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Show that the last two decimal digits of a perfect square must be one of the following pairs. Show that the last two decimal digits of a perfect square must be one of the following pairs:
$00, e1, e4, 25, o6, e9$ ($e$ stand for even digit, $o$ for odd).
Solution is given, but I don't understand it at all.
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So the last two digits of $n^2$ clearly depend on the last two digits of $n$. There are $100$ possibilities.
Now $(50+n)^2=2500+100n+n^2=100(25+n)+n^2$ and the multiple of $100$ shows that this part can't affect the last two digits, which are therefore the same as the last two digits of $n^2$. So numbers which differ by $50$ produce the same final pair of digits when squared. Since the cycle repeats after $50$ numbers there are now only $50$ possibilities to consider say the integers $0$ to $49$.
Then $(50-n)^2=2500-100n+n^2=100(25+n)+n^2$ so, for the same reason, $n$ and $50-n$ have the same final digits. In other words two numbers which sum to $50$ have the same final digits. We therefore pair $(1,49); (2,48) \dots (24,26)$ and therefore only have to consider the $26$ numbers $0$ to $25$.
Now consider $10a+5\pm b$ where $0\le a\le 2$ and $0\le b \le 5$
$$(10a+5\pm b)^2=100a^2+25+b^2+100a\pm 20ab\pm 10b$$
If we work modulo $20$ we have that the final digit and the parity of the penultimate digit are determined by that of $(b-5)^2$. Taking $b=0$ and $b=5$ as special cases giving $25$ and $00$ respectively. we need only test $|b-5|=1,2,3,4$ giving $e1, e4, e9, o6$
All the cases do in fact occur, as we find by squaring all the numbers $0 \to 25$
|
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|
Solving for $x$ an $x\log(x)$ kind of equation Given the following:
$$\frac{x \log\left(\frac{4x}{d}\right)}{y_0} = 10$$
where $d$ and $y_0$ are constants, how can I solve for $x$?
|
For the equation
$$\frac{x \log\left(\frac{4x}{d}\right)}{y_0} = 10$$
multiply both sides by $4 y_{0}/d$ to obtain
$$\frac{4 x}{d} \, \log\left(\frac{4 x}{d}\right) = \frac{40 y_{0}}{d}.$$
Now using the Lambert W-function defined by $x e^{x} = W(x)$ with the property $e^{W(x)} = \frac{x}{W(x)}$ then
\begin{align}
\frac{4 x}{d} \, \log\left(\frac{4 x}{d}\right) &= \frac{40 y_{0}}{d} \\
t \, \log(t) &= \frac{40 y_{0}}{d} \hspace{15mm} t = \frac{4 x}{d} \\
u \, e^{u} &= \frac{40 y_{0}}{d} \hspace{15mm} u = \log(t) = \log\left(\frac{4 x}{d}\right) \\
u &= W\left(\frac{40 y_{0}}{d}\right) \\
\log\left(\frac{4 x}{d}\right) &= W\left(\frac{40 y_{0}}{d}\right) \\
\frac{4 x}{d} &= e^{W\left(\frac{40 y_{0}}{d}\right)} \\
x &= \frac{\frac{d}{4} \, \frac{40 y_{0}}{d} }{W\left(\frac{40 y_{0}}{d}\right)} = \frac{10 \, y_{0}}{W\left(\frac{40 y_{0}}{d}\right)}.
\end{align}
|
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|
Miscellaneous Olympiad Problem Two brothers sold a herd of sheep which they owned. for each sheep, they sold they received as many rubles as the number of sheep originally in the herd. The money was then divided in the following manner. First, the older brother took ten rubles followed by the younger brother taking ten and so on. At the end of the division, it was the younger brothers turn who found that there were fewer than ten rubles so he took what remained. The older brother to make the division fair gave the younger brother his penknife.
We have to find the value of the penknife.
The only thing I was able to determine is that the total money must be a two digit number so it can be represented as 10a+b, and the whole square( since each sheep was sold for the number of sheep originally in the herd) is 100a^2+20ab+b^2.
However, I am not able to do much after this. Help would be much appreciated.
|
If $N$ represents the total number of sheep that were originally in the flock, then the brothers were paid a total of $N^2$ rubles. From the description of how the proceeds were divided, we know that if $N^2$ is divided by 20, the remainder will be between 10 and 19 (inclusive). That is, there are nonnegative integers $q$ and $r$ such that
$$ N^2 = 20q + r
\qquad\text{and}\qquad
10 \le r \le 19. \tag{$\ast$}$$
While there are likely cleverer ways of approaching the problem at this point (the term "quadratic residues" comes to mind...), we can just brute force this problem: listing the remainders when square integers are divided by 20, we obtain
\begin{align}
1^2 &\equiv 1 \pmod{20} \\
2^2 &\equiv 4 \pmod{20} \\
3^2 &\equiv 9 \pmod{20} \\
4^2 &\equiv 16 \pmod{20} \\
5^2 &\equiv 5 \pmod{20} \\
6^2 &\equiv 16 \pmod{20} \\
7^2 &\equiv 9 \pmod{20} \\
8^2 &\equiv 4 \pmod{20} \\
9^2 &\equiv 1 \pmod{20} \\
10^2 &\equiv 0 \pmod{20} \\
11^2 &\equiv 1 \pmod{20} \\
12^2 &\equiv 4 \pmod{20} \\
13^2 &\equiv 9 \pmod{20} \\
14^2 &\equiv 16 \pmod{20} \\
15^2 &\equiv 5 \pmod{20} \\
16^2 &\equiv 16 \pmod{20} \\
17^2 &\equiv 9 \pmod{20} \\
18^2 &\equiv 4 \pmod{20} \\
19^2 &\equiv 1 \pmod{20} \\
20^2 &\equiv 0 \pmod{20},
\end{align}
at which point the pattern will repeat. Note that there is exactly one term that satisfies our requirements: 16. That is, if ($\ast$) holds, then $r=16$ is the only possibility. Therefore there were 6 rubles left over when the process finished.
Hence in the last division, the younger brother took 6 rubles. The older brother had just taken 10 rubles, which means that the older brother was up by 4. To make things even, the older brother would have to give the younger brother 2 rubles (subtracting two from his own total, and adding two to the total of the younger brother, netting each brother 8 rubles in the last division), therefore the penknife must be worth two rubles.
|
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prove factored polynomial has no real roots the polynomial $6x^3-18x^2-6x-6$ can be factored as $6(x-r)(x^2+ax+b)$ for some $a,b \in \Bbb{R}$ and where $r$ is a real root fo the polynomial. How would you prove that the polynomial $x^2+ax+b$ has no real roots. I know that you can do polynomial long division to get concrete values for $a$ and $b$ but $r$ is a decimal so the long division would be messy and inaccurate, but other then that method I have no idea how i would prove that it has no real roots.
|
A non-calculus approach is possible, although not necessarily the most efficient.
Consider $$f(x) = x^3 - 6x^2 - x - 1.$$ Let $r$ be a real root of $f$, which we know must exist. Then long division yields $$\frac{f(x)}{x-r} = x^2 + (r-6)x + (r^2 - 6r - 1) + \frac{r^3 - 6r^2 - r - 1}{x-r},$$ and since $r$ is a root, the remainder term is zero, yielding the factorization $$f(x) = (x-r)(x^2 + (r-6)x + (r^2 - 6r - 1)).$$ Then the quadratic term has discriminant $$(r-6)^2 - 4(r^2 - 6r - 1) = -3r^2 + 12r + 40.$$ This discriminant is nonnegative when $L = \frac{2}{3}(3 - \sqrt{39}) \le r \le \frac{2}{3}(3 + \sqrt{39}) = U$, and negative otherwise. To determine where $r$ lies relative to the boundaries $L, U$, we evaluate $$f(L) = \frac{-171 - 26\sqrt{39}}{9}, \quad f(U) = \frac{-171+26\sqrt{39}}{9}.$$ Both are negative. Since $U < 2(3+\sqrt{49})/3 = 20/3$, we evaluate $f(20/3) = 593/27 > 0$, meaning there is a sign change in the interval $(U, 20/3)$, and $r$ must be in this region; thus the discriminant is negative and the quadratic factor has no real roots. We knew to search for $r > U$ because once we determined $f(U) < 0$, the sign of the cubic term of $f$ being positive assures that for some $r > U$, $f(r) > 0$.
The proof that $f(U) < 0$ amounts to showing that $39 < (\frac{171}{26})^2$, which is elementary arithmetic.
|
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|
I need help with geometry Let $ABCD$ be a convex quadrilateral. Let $|\measuredangle ABC|$=$|\measuredangle ACD|$ and $|\measuredangle ACB|$=$|\measuredangle ADC|$. Let $O$ be the circumcenter of triangle $BCD$, distinct from $A$. Prove, that $|\measuredangle OAC|$ is a right angle.
I'm supposed to use spiral similarity with center A and midpoints of sides $BC$ and $CD$ (let them be $M_A$ and $M_B$, respectively) to show that $AOM_ACM_B$ is a cyclic pentagon, but i don't know how to proceed with it. How do I do this?
|
Particularly in the absence of better ideas, I like coordinates. Without loss of generality (as the setup is invariant under similarity transformations), you can use $A$ as the origin and $\overrightarrow{AB}$ as first unit vector. So you get
$$A=\begin{pmatrix}0\\0\end{pmatrix}\qquad
B=\begin{pmatrix}1\\0\end{pmatrix}\qquad
C=\begin{pmatrix}x\\y\end{pmatrix}$$
for some $x,y\in\mathbb R$. This defines the spiral similarity transformations, and applying that similarity to $C$ you get
$$D=\begin{pmatrix}x&-y\\y&x\end{pmatrix}\cdot
\begin{pmatrix}x\\y\end{pmatrix}=
\begin{pmatrix}x^2-y^2\\2xy\end{pmatrix}\;.$$
The midpoints are thus
$$M_A=\tfrac12(B+C)=\frac12\begin{pmatrix}x+1\\y\end{pmatrix}\qquad
M_B=\tfrac12(C+D)=\frac12\begin{pmatrix}x+x^2-y^2\\y+2xy\end{pmatrix}$$
and intersecting the orthogonal lines in these points you find that the circumcenter is
$$O=\frac{x^2+y^2-1}{2y}\begin{pmatrix}-y\\x\end{pmatrix}\;.$$
To show that $\overrightarrow{AC}$ and $\overrightarrow{AO}$ are orthogonal, you check that the scalar product between these two is zero:
$$\langle C,O\rangle=\frac{x^2+y^2-1}{2y}\left\langle
\begin{pmatrix}x\\y\end{pmatrix},
\begin{pmatrix}-y\\x\end{pmatrix}
\right\rangle=0\;.$$
|
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|
Prove that $7^{100} - 3^{100}$ is divisible by $1000$
Prove that $7^{100} - 3^{100}$ is divisible by $1000$
Equivalently, we want to show that $$7^{100} = 3^{100} \pmod {1000}$$
I used WolframAlpha (not sure if that's the right way though) and found that $\varphi (250) = 100$.
So by Euler's theorem: $$7^{100} \equiv 7^{\varphi(250)} \equiv 1 \pmod {250} \\ 3^{100} \equiv 3^{\varphi(250)} \equiv 1 \pmod {250}$$
but of course, we want $\pmod {1000}$.
Is that what I'm intended to do in this exercise (how to proceed if so)? Is there a solution without the need to use WolframAlpha?
Thanks!
|
\begin{eqnarray}
7^{100}-3^{100} &=& (10-3)^{100}-3^{100}\\
&=& \underbrace{{100\choose 0}10^{100}-{100\choose 1}10^{99}\cdot 3+...-{100\choose 97}10^3 \cdot 3^{97}}_{10^3k}+{100\choose 98}10^2 \cdot 3^{98} -{100\choose 99}10 \cdot 3^{99}+3^{100}-3^{100}\\
&=&1000k +50\cdot 99\cdot10^2 \cdot 3^{98} -100\cdot 10 \cdot 3^{99}
\end{eqnarray}
|
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solution using synthetic geometry I managed to solve this problem only using complex numbers but I'd like to solve it using synthetic geometry and I can't.
Can someone help me to solve this problem using synthetic geometry?
Let $ABC$ an acute triangle with $AB > AC$ . Let $O$ its circumcenter and let $D$ the midpoint of $BC$. The circle of diameter $AD$ intersects again $AB$ and $AC$ in $E$ and in $F$, respectively. Let $M$ the midpoint of $EF$.
Prove that $MD$ is parallel to $AO$.
This is my solution. But, as I wrote above, I'd like to solve it using synthetic geometry and I can't.
Setting the origin of the plane in O and the points A, B and C on the circumference of unit radius, we have that
$ a \bar{a} = 1 \text{;} \ b \bar{b} = 1 \text{;} \ c \bar{c} = 1 $.
Because $ D $ is the midpoint of $ BC $, we can write
1) $ d = \dfrac{b + c}{2} $
and because $ AD $ is a diamtere of the new circle then said $ Q $ his midpoint we have
2) $ q = \dfrac{a + d}{2} = \dfrac{2a + b + c}{4}$
Said $ M_{1} $ the projection of $ Q $ on $ AB $, then
$ m_{1} = \frac{1}{2} \left[ \left( \dfrac{\bar{q} - \bar{a}}{\bar{b} - \bar{a}}\right) (b - a) + a + q \right] $
but
$\dfrac{1}{\bar{b} - \bar{a}} = \dfrac{1}{\dfrac{1}{b} - \dfrac{1}{a}} = \dfrac{ab}{a – b} $
and so
$ m_{1} = \frac{1}{2} \left[ \bar{q} ab (-1) - \dfrac{ab}{a} (- 1) + a + q \right] \Rightarrow $
$m_{1} = \frac{1}{2} \left( a + b + q - ab\bar{q} \right) $
In the same way, said $ M_{2} $ the projection of $ Q $ on $ AC $ we have
$ m_{2} = \frac{1}{2} \left[ \dfrac{\bar{q} - \bar{a}}{\bar{c} - \bar{a}} (c - a) + a + q \right] \Rightarrow $
$ m_{2} = \frac{1}{2} \left( a + c + q - ac\bar{q} \right) $
Because $ Q $ is the center of the new circle passing through $ A \text{,} D \text{,} E \text{,} F $ we have that $ M_{1}Q $ is axes of $ AE $ and that $ M_{2}Q $ in axes of $ AF $.
So we have that $ M_{1} $ is midpoint of $ AE $ and that $ M_{2} $ is midpoint of $ AF $.
So we can write
$ m_{1} = \dfrac{a + e}{2} \ \Rightarrow \ e = 2m_{1} – a $
and also
$ m_{2} = \dfrac{a + f}{2} \ \Rightarrow \ f = 2m_{2} - a$
The point $ M $ is defined as midpoint of $ EF $ so we have
$ m = \dfrac{e + f}{2} = \dfrac{2m_{1} + 2m_{2} - 2a}{2} = m_{1} + m_{2} - a$
therefore replacing $ m_{1} $ and $ m_{2} $ we have
$m = \frac{1}{2} a + \frac{1}{2} c + \frac{1}{2} q + \frac{1}{2} a + \frac{1}{2} b + \frac{1}{2} q - \frac{ab\bar{q}}{2} - \frac{ac\bar{q}}{2} - a \ \Rightarrow $
$m = \dfrac{b + c}{2} + q - a\bar{q} \dfrac{b + c}{2}$
We know, furthemore, that $ \bar{q} $ is:
3) $\bar{q} = \dfrac{2\bar{a} + \bar{b} + \bar{c}}{4} = \frac{1}{4} \left( \frac{2}{a} + \frac{1}{b} + \frac{1}{c} \right) = \dfrac{2bc + ab + ac}{4abc}$
Now we write the equation of the parallel line through $D$ parallel to $AO$:
let $z$ be a generic point of this line it is possible to write
4) $ \dfrac{z - d}{a - 0} = \dfrac{\bar{z} - \bar{d}}{\bar{a} – 0} $
Replacing $d$ with the expression of 1) and taking into account that $ \bar {a} = \frac{1}{a} $, we get
5) $z - \dfrac{b + c}{2} = a^{2} \left( \bar{z} - \dfrac{\bar{b} + \bar{c}}{2} \right) = a^{2}\bar{z} - \dfrac{a^{2} (\bar{b} + \bar{c})}{2}$
If the $ M $ point belongs to this line, $ m $ must satisfy equation 5). Substituting the value of $ m $ we have
$\dfrac{b + c}{2} + q - a \bar{q} \cdot \left( \dfrac{b + c}{2} \right) - \dfrac{b + c}{2} = a^{2}\left( \dfrac{\bar{b} + \bar{c}}{2} + \bar{q} - \bar{a}q \cdot \dfrac{\bar{b} + \bar{c}}{2} \right) - \dfrac{a^{2} (\bar{b} + \bar{c})}{2} \Rightarrow $
$ q - a \bar{q} \cdot \left( \dfrac{b + c}{2} \right) = a^{2}\bar{q} - \dfrac{aq( b + c)}{2bc} \ \Rightarrow $
$ q \left( 1 + \dfrac{ab + ac}{2bc} \right) = \bar{q}a \left( a + \dfrac{b + c}{2} \right)$
Substituting the values of $ q $ in 2) and the value of $ \bar{q} $ in 3) we have
$ \dfrac{(2a + b + c)(2bc + ab + ac)}{8bc} = \dfrac{a (2bc + ab + ac) (2a + b + c)}{a \cdot 8bc} $
which is, obviously, an identity, so $ MD $ is parallel to $ AO $, as we wanted to prove.
|
Hints:
Construct squares $EDPR$ and $FDQS$ outwardly on the sides $ED$ and $FD$ of $\triangle DEF$. Let the line $PQ$ intersect lines $AB, AC$ at $U, V$ respectively .
First show that $MD \perp \operatorname{line} PQ = \operatorname{line}\ UV$. Therefore if $l$ is the line through $A$ that is parallel to $MD$ then $l \perp \operatorname{line} UV$.
Using $\triangle DEF \cong \triangle ABC$ (use ratio $AB/AC$ and $\angle A$) and $\operatorname{line} DP \parallel \operatorname{line} AB$ show that $\angle AUV = \angle ACB$. Since $\angle OAB = \frac\pi 2 - \angle ACB$, conclude that $\operatorname{line} AO \perp UV$.
Thus lines $AO$ and $l$ through $A$ are both perpendicular to the line $UV$, which implies that $\operatorname{line} AO = l \parallel \operatorname{line} MD$.
|
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I want to know the limits and convergence/divergence of $a_{n+1} = (a_n+3)/(a_n+4)$ I`ve the following iteratively defined sequence:
$a_1=1$ and for $n\ge 1$,
$$a_{n+1} = \frac{a_{n}+3}{a_n+4}$$
I want to know the limits and if it is convergent or divergent, but I don't know how to handle two $a_n$'s in the formula.
If I have got one $a_n$ I would find it out by proving monotony and boundedness. I think that I can use the same technique here but I got stuck. Thanks in advance.
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After testing a couple $a_n$ by hand it appears to converge. So let's try to solve:
$$x = \frac{x+3}{x+4}$$
$$x(x + 4) = x+3$$
$$x^2 +3x - 3 = 0$$
$$x = \frac{1}{2}(\sqrt{21} - 3)$$
But if we want to be rigorous and prove its convergence:
$$a_n = \frac{p_n}{q_n}$$
$$\frac{p_{n+1}}{q_{n+1}} = \frac{\frac{p_n}{q_n}+3}{\frac{p_n}{q_n}+4} = \frac{p_n + 3q_n}{p_n + 4q_n}$$
$$P(x) = \sum_{n=1}^\infty p_nx^n \quad Q(x) = \sum_{n=1}^\infty q_nx^n$$
$$P(x) = x + x\sum_{n=1}^\infty p_{n+1}x^n = x + x\sum_{n=1}^\infty (p_n + 3q_n)x^n = x + xP(x) + 3xQ(x)$$
$$Q(x) = x + x\sum_{n=1}^\infty q_{n+1}x^n = x + x\sum_{n=1}^\infty (p_n + 4q_n)x^n = x + xP(x) + 4xQ(x)$$
$$Q(x) = \frac{x + xP(x)}{1 - 4x}$$
$$P(x) = x + xP(x) + 3x\cdot \frac{x + xP(x)}{1 - 4x} = \frac{x(1 - x)}{x^2 - 5x + 1}$$
After plugging $P(x)$ back in we find that $Q(x)$ is quite simple:
$$Q(x) = \frac{x}{x^2 -5x + 1} = \frac{1}{\beta - \alpha}\left(\frac{\alpha}{\alpha - x} - \frac{\beta}{\beta - x}\right)$$
Where $x^2 - 5x + 1 = (x - \alpha)(x - \beta)$ and thus $\alpha = \frac{1}{2}(5 + \sqrt{21})$ and $\beta = \frac{1}{2}(5 - \sqrt{21})$.
Knowing the formula for the inverse geometric series $\displaystyle \sum_{n=1}^\infty \alpha^{-n}x^n = \frac{a}{a-x}$ we get:
$$Q(x) = \frac{1}{\beta - \alpha}\left(\sum_{n=1}^\infty (\alpha^{-n} - \beta^{-n})x^n\right)$$
$$q_n = \frac{\alpha^{-n} - \beta^{-n}}{\beta - \alpha}$$
But subtracting the original numerator from the denominator we find $q_{n+1} - p_{n+1} = q_n$. Thus
$$\dfrac{p_{n}}{q_{n}} = 1 - \dfrac{q_{n-1}}{q_{n}} = 1 - \frac{\alpha^{1-n} - \beta^{1-n}}{\alpha^{-n} - \beta^{-n}}$$
Finally, for large $n$ we find that $\alpha^{-n} \to 0$ as $\alpha > 1$, thus:
$$\lim_{n\to\infty} \dfrac{p_{n}}{q_{n}} = 1 - \frac{\beta^{1-n}}{\beta^{-n}} = 1 - \beta \approx 0.791288$$
As a fun fact we have:
$$p_n = \frac{(\alpha - 1)\alpha^{-n} - (\beta- 1)\beta^{-n}}{\alpha - \beta}$$
|
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How to find matrix exponential $e^A$ I have the matrix $$A =\begin{pmatrix} 0 & 1 \\ - 1 & 0 \end{pmatrix}$$
and I have to find $e^A$
I've found two complex-conjugate eigenvalues $\lambda_{1,2} = \pm i$
so substracting $\lambda_1 = i$ from the matrix's diagonal I got:
$$A_1 = \begin{pmatrix} -i & 1 \\-1 & i \end{pmatrix}$$
and therefore. to find eigenvector I have to solve the system:
$$A_1 = \begin{pmatrix} -i & 1 \\-1 & i \end{pmatrix} \begin{pmatrix} x_1 \\ x_2 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix}$$
so the first eigenvector is $h_1 = \begin{pmatrix}1 \\ i\end{pmatrix}$
and the second one is $h_2 = \begin{pmatrix}1 \\ -i\end{pmatrix}$
so the general solution is $$x(t) = C_1e^{it}\begin{pmatrix} 1 \\i\end{pmatrix} + C_2e^{-it}\begin{pmatrix} 1 \\-i\end{pmatrix}$$
I know that now I have to solve two Cauchy's problems for the standard basis $\mathbb{R}^2$ with vectors $v_1 = (1, 0)$ and $v_2 = (0,1)$ But I do not know how to approach it for complex numbers
|
Answer
Let a,b $\in \mathbb{R}$ and
$$
\begin{align}
I &=
\begin{pmatrix}
1 & 0 \\
0 & 1 \\
\end{pmatrix} \\
J &=
\begin{pmatrix}
0 & 1 \\
-1 & 0 \\
\end{pmatrix}
\end{align}
$$
Then the following formula is true,
$$ \exp(aI + bJ) = \exp(a)[I \cos(b) + J \sin(b)]$$
In the special case where a = 0 and b = 1,
$$ \exp(J) =
\begin{pmatrix}
\cos(1) & \sin(1) \\
-\sin(1) & \cos(1) \\
\end{pmatrix}
$$
Proof
We will need the following lemmas to go through the chain of equalities. The lemmas are not going to be proved or explained here.
Let n $\in \mathbb{N}$, x $\in \mathbb{R}$, A and B two matrices
*
*The product of $aI$ and $bJ$ is commutative
*$\exp(A + B) = \exp(A)\exp(B)$ if $AB = BA$
*$J^{2n} = (-1)^n I$
*$J^{2n+1} = (-1)^n J$
Therefore,
$$
\begin{align}
\exp(aI + bJ)
&= \exp(aI)\exp(bJ) \\
&= \exp(a)[I + bJ + (bJ)^2/2! + (bJ)^3/3! + \cdots] \\
&= \exp(a)[I(1 - b^2/2! + b^4/4! - \cdots) + J(b - b^3/3! + b^5/5! - \cdots)] \\
&= \exp(a)[I\cos(b) + J \sin(b)]
\end{align}
$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Evaluate $\int_0^a \sqrt{\frac{x^3}{a^3-x^3}} dx$. Evaluate following in terms of Gamma function:
$$\int_0^a \sqrt{\frac{x^3}{a^3-x^3}} dx$$
I don't know how to proceed. So, please tell the intuition behind the solution.
|
Set $$x =a\sin^{\frac{2}{3}}(t) \implies dx = \frac{2a}{3}\frac{\cos t}{\sqrt[3]{\sin t}} dt$$ giving us: $$I = \int_{0}^{a}\sqrt{\frac{x^3}{a^3-x^3}} dx = \frac{2a}{3}\int_{0}^{\frac{\pi}{2}}\sqrt{\frac{a^3\sin^2t}{a^3\cos^2t}}\frac{\cos t}{\sqrt[3]{\sin t}} dt$$ $$=\frac{2a}{3}\int_{0}^{\frac{\pi}{2}}\sin^{\frac{2}{3}}(t) dt$$
Using the result: $$\int_{0}^{\frac{\pi}{2}}\sin^p x \cos^q x dx = \frac{1}{2}B(\frac{1+p}{2}, \frac{1+q}{2})$$ we get, $$I = \frac{2a}{3}\frac{1}{2}B(\frac{5}{6},\frac{1}{2})= \frac{a}{3}B(\frac{1}{2},\frac{5}{6})$$ and since, $$B(l,m) = \frac{\Gamma(l)\Gamma(m)}{\Gamma(l+m)}$$ we have, $$\boxed{I =\frac{a}{3}\frac{\Gamma(\frac{1}{2})\Gamma(\frac{5}{6})}{\Gamma(\frac{4}{3})}}$$
|
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|
Prove $5^n + 2 \cdot 3^n - 3$ is divisible by 8 $\forall n\in \mathbb{N}$ (using induction)
Prove $5^n + 2 \cdot 3^n - 3$ is divisible by 8 $\forall n\in \mathbb{N}$
Base case $n = 1\to 5 + 6 - 3 = 8 \to 8 \mid 8 $
Assume that for some $n \in \mathbb{N}\to 8 \mid 5^n + 2 \cdot 3^n - 3$
Showing $8 \mid 5^{n+1} + 2 \cdot 3^{n+1} - 3$
$$5^{n+1} + 2 \cdot 3^{n+1} - 3$$ $$5\cdot 5^n + 2\cdot 3\cdot 3^n - 3$$ $$ (5^n + 2\cdot 3^n - 3) + 4\cdot 5^n + 2\cdot 2\cdot3^n $$ $$ 5\cdot(5^n + 2\cdot 3^n - 3) + 4\cdot 5^n + 2\cdot 2\cdot3^n - 4\cdot(5^n + 2\cdot 3^n - 3)$$
$$ [5\cdot(5^n + 2\cdot 3^n - 3)] - [4\cdot 3^n - 12]$$
The first term divides by 8 but I am not sure how to get the second term to divide by 8.
|
Hint: Try reducing your expression mod 8. For example, what is $5^n$ mod 8? Since $5 \equiv 5$ and $5^2 = 25 \equiv 1$,
$5^n \equiv \begin{cases} 5, n \text{ odd} \\ 1, n \text{ even} \end{cases}$.
Now do the same for $3^n$, and add.
(This solution doesn't use induction.)
|
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|
Find the relation of the sides. How can I find the following relation $\frac{AO}{r}$?
$BO = OC, AB = AC, AD = BD$
I tried to do the following $r = \frac{AO + OC + \sqrt{{OC}^{2}+{AO}^{2}}}{2} $ but I was not able to infer the relation.
$CD$ is touched by the inner circle.
|
Let's define, as in the diagram below: $a=AG$, $b=GC$, $c=CH$. From Pythagoras' theorem applied to triangle $AOC$ we have:
$(a+r)^2+(b+r)^2=(a+b)^2$, that is:
$$
\tag{1}
ar+br+r^2=ab.
$$
As $AD=BD$, we also have $AH=BL=BJ=2OC-JC$, that is: $a+b+c=2b+2r-c$, or: $2c=2r+b-a$. On the other hand, triangles $EGC$ and $FCH$ are similar (because angle bisectors $CE$ and $CF$ are perpendicular), which gives:
$b:r=r:c$, or: $c=r^2/b$. Combining these two expressions for $c$ we get:
$$
\tag{2}
2r^2=2rb+b^2-ab.
$$
From equations $(1)$ and $(2)$ we can obtain $a$ and $b$ as a function of $r$. If we set $x=a/r$ and $y=b/r$ these equations can be rewritten as
$$
\cases{
x+y+1=xy\\
2+xy=2y+y^2
}.
$$
By eliminating $xy$ we get $x=y^2+y-3$ and plugging this into the first equation yields $y^3-5y+2=0$, which can be factored as $(y-2)(y^2+2y-1)=0$. The solutions are then $y=2$ and $y=-1\pm\sqrt2$, but only the first one leads to positive values for both $x$ and $y$. We obtain then $y=2$ and $x=3$, so that:
$$
{AO\over r}={a+r\over r}=x+1=4.
$$
|
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|
Show that $\int_{0}^{\pi}{\sin^5(x)\over 1+\cos^3(x)}\mathrm dx =\ln{3}?$
$$\int_{0}^{\pi}{\sin^5(x)\over 1+\cos^3(x)}\mathrm dx =\ln{3}\tag1$$
How can we show that $(1)=\ln{3}$
$u=\sin^3{x}$ then $du=3\sin^2{x}\cos{x}dx$
$${1\over 3}\int{u\mathrm du\over \cos{x}+\cos^4{x}}\tag2$$
This is not a good substitution
|
Without using integration by substitution or by parts:
Trigonometric algebra
We use the identity that $(1-\cos x)(1+ \cos^3x)=(1-\cos^2x)(1-\cos x+\cos^2x)$ so $$\frac{\sin^2x}{1+\cos^3x}=\frac{1-\cos x}{1-\cos x + \cos^2x} \implies \frac{\sin^5x}{1+\cos^3x}=\frac{\sin^3x(1-\cos x)}{1-\cos x + \cos^2x}$$ Now RHS is equivalent to $$\frac{\sin x(1-\cos^2x)(1-\cos x)}{1-\cos x + \cos^2x}=\frac{\sin x-\sin x \cos x+\sin x \cos^2x(\cos x-1)}{1-\cos x + \cos^2x}$$ or $$\frac{\sin x(1-2\cos x)+\sin x \cos x(1-\cos x + \cos^2x)}{1-\cos x + \cos^2x}=\frac{2\sin x(1-\cos^2x)}{2-2\cos x + 2\cos^2x}+\sin x \cos x$$ We do this so that we can replace the square term with $\cos 2x$. Continuing, we have $$\frac{2\sin x - 4 \sin x \cos x}{3-2\cos x + \cos 2x}+\frac12 \sin 2x=\frac{-2(-\sin x)+2(-\sin 2x)}{3-2\cos x + \cos 2x}-\frac12(-\sin 2x)$$
Notice that the numerator is the derivative of the denominator! Thus the integral is $$\int \frac{\sin^5x}{1+\cos^3x} dx=\ln|3-2\cos x + \cos 2x|-\frac14\cos 2x + c$$ where $c$ is a constant.
Substituting the upper and lower limits of $\pi$ and $-\pi$ respectively yields $\ln 3$.
|
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|
How to Simplify $ \frac{\sin(3x)+\sin^3(x)}{\cos(3x)-\cos^3(x)} $
Simplify $$ \frac{\sin(3x)+\sin^3(x)}{\cos(3x)-\cos^3(x)}.$$
The solution is : $-\cot(x)$
I tried to: $$\frac{\sin(2x)\cos(x)+\cos(2x)\sin(x)+\sin^3(x)}{\cos(2x)\cos(x)+\sin(2x)\sin(x)-\cos^3(x)}.$$
|
Note that
$$\begin{align}\cos(3x)+i\sin(3x)&=(\cos(x)+i\sin(x))^3
\\&=\cos^3(x)+3i\cos^2(x)\sin(x)
-3\cos(x)\sin^2(x)-i\sin^3(x).
\end{align}$$
Therefore, after separating real and complex parts, we obtain
$$
\cos(3x)=\cos^3(x)
-3\cos(x)\sin^2(x),\quad
\sin(3x)=3\cos^2(x)\sin(x)
-\sin^3(x).
$$
Finally
$$\frac{\sin(3x)+\sin^3(x)}{\cos(3x)-\cos^3(x)}=\frac{3\cos^2(x)\sin(x)}{-3\cos(x)\sin^2(x)}=-\frac{\cos(x)}{\sin(x)}=-\cot(x).$$
|
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|
Distance from center of a spiral Consider the following curve:
*
*We start at the center and take a step of 1 unit length
*Then turn to the left by $\pi/3$ and take a step of $9/10$ units in length
*Then turn to the left by $\pi/3$ and take a step of $81/100$ units in
length
*Then turn to the left by $\pi/3$ and take a step of $729/1000$ units in length
*Repeat the left turns by $\pi/3$ with the same ratio of length.
How far from the center do we go?
If the question were the total distance this would be an infinite series. Because the question is the distance from the starting point it seems to me I somehow must use complex numbers and do an infinite sum of some kind. But not sure
|
Short answer: take advantage of the periodicity: steps taken three turns apart are parallel.
With more detail: each step you take corresponds to adding a term of the form
\begin{equation}
\left(\frac{9}{10}\right)^ke^{ik\frac{\pi}{3}},
\end{equation}
starting from $k=0$. The terminal point of our spiral is then
\begin{equation}
\sum_{k=0}^\infty \left(\frac{9}{10}\right)^ke^{ik\frac{\pi}{3}}\in\mathbb{C},
\end{equation}
assuming this series converges. This seems an unwieldy modulus to compute at first, but we can take advantage of some periodicity. Because $\exp(i(k+3)\frac{\pi}{3})=-\exp(ik\frac{\pi}{3})$ for every $k$, we have
\begin{align}
\sum_{k=0}^\infty \left(\frac{9}{10}\right)^ke^{ik\frac{\pi}{3}} &= \sum_{k=0}^\infty (-1)^k\left(\frac{9}{10}\right)^{3k}e^{0} + \sum_{k=0}^\infty (-1)^k\left(\frac{9}{10}\right)^{3k+1}e^{i\frac{\pi}{3}} + \sum_{k=0}^\infty (-1)^k\left(\frac{9}{10}\right)^{3k+2}e^{2i\frac{\pi}{3}}\\
&= \left(1 + \frac{9}{10}e^{i\frac{\pi}{3}} + \frac{81}{100}e^{2i\frac{\pi}{3}}\right)\sum_{k=0}^\infty(-1)^k\left(\frac{9^3}{10^3}\right)^k.
\end{align}
Because the alternating geometric series we see here will converge to
\begin{equation}
\sum_{k=0}^\infty(-1)^k\left(\frac{9^3}{10^3}\right)^k = \frac{1}{1+9^3/10^3},
\end{equation}
our spiral will converge in $\mathbb{C}$. We can compute the modulus of the complex number that's left out front:
\begin{equation}
\bigg\vert1 + \frac{9}{10}e^{i\frac{\pi}{3}} + \frac{81}{100}e^{2i\frac{\pi}{3}}\bigg\vert = \frac{19\sqrt{91}}{100},
\end{equation}
meaning that the modulus of our terminal point is
\begin{equation}
\frac{1}{1+9^3/10^3}\frac{19\sqrt{91}}{100} = \frac{190\sqrt{91}}{1729}.
\end{equation}
|
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|
let $\forall x \neq 0 \ \ \ f(x+\frac{1}{x})=x^2-\frac{1}{x^2}$ then find the $f(x)$
let $\forall x \neq 0 \ \ \ f(x+\frac{1}{x})=x^2-\frac{1}{x^2}$ then find the $f(x)$
My try :
$$x^2-\frac{1}{x^2}=(x+\frac{1}{x})(x-\frac{1}{x})$$
And $$(x-\frac{1}{x})^2 =(x+\frac{1}{x})^2-4$$
so we have :
$$f(t)=\pm t \sqrt{t^2-4}$$
it is right ?
|
Such function cannot exist. Take $x=2$, then we have $$f(2+1/2)= 4-1/4= 15/4$$ Now take $x=1/2$, then we have $$f(2+ 1/2)=1/4-4=-15/4$$But $-15/4\neq 15/4$. Hence such function cannot exist.
|
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|
How can i solve $5^{2x-\frac{1}{3}x^2} < 5^{2-2x} * (5^\frac{1}{3})^{x^2}+24$? How can i solve it?
$$5^{2x-\frac{1}{3}x^2} < 5^{2-2x} * (5^\frac{1}{3})^{x^2}+24$$
I don't have idea how to solve it..
|
We have,
$$5^{2x-\frac{1}{3}x^2} < 5^{2-2x} * (5^\frac{1}{3})^{x^2}+24$$
$$5^{2x-\frac{1}{3}x^2} - 5^{2-2x} \cdot (5^\frac{1}{3})^{x^2}- 24<0 $$
$$5^{2x-\frac{1}{3}x^2} - 5^{2-2x + \frac{1}{3} x^2 } - 24 < 0 $$
Substitute $ 5^{2x - \frac{1}{3}x^2} = t $,
$$t - \frac{5^2}{t} - 24 < 0 $$
$$\frac{(t-25)(t+1)}{t} < 0 $$
$$ t \in (-∞,-1) \cup (0,25) $$
Since $ t$ is a constant raised to some exponent , it can't be negative.
So, $$ t \in (0,25) $$
Which implies,
$$5^{2x-\frac{1}{3}x^2} < 25 $$
To get $x$ taking logarithm with base as $5$,
$$\log_5{5^{2x-\frac{1}{3}x^2} } < \log_5 {5^2} $$
$$ 2x- \frac{1}{3}x^2 < 2 $$
$$ - \frac{1}{3}x^2 + 2x -3 < 0 $$
$$ \frac{1}{3}x^2 -2x + 3 > 0 $$
$$ x^2- 6x + 6 > 0 $$
$$ x = 3+√3, x= 3-√3 $$
|
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|
Wrong Wolfram Alpha result for $\lim_{(x,y)\to(0,0)}\frac{xy^4}{x^4+x^2+y^4}$?
I'm trying to solve this limit:
$$
\lim_{(x,y)\to(0,0)}\frac{xy^4}{x^4+x^2+y^4}
$$
Here's my attempt:
$$0 \le |\frac{xy^4}{x^4+x^2+y^4} - 0| = \frac{|x|y^4}{x^4+x^2+y^4},$$ and since $x^4+x^2 \ge0$ then $\frac{y^4}{x^4+x^2+y^4} \le 1$ so
$$
\frac{|x|y^4}{x^4+x^2+y^4} \le |x|,$$ so
$$
0 \le \lim_{(x,y)\to(0,0)}|\frac{xy^4}{x^4+x^2+y^4} - 0| \le \lim_{(x,y)\to(0,0)} |x| = 0,
$$ and using the squeeze theorem the limit is $0$.
But if I input the limit in wolfram alpha, it says that the limit doesn't exist. Here is the link to the limit in Wolfram Alpha.
|
For
\begin{align*}
\lim_{(x,y)\rightarrow(0,0)}\dfrac{xy^{4}}{x^{4}+x^{2}+y^{2}},
\end{align*}
one does the following step which is similar to your technique:
\begin{align*}
\left|\dfrac{xy^{4}}{x^{4}+x^{2}+y^{2}}\right|\leq|xy^{2}|\cdot\dfrac{y^{2}}{x^{4}+x^{2}+y^{2}}\leq|x|\cdot y^{2}\rightarrow 0
\end{align*}
as $(x,y)\rightarrow(0,0)$.
|
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|
Evaluate $\int_0^\infty \frac {(\log x)^4dx}{(1+x)(1+x^2)}$ Evaluate $$\displaystyle\int_0^\infty \frac {(\log x)^4dx}{(1+x)(1+x^2)}$$
This is a past final term exam problem of a complex analysis course at my university. I am studying for this year’s exam and I found this problem. The examiner assumes us to use residue calculus. Could you please give your valuable suggestions on how to proceed ?
|
For $-1\lt a\lt0$,
$$
\begin{align}
&\int_0^\infty\frac{x^a}{(1+x)(1+x^2)}\,\mathrm{d}x\\
&=\int_0^\infty\frac{x^a}2\left(\frac1{1+x}+\frac{1-x}{1+x^2}\right)\mathrm{d}x\\
&=\frac12\int_0^\infty\frac{x^a}{1+x}\,\mathrm{d}x+\frac14\int_0^\infty\frac{x^{\frac{a-1}2}}{1+x}\,\mathrm{d}x-\frac14\int_0^\infty\frac{x^{\frac{a}2}}{1+x}\,\mathrm{d}x\\
&=\frac12\Gamma(1+a)\Gamma(-a)+\frac14\Gamma\!\left(\frac{1+a}2\right)\Gamma\!\left(\frac{1-a}2\right)-\frac14\Gamma\!\left(\frac{2+a}2\right)\Gamma\!\left(-\frac{a}2\right)\\
&=-\frac12\frac\pi{\sin(\pi a)}+\frac14\frac\pi{\cos\left(\frac\pi2a\right)}+\frac14\frac\pi{\sin\left(\frac\pi2a\right)}\\
&=\frac\pi4\frac{1+\cos\left(\frac\pi2a\right)-\sin\left(\frac\pi2a\right)}{\left(1+\cos\left(\frac\pi2a\right)\right)\cos\left(\frac\pi2a\right)}\\
&=\frac\pi2\frac1{1+\cos\left(\frac\pi2a\right)+\sin\left(\frac\pi2a\right)}\\
\end{align}
$$
This can be analytically continued to $-1\lt a\lt 2$.
Taking $4$ derivatives gives
$$
\begin{align}
&\int_0^\infty\frac{\log(x)^4x^a}{(1+x)(1+x^2)}\,\mathrm{d}x\\
&=\frac{\pi^5}{32}\frac{\scriptsize105+45\left(\sin\left(\frac\pi2a\right)+\cos\left(\frac\pi2a\right)\right)-54\sin(\pi a)-11\left(\sin\left(\frac{3\pi}2a\right)-\cos\left(\frac{3\pi}2a\right)\right)-\cos(2\pi a)}{2\left(1+\cos\left(\frac\pi2a\right)+\sin\left(\frac\pi2a\right)\right)^5}
\end{align}
$$
and evaluating at $a=0$ gives
$$
\int_0^\infty\frac{\log(x)^4}{(1+x)(1+x^2)}\,\mathrm{d}x
=\frac{5\pi^5}{64}
$$
|
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|
Why is function domain of fractions inside radicals not defined for lower values than those found by searching for domain of denominator in fraction? Consider function $y = \sqrt{\frac{1-2x}{2x+3}}$. To find the domain of this function we first find the domain of denominator in fraction:
$2x+3 \neq 0$
$2x \neq -3$
$x \neq - \frac{3}{2}$
So, the domain of $x$ (for fraction to be valid) is $x \in \left(- \infty, - \frac{3}{2}\right) \cup \left(- \frac{3}{2}, + \infty\right)$.
Then we find the domain for whole fraction:
$\frac{1-2x}{2x+3} \ge 0$
$1-2x \ge 0$
$-2x \ge -1$
$x \le \frac{1}{2}$
My textbook says that the (real) domain of the whole $y$ function is $x \in \left(- \frac{3}{2}, \frac{1}{2}\right]$.
I understand why the function is not defined in values larger than $\frac{1}{2}$ (because condition is $x \le \frac{1}{2}$), but I don't understand why it can't be have values less than $- \frac{3}{2}$ (because condition says only $x \neq - \frac{3}{2}$).
I checked the domain of this function and the domain given in the textbook is correct. Function has imaginary values for $x$ values less than $- \frac{3}{2}$ or bigger than $\frac{1}{2}$. It is undefined in $- \frac{3}{2}$.
Real values only in $\left(- \frac{3}{2}, \frac{1}{2}\right]$ domain.
|
Find the domain of $y=\sqrt{\frac{1-2x}{2x+3}}$
Draw a simple table:
$$
\begin{array}{c|ccccc}
x & \text{under $-3/2$} & \text{$-3/2$} & \text{$]-3/2;1/2[$} & \text{$1/2$} & \text{over $1/2$} \\
\hline
1-2x & + & + & + & 0 & - \\
2x+3 & - & 0 & + & + & + \\
\frac{1-2x}{2x+3} & - & /// & + & 0 & - \\
\sqrt{\frac{1-2x}{2x+3}} & /// & /// & + & 0 & /// \\
\end{array}
$$
In the above table : /// means "not defined" (easy to sketch).
$y$ is defined for any $x$ in $]\frac{-3}{2};\frac{1}{2}]$.
|
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|
Minimum distance from $2x^2 + 2xy + 4yz + z^2 = 1 $ to the origin Let Σ be the surface in $R^3$ given by
$2x^2 + 2xy + 4yz + z^2 = 1 $
By writing this equation as
$x^TAx$ = 1,
with A a real symmetric matrix, show that there is an orthonormal basis such that, if we use coordinates
$(u, v, w)$ with respect to this new basis, Σ takes the form
$λu^2 + µv^2 + νw^2 = 1.$
Find $λ$, $µ$ and $ν$ and hence find the minimum distance between the origin and Σ.
Hint: it is not
necessary to find the basis explicitly.
I found matrix $A$ = $\begin{pmatrix}
2& 1 & 0 \\
1 & 0 & 2 \\
0 & 2 & 1
\end{pmatrix} $ , and figured that $λ$, $µ$ and $ν$ should be its eigenvalues, after changing the basis to the eigenvectors. The eigenvalues are $ 3, \sqrt{3}, -\sqrt{3} $, so I have $3u^2 + \sqrt{3}^2 -\sqrt{3}w^2 = 1.$
I feel like I have made a mistake somewhere, I don't know how to proceed.
|
Apply Lagrange Multipliers Method to minimize $f(x,y,z)= x^2+y^2+z^2$ subject to $g(x,y,z)= 2x^2+2xy+4yz+z^2-1$, results in $x=y=z=\frac13$ or $x=y=z=-\frac13$ with the minimum value of $f(x,y,z)=\frac13$. Therefor the minimum distance is $\sqrt{\frac13}$.
|
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|
Relatively prime factors of $24500$
Let $N=24500$, then find the number of ways by which $N$ can be resolved into two coprime factors?
My tries:
$N=24500=2^2\cdot 5^3\cdot 7^2$, for co prime no those two factors of $24500$ should share something in common, so the two factors should be such that either one of the two should use all of $2's,5's,7's$, hence picking up any two from $2^2\cdot 5^3\cdot 7^2\ \ ,\ \# ={3\choose {2}}=3$, then multiplying them together to get new number, for example $2^2\cdot 7^2=196$, then we are left with $5^3$, hence one such possible resolution should be $196\cdot 5^3$. So, ${3\choose {2}}=3$ should count all answer, but book says answer should be $4$.
Please help.
|
For integer $a,b,c>0$
$$2^a\cdot5^b\cdot7^c=(1\cdot2^a)(1\cdot5^b)(1\cdot7^c)$$
$=((1\cdot2^a5^b)$ or $(2^a\cdot5^b))(1\cdot7^c)$
$=(1\cdot2^a5^b7^c)$ or $(7^c\cdot2^a5^b)$ or $(2^a7^c,5^b)$ or $(2^a,5^b7^c)$
|
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Weird result proving : $1+2+3+...+n<\frac{(2n+1)²}{8} $
$$1+2+3+...+n<\frac{(2n+1)²}{8} $$
So, I'm going through Apostol's Calculus, and solving this:
I want to prove the aforementioned inequality, and proceed like this :
*
*$A(1) = 1 < \frac{9}{8}$
$A(2) = 1+2=3 < \frac{25}{8}$
$...$
*
*$A(k)=1+2+3+...+k<\frac{(2k+1)²}{8} | k≥1 $
*$A(k+1)=1+2+3+...+k+(k+1)<\color{red}{\frac{(2(k+1)+1)²}{8}} $
*$1+2+3+...+k+(k+1)<\color{red}{\frac{(2k+1)²}{8}+(k+1)} $
Now, following the way how Apostol proceeded from this point in other proofs like this one in the book, I have to prove:
$\frac{(2k+1)²}{8}+(k+1)<\frac{(2(k+1)+1)²}{8}$
But
*
*$\frac{(2k+1)²}{8}+(k+1)=\frac{4k²+12k+9}{8}$
*$\frac{(2(k+1)+1)²}{8}=\frac{4k²+12k+9}{8}$
So $\color{blue}{\frac{(2k+1)²}{8}+(k+1)<\frac{(2(k+1)+1)²}{8}}$ is false, but according to this other proof in the book
$\color{blue}{\frac{(2k+1)²}{8}+(k+1)<\frac{(2(k+1)+1)²}{8}}$ should be true, for the inequality to hold, but the result I'm getting tells me
$$1+2+3+...+n=\frac{(2n+1)²}{8} $$
which seems to be false according to
$A(1) = 1 < \frac{9}{8}$
$A(2) = 1+2=3 < \frac{25}{8}$
and
So, what am I doing wrong?
|
You are almost done. Note that you should show that
$$1+2+3+\dots+k+(k+1)<\frac{(2k+1)^2}{8}+(k+1)\color{red}{\leq }\frac{(2(k+1)+1)^2}{8}.$$
which implies that
$$1+2+3+\dots+k+(k+1)< \frac{(2(k+1)+1)^2}{8}.$$
So showing that
$$\frac{(2k+1)^2}{8}+(k+1)=\frac{(2(k+1)+1)^2}{8}$$
concludes the proof.
|
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Solve $x^3 +y^3 + z^3 =57$ How can we solve $x^3 + y^3 + z^3 =57$ efficiently in a shorter way. $x$ $y$ and $z$ are integers. Given that modulus of $x$ $y$ and $z$ is less than or equal to five. We can of course do by hit and trial but what is the method of solving such questions. I actually stumbled upon this equation while solving a determinant. How to proceed. Pls help
|
With some forethought, you can bring it down to one case to check directly, which turns out to be a solution. A previous version of this answer examined the equation mod $7$, which led to $16$ cases to directly examine. This answer examines mod $9$, which works out even better. (The reason $7$ and $9$ are good moduli to consider is because there are relatively few cubes mod these numbers.)
Mod $9$, the only cubes are $0$, $1$, and $8$. For solutions to $X+Y+Z\equiv57\equiv3$, the only solution is $1+1+1\equiv3$. This means $x^3\equiv y^3\equiv z^3\equiv1$ mod $9$, which means $x\equiv y\equiv z\equiv1$ mod $3$. So all of $x$, $y$, and $z$ are among $\{-5,-2, 1, 4\}$.
As in @RossMillikan's answer, we can assume $x, y\leq z$ and $z$ must equal $4$. So now we must solve $x^3+y^3=-7$, with $x\leq y$ among $\{-5,-2, 1, 4\}$.
Consider this equation mod $7$, where the cubes of $\{-5,-2, 1, 4\}$ are equivalent to $\{1,6, 1, 1\}$ respectively. The only way to sum two of these and get $-7\equiv0$ is using $6$ and $1$. Therefore one of $x,y$ is $-2$. Wlog, assume it's $x$, and examining $(-2)^3+y^3=-7$, we have that $y=1$ makes a complete solution $(x,y,z)=(-2,1,4)$.
|
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Property of Medians and Cirumcircle Let $ABC$ be a non-isosceles triangle. Medians of $\triangle ABC$ intersect the circumcircle in points $L,M,N$. If $L$ lies on the median of $BC$ and $LM=LN$, then prove that $2a^2=b^2+c^2$.
My Attempt:
Let $G$ be the centroid of $\triangle ABC$ and $D$ be the mid-point of $BC$.
Since $LM=LN$, therefore $LM$ and $LN$ will subtend equal angles on the circumference of circumcircle
$\Rightarrow \angle GBL=\angle LCG$
$GL=GL$
Altitude of $\triangle BGL$ from $B$ to $GL$=Altitude of $\triangle LGC$ from $C$ to $GL$
Area($\triangle BGL$)=Area($\triangle LGC$)
Now, based on this ,can it be said that $\square GBLC$ is a parallelogram. If yes then
$$DL=GD=\frac{m_{a}}{3}$$
$\Rightarrow AD.DG=BD.DC$
$$m_{a}.\frac{m_{a}}{3}=\frac{a^2}{4}$$
$\Rightarrow 4m^2_{a}=3a^2$
$\Rightarrow 2b^2+2c^2-a^2=3a^2$
$\Rightarrow b^2+c^2=2a^2$
I am not sure whether this justification is sufficient(the one that I have written in bold). What more can be added to seal the issue
|
Consider $\triangle AGB$ and $\triangle MGL$
$$\angle BGA=\angle LGM$$
$$\angle GAB=\angle GML$$
$$\triangle AGB \sim \triangle MGL$$
$$\frac{GB}{GL}=\frac{AG}{GM}=\frac{AB}{LM}$$
$$AG=\frac{c}{LM}GM$$
Similarly $\triangle AGC \sim \triangle NGL$
$$\frac{AG}{GN}=\frac{GC}{LG}=\frac{AC}{LN}$$
$$AG=\frac{b}{LN}GN$$
Thus$$AG=\frac{b}{LN}GN=\frac{c}{LM}GM$$
$$\Rightarrow c.GM=b.NG$$
Also, we have $$BG.GM=CG.GN$$
$$\Rightarrow \frac{2}{3}m_{b}.GM=\frac{2}{3}m_{c}.NG$$
$$\Rightarrow \frac{2}{3}m_{b}.\frac{b}{c}.NG=\frac{2}{3}m_{c}.NG$$
$$\Rightarrow b.m_{b}=c.m_{c}$$
$$\Rightarrow b\sqrt{2c^2+2a^2-b^2}=c\sqrt{2a^2+2b^2-c^2}$$
$$\Rightarrow 2b^2c^2+2a^2b^2-b^4=2a^2c^2+2b^2c^2-c^4$$
$$\Rightarrow 2a^2(b^2-c^2)=b^4-c^4$$
$$\Rightarrow 2a^2=b^2+c^2$$
|
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Two fair coins are tossed until both turn up heads A penny and a dime are tossed together until both turn up heads, after which no more tosses are made. Find the expected number of times the penny comes up heads.
What I've tried:
Let $X$ and $Y$ be the number of times the penny and dime come up heads, respectively. Then, for $x = 1,2,3,\ldots$
$$P(X = x) = \sum_{y=1}^\infty P(X = x, Y = y) \\ = \sum_{y=1}^x P(X=x,Y=y) + \sum_{y=x+1}^\infty P(X=x, Y=y) \\ = \sum_{y=1}^x \sum_{n=x}^\infty \left( \frac{1}{4} \right) \binom{n-1}{x-1} \left( \frac{1}{2} \right)^{n-1} \binom{n-1}{y-1} \left( \frac{1}{2} \right)^{n-1} + \\ \sum_{y=x+1}^\infty \sum_{n=y}^\infty \left( \frac{1}{4} \right) \binom{n-1}{x-1} \left( \frac{1}{2} \right)^{n-1} \binom{n-1}{y-1} \left( \frac{1}{2} \right)^{n-1} \\ = \sum_{y=1}^x \sum_{n=x}^\infty \left( \frac{1}{4} \right)^n \binom{n-1}{x-1} \binom{n-1}{y-1} + \sum_{y=x+1}^\infty \sum_{n=y}^\infty \left( \frac{1}{4} \right)^n \binom{n-1}{x-1} \binom{n-1}{y-1}$$
Is there a way to simplify the above expression? Or is there an easier approach that I'm missing?
|
Yes, there is an easier way. Suppose the expected number of times the penny comes up heads is $x$. Then consider the outcome of the first pair of tosses: if both come up heads, $x = 1$, with probability $1/4$.
With probability $1/2$, the penny comes up tails. Regardless of the outcome of the dime, we don't stop; but since the penny came up tails, this toss also doesn't count toward the total number of times the penny comes up heads--so in this case, it is as if the toss never occurred.
In the remaining case, with probability $1/4$, the penny comes up heads but the dime comes up tails. We don't stop, but now the penny has shown a head and for the remaining tosses, we can reason that the expected number of heads shown by the penny is still $x$, because the outcome of pairs of tosses are independent of previous tosses--the coins don't "remember" what they did before.
To summarize, we must have $$x = \frac{1}{4}(1) + \frac{1}{2}x + \frac{1}{4}(x+1).$$ This gives $x = 2$ as the expected number of heads seen from the penny.
For your own benefit, you should consider how this problem generalizes to the case where the probability of the penny showing heads is $p_1$, and the dime's probability of showing heads is $p_2$. What is the expectation in terms of these probabilities?
For even more understanding, can you compute the probability distribution of the random number of heads obtained from the penny (in the fair case, $p_1 = p_2 = 1/2$)? What is it? Why does it make intuitive sense? Can you use this result to formulate an alternate method of solution?
|
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|
Compute $e^{Ax}$ for the $2 \times 2$ matrix $A$ Consider the real $2 \times 2$ matrix $A$ $$A=\begin{pmatrix} 1 & 0 \\ 0 & -1\end{pmatrix}$$
I want to compute $e^{Ax}$.
I can see that $A^2=I$ which looks like it should be useful.
Using the definition of the matrix exponential $$e^{Ax}=I+Ax+\frac{(Ax)^2}{2!}+ \cdots$$ I find that $$e^{Ax}=I\left(1+\frac{x^2}{2!}+ \cdots\right)+A\left(x+\frac{x^3}{3!}+ \cdots\right)$$
Is there anything I can do from here?
|
We have $$e^{Ax}=\begin{pmatrix} 1 & 0 \\ 0 & 1\end{pmatrix}\left(1+\frac{x^2}{2!}+ \cdots\right)+\begin{pmatrix} 1 & 0 \\ 0 & -1\end{pmatrix}\left(x+\frac{x^3}{3!}+ \cdots\right)$$ This can be written as $$\begin{pmatrix} 1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+... & 0 \\ 0 & 1+(-x)+\frac{x^2}{2!}+\frac{(-x)^3}{3!}+...\end{pmatrix}$$ so $$\boxed{e^{Ax}=\begin{pmatrix} e^x & 0 \\ 0 & e^{-x}\end{pmatrix}}$$
|
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|
Check if the sequence converges using Cauchy criterion I need to check if the sequence of $x_n = \dfrac1{\sqrt1} + \dfrac1{\sqrt2} ... + \dfrac1{\sqrt n}$ converges using the Cauchy criterion.
Obviously, first I have to do is to somehow check if $|x_n - x_{n+p}| $ (where $p$ is any positive whole number) is convergent, but I have no idea even how to start. How can it be solved?
|
For any $n \in \mathbb{N}$ we have:
\begin{align}
x_{n^2} - x_n &= \left(\frac{1}{\sqrt{1}} +\ldots + \frac{1}{\sqrt{n^2}}
\right) - \left(\frac{1}{\sqrt{1}} +\ldots + \frac{1}{\sqrt{n}}
\right)\\
&= \frac{1}{\sqrt{n+1}} + \frac{1}{\sqrt{n+2}} + \ldots + \frac{1}{\sqrt{n^2}}\\
&\ge \underbrace{\frac{1}{\sqrt{n^2}} + \frac{1}{\sqrt{n^2}} + \ldots + \frac{1}{\sqrt{n^2}}}_{n^2 - n \text{ terms}}\\
&= \frac{n^2 - n}{\sqrt{n^2}}\\
&= \frac{n^2 - n}{n}\\
&= n - 1\\
\end{align}
Therefore, $(x_n)_{n=1}^\infty$ cannot be Cauchy since $|x_{n^2} - x_n| \ge n-1$ for all $n \in \mathbb{N}$.
We conclude that $(x_n)_{n=1}^\infty$ does not converge.
|
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|
Factoring $x^7+3x^6+9x^5+27x^4+81x^3+243x^2+729x+2187$
Question: How would you factor$$P(x)=x^7+3x^6+9x^5+27x^4+81x^3+243x^2+729x+2187$$
I thought for a while and realized that the coefficients are in powers of $3$, so $x=-3$ is a factor. Taking that factor out, we see that the septic is equal to$$P=(x+3)(x^2+9)(x^4+81)$$I'm wondering, however, if there is a quicker way to factor it because the original method was pretty tedious.
|
For the sake of an alternative, less clever approach: pretending to not notice the pattern of increasing powers of $3$, the root $x=-3$ can also be found by brute force using the rational root theorem. Quite obviously, the polynomial has no positive roots, so it's enough to try the negative divisors of $2187=3^7$, which finds $-3$ pretty quickly.
Then, dividing by the factor of $x+3$ using (for example) polynomial long division gives:
$$
P(x)=(x+3)(x^6 + 9 x^4 + 81 x^2 + 729)
$$
The sextic that remains to be factored is a cubic in $y=x^2$:
$$
Q(y) = y^3+9y^2+81y+729
$$
Using the rational root theorem again, $y=-9$ is a root, then dividing by $y+9$ gives:
$$
Q(y) = (y+9)(y^2+81)
$$
So in the end $P(x)=(x+3)Q(x^2)=(x+3)(x^2+9)(x^4+81)\,$.
|
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How to algebraically solve $\frac{1}{x} < 5$ inequality to obtain two solutions? Given the inequality:
$\frac{1}{x} < 5$
In order to find a solution, I would normally multiply both sides by $x$:
$1 < 5x$
Then I would divide by $5$
$\frac{1}{5} < x$
To obtain the solution: $x > \frac{1}{5}$.
Now, the thing is, the solutions are actually two: $x > \frac{1}{5}$ and $x < 0$
How am I supposed to reach this conclusion algebraically? It seems I'm not able to obtain the second solution ($x < 0$).
Thanks!
|
Note that by going from $\dfrac{1}{x} < 5$ to $1 < 5x$, you are assuming that $x > 0$. You see that by assuming $x > 0$, you obtain $x > \dfrac{1}{5}$.
Now assume that $x < 0$. Then $\dfrac{1}{x} < 5 \implies 1 > 5x\implies x < \dfrac{1}{5}$ (flipping the inequality). But, remember, we assumed that $x < 0$. Thus, if $x < \dfrac{1}{5}$ and $x < 0$, we can succinctly write this as $x < 0$.
|
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|
Find all real $x$ for which $\log_{10}(2^{x-1}+3^{x+1})=2x-\log_2{3^x}$ My attempt:
$$\log_{10}(2^{x-1}+3^{x+1})=x\cdot(2-\log_2{3})$$
I am stuck after this. If the terms inside LHS were a product, this problem would have been a piece of cake. The domain of $x$ is all real here so I can't limit the range of $x$ for a hit and trial solution.
Any starting steps please?
|
Hint: power each side by 2.
$$A = 2^{x-1} + 3^{x+1} = 2^{2x-xlog_2(3)}$$
$$= (2^{2-log_2(3)})^x = \left(\frac{4}{3}\right)^x$$
$$\Rightarrow 2^{x-1}3^x+3^{2x+1} = 2^{2x}$$
Suppose $X = 2^x$, $Y=3^x$:
$$\frac{1}{2}XY+ 3Y^2 = X^2$$
$$ X^2 - 3Y^2 - \frac{1}{2}XY = 0$$
|
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To find the sum of $1+\frac13-\frac12-\frac14-\frac16+\frac15+\frac{1}{7}-\frac{1}{8}-\frac{1}{10}-\frac{1}{12}+\ldots$ I have to find the sum of:
$$1+\frac13-\frac12-\frac14-\frac16+\frac15+\frac{1}{7}-\frac{1}{8}-\frac{1}{10}-\frac{1}{12}+\ldots$$
My attempt:
$$\left( 1+\frac13-\frac12-\frac14-\frac16\right)+\left(\frac15+\frac{1}{7}-\frac{1}{8}-\frac{1}{10}-\frac{1}{12}\right)+\ldots$$
\begin{eqnarray*}\sum_{n=0}^{+\infty}\left(\frac{1}{4n+1}+\frac{1}{4n+3}-\frac{1}{6n+2}-\frac{1}{6n+4}-\frac{1}{6n+6}\right)&=&\int_{0}^{1}\frac{1+x^2}{1-x^4}-\frac{x+x^3+x^5}{1-x^6}\,dx\\&=&\log2\end{eqnarray*}
But the answer is given as $\frac12\log{\frac83}$.Where did I make a mistake?
Also,could I get the solution using $\log 2=1-\frac12+\frac13-\frac14+\frac15-\frac16+\ldots$?If so,how?
|
The series is not absolutely convergent; therefore, you cannot move around terms freely.In other words,
\begin{eqnarray*}\sum_{n=0}^{+\infty}\left(\frac{1}{4n+1}+\frac{1}{4n+3}-\frac{1}{6n+2}-\frac{1}{6n+4}-\frac{1}{6n+6}\right)\neq \sum_{n=0}^{+\infty}\left(\frac{1}{4n+1}+\frac{1}{4n+3}\Bigg)-\\-\sum_{n=0}^{\infty}\Bigg(\frac{1}{6n+2}+\frac{1}{6n+4}+\frac{1}{6n+6}\right) \end{eqnarray*}
In fact, both terms on the right hand side of the above equation are both diverging to $\infty$, which should make your mistake even obvious.
If you want a hint, then consider $$f(t) = \sum_{n=0}^{+\infty}\left(\frac{t^{4n+1}}{4n+1}+\frac{t^{4n+3}}{4n+3}-\frac{t^{6n+2}}{6n+2}-\frac{t^{6n+4}}{6n+4}-\frac{t^{6n+6}}{6n+6}\right)$$
on $[0,1)$ and use Abel's continuity theorem and differentiation.
|
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|
Count rational numbers in lowest terms in $[0,1]$ I was wondering if there is a formula for the count of rational numbers $\frac{p}{q}$ in lowest terms in $[0,1]$ similar to the sequence in this answer.
For example:
*
*For $q = 1$ we have two: $\{\frac{0}{1}, \frac{1}{1}\}$
*For $q = 2$ we have one: $\{\frac{1}{2}\}$
*For $q = 3$ we have two: $\{\frac{1}{3}, \frac{1}{2}\}$
*For $q = 4$ we have two: $\{\frac{1}{4}, \frac{3}{4}\}$
*For $q = 5$ we have four: $\{\frac{1}{5}, \frac{2}{5}, \frac{3}{5}, \frac{4}{5}\}$
*For $q = 6$ we have two: $\{\frac{1}{6}, \frac{5}{6}\}$
*For $q = 7$ we have six: $\{\frac{1}{7}, \frac{2}{7}, \frac{3}{7}, \frac{4}{7}, \frac{5}{7}, \frac{6}{7}\}$
*and so on...
So we have a total of $19$ rational numbers in lowest terms with $q \le 7$. I'd be happy with the total count for all $q \le N$ too.
|
Your question is related to Euler's totient function. In particular, the total number of fractions $p/q$, where $1\le p < q$, and $\gcd(p,q)=1$ equals $\varphi(q)$, where $\varphi(\cdot)$ is the Euler's totient function. Consequently, the required number of fractions is
$$2+\sum_{q=2}^{N} \varphi(q).$$
|
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Is 2018 special because of these properties? I discovered that: $$2018=(6^2)^2+(5^2)^2+(3^2)^2+(2^2)^2$$
We also have: $$13^2+43^2=2018$$
And we have:
$$2018=44^2+9^2+1^2$$
I somehow tend to believe that there could be a finite number of these numbers that are sum of two squares, three squares and four fourth powers.
So we have a system of three Diophantine equations:
$$n=a^2+b^2$$
and $$n=c^2+d^2+e^2$$
and $$n=f^4+g^4+h^4+i^4$$
where, $n,a,b,c,d,e,f,g,h,i \in \mathbb N$.
Is there a finite number of these numbers?
Edit : Also, it is $$2018=35^2+26^2+8^2+7^2+2^2$$ a sum of five squares.
And of $$2018=11^3+7^3+7^3+1^3$$ four cubes.
|
One infinite set is $2018k^4$ for any natural $k$. I strongly suspect that there are plenty more. Numbers that are a sum of three squares are very common. Numbers that are a sum of two squares are not so rare, so I would just start picking sums of four cubes and try to satisfy the other two. Another example is $$1^4+2^4+3^4+6^4=1394=2\cdot 17 \cdot 41 \equiv 4\pmod 8$$
so is a sum of two and three squares.
|
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Given $3\cos x - 4 \sin x = 2$, find $3 \sin x + 4 \cos x$ without first solving for $x$ If $$3\cos{x}-4\sin{x}=2$$
find $$3\sin{x} +4\cos{x} $$
I have solved the equation for $x$, then calculated the required value, but I think there is a direct solution without solving the equation.
|
Hint:
Using Brahmagupta-Fibonacci Identity,
$$(a\cos x-b\sin x)^2+(a\sin x+b\cos x)^2=(a^2+b^2)(\cos^2x+\sin^2x)=?$$
|
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Inequality: $\frac{1+x}{1-x} > e^{2x}$ for $0 < x < 1$ I was asked to prove the inequality $\frac{1+x}{1-x} > e^{2x}$ for $0 < x < 1$
My approach would be with the differential equation.
$f(x) = e^{2x} - \frac{1+x}{1-x} $
$f(x)' = 2e^{2x} − \frac{x+1}{(1−x)^2} − \frac{1}{1−x} = 2e^{2x}−\frac{2}{(1−x)^2} $
Now it would suffice to say that $f(x)' > 0$ for $0 < x < 1$, but $f(x)' < 0$ for $0 < x < 1$
Could you please tell me how to solve this inequality correctly (with the differential equation)?
|
$$\frac{1+x}{1-x} > e^{2x}\iff \ln\left(\frac{1+x}{1-x}\right)>2x$$
$$\ln\left(\frac{1+x}{1-x}\right)=\ln(1+x)-\ln (1-x)>x-\frac{x^2}2-\left(-x-\frac{x^2}2 \right)=2x$$
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|
Determine exponential generating function of the case given I have a problem to determine EGF that show how many ways we can distribute r different people into n different rooms such that every room has at least 2 people and no more than 5 people. Of course, we know that this is a permutation case (and that's why it leads to EGF).
In ordinary generating function, we can manipulate the expression $(x^2 + x^3 + x^4 + x^5)$ as $x^2(1 + x + x^2 + x^3) = x^2(1-x^4)(1 + x + x^2 + \cdots)$, but how about in exponential generating function which lets to
$\left(\dfrac{x^2}{2!} + \dfrac{x^3}{3!} + \dfrac{x^4}{4!} + \dfrac{x^5}{5!}\right)$?
Could anyone help me? Your help will be appreciated. Thank you.
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Using the coefficient of operator $[x^j]$ to denote the coefficient of $x^j$ in a series the wanted number is
\begin{align*}
r![x^r]&\left(\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}+\frac{x^5}{5!}\right)^n\\
&=r![x^r]\frac{x^{2n}}{5!}\left(60+20x+5x^2+x^3\right)^n\\
&=\frac{r!}{5!}[x^{r-2n}]\left(60+20x+5x^2+x^3\right)^n\tag{1}\\
\end{align*}
Here we have $r\geq 2n$ since we need at least $2n$ people to arrange them into $n$ rooms.
Comment: In (1) we apply the rule $[x^{p-q}]A(x)=[x^p]x^qA(x)$.
Hint:
*
*The paper Bracket notation for the "coefficient of" operator by D.E. Knuth might be instructive.
*This MSE post provides an easy application of the coefficient of operator.
*... and this one a somewhat more demanding.
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|
How to obtain the following definite integral? I have no idea how to obtain the answer to the following problem:
$$I=\int_{0}^{1}\frac{dx}{\sqrt{8-x^2-x^3}}$$
The answer has been given as
$$\sin^{-1} \frac{1}{2\sqrt{2}}< I < \frac{1}{\sqrt{2}}\sin^{-1} \frac{1}{2}$$
Any help will be appreciated. Thanks in advance.
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$I$ is an elliptic integral, accurate approximations can be deduced from convexity. Over $(0,1)$ we clearly have $8-x^2-x^3 > 8-2x^2$, hence
$$ I < \int_{0}^{1}\frac{dx}{\sqrt{8-2x^2}} = \frac{\pi}{6\sqrt{2}} \tag{Ub}$$
but we also have $\frac{1}{\sqrt{8-x^2-x^3}}>\frac{1}{2\sqrt{2}}\left(1+\frac{x^2+x^3}{16}\right)$, hence
$$ I > \int_{0}^{1}\frac{1}{2\sqrt{2}}\left(1+\frac{x^2+x^3}{16}\right)\,dx = \frac{199}{384\sqrt{2}}\tag{Lb}$$
and by combining the upper bound and the lower bound
$$ 0.3664 < I < 0.3703. \tag{B} $$
$(\text{Lb})$ can be improved through Jensen's inequality:
$$ I > \frac{1}{2\sqrt{2}\sqrt{1-\int_{0}^{1}\frac{x^2+x^3}{8}\,dx}}=\sqrt{\frac{12}{89}}>0.36719 $$
and $(\text{Ub})$ can be improved through the Cauchy-Schwarz inequality:
$$ I < \sqrt{\int_{0}^{1}\frac{dx}{8-x^2-x^3}}<0.36841 $$
such that $I=\color{green}{0.36}77\pm 0.0008$.
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|
Solve $y^{(4)}-2y^{(3)}+2y'-y=xe^x.$ Solve $y^{(4)}-2y^{(3)}+2y'-y=xe^x.$
The characteristic equation is $(r-1)^3(r+1)\Rightarrow y_h=(C_1+C_2x+C_3x^2)e^x+C_4e^{-x}.$
The problem is the particular equation. Why doesn't it work with the ansatz
$$y=(ax+b)e^x?$$
I get
\begin{array}{lcl}
y & = & e^x(ax+b) \\
y' & = & e^x(a(x+1)+b)\\
y'' & = & e^x(a(x+2)+b) \\
y^{(3)} & = & e^x(a(x+3)+b) \\
y^{(4)} & = & e^x(a(x+4)+b)
\end{array}
Setting these in i get $$e^x[((a(x+4)+b))-2((a(x+3)+b))+2(e^x(a(x+1)+b))-(e^x(ax+b))] = e^x\cdot 0=0.$$
So this doesn't work. Is it because I already have corresponding powers of $x$
in the homogenous solution? How can i fix my ansatz?
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Solve step by step the equation $$(D-1)^3(D+1)y=xe^x$$ (where $D$ is differentiation operator) and there is no need to worry about guessing particular solution. Let $z=(D-1)^3y$ so that $$(D+1)z=xe^x$$ or $$D(e^xz) =xe^{2x}$$ or $$ze^x=\frac{xe^{2x}}{2}-\frac{e^{2x}}{4}+a$$ or $$z=\frac{xe^x} {2}-\frac{e^x}{4}+ae^{-x}$$ Next put $t=(D-1)^2y$ so that $$(D-1)t=z$$ or $$D(te^{-x})= ze^{-x} =\frac{x} {2}-\frac{1}{4}+ae^{-2x}$$ or $$t=e^x\left(\frac{x^2}{4}-\frac{x}{4}\right) -\frac{ae^{-x}}{2}+be^x$$ Since $a$ is arbitrary constant one can replace $-a/2$ by $a$ to get $$t=e^x\left(\frac{x^2}{4}-\frac{x}{4}\right)+ae^{-x}+be^x$$ Going further let $u=(D-1)y$ so that $$(D-1)u=t$$ or $$D(ue^{-x}) =te^{-x} =\frac{x^2}{4}-\frac{x}{4}+ae^{-2x}+b$$ or $$u=e^{x} \left(\frac{x^3}{12}-\frac{x^2}{8}\right)-\frac{ae^{-x}}{2}+bxe^x+ce^x$$ Replacing $-a/2$ by $a$ we get $$u=e^x\left(\frac{x^3}{12}-\frac{x^2}{8}\right)+ae^{-x}+bxe^x+ce^x$$ and in similar manner solving the final equation $(D-1)y=u$ we get $$y=e^x\left(\frac{x^4}{48}-\frac{x^3}{24}\right)+ae^{-x}+bx^2e^x+cxe^x+de^x$$ which is the desired solution.
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|
Evaluate a limit at infinity Find the following limit without Lospital rule nor series
$$\lim_{n\to \infty}\left(\frac{1}{a^n}+\frac{1}{b^n}+\frac{1}{c^n}\right)^{\frac{1}{n}}$$
I tried to take log to both sides to be
$$\ln L=\lim_{n\to \infty}\frac{\ln\left((ab)^n +(bc)^n+(ac)^n\right)}{n}-\lim_{x\to \infty}\frac{n\ln(abc)}{n}$$
$$=\lim_{n\to \infty}\frac{\ln\left((ab)^n +(bc)^n+(ac)^n\right)}{n}-\ln(abc)$$
But i could not solve the first limit? $$0<a<b<c$$
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Assume $a,b,c>0$ and wlog $\frac1a=max\{\frac1a,\frac1b\frac1c\}$ thus
$$\left(\frac{1}{a^n}+\frac{1}{b^n}+\frac{1}{c^n}\right)^{\frac{1}{n}}=e^{\frac{\log{\left(\frac{1}{a^n}+\frac{1}{b^n}+\frac{1}{c^n}\right)}}{n}}=e^{\frac{\log{\left(\frac{1}{a^n}\right)}+\log{\left(1+\frac{a^n}{b^n}+\frac{a^n}{c^n}\right)}}{n}}\to\frac1a$$
indeed
$$\frac{\log{\left(\frac{1}{a^n}\right)}+\log{\left(1+\frac{a^n}{b^n}+\frac{a^n}{c^n}\right)}}{n}=\frac{n\log{\left(\frac{1}{a}\right)}+\log{\left(1+\frac{a^n}{b^n}+\frac{a^n}{c^n}\right)}}{n}=$$
$$=\log{\left(\frac{1}{a}\right)}+\frac{\log{\left(1+\frac{a^n}{b^n}+\frac{a^n}{c^n}\right)}}{n}\to \log{\left(\frac{1}{a}\right)}+0=\log{\left(\frac{1}{a}\right)}$$
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|
Find $4762^{5367} \pmod{13}$
Find $4762^{5367} \pmod{13}$
So I started off and got:
$$4762 = 4 \pmod{13}$$
Therefore, $$4762^{5367} = 4^{5367} \pmod{13}$$
But I did not know how to proceed after this.
Any help?
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For a small modulus like $13$, it's worth just looking for the pattern that $4$ follows on exponentiation:
$\bmod 13: \\
\begin{align}
\qquad
4^0 &\equiv 1\\
4^1 &\equiv 4\\
4^2 &\equiv 16 \equiv 3\\
4^3 &\equiv 12 \equiv -1\\
4^4 &\equiv -4 \equiv 9 \\
4^5 &\equiv -3 \equiv 10\\
4^6 &\equiv -1^2 \equiv 1\\
\end{align}$
And what we're after is closing the repeat, which we can be sure will happen because there are a limited number of values $(\phi(13)=12)$ that the values can take.
We could be sure in advance, in fact, that $4^6\equiv 1\bmod 13$ because Fermat's little theorem gives $a^{12}\equiv 1 \bmod 13$ (for $13\nmid a$) and thus $2^{12}=4^6\equiv 1 \bmod 13$.
Then $4^{5364} = 4^{6\times 894} \equiv 1^{894}\equiv 1 \bmod 13$. So $4^{5367}\equiv 4^3\equiv 64\equiv 12 \bmod 13$
|
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|
Which graph corresponds to which equation? The figure below represents the graphs of parametric curve $x(t) = \sin(t)$, $y(t) = \frac{1}{2}\sin(2t)$, $0 \le t \le 2\pi$ and polar curve $r^2 = \cos(2\theta)$, $0\le \theta \le 2\pi$. Which curve corresponds to which equation?
I simply have no idea how to approach this, as it seems that $t$ doesn't have a direct relationship to $\theta$.
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The point $(x,y)=(1,0)$ belongs to both curves, at $t=\pi/2$ for the parametric curve and at $\theta=0$ for the polar curve, so let us compare the behaviour of the two curves in a neighborhood of this point.
Our only prerequisite shall be the one term Taylor expansion $$\sqrt{1-2z}=1-z+o(z^2)$$ when $z\to0$, which (if need be) follows from the double inequality, with an elementary proof, stating that, for every $|z|\leqslant\frac12$, $$1-z-2z^2\leqslant\sqrt{1-2z}\leqslant1-z$$
*
*On the polar curve, $r^4=r^2(\cos^2\theta-\sin^2\theta)$ hence $(x^2+y^2)^2=x^2-y^2$. Solving this quadratic identity in $x^2$ yields $2x^2=1-2y^2\pm\sqrt{1-8y^2}$.
Thus, in a neighborhood of $(x,y)=(1,0)$, on the polar curve, $$2x^2=1-2y^2+\sqrt{1-8y^2}$$
The limited expansion when $y\to0$ is $$2x^2=2-6y^2+o(y^2)$$
*On the parametric curve, $y^2=\sin^2t\cos^2t=x^2(1-x^2)$ hence $x^4-x^2+y^2=0$. Solving this quadratic identity in $x^2$ yields $2x^2=1\pm\sqrt{1-4y^2}$.
Thus, in a neighborhood of $(r,\theta)=(1,0)$, which is also $(x,y)=(1,0)$, on the parametric curve, $$2x^2=1+\sqrt{1-4y^2}$$
The limited expansion when $y\to0$ is $$2x^2=2-2y^2+o(y^2)$$
This proves that the parametric curve leaves $(1,0)$ by staying slightly farther from $(0,0)$ than the polar curve, thus the parametric curve is drawn in red and the polar one in green.
Edit: For a non asymptotic approach, one should study the sign of $$u(y)=\left(1+\sqrt{1-4y^2}\right)-\left(1-2y^2+\sqrt{1-8y^2}\right)$$ which is also $$u(y)=2y^2+\left(\sqrt{1-4y^2}-\sqrt{1-8y^2}\right)$$ The $2y^2$ term and the term in the parenthesis are both nonnegative hence $u(y)\geqslant0$ everywhere. This shows that the polar curve is always "inside" the parametric curve, extending the validity of the conclusion above from a neighborhood of $(1,0)$ to the whole curves.
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How to prove that this triangle is equilateral? Question: If $\cos A +\cos B +\cos C=\frac{3}{2}$, prove that the triangle is equilateral.
My attempt: According to cosine rule, $\cos A=\frac{b^2+c^2-a^2}{2bc}$, $\cos B=\frac{c^2+a^2-b^2}{2ca}$ and $\cos C=\frac{a^2+b^2-c^2}{2ab}$.
$\cos A +\cos B +\cos C=\frac{3}{2}$
$\therefore\frac{b^2+c^2-a^2}{2bc}+\frac{c^2+a^2-b^2}{2ca}+\frac{a^2+b^2-c^2}{2ab}=\frac{3}{2}$
$\implies\frac{ab^2+ac^2-a^3+bc^2+ba^2-b^3+ca^2+cb^2-c^3}{2abc}=\frac{3}{2}$
I have tried simplifying the given equation using cosine rule but could not get far. Please help.
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A slick way is to combine Carnot's theorem with Euler's inequality $R\geq 2r$.
$$ R\cos A+R\cos B + R\cos C $$
is the sum of the signed distances of the circumcenter from the triangle sides, and it equals $R+r$. Since $r<\frac{R}{2}$ unless $ABC$ is equilateral,
$$ R\cos A+R\cos B + R\cos C = \frac{3}{2}R$$
implies that $ABC$ is equilateral.
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How can I prove that $7+7^2+7^3+...+7^{4n} = 100*a$ (while a is natural number)? How can I prove that $7+7^2+7^3+...+7^{4n} = 100*a$ (while a is entire number) ?
I thought to calculate $S_{4n}$ according to:
$$ S_{4n} = \frac{7(7^{4n}-1)}{7-1} = \frac{7(7^{4n}-1)}{6} $$
But know, I don't know how to continue for get what that rquired.
I will be happy for help or hint.
After beautiful ideas for solving this question, someone know how to do it with induction too?
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It remains to show that $7^{4n} - 1$ is a multiple of $600$.
Since $600 = (2^3)(3)(5^2)$, the goal is equivalent to showing that the three congruences
\begin{align*}
7^{4n} &\equiv 1\;(\text{mod}\;2^3)\\[4pt]
7^{4n} &\equiv 1\;(\text{mod}\;3)\\[4pt]
7^{4n} &\equiv 1\;(\text{mod}\;5^2)\\[4pt]
\end{align*}
hold for all positive integers $n$.
Now simply note that
\begin{align*}
7 &\equiv -1\;(\text{mod}\;8)\\[4pt]
7 &\equiv 1\;(\text{mod}\;3)\\[4pt]
7^2 &\equiv -1\;(\text{mod}\;25)\\[4pt]
\end{align*}
Can you finish it?
For an inductive approach, note that
\begin{align*}
S_{4(n+1)}-S_{4n}
&=
\frac{7(7^{4(n+1)}-1)}{6}-\frac{7(7^{4n}-1)}{6}
\\[4pt]
&=
\frac{7(7^{4(n+1)}-7^{4n})}{6}
\\[4pt]
&=
\frac{7^{4n+1}(7^4-1)}{6}
\\[4pt]
&=
\frac{7^{4n+1}(2400)}{6}
\\[4pt]
&=7^{4n+1}(400)
\\[4pt]
\end{align*}
hence, if $S_{4n}$ is a multiple of $100$, then so is $S_{4(n+1)}$.
Since $S_{4n}$ is a multiple of $100$ when $n=1$, it follows (by induction on $n$), that $S_{4n}$ is a multiple of $100$, for all positive integers $n$.
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Range of $S,T$ in $2$ sums
If $\displaystyle S = \sum^{4n-1}_{k=3n}\bigg(\frac{k^2-7kn+13n^2}{n^3}\bigg)$ and $\displaystyle T= \sum^{4n}_{k=3n+1}\bigg(\frac{k^2-7kn+13n^2}{n^3}\bigg)$. then which one is/are
right $\; (a)\displaystyle \; S<\frac{5}{6}\; (b)\; T<\frac{5}{6}\; (c)\; S>\frac{5}{6}\; (d)\; T>\frac{5}{6}$
Try: assuming for $$\lim_{n\rightarrow \infty}S = \lim_{n\rightarrow \infty}\sum^{4n-1}_{k=3n}\bigg(\frac{k^2-7kn+13n^2}{n^3}\bigg) =\lim_{n\rightarrow \infty}\frac{1}{n} \sum^{4n-1}_{k=3n}\bigg[\bigg(\frac{k}{n}\bigg)^2-7\bigg(\frac{k}{n}\bigg)+13\bigg]$$
conversion into definite integration by substituting $\displaystyle \frac{k}{n}=x$ and $\displaystyle \frac{1}{n}=dx$
$$ = \int^{4}_{3}(x^2-7x+13)dx = \frac{5}{6}$$
could some help me how to estimate $S$ and $T$ , thanks
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First,\begin{align*}
S - T &= \sum_{k = 3n}^{4n - 1} \frac{1}{n^3} (k^2 - 7kn + 13n^2) - \sum_{k = 3n + 1}^{4n} \frac{1}{n^3} (k^2 - 7kn + 13n^2)\\
&= \frac{1}{n^3} ((3n)^2 - 7 \cdot 3n \cdot n + 13n^2) - \frac{1}{n^3} ((4n)^2 - 7 \cdot 4n \cdot n + 13n^2)\\
&= \frac{1}{n} - \frac{1}{n} = 0.
\end{align*}
Now,\begin{align*}
S &= \sum_{k = 3n}^{4n - 1} \frac{1}{n^3} (k^2 - 7kn + 13n^2) = \frac{1}{n^3} \sum_{k = 3n}^{4n - 1} k^2 - \frac{7}{n^2} \sum_{k = 3n}^{4n - 1} k + 13\\
&\geqslant \frac{1}{n^3} \int_{3n}^{4n} x^2 \,\mathrm{d}x - \frac{7}{n^2} \sum_{k = 3n}^{4n - 1} k + 13\\
&= \frac{1}{n^3} \cdot \frac{1}{3} ((4n)^3 - (3n)^3) - \frac{7}{n^2} \cdot \frac{1}{2} (3n + (4n - 1)) n + 13\\
&= \frac{37}{3} - \frac{49n - 7}{2n} + 13 > \frac{37}{3} - \frac{49}{2} + 13 = \frac{5}{6}.
\end{align*}
Therefore,$$
S = T > \frac{5}{6}.
$$
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Show that if $x+y+z=x^2+y^2+z^2=x^3+y^3+z^3=1$ then $xyz=0$ For any real numbers $x,y,z$ such that $x+y+z = x^2+y^2+z^2 = x^3+y^3+z^3 =1\\$ show that $x \cdot y \cdot z=0$.
I think that because $x \cdot y \cdot z = 0$ at least one of the three number should be equal to zero, but I'm stuck into relating the other things.
|
Consider $f(w)=(w-x)(w-y)(w-z).$ We have $f(w)=w^3-Aw^2+Bw-C$ where $A=x+y+z$ and $B=xy+yz+zx$ and $C=xyz.$
Suppose $x+y+z=x^2+y^2+z^2=x^3+y^3+z^3=1.$ We have $0=f(x)=f(y)=f(z).$ Hence $$0=f(x)+f(y)+f(z)=$$ $$=(x^3+y^3+z^3)-A(x^2+y^2+z^2)+B(x+y+z)-3C=$$ $$=1-A+B-3C=$$ $$=1-(x+y+z)+(xy+yz+zx)-3C=$$ $$=1-1+(xy+yz+zx)-3C=$$ $$=(xy+yz+zx)-3C.$$ Now $(xy+yz+zx)=\frac {1}{2} ((x+y+z)^2-(x^2+y^2+z^2))=\frac {1}{2} (1^2-1)=0 .$
Therefore $0=(xy+yz+zx)-3C=0-3C=0-3xyz.$
|
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Span of Integer Linear Combinations This question relates to linear combinations of vectors; however, vectors can only be scaled by integer values. This means that if we have two vectors $\binom{a}{b}$ and $\binom{c}{d}$, we can only have $$x\binom{a}{b} + y\binom{c}{d},$$ where $a, b, c, d, x,y\in \mathbb{Z}.$
I was wondering how to determine when linear combinations of two vectors will span $\mathbb{Z^2}$. I know that if the second vector is a multiple of the first, this will not be true, but I'm not sure about the other cases.
|
Claim: $\begin{pmatrix}a\\b\end{pmatrix}$ and $\begin{pmatrix}c\\d\end{pmatrix}$ span the whole $\mathbb Z^2=M_{12}(\mathbb Z)$ if and only if $\det\left[\begin{matrix}a&c\\b&d\end{matrix}\right]=ad-bc=\pm 1$.
Proof sketch: basically, the span is the whole of $\mathbb Z^2$ if and only if $\begin{pmatrix}1\\0\end{pmatrix}$ and $\begin{pmatrix}0\\1\end{pmatrix}$ can be represented as linear combinations of $\begin{pmatrix}a\\b\end{pmatrix}$ and $\begin{pmatrix}c\\d\end{pmatrix}$. However:
$$\begin{pmatrix}1\\0\end{pmatrix}=x\begin{pmatrix}a\\b\end{pmatrix}+y\begin{pmatrix}c\\d\end{pmatrix}$$
$$\begin{pmatrix}0\\1\end{pmatrix}=z\begin{pmatrix}a\\b\end{pmatrix}+t\begin{pmatrix}c\\d\end{pmatrix}$$
is equivalent to:
$$\left[\begin{matrix}a&c\\b&d\end{matrix}\right]\cdot\left[\begin{matrix}x&z\\y&t\end{matrix}\right]=\left[\begin{matrix}1&0\\0&1\end{matrix}\right]=I$$
so existence of $\left[\begin{matrix}x&z\\y&t\end{matrix}\right]$ implies that $\det\left[\begin{matrix}a&c\\b&d\end{matrix}\right]=\pm 1$, and, conversely, if $\det\left[\begin{matrix}a&c\\b&d\end{matrix}\right]=\pm 1$, then the inverse $\left[\begin{matrix}a&c\\b&d\end{matrix}\right]^{-1}$ is easily shown to be an integral matrix.
Note: Notice that, using the same logic, one can conclude that $\begin{pmatrix}a\\b\end{pmatrix}$ and $\begin{pmatrix}c\\d\end{pmatrix}$ are linearly independent if and only if $\det\left[\begin{matrix}a&c\\b&d\end{matrix}\right]=ad-bc\ne 0$. Thus, in the $\mathbb Z$-module $\mathbb Z^2$ there are pairs of "vectors" which are linearly independent but don't span the whole module.
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Determinant equal to zero : elements are angles Let $\alpha, \beta, \gamma$ be internal angles of an arbitrary triangle.
I want to show that $$\det \begin{pmatrix}\cos \beta & \cos \alpha&-1 \\ \cos\gamma & -1 & \cos\alpha \\ -1 & \cos\gamma & \cos\beta\end{pmatrix}=0$$
We have the following:
\begin{align*}&\det \begin{pmatrix}\cos \beta & \cos \alpha&-1 \\ \cos\gamma & -1 & \cos\alpha \\ -1 & \cos\gamma & \cos\beta\end{pmatrix}=\cos \beta \cdot \det \begin{pmatrix} -1 & \cos\alpha \\ \cos\gamma & \cos\beta\end{pmatrix}-\cos \alpha \cdot \det \begin{pmatrix} \cos\gamma & \cos\alpha \\ -1 & \cos\beta\end{pmatrix}+(-1)\cdot \det \begin{pmatrix} \cos\gamma & -1 \\ -1 & \cos\gamma \end{pmatrix} \\ & = \cos \beta \cdot \left (-\cos \beta-\cos\gamma\cdot \cos\alpha\right )-\cos \alpha \cdot \left (\cos\gamma\cdot \cos\beta-(-1)\cdot \cos\alpha\right )-\left (\cos^2 \gamma -(-1)^2\right ) \\ & =-\cos^2 \beta-\cos\gamma\cdot \cos\alpha\cdot \cos \beta-\cos\alpha\cdot \cos\gamma\cdot \cos\beta- \cos^2\alpha-\cos^2 \gamma +1 \\ & =1-2\cdot \cos\alpha\cdot \cos\beta\cdot \cos\gamma - \cos^2\alpha-\cos^2 \beta-\cos^2 \gamma \end{align*}
Do we use here the Law of cosines?
\begin{align*}&c^2=a^2+b^2-2ab\cdot \cos \gamma \Rightarrow \cos \gamma=\frac{a^2+b^2-c^2}{2ab} \\ &b^2=a^2+c^2-2ac\cdot \cos \beta \Rightarrow \cos \beta=\frac{a^2+c^2-b^2}{2ac} \\ &a^2=b^2+c^2-2bc\cdot \cos \alpha \Rightarrow \cos\alpha=\frac{b^2+c^2-a^2}{2bc}\end{align*}
Or how could we continue?
|
the product of the three cosines is given by $$-\frac{\left(a^2-b^2-c^2\right)
\left(a^2+b^2-c^2\right) \left(a^2-b^2+c^2\right)}{2
a^2 b^2 c^2}$$
for your second sum we get
$$2 \left(\frac{\left(a^2+b^2-c^2\right)^2}{4 a^2
b^2}+\frac{\left(a^2-b^2+c^2\right)^2}{4 b^2
c^2}+\frac{\left(-a^2+b^2+c^2\right)^2}{4 b^2
c^2}\right)=\frac{2 a^6-4 a^4 b^2+a^4 c^2+2 a^2 b^4+2 a^2 b^2
c^2+b^4 c^2-2 b^2 c^4+c^6}{2 a^2 b^2 c^2}$$you must only multiply out the first sum
|
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "2",
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|
How to find major & minor axes of the ellipse $10x^2+14xy+10y^2-7=0$?
What are major & minor axes of the ellipse: $10x^2+14xy+10y^2-7=0$ ?
My trial:
from given equation: $10x^2+14xy+10y^2-7=0$
$$10x^2+14xy+10y^2=7$$
$$\frac{x^2}{7/10}+\frac{xy}{7/14}+\frac{y^2}{7/10}=1$$
I know the standard form of ellipse: $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$
But the term of $xy$ is a bottleneck.
From here I can't proceed.
Can somebody please help me solve this problem?
Thank you.
|
Yet another method is with matrices and linear algebra. If we stuff $(x,y,1)^T$ into a vector, then the matrix
$${\bf M}=\left[\begin{array}{ccc}10&7&0\\7&10&0\\0&0&-7\end{array}\right]$$
Can be used to express the ellipse as a scalar product with matrix multiplication:
$$\left[\begin{array}{ccc}x&y&1\end{array}\right] \left[\begin{array}{ccc}10&7&0\\7&10&0\\0&0&-7\end{array}\right] \left[\begin{array}{c}x\\y\\1\end{array}\right]$$
A linear coordinate change can be expressed like this:
$$\left[\begin{array}{c}x_1\\y_1\\1\end{array}\right]= {\bf T}_{0\to 1}\left[\begin{array}{c}x_0\\y_0\\1\end{array}\right]$$
By making an eigenvalue decomposition of $\bf M$ we can find the coordinate system to transform to:
$${\bf M} = \left[\begin{array}{ccc}0&-1/\sqrt 2&1/\sqrt 2\\0&1/\sqrt 2&1/\sqrt 2\\1&0&0\end{array}\right]\left[\begin{array}{ccc}-7&0&0\\0&3&0\\0&0&17\end{array}\right]\left[\begin{array}{ccc}0&-1/\sqrt 2&1/\sqrt 2\\0&1/\sqrt 2&1/\sqrt 2\\1&0&0\end{array}\right]^{-1}$$
Here the coefficients of the new expression appears in the diagonal: $-7,3, 17$. The rest of the steps are similar to @user45914123 answer.
The main benefit of this method is that we can bake translation into the base change. For example if you had $X^2+2X+1+Y^2=R^2$, you would not know what to do with the $2X$ if you could only do linear transformations on the $[X,Y]^T$ vector.
|
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"timestamp": "2023-03-29T00:00:00",
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|
How can I calculate $\lim\limits_{x\rightarrow 0} \frac{(1+x)^{\frac{1}{x}}-e}{x}$?
How can I calculate this limit?
$$\lim_{x\rightarrow 0} \frac{(1+x)^{\frac{1}{x}}-e}{x}$$
I thought about L'Hospital because case of $\frac{0}{0}$, but I don't know how to contiune from this point..
|
By inequality
$$x-\frac{x^2}{2}\leq \log (1+x) \leq x-\frac{x^2}{2}+\frac{x^3}{3}$$
$$1-\frac{x}{2}\leq \frac{\log (1+x)}{x} \leq 1-\frac{x}{2}+\frac{x^2}{3}$$
we have that
$$e^{1-\frac{x}{2}}\leq (1+x)^\frac{1}{x}=e^{\frac{\log (1+x)}{x}} \leq e^{1-\frac{x}{2}+\frac{x^2}{3}} $$
thus
$$-\frac{e}2\le\frac{e^{1-\frac{x}{2}}-e}{x}\leq \frac{(1+x)^{\frac{1}{x}}-e}{x} \leq \frac{ e^{1-\frac{x}{2}+\frac{x^2}{3}}-e}{x}\to -\frac{e}2$$
indeed
$$ \frac{ e^{1-\frac{x}{2}+\frac{x^2}{3}}-e}{x} =e \cdot \frac{ e^{-\frac{x}{2}+\frac{x^2}{3}}-1}{ -\frac{x}{2}+\frac{x^2}{3} }\cdot\frac{-\frac{x}{2}+\frac{x^2}{3}}{x}\to e\cdot1\cdot -\frac12=-\frac{e}2$$
therefore for squeeze theorem
$$\lim_{x\rightarrow 0} \frac{(1+x)^{\frac{1}{x}}-e}{x}=-\frac{e}2$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Proof verification: $\forall x \in [\frac{\pi}{2}, \pi], sin(x) - cos(x) \geq 1$. Would someone be willing to verify the following proof by contradiction?
Theorem: $\forall x \in [\frac{\pi}{2}, \pi], sin(x) - cos(x) \geq 1$.
Suppose, for the sake of contradiction, that this statement is not true. Then, $\exists x \in [\frac{\pi}{2}, \pi], sin(x) - cos(x) < 1$.
$sin(x) - cos(x) < 1 \Longrightarrow sin^2(x) - 2sin(x)cos(x) + cos^2(x) < 1$
$\Longrightarrow sin^2(x) + cos^2(x) < 1 + 2sin(x)cos(x) \Longrightarrow 0 < sin(x)cos(x)$
On the interval $(\frac{\pi}{2}, \pi), sin(x) > 0$, and $cos(x) < 0$. Therefore, $sin(x)cos(x) < 0$, which is a contradiction.
|
Use that
$$\cos x=\sin \left( \frac{\pi}{2} -x\right)$$
and the sum-to-product formulas.
Notably from
$$\sin \theta - \sin \varphi = 2 \sin\left( \frac{\theta - \varphi}{2} \right) \cos\left( \frac{\theta + \varphi}{2} \right)$$
we obtain
$$\sin x - cos x=\sin x -\sin \left( \frac{\pi}{2} -x\right)=2\sin\left(x-\frac{\pi}{4}\right) \cdot\cos \frac{\pi}{4}=\sqrt2\cdot \sin\left(x-\frac{\pi}{4}\right) \geq 1$$
whic is true since
$$\frac{\pi}{2}\le x\le\pi\iff\frac{\pi}{4}\le x-\frac{\pi}{4}\le\frac{3\pi}{4}\implies \sin\left(x-\frac{\pi}{4}\right)\ge\frac{\sqrt2}{2} $$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Evaluating: $\int \frac {1+\sin (x)}{1+\cos (x)} dx$
Evaluate: $\int \dfrac {1+\sin (x)}{1+\cos (x)} dx$
My Attempt:
$$=\int \dfrac {1+\sin (x)}{1+\cos (x)} dx$$
$$=\int \dfrac {(\sin (\dfrac {x}{2}) + \cos (\dfrac {x}{2}))^2}{2\cos^2 (\dfrac {x}{2})} dx$$
$$=\dfrac {1}{2} \int (\dfrac {\sin (\dfrac {x}{2}) + \cos (\dfrac {x}{2})}{\cos (\dfrac {x}{2})})^2 dx$$
$$=\dfrac {1}{2} \int (\tan (\dfrac {x}{2}) +1)^2 dx$$
How do I continue?
|
You were almost there, just substitute $t=\tan(\frac x2)$ now.
$\displaystyle\dfrac 12\int \left(1+\tan(\dfrac x2)\right)^2\mathop{dx}=\int \dfrac{(1+t)^2}2\times\dfrac{2\mathop{dt}}{1+t^2}=\int \left(1+\dfrac{2t}{1+t^2}\right)\mathop{dt}=t+\ln(1+t^2)+C$
Remark that you can substitute directly without intermediate trigonometric changes :
$\dfrac{1+\sin(x)}{1+\cos(x)}=\dfrac{1+\frac{2t}{1+t^2}}{1+\frac{1-t^2}{1+t^2}}=\dfrac{1+2t+t^2}{2}$
|
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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|
System of equations involving complex numbers I'm getting confused to figure this work out. The only thing that came into my head was using AM-GM inequality, but i just get stuck. Here's the problem:
Let $a,b,c,$ be the complex numbers such that $abc=1$. and
\begin{cases}
a^{20}+b^{20} + c^{20} &= \frac{1}{a^{20}} + \frac{1}{b^{20}} + \frac{1}{c^{20}} \\
a^{17}+b^{17}+c^{17} &= \frac{1}{a^{17}} + \frac{1}{b^{17}} + \frac{1}{c^{17}}\\
a^{2017} + b^{2017}+c^{2017} &= \frac{1}{a^{2017}} + \frac{1}{b^{2017}} + \frac{1}{c^{2017}}\\
\end{cases}
show that $1 \in \{a,b,c\}$.
Please help me to solve this question, any thought would be helpful.
|
Actually,
Lemma 1: if $abc=1$, and if $a+b+c=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}$, then one of $a,b,c$ is 1.
Proof: multiply by $abc=1$:
$$a+b+c=ab+bc+ac\Leftrightarrow abc-ab-ac-bc+a+b+c-1=0\Leftrightarrow (a-1)(b-1)(c-1)=0$$
Lemma 2: If $\gcd(u,v)=1$, and $a^u=a^v=1$, then $a=1$.
Proof: $1=mu+nv$ for some $m,n\in\mathbb Z$, and so $a=a^{mu+nv}=(a^u)^m(a^v)^n=1$.
We can now apply Lemma 1 to the triplets $(a^{17}, b^{17}, c^{17})$, $(a^{20}, b^{20}, c^{20})$ and $(a^{2017}, b^{2017}, c^{2017})$. Note also that $17, 20, 2017$ are pairwise coprime. We have two cases:
*
*Either the same variable ($a$, $b$ or $c$) raised to two different exponents is 1. WLOG, let's assume $a^{17}=1$ and $a^{20}=1$. Then $a=1$ by Lemma 2. The same proof goes if the exponents are $20,2017$ or $17,2017$.
*Or, it is all three variables. WLOG, assume $a^{17}=b^{20}=c^{2017}=1$. Thus $a^{17\cdot 20}b^{17\cdot 20}=1$ so $c^{17\cdot 20}=1$, but because also $c^{2017}=1$ and $\gcd(17\cdot 20, 2017)=1$ we then have $c=1$ by Lemma 2.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Computing the definite integral $\int _0^a \:x \sqrt{x^2+a^2} \,\mathrm d x$
Compute the following definite integral $$\int _0^a \:x \sqrt{x^2+a^2} \,\mathrm d x$$
This is what I did:
$u = x^2 + a^2 $
$du/dx = 2x$
$du = 2xdx$
$1/2 du = x dx$
$\int _0^a\:\frac{1}{2}\sqrt{u}du = \frac{1}{2}\cdot \frac{u^{\frac{3}{2}}}{\left(\frac{3}{2}\right)}$ from $0$ to $a$.
$\frac{1}{3}\cdot \left(x^2+a^2\right)^{\frac{3}{2}}$ from $0$ to $a$.
I eventually got:
$\frac{1}{3}\left(81+a^2\right)^{\frac{3}{2}}-\frac{1}{3}\left(a^2\right)^{\frac{3}{2}}$
but this was incorrect.
The correct answer was:
$\frac{1}{3}\left(2\sqrt{2}-1\right)a^3$
Any help?
|
Hint
You made mistakes here:
$$\frac{1}{3}\cdot \left(x^2+a^2\right)^{\frac{3}{2}}\big |_0^a=\frac 1 3 (2a^2)^{\frac 3 2}-\frac {a^3}{3}=\frac 1 3 (\sqrt{2^3}|a|^3)-\frac {a^3}{3}=....$$
You were almost done...
|
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"timestamp": "2023-03-29T00:00:00",
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|
The number of ways in which one can choose three distinct numbers from the set so that the product of the chosen numbers is divisible by $9$
Consider the set $A=\{1,2,3,...,30\}$
. The number of ways in which one can choose three
distinct numbers from $A$ so that the product of the chosen numbers is divisible by $9$ is $X$
Find $X$
This question is from KVPY SA 2017.
My working:
Nos in set divisible by $3$(but not by $9$)-$3,6,12,15,21,24,30$-Total $7$
Nos in set divisible by $9$-$9,18,27$-Total $3$
Therefore, the ans should be $7.7.20+7.7.7+7.7.3+7.3.3+3.3.3+3.20.20+3.3.20+3.7.20$
Where $.$ is multiplication
Is this correct?
|
You haven't adressed "three distinct numbers" correctly. For instance, you aren't allowed to pick $3, 3, 5$, so the first term shouldn't be $7\cdot 7\cdot 20$, it should be $7\cdot 6\cdot 20$. And then, still, you count $3, 6, 5$ as a separate choice from $6, 3, 5$. Therefore, what you really want is $\frac{7\cdot 6}2\cdot 20$ for that term.
With these two corrections, the final answer is
$$
\frac{7\cdot 6}{2}\cdot 20 + \frac{7\cdot 6\cdot 5}{6} + \frac{7\cdot 6}{2}\cdot 3 + 7\cdot\frac{3\cdot 2}{2} + \frac{3\cdot 2\cdot 1}6 + 3\cdot \frac{20\cdot 19}2 + \frac{3\cdot 2}2\cdot 20 + 3\cdot 7\cdot 20
$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Computing $\lim\limits_{x\to0}\int_0^x\frac{t^2}{(x-\sin x)\sqrt{a+t}}\,dt$ without L'Hopital I am computing the following limit
$$\lim_{x\to0}\int_0^x\frac{t^2}{(x-\sin x)\sqrt{a+t}}\,dt$$
where $a$ is a parameter. the case $a= 0$ has been resolved. But yet, for the case $ a\neq 0$ I want to compute it without using L'Hopital rule.
|
With Maclaurin series:
$$\lim_{x\to0}\int_0^x\frac{t^2}{(x-\sin x)\sqrt{a+t}}\,dt=\lim_{x\to0}\frac{1}{(x-\sin x)}\int_0^x\frac{t^2}{\sqrt{a+t}}\,dt=\lim_{x\to0}\frac{1}{(x-\sin x)}\left[\frac{2}{15}\sqrt{a+t}(8a^2-4at+3t^2)\right]_{0}^{x}=\lim_{x\to0}\frac{1}{(x-\sin x)}\frac{2}{15}\left[\sqrt{a+x}(8a^2-4ax+3x^2)-\sqrt{a}(8a^2)\right]=\frac{2\sqrt{a}}{15}\lim_{x\to0}\frac{\sqrt{1+\frac{x}{a}}(8a^2-4ax+3x^2)-8a^2}{x-\sin x}=\frac{2\sqrt{a}}{15}\lim_{x\to0}\frac{(1+\frac{x}{2a}-\frac{x^2}{8a^2}+\frac{x^3}{16a^3}+O(x^4))(8a^2-4ax+3x^2)-8a^2}{x-x+\frac{x^3}{6}+O(x^4)}=\frac{2\sqrt{a}}{15}\lim_{x\to0}\frac{\frac{3x^3}{2a}+\frac{x^3}{2a}-\frac{5x^3}{16a}+O(x^4)}{\frac{x^3}{6}+O(x^4)}=\frac{2\sqrt{a}}{15}\frac{5}{2a}\frac{1}{\frac{1}{6}}=\frac{2}{\sqrt{a}}$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
The Arithmetic Mean (A.M) between two numbers exceeds their Geometric Mean (G.M.) The Arithmetic Mean (A.M) between two numbers exceeds their Geometric Mean (G.M.) by $2$ and the GM exceeds the Harmonic Mean (H.M) by $1.6$. Find the numbers.
My Attempt:
Let the numbers be $a$ and $b$. Then,
$$A.M=\dfrac {a+b}{2}$$
$$G.M=\sqrt {ab}$$
$$H.M=\dfrac {2ab}{a+b}$$
According to question:
$$\dfrac {a+b}{2} =\sqrt {ab}+2$$
$$\dfrac {a+b}{2}-2=\sqrt {ab}$$
$$a+b-4=2\sqrt {ab}$$
Also,
$$\sqrt {ab}=\dfrac {2ab}{a+b} + 1.6$$
Then,
$$a+b-4=2(\dfrac {2ab}{a+b} + 1.6)$$
$$a+b-4=\dfrac {4ab+3.2(a+b)}{a+b}$$
$$(a+b-4)(a+b)=4ab+3.2(a+b)$$
$$(a+b)^2-4(a+b)=4ab+3.2a+3.2b$$
How do I solve further?
|
I don't know
why I do this,
but here is
the general case.
Suppose
$am = gm+u$
and
$gm = hm+v$
with
$u, v \ne 0$
and
$u \ne v$.
Since
$hm \le gm
\le am
$,
$u \ge 0$
and
$v \ge 0$.
Since
$gm^2 = am\cdot hm$,
$gm^2
=(gm+u)(gm-v)
=gm^2+gm(u-v)-uv
$
so
$gm(u-v) =uv$.
Therefore
$u > v$
and
$gm
=\dfrac{uv}{u-v}
$.
Then
$am
=gm+u
=\dfrac{uv}{u-v}+u
=\dfrac{uv+u(u-v)}{u-v}+u
=\dfrac{u^2}{u-v}
$
and
$hm
=gm-v
=\dfrac{uv}{u-v}-v
=\dfrac{uv-v(u-v)}{u-v}
=\dfrac{v^2}{u-v}
$.
If there are only
two values,
$a$ and $b$,
then
$\dfrac{a+b}{2}
=\dfrac{u^2}{u-v}
$
and
$\sqrt{ab}
=\dfrac{uv}{u-v}
$
so
$b
=\dfrac{u^2v^2}{a(u-v)^2}
$
and
$\dfrac{u^2}{u-v}
=\dfrac{a+\dfrac{u^2v^2}{a(u-v)^2}}{2}
=\dfrac{a^2(u-v)^2+u^2v^2}{2a(u-v)^2}
$
or
$2au^2(u-v)
=a^2(u-v)^2+u^2v^2
$.
Solving
$a^2(u-v)^2-2u^2(u-v)a+u^2v^2
=0$,
$\begin{array}\\
a
&=\dfrac{2u^2(u-v)\pm\sqrt{4u^4(u-v)^2-4(u-v)^2u^2v^2}}{2(u-v)^2}\\
&=\dfrac{u^2(u-v)\pm u(u-v)\sqrt{u^2-v^2}}{(u-v)^2}\\
&=\dfrac{u^2\pm u\sqrt{u^2-v^2}}{u-v}\\
&=u\dfrac{u\pm \sqrt{u^2-v^2}}{u-v}\\
\text{and}\\
b
&=\dfrac{2u^2}{u-v}-a\\
&=\dfrac{2u^2}{u-v}-\dfrac{u^2\pm u\sqrt{u^2-v^2}}{u-v}\\
&=\dfrac{2u^2-u^2\mp u\sqrt{u^2-v^2}}{u-v}\\
&=\dfrac{u^2\mp u\sqrt{u^2-v^2}}{u-v}\\
\end{array}
$
For this case,
$u=2$
and
$v=1.6$.
$gm
=\dfrac{2\cdot 1.6}{2-1.6}
=\dfrac{3.2}{.4}
=8
$.
To get $a$ and $b$,
$\sqrt{u^2-v^2}
=\sqrt{4-2.56}
=\sqrt{1.44}
=1.2
$
so
$a
=2\dfrac{2\pm 1.2}{.4}
=2\dfrac{3.2, .8}{.4}
=2(8, 2)
=(16, 4)
$
and
$b
=(4, 16)
$.
|
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|
Equation System with 4 real variables I need to solve the next equation system:
Find all real numbers $a,b,c,d$ such that:
$$
\left\{
\begin{array}{c}
a+b+c+d=20 \\
ab+ac+ad+bc+bd+cd=150 \\
\end{array}
\right.
$$
I tried something like this:
$b+c+d=20-a$
And i put the second equation like this
$a(b+c+d) + bc+bd+cd=150$
Getting
$20a-a^2 +bc+bd+cd=150$
But i see that this is useless, so i don't know how to start this problem.
|
$$0=3(a+b+c+d)^2-8(ab+ac+ad+bc+bd+cd)=$$
$$=(a-b)^2+(a-c)^2+(a-d)^2+(b-c)^2+(b-d)^2+(c-d)^2\geq0,$$
where the equality occurs for $a=b=c=d.$
Thus, $a=b=c=d=5.$
|
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|
Minimum value of $(a+1)(b+1)$
If $a,b$ are the roots of the equation $$(\arcsin x+\arctan x)p^2+25p\pi+2(\arccos x+\operatorname{arccot} x)=0$$
Then minimum of $(a+1)(b+1)$
Try: $$a+b=\frac{25\pi}{\arcsin x+\arctan x}$$ and $$ab=\frac{2(\arccos x+\operatorname{arccot} x)}{\arcsin x+\arctan x}$$
So $$ab+a+b+1=\frac{25\pi+2(\arccos x+\operatorname{arccot} x)}{\arcsin x+\arctan x}$$
Could some help me to solve it, thanks
|
$x\neq0$, otherwise our equation has unique root.
Also, $$\left(25\pi\right)^2-8(\arcsin{x}+\operatorname{arctan}x)(\arccos{x}+\operatorname{arccot}x)\geq$$
$$\geq\left(25\pi\right)^2-8\left(\frac{\arcsin{x}+\operatorname{arctan}x+\arccos{x}+\operatorname{arccot}x}{2}\right)^2=625\pi^2-2\pi^2>0,$$
which says that our equation has two real roots for all $x\neq0$ and $-1\leq x\leq 1.$
Now, $$(a+1)(b+1)=\frac{-25\pi+2(\arccos x+\operatorname{arccot} x)}{\arcsin x+\arctan x}+2-1=$$
$$=\frac{-25\pi+2(\arcsin{x}+\arccos x+\operatorname{arctan} x+\operatorname{arccot} x)}{\arcsin x+\arctan x}-1=$$
$$=\frac{-25\pi+2\left(\frac{\pi}{2}+\frac{\pi}{2}\right)}{\arcsin x+\arctan x}-1\rightarrow-\infty$$ for $x\rightarrow0^+$, which says that the minimum does not exist.
|
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"timestamp": "2023-03-29T00:00:00",
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|
How to get first derivative? My question is to find the first derivative using product and/or quotient rule
$$f(x) = {(x^2+1)(x^2-2) \over 3x+2}.$$
The solution to the problem is:
$$f'(x) = {[(2x)(x^2-2)+(x^2+1)(2x)](3x+2)-(3)[(x^2+1)(x^2-2)] \over (3x+2)^2}.$$
I'm having problems getting the same solution would someone be able to help me by showing me the correct way of getting the first derivative of this question
|
When you have expressions which just contains products, quotients and powers, logarithmic differentiation makes life easier.
For your case
$$f = {(x^2+1)(x^2-2) \over 3x+2}\implies \log(f)=\log(x^2+1)+\log(x^2-2)-\log(3x+2)$$ Differentiate both sides
$$\frac{f'}f=\frac{2x}{x^2+1}+\frac{2x}{x^2-2}-\frac 3{3x+2}=\frac{9 x^4+8 x^3-3 x^2-4 x+6 }{(x^2+1 )(x^2-2 ) (3x+2 ) }$$
Now
$$f'=f \times \frac{f'}f={(x^2+1)(x^2-2) \over 3x+2}\times \frac{9 x^4+8 x^3-3 x^2-4 x+6 }{(x^2+1 )(x^2-2 ) (3x+2 ) }$$ Now, simplify.
|
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|
Value of $S_{10}+I_{10}$ Let $S_n=\sum_{i=1}^n \frac{1}{k}$ and$$I_n=\int_{1}^{n} \frac{x-[x]}{x^2} dx$$ then what is the value of $S_{10}+I_{10}$ ?
$S_{10}=1+1/2+1/3+.....+1/10$ and $$I_{10}=\int_{1}^{10} \frac{x-[x]}{x^2} dx$$
$$=ln 10-\int_{1}^{10} \frac{[x]}{x^2} dx$$
But I am stuck here.How to proceed further?
|
You just have to use
$$
\begin{align*}
\int_{1}^{10} \frac{[x]}{x^2} dx & =\int_1^2\frac{1}{x^2} dx+2\int_2^3\frac{1}{x^2} dx +3\int_3^4\frac{1}{x^2} dx+\cdots +9\int_9^{10}\frac{1}{x^2} dx \\
& =\sum_{k=1}^{9} k \int_k^{k+1}\frac{1}{x^2} dx \\
& = \sum_{k=1}^{9} k [-\tfrac{1}{u}]_k^{k+1} \\
&= \sum_{k=1}^{9} k\big( \tfrac{1}{k} -\tfrac{1}{k+1}\big) \\
&=\sum_{k=1}^{9} 1 -(1 -\tfrac{1}{k+1})\\
&= \sum_{k'=2}^{10}\tfrac{1}{k'} = S_{10}-1
\end{align*}
$$
Since $[x]=k$ on each interval $[k,k+1)$ ; you see that it is easy to generalize for other indices values than 10... (and $S_n+I_n=S_N+(\ln(n)-S_n+1)=\ln(n)+1)$.
|
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|
Find the volume of the following solid. I need to find the volume of that solid:
$$(x^2+y^2+z^2)^3=4y^2z^2$$
I obviously tried spherical coordinates, but that simply led me to nowhere. I used:
$$\begin{cases}x=r\cos\alpha \cos\beta\\[2ex]y=r\sin\alpha \cos\beta\\[2ex]z=r\sin\beta\end{cases}$$
and it got me to the point where I have:
$$r^2=4\cos\alpha\sin\alpha\cos^2\beta.$$
and honestly I don't know what to do next.
|
In spherical coordinates $(R,\theta,\phi)$ the Jacobian of the spherical coordinates is $R^2\sin\theta$. Also the equation of the surface enveloping the solid is:$$R^2=4\sin^2\theta\sin^2\phi\cos^2\phi=\sin^2\theta\sin^2(2\phi)$$
For calculating the volume we have
$$\iiint_VR^2\sin\theta dR d\theta d\phi=
\int_{0}^{\pi}\int_{0}^{2\pi}\int_{0}^{\sqrt{\sin^2\theta\sin^2(2\phi)}}R^2\sin\theta dR d\phi d\theta=
\int_{0}^{\pi}\int_{0}^{2\pi}\dfrac{R^3}{3}|_{0}^{\sqrt{\sin^2\theta\sin^2(2\phi)}}\sin\theta d\phi d\theta=
\int_{0}^{\pi}\int_{0}^{2\pi}\dfrac{|\sin\theta|^3|\sin(2\phi)|^3}{3}\sin\theta d\phi d\theta$$
since $\sin\theta\ge0$ when $0\le\theta\le\pi$, the period of $|\sin(2\phi)|^3$ is $\dfrac{\pi}{4}$ and $\sin(2\phi)\ge0$ when $0\le\phi\le\dfrac{\pi}{2}$ we can write the integral formula as following:
$$I=4\int_{0}^{\pi}\int_{0}^{\frac{\pi}{2}}\dfrac{\sin^4\theta}{3}\sin^3(2\phi) d\phi d\theta$$
also
$$\int_{0}^{\frac{\pi}{2}}4\sin^3(2\phi) d\phi=\int_{0}^{\frac{\pi}{2}}3\sin(2\phi)-\sin(6\phi) d\phi=\dfrac{8}{3}$$
therefore the integral can be simplified as follows:
$$I=\dfrac{8}{9}\int_{0}^{\pi}\sin^4\theta d\theta$$
which by substituting $\sin^2p=\dfrac{1-\cos(2p)}{2}$ gives us the final result:
$$I=\dfrac{\pi}{3}$$
|
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|
how can I calculate: $\lim_{n \to \infty} \frac{\frac{1}{n}-\ln(1+\frac{1}{n})}{n^{\frac{1}{n}}-1} $?
how can I calculate:
$$\lim_{n \to \infty} \frac{\frac{1}{n}-\ln(1+\frac{1}{n})}{n^{\frac{1}{n}}-1} $$
I tried with Hospital and it's not working. Can help please ?
|
Using Taylor expansion
We have $$\frac{1}{n}-\ln\left(1+\frac{1}{n}\right)= \frac{1}{2n^2} +o\left(\frac{1}{n^2}\right)$$
since $$\ln(x+1) = x-\frac{x^2}{2} +o(x^2)$$
and similarly we have,
$$n^{1/n} = \exp(n\ln (1/n)) =\exp\left(n\left(\frac{1}{n}-\frac{1}{2n^2}+o\left(\frac{1}{n^2}\right)\right)\right) = 1 -\frac{1}{2n} +o\left(\frac{1}{n}\right)$$
hence, $$ \frac{\frac{1}{n}-\ln(1+\frac{1}{n})}{n^{\frac{1}{n}}-1}= \frac{\frac{1}{2n^2} +o\left(\frac{1}{n^2}\right)}{-\frac{1}{2n} +o\left(\frac{1}{n}\right)} =-\frac{1}{n}+o\left(\frac{1}{n}\right) \to 0~~ as~~~ n\to\infty$$
$$\lim_{n \to \infty} \frac{\frac{1}{n}-\ln(1+\frac{1}{n})}{n^{\frac{1}{n}}-1} =0$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Formatting Functions To Avoid Loss Of Significant
Rewrite the following to avoid loss of significant
*
*$\ln(x+1)-\ln(x)$ where $x>>1$
*$\cos^2(x)-\sin^2(x)$ where $x\approx \frac{\pi}{4}$
*$\sqrt{x^2+1}-x$ where $x>>1$
*$\sqrt{\frac{1+\cos x}{2}}$
*
*Using taylor expansion we get $$x-\frac{x^2}{2}+\frac{x^3}{3}-[(x-1)-\frac{(x-1)^2}{2}+\frac{(x-1)^3}{3}]=1+\frac{(x-1)^2-x^2}{2}+\frac{x^3-(x-1)^3}{3}$$
*Using taylor expansion we get
$$(1-\frac{x^2}{2!}+\frac{x^4}{4!})^2-(x-\frac{x^3}{3!}+\frac{x^5}{5!})^2$$
*$$\sqrt{x^2+1}-x=(\sqrt{x^2+1}-x)\cdot (\frac{\sqrt{x^2+1}+x}{\sqrt{x^2+1}+x})=\frac{1}{\sqrt{x^2+1}+x}$$
*$$\sqrt{\frac{1+\cos x}{x}}\approx \sqrt{\frac{1+1-\frac{x^2}{2!}+\frac{x^4}{4!}}{2}}$$
Is this valid?
|
You don't want to use the Taylor series when $x$ is large. Better to use the law of exponents to write $\log(x+1) - \log (x)=\log(1+\frac 1x)$ and use the Taylor series from there.
This is the usual approach. You want to analytically subtract the large parts of the two numbers, reducing the cancellation.
|
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|
Find all $x \in \mathbb R$ such that $x + \sqrt{3}$ and $x^2 + \sqrt{3}$ are rational. Find all $x \in \mathbb R$ such that $x + \sqrt{3}$ and $x^2 + \sqrt{3}$ are rational. I have started by assuming that $x + \sqrt{3} = \frac{a}{b}$ and substituting $x = \frac{a}{b} - \sqrt{3}$ into $x^2 + \sqrt{3}$. It led me to $\frac{a^2}{b^2} + 3 + \sqrt{3} \cdot \frac{b - 2a}{b}$. I don't know how to continue.
|
Let $x+\sqrt3=r$ and $x^2+\sqrt3=q$.
Thus, $$(r-\sqrt3)^2=q-\sqrt3$$ or
$$r^2-2\sqrt3r+3=q-\sqrt3,$$ which gives $2r=1$ and $r^2+3=q$ and $x=\frac{1}{2}-\sqrt3.$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Point in triangle Inside a random triangle, consider a point P and its distances PM, PN and PQ to the 3 sides a, b and c.
For which location of the point P the below sum gets minimal?
$$\frac{a}{PM} + \frac{b}{PN} + \frac{c}{PQ}$$
I think it is the incenter but how do we prove it?
|
Note that
\begin{align}
a\cdot x+b\cdot y+c\cdot z&=2\,S=\mathrm{const}\quad\text{for fixed }a,b,c.
\end{align}
\begin{align}
\operatorname*{argmin}_{P}\left(\frac{a}x + \frac{b}y + \frac{c}z\right)
&=
\operatorname*{argmin}_{P}2\,S\cdot\left(\frac{a}x + \frac{b}y + \frac{c}z\right)
\\
&=
\operatorname*{argmin}_{P}
\left(
(\tfrac{x}y+\tfrac{y}x)\,ab+(\tfrac{y}z
+\tfrac{z}y)bc+(\tfrac{z}x+\tfrac{x}z)\,ca
+\underbrace{(a^2+b^2+c^2)}_{\mathrm{const}}
\right)
\\
&=
\operatorname*{argmin}_{P}
\left(
(\tfrac{x}y+\tfrac{y}x)\,ab+(\tfrac{y}z
+\tfrac{z}y)bc+(\tfrac{z}x+\tfrac{x}z)\,ca
\right)
,\\
\end{align}
\begin{align}
\left(
(\tfrac{x}y+\tfrac{y}x)\,ab+(\tfrac{y}z
+\tfrac{z}y)bc+(\tfrac{z}x+\tfrac{x}z)\,ca
\right)
&\overset{\text{AM-GM}}{\ge}
\left(
3\,
\sqrt[3]{
a^2\,b^2\,c^2\,
(\tfrac{x}y+\tfrac{y}x)
(\tfrac{y}z+\tfrac{z}y)
(\tfrac{z}x+\tfrac{x}z)
}
\right)
,\\
\operatorname*{argmin}_{P}
\left(
3\,
\sqrt[3]{
a^2\,b^2\,c^2\,
(\tfrac{x}y+\tfrac{y}x)
(\tfrac{y}z+\tfrac{z}y)
(\tfrac{z}x+\tfrac{x}z)
}
\right)
&=
\operatorname*{argmin}_{P}
\left(
(\tfrac{x}y+\tfrac{y}x)
(\tfrac{y}z+\tfrac{z}y)
(\tfrac{z}x+\tfrac{x}z)
\right)
,\\
(\tfrac{x}y+\tfrac{y}x)
(\tfrac{y}z+\tfrac{z}y)
(\tfrac{z}x+\tfrac{x}z)
&\overset{\text{AM-GM}}{\ge}
2\cdot2\cdot2=8
,\\
\operatorname*{min}_{P}
\left(
(\tfrac{x}y+\tfrac{y}x)
(\tfrac{y}z+\tfrac{z}y)
(\tfrac{z}x+\tfrac{x}z)
\right)
&=8,\quad P:x=y=z
.
\end{align}
That is, $x=y=z=r$ and $P$
coincides with the incenter of the triangle.
|
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|
Find $g(2002)$ given $f(1)$ and two inequalities It is given that $f(x)$ is a function defined on $\mathbb{R}$, satisfying $f(1)=1$ and for any $x\in \mathbb{R}$,
$$f(x+5)\geq f(x)+5,$$
$$f(x+1)\leq f(x)+1.$$
If $g(x)=f(x)+1-x$ then find $g(2002)$.
Here,
$$f(x+5)\leq f(x+4) +1,$$
I didn't get any idea..
|
Rewriting the given inequalities in terms of $\,f(x)=g(x)+x-1\,$:
$$\require{cancel}
\begin{align}
f(x+5)\geq f(x) + 5 \quad&\iff\quad g(x+5)+\bcancel{x}+\cancel{5}- \xcancel{1} \ge g(x) + \bcancel{x} -\xcancel{1} + \cancel{5} \\
&\iff\quad g(x+5) \ge g(x) \\
f(x+1)\leq f(x)+1 \quad&\iff\quad g(x+1)+\bcancel{x}+\cancel{1}- \xcancel{1} \le g(x) + \bcancel{x} -\xcancel{1} + \cancel{1} \\
&\iff\quad g(x+1) \le g(x) \\
\end{align}
$$
It follows that $\,\color{blue}{g(x)} \ge g(x+1) \ge g(x+2) \ge g(x+3) \ge g(x+4) \ge g(x+5) \ge \color{blue}{g(x)}\,$, so equalities must hold throughout, then $\,g(x+1)=g(x)\,$, and $\,g(2002)=g(2001)=\ldots =g(1)\,$.
|
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|
Proof of inequalities using binomial theorem
How do they achieve $$1+1+\frac{1}{2}\left(1-\frac{1}{n}\right)?$$ If I consider the first three terms of the binomial expansion, I get the first two terms, but can't factor the third to match this.
|
$(1+x)^n = 1 + n x + \frac {n(n-1)}{2} x^2 + \cdots$
All the terms in the "$\cdots$" are positive if $x$ is positive.
If $n \ge 2$
$(1+x)^n \ge 1 + n x + \frac {n(n-1)}{2} x^2 $
$(1+\frac {1}{n})^n$ replace $x$ above with $\frac 1n$
$(1+\frac 1n)^n \ge 1 + n \frac 1n + \frac {n(n-1)}{2} (\frac 1n)^2 $
and distribute
$(1+\frac 1n)^n \ge 1 + 1 + \frac {n^2-n}{2n^2} = 2+\frac {n-1}{2n} = 2+\frac 12 - \frac {1}{2n} $
|
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|
Determine the whether the sequence $a_{n+1} = 3 - \frac{1}{a_n} \text{ for n > 1}$ is convergent or divergent. Consider the recursively defined sequence $a_n = 1$
$$a_{n+1} = 3 - \frac{1}{a_n} \text{ for n > 1}$$
Is the sequence convergent?
This is my attempt:
First, we prove that the sequence is positive, and monotonically increasing, using induction.
$\textbf{Base Case:}$ $a_1 = 1$ and $a_2 = 2$. $a_1 < a_2$ and $a_1,a_2 > 0$.
$\textbf{Inductive Hypothesis:}$ $a_n < a_{n+1}$ where $a_n, a_{n+1}$.
$\textbf{Inductive Step:}$ We prove that $a_n < a_{n+1} \Rightarrow a_{n+1} < a_{n+2}$.
By our induction hypothesis:
$$a_n < a_{n+1}$$
$$-a_n > -a_{n+1}$$
$$-\frac{1}{a_n} < -\frac{1}{a_{n+1}}$$ (True by our IH since $a_n, a_{n+1} > 0$ and thus, $-a_n, -a_{n+1}$ share the same sign).
$$3-\frac{1}{a_n} < 3-\frac{1}{a_{n+1}}$$
$$a_{n+1} < a_{n+2}$$.
Note that $a_{n+1} > 0$ by our IH, so $a_{n+2} > 0$.
Now we prove that the sequence is bounded.
Observe that $a_n$ is monotonically increasing, which means that $ - \frac{1}{a_n}$ is monotonically increasing as well and it is upped bounded by $0$. Thus, $3-\frac{1}{a_n}$ is upper bounded by $3$.
We have a sequence that is monotone and bounded. Hence, by the Monotone Convergence Theorem, This sequence converges.
I was wondering if this method is correct.
|
A shorter approach, look at the function $f(x)=3-\frac{1}{x}$ because this function "generates" the sequence, i.e. $f(a_n)=a_{n+1}$. This function is ascending, because $f'(x)=\frac{1}{x^2}$ and $\frac{1}{3}<a_1<a_2 \Rightarrow 0< \color{red}{f(a_1)\leq f(a_2)} \Rightarrow \frac{1}{3}<a_1<\color{red}{a_2 \leq a_3}$ and by induction $\frac{1}{3}< a_{n} \leq a_{n+1}$. So, the sequence is ascending and positive.
Given $a_n > \frac{1}{3} > 0, \forall n$, then $\frac{1}{a_n}>0 \Rightarrow a_{n+1}=3-\frac{1}{a_n} < 3 \Rightarrow 0<a_{n+1} < 3, \forall n$. So the sequence in bounded.
|
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|
A triangle has the maximum possible area on the parabola $y = x^2$ Given points $P(−1, 1)$ and $Q(2, 4)$ on the parabola $y = x^2$, where should the point $R$ be on the parabola (between P and Q) so that the triangle $PQR$ has the maximum possible area?
How should I do this question?
|
Let $R(x,x^2),$ where $-1\leq x\leq2.$
We have
$$m_{PQ}=1,$$ which gives the equation of $PQ:$
$$y-1=1(x+1)$$ or $$x-y+2=0.$$
Thus, the altitude of the triangle from $R$ is
$$\frac{|x-x^2+2|}{\sqrt2}$$ and since $$PQ=\sqrt{3^2+3^2}=\sqrt{18},$$ by AM-GM we obtain:
$$S_{\Delta PQR}=\frac{\sqrt{18}\cdot\frac{|x-x^2+2|}{\sqrt2}}{2}=\frac{3}{2}\cdot(2-x)(x+1)\leq\frac{3}{2}\left(\frac{2-x+x+1}{2}\right)^2=\frac{27}{8}.$$
The equality occurs for $x+1=2-x$, which says that we got a maximal value.
|
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|
Determinant of a matrix after changes If $\det \begin{pmatrix}a&1&d\\ b&1&e\\ c&1&f\end{pmatrix}=1$ and $\det \begin{pmatrix}a&1&d\\ b&2&e\\ c&3&f\end{pmatrix}=1$, what is $\det \begin{pmatrix}a&-4&d\\ b&-5&e\\ c&-6&f\end{pmatrix}$?
So I am aware about all the different operations and what changes they bring to the value of the determinant, but I am not exactly sure which one of them is being applied here. There is no constant multiplication or an addition or subtraction by a row. I have tried to add and subtract various multiples of the matrices from each other as well but not to any avail. I believe I am just not spotting something.
Any help?
|
By the multilinearity of determinant,
$$\begin{aligned}
& \begin{vmatrix}a&-4&d\\ b&-5&e\\ c&-6&f\end{vmatrix} \\
&= \begin{vmatrix}a&-1&d\\ b&-2&e\\ c&-3&f\end{vmatrix}+\begin{vmatrix}a&-3&d\\ b&-3&e\\ c&-3&f\end{vmatrix} \\
&= -\begin{vmatrix}a&1&d\\ b&2&e\\ c&3&f\end{vmatrix}-3\begin{vmatrix}a&1&d\\ b&1&e\\ c&1&f\end{vmatrix} \\
&= -1 -3(1) =-4.
\end{aligned}
$$
|
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|
Solve this initial value problem $y'=\tan^2(5x+5y)$, with the initial condition $y(0)=0$
So I'm about to lose my hair, I already have 60 attempts done for this problem.
I took $u= 5x+5y$, which gave me $u'= 5+5y'$.
Then I isolated $y'= (u'-5)/5$, which would give me
$5\tan^2(u)+5 = u'$ and then I have integral of $dx = \int \frac{du}{5\tan^2(u)+5}$.
And after by using an integral calculator I get that $x +C= \frac{\tan(u)}{10\tan^2(u) +10} +\frac u{10}$.
By replacing the $u$ back and using the initial condition $y(0)=0$, I get $C=0$ and the solution that I found was $\frac{\tan(5(x+y))}{10\tan^2(5(x+y) )+10} +\frac{5x+5y}{10} -x$.
|
Let us observe that
\begin{align}
y' =\tan^2(5x+5y) \ \ \implies& \ \ \cos^2(5x+5y)y' = \sin^2(5x+5y) = 1-\cos^2(5x+5y)\\
\implies& \ \ \cos^2(5x+5y)(1+y') = 1.
\end{align}
Next, notice if we define $u(x) =5y(x)+5x$, then it follows
\begin{align}
\cos^2(5x+5y)(1+y') = 1 \ \ \implies \ \ \cos^2(u)u' = 5 \ \ \implies \ \ \frac{1+\cos(2u)}{2} du = 5 dx.
\end{align}
Hence the general implicit solution is given by
\begin{align}
\frac{1}{2}u + \frac{1}{4}\sin(2u)= 5x+C \ \ \implies \ \ \frac{5}{2}(y+x) + \frac{1}{4}\sin(10(x+y))= 5x+C.
\end{align}
Plugging in the initial condition yields
\begin{align}
F(x, y) = \frac{5}{2}(y-x) + \frac{1}{4}\sin(10(x+y))+1 = 1.
\end{align}
Edit: The problem looks ill-posed since
\begin{align}
F(x, y) = \frac{1}{2}(y-x) + \frac{1}{20}\sin(10(x+y))+1 = 1.
\end{align}
also works.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Congruence equation doubt Given the congruence equation:
$$70x\equiv 36 \pmod{198}$$
I have to find the smallest positive solution. I solved this by attempts (dividing the multiples of $70$ by $198$ and checking if the remainder is $36$) and I've got $x=9$ as a solution. Is my method incorrect?
|
$$70x-36 = 198k$$
Divide it $2$,
$$35x-18=99k$$
We want to find the smallest positive integer $x$ such that an integer $k$ exists.
I would be great if we can find $35^{-1} \pmod{99}$, we know it exists since $35$ and $99$ are corprime.
Let's use Euclidean algorithm:
\begin{align}99&=2(35)+29\\
35&=1(29)+6\\
29&=4(6)+5\\
6&=1(5)+1 \end{align}
Hence \begin{align}
1&=6-1(5)\\
&=6-(29-4(6))\\
&=5(6)-29 \\
&=5(35-29)-29\\
&=5(35)-6(29) \\
&=5(35)-6(99-2(35)) \\
&=17(35)-6(99) \end{align}
Hence $$17(35) \equiv 1 \pmod{99}$$
Our goal was to solve for $$35x \equiv 18 \pmod{99}$$
Multiply both sides by $17$, we have $$x \equiv 18(17) \pmod{99}$$
$$x \equiv 306 \equiv 3(99+1)+6\equiv 9 \pmod{99}$$
Hence the smallest positive solution is $9$.
|
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|
Proving $ \frac {(a+b)^2 }{\sqrt{(a+b)^2 +1}} < 2 \left( \frac {a^2 }{\sqrt{ a^2 +1}} + \frac{b^2 } {\sqrt{ b^2 +1 }} \right) $
For positive real $a, b$, $$ \frac {(a+b)^2 }{\sqrt{(a+b)^2 +1}} < 2 \left( \frac {a^2 }{\sqrt{ a^2 +1}} + \frac{b^2 } {\sqrt{ b^2 +1 }} \right).$$
I know the inequality $ (a+b)^2 \le 2(a^2 +b^2 ) $, but this is not useful for me. Then I think the function $$f(x) = \frac{ x^2 } {\sqrt{x^2 +1 }},$$
and the inequality is $$f(a+b) <2 ( f(a) + f(b)).$$
But I meet a wall and I don't know if this inequality is true. I want some hints. Thank you.
|
By C-S we obtain:
$$2\left(\frac {a^2}{\sqrt{a^2 + 1}} + \frac{b^2}{\sqrt{b^2 + 1}}\right)\geq\frac{2(a+b)^2}{\sqrt{a^2+1}+\sqrt{b^2+1}}\geq$$
$$\geq\frac{2(a+b)^2}{\sqrt{(1^2+1^2)(a^2+1+b^2+1)}}=\frac{(a+b)^2}{\sqrt{\frac{a^2+b^2}{2}+1}}\geq\frac{(a+b)^2}{\sqrt{(a+b)^2+1}}.$$
|
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|
On the conjecture that if $\,\forall\{x, y, z\}\subset \Bbb{Z^+}$ squarefree, and $\ x^3 + y^3 = z^2$, then $\gcd(3, x, y) = 1$. I made a conjecture that if $\,\forall\{x, y, z\}\subset\mathbb{Z}^+$ with $x, y, z$ squarefree (not raised to a power greater than $1$; not a perfect power), and $$x^3 + y^3 = z^2,$$ then $\gcd(3, x, y) = 1$. I was able to prove my conjecture for all $z$ prime.
Proof: Let $z$ be a prime number.
Since $(x+y)^3 = x^3 + y^3 + 3xy(x+y),$ then $$\begin{align} x^3 + y^3 &= (x+y)^3 - 3xy(x+y) \\ &= (x+y)\big((x+y)^2-3xy\big) \\ &=(x+y)(x^2 + y^2 - xy).\tag1\end{align}$$ Therefore, we need integer solutions such that $(*)\,\, x + y = x^2 + y^2 - xy$ because $(1) = z^2$, follows that $$\begin{align} 0 &= x^2 + y^2 - xy - x - y \\ &= x(x - 1) + y(y - 1) - xy \\ \Leftrightarrow 3xy &= x(x-1) + y(y-1) + 2xy.\end{align}$$ Since $3xy$ is not a square number, this means that $x \neq 3y$, $y\neq 3x$ and $3\neq xy$.
Therefore $\gcd(3, x, y) = 1$ for this case, but there is also another case where $(**)\,\, x + y = z$ and $x^2 + y^2 - xy = 1$. Of course $x + y\neq 1$ because $x, y\in\mathbb{Z}^+$, thus we need to consider statement $(**)$. $$\begin{align} x^3 + y^3 &= (x+y)^2 \\ \Leftrightarrow (x+y)(x^2 + y^2 - xy) &= (x+y)^2 \\ \Leftrightarrow x^2 + y^2 - xy &= x + y\tag*{$\unicode{x21af}$ (Contradiction)}\end{align}$$ Therefore, statement $(*)$ must always be true, i.e. $\gcd(3, x, y) = 1$.$\quad\quad\quad\quad\qquad\qquad\qquad\quad\Box$
Of course if $z$ is not prime, then $x+y$ does not necessarily have to equal $x^2 + y^2 - xy$. Take $z = 6$ for example, then $z^2 = 6^2 =6\times 6$ but it also equals $9\times 4$, etc.
Could somebody prove/disprove my conjecture for all $z\in\mathbb{Z^+}$? Along with that, is my proof valid?
Thank you in advance.
Edit:
The symbol $\unicode{x21af}$ (Harry Potter's Scar) is used to symbolise a contradiction in this case. Most people prefer $\Rightarrow\Leftarrow$, so if you prefer that instead, treat the Scar as if it were the arrows.
|
As stated your conjecture is false.
I supposed that $x,y$ are squarefree integers.
The following paramertization holds,
$$ x = s^{4} + 6 \, s^{2} t^{2} - 3 \, t^{4}, \
y = -s^{4} + 6 \, s^{2} t^{2} + 3 \, t^{4},\
z = 6 \, {\left(s^{4} + 3 \, t^{4}\right)} s t$$
(See for instance lemma 3.2.6 of Zagier http://www.cecm.sfu.ca/~nbruin/thesis.pdf).
Then we get many solutions with $\gcd(x,y,3)>1$ and $x,y,z>0.$
For instance $s=-3,t=2$ provides $(x,y,z)=(249, 183, 4644)$
or $s=-3,t=4$ gives $(x,y,z)=(177, 1551, 61128).$
In your proof, there are more conditions except $x+y=x^2+y^2-xy$ that implies $(1)=z^2$, e.g. $x+y = ab, x^2+y^2-xy=abc^2.$
|
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|
How to calculate the integral $\int\frac{1}{\sqrt{(x^2+8)^3}}dx$? I need to solve something like this
$$\int\frac{1}{\sqrt{(x^2+8)^3}}dx$$
Wolfram alpha says the solution is $$\frac{x}{8\sqrt{x^2+8}} + c$$
The problem is that the integrand is obtained by the quotient rule:
$$\bigg(\frac{g(x)}{h(x)}\bigg)'=\frac{g'(x)h(x)-g(x)h'(x)}{h^2(x)}$$
$$\bigg(\frac{x}{8\sqrt{x^2+8}}\bigg)'=\frac{1}{8}\cdot\frac{\sqrt{x^2+8}-\frac{x^2}{\sqrt{x^2+8}}}{x^2+8}=\frac{1}{8}\cdot\frac{\frac{x^2+8-x^2}{\sqrt{x^2+8}}}{x^2+8}=\frac{1}{8}\cdot\frac{8}{\sqrt{(x^2+8)^3}}=\frac{1}{\sqrt{(x^2+8)^3}}$$
It there a way to extract the solution from these types of integrals which argument is born from the easy quotient rule?
|
It there a way to extract the solution from these types of integrals which argument is born from the easy quotient rule?
None that I know of. You just see it or maybe make a lucky substitution.
Alternatively, you can take a look at Euler substitutions, which will work in this case and can help for quite a lot of hand-made integrals for homeworks.
|
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|
Inductive proof of sequence defined by recurrence relation I'm working on a proof that shows for any $a \in \mathbb Q$ such that $a < 2$, there exists some $b \in \mathbb Q$ such that $a < b^2 < 2$. It was suggested that as part of the proof, I should consider the series defined by $a_1=1, a_{n+1}=\frac{2a_n+2}{a_n+2}$ and prove by induction that $2 - a_n^2 \le \frac{1}{n}$. I can then use this fact to help pick some $b^2$ larger than $a$. I have a decent idea on what to do with the result, but I seem to be struggling with the induction. Here is what I have so far:
First, we consider our base inductive case. Let $n_0=1$. We have $$2-a_1^2=2-(1)^2=2-1=1 \le1$$ so our hypothesis holds. Now, assume that it holds for some $k > n_0$. We want to show that $2-a_{k+1}^2 \le \frac{1}{k+1}$. We obtain the following:
\begin{align}
\\2-a_{k+1}^2 & = 2-\left( \frac{2a_k+2}{a_k+2} \right) ^2 \\
& = 2 - \frac{(2a_k+2)^2}{(a_k+2)^2} \\
& = 2 - \frac{4a_k^2+8a_k+4}{(a_k+2)^2} \\
& = \frac{2(a_k+2)^2-(4a_k^2+8a_k+4)}{(a_k+2)^2} \\
& = \frac{2a_k^2+8a_k+8-4a_k^2-8a_k-4}{(a_k+2)^2} \\
& = \frac{-2a_k^2+4}{(a_k+2)^2}
\end{align}
It isn't totally clear to me how to proceed from here. From our initial induction hypothesis, we have $2-a_k^2 \le \frac{1}{k}$, so $-a_k^2 \le \frac{1}{k} -2$. Returning to where we left off, we can see:
\begin{align}
\\ \frac{-2a_k^2+4}{(a_k+2)^2} & = \frac{2(-a_k^2+2)}{(a_k+2)^2} \\
& \le \frac{2\left((\frac{1}{k}-2)+2\right)}{(a_k+2)^2} \\
& = \frac{2}{k(a_k+2)^2}
\end{align}
Unfortunately, I don't really see how this can help me reach what I'm trying to show. Any suggestions on where to go from here (or where I went wrong previously) would be more than welcome. I'm not sure if this is relevant, but I'd like to reiterate that this is going to be used as a lemma in a proof regarding the rationals and not the reals, so I'd like to avoid taking the square root of anything that may not result in a rational number.
|
$2-a_{k+1}^2 = \dfrac{-2a_k^2+4}{(a_k+2)^2}$
You are essentially done at this point. All that's left to note is that $\,a_k \ge 0\,$, and therefore:
$$
\\2-a_{k+1}^2 \,=\, \frac{2}{(a_k+2)^2} \cdot (2-a_k^2) \;\le\; \frac{2}{2^2} \cdot \frac{1}{k} \;\le\; \frac{1}{k+1}
$$
|
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|
How do I find the singular solution of the differential equation $y' = \frac{y^2 + 1}{xy + y}$? I start out with the separable differential equation,
$$y' =\frac{dy}{dx} = \frac{y^2 + 1}{xy + y} = \frac{y^2 + 1}{y(x+1)}$$
Thus, $\frac{1 }{x+1}dx = \frac{y }{y^2 + 1}dy$.
Then integrating both sides of the equation, I get
$$\ln(x+1) = \frac{1}{2}\ln(y^2 +1) + C$$
Now, $e^{\ln(x+1)}$ = $e^{\frac{1}{2}\ln(y^2 +1) + C}$. So...
$$(x+1) = e^C(y^2 + 1)^{\frac{1}{2}}$$
I kind of wanted to know if this is indeed the correct general formula. And also, how would I find the singular solution, if there happens to be one in this case.
|
You should get $x+1=C_1\sqrt{y^2+1}$, where $C_1=e^C$.
|
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|
When is the following matrix diagonalizable? $\begin{pmatrix}8&k\\ 0&4\end{pmatrix}$
So I know that for a matrix to be diagonalizable, it needs to have 2 distinct real eigenvalues.
So I calculated the characteristic polynomial to be:
$x^2-12x+32-k$
Therefore, for the discriminant to be greater than $0$, I got the inequality:
$k>-4$
Therefore, I thought that it would diagonalizable for all $k$ except $k=-4$, but this is not correct.
Any help?
|
It is more concrete to just go ahead and see what happens. With the eigenvalues $8,4$ we can get a matrix with eigenvectors as columns from
$$
R =
\left(
\begin{array}{rr}
1 & -k \\
0 & 4
\end{array}
\right).
$$
We can confirm this with
$$
\left(
\begin{array}{rr}
8 & k \\
0 & 4
\end{array}
\right)
\left(
\begin{array}{rr}
1 & -k \\
0 & 4
\end{array}
\right) =
\left(
\begin{array}{rr}
8 & -4k \\
0 & 16
\end{array}
\right)
$$
Then
$$
R^{-1} =
\left(
\begin{array}{rr}
1 & -\frac{k}{4} \\
0 & \frac{1}{4}
\end{array}
\right)
$$
and
$$
\left(
\begin{array}{rr}
1 & -\frac{k}{4} \\
0 & \frac{1}{4}
\end{array}
\right)
\left(
\begin{array}{rr}
8 & k \\
0 & 4
\end{array}
\right)
\left(
\begin{array}{rr}
1 & -k \\
0 & 4
\end{array}
\right) =
\left(
\begin{array}{rr}
8 & 0 \\
0 & 4
\end{array}
\right)
$$
Whatever the appearances, the value of $k$ does not matter. Since someone mentioned it, we are taking the characteristic of the field to not be two.
|
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|
How to extract $(x+y+z)$ or $xyz$ from the determinant Prove
$$\color{blue}{
\Delta=\begin{vmatrix}
(y+z)^2&xy&zx\\
xy&(x+z)^2&yz\\
xz&yz&(x+y)^2
\end{vmatrix}=2xyz(x+y+z)^3}
$$
using elementary operations and the properties of the determinants without expanding.
My Attempt
$$
\Delta\stackrel{C_1\rightarrow C_1+C_2+C_3}{=}\begin{vmatrix}
xy+y^2+yz+zx+yz+z^2&xy&zx\\
x^2+xy+xz+xz+yz+z^2&(x+z)^2&yz\\
x^2+xy+xz+xy+y^2+yz&yz&(x+y)^2\\
\end{vmatrix}\\
=\begin{vmatrix}
y(x+y+z)+z(x+y+z)&xy&zx\\
x(x+y+z)+z(x+y+z)&(x+z)^2&yz\\
x(x+y+z)+y(x+y+z)&yz&(x+y)^2\\
\end{vmatrix}\\
=(x+y+z)\begin{vmatrix}
y+z&xy&zx\\
x+z&(x+z)^2&yz\\
x+y&yz&(x+y)^2\\
\end{vmatrix}\\
\stackrel{R_1\rightarrow R_1+R_2+R_3}{=} \\(x+y+z)\begin{vmatrix}
2(x+y+z)&x^2+xy+xz+xz+yz+z^2&x^2+xy+zx+xy+y^2+yz\\
x+z&(x+z)^2&yz\\
x+y&yz&(x+y)^2\\
\end{vmatrix}\\
=(x+y+z)^2\begin{vmatrix}
2&x+z&x+y\\
x+z&(x+z)^2&yz\\
x+y&yz&(x+y)^2\\
\end{vmatrix}\\
=2(x+y+z)^2\begin{vmatrix}
1&\frac{x+z}{2}&\frac{x+y}{2}\\
x+z&(x+z)^2&yz\\
x+y&yz&(x+y)^2
\end{vmatrix}
\stackrel{R_2\rightarrow R_2-(x+z)R_1|R_2\rightarrow R_3-(x+y)R_1}{=}
2(x+y+z)^2\begin{vmatrix}
1&\frac{x+z}{2}&\frac{x+y}{2}\\
0&\frac{(x+z)^2}{2}&\frac{yz-x^2-xy-xz}{2}\\
0&\frac{yz-x^2-xy-xz}{2}&\frac{(x+y)^2}{2}
\end{vmatrix}
$$
How do I extract $x+y+z$ or $xyz$ from the determinant to find the solution ?
Note: In a similar problem How to solve this determinant there seems to be solutions talking about factor theorem and polynomials. I am basically looking for extracting the terms $(x+y+z)$ or $xyz$ from the given determinant to solve it only using the basic properties of determinants.
|
The determinant must be a polynomial in $x,y,z$ of degree $6$.
If you set $x=0$,
$$
\begin{vmatrix}
(y+z)^2&0&0\\
0&z^2&yz\\
0&yz&y^2
\end{vmatrix}=0
$$
so that $x$ is a factor. And by symmetry, $xyz$ as well.
Then with $x=-y-z$,
$$
\begin{vmatrix}
(y+z)^2&xy&zx\\
xy&y^2&yz\\
xz&yz&z^2
\end{vmatrix}=0
$$ (the bottom six minors are zero), and $x+y+z$ is a factor.
To reach the sixth degree, we are missing two other linear factors. But by symmetry, they cannot be other than $x+y+z$ both (two factors alone cannot sustain a permutation of the variables).
Remains to find the global factor, for example from
$$
\begin{vmatrix}
4&1&1\\
1&4&1\\
1&1&4
\end{vmatrix}=\lambda\cdot1\cdot1\cdot1\cdot(1+1+1)^3.$$
|
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|
Find the kernel of a 4x4 matrix $$
\begin{pmatrix}
1 & 2 & 3 & 4\\
5 & 6 & 7 & 8\\
9 & 10 & 11 & 12\\
13 & 14 & 15 & 16\\
\end{pmatrix}
$$
I am asked to find the kernel of the matrix $M$. After doing some row operation I get to
$$
\begin{pmatrix}
1 & 2 & 3 & 4\\
0 & -4 & -8 & -12\\
0 & 0 & 0 & 0\\
0 & 0 & 0 & 0\\
\end{pmatrix}
$$
and for $x$ I find $x = \alpha + 2\beta$, whereas $y = -2\alpha -3\beta$
Therefore,
$$
\begin{pmatrix}
\alpha + 2\beta\\
-2\alpha - 3\beta\\
\alpha\\
\beta\\
\end{pmatrix}
$$
When we take outside alpha and beta:
we get two vectors:
$$
\begin{pmatrix}
1\\
-2\\
1\\
0\\
\end{pmatrix}
$$
and
$$
\begin{pmatrix}
2\\
-3\\
0\\
1\\
\end{pmatrix}
$$
which are linearly independent and form a basis of this $ker(M)$
Could you please confirm with me whether you get the same result? Thank you.
|
Yes it looks good great. However, you should precise why the dimension of Kernel is $2$. For example by extracting the $2 \times 2$ determinant
$$
\begin{vmatrix}
3&4 \\
7&8
\end{vmatrix}=24-28 \ne 0
$$
So the dimesion of the image is at least $2$. Then you found two vectors in it, so the dimension of the image cannot exceed two so it values $2$. as well as the dimension of the Kernel.
|
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|
Find $\cos2B+\cos2C$ in a triangle In triangle $ABC$ we have $2a^2=b^2+c^2$. The measure of angle $A$ is $30^{\circ}$. Find $\cos2B+\cos2C$
|
By the law of cosines we obtain:
$$a^2=b^2+c^2-2bc\cdot
\frac{\sqrt3}{2},$$
which gives $$2(b^2+c^2-\sqrt3bc)=b^2+c^2$$ or $$b^2+c^2=2\sqrt3bc.$$
Also, we have $$a^2=\sqrt3bc.$$
Thus, $$\cos2\beta+\cos2\gamma=2\left(\frac{a^2+c^2-b^2}{2ac}\right)^2+2\left(\frac{a^2+b^2-c^2}{2ab}\right)^2-2=$$
$$=2\left(\frac{\frac{b^2+c^2}{2}+c^2-b^2}{2ac}\right)^2+2\left(\frac{\frac{b^2+c^2}{2}+b^2-c^2}{2ab}\right)^2-2=$$
$$=\frac{(3c^2-b^2)^2}{4c^2(b^2+c^2)}+\frac{(3b^2-c^2)^2}{4b^2(b^2+c^2)}-2=$$
$$=\frac{b^6+c^6+3b^4c^2+3c^4b^2}{4b^2c^2(b^2+c^2)}-2=\frac{(b^2+c^2)^2}{4b^2c^2}-2=3-2=1.$$
|
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|
Prove that $n^2 - 2n + 7$ is even then $n + 1$ is even I have to use Proof by contradiction to show what if $n^2 - 2n + 7$ is even then $n + 1$ is even.
Assume $n^2 - 2n + 7$ is even then $n + 1$ is odd. By definition of odd integers, we have $n = 2k+1$.
What I have done so far:
\begin{align}
& n + 1 = (2k+1)^2 - 2(2k+1) + 7 \\
\implies & (2k+1) = (4k^2+4k+1) - 4k-2+7-1 \\
\implies & 2k+1 = 4k^2+1-2+7-1 \\
\implies & 2k = 4k^2 + 4 \\
\implies & 2(2k^2-k+2)
\end{align}
Now, this is even but I wanted to prove that this is odd(the contradiction). Can you some help me figure out my mistake?
Thank you.
|
$n^2-2n+7
=n^2+2n+1-(4n+6)
=(n+1)^2-2(2n+3)
$.
If this is even then,
since $2(2n+3)$ is even,
their sum is even,
so $(n+1)^2$ is even
so $n+1$ is even.
|
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|
Compute the volume bounded by the parabolic cylinders $x^2=4-4z, \quad y^2=4-4z$ and the $xy-$plane.
Compute the volume bounded by the parabolic cylinders $x^2=4-4z, \quad
> y^2=4-4z$ and the $xy-$plane.
Say the $xy-$plane is the ground, the two cylindrcal paraboloids then make upp the roof, where the base is simply a square. Looking at this from above and onto the $xy-$ plane, the following image should emerge:
Hence the volume of this region should easily be computed by
$$V=\iiint_K1 \ dxdydz = 8\iint_D\left(\int_0^{1-x^2/4}1\right)\ dxdy=8\iint_D1-\frac{x^2}{4} \ dydx=$$
$$8\int_{0}^{1}\left(1-\frac{x^2}{4}\right)\int_0^x1 \ dydx= 8\int_0^1x-\frac{x^3}{4} \ dx =\frac{7}{2}.$$
Correct answer: $V=8.$ Why is this?
|
Note that in the integral $x$ varies between $0$ and $2$, thus
$$8\int_0^2 x-\frac{x^3}{4} \ dx = \left[\frac{x^2}2-\frac{x^4}{16}\right]_0^2=8(2-1)=8$$
|
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|
Diophantine Equation $a^6+b^6+c^6=d^6$ Within the literature on Diophantine equations there seems to be very little on the $6,1,3$ equation:
$$a^6+b^6+c^6=d^6\quad\quad(1)$$
Mathworld, for example, simply records that there are no known solutions of $6.1.n$ equations for $n \leq 6$. The former Euler Project searching for solutions of equations in equal sums of like powers showed $6,1,5$ at the top of its most wanted list but did not mention $6,1,3$.
One way of looking at the $6,1,3$ equation is as a special case of the much more familiar $3,1,3$ equation:
$$w^3+x^3+y^3=z^3\quad\quad(2)$$
in which each of the terms is also a square. It may be considered relevant that there are known solutions of $(2)$ in which two or three of the terms are square, the smallest respectively being:
$$(1^2)^3 + 6^3 + 8^3 = (3^2)^3\quad\quad(3)$$
$$118^3 + (15^2)^3 + (18^2)^3 = (19^2)^3\quad\quad(4)$$
Question: Are there any discussions in the literature of either:
*
*direct strategies for searching for solutions of $(1)$;
*research seeking a proof that $(1)$ has no non-trivial solutions?
By a direct strategy I mean one which addresses 6,1,3 itself, rather than one which searches for $6,1,n$ for higher $n$ with the remote possibility of finding a solution in which some of the terms are zero. An example of a direct strategy would be to take the following parametric solution of $(2)$:
$$(9s^4)^3 + (3s(t^3-3s^3))^3 + (t(t^3-9s^3))^3 = (t^4)^3$$
which already has two square terms, and to try to find values of $s$ and $t$ such that the other two terms are also square.
|
COMMENT (This is not an answer!)
The identity $(2xz)^2+(2yz)^2+(z^2-x^2-y^2)^2=(x^2+y^2+z^2)^2$ gives the general solution of $X^2+Y^2+Z^2=W^2$ so one has the necessary condition with parameters $x,y,z$
$$\begin{cases}a^3=2xz\\b^3=2yz\\c^3=z^2-x^2-y^2\\d^3=x^2+y^2+z^2\end{cases}$$ This could be a good first step to get an answer.
(Follow the COMMENT).- By the way, if
$$A_1=a^5+10a^4-8a^3+16a^2+64a-32\\A_2=a^5-10a^4-8a^3-16a^2+64a+32\\A_3=-a^5+8a^4+8a^3-16a^2+80a+32\\A_4=-a^5-8a^4+8a^3+16a^2+80a-32$$ then we have for all real parameter $a$ the identity
$$a(6a^4+24a^2+96)^3=A_1^3+A_2^3+A_3^3+A_4^3$$ Consequently, making $a=n^3$, one has for all natural $n$ an identity:
$$X_1^3+X_2^3+X_3^3+X_4^3=X_5^3$$ where each $X_i$ has degree $15$.
|
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|
Solve the following differential equations 2x2 with complex eigenvert
Solve the following equation:
$$X'=\pmatrix { 2 & 5
\\ -5 & 8}X$$
$$x(0)=(5,5)^t$$
I already got out the followings:
$$\lambda_{1,2} = 5 \pm 4i$$
$$v_{1} = \begin{pmatrix} 3 \\ 5 \end{pmatrix} - i\cdot \begin{pmatrix} 4 \\ 0 \end{pmatrix}$$
$$v_{1} = \begin{pmatrix} 3 \\ 5 \end{pmatrix} + i\cdot \begin{pmatrix} 4 \\ 0 \end{pmatrix}$$
And so far I have this(Sry its messy):
$$\begin{align*}\vec{x}_{1}(t) = (3-4i)e^{5t} \begin{pmatrix}a\cos(4t)+b\sin(4t)\end{pmatrix}\end{align*}$$
$$\begin{align*}\vec{x}_{2}(t) = 5\cdot e^{5t} \begin{pmatrix}c\cos(4t)+d\sin(4t)\end{pmatrix}\end{align*}$$
I dont know if this is right so far... but I also tried to sum up x_1:
My attend to solve x_1
This is the right answer in the end:$$\begin{align*}\vec{x}(t) = e^{5t} \begin{pmatrix}5\cos(4t)+\frac{5}{2}\sin(4t)\\ 5\cos(4t)-\frac{5}{2}\sin(4t)\end{pmatrix}\end{align*}$$
tyx
|
Since we have complex conjugate eigenvalues/eigenvectors, we only need to use one of them.
For $\lambda_1 = 5 + 4 i, v_1 = \begin{pmatrix} 3 -4i\\ 5 \end{pmatrix} $, we have
$\begin{align} e^{(5 + 4i)t} \begin{pmatrix} 3 -4i\\ 5 \end{pmatrix} \\ = e^{5t}e^{4it} \begin{pmatrix} 3 -4i\\ 5 \end{pmatrix}\end{align} \\
= e^{5t}(\cos 4t + i \sin 4t)\begin{pmatrix} 3 -4i\\ 5 \end{pmatrix} \\
= e^{5t}\begin{pmatrix} 3 \cos 4t + 4 \sin 4t +i(3 \sin 4t - 4 \cos 4t)\\ 5 \cos 4 t + i( 5 \sin 4t )\end{pmatrix}$
We know that the real and imaginary parts are both solutions, so our general solution is
$$\vec{x}(t) = e^{5t}\left(c_1\begin{pmatrix} 3 \cos 4t + 4 \sin 4t \\ 5 \cos 4 t \end{pmatrix} + c_2 \begin{pmatrix} 3 \sin 4t - 4 \cos 4t\\ 5 \sin 4t \end{pmatrix} \right)$$
We now use our initial condition $x(0)=(5,5)^t$
$$c_1\begin{pmatrix} 3 \\ 5 \end{pmatrix} + c_2 \begin{pmatrix} - 4 \\ 0 \end{pmatrix} = \begin{pmatrix} 5 \\ 5 \end{pmatrix}$$
You can use many methods to solve this $2 \times 2$ and it yields $c_1 = 1, c_2 = -\dfrac{1}{2}$ and a final result
$$\begin{align*}\vec{x}(t) = e^{5t} \begin{pmatrix}5\cos 4t+\dfrac{5}{2}\sin 4t\\ 5\cos 4 t-\dfrac{5}{2}\sin 4t\end{pmatrix}\end{align*}$$
|
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|
Calculating the summation $\sum_{n=1}^{\infty}\frac{H_{n+1}}{n(n+1)}$ I need to find explicitly the following summation
$$\sum_{n=1}^{\infty}\frac{H_{n+1}}{n(n+1)}, \quad
H_{n}=1+\frac{1}{2}+\cdots+\frac{1}{n}$$
From Mathematica, I checked that the answer is $2$. The same result is returned by WolframAlpha.
A thought from afar:
$$
\begin{align}
\frac{H_{n+1}}{n(n+1)}
&=H_{n+1}\left(\frac1n-\frac1{n+1}\right)\tag1\\
&=\color{#C00}{H_{n+1}}\color{#090}{(H_n-H_{n-1})}-H_{n+1}(H_{n+1}-H_n)\tag2\\
&=\color{#C00}{\frac1{n+1}}\color{#090}{\frac1n}+\color{#C00}{H_n}\color{#090}{(H_n-H_{n-1})}-H_{n+1}(H_{n+1}-H_n)\tag3\\
&=\frac1n-\frac1{n+1}+\frac{H_n}{n}-\frac{H_{n+1}}{n+1}\tag4\\
&=\frac{H_n+1}{n}-\frac{H_{n+1}+1}{n+1}\tag5
\end{align}
$$
What was done:
$(1)$: partial fractions
$(2)$: used $\frac1n=H_n-H_{n-1}$ and $\frac1{n+1}=H_{n+1}-H_n$
$(3)$: used $\color{#C00}{H_{n+1}=\frac1{n+1}+H_n}$ and $\color{#090}{\frac1n=H_n-H_{n-1}}$
$(4)$: partial fractions and $\frac1n=H_n-H_{n-1}$ and $\frac1{n+1}=H_{n+1}-H_n$
$(5)$: combined terms
Now this looks like Clement C's hint.
|
(Big) hint:
$$\frac{H_{n+1}}{n(n+1)}
= \frac{H_{n+1}}{n}-\frac{H_{n+1}}{n+1}
= \frac{1}{n(n+1)} + \frac{H_{n}}{n}-\frac{H_{n+1}}{n+1}$$
so that you can get a telescopic series: for any $N\geq 1$,
$$
\sum_{n=1}^N \frac{H_{n+1}}{n(n+1)}
= \sum_{n=1}^N \frac{1}{n(n+1)} + \sum_{n=1}^N \frac{H_{n}}{n} - \sum_{n=2}^{N +1}\frac{H_{n}}{n}
$$
Spoiler:
The value you should arrive at is $2$, which is $\sum_{n=1}^\infty \frac{1}{n(n+1)} + 1$.
|
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real value of $k$ inirrational equation Find value of $k$ for which the equation $\sqrt{3z+3}-\sqrt{3z-9}=\sqrt{2z+k}$ has no solution.
solution i try $\sqrt{3z+3}-\sqrt{3z-9}=\sqrt{2z+k}......(1)$
$\displaystyle \sqrt{3z+3}+\sqrt{3z-9}=\frac{12}{\sqrt{2z+k}}........(2)$
$\displaystyle 2\sqrt{3z+3}=\frac{12}{\sqrt{2z+k}}+\sqrt{2z+k}=\frac{12+2z+k}{\sqrt{2z+k}}$
$\displaystyle 2\sqrt{6z^2+6z+3kz+3k}=\sqrt{12+2z+k}$
$\displaystyle 4(6z^2+6z+3kz+3k)=144+4z^2+k^2+48z+4kz+24k$
Help me how i solve it after that point
|
Starting from your last step,
$$4(6z^2+6z+3kz+3k)=144+4z^2+k^2+48z+4kz+24k$$
is equivalent to
$$28z^2+(24+12k+48+4k)z-(k^2+12k-144)=0 $$
and this as no solution (since $z$ is real, for the squared roots be defined) if and only if
$$ \Delta = (24+12k+48+4k)^2-4\times 28 \times -(k^2+12k-144) <0 $$
So it remain to study a 2de order equation degree in $k$.
|
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Show that $\frac{1}{a^n+b^n+c^n} = \frac{1}{a^n} + \frac{1}{b^n} + \frac{1}{c^n}$ Let $n \in N, n=2k+1, and \text{ } \frac{1}{a+b+c} = \frac{1}{a} + \frac{1}{b} + \frac{1}{c}$.
Show that $$\frac{1}{a^n+b^n+c^n} = \frac{1}{a^n} + \frac{1}{b^n} + \frac{1}{c^n}$$
I have tried, but I don't get anything. Can you please give me a hint?
|
Hint
*
*At the first step, show that $(a+b)(a+c)(b+c)=0$
*Next, show that two of the three numbers are opposite.
|
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$36 \leq 4(a^3+b^3+c^3+d^3) - (a^4+b^4+c^4+d^4)\leq4 8.$ Let $a,b,c,d \in \Bbb R$, $a+b+c+d=6$ and $a^2+b^2+c^2+d^2=12$. Then
$$36 \leq 4(a^3+b^3+c^3+d^3) - (a^4+b^4+c^4+d^4)\leq4
8.$$
I have found only two bounds: $216 \geq a^3+b^3+c^3+d^3$ and $144 \geq a^4+b^4+c^4+d^4$.
How to prove this inequality?
|
For upper bound, we have
$$ x^3(4-x) \leq 4x^2\;\; \forall x$$
and we are done.
For lower bound:
Let $a+b+c =x$, then $d=6-x$ and by Cauchy we have $a^2+b^2+c^2\geq x^2/3$
so $$12 \geq {x^2\over 3} +(6-x)^2$$ so $x\in (3,6) $ and so $0\leq d\leq 3$. But this holds also for $a,b,c$.
Idea for finish:
Let $f(x) = 4x^3-x^4$, for $x\in[0,3]$ we have $4x^3-x^4\geq x^3$, so we have by Cauchy
$$ f(a)+f(b)+f(c)\geq a^3+b^3+c^3\geq {(a^2+
b^2+c^2)^2\over a+b+c} = {(12-d^2)^2\over 6-d}$$
So we have to check if following holds:
$$ f(d)+ {(12-d^2)^2\over 6-d}\geq 36$$
|
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How to calculate $ \int \frac{\sin^{6}(x)}{\sin^{6}(x) + \cos^{6}(x)} dx? $ How to calculate
$$ \int \frac{\sin^{6}(x)}{\sin^{6}(x) + \cos^{6}(x)} dx? $$
I already know one possible way, that is by :
$$ \int \frac{\sin^{6}(x)}{\sin^{6}(x) + \cos^{6}(x)} dx = \int 1 - \frac{\cos^{6}(x)}{\sin^{6}(x) + \cos^{6}(x)} dx $$
$$= x-
\int \frac{1}{1+\tan^{6}(x)} dx $$
Then letting $u=\tan(x)$, we must solve
$$\int \frac{1}{(1+u^{6})(1+u^{2})} du $$
We can reduce the denominator and solve it using Partial Fraction technique. This is quite tedious, I wonder if there is a better approach.
Using same approach, for simpler problem, I get
$$\int \frac{\sin^{3}(x)}{\sin^{3}(x)+\cos^{3}(x)} dx = \frac{x}{2} - \frac{\ln(1+\tan(x))}{6} + \frac{\ln(\tan^{2}(x)- \tan(x)+1)}{3} - \frac{\ln(\sec(x))}{2} + C$$
|
An alternative would be to do a load of trig manipulation:
Starting with $$\frac{s^6}{s^6+c^6}=1-\frac{c^6}{s^6+c^6}$$
$$=1-\frac{\frac 18(1+\cos2x))^3}{ \frac 18(1+\cos2x))^3+\frac 18(1-\cos2x))^3}$$
= a few more lines of algebra = $$=\frac 12+\frac 12\cos 2x\left(\frac{3+\cos^2 2x}{1+3\cos^2 2x}\right)$$
$$=\frac 12+\frac 12\cos 2x\left(\frac{4-\sin^2 2x}{4-3\sin^2 2x}\right)$$
Now integrate, and for the second part us the substitution $$u=\frac{\sqrt{3}}{2}\sin 2x$$
and you end up with $$\frac 12 x+\frac 16\sin x\cos x+\frac{1}{3\sqrt{3}}\operatorname{artanh}(\sqrt{3}\sin x\cos x)+c$$
|
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|
Find a pattern and prove it by mathematical induction: $1 = 1$
$3 + 5 = 8$
$7 + 9 + 11 = 27$
$13 + 15 + 17 + 19 = 64$
Etc...
I am having trouble seeing a pattern with this, I know it is relatable with Fibonacci Numbers but I am having trouble grasping this topic
|
Let $a_n$ the $n$-th term of the sequence.
The sum of the $n$ first terms is the sum of the first $\frac{n(n+1)}{2}$ odd integers, which is $\left( \frac{n(n+1)}{2} \right)^2$.
So that:
\begin{equation}
a_1+\ldots +a_n = \left( \frac{n(n+1)}{2} \right)^2
\end{equation}
\begin{equation}
a_1+\ldots +a_{n-1} = \left( \frac{(n-1)n}{2} \right)^2
\end{equation}
Then
$$a_n = \left( \frac{n(n+1)}{2} \right)^2 - \left( \frac{(n-1)n}{2} \right)^2 = \\
\left( \frac{n(n+1)}{2}+\frac{(n-1)n}{2} \right)\left( \frac{n(n+1)}{2}-\frac{(n-1)n}{2} \right) =
\left( \frac{n^2+n}{2}+\frac{n^2-n}{2} \right)\left( \frac{n^2+n}{2}-\frac{n^2-n}{2} \right) =
n^2 \cdot n = n^3
$$
|
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|
Sketch the region in the plane consisting of all points (x,y) such that $2xy\le |x-y|\le x^2+y^2$ Question: Sketch the region in the plane consisting of all points (x,y) such that $2xy\le |x-y|\le x^2+y^2$
My attempt:
If $x>y$ then
$|x-y|=x-y$
Thus now,
$2xy\le x-y\le x^2+y^2$
If $x-y\geq 2xy$ then let $x-y=k$ ($k$ is some real number)
then $y=x-k \tag{1}$
Thus the straight line (1) has a negative slope.
After this how to proceed?
|
we have $$2xy\le |x-y|\le x^2+y^2$$ now we assume $$x\geq y$$ then we get
$$2xy\le x-y$$ and $$x-y\le x^2+y^2$$
the right inequality Can be written as $$\frac{1}{2}\le (x-\frac{1}{2})^2+(y-\frac{1}{2})^2$$
the left Hand side is:
$$y(1+2x)\le x$$
$$x=-\frac{1}{2}$$ gives a contradiction
if $$x<-\frac{1}{2}$$ we get $$y\geq \frac{x}{1+2x}$$
if $$x>-\frac{1}{2}$$ we get $$y\le \frac{x}{1+2x}$$
Can you finish now?
|
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|
Greatest Common Divisor of two Polynomials. Find the $\gcd(x^3-6x^2+14x-15, x^3-8x^2+21x-18)$ over $\mathbb{Q}[x]$. Then find two polynomials $a(x),b(x) \in \mathbb{Q}[x]$ such that, $$a(x)(x^3-6x^2+14x-15) + b(x)(x^3-8x^2+21x-18)=\gcd(x^3-6x^2+14x-15, x^3-8x^2+21x-18)$$
I have managed to find,
$$x^3-6x^2+14x-15=(x-3)(x^2-3x+5)$$
$$x^3-8x^2+21x-18=(x-3)(x-3)(x-2)$$
Now since $x^2-3x+5$ is irreducible over $\mathbb{Q}[x]$ and so the greatest common divisor is $(x-3)$. Now to find $a(x)$ and $b(x)$ I have no clue how to do that. I have looked online and it seems there is extended euclidean algorithm for polynomials but I haven't formally learned it in my class yet, so I was wondering if there is another efficient way to find these polynomials. Any help is appreciated, thanks!
|
This is just the Extended Euclidean Algorithm. Instead of back-substitution, I have always preferred to write the construction steps in the style of continued fractions. Furthermore, I have always depended on the kindness of strangers.
$$ \left( x^{3} - 6 x^{2} + 14 x - 15 \right) $$
$$ \left( x^{3} - 8 x^{2} + 21 x - 18 \right) $$
$$ \left( x^{3} - 6 x^{2} + 14 x - 15 \right) = \left( x^{3} - 8 x^{2} + 21 x - 18 \right) \cdot \color{magenta}{ \left( 1 \right) } + \left( 2 x^{2} - 7 x + 3 \right) $$
$$ \left( x^{3} - 8 x^{2} + 21 x - 18 \right) = \left( 2 x^{2} - 7 x + 3 \right) \cdot \color{magenta}{ \left( \frac{ 2 x - 9 }{ 4 } \right) } + \left( \frac{ 15 x - 45 }{ 4 } \right) $$
$$ \left( 2 x^{2} - 7 x + 3 \right) = \left( \frac{ 15 x - 45 }{ 4 } \right) \cdot \color{magenta}{ \left( \frac{ 8 x - 4 }{ 15 } \right) } + \left( 0 \right) $$
$$ \frac{ 0}{1} $$
$$ \frac{ 1}{0} $$
$$ \color{magenta}{ \left( 1 \right) } \Longrightarrow \Longrightarrow \frac{ \left( 1 \right) }{ \left( 1 \right) } $$
$$ \color{magenta}{ \left( \frac{ 2 x - 9 }{ 4 } \right) } \Longrightarrow \Longrightarrow \frac{ \left( \frac{ 2 x - 5 }{ 4 } \right) }{ \left( \frac{ 2 x - 9 }{ 4 } \right) } $$
$$ \color{magenta}{ \left( \frac{ 8 x - 4 }{ 15 } \right) } \Longrightarrow \Longrightarrow \frac{ \left( \frac{ 4 x^{2} - 12 x + 20 }{ 15 } \right) }{ \left( \frac{ 4 x^{2} - 20 x + 24 }{ 15 } \right) } $$
$$ \left( x^{2} - 3 x + 5 \right) \left( \frac{ 2 x - 9 }{ 15 } \right) - \left( x^{2} - 5 x + 6 \right) \left( \frac{ 2 x - 5 }{ 15 } \right) = \left( -1 \right) $$
$$ \left( x^{3} - 6 x^{2} + 14 x - 15 \right) = \left( x^{2} - 3 x + 5 \right) \cdot \color{magenta}{ \left( x - 3 \right) } + \left( 0 \right) $$
$$ \left( x^{3} - 8 x^{2} + 21 x - 18 \right) = \left( x^{2} - 5 x + 6 \right) \cdot \color{magenta}{ \left( x - 3 \right) } + \left( 0 \right) $$
$$ \mbox{GCD} = \color{magenta}{ \left( x - 3 \right) } $$
$$ \left( x^{3} - 6 x^{2} + 14 x - 15 \right) \left( \frac{ 2 x - 9 }{ 15 } \right) - \left( x^{3} - 8 x^{2} + 21 x - 18 \right) \left( \frac{ 2 x - 5 }{ 15 } \right) = \left( - x + 3 \right) $$
............
|
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|
$\lim_{x\rightarrow 0} \frac{2\cdot \ln(1+x)+x^2-2x}{x^3}$ without 'Hopital Rule Compute $$\lim_{x\rightarrow 0} \frac{(x+2)\cdot \ln(1+x)-2x}{x^3}$$ without L'Hopital Rule.
I proved before that that $$\lim_{x\rightarrow 0} \frac{ \ln(1+x)-x}{x^2}=-\frac{1}{2}$$
I tried to use it and I have to compute $$\lim_{x\rightarrow 0} \frac{2\cdot \ln(1+x)+x^2-2x}{x^3}$$
This limit looks easier to compute.
|
As shown, without l'Hopital, the limit can be solved easily by Taylor's expansion that is given the two following limits
*
*$\frac{\ln(1+x)-x}{x^2}\to -\frac12$ (as already mentioned)
and
*
*$\frac{\ln(1+x)-x+\frac{x^2}2}{x^3}\to \frac13$ (which correspond to to your second limit)
indeed by those limits, as you noticed, it follow that
$$\frac{(x+2)\cdot \ln(1+x)-2x}{x^3}=\frac{\ln(1+x)-x}{x^2}+\frac{2\cdot \ln(1+x)+x^2-2x}{x^3}\to-\frac12+\frac23=\frac16$$
Note that I've also tried by various manipulation as the following but I can't reach the result.
$$\frac{2\cdot \ln(1+x)+x^2-2x}{x^3}=\frac23\frac{3\ln(1+x)+\frac32x^2-3x}{x^3}=\frac23\frac{\ln(1+3x+3x^2+x^3)-(3x+3x^2+x^3)}{(3x+3x^2+x^3)^2}\frac{(3x+3x^2+x^3)^2}{x^3}+\frac23\frac{x^3+\frac92x^2}{x^3}$$
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"timestamp": "2023-03-29T00:00:00",
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|
Simplification algebraic of a cube root. I am trying to simplify this:
$$\frac{1000}{\pi \cdot (\frac{500}{\pi})^{\frac{2}{3}}}$$
and I think it becomes:
$$2 \cdot \sqrt[3]{\frac{500}{\pi}}$$
I basically thought we cube root the $\frac{500}{\pi}$ and then multiply the denominator by $\frac{500}{\pi}$ which could cancel out some stuff. This is how I thought about it:
$$\frac{1000}{\pi \cdot \sqrt[3]{\frac{500}{\pi}} \cdot \frac{500}{\pi}}$$
$$ = \frac{1000 \cdot \pi}{\pi \cdot \sqrt[3]{\frac{500}{\pi}} \cdot 500}$$
Is there a better way to do this cancellation?
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$$\frac{1000}{ \pi \cdot (\frac{500}{\pi})^{\frac{2}{3}} }
=\frac{1000}{\pi^{\frac{1}{3}}\cdot 2^{-\frac{2}{3}}\cdot 2^{\frac{2}{3}}\cdot 500^{\frac{2}{3}}}
=\frac{1000}{\pi^{\frac{1}{3}}\cdot 2^{-\frac{2}{3}}\cdot 1000^{\frac{2}{3}}}
=\frac{2\cdot 1000^{\frac{1}{3}}}{(2\pi)^{\frac{1}{3}}}
=\frac{20}{\sqrt[3] {2\pi} }=\frac{10\sqrt[3] {4\pi^2} }{\pi}=10\sqrt[3]{\frac{4}{\pi}}$$
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{
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"url": "https://math.stackexchange.com/questions/2683213",
"timestamp": "2023-03-29T00:00:00",
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|
Find $\int_0^{\sqrt{R^2-1}}g(x)-f(x)\,\mathrm dx$
How do you evaluate this integral?
$$\int_0^{\sqrt{R^2-1}}g(x)-f(x)\,\mathrm dx$$
where $f(x)=1$ and $g(x)=\sqrt{R^2-x^2}$.
Wolfram tells me I exceeded my computational limit. Mathematica gives me a long answer which is very difficult to read (for me).
EDIT: I was told to consider instead $x=f(y)$, so:
$$\int_0^{1}\sqrt{R^2-y^2}\,\mathrm dy$$
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Using linearity, you can split the integral
$$\begin{align*}\int_{0}^{\sqrt{R^2 - 1}} g(x) - f(x)~\mathrm{d}x &= \int_{0}^{\sqrt{R^2 - 1}} \sqrt{R^2 - x^2} - 1~\mathrm{d}x \\ &= \int_{0}^{\sqrt{R^2 - 1}} \sqrt{R^2 - x^2}~\mathrm{d}x - \int_{0}^{\sqrt{R^2 - 1}} 1~\mathrm{d}x\tag{1}\end{align*}$$
Let's first solve the first part of the integral using the substituion $$x = R\cdot\sin(u) \implies \mathrm{d}x = R\cdot\cos(u)~\mathrm{d}u$$
The limits change as follows
$$0 = R\cdot\sin(u)\implies u = 0$$ and $$\sqrt{R^2 - 1} = R\cdot\sin(u)\implies u = \arcsin\left(\frac{\sqrt{R^2 - 1}}{R}\right)$$
$$\begin{align*}\int_{0}^{\sqrt{R^2 - 1}} \sqrt{R^2 - x^2}~\mathrm{d}x &= \int_{0}^{\arcsin\left(\frac{\sqrt{R^2 - 1}}{R}\right)} R\cdot\cos(u)\sqrt{R^2 - R^2\cdot\sin^2(u)}~\mathrm{d}u \\ &= \int_{0}^{\arcsin\left(\frac{\sqrt{R^2 - 1}}{R}\right)} R\cdot\cos(u)\sqrt{R^2\cdot\cos^2(u)}~\mathrm{d}u \\ &= R^2\int_{0}^{\arcsin\left(\frac{\sqrt{R^2 - 1}}{R}\right)} \cos^2(u)~\mathrm{d}u \\ &= R^2\left(\frac{\cos(u)\cdot\sin(u)}{2} + \frac{u}{2}\right)\Bigg|_{0}^{\arcsin\left(\frac{\sqrt{R^2 - 1}}{R}\right)} \\ &=\frac{R^2}{2}\left(\left[\cos\Big({\arcsin\left(\frac{\sqrt{R^2 - 1}}{R}\right)\Big)}\cdot\sin\Big({\arcsin\left(\frac{\sqrt{R^2 - 1}}{R}\right)\Big)} \\ + \arcsin\left(\frac{\sqrt{R^2 - 1}}{R}\right)\right]\right) \\
&= \frac{\sqrt{R^2 - 1}}{2} + \frac{R^2}{2}\cdot\arcsin\left(\frac{\sqrt{R^2 - 1}}{R}\right)\tag{2}\end{align*}$$
And secondly,
$$\int_{0}^{\sqrt{R^2 - 1}} 1~\mathrm{d}x = \sqrt{R^2 - 1}\tag{3}$$
Substitute $(2)$ and $(3)$ in $(1),$
$$\int_{0}^{\sqrt{R^2 - 1}} \sqrt{R^2 - x^2}~\mathrm{d}x - \int_{0}^{\sqrt{R^2 - 1}} 1~\mathrm{d}x = \frac{R^2}{2}\cdot\arcsin\left(\frac{\sqrt{R^2 - 1}}{R}\right) - \frac{\sqrt{R^2 - 1}}{2}$$
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"timestamp": "2023-03-29T00:00:00",
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|
Area of largest inscribed rectangle in an ellipse. Can I take the square of the area to simplify calculations? So say I have an ellipse defined like this:
$$\frac{x^2}{9} + \frac{y^2}{4} = 1$$
I have to find the largest possible area of an inscribed rectangle.
So the area ($A$) of a rectangle is $2x2y=4xy$. Also we can redefine $y$ in terms of $x$:
$$\frac{y^2}{4} = 1 - \frac{x^2}{9}$$
$$y^2 = 4 - \frac{4x^2}{9}$$
$$y = \sqrt{4 - \frac{4x^2}{9}}$$
So the area function is now:
$$A=4x \cdot \sqrt{4 - \frac{4x^2}{9}}$$
$$A' = \frac{4x}{2 \cdot \frac{-8x}{9}} + \sqrt{4 - \frac{4x^2}{9}} \cdot 4$$
So this track seems too difficult, so I'd like to find another approach. Can I square the area first, find the derivative of that to solve for $x$?
So the $\text{Area} = 4x \cdot \sqrt{4 - \frac{4x^2}{9}}$
Is this valid?
$$\text{Area}^2 = 16x^2 \cdot \left(4 - \frac{4x^2}{9}\right)$$
$$= 64x^2 - \frac{64x^4}{9}$$
Derivative:
$$ \frac{d}{dx} \text{Area}^2 = 128x - \frac{256x^3}{9}$$
$$128x\left(1-\frac{2x^2}{9}\right)$$
So critical values: $x = 0, \frac{3}{\sqrt{2}}$
because the derivative equals $0$ when:
$$2x^2 = 9$$
$$x = \frac{3}{\sqrt{2}}$$
Plugging this value of $x$ into $y$ we get that $y = \sqrt{2}$ so the $\text{Area}$ is $3$.
Is this valid? If so why? Does squaring not cause any problems?
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I agree with José Carlos Santos, yet I don't agree with your $A^2$ derivative. So, I like to show my way of solving with $A^2$.
$$Area =A= 4x \cdot \left(4 - \frac{4x^2}{9}\right)^{\frac12}$$
$$A^2= 16x^2 \cdot \left(4 - \frac{4x^2}{9}\right) = 64x^2 - \frac{64x^4}{9}$$
Thus, derivative of $A^2$: $$ \frac1A\cdot\frac{dA}{dx} = 128x - \frac{256x^3}{9}=128x\left(1-\frac{2x^2}{9}\right)$$ Thus,
$$ \frac{dA}{dx} =A\cdot 128x\cdot \left(1-\frac{2x^2}{9}\right)$$
Since $A\ne 0$, the critical values are: $x = 0$ or $x=\pm \frac{3}{\sqrt{2}}$ because $\frac{dA}{dx}$ equals $0$ at critical points. Note that, as Professor Vector pointed out, "the rectangle with the largest area among those with sides parallel to the axes of the ellipse." There are two sides parallel to the $y$-axis, and the points of $x=+\frac{3}{\sqrt{2}}$ and $x=-\frac{3}{\sqrt{2}}$ are where these parallel sides meet the $x$-axis.
Consequently, the maximum area of rectangle is: $$A_{max}=$4\left(\frac{3}{\sqrt{2}}\right) \cdot \left(4 - \frac{4\left(\frac{3}{\sqrt{2}}\right)^2}{9}\right)^{\frac12}=12$$
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"timestamp": "2023-03-29T00:00:00",
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|
On the series $\sum \limits_{n=1}^{\infty} \frac{1}{n^2-3n+3}$ and $\sum\limits_{n=-\infty}^{\infty} \frac{1}{n^2-3n+3}$ Wolfram Alpha says that
$$\sum_{n=1}^{\infty} \frac{1}{n^2-3n+3} = 1 + \frac{\pi \tanh \left ( \frac{\sqrt{3}\pi}{2} \right )}{\sqrt{3}}$$
However I am unable to get it. It is fairly routine to prove that
$$\sum_{n=-\infty}^{\infty} \frac{1}{n^2-3n+3} = \frac{2\pi \tanh \left ( \frac{\sqrt{3}\pi}{2} \right )}{\sqrt{3}}$$
by using complex analysis ( contour integration ) but honestly I am stuck how to retrieve the original sum. Split up , the last sum gives:
\begin{align*}
\sum_{n=-\infty}^{\infty} \frac{1}{n^2-3n+3} &= \sum_{n=-\infty}^{-1} \frac{1}{n^2-3n+3} + \frac{1}{3} + \sum_{n=1}^{\infty} \frac{1}{n^2-3n+3} \\
&=\frac{1}{3} +\sum_{n=1}^{\infty} \frac{1}{n^2+3n+3} + \sum_{n=1}^{\infty} \frac{1}{n^2-3n+3} \\
&=\frac{1}{3}+ \sum_{n=1}^{\infty} \left [ \frac{1}{n^2-3n+3} + \frac{1}{n^2+3n+3} \right ]
\end{align*}
Am I overlooking something here?
P.S: Working with digamma on the other hand I am not getting the constant. I'm getting $\frac{1}{3}$ instead.
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$$\sum_{n\geq 0}\frac{1}{(n+a)(n+b)}=\frac{\psi(a)-\psi(b)}{a-b}\tag{1}$$
for any $a\neq b$ in the half-plane $\text{Re}(s)>0$ is fairly routine, too. Here we have to find
$$ \sum_{n\geq 0}\frac{1}{n^2-n+1}=1+\sum_{n\geq 0}\frac{1}{n^2+n+1}=1+\frac{\psi\left(\frac{1+i\sqrt{3}}{2}\right)-\psi\left(\frac{1-i\sqrt{3}}{2}\right)}{i\sqrt{3}}\tag{2}$$
which (by the reflection formula for the $\psi$ function) simplifies into
$$ 1+\frac{-\pi\cot\left(\frac{\pi}{2}(1+i\sqrt{3})\right)}{i\sqrt{3}}=1+\frac{\pi}{\sqrt{3}}\tanh\left(\frac{\pi\sqrt{3}}{2}\right)\approx 2.79814728\tag{3}$$
as wanted.
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{
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"url": "https://math.stackexchange.com/questions/2686931",
"timestamp": "2023-03-29T00:00:00",
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|
Hyperbolic functions proof using $\cosh x$
Prove that $\cosh x-1\equiv \frac12(e^{0.5x}-e^{-0.5x})^2$
I'm stuck on what appears to be the last step, please could someone explain where I have made a mistake?
\begin{align}
\frac12(e^{0.5x}-e^{-0.5x})^2 & \equiv \frac12(\sinh(0.5x))^2 \\
& \equiv \frac12\sinh^2(0.5x) \\
& \equiv \frac12(\cosh^2(0.5x) -1) \\
& \equiv \frac12((\frac{e^{0.5x}+e^{-0.5x}}{2})^2 -1) \\
& \equiv \frac12((\frac{e^x+e^{-x}+2}4) -1) \\
& \equiv \frac14((\frac{e^x+e^{-x}+2}2)-2) \\
& \equiv \frac14(\frac{e^x+e^{-x}}2+\frac22-2) \\
& \equiv \frac14(\frac{e^x+e^{-x}}{2}-1) \\
& \equiv \frac14(\cosh x-1) \\
\end{align}
which is a quarter of what I am trying to prove?
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Note that $$\begin{align}
\frac12(e^{0.5x}-e^{-0.5x})^2 & \equiv \frac12(2\sinh(0.5x))^2 \\
& \equiv 2\sinh^2(0.5x) \\
& \equiv 2(\cosh^2(0.5x) -1) \\
& \equiv 2((\frac{e^{0.5x}+e^{-0.5x}}{2})^2 -1) \\
& \equiv 2((\frac{e^x+e^{-x}+2}4) -1) \\
& \equiv ((\frac{e^x+e^{-x}+2}2)-2) \\
& \equiv (\frac{e^x+e^{-x}}2+\frac22-2) \\
& \equiv (\frac{e^x+e^{-x}}{2}-1) \\
& \equiv (\cosh x-1) \\
\end{align}$$ Which is what you wanted to prove.
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"timestamp": "2023-03-29T00:00:00",
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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.