone could tell me how my method went wrong, although since my solution is kinda long (but it's mostly algebra) I would welcome an alternative solution.
$$
x^2+y^2=100\\
(x-11)^2+(y-4)^2=9
$$
and the centered at $(22,8)$.
The way I went about solving this was first starting with the family of curves $F(x,y,h,k)=0$ which go through the point of intersection
$$
h(x^2+y^2-100)+k(x-11)^2+(y-4)^2-9)=0\\
\Rightarrow (h+k)(x^2+y^2)-22kx-8ky+121k+16k-9k-100h=0\\
\Rightarrow (h+k)(x^2+y^2)-22kx-8ky+128k-100h=0\\
\Rightarrow x^2+y^2-\frac{22k}{h+k}x-\frac{8k}{h+k}y+\frac{128k}{h+k}-\frac{100h}{h+k}=0\\
\Rightarrow(x-\frac{11k}{h+k})^2+(y-\frac{4k}{h+k})^2
+\frac{128k+100h}{h+k}=\frac{121k^2+16k^2}{(h+k)^2}=\frac{137k^2}{(h+k)^2}\\
\Rightarrow (x-\frac{11k}{h+k})^2+(y-\frac{4k}{h+k})^2=\frac{137k^2}{(h+k)^2}-\frac{128k+100h}{h+k}
$$
Then imposing the restriction that this circle is centered at $(22,8)$ we have the system
$$
\frac{11k}{h+k}=22\\
\frac{4k}{h+k}=8\Rightarrow 2h=-k
$$
Which yields a potential solution of $(h,k)=(1,-2)$ and a radius of $392$.
edit: I fixed an algebraic mistake at the end, would still be curious to hear alternative solutions!
|
First of all, it would be slightly easier if we write the resultant circle as:-
$((x-11)^2+(y-4)^2-9) + j(x^2+y^2-100) = 0$
because only one variable is involved.
Secondly, the painful completing square process can be avoided.
Therefore the equation of the required circle is $x^2 + y^2 + 2(\dfrac {-11}{1 + j})x + 2(\dfrac {-4}{1 + j})y + \dfrac {121 + 16 – 9 -100j}{1 + j} = 0$
Correction:-
From $\dfrac {-11}{1 + j} = -22$, we get $j = -0.5 (= h/k)$
Then, the radius $= \sqrt {22^2 + 8^2 - 356}$
|
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|
Prove $\frac{a}{b+c}+\frac{b}{c+d}+\frac{c}{d+a}+\frac{d}{a+b} \ge \frac{1}{2}$ where $a,b,c,d \ge 0$ Prove $\frac{a}{b+c}+\frac{b}{c+d}+\frac{c}{d+a}+\frac{d}{a+b} \ge \frac{1}{2}$ where $a,b,c,d \ge 0$
My attempt:I used two ways but I get to a wrong answer.
My first way:We know that $\frac{a}{b}+\frac{b}{a} \ge 2$ where $a,b \ge 0$
Then:
$\frac{a}{b+c}+\frac{b+c}{a}+\frac{b}{c+d}+\frac{c+d}{b}+\frac{c}{d+a}+\frac{d+a}{c}+\frac{d}{a+b}+\frac{a+b}{d} \ge 8$
And:
$\frac{b+c}{a}+\frac{c+d}{b}+\frac{d+a}{c}+\frac{a+b}{d}=\frac{b}{a}+\frac{c}{a}+\frac{c}{b}+\frac{d}{b}+\frac{d}{c}+\frac{a}{c}+\frac{a}{d}+\frac{b}{d}=$
$\frac{a}{c}+\frac{c}{a}+\frac{b}{d}+\frac{d}{b} \ge 4$
$+$
$\frac{b}{a}+\frac{c}{b}+\frac{d}{c}+\frac{a}{d}\ge 4\sqrt{\frac{b}{a}*\frac{c}{b}*\frac{d}{c}*\frac{a}{d}}=4$
Then:
$\frac{a}{b+c}+\frac{b+c}{a}+\frac{b}{c+d}+\frac{c+d}{b}+\frac{c}{d+a}+\frac{d+a}{c}+\frac{d}{a+b}+\frac{a+b}{d} \ge 8$
And:
$\frac{b+c}{a}+\frac{c+d}{b}+\frac{d+a}{c}+\frac{a+b}{d}=\frac{b}{a}+\frac{c}{a}+\frac{c}{b}+\frac{d}{b}+\frac{d}{c}+\frac{a}{c}+\frac{a}{d}+\frac{b}{d}\ge 8$
Then we will get:
$\frac{a}{b+c}+\frac{b}{c+d}+\frac{c}{d+a}+\frac{d}{a+b} \ge 0$
Which is not true.
My second way:I don't have enough time then I just explain it.
I used caushy-shuartz and I get:
$\frac{a}{b+c}+\frac{b}{c+d}+\frac{c}{d+a}+\frac{d}{a+b} \ge 2$
|
Let
$$P=\frac{a}{b+c}+\frac{b}{c+d}+\frac{c}{d+a}+\frac{d}{a+b}$$
$$Q=\frac{b}{b+c}+\frac{c}{c+d}+\frac{d}{d+a}+\frac{a}{a+b}$$
$$R=\frac{c}{b+c}+\frac{d}{c+d}+\frac{a}{d+a}+\frac{b}{a+b}$$
We have $$Q+R=4\tag{1}$$
$$P+Q=\frac{a+b}{b+c}+\frac{b+c}{c+d}+\frac{c+d}{d+a}+\frac{d+a}{a+b} \overbrace{\geq}^{\color{red}{\text{AM} \geq \text{GM}}} 4\tag{2}$$
$$\begin{align} P+R=\frac{a+c}{b+c}+\frac{b+d}{c+d}+\frac{c+a}{d+a}+\frac{d+b}{a+b}\\ = \left(\frac{a+c}{b+c}+\frac{a+c}{d+a}\right)+ \left(\frac{b+d}{c+d}+\frac{b+d}{a+b}\right) \\
\overbrace{\geq}^{\color{blue}{\text{Titu's Lemma}}} \frac{4(a+c)}{a+b+c+d}+\frac{4(b+d)}{a+b+c+d}=4\tag{3}
\end{align}
$$
Using $(1),(2)$ and $(3)$, we get the desired result.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Area enclosed by the curve $5x^2+6xy+2y^2+7x+6y+6=0$ We have to find the area enclosed by the curve
$$5x^2+6xy+2y^2+7x+6y+6=0.$$
I tried and I got that it is an ellipse, and I know its area is $\pi ab$ where $a$ and $b$ are the semiaxis lengths of the ellipse.
But I am unable to find the value of $a$ and $b$.
|
The answer by Doug M solves your problem. I just want to expand on it to the general ellipse.
Assume $$ax^2+2bxy+cy^2+dx+ey+f=0$$ is an ellipse ($ac-b^2>0, a>0$). It has center $x=\frac{eb-cd}{2(ac-b^2)}, y=\frac{bd-ae}{2(ac-b^2)}$, which means that when the equation transforms to $$ax'^2+2bx'y'+cy'^2+f-\frac{ae^2-2bde+cd^2}{4(ac-b^2)}=0,$$ that is $f-\frac{ae^2-2bde+cd^2}{4(ac-b^2)}$ is negative (if not the conic is just a point or doesn't have real points). We need to multiply this by the $g>0$ so that $g(f-\frac{ae^2-2bde+cd^2}{4(ac-b^2)})=-1$. Then $agx'^2+2bgx'y'+cgy'^2-1=0$ and the area is $\frac{\pi}{g\sqrt{ac-b^2}}$ by Doug M's method.
For your equation $g=2$ and $ac-b^2=1$ so $A=\frac{\pi}{2}$.
|
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Challenge: Prove $\displaystyle \sum_{n \in \mathbb{N}} \frac{n!}{(2n+1)!}=e^{1/4}\sqrt{\pi}\ \ \mathrm{erf}(\frac{1}{2})$ I stumbled upon this cute sum while messing about, and I want to see what other solutions people propose before I put forward my own (which may be unnecessarily complicated).
You can use any maths you like.
NB: $\displaystyle \mathrm{erf}(x)=\frac{2}{\sqrt{\pi}}\int_0^xe^{-t^2}dt$
|
Outline:
*
*We know that $$e^{1/4} = \sum_{n=0}^\infty \frac{1}{2^{2n} n!}$$
*and that $$\begin{align}
\sqrt{\pi}\operatorname{erf}\left(\frac{1}{2}\right) &= \int_0^{1/2} e^{-t^2}dt
= \int_0^{1/2} \sum_{n=0}^{\infty} \frac{(-1)^n t^{2n}}{n!} dt
= \sum_{n=0}^{\infty} \frac{(-1)^n}{n!}\int_0^{1/2} t^{2n} dt
\\
&= \sum_{n=0}^{\infty} \frac{(-1)^n}{n! 2^{2n}} \frac{1}{2n+1}
\end{align}$$
so that, writing it as a product of series and using the Cauchy product of these,
$$\begin{align}
\sqrt{\pi}\operatorname{erf}\left(\frac{1}{2}\right) e^{1/4}
&= \sum_{n=0}^{\infty} \frac{(-1)^n}{n! 2^{2n}} \frac{1}{2n+1}\sum_{n=0}^\infty \frac{1}{2^{2n} n!}
= \sum_{n=0}^\infty \sum_{k=0}^n \frac{(-1)^k}{k! 2^{2k} (2k+1)}\frac{1}{(n-k)!2^{2n-2k}} \\
&= \sum_{n=0}^\infty \frac{1}{2^{2n}n!}\sum_{k=0}^n \binom{n}{k} \frac{(-1)^k}{k! (2k+1)}
= \sum_{n=0}^\infty \frac{1}{2^{2n}n!}\frac{(2n)!!}{(2n+1)!!}
\end{align}$$
and from there it is not hard to conclude from the properties of the double factorial.
(If needed, I can fill in the remaining steps later.)
Finishing it:
$$\begin{align}
\frac{(2n)!!}{(2n+1)!!}
= \frac{2^n n!}{(2n+1)!!}
= \frac{2^n n!\sqrt{\pi}}{2^{n+1}\Gamma(\frac{1}{2}+n+1)}
\end{align}$$
from which, using the properties of the Gamma function,
$$\begin{align}
\frac{(2n)!!}{2^{2n}n!(2n+1)!!}
&= \frac{\sqrt{\pi}}{2^{2n+1}\Gamma(\frac{1}{2}+n+1)}
= \frac{\sqrt{\pi}2^{2n+2}(n+1)!}{2^{2n+1}\sqrt{\pi}(2n+2)!}\\
&= \frac{2 (n+1)n!}{2(n+1)(2n+1)!}
= \frac{n!}{(2n+1)!}
\end{align}$$
and we are done.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Prove that $\frac{a}{b} < \frac{a+c}{b+d} <\frac{ c}{d}$ I'm reading a introductory book on mathematical proofs and I am stuck on a question.
Let $a, b, c, d$ be positive real numbers, prove that if $\frac{a}{b} < \frac{c}{d}$, then $\frac{a}{b} < \frac{a+c}{b+d} <\frac{ c}{d}$.
|
$$\frac{a}{b}=\frac{a(1+\frac{d}{b})}{b(1+\frac{d}{b})}=\frac{a+d(\frac{a}{b})}{b+d}<\frac{a+d(\frac{c}{d})}{b+d}=\frac{a+c}{b+d}$$
and similarly for the other inequality.
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
$\frac{1}{1+x^2}+\frac{1}{1+y^2}+\frac{1}{1+xy}≥\frac{3}{1+(\frac{x+y}{2})^2}$ if $x^2+y^2=1$. Show that $\frac{1}{1+x^2}+\frac{1}{1+y^2}+\frac{1}{1+xy}≥\frac{3}{1+(\frac{x+y}{2})^2}$. It is given that $x^2+y^2=1$. $x,y$ are positive real numbers.
[From a Regional Mathematical Olympiad, 2013 in India]
|
$$\frac{1}{1+x^2}+\frac{1}{1+y^2}+\frac{1}{1+xy}\ge \frac{4}{1+x^2+1+y^2} + \frac{1}{1+xy}$$
Use of AM-HM Inequality for $n=2$
$$=\frac{4}{3}+\frac{1}{1+xy} \ge \frac{1}{\frac{5}{4}+\frac{xy}{2}}$$
Expanding out
$$\frac{7+4xy}{3(1+xy)} \ge \frac{4}{5+2xy}$$
Common Denominators
$$(7+4xy)(5+2xy) \ge 12(1+xy)$$
Justified through positive terms
$$(7+4u)(5+2u) \ge 12(1+u)$$
$$35+34u+8u^2 \ge 12+12u$$
$$8u^2 + 22u + 23 \ge 0$$
But the global minimum of the LHS is $\frac{63}{8}$ for all real values of $u$. So the inequality is strictly true for all valid $x$ and $y$.
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
A quadratic equation Find all values of a for which the quadratic $$\cos^2x - (a^2 + a + 5) |\cos x| + (a^3 + 3a^2 + 2a + 6) = 0$$ has real solution(s)
A. $a=-3$, B. $a=-2$, C. $a=-1$, D. $a=0$
I have solved this question by putting the values of the given options :
Let $\cos x = t$
then it will look like $$t^2 - (a^2 + a + 5) |t| + (a^3 + 3a^2 + 2a + 6)$$
Comparing this quadratic with the standard quadratic and then finding out the discriminant for all four values of $a$ we get $D>0$ for all four values while the answer is given only $A$ and $B$
Kindly help.
|
You have to check if for at least one root $t$ of the equation
$$t^2 - (a^2 + a + 5) t + (a^3 + 3a^2 + 2a + 6)=
t^2 - (a(a+1)+ 5) t + (a(a+1)(a+2) + 6)
=0$$
there is a real $x$ such that $|\cos(x)|=t$.
Since $|\cos(x)|=t$ is solvable as soon as $t\in [0,1]$, just see if there is a root $t\in [0,1]$.
i) If $a=-3$ then $t^2 - 11t=0$ is solved by $t=0$ and $t=11$.
$0\in[0,1]$ so the answer is YES.
ii) If $a=-2$ then $t^2 - 7t+6=0$ is solved by $t=6$ and $t=1$.
$1\in[0,1]$ so the answer is YES.
iii) and iv) If $a=-1$ or $a=0$ then $t^2 - 5t+6=0$ is solved by $t=2$ and $t=3$.
No roots in in $[0,1]$ so the answer is NO.
|
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|
A difficult functional equation Is it possible to solve the following functional equation-
Determine all functions $f: \mathcal{R} \rightarrow \mathcal{R}$ such that $f(f(x)-f(y))=f(f(x))-2x^2f(y)+f(y^2)$ for all reals $x,y$
Here , $\mathcal{R}$ denotes the set of all reals.
|
I suppose that $f$ is not $0$.
1) I put first $a=f(0)$. Then $x=y=0$ gives $f(a)=0$. Putting $y=a$ gives $f(a^2)=0$, and replacing $x$ by $a$ gives $$f(-f(y))=a-2a^2f(y)+f(y^2)$$
2) Let $b$ such that $f(b)=0$. Replacing $x$ by $b$ gives
$$f(-f(y))=a-2b^2f(y)+f(y^2)$$. By the above, this is true $b=a^2$. If $a^4\not =a^2$, then the two relations gives $f=0$. Hence we must have $a=0$ or $a=\pm 1$.
3) I suppose now that $a=f(0)=0$. We have hence $$f(-f(y))=f(y^2)\quad (1)$$ In addition, if $f(b)=0$, we must have $b=0$. If we put $x=y$, we get
$$f(f(x)=2x^2f(x)-f(x^2)\quad (2)$$
a) Let $x,y$ not zero, and suppose that $f(x)=f(y)$. Then we get that $f(x^2)=f(-f(x))=f(-f(y))=f(y^2)$, and $f(f(x))=f(f(y))$. Hence by (2), we have $f(x)(y^2-x^2)=0$. As $f(x)\not =0$, we find $y=\pm x$.
b) From (1), we get that $f(y)=\pm y^2$ for all $y$.
Suppose that $f(x)=-x^2$ By using (2), we get that $f(x^2)+f(-x^2)=-2x^4$. But as $f(\pm x^2)=\pm x^4$, we get that $f(x^2)=f(-x^2)=-x^4$.
Now if $f(y)=y^2$, by using (2) again, we show that $f(y^2)=y^4$.
c) Now suppose that there exists $x,y$, not zero, with $f(x)=-x^2$ and $f(y)=+y^2$. The original equation show that
$$f(-x^2-y^2)=-x^4-2x^2y^2+y^4$$
But we must have $f(-x^2-y^2)=\pm (x^2+y^2)^2$, and it is easy to see that this is not the case.
d) Hence we must have $f(x)=-x²$ for all $x\not =0$, (hence for all $x$) or $f(x)=x^2$ for all $x$. It is easy to see that they both are solutions.
4) Now if $a=f(0)=\pm 1$, I think that we can follow the same way, If $f(0)=1$, then $f(1)=0$, if $f(b)=0$, then $b=\pm 1$ in the same way, etc.
I must say that I have not completed the computations. If I am not wrong, this leads to two other solutions, namely $f(x)=1-x^2$ and $f(x)=-1+x^2$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
What is the remainder when the product of the primes between 1 and 100 is divided by 16? The product of all the prime numbers between 1 and 100 is equal to $P$. What is the remainder when $P$ is divided by 16?
I have no idea how to solve this, any answers?
|
If the product of the odd primes is $8k+r$ then the product of all of them is $16+2r$. So we only need to work $\bmod 8$. We only need to find primes with residues $3,5$ and $7$. We do it as follows:
$$\begin{pmatrix}
3 && 5 && 7\\
11 && 13 && 15 \\
19 && 21 && 23\\
27 && 29 && 31\\
35 && 37 && 39\\
43 && 45 && 47\\
51 && 53 && 55\\
59 && 61 && 63\\
67 && 69 && 71\\
75 && 77 && 79\\
83 && 85 && 87\\
91 && 93 && 95\\
99\\
\end{pmatrix}$$
Then take out non-prime multiples of $3,5,7$
$$
\require{cancel}
\begin{pmatrix}
3 && 5 && 7\\
11 && 13 && \cancel{15} \\
19 && \cancel{21} && 23\\
\cancel{27} && 29 && 31\\
\cancel{35} && 37 && \cancel{39}\\
43 && \cancel{45} && 47\\
\cancel{51} && 53 && \cancel{55}\\
59 && 61 && \cancel{63}\\
67 && \cancel{69} && 71\\
\cancel{75} && \cancel{77} && 79\\
83 && \cancel{85} && \cancel{87}\\
\cancel{91} && \cancel{93} && \cancel{95}\\
\cancel{99}\\
\end{pmatrix}$$
The only column with an odd number of remaining elements is $3\bmod 8$, so the answer is $6$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Constant when integrating I was integrating $(t+2)^2$ by using the substitution method and by expanding the function. By expanding it then integrating I got the same answer as the book ($\frac{1}{3}(t^3+6t^2+12t)+C$), but when I substituted I got $\frac{1}{3}\cdot(t+2)^3 + C = \frac{1}{3} \cdot (t^3+6t^2+12t+8) + C$.
Am I missing something here? Got an 8 at the end when substituting, but no constant when expanding.
|
Your answers differ by a constant factor, so the values of $C$ will be different in these 2 methods.
In other words,
$$
\frac{t^3+6t^2+12t}{3} + C= \frac{t^3 + 6t^2 + 12t +8}{3} + K
$$
just $C = K + 8/3$...
|
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"timestamp": "2023-03-29T00:00:00",
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|
$(a+2)^3+(b+2)^3+(c+2)^3 \ge 81$ while $a+b+c=3$ I need to show that $(a+2)^3+(b+2)^3+(c+2)^3 \ge 81$ while $a+b+c=3$ and $a,b,c > 0$ using means of univariate Analysis.
It is intuitively clear that $(a+2)^3+(b+2)^3+(c+2)^3$ is at its minimum (when $a+b+c=3$) if $a,b,c$ have "equal weights", i.e. $a=b=c=1$.
To show that formally one can firstly fix $0<c \le 3$, introduce a variable $0\le\alpha\le1$ and express $a$ and $b$ through $\alpha$
$$a= \alpha(3-c)$$ $$b=(1-\alpha)(3-c)$$
Then we minimize $(a+2)^3+(b+2)^3+(c+2)^3$ w.r.t $\alpha$ treating $c$ as a parameter. We get $\alpha=\frac{1}{2}(3-c)$. Further we maximize $(a+2)^3+(b+2)^3+(c+2)^3$ one more time w.r.t. $c$.
That is quite tedious. I am wondering whether their is a more elegant way. Probably using idea of norms. Basically we need to show that $\lVert(a,b,c)+(2,2,2)\rVert_3 \ge (81)^{1/3}$ while $\lVert(a,b,c)\rVert_1=3$ and $a,b,c>0$.
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Let $f(x) = x^3 $ is a convex function. Using Jenson we get :
$$ \frac{f(a+2)+f(b+2)+f(c+2)}{3} \geq f(\frac{a+b+c+6}{3})=f(3)=27$$
|
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|
Basic Algebra combining exponent fractions/simplifying (George F Simmons "Precalculus in Nutshell") From George F. Simmons 'Precalculus' book, Algebra section, 5(d);
Combine and Simplify
$$\frac{x}{xy^2} + \frac{y}{x^2y}$$
Combine:
= $$\frac{x(x^2y) + y(xy^2)}{(xy^2)(x^2y)}$$
Simplify: = $$\frac{x(x^2y) + y(xy^2)}{xy(x^2y^2)}$$
The given answer, which I can't figure out how he arrives at it is: $$\frac{x^2 +y^2}{x^2y^2}$$
|
$$\frac{x}{xy^2}+\frac{y}{x^2y}=\frac{x}{x}\cdot\frac{x}{xy^2}+\frac{y}{y}\cdot\frac{y}{x^2y}=\frac{x\cdot x}{xy^2\cdot x}+\frac{y\cdot y}{x^2y\cdot y}=\frac{x^2}{x^2y^2}+\frac{y^2}{x^2y^2}=\frac{x^2+y^2}{x^2y^2}$$
|
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|
Why is minimizing least squares equivalent to finding the projection matrix $\hat{x}=A^Tb(A^TA)^{-1}$? I understand the derivation for $\hat{x}=A^Tb(A^TA)^{-1}$, but I'm having trouble explicitly connecting it to least squares regression.
So suppose we have a system of equations: $A=\begin{bmatrix}1 & 1\\1 & 2\\1 &3\end{bmatrix}, x=\begin{bmatrix}C\\D\end{bmatrix}, b=\begin{bmatrix}1\\2\\2\end{bmatrix}$
Using $\hat{x}=A^Tb(A^TA)^{-1}$, we know that $D=\frac{1}{2}, C=\frac{2}{3}$. But this is also equivalent to minimizing the sum of squares: $e^2_1+e^2_2+e^2_3 = (C+D-1)^2+(C+2D-2)^2+(C+3D-2)^2$.
I know the linear algebra approach is finding a hyperplane that minimizes the distance between points and the plane, but I'm having trouble understanding why it minimizes the squared distance. My intuition tells me it should minimize absolute distance, but I know this is wrong because it's possible for there to be non-unique solutions.
Why is this so? Any help would be greatly appreciated. Thanks!
|
The matrix has full column rank; we are guaranteed a unique solution.
Problem statement
$$
\begin{align}
\mathbf{A} x &= b \\
\left[
\begin{array}{cc}
1 & 1 \\
1 & 2 \\
1 & 3 \\
\end{array}
\right]
%
\left[
\begin{array}{cc}
x_{1} \\
x_{2} \\
\end{array}
\right]
&=
\left[
\begin{array}{cc}
1 \\
2 \\
2 \\
\end{array}
\right]
\end{align}
$$
Normal equations
$$
\begin{align}
\mathbf{A}^{*} \mathbf{A} x &= \mathbf{A}^{*} b \\
%
\left[
\begin{array}{ccc}
1 & 1 & 1 \\
1 & 2 & 3 \\
\end{array}
\right]
%
\left[
\begin{array}{cc}
1 & 1 \\
1 & 2 \\
1 & 3 \\
\end{array}
\right]
%
\left[
\begin{array}{cc}
x_{1} \\
x_{2} \\
\end{array}
\right]
&=
\left[
\begin{array}{ccc}
1 & 1 & 1 \\
1 & 2 & 3 \\
\end{array}
\right]
%
\left[
\begin{array}{cc}
1 \\
2 \\
2 \\
\end{array}
\right] \\[3pt]
%
\left[
\begin{array}{cc}
3 & 6 \\
6 & 14 \\
\end{array}
\right]
%
\left[
\begin{array}{cc}
x_{1} \\
x_{2} \\
\end{array}
\right]
%
&=
%
\left[
\begin{array}{cc}
5 \\
11 \\
\end{array}
\right]
%
\end{align}
$$
Least squares solution
$$
\begin{align}
x_{LS} &= \left( \mathbf{A}^{*} \mathbf{A} \right)^{-1} \mathbf{A}^{*} b \\
%
\left[
\begin{array}{cc}
x_{1} \\
x_{2} \\
\end{array}
\right]
&=
\frac{1}{6}
\left[
\begin{array}{cc}
14 & -6 \\
-6 & 3 \\
\end{array}
\right]
%
\left[
\begin{array}{cc}
5 \\
11
\end{array}
\right] \\
%
&=
\frac{1}{6}
\left[
\begin{array}{cc}
4 \\
3
\end{array}
\right]
%
\end{align}
$$
|
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|
find integers a and b such $x^2-x-1$ divides $ax^{17}+bx^{16}+1 = 0$
find integers a and b such $x^2-x-1$ divides $ax^{17}+bx^{16}+1 = 0$
By really long division i got :-
$$Q=ax^{15} + (a+b)x^{14} + \dots +(610a + 377b) $$
$$R = x(987a+610b)+1+610a+377b$$
since remainder is $0 $,
$$987a+610b = 0$$
$$1+610a+377b = 0$$
from which i got $a = -610, b = 987$
but from wolfram alpha the remainder of $-610x^{17}+987x^{16}+1 \over x^2-x-1$ is $x-1$
Somebody please show me where i went wrong ?
Thanks.
|
$$x^2-x-1=
\left( x-\frac{1+\sqrt{5}}{2} \right)\left( x-\frac{1-\sqrt{5}}{2} \right)$$
By factor theorem,
\begin{align*}
a\left( \frac{1+\sqrt{5}}{2} \right)^{17}+
b\left( \frac{1+\sqrt{5}}{2} \right)^{16}+1 &=0 \quad \cdots \cdots \: (1) \\
a\left( \frac{1-\sqrt{5}}{2} \right)^{17}+
b\left( \frac{1-\sqrt{5}}{2} \right)^{16}+1 &=0 \quad \cdots \cdots \: (2)
\end{align*}
$\left( \frac{1-\sqrt{5}}{2} \right)^{16} \times (1)
-\left( \frac{1+\sqrt{5}}{2} \right)^{16} \times (2)$,
$$a\sqrt{5}
=\left( \frac{1+\sqrt{5}}{2} \right)^{16}
-\left( \frac{1-\sqrt{5}}{2} \right)^{16}$$
$$a=F_{16}=987$$
$\left( \frac{1-\sqrt{5}}{2} \right)^{17} \times (1)
-\left( \frac{1+\sqrt{5}}{2} \right)^{17} \times (2)$,
$$-b\sqrt{5}
=\left( \frac{1+\sqrt{5}}{2} \right)^{17}
-\left( \frac{1-\sqrt{5}}{2} \right)^{17}$$
$$b=-F_{17}=-1597$$
In general,
$$
\frac{F_{n} x^{n+1}-F_{n+1} x^n+(-1)^n}{x^2-x-1}=
F_{n} x^{n-1}-F_{n-1} x^{n-2}+\ldots+(-1)^{n-1}F_{1}$$
|
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"url": "https://math.stackexchange.com/questions/1900326",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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|
Find value of x, where $\frac{3+\cot 80^{\circ} \cot 20^{\circ}}{\cot 80^{\circ}+\cot 20^{\circ}}=\cot x^{\circ}$ $$\frac{3+\cot 80^{\circ}\cot 20^{\circ}}{\cot 80^{\circ}+\cot 20^{\circ}}=\cot x^{\circ}$$ Then find $x$
My Try:
Using $\cot 80=\tan 10$ and $\cot 20=\frac{1}{\tan 20}$
we have LHS as $$\frac{3+\frac{\tan 10^{\circ}}{\tan 20^{\circ}}}{\tan 10^{\circ}+\frac{1}{\tan 20^{\circ}}}$$
assuming $y=10$ and using $\tan 2y=\frac{2 \tan y}{1-\tan ^2 y}$ we get
LHS as
$$\frac{3+\frac{\tan y}{\tan 2y}}{\tan y+\frac{1}{\tan 2y}}=\frac{\tan y(7-\tan ^2 y)}{1+\tan ^2 y}$$
how to proceed further?
|
Changing in terms of $$\sin \text {and} \cos ,3=2+1$$
$$L.H.S=\frac {2\sin 80^{\circ}.\sin20^{\circ} +\cos (80^{\circ}-20^{\circ})}{\sin(80^{\circ}+20^{\circ})} $$
$$=\frac {\cos(80^{\circ}-20^{\circ}) -\cos(80^{\circ}+20^{\circ})+\cos 60^{\circ}}{\sin 100^{\circ}} $$
$$=\frac {1-\cos 100^{\circ}}{\sin100^{\circ}}=\tan 50^{\circ}=\cot 40^{\circ}$$
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
Prove this inequality $\sum\limits_\text{cyc}\sqrt{1-xy}\ge 2$ Let $x,y,z\ge 0$, and $x+y+z=2$, show that
$$\sqrt{1-xy}+\sqrt{1-yz}+\sqrt{1-xz}\ge 2.$$
Mt try: $$\Longleftrightarrow 3-(xy+yz+zx)+2\sum_\text{cyc}\sqrt{(1-xy)(1-yz)}\ge 4$$
or
$$\sum_\text{cyc}\sqrt{(1-xy)(1-yz)}\ge\dfrac{1}{2}(1+xy+yz+zx)$$
|
After squaring of the both sides we need to prove that
$$2\sum\limits_{cyc}\sqrt{(1-xy)(1-xz)}\geq1+xy+xz+yz$$ or
$$2\sum\limits_{cyc}\sqrt{((x+y+z)^2-4xy)((x+y+z)^2-4xz)}\geq(x+y+z)^2+4(xy+xz+yz)$$ or
$$2\sum\limits_{cyc}\sqrt{((x-y)^2+z^2+2xz+2yz)((x-z)^2+y^2+2xy+2yz)}\geq\sum\limits_{cyc}(x^2+6xy)$$
and after using C-S it remains to prove that
$$2\sum_{cyc}((x-y)(x-z)+xy+2x\sqrt{yz}+2xy)\geq\sum\limits_{cyc}(x^2+6xy)$$ or
$$\sum_{cyc}(x^2-2xy+4x\sqrt{yz})\geq0$$ or
$$\sum\limits_{cyc}\left(x^2-\sqrt{x^3y}-\sqrt{x^3z}+x\sqrt{yz}\right)+\sum_{cyc}\left(\sqrt{x^3y}+\sqrt{x^3z}-2xy\right)+3\sum_{cyc}x\sqrt{yz}\geq0$$
which is Schur and Muirhead.
|
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|
vectorial equation i need help to Calculate the vectorial equation of the line $$ g = E_{1}\cap E_{2} $$ you get by intersecting the planes$$ E_{1}:\overrightarrow{x}=\left(\begin{array}{c}1\\ 0\\1\end{array}\right)+r.\left(\begin{array}{c}1\\ 1\\0\end{array}\right)+s.\left(\begin{array}{c}0\\ 1\\1\end{array}\right);r,s\in R $$ $$ E_{2}:\overrightarrow{x}=\left(\begin{array}{c}1\\ 1\\1\end{array}\right)+v.\left(\begin{array}{c}1\\ 1\\1\end{array}\right)+w.\left(\begin{array}{c}1\\ 0\\0\end{array}\right);v,w\in R $$
And to determine the vectorial equation of the line ( ℎ ), so that ( ℎ ) is perpendicular to ( E1 ) and that the point A(1|0|1) is located on ℎ $$ ((h \bot E_{1}) ; A(1|0|1)\in h) $$
|
The normal vectors to planes $E_1$ and $E_2$ are:
$$\vec{N_1}=\begin{pmatrix}1\\1\\0\end{pmatrix} \times \begin{pmatrix}1\\1\\0\end{pmatrix}=\begin{pmatrix}1\\-1\\1\end{pmatrix} \ \text{and} \ \vec{N_2}=\begin{pmatrix}1\\1\\1\end{pmatrix} \times \begin{pmatrix}1\\0\\0\end{pmatrix}=\begin{pmatrix}0\\1\\-1\end{pmatrix} \ \text{resp.}$$
(They are attached to the vectorial planes associated with these affine planes).
Thus, a directing vector of the intersection line is:
$$\vec{N}:=\vec{N_1}\times\vec{N_2}=\begin{pmatrix}0\\1\\-1\end{pmatrix}$$
Now let us find a common point of the affine planes. This can be done by inspection: $A=\begin{pmatrix}2\\1\\1\end{pmatrix}$ is such a point (take $r=1,s=0,v=0,w=1$).
Thus a parametric representation of the looked for intersection line :
$$M=A+t\vec{N}=\begin{pmatrix}2\\1\\1\end{pmatrix}+t\begin{pmatrix}0\\1\\-1\end{pmatrix}=\begin{pmatrix}2\\1+t\\1-t\end{pmatrix} \ \ t \in \mathbb{R}$$
Edit: it is interesting to check the answer by writing that intersection line $g$ lies in $E_1$ for example:
$$\begin{cases}2&=&1+r\\1+t&=&r+s\\1-t&=&1+s\end{cases}$$
which are compatible with $r=1$ and $s=-t$. In particular $r=1$ can be considered as the equation of $g$ in plane $E_1$.
|
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|
Sum of the series $1\cdot3\cdot2^2+2\cdot4\cdot3^2+3\cdot5\cdot4^2+\cdots$ Find the sum of $n$ terms of following series:
$$1\cdot3\cdot2^2+2\cdot4\cdot3^2+3\cdot5\cdot4^2+\cdots$$
I was trying to use $S_n=\sum T_n$, but while writing $T_n$ I get a term having $n^4$ and I don't know $\sum n^4$. Is there any other way find the sum?
|
We have
$$ n(n+1)^2(n+2) = 24\binom{n+2}{4}+12\binom{n+2}{3}\tag{1}$$
hence
$$ \sum_{n=1}^{N}n(n+1)^2(n+2) = 24\binom{N+3}{5}+12\binom{N+3}{4} \tag{2}$$
is a consequence of the Hockey Stick identity, leading to:
$$ \sum_{n=1}^{N}n(n+1)^2(n+2) = \color{red}{\frac{1}{10} N (1+N) (2+N) (3+N) (3+2 N)}.\tag{3}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1903470",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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|
Does this system of Diophantine equations have a solution?
Are there natural numbers $a,b,c,d,e,f$ such that we have $a \neq b$ and $a \neq c$ and $b \neq c$ and that they are solution of this system of equations:
$9ab-3a-3b+1=d^2$
$9ac-3a-3c+1=e^2$
$9bc-3b-3c+1=f^2$
|
There are infinitely many. Let,
$$a = (2p+1)^2+2p^2\\
b = (2q+1)^2+2q^2\\
c = (2r+1)^2+2r^2$$
then,
$$d =4(3p+1)^2(3q+1)^2\\
e =4(3p+1)^2(3r+1)^2\\
f =4(3q+1)^2(3r+1)^2$$
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
cardinality of cartesian product of the infinite set of natural numbers One of the problems in my discrete math course states that we need to prove that $\mathcal{N}\times\mathcal{N}$ is countable specifically when there's a function $f:\mathcal{N}\times\mathcal{N}\to\mathcal{N}$ defined as follows:
$$f(a,b)=\frac{1}{2}(a+b+1)(a+b)+a$$
where $a,b \in \mathcal{N}$. The solution uses the function definition in order to prove that the function is bijective as shown below:
$$f(a,b+1)=\frac{1}{2}(a+b+1+1)(a+b+1)+a=$$
$$=\frac{1}{2}(a+b+1)(a+b)+a+\frac{1}{2}(a+b+1)\cdot2$$
How is the transition achieved? I tried all kinds of arithmetics and couldn't arrive to the expression after the equal sign.
|
$\frac{1}{2}(a+b+1+1)(a+b+1)+a=\frac{a^2+ab+a+ba+b^2+b+a+b+1+a+b+1}{2}+a=\frac{a^2+b^2+2ab+3a+3b+2}{2}+a=\frac{a^2+b^2+2ab+5a+3b+2}{2}$
$\frac{1}{2}(a+b+1)(a+b)+a+\frac{1}{2}(a+b+1)\cdot 2=\frac{a^2+ab+ba+b^2+a+b}{2}+a+\frac{2a+2b+2}{1}=\frac{a^2+b^2+2ab+5a+3b+2}{2}$
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
Contradictory result when testing Linear independence using Gaussian elimination Consider a set of vectors - (2,3,1) , (1,-1,2) and (7,3,8).
I want to find if its linearly dependent or independent.
Putting it as:
\begin{equation}
2a + b + 7c = 0 \\
3a - b + 3c = 0 \\
a + 2b + 8c = 0
\end{equation}
If I use Gaussian elimination of equations to calculate row echelon form of the matrix:
\begin{bmatrix}
2&1&7&0 \\
3&-1&3&0 \\
1&2&8&0 \\
\end{bmatrix}
I get:
\begin{bmatrix}
1&0&0&0 \\
0&1&0&0 \\
0&0&1&0 \\
\end{bmatrix}
which means a = b = c = 0
*
*So, as per
Determine if vectors are linearly independent
the vectors are linearly independent.
*But I know that they are linearly dependent, with a = 2, b = 3, c =
-1
Why is there such a contradiction? did i do something procedurally wrong?
Note: I just want to understand the mistake in above procedure, dont want an alternate solution like using determinant etc (unless this procedure itself is totally wrong!)
|
First note that
$$
\left[\begin{array}{ccc|c}
2&1&7&0 \\
3&-1&3&0 \\
1&2&8&0 \\
\end{array}\right]
\Rightarrow
\left[\begin{array}{ccc|c}
1&0&2&0 \\
0&1&3&0 \\
0&0&0&0 \\
\end{array}\right]
$$
Using matrix equations
We have
$$
a\begin{bmatrix}
1\\
0\\
\end{bmatrix}
+b\begin{bmatrix}
0\\
1\\
\end{bmatrix}
+c\begin{bmatrix}
2\\
3\\
\end{bmatrix}
=\begin{bmatrix}
0\\
0\\
\end{bmatrix}
$$
$$
a\begin{bmatrix}
1\\
0\\
\end{bmatrix}
+b\begin{bmatrix}
0\\
1\\
\end{bmatrix}
=-c\begin{bmatrix}
2\\
3\\
\end{bmatrix}
$$
$$
\begin{bmatrix}
a\\
b\\
\end{bmatrix}
=-c\begin{bmatrix}
2\\
3\\
\end{bmatrix}
\neq\begin{bmatrix}
0\\
0\\
\end{bmatrix}
$$
Therefore, the three vectors are linearly dependent.
Using determinants
We have $$\begin{vmatrix}
2&1&7 \\
3&-1&3 \\
1&2&8\\
\end{vmatrix}=
\begin{vmatrix}
1&0&2 \\
0&1&3 \\
0&0&0 \\
\end{vmatrix}=0$$
Therefore, the three vectors are linearly dependent.
|
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|
How should I have derived the extra solution my book lists? I am working through some geometry exercises, one of which is to find the point(s) where the circle defined by $x^2 + y^2 = 16$ touches the line defined by $x - 2y = 4$.
I defined $x$ in terms of $y$ starting from the equation for the line: $x = 4 + 2y$.
Substituted in the equation for the circle this yields $(4 + 2y)^2 + y^2 = 16$. The abc-formula gives $y = \frac{-20 \pm 20}{2} \implies y = 0 \lor y = -20$. $-20$ can't be a solution since $x^2 + 400 = 16$ has no solutions, so one solution that satisfies both equations is $(4, 0)$.
Then I checked my answer to see that, although my answer was correct, my book listed another solution: $(-\frac{12}{5}, -\frac{16}{5})$. Substitution shows that this indeed satisfies both equations, yet I never got there. How should I have derived this second solution?
TL;DR
I only found one solution to the following system of equations - $(4,0)$ -, yet my book lists two:
$$
\left\{
\begin{array}{c}
x^2 + y^2 = 16 \\
x - 2y = 4
\end{array}
\right.
$$
How should I have derived the second solution - $(-\frac{12}{5}, -\frac{16}{5})$ -?
|
plugging $$x=4+2y$$ in $$x^2+y^2=16$$ we get $$(4+2y)^2+y^2=16$$ from here we get
$$16+4y^2+16y+y^2=16$$ or equivalent to $$y(5y+16)=0$$ can you proceed?
|
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|
Find the angles using the given conditions
I got stuck in last step. What should I do next? Please help me
|
$$
\text{Let } a = \frac{\alpha}{2}, b = \frac{\beta}{2}, c = \frac{\gamma}{2} \\
\sin{\frac{\alpha-\beta}{2}}+\sin{\frac{\alpha-\gamma}{2}}+\sin{\frac{3\alpha}{2}} = \frac{3}{2} \\
\sin\left(a-b\right)+\sin\left(a-c\right)+\sin{3a} = \frac{3}{2}
\\
\sin\left(a-b\right)+\sin\left(a-c\right)+\sin\left(a-b+a-c+a-a+a+b+c\right) = \frac{3}{2}\\
\sin\left(a-b\right)+\sin\left(a-c\right)+\sin\left(a-b+a-c+90^{\circ}\right) = \frac{3}{2}\\
\sin\left(a-b\right)+\sin\left(a-c\right)+\cos\left(a-b+a-c\right) = \frac{3}{2}\\
\text{So, } a-b=30^{\circ},a-c=30^{\circ} \text{(messy proof below)}
$$
Do the rest and you should get $\alpha=100^{\circ}, \beta=40^{\circ}, \gamma=40^{\circ}$
Proof that $\sin x +\sin y + \cos(x+y) = \frac{3}{2} \implies x = y= 30^{\circ}$
It actually turns out that $\frac{3}{2}$ is the maximum of that two dimensional function. Finding the partial derivatives wrt $x$ and equating it to zero,
$$
\cos x = \sin(x+y) \\
\sin\left(\frac{\pi}{2}-x\right) = \sin(x+y) \\
y = \frac{\pi}{2}-2x
$$
Note that in general, $\left(\frac{\pi}{2}-x\right) = n\pi+(-1)^n(x+y)$. But for the range($0\leq x,y< 90^{\circ}$) we are working here this is the only possible condition. Similarly, the other partial derivative gives,
$$
x = \frac{\pi}{2}-2y
$$
Solving we get that $\sin x +\sin y + \cos(x+y) = \frac{3}{2}, 0\leq x,y< 90^{\circ}$ only if $x = y = 30^{\circ}$.
I know this is not the most elegant proof, but it works and if you know a better way, please comment/answer.
|
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|
Simplifying a rational function with infinite series in numerator and denominator We're working with Taylor Series and I have to simplify the rational expression
$$ \frac{x-2\frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^5}{5!} - 2\frac{x^6}{6!} + \cdots}{x- \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \cdots}.$$
I'm not sure how to complete the problem -- the answer is
$$ 1 - x + (x^2/3) - (x^3/6) + \cdots $$
but I'm not sure how to arrive at the solution. Thank you for the help.
|
The long division mentioned in the comments is likely the way you are 'supposed' to answer the question. Here's the $\operatorname{csc}$ method:
$$\begin{align}R & = \frac{x - 2\frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^5}{5!} - 2 \frac{x^6}{6!} + \ldots}{x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \ldots} \\
& = \operatorname{csc} x \times \sum_{n=0}^\infty \left[\frac{x^{2n+1}}{(2n+1)!}- (1 + [-1]^n) \frac{x^{2n}}{(2n)!}\right] \\
& = \left(\sum_{k=0}^\infty \frac{(-1)^{k-1} 2 (2^{2k-1} - 1) B_{2k}}{(2k)!} x^{2k-1}\right)\left(\sum_{n=0}^\infty \left[\frac{x^{2n+1}}{(2n+1)!}- (1 + [-1]^n) \frac{x^{2n}}{(2n)!}\right]\right) \\
& = \sum_{n,k=0}^\infty\left[\left(\frac{(-1)^{k-1} 2 (2^{2k-1} - 1) B_{2k}}{(2k)!}\right)\left(\frac{1}{(2n+1)!}\right) x^{2(n+k)} \right.\\
& \hphantom{= \sum_{n,k=0}^\infty} \left.- \left(\frac{(-1)^{k-1} 2 (2^{2k-1} - 1) B_{2k}}{(2k)!}\right) \left(\frac{1-[-1]^n}{(2n)!}\right)x^{2(n+k)-1} \right] .\end{align}$$
Renumbering to group the coefficients of particular powers of $x$ using $k = j - n$:
$$\begin{align}R& = \sum_{j=0}^\infty \left[ \left\{ \sum_{n=0}^j \left(\frac{(-1)^{j-n-1} 2 (2^{2[j-n]-1} - 1) B_{2[j-n]}}{(2[j-n])!}\right)\left(\frac{1}{(2n+1)!}\right)\right\} x^{2j} \right.\\
& \hphantom{= \sum_{n,k=0}^\infty} \left.- \left\{ \sum_{n=0}^j \left(\frac{(-1)^{j-n-1} 2 (2^{2[j-n]-1} - 1) B_{2[j-n]}}{(2[j-n])!}\right) \left(\frac{1-[-1]^n}{(2n)!}\right) \right\} x^{2j-1} \right].\end{align}$$ Granted, the expression is pretty complicated, and could do with some simplification, particularly the sums in the curly braces. Assuming I didn't make any mistakes, though, the expressions are correct.
|
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|
Determine the diameter of a circle given its length is a two digit number and when the number is reversed the length of a chord is obtained. Problem:
The length of diameter AB is a two digit integer. Reversing the digits gives the length of a perpendicular chord CD.The distance from their intersection point H to the center O is a positive rational number. Determine the length of AB.
My attempt:
Let the length of AB be 10a+b. Then the length of CD is 10b+a. Now since AO is a radius, H bisects CD. Therefore, HD=CD/2=(10b+a)/2 and OD=AO=AB/2=(10a+b)/2. In triangle OHD,
$OH^2+HD^2=OD^2$,
$OH^2=OD^2-HD^2$
$OH^2=(\frac{10a+b}{2})^2-(\frac{10b+a}{2})^2$
$=>OH^2=\frac{(10a+b)^2}{4}-\frac{(10b+a)^2}{4}$
$=>OH^2=\frac{100a^2+b^2+20ab-100b^2-a^2-20ab}{4}$
$=>OH^2=\frac{99a^2-99b^2}{4}$
$=>OH^2=\frac{99}{4}*(a^2-b^2)$
$=>OH=\frac{3\sqrt{11}}{2}*\sqrt{(a^2-b^2)}$
Since it is given that OH is a positive rational number, $\sqrt{(a^2-b^2)}$
should be of the form $x\sqrt{11}$ where $x$ is an integer. Therefore $a^2-b^2$ must be of the form $11x^2$. $a^2-b^2=(a-b)(a+b)$. Since $0<a,b\leq 9$ and both cannot be $0$, we have only two cases when $a^2-b^2$ is a multiple of $11$:
1)$a-b=0$ and
2) $a+b=11$
If the first case is true, OH=0-which is not possible because OH is a positive rational number. Therefore $a+b=11$. Considering all a,b such that $0\leq a,b \leq 9$ and such that $a+b=11$, the only a,b which makes $a-b$ a perfect square is $(6,5)$. Hence the length of $AB=10a+b=65$.
Is my attempt to the problem correct?
|
Yes, your attempt to the problem is correct. here is another way of solving it.
The equation of a circle is $$x^2 + y^2 = R^2\Rightarrow x^2 + y^2 = (\frac{d}2)^2$$
where, R and d are the radius and the diameter of the circle respectively.
So, the equation for the top semi circle is $$y = \sqrt{(\frac{d}2)^2 - x^2}$$
Let, the diameter of the circle be of the form $$d = 10a + b$$
where, $a$ and $b$ are whole numbers between 0 and 9 inclusive.
So, $a$ and $b$ are the digits of the two digit number.
Also, it is given that the length of cord CD is obtained by reversing the digits of the diameter. So, $$CD = 10b + a$$
Let us define a function $C(x)$ which gives us the length of a perpendicular chord, whose input is the distance of the center of the chord from the center of the circle.
We know that the length of half of the perpendicular chord is given by $$\sqrt{(\frac{d}2)^2 - x^2}$$
So,$$C(x) = 2\sqrt{(\frac{d}2)^2 - x^2}\Rightarrow 2\sqrt{(\frac{10a + b}2)^2 - x^2}$$
Now, let the distance of CD from the center of the circle be a rational number of the form $\frac{p}q$, where $p$ and $q$ are relatively prime integers.
Hence, the length of CD is given by $$C(\frac{p}q) = 2\sqrt{(\frac{10a + b}2)^2 - (\frac{p}q)^2}$$$$\Rightarrow10b + a = 2\sqrt{(\frac{10a + b}2)^2 - (\frac{p}q)^2}$$
Simplifying the equation will give us,$$(\frac{2p}{3q})^2 = 11(a^2-b^2)$$Since, the LHS is a perfect square, RHS must also be a perfect square. As, the RHS is already a multiple of $11$, $ a^2-b^2$ must be an odd power of 11, like $11^1, 11^2, 11^3$, etc.Since, $a$ and $b$ are whole numbers between 0 and 9, the maximum value of $a^2-b^2$ will be when $a = 9$ and $b = 0$ , i.e. $81$.
Because, all odd powers of $11$ are greater than $81$ except $11$ itself, $a^2 - b^2$ must be $11$.So, when we see a list of perfect squares from $1^2$ to $9^2$ $$1, 4, 9, 16, 25, 36, 49, 64, 81$$we see that only $36\;(i.e.\;6^2)$ and $25\;(i.e.\;5^2)$ are $11$ apart. Hence, we can choose, $a = 6$ and $b = 5$, to satisfy the equation $a^2 - b^2 = 11$.
Therefore, the length of the diameter of the circle is $65$ units.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Inequality with $a+b+c=1$ Let $a,b,c\in\mathbb{R^+}$ such that $a+b+c=1$. Prove that $$\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}+3(ab+bc+ca)\geq\frac{11}{2}.$$
I am trying to resolve this problem but actually i found some issues:
$$\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}=3+\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\geq \frac{9}{2}$$
I only use Nesbitt's inequality. Then we only need that $3(ab+bc+ca)\geq 1$, but it is not correct, because $3(ab+bc+ca)\leq 1$.
Maybe some ideas will be more that grateful.
|
$$\sum\frac{1}{a+b}+3(ab+bc+ca)\geq\frac{11}{2}$$
$$\sum\frac{(a+b+c)^2}{a+b}+\frac{3}{2}(1-\sum a^2)\geq\frac{11}{2}$$
$$\sum\frac{a^2}{b+c} + 4 +\frac{3}{2}(1-\sum a^2)\geq\frac{11}{2}$$
$$2\sum\frac{a^2(a+b+c)}{b+c}\geq3\sum a^2$$
$$2\sum\frac{a^3}{b+c}\geq\sum a^2$$
Using CSB we get:
$$(\sum\frac{a^3}{b+c})(\sum a(b+c))\geq(\sum a^2)^2 \geq (\sum a^2)(\sum ab)$$
and from that follows the inequality from above.
|
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|
linear transformation eigenvectors and eigenvalues
Consider the operator defined by $T(x, y, z) = (-x+2y, 3y, 0)$.
*
*Find the eigenvalues of T and all corresponding eigenvectors.
*Find each generalised eigenvector corresponding to each eigenvalue.
So for 1, I found the matrix with respect to the standard basis $(1,0,0), (0,1,0), (0,0,1)$. It was upper triangular so I read the eigenvalues from the diagonal entries, which gave me eigenvalues of $-1, 3$ and $0$. I'm just so lost with how to find the corresponding eigenvectors, I feel like this should be really simple.
I know once I have found these eigenvectors for 2, I can simply compute the generalized eigenvectors by solving the equation $(T- \lambda I)^j (v) = 0$ where $v$ is my particular eigenvector with corresponding eigenvalue $\lambda$.
Thanks in advance!
|
Operator
$$
\mathbf{T} =
\left[
\begin{array}{rrr}
-1 & 2 & 0 \\
0 & 3 & 0 \\
0 & 0 & 0 \\
\end{array}
\right]
$$
Eigenvalues
To find the eigenvalues, compute $p(\lambda)$, the characteristic polynomial.
$$
p (\lambda ) = \det \left( \mathbf{T} - \lambda \mathbf{I}_{3} \right) = \det
\left[
\begin{array}{ccr}
-\lambda -1 & 2 & 0 \\
0 & 3-\lambda & 0 \\
0 & 0 & -\lambda \\
\end{array}
\right]
=
-\lambda \left( 3 - \lambda \right) \left( -1 - \lambda \right)
$$
The roots $p(\lambda) = 0$ are the eigenvalues: $\lambda = \left\{ 3, -1, 0 \right\}$.
Eigenvectors
Solve
$$
\mathbf{T} v_{k} = \lambda_{k} v_{k} \qquad \Rightarrow \qquad \left( \mathbf{T} - \lambda_{k} \mathbf{I}_{3} \right) v_{k} = \mathbf{0}
$$
$\lambda = 3$
$$
\begin{align}
\left( \mathbf{T} - 3 \mathbf{I}_{3} \right) v_{1} &= \mathbf{0} \\
\left[
\begin{array}{rrr}
-4 & 2 & 0 \\
0 & 0 & 0 \\
0 & 0 & -3 \\
\end{array}
\right]
%
\left[
\begin{array}{c}
v_{1_{1}} \\
v_{1_{2}} \\
v_{1_{3}}
\end{array}
\right]
&=
\left[
\begin{array}{c}
0 \\
0 \\
0
\end{array}
\right]
%
\qquad \Rightarrow \qquad
v_{1} =
\left[
\begin{array}{c}
1 \\
2 \\
0
\end{array}
\right]
%
\end{align}
$$
$\lambda = -1$
$$
\begin{align}
\left( \mathbf{T} + \mathbf{I}_{3} \right) v_{2} &= \mathbf{0} \\
\left[
\begin{array}{rrr}
0 & 2 & 0 \\
0 & 4 & 0 \\
0 & 0 & 1
\end{array}
\right]
%
\left[
\begin{array}{c}
v_{2_{1}} \\
v_{2_{2}} \\
v_{2_{3}}
\end{array}
\right]
&=
\left[
\begin{array}{c}
0 \\
0 \\
0
\end{array}
\right]
%
\qquad \Rightarrow \qquad
v_{2} =
\left[
\begin{array}{c}
1 \\
0 \\
0
\end{array}
\right]
%
\end{align}
$$
$\lambda = 0$
$$
\begin{align}
\left( \mathbf{T} + 0 \mathbf{I}_{3} \right) v_{3} &= \mathbf{0} \\
\left[
\begin{array}{rrr}
-1 & 2 & 0 \\
0 & 3 & 0 \\
0 & 0 & 0 \\
\end{array}
\right]
%
\left[
\begin{array}{c}
v_{3_{1}} \\
v_{3_{2}} \\
v_{3_{3}}
\end{array}
\right]
&=
\left[
\begin{array}{c}
0 \\
0 \\
0
\end{array}
\right]
%
\qquad \Rightarrow \qquad
v_{3} =
\left[
\begin{array}{c}
0 \\
0 \\
1
\end{array}
\right]
%
\end{align}
$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Range of $\tan^2\frac A2+\tan^2\frac B2+\tan^2 \frac C2$ with $A,B,C$ angles in a triangle
What is the range of $\tan^2\frac A2+\tan^2\frac B2+\tan^2 \frac C2$ if $A,B,C$ are angles in a triangle?
For $\tan^2\frac A2+\tan^2\frac B2+\tan^2 \frac C2$, I know we have to apply the AM–GM inequality. So
$$\begin{align}\frac12\left(\tan^2\frac A2+\tan^2\frac B2+\tan^2\frac C2\right)&\ge\sqrt{\tan^2\frac A2\tan^2\frac B2\tan^2\frac C2}\\
&\ge\tan\frac A2\tan\frac B2\tan\frac C2\\
\tan^2\frac A2+\tan^2\frac B2+\tan^2\frac C2&\ge2\tan\frac A2\tan\frac B2\tan\frac C2
\end{align}$$
After this, the LHS is of the required form, but the RHS is having half-angle terms. Then again, if $A+B+C=\pi$, $\tan\frac{A+B+C}2=\tan90^\circ=\infty$. Now I am stuck; how to solve further?
The given options are:
*
*$>1$
*$<1$
*$\ge1$
*$\le1$
|
Let $a+b-c=z$, $a+c-b=y$ and $b+c-a=x$.
Hence, by C-S $\sum\limits_{cyc}\tan^2\frac{\alpha}{2}=\sum\limits_{cyc}\frac{yz}{x(x+y+z)}=\sum\limits_{cyc}\frac{y^2z^2}{xyz(x+y+z)}\geq\frac{(xy+xz+yz)^2}{3xyz(x+y+z)}\geq1$.
The equality occurs for $x=y=z$, id est, for $a=b=c$.
In another hand, our sum $\rightarrow+\infty$ for $\alpha\rightarrow180^{\circ}$.
Id est, the answer is $[1,+\infty)$.
|
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|
proof using the mathematical induction Prove $$\frac{k^7}{7}+\frac{k^5}{5}+\frac{2k^3}{3}-\frac{k}{105}$$ is an integer for every positive integer k.
In proving for (n+1) integer,the expression is integer,I found $(n+1)^7$ term.I use binomial theorem to expand but finally it won't work(some term become integer While some other remaining as rational)
Any hint to solve the problem is appreciated.Thanks.
|
Based on other similar questions, you might think to first prove that
$$ \frac{k^7 - k}{7} $$
is always an integer. Once you have done so, you can simplify
$$\frac{k^7}{7}+\frac{k^5}{5}+\frac{2k^3}{3}-\frac{k}{105}
=
\frac{k^7 - k}{7}
+\frac{k^5}{5}+\frac{2k^3}{3}+\frac{2k}{15}
$$
You can do the same with $5$ and $3$, leaving you with
$$\frac{k^7}{7}+\frac{k^5}{5}+\frac{2k^3}{3}-\frac{k}{105}
=
\frac{k^7 - k}{7} + \frac{k^5 - k}{5} + 2 \frac{k^3 - k}{3} + k
$$
Another path to the same approach might be to use a variant of partial fractions to separate the denominator into its prime components:
$$ \frac{1}{105} = \frac{1}{7} + \frac{1}{5} + \frac{2}{3} - 1$$
and then group like denominators together.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Three digit number such that $A^2+B^2+C^2$ is divisible by $26$. Find all three digit natural numbers $ABC,(A \neq 0)$ such that $A^2+B^2+C^2$ is divisible by $26$.
Could someone give me some hint as how to approach this question as I am not able to initiate?
|
As $26=2\cdot13$
and $x^2\equiv x\pmod2\implies A^2+B^2+C^2\equiv A+B+C\pmod2$
So, there should be even number of odd values among $\{A,B,C\}\ \ \ \ (1)$
Again, as $A,B,C$ are decimal digits, $$0\le A,B,C\le9\ \ \ \ (2)$$
Now for any integer $y, y^2\equiv0,\pm1,\pm4,\pm3\pmod{13}$
For $13\mid(A^2+B^2+C^2),$
we need $\{A^2,B^2,C^2\}\equiv$ to be one of $\{0,0,0\};\{0,\pm1\};\{0,\pm4\};\{0,\pm3\};\{4,-1,-3\};\{-4,1,3\}$
$\implies\{A,B,C\}\equiv$ to be one of
$\{0,0,0\};$
$\{0,\{1,-1\equiv12\text{(which is unacceptable as }>10\},\{5,-5\equiv8\}\};$
$\{0,2,3\};$
$\{0,\{4,-4\equiv9\},\{7,-7\equiv6\}\};$
$\{2,\{5,-5\equiv8\},\{7,-7\equiv6\}\};$
By $(1),(2);$ we can accept $$\{0,1,5\};\{0,4,6\};\{0,9,7\};\{2,5,7\};\{2,8,6\}$$
|
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|
Minimum value of $a^2 + b^2$ so that the quadratic $x^2 + ax + (b+2) = 0$ has real roots
The equation $ x^2 + ax + (b+2) = 0 $ has real roots, where $a$ and $b$ are real numbers.
How would I find the minimum value of $a^2 + b^2$ ?
|
We get the solutions
$$
0 = x^2 + a x + b + 2 = (x + a/2)^2 - a^2/4 + b + 2 \iff \\
x = \frac{-a \pm \sqrt{a^2 - 4b - 8}}{2}
$$
which are real for
$$
a^2 - 4b - 8 \ge 0
$$
We now want to solve the optimization problem
\begin{matrix}
\min & a^2 + b^2 \\
\text{w.r.t} & a^2 - 4b - 8 \ge 0
\end{matrix}
The above image shows the function
$$
z = f(x,y) = x^2 + y^2
$$
(red surface). It is rotational symmetric around the $z$-axis and has circles in the $x$-$y$-plane with center $(0,0)$ and radius $\sqrt{c}$ as isolines $f(x,y) = c$ (yellow).
The curve
$$
x^2 - 4y - 8 = 0 \iff \\
y = (1/4) x^2 - 2
$$
(green) is the border of the feasible area $x^2 - 4y -8 \ge 0$, shown in light blue. Point $A=(2,0)$ (pink) is not feasible with the condition, $B=(2,-2)$ (blue) is a feasible point.
Going from inner to outer isolines, we see that the isoline with radius $r=2$, thus $x^2 + y^2 = 2^2 = 4 = c$ is the first one to be part of the feasible area. Thus the minimum is $4$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Stars and Bars with bounds This question is related to Error solving “stars and bars” type problem
I have what I thought is a fairly simple problem: Count non-negative
integer solutions to the equation
$$x_1 + x_2 + x_3 + x_4 + x_5 = 23$$
such that $0 \leq x_1 \leq 9$.
The difference here is on the constraint. It bounds all the $x_i$ under 10 : $\forall i\le5$ , $0 \leq x_i \leq 9$.
$8+8+0+0+7=23$ is accepted but not $18+3+0+0+2=23$ or $11+0+0+0+12=23$ . In other words all the $x_i$ must be usual digits.
Note that bad solutions may include one or two bad $x_i$. It is the main difficulty.
What is the count of combinations ?
Edit : this question includes a double bounds and a supplemental difficulty to find the right solution.
|
We cannot have $3$ $x_i= > 9$ because the result might be $>= 30$
We can have two $x_i > 9$ because the result may be in the bounds , example $10+10+1+1+1=23$ and $11+10+1+1+0=23$
1 ) let's see the case where there is one or two $x_i > 9$
We can have one $x_i > 9$ because the result may be in the bounds , example $10+9+2+1+1=23$.
The trick would be to remove $10$ and to focalize on the distribution of the remaining $13$ ( which will include a case of twice overflow ) :
Ways to chose the overflow candidates $ = \binom {5}{1} \begin{pmatrix}23 + (5-1) -10 \\ (5-1)\end{pmatrix} = \binom {5}{1} \begin{pmatrix}17 \\ 4\end{pmatrix} = 11900$
2 ) let's see the case where there are two $x_i > 9$ .
Again, the trick is to remove twice 10 and to focalize on the distribution of the remaining $3$.
Ways to chose the twice overflow candidates $ = \binom {5}{2} \binom {23 + (5-1) -10-10}{5-1} = \binom {5}{2} \binom {7}{4} = 350$
These last combinations had been counted twice and then we must deduced it once.
3 )
Now ignoring the bounds limitation, we have a total of combinations of
$\begin{pmatrix}23 + (5-1) \\ (5-1)\end{pmatrix} = \begin{pmatrix}27 \\ 4\end{pmatrix} = 17550$
The result must be $17550 - (11900 - 350) =6000$
|
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|
Find all functions satisfying the functional equation $ xf(x) + f(1-x) = x^3 - x $
Find all functions, for all real x, that satisfy the following functional equation:
$$ xf(x) + f(1-x) = x^3 - x $$
|
We have
\begin{align*}
xf(x) + f(1-x) &= x^3-x\\
(1-x)f(1-x) + f(1-(1-x)) &= (1-x)^3 - (1-x)
\end{align*}
and hence
\begin{align*}
f(x) + (1-x)f(1-x) = (1-x)^3 - (1-x)
\end{align*}
Multiplying the first equation by $1-x$ and subtracting from the third equation, we get
\begin{align*}
(x(1-x) - 1)f(x) &= (x^3-x)(1-x) - (1-x)^3 + (1-x)\\
(-x^2+x-1)f(x) &= x^3-x - x^4 + x^2 - (1-3x+3x^2-x^3) + 1-x \\
&= -x^4 + 2x^3-2x^2 +x
\end{align*}
Hence
\begin{align*}
f(x) = \frac{-x^4 + 2x^3-2x^2 +x }{-x^2+x-1} = x^2 - x = x(x-1)
\end{align*}
|
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|
Expected Run before Absorption Consider a $4$ state Markov Chain that includes the row-stochastic
transition matrix
$$P = \left(\begin{array}{cccc} 1/4&1/4&1/4&1/4 \\ 1/2&0&0&1/2 \\
1/3&1/3&1/3&0 \\ 0&0&0&1 \end{array}
\right).$$
Assume that we start the chain in state $1$. I would like to find the expected
length of time intervals until we are absorbed (as in, before we reach state 4).
To me, this seems like a problem where we have a large number of potential
transitions, and I need some help on considering a clean way to calculate the
terms for the probability that the expected length is $\geq 3$. I see that for
the first term of this expectation, we have
$1 \times P_{1,4} = \frac{1}{4},$ and for the second term, we have
$$2 \times \sum_{i=1}^3 P_{1,i}P_{i,4}
= 2 \times (\frac{1}{4} \times \frac{1}{4} + \frac{1}{4} \times \frac{1}{2}
+ \frac{1}{4} \times 0)
= 2 \times (\frac{3}{16}) = \frac{3}{8}.$$
However, I see how this computation becomes much more complicated given the
length of the chain. Is there potentially a more efficient way to calculate
the individual terms of this expectation?
|
Considering this,
assume $$Q = \left(\begin{array}{cccc} 1/4&1/4&1/4 \\ 1/2&0&0 \\
1/3&1/3&1/3 \end{array}
\right)$$
Then the expected number of transitions from state $i$ to $j$ is
$$N=(I-Q)^{-1}=\left(\begin{array}{cccc} 2.2857 &0.8571&0.8571 \\ 1.1429 & 1.4286 & 0.4286\\
1.7143 & 1.1429 & 2.1429\end{array}
\right)$$
The $i$th element of the vector $\mathbf{t}=N\mathbf{1}$ gives the expected number of transitions until absorption when beginning from state $i$.
For this case, $$\mathbf{t}=\left(\begin{array}{cccc} 2.2857 &0.8571&0.8571 \\ 1.1429 & 1.4286 & 0.4286\\
1.7143 & 1.1429 & 2.1429\end{array}
\right)\begin{pmatrix}
1\\
1\\
1
\end{pmatrix}=\begin{pmatrix}
4\\
3\\
5
\end{pmatrix}$$
So the answer to your question is $\color{red}{4}$.
|
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|
Solve the equation: $x=\sqrt{a-\sqrt{a+x}},(a\geq 1)$
Solve the equation: $x=\sqrt{a-\sqrt{a+x}},(a\geq 1)$
We have: $x^2=a-\sqrt{a+x}$ , if we take both sides to the power of 2 we will get a 4th order equation which I don't know of!
Please help.
|
Note that $0 \le x\le a(a-1)$,
\begin{align*}
x &= \sqrt{a-\sqrt{a+x}} \\
x^2 &= a-\sqrt{a+x} \\
a+x &= (a-x^2)^2 \\
a+x &= a^2-2ax^2+x^4 \\
x^4-2ax^2-x+a^2-a &= 0 \\
(x^2-x-a)(x^2+x-a+1) &= 0 \\
\end{align*}
$$x=\frac{1 \pm \sqrt{4a+1}}{2} \quad \text{or} \quad \frac{-1 \pm \sqrt{4a-3}}{2}$$
For $a\ge 1$,
$$x=\frac{-1+\sqrt{4a-3}}{2}$$
Alternative solution from problem $168$ in The USSR Olympiad Problem Book
Let $y=\sqrt{a+x}$, then $x=\sqrt{a-y}$
\begin{align*}
a+x &= y^2 \\
a-y &= x^2 \\
x+y &= y^2-x^2 \\
x^2-y^2+x+y &= 0 \\
(x+y)(x-y+1) &= 0 \\
(x+\sqrt{a+x})(x-\sqrt{a+x}+1) &= 0
\end{align*}
Rejecting $x+\sqrt{a+x}=0$,
$$x^2+x-a+1=0$$
$$x=\frac{-1+\sqrt{4a-3}}{2}$$
|
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|
$\int \frac{\sqrt{1-36 x^2}}{x^2}\,dx$ I need help integrating
$$\int \frac{\sqrt{1-36x^2}}{x^2} \ dx$$
using trigonometric substitution.
My first step was simplifying the integral down to $$\int \frac{\sqrt{36(\frac{1}{36}-x^2)}}{x^2} \ dx$$
and use $x=\frac{1}{6} \sin \theta$ to perform trigonometric substitution. I then perform trigonometric substitution as so
$$6\int \frac{\sqrt{\frac{1}{36}(1-\sin^2\theta)}}{\frac{1}{36}\sin^2\theta} \ d\theta$$
Could someone please perform the next couple steps so I can find my error? I keep getting the incorrect answer after this last step. Thank you.
|
Consider the following substitution $x=\frac{\sin (u)}{6}$ and $dx=\frac{\cos(u)}{6}$. Then, $\sqrt{1-36x^2}=\sqrt{1-\sin^2(u)}=\cos(u)$ and $u=\sin^{-1}(6x)$. Hence,
$$
\int \frac{\sqrt{1-36x^2}}{x^2} \,dx=6\int \cot^2(u) \,du=6\int (\csc^2(u)-1) \,du=-6\cot(u)-6u.
$$
Substituting back, we get
$$
-6\cot (\sin^{-1}(6x))-6\sin^{-1}(6x).
$$
Finally, using the identity $\cot(\sin^{-1}(z))=\frac{\sqrt{1-z^2}}{z}$, we get the simplified expression
$$
\int \frac{\sqrt{1-36x^2}}{x^2} \,dx=-\frac{\sqrt{1-36x^2}+6x\sin^{-1}(6x)}{x}.
$$
|
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}
|
Integration of $\int\frac{x-1}{\sqrt{1+\sqrt{1-x^2}}}dx$ Integral of $\int\frac{x-1}{\sqrt{1+\sqrt{1-x^2}}}dx$
This is the answer, but I'm not sure how to get to it...
http://www.wolframalpha.com/input/?i=Integral+of+(x-1)%2F((sqrt(1%2Bsqrt(1-x%5E2))))dx
|
$$I=\int\frac{x-1}{\sqrt{1+\sqrt{1-x^2}}}dx$$
Start from substitution $t^2=1-x^2,\quad x=\sqrt{1-t^2},\quad dx=-\frac{t}{\sqrt{1-t^2}}dt$
$$
I=\int\frac{t(1-\sqrt{1-t^2})}{\sqrt{1+t}\sqrt{1-t^2}}dt=\int\frac{t}{\sqrt{1+t}\sqrt{1-t^2}}dt-\int\frac{t}{\sqrt{1+t}}dt=I_1-I_2+const
$$
$I_2$ is trivial and equal to $\frac{2}{3}(t-2)\sqrt{t+1}$, $I_1 $ requires further transformation
$$
I_1=\int\frac{t}{\sqrt{1+t}\sqrt{1-t^2}}dt=\int\frac{t}{(1+t)\sqrt{1-t}}dt\\
I_1=\int\frac{1}{\sqrt{1-t}}dt-\int\frac{1}{(1+t)\sqrt{1-t}}dt
$$
First integral is trivial, second is also can be considered as a table one (alternatively use substitution $z^2=1-t$).
$$
I_1=(-2\sqrt{1-t})-\left(-\sqrt{2}\tanh^{-1}\left(\frac{1-t}{2}\right)\right)
$$
Hence,
$$
I=-2\sqrt{1-t}+\left(\sqrt{2}\tanh^{-1}\left(\frac{1-t}{2}\right)\right)-\frac{2}{3}(t-2)\sqrt{1+t}+const
$$
Or in initial variables
$$
I=-2\sqrt{1-\sqrt{1-x^2}}+\left(\sqrt{2}\tanh^{-1}\left(\frac{1-\sqrt{1-x^2}}{2}\right)\right)-\frac{2}{3}(\sqrt{1-x^2}-2)\sqrt{1+\sqrt{1-x^2}}+const
$$
|
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|
Lengthy partial fractions? $$\frac{1}{(x+3)(x+4)^2(x+5)^3}$$
I was told to integrate this, I see partial fractions as a way, but this absurd! Is there an easier way?
|
Lazy people like me will do this as follows. The function
$$f(x) = \frac{1}{(x+3)(x+4)^2(x+5)^3}$$
has a simple pole at $x = -3$. The principal part of the Laurent expansion around this point (i.e. the sum of the singular terms in the expansion) is:
$$s_1(x) = \frac{1}{8}\frac{1}{x+3}$$
If you calculate the principal parts of the Laurent expansion around the other singular points and add them up, then that's the partial fraction expansion, because the difference between that sum and $f(x)$ won't have any singularities left. But this is then a rational function without any singularities which is therefore a polynomial. But it will tend to zero in the limit of $x$ to infinity, therefore that polynomial must be identical to zero.
Now, we can save the work needed to actually calculate the Laurent expansion around $x = -4$ and $x = -5$. We can instead exploit the fact that $f(x)$ for large $x$ tends to zero as $x^{-6}$. This means that terms of lower order in $x^{-1}$ must cancel out in the partial fraction expansion. And instead of negative powers of $x$ we can just as well consider expanding in negative powers of $x+a$ for some suitable $a$.
Then directly calculating the most singular terms of the principal parts amounts to a trivial calculation, see eq. (2) in Jack D'Aurizio's answer. Let's take from there that the principal part of the Laurent expansion around $x = -4$ is of the form:
$$s_2(x) = \frac{B}{x+4} -\frac{1}{(x+4)^2}$$
If we use the fact that the coefficient of $(x+5)^{-4}$ in the large $x+5$ expansion of $f(x)$ is zero, then since the principal part of the Laurent expansion around $x = -5$ doesn't yield any contributions to this term, we get an equation involving only $B$. The principal part of the Laurent expansion around $x = -4$, $s_2(x)$, yields two terms, one comes from re-expanding the term
$$-\frac{1}{(x+4)^2} = -\frac{1}{(x+5)^2}\frac{1}{\left(1-\frac{1}{x+5}\right)^2}=-\frac{1}{(x+5)^2} - \frac{2}{(x+5)^3}- \frac{3}{(x+5)^4}+\cdots$$
which therefore yields the contribution:
$$-\frac{3}{(x+5)^4}\tag{1}$$
Then we have a contribution coming from the unknown term of the same principal part:
$$\frac{B}{x+4} = \frac{B}{x+5} \frac{1}{1 - \frac{1}{x+5}} = \frac{B}{x+5} +\frac{B}{(x+5)^2} + \frac{B}{(x+5)^3}+ \frac{B}{(x+5)^4}+\cdots $$
So, we get the contribution:
$$\frac{B}{(x+5)^4}\tag{2}$$
Finally, we have a contribution coming from $s_1(x)$:
$$\frac{1}{8}\frac{1}{x+3} = \frac{1}{8}\frac{1}{x+5}\frac{1}{1-\frac{2}{x+5}} = \frac{1}{8}\frac{1}{x+5} + \frac{1}{4}\frac{1}{(x+5)^2}+ \frac{1}{2}\frac{1}{(x+5)^3}+\frac{1}{(x+5)^4}\cdots$$
This thus yields the contribution:
$$\frac{1}{(x+5)^4}\tag{3}$$
Adding up the contributions (1), (2), (3), and (3) must yield zero, therefore we have $B = 2$.
We can then similarly proceed to calculate the principal part of the Laurent expansion around $x=-5$ by considering the coefficient of $(x+5)^{-3}$, $(x+5)^{-2}$, and $(x+5)^{-1}$, but we don't need to do the first one as we can obtain this from a straightforward limit computations, see eq.(2) in Jack D'Aurizio's answer. Let's take that result from there and write the principal part as:
$$s_3(x) = \frac{C}{x+5} + \frac{E}{(x+5)^2} -\frac{1}{2}\frac{1}{(x+5)^3}$$
To calculate $E$ we can then consider the coefficient of $(x+5)^{-2}$ in the large $x+5$ expansion of $f(x)$ and equate that to zero. Then $s_3(x)$ yields the contribution:
$$\frac{E}{(x+5)^2} \tag{4}$$
And $s_2(x)$ yields the contribution:
$$\frac{1}{(x+5)^2} \tag{5}$$
And $s_1(x)$ yields:
$$\frac{1}{4}\frac{1}{(x+5)^2} \tag{6}$$
Adding up (4), (5), and (6) and equating to zero yields $E = -\frac{5}{4}$
Finally $C$ follows from equating the coefficient of $(x+5)^{-1}$ to zero. The contribution from $s_3(x)$ is:
$$\frac{C}{x+5} \tag{7}$$
The contribution from $s_2(x)$ is:
$$\frac{2}{x+5} \tag{8}$$
The contribution from $s_1(x)$ is:
$$\frac{1}{8}\frac{1}{x+5} \tag{9}$$
Adding up (7), (8), and (9) and equating the result to zero yields $C = -\frac{17}{8}$.
|
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|
CDF for Negative Binomial Distribution I am trying to show that the following statement is true.
$$
\sum_{x = r}^{X}\binom{x-1}{r-1}p^r(1-p)^{x-r} =
\sum_{x = r}^{X}\binom{X}{x}p^x(1-p)^{X-x}
$$
Where $X$ and $r$ and $p$ are constants, with $X \geq r$, and $ 0 \leq p \leq 1.$
How did I get there? Well, this is the story:
Consider a sequence of independent binomial trials, each one producing the result success or failure, with probabilities $p$, and $1-p$, respectively.
Let $x$ be the total number of trials which must be carried out in order to attain exactly $r$ successes.
Knowing that the probability mass function for this Negative Binomial Distribution is as follows,
$P(x=X)=\binom{X-1}{r-1}p^r(1-p)^{X-r}$, (for $X \geq r$),
I was trying to prove the following about the corresponding Cumulative Distribution Function.
$P(x \leq X)=\sum_{x=r}^{x=X}\binom{x-1}{r-1}p^r(1-p)^{x-r}=1-\sum_{x=0}^{x=r-1}\binom{X}{x}p^x(1-p)^{X-x}$
I started out with the following:
$\sum_{x=r}^{\infty}\binom{x-1}{r-1}p^r(1-p)^{x-r}=1$,
which can be recast as below. (Relation I)
$\sum_{x=r}^{x=X}\binom{x-1}{r-1}p^r(1-p)^{x-r}+\sum_{x=X+1}^{\infty}\binom{x-1}{r-1}p^r(1-p)^{x-r}=1$
In addition, from binomial theorem, we have:
$\left ( p+(1-p) \right )^X=\sum_{x=0}^{x=X}\binom{X}{x}p^x(1-p)^{X-x}=1$
Which can be restated in Relation II as below.
$\sum_{x=0}^{x=r-1}\binom{X}{x}p^x(1-p)^{X-x}+\sum_{x=r}^{x=X}\binom{X}{x}p^x(1-p)^{X-x}=1$
Comparing the relations I and II with the expression for the CDF, the proof boils down to verification of the following:
$\sum_{x=r}^{x=X}\binom{x-1}{r-1}p^r(1-p)^{x-r}=\sum_{x=r}^{x=X}\binom{X}{x}p^x(1-p)^{X-x}$
Any idea how to continue form this point onward?
Thanks.
|
Suppose we seek to prove that
$$\sum_{k=0}^m {n+k-1\choose k} q^k
= \sum_{k=0}^m {m+n\choose m-k} q^{m-k} (1-q)^k.$$
The RHS is
$$\sum_{k=0}^m {m+n\choose k} q^{k} (1-q)^{m-k}.$$
Extracting the coefficient on $q$ on the RHS we get
$$[q^p] \sum_{k=0}^m {m+n\choose k} q^{k} (1-q)^{m-k}
= \sum_{k=0}^m {m+n\choose k} [q^p] q^{k} (1-q)^{m-k}
\\ = \sum_{k=0}^m {m+n\choose k} [q^{p-k}] (1-q)^{m-k}.$$
Now
$$[q^{p-k}] (1-q)^{m-k} =
\frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{p-k+1}} (1-z)^{m-k} \; dz.$$
Observe that this vanishes when $p\lt k$ as needed. Now by inspection
of the original RHS we see that we may assume $p\le m.$ Therefore when
$k\gt m$ the integral vanishes and we may extend the range of the sum
to $m+n$, getting
$$\frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{p+1}} (1-z)^{m}
\sum_{k=0}^{m+n} {m+n\choose k} \frac{z^k}{(1-z)^k}
\; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{p+1}} (1-z)^{m}
\left(1+\frac{z}{1-z}\right)^{m+n}
\; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{p+1}} (1-z)^{m}
\frac{1}{(1-z)^{m+n}}
\; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{p+1}}
\frac{1}{(1-z)^{n}}
\; dz
\\ = {p+n-1\choose p}.$$
This is the claim.
|
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|
How to solve $z^2 + \left( 2i - 3 \right)z + 5-i = 0$ in $\mathbb{C}$
Solve the equation $z^2 + \left( 2i - 3 \right)z + 5-i = 0$ in $\mathbb{C}$
I was thinking on it for a few minutes and came up with a few ideas (none of them worked).
My first idea: Use the quadratic formula. Is that allowed? If so, I got to this:
$$
z = \frac{-(2i-3) \pm \sqrt{(2i-3)^2-4(5-i)}}{2} = \cdots = \frac{3-2i \pm \sqrt{-8i-15}}{2}
$$
I am guessing I cannot go further than this.
Other idea: Being $z = ai+b$,
$$
(ai+b)^2 + (2i-3)(ai+b) + 5 - i = 0
$$
Which will give me a system of two equations and two variables
$$
\begin{cases}
-a^2-2a+b^2-3b = -5\\
2ab+2b-3a = 1
\end{cases}
$$
which seems almost impossible to solve.
What am I missing? Is there a simpler solution to this that I am not thinking about?
|
By solving $(a+ib)^2=8i+15$ we get
$a^2-b^2=15$, $2ab=8$ which are satisfied by $a=4$ and $b=1$.
Hence we can go on with the quadratic formula:
$$z = \frac{-(2i-3) \pm \sqrt{(2i-3)^2-4(5-i)}}{2} = \frac{3-2i \pm \sqrt{-8i-15}}{2}\\
= \frac{3-2i \pm i\sqrt{8i+15}}{2}=\frac{3-2i \pm i(4+i)}{2}=\frac{(3\mp 1)+ i(-2\pm 4)}{2}$$
which implies that the solutions are $z_1=1+i$ and $z_2=2-3i$.
|
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|
I don't know what is wrong with my graph for this problem - at most how many students take none of these courses? Is my graph wrong? If it is, what is the correct one? I used this graph to solve the problem and got a wrong answer. It is highly appreciated if one can show me the correct graph.
Of the 60 students at Hope Middle School, 40 take algebra, 45 take geometry, 48 take trigonometry. Furthermore, 22 take all three courses. At most how many students take none of these courses?
According to this graph, I have:
$$a+b=45-22$$
$$b+c=48-22$$
$$c+a=40-22$$
$$x=60-(40+45+48-a-b-c-22)$$
$x$ is many students taking none of these courses.
|
Total of all the classes
$$
a+b+c+2d+2e+2f=133-66=67\tag{1}
$$
Algebra
$$
a+d+e=40-22=18\tag{2}
$$
Trigonometry
$$
b+d+f=48-22=26\tag{3}
$$
Geometry
$$
c+e+f=45-22=23\tag{4}
$$
Number of students in one of these classes
$$
t=a+b+c+d+e+f+22\tag{5}
$$
Number of students not in any class
$$
\begin{align}
60-t
&=38-a-b-c-d-e-f\\
&=38-\frac{67+a+b+c}2\\
&=\frac{9-(a+b+c)}2\tag{6}
\end{align}
$$
To maximize $(6)$, we want to minimize a+b+c. We cannot have $a+b+c=0$ since $(1)$ would then imply that $2(d+e+f)=67$. Thus, the minimum would be $a+b+c=1$. There are three cases:
$a=1$: $b=c=0$, $d=10$, $e=7$, $f=16$
$b=1$: $a=c=0$, $d=10$, $e=8$, $f=15$
$c=1$: $a=b=0$, $d=11$, $e=7$, $f=15$
Each of these cases leaves us with $4$ students not taking any classes.
|
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|
Solutions to ceiling equation system
Prove that there does not exist an $x$ with $1000 \leq x \leq 1990$ that can be expressed in the forms $$\dfrac{10000}{x} = \left \lceil\dfrac{10000}{x} \right \rceil -\dfrac{1}{m} \quad \text{and} \quad \dfrac{10000}{x-1} = \left \lceil\dfrac{10000}{x-1} \right \rceil-\dfrac{1}{k}$$ where $m$ and $k$ are positive integers and $\left \lceil\dfrac{10000}{x} \right \rceil = \left \lceil\dfrac{10000}{x-1} \right \rceil$.
I wasn't sure how to go about solving this question. I found an example that worked but instead of $10000$ it was $1000$. We have $$\dfrac{1000}{144} = 7-\dfrac{1}{18} \quad \text{and} \quad \dfrac{1000}{143} = 7-\dfrac{1}{143}.$$ How do we go about finding solutions here?
|
Some necessary conditons can be found, for example both $x \cdot 10000 \bmod x$ and $(x-1) \cdot 10000 \bmod (x-1)$ must be even multiples of triangular numbers. Unfortunately I did not see how to use those for a systematic approach to find solutions. (Not that it matters, but I am curious what the source or background of the problem is.)
By brute force, there are no solutions $1000 \le x \le 1990$. Dropping the range constraint, however, there exists a set of solutions that work for arbitrary $N = 10^n$.
*
*$n = 2$
$$\frac{100}{34} = 3 - \frac{1}{17}, \;\;\; \frac{100}{35} = 3 - \frac{1}{7}$$
*$n = 3$
$$\frac{1000}{334} = 3 - \frac{1}{167}, \;\;\; \frac{1000}{335} = 3 - \frac{1}{67}, \;\;\; \frac{1000}{336} = 3 - \frac{1}{42}$$
*$n = 4$
$$\frac{10000}{3334} = 3 - \frac{1}{1667}, \;\;\; \frac{10000}{3335} = 3 - \frac{1}{667}, \;\;\; \frac{10000}{3336} = 3 - \frac{1}{417}$$
*$...$
The explicit solution is $x = \frac{10^n + 5}{3}$ for $n \ge 2$, and $(x+1)$ is a solution as well for $n \ge 3$.
The proof follows by simply substituting the values and noting that $5 \,|\, x$, $2 \,|\,x-1$, and (for $n \ge 3$) also $8 \,|\, x+1$.
$$\frac{10^n}{x-1} = \frac{3 \cdot 10^n}{10^n + 2} = 3 - \frac{6}{10^n + 2} = 3 - \frac{1}{(\frac{x-1}{2})}
$$
$$\frac{10^n}{x} = \frac{3 \cdot 10^n}{10^n + 5} = 3 - \frac{15}{10^n + 5} = 3 - \frac{1}{(\frac{x}{5})}
$$
$$\frac{10^n}{x+1} = \frac{3 \cdot 10^n}{10^n + 8} = 3 - \frac{24}{10^n + 8} = 3 - \frac{1}{(\frac{x+1}{8})}
$$
|
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|
Generalizing proof of AM-GM inequality? For two positive real numbers $x$ and $y$, we can prove the AM-GM inequality between them.
$$\left(\sqrt{x}-\sqrt{y}\right)^2=x+y-2\sqrt{xy}\geq 0$$
$$\implies \frac{x+y}{2}\geq\sqrt{xy}$$
Can this be generalized easily to the case of more numbers?
|
If $n=2^k$, where $k\in\mathbb N$, so we can use $\frac{x+y}{2}\geq\sqrt{xy}$ for non-negatives $x$ and $y$.
If $n>2^k$ we can use a similar to the following reasoning.
We'll prove that $\frac{x+y+z}{3}\geq\sqrt[3]{xyz}$ for non-negatives $x$, $y$ and $z$.
Indeed, by AM-GM for four variables we obtain:
$$\frac{x+y+z+\sqrt[3]{xyz}}{4}\geq\sqrt[4]{xyz\sqrt[3]{xyz}}=\sqrt[3]{xyz}$$
and we are done!
For five non-negative variables we have
$$\frac{a+b+c+d+e+3\sqrt[5]{abcde}}{8}\geq\sqrt[8]{abcde\sqrt[5]{a^3b^3c^3d^3e^3}}=\sqrt[5]{abcde}$$
which gives
$$\frac{a+b+c+d+e}{5}\geq\sqrt[5]{abcde}$$
|
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|
Prove $ \frac 1 2 \cdot \frac 3 4 \cdot \frac 5 6 \cdots \frac{2n-1}{2n} < \frac 1 {(2n+1)^{0.5}} $ . Prove
$$
\frac 1 2 \cdot \frac 3 4 \cdot \frac 5 6 \cdots \frac{2n-1}{2n} < \frac 1 {(2n+1)^{0.5}}
$$
Can this be done by induction using the pi function.
If no, why not.
|
Hint. If you want to prove it by induction, you may observe that, for $n\ge1$,
$$
(2n+2)^2=4n^2+8n+4>4n^2+5n+3=(2n+1)(2n+3)
$$ giving
$$
\frac1{2n+2}<\frac1{\sqrt{2n+1}\cdot \sqrt{2n+3}}, \qquad n\ge1,
$$ which is equivalent to the inductive step:
$$
\frac1{\sqrt{2n+1}}\cdot \frac{2n+1}{2n+2}<\frac1{\sqrt{2n+3}}.
$$
|
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|
Taylor Series of $ \cfrac{x}{1+x} $ at $ x = -2 $ I am trying to find the taylor series for the following:
$$ \cfrac{x}{1+x} \ \text{ at } \ x = -2 $$
The following are my steps:
$$ \cfrac{x}{1+x} = 1 - \cfrac{1}{1+x} \ \text{ (by long division) } $$
Taylor series of $ \cfrac{1}{1+x} $ at $ x = 0, $
$$ \begin{align} \cfrac{1}{1+x} = \displaystyle\sum_{n=0}^{\infty}{(-1)^n(x)^n}\end{align} $$
Taylor series of $ \cfrac{1}{1+x} $ at $ x = -2, $
$$ \begin{align} \cfrac{1}{1+(x+2)} = \displaystyle\sum_{n=0}^{\infty}{(-1)^n(x+2)^n}\end{align} $$
Taylor series of $ \cfrac{x}{1+x} $ at $ x = -2, $
$$ \begin{align}
\cfrac{x}{1+x}
&= 1 - \cfrac{1}{1+x} \\
&= 1 - \displaystyle\sum_{n=0}^{\infty}{(-1)^n(x+2)^n} \\
&= 1 - \left(\cfrac{1}{1+(-2)} + \displaystyle\sum_{n=1}^{\infty}{(-1)^n(x+2)^n}\right) \\
&= 2 - \displaystyle\sum_{n=1}^{\infty}{(-1)^n(x+2)^n}
\end{align}
$$
Somehow, the answer is:
$ 2 + \displaystyle\sum_{n=1}^{\infty}{(x+2)^n} $
I know that my last block of working seems wrong. But I just cannot think of anything else. Could someone please advise me?
|
Hint. Note that
$$\frac{x}{1+x}=\frac{x}{-1+x+2}=\frac{-(x+2)+2}{1-(x+2)}=1+\frac{1}{1-(x+2)}.$$
Now use the result: for $|z|<1$ then
$\displaystyle \cfrac{1}{1-z} = \displaystyle\sum_{n=0}^{\infty}{ z^n}$.
|
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|
Showing that $\lim_{x \to 1} \left(\frac{23}{1-x^{23}}-\frac{11}{1-x^{11}} \right)=6$ How does one evaluate the following limit?
$$\lim_{x \to 1} \left(\frac{23}{1-x^{23}}-\frac{11}{1-x^{11}} \right)$$
The answer is $6$.
How does one justify this answer?
Edit: So it really was just combine the fraction and use L'hopital's rule twice (because function and its first derivative are of indeterminate form at $x=1$). This problem is more straightforward than it seems at first.
|
$\lim_\limits{x\to1}\frac {23(1-x^{11})-11(1-x^{23})}{(1-x^{11})(1-x^{23})}$
$\lim_\limits{x\to1}\frac {12-23x^{11}+ 11x^{23}}{(1-x^{11})(1-x^{23})}$
Now we could apply L'Hopitals at this point, or we can use algebra.
Using algebra, numerator and denominator both divide by $(x-1)^2$
$1-x^{11} = (1-x)\sum_\limits{i=0}^{10} x^i\\
1-x^{23} = (1-x)\sum_\limits{i=0}^{22} x^i$
$11x^{23} -23x^{11}+ 12 = (x-1)(11 x^{22}\cdots 11x^{11} - 12x^{10}\cdots +12)\\ =(x-1)^2 (11x^{21} + 2\cdot11 x^{20} + 3\cdot11 x^{19}\cdots +12\cdot 11 x^{10} + 12\cdot 10 x^9\cdots +12\cdot 2x + 12 $
Evaluated at 1.
The denominator:
$\sum_\limits{i=0}^{10} x^i = 11, \sum_\limits{i=0}^{22} x^i = 23$
the numerator:
$11 \sum_\limits{i=1}^{11} i + 12 \sum_\limits{i=1}^{11} i = (23)(11)(12)/2$
and the ratio $= 6$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Finding $m$ so that the roots of $x^2 - (3m - 1)x + m^2 - 2 = 0$ lie inside the interval $(1, 5)$
Find $m$ so that the roots of the equation $x^2 - (3m - 1)x + m^2 - 2 = 0$ lie inside the interval $(1, 5)$.
Having $x_1, x_2 \in (1, 5)$, we get $x_1 + x_2 \in (2, 10)$ and $x_1x_2 \in (1, 25)$. By solving these two inequations I got $m \in \left( \sqrt3, \frac{11}{3} \right)$. However, this result is wrong. By substituting $m$ with $2$, for example, the given conditions are not reached.
So, please, explain what I did wrong and how to get the right solution.
Thank you in advance!
|
Notice that $\Delta = 5m^2 -6m +9 > 0, \forall m \in \mathbb{R}.$ Hence,
$x = \dfrac{3m-1 \pm \sqrt{5m^2 -6m +9}}{2} \in \mathbb{R}.$
$(1):~~~\dfrac{3m-1 - \sqrt{5m^2 -6m +9}}{2} > 1 \implies (3m-3)^2 > 5m^2 -6m +9 \implies 4m(m -3) > 0 \implies m < 0 ~~\text{or}~~ m > 3.$ But, only $ m > 3$ satisfies the condition (we squared the inequality!).
$(2):~~~ \dfrac{3m-1 + \sqrt{5m^2 -6m +9}}{2} < 5 \implies 5m^2 -6m +9 <(-3m+11)^2 \implies m^2 -15m +28 > 0 \implies m < \dfrac{15-\sqrt{113}}{2} \approx 2.185 ~~\text{or}~~ m > \dfrac{15+\sqrt{113}}{2} \approx 12.185.$ However, only $m < \dfrac{15-\sqrt{113}}{2}$ is solution of the inequality $(2)$.
Thus, combine the results of $(1)$ and $(2)$, we notice that there is no $m$ such that $x \in (1,5)$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Does this trig equation have no solution? $\sqrt 2\sin \left(\sqrt 2x\right)+\sin (x)=0$
Solve the following trig. equation for $x$
$$\sqrt 2\sin \left(\sqrt 2x\right)+\sin (x)=0$$
My try:
I divided by $\sqrt{\left(\sqrt2\right)^2+1^2}=\sqrt 3$,
$$\frac{\sqrt 2}{\sqrt 3}\sin\left(\sqrt 2x\right)+\frac{1}{\sqrt 3}\sin (x)=0$$
$$\sqrt{\frac{2}{3}}\sin\left(\sqrt 2x\right)+\frac{1}{\sqrt 3}\sin (x)=0$$
I got stuck here, I have no clue to proceed to find the values of $x$.
|
$\sqrt{2} \sin(\sqrt{2}x)$ goes from $-1$ to $1$ on the interval $[(2n-1/4) \pi/\sqrt{2}, (2n+1/4)\pi/\sqrt{2}]$ and from $1$ to $-1$ on the interval $[(2n+3/4)\pi/\sqrt{2}, (2n+5/4)\pi/\sqrt{2}]$. On each of these intervals
the derivative of $\sqrt{2}\sin(\sqrt{2}x)$ has absolute value at least $\sqrt{2}$. Thus $\sqrt{2} \sin(\sqrt{2}x) + \sin(x)$ is monotone and has exactly one zero in each of these intervals.
|
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|
Find the set of $n\in\Bbb Z^+$ with $M=\{n,n+1,n+2,n+3,n+4,n+5\}$ partitionable into two sets
Find the set of all positive integers $n$ with the property that the set $M=\{n, n + 1,n + 2,n + 3,n + 4,n + 5\}$ can be partitioned into two sets such that the product of the numbers in one set equals the product of the numbers in the other set.
If $n=1$ them $M=\{1,2,3,4,5,6\}$ and there is no such partition, so $n \ge 2$.
If the prime $p|n$ then either $p|2$ or $p|3$ or $p|5$ which means either $p=2$ or $p=3$ or $p=5$.
Suppose $n=2k$. Then $M=\{2k, 2k + 1,2k + 2,2k + 3,2k + 4,2k + 5\}$. I have no idea how to proceed.
|
Hint 1: Start by proving that each set of the partition must have 3 elements. To do this just argue that, for $n \geq 2$, the product of the smallest 4 elements is higher than the product of the largest 2.
Hint 2: For all $n >0 $ we have
$$(n+3)(n+4)(n+1)>(n+5)(n+2)n$$
Hint 3 one of the sets of your partition must contain two of $n+3, n+4, n+5$. Now use
$$(n+3)(n+4)(n+1)>(n+5)(n+2)n$$
to argue that the remaining number must be $n$.
Therefore, one of your set contains $n$ and two of $n+3, n+4, n+5$.
In each of the three cases, you get a quadratic equation.
|
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|
Prove by induction that $\left [ \left (3+\sqrt{5} \right )^{n} + \left (3-\sqrt{5} \right )^{n} \right ]$ is multiple of $2^{n}$ Prove by induction that :
$\left [ \left (3+\sqrt{5} \right )^{n} + \left (3-\sqrt{5} \right )^{n} \right ]$ is multiple of $ 2^{n}$
My proof is :
At $n=1$
$$\frac{\left ( 3+\sqrt{5} \right )+\left ( 3-\sqrt{5} \right )}{2} = \frac{6}{2} = 3 \in \mathbb{Z}$$
Assume $P(k)$ is true
$$\frac{\left ( 3+\sqrt{5} \right )^{k}+\left ( 3-\sqrt{5} \right )^{k}}{2^{k}} = m \in \mathbb{Z}$$
$$\Rightarrow \left ( 3+\sqrt{5} \right )^{k}+\left ( 3-\sqrt{5} \right)^{k} = 2^{k}.m \rightarrow (*)$$
At $n=k+1$
$$\left ( 3+\sqrt{5} \right )^{k+1}+\left ( 3-\sqrt{5} \right )^{k+1} = \left [ \left ( 3+\sqrt{5} \right )^{k}+\left ( 3-\sqrt{5} \right )^{k} \right ]\left [ \left ( 3+\sqrt{5} \right )+\left ( 3-\sqrt{5} \right ) \right ]-\left [ \left ( 3+\sqrt{5} \right )^{k}\left ( 3-\sqrt{5} \right )+\left ( 3-\sqrt{5} \right )^{k}\left ( 3+\sqrt{5} \right ) \right ]$$
from $(*) $:
$$\left ( 3+\sqrt{5} \right )^{k+1}+\left ( 3-\sqrt{5} \right )^{k+1} =2^{k}.m.6-\left ( 3+\sqrt{5} \right )\left ( 3-\sqrt{5} \right ) \left [ \left ( 3+\sqrt{5} \right )^{k-1}+\left ( 3-\sqrt{5} \right )^{k-1} \right ] $$
but i can't resume my proof .. i can't do any thing with $\left [ \left ( 3+\sqrt{5} \right )^{k-1}+\left ( 3-\sqrt{5} \right )^{k-1} \right ]$ .
|
An idea for you assuming the claim's true for exponents up to $\;k\;$ . Observe that I begin from the middle of your work:
$$\left ( 3+\sqrt{5} \right )^{k+1}+\left ( 3-\sqrt{5} \right )^{k+1} = \left [ \left ( 3+\sqrt{5} \right )^{k}+\left ( 3-\sqrt{5} \right )^{k} \right ]\left [ \left ( 3+\sqrt{5} \right )+\left ( 3-\sqrt{5} \right ) \right ]-$$
$$-\left [ \left ( 3+\sqrt{5} \right )^{k}\left ( 3-\sqrt{5} \right )+\left ( 3-\sqrt{5} \right )^{k}\left ( 3+\sqrt{5} \right ) \right ]=$$
$$=2^km\cdot6-(3+\sqrt5)(3-\sqrt5)\overbrace{\left[(3+\sqrt5)^{k-1}+(3-\sqrt5)^{k-1}\right]}^{=2^{k-1}\cdot r\,,\,\,r\in\Bbb Z}=$$
$$=2^{k+1}\cdot3m-(9-5)\cdot2^{k-1}\cdot r=2^{k+1}(3m+r).\;\;\;\;\;\square$$
|
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|
Proving $E(X)$ and $Var(X)$ for a random variable that has a discrete uniform distribution problem, what am I doing wrong I am having a problem proving $E(X)$ and $Var(X)$ for a random variable that has a discrete uniform distribution. The question and my current workings are below, can someone please indicate what I am doing wrong and if possible correct me?
Statement: let $X=a,a+1,\dots,a+(s-1)$ be a random variable that has a discrete uniform distribution.
a. show that $E(X)=a+\frac{s-1}2$
method 1:
$$E(X)=\sum_{k=1}^{a+(s-1)}k\cdot P(X=x)=\sum_{k=1}^{a+(s-2)}k\cdot\left(a+\frac1{(s-2)}\right)\\
=a+\frac1{(s-2)}\sum_{k=1}^{(s-2)}k=a+\frac1{(s-2)}\left(\frac{(s-2)((s-2)+1)}{2}\right)\\
=a+\frac{(s-2)}{2}$$
method 2:
$$E(X)=\sum_{k=1}^{a+(s-1)}k\cdot P(X=x)=\sum_{k=1}^{a}k\cdot P(X=x)+\sum_{k=1}^{(s-1)}k\cdot P(X=x)\\
=\sum_{k=1}^a k\cdot\frac1a+\sum_{k=1}^{(s-1)} k\cdot\frac1{(s-1)}\\
=\frac1a\left(\frac{a(a+1)}{2}\right)+\frac1{(s-1)}\left(\frac{(s-1)((s-1)+1)}{2}\right)=\frac{a+1}{2}+\frac{s}{2}$$
b. show that $Var(X)=\frac{s^2-1}{12}$
$$Var(X)=E(X^2)-(E(X))^2$$
$$E(X^2)=\sum_{k=1}^{a+(s-1)}k^2\cdot P(X=x)=\sum_{k=1}^{a+(s-1)}k^2\cdot\frac{1}{a+(s-1)}=\frac{1}{a+(s-1)}\sum_{k=1}^{a+(s-1)}k^2\\
=\frac{1}{a+(s-1)}\left(\frac{(a+(s-1))(a+(s-1)+1)(2a+2(s-1)+1)}{6}\right)\\
=\frac{(a+s)(2a+2s+1}{6}=\frac{2a^2+4as-a+2s^2-s}{6}$$
Now
$$Var(X)=E(X^2)-(E(X))^2=\frac{2a^2+4as-a+2s^2-s}{6}-\left(a+\frac{s-1}{2}\right)^2\\
=\frac{2a^2+4as-a+2s^2-s}{6}-\left(a^2+a(s-1)+\frac{(s^2-2s-1)}{4}\right)\\
=\frac{2a^2+4as-a+2s^2-s}{6}-\left(\frac{4a^2+4as-4a+(s^2-2s-1)}{4}\right)\\
=\frac{4a^2+8as-2a+4s^2-2s-12a^2-12as+12a-3s^2+6s+3}{12}\\
=\frac{-8a^2-4as+10a+s^2+4s+3}{12}
$$
|
$$E(X^2) = \sum_{k=a}^{a+s-1} k^2P(X=k) = \frac 1{a+s-1}\sum_{k=a}^{a+s-1}k^2 = \frac{ (s-1)(6a^2 + 6as-6a + 2s^2 + 1 - 3s)}{6(a+s-1)}$$
$$Var(X) = E(X^2) - E(X)^2 = \frac{ (s-1)(6a^2 + 6as-6a + 2s^2 + 1 - 3s)}{6(a+s-1)} - a^2 - \frac {s^2 - 2s + 1}4 - a(s-1)$$
Now you could go ahead and simplify this, or using arguments similar to the one @BGM posted in the comments you can see that the above does not depend on $a$, so choose, say, $a = 0$ to find
$$Var(X) = \frac{(s-1)(2s^2 - 3s + 1)}{6(s-1)} - \frac{s^2 - 2s + 1}{4} = \frac {s^2}3 - \frac s2 + \frac 16 - \frac{s^2}4 + \frac s2 - \frac 14 = \frac {s^2}{12}-\frac 1{12} $$
|
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Why do some divisibility rules work only in base 10? Aside from the zero rule (a number ending with zero means it is divisible by the base number and its factors), why is it that other rules cannot apply in different bases?
In particular for 3 one can just sum all digits and see if it is divisible. What relation exists between 10 and 3 to have this property? Does this property exist in another base?
Also:
Are there divisibility rules for base 2?
Are there general divisibility formulae for all bases?
|
There are other rules for other bases.
To understand why the rule for $3$ works in base $10$, you need to write your number $x=d_n\times 10^n+\dots+d_1\times 10^1+d_0$ where the $d_i$ are the digits of $x$ in base $10$.
You can notice that $10=3\times 3+1$, $10^2=3\times 33+1$, $10^n=3...3\times 3+1$ (where $3...3$ has $n-1$ digits equal to $3$).
So $$\begin{align}x &=d_n\times (3...3\times 3+1)+\dots+d_1\times (3\times 3+1)+d_0 \\ &= (d_n\times 3...3 + \dots+d_1\times 3)\times 3 +d_n+\dots+d_1+d_0\end{align}$$
As you can see, the first term is divisible by $3$, so $x$ is divisible by $3$ if and only if $d_n+\dots+d_1+d_0$ is divisible by $3$.
So the reason that this rule works is the fact that the rest of the division of a power of $10$ by $3$ is $1$.
In base $5$, the same rule applies with $2$ (because the powers of $5$ are odd). For example, in base $5$, $12$ is not divisible by $2$ ($1+2=3$ is not divisible by $2$) but $13$ and $431$ are.
To find rules like this by yourself for a base $b$ and a divisor $d$, you need to study how the powers of $b$ behave when divided by $d$.
|
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|
showing $\frac{x}{e^{x}-1} = \sum_{n=1}^{\infty} \frac{(1-e^x)^{n-1}}{n} $ I want to show
$\frac{x}{e^{x}-1} = -\sum_{k=0}^{\infty} x e^{kx} = \frac{\ln(e^x)}{e^x-1} = -\frac{\ln[1-(1-e^x)]}{1-e^x} = \sum_{n=1}^{\infty} \frac{(1-e^x)^{n-1}}{n} $
The first step seems trivial, what i am confused is the process of
\begin{align}
-\sum_{k=0}^{\infty}xe^{kx} = \frac{\ln(e^x)}{e^x-1}
\end{align}
and the last step
\begin{align}
-\frac{\ln[1-(1-e^x)]}{1-e^x} = \sum_{n=1}^{\infty} \frac{(1-e^x)^{n-1}}{n}
\end{align}
Can you give me some idea for these steps?
|
In order to show the identity
\begin{align*}
\frac{x}{e^x-1}=\sum_{n=1}^\infty \frac{\left(1-e^x\right)^{n-1}}{n}\qquad\qquad |1-e^x|<1\tag{1}
\end{align*}
we recall that
\begin{align*}
\ln(1-x)=-\sum_{n=1}^\infty\frac{x^n}{n}\qquad\qquad\qquad |x|<1
\end{align*}
We obtain
\begin{align*}
x&=\ln(e^x)=\ln(1-(1-e^x))\\
&=-\sum_{n=1}^\infty \frac{1}{n} (1-e^x)^n\qquad\qquad\qquad\qquad\qquad |1-e^x|<1\\
\end{align*}
and after division by $1-e^x$ with $x \neq 0$ we get (1)
\begin{align*}
\frac{x}{e^x-1}=\sum_{n=1}^\infty \frac{1}{n} (1-e^x)^{n-1}\qquad\qquad\qquad\qquad x<\ln 2,x\neq 0
\end{align*}
Note, it is not necessary to consider a geometric series expansion. In fact the representation in this case is different from the series representation in (1) since the radius of convergence is different.
\begin{align*}
\frac{x}{e^x-1}=\sum_{n=0}^\infty xe^{kx}\qquad\qquad |e^x|<1\quad\text{resp.}\quad x<0
\end{align*}
|
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|
Challenging Integral [indefinite] Integrate: $$\int \frac{x^2+n(n-1)}{(x\sin x+n\cos x )^2}dx$$
I've been beating my head around this problem for quite some time now, but I've got nowhere. I'd request the person writing the solution to please explain his thought process because I would like to learn how to appraoch such Integrals in the future.
|
Using the identities $n^2+x^2=(n\sin x-x\cos x)^2+(n\cos x+x\sin x)^2$ and
$n=\sin x(n\sin x-x\cos x)+\cos x(n\cos x+x\sin x)$, we can rewrite the integral as
$\displaystyle \int\frac{(n\sin x-x\cos x)^2+(n\cos x+x\sin x)^2-[\sin x(n\sin x-x\cos x)+\cos x(n\cos x+x\sin x)]}{(n\cos x+x\sin x)^2} dx$
$\displaystyle=\int\frac{(n\cos x+x\sin x)[n\cos x+x\sin x-\cos x]-(n\sin x-x\cos x)[-n\sin x+x\cos x+\sin x]}{(n\cos x+x\sin x)^2} dx$
$\displaystyle=\int\bigg(\frac{n\sin x-x\cos x}{n\cos x+x\sin x}\bigg)^{\prime} dx$
$\;\;\;\;\displaystyle=\frac{n\sin x-x\cos x}{n\cos x+x\sin x}.$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Roots of $x^2-2ax+b=0,~a^2>>b$, stable algorithm
Find a stable algorithm for the computation of the roots of the equation $x^2-2ax+b=0$ if $a,~b>0$ and $a^2>>b$.
Attempt. The roots are $x_1=a-\sqrt{a^2-b}$ and $x_2=a+\sqrt{a^2-b}$, where we have a cancelation error for the computation of $x_1$, since $a^2-b\cong a^2$. So we compute $\displaystyle x_1=\frac{b}{a+\sqrt{a^2-b}}$ (stable algorithm).
My question: is it correct to say that $\displaystyle \frac{b}{2a}$ is a "good" approximation for $x_1$? Or should we remain on the algorithm $\displaystyle x_1=\frac{b}{a+\sqrt{a^2-b}}$?
Thanks in advance!
|
*
*About one part of your question, no purely mathematical answer can be given because $\displaystyle x_1=\frac{b}{a+\sqrt{a^2-b}}$ is strictly the same as $a-\sqrt{a^2-b}.$ It is a floating point issue.
*About the other part, yes, $\frac{b}{2a}$ is a good approximation.
This can be attested by applying Taylor expansion :
$$\sqrt{1-x}=1-\frac{1}{2}x-\frac{1}{8}x^2-\cdots$$
in the following way:
$$x_1=a-\sqrt{a^2-b}=a-a\sqrt{1-\frac{b}{a^2}}=a-a\left(1-\frac{1}{2}\frac{b}{a^2}-\frac{1}{8}\frac{b^2}{a^4}- \cdots\right)$$
Giving :
$$x_1 \ \approx \ \frac{b}{2a}+\frac{b^2}{8a^3}$$
Thus, the order of the error is approximately: $\frac{b^2}{8a^3}$,
which is very small because $b << a^2.$
|
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|
Binomial identity in a Change of Basis Matrix I need the following identity in my research and have a rather ugly proof but cannot find anything more elegant.
If $L\colon\mathbb{Q}[m]\to\mathbb{Q}[m]$ is a linear map which satisfies
$L\left(\binom{m}{c+d}\right)=L\left(\binom{m}{c}\right)L\left(\binom{m}{d}\right)$
for all integer $c,d\geq0$, then $L$ also satisfies
$L\left(\binom{m+a+b}{c+d}\right)=L\left(\binom{m+a}{c}\right)L\left(\binom{m+b}{d}\right)$
for all integer $a,b,c,d\geq0$.
We shall proceed by induction on $a+b$.
For the base case where $a+b=0$, $a=b=0$ and the result is a re-indexed form of the assumption on $L$.
Now suppose that the result holds for all $a+b=n$ and that $a+b=n+1$.
Without loss of generality, $a\geq1$.
Then by Pascal's identity and the inductive assumption,
\begin{align*}
L\left(\binom{m+a+b}{c+d}\right)&=L\left(\binom{m+(a-1)+b}{(c-1)+d}\right)+L\left(\binom{m+(a-1)+b}{c+d}\right)\\
&=L\left(\binom{m+(a-1)}{c-1}\right)L\left(\binom{m+b}{d}\right)+L\left(\binom{m+(a-1)}{c}\right)L\left(\binom{m+b}{d}\right)\\
&=L\left(\binom{m+(a-1)}{c-1}+\binom{m+(a-1)}{c}\right)L\left(\binom{m+b}{d}\right)\\
&=L\left(\binom{m+a}{c}\right)L\left(\binom{m+b}{d}\right)
\end{align*}
which shows the inductive step and completes the proof. Does anyone have a better proof of this? Furthermore, is this the sort of result one would include in a paper or exclude as obvious enough for the reader to conclude?
|
Vandermonde's identity and diagonal summation can be used to give a non-inductive proof:
\begin{align*}
L\left(\binom{m+a}{c}\right)L\left(\binom{m+b}{d}\right)&=L\left(\sum_{i=0}^c\binom{m}{i}\binom{a}{c-i}\right)L\left(\sum_{j=0}^d\binom{m}{j}\binom{b}{d-j}\right)\\
&=\sum_{i=0}^c\sum_{j=0}^d\binom{a}{c-i}\binom{b}{d-j}L\left(\binom{m}{i}\right)L\left(\binom{m}{j}\right)\\
&=\sum_{k=0}^{c+d}L\left(\binom{m}{k}\right)\sum_{i=0}^c\binom{a}{c-i}\binom{b}{d-(k-i)}\\
&=L\left(\sum_{k=0}^{c+d}\binom{m}{k}\binom{a+b}{c+d-k}\right)\\
&=L\left(\binom{m+a+b}{c+d}\right).
\end{align*}
|
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|
Find the volume of the solid obtained by rotating the region bounded by the given curves about the $x$-axis. $y=x^{3/2}$, $y=8$, $x=0$ Use the method of cylindrical shells to find the volume of the solid obtained by rotating the region bounded by the given curves about the $x$-axis.
$y=x^{3/2}$, $y=8$, $x=0$
What I did:
$y=x^{3/2}\Rightarrow x=y^{2/3}$
$Circumference= 2\pi(8-y)$
$Height=(4-y^{2/3})$
$Surface Area =2\pi(8-y)(4-y^{2/3})$
$Volume=2\pi\int _{ 0 }^{ 8 }{ (8-y)(4-y^{2/3})dy} $
$=2\pi\int _{ 0 }^{ 8 }{ 32-8y^{2/3}-4y+y^{5/3}dy}$
$=2\pi[32y-\frac{24y^{5/3}}{5}-2y^2+\frac{3y^{8/3}}{8}]_{0}^{8}$
$=2\pi[32(8)-\frac{24(8)^{5/3}}{5}-2(8)^2+\frac{3(8)^{8/3}}{8}]-2\pi[0]$
$=2\pi[256-\frac{768}{5}-128+\frac{768}{8}]$
$=2\pi(\frac{352}{5})$
$=\frac{704\pi}{5}$
My final answer seems to be wrong, but I am not sure where I went wrong in my (wrong) solution for this problem. Any help/guidance would be greatly appreciated.
|
Since the revolution is around the $x$-axis, each cylindrical shell at $x$ for $x\in [0,4]$ (where $4=8^{3/2}$) is actually an annulus with inner radius $r(x)=x^{3/2}$ and outer radius $R(x)=8$.
The volume of this shell is $(\pi R^2(x) -\pi r^2(x))\cdot dx$. Hence the total volume can be obtained by "summing" the volumes of all these thin shells,
$$\int_{x=0}^4 (\pi R^2(x) -\pi r^2(x))\cdot dx=\pi\int_{0}^4 (8^2 -x^3)\cdot dx=\pi \left[64x-\frac{x^4}{4}\right]_0^4= 192\pi.$$
Note that if the revolution is around the $y$-axis, the solid is different and its volume is
$$\int_{y=0}^8 \pi (y^{2/3})^2\cdot dy=\pi \left[\frac{y^{7/3}}{7/3}\right]_0^8=\frac{384\pi}{7}.$$
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"language": "en",
"url": "https://math.stackexchange.com/questions/1956151",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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