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Transformation of general equation of straight line to normal form Consider the general form of equation of straight line.
$$ax + by + c = 0$$
$$\Rightarrow ax + by = -c$$
$$\Rightarrow \frac{ax}{-c} + \frac{by}{-c} = 1$$
$$\Rightarrow \frac{x}{\frac {-c}{a}} + \frac{y}{\frac {-c}{b}} =1 $$
this is equation of line in intercept form with x-intercept $\frac {-c}{a}$ and y-intercept $ \frac {-c}{b}$
In above fig. $\overline {OD}\bot\overline {AB}$
Now consider $\Delta$OAD
$$Cos\alpha =\frac {\overline{OD}}{\overline{OA}} = \frac{P}{\frac {-c}{a}} =\frac {-aP}{c} $$
In $\Delta$OBD
$$Cos(90-\alpha)=\frac {\overline{OD}}{\overline{OB}} = \frac{P}{\frac {-c}{b}} =\frac {-bP}{c} $$
Since Cos($90-\alpha$)= Sin $\alpha$ Therefore,
$$Sin\alpha = \frac {-bP}{c}$$
Now
$$tan\alpha =\frac {sin \alpha}{cos\alpha} = (\frac {-bP}{c})(\frac{-c}{aP})$$
$$tan \alpha = \frac {b}{c}=> \frac {perpendicular}{base}$$
$$\Rightarrow hyp = \pm\sqrt{a^2+b^2} (by Pythagorean theorem).........(a)$$
$$\Rightarrow cos\alpha = \frac {a}{\pm\sqrt {a^2+b^2}},sin\alpha = \frac {b}{\pm\sqrt {a^2+b^2}}...............(1)$$
Now for transforming general equation of straight line to normal form:
Consider the general form of equation of straight line
$$ax+by+c=0$$
$$\Rightarrow ax+by=-c$$
Now in order to to have Normal form of equation of straight line it is said that divide both sides by $\pm\sqrt{a^2+b^2}$, then we get
$$\frac {a}{\pm\sqrt{a^2+b^2}}x+\frac {b}{\pm\sqrt{a^2+b^2}}y=\frac {-c}{\pm\sqrt{a^2+b^2}}$$
Using equations (1) in above we get
$$xcos\alpha+ysin\alpha=\frac {-c}{\pm\sqrt{a^2+b^2}}$$
Now comparing with normal equation of straight line i.e.$xcos\alpha+ysin\alpha=P$
$$\implies P=\frac {-c}{\pm\sqrt{a^2+b^2}}$$
Now the question is:
1: In equation (a) $hyp = \pm\sqrt{a^2+b^2}$ but according to figure in $\Delta OAD hyp=\overline{OA}= \frac{-c}{a}$
How $\pm\sqrt{a^2+b^2}=\frac{-c}{a}$?
|
The general equation $ax+by+c=0$ of a straight line in $\mathbb R^2$ can be rewrtten as $\mathbf n\cdot\mathbf x=-c$, with $\mathbf n\ne0$. This form of the equation tells us that a line can be characterized as the set of points that have the same dot product with some fixed vector $\mathbf n$. If we normalize $\mathbf n$ by dividing both sides of the equation by its length, we have $${\mathbf n\over\|\mathbf n\|}\cdot\mathbf x=-{c\over\|\mathbf n\|}.$$ This is the so-called normal form of the equation for the line, characterized by $\mathbf n$ being a unit vector. The left-hand side of this equation is the (signed) length of the orthogonal projection of $\mathbf x$ onto $\mathbf n$, so an equivalent way to charactere the line is as the set of vectors $\mathbf x$ that have the same projection $\mathbf x_\parallel$ onto $\mathbf n$. The remainder $\mathbf x-\mathbf x_\parallel$ is perpendicular to $\mathbf n$, from which we can see that $\mathbf n$ is perpendicular to the line, and so the (perpendicular) distance of the line from the origin is ${|c|\over\|\mathbf n\|}$. Substituting the coefficients from the original equation yields the normal equation $${a\over\sqrt{a^2+b^2}}x+{b\over\sqrt{a^2+b^2}}y={-c\over\sqrt{a^2+b^2}}.$$ Since we have a unit normal, we can replace the coefficients on the left-hand side: $$x\cos\alpha+y\sin\alpha={-c\over\sqrt{a^2+b^2}},$$ where $\alpha$ is the angle made by the normal vector with the $x$-axis.
Now, as far as your equation (a) is concerned, remember that in computing $\tan\alpha$, you canceled a constant of proportionality that appeared in both the numerator and denominator, so the most that you can conclude from it is that the length of the hypotenuse of $\triangle{OAD}$ is proportional to $\sqrt{a^2+b^2}$. Indeed, in your diagram you have two other triangles that are similar to $\triangle{OAD}$: $\triangle{BOD}$ and $\triangle{BOA}$, so we have $$\tan\alpha = {\overline{AD}\over\overline{OD}} = {\overline{OD}\over\overline{BD}}={\overline{OA}\over\overline{OB}}.$$ The legs of the largest of these are the hypotenuses of the smaller ones, so there are at least two different lengths of hypotenuse in play. Moreover, it might be that none of the lengths is equal to $\sqrt{a^2+b^2}$!
|
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|
Finding $\lim_{ n \to \infty }(1-\tan^2\frac{x}{2})(1-\tan^2\frac{x}{4})(1-\tan^2\frac{x}{8})...(1-\tan^2\frac{x}{2^m})=?$ Find the limit :
$$\lim_{ n \to \infty }(1-\tan^2\frac{x}{2})(1-\tan^2\frac{x}{4})(1-\tan^2\frac{x}{8})...(1-\tan^2\frac{x}{2^n})=?$$
My try :
$$1-\tan^2 y = \frac{2\tan y }{\tan(2y)}$$
$$\lim_{ n \to \infty }\left( \frac{2\tan\frac{x}{2} }{\tan(x)}\right)( \frac{2\tan\frac{x}{4} }{\tan(\frac{x}{2})})( \frac{2\tan\frac{x}{8} }{\tan(\frac{x}{4})})...( \frac{2\tan\frac{x}{2^n} }{\tan(\frac{x}{2^{n-1}})})=?$$
Now?
|
Note that the numbers in the deonominator and numerator cancel, leaving only $$\lim_{n \to \infty} 2^{n} \frac{\tan \frac{x}{2^n}}{\tan x}$$
In your original equation. However, as $\lim\limits_{n \to \infty} \frac{x}{2^n}=0$, we have that $$\lim_{n \to \infty} 2^n \tan \frac{x}{2^n}=\lim_{n \to \infty}x \times \dfrac{ \tan \frac{x}{2^n}}{\frac{x}{2^n}} =x$$
Using the fact that $\lim\limits_{a \to 0}\frac{\tan a}{a}=1$. So the limit becomes $$\lim_{n \to \infty} 2^n \times {\tan \frac{x}{2^n}} \times \frac{1}{\tan x}= \frac{x}{\tan x}=x \cot x$$
The answer is $x \cot x$.
|
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|
Which of the following numbers is greater? Which of the following numbers is greater?
Without using a calculator and logarithm.
$$7^{55} ,5^{72}$$
My try
$$A=\frac{7^{55} }{5^{55}×5^{17}}=\frac{ 7^{55}}{5^{55}}×\frac{1}{5^{17}}= \left(\frac{7}{5}\right)^{55} \left(\frac{1}{5}\right)^{17}$$
What now?
|
Observe that $7^2=49<2\cdot 5^2$. In this case,
$$
7^{55}=7\cdot 7^{54}=7\cdot(7^2)^{27}<7\cdot (2\cdot 5^2)^{27}=7\cdot 2^{27}5^{54}.
$$
Observe that $2^3<10=2\cdot 5$. In this case,
$$
7\cdot 2^{27}5^{54}=7\cdot (2^3)^95^{54}<7\cdot(2\cdot 5)^95^{54}=7\cdot 2^9\cdot 5^{63}
$$
Using that $2^3<10=2\cdot 5$ again, we get
$$
7\cdot 2^9\cdot 5^{63}=7\cdot (2^3)^3\cdot 5^{63}<7\cdot (2\cdot 5)^3\cdot 5^{63}=7\cdot 2^3\cdot 5^{66}.
$$
Since $7\cdot 2^3=7\cdot 8=56<125=5^3$, we get
$$
7\cdot 2^3\cdot 5^{66}<5^{69}<5^{72}.
$$
|
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|
Find the limit : $\lim_{ x \to 1}\frac{\sqrt[n]{x^n-1}}{\sqrt[n]{nx}-\sqrt[n]{n}-\sqrt[n]{nx-n}}$ Find the limit: Without the use of the L'Hôspital's Rule
$$\lim_{ x \to 1}\frac{\sqrt[n]{x^n-1}}{\sqrt[n]{nx}-\sqrt[n]{n}-\sqrt[n]{nx-n}}$$
My try:
$u=x-1$
Now:
$$\lim_{ x \to 1}\frac{\sqrt[n]{(u+1)^n-1}}{\sqrt[n]{n(u+1)}-\sqrt[n]{n}-\sqrt[n]{n(u+1)-n}}$$
|
Let us assume $n > 1$. We must have $x \to 1^{+}$ to ensure that the roots are well defined for all $n > 1$. We can proceed as follows
\begin{align}
L &= \lim_{x \to 1^{+}}\frac{\sqrt[n]{x^{n} - 1}}{\sqrt[n]{nx} - \sqrt[n]{n} - \sqrt[n]{nx - n}}\notag\\
&= \frac{1}{\sqrt[n]{n}}\lim_{x \to 1^{+}}\frac{\sqrt[n]{x^{n} - 1}}{\sqrt[n]{x} - 1 - \sqrt[n]{x - 1}}
&= \frac{1}{\sqrt[n]{n}}\lim_{x \to 1^{+}}\dfrac{\sqrt[n]{\dfrac{x^{n} - 1}{x - 1}}}{\dfrac{\sqrt[n]{x} - 1}{\sqrt[n]{x - 1}} - 1}\notag\\
&= \lim_{x \to 1^{+}}\dfrac{1}{\dfrac{\sqrt[n]{x} - 1}{x - 1}\cdot (x - 1)^{1 - 1/n} - 1}\notag\\
&= \dfrac{1}{\dfrac{1}{n}\cdot 0 - 1}\notag\\
&= -1\notag
\end{align}
|
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|
Find the value of $\arctan(1/3)$ How can I calculate $\arctan\left({1\over 3}\right)$ in terms of $\pi$ ? I know that $\tan^2(\frac{\pi}{6})= {1\over3}$ but don't know if that helps in any way.
|
For the evaluation of $\tan^{-1}(x)$, you can also use Padé approximants such as
$$\frac{ x+\frac{4 x^3}{15}} { 1+\frac{3 x^2}{5}}\tag 1$$
$$\frac{x+\frac{11 x^3}{21} } {1+\frac{6 x^2}{7}+\frac{3 x^4}{35} } \tag 2$$
$$\frac{x+\frac{7 x^3}{9}+\frac{64 x^5}{945}} {1+\frac{10 x^2}{9}+\frac{5 x^4}{21} } \tag 3$$
$$\frac{x+\frac{34 x^3}{33}+\frac{x^5}{5} } { 1+\frac{15 x^2}{11}+\frac{5 x^4}{11}+\frac{5 x^6}{231}} \tag 4$$
Using the above formulae, you would get $\frac{139}{432}$, $\frac{250}{777}$, $\frac{20806}{64665}$, $\frac{19593}{60895}$ which are $\approx 0.3217592593$, $\approx 0.3217503218$, $\approx 0.3217505606$, $\approx 0.3217505542$ while the "exact" value would be $\approx 0.3217505544$.
|
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|
If the points $(0,0),(a,11),(b,37)$ are vertices of equilateral triangle what is the value of ab? If the points $(0,0),(a,11),(b,37)$ are vertices of equilateral triangle, find the value of $ab$
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I'd like to propose an exotic solution using complex numbers. Let the three vertices be represented on the Argand diagram by
$$z_0 = 0 + 0i = 0$$
$$z_1 = a + 11i$$
$$z_2 = b + 37i$$
Then we know that $z_1-z_0 = z_1$ and $z_2-z_0 = z_2$ are $\frac{\pi}{3}$ apart. That is
$$z_1 = z_2e^{\pm i\frac{\pi}{3}}$$
$$a + 11i = (b + 37i)\left(\frac{1}{2} \pm i \frac{\sqrt{3}}{2}\right)$$
$$a + 11i = \left(\frac{1}{2}b \pm\frac{37\sqrt{3}}{2}\right) + \left(\mp\frac{\sqrt{3}}{2}b + \frac{37}{2} \right)i$$
Do note that the $\pm$ and $\mp$ signs are alternate.
Next, comparing coefficients,
$$2a = b \pm 37\sqrt{3}~~,~~22 = \mp\sqrt{3}b + 37$$
$$2a = b \pm 37\sqrt{3}~~,~~\mp\sqrt{3}b = -15$$
$$2a = b \pm 37\sqrt{3}~~,~~b = \pm\frac{15}{\sqrt{3}} = \pm5\sqrt{3}$$
Hence,
$$\begin{align}2ab &= b^2 \pm37\sqrt{3}b\\
&=75 + 555\\
&=630\end{align}$$
$$ab = 315$$
|
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|
Solving for a matrix $$
\begin{bmatrix}1&2&3\\4&5&6\\7&8&9\end{bmatrix}X = \begin{bmatrix}3&2&1\\6&5&4\\9&8&7\end{bmatrix}.
$$
I know how to do this normally, but I'm confused with this case. It seems like an inverse does not exist for the matrices, and the second is the first matrix in reverse.
|
Recall that, when we multiply matrices $A$ and $B$, each column of $AB$ is a linear combination of the columns of $A$, with weights given by entries in the corresponding column of B.
In this case, you want the first column of $X$ to be something that says, "ignore the first two columns of $A$, and take 1 times the third. That column should be $\left[\begin{matrix} 0 \\ 0 \\ 1 \end{matrix}\right]$. Can you take it from there?
[EDIT]
This solution won't be unique, since, as we've noted, the columns of $A$ are linearly dependent, so there's more than one way to get column 3.
[EDIT]
To obtain a general solution, we could work with the augmented matrix $[A|AX]$, and see what happens:
$\begin{align}\left[\begin{array}{ccc|ccc}1 & 2 & 3 & 3 & 2 & 1 \\ 4 & 5 & 6 & 6 & 5 & 4 \\ 7 & 8 & 9 & 9 & 8 & 7\end{array}\right] &\sim \left[\begin{array}{ccc|ccc}1 & 2 & 3 & 3 & 2 & 1 \\ 4 & 5 & 6 & 6 & 5 & 4 \\ 0 & 0 & 0 & 0 & 0 & 0\end{array}\right] \\ &\sim \left[\begin{array}{ccc|ccc}1 & 2 & 3 & 3 & 2 & 1 \\ 0 & 1 & 2 & 2 & 1 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0\end{array}\right] \\ &\sim \left[\begin{array}{ccc|ccc}1 & 0 & -1 & -1 & 0 & 1 \\ 0 & 1 & 2 & 2 & 1 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0\end{array}\right]\end{align}$
Now we just have to interpret this result. For each column we get a free variable. Call one free variable $t_1$. Then the first column of $X$ is $\left[\begin{matrix}-1+t_1 \\ 2- 2t_1 \\ t_1 \end{matrix}\right]$. Similarly, the second column is $\left[\begin{matrix}t_2 \\ 1- 2t_2 \\ t_2 \end{matrix}\right]$.
Do you see how this is working?
|
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|
Show that $a+b+c=0$ implies that $32(a^4+b^4+c^4)$ is a perfect square. There are given integers $a, b, c$ satysfaying $a+b+c=0$. Show that $32(a^4+b^4+c^4)$ is a perfect square.
EDIT: I found solution by symmetric polynomials, which is posted below.
|
It suffices to show that $2(a^4+b^4+c^4)$ is a perfect square.
\begin{align}
& 2(a^4+b^4+c^4) \\
=& 2((b+c)^4+b^4+c^4) \\
=& 2(2b^4+4b^3c+6b^2c^2+4bc^3+2c^4) \\
=& 4(b^4+\bbox[2px, border:1px solid]{2b^3c}+\bbox[2px, border:1px dashed]{3b^2c^2}+2bc^3+c^4) \\
=& 4((b^4+\bbox[2px, border:1px solid]{b^3c}+\bbox[2px, border:1px dashed]{b^2c^2})+(\bbox[2px, border:1px solid]{b^3c}+\bbox[2px, border:1px dashed]{b^2c^2}+bc^3)+(\bbox[2px, border:1px dashed]{b^2c^2}+bc^3+c^4)) \\
=& 2^2 (b^2(b^2+bc+c^2)+bc(b^2+bc+c^2)+c^2(b^2+bc+c^2))\\
=& 2^2(b^2+bc+c^2)^2
\end{align}
|
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|
Find matrix $B$ such that $BA=4A$ Find the unique $B \in \mathbb{R}^{3 \times 3}$ such that for every $A \in \mathbb{R}^{3 \times 3}$ we have
i) $BA=4A$
ii) The 1, 2 and 3 rows of $BA$ are the 3,2 and 1 rows of A.
This problem is in the section where they define matrix multiplication. My only idea was to set up a giant system of equations, but well...it was too giant. Is there a smart way to solve it?
|
Taking them to be separate questions,
You should start by looking at what $4A$ is equal to as an arbitrary matrix. Then, seeing what matrix can a multiply by in order to get this this product? In the second case, the same reasoning follows. See below. Let $A$ be the matrix with a standard representation.
*
*$4A = \begin{pmatrix} 4a_{11} &4a_{12} &4a_{13} \\ 4a_{21} &4a_{22} &4a_{23} \\ 4a_{31} &4a_{32} &4a_{33} \end{pmatrix} = \begin{pmatrix} 4&0 &0 \\0&4 &0 \\0&0 &4 \\\end{pmatrix} \begin{pmatrix} a_{11} &a_{12} &a_{13} \\ a_{21} &a_{22} &a_{23} \\ a_{31} &a_{32} &a_{33} \end{pmatrix} = BA$ so that $B = \begin{pmatrix} 4&0 &0 \\0&4 &0 \\0&0 &4 \\\end{pmatrix}$.
*$BA = \begin{pmatrix} a_{31} &a_{32} &a_{33} \\ a_{21} &a_{22} &a_{23} \\ a_{11} &a_{12} &a_{13} \\ \end{pmatrix}= \begin{pmatrix} 0&0&1\\ 0&1&0\\ 1&0&0 \end{pmatrix} \begin{pmatrix} a_{11} &a_{12} &a_{13} \\ a_{21} &a_{22} &a_{23} \\ a_{31} &a_{32} &a_{33} \end{pmatrix}\ $ so that $B = \begin{pmatrix} 0&0&1\\ 0&1&0\\ 1&0&0 \end{pmatrix} $.
|
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|
Closed form for the series $\sum_{k=1}^\infty (-1)^k \ln \left( \tanh \frac{\pi k x}{2} \right)$
Is there a closed form for: $$f(x)=\sum_{k=1}^\infty (-1)^k \ln \left( \tanh \frac{\pi k x}{2} \right)=2\sum_{n=0}^\infty \frac{1}{2n+1}\frac{1}{e^{\pi (2n+1) x}+1}$$
This sum originated from a recent question, where we have:
$$f(1)= -\frac{1}{\pi}\int_0^1 \ln \left( \ln \frac{1}{x} \right) \frac{dx}{1+x^2}=\ln \frac{\Gamma (3/4)}{\pi^{1/4}}$$
If we differentiate w.r.t. $x$, we obtain:
$$f'(x)=\sum_{k=1}^\infty (-1)^k \frac{\pi k}{\sinh \pi k x}$$
There is again a closed form for $x=1$ (obtained numerically):
$$f'(1)=-\frac{1}{4}$$
So, is there a closed form or at least an integral definition for arbitrary $x>0$?
The series converges absolutely (numerically at least):
$$\sum_{k=1}^\infty \ln \left( \tanh \frac{\pi k x}{2} \right)< \infty$$
Thus, this series can also be expressed as a logarithm of an infinite product:
$$f(x)=\ln \prod_{k=1}^\infty \tanh (\pi k x) - \ln \prod_{k=1}^\infty \tanh \left( \pi (k-1/2) x \right)$$
$$e^{f(x)}= \prod_{k=1}^\infty \frac{\tanh (\pi k x)}{\tanh \left( \pi (k-1/2) x \right)}$$
This by the way leads to:
$$\prod_{k=1}^\infty \frac{\tanh (\pi k)}{\tanh \left( \pi (k-1/2) \right)}=\frac{\pi^{1/4}}{\Gamma(3/4)}$$
I feel like there is a way to use the infinite product form for $\sinh$ and $\cosh$:
$$\sinh (\pi x)=\pi x \prod_{n=1}^\infty \left(1+\frac{x^2}{n^2} \right)$$
$$\cosh (\pi x)=\prod_{n=1}^\infty \left(1+\frac{x^2}{(n-1/2)^2} \right)$$
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Let’s use $~\displaystyle\prod\limits_{k=1}^\infty (1+z^k)(1-z^{2k-1}) =1~$ . $\enspace$ (It's explained in a note below.)
For $~z:=q^2~$ and $~q:=e^{-\pi x}~$ with $~x>0~$ we get
$\displaystyle e^{f(x)} = \prod\limits_{k=1}^\infty\frac{\tanh(k\pi x)}{\tanh((k-\frac{1}{2})\pi x)} = \prod\limits_{k=1}^\infty\frac{ \frac{q^{-k}-q^k}{q^{-k}+q^k} }{ \frac{q^{\frac{1}{2}-k}-q^{k-\frac{1}{2} }}{q^{\frac{1}{2}-k}+q^{k-\frac{1}{2}}} } = \prod\limits_{k=1}^\infty\frac{(1-q^{2k})(1+q^{2k-1})}{(1+q^{2k})(1-q^{2k-1})} =$
$\displaystyle = \prod\limits_{k=1}^\infty (1-q^{2k})(1+q^{2k-1})^2 = \sum\limits_{k=-\infty}^{+\infty} q^{k^2} = \vartheta(0;ix)$
The “closed form” for $\,f\,$ is:
$$f(x) = \ln\vartheta(0;ix)$$
Please see e.g. Theta function .
Note:
$\displaystyle\prod\limits_{k=1}^\infty (1+z^k)(1-z^{2k-1}) =1$
$\Leftrightarrow\hspace{2cm}$ (logarithm)
$\displaystyle \sum\limits_{k=1}^\infty \sum\limits_{v=1}^\infty \frac{(-1)^{v-1}z^{kv}}{v} = \sum\limits_{k=1}^\infty \ln(1+z^k) = -\sum\limits_{k=1}^\infty \ln(1-z^{2k-1}) = \sum\limits_{k=1}^\infty \sum\limits_{v=1}^\infty \frac{z^{(2k-1)v}}{v}$
$\Leftrightarrow\hspace{2cm}$ (exchanging the sum symbols which is valid for $~|z|<1~$
$\hspace{2.7cm}$ and using $~\displaystyle\frac{x}{1-x}=\sum\limits_{k=1}^\infty x^k~$)
$\displaystyle\sum\limits_{v=1}^\infty \frac{(-1)^{v-1}}{v}\frac{z^v}{1-z^v} = \sum\limits_{v=1}^\infty \frac{1}{v}\frac{z^v}{1-z^v} - 2\sum\limits_{v=1}^\infty \frac{1}{2v}\frac{z^{2v}}{1-z^{2v}} = \sum\limits_{v=1}^\infty \frac{1}{v}\frac{z^v}{1-z^{2v}}$
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|
Solving: $(b+c)^2=2011+bc$
Solve $$(b+c)^2=2011+bc$$ for integers $b$ and $c$.
My tiny thoughts:
$(b+c)^2=2011+bc\implies b^2+c^2+bc-2011=0\implies b^2+bc+c^2-2011=0$
Solving in $b$ as Quadratic.$$\implies b=\frac{-c\pm \sqrt{8044-3c^2}} {2}.$$
So $8044-3c^2=k^2$, as $b$ and $c$ are integers. We also have inequalities: $8044>3c^2,8044>3b^2\\ \ \ \ \ 51>c\ \ \ \ , \ \ \ \ 51>b$
How to proceed further. Help.
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We can assume that $b\geq0$ and by your starting reasoning we ca get $b\in\{10,39,49\}$,
which gives all 12 solutions.
|
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|
Quadratic equation find all the real values of $x$ Find all real values of $x$ such that
$\sqrt{x - \frac{1}{x}} + \sqrt{1 - \frac{1}{x}} = x$
I tried sq both sides by taking 1 in RHS but it didn't worked out well...
|
HINT: after squaring one times we get
$$2\sqrt{x-\frac{1}{x}}\sqrt{1-\frac{1}{x}}=x^2-x+\frac{2}{x}-1$$
can you finish this?
squaring this one more times we get
$$4\left(x-\frac{1}{x}\right)\left(1-\frac{1}{x}\right)=\left(x^2-x+\frac{2}{x}-1\right)^2$$
expanding the left Hand side we obtain $$4\,x-4-4\,{x}^{-1}+4\,{x}^{-2}$$ and the right Hand side is given by $${x}^{4}-2\,{x}^{3}+6\,x-{x}^{2}-3+4\,{x}^{-2}-4\,{x}^{-1}$$ have you got this?
Bringung all together we obtain this equation
$$0=x^4-2x^3-x^2+2x+1$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Inequality for positive real numbers less than $1$: $8(abcd+1)>(a+1)(b+1)(c+1)(d+1)$ If $a,b,c,d$ are positive real numbers, each less than 1, prove that the following inequality holds:
$$8(abcd+1)>(a+1)(b+1)(c+1)(d+1).$$
I tried using $\text{AM} > \text{GM}$, but I could not prove it.
|
Given that $0<a<1$, $0<b<1$, $0<c<1$, $0<d<1$
$\therefore$ $-1<a-1<0$ and $-1<b-1<0$
Now
\begin{align*}
&(a-1)(b-1)>0 \\
\Rightarrow& ab - (a+b) +1 >0 \\
\Rightarrow& (ab+1) > a+b \\
\Rightarrow& (ab+1)+(ab+1) > (ab+1)+(a+b) \\
\Rightarrow& 2(ab+1) > (ab+1)+(a+b) \\
\Rightarrow& 2(ab+1) > (a+1)(b+1) \tag{I}
\end{align*}
Similarly
$$2(cd+1) > (c+1)(d+1)$$
$\therefore$ $0<a<1$ and $0<b<1$ $\Rightarrow$ $0<ab<1$.
Similarly $0<cd<1$.
Now,
\begin{align*}
4(ab+1)(cd+1) &> (a+1)(b+1)(c+1)(d+1) \\
\Rightarrow (a+1)(b+1)(c+1)(d+1) &< 4(ab+1)(cd+1) \tag{II}
\end{align*}
As $(I)$
\begin{align*}
2(abcd+1) &> (ab+1)(cd+1) \\
\Rightarrow (ab+1)(cd+1) &< 2(abcd+1) \tag{III}
\end{align*}
Putting these value in $(II)$,we get
\begin{align*}
(a+1)(b+1)(c+1)(d+1) &< 4\left\{2(abcd+1)\right\} \\
(a+1)(b+1)(c+1)(d+1) &< 8(abcd+1)
\end{align*}
answered by SAROJ GHOSH .
|
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|
Let $a,b,c\in \Bbb R^+$ such that $(1+a+b+c)(1+\frac{1}{a}+\frac{1}{b}+\frac{1}{c})=16$. Find $(a+b+c)$ Let $a,b,c\in \Bbb R^+$ such that $(1+a+b+c)(1+\frac{1}{a}+\frac{1}{b}+\frac{1}{c})=16$.
Find $(a+b+c)$.
I computed the whole product ;If $(a+b+c)=x\implies (1+x)(1+\frac{bc+ca+ab}{abc})=16$. Unable to view how to proceed further.
Please help.
|
Since scaling by a constant
leaves the product unchanged,
we can assume that
$a=1$.
This becomes
$(1+b+c)(1+1/b+1/c)
=16$.
Looking for a special solution,
assume $b = c$.
This becomes
$16
=(1+2b)(1+2/b)
=1+2b+2/b+4
$
or
$11
=2b+2/b
$
or
$2b^2-11b+2 = 0$.
Solving this,
$b
=\dfrac{11\pm \sqrt{11^2-16}}{4}
=\dfrac{11\pm \sqrt{105}}{4}
$.
So these are two solutions.
Going back to
the general case,
$16
=(1+b+c)(1+1/b+1/c)
=1+b+c+1/b+1/c+(b+c)(1/b+1/c)
=1+b+c+1/b+1/c+2+b/c+c/b
$
or
$13
=b+c+1/b+1/c+b/c+c/b
$.
If
$c = rb$,
this becomes
$13
=b+rb+1/b+1/(rb)+r+1/r
$
or
$13r
=br+r^2b+r/b+1/(b)+r^2+1
$
or
$(b+1)r^2+(b-13+1/b)r+1+1/b
=0
$.
The discriminant of this is
$d^2
=(b-13+1/b)^2-4(b+1)(1+1/b)
=(b^4 - 30 b^3 + 163 b^2 - 30 b + 1)/b^2
$
(according to Wolfy).
This has real roots
$b
= \dfrac{7 \pm 3 \sqrt{5}}{2},
2/(23 + 5 \sqrt(21)),
23/2 + (5 \sqrt(21))/2
$
with approximate values
$0.14590,
6.8541,
0.043561,
22.956
$.
The plot shows that
for $b$ from
0.14590 to 6.8541
and $b > 22.956$
this is positive,
so there are real roots
for these $b$.
By taking the positive square root,
we can get a positive $r$.
I'll stop here.
|
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|
Multiple choice question of indefinite integral, $\int \frac{x + 9}{x^3 + 9x} dx$. If $\int \frac{x + 9}{x^3 + 9x} dx = k\arctan(mx) + n\ln (x) + p \ln (x^2 + 9) + c$, then $(m+n)/(k+p) = $
(A) 6
(B) -8
(C) -3
(D) 4
I tried solving it by differentiating the R.H.S. but couldn't arrive at the answer.
|
\begin{align}
\int \frac{x + 9}{x^3 + 9x} dx &= \int \frac{x}{(x^2 + 9)x} dx + \int \frac{9}{(x^2 + 9)x} dx
\\
\\
&= \int \frac{dx}{x^2 + 9} + \int \left(\frac{A}{x} + \frac{Bx+C}{x^2 + 9}\right) dx
\\
\\
&= \int \frac{dx}{x^2 + 9} + \int \left(\frac{1}{x} + \frac{-x}{x^2 + 9}\right) dx
\\
\\
&= \frac{1}{9}\int \frac{dx}{\left(x/3\right)^2 + 1} + \int \frac{dx}{x} - \int\frac{x}{x^2 + 9}dx
\\
\\
&= \frac{1}{3}\arctan(x/3) + \ln x - \frac{1}{2}\ln(x^2+9) + c
\end{align}
$$\int \frac{x + 9}{x^3 + 9x} dx = k\arctan(mx) + n\ln (x) + p \ln (x^2 + 9) + c$$
$$\frac{m+n}{k+p} = \frac{1/3+1}{1/3-1/2}=\frac{4/3}{-1/6}=-8$$
|
{
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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|
Show that $\sqrt{2-\sqrt{3}}=\frac{1}{2}(\sqrt{6}-\sqrt{2})$
Prove the following:
$\sqrt{2-\sqrt{3}}=\frac{1}{2}(\sqrt{6}-\sqrt{2})$
How do you go about proving this?
|
\begin{align}\sqrt{2-\sqrt{3}} &=\frac{\sqrt{2}\sqrt{2-\sqrt{3}}}{\sqrt{2}}\\
&=\frac {\sqrt { 4-2\sqrt{3}}}{\sqrt{2}} \\
&=\frac {\sqrt{1-2\sqrt{3} +{ \left( \sqrt{3}\right)}^{2} } }{ \sqrt {2} } \\
&=\frac{\sqrt { { \left( 1-\sqrt{3}\right)}^{2}}}{ \sqrt{2}}\\
&=\frac{\sqrt {3} -1 }{ \sqrt {2} }\\
& = \frac{\sqrt{6}-\sqrt{2}}{2} \end{align}
|
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"timestamp": "2023-03-29T00:00:00",
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|
If $\sin A+\sin^2 A=1$ and $a\cos^{12} A+b\cos^{8} A+c\cos^{6} A-1=0$ If $\sin A+\sin^2 A=1$ and $a\cos^{12} A+b\cos^{8} A+c\cos^{6} A-1=0$.
Find the value of $2b + \dfrac {c}{a}$.
My Attempt:
$$\sin A+\sin^2 A=1$$
$$\sin A + 1 - \cos^2 A=1$$
$$\sin A=\cos^2 A$$
Now,
$$a(\cos^2 A)^{6}+b(\cos^2 A)^{4}+c(\cos^2 A)^{3}=1$$
$$a\sin^6 A+ b\sin^4 A+c\sin^3 A=1$$
How do I proceed further?
|
$$\sin A= \cos ^2A→\sin^2A=\cos^4A→1-\cos^2A=\cos^4A\\
(\cos^2A)^2+(\cos^2A)-1=0→\cos^2A=\frac{-1+ \sqrt{5}}{2}$$
So,
$$a\left(\frac{-1+ \sqrt{5}}{2}\right)^6+b\left(\frac{-1+ \sqrt{5}}{2}\right)^4+c\left(\frac{-1+ \sqrt{5}}{2}\right)^3=1\\
a(9-4\sqrt{5})+b\left(\frac{7-3\sqrt{5}}{2}\right)+c(\sqrt{5}-2)=1\\
(9a+7b/2-2c)+(-4a-3b/2+c)\sqrt{5}=1$$
Maybe the question want:
$$9a+\frac{7b}{2}-2c=1→18a+7b-4c=2\\
-4a-\frac{3b}{2}+c=0→8a+3b=2c$$
But on that case we get:
$$b=2-2a\\
c=a+3$$
And then
$$2b+\frac{c}{a}=5-4a+\frac{3}{a}$$
and the result depends on $a$. Otherwise I don't have another guess.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Find radius of convergence and interval of converges of the series; $\displaystyle\sum_{n=0}^{\infty}\frac{1}{(-3)^{2+n}(n^2+1)}(4x-12)^n$ $\displaystyle\sum_{n=0}^{\infty}\frac{1}{(-3)^{2+n}(n^2+1)}(4x-12)^n$
My try:$$\displaystyle\lim_{n\to\infty}\left(\frac{1}{(-3)^{2+n}(n^2+1)}(4x-12)^n\right)^{\frac{1}{n}}\leq1$$ $$\displaystyle\lim_{n\to\infty}\frac{1}{(-3)^{2/n+1}(n^2+1)^{1/n}}(4x-12)\leq1$$ $$\frac{1}{(-3)\displaystyle\lim_{n\to\infty}(n^2+1)^{1/n}}(4x-12)\leq1$$
Here I found $|4x-12|\leq3$, then what is radius of convergence?
Thank you.
|
I personally prefer the ratio test. If a series converges, then:
$$\lim_{k\to \infty}\left|\frac{a_{k+1}}{a_k}\right|<1$$
Substituting gives:
$$\lim_{k\to \infty}\left|\frac{\left(\frac{(4x-12)^{k+1}}{(-3)^{3+k}((k+1)^2+1)}\right)}{\left(\frac{(4x-12)^{k}}{(-3)^{2+k}(k^2+1)}\right)}\right|<1$$
$$\lim_{k\to \infty}\left|\frac{(4x-12)^{k+1}}{(4x-12)^k}\cdot \frac{(-3)^{2+k}}{(-3)^{3+k}}\cdot \frac{k^2+1}{(k+1)^2+1}\right|<1$$
$$\lim_{k\to \infty}\left|(4x-12)\cdot \frac{-1}{3} \cdot \frac{k^2+1}{(k+1)^2+1}\right|<1$$
Taking limits gives:
$$\left|(4x-12)\cdot \frac{-1}{3}\cdot 1\right|<1$$
$$|4x-12|<3$$
To find the radius of convergence, put it in the form $|x-a|<R$ where $R$ is the radius of convergence. Now, continuing from what you've done:
$$4|x-3|<3$$
$$|x-3|<\frac{3}{4}$$
This gives the radius of convergence $R=\frac{3}{4}$.
Therefore, we obtain:
$$\frac{9}{4}<x<\frac{15}{4} \tag{1}$$
To find the interval of convergence, test the lower and upper bounds of the inequality by substituting them into your sum (i.e. $x=\frac{9}{4}$ and $x=\frac{15}{4}$) and check if they converge. If one of these bounds converge, put the $\leq$ sign next to that bound. Otherwise, leave it as $<$.
|
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|
Can this symmetric matrix be an orthogonal matrix? So the question is to prove whether this matrix M is orthogonal?
$$
M = \begin{bmatrix}
a & k & k \\
k & a & k \\
k & k & a \\
\end{bmatrix}
$$
My attempt to find the inverse of M :
$$det(M)=(a+2k)(a-k)(a-k)$$
The inverse of $M$ is :
$$
\begin{matrix}
(a+k)/((a-k)(a+2k)) & (-k)/((a-k)(a+2k)) & (-k)/((a-k)(a+2k))\\
(-k)/((a-k)(a+2k)) & (a+k)/((a-k)(a+2k)) & (-k)/((a-k)(a+2k))\\
(-k)/((a-k)(a+2k)) & (-k)/((a-k)(a+2k)) & (a+k)/((a-k)(a+2k))\\
\end{matrix}
$$
$M$ is orthogonal if $M^T = M^{-1}$
So I end up with a system of 2 equations to solve :
$a=\dfrac{a+k}{(a-k)(a+2k)}$ and $k=\dfrac{-k}{(a-k)(a+2k)}$
and now I am stucked.
I would really appreciate feedbacks about whether my steps are all correct and if yes, what to do next ?
|
Since $M$ is symmetric, $M=M^T$, so to check orthogonality, we compute $MM^T=M^2$ and get
$$
\begin{bmatrix}
a^2+2k^2&2ak+k^2&2ak+k^2\\
2ak+k^2&a^2+2k^2&2ak+k^2\\
2ak+k^2&2ak+k^2&a^2+2k^2
\end{bmatrix}
$$
Therefore, you need $2ak+k^2=0$ and $a^2+2k^2=1$. Factoring the first equation, we have $k(2a+k)=0$. Therefore, either $k=0$ or $k=-2a$.
*
*When $k=0$, the second equation simplifies to $a^2=1$, so $a=\pm 1$ gives orthogonality.
*When $k=-2a$, the second equation simplifies to $9a^2=1$, so $a=\pm\frac{1}{3}$ with $k=\mp \frac{2}{3}$ gives orthogonality.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Solving $\frac{x^2 + k^2 - 2xk}{x^2 + c^2 - 2xc} = e^{-2yg}\frac{k^2}{c^2}$ $$\frac{x^2 + k^2 - 2xk}{x^2 + c^2 - 2xc} = e^{-2yg}\frac{k^2}{c^2}$$
I am attempting to solve for $x$. I've been at it for a while and have reached to above equation, but now it seems to get quite messy. Any help?
Important:
$k = \frac{a+g}{2E}$,
$c = \frac{a-g}{2E}$,
$g = \sqrt{a^2 + 4E}$.
What has yet to be defined are just real constants (i.e, $E$, $a$, $y$)
|
It helps to note that everything is pretty much a constant, hence we may reduce as follows:
$$\frac{x^2-2kx+k^2}{x^2-2cx+c^2}=P$$
where $P=e^{-2yg}\frac{k^2}{c^2}$. Then notice some perfect squares:
$$x^2-2kx+k^2=(x-k)^2\\x^2-2cx+c^2=(x-c)^2$$
Thus, the problem further reduces:
$$\frac{x-k}{x-c}=\pm\sqrt P$$
Multiply both sides by $x-c$ to get
$$x-k=\pm\sqrt P(x-c)$$
$$\mp c\sqrt P-k=(\pm\sqrt P-1)x$$
$$x=\frac{\mp c\sqrt P-k}{\pm\sqrt P-1}$$
(watch the signs, as $\pm$ is always the opposite sign of $\mp$)
|
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|
Is $f_n(z) = \dfrac {2nz - 1} {z + n^2}$ uniformly convergent on the disc $|z| <1$? The pointwise limit of $f_n(z) = \dfrac {2nz - 1} {z + n^2}$ is $0$.
By the triangle inequality,
$\Bigg|\dfrac {2nz - 1} {z + n^2}\Bigg| \le \Bigg|\dfrac {2nz} {z + n^2}\Bigg| + \Bigg|\dfrac {1} {z + n^2}\Bigg|$
On the disc $|z| <1$, $\Bigg|\dfrac {1} {z + n^2}\Bigg|\le \dfrac {1} {n^2} \le \dfrac {1} {n}$. However I cannot seem to find bounds for $\Bigg|\dfrac {2nz} {z + n^2}\Bigg|$.
I'm beginning to think that the sequence is only pointwise convergent in the specified disc.
|
No, $|1/(z+n^2)| \le 1/n^2$ fails. Try $z=-1/2$ for example. Instead note that for $n>1,$
$$\left |\frac{2nz-1}{z+n^2}\right | \le \frac{|2nz|+1}{n^2 - |z|} \le \frac{2n+1}{n^2 - 1}.$$
|
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|
Determine: $S = \frac{2^2}{2}{n \choose 1} + \frac{2^3}{3}{n \choose 2} + \frac{2^4}{4}{n \choose 3} + \cdots + \frac{2^{n+1}}{n+1}{n \choose n}$ We are given two hints: consider $(n+1)S$; and use the Binomial Theorem. But we are not to use calculus.
My consideration of $(n+1)S$ goes like this:
\begin{align*}
\sum\limits_{k=1}^{n}\frac{2^{k+1}}{k+1}{n \choose k}
&= \frac{1}{n+1}\sum\limits_{k=1}^{n}(n+1)\frac{2^{k+1}}{k+1}{n \choose k} \\
&= 2\frac{1}{n+1}\sum\limits_{k=1}^{n}2^k{n+1 \choose k+1} \\
&= 2\frac{1}{n+1}\sum\limits_{k=1}^{n}(1+1)^k{n+1 \choose k+1} \\
\end{align*}
Now I think I'm in a position to use the Binomial Theorem, giving
\begin{equation*}
2\frac{1}{n+1}\sum\limits_{k=1}^{n}\sum\limits_{i=0}^{k}{k \choose i}{n+1 \choose k+1}
\end{equation*}
I don't know if I am on the right track, but I do know that I'm stuck. Can anyone offer any advice on how to proceed?
|
Note that
$$\sum_{k=1}^n2^{k+1}{n+1\choose k+1}=\sum_{m=2}^{n+1}2^m{n+1\choose m}$$
So
$$\sum_{k=1}^n\frac{2^{k+1}}{k+1}{n\choose k}=\frac{1}{n+1}\sum_{m=2}^{n+1}2^m{n+1\choose m}=\frac{1}{n+1}\left((1+2)^{n+1}-1-2(n+1)\right)=\frac{3^{n+1}-3-2n}{n+1}$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
why is $2^{\log^{2}\frac{1}{2}} = 1/2$? I have no understanding of $\log^2n$, wouldn't $2^{\log^{2}\frac{1}{2}} = 1/4$ since $2^{\log\frac{1}{2}} = 1/2$?
$\log$ here has the base of $2$.
|
$\textbf{Remark}$: Note that we have that $(a^b)^2 = a^{2b}$ and $a^{(b^2)} = a^{b^2}$, but in general we have that
$$a^{b^2} \neq a^{2b}$$ (unless $b^2 = 2b$, which is only true for $b = 2$ and $b = 0$).
(Consider for example $(2^3)^2 = 2^(3 \cdot 2) = 2^6 = 64$, whereas $2^{(3^2)} = 2^9 = 512$).
In your case, we have $a^{b^2}$ and with this in mind, we find
$$\log(\frac{1}{2}) = \log(1) - \log(2) = 0 -1 = -1,$$
since we work in base $2$ and therefore
$$2 ^{\log^2(\frac{1}{2})} = 2^{(-1)^2} = 2^1 = 2.$$
|
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|
How to show that $\sqrt{3 - \sqrt{2}} \notin \mathbb{Q}(\sqrt{3 + \sqrt{2}})$ Is there a simple way to show that $\sqrt{3 - \sqrt{2}} \notin \mathbb{Q}(\sqrt{3 + \sqrt{2}})$ (assuming this is true), using Galois theory? I have tried to show this by computational means but the calculations are horrible.
|
Let $\,K\,$ denote $\,\mathbb Q(\sqrt{3+\sqrt2})\,$ for convenience, then observe that
$$x=\sqrt{3+\sqrt2}$$
$$\Rightarrow\quad x^2-3=\sqrt2\qquad\ \ \ $$
$$\Rightarrow\quad x^4-6x^2+7=0\quad$$
$$\Rightarrow\quad[\,K:\mathbb Q\,]=4\qquad\ \ \ $$
First, $\,(\sqrt{3+\sqrt2})^2-3=\sqrt2\in K$
Now if $\,\sqrt{3-\sqrt2}\in K,$ then $\,\sqrt{3+\sqrt2}\cdot\sqrt{3-\sqrt2}=\sqrt7\in K$
Thus $\,K=\mathbb Q(\sqrt2,\sqrt7)$,$\,$ and this means that
$$\qquad\qquad\exists\ a,b,c,d\in\mathbb Q\ :\ a+b\sqrt2+c\sqrt7+d\sqrt{14}=\sqrt{3+\sqrt2}\qquad(\#)$$
We must have $\,c=d=0\quad(\%)$
Hence $\,\sqrt{3+\sqrt2}\in\mathbb Q(\sqrt2)\,$, which is impossible because $$[\mathbb Q(\sqrt2):\mathbb Q]=2<\mathbb [\mathbb Q(\sqrt{3+\sqrt2}):\mathbb Q]$$
As a result, $\,\sqrt{3-\sqrt2}\notin K$
$\left.\right.$
$\textbf{Proof}$ of $(\%):$
Square both sides of $(\#)$, we get
$$(a^2+2b^2+7c^2+14d^2-3)+2(ab+7cd-1)\sqrt2+2(ac+2bd)\sqrt7+2(ad+bc)\sqrt{14}=0$$
Now we must have
\begin{align*}
W&=a^2+2b^2+7c^2+14d^2-3=0\\
X&=ab+7cd-1=0\\
Y&=ac+2bd=0\\
Z&=ad+bc=0
\end{align*}
First, if $\,c=0,\,d\neq0$, we have $\,a=b=0\,$ by $Y$ and $Z$. Then $X$ will become $\,-1=0$, which is obviously a contradiction.
Next, if $\,c\neq0,\,d=0$, the same argument as above will lead to a contradiction.
Hence suppose $\,c,d\neq0$, then we have the following two cases:
(1) $b\neq0:$
$$c=-\frac{ad}b\ \ (\text{by }Z)\ \ \Rightarrow\ \ \frac db(a^2-2b^2)=0\ \ (\text{by }Y)\ \ \Rightarrow\ \ a=\pm\sqrt2b\notin\mathbb Q\ \ (\text{contradiction})$$
(2) $b=0:$
$$a=0\ \ (\text{by }Z)\ \ \Rightarrow\ \ c=\frac1{7d}\ \ (\text{by }X)\ \ \Rightarrow\ \ 98d^4-21d^2=-1\ \ (\text{by }W)\ \ \Rightarrow\ \ d\notin\mathbb Q\ \ (\text{contradiction})$$
As a result, we must have $\,c=d=0$
$\left.\right.$
(This answer has been improved by the suggestions from Kenny Wong)
|
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|
Finding $a,b,c$ using the equation provided Provided with the equation
$$\sin^2 (n+1)x - \sin^2 (n-1)x = \sin2nx \sin2x$$
Find the values of $a,b,c$ if
$$ \sin^2 3x - \sin^2 x = 8\cos^2 x (a\cos^4 x + b\cos^2 x + c)$$
What I have tried
$(1 - \cos^2 3x)- (1- \cos^2 x)=8\cos^2 x (a\cos^4 x + b\cos^2 x + c)$
Expanding out the LHS
$(1 - (\cos 2x \cos x - \sin 2x \sin x )^2) + \cos^2 x -1 $
$=( \cos x( 2\cos^2 x -1)-2\sin^2 x \cos x)^2$
$=\cos^2 x((2\cos^2x -1) -2(1-\cos^2 x))^2 $
$=\cos^2 x(4\cos^2 x -3)^2 $
$=16\cos^6 x -24\cos^4 x + 9\cos^2 x$
By comparing,
$a=2$
$b=-3$
$c=\frac{9}{8}$
Is there a way to do the question using the equation they provided? I can't seem to see it.Any tips would be useful.Thank you!
|
Friend, your answer is wrong,
Correct answer is $a=-2$, $b=3$ and $c=-1$
For a quick verification, put $x=0$ ;
LHS = $0$
thus RHS must be $0$, i.e. $a+b+c=0$. Which isn't true in your case.
Proper proof using given equation :
$\sin^2(n+1)x−\sin^2(n−1)x=\sin2nx\sin2x$
Put $n=2$
$\sin^23x−\sin^2x= \sin4x\sin2x$
$= 2\sin^22x\cos2x$
$= 2(1-(2\cos^2x-1)^2)(2\cos^2x-1)$
$= 2(4\cos^2x-\cos^4x)(2\cos^2x-1)$
$= 8\cos^2x(1-\cos^2x)(2cos^2x-1)$
$= 8\cos^2x(-2\cos^4x+3\cos^2x-1)$
Thus ;
$a=-2$ ,
$b=3$ , and
$c=-1$
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/2179300",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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|
Why is there another transformation matrix for the bases of the image and of the preimage of this mapping? There is the following Matrix:
\begin{pmatrix}1&1&1\\ a&b&c\\ a^2&b^2&c^2\end{pmatrix}
At a point it is needed to calculate the determinant of the matrix. In the official solution it is written:
$det\begin{pmatrix}1&1&1\\ a&b&c\\ a^2&b^2&c^2\end{pmatrix} = (c-b)(c-a)(b-a)$
And I don't see how they get this. If I calculate the determinant I am always getting this:
$det\begin{pmatrix}1&1&1\\ a&b&c\\ a^2&b^2&c^2\end{pmatrix} = (bc^2-b^2c)-(ac^2-a^2c)+(ab^2-a^2b)=c(b(c-b)-a(c-a)+ab(b-a)).$
But after that point I don't know how to proceed and get the form above. Can you help me?
|
Subtracting the first colum from the second and third columns we get
$$
\det\begin{pmatrix}1&1&1\\ a& b& c\\a^2& b^2& c^2\end{pmatrix}=\det\begin{pmatrix}1&0&0\\a&b-a&c-a\\ a^2&b^2-a^2&c^2-a^2\end{pmatrix}
$$
It follows that
\begin{eqnarray}
\det\begin{pmatrix}1&1&1\\ a& b& c\\a^2& b^2& c^2\end{pmatrix}&=&\det\begin{pmatrix}1&0&0\\a&b-a&c-a\\ a^2&b^2-a^2&c^2-a^2\end{pmatrix}=\det\begin{pmatrix}b-a&c-a\\ b^2-a^2&c^2-a^2\end{pmatrix}\\
&=&\det\begin{pmatrix}b-a&c-a\\ (b-a)(b+a)&(c-a)(c+a)\end{pmatrix}\\
&=&(b-a)(c-a)\det\begin{pmatrix}1&1\\ b+a&c+a\end{pmatrix}=(b-a)(c-a)(c+a-b-a)\\
&=&(b-a)(c-a)(c-b)
\end{eqnarray}
|
{
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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|
Find minimum of $y=\sqrt{-x^2+4x+12}-\sqrt{-x^2+2x+3}$
Find minimum of $y=\sqrt{-x^2+4x+12}-\sqrt{-x^2+2x+3}$
My work so far:
1) $$y =\frac{2x+9}{\sqrt{-x^2+4x+12}+\sqrt{-x^2+2x+3}} $$
2) I used a derivative and found the answer ($y=\sqrt3$ at $x=0$). Is there any other way?
|
Rewriting the function that way only complicates things.
First let's determine the domain:
\begin{cases}
-x^2+4x+12\ge0\\
-x^2+2x+3\ge0
\end{cases}
reduces to $-1\le x\le3$.
The derivative is
$$
y'=
\frac{-x+2}{\sqrt{-x^2+4x+12}}-
\frac{-x+1}{\sqrt{-x^2+2x+3}}
$$
(which is undefined at $-1$ and $3$) and we want to see where it vanishes, that is,
$$
(2-x)\sqrt{-x^2+2x+3}=(1-x)\sqrt{-x^2+4x+12}
$$
We need either $-1<x<1$ or $2<x<3$ so that both terms are either positive or negative. Now we can square safely:
$$
(2-x)^2(-x^2+2x+3)=(1-x)^2(-x^2+4x+12)
$$
becomes $12x^2-16x=0$, so the only critical point is $0$ (because $4/3$ doesn't satisfy the above limitations).
We also have
$$
f(-1)=\sqrt{7},\qquad
f(0)=\sqrt{3},\qquad
f(3)=\sqrt{15}
$$
Since maxima and minima are at critical points or at the extremes of the domain, we can conclude.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2180644",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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|
A simple algebra question involving a sum of a series I'm sorry if the question is silly, but I wondered if the next equation is right:
$$\sum\limits_{k = 1}^n {2^{n-k}}k = \sum\limits_{k = 1}^n k \sum\limits_{k = 1}^n {2^k}$$
And if not, is there a way to represent it without the sum $\sum\limits_{k = 1}^n {2^{n-k}}k$ ?
|
At first note that left-hand side and right-hand side of the equation are different.
The LHS is
\begin{align*}
\sum_{k=1}^nk\sum_{k=1}^n2^k&=\left(\sum_{k=1}^nk\right)\left(\sum_{k=1}^n2^k\right)\\
&=(1+2+3+\cdots+n)(2^1+2^2+2^3+\cdots+2^n)
\end{align*}
whereas the RHS is
\begin{align*}
\sum_{k=1}^n 2^{n-k}k=1\cdot 2^{n-1}+2\cdot 2^{n-2}+3\cdot 2^{n-3}+\cdots+n\cdot 2^0
\end{align*}
which are quite different, already when $n=1$.
We can derive a closed formula as follows:
\begin{align*}
\sum_{k=1}^nk2^{n-k}&=2^n\left(\sum_{k=1}^nk\frac{1}{2^k}\right)=2^n\left.\left(\sum_{k=1}^nkx^k\right)\right|_{x=\frac{1}{2}}\\
&=2^n\left.\left(x\sum_{k=1}^nkx^{k-1}\right)\right|_{x=\frac{1}{2}}\tag{1}\\
&=2^{n-1}\left.\left(\frac{d}{dx}\sum_{k=1}^nx^k\right)\right|_{x=\frac{1}{2}}\tag{2}\\
&=2^{n-1}\left.\frac{d}{dx}\left(\frac{x-x^{n+1}}{1-x}\right)\right|_{x=\frac{1}{2}}\tag{3}\\
&=2^{n-1}\left.\left(\frac{nx^{n+1}-(n+1)x^n+1}{(1-x)^2}\right)\right|_{x=\frac{1}{2}}\tag{4}\\
&=2^{n+1}-n-2\tag{5}
\end{align*}
Comment:
*
*In (1) we write the sum as series in $x$ evaluated at $x=\frac{1}{2}$ and factor out $x$ to prepare it for differentiation.
*In (2) we write the sum as derivative of a finite geometric series in $x$ and evaluated the single factor $x$ left from the sum already at $x=\frac{1}{2}$.
*In (3) we apply the formula for the finite geometric series.
*In (4) we differentiate the expression.
*In (5) we finally evaluate the expression at $x=\frac{1}{2}$ in order to get a closed formula.
Note: If this derivation is at the time too complicated, we could also claim the final result (5) and prove it using mathematical induction.
|
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|
Evaluate the sum of series $\sum_{k=1}^{+\infty}(-1)^{k-1}\frac{1}{k(4k^2-1)}$ Evaluate the sum of series
$$\sum_{k=1}^{+\infty}(-1)^{k-1}\frac{1}{k(4k^2-1)}$$
I have tried two methods:
1) using power series
2) using partial sums
but I can't find the sum.
1) Using power series:
$$\sum_{k=1}^{+\infty}(-1)^{k-1}\frac{1}{k(4k^2-1)}x^{k(4k^2-1)}$$
$$f(x)=\sum_{k=1}^{+\infty}(-1)^{k-1}\frac{1}{k(4k^2-1)}x^{k(4k^2-1)}$$
After derivation:
$$f'(x)=\sum_{k=1}^{+\infty}(-1)^{k-1}x^{4k^3-k-1}$$
The problem here is that:
$$\sum_{k=1}^{+\infty}(-1)^{k-1}x^{4k^3-k-1}=x^2-x^{29}+x^{104}-...$$
Is it possible to find the closed form for the last series?
2) Using partial sums:
$$S_n=\sum_{k=0}^{n}(-1)^{k}\frac{1}{(k+1)(4(k+1)^2-1)}$$
Now, using the formula:
$$S_n+a_{n+1}=a_0+\sum_{k=1}^{n+1}(-1)^{k}\frac{1}{(k+1)(4(k+1)^2-1)}\Rightarrow$$
$$S_n+(-1)^{n+1}\frac{1}{(k+2)(4(k+2)^2-1)}=\frac{1}{3}+\sum_{k=1}^{n+1}(-1)^{k}\frac{1}{(k+1)(4(k+1)^2-1)}$$
$$S_n=\frac{1}{3}-\frac{1}{30}+...+(-1)^{n}\frac{1}{(n+1)(4(n+1)^2-1)}$$
$$\sum_{k=1}^{n+1}(-1)^{k}\frac{1}{(k+1)(4(k+1)^2-1)}=T_n=-\frac{1}{30}+...+(-1)^{n+1}\frac{1}{(n+2)(4(n+2)^2-1)}$$
$$T_n=S_n-\frac{1}{3}+(-1)^{n+1}\frac{1}{(n+2)(4(n+2)^2-1)}$$
Going back to the formula $$S_n+a_{n+1}=a_0+\sum_{k=1}^{n+1}(-1)^{k}\frac{1}{(k+1)(4(k+1)^2-1)}$$
we have that $S_n$ cancels, so we can't determine partial sums using this method?
$$S_n+(-1)^{n+1}\frac{1}{(k+2)(4(k+2)^2-1)}=\frac{1}{3}+T_n$$
$$S_n+(-1)^{n+1}\frac{1}{(k+2)(4(k+2)^2-1)}=\frac{1}{3}+S_n-\frac{1}{3}+(-1)^{n+1}\frac{1}{(n+2)(4(n+2)^2-1)}$$
Question: How to find the sum of this series?
|
The way to proceed is to exploit the telescoping series as illustrated by @MarkusScheuer. But, ...
If one really wants to use the method proposed in the OP, then writing
$$\frac{1}{k(4k^2-1)}=\int_0^1 x^{k(4k^2-1)-1}\,dx$$
won't lead to tractable way forward since $\sum_{k=1}^K x^{k(4k^2-1)-1}$ is not a geometric series.
However, we can proceed by writing
$$\begin{align}
\frac{1}{k(4k^2-1)}&=\frac{1}{2k+1}+\frac{1}{2k-1}-\frac1k\\\\
&=\int_0^1 (x^{2k}+x^{2k-2}-x^{k-1})\,dx \tag 1
\end{align}$$
Then, using $(1)$ it is easy to see that
$$\begin{align}
\sum_{k=1}^K (-1)^{k-1}\frac{1}{k(4k^2-1)}&=-\int_0^1 \sum_{k=1}^K((-x^2)^k+x^{-2}(-x^2)^k-x^{-1}(-x)^k)\,dx\\\\
&=\int_0^1 (1 -(-1)^Kx^{2K}) \,dx-\int_0^1 \frac{1-(-1)^Kx^{K}}{1+x}\,dx \\\\
&=1-\frac{(-1)^K}{2K+1}-\log(2)+(-1)^K\int_0^1 \frac{x^K}{1+x}\,dx\tag 2
\end{align}$$
whence letting $K\to \infty$ in $(2)$ yields the coveted result
$$\bbox[5px,border:2px solid #C0A000]{\sum_{k=1}^\infty (-1)^{k-1}\frac{1}{k(4k^2-1)}=1-\log(2)}$$
as expected!
NOTE:
There are a number of ways to show that the integral on the right-hand side of $(2)$ approaches $0$ as $K\to \infty$.
One way is to apply the Dominated Convergence Theorem with dominating function $\frac{1}{1+x}$.
A second way is to integrate by parts with $u=\frac1{1+x}$ and $v=\frac{x^{K+1}}{K+1}$.
A third way is to note that $0\le \int_0^1 \frac{x^K}{1+x}\,dx\le \int_0^1x^K\,dx=\frac{1}{K+1}$ and apply the squeeze theorem.
|
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|
How can I derive the conditions of Positive semidefinite cone in $2\times2$ matrix. By the definition, in order for $X$ to be positive semidefinite cone in $S^2$, it should satisfy that
\begin{equation}
X=\left[
\begin{array}{cc}
x & y \\
y & z
\end{array}
\right]\in S_+^2 \quad\Longleftrightarrow\quad x\ge0,\quad z\ge0, \quad xz\ge y^2,\tag{1}
\end{equation}
where
$$
S_+^2 = \left\{X\in S^2 | X \succeq 0\right\}.
$$
I have failed to prove the $(1)$.
$$
\begin{align}
\forall \alpha,\beta,\quad
\left[
\begin{array}{cc}
\alpha & \beta
\end{array}
\right]
\left[
\begin{array}{cc}
x & y \\
y & z
\end{array}
\right]
\left[
\begin{array}{cc}
\alpha \\
\beta
\end{array}
\right] &\ge 0\\
\alpha^2x + 2\alpha\beta y + \beta^2z &\ge 0\\
\left(\alpha\sqrt{x} + \beta\sqrt{z}\right)^2 + 2\alpha\beta(y-\sqrt{xz}) &\ge 0
\\\therefore \alpha\beta(y-\sqrt{xz})\ge0\\
\text{if}\quad\alpha\beta\ge0,\quad \text{then} \quad y^2\ge xz\\
\text{if}\quad\alpha\beta\le0,\quad \text{then} \quad y^2\le xz
\end{align}
$$
How can I reach to
$$
x\ge0,\quad z\ge0,\quad xz\ge y^2\qquad?
$$
|
This is essentially the implication from the Sylvester's criterion extended to the case of positive semi-definite matrices.
All of the leading principal minors must be nonnegative. So you immediately get $x\ge 0, x z\ge y^2, z\ge 0$.
Following your proof for $2\times2$ matrices, take consequently $\alpha = 0$ and $\beta = 0$ to get $x\ge 0, z\ge 0$. The last step is to take $\alpha = \beta = 1$
$$2y \ge -x-z\Rightarrow 4y^2 \ge (x+z)^2 \ge 4xz$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
How do I solve the differential equation: $ \frac{dy}{dx} = \frac{x+y}{x-y} $? $$ \frac{dy}{dx} = \frac{x+y}{x-y} $$
I have tried this problem so long... such as
please help..
|
This is a dimensionally homogenous ODE.
Let $u = \frac{y}{x}$
Then we have $\frac{du}{dx} = \frac{x\frac{dy}{dx} - y}{x^{2}}$
So $\frac{du}{dx} = \frac{\frac{dy}{dx} - u}{x}$
$\frac{dy}{dx} = x\frac{du}{dx} + u$
Substituting this in:
$x\frac{du}{dx} + u = \frac{x + ux}{x-ux}$
$x\frac{du}{dx} + u = \frac{1+u}{1-u}$
$x\frac{du}{dx} = \frac{1+u - u + u^{2}}{1-u}$
$x\frac{du}{dx} = \frac{u^{2} + 1}{1-u}$
$\int \frac{1-u}{u^{2}+1}du = \int \frac{1}{x} dx$
Can you solve from here?
|
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|
Evaluate: $\int_0^1 \frac{\left(1-x^4\right)^{3/4}}{\left(x^4+1\right)^2} \, dx$ Here is what I did:
$\int_0^1 \frac{\left(1-x^4\right)^{3/4}}{\left(x^4+1\right)^2} \, dx$
Put $x^2=\tan (\theta )$
Then: $x=\sqrt{\tan (\theta )}$ and $dx=\frac{1}{2} \tan ^{-\frac{1}{2}}(\theta ) \sec ^2(\theta )d\theta$
So the integral becomes $\int_0^{\frac{\pi }{4}} \frac{\left(1-\tan ^2(x)\right)^{3/4} \tan ^{-\frac{1}{2}}(x) \sec ^2(x)}{2 \left(\tan ^2(x)+1\right)^2} \, dx$
= $\frac{1}{2} \int_0^{\frac{\pi }{4}} \frac{\sin ^{-\frac{1}{2}}(x) {\cos^{\frac{1}{2}} (x)} \sec ^2(x) \left(\frac{\cos ^2(x)-\sin ^2(x)}{\cos ^2(x)}\right)^{3/4}}{\sec ^4(x)} \, dx$
= $\frac{1}{2} \int_0^{\frac{\pi }{4}} \sin ^{-\frac{1}{2}}(x) \cos (x) \cos ^{\frac{3}{4}}(2 x)\, dx$
Now how do I proceed?
I tried converting it to the form $\sin ^p(2 x) \cos ^q(2 x)$ but that proved impossible.
The textbook gives the answer as $\frac{1}{2^{9/2}}B\left(\frac{7}{4},\frac{1}{4}\right)$ which WolframAlpha tells me is 0.1472... but if ask it to evaluate the integral, I get the answer as 0.700... Who is right?
|
$\displaystyle J=\int_0^1 \frac{\left(1-x^4\right)^{3/4}}{\left(x^4+1\right)^2} \, dx$
Perform the change of variable $y=\dfrac{1-x^4}{1+x^4}$,
$\displaystyle J=\dfrac{1}{2^{\tfrac{9}{4}}}\int_0^1 \dfrac{x^{\tfrac{3}{4}}}{(1-x)^{\tfrac{3}{4}}}dx=\boxed{\dfrac{1}{2^{\tfrac{9}{4}}}\text{B}\left(\dfrac{1}{4},\dfrac{7}{4}\right)}$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Did I do it correctly? What's the image and the basis of the image of this linear mapping?
$A$ is the linear mapping $f(x)= Ax,\mathbb{R} \rightarrow \mathbb{R}$
$$f\left( \begin{pmatrix} x_{1}\\ x_{2}\\ x_{3}
\end{pmatrix}\right)= \begin{pmatrix} x_{2}-x_{3}\\
x_{1}+3x_{2}-2x_{3}\\ x_{1}-4x_{2}+5x_{3} \end{pmatrix}$$
What's the image $\text{Im}(f)$? What's the basis of the image?
So far I have only calculated the image / basis of matrix but I hope / think it's actually the same for these mappings.
So I form $f$ to a matrix $\begin{pmatrix}
0 & 1 & -1\\
1 & 3 & -2\\
1 & -4 & 5
\end{pmatrix}$
Then transpose this matrix: $\begin{pmatrix}
0 & 1 & 1\\
1 & 3 & -4\\
-1 & -2 & 5
\end{pmatrix}$
Now get as many zero lines as possible using Gauss:
Take third line and add it to second line:
$\begin{pmatrix}
0 & 1 & 1\\
0 & 1 & 1\\
-1 & -2 & 5
\end{pmatrix}$
Now make second line negative by $\cdot(-1)$ and add it to first line:
$\begin{pmatrix}
0 & 0 & 0\\
0 & -1 & -1\\
-1 & -2 & 5
\end{pmatrix}$
Transpose back we have $\begin{pmatrix}
0 & 0 & -1\\
0 & -1 & -2\\
0 & -1 & 5
\end{pmatrix}$
Thus $\text{Im}(f)= \text{span} \left(\left\{ \begin{pmatrix}
0\\
-1\\
-1
\end{pmatrix},\begin{pmatrix}
-1\\
-2\\
5
\end{pmatrix} \right\}\right )$ and basis $B= \left\{ \begin{pmatrix}
0\\
-1\\
-1
\end{pmatrix},\begin{pmatrix}
-1\\
-2\\
5
\end{pmatrix} \right\}$
Can you please tell me if I did it correctly and if not how to do it correctly?
|
You start with the matrix
$$
f =
\left[
\begin{array}{rrr}
0 & 1 & -1 \\
1 & 3 & -2 \\
1 & -4 & 5 \\
\end{array}
\right]=
\left[
\begin{array}{rrr}
f_{1} & f_{2} & f_{3}
\end{array}
\right].
$$
This looks like an academic exercise, so we look for trivial combinations of rows or columns. We see the column property
$$
f_{1} - f_{2} = f_{3}.
$$
The image of $f$ is the combination of all independent column vectors:
$$
\text{Im} \left( f \right) = a_{1} f_{1} + a_{2} f_{2} =
a_{1}
\left[
\begin{array}{r}
0 \\
1 \\
1
\end{array}
\right]
+ a_{2}
\left[
\begin{array}{r}
1 \\
3 \\
-4
\end{array}
\right]
%
=
%
\left[
\begin{array}{r}
a_{2} \\
a_{1} + 3 a_{2} \\
a_{1} - 4 a_{2}
\end{array}
\right].
$$
The column combination formula provides the kernel
$$
\text{Ker} \left( f \right) =
%
\left[
\begin{array}{r}
1 \\
-1 \\
-1
\end{array}
\right].
$$
A minimal basis is any two linearly independent vectors from
$$
\text{span} \left\{ \,
\left[
\begin{array}{r}
0 \\
1 \\
-1
\end{array}
\right]
, \,
\left[
\begin{array}{r}
1 \\
3 \\
-4
\end{array}
\right] \,
\right\}
$$
For example $v_{1} = \left[
\begin{array}{r}
0 \\
1 \\
-1
\end{array}
\right]$, and $v_{2} =
\left[
\begin{array}{r}
1 \\
3 \\
-4
\end{array}
\right].$
You could provide an orthogonal basis using the process of Gram and Schmidt:
$u_{1} = \left[
\begin{array}{r}
0 \\
1 \\
-1
\end{array}
\right]$, and $u_{2} =
\left[
\begin{array}{r}
1 \\
3 \\
-4
\end{array}
\right].$
|
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|
The probability that $a$ and $b$ satisfy the inequality $a-2b+10>0$? Nine identical balls are numbered $1,2,3,.........,9$ are put in a bag.$A$ draws a ball and gets the number $a$ and puts back in the bag. Next $B$ draws a ball and gets the number $b$.
The probability that $a$ and $b$ satisfy the inequality $a-2b+10>0$ ?
My Try :-
Total pairs of $(a,b)$ possible are $81$ .
*
*If $a=1$, then $b = 1,2,3,4,5$. Similarly for $a=2$.
*If $a=3$, then $b = 1,2,3,4,5,6$. Similarly for $a=4$.
*If $a=5$, then $b = 1,2,3,4,5,6,7$. Similarly for $a=6$.
*If $a=7$, then $b = 1,2,3,4,5,6,7,8$. Similarly for $a=8$.
*If $a=9$, then $b = 1,2,3,4,5,6,7,8,9$.
Total favourable pairs are $61$.
Hence, Total Probability = $\frac{61}{81}$
However, I don't have an answer for this. Am I right or missing something ?
|
You seek the probability that $~2b < 10+a$ when selecting the values with replacement and no bias.
Let us see: You have listed values of $b$ that satisfy this for every $a$, counted them, and compared as a ratio of the size of the sample space.
Yes, that is okay; you have arrived at the correct answer by a valid process and made no errors.
|
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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|
Congruence Modulo Powers of Primes I need help with the following:
Let $p$ be an odd prime and $x, y \in \mathbb{Z}$. Then $x \equiv y$ mod $p^k \implies x^p \equiv y^p$ mod $p^{k+1}$.
I know that I can write $x^p - y^p$ as $(x - y)F(x,y)$ for some ugly polynomial $F$, but I'm not sure whether that's too much help.
Ta
|
Suppose $x \equiv y \pmod {p^k}$, for some positive integer $k$.
Note the identity
\begin{align*}
x^p - y^p &= (x - y)
\left(
x^{p-1} + x^{p-2}y + \cdots + xy^{p-2} + y^{p-1}
\right)\\[4pt]
&=(x - y)
\left(
\sum_{i=0}^{p-1}x^{p-1-i}y^i
\right)
\\[4pt]
\end{align*}
Then
\begin{align*}
&x \equiv y \pmod{p^k}\\[4pt]
\implies\; &x \equiv y \pmod{p}\\[4pt]
\implies\; &x^i \equiv y^i \pmod{p},\;\;\text{for all nonnegative integers $i$}\\[4pt]
\end{align*}
Hence
\begin{align*}
\sum_{i=0}^{p-1}x^{p-1-i}y^i &\equiv \sum_{i=0}^{p-1}x^{p-1-i}x^i \pmod{p}\\[4pt]
&\equiv \sum_{i=0}^{p-1}x^{p-1} \pmod{p}\\[4pt]
&\equiv px^{p-1} \pmod{p}\\[6pt]
&\equiv 0 \pmod{p}\\[4pt]
\end{align*}
Thus, $\displaystyle{\sum_{i=0}^{p-1}x^{p-1-i}y^i}$ is a multiple of $p$.
But also, by hypothesis, $x - y$ is a multiple of $p^k$, hence
\begin{align*}
&p^k\,|\,(x-y)\;\;\;\text{and}\;\;\, p\;{\large{|}}\left(\sum_{i=0}^{p-1}x^{p-1-i}y^i\right)\\[6pt]
\implies\; &p^{k+1}\;{\Large{|}} \left((x - y)
\left(
\sum_{i=0}^{p-1}x^{p-1-i}y^i
\right)\right)\\[6pt]
\implies\; &x^p - y^p \equiv 0 \pmod {p^{k+1}}\\[6pt]
\end{align*}
as was to be shown.
A point of interest: We never used the fact that $p$ is an odd prime, hence the same result holds for any positive integer $p$.
|
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|
Showing that $m^2-n^2+1$ is a square Prove that if $m,n$ are odd integers such that $m^2-n^2+1$ divides $n^2-1$ then $m^2-n^2+1$ is a square number.
I know that a solution can be obtained from Vieta jumping, but it seems very different to any Vieta jumping problem I've seen.
To start, I chose $m=2a+1$ and and $n=2b+1$ which yields: $$ 4ka^2+4ka-4kb^2-4kb+k = b^2+b$$
Then suppose that $B$ is a solution, and $B_0$ is another solution. Then using Vieta jumping we get (with a bit of algebra) that $B+B_0 = -1$ and $B_0 = \frac {-k(2a+1)^2}{B(4k+1)}$.
But I'm not sure these final equalities are particularly helpful; I can't find any way to yield more solutions from them. How can I solve the problem? A solution without Vieta jumping is probably also possible
|
Under the assumption that the integer ratio is positive:
=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=
LEMMA
Given integers $$ M \geq m > 0, $$ along with positive integers $x,y$ with
$$ x^2 - Mxy + y^2 = m. $$
Then $m$ is a square.
PROOF.
First note that we cannot have integers $xy < 0$ with $ x^2 - Mxy + y^2 = m, $ since then
$ x^2 - Mxy + y^2 \geq 1 + M + 1 = M + 2 > m.$ If we have a solution with $x > 0$ and $xy \leq 0,$ it follows that $y=0.$
This is the Vieta jumping part, with some extra care about inequalities.
Case I: We begin with integers $$ y > x > 0 $$ and the stronger $$y > Mx.$$
Then we get a new solution by jumping
$$ (x,y) \mapsto (Mx - y,x). $$
However, the assumption $y > Mx$ means $Mx-y < 0,$ we cannot have a solution with one variable positive and the other negative. This case cannot occur.
Case II. $y > x > 0$ and $y = Mx.$ But then $x^2 - Mxy + y^2 = x^2 - M^2 x^2 + M^2 x^2 = x^2.$ Therefore $x^2 = m$ which is a square.
Case III.
$$ y > x $$ and
$$ y < Mx. $$ We have
$$ x^2 - Mxy + y^2 > 0, $$
$$ x^2 > Mxy - y^2 = y(Mx - y) > x(Mx-y), $$
$$ x > Mx - y > 0. $$
That is, the jump
$$ (x,y) \mapsto (Mx - y,x) $$
takes us from one ordered solution to another ordered solution while strictly decreasing $x+y.$
Within a finite number of such jumps we violate the conditions we were preserving; we reach a solution $(x,y)$ with $y \geq Mx,$ that is
$x > 0$ but $Mx-y \leq 0.$ Since $(Mx - y,x) $ is another solution we know that $Mx-y = 0.$ Therefore $x^2 = m$ and $m$ is a square.
Graph for $x^2 - 5xy + y^2 = 3$
=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=
Suppose we have odd integers $m,n > 0$ such that
$$ \frac{n^2 - 1}{m^2 - n^2 + 1} = k > 0 $$
is an integer. Then
$$ k+1 = \frac{m^2 }{m^2 - n^2 + 1}. $$
Name $w = 1 + k,$ so
$$ w = \frac{m^2 }{m^2 - n^2 + 1}. $$
We are sticking with positive $w$ so we may take $m \geq n >0.$ When we write
$$ m-n = 2x, $$
$$ m+n = 2y, $$
we are introducing positive variables. Then $m=x+y,$ $n = y - x,$ and
$$ w = \frac{x^2 + 2xy + y^2}{4xy+1}, $$
$$ x^2 + 2xy+ y^2 = 4wxy + w, $$
$$ x^2 - (4w-2)xy + y^2 = w. $$
From the LEMMA, we find that $w$ is a square. From
$$ w = \frac{m^2 }{m^2 - n^2 + 1} $$
we see that
$$ m^2 - n^2 + 1 $$ is also a square.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Polynomial form of $\det(A+xB)$
Let $A$ and $B$ be two $2 \times 2$ matrices with integer entries. Prove that $\det(A+xB)$ is an integer polynomial of the form $$P(x) = \det(A+xB) = \det(B)x^2+mx+\det(A).$$
I tried expanding the determinant of $\det(A+xB)$ for two arbitrary matrices, but it got computational. Is there another way?
|
$$\begin{pmatrix}a_{11}&&a_{12}\\a_{21}&&a_{22}\end{pmatrix}+x\begin{pmatrix}b_{11}&&b_{12}\\b_{21}&&b_{22}\end{pmatrix}=\begin{pmatrix}a_{11}+xb_{11}
&&a_{12}+xb_{12}\\a_{21}+xb_{21}&&a_{22}+xb_{22}\end{pmatrix}=C$$
$$\det(C)=
a_{11}a_{22}+a_{11}xb_{22}+a_{22}xb_{11}+x^2b_{11}b_{22}-
a_{21}a_{12}-a_{21}xb_{12}-a_{12}xb_{21}-x^2b_{21}b_{12}
$$
$$=\det(A)+x\left[\det\begin{pmatrix}a_{11}&&a_{12}\\b_{21}&&b_{22}\end{pmatrix}+\det\begin{pmatrix}b_{11}&&b_{12}\\a_{21}&&a_{22}\end{pmatrix}\right]+x^2\det(B)$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Uniform continuity : stuck at an intermediate step $$f(x)= \frac{x}{1+x^2}, \, x \in \mathbb{R}$$
I have proved till the following step. I can't understand how do I proceed further. Please help $$f(x)\le|(x-y)||1-xy|$$
|
For all $x \in \mathbb{R},$ we have
$$f'(x) = \frac{1 - x^2}{(1+x^2)^2} $$
For $|x| \leqslant 1$ we have
$$|f'(x)| = \frac{1 - x^2}{(1+x^2)^2} \leqslant \frac{1}{(1+x^2)^2} \leqslant 1,$$
and for $|x| > 1$ we have $|f'(x)| \to 0$ as $x \to \pm \infty$ and local maxima at $x = \pm \sqrt{3}$ where $f'(\pm \sqrt{3}) = 1/8.$
Since the derivative is bounded, $f$ is uniformly continuous.
Alternatively, following your approach
$$\left|\frac{x}{1+x^2} - \frac{y}{1+y^2}\right| = |x-y|\frac{|1 - xy|}{1 + x^2 + y^2 +x^2y^2} \leqslant |x-y|\frac{1 + |xy|}{1 + x^2 + y^2 +x^2y^2}. $$
I'll leave it to you to show that the RHS is bounded by $C |x - y|$ where $C$ is a constant.
Hint:
If $|xy| < 1$ then
$$\frac{1 + |xy|}{1 + x^2 + y^2 +x^2y^2} < \frac{2}{1 + x^2 + y^2 +x^2y^2}$$
If $|xy| > 1$ then
$$\frac{1 + |xy|}{1 + x^2 + y^2 +x^2y^2} < \frac{2|x||y|}{x^2 + y^2}$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Three Distinct Points and Their Normal Lines
Suppose That three points on the graph of $y=x^2$ have the property that their normal lines intersect at a common point. Show that the sum of their $x$-coordinates is $0$.
I have a lot going but can not finish it.
Proof:
Let $(a,a^2)$, $(b,b^2)$, and $(c,c^2)$ be three distinct points on $y=x^2$ such that $a\not=b\not=c$.
Find the tangent slope and the slope of their normal lines.
$$(a,a^2) \hspace{4mm}m_{tan}=2a, \hspace{4mm} m_{norm}=\frac{-1}{2a}$$
$$(b,b^2) \hspace{4mm}m_{tan}=2b, \hspace{4mm} m_{norm}=\frac{-1}{2b}$$
$$(c,c^2) \hspace{4mm}m_{tan}=2c, \hspace{4mm} m_{norm}=\frac{-1}{2c}$$
Normal Line $(a,a^2)$
$y-a^2=-\frac{1}{2a}(x-a) \implies y=-\frac{1}{2a}x+\frac{1}{2}+a^2$
Normal Line $(b,b^2)$
$y-b^2=-\frac{1}{2b}(x-b) \implies y=-\frac{1}{2b}x+\frac{1}{2}+b^2$
Normal Line $(c,c^2)$
$y-c^2=-\frac{1}{2c}(x-c) \implies y=-\frac{1}{2c}x+\frac{1}{2}+c^2$
Find the $x$-interception points between the normal lines of $(a,a^2)$ and $(b,b^2)$.
$-\frac{1}{2a}x+\frac{1}{2}+a^2=-\frac{1}{2b}x+\frac{1}{2}+b^2 \implies x=-(b+a)2ab$
Find the $x$-interception points between the normal lines of $(a,a^2)$ and $(c,c^2)$.
$-\frac{1}{2a}x+\frac{1}{2}+a^2=-\frac{1}{2c}x+\frac{1}{2}+c^2 \implies x=-(c+a)2ac$
Find the $x$-interception points between the normal lines of $(b,b^2)$ and $(c,c^2)$.
$-\frac{1}{2b}x+\frac{1}{2}+b^2=-\frac{1}{2c}x+\frac{1}{2}+c^2 \implies x=-(c+b)2bc$
Show that $a+b+c=0$
$$\begin{align}
....\\
....\\
....\\
\end{align}$$
I do not know how to show that $a+b+c=0$. Any advice on how to continue? Thanks in advance!
|
Since all normal lines intersect at a common point you have
$$
-(b+a)2ab=-(c+a)2ac=-(c+b)2bc,
$$
which means
$$
b^2a+a^2b=c^2a+a^2c=c^2b+b^2c.
$$
Now if $a=0$
$$
c^2b+b^2c=0\Rightarrow c=-b.
$$
If $a\neq0$,since $b\neq c$, from the first equality
$$
(b^2-c^2)a=-a^2(b-c) \Rightarrow b+c=-a.
$$
|
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|
What is the probability that if $3$ of $12$ lockers are selected at random that at least two of the selections are consecutive? Three persons randomly choose a locker among 12 consecutive lockers.
What is the probability that at least two of the lockers are consecutive?
This is what I came up with:
$|S| = C(12,3)$ all possibilities
$|E| = C(11,2) \times C(10,1) $
Thus, the probability should be
$ P(E) = $ $C(11,2) \times C(10,1) \over C(12,3)$
Is my answer correct?
Thanks!
|
Your answer is incorrect. Notice that
$$\frac{\binom{11}{2}\binom{10}{1}}{\binom{12}{3}} = \frac{55 \cdot 10}{220} = \frac{550}{220} = \frac{5}{2} > 1$$
Method 1: There are $12 \cdot 11 \cdot 10 = 1320$ possible ways for three people to choose their lockers.
Now, let's count arrangements in which at least two people select consecutive lockers.
There are $\binom{3}{2}$ ways to select two of the three people to have consecutive lockers. There are $11$ possible starting positions for the block of two consecutive lockers and $2!$ ways to order the people within that block. That leaves $10$ ways for the third person to select his or her locker. Hence, there are
$$\binom{3}{2} \cdot 11 \cdot 2! \cdot 10 = 660$$
arrangements in which at least two people have selected consecutive lockers.
However, we have counted selections in which all three people have counted consecutive lockers twice, once when we counted the leftmost pair as having two consecutive lockers and once when we counted the rightmost pair as having two consecutive lockers. We only want to count such selections once. Thus, we must subtract these selections from the total.
There are $10$ possible starting positions for a block of three consecutive lockers and $3!$ orders for the people within that block. Hence, there are
$$10 \cdot 3! = 60$$
locker selections in which all three selected lockers are consecutive.
Hence, the number of selections in which at least two lockers are consecutive is
$$\binom{3}{2} \cdot 11 \cdot 2! \cdot 10 - 10 \cdot 3! = 660 - 60 = 600$$
which yields a probability of
$$\frac{600}{1320} = \frac{5}{11}$$
that at least two consecutive lockers are selected by the three people.
Method 2: There are $$\binom{12}{3}$$ ways to select three of the twelve lockers. We count the number of selections in which no two lockers are consecutive, then subtract that from the total to determine the number of selections in which at least two selections are consecutive.
Line up nine blue balls in a row, leaving spaces between them and at the ends of the row in which to place three green balls.
$$\square \color{blue}{\bullet} \square \color{blue}{\bullet} \square \color{blue}{\bullet} \square \color{blue}{\bullet} \square \color{blue}{\bullet} \square \color{blue}{\bullet} \square \color{blue}{\bullet} \square \color{blue}{\bullet} \square \color{blue}{\bullet} \square$$
That gives ten spaces in which to place the three green balls. If we select three of those ten spaces, no two of the three green balls will be consecutive. There are $$\binom{10}{3}$$ ways to select those three spaces. If we now number the twelve balls from left to right, the numbers on the green balls represent a selection in which no two of the three selected lockers are consecutive. For instance,
$$\color{green}{\bullet} \color{blue}{\bullet} \color{blue}{\bullet} \color{blue}{\bullet} \color{blue}{\bullet} \color{green}{\bullet} \color{blue}{\bullet} \color{blue}{\bullet} \color{green}{\bullet} \color{blue}{\bullet} \color{blue}{\bullet} \color{blue}{\bullet}$$
represents the selection of the first, sixth, and ninth lockers. Hence, there are
$$\binom{12}{3} - \binom{10}{3}$$
selections in which at least two consecutive lockers are selected. Therefore, the probability that at least two consecutive lockers are selected is
$$\frac{\binom{12}{3} - \binom{10}{3}}{\binom{12}{3}} = 1 - \frac{\binom{10}{3}}{\binom{12}{3}} = 1 - \frac{120}{220} = 1 - \frac{6}{11} = \frac{5}{11}$$
|
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|
LU-decomposition of A I have:
$A=\begin{bmatrix}
2 & -1 & 2 & 3 & 4
\\
4 & -2 & 7 & 7 & 6
\\
2 & -1 & 20 & 9 & -8
\end{bmatrix}$
and I'm asked to LU-decomposition A, then solve $Ax=0$.
What I did:
With these steps:
$1)$ $R2=R2-2\times R1$
$2)$ $R3=R3-R1$
I got:
$U=\begin{bmatrix}
2 & -1 & 2 & 3 & 4
\\
0 & 0 & 3 & 1 & -2
\\
0 & 0 & 18 & 6 & -12
\end{bmatrix}$
$L=\begin{bmatrix}
1 & 0 & 0
\\
2 & 1 & 0
\\
1 & 0 & 1
\end{bmatrix}$
And because:
$Ax=b\hspace{12mm}L(Ux)=b$
$Ly=b\hspace{12mm}Ux=y$
I got:
$Ly=0$
$
\begin{bmatrix}
1 & 0 & 0
\\
2 & 1 & 0
\\
1 & 0 & 1
\end{bmatrix}
\begin{bmatrix}
y_1
\\
y_2
\\
y_3
\end{bmatrix}
=
\begin{bmatrix}
0
\\
0
\\
0
\end{bmatrix}
$
Using this step:
$1)$ $R2=R2-2\times R1$
I got $y=0$
Then solving:
$Ux=y$
$
\begin{bmatrix}
2 & -1 & 2 & 3 & 4
\\
0 & 0 & 3 & 1 & -2
\\
0 & 0 & 18 & 6 & -12
\end{bmatrix}
\begin{bmatrix}
x_1
\\
x_2
\\
x_3
\\
x_4
\\
x_5
\end{bmatrix}
=
\begin{bmatrix}
0
\\
0
\\
0
\end{bmatrix}
$
And using this steps:
$1)$ $R1=R1+2\times R2$
$2)$ $R3=\frac{R3}{6}$
$3)$ $R3=R3-R2$
I got:
$
\begin{cases}
x_2=2x_1+8x_3+5x_4
\\
x_4=-3x_3+2x_5
\\
x_1\,,\,x_3\,,\,x_5\quad free
\end{cases}
$
And I chose free variables to be $x_1=1$ , $x_3=1$ , $x_5=1$ so I got:
$
\begin{cases}
x_1=1
\\
x_2=5
\\
x_3=1
\\
x_4=-1
\\
x_5=1
\end{cases}
$
Just wondering if I have done it right? And if not, then how should I do it?
Thanks in advance.
|
$A=\begin{bmatrix}
2 & -1 & 2 & 3 & 4
\\
4 & -2 & 7 & 7 & 6
\\
2 & -1 & 20 & 9 & -8
\end{bmatrix}$
With these steps:
$1)$ $R2=R2-2\times R1$
$2)$ $R3=R3-R1$
$3)$ $R3=R3-6\times R2$
I got:
$U=\begin{bmatrix}
2 & -1 & 2 & 3 & 4
\\
0 & 0 & 3 & 1 & -2
\\
0 & 0 & 0 & 0 & 0
\end{bmatrix}$
$L=\begin{bmatrix}
1 & 0 & 0
\\
2 & 1 & 0
\\
1 & 6 & 1
\end{bmatrix}$
Solving $x$ from $Ax=0$:
$Ly=0$
$
\begin{bmatrix}
1 & 0 & 0
\\
2 & 1 & 0
\\
1 & 0 & 1
\end{bmatrix}
\begin{bmatrix}
y_1
\\
y_2
\\
y_3
\end{bmatrix}
=
\begin{bmatrix}
0
\\
0
\\
0
\end{bmatrix}
\Longrightarrow
\begin{cases}
y_1=0
\\
2y_1+y_2=0
\\
y_1+6y_2+y_3=0
\end{cases}
\Longrightarrow
\begin{cases}
y_1=0
\\
y_2=0
\\
y_3=0
\end{cases}
\Longrightarrow
y=
\begin{bmatrix}
0
\\
0
\\
0
\end{bmatrix}
=0
$
$Ux=y$
$Ux=0$
$
\begin{bmatrix}
2 & -1 & 2 & 3 & 4
\\
0 & 0 & 3 & 1 & -2
\\
0 & 0 & 0 & 0 & 0
\end{bmatrix}
\begin{bmatrix}
x_1
\\
x_2
\\
x_3
\\
x_4
\\
x_5
\end{bmatrix}
=
\begin{bmatrix}
0
\\
0
\\
0
\end{bmatrix}
$
And using this steps (to get echelon form):
$1)$ $R1=\frac{R1}{2}$
$2)$ $R1=R1+R2$
$3)$ $R2=\frac{R2}{3}$
And since the pivot columns of the matrix are $1$ and $3$, so the basic variables are $x_1$ and $x_3$. The remaining variables, $x_2\,,\,x_4$ and $x_5$ must be free.
I got:
$
\begin{cases}
x_1=\frac{1}{2}x_2-4x_3-\frac{5}{2}x_4
\\
x_3=-\frac{1}{3}x_4+\frac{2}{3}x_5
\\
x_2\,,\,x_4\,,\,x_5\quad free
\end{cases}
$
And I chose free variables to be $x_2=2$ , $x_4=6$ , $x_5=3$ so I got:
$
\begin{cases}
x_1=-14
\\
x_2=2
\\
x_3=0
\\
x_4=6
\\
x_5=3
\end{cases}
\Longrightarrow
x=\begin{bmatrix}
-14
\\
2
\\
0
\\
6
\\
3
\end{bmatrix}
=\begin{bmatrix}
-14 & 2 & 0 & 6 & 3
\end{bmatrix}^T
$
And I check it out by calculating:
$\Rightarrow L\times U$ and I got the $A$ so L and U matrices are correct
$\Rightarrow A\times x$ and I got $0$ so it $x$ values are correct, since it is true when $0=0$
Here is a photo of what I have done with the coefficients of $x_3$ and $x_4$:
|
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|
Find closed form of Wallis's product type $\prod_{n=1}^{\infty}\left(\prod_{k=0}^{m}(n+k)^{{m\choose k}(-1)^k}\right)^{(-1)^n}$ Wallis's product
$${2\over 1}\cdot{2\over 3}\cdot{4\over 3}\cdot{4\over 5}\cdots={\pi\over 2}\tag1$$
Generalised of Wallis's product type
$$\prod_{n=1}^{\infty}\left(\prod_{k=0}^{m}(n+k)^{{m\choose k}(-1)^k}\right)^{(-1)^n}\tag2$$
Where $m\ge 1$
Setting $m=1$, we get $(1)$
$$\prod_{n=1}^{\infty}\left({n\over n+1}\right)^{(-1)^n}={\pi\over
2}\tag3$$
and $m=2$ we got
$$\prod_{n=1}^{\infty}\left(n\cdot{n+2\over (n+1)^2}\right)^{(-1)^n}={\pi^2\over 8}\tag4$$
How can we find the closed form for $(2)$? Conjecture closed form may take the form of $${\pi^{2^{m-1}}\over F(m)}$$
|
Using $~\displaystyle \prod\limits_{n=1}^M (x+n) \approx \frac{M!M^x}{\Gamma(x+1)}~$ for large $M$ we get with $~m\geq 1~$ :
$$\prod_{n=1}^{\infty}\left(\prod_{k=0}^{m}(n+k)^{{m\choose k}(-1)^k}\right)^{(-1)^n} = 2^{\delta_{m,1}} \prod\limits_{k=1}^m {\binom {k}{k/2}}^{(-1)^k {\binom m k}}$$
where $~\delta_{i,j}~$ is the Kronecker Delta, see here
$\,$
$\displaystyle m:=1 :\hspace{1cm} 2^{\delta_{1,1}} \prod\limits_{k=1}^1 {\binom {k}{k/2}}^{(-1)^k {\binom 1 k}} = 2{\binom {1}{1/2}}^{-1}=\frac{\pi}{2}$
$\displaystyle m:=2 :\hspace{1cm} 2^{\delta_{2,1}} \prod\limits_{k=1}^2 {\binom {k}{k/2}}^{(-1)^k {\binom 2 k}} = {\binom {1}{1/2}}^{-2} {\binom {2}{1}}^{+1} = \frac{\pi^2}{2^3}$
$\displaystyle m:=3 :\hspace{1cm} 2^{\delta_{3,1}} \prod\limits_{k=1}^3 {\binom {k}{k/2}}^{(-1)^k {\binom 3 k}} = {\binom {1}{1/2}}^{-3} {\binom {2}{1}}^{+3} {\binom {3}{3/2}}^{-1} = \frac{3\pi^4}{2^8}$
$\displaystyle m:=4 :\hspace{1cm} 2^{\delta_{4,1}} \prod\limits_{k=1}^4 {\binom {k}{k/2}}^{(-1)^k {\binom 4 k}} = {\binom {1}{1/2}}^{-4} {\binom {2}{1}}^{+6} {\binom {3}{3/2}}^{-4} {\binom {4}{2}}^{+1} = \frac{3^5\pi^8}{2^{21}}$
...
|
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|
Find the last two digits of $47^{89}$ Find the last two digits of the number $47^{89}$
I applied the concept of cyclicity
$47\cdot 47^{88}$
I basically divided the power by 4 and then calculated $7^4=2401$ and multplied it with 47 which gave me the answer $47$ but the actual answer is $67$. How?
|
\begin{align}
47^2 &= 40^2 + 2\times 40\times 7 + 7^2 \equiv 9 \pmod{100},\\
47^4 &\equiv 9^2 \equiv 81 \pmod{100}, \\
47^8 &\equiv 81^2 \equiv 61 \pmod{100}, \\
47^{16} &\equiv 61^2 \equiv 21 \pmod{100}. \\
\end{align}
At this point, if we recognize how convenient the results
$47^4 \equiv 81 \pmod{100}$ and $47^{16} \equiv 21 \pmod{100}$ are,
we might continue with
\begin{align}
47^{20} &= 47^{16} \times 47^4 \equiv 21\times 81 \equiv 1 \pmod{100}, \\
46^{80} &\equiv 1 \pmod{100}, \\
47^9 &\equiv 61 \times 47 \equiv 67 \pmod{100}, \\
47^{89} &= 47^{80} \times 47^9 \equiv 67 \pmod{100}. \\
\end{align}
If we did not notice that $47^{20}\equiv 1 \pmod{100},$ however, we
could continue squaring to find $47^{32} \equiv 41 \pmod{100}$
and $47^{64} \equiv 81 \pmod{100},$
and then use the fact that
$47^{89} = 47^{64} \times 47^{16} \times 47^{8} \times 47.$
|
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|
How can I prove this trigonometric equation with squares of sines? Here is the equation:
$$\sin^2(a+b)+\sin^2(a-b)=1-\cos(2a)\cos(2b)$$
Following from comment help,
$${\left(\sin a \cos b + \cos a \sin b\right)}^2 + {\left(\sin a \cos b - \cos a \sin b\right)}^2$$
$$=\sin^2 a \cos^2b + \cos^2 a \sin^2 b + \sin^2 a \cos^2 b + \cos^2 a \sin^2 b$$
I am stuck here, how do I proceed from here?
Edit: from answers I understand how to prove,but how to prove from where I am stuck?
|
On RHS we have,
$$
1-\cos(2a)\cos(2b) = 1-\cos^2a\cos^2b+\cos^2a\sin^2b+\sin^2a\cos^2b-\sin^2a\sin^2b
= (\cos^2a\sin^2b+\sin^2a\cos^2b)+1-\cos^2a+\sin^b\cos^2a-\sin^2a\sin^2b
= (\cos^2a\sin^2b+\sin^2a\cos^2b)+\sin^2b\cos^2a+\sin^2a-\sin^2a\sin^2b
= 2(\cos^2a\sin^2b+\sin^2a\cos^2b)
= LHS
$$
Hence proved
|
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|
Tan function and isosceles triangles I have a non-right-angled isosceles triangle with two longer sides, X, and a short base Y.
I know the length of the long sides, X.
I also know the acute, vertex angle opposite the base Y, let's call it angle 'a'
I have been told I can calculate the length of the base Y by:
Y = tan(a) x X
I've sketched this out with a few hand drawn triangles and it does seem to work...... But why?
I can't derive that formula from any of the trigonometry I know. What am I missing?
|
bisect angle $a$
$\sin \frac a2 = \frac {y}{2x}\\
\cos \frac a2 = \frac {\sqrt{4x^2-y^2}}{2x}\\
\tan \frac a2 = \frac {y}{\sqrt{4x^2-y^2}}\\
\tan a = \frac {2\tan\frac a2}{1-\tan^2 \frac a2}\\
\tan a = \frac {2y}{\sqrt{4x^2-y^2}(1-\frac {y^2}{4x^2-y^2})}\\
\tan a = \frac {y\sqrt{4x^2-y^2}}{2x^2-y^2}$
$\frac yx$ is small, $\frac{\sqrt{4x^2-y^2}}{2x^2-y^2} \approx \frac {1}{x}$
and $\tan a \approx \frac yx$
|
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|
Solve $f(x) = f(\lfloor x/2 \rfloor) + f(\lfloor x/3 \rfloor)$
Solve the following recurrence relation
$$f(1)=1, f(2)=2\\
f(x) = f\left(\left \lfloor \frac{x}{2} \right \rfloor\right) + f\left(\left\lfloor \frac{x}{3} \right\rfloor\right), \forall x \in \mathbb{N}, x \geq 3 $$
I tried the simple ways to solve a recurrence relation but got things messed up. Any Hint will be helpful.
|
Is $x$ constrained in some particular set?
Note that:
$$
\begin{aligned}
f(2) &= f\left(\left\lfloor \frac{2}{2} \right\rfloor\right) + f\left(\left\lfloor \frac{2}{3} \right\rfloor\right)\\
&=f(1) + f(0)
\end{aligned}
$$
Therefore, $f(0) = f(2) - f(1) = 1$. At the same time, for $x=1$,
$$
\begin{aligned}
f(1) &= f\left(\left\lfloor \frac{1}{2} \right\rfloor\right) + f\left(\left\lfloor \frac{1}{3} \right\rfloor\right)\\
&=2f(0),
\end{aligned}
$$
therefore, $f(x) = \tfrac{1}{2}f(1) = \tfrac{1}{2}$, which is a contradition. Therefore, $f$ cannot be defined (at least) at $x=0$.
|
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|
Convergence, absolute convergence of an integral $\int_{0}^{\infty}\frac{(x^{4}+2017)*(sin x)^{3}}{x^{5}+2018*x^{3}+1}dx$
How do I prove that it converges and that it does not converge absolutely?
|
Recall that $\sin^3(x)=\frac34\sin(x)-\frac14\sin(3x)$. Furthermore, $\left|\int_0^L \sin(x)\,dx\right|\le 2$ for all $L$.
Then, note that for sufficiently large $x$, $\frac{x^4+2107}{x^5+2018x^3+1}$ monotonically decreases to $0$.
Abel's Test (Dirichlet's Test) for improper integrals guarantees convergence of the integral
$$\begin{align}\int_0^\infty \frac{(x^4+2107)\sin^3(x)}{x^5+2018x^3+1}\,dx&=\frac34\int_0^\infty \frac{(x^4+2107)\sin(x)}{x^5+2018x^3+1}\,dx-\frac14\int_0^\infty \frac{(x^4+2107)\sin(3x)}{x^5+2018x^3+1}\,dx\end{align}$$
To show that we have conditional convergence only, we have for $x>16\pi$,
$$\left|\frac{(x^4+2107)\sin^3(x)}{x^5+2018x^3+1}\right|\ge \frac{|\sin^3(x)|}{x}$$
Then, we can write
$$\begin{align}
\int_{16\pi}^{(n+1)\pi}\frac{|\sin^3(x)|}{x}\,dx&=\sum_{k=16}^n\int_{k\pi}^{(k+1)\pi}\frac{|\sin^3(x)|}{x}\,dx\\\\
&\ge \sum_{k=16}^n \frac{1}{(k+1)\pi}\int_0^\pi \sin^3(x)\,dx\\\\
&=\sum_{k=16}^n \frac{4}{3(k+1)\pi}\tag 1
\end{align}$$
Since, the series in $(1)$ diverges by comparison with the harmonic series, then the integral of interest does not absolutely converge.
|
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|
Prove that $\displaystyle \prod_{j=1}^{2n}\left(j^j(2n+j)^{j-1}\right)$ is a perfect square
For any positive integer $n$, prove that $\displaystyle \prod_{j=1}^{2n}\left(j^j(2n+j)^{j-1}\right)$ is a perfect square.
Let $f(n) = \prod_{j=1}^{2n}\left(j^j(2n+j)^{j-1}\right)$. Then $$f(1) = 2^4, \quad f(2) = 2^{20} \cdot 3^4 \cdot 7^2, \quad f(3) = 2^{32} \cdot 3^{18} \cdot 5^8 \cdot 11^4.$$ I didn't see a way to prove the expression is a perfect square for all $n$.
|
Extracting even powers, we see that, up to a perfect square, this expression is equal to
$$
\prod_{\text{odd }j\le 2n} j \cdot \prod_{\text{even }j\le 2n}(2n+j) = (2n-1)!! \cdot \prod_{k=1}^n(2n+2k) = (2n-1)!! \cdot 2^n \frac{(2n)!}{n!} \\
= (2n-1)!! \cdot 2^{2n} \frac{(2n)!}{(2n)!!} = \bigl(2^n (2n-1)!!\bigr)^2.
$$
|
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|
Show $\sum_{n=1}^{\infty}\frac{1}{n^2}=\frac{\pi^2}{6}$ I know there are various methods showing that $\sum_{n=1}^{\infty}\frac{1}{n^2}=\frac{\pi^2}{6}$, but I want to know how to derive it from letting $t\rightarrow 0^{+}$ for the following identity:
$$\sum_{n=-\infty}^{\infty}\frac{1}{t^2+n^2}=\frac{\pi}{t}\frac{1+e^{-2\pi t}}{1-e^{-2\pi t}}$$
|
Note we can write:
$$\frac{\pi}{t}-\frac{1}{t^2}\frac{(-2\pi t)e^{-2\pi t}}{1-e^{-2\pi t}}=\frac{\pi}{t}+\frac{1}{t}\frac{2\pi e^{-2\pi t}}{1-e^{-2\pi t}}=\frac{\pi}{t}\left(1+\frac{2e^{-2\pi t}}{1-e^{-2\pi t}}\right)=\frac{\pi}{t}\frac{1-e^{-2\pi t}+2e^{-2\pi t}}{1-e^{-2\pi t}}\\=\frac{\pi}{t}\frac{1+e^{-2\pi t}}{1-e^{-2\pi t}}=\sum_{n=-\infty}^{\infty}\frac{1}{t^2+n^2}=\frac{1}{t^2}+2\sum_{n=1}^{\infty}\frac{1}{t^2+n^2}=\frac{1}{t^2}+2\sum_{n=1}^{\infty}\frac{1/n^2}{(t/n)^2+1}\\=\frac{1}{t^2}-\frac{2}{t^2}\sum_{n=1}^{\infty}\frac{(ti/n)^2}{(ti/n)^2-1}=\frac{1}{t^2}-\frac{2}{t^2}\sum_{n=1}^{\infty}\left(\sum_{k=1}^{\infty}\frac{(ti)^{2k}}{n^{2k}}\right)=\frac{1}{t^2}-\frac{2}{t^2}\sum_{k=1}^{\infty}\left(\sum_{n=1}^{\infty}\frac{(ti)^{2k}}{n^{2k}}\right)\\\implies \frac{\pi}{t}-\frac{1}{t^2}\frac{(-2\pi t)e^{-2\pi t}}{1-e^{-2\pi t}}=\frac{1}{t^2}-\frac{2}{t^2}\sum_{k=1}^{\infty}(ti)^{2k}\zeta(2k)=\frac{1}{t^2}-\frac{2}{t^2}\sum_{k=1}^{\infty}t^{2k}(-1)^k\zeta(2k)\\\implies 1-\pi t+\frac{(-2\pi t)e^{-2\pi t}}{1-e^{-2\pi t}}=\sum_{k=1}^{\infty}t^{2k}\left(2(-1)^k\zeta(2k)\right)$$
Thus if we let $\frac{ze^{z}}{1-e^{z}}=f(z)=\sum_{k=0}^{\infty}\frac{f^{(k)}(0)}{k!}z^k$ for appropriately chosen $z$, we get that:
$$1-\pi t+\sum_{k=0}^{\infty}\frac{f^{(k)}(0)(-2\pi)^k}{k!}t^k=1-\pi t+f(-2\pi t)=\sum_{k=1}^{\infty}t^{2k}\left(2(-1)^k\zeta(2k)\right)$$
However since the summands on the right hand-side consist of only those terms at even indices we must have that all the terms of odd index on the left hand-side vanish as well, which gives us:
$$\sum_{k=0}^{\infty}\frac{f^{(2k)}(0)(2\pi)^{2k}}{(2k)!}t^{2k}=\sum_{k=1}^{\infty}t^{2k}\left(2(-1)^k\zeta(2k)\right)\implies \frac{f^{(2n)}(0)(-1)^n(2\pi)^{2n}}{2(2n)!}=\zeta(2n)$$
For all $n\in \mathbb{N}$ and thus in particular at $n=1$ we get $f^{(2)}(0)=-\frac{1}{6}$ so that $\zeta(2)=\frac{\pi^2}{6}$.
|
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|
What is wrong with this way of solving trig equations? Let's suppose I have to find the values of $\theta$ and $\alpha$ that satisfy these equations:
*
*$\cos^3 \theta$ = $\cos \theta $
*$3\tan^3 \alpha = \tan \alpha$
on the interval $[0; 2 \pi]$.
If I try to solve, for instance, the first equation like this:
$$\cos^3 \theta = \cos \theta $$
$$\cos^2 \theta * \cos \theta = \cos \theta$$
$$\cos^2 \theta = \cos \theta \div \cos \theta $$
$$\cos^2 = 1$$
$$\cos \theta = \pm \sqrt {1}$$
$$\cos \theta = \pm 1$$
I end up getting only:
$\theta = 0 ; \pi ; 2 \pi$
But I know that $\pi /2$ and $3\pi /2$ would also make the equation true since $\cos^3 (\pi /2) = cos (\pi /2) = 0$ and $\cos^3 ( 3 \pi /2) = cos (3 \pi /2) = 0$.
The same problem arises when I try to solve the second equation:
$$3\tan^3 \alpha = \tan \alpha$$
$$3\tan^2 \alpha * \tan \alpha = \tan \alpha$$
$$3\tan^2 \alpha = \tan \alpha \div \tan \alpha$$
$$3\tan^2 \alpha =1$$
$$\tan^2 \alpha =1/3$$
$$\tan \alpha =\pm \sqrt{1/3}$$
$$\tan \alpha =\pm 1/\sqrt{3}$$
The values of $\alpha$ that make the tangent equal to $\pm 1/\sqrt{3}$ between 0 and $2 \pi$ are only these:
$ \alpha = \pi / 6 ; 5 \pi /6 ; 7 \pi / 6 ; 11 \pi /6$
However, I should also find $0$ and $\pi$ because $3\tan^3 (0) = \tan (0) = 3 tan^3 (\pi) = tan ( \pi ) = 0 $.
Something similar happens when I'm looking for the local minima and maxima of this function:
$$ f(x) = \sin^2(x) + \cos(x)$$
on the interval $[0; 2 \pi]$
$f'(x) = 2 \sin(x) \cos(x) - \sin(x)$
$ 0 = 2 \sin(x) \cos(x) - \sin(x)$
$ \sin(x) = 2 \sin(x) \cos(x)$
$ \sin(x) / \sin(x) = 2 \cos(x)$
$1 = 2 \cos(x)$
$1/2 = \cos(x)$
$x = \pi / 3 ; 5 \pi /3$
And again, plugging $x = 0$ or $x= \pi$ or $x = 2 \pi$ also make the derivative 0.
I've noticed that in all of these 3 cases I have the same trigonometric function on both sides of the equation and I'm dividing both sides by that function. This is making one side of the equation equal to 1 at some point. What is wrong or incomplete with this method? Why am I missing some results when I do this?
|
Clearly, when $\cos \theta = 0$, then $\cos^3 \theta = \cos \theta$.
So, in the interval $[0, 2\pi]$, we get $\theta \in \left\{\frac 12 \pi, \frac 32 \pi \right\}$
If $\cos \theta \ne 0$, then we can divide both sides by $\cos \theta$...
|
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|
The integral $\int_{-\infty}^\infty \frac{e^{-x^2}}{(1+e^x)^2} dx$. Let
$$T(n) = \int_{-\infty}^\infty \frac{e^{-x^2}}{(1+e^{x})^n} dx.$$
We have that
$$ T(0) = \sqrt{\pi} \text{ and } T(1) = \frac{\sqrt{\pi}}{2}$$
and also that
$$ T(3) = \tfrac{3}{2} T(2) - \frac{\sqrt{\pi}}{4}.$$
Can we find a closed form for $T(2)$? It would also give us $T(3)$. Perhaps something in terms of special functions?
|
An expression for $T(2)$ as a series of Bernoulli numbers:
\begin{align}
T(2) &= \int_{-\infty}^\infty \frac{e^{-x^2}}{(1+e^{x})^2} d\,dx\\
&=\frac{1}{4}\int_{-\infty}^\infty \frac{e^{-x^2-x}}{\cosh^2 \frac{x}{2}} \,dx\\
&=\frac{1}{2}\int_{-\infty}^\infty \frac{e^{-4x^2-2x}}{\cosh^2 x} \,dx\\
&=\frac{1}{2}\int_{-\infty}^\infty \frac{e^{-4x^2+2x}}{\cosh^2 x} \,dx
\end{align}
Summing the last two expressions and using the formula $\cosh 2x=2\cosh^2x-1$, it comes
\begin{align}
T(2)&=\frac{1}{2}\int_{-\infty}^\infty \frac{\cosh 2x}{\cosh^2 x}e^{-4x^2}\,dx\\
&=\int_{-\infty}^\infty e^{-4x^2}\,dx-\frac{1}{2}\int_{-\infty}^\infty \frac{e^{-4x^2}}{\cosh^2 x}\,dx\\
&=\frac{\sqrt{\pi}}{2}-\frac{1}{2}\int_{-\infty}^\infty \frac{e^{-4x^2}}{\cosh^2 x}\,dx
\end{align}
We integrate by parts,
\begin{equation}
T(2)=\frac{\sqrt{\pi}}{2}-4\int_{-\infty}^\infty xe^{-4x^2}\tanh x\,dx
\end{equation}
Now, using the series expansion for $\tanh$ DLMF:
\begin{equation}
\tanh z=\sum_{n=1}^\infty\frac{2^{2n}(2^{2n}-1)B_{2n}}{(2n)!}z^{2n-1}
\end{equation}
one can express,
\begin{align}
T(2)&=\frac{\sqrt{\pi}}{2}-4\sum_{n=1}^\infty\frac{2^{2n}(2^{2n}-1)B_{2n}}{(2n)!}\int_{-\infty}^\infty x^{2n}e^{-4x^2}\,dx\\
&=\frac{\sqrt{\pi}}{2}-2\sqrt{\pi}\sum_{n=1}^\infty\frac{2^{2n}-1}{2^{2n}}\frac{B_{2n}}{n!}
\end{align}
where we have used the tabulated formula
\begin{equation}
\int_0^\infty x^{2n}e^{-px^2}\,dx=\frac{(2n-1)!!}{2(2p)^n}\sqrt{\frac{\pi}{p}}
\end{equation}
|
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|
Proving $\cos(4\theta) - 4\cos(2\theta) \equiv 8\sin^4\theta - 3$.
Prove the following identity:
$$\cos(4\theta) - 4\cos(2\theta) \equiv 8\sin^4\theta - 3$$
How can I express $\cos(4\theta) $ in other terms?
|
$$\cos(4\theta)=2\cos^2(2\theta)-1\to\\ \cos(4\theta) - 4\cos(2\theta)=2\cos^2(2\theta)- 4\cos(2\theta)-1\to\\
\cos(4\theta) - 4\cos(2\theta)=2(1-2\sin^2(\theta))^2-4(1-2\sin^2(\theta))-1=8\sin^4(\theta)-3$$
|
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|
Need direction for solving integral
Evaluate: $$\int \frac{\sin^2x}{\cos^2x +4}dx.$$
I tried to this things:
First
$$\tan\left(\frac{x}{2}\right) = t$$
$$dx = \frac{2~dt}{1+t^2}$$
$$\sin x= \frac{2t}{1+t^2}$$
$$\cos x= \frac{1-t^2}{1+t^2}$$
Tried also but this is same thing?
$$\tan\left(\frac{x}{2}\right) = t$$
There I don't have to place $2$ left to integral ( same thing )
Second
I tried trigonometric transformations like
$$\sin^2x \equiv 1-\cos^2x$$
$$\int \frac{1-\cos^2x}{\cos^2x +4}dx$$
$$-1\cdot \int \frac{\cos^2x-1}{\cos^2x +4}~dx$$
$$-1\cdot \int \frac{\cos^2x+4-5}{\cos^2x +4}~dx$$
But I get stucked in this part
$$5\int \frac{1}{\cos^2x +4}dx.$$
And mix something from first step
Third
I tried to divide everything with $\sin^2x$, but didn't succeed to solve + mix something from first step
Can anyone give me direction how to solve this?
sorry, for late update....
My goal is to complete this task without trigonometry function sec
|
substituting $t=\tan x$, with $dx=\frac{dt}{1+t^2}$ gives $$\int\frac{t^2}{(5+4t^2)(1+t^2)}dt=\int\frac{5}{5+4t^2}-\frac{1}{1+t^2}dt$$
So you end up with $$\frac{\sqrt{5}}{2}\arctan\left(\frac{2}{\sqrt{5}}\tan x\right)-x+c$$
|
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|
How to know multipliers in solving PDE Equations? Solve $$\frac{dx}{y+z}=\frac{dy}{z+x}=\frac{dz}{x+y}.$$ Solve the similtaneous equation by using method of multipliers. How can we choose these multipliers? Is there any specific method to know these multipliers?
|
Try to find triplets (multipliers) $l,m,n$ such that in,
$$\frac{dx}{y+z}=\frac{dy}{z+x}=\dfrac{dz}{x+y}=\dfrac{l dx+m dy+n dz}{l(y+z)+m(z+x)+n (x+y)}$$
make the denominator vanish
Set $l=y-z,m=z-x,n=x-y$
Checking for the denominator to vanish: $(y-z)(y+z)+(z-x)(z+x)+(x-y)(x+y)=y^2-z^2+z^2-x^2+x^2-y^2=0$
$$\frac{dx}{y+z}=\frac{dy}{z+x}=\dfrac{dz}{x+y}=\dfrac{(y-z)dx+(z-x)dy+(x-y)dz}{0}$$
We have to solve $(y-z)dx+(z-x)dy+(x-y)dz=0$ Rearranging,
$(y-z)dx+(z-y+y-x)dy+(x-y)dz=0$
$(y-z)dx-(y-z)dy-(x-y)dy+(x-y)dz=0$
$(y-z)(dx-dy)-(x-y)(dy-dz)=0$
$\dfrac{(dx-dy)(y-z)-(x-y)(dy-dz)}{(y-z)^2}=0$
Finally
$\dfrac{x-y}{y-z}=c_1$
Choosing another triplet in a similar fashion, you can get the second equation.
The specifity of the method is the posing of the quotient. With respecto to something more systematic, I don't think there is an algorithm to find them.
|
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|
What is the minimum value of $|z+\frac{1}{2}|$ for $|z|\geq 2$ Find the minimum value of $|z+\frac{1}{2}|$ for $|z|\geq 2$.
Is the following approaches correct ?
$\left|z+\frac{1}{2} \right|=\sqrt{(x+\frac{1}{2})^2+y^2}=\sqrt{x^2+y^2+x+\frac{1}{4}}$. What will be the next?
Update:
$|z|\geq 2 \implies x^2+y^2\geq 4\implies x^2+y^2+x+\frac{1}{4}\geq 4+\frac{1}{4}+x\implies \sqrt{x^2+y^2+x+\frac{1}{4}} \geq \sqrt{\frac{17}{4}+x}$.
|
Use the triangle inequality:
$$||a|-|b|| \leq |a+b| \leq |a|+|b|.$$
You need only the first part with $a=z$ and $b=1/2$. So
$$ |z+\frac12| \geq ||z|-\frac12| = |z| - \frac12 \geq 2 - 0.5 = 1.5$$
It is enough to notice, that for $z=-2$ we have $|z|=2$ and $|z+\frac12| = 1.5$.
More about triangle inequality:
https://en.wikipedia.org/wiki/Triangle_inequality
|
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|
Evaluate $\int_0^{2\pi}\frac{1}{1 + A\sin(\theta) + B\cos(\theta)}\, \mathrm{d}\theta \ \text{ for }\ A,B <<1$ I need to evaluate the definite integral $$\int_0^{2\pi}\frac{1}{1 + A\sin(\theta) + B\cos(\theta)}\, \mathrm{d}\theta \ \text{ for various}\ A,B \text{; with}\ A,B<<1.$$
Wolfram Alpha provides the following indefinite general solution:-
$$\int \frac{1}{1 + A\sin(\theta) + B\cos(\theta)}\, \mathrm{d}\theta = -(2/K) \tanh^{-1} ( \frac{A-(B-1)\tan(\theta/2)}{K}) $$
where $K = \sqrt{A^2 + B^2 -1}$.
But I am having trouble checking it for the simple case when $A=B=0$ when I would expect the answer to be given by:-
$$\int_0^{2\pi}\frac{1}{1 + 0 + 0}\, \mathrm{d}\theta = 2\pi.$$
I have approached the Wolfram Alpha solution thus:-
$$ -(2/K) \tanh^{-1} ( \frac{\tan(2\pi/2)}{K}) +(2/K) \tanh^{-1} ( \frac{\tan(0/2)}{K}) $$
$$ -(2/K) \tanh^{-1} ( \frac{\tan(\pi)}{K}) +(2/K) \tanh^{-1} ( \frac{\tan(0)}{K}) $$
$$ -(2/K) \tanh^{-1} ( \frac{0}{K}) +(2/K) \tanh^{-1} ( \frac{0}{K}) $$
which gives the result of zero.
I presume this error comes from trying to integrate across the range $0, 2\pi$ where the $\tan$ function has singularities at $\pi/2$ and $3\pi/2$.
However when I try and break the integration into the three continuous ranges $0,\pi/2$ and $\pi/2,3\pi/2$ and $3\pi/2,2\pi$ I am still getting a result of zero thus:-
$$ -(2/K) \tanh^{-1} ( \frac{\tan(2\pi/2)}{K}) +(2/K) \tanh^{-1} ( \frac{\tan(3\pi/4)}{K}) + $$
$$ -(2/K) \tanh^{-1} ( \frac{\tan(3\pi/4)}{K}) +(2/K) \tanh^{-1} ( \frac{\tan(\pi/4)}{K}) +$$
$$ -(2/K) \tanh^{-1} ( \frac{\tan(\pi/4)}{K}) +(2/K) \tanh^{-1} ( \frac{\tan(0)}{K}) $$
leading to
$$ -(2/K) \tanh^{-1} ( \frac{0}{K}) +(2/K) \tanh^{-1} ( \frac{-1}{K}) + $$
$$ -(2/K) \tanh^{-1} ( \frac{-1}{K}) +(2/K) \tanh^{-1} ( \frac{1}{K}) +$$
$$ -(2/K) \tanh^{-1} ( \frac{1}{K}) +(2/K) \tanh^{-1} ( \frac{0}{K}) $$
which gives the same result of zero.
I would be grateful if somebody could tell me where I am going wrong here?
EDIT 1: I have accepted the solution provided kindly by Dr. MV.
I have posted a related question which seeks to understand where my original evaluation of the definite integrand goes wrong.
EDIT 2: In the related question comments from user mickep pointed out that the wrong partitions had been used in the original evaluation. Using the correct partitions ($0...\pi$) and ($\pi...2\pi$) leads to the correct answer, for $A=B=0$, of $2\pi$ (as described in my self-answer to that same question).
EDIT 3: It was pointed out by user mickep in comments to the related question that the Wolfram Alpha solution
$$ -\left(\frac{2}{K1}\right) {\arctan}h \left( \frac{A-(B-1)\tan(\theta/2)}{K1}\right) $$
where $K1= \sqrt{A^2 + B^2 -1}$.
is not as friendly as an alternative solution (reported by user mickep) which is:
$$ +\left(\frac{2}{K2}\right) \arctan \left( \frac{A+(1-B)\tan(\theta/2)}{K2}\right) $$
where $K2 = \sqrt{1 - A^2 - B^2}$.
|
When one of $A$ and $B$ is non-zero, then one way to figure this out for yourself is to put
$$A \colon= r \cos \beta \ \mbox{ and } \ B \colon= r \sin \beta,$$
where
$$r = \sqrt{A^2 + B^2} \ \mbox{ and } \ B \tan \beta = A.$$
Then
$$
\begin{align}
\int_0^{2\pi} \frac{1}{1+ A \sin \theta + B \cos \theta } \mathrm{d} \theta &= \int_0^{2\pi} \frac{1}{1+r\sin(\beta + \theta) } \mathrm{d} \theta.
\end{align}
$$
Can you take it from here?
As one trick, you can put
$$z = \tan \frac{\beta + \theta}{2}.$$
Then $$\sin (\beta+\theta) = \frac{2z}{1+z^2} \ \mbox{ and } \ \mathrm{d} \theta = \frac{2 \mathrm{d} z}{1+z^2}.$$
When $\theta = 0$, $z= \tan \beta/2$, and when $\theta = 2\pi$, $z= \tan \left( \pi+ \beta/2 \right) = \tan \beta/2$.
So,
$$
\begin{align}
\int_0^{2\pi} \frac{1}{1+ A \sin \theta + B \cos \theta } \mathrm{d} \theta &= \int_0^{2\pi} \frac{1}{1+r\sin(\beta + \theta) } \mathrm{d} \theta \\
&= 2 \int_{\tan \beta/2}^{\tan \beta/2} \frac{1}{1+2rz+z^2} \mathrm{d} \theta \\
&= 0
\end{align}
$$
For $A = B=0$, the answer is $2\pi$, as you've correctly found out.
|
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|
Absolute value distance Let $a$ and $b$ be in the interval $[0, 10]$, and let $a_0$ have a distance inferior to $10^{-3}$ from $a$ and $b_0$ a distance inferior to $10^{-3}$ from $b$. How close is $a_0 b_0$ from $ab$?
This has got me stuck. How can I verify this through inequalities?
|
Partial answer: Upper bound for $a_0b_0-ab.$
For $a\leq 10-10^{-3}$ and $b\leq 10-10^{-3}$ we have $$a_0b_0<(a+10^{-3})(b+10^{-3})=ab+10^{-3}(a+b)+10^{-6}\leq$$ $$\leq ab+10^{-3}(20-2\cdot 10^{-3})+10^{-6}=$$ $$=ab+10^{-3}(20-10^{-3}).$$ Observe that in this case $a_0b_0-ab$ can be arbitrarily close to $10^{-3}(20-10^{-3}).$
For $a\leq 10-10^{-3}$ and $10-10^{-3}<b\leq 10$ we have $$a_0b_0<(a+10^{-3})10=(a+10^{-3})(b+(10-b))=$$ $$= ab+a(10-b)+10^{-3}(10)< ab+a(10^{-3})+10^{-3}(10)\leq $$ $$\leq ab+(10-10^{-3})10^{-3}+10^{-3}(10)=$$ $$=ab+10^{-3}(20-10^{-3}).$$For $10-10^{-3}<a<10$ and $10-10^{-3}<b\leq 10$ we have $$a_0b_0\leq 10^2=(a+(10-a))(b+10-b)=$$ $$=ab +a(10-b)+b(10-a)+(10-a)(10-b)<$$ $$<ab+10(10^{-3})+10(10^{-3})+10^{-6}=ab+10^{-3}(20-10^{-3}).$$
|
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|
Cubic equations and relationship between their roots If there exists a cubic equation $x^3 + 2x^2 +3x + 1 = 0$ has the roots $a,b,c$
And it is given that
*
*$\frac{1}{a^3} + \frac{1}{b^3} - \frac{1}{c^3}$
*$\frac{1}{a^3} + \frac{1}{c^3} - \frac{1}{b^3}$
*$\frac{1}{c^3} + \frac{1}{b^3} - \frac{1}{a^3}$
Are the roots of another cubic
$px^3 + qx^2 + rx + s$
And then what will the value of
$p+q-12r+s$
be equal to?
my work so far
I initially tried establishing the relations of the sum of the roots taken 1,2 and 3 at a time and the coefficients that exists for any polynomial, I did this for the first and second equation but as I thought not much was simplified in the second equation which left me wondering if there was another shorter way, perhaps a trick or an observation that I am not able to make. I also tried expressing the cubic relation of $a,b, c$ in different ways by using the formulas for $(a+b+c)^3$ but that didn't result in much. I don't think I missed anything in the mentioned attempts that could be of significance but do feel free to check.
the answer, p+q-12r+s=5
I wasn't even able to get close to an answer so.
All help is greatly appreciated
|
I have to sleep, so this is all I have right now. Perhaps I'll update tomorrow? It's too long for a comment anyways.
So, if we assume it's monic, obviously $p$ will be 1. Now, by Vieta's formulas, $-s$ will be the sum of the roots. And the sum of the roots turns out to be:
$\frac{1}{a^3} + \frac{1}{b^3} + \frac{1}{c^3}$
We want to express this in elementary symmetric polynomials, so I used Mathematica's SymmetricReduction function and got that it is equivalent to:
$$-\frac{3 a^2 b^2 c^2+(a b+a c+b c)^3-3 a b c (a+b+c) (a b+a c+b c)}{(abc)^3}$$
Use Vieta's formulas to get this equal to 12. So $s = 12$.
|
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|
Find the value of $\sin \dfrac {\pi}{10} - \sin \dfrac {3\pi}{10}$ Find the value of $\sin \dfrac {\pi}{10} - \sin \dfrac {3\pi}{10}$
My Attempt:
$$\sin \dfrac {\pi}{10} - \sin \dfrac {3\pi}{10}$$
$$\sin \dfrac {180}{10} - \sin \dfrac {3\times 180}{10}$$
$$\sin 18^\circ - \sin 54^\circ$$
Now,
Let $A=18^\circ$.
$$5A=90^\circ$$
$$2A+3A=90^\circ$$
$$3A=90^\circ - 2A$$
Taking 'sin' on the both sides,
$$\sin 3A=\sin (90^\circ - 2A)$$
$$3\sin A-4\sin^3 A=\cos 2A$$
What should I do further?
|
$$\sin \dfrac {\pi}{10} - \sin \dfrac {3\pi}{10}=\cos72^{\circ}+\cos144^{\circ}=$$
$$=\frac{2\sin36^{\circ}\cos72^{\circ}+2\sin36^{\circ}\cos144^{\circ}}{2\sin36^{\circ}}=$$
$$=\frac{\sin108^{\circ}-\sin36^{\circ}+\sin180^{\circ}-\sin108^{\circ}}{2\sin36^{\circ}}=-\frac{1}{2}.$$
Done!
|
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|
Prove inequality $\sum\frac{{{a^3} - {b^3}}}{{{{\left( {a - b} \right)}^3}}}\ge \frac{9}{4}$ Given $a,b,c$ are positive number. Prove that $$\frac{a^3-b^3}{\left(a-b\right)^3}+\frac{b^3-c^3}{\left(b-c\right)^3}+\frac{c^3-a^3}{\left(c-a\right)^3}\ge \frac{9}{4}$$
$$\Leftrightarrow \sum \frac{3(a+b)^2+(a-b)^2}{(a-b)^2} \ge 9$$
$$\Leftrightarrow \frac{(a+b)^2}{(a-b)^2}+\frac{(b+c)^2}{(b-c)^2}+ \frac{(c+a)^2}{(c-a)^2}\ge 2$$
Which $$\frac{a+b}{a-b}.\frac{b+c}{b-c}+\frac{b+c}{b-c}.\frac{c+a}{c-a}+\frac{c+a}{c-a}.\frac{a+b}{a-b} =-1$$
Use $x^2+y^2+z^2 \ge -2(xy+yz+zx)$ then inequality right
I don't know why use $x^2+y^2+z^2 \ge -2(xy+yz+zx)$ then inequality right?
P/s: Sorry i knowed, let $x=\frac{a+b}{a-b}$
We have $(x+1)(y+1)(z+1)=(x-1)(y-1)(z-1) => xy+yz+zx=-1 $
|
$$a^2-2ab+b^2\le a^2+ab+b^2$$
$$1 \le \frac{a^2+ab+b^2}{(a-b)^2}$$$$1\le\frac{a^3-b^3}{(a-b)^3}$$$$\frac{9}{4}<3\le\sum_{cyc}\frac{a^3-b^3}{(a-b)^3}$$
|
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|
solve for $x$ and $y$ in the following equation $x^2 +y^2 -3=3xy$ and $2x^2 +y^2 =6$ Solve for $x$ and $y$ in the following equations: $x^2 +y^2 -3=3xy$ and $2x^2 +y^2 =6$. I made $y^2$ the subject of the formula in eqn 2.
This gives $y^2 = 6 -2x^2$.
I substitute this into the first eqn.
This gives $x^2+3xy -3$. There's where am stuck.
Can someone pull me out.
|
Hint....From what you have obtained, $x^2+3xy=3$ make $y$ the subject and substitute this into the equation $2x^2+y^2=6$ and you will obtain a quadratic in $x^2$
|
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|
Find the rational canonical form of a matrix from its minimal and characteristic polynomials
What is the rational canonical form of $A$?
$$A=\begin{bmatrix}
1 & 1 & 0 & 0 \\
0 & 1 & 0 & 0 \\
2 & 3 & -1 & 4\\
1 & 1 & -1 & 3\\
\end{bmatrix}$$
I found that the minimal polynomial $m_A(x)=(x-1)^2$ and the characteristic polynomial $c_A(x)=(x-1)^4$. Therefore the invariant factors can be
$$x-1,x-1,(x-1)^2$$
or
$$(x-1)^2,(x-1)^2$$
Therefore the rational canonical form may be
$$\begin{bmatrix}
1 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 \\
0 & 0 & 0 & -1\\
0 & 0 & 1 & 2\\
\end{bmatrix}$$
or
$$\begin{bmatrix}
0 & -1 & 0 & 0 \\
1 & 2 & 0 & 0 \\
0 & 0 & 0 & -1\\
0 & 0 & 1 & 2\\
\end{bmatrix}$$
How do I quickly figure out which one is the correct one?
|
If the first matrix is the rational form of $A$, we should have $\dim \ker (A - I) = 3$ (because this is true for the rational form and so it should be true for $A$ as well) while if the second matrix is the rational form of $A$, we should have $\dim \ker(A - I) = 2$. Just check which of those two options holds for $A$ by computing the rank of $A - I$.
|
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|
Finding $\int x \sqrt{1- x^2 \over 1 + x^2}dx$.
$$J =\int x \sqrt{1- x^2 \over 1 + x^2}dx$$
Substituting $u = \sqrt{1 + x^2}$
$$J = \int \sqrt{2 - u^2} du = \sqrt{2}\int \sqrt{1 - \left({u \over \sqrt{2}}\right)^2} du$$
Now substituting $\sin z = u/\sqrt{2}$
$$J = 2\int \cos^2 z dz = z + (\sin 2z)/2 + C = \arcsin\left(\sqrt{1+ x^2}\over 2\right) + {\sqrt{1-x^4}\over 2} + C $$.
The give answer is $\displaystyle \arcsin(x^2) + {\sqrt{1-x^4}\over 2} + C$.
What went wrong in my attempt ?
|
Note that $$\frac{d}{dx}\arcsin(x^2)=\frac{2x}{\sqrt{1-x^4}}$$ and $$\frac{d}{dx}2\arcsin\left(\frac{\sqrt{1+x^2}}{\sqrt 2}\right)=\frac{2x}{\sqrt{1-x^4}}$$
So, the error in the OP is in the term $\arcsin\left(\frac{\sqrt{1+x^2}}{2}\right)$, which should be instead $\arcsin\left(\frac{\sqrt{1+x^2}}{\sqrt 2}\right)$ and the "given answer" needs to have $\frac12 \arcsin(x^2)$ instead of $\arcsin(x^2)$.
The resolution to the apparent discrepancy comes from the identity
$$2\arcsin\left(\sqrt{1+x^2}\over\sqrt{2}\right) = \arcsin(x^2)+\pi/2\tag 1$$
which can be verified directly by taking the sine of both sides of $(1)$ and using $\arcsin(1)=\pi/2$. We have used instead that the two sides of $(1)$ have equal derivatives and are equal at $x=1$.
|
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|
Decide if the series $\sum\frac{4^{n+1}}{3^{n}-2}$ converges or diverges and, if it converges, find its sum
Decide if the series $$\sum_{n=1}^\infty\frac{4^{n+1}}{3^{n}-2}$$ converges or diverges and, if it converges, find its sum.
Is this how you would show divergence attempt:
For $n \in [1,\infty), a_n = \frac{4^{n+1}}{3^n -2} \geq 0$
For $n \in [1,\infty), a_n = \frac{4^{n+1}}{3^n-2} \geq \frac{4^{n+1}}{3^n} = b_n$
Since $\sum_{n=1}^{\infty} \frac{4^{n+1}}{3^n}$ is a geometric series with $r = \frac{4}{3} > 1$. Therefore it diverges by the geometric series test and by the comparison test $\sum a_n$ diverges too.
|
For $n\geq 1$ we have $3^n>3^n-2>0$ so $0<3^{-n}<(3^n-2)^{-1}$ so $4^{n+1}(3^n-2)^{-1}>4^{n+1}3^{-n}=4 (4/3)^n>4.$ If the terms of the series do not converge to $0$ then the sum cannot be convergent.
|
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|
How do I compute the taylor series for $(1+x+x^2)^\frac{1}{x}$ I tried rewriting $(1+x+x^2)^\frac{1}{x}$ as $e^{\frac{1}{x}\ln(1+x+x^2)}$ and then computing the taylor series of $\frac{1}{x}$ and $\ln(1+x+x^2)$ but I'm still not getting the correct answer..
|
This appears to be the approach you used, but without seeing your work, it is hard to tell where you were having trouble.
Use the series $\log(1+x)=x-\frac{x^2}2+\frac{x^3}3+\cdots$
$$
\begin{align}
\frac1x\log\left(1+x+x^2\right)
&=\frac1x\left(x+x^2-\frac{x^2+2x^3+x^4}2+\frac{x^3+3x^4+3x^5+x^6}3+\cdots\right)\\
&=1+\frac12x-\frac23x^2+\cdots
\end{align}
$$
Then use the series $e^x=1+x+\frac{x^2}2+\cdots$
$$
\begin{align}
\left(1+x+x^2\right)^{1/x}
&=e^{1+\frac12x-\frac23x^2+\cdots}\\
&=e\left(1+\left(\frac12x-\frac23x^2+\cdots\right)+\frac12\left(\frac12x-\frac23x^2+\cdots\right)^2+\cdots\right)\\
&=e\left(1+\frac12x-\frac{13}{24}x^2+\cdots\right)
\end{align}
$$
|
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|
What is $x$ in the formula? What is $x$ (assume it's integer) in $512p+ 1 = x^3$, where $p$ is a prime number.
Attempt:
$$a^3 - b^3 = (a-b) (a^2 + ab+b^2) \Longrightarrow 512p = x^3 -1 = (x-1)(x^2+x+1).$$
Here I got stuck. Am I suppose to plug in $p$ and try it one by one?
What about $16p + 1 = x^3$??
|
$512p + 1 = x^3$
$512p = x^3 -1$
$512p = (x-1)(x^2 + x + 1)$
$8^3p = (x-1)(x^2 + x + 1)$
If $x$ is even then $x^2 + x + 1$ is odd, and $x-1$ is odd and LHS is odd. That's impossible.
So $x$ is odd. So $x^2 + x + 1$ is odd. So $512|x-1$
$p = \frac {x-1}{512}(x^2 + x + 1)$.
If $x \le 0$ then $ \frac {x-1}{512}(x^2 + x + 1) < 0$ so that's a contradiction, so $x^2 + x + 1 > 1$
$p$ is prime. So it only has $1$ and $p$ as factors. So $x^2 + x + 1 = p$ or $x^2 + x+1=1$. The later is impossible so $\frac {x-1}{512} = 1$ and $x^2 + x+1 = p$... That is, if there is any answer at all.
So $x = 513$ and $p =513^2 + 513 +1$... if, such a number is prime. If not, there is no solution.
|
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|
Proving $\lim_{(x,y)\to (0,0)} (x^4+y^4)/(x^2+y^2)=0$ by definition I need to prove by definition (and nothing else) that
$$\lim_{(x,y) \to (0,0)}\frac{x^4+y^4}{x^2+y^2} = 0.$$
I've been stuck on this for almost an hour with no luck, and ran out of ideas.
Can anyone help or give a hint?
|
These things are always nicer in polar form...
$$\frac{x^4+y^4}{x^2+y^2} = \frac{r^4 (\sin^4 \theta + \cos^4 \theta)} {r^2} = r^2 (\sin^4 \theta + \cos^4 \theta) \leq 2r^2$$
Let's assume $\|(x,y)\| < \delta$, that is $(x,y)$ is inside the open disk around $0$ with a radius of $\delta$.
Then using the above inequality we can guarantee that
$$\left\| \frac{x^4+y^4}{x^2+y^2} \right\| \leq 2\delta^2,$$
which in turn means that the image of $(x,y)$ is inside the open disk around $0$ with a radius of $2\delta^2$.
Therefore if you get an $0 < \epsilon$, and you pick $\delta$, such that $2\delta^2 < \epsilon$, then
$$\|(x,y)\| < \delta \Rightarrow \left\| \frac{x^4+y^4}{x^2+y^2} \right\| \leq 2\delta^2 < \epsilon.$$
|
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|
Prove $\lfloor 2x \rfloor + \lfloor 2y \rfloor \geq \lfloor x \rfloor + \lfloor y\rfloor+\lfloor x+y\rfloor$
Prove that $\lfloor 2x \rfloor + \lfloor 2y \rfloor \geq \lfloor x \rfloor + \lfloor y \rfloor + \lfloor x+y \rfloor$ for all real $x$ and $y$.
If anybody could post a simple solution (no complicated abstract theories or calc :D) to the above question, I would greatly appreciate it. Thanks!
Also, if possible, the solution shouldn't wander too much away from floor, ceiling, fraction functions, and other related functions.
|
HINT
Let $x = n + r$ with $n$ an integer and $0 \le r <1$
Similarly, $y = m + s$ with $m$ an integer and $0 \le s < 1$
Now there are 4 cases to consider:
*
*$0 \le r < \frac{1}{2}$ and $0 \le s < \frac{1}{2}$
*$0 \le r < \frac{1}{2}$ and $\frac{1}{2} \le s <1$
*$\frac{1}{2} \le r < 1 $ and $0 \le s < \frac{1}{2}$
*$\frac{1}{2} \le r < 1$ and $\frac{1}{2} \le s <1$
Just to show case 2:
$\lfloor 2x \rfloor + \lfloor 2y \rfloor = \lfloor 2n + 2r \rfloor + \lfloor 2m + 2s \rfloor = 2n + 0 + 2m + 1$
$\lfloor x \rfloor + \lfloor y \rfloor + \lfloor x + y\rfloor = n + m + n + m + \lfloor s +r \rfloor$ where $\lfloor s + r \rfloor \le 1$
|
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|
Show that if $p$ and $q$ are primes $\equiv 3$(mod $4$) then at least one of the equations $px^{2}-qy^{2} = \pm 1$ is soluble in integers $x, y$. So far we have only talked about equations in the form $x^{2} - dy^{2} = \pm 1$ and I'm unsure how to handle the specific coefficients. We have normally found $\sqrt{d}$ and looked at the convergents of the continued fraction to find solutions. I was just wondering how to get started on this type of problem.
|
You have a primitive solution to $x^2-pqy^2=1$ ($x$, $y$ positive, $y$ minimal). Then $x$ is odd and $y$ is even (think modulo $4$). Also
$$x^2-1=(x+1)(x-1)=pqy^2$$
so that
$$\frac{x+1}2\frac{x-1}2=pq\left(\frac{y}{2}\right)^2.$$
As $(x\pm 1)/2$ are coprime integers then we have one of the following
*
*$(x+1)/2=u^2$, $(x-1)/2=pqv^2$
*$(x+1)/2=pu^2$, $(x-1)/2=qv^2$
*$(x+1)/2=qu^2$, $(x-1)/2=pv^2$
*$(x+1)/2=pqu^2$, $(x-1)/2=v^2$
We succeed with (2) and (3) (write $1=(x+1)/2-(x-1)/2$). We eliminate
(1) by minimality to solution of Pell and (4) gives $v^2-pqu^2=-1$
which is impossible modulo $p$.
|
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|
Prove by induction that $\forall n\geq 1,\ 7\mid 3^{2n+1} + 2^{n-1}$ Prove by induction that $$7 \mid 3^{2n+1} + 2^{n-1},\ \forall n\geq 1$$
Base case $n=1$:
$$3^{2 × 1+1} + 2^{1-1} = 28.$$
Induction:
$$P(k): 3^{2k+1} + 2^{k-1},\ P(k+1): 3^{2(k+1)+1} + 2^{(k+1)-1}.$$
$$3^{2k+3} + 2^k = 9 \times 3^{2k+1} + 2^{k-1} \times 2 = 7 \times 3^{2k+1} + 2 \times 3^{2k+1} + 2^{k-1} \times 2.$$
Where do I go from here?
|
$$P(k)=3^{2k+1}+2^{k-1}=7 \lambda (let)$$
Now, For $(k+1)$,
$$9*3^{2k+1}+2^{k-1}.2$$
Put $2^{k-1}$ form first equation i wrote
$$7.(3^{2k+1}+2\lambda)$$
Hope your doubt is solved?
|
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|
Showing $\frac{1}{x^3+x}$ is continuous at $x=1$ I've been attempting to use the definition of continuity and felt a little uncertain about my working on this question.
We define $\epsilon>0$ and choose $\delta $ s.t .....
Let $\lvert x-1\rvert < 1 $:
$\lvert \frac{1}{x^3+x}-\frac{1}{2} \rvert = \lvert \frac{2-x^3-x}{2x^3+2x}\rvert = \lvert (x-1)\rvert\lvert\frac{-x^2-x-2}{2x^3+2x}\rvert$
Then we restrict $\delta < 1$ so now $ 0<x<2$ which tells us that
$\lvert\frac{-x^2-x-2}{2x^3+2x}\rvert<-\frac{2}{5}$
Hence $\lvert (x-1)\rvert\lvert\frac{-x^2-x-2}{2x^3+2x}\rvert < -\frac{2\delta}{5}<\epsilon$.
So we must choose $\delta > -\frac{2\epsilon}{5}$.
However I am used to finding something of the form $0<\delta<$min{$a,b$}
Does this mean the solution is $-\frac{2\epsilon}{5}<\delta<1$ ?
Apologies if this is really trivial/simple, I have just never seen a solution like that so wondered if perhaps I made an error somewhere.
|
Note that we have
$$\left|\frac{1}{x(x^2+1)}-\frac12\right|=\left|\frac{x^2+x+2}{2x(x^2+1)}\right|\,|x-1|$$
Now, we need to bound $x$ away from $0$. So, let us first restrict $x$ so that $1/2<x<3/2$. Then, we see that
$$\left|\frac{x^2+x+2}{2x(x^2+1)}\right|\le \frac{(3/2)^2+(3/2)+2}{((1/2)^2+1)}=\frac{23}{5}$$
Hence, we see that
$$\left|\frac{1}{x(x^2+1)}-\frac12\right|\le \frac{23}5|x-1|<\epsilon$$
whenever $|x-1|<\delta=\min\left(1/2,\frac{5}{23}\epsilon\right)$.
|
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|
why$2 (-1 + 2^{1 + n})$ is the answer to the recurrence relation $a_{n}=2a_{n-1}+2$? $a_{0}=2$
$a_{1}=2(2)+2$
$a_{2}=2(2(2)+2)+2$
$a_{3}=2(2(2(2)+2)+2)+2$
$a_{4}=2(2(2(2(2)+2)+2)+2)+2$
$a_{5}=2(2(2(2(2(2)+2)+2)+2)+2)+2$
To simplifiy
$a_{6}=2^{6}+2^{5}...2^{1}$
so my answer is
$a_{n}=2^{n+1}+2^{n}+...2^{1}$
The correct answer is $2 (-1 + 2^{1 + n})$
How do I make this transition?
|
Let $a_n=2^nb_n$. Then
$$
a_{n}=2a_{n-1}+2\\\Downarrow\\2^nb_n=2^nb_{n-1}+2\\\Downarrow\\b_n=b_{n-1}+2^{1-n}
$$
Therefore,
$$
\begin{align}
b_n
&=b_0+\sum_{k=0}^{n-1}2^{-k}\\
&=b_0+\frac{1-2^{-n}}{1-2^{-1}}\\[9pt]
&=b_0+2-2^{1-n}
\end{align}
$$
and
$$
2^{-n}a_n=a_0+2-2^{1-n}\\
\Downarrow\\
\begin{align}
a_n
&=2^na_0+2^{n+1}-2\\[9pt]
&=2^{n+2}-2
\end{align}
$$
|
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|
Stars and Bars with a maximum number per tuple I have 4-tuples of numbers, each consisting of 1 to 10. How do I find the number of permutations possible to reach a total sum of 21? I tried using the formula given in https://en.wikipedia.org/wiki/Binomial_coefficient and plotted it against a graph, problem is, my graph should look like an n-shaped graph, but instead, I got an exponential graph instead.
|
An approach using generating functions:
In order to count up the number of solutions that sum to $n$, we can look at the coefficient of $x^n$ in the generating function $f(x)$:
$$[x^{n}]f(x) = [x^{n}]\left(\sum_{n=1}^{10}x^n\right)^4$$
The $[x^n]$ operator refers to the coefficient of $x^n$.
Use the identity $\displaystyle \sum_{n=0}^{k}x^n = \frac{1-x^{k+1}}{1-x}$ to convert this expression:
$$[x^{n}]f(x) = [x^{n}]\left(\frac{1-x^{11}}{1-x} - x^0\right)^4 = [x^{n}]\left(\frac{x-x^{11}}{1-x}\right)^4$$
Simplify:
$$[x^{n}]f(x) = [x^{n}]\left(\frac{x^{44} - 4 x^{34} + 6 x^{24} - 4 x^{14} + x^4}{(1-x)^4}\right)$$
Factor out $\dfrac{1}{(1-x)^4}$ and use the identity $\displaystyle \frac{1}{(1-x)^{m+1}} = \sum_{n=0}^{\infty}\binom{n+m}{m}x^n$:
$$[x^{n}]f(x) = [x^{n}]\left(x^{44} - 4 x^{34} + 6 x^{24} - 4 x^{14} + x^4\right)\sum_{n=0}^{\infty}\binom{n+3}{3}x^n$$
Distribute:
$$[x^{n}]f(x) = [x^{n}]\left(\sum_{n=0}^{\infty}\binom{n+3}{3}x^{n+44}-4\sum_{n=0}^{\infty}\binom{n+3}{3}x^{n+34}+6\sum_{n=0}^{\infty}\binom{n+3}{3}x^{n+24}-4\sum_{n=0}^{\infty}\binom{n+3}{3}x^{n+14} + \sum_{n=0}^{\infty}\binom{n+3}{3}x^{n+4}\right)$$
Shift indices:
$$[x^{n}]f(x) = [x^{n}]\left(\sum_{n=44}^{\infty}\binom{n-41}{3}x^{n}-4\sum_{n=34}^{\infty}\binom{n-31}{3}x^{n}+6\sum_{n=24}^{\infty}\binom{n-21}{3}x^{n}-4\sum_{n=14}^{\infty}\binom{n-11}{3}x^{n} + \sum_{n=4}^{\infty}\binom{n-1}{3}x^{n}\right)$$
If we look at the lower bounds of the indices, the smallest $x^n$ that we could get the coefficient for is $x^{4}$. This is because $4$ is the smallest possible value of $n$ under the $x_i$ constraints, which occurs when each value of the tuple is $1$. The coefficients for all $n<4$ are simply $0$.
Now we can invoke the coefficient operator:
$$[x^n]f(x) = \binom{n-41}{3}-4\binom{n-31}{3}+6\binom{n-21}{3}-4\binom{n-11}{3}+\binom{n-1}{3}$$
Set $n=21$ to get the number of solutions to the original problem:
$$[x^{21}]f(x) = 660$$
|
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|
Remainder when dividing power series I know how long division of polynomials works. For example I can use it to divide $x^2+1$ into $x$ to get
$$x=(x^2+1)\frac{1}{x}-\frac{1}{x}$$
with remainder $R=-1/x$, so that
$$\frac{x}{x^2+1}=\frac{1}{x}-\frac{1/x}{x^2+1}.$$
My question is, suppose $f$ and $g$ have infinite power series. I have seen people perform long division to compute $f/g$ up to the term, say, $x^4$, and what they say is that, up to that term, $\frac{f}{g}$ is whatever the quotient is up to that order. But what about the remainder? Is that zero?
|
No, the remainder is not $0$, but another power series. For example,
$$
\begin{align}
\frac{\sin(x)}{\cos(x)}
&=\frac{x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+O\!\left(x^9\right)}{1-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}+O\!\left(x^8\right)}\\[6pt]
&=x+\frac{x^3}{3}+\frac{2x^5}{15}+\frac{17x^7}{315}+O\!\left(x^9\right)
\end{align}
$$
where the latter $O\!\left(x^9\right)$ is the series for
$$
\tan(x)-\left(x+\frac{x^3}{3}+\frac{2x^5}{15}+\frac{17x^7}{315}\right)
$$
However, in the terminology for long division, the remainder would be
$$
\sin(x)-\cos(x)\left(x+\frac{x^3}{3}+\frac{2x^5}{15}+\frac{17x^7}{315}\right)
$$
which is another power series.
However, when dealing with power series, it is often useful, and usually simpler, to use Landau "big-O" notation.
|
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|
Prove that $T$ is a Linear Transformation , if $T$ is defined by $T(A)=XA-AX$ Let $T : M_{2\times 2} \to M_{2\times 2}$ be defined by
$$
T(A) =
\begin{pmatrix} 2 & 1 \\ 1 & 2 \end{pmatrix} A
− A \begin{pmatrix} 2 & 1 \\ 1 & 2 \end{pmatrix}.
$$
How do I prove that $T$ is a linear transformation and then find the basis for the kernel of $T$. The $2\times 2$ matrix multiplied by $A$ confuses me.
|
Linearity is easy to prove.
Now let $A=\begin{pmatrix} a & b\\ c & d \end{pmatrix}$ then $\begin{pmatrix} a & b \\ c & d \end{pmatrix} \begin{pmatrix} 2 & 1 \\ 1 &2\end{pmatrix}=\begin{pmatrix} 2a+b & a+2b \\ 2c+d & c+2d \end{pmatrix}=D_1$
And $\begin{pmatrix} 2 & 1 \\ 1 & 2 \end{pmatrix}A=\begin{pmatrix} 2a+c & 2b+d \\ a+2c & b+2d \end{pmatrix}=D_2$
Thus $T(A)=D_1-D_2=\begin{pmatrix} b-c & a-d \\ d-a &c-b \end{pmatrix}$
Now let $T(A)=\mathbb{O}_{2\times2}$ then we have that $b=c$ and $a=d$ thus $A \in ker(T)$ iff $A=\begin{pmatrix} a & b \\ b & a \end{pmatrix}$ for $a,b \in \mathbb{R}$.
Thus $Ker(T)=\{a\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}+b\begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}|a,b \in \mathbb{R}\}$ thus $Ker(T)=span(\{\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix},\begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}\})$
And you can easily see that $\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$ and $\begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}$ are linearly independent.
|
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|
Floors Complicated Proof Problem:
Prove that $\lfloor 2x \rfloor + \lfloor 2y \rfloor \geq \lfloor x \rfloor + \lfloor y \rfloor + \lfloor x+y \rfloor$ for all real $x$ and $y$.
I have my proof down below. Its a little complicated though.
This is not a duplicate of the same question on another page on math.stackexchange.com because I provide my answer.
|
There is a simple solution to this, and it involves a little casework.
Case 1: The fractional parts of $x$ and $y$ are both less than $0.5$. Then
$$\lfloor 2x\rfloor=2\lfloor x\rfloor$$
$$\lfloor 2y\rfloor=2\lfloor y\rfloor$$
$$\lfloor x+y\rfloor=\lfloor x\rfloor + \lfloor y\rfloor$$
and the two sides of the inequality are equal, making it trivially true.
Case 2: The fractional parts of $x$ and $y$ are both greater than $0.5$. Then
$$\lfloor 2x\rfloor=2\lfloor x\rfloor+1$$
$$\lfloor 2y\rfloor=2\lfloor y\rfloor+1$$
$$\lfloor x+y\rfloor=\lfloor x\rfloor+\lfloor y\rfloor+1$$
And the two sides of the equation look like
$$2\lfloor x\rfloor+2\lfloor y\rfloor+2\ge2\lfloor x\rfloor+2\lfloor y\rfloor+1$$
Which is always true.
Case 3: The fractional part of $x$ is less than $0.5$ and the fractional part of $y$ is greater than $0.5$ (or vice versa, without loss of generality). Then
$$\lfloor 2x\rfloor=2\lfloor x\rfloor$$
$$\lfloor 2y\rfloor=2\lfloor y\rfloor+1$$
And the quantity $\lfloor x+y\rfloor$ is either $0$ or $1$ more than the quantity $\lfloor x\rfloor+\lfloor y\rfloor$, and we have one of these two inequalities:
$$2\lfloor x\rfloor+2\lfloor y\rfloor+2\ge2\lfloor x\rfloor+2\lfloor y\rfloor+1$$
or
$$2\lfloor x\rfloor+2\lfloor y\rfloor+2\ge2\lfloor x\rfloor+2\lfloor y\rfloor$$
Both of which are always true.
This is true for all cases, so it is always true.
|
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|
On the differentiability of $\frac{\sin(x^2)+\cos(y^2)}{\sqrt{x^2+y^2}}$ Does this function $f:\mathbb{R^2}\to\mathbb{R}$ defined by:
$$\frac{\sin(x^2)+\cos(y^2)}{\sqrt{x^2+y^2}}$$
differentiable at the origin? Thank you
|
The function
$$
f(x) = \frac{\sin \left(x^2\right)+\cos \left(y^2\right)}{\sqrt{x^2+y^2}}
$$
is not differentiable at the origin.
The iterated limit demonstrates the problem.
$$
\begin{align}
\lim_{\color{red}{x}\to 0} f(x,y) &= \frac{\cos \left(y^2\right)}{\sqrt{y^2}} \\[3pt]
\lim_{\color{blue}{y}\to 0} \left( \lim_{\color{red}{x}\to 0} f(x,y) \right) &= \infty \\
\end{align}
$$
$$
\begin{align}
\lim_{\color{blue}{y}\to 0} f(x,y) &= \frac{\sin \left(x^2\right)+1}{\sqrt{x^2}} \\[3pt]
\lim_{\color{red}{x}\to 0} \left( \lim_{\color{blue}{y}\to 0} f(x,y) \right) &= \infty \\
\end{align}
$$
The function is growing rapidly near the origin.
Hint: Maclaurin series
$$
\sin x^{2} = x^{2} - \frac{x^{2}}{6} + \mathcal{O}\left( x^{8} \right)
$$
|
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|
Solve the Differential equation $x^3 \frac{dy}{dx}=y^3+y^2\sqrt{x^2+y^2}$ Solve the Differential equation $$x^3 \frac{dy}{dx}=y^3+y^2\sqrt{x^2+y^2}$$
i reduced the equation as
$$x^3\frac{dy}{dx}=y^3\left(1+\sqrt{1+\left(\frac{x}{y}\right)^2}\right)$$ $\implies$
$$\frac{x^3}{y^3}\frac{dy}{dx}=\left(1+\sqrt{1+\left(\frac{x}{y}\right)^2}\right) \tag{1}$$ Next put $$\frac{x}{y}=v$$ we get
$$ x=vy$$ then
$$ \frac{dx}{dy}=v+y\frac{dv}{dy}$$ Then $(1)$ becomes
$$\frac{v^3}{v+y\frac{dv}{dy}}=1+\sqrt{1+v^2}$$ Reciprocating we get
$$\frac{v+y\frac{dv}{dy}}{v^3}=\frac{1}{\sqrt{1+v^2}+1}$$ Rationalizing RHS we get
$$\frac{v+y\frac{dv}{dy}}{v^3}=\frac{\sqrt{1+v^2}-1}{v^2}$$ Rearranging we get
$$\frac{dv}{v \times \left(\sqrt{1+v^2}-2\right)}=\frac{dy}{y}$$
EDIT: i am posting here the clue given by paul using Substitution $v=\tan z$:
$$\int\frac{dv}{v \times \left(\sqrt{1+v^2}-2\right)}=\int\frac{\sec^2 z\: dz}{\tan z(\sec z-2)}=\int\frac{dz}{\sin z(1-2\cos z)}$$ So
$$\int\frac{dz}{\sin z(1-2\cos z)}=\int \frac{\sin z\: dz}{\sin^2 z(1-2\cos z)}$$
Put $\cos z=t$ and use partial fractions
|
The solution of the ODE is presented below on parametric form. The explicit form can be obtained but involves the roots of a cubic polynomial equation which would lead to a big formula.
|
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|
Measure of a subset of $\mathbb{R}^2$ I'd like to know whether the following integral is right or wrong. I have to calculate the measure of $A$, where
$$
A=\left \{ (x,y): \frac{x^2}{3}+y^2\leqslant 1, \, x\geqslant 0, \, -x\leqslant y\leqslant x \right \}
$$
using polar coordinates $\Phi:\left \{ x= \sqrt{3}\rho \cos \theta, y= \rho \sin \theta \right \}$, the polar region is
$B=\left \{ (\rho, \theta): 0 \leqslant \rho \leqslant 1 , -\frac{\pi}{4} \leqslant \theta \leqslant \frac{\pi}{4} \right \}$, $A=\Phi (B)$ and $|\det J_\Phi|=\sqrt{3} \rho$. Then,
$$
m(A)=2\int_{0}^{\frac{\pi}{4}}\int_{0}^{1}\sqrt{3}\rho \, d\rho d\theta=\int_{0}^{\frac{\pi}{4}}\sqrt{3} \, d \theta=\frac{\sqrt{3}}{4}\pi
$$
The answer would be $\frac{\pi}{\sqrt{3}}$ but I think it's incorrect.
|
I don't believe $B$ is correctly defined. $\theta$ should run between $-\pi/3$ and $\pi/3$. Notice that for $\theta=\pi/3$, $x=\frac{\sqrt{3}}{2}\rho$ and $y=\frac{\sqrt{3}}{2}\rho$.
Here is another approach to verify the answer. You want to intersect two regions
1)the interior of the ellipse $x^2/3+y^2=1$
2)the cone about the positive $x$-axis bounded by $y=x$ and $y=-x$
The region is symmetric about the $x$-axis, so we can focus on the first quadrant.
Considering $y$ as the independent variable, the ``top curve'' is the ellipse
$x=\sqrt{3-3y^2}$ and the bottom curve is $x=y$.
Therefore the area is
\begin{align}
m(A)
&=
2\int_{0}^{\sqrt{3}/2} \sqrt{3-3y^2} - y\ dy\\
&=
2\sqrt{3}\int_{0}^{\sqrt{3}/2} \sqrt{1-y^2}\ dy - 2\int_{0}^{\sqrt{3}/2} y\ dy\\
&=
2\sqrt{3}\int_{0}^{\sqrt{3}/2} \sqrt{1-y^2}\ dy - \frac{3}{4}
\end{align}
After a trig substitution $y=\sin(t)$, we have
\begin{align}
m(A)
&=
2\sqrt{3}\int_{0}^{\pi/3} \sqrt{1-\sin^2(t)}\cos(t)\ dt - \frac{3}{4}
\\
&=
2\sqrt{3}\int_{0}^{\pi/3} \cos^2(t)dt- \frac{3}{4}\\
&=
\left(\frac{\pi}{\sqrt{3}} + \frac{3}{4} \right) -\frac{3}{4}\\
&=
\frac{\pi}{\sqrt{3}}
\end{align}
|
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|
Equation of common tangent Equation of common tangent of circle $x^2 + y^2 -8x =0$ and hyperbola with $x^2/9 - y^2/4 =1.$
I tried this but was left with lengthy solutions can u please explain with easy solutions.
|
Sketch (this answer is to find common tangent lines between the two objects). If you just want the intersections, substitute $y^2$ from the second equation into the first equation:
*
*First, observe that by implicit differentiation on the hyperbola's equation, we get
$$
\frac{2}{9}x-\frac{1}{2}y\frac{dy}{dx}=0.
$$
Therefore,
$$
\frac{dy}{dx}=\frac{4x}{9y}
$$
Observe that the tangent line is a vertical line when $y=0$; in this case, $x=\pm 3$.
*On the other hand, completing the square on the circle, we get
$$
(x-4)^2+y^2=16
$$
The vertical tangents for the circle are at $y=0$ and $x=0,8$, so they don't agree with the vertical tangents for the hyperbola, so the circle and hyperbola don't have common vertical tangents. Throughout the rest of this answer, we assume the tangent line isn't vertical.
*Suppose that $y=mx+b$ is tangent to the circle. Then when we substitute $y=mx+b$ into the equation for a circle, we get
$$
x^2-8x+m^2x^2+2mbx+b^2=0.
$$
For this line to be tangent, the discriminant of this quadratic must be zero. (This happens because otherwise, the line intersects the circle in two positions instead of one point, twice). In other words,
$$
(2mb-8)^2-4(1+m^2)b^2=0.
$$
Simplifying a bit, we get
$$
-32mb+64-b^2=0.
$$
*If we also try this with the hyperbola, we get
$$
x^2/9-m^2x^2/4-2mbx/4-b^2/4=1
$$
Clearing fractions, we get
$$
4x^2-9m^2x^2-18mbx-9b^2-36=0.
$$
Again, to be tangent, the discriminant must be zero so
$$
(-18mb)^2-4(4-9m^2)(-36-9b^2)=0
$$
Simplifying a bit, we get
$$
576+144b^2-1296m^2=0.
$$
*At this point, we only need to solve the simultaneous system of equations
\begin{align*}
576+144b^2-1296m^2&=0\\
-32mb+64-b^2&=0
\end{align*}
We can make this a little more explicit by writing
$$
32mb=64-b^2
$$
and squaring both sides to get
$$
1024m^2b^2=4096-128b^2+b^4.
$$
Substituting
$$
\frac{4}{9}+\frac{1}{9}b^2=m^2
$$
we get
$$
1024\left(\frac{4}{9}+\frac{1}{9}b^2\right)b^2=4096-128b^2+b^4
$$
This is quadratic in $b^2$, so let $c=b^2$, use the quadratic formula and work backwards from there.
Please do check the calculations (and make sure to throw out complex solutions).
|
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|
$A$ and $B$ are the positive acute angles satisfying the equations $3\cos^2 A + 2\cos^2 B=4$ $A$ and $B$ are the positive acute angles satisfying the equations $3\cos^2 A + 2\cos^2 B=4$ and $\dfrac {3\sin A}{\sin B}=\dfrac {2\cos B}{\cos A}$. Then $A+2B$ is equal to:
$1$. $\dfrac {\pi}{4}$
$2$. $\dfrac {\pi}{3}$
$3$. $\dfrac {\pi}{6}$
$4$. $\dfrac {\pi}{2}$.
My Attempt
$$\dfrac {3\sin A}{\sin B}=\dfrac {2\cos B}{\cos A}$$
$$3\sin A.\cos A= 2\cos B.\sin B$$
$$\dfrac {3}{2} \sin 2A=2\sin 2B$$.
How do I proceed further?
|
The first condition gives $3\cos2A+2\cos2B=3$.
The second condition gives $3\sin2A=2\sin2B$ or
$$9\sin^22A=4\sin^22B$$ or
$$9(1-\cos^22A)=4(1-\cos^22B)$$ or
$$9\cos^22A-4\cos^22B=5$$ or
$$(3\cos2A+2\cos2B)(3\cos2A-2\cos2B)=5$$ or
$$3(3\cos2A-2\cos2B)=5$$ or
$$3\cos2A-2\cos2B=\frac{5}{3},$$
which after summing with $$3\cos2A+2\cos2B=3$$ gives $6\cos2A=\frac{14}{3}$,
which says $\cos2A=\frac{7}{9}$ and $\cos2B=\frac{1}{3}$.
Thus,
$$\sin(A+2B)=\sin{A}\cos2B+\cos{A}\sin2B=$$
$$=\sqrt{\frac{1-\frac{7}{9}}{2}}\cdot\frac{1}{3}+\sqrt{\frac{1+\frac{7}{9}}{2}}\cdot\sqrt{1-\frac{1}{9}}=1,$$
which gives $A+2B=\frac{\pi}{2}$
|
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|
Exercise 5.3 in Calculus Made Easy: are these answers equivalent? Exercise 5.3 in Calculus Made Easy, by Silvanus Thompson, is to find $\mathrm d\over \mathrm dx$ when the following relationship holds ($a$ and $b$ are both constants):
$$ay + bx = by - ax + (x + y)\sqrt{a^2 - b^2}$$
I tried differentiating both sides of this equation with respect to $x$ as follows:
$$\begin{align}
{\mathrm d\over \mathrm dx}(ay + bx) &= {\mathrm d\over \mathrm dx}\left(by - ax + (x + y)\sqrt{a^2 - b^2}\right)\\
a{\mathrm dy\over \mathrm dx} + b &= b{\mathrm dy\over \mathrm dx} - a + \left(1 + {dy\over dx}\right)\sqrt{a^2 - b^2}\\
\left(a - b - \sqrt{a^2 - b^2}\right){\mathrm dy\over \mathrm dx} &= \sqrt{a^2 - b^2} - a - b\\
{\mathrm dy\over \mathrm dx} &= {\sqrt{a^2 - b^2} - a - b \over a - b - \sqrt{a^2 - b^2}}
\end{align}$$
The author has considerably more finess and first does some algebraic manipulation to get the equation into better shape. First he squares both sides, then pushes some things around, then takes the square root of both sides:
$$\begin{align}
(a - b)^2y^2 + (a + b)^2x^2 + 2(a + b)(a - b)xy &= (x^2 + y^2 + 2xy)(a^2 - b^2)\\
\left[(a - b)^2 - \left(a^2 - b^2\right)\right]y^2 &= \left[\left(a^2 - b^2\right) - (a + b)^2\right]x^2\\
2b(b - a)y^2 &= -2b(b + a)x^2\\
y &= \sqrt{a + b \over a - b}x\\
{\mathrm dy\over \mathrm dx} &= \sqrt{a + b \over a - b}
\end{align}$$
I believe that both answers are correct, but if that is true, it should be possible to convert one to the other algebraically. For the life of me, I can't see how to do this. So my question is: are these two answers really equivalent?
|
$${\sqrt{a^2 - b^2} - a - b \over a - b - \sqrt{a^2 - b^2}}=\frac{\sqrt{a+b}\left(\sqrt{a-b}-\sqrt{a+b}\right)}{\sqrt{a-b}\left(\sqrt{a-b}-\sqrt{a+b}\right)}=\frac{\sqrt{a+b}}{\sqrt{a-b}}$$
|
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|
calculate polynomial equation from roots I want to derive a polynomial equation from the following roots:
$$\left\{-1,1+\sqrt5,1-\sqrt5\right\}$$
Below is my attempt.
\begin{align}f(x)&=(x+1)( (x-1) - \sqrt5 ) ( (x-1) + \sqrt5 )\\
&=(x+1)( (x-1)^2 - 5)\\
&=(x+1)(x^2 - 2x - 4)\end{align}
And the answer in the book is $(x+1)(x^2 + 2x -4)$. I'm not sure what's correct?
I'm taking a Math correspondence. The book has very little explanation and few errors. My hunch is that, the answer is wrong. Please advice.
EDIT
The first root is $-1$ instead of $1$. Sorry about typo.
|
When we have the roots $\{a,b,c\}$, we can write $$f(x)=(x-a)(x-b)(x-c)$$We have the roots $\{-1,1+\sqrt 5,1−\sqrt5\}$ so we can write the equation in the following way:
\begin{align}f(x)&=(x-(-1))(x-(1+\sqrt5))(x-(1-\sqrt5))\\
&=(x+1)(x-1-\sqrt5)(x-1+\sqrt5)\\
&=(x+1)(x^2-x+x\sqrt5-x+1-\sqrt5-x\sqrt5+\sqrt5-5)\\
&=(x+1)(x^2-(1-\sqrt5+1+\sqrt5)x+(1+-\sqrt5+\sqrt5-5))\\
&=(x+1)(x^2-2x-4)\\
&=x^3-2x^2-4x+x^2-2x-4\\
&=x^3-x^2-6x-4\end{align}
We can verify this with WolframAlpha and see that it is correct
|
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|
How to prove this property involving the floor function? How does one prove that $$\lfloor x \rfloor = \lfloor x/2 \rfloor + \lfloor (x+1)/2 \rfloor $$
where $\lfloor x \rfloor$ represents the greatest integer less than or equal to $x$?
I tried to do this by using inequalities but end up with $\lfloor x \rfloor + 1/2$ on the lhs. Can someone help?
|
Let $x = a + b$ where $a \in \mathbb{N}$ and $0 \leqslant b < 1$
We need to prove that
$$\lfloor a + b \rfloor = \left\lfloor \frac{a+b}{2} \right\rfloor + \left\lfloor \frac{a+b +1}{2} \right\rfloor $$
The L.H.S. will be just $a$
Now the R.H.S.
$$\left\lfloor \frac{a+b}{2} \right\rfloor + \left\lfloor \frac{a+b +1}{2} \right\rfloor = \left\lfloor \frac{a}{2} \right\rfloor + \left\lfloor \frac{a+1}{2} \right\rfloor $$
If $a$ is even,
$$\left\lfloor \frac{a}{2} \right\rfloor + \left\lfloor \frac{a+1}{2} \right\rfloor = \frac{a}{2} + \frac{a}{2} = a $$
If $a$ is odd,
$$\left\lfloor \frac{a}{2} \right\rfloor + \left\lfloor \frac{a+1}{2} \right\rfloor = \frac{a-1}{2} + \frac{a+1}{2} = a $$
|
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|
When is $n\cdot m\cdot [(n-8)\cdot m+4]-4$ an integer square? For which odd integers $n$ and $m$, with $n \ge m$ is $$n\cdot m\cdot [(m-8)\cdot n+4]-4$$ a square?
Part of my efforts so far:
The product of the three factors above is a sum of two squares (the square we wish to find plus $2^2$), thus all three factors are a sum of two squares too (since their prime factorization is of the right form).
But that is not enough, as their product minus 4 needs to be a square too.
So $n=17$ and $m=13$ look to be OK at first sight, as the third factor then is $89=8^2+5^2$, but the product $13\cdot 17\cdot 89$ is not a square plus 4.
On the other hand, $n=17$ and $m=17$ is OK, as the third factor now is $157=11^2+6^2$ and also $17\cdot 17 \cdot 157=213^2+2^2$.
Taking the sum of two squares to be the determinant of a $2x2$ matrix
$$\left(\begin{array}{cc}a&b\\-b&a \end{array}\right)$$
it is easy to see that $n$, $m$ and $(m-8)\cdot n+4$ then correspond to matrices whose product is a matrix as above with $b=\pm2$.
For example
$$\left(\begin{array}{cc}4&1\\-1&4 \end{array}\right)\cdot \left(\begin{array}{cc}4&1\\-1&4 \end{array}\right)\cdot \left(\begin{array}{cc}11&-6\\6&11 \end{array}\right)=\left(\begin{array}{cc}213&-2\\2&213 \end{array}\right)$$
Not sure if this helps finding the solution, though.
A couple of the solutions I found are
$$
\begin{array}{l|l|c|l}
\text{n} & \text{m} & (m-8)\cdot n+4 & \text{product} \\
\hline
17=4^2+1^2 & 17=4^2+1^2 & 157=11^2+6^2 & 45373=213^2+2^2 \\
29=5^2+2^2 & 13=3^2+2^2 & 149=10^2+7^2 & 56173=237^2+2^2 \\
3277=54^2+19^2 & 17=4^2+1^2 & 29497=171^2+16^2 & 1643248373=40537^2+2^2 \\
\end{array}
$$
What I would like to have is a general expression for all solutions, or an algorithm to find the next solution.
|
For every $m$ such that $-1$ is quadratic residue modulo $m$ and $m>5$ you will have infinitely many $n$ which satisfy that $n m ((m-8) n+4)-4$ is perfect square.
For Example: for $m=13$ we get that $n=\{29,125,1956797,8346845,130248348509,555582723389,8669590571525885,3698069722589
8397,577065287491657635485,2461509168194666404349,38410619657350124961653309,1
63842973216392687980406365,\cdots\}$
Note that the above sequence follow an explicit formula, so for any fixed $m$ which fulfill the conditions above, you can have a general formula, but it will be correct just for the fixed $m$, a general formula for $n,m$ without any one be fixed might be hard to come by if not impossible.
|
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|
Find the value of $A$ for the inequality if $x,y,z$ are positive reals and
$$S=\frac{x}{(x+y)(x+z)}+\frac{y}{(y+x)(y+z)}+\frac{z}{(z+x)(z+y)} \le A$$ find $A$
i have taken L.C.M getting
$$S=\frac{2(xy+yz+zx)}{(x+y)(y+z)(z+x)}$$ dividing by $xyz$ we get
$$S=\frac{2\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)}{xyz\left((\frac{1}{x}+\frac{1}{y})(\frac{1}{y}+\frac{1}{z})(\frac{1}{z}+\frac{1}{x})\right)}$$
Letting $a=\frac{1}{x}$, $b=\frac{1}{y}$ and $c=\frac{1}{z}$ we get
$$S=\frac{2abc(a+b+c)}{(a+b)(b+c)(c+a)}$$ $\implies$
$$S=\frac{abc(a+b+b+c+c+a)}{(a+b)(b+c)(c+a)}$$ $\implies$
$$S=abc \times \left(\frac{1}{(a+b)(b+c)}+\frac{1}{(b+c)(c+a)}+\frac{1}{((a+b)(c+a)}\right)$$
Now by AM-GM inequality
$$(a+b) \ge 2 \sqrt{ab}$$ and $(b+c) \ge 2 \sqrt{bc}$ so
$$(a+b)(b+c) \ge 4c\sqrt{ab}$$ so
$$S \le \frac{abc}{4c\sqrt{ab}}+\frac{abc}{4b\sqrt{ac}}+\frac{abc}{4a\sqrt{bc}}$$
so
$$S \le \frac{\sqrt{ab}+\sqrt{bc}+\sqrt{ca}}{4} \le \frac{a+b+c}{4}$$ by cauchy inequality
but how can we get $A$ from here
|
Without loss of generality, we can consider $x\le y \le z$ or $x \le ax \le abx$ with $a,b\ge1$.
The given inequality becomes equivalent to:
$$S=\frac{x}{x(a+1)\cdot x(ab+1)}+\frac{ax}{x(b+1) \cdot ax(b+1)}+\frac{abx}{x(ab+1) \cdot ax(b+1)} \le A$$
If $x\to0^+$, the LHS $\to+\infty=A$.
|
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|
Finding $\gcd(x^3, x^3+x+1)$ Suppose $f(x) = x^3$ and $g(x) = x^3+x+1$. I am trying to show that the $\gcd(f,g)= 1$ but am running into some trouble...
My attempt: (Division algorithm)
$x^3+x+1 = (1)(x^3)+(x+1)$
$x^3 = x^2(x+1)-x^2$
But I cannot go further than this, what is going wrong here?
|
Just keep going. You found that
$$
x^3+x+1 = (1)(x^3) + (x+1)
$$
Now observe that
$$
x^3 = (x^2-x+1)(x+1) - (1)
$$
What can you conclude?
|
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|
Show that the geometric images of any complex number defined by $cis \theta$ belong to a circumference of radius 1
Show that the geometric images of any complex number defined by $cis
\theta$ belong to the circumference centered at the origin with radius
1.
My book states that this is proven because $|cis \theta|= 1$ . I tried:
$$|cis \theta| = \sqrt{\cos^2\theta+\sin^2\theta\cdot i^2} = \sqrt{\cos^2\theta-\sin^2\theta} = \sqrt{(\cos(2\theta)} = ??$$
What do I do next?
|
The modulus or norm of any complex number $z = a + bi$, $a, b \in \Bbb R$, is given by
$\vert z \vert = \sqrt{a^2 + b^2}; \tag{1}$
using the identity
$x^2 - y^2 = (x + y)(x - y), \tag{2}$
which holds in any field (actually, in any commutative ring), and taking $x = a$, $y = bi$, and recalling that $i^2 = -1$, we find
$a^2 + b^2 = a^2 - (-b^2) = a^2 - (ib)^2 =$
$(a + ib)(a - ib) = z \bar z; \tag{3}$
this may also be written
$\vert z \vert ^2 = z \bar z. \tag{4}$
We apply the formula (4) to $cis \theta$:
$cis \theta = \cos \theta + i \sin \theta, \tag{5}$
whence
$\vert cis \theta \vert^2 = cis \theta \overline{cis \theta}= (\cos \theta + i \sin \theta)(\cos \theta - i \sin \theta) = \cos^2 \theta + \sin^2 \theta = 1, \tag{6}$
from which
$\vert cis \theta \vert= 1 \tag{7}$
readily follows.
|
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|
why $x^2-7y^2 = 11$ cannot have integer solutions? I have a quadratic equation $x^2-7y^2 = 11$, I understand it's not a pell equation, because $7/11$ is not an integer, but how can I prove it has no integer solutions?
|
If there were integers $x,y$ such that
$$ x^2 - 7y^2 = 11 $$
we see that it would follow that:
\begin{align*}
x^2 - 7y^2 &\equiv 11 \pmod{4} \\
x^2 + y^2 &\equiv 3 \pmod{4}
\end{align*}
We see that the congruence classes mod 4 are $\{0,1,2,3\}$. Hence, the quadratic residues are $\{0,1\}$. Since $x^2, y^2 \in \{0,1 \} \pmod{4}$, it follows that:
$$ x^2 + y^2 \not\equiv 3 \pmod{4} $$
By contradiction,
$$ \text{there exists no integer solutions to } x^2 - 7y^2 = 11$$
|
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|
To prove $\frac{1}{2} \times \frac{3}{4} \times \frac{5}{6} \cdots \frac{(2n-1)}{2n} \le \frac{1}{\sqrt{3n+1}}$ To prove $$P=\frac{1}{2} \times \frac{3}{4} \times \frac{5}{6} \cdots \frac{(2n-1)}{2n}\le \frac{1}{\sqrt{3n+1}}$$
i have written $P$ as
$$P=\frac{(2n)!}{2^{2n}(n!)^2}=\frac{(2n)!}{4^{n}(n!)^2}=\frac{\binom{2n}{n}}{4^n}$$
Now
$$P=\frac{\binom{2n}{n}}{(1+3)^n} \lt \frac{\binom{2n}{n}}{1+3n}$$ since
$$(1+3)^n=1+3n+\binom{n}{2}3^2+\cdots$$
Any help here..
|
We can use simple induction to find the answer. First, at $n=1$, the inequality holds true. Now let us assume the above for n and prove it for $n+1$.
Let us divide the LHS of the inequality for $n+1$ by LHS of the inequality for n, and do the same with the RHS for $n+1$ and $n$.
$(2n+1)/(2n+2)$ must be lesser than or equal to $\sqrt{(3n+1)}/\sqrt{(3n+3)}$. We can square on both sides.
$(4n^2+4n+1)/(4n^2+8n+4)$ must be lesser than or equal to $(3n+1)/(3n+3)$.
$(4n^2+4n+1)(3n+3)$ must be lesser than or equal to $(4n^2+8n+4)(3n+1)$. This can be easily verified by multiplying the polynomials. Hence, for all '$n$', the inequality holds
|
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|
If $\sec \theta + \tan \theta =x$, then find the value of $\sin \theta$. If $\sec \theta + \tan \theta =x$, then find the value of $\sin \theta$.
$$\sec \theta + \tan \theta = x$$
$$\dfrac {1}{\cos \theta }+\dfrac {\sin \theta }{\cos \theta }=x$$
$$\dfrac {1+\sin \theta }{\sqrt {1-\sin^2 \theta }}=x$$
$$1+\sin \theta =x\sqrt {1-\sin^2 \theta }$$
$$1+2\sin \theta + \sin^2 \theta = x^2-x^2 \sin^2 \theta $$
$$x^2 \sin^2 \theta + \sin^2 \theta + 2\sin \theta = x^2-1$$
$$\sin^2 \theta (x^2+1) + 2\sin \theta =x^2-1$$
|
Here is a different approach: Since $1 + \tan^2\theta = \sec^2\theta$, we have
$$\sec^2\theta - \tan^2\theta = 1$$
Factoring yields
$$(\sec\theta + \tan\theta)(\sec\theta - \tan\theta) = 1$$
Since we are given that $\sec\theta + \tan\theta = x$, we obtain
$$x(\sec\theta - \tan\theta) = 1$$
Therefore,
$$\sec\theta - \tan\theta = \frac{1}{x}$$
This yields the system of equations
\begin{align*}
\sec\theta + \tan\theta & = x \tag{1}\\
\sec\theta - \tan\theta & = \frac{1}{x} \tag{2}
\end{align*}
Adding equations 1 and 2 and solving for $\sec\theta$ yields
\begin{align*}
2\sec\theta & = x + \frac{1}{x}\\
2\sec\theta & = \frac{x^2 + 1}{x}\\
\sec\theta & = \frac{x^2 + 1}{2x}
\end{align*}
Therefore,
$$\cos\theta = \frac{1}{\sec\theta} = \frac{2x}{x^2 + 1}$$
Subtracting equation 2 from equation 1 and solving for $\tan\theta$ yields
\begin{align*}
2\tan\theta & = x - \frac{1}{x}\\
2\tan\theta & = \frac{x^2 - 1}{x}\\
\tan\theta & = \frac{x^2 - 1}{2x}
\end{align*}
Thus,
$$\sin\theta = \tan\theta\cos\theta = \frac{x^2 - 1}{2x} \cdot \frac{2x}{x^2 + 1} = \frac{x^2 - 1}{x^2 + 1}$$
|
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|
How to solve this integral using the method of residues? Using the method of residues, verify the following.
$$\int_0^{\pi} \frac{d \theta}{(3+2cos \theta)^2} = \frac{3 \pi \sqrt{5}}{25}$$
I tried doing this but can't get the correct answer, here is my attempt:
first I substituted $cos \theta = \frac{z+ \frac{1}{z}}{2}$ and $d \theta = \frac{dz}{iz}$ and got $\int_0^{\pi} \frac{dz}{iz(3+z+\frac{1}{z})^2}$
then I multiplied top and bottom my $iz$ so that the $i$ moves to the top and on the bottom I'll have $z^2$ so I can distribute it into the rest of it and get $-i \int_0^{\pi} \frac{zdz}{(3z+z^2+1)^2}$
the denominator has two zeroes at $-\frac{3}{2} \pm \sqrt{\frac{5}{4}}$, since the denominator is squared the function has poles here of order two.
The I used Cauchy's Residue theorem but first I needed to find the residue at $-\frac{3}{2} + \sqrt{\frac{5}{4}}$ because only this pole is inside the unit circle.
I found the residue using theorem 1 on pdf page 324/579 from this textbook
http://english-c.tongji.edu.cn/_SiteConf/files/2014/05/05/file_53676237d7159.pdf
After finding the residue using that formula and applying cauchy's residue theorem I do not get an answer close to $\frac{3 \pi \sqrt{5}}{25}$
I am not sure how to handle the fact that the integral is from $0$ to $\pi$ rather then from $0$ to $2\pi$, and I am not sure if I made any mistakes when finding the poles and residues.
|
Note
$$\int_0^{\pi} \frac{d \theta}{(3+2\cos \theta)^2} = \frac12\int_0^{2\pi} \frac{d \theta}{(3+2\cos \theta)^2}.$$
Let $z=e^{i\theta}$ and hence one has
\begin{eqnarray}
&&\int_0^{\pi} \frac{d \theta}{(3+2\cos \theta)^2}\\
&=&\frac12\int_0^{2\pi} \frac{d \theta}{(3+2\cos \theta)^2}\\
&=&\frac12\int_{|z|=1}\frac{1}{(3+z+z^{-1})^2}\frac{dz}{iz}\\
&=&\frac1{2i}\int_{|z|=1}\frac{z}{(z^2+3z+1)^2}dz\\
&=&\pi\text{Res}(f(z),z=\frac{1}{2}(-3+\sqrt5))\\
&=& \frac{3 \pi \sqrt{5}}{25}
\end{eqnarray}
where $f(z)=\frac{z}{(z^2+3z+1)^2}$ has only one pole $z=\frac{1}{2}(-3+\sqrt5)$ inside $|z|=1$.
|
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|
Prove $\frac{1}{r} = \frac{1}{a} + \frac{1}{b}$ for a semicircle tangent within a right triangle
I have a right angled triangle with the sides which are not hypotenuses $a$ and $b$. There is also a semi-circle radius $r$ whose diameter lies on the hypotenuse of the right angles triangle, and sides $a$ and $b$ are tangents of the semicircle.
Prove that $$\frac{1}{r} = \frac{1}{a} + \frac{1}{b}$$
My attempt: First I drew the diagram:
I marked all areas of significance:
The vertices opposite side $a$ is $A$ , $b$ to $B$ and $r$ to $C$. Where the semi circle touches $a$, I called $P$, and where it touched $b$ I called $Q$. The centre of the semicircle I call $O$.
But then, I couldn't proceed. What do I do after this?
|
We can write the area of $\triangle ABC$ in two ways: $$Area = \frac{1}{2}\cdot ab$$ $$Area = \frac{1}{2}\cdot a\cdot OP+\frac{1}{2}\cdot b\cdot OQ = \frac{1}{2}\cdot (ar+br)$$
Equate these two equations and divide both sides by $\frac{1}{2}\cdot abr$ to get your result:
$$\frac{1}{2}\cdot ab=\frac{1}{2}\cdot (ar+br) \implies \frac{1}{r}=\frac{1}{a}+\frac{1}{b}$$
|
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|
GCD of polynomials in $\mathbb Z_5$
I have to find the $\gcd(p(x),h(x))$ where $p(x)=1+2x+x^2+3x^3$ and $h(x)=2+x+x^5+x^6$ are in $\mathbb Z_5$.
To find the solution, I divide $h(x)/p(x)$ and I get $q_1(x)=2x^3+3x^2-1$ and $r_1(x)=3x^2+3x+3$
Should I divide $p(x)/r_1(x)$ ?
I think $r_1(x)$ is irreducible but I don't know what to conclude from this .
Also the answer of the exercise is $\gcd(p(x),h(x))=2$ but I don't know how to get $2$.
|
The GCD doesn't change if you multiply $p$ by $2$, so to get
$$
2+4x+2x^2+x^3
\qquad\text{and}\qquad
2+x+x^5+x^6
$$
The first remainder is $3x^2+x$, but we can multiply it by $2$:
$$
2x+x^2
\qquad\text{and}\qquad
2+4x+2x^2+x^3
$$
The remainder is $4x+2$, but we can multiply it by $4$:
$$
x+3
\qquad\text{and}\qquad
2x+x^2
$$
The remainder is $3$.
Thus the two polynomials are coprime. Answering $3$, $2$ or $1$ for the GCD is the same, as it is determined up to an invertible element.
|
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|
Prove that $n^5+n^4+1$ is composite for $n>1.$
Prove that $f(n)=n^5+n^4+1$ is composite for $n>1, n\in\mathbb{N}$.
This problem appeared on a local mathematics competition, however it looks like there is no simple method to solve it. I tried multiplying it by $n+1$ or $n-1$ and then tried factorizing it, but it was too tough for me.
Any help will be appreciated.
|
If $\omega$ is a (complex) root of $x^2+x+1$, then $\omega^3 - 1 = (\omega-1)(\omega^2+\omega+1) = (\omega-1)\cdot 0 = 0$, and hence $\omega^3 = 1$.
Observe then that $\omega^5 = \omega^3 \cdot \omega^2 = 1 \cdot \omega^2 = \omega^2$ and $\omega^4 = \omega^3 \cdot \omega^1 = 1 \cdot \omega = \omega$, so that $\omega^5+\omega^4 +1 = \omega^2 +\omega + 1 = 0$, hence $\omega$ is a root of $x^5+x^4+1$. Hence both roots of $x^2+x+1$ are roots of $x^5+x^4+1$ and thus $(x^2 + x + 1) \mid (x^5+x^4+1)$.
Note that by a similar argument we can show that $x^2+x+1$ is a factor of $x^8+x^7 + 1$ or even of $x^{101}+x^4+1$. All that matters is that the exponents give all the different residues modulo $3$.
This problem appears as E$13$ in the Number Theory Chapter of Arthur Engel's Problem Solving Strategies.
|
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|
$a,b,c,d$ are real numbers such that....
Suppost that $a,b,c,d$ are real numbers such that $$a^2+b^2=1$$ $$c^2+d^2=1$$ $$ac+bd=0$$
I've to show that $$a^2+c^2=1$$ $$b^2+d^2=1$$ $$ab+cd=0$$
Basically,I've no any idea or tactics to tackle this problem. Any methods? Thanks in advance.
EDITED.
The given hint in the books is $S:=(a^2+c^2-1)^2+(b^2+d^2-1)^2=(ac+bd)^2$
|
What if $b=0?$
Else $abcd\ne0,$
we have $\dfrac ab=-\dfrac dc=k$(say)
$\implies a=bk,d=-ck$
$$1=a^2+b^2=b^2(1+k^2)\implies b^2=?,a^2=(bk)^2=?$$
$$1=c^2+d^2=c^2(1+k^2)\implies c^2=?,d^2=(-ck)^2=?$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
How can we prove the equality of the following determinants? Assume that $A$ is an $n\times n$ real matrix whose entries are all $1$. Then how can we show the following determinant equality for any $x$?
$\det(A-xI)$=\begin{vmatrix}
1 -x & 1 & 1 & \cdots & 1 \\
1 & 1 -x & 1 & \cdots & 1 \\
1 & 1 & 1 -x & \cdots & 1\\
\vdots & \vdots & \vdots & \ddots & \vdots\\
1 & 1 & 1 & \cdots & 1 -x
\end{vmatrix}=
\begin{vmatrix}
n-x & n-x & n-x & \cdots & n -x \\
1 & 1 -x & 1 & \cdots & 1 \\
1 & 1 & 1 -x & \cdots & 1\\
\vdots & \vdots & \vdots & \ddots & \vdots\\
1 & 1 & 1 & \cdots & 1 -x
\end{vmatrix}
Thanks.
|
Just add row 2, row 3,... up to row $n$ to row 1.
|
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|
Solving a rational inequality with an unknown using cases So the question is,
For what values of x is $\dfrac{1}{1 + x}> -1 $ ?
Now, one way to do it is write the inequality relative to zero, and then find the behavior of the graph relative to the zeroes of the numerator and denominator:
$\dfrac{1}{1 + x}> -1 \implies \dfrac{1}{1+x} + 1> 0 \implies \dfrac{1}{1+x} + \dfrac{1+x}{1+x} > 0 $
$\implies \dfrac{2+x}{1+x} > 0$, and we simply find what happens when $x < -2, -2 < x < -1$, and $x > -1$.
What I don't understand is why I can't answer this by multiplying through by $1 +x$ if I consider both a) $1 +x >0$ and b) $1+x < 0$.
Here's the attempt:
Case a):
$1 + x > 0 \implies x > -1$, so
$\dfrac{1}{1+x} > -1$
$\implies (1+x) \dfrac{1}{1+x} > -1 (1+x) $ (we can multiply both sides by 1+x because we assume 1+x is positive)
$\implies x > -2 $.
Hence, the inequality is true when $x > -2$ and $x > -1$--so $x > -1$.
Case b): $1+x < 0 \implies x < -1$, so
$\dfrac{1}{1+x} > -1$
$\implies (1+x) \dfrac{1}{1+x} < -1 (1+x) $ (we assume $1+x$ is negative, so we reverse the sign)
$\implies 1 < 1 + x$
$\implies x > 0$.
Hence, the inequality is true when $x < -1$ and $x > 0$--this is never true.
So the real answer is $x > -1$ or $x <-2$, and I think the reason my attempt at multiplying through by an unknown doesn't work is that, in my assumption $x\in (-1, \infty)$ clearly includes both negative and positive values, and that leads the answer to be wrong.
But I'm not 100 percent certain it's impossible, and would like some help!
|
I think the question has been well-answered, but in case you were looking for a way to multiply through and not have to worry about these cases, try this:
$$\frac{1}{1+x} > -1$$ Multiplying through by the positive quantity $(1+x)^2$ gives, $$ (1+x)^2 \frac{1}{1+x} > -(1+x)^2$$ $$ 1+ x > -(1+x)^2$$ $$(1+x) + (1+x)^2 > 0 $$$$ (1+x) [1+ (1+x)]>0$$ $$(1+x) (2+x)>0$$ which has solutions of $x<-2$ or $x>-1$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
The distance from the incenter to an acute vertex of a right triangle I'm seeking an alternative proof of this result:
Given $\triangle ABC$ with right angle at $A$. Point $I$ is the intersection of the three angle lines. (That is, $I$ is the incenter of $\triangle ABC$.) Prove that
$$|CI|^2=\frac12\left(\left(\;|BC|-|AB|\;\right)^2+|AC|^2\right)$$
My Proof. Draw $ID \perp AB$, $IE\perp BC$, and $IF\perp CE$.
We have $|ID|=|IE|=|IF|=x$. Since $\triangle ADI$ is right isosceles triangle, we also have that $|AD|=|ID|=x$. Respectively, we have: $$|ID|=|IF|=|IE|=|AD|=|AF|=x$$
$\triangle BDI=\triangle BEI \Rightarrow |BD|=|BE|=y$. And $|CE|=|CF|=z$
We have:
$$|CI|^2=|CE|^2+|IE|^2=x^2+z^2 \tag{1}$$
And
$$\begin{align}\frac12\left(\left(|BC|-|AB|\right)^2+|AC|^2\right) &=\frac12\left(\left(\;\left(y+z\right)-\left(x+y\right)\;\right)^2+\left(x+z\right)^2\right) \\[4pt]
&=\frac12\left(\left(x-z\right)^2+\left(x+z\right)^2\right) \\[4pt]
&=\frac22\left(x^2+z^2\right) \\[4pt]
&=x^2+z^2
\tag{2}\end{align}$$
From $(1);(2)$ we are done. $\square$
|
Alternative proof: by Stewart's theorem the length $\ell_c$ of the angle bisector through $C$ is given by
$$ \ell_c^2 = \frac{ab}{(a+b)^2}\left[(a+b)^2-c^2\right] $$
and by Van Obel's theorem and the bisector theorem $\frac{CI}{\ell_c}=\frac{a+b}{a+b+c}$. It follows that, in general:
$$ CI^2 = \frac{ab}{(a+b+c)^2}\left[(a+b)^2-c^2\right]=\frac{ab(a+b-c)}{(a+b+c)} $$
and it is enough to prove that the given hypothesis ($a^2=b^2+c^2$) ensure
$$ \frac{ab(a+b-c)}{(a+b+c)} = \frac{(a-c)^2+b^2}{2} $$
or
$$ 2ab(a+b-c) = (a+b+c)(a^2+b^2+c^2-2ac). $$
That is trivial since the difference of the RHS and the LHS of the last expression is
$$ (b+c-a)(a^2-b^2-c^2). $$
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
Find the limit of $(1-\cos x)/(x\sin x)$ as $x \to 0$ Can you please help me solve:
$$\lim_{x \rightarrow 0} \frac{1- \cos x}{x \sin x}$$
Every time I try to calculate it I find another solution and before I get used to bad habits, I'd like to see how it can be solved right, so I'll know how to approach trigonometric limits.
I tried to convert $\cos x$ to $\sin x$ by $\pi -x$, but I think it's wrong. Should I use another identity?
|
Just taking terms of the fraction in question leads to:
\begin{align}
\frac{1-\cos(x)}{x \, \sin(x)} &= \frac{1 - \left(1 - \frac{x^2}{2!} + \frac{x^4}{4!} + \cdots\right)}{x^2 \, \left(1 - \frac{x^2}{3!} + \frac{x^4}{5!} - \cdots \right)} \\
&= \frac{\frac{1}{2!} - \frac{x^2}{4!} + \cdots}{1 - \frac{x^2}{3!} + \frac{x^4}{5!} - \cdots } \\
&= \frac{1}{2!} + \frac{x^2}{4!} + \frac{3 \, x^4}{6!} + \mathcal{O}(x^6)
\end{align}
which leads to
$$\lim_{x \to 0} \frac{1 - \cos(x)}{x \, \sin(x)} = \frac{1}{2}.$$
Alternatively, L'Hospital's rule applies.
\begin{align}
\lim_{x \to 0} \frac{1 - \cos(x)}{x \, \sin(x)} \to \frac{0}{0}
\end{align}
which leads to
\begin{align}
\lim_{x \to 0} \frac{1 - \cos(x)}{x \, \sin(x)} = \lim_{x \to 0} \frac{\sin(x)}{x \, \cos(x) + \sin(x)} \to \frac{0}{0}
\end{align}
and finally becomes
\begin{align}
\lim_{x \to 0} \frac{1 - \cos(x)}{x \, \sin(x)} = \lim_{x \to 0} \frac{\sin(x)}{x \, \cos(x) + \sin(x)} = \lim_{x \to 0} \frac{\cos(x)}{2 \, \cos(x) - x \, \sin(x)} = \frac{\cos(0)}{2 \, \cos(0)} = \frac{1}{2}.
\end{align}
|
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Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.