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Solving for $x$ in $x + \frac{s}{s^2+4} = \frac{2x}{s^2+4}$ How would I go about solving for $x$ in this equation?
$$x + \frac{s}{s^2+4} = \frac{2x}{s^2+4}$$
|
$$x+\frac{s}{s^2+4}=\frac{2x}{s^2+4}$$
$$\frac{s^2x+4x+s}{s^2+4}=\frac{2x}{s^2+4}$$
For real $s$, $s^2+4\ne0$. Hence,
$$s^2x+4x+s=2x$$
$$x(s^2+2)+s=0$$
$$x=-\frac{s}{s^2+2}$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Let $A$ be the $n\times n$ matrix with a $1$ in every entry. What are the eigenvalues of $A$? Is $A$ diagonalizable? Let $A$ be the $n\times n$ matrix with a $1$ in every entry. What are the eigenvalues of $A$ and a basis for each eigenspace? Is $A$ diagonalizable?
Having some trouble with this one. I tried using the fact that $\lambda$ is an eigenvalue of $A$ iff there exist non-zero solutions to $Ax=\lambda x$. Well, clearly the $1\times n$ non-zero vector $x$ with all entries equal is an eigenvector of $A$, with corresponding eigenvalue $\lambda=n$, since $Ax=nx$. But I can't seem to get much further than that in terms of finding eigenvalues for $A$, any hints/suggestions?
|
Consider the matrix $B = A - \lambda I$, where $ \lambda \in \text{Spec}(A)$.
Then $\lambda_{1} = n - \lambda$ is an eigenvalue of $B$ with an associated eigenvector $v_{1} = \begin{pmatrix} 1 \\ 1 \\ 1\\ \cdot \\ \cdot \\ \cdot \\ 1 \\1 \\1\\ \end{pmatrix}$.
The remaining eigenvalues of $B$ are $\lambda_{2} = \lambda_{3} =...=\lambda_{n} = -\lambda$ and the corresponding eigenvectors are:
$v_{2} = \begin{pmatrix} 1 \\ -1 \\ 0\\ \cdot \\ \cdot \\ \cdot \\ 0 \\0 \\0\\ \end{pmatrix}$, $v_{3} = \begin{pmatrix} 0 \\ 1 \\ -1\\ \cdot \\ \cdot \\ \cdot \\ 0 \\0 \\0\\ \end{pmatrix}$, ..., $v_{n} = \begin{pmatrix} 0 \\ 0 \\ 0\\ \cdot \\ \cdot \\ \cdot \\ 0 \\1 \\-1\\ \end{pmatrix}$.
Now, it is easier to see that $v_{1}, ..., v_{n}$ are linearly independent and they are eigenvectors of $A$ with $v_{1}$ associated with the eigenvalue $\lambda =n$ and $v_{j}$ for $j \geq 2$ associated with $\lambda = 0$. From the linearly independence of these eigenvectors we see $A$ is diagonalizable.
|
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|
The sum of k times the kth power of a is given analytically by? I was wondering how would someone derive (Not prove) the result in terms of n and a of the following sum:
$$\sum_{k=1}^n ka^k$$
My question basically is, given that summation how would you tackle the problem in order to get to a solution such as, for example, this:
$\sum_{i=0}^n i^2 = \frac{(n^2+n)(2n+1)}{6}$
|
The trick is to eliminate the $k$ in the sum by integrating:
$$\begin{align*}
\sum_{k = 1}^n ka^k & ~=~ a \sum_{k = 1}^n k a^{k - 1} \\
& ~=~ a \sum_{k = 1}^n \frac{d}{da} \left ( \int ka^{k - 1} \ da\right ) \\
& ~=~ a \frac{d}{da} \left ( \sum_{k = 1}^n \int ka^{k - 1} \ da \right ) \\
& ~=~ a \frac{d}{da} \left ( \sum_{k = 1}^n \left ( a^k + h \right ) \right ) \\
& ~=~ a \frac{d}{da} \left ( \frac{a^{n + 1} - a}{a - 1} + hn \right ) \\
& ~=~ a \left ( \frac{na^{n + 1} - na^n - a^n + 1}{(a - 1)^2}\right ) \\
& ~=~ \frac{na^{n + 2} - na^{n + 1} -a^{n + 1} + a}{(a - 1)^2}
\end{align*}$$
Alternatively, you can differentiate the geometric sum:
$$\begin{align*}
\sum_{k = 1}^n a^k = \frac{a^{n + 1} - a}{a - 1}
& \rightarrow ~ \frac{d}{da} \left ( \sum_{i = 1}^k a^k \right ) = \frac{d}{da} \left ( \frac{a^{n + 1} - a}{a - 1} \right )\\
& \rightarrow ~ \sum_{i = 1}^k ka^{k - 1} = \frac{na^{n + 1} - na^n - a^n + 1}{(a - 1)^2} \\
& \rightarrow ~ \sum_{i = 1}^k ka^k = \frac{na^{n + 2} - na^{n + 1} - a^{n + 1} + a}{(a - 1)^2}
\end{align*}$$
|
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|
Quick Question on Zero Matrix Find all $2\times2$ matrices
$$B =\begin{pmatrix} \alpha & \beta\\ \gamma & \delta\\ \end{pmatrix}$$
such that $B^2$ is the zero matrix.
Any help would be greatly appreciated. Thanks
|
$B^2=0$ means $\operatorname{im}(B)\subseteq\ker(B)$. In particular, $B$ needs to be of rank at most one: If $\dim\operatorname{im}(B)=2$, then $\dim\ker(B)=0$ and the above inclusion is impossible. Furthermore, $B^2=0$ is equivalent to $(SBS^{-1})^2=0$, so we can change bases as we please. Let us assume $B\ne0$ and $B^2=0$, then the kernel and image of $B$ are both of dimension one. We change bases so that $\ker(B)$ is spanned by
$$e_1=\begin{pmatrix}1\\0\end{pmatrix}.$$
Then,
$$\exists a,b\colon\quad B=\begin{pmatrix} 0 & a \\ 0 & b \end{pmatrix}.$$
Since $\operatorname{im}(B)\subseteq\ker(B)$, we have $b=0$. As $B\ne0$, we have $a\ne 0$. By a change of bases, we can achieve that $a=1$, because
$$
\begin{pmatrix} a^{-1} & 0 \\ 0 & 1 \end{pmatrix}
\begin{pmatrix} 0 & a \\ 0 & 0 \end{pmatrix}
\begin{pmatrix} a & 0 \\ 0 & 1 \end{pmatrix}
=
\begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}.
$$
Hence, aside from the zero matrix, every matrix $B$ with $B^2=0$ is of the form
$$S \cdot \begin{pmatrix} 0 & 1 \\ 0 &0\end{pmatrix}\cdot S^{-1}$$
for an invertible matrix $S$.
|
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|
Solving Recurrence Relation problem I am trying to solve the recurrence relation:
$$G_n = \frac{1-G_{n-1}}{4}$$
$$G_0 = 0$$
$$G_1 = \frac{1}{4}$$
I am told that the answer is
$$G_n = \frac{1}{5}\left(1+\left(\frac{-1}{4}\right)^{n+1}\right)$$
I have found the characteristic equation to be
$$G_n = B\left(\frac{-1}{4}\right)^{n}$$
But then I am not sure how to continue.
|
By rewriting the recurrence formula we have $G_n=\dfrac{1}{4}-\dfrac{1}{4}G_{n-1}$.
If we write the first elements of the sequence we get:
$$\begin{align*}
G_0&=0, G_1=\dfrac{1}{4} \\
G_2&=\dfrac{1}{4}-\dfrac{1}{4}G_1 =\dfrac{1}{4}-\left(\dfrac{1}{4}\right)^2\\
G_3&=\dfrac{1}{4}-\dfrac{1}{4}G_2 =\dfrac{1}{4}-\dfrac{1}{4}\left(\dfrac{1}{4}-\left(\dfrac{1}{4}\right)^2\right)\\
& \hspace{2.3cm}=\dfrac{1}{4}-\left(\dfrac{1}{4}\right)^2+\left(\dfrac{1}{4}\right)^3\\
\vdots & \hspace{3cm}\vdots\\
G_n&=^{(?)}\sum_{k=1}^n (-1)^{k+1}\dfrac{1}{4^k}=(-1)\cdot \sum_{k=1}^n \left(\dfrac{-1}{4}\right)^k
\end{align*}$$
Once you have proved the recurrence formula then by properties of geometric series you get
$$G_n=(-1)\cdot \sum_{k=1}^n \left(\dfrac{-1}{4}\right)^k=(-1)\left(\dfrac{1-(-\frac{1}{4})^{n+1}}{1-(-\frac{1}{4})}-1\right)=\dfrac{-1}{1+\frac{1}{4}}\left(1-\left(-\frac{1}{4}\right)^{n+1}-\dfrac{5}{4}\right)=\dfrac{-4}{5}\left(-\dfrac{1}{4}-\left(-\frac{1}{4}\right)^{n+1}\right)=\dfrac{1}{5}\left(1-\left(-\frac{1}{4}\right)^{n}\right)$$
|
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|
Find $\lim_{x\to0}\frac{\ln(1+\sin^3x \cos^2x)\cot(\ln^3(1+x))\tan^4x}{\sin(\sqrt{x^2+2}-\sqrt{2})\ln(1+x^2)}$ $$\lim_{x\to0}\frac{\ln(1+\sin^3x \cos^2x)\cot(\ln^3(1+x))\tan^4x}{\sin(\sqrt{x^2+2}-\sqrt{2})\ln(1+x^2)}$$
I don't think L'hospital's rule will make the problem easy. (I am afraid to differentiate the numerator). The given limit has a $\frac{0}{0}$ form. I tried using taylor series but the it made the problem more complicated.
|
Use equivalents:
$$\ln(1+u)\sim_0 u, \enspace\sqrt{1+u}-1\sim_0 \dfrac u2, \enspace \sin u\sim_0 u, \enspace \cos u\sim_0 1, \enspace\cot u\sim_0 \dfrac1u, \enspace \tan u\sim_0 u$$
hence $$\frac{\ln(1+\sin^3x \cos^2x)\cot(\ln^3(1+x))\tan^4x}{\sin(\sqrt{x^2+2}-\sqrt{2})\ln(1+x^2)}\sim_0 \frac{x^3\cfrac1{x^3}x^4}{\sqrt2\Bigl(\sqrt{1+\frac{x^2}2}-1\Bigr)x^2}\sim_0 \frac{x^2}{\sqrt2\dfrac{x^2}{4}}=2\sqrt2.$$
|
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|
$f(z) = \frac{1}{z^2-2z+2}$ - $|f^{n}(0)| \leq n!$ - Cauchy inequality Let the function $f(z) = \frac{1}{z^2-2z+2}$. Show that for each integer $n \geq 0$, we have $|f^{n}(0)| \leq n!$.
Honnestly, I don't know how to do that problem, maybe in using the Cauchy inequality (complex analysis).
Is anyone could help me at this point?
|
We'll use the following result from geometric series to compute the power series of $f$.
Let $c\neq 0$, the power series of $\frac{1}{z-c}$ around $0$ is $\frac{1}{z-c}=\frac{\frac{-1}{c}}{1-\frac{z}{c}}=\frac{-1}{c} (1+(\frac{z}{c})+(\frac{z}{c})^2+(\frac{z}{c})^3+...)$, radius of convergence is $c$.
$f(z)=\frac{1}{(z-a)(z-b)}$, $a=1+i=\sqrt{2}e^{\pi i/4}$, $b=1-i=\sqrt{2}e^{-\pi i/4}$
$$f(z)\\=\frac{1}{z-a}\frac{1}{z-b}\\=\frac{-1}{a}\frac{-1}{b}(1+(\frac{z}{a})+(\frac{z}{a})^2+(\frac{z}{a})^3+...)(1+(\frac{z}{b})+(\frac{z}{b})^2+(\frac{z}{b})^3+...)\\=\frac{1}{2}(1+(\frac{z}{a})+(\frac{z}{a})^2+(\frac{z}{a})^3+...)(1+(\frac{z}{b})+(\frac{z}{b})^2+(\frac{z}{b})^3+...)$$
Because $\frac{f^{(n)}(0)}{n!}=c_n=$ the coefficient in front of $z^n$ in the expansion of $f(z)$. It's sufficient to prove $|c_n|\leq 1$ for each $n$.
Not hard to see from the above equation $c_n=\frac{1}{2}(b^{-n}+a^{-1}b^{-n+1}+...+a^{-n})=\frac{b^{-n-1}-a^{-n-1}}{2(b^{-1}-a^{-1})}=\frac{1}{2(\sqrt2)^n}\frac{\cos((n+1)\frac{\pi}{4})}{\cos(\frac{\pi}{4})}\leq \frac{1}{2\sqrt{2}^{n-1}}\leq 1$.
|
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Find a Cartesian equation for the curve and identify it. $ r^2 \cos 2\theta = 1$
Find a Cartesian equation for the curve and identify it. $$ r^2 \cos 2\theta = 1$$
I'm confused by the $2\theta.$
I isolated $r^2$ to get $r^2 = \frac{1}{\cos2\theta}$
Now, normally if it was just a $\cos \theta$ I would multiply both sides by $\frac{1}{r}$ and then substitute $r$ for $\sqrt {x^2+y^2}$ and then substitute $r\cos\theta$ for $x$. But the $2\theta$ doesn't allow for that to happen.
|
Since $\cos 2 \theta = 2 \cos^2 \theta -1$, then $$r^2 \cos 2 \theta = 2 r^2 \cos^2 \theta - r^2 = 2(r \cos \theta)^2 - r^2 = 2x^2 - (x^2 + y^2) = x^2 - y^2 ,$$ so your curve is $x^2 - y^2 = 1$ which is known to be a hyperbola having the lines $y = \pm x$ as asymptotes.
|
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|
Simplifying Fractions involving negative numbers I want to simplify $$\frac{\frac{7}{-10} \times \frac{-15}{6}}{\frac{7}{-19} + \frac{-17}{-8}}$$
I really don't understand how to do this, or even how to start?
Negative numbers make it even harder for me to try and simplify it.
|
A few important facts to know are the following. For all complex numbers $a,b, n$:
$$ \frac{-a}{b} = \frac{a}{-b} = -\left(\frac{a}{b}\right) $$
$$ \frac{a}{b} + \frac{c}{b} = \frac{a+c}{b} $$
$$ \frac{n\times a}{n \times b} = \frac{a}{b} $$
$$ \frac{a}{b} = a \times \frac{1}{b} $$
In particular, by the last identity above:
$$ \frac{a}{\frac{b}{c}} = a \times \frac{c}{b} $$
Now it follows that:
\begin{align}
\frac{\frac{7}{-10} \times \frac{-15}{6}}{\frac{7}{-19} + \frac{-17}{-8}} & = \frac{\frac{7}{10} \times \frac{15}{6}}{\frac{-7}{19} + \frac{17}{8}} \\
& = \frac{\frac{7}{4}}{\frac{-7\times8 + 17\times19}{19\times8}} \\
& = \frac{\frac{7}{4}}{\frac{267}{152}} \\
& = \frac{7}{4} \times \frac{152}{267} \\
& = \frac{1064}{1068}
\end{align}
You can simplify this last expression by yourself if you wish.
|
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In how many ways can $10$ chocolates be distributed to $3$ people such that no one gets more than $4$? $10$ chocolates are distributed to $3$ people such that no one gets more than $4$ or $A,B,C \leq 4$. How many ways can this be done?
I know how to do this if the condition is every person gets at least one. Here the condition is different. So need help.
|
drhab has provided you with an elegant solution to this problem based on the assumption that the chocolates are indistinguishable. I will make the same assumption.
We wish to distribute ten chocolates to three people so that no person receives more than four chocolates. The number of ways we can do this is equal to the number of solutions of the equation
$$x_1 + x_2 + x_3 = 10 \tag{1}$$
in the non-negative integers subject to the restrictions that $x_1, x_2, x_3 \leq 4$.
The number of ways ten chocolates can be distributed to three people is the number of solutions of equation 1 in the non-negative integers. A particular solution corresponds to the placement of two addition signs in a row of ten ones. For instance,
$$+ 1 1 1 1 1 1 + 1 1 1 1$$
corresponds to the solution $x_1 = 0$, $x_2 = 6$, and $x_3 = 4$, while
$$1 1 1 1 1 + 1 1 1 + 1 1$$
corresponds to the solution $x_1 = 5$, $x_2 = 3$, and $x_4 = 2$. Thus, the number of solutions of equation 1 in the non-negative integers is the number of ways two addition signs can be inserted into a row of ten ones, which is
$$\binom{10 + 2}{2} = \binom{12}{2}$$
since we must choose which two of the twelve symbols (ten ones and two addition signs) will be addition signs.
However, we have counted solutions in which at least one person receives more than four chocolates. We must exclude those cases.
Notice that $2 \cdot 5 = 10$, so at most two of the people could receive more than four chocolates.
Suppose $x_1 > 4$. Since $x_1$ is an integer, then $x_1 \geq 5$. Let $y_1 = x_1 - 5$. Then $y_1$ is a non-negative integer. Substituting $y_1 + 5$ for $x_1$ in equation 1 yields
\begin{align*}
y_1 + 5 + x_2 + x_3 & = 10\\
y_1 + x_2 + x_3 & = 5 \tag{2}
\end{align*}
Equation 2 is an equation in the non-negative integers with
$$\binom{5 + 2}{2} = \binom{7}{2}$$
solutions. There are $\binom{3}{1}$ ways to select a person to receive more than four chocolates. Hence, the number of ways of distributing ten chocolates to three people so that one person receives more than four chocolates is
$$\binom{3}{1}\binom{7}{2}$$
However, if we subtract this number from the total number of ways of distributing ten chocolates, we will have subtracted those solutions in which two people received more than four chocolates twice. Suppose $x_1, x_2 > 4$. Let $y_1 = x_1 - 5$; let $y_2 = x_2 - 5$. Then $y_1, y_2$ are non-negative integers. Substituting $y_1 + 5$ for $x_1$ and $y_2 + 5$ for $x_2$ in equation 1 yields
\begin{align*}
y_1 + 5 + y_2 + 5 + x_3 & = 10\\
y_1 + y_2 + x_3 & = 0 \tag{3}
\end{align*}
Equation 3 is an equation in the non-negative integers with
$$\binom{0 + 2}{2} = \binom{2}{2} = 1$$
solution (namely $y_1 = y_2 = x_3 = 0$). Since there are $\binom{3}{2}$ ways of selecting two people to receive five chocolates each, the number of ways two people could receive more than four chocolates is
$$\binom{3}{2}\binom{2}{2}$$
By the Inclusion-Exclusion Principle, the number of ways ten chocolates can be distributed to three people so that no person receives more than chocolates is
$$\binom{12}{2} - \binom{3}{1}\binom{7}{2} + \binom{3}{2}\binom{2}{2} = 6$$
As a check, they are $(2, 4, 4), (4, 2, 4), (4, 4, 2), (3, 3, 4), (3, 4, 3), (4, 3, 3)$.
|
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positive definite of a matrix(diagonal decay) Is this matrix positive definite:
$$\begin{bmatrix}a & b &c & d\\b & b & c & d\\c & c & c & d\\d&d&d&d\end{bmatrix}$$
when $a>b>c>d>0$?
I tried several $a,b,c,d$, the result seems to be true, but I cannot prove it. Thanks!
|
Thanks to @Friedrich Philipp and @Omnomnomnom's comments, I notice huge defects in my original proof. Instead, I would like to present an another
proof followed by @Omnomnomnom's method.
Denote the given matrix by $A$. Observe that
\begin{align}
A=\begin{bmatrix}a & b &c & d\\b & b & c & d\\c & c & c & d\\d&d&d&d\end{bmatrix}
&=
\begin{bmatrix}a-b & 0 &0 &0 \\0 & 0& 0 &0 \\ 0& 0 & 0 & 0\\0& 0& 0&0\end{bmatrix}
+
\begin{bmatrix}b-c & b-c &0 &0 \\b-c & b-c& 0 &0 \\ 0& 0 & 0 & 0\\0& 0& 0&0\end{bmatrix}\\
&+
\begin{bmatrix}c-d & c-d &c-d &0 \\c-d & c-d& c-d &0 \\ c-d& c-d & c-d & 0\\0& 0& 0&0\end{bmatrix}
+\begin{bmatrix}d & d &d &d\\d& d& d &d \\ d& d & d & d\\d& d& d&d\end{bmatrix}\\
&=A_1+A_2+A_3+A_4,
\end{align}
where $A_i$ denotes the above corresponding decomposed matrix. Then given
$x=(x_1,x_2,x_3,x_4)\in{\sf R}^4$, we see that
\begin{align}
x^tAx&=x^tA_1x+x^tA_2x+x^tA_3x+x^tA_4x\\
&=(a-b)x_1^2+(b-c)(x_1+x_2)^2+(c-d)(x_1+x_2+x_3)^2+d(x_1+x_2+x_3+x_4)^2.
\end{align}
That is, $x^tAx$ is the sum of the four non-zero terms. Now, if $x\neq{\it 0}$, we discuss the following cases.
*
*$x_1\neq 0$: It is clear that
$x^tAx\geq(a-b)x_1^2>0$.
*$x_1=0$ and $x_2\neq 0$: It follows that $x^tAx\geq(b-c)x_2^2>0$.
*$x_1=x_2=0$ and $x_3\neq 0$: It follows that $x^tAx\geq(c-d)x_3^2>0$.
*$x_1=x_2=x_3=0$ and $x_4\neq 0$: It follows that $x^tAx= dx_4^2>0$.
Hence we conclude that $x^tAx>0$ for all $x\neq{\it 0}$, that is, $A$ is positive definite.
|
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Trigonometric inequality - x domain the following equality is given:
$$2 \sqrt2\sin{}x+\sqrt2\cos{x}=\sqrt{-\sin2x}$$
I managed to solve it, but I have a problem with this inequality to define x domain:$$2 \sqrt2\sin{}x+\sqrt2\cos{x}\ge0$$
Wolfram says:
$$x \in\left\langle 2\arctan(2- \sqrt{5}) + 2k \pi, 2\arctan(2+ \sqrt{5} ) + 2k \pi \right\rangle$$
but I don't know how to get to this point. I've tried some transformations and came up with this:
$$3\cos(\frac{\pi}{4}-x)-\cos(x+\frac{\pi}{4})\ge0$$
and still can't go any further.
Any help will be appreciated :)
|
Since $\sqrt{(2\sqrt 2)^2+(\sqrt 2)^2}=\sqrt{10}$,
we have$$\begin{align}2\sqrt 2\sin x+\sqrt 2\cos x&=\sqrt{10}\left(\frac{2\sqrt{2}}{\sqrt{10}}\sin x+\frac{\sqrt 2}{\sqrt{10}}\cos x\right)\\&=\sqrt{10}\left(\frac{2}{\sqrt 5}\sin x+\frac{1}{\sqrt 5}\cos x\right)\\&=\sqrt{10}(\cos\theta\sin x+\sin \theta\cos x)\\&=\sqrt{10}\sin(x+\theta)\end{align}$$
where
$$\cos\theta=\frac{2}{\sqrt 5},\quad \sin\theta=\frac{1}{\sqrt 5}\quad\Rightarrow \quad \tan\theta=\frac 12\quad\Rightarrow \quad \theta=\arctan(1/2)$$
Thus, for $k\in\mathbb Z$,
$$\begin{align}&2\sqrt 2\sin x+\sqrt 2\cos x\ge 0\\&\iff \sqrt{10}\sin(x+\arctan(1/2))\ge 0\\&\iff 0+2k\pi\le x+\arctan(1/2)\le \pi+2k\pi\\&\iff -\arctan(1/2)+2k\pi\le x\le \pi-\arctan(1/2)+2k\pi\end{align}$$
We can see that this is the same as
$$2\arctan(2-\sqrt 5)+2k\pi\le x\le 2\arctan(2+\sqrt 5)+2k\pi$$
because
$$\begin{align}&2\arctan(2-\sqrt 5)+\arctan(1/2)\\&=\arctan(2-\sqrt 5)+\arctan(2-\sqrt 5)+\arctan(1/2)\\&=\arctan(2-\sqrt 5)+\arctan\left(\frac{2-\sqrt 5+(1/2)}{1-(2-\sqrt 5)/2}\right)\\&=\arctan(2-\sqrt 5)+\arctan(\sqrt 5-2)\\&=0\end{align}$$
and
$$\begin{align}&2\arctan(2+\sqrt 5)+\arctan(1/2)\\&=\arctan(2+\sqrt 5)+\arctan(2+\sqrt 5)+\arctan(1/2)\\&=\arctan(2+\sqrt 5)+\pi+\arctan\left(\frac{2+\sqrt 5+(1/2)}{1-(2+\sqrt 5)/2}\right)\\&=\arctan(2+\sqrt 5)+\pi+\arctan(-2-\sqrt 5)\\&=\pi\end{align}$$
|
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|
Integrate $\int_0^{2 \pi} \frac{2}{\cos^{6}(x) + \sin^{6}(x)} dx$ Integrate $$\int_0^{2 \pi} \frac{2}{\cos^{6}(x) + \sin^{6}(x)} dx$$
I have tried to do Tangent substitution, but there is huge power.
What method is better to use here?
|
$$
\begin{aligned}
\int_0^{2 \pi} \frac{2}{\cos^6 x + \sin^6 x} dx
&=8\int_0^{\frac{\pi}{2}} \frac{1}{\cos^6 x + \sin^6 x} dx\\
&=8\int_0^{\frac{\pi}{2}} \frac{1}{(\cos^2 x + \sin^2 x)(\cos^4 x-\cos^2 x\sin^2 x+\sin^4 x)} dx\\
&=8\int_0^{\frac{\pi}{2}} \frac{1}{\cos^4 x-\cos^2 x\sin^2 x+\sin^4 x} dx\\
&=8\int_0^{\frac{\pi}{2}} \frac{1}{\cos^4 x}\frac{1}{1-\tan^2x+\tan^4 x} dx\\
&=8\int_0^{\frac{\pi}{2}} \frac{dx}{\cos^2 x}\frac{1}{\cos^2 x}\frac{1}{1-\tan^2x+\tan^4 x}\\
&=8\int_0^{\frac{\pi}{2}} d(\tan x)\left(1+\tan^2 x\right)\frac{1}{1-\tan^2x +\tan^4 x}\\
&=8\int_0^{+\infty} \frac{(1+t^2)}{1-t^2+t^4} dt\\
&=8\int_0^{+\infty} \left(\frac{1}{t^2}+1\right)\frac{1}{\frac{1}{t^2}-1+t^2} dt\\
&=8\int_0^{+\infty} \frac{d\left(t-\frac{1}{t}\right)}{\left(t-\frac{1}{t}\right)^2+1}\\
&=8\int_{-\infty}^{+\infty} \frac{ds}{s^2+1}\\
&=8\operatorname{arctan} s|_{-\infty}^{+\infty}\\
&=8\pi
\end{aligned}
$$
|
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|
How do i find sequences using linear algebra? My task is this;
Find two sequences $\{x_n\}$ and $\{y_n\}$ such that $$x_{n+1} = x_n + 3y_n\\ y_{n+1} = 2x_n + 2y_n$$
When $x_0 = 5, y_0 = -5$.
My work so far (not sure if this is the right approach):
We are interested in finding eigenvectors and values to this system. Let $A = \begin{pmatrix}1 & 3\\2 & 2\end{pmatrix}$, $r_{n+1} = \begin{pmatrix}x_{n+1}\\y_{n+1}\end{pmatrix}$, then $r_{n+1} = Ar_n = A^nr_0$. Finding eigenvalues/vectors to $A$ we use $det(\lambda I_2-A) = 0 \to \begin{vmatrix}\lambda - 1 & -3\\-2& \lambda - 2\end{vmatrix} = (\lambda -1)(\lambda - 2) - 6 =\\ \lambda^2 - 3\lambda -4 = (\lambda +1)(\lambda - 4) \to \lambda = \{-1, 4\}$
Solving for each eigenvalue i.e. $Av_1 = -v_1$ and $Av_2 = 4v_2$ gives us eigenvectors $v_1 =\begin{pmatrix}3&-2\end{pmatrix}, v_2 = \begin{pmatrix}1&1\end{pmatrix}$. If i did this right.
Now i'm not sure if this is the right way and certainly not to proceed so any hints and good explanations would be more than welcome.
Thanks in advance!
|
So now you know
$$A=\begin{pmatrix}1&3\\2&2\end{pmatrix}=\overbrace{\begin{pmatrix}3&1\\\!\!-2&1\end{pmatrix}}^{=P}\begin{pmatrix}\!\!-1&0\\0&4\end{pmatrix}\overbrace{\begin{pmatrix}\frac15&\!\!-\frac15\\\frac25&\frac35\end{pmatrix}}^{=P^{-1}}\implies$$
$$A^n=P\begin{pmatrix}(-1)^n&0\\0&4^n\end{pmatrix}P^{-1}=\frac15\begin{pmatrix}3(-1)^n+2\cdot4^n&3(-1)^{n+1}+3\cdot4^n\\2(-1)^{n+1}+2\cdot4^n&2(-1)^n+3\cdot4^n\end{pmatrix}\implies$$
The sequences are then
$$\binom{x_{n+1}}{y_{n+1}}=A^n\binom{\;5}{\!-5}=\frac15\begin{pmatrix}3(-1)^n+2\cdot4^n&3(-1)^{n+1}+3\cdot4^n\\2(-1)^{n+1}+2\cdot4^n&2(-1)^n+3\cdot4^n\end{pmatrix}\binom{\;5}{\!-5}=$$$${}$$
$$=\begin{pmatrix}6(-1)^n-4^n\\{}\\4(-1)^{n-1}-4^n\end{pmatrix}$$
|
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|
Finding when $y=1+2\sin(x)$ and $y=2\sin(\frac{x}{2})+2\cos(\frac{x}{2})$ are equal How would you solve the following equation?
$$1 +2\sin(x) = 2\sin(\tfrac{x}{2}) +2\cos(\tfrac{x}{2}) \quad\text{for }−2\pi\leq x \leq 2\pi.$$
Steps I tried:
\begin{align}
1+2\sin(x) &= 2\sin(\tfrac{x}{2}) +2\cos(\tfrac{x}{2}) \\
2\sin(x) -2\sin(\tfrac{x}{2}) &= 2\cos(\tfrac{x}{2}) -1 \\
2[\sin(x) -\sin(\tfrac{x}{2})] &= 2\cos(\tfrac{x}{2}) -1 \\
\sin(x) -\sin(\tfrac{x}{2}) &= \cos(\tfrac{x}{2}) -\tfrac{1}{2} \\
\sin(x) +\tfrac{1}{2} &= \cos(\tfrac{x}{2}) +\sin(\tfrac{x}{2})
\end{align}
Now, squaring both sides,
$$\sin^2(x) +\tfrac{1}{4} = \cos^2(\tfrac{x}{2}) +\sin^2(\tfrac{x}{2}).$$
Using $\sin^2(x) +\cos^2(x) = 1$,
\begin{align}
\sin^2(x) &= 1 -\tfrac{1}{4} \\
\sin(x) &= \mp\sqrt\frac{3}{4}.
\end{align}
When I follow through, I get the solutions below the $x$-axis but not above. Can anyone please explain what's wrong with my solution?
|
HINT:
Avoid squaring as it immediately introduces extraneous root.
Using $\sin x=\sin2\cdot\dfrac x2=2\sin\dfrac x2\cos\dfrac x2$
$$1+2\sin x=2\left(\sin\dfrac x2+\cos\dfrac x2\right)$$
$$\iff\left(2\sin\dfrac x2-1\right)\left(2\cos\dfrac x2-1\right)=0$$
Hope you can take it from here!
|
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|
Maclaurin serie of $\int_0^x\frac{sin(t)}{t}$ If $f(x)=\int_0^x\frac{\sin(t)}{t}$. Show that
$$f(x)=x-\frac{x^3}{3*3!}+\frac{x^5}{5*5!}-\frac{x^7}{7*7!}+...$$
Calculate f(1) to three decimal places.
Would you mind showing how to build this Maclaurin serie?
|
$$\sin x = \sum^{\infty}_{n=0} \frac{(-1)^n}{(2n+1)!} x^{2n+1} = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \cdots$$
Now, we divide by $x$ to get that
$$\frac{\sin x}{x} = \sum^{\infty}_{n=0} \frac{(-1)^n}{(2n+1)!} x^{2n} = 1 - \frac{x^2}{3!} + \frac{x^4}{5!} - \cdots$$
Now, we write
$$f(x)=\int_0^x\frac{\sin tdt}{t}=\int_0^x\sum^{\infty}_{n=0} \frac{(-1)^n}{(2n+1)!} t^{2n}=\sum^{\infty}_{n=0} \int_o^x\frac{(-1)^n}{(2n+1)!} t^{2n}dt=\sum^{\infty}_{n=0} \frac{(-1)^n}{(2n+1)(2n+1)!} t^{2n+1}\Big|_0^x=\sum^{\infty}_{n=0} \frac{(-1)^n}{(2n+1)(2n+1)!} x^{2n+1}=x-\frac{x^3}{3*3!}+\frac{x^5}{5*5!}-\frac{x^7}{7*7!}+\cdots$$
To calculate $f(1)$, subsitute $1$ for $x$ into the derived series and hammer through some arithmetic until you find the required accuracy.
|
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|
Solve: $\int x^3\sqrt{4-x^2}dx$ $$\int x^3\sqrt{4-x^2}dx$$
I tried changing the expression like this:
$$\int x^3\sqrt{2^2-x^2}dx=\int x^3\sqrt{(2-x)(2+x)}dx$$
And that's where I got stuck. I tried integration by parts but it doesn't simplify the expression.
|
Here's yet another way. First write: $$\int x^3\sqrt{4-x^2} \, dx = \int x^2 \sqrt{4-x^2} \, \cdot x \, dx$$
Now let $u = 4-x^2$, so then $du = -2x\, dx$ (meaning $x\,dx = -\frac{1}{2}du$), and $x^2 = 4 - u$. Then you get:
\begin{align}
\int x^2 \sqrt{4-x^2} \, \cdot x \, dx
&= \int (4-u)\sqrt{u} \, \cdot \left(-\frac{1}{2}\right) \, du\\[0.3cm]
&= -\frac{1}{2} \int (4u^{1/2} - u^{3/2}) \, du\\[0.3cm]
&= -\frac{1}{2}\left(4 \cdot \frac{2}{3} u^{3/2} - \frac{2}{5}u^{5/2}\right) + C\\[0.3cm]
&= \frac{1}{5}(4-x^2)^{5/2} - \frac{4}{3}(4-x^2)^{3/2} + C
\end{align}
|
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|
Possible number of terms in an Arithmetic Progression The sum of the first $n$ $(n>1)$ terms of the A.P. is $153$ and the common difference is $2$. If the first term is an integer , then number of possible values of $n$ is
$a)$ $3$
$b)$ $4$
$c)$ $5$
$d)$ $6$
My approach : I used the formula for the first $n$ terms of an A.P. to arrive at the following quadratic equation
$n^2 + n(a-1) -153 = 0 $
Next up I realised that since we are talking about the number of terms , thus the
possible values which n can take must be whole numbers. That is the discriminant of the above quadratic should yield a whole number in other words
$ (a-1)^2 + 612 = y^2 $ for some y .
However I am stuck at this point , as from here I am unable to figure out the number of such a's ( i.e. the initial terms of an AP ) which will complete the required pythagorean triplet
The answer mentioned is $5$
Please let me know , if I am doing a step wrong somewhere . Or If you have a better solution , that will be welcomed too.
|
\begin{align}
n^2 +(a-1)n - 153 &= 0 \\
(n+u)(n-v) &= 0 \\
n^2 + (u-v)n - uv &= 0
\end{align}
So you need two positive integers, say $u>v>0$, such that $uv=153$ and $u-v=a-1$. Which will have solution $n=v$.
There aren't that many possibilities
\begin{array}{|rr|r|rr|}
\hline
u & v & u-v & a & n \\
\hline
153 & 1 & 152 & 153 & 1 \\
51 & 3 & 48 & 49 & 3 \\
17 & 9 & 8 & 9 & 9 \\
\hline
\end{array}
Above I was assuming $a > 1$. If you allow $a$ to be a negative number, then the analysis changes a bit.
$$n^2 +(a-1)n - 153 = 0$$
The divisors of $153$ are $1,3,9,17,51,153$ We find the following.
\begin{array}{|l|rr|}
\hline
\text{factors} & a & n\\
\hline
(n-1)(n+153) = n^2+152n - 153 & 153 & 1 \\
(n-3)(n+ 51) = n^2 +48n - 153 & 49 & 3 \\
(n-9)(n+17) = n^2 +8n - 153 & 9 & 9 \\
(n-17)(n+9) = n^2 -8n - 153 & -7 & 17 \\
(n-51)(n+3) = n^2 -48n - 153 & -47 & 51 \\
(n-153)(n+1) = n^2-152n - 153 & -151 & 153 \\
\hline
\end{array}
|
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|
How to solve the congruence $y^{31}\equiv 3 \mod{100}$ $\phi (100) = 40$
Hence:
$y^{31}\equiv y^{-9} \equiv 3\mod{100}$
$y^{9}\equiv 67 \mod{100}$
However I do not know where to go from here.
|
This is a kind of tedious solution:
$$x^{31}\equiv 3\pmod{100}\iff \begin{cases}x^{31}\equiv 3\pmod{25}\\ x^{31}\equiv 3\pmod{4}\end{cases}$$
$$\iff \begin{cases}x^{31}\equiv 3\pmod{25}\\ x\equiv 3\pmod{4}\end{cases}$$
Clearly $\gcd(x,25)=1$, so by Euler's theorem:
$$\iff \begin{cases}x^{11}\equiv 3\pmod{25}\\ x\equiv 3\pmod{4}\end{cases}$$
$2$ is a primitive root mod $5^2$, because $\gcd(2,25)=1$ and $\phi(25)=20=2^2\cdot 5$ and
$2^{\frac{20}{2}}\equiv 2^{10}\equiv 1024\equiv -1\not\equiv 1\pmod{25}$
and $2^{\frac{20}{5}}\equiv 2^{4}\equiv 16\not\equiv 1\pmod{25}$.
Let $x\equiv 2^k\pmod{25}$ for some $k\in\mathbb Z^+$.
$$2^{11k}\equiv 3\equiv 2^7\pmod{25}\iff 11k\equiv 7\equiv -33\pmod{20}$$
$$\stackrel{:11}\iff k\equiv -3\equiv 17\pmod{20}$$
Therefore $x\equiv 2^{17}\equiv 2^{10}2^{7}\equiv (1024)(128)\equiv (-1)(3)\equiv -3\equiv 22\pmod{25}$,
so by Chinese Remainder Theorem $x\equiv 47\pmod{100}$.
|
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|
integral $\int_0^1\frac{1}{\sqrt{1+x^4}}\text{d}x$
How to prove that $$0.78<\int_0^1\frac{1}{\sqrt{1+x^4}}\text{d}x<0.93$$
approch: $x^2<\sqrt{1+x^4}<x^2+\frac{1}{2x^2}$
Any hint would be appreciated.
|
I know this doesn't constitute as a proof, but I thought the following was interesting. It turns out we can use the Beta function to express your integral in terms of the Gamma function by taking advantage of the fact that
\begin{align*}
B(x,y) = \int_0^\infty \frac{t^{x-1}}{(1+t)^{x+y}}dt.
\end{align*}
First, substituting $x=t^4$, we see that
\begin{align*}
\int_0^1\frac{dx}{\sqrt{1+x^4}}= \frac{1}{4}\int_0^1 \frac{t^{-3/4}}{(t+1)^{1/2}} \, dt.
\end{align*}
By making a substitution such as $u=1/t$, after simplification one ends up with an identical integral but instead over $(1,\infty)$, so we can deduce that
\begin{align*}
B(1/4,1/4)=\int_0^\infty \frac{t^{-3/4}}{(t+1)^{1/2}} \, dt=2\int_0^1 \frac{t^{-3/4}}{(t+1)^{1/2}} \, dt.
\end{align*}
It follows that
\begin{align*}
\frac{1}{4}\int_0^1 \frac{t^{-3/4}}{(t+1)^{1/2}}dt &= \frac{B(1/4,1/4)}{8}\\
&= \frac{\Gamma(1/4)\Gamma(1/4)}{8\Gamma(1/2)}\\
&=\frac{\Gamma(1/4)^2}{8\sqrt \pi}.
\end{align*}
From here there are probably ways of getting algebraic bounds that would be better for a proof. This particular value of Gamma has been approximated to death however, and so we see that your original integral is
\begin{align*}
\int_0^1\frac{dx}{\sqrt{1+x^4}} = \frac{\Gamma(1/4)^2}{8\sqrt \pi} \approx 0.927037.
\end{align*}
|
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|
Prove that $36|n^{12k-6}-n^6$ Prove that if gcd(n,6)=1 and k>0, then $36|n^{12k-6}-n^6$.
Idea: I want to show that $2|n^{12k-6}-n^6$ and $3|n^{12k-6}-n^6$.
For the first part, since gcd(n,6)=1, n must be an odd number, so it is easy to show that $n^{12k-6}-n^6$ is divisible by 2. I am stuck by the second part and i think that I need to use Fermat's little theorem somehow. Anyone can give some hint?
|
Let $n$ and $k>0$ integers. Notice that
\begin{align*}
n^{12k-6}-n^6&=\left(n^{6k-3}-n^3\right)\left(n^{6k-3}+n^3\right)\\
&=\left(n^{2k-1}-n\right)\left(n^{4k-2}+n^{2k}+n^2\right)\left(n^{2k-1}+n\right)\left(n^{4k-2}-n^{2k}+n^2\right)\\
&=n^6\left(n^{2k-2}-1\right)\left(n^{4k-4}+n^{2k-2}+1\right)\left(n^{2k-2}+1\right)\left(n^{4k-4}-n^{2k-2}+1\right)
\end{align*}
*
*If $n$ is even, then $2^6=64$ divides $n^{12k-6}-n^6$.
*For $n$ odd we have $4|(n^{2k-2}-1)$ and $2|(n^{2k-2}+1)$, so $8$ divides $n^{12k-6}-n^6$.
*If $3|n$, then $729$ divides $n^{12k-6}-n^6$.
*If $3$ does not divide $n$, then the squares $n^{4k-4}$ and $n^{2k-2}$ gives remainder $1$ under division by $3$, then $3$ divides $n^{2k-2}-1$ and $n^{4k-4}+n^{2k-2}+1$, so $9$ divides $n^{12k-6}-n^6$.
Then, $72$ divides $n^{12k-6}-n^6$ for any integer $n$.
|
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|
None of $3,5,7$ can divide $r^4+1$ Let $n=r^4+1$ for some $r$. Show that none of $3,5,$ and $7$ can divide $n$.
I am thinking to use a corollary that "each prime divisor p of an integer of the form $(2m)^4+1$ has the form $8k+1$", but I failed. Anyone can give some hint?
|
Use Fermat little theorem. If $r $ isn't divisible by 3, $r^2 \equiv 1 \mod 3$. If $r$ is divisible by 3 then $r^2 \equiv 0 \mod 3$. So $r^4 \not \equiv -1 \mod 3$ so $r^4 + 1$ is not divisible by 3.
Similarly $r^4 \equiv 1,0 \mod 5$ so $r^4 + 1$ is not divisible by 5.
If $r$ is not divisible by 7 then $r^6 \equiv 1 \mod 7$. So if $r^4 \equiv -1 \mod 7$ then $r^8 \equiv r^2 \equiv 1 \mod 7$. So, contradictory, $r^4 \equiv 1 \mod 7$. So $r^4 \not \equiv -1 \mod 7$. So $r^4 + 1$ is never disible by 7.
|
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|
How to find whether equation of angle bisector represents the obtuse or acute angle bisector of two given straight lines? Two lines: $a_1x + b_1y + c_1 = 0$ and $a_2x + b_2y + c_2 = 0$ are given. I know that the equation of its bisectors is ${a_1x + b_1y + c_1 \over \sqrt{(a_1^2 + b_1^2)}} = \pm {a_2x + b_2y + c_2 \over\sqrt{ (a_2^2 + b_2^2)}}$ But I intend to find which one is the obtuse angle bisector and which one is the acute angle bisector. I want to find a general formula Assuming $c_1 , c_2$ both are of same sign, I know if $a_1a_2 + b_1b_2 > 0$ and if we take the positive sign we get the obtuse angle bisector and vice versa. But I want to prove it using general equation of line, I tried to find the angle between bisector and original line i.e. $tan θ = {m_1 - m_2 \over 1+ m_1m_2}$ and then if it is greater than one it will be of obtuse angle but calculations are tough if we use general equation of line. May anyone give a simple proof of the following statement: "Assuming $c_1 , c_2$ both are of same sign IF $a_1a_2 + b_1b_2 > 0 $then if we take positive sign we get the obtuse angle bisector".
|
We have two lines :
$$L_1 : a_1x+b_1y+c_1=0,\quad L_2 : a_2x+b_2y+c_2=0$$
and the angle bisectors :
$$L_{\pm} : \frac{a_1x+b_1y+c_1}{\sqrt{a_1^2+b_1^2}}=\pm\frac{a_2x+b_2y+c_2}{\sqrt{a_2^2+b_2^2}}$$
If we let $\theta$ be the (smaller) angle between $L_+$ and $L_1$, then we have
$$\cos\theta=\frac{\left|a_1\left(\frac{a_1}{\sqrt{a_1^2+b_1^2}}-\frac{a_2}{\sqrt{a_2^2+b_2^2}}\right)+b_1\left(\frac{b_1}{\sqrt{a_1^2+b_1^2}}-\frac{b_2}{\sqrt{a_2^2+b_2^2}}\right)\right|}{\sqrt{a_1^2+b_1^2}\sqrt{\left(\frac{a_1}{\sqrt{a_1^2+b_1^2}}-\frac{a_2}{\sqrt{a_2^2+b_2^2}}\right)^2+\left(\frac{b_1}{\sqrt{a_1^2+b_1^2}}-\frac{b_2}{\sqrt{a_2^2+b_2^2}}\right)^2}}$$
$$=\frac{\left|\sqrt{a_1^2+b_1^2}-\frac{a_1a_2+b_1b_2}{\sqrt{a_2^2+b_2^2}}\right|}{\sqrt{a_1^2+b_1^2}\sqrt{2-2\frac{a_1a_2+b_1b_2}{\sqrt{(a_1^2+b_1^2)(a_2^2+b_2^2)}}}}\times\frac{2\frac{1}{\sqrt{a_1^2+b_1^2}}}{2\frac{1}{\sqrt{a_1^2+b_1^2}}}=\sqrt{\frac{1-\frac{a_1a_2+b_1b_2}{\sqrt{(a_1^2+b_1^2)(a_2^2+b_2^2)}}}{2}}$$
Hence, we can see that
$$\begin{align}a_1a_2+b_1b_2\gt 0&\iff\cos\theta\lt 1/\sqrt 2\\&\iff \theta\gt 45^\circ\\&\iff \text{$L_+$ is the obtuse angle bisector}\end{align}$$
as desired.
(Note that "$c_1,c_2$ both are of same sign" is irrelevant.)
|
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|
How to evaluate $\lim _{x\to \infty }\:\frac{\left(\sqrt{1+\frac{x^3}{x+1}}-x\right)\ln x}{x\left(x^{\frac{1}{x}}-1\right)+\sqrt{x}\ln^2x}$? I have a problem with this limit, I have no idea how to compute it. Can you explain the method and the steps used(without L'Hopital if is possible)? Thanks
$$\lim _{x\to \infty }\:\frac{\left(\sqrt{1+\frac{x^3}{x+1}}-x\right)\ln x}{x\left(x^{\frac{1}{x}}-1\right)+\sqrt{x}\ln^2x}$$
The result should be $-\frac{1}{2}$, but wolfram says that is $0$
|
$x^{\frac{1}{x}}-1 = \exp\left(\frac{\log x}{x}\right)-1 = \frac{\log x}{x}+O\left(\frac{\log^2 x}{x^2}\right)$ for $x\to +\infty$, so the denominator behaves like $x^{1/2}\log(x)+O(\log x)$. On the other hand:
$$ \sqrt{1+\frac{x^3}{x+1}}-x = \frac{1+\frac{x^3}{x+1}-x^2}{x+\sqrt{1+\frac{x^3}{x+1}}}=\frac{x+1+x^3-x^2(x+1)}{x(x+1)+\sqrt{(x+1)^2+(x+1)x^3}}$$
behaves like $-\frac{1}{2}+O\left(\frac{1}{x}\right)$ for $x\to +\infty$, hence the wanted limit is just $\color{red}{0}$.
|
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|
Determine if the set of vectors are linearly independent or linearly dependent $$u=(1,1,1,3)$$
$$v=(1,2,1,3)$$
$$w=(1,2,3,2)$$
I need help understanding the method of how to do solve this type of problem. I understand that the concept is just to find out if the constants $k_n$ in $k_1u+k_2v+k_3w=0$ all equal zero or not.
Linear independence : this means that $k_1$, $k_2$, and $k_3$ are all equal to zero and that these are the only values that will make the overall equation equal to zero.
Linear dependence : this means that at least one of the $k$ values is not equal to zero.
This is how I tried solving this:
*
*Construct an augmented matrix $A$ out of the vectors.
$$A = \left[\begin{array}{ccc|c}1 & 1 & 1 & 0\\1 & 2 & 2 & 0\\ 1 & 1 & 3 & 0 \\ 3 & 3 & 2 & 0\end{array}\right]$$
*Use row operations as much as possible to get to row-echelon form
$$A = \left[\begin{array}{ccc|c}1 & 1 & 1 & 0\\0 & 1 & 1 & 0\\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0\end{array}\right]$$
*We now have the following:
$$a_1+a_2+a_3=0$$
$$a_2+a_3=0$$
$$a_3=0$$
This is where I'm confused. I don't know what these $a$ values are supposed to represent. They can't be the vector values because we had those then changed them. Are they the $k$ values I'm looking for?
I would conclude that this system of equations is linearly independent but that's assuming I even understand what the $a_n$ values are...
|
To answer where those $a$'s come from, and what they represent, when you construct an augmented matrix like you did, what you really have is,
$$[A|0] = \left[\begin{array}{ccc|c} 1 & 1 & 1 & 0 \\ 1 & 2 & 2 & 0 \\ 1 & 1 & 3 & 0 \\ 3 & 3 & 2 & 0 \\ \end{array} \right].$$
Which is equivalent to saying,
$$A\vec{x} = \vec{0},$$
or,
$$\left[\begin{array}{ccc} 1 & 1 & 1 \\ 1 & 2 & 2 \\ 1 & 1 & 3 \\ 3 & 3 & 2 \\ \end{array} \right] \left[\begin{array}{c} a_{1} \\ a_{2} \\ a_{3} \end{array} \right] = \left[\begin{array}{c} 0 \\ 0 \\ 0 \end{array} \right], $$
which describes the system,
$$\begin{align}
1\cdot a_{1} + 1\cdot a_{2} + 1\cdot a_{3} &= 0\\
1\cdot a_{1} + 2\cdot a_{2} + 2\cdot a_{3} &= 0\\
1\cdot a_{1} + 1\cdot a_{2} + 3\cdot a_{3} &= 0\\
3\cdot a_{1} + 3\cdot a_{2} + 2\cdot a_{3} &= 0\\
\end{align}$$
But once row reduced,
$$\begin{align}
1\cdot a_{1} + 1\cdot a_{2} + 1\cdot a_{3} &= 0\\
0\cdot a_{1} + 1\cdot a_{2} + 1\cdot a_{3} &= 0\\
0\cdot a_{1} + 0\cdot a_{2} + 1\cdot a_{3} &= 0\\
0\cdot a_{1} + 0\cdot a_{2} + 0\cdot a_{3} &= 0.\\
\end{align}$$
Which is just
$$\begin{align}
1\cdot a_{1} + 1\cdot a_{2} + 1\cdot a_{3} &= 0\\
1\cdot a_{2} + 1\cdot a_{3} &= 0\\
1\cdot a_{3} &= 0.\\
\end{align}$$
But as Brian mentioned, continuing to RREF form we have
$$\begin{align}
1 \cdot a_{1} &= 0\\
1 \cdot a_{2} &= 0\\
1 \cdot a_{3} &= 0.
\end{align}$$
Thus the only solution to your original equation is the vector,
$$ \left[ \begin{array}{c} a_{1} \\ a_{2} \\ a_{3} \end{array} \right] = \left[ \begin{array}{c} 0 \\ 0 \\ 0 \end{array} \right]$$
Which is the definition of linear independence.
|
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|
Find a polynomial f(x) of degree 5 such that 2 properties hold. I have been trying to find a polynomial $f(x)$ such that these $2$ properties hold:
*
*$f(x)-1$ is divisible by $(x-1)^3$
*$f(x)$ is divisible by $x^3$
To start, I set $f(x) =ax^5 + bx^4 + cx^3 + dx^2 + ex + f$.
This is divisible by $x^3$, so $d, e, f $ must be $0$. So polynomial $f(x) = ax^5 + bx^4 + cx^3$
This minus $1$ is divisible by $(x-1)^3$. So I used synthetic division and got that the remainder of $ax^5 + bx^4 + cx^3$ divided by $(x-1)$ is $a+b+c-1$. But it should divide out evenly, so $a+b+c-1 = 0$.
From here, I can't find any other equations involving the three variables. Could someone have any suggestions for how to continue? Thanks in advance. :)
|
Since $x^2$ and $(x-1)^2$ divide $p'(x)$ we have $$p'(x) =ax^2(x-1)^2 =ax^4-2ax^3+ax^2,$$ thus $$p(x) = {a\over 5}x^5-{a\over 2}x^4+{a\over 3}x^3+c$$
Since $p(0)=0$ we have $c=0$ and since $p(1)=1$ we have ${a\over 5}-{a\over 2}+{a\over 3}=1$ so $a=30$.
|
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|
If $x+y+z=0$, prove that $\frac{x^2}{2x^2+yz}+\frac{y^2}{2y^2+zx}+\frac{z^2}{2z^2+xy}=1$ A problem in my homework had asked me:
When $x+y+z=0$, evaluate$$\frac{x^2}{2x^2+yz}+\frac{y^2}{2y^2+zx}+\frac{z^2}{2z^2+xy}$$
Without too much difficulty, one can see that the value should be $1$ using $(x,y,z)=(1,0,-1)$.
I decided to use $x=-y-z$, which turned out not to be as difficult as initially thought. However, would someone care to enlighten me to some other methods of doing this?
|
HINT:
$$2x^2+yz=2(y+z)^2+yz=(2y+z)(y+2z)$$
Now $2y+z=x+y+z+y-x=y-x$
|
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|
If $f$ is an even function defined on the interval $(-5,5)$ then four real values of $x$ satisfying the equation $f(x)=f(\frac{x+1}{x+2})$ are? If $f$ is an even function defined on the interval $(-5,5)$ then four real values of $x$ satisfying the equation $f(x)=f(\frac{x+1}{x+2})$ are?
I thought that $(x+1)/(x+2)=-x$.But I'm getting only two values by solving this.How do I get the other two values?
And if i put $(x+1)/(x+2)=x$ then the answer is not matching.I dont know why.
|
Don't forget the possibility $(x+1)/(x+2)=+x$.
Setting $\frac{x+1}{x+2}=-x$ leads to $x+1=-x^2-2x$, i.e., $x^2+3x+1=0$, $x=\frac{-3\pm\sqrt{5}}2$. Setting $\frac{x+1}{x+2}=+x$ leads to $x+1=x^2+2x$, i.e., $x^2+x-1=0$, $x=\frac{-1\pm\sqrt{5}}2$.
All four values are well within $(-5,5)$.
|
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|
$\frac{1}{{{a_1}}} + \frac{1}{{{a_2}}} + \cdots + \frac{1}{{{a_n}}} < 2$
If ${{a}_{1}},{{a}_{2}},\ldots ,{{a}_{n}}$ are distinct odd natural
numbers not divisible by any prime greater than 5, then show that
$\frac{1}{{{a_1}}} + \frac{1}{{{a_2}}} + \cdots + \frac{1}{{{a_n}}}
< 2$.
First I have noted that $a_i$s are in the form $3^a5^b$ where $a,b$ are non-negative integers.Now let ${a_1} < {a_2} < \cdots < {a_n}$. As the inequality involves 2, I thought of the geometric series $1,\frac{1}{2},\frac{1}{{{2^2}}}, \cdots $ and I thought that each $\frac{1}{{{a_i}}}$ would be less than or equal to $\frac{1}{{{2^{i - 1}}}}$ and this works for some $a_i$s such as $\frac{1}{{{a_1}}} = 1 \leqslant 1, \frac{1}{{{a_2}}} = \frac{1}{3} < \frac{1}{2},\frac{1}{{{a_3}}} = \frac{1}{5} < \frac{1}{{{2^2}}}, \frac{1}{{{a_4}}} = \frac{1}{9} < \frac{1}{{{2^3}}}$. But this do not work for $i\ge 5$. I would appreciate any suggestion or idea on this problem.
|
Look at the product
$$\left(1+\frac{1}{3}+\frac{1}{3^2}+...\right)\left(1+\frac{1}{5}+\frac{1}{5^2}+...\right) = \frac{15}{8}$$
and notice it contains every possible $\frac{1}{3^a 5^b}$, hence it gives you upper bound for all of your finite sums.
|
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|
If $a,b,c,d > 0$ and $a+b+c+d=1$ , find the maximum value of $M=abc+abd+acd+bcd-abcd$
If $a,b,c,d > 0$ and $a+b+c+d=1$ , find the maximum value of $$M=abc+abd+acd+bcd-abcd$$
I don't have any idea how to simplify this equation. I have tried for $a,b,c>0$ and $a+b+c=1$ and I got $ab+ac+bc-abc=(1-a)(1-b)(1-c)<= 8/27$.
Thanks.
|
Let $\mathcal{T}$ be the regular tetrahedron of side $\sqrt{2}$ with vertices at
$\begin{cases}
A &= (+1,+1,+1),\\
B &= (+1,-1,-1),\\
C &= (-1,+1,-1),\\
D &= (-1,-1,+1).
\end{cases}$
For every point $X \in \mathcal{T}$, its baricentric coordinate is an unique $4$-tuple $(a,b,c,d)$ such that
$$\begin{cases}
a, b, c, d \ge 0, a + b + c + d = 1\\
X = aA + bB + cC +dD
\end{cases}
$$
Conversely, any $4$-tuple $(a,b,c,d)$ satisfies the first condition
define a point $X$ on $\mathcal{T}$ by the second formula.
Let $X = (x,y,z)$, it is easy to work out
$$\begin{cases}
a &= \frac14 (+x + y + z + 1)\\
b &= \frac14 (+x - y - z + 1)\\
c &= \frac14 (-x + y - z + 1)\\
d &= \frac14 (-x - y + z + 1)
\end{cases}$$
In terms of $(x,y,z)$, we have
$$abc+bcd + cda + dab - abcd = \frac{15-M(x,y,z)}{256}$$
where
$$M(x,y,z) = (14-(x^2+y^2+z^2))(x^2+y^2+z^2)+2(x^4+y^4+z^4)-24xyz$$
Let $r^2 = x^2 + y^2 + z^2$. Using $AM \ge GM$, we have
$$|xyz| \le \left(\frac{x^2+y^2+z^2}{3}\right)^{3/2} = \frac{r^3}{3\sqrt{3}}
\implies
M(x,y,z) \ge (14-r^2)r^2 - \frac{8r^3}{\sqrt{3}}
$$
As $X = (x,y,z)$ varies over $\mathcal{T}$, $r$ take values from the interval
$[0,\sqrt{3}]$. It is easy to check the polynomial on RHS is non-negative over
such a interval. This means $M(x,y,z) \ge 0$ for $X \in \mathcal{T}$.
As a result, we have
$$abc + bcd + cda + dab - abcd \le \frac{15}{256}$$
Since $M(0,0,0) = 0$, the equality is achieved at $a = b = c = d = \frac14$. This implies the maximum value we seek is $\frac{15}{256}$.
|
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|
How to compute $\int \frac{1}{(x^2+1)^2}dx$? Suppose we know $\int \frac{1-x^2}{(x^2+1)^2}=\frac{x}{x^2+1}+C$
How to compute $\int \frac{1}{(x^2+1)^2}dx$?
I tried writing it as $\frac{1+x^2-x^2}{(x^2+1)^2}=\frac{1-x^2}{(x^2+1)^2}+\frac{x^2}{(x^2+1)^2}$. But then how do you deal with $\frac{x^2}{(x^2+1)^2}$?
Some ideas?
|
So your problem reduces to
$$\int\frac{x^2dx}{(x^2+1)^2}$$
Integrate by parts: let $u=x$ and $dv=\frac{x}{(x^2+1)^2}dx$. Then $du=dx$ and $v=-\frac{1}{2}(x^2+1)^{-1}$ so we have
$$-\frac{1}{2}x(x^2+1)^{-1}+\frac{1}{2}\int\frac{dx}{x^2+1}$$
and that last integral may jump out at you as a particular trig derivative
|
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|
Sum of number-of-divisors function equals $\sum_{j=1}^{n} \lfloor n/j \rfloor$. I am trying to prove the identity
$$t(1) + t(2) + \cdots + t(n) = \Big\lfloor \dfrac{n}{1} \Big\rfloor + \Big\lfloor \dfrac{n}{2} \Big\rfloor + \cdots + \Big\lfloor \dfrac{n}{n} \Big\rfloor,$$
where $t(j)$ is the number of divisors of $j$.
Attempt: The highest power of $k \in \{2, 3, \ldots, n\}$ that divides $n!$ is given by
$$H_{k} := \Big\lfloor \dfrac{n}{k}\Big\rfloor + \Big\lfloor \dfrac{n}{k^{2}}\Big\rfloor + \cdots + \Big\lfloor\dfrac{n}{k^{N_{c}}}\Big\rfloor,$$
where $N_{c}$ is the maximal integer satisfying $k^{N_{c}} \leq n$. With a little thought, one can see that
\begin{equation*}
\begin{split}
\Big\lfloor\dfrac{n}{1}\Big\rfloor + \Big\lfloor\dfrac{n}{2}\Big\rfloor + \cdots + \Big\lfloor\dfrac{n}{n}\Big\rfloor &= H_{2} + H_{3} + \cdots + H_{n} + \Big\lfloor\dfrac{n}{1}\Big\rfloor \\
&= H_{2} + H_{3} + \cdots + H_{n} + n.
\end{split}
\end{equation*}
Each $H_{k}$ is a counted in the sum $t(1) + t(2) + \cdots + t(n)$, since each $t(k^{j})$ is a summand for $j \in \{1, 2, \ldots, N_{k}\}$, and also each $t(m)$ is a summand, where $m$ is a multiple of $k$ less than or equal to $n$.
This is where I'm a little stuck: I think that the only other summand in $t(1) + t(2) + \cdots + t(n)$ would be $1 + 1 + \cdots + 1 = n$, since each $t(j)$ counts $1$ as a divisor. My problem is showing that these are the only extra terms in $t(1) + t(2) + \cdots + t(n)$, which would show equality.
Any hints would be appreciated, or even suggestions of going about it a different way.
|
For the sake of completeness I would like to point out that this is
usually done by induction. We seek to show that
$$\sum_{k=1}^n \tau(k)
= \sum_{k=1}^n \bigg\lfloor \frac{n}{k} \bigg\rfloor.$$
It holds for $n=1:$ $$\tau(1) = \lfloor 1\rfloor.$$
In the induction step we start from
$$\sum_{k=1}^n \tau(k)
= \sum_{k=1}^n \bigg\lfloor \frac{n}{k} \bigg\rfloor.$$
to get
$$\tau(n+1) + \sum_{k=1}^n \tau(k)
= \tau(n+1) + \sum_{k=1}^n \bigg\lfloor \frac{n}{k} \bigg\rfloor.$$
Now $$\tau(n+1) = \sum_{k=1}^{n+1}
\left(\bigg\lfloor \frac{n+1}{k} \bigg\rfloor
- \bigg\lfloor \frac{n}{k} \bigg\rfloor\right)$$
since
$$\bigg\lfloor \frac{n+1}{k} \bigg\rfloor
- \bigg\lfloor \frac{n}{k} \bigg\rfloor
= \begin{cases} 1 \quad\text{if}\quad k|n+1
\\ 0 \quad\text{otherwise}.\end{cases}$$
Therefore we have
$$\sum_{k=1}^{n+1} \tau(k)
= \sum_{k=1}^{n+1}
\left(\bigg\lfloor \frac{n+1}{k} \bigg\rfloor
- \bigg\lfloor \frac{n}{k} \bigg\rfloor\right)
+ \sum_{k=1}^n \bigg\lfloor \frac{n}{k} \bigg\rfloor
\\ = \sum_{k=1}^{n+1}
\left(\bigg\lfloor \frac{n+1}{k} \bigg\rfloor
- \bigg\lfloor \frac{n}{k} \bigg\rfloor\right)
+ \sum_{k=1}^{n+1} \bigg\lfloor \frac{n}{k} \bigg\rfloor
\\ = \sum_{k=1}^{n+1}
\bigg\lfloor \frac{n+1}{k} \bigg\rfloor$$
and the induction is complete.
|
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|
arithmetic problem concerning equilateral triangle An equilateral triangle exists with vertices $(0,0), (a,11)$ and $(b,37)$. Using the distance formula three times, I eventually arrive to:
$$a^2 + 121 = b^2 + 1369 = a^2 + b^2 - 2ab + 676$$
The only problem is that since none of the equations are actually equal to anything, I can't figure it out. How can I solve this?
|
You actually have two equations: for instance,
\begin{align}
a^2+121&=b^2+1369\\
b^2+1369&=a^2+b^2-2ab+676.
\end{align}
It is easy to see that $b=0$ gives no solution: the second equation becomes $121=676$. So we assume $b\ne0$. We rewrite the equations as
\begin{align}
a^2-b^2&=1248\\
a^2-2ab&=693.
\end{align}
If you subtract both equations, you get
$$
-b^2+2ab=555.
$$
Thus,
$$
a=\frac{555+b^2}{2b}.$$ If we plug this into $a^2-b^2=1248$,
$$
1248=\frac{555^2+b^4+1110b^2}{4b^2}-b^2
=\frac{555^2-3b^4+1110b^2}{4b^2},
$$
so
$$
3b^4+3882b^2-555^2=0.
$$
Now
\begin{align}
b^2&=\frac{-3882\pm\sqrt{3882^2+4\times3\times555^2}}{6}
=\frac{-3882\pm\sqrt{2^4\times 3^2\times 19^4}}{6}\\ &=\frac{-3882\pm2^2\times3\times 19^2}{6}
=647\pm722,
\end{align}
so $b^2=75$ (the negative solution is not acceptable because it won't give $b$ real), and then $b=\pm5\sqrt3$.
Correspondingly,
$$
a=\pm\sqrt{1248+b^2}=\pm\sqrt{1248+75}=\pm\sqrt{1323}=\pm21\sqrt3.
$$
|
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|
Solve equation $8\cos x - 6\sin x = 5$ where $0 \le x \le 360$ Solve equation $8\cos x - 6\sin x = 5$ where $0 \le x \le 360$.
I am not asked to use any form, so I am going to use $k\cos(x-\alpha)$.
$8\cos x - 6\sin x = k\cos(x-\alpha)$
$$=k(\cos x\cos\alpha + \sin x\sin\alpha)$$
$$=k\cos\alpha\cos x + k\sin\alpha\sin x$$
equating coefficients:
$k\cos\alpha = 8$
$k\sin\alpha = 6$, or is it $k\sin\alpha = -6$? I find this really confusing
$k = \sqrt{8^2 + 6^2} = 10$
$\alpha$ is in the 1st quadrant where both sin and cos are positive.
$\alpha = \arctan\frac{6}8 = 36.8^{\circ}$
$\therefore10\cos(x - 36.8) = 5$
I have a maximum and minimum value of 10
From here I do not know how to finish solving for x.
|
The general method of solving $A\cos x+B\sin x=C$ when $A^2+B^2\ne 0$ is to take any $y$ such that $$(1)\quad A/\sqrt {A^2+B^2}=\cos y\;\land \;B/\sqrt {A^2+B^2}=\sin y.$$ $$\text {Then } C/\sqrt {A^2+B^2}=\cos y \cos x+\sin y \sin x =\cos (x-y).$$ So let $x=y+z$ for any $z$ such that $\cos z=C/\sqrt {A^2+B^2}.$
If we are strictly in the real numbers we require $|C/\sqrt {A^2+B^2}|\leq 1......$ Let $y_0$ satisfy (1) with $y_0\in [0,\pi].$ Let $z_0=\cos^{-1}(C/\sqrt {A^2+B^2})\in [0,\pi].$ Then $x=2\pi n \pm (y_0+z_0)$ for some (any) integer $n.$
|
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|
calculate arbitrary points from a plane equation I understand how one can calculate a plane equation (ax+by+cz=d) from three points but how can you go in reverse?
How can you calculate arbitrary points from a plane equation?
|
From your comment I finally understood what you are looking for:
If you have a plane defined by $a x + b y + c z = d$ then you also have the following properties:
*
*Plane normal direction:
$$\hat{n} = \begin{pmatrix}
\frac{a}{\sqrt{a^2+b^2+c^2}} \\
\frac{b}{\sqrt{a^2+b^2+c^2}} \\
\frac{c}{\sqrt{a^2+b^2+c^2}} \end{pmatrix}$$
*Point on plane closest to the origin (position of plane)
$$ \vec{r} = \begin{pmatrix}
\frac{a d}{a^2+b^2+c^2} \\
\frac{b d}{a^2+b^2+c^2} \\
\frac{c d}{a^2+b^2+c^2} \end{pmatrix} $$
*Distance of plane from the origin
$$ r = \frac{d}{\sqrt{a^2+b^2+c^2}} $$
*Directions along the plane (not unit vectors), and perpendicular to $\hat{n}$.
$$ \begin{align}
\hat{e}_1 & = \begin{pmatrix} c-b \\ a-c \\ b-a \end{pmatrix} &
\hat{e}_2 & = \begin{pmatrix} a (b+c)-b^2-c^2 \\ b (a+c) -a^2-c^2 \\ c (a+b)-a^2 - b^2 \end{pmatrix} \end{align} $$
You can verify that $\hat{e}_1 \cdot \hat{n} =0$, $\hat{e}_2 \cdot \hat{n}=0$ and $\hat{e}_1 \cdot \hat{e}_2 =0$, where $\cdot$ is the dot (inner) product.
Confirmation via GeoGebra
NOTES: Please edit the equation to make it clear you are looking for the plane properties when given a plane in equation form.
|
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Real solution of the equation $\sqrt{a+\sqrt{a-x}} = x\;,$ If $a>0$
For a real number $a>0\;,$ How many real solution of the equation $\sqrt{a+\sqrt{a-x}} = x$
$\bf{My\; Try::}$ We can Write $\sqrt{a+\sqrt{a-x}} = x$ as $a+\sqrt{a-x}=x^2$
So we get $(x^2-a)=\sqrt{a-x}\Rightarrow (x^2-a)^2 = a-x\;,$ Where $x<a$
So we get $x^4+a^2-2ax^2=a-x\Rightarrow x^4-2ax^2+x+a^2-a=0$
Now How can i solve it after that, Help me
Thanks
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Firstly, let's say $f(x) = \sqrt{a+\sqrt{a-x}} - x$. If $f$ has real solutions, then it must hold that $x \in [0,a]$. Now, taking the derivative of $f$, we have:
$$f'(x) = -1-\dfrac{1}{4\sqrt{a+\sqrt{a-x}}\cdot (\sqrt{a-x})}.$$
Clearly, $f'(x)<0, \forall x \in [0,a).$ Thus, $f$ is strictly decreasing in $[0,a]$.
Also, we have that $f(0) = \sqrt{a+\sqrt{a}}>0, \, \forall a>0$. Moreover, $f(a) = \sqrt{a} - a$. Thus, everything depends on the value of $a.$
Thus, there will be either zero roots or one root of $f$.
|
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Is it true that $log(i) = \frac\pi2i$ ? If so, are both of these legitimate proofs? They seem too beautiful not to be... Sorry if this is a naive question. I have not yet taken any upper level math courses involving complex numbers. However, in preparation for those courses, together with utilizing the knowledge that mathematicians such as Euler would just play around with infinite series, led me to this result. This seems too beautiful not to be true. If it is true, could someone explain the general framework from which this is derived?
Statement: $$log(i) = \frac\pi2i$$
Proof One: By Euler's Identity $e^{ix} = cos{x} + isin{x}$
So that $$e^{\frac\pi2i} = cos\frac\pi2 + isin\frac\pi2 = i$$
But $$e^{log(i)} = i$$
Thus $$e^{log(i)} = e^{\frac\pi2i}$$
So that $$log(i) = \frac\pi2i$$
Proof Two: Using Power Series
$$log(i) = log(\frac{2i}2) = log\left(\frac{(1+i)(1+i)}{(1+i)(1-i)}\right) = log\left(\frac{1+i}{1-i}\right) = log(1+i) - log(1-i)$$
But $$ log(1+i) = i - \frac{i^2}2 + \frac{i^3}3 - \frac{i^4}4 + ...$$
And $$-log(1-i) = i + \frac{i^2}2 + \frac{i^3}3 + \frac{i^4}4 + ...$$
Thus $$log(1+i) - log(1-i) = 2\{i + \frac{i^3}3 + \frac{i^5}5 + \frac{i^7}7 + ...\} = 2i\{1 - \frac13 + \frac15 - \frac17 + ...\} = 2i\frac\pi4 = \frac\pi2i$$
Therefore $$log(i) = \frac\pi2i$$
Added June 4, 2016: I found a delightful identity for logarithms of complex quantites in general as opposed to the case $a+bi = i$. Since the proof is essentially a generalization of my proof that $log(i) = \frac{\pi}{2}i$ as well as being aesthetically pleasing I thought it was worth posting.
Statement: $$log\left(a+bi\right) = \frac{1}{2}log\left(a^{2}+b^{2}\right) + itan^{-1}\frac{b}{a}$$
Proof: $$log\left(a+bi\right) = log\left({\sqrt{a^{2}+b^{2}}}{\frac{a+bi} {\sqrt{a^{2}+b^{2}}}}\right) = \frac{1}{2}log\left(a^{2}+b^{2}\right) + log\left({\frac{a+bi}{\sqrt{a^{2}+b^{2}}}}\right)$$
Now, $$log\left({\frac{a+bi}{\sqrt{a^{2}+b^{2}}}}\right) = log\left({\frac{1+i\frac{b}{a}}{\sqrt{1+\frac{b^{2}}{a^2}}}}\right) = log\left({1+i\frac{b}{a}}\right) - \frac{1}{2}log\left(1+\frac{b^{2}}{a^2}\right)$$
But, $$log\left(1+i\frac{b}{a}\right) = \left(i\frac{b}{a}\right)-\frac{1}{2}{\left(i\frac{b}{a}\right)}^{2} + \frac{1}{3}{\left(i\frac{b}{a}\right)}^{3} - \frac{1}{4}{\left(i\frac{b}{a}\right)}^{4} +... =
i\left( \left(\frac{b}{a}\right)-\frac{1}{3}{\left(\frac{b}{a}\right)}^{3} + \frac{1}{5}{\left(\frac{b}{a}\right)}^{5} - \frac{1}{7}{\left(\frac{b}{a}\right)}^{7} +... \right) +
\\ \frac{1}{2}\left(\frac{b}{a}\right)-\frac{1}{4}{\left(\frac{b}{a}\right)}^{4} + \frac{1}{6}{\left(\frac{b}{a}\right)}^{6} - \frac{1}{8}{\left(\frac{b}{a}\right)}^{8} +...$$
And,$$ \frac{1}{2}log\left(1+\frac{b^2}{a^2}\right) = \frac{1}{2}\left(\left({\frac{b^{2}}{a^{2}}}\right) - \frac{1}{2}\left({\frac{b^{2}}{a^{2}}}\right)^{2} + \frac{1}{3}\left({\frac{b^{2}}{a^{2}}}\right)^{3} - \frac{1}{4}\left({\frac{b^{2}}{a^{2}}}\right)^{4} +...\right) = \\ \frac{1}{2}\left(\frac{b}{a}\right)^{2}-\frac{1}{4}{\left(\frac{b}{a}\right)}^{4} + \frac{1}{6}{\left(\frac{b}{a}\right)}^{6} - \frac{1}{8}{\left(\frac{b}{a}\right)}^{8} +...$$
So that, $$log\left({1+i\frac{b}{a}}\right) - \frac{1}{2}log\left(1+\frac{b^{2}}{a^2}\right) = i\left( \left(\frac{b}{a}\right)-\frac{1}{3}{\left(\frac{b}{a}\right)}^{3} + \frac{1}{5}{\left(\frac{b}{a}\right)}^{5} - \frac{1}{7}{\left(\frac{b}{a}\right)}^{7} +... \right) = itan^{-1}{\frac{b}{a}}$$
Therefore, $$log(a+bi) = \frac{1}{2}log(a^{2}+b^{2}) + itan^{-1}\frac{b}{a}$$
Example: Let a=0 and b=1 and we have
$$log(i) = itan^{-1}\infty = \frac{\pi}{2}i + 2\pi mi$$
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It depends on how you define "logarithm" in the first place. But the simplest and most straightforward approach would be to define
Definition. $\log x$ means the number $y$ such that $e^y=x$.
With this definition the first two lines of your first proof is a complete, direct proof that $\log i = \frac12\pi i$.
The only problem is that with this reasoning $\log i$ can also be shown to be $\frac52\pi i$ and $\frac 92\pi i$ and $\frac{13}2\pi i\ldots$ because the exponential function of those numbers are also $i$.
This is not just a minor deficiency of the definition, though -- the complex logarithm is inherently multivalued in the precise sense that it is not possible to define a single function that is continuous on all of $\mathbb C\setminus\{0\}$ such that $e^{f(z)}=z$ everywhere.
The usual fudging one does in order to overcome this is to speak of the "principal logarithm", notated $\operatorname{Log}$ with a capital L, and define
$\operatorname{Log} z$ means the number $y$ such that $e^y=x$ and the imaginary part of $y$ is in $(-\pi,\pi]$.
This is, unfortunately, discontinuous on the negative real axis, but we can't really do better.
Your second proof is an ingenious manipulation, but of a kind that will easily lead to nonsense if you're not careful. Basically you're taking the power series for $\log(1+z)$ which will only work for $|z|\le 1, z\ne -1$ and then just assuming that this extends to a function on all of $\mathbb C\setminus 0$ that satisfies the usual logarithm rules everywhere. But this is not the case -- with the same reasoning we could say
$$ 1 = \Bigl(\frac{1+i}{\sqrt2}\Bigr)^8 $$
so
$$ 0 = \log 1 = 8\log\Bigl(\frac{1+i}{\sqrt2}\Bigr) $$
which requires the logarithm on the right to be $0$ -- but actually the series will yield $\frac14\pi i$.
This is the multi-valuedness of the logarithm that strikes again -- it is not possible for any nontrivial differentiable function on $\mathbb C\setminus\{0\}$ to satisfy $f(ab)=f(a)+f(b)$ everywhere.
|
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Solve $\log_9(x-4) - \log_9(x-8)= \frac{1}2$
Solve $\log_9(x-4) - \log_9(x-8)= \frac{1}2$
$(x-4) - (x-8)= 9^\frac{1}2$
$(x-4) - (x-8)= 3$
The answer is 10 but I am not sure how that was obtained.
|
$\log_9(x-4)-\log_9(x-8)=\log_9(\frac{x-4}{x-8})=\frac{1}{2}$
$\frac{x-4}{x-8}=9^{1/2}=3$
$x-4 = 3(x-8)=3x-24$
$x-4=3x-24$
$2x=20$
$x=10$
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Solve the system of equations: $a+b+c=2$, $a^2+b^2+c^2=6$, $a^3+b^3+c^3=8$ If we have \begin{cases} a+b+c=2 \\ a^2+b^2+c^2=6 \\ a^3+b^3+c^3=8\end{cases} then what is the value of $a,b,c$?
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Note this convenient fact: $$(a+b+c)^3-(a^3+b^3+c^3) = 3(a+b)(b+c)(c+a)$$
Since we know the left-hand side is zero, we also know the right-hand side is zero; this tells us that at least one of these is true:
$$a+b=0 \qquad b+c=0 \qquad c+a=0 \tag{$\star$}$$
We'll go ahead and take the middle one (the others will work out similarly), so that $c = -b$. Then, we have
$$2 = a + b + c = a + 0 = a$$
which gives us $a$. Also
$$6 = a^2 + b^2 + c^2 = 2^2 + b^2 + (-b)^2 = 4 + 2 b^2$$
which gives us $b^2 = 1$, so that $b$ is either $+1$ or $-1$, and $c$ is its opposite.
If we'd made a different choice at $(\star)$, we'd have gotten the same values, $2$, $1$, $-1$, assigned to different variables. That's to be expected, since the original equations are symmetric in $a$, $b$, $c$; the letters are interchangeable. Therefore, the best we can say here is that
$$\{ a, b, c\} = \{2, 1, -1\}$$
|
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Find all integral solutions for the Diophantine Equations $x^4 - x^2y^2 + y^4 = z^2$ and $x^4 + x^2y^2 + y^4 = z^2$.
Find all integral solutions for the Diophantine Equations $$x^4 - x^2y^2 + y^4 = z^2$$ and $$x^4 + x^2y^2 + y^4 = z^2$$
I basically think that to solve these equations we need to use the fact that all Pythagoras Triplets are like $(k \cdot 2mn, k \cdot (m^2 - n^2), k \cdot (m^2 + n^2))$. The above equations can be modified a little bit to make all the terms perfect squares. Then the first equation would be $$(x^2 - y^2)^2 + (xy)^2 = z^2$$ and the second one would be $$z^2 + (xy)^2 = (x^2 + y^2)^2$$
I have found no other clues. Please help me proceed.
Thanks.
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COMMENT.- I feel but I have no proof that there are only the trivial solutions $(x,y,z)= (t,0,t^2),(0,t,t^2)$. I give here an outline of what could perhaps lead to a proof.
$$x^4-x^2y^2+y^4=z^2\iff (x^2-y^2)^2+x^2y^2=z^2\qquad (1)$$ Hence, as it is well known,
$$\begin{cases}x^2-y^2=t^2-s^2\\xy=2ts\\z=t^2+s^2\end{cases}\qquad (2)$$
It follows in particular $$x^2+s^2=t^2+y^2\qquad (3)$$ The general solution of $(3)$, which likely as for Pithagorean triples comes from an identity easily verified, is given by
$$\begin{cases}2x=aX+bY\\2s=aY-bX\\2t=aX-bY\\2y=aY+bX\end{cases}\qquad (4)$$
where $a,b$ are arbitrary integers and $X,Y$ two parameters.
In any case, because of $(3)$, $x$ and $y$ must have the form, given by $(4)$, for certain integer values $a,b,X,Y$. However, in order to use parameters $t$ and $s$ we need besides to fit our values with the constraint $xy=2ts$ so we get the condition
$$(aX+bY)(aY+bX)=2(aX-bY)(aY-bX)\iff 3ab(X^2+Y^2)=(a^2+b^2)XY$$
It follows $$\left(\frac XY\right)^2-\left(\frac{a^2+b^2}{3ab}\right)\left(\frac XY\right)+1=0$$
Hence $$ \frac{6abX}{Y}= a^2+b^2\pm \sqrt{ (a^2+b^2)^2-(6ab)^2}$$ so the equation $$(a^2+b^2)^2-(6ab)^2=c^2\iff (a^2+b^2+6ab)(a^2+b^2-6ab)=c^2$$
It could be useful perhaps for someone. I stop here the comment.
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What is the coefficient and constant term in the following sequence defined recursively? Let $f_n(x)$ be a sequence of polynomials defined inductively as
$f_1(x) = (x - 2)^2$
$f_{n+1}(x) = (f_n(x) - 2)^2$ $; n \ge 1$
Let $a_n$ and $b_n$ respectively denote the constant term and the coefficient of $x$ in $f_n(x)$. Then
$(A) a_n = 4, b_n = -4^n (B) a_n = 4, b_n = -4n^2
(C) a_n = 4^{(n-1)!} , b_n = -4^n
(D) a_n = 4^{(n-1)!}, b_n = -4n^2.$
By simpllifying the expression, we can derive the term, ie, by expanding out the powers with $(a+b)^2$ formula recursively. But is there an analytical approach to solve this problem?
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A straightforward approach is to see how $a_n$ and $b_n$ change when $n$ is increased by $1$.
The constant term of $f_1(x)$ is $4$. Suppose that $f_n(x)$ for some $n$; then $f_n(x)=xg_n(x)+4$ for some polynomial $g_n(x)$ (why?), so
$$f_{n+1}(x)=\left(\big(xg_n(x)+4\big)-2\right)^2=\big(xg_n(x)+2\big)^2=x^2\big(g_n(x)\big)^2+4xg_n(x)+4\;.$$
The first two terms of $x^2\big(g_n(x)\big)^2+4xg_n(x)+4$ are multiples of $x$, so the constant term of $f_{n+1}(x)$ must be $4$. Thus, $a_n=4$ for each $n\in\Bbb Z^+$. (Technically this is a proof by mathematical induction.)
Now let’s look at the $x$ term. We can do pretty much the same thing: we know that $$f_n=x^2h_n(x)+b_nx+4$$ for some polynomial $h_n(x)$ (why?), so
$$\begin{align*}
f_{n+1}(x)&=\big(x^2h_n(x)+b_nx+2\big)^2\\
&=\Big(x\big(xh_n(x)+b_n\big)+2\Big)\\
&=x^2\big(xh_n(x)+b_n\big)^2+4x\big(xh_n(x)+b_n\big)+4\\
&=x^2\Big(x\big(xh_n(x)+b_n\big)^2+4h_n(x)\Big)+4b_nx+4\;.
\end{align*}$$
The first of those three terms is a multiple of $x^2$, so $4b_nx$ is the $x$ term of $f_{n+1}(x)$, and we see that $b_{n+1}=4b_n$. Thus, $b_2=4b_1$, $b_3=4b_2=4^2b_1$, $b_4=4b_3=4^3b_1$, and in general to get from $b_1$ to $b_n$ we must multiply by $4$ a total of $n-1$ times:
$$b_n=4^{n-1}b_1=4^{n-1}(-4)=-4^n\;.$$
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If $a,A,b,B,c,C$ are non negative reals such that $a+A=b+B=c+C=k$ Prove that $aB+bC+cA \le k^2$ If $a,A,b,B,c,C$ are non negative reals such that $a+A=b+B=c+C=k$ Prove that $aB+bC+cA \le k^2$
I substituted $B=k-b,C=k-c,A=k-a$ and plugged them to get a quadratic of $k$ which I had to show positive .SO, the discriminant should be $<0$ but actually its not coming
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From
$$
ABC = (k-a)(k-b)(k-c) \\
= k^3 - (a+b+c)k^2 + (ab + bc + ca)k - abc \\
= k \bigl( k^2 - (a+b+c)k + (ab + bc +ca) \bigr) - abc \\
= k \bigl( k^2 - (a B + b C + c A ) \bigr) - abc
$$
it follows that
$$
a B + b C + c A = k^2 - \frac{abc + ABC}{k}
\le k^2
$$
Equality holds if and only if
$$
abc = ABC = 0
$$
which means that one of $a, b, c$ is zero and another
one equal to $k$.
|
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What is the probability to fill rows of a cinema hall? This is the problem I'm trying to solve, but I'm not sure I'm on the correct path! would appreciate your feedback guidence and help.
So the problem is: there're 3 rows in a cinema hall. the first one is of 5 chairs, the second one with 7 chairs and the third one with 10 chairs.
There's a group of 18 people (15 men and 3 women) arrive to the hall and sit randomly at the 22 chairs.
*
*what is the probability that the first row is full?
*what is the probability that the first row is full if we know that the second row isn't full?
$$$$
This is what I've got so far:
1.
*
*Our probability space would be: $C_{22}^{18}18!$ (for choosing 18 places for the people, and in-ordering them in the chosen chairs)
*There're $(C_{15}^{5}+C_{15}^{4}C_{3}^{1}+C_{15}^{3}C_{3}^{2}+C_{15}^{2})$ ways to choose a group of 5 people (from both men and women).
*There are $C_{22-5}^{13}13!$ ways to choose chairs for the rest and $5!$ ways to order the 5 people of the first row.
So finally, the probability $P(first-row-is-full)=\frac{C_{22-5}^{13}13!5!(C_{15}^{5}+C_{15}^{4}C_{3}^{1}+C_{15}^{3}C_{3}^{2}+C_{15}^{2})}{C_{22}^{18}18!}$
*Let's mark:
*
*A - first row is full
*B - second row isn't full
then:
$P(A|B)=\frac{P(A\cap B)}{P(B)}=\frac{P(A)-P(A\cap \bar B)}{1-P(\bar B)}$
$$$$
Is it correct what I did? mostly I'm not sure about 2 because I get negative numbers.
Thank you!
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What matters here is which seats are occupied, not who sits in which seat. Therefore, we can take our sample space to be the
$$\binom{5 + 7 + 10}{18} = \binom{22}{18}$$
ways the patrons can occupy $18$ of the $22$ available chairs.
What is the probability the first row is full?
If the first row is full, $5$ of the $18$ people sit in the first row, which leaves $13$ people to sit in the $7 + 10 = 17$ seats available in the second and third row. Hence, the probability that the first row is full is
$$\frac{\dbinom{17}{13}}{\dbinom{22}{18}}$$
What is the probability the first row is full if the second row is not full?
Method 1: In this case, at least one of the four empty seats must be in the second row. Hence, the possible seating arrangements in which the first row is full and the second row is not are $(5, 3, 10)$, $(5, 4, 9)$, $(5, 5, 8)$, and $(5, 6, 7)$, where the ordered triple $(f, s, t)$ represents the number of people sitting in the first, second, and third rows, respectively. Since there are five seats in the first row, seven seats in the second row, and ten seats in the third, the probability that $f$ people sit in the first row, $s$ people sit in the second row, and $t$ people sit in the third row is
$$\binom{5}{f}\binom{7}{s}\binom{10}{t}$$
Adding up the possible cases yields
$$\binom{5}{5}\binom{7}{3}\binom{10}{10} + \binom{5}{5}\binom{7}{4}\binom{10}{9} + \binom{5}{5}\binom{7}{5}\binom{10}{8} + \binom{5}{5}\binom{7}{6}\binom{10}{7}$$
If the second row were full, then seven of the $22$ seats would be occupied, leaving $15$ seats in the first and third rows for the remaining $11$ patrons. Thus, the number of seating arrangements in which the second row is full is $\binom{15}{11}$. Hence, the number of seating arrangements in which the second row is not full is
$$\binom{22}{18} - \frac{15}{11}$$
Therefore, the conditional probability that the first row row is full given that the second row is not full is
$$\frac{\dbinom{5}{5}\dbinom{7}{3}\dbinom{10}{10} + \dbinom{5}{5}\dbinom{7}{4}\dbinom{10}{9} + \dbinom{5}{5}\dbinom{7}{5}\dbinom{10}{8} + \dbinom{5}{5}\dbinom{7}{6}\dbinom{10}{7}}{\dbinom{22}{18} - \dbinom{15}{11}}$$
Method 2: Note that the probability that the first row is full and the second row is not full is the probability that the first row is full minus the probability that the second row is also full. If the second row is also full, then only six of the ten seats in the third row are occupied. Hence, the probability that the first row is full and the second row is not full is
$$\frac{\dbinom{17}{13} - \dbinom{10}{6}}{\dbinom{22}{18} - \dbinom{15}{11}}$$
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Prove the inequality $\frac{a+c}{a+b}+\frac{b+d}{b+c}+\frac{c+a}{c+d}+\frac{d+b}{d+a}\geq 4$ $a, b, c, d$ are positive reals.
How would I prove the inequality
$$\frac{a+c}{a+b}+\frac{b+d}{b+c}+\frac{c+a}{c+d}+\frac{d+b}{d+a} \geq 4$$
I have tried using the rearrangement inequality with $a\leq b\leq c\leq d$
But it doesn't seem to work well.
Any hints please?
|
By C-S
$$\sum_{cyc}\frac{a+c}{a+b}=(a+c)\left(\frac{1}{a+b}+\frac{1}{c+d}\right)+(b+d)\left(\frac{1}{b+c}+\frac{1}{d+a}\right)\geq$$
$$\geq\frac{4(a+c)}{a+b+c+d}+\frac{4(b+d)}{b+c+d+a}=4.$$
|
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Is this an acceptable trig-sub and reversion from trig at the answer? Today I had to take the indefinite integral $\int x^3 \sqrt{x^2-1} \, dx$
My steps:
*
*$x=\sec\theta$, $dx=\sec\theta\tan\theta\, d\theta$
*$\displaystyle \int \sec^3\theta \sqrt{\sec^2\theta - 1} \, \sec\theta\tan\theta\, d\theta$
*${\displaystyle \int \sec^3\theta \sqrt{\tan^2\theta}\sec\theta\tan\theta d\theta}$
*${\displaystyle \int \sec^3\theta \tan\theta \cdot \sec\theta \tan\theta d\theta}$
*${\displaystyle \int \sec^4\theta \cdot \tan^2\theta d\theta}$
*${\displaystyle \int \sec^2\theta \sec^2\theta\tan^2\theta d\theta}$
*$u=\tan\theta$, $du= \sec^2\theta d\theta$
*${\displaystyle \int u^2(u^2 + 1)}$
*${\displaystyle \int u^4 + u^2}$
*${\displaystyle = \frac{1}{5}u^5 + \frac{1}{3}u^3 + C}$
here is where I'm even more shaky:
*${\displaystyle \frac{1}{5}\tan^5 + \frac{1}{3}\tan^3 + C}$
*since $\tan= \sin/\cos$, $\tan$ also $= \sec/\csc$, and if $\sec\theta=x$, per the substitution, couldn't $\tan$ also be written $x/(1/x)=x^2$, right?
*I replaced all $\tan$s in the final answer with $x^2$, leaving a final answer of ${\displaystyle \frac{1}{5}x^{10}+\frac{1}{3}x^6+C}$
|
This would have been lots easier with $x=\cosh\theta$:
$$\begin{align}\int x^3\sqrt{x^2-1}dx&=\int\cosh^3\theta\sinh^2\theta\,d\theta\\
&=\int(\sinh^2\theta+1)\sinh^2\theta\cosh\theta\,d\theta\\
&=\frac15\sinh^5\theta+\frac13\sinh^3\theta+C\\
&=\frac15(x^2-1)^{5/2}+\frac13(x^2-1)^{3/2}+C\\
&=\left(\frac15x^2-\frac15+\frac13\right)(x^2-1)^{3/2}+C\\
&=\frac1{15}(3x^2+2)(x^2-1)^{3/2}+C\end{align}$$
But I don't know if hyperbolic substitution counts as trigonometric substitution in this context.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Factoring a 4th degree trinomial I am trying to factor $3x^4-8x^3+16$, but I have no idea how to even start. I put into Wolfram Alpha, and it said that the answer was $(x-2)^2 (3 x^2+4 x+4)$. How would you factor something like this by hand? Is there any way without using the quartic formula in this case?
|
You can factor it if you know its roots. In general if $r$ is a root of a polynomial $f(x)$, then $(x-r)$ divides it.
For example, if you see that $f(x)=3 x^4−8 x^3+16$ has the root $2$, that is, $3(2)^4-8(2)^3+16=48-64+16=0$, then you know that $(x-2)$ divides it. You can then use long division to show that $3 x^4−8 x^3+16= (x-2)(3x^3-2x^2-4x-8) $. Now noticing that $2$ is also a root of $3x^3-2x^2-4x-8$ we have that $(x-2)$ divides that also. By long division $3x^3-2x^2-4x-8 = (x-2)(3x^2+4x+4)$. Now $f(x)=(x-2)^2(3x^2+4x+4)$.
So what you can do is first look for some roots using various methods.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Cauchy but not rapidly Cauchy I want to show that the sequence $\{\frac{(-1)^n}{n}\}$ is Cauchy but not rapidly Cauchy. Here is the work I done so far. I am curiously if I made any errors.
Consider the normed linear space $\mathbb{R}$ with the norm of the absolute value. Let
$$
\{f_n\}=\left\{\frac{(-1)^n}{n}\right\}=\left\{-1,\frac{1}{2},\frac{-1}{3},\ldots\right\}.
$$
Suppose $\{f_n\}\rightarrow 0$. Let $\epsilon>0$ and $N\in\mathbb{N}$ such that $n\geq N$. We want to show $|f_n-0|<\epsilon$ in order to prove the convergence of $\{f_n\}\rightarrow 0$.
Consider
\begin{eqnarray*}
\left|\frac{(-1)^n}{n}-0\right| &<& \epsilon\\
\left|\frac{(-1)^n}{n}\right| &<& \epsilon\\
\left| \frac{1}{n} \right| &<& \epsilon\\
\frac{1}{n}&<& \epsilon
\end{eqnarray*}
Therefore $n>\frac{1}{\epsilon}$. Let $N=\frac{1}{\epsilon}$. Then $n>\frac{1}{\epsilon}$ implies $|f_n-0|<\epsilon$, meaning $\{f_n\}\rightarrow 0$. Since $\{f_n\}$ converges in $\mathbb{R}$, $\{f_n\}$ is Cauchy.
Now we want to show that $\{f_n\}$ is not rapidly Cauchy. In other words, we want to show that for $\{f_n\}$ there is not a convergent series of positive numbers $\sum_{k=1}^\infty \epsilon_k$ for which
$$
|f_{k+1}-f_k|\leq \epsilon_k^2.
$$
Consider $|f_{k+1}-f_k|\leq \epsilon_k^2$.
\begin{eqnarray*}
|f_{k+1}-f_k| = \left| \frac{(-1)^{k+1}}{k+1}-\frac{(-1)^k}{k} \right| &\leq& \epsilon_k^2\\
\left|\frac{k(-1)^{k+1}-(k+1)(-1)^k}{k(k+1)} \right|&\leq& \epsilon_k^2
\end{eqnarray*}
Suppose $k$ is even. Then
$$
\left|\frac{-k-(k+1)}{k(k+1)} \right| = \left|\frac{-2k-1}{k^2+k} \right| \leq \epsilon_k^2
$$
Suppose $k$ is odd. Then
$$
\left|\frac{k+(k+1)}{k(k+1)} \right| = \left|\frac{2k+1}{k^2+k} \right| \leq \epsilon_k^2
$$
Therefore
$$
\frac{2k+1}{k^2+k} \leq \epsilon_k^2
$$
So $\epsilon_k=\sqrt{\frac{2k+1}{k^2+k}}$. Consider $\sum\limits_{k=1}^\infty \sqrt{\frac{2k+1}{k^2+k}}$. Notice $\sqrt{\frac{2k}{k^2}}=\frac{\sqrt{2}}{k}\leq\sqrt{\frac{2k+1}{k^2+k}}$ and $\sqrt{2}\sum\limits_{k=1}^\infty\frac{1}{k}$ diverges by the $p$-series since $p=1$. Therefore by the comparison test, $\sum\limits_{k=1}^\infty \sqrt{\frac{2k+1}{k^2+k}}$ diverges. Thus $\{f_n\}$ is not rapidly Cauchy.
|
Your argument looks correct, although it's a bit long and drawn out. I might argue this way: From your definition we have $f_k$ rapidly Cauchy iff
$$\sum_{k=1}^{\infty} |f_{k+1} - f_k|^{1/2} < \infty.$$
(Never heard of "rapidly Cauchy" before). Now
$$f_{k+1} - f_k = \frac{(-1)^{k+1}}{k+1} - \frac{(-1)^{k} }{k} = (-1)^k\left ( -\frac{1}{k+1} -\frac{1}{k} \right ).$$
So $|f_{k+1} - f_k| = 1/(k+1) + 1/k > 1/k.$ Thus $|f_{k+1} - f_k|^{1/2}> 1/\sqrt k.$ Since $\sum_k 1/\sqrt k = \infty,$ $f_k$ is not rapidly Cauchy
|
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"url": "https://math.stackexchange.com/questions/1771631",
"timestamp": "2023-03-29T00:00:00",
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|
Use generating functions to determine the number of ways Use generating functions to determine the number of different ways $12$ identical action figures can be given to $5$ children so that each child receives at most $3$ action figures
So far I have come up with the generating function of: $$(1 + x + x^2 + x^3)^5$$
To correspond to each child having either $0, 1, 2, $ or $3$ action figures with each child representing a factor. But, I am confused as to how to expand this. Any help is appreciated, thanks!
|
It is convenient to use the coefficient of operator $[x^n]$ to denote the coefficient of $x^n$ of a series.
We obtain
\begin{align*}
[x^{12}]&(1+x+x^2+x^3)^5\\
&=[x^{12}]\left(\frac{1-x^4}{1-x}\right)^5\tag{1}\\
&=[x^{12}]\left(\sum_{k=0}^{5}\binom{5}{k}(-1)^kx^{4k}\right)\left(\sum_{n=0}^\infty\binom{-5}{n}(-x)^n\right)\tag{2}\\
&=\left(\binom{5}{0}[x^{12}]-\binom{5}{1}[x^8]+\binom{5}{2}[x^4]-\binom{5}{3}[x^0]\right)
\sum_{n=0}^\infty\binom{n+4}{4}x^n\tag{3}\\
&=\binom{5}{0}\binom{16}{4}-\binom{5}{1}\binom{12}{4}+\binom{5}{2}\binom{8}{4}-\binom{5}{3}\binom{4}{4}\tag{4}\\
&=1\cdot 1820-5\cdot 495+10\cdot 70-10\cdot 1\\
&=35
\end{align*}
Comment:
*
*In (1) we use the formula for the finite geometric series
*In (2) we use the binomial series expansion for the denominator
*In (3) we use the linearity of the coefficient of operator and the rule $[x^{p-q}]A(x)=[x^{p}]x^{q}A(x)$. Note that we only need to respect $4$ of the $6$ summands of the series, since terms with exponents greater than $12$ do not contribute. We also use in the right series the binomial identity
\begin{align*}
\binom{-p}{q}=\binom{p+q-1}{p-1}(-1)^q
\end{align*}
*In (4) we select the coefficients from the right series accordingly to the coefficient of operator
|
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"timestamp": "2023-03-29T00:00:00",
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|
finding the number of factors $2^{15}\times3^{10}\times5^6$ The number of factors of $2^{15}\times3^{10}\times5^6$ which are either perfect square or perfect cubes(or both)
I don't know how to start this even!
Plz solve this!
|
A factor can be described by a tuple of exponents
$f \in
\{ 0, \dotsc, 15 \}
\times
\{ 0, \dotsc, 10 \}
\times
\{ 0, \dotsc, 6 \}
$.
As
$$
(2^{f_1} 3^{f_2} 5^{f_3})^{1/n} =
2^{f_1/n} \, 3^{f_2/n} \, 5^{f_3/n}
$$
we need $n$ to divide the exponents $f_i$.
The perfect squares ($n=2$) are combined from
$
S = \{ 0, 2, 4, 6, 8, 10, 12, 14 \} \times
\{ 0, 2, 4, 6, 8, 10 \} \times
\{ 0, 2, 4, 6 \}
$, the perfect cubes ($n=3$) from
$
C = \{ 0, 3, 6, 9, 12, 15 \} \times
\{ 0, 3, 6, 9 \} \times
\{ 0, 3, 6 \}
$.
Then
$S \cap C =
\{ 0, 6, 12 \} \times \{ 0, 6 \} \times \{ 0, 6 \}
$.
This gives
$$
\lvert S \rvert + \lvert C \rvert - \lvert S \cap C \rvert
= 8 \cdot 6 \cdot 4 + 6 \cdot 4 \cdot 3 - 3 \cdot 2 \cdot 2
= 252
$$
|
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"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Evaluate the improper integral with residues.
$$\int_0^{\infty} \frac{x^2+1}{x^4+1}dx$$
What i've found are the singularities at: $e^{\pi/4+\pi/2n}$ for $n=0,1,2,3$. But i'm unsure how to calculate the integral since I don't want to include the singularity at $x_2=\frac{-1+i}{\sqrt{2}}$ since then I would be calculating the integral from negative to positive infinity.
My idea was to construct a contour $\gamma : x=ti$ where $R<t\leq0$ and $\lim_{R\rightarrow \infty}$. My reckoning was that since:
$$\int_{\gamma}\frac{x^2+1}{x^4+1}dx=i\int_R^0\frac{1-t^2}{t^4+1}dt$$
doesn't have any singularities along the positive real numberline, it is therefore bounded and thus:
$$i\int_R^0\frac{1-t^2}{t^4+1}dt\leq \Big| i\int_R^0\frac{1-t^2}{t^4+1}dt\Big|\leq\int_R^0\frac{|1-t^2|}{|t^4+1|}dt\leq\frac{|R^2|+1}{|R^4|-1}\rightarrow0 \text{ when } R\rightarrow \infty.$$
The answer would then be:
$$\lim_{R\rightarrow \infty}\int_0^{R}\frac{x^2+1}{x^4+1}dx=2\pi iRes(f;e^{\pi/4})-\int_0^{\pi/2}f(x)dx-i\int_R^0f(x)dx=\frac{\pi}{2\sqrt{2}}.$$
But sadly I am mistaken, what is wrong?
|
Let $I$ be the integral defined by
$$I=\int_0^\infty\frac{x^2+1}{x^4+1}\,dx \tag 1$$
We proceed to evaluate the integral in $(1)$ by analyzing the contour integral $J$
$$J=\oint_C \frac{z^2+1}{z^4+1}\,dz$$
where $C$ is the quarter circle in the first quadrant, centered at the origin, with radius $R$. Then, we have
$$J=\int_0^R \frac{x^2+1}{x^4+1}\,dx+i\int_0^R \frac{y^2-1}{y^4+1}\,dy+\int_0^{\pi/2}\frac{R^2e^{i2\phi}+1}{R^4e^{i4\phi}+1}\,iRe^{i\phi}\,d\phi \tag 2$$
From the residue theorem, we find that
$$\begin{align}
J&=2\pi i \text{Res}\left(\frac{z^2+1}{z^4+1}, z=e^{i\pi/4}\right)\\\\
&=2\pi i \frac{e^{i\pi/2}+1}{4e^{i3\pi/4}}\\\\
&=\frac{\sqrt{2}\pi}{2}
\end{align}$$
As $R\to \infty$, the third integral on the right-hand side of $(2)$ vanishes and we find
$$\int_0^\infty \frac{x^2+1}{x^4+1}\,dx+i\int_0^\infty \frac{y^2-1}{y^4+1}\,dy=\frac{\sqrt{2}\pi}{2} \tag 3$$
Equating the real parts of $(3)$, we find
$$\bbox[5px,border:2px solid #C0A000]{\int_0^\infty \frac{x^2+1}{x^4+1}\,dx=\frac{\sqrt{2}\pi}{2}}$$
As a bonus, we find that
$$\int_0^\infty \frac{x^2-1}{x^4+1}\,dx=0$$
which one can verify directly by enforcing the substitution $x\to 1/x$.
|
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|
$x,y,z \geqslant 0$, $x+y^2+z^3=1$, prove $x^2y+y^2z+z^2x < \frac12$
$x,y,z \geqslant 0$, $x+y^2+z^3=1$, prove
$$x^2y+y^2z+z^2x < \frac12$$
This inequality has been verified by Mathematica. $\frac12$ is not the best bound. I try to do AM-GM for this one but not yet success. The condition $x+y^2+z^3$ is very weird.
|
Let $x=0$.
Then $y^2+z^3=1 \Rightarrow z^3=1-y^2$
Seek to maximise $p=y^2z=y^2(1-y^2)^{\frac 13}$
WLOG let $u=y^2$ so that $p=u(1-u)^{\frac 13}$
$\frac {dp}{du}=(1-u)^{\frac 13}-u{\frac 13}(1-u)^{-\frac 23}$
$\frac {dp}{du}=\frac 13(3-4u)(1-u)^{-\frac 23}$
Maximum is at $u=\frac 34$
Thus maximum is $p=\frac 34(\frac 14)^{\frac 13}$
$p \le \frac 34(\frac 14)^{\frac 13}$
$p^3 \le \frac {27}{64}\frac 14$
$p^3 \le \frac {27}{256} < \frac {32}{256}$
$p^3 < \frac {1}{8}$
$p < \frac {1}{2}$
We then have to demonstrate that any increase in $x$ creates corresponding decreases in $y^2$ and $z^2$ such that the other two terms in the expression $x^2y+y^2z+z^2x$ increase more slowly than the decrease in $p$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Olympiad Inequality $\sum\limits_{cyc} \frac{x^4}{8x^3+5y^3} \geqslant \frac{x+y+z}{13}$
$x,y,z >0$, prove
$$\frac{x^4}{8x^3+5y^3}+\frac{y^4}{8y^3+5z^3}+\frac{z^4}{8z^3+5x^3} \geqslant \frac{x+y+z}{13}$$
Note:
Often Stack Exchange asked to show some work before answering the question. This inequality was used as a proposal problem for National TST of an Asian country a few years back. However, upon receiving the official solution, the committee decided to drop this problem immediately. They don't believe that any students can solve this problem in 3 hour time frame.
Update 1: In this forum, somebody said that BW is the only solution for this problem, which to the best of my knowledge is wrong. This problem is listed as "coffin problems" in my country. The official solution is very elementary and elegant.
Update 2: Although there are some solutions (or partial solution) based on numerical method, I am more interested in the approach with "pencil and papers." I think the approach by Peter Scholze in here may help.
Update 3: Michael has tried to apply Peter Scholze's method but not found the solution yet.
Update 4: Symbolic expanding with computer is employed and verify the inequality. However, detail solution that not involved computer has not been found. Whoever can solve this inequality using high school math knowledge will be considered as the "King of Inequality".
|
This is too long to fit into a comment. I wanted to ask a question about my proof on this problem. (It might help discover another proof)
This proof has a flaw -- From $AB \ge C$ and $A \ge D$, I wrongly implied that $DB \ge C$.
Is there a way to slightly modify it such that it can prove the statement or is it completely wrong?
Seeing that the inequality is homogeneous (meaning that the transformation
$(x, y, z) \mapsto (kx, ky, kz)$ does not change anything), it is natural to impose a constraint on it. So let us assume
without the loss of generality that $xyz=1$.
From Cauchy-Schwarz Inequality,
$$([8x^3+5y^3]+[8y^3+5z^3]+[8z^3+5x^3])(\frac{x^4}{8x^3+5y^3}+\frac{y^4}{8y^3+5z^3}+\frac{z^4}{8z^3+5x^3})\geqslant (x^2+y^2+z^2)^2$$
Since (By AM-GM) $$[8x^3+5y^3]+[8y^3+5z^3]+[8z^3+5x^3] = 13(x^3+y^3+z^3) \geqslant 13(3 \sqrt[3]{(xyz)^3}) = 13(3)$$
Therefore
$([8x^3+5y^3]+[8y^3+5z^3]+[8z^3+5x^3])(\frac{x^4}{8x^3+5y^3}+\frac{y^4}{8y^3+5z^3}+\frac{z^4}{8z^3+5x^3}) \geqslant (13)(3)(\frac{x^4}{8x^3+5y^3}+\frac{y^4}{8y^3+5z^3}+\frac{z^4}{8z^3+5x^3}) \geqslant (x^2+y^2+z^2)^2$
Therefore
$$\frac{x^4}{8x^3+5y^3}+\frac{y^4}{8y^3+5z^3}+\frac{z^4}{8z^3+5x^3} \geqslant \frac{(x^2+y^2+z^2)^2}{(13)(3)}$$
Now it reamins to prove that $\frac{(x^2+y^2+z^2)^2}{(13)(3)} \geqslant \frac{x+y+z}{13}$, i.e.
$$(x^2+y^2+z^2)(x^2+y^2+z^2)\geqslant 3(x+y+z)$$
which is straightforward by AM-GM:
Notice that for all $xyz=1$
$$(x - 1)^2 + (y-1)^2 + (z - 1)^2 \ge 0$$
$$x^2 + y^2 + z^2 - 2a - 2b - 2c + 3 \ge 0$$
$$x^2 + y^2 + z^2 \ge -3 + (x + y + z) + (x + y + z)$$
But by AM-GM, $x + y + z \ge 3\sqrt[3]{xyz} = 3$. So,
$$x^2 + y^2 + z^2 \ge -3 + 3 + (x + y + z)$$
$$x^2 + y^2 + z^2 \ge x + y + z \ge 3$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
If $x^2+ax-3x-(a+2)=0\;,$ Then $ \min\left(\frac{a^2+1}{a^2+2}\right)$
If $x^2+ax-3x-(a+2)=0\;,$ Then $\displaystyle \min\left(\frac{a^2+1}{a^2+2}\right)$
$\bf{My\; Try::}$ Given $x^2+ax-3x-(a+2)=0\Leftrightarrow ax-a = -(x^2-3x-2)$
So we get $$a=\frac{x^2-3x-2}{1-x} = \frac{x^2-2x+1+1-x-4}{1-x} = \left[1-x-\frac{4}{1-x}+1\right]$$
Now $$f(a) = \frac{a^2+1}{a^2+2} = \frac{a^2+2-1}{a^2+2} = 1-\frac{1}{a^2+2}$$
So $$f(x) = 1-\frac{1}{\left[(1-x)-\frac{4}{1-x}+1\right]^2+2}$$
Now put $1-x=t\;,$ Then we get $$f(t) =1- \frac{1}{\left(t-\frac{4}{t}+1\right)^2+2}$$
Now How can I maximize $\displaystyle \frac{1}{\left[(1-x)-\frac{4}{1-x}+1\right]^2+2, }\;,$ Help Required, Thanks
|
Minimizing $1 - \frac{1}{(t - \frac 4t +1)^2 + 2}$ is equivalent to minimizing $(t - \frac 4t +1)^2 + 2$. But obviously the minimal value for this is $2$, as the square of a number is always bigger than $0$. To find the value which minimizes it just solve $t- \frac4t + 1 = 0$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Inequality $\sqrt{1-abc}(3-a-b-c)\geq |(1-a)(1-b)(1-c)|$ Let $a,b,c>0$ and $a^2+b^2+c^2+abc=4$. Prove that
$$\sqrt{1-abc}(3-a-b-c)\geq |(1-a)(1-b)(1-c)|.$$
Note that equality holds trivially when $a=b=c=1$, but it is not the only case. It also holds more generally when $a=x, b=x, c=2-x^2$, where $0<x^2<2$.
|
squaring the given inequality we get
$$- \left( ac+b-2 \right) \left( bc+a-2 \right) \left( ab+c-2 \right) \geq 0$$
from $$a^2+b^2+c^2+abc=4$$ we get
$$(b-2)(b+2)=-(a^2+c^2+abc)$$
$$b-2=-\frac{a^2+c^2+abc}{b+2}$$
$$ac+b-2=ac-\frac{a^2+c^2+abc}{abc}$$
$$ac+b-2=-\frac{(a-c)^2}{b+2}<0$$
etc. Thus our product above is non negativ.
|
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|
Inequality $\sum\limits_{cyc}\frac{a^3}{13a^2+5b^2}\geq\frac{a+b+c}{18}$
Let $a$, $b$ and $c$ be positive numbers. Prove that:
$$\frac{a^3}{13a^2+5b^2}+\frac{b^3}{13b^2+5c^2}+\frac{c^3}{13c^2+5a^2}\geq\frac{a+b+c}{18}$$
This inequality is strengthening of the following Vasile Cirtoaje's one, which he created in 2005.
Let $a$, $b$ and $c$ be positive numbers. Prove that:
$$\frac{a^3}{2a^2+b^2}+\frac{b^3}{2b^2+c^2}+\frac{c^3}{2c^2+a^2}\geq\frac{a+b+c}{3}.$$
My proof of this inequality you can see here: https://artofproblemsolving.com/community/c6h22937p427220
But this way does not help for the starting inequality.
A big problem we have around the point $(a,b,c)=(0.785, 1.25, 1.861)$ because the difference between the LHS and the RHS in this point is $0.0000158...$.
I tried also to use Cauchy-Schwarz, but without success.
Also, I think the BW (see here https://math.stackexchange.com/tags/buffalo-way/info I tryed!) does not help.
|
A bit algebra shows that the inequality is equivalent to
$$
5 \left(5 a^5 \left(13 b^2+5 c^2\right)+13 a^4 \left(5 b^3-13 b^2 c-5 b c^2-5 c^3\right)+a^3 \left(-65 b^4+144 b^2 c^2+65 c^4\right)+a^2 \left(25 b^5-65 b^4 c+144 b^3 c^2+144 b^2 c^3-169 b c^4+65 c^5\right)-13 a \left(13 b^4 c^2+5 b^2 c^4\right)+5 b^2 c^2 \left(13 b^3+13 b^2 c-13 b c^2+5 c^3\right)\right) \ge0
$$
The left hand side is a polynomial, so this can be solved by Cylindrical Algebra Decomposition -- https://en.wikipedia.org/wiki/Cylindrical_algebraic_decomposition
The following code in Mathematica does the job.
ex1 = a^3/(13 a^2 + 5 b^2) + b^3/(13 b^2 + 5 c^2) + c^3/(5 a^2 + 13 c^2) >= 1/18 (a + b + c);
ex2 = ex1[[1]] - ex1[[2]] // Together // Numerator // Simplify;
ex3 = ForAll[{a, b, c}, And @@ {a >= 0, b >= 0, c >= 0}, ex2 >= 0];
CylindricalDecomposition[ex3, {}]
Note this may take a few minutes to run.
|
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|
Show that, $2\arctan\left(\frac{1}{3}\right)+2\arcsin\left(\frac{1}{5\sqrt2}\right)-\arctan\left(\frac{1}{7}\right)=\frac{\pi}{4}$ Show that,
$$2\arctan\left(\frac{1}{3}\right)+2\arcsin\left(\frac{1}{5\sqrt2}\right)-\arctan\left(\frac{1}{7}\right)=\frac{\pi}{4}$$
There is a mixed of sin and tan, how can I simplify this to $\frac{\pi}{4}$
We know the identity of $\arctan\left(\frac{1}{a}\right)+\arctan\left(\frac{1}{b}\right)=\arctan\left(\frac{a+b}{ab-1}\right)$
|
Explanation for your question here
$2arctan(\frac{1}{3})$ = $arctan(\frac{3}{4})$ as you already understand
$arcsin(x) = arctan(\frac{x}{\sqrt{1-x^2}})$Inverse trigonometric functions
This simplifies $arcsin(\frac{1}{5\sqrt2}) = arctan (\frac{1}{7})$ which makes
$2arcsin(\frac{1}{5\sqrt2}) = 2arctan(\frac{1}{7})$
Now the solution is
$2arctan(\frac{1}{3})+ 2arcsin(\frac{1}{5\sqrt2})-arctan(\frac{1}{7})$
simplifies into
$arctan(\frac{3}{4})+ 2arctan(\frac{1}{7})-arctan(\frac{1}{7})$
$arctan(\frac{3}{4})+arctan(\frac{1}{7})$
$arctan(\frac{(21+4)}{(28-3)}) = arctan(1) = \frac{\pi}{4}$
|
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|
How to compute $\int_0^1\frac{\ln(x)}{1+x^5}dx$?
Let $\phi$ denote the golden ratio $\phi=\frac{1+\sqrt5}{2}$.
How can I prove this sum?
$$\sum_{n=0}^{\infty}(-1)^n\left[\frac{\phi}{(5n+1)^2}+\frac{\phi^{-1}}{(5n+2)^2}-\frac{\phi^{-1}}{(5n+3)^2}-\frac{\phi}{(5n+4)^2}\right]=\left(\frac{2\pi}{5}\right)^2$$
My try:
Change the sums into an integrals:
$\sum_{n=0}^{\infty}\frac{(-1)^n}{(5n+1)^2}=\int_0^1\frac{-\ln(x)}{1+x^5}dx$
Can somebody give a hint how on to integrate this integral.
Try substitution by letting $u=\ln(x)$ is not working and integration by part is making it more complicated than before. What kind of substitution should I be using?
|
Let's define the sum we are looking for with $S$.
Furthermore, notice that ($k \in [1,4] $)
$$
S_k=\sum_{n=0}^{\infty}\frac{(-1)^n}{(5n+k)^2}=\sum_{n=0}^{\infty}\frac{1}{(5(2n)+k)^2}-\sum_{n=0}^{\infty}\frac{1}{(5(2n+1)+k)^2}=\\
\frac{1}{5^22^2}\left[\sum_{n=0}^{\infty}\frac{1}{(n+k/10)^2}-\sum_{n=0}^{\infty}\frac{1}{(n+(k+5)/10)^2}\right]=\\
\frac{1}{5^22^2}\left[\psi^{(1)}(k/10)-\psi^{(1)}((k+5)/10)\right]
$$
where we used the definition of the Trigamma function
Taking now for example $\Delta_{14}=S_1-S_4$
$$
\Delta_{14} =\frac{1}{5^22^2}\left[\psi^{(1)}(1/10)-\psi^{(1)}(6/10)\right]-\frac{1}{5^22^2}\left[\psi^{(1)}(4/10)-\psi^{(1)}(9/10)\right]=
\frac{1}{5^22^2}(\psi^{(1)}(1/10)+\psi^{(1)}(9/10))-\frac{1}{5^22^2}(\psi^{(1)}(4/10)+\psi^{(1)}(6/10))
$$
Trigamma reflection $\psi^{(1)}(1-z)+\psi^{(1)}(z)=\frac{\pi^2}{\sin^2(\pi z)}$ yields the massive simplification
$$
\Delta_{14}=\frac{\pi^2}{100}\left(\frac{1}{\sin^2(\pi/10)}-\frac{1}{\sin^2(4\pi/ 10)}\right)=\\
\frac{\pi^2}{125}(5+3\sqrt{5})
$$
Playing the same game with $\Delta_{23}=S_2-S_3$ we obtain
$$
\Delta_{23}=\frac{\pi^2}{125}(-5+3\sqrt{5})
$$
Doing the algebra yields
$$
S=\phi\Delta_{14}+\frac{\Delta_{23}}{\phi}=\frac{4 \pi^2}{25}\\
\bf{QED}
$$
|
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|
Prove that $\displaystyle \sum_{k=1}^n \bigg(\dfrac{1}{k}+\dfrac{2}{k+n}\bigg ) \leq \ln(2n) + 2 -\ln(2)$
Prove that $$\displaystyle \sum_{k=1}^n \bigg(\dfrac{1}{k}+\dfrac{2}{k+n}\bigg ) \leq \ln(2n) + 2 -\ln(2).$$
I was thinking of using mathematical induction for this. That is,
We prove by induction on $n$. The case $n=1$ holds trivially since $2 \leq 2$. Now assume the result holds for some $m$. Then by assumption we know that $$\displaystyle \sum_{k=1}^{m+1} \bigg(\dfrac{1}{k}+\dfrac{2}{k+m}\bigg ) \leq \ln(2m) + 2 -\ln(2)+\dfrac{1}{m+1}+\dfrac{2}{2m+1}. $$
We must relate this somehow to $\ln(2(m+1)) + 2 -\ln(2)$.
|
If you want to do a proof by induction.
you have covered the base case.
Inductive hypothesis:
$\sum_\limits{k=1}^n \bigg(\dfrac{1}{k}+\dfrac{2}{k+n}\bigg ) \leq \ln(2n) + 2 -\ln(2)$.
We need to show that:
$\sum_\limits{k=1}^{n+1} \bigg(\dfrac{1}{k}+\dfrac{2}{k+n+1}\bigg ) \leq \ln(2n) + 2 -\ln(2)$
thoughts:
$\ln(2n) + 2 -\ln(2) = \ln (n) + 2$
$\sum_\limits{k=1}^{n+1} \bigg(\dfrac{1}{k}+\dfrac{2}{k+n+1}\bigg ) = \sum_\limits{k=1}^{n} \bigg(\dfrac{1}{k}+\dfrac{2}{k+n}\bigg ) + \dfrac{2}{2n+1}$
$\sum_\limits{k=1}^{n} \bigg(\dfrac{1}{k}+\dfrac{2}{k+n}\bigg )\leq \ln n + 2$ by the inductive hypothesis.
So, if $\dfrac{2}{2n+1}-\ln (n) + 2 \leq \ln(n+1) + 2$, we are done.
$\dfrac{2}{2n+1} < \ln(1+\frac{1}{n})$
$\sum_\limits{k=1}^{n+1} \bigg(\dfrac{1}{k}+\dfrac{2}{k+n+1}\bigg ) \leq \ln(2n) + 2 -\ln(2)$
QED
|
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|
How to solve this equation algebraically Solve the following simultaneous equations on the set of real numbers:
\begin{cases}x^2 + y^3 = x+1 \\ x^3+y^2=y+1\end{cases}
Thanks for helping!
|
Hint:
Eliminate one of the unknowns in a way to get a polynomial.
From the second equation,
$$y^2-y+1=2-x^3$$ and multiplying by $y+1$ and using the first equation,
$$(2-x^3)(y+1)=(y^2-y+1)(y+1)=y^3+1=2+x-x^2,$$
and
$$y=\frac{2+x-x^2}{2-x^3}-1=\frac{x-x^2+x^3}{2-x^3}.$$
Then
$$\left(x-x^2+x^3\right)^3=(1+x-x^2)(2-x^3)^3.$$
|
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|
Calculate $\sum_{n=0}^\infty(n+2)x^n$ I am trying to calculate $\sum_{n=0}^\infty(n+2)x^n$.
I was thinking it is like the second derivative of $x^{n+2}/(n+1)$ but I am not sure how to go about calculating it. Any hints?
|
Hint: It is sometimes convenient to work with operators. Since
\begin{align*}
\left(x\frac{d}{dx}\right)\sum_{n=0}^\infty a_nx^n=x\sum_{n=0}^\infty na_nx^{n-1}=\sum_{n=0}^\infty na_nx^{n}
\end{align*}
we consider $xD:=\left(x\frac{d}{dx}\right)$ as operator and obtain for $k\geq 0$
\begin{align*}
\left(xD\right)^k\sum_{n=0}^{\infty}a_nx^n=\sum_{n=0}^\infty n^ka_nx^{n}
\end{align*}
We conclude
\begin{align*}
\sum_{n=0}^{\infty}(n+2)x^n&=\left(xD+2\right)\sum_{n=0}^{\infty}x^n\\
&=\left(xD+2\right)\frac{1}{1-x}\\
&=x\left(\frac{1}{1-x}\right)^\prime+\frac{2}{1-x}\\
&=\ldots
\end{align*}
|
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|
Number of polynomials which are divisible by $x+1$ Let $a,b,c,d$ be four integers (not necessarily distinct) in the set ${1,2,3,4,5}$ . The number of polynomials $f(x)=x^4+ax^3+bx^2+cx+d$ which are divisible by $x+1$ are:
$(A)$ Between 55 and 65
$(B)$ Between 65 and 85
$(C)$ Between 86 and 105
$(D)$ More than 105
I see that it will be divisible when $f(-1)=0$
i.e. $1-a+b-c+d=0$ but how should I proceed further ?
|
Your approach of looking for the number of solutions to $1-a+b-c+d=0$ is excellent.
Let $f(n)$ be the number of ways of choosing two (possibly identical) numbers $h,k$ from $\{1,2,3,4,5\}$ so that $h-k=n$. It is easy to check that $f(n)=1,2,3,4,5,4,3,2,1$ for $n=-4,-3,-2,\dots,3,4$ and 0 for all other $n$.
So $1-a+b$ must be one of $-3,-2,\dots,4,5$. The number of $(a,b)$ giving each of these is $1,2,3,4,5,4,3,2,1$. The number of $(c,d)$ giving minus those values is $2,3,4,5,4,3,2,1,0$. So the total number of solutions is $1\cdot2+2\cdot3+3\cdot4+4\cdot5+5\cdot4+4\cdot3+3\cdot2+2\cdot1=80$.
So the answer to the multiple choice question is (B).
|
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|
Find all positive integers $n$ such that $n^2+n+43$ becomes a perfect square Find all positive integers $n$ such that $n^2+n+43$ becomes a perfect square.
Since $n^2+n+43$ is odd,if it's a perfect square it can be written as: $8k+1$,then:
$$n^2+n+43=8k+1\Rightarrow\ n^2+n+42=8k\ \Rightarrow\ n(n+1)\equiv6\pmod8$$
So $n$ becomes $2,5,...$ but $5$ doesn't work!!
|
Trying to solve: $n^2+n+43=m^2$, multiply by $4$ and complete the square, and you get:
$$(2n+1)^2+4\cdot 43-1 = (2m)^2.$$
Subtract and you are trying to solve $(2m)^2-(2n+1)^2=171$.
The difference of two squares means you need:
$$(2(m-n)-1)(2(m+n)+1)=171=9\cdot 19$$
So you need to factor $171$ as two odd numbers whose sum is a multiple of $4$.
There will only be finitely many solutions.
Alternatively, you know that if $n>42$ then $n^2<n^2+n+43<n^2+2n+1=(n+1)^2$. So you only have to check finitely many values $n$.
|
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|
Show that $n^2+11n+2$ is not divisible by $113^2$ ( n is integer) Show that $n^2+11n+2$ is not divisible by $113^2$ ( n is integer)
It's obvious that if we show $113$ doesn't divide $n^2+11n+2$ we are done...
|
A good start would be to find if the equation $n^2 + 11n + 2 \equiv 0$ has any solutions modulo 113. We can do this by completing the square, so we want to find $a,b \in \mathbb{Z}/113\mathbb{Z}$ such that
$$
n^2 + 11n + 2 \equiv (n-a)^2 + b \pmod{113}.
$$
By expanding we see that $-2a \equiv 11 \pmod{113}$, so $a \equiv (-2)^{-1}\cdot 11 \equiv 51 \pmod{113}$. If we fill this in we find that actually $b \equiv 0 \pmod{113}$ so
$$
n^2 + 11n + 2 \equiv (n-51)^2 \pmod{113}.
$$
This shows that $n\equiv51 \pmod{113}$ is the only solution to the equation $$n^2 + 11n + 2 \equiv 0 \pmod{113}.$$
Now let $n = 51 + 113k$, we know the expression is divisible by $113$ for all $k$, which we can also calculate:
$$
\frac{(51 + 113k)^2 + 11(51 + 113k) + 2}{113} = 113 k^2+113 k+28 \equiv 28 \pmod{113}.
$$
So we see that $n^2 + 11n + 2$ is never divisible by $113$ twice.
|
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|
Show that $1-\frac{1}{2}-\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}-\frac{1}{7} -\frac{1}{8}-\frac{1}{9}-\frac{1}{10} -\cdots $ converge $$1-\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{6}-\dfrac{1}{7}
-\dfrac{1}{8}-\dfrac{1}{9}-\dfrac{1}{10} ... $$
I added parentheses for each sub-sequence with the same sing.
so i got :
$$1-(\dfrac{1}{2}+\dfrac{1}{3})+(\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{6})-(\dfrac{1}{7}
+\dfrac{1}{8}+\dfrac{1}{9}+\dfrac{1}{10}) ... $$
I want to show that the new sequence is a leibniz sequence and by that conclude that is converge.
I managed to show that each pair of parentheses is greater than:
$$\dfrac{2}{n+1} $$
Cant find a way to proceed.
Thanks for helping.
|
Michael's grouping answer works in general. Note that the group with $n$ terms is of the form
$${1\over k+1}+{1\over k+2}+\cdots+{1\over k+n}$$
so the next group is
$${1\over k+n+1}+{1\over k+n+2}+\cdots{1\over k+2n}+{1\over k+2n+1}$$
It's easy to see that
$${1\over k+j}-{1\over k+n+j}={n\over(k+j)(k+n+j)}\gt{n\over(k+n)(k+2n+1)}$$
for each $1\le j\le n$, and so to show that the group with $n$ terms is bigger than the next group, it suffices to show
$${n^2\over(k+n)(k+2n+1)}\ge{1\over k+2n+1}$$
which is to say,
$$n^2\ge k+n$$
But this is easy: The number that ends the group with $n$ terms, $k+n$, is the triangular number $n(n+1)\over2$. And $n^2\ge{n(n+1)\over2}$ for all $n\ge1$.
|
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|
Prove that $1^8+2^8+\cdots+99^8 \equiv 1^4+2^4+\cdots+99^4 \pmod{25}.$
Prove that $$1^8+2^8+\cdots+99^8 \equiv 1^4+2^4+\cdots+99^4 \pmod{25}.$$
Attempt:
We can easily show that an eighth power can be expressed as a fourth power since $x^8 = (x^2)^4$. Conversely, by Fermat's Little Theorem, $x^{\phi(25)} = x^{20} \equiv 1 \pmod{25}$ if $\gcd(x,25) = 1$ and thus $x^4 = x \cdot x^3 \equiv x^{21} \cdot x^3 \equiv x^{24} \equiv (x^3)^8 \pmod{25}$. $\square$
I am not sure if the above proves the result, but it does show that if we have an $8$th power of an integer modulo $25$ and it is relatively prime to $25$, then it is equal to a fourth power of an integer modulo $25$ and vice-versa. Does that therefore mean they are equivalent?
|
$\mathbb{Z}/(25\mathbb{Z})^*$ is a cyclic group generated by $2$, hence
$$ \sum_{\substack{1\leq n \leq 25\\ 5\nmid n}}n^4 \equiv \sum_{r=1}^{20}2^{4r}\equiv \frac{16^{24}-16^4}{16-1}\equiv 20\pmod{25} $$
and
$$ \sum_{\substack{1\leq n \leq 25\\ 5\nmid n}}n^8 \equiv \sum_{r=1}^{20}2^{8r}\equiv \frac{256^{168}-256^8}{256-1}\equiv 20\pmod{25} $$
prove our claim.
To compute $2^N\pmod{125}$ is not difficult since $2$ is a generator of $\mathbb{Z}/(125\mathbb{Z})^*$, too.
|
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|
The function $h(x) = x^3 + bx^2 + d$ has a critical point at $(2, -4)$. Determine the constants $b$ and $d$ and find the equation of $h(x)$. The function $h(x) = x^3 + bx^2 + d$ has a critical point at $(2, -4)$. Determine the constants $b$ and $d$ and find the equation of $h(x)$. Please help me answer this question.
|
The critical points occur where the derivative is equal to zero.
\begin{align*}
h(x) & = x^3 + bx^2 + d\\
h'(x) & = 3x^2 + 2bx\\
& = x(3x + 2b)
\end{align*}
Setting the derivative equal to zero yields
\begin{align*}
x & = 0 & 3x + 2b & = 0\\
& & 3x & = -2b\\
& & x & = -\frac{2b}{3}
\end{align*}
We know that a critical point occurs when $x = 2$. Hence,
$$2 = -\frac{2b}{3} \implies b = -3$$
Thus,
$$h(x) = x^3 - 3x^2 + d$$
We also know that $h$ has the critical point $(2, -4)$. Hence,
$$h(2) = 2^3 - 3 \cdot 2^2 + d = 8 - 12 + d = -4 + d = -4$$
Thus, $d = 0$.
As a check, you should verify that $h(x) = x^3 - 3x^2$ has a critical point at $(2, -4)$.
|
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|
Radical equation $\sqrt{x+1}+\sqrt{x-1}-\sqrt{x^2 -1}=x$ The Question: $\sqrt{x+1}+\sqrt{x-1}-\sqrt{x^2 -1}=x$
Only thing I can take from this is that $x^2 -1=(x+1)(x-1)$, but I don't think that would help in any way.
I know the answer, but I don't know how to use that to work backwards.
Answer:
$x= \frac54$
|
You have already seen a solution which is much more clever. But let us try whether this can be done with repeated squaring - which is usually one of the first thing which comes to mind in problems like this.
Do not forget that when you square an equation, you can get an extraneous solutions. (See here, here or here.)
We first move $\sqrt{x^2+1}$ to the RHS. We hope to get nicer expressions if we do not square something containing three square roots.
\begin{align*}
\sqrt{x+1}+\sqrt{x-1}&=x+\sqrt{x^2-1}\\
(\sqrt{x+1}+\sqrt{x-1})^2&=(x+\sqrt{x^2-1})^2\\
(x+1)+2\sqrt{(x+1)(x-1)}+(x-1)&=x^2+2x\sqrt{x^2-1}+(x^2-1)\\
2x+2\sqrt{x^2-1}&=2x^2-1+2x\sqrt{x^2-1}\\
2(1-x)\sqrt{x^2-1}&=2x^2-2x-1
\end{align*}
Let me also mention that if we look at the LHS, we can see that $(\sqrt{x+1}+\sqrt{x-1})^2$ is precisely twice the expression $x+\sqrt{x^2-1}$. So we got some kind of relation between LHS and the RHS of the equation we started with. Perhaps this is one of possible ways to discover the clever substitution from Dr. MV's answer. But if we do not notice this, we can simply try to continue with squaring.
Now we only have one square root in our equation. We rearrange it in such way that all terms containing the square root are on one side of the equation and the remaining terms are on the other sited.
\begin{align*}
2(1-x)\sqrt{x^2-1}&=2x^2-2x-1\\
4(1-x)^2(x^2-1)&=(2x^2-2x-1)^2\\
4(x-1)^2(x-1)(x+1)&=(2x(x-1)-1)^2
\end{align*}
It looks that at this moment the substitution $x-1=t$ might simplify things a bit. (Of course, we could also continue without this substitution. You can try it that way, if you prefer.)
\begin{align*}
4t^3(t+2)&=(2t(t+1)-1)^2\\
4t^3(t+2)&=(2t^2+2t-1)^2\\
4t^4+8t^3&=4t^4+8t^3-4t+1\\
0&=4t+1\\
t&=\frac14\\
x&=t+1=\frac54
\end{align*}
So we get $$x=\frac54.$$
Since we used squaring in the process, we also have to check whether $5/4$ fulfills the original equation. And we find out that it does.
The LHS is $\sqrt{\frac94}+\sqrt{\frac14}-\sqrt{\frac{25}{16}-1}=\sqrt{\frac94}+\sqrt{\frac14}-\sqrt{\frac{9}{16}}=\frac32+\frac12-\frac34=2-\frac34=\frac54$. And we get $\frac54$ on the RHS, too.
|
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|
Why is $5\tan(54^\circ) = \sqrt{25 + 10\sqrt{5}}$ and $\tan\left(\frac{\pi}{5}\right) = \sqrt{5 - 2\sqrt{5}}$? On the Wikipedia Page about Pentagons, I noticed a statement in their work saying that $\sqrt{25+10\sqrt{5}}=5\tan(54^{\circ})$ and $\sqrt{5-2\sqrt{5}}=\tan(\frac {\pi}{5})$
My question is: How would you justify that? My goal is to simplify $\sqrt{25+10\sqrt{5}}$ and $\sqrt{5-2\sqrt{5}}$ without knowledge on the detested radical!
|
You surely know that $z=\cos\frac{2\pi}{5}+i\sin\frac{2\pi}{5}$ is the root in the first quadrant of $z^5-1=0$ so it satisfies
$$
z^4+z^3+z^2+z+1=0
$$
that can also be written as
$$
z^2+\frac{1}{z^2}+z+\frac{1}{z}+1=0
$$
or, by noting that $z^2+1/z^2=(z+1/z)^2-2$,
$$
\left(z+\frac{1}{z}\right)^2+\left(z+\frac{1}{z}\right)-1=0
$$
Therefore,
$$
z+\frac{1}{z}=2\cos\frac{2\pi}{5}=\frac{\sqrt{5}-1}{2}
$$
Thus
$$
\cos\frac{2\pi}{5}=\frac{\sqrt{5}-1}{4},
\qquad
\sin\frac{2\pi}{5}=\sqrt{1-\frac{5-2\sqrt{5}+1}{16}}=
\frac{\sqrt{10+2\sqrt{5}}}{4}
$$
Now use that
$$
\tan\frac{x}{2}=\frac{\sin x}{1+\cos x}
$$
and you get
$$
\tan\frac{\pi}{5}=\frac{\sqrt{10+2\sqrt{5}}}{3+\sqrt{5}}=
\frac{\sqrt{(10+2\sqrt{5})(3-\sqrt{5})^2}}{4}=
\frac{\sqrt{80-32\sqrt{5}}}{4}=\sqrt{5-2\sqrt{5}}
$$
Since $54^\circ=3\pi/10=\pi/2-\pi/5$, we have
$$
\tan\frac{3\pi}{10}=\cot\frac{\pi}{5}=\frac{1}{\sqrt{5-2\sqrt{5}}}=
\sqrt{\frac{5+2\sqrt{5}}{25-20}}=\frac{\sqrt{25+10\sqrt{5}}}{5}
$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Sum of real values of $x$ satisfying the equation $(x^2-5x+5)^{x^2+4x-60}=1$ I have this equation from this paper (Q.63)
Find the sum of all real values of $x$ satisfying the equation-$(x^2-5x+5)^{x^2+4x-60}=1$.
My attempt-
$(x^2-5x+5)^{x^2+4x-60}=(x^2-5x+5)^{(x-6)(x+10)}$
So, $x=6$ and $x=-10$ makes the power $0$ and hence the equation becomes one. So, the sum of the real values of $x$ satisfying the equation is $6+(-10)=-4$. But answer is $+3$.
Where have I gone wrong?
|
You missed some cases.
The cases are: $\text{number}^0,1^\text{number},{-1}^{\text{even}}$
You have only considered the first case.
*
*For first case, $$x^2+4x-60=0$$ the roots are $6,-10$.
*For the second case, $$x^2-5x+5=1$$ the roots are $1,4$.
*For the third case, $$x^2-5x+5=-1$$ $x$ can be $2$ or $3$. If you substitute the value $2$ on the equation $x^2+4x-60$, you get $-48$ which satisfies the condition since $-1^{-48}=1$. If you substitute $x=3$, you get $-39$ which does not satisfy the given condition since $-1^{-39}\neq 1$. So only $x=2$ satisfies the condition.
The real values of $x$ are therefore: $1,2,4,6,-10$. The sum of real values of $x$ is therefore $3$.
|
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|
Find the limit $\lim_{x \to 1} \left(\frac{p}{1-x^p} - \frac{q}{1-x^q}\right) $ $p ,q >0$ I Know series expansion and L'Hospital's rule . But here both of them are not of any help.
|
Here's a rather standard series expansion method: rewriting $x$ as $1+(x-1)$, and using Newton's generalised binomial expansion,
\begin{align}
\frac{p}{1-x^p}-\frac{q}{1-x^q} & = \frac{p}{1-(1+(x-1))^p}-\frac{q}{1-(1+(x-1))^q} \\
& = \frac{p}{1-(1+p(x-1)+\frac{p(p-1)}{2}(x-1)^2+O(|x-1|^3))} \\
&-\frac{q}{1-(1+q(x-1)+\frac{q(q-1)}{2}(x-1)^2+O(|x-1|^3))} \\
& = \frac{1}{(x-1)+\frac{q-1}{2}(x-1)^2+O(|x-1|^3)} \\
&-\frac{1}{(x-1)+\frac{p-1}{2}(x-1)^2+O(|x-1|^3)} \\
& = \frac{1}{x-1}(\frac{1}{1+\frac{q-1}2(x-1)+O(|x-1|^2)}-\frac{1}{1+\frac{p-1}2(x-1)+O(|x-1|^2)})\\
& = \frac1{x-1}(\frac{p-1}2(x-1)-\frac{q-1}{2}(x-1)+O(|x-1|^2)) \\
& = \frac{p-q}2.
\end{align}
Note that this requires no condition on $p,q$. They can even be negative.
|
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|
How can I compute $\tan(.5\arctan(x))$? The plot for this function appears to be in the form of $\alpha*\arctan(\beta*x)$ but I've no clue how to go about simplifying the expression.
|
You have to compute $y=\tan\left( \frac{\arctan{(x)}}{2} \right)$. Let's call $\theta=\arctan(x)$, so you have to compute $y=\tan\left(\frac{\theta}{2}\right)$.
You should write $\tan\left(\frac{\theta}{2}\right)$ in terms of $\tan\left(\theta\right)$, say you use:
$$\tan\left(\frac{\theta}{2}\right)=\frac{1-\cos\left(\theta\right)}{\sin\left(\theta\right)} = \frac{1}{\sin\left(\theta\right)}-\frac{1}{\tan\left(\theta\right)}$$
but from $$\cos^2\left(\theta\right)+\sin^2\left(\theta\right)=1$$ we get that $$\frac{1}{\tan^2\left(\theta\right)}+1 = \frac{1}{\sin^2\left(\theta\right)}$$
which is the same of
$$
\frac{1}{\sin\left(\theta\right)}=\sqrt{1+\frac{1}{\tan^2\left(\theta\right)}}\quad 0\lt\theta\lt\pi \Rightarrow x \gt 0 \\
$$
$$
\frac{1}{\sin\left(\theta\right)}=-\sqrt{1+\frac{1}{\tan^2\left(\theta\right)}}\quad -\pi\lt\theta\lt 0 \Rightarrow x \lt 0\\
$$
so finally we can write
$$\tan\left(\frac{\theta}{2}\right)= \frac{1}{\sin\left(\theta\right)}-\frac{1}{\tan\left(\theta\right)}=\pm\sqrt{1+\frac{1}{\tan^2\left(\theta\right)}}-\frac{1}{\tan\left(\theta\right)}$$
This is very helpfull because $\theta=\arctan(x)$ so $\tan(\arctan(x))=x$ and
$$y=\tan\left( \frac{\arctan{(x)}}{2} \right)=+\sqrt{1+\frac{1}{x^2}}-\frac{1}{x}\quad x\gt 0$$
$$y=\tan\left( \frac{\arctan{(x)}}{2} \right)=-\sqrt{1+\frac{1}{x^2}}-\frac{1}{x}\quad x\lt 0$$
|
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|
Find the condition such that the roots of the polynomial are in AP
$f(x)=x^3+3px^2+3qx+r$ has roots in AP.Find the relation between $p,q$ and $r$.
[Answer:$-2p^2-3pq+r=0$]
My attempt:-
Taking $d$ as the common difference of the roots in AP we have $f(x)=(x-(a-d))(x-a)(x-(a+d))=x^3-3x^2a+(3a^2-d^2)x-a(a^2-d^2)$.
Comparing this with,the given equation,we get,
$(p=-a),q=(\frac{3a^2-d^2}{3}),r=-a(a^2-d^2)$.
But,after this I have no idea how to eliminate $a$ and $d$ from the equation and find a relation between $p,r$ and $r$.
Thanks for any help!!
|
As you suggest, take the roots as $a-d,a,a+d$, then you get $p=-a,3q=3a^2-d^2,r=-a^3+ad^2$. Substituting the first and third in the second we get $$-2p^3+3pq-r=0$$ Check: take roots 1,2,3. The equation is $(x-1)(x-2)(x-3)=x^3-6x^2+11x-6$, so $p=-2,3q=11,r=-6$. The relation $-2p^3+3pq-r=0$ holds. The relation given in the question of $-2p^2-3pq+r$ does not.
|
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|
how to verify $\frac{\sin(x)\cos(x)}{\cos^2(x)-\sin^2(x)}=\frac{\tan(x)}{1-\tan^2(x))}$? How would I verifty the following trig identity?
$$
\frac{\sin(x)\cos(x)}{\cos^2(x)-\sin^2(x)}=\frac{\tan(x)}{1-\tan^2(x)}
$$
I am not sure how to start.
|
$$\frac{\sin(x)\cos(x)}{\cos^2(x)-\sin^2(x)}=\frac{\tan(x)}{1-\tan^2(x)}\Longleftrightarrow$$
$$\frac{\sec^2(x)}{\sec^2(x)}\cdot\frac{\sin(x)\cos(x)}{\cos^2(x)-\sin^2(x)}=\frac{\tan(x)}{1-\tan^2(x)}\Longleftrightarrow$$
$$\frac{\tan(x)}{\sec^2(x)\left(\cos^2(x)-\sin^2(x)\right)}=\frac{\tan(x)}{1-\tan^2(x)}$$
Now, use $\sec(x)=\frac{1}{\cos(x)}$:
$$\sec^2(x)\left(\cos^2(x)-\sin^2(x)\right)=\cos^2(x)\sec^2(x)-\sin^2(x)\sec^2(x)=$$
$$\left(\frac{1}{\cos(x)}\right)^2\cos^2(x)-\left(\frac{1}{\cos(x)}\right)^2\sin^2(x)=1-\frac{\sin^2(x)}{\cos^2(x)}=1-\tan^2(x)$$
So, we can conclude:
$$\frac{\tan(x)}{\sec^2(x)\left(\cos^2(x)-\sin^2(x)\right)}=\frac{\tan(x)}{1-\tan^2(x)}$$
|
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|
If $f$ is differentiable in $(a,b)$ then $\frac{1}{f}$ is differentiable at $(a,b)$, provided $f(a,b)\neq0$ "Suppose that $f$ is a differentiable function at $(a,b)$. Prove that $\frac{1}{f}$ is differentiable in $(a,b)$, provided $f(a,b)\neq0$"
We were given the following definition of differentiability: a function is differentiable in $(a,b)$ if there exists a linear map $L$ such that $f(x,y)=f(a,b)+L_{(a,b)}\cdot(x-a,y-b)+r(x,y)$ with $\frac{r(x,y)}{|(x-a,y-b)|}\to0$. However, I still have no idea how to use this to prove the theorem. I guess you want to end up with something like $\frac{1}{f(x,y)}=\frac{1}{f(a,b)}+L_{(a,b)}\cdot(x-a,y-b)+r(x,y)$ with $\frac{r(x,y)}{|(x-a,y-b)|}\to0$, but I don't see how you would get there.
Please help. Thanks in advance.
|
To be able to prove differentiability, you first need to guess what the linear map is. Usually one does this by using differentiation rules (derivative of a product, chain rule) that have appropriate extensions to the multivariable setting. But let us work this directly.
A decent rule of thumb is to mimic the one-variable rules. So we may guess, from $$ \left(\frac1f\right)'(t)=-\frac{f'(t)}{f(t)^2},$$ that the linear map $M$ for $1/f$ should be
$$
M_{(a,b)}(x,y)=-\frac{L_{(a,b)}(x,y)}{f(a,b)^2}.
$$
Let us give it a try: because $f(a,b)\ne0$ and $f$ is continuous (being differentiable), there exist $c,d>0$ with $d\geq|f(a+x,b+y)|\geq c$ for $x,y$ small enough. Then
\begin{align}
\frac{\left|\displaystyle\frac1{f(a+x,b+y)}-\frac1{f(a,b)}+\frac{L_{(a,b)}(x,y)}{f(a,b)^2}\right|}{(x^2+y^2)^{1/2}}
&=
\frac{\left|\displaystyle{f(a,b)^2}{}-{f(a+x,b+y)f(a,b)}{}+{L_{(a,b)}(x,y)f(a+x,b+y)}\right|}{(x^2+y^2)^{1/2}|f(a+x,b+y)|f(a,b)^2}\\ \ \\
&\leq
\frac{\left|\displaystyle{f(a,b)^2}{}-{f(a+x,b+y)f(a,b)}{}+{L_{(a,b)}(x,y)f(a+x,b+y)}\right|}{c^3(x^2+y^2)^{1/2}}\\ \ \\
&=
\frac{\left|\displaystyle{f(a,b)}(f(a,b)-{f(a+x,b+y))}{}+{L_{(a,b)}(x,y)f(a+x,b+y)}\right|}{c^3(x^2+y^2)^{1/2}}\\ \ \\
&=
\frac{\left|\displaystyle{f(a,b)}(-L_{(a,b)}(x,y)-r(x,y))+{L_{(a,b)}(x,y)f(a+x,b+y)}\right|}{c^3(x^2+y^2)^{1/2}}\\ \ \\
&=
\frac{\left|\displaystyle(f(a+x,b+y)-f(a,b))\,L_{(a,b)}(x,y)-f(a,b)\,r(x,y))\right|}{c^3(x^2+y^2)^{1/2}}\\ \ \\
&=
\frac{\left|\displaystyle(L_{(a,b)}(x,y)+r(x,y))\,L_{(a,b)}(x,y)-f(a,b)\,r(x,y))\right|}{c^3(x^2+y^2)^{1/2}}\\ \ \\
&\leq
\frac{\displaystyle(L_{(a,b)}(x,y)^2}{c^3(x^2+y^2)^{1/2}}+\frac{|r(x,y))\,L_{(a,b)}(x,y)|}{c^3(x^2+y^2)^{1/2}}+\frac{|f(a,b)\,r(x,y))|}{c^3(x^2+y^2)^{1/2}}\\ \ \\
&\leq
\frac{\displaystyle|L_{(a,b)}|^2(x^2+y^2)}{c^3(x^2+y^2)^{1/2}}+\frac{|r(x,y)|\,|L_{(a,b)}|^2(x^2+y^2)}{c^3(x^2+y^2)^{1/2}}+\frac{d\,|r(x,y))|}{c^3(x^2+y^2)^{1/2}}\\ \ \\
&=
\frac{\displaystyle|L_{(a,b)}|^2(x^2+y^2)^{1/2}}{c^3}+\frac{|r(x,y)|\,|L_{(a,b)}|^2(x^2+y^2)^{1/2}}{c^3}+\frac{d\,|r(x,y))|}{c^3(x^2+y^2)^{1/2}}\\ \ \\
\end{align}
and now it is clear that the three terms go to zero as $(x,y)\to0$, so $1/f$ is differentiable at $(a,b)$.
|
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|
Most natural way to prove $\sum_{n=1}^{\infty}\frac{1}{n+2}$ diverges I don't know how my teacher wants me to prove that
$$\sum_{n=1}^{\infty}\frac{1}{n+2}$$
diverges. All I know is that I have to use the $a_n>b_n$ criteria and prove that $b_n$ diverges.
I tried this:
$$\sum_{n=1}^{\infty} \frac{1}{n+2} =
\frac{1}{1+2}+\frac{1}{2+2}+\frac{1}{3+2} + \cdots + \frac{1}{n+2} = \left(\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n}\right)-\left(\frac{1}{1}+\frac{1}{2}\right) = \left(\sum_{n=1}^{\infty}\frac{1}{n}\right)-\frac{3}{2}$$
but I can't get a relationship between $a_n$ and $b_n$ of these series.
If I go to the root of the comparsion criterion, I know that:
$$\sum_{n=1}^{\infty} \frac{1}{n+2} =
\frac{1}{1+2}+\frac{1}{2+2}+\frac{1}{3+2} + \cdots + \frac{1}{n+2} = \left(\frac{1}{3}+\frac{1}{4}\right)+\left(\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}\right)+\cdots+\frac{1}{n}> \frac{2}{4}+\frac{4}{8}+\cdots+g_n$$ for some $g_n$ that I'm lazy to calculate.
Therefore, by the properties of limits, since the righthand sum $p_n$ diverges, the lefthand sum $s_n$ diverges too, because $s_n>p_n$. But I don't know if my teacher would accept that, I think she would only accept that I use the argument for the $a_n$ of the sum, not for the partial sum.
I also thought about rewriting:
$$\sum_{n=1}^{\infty} \frac{1}{n+2} = \sum_{n=3}^{\infty} \frac{1}{n}$$
but I don't see how it helps because the indexes are different.
All I need is an argument that will work with the $a_n$ of the $\sum_{n=1}^{\infty}a_n$, in relation with $b_n$ from $\sum_{n=1}^{\infty} b_n$
|
$$
\sum_{n=1}^N \frac{1}{n+2} =
\sum_{n=3}^{N+2} \frac{1}{n}
$$
So your series differs just by finite many terms from the harmonic series, which diverges.
|
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|
If $9 \mid 2^b-2^a$, then $7\mid2^b-2^a$
Prove that if $9 \mid 2^b-2^a$, then $7\mid2^b-2^a$.
I am not sure how to prove this statement, but it seems that from $9 \mid 2^b-2^a$ we have $b-a = 6n$. Then what should I do from here to prove the statement?
|
$9 \mid 2^b-2^a$
Let $a \leq b$, otherwise just exchange $a$ and $b$ and note that if $ x \mid -y \Rightarrow x \mid y$
$2^b -2^a=2^a(2^{b-a}-1)$
So, $9 \mid (2^{b-a}-1)$, as $9 \nmid 2^a$, because $9=3^2$ and so they don't have any common prime factors.
$2^{b-a} \equiv 1 \pmod 9$ and $\phi(9)=6$
$2$ is a primitive root modulo $9$, as all the reminders obtained on raising $2$ to a power are co-prime to $9$.
$$\begin{array}\\
2^1=2 \equiv 2 \pmod 9\\
2^2=4 \equiv 4 \pmod 9\\
2^3=8 \equiv 8 \pmod 9\\
2^4=16 \equiv 7 \pmod 9\\
2^5=32 \equiv 5 \pmod 9\\
2^6=63 \equiv 1 \pmod 9\\
\end{array}$$
$\Rightarrow b-a=6n$, where $n \in \mathbb{Z}$
Since $7 \nmid 2^a$, $7 \mid 2^b-2^a \Rightarrow 7 \mid (2^{b-a}-1)$, or $2^{b-a} \equiv 1 \pmod 7$
Now, since $\phi(7)=6$ and $b-a=6n$
$\Rightarrow 2^{6n} \equiv (2^6)^n \equiv (1)^n \equiv 1 \pmod 7$
So $7 \mid 2^b-2^a$, iff $9 \mid 2^b-2^a$
|
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|
$6 \times 6$ and $7 \times 7$ integer matrices
Can one fill a $6 \times 6$ matrix with integers so that the sum of all the numbers in each $3 \times 3$ square equals $2016$ and the sum of all the numbers in each $5 \times 5$ square equals $2015$?
Solve the same problem for the $7 \times 7$ case.
My work so far:
I solved the problem for table $6\times6$:
\begin{bmatrix}
2017 & -1 & 0 & 0 & 0 & 2017 \\
0 & -6050 & 4033 & 4034 & -6050 & -1 \\
0 & 4034 & -2017 & -2017 & 4033 & 0 \\
0 & 4033 & -2017 & -2017 & 4034 & 0 \\
-1 & -6050 & 4034 & 4033 & -6050 & 0 \\
2017 & 0 & 0 & 0 & -1 & 2017
\end{bmatrix}
Addition:
I proved that $x_1+x_2+x_3+x_4=4\cdot 2017$
$y_1+y_2+y_3+y_4=-4$
$x_1+x_3=x_2+x_4$
$S=0$
Further
Q: How to prove or disprove the existence of the table $7\times7$?
|
Hint: in the $7 \times 7$ case, how many $3 \times 3$ squares and how many $5 \times 5$ squares contain each entry of the table?
EDIT:
OK, time's up... each entry of the $7 \times 7$ table is in the same number of $3 \times 3$ squares as it is in $5 \times 5$ squares. So i you take the sums of the numbers in each $3 \times 3$ square and add them up, you should get the same result as if you take the sum of the numbers in each $5 \times 5$ square and add them up. Now, what does that tell you?
|
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|
Permutations conjugated
Show that the permutations:
$\alpha=
\begin{pmatrix}
1 & 2 & 3 & 4 & 5 & 6 \\
2 & 5 & 3 & 6 & 1 & 4 \\
\end{pmatrix}
$
and
$\beta=
\begin{pmatrix}
1 & 2 & 3 & 4 & 5 & 6 \\
5 & 3 & 4 & 2 & 1 & 6 \\
\end{pmatrix}
$
Are conjugated in $S_6$ and you hrite all permutations $\gamma\in S_6$ such that $\gamma\alpha\gamma^{-1}=\beta$
I know that $\alpha=(1,2,5)(4,6)(3)$ and $\beta=(2,3,4)(1,5)(6)$ also if $\sigma_1=\bigl( \begin{smallmatrix} 1 & 2 & 3 & 4 & 5 & 6 \\ 2 & 3 & 1 & 5 & 4 & 6 \end{smallmatrix} \bigr)$ then $\sigma_1(1,2,5)\sigma_1^{-1}=(2,3,4)$; if $\sigma_2=\bigl( \begin{smallmatrix} 1 & 2 & 3 & 4 & 5 & 6 \\ 2 & 3 & 4 & 1 & 6 & 5 \end{smallmatrix} \bigr)$ then $\sigma_2(4,6)\sigma_2^{-1}=(1,5)$ and if $\sigma_3=(3,6)$ then $\sigma_3(3)\sigma_3^{-1}=(6)$.
But I don't know how to build $\gamma$ such that $\gamma\alpha\gamma^{-1}=\beta$. Can you help me, please?
|
Note that $S_6$ is the symmetric group of $6$ symbols which consists of $6!$ element.
We can use the theorem that two permutations are conjugate iff they have the same cycle type.
As you have already simplified the permutations into the product of disjoint cycles, it is easy to see that the two permutations have the same cycle type. Fortunately, using the theorem, no further calculations are needed.
|
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|
How can you prove that $1+ 5+ 9 + \cdots +(4n-3) = 2n^{2} - n$ without using induction? Using mathematical induction, I have proved that
$$1+ 5+ 9 + \cdots +(4n-3) = 2n^{2} - n$$
for every integer $n > 0$.
I would like to know if there is another way of proving this result without using PMI. Is there any geometric solution to prove this problem? Also, are there examples of problems where only PMI is our only option?
Here is the way I have solved this using PMI.
Base Case: since $1 = 2 · 1^2 − 1$, the formula holds for $n = 1$.
Assuming that the
formula holds for some integer $k ≥ 1$, that is,
$$1 + 5 + 9 + \dots + (4k − 3) = 2k^2 − k$$
I show that
$$1 + 5 + 9 + \dots + [4(k + 1) − 3] = 2(k + 1)^2 − (k + 1).$$
Now if I use hypothesis I observe.
$$
\begin{align}
1 + 5 + 9 + \dots + [4(k + 1) − 3]
& = [1 + 5 + 9 + \dots + (4k − 3)] + 4(k + 1) −3 \\
& = (2k^2 − k) + (4k + 1) \\
& = 2k^2 + 3k + 1 \\
& = 2(k + 1)^2 − (k + 1)
\end{align}
$$
$\diamond$
|
Let
$$a_k := 4 k - 3 \qquad \qquad \qquad b_k := k - 1$$
Using summation by parts,
$$\begin{array}{rl} \displaystyle\sum_{k=1}^n a_k &= \displaystyle\sum_{k=1}^n a_k (b_{k+1} - b_k)\\ &= \left(a_k \, b_k \,\bigg|_1^{n+1}\right) - \displaystyle\sum_{k=1}^n (a_{k+1} - a_k) \, b_{k+1}\\ &= a_{n+1} \, b_{n+1} - a_1 \, b_1 - \displaystyle\sum_{k=1}^n 4 k\\ &= a_{n+1} \, b_{n+1} - \left(\displaystyle\sum_{k=1}^n (4 k - 3)\right) - 3n\\ &= (4 n + 1) n - 3 n - \displaystyle\sum_{k=1}^n a_k\\ &= 4 n^2 - 2n - \displaystyle\sum_{k=1}^n a_k\end{array}$$
Hence,
$$\displaystyle\sum_{k=1}^n a_k = 2 n^2 - n$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1802846",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "53",
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"answer_id": 2
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|
Household reflector or transformation
Let $A\in\mathbb{R}^{n\times k}$, $n\geq k$, and $rank(A) = k$. Consider the use of Household reflectors, $H_i$, $1\leq i\leq k$, to transform $A$ to upper trapezoidal form, i.e., $$H_{k}H_{k-1}\ldots H_2 H_1 A = \begin{pmatrix}R\\0\end{pmatrix}$$
where $R\in\mathbb{R}^{k\times k}$ nonsingular upper triangular. We can define $$H_i = I + \alpha_i x_i x_i^{T}$$
$$\alpha_i = \frac{-2}{\|x_i\|_{2}^{2}}$$
not sure what $x_i$ is though don't see it in my notes anywhere.
So suppose $$A = \begin{pmatrix}
2 & 1\\
1 & 2\\
1 & 2\\
\end{pmatrix}$$
and I want to get $R$. I tried to apply the formula but my $R$ is not upper trapezoidal. So there must be something I am doing wrong. Could someone kindly show the steps so I know how to do this?
Attempt - $$H_1 = I + \frac{-2}{6}\begin{pmatrix}
2\\
1\\
1\\
\end{pmatrix}\begin{pmatrix} 2 & 1 & 1\end{pmatrix}$$ thus we have $$\begin{pmatrix}
1 & 0 & 0\\
0 & 1 & 1\\
0 & 0 & 0\\
\end{pmatrix} - \frac{1}{3}\begin{pmatrix}
4 & 2 & 2\\
2 & 1 & 1\\
2 & 1 & 1\\
\end{pmatrix} = \begin{pmatrix}
-1/3 & -2/3 & -2/3\\
-2/3 & 2/3 & -1/3\\
-2/3 & -1/3 & 2/3\\
\end{pmatrix}$$
Thus $$H_1 A = \begin{pmatrix}
-2 & -3\\
-1 & 0\\
-1 & 0\\
\end{pmatrix}$$
thus I have a problem, please let me know if there is an error in my calculations.
|
The point of $H_1$ is to make the first column a column with zeros except in the top entry.
To find $x_1$, note that $\|(2,1,1)\| = \sqrt{6}$, so we should aim to get $(2,1,1)$ to $\sqrt{6}(1,0,0)$. Take
$$
x_1 = (2,1,1) - (\sqrt{6},0,0) = (2-\sqrt{6},1,1)
$$
(why should this work? Try to figure it out geometrically).
In the next step, since the first column is already where we want it, we apply a Householder transformation for the $2 \times 2$ submatrix that comes from deleting the first row and column. In this way, our upper-triangularization process is recursive.
|
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"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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|
Curve sketching without a computer program How to sketch the curve x^6 + y^6 = (x^4)*y without using a computer program ? Could someone give me the step by step ?
|
$\dfrac{y^6}{x^4} = y - x^2$, so there are no solutions whenever $y - x^2 < 0$. Whenever $y - x^2 = 0$, we find (by substitution) that the only solution is the origin. Notice that the curve is symmetric with respect to the y-axis, that is $(-x,y)$ is a solution if $(x,y)$ is a solution. Therefore, it is only necessary to graph the curve in the region bounded by the parabola $y = x^2$ and the y-axis. You may also want to ask if the curve is bounded in its y-component. In fact, if $y = 1$, then $x^6 - x^4 - 1 = 0$. The polynomial $f(x) = x^6 - x^4$ is pretty simple to graph with the three roots $(-1,0), (0,0),$ and $(1,0)$. By translating the graph of $x^6 - x^4$ down by one unit, to obtain the graph of $x^6 - x^4 - 1$, we see that there are two roots: one whose x-component is less than -1 and the other whose x-component is greater than 1, yet these points do not belong in the region that we described. This may be generalized for any $y \geq 1$, so the curve must be bounded above by $y = 1$.
In polar coordinate form, recall $x = r\cos(\theta)$ and $y = r\sin(\theta)$.
$x^6 + y^6 = r^6\cos^6(\theta) + r^6\sin^6(\theta) = r^6[\cos^2(\theta) + \sin^2(\theta)][\cos^4(\theta) - \cos^2(\theta)\sin^2(\theta)+\sin^4(\theta)] = r^6[\cos^4(\theta) - \cos^2(\theta)\sin^2(\theta)+\sin^4(\theta)]$
The equation $x^6 + y^6 = x^4y$ becomes
$r^6[\cos^4(\theta) - \cos^2(\theta)\sin^2(\theta)+\sin^4(\theta)] = r^5\cos^4(\theta)\sin(\theta).$ We may divide appropriately since we've already discussed the points at which division would not be well-defined.
$r = \dfrac{\sin(\theta)}{1 - \tan^2(\theta) + \tan^4(\theta)}$
You may plot a few points of the form $(r,\theta)$ by way of a trigonometric table: $\big(\frac{9}{14}, \frac{\pi}{6}\big), \big(\frac{1}{\sqrt{2}},\frac{\pi}{4}\big), \big(\frac{\sqrt{3}}{14}, \frac{\pi}{3}\big)$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1804084",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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|
Integral inequality :$\int_0^1(f'(x))^2dx\geq 32\int_0^1(f(x))^2dx + 16\left(\int_0^{\frac{1}{2}}f(x)dx-\int_{\frac{1}{2}}^1f(x)dx\right)^2$ Assume $f:[0,1]\to \mathbb{R}$ is differentiable and $f'$ is integrable. Given $f\left(\frac{1}{4}\right)=f\left(\frac{3}{4}\right)=f(1)-f(0)=0$, then prove that $$\int_0^1(f'(x))^2dx\geq 32\int_0^1(f(x))^2dx + 16\left(\int_0^{\frac{1}{2}}f(x)dx-\int_{\frac{1}{2}}^1f(x)dx\right)^2$$ My attempt: I tried to make the most out of the given information. Consider the integrals $$I_1=\int_0^{\frac{1}{4}}\left(x-\frac{1}{8}\right)f'(x)dx=\frac{f(0)}{8} - \int_0^{\frac{1}{4}}f(x)dx$$ $$I_2=\int_{\frac{1}{4}}^{\frac{1}{2}}\left(x-\frac{3}{8}\right)f'(x)dx=\frac{f\left(\frac{1}{2}\right)}{8} - \int_{\frac{1}{4}}^{\frac{1}{2}}f(x)dx$$ $$I_3=\int_{\frac{1}{2}}^{\frac{3}{4}}\left(x-\frac{5}{8}\right)f'(x)dx=\frac{f\left(\frac{1}{2}\right)}{8} - \int_{\frac{1}{2}}^{\frac{3}{4}}f(x)dx$$ $$I_4=\int_{\frac{3}{4}}^{1}\left(x-\frac{7}{8}\right)f'(x)dx=\frac{f\left(1\right)}{8} - \int_{\frac{3}{4}}^{1}f(x)dx$$ Notice that $$I_1+I_2-I_3-I_4=-\left(\int_0^{\frac{1}{2}}f(x)dx-\int_{\frac{1}{2}}^1f(x)dx\right)$$ Hence $$\left|\int_0^{\frac{1}{2}}f(x)dx-\int_{\frac{1}{2}}^1f(x)dx\right|^2\leq \left(I_1+I_2-I_3-I_4\right)^2\leq 4\left(|I_1|^2+|I_2|^2+|I_3|^2+|I_4|^2\right)$$ Using C-S inequality, we get $$\left|I_k\right|^2\leq \frac{1}{2^8\times 3}\int_{\frac{k}{4}}^{\frac{k+1}{4}}(f'(x))^2dx ,\quad k\in \{0,1,2,3\}$$ Hence $$\left(\int_0^{\frac{1}{2}}f(x)dx-\int_{\frac{1}{2}}^1f(x)dx\right)^2\leq \frac{4}{2^8\times 3}\int_0^1(f'(x))^2dx$$ Upto this seems fine , but i am having trouble estimating the second term i.e $\int_0^1(f(x))^2dx$. I tried considering integrals like $\int_0^{\frac{1}{4}}xf'(x)f(x)dx,\quad \int_{\frac{1}{4}}^{\frac{1}{2}}\left(x-\frac{1}{2}\right)f'(x)f(x)dx$ etc, but they are not giving satisfactory results. Please see if i can improve this or is there some other approach.
|
Thanks for mentioning Wirtinger's inequality, i was not aware of it( have not studied fourier analysis yet). I will use two of its applications : $1)$ If $f(a)=f(b)=0$ then $$\frac{(b-a)^2}{\pi^2}\int_a^b(f'(x))^2dx\geq\int_0^a(f(x))^2dx$$ $2)$ If either one of $f(a)$ or $f(b)$ is zero then $$\frac{4(b-a)^2}{\pi^2}\int_a^b(f'(x))^2dx\geq\int_a^b(f(x))^2dx$$ Now using $(1)$ for the interval $\left[\frac{1}{4},\frac{3}{4}\right]$ we get $$\frac{1}{4\pi^2}\int_{\frac{1}{4}}^{\frac{3}{4}}(f'(x))^2dx\geq\int_{\frac{1}{4}}^{\frac{3}{4}}(f(x))^2dx$$ and using $(2)$ for intervals $\left[0,\frac{1}{4}\right]$ and $\left[\frac{3}{4},1\right]$ $$\frac{1}{4\pi^2}\int_{0}^{\frac{1}{4}}(f'(x))^2dx\geq\int_{0}^{\frac{1}{4}}(f(x))^2dx$$
$$\frac{1}{4\pi^2}\int_{\frac{3}{4}}^{1}(f'(x))^2dx\geq\int_{\frac{3}{4}}^{1}(f(x))^2dx$$ adding these three $$\frac{1}{4\pi^2}\int_{0}^{1}(f'(x))^2dx\geq\int_{0}^{1}(f(x))^2dx$$ and i have already established that $$ \frac{4}{2^8\times 3}\int_0^1(f'(x))^2dx\geq\left(\int_0^{\frac{1}{2}}f(x)dx-\int_{\frac{1}{2}}^1f(x)dx\right)^2$$ Hence $$32\int_0^1(f(x))^2dx + 16\left(\int_0^{\frac{1}{2}}f(x)dx-\int_{\frac{1}{2}}^1f(x)dx\right)^2\leq\left(\frac{1}{12}+\frac{8}{\pi^2}\right)\int_0^1(f'(x))^2dx$$ and $\left(\frac{1}{12}+\frac{8}{\pi^2}\right)<1$
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/1804720",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
How to find the Laplacian Eigen Values of the given graph
Find the Laplacian eigen-values of the of the graph on $N$ vertices whose edge set is given by $\{(i,i+1),1\le i<n\}$ and the edge $(1,n)$ .
The answer is given to be $2-2\cos{\frac{2k\pi}{n}}$ .
My try:
I tried to proceed by induction.
For the case $n=3$ ;I got the matrix as $L$=
\begin{array}{cccc}
2& -1 &-1\\
-1& 2 &-1\\
-1 & -1 &2
\end{array}
I got the eigen values as $0,3,3$ which are nowhere near the given the eigen values.
How should I do it? Please give some ways to solve this problem.
|
Consider the Laplacian matrix of the cycle on $n$ vertices, given by
\begin{equation*}
L = \begin{bmatrix}
2 & - 1 & 0 & \cdots & 0 & -1\\
-1 & 2 & -1 & \cdots & 0 & 0\\
0 & -1 & 2 & \cdots & 0 & 0\\
\vdots & \vdots & \vdots & \ddots & \vdots & \vdots\\
0 & 0 & 0 & \cdots & 2 & -1\\
-1 & 0 & 0 & \cdots & -1& 2\\
\end{bmatrix}.
\end{equation*}
$L$ is an example of a circulant matrix. Any circulant matrix is fully determined by its first row $[a_0\ a_1\ \cdots\ a_{n-1}]$, and its eigenvalues are given by a general formula. However, I will demonstrate the method of computation here (for the special case $L$).
Let $\omega = \exp\left( \dfrac{2i\pi}{n} \right)$, the primitive $n$-th root of unity, and let $v_1 = [1\ \omega\ \omega^2\ \cdots \omega^{n-1}]^T$. Then,
\begin{align*}
Lv_1 = \begin{bmatrix}
2 - \omega - \omega^{n-1}\\
2 \omega - \omega^2 - 1\\
2 \omega^2 - \omega^3 - \omega\\
\vdots\\
2\omega^{n-1} - 1 - \omega^{n-2}
\end{bmatrix} = (2 - \omega - \omega^{n-1})\begin{bmatrix}
1 \\ \omega \\ \omega^2\\ \vdots\\ \omega^{n-1}
\end{bmatrix},
\end{align*}
(keeping in mind that $\omega^n = 1$).
Thus, $2 - \omega - \omega^{n-1} = 2 - \omega - \omega^{-1}$ is an eigenvalue of $L$, with eigenvector $v_1$.
Similarly, for $k = 2, \cdots, n$, define $v_k = [1\ \omega^k\ \cdots\ \omega^{k(n-1)}]^T$, the vector obtained by taking the $k$-th power of each entry of $v_1$. Note that $v_n$ is the all-ones vector. Then, a computation exactly similar to the above one shows that $v_k$ is an eigenvector of $L$ for the eigenvalue $2 - \omega^k - \omega^{-k} = 2 - (\omega^k + \omega^{-k})$.
Simplifying, $\omega^k + \omega^{-k} = \exp\left( \dfrac{2ki\pi}{n} \right) + \exp\left( -\dfrac{2ki\pi}{n} \right) = 2 \cos \left( \dfrac{2k\pi}{n} \right)$. Thus, the eigenvalues of $L$ are
\begin{align*}\large\boxed{
2 \left( 1 - \cos \dfrac{2k\pi}{n} \right), \quad k = 1, 2, \ldots, n.}
\end{align*}
Alternatively, you can find the eigenvalues $\lambda_k$ of the adjacency matrix $A$ (which is also circulant), and use the fact that $L = 2I - A$ to get the eigenvalues $2 - \lambda_k$ of $L$.
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/1809210",
"timestamp": "2023-03-29T00:00:00",
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|
Show that the equation $x^2+y^2+z^2=x^2y^2$ has no integer solution,except $x=y=z=0$ Show that the equation $x^2+y^2+z^2=x^2y^2$ has no integer solution,except $x=y=z=0.$
Let one of the $x,y,z$ be even number.Let $x=2p$
$x^2+y^2+z^2=x^2y^2$
This gives $y^2+z^2$ is also even,which means either $y,z$ are both even or $y,z$ are both odd.
So either $x=2p,y=2q,z=2r$ is the solution or $x=2p,y=2q+1,z=2r+1$ is the solution.I put $x=2p,y=2q+1,z=2r+1$ in the equation $x^2+y^2+z^2=x^2y^2$ and see that it does not satisfy both sides,so it is not the solution.
When I put $x=2p,y=2q,z=2r$,i get $p^2+q^2+r^2=4p^2q^2$
I am confused here how to prove that $p^2+q^2+r^2=4p^2q^2$ does not hold true.
Also i cannot solve when $x$ is odd which is the second case.
Is my method correct?Please help me complete the solution.
|
$$ z^2 + 1 = (x^2 - 1)(y^2 - 1) $$
From quadratic reciprocity, $z^2 + 1$ is not divisible by $4$ or by any prime $q \equiv 3 \pmod 4.$
If $x$ or $y$ is odd, then $z^2 + 1$ is divisible by $4.$ Nope.
If $x$ or $y$ is $0$ then all $x,y,z = 0.$
Otherwise, if $x$ or $y$ is even, say it is $x,$ then one of $x-1,x+1$ is $1 \pmod 4$ but the other is $3 \pmod 4,$ the latter therefore divisible by some prime $q \equiv 3 \pmod 4.$ Nope.
|
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|
Find dimension from subspace equations In order to calculate its dimension, I need the basis of the subspace. How can I get it from the equations?
Find the dimension of the following subspaces of $\mathbb{R}^5$:
$$ U = \{(x_1,x_2,x_3,x_4,x_5) \ | \ 2x_1 - x_2 - x_3 = 0, x_4-3x_5=0 \}\\
V = \{(x_1,x_2,x_3,x_4,x_5) \ | \ 2x_1 - x_2 + x_3 + 4x_4 + 4x_5 = 0 \}.$$
Let $W$ be the subspace satisfying all 3 equations. Is it true that dim($W) = 5$?
|
If $(x_1,x_2,x_3,x_4,x_5) \in U$, then $x_1 = \frac{1}{2}x_2 + \frac{1}{2}x_3$ and $x_4 = 3x_5$.
So
$$\begin{pmatrix}
x_1\\
x_2\\
x_3\\
x_4\\
x_5\\
\end{pmatrix} = \begin{pmatrix}
\frac{1}{2}x_2 + \frac{1}{2}x_3\\
x_2 \\
x_3 \\
3x_5 \\
x_5
\end{pmatrix}= x_2\begin{pmatrix}
\frac{1}{2}\\
1\\
0\\
0\\
0\\
\end{pmatrix}+ x_3 \begin{pmatrix}
\frac{1}{2}\\
0\\
1\\
0\\
0\\
\end{pmatrix} + x_5 \begin{pmatrix}
0\\
0\\
0\\
3\\
1\\
\end{pmatrix}$$
So we see that these three vectors form a basis for $U$. Hence dim($U) = 3$.
Now if $(x_1,x_2,x_3,x_4,x_5) \in V$, then $x_2 = 2x_1 + x_3 + 4x_4 + 4x_5$.
So
$$\begin{pmatrix}
x_1\\
x_2\\
x_3\\
x_4\\
x_5\\
\end{pmatrix} = \begin{pmatrix}
x_1\\
2x_1 + x_3 + 4x_4 + 4x_5\\
x_3\\
x_4\\
x_5\\
\end{pmatrix}= x_1 \begin{pmatrix}
1\\
2\\
0\\
0\\
0\\
\end{pmatrix} + x_3 \begin{pmatrix}
0\\
1\\
1\\
0\\
0\\
\end{pmatrix} + x_4 \begin{pmatrix}
0\\
4\\
0\\
1\\
0\\
\end{pmatrix} + x_5 \begin{pmatrix}
0\\
4\\
0\\
0\\
1\\
\end{pmatrix}$$
So we see that these four vectors form a basis for $V$. Hence dim($V)=4$.
For $W$, rather than go through the same sort of computation, note that $W = U \cap V$. So $dim(W) \leq \min\{dim(U),dim(V)\} = 3$. So $dim(W) \neq 5$.
|
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "2",
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|
Solution of equation If $f(x) = x^2 - 2ax + a(a+1)$ , $f:[ a, \infty] \to [a,\infty]$ . If one of the solution of the equation $f(x)=f^{-1}(x)$ is $5049$ , then what may be the other solution ?
My WORK:
I found the inverse, but when I equate them , it is very difficult to solve . Please help me with some method to solve it.
|
Your method of equating the function and its inverse directly is perfectly alright.
$f(x) = x^2 - 2ax + a^2 + a$
or $f(x) = (x-a)^2 + a$
or $f(x) - a = (x-a)^2$
$x = a \pm \sqrt{f(x) - a}$
Here the negative sign is rejected as $x \in [a,\infty]$
So, we have $f^{-1}(x) = a + \sqrt{x - a}$
Now, $f(x) = f^{-1}(x)$
$(x-a)^2 + a = a + \sqrt{x-a}$
$(x-a)^2 = \sqrt{x-a}$
$(x - a) ^ {\frac{1}{2}} ((x - a) ^ {\frac{3}{2}} - 1) = 0$
Therefore,
$x - a = 0$ or $(x - a)^{\frac{3}{2}} = 1$
$x = a$, or $x = a + 1$
|
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"url": "https://math.stackexchange.com/questions/1818134",
"timestamp": "2023-03-29T00:00:00",
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|
To show that two quadratic forms are equivalent over $\mathbb{C}$ but not over $\mathbb{R}$ Consider the quadratic forms $p$ and $q$ given by $$q(x,y,z,w)=x^2+y^2+z^2+bw^2$$ and $$p(x,y,z,w)=x^2+y^2+czw.$$
Then we can write their respective matrices as
$$
A = \pmatrix{1\\&1\\&&1\\&&&b}, \quad
B = \pmatrix{1\\&1\\&&0&c/2\\&&c/2&0}
$$
To Prove: If $b$ and $c$ are non-zero than $p$ and $q$ are equivalent over $\mathbb{C}$ (need not be equivalent over $\mathbb{R}$).
They are equivalent over $\mathbb{R}$ only if $b,c$ are non-zero and $b$ is negative.
Now that characteristic polynomial of $A$ and $B$ are $(x-1)^3(x-b)$ and $(x-1)^2(x^2-\frac{c^2}{4})$ respectively. So I can see from Sylvester's theorem why they are equivalent over $\mathbb{R}$ only if $b,c \neq 0 \ \& \ b<0$. But I can't prove the part over $\mathbb{C}$ I am trying to work with the inertia of matrices described here but I need help.
|
Write the coordinates in $q(x,y,z,w)$ as $\widetilde z:= \frac1{\sqrt c}\cdot (z - i\sqrt bw)$ and $\widetilde w:= \frac1{\sqrt c}\cdot (z+i\sqrt bw)$ (i. e. make a base change). Then you have
\begin{align*}
q(x,y,z,w) &= x^2+y^2+z^2+bw^2\\
&= x^2+y^2 + (z-i\sqrt bw)(z+i\sqrt bw)\\
&= x^2+y^2 + c\cdot \left(\frac{1}{\sqrt c}\cdot (z-i\sqrt bw)\right)\cdot \left(\frac1{\sqrt c}\cdot (z+i\sqrt bw)\right)\\
&= x^2+y^2 + c\widetilde z\widetilde w\\
&= p(x,y,\widetilde z,\widetilde w).
\end{align*}
Therefore, $p$ and $q$ are equivalent over $\mathbb C$.
If you want to calculate with matrices, notice that
$$\tag{$*$}
\begin{pmatrix}x\\y\\z\\w \end{pmatrix} = \begin{pmatrix} 1 & 0 & 0 & 0\\ 0 & 1
& 0 & 0\\ 0 & 0 & \frac1{\sqrt c} & -i\sqrt{\frac bc}\\
0 & 0 & \frac1{\sqrt c} & i\sqrt{\frac bc} \end{pmatrix}\cdot \begin{pmatrix}
x\\ y\\ \widetilde z\\ \widetilde w\end{pmatrix}.
$$
Therefore,
\begin{align*}
q(x,y,z,w) &= \begin{pmatrix}x & y & z & w\end{pmatrix}\cdot \begin{pmatrix} 1 &
0 & 0 & 0\\ 0 & 1 & 0 & 0\\ 0 & 0 & 0 & \frac c2\\ 0 & 0 & \frac c2 &
0\end{pmatrix}\cdot \begin{pmatrix}x\\y\\z\\w\end{pmatrix}\\
&=\begin{pmatrix}x & y & \widetilde z & \widetilde w\end{pmatrix} \cdot
\begin{pmatrix} 1 & 1 & 0 & 0\\ 0 & 1 & 0 & 0\\ 0 & 0 & \frac1{\sqrt c} &
\frac1{\sqrt c}\\ 0 & 0 & -i\sqrt{\frac bc} & i\sqrt{\frac bc} \end{pmatrix}
\cdot \begin{pmatrix} 1 & 0 & 0 & 0\\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & \frac c2\\ 0
& 0 & \frac c2 & 0\end{pmatrix}\cdot \begin{pmatrix}1 & 0 & 0 & 0 \\ 0 & 1 & 0 &
0\\ 0 & 0 & \frac1{\sqrt c} & -i\sqrt{\frac bc}\\ 0 & 0 & \frac1{\sqrt c} &
i\sqrt{\frac bc}\end{pmatrix} \cdot \begin{pmatrix}x\\y\\ \widetilde z\\
\widetilde w\end{pmatrix}\\
&= \begin{pmatrix}x & y & \widetilde z & \widetilde w\end{pmatrix} \cdot
\begin{pmatrix} 1 & 0 & 0 & 0\\ 0 & 1 & 0 & 0\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 &
b\end{pmatrix} \cdot \begin{pmatrix} x\\ y\\ \widetilde z\\ \widetilde
w\end{pmatrix}\\
&= p(x,y,\widetilde z, \widetilde w).
\end{align*}
To find the base change matrix, you could apply symmetric Gauss elimination to
the matrix of $q$: Use column operations (followed directly by the corresponding
row operation) to transform the matrix of $q$ into the matrix of $p$. Also, at
every step, keep track of the column operations (e. g. by applying them to the
identity matrix): the resulting matrix is your base change matrix (in the sense of ($*$)).
|
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|
Two dice are rolled and the sum of the face values is six. What is the probability that at least one of the dice came up a three?
Two dice are rolled and the sum of the face values is six. What is the probability that at least one of the dice came up a three?
Attempt: I feel I'm having trouble formalizing the expression for what I believe to be happening. This is a question concerning conditional probabilities so as such I will need first the probability of obtaining a sum of $6$:
$$P(6) = P(6|(4,2))P(4,2) + P(6|(5,1))P(5,1) + P(6|(3,3))P(3,3)$$
So in terms of notation, the ordered pairs represent the way in which a sum of $6$ could be attained (ex: $(4,2)$, where you roll a $4$ on the first die and $2$ on the second). With that being taken into account:
$$\left(\frac{2}{5}\right)\left(\frac{2}{36}\right) + \left(\frac{2}{5}\right)\left(\frac{2}{36}\right) + \left(\frac{1}{5}\right)\left(\frac{1}{36}\right) = 0.0495$$
Now what was requested was the probability of a $3$ coming up given the sum of 6:
\begin{align}P((3,3)|6) &= \frac{P(6|(3,3))P(3,3)}{P(6) = P(6|(4,2))P(4,2) + P(6|(5,1))P(5,1) + P(6|(3,3))P(3,3)}\\\\
&= \frac{0.0056}{0.0495}\\\\
&= 0.113\end{align}
But I still feel I'm not treating the condition on the sum of $6$ properly.. Is this correct?
|
I would use a table to first see this intuiutively
$$\begin{array}{|cc|cccccc|}\hline
&&&&&D_1\\
&&1&2&3&4&5&6\\
\hline
&1&2&3&4&5&\color{red}6&7\\
&2&3&4&5&\color{red}6&7&8\\
D_2&3&4&5&\color{blue}6&7&8&9\\
&4&5&\color{red}6&7&8&9&10\\
&5&\color{red}6&7&8&9&10&11\\
&6&7&8&9&10&11&12\\
\hline\end{array}$$
The red numbers show all combinations where the dice sum to $6$ and the blue shows the one event we are interested in, where at least one dice shows a $3$
So we have $\dfrac 15$ elements which satisfy the given constraints
Now consider it mathematically
We know that $P(\text{sum}=6)=\dfrac 5{36}$
We also know that $P((3,3))=\dfrac 1{36}$
We can therefore say that \begin{align}P(\text{sum}=6\text{ AND one dice shows }3)&=\frac{P((3,3))}{P(\text{sum}=6)} \\\\
&=\frac{\frac1{36}}{\frac5{36}}\\\\
&=\frac1{36}\cdot\frac{36}5\\\\
&=\frac 15\end{align}
|
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|
Diophantine equations using Euclidean algorithm I solved two systems of Diophantine equations using the Euclidean algorithm and I can't figure out where I went wrong because the solutions I test aren't working but I have rechecked my work several times.
a) $56x+72y=40$
using the algorithm:
$72=1*56+16$
$56=3*16+$
$16=2*8+$
so $\gcd(56, 72)=8$. Then,
$8=56-3*16$
$=56-3(72-56)$
$=-3(72)+4(56)$
since $c/\gcd=40/8=5$, $x_0=4(5)=20$ and $y_0=-3(5)=-15$ so all solutions should be $(20+9t,-15+7t)$ since $9=b/\gcd$ and $7=a/\gcd$. Everything looks right to me and matches my notes and all the equations seem to make sense, but when I try solutions with different values of t, the only one that works and actually $=40$ is when $t=0$. I have the same problem with the next system:
b) $101x+40y=1$
using the algorithm the same way as above, I find $\gcd(101,40)=1$, and
$1= 2-1(1)= 2-(19-9*2)= 10(2)-19$
$= 10(21-19)-19 = 10(21)-11(19)$
$= 10(101-40*2)-11(40-21)$
$= 10(101)-20(40)-11(40)+11(101-40*2) = 101(21)+40(-53)$
since $\gcd=1$, $x_0=21$ and $y_0=-53$ so the set of all equations is $(21+40t,-53+101t)$
|
Let $d = \gcd(a, b)$. The linear Diophantine equation $ax + by = c$ has a solution if and only if $d \mid c$. If $(x_0, y_0)$ is a particular solution of the equation $ax + by = c$, then the general solution is
$$x = x_0 + \frac{b}{d}t \qquad y = y_0 - \frac{a}{d}t$$
where $t \in \mathbb{Z}$.
In both problems, you incorrectly set $y = y_0 \color{red}{+} \dfrac{a}{d}t$.
$56x + 72y = 40$
You correctly found the particular solution $(20, -15)$. Let's see what happens when we substitute your general solution $(20 + 9t, -15 + 7t)$ into the equation $56x + 72y = 40$.
\begin{align*}
56x + 72y & = 56(20 + 9t) + 72(-15 + 7t)\\
& = 1120 + 504t - 1008 + 504t\\
& = 40 + 1008t
\end{align*}
which only yields a correct solution when $t = 0$. We want the $504t$ terms to cancel, which we can achieve by changing the sign of $7t$ to obtain the general solution $(20 + 9t, -15 - 7t)$.
A recommendation: Divide by $\gcd(56, 72, 40) = 8$ first to obtain the equivalent equation $7x + 9y = 5$.
Apply the extended Euclidean algorithm.
\begin{align*}
9 & = 1 \cdot 7 + 2\\
7 & = 3 \cdot 2 + 1\\
2 & = 2 \cdot 1
\end{align*}
Working backwards, we obtain
\begin{align*}
1 & = 7 - 3 \cdot 2\\
& = 7 - 3(9 - 7)\\
& = 4 \cdot 7 - 3 \cdot 9
\end{align*}
Multiplying both sides of the equation $4 \cdot 7 - 3 \cdot 9 = 1$ by $5$ yields $20 \cdot 7 - 15 \cdot 9 = 5$. Hence, a particular solution of the equation is $(20, -15)$, as you found above.
We have $a = 9$, $b = 7$, $x_0 = 20$, $y_0 = -15$, and $d = \gcd(a, b) = \gcd(9, 7) = 1$, so the general solution is
\begin{align*}
x & = x_0 + \frac{b}{d}t & y & = y_0 - \frac{a}{d}t\\
& = 20 + \frac{9}{1}t & & = -15 - \frac{7}{1}t\\
& = 20 + 9t & & = -15 - 7t
\end{align*}
which you can verify by substituting $20 + 9t$ for $x$ and $-15 - 7t$ for $y$ in the equation $56x + 72y = 40$.
$101x + 40y = 1$
You correctly found that $(21, -53)$ is a particular solution. If you correct the sign error in the formula for the general solution, you should obtain $(21 + 40t, -53 - 101t)$.
|
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|
Can $b+c$ in the pythagorean triplets $(a, b, c)$ be a prime number? If $a^2 + b^2 = c^2$, can $b + c$ be a prime number?
|
Hint: $a^2+b^2=c^2\implies b^2-c^2=-a^2\implies (b+c)(b-c)=-a^2 \implies b+c=\frac{a^2}{c-b}$ $ \implies b+c=a^2 \times \frac{1}{c-b}$
|
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|
Uniform convergence of following series Prove that $\sum_{n=1}^\infty \frac{x^{2n}}{(1 + x + \dots + x^{2n})^2}$ converges uniformly when $x \geq 0$.
|
$\sum \dfrac {x^{2n}}{(1+x+...x^{2n})^2}$
Suppose $x< 1$
$\sum \dfrac {(1-x)^2x^{2n}}{(1-x^{2n+1})^2}\\
\sum \big(\dfrac {(1-x)x^{n}}{1-x^{2n+1}}\big)^2\\
\big(\dfrac {(1-x)x^{n}}{1-x^{2n+1}}\big)^2<x^{2n}$
$\sum x^{2n}$ converges when $x<1$ and the series converges by the comparison test.
Suppose $x>1$
$\sum \big(\dfrac {(x-1)x^n}{(x^{2n+1}-1)}\big)^2\\
\sum \big(\dfrac {1-x^{-1}}{x^n-x^{-n-1}}\big)^2\\$
$\big(\dfrac {1-x^{-1}}{x^n-x^{-n-1}}\big)^2<\frac{1}{x^{2n}}$
$\sum \frac{1}{x^{2n}}$ converges when $x>1$ and the series converges by the comparison test.
Suppose $x=1,$
$\sum \dfrac {1}{(2n+1)^2}$ converges.
|
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|
How to prove this inequality? $a^{2}+b^{2}+c^{2}\leq 3$ How to prove this inequality?
If $a^{2}+b^{2}+c^{2}\leq 3$ and $a,b,c\in \Bbb R^+$, then
$$\left( a+b+c\right) \left( a+b+c-abc\right)\geq 2\left( a^{2}b+b^{2}c+c^{2}a\right) $$
I tried AM>GM but I couldn't get result
|
Assume w.l.o.g that $a \leq b \leq c$.
Then we have : $(a+b+c)^3 \geq 7(a^2b+b^2c+c^2a)+6abc$; thus it is sufficient to prove: $(a+b+c)(a+b+c-abc) \geq \dfrac{2}{7}((a+b+c)^3-6abc)$.
Let $x=a+b+c$ and $y=abc$. The last inequality is then: $x^2-xy \geq \dfrac{2}{7}(x^3-6y)$, which is equivalent to: $x^2+\dfrac{12y}{7} \geq \dfrac{2x}{7}(x^2+\dfrac{7y}{2})$ and since $x \leq 3$ (using Cauchy-Schwarz on the original inequality), it is sufficient to prove: $x^2+\dfrac{12y}{7} \geq \dfrac{6}{7}(x^2+\dfrac{7y}{2})$, which is equivalent to: $x^2 \geq 9y$, which follows from A.M.-G.M. ($x \geq 3y^{1/3} \geq 3y^{1/2}$) where we used $y=abc\leq 1$.
|
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|
Prove $a^2+b^2+c^2 \ge a+b+c$ if $abc=1$, and $a$, $b$, $c$ are positive real numbers
Prove $a^2+b^2+c^2 \ge a+b+c$ if $abc=1$, and $a$, $b$, $c$ are positive real numbers
It is in the exercises of the AM-GM inequality chapter of a book, and that is why I believe it will be solved by that. Can anyone give me a proof using that or otherwise, too?
|
The previous answer was not a full answer, let me offer my thoughts and let me expand a little (using Cauchy-Schwartz):
Using the Cauchy-Schwarz inequality we get
\begin{align}
a+b+c=a\cdot 1+b\cdot 1+c\cdot 1+\leq\sqrt{a^2+b^2+c^2}\sqrt{1^2+1^2+1^2}\tag{1}
\end{align}
From the AM-GM inequality we obtain $a^2+b^2+c^2\geq 3\sqrt[3]{a^2b^2c^2}=3$, so
\begin{align}
\sqrt{3}\leq\sqrt{a^2+b^2+c^2}\tag{2}
\end{align}
From $(1)$ and $(2)$ it clearly follows that
\begin{align}
a+b+c\leq\sqrt{a^2+b^2+c^2}\sqrt{3}\leq\sqrt{a^2+b^2+c^2}\sqrt{a^2+b^2+c^2}=a^2+b^2+c^2
\end{align}
Further Edit using only AM-GM
Reduce the problem to elementary algebra by considering
$$a^2 \ge 2a -1 \tag3$$
$$b^2 \ge 2b-1 \tag4$$
$$c^2 \ge 2c-1 \tag5$$
Then add up $(3), (4)$ & $(5)$ and get:
$$a^2+b^2+c^2 \ge a+b+c +a+b+c -3 \tag6$$
By AM-GM we have
$$\frac{a+b+c}{3} \ge (abc)^\frac{1}{3}=1 $$
$$ a+b+c \ge 3 \tag7 $$
Finally, from $(6)$ and $(7)$ we obtain the required inequality:
$$a^2+b^2+c^2 \ge a+b+c +a+b+c -3 \ge a+b+c $$
|
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|
Limit of the sequence $a_{n+1}=\frac{1}{2} (a_n+\sqrt{\frac{a_n^2+b_n^2}{2}})$ - can't recognize the pattern Consider the sequence:
$$a_0=x,~~~b_0=y$$
$$a_{n+1}=\frac{1}{2} \left(a_n+\sqrt{\frac{a_n^2+b_n^2}{2}} \right),~b_{n+1}=\frac{1}{2} \left(b_n+\sqrt{\frac{a_n^2+b_n^2}{2}}\right)$$
$$\lim_{n \to \infty} a_n=\lim_{n \to \infty} b_n=l(x,y)$$
I can't pin down the pattern for the limit. Numerically, I've got the following result:
$$\frac{1}{l(x,y)}=\arctan (f(x,y))$$
What is the explicit expression for $f(x,y)$?
Numerical examples:
$$\begin{array}( x & y & \frac{1}{l(x,y)} \\ 1 & 2 & \arctan \left( \frac{3}{4} \right) \\ 1 & 3 & \arctan \left( \frac{1}{2} \right) \\ 2 & 3 & \arctan \left( \frac{5}{12} \right) \\ 3 & 5 & \arctan \left( \frac{1}{4} \right) \\ 1 & 5 & \arctan \left( \frac{\sqrt{13}-3}{2} \right) \\ 4 & 5 & \arctan \left( \frac{9}{40} \right) \\ 3 & 4 & \arctan \left( \frac{7}{24} \right) \\ 3 & 7 & \arctan \left( \frac{\sqrt{29}-5}{2} \right) \end{array}$$
Edit
To summarize @IvanNeretin's comment (and adding my own numerical results), so far we have:
$$l(x,0)=\frac{2x}{\pi}$$
$$f(x,x+1)=\frac{2x+1}{2x(x+1)}$$
$$f(x,x+2)=\frac{1}{x+1}$$
$f(x,x+3)$ - see my answer below
$$f(x,x+4)=\frac{\sqrt{x^2+4x+8}-(x+2)}{2}$$
Here $x \in \mathbb{R}^{+}$, $x \neq 0$
Also by symmetry we have of course:
$$f(x,y)=f(y,x)$$
At this point it's obvious that no ordinary function works here.
My hope is to connect $f(x,y)$ to some special function or an integral.
|
Let $z_i=a_i + ib_i$. Then your recursion is
$$
z_{n+1}=a_{n+1}+ib_{n+1}=\frac{1}{2}(a_n+ib_n)+\frac{1}{2}(1+i)\sqrt{\frac{a_n^2+b_n^2}{2}}=\frac{1}{2}z_n+\frac{1+i}{2\sqrt{2}}|z_n|=\frac{1}{2}z_n+\frac{1}{2}e^{i\pi/4}|z_n|.
$$
Now, let $z_n=e^{i(\theta_n+\pi/4)}y_n$, where the $y_n$ and $\theta_n$ are real. Then
$$
y_{n+1}e^{i\theta_{n+1}}=\frac{1}{2}y_{n}e^{i\theta_{n}}+\frac{1}{2}y_n=\frac{1}{2}y_n(1+e^{i\theta_n}),
$$
so
$$
y_{n+1}=\frac{1}{2}y_n\sqrt{2+2\cos\theta_n}=y_n\cos\left(\theta_n/2\right)
$$
and
$$
\cos(\theta_{n+1})=\frac{y_n}{2y_{n+1}}(1+\cos\theta_n)=\frac{1+\cos\theta_n}{2\cos(\theta_n/2)}=\cos(\theta_n/2),$$
or
$$
\theta_{n+1}=\frac{1}{2}\theta_n.
$$
This just gives $\theta_{n}=2^{-n}\theta_{0}$, so $\theta_{\infty}=0$, and
$$
y_{\infty}=y_0\prod_{i=0}^{\infty}\cos(2^{-i}\theta_0/2)=y_0\frac{\sin(\theta_0)}{\theta_0}.
$$
The limit of the real and imaginary parts of $z_n$ is
$$
l(x,y)=\frac{1}{\sqrt{2}}y_{\infty}=\frac{y_0}{\sqrt{2}}\frac{\sin\theta_0}{\theta_0}=\sqrt{\frac{x^2+y^2}{2}}\frac{\sin\theta_0}{\theta_0},
$$
where
$$
\theta_0 = \tan^{-1}(y/x) - \frac{\pi}{4}=\tan^{-1}(y/x)-\tan^{-1}(1)=\tan^{-1}\left(\frac{y/x-1}{1+y/x}\right)=\tan^{-1}\left(\frac{y-x}{y+x}\right).
$$
Then
$$
l(x,y)=\frac{y-x}{2\tan^{-1}\left(\frac{y-x}{y+x}\right)}.
$$
For $(x,y)=(1,2)$, for instance, we have
$$
l(1,2)=\frac{1}{2\tan^{-1}(1/3)}=\frac{1}{\tan^{-1}(3/4)}.
$$
|
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|
Prove that $\Gamma(-k+\frac{1}{2})=\frac{(-1)^k 2^k}{1\cdot 3\cdot 5\cdots(2k-1)}\sqrt{\pi}$. I was able to prove that
$$
\Gamma\left (k+\frac{1}{2} \right )=\frac{1\cdot 3\cdot 5\cdots(2k-1)}{2^k}\sqrt{\pi}.\tag{$k\geq 1$}$$
using the Legendre's duplication formula. But I can't do the same to
$$\Gamma\left ( -k+\frac{1}{2} \right )=\frac{(-1)^k 2^k}{1\cdot 3\cdot 5\cdots(2k-1)}\sqrt{\pi}.\tag{$k\geq 1$}$$
If possible, I'd like you to give a hint. If it is not possible to use Legendre's duplication formula, then I tried this way, for $n\geq 1$,
\begin{align*}
\Gamma\left ( -n+\frac{1}{2} \right )&=\left ( -n-\frac{1}{2} \right )\Gamma\left ( -n-\frac{1}{2} \right )\\
&=\dots\\
&=\left ( \frac{1}{2} \right )\left ( \frac{3}{2} \right )\cdots\left ( \frac{-2n-3}{2} \right )\left ( \frac{-2n-1}{2} \right )\Gamma\left ( \frac{1}{2} \right )
\end{align*}
and not anymore. How do I count how many factors there are to right side beside $\Gamma(1/2)$?
|
Is this Possible?
Since,
$\Gamma(a)=\frac{\Gamma(a+n)}{(a)_n}$ You can deduce that
\begin{eqnarray*}
\Gamma(-m+\frac{1}{2})&=&\frac{\Gamma(\frac{1}{2})}{(-m+\frac{1}{2})_m}\\
&=&\frac{\sqrt{\pi}}{\left(\frac{-2m+1}{2}\right)\cdot\left(\frac{-2m+3}{2}\right)\cdot\left(\frac{-2m+5}{2}\right)\cdots\left(\frac{1}{2}\right)}\\
&=&\frac{(-1)^m\cdot 2^m\cdot\sqrt{\pi}}{(2m-1)\cdot(2m-3)\cdots 3\cdot 2\cdot 1}
\end{eqnarray*}
|
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|
Sum of $\frac{1}{5}-\frac{2}{5^2}+\frac{3}{5^3}-\frac{4}{5^4}+....$ Find the sum of following series:
$S=\frac{1}{5}-\frac{2}{5^2}+\frac{3}{5^3}-\frac{4}{5^4}+....$ upto infinite terms
Could someone give me slight hint to solve this question?
|
$$S=\frac 15-\frac2{5^2}+\frac 3{5^3}-\dots\tag 1$$
and $$\frac S5=\frac 1{5^2}-\frac2{5^3}+\frac 3{5^4}-\dots\tag 2$$
Now, $(1)+(2)$ yields $$S+\frac S5=\frac 15-\frac1{5^2}+\frac1{5^3}+\dots\\\implies \frac{6S}5=\frac 15-\frac1{5^2}+\frac1{5^3}+\dots=\frac 16\\\implies S=\frac5{36}.$$
|
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|
Expected Number of flips for alternating Heads/Tails 10 times What is the expected number of flips needed to flip a coin 10 times and have the outcome be alternating heads/tails (starting with heads, then tails, then heads etc...).
I wrote a c++ program and it gives me 2,730. Is this correct and how would you do this mathematically?
|
Let $x$ be the expected number of flips to get $(HT)^5$ and $y$ the expected number of further tosses required if we are starting with $XH$ where we cannot use part of $X$ to form the first part of the required run (eg $X$ is empty or ends in $H$).
The first toss is $H$ or $T$, so $x=\frac{1}{2}(1+y)+\frac{1}{2}(1+x)$. Hence $y=x-2$.
How consider the outcomes: $T,HH,(HT)T,(HT)HH,(HT)^2T,(HT)^2HH,\dots,(HT)^4T,(HT)^4HT,(HT)^4HH,(HT)^5$.
They give $$x=\frac{1}{2}(x+1)+\frac{1}{2^2}(y+2)+\frac{1}{2^3}(x+3)+\dots+\frac{1}{2^9}(x+9)+\frac{1}{2^{10}}(y+10)+\frac{1}{2^{10}}10$$ $$=\frac{1}{2}(x+1)+\frac{1}{2^2}x+\frac{1}{2^3}(x+3)+\dots+\frac{1}{2^9}(x+9)+\frac{1}{2^{10}}(x+8)+\frac{1}{2^{10}}10$$ $$=x\left(1-\frac{1}{1024}\right)+\left(\frac{1}{2}+\frac{3}{2^3}+\frac{2}{2^4}+\frac{5}{2^5}+\frac{4}{2^6}+\frac{7}{2^7}+\frac{6}{2^8}+\frac{9}{2^9}+\frac{18}{2^{10}}\right)$$ $$=x\left(1-\frac{1}{1024}\right)\frac{341}{256}$$ So finally $x=1364$.
$\textbf{Comments}$
This is easy to get wrong. I got it wrong, but fortunately @Did picked up the error (see comment below).
The video the OP refers to seems to be about 10 consecutive heads. That case is different because a single $T$ wrecks the run and puts you back to the start. In that case you can consider the outcomes $T,HT,HHT,HHHT,\dots,HHHHHHHHHT,HHHHHHHHHH$. The all but the last (which is a run of 10 heads) put you back to the start. So we have $$x=\frac{1}{2}(x+1)+\frac{1}{4}(x+2)+\dots+\frac{1}{1024}(x+10)+\frac{10}{1024}$$ giving $$\frac{1}{1024}x=\frac{1}{2}+2\left(\frac{1}{2}\right)^2+3\left(\frac{1}{2}\right)^3+\dots+10\left(\frac{1}{2}\right)^{10}+\frac{10}{1024}$$ giving $$x=2046$$
In both cases it is not too difficult to generalise to the case of $n$ flips.
Note that a run of 10 heads or a run of 10 tails are the hardest runs to get because a single wrong flip puts you back to the start, whereas if you are aiming for alternate heads and tails, an unwanted $H$ knocks out your run but gives the first element of a new run. So the 2730 from the $C^{++}$ must be some kind of error.
|
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|
Find the smallest positive integer which can be written as the sum of the squares of two positive integers in two different ways
Find the smallest positive integer which can be written as the sum of the squares of two positive integers in two different ways.
I took extremely long to solve this
I got
$50= 7^2 + 1^2 $
$50= 5^2 + 5 ^2 $
I did it with the trial and error method.
I'm just curious if this came out during my exams, is there a quicker way to find the answer?
|
Consider the equation $x^2+y^2=z^2+w^2=N.$
This is equivalent to: $x^2-z^2=w^2-y^2=D.$
and we need $D$ to have at least two different factorizations with the factors having the same parity. One approach is thus to find the smallest values of $D$ which satisfy this property. Another approach is to find parametric solutions.
Let $(x-z)=ab$ and $(x+z)=cd$ with $(w-y)=ac$ and $(w+y)=bd$,
which gives the parametric solution: $\dfrac{1}{2}(ab+cd,bd-ac,cd-ab,ac+bd)$.
We have $N=\dfrac{1}{4}(a^2b^2+c^2d^2+b^2d^2+a^2c^2)$.
Now from the fact that $cd-ab$ and $bd-ac$ are distinct positive integers, we can see that no three of $a,b,c,d$ can be the same.
Thus, the lexicographically smallest value for the tuple $(a,b,c,d)$ is $(1,1,2,3)$, which gives the solution $(7,1,5,5)$ and $N=50$. Once we have this solution, it is easily checked that other tuples of $a,b,c,d$ give larger values of $N$.
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If $\ln(1+x) \approx A+Bx+Cx^2$, differentiate twice both sides and show that $\ln(1+x) \approx x-\frac{1}{2}x^2$
Question: $\ln(1+x) \approx A+Bx+Cx^2$, for $-1<x\leq1$, where $A,B,C$ are constants.
Differentiate twice both sides of the approximation above and hence show that $$ \ln(1+x) \approx x-\frac{1}{2}x^2$$
Using logs, estimate the smallest value of the positive integer $n$ for which $$ \left(1+\frac{1}{2n} \right)^{n+3} < \left(1+\frac{1}{n} \right)^{n-1}$$
My working:
For the first part:
$$\ln(1+x) \approx A+Bx+Cx^2$$
$$ \Leftrightarrow \frac{1}{1+x} \approx B+2Cx$$
$$ \Leftrightarrow \frac{-1}{(1+x)^2} \approx 2C $$
But now I do not know what to do with this....
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@Michael Hardy offered a clear solution to the first part.
Second Part solution
The second part of the problem can be solved by applying the solution from the first part:
\begin{align*}
\left(1+\frac{1}{2n}\right)^{n+3} &< \left(1+\frac{1}{n}\right)^{n-1} \\
\ln \left(1+\frac{1}{2n}\right)^{n+3} &< \ln \left(1+\frac{1}{n}\right)^{n-1} &\text{($\ln$ is increasing function)} \\
(n+3) \ln \left(1+\frac{1}{2n}\right) &< (n-1) \ln \left(1+\frac{1}{n}\right) \\
(n+3) \left(\frac{1}{2n}-\frac{\left(\frac{1}{2n}\right)^2}{2}\right) &< (n-1) \left(\frac{1}{n}-\frac{\left(\frac{1}{n}\right)^2}{2}\right) \\
\frac{n+3}{2n}-\frac{n+3}{8n^2} &< \frac{n-1}{n}-\frac{n-1}{2n^2} \\
(4n^2+12n)-(n+3) &< (8n^2-8n)-(4n-4) \\
0 &< 4n^2-23n+7 \\
\end{align*}
(Algebra in $\LaTeX$. Yuck!)
We solve the quadratic for zeros (alternatively, plot it and eyeball the solution since we're estimating anyways) to get
$n \ge \frac{23 + \sqrt{417}}{8} \approx 5.428 \text{ or } n \le \frac{23 - \sqrt{417}}{8} \approx 0.322$. Since $n$ must be a positive integer, we solve $$n \ge 6$$
Additional Notes
Wolfram Alpha gives an "exact approximation" of $n \ge 5.357$ for the intersection point of the two functions. This value is, as we see, quite close to our polynomial-based approximation, because, based on a plot or a quick calculation, the second-degree approximation $\ln(1+x)\approx x-\frac{x^2}{2}$ is pretty darn close arount $x=0.2$.
If we try to solve this with the first-degree approximation $\ln(1+x) \approx x$, we get
\begin{align*}
(n+3)\left(\frac{1}{2n}\right) &< (n-1)\left(\frac{1}{n}\right) \\
\frac{5}{2n} &< \frac{1}{2} \\
5 &< n \\
n &\ge 6
\end{align*}
which, amusingly, is just enough accuracy for me to get the right answer. Should've done that, maybe, and saved myself from having to use the quadratic equation. A quick calculation shows that the first-degree approximation is not as good, but still somewhat good, as the second-degree approximation.
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Prove: if $|x-1|<\frac{1}{10}$ so $\frac{|x^2-1|}{|x+3|}<\frac{1}{13}$
Prove: $$|x-1|<\frac{1}{10} \rightarrow \frac{|x^2-1|}{|x+3|}<\frac{1}{13}$$
$$|x-1|<\frac{1}{10}$$
$$ -\frac{1}{10}<x-1<\frac{1}{10}$$
$$ \frac{19}{10}<x+1<\frac{21}{10}$$
$$|x+1|<\frac{19}{10}$$
Adding 4 to both sides of $$ -\frac{1}{10}<x-1<\frac{1}{10}$$ gives:
$$\frac{39}{10}<x+3<\frac{41}{10}$$
$$|x+3|<\frac{39}{10}$$
Plugging those results in $$\frac{|x-1|*|x+1|}{|x+3|}<\frac{1}{13}$$
We get: $$\frac{\frac{1}{10}*\frac{19}{10}}{\frac{39}{10}}<\frac{1}{13}$$
$$\frac{19}{390}<\frac{1}{13}$$ Which is true, is this proof is valid as I took the smallest intervals, like $|x+3|<\frac{39}{10}$ and not $|x+3|<\frac{41}{10}$?
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You are very close but you have actually made two mistakes that have 'cancelled' each other out to get the right result.
Like you pointed out the inequality you used was wrong. In fact, you have the reverse of this inequality. However, this is exactly what you need since you have to remember that dividing reverses the direction of the inequality as well. Hence, the rest of the proof is correct.
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