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Find the exact value of integration $ \int_0^1 \frac{1}{\sqrt{1-x}+\sqrt{x+1}+2} dx$ Can you help me find the exact value for integration with the given steps?
$$ \int_0^1 \frac{1}{\sqrt{1-x}+\sqrt{x+1}+2} dx$$
Some of my attempts as indefinite Integral
$$
\int \frac{1}{2+\sqrt{1-x}+\sqrt{1+x}} \, dx\approx \left(\sqrt{x+1}+\left(-\frac{1}{\sqrt{x+1}+1}-1\right) \sqrt{1-x}+\frac{1}{\sqrt{x+1}+1}-\frac{2 \left(0.707107 \sqrt{x+1}\right)}{\sin }\right)+C
$$
Is it considered Improper Integral?
|
Hint:
$$
\begin{aligned}
& \int\frac{dx}{\sqrt{1-x}+\sqrt{x+1}+2}\\
& \stackrel{x\to\cos2\phi}=
\int\frac{\sin{2\phi}\,d\phi}{1+\frac1{\sqrt2}(\sin\phi+\cos\phi)}
=\int\frac{\sin{2\phi}\,d\phi}{1+\sin(\phi+\frac\pi4)}\\
&\stackrel{\phi\to\theta+\frac\pi4}
=\int\frac{\cos{2\theta}\,d\theta}{1+\cos\theta}=\int\frac{2\cos^2{\theta}-1}{1+\cos\theta}\,d\theta\\
&\stackrel{\theta\to2\arctan t}=\int\left[2\left(1-\frac{2}{t^2+1}\right)^2-1\right]\,dt.
\end{aligned}
$$
The rest should not be complicated.
|
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|
What is the derivative of $\mathbf{a}^TX^2\mathbf{a}$ with respect to the symmetric matrix $X$? Given a constant vector $\mathbf{a}\in{\rm I\!R}^n$ and a real symmetric matrix $X\in{\rm I\!R}^{n\times n}$, what is the derivative of $\mathbf{a}^T X^2 \mathbf{a}$ with respect to $X$?
I tried a simple example using $n=2$, with the following vector and matrix :
$$
\mathbf{a} = \begin{bmatrix}a\\b\end{bmatrix} \qquad X = \begin{bmatrix} x & z \\ z & y \end{bmatrix}
$$
What we want to differentiate is
$$
\mathbf{a}^T X^2 \mathbf{a} = a^2 (a^2 + z^2) + 2 ab (xz+yz) + b^2 (z^2 + y^2)
$$
Which gives the following result
$$
\frac{\partial \mathbf{a}^T X^2 \mathbf{a}}{\partial X} =
\begin{bmatrix}
\frac{\partial \mathbf{a}^T X^2 \mathbf{a}}{\partial x} & \frac{\partial \mathbf{a}^T X^2 \mathbf{a}}{\partial z} \\
\frac{\partial \mathbf{a}^T X^2 \mathbf{a}}{\partial z} & \frac{\partial \mathbf{a}^T X^2 \mathbf{a}}{\partial y}
\end{bmatrix}
=
2
\begin{bmatrix}
a^2 x + abz & a^2z + abx + aby + b^2z \\
a^2z + abx + aby + b^2z & b^2 y + abz
\end{bmatrix}
$$
I tried to re-write this to end up with something meaningful, but I could only write
$$
\frac{\partial \mathbf{a}^T X^2 \mathbf{a}}{\partial X} = 2X\mathbf{a}\mathbf{a}^T + 2
\begin{bmatrix}
0 & a^2 z + ab y \\
abx + b^2z & 0
\end{bmatrix}
$$
and I do not know what to do with the matrix on the right...
What I would like is a valid expression for any $n > 1$ involving only $X$ and $\mathbf{a}$.
|
Hint
Fréchet derivative of $f(X) =\mathbf{a}^TX^2\mathbf{a}$ is given by
$$\partial_{X_0}f(h) = \mathbf{a}^TX_0 h\mathbf{a} + \mathbf{a}^Th X_0 \mathbf{a}$$
|
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|
Area of square that is inscribed in a circle that is also inscribed in a rhombus
A circle is inscribed in a rhombus whose diagonals are $17 cm$ and $27 cm$. What is the area of square inscribed on the same circle?
Solution:
Centred at the origin, one side can lie on the line $\frac{x}{(13.5)} + \frac{y}{(8.5)} = 1$.
The square of the distance from the origin to the line/side $\frac{x}{(13.5)} + \frac{y}{(8.5)} = 1$ is $\frac{1} {\frac{1}{(13.5)^2}} + \frac{4}{(8.5)^2}$ which is $\frac{210681}{4072}$ which is half the area of the inscribed square is which is $\frac{210681}{2036}cm²$.
Question:
Is $\frac{210681}{2036}cm²$ correct? Or is there something wrong with my solution?
|
Let $r$ be the circle of the radius. Then,
$$r=\frac{a_1a_2}{\sqrt{a_1^2+a_2^2}}$$
with $a_1$ and $a_2$ being the half diameters of the rhombus. Then, the area of the square is
$$ Area = 2r^2 = \frac{2\left( \frac{17}2 \cdot\frac{27}2\right)^2}{{\left(\frac{17}2\right)^2+\left(\frac{27}2\right)^2}} = \frac12\cdot\frac{(17\cdot27)^2}{17^2+27^2}$$
|
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|
Express the matrix of $f$ with respect to the basis $\{1,x+1,x^{2}+x\}$. Let $\textbf{P}_{2}(\textbf{R})$ be the real vector space of polynomials of degree less than or equal to $2$.
Let $f:\textbf{P}_{2}(\textbf{R})\rightarrow \textbf{P}_{2}(\textbf{R})$ be the linear map given by differentiation, i.e., $f(p(x)) = p'(x)$.
Compute the matrix of $f$ with respect to the basis $\{1, x + 1, x^2 + x\}$.
MY ATTEMPT
So far I've done $f(1) = 0$, $f(x+1) = 1$ and $f(x^2 + x) = 2x + 1$.
|
Since we are dealing with a linear operator, we can assume the same basis for the domain and the counterdomain.
Given the basis $\mathcal{B} = \{1,x+1,x^{2}+x\}$, we have the following system of equations to solve
\begin{align*}
\begin{cases}
a_{11} + a_{12}(x+1) + a_{13}(x^{2}+x) = f(1) = 0\\\\
a_{21} + a_{22}(x+1) + a_{23}(x^{2} + x) = f(x+1) = 1\\\\
a_{31} + a_{32}(x+1) + a_{33}(x^{2} + x) = f(x^{2}+x) = 2x + 1
\end{cases}
\end{align*}
whence we conclude $f(1) = (0,0,0)_{\mathcal{B}}$, $f(x+1) = (1,0,0)_{\mathcal{B}}$ and $f(x^{2}+x) = (-1,2,0)_{\mathcal{B}}$.
Finally, one has that
\begin{align*}
[f]_{\mathcal{B}} =
\begin{bmatrix}
0 & 1 & -1\\
0 & 0 & 2\\
0 & 0 & 0
\end{bmatrix}
\end{align*}
and we are done.
Hopefully this helps.
|
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|
Range of $f(x)=ax^2-c$ The function $f(x)=ax^2-c$ satisfies $-4\le f(1) \le -1$ and $-1\le f(2) \le 5$. Which of the following statement is true ?
(1)$-7\le f(3) \le 26$
(2)$-4\le f(3) \le 15$
(3)$-1\le f(3) \le 20$
(4)$-\frac {28}{3}\le f(3) \le \frac {35}{3}$
My approach is as follow
$f(1)=a-c$
$f(2)=4a-c$
$f(3)=9a-c$
$-4\le a-c \le -1$ and $-1\le 4a-c \le 5$
We need to find $a'\le 9a-c \le b'$, how do I proceed.
|
First, consider how to express $f(3)$ as a linear combination of $f(1)$ and $f(2)$ so you can effectively use their adjusted limits to determine the limits for $f(3)$. To do this, for some real constants $d$ and $e$ you have
$$\begin{equation}\begin{aligned}
9a - c & = d(a - c) + e(4a - c) \\
& = (d)a - (d)c + (4e)a - (e)c \\
& = (d + 4e)a + (-d - e)c
\end{aligned}\end{equation}\tag{1}\label{eq1A}$$
Matching the coefficients of $a$ and $c$ between the left & right sides gives
$$d + 4e = 9 \tag{2}\label{eq2A}$$
$$-d - e = -1 \tag{3}\label{eq3A}$$
Adding \eqref{eq2A} and \eqref{eq3A} gives $3e = 8 \implies e = \frac{8}{3}$ which, substituting back into \eqref{eq3A} gives $-d - \frac{8}{3} = -1 \implies d = -\frac{5}{3}$. This now gives
$$\begin{equation}\begin{aligned}
f(3) & = 9a - c \\
& = - \frac{5}{3}(a - c) + \frac{8}{3}(4a - c) \\
& = - \frac{5}{3}f(1) + \frac{8}{3}f(2)
\end{aligned}\end{equation}\tag{4}\label{eq4A}$$
Next, you've got
$$-4\le f(1) \le -1 \implies \frac{20}{3} \ge -\frac{5}{3}f(1) \ge \frac{5}{3} \tag{5}\label{eq5A}$$
$$-1\le f(2) \le 5 \implies -\frac{8}{3} \le \frac{8}{3}f(2) \le \frac{40}{3} \tag{6}\label{eq6A}$$
Adding \eqref{eq5A} and \eqref{eq6A} gives, from using \eqref{eq4A}, that
$$\frac{5}{3} - \frac{8}{3} \le - \frac{5}{3}f(1) + \frac{8}{3}f(2) \le \frac{20}{3} + \frac{40}{3} \implies -1 \le f(3) \le 20 \tag{7}\label{eq7A}$$
This matches option ($3$).
|
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|
Proving Proposition 4.1.8. from Terence Tao's Analysis I Let $a$ and $b$ be integers such that $ab = 0$. Then either $a = 0$ or $b = 0$ (or both $a=b=0$).
MY ATTEMPT
Let us consider that $a = m - n$ and $b = p - q$, where $m,n,p,q$ are natural nubmers.
Then we have that
\begin{align*}
ab = (m-n)(p-q) & = (mp + nq) - (mq + np) = 0
\end{align*}
but then I get stuck. Can somebody help me out?
|
You can try and prove the contrapositive, namely, that $a \neq 0$ and $b \neq 0$ implies $ab \neq 0$. Use trichotomy of integers for $a$ and $b$ and Lemma 2.3.3.
Or you can use contradiction and assume that $a \neq 0$ and $b \neq 0$. Then, use trichotomy of integers and see that it contradicts Lemma 2.3.3.
Or if you prefer the direct approach to exhaust all possible cases:
Let $a$ and $b$ be integers. Suppose that $ab = 0$. By Lemma 4.1.5 (Trichotomy of integers) on p. 77 we have for $a$ three cases, namely, $a = 0$, $a = n$ for positive natural number $n$, $a = -n$ for positive natural number $n$. Similarly, for $b$ we have three cases $b = 0$, $b = m$ for positive natural number $m$, $b = -m$ for positive natural number $m$. Then, check cases.
(1) $a = 0$ and $b = 0$.
(2) $a = 0$ and $b = m$.
(3) $a = 0$ and $b = -m$.
(4) $a = n$ and $b = 0$.
(5) $a = n$ and $b = m$. Hint use Lemma 2.3.3 on p. 30.
(6) $a = n$ and $b = -m$.
(7) $a = -n$ and $b = 0$.
(8) $a = -n$ and $b = m$.
(9) $a = -n$ and $b = -m$.
|
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|
Conjecture Prove that : $\sum_{cyc}\frac{a}{a^n+1}\leq \sum_{cyc}\frac{a}{a^2+1}\leq \frac{3}{2}$ Conjecture, Prove that :
$$\sum_{cyc}\frac{a}{a^n+1}\leq \sum_{cyc}\frac{a}{a^2+1}\leq \frac{3}{2}$$
Under the assumptions $a\geq b\geq 1\geq c>0$ such that $abc=1$ and $\frac{c}{c^n+1}\geq \frac{b}{b^n+1}\geq \frac{a}{a^n+1}$ and $\frac{a}{a+b}\geq \frac{b}{b+c}\geq \frac{c}{c+a}$ and finally $n\geq 10$ a natural number.
My work
We start with the following expression :
$$\Big(\sum_{cyc}\frac{a}{a^n+1}\Big)\Big(\sum_{cyc}\frac{a}{a+b}\Big)$$
Using Tchebytchev's inequality and the order we have :
$$\Big(\sum_{cyc}\frac{a}{a^n+1}\Big)\Big(\sum_{cyc}\frac{a}{a+b}\Big)\leq 3\Big(\frac{a}{a^n+1}\frac{c}{c+a}+\frac{b}{b^n+1}\frac{b}{b+c}+\frac{a}{a+b}\frac{c}{c^n+1}\Big)$$
Now we study the following expression :
$$3\Big(\frac{ac+a^n}{(a^n+1)(c+a)}+\frac{b^2+b^{n-1}c}{(b^n+1)(b+c)}+\frac{ac+c^{n-1}b}{(a+b)(c^n+1)}\Big)$$
Take one element like :
$$\frac{ac+c^{n-1}b}{(a+b)(c^n+1)}$$
We prove that :
$$\frac{ac+c^{n-1}b}{(a+b)(c^n+1)}\leq \frac{c}{c^2+1}\leq \frac{1}{2}$$
Or :
$$\frac{a+c^{n-2}b}{(a+b)(c^n+1)}\leq \frac{1}{c^2+1}$$
Or :
$$\frac{(a+c^{n-2}b)(c^2+1)}{(a+b)(c^n+1)}\leq 1$$
Or :
$$(a+c^{n-2}b)(c^2+1)\leq (a+b)(c^n+1)$$
Or :
$$ac^2+a+c^nb+c^{n-2}b\leq ac^n+a+bc^n+b$$
Or :
$$ac^2+c^{n-2}b\leq ac^n+b$$
Using the fact that $abc=1$ we have :
$$ac^2+\frac{c^{n-2}}{ac}\leq ac^n+\frac{1}{ac}$$
Or :
$$a^2c^3+c^{n-2}\leq a^2c^{n+1}+1$$
Wich have the form :
$$x+y\leq xy+1$$
Wich can be solved with tangent hyperbolic .
Simpler if we have $a+c\geq 2$ and $ac\leq 1$ we deduce that :
$$\frac{ac+a^n}{(a^n+1)(c+a)}\leq \frac{1}{2}$$
Similar method conducts to :
$$\frac{b^2+b^{n-1}c}{(b^n+1)(b+c)}\leq \frac{1}{2}$$
Summing each elements we have :
$$3\Big(\frac{ac+a^n}{(a^n+1)(c+a)}+\frac{b^2+b^{n-1}c}{(b^n+1)(b+c)}+\frac{ac+c^{n-1}b}{(a+b)(c^n+1)}\Big)\leq \frac{9}{4}$$
But :
$$\frac{a^n}{(a^n+1)(c+a)}=\frac{\frac{1}{a}\frac{1}{c}}{(\frac{1}{a^n}+1)(\frac{1}{a}+\frac{1}{c})}$$
The same trick apply to the other elements conducts to :
$$3\Big(\frac{a}{a^n+1}\frac{c}{c+a}+\frac{b}{b^n+1}\frac{b}{b+c}+\frac{a}{a+b}\frac{c}{c^n+1}\Big)\leq \frac{9}{4}$$
And we deduce that :
$$\Big(\sum_{cyc}\frac{a}{a^n+1}\Big)\Big(\sum_{cyc}\frac{a}{a+b}\Big)\leq \frac{9}{4}$$
But with the assumptions we have :
$$\sum_{cyc}\frac{a}{a+b}\geq \frac{3}{2}$$
And so :
$$\sum_{cyc}\frac{a}{a^n+1}\leq \frac{3}{2}$$
My questions
Can someone correct it if it's wrong ?
With all this elements of proof can someone achieve or complete my proof ?
Have you other method to learn ?
Any helps is greatly appreciated .
Thanks in advance !
Ps : if you have a counter-example say where I'm wrong in my proof please .
PPs:I add the tag "contest-maths" to see if there is an elegant hand proof other than mine .
Update :
My trick is false because it works only for $a=b=c=1$ so the idea is to prove
$$3\Big(\frac{ac-a^n}{(a^n+1)(c+a)}+\frac{b^2-b^{n-1}c}{(b^n+1)(b+c)}+\frac{ac-c^{n-1}b}{(a+b)(c^n+1)}\Big)\leq 0$$
Under some assumptions .
The idea is to use the method A special case of Karamata's inequality to solve one or more Olympiad inequality? . I will develop it later but with this method it's easy the only problem is the order .
|
I will prove the second inequality $\sum_{cyc}\frac{a}{a^2+1}\leq \frac{3}{2}$ for you.
Note that for $x\gt0,$
$$\dfrac{x}{x^2+1}=\left(x+\dfrac1x\right)^{-1}$$ and by AM-GM inequality $$x+\dfrac1x\ge2$$ with equality occurs at $x=1.$ Hence the result follows.
|
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|
Linear combinations problem The vectors $\dbinom{3}{2}$ and $\dbinom{-4}{1}$ can be written as linear combinations of $\mathbf{u}$ and $\mathbf{w}$:
\begin{align*}
\dbinom{3}{2} &= 5\mathbf{u}+8\mathbf{w} \\
\dbinom{-4}{1} &= -3\mathbf{u}+\mathbf{w} .
\end{align*}The vector $\dbinom{5}{-2}$ can be written as the linear combination $a\mathbf{u}+b\mathbf{w}$. Find the ordered pair $(a,b)$.
I've tried to eliminate $\mathbf{u}$ by multiplying the first equation by 3, the second equation by 5, then adding, but it only leads to $\mathbf{w}=\dbinom{-\frac{11}{29}}{\frac{11}{29}}$. I feel like the algebra from here would be too complicated for what the people who wrote the problem were intending, so perhaps I'm going down the wrong path. Would there instead be a convenient way to manipulate the terms to eventually get $\dbinom{5}{-2}$ on the LHS?
|
Here is how I would do it.
$\pmatrix{3\\2} - 2\pmatrix{-4\\1} = \pmatrix{11\\0}$
Allowing us to find one of the princicipal component vectors of the standard basis in terms of the $\{u,w\}$ basis.
$5u + 8w - 2(-3u+w) = \pmatrix{11\\0}\\
11u + 6w = \pmatrix{11\\0}\\
\pmatrix{1\\0} = u + \frac {6}{11} w$
With that we can say:
$\pmatrix{5\\-2} = 8\pmatrix{1\\0}-\pmatrix{3\\2}$
And write our vector in terms of this basis.
$\pmatrix{5\\-2} = 8(u+\frac 6{11}w) - (5u+8w) = 3u- \frac {40}{11}w$
|
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Solve the equation $y-3px+ayp^2=0$ where $p=\frac {dy}{dx}$ Solve the equation $y-3px+ayp^2=0$ where $p=\frac {dy}{dx}$
My Attempt:
Given
$$y-3px+ayp^2=0$$
$$3px=y+ayp^2$$
$$x=\frac {1}{3} \cdot \frac {y}{p} + \frac {a}{3} \cdot yp$$
This is solvable for x. Differentiating both sides with respect to $y$
$$\frac {dx}{dy}=\frac {1}{3} \cdot \frac {p-y\frac {dp}{dy}}{p^2} + \frac {a}{3} (y\cdot \frac {dp}{dy} +p)$$
$$\frac {dx}{dy} = \frac {1}{3p} - \frac {y}{3p^2} \cdot \frac {dp}{dy} + \frac {ay}{3} \cdot \frac {dp}{dy} + \frac {ap}{3}$$
$$\frac {1}{p} - \frac {1}{3p} -\frac {ap}{3} = \frac {y}{3} \cdot \frac {dp}{dy}\cdot \frac {ap^2-1}{p^2}$$
$$\frac {3-1-ap^2}{3p} = \frac {y}{3} \cdot \frac {ap^2-1}{p^2} \cdot \frac {dp}{dy}$$
$$(2-ap^2) = \frac {y}{p} \cdot (ap^2-1) \cdot \frac {dp}{dy}$$
$$\frac {dy}{y} = \frac {ap^2-1}{p(2-ap^2)} dp$$
How do I integrate the RHS of above equation?
|
Hint 1:
$$\frac{ap^2-1}{p(2-ap^2)} = -\frac{1-ap^2}{p(2-ap^2)} = - \frac{(2-ap^2) - 1}{p(2-ap^2)} = - \frac{1}{p} + \frac{1}{p(2-ap^2)}$$
Hint 2: The second expression involves two simple logarithmic expressions and can be and explicitly calculated.
|
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|
Solve $x+yp=ap^2$ where $p=\frac {dy}{dx}$ Solve $x+yp=ap^2$ where $p=\frac {dy}{dx}$
My Attmept:
$$x+yp=ap^2$$
$$x=ap^2-yp$$
This is solvable for $x$ so differentiating both sides w.r.t $y$
$$\frac {dx}{dy} = 2ap\cdot \frac {dp}{dy} - y\cdot \frac {dp}{dy} -p$$
$$\frac {1}{p} + p = (2ap-y) \cdot \frac {dp}{dy}$$
$$\frac {1+p^2}{p} = (2ap-y) \cdot \frac {dp}{dy}$$
$$\frac {dy}{dp} = \frac {2ap^2}{1+p^2} - \frac {yp}{1+p^2}$$
$$\frac {dy}{dp} + \frac {p}{1+p^2} \cdot y = \frac {2ap^2}{1+p^2}$$
$$\textrm {This is Linear so}$$
$$\textrm {Integrating Factor}=e^{\int \frac {p}{1+p^2} dp}=\sqrt {1+p^2}$$
So we have
$$y\sqrt {1+p^2} = \int \frac {2ap^2}{1+p^2} \times \sqrt {1+p^2} dp + c$$
$$y\sqrt {1+p^2} = 2a \int \frac {p^2}{\sqrt {1+p^2}} dp + c $$
I could not solve further from here.
|
Yes go ahead $$y\sqrt{1+p^2}=ap\sqrt{1+p^2}-a\sinh^{-1}{p}+C \implies y(p)=ap+\frac{C-\sinh^{-1}p}{\sqrt{1+p^2}}.$$
put this in $$x(p)=ap^2-py(p)$$ to get $x(p)$.
Finally
$x(p)$ and $y(p)$ constitue the parametric solution of the ODE, where $p$ acts as a parameter only.$C$ is the integration-constant,
Note that $$\int \frac{p^2}{\sqrt{1+p^2}}dp= \int \left(\sqrt{1+p^2}-\frac{1}{\sqrt{1+p^2}}\right)dp=\frac{1}{2}\left( p\sqrt{1+p^2}-\sinh^{-1} p\right)$$
Both the integrals are standard ones.
|
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Ramanujan's rational elementary results on $A^3+B^3=C^2$. In the bottom of one of the pages of Ramanujan's notebooks (see below), he writes down the following equations: $$\left(11\frac 12\right)^3+\left(\frac 12\right)^3=39^2$$ $$\left(3\frac 17\right)^3-\left(\frac 17\right)^3=\left(5\frac 47\right)^2$$ $$\left(3-\frac 1{105}\right)^3+\left(\frac 1{105}\right)^3=\left(5\frac 6{35}\right)^2$$ $$\left(3\frac 1{104}\right)^3-\left(\frac 1{104}\right)^3=\left(5\frac{23}{104}\right)^2$$ It is clear he was studying the equation $A^3+B^3=C^2$ just like how Euler did, but he also took it one step further and extended the solutions to the set of rational numbers. Now Ramanujan was particularly fond of finding remarkable general identities, but often his most curious ones was he going to remember the most. So he never really wrote them down, and predominantly wrote special cases of them. From the elegant form of the equations above, I presume he had a general identity.
Empirically I have found a near solution (my best one).
$$\left(14+\frac 13\right)^3+\left(\frac 13\right)^3\simeq \left(54+\frac{442}{1665}\right)^2$$
And some solutions with irrationals. $$\left(6-\frac 4{3\pm \sqrt 5}\right)^3+\left(\frac 4{3\pm \sqrt 5}\right)^3=12^2$$
$$\left(1\pm\frac 2{\sqrt 5-1}\right)^3\mp\left(\frac 2{\sqrt 5-1}\right)^3=2^2$$
Anybody know how Ramanujan got these results? I just factored out $(a+b)^3-b^3$ since the $b^3$ terms cancelled out, leaving me with quadratics, which is fairly simple. I think if Ramanujan had a general identity for this, it would do the same. Note that $(-1\frac 12)^3+(\frac 12)^3=0^2$ which is another solution to his first equation, and both solutions can be written as $\frac{10\pm 12}{2}$ and $\frac{39\pm 39}{2}$ which looks quadratic.
Any thoughts?
Thanks.
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Ramanujan could not have had a general formula with the machinery of the time. Note that all of the listed results can be rewritten to have the same denominator $A$ in each number, so after multiplying by $A^3$ we have
$$x^3+y^3=Az^2$$
where $x,y,z$ are integers.
The given solutions correspond to the following integral relations:
$$23^3+1^3=2\cdot78^2$$
$$22^3-1^3=7\cdot39^2$$
$$314^3+1^3=105\cdot543^2$$
$$313^3-1^3=104\cdot543^2$$
Also note that the second term is always $\pm1$. So Ramanujan was merely looking for factorisations of incremented and decremented cubes with small squarefree parts.
From this framework other solutions arise, like
$$\left(\frac47\right)^3-\left(\frac17\right)^3=\left(\frac37\right)^2$$
$$\left(\frac{31}{38}\right)^3+\left(\frac1{38}\right)^3=\left(\frac{14}{19}\right)^2$$
|
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|
Transform Collatz sequence to a strictly decreasing sequence While playing with numbers, I found that every Collatz sequence $n, T(n), T^2(n), \ldots, 1$ can be associated with a strictly decreasing sequence of integers.
The Collatz conjecture asserts that a sequence defined by repeatedly applying the Collatz function
\begin{align*}
T(n) = \begin{cases}
(3n+1)/2 &\text{ if $n \equiv 1 \pmod{2}$, or}\\
n/2 &\text{ if $n \equiv 0 \pmod{2}$}
\end{cases}
\end{align*}
will always converge to the cycle passing through the number 1 for arbitrary positive integer $n$.
Note that multiplying the $n$ by positive odd integer $a$ does not affect the result of the modulo 2 operation.
By multiplying the Collatz function by an odd integer $a$, and tracking the $m = an$ rather than $n$, we get
\begin{align*}
S(m) = \begin{cases}
(3m+a)/2 &\text{ if $m \equiv 1 \pmod{2}$, or}\\
m/2 &\text{ if $m \equiv 0 \pmod{2}$,}
\end{cases}
\end{align*}
where each iterate $S^i(m) = a \, T^i(n)$.
Now we can choose a sufficiently large positive integer $A$ and track $m = 3^A n$.
But we do a little trick.
Instead of multiplying the $m$ by 3 in the "odd" branch, we just replace $3^A$ with $3^{A-1}$, and track the $3^{A-1}$ from that moment on (the effect is the same).
We get the following algorithm:
It can be shown that every next $m$ is strictly less than the previous $m$.
Since every next $m$ is smaller than its predecessor, we must hit $m = 1$ at the end.
Since we track $m = 3^A n$, once the $m = 1$, then the $A = 0$ and $n = 1$.
This implies that for arbitrary positive integer $n$, the sequence $n, T(n), T^2(n), \ldots$ leads to one.
Note that once the $m = 3^A$, then the $n = 1$.
I am however stuck to show that there is always the sufficiently large $A$ for a given $n$.
Is it possible to show this?
I found out that the sufficiently large $A$ does not always exist for the $3x-1$ problem.
Example
The trajectory starting at $n=19$ with $A=9$ (termination at $m = 1$):
$$\begin{matrix}
n & m & A \\
\hline
19 & 373977 & 9 \\
29 & 190269 & 8 \\
44 & 96228 & 7 \\
22 & 48114 & 7 \\
11 & 24057 & 7 \\
17 & 12393 & 6 \\
26 & 6318 & 5 \\
13 & 3159 & 5 \\
20 & 1620 & 4 \\
10 & 810 & 4 \\
5 & 405 & 4 \\
8 & 216 & 3 \\
4 & 108 & 3 \\
2 & 54 & 3 \\
1 & 27 & 3 \\
2 & 18 & 2 \\
1 & 9 & 2 \\
2 & 6 & 1 \\
1 & 3 & 1 \\
2 & 2 & 0 \\
1 & 1 & 0 \\
\end{matrix}$$
|
It suffers from the same pitfall as other representations which relies on the fact that the sequence reaches 1.
e.g. in the Collatz tree, you pick a number, and it does not matter if it seems to rise, in the tree it is a step closer to the root.
another one is the "inverse Collatz" representation of a number:
$7 = \frac{2^5}{3^5}\cdot 2^{(3+2+1+0+0)} - \frac{2^4}{3^5}\cdot 2^{(2+1+0+0)} - \frac{2^3}{3^4}\cdot 2^{(1+0+0)} - \frac{2^2}{3^3}\cdot 2^{(0+0)} - \frac{2^1}{3^2}\cdot 2^{(0)} - \frac{2^0}{3^1}\\
11 = \frac{2^4}{3^4}\cdot 2^{(3+2+1+0)} - \frac{2^3}{3^4}\cdot 2^{(2+1+0)} - \frac{2^2}{3^3}\cdot 2^{(1+0)} - \frac{2^1}{3^2}\cdot 2^{(0)} - \frac{2^0}{3^1}\\
17 = \frac{2^3}{3^3}\cdot 2^{(3+2+1)} - \frac{2^2}{3^3}\cdot 2^{(2+1)} - \frac{2^1}{3^2}\cdot 2^{(1)} - \frac{2^0}{3^1}\\
13 = \frac{2^2}{3^2}\cdot 2^{(3+2)} - \frac{2^1}{3^2}\cdot 2^{(2)} - \frac{2^0}{3^1}\\
5 = \frac{2^1}{3^1}\cdot 2^{(3)} - \frac{2^0}{3^1}\\
1 = \frac{2^0}{3^0}$
it does not matter if 7 rises to 11. In it's representation, at each step, the exponent decreases, as well as the length of the representation.
|
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|
If $f : \mathbb R \rightarrow \mathbb R $ such that $f(x^2+x)+2f(x^2-3x+2) = 9x^2-15x$. Find $f(2016)$.
Determine all $f : \mathbb R \rightarrow \mathbb R $ such that $$f(x^2+x)+2f(x^2-3x+2) = 9x^2-15x$$ for all $x$. Find $f(2016)$.
Similar problem appeared on this site before: $f(x^2 + x)+2f(x^2 - 3x + 2)=9x^2 - 15x$ then find $f(2016)$. (The question is now deleted.) The same problem with finding $2011$ (instead of $2016$) appeared in 2011 Singapore Mathematical Olympiad as problem 17 (Wayback Machine).
I’ve tried put $x=0,1$ and got \begin{align*}
f(0)+2f(2)&=0\\
f(2)+2f(0)&=-6
\end{align*}
which gives me $f(0)=-4$, $f(2)=2$.
Similarly, if we notice that $x^2+x=x^2-3x+2$ holds for $x=\frac12$, we can find the value at the point $\frac34=\left(\frac12\right)^2+\frac12$.
But the above doesn’t seem to help for other values.
Thank you very much for helping.
|
Denote: $x^2+x=a$. Then:
$$f(a)+2f(a-4x+2)=9a-24x.$$
Plug $x=\frac12$ to get:
$$f(a)+2f(a)=9a-12 \Rightarrow f(a)=3a-4.$$
Hence:
$$f(2016)=3\cdot 2016-4=6044.$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Evaluate $\frac{q}{1+q^2}+\frac{q^2}{1+q^4}+\frac{q^3}{1+q^6}$, where $q^7=1$ and $q\neq 1$.
Let $q$ be a complex number such that $q^7=1$ and $q\neq 1$. Evaluate $$\frac{q}{1+q^2}+\frac{q^2}{1+q^4}+\frac{q^3}{1+q^6}.$$
The given answer is $\frac{3}{2}$ or $-2$. But my answer is $\pm 2$.
At first, I tried to evaluate it directly. And the LHS equals to
\begin{align}
\frac{q}{1+q^2}+\frac{q^2}{1+q^4}+\frac{q^3}{1+q^6}
& = \frac{q}{1+q^2}+\frac{q^2}{1+q^4}\cdot\frac{q^3}{q^3}+\frac{q^3}{1+q^6}\cdot\frac{q}{q} \\
& = \frac{q}{1+q^2}+\frac{q^5}{1+q^3}+\frac{q^4}{1+q} \\
& = q\cdot\frac{(1+q)(1+q^3)+q^4(1+q)(1+q^2)+q^3(1+q^2)(1+q^3)}{(1+q)(1+q^2)(1+q^3)} \\
& = q\cdot\frac{1+q+q^3+q^4+q^4+q^5+q^6+1+q^3+q^5+q^6+q}{(1+q)(1+q^2)(1+q^3)} \\
& = \frac{-2q^3}{(1+q)(1+q^2)(1+q^3)} \\
\end{align}
And
$$(x-q)(x-q^2)(x-q^3)(x-q^4)(x-q^5)(x-q^6)=x^6+x^5+x^4+x^3+x^2+x+1$$
Let $x=-1$ I get that
$$(1+q)(1+q^2)(1+q^3)(1+q^4)(1+q^5)(1+q^6)=1$$
and
$$(1+q)(1+q^2)(1+q^3)\cdot q^4(q^3+1)\cdot q^5(q^2+1)\cdot q^6(q+1)=1$$
therefore
$$\left[(1+q)(1+q^2)(1+q^3)\right]^2=\frac{1}{q^{15}}=\frac{1}{q}$$
hence
$$\left[\frac{-2q^3}{(1+q)(1+q^2)(1+q^3)}\right]^2=\frac{q}{1}\cdot 4q^6=4$$
$$\frac{-2q^3}{(1+q)(1+q^2)(1+q^3)}=\pm 2$$
And I try for a solution as a polar-form method$.\\$Suppose $q=\cos\frac{2j\pi}{7}+i\sin\frac{2j\pi}{7}$
$$\frac{q^k}{1+q^{2k}}=\frac{\cos\frac{2jk\pi}{7}+i\sin\frac{2jk\pi}{7}}{2\cos\frac{2jk\pi}{7}\left(\cos\frac{2jk\pi}{7}+i\sin\frac{2jk\pi}{7}\right)}=\frac{1}{2\cos\frac{2jk\pi}{7}}$$
Am I going to the right direction? How I finish it? And please help to figure out what's wrong with my calculation at the first part. I appreciate for your help.
|
Let $7x=2k\pi$ where $k=\pm1,\pm2,\pm3$
like Prove that $\cos(\pi/7)$ is root of equation $8x^3-4x^2-4x+1=0$
$q_k=2\cos\dfrac{2k\pi}7; k=1,2,3$ are the roots of $$c^3+c^2-2c-1=0$$
Use Veita's formula $$\sum_{k=1}^3\dfrac1{q_k}=\dfrac{q_1q_2+q_2q_3+q_3q_1}{q_1q_2q_3}=\dfrac{-\dfrac21}{\dfrac11}$$
|
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|
A parabola touches the bisectors of the angles formed by lines $x+2y+3=0$ and $2x+y+3=0$ at $(1,1)$ and $(0,-2)$. Find its focus and directrix.
A parabola touches the bisectors of the angles obtained by the lines $x+2y+3=0$ and $2x+y+3=0$ at the points $(1,1)$ and $(0,-2)$. Then find its focus and the equation of the directrix.
My approach is as follows:
The equation of bisector is
$$\frac{x+2y+3}{\sqrt{5}}= \pm \frac{2x+y+3}{\sqrt{5}}$$
We get the required bisectors as $x-y=0$ and $3x+3y+6=0$, or $x+y+2=0$.
$x-y=0$ is tangent to the parabola at $(1,1)$, whereas $x+y+2=0$ is tangent to the parabola at $(0,-2)$.
From here, how do I proceed?
|
Since angular bisectors are perpendicular to each other and are tangent to the parabola at $A(1,1)$ and $B(0,2)$, the point of intersection of the two given lines $(-1,-1)$ is a point on the directrix of the parabola.
The median through $P$ to the Archimedes triangle $PAB$ is known to be parallel to axis, which gives the slope of directrix as $-3$. So foot of perpendicular from $A$ on to directrix is $\left(-\dfrac{7}{5},\dfrac{1}{5}\right)$. This is also the image of the focus in the tangent $PA \equiv x-y=0$ which is easily found as $\boxed{\left(\dfrac{1}{5},-\dfrac{7}{5}\right)}$
|
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|
If $a+b+c=3$ Prove that $a^{2}+b^{2}+c^{2}\geq\frac{2+a}{2+b}+\frac{2+b}{2+c}+\frac{2+c}{2+a}$ Question -
Let $a, b, c$ be positive real numbers such that $a+b+c=3 .$ Prove that
$$
a^{2}+b^{2}+c^{2} \geq \frac{2+a}{2+b}+\frac{2+b}{2+c}+\frac{2+c}{2+a}
$$
My try -
i tried putting $a+2 = x, b+2=y , c+2=z$
then we get $x+y+z=9$ and after simplification we have to prove that
$3>x/y + y/z + z/x$ which i am not able to prove...
i also tried C-S,Chebyshev,rearrangement..etc but none of them working
any hints ???
thankyou
|
Here's another way. Notice that
$$\frac{2+a}{2+b} = \frac{5a+2b+2c}{2a+5b+2c} = \frac52 -\frac32\cdot\frac{7b+2c}{2a+5b+2c} $$
Also using CS inequality ($\sum $ representing cyclic sums):
$$\sum \frac{7b+2c}{2a+5b+2c} \geqslant \frac{\left( \sum (7b+2c)\right)^2}{\sum (7b+2c)(2a+5b+2c)}= \frac{81(a+b+c)^2}{18\sum a^2+21(a+b+c)^2} \\= \frac{729}{18\sum a^2+189}$$
By the above bound, with $x = \sum a^2 \geqslant 3$, it is enough to show
$$x \geqslant \frac{15}2- \frac32\cdot\frac{729}{18x + 189} \iff \frac{2(x+6)(x-3)}{2x+21}\geqslant 0$$
|
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|
Find limit (Riemann sum) $\displaystyle a_n=\sum^{n^2}_{k=1}\sqrt{\frac{k^2-k}{n^8+k^2n^4-kn^4}}$.
I tried solving it by changing it a Riemann sum then integrating, however I couldn't manipulate the algebra to its form.
|
We have:
\begin{align*}
\sum^{n^2}_{k=1}\sqrt{\frac{k^2-k}{n^8+k^2n^4-kn^4}} &= \frac{1}{n^2} \sum^{n^2}_{k=1}\sqrt{\frac{k^2-k}{n^4+k^2-k}} \\ &= \frac{1}{n^2} \sum^{n^2}_{k=1}\sqrt{1 - \frac{n^4}{n^4+k(k-1)}} \\ &= \frac{1}{n^2} \sum^{n^2}_{k=1}\sqrt{1 - \frac{1}{1+\frac{k}{n^2}\frac{k-1}{n^2}}}\end{align*}
|
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|
Find any two integers that satisfy $19x+47y = 1$ How can I find two integers that satisfy $19x+47y = 1$? Is there some technique to finding 2 numbers? I can't find any 2 numbers in $\mathbb{Z}$ that make this work.
Thank you!
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Consider the continued fraction of $\frac{47}{19}$:
$$ \frac{47}{19}=2+\frac{1}{\frac{19}{9}}=2+\frac{1}{2+\frac{1}{9}}=[2;2,9] $$
truncate it and expand it back:
$$ [2;2] = 2+\frac{1}{2} = \frac{5}{2}. $$
We have that $\frac{5}{2}$ and $\frac{47}{19}$ are consecutive convergents of the same continued fraction, hence their difference is $\pm\frac{1}{2\cdot 19}$.
Indeed
$$ \frac{47}{19}-\frac{5}{2}=\frac{94-95}{38}=\frac{-1}{38} $$
leads to
$$ 47\cdot 2 - 5\cdot 19 = -1 $$
$$ 47\cdot(-2) + 19\cdot 5 = 1 $$
$$ 47(19-2)+19(5-47) = 1 $$
$$ \color{red}{47\cdot 17 - 19\cdot 42 = 1.}$$
|
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|
There are two positive integers of the form $p-n^2$ such that one divides the other.
Let $p>3$ be a prime number. Consider the numbers of the form $p - n^2$ where $n = 1 , 2 , 3 , ... , \lfloor\sqrt{p}\rfloor$. Show that there exist two such numbers, $a$ and $b$, that $a | b$.
I tried show that $p - n_{max}^2 | p - 1$. But I’ve found that $p = 107$ is a counter-example.
Edit : $p - n_1^2$ is a and $p - n_2^2$ is b for some $n_1,n_2$
Please give me hints or answers. Thank you!
|
Let $m:=\lfloor \sqrt{p}\rfloor$. If $p=m^2+1$, then obviously, we have $a\mid b$ if $a:=p-m^2$ and $b:=p-(m-1)^2$. We now suppose that $p-m^2>1$.
We shall prove that, if $a:=p-m^2$, then there exists an integer $k$, where $1\le k<m$, such that $a\mid p-k^2$. First, we note that $p< (m+1)^2-1$; therefore, $p\leq (m+1)^2-2=m^2+2m-1$, so
$$a=p-m^2\leq 2m-1\,.\tag{*}$$
Observe that $a\neq m$; otherwise, $p=m^2+a=m^2+m=m(m+1)$, whence $p=2$, but we assume that $p\geq 3$.
Let $X$ be the set of integers $x$ such that $x\equiv -m\pmod{a}$ and $x<m$. Clearly, $X$ is nonempty. Take $k$ to be the largest element of $X$. Then, $k<m$ is given.
Now, if $k\leq 0$, then $k+a\equiv k\equiv -m\pmod{a}$ and $k+a>k$. By maximality of $k$, we must have $k+a\geq m$. Note that
$$a\geq m-k\geq m\,.\tag{#}$$
We know that $a\neq m$. That is, $a>m$. Because $k\equiv -m\pmod{a}$ and $k\leq 0$, we must have $k\leq -m$. Thus, by (#), we get
$$a\geq m-k \geq 2m\,.$$
This contradicts (*). Hence,
$$0<k<m\,.$$
Now, $$p-k^2\equiv p-(-m)^2= p-m^2=a\equiv0\pmod{a}\,,$$
whence $a\mid b$ if $b:=p-k^2$.
For example, when $p=107$, we see that $m=10$ and $a=7$. Thus, if $x$ is an integer such that $x\equiv -m\pmod{a}$, then $x\equiv -10\pmod{7}$, or $x\equiv 4\pmod{7}$. Ergo, $k=4$ works. We can then take $b:=p-k^2=107-16=91=7\cdot 13$, which is divisible by $a=7$.
Remark. This solution does not really use primality of $p$. We can take $p$ to be any positive integer such that $p$ is not of the form $q^2$, $q(q+1)$, and $q(q+2)$ for any integer $q$. For example, if $p=1243=11\cdot 113$, then you can take $m=35$ and $a=18$. Thus, if $x$ is an integer such that $x\equiv -m\pmod{a}$, then $x\equiv -35\pmod{18}$, or $x\equiv 1\pmod{18}$. Ergo, $k=19$ works. We can then take $b:=p-k^2=1243-19^2=882=18\cdot 49$, which is divisible by $a=18$.
|
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|
Simplifying $\sqrt{(\cos(\arctan x)+x\sin(\arctan x))^2-1}$
The expression $\sqrt{(\cos(\arctan x)+x\sin(\arctan x))^2-1}$, where $0<x<1$, is equal to $x$ or $\sqrt{(1+x^2)}$ or $\frac1{\sqrt{(1+x^2)}}$ or $\frac x{\sqrt{(1+x^2)}}$? (one of these 4 is correct).
My attempt: $$0<x<1\implies 0<\arctan x<\frac{\pi}{4}\implies\frac{1}{\sqrt2}<\cos(\arctan x)<1$$ and $$0<x\sin(\arctan x)<\frac{x}{\sqrt2}$$$$\implies\frac{1}{\sqrt2}<\cos(\arctan x)+x\sin(\arctan x)<1+\frac{x}{\sqrt2}$$$$\implies\frac{1}{2}<(\cos(\arctan x)+x\sin(\arctan x))^2<(1+\frac{x}{\sqrt2})^2$$$$\implies\sqrt\frac{-1}{2}<\sqrt{(\cos(\arctan x)+x\sin(\arctan x))^2-1}<\sqrt{(1+\frac{x}{\sqrt2})^2-1}$$$$\implies\sqrt\frac{-1}{2}<\sqrt{(\cos(\arctan x)+x\sin(\arctan x))^2-1}<\sqrt{\frac{x^2}{2}-\sqrt2x}$$ Not sure if I have reached anywhere.
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Since $x>0$ and $$1+\tan^2\alpha=\frac{1}{\cos^2\alpha},$$ we obtain: $$\sqrt{(\cos(\arctan x)+x\sin(\arctan x))^2-1}=\sqrt{\left(\frac{1}{\sqrt{1+x^2}}+x\cdot\frac{x}{\sqrt{1+x^2}}\right)^2-1}=$$
$$=\sqrt{\left(\sqrt{1+x^2}\right)^2-1}=\sqrt{x^2}=x.$$
|
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|
If $z_1,\dotsc,z_n$ are the vertices of a regular $n$-gon, with $z_3 + z_n = Az_1 + \bar{A}z_2$, find $\lfloor |A| \rfloor$. The question goes:
The complex numbers $z_1, z_2, z_3 \ldots z_n, \sum_{i=1}^{n} z_i \neq 0$ represent the vertices of a regular polygon of $n$ sides in order, inscribed in a circle of unit radius such that $z_3 + z_n =A\cdot z_1 + \bar A\cdot z_2$ Find $\lfloor |A|\rfloor$ when $n=4,6,8,12$.
In my attempt I tried to generalize over $n$. My attempt:
Suppose $z_1=e^{i\phi}$. Now, $$z_j=e^{i((j-1)\frac{2\pi}{n} + \phi)}$$
On substitution, I obtained
$$e^{i\frac{4\pi}{n}}+e^{i(n-1)\frac{2\pi}{n}}=A+\bar A\cdot e^{i\frac{2\pi}{n}}=A+i\bar A $$
Then I put $A=x +iy$ and $n=4$ to end up with $-1-i=(x+y)+i(x+y)$ giving me only $x+y=-1$.
If I'm going right, how should I proceed? Thanks in advance.
Edit:
The answers are $2,2,1,1$ respectively.
Edit 2: The original question.
Answer is A-R,B-R,C-Q,D-Q.
Edit 3: The solution as given by the problem setter.
|
Let $n \geq 4$. We are given that $z_1, \dotsc, z_n$ are the vertices of a regular $n$-gon inscribed in a circle in the complex plane, with $\sum_i z_i \neq 0$. So, letting $w := x_0 + iy_0$ denote the centre of the circle, we have $w \neq 0$.
Let $\theta \in [0,2\pi)$ such that $z_1 = w + e^{i \theta}$. Then, $$z_j = w + e^{i(\theta + 2\pi(j-1)/n)} = \left[x_0 + \cos\left(\theta + \frac{2\pi(j-1)}{n}\right)\right] + i\left[y_0 + \sin\left(\theta + \frac{2\pi(j-1)}{n}\right)\right]$$ for all $j = 1, \dotsc, n$.
Hence,
$$
z_3 + z_n = \left[2x_0 + \cos\left(\theta + \frac{4\pi}{n}\right) + \cos\left(\theta + \frac{2(n-1)\pi}{n}\right)\right] + i\left[2y_0 + \sin\left(\theta + \frac{4\pi}{n}\right)+\sin\left(\theta + \frac{2(n-1)\pi}{n}\right)\right].
$$
Let $A = p + iq$. Then,
$$
Az_1 + \bar{A}z_2 = \left\{\left[ 2x_0 + \cos( \theta ) + \cos\left( \theta + \frac{2\pi}{n} \right)\right]p + \left[-\sin( \theta ) + \sin\left( \theta + \frac{2\pi}{n} \right) \right]q \right\} + i \left\{\left[ 2y_0 + \sin( \theta ) + \sin\left( \theta + \frac{2\pi}{n} \right)\right]p + \left[\cos( \theta ) - \cos\left( \theta + \frac{2\pi}{n} \right) \right]q\right\}.
$$
We are given that $z_3 + z_n = Az_1 + \bar{A}z_2$. So, equating the real and imaginary parts of the two sides, we get a system of two linear equations in two unknowns:
$$
\begin{bmatrix}
2x_0 + \cos( \theta ) + \cos\left( \theta + \frac{2\pi}{n} \right) & -\sin( \theta ) + \sin\left( \theta + \frac{2\pi}{n} \right)\\
2y_0 + \sin( \theta ) + \sin\left( \theta + \frac{2\pi}{n} \right) & \cos( \theta ) - \cos\left( \theta + \frac{2\pi}{n} \right)
\end{bmatrix}
\begin{bmatrix}
p \\
q
\end{bmatrix}
=
\begin{bmatrix}
2x_0 + \cos\left(\theta + \frac{4\pi}{n}\right) + \cos\left(\theta + \frac{2(n-1)\pi}{n}\right) \\
2y_0 + \sin\left(\theta + \frac{4\pi}{n}\right)+\sin\left(\theta + \frac{2(n-1)\pi}{n}\right)
\end{bmatrix}
$$
Now, the determinant of the matrix of coefficients is
$$
\Delta = 4 \sin \left( \frac{\pi}{n} \right) \left[ -y_0 \cos\left( \theta + \frac{\pi}{n} \right) + x_0 \cos\left( \theta + \frac{\pi}{n} \right) \right],
$$
so $\Delta = 0 \iff w$ lies on the line with slope $\tan\left(\theta + \frac{\pi}{n} \right)$. Note: here we mean that $w$ lies on the imaginary axis if $\theta + \pi/n = \pi/2$ or $3\pi/2$.
Suppose that $\Delta \neq 0$. Then, left multiplying by the inverse of the matrix of coefficients, we get
$$
\begin{bmatrix}
p\\
q
\end{bmatrix}
=
\frac{1}{\Delta}
\begin{bmatrix}
2x_0\left[ \cos( \theta ) - \cos\left( \theta + \frac{2\pi}{n} \right) \right] + 2y_0\left[ \sin(\theta) - \sin\left( \theta + \frac{2\pi}{n} \right) \right] \\
\\
2x_0\left[ - \sin(\theta) - \sin\left( \theta + \frac{2\pi}{n} \right) + \sin\left( \theta + \frac{4\pi}{n} \right) + \sin\left( \theta + \frac{2(n-1)\pi}{n} \right)\right] + 2y_0\left[ \cos(\theta) + \cos\left( \theta + \frac{2\pi}{n} \right) -\cos\left( \theta + \frac{4\pi}{n} \right) - \cos\left( \theta + \frac{2(n-1)\pi}{n} \right)\right]
\end{bmatrix}
=
\frac{1}{\Delta}
\begin{bmatrix}
\Delta\\
-2\Delta\sin\left( \frac{2\pi}{n} \right)
\end{bmatrix}.
$$
Hence, $$A = 1 - 2i\sin\left( \frac{2\pi}{n} \right).$$
Since $\sin(\pi - x) = \sin(x)$ for all $x$, and $\pi - 2\pi/n$ is the interior angle of the polygon, our solution for $A$ matches the one in the "official" solution.
Now, the given solution is damn clever and makes my calculations look caveman-like. But, just so I can salvage some of my previous attempt, I will note that the "official" solution actually does not make use of the fact that $\sum_i z_i \neq 0$ anywhere! This suggests that they are implicitly dealing with the non-degenerate case only.
Since we've come so far, let's complete the degenerate case as well. :)
Firstly, our analysis tells us that the correct criterion for degeneracy is $\Delta = 0$. This is the condition that $w$ lies on the line with slope $\tan\left( \theta + \frac{\pi}{n} \right)$. This is stronger than the given condition, which just says that $w$ must not be the origin.
So, let's assume that $\Delta = 0$. In this case, the system of linear equations looks like:
$$
\begin{bmatrix}
\pm 2\lvert w \rvert \cos\left( \theta + \frac{\pi}{n} \right) + \cos( \theta ) + \cos\left( \theta + \frac{2\pi}{n} \right) & -\sin( \theta ) + \sin\left( \theta + \frac{2\pi}{n} \right)\\
\pm 2\lvert w \rvert \sin\left( \theta + \frac{\pi}{n} \right) + \sin( \theta ) + \sin\left( \theta + \frac{2\pi}{n} \right) & \cos( \theta ) - \cos\left( \theta + \frac{2\pi}{n} \right)
\end{bmatrix}
\begin{bmatrix}
p \\
q
\end{bmatrix}
=
\begin{bmatrix}
\pm 2\lvert w \rvert \cos\left( \theta + \frac{\pi}{n} \right) + \cos\left(\theta + \frac{4\pi}{n}\right) + \cos\left(\theta + \frac{2(n-1)\pi}{n}\right) \\
\pm 2\lvert w \rvert \sin\left( \theta + \frac{\pi}{n} \right) + \sin\left(\theta + \frac{4\pi}{n}\right)+\sin\left(\theta + \frac{2(n-1)\pi}{n}\right)
\end{bmatrix}
$$
which can be simplified to
$$
\begin{bmatrix}
2 \cos\left( \theta + \frac{\pi}{n} \right) \left[ \pm \lvert w \rvert + \cos\left( \frac{3\pi}{n} \right)\right] & 2 \sin\left( \frac{\pi}{n} \right) \cos\left( \theta + \frac{\pi}{n} \right) \\
2 \sin\left( \theta + \frac{\pi}{n} \right) \left[ \pm \lvert w \rvert + \cos\left( \frac{3\pi}{n} \right)\right] & 2 \sin\left( \frac{\pi}{n} \right) \sin\left( \theta + \frac{\pi}{n} \right)
\end{bmatrix}
\begin{bmatrix}
p \\
q
\end{bmatrix}
=
\begin{bmatrix}
2 \cos\left( \theta + \frac{\pi}{n} \right) \left[ \pm \lvert w \rvert + \cos\left(\frac{3\pi}{n}\right)\right] \\
2 \sin\left( \theta + \frac{\pi}{n} \right) \left[ \pm \lvert w \rvert + \cos\left(\frac{3\pi}{n}\right)\right] \\
\end{bmatrix}.
$$
If $\theta = \pi - \pi/n$ or $2\pi - \pi/n$, then the second equation is identically zero; if $\theta = \pi/2 - \pi/n$ or $3\pi/2 - \pi/n$, then the first equation is identically zero; otherwise, we can subtract $\tan\left( \theta + \frac{\pi}{n} \right)$ times the first row from the second row. In each case, we see that the system is consistent and there is one nontrivial equation.
So, every complex number on the line determined by that equation is a valid choice for $A$, and $\lfloor \lvert A \rvert \rfloor$ is not uniquely determined!
Lastly, one can check that the point $1 - 2i\sin\left( \frac{2\pi}{n} \right)$ lies on this degenerate line so there is no contradiction in the given solution. It is only that the problem setter missed some additional solutions in this case.
I made a visualization of the solutions in GeoGebra:
*
*Non-degenerate case
*Degenerate case
I give some screenshots below:
|
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|
How to find the sum of this complex series? Question says:
The convergent infinite series C and S are defined as
$$
C=1+\frac{1}{4}\cos(2\theta)+\frac{1}{16}\cos(4\theta)+\frac{1}{64}\cos(6\theta)+...
$$
$$
S=\frac{1}{4}\sin(2\theta)+\frac{1}{16}\sin(4\theta)+\frac{1}{64}\sin(6\theta)+...
$$
Show that $C+iS=\frac{k}{k-e^{2i\theta}}$ where $k$ is an integer to be determined.
(Side note: Sorry if there is a formatting problem; first time using LaTeX)
|
Add the corresponding terms of C and iS
You get
$$1+\frac{1}{4}\left(\cos 2\theta +i\sin 2\theta \right)+\frac{1}{16}\left(\cos 4\theta +i\sin \theta \right)...$$
use the identity $\begin{array}{l}e^{i\theta }=\cos \theta +i\sin \theta \end{array}$
$$C+iS=\left(\frac{e^{i2\theta}}{4}\right)^0+\left(\frac{e^{i2\theta}}{4}\right)^1+\left(\frac{e^{i2\theta}}{4}\right)^2...$$
Now use sum of terms in a GP formula to get
$$C+iS=\frac{4}{4-e^{2i\theta }}$$
Therefore k=4
|
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"timestamp": "2023-03-29T00:00:00",
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|
Investigate the convergence of the integral Investigate the convergence of the integral
$$\int_{0}^{1} \left\vert \sin\frac{1}{x} \right\vert\ \frac{1}{x^{\alpha}} dx \ \text{where $\alpha \geq 1$ }$$
I assume that it diverges, so I rated function from below
$|sin\frac{1}{x}| \ \frac{1}{x^{\alpha }} \geq \sin^2(\frac{1}{x}) \ \frac{1}{x^\alpha} = \frac {(1-cos(\frac{2}{x}))}{2x^\alpha } = \frac{1}{2x^\alpha} - \frac {cos(\frac{2}{x})}{x^\alpha}$
The $\int_{0}^{1} \frac{1}{2x^{\alpha}} dx$ diverges
How to show that the integral $\int_{0}^{1}\frac {cos(\frac{2}{x})}{x^\alpha}$ converges?
|
Using integration by substitution, we have to study the convergence of the integral
$$I_\alpha=\int_1^\infty x^{\alpha-2} \left\vert \sin x \right\vert \ dx.$$
For $k \in \mathbb N$, and $x \in [k \pi + \frac{\pi}{6}, k\pi + \frac{5\pi}{6}]$, you have
$$\left\vert \sin x \right\vert \ge \frac{1}{2}$$ hence
$$g(x) = x^{\alpha-2}\left\vert \sin x \right\vert \ge \frac{x^{\alpha-2}}{2}.$$
Based on that, you get
$$I_\alpha \ge \frac{\pi}{3} \sum_{k=1}^\infty \frac{1}{\left(k\pi + \frac{5\pi}{6}\right)^{2-\alpha}}.$$
As the right end series is diverging for $2- \alpha \le 1$, i.e. $\alpha \ge 1$, we finally get that the integral is also diverging for $\alpha \ge 1$.
|
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|
When $ab/(a+b)$ is an integer, where $a,b$ are positive integers. When $ab/(a+b)$ is an integer, where $a,b$ are positive integers?
clear;
maxn:=30;
for a in [1..maxn] do
for b in [a..maxn] do
q1:=a*b; q2:=a+b;
if q1 mod q2 eq 0 then
print a,b,q1 div q2;
end if;
end for;
end for;
the Magma code given above outputs the following.
2 2 1
3 6 2
4 4 2
4 12 3
5 20 4
6 6 3
6 12 4
6 30 5
8 8 4
8 24 6
9 18 6
10 10 5
10 15 6
12 12 6
12 24 8
14 14 7
15 30 10
16 16 8
18 18 9
20 20 10
20 30 12
21 28 12
22 22 11
24 24 12
26 26 13
28 28 14
30 30 15
So I conjectrue that $\frac{ab}{a+b}$ is an integer if and only if $\frac{a}{d}+\frac{b}{d} \mid d$, where $d=\gcd(a,b)$. It is true if $a=b$.
If $\frac{a}{d}+\frac{b}{d} \mid d$, then
$$\frac{ab}{a+b}=\frac{a}{d} \cdot \frac{b}{d}\cdot \frac{d}{\frac{a}{d}+\frac{b}{d}}$$
is a product of three positive integers and hence is an integer.
Conversely, i.e., $\frac{ab}{a+b}$ is an integer. Let $b=\frac{p}{q}\cdot a$ with $\gcd(p,q)=1$ where $p=\frac{b}{d}$, $q=\frac{a}{d}$ and $d=\gcd(a,b)$. Then
$$\frac{ab}{a+b}=\frac{a\cdot \frac{p}{q}\cdot a }{(1+\frac{p}{q})a}=\frac{ap}{p+q}$$
Since $\gcd(p,p+q)=\gcd(p,q)=1$, it has $p+q \mid a$. Similarly, it has $p+q \mid b$. As a result, $p+q \mid \gcd(a,b)$, i.e. $\frac{a}{d}+\frac{b}{d} \mid d$.
It completes the proof.
|
We can do way better.
Thm. We have $a+b\mid ab$ if and only if $a=\dfrac{u+v+w}{2}$ and $b=\dfrac{-u+v+w}{2}$, where $u,v,w\in\mathbb{Z}$ such that $v$ is even and $u^2+v^2=w^2.$
Since we know how to describe completely Pythagorean triples, this gives a full description of the solutions.
Proof. If $a+b\mid ab$, there exists $m\in\mathbb{Z}$ are integers such that $ab=ma+mb$.
Hence $(a-m)(b-m)=m^2$, that is $(a+b-2m)^2-(a-b)^2=4m^2$. Now set $u=a-b,v=2m,w=a+b-2m$.
Notice that $a=\dfrac{u+v+w}{2}$ and $b=\dfrac{-u+v+w}{2}$
Conversely, if $u,v,w$ satisfies the conditions of the theorem, then $u$ and $w$ necessarily have smae parity, and $a=\dfrac{u+v+w}{2}$ and $b=\dfrac{-u+v+w}{2}$ are then integers (since $v$ is even). Moreover, $a+b=v+w$ and $ab=\dfrac{(v+w)^2-u^2}{4}$.
Now $(v+w)^2-u^2=v^2+w^2+2vw-u^2=2v^2+2vw=2v(v+w)$. Hence $ab=\dfrac{v}{2}(v+w)=\dfrac{v}{2}(a+b)$. Thus, $a+b\mid ab$ since $v$ is even.
|
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|
Sum of the squares of twelve real numbers which add to $1.$ Let $a, b, c,\cdots, l$ be real numbers such that $a + b + c \dots + l = 1.$ Find the minimum value of $$a^2 + b^2 + c^2 + \dots + l^2.$$
My first glance at this problem, I would assume the minimum would be $1,$ gotten by a bunch of zeros, and then a $1.$ I'm not sure if this is correct, and is there a rigorous proof directed toward this?
|
Minimum occurs when $a_1 = a_2 = a_3 = \ldots = a_{12} = \frac{1}{12}$, and the minimum such value is $\frac{1}{12}$. To prove this, consider shifting the variables by setting $b_i = a_i - \frac{1}{12}$. We are then trying to minimise
$$\left(b_1 + \frac{1}{12}\right)^2 + \left(b_2 + \frac{1}{12}\right)^2 + \ldots + \left(b_{12} + \frac{1}{12}\right)^2$$
subject to $b_1 + b_2 + \ldots + b_{12} = 0$. We then have,
\begin{align*}
&\left(b_1 + \frac{1}{12}\right)^2 + \left(b_2 + \frac{1}{12}\right)^2 + \ldots + \left(b_{12} + \frac{1}{12}\right)^2 \\
= \; &b_1^2 + b_2^2 + \ldots + b_{12}^2 + \frac{1}{6}(b_1 + b_2 + \ldots + b_{12}) + \frac{1}{12} \\
= \; &b_1^2 + b_2^2 + \ldots + b_{12}^2 + \frac{1}{12}.
\end{align*}
Clearly the above function has a minimum of $\frac{1}{12}$, and is achieved if and only if $b_1 = b_2 = \ldots = b_{12} = 0$ (which satisfies the constraint), i.e. when $a_1 = a_2 = \ldots = a_{12} = \frac{1}{12}$.
|
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|
How can we prove that $\gcd((n^4) + (n+1)^4 , (n+1)^4 + (n+2)^4) = 1$? I have found through experimentation that the consecutive sums of consecutive natural numbers raised to certain powers$(1,2,4)$ are always coprime. I was looking for modular multiplicative inverses and came across this facet.
If you define a sequence $a(n): n^k + (n+1)^k$ , then $a(n)$ and $a(n+1)$ are coprime when $k=1,2$ or $4$ but no others. How can we prove this?
Since $a(n)$ and $a(n+1)$ are coprime when $k=1,2$ or $4$, their modular multiplicative inverse exists, and are sequences in the OEIS already.
I sincerely hope that this is moderately worthy and not cringe.
|
Suppose $d$ is a common factor of $n^4+(n+1)^4$ and $(n+1)^4+(n+2)^4$. Then
$$d\bigg|(n+2)^4-n^4\\
\implies d\bigg|4(n+1)(n^2+(n+2)^2)$$
But note that $4(n+1)$ and $d$ are coprime (since both of the given numbers are odd, hence $d$ is also odd, also, if $k$ is a common factor of $d$ and $n+1$, then $k$ also divides $n^4$ which is a contradiction, since, $k$ and $n$ are coprime). Hence,
$$d\bigg|n^2+(n+2)^2\tag1\\
\implies d\bigg|n^4+2(n+1)^4+(n+2)^4-\Big(n^2+(n+2)^2\Big)^2\\
\implies d\bigg|2(n+1)^4-2n^2(n+2)^2\\
\implies d\bigg|\Big((n+1)^2-n(n+2)\Big)\Big((n+1)^2+n(n+2)\Big)$$
$$\implies d\bigg|2n^2+4n+1\tag2$$
Subtracting the second equation from the first one, we get,
$$d\bigg|3\tag3$$
Hence, $d$ is either $1$ or $3$. Since exactly one of $n, n+1$ and $n+2$ is divisible by $3$, we are done.
EDIT
As N.S. pointed out in comments, after $(1)$, if one observes that
$$\Big(n^2+1\Big)\Big(n^2+(n+2)^2\Big)-\Big(n^4+(n+1)^4\Big)=3$$
then $d$ being a common factor of both terms in LHS, is also a factor of RHS and we again reach $(3)$.
|
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|
Finding the area of $\triangle ABC$ with $B$ and $C$ lying on the line $\frac{x-2}{3}=\frac{y–1}{0}=\frac{z}{4}$
The vertices $B$ and $C$ of a $\triangle ABC$ lie on the line, $\frac{x-2}{3}=\frac{y–1}{0}=\frac{z}{4}$ such that $BC = 5$ units. Then the area (in sq. units) of this triangle, given that the point $A(1, –1, 2)$, is: $5 \sqrt{17} /\sqrt{34}/ 6 / 2\sqrt{34}?$
My attempt: Let $B$ be $(3k+2,1,4k)$ and $C$ be $(3l+2,1,4l)$. So, using distance formula between $B$ and $C$, I get $k-l=1$. Now, using determinant, where the first row is coordinates of $A$, the second is of $B$, and the third $C$, I get area to be $5$. But the answer is given as $\sqrt{34}$. What is wrong in my method?
|
Let $AD$ be an altitude of the triangle.
Thus, $$D(2+3t,1,4t)$$ and since $$\vec{AD}\cdot\vec{(3,0,4)}=0,$$ we obtain:
$$(1+3t,2,4t-2)(3,0,4)=0,$$ which gives $$t=\frac{1}{5}.$$
Can you end it now?
|
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|
$\sum(-1)^{k}k^{2}C_{n}^{k}$ I'm asked to compute $\sum^{n}_{k=1}(-1)^{k}k^{2}C_{n}^{k}$ for $n\geqslant 1$. I tried to use generating functions to solve the problem: $A(x):=\sum^{n}_{k=1}k^{2}C_{n}^{k}x^{k}$. So, I need to compute $A(-1)$.
For $n=1$ $A(-1)=-1$, for $n=2$ $A(-1)=2$, for $n=2$ $A(-1)=2$, for $n=3,4,5,6,7,...(?)$ $A(-1)=0$. It seems that the answer is $0$ for $n\geqslant 3$...
Thank you in advance!
|
Computing
$$\sum_{k=0}^n {n\choose k} (-1)^k k^2$$
we write
$$\sum_{k=0}^n {n\choose n-k} (-1)^k k^2
= [z^n] (1+z)^n \sum_{k=0}^n z^k (-1)^k k^2.$$
Here the coefficient extractor controls the range and we find
$$[z^n] (1+z)^n \sum_{k\ge 0} z^k (-1)^k k^2
= [z^n] (1+z)^n \frac{z(z-1)}{(1+z)^3}
\\ = [z^n] (1+z)^{n-3} (z^2-z).$$
This is zero when $n\ge 3$ because we are extracting the coefficient
$[z^n]$ from a polynomial in $z$ of degree $n-1.$
Continuing, it becomes
$$[z^n] z^2 (1+z)^{n-3} - [z^n] z (1+z)^{n-3}
= {n-3\choose n-2} - {n-3\choose n-1}.$$
This is again zero by inspection when $n\ge 3$ because
$(n-3)^\underline{n-2} = 0$ and $(n-3)^\underline{n-1} = 0.$
For $n=2$ we get
${-1\choose 0} - {-1\choose 1} = 1 - (-1) = 2$
and for $n=1$ we find
$- {-2\choose 0} = -1.$
|
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|
Is there missing information to answer this question! I have two quantities $A_1$ and $A_2$, and I would like to compare them in order to know which one is bigger.
Knowing that $x$, $y$ are two constants such that $y \leq \frac{x}{2}$
$$A_1 = \sum_{z=1}^y (x-y+z)!(y-z+1)! (z+1)$$
$$A_2 = \sum_{z=1}^y (x-y+z+2)! (y-z)! (z+3)$$
Simply I want to specify if $A_1 > A_2$ or $A_1 < A_2$ ?
I guess that we can say $A_1 > A_2$ $n$ increases because the factorial increases faster, but which constraints I should put here?
|
We have $x-y\ge y.$ So, for each given $z,$ the ratio of the "$z$-term" in $A_2$ to the "$z$-term" in $A_1$ is $$\frac {((x-y)+z+2)((x-y)+z+1)}{y-z+1}\cdot \frac {z+3}{z+1}\ge \frac {(y+z+2)(y+z+1)}{y-z+1}\cdot \frac {z+3}{z+1}=$$ $$=(y+z+2)\cdot \frac {(y+1)+z}{(y+1)-z}\cdot \frac {z+3}{z+1}>$$ $$>(y+z+2)\cdot 1 \cdot 1\ge 4.$$ So $A_2>4A_1>0.$
|
{
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Calculation of Complex Limit When trying to calculate a residue, I came across this limit:
$$L:=\lim_{z\to \pi k} \frac{z^3-2z^2}{(1-\mathrm e^{\mathrm iz})\sin(z)}\left[\frac{(3z^2-4z)(z-\pi k)^2}{z^3-2z^2}+2(z-\pi k)-\frac{(z-\pi k)^2\cos(z)}{\sin(z)}+\frac{(z-\pi k)^2\mathrm i\,\mathrm e^{\mathrm i z}}{1-\mathrm e^{\mathrm iz}}\right]$$
where $0\neq k\in\mathbb Z$ and $k$ is even.
Since the limits of the individual summands do not exist, it seems to be very hard to apply l'Hopital to the whole fraction.
Is there any way to simplify this in order to calculate the limit?
(The original problem was to find the residue of $z\mapsto\frac{z^2(z-2)}{(1-\exp(\mathrm i z))\sin(z)}$ at $z=k\pi$ where $0\neq k$ is an even integer.)
Thank you.
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We can calculate the limit using series expansion at $z=\pi k$ where $0\neq k\in\mathbb Z, k \text{ even}$. We recall
\begin{align*}
\cos(z)&=1+\mathcal{O}\left((z-\pi k)^2\right)\\
e^{iz}&=1+i(z-\pi k)+\mathcal{O}\left((z-\pi k)^2\right)\\
\frac{1}{\sin(z)}&=\frac{1}{z-\pi k}+\mathcal{O}\left(z-\pi k\right)\\
\frac{1}{1-e^{iz}}&=\frac{i}{z-\pi k}+\frac{1}{2}+\mathcal{O}\left(z-\pi k\right)
\end{align*}
where we expand the series up to terms we need for calculation and put everything else into big-$\mathcal{O}$'s.
We obtain
\begin{align*}
\color{blue}{\lim_{z\to\pi k}}&\color{blue}{\frac{z^3-2z^2}{\left(1-e^{iz}\right)\sin z}\left[\frac{(3z^2-4z)(z-\pi k)^2}{z^3-2z^2}+2(z-\pi k)\right.}\\
&\qquad\qquad\color{blue}{\left.-\frac{(z- \pi k)^2\cos z}{\sin z}+\frac{(z-\pi k)^2ie^{iz}}{1-e^{iz}}\right]}\\
&=\lim_{z\to \pi k}(3z^2-4z)(z-\pi k)^2\left(\frac{1}{z-\pi k}+\mathcal{O}\left(z-\pi k\right)\right)\\
&\qquad\qquad\cdot\left(\frac{i}{z-\pi k}+\frac{1}{2}+\mathcal{O}\left(z-\pi k\right)\right)\\
&\qquad+\lim_{z\to\pi k}\left(z^3-2z^2\right)\left[2(z-\pi k)-\frac{(z- \pi k)^2\cos z}{\sin z}+\frac{(z-\pi k)^2ie^{iz}}{1-e^{iz}}\right]\\
&\qquad\qquad\cdot\left(\frac{1}{z-\pi k}+\mathcal{O}\left(z-\pi k\right)\right)
\left(\frac{i}{z-\pi k}+\frac{1}{2}+\mathcal{O}\left(z-\pi k\right)\right)\tag{1}\\
&=\left(3\pi^2k^2-4\pi k\right)i+\left(\pi^3k^3-2\pi^2k^2\right)\lim_{z\to \pi k}\left[\frac{2i}{z-\pi k}\right.\\
&\qquad\left.-i\frac{\cos z}{\sin z}-\frac{e^{iz}}{1-e^{iz}}
+1-\frac{1}{2}\frac{(z-\pi k)\cos z}{\sin z}+\frac{i}{2}\frac{(z-\pi k)e^{iz}}{1-e^{iz}}\right]\tag{2}\\
&=\left(3\pi^2k^2-4\pi k\right)i+\left(\pi^3k^3-2\pi^2k^2\right)\lim_{z\to \pi k}\left[\frac{2i}{z-\pi k}+1\right.\\
&\qquad\left.-\left(\frac{z-\pi k}{2}+i\right)\frac{\cos z}{\sin z}+\left(i\frac{z-\pi k}{2}-1\right)\frac{e^{iz}}{1-e^{iz}}\right]\tag{3}\\
&=\left(3\pi^2k^2-4\pi k\right)i+\left(\pi^3k^3-2\pi^2k^2\right)\lim_{z\to \pi k}\left[\frac{2i}{z-\pi k}+1\right.\\
&\qquad-\left(\frac{z-\pi k}{2}+i\right)\left(\frac{1}{z-\pi k}+\mathcal{O}\left(z-\pi k\right)\right)\\
&\qquad\left.+\left(i\frac{z-\pi k}{2}-1\right)\left(\frac{i}{z-\pi k}-\frac{1}{2}+\mathcal{O}\left(z-\pi k\right)\right)\right]\tag{4}\\
&=\left(3\pi^2k^2-4\pi k\right)i+\left(\pi^3k^3-2\pi^2k^2\right)\lim_{z\to \pi k}\left[\frac{2i}{z-\pi k}+1\right.\\
&\qquad\left.-\frac{1}{2}-\frac{i}{z-\pi k}-\frac{1}{2}-\frac{i}{z-\pi k}-\frac{i}{4}\left(z-\pi k\right)+\frac{1}{2}\right]\tag{5}\\
&\,\,\color{blue}{=\frac{1}{2}\pi^3k^3-\pi^2k^2+\left(3\pi^2k^2-4\pi k\right)i}
\end{align*}
Comment:
*
*In (1) we separate the first bracketed term by a limit of its own, since it can be calculated separately from the other terms. We cancel $z^3-2z^2$ and use the series expansion of $\frac{1}{1-e^{iz}}$ and $\frac{1}{\sin z}$.
*In (2) we calculate the first limit. We factor out $z^3-2z^2$ and evaluate it at $z=\pi k$. We multiply out within in the limit and skip terms which do not contribute.
*In (3) we collect related terms.
*In (4) we expand $\frac{\cos z}{\sin z}$ and $\frac{e^{iz}}{1-e^{iz}}$ at $z=\pi k$.
*In (5) we multiply out and cancel again terms which do not contribute. Now we are ready to cancel away some terms in the last step and do a final limit calculation.
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|
Do we reduce the problem into the known $3^x+4^x=5^x $? I want to solve the following equations: \begin{align*}&(i) \ \ \ \ \ 3^x+4^x=5^x \\ &(ii) \ \ \ \ \ (x+1)^x+(x+2)^x=(x+3)^x, \ x>-1 \\ &(iii) \ \ \ \ \ (x+1)^x+(x+2)^x=(x+4)^x, \ x>-1\end{align*}
$$$$
I have done the following:
For (i):
We have that $$3^x+4^x=5^x\Rightarrow \left (\frac{3}{5}\right )^x+\left (\frac{4}{5}\right )^x=1$$ This equation of the form $f(x)=a^x+b^x$ with $0<a<b<1$. For this equation we have that: Since $0<a,b<1$ the function $a^x$ and $b^x$ and so also $f(x)$ are decreasing.
So there is at most one solution.
Since $a<b$ so $a^x<b^x$ for positive $x$ then we get $2a^x<f(x)<2b^x$. Both the lower and upper bound for $f(x)$ pass through 1, we know that f(x) must also pass through 1, so there is exactly one solution,.
We symbolize by $c$ the unique solution. From $2a^c<1<2b^c$ we get the inequality $-\log_a2<c<-\log_b2$. For $a=\frac{3}{5}$ and $b=\frac{4}{5}$ we get $1.35<c<3.10$. We consider the natural numers between $1.35$ and $3,10$.
We see that for $c=2$ the equation is satisfied, and so $x=2$ is the only solution.
Is everything correct?
As for (ii) and (iii) do we have to find an appropriate $x$ so that we can reduce it to a problem as in (i) ?
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Another approach
$x=2$ is an obvious solution to the problem. So consider the function
$$f(x)=\left(\frac{3}{5}\right)^x+\left(\frac{4}{5}\right)^x-1.$$
We know $f(2)=0$. Also $f'(x)=\left(\frac{3}{5}\right)^x\ln\left(\frac{3}{5}\right)+\left(\frac{4}{5}\right)^x \ln\left(\frac{4}{5}\right) <0$ for all $x$. Should this $f$ had another zero, say $f(c)=0$, then by Rolle's theorem, there exists a point $k \in (c,2)$ (or $(2,c)$) such that $f'(k)=0$ but thay contradicts the sign of the derivative we have shown above. Thus only one solution.
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|
Recurrence relations: prove that if $a_i = a_j$, then $i=j$
let $$a_n = \frac{n^2+10}{10^4}a_{n-1}, \space a_1=99$$
Prove (or disprove) that $a_i = a_j \implies i = j$
My proof:
The above sequence is non-monotonic: $\frac {a_n}{a_{n-1}} > 1$ when $n^2 > 9900$. Therefore the least element will be $a_{99}$. Since a difinite minimum element exists, it is possible that if $a_i = a_j$, then $i=j$ exists such that $i<99, j>99$.
This proof is not only weak but also wrong: my book says that this property is true. Please help me in proving so.
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Well, this is the same as proving that $f:\mathbb{Z^{+}}\rightarrow\mathbb{Z^{+}}$ is injective $(f(a)=f(b) \implies a=b)$
while $f(1)=99$ and
$$f(x)=\frac{x^2+10}{10^4}\cdot f(x-1)$$
$$f(x)=\frac{x^2+10}{10^4}\cdot\frac{(x-1)^2+10}{10^4}\cdot \dots \cdot\frac{(x-k)^2+10}{10^4} \cdot f(x-k-1) \tag{1}$$
for all positive integers $k \le x-2$
Assume for the sake of contradiction that there exists positive integers $a,b$ such that $f(a)=f(b) \ne0$ while $a \ne b$, and assume WLOG that $a>b$, so that we can choose $k$ such that $b=a-k-1$, we get:
$$f(a)=\frac{a^2+10}{10^4}\cdot\frac{(a-1)^2+10}{10^4}\cdot \dots \cdot\frac{(b+1)^2+10}{10^4} \cdot f(b)$$
$$\iff(a^2+10)((a-1)^2+10)\dots((b+1)^2+10)=1$$
which is, of course, impossible.
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|
Does $a^2 + b^2 = 2 c^2$ have any integer solution?
Does the equation $a^2 + b^2 = 2 c^2$ have any integer solution with $|a| \neq |b|$?
I think not, because of these equations for pythagorean triplets:
$$\left(\frac{a}{c}\right)^2+\left(\frac{b}{c}\right)^2=1$$
$$x=\frac{1-t^2}{1+t^2}, y=\frac{2t}{1+t^2}$$
I think I would need to multiply $x$ and $y$ by $\sqrt{2}$ and they would never be rational numbers.
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Suppose $(x,y,z)$ is any Pythagorean triple. Then:
$$(x-y)^2+(x+y)^2=2x^2+2y^2=2z^2$$
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|
Pair of real number satisfying $56x+33y=-\frac{y}{x^2+y^2}$ and $33x+56y=\frac{x}{x^2+y^2}$. Let (x,y) be a pair of real number satisfying $56x+33y=-\frac{y}{x^2+y^2}$ and $33x+56y=\frac{x}{x^2+y^2}$. If $|x|+|y|=\frac{p}{q}$ (where p and q are relatively prime), then find the value (6p – q).
I used the concept $\frac{x}{y}=t$ while solving i get $56t+33=-\frac{1}{t^2+1}$ and $33t+56=\frac{t}{t^2+1}$ on dividing the two equation i end up getting quadratic equation but that is not helping me
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Both of the existing answers posted here are incorrect, so here I will start with @Aditya Dwivedi's solution and go from there.
Let $ z = x - yi $
Then, subtract the first equation times $i$ from the second equation.
\begin{align*}
&33x - 56y - 56xi - 33yi = \frac{x + yi}{x^2 + y^2} \\
\implies &33(x-yi) - 56(y + xi) = \frac{1}{x - yi} \\
&\text{you can see that $y + xi = i \cdot (x - yi) $, so} \\
\implies & z(33 - 56i) = \frac{1}{z} \implies z^2 = \frac{1}{33-56i}
\end{align*}
$33 - 56i = (7 - 4i)^2$, so $ z = \pm \frac{1}{7 - 4i} = \pm(\frac{7}{65} + \frac{4}{65} i) \implies (x, y) = (\pm \frac{7}{65}, \mp \frac{4}{65}) $. Thus, $|x| + |y| = \boxed{\frac{11}{65}} \hspace{.5em} \blacksquare $
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How does $x - 4y = 12$ result in $y = \frac{1}{4}x - 3$ instead of $y = -\frac{1}{4}x - 3$? I'm learning math. That's why I'm asking a simple explanation from a math expert.
When I solve this equation it results in $y = -\frac{1}{4}x - 3$, but the right answer is $y = \frac{1}{4}x - 3$.
Here is my solution:
$$x-4y = 12$$
$$x-x-4y = 12-x$$
$$-4y = 12-x $$
$$\frac{-4}{-4}y = -\frac{12}{4}-\frac{x}{4}$$
$$y = -3 - \frac{x}{4} $$
$$y = - \frac{1}{4}x - 3$$
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The error is that when dividing both sides by $-4$ you went from $-x$ to $-\frac x4$; it should go to $-\frac{x}{-4}=+\frac{x}{4}$.
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Prove by induction that for all $n\in\mathbb N, (\sqrt3+i)^n+(\sqrt3-i)^n=2^{n+1}\cos(\frac{n\pi}6)$ I want to prove by induction that for all $n \in \mathbb{N}$, $$(\sqrt{3} + i)^n + (\sqrt{3} - i)^n = 2^{n+1} \cos\left(\frac{n\pi}{6} \right)$$
I can prove the identity using direct complex number manipulation and de Movire (which gets it for all integers), but I'm getting stuck when trying induction out of interest instead.
For the inductive step, I have $$(\sqrt{3} + i)^{k+1} + (\sqrt{3} - i)^{k+1} $$ but what can I do with that? If I separate the terms into $(\sqrt{3} + i)^{k+1} = (\sqrt{3} + i)^{k} (\sqrt{3} + i)$ etc, the minus sign from the second term is causing me troubles.
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You need to induct the two statements below simultaneously
$$(\sqrt{3} + i)^n + (\sqrt{3} - i)^n = 2^{n+1} \cos\frac{n\pi}{6} \\
(\sqrt{3} + i)^n - (\sqrt{3} - i)^n = i 2^{n+1} \sin\frac{n\pi}{6} $$
Note
\begin{align}
&(\sqrt{3} + i)^{n+1}+ (\sqrt{3} - i)^{n+1}
\\
=& \sqrt3[ (\sqrt{3} + i)^{n}+ (\sqrt{3} - i)^{n}]+i [(\sqrt{3} +i)^{n}- (\sqrt{3} - i)^{n}]\\
=& 2^{n+2} \left( \frac{\sqrt3}2\cdot\cos\frac{n\pi}{6} - \frac12\cdot\sin\frac{n\pi}{6} \right)\\
=& 2^{n+2} \cos\frac{(n+1)\pi}{6}
\end{align}
Similarly
\begin{align}
&(\sqrt{3} + i)^{n+1}-(\sqrt{3} - i)^{n+1}=i 2^{n+2} \sin\frac{(n+1)\pi}{6}
\end{align}
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$f:\mathbb N_0\to\mathbb N_0$ with $2f\left(m^2+n^2\right)=f(m)^2+f(n)^2$ and $f\left(m^2\right)\geqslant f\left(n^2\right)$ when $m\geqslant n$
Determine all $f : \mathbb N_0 \to \mathbb N_0$ that satisfy $2f\left(m^2 + n^2\right) = f(m)^2 + f(n)^2$ and $f\left(m^2\right) \geqslant f\left(n^2\right)$ when $m \geqslant n$.
I managed to prove that $f$ is increasing, but I don't know what to do next. Can anyone give some hints or solution please. Thank you very much!
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Observe the following things :
*
*For any $m, n \in \mathbb{N}_0$, $f(m)^2 + f(n)^2 $ is an even number, so the values of $f(x)$ are all odd or all even.
*$2f(0^2+ 0^2) = 2f(0)^2$ so we have $f(0) = 0$ or $1$.
*$2f(1^2 + 0^2) = f(1)^2 + f(0)$.
First assume that $f(x)$ values are all odd. then we have $f(0 ) = 1$ and $(f(1)-1)^2 = 0$, $f(1) = 1$. Again $f(2) = f(1)^2 = 1$. If $f(a) = 1$, then $2f(a^2) =f(a)^2 + f(0) = 2$, i.e. $f(a^2) = 1$. This leads us to $f(2) = f(4) = f(16) = \cdots = 1$, and from the fact that $f$ is increasing (I didn't check this but you said you proved this, so, ) we can see that $f(x) = 1$ for all $x \in \mathbb{N}_0$.
For the second case we assume that $f(x)$ values are all even. $f(0) = 0$, and $f(1) = 0$ or $2$. If $f(1) = 0$, we have $f(2) = 0$ and $f(4) = f(16) = \cdots = 0$. One can verify this as same as $f$ of odd values case.
The only remaining case is that $f$ has even values, $f(0) = 0$, $f(1) =2$.
From now on, for simplicity, Let $g(x) = f(x)/2$. Then $g$ is also an integer function and satisfies $g(n^2+m^2) = g(n)^2 + g(m)^2$. From this step, we can refer to this paper. I think the additional conditions we have would make this problem much easier, but I will proceed after the paper.
Note that $(xy + zw)^2 + (xw-yz)^2 = (xy - zw)^2 + (xw + yz)^2$, so $g(xy + zw)^2 + g(xw-yz)^2 = g(xy - zw)^2 + g(xw + yz)^2$ whenever the terms are nonnegative.
Putting $(x, y, z, w) = (k, 2, 1, 1)$ lead us to
$$ g(2k+1)^2 = g(2k-1)^2 + g(k+2)^2 - g(k-2)^2.$$ If $g(x) = x$ for $x \le 2k-1$, one can wee that $g(2k+1) = (2k-1)^2 +(k+2)^2 - (k-2)^2 = 4k^2 + 1+ 4k = (2k+1)^2$, i.e. $g(2k + 1) = 2k+ 1$.
Putting $(x, y, z, 2) = (k-1, 2, 2, 1)$ lead us to the similar induction step for even $x$ case.
So, to show that $g(x) = x$, i.e. $f(x) = 2x$, it is enough to show for some initial cases. Refer to the paper I linked for remaining details.
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|
Vieta's formulas for $x^2+px+1$ and $x^2+qx+1$.
Let $\alpha$ and $\beta$ be the roots for $x^2+px+1$ and let $\gamma$ and $\delta$ be the roots
for $x^2+qx+1$. Show that $$(\alpha-\gamma)(\beta-\gamma)(\alpha+\delta)(\beta+\delta)=q^2-p^2.$$
This seemed to be rather peculiar. It should be a simple application of Vieta's formulas, but I couldn't get it to the form they wanted...
We have $\alpha+\beta=-p$, $\alpha\beta=1$ and $\gamma+\delta=-q$, $\gamma\delta=1$.
Also $\alpha^2+p\alpha+1=0 \Rightarrow p = -\alpha-\frac{1}{\alpha} \Rightarrow p^2 = \alpha^2+2+ \frac{1}{\alpha^2}$
and $\gamma^2+q\gamma+1=0 \Rightarrow p = -\gamma-\frac{1}{\gamma} \Rightarrow q^2 = \gamma^2+2+ \frac{1}{\gamma^2}$.
This should imply that $q^2-p^2 = (\gamma^2+2+\frac{1}{\gamma^2})-(\alpha^2+2+ \frac{1}{\alpha^2}) = \gamma^2+\frac{1}{\gamma^2} - \alpha^2-\frac{1}{\alpha^2}$.
What should I do here?
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Note $x^2+px+1= (x-\alpha)(x-\beta)$. Then
\begin{align}
& (\alpha-\gamma)(\beta-\gamma)(\alpha+\delta)(\beta+\delta)\\
= &(\gamma^2+p\gamma +1)(\delta^2-p\delta +1)\\
= &(\gamma\delta)^2+\delta^2 +\gamma^2+1-(\gamma\delta)p^2\\
= &(\delta+\gamma)^2-2\gamma\delta+2-p^2\\
=&q^2-p^2
\end{align}
where $\gamma+\delta=-q $ and $\gamma\delta=1$.
|
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|
Finding the sum of $1 \cdot 2x + 2 \cdot 3x^{2} + 3 \cdot 4x^{3} + \dots$ . What is the sum of $1 \cdot 2x + 2 \cdot 3x^{2} + 3 \cdot 4x^{3} + \dots$ ?
I have been stuck on this problem with no direction. I have tried multiplying the sequence with $x$ and trying out $S-Sx$ but have gotten nowhere. Any help?
Thanks.
|
Expanding my hint, you have:
$$\frac{s}{x} = \frac{\text{d}^2}{\text{d} x^2}\left[ x^2 + x^3 + x^4 + x^5 + \cdots \right] = \frac{\text{d}^2}{\text{d} x^2}\left[ 1 + x + x^2 + x^3 + x^4 + x^5 + \cdots \right] = \frac{\text{d}^2}{\text{d} x^2}\left[ \frac{1}{1 - x} \right] = \frac{2}{(1 - x)^3}$$
(you can add $1+x$ to the sum in square brackets because $1+x$ has second derivative equal to $0$), hence:
$$ s = \frac{2x}{(1 - x)^3}\;.$$
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|
Finding the general formula for the sequence with $d_0=1$, $d_1=-1$, and $d_k=4 d_{k-2}$
Suppose that we want to find a general formula for the terms of the sequence
$$d_k=4 d_{k-2}, \text{ where } d_0=1 \text{ and } d_1=-1$$
I have done the following:
\begin{align*}d_k=4d_{k-2}&=2^2d_{k-2} \\ &=2^2\left (2^2d_{(k-2)-2}\right )=2^4d_{k-4} \\ & =2^4\left (2^2d_{(k-4)-2}\right )=2^6d_{k-6} \\ & = 2^6\left (2^2d_{(k-6)-2}\right )=2^8d_{k-8} \\ & = \ldots \\ & = 2^id_{k-i}\ , \ \ i \text{ even}\end{align*}
If $k$ even, then at the last step we have for $i=k$ (since $k$ is the maximum even number $\leq k$) : $d_k=2^kd_{k-k}=2^kd_0=2^k$.
If $k$ odd, then at the last step we have for $i=k-1$ (since $k-1$ is the maximum even number $\leq k$) : $d_k=2^{k-1}d_{k-(k-1)}=2^{k-1}d_1=-2^{k-1}$.
How can we find the general form for the terms of the recurrence relation? Or do we distinguish cases when $k$ is even and odd?
I am interested to find the general formula without using the characteristic equation.
|
You found $d_k=2^k$ when $k$ is even and $d_k=-2^{k-1}$ when $k$ is odd.
To put this in one formula, note that $\dfrac{1+(-1)^n}2$ is $0$ when $n$ is odd and $1$ when $n$ is even,
whereas $\dfrac{1-(-1)^n}2$ is $1$ when $n$ is odd and $0$ when $n$ is even.
So you could say $d_k=2^k\dfrac{1+(-1)^k}2-2^{k-1}\dfrac{1-(-1)^k}2,$
which simplifies to $2^{k-2}\left[1+3(-1)^k\right]$.
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How to find the intergral $I_{A}=\int_{0}^{2\pi}\frac{\sin^2{x}}{(1+A\cos{x})^2}dx$ Let $A\in (0,1)$be give real number ,find the closed form intergral
$$I_{A}=\int_{0}^{2\pi}\dfrac{\sin^2{x}}{(1+A\cos{x})^2}dx$$
This integral comes from a physical problem，following is my try:
since
$$I_{A}=\int_{0}^{2\pi}\dfrac{\sin^2{x}}{(1+A\cos{x})^2}dx=I_{1}+I_{2}$$
Where $$I_{1}=\int_{0}^{\pi}\dfrac{\sin^2{x}}{(1+A\cos{x})^2}dx,I_{2}=\int_{\pi}^{2\pi}\dfrac{\sin^2{x}}{(1+A\cos{x})^2}dx$$
For $I_{2}$ Let $x=\pi+t$,then we have
$$I_{2}=\int_{0}^{\pi}\dfrac{\sin^2{x}}{(1-A\cos{x})^2}dx$$
so
$$I_{A}=I_{1}+I_{2}=2\int_{0}^{\pi}\dfrac{\sin^2{x}(1+A^2\cos^2{x})}{(1-A^2\cos^2{x})^2}dx=4\int_{0}^{\frac{\pi}{2}}\dfrac{\sin^2{x}(1+A^2\cos^2{x})}{(1-A^2\cos^2{x})^2}dx$$
Then I fell ugly, so how to prove it? Thank you
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Most probably you would like to have a formula depending on $A$.
Using complex residues we need to find the residues of
$$f(z) = -\frac{(z^2-1)^2}{A^2 z \left(1 + \frac{2z}{A}+ z^2\right)^2}$$
within the unit disc.
$f(z)$ has a single pole at $z=0$ and for $0<A<1$ a pole of order $2$ at $z_A=\frac{\sqrt{1-A^2}-1}{A}$.
While the residue at $z=0$ is easy to calculate, the residue at $z_A$ is a bit ugly. So, I used Mathematica to calculate the residue at $z_A$ and will give only the final result without further simplifying:
$$I_A = 2\pi\left(Res_{z=0}f(z) + Res_{z=z_A}f(z)\right)$$
$$= 2 \pi \left(-\frac{1}{A^2} + \frac{A^2+2 \sqrt{1-A^2}-2}{A^2 \left(\left(\sqrt{1-A^2}-2\right) A^2-2 \sqrt{1-A^2}+2\right)}\right)$$
Of course, I tested it numerically and the formula produces nicely the searched for integral:
$$\left(
\begin{array}{ccc}
\text{A} & I_A \text{numerical} & I_A \text{ via residues} \\
0.1 & 3.16535 & 3.16535 \\
0.2 & 3.2391 & 3.2391 \\
0.3 & 3.37092 & 3.37092 \\
0.4 & 3.57707 & 3.57707 \\
0.5 & 3.88805 & 3.88805 \\
0.6 & 4.36332 & 4.36332 \\
0.7 & 5.13272 & 5.13272 \\
0.8 & 6.54498 & 6.54498 \\
0.9 & 10.0388 & 10.0388 \\
\end{array}
\right)$$
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|
Prove $3(x+y)(x+z)(y+z)\neq a$ cubic when $x,y,z$ are different co-primal positive integers Prove $3(x+y)(x+z)(y+z)\neq a$ cubic when $x,y,z$ are different co-primal positive integers.
I believe you can look at the prime factors of $x,y$ and $z$. As for the equation to equal a cubic, the prime factors must all be present in multiples of $3$'s.
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What you're asking to prove is not always true. One specific counter-example, among infinitely many, is
$$x = 19 \tag{1}\label{eq1A}$$
$$y = 324 = 2^2 \times 3^4 \tag{2}\label{eq2A}$$
$$z = 4\text{,}589 = 13 \times 353 \tag{3}\label{eq3A}$$
These are all distinct, co-primal positive integers. You next have
$$x + y = 343 = 7^3 \tag{4}\label{eq4A}$$
$$x + z = 4\text{,}608 = 3^2 \times 2^9 \tag{5}\label{eq5A}$$
$$y + z = 4\text{,}913 = 17^3 \tag{6}\label{eq6A}$$
This gives
$$\begin{equation}\begin{aligned}
3(x + y)(x + z)(y + z) & = 3(343)(4\text{,}608)(4\text{,}913) \\
& = 3(7^3 \times (3^2 \times 2^9) \times 17^3) \\
& = (2^3)^3 \times 3^3 \times 7^3 \times 17^3 \\
& = (8 \times 3 \times 7 \times 17)^3 \\
& = (2\text{,}856)^3
\end{aligned}\end{equation}\tag{7}\label{eq7A}$$
Are there perhaps some other required conditions which were not mentioned?
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|
Problem 1, Exercise 3.4, Linear Algebra, Hoffman and Kunze The fundamental question is this: Let $V$ be a n-dimensional F-vector space. Is the matrix $A \in F^{n \times n}$ of $T \in L(V,V)$ relative to $\mathcal{B}$ is unique upto row-equivalence or not. With this out, let me discuss what is my issue with the problem.
The problem says: Let $T \in L(\mathbb{C}^2,\mathbb{C}^2)$ be defined as $T(x_1,x_2)=(x_1,0)$. Let $\mathcal{B}=\{e_1, e_2\}$ with $e_1 = \begin{pmatrix}1 \\0 \end{pmatrix}$ and $e_2 = \begin{pmatrix} 0 \\1 \end{pmatrix}$.Let $\mathcal{B}' = \{\alpha_1, \alpha_2\}$ with $\alpha_1 = \begin{pmatrix}1 \\i \end{pmatrix}$ and $\alpha_2 = \begin{pmatrix} -i \\2 \end{pmatrix}$. Then:
a) What is the matrix T relative to the pair $\mathcal{B}, \mathcal{B}'$.
My approach: We know $Te_j = A_{ij} \alpha_i$. So, columns of A are $A_1, ... , A_n$ such that $A_i = [Te_i]_{\mathcal{B}'}$. Now, we can find a $2 \times 2$ matrix Q such that $\begin{pmatrix} \alpha_1 \\ \alpha_2 \end{pmatrix} = Q \begin{pmatrix} e_1 \\ e_2 \end{pmatrix}$ and then $P := Q^{-1}$ gives $X = P X'$, where X is the coordinates of a vector in $\mathbb{C}^2$ wrt the basis $\begin{pmatrix} e_1 \\ e_2 \end{pmatrix}$ and $X'$ is the coordinates of the same vector in $\mathbb{C}^2$ wrt the basis $\begin{pmatrix} \alpha_1 \\ \alpha_2 \end{pmatrix}$ (based on section 2.5-6 of H&K).
Then, $A_1 = \begin{pmatrix} 1 \\ 0 \end{pmatrix}_{\mathcal{B}'}=P\begin{pmatrix} 1 \\ 0 \end{pmatrix} = \begin{pmatrix} 2 \\ -i \end{pmatrix}$. Similarly, $A_2 = \begin{pmatrix} 0 \\ 0 \end{pmatrix}$. So the answer is \begin{pmatrix} 2 & 0 \\ -i & 0 \end{pmatrix}
This is all right, as it matches with [1,2]. The problem is with b, which says:
b) What is the matrix T relative to the pair $\mathcal{B}', \mathcal{B}$.
My approach: $A_i = [T \alpha_i]_{\mathcal{B}}$, then $A_1 = \begin{pmatrix} 1 \\ 0 \end{pmatrix}_{\mathcal{B}}=Q\begin{pmatrix} 1 \\ 0 \end{pmatrix} = \begin{pmatrix} 1 \\ i \end{pmatrix}$. Similarly, $A_2 = \begin{pmatrix} -i \\ 1 \end{pmatrix}$. So the answer is $Matrix_1 = \begin{pmatrix} 1 & -i \\ i & 1 \end{pmatrix}$.
Here is the confusion: the answer I got from online solutions (See [1], [2]) is $Matrix_2 = \begin{pmatrix} 1 & -i \\ 0 & 0 \end{pmatrix}$.
$Matrix_1$ and $Matrix_2$ are row equivalent but not the same. So where have I gone wrong?
I got similar results for c) and d), which asks: c) T relative to $\mathcal{B}'$ and d) T relative to $\{ \alpha_2, \alpha_1 \}$
My approach: c) $A_1 = [T \alpha_1]_{\mathcal{B}'} = \begin{pmatrix} 1 \\ 0 \end{pmatrix}_{\mathcal{B}'}= I \begin{pmatrix} 1 \\ 0 \end{pmatrix} = \begin{pmatrix} 1 \\ 0 \end{pmatrix}$. Similarly, $A_2 = \begin{pmatrix} -i \\ 0 \end{pmatrix}$
Thus my answer is: $Matrix_3 = \begin{pmatrix} 1 & -i \\ 0 & 0 \end{pmatrix} $. reference gives $Matrix_4 = \begin{pmatrix} 2 & -2i \\ -i & -1 \end{pmatrix} $ which is row-equivalent to $Matrix_3$. Again, the same question?
d) Similar approach gives me: $Matrix_5 = \begin{pmatrix} -i & 1 \\ 0 & 0 \end{pmatrix} $. reference gives $Matrix_6 = \begin{pmatrix} -1 & -i \\ -2i & 2 \end{pmatrix} $ which is row-equivalent to $Matrix_5$. Again, the same question?
References:
[1] G. Grant, Solutions to Linear Algebra H&K, p64,
http://greggrant.org/hoffman_and_kunze.pdf
[2] https://linearalgebras.com/solution-hoffman-kunze.html
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I have figured it out. I was wrong and the right answer is:
$T = [T]_{\mathcal{B}}=\begin{bmatrix} 1&0\\0&0 \end{bmatrix}$
Let $\alpha = \begin{bmatrix} \alpha_1 \\ \alpha_2 \end{bmatrix} $ and $e = \begin{bmatrix} e_1 \\ e_2 \end{bmatrix} $. Then $\alpha = P e$ where, $P = \begin{bmatrix}1 & -i\\ i&2 \end{bmatrix} $ and $P^{-1} = \begin{bmatrix}2 & i\\ -i&1 \end{bmatrix} $.
a) $[Te_1]_{\mathcal{B}} = \begin{bmatrix} 1\\0 \end{bmatrix}$ and $[Te_2]_{\mathcal{B}} = \begin{bmatrix} 0\\0 \end{bmatrix}$
$[Te_1]_{\mathcal{B}'}=P^{-1} [Te_1]_{\mathcal{B}} = \begin{bmatrix} 2\\-i \end{bmatrix}$ and similarly, $[Te_1]_{\mathcal{B}'}= \begin{bmatrix} 0\\0 \end{bmatrix}$. So, $A =\begin{bmatrix}2&0\\-i&0 \end{bmatrix} $
Also note, $Te_1 = \begin{bmatrix} 1 \\0 \end{bmatrix} = 2 \alpha_1 - i \alpha_2$ and $Te_2 = \begin{bmatrix} 0 \\0 \end{bmatrix} = 0 \alpha_1 - 0 \alpha_2$. So the answer is correct and this is the alternative way of finding the matrix.
b) $T\alpha_1 = \begin{bmatrix} 1 \\0 \end{bmatrix} = 1 e_1 + 0 e_2$ and $T\alpha_2 = \begin{bmatrix} -i \\0 \end{bmatrix} = -i e_1 - 0 e_2$. So, $A =\begin{bmatrix}1&-i\\0&0 \end{bmatrix} $
c) $[T]_{\mathcal{B}'}=P^{-1}[T]_{\mathcal{B}}P$
d) $\begin{bmatrix} \alpha_2 \\ \alpha_1 \end{bmatrix} = \begin{bmatrix} -i&i\\2&i \end{bmatrix} e$, So, $P=\begin{bmatrix} -i&i\\2&i \end{bmatrix}$ and $P^{-1} = \begin{bmatrix} -i&1\\2&i \end{bmatrix}$.
$[T]_{\mathcal{B}''}=P^{-1}[T]_{\mathcal{B}}P$
all answers match the references.
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What are the steps to factor $x^2 - 1$ into $(x+1)(x-1)$? Does $(x+1)(x-1) = x^2+1x-1x-1$? If so where are the $+1x$ and the $-1x$ when it is being factored from $x^2-1$ into $(x+1)(x-1)$?
What exactly are we dividing $x^2-1$ by to get $(x+1)(x-1)$ and how did you know what to divide it by?
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Let $y:=x-1$. Then
$$x^2-1=(y+1)^2-1=y^2+2y$$ which obviously factors as $$(y+2)y=(x+1)(x-1).$$
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Strategy to calculate $ \frac{d}{dx} \left(\frac{x^2-6x-9}{2x^2(x+3)^2}\right) $. I am asked to calculate the following: $$ \frac{d}{dx} \left(\frac{x^2-6x-9}{2x^2(x+3)^2}\right). $$
I simplify this a little bit, by moving the constant multiplicator out of the derivative:
$$ \left(\frac{1}{2}\right) \frac{d}{dx} \left(\frac{x^2-6x-9}{x^2(x+3)^2}\right) $$
But, using the quotient-rule, the resulting expressions really get unwieldy:
$$ \frac{1}{2} \frac{(2x-6)(x^2(x+3)^2) -(x^2-6x-9)(2x(2x^2+9x+9))}{(x^2(x+3)^2)^2} $$
I came up with two approaches (3 maybe):
*
*Split the terms up like this: $$ \frac{1}{2}\left( \frac{(2x-6)(x^2(x+3)^2)}{(x^2(x+3)^2)^2} - \frac{(x^2-6x-9)(2x(2x^2+9x+9))}{(x^2(x+3)^2)^2} \right) $$
so that I can simplify the left term to $$ \frac{2x-6}{x^2(x+3)^2}. $$
Taking this approach the right term still doesn't simplify nicely, and I struggle to combine the two terms into one fraction at the end.
*The brute-force-method. Just expand all the expressions in numerator and denominator, and add/subtract monomials of the same order. This definitely works, but i feel like a stupid robot doing this.
*The unofficial third-method. Grab a calculator, or computer-algebra-program and let it do the hard work.
Is there any strategy apart from my mentioned ones? Am I missing something in my first approach which would make the process go more smoothly?
I am looking for general tips to tackle polynomial fractions such as this one, not a plain answer to this specific problem.
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Logarithmic differentiation can also be used to avoid long quotient rules. Take the natural logarithm of both sides of the equation then differentiate:
$$\frac{y'}{y}=2\left(\frac{1}{x-3}-\frac{1}{x}-\frac{1}{x+3}\right)$$
$$\frac{y'}{y}=-\frac{2\left(x^2-6x-9\right)}{x(x+3)(x-3)}$$
Then multiply both sides by $y$:
$$y'=-\frac{{\left(x-3\right)}^3}{x^3{\left(x+3\right)}^3}$$
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Solve $\sqrt{x^2+8x+7}+\sqrt{x^2+3x+2}=\sqrt{6x^2+19x+13}$ I tried squaring both sides but doesn't seem like a good idea.
$$x^2+8x+7+\sqrt{(x^2+8x+7)(x^2+3x+2)}+x^2+3x+2=6x^2+19x+13$$
$$\sqrt{(x^2+8x+7)(x^2+3x+2)}=4x^2+8x+4$$
$$\sqrt{(x^2+8x+7)(x^2+3x+2)}=4(x+1)^2$$
Is there a better way for solving this equation?
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I would factor both sides first so that:
$$\sqrt{(x+7)(x+1)}+\sqrt{(x+2)(x+1)}=\sqrt{(6x+13)(x+1)}$$
$$\sqrt{x+1}(\sqrt{x+7}+\sqrt{x+2})=\sqrt{x+1}\sqrt{6x+13}$$
$$\sqrt{x+7}+\sqrt{x+2}=\sqrt{6x+13}$$
$$2x+9+2\sqrt{(x+2)(x+7)}=6x+13$$
$$\sqrt{(x+2)(x+7)}=2x+2$$
$$(x+2)(x+7)=4x^2+8x+4$$
$$3x^2-x-10=0$$
$$(3x+5)(x-2)=0$$
So $x=-\frac{5}{3}$ or $x=2$. However, you find that only $x=2$ works by putting the value into the equation. Do note that while this method may be easier, another solution, $x=-1$ has not been found which would be found by squaring both sides: https://www.symbolab.com/solver/step-by-step/%5Csqrt%7Bx%5E%7B2%7D%2B8x%2B7%7D%2B%5Csqrt%7Bx%5E%7B2%7D%2B3x%2B2%7D%3D%5Csqrt%7B6x%5E%7B2%7D%2B19x%2B13%7D. While this may not be a complete answer, I hope this helped!
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Prove that if $\mathcal{B}_{1}\cup\mathcal{B}_{2}$ is a basis for $V$, then $V = W_{1}\oplus W_{2}$. (a) Let $W_{1}$ and $W_{2}$ be subspaces of a vector space $V$ such that $V = W_{1}\oplus W_{2}$. If $\mathcal{B}_{1}$ and $\mathcal{B}_{2}$ are bases for $W_{1}$ and $W_{2}$, respectively, show that $\mathcal{B}_{1}\cap\mathcal{B}_{2} = \varnothing$ and $\mathcal{B}_{1}\cup\mathcal{B}_{2}$ is a basis for $V$.
(b) Conversely, let $\mathcal{B}_{1}$ and $\mathcal{B}_{2}$ be disjoint bases for subspaces $W_{1}$ and $W_{2}$, respectively, of a vector space $V$. Prove that if $\mathcal{B}_{1}\cup\mathcal{B}_{2}$ is a basis for $V$, then $V = W_{1}\oplus W_{2}$.
MY ATTEMPT
(a) Let $\mathcal{B}_{1} = \{\alpha_{1},\alpha_{2},\ldots,\alpha_{m}\}$ and $\mathcal{B}_{2} = \{\beta_{1},\beta_{2},\ldots,\beta_{n}\}$ where $\dim W_{1} = m$ and $\dim W_{2} = n$.
Let $v\in V = W_{1}\oplus W_{2}$. Then $v = w_{1} + w_{2}$ where $w_{1}\in W_{1}$ and $w_{2}\in W_{2}$.
Consequently, there are scalars $a_{1},a_{2},\ldots,a_{m}$ and $b_{1},b_{2},\ldots,b_{n}$ such that
\begin{align*}
\begin{cases}
w_{1} = a_{1}\alpha_{1} + a_{2}\alpha_{2} + \ldots + a_{m}\alpha_{m}\\\\
w_{2} = b_{1}\beta_{1} + b_{2}\beta_{2} + \ldots + b_{n}\beta_{n}
\end{cases}
\end{align*}
Thence we conclude that
\begin{align*}
v = w_{1} + w_{2} = a_{1}\alpha_{1} + a_{2}\alpha_{2} + \ldots + a_{m}\alpha_{m} + b_{1}\beta_{1} + b_{2}\beta_{2} + \ldots + b_{n}\beta_{n}
\end{align*}
Thus $\mathcal{B}_{1}\cup\mathcal{B}_{2}$ spans $V$. Besides that, $\mathcal{B}_{1}\cap\mathcal{B}_{2} = \varnothing$. Indeed, if it were not the case, we would have $b\in\mathcal{B}_{1}\cap\mathcal{B}_{2}\subseteq W_{1}\cap W_{2}$ such that $b\neq 0$, which contradicts the fact that $W_{1}\cap W_{2} = \{0\}$.
Finally, let us prove that $\mathcal{B}_{1}\cup\mathcal{B}_{2}$ is LI. Indeed, if
\begin{align*}
c_{1}\alpha_{1} + c_{2}\alpha_{2} + \ldots + c_{m}\alpha_{m} + d_{1}\beta_{1} + d_{2}\beta_{2} + \ldots + d_{n}\beta_{n} = 0
\end{align*}
then we should have
\begin{align*}
c_{1}\alpha_{1} + c_{2}\alpha_{2} + \ldots + c_{m}\alpha_{m} = -d_{1}\beta_{1} - d_{2}\beta_{2} - \ldots - d_{n}\beta_{n}
\end{align*}
which implies that
\begin{align*}
c_{1}\alpha_{1} + c_{2}\alpha_{2} + \ldots + c_{m}\alpha_{m}\in W_{1}\cap W_{2} = \{0\}
\end{align*}
whence we conclude that $c_{1} = c_{2} = \ldots = c_{m} = 0$. Similar reasoning shows that $d_{1} = d_{2} = \ldots = d_{n} = 0$, and the result holds.
(b) Based on the same notation as previously established, let $v\in V$. According to the given assumptions, there are scalars $a_{1},a_{2},\ldots,a_{m}$ as well as $b_{1},b_{2},\ldots,b_{n}$ such that
\begin{align*}
v = a_{1}\alpha_{1} + a_{2}\alpha_{2} + \ldots + a_{m}\alpha_{m} + b_{1}\beta_{1} + b_{2}\beta_{2} + \ldots + b_{n}\beta_{n} = w_{1} + w_{2}
\end{align*}
where $w_{1}\in W_{1}$ and $w_{2}\in W_{2}$. Thus $V = W_{1}+W_{2}$. It remains to prove that $W_{1}\cap W_{2} = \{0\}$.
Let us assume that $w\in W_{1}\cap W_{2}$. Then we conclude that
\begin{align*}
w = c_{1}\alpha_{1} + c_{2}\alpha_{2} + \ldots + c_{m}\alpha_{m} = d_{1}\beta_{1} + d_{2}\beta_{2} + \ldots + d_{n}\beta_{n}
\end{align*}
Rearranging this relation, it results that
\begin{align*}
c_{1}\alpha_{1} + c_{2}\alpha_{2} + \ldots + c_{m}\alpha_{m} - d_{1}\beta_{1} - d_{2}\beta_{2} - \ldots - d_{n}\beta_{n} = 0
\end{align*}
thence $a_{1} = a_{2} = \ldots = a_{m} = b_{1} = b_{2} = \ldots = b_{n} = 0$, and we are done.
Are the provided proofs correct? Is there a neater way to rephrase my arguments? Any comments are appreciated.
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I prefer denoting a basis as a list and not a set.
(a)
Let $B_{1}= (v_{i}:i\in I), B_{2}= (u_{j}:j\in J)$ (where $I, J$ are arbitrary index sets).
By hypothesis $V = W_{1} + W_{2}, W_{1}\cap W_{2}=\{0\}$. Suppose $B_{1}\cap B_{2} \neq \emptyset,$ then $\exists x\in B_{1}\cap B_{2},$ then $x\neq 0$ (since $B_{1}, B_{2}$ are l.i.). Thus $W_{1}\cap W_{2}\neq \{0\}$.
Additionally, we have $B_{1}\cup B_{2}=(v_{i}, u_{j}: i\in I, j\in J)$, then
let's consider an arbitrary finite null combination of the elements of $B_{1}\cup B_{2}$ as $\sum\alpha_{i}v_{i} + \sum\beta_{j}u_{j} = 0,$ then by hypothesis $\sum\alpha_{i}v_{i} = \sum\beta_{j}u_{j} = 0$. So, $\alpha_{i} = \beta_{j}=0, \forall i,j$. This proves the linear independence of $B_{1}\cup B_{2}=(v_{i}, u_{j}: i\in I, j\in J)$. And it is trivial the spanning property.
(b)
Let $v\in V$, then $v = \sum_{finite}\alpha_{i}v_{i} + \sum_{finite}\beta_{j}u_{j}$ (since $B_{1}\cup B_{2}=(v_{i}, u_{j}: i\in I, j\in J)$ is basis of $V$). Thus $v\in W_{1} + W_{2}$.
Suppose $\exists x\in W_{1}\cap W_{2}: x\neq 0$, then $x = \sum_{finite}\alpha_{i}v_{i} = \sum_{finite}\beta_{j}u_{j},$ for some $\alpha_{i}'s,\beta_{j}'s$. Then $\sum_{finite}\alpha_{i}v_{i} + \sum_{finite}-\beta_{j}u_{j} = 0$, thus $\alpha_{i} = \beta_{j}=0, \forall i,j$ which proves $W_{1}\cap W_{2} = \{0\}$
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A Systematic way to solve absolute value inequalities? So, I had to solve this problem: $\left\vert \dfrac{x^2-5x+4}{x^2-4}\right\vert \leq 1$
I factored it in the form: $\left\vert \dfrac{(x-4)(x-1)}{(x-2)(x+2)} \right\vert \leq 1$.
After that I found the intervals in which the expression is positive: $x \in(-\infty, -2) \cup [1,2)\cup[4, \infty) $. And when expression is negative: $x \in (-2,1) \cup(2,4) $.
So, then I went ahead and solved the equations, when the expression is positive and when it's negative. The answer I got for positive is: $ x \in (-2,8/5] \cup(2, \infty).$ I'm not sure how to proceed from here, I figured that I'd find the intersection of the answer I got (when solved for positive expression) and when x is positive. But that doesn't seem to be the correct way to proceed. Since my answer is way off.
I've tried searching how to solve these kinds of problems, but couldn't find anything close to this. So I'd really like to know how to solve this problem and how to approach problems like these in general. I really lack intuition and would like to understand it more. I'd also appreciate any textbook / information about how to solve more complex problems in general.
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I can think of two methods to solve these kinds of equations, so I'll write them both down and you can see which you prefer. The first is a lot more algebraic, the second you can argue is more geometric. It may also not fly in your class, depending on the syllabus and professor.
First, we have $-1\leq\frac{x^2-5x+4}{x^2-4}\leq 1$, so we've gotten rid of the absolute value. If this was an equation rather than two inequalities we'd definitely want to multiply by the denominator, and here is no different. The problem is we don't know if the denominator is positive or negative. So we first assume $x^2-4=(x-2)(x+2)>0$. Then multiplying by the denominator gives us
\begin{align*}
&\left\{\begin{array}{ll}-x^2+4\leq x^2-5x+4,\\ x^2-5x+4\leq x^2-4\end{array}\right. \\
\Rightarrow &\left\{\begin{array}{ll}0\leq 2x^2-5x,\\ 0\leq 5x-8\end{array}\right. \\
\Rightarrow &\left\{\begin{array}{ll}0\leq x(2x-5),\\ 0\leq 5x-8.\end{array}\right. \\
\end{align*}
So if $x^2-4=(x-2)(x+2)>0$, we need these equations to be true as well. Now looking at the intersection of the sets that cause each of the equations to be true, we have that the original absolute inequality is true in the following set, where I use curly braces rather than parentheses to show order:
\begin{align*}
&\{(-\infty,-2)\cup(2,\infty)\}\bigcap\left\{(-\infty,0]\cup\left[\frac{5}{2},\infty\right)\right\}\bigcap \left[\frac{8}{5},\infty\right) \\
=&\left\{(-\infty,-2)\bigcap\left\{(-\infty,0]\cup\left[\frac{5}{2},\infty\right)\right\}\bigcap \left[\frac{8}{5},\infty\right)\right\}\bigcup\left\{(2,\infty)\bigcap\left\{(-\infty,0]\cup\left[\frac{5}{2},\infty\right)\right\}\bigcap \left[\frac{8}{5},\infty\right)\right\} \\
=&\emptyset\bigcup \left[\frac{5}{2},\infty\right) \\
=&\left[\frac{5}{2},\infty\right).
\end{align*}
If you repeat the above steps yourself, assuming instead that $x^2-4<0$, you should get that $x$ needs to be in the set $\left[0,\frac{8}{5}\right]$. So our overall answer is $\left[0,\frac{8}{5}\right]\cup\left[\frac{5}{2},\infty\right)$.
I've made one assumption here which is that you know how to solve inequalities like $0<(x-2)(x+2)$, let me know if that's not the case and we can go through that as well.
Now, the second method. Your function inside the absolute value is continuous everywhere except the asymptotes, which is where the denominator is equal to 0. So lets first look at the boundaries of the set where the equation is satisfied.
\begin{align*}
&\left\vert\frac{x^2-5x+4}{x^2-4}\right\vert=1 \\
\Rightarrow&\frac{x^2-5x+4}{x^2-4}=\pm 1. \\
&\left\{\begin{array}{ll}\frac{x^2-5x+4}{x^2-4}=1\\ \frac{x^2-5x+4}{x^2-4}=-1\end{array}\right. \\
\Rightarrow&\left\{\begin{array}{ll}x^2-5x+4=x^2-4\\ x^2-5x+4=-x^2+4\end{array}\right. \\
\Rightarrow&\left\{\begin{array}{ll}-5x+8=0\\ 2x^2-5x=0\end{array}\right. \\
\Rightarrow&\left\{\begin{array}{ll}-5x+8=0\\ x(2x-5)=0\end{array}\right. \\
\Rightarrow &x=0,\frac{8}{5},\frac{5}{2}.
\end{align*}
At these points, the function $\left\vert\frac{x^2-5x+4}{x^2-4}\right\vert$ potentially switches from being less than 1 to being greater than 1, or from being greater than 1 to being less than 1. This can also happen when the denominator is equal to 0, which is when $x=-2,2$, as the function can switch from being close to $-\infty$ to being close to $+\infty$ for example. These values together are then the only times the inequality $\left\vert\frac{x^2-5x+4}{x^2-4}\right\vert\leq 1$ potentially changes from being true to not true. Therefore in between these points, if the inequality is true (or false) for one value it's true (or false) for all values. So we just need to pick values in the intervals \begin{align*}
(-\infty,-2),(-2,0],\left(0,\frac{8}{5}\right],\left(\frac{8}{5},2\right),\left(2,\frac{5}{2}\right],\left(\frac{5}{2},\infty\right)
\end{align*}
to see if the inequality is true in the interval. Note when I use open and closed ends in the intervals, it doesn't matter in which interval you include $x=\frac{5}{2}$, but you don't include $x=2$ in either interval because the function is not defined there.
So for $(-\infty,-2)$ let $x=-3$. Then $\left\vert\frac{x^2-5x+4}{x^2-4}\right\vert=\left\vert\frac{9+15+4}{9-4}\right\vert=\left\vert\frac{28}{5}\right\vert=\frac{28}{5}> 1$. So the inequality is false when $x=-3$ and so is false in the entire interval $(-\infty,-2)$. Repeat for the other intervals to get the same answers as before.
|
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|
Determining $\arg(u-\sqrt{3})$ in exact form. Let $u$ be the solution to $z^5=-9\sqrt{3}i$ so that $\frac{\pi}{2}\le arg(u) \le \pi$.
Determine $\arg(u-\sqrt{3})$ in exact form.
How would I go about completing this question?
|
Let $z=r(\cos\theta+i\sin\theta)$. We know $-9i\sqrt3=9\sqrt3(\cos-\frac{\pi}{2}+i\sin-\frac{\pi}{2} )$
This means (using De Moivre's Theorem)
$$r^{5}(\cos5\theta+i\sin5\theta)=9\sqrt3(\cos-\frac{\pi}{2}+i\sin-\frac{\pi}{2})=9\sqrt3(\cos(-\frac{\pi}{2}+2k\pi)+i\sin(-\frac{\pi}{2}+2k\pi)$$
So $r^{5}=9\sqrt3$ so $r=\sqrt3$.
We also have $5\theta=-\frac{\pi}{2}+2k\pi$ so $\theta=-\frac{\pi}{10}+\frac{2k\pi}{5}$
For different values of $k$:
$$k=0: \theta=-\frac{\pi}{10}$$
$$k=1: \theta=-\frac{\pi}{10}+\frac{2\pi}{5}=\frac{3\pi}{10}$$
$$k=-1: \theta=-\frac{\pi}{10}-\frac{2\pi}{5}=-\frac{\pi}{2}$$
$$k=2: \theta=-\frac{\pi}{10}+\frac{4\pi}{5}=\frac{7\pi}{10}$$
$$k=-2: \theta=-\frac{\pi}{10}-\frac{4\pi}{5}=-\frac{9\pi}{10}$$
Ignore all values of $\theta<\frac{\pi}{2}$ as this is what you specified. This leaves us with : $$\theta= \frac{7\pi}{10}$$
So (using Euler's relation, that $e^{i\theta}=\cos\theta+i\sin\theta)$:
$$ u=\sqrt3 e^{\frac{7i\pi}{10}} $$
Can you do it from there?
Hope that helped :)
|
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|
Value of $\alpha$ for which $x^5+5\lambda x^4-x^3+(\lambda\alpha-4)x^2-(8\lambda+3)x+\lambda\alpha-2=0$ has roots independent of $\lambda$
Consider the equation $$x^5 + 5\lambda x^4 -x^3 + (\lambda \alpha -4)x^2 - (8\lambda +3)x + \lambda\alpha - 2 = 0$$ The value of $\alpha$ for which the roots of the equation are independent of $\lambda$ is _______
My approach: The equation can be rewritten as:
$$\underbrace{(x-2)(x^4 + 2x^3 + 3x^2 + 2x + 1)}_{f(x)} + \lambda\underbrace{(5x^4 + \alpha x^2 -8x + \alpha)}_{g(x)} = 0$$
For this equation to be valid independent of $\lambda$, $f(x) = g(x) = 0$. $f(x)$ has $2$ as one of it's roots. Solving $g(2) = 0$, the value of $\alpha$ comes out to be
$$\alpha = -\frac{64}{5}$$
which is unfortunately not the correct answer. Where is my approach breaking down?
|
The problem with your answer is that it's true that $2$ would be a root independent on $\lambda$ but there may be other roots which depends on $\lambda$.
Indeed with $\alpha=-\frac{64}{5}$ your equation becomes
$$\frac{1}{5}(x-2)(5(1+x+x^2)^2+\lambda(5x+8)(4+x(2+5x))$$
which may have solutions different from $2$ and $\lambda$-dependent.
However I don't see how could you get all the solution independent of $\lambda$, unless you allow $\alpha$ to be $x$-dependent.
Indeed I would say that the only way to get all roots independent of $\lambda$ is to ask $g(x) = 0$ or $g(x) = f(x)$. Ando both require a dependence of $\alpha$ on $x$.
For $g(x) = 0$ you must have
$$\alpha = -\frac{5 x^4 - 8x}{1+x^2}\,.$$
For $g(x)=f(x)$ things look even worse...
$$\alpha = \frac{-2 + 5 x - 4 x^2 - x^3 - 5 x^4 + x^5}{1 + x^2},.$$
|
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|
The number of critical points of the function $f(x,y)=(x^2+3y^2)e^{-(x^2+y^2)}$ is
The number of critical points of the function $$f(x,y)=(x^2+3y^2)e^{-(x^2+y^2)}$$ is _____.
My attempt:
$f_x=-(2x^3+(6y^2-2)x)e^{-(x^2+y^2)}$ and $f_y=-(6y^3+(2x^2-6)y)e^{-(x^2+y^2)}$
Clearly, $(0,0)$ is a critical point since at $(0,0)$ we have: $f_x=0, f_y=0$. Are there anymore, if yes how to find them?
|
Clearly, you can factor:
\begin{align}
f'x&=-2x(x^2+3y^2-1)\mathrm e^{-(x^2+y^2)} \\
f'y&=-2y(x^2+3y^2-3)\mathrm e^{-(x^2+y^2)}
\end{align}
and observe that $x^2+3y^2-1$ and $x^2+3y^2-3$ can also be $0$, but not simultaneously. Therefore you also have the solutions of the systems
$$
\begin{cases}
x=0\\x^2+3y^2-3=0
\end{cases}\quad\text{ and }\quad
\begin{cases}
y=0\\x^2+3y^2-1=0
\end{cases}
$$
|
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What is the solution of this summation? $$S(x) = \frac{x^4}{3(0)!} + \frac{x^5}{4(1)!}+\frac{x^6}{5(2)!}+.....$$
If the first term was $$x^3$$ and the next terms were $$x^{3+i}$$ then differentiating it would have given $$x^2.e^x$$ and then it was possible to integrate it. But how to solve this one?
|
Your idea is right, just let $x^3$ appear and differentiate
$$\left(\frac{S(x)}x\right)'=\left(\frac{x^3}{3\cdot0!}+\frac{x^4}{4\cdot1!}+\frac{x^5}{5\cdot2!}+\cdots\right)'=\frac{x^2}{0!}+\frac{x^3}{1!}+\frac{x^4}{2!}+\cdots=x^2e^x.$$
Then by integration,
$$\color{green}{\frac{S(x)}x=(x^2-2x+2)e^x-2}$$ (the constant of integration is drawn from $\left.\dfrac{S(x)}x\right|_{x\to0}=0$).
|
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|
Integrate $\int_0^{\frac{\pi}{2}} \frac{dx}{{\left(\sqrt{\sin{x}}+\sqrt{\cos{x}}\right)}^4} $ I found a challenge problem and am confused$$\int_0^{\frac{\pi}{2}} \frac{dx}{{\left(\sqrt{\sin{x}}+\sqrt{\cos{x}}\right)}^4} $$
$u=\frac{\pi}{2}-x$ is no good and square or 4th power the denominator does not help? Suggestion?
|
\begin{align}
\int_0^{\frac{\pi}{2}} \frac{1}{{\left(\sqrt{\sin{x}}+\sqrt{\cos{x}}\right)}^4}{\rm d}x &= \int_0^{\pi/2}\dfrac{\sec^2 x}{(\sqrt{\tan x} + 1)^4}{\rm d}x.
\end{align}
Denote the upper integral by $I$.
Put $u = \tan x$ to get
$$I = \int_0^\infty \dfrac{1}{(1 + \sqrt u)^4}{\rm d}u.$$
Put $u = t^2$ to get
\begin{align}
I &= \int_0^\infty\dfrac{2t}{(1 + t)^4}{\rm d}t.
\end{align}
Puting $v = t+1$ to get
\begin{align}
I &= \int_1^\infty \dfrac{2(v - 1)}{v^4}{\rm d}v\\
&= 2\int_1^\infty \left(\dfrac{1}{v^3} - \dfrac{1}{v^4}\right){\rm d}v\\
&= 2\left(\dfrac{1}{2} - \dfrac{1}{3}\right)\\
&= \boxed{\dfrac{1}{3}}.
\end{align}
|
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|
Evaluation of integral $\int_{S^2} \frac{dS}{((x-a)^2 +y^2+z^2)^{1/2}}$, where $a>1$ and $S$ is the unit sphere. I want to evaluate$$\int_{S^2} \frac{dS}{((x-a)^2 +y^2+z^2)^{1/2}}$$
where $a >1$ and $S$ is the unit sphere.
I'm not sure how to do this using only multivariable calculus techniques. My only idea was to use the fact that the function $(x^2+y^2+z^2)^{-1/2}$ is harmonic away from the origin and use the mean value formula.
|
We can use a few symmetry conditions to make our lives easier. Note that your proposed integral is equivalent to
$$\iint_{(x+a)^2+y^2+z^2=1} \frac{1}{\sqrt{x^2+y^2+z^2}}\:dS = \iint_{(x+a)^2+y^2+z^2=1} \frac{(x+a,y,z)\cdot(x+a,y,z)}{\sqrt{x^2+y^2+z^2}}\:dS$$
Thus we can use the divergence theorem
$$= \iint_{(x+a)^2+y^2+z^2\leq 1} \frac{2}{\sqrt{x^2+y^2+z^2}}-\frac{ax}{(x^2+y^2+z^2)^{\frac{3}{2}}}\:dV$$
In spherical coordinates centered around the $x$ axis, the boundary can be rewritten as
$$x^2+y^2+z^2+2ax+a^2-1 = 0 \implies \rho^2+2a\rho\cos\phi + a^2-1 = 0$$
The best way to set up the integral is with $\phi$ first like so:
$$\int_0^{2\pi} \int_{a-1}^{a+1} \int_{\cos^{-1}\left(\frac{1-a^2}{2a\rho}-\frac{\rho}{2a}\right)}^\pi 2\rho\sin\phi - a\cos\phi\sin\phi \:d\phi\:d\rho\:d\theta$$
$$ = 2\pi\int_{a-1}^{a+1} 2\rho + \frac{5-3a^2}{4a} - \frac{9\rho^2}{8a} -\frac{(1-a^2)^2}{8a\rho^2}\:d\rho$$
$$= 2\pi \left[4a + \frac{5-3a^2}{2a}-\frac{3}{8a}(6a^2+2) - \frac{a^2-1}{4a}\right] = \frac{4\pi}{a}$$
where you can see the answer was more easily retrieved by the Mean Value Property of harmonic functions (i.e. $\int_{S^2} f = 4\pi f(0)$ due to the symmetries in the problem), but this is a purely multivariable calculus integration way of proving the result without resorting to PDEs.
|
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|
For real values with $abc\neq0$, if $\frac{xb+(1-x)c}{a}=\frac{xc+(1-x)a}{b}=\frac{xa+(1-x)b}{c}$, show that $x^3 = -1$ and $a=b=c.$
Let $a,b,c$ and $x$ are real numbers such that $abc \neq 0$ and $$\frac {xb+(1-x)c} {a} = \frac {xc + (1-x)a}{b} = \frac {xa+(1-x)b}{c}.$$ Prove that $x^3=-1$ and $a=b=c.$
My attempt $:$ If $a+b+c \neq 0$ then $$\frac {xb+(1-x)c} {a} = \frac {xc + (1-x)a}{b} = \frac {xa+(1-x)b}{c} = \frac {\left [\{xb+(1-x)c\} + \{xc + (1-x)a\} + \{xa+(1-x)b\} \right]} {a+b+c} =1.$$ Therefore $$x = \frac {a-c}{b-c} = \frac {b-a} {c-a} = \frac {c-b} {a-b}.$$
Comparing the first two expressions of $x$ and simplifying we get \begin{align*} a^2+b^2+c^2 - ab - bc -ca & = 0 \\ \implies (a-b)^2+(b-c)^2 +(c-a)^2 & = 0. \end{align*} Therefore we have $a=b=c.$ But then $x$ would be an indeterminant form. Does it imply that $a+b+c = 0$? How to proceed further? Any help will be highly appreciated. Thank you very much.
|
$$bc((b-c)x+c)=ca((c-a)x+a)=ab((a-b)x+b)$$
has a solution in $x$ iff
$$\begin{vmatrix}bc(b-c)&bc^2&1\\ca(c-a)&ca^2&1\\ab(a-b)&ab^2&1\end{vmatrix}=-abc(3abc-a^3-b^3-c^3)\\=-abc(a+b+c)(a+\omega b+\omega^2c)(a+\omega^2b+\omega c)=0$$
where $\omega$ is a cube root of unity.
There are two cases:
*
*one of the complex factors vanishes when $a=b=c$, and $x$ is indeterminate;
*$a+b+c=0$ and the equations reduce to
$$2x=1.$$
The correct answer is
$$\color{green}{a=b=c\lor \left(a+b+c=0\land x=\frac12\right)}.$$
|
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How does the square root disappear when differentiating $y=\frac{\sqrt{2x^2}}{\cos x}$? Finding the derivative of $$y=\frac{\sqrt{2x^2}}{\cos x}$$ I am going through the steps and having trouble using the quotient rule. I have seen the final answer, and I've had no trouble using the quotient rule in the past, but this one is giving me trouble in terms of figuring out where all the fractions end up going. Below is what I have so far:
$$\frac{dy}{dx}=\frac{\dfrac{4x\cos x}{2\sqrt{2x^2}}+\sqrt{2x^2}\sin x}{(\cos x)^2}$$
I assume I can get rid of the 2 in the denominator in the first term in the numerator, leaving $\dfrac{2x\cos x}{\sqrt{2x^2}}$ and after seeing the final answer, I believe the denominator can be moved down so that the final answer's denominator is $\sqrt{2x^2}(\cos x)^2$. However, in the final answer the numerator is $2x\cos x +2x^2\sin x$. Where does the $2x^2$ come from? How could the square root have gone away? For reference, here is the final answer I'm supposed to get:
$$\frac{2x\cos x+2x^2\sin x}{\sqrt{2x^2}(\cos x)^2}$$
|
You have
$$\frac{\frac{4x\cos x}{2\sqrt{2x^2}}+\sqrt{2x^2}\sin x}{(\cos x)^2}$$
You correctly mentioned that this can be simplified to
$$\frac{\frac{2x\cos x}{\sqrt{2x^2}}+\sqrt{2x^2}\sin x}{(\cos x)^2}$$
To simplify this you can remove the compound fraction. To get rid of the fraction in the numerator, multiply numerator and denominator of the big fraction by the denominator of the small fraction. That is, multiply the big fraction by $\frac{\sqrt{2x^2}}{\sqrt{2x^2}}$, giving:
$$\frac{\frac{2x\cos x}{\sqrt{2x^2}}+\sqrt{2x^2}\sin x}{(\cos x)^2}\times\frac{\sqrt{2x^2}}{\sqrt{2x^2}}$$
The $\sqrt{2x^2}$ term in the denominator of the fraction in the numerator is cancelled, and $\sqrt{2x^2}\times\sqrt{2x^2}=2x^2$, this is how the square root disappears. So it simplifies to
$$\frac{2x\cos x+2x^2\sin x}{\sqrt{2x^2}(\cos x)^2}$$
as you mentioned.
|
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|
Second system of equations I've solved that system and recieved $y=\frac{1}{4}$, $x = -\frac{4}{5}$ but there are one more pair of $(x, y)$ in book and I don't know what I have to do to find its.
\begin{cases} 2x - 3xy + 4y=0 \\ x + 3xy -3x = 1 \end{cases}
In book there are two answers and the second answer is $(1, -2)$.
I've reached $y=\frac{1}{4}$, $x = -\frac{4}{5}$ through that way:
\begin{array}{lcl}2x - 3xy + 4y = 0\\2x - 3xy = -4y\\x(2 - 3y) = -4y\end{array}
2)
\begin{array}{lcl}x + 3xy - 3x = 1\\-2x + 3xy = 1\\-x(2 - 3y) = 1\end{array}
Get new system:
\begin{cases} x(2 - 3y) = -4y \\ -x(2 - 3y) = 1 \end{cases}
Then I divide first to second and get
\begin{array}{lcl}-1 = -4y\\y = \frac{1}{4}\end{array}
Next solve one of equation and get \begin{array}{lcl} x = -\frac{4}{5}\end{array}
But how to get $x = 1, y = -2$?
|
We have
$$2x-3xy+4y\quad=x+3xy-3x-1=0\implies 4x-6xy+4y+1=0$$
$$\implies\quad x = \frac{(4 y + 1)}{(6 y - 4)} \land 3 y\ne 2\qquad y = \frac{(4 x + 1)}{(6 x - 4)} \land 3 x\ne2$$
so the solution for one variable in terms of the other yields the same results in that $1)$ either independent variable can be any of an infinity of numbers except $\frac{2}{3}$ and, $2)$ both variables cannot be integers and the same time.
examples
$$x=\frac{4(1)+1}{6(1)-4}=\frac{5}{2}\rightarrow (\frac{5}{2},1)\land (1,\frac{5}{2})$$
$$y=\frac{4(2)+1}{6(2)-4}=\frac{9}{8}\rightarrow (2,\frac{9}{8})\land (\frac{9}{8},2)$$
|
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Proof for the general formula for $a^n+b^n$. Based on the following observations. That is
$$a+b = (a+b)^1 \\ a^2+b^2 = (a+b)^2-2ab \\ a^3+b^3 = (a+b)^3-3ab(a+b) \\ a^4+b^4= (a+b)^4-4ab(a+b)^2+2(ab)^2\\ a^5+b^5
= (a+b)^5 -5ab(a+b)^3+5(ab)^2(a+b)\\\vdots$$
I came to make the following conjecture as general formula.
$$ a^n +b^n =\sum_{k=0}^{n-1}(-1)^k \frac{n\Gamma(n-k)}{\Gamma(k+1)\Gamma(n-2k+1)}(a+b)^{n-2k}(ab)^k $$ where $\Gamma(.) $ is gamma function.
I tried up proving the result using binomial theorem $\displaystyle (a+b)^n=\sum_{r=0}^n a^{n-r}b^r$ for positive integers $a,b$ however, I didn't find any elegance in the work. So in the expect of some beautiful proofs, I wish to share general formula here.
Thank you
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You want to express $x^n+b^n$ in terms of their sum, or (for convenience) half-sum $s:=\dfrac{a+b}2$, and geometric mean $p:=\sqrt{ab}$.
We have
$$2as=a^2+ab=a^2+p^2,$$ giving
$$a=s\pm\sqrt{s^2-p^2},b=s\mp\sqrt{s^2-p^2}.$$
Now
$$a^n+b^n=\sum_{k=0}^n\binom nk\left(s^{n-k}(s^2-p^2)^{k/2}+(-1)^ks^{n-k}(s^2-p^2)^{k/2}\right)
\\=2\sum_{j=0}^{2j\le n}\binom n{2j}s^{n-2j}(s^2-p^2)^j.$$
This further expands as
$$2\sum_{j=0}^{2j\le n}\binom n{2j}s^{n-2j}(s^2-p^2)^j
=2\sum_{j=0}^{2j\le n}\binom n{2j}s^{n-2j}\sum_{i=0}^j\binom ji(-1)^is^{2(j-i)}p^{2i}
\\=2s^n\sum_{j=0}^{2j\le n}\sum_{i=0}^j(-1)^i\binom n{2j}\binom ji\left(\frac ps\right)^{2i}.$$
Some more work is needed to regroup the terms by equal $i$.
|
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Proving $(a+b+c)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)+3\ge 4\cdot \frac{a+b+c}{\sqrt[3]{abc}}$ for positive $a$, $b$, $c$
For $a,b,c>0$ Prove that $$(a+b+c)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)+3\ge 4\cdot \frac{a+b+c}{\sqrt[3]{abc}}$$
My attempt: By AM-GM we obtain $$\frac{a}{b}+\frac{a}{b}+\frac{b}{c}\ge 3\sqrt[3]{\frac{a^2}{bc}}=\frac{3a}{\sqrt[3]{abc}}$$
Thus $$\sum \frac{a+c}{b}\ge \frac{2(a+b+c)}{\sqrt[3]{abc}}$$
So it suffices to show that $$6\ge \frac{2(a+b+c)}{\sqrt[3]{abc}}\Leftrightarrow 3\sqrt[3]{abc}\ge a+b+c$$
Which is clearly wrong. :"(
Thank you very much.
|
Suppose the inequality is true,then,
$$(a+b+c)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)+3\ge 4\cdot \frac{a+b+c}{\sqrt[3]{abc}}$$$$\Rightarrow \left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)+\frac3{a+b+c}\ge 4\cdot \frac{1}{\sqrt[3]{abc}}$$
$$\Rightarrow \left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)+\frac3{a+b+c}\ge 4\cdot \frac{1}{\sqrt[3]{abc}}\geq \frac{12}{a+b+c}$$
$$\Rightarrow \left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\geq \frac {9}{a+b+c} $$$$\Rightarrow \frac {a+b+c}{3}\geq\frac{3}{\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)}$$Which is obviously true ($AM\geq HM $).
|
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"timestamp": "2023-03-29T00:00:00",
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|
Proofs of the Reflection Rules I couldn't find a formal proof for the rule:
when a point $(a,b)$ is reflected along $y=x$, it becomes $(b,a)$.
I tried to prove it by sketching out the situation:
However, I still don't know how to prove that $b'=b, a'=a$.
Furthermore, I just want to make sure, for the following two rules:
*
*Reflection Across Y-Axis. $(x,y)\to(-x,y)$
*Reflection Across X-Axis. $(x,y)\to(x,-y)$.
Do they have formal proofs or do we just prove them by visualizing where a point ends up to be on a cartesian plane?
|
To find the coordinates of the reflected point $P'$, let us first find the intersection point of the line $y=x$ and the line perpendicular to that line and passing through the point $P=(a,b)$.
As we know, the equation of the line perpendicular to the line $y=x$ and passing through the point $P=(a,b)$ is$$y=-(x-a)+b.$$So, the intersection point can be obtained by solving the following system of equations as follows.$$\begin{cases} y=x \\ y=-(x-a)+b \end{cases} \quad \Rightarrow \quad M=\left ( \frac{a+b}{2}, \frac{a+b}{2} \right ).$$According to the definition of reflection, the point $M$ is the midpoint of the segment $\overline{PP'}$. So the reflected point $P'$ can be obtained by the following vector addition:$$\overrightarrow{OP'}=\overrightarrow{OP}+ 2 \overrightarrow{PM},$$where $O=(0,0)$ is the origin.
So, we need to do some vector algebra as follows.$$\overrightarrow{PM}=\left ( \frac{a+b}{2}, \frac{a+b}{2} \right ) - \left ( \vphantom{\frac{a}{b}} a,b \right )= \left ( \frac{b-a}{2}, \frac{a-b}{2} \right )$$$$\Rightarrow \quad \overrightarrow{OP'}= \left ( \vphantom{\frac{a}{b}} a,b \right )+ 2 \left ( \frac{b-a}{2}, \frac{a-b}{2} \right )=(b,a).$$Thus, the coordinates of the reflected point $P'$ is$$P'=(b,a).$$
Addendum
We can also find the coordinates of the reflected point by equating the distances of the points $P$ and $P'$ from the Point $M$ as follows (Please note that the point $P'$ lies on the line $y=-(x-a)+b$).$$d_{P',M}=d_{P,M}$$$$\Rightarrow \quad \sqrt{\left ( x- \frac{a+b}{2} \right )^2+ \left ( (-x+a+b) - \frac{a+b}{2} \right )^2}= \sqrt{ \left ( a- \frac{a+b}{2} \right )^2 + \left ( b - \frac{a+b}{2} \right )^2}$$$$ \Rightarrow \quad x=a \quad \text{ or } \quad x=b$$ $x=a$ corresponds to the point $P$. Thus, the coordinates of the reflected point $P'$ is$$P'=(b,a).$$
|
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|
$\frac{1}{1-x} = (1+x)(1+x^2)(1+x^4)(1+x^8)...$ How can I prove the identity $\frac{1}{1-x} = (1+x)(1+x^2)(1+x^4)(1+x^8)\ldots$ for $|x|<1$? I am preferably looking for a derivation rather than using the RHS. I have tried using binomial expansion, but it only seems to give the LHS back. I also tried taking the logarithm of $\frac1{1-x}$ on seeing a product and using the Taylor series of $\ln{(1+x)}$, but this appears to be a dead end.
|
Use $(P+Q)((P-Q)=P^2-Q^2$, repeatedly:
$$F=(1-x)(1+x)(1+x^2)(1+x^4)(1+x^8)......=(1-x^2)(1+x^2)(1+x^4)(1+x^8).....=(1-x^4)(1+x^4)(1+x^8)(1+x^{16})....=(1-x^8)(1+x^8)(1+x^{16})...(1+x^{2^n})=(1-x^{2^{n+1}})$$
When $|x|<1$ and $n \to \infty$, then $F =1$,
$$\implies \frac{1}{1-x}=(1+x)(1+x^2)(1+x^4)(1+x^8)....., |x|<1.$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3748821",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Continuity of $a^x+b$ with $a, b \in \mathbb R$ Let $a,b \in \mathbb{R}$ with $a > 0$. find $a$, $b$ so the function would be continuous
$$
f(x) = \begin{cases} a^x + b, & |x|<1 \\
x, & |x| \geq 1 \end{cases}
$$
I got $b = -a^x+x$ as my answer, but I'm unsure.
|
Just do the definitions. $x$, and $a^x + b$ are continuous so the the only possible point of discontinuity is as $|x| = 1$.
If $f$ is continuous at $x = 1$ then $\lim_{x\to 1^-} f(x) = \lim_{x\to 1^-}a^x + b = a+b$ must equal $f(1) = x|_1 = 1$ which must equal $\lim_{x\to 1^+} f(x) = \lim_{x\to 1^+} x = 1$. So we must have $a+b = 1$
If $f$ is continuous at $x =-1$ then $\lim_{x\to -1^+} f(x) = \lim_{x\to -1^+} a^x + b = \frac 1a + b$ must equal $f(-1) = x|_{-1} = -1$ which must equal $\lim_{x\to -1^-}f(x) = \lim_{x\to -1^-}x = -1$. So we must have $\frac 1a+b =-1$.
So $a+b =1$ and $\frac 1a + b = -1$. So $b= 1-a=-1-\frac 1a$ so $a \ne 0$ and $1-a=-1-\frac 1a$ so $a-a^2=-a-1$ so $a^2-2a-1=0$ so $a =\frac {2 \pm\sqrt{4+4}}2=1\pm \sqrt 2$.
So $b =1-(1\pm \sqrt 2)=\mp \sqrt 2$ [Note: $b= -1-\frac 1{1\pm\sqrt 2} = -1-\frac {1\mp \sqrt 2}{(1\pm \sqrt 2)(1\mp \sqrt 2)} = -1-\frac {1\mp \sqrt 2}{1-2}=-1+(1\mp \sqrt 2= \mp \sqrt 2$]
so we can have $a = 1+\sqrt 2$ and $b= -\sqrt 2$ (then $\lim_{x\to 1^-} f(x) = (1+\sqrt 2)^1 + (-\sqrt 2) = 1+\sqrt 2 -\sqrt 2 = 1$ and $\lim_{x\to -1^+} f(x) = \frac 1{1+\sqrt 2} -\sqrt 2 =\frac 1{1+\sqrt 2}\frac {1-\sqrt 2}{1-\sqrt 2} - \sqrt 2=\frac {1-\sqrt 2}{1-2} - \sqrt 2 = (-1+\sqrt 2)-\sqrt 2 = -1$)
or we can have $a=1-\sqrt 2$ and $b = \sqrt 2$ (then $\lim_{x\to 1^-} f(x) = (1-\sqrt 2)^1 + \sqrt 2 = 1$ and $\lim_{x\to -1^+} f(x) = \frac 1{1-\sqrt 2} +\sqrt 2 =\frac 1{1-\sqrt 2}\frac {1+\sqrt 2}{1+\sqrt 2} + \sqrt 2=\frac {1+\sqrt 2}{1-2} + \sqrt 2 = (-1-\sqrt 2)+\sqrt 2 = -1$)
|
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"timestamp": "2023-03-29T00:00:00",
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|
Find the value of $\frac{1}{\sin^3\alpha}-\frac{1}{\cos^3\alpha}$ given that $\sin\alpha-\cos\alpha=\frac12$ Given that $\sin\alpha-\cos\alpha=\frac12$. What is the value of $$\frac{1}{\sin^3\alpha}-\frac{1}{\cos^3\alpha}?$$
My work:
$$\sin\alpha-\cos\alpha=\frac12$$
$$\sin\alpha\frac1{\sqrt2}-\cos\alpha\frac1{\sqrt2}=\frac1{2\sqrt2}$$
$$\sin\left(\alpha-\frac{\pi}{4}\right)=\frac1{2\sqrt2}$$
$$\alpha-\frac{\pi}{4}=\sin^{-1}\left(\frac{1}{2\sqrt{2}}\right)$$
I calculated the value of $\sin^{-1}\left(\frac{1}{2\sqrt{2}}\right)\approx 20.705^\circ$,
so I got $\alpha\approx 45^\circ+20.705^\circ=65.705^\circ$
I calculated
$$\frac{1}{\sin^3\alpha}-\frac{1}{\cos^3\alpha}=\frac{1}{\sin^365.705^\circ}-\frac{1}{\cos^3 65.705^\circ}\approx -13.0373576$$
My question: Can I find the value of above trigonometric expression without using calculator? Please help me solve it by simpler method without solving for $\alpha$. Thanks
|
Hint:
$$\left(\dfrac12\right)^2=(\sin\alpha-\cos\alpha)^2=?$$
So, we know $\sin\alpha\cos\alpha=?$
$$\dfrac1{\sin\alpha}-\dfrac1{\cos\alpha}=\dfrac{?}{\sin\alpha\cos\alpha}=?$$
Finally use $$\left(\dfrac1{\sin\alpha}-\dfrac1{\cos\alpha}\right)^3=\dfrac1{\sin^3\alpha}-\dfrac1{\cos^3\alpha}-\dfrac3{\sin\alpha\cos\alpha}\left(\dfrac1{\sin\alpha}-\dfrac1{\cos\alpha}\right)$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3750756",
"timestamp": "2023-03-29T00:00:00",
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|
Generalized repetitions of letters with limited amount of adjacent letters Say I have the first $x$ letters of the alphabet, and I want to generate a sequence of length $y$, such that there are no more than $z$ adjacent repeated letters. For example, if $x = 2$, $y = 3$ and $z = 2$, here are all the valid sequences:
AAB ABA ABB BAA BAB BBA
How many total possible sequences are there?
Without the restriction of z, the question devolves to $x^y$, but I am stuck as to how to proceed further.
The case I have described earlier is also not too hard, I think it is $x^y-x$ since $z = y-1$.
Would it be better to count all cases and subtract (like I did for the $z = y-1$ case) or should I try and count upwards?
To be clear, you are allowed to repeat letters as many times as you want so long as they never lie next to each other more than $z$ times.
Thanks for the help!
|
The following answer is based upon the Goulden-Jackson Cluster Method which provides a convenient technique to solve problems of this kind. We consider the set of words of length $y\geq 0$ built from an alphabet $$\mathcal{V}=\{A_1,A_2,\ldots A_x\}$$ and the set $B=\{A_1^{z+1},A_2^{z+1},\ldots,A_x^{z+1}\}$ of bad words each of length $z+1$, which are not allowed to be part of the words we are looking for. We derive a generating function $f(t)$ with the coefficient of $t^y$ being the number of wanted words of length $y$.
According to the paper (p.7) the generating function $f(t)$ is
\begin{align*}
f(t)=\frac{1}{1-xt-\text{weight}(\mathcal{C})}\tag{1}
\end{align*}
with $x=|\mathcal{V}|$ the size of the alphabet and $\mathcal{C}$ the weight-numerator of bad words with
\begin{align*}
\text{weight}(\mathcal{C})=\text{weight}(\mathcal{C}[A_1^{z+1}])+\text{weight}(\mathcal{C}[A_2^{z+1}])
+\cdots+\text{weight}(\mathcal{C}[A_x^{z+1}])\tag{2}
\end{align*}
We calculate according to the paper for $1\leq j\leq x$:
\begin{align*}
\text{weight}(\mathcal{C}[A_j^{z+1}])&=-t^{z+1}-(t+t^2+\cdots+t^{z})\text{weight}(\mathcal{C}[A_j^{z+1}])\tag{3}\\
\end{align*}
where the long expression at the right-hand side of (3) accounts for overlappings of $A_{j}^{z+1}$ with itself. We obtain from (3):
\begin{align*}
\text{weight}(\mathcal{C}[A_j^{z+1}])&=-\frac{t^{z+1}}{1+t+\cdots t^{z}}\\
&=-\frac{t^{z+1}(1-t)}{1-t^{z+1}}\qquad\qquad 1\leq j \leq x\\
\end{align*}
It follows from (1) - (3):
\begin{align*}
\color{blue}{f(t)}&=\frac{1}{1-xt-\text{weight}(\mathcal{C})}\\
&=\frac{1}{1-xt+x\frac{t^{z+1}(1-t)}{1-t^{z+1}}}\\
&\,\,\color{blue}{=\frac{1-t^{z+1}}{1-xt-(1-x)t^{z+1}}}\tag{4}
\end{align*}
and conclude the number of valid words of length $y\geq 0$ is the coefficient of $t^y$ in (4). We use the coefficient of operator $[t^y]$ to denote the coefficient of $y^n$ of a series. Applying standard techniques to extract the coefficient $[t^y]$ from $f(t)$ we obtain
\begin{align*}
[y^t]f(t)&=\sum_{k=0}^{\left\lfloor\frac{y}{z}\right\rfloor}\binom{y-zk}{k}(1-x)^kx^{y-(z+1)k}\\
&\qquad- \sum_{k=0}^{\left\lfloor\frac{y-z-1}{z}\right\rfloor}\binom{y-z-1-zk}{k}(1-x)^kx^{y-z-1-(z+1)k}\tag{5}
\end{align*}
Example: $x=2,y=3,z=2$
We look at OPs example where the alphabet $\mathcal{V}=\{A,B\}$ consists of two characters. Invalid words are those containing substrings which are elements from the set of $\{AAA,BBB\}$ of bad words. According to (4) the generating function is given as
\begin{align*}
f(t)&=\left.\frac{1-t^{z+1}}{1-xt-(1-x)t^{z+1}}\right|_{x=2,z=2}\\
&=\frac{1-t^3}{1-2t+t^3}\\
&=1 + 2 t + 4 t^2 + 6t^3 + \color{blue}{10} t^4 + 16 t^5\\
&\qquad + 26 t^6 + 42t^7 + 68 t^8 +\cdots\\
\end{align*}
The last line was calculated with some help of Wolfram Alpha. Note, the coefficient $6$ of $t^{3}$ according to the six valid words stated by OP. The blue marked coefficient of $t^4$ is $10$ and the $2^4-10=6$ invalid words are
\begin{align*}
&\rm{\color{blue}{aaa}a}&\rm{a\color{blue}{bbb}}\\
&\rm{\color{blue}{aaa}b}&\rm{b\color{blue}{bbb}}\\
&\rm{b\color{blue}{aaa}}&\rm{\color{blue}{bbb}a}\\
\end{align*}
Applying (5) we find
\begin{align*}
\color{blue}{[t^3]f(t)}&=\sum_{k=0}^1\binom{3-2k}{k}(-1)^k2^{3-3k}-\sum_{k=0}^0\binom{0-2k}{k}(-1)^k2^{0-3k}\\
&=\binom{3}{0}(-1)^02^3+\binom{1}{1}(-1)^12^0-\binom{0}{0}(-1)^02^0\\
&=8-1+1\\
&\,\,\color{blue}{=6}\\
\\
\color{blue}{[t^4]f(t)}&=\sum_{k=0}^2\binom{4-2k}{k}(-1)^k2^{4-3k}-\sum_{k=0}^0\binom{0-2k}{k}(-1)^k2^{1-2k}\\
&=\binom{4}{0}(-1)^02^4+\binom{2}{1}(-1)^12^1+\binom{0}{2}(-1)^22^{-2}-\binom{0}{0}(-1)^02^1\\
&=16-4+0-2\\
&\,\,\color{blue}{=10}
\end{align*}
in accordance with the results above.
|
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|
Solve the following logarithmic equation over real numbers
Solve the equation:
$$\log_{2020} {(x^{10} + x^9 + x^8 + x^7 + x^6+ x^5+ x^2 )}=\log_2 x$$
over real numbers.
I found out that $x=2$ is a solution and I suspect is the only one, but cannot prove it.
|
Solve for $x$ in $\log_{2020} {(x^{10} + x^9 + x^8 + x^7 + x^6+ x^5+ x^2 )}=\log_2 x$.
$$\log_2{2020} =\log_x{(x^{10} + x^9 + x^8 + x^7 + x^6+ x^5+ x^2 )}$$
$x=2$ is a solution on checking.
To check if it is the only solution, let $t=\log_2 2020$ and consider
\begin{align*}
f(x)&=\ln(x^{10}+x^9+\cdots+x^2)-t\ln x\\
f^\prime(x)&=\frac{(10-t)x^{10}+(9-t)x^9+\cdots+(2-t)x^2}{x(x^{10}+x^9+\cdots+x^2)}<0\ \forall\ x\in[0,+\infty)&(\because 10<t)
\end{align*}
Since $f(x)$ is strictly decreasing, it cuts the $x-$axis at $x=2$ only.
|
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|
How can I integrate $\int \frac{u^3}{(u^2+1)^3}du?$ How to integrate following
$$\int\frac{u^3}{(u^2+1)^3}du\,?$$
What I did is here:
Used partial fractions
$$\dfrac{u^3}{(u^2+1)^3}=\dfrac{Au+B}{(u^2+1)}+\dfrac{Cu+D}{(u^2+1)^2}+\dfrac{Au+B}{(u^2+1)^3}$$
After solving I got
$A=0, B=0, C=1, D=0, E=-1, F=0$
$$\dfrac{u^3}{(u^2+1)^3}=\dfrac{u}{(u^2+1)^2}-\dfrac{u}{(u^2+1)^3}$$
Substitute $u^2+1=t$, $2u\ du=dt$, $u\ du=dt/2$
$$\int\frac{u^3}{(u^2+1)^3}du=\int \frac{dt/2}{t^2}-\int \frac{dt/2}{t^3}$$
$$=\frac12\dfrac{-1}{t}-\frac{1}{2}\dfrac{-1}{2t^2}$$
$$=-\dfrac{1}{2t}+\dfrac{1}{4t^2}$$
$$=-\dfrac{1}{2(u^2+1)}+\dfrac{1}{4(u^2+1)^2}+c$$
My question: Can I integrate this with suitable substitution? Thank you
|
One alternate method is thus: make the substitution $t=u^2$ which gives $dt = 2u\; du \iff du = dt/2 \sqrt t$. Then
$$\mathcal I := \int \frac{u^3}{(u^2+1)^3} du = \int \frac{t \sqrt t}{(t+1)^3} \cdot \frac{dt}{2 \sqrt t} = \frac 1 2\int\frac{t}{(t+1)^3}dt$$
Now let $w = t+1 \implies dw = dt$. Then
$$\mathcal I = \frac 1 2 \int \frac{w-1}{w^3}dw = \frac 1 2 \left( \int \frac 1 {w^2} dw - \int\frac{1}{w^3} dw \right)$$
I imagine you can finish things up from here.
|
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|
For $x≠y$ and $2005(x+y) = 1$; Show that $\frac{1}{xy} = 2005\left(\frac{1}{x} + \frac{1}{y}\right)$ Problem:
Let $x$ and $y$ two real numbers such that $x≠0$ ; $y≠0$ ; $x≠y$ and $2005(x+y) = 1$
*
*Show that $$\frac{1}{xy} = 2005\left(\frac{1}{x} + \frac{1}{y}\right)$$
*Calculate $l$:
$$l = \frac{y}{y-x} - \frac{y-x}{y} - \frac{x}{y-x} - \frac{y-x}{x} + \frac{y}{x} - \frac{x}{y} +2 $$
For the first question, I tried to work it out with algebra; I solved for x through the equation given, then multiplied it by y and I got the value of $\frac{1}{xy} = 2005\left(\frac{1}{y-2005y^2}\right) $. Then I tried proving that $\frac{1}{y-2005y^2} =\frac{1}{x} + \frac{1}{y} $ but I failed at this.
|
According to the problem, $x, y$ $\in$ $\mathbb{R}$ and $x \neq y$.
$\bullet$ For the first part,
(I) From the given, we have that
\begin{align*}
&2005(x + y) = 1\\
\implies & 2005 \cdot \frac{(x + y)}{xy} = \frac{1}{xy}\\
\implies & 2005 \cdot \bigg( \frac{1}{x} + \frac{1}{y} \bigg) = \frac{1}{xy}
\end{align*}
Hence, done!
$\bullet$ For the second part,
(II) According to question,
\begin{align*}
l =& ~\frac{y}{y-x} - \frac{y - x}{y} - \frac{x}{y - x} - \frac{y - x}{x} + \frac{y}{x} - \frac{x}{y} + 2\\
= & ~ \bigg[ \frac{y}{y-x} - \frac{x}{y - x} \bigg] - \bigg[ \frac{y - x}{y} + \frac{x}{y} \bigg] + \bigg[ \frac{y}{x} - \frac{y - x}{x} \bigg] + 2\\
= & ~ 1 - 1 + 1 + 2\\
= &~ 3
\end{align*}
Hence, the value of $l$ is $3$.
|
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|
Solve second order linear ODE if particular solution of the homogenous part is known Here's this ODE: $$x(x-1)y'' -(2x-1)y' + 2y = 2x^3 -3x^2$$
and $$y_1 =x^2$$
I know that I have to consider the homogenous part of the ODE first, which is $$x(x-1)y'' -(2x-1)y' + 2y = 0$$
If one solution is already known, then the second one can be calculated as:
$$y_2(x) = y_1(x)\int \frac{e^{-\int P(x)dx}}{y_1^2(x)}dx$$, after which the solution of the homogenous part is $$C_1y_1(x) + C_2y_2(x)$$
After I divided my equation by $x(x-1)$ I solved the first part and got $$C_1x^2 + C_2(-x+\frac{1}{2})$$
Now I need to find the rest of the solution (By the way, the full solution is: $$C_1x^2 + C_2(-x+\frac{1}{2}) +x^3 -\frac{x^2}{2}+x-\frac{1}{2}$$
I tried the variation of constants by forming the following system of equations:
$$C_1'(x)x^2 + C_2'(x)(-x+\frac{1}{2})=0$$
$$C_1'(x)2x - C_2'(x)=\frac{x(2x-3)}{x-1}$$
I used Cramer's method of solving the system and got that
$C_1' = \frac{-(2x-3)(-2x+1)}{2(x-1)^2}$ and
$C_2' = \frac{x^2(2x-3)}{(x-1)^2}$
Now if I integrate both I get:
$C_1 = 2x + \frac{1}{2(x-1)} -2$
and
$C_2 = \frac{x^3 -3x +3}{x-1}$
However, when I add these two $C_1$ and $C_2$ I don't get anything that resembles the $x^3 -\frac{x^2}{2}+x-\frac{1}{2}$ part of the solution.
Can anyone please help me with this? Have I correctly calculated $C_1$ and $C_2$? What do I do with them?
|
First, after considering $y_1 = c_1 x^2$ as a particular solution for the homogeneous
$$
x(x-1)y_h'' -(2x-1)y_h' + 2y_h = 0
$$
we have by constants variation (Lagrange)
$$
y_h = \left(\frac{c_2 (2 x-1)}{2 x^2}+c_3\right)x^2 = \frac{c_2}{2}(2x-1)+c_3 x^2
$$
now, using the Lagrange method again we propose for the complete ODE a particular solution with the structure
$$
y_p = \frac{c_2(x)}{2}(2x-1)+c_3(x) x^2
$$
and after substitution into
$$
x(x-1)y_p'' -(2x-1)y_p' + 2y_p = 2x^3 -3x^2
$$
we obtain
$$
\frac{1}{2} x (x (2 x-3)+1) c_2''(x)-\frac{c_2'(x)}{2}+(x-1) x^3 c_3''(x)+(2 x-3) x^2 c_3'(x)+(3-2 x) x^2 = 0
$$
as long as $c_2(x), c_3(x)$ are independent functions we can establish
$$
\cases{
\frac{1}{2} x (x (2 x-3)+1) c_2''(x)-\frac{c_2'(x)}{2}=0\\
(x-1) x^3 c_3''(x)+(2 x-3) x^2 c_3'(x)+(3-2 x) x^2 = 0
}
$$
We are looking for a particular solution then $c_2(x) = 0$ satisfies this condition. Regarding $c_3(x)$ we find
$$
c_3(x) = -\frac{3}{2 x^2}+x+\frac{3}{x}
$$
and finally
$$
y = y_h+y_p = \frac{c_2}{2}(2x-1)+c_3 x^2 + x^3+3 x-\frac{3}{2}
$$
The constants are $c_2, c_3$
|
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "2",
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|
Solving $\int \frac{x^3}{(4x^2 + 9)^\frac{3}{2}} dx$ The integral I must solve is this:
$$\begin{equation} \int \frac{x^3 }{(4x^2 + 9)^\frac{3}{2}} dx \tag{1}\end{equation}$$
Now I know that I'm supposed to convert the denominator into some form of $\sqrt{a^2 + x^2}$ in order to apply the substitution $x = a \ tan \ \theta$. Now I can change the form of the denominator to become easy enough to substitute as follows:
$$\bigg (4 \bigg (\ \frac{9}{4}\ +x^2\bigg)\bigg) ^\frac{3}{2} $$
Which makes it clear that $x$ needs to be substituted as $x = \frac{3}{2} \tan \theta $, and $dx = \frac{3}{2} \sec^2 \theta $
for later use. At this point I can represent $(1)$ in terms of my substituted trignometric function. The only problem comes with the denominator where I get stuck on the power. Here is how I went about solving it:
$$\bigg (4 \bigg (\ \frac{9}{4}\ +\left(\frac{3}{2} \tan \theta\right)^2 \bigg)\bigg) ^\frac{3}{2} $$
$$\bigg (4 \bigg (\ \frac{9}{4}\ +\frac{9}{4} \tan^2 \theta) \bigg)\bigg) ^\frac{3}{2} $$
$$\bigg (4 \frac{9}{4}\bigg (\ 1 + \tan^2 \theta) \bigg)\bigg) ^\frac{3}{2} $$
$$ 9^\frac{3}{2}\bigg( \ 1 + \tan^2 \theta \bigg) ^\frac{3}{2} $$
$$ 27\ ( \sec^2 \theta ) ^\frac{3}{2} $$
Now I have no idea how to evaluate this power of $sec$. The author says that it changes into $sec^3 \theta$ but I just can't fathom how that would go about. If what I understand is correct, the power it is raised to would be added to it's own making it $ \sec^\frac{7}{2} \theta$. My question is that how exactly is my reasoning wrong here?
|
Hint: Use substitution $\space 9+4x^2 = t^2. \space$ New integral shall be the integral of rational function...
|
{
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 7,
"answer_id": 2
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|
Evaluate $\lim_{x \to 0} \frac{\sqrt{1 + x\sin x} - \sqrt{\cos x}}{x\tan x}$ What I attempted thus far:
Multiply by conjugate
$$\lim_{x \to 0} \frac{\sqrt{1 + x\sin x} - \sqrt{\cos x}}{x\tan x} \cdot \frac{\sqrt{1 + x\sin x} + \sqrt{\cos x}}{\sqrt{1 + x\sin x} + \sqrt{\cos x}} = \lim_{x \to 0} \frac{1 + x\sin x - \cos x}{x\tan x \cdot(\sqrt{1 + x\sin x} + \sqrt{\cos x})}$$
From here I can’t see any useful direction to go in, if I even went in an useful direction in the first place, I have no idea.
|
@GregMartin's hint is to compute the numerator and denominator each to $O(x^2)$, respectively as $1+\tfrac12x^2-(1-\frac14x^2)=\tfrac34x^2$ and $x^2$, so the limit is $\tfrac34$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3758133",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 4
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|
in the triangle ABC on the AC side, points M and N are chosen such that ABM = MBN = NBC in the triangle ABC on the AC side, points M and N are chosen such that <ABM = <MBN = <NBC It turned out that NB = BC. On the side AB, a point K was marked such that BK = BM. Prove that AK + NC> AM.
I tried to get a triangle with sides AK NC and AM.
But I couldn't. So I don't know how to prove this
|
let $BC=a$ and $\measuredangle ABC=3\beta$.
Thus, $$\measuredangle ACB=90^{\circ}-\frac{\beta}{2},$$
$$\measuredangle ANC=90^{\circ}+\frac{\beta}{2},$$
$$\measuredangle AMB=90^{\circ}+\frac{3\beta}{2}$$ and
$$\measuredangle BAC=180^{\circ}-\left(90^{\circ}+\frac{5\beta}{2}\right),$$ which says
$$90^{\circ}+\frac{5\beta}{2}<180^{\circ},$$ which gives $$0^{\circ}<\beta<36^{\circ}.$$
Now, $$NC=2a\sin\frac{\beta}{2}$$ and by law of sines we can show that:
$$AM=\frac{a\cos\frac{\beta}{2}\sin\beta}{\cos\frac{5\beta}{2}\cos\frac{3\beta}{2}}$$ and
$$AK=AB-BK=\frac{a\cos\frac{\beta}{2}}{\cos\frac{5\beta}{2}}-\frac{a\cos\frac{\beta}{2}}{\cos\frac{3\beta}{2}}=\frac{a\sin\beta\sin2\beta}{\cos\frac{5\beta}{2}\cos\frac{3\beta}{2}}.$$
Id est, we need to prove that:
$$\frac{a\sin\beta\sin2\beta}{\cos\frac{5\beta}{2}\cos\frac{3\beta}{2}}+2a\sin\frac{\beta}{2}>\frac{a\cos\frac{\beta}{2}\sin\beta}{\cos\frac{5\beta}{2}\cos\frac{3\beta}{2}}$$ or
$$\frac{\cos\frac{\beta}{2}\sin2\beta}{\cos\frac{5\beta}{2}\cos\frac{3\beta}{2}}+1>\frac{\cos^2\frac{\beta}{2}}{\cos\frac{5\beta}{2}\cos\frac{3\beta}{2}}$$ or $$2\cos\frac{\beta}{2}\sin2\beta+\cos4\beta+\cos\beta>1+\cos\beta$$ or
$$\cos\frac{\beta}{2}\sin2\beta>2\sin^22\beta$$ or
$$1>4\sin\frac{\beta}{2}\cos\beta$$ or
$$1>4\sin\frac{\beta}{2}\left(1-2\sin^2\frac{\beta}{2}\right)$$ or
$$1-2\sin\frac{\beta}{2}>2\sin\frac{\beta}{2}\left(1-4\sin^2\frac{\beta}{2}\right)$$
$$1>2\sin\frac{\beta}{2}\left(1+2\sin\frac{\beta}{2}\right)$$ or
$$\sin\frac{\beta}{2}<\frac{\sqrt5-1}{4},$$ which we got by looking for domain of $\beta$.
|
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
if $f'(x)\ne0, g(x)>0 \forall x \in 0$, prove that $|f(x)| < 5$
if $f(x)$ is a twice differentiable real-valued function satisfying $f'(x)\ne0, f(0) = -3, f'(0) = 4$, such that $f(x) + f''(x) = -xg(x)f'(x), \space g(x) > 0$ for all $x > 0$. Prove that $|f(x)| \le 5$
There was a hint in the problem to first prove that $(f(x))^2 + (f'(x))^2$ is decreasing for all $x \in R^+$ and to use this result to prove the above statement. This relation can be easily proved since
$$\frac{d}{dx}((f(x))^2 + (f'(x))^2) = 2f'(x)(f(x) + f''(x)) \\
= -2xg(x)(f'(x))^2 \\
< 0 \space \space \forall \space x \in R^+$$
I could not use this result to prove that $|f(x)| < 5$. Any hints/solutions in solving this problem are appreciated!
|
$\boxed{\textit{Solution:}}~$We consider 2 cases.
$\bullet~$When $x > 0$.
as we know that $g(x) \geqslant 0$, for $x > 0$, we have for the interval $[0, t]$ for $t > 0$.
\begin{align*}
&f(x) + f''(x) = - x g(x) f'(x)\\
\implies & f(x) f'(x) + f'(x) f''(x) = - x g(x) \{ f(x) \}^{2} \leqslant 0 \quad [\text{ as } x \geqslant 0 \text{ } ] \\
\implies & \int_{0}^{t} f(x) f'(x) dx + \int_{0}^{t} f'(x) f''(x) dx \leqslant 0\\
\implies & \bigg[ \frac{[f(t)]^2}{2} - \frac{[f(0)]^2}{2} \bigg] + \bigg[ \frac{[f'(t)]^2}{2} - \frac{[f'(0)]^2}{2} \bigg] \leqslant 0\\
\implies & \bigg[ \frac{[f(t)]^2}{2} + \frac{[f'(t)]^2}{2} \bigg] \leqslant \bigg[ \frac{[f(0)]^2}{2} + \frac{[f'(0)]^2}{2} \bigg]\\
\implies &\frac{[f(t)]^2}{2} \leqslant \bigg[ \frac{[f(t)]^2}{2} + \frac{[f'(t)]^2}{2} \bigg] \leqslant \frac{25}{2} ~; [\text{ where } \bigg[ \frac{[f(0)]^2}{2} + \frac{[f'(0)]^2}{2} \bigg] = \frac{3^2}{2} + \frac{4^2}{2} = \frac{25}{2} > 0 \text{ }]\\
\implies & \frac{[f(t)]^2}{2} \leqslant \frac{25}{2}\\
\implies & {[f(t)]^2} \leqslant 25 \\
\implies & \lvert f(t) \rvert \leqslant \sqrt{25} = 5 \quad
%\text{ hence } 5 \text{ be the bound we require for all } t > 0
\end{align*}
$\bullet$ When $x < 0$.
as we know that $g(x) \geqslant 0$, for $x < 0$, we have for the interval $[t, 0]$ for $t < 0$.
\begin{align*}
&f(x) + f''(x) = - x g(x) f'(x)\\
\implies & f(x) f'(x) + f'(x) f''(x) = - x g(x) \{ f(x) \}^{2} \geqslant 0 \quad [\text{ as } x \leqslant 0 \text{ } ] \\
\implies & \int_{t}^{0} f(x) f'(x) dx + \int_{t}^{0} f'(x) f''(x) dx \geqslant 0\\
\implies & \bigg[ \frac{[f(0)]^2}{2} - \frac{[f(t)]^2}{2} \bigg] + \bigg[ \frac{[f'(0)]^2}{2} - \frac{[f'(t)]^2}{2} \bigg] \geqslant 0\\
\implies & \bigg[ \frac{[f(t)]^2}{2} + \frac{[f'(t)]^2}{2} \bigg] \leqslant \bigg[ \frac{[f(0)]^2}{2} + \frac{[f'(0)]^2}{2} \bigg]\\
\implies & \frac{[f(t)]^2}{2} \leqslant \bigg[ \frac{[f(t)]^2}{2} + \frac{[f'(t)]^2}{2} \bigg] \leqslant \frac{25}{2} ~; [\text{ where }\bigg[ \frac{[f(0)]^2}{2} + \frac{[f'(0)]^2}{2} \bigg] = \frac{3^2}{2} + \frac{4^2}{2} = \frac{25}{2} > 0 \text{ }]\\
\implies & \frac{[f(t)]^2}{2} \leqslant \frac{25}{2}\\
\implies & {[f(t)]^2} \leqslant 25\\
\implies & \lvert f(t) \rvert \leqslant \sqrt{25} = 5
%\quad \text{ hence } M_{2} \text{ be the bound we require for all } t < 0
\end{align*}
Therefore from the values of $f(0)$ and $f'(0)$ given we have obtained
\begin{align*}
\lvert f(x) \rvert \leqslant 5\ \quad \text{for all } x \in \mathbb{R}
\end{align*}
Hence we have proved that the function $f$ is bounded.
|
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"timestamp": "2023-03-29T00:00:00",
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|
How many roots does $(x+1)\cos x = x\sin x$ have in $(-2\pi,2\pi)$? So the nonlinear equation that I need to find the number of its roots is
$$(x+1)\cos x = x\sin x \qquad \text{with } x\in (-2\pi,2\pi)$$
Using the intermediate value theorem I know that the equation has at least one root on this interval, and if I use drawing I see that $x\sin x$ and $(x+1)\cos x$ intersect in three points, but from drawing I can't know if they might intersect again somewhere.
And the problem is that the number of zeroes is definitely not 3, the options are 4, 5, 6, 7 based on my textbook.
I tried the Fixed point method but $\{x\}$ didn't converge, either my starting point or the function I chose were inappropriate.
Can you help?
|
First we can manipulate the expression given:
$(x+1)\cos x= x\sin x$
divide both sides by $x\cos x$
Yielding: $(\frac{x+1}{x})= \tan x $
The Inverse tangent of both sides yields:
$\operatorname{arctan}(\frac{x+1}{x})=x$
We can state without requiring a proof, although can be proved, that:
$\frac{d}{dz} \operatorname{arctan}(z) = \frac{1}{1+z^2}$
In our example we could let
$z=\frac{x+1}{x}$
So:
$\frac{d}{dx}\operatorname{arctan}(z)= (\frac{d}{dz}\operatorname{arctan}(z)\cdot \frac{dz}{du})$
By the Chain Rule !
Right Hand Side (RHS) of the above expression yields
$\frac{1}{1+z^2} \cdot \frac{-1}{x^2}$
Substitute value of $z$ in terms of $x$ into the RHS to yield an overall equation:
$\frac{d}{dx}\operatorname{arctan}(z)= \frac{1}{1+(\frac{x+1}{x})^2}\cdot\frac{-1}{x^2}$
$= \frac{-1}{x^2+(x+1)^2}$
$ \therefore , \operatorname{arctan}(z)=\int \frac{-1}{x^2+(x+1)^2}$
and as: $\operatorname{arctan}(z)= x$. Then:
$x=\int \frac{-1}{x^2+(x+1)^2}$
Hopefully you can solve this integral now to find all the solutions for x
|
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|
Sum of squares and linear sum
For which positive integer $n$ can we write $n=a_1+a_2+\dots+a_k$ (for some unfixed $k$ and positive integers $a_1,a_2,\ldots,a_k$) such that $\sum_{i=1}^k a_i^2 = \sum_{i=1}^k a_i + 2\sum_{i<j}a_ia_j$?
When $k=1$, the equation is $a_1^2=a_1$, so $a_1=1$ and $n=1$ is the only possibility.
When $k=2$, we get $a_1^2+a_2^2 = a_1+a_2+2a_1a_2$, or $(a_1-a_2)^2 = a_1+a_2$, hence $n$ must be a perfect square. For perfect square $n=r^2$, we can solve $a_1+a_2=r^2$ and $a_1-a_2=r$, which must have a solution because $r^2\equiv r\pmod 2$. So all perfect squares work.
|
Here we prove an auxiliary claim for WhatsUp’s answer.
Let $S$ be the set of all pairs $(a,b)$ of natural numbers such that a system
$$\cases{
c_1 + 2c_2 + 3c_3 + 4c_4 + 5c_5 + 6c_6=x\\
c_1 + 4c_2 + 9c_3 + 16c_4 + 25c_5 + 36c_6=y}$$
has a solution in non-negative integers.
The set $S$ can be constructed recursively as follows. Put $(0,0)\in S$ and if $(x,y)\in S$ then put to $S$ pairs $(x+1,y+1)$, $(x+2,y+4)$, $(x+3,y+9)$, $(x+4,y+16)$, $(x+5,y+25)$, and $(x+6,y+36)$. If $(x,y)\in S$ then $x\le y \le 6x$. Moreover, $y-x=4c_2 + 6c_3 + 12c_4 + 20c_5 + 30c_6$, so $x$ and $y$ have the same parity.
Black points are pairs $(x,y)\in S$ for $0\le x, y\le 255$. Red points are pairs $(x,y)\not\in S$ for $x\le y \le 6x$ and $x,y$ of the same parity.
We claim that if $16\le x\le y\le 5x$ and $x$ and $y$ have the same parity then $(x,y)\in S$. We shall prove the claim by indution with respect to $x$. For $16\le x\le 20$ the claim can be checked directly, see the graph below. Assume that the claim is already proved for $x\ge 20$ and $x+1\le y\le 5(x+1)$. If $x+1\le y-20$ then $(x-4,y-25)\in S$ by the induction hypoteses and so $(x+1,y)=(x-4,y-25)+(5,25)\in S$ by the construction of $S$. If $x+1>y-20$ then $(x,y-1)\in S$ by the induction hypoteses and so $(x+1,y)=(x,y-1)+(1,1)\in S$ by the construction of $S$.
|
{
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"url": "https://math.stackexchange.com/questions/3760391",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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|
Integral: $\int \dfrac{dx}{(x^2-4x+13)^2}$?
How can I integrate $$\int \dfrac{dx}{(x^2-4x+13)^2}?$$
Here is my attempt:
$$\int \dfrac{dx}{(x^2-4x+13)^2}=\int \dfrac{dx}{((x-2)^2+9)^2}$$
Substitute $x-2=3\tan\theta$, $\ dx=3\sec^2\theta d\theta$
\begin{align*}
&=\int \dfrac{3\sec^2\theta d\theta}{(9\tan^2\theta+9)^2}\\
&=\int \dfrac{3\sec^2\theta d\theta}{81\sec^4\theta}\\
&=\dfrac{1}{27}\int \cos^2\theta d\theta\\
&=\dfrac{1}{27}\int \frac{1+\cos2\theta}{2} d\theta\\
&=\dfrac{1}{54}\left(\theta+\frac{\sin2\theta}{2}\right)+C
\end{align*}
This is where I got stuck. How can I get the answer in terms of $x$?
Can I solve it by other methods?
|
By setting
$$
\frac{1}{\left(x^2-4 x+13\right)^2}=\frac{A (2 x-4)+B}{x^2-4 x+13}+\frac{d}{dx}\left(\frac{C x+D}{x^2-4 x+13}\right)
$$
you get
\begin{align}
A &= 0,\\
B &= \frac{1}{18},\\
C &= \frac{1}{18},\\
D &= -\frac{1}{9}
\end{align}
so that
\begin{align}
\int\frac{1}{\left(x^2-4 x+13\right)^2}dx
&= \frac{1}{18}\int\frac{1}{x^2-4 x+13}dx+\frac{x-2}{18(x^2-4 x+13)}=\\
&= \frac{1}{54} \arctan\left(\frac{x-2}{3}\right)+\frac{x-2}{18(x^2-4x+13)}+c
\end{align}
|
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"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 9,
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|
Find the value of $\lim _{a \to \infty} \frac{1}{a} \int_{0}^{\infty} \frac{x^{2}+a x+1}{1+x^{4}} \cdot \tan ^{-1}\left(\frac{1}{x}\right) \,d x $ Find the value of : $$
\lim _{a \rightarrow \infty} \frac{1}{a} \int_{0}^{\infty} \frac{x^{2}+a x+1}{1+x^{4}} \cdot \tan ^{-1}\left(\frac{1}{x}\right) \,d x
$$ I have tried to evaluate this integral by L'Hospitals rule, by separately diffrentiating the numerator and the denominator in the following steps: $$\int_{0}^{\infty} \frac{x^{2}+a x+1}{1+x^{4}} d x\cdot\dfrac{\pi}{2} (\text{by taking x=1/t and adding the integrands)}
$$ I end up converting the above integral in the form $(\text{by Leibnitz's integral rule})$ $$ \dfrac{\pi}{2}
\left(\int_{0}^{\infty} \frac{x}{1+x^{4}} d x\right)
$$ please provide an approach to this problem after this step.
|
Your solution attempt honestly works just fine as well (and it's also the first thing I thought of when I saw the question!). Indeed from ${x=\frac{1}{t}}$ you get
$${\int_{0}^{\infty}\frac{x^2 + ax + 1}{1+x^4}\arctan\left(\frac{1}{x}\right)dx=\int_{0}^{\infty}\frac{\left(\frac{1}{t}\right)^2 + \frac{a}{t} + 1}{1 + \left(\frac{1}{t}\right)^4}\arctan(t)\frac{1}{t^2}dt}$$
$${=\int_{0}^{\infty}\frac{\left(\frac{1}{t}\right)^4 + \frac{a}{t^3} + \frac{1}{t^2}}{1 + \left(\frac{1}{t}\right)^4}\arctan(t)dt=\int_{0}^{\infty}\frac{1 + at + t^2}{t^4 + 1}\arctan(t)dt}$$
And so
$${2I = \int_{0}^{\infty}\frac{x^2 + ax + 1}{1 + x^4}\left(\arctan(x) + \arctan\left(\frac{1}{x}\right)\right)dx=\frac{\pi}{2}\int_{0}^{\infty}\frac{x^2 + ax + 1}{1 + x^4}dx}$$
So
$${I = \frac{\pi}{4}\int_{0}^{\infty}\frac{x^2 + ax + 1}{x^4 + 1}dx}$$
Now, we don't actually need to use L'hopitals rule. Just notice that
$${\frac{I}{a} = \frac{1}{a}\int_{0}^{\infty}\frac{\frac{\pi}{4}x^2}{x^4 + 1}dx + \frac{\pi}{4}\int_{0}^{\infty}\frac{x}{x^4 + 1}dx + \frac{1}{a}\int_{0}^{\infty}\frac{\frac{\pi}{4}}{x^4 + 1}dx}$$
(We can split the integrals up like this because the integrals all converge separately). As ${a\rightarrow \infty}$, the left-most and right-most terms will obviously go to zero, and the only one that survives is the middle term. Hence we get that the limit is
$${=\frac{\pi}{4}\int_{0}^{\infty}\frac{x}{1+x^4}dx}$$
Now just make the substitution ${u=x^2}$:
$${\Rightarrow \frac{\pi}{8}\int_{0}^{\infty}\frac{1}{1+u^2}du}$$
And finally we are at a standard integral. We know
$${\int_{0}^{\infty}\frac{1}{1+x^2}dx=\frac{\pi}{2}}$$
And so finally we get the limit is
$${\Rightarrow \lim_{a\rightarrow\infty}\frac{I}{a}=\frac{\pi^2}{16}}$$
|
{
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Evaluate $\lim _{n \rightarrow \infty} \sum_{i=1}^{n} \frac{1}{i-3 n}$ Evaluate $\lim _{n \rightarrow \infty} \sum_{i=1}^{n} \frac{1}{i-3 n}$
Here $\sum_{i=1}^{n} \frac{1}{i-3 n}=\frac{1}{1-3 n}+\frac{1}{2-3 n}+\cdots+\frac{1}{n-3 n}$.
I tried to make the sum squeezed
between Convergent Sequences, but failed by getting
$\frac{-1}{2} \leq \frac{1}{-3 n+1}+\frac{1}{-3 n+2}+\cdots+\frac{1}{-3 n+n} \leq \frac{-1}{3}$
|
One approach is to use an expansion of harmonic numbers where $H(n)= \sum \limits_{i=1}^n \frac1n = \log_e(n) + \gamma + \frac{1}{2n}+O(n^{-2})$
You then have $\sum\limits_{i=1}^{n} \frac{1}{i-3 n} = H(2n-1)-H(3n-1) = \log_e(\frac{2n-1}{3n-1}) + \frac{n}{2(2n-1)(3n-1)}+O(n^{-2})$
and so $\lim\limits _{n \rightarrow \infty} \sum\limits_{i=1}^{n} \frac{1}{i-3 n} =\log_e(\frac{2}{3}) =\log_e(2)-\log_e(3) \approx-0.405$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Rolling Dice Game, Probability of Ending on an Even Roll The game is described as follows. $A$ and $B$ take turns rolling a fair six sided die. Say $A$ rolls first. Then if $A$ rolls {1,2} they win. If not, then $B$ rolls. If $B$ rolls {3,4,5,6} then they win. This process repeats until $A$ or $B$ wins, and the game stops.
What is the probability that the game ends on an even turn when $A$ rolls first?
Now the book gives the answer as $\frac{4}{7}$, however, when try to calculate I end up with $\frac{2}{11}$.
Below is my work:
To calculate this probability, we decompose the event into two disjoint events, (a) the event where $A$ wins on an even roll, and (b) the event where $B$ wins on an even roll.
(a) Now, the probability $A$ wins can be calculated as follows
\begin{align*}
\biggr(\frac{2}{3} \cdot \frac{1}{3}\biggr)\biggr(\frac{1}{3}\biggr) + \biggr(\frac{2}{3} \cdot \frac{1}{3}\biggr)\biggr(\frac{2}{3} \cdot \frac{1}{3}\biggr)\biggr(\frac{2}{3} \cdot \frac{1}{3}\biggr)\biggr(\frac{1}{3}\biggr) + \dots = \sum_{k=0}^\infty \biggr(\frac{2}{9}\biggr)^{2k+1}\frac{1}{3}\\
= \sum_{k=0}^\infty \frac{2}{27}\biggr(\frac{2}{9}\biggr)^{2k} = \sum_{k=0}^\infty \frac{2}{27}\biggr(\frac{4}{81}\biggr)^k = \frac{2}{27}\cdot \frac{1}{1- \frac{4}{81}} = \frac{6}{77}.
\end{align*}
(b) Similarly we calculate the probability $B$ wins on an even roll as
\begin{align*}
\biggr(\frac{2}{3} \cdot \frac{1}{3}\biggr)\biggr(\frac{2}{3}\cdot \frac{2}{3}\biggr) + \biggr(\frac{2}{3} \cdot \frac{1}{3}\biggr)\biggr(\frac{2}{3} \cdot \frac{1}{3}\biggr)\biggr(\frac{2}{3} \cdot \frac{1}{3}\biggr)\biggr(\frac{2}{3}\cdot\frac{2}{3}\biggr) + \dots = \sum_{k=0}^\infty \biggr(\frac{2}{9}\biggr)^{2k+1}\frac{4}{9}\\
= \sum_{k=0}^\infty \frac{8}{81}\biggr(\frac{2}{9}\biggr)^{2k} = \sum_{k=0}^\infty \frac{8}{81}\biggr(\frac{4}{81}\biggr)^k = \frac{8}{81}\cdot \frac{1}{1- \frac{4}{81}} = \frac{8}{77}.
\end{align*}
Therefore, it follows that the probability of the game ending on an even number of rolls is
\begin{equation*}
\frac{6}{77} + \frac{8}{77} = \frac{2}{11}.
\end{equation*}
Am I missing something?
|
The answer = 1/2
The game has to end by either A winning or B winning
Let's say A wins. He is just as likely to roll a 1 or a 2 on the last roll. Therefore in a game that A wins, probability of an even roll ending the game is 1/2, as 1(odd) and 2(even) are equally likely.
Let's say B wins. He is just as likely to roll a 3/4/5/6 on the last roll. Therefore in a game that B wins, probability of an even roll ending the game is 1/2, as 4 and 6 are favourable outcomes.
P.S. I have assumed that "ending on an even roll" as written in the title means the die outputs an even number. I agree that while the body of the question seems to suggest an even turn, this seems like the correct interpretation to me.
|
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|
How to solve $\int_0^1dx\int_0^1\frac{x^2-y^2}{(x^2+y^2)^2}\,dy$ The original question is:
Prove that:$$\begin{aligned}\\
\int_0^1dx\int_0^1\frac{x^2-y^2}{(x^2+y^2)^2}\,dy\neq\int_0^1dy&\int_0^1\frac{x^2-y^2}{(x^2+y^2)^2}\,dx\\
\end{aligned}\\$$
But I can't evaluate the integral $$\int_0^1dx\int_0^1\frac{x^2-y^2}{(x^2+y^2)^2}\,dy$$
At first, I assumed $x^2+y^2=z^2$. But, it is so complicated. Then, I assumed $x=r\cos\theta$ and $y=r\sin\theta$. But, I can't calculate the limits. Solving the equations I got three values of $\theta$ i.e. $\theta=0$, $\theta=\frac{\pi}{4}$ and $\theta=\frac{\pi}{2}$. I am just confused. Please help.
|
Hint:
$$\int_0^1\frac {x^2-y^2}{(x^2+y^2)^2}\, dy = \int_0^1\frac{\partial}{\partial y} \left(\frac{y}{x^2 + y^2}\right)\, dy = \frac{1}{1+x^2} $$
|
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"timestamp": "2023-03-29T00:00:00",
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|
System of congurences and the Chinese Remainder Theorem I have the following system of congruences:
\begin{align*}
x &\equiv 1 \pmod{3} \\
x &\equiv 4 \pmod{5} \\
x &\equiv 6 \pmod{7}
\end{align*}
I tried solving this using the Chinese remainder theorem as follows:
We have that $N = 3 \cdot 5 \cdot 7 = 105$ and $N_1=35, N_2=21, N_3=15$.
From this, we get the following
\begin{align*}
35x_1 &\equiv 1 \pmod{3} \\
21x_2 &\equiv 1 \pmod{5} \\
15x_3 &\equiv 1 \pmod{7}
\end{align*}
and this will result in
\begin{align*}
2x_1 &\equiv 1 \pmod{3} \\
x_2 &\equiv 1 \pmod{5} \\
x_3 &\equiv 1 \pmod{7}
\end{align*}
so from CRT $x =x_1N_1b_1 + x_2N_2b_2 + x_3N_3b_3 = 2 \cdot 35 \cdot3 + 1 \cdot 21 \cdot 5 + 1 \cdot 15 \cdot7 = 420 $.
However $420$ doesn't seem to satisfy the given system, what would be the problem here?
|
Here's how I would do it: using Bezout coefficients, we get $2\cdot5-3\cdot3=1$. So the solution to $\begin{cases}x\cong1\pmod3\\x\cong4\pmod5\end{cases}$ is $x=1\cdot{10}-4\cdot9=-26\pmod{15}=4\pmod{15}$.
Next we solve $\begin{cases} x\cong{4}\pmod{15}\\x\cong6\pmod7\end{cases}$.
Since $-6\cdot15+13\cdot7=1$, we get $x=6\cdot{-90}+4\cdot91=-540+364=-176\pmod{105}=34\pmod{105}$.
Hence $x\cong 34\pmod{105}$.
|
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|
Evaluate $\lim\limits_{x \to \infty} \sqrt[n]{(1+x^2)(2+x^2)...(n+x^2)}-x^2 $ I'm trying to calculate:
$$T = \lim\limits_{x \to \infty} \sqrt[n]{(1+x^2)(2+x^2)...(n+x^2)}-x^2$$
Here is my attempt.
Put $x^2=\dfrac{1}{t}$ so when $x\to \infty, t \to 0$ and the limit become
\begin{align*}
T &= \lim\limits_{t \to 0} \sqrt[n]{\left(1+\dfrac{1}{t}\right)\left(2+\dfrac{1}{t}\right)...\left(n+\dfrac{1}{t}\right)}-\dfrac{1}{t}\\
&=\lim\limits_{t \to 0} \sqrt[n]{\left(\dfrac{t+1}{t}\right)\left(\dfrac{2t+1}{t}\right)...\left(\dfrac{nt+1}{t}\right)}-\dfrac{1}{t} \\
&=\lim\limits_{t \to 0} \dfrac{\sqrt[n]{(t+1)(2t+1)...(nt+1)}-1}{t}
\end{align*}
My idea is to use $\lim\limits_{x\to0}\dfrac{(ax+1)^{\beta}-1}{x} =a\beta .$ But after some steps (above), now I'm stuck.
Thanks for any helps.
|
By definition of pochammer symbol $$(x^2+1)^{(n)}=(1+x^2)(2+x^2)\cdots (n+x^2)=\frac{\Gamma(x^2+1+n)}{\Gamma(x^2+1)}\sim x^{2n}\left(1+\frac{n(n+1)}{2x^2}+O(x^{-4})\right)$$ thus $$\sqrt[n]{(x^2+1)^{(n)}}-x^2 =x^2\left(1+\frac{n(n+1)}{2x^2}\right)^{\frac{1}{n}}-x^2$$ using the fractional binomial theorem we have limit $$\lim_{x\to \infty}\left(\sqrt[n]{(x^2+1)^{(n)}}-x^2\right)= x^2\left(1+\frac{n(n+1)}{2n} x^{-2} +O(x^{-4}) -x^2\right)=\frac{n+1}{2}$$
Notation: $O(.)$ is Big O notation.
|
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|
Proving that $ \sum_{k=1}^\infty \frac{k^{8} + 2^{k} }{3^{k} - 2^{k}} $ converges by the comparison test I would like to prove that the following series converges:$ \sum_{k=1}^\infty \frac{k^{8} + 2^{k} }{3^{k} - 2^{k}} $ by comparing it with a series that I already know converges. One such series could be the geometric series $ \sum_{k=1}^\infty \frac{2^{k}}{3^{k}} $. Since I know that $ \sum_{k=1}^\infty \frac{2^{k}}{3^{k}} $ converges, the only thing left to prove is that
$$ L = \lim_{k \to \infty}\frac{a_k}{b_k} < +\infty ,$$where $a_k = \frac{k^{8} + 2^{k} }{3^{k} - 2^{k}} $ and $ b_k = \frac{2^{k}}{3^{k}} $. However, when I try to prove this I get the following problem:
$$ \lim_{k \to \infty}\frac{3^{k}(k^{8}+2^{k})}{2^{k}(3^{k} - 2^{k})} $$
And I don't know how to solve this limit problem, and show that the above limit is less than $ +\infty$ so I'm kinda stuck and would appreciate any help!
|
$$\frac{k^{8} + 2^{k} }{3^{k} - 2^{k}}= \left(\frac{2}{3}\right)^{k} \cdot \frac{1+ \frac{k^8}{2^{k}} }{1-\left( \frac{2}{3}\right)^{k}}$$
As $\frac{1+ \frac{k^8}{2^{k}} }{1-\left( \frac{2}{3}\right)^{k}} \to 1$, then we can say, that $\exists N \in \mathbb{N}$ such that for $k>N$ holds $\frac{k^{8} + 2^{k} }{3^{k} - 2^{k}}< \frac{3}{2} $, so $$\frac{k^{8} + 2^{k} }{3^{k} - 2^{k}}< \left(\frac{2}{3}\right)^{k-1}$$
for $k>N$.
|
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|
Greatest Common Divisor Problem: Prove that $\gcd(\frac{a^3+b^3}{a+b}, a+b) = \gcd(a+b, 3ab)$ I've been stuck in this problem for some time now. Currently what I have accomplished is, using the propriety $\gcd(a,b) = \gcd(b,a \bmod(b))$ to get in the equation
$$ \gcd(a+b, \frac{a^3+b^3}{a+b}\bmod(a+b)) $$
but I don't know where to go anymore or even if I'm in the correct path. Any tips or solutions would be great appreciated.
|
We have
$$
\begin{pmatrix}
\frac{a^3+b^3}{a+b} \\ a+b
\end{pmatrix}
=
\begin{pmatrix}
a^2-ab+b^2 \\ a+b
\end{pmatrix}
=
\begin{pmatrix}
-1 & a+b \\ 0 & 1
\end{pmatrix}
\begin{pmatrix}
3ab \\ a+b
\end{pmatrix}
$$
The result follows because the matrix has determinant $-1$ and so has an integer inverse.
|
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|
Find $a$, $b$ such that $x^2 - x -1$ is a factor of $ax^9 + bx^8 + 1$
Find $a$, $b$ such that $x^2 - x -1$ is a factor of $ax^9 + bx^8 + 1$
The second polynomial can be rewritten as
$$ax^9 + bx^8 + 1 = f(x)(x^2 - x - 1)$$
The roots of this polynomial are $\frac{1 \pm \sqrt 5}{2}$. Substituting one of these roots in this equation gives us:
$$a\left( \frac{1 + \sqrt5}{2}\right)^9 + b\left( \frac{1 + \sqrt 5}{2}\right)^8 + 1 = 0$$
I was able to solve this far, but I gave up because the calculation past this point gets too tedious. The textbook has gone ahead and simplified this to
$$2^9 a + 2^8b(\sqrt 5 - 1) + (\sqrt5 - 1)^9 = 0$$
after which it simplifies to (divide by $2^8$ and solve the binomial expression)
$$2a + b(\sqrt 5 -1) = 76 - 34\sqrt5$$
Is there a more elegant way to solve this problem? Preferably one that does not include the magical use of a calculator or the evaluation of that ugly binomial expansion?
|
Consider
$$\frac{a x^9+bx^8+1}{x^2-x-1}$$ perform long division to get
$$-1+x-2 x^2+3 x^3-5 x^4+8 x^5-13 x^6+21 x^7-(b+34) x^8+x^9
(-a+b+55)+O\left(x^{10}\right)$$
So $b=-34$ and $a=21$ and the result is
$$\frac{a x^9+bx^8+1}{x^2-x-1}=-1+x-2 x^2+3 x^3-5 x^4+8 x^5-13 x^6+21 x^7$$
|
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|
Evaluate integral $\int (x^2-1)(x^3-3x)^{4/3} \mathop{dx}$ How can I evaluate this integral $$\int (x^2-1)(x^3-3x)^{4/3} \mathop{dx}=\;\;?$$
My attempt:
I tried using substitution $x=\sec\theta$, $dx=\sec\theta\ \tan\theta d\theta$,
$$\int (\sec^2\theta-1)(\sec^3\theta-3\sec\theta)^{4/3}
\sec\theta\ \tan\theta d\theta $$
$$=\int \tan^2\theta \sec^4\theta(1-3\cos^2\theta)^{4/3} \sec\theta\ \tan\theta d\theta $$
$$=\int \tan^3\theta \sec^5\theta(1-3\cos^2\theta)^{4/3}\ d\theta $$
$$=\int\dfrac{ \sin^3\theta}{ \cos^8\theta}(1-3\cos^2\theta)^{4/3}\ d\theta $$
I can't see if this substitution will work or not. This has become so complicated.
Please help me solve this integral.
|
Let $x^3-3x=t\implies (3x^2-3)dx=dt$ or $(x^2-1)dx=\frac{dt}{3}$
$$\int (x^2-1)(x^3-3x)^{4/3} \mathop{dx}=\int t^{4/3}\frac{dt}{3}$$
$$=\frac13\frac{t^{7/3}}{7/3}+C$$$$=\frac{(x^3-3x)^{7/3}}{7}+C$$
or alternatively,
$$\int (x^2-1)(x^3-3x)^{4/3}\ dx=\frac13\int (3x^2-3)(x^3-3x)^{4/3}\ dx$$
$$=\frac13\int (x^3-3x)^{4/3}\ d(x^3-3x)$$
$$=\frac13\frac{(x^3-3x)^{7/3}}{7/3}+C$$
$$=\frac{(x^3-3x)^{7/3}}{7}+C$$
|
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|
Calculus of $ \lim_{(x,y)\to (0,0)} \frac{8 x^2 y^3 }{x^9+y^3} $ By Wolfram Alpha I know that the limit
$$
\lim_{(x,y)\to (0,0)} \dfrac{8 x^2 y^3 }{x^9+y^3}=0.
$$
I have tried to prove that this limit is $0$, by using polar coordinate, the AM–GM inequality and the change of variable $ x^9= r^2 \cos^2(t) $ and $y^3= r^2 \sin^2(t)$, but these attempts were unsatisfactory.
I also have reviewed the similar questions and their answers but there are difference between those functions and mine one, I think the principal difference is that the powers of the denominators are odd.
|
As suggested in the comments, let consider the following path "near" the problematic points $y=-x^3$ for which denominator vanishes:
*
*$x=t$
*$y=-t^3+t^5$
then we have
$$\frac{8 x^2 y^3 }{x^9+y^3}
=\dfrac{8 t^2 (-t^3+t^5)^3 }{t^9+(-t^3+t^5)^3}
=\frac{8 t^2(-t^{9}+3t^{11}-3t^{13}+t^{15})}{t^9+(-t^{9}+3t^{11}-3t^{13}+t^{15})}=\\
=8\frac{-t^{11}+3t^{13}-3t^{15}+t^{17}}{3t^{11}-3t^{13}+t^{15}}=8\frac{-1+3t^2-3t^{4}+t^{6}}{3-3t^{2}+t^{4}}$$
For the general strategy, see also the related:
*
*Find $\lim_{(x,y)\rightarrow (0,0)}\frac{\sin (xy)}{x+y}$
|
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|
If $a$, $b$, $c$, $d$ are positive reals so $(a+c)(b+d) = 1$, prove the following inequality would be greater than or equal to $\frac {1}{3}$.
Let $a$, $b$, $c$, $d$ be real positive reals with $(a+c)(b+d) = 1$. Prove that $\frac {a^3}{b + c + d} + \frac {b^3}{a + c + d} + \frac {c^3}{a + b + d} + \frac {d^3}{a + b + c} \geq \frac {1}{3}$.
So I've been trying to solve this problem, and I've been trying to find a way to modify it into using AM-GM. The issue is that the $(a+c)(b+d) = 1$ is really throwing me off, as I haven't dealt with any inequalities that have used that as a condition yet (most other conditions I have seen go along the lines of $abcd = 1$ or something like that), and I'm not sure how exactly to deal with this inequality. Does anyone have any ideas?
|
By C-S $$\sum_{cyc}\frac{a^3}{b+c+d}=\sum_{cyc}\frac{a^4}{ab+ac+ad}\geq\frac{(a^2+b^2+c^2+d^2)^2}{\sum\limits_{cyc}(ab+ac+ad)}\geq$$
$$\geq\frac{a^2+b^2+c^2+d^2}{\sum\limits_{cyc}(ab+ac+ad)}\geq\frac{1}{3},$$
where the last inequality it's $$\sum_{sym}(a-b)^2\geq0.$$
|
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|
How do I integrate $\frac1{x^2+x+1}$? I have tried this:
$$\frac1{x^2+x+1} = \frac1{\left( (x+\frac12)^2+\frac34\right)}$$
Now $u = x+\frac12$
$$\frac1{ u^2+\frac34 }$$ Now multiply by $ \frac34$
$$\frac1{ \frac43 u^2 + 1}$$
Now put the $\frac43$ outside the integral
$$\frac34 \int \frac1{u^2+1}\,du=\frac34\arctan(u)=\frac34\arctan(x+1/2)$$
But the result is not the same result calculated by computers.
What did I do wrong?
Could someone please help me with this?
I don't know where my wrong calculation is. The way should be correct to get to the result.
Edit:
So now
$\int \frac{1}{x^2+x+1}=\int \frac{1}{(x+\frac{1}{2})^2}= \int \frac{4}{3} \frac{1}{\frac{4}{3}(x+\frac{1}{2})^2+1}$
u=x^2+1/2
$\int \frac{4}{3} \frac{1}{\frac{4}{3}(u)^2+1}$
$\int \frac{4}{3} \frac{1}{\frac{4u^2}{3}+1}$
$\int \frac{4}{3} \frac{1}{\frac{2u^2}{\sqrt{3}}+1}$
Now it is:
$\frac{4}{3} \int \frac{1}{\frac{2u^2}{\sqrt{3}}+1}$
$=\frac{4}{3} * arctan(2*(x^2+1/2)/(\sqrt{3}))$
Why is this still not the same as the computer calculated solution?
|
Note that $$\frac{1}{u^2+3/4}=\frac{1}{\frac34((2u/\sqrt 3)^2+1)}=\frac43 \frac{1}{v^2+1}$$
where $v=2u/\sqrt 3$. Can you finish now?
|
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|
Prove that $(x-1)\log(1 - 2 x) -2x = \sum_{n=2}^{\infty} \frac{2^n (n-1) x^{n+1}}{n^2 +n} \quad \text{for} \; 2|x|<1$ Prove that $$(x-1)\log(1 - 2 x) -2x = \sum_{n=2}^{\infty} \frac{2^n (n-1) x^{n+1}}{n^2 +n} \quad \text{for} \; 2|x|<1$$
First of all, I don't really know if by proving it means finding the function Sum of $\sum_{n=2}^{\infty} \frac{2^n (n-1) x^{n+1}}{n^2 +n}$ and concluding that $f(x)=(x-1)\log(1 - 2 x) -2x$ or to replace $\log(1 - 2 x)$ for it's Taylor Expansion and resaulting in $ \sum_{n=2}^{\infty} \frac{2^n (n-1) x^{n+1}}{n^2 +n}$. I've tried both ways, starting from the Taylor Expansion of $\log(1 - 2 x)=\sum_{n=1}^{\infty} -\frac{2^n x^{n}}{n}$ but I've failed both ways. Any hints on how to prove this? Thanks in advance.
Edit: What I did
$$\log(1 - 2 x)=\sum_{n=1}^{\infty} -\frac{2^n x^{n}}{n} = -2x + \sum_{n=2}^{\infty} -\frac{2^n x^{n}}{n} \\ \rightarrow (x-1)\log(1 - 2 x)=-2x^2 + x\sum_{n=2}^{\infty} -\frac{2^n x^{n}}{n} + 2x +\sum_{n=2}^{\infty} \frac{2^n x^n}{n} \\ \rightarrow (x-1)\log(1 - 2 x) -2x =-2x^2 + \sum_{n=2}^{\infty} -\frac{2^n x^{n+1}}{n} + \sum_{n=2}^{\infty} \frac{2^n x^n}{n} \\ = \sum_{n=1}^{\infty} -\frac{2^n x^{n+1}}{n} + \sum_{n=2}^{\infty} \frac{2^n x^n}{n} \\ = \sum_{n=1}^{\infty} -\frac{2^{n} x^{n+1}}{n} + \sum_{n=1}^{\infty} \frac{2^{n+1} x^{n+1}}{n+1} \\ = \sum_{n=1}^{\infty} \frac{-2^{n} x^{n+1} (n+1) + 2^{n+1} x^{n+1} n}{n(n+1)} \\ = \sum_{n=1}^{\infty} \frac{-2^{n} x^{n+1}[(n+1)-2n]}{n(n+1)} \\ = \sum_{n=1}^{\infty} \frac{-2^{n} x^{n+1}(1-n)}{n(n+1)} \\ = \sum_{n=1}^{\infty} \frac{2^{n} x^{n+1}(n-1)}{n(n+1)} \\ = 0 + \sum_{n=2}^{\infty} \frac{2^{n} x^{n+1}(n-1)}{n(n+1)} \\ = \sum_{n=2}^{\infty} \frac{2^{n} x^{n+1}(n-1)}{n(n+1)}$$
Which is what I was trying to prove.
|
Another approach.
$$f(x)=(x-1)\log(1 - 2 x) -2x \implies f''(x)=\frac{4 x}{(1-2 x)^2}$$ Let $t=2x$ to make
$$g''(t)=\frac{2 t}{(1-t)^2}=2\sum_{n=1}^\infty n t^n\implies g'(t)=2\sum_{n=1}^\infty\frac{ n t^{n+1}}{n+1}\implies g(t)=2\sum_{n=1}^\infty\frac{ n t^{n+2}}{(n+1) (n+2)}$$ Replace $t$ by $2x$ and shift the index by $1$
|
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|
How do we solve pell-like equations? I need to find all solutions $(x,y)∈Z^2$ to the Pell-like equation $x^2-21y^2= 4$
Method I used to solve above problem:-
I solved the pell-equation $x^2-21y^2= 1$ and calculated the possible solutions to the equation and further multiplied the above equation with the initial equation, i.e, $x^2-21y^2= 4$.
But I am still not able to figure out what should I do next? Could someone help me out in this problem?
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Given
$x^2-21y^2= 4$ we can see $(5,1)$ as an easy solution where $5^2-21= 4$.
Another observation is
$$x^2-21y^2= 4\implies \frac{x^2-4}{21}=y^2=\frac{x-2}{p}\cdot\frac{x+2}{q}\quad \text{ where }\quad p|x-2\quad\land\quad q|x+2$$
The factors of $21$ are $1,3,7,21$ and trying the cofactors $(1,21)$ we get conflicting answers for what is x.
$$x-2=1\implies x=3\quad \land \quad x+2=21\implies x=19 \lor\\
x-2=21\implies x=23\quad \land \quad x+2=1\implies x=-1$$
but the other two cofactors do yield consistent results for what is x.
$$x-2=3\implies x=5\quad \land \quad x+2=7\implies x=5$$
and this fits our desire to have integers that, multiplied, yield a square.
$$\frac{x-2}{3}\cdot\frac{x+2}{7}=\frac{5-2}{3}\cdot\frac{5+2}{7}
=\frac{3}{3}\cdot\frac{7}{7}=\frac{21}{21}=1=y^2$$
Only positive integers have been used in this demonstration but the results are the same with $(-5,-1)$ because, multiplied, they become positive.
$$\therefore x^2-21y^2= 4\implies x=\pm5\quad y=\pm 1 $$
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Show that by means of the transformation $w=\frac 1z$ the circle $C$ given by $|z-3|=5$ is mapped into the circle $|w+\frac{3}{16}|=\frac{5}{16}$
Show that by means of the transformation $w=\frac 1z$ the circle C given by $|z-3|=5$ is mapped into the circle $\left|w+\frac{3}{16}\right|=\frac{5}{16}$
My try:
$$\begin{align}\\
&w=\frac 1z\\
&\implies 3w=\frac 3z\\
&\implies1-3w=1-\frac 3z=\frac{z-3}{z}\\
&\implies|1-3w|=\left|\frac{z-3}{z}\right|\\
&\implies|1-3w|=|z-3|\left|\frac 1z\right|\\
&\implies|1-3w|=5|w|\\
\end{align}\\$$
But, I can't understand what to do next and how can I proof that. Any suggestions? or, you can add an answer.
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$$|z-3|=5.$$
Let $\displaystyle w= \frac{1}{z}$ and let $w^\star$ be the conjugate of $w$.
$$\begin{aligned}
\left(\frac{1}{w} -3 \right) \left(\frac{1}{w^\star} -3 \right) &= 25 \\
(1 -3 w^\star) \left( 1 - 3w \right) &= 25 w w^\star\\
1-3(w+w^\star) + 9 w w^\star &= 25 w w^\star \\
16 w w^\star + 3(w + w^\star) -1 &= 0\\
w w^\star + \frac{3}{16} (w + w^\star) &= \frac{1}{16}\\
\left( w + \frac{3}{16}\right) \left( w^\star + \frac{3}{16} \right) &= \frac{9}{16^2} + \frac{1}{16} = \left( \frac{5}{16}\right)^2\\
\left| w + \frac{3}{16} \right| &= \frac{5}{16}.
\end{aligned}$$
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Calculate the minimum value of $\left|\frac{a^2 - bc}{b - c}\right| + \left|\frac{b^2 - ca}{c - a}\right| + \left|\frac{c^2 - ab}{a - b}\right|$.
Given positives $a, b, c$ such that $abc = 1$, if possible, calculate the minimum value of $$\left|\frac{a^2 - bc}{b - c}\right| + \left|\frac{b^2 - ca}{c - a}\right| + \left|\frac{c^2 - ab}{a - b}\right|$$
Without loss of generalisation, assume that $a \le b \le c$.
We have that $$\left|\frac{a^2 - bc}{b - c}\right| + \left|\frac{b^2 - ca}{c - a}\right| + \left|\frac{c^2 - ab}{a - b}\right| \ge \frac{c^2 - ba}{b - a} + \frac{a^2 - bc}{b - c}$$
$$ = \frac{(c + a)(a^2 + b^2 + c^2 - bc - ca - ab)}{(c - b)(b - a)} \ (1)$$
Let $c' = b - a, a' = c - b \iff c = a' + b, a = b - c'$, $(1)$ becomes $$\frac{(2b - c' + a')(c'^2 + c'a' + a'^2)}{c'a'}$$
and $(b - c')b(b + a') = b^3 - (c' - a')b^2 - c'a'b = 1$
$$\iff (2b - c' + a')b^2 = b^3 + c'a'b + 1 \iff 2b - c' + a' = \frac{b^3 + c'a'b + 1}{b^2}$$
Another idea I had was that $\left|\dfrac{a^2 - bc}{b - c}\right| + \left|\dfrac{b^2 - ca}{c - a}\right| + \left|\dfrac{c^2 - ab}{a - b}\right|$
$$ = \frac{1}{2}\sum_{\text{cyc}}\left(|c - a|\left|\frac{2(b^2 - ca)}{(c - a)^2}\right|\right) = \frac{1}{2}\sum_{\text{cyc}}\left(|c - a|\left|\frac{2b^2 - c^2 - a^2}{(c - a)^2} + 1\right|\right)$$
$$ = \frac{1}{2}\left[(c - b)\left(\left|\frac{2(b^2 - a^2)}{(b - c)^2} + \frac{c^2 - b^2}{(b - c)^2} - 1\right| + \left|\frac{b^2 - a^2}{(c - a)^2} - \frac{c^2 - b^2}{(c - a)^2} + 1\right|\right)\right.$$
$$\left. + (b - a)\left(\left|\frac{b^2 - a^2}{(c - a)^2} - \frac{c^2 - b^2}{(c - a)^2} + 1\right| + \left|\frac{b^2 - a^2}{(a - b)^2} + \frac{2(c^2 - b^2)}{(a - b)^2} + 1\right|\right)\right]$$
$$ \ge \frac{1}{2}\left[(c - b)\left(\frac{2(b^2 - a^2)}{(b - c)^2} + \frac{c^2 - b^2}{(b - c)^2} + \frac{b^2 - a^2}{(c - a)^2} - \frac{c^2 - b^2}{(c - a)^2}\right)\right.$$
$$\left. + (b - a)\left(\frac{b^2 - a^2}{(c - a)^2} - \frac{c^2 - b^2}{(c - a)^2} + \frac{b^2 - a^2}{(a - b)^2} + \frac{2(c^2 - b^2)}{(a - b)^2} + 2\right)\right]$$
$$ = \frac{1}{2}\left[(c^2 - b^2)\left(\frac{1}{c - b} - \frac{1}{c - a} + \frac{2}{b - a}\right) + (b^2 - a^2)\left(\frac{2}{c - b} + \frac{1}{c - a} + \frac{1}{b - a}\right)\right] + (b - a)$$
$$ = \frac{1}{2}\left(\frac{b^2 + c^2 - 2a^2}{c - b} + \frac{2b^2 - c^2 - a^2}{c - a} + \frac{2c^2 - a^2 - b^2}{b - a}\right) + (b - a)$$
There must have been something wrong, but that's all I have for now.
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Answer to the question: $\lim_{h \to 0}6+h$
The given expression$$\left|\frac{a^2 - bc}{b - c}\right| + \left|\frac{b^2 - ca}{c - a}\right| + \left|\frac{c^2 - ab}{a - b}\right| $$
upon condition that $abc=1$
This can be rewritten as
$$\left|\frac{a^2 - \frac{1}{a}}{b - c}\right| + \left|\frac{b^2 - \frac{1}{b}}{c - a}\right| + \left|\frac{c^2 - \frac{1}{c}}{a - b}\right| $$ which is further reducable to $$\left|\frac{a-1}{b - c}(a+\frac{1}{a}+1)\right| + \left|\frac{b-1}{c-a}(b+\frac{1}{b}+1)\right| + \left|\frac{c-1}{a-b}(c+\frac{1}{c}+1)\right| $$
if $a,b,c >0$
this reduces to $$3\big(\left|\frac{a-1}{b - c}\right| + \left|\frac{b-1}{ c-a}\right| + \left|\frac{c-1}{a-b}\right|\big) $$
Note: The reduction for $a+\frac{1}{a} +1>3$ and not $\geq 3$ because doing so would imply all variables equal $1$
The least value for this expression I could yield using $a$ fixed at 1 and manipulating $b,c$ was 6
Since there's a modulus operation, the least we can get out of |x| is 0, so set $a=1$, by doing that we are left with
$$\left|\frac{b-1}{ c-1}\right| + \left|\frac{c-1}{b-1}\right|\ $$
which is definitely $$> 2$$ (remember $b$ and $c$ cannot be equal)
Hence the answer.
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Proving $\int_0^1\int_0^1\frac{-x\ln(xy)}{1-x^2y^2}\,dx\,dy=\frac{\pi^2}{12}$
Proving $\displaystyle\int_0^1\int_0^1\frac{-x\ln(xy)}{1-x^2y^2}\,dx\,dy=\frac{\pi^2}{12}$
My atempt:
\begin{align*}
\int_0^1 \int_0^1\frac{-x\ln(xy)}{1-x^2y^2} \, dx \, dy &=\int_0^1I_x(y)\,dy\\[6pt]
\text{where }I_x(y)=\int_0^1\frac{-x\ln(xy)}{1-x^2y^2} \, dx
\end{align*}
\begin{align*}
I_x(y)&=\int_0^1\frac{-x\ln(xy)}{2(1-xy)}-\frac{x\ln(xy)}{2(1+xy)} \, dx\\[6pt]
&=\frac{1}{2}\int_0^1\frac{-x\ln(xy)}{1-xy}\, dx+\frac{1}{2} \int_0^1\frac{-x\ln(xy)}{1+xy} \, dx\\[6pt]
&=\frac{1}{2} \sum_{n=0}^\infty-y^n \int_0^1x^{n+1}\ln(xy)\,dx+\frac{1}{2} \sum_{n=0}^\infty(-1)^{n+1} \int_0^1y^nx^{n+1}\ln(xy) \, dx\\[6pt]
&=\frac{1}{2} \sum_{n=0}^\infty-y^n\left(\frac{\ln(y)}{n+2}-\int_0^1 \frac{x^{n+1}}{(n+2)y} \, dx\right)+\frac{1}{2} \sum_{n=0}^\infty (-1)^{n+1} y^n \left(\frac{\ln(y)}{n+2}-\int_0^1\frac{x^{n+1}}{(n+2)y} \, dx\right)\\[6pt]
&=\frac{1}{2} \sum_{n=0}^\infty \frac{y^n\ln(y)}{n+2} ((-1)^{n+1}-1) +\frac{1}{2} \sum_{n=0}^\infty \frac{y^{n-1}}{(n+2)^2}(1-(-1)^{n+1})
\end{align*}
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I thought it might be instructive to present an approach that uses elementary calculus tools to reduce the double integral to a single integral.
Then, after a standard use of a Taylor series expansion of $\log(1+y)$, the answer is expressed as the familiar alternating series of reciprocal squares $\sum_{n=1}^\infty\frac{(-1)^n}{n^2}=-\frac{\pi^2}{12}$.
To this end, we now proceed.
Let $I$ be the double integral of interest given by
$$I=\int_0^1\int_0^1\frac{-x\ln(xy)}{1-x^2y^2}\,dy\,dx$$
Enforcing the substitution $y\mapsto y/x$ in the inner integral reveals
$$\begin{align}
I&\overbrace{=}^{y\mapsto y/x}-\int_0^1 \int_0^x \frac{\log(y)}{1-y^2}\,dy\,dx\tag1
\end{align}$$
We change the order or integration in $(1)$ to find that
$$\begin{align}
I&=-\int_0^1 \int_0^x \frac{\log(y)}{1-y^2}\,dy\,dx\\\
&=-\int_0^1 \int_y^1 \frac{\log(y)}{1-y^2}\,dx\,dy\\\\
&=-\int_0^1 \frac{\log(y)}{1+y}\,dy\tag2
\end{align}$$
It is interesting to note that the integral on the right-hand side of $(2)$ can be recognized as $-\text{Li}_2(-1)=\frac{\pi^2}{12}$ and we are done!
Integrating by parts the integral on the right-hand side of $(2)$, we find that
$$\begin{align}I&=-\int_0^1 \frac{\log(y)}{1+y}\,dy\\\\
&\overbrace{=}^{\text{IBP}}\int_0^1 \frac{\log(1+y)}{y}\,dy\tag3
\end{align}$$
Using the Taylor series for $\log(1+y)=\sum_{n=1}^\infty \frac{(-1)^{n-1}y^n}{n}$ in $(3)$ yields
$$\begin{align}
I&=\int_0^1 \frac{\log(1+y)}{y}\,dy\\\\
&=\sum_{n=1}^\infty \frac{(-1)^{n-1}}{n^2}\\\\
&=\frac{\pi^2}{12}
\end{align}$$
as was to be shown!
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|
What's the problem with differentiating $y = \sin(x^2)$ by applying the limit definition of a derivative directly? I was taking the derivative of $y = \sin(x^2)$. I know that we can solve it by applying chain rule, but i tried without any rules, just like a normal method. This is what i did:
$$\frac{\sin((x + h)^2) - \sin((x)^2)}{h}$$
Is this method correct? If not, then why? Because wherever I search for the derivative of $y = \sin(x^2)$, none did like this. And also I am not able to come to the proper answer which is $2x\cos(x^2)$ through that method.
Can somebody help me!
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This is OK,
but the chain rule is easier.
Continue like this,
using $\sin(x) = x+O(x^3),
\cos(x) = 1-x^2/2+O(x^2)$
for small $x$:
$\begin{array}\\
\Delta_h(\sin(x^2))
&=\dfrac{\sin((x + h)^2) - \sin((x)^2)}{h}\\
&=\dfrac{\sin(x^2+2hx+h^2) - \sin(x^2)}{h}\\
&=\dfrac{\sin(x^2)\cos(2hx+h^2)+\cos(x^2)\sin(2hx+h^2) - \sin(x^2)}{h}\\
&=\dfrac{\sin(x^2)(\cos(2hx+h^2)-1)+\cos(x^2)\sin(2hx+h^2)}{h}\\
&=\dfrac{\sin(x^2)(\cos(2hx+h^2)-1)}{h}+\dfrac{\cos(x^2)\sin(2hx+h^2)}{h}\\
&\approx\dfrac{\sin(x^2)((1-(2hx+h^2)^2/2-1)}{h}+\dfrac{\cos(x^2)(2hx+O(h^2)}{h}\\
&=\dfrac{-\sin(x^2)((2hx+h^2)^2/2}{h}+\dfrac{\cos(x^2)(2hx+O(h^2)}{h}\\
&=-\sin(x^2)(h(2x+h)^2/2+2x\cos(x^2)+O(h)\\
&\to 2x\cos(x^2)\\
\end{array}
$
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Proving: $\int_0^1 \int_0^1\frac{\ln^4(xy)}{(1+xy)^2}dxdy=\frac{225}{2}\zeta(5)$
Proving:$$\displaystyle\int_0^1\displaystyle\int_0^1\frac{\ln^4(xy)}{(1+xy)^2}dxdy=\frac{225}{2}\zeta(5)$$
I tried using variable switching
$\ln(xy)=t$ But I did not reach any results after the calculation
\begin{align*}
k&=\displaystyle\int_0^1\displaystyle\int_0^1\frac{\ln^4(xy)}{(1+xy)^2}dx\\
&=\displaystyle\int_0^1\displaystyle\int_{-\infty}^{\ln(y)}\frac{t^4e^t}{(1+e^t)^2y^2}dtdy\\
&=\displaystyle\int_0^1\displaystyle\int_{-\infty}^{\ln(y)}\frac{t^4e^t}{y^2(1+e^t)}\displaystyle\sum_{n=0}^{\infty}(-e^t)dtdy\\
&=\displaystyle\int_0^1\frac{1}{y^2}\left(\displaystyle\sum_{n=0}^{\infty}\displaystyle\int_{-\infty}^{\ln(y)}\frac{t^4(-e^{2t})}{1+e^t}dt\right)dy\\
&=\displaystyle\int_0^1\frac{1}{y^2}\left(\displaystyle\sum_{n=0}^{\infty}\displaystyle\int_{\ln(y)}^{\infty}\frac{t^4e^{2t}}{1+e^t}dt\right)dy\\
\end{align*}
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Starting off after your first substitution, notice that $$\sum_{n=1}^{\infty} nx^n = \frac{x}{(1-x)^2}, \text{ for } x \in(-1,1)$$
Since the domain of $x,y$ is $(0,1)$, we can write $$\frac{e^t}{(1+e^t)^2}=-\sum_{n=1}^{\infty} {(-1)}^n n e^{tn}$$
In addition, you made a slight error when calculating $dt$ I suppose. It should be $y$ not $y^2$ in the denominator.
$$\int_0^1 \int_0^{\ln{y}} \frac{t^4}{y}\left( -\sum_{n=1}^{\infty} {(-1)}^n n e^{tn}\right) \; dt \; dy$$
Because the summation converges, we can interchange the summation and integral sign from Fubini's theorem:
\begin{align}
k &= -\sum_{n=1}^{\infty} {(-1)}^n n \int_0^1 \frac{1}{y} \int_0^{\ln{y}} t^4 e^{tn}\; dt \; dy \\
&\overset{\text{IBP}}= -\sum_{n=1}^{\infty} {(-1)}^n n \int_0^1 \frac{y^{n-1} \left(n^4 \ln^4{y}-4n^3\ln^3{y}+12n^2\ln^2{y}-24n\ln{y}+25\right)}{n^5} \; dy \\
&\overset{\text{IBP}}= -\sum_{n=1}^{\infty} {(-1)}^n n \cdot \frac{120}{n^6} \\
&= 120\sum_{n=1}^{\infty} \frac{{(-1)}^{n+1}}{n^5} \\
&= \boxed{\frac{225}{2}\zeta(5)} \\
\end{align}
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Find all values of $a$ for which the maximum value of $f(x)=\frac{ax-1}{x^4-x^2+1}$equals $1$. Find all values of $a$ for which the maximum value of $$f(x)=\frac{ax-1}{x^4-x^2+1}$$equals $1$.
I equated $f(x)$ equal to $1$ to obtain a polynomial $x^4-x^2-ax+2=0$. Now this must have at least one repeated root.But I could not get any further.
Is there a calculus way to do it
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Compute the polynomial GCD of your polynomial and its derivative.
A repeated root of a polynomial is a root of its derivative. So we want to find $a$ such that $g = \gcd(x^4 - x^2 - a x+2, 4 x^3 - 2x - a) \neq 1$.
Taking $4$ of the left and $-x$ of the right, as well as the right, we have
$$ g = \gcd(-2x^2-3ax+8, 4 x^3 - 2x - a ) \text{.} $$
Taking $2x$ of the left and $1$ of the right, as well as the left, we have
$$ g = \gcd(-6ax^2+14x-a, -2x^2-3ax+8) \text{.} $$
Taking $1$ of the left and $-3a$ of the right, as well as the right, we have
$$ g = \gcd(9a^2x+14x-25a, -2x^2-3ax+8) \text{.} $$
We started with two polynomials equal to zero and have constructed sums of multiples of them. Since sums of multiples of zeroes are zero, we solve the left of these: from $9a^2x+14x-25a = 0$, $x = \frac{25a}{9a^2 + 14}$. Notice this new denominator is always positive. We substitute this into the right member of our most recent pair of polynomials to find
$$ \frac{27a^4 + 284a^2 - 1568}{9a^2 + 14} = 0 \text{.} $$
Since the denominator is always positive, this is zero only when the numerator is zero. By applying the quadratic formula, we find $a^2 = 4$, so $a = \pm 2$ or $a^2 = \frac{-392}{27}$, so $a = \pm \frac{14 \mathrm{i} \sqrt{2}}{3\sqrt{3}}$. There is no sensible meaning of "maximum" if we allow $a$ to be non-real, so we restrict to $a = \pm 2$.
Continuing our GCDs,
\begin{align*}
g_{a = 2} &= \gcd(50x-50,-2x^2-6x+8) \\
&= \gcd(50x-50, x(50x-50) + 25(-2x^2-6x+8)) \\
&= \gcd(50x-50, -200x+200) \\
&= \gcd(50(x-1), -200(x-1)) \\
&= 50(x-1) \text{.}
\end{align*}
So when $a = 2$, the double root occurs at $x = 1$. Here, $f''(1) = -10$, so this is a local maximum. The first derivative has one other zero at the real root of $3x^3+x^2+1$ (the factor of the numerator of $f'$ left after the factors $-2$ and $x-1$), but the second derivative is positive there, so this corresponds to a local minimum. The only other potential critical point is any zero of $x^4 - x^2 + 1$, but this has no real roots, so there are two critical points, we have characterized them, there is one local maximum, and it is necessarily the global maximum, $\left.f(1)\right|_{a = 2} = 1$.
The analysis for $a = -2$ parallels the above, with maximum $1$ occurring at the critical point $x = -1$.
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|
How to solve double absolute value inequality? This question comes from Spivak Calculus Chapter 1.
How can we algebraically solve $|x − 1|+|x − 2| > 1?$
I know that if we 2 absolute values and no constants, we can square both sides, but I'm pretty sure this is not the case here. My attempt was to split this into different sections:
$|x − 1|+|x − 2| > 1 \rightarrow |x − 1| > 1 - |x − 2|$.
So we would have:
$x − 1 > 1 - |x − 2|$
$x − 1< -1 +| x − 2|$
Then we can split this into 4 equations more equations based on the absolute value on $(x-2)$.
However, after doing this, I obtained conflicting solutions and unsolvable expressions (i.e $2<-2$).
That being said, how would I go about algebraically solving this inequality?
Thanks!
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"However, after doing this, I obtained conflicting solutions and unsolvable expressions"
Those are cases with no solutions. Nothing wrong with that.
Do cases be keep track of you initial assumptions.
Case 1: $x-1 \ge 0; x-2 \ge 0$. Thus $x\ge 1$ and $x \ge 2$. This is the case that $x \ge 2$.
Okay $|x-1| + |x-2|> 1$ so
$(x-1) + (x-2) > 1$ so
$2x - 3 > 1$ so $2x > 4$ and $x >2$. And we restrict this to $x \ge 2$ to get
$x > 2$ AND $x \ge 2$ so
Conclusion $x > 2$.
Case 2: $(x-1) \ge 0$ and $(x-2) < 0$. That is $x \ge 1$ and $x < 2$ so this is the case that $1 \le x < 2$.
We get $(x-1) -(x-2) > 1$ so
$1 > 1$. This is never the case so there are no solutions where $1 \le x < 2$.
If we want to be thurough we would say.
We must restrict to where $1 > 1$ AND $1\le x < 2$. There are no cases where both are true.
Case 3: $(x-1) < 0$ and $x -2 \ge 0$. This means $x < 1$ and $x \ge 2$. This is impossible. There are no such $x$ and so no such $x$ can be a solution (as there are no such $x$!).
If we want to be thorough (which we don't but let's pretend we do) we would solve
$-(x-1) + (x-2) > 1$ so $-1 > 1$ and or solution occurs when $-1 > 1$ and $x< 1$ and $x \ge 2$. As those three conditions are never concurrently true we have no solution in this interval which doesn't exist in the first place.
Case 4: $(x-1) < 0$ and $(x-2) < 0$. This means $x < 1$ and $x < 2$ so is the case when $x < 1$.
So $-(x-1) -(x-2) > 1$ so $-2x + 3> 1$ so $-2x > -2$ so $x < 2$.
So these solutions occur when $x < 2$ AND $x < 1$
Conclusion: so these solutions occur whenever $x < 1$
Combining Case 1, and Case 4 (and 2 and 3 although those had no result) we have final solution
$|x-1| + |x-2| >1 $ if
$x >2$ OR $x < 1$ or $x \in (-\infty, 1)\cup (2, \infty)$.
If we want to be thorough (which be now you should know we don't)
We could so we have solutions when:
$x > 2$ OR $1 < 1$ OR ($x < 1$ AND $x\ge 2$) OR $x < 1$ or
$x \in (2, \infty) \cup \emptyset \cup \emptyset \cup (-\infty, 1)=$
$(-\infty, 1)\cup (2, \infty)$.
=====
Familiarity and common sense and we can allow ourselve to consider then intervals $(-\infty, 1], [1,2],$ and $[2,\infty)$.
If $x \in (-\infty 1]$ then $(x-1)\le 0; x-2 < 0$ so $|x-1|+|x-2|=-(x-1)-(x-2)=-2x+3 > 1$ so $x < 1$.
If $x \in [1,2]$ then $x-1 \ge 0$ and $x-2\le 0$ so $|x-1|+|x-2| = (x-1)-(x-2) = 1 > 1$ which is impossible.
If $x \in [2,\infty)$ then $x-1>0$ and $x -2\ge 0$ so $|x-1| + |x-2| = x-1 + x-2=2x -3 >1$ so $x > 2$.
So $x< 1$ or $x > 2$ and $x \in (-\infty,1)\cup (2, \infty)$.
....
this way we know $x-1 <0$ while $x-2 \ge 0$ was absurd from the start and never needed to be considered in the first place.
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{
"language": "en",
"url": "https://math.stackexchange.com/questions/3786586",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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|
Evaluate $\frac{1+3}{3}+\frac{1+3+5}{3^2}+\frac{1+3+5+7}{3^3}+\cdots$ It can be rewritten as
$$S = \frac{2^2}{3}+\frac{3^2}{3^2}+\frac{4^2}{3^3}+\cdots$$
When $k$ approaches infinity, the term $\frac{(k+1)^2}{3^k}$ approaches zero. But, i wonder if it can be used to determine the value of $S$. Any idea?
Note: By using a programming language, i found that the value of $S$ is $3.5$
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If we assume $z\in(-1,1)$ and start with
$$ \sum_{n\geq 0} (2n+1) z^n = \frac{1+z}{(1-z)^2} \tag{1}$$
granted by stars&bars
$$ \frac{1}{(1-z)^{m+1}}=\sum_{n\geq 0}\binom{n+m}{m}z^n\tag{2} $$
by convolution it follows that
$$ \sum_{n\geq 0}\left(\sum_{k=0}^{n}(2k+1)\right) z^n = \frac{1+z}{(1-z)^3}\tag{3} $$
and by evaluating both sides of $(3)$ at $z=\frac{1}{3}$ we get
$$ \sum_{n\geq 0}\frac{(n+1)^2}{3^n} = \frac{9}{2}\tag{4} $$
so
$$ \frac{1}{3}+\frac{1+3}{3^2}+\frac{1+3+5}{3^3}+\frac{1+3+5+7}{3^4}+\ldots = \color{red}{\frac{7}{2}}.\tag{5} $$
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{
"language": "en",
"url": "https://math.stackexchange.com/questions/3788078",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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|
Check the convergence of the series : Which test do we apply? Check the convergence of the following series:
*
*$\displaystyle{\sum_{n=1}^{+\infty}\frac{n^2+1}{n^4+n}}$
*$\displaystyle{\sum_{n=1}^{+\infty}\frac{n^{n^2}}{(n+1)^{n^2}}}$
For the first series do we use the comparison test? But with which sequence do we compare $\frac{n^2+1}{n^4+n}$ ? Do we maybe do the following? $$\frac{n^2+1}{n^4+n}<\frac{n^2+n^2}{n^4+n}<\frac{n^2+n^2}{n^4}=\frac{2n^2}{n^4}=\frac{2}{n^2}$$
As for the second series I thought that we could use the ratio test but I am not sure if that works because we get $$\frac{a_{n+1}}{a_n}=\frac{\frac{(n+1)^{(n+1)^2}}{(n+2)^{(n+1)^2}}}{\frac{n^{n^2}}{(n+1)^{n^2}}}=\frac{(n+1)^{(n+1)^2}(n+1)^{n^2}}{(n+2)^{(n+1)^2}n^{n^2}}=\frac{(n+1)^{(n+1)^2+n^2}}{(n+2)^{(n+1)^2}n^{n^2}}$$ So is it better to apply an other test here?
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The first one is fine, for the second one, by $\left(1+\frac1n\right)^{n}\ge 2$, we have
$$\frac{n^{n^2}}{(n+1)^{n^2}}=\frac1{\left(1+\frac1n\right)^{n^2}}\le\frac1{2^n}$$
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"url": "https://math.stackexchange.com/questions/3788940",
"timestamp": "2023-03-29T00:00:00",
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|
Evaluate $\int\limits_{0}^{\infty}\frac{n \sin x}{1+n^2x^2}dx$ Notice that $\frac{n \sin x}{1+n^2x^2}\to 0$ pointwise.
And we have,$$\int\limits_{0}^{\infty}\frac{n \sin x}{1+n^2x^2}dx=\int\limits_{0}^{1}\frac{n \sin x}{1+n^2x^2}dx+\int\limits_{1}^{\infty}\frac{n \sin x}{1+n^2x^2}dx$$
Then for $\int\limits_{1}^{\infty}\frac{n \sin x}{1+n^2x^2}dx$ portion we have,
$$\left|\frac{n \sin x}{1+n^2x^2}\right|\leq\left|\frac{n}{n^2x^2}\right|=\left|\frac{1}{nx^2}\right|\leq\frac{1}{x^2}$$
And $\frac{1}{x^2}$ is integrable on $(1,\infty)$
So by the dominated convergence theorem:
$$\int\limits_{1}^{\infty}\frac{n \sin x}{1+n^2x^2}dx\to\int\limits_0^{\infty}0=0$$
But how should I do the $\int\limits_{0}^{1}\frac{n \sin x}{1+n^2x^2}dx$?
Appreciate your help
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The change fo variable $u=nx$ gives
$$
\int^\infty_0\frac{\sin(u/n)}{1+u^2}\,du
$$
The integrand is dominated by $\frac{1}{1+u^2}$ which is integrable. Then, by dominated convergence
$$
\lim_{n\rightarrow\infty}\int^\infty_0\frac{n\sin x}{1+n^2x^2} =\lim_{n\rightarrow\infty}\int^\infty_0\frac{\sin(u/n)}{1+u^2}\,du=
\int^\infty_0\lim_{n\rightarrow\infty}\frac{\sin(u/n)}{1+u^2}\,du=0
$$
for $\sin(u/n)\xrightarrow{n\rightarrow\infty}0$ for all $u$.
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{
"language": "en",
"url": "https://math.stackexchange.com/questions/3789989",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.