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Solving 2nd Order ODE w/Laplace Transforms I am having difficulty with this problem:
*Note: The Delta3(t) is the delta dirac function, also the answer in the image is WRONG.
Attempt at solution :
*
*Let Laplace{y(t)}=Y
*Take Laplace of LHS and RHS.
*Solve for Y.
*Take inverse Laplace of Y, giving me a function of y(t)= some function
Now, are my steps correct for this type of problem?
How do I write this answer as a Heaviside function, as posed in the question?
|
Define the Laplace transform as
\begin{align}
f(s) = \int_{0}^{\infty} e^{-s t} \, y(t) \, dt
\end{align}
then for the differential equation
\begin{align}
y'' - 3 y' -4 y = -4 t + \delta(t),
\end{align}
where $y(0) = -2$ and $y'(0) = -1$ then
\begin{align}
s^{2} f(s) - y'(0) - s y(s) - 3s f(s) + 3 y(0) - 4 f(s) = - \frac{4}{s^{2}} + 1
\end{align}
which simplifies to
\begin{align}
(s+1)(s-4) f(s) = 6 - 2s - \frac{4}{s^2}.
\end{align}
Solving for $f$ leads to
\begin{align}
f(s) &= \frac{6}{(s+1)(s-4)} - \frac{2 s}{(s+1)(s-4)} - \frac{4}{s^{2}(s+1)(s-4)} \\
&= \frac{6}{5} \left( \frac{1}{s-4} - \frac{1}{s+1} \right) - \frac{2}{5} \left( \frac{1}{s+1} + \frac{4}{s-4} \right) + \frac{1}{ s^2} - \frac{3}{4 s} + \frac{4}{5 (s+1)} - \frac{1}{20 (s-4)} \\
&= \frac{1}{s^{2}} - \frac{3}{4 s} - \frac{9}{20} \frac{1}{s-4} - \frac{4}{5} \frac{1}{s+1}.
\end{align}
Inversion provides
\begin{align}
y(t) = t - \frac{3}{4} - \frac{4}{5} e^{-t} - \frac{9}{20} e^{4t}.
\end{align}
The fraction involving $s^{2} (s+1)(s-4)$ is separated as follows
\begin{align}
\frac{1}{s^{2}(s+1)(s-4)} &= \frac{a}{s^{2}} + \frac{b}{s} + \frac{c}{s+1} + \frac{d}{s-4} \\
&= \frac{a + b s}{s^{2}} + \frac{c}{s+1} + \frac{d}{s-4}.
\end{align}
Now
\begin{align}
d &= \left.\frac{(s-4)}{s^2 (s+1)(s-4)}\right|_{s=4} = \frac{1}{80} \\
c &= \left. \frac{s+1}{s^2 (s+1)(s-4)} \right|_{s=-1} = - \frac{1}{5} \\
a &= \left. \frac{s^2}{s^2 (s+1)(s-4)} \right|_{s=0} = - \frac{1}{4} \\
b &= \frac{3}{16}
\end{align}
for which
\begin{align}
\frac{1}{s^{2}(s+1)(s-4)} = - \frac{1}{s^{2}} + \frac{3}{16 \, s} - \frac{1}{5(s+1)} + \frac{1}{80 \, (s-4)}
\end{align}
|
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|
Use Euclid's algorithm to find the multiplicative inverse $11$ modulo $59$ I was wondering if this answer would be correct
the multiplicative of $11$ modulo $59$ would be $5$ hence $5\cdot11 \equiv 4 \pmod{59}$.
Is this correct?
|
No. A multiplicative inverse $a$ has $5 \cdot a \equiv 1 \pmod{59}$ and $11$ has $5 \cdot 11 \equiv -4 \not\equiv 1\pmod{59}$. If you want to find the inverse, use the Euclidean algorithm, as you said:
\begin{align*}
59 &= 5 \cdot 11 + 4\\
11 &= 2 \cdot 4 + 3\\
4 &= 1 \cdot 3 + 1\\
\end{align*}
Hence
\begin{align*}
1 &= -1 \cdot 3 + 4\\
&= -1 \cdot (11 - 2 \cdot 4) + 4\\
&= -1 \cdot 11 + 3 \cdot 4\\
&= -1 \cdot 11 + 3 \cdot (59 - 5 \cdot 11)\\
&= -16 \cdot 11 + 3 \cdot 59
\end{align*}
Hence, $\pmod{59}$, we have
$$ 1 \equiv -16 \cdot 11 \equiv 43 \cdot 11 \pmod{59} $$
So, the multiplicative inverse is $43$.
|
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|
If the equation $|x^2+4x+3|-mx+2m=0$ has exactly three solutions then find value of m. Problem :
If the equation $|x^2+4x+3|-mx+2m=0$ has exactly three solutions then find value of $m$.
My Approach:
$|x^2+4x+3|-mx+2m=0$
Case I : $x^2+4x+3-mx+2m=0$
$\Rightarrow x^2+ x (4-m) + 3+2m=0 $
Discriminant of above qudratic is
$D = (4-m)^2 -4(3+2m) \geq 0$
$D = 16+m^2-8m-12-8m$
Solving for $m$ we get the values $-8 \pm 2\sqrt{15}$
Case II :
Similarly solving for the given equation taking negative sign of modulus we get the solution
for $m =$$8 \pm 2\sqrt{15}$
Can we take all the values of m to satisfy the given condition of the problem , please suggest which value of m should be neglected in this. Thanks.
|
If you recognize that $x^2 + 4x + 3 = (x+2)^2 - 1,$
then it can quickly be seen that the graph of that function is an
upward-opening parabola with its minimum at $x=-2$, where the value of the
function is $-1$.
Clearly for values of $x$ far from $-2$, the function is positive,
those parts of the graph of the function are preserved when
we take the absolute value, $|x^2 + 4x + 3|,$
but the part near $x=-2$ (bounded by the points where the parabola crosses
the $x$ axis to the left and right of $x=-2$) is flipped over the $x$ axis;
that is, the section of that parabola below the $x$ axis is replaced
by the section of the parabola $1-(x+2)^2$ above the $x$ axis.
So the graph of $|x^2 + 4x + 3|$ starts at large positive $y$ and large
negative $x$, goes down to the $x$ axis, then goes up over a "hill"
and down again until it arrives at the $x$ axis again, and finally goes up
indefinitely.
The solutions of $|x^2+4x+3|-mx+2m=0$ are the points where
$$|x^2 + 4x + 3| = m(x - 2),$$ so we can look for a line that intercepts the
$x$ axis at $x=2$ with slope $m$ and that intersects the graph of
$|x^2 + 4x + 3|$ in exactly three points.
It is clear that a positive $m$ will not work, and for zero or negative $m$
it is easy to get either two intersection points or four;
but the only way to get three intersection points is for the line to be just
tangent to the "hill" in the middle of the graph of $|x^2 + 4x + 3|$.
(A slightly shallower slope gives four intersections; steeper gives two.)
So you need there to be a unique solution to $m(x-2) = 1-(x+2)^2$.
Rather than do much more algebra on that form of the equation,
we can observe that $1-(x+2)^2$ describes the part of the graph of
$|x^2 + 4x + 3|$ where the $|\cdot|$ operation changed the sign of the formula,
so we really want there to be one (double root) solution of
$$-(x^2+4x+3)-mx+2m = -x^2 -(4+m)x - (3-2m)=0.$$
Using the $b^2-4ac$ formula for the discriminant,
the discriminant is $(m+4)^2 - 4(3-2m)$
or simplified, $m^2 + 16m + 4$.
This has two possible solutions:
the larger (steeper) value of $m$ corresponds to a line that
is tangent to the parabola $-(x^2+4x+3)$ somewhere to the right of $x=2$
and far below the $x$-axis, in other words, nowhere near the graph of
$|x^2 + 4x + 3|$;
the smaller (shallower) slope $m$
is the solution you want.
|
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|
Prove by strong induction that $2^n$ divides $p_n$ for all integers n ≥ 1
Let $p_1 = 4$, $p_2 = 8$, and $p_n = 6p_{n−1} − 4p_{n−2}$ for each integer $n ≥ 3$.
Prove by strong induction that $2^n$ divides $p_n$ for all integers $n ≥ 1$
I got up to the base step where by you prove for $p_3$ but unsure about the strong induction step
|
If $2^n$ divides $p_n$ then we have that $p_n = \lambda_n 2^n$ for some integer $\lambda_n$. Therefore, the inductive hypothesis states that:
$$
p_n = \lambda_n2^n = 6p_{n - 1} - 4p_{n - 2}
$$
If we assume that it holds for both $n - 1$ and $n - 2$ (this is the strong induction) then we get that: $p_{n - 1} = \lambda_{n - 1}2^{n - 1}$ and $p_{n - 2} = \lambda_{n - 2}2^{n - 2}$. Putting this together we find that:
\begin{align}
6p_{n - 1} - 4p_{n - 2} =&\ 6\lambda_{n - 1}2^{n - 1} - 4\lambda_{n - 2}2^{n - 2}\\
=&\ 2^n \left(\frac{6\lambda_{n - 1}}{2^1} - \frac{4\lambda_{n - 2}}{2^2}\right) \\
=& 2^n \left(\frac{6}{2}\lambda_{n - 1} - \frac{4}{4}\lambda_{n - 2}\right) \\
=&\ 2^n \left(3\lambda_{n - 1} - \lambda_{n - 2}\right)
\end{align}
Therefore $2^n$ divides $p_n$ by a factor of $3\lambda_{n - 1} - \lambda_{n - 2}$ (which are both assumed to be integers through the inductive hypothesis).
|
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|
Prove the set $\mathcal{O}_d := \left\{\frac{a + b\sqrt{d}}{2}:a,b \in \mathbb{Z}, a \equiv b \mod 2 \right\}$ is a ring.
Prove that if $d$ is a non-square integer with $d \equiv 1 \mod 4$ then the set $$\mathcal{O}_d := \left\{\frac{a + b\sqrt{d}}{2}:a,b \in \mathbb{Z}, a \equiv b \mod 2 \right\}$$ is a ring, and in particular a integral domain.
Little bit stuck here.
For $x,y \in \mathcal{O}_d,$ we define $+: \mathcal{O}_d \times \mathcal{O}_d \to \mathcal{O}_d$ by $(x + y) = \dfrac{a + b \sqrt{d}}{2} + \dfrac{a' + b'\sqrt{d}}{2} = \dfrac{(a+a') + (b+b')\sqrt{d}}{2} = \dfrac{a'+b'\sqrt{d}}{2} + \dfrac{a + b\sqrt{d}}{2} = (y + x) \in \mathcal{O}_d.$
We have that $0 = \dfrac{0 + 0\sqrt{d}}{2} \in \mathcal{O}_d$ and $\dfrac{a + b \sqrt{d}}{2} + \dfrac{-a + -b\sqrt{d}}{2} = \dfrac{(a-a) + (b-b)\sqrt{d}}{2} = 0 .$ Hence $-a \in \mathcal{O}_d.$ Also $$\begin{align} (x + y) + z &= \left(\dfrac{a + b\sqrt{d}}{2} + \dfrac{a' + b'\sqrt{d}}{2}\right) + \dfrac{a'' + b''\sqrt{d}}{2} \\&= \dfrac{(a + a') + (b + b')\sqrt{d}}{2} + \dfrac{a'' + b''\sqrt{d}}{2} \\&= \dfrac{(a + a' + a'') + (b + b' + b'')\sqrt{d}}{2} \\&= \dfrac{a + b\sqrt{d}}{2} + \dfrac{(a'+a'')+(b+b'')\sqrt{d}}{2} \\&= \dfrac{a + b\sqrt{d}}{2} + \left(\dfrac{a'+b'\sqrt{d}}{2} + \dfrac{a''+b''\sqrt{d}}{2}\right) \\&= x + (y+z) \end{align}$$
for all $x,y,z \in \mathcal{O}_d.$
Hence $\mathcal{O}_d$ is a group under addition.
I've had difficulty showing the multiplicative operation $\circ: \mathcal{O}_d \times \mathcal{O}_d \to \mathcal{O}_d$ stays in the set. How do we define the operation so that $x \circ y \in \mathcal{O}_d?$ When I multiply two elements from $\mathcal{O}_d$ I get $x \circ y = \left(\dfrac{a + b \sqrt{d}}{2} \right)\left(\dfrac{a' + b' \sqrt{d}}{2} \right) = \dfrac{(a + b \sqrt{d})(a' + b'\sqrt{d})}{4} = \dfrac{aa' + ab'\sqrt{d} +ba'\sqrt{d} + bb'd}{4}= \dfrac{aa' + bb'd + (ab' +ba')\sqrt{d}}{4}$ which doesn't seem to be an element of $\mathcal{O}_d.$
|
Hint $\ $ Show that $\,{\cal O}_d = \Bbb Z\left[w\right],\ w = (1\!+\!\sqrt d)/2,\,$ using $\ w^2\! -w + n = 0,\,\ n =(1\!-\!d)/4$
|
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|
Consider the equation $|z + 3i|=3|z|$ for complex z and give a geometric description of the set S of all solutions. Writing $z$ in the form $a+ib$ and then rearranging gives $-8a^2-8b^2+6b+9=0$. The most promising form I could manage from this is $(b-\frac{3}{8})^2=(\frac{9}{8}-a)(\frac{9}{8}+a)$ but I still do not know how this looks like.
|
If $z=x+iy$ then
\begin{align*}
|z+3i|&=3|z|&\;\iff\;&&x^2+(y+3)^2&=9(x^2+y^2)\\
& &\;\iff\;&&8x^2+8y^2-6y-9&=0\\
& &\;\iff\;&& x^2+ y^2-\frac{3}{4}y-\frac{9}{8}&=0\\
& &\;\iff\;&& x^2+ \left(y-\frac{3}{8}\right)^2&=\frac{81}{64}\\
\end{align*}
Therefore the set $S$ is a circle with center in $z_0=\frac{3}{8}i\;$ and radius $\frac{9}{8}$.
|
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|
prove Inequalities for integrals prove
$\frac{\pi}{6}+\frac{1}{3}\leq \int_0^\frac{\pi}{2}\frac{1+\cos(x)}{2+\sin(x)}dx \leq \frac{\pi}{4}+\frac{1}{2}$
I got to the point where $\frac{1}{3} \leq f(x) \leq 1$,
so $\frac{\pi}{6} \leq \int_0^\frac{\pi}{2}f(x)dx \leq \frac{\pi}{2}$ and where do I go from here , thanks
|
The upper bound is easy:$$\int_0^{\frac{\pi}{2}}\frac{1+\cos x}{2+\sin x}\mathrm{d}x\leq \int_2^{\frac{\pi}{2}}\frac{1}{2}(1+\cos x)\mathrm{d}x=\pi/4+1/2.$$
For the lower bound, $$\int_0^{\frac{\pi}{2}}\frac{1+\cos x}{2+\sin x}\mathrm{d}x=\int_0^{\frac{\pi}{2}}\frac{1}{2+\sin x}\mathrm{d}x+\int_0^{\frac{\pi}{2}}\frac{\cos x}{2+\sin x}\mathrm{d}x.$$ The first term on the right hand side is $\geq \pi/6$. The second term $$\int_0^{\frac{\pi}{2}}\frac{\cos x}{2+\sin x}\mathrm{d}x=\int_0^1 \frac{1}{2+t}\mathrm{d}t=\ln 3-\ln 2 \geq 1/3.$$
|
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|
Use the generating function to solve a recurrence relation We have the recurrence relation $\displaystyle a_n = a_{n-1} + 2(n-1)$ for $n \geq 2$, with $a_1 = 2$.
Now I have to show that $\displaystyle a_n = n^2 - n +2$, with $n \geq 1$ using the generating function.
The theory in my book is scanty, so with the help of the internet I have the following:
$\displaystyle \sum_{n = 2} ^\infty (a_n - a_{n-1}) x^n = \sum_{n = 2} ^\infty a_n x^n - \sum_{n = 2} ^\infty a_{n-1} x^n = \sum_{n = 2} ^\infty a_n x^n - x \sum_{n = 2} ^\infty a_{n-1} x^{n-1} = (a(x) - a_1 x) - x(a(x)) = (a(x) - 2 x) - x(a(x)) = a(x) (1-x) - 2x$
But how I have to work out $\displaystyle \sum_{n = 2} ^\infty 2(n-1) x^n$ ?
If I have this, an expression for $a(x)$ can be found. How should $a_n$ be found from $a(x)$?
|
another observation is :$$a_n=a_{n-1}+2(n-1) ,\space \space \space a_1=2$$ $$\rightarrow a_n-a_{n-1}=2(n-1)\\$$ put $n=1,2,3,..(n-1)$ $$a_2-a_1=2(2-1
)=2(1)\\ a_3-a_2=2(3-1)=2(2)\\a_4-a_3=2(4-1)=2(3)\\...\\a_n-a_{n-1}=2(n-1)=2(n-1)\\$$no look at sum of them $$a_n-a_1=2(1)+2(2)+2(3)+...2(n-1)=\\2(1+2+3+4+...+(n-1))=2 \frac{(n-1)(n-1+1)}{2} \\so\\a_n-2=2\frac{n^2-n}{2}\\a_n=n^2-n+2$$
|
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|
Magic square with not distinct numbers There's a 4x4 magic square:
4 0 1 0
3 0 2 0
0 3 0 2
0 4 0 1
Where 0s are different numbers, 1=1, 2=2, 3=3, 4=4.
Only the rows and the columns have the same sum, the diagonals don't.
Question: If rotating or reflecting the magic square counts as a different solution, how many different magic squares is it possible to build?
I suppose if all the numbers were different, the solution would be 2 * 4! * 4!.
But I just can't figure it out how to deal with the repeated numbers.
|
Let's start by giving names to the unknown quantities
$$
\begin{bmatrix}
4 & a & 1 & e \\
3 & b & 2 & f \\
c & 3 & g & 2 \\
d & 4 & h & 1
\end{bmatrix}
$$
and call the total sum of the rows and columns $T$. We can now subtract the 1st row from the 2nd column as follows
$$(a+b+3+4)-(4+a+1+e)=T-T=0$$
$$|| \quad\quad\quad\quad\quad\quad$$
$$\quad\quad\quad\ \ \ b-e+2 \quad\quad\Rightarrow\quad\quad e=b+2$$
That column and row both contained $a$. This trick works whenever you choose a row and column that contain the same variable. When we do the same trick for the other variables, we get 3 more equalities for 4 total:
$$e=b+2 \quad\quad f=a+2 \quad\quad g=d+2 \quad\quad h=c+2$$
Substituting these into the square we get
$$
\begin{bmatrix}
4 & a & 1 & b+2 \\
3 & b & 2 & a+2 \\
c & 3 & d+2 & 2 \\
d & 4 & c+2 & 1
\end{bmatrix}
$$
Most importantly, for some particular choice of $a,b,c,$ and $d$ this magic square is valid exactly when
$$a+b=c+d$$
So if you're counting how many valid magic squares of this form are possible, there are $\bf{infinitely\ many}$. For example, choose $a=7$, $b=8$, $c=9$, $d=6$ and a magic square pops out
$$
\begin{bmatrix}
4 & 7 & 1 & 10 \\
3 & 8 & 2 & 9 \\
9 & 3 & 8 & 2 \\
6 & 4 & 11 & 1
\end{bmatrix}
$$
If however you wanted to know, for some particular choice of $a,b,c$, and $d$, how many distinct squares can be made by rotations and reflections, then the answer is $\textbf{8}$. This is because there are 4 rotations and their corresponding reflections. (Note: a reflection of a reflection creates either the original square, or a rotation of the original square)
$\quad$ And finally if you wanted to also count exchanging rows and columns as part of "building" new squares, then their are two choices of $a,b,c$, and $d$ that affect the answer. Specifically
$$(1)\quad\quad a=1 \quad b=2 \quad c=1 \quad d=2 \quad\quad\quad$$
$$(2)\quad\quad a=2 \quad b=1 \quad c=2 \quad d=1 \quad\quad\quad$$
With either of these two choices, two of the columns will be equal. (It's impossible for two rows to ever be equal, so we don't need to worry about that).
In conclusion, for most choices of $a,b,c,$ and $d$ are $8\cdot 4! \cdot 4! = 4608$ distinct squares.
Except when the choice of $a,b,c,$ and $d$ is $(1)$ or $(2)$, in which case there are $8 \cdot \frac{4!}{2} \cdot 4! = 2304$ distinct squares.
|
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|
Determining the Laurent Series I need to determine the Laurent series of this function:
$$\frac{1}{(z-1)(z+5)}$$ Inside the annulus: $$\left\{z|1<|z-2|<6\right\}$$ Any help appreciated.
|
Here is just a set up: $\frac{1}{1-x}=1+x+x^2+x^3+x^4+.....$,then
$\frac{1}{1+x}=1-x+x^2-x^3+x^4-.....$ then $\frac{1}{1+x/5}=\frac{5}{5+x}=1-(x/5)+(x/5)^2-(x/5)^3+(x/5)^4-.....$ Now divide by 5 to get $\frac{1}{5+x}=1/5-(x/5)/5+(x/5)^2/5-(x/5)^3/5+...$ And then integrate. Can you now work out the details?
|
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|
Draw the line segment joining the centers of two circles. Where does it meet the circles? I'm trying to construct a line segment between two circles. Given each radius and $x$, $y$ center of each circle, how can I find the endpoints for the blue line segment?
|
Let the centres of the two circles be $(x_1, y_1)$ and $(x_2, y_2)$, where either $x_1 < x_2$ or $x_1 = x_2$ and $y_1 < y_2$, and radii $r_1$, $r_2$.
Suppose $x_1 = x_2$. In this case, the blue line segment is vertical, and its endpoints can easily be seen to be $(x_1, y_1 + r_1)$ and $(x_2, y_2 - r_2) = (x_1, y_2 - r_2)$.
Now suppose $x_1 < x_2$. By shifting horizontally and vertically, we can assume the two circles are centred at $(0, 0)$ and $(a, b)$ $a = x_2 - x_1$ and $b = y_2 - y_1$. The slope of the line passing through the two centres is $b/a$; note, $a \neq 0$ as $x_1 \neq x_2$. As the $y$-intercept of the line is $0$, the equation of the line which passes through them is $y = \frac{b}{a}x$; note that the endpoints of the blue line segment lie on this line.
As can be see from the right-angled triangle in the image below, $\sin\theta = \frac{b}{\sqrt{a^2+b^2}}$, $\cos\theta=\frac{a}{\sqrt{a^2+b^2}}$, and $\tan\theta = \frac{b}{a}$ where $\theta$ is the angle the line $y = \frac{b}{a}x$ makes with the positive $x$-axis.
$\hspace{37mm}$
In order to determine the first endpoint of the blue line segment, consider the following image.
$\hspace{37mm}$
The coordinates of the first endpoint are $(a_1, b_1)$; we just need to determine what $a_1$ and $b_1$ are. Note that $R_1 = r_1$, so using the small right-angled triangle, we see that $\cos\theta = \frac{a_1}{r_1}$ so $a_1 = r_1\cos\theta$; likewise, $b_1 = r_1\sin\theta$.
In order to determine the second endpoint of the blue line segment, consider the following image.
$\hspace{37mm}$
The coordinates of the second endpoint are $(a_2, b_2)$; we just need to determine what $a_2$ and $b_2$ are. Note that $R_2 = \sqrt{a^2 + b^2} - r_2$, so using the smaller of the two right-angled triangles, we get the following expressions for $a_2$ and $b_2$ as we did above: $a_2 = (\sqrt{a^2 + b^2} - r_2)\cos\theta$, $b_2 = (\sqrt{a^2 + b^2} - r_2)\sin\theta$.
Combining the information deduced from the three images above, we see that, after a little bit of algebra, the endpoints of the blue segment are
$$\left(\frac{ar_1}{\sqrt{a^2 + b^2}}, \frac{br_1}{\sqrt{a^2+b^2}}\right)\ \text{and}\ \left(a\left(1 - \frac{r_2}{\sqrt{a^2 + b^2}}\right), b\left(1 - \frac{r_2}{\sqrt{a^2 + b^2}}\right)\right).$$
We can undo the original horizontal and vertical shifts we did at the beginning and well as express $a$ and $b$ in terms of the original data to obtain the endpoints
$$\left(x_1 + \frac{(x_2 - x_1)r_1}{\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}}, y_1 + \frac{(y_2 - y_1)r_1}{\sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}}\right)$$
$$\text{and}$$
$$\left(x_2 - \frac{(x_2 - x_1)r_2}{\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}}, y_2 - \frac{(y_2 - y_1)r_2}{\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}}\right).$$
Note, these formulae give the correct endpoints even in the case where $x_1 = x_2$.
|
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Precalculus unit circle with imaginary axis. (a) Suppose $p$ and $q$ are points on the unit circle such that the line through $p$ and $q$ intersects the real axis. Show that if $z$ is the point where this line intersects the real axis, then $z = \dfrac{p+q}{pq+1}$.
(b) Let $P_1 P_2 \dotsb P_{18}$ be a regular 18-gon. Show that $P_1 P_{10}$, $P_2 P_{13}$, and $P_3 P_{15}$ are concurrent.
I have gotten nowhere on this problem, but I have a hint:
One of those three segments is more interesting than the other two. Which one, and why? And how can you use that fact to make part (a) relevant?
Any help is appreciated!
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(b) Let $\omega = e^{2 \pi i/18}$, a primitive $18^{\text{th}}$ root of unity. Then $\omega^{18} = 1$. Also, $\omega^{18} - 1 = 0$, which factors as
$(\omega^9 - 1)(\omega^9 + 1) = 0.$Since $\omega^9 = e^{2 \pi i/2} = e^{\pi i} = -1 \neq 1$, $\omega$ must satisfy the equation $\omega^9 + 1 = 0$. We can factor this equation as
$(\omega^3 + 1)(\omega^6 - \omega^3 + 1) = 0.$Since $\omega^3 = e^{2 \pi i/6} = e^{\pi i/3} \neq -1$, $\omega$ must satisfy the equation
$\omega^6 - \omega^3 + 1 = 0.$
We place regular 18-gon $P_1 P_2 \dotsb P_{18}$ in the complex plane so that point $P_1$ goes to 1, point $P_2$ goes to $\omega$, and so on.
To prove that all three lines concur, we are going to find the point where the line joining $\omega$ and $\omega^{12}$ intersects the real axis, and the point where the line joining $\omega^2$ and $\omega^{14}$ intersects the real axis. We will show that these points coincide.
We want to compute the point where the line joining $\omega$ and $\omega^{12}$ intersects the real axis. Since $\omega^9 = -1$, we can write $\omega^{12} = -\omega^3$. Then from our formula in part (a) above, the intersection is
$\frac{\omega - \omega^3}{-\omega^4 + 1}.$
Similarly, the point where the line joining $\omega^2$ and $\omega^{14} = -\omega^5$ intersects the real axis is
$\frac{\omega^2 - \omega^5}{-\omega^7 + 1}.$Hence, we have reduced the problem to showing that
$\frac{\omega - \omega^3}{-\omega^4 + 1} = \frac{\omega^2 - \omega^5}{-\omega^7 + 1}.$Cross-multiplying, we get
$\omega^{10} - \omega^8 - \omega^3 + \omega = \omega^9 - \omega^6 - \omega^5 + \omega^2.$Using the fact that $\omega^9 = -1$, this equation simplifies to
$-\omega - \omega^8 - \omega^3 + \omega = -1 - \omega^6 - \omega^5 + \omega^2,$or
$\omega^8 - \omega^6 - \omega^5 + \omega^3 + \omega^2 - 1 = 0.$We can write this equation as
$(\omega^2 - 1) \omega^6 - (\omega^2 - 1) \omega^3 + (\omega^2 - 1) = 0,$which factors as
$(\omega^2 - 1)(\omega^6 - \omega^3 + 1) = 0.$We know that $\omega^6 - \omega^3 + 1 = 0$. Therefore, our two points of intersection with the real axis coincide, which means that all three lines are concurrent.
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|
Prove that $f : [-1, 1] \rightarrow \mathbb{R}$, $x \mapsto x^2 + 3x + 2$ is strictly increasing.
Prove that $f : [-1, 1] \rightarrow \mathbb{R}$, $x \mapsto x^2 + 3x + 2$ is strictly increasing.
I do not have use derivatives, so I decided to apply the definition of being a strictly increasing function, which should be:
If we pick 2 numbers $a$ and $b$ from the domain of a function $f$, where $a < b$, then $f(a) < f(b)$.
Now (if that is a correct definition), I have tried to apply it to my case:
Let $a$, $b \in [-1, 1]$ and $a < b$. We want to show that $f(a) < f(b)$ or that $f(b) - f(a) > 0$.
We know that $x^2 + 3x + 2 = (x + 2)(x + 1)$, thus we have that $f(a) = (a + 2)(a + 1)$ and $f(b) = (b + 2)(b + 1)$ , therefore we need to show that:
$$(b + 2)(b + 1) - (a + 2)(a + 1) > 0$$
We can see that $(b + 2)$ and $(a + 2)$ will always be positive, and that $(b + 2) > (a + 2)$ , since $b > a$ (by assumption).
Since $b > a$, we know that $b > -1$ (otherwise $a \geq b$), thus $(b + 1) > 0$ (so we know that $(b + 2)(b + 1) > 0$ . We also have at most $(a + 1) = 0$, and thus we have that $(a + 2)(a + 1) \ge 0$.
So, here is the proof that $f(b) - f(a) > 0$ .
Am I correct? If yes, what can I improve it? If not, where are the erros and possible solutions?
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Yes, to see more clearly, let $x=a+2, y =a+1, x+c = b+2, y+c = c+2$, compute $(x+c)(y+c)- xy$. Notice that $x, c$ are positive, $y$ is non-negative.
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|
The implication $\sqrt{(x-2)^2+(y-1)^2+(z-1)^2+(w-3)^2}Given a real number $a>0$ find a $b>0$ such that
$\sqrt{(x-2)^2+(y-1)^2+(z-1)^2+(ω-3)^2}<b\Longrightarrow |xyzω-6|<a$
I tried the procedure followed in another one of my questions, but it doesn't work.
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It seems the following.
The question seems to be a numerical adjustment of the continuity of multiplication. Assume that $$\sqrt{(x-2)^2+(y-1)^2+(z-1)^2+(w-3)^2}<b.$$ Then $|x-2|<b$, $|y-1|<b$, $|z-1|<b$, and $|w-3|<b$.
We have $2\cdot 1\cdot 1\cdot 3=6$ and
$$|xyzw-6|\le$$ $$|xyzw-xyz\cdot 3|+|xyz\cdot 3-xy\cdot 2\cdot 3|+$$ $$|xy\cdot 2\cdot 3 - x\cdot 1\cdot 2\cdot 3|+
|x\cdot 1\cdot 2\cdot 3- 1\cdot 1\cdot 2\cdot 3|=$$ $$
|xyz|\cdot|w-3|+|xy||z-2|\cdot 3+|x||y-1|\cdot 2\cdot 3+
|x- 1|\cdot 1\cdot 2\cdot 3<$$ $$|xyz|\cdot b+|xy|\cdot b\cdot 3+|x|\cdot b \cdot 6+
b \cdot 6.$$
If we additionally assume that $b<1$ then $1<x<3$, $0<y<2$, $0<z<2$, and $2<w<4$. Hence
$$|xyz|\cdot b+|xy|\cdot b\cdot 3+|x|\cdot b \cdot 6+b \cdot 6<
3\cdot 2\cdot 2\cdot b+3\cdot 2\cdot b\cdot 3+3\cdot b \cdot 6+b \cdot 6=
54b.$$
Thus it suffices to put $b=\min\{1, a/54\}$.
Update. Such breaking down is a standard trick, used in analysis and based on the triangle inequality.
The usage of $\min$ is a standard and simple act to show that $b$ should be small. The smallness is typical here, because the continuity of function in a point is based on its behavior in small neighborhoods of this point. If we reject this approach, we may obtain a formula for $b$, which uses a polynomials or roots of $a$ and which is more complex than the minimum of a constant and a linear function.
For instance, we have $|x|<2+b$, $|y|<1+b$, $|z|<1+b$, and $|w|<3+b$. Then
$$|xyz|\cdot b+|xy|\cdot b\cdot 3+|x|\cdot b \cdot 6+b \cdot 6<$$
$$(2+b)(1+b)^2\cdot b+(2+b)(1+b)\cdot b\cdot 3+(2+b)\cdot b \cdot 6+b \cdot 6=$$ $$
b(b^3+13b^2+23b+17)<b(b^3+15b^2+75b+125) = b(b+5)^3.$$
Then for the given $a$ we have to find $b$ such that $b(b+5)^3<a$. It suffices to put $b=\frac a{a+6^3}$. Indeed,
$$\frac a{a+6^3}\left(\frac a{a+6^3}+5\right)^3<\frac a{a+6^3}6^3<a.$$
We can deal with your next inequality as follows. Put $$f=\sqrt{(x-2)^2+(y-1)^2+(z-1)^2}.$$ For given $a>0$ similarly to the above we find positive $b_1, b_2, b_3$ such that
$f<b_1$ implies $|x^2yz−4|<a/\sqrt{3},$
$f<b_2$ implies $|xy^2z−2|<a/\sqrt{3},$
$f<b_3$ implies $|xyz^2−2|<a/\sqrt{3}.$
Then if $f<\min\{b_1, b_2, b_3\}$ we have $$\sqrt{(x^2yz−4)^2+(x y^2z−2)^2+(xyz^2−2)^2}<\sqrt{3\cdot a^2/3}=a.$$
If you insist on avoiding $\min$, you can put $b=\frac{b_1b_2b_3}{b_1b_2+b_1b_3+b_2b_3}<\min\{b_1, b_2, b_3\}$.
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Positivity of power function. Prove that
$6^a-7^a+2\cdot 4^a-3^a-5^a\ge0$ for $-\frac{1}{2}\le a\le0$.
I tried to do it by first derivative test but derivative become more complicated (same with 2nd derivative for convexity).
Here is it's plotting.
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For $a\in[-\frac12,0)$ and $x>0$ let $$g_a(x)=x^a-(x+1)^a-(x+3)^a$$
Then $$f(a)=g_a(4)-g_a(3)$$
By the Mean Value Theorem this equals $g_a'(\theta)=ag_{a-1}(\theta)$ for some $\theta\in [3,4]$. We have
$$\begin{align}\frac{g_{a-1}(\theta)}{\theta^{a-1}}&=1-(1+\tfrac1\theta)^{a-1}-(1+\tfrac3\theta)^{a-1}\\&\le 1-(1+\tfrac13)^{a-1}-2^{a-1}&\text{(negative exponent $\Rightarrow$ maximize $\tfrac1\theta$)}\\
&\le 1-(\tfrac43)^{-3/2}-2^{-3/2}&\text{(bases $>1$ $\Rightarrow$ minimize neg. exponent)}\\
&=\tfrac18\left(8-3\sqrt3-2\sqrt 2\right).\end{align} $$
We have $5<3\sqrt 3<6$ and $2<1\sqrt 2<3$, therefore the sign of the expression in parenthesis is not immediately clear (without using a calculator). But the conjugates $8+3\sqrt3-2\sqrt 2$, $8-3\sqrt3+2\sqrt 2$, $8+3\sqrt3+2\sqrt 2$ are all definitely positive (in fact $>8+2-6=4$).
The product of all four conjugates however is
$$\begin{align}&(8-3\sqrt3-2\sqrt 2)(8-3\sqrt3+2\sqrt 2)(8+3\sqrt3-2\sqrt 2)(8+3\sqrt3+2\sqrt 2)\\=\:&\bigl((8-3\sqrt 3)^2-8\bigr)\bigl((8+3\sqrt 3)^2-8\bigr)\\=\:&(83-48\sqrt 3)(83+48\sqrt 3)\\=\:&83^2-3\cdot 48^2\\
=\:&{-23}.\end{align} $$
We conclude $8-3\sqrt 3-2\sqrt 2<0$, so that
$$f(a)=ag_{a-1}(\theta)\ge a\theta^{a-1}\tfrac18\left(8-3\sqrt3-2\sqrt 2\right)>0 $$
for $-\frac12\le a<0$.
|
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Solution Invariant Explanation Trick
Suppose not all 4 integers, $a,b,c,d$ are equal. Start with $(a,b,c,d)$ and repeatedly replace $(a,b,c,d)$ by $(a−b,b−c,c−d,d−a)$. Then show that at least one number of the quadruple will become arbitrarily large.
Solution:
We have
$$(a_{n+1},b_{n+1},c_{n+1},d_{n+1}) = (a_n-b_n,b_n-c_n,c_n-d_n,d_n-a_n)$$
As you have rightly observed, the invariant is $a_n+b_n+c_n+d_n = 0$ for all $n \geq 1$, where $a_0 = a$, $b_0=b$, $c_0 = c$ and $d_0 = d$. Let us now look at $a_{n+1}^2+b_{n+1}^2 + c_{n+1}^2 + d_{n+1}^2$. We have
\begin{align}
a_{n+1}^2+b_{n+1}^2 + c_{n+1}^2 + d_{n+1}^2 & = (a_n-b_n)^2+(b_n-c_n)^2+(c_n-d_n)^2 + (d_n-a_n)^2\\
& = 2(a_n^2+b_n^2+c_n^2+d_n^2) - 2(a_nb_n+b_nc_n+c_nd_n+d_na_n) & (\spadesuit)
\end{align}
We shall now show that $ - 2(a_nb_n+b_nc_n+c_nd_n+d_na_n)$ is non-negative for $n \geq 1$. Since $a_n+b_n+c_n+d_n=0$, we have
\begin{align}
0 & = (a_n+b_n+c_n+d_n)^2 = (a_n+c_n)^2 + (b_n+c_n)^2 + 2(a_n+c_n)(b_n+d_n)\\
& = (a_n+c_n)^2 + (b_n+c_n)^2 + 2a_nb_n + 2b_nc_n + 2c_nd_n + 2d_na_n
\end{align}
This gives us
$$-2(a_nb_n+b_nc_n+c_nd_n+d_na_n) = (a_n+c_n)^2 + (b_n+c_n)^2 \geq 0 \,\,\, (\clubsuit)$$
Making use of $(\clubsuit)$ in $(\spadesuit)$, we obtain that
$$a_{n+1}^2+b_{n+1}^2 + c_{n+1}^2 + d_{n+1}^2 \geq 2(a_n^2+b_n^2+c_n^2+d_n^2)$$
This gives us
$$a_{n+1}^2+b_{n+1}^2 + c_{n+1}^2 + d_{n+1}^2 \geq 2^n(a_1^2+b_1^2+c_1^2+d_1^2)$$
Since $a, b,c,d$ are distinct, one of the terms $a_1,b_1,c_1$ or $d_1$ is non-zero. Hence, $a_{n+1}^2+b_{n+1}^2 + c_{n+1}^2 + d_{n+1}^2$ grows unbounded, which means the largest of the terms must keep growing unbounded.
I understand everything until the end. I got:
$$a_{n+1}^2 + b_{n+1}^2 + c_{n+1}^2 + d_{n+1}^2 \ge 2(a_n^2 + b_n^2 + c_n^2 + d_n^2)$$
How do you get to:
$$a_{n+1}^2+b_{n+1}^2 + c_{n+1}^2 + d_{n+1}^2 \geq 2^n(a_1^2+b_1^2+c_1^2+d_1^2)$$
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Set $x_n=a_n^2+b_n^2+c_n^2+d_n^2$. The relationship, to be used repeatedly, is $$\color{red}{ x_{n+1}\ge 2 x_n,\text{ for all }n\in\mathbb{N}}.$$ Hence $$x_4\ge 2x_3\ge 2(2x_2)\ge 2(2(2x_1))=2^3x_1$$
We can prove by induction that $x_{n+1}\ge 2^n x_1$.
Details of the induction, as requested:
The base case is $n=1$, and $x_2\ge 2^1x_1$.
Suppose now that $x_{n+1}\ge 2^n x_1$. By the relationship property we are given, $x_{n+2}\ge 2x_{n+1}$. Combining with the inductive hypothesis we have $$x_{n+2}\ge 2x_{n+1}\ge 2(2^n x_1)=2^{n+1}x_1$$
This establishes the induction.
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Find the equation whose roots are each six more than the roots of $x^2 + 8x - 1 = 0$
Find the equation whose roots are each six more than the roots of $x^2 + 8x - 1 = 0$
I must use Vieta's formulas in my solution since that is the lesson we are covering with our teacher.
My solution:
Let p and q be the roots of the quadratic.
$$\begin{align} p + q = & -8 \\ pq = & -1 \end{align}$$
If the roots are each six more than the roots of the quadratic, then we will have:
$$\begin{align} p + q + 12 = & -8 \\
p + q = & -20 \\
pq = & - 3 \end{align}$$
Also,
$$ \begin{align} (p+6)(q+6) = & pq + 6p + 6q + 36\\
=&pq + 6(p+q) + 36 \\
=&-3 + 6(-8) + 36 \\
=&-3 - 48 + 36 \\
&36 - 51 = -15 \end{align} $$
Hence, -15 = constant.
Thus, the quadratic equation is $x^2 + 20x - 15$
My worksheet gives an answer of $x^2 - 4x - 13$, how am I wrong?
Thanks!
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Let $x_1,x_2$ be the roots of the equation: $x^2+8x-1 = 0$, and let $y_1 = x_1 +6, y_2 = x_2+6 \Rightarrow y_1+y_2 = x_1+x_2+12=-8+12 = 4, y_1y_2 = (x_1+6)(x_2+6)=x_1x_2+6(x_1+x_2)+36=-1+6\cdot (-8) + 36=-13\Rightarrow y^2-4y-13=0$ is the sought after equation.
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Compute the largest root of $x^4-x^3-5x^2+2x+6$ I want to calculate the largest root of $p(x)=x^4-x^3-5x^2+2x+6$. I note that $p(2) = -6$ and $p(3)=21$. So we must have a zero between two and three. Then I can go on calculating $p(\tfrac52)$ and see that the zero must lie in the interval $]2,\tfrac52[$. The answer is $\frac{1+\sqrt{13}}{2}$. But how can I derive that exact result?
Furthermore, to know that it is the largest root, I must find the other two roots, must not I? By sign shift I know one root between $1$ and $2$, but this method cannot give me the third root (the graph touches the $x$-axis). Maybe calculus? Using calculus we get a degree polynomial of degree which also is hard to find roots to. We get $p'(-1)=5$ and $p'(-2)=-22$ so we have a stationary point in $]-2,-1[$ and similarly we get one in $]1,2[$. Since it is two stationary points it must have one maximum and one minimum point. But how do I know which is which, I mean I cannot plug in a value in the second derivative.
Can somebody help me?
Thanks in advance
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You've : $p(x) = x^4 - x^3 - 5x^2 + 2x + 6 $ .
Rearrange the given polynomial as : $$ p(x) =(x^4 - 5x^2 + 6) - x(x^2 - 2) $$
For the first term, let us factor it separately. Say, $x^2 = t$ . So, you've :
$$\begin{align} f(x) =&\ x^4 - 5x^2 + 6\\
f(x) =&\ t^2 - 5t + 6 \end{align}$$
Roots of $f(x)$ are : $t = 3 \ \text{OR} \ 2 $ . Now, plugging-in $t = x^2$ back, we get: $x = \pm \sqrt{3},\pm \sqrt{2}$ .
So, we can write $f(x)$ as : $f(x) = (x-\sqrt{3})(x+\sqrt{3})(x-\sqrt{2})(x+\sqrt{2})$
And thus, $p(x)$ can be written as:
$$\begin{align} p(x) =& (x-\sqrt{3})(x+\sqrt{3})(x-\sqrt{2})(x+\sqrt{2}) - x(x^2 -2) \\
=& (x^2 - 3)(x^2 -2 ) -x(x^2-2) \\
=& (x^2-2)(x^2 - x - 3) \end{align}$$
Therefore, you've : $p(x) = (x^2 - 2)(x^2 - x - 3)$
You can proceed from here, I guess. Hope it helps.
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I cannot solve this limit $$
\lim_{n\to\infty}\frac{(\frac{1}{n}+1)^{bn+c+n^2}}{e^n}=e^{b-\frac{1}{2}}
$$
I am doing it like this, and I cannot find the mistake:
$$
\lim_{n\to\infty}\frac{1}{e^n}e^{n+b+c/n}=
\lim_{n\to\infty}e^{n+b-n+c/n}=e^b
$$
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I would say that $$\log_e\left(\frac{(\frac{1}{n}+1)^{bn+c+n^2}}{e^n}\right) $$ $$ = (bn+c+n^2)\log_e\left(\frac{1}{n}+1\right) - n$$ $$= (bn+c+n^2)\left(\frac{1}{n}-\frac{1}{2\,{n}^{2}}+\frac{1}{3\,{n}^{3}}-\frac{1}{4\,{n}^{4}}+\cdots\right) - n$$ $$= b - \frac{b}{2n} + o(1/n) +\frac{c}{n} - o(1/n) +n -\frac{1}{2} + \frac{1}{3n} -o(1/n) -n$$ $$= b -\frac{1}{2} +\frac{-3b+6c+2}{6n} + o(1/n)$$
making the limit of the logarithm $b-\frac{1}{2}$ as $n\to \infty$ and so the original limit $\displaystyle e^{b-\frac{1}{2} }$.
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|
Integration of $\int\frac{x^{4}}{x+1}dx$ Could you help me with the following integral:
$$\int\frac{x^{4}}{x+1}dx$$
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HINT: $$x^4=x^4-1+1=(x-1)(x+1)(x^2+1)+1$$
Generalization : For integer $n\ge1,$
$$x^n-1=(x-1)(x^{n-1}+x^{n-2}+\cdots+x+1)$$
For $x=y^2,y^{2n}-1=(y^2-1)(y^{2(n-1)}+y^{2(n-2)}+\cdots+y^2+1)$
$\iff y^{2n}-1=(y-1)(y+1)(y^{2n-2}+y^{2n-4}+\cdots+y^2+1)$
|
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|
To find the smallest integer with $n$ distinct divisors For example, if $n=20$, how can I find the smallest integer which has exactly $20$ distinct divisors? Can someone give me some hints?
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If $m=p_1^{a_1}p_2^{a_2}\ldots p_k^{a_k}$, we have number of divisors to be
$$(1+a_1)(1+a_2)\ldots(1+a_k)$$
For this to be $20$, we want
$$(1+a_1)(1+a_2)\ldots(1+a_k) = 20 = 2 \cdot 2 \cdot 5$$
This means $m$ can have at most $3$ distinct prime divisors.
*
*$3$ prime divisors. The smallest possible $m$ is $2^4 \cdot 3 \cdot 5 = 240$.
*$2$ prime divisors. The smallest possible $m$ is of the form $2^a \cdot 3^b$. This means we need $(1+a)(1+b) = 20$, where $a\geq b$.
*
*$a=9,b=1$, gives $m=2^9 \cdot 3 = 1536$.
*$a=4,b=3$, gives $m=2^4\cdot3^3 = 432$.
*$1$ prime divisor. The smalles possible $m$ is of the form $2^{19} = 524288$.
Hence, the smallest $m$ is $240$.
|
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|
Solve this logarithmic equation: $2^{2-\ln x}+2^{2+\ln x}=8$ Solve this logarithmic equation: $2^{2-\ln x}+2^{2+\ln x}=8$. I thought to write
$$\dfrac{2^2}{2^{\ln(x)}} + 2^2 \cdot 2^{\ln(x)} = 2^3 \implies \dfrac{2^2 + 2^2 \cdot \left(2^{\ln(x)}\right)^2}{2^{\ln(x)}} = 2^3$$
What should I do else?
|
Since $2^2+2^2 = 4+4 = 8$, can you just let $\ln x = 0$ and find $x = 1$?
ETA: This can be shown to be unique by finding the derivative with respect to $x$:
$$
\frac{d}{dx} 2^{2-\ln x} + 2^{2+\ln x}
= \frac{\ln 2}{x} \left(2^{2+\ln x}-2^{2-\ln x}\right)
$$
and provided that $x > 0$, this only equals $0$ at $\ln x = 0$.
|
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|
Prove that number of zeros at the right end of the integer $(5^{25}-1)!$ is $\frac{5^{25}-101}{4}.$
Prove that number of zeros at the right end of the integer $(5^{25}-1)!$ is $\frac{5^{25}-101}{4}.$
Attempt:
I want to use the following theorem:
The largest exponent of $e$ of a prime $p$ such that $p^e$ is a divisor of $n!$ is given by $$e=[\frac{n}{p}]+[\frac{n}{p^2}]+[\frac{n}{p^3}]+\cdots$$.
The number of times the prime divisor $5$ is repeated in $(5^{25}-1)!$ equals the greatest exponent of $5$ contained in $(5^{25}-1)!$, which is $e_1=[\frac{(5^{25}-1)}{5}]+[\frac{(5^{25}-1)}{5^2}]+[\frac{(5^{25}-1)}{5^3}]+[\frac{(5^{25}-1)}{5^4}]+\cdots$
The number of times the prime divisor $2$ is repeated in $(5^{25}-1)!$ equals the greatest exponent of $2$ contained in $(5^{25}-1)!$, which is $e_2=[\frac{(5^{25}-1)}{2}]+[\frac{(5^{25}-1)}{2^2}]+[\frac{(5^{25}-1)}{2^3}+[\frac{(5^{25}-1)}{2^4}]+\cdots$
Therefore, the number of zeros at the right end equals the greatest exponent of $10$ contained in $(5^{25}-1)!$$=\min{(e_1, e_2)}$.
I am unable to simplify $e_1, e_2$. Please help me.
|
As shown in this answer, the number of factors of a prime $p$ in $n!$ is
$$
\frac{n-\sigma_p(n)}{p-1}
$$
where $\sigma_p(n)$ is the sum of the digits in the base-$p$ expansion of $n$.
The base-$5$ expansion of $5^{25}-1$ consists of $25\,\,4$s, thus
$$
\sigma_{5}(5^{25}-1)=25\cdot4=100
$$
Thus, the number of factors of $5$ in $(5^{25}-1)!$ is
$$
\frac{5^{25}-1-100}{5-1}=\frac{5^{25}-101}{4}
$$
The binary representation of $5^{25}-1$ has at most $3\cdot25=75$ bits since $5\lt2^3$. Thus, the number of factors of $2$ in $(5^{25}-1)!$ is at least
$$
\frac{5^{25}-1-75}{2-1}=5^{25}-76\gt\frac{5^{25}-101}{4}
$$
Thus, the number of factors of $10$ in $(5^{25}-1)!$ is $\dfrac{5^{25}-101}{4}$.
|
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|
$f\left( x-1 \right) +f\left( x+1 \right) =\sqrt { 3 } f\left( x \right)$ Let f be defined from real to real
$f\left( x-1 \right) +f\left( x+1 \right) =\sqrt { 3 } f\left( x \right)$
Now how to find the period of this function f(x)?
Can someone provide me a purely algebraic method to solve this problem please?
Update:My method
An elementary algebraic approach to the problem :
$f(x-1)+f(x+1)=\sqrt { 3 } f(x)$
Replace $x$ with $x+1$ and $x-1$ respectively.
We get $f(x)+f(x+2)=\sqrt { 3 } f(x+1)$ and
$f(x-2)+f(x)=\sqrt { 3 } f(x-1)$
From these three equations we get $f(x-2)+f(x+2)=0$
Putting $x=x+2$ and adding with last equation we get $f(x-2)+f(x+4)=0$....(1)
Similarly $f(x-4)+f(x+2)=0$.....(2)
Put $x=x-6$ in (1)
We get $f(x-8)+f(x-2)=0$.....(3)
From (1) and (3) we get $f(x-8)=f(x+4)$
So the period of $f(x)$ is 12
|
Algebraic (non-coanstant)? NO.
But, for example
$$
f(x) = \sin\frac{\pi x}{6}
$$
is a solution.
|
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|
How does $-\frac{1}{x-2} + \frac{1}{x-3}$ become $\frac{1}{2-x} - \frac{1}{3-x}$ I'm following a solution that is using a partial fraction decomposition, and I get stuck at the point where $-\frac{1}{x-2} + \frac{1}{x-3}$ becomes $\frac{1}{2-x} - \frac{1}{3-x}$
The equations are obviously equal, but some algebraic manipulation is done between the first step and the second step, and I can't figure out what this manipulation could be.
The full breakdown comes from this solution
$$
\small\begin{align}
\frac1{x^2-5x+6}
&=\frac1{(x-2)(x-3)}
=\frac1{-3-(-2)}\left(\frac1{x-2}-\frac1{x-3}\right)
=\bbox[4px,border:4px solid #F00000]{-\frac1{x-2}+\frac1{x-3}}\\
&=\bbox[4px,border:4px solid #F00000]{\frac1{2-x}-\frac1{3-x}}
=\sum_{n=0}^\infty\frac1{2^{n+1}}x^n-\sum_{n=0}^\infty\frac1{3^{n+1}}x^n
=\bbox[4px,border:1px solid #000000]{\sum_{n=0}^\infty\left(\frac1{2^{n+1}}-\frac1{3^{n+1}}\right)x^n}
\end{align}
$$
Original image
|
We have $- \frac{1}{x-2} + \frac{1}{x-3} = \frac{1}{-1} \times \frac{1}{x-2} + \frac{-1}{-1} \times \frac{1}{x-3} = \frac{1}{(-1)\times (x-2)} + \frac{(-1)\times 1}{(-1)\times (x-3)} = \frac{1}{-x+2} + \frac{-1}{-x+3}= \frac{1}{-x+2} - \frac{1}{-x+3} $. And we have the result by just nature of your own question.
|
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|
Generating Series and Recurrence Relation and Closed Form We have the following recurrence relation:
$b_n=2b_{n-1}+b_{n-2}$
and initial conditions $b_0=0, b_1=2$
I use the generating series method to solve as following:
Let
$B(x)=b_0+b_1x+b_2x^2+...+b_nx^n+...$
$-xB(x)=-b_0x-b_1x^2-...-b_{n-1}x^n-...$
$-x^2B(x)=-b_0x^2-b_1x^3+...-b_{n-2}x^n+...$
After cancelling terms I get
$(1-x-x^2)B(x)=b_0+(b_1-2b_0)x$
And using initial conditions and isolating $B(x)$
$B(x)={2x\over (1-x-x^2)}$
When I factorise the quadratic in the denominator I get roots:
$x_1={-1-\sqrt{5}\over 2}$ and $x_2={-1+\sqrt{5}\over 2}$
I'm having trouble completing the question using partial fractions from this point, I would appreciate if someone could help me out
I am required to use the Generating Series method
|
Shift indices to get:
$$
b_{n + 2} = 2 b_{n + 1} + b_n
$$
Define the generating function:
$$
B(z) = \sum_{n \ge 0} b_n z^n
$$
Multiply the recurrence by $z^n$, sum over $n \ge 0$ and recognize the resulting sums:
$$
\frac{B(z) - b_0 - b_1 z}{z^2}
= 2 \frac{B(z) - b_0}{z} + B(z)
$$
As partial fractions:
$\begin{align}
B(z)
&= \frac{2z}{1 - 2 z - z^2} \\
&= \frac{\sqrt{2}}{2} \cdot \frac{1}{1 - (1 + \sqrt{2}) z}
- \frac{\sqrt{2}}{2} \cdot \frac{1}{1 - (1 - \sqrt{2}) z}
\end{align}$
This is just two geometric series:
$\begin{align}
b_n
&= \frac{\sqrt{2}}{2}
\left(
(1 + \sqrt{2})^n
- (1 - \sqrt{2})^n
\right)
\end{align}$
|
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|
How to compute $\sum_{n=0}^{\infty}{\frac{3^n(n + \frac{1}{2})}{n!}}$? I have to compute the series $\displaystyle\sum_{n=0}^{\infty}{\frac{3^n(n + \frac{1}{2})}{n!}}$.
$$\displaystyle\sum_{n=0}^{\infty}{\frac{3^n(n + \frac{1}{2})}{n!}} = \sum_{n=0}^{\infty}{\frac{3^n\frac{1}{2}}{n!}} + \sum_{n=0}^{\infty}{\frac{3^nn}{n!}} = \frac{e^3}{2} + \sum_{n=0}^{\infty}{\frac{3^nn}{n!}},$$ but how to compute the $\displaystyle\sum_{n=0}^{\infty}{\frac{3^nn}{n!}}$?
|
Hint:
$$\sum_{n=0}^{\infty}{\frac{3^nn}{n!}}=0+3\sum_{n=1}^{\infty}{\frac{3^{n-1}}{(n-1)!}}$$
|
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|
Find the domain of $x$ in $4\sqrt{x+1}+2\sqrt{2x+3}\leq(x-1)(x^2-2)$
Solve this equation for $x$: $4\sqrt{x+1}+2\sqrt{2x+3}\leq(x-1)(x^2-2)$
I have no idea to solve that, but I know solutions are $x=-1$ or $x\ge 3$.
|
It seems the following.
We can straightforwardly verify your answer with Mathcad as follows:
Since $\sqrt{x+1}$ exists, $x\ge -1$. Since the left-hand side of the inequality is non-negative, its right-hand side is non-negative too, that is $x\ge\sqrt{2}$ or $-1\le x\le 1$. Squaring both sides of the inequality, we obtain:
$$16(x+1)+4(2x+3)+16\sqrt{(x+1)(2x+3)} \leq(x-1)^2(x^2-2)^2$$
$$16\sqrt{(x+1)(2x+3)}\le x^6-2x^5-3x^4+8x^3-32x-24$$
Mathcad shows that the right-hand side has two real roots: $x_1=-1$ and $x_2\simeq 2.7472$.
So $x\ge x_2.$
Squaring both sides of the inequality again, we obtain:
$$0 \le x^{12}-4x^{11}-2x^{10}+28x^9-23x^8-112x^7+144x^6+288x^5-368x^4-384x^3+512x^2+256x-192$$
$$0 \le (x-3)(x+1)g(x)$$
where $$g(x)=x^{10}-2x^9-3x^8+16x^7-64x^5+16x^4+128x^3-64x^2-128x+64.$$
Mathcad shows that the function $g$ has two real roots $x’_1\simeq 0.4917$ and $x_2’\simeq 1.9309$, both are less than $x_2$.
Therefore the value of $(x-3)(x+1)g(x)$ is non-positive, if $x_2\le x\le 3$ and non-negative, if $x\ge 3$.
|
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|
Solve $\lim \limits_{x \to 0} (\frac{\sin x}{x})^\frac{1}{1-\cos x}$ without L'Hospital's Rule or Taylor expansion. I solved it with L'Hospital's Rule, but I had to use it 4 times! Is there any way to avoid it?
|
Can we use some basics: $\lim \limits_{x \to 0} (1+x)^\frac{1}{x} = e$ and $\lim \limits_{x \to 0} \frac{\sin x}{x} = 1$?
So, we separate $1$: $$\large\lim \limits_{x \to 0} \left(\frac{\sin x}{x}\right)^\frac{1}{1-\cos x} = \lim \limits_{x \to 0} \left(1+\frac{\sin x}{x}-1\right)^\frac{1}{1-\cos x}
= \lim \limits_{x \to 0} \left(\left(1+\frac{\sin x}{x}-1\right)^\frac{1}{\frac{\sin x}{x}-1}\right)^\frac{\frac{\sin x}{x}-1}{1-\cos x}=e^{\lim \limits_{x \to 0}\frac{\frac{\sin x}{x}-1}{1-\cos x}}, $$
$$\lim \limits_{x \to 0}\frac{\frac{\sin x}{x}-1}{1-\cos x} =
\lim \limits_{x \to 0}\frac{\frac{\sin x}{x}-1}{2\left(\frac{x}{2}\right)^2}\cdot \frac{2\left(\frac{x}{2}\right)^2}{2\sin^2\left(\frac{x}{2}\right)} = 2\lim \limits_{x \to 0}\frac{\sin x-x}{x^3},$$
$\lim \limits_{x \to 0}\frac{\sin x-x}{x^3}$ is here.
|
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|
Calculating all possible sums of the numbers $2^0, 2^1, \ldots, 2^{(n-1)}$ Using the simple equation $2^{n-1}$ you get values such as:
$1,2,4,8,16,32,64,128,256,$ etc.
How can I find all possible number combinations within this range? For example, the numbers $1,2,4,8$ give:
\begin{align}
1 &= 1 \\
2 &= 2 \\
1 + 2 &= 3 \\
4 &= 4 \\
1 + 4 &= 5 \\
2 + 4 &= 6 \\
1 + 2 + 4 &= 7 \\
8 &= 8 \\
1 + 8 &= 9 \\
2 + 8 &= 10 \\
1 + 2 + 8 &= 11 \\
4 + 8 &= 12 \\
1 + 4 + 8 &= 13 \\
2 + 4 + 8 &= 14 \\
1 + 2 + 4 + 8 &= 15
\end{align}
The numbers calculable are: $1,2,3,4,5,6,7,8,9,10,11,12,13,14,15$.
Background I'm a computer programmer trying to get a unique number which represents a series of other numbers without the unique IDs being lost.
If I assign $5$ people with the id's, $1,2,4,8,16$, then in order to pass the list of them I don't need to list them all, just the mathematical equation to work it out. For example, if I wanted people $1,2,4$ then I could just pass the number $7$ and therefore it would HAVE to be people $1,2,4$ as there is no other possible combination to make this.
Please let me know if my logic is flawed somewhere!
Thanks,
Nick
|
Computing the decomposition of a nonnegative integer into sums of powers of $2$ is essentially determining the binary (base-$2$)-representation of a number.
For example, we can decompose $42$ as
$$42 = 32 + 8 + 2 = 1 \cdot 2^5 + 0 \cdot 2^4 + 1 \cdot 2^3 + 0 \cdot 2^2 + 1 \cdot 2^1 = 0 \cdot 2^0,$$
so in binary we write $42$ in bits (the binary analogue of digits) as
$$101010_2$$
(the subscript $_2$ here just reminds us that this should be read as a binary expression and not as a decimal one). Compare this decomposition with the decimal decomposition $$42 = 4 \cdot 10^1 + 2 \cdot 10^0 ;$$ we could thus write $42$ as $42_{10}$ but generally suppress a subscript $_{10}$ unless we're writing about multiple bases at once.
More generally, given the ($n + 1$) numbers $1, 2, 4, \ldots, 2^n$, the numbers expressible as sums of these are precisely the nonnegative integers with at most $n$ bits in their binary representations. (As usual, we take the empty sum to have value $0$.) These are precisely the nonnegative integers no larger than $${\underbrace{1 \cdots 1}_{n + 1}}{}_2 = 2^n + 2^{n - 1} + \cdots + 2 + 1 = 2^{n + 1} - 1.$$
By construction, this representation of any such number is unique (for the same reason that any given positive integer can only be written in decimal in a single way).
Then, in the encoding you describe, the binary representation $a_n \cdots a_0$ encodes the subset of $\{0, \ldots, n\}$ consisting precisely of the numbers $k$ for which $a_k = 1$. (In fact, this prescribes an easy algorithm for this subset from a given value of an $\texttt{integer}$-type variable.)
|
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|
Find $\int \frac{1}{x^4+x^2+1} \,\, dx$ Find $$\int \frac{1}{x^4+x^2+1} \,\, dx$$
I tried to find like that:
$\int \frac{1}{x^4+x^2+1} = \int \frac{\frac{1}{2}x + \frac{1}{2}}{x^2+x+1} \,\, dx + \int \frac{-\frac{1}{2}x + \frac{1}{2}}{x^2-x+1} \,\, dx = \frac{1}{2} \Big(\int \frac{2x + 1}{x^2+x+1} - \int \frac{x}{x^2+x+1} \Big) - \frac{1}{2} \Big( \int \frac{2x - 1}{x^2-x+1} \,\, dx - \int \frac{x}{x^2-x+1} \,\, dx \Big)$
but then I don't know how to find integrals:
$\, \int \frac{x}{x^2-x+1} \,\, dx \,$ and $\, \int \frac{x}{x^2+x+1} \,\, dx \,$
Is there another way to integrate this function or way to end my calculations?
|
The above method is a good way to solve it, however I think you are trying to solve the wrong question because its integral is very complicated.Take a look at this: http://www.wolframalpha.com/input/?i=integral+1%2F%28x%5E4%2Bx%5E2%2B1%29
|
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|
Show that $f:= \frac{1 - 2x}{x^2 - 1}$ is monotonically increasing in the open interval $(-1, 1)$ Show that $f:= \frac{1 - 2x}{x^2 - 1}$ is monotonically increasing in the open interval $(-1, 1)$
To show a function is monotonically increasing, I started by saying that:
A function $f$ is monotonically increasing in an interval $(a, b)$ if for all $x, y \in (a, b)$, with $x$ < $y$, we have that $f(x) \leq f(y)$.
So I started by tryin to show:
Let $x$, $y$ be arbitrary fixed numbers in $(-1, 1)$, with $x < y$, we want to show that $f(x) \leq f(y)$:
$$f(x) = \frac{1 - 2x}{x^2 - 1} \leq \frac{1 - 2y}{y^2 - 1} = f(y)$$
We multiply the left side by $\frac{y^2 - 1}{y^2 - 1}$, which is positive because $y^2$ is always less than $1$, so we don't need to change the sign:
$$\frac{1 - 2x}{x^2 - 1}\cdot \frac{y^2 - 1}{y^2 - 1} \leq \frac{1 - 2y}{y^2 - 1}$$
We know multiply by $\frac{x^2 - 1}{x^2 - 1}$ the right side:
$$\frac{1 - 2x}{x^2 - 1}\cdot \frac{y^2 - 1}{y^2 - 1} \leq \frac{1 - 2y}{y^2 - 1} \cdot \frac{x^2 - 1}{x^2 - 1}$$
So we can simplify:
$$(1 - 2x)\cdot (y^2 - 1) \leq (1 - 2y) \cdot (x^2 - 1)$$
Now, how do we proceed? Is this correct?
|
A simpler solution:
$$ f(x)=\frac{1-2x}{x^2-1}=\frac{1}{2}\left(\frac{1}{1-x}-\frac{3}{x+1}\right)\tag{1}$$
and both $\frac{1}{1-x}$ and $\frac{-1}{x+1}$ are increasing over $(-1,1)$.
|
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|
Computing a limit similar to the exponential function I want to show the following limit:
$$
\lim_{n \to \infty}
n
\left[
\left( 1 - \frac{1}{n} \right)^{2n}
- \left( 1 - \frac{2}{n} \right)^{n}
\right]
= \frac{1}{e^{2}}.
$$
I got the answer using WolframAlpha, and it seems to be correct numerically, but I am having trouble proving the result. My first instinct was to write the limit as
$$
\lim_{n \to \infty}
\frac
{
\left( 1 - \frac{1}{n} \right)^{2n}
- \left( 1 - \frac{2}{n} \right)^{n}
}
{1/n}.
$$
Then, I tried applying l'Hopital's rule, and I got
$$
\lim_{n \to \infty}
\frac
{
\left( 1 - \frac{1}{n} \right)^{2n}
\left( 2 \log\left( 1 - \frac{1}{n} \right) + \frac{2}{n-1} \right)
-
\left( 1 - \frac{2}{n} \right)^{n}
\left( \log\left( 1 - \frac{2}{n} \right) + \frac{2}{n-2} \right)
}
{-1/n^{2}}.
$$
This does not seem to have gotten me anywhere. My second attempt was to use the binomial theorem:
$$
\begin{align*}
n
\left[
\left( 1 - \frac{1}{n} \right)^{2n}
- \left( 1 - \frac{2}{n} \right)^{n}
\right]
&
=
n
\left[
\sum_{k=0}^{2n} \binom{2n}{k} \frac{(-1)^{k}}{n^{k}}
- \sum_{k=0}^{n} \binom{n}{k} \frac{(-1)^{k} 2^{k}}{n^{k}}
\right]
\\ &
=
\sum_{k=2}^{n} \left[ \binom{2n}{k} - \binom{n}{k} 2^{k} \right] \frac{(-1)^{k}}{n^{k-1}}
+ \sum_{k=n+1}^{2n} \binom{2n}{k} \frac{(-1)^{k}}{n^{k-1}}.
\end{align*}
$$
At this point I got stuck again.
|
In the same spirit as Lucian, you can consider the general expansion for large values of $n$ $$ \left( 1 - \frac{a}{n} \right)^{b\,n}= \left(1-\frac{a^2 b}{2 n}+\frac{a^3 b (3 a b-8)}{24 n^2}-\frac{a^4 b
\left(a^2 b^2-8 a b+12\right)}{48 n^3}+\cdots\right)e^{-a b} $$ Applied to your case $$\left( 1 - \frac{1}{n} \right)^{2n}=\left(1-\frac 1n-\frac1 {6n^2}+\cdots\right)e^{-2}$$ $$\left( 1 - \frac{2}{n} \right)^{n}=\left(1-\frac 2n-\frac2 {3n^2}+\cdots\right)e^{-2}$$ which finally make $$n
\left[
\left( 1 - \frac{1}{n} \right)^{2n}
- \left( 1 - \frac{2}{n} \right)^{n}
\right]=\left( 1 +\frac{1}{2n}+\cdots \right)e^{-2}$$
In a more general manner,$$\left[
\left( 1 - \frac{a}{n} \right)^{bn}
- \left( 1 - \frac{b}{n} \right)^{an}
\right]=(a-b)\left(-\frac{a b }{2 n}+\frac{a b (a+b) (3 a b-8)}{24 n^2}\right)e^{-ab}$$
|
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|
Integer solutions of $x^3 = 7y^3 + 6 y^2+2 y$? Does the equation $$x^3 = 7y^3 + 6 y^2+2 y\tag{1}$$ have any positive integer solutions? This is equivalent to a conjecture about OEIS sequence A245624.
Maple tells me this is a curve of genus $1$, and its Weierstrass form is $s^3 + t^2 + 20 = 0$, with $$ \eqalign{ s = \dfrac{-2(7 y^2 + 6 y + 2)}{x^2},& \
t = \dfrac{-2(3 x^3 + 14 y^2 + 12 y + 4)}{x^3}\cr
x = \dfrac{-2s(t-6)}{s^3+56},&\ y = \dfrac{4t-24}{s^3+56}}$$
So I can find rational points on both curves, but I haven't been able to find integer points on (1) other than the trivial $(0,0)$.
|
$$ \gcd(y, 7y^2 + 6 y + 2) = 1,2 $$
The first case is odd $y,$ so that $7y^2 + 6y+2$ is odd and $\gcd(y, 7y^2 + 6 y + 2) = 1.$ Both $y$ and $7y^2 + 6 y + 2$ must be cubes. Take $y = n^3.$ We want $7n^6 + 6 n^3 + 2$ to be a cube. Cubes are $1,0,-1 \pmod 9.$ If $n \equiv 0 \pmod 3,$ then $7n^6 + 6 n^3 + 2 \equiv 2 \pmod 9$ and is not a cube. If $n^3 \equiv 1 \pmod 9,$ then $7n^6 + 6 n^3 + 2 \equiv 6 \pmod 9$ and is not a cube. If $n^3 \equiv -1 \pmod 9,$ then $7n^6 + 6 n^3 + 2 \equiv 3 \pmod 9$ and is not a cube.
Next
$\gcd(y, 7y^2 + 6 y + 2) = 2.$ Both $y= 4n^3$ and $7y^2 + 6 y + 2 = 2 w^3$ .
give me a minute, it is not guaranteed to be easy just because the other case was. Hmmm, it is possible both mod 9 and mod 7, which reflects the solution with my $n=0, w=0.$ Sigh. Just taking $y=4u,$ there may be a tractable way to deal with $56u^2 + 12 u + 1 = w^3.$
Monday: computer suggests the only integer point on $56u^2 + 12 u + 1 = w^3$ is $(0,1),$ which would finish the problem if confirmed. CONFIRMED: see Integral solutions to $56u^2 + 12 u + 1 = w^3$
|
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|
Trigonometric root of a polynomial
If $4\cos^2 \left(\dfrac{k\pi}{j}\right)$ is the greatest root of the equation
$$x^3-7x^2+14x-7=0$$
where $\gcd(k,j)=1$
Evaluate $k+j$
I tried factorizing the equation but it wasn't of much help.
Btw, the answer given in my book is $k=1$ and $j=14$
Any help will be appreciated.
Thanks!
|
Let $z=\frac{k\pi}{j}$. If $4\cos^2z$ is the root of $x^3−7x^2+14x−7=0$, then
$2\cos z$ is the root of $x^6−7x^4+14x^2−7=0$.
Let's plug $x=2\cos z=2\frac{e^{iz}+e^{-iz}}{2}$ in and see what happens.
The expression comes out to be $e^{-6 i z} \left(-e^{2 i z}+e^{4 i z}-e^{6 i z}+e^{8 i z}-e^{10 i z}+e^{12 i z}+1\right)=0$.
Let's change the variables again, $-e^{2 i z}=t$, so $-t^{-3}\left(1+t+t^2+t^3+t^4+t^5+t^6\right)=0 \iff \frac{t^7-1}{t^3(t-1)}=0$, so $-e^{2 i z}$ is the $7$th-root of unity: $-e^{2iz}=e^{i\frac{2\pi k}{7}}, \gcd(k,7)=1$.
$$2iz+\pi i=i\frac{2\pi k}{7}$$
$$z=-\frac{\pi}{2}+\frac{\pi k}{7}$$
Now we need $Re(t)$ to be maximal, so we select $k$ to $-\frac{\pi}{2}+\frac{\pi k}{7}$ be the closest to $0$: $z=\pm\frac{\pi}{14}$.
The answer is either $13$ or $15$.
|
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|
Roots of a polynomial whose coefficients are ratios of binomial coefficients
Prove that $\left\{\cot^2\left(\dfrac{k\pi}{2n+1}\right)\right\}_{k=1}^{n}$ are the roots of the equation
$$x^n-\dfrac{\dbinom{2n+1}{3}}{\dbinom{2n+1}{1}}x^{n-1} + \dfrac{\dbinom{2n+1}{5}}{\dbinom{2n+1}{1}}x^{n-2} - \ldots \ldots \ldots + \dfrac{(-1)^{n}}{\dbinom{2n+1}{1}} =0 $$
Hence prove that
$$\sum_{r=1}^{\infty} \dfrac{1}{r^2}=\dfrac{\pi^{2}}{6}$$
I was stumped on the first sight. Then I tried using complex numbers but it was in vein. I further tried simplifying the equation, but since it contains only half of the binomial coefficients, I wasn't able to get a simpler equation.
Any help will be appreciated.
Thanks.
|
We have
$$\frac{e^{m i \theta}}{\sin^m \theta} = \frac{(\cos \theta + i \sin \theta)^m}{\sin^m \theta} = (\cot \theta + i)^m $$
for $m \in \mathbb{N}$ and $\theta \in \mathbb{R}$. Take $m=2n+1$ and $\theta = k\pi/(2n+1)$ to get
$$\frac{(-1)^k}{\sin^{2n+1}\theta} = \sum_{r=0}^{2n+1} \binom{2n+1}{r}i^r \cot^{2n+1-r} \theta . $$
The imaginary part comes from the summands for which $r$ is odd, hence
$$0 = \sum_{s=0}^n \binom{2n+1}{2s+1} (-1)^s (\cot^2\theta)^{n-s}. $$
Equivalently, the polynomial
$$\sum_{s=0}^n \binom{2n+1}{2s+1} (-1)^s x^{n-s} $$
has $\cot^2 \theta$ as a root. Divide by $2n+1$ to get the relation in the question. For the second part, take the coefficient of $x^{n-1}$ to get
$$\sum_{k=0}^n (\cot \frac{k\pi}{2n+1})^2 = \frac{1}{2n+1}\binom{2n+1}{3}. $$
Now divide by $2n(2n-1)$ and take a limit using $\cot \theta = 1/\theta + O(\theta)$.
|
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|
Determinant of symmetric matrix $(A-\lambda I)$ If we have a matrix $(A-\lambda I)$ which is:
$\left(
\begin{array}{ccc}
1-\lambda & -1 & 2 \\
-1 & 1-\lambda & 2 \\
2 & 2 & 2-\lambda \\
\end{array}
\right)
$
Then it's determinant can be written as : $(-1)^n(\lambda-\lambda_1)(\lambda-\lambda_2)(\lambda-\lambda_3)$. In this case what will $\lambda_1$,$\lambda_2$ and $\lambda_3$ be equal to? And how do we determine it's value given that the matrix is symmetric?
|
$\left( \begin{array}{ccc}
1-\lambda & -1 & 2 \\
-1 & 1-\lambda & 2 \\
2 & 2 & 2-\lambda \\
\end{array} \right)
$ = $\left( \begin{array}{ccc}
2-\lambda & -2+\lambda & 0 \\
-1 & 1-\lambda & 2 \\
2 & 2 & 2-\lambda \\
\end{array} \right)
=(-2+\lambda)
\left( \begin{array}{ccc}
-1 & 1 & 0 \\
-1 & 1-\lambda & 2 \\
2 & 2 & 2-\lambda \\
\end{array} \right) =
(-2+\lambda)
\left( \begin{array}{ccc}
-1 & 0 & 0 \\
-1 & -\lambda & 2 \\
2 & 4 & 2-\lambda \\
\end{array} \right)= (-1)(-2+\lambda)
\left( \begin{array}{ccc}
-\lambda & 2 \\
4 & 2-\lambda \\
\end{array} \right)
= (-1)(-2+\lambda)
\left( \begin{array}{ccc}
4 -\lambda & 4 -\lambda \\
4 & 2-\lambda \\
\end{array} \right)=
(-2+\lambda)(\lambda-4)
\left( \begin{array}{ccc}
1 & 1\\
4 & 2-\lambda \\
\end{array} \right)
=
(-2+\lambda)(\lambda-4)
\left( \begin{array}{ccc}
0 & 1\\
2+\lambda & 2-\lambda \\
\end{array} \right)=(-2+\lambda)(\lambda-4)(2-\lambda)=(-1)(\lambda-2)(\lambda-4)(\lambda-2)
$ Therefore,
$\lambda_1=2,\lambda_2=2,\lambda_3=4$.
|
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|
Cannot understand an Integral $$\displaystyle \int _{ \pi /6 }^{ \pi /3 }{ \frac { dx }{ \sec x+\csc x } } $$
I had to solve the integral and get it in this form.
My attempt:
$$\int _{ \pi /6 }^{ \pi /3 }{ \frac { dx }{ \sec x+\csc x } } $$
$$=\int _{\frac{\pi}{6}}^{ \frac{\pi}{3}} \dfrac{\sin x \cos x }{ \sin x+\cos x }dx $$
Substituting $t=\tan(\frac{x}{2})$,
$$\int_{\tan(\frac{\pi}{12})}^{\tan(\frac{\pi}{6})} \dfrac{2t}{1+t^2}\times\dfrac{1-t^2}{1+t^2}\times\dfrac{2}{1+t^2}dt$$
$$2\int_{2-\sqrt{3}}^{\frac{1}{\sqrt{3}}} \dfrac{2t(1-t^2)}{(1+t^2)^3}dt$$
Substituting $u=1+t^2$, $2t dt=du$, $1-t^2 = 2-u$
$$2\int_{8-4\sqrt{3}}^\frac{4}{3} \dfrac{(2-u)}{u^3}du$$
$$\displaystyle 4\int_{8-4\sqrt{3}}^\frac{4}{3} \dfrac{du}{u^3} \displaystyle -2\int_{8-4\sqrt{3}}^\frac{4}{3} \dfrac{du}{u^2}$$ Could somebody please tell me where I have gone wrong? Also could someone please tell me how to change the limits of the definite integral throughout?
|
Integrate directly as follows
\begin{align}
\int _{ \pi /6 }^{ \pi /3 }{ \frac { dx }{ \sec x+\csc x } }
& =\frac12 \int _{ \pi /6 }^{ \pi /3 } \frac{(\sin x +\cos x)^2-1 }{ \sin x+\cos x }dx \\
&= \frac{1}2\int _{ \pi /6 }^{ \pi /3 } \left(\sqrt2\cos(\frac\pi4-x)
-\frac1{\sqrt2\cos(\frac\pi4-x)}\right)dx\\
&= \frac12\left[{\sqrt2}\sin(\frac\pi4-x)dx
-\frac1{\sqrt2} \ln \cot(\frac\pi8-\frac x2)\right]_{\pi/6}^{\pi/3}\\
&= \frac{\sqrt3-1}2-\frac1{2\sqrt2}\ln \frac{\cot\frac{5\pi}{24}}{\cot\frac{7\pi}{24}}
\end{align}
|
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|
Find the diameter of iron ball.
Iron weights $8$ times the weight of oak, Find the diameter of an iron ball whose weight is equal to that of a ball of oak $18$ cm in diameter.
$a)\quad 4.5\quad cm \\\color{green}{b)\quad 9 \quad cm}\\c)\quad 12\quad cm\\d) \quad 15 \quad cm$
I tried
$\dfrac43\times \pi\times (\dfrac{d}{2})^3 =8\times \dfrac43\times \pi\times (9)^3\\~\\
d=36\quad cm$
what is the mistake ?
|
Since iron weighs $8$ times what oak weighs, you need the iron ball to have $\frac{1}{8}$ the volume of the oak ball.
$$\begin{align}
\frac{4}{3} \times \pi \times \left(\frac{d}{2}\right)^3 &=
\color{red}{\frac{1}{8}} \times \frac{4}{3} \times \pi \times (9)^3 \\
\frac{d^3}{8} &= \frac{1}{8} \times (9)^3 \\
d^3 &= 9^3 \\
d &= 9
\end{align}$$
|
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|
Find the coordinates for $v$ in the subspace $W$ spanned by the following vectors:
I'm confused as to how I solve this. I was told to project each $u$ onto $v$ but how would that get me to the answer that's required?
|
Resolve the data vector into its components in the $u$ coordinate system. because the $u$ vectors are orthogonal, we can solve for the projection terms independently.
$$
\begin{array}{cccccccccc}
v & = &
\frac{\langle v,u_{1} \rangle}{\langle u_{1},u_{1} \rangle} & u_{1} & + & \frac{\langle v,u_{2} \rangle}{\langle u_{2},u_{2} \rangle} & u_{2} & + &
\frac{\langle v,u_{3} \rangle}{\langle u_{3},u_{3} \rangle} & u_{3} & \\
%
\left[
\begin{array}{r}
5 \\ -2 \\ 9 \\ -5
\end{array}
\right]
& = &
\frac{27}{22} & \left[
\begin{array}{r}
3 \\3 \\ 2 \\ 0
\end{array}
\right]
& + &
-\frac{11}{7} & \left[
\begin{array}{r}
2 \\ 0 \\ -3 \\ 1
\end{array}
\right]
& + &
-\frac{65}{488} & \left[
\begin{array}{r}
-6 \\ 2 \\ 6 \\ 30
\end{array}
\right] \\
%
&=&
\frac{1}{18788 } & \left[
\begin{array}{r}
-68799 \\ 101745 \\ -49419 \\ -10659
\end{array}
\right]
%
\end{array}
$$
|
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|
Derivative of $f(x) = \frac{\cos{(x^2 - 1)}}{2x}$ Find the derivative of the function $$f(x) = \frac{\cos{(x^2 - 1)}}{2x}$$
This is my step-by-step solution: $$f'(x) = \frac{-\sin{(x^2 - 1)}2x - 2\cos{(x^2 -1)}}{4x^2} = \frac{2x\sin{(1 - x^2)} - 2\cos{(1 - x^2)}}{4x^2} = \frac{x\sin{(1 - x^2)} - \cos{(1 - x^2)}}{2x^2} = \frac{\sin{(1 - x^2)}}{2x} - \frac{\cos{(1-x^2)}}{2x^2}$$
and this is the output of WolphramAlpha: http://www.wolframalpha.com/input/?i=derivative+cos(x^2+-+1)%2F(2x)
Where is the mistakes?
|
In the first step, when you take the derivative of $\cos(x^2 - 1)$ in the quotient rule calculation, you're getting $-\sin(x^2-1)$, when you should be getting $-2x\sin(x^2 - 1)$ by the chain rule. This is then multiplied by the $2x$ from the denominator, giving $-4x^2\sin(x^2 - 1)$.
This is the only error--you'll notice that the difference between your solution and WolframAlpha's is the factor of $2x$ on the $\sin$ term.
|
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|
Properties of Jacobi symbol Let $\left(\frac{a}{n}\right) $ be Jacobi symbol .
It is well known that Jacobi symbol for $a=-1$ and $a=2$ satisfies the following:
$\left(\frac{-1}{n}\right) =
\begin{cases}
1, & \text{if } n \equiv 1 \pmod{4} \\
-1, & \text{if } n \equiv 3 \pmod{4}
\end{cases}$
and ,
$\left(\frac{2}{n}\right) =
\begin{cases}
1, & \text{if } n \equiv 1,7 \pmod{8} \\
-1, & \text{if } n \equiv 3,5 \pmod{8}
\end{cases}$
Are there similar properties for other values of $a$ ?
|
Yes, it does for all $a$ - that's arguably the point of quadratic reciprocity.
For one example,
$$\left(\frac{5}{n}\right)=\left(\frac{n}{5}\right)(-1)^{\frac{n-1}{2}\cdot\frac{5-1}{2}}=\left(\frac{n}{5}\right)=\begin{cases}
\hphantom{-}1&\text{if }n\equiv 1,4\bmod 5\\
\hphantom{-}0&\text{if }n\equiv 0\bmod 5\\
-1 & \text{if }n\equiv 2,3\bmod 5
\end{cases}$$
Even if $a$ is negative or even, you factor $-1$ and any $2$'s out and still get a breakdown in terms of residues of $n$. For example,
$$\left(\frac{-5}{n}\right)=\left(\frac{-1}{n}\right)\left(\frac{5}{n}\right)=\left(\frac{-1}{n}\right)\left(\frac{n}{5}\right)(-1)^{\frac{n-1}{2}\cdot\frac{5-1}{2}}=\left(\frac{-1}{n}\right)\left(\frac{n}{5}\right)$$
$$=\left.\begin{cases}
\hphantom{-}1&\text{if }n\equiv 1\bmod 4\\
-1&\text{if }n\equiv 3\bmod 4
\end{cases}\right\}\left.\begin{cases}
\hphantom{-}1&\text{if }n\equiv 1,4\bmod 5\\
\hphantom{-}0&\text{if }n\equiv 0\bmod 5\\
-1 & \text{if }n\equiv 2,3\bmod 5
\end{cases}\right\}$$
$$=\begin{cases}
\hphantom{-}1 & \text{if }n\equiv 1,3,7,9\bmod 20\\
\hphantom{-}0 & \text{if }n\equiv 5,15\bmod 20\\
-1 & \text{if }n\equiv 11,13,17,19\bmod 20
\end{cases}$$
|
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|
Show that there exists a vector $v$ such that $Av\neq 0$ but $A^2v=0$ Let $A$ be a $4\times 4$ matrix over $\mathbb C$ such that $rank A=2$ and $A^3=A^2\neq 0$.Suppose that $A$ is not diagonalisable. Then
Show that there exists a vector $v$ such that $Av\neq 0$ but $A^2v=0$
My try:$\dim \operatorname{Im}(A)+\dim \ker (A)=4$ so $\dim(\ker A)=\dim (\operatorname{Im}A)=2$.
Now $A$ satisfies $x^3=x^2\implies x^2(x-1)=0$ thus $0,1$ are the only eigen values of $A$. Since $A$ has rank $2$ so geometric multiplicity of $0$ is $2$ but $A$ is not diagonalizable thus algebraic multiplicity of $0$ is $3$.
So the characteristic polynomial will be $x^4-x^3=0$
Obviously $A$ will have a Jordan block of size $2$ corresponding to 0
$$\ A \text{ will have Jordan form as } \pmatrix{1&0&0&0\\0&0&1&0\\0&0&0&0\\0&0&0&0}
$$
How should I use these information to conclude my result??
|
There are only two possible Jordan forms:
$$\pmatrix{0&1&0&0\\0&0&0&0\\0&0&0&0\\0&0&0&1}, \quad \pmatrix{0&0&0&0\\0&0&0&0\\0&0&1&1\\0&0&0&1}$$ only the first first one satisfies $A^2 = A^3.$
$\bf p.s.$
the nonzero vector $v$ you are looking for is in $\ker(A^2)\setminus \ker(A)$ we have $dim(\ker(A^2) = 3,dim(\ker(A) = 2.$
step 1. find a basis of $3$ vector for $ker(A^2)$ by row reducing $A$ if you must.
step 2. find the vector $v$ in the basis so that $Av \neq 0.$
|
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|
Find the number of possible values of $a$
Positive integers $a, b, c$, and $d$ satisfy $a > b > c > d, a + b + c + d = 2010$,
and $a^2 − b^2 + c^2 − d^2 = 2010$. Find the number of possible values of $a.$
Obviously, factoring,
$$(a-b)(a+b) + (c-d)(c+d) = 2010$$
$$a + b + c + d = 2010$$
Substituting you get:
$$(a-b)(a+b) + (c-d)(c+d) - (a + b) - (c + d) = 0$$
$$\implies (a - b - 1)(a + b) + (c - d- 1)(c+d) = 0$$
But I dont see anything else.
|
Denote $$b=a-x, \quad d=c-y. \qquad(x,y\in\mathbb{N}).\tag{1}$$
Then
$$
a+b+c+d = 2a-x+2c-y = 2010,\\
a^2-b^2+c^2-d^2=(2a-x)x+(2c-y)y=2010.\tag{2}
$$
Since $x,y\in\mathbb{N}$, then unique solution of $(2)$ has $$x=y=1.\tag{3}$$
So, solution has form
$$
(a,\;a-1,\;c,\;c-1).\tag{4}
$$
$(2),(4)\Rightarrow$
$$a+c-1=1005,$$
$$c=1006-a.$$
So, solution has form
$$
(a,\;a-1,\;1006-a,\;1005-a).\tag{5}
$$
Smallest possible $a$: $a_{min}=504 \rightarrow (a,b,c,d)=(504,503,502,501)$.
Largest possible $a$: $a_{max}=1004 \rightarrow (a,b,c,d)=(1004,1003,2,1)$.
Number of possible values of $a$: $501$.
|
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|
What mistake have I made when trying to evaluate the limit $\lim \limits _ {n \to \infty}n - \sqrt{n+a} \sqrt{n+b}$? Suppose $a$ and $b$ are positive constants.
$$\lim \limits _ {n \to \infty}n - \sqrt{n+a} \sqrt{n+b} = ?$$
What I did first:
I rearranged $\sqrt{n+a} \sqrt{n+b} = n \sqrt{1+ \frac{a}{n}} \sqrt{1+ \frac{b}{n}}$ and so: $$\lim \limits _ {n \to \infty} n - n \sqrt{1+ \frac{a}{n}} \sqrt{1+ \frac{b}{n}} = 0$$
Because both $\frac{a}{n}$ and $\frac{b}{n}$ tend to $0$.
What would give a correct answer:
Plotting the function $$f(x) = x - \sqrt{x+a} \sqrt{x+b}$$
Clearly indicates that it has an asymptote in $- \frac{a+b}{2}$. This result can be obtained multiplying the numerator and the denominator by $n + \sqrt{n+a} \sqrt{n+b}$:
$$\lim \limits _ {n \to \infty}n - \sqrt{n+a} \sqrt{n+b} = $$
$$-\lim \limits _{n \to \infty} \frac {n(a+b)}{n + \sqrt{n+a} \sqrt{n+b}} - \lim \limits _{n \to \infty} \frac {ab}{n + \sqrt{n+a} \sqrt{n+b}}$$
The second limit is clearly $0$ and the first one gives the correct answer (dividing the numerator and denominator by $n$).
Why the first way I tried is wrong? I might have done something silly but I cannot find it.
|
Another approach to solving this is via L'Hopital's rule. For this, we first need to do a bit of rearranging...
$$
\lim_{n\to\infty} \frac{1-\sqrt{1+a/n}\sqrt{1+b/n}}{1/n}
$$
which, letting $m=1/n$, gives us
$$
\lim_{m\to0} \frac{1-\sqrt{1+am}\sqrt{1+bm}}{m}
$$
It is from here that L'Hopital's rule (in this case, it's equivalent to evaluating the derivative of $1-\sqrt{1+ax}\sqrt{1+bx}$ at $x=0$) can be applied, namely
$$
\lim_{m\to0} \frac{-\frac12\left(\frac{a\sqrt{1+bm}}{\sqrt{1+am}}+\frac{b\sqrt{1+am}}{\sqrt{1+bm}}\right)}{1} = -\frac{a+b}2
$$
|
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"source": "stackexchange",
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|
Solving differential equation $x'=\frac{x+2t}{x-t}$ I am trying to solve the following differential equation:
$$x'=\frac{x+2t}{x-t}$$
with initial value condition: $x(1)=2$
This is what I have so far:
Substitution: $u=\frac{x}{t}$
$$\implies u't+u=\frac{2t+tu}{-t+tu}$$
Separation of variables:
$$\implies \frac{u'(u-1)}{-u^2+2u+2}=\frac{1}{t}$$
Integrate both sides:
$$\implies \int \frac{u'(u-1)}{-u^2+2u+2} dt=\int \frac{1}{t} dt \iff -\frac{1}{2}\ln(-u^2+2u+2)=\ln(t)+C$$
Solve for "$u$":
$$\implies u=1 \pm\sqrt{\frac{-1+3e^{2C}t^2}{e^{2C}t^2}}$$
Resubstitution: $u=\frac{x}{t}$
$$\implies x=t \pm t\sqrt{\frac{-1+3e^{2C}t^2}{e^{2C}t^2}}$$
Solving for "$C$" using $x(1)=2$:
$$x(1)=2=1\pm 1\sqrt{\frac{-1+3e^{2C}1^2}{e^{2C}1^2}} \implies C=-\frac{\ln(2)}{2}$$
Substituting "$C$" into the solution:
$$x=t \pm t\sqrt{\frac{-1+\frac{3}{2}t^2}{\frac{1}{2}t^2}}=t \pm t\sqrt{3t^2-2}$$
Plugging "$x$" into the differential equation:
$$1 \pm (\sqrt{3t^2-2}+\frac{6t^2}{\sqrt{3t^2-2}}) \not = \frac{3t \pm t\sqrt{3t^2-2}}{\pm t\sqrt{3t^2-2}}$$
Where did I go wrong?
|
here is another way to do this. we have $$\frac{dx}{dt} = \frac{x+2t}{x-t}. $$ we can rewrite this as an exact differential in the form $$0=(t-x) dx + (x+2t) dt = f_x\, dx + f_t \, dt $$ we have $$f_{xt} = f_{tx} =1 $$
we can integrate $$ f_x = t-x \to f = tx-\frac12x^2 + C(t) \to f_t = x + C'=x+2t\to C= t^2 $$ to therefore the solution is $$tx - \frac12x^2 + t^2 = 1\times 2-2+1= 1. $$
|
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|
Yamabe's equation This is PDE Evans, 2nd edition: Chapter 9, Exericse 8(a).
*(a) Assume $n \ge 3$. Find a constant $c$ such that $$u(x) := (1+|x|^2)^{\frac{2-n}2}$$ solves Yamabe's equation $$-\Delta u = cu^{\frac{n+2}{n-2}} \quad \text{in }\mathbb{R}^n.$$ Note the appearance of the critical exponent $\frac{n+2}{n-2}$.
My work so far:
Given $u(x):= (1+|x|^2)^{\frac{2-n}2}$, we find that $$u^{\frac{n+2}{n-2}}=(1+|x|^2)^{-\frac{n+2}2}.$$
Also, I differentiate $u(x)$ to find $$u_{x_i}=(2-n)(1+|x|^2)^{-\frac n2} x_i.$$ I differentiate again to find \begin{align}
u_{x_i x_i} &= (2-n)\left[-n(1+|x|^2)^{-\frac{n+2}2} x_i^2 + (1+x^2)^{-\frac n2}\right] \\
&= (2-n)[-nx_i^2 +(1+x^2)](1+|x|^2)^{-\frac{n+2}2}.
\end{align}
Therefore, the negative Laplacian is
\begin{align}
-\Delta u = -\sum_{i=1}^n u_{x_i x_i}
&= (n-2)[-n|x|^2 +n(1+x^2)](1+|x|^2)^{-\frac{n+2}2} \\
&= (n-2)n(1+|x|^2)^{-\frac{n+2}2}
\end{align}
Putting these altogether, Yamabe's PDE $-\Delta u=cu^{\frac{n+2}{n-2}}$ becomes $$(n-2)n(1+|x|^2)^{-\frac{n+2}2}=c(1+|x|^2)^{-\frac{n+2}2}.$$
This would mean that $$c=n(n-2).$$
I had fixed my work acordingly (see my revision history), so this is actually the answer. I do have a follow-up question here, though.
|
You should have $$u_{x_i x_i} = (2-n)\left[-n(1+|x|^2)^{-\frac{n+2}2} x_i^2 + (1+|x|^2)^{-\frac n2}\right].$$
That is, not $x_i$, but $x_i^2$ in the squred bracket. Therefore, we have
$$-\Delta u=-\sum_{i=1}^nu_{x_i x_i} = -(2-n)\left[-n(1+|x|^2)^{-\frac{n+2}2} |x|^2 + n(1+|x|^2)^{-\frac n2}\right]=n(n-2)(1+|x|^2)^{-\frac{n+2}2}$$
where the second equality follows from $|x|^2=\sum_{i=1}^nx_i^2$.
As you have calculated, $u^{\frac{n+2}{n-2}}=(1+|x|^2)^{-\frac{n+2}2}$. We have
$$-\Delta u=n(n-2)u^{\frac{n+2}{n-2}}.$$
|
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|
For every $n$ and $x$ there is an $m$ such that $3^n\mid 4^mx + (4^m-1)/3$ I am interested to know if for any $x\in 2\mathbb N +1$ and $n\in\mathbb N$ there exists a number $m\in \mathbb N\cup\{0\}$ such that
$3^n$ divides $$4^mx + \frac{4^m-1}{3}$$
I can easily show this if $x\equiv 1 \mod 3^n$, however I don't seem to be able to use this extra information to do it for any $x$. Here is the proof for $x\equiv 1\mod 3^n$:
Proof is by induction. If $n=1$, we can write $x=3y+1$ then with $m=2$ we have $3$ divides $4^2(3y+1)+(4^2-1)/3 = 21+3(16y)$. Let $n>1$ and suppose that the result is true for this $n$. Suppose also that $x\equiv 1 \mod 3^{n+1}$, then clearly $x\equiv 1 \mod 3^n$ so there is an $m\in\mathbb N$ such that $3^n$ divides $4^mx + \displaystyle\frac{4^m-1}{3}$. We can write $x=3^{n+1}y+1$ and so $3^n$ divides
$$4^m(3^{n+1}y+1) + \frac{4^m-1}{3}$$
so because of the formula for geometric series, $3^n$ divides $\frac{4^{m+1}-1}{3}$.
Consider now the fact that
$$4^{3(m+1)}-1 = (4^{2(m+1)}+4^{m+1} + 1)(4^{m+1}-1)$$
The left factor on the right hand side is a multiple of 3 (because $4 \equiv 1 \mod 3$), and so $3^{n+1}$ divides
$$\frac{4^{3(m+1)}-1}{3}$$
Thus $3^{n+1}$ divides
$$4^{3(m+1)-1}(3^{n+1}y+1) + \frac{4^{3(m+1)-1}-1}{3}$$
so we choose $3m+2$ for $x$.
Maybe someone can use this to help me figure this out for any $x$ independent of $n$?
|
Generalizing slightly: Let $p$ be an odd prime (in the question we have $p=3$) and $n$ be a positive integer. We want to show that for every integer $x$ there exists an $m\in\mathbb N$ such that
$$p^n\mid (p+1)^mx+\dfrac{(p+1)^m−1}p$$
We multiply this by $p$ on both sides to get
$$p^{n+1}\mid (p+1)^m(px+1)−1$$
or in other words
$$\tag{*} (p+1)^m(px+1)\equiv 1\pmod{p^{n+1}}.$$
At this stage we can see that the question's restriction to odd $x$ doesn't matter -- an $m$ that works for some $x$ will also work for $x+p^n$, so if there are $m$ for every odd $x$, there is also an $m$ that works for every even $x$, namely the one we get for $x+p^n$.
On the other hand, the values of $px+1$ for $x\in\{0,1,2,\ldots,p^n-1\}$ are all coprime to $p^{n+1}$ and different modulo $p^{n+1}$. In fact they are exactly the elements of a group $G$ of order $p^n$ under multiplication modulo $p^{n+1}$.
What we need is that $(px+1)^{-1}$ is always a power of $p+1$, which is the same as saying that $p+1$ generates $G$.
The multiplicative group modulo a power of an odd prime is always cyclic, and therefore in particular its subgroup $G$ is cyclic. And in a cyclic group of order $p^n$, everything that is not a generator is the $p$th power of something. However, by the binomial theorem,
$$(px+1)^p = 1+\binom{p}{1}px+(\text{terms involving }p^2) \equiv 1 \pmod{p^2} $$
So in particular $p+1$ cannot be a $p$th power, so it does generate $G$, and therefore $(px+1)^{-1}$ is always a power of $p+1\pmod{p^{n+1}}$.
An aside: The case where $x\equiv 1\pmod{p^n}$ (which was proved in the question) can be handled directly from $\text{(*)}$. Namely if $x=p^nj+1$ then $px+1=p^{n+1}j+p+1\equiv p+1\pmod{p^{n+1}}$, and then simply because $p+1$ is coprime to $p^{n+1}$, some positive power of it must be $1$ modulo $p^{n+1}$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Remainder of $(1+x)^{2015}$ after division with $x^2+x+1$
Remainder of $(1+x)^{2015}$ after division with $x^2+x+1$
Is it correct if I consider the polynomial modulo $5$
$$(1+x)^{2015}=\sum\binom{2015}{n}x^n=1+2015x+2015\cdot1007x^2+\cdots+x^{2015}$$
RHS stays the same and then The remainder must be of the form $Ax+B$
$$x^{2015}+1\equiv Ax+B\pmod{1+x+x^2}$$
plug in $x=0\implies B=0$
plug in $x=1\implies A=-1\implies $ the remainder is $-x$
Is this a good way to solve the problem or were we lucky ?
|
Method $\#1:$
Observe that $$1+x\equiv-x^2\pmod{x^2+x+1}\text{ and }x^3\equiv1$$
$$\implies(1+x)^{3n+2}\equiv(-x^2)^{3n+2}\equiv(-1)^{3n+2}(x^3)^{2n+1}\cdot x\equiv(-1)^nx$$ as $3n+2\equiv n\pmod2$
Method $\#2:$
Observe that $$(1+x)^2\equiv x\pmod{x^2+x+1}\text{ and }x^3\equiv1$$
$$\implies(1+x)^{6n+5}=\{(1+x)^2\}^{3n+2}\cdot(1+x)\pmod{x^2+x+1}$$
$$\equiv x^{3n+2}\cdot(x+1)$$
$$\equiv(x^3)^n\cdot x^2(x+1)$$
$$\equiv1^n\cdot(x^3+x^2)$$
$$\equiv1+x^2\equiv-x$$
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
Determine whether or not the limit exists: $\lim_{(x,y)\to(0,0)}\frac{(x+y)^2}{x^2+y^2}$
Determine whether or not the limit $$\lim_{(x,y)\to(0,0)}\frac{(x+y)^2}{x^2+y^2}$$
exists. If it does, then calculate its value.
My attempt:
$$\begin{align}\lim \frac{(x+y)^2}{x^2+y^2} &= \lim \frac{x^2+y^2}{x^2+y^2} + \lim \frac {2xy}{x^2+y^2} =\\&= 1 + \lim \frac 2{xy^{-1}+yx^{-1}} = 1+ 2\cdot\lim \frac 1{xy^{-1}+yx^{-1}}\end{align}$$
But $\lim_{x\to 0^+} x^{-1} = +\infty$ and $\lim_{x\to 0^-} x^{-1} = -\infty$
Likewise, $\lim_{y\to 0^+} y^{-1} = +\infty$ and $\lim_{y\to 0^-} y^{-1} = -\infty$
So the left hand and right hand limits cannot be equal, and therefore the limit does not exist.
|
Here's another way
$$ \lim\limits_{(x,y)\to(0,0)} \frac{(x+y)^2}{x^2+y^2} $$
Using polar coordinates, we have
$$ \lim\limits_{r\to 0^+} \frac{\left(r\cos\phi+r\sin\phi\right)^2}{r^2\cos^2\phi + r^2\sin^2\phi} $$
$$ = \lim\limits_{r\to 0^+} \frac{r^2\left(\cos\phi+\sin\phi\right)^2}{r^2\left(\cos^2\phi + \sin^2\phi\right)} $$
$$ = \lim\limits_{r\to 0^+} \frac{\left(\cos\phi+\sin\phi\right)^2}{\cos^2\phi + \sin^2\phi} $$
$$ = \lim\limits_{r\to 0^+} \left(\cos\phi+\sin\phi\right)^2 $$
$$ = \sin(2\phi)+1 $$
This limit is clearly dependent on $\phi$. Therefore
$$ \lim\limits_{(x,y)\to(0,0)} \frac{(x+y)^2}{x^2+y^2}\ \mbox{does not exist} $$
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/1307149",
"timestamp": "2023-03-29T00:00:00",
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|
Linear Transformation Matrix with polynomials A linear transformation $T : P_2 \to P_2$ has matrix with respect to $S$ given by:
$$[T]\,( S) = \begin{bmatrix}
1/2&-3&1/2\\
-1&4&-1\\
1/2&2&1/2\\
\end{bmatrix}
$$
How do you find $T(a+bx+cx^2)$?
Thank you!!
|
Every polynomial $S$ of degree $2$ s.t. $S \in P_2$ can be represented as a vector in three dimensional space:
$$
S = a + bx + cx^2 \quad \iff \quad
\begin{bmatrix}
a \\ b \\ c \\
\end{bmatrix}
\cdot
\begin{bmatrix}
1 \\ x \\ x^2 \\
\end{bmatrix},
$$
therefore we can associate $S$ with a 3D vector
$$
S \longleftrightarrow
\begin{bmatrix}
a \\ b \\ c \\
\end{bmatrix}.
$$
When you have the polynomial represented as a vector, applying linear transformation $T$, given in form of a matrix, is a piece of cake:
$$
[T]\,( S) = \begin{bmatrix}
1/2&-3&1/2\\
-1&4&-1\\
1/2&2&1/2\\
\end{bmatrix}
\begin{bmatrix}
a \\ b \\ c \\
\end{bmatrix}
=
\begin{bmatrix}
\frac{1}{2}a - 3 b + \frac{1}{2}c \\
-a + 4 b -c \\
\frac{1}{2}a +2 b + \frac{1}{2}c \\
\end{bmatrix}
,
$$
which will correspond to the polynomial
$$
T\left( a+bx+cx^2\right) =
\left(\frac{1}{2}a - 3 b + \frac{1}{2}c\right) +
\left( -a + 4 b -c \right) x +
\left(\frac{1}{2}a +2 b + \frac{1}{2}c \right)x^2
$$
|
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|
find the minimum value of $\sqrt{x^2+4} + \sqrt{y^2+9}$ Question:
Given $x + y = 12$, find the minimum value of $\sqrt{x^2+4} + \sqrt{y^2+9}$?
Key:
I use $y = 12 - x$ and substitute into the equation, and derivative it.
which I got this
$$\frac{x}{\sqrt{x^2+4}} + \frac{x-12}{\sqrt{x^2-24x+153}} = f'(x).$$
However, after that. I don't know how to do next in order to find the minimum value. Please help!
|
Setting $f'(x)=0$ gives
$$\frac{x}{\sqrt{x+4}}+\frac{x-12}{\sqrt{x^2-24x+153}}=0 $$
or
$$\frac{x}{\sqrt{x+4}}=-\frac{x-12}{\sqrt{x^2-24x+153}}\tag1$$
Squaring both sides of $(1)$ gives
$$\left(\frac{x}{\sqrt{x^2+4}}\right)^2=\left(\frac{x-12}{\sqrt{x^2-24x+153}}\right)^2$$
which can be rewritten as
$$\frac{x^2}{x^2+4}=\frac{(x-12)^2}{x^2-24x+153} \tag2$$
Thus, writing the numerator of the left-hand side as $x^2=x^2+4-4$ and writing the numerator of the right-hand side as $(x-12)^2=(x-12)^2+9-9$ simplifies to $(2)$
$$\frac{4}{x^2+4}=\frac{9}{x^2-24x+153} \tag3$$
Can you finish from here?
|
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|
$5x\big(1+\frac{1}{x^2 +y^2}\big)=12$ ; $5y\big(1-\frac{1}{x^2 +y^2}\big)=4$ find $x$ and $y$ I already tried to solve using substitution and cross multiplication method .
I got the first simplified
(1)$$\frac{12}{5x}=1+ \frac{1}{x^2} +y^2$$
(2)
$$\frac{4}{5y}=1- \frac{1}{x^2+y^2}$$
Adding (1) and (2)
$$12y +4x= 50xy$$
I am pretty sure that it is irrelevant as there has to be at be $2$ solutions for $x$ and $y$. Please help
Thanks in advance
|
Let $x=r\cos\varphi,\, y=r\sin\varphi$
$$\begin{cases}
\frac{12}{5r\cos\varphi}=\frac{r^2+1}{r^2}\\
\frac{4}{5r\sin\varphi}=\frac{r^2-1}{r^2}
\end{cases}$$
$$\begin{cases}
\frac{5\cos\varphi}{12}=\frac{r}{r^2+1}\\
\frac{5\sin\varphi}{4}=\frac{r}{r^2-1}
\end{cases}$$
$$\begin{cases}
\cos\varphi=\frac{12}{5}\cdot\frac{r}{r^2+1}\\
\sin\varphi=\frac{4}{5}\cdot\frac{r}{r^2-1}
\end{cases}$$
$$\left(\frac{12}{5}\cdot\frac{r}{r^2+1}\right)^2+\left(\frac{4}{5}\cdot\frac{r}{r^2-1}\right)^2=1,$$
which yields $\frac{5 r^4-26 r^2+5}{r^2-1} = 0$
$r^2=5$ or $r^2=\frac{1}{5}$, substituting $r^2$ back, we obtain $x$ and $y$
|
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|
Showing that $\sqrt[105]{105}>\sqrt[106]{106}$ How can one prove that
$$\sqrt[105]{105}>\sqrt[106]{106}\text{ ?}$$
Induction on the statement $$\sqrt[n]{n}>\sqrt[n+1]{n+1} \text{ for } n \in \mathbb{N}| n>2$$ would yield $\sqrt[3]{3}>\sqrt[4]{4}$ at the base step $n=3$, which we cannot assume.
So, based on the properties of powers and square roots alone, can we prove the first statement?
EDIT: No calculus, no functions, not even logs.
It's more of a riddle that anything else.
|
All you need here is the binomial theorem for positive integer exponents:
$$(1+x)^n = \sum_{r=0}^n \binom{n}{r}x^r$$
Fix $n \ge 3$, and put $x = \frac{1}{n}$:
$$\left(1+\frac{1}{n}\right)^n = \sum_{r=0}^n \binom{n}{r}n^{-r}$$
Now, $\binom{n}{r} \le n^r$ for all $r=0,\ldots,n$ (easily proven by induction; or just look at the expression $\binom{n}{r} = \frac{n}{r}\frac{n-1}{r-1}\cdots\frac{n-r+1}{1}$). So
$$\left(1+\frac{1}{n}\right)^n \le \sum_{r=0}^n n^rn^{-r} = n+1$$
This is not quite good enough for our purposes, but we can improve the estimate by considering the last two terms separately:
$$\begin{align}
\left(1+\frac{1}{n}\right)^n &\le \sum_{r=0}^{n-2} n^rn^{-r} + \binom{n}{n-1}n^{-(n-1)}+\binom{n}{n}n^{-n} \\
&= n - 1 + n^{-(n-2)}+n^{-n} \\
&< n - 1 + 2n^{-(n-2)} \\
&< n
\end{align}$$
because $n^{n-2} > 2$ if $n \ge 3$, so $n^{-(n-2)} < \frac12$.
Therefore $n > (1+\frac{1}{n})^n$ for all $n \ge 3$. Multiplying both sides by $n^n$ gives $n^{n+1} > (n+1)^n$. Taking the $n(n+1)$-th root gives, for all $n \ge 3$,
$$\sqrt[n]{n} > \sqrt[n+1]{n+1}$$
|
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|
What is the most unusual proof you know that $\sqrt{2}$ is irrational? What is the most unusual proof you know that $\sqrt{2}$ is irrational?
Here is my favorite:
Theorem: $\sqrt{2}$ is irrational.
Proof:
$3^2-2\cdot 2^2 = 1$.
(That's it)
That is a corollary of
this result:
Theorem:
If $n$ is a positive integer
and there are positive integers
$x$ and $y$ such that
$x^2-ny^2 = 1$,
then
$\sqrt{n}$ is irrational.
The proof is in two parts,
each of which
has a one line proof.
Part 1:
Lemma: If
$x^2-ny^2 = 1$,
then there are arbitrarily large integers
$u$ and $v$ such that
$u^2-nv^2 = 1$.
Proof of part 1:
Apply the identity
$(x^2+ny^2)^2-n(2xy)^2
=(x^2-ny^2)^2
$
as many times as needed.
Part 2:
Lemma: If
$x^2-ny^2 = 1$
and
$\sqrt{n} = \frac{a}{b}$
then
$x < b$.
Proof of part 2:
$1
= x^2-ny^2
= x^2-\frac{a^2}{b^2}y^2
= \frac{x^2b^2-y^2a^2}{b^2}
$
or
$b^2
= x^2b^2-y^2a^2
= (xb-ya)(xb+ya)
\ge xb+ya
> xb
$
so
$x < b$.
These two parts
are contradictory,
so
$\sqrt{n}$
must be irrational.
Two things to note about
this proof.
First,
this does not need
Lagrange's theorem
that for every
non-square positive integer $n$
there are
positive integers $x$ and $y$
such that
$x^2-ny^2 = 1$.
Second,
the key property of
positive integers needed
is that
if $n > 0$
then
$n \ge 1$.
|
This proof will look much better if someone can add a diagram to it.
Suppose $\sqrt2=a/b$ with $a$ and $b$ positive integers chosen as small as possible. Draw a square of side $a$. In the upper left corner, place a square of side $b$, and another one in the lower right corner. The two $b$-squares have total area $2b^2=a^2$, the same as the area of the $a$-square. Thus, the overlap of the two $b$-squares, which is a square we'll call a $c$-square, must have area equal to that of the two corners of the $a$-square not covered by the $b$-squares. Those corners are squares, call them $d$-squares. Then $c^2=2d^2$, so $\sqrt2=c/d$, where $c$ and $d$ are integers (indeed, $c=2b-a$, $d=a-b$) and are less than $a$ and $b$. Contradiction!
You can see this, with diagram, here. It also appears, with many other proofs, here.
|
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|
Evaluate the directional derivative of $f$ for the points and directions specified Evaluate the directional derivative of $f$ for the points and directions specified:
(a) $f(x,y,z)=3x-5y+2z$ at $(2,2,1)$ in the direction of the outward normal to the sphere $x^2+y^2+z^2=9.$
(b) $f(x,y,z)=x^2-y^2$ at a general point of the surface $x^2+y^2+z^2=4$ in the direction of the outward normal at that point.
(c) $f(x,y,z)=x^2+y^2-z^2$ at $(3,4,5)$ along the curve of intersection of the two surfaces $2x^2+2y^2-z^2=25$ and $x^2+y^2=z^2.$
The answer for the problems are (a) -2/3 (b) $x^2-y^2$ (c) $0$.
For (a) and (b) I thought the direction of the outward normal was the gradient of the level surface, which would be $(2x,2y,2z)$ in both cases. So to get the directional derivative I would simply have to compute the dot product of grad$f$ and $(2x,2y,2z)$ at the corresponding points, however, the result does not match the answers.
For (c), I'm stuck on finding the curve of intersection of the two surfaces.
I would greatly appreciate it if anyone could help me out.
|
(a)
The outward normal of the sphere given is $\left(\frac{x}{3},\frac{y}{3}, \frac{1}{3}\sqrt{9-x^2-y^2}\right)$. The same at the given point is $\overline n=\left(\frac{2}{3},\frac{2}{3}, \frac{1}{3}\right).$ The vector of the partial derivatives is $\nabla f =(3,-5,2).$ The directional derivative is then
$$\nabla f \cdot \overline n=\left(\frac{2}{3},\frac{2}{3}, \frac{1}{3}\right)\cdot (3,-5,2)=2-\frac{10}{3}+\frac{2}{3}=-\frac{2}{3}.$$
(b)
The outward normal of the sphere given is $\left(\frac{x}{2},\frac{y}{2}, \frac{1}{2}\sqrt{4-x^2-y^2}\right)$.The vector of the partial derivatives is $\nabla f =(2x,-2y,0)$. So,
$$\nabla f \cdot \overline n=(2x,-2y,0)\cdot\left(\frac{x}{2},\frac{y}{2}, \frac{1}{2}\sqrt{4-x^2-y^2}\right)=x^2-y^2.$$
(c)
We have two surfaces this time: $2x^2+2y^2-z^2=25$ and $x^2+y^2=z^2.$ These intersect above the circle $x^2+y^2=25$ at $z=\pm 5$. (The intersection line was given by setting $2x^2+2y^2-25=x^2+y^2$. Also, where $x^2+y^2=25$, there $z=\pm 5$.) Now, the vectors pointing to the intersection lines are of the form $(x,\sqrt{25-x^2},\pm 5)$; at $(3,4,5)$ it gives $(3,4,5)$. The vector of the partial derivatives: $(2x,2y,-2z)$, which at the given point is $(6,8,-10)$. So,
$$\nabla f \cdot \overline n= (3,4,5)\cdot(6,8,-10)=0.$$
(c')
I've seen that the other people answering considered the tangent vector to the intersection line. That approach gave the same result at the given point. If "along" means "tangent to" then my answer was only accidentally correct.
|
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|
Generating function for $l_{n+1} = 3l_n+1$. I have the sequence $l_{n+1} = 3l_n+1$ for $l_0 = 0$ or $1$ (This just shifts the sequence one index back or forward.) The first terms are $(0),1,4,13,40,121,364,\ldots$
So I am looking for an expression for $S(x) = \sum\limits_{n=0}^\infty l_n x^n$ I tried to standard approach:
$$\begin{align}l_{n+1} &= 3l_n +1 \\
\implies \sum\limits_{n=0}^\infty x^{n+1}l_{n+1} &= \sum\limits_{n=0}^\infty 3l_n x^{n+1}+x^{n+1} \\
\implies S(x)-l_0 &= 3xS(x)+\left(\frac 1 {1-x} -1\right) \\
&= 3xS(x)+\frac{1+x}{1-x}\\
\implies (1-3x)S(x) &= \frac{1+x}{1-x}+l_0 \\
\implies S(x) &= \frac{1+x}{(1-3x)(1-x)}+\frac{l_0}{1-3x} \end{align} $$
But the Taylor series of $S$ for $l_0 = 1$ yields:
$2+8 x+26 x^2+80 x^3+242 x^4+728 x^5+\ldots$ which has a factor $2$ compared to what is desired. (Curiously the same series appears for $l_0 = -1$ , but just a side note.) When we simplify $S(x)$ for $l_0 = 1$ we get $\frac{2}{(1-3x)(1-x)}$ so obviously the right expression for $S$ should be $S(x) = \frac{1}{(1-3x)(1-x)}$
For $l_0 = 0$ we should obviously get the same series multiplied by $x$, but this is not the case, we get:
$1+5 x+17 x^2+53 x^3+161 x^4+485 x^5+\ldots$
This leads me to the conclusion that I made an error when derivin $S$ but I just could not find it. Please help me find the error! Thank you very much in advance!
|
Your mistake is that $\sum_{n=0}^{\infty} x^{n+1} = \frac{x}{1-x}$, not $\frac{1+x}{1-x}$.
Fixing that, the rest of the steps are:
$$S(x) = \frac{l_0}{1 - 3x} + \frac{x}{(1-x)(1-3x)}$$
With $l_0 = 0$, that's:
$$\begin{split}
S(x) &= \frac{-\frac12}{1-x} + \frac{\frac12}{1-3x} \\
&= -\frac12\sum_{n=0}^{\infty}x^n + \frac12\sum_{n=0}^{\infty}(3x)^n \\
&= \sum_{n=0}^{\infty}\frac{3^n-1}2x^n
\end{split}$$
So your series is $l_n = \frac{3^n - 1}2$.
If $l_0 = 1$, the coefficient of the $\frac1{1-3x}$ term becomes $\frac32$ , instead of $\frac12$, so we'd end up with $l_n = \frac{3^{n+1} - 1}2$ instead.
|
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|
Is this a valid partial fraction decomposition?
Write $\dfrac{4x+1}{x^2 - x - 2}$ using partial fractions.
$$ \frac{4x+1}{x^2 - x - 2} = \frac{4x+1}{(x+1)(x+2)} = \frac{A}{x+1} + \frac{B}{x-2} = \frac{A(x-2)+B(x+1)}{(x+1)(x-2)}$$
$$4x+1 = A(x-2)+B(x+1)$$
$$x=2 \Rightarrow 4 \cdot2 + 1 = A(0) + B(3) \Rightarrow B = 3$$
$$x = -1 \Rightarrow 4(-1) +1 = A(-3)+ B(0) \Rightarrow A = 1$$
Thus,
$$\frac{4x+1}{x^2-x-2} = \frac{1}{x+1} + \frac{3}{x-2}\textrm{.}$$
The substitution of $x$ ($x = 2, -1$) is a common method to find out the coefficient of the partial fractions. However, the equation on the third line is obtained by multiplying $(x+1)(x-2)$, which is assumed to be nonzero. Here we have a contradiction. Furthermore, the original function is not defined at $x=-1,2$.
How can we substitute these value for $x$? So is this method valid and rigorous? How to modify it so that it is rigorous?
|
$A(x-2)+B(x+1) = 4x+1$ gives us a system of two equations:
$$\begin{split}
A+B &= 4 \\
-2A + B &= 1
\end{split}$$
But really the first equation as really $(A+B)x = 4x$, so the substitution $x=2$ simply gives us the system:
$$\begin{split}
2A+2B &= 8 \\
-2A + B &= 1 \\
\end{split}$$
Which conveniently sums to $3B = 9$, $B=3$. It's just a choice of substitution that makes it easy to solve the system. Any such substitution is valid, just that $x=2$ and $x=-1$ happen to be particularly good choices.
|
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|
Prove $(a^2+b^2+c^2+d^2)^2≥(a+b)(b+c)(c+d)(d+a)$ I've been unsuccessfully trying to solve this contest-style problem for a while.
Tried different substitutions and the such, but nothing helped. I presume the solution is related to Cauchy-Schwarz? Anyway, any hints would be appreciated!
Given $a, b, c, d \in \mathbb{R^+}$, prove that
$$(a^2 + b^2 + c^2 + d^2)^2 \ge (a+b)(b+c)(c+d)(d+a)$$
Note- I'm new on the site, sorry if I'm breaking some rule or convention.
|
By the AM-GM inequality:
$$ (a+b)(c+d) \leq \frac{1}{4}(a+b+c+d)^2, $$
$$ (b+c)(d+a) \leq \frac{1}{4}(a+b+c+d)^2, $$
and by the AM-QM inequality:
$$ \frac{1}{4}(a+b+c+d)^2 \leq (a^2+b^2+c^2+d^2).$$
The claim easily follows.
|
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|
Simple mod problem It’s kind of a silly question but I can't find a simple way for finding the value of variable $d$ .
$(5*d) \mod 8 = 1$
I normally just do this recursively by saying $d=d+1$ until I get the right answer. Is there a better way of solving this?
|
To find the multiplicative inverse of $5$ modulo $8$, use the Extended Euclidean Algorithm.
First, solve for the greatest common divisor of $5$ and $8$, which is $1$ since they are relatively prime.
\begin{align*}
8 & = 1 \cdot 5 + 3\\
5 & = 1 \cdot 3 + 2\\
3 & = 1 \cdot 2 + 1\\
2 & = 2 \cdot 1
\end{align*}
Now, work backwards to solve for $1$ as a linear combination of $5$ and $8$.
\begin{align*}
1 & = 3 - 2\\
& = 3 - (5 - 3)\\
& = 2 \cdot 3 - 5\\
& = 2 \cdot (8 - 5) - 5\\
& = 2 \cdot 8 - 3 \cdot 5
\end{align*}
Since $2 \cdot 8 - 3 \cdot 5 = 1$, $-3 \cdot 5 = 1 + 2 \cdot 8$, so
$$-3 \cdot 5 \equiv 1 \pmod{8}$$
Hence, $-3$ is in the residue class of the multiplicative inverse of $5$ modulo $8$. To express the multiplicative inverse as one of the residues $\{0, 1, 2, 3, 4, 5, 6, 7\}$, we add $8$ to $-3$ to obtain $5$. Thus, $5$ is the multiplicative inverse of $5 \pmod{8}$.
Check: $5 \cdot 5 \equiv 25 \equiv 1 + 3 \cdot 8 \equiv 1 \pmod{8}$.
|
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|
When is the series converges?
Let the series $$\sum_{n=1}^\infty \frac{2^n \sin^n x}{n^2}$$. For $x\in (-\pi/2, \pi/2)$, when is the series converges?
By the root-test:
$$\sqrt[n]{a_n} = \sqrt[n]{\frac{2^n\sin^n x}{n^2}} = \frac{2\sin x}{n^{2/n}} \to 2\sin x$$
Thus, the series converges $\iff 2\sin x < 1 \iff \sin x < \frac{1}{2}$
Is that right?
|
I like the ratio test here:
$$\begin{split}
L &= \lim_{n\to\infty} \left|\frac{2^{n+1} \sin^{n+1} x}{(n+1)^2} \frac{n^2}{2^n\sin^n{x}} \right| \\
&=\lim_{n\to\infty} \left|\frac{2n^2\sin{x}}{(n+1)^2}\right| \\
&=\lim_{n\to\infty} \left|\frac{2\sin{x}\cdot n^2}{n^2}\right| \\
&= |2\sin{x}|
\end{split}$$
$L < 1 \iff |2\sin x| < 1 \iff |\sin x| < \frac12 \iff x \in (-\frac{\pi}6,\frac{\pi}6)$
We can test the boundary separately. If $\sin x = \pm\frac12$, then the series becomes $\sum_{n=0}^{\infty}\frac{(\pm1)^n}{n^2}$, which clearly converges. Thus, the solution is the closed interval, $x \in [-\frac{\pi}6,\frac{\pi}6]$
|
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|
What two numbers when multiplied gives $-25$ and when added, $-10$? Factors of $25$ are; $1, 5, 25$
$$5 \times 5 = 25$$
$$5 + 5 = 10$$
$$-5 \times 5 = -25$$
$$-5 + 5 = 0$$
How can I solve this?
Thanks
|
Well one cannot solve this if only integers may be used.
The answer is:
$$(a,b) = (5(\sqrt{2} - 1), -5(1+\sqrt{2}))$$
We can see this as we have
$$a*b = -25,\qquad a+b =-10.$$
From this it follows that $a = -10-b$, thus
$$-10b -b^2 = -25, \Rightarrow b^2+10b-25 = 0.$$
This can be solved quite easily, as this gives
$$b = \frac{-10\pm \sqrt{10^2-4*-25}}{2} = -5 \pm 5\sqrt{2}.$$
|
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|
Help with solving this Recurrence Relation I really need help with this question
Would anyone please give a simple step-by-step on how to solve this Recurrence Relation??
$a_n = 2a_{n-1} - 2a_{n-2}$ where $a_0 = 1$ and $a_1 = 3$
It would really be great if someone could explain how to solve this
So far I made it into the characteristic equation:
$r^2 - 2r + 2 = 0$
But not sure where to go or if I am on the right track
Thanks
|
A generatingfunctionological solution would be to define $A(z) = \sum_{n \ge 0} a_n z^n$, write the recurrence shifting indices as:
$$
a_{n + 2} = 2 a_{n + 1} - 2 a_n
$$
Multiply the recurrence by $z^n$, sum over $n \ge 0$ and recognize some sums:
$$
\frac{A(z) - a_0 - a_1 z}{z^2}
= 2 \frac{A(z) - a_0}{z} - 2 A(z)
$$
Plug in $a_0 = 1, a_1 = 3$ and solve for $A(z)$ to get
$$
A(z) = \frac{1 + z}{1 - 2 z + 2 z^2}
$$
Write this as partial fractions, factoring the denominator as $1 - 2 z + 2 z^2 = (1 - (1 - \mathrm{i}) z) (1 - (1 + \mathrm{i}) z)$:
$$
A(z) = \frac{1 + 2 \mathrm{i}}{2} \cdot \frac{1}{1 - (1 + \mathrm{i}) z}
+ \frac{1 - 2 \mathrm{i}}{2} \cdot \frac{1}{1 - (1 - \mathrm{i}) z}
$$
This is two geometric series, so we can read off the coefficients:
$$
a_n = \frac{1 + 2 \mathrm{i}}{2} \cdot (1 + \mathrm{i})^n
+ \frac{1 - 2 \mathrm{i}}{2} \cdot (1 - \mathrm{i})^n
$$
This is the sum of a complex number and it's conjugate, so it is just twice the real part of the first one:
$$
a_n = 2 \Re\left(\frac{1 + 2 \mathrm{i}}{2} \cdot (1 + \mathrm{i})^n\right)
$$
Write the numbers in polar form:
$\begin{align}
a_n &= 2 \Re\left( \sqrt{5} \cdot \exp(\mathrm{i} \arctan 2)
\cdot \sqrt{2}^n \cdot \exp(\pi n \mathrm{i} / 4) \right) \\
&= 2 \sqrt{5} \cdot \sqrt{2}^n
\cdot \Re\left(
\exp\left(
\left(
\frac{n \pi}{4} + \arctan 2
\right) \mathrm{i}
\right)
\right) \\
&= 2 \sqrt{5} \cdot \sqrt{2}^n
\cdot \cos \left(
\frac{n \pi}{4} + \arctan 2
\right)
\end{align}$
|
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|
solving difficult complex number proving if $z= x+iy$ where $y \neq 0$ and $1+z^2 \neq 0$, show that the number $w= z/(1+z^2)$ is real only if $|z|=1$
solution :
$$1+z^2 = 1+ x^2 - y^2 +2xyi$$
$$(1+ x^2 - y^2 +2xyi)(1+ x^2 - y^2 -2xyi)=(1+ x^2 - y^2)^2 - (2xyi)^2$$
real component
$$(1+ x^2 - y^2)x - yi(2xyi) = x + x^3 + xy^2$$
imaginary component
$$-2yx^2 i +yi + x^2 yi - y^3 i=0i$$
$$-2yx^2 +y + x^2 y - y^3 =0$$
...
can't solve this question
|
It's easier to work with $w=\dfrac{1+z^2}z=\dfrac1z+z$ (that doesn't change the claim).
Taking the imaginary part,
$$-\frac y{x^2+y^2}+y=y\left(1-\frac1{x^2+y^2}\right)=0,$$
i.e.
$$x^2+y^2=1.$$
Or, in polar coordinates,
$$-\frac1r\sin(\phi)+r\sin(\phi)=0\implies r=1.$$
|
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|
Solve $p_{n+1} + \frac 16 p_n = \frac 1 2 (\frac 5 6 ) ^{n-1}$ I'm trying to solve:
$$p_{n+1} + \frac 16 p_n = \frac 1 2 \left(\frac 5 6 \right) ^{n-1}$$
with initial condition: $p_1 = 1$.
First, I search particular solution of the form $p_n^* = \lambda (\frac 5 6 ) ^n$. I found $\lambda = \frac 3 5$.
Next, I know that $p_n = \alpha (-\frac 1 6)^n + \frac 3 5 (\frac 5 6 )^n$, and using $p_1 = 1$, I found: $\alpha = -3$.
Which leads to: $p_n = -3(-\frac 1 6)^n + \frac 3 5 (\frac 5 6 )^n$.
Well... But thanks to the recurrence formula, I have $p_2 = \frac 1 4$ and with this new formula, I have $p_2 = \frac 1 3$.
Where is my mistake?
|
Your answer is correct! Here's a full solution to the recurrence.
We have the following:
$$
\begin{align*}
p_{n+1} + \dfrac{1}{6} p_n &= \dfrac{1}{2}\left(\dfrac{5}{6}\right)^{n-1} \\
p_{n} + \dfrac{1}{6} p_{n-1} &= \dfrac{1}{2}\left(\dfrac{5}{6}\right)^{n-2} \\
\end{align*}
$$
Dividing the two, we have:
$$
\begin{align*}
\dfrac{6p_{n+1} + p_n}{6p_{n} + p_{n-1}} &= \dfrac{5}{6} \\
36p_{n+1} + 6p_n &= 30p_n + 5p_{n-1} \\
36p_{n+1} - 24p_n - 5p_{n-1} &= 0
\end{align*}
$$
The roots of the characteristic polynomial are $-\dfrac{1}{6}$ and $\dfrac{5}{6}$. Thus, the solution is in the form
$$
p_n = \alpha \left(-\dfrac{1}{6}\right)^n + \beta \left(\dfrac{5}{6}\right)^n
$$
Plugging in $p_{1} = 1$ and $p_2 = \dfrac{1}{3}$, we find that $\alpha = -3$, $\beta = \dfrac{3}{5}$.
Thus, the solution is
$$
p_n = -3 \cdot \left(-\dfrac{1}{6}\right)^n + \dfrac{3}{5} \cdot \left(\dfrac{5}{6}\right)^n
$$
|
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|
System of equations with complex numbers-circles The system of equations
\begin{align*}
|z - 2 - 2i| &= \sqrt{23}, \\
|z - 8 - 5i| &= \sqrt{38}
\end{align*}
has two solutions $z_1$ and $z_2$ in complex numbers. Find $(z_1 + z_2)/2$.
So far I have gotten the two original equations to equations of circles,
$(a-2)^2 +(b-2)^2=23$ and $(a-8)^2+(b-5)^2=38$.
From here how do I find the solutions?
Thanks.
|
You're given the distances of the solution points to $A = 2+2i$ and $B = 8+5i.$
One of the solutions and these two complex numbers give you a triangle in the complex plane, and you know the lengths of all the sides.
We can use Law of Cosines to find the angle $\theta$ that has $2+2i$ as its vertex:
$$38 = 45 + 23 - 2 \sqrt{45} \sqrt{23} \cos \theta \to \cos \theta \approx. 0.466252.$$
The projection of $AZ_1$ onto $AB$ is then $\sqrt{23} \cdot 0.466252 \approx 2.23606$.
Why calculate this? Because the other solution is symmetric on the other side of $AB$, and what you're asked to find, in essence, is the average of the two solutions, which lies on $AB$.
The unit vector from $A$ to $B$ is $(6 + 3i)/\sqrt{45} = (2+i)/\sqrt{5}.$
So, the solution is
$$(Z_1 + Z_2)/2 = 2 + 2i + 2.23606 \cdot (6 + 3i)/\sqrt{45} \approx 4 + 3i.$$
(The fact that this worked out so nicely makes me think there was an easier way, but oh well ...)
|
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|
High School Trigonometry ( Law of cosine and sine) I am preparing for faculty entrance exam and this was the question for which I couldn't find the way to solve (answer is 0). I guess they ask me to solve this by using the rule of sine and cosine:
Let $\alpha$, $\beta$ and $\gamma$ be the angles of arbitrary triangle with sides a, b and c respectively. Then $${b - 2a\cos\gamma \over a\sin\gamma} + {c-2b\cos\alpha \over b\sin\alpha} + {a - 2c\cos\beta \over c\sin\beta}$$ is equal to (answer is zero but I need steps).
|
Consider the first summand, to begin with:
$$
C=\frac{b-2a\cos\gamma}{a\sin\gamma}
$$
The law of cosines tells you that
$$
c^2=a^2+b^2-2ab\cos\gamma
$$
so we can write
$$
b-2a\cos\gamma=\frac{c^2-a^2}{b}
$$
and so we have
$$
C=\frac{c^2-a^2}{ab\sin\gamma}
$$
The law of sines is
$$
\frac{a}{\sin\alpha}=\frac{b}{\sin\beta}=\frac{c}{\sin\gamma}=2R
$$
where $R$ is the radius of the circumscribed circle. In particular,
$$
\sin\gamma=\frac{c}{2R}
$$
and so we have
$$
C=\frac{2R(c^2-a^2)}{abc}
$$
Similarly,
$$
A=\frac{c-2b\cos\alpha}{b\sin\alpha}=\frac{a^2-b^2}{abc}\\
B=\frac{a-2c\cos\beta}{c\sin\beta}=\frac{b^2-c^2}{abc}
$$
and
$$
C+A+B=\frac{c^2-a^2+a^2-b^2+b^2-c^2}{abc}=0
$$
|
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Solution to Fibonacci Recursion Equations Let the sequence $(a_n)_{n\geq0}$ of the fibonacci numbers: $a_0 = a_1 = 1, a_{n+2} = a_{n+1} + a_n, n \geq 0$
Show that:
i) $$a^2_n - a_{n+1}a_{n-1} = (-1)^n \text{ for }n\geq1$$
I try to show this with induction:
$n=1:$
\begin{align*}
a^2_1 - a_{2}a_{0}&=1 - a_2a_0\\
&= 1- (a_1 + a_0 )a_0\\
&= 1-(1+1)1 = -1 = (-1)^1\\
\end{align*}
Assume that: $a^2_n - a_{n+1}a_{n-1} = (-1)^n ,\forall n\geq1$
Inductive step:
\begin{align*}
a^2_{n+1} - a_{n+2}a_{n}
&=(a_n + a_{n-1}) ^2 - (a_{n+1}+a_{n})(a_{n-1}-a_{n-2})\\
&=a^2_{n}+2a_na_{n-1}+a^2_{n-1}-a_{n+1}a_{n-1}+a_{n+1}a_{n-2}-a_{n}a_{n-1}+a_{n}a_{n-2}\\
&=(a^2_{n}-a_{n+1}a_{n-1})+(a^2_{n-1}-a_{n}a_{n-1})+2a_na_{n-1}+a_{n+1}a_{n-2}+a_{n}a_{n-2}\\
&=\underbrace{(-1)^n + (-1)^{n-1}}_{=0}+2a_na_{n-1}+a_{n+1}a_{n-2}+a_{n}a_{n-2}\\
&=2a_na_{n-1}+a_{n+1}a_{n-2}+a_{n}a_{n-2}\\
&=??\\
\end{align*}
ii) $$\sum \limits_{i=0}^na_i=a_{n+2}-1 , n\geq0$$
iii) $$a^2_{n-1}+a^2_n=a_{2n} \text{ and } a_{n-1}a_n+a_na_{n+1}=a_{2n+1}, n\geq1$$
Thank you for any help
|
Since the first term of the induction process for part one has been established the t remains to be seen for general $n$. Now,
\begin{align}
a_{n+1}^{2} - a_{n} \, a_{n+2} &= a_{n+1}^{2} - a_{n} \, ( a_{n+1} + a_{n} )\\
&= a_{n+1} ( a_{n+1} - a_{n} ) - a_{n}^{2} \\
&= a_{n-1} \, a_{n+1} - a_{n}^{2} = (-1)^{n+1}.
\end{align}
This provides $a_{n}^{2} - a_{n-1} \, a_{n+1} = (-1)^{n}$.
For the second part:
Using $a_{n} = a_{n+2} - a_{n+1}$ then
\begin{align}
\sum_{i=0}^{n} a_{i} &= \sum_{i=0}^{n} ( a_{i+2} - a_{i+1} ) = a_{n+2} - a_{1}.
\end{align}
For the third part:
Using the solution of the difference equation in the form $(\alpha - \beta) a_{n} = \alpha^{n} - \beta^{n}$ where $2 \alpha = 1 + \sqrt{5}$ and $2 \beta = 1 - \sqrt{5}$. Now,
\begin{align}
a_{n}^{2} + a_{n-1}^{2} &= \frac{1}{5} \left[ (\alpha^{2n} - 2(-1)^{n} + \beta^{2n}) + (\alpha^{2n-2} + 2 (-1)^{n} + \beta^{2n-2}) \right] = a_{2n-1}.
\end{align}
For the fourth part:
\begin{align}
a_{n} ( a_{n+1} + a_{n-1} ) &= \frac{a_{n}}{\alpha - \beta} \, \left[ \alpha^{n} \left( \alpha + \frac{1}{\alpha} \right) - \beta^{n} \left( \beta + \frac{1}{\beta} \right) \right] \\
&= \frac{(\alpha^{n} - \beta^{n})(\alpha^{n} + \beta^{n})}{\alpha - \beta} \\
&= a_{2n}.
\end{align}
|
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"timestamp": "2023-03-29T00:00:00",
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|
Given $x^2 + 4x + 6$ as factor of $x^4 + ax^2 + b$, then $a + b$ is I got this task two days ago, quite difficult for me, since I have not done applications of Vieta's formulas and Bezout's Theorem for a while. If can someone solve this and add exactly how I am supposed to use these two theorem's on this task, I would be thankful.
Given $x^2 + 4x + 6$ as factor of $x^4 + ax^2 + b$, then $a + b$ is equal to?
|
If $x^2 + 4x + 6$ is factor of $x^4 + ax^2 + b$, then it must satify
$$x^4 + ax^2 + b=(x^2 + 4x + 6)Q(x)$$ where $Q(x)$ is a second degree polynominal. You do not need to find $Q(x)$
$x^2 + 4x + 6=0$
$x^2=-(4x + 6)$
$$x^4 + ax^2 + b=(x^2 + 4x + 6)Q(x)$$
$$(-(4x + 6))^2 - a(4x + 6)+ b=0$$
$$16x^2+48x+36 - 4ax -6a+ b=0$$
$$-16(4x + 6))+48x+36 - 4ax -6a+ b=0$$
$$-64x -96+48x+36 - 4ax -6a+ b=0$$
$$-16x -60 - 4ax -6a+ b=0$$
$$-16-4a=0 $$
$$a=-4$$
$$-60 -6a+ b=0$$
$$-60 +24+ b=0$$
$$b=36$$
then $a+b=32$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Comparing two large numbers Can you compare two large exponential numbers, like $5^{44}$ and $4^{53}$ without taking their logs?
|
One approach is to figure that, roughly, $2^{10} \approx 10^3$, and $5^9 \approx 2,000,000 = 2 \cdot 10^6$.
Then,
\begin{align}
4^{53} &= (2^2)^{53} \\
&= 2^{106} \\
&= 2^{100} \cdot 2^6 \\
&= (2^{10})^{10} \cdot 2^6 \\
&\approx (10^3)^{10} \cdot 2^6 \\
&= 10^{30} \cdot 2^6
\end{align}
By way of comparison,
\begin{align}
5^{44} &= 5^{45} \div 5 \\
&= (5^9)^5 \div 5 \\
&\approx (2 \cdot 10^6)^5 \div 5 \\
&= 2^5 \cdot (10^6)^5 \div 5 \\
&= 10^{30} \cdot (2^5 \div 5).
\end{align}
Now, $2^6 > 2^5 \div 5$, so one might suppose $4^{53} > 5^{44}$.
|
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|
Integration of $\int \frac {x\arctan x}{(\sqrt{1+x^2})^3}$ How can I evaluate
$$\int \frac{x \arctan x}{(\sqrt{1+x^2})^3}\,dx$$
I think it must be done by parts, but I can't get any appropriate substitution or such.
|
There is no magic in this calculation, the key is familiarity, familiar with $\int\frac{dx}{1+x^2}=\arctan(x)+C$.
Let $\theta=\arctan(x)$, then we have $d\theta=\frac{dx}{1+x^2}$, $\frac{x}{\sqrt{1+x^2}}=\sin\theta$ and $\frac{1}{\sqrt{1+x^2}}=\cos\theta$, so
$$
I=\int\frac{x\theta}{\sqrt{1+x^2}}\centerdot\frac{dx}{1+x^2}=\int\theta(\sin\theta) d\theta = -\int\theta d\cos\theta =
$$
$$
-\theta\cos\theta + \int(\cos\theta) d\theta = -\theta\cos\theta+\sin\theta = \frac{-\arctan(x)+x}{\sqrt{1+x^2}}+C
$$
|
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|
Find the cubic equation of $x=\sqrt[3]{2-\sqrt{3}}+\sqrt[3]{2+\sqrt{3}}$ Find the cubic equation which has a root $$x=\sqrt[3]{2-\sqrt{3}}+\sqrt[3]{2+\sqrt{3}}$$
My attempt is
$$x^3=2-\sqrt{3}+3\left(\sqrt[3]{(2-\sqrt{3})^2}\right)\left(\sqrt[3]{(2+\sqrt{3})}\right)+3\left(\sqrt[3]{(2-\sqrt{3})}\right)\left(\sqrt[3]{(2+\sqrt{3})^2}\right)+2-\sqrt{3}$$
$$x^3=4+\left(\sqrt[3]{(2-\sqrt{3})^2}\right)\left(\sqrt[3]{(2+\sqrt{3})}\right)+3\left(\sqrt[3]{(2-\sqrt{3})}\right)\left(\sqrt[3]{(2+\sqrt{3})^2}\right)$$
then what I will do??
|
Probably it is better to notice, for first, that the product between $2-\sqrt{3}$ and $2+\sqrt{3}$ is one. After that, the minimal polynomial of $a=\sqrt[3]{2+\sqrt{3}}$ over $\mathbb{Q}$ is quite trivially:
$$ p(x)=(x^3-2)^2-3 = x^6-4x^3+1 $$
hence $a$ is a root of:
$$ x^3+x^{-3}-4 = \left(x+\frac{1}{x}\right)^3 - 3\left(x+\frac{1}{x}\right)-4$$
and $a+\frac{1}{a}$ is a root of:
$$ q(x) = x^3-3x-4.$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Find the remainder when $f(x)$ is divided by $x^2+x-2$
When $f(x)$ is divided by $x-1$ and $x+2$, the remainders are $4$ and $-2$ respectively. Find the remainder when $f(x)$ is divided by $x^2+x-2$.
Please help. The answer is $2x+2$.
I tried to understand the link between the first sentence and the second sentence but all I could make out was $f(1)=4$ ($4$ being the remainder) and $f(-2)=-2$.
I also broke down $x^2+x-2$ into $(x-1)(x+2)$ but after that, I didn't know how to continue. It seemed like there was a link as both have $x-1$ and $x+2$. I tried to attempt it myself for quite a long time but I couldn't make out anything. I'm not sure what I made out was even on the correct track though.
|
Use the Chinese Remainder theorem (Wikipedia link) in the ring of polynomials.
You'll want to look at the polynomials $g=\frac{1}{3}(x+2)$ and $h=-\frac{1}{3}(x-1)$, which satisfy
$$\begin{align*}
g&\equiv 1\bmod (x-1) & h&\equiv 0\bmod(x-1)\\
g&\equiv 0\bmod (x+2) & h&\equiv 1\bmod (x+2)
\end{align*}$$
Thus, for any $a$ and $b$,
$$\begin{align*}
ag+bh&\equiv a\bmod (x-1)\\
ag+bh&\equiv b\bmod (x+2)
\end{align*}$$
In particular, the polynomial
$$k=4g-2h=(\tfrac{4}{3}x+\tfrac{8}{3})+(\tfrac{2}{3}x-\tfrac{2}{3})=2x+2$$
has the property that
$$\begin{align*}
k&\equiv \hphantom{-}4\bmod (x-1)\\
k&\equiv -2\bmod (x+2)
\end{align*}$$
The Chinese remainder theorem then guarantees that $f\equiv k\bmod (x^2+x-2)$ for any $f$ such that $f(1)=4$ and $f(-2)=-2$.
|
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|
Triple fractions I've got this simple assignment, to find out the density for a give sphere with a radius = 2cm and the mass 296g. It seems straightforward, but it all got hairy when i've got to a fraction with three elements(more precisely a fraction divided by a number actually this was wrong, the whole point was that the number is divided by a fraction, and it's different than a fraction being divided by a number.). I tend to solve these by dividing the element on the bottom by 1, and extracting from that 2 fraction division like this :
$$
\frac{a}{\frac{b}{c}} \Rightarrow \frac{\frac{a}{b}}{\frac{c}{1}} \Rightarrow \frac{a}{b} \div \frac{c}{1} => \frac{a}{b} \cdot \frac{1}{c} \Rightarrow \frac {a} {b \cdot c}
$$
And it used to work, though for the next example it doesn't seem to, it looks like another technique is used:
$$
\frac{a}{\frac{b}{c}} \Rightarrow a \div \frac{b}{c} \Rightarrow a \cdot \frac{c}{b} \Rightarrow \frac {a \cdot c}{ b}
$$
For the example below cleary the second method is used/needed, to get the right response. But i'm confused when to use each, as i've use both before, and both gave correct asnwers(matching with the answers at the end of the book).
$$
v = \frac43\pi r^3
$$
$$
d = \frac mv
$$
$$
m = 296g
$$
$$
r=2cm
$$
$$
v = \frac43\pi 2^3 \Rightarrow \frac{32\pi}{3}
$$
$$
d = \frac{m}{v} \Rightarrow \frac{296}{\frac{32\pi}{3}} \Rightarrow \frac {296}{32\pi} \div \frac31 \Rightarrow \frac{296}{32\pi} \cdot \frac{1}{3} \Rightarrow \frac{296}{96\pi} \approx 0.9814\frac{g}{cm^3}
$$
$$
d_{expected} = 8.8 \frac{g}{cm^3}
$$
I am, clearly, missing something fundamental about the use of these.
Can anyone enlighten me please?
Can't quite find a good explanation online.
|
$$\frac{a}{\frac{b}{c}}\ne\frac{\frac{a}{b}}{c} \tag 1$$
The left-hand side of $(1)$ can be written as
$$\frac{a}{\frac{b}{c}}=\frac{ac}{b}$$
whereas the right-hand side of $(1)$ can be written as
$$\frac{\frac{a}{b}}{c}=\frac{a}{bc}$$
Let's look at an example: Suppose $a=3$, $b=6$, and $c=2$. Then, we have
$$\frac{a}{\frac{b}{c}}=\frac{3}{\frac{6}{2}}=\frac{3}{3}=1$$
but
$$\frac{\frac{a}{b}}{c}=\frac{\frac{3}{6}}{2}=\frac{1/2}{2}=\frac{1}{4}$$
|
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|
Distance between orthocenter and circumcenter.
Let $O$ and $H$ be respectively the circumcenter and the orthocenter of triangle $ABC$. Let $a$, $b$ and $c$ denote the side lengths. We are given that $a^2+b^2+c^2=29$ and the circumradius is $R=9$. We need to find $OH^2$.
I know that there is formula $OH^2=9R^2-(a^2+b^2+c^2)$, but I cannot use it unless I prove it.
I tried placing ABC triangle in a Cartesian plane and found coordinates of $O$ and $H$, but the expressions are not nice and I didn't manage to simplify them to required result.
Maybe vectors can be used? Please, note that I cannot use sophisticated vector knowledge in the solution. Any suggestions or ideas would be appreciated.
|
since
$$\angle HAO=|B-C|.OA=R,AH=2R\cos{A}$$
use cosin theorem we have
\begin{align*}
OH^2&=AH^2+AO^2-2AH\cdot AO\cos{|B-C|}
=4R^2\cos^2{A}+R^2-4R^2\cos{A}\cos{(B-C)}\\
&=5R^2-4R^2\sin^2{A}+4R^2\cos{(B+C)}\cos{(B-C)}\\
&=9R^2-4R^2(\sin^2{A}+\sin^2{B}+\sin^2{C})\\
&=9R^2-(a^2+b^2+c^2)
\end{align*}
|
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|
How to find sum of the infinite series $\sum_{n=1}^{\infty} \frac{1}{ n(2n+1)}$ $$\frac{1}{1 \times3} + \frac{1}{2\times5}+\frac{1}{3\times7} + \frac{1}{4\times9}+\cdots $$
How to find sum of this series?
I tried this: its $n$th term will be = $\frac{1}{n}-\frac{2}{2n+1}$; after that I am not able to solve this.
|
First, we can rewrite the partial sum as an integral
$$\sum_{n=1}^N \frac{1}{n(2n+1)} = 2\sum_{n=1}^N \left(\frac{1}{2n} - \frac{1}{2n+1}\right)
= 2\sum_{n=1}^N \int_0^1 (z^{2n-1} - z^{2n}) dz\\
= 2 \int_0^1 z(1-z)\left(\sum_{n=0}^{N-1} z^{2n}\right) dz
= 2 \int_0^1 \frac{z}{1+z}( 1 - z^{2N} ) dz
$$
Notice the $N$ dependence piece on RHS can be bounded from above
$$\left| 2 \int_0^1 \frac{z}{1+z} z^{2N} dz \right|
< 2 \int_0^1 z^{2N} dz = \frac{2}{2N+1} \to 0
\quad\text{ as }\quad N \to \infty
$$
We have
$$\sum_{n=1}^\infty \frac{1}{n(2n+1)}
=\lim_{N\to\infty}\sum_{n=1}^N \frac{1}{n(2n+1)}
= 2 \int_0^1 \frac{z}{1+z} dz = 2 (1 - \log 2)$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Simplify P(n), where n is a positive integer : $ P(x)=\sum \limits_{k=1}^\infty \arctan\left(\frac{x-1}{(k+x+1)\sqrt{k+1}+(k+2)\sqrt{k+x}}\right). $ This is what I have tried, but I don't know what to do next, so I need help
:
$ P(x)=\sum \limits_{k=1}^\infty \arctan\left(\frac{x-1}{(n+x+1)\sqrt{n+1}+(n+2)\sqrt{n+x}}\right). $
$ P(1)=\arctan(\frac{x-1}{(1+x+1)\sqrt{1+1}+(1+2)\sqrt{1+x}})$
$=\arctan(\frac{x-1}{(2+x)\sqrt{2}+3\sqrt{1+x}})$
$ P(2)=\arctan(\frac{x-1}{(2+x+1)\sqrt{2+1}+(2+2)\sqrt{2+x}})$
$=\arctan(\frac{x-1}{(3+x)\sqrt{3}+4\sqrt{2+x}})$
$ P(3)=\arctan(\frac{x-1}{(3+x+1)\sqrt{3+1}+(3+2)\sqrt{3+x}})$
$=\arctan(\frac{x-1}{(4+x)\sqrt{4}+5\sqrt{3+x}})$
$ P(4)=\arctan(\frac{x-1}{(4+x+1)\sqrt{4+1}+(4+2)\sqrt{4+x}})$
$=\arctan(\frac{x-1}{(5+x)\sqrt{5}+6\sqrt{4+x}})$
$ P(5)=\arctan(\frac{x-1}{(5+x+1)\sqrt{5+1}+(5+2)\sqrt{5+x}})$
$=\arctan(\frac{x-1}{(6+x)\sqrt{6}+7\sqrt{5+x}})$
I tried to calculate it for a few numbers to see if there will be any order to conclude something. Only that I conclude is that the series is diverging.
|
I'll assume $x$ is an positive integer. Define
$$\begin{cases}
a_n &= \frac{x-1}{(n+x+1)\sqrt{n+1}+(n+2)\sqrt{n+x}}\\
u_n &= \sqrt{n+x}\\
v_n &= \sqrt{n+1}
\end{cases}$$
We have
$$a_n = \frac{u_n^2-v_n^2}{(u_n^2+1)v_n + (v_n^2+1)u_n} = \frac{u_n-v_n}{1 + u_nv_n}
$$
This leads to
$$\tan^{-1}a_n
= \tan^{-1}\left(\frac{u_n-v_n}{1+u_n v_n}\right)
= \left(\tan^{-1}u_n - \tan^{-1}v_n\right) + \pi N
$$
for some integer $N$ to be determined. By throwing in some explicit numbers, it is not hard to see $N = 0$ in this case. As a result, we have
$$\begin{align}
P(x) = \sum_{n=1}^\infty \tan^{-1} a_n
&= \sum_{n=1}^\infty \left(\tan^{-1}\sqrt{n+x} - \tan^{-1}\sqrt{n+1}\right)\\
&= \sum_{n=1}^\infty \left(
\tan^{-1}\frac{1}{\sqrt{n+1}} - \tan^{-1}\frac{1}{\sqrt{n+x}}\right)\\
&= \sum_{n=1}^{x-1} \tan^{-1}\frac{1}{\sqrt{n+1}}
\end{align}
$$
|
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|
Let $z \in \Bbb Z_m$, when is $z^2 \equiv 1$? Let $z \in \Bbb Z_m$. When is $z^2=1, (z\neq1)$?
I know that for $m$ prime, $z=p-1$ is it's own inverse, but what about nonprime $m$?
Is $p-1$ the only self inverse element in $\Bbb Z_p$ ?
|
Let $p$ be an odd prime throughout.
Claim 1: In $\mathbb{Z}/p\mathbb{Z}$, the only numbers $n$ such that $n^2 \equiv 1$ are $\pm 1$.
Proof. Suppose $n$ is such that $n^2 \equiv 1 \pmod p$. Then we must have that $p \mid n^2 - 1 = (n+1)(n-1)$. Since $p$ is prime, we know that we then have $p \mid n+1$ or $p \mid n-1$. In the first case, we have that $n \equiv -1 \pmod p$, and in the second case we have that $n \equiv 1 \pmod p$. So we've proven the claim. $\clubsuit$
Claim 2: In $\mathbb{Z}/p^n\mathbb{Z}$, the only numbers $n$ such that $n^2 \equiv 1$ are $\pm 1$.
Proof. This is not actually conceptually any different, but I present this separately. Suppose $n$ is such that $n^2 \equiv 1 \pmod {p^n}$. The in particular $p^n \mid n^2 - 1 = (n+1)(n-1)$ as above. And in particular, $p \mid (n+1)(n-1)$, so that either $p \mid n+1$ or $p \mid n-1$. Since $p > 2$, we cannot have that $p$ divides both, so $p$ divides exactly one of $n+1$ or $n-1$. Thus $p^n$ divides exactly one of $n+1$ or $n-1$. In the former case, we have that $n \equiv -1 \pmod {p^n}$. In the latter, $n \equiv 1 \pmod {p^n}$. So we've proven the claim. $\clubsuit$
Claim 3: In $\mathbb{Z}/2\mathbb{Z}$ and $\mathbb{Z}/4\mathbb{Z}$, there are $1$ and $2$ solutions to $x^2 \equiv 1$, respectively.
Proof. You can simply do it, checking manually. $\spadesuit$
Claim 4: In $\mathbb{Z}/2^n\mathbb{Z}$ for $n \geq 3$, there are exactly $4$ solutions to $x^2 \equiv 1$. They are given by $x \equiv \pm 1 \pmod {2^{n-1}}$.
Proof. This is similar to the above proofs, but slightly more annoying. Again suppose that $x$ satisfies $x^2 \equiv 1 \pmod {2^n}$. Then $2^n \mid (x+1)(x-1)$. But since the difference between $x+1$ and $x-1$ is $2$, we will have that one will be divisible by $2$ and the other divisible by $2^{n-1}$ (note that they cannot both be divisible by $4$ for instance). If $2^{n-1} \mid x+1$, then we have that $x \equiv -1 \pmod{2^{n-1}}$. If $2^{n-1} \mid x-1$, then we have that $x \equiv 1 \pmod{2^{n-1}}$. When raised to $2^n$, these give the $4$ solutions mod $2^n$. $\clubsuit$
We now appeal to the Chinese Remainder Theorem to patch together different congruences mod different primes. In particular, we get $2$ choices of congruence class per odd prime and either $1, 2,$ or $4$ choices for powers of $2$. So if $n = 2^{a_0} p_1^{a_1} \ldots p_k^{a_k}$, then we have $2^k \cdot T(a_0)$ choices of congruences classes from the Chinese Remainder Theorem, where $T(0) = T(1) = 1, T(2) = 2$, and $T(a) = 4$ for $a \geq 3$.
And so this is the number of square roots of $1$ mod $n$. $\diamondsuit$
|
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|
Prove that $\sin{\frac{2\pi x}{x^2+x+1}}=\frac{1}{2}$ has no rational roots.
Show that the following equation has no rational roots.
$$\sin{\frac{2\pi x}{x^2+x+1}}=\frac{1}{2}$$
This is what I've tried:
$$\left ( \frac{2\pi x}{x^2+x+1}=\frac{\pi}{6}+2k\pi \right)\lor\left (\frac{2\pi x}{x^2+x+1}=\frac{5\pi}{6}+2l\pi \right ), \; \left (k,l \in \mathbb{Z} \right ) $$
First case only:
$$ \frac{2\pi x}{x^2+x+1}=\frac{\pi}{6}+2k\pi \iff \frac{2\pi x}{x^2+x+1}=2\pi \left ( \frac{1}{12}+k \right )$$
$$\left ( 12k+1 \right )x^2+\left (12k-11 \right)x+(12k+1)=0 \tag{$\bigstar$}$$
Now $$\Delta=96k^2-316k+117\geqslant 0 $$
For case $\Delta=0$:
$$k_{1,2}=\frac{316\pm \sqrt{54928}}{2\cdot96} \notin \mathbb{Z}$$
But it means nothing 'cause I still have entire interval $\left ( -\infty,k_{-} \right ) \cup \left ( k_{+}, +\infty\right )$ to check.
I've rewritten $\left ( \bigstar \right )$:
$$x^2+x\left (\frac{12k-11}{12k+1} \right)+1=0$$
By using Vieta's formulas: $$x_{1}+x_{2}=\frac{11-12k}{12k+1} \qquad \land \qquad x_{1}x_{2}=1$$
Since I have to prove that there are no rational roots, I wanted to give it a shot with contradiction, so I assumed:
$$x_{1}=\frac{p}{q}, \;\; x_{2}=\frac{r}{s}, \;\; \left (p,q \right)=\left (r,s \right)=1$$
Since $\left (p,q \right)=\left (r,s \right)=1$ we obtain from $x_{1}x_{2}=1$ that $p=s$ and $q=r$.
Now we have: $$x_{1}+x_{2}=\frac{p}{q}+\frac{r}{s}=\frac{p}{q}+\frac{q}{p}$$
$$x_{1}+x_{2}=\frac{p^2+q^2}{pq}=-\frac{12k+1-12}{12k+1}=\frac{12}{12k+1}-1$$
Now if I could prove: $$\left (p,q \right)=1 \implies \frac{p^2+q^2}{pq}\in \mathbb{Z}$$
...then I would show that my assumption is wrong i.e. there are no rational roots of the given equation.
But even if that is true, I'd rather hear some solution that doesn't involve number theory, not even elementary one.
|
You are almost there. If the equation has rational roots, then $\Delta$ must be a perfect square (why?).
To show that $\Delta$ is not a perfect square just observe that
$$\Delta=96k^2-316k+11 \equiv 3 \pmod{4}$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Integral of binomial coefficients Let the integral in question be given by
\begin{align}
f_{n}(x) = \int_{1}^{x} \binom{t-1}{n} \, dt.
\end{align}
The integral can also be seen in the form
\begin{align}
f_{n}(x) = \frac{1}{n!} \, \int_{1}^{x} \frac{\Gamma(t) \, dt}{\Gamma(t-n)} = \frac{(-1)^{n+1}}{n!} \int_{1}^{1-x} (u)_{n} \, du.
\end{align}
The first few values are
\begin{align}
f_{0}(x) &= x-1 \\
f_{1}(x) &= \frac{(x-1)^{2}}{2!} \\
f_{2}(x) &= \frac{2}{4!} \, (2 x^{3} - 9 x^{2} + 12 x - 5).
\end{align}
The question then becomes:
*
*What is the general series form of $f_{n}(x)$ ?
*Is there a connection to known special functions? In particular can $f_{n}(x)$ be related to the Bernoulli numbers of the second kind?
|
Here's another variation of the theme based upon Stirling Numbers.
Starting from
\begin{align*}
f_n(x)&=\int_1^x\binom{t-1}{n}dt
=\frac{1}{n!}\int_1^x{(t-1)}_ndt
=\frac{1}{n!}\int_0^{x-1}{(u)}_ndu
\end{align*}
we can use the Stirling Numbers of the first kind $s(n,k)$ which can be defined for $n\geq 0$ and $0\leq k \leq n$ by
\begin{align*}
{(u)}_n=\sum_{k=0}^ns(n,k)u^k\tag{1}
\end{align*}
Using this relationship we obtain for $n\geq 0$
\begin{align*}
f_n(x)&=\frac{1}{n!}\int_0^{x-1}\sum_{k=0}^ns(n,k)u^kdu\\
&=\frac{1}{n!}\sum_{k=0}^ns(n,k)\int_0^{x-1}u^kdu\\
&=\frac{1}{n!}\sum_{k=0}^ns(n,k)\left.\frac{1}{k+1}u^{k+1}\right|_0^{x-1}\\
&=\frac{1}{n!}\sum_{k=0}^ns(n,k)\frac{1}{k+1}(x-1)^{k+1}
\end{align*}
We conclude
\begin{align*}
f_n(x)=
\frac{1}{n!}\sum_{k=0}^n\frac{s(n,k)}{k+1}(x-1)^{k+1}&\qquad n\geq 0\tag{2}\\
\end{align*}
An exponential generating function $\sum_{n=0}^{\infty}f_n(x)\frac{z^n}{n!}$ can be derived from the generating function of the Stirling Numbers of the first kind.
Generating function: In accordance to the answer of @SylvainL. we claim
The following is valid
\begin{align*}
\sum_{n=0}^{\infty}f_n(x)\frac{z^n}{n!}=\frac{(1+z)^{x-1}-1}{\log(1+z)}
\end{align*}
We obtain from (1)
\begin{align*}
(1+z)^u&=\sum_{n=0}^{\infty}\binom{u}{n}z^n\\
&=\sum_{n=0}^{\infty}{(u)}_n\frac{z^n}{n!}\\
&=\sum_{n=0}^{\infty}\left(\sum_{k=0}^ns(n,k)u^k\right)\frac{z^n}{n!}\tag{3}
\end{align*}
In the following we consider $f_n(x+1)$ instead of $f_n(x)$.
We observe
\begin{align*}\sum_{n=0}^{\infty}f_n(x+1)\frac{z^n}{n!}
&=\sum_{n=0}^{\infty}\left(\sum_{k=0}^{n}\frac{s(n,k)}{k+1}x^{k+1}\right)\frac{z^n}{n!}\tag{4}\\
&=\int_0^x\sum_{n=0}^{\infty}\left(\sum_{k=0}^{n}s(n,k)u^{k}\right)\frac{z^n}{n!}du\\
&=\int_0^{x}(1+z)^udu\tag{5}\\
&=\int_0^xe^{u\log(1+z)}du\\
&=\left.\frac{1}{\log(1+z)}e^{u\log(1+z)}\right|_0^x\\
&=\frac{(1+z)^x-1}{\log(1+z)}\\
&\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\Box
\end{align*}
and the claim follows.
Comment:
*
*In (4) we use the representation (2) of $f_n(x)$
*In (5) we use the generating function of $s(n,k)$ from (3)
Calculation for small values $n=0,1,2$:
Since $$s(0,0)=1,s(1,1)=1,s(2,1)=-1,s(2,2)=1$$
we obtain
\begin{align*}
f_0(x)&=s(0,0)(x-1)=x-1\\
f_1(x)&=s(1,1)\frac{1}{2}(x-1)^2=\frac{1}{2}(x-1)^2\\
f_2(x)&=\frac{1}{2}\left(s(2,1)\frac{1}{2}(x-1)^2+s(2,2)\frac{1}{3}(x-1)^3\right)\\
&=-\frac{1}{4}(x-1)^2+\frac{1}{6}(x-1)^3\\
&=\frac{1}{12}(2x^3-9x^2+12x-5)
\end{align*}
Note: We don't need analytical considerations for the calculations above, since we can state that all calculations are done within the ring of formal power series.
|
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|
Compute definite integral Question: Compute
$$\int_0^1 \frac{\sqrt{x-x^2}}{x+2}dx.$$
Attempt: I've tried various substitutions with no success. Looked for a possible contour integration by converting this into a rational function of $\sin\theta$ and $\cos \theta$.
|
$\displaystyle\int_0^1\frac{x^{1/2}(1-x)^{1/2}}{x+2}dx=\int_0^1\sum_{n=0}^{\infty}(-1)^n\frac{x^n}{2^{n+1}}x^{1/2}(1-x)^{1/2}dx=\sum_{n=0}^{\infty}\frac{(-1)^n}{2^{n+1}}\int_0^1x^{n+\frac{1}{2}}(1-x)^{\frac{1}{2}}dx$
$\displaystyle=\sum_{n=0}^{\infty}\frac{(-1)^n}{2^{n+1}}\beta(n+\frac{3}{2},\frac{3}{2})=\sum_{n=0}^{\infty}\frac{(-1)^n}{2^{n+1}}\frac{\Gamma(n+\frac{3}{2})\Gamma(\frac{3}{2})}{\Gamma(n+3)}=\sum_{n=0}^{\infty}\frac{(-1)^n}{2^{n+3}(n+2)!}(2n+1)\bigg(\frac{(2n)!}{4^nn!}\sqrt{\pi}\bigg)(\sqrt{\pi})$
$\displaystyle=\frac{\pi}{2}\sum_{n=0}^{\infty}(-1)^n\frac{(2n+1)!}{2^{3n+2}(n+2)!n!}=-\frac{\pi}{2}\sum_{n=0}^{\infty}\left(-\frac{1}{8}\right)^{n+1}\frac{1}{n+2}\binom{2n+2}{n+1}$.
Using the generating function for the Catalan numbers, this equals
$\displaystyle-\frac{\pi}{2}\left[\frac{1-\sqrt{1-4x}}{2x}-1\right]_{x=-\frac{1}{8}}=-\frac{\pi}{2}\left[\frac{1-\sqrt{\frac{3}{2}}}{-\frac{1}{4}}-1\right]=\frac{\pi}{2}\left[5-2\sqrt{6}\right]$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Concluding three statements regarding $3$ real numbers.
$\{a,b,c\}\in \mathbb{R},\ a<b<c,\ a+b+c=6 ,\ ab+bc+ac=9$
Conclusion $I.)\ 1<b<3$
Conclusion $II.)\ 2<a<3$
Conclusion $III.)\ 0<c<1$
Options
By the given statements
$\color{green}{a.)\ \text{Only conclusion $I$ can be derived}}$.
$b.)\ $ Only conclusion $II$ can be derived.
$c.)\ $ Only conclusion $III$ can be derived.
$d.)\ $ Conclusions $I,\ II,\ III$ can be derived.
$e.)\ $ None of the three conclusions can be derived.
$\quad\\~\\$
I tried $(a+b+c)^2=36 \implies a^2+b^2+c^2=18$
and found that $(a,b,c)\rightarrow \{(-1,1,4),(-3,0,3)\}$ satisfies the two conditions
$a^2+b^2+c^2,\ a<b<c $ but not this one $a+b+c=6$
I thought a lot but can't find any suitable pairs.
I look for a simple and short way.
I have studied maths upto $12$th grade.
|
I shall show that only Conclusion I is correct. For a fixed $b$, we have
$a+c=6-b$ and $ac=9-(a+c)b=9-(6-b)b=(b-3)^2$. Hence, the quadratic polynomial
$x^2-(6-b)x+(b-3)^2$ has two distinct real roots $x=a$ and $x=c$. Therefore,
the discriminant $(6-b)^2-4(b-3)^2=3b(4-b)$ of this quadratic is strictly
positive (hence, $0<b<4$). Furthermore, the roots of the quadratic are
$a=\dfrac{(6-b)-\sqrt{3b(4-b)}}{2}$ and $c=\dfrac{(6-b)+\sqrt{3b(4-b)}}{2}$. As
$a<b<c$, we must have $\dfrac{(6-b)-\sqrt{3b(4-b)}}{2}<b<\dfrac{(6-b)+\sqrt{3b(4-b)}}{2} $, or equivalently,
$-\sqrt{3b(4-b)}<3b-6<+\sqrt{3b(4-b)}$. Hence,
$\sqrt{3}|b-2|<\sqrt{b(4-b)}$, or $3(b-2)^2<b(4-b)$. Ergo, $4(b-1)(b-3)<0$.
That means $1<b<3$.
Note that $b=2$ gives a solution $(a,b,c)=(2-\sqrt{3},2,2+\sqrt{3})$.
Consequently, Conclusions II and III are false. In fact, we can prove that
$0<a<1<b<3<c<4$. It might be a good exercise for you to show that $0<abc<4$.
|
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|
What is the last digit of $2003^{2003}$?
What is the last digit of this number?
$$2003^{2003}$$
Thanks in advance.
I don't have any kind of idea how to solve this. Except that $3^3$ is $27$.
|
When calculating the last digit of $a\cdot b$ you only need to find the product of the last digit of $a$ and the last digit of $b$.
so $2003^{2003}$ ends in the same digit as $3^{2003}$
$3^1$ ends in $3$
$3^2$ ends in $9$
$3^3$ ends in $7$
$3^4$ ends in $1$
$3^5$ ends in $3$
$3^6$ ends in $9$
$3^7$ ends in $7$
$3^8$ ends in $1$.
So $3^{4k}$ ends in $1$.
From here $3^{2000}$ ends in $1$ (since $2000$ is $4\cdot 500$).
Therefore $3^{2001}$ ends in $3$, $2^{2002}$ ends in $9$ and $3^{2003}$ ends in $7$
|
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|
How to derive $\sum_{n=0}^\infty 1 = -\frac{1}{2}$ without zeta regularization On Wikipedia we find $$\displaystyle \bbox[5px,border:1px solid #F5A029]{1 + 1 + 1+\dots =\sum_{n=0}^\infty 1 = -\frac{1}{2}}$$ using (the rather complicated) zeta-function regularization. I asking for an elementary derivation possibly based on the idea that on average
$$ \sum_{n=0}^\infty (-1)^n =
\begin{cases}
1 & \text{ if }n\text{ is odd} \\
0 & \text{ if }n\text{ is even} \\
\end{cases} = \frac{1}{2} $$
There is a blog discussion that uses very general "cutoff" functions, but I am having a hard time specializing to the case at hand.
A serious question is to qualify where the basic properties of addition break down:
*
*$(a+b)+c = a+(b+c)$ associativity
*$a+b = b+a$ commutativity
*$a + 0 = a = 0 + a$ addition by zero
This example obviously shows we can't use these axioms infinitely many times without generating contradictions. A related question even has $\sum 1 = 0$.
|
For $f(x)=\sum_0^\infty a_n x^n$, a real number $R$, $R\neq 1$, and
$g(x)=f(x)-Rf(x^2)$, if $g(1)$ is Abel summable, that is,
$g(1)=\lim_{x\to 1^-} g(x)$, an elementary Ramanujan sum of $f(1)$ is
defined by
$$
f(1)=\frac{g(1)}{1-R}.
$$
Inversely, for an Abel summable series $g(1)=\sum_0^\infty g_n$ and
a real number $R$, $R\neq 1$, the corresponding elementary Ramanujan
summable series $f(1)$ is defined by
$$
a_0=\frac{g_0}{1-R}~,~~a_{2n+1}=g_{2n+1}~,~~a_{2(n+1)}=g_{2(n+1)}+Ra_{n+1}~.
$$
$f(1)$ is an element of the linear space of elementary Ramanujan summable
series corresponding to $R$, the Ramanujan class $R$.
If $f(1)$ is an element of two distinct classes, $f(1)$ is Abel summable,
that is, an element of the class $R=0$.
One should not naively sum series of distinct classes (here).
For $F_k(x)=f(x^k)$, $F_k(1)=f(1)$ since $\lim_{x\to 1^-} g(x^k)=g(1)$.
$F_k(1)=a_0+0+\cdots +0+a_1+0+\cdots +0+a_2+\cdots$, with $k-1$ zeros between
$a_n$ and $a_{n+1}$. $F_k(1)$ is an element of the class $R$, as $f(1)$.
For $f'(x)=x\frac{df(x)}{dx}$, $f'(1)$ is an element of the class $2R$.
Consider $f(x)=\frac{x}{1-x}=x+x^2+x^3+\cdots$,
$g(x)=\frac{x}{1+x}=x-x^2+x^3-\cdots$ and $R=2$.
$g(1)$ is Abel summable to 1/2 and the elementary sum of $f(1)$
is -1/2.
$f'(1)=1+2+3+\cdots = -1/12$ is an element of the class $R=4$ and
$f''(1)=1+4+9+\cdots = 0$ is an element of the class $R=8$ etc.
Consider now $f(x)=x+2x^2+4x^4+\cdots$, $g(x)=x$ and $R=2$.
The elementary sum of $f(1)=\sum_1^\infty a_n$, with $a_n=n$
if $n$ is a power of 2 and $a_n=0$ if not, is $-1$.
$f'(1)=\sum_1^\infty a_n^2=-1/3$, $f''(1)=\sum_1^\infty a_n^3=-1/7$ etc.
To verify that $1-1+2-2+3-3+\cdots=1/8$,
consider
$$
f(x)=x-x^2+2x^3-2x^4+3x^5-3x^6+\cdots=\frac{1}{(1+x)^2}\frac{x}{1-x}~,
$$
$$
g(x)=\frac{x-x^2-x^3-x^4}{(1+x)^2(1+x^2)^2}
$$
and $R=2$.
Edit.
The consistency of (iterated) elementary Ramanujan summation with analytic continuation of Dirichlet series is shown here (this question is deleted!), here (this question is deleted!) and here.
|
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|
How to determine a basis and the dimension for this vectorspace? Determine a basis and the dimension for the following vectorspace:
\begin{align*} W = \left\{A \in \mathbb{R}^{3 \times 3} \mid A \ \text{is a diagonal matrix and} \ \sum_{i=1}^3 A_{ii} = 0\right\} \end{align*} I know the dimension of all $(3 \times 3)$-diagonal matrices is $3$. A basis for that space would be: \begin{align*} \left\{ \begin{pmatrix} 1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix}, \begin{pmatrix} 0 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 0 \end{pmatrix}, \begin{pmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 1 \end{pmatrix} \right\} \end{align*} But now there is another condition: the trace should be zero. How should I bring that into account? Will the basis be smaller?
|
Note that $A\in W$ if and only if
$$
A=\begin{bmatrix}a&0&0\\0&b&0\\0&0&c\end{bmatrix}
$$
where $a+b+c=0$. It follows that
$$
A=\begin{bmatrix}a&0&0\\0&b&0\\0&0&-a-b\end{bmatrix}
=a\begin{bmatrix}1&0&0\\0&0&0\\0&0&-1\end{bmatrix}+
b\begin{bmatrix}0&0&0\\0&1&0\\0&0&-1\end{bmatrix}
$$
This proves that
$$
\left\{
\begin{bmatrix}1&0&0\\0&0&0\\0&0&-1\end{bmatrix},
\begin{bmatrix}0&0&0\\0&1&0\\0&0&-1\end{bmatrix}
\right\}
$$
spans $W$. Can you prove that these two matrices are linearly independent? Once this is done what can we conclude?
|
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|
How do you multiply this How can you multiply these ordinal numbers:
$(\omega+1)(\omega+1)(\omega2+2)$
I tried and have gotten to this:
$(\omega^2+1)(\omega2+2)$
Is that the correct way, or did i made a mistake?
|
Ordinal multiplication is distributive in the rightmost element. Therefore
$$(\omega+1)(\omega+1)=(\omega+1)\omega+(\omega+1)=\omega^2+\omega+1$$
Therefore:
\begin{align*}
(\omega+1)(\omega+1)(\omega\cdot2+2) &=(\omega+1)(\omega+1)\omega\cdot2+(\omega+1)(\omega+1)2 \\ &=(\omega+1)(\omega+1)\omega+(\omega+1)(\omega+1)\omega+(\omega+1)(\omega+1)2 \\ &= (\omega^2+\omega+1)\omega+(\omega^2+\omega+1)\omega+(\omega^2+\omega+1)+(\omega^2+\omega+1) \\ &= \omega^3+\omega^3+\omega^2+\omega+1+\omega^2+\omega+1 \\ &=\omega^3 \cdot 2+ \omega^2 \cdot 2 + \omega + 1 \end{align*}
|
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|
This general inequality maybe is true? $\sum_{i=1}^{n}\frac{i}{1+a_{1}+\cdots+a_{i}}<\frac{n}{2}\sqrt{\sum_{i=1}^{n}\frac{1}{a_{i}}}$
Let $a_{1},a_{2},\ldots,a_{n}>0$ and prove or disprove
$$\dfrac{1}{1+a_{1}}+\dfrac{2}{1+a_{1}+a_{2}}+\cdots+\dfrac{n}{1+a_{1}+a_{2}+\cdots+a_{n}}\le\dfrac{n}{2}\sqrt{\dfrac{1}{a_{1}}+\dfrac{1}{a_{2}}+\cdots+\dfrac{1}{a_{n}}}\tag{1}$$
This problem from when I solve this two variable inequality
since $n=1$ it is clear
$$\dfrac{1}{1+a_{1}}\le\dfrac{1}{2}\sqrt{\dfrac{1}{a_{1}}}$$
because $1+a_{1}\ge 2\sqrt{a_{1}}$
$n=2$ case,can see this links my answer.
For general
simaler this two variable inequality methods, then I use Cauchy-Schwarz inequality we have
$$\left(\sum_{i=1}^{n}\dfrac{1}{1+a_{1}+\cdots+a_{i}}\right)^2\le\left(\sum_{i=1}^{n}\dfrac{1}{a_{i}}\right)\cdot\left(\sum_{i=1}^{n}\dfrac{i^2a_{i}}{(1+a_{1}+a_{2}+\cdots+a_{i})^2}\right)$$
it suffices to show that
$$\sum_{i=1}^{n}\dfrac{i^2a_{i}}{(1+a_{1}+\cdots+a_{i})^2}\le\dfrac{n^2}{4}\tag{2}$$
it seem hard.
because I tried following also fail;
$$\sum_{i=1}^{n}\dfrac{i^2a_{i}}{(1+a_{1}+\cdots+a_{i})^2}<\sum_{i=1}^{n}i^2\left(\dfrac{1}{1+a_{1}+\cdots+a_{i-1}}-\dfrac{1}{1+a_{1}+a_{2}+\cdots+a_{i}}\right)$$
and use Abel transformation.not can to prove $(2)$,
Note $(1)$ Left side hand was simaler Hardy's inequality when $p=-1$,But there are different problem.
EDIT:Numerical tests $(2)$ is not right.so my idea can't works
|
ONLY AN IDEA (and no solution, or at least a partial solution):
From this question (thx to the comments) we have
\begin{align*}
\sum_{k=1}^n\frac{k}{1+a_1+\ldots+a_k}\leq
\sum_{k=1}^n\frac{k}{(1/n+a_1)+\ldots+(1/n+a_k)}\leq2\left(\frac{1}{1/n+a_1}+\ldots+\frac{1}{1/n+a_n}\right)=2x,
\end{align*}
where $x:=\sum_{k=1}^n\frac{1}{1/n+a_k}$. Since
\begin{align*}
\sum_{k=1}^n\frac{1}{1/n+a_k}\leq\sum_{k=1}^n\frac{1}{1/n}=n^2,
\end{align*}
we have $x/n^2\leq 1$ and thus $x/n^2\leq\sqrt{x/n^2}$. Consequently, $x\leq n\sqrt{x}$, and this implies
\begin{align*}
\sum_{k=1}^n\frac{k}{1+a_1+\ldots+a_k}\leq2n\cdot\sqrt{\sum_{k=1}^n\frac{1}{1/n+a_k}}\leq2n\cdot\sqrt{\sum_{k=1}^n\frac{1}{a_k}}.
\end{align*}
|
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|
Solve integrals using residue theorem? $$\int_{0}^{\pi}\frac{d\theta }{2+\cos\theta}$$
$$\int_{0}^{\infty}\frac{x }{(1+x)^6} dx$$
My problem is that I don't know how to start solving these integrals, or how to convert them into usual types that can be solved.
|
Hint for the First one First compute
$$
I=\int_0^{2\pi} \frac{d\theta}{2 + \cos\theta}=\int_0^{2\pi}R(\cos(\theta), \sin(\theta)) d\theta
$$
Where $R$ is the rational function given by $$R(x,y)=\frac{1}{2+x}$$
How to do this using the residue theorem? Put $z=e^{it}=\cos(t)+i\sin(t)$, thus
$$
\cos(t)=Re(z)=\frac{z+z^{-1}}{2}, \ \\ \sin(t)=Im(z)=\frac{z-z^{-1}}{2i}, \\ dz=ie^{it} dt=iz \ dt \Longrightarrow dt=\frac{1}{iz}dz
$$
Then $I$ can be seen as a contour integral, solve it by using residues
$$
I=\int_0^{2\pi}R(\cos(t), \sin(t)) dt= \int_{|z|=1} R\left(\frac{z+1/z}{2}, \frac{z-1/z}{2i} \right)\frac{1}{iz}dz
$$
Hence in your case the integral you will compute is
$$
I= \int_{|z|=1} \left(\frac{1}{2+\frac{z+1/z}{2}} \right)\frac{1}{iz}dz = \frac{2}{i} \int_{|z|=1} \frac{dz}{z^2+4z+1}
$$
which can be easily obtain by the residue theorem!
Finally: Note that
$$
\int_0^{\pi} \frac{d\theta}{2 + \cos\theta}=\frac{I}{2}
$$
and hence your result follows by computing the next integral
$$
\int_0^{\pi} \frac{d\theta}{2 + \cos\theta} = \frac{1}{i} \int_{|z|=1} \frac{dz}{z^2+4z+1}
$$
Spoiler Solution
Since $z^2+4z+1=(z-(\sqrt{3}-2))(z-(-\sqrt{3}-2))$ and $z_0=\sqrt{3}-2$ is the only root inside the contour $|z|=1$, then by the residue theorem we have $\displaystyle \int_{|z|=1} \frac{dz}{z^2+4z+1} = 2\pi i \left( \frac{1}{2\sqrt{3}}\right)=i \frac{\pi}{\sqrt{3}}$. Hence we get that the final result is $\displaystyle \int_0^{\pi} \frac{d\theta}{2 + \cos\theta} =\frac{1}{i} \times i \frac{\pi}{\sqrt{3}}=\frac{\pi}{\sqrt{3}}$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1344724",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Convert the equation to rectangular form $r = \frac {6}{1-\sinθ}$ Convert the equation to rectangular form $r = \frac {6}{1-\sinθ}$
The answer should be: $y = \frac{1}{12} x^2 -3$
But how to arrive at the answer?
I tried replacing r with $\sqrt{x^2 + y^2}$, then $\sinθ$ as $\frac{y}{r}$ but to no avail.
I also ended up with
$r = \frac{6}{\frac{r-y}{r}} $ ->
$ 1 = \frac{6}{\sqrt{x^2+y^2}-y}$ -> x = +- 6
but that's not the answer...
|
Write
$$6=r(1-\sin\theta)=r-r\sin\theta=\sqrt{x^2+y^2}-y$$
Then
$$6+y=\sqrt{x^2+y^2}\implies12y+36=x^2$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1345356",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Solving a system of non-linear equations Let
$$(\star)\begin{cases}
\begin{vmatrix}
x&y\\
z&x\\
\end{vmatrix}=1, \\
\begin{vmatrix}
y&z\\
x&y\\
\end{vmatrix}=2, \\
\begin{vmatrix}
z&x\\
y&z\\
\end{vmatrix}=3.
\end{cases}$$
Solving the above system of three non-linear equations with three unknowns.
I have a try.
Let$$A=\begin{bmatrix}
1& 1/2& -1/2\\
1/2& 1& -1/2\\
-1/2& -1/2& -1
\end{bmatrix}$$
We have $$(x,y,z)A\begin{pmatrix}
x\\
y\\
z
\end{pmatrix}=0.$$
There must be a orthogonal matrix $T$,such that $T^{-1}A T=diag \begin{Bmatrix}
\frac{1}{2},\frac{\sqrt{33}+1}{4},-\frac{\sqrt{33}-1}{4}
\end{Bmatrix}.$
$$\begin{pmatrix}
x\\
y\\
z
\end{pmatrix}=T\begin{pmatrix}
x^{'}\\
y^{'}\\
z^{'}
\end{pmatrix}\Longrightarrow\frac{1}{2} {x'}^{2}+\frac{\sqrt{33}+1}{4} {y'}^{2}-\frac{\sqrt{33}-1}{4}{z'}^{2}=0.$$
But even if we find a $\begin{pmatrix}
x_0^{'}\\
y_0^{'}\\
z_0^{'}
\end{pmatrix} $ satisfying $\frac{1}{2} {x_0'}^{2}+\frac{\sqrt{33}+1}{4} {y_0'}^{2}-\frac{\sqrt{33}-1}{4}{z_0'}^{2}=0,\begin{pmatrix}
x_0\\
y_0\\
z_0
\end{pmatrix}=T\begin{pmatrix}
x_0^{'}\\
y_0^{'}\\
z_0^{'}
\end{pmatrix}$ may not be the solution of $(\star)$
If you have some good ideas,please give me some hints. Any help would be appreciated!
|
The resultant of $x^2-yz-1$ and $y^2-xz-2$ with respect to $z$ is $x^3-y^3-x+2y$. The resultant of $x^2 - yz - 1$ and $z^2 - xy - 3$ with respect to $z$ is $x^4-x y^3-2 x^2-3 y^2+1$. The resultant of $x^3-y^3-x+2y$ and
$x^4-x y^3-2 x^2-3 y^2+1$ with respect to $x$ is $-y^4(18 y^2-1)$. So either
$y = 0$ or $y = \pm 1/\sqrt{18}$.
With $y=0$ we get $x^2 - 1 = 0$, $-xz-2 = 0$ and $z^2 - 3 = 0$, which clearly will not work.
With $y = \pm 1/\sqrt{18} = \sqrt{2}/6$ we do get
solutions: $x = \mp 5 \sqrt{2}/6$, $z = \pm 7 \sqrt{2}/6$.
|
{
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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|
value of an $\sum_3^\infty\frac{3n-4}{(n-2)(n-1)n}$ I ran into this sum $$\sum_{n=3}^{\infty} \frac{3n-4}{n(n-1)(n-2)}$$
I tried to derive it from a standard sequence using integration and derivatives, but couldn't find a proper function to describe it.
Any ideas?
|
Setting
$$\frac{3n-4}{n(n-1)(n-2)}=\frac{A(n-1)-B}{(n-2)(n-1)}-\frac{An-B}{(n-1)n}$$
gives you $A=3,B=2$, i.e.
$$\frac{3n-4}{n(n-1)(n-2)}=\frac{3(n-1)-2}{(n-2)(n-1)}-\frac{3n-2}{(n-1)n}.$$
Hence, we have
$$\begin{align}\sum_{n=3}^{\infty}\frac{3n-4}{n(n-1)(n-2)}&=\lim_{m\to\infty}\sum_{n=3}^{m}\left(\frac{3(n-1)-2}{(n-2)(n-1)}-\frac{3n-2}{(n-1)n}\right)\\&=\lim_{m\to\infty}\left(\frac{3\cdot 2-2}{1\cdot 2}-\frac{3m-2}{(m-1)m}\right)\\&=2\end{align}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1348046",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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|
Expanding $\frac{2x^2}{1+x^3}$ to series So I was doing some series expansion problems and stumbled upon this one ( the problem is from Pauls Online Notes )
$$f(x) = \frac{2x^2}{1+x^3}$$
The actual solution to this problem uses a different method then mine. Since this function is really seams nice for integration, i thought i could do:
$$h(x) = \int f(x) dx =\int \frac{2x^2}{1+x^3}dx = \frac{2}{3} \ln({1+x^3})$$
If we let $t=x^3$, and use Maclaurin series for $\ln{(1+t)}$ we have
$$\frac{2}{3} \ln({1+t}) = \frac{2}{3} \sum_{n=1}^{+\infty}(-1)^{n-1} \frac{t^n}{n} $$
No we can take the derivative part by part in the radius of convergence we will get the original function series expansion:
$$f(x)= \frac{2}{3} \sum_{n=1}^{+\infty}(-1)^{n-1} t^{n-1} = \frac{2}{3} \sum_{n=1}^{+\infty}(-1)^{n-1} (x^3)^{n-1}$$
But the solution from the course is :
$$\sum_{n=0}^{+\infty}2(-1)^{n} x^{3n+2}$$
Where did i go wrong? :(
Thanks.
|
Intelligent question. The problem is that you first differentiate with respect to $t$, then substitute $t$ back for $x^3$. The correct approach is the reverse order: replace $t$ by $x^3$ and derive with respect to $x$. Otherwise, your solution is nice. Below you will find a slightly more direct one.
Assume $x \ne 0$. Then
$$\frac {2 x^2} {1+x^3} = \frac 1 x \frac {2 x^3} {1+x^3} = 2 - \frac 2 {1+x^3} = \frac 2 x - \frac 2 x \sum \limits _{n=0} ^\infty (-1)^n x^{3n} = - \frac 2 x \sum \limits _{n=1} ^\infty (-1)^n x^{3n} = \\ -\frac 2 x \sum \limits _{m=0} ^\infty (-1)^{m+1} x^{3m+3} = 2 \sum \limits _{m=0} ^\infty (-1)^m x^{3m+2} .$$
All computations are assumed to happen where they are defined (i.e. inside the interval of convergence).
Finally, you can see the this equality is also valid for $x=0$.
|
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"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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|
Solving for y' in a fraction Given the equation $x+xy^2 = \tan^{-1}(x^2y)$ find $y'$.
I have tried doing this but solving for $y'$ I need some help and would like your advice.
Work so far...
$$1+y^2+2xy\left(\frac{dy}{dx}\right)= \frac{2xy+x^2\left(\frac{dy}{dx}\right)}{1+x^4y^2}$$
What can be done now to solve for $y'$?
|
So you have the expression
$$ 1 + y^{2} + 2xy(\dfrac{dy}{dx}) = \dfrac{2xy + x^{2}(\dfrac{dy}{dx})}{1 + x^{4}y^{2}}$$
And we can break up the term on the left hand side in this way: $$\dfrac{2xy + x^{2}(\dfrac{dy}{dx})}{1 + x^{4}y^{2}} = \dfrac{2xy }{1 + x^{4}y^{2}} + \dfrac{x^{2}(\dfrac{dy}{dx})}{1 + x^{4}y^{2}} $$
Then, subtracting $\dfrac{2xy }{1 + x^{4}y^{2}}$ from both sides of the original equation, and subtracting $2xy(\dfrac{dy}{dx})$ from both sides, the first equation becomes:
$$ 1 + y^{2} - \dfrac{2xy }{1 + x^{4}y^{2}} = \dfrac{x^{2}(\dfrac{dy}{dx})}{1 + x^{4}y^{2}} - 2xy(\dfrac{dy}{dx})$$
Finally, you can factor $\dfrac{dy}{dx}$ out from the right hand side, and divide both sides by the other factor to solve for $\dfrac{dy}{dx}$.
You should get:
$$ \dfrac{1 + y^{2} - \dfrac{2xy }{1 + x^{4}y^{2}}}{\dfrac{x^{2}}{1 + x^{4}y^{2}} - 2xy} = \dfrac{dy}{dx} $$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1351689",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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|
Stability analysis of Ralston's method Ralston's method is given by:
$$y_{n+1} = y_n + \frac{h}3(f(t_n,y_n)+2f(t_n+\frac34h, y_n + \frac34h f(t_n,y_n)))$$
carry out a stability analysis of this method to determine the condition for stability based on the test problem:
$$\frac{dy}{dt} -\lambda y$$
From what I understand I need to use the test problem to determine rewrite the method without the $f$ function. I have that
$f(t_n,y_n) = -\lambda y_n$
$f(t_n+\frac34h, y_n + \frac34h f(t_n,y_n)) = f(t_n+\frac34h, y_n + \frac34h (-\lambda y_n))$
I'm not sure how to simplify this second $f$ equation though.
Any help would really be appreciated!
|
Usually, stability of a method is studied on a test ODE
$$
y' = f(t,y) \quad\Leftrightarrow\quad y' - f(t,y) = 0
$$
with $f(t,y) = \lambda y,\; \lambda \in \mathbb{C}^{-}$.
So
$$
f(t_n, y_n) = \lambda y_n\\
f(t_n + \tfrac{3}{4}h, y_n + \tfrac{3}{4}h f(t_n, y_n)) =
f(t_n + \tfrac{3}{4}h, y_n + \tfrac{3}{4}h \lambda y_n) =
\lambda( y_n + \tfrac{3}{4}h \lambda y_n) =
\lambda y_n(1 + \tfrac{3}{4}h \lambda)
$$
Now
$$
y_{n+1} = y_n + \tfrac{h}{3}\left(\lambda y_n + 2 \lambda y_n(1 + \tfrac{3}{4}h\lambda)\right) = y_n\left(1 + \lambda h + \frac{(\lambda h)^2}{2}\right) = y_n r(\lambda h)\\
r(z) = 1 + z + \frac{z^2}{2}.
$$
The function $r(z)$ is the stability function of the method.
Also, one can note that the given method falls in Runge-Kutta family with following Butcher's tableau
$$
\begin{array}{c|c}
\mathbf{c} & \mathbf{A}\\
\hline
& \mathbf{b}^\top
\end{array}
=
\begin{array}{c|cc}
0 & 0 & 0 \\
\tfrac{3}{4} & \tfrac{3}{4} & 0\\
\hline
& \tfrac{1}{3} & \tfrac{2}{3}
\end{array}
$$
and for Runge-Kutta methods the stability function can be found directly from their Butcher's tableau by
$$
r(z) = \frac{\operatorname{det}(\mathbf{E} - z\mathbf{A} + z\mathbf{e}\mathbf{b}^\top)}{\operatorname{det}(\mathbf{E} - z\mathbf{A})} =
\frac{\begin{vmatrix}1 +\tfrac{z}{3} & \tfrac{2z}{3}\\-\tfrac{3z}{4}+\tfrac{z}{3}& 1+\tfrac{2z}{3}\end{vmatrix}}{\begin{vmatrix}1 & 0\\-\tfrac{3z}{4}& 1\end{vmatrix}} = \left(1 + \frac{z}{3}\right)\left(1 + \frac{2z}{3}\right) + \frac{5z}{12} \frac{2z}{3} = \\ = 1 + z + \frac{2z^2}{9} + \frac{5z^2}{18} = 1 + z + \frac{z^2}{2}
$$
It is easy to check that this RK method is an explicit two-stage second order method. From the former one can deduce that $r(z)$ is some quadratic polynomial in $z$, and from the latter
$$
r(z) = e^z + o(z^2).
$$
That gives the only possibility for $r(z) = 1 + z + \tfrac{z^2}{2}$ which is the stability function of every explicit two-stage second order Runge-Kutta method.
|
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"url": "https://math.stackexchange.com/questions/1354505",
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|
System of Congruences with Special Symmetry Show that the following system of congruences
\begin{align}
\begin{cases}
3 x^4 - 7 x^2 y^2 - 7 x^2 z^2 - 35 y^2 z^2 \equiv 0 \pmod{p} \\
3 y^4 - 7 x^2 y^2 - 7 y^2 z^2 - 35 x^2 z^2 \equiv 0 \pmod{p} \\
3 z^4 - 7 y^2 z^2 - 7 x^2 z^2 - 35 x^2 y^2 \equiv 0 \pmod{p}
\end{cases}
\end{align}
has only the trivial solution when $p$ is a prime distinct from $2, 3, 5,$ and $23$.
Note: To obtain the restriction $p \neq 2, 23$, suppose that the system has a solution in which $x = y = z$. Then
\begin{align}
3 x^4 - 7 x^2 y^2 - 7 x^2 z^2 - 35 y^2 z^2 \equiv -46 x^4 \equiv 0 \pmod{p}
\end{align}
has nontrivial solutions only when $p = 2, 23$.
I suspect that the symmetry in this system will help in showing that it has only the trivial solution at the other primes.
|
The symmetries are, indeed, the key. I let the variables range over the field $\Bbb{Z}_p$ to avoid having to write $\pmod p$ all the time.
Assume first that the squares $x^2,y^2,z^2$ are all pairwise non-congruent modulo $p$. Then subtracting your second equation from the first gives
$$
0=3x^4-3y^4+28x^2z^2-28y^2z^2=(x^2-y^2)(3x^2+3y^2+28z^2).
$$
Given our assumption we can cancel the factor $x^2-y^2$ and end up with the congruence
$$
3x^2+3y^2+28z^2=0.\qquad(*)
$$
Similarly operations with other pairs lead to the congruences, where the variables $x,y,z$ switch roles.
We thus arrive at a system of three congruences. These are "linear" in the unknowns $u=x^2$, $v=y^2$ and $w=z^2$. Let us denote the $3\times3$ matrix
$$
A=\left(\begin{array}{ccc}28&3&3\\3&28&3\\3&3&28\end{array}\right).
$$
We have
$$
\det A=21250=2\cdot5^4\cdot17,
$$
so the system
$$
A\left(\begin{array}{c}u\\v\\w\end{array}\right)=\left(\begin{array}{c}0\\0\\0\end{array}\right)
$$
has no non-trivial solutions unless $p\in\{2,5,17\}$. Two first of these possibilities were excluded at the starting gate. If $p=17$, then $A$ has rank $2$, so the solution space is 1-dimensional and clearly consists of the vectors $u=v=w$. But it is easy to check that the original system as no solutions with $x^2=y^2=z^2$ when $p=17$ (you already did that).
The remaining possibility is that some pair of squares of the variables are equal. By symmetry it suffices to handle the case $x^2=y^2$. In that case the original system becomes the pair
$$
\begin{cases}
-4 x^4 - 42 x^2 z^2 =0 \\
3 z^4 - 14 x^2 z^2 - 35 x^4 =0
\end{cases}
$$
If $x^2=0$ then the latter congruence implies $z=0$ also. Otherwise we can solve from the first congruence that $2x^2=-21z^2$. Multiplying the second congruence by four and plugging this in yields $14835z^4=0$. The original system has no non-trivial solutions with $x^2=y^2$ when $z=0$ so we are left with the case
$$
14835=3\cdot5\cdot7\cdot137=0.
$$
In the case $p=7$ the original system immediately tells that only the trivial solution is there. The case $p=137$ remains. But if $p=137$ then
$$
2x^2\equiv -21 z^2\equiv116z^2\implies x^2\equiv 58z^2.
$$
This has non-trivial solutions only if $58$ is a quadratic residue modulo $137$. It is easy to verify that this is not the case. Clearly $58=2\cdot29$. Because $137\equiv1\pmod 8$ we know that $2$ is a quadratic residue modulo $137$. With $29$ we use the law of quadratic reciprocity
$$
\left(\frac{29}{137}\right)=\left(\frac{137}{29}\right)=\left(\frac{21}{29}\right).
$$
As $29\equiv1\pmod7$ another round of reciprocity gives that $\left(\dfrac 7{29}\right)=1$. Similarly because $29\equiv2\pmod3$ we see that $\left(\dfrac 3{29}\right)=-1.$ The claim follows from this.
|
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"timestamp": "2023-03-29T00:00:00",
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|
$ \lim_{x \to \infty} x(\sqrt{2x^2+1}-x\sqrt{2})$ Any ideas fot evaluating:
$$ \lim_{x \to \infty} x(\sqrt{2x^2+1}-x\sqrt{2})$$
thanks.
|
While one way to proceed here is to write the term
$\sqrt{2x^2+1}-\sqrt{2}x=\frac{1}{\sqrt{2x^2+1}+\sqrt{2}x}$
a second way is to write
$$\sqrt{2x^2+1}=\sqrt{2}x\left(1+\frac{1}{4x^2}+O\left(\frac{1}{x^4}\right)\right)$$
so that
$$\begin{align}
x\left(\sqrt{2x^2+1}-\sqrt{2}x\right)&=\sqrt{2}x^2\left(1+\frac{1}{4x^2}+O\left(\frac{1}{x^4}\right)\right)-\sqrt{2}x^2\\\\
&=\frac{\sqrt{2}}{4}+O\left(\frac{1}{x^2}\right)\\\\
&\to \frac{\sqrt{2}}{4}
\end{align}$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Shortest distance of the point $(0,c)$ from the parabola $y=x^2$ Shortest distance of the point $(0,c)$ from the parabola $y=x^2$ ? (Where $0\le c \le5$)
My approach: I wrote the distance formula by taking parametric coordinates as $(t,t^2)$ and then differentiated the equation.I got the extremum as $x=\sqrt (c)$,but that's the wrong answer? I can't figure out my mistake.
Please help!
|
Given is
$$
y = x^2
$$
So the normal for point $(x_o,x_o^2)$ can be written as
$$
(x,y) = (x_o - 2\alpha x_o, x_o^2 + \alpha )
$$
The length of the normal from point $(x_o,x_o^2)$ is given by
$$
\ell = \alpha \sqrt{4 x_o^2 + 1}
$$
We need to solve
$$
(0,c) = (x_o - 2\alpha x_o, x_o^2 + \alpha )
$$
or
$$
\left[
\begin{array}{rcl}
x_o - 2 \alpha x_o &=& 0\\
c &=& x_o^2 + \alpha
\end{array}
\right.
$$
From
$$
x_o - 2 \alpha x_o = 0
$$
follows $x_o$ or $\alpha = \tfrac{1}{2}$.
Case $x_o=0$
We obtain
$$
\left[
\begin{array}{rcl}
0 &=& 0\\
c &=& \alpha
\end{array}
\right.
$$
Thus $\alpha = c$ and $x_o=0$. So the length is given by
$$
\ell = c
$$
Case $\alpha = \tfrac{1}{2}$
We obtain
$$
\left[
\begin{array}{rcl}
0 &=& 0\\
c &=& x_o^2 + \frac{1}{2}
\end{array}
\right.
$$
Thus $\alpha = \tfrac{1}{2}$ and $x_o = \sqrt{c - \tfrac{1}{2}}$. The case $c<\tfrac{1}{2}$ givens the $x_o=0$ solution.
The case $c \ge \frac{1}{2}$ gives the length
$$
\ell = \sqrt { c^2 - \tfrac{1}{4} }
$$
Note that $c \ge \sqrt{ c^2 - \tfrac{1}{4}}$ for $c \ge \frac{1}{2}$, so the shortest distance is given by
$$
c \textrm{ for $c \le \tfrac{1}{2}$}
$$
or
$$
\sqrt{c^2 - \tfrac{1}{4}} \textrm{ for $c \ge \tfrac{1}{2}$}
$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Probability that team $A$ has more points than team $B$
Seven teams play a soccer tournament in which each team plays every other team exactly once. No ties occur, each team has a $50\%$ chance of winning each game it plays, and the outcomes of the games are independent. In each game, the winner is awarded a point and the loser gets 0 points. The total points are accumulated to decide the ranks of the teams. In the first game of the tournament, team $A$ beats team $B.$ The probability that team $A$ finishes with more points than team $B$ is $m/n,$ where $m$ and $n$ are relatively prime positive integers. Find $m+n.$
I got that since, Team $B$ already has one loss, it doesnt matter how many games team $B$ wins.
We must find the probability that team $A$ wins the rest $5$ games.
Since it says: "outcome of games is independent, I am confused."
My first approach was:
$$P(\text{A Wins 5}) = \frac{1}{32} \implies m + n = 33$$
This was wrong.
Second approach.
Suppose $A$ has a match with Team $C$.
$$P(\text{Team A wins, Team C loses}) = \frac{1}{2}\frac{1}{2} = \frac{1}{4}$$
But then overall: $\frac{1}{4^5} > 1000$ too big of an answer ($m + n < 1000$ requirement).
HINTS ONLY PLEASE!!
EDIT:
I did some casework and the work is very messy and I don't think I got the right answer anyway. I have to find:
$$\binom{5}{1} = 5, \binom{5}{2} = 10, \binom{5}{3} = 10, \binom{5}{4} = 5, \binom{5}{5} = 1.$$
Let $A = x$
$$P(B=1, x \ge 1) + P(B = 2, x\ge 2) + P(B=3, x\ge 3) + ... + P(B=5, x = 5)$$
$$P(B=1, x \ge 1) = \binom{5}{1}(0.5)^{5} \cdot \bigg(\binom{5}{1} (0.5)^5 + \binom{5}{2} (0.5)^5 + ... + (0.5)^5\binom{5}{1} \bigg) = \frac{5}{1024} \cdot \bigg(31\bigg) = \frac{155}{1024} $$
$$P_2 = \frac{10}{1024} \bigg(\binom{5}{2} + ... + \binom{5}{5}\bigg) = \frac{260}{1024}$$
$$P_3 = \frac{10}{1024} \bigg(\binom{5}{3} + ... + \binom{5}{5} \bigg) = \frac{160}{1024}$$
$$P_4 = \frac{5}{1024} \bigg( \binom{5}{4} + \binom{5}{5}\bigg) = \frac{60}{1024}$$
$$P_5 = \frac{1}{1024} \bigg( 1\bigg) = \frac{1}{1024}$$
$$P(\text{Total}) = \frac{636}{1024} = \frac{318}{512} = \text{wrong}$$
What is wrong with this method?
|
$A$ and $B$ have played one game against each pther, and $B$ lpst.
Bpth teams have each to play against the remaining five teams (independently). Let $X$ be the count of those games $A$ wins (ie: the points), and let $Y$ be the count of games $B$ wins. Since $A$ has at least one point from the game between the two teams, you wish to find: $\;\mathsf P(X+1>Y)\;$.
Independence means that how many and which teams $A$ wins against has no influence on the probabilities of $B$ winning against any team.
Hint: The random variables $X$, $Y$ have iid binomial distributions, parameters $p=0.5, n=5$. $$X,Y\mathop{\sim}^{iid}\mathcal{Bin}(5, 0.5)$$
$$\begin{align}
m/n & = \mathsf P(X+1> Y)
\\[1ex] & = \sum_{x=0}^5 \mathsf P(X=x \cap Y< x+1)
\\[1ex]
& = \sum_{x=0}^5 \mathsf P(X=x) \sum_{y=0}^{x}\mathsf P(Y=y)
\end{align}$$
The rest is left to you.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1358776",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
}
|
Integer solutions for $x^2+y^2=208$ Which steps I can follow to find the integer solutions for the equation $x^2 + y^2 = 208$?
|
A positive integer is the sum of two squares iff primes $\equiv 3\pmod 4$ (following the Fermat's theorem on sums of two squares) of its factorization are raised to even exponents.
$$208=2^4\cdot 13$$
So, $208$ meets the condition.
Now, $13=3^2+2^2$, so $208=4^2(3^2+2^2)=12^2+8^2$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1359674",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
}
|
Finding the area between : $\frac{\pi}{3}\leq \theta \leq \frac{2 \pi}{3}, 0 \leq r \leq 1$ using polar coordinates
I need to find the area between : $\frac{\pi}{3}\leq \theta \leq \frac{2 \pi}{3}, 0 \leq r \leq 1$ using polar coordinates $x=r\cos \theta , \, y=r\sin \theta$
$$x^2+y^2=r \Longrightarrow r=1$$
My attempt:
$$\iint (r\cdot J )\, dr \,d\theta$$
Using double integral:
$$\color{blue}{J=r=1}:$$
$$\int\limits_{y=\frac{\pi}{3}}^{y=\frac{2\pi}{3}}\left[~\int\limits_{x=0}^{x=1} r^2\,dr \right]d\theta=\int\limits_{y=\frac{\pi}{3}}^{y=\frac{2\pi}{3}} \frac{r^3}{3} \,d\theta=\frac{r^4}{12}\bigg|_\frac{\pi}{3}^\frac{2\pi}{3}=\boxed{\color{blue} {\frac{\pi}{36}}} $$
The answer should be : $\boxed{\color{red} {\frac{\pi}{6}}} $
Where am I wrong?
|
Here's how it should look:
$$\int_{\pi/3}^{2 \pi / 3} \int_0^1 \color{blue}{r} \, dr \, d \theta = \int_{\pi/3}^{2 \pi/3} \left. \color{blue}{\frac{r^2}{2}} \right|_0^1 d \theta$$
$$= \int_{\pi/3}^{2 \pi/3} \frac{1}{2} \, d \theta = \left. \frac{1}{2} \theta \right|_{\pi/3}^{2\pi/3} = \color{red}{\frac{\pi}{6}}$$
You did two things wrong: One, you had $r^2$ instead of $r$.
Second, you did not evaluate inner definite integral; instead you took the indefinite integral and then used that result in the outer integral.
Visually, the inner integral represents a straight line from the origin that is $1$ unit long. Then, your $d \theta$ integral takes that straight line and "sweeps" along the origin at angle $\frac{\pi}{3}$ to the angle $\frac{2 \pi}{3}$. This is how the area of the region is found.
I am unfamiliar with the $J$ notation. If you could explain it I could tell you why it gave you a wrong answer.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1359921",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
}
|
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