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A limit involving $\cos x$ and $x^2$ The question is finding the value of $L = \lim_{x \to 0}\frac{1 - \cos x \cos 2x \cos 3x}{x^2}$ if it exists . I've found the answer using taylor's formula but I'm looking for other solutions maybe using trigonometric identities .
|
Observe you have
\begin{align}
\cos x \cos 2x \cos 3x= \cos x(\cos 5x+\sin 2x \sin 3x)
\end{align}
and
\begin{align}
\cos x \cos 5x =\cos 6x+ \sin x\sin 5x
\end{align}
which means
\begin{align}
\frac{1-\cos x \cos 2x \cos 3x}{x^2} =&\ \frac{1-\cos 6x}{x^2}-\frac{\sin x\sin 5x}{x^2}-\frac{\cos x \sin 2x\sin 3x}{x^2}\\
=&\ 18\frac{\sin^2 3x}{(3x)^2}-5\frac{\sin x\sin 5x}{x(5x)}-6\frac{\cos x \sin 2x\sin 3x}{(2x)(3x)}
\end{align}
In the limit, we have
\begin{align}
18-5 - 6= 7
\end{align}
|
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"timestamp": "2023-03-29T00:00:00",
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|
British Maths Olympiad (BMO) 2004 Round 1 Question 1 alternative approaches? The questions states:
Solve the simultaneous equations (which I respectively label as $
> \ref{1}, \ref{2}, \ref{3}, \ref{4}$)
$$\begin{align} ab + c + d &= 3 \tag{1} \label{1} \\ bc + d + a &= 5
\tag{2} \label{2} \\ cd + a + b &= 2 \tag{3} \label{3} \\ da + b + c
&= 6 \tag{4} \label{4} \end{align}$$
where $a,b,c,d$ are real numbers.
I solved this system after quite a while by taking
$eqns$ 1 - 3 = $eqns$ 4 - 2
which yields $a + c = 2$
You can then substitute that in and find the other variables
I also noticed that $(a+1)(b+1) + (a+1)(d+1) + (c+1)(b+1) + (c+1)(d+1) = 20$ but that line didnt really help me.
I'm interested in seeing the other approaches people can take with this system.
Additionally, is there a sufficient enough hint to take another route? Did I miss an easy solution?
|
My approach was to set $A=a-1,B=b-1,C=c-1,D=d-1$ so that we get
\begin{align}
AB + A + B + C + D &= 0 \\
BC + A + B + C + D &= 2 \\
CD + A + B + C + D &= -1 \\
AD + A + B + C + D &= 3.
\end{align}
Letting $S = A+B+C+D$, we have $ABCD = (-S)(-1-S) = (2-S)(3-S)$, so $S=1$, and therefore
\begin{align}
AB &= -1 \\
BC &= 1 \\
CD &= -2 \\
AD &= 2.
\end{align}
This gives us $A = -\frac1B$, $C = \frac1B$, and $D = -\frac2C = -2B$.
From $A+B+C+D=1$, we have $-\frac1B + B + \frac1B - 2B = 1$, or $B = -1$. Then we can solve for $A,C,D$ and finally get $a,b,c,d$.
|
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|
How to find the particular solution of the differential equation? $$\frac{dy}{dx} = 1 + \frac{y}{x} + \left(\frac{y}{x}\right)^2 , ~~ y(1)=1$$
Thank you
|
We start by setting a variable, $z$ equal to $\frac{y}{x}$
$$\frac{{\rm d}y}{{\rm d}x} = 1 + z + z^2$$
Next, find $\frac{{\rm d}z}{{\rm d}x}$.
$$\frac{{\rm d}z}{{\rm d}x} = \frac{x \frac{{\rm d}y}{{\rm d}x} - y}{x^2} = \frac{1}{x} \frac{{\rm d}y}{{\rm d}x} - \frac{y}{x^2} = \frac{1}{x}\left(\frac{{\rm d}y}{{\rm d}x} - z\right)$$
Now, solve for $\frac{{\rm d}y}{{\rm d}x}$.
$$\frac{{\rm d}z}{{\rm d}x} = \frac{1}{x}\left(\frac{{\rm d}y}{{\rm d}x} - z\right)$$
$$x \frac{{\rm d}z}{{\rm d}x} = \frac{{\rm d}y}{{\rm d}x} - z$$
$$\frac{{\rm d}y}{{\rm d}x} = z + x \frac{{\rm d}z}{{\rm d}x}$$
Substitute this into the original equation.
$$z + x \frac{{\rm d}z}{{\rm d}x} = 1 + z + z^2$$
$$x \frac{{\rm d}z}{{\rm d}x} = 1 + z^2$$
Now, solve this as a separable differential equation.
$$\frac{{\rm d}z}{1 + z^2} = \frac{{\rm d}x}{x}$$
$$\int\frac{{\rm d}z}{1 + z^2} = \int\frac{{\rm d}x}{x}$$
$$\arctan(z) = \ln(x) + C$$
$$z = \tan(\ln(x) + C)$$
Now, we finally substitute $\frac{y}{x}$ for $z$.
$$\frac{y}{x} = \tan(\ln(x) + C)$$
$$y = x\tan(\ln(x) + C)$$
To find $C$, we substitute $x=1$ and $y=1$.
$$1=1\cdot\tan(\ln(1) + C)$$
$$\arctan 1=C$$
$$C=\frac{\pi}{4}$$
Therefore,
$$y = x\tan\left(\ln(x) + \frac{\pi}{4}\right)$$
|
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|
Solving $x^2u_x + y^2u_v + z(x+y)u_z = 0$ Exercise :
Solve the PDE
$$x^2u_x + y^2u_v + z(x+y)u_z = 0$$
Attempt :
We have to solve the problem :
$$\frac{dx}{x^2} = \frac{dy}{y^2} = \frac{dz}{z(x+y)}$$
Choosing the first two fractions, we yield :
$$\frac{dx}{x^2} = \frac{dy}{y^2} \Rightarrow \int \frac{dx}{x^2} = \int \frac{dy}{y^2} \Leftrightarrow -\frac{1}{x} = \frac{1}{y} + c $$
$$\implies $$
$$u_1 = \frac{1}{x} - \frac{1}{y}$$
Now, choosing the first and the last fraction, we yield :
$$\frac{dx}{x^2} = \frac{dz}{z(x+y)} \Rightarrow \int z(x+y)dx = \int x^2dz \Leftrightarrow z\frac{x^2}{2} + zxy = x^2z + c' $$
but this is a different result for $u_2$ and eventually the general solution $u(x,y,z) = F(u_1,u_2)$ than Wolfram Alpha calculates here. Why is this the case and where am I wrong with my solution ?
|
For the first equation
$$\frac{dx}{x^2} = \frac{dy}{y^2} \Rightarrow \int \frac{dx}{x^2} = \int \frac{dy}{y^2} \Leftrightarrow -\frac{1}{x} = -\frac{1}{y} + C$$
$$ \implies K=\frac 1x- \frac 1y $$
For the last equation use this trick @Rebellos
$$\frac{dx}{x^2} = \frac{dy}{y^2} = \frac{dz}{z(x+y)}$$
$$\frac{dx-dy}{x^2-y^2} = \frac{dz}{z(x+y)}$$
$$\frac{d(x-y)}{x-y} = \frac{dz}{z}$$
I let you finish this last equation
|
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|
Determine the region of the phase plane in which all phase paths are periodic orbits Sketch the phase portrait of the dynamical system
\begin{align} \dot{x}&=y+2xy \\ \dot{y}&=-x+x^2-y^2\end{align}
and determine the region of the phase plane in which all phase paths are periodic orbits.
I have found that the equilibrium points $(0,0),(1,0), (-\frac{1}{2},\frac{\sqrt{3}}{2}),(-\frac{1}{2},-\frac{\sqrt{3}}{2})$ are respectively a centre and three saddle points,
and that
the horizontal isocline is $(x-\frac{1}{2})^2-y^2=\frac{1}{4}$ and the vertical isoclines are $y=0$ and $x=-\frac{1}{2}$.
How to determine the region of the phase plane in which all phase paths are periodic orbits?
|
Note that the system is Hamiltonian since
$$\frac{\partial}{\partial x} (y+2xy) = 2y = - \frac{\partial}{\partial y} (-x+x^2 -y^2). $$
Indeed a Hamiltonian function is given by $H(x, y) = \frac{1}{2}x^2 -\frac{1}{3} x^3 + \frac{1}{2} y^2 + xy^2$. The flow lines are given by the level curve of $H$.
Now we consider the level set which contains the line $x = -1/2$ (From your analysis, we know that this line is part of the vertical nullclines, thus is part of a solution curve). Since $H(-1/2, 0) = 1/6$, after some calculations,
$$ \frac{1}{2}x^2 - \frac{1}{3} x^3 + \frac{1}{2} y^2 + xy^2= \frac 1{6}\Rightarrow \left( x+\frac 12\right)\left(y+ \frac{1}{\sqrt 3}(x-1)\right)\left(y - \frac{1}{\sqrt 3}(x-1)\right)=0.$$
So luckily this level curves is a union of three lines. The triangle bounded by these three lines are exactly where all the periodic orbit are located.
|
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|
show this number is not a prime number The following problem is a special case Show this number always is composite number? , but I think this special case is relatively difficult to deal with, that is, if this is solved, it may solve the general situation.
show this number
$$A=2004^{2005}+1002^{2005}\cdot 2005^{1002}+2005^{2004}$$is not a prime number.
This is a question from an eighth-grade math contest exercise, (this .：see:at last page problem 20
I'm thinking mod 3, mod 5, mod 7. It doesn't seem to work.
because
$$A\equiv 0+0+1=1\pmod 3$$
$$A\equiv -1+0+0=4\pmod 5$$
Use Fermat's little theorem
$$A\equiv 2+1+3^{0}=4\pmod 7$$
|
$$2004^{2005}+1002^{2005} 2005^{1002}+2005^{2004}$$
$$x=2004$$
$$x^{x+1}+\left(\frac{x}{2}\right)^{x+1} (x+1)^{x/2}+(x+1)^x=2^{-x-1} \left(x^{x+1} (x+1)^{x/2}+2^{x+1} x^{x+1}+2^{x+1} (x+1)^x\right)$$
which should never be prime when $x$ is even:
Consider the portion (ignoring the extra $2^{x+1}$ for now):
$$\left(x^{x+1}+(x+1)^{x}\right)$$
Then, let $x={1,2,3,...}$, but don't evaluate to see:
$$x^2+x+1$$
$$x^3+x^2+2 x+1$$
$$x^4+x^3+3 x^2+3 x+1$$
...
Then the other portion:
$$\left(x^{x+1}(x+1)^{x/2}\right)$$
$$x^4+x^3$$
$$x^7+2 x^6+x^5$$
$$x^{10}+3x^9+3x^8+x^7$$
...
So the inside part of the polynomial is the sum of two different variations of binomial expansions with different powers for $x$, so it cannot be prime when $x$ is even.
One other note: it's almost like $2^2(x^2+x+1)+x^4+x^3$ characterizes the entire function.
I'm pretty sleepy right now, if I missed something let me know. (nicely, please)
|
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|
Find the altitude of a tetrahedron whose faces are congruent triangles Any set of four identical triangles can be arranged to form a four-surface solid. If these four triangles are not equilateral, e.g. 3,4,5, how do I find the altitude of from the one used for the bottom to the apex formed by the other three.
|
Elaborating on my comment ...
Let the triangle have side-lengths $a$, $b$, $c$, and area $T$. The Cayley-Menger determinant for area ---equivalently Heron's formula--- tells us
$$16 T^2 = (-a+b+c)(a-b+c)(a+b-c)(a+b+c) \tag{1}$$
The CM determinant for volume gives
$$288V^2 = \left|\begin{matrix}
0 & 1 & 1 & 1 & 1 \\
1 & 0 & a^2 & b^2 & c^2 \\
1 & a^2 & 0 & c^2 & b^2 \\
1 & b^2 & c^2 & 0 & a^2 \\
1 & c^2 & b^2 & a^2 & 0
\end{matrix}\right| = 4 (-a^2 + b^2 + c^2) (a^2 - b^2 + c^2) (a^2 + b^2 - c^2) \tag{2}$$
Observe that, if the triangle has a right angle, then $(2)$ vanishes; the tetrahedron has zero volume. Moreover, if the triangle has an obtuse angle, then $(2)$ is negative, making $V$ imaginary: in this case, there must be no tetrahedron at all.
Since altitude $h$ satisfies $3 V = h T$,
$$h^2 = \frac{9 V^2}{T^2} = \frac{2(-a^2+b^2+c^2)(a^2-b^2+c^2)(a^2+b^2-c^2)}{(-a+b+c)(a-b+c)(a+b-c)(a+b+c)} \tag{$\star$}$$
For something more angle-related ... By the Law of Cosines, we have
$$a^2 = b^2 + c^2 - 2 b c \cos A \qquad\text{etc}$$
so that
$$V^2 = \frac{1}{72}\cdot 2 b c \cos A \cdot 2c a \cos B \cdot 2 a b \cos C = \frac{1}{9} a^2 b^2 c^2 \cos A \cos B \cos C$$
By the Law of Sines,
$$a = d \sin A \qquad\text{etc}$$
where $d$ is the circumdiameter of the triangle, so that
$$T = \frac12 a b \sin C = \frac12 d^2 \sin A \sin B \sin C \qquad
V = \frac13 d^3\sin A \sin B \sin C\sqrt{\cos A \cos B \cos C} $$
Hence,
$$h = \frac{3V}{T} = 2d\sqrt{\cos A \cos B \cos C} \tag{$\star\star$}$$
|
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|
What am I doing wrong in finding the $\gcd(x^3-2x^2-x+2, x^3-4x^2+3x)$? Could someone tell me what I am doing wrong in the process of finding the gcd with Euclid's Algorithm? The gcd is supposedly $(x-1)$ but I only get $-2(x-1)$.
$f(x) = x^3-2x^2-x+2$
$g(x) = x^3-4x^2+3x$
$f(x) = 1*g(x)+2x^2-4x+2$
$g(x) = 1/2x*r_1 -(2x^2-2x)$
$r_1 = (-1)*r_2 -2x+2$
$r_2 = x*(r_3)$
$\gcd(f(x),g(x)) = -(2x+2) \implies -2(x-1)$
Thanks in advance!
|
$$ \left( x^{3} - 2 x^{2} - x + 2 \right) $$
$$ \left( x^{3} - 4 x^{2} + 3 x \right) $$
$$ \left( x^{3} - 2 x^{2} - x + 2 \right) = \left( x^{3} - 4 x^{2} + 3 x \right) \cdot \color{magenta}{ \left( 1 \right) } + \left( 2 x^{2} - 4 x + 2 \right) $$
$$ \left( x^{3} - 4 x^{2} + 3 x \right) = \left( 2 x^{2} - 4 x + 2 \right) \cdot \color{magenta}{ \left( \frac{ x - 2 }{ 2 } \right) } + \left( - 2 x + 2 \right) $$
$$ \left( 2 x^{2} - 4 x + 2 \right) = \left( - 2 x + 2 \right) \cdot \color{magenta}{ \left( - x + 1 \right) } + \left( 0 \right) $$
$$ \frac{ 0}{1} $$
$$ \frac{ 1}{0} $$
$$ \color{magenta}{ \left( 1 \right) } \Longrightarrow \Longrightarrow \frac{ \left( 1 \right) }{ \left( 1 \right) } $$
$$ \color{magenta}{ \left( \frac{ x - 2 }{ 2 } \right) } \Longrightarrow \Longrightarrow \frac{ \left( \frac{ x }{ 2 } \right) }{ \left( \frac{ x - 2 }{ 2 } \right) } $$
$$ \color{magenta}{ \left( - x + 1 \right) } \Longrightarrow \Longrightarrow \frac{ \left( \frac{ - x^{2} + x + 2 }{ 2 } \right) }{ \left( \frac{ - x^{2} + 3 x }{ 2 } \right) } $$
$$ \left( x^{2} - x - 2 \right) \left( \frac{ x - 2 }{ 4 } \right) - \left( x^{2} - 3 x \right) \left( \frac{ x }{ 4 } \right) = \left( 1 \right) $$
$$ \left( x^{3} - 2 x^{2} - x + 2 \right) = \left( x^{2} - x - 2 \right) \cdot \color{magenta}{ \left( x - 1 \right) } + \left( 0 \right) $$
$$ \left( x^{3} - 4 x^{2} + 3 x \right) = \left( x^{2} - 3 x \right) \cdot \color{magenta}{ \left( x - 1 \right) } + \left( 0 \right) $$
$$ \mbox{GCD} = \color{magenta}{ \left( x - 1 \right) } $$
$$ \left( x^{3} - 2 x^{2} - x + 2 \right) \left( \frac{ x - 2 }{ 4 } \right) - \left( x^{3} - 4 x^{2} + 3 x \right) \left( \frac{ x }{ 4 } \right) = \left( x - 1 \right) $$
........................
|
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Finding the residue of $\frac{1}{z^3 \sin z}$ at z = 0 Given the function $f(z) = \frac{1}{z^3 \sin{(z)}}$, what is the residue of $f(z)$ at $z = 0$?
I want to find $b_1$ from the Laurent expansion. So I did the following:
\begin{align*}
\frac{1}{z^3 \sin{(z)}}
&= ( \dots \frac{b_5}{z^5} + \frac{b_4}{z^4} + \frac{b_3}{z^3} + \frac{b_2}{z^2} + \frac{b_1}{z} + a_0 + a_1z + \dots )\\
1& = \Big ( \dots \frac{b_5}{z^5} + \frac{b_4}{z^4} + \frac{b_3}{z^3} + \frac{b_2}{z^2} + \frac{b_1}{z} + a_0 + a_1z + \dots \Big ) \cdot \Big ( z^3 \sin{(z)} \Big )\\
&= \Big ( \dots \frac{b_5}{z^2} + \frac{b_4}{z} + b_3 + b_2z + b_1z^2 + a_0z^3 + a_1z^4 + \dots \Big ) \cdot \Big ( \sin{(z)} \Big )\\
&= \Big ( \dots \frac{b_5}{z^2} + \frac{b_4}{z} + b_3 + b_2z + b_1z^2 + a_0z^3 + a_1z^4 + \dots \Big ) \cdot \Big ( z - \frac{z^3}{3!} + \frac{z^5}{5!} - \dots \Big )\\
\end{align*}
After some more thought...
Is it true to say that because f(z) has a pole of order 4 at $z=0$ that our $b_n$'s only go out to the 4th term? Meaning there are no $b_5$, $b_6$, etc like how wrote previously. That is,
$\frac{1}{z^3 \sin{(z)}}
= \Big ( \frac{b_4}{z^4} + \frac{b_3}{z^3} + \frac{b_2}{z^2} + \frac{b_1}{z} + a_0 + a_1z + \dots \Big )\\$
followed by
\begin{align*}
1
&= \Big ( \frac{b_4}{z} + b_3 + b_2z + b_1z^2 + a_0 + \dots \Big ) \cdot \Big ( z - \frac{z^3}{3!} + \frac{z^5}{5!} - \dots\Big )\\
\end{align*}
Which then when multiplying out $b_1z^2$ with each term from sin(z)'s Laurent expansion will never yield a $\frac{1}{z}$ term, concluding that the coefficient $b_1 = 0$?
|
It's much simpler than what you do, using asymptotic analysis:
\begin{align}
\frac1{\sin z}&=\frac 1{z-\cfrac{z^3}6+\cfrac{z^5}{120}+O(z^7)}\\
&=\frac1z\cdot\frac1{1-\biggl(\underbrace{\cfrac{z^2}6-\cfrac{z^4}{120}+O(z^6)}_{=\,u}\biggr)} \\
&=\frac1z\biggl[1+
\cfrac{z^2}6-\cfrac{z^4}{120}+\biggl( \cfrac{z^2}6-\cfrac{z^4}{120}\biggr)^{\!\!2}+O(z^6)\biggr] \\
&=\frac1z\biggl[1+
\cfrac{z^2}6-\cfrac{z^4}{120}+ \cfrac{z^4}{36}+O(z^6)\biggr] \\
&=1+ \cfrac{z^2}6+\frac{7z^4}{360}+O(z^6)\\
\text{so that }\qquad
\frac1{z^3}\frac1{\sin z}&=\frac1{z^4}\biggl[1+
\cfrac{z^2}6+ \cfrac{7z^4}{360}+O(z^6)\biggr] \\
&=\frac1{z^4}+\frac1{6z^2}+ \cfrac{7}{360}+O(z^2).
\end{align}
Finally, $\;\operatorname{Res}(f,0)=0$.
In this case, it could have been anticipated: the function $\dfrac1{z^3\sin z}$ is even, and therefore, its Laurent expansion around $0$ has only terms of even degree, so $a_{-1}=0$.
|
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|
Finding every solution for equation of complex numbers I need to find every solution for:
$\ z^{3} + 3i \overline z = 0 $
So I tried was just to compare imaginary and complex part of $\ z^{3} $ and $\ 3i\overline z$
Ill spare you the alegbra, here is the result:
$$\ a^{3} - 3ab^{2} + i(3a^{2}b-b^{3}) = -3b -3ai \\a^{3} - 3ab^{2} = -3b \\3a^{2}b - b^{3} = -3a$$ and so $$\ a^{3} -3ab^{2} + 3b = 0 \\ b^{3} -3a^{2}b-3a=0
$$
but I'm pretty stuck here. not sure what do next. I also tried using eulers rule so
$$\ z^{3} = -3i\overline z \\ r^{3}e^{{i\theta}^{3}} = -3i \times re^{-i\theta} \\ r^{3}e^{{i\theta}^{3}} = -3i \times re^{2\pi-i\theta} \\ 3\theta = 2\pi - \theta +2\pi k \\ 4\theta = 2\pi + 2\pi k \\ \theta = \frac{\pi}{2} + \frac{\pi k }{2}$$
and
$$\ r^{3} = -3ir \\ r^{2} = -3i$$
|
there is the trivial solution $z = 0$
or
$z^3 + \bar z 3i = 0\\
z(z^3 + \bar z 3i) = 0\\
z^4 + z\bar z 3i = 0\\
z^4 + |z|^2 3i = 0\\
\frac {z^4}{|z|^2} = -3i \\
\arg z^4 = \frac {-\pi}{2}+2n\pi\\
\arg z = \frac {-\pi}{8}, \frac {-5\pi}{8},\frac {3\pi}{8},\frac {7\pi}{8}\\
|\frac {z^4}{|z|^2}| = |z^2| = 3\\
|z| = \sqrt 3\\
z = \sqrt 3 e^{\frac {-\pi}{8}i},z = \sqrt 3 e^{\frac {-5\pi}{8}i},z = \sqrt 3 e^{\frac {3\pi}{8}i},z = \sqrt 3 e^{\frac {7\pi}{8}i}$
|
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|
Find integer triples $(x,y,z)$ such that $2018^x=y^2+z^2+1$ To find the triples, I had some tries.
First , consider $x\ge 2$, then $2018^x=0\pmod 4$. But for the right side $y^2+z^2+1$. Consider 3 situations.
*
*case 1. Both $y$ and $z$ are even. Then $y^2+z^2+1=1\pmod 4$.
*case 2. One of them are even, the other is odd. Then $y^2+z^2+1=2\pmod 4$.
*case 3. Both $y$ and $z$ are odd. Then $y^2+z^2+1=3\pmod 4$.
Thus for $x\ge 2$, there is no solution. Then $x=0$ or $x=1$.
Now for $x=0$, $2018^0=y^2+z^2+1$ ,then $y^2+z^2=0$, hence we have $y=0$ and $z=0$.
For $x=1$, $2018^1=y^2+z^2+1$, then my question is how to solve $y^2+z^2=2017$.
Also since I didn't learn Number Theory systematically, I don't know if I was proceeding in the right way for this question.
Lastly, Are there more general methods for problems like this?
|
You've done most of the work already; you might also want to explicitly exclude negative $x$. One way to find all solutions with $x=1$ without any advanced machinery is as follows:
From any solution of $2017=y^2+z^2$ we get another solution by changing the signs of $y$ and $z$ and by interchanging $y$ and $z$. So let's restrict our attention to nonnegative $y$ and $z$, where $y=2a$ is even and $z=2b+1$ is odd. Then
$$2017=y^2+z^2=(2a)^2+(2b+1)^2=4a^2+4b^2+4b+1,$$
which tells us that
$$a^2+b^2+b=504,$$
where $a$ and $b$ are nonnegative. Note that $b^2+b$ must be even, so also $a$ is even. This shows that
$$a\leq\sqrt{504}<23\qquad\text{ and }\qquad b<\sqrt{b^2+b}\leq\sqrt{504}<23.$$
This leaves only $12$ values for $a$ to check. We can also continue by noting that $a^2\equiv0\pmod4$ and so $b^2+b\equiv0\pmod{4}$, which implies that $b$ is congruent to either $0$ or $3$ mod $4$, leaving only $11$ values for $b$ to check.
|
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|
Find the values of $(2\sin x-1)(\cos x+1)=0$ How many different value of x from 0° to 180° for the equation $(2\sin x-1)(\cos x+1) = 0$?
The solution shows that one of these is true:
$\sin x = \frac12$ and thus $x = 30^\circ$ or $120^\circ$
$\cos x = -1$ and thus $x = 180^\circ$
Question: Inserting the $\arcsin$ of $1/2$ will yield to $30°$, how do I get $120^\circ$? and what is that $120^\circ$, why is there $2$ value but when you substitute $\frac12$ as $x$, you'll only get $1$ value which is the $30^\circ$?
Also, when I do it inversely: $\sin(30^\circ)$ will result to 1/2 which is true as $\arcsin$ of $1/2$ is $30^\circ$. But when you do $\sin(120^\circ)$, it will be $\frac{\sqrt{3}}{2}$, and when you calculate the $\arcsin$ of $\frac{\sqrt{3}}{2}$, it will result to $60^\circ$ and not $120^\circ$. Why?
|
As you noted $x=120°$ is not a solution indeed
*
*$(2\sin 120°-1)(\cos 120°+1)=\left(2\frac{\sqrt3}2-1\right)\left(-\frac12+1\right)\neq 0$
but
*
*$2\sin x-1=0 \implies x=\frac16 \pi + 2k\pi,\,x=\frac56 \pi + 2k\pi$
*$\cos x+1 = 0 \implies x=\pi + 2k\pi=(2k+1)\pi$
|
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|
Testing $\sum\limits_{k=1}^∞(\frac{k+1}k)^{k^2}3^{-k}$ for convergence and absolute convergence
Test $$\sum_{k=1}^{\infty}\left(\frac{k+1}{k}\right)^{k^2}3^{-k}$$ for convergence and absolute convergence.
We apply the ratio test for $\displaystyle \sum_{k=1}^{\infty}\left|\left(\frac{k+1}{k}\right)^{k^2}3^{-k}\right|$:
$$
\left|\frac{\left(\dfrac{k+2}{k+1}\right)^{(k+1)^2}3^{-(k+1)}}{\left(\dfrac{k+1}{k}\right)^{k^2}3^{-k}}\right|=\left|\frac{\left(\left(\dfrac{k+2}{k+1}\right)^{k}\right)^2\left(\dfrac{k+2}{k+1}\right)^{2k+1}3^{-k}\dfrac{1}{3}}{\left(\dfrac{k+1}{k}\right)^{k^2}3^{-k}}\right|→\frac{e^2\cdot e^2\cdot\dfrac13}{e^2}=\frac{1}{3}e^2.
$$
Since $\dfrac{1}{3}e^2 \geq 1$, the series diverges. Is this correct? I feel like I made a mistake somewhere that I cannot pin down.
|
Too long for a comment.
Consider
$$a_k=\left(\frac{k+1}{k}\right)^{k^2}3^{-k}$$ and you want to analyze $\frac{a_{k+1}}{a_k}$. It is convenient to consider first
$$\log\left(\frac{a_{k+1}}{a_k}\right)=\log\left({a_{k+1}}\right)-\log\left({a_{k}}\right)$$ where
$$\log\left({a_{k}}\right)=k^2 \log \left(\frac{k+1}{k}\right)-k\log (3)$$ making
$$\log\left(\frac{a_{k+1}}{a_k}\right)=(k+1)^2
\log \left(\frac{k+2}{k+1}\right)-k^2 \log \left(\frac{k+1}{k}\right)+(k+1)\log(3)-k \log (3)$$ Now, use Taylor series for large values of $k$ to get
$$\log\left(\frac{a_{k+1}}{a_k}\right)=\log \left(\frac{e}{3}\right)-\frac{1}{3
k^2}+O\left(\frac{1}{k^3}\right)$$ Take now the exponential of both sides and continue with Taylor and get
$$\frac{a_{k+1}}{a_k}=\frac{e}{3}-\frac{e}{9 k^2}+O\left(\frac{1}{k^3}\right)$$
|
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|
Integrate $\int x\sin^2 (x) dx$ Integrate $\int x\sin^2 (x) dx$
My attempt:
$$=\int x\sin^2 (x) dx\\
=x^2\sin^2 (x) - \int 2\sin (x)\cos (x)x^2 dx\\
=x^2\sin^2 (x) - \int \sin (2x) x^2 dx.$$
|
Take $\int x\sin ^2(x)dx = \int udv$, where $u = x\sin x$ and $dv = \sin (x)dx$. Then by integrating by parts we get
$$ \int x \sin ^2(x) dx = -x\sin (x)\cos x + \int (\sin x + x\cos x)\cos (x)dx =\ldots $$
Now, with $\cos ^2x = 1-\sin ^2x$ we get
$$\ldots = -x\sin (x)\cos x + \int \sin (x)\cos (x) dx + \int x\left (1-\sin ^2x\right )dx +C.$$
Collect same quantities to get
$$2\int x\sin ^2(x) dx = -x\sin (x)\cos x + \frac{\sin ^2x}{2} + \frac{x^2}{2}+C.$$
For a more pretty result, we can multiply both sides by $2$, then
$$4\int x\sin ^2(x)dx = \sin ^2x + x^2 - x\sin (2x) + C.$$
|
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|
solving differential equation $\frac{d^3y}{dx^3}+\frac{d^2y}{dx^2}=x^2+2x+2$ How would you solve this third order differential equation:
$$\frac{d^3y}{dx^3}+\frac{d^2y}{dx^2}=x^2+2x+2$$
My first thought was to take a double integral:
$$\iint\frac{d^3y}{dx^3}+\frac{d^2y}{dx^2}dxdx=\iint{x^2+2x+2}dxdx$$
so:
$$y+\frac{dy}{dx}=\frac{x^4}{12}+\frac{x^3}{3}+x^2+c_1x+c_2$$
This is what I got to but I am unsure beyond here
|
Here is another simple way
$$\frac{d^3y}{dx^3}+\frac{d^2y}{dx^2}=x^2+2x+2$$
Multiply both side by $e^x$ We get
$$(y''e^x)'= (e^x(x^2+2))'$$
we can reduce the order by direct integration
$$y''e^x= e^x(x^2+2)+K_1$$
$$y"=x^2+2+K_1e^{-x}$$
Integrate
$$y'=\frac {x^3}3+2x+K_1e^{-x}+K_2$$
Integrate again
$$y=\frac {x^4}{12}+x^2+K_1e^{-x}+K_2x+K_3$$
|
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|
Grade $6$ Math Problem
$10$ players are playing in a card game in which the winner is the player having the most number of cards. There are $230$ cards in total. What is the smallest number of cards the winning player could have collected, assuming that each player collected a different number of cards?
When I attempted the question, I tried the highest consecutive numbers the other $9$ players could have and then added them and took that number away from $230$.
So
$$26+25+24+ \cdots + 18 = 198$$
The winner has then $230-198= 32$ cards.
|
Extending this to $n$ cards and $k$ players, let $m$ be the minimum possible sum with $k$ players i.e. $m=\sum_{i=1}^ki$. The solution to the problem is then $\lceil \frac {n-m}k \rceil+k$.
e.g. $n=230$, $k=10$, then $m=55$. Solutions is:
$\lceil \frac {230-55}{10} \rceil+10=\lceil \frac {175}{10} \rceil+10=\lceil 17.5 \rceil+10=17+10=28$
Proof: $28+27+26+25+24+22+21+20+19+18=230$
e.g. $n=21$, $k=4$, then $m=10$. Solutions is:
$\lceil \frac {21-10}{4} \rceil+4=\lceil \frac {11}{4} \rceil+4=\lceil 2.75 \rceil+4=3+4=7$
Proof: $3+5+6+7=21$
e.g. $n=23$, $k=4$, then $m=10$. Solutions is:
$\lceil \frac {23-10}{4} \rceil+4=\lceil \frac {13}{4} \rceil+4=\lceil 3.25 \rceil+4=4+4=8$
Proof: $4+5+6+8=23$
|
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|
Generating function for sequence $a_i={n+im-1 \choose im}$ It is well known that $\sum_{i=0}^\infty {n+i-1 \choose i}x^i=\frac{1}{(1-x)^n},$ i.e. $\frac{1}{(1-x)^n}$ is generating function for sequence $a_i={n+i-1 \choose i.}$
But I want to find generating function for sequence $a_i={n+im-1 \choose im.}$
Using The On-Line Encyclopedia of Integer Sequences, I understood that generating function for sequence $a_i={n+2i-1 \choose 2i.}$ is $\frac{1+(n-1)x}{(1-x)^2}.$ But I cann't recieve generating function if $m>2.$ I found two formulas for multiset formula. But they didn't help me.
Thanks a lot in advance for any help!
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The $m$-th roots of unity $\omega_j=\exp\left(\frac{2\pi ij}{m}\right), 0\leq j<m$ have the nice property to filter elements. For $m,N>0$ we have
\begin{align*}
\frac{1}{m}\sum_{j=0}^{m-1}\exp\left(\frac{2\pi ij N}{m}\right)=
\begin{cases}
1&\qquad m\mid N\\
0& \qquad otherwise
\end{cases}
\end{align*}
If $A(x)=\sum_{k=0}^\infty a_kx^k$ then
\begin{align*}
A(x)+A(\omega_1x)+\cdots+A(\omega_{m-1}x)=\sum_{k=0}^\infty a_{mk}x^{mk}\tag{1}
\end{align*}
We set
\begin{align*}
A_1(x)=\sum_{k=0}^\infty \binom{n+k-1}{k}x^k=\frac{1}{(1-x)^n}
\end{align*}
According to (1) we obtain
\begin{align*}
\color{blue}{A_m(x)}&\color{blue}{=\sum_{k=0}^\infty\binom{n+mk-1}{mk}x^k}\\
&=A_1\left(x^{1/m}\right)+A_1\left(\omega_1 x^{1/m}\right)+\cdots+A_{m-1}\left(\omega_{m-1} x^{1/m}\right)\\
&=\sum_{j=0}^{m-1}A_1\left(\omega_j x^{1/m}\right)\\
&\,\,\color{blue}{=\sum_{j=0}^{m-1}\frac{1}{\left(1-\omega_j x^{1/m}\right)^n}}
\end{align*}
Some special cases:
Case $m=2$:
We have $\{\omega_0,\omega_1\}=\{1,e^{\pi i}\}=\{1,-1\}$
\begin{align*}
A_2(x)&=\sum_{k=0}^\infty \binom{n+2k-1}{2k}x^k\\
&=\frac{1}{(1-\sqrt{x})^n}+\frac{1}{(1+\sqrt{x})^n}
\end{align*}
Case $m=3$:
We have $\{\omega_0,\omega_1,\omega_2\}=\{1,e^{\frac{2\pi i}{3}},e^{\frac{4\pi i}{3}}\}=\{1,\frac{1}{2}(-1+\sqrt{3}),\frac{1}{2}\left(-1-\sqrt{3}\right)\}$
\begin{align*}
A_3(x)&=\sum_{k=0}^\infty \binom{n+3k-1}{3k}x^k\\
&=\frac{1}{(1-\sqrt[3]{x})^n}+\frac{1}{(1+\frac{1}{2}(1-\sqrt{3})\sqrt[3]{x})^n}+\frac{1}{(1+\frac{1}{2}(1+\sqrt{3})\sqrt[3]{x})^n}
\end{align*}
Case $m=4$:
We have $\{\omega_0,\omega_1,\omega_2,\omega_3\}=\{1,e^{\frac{\pi i}{2}},e^{\pi i},e^{\frac{3\pi i}{2}}\}=\{1,i,-1,-i\}$
\begin{align*}
A_4(x)&=\sum_{k=0}^\infty \binom{n+4k-1}{4k}x^k\\
&=\frac{1}{(1-\sqrt[4]{x})^n}+\frac{1}{(1-i\sqrt[4]{x})^n}+\frac{1}{(1+\sqrt[4]{x})^n}+\frac{1}{(1+i\sqrt[4]{x})^n}
\end{align*}
|
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|
Implicit derivative problem has two algebraically unequal answers with the exact same graph (??) In my math class we were assigned to find the implicit derivative of
$\dfrac yx + \dfrac xy= 2x$
And most of the class found the answer to be
$\text{(student answer)} \to\dfrac{dy}{dx} = \dfrac{(y^3-x^2y)}{(xy^2 -x^3 -2x^2y^2)}$
But the teacher had the answer
$\text{(teacher answer)}\to\dfrac{dy}{dx} = \dfrac{(y^2-x)}{(y-2xy)}$
She says that she sees how all of her students got the their answers, but she didn't understand why both answers were valid even when they are not algebraically equivalent. She says that she suspects it has something to do with variables acting as zeros in the denominator of the original function.
Can anyone explain why this happens?
|
I'm not sure about how the two answers were reached: either you differentiate as is, getting
$$ \frac{1}{x} \frac{dy}{dx} - \frac{y}{x^2} + \frac{1}{y} - \frac{x}{y^2} \frac{dy}{dx} - 2 = 0, $$
which can be rearranged to
$$ \frac{dy}{dx} = \frac{2+y/x^2-1/y}{1/x-x/y^2} = \frac{2x^2y+y^2-x^2}{y(y^2-x^2)} $$
or you clear denominators to start with
$$ x^2+y^2 - 2x^2y =0, $$
of which differentiation gives
$$ 2x + 2y \frac{dy}{dx} - 4xy - 2x^2\frac{dy}{dx} = 0, $$
and can be rearranged into
$$ \frac{dy}{dx} = \frac{2xy-x}{y-x^2}. $$
How are these two equivalent? It becomes slightly more apparent if we multiply numerator and denominator of the first one by the denominator of the original formula $xy$:
$$ \frac{2+y/x^2-1/y}{1/x-x/y^2} = \frac{2xy + y^2/x - x}{y-x^2/y} = \frac{(2xy-x) + y(y/x)}{y-x^2 +x(x-x/y)} $$
But now we can use the original formula $x/y+y/x=2x$:
$$ \frac{(2xy-x) + y(y/x)}{y-x^2 +x(x-x/y)} = \frac{(2xy-x) + y(2x-x/y)}{y-x^2 +x(-x+y/x)} = \frac{(2xy-x) + (2xy-x)}{y-x^2 +(y-x^2)} = \frac{2xy-x}{y-x^2}. $$
What's actually going on here? Suppose we have $f(x,y)/g(x,y) = 0$. Differentiating gives
$$ \frac{f_x}{g} - \frac{fg_x}{g^2} + \left( \frac{f_y}{g} - \frac{fg_y}{g^2} \right) \frac{dy}{dx}, $$
where $f_x$ is the partial derivative $\partial f/\partial x$ (i.e. the derivative holding $y$ constant), which rearranges to
$$ \frac{dy}{dx} = \frac{-f_x+(f/g)g_x}{f_y-(f/g)g_y} = \frac{-gf_x+fg_x}{gf_y-fg_y}. $$
But $f=0$, so this cancels to $dy/dx = -f_x/f_y$ as it should be. This is obscured when we have actual expressions for $f$ and $g$, so that there can be cancellations between the two terms in the numerator and/or denominator, hence why it was helpful to take a step back from the fully simplified version.
|
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|
How to solve the problem on number theory Find the number of positive integer pairs $x,y$ such that
$$xy+\dfrac{(x^3+y^3)}3=2007.$$
I solved the question by using factorization and further checking possible values of $x$ and $y$. But it was very lengthy as I had to check many cases for $x$ and $y$. Is there any possible other method?
|
Denote: $$\begin{cases}x+y=a \\ x-y=b\end{cases} \Rightarrow \begin{cases} x=\frac{a+b}{2}\\ y=\frac{a-b}{2}\end{cases}$$ Then:
$$xy+\dfrac{x^3+y^3}3=2007 \Rightarrow (a+2)^2+3b^2=\frac{24080}{a-1}.$$
Note that $a>1$ and $a$ can be $2,3, 5,6,8,9,11, 15,17,21$. Note that $a\ge 28 \Rightarrow b^2<0$. So, among them only $21$ suits. Hence $b=15$. Hence: $(x,y)=(18,3)$. Since the equation is symmetric, $(x,y)=(3,18)$.
|
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|
What is the coefficient of $x^{11}$ in the power series expansion of $\frac{1}{1-x-x^4}$? I am really stuck on this problem. I don't really understand power series expansions. However, I think this has to do with generating functions.
|
$\frac{1}{1-x-x^4}=\frac{1}{1-(x+x^4)}=1+(x+x^4)+(x+x^4)^2+(x+x^4)^3+(x+x^4)^4+(x+x^4)^5+(x+x^4)^6+(x+x^4)^7+(x+x^4)^8+(x+x^4)^9+(x+x^4)^{10}+(x+x^4)^{11}+\cdot\cdot\cdot$.
Checking how many $x^{11}$s are contributed by each term and adding up, we see the answer is $0+0+0+0+0+\binom{5}{2}+0+0+\binom{8}{1}+0+0+\binom{11}{0}=10+8+1$.
So the answer is 19.
|
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|
3D vector projection If I have an arbitrary vector in the 3D coordinate plane with given azimuth angle and polar angle with radius r=1, how could I use this information to project this vector onto the three 2D planes?(xy,yz,xz) There are no specific number available because I need to represent these "component" vectors symbolically rather than numerically.
|
If your vector is $$\vec{p} = \left[\begin{matrix}x\\y\\z\end{matrix}\right] = \left[\begin{matrix}r \cos\theta \cos\varphi \\
r \cos\theta \sin\varphi \\
r \sin\theta \end{matrix}\right]$$
where $-180° \le \varphi \le +180°$ is the azimuth angle, and $-90° \le \theta \le +90°$ the polar angle, then its projection on the $xy$, $xz$, and $yz$ planes are
$$\vec{p}_{xy} = \left[\begin{matrix}x\\y\\0\end{matrix}\right], \quad
\vec{p}_{xz} = \left[\begin{matrix}x\\0\\z\end{matrix}\right], \quad
\vec{p}_{yz} = \left[\begin{matrix}0\\y\\z\end{matrix}\right]$$
respectively.
In spherical coordinates,
$$\left\lbrace\begin{aligned}
r_{xy} &= r \cos\theta \\
\varphi_{xy} &= \varphi \\
\theta_{xy} &= 0 \\
\end{aligned}\right.$$ $$\left\lbrace\begin{aligned}
r_{xz} &= r \sqrt{1 - (\cos\theta)^2(\cos\varphi)^2} \\
\varphi_{xz} &= \begin{cases}
0°, & \cos\varphi \ge 0 \\
180°, & \cos\varphi \lt 0 \\
\end{cases} \\
\theta_{xz} &= \arcsin\left(\frac{\sin\theta}{\sqrt{1 - (\cos\theta)^2(\sin\varphi)^2}}\right) \\
\end{aligned}\right.$$ $$\left\lbrace\begin{aligned}
r_{yz} &= r \sqrt{1 - (\cos\theta)^2(\sin\varphi)^2} \\
\varphi_{yz} &= \begin{cases}
+90°, & \cos\varphi \ge 0 \\
-90°, & \cos\varphi \lt 0 \\
\end{cases} \\
\theta_{yz} &= \arcsin\left(\frac{\sin\theta}{\sqrt{1 - (\cos\theta)^2(\cos\varphi)^2}}\right)
\end{aligned}\right.$$
In general, if you have a plane through origin with unit normal vector $\hat{n}$, $\lVert\hat{n}\rVert = 1$, then the projection of $\vec{p}$ on that plane in Cartesian coordinates is $\vec{p}_n$,
$$\vec{p}_n = \vec{p} - \vec{p}\cdot\hat{n}$$
|
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|
If $a,b\in R$ and $a \neq 0$ and quadratic $ax^2-bx+c=0$ If $a,b\in R$ and $a \neq 0$ and quadratic $ax^2-bx+c=0$ has imaginary roots then $a+b+1$
why a+b+1 should be positive?
|
The roots of $ax^2-bx+c$ are given by
$$
\frac{b\pm\sqrt{b^2-4ac}}{2},
$$
which are both complex if and only if $b^2-4ac<0$.
|
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|
Integral $\int_0^{\pi} \frac{\cos(2018x)}{5-4\cos{x}}dx$ I wish to evaluate $$I(2018)=\int_{0}^{\pi}\frac{\cos(2018x)}{5-4\cos x} dx$$ Considering $$X=I(k)+iJ(k)=\int_{-\pi}^{\pi}\frac{\cos{kx}}{5-4\cos x} dx +i\int_{-\pi}^{\pi}\frac{\sin{kx}}{5-4\cos x} dx=\int_{-\pi}^{\pi}\frac{e^{ikx}}{5-4\cos x} dx$$ let us substitute $$e^{ix}=z\rightarrow dx=\frac{dz}{iz} \, ,|z|=1$$ Due to Euler's formula we can rewrite $$\cos x=\frac{z^2+1}{2z}$$ $$X=\oint_{|z|=1} \frac{z^k}{5-4\frac{z^2+1}{2z}}\frac{dz}{iz}=\frac{1}{i}\oint_{|z|=1} \frac{z^k}{-2z^2+5z-2}dz$$ $$-2z^2+5z-2=-\frac{1}{2}((2z)^2-5(2z)+4)=-\frac{1}{2}(2z-4)(2z-1)=-2(z-2)(z-\frac{1}{2})$$ Now let us notice that in our contour $|z|=1\,$ only the pole $z_2=\frac{1}{2}$ is found. Thus the integral we seek to evaluate is $$\frac{1}{i} \cdot 2\pi i \, \text{Res} (f(z),z_2)$$ where $f(z)=\frac{z^k}{-2(z-2)(z-\frac{1}{2})}$ $$X=2\pi \lim_{z\to z_2} (z-z_2)\frac{z^k}{-2(z-2)(z-z_2)}=\frac{2}{3}\pi \frac{1}{2^k}$$ therefore $$I(k)=\Re (X) =\frac{2\pi}{3}\frac{1}{2^k}$$ And $$\int_{0}^{\pi}\frac{\cos(2018 x)}{5-4\cos x} dx=\frac{\pi}{3}\cdot\frac{1}{2^{2018}}.$$ Now someone told me that the answer is $0$ and I am wrong (also wolfram gives $0$ as an answer). Could you please clarify? Or maybe give another solution to this integral if it's $0$ or another answer?
|
You are correct! This is an elementary approach without complex analysis.
Since $$\cos((n+1)x)+\cos((n-1)x)=2\cos(nx)\cos(x)$$ then for $n\geq 1$, we have the linear recurrence
$$\begin{align}
I(n-1)+I(n+1)&=\int_{0}^{\pi}\frac{2\cos(nx)\cos(x)}{5-4\cos(x)} dx\\
&=
-\frac{1}{2}\int_{0}^{\pi}\frac{\cos(nx)(-5+5-4\cos(x))}{5-4\cos(x)} dx\\
&=
\frac{5}{2}I(n)-\frac{1}{2}\int_{0}^{\pi}\cos(nx) dx=\frac{5}{2}I(n).
\end{align}$$
Then
$$I(n)=A2^n+\frac{B}{2^n}$$
for some constants $A$ and $B$. Since $I(n)$ is bounded it follows that $A=0$ and
$$I(n)=\frac{I(0)}{2^n}=\frac{1}{2^n}\int_{0}^{\pi}\frac{dx}{5-4\cos(x)} =\frac{1}{2^n}\left[\frac{2\arctan(3\tan(x/2))}{3}\right]_{0}^{\pi}=\frac{\pi/3}{2^n}.$$
P.S. In order to show that $I(n)$ is not zero for any $n$ it suffices to say that $I(0)>0$ since it is the integral of a positive continuous function.
|
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|
If for a real $x, x +\frac1x$ is an integer, prove $x^{2017}+\frac1{x^{2017}}$ is also. For some real $x$, let $y=x +\frac1x$, with $y\in \mathbb{Z}$.
$y=x +\frac1x\implies x^2 -xy +1 =0\implies x= \frac{y\pm \sqrt{y^2 -4}}2$, so $x+\frac1x = \frac{y\pm \sqrt{y^2 -4}}2+ \frac2{y\pm \sqrt{y^2 -4}}$
This implies: $ y = \frac{(y\pm \sqrt{y^2 -4})^2+4}{2\cdot (y\pm \sqrt{y^2 -4})}$
Not sure of proper logic (request guidance on this part), but hope that can take positive & negative signs for $ \sqrt{y^2 -4}$ alternately, in both numerator & denominator simultaneously.
Case (a) : positive $ \sqrt{y^2 -4}$ :
$ y = \frac{(y+ \sqrt{y^2 -4})^2+4}{2\cdot (y+ \sqrt{y^2 -4})}=\frac{y^2 + y\sqrt{y^2-4}}{y+\sqrt{y^2-4}}\implies y$
Case (b) : negative $ \sqrt{y^2 -4}$ :
$ y = \frac{(y- \sqrt{y^2 -4})^2+4}{2\cdot (y-\sqrt{y^2 -4})}=\frac{y^2 - y\sqrt{y^2-4}}{y-\sqrt{y^2-4}}\implies y$
There can be formed no conclusion with above, except vindicating that the roots are correct.
Now to show further that $y'= x^{2017}+\frac1{x^{2017}}$ is also an integer, there seems no way 'algebraically (i.e., direct multiplication) ' except possibly by modulus arithmetic. The direct route is not clear to me, as modulus is to be different (higher with each step) powers of $x^i+\frac1{x^i}, i\in \mathbb{Z+}$.
A variant of the above idea can be with strong induction, which considers all powers of $x, \frac1x$ till $2017$, or any positive integral power $i$ as follows:
Step 1: Base case of $n=1$ holds true, i.e. for some real $x, y = x+\frac1x$ is an integer.
Step 2: Suppose hypothesis holds for all $i\le n$ for the induction hypothesis step, i.e. :$x^n+\frac1{x^n}$ is an integer too.
Step 3: Need prove that for $i = n+1, x^i +\frac 1{x^i}$ is also an integer.
$$(x^n+\frac1{x^n})(x+\frac1x)=x^{n+1}+\frac1{x^{n-1}}+x^{n-1}+\frac1{x^{n+1}}=(x^{n+1}+\frac1{x^{n+1}})+ (\frac1{x^{n-1}}+x^{n-1})$$
so
$$x^{n+1}+\frac1{x^{n+1}}=(x^n+\frac1{x^n})(x+\frac1x)-(x^{n-1}+1/x^{n-1})$$
with the r.h.s. being an integer, hence proved.
Have two doubts:
1. Is it possible to prove by finding the complex roots of $x^i +\frac1{x^i}$ and proving using finding roots, by changing / extending domain of $x$ to be in $\mathbb{C}$.
2. No idea for negative integer values of powers of $x, \frac1x$ by the strong induction approach.
|
Alt. hint: $\;a=x\,$ and $\,b=\dfrac{1}{x}\,$ are roots of the quadratic $\,z^2-y z + 1\,$ where $\,y = a+b\,$.
If $\,y=x+\dfrac{1}{x}\,$ is an integer, then the quadratic is a monic polynomial with integer coefficients, and therefore $\,a^n+b^n=x^n+\dfrac{1}{x^n}\,$ is an integer for all $\,\forall n\,$ by Newton's identities.
|
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|
Rank of a matrix in $Z_p^{r\times n}$. Let $p$ be a prime number and $r$ even such that $0<r<n<p$. Calculate rank of matrix$\begin{pmatrix}1&1&1&...&1\\
0&1&2&...&n-1\\
0^2&1^2&2^2&...&(n-1)^2\\
\vdots&\vdots&\vdots&&\vdots\\
0^{r-1}&1^{r-1}&2^{r-1}&...&(n-1)^{r-1}
\end{pmatrix} \in \mathbb Z_p^{r\times n}$
Here is an example for $0<2<3<7$ we have
$\begin{pmatrix}1&1&1\\
0&1&2\\
\end{pmatrix}$. It's rank is $2=r$.
Another $0<4<5<7 $ :$\begin{pmatrix}1&1&1&1&1\\
0&1&2&3&4\\
0^2&1^2&2^2&3^2&4^2\\
0^{3}&1^{3}&2^{3}&3^3&4^{3}
\end{pmatrix}=\begin{pmatrix}1&1&1&1&1\\
0&1&2&3&4\\
0&1&4&2&2\\
0&1&1&6&1
\end{pmatrix}\rightarrow \begin{pmatrix}1&0&0&0&6\\
0&1&0&0&4\\
0&0&1&0&1\\
0&0&0&1&4
\end{pmatrix}$ has full rank $4=r$.
Does it hold in general? If so, how do I proove it?
|
As already observed in the comments, this matrix is related to a Vandermonde matrix. Specifically, the $(1,1)$ minor is the determinant of a matrix that is obtained from a Vandermonde matrix with distinct factors by multiplying each column with a non-zero factor. Thus, the Laplace expansion for the first column yields a non-zero determinant; hence the matrix has full rank.
|
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|
Finding $C_n=-C_{n-1}+C_{n-2}$ $$\begin{cases}
C_n=-C_{n-1}+C_{n-2}\\
C_0=1\\
C_1=0
\end{cases} $$
$
\begin{pmatrix}
C_n \\
C_{n-1}
\end{pmatrix}
\begin{pmatrix}
-1 && 1 \\
1 && 0
\end{pmatrix}^n
\begin{pmatrix}
1\\
0
\end{pmatrix}$
$\begin{vmatrix}
\lambda+1 && -1 \\
-1 && \lambda
\end{vmatrix}=\lambda^2+\lambda-1$ So $\lambda_{1,2}=-\frac{1}{2}\pm \frac{ \sqrt{5}}{2}$
$$C_n=a(-\frac{1}{2}+ \frac{ \sqrt{5}}{2})^nC_{n-1}+b(\frac{1}{2}- \frac{ \sqrt{5}}{2})^nC_{n-2}$$
$$\begin{cases}
1=a+b\\
0=a(-\frac{1}{2}+ \frac{ \sqrt{5}}{2})+b(\frac{1}{2}- \frac{ \sqrt{5}}{2})
\end{cases}$$
$C_n=\frac{1}{2}(-\frac{1}{2}+\frac{\sqrt{5}}{2})^n+\frac{1}{2}(\frac{1}{2}- \frac{ \sqrt{5}}{2})^n$
Is it correct?
|
Apart from an offset (shift of n to n-1) this is https://oeis.org/A039834 in the Online Encyclopedia of Integer Sequences, which provides an equivalent formula.
|
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|
Uniform Convergence of $\frac{2x}{1+x^2}+\frac{4x^3}{1+x^4}+\frac{8x^7}{1+x^8}+...$ Show that following series is uniformly convergent in $(-1,1)$.
$$\frac{2x}{1+x^2}+\frac{4x^3}{1+x^4}+\frac{8x^7}{1+x^8}+...$$
I had previously solved this problem using Weierstrass M test by taking
$$|f_n| = |\frac{2^n.x^{2^n-1}}{1+x^{2^n}}| \leq 2^n.k^{2^n-1}, k\in(x,1)$$
Thus, $\frac{M_n}{M_{n+1}}\to \infty$ hence the series is uniformly convergent. But now I think this is incorrect as $k$ shouldn't be dependent on $x$. How should I proceed then?
|
The series $\sum \frac{2^n.x^{2^n-1}}{1+x^{2^n}}$ is not uniformly convergent on $(-1,1)$. If it was $v_n = \sup\limits_{x \in (-1,1)} \left\vert \frac{2^n.x^{2^n-1}}{1+x^{2^n}} \right\vert$ would be such that $v_n \to 0$.
But
$$\left\vert \frac{2^n.x^{2^n-1}}{1+x^{2^n}} \right\vert \ge 2^{n-1} \vert x \vert^{2^n-1}$$ and the right hand side of the inequality is unbounded in $(-1,1)$.
|
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Consider the next succession and prove by induction The exercise says:
Knowing the next succession,
$$a_1=1$$
$$a_{n+1}=\frac{2a_n}{(2n+2)(2n+1)}, n>=1$$
Prove by induction that $a_n=\frac{2^n}{(2n)!}$
What I've done so far is prove for $n=1$
For $n=1$:
$$a_1=\frac{2^1}{(2\times1)!}=\frac{2}{2}=1$$
Which is correct.
Therefore, to prove the induction:
$$a_{n+1}=\frac{2\times a_{n+1}}{(2(n+1)+2)(2n+1)}=$$
$$=\frac{2}{(2(n+1)+2)(2n+1)}\frac{2^{n+1}}{(2(n+1))!}=$$
$$=\frac{2^{n+2}}{(2n+4)(2n+3)}\times\frac{1}{(2(n+1))!}=$$
$$=\frac{2^{n+2}}{4n^2+6n+18n+12}=$$
$$=\frac{2^{n+2}}{4n^2+24n+12}=$$
$$=\frac{2\times2^{n+1}}{2(2n^2+12n+6)}=$$
$$=\frac{2^{n+1}}{2n^2+12n+6}$$
Is this correct? Do I have to simplify even more?
Thanks
|
When you wrote that$$a_{n+1}=\frac{2\times a_{n+1}}{(2(n+1)+2)(2n+1)},$$I suppose that you meant to write $a_{n+2}$ as the LHS.
Anyway, this is inconclusive. If you were assuming that $a_{n+1}=\frac{2^{n+1}}{(2(n+1))!}$, your conclusion should have been that $a_{n+2}=\frac{2^{n+2}}{(2(n+2))!}$.
Note that$$\frac{a_{n+1}}{a_n}=\frac2{(2n+2)(2n+1)}\tag1$$and that$$\frac{\frac{2^{n+1}}{(2(n+1))!}}{\frac{2^n}{(2n)!}}=\frac{2\times(2(n+1))!}{(2n)!}=\frac2{(2n+2)(2n+1)}=(1).$$Can you take it from here?
|
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|
Proving $\sum_{k=0}^{n-1} \text{cos}\left(\frac{2\pi k}{n} \right) = \sum_{k=0}^{n-1} \text{sin}\left(\frac{2\pi k}{n} \right) = 0$ I would like to prove the following, $$\sum_{k=0}^{n-1} \text{cos}\left(\frac{2\pi k}{n} \right) = \sum_{k=0}^{n-1} \text{sin}\left(\frac{2\pi k}{n} \right) = 0$$ This is equivalent to showing that if we have a regular n-gon, all of the vectors in $\mathbb{R}^2$ pointing to every vertex cancel eachother out to $\textbf{0}$. It seems obvious enough and i've tried a number of examples for small $n$. I am not sure how to start a proof though.
|
I assume $n \ge 2$.
Set
$\omega = \exp \left ( \dfrac{2\pi i}{n} \right ); \tag 1$
then
$\omega^n = \left ( \exp \left ( \dfrac{2\pi i}{n} \right ) \right )^n = \exp \left ( \dfrac{2 \pi i n}{n} \right ) = \exp ( 2 \pi i ) = 1; \tag 2$
we have
$( \omega - 1) \displaystyle \sum_0^{n - 1} \omega^k = \omega^n - 1 = 0; \tag 3$
since
$\omega \ne 1, \tag 4$
(3) yields
$ \displaystyle \sum_0^{n - 1} \omega^k = 0; \tag 5$
from (1)
$\omega^k = \exp \left ( \dfrac{2\pi i k}{n} \right ) = \cos \left ( \dfrac{2 \pi k}{n} \right ) + i \sin \left ( \dfrac{2 \pi k}{n} \right ) ; \tag 6$
substituting (6) into (5) and separating out the real and imaginary parts gives us
$\displaystyle \sum_0^{n - 1} \cos \left ( \dfrac{2 \pi k}{n} \right ) + i \sum_0^{n - 1} \sin \left ( \dfrac{2 \pi k}{n} \right ) = 0, \tag 7$
from which we immediately have
$\displaystyle \sum_0^{n - 1} \cos \left ( \dfrac{2 \pi k}{n} \right ) = \sum_0^{n - 1} \sin \left ( \dfrac{2 \pi k}{n} \right ) = 0. \tag 8$
|
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|
Proving the product of four consecutive integers, plus one, is a square I need some help with a Proof:
Let $m\in\mathbb{Z}$. Prove that if $m$ is the product of four consecutive integers, then $m+1$ is a perfect square.
I tried a direct proof where I said:
Assume $m$ is the product of four consecutive integers.
If $m$ is the product of four consecutive integers, then write $m=x(x+1)(x+2)(x+3)$ where $x$ is an integer.
Then $m=x(x+1)(x+2)(x+3)=x^4+6x^3+11x^2 +6x$.
Adding $1$ to both sides gives us:
$m+1=x^4+6x^3+11x^2+6x+1$.
I'm unsure how to proceed. I know I'm supposed to show $m$ is a perfect square, so I should somehow show that $m+1=a^2$ for some $a\in\mathbb{Z}$, but at this point, I can't alter the right hand side of the equation to get anything viable.
|
\begin{eqnarray*}
m=n(n+1)(n+2)(n+3) =n^4+6n^3+11n^2+6n.
\end{eqnarray*}
So
\begin{eqnarray*}
m+ 1 =n^4+6n^3+11n^2+6n+1=(n^2+3n+1)^2.
\end{eqnarray*}
|
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|
Proving $x^2+x+1\gt0$ I was doing a question recently, and it came down to proving that $x^2+x+1\gt0$. There are of course many different methods for proving it, and I want to ask the people here for as many ways as you can think of.
My methods:
*
*$x^2+x+1=(x+\frac12)^2+\frac34$, which is always greater than $0$.
*Let it be $0$ for some $x=k$. Then $x^2+x+1=0$ has a real solution. But since $1^2\not\gt4$, this has no real solution. Therefore it is more than $0$.
|
If $x$ is positive, then $x^2+x+1$ is clearly positive.
If $x$ is negative then $x^2-x+1$ is certainly positive. Now $$(x^2+x+1)(x^2-x+1)=x^4+x^2+1$$ is certainly positive, so $x^2+x+1$ must also be positive in this case.
If $x$ is zero then $x^2+x+1=1\gt 0$
|
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Binomial coefficient identity found in Apéry's theorem In Apéry's proof of the irrationality of $\zeta(3)$, while proving the formula for the fast-converging series $\zeta(3)=\frac{5}{2} \sum_{k=1}^{\infty}{\frac{ (-1)^{k-1}} {k^{3}\binom {2k}{k}}}$ there is the following identity:
$$
\frac{(-1)^{n-1}(n-1)!^2}{n^2(n^2-1^2)\ldots(n^2-(n-1)^2)} = \frac{2(-1)^{n-1}}{n^2 \binom{2n}{n}},
$$
for positive integer $n>0$. I can't for the life of me figure out how to prove this is true. By eliminating on both sides the $(-1)^{n-1}$ in the numerator and the $n^2$ in the denominator, we are left with:
$$
\frac{(n-1)!^2}{(n^2-1^2)\ldots(n^2-(n-1)^2)} = \frac{2}{ \binom{2n}{n}}.
$$
But this already seems absurd! Using Wolfram Alpha shows that LHS-RHS is not zero, therefore the "identity" does not hold.
What am I missing? Note that I am using Van der Poorten's paper to understand Apéry's proof. The identity in question can be seen on the page labelled "197", on the bottom of the left column.
|
For each integer $k$, we have $n^2 - k^2 = \left(n-k\right) \left(n+k\right)$. Thus,
\begin{align*}
\prod\limits_{k=1}^{n-1} \left(n^2-k^2\right) &= \prod\limits_{k=1}^{n-1} \left(\left(n-k\right) \left(n+k\right)\right) = \underbrace{\left(\prod\limits_{k=1}^{n-1} \left(n-k\right) \right)}_{=\left(n-1\right)!} \underbrace{\left(\prod\limits_{k=1}^{n-1} \left(n+k\right)\right)}_{= \left(2n-1\right) \left(2n-2\right) \cdots \left(n+1\right)} \\
&= \left(n-1\right)! \left( \left(2n-1\right) \left(2n-2\right) \cdots \left(n+1\right) \right) .
\end{align*}
Hence,
\begin{align*}
\dfrac{\prod\limits_{k=1}^{n-1} \left(n^2-k^2\right)}{\left(n-1\right)!^2}
&= \dfrac{\left(n-1\right)! \left( \left(2n-1\right) \left(2n-2\right) \cdots \left(n+1\right) \right)}{\left(n-1\right)!^2} \\
&= \dfrac{\left(2n-1\right) \left(2n-2\right) \cdots \left(n+1\right) }{\left(n-1\right)!} \\
&= \dfrac{\left(2n\right) \cdot \left( \left(2n-1\right) \left(2n-2\right) \cdots \left(n+1\right) \right)}{\left(2n\right) \cdot \left(n-1\right)!} \\
&= \dfrac{\left(2n\right) \left(2n-1\right) \left(2n-2\right) \cdots \left(n+1\right) }{2 n \cdot \left(n-1\right)!} \\
&= \dfrac{\left(2n\right) \left(2n-1\right) \left(2n-2\right) \cdots \left(n+1\right) }{2 n!}
\qquad \left(\text{since } n \cdot \left(n-1\right)! = n! \right) \\
&= \dfrac{1}{2} \cdot \underbrace{\dfrac{\left(2n\right) \left(2n-1\right) \left(2n-2\right) \cdots \left(n+1\right) }{n!}}_{\substack{=\dbinom{2n}{n}\\ \text{(by the definition of }\dbinom{2n}{n}\text{)}}} \\
&= \dfrac{1}{2} \dbinom{2n}{n}.
\end{align*}
Taking the reciprocal of both sides, we obtain
\begin{align*}
\dfrac{\left(n-1\right)!^2}{\prod\limits_{k=1}^{n-1} \left(n^2-k^2\right)}
= \dfrac{2}{\dbinom{2n}{n}} .
\end{align*}
In other words,
\begin{align*}
\dfrac{\left(n-1\right)!^2}{\left(n^2-1^2\right)\left(n^2-2^2\right)\cdots\left(n^2-\left(n-1\right)^2\right)}
= \dfrac{2}{\dbinom{2n}{n}} .
\end{align*}
|
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|
Solving a congruence of the form $a^x = b \pmod m$ without indices or primitive roots Consider $9^x\equiv 7 \mod 19$. So $9^x\equiv 26 \equiv 45$, $9^{x-1} \equiv 5 \equiv 24 \equiv 43 \equiv 62 \equiv 81$, so $x=3$, and $19 \mid 722$.
But what I really want to solve is $12^x\equiv 17 \mod 25$. Using the same method, $$2^{2x}\cdot3^x \equiv 17 \equiv 42,$$ $$ 2^{2x- 1}3^{x-1}\equiv 7 \equiv 32,$$ $$2^{2x-6}3^{x-1}\equiv 1\equiv 26, $$ $$ 2^{2x-7}3^{x-1}\equiv 13 \equiv38,$$ $$ 2^{2x-8}3^{x-1} \equiv 19\equiv 44, $$ $$2^{2x-10}3^{x-1}\equiv 11 \equiv 36 \equiv (2^2)(3^2),$$ so $2x-10=2$, hence $x=6$, and $x-1=2$, so $x=3$.
Why doesn't this work? Is it because $12$ is not the power of a prime and $9$ is? Any help is appreciated!
|
It is probably quickiest & easiest to calculate the powers of $12$ modulo $25$ ..
\begin{eqnarray*}
12,19,3,11,7,9,8,21,2,24 \\
13,6,22,14,18,16,\color{red}{17} \cdots
\end{eqnarray*}
So
\begin{eqnarray*}
12^{17}=17 \pmod{25}.
\end{eqnarray*}
|
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|
Determining an orthonormal set of basis vectors for the linear space The following is example C.4 from Appendix C (Linear Spaces Review) of Introduction to Laplace Transforms and Fourier Series, Second Edition, by Phil Dyke:
Example C.4 Determine an orthonormal set of vectors for the linear space that consists of all real linear functions: $$\{a+bx:a,b\in\mathbb{R}\ 0\leq x\leq1\}$$ using an inner product $$\langle f,g\rangle=\int_0^1f g\,dx.$$ Solution The set $\{1,x\}$ forms a basis, but is is not orthogonal. Let $a+bx$ and $c+dx$ be two vectors. In order to be orthogonal we must have $$\langle a+bx,c+dx\rangle=\int_0^1(a+bx)(c+dx)\,dx=0.$$ Performing the elementary integration gives the following condition on the constants $a,b,c$ and $d$ $$ac+\frac{1}{2}(bc+ad)+\frac{1}{3}bd=0.$$ In order to be orthonormal too we also need $$\|a+bc\|=1\,\text{ and }\,\|c+dx\|=1$$ and these give, additionally, $$a^2+b^2=1,\ c^2+d^2=1.$$ There are four unknowns and three equations here, so we can make a convenient choice. Let us set $$a=-b=\frac{1}{\sqrt{2}}$$ which gives $$\frac{1}{\sqrt{2}}(1-x)$$
as one vector. The first equation now gives $3c=-d$ from which $$c=\frac{1}{\sqrt{10}},\ \ d=-\frac{3}{\sqrt{10}}.$$ Hence the set $\{(1-x)/\sqrt{10},(1-3x)/\sqrt{10}\}$ is a possible orthonormal one.
$\ $ $ \ $ Of course there are infinitely many possible orthonormal sets, the above was one simple choice. The next definition follows naturally.
I have the following questions:
*
*How do we determine $a^2 + b^2 = 1$ and $c^2 + d^2 = 1$ from $\|a + bx\| = 1$ and $\|c + dx\| = 1$? This seems similar to the norm of a complex number $a + bi$, but we're not dealing with complex numbers in this case, since we're dealing with the space of all real linear functions, so I'm not sure how these are being derived?
*If we have $a = -b = \dfrac{1}{\sqrt{2}}$ and $3c = -d$, then we have the following: $$\begin{align}
\dfrac{1}{\sqrt{2}}c + \dfrac{1}{2} \left[ \left( \dfrac{-1}{\sqrt{2}} \right)c + \left( \dfrac{1}{\sqrt{2}} \right) (-3c) \right] + \dfrac{1}{3} \left( \dfrac{-1}{\sqrt{2}} \right)(-3c) = 0
\\
\rightarrow \dfrac{c}{\sqrt{2}} - \dfrac{c}{2 \sqrt{2}} - \dfrac{3c}{2\sqrt{2}} + \dfrac{c}{\sqrt{2}} = 0
\\
\rightarrow \dfrac{2c}{\sqrt{2}} - \dfrac{4c}{2\sqrt{2}} = 0
\\
\rightarrow 0 = 0\ ?\end{align}$$
Have I made an error? Where does the $c = \dfrac{1}{\sqrt{10}}$ and $-\dfrac{3}{\sqrt{10}}$ come from?
I would greatly appreciate it if people could please take the time to clarify these.
EDIT:
The following is proved in the textbook:
Example C.3 Prove that $\|a\|=\sqrt{\langle\mathbf{a}.\mathbf{a}\rangle}\in V$ is indeed a norm for the vector space $V$ with inner product $\langle,\,\rangle$.
Which seems to suggest that $\|x\| = \sqrt{\langle x,x\rangle}$?
|
By definition of norm induced from inner product, $||x||^2 = \langle x,x\rangle$.
That is, if $||a+bx|| = 1$ then $\langle a+bx,a+bx\rangle = 1$ ,so writing this down:
$$
\int_{0}^1 (a+bx)(a+bx) dx = 1 \implies \int_0^1 (a^2 + 2abx + b^2x^2) dx = 1 \\ \implies a^2 + ab + \frac {b^2}3 = 1
$$
therefore, the statement that $a^2 + b^2 = 1$ is FALSE.
This is actually quite clear with an example : $a=0 , b=1$ satisfies $a^2+b^2 = 1$, and gives the polynomial $x$, but $||x||^2 = \int_0^1 x^2 = \frac 13$, so $||x|| \neq 1$. Instead, the other statement given is correct. Replacing $a$ and $b$ by $c$ and $d$ gives you the other analogously correct statement.
Once this happens, you may set $a,b$ to any suitable values, and check what happens to $c$ and $d$.
For example, set $b = 0$ : from the above equation, this forces $a = \pm 1$, we will take $a =1$.
From the equation that $\langle a+bx,c+dx\rangle = 0$ that the author has derived in your question above, substituting (and cancelling $b$) and rearranging gives $2c+d = 0$, so $d = -2c$.
This must be combined with $c^2 + cd + \frac{d^2}{3} = 1$. Setting it, we get $c^2(1 - 2 + \frac 43) = 1$, so $c^2 = 3$. Just take $c = + \sqrt 3$, so $d = -2\sqrt 3$.
In this manner, we may verify that the polynomial $a+bx = 1$ and $c + dx = \sqrt 3(1 - 2x)$ form an orthonormal basis for the space.
|
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"url": "https://math.stackexchange.com/questions/2839288",
"timestamp": "2023-03-29T00:00:00",
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|
Prove or disprove that $ \sum\limits_{k = 1 }^T f(k)=0 $ where $f(m)=\sum\limits_{n = 1 }^ m (-1)^n \sin(\frac{n(n+1)(2n+1)}{6}x) $ $$f(m)=\sum\limits_{n = 1 }^ m (-1)^n \sin\left(\frac{n(n+1)(2n+1)}{6} \frac{a \pi}{b}\right) \tag 1 $$
Where $a,b,m$ positive integers.
I have tested in WolframAlpha for many $a$ and $b$ values.
I conjecture (1) without proof that $f(m)$ function is periodic when $a,b,m$ positive integers and the sum of $f(m)$ is $0$ between period.
Edit: In other way to express my claim above in my conjecture ($1$) that
$ \sum\limits_{k = 1 }^T f(k)=0 $
where ($T$) is the period value.
The wolframalpha link for testing some $a,b$ values
I also conjecture (2) without proof that the sum of $f(m)$ should be zero if $x$ is any real number.
$$f(m)=\sum\limits_{n = 1 }^ m (-1)^n \sin\left(\frac{n(n+1)(2n+1)}{6}x\right) \tag 2 $$
$$ \lim\limits_{n \to \infty}\sum\limits_{k = 1 }^ n f(k)=0 \tag 3 $$
*
*What is the period formula when $a,b$ are positive integers?
*Please help me to prove my conjectures 1 and 2 or disprove .
Note that:$$\sum\limits_{k = 1 }^ n k^2= \frac{n(n+1)(2n+1)}{6} $$
EDIT:
The period value is ($T$) and $f(m)$ satisfies $f(m)=f(m+kT)$ relation where $k$ is non-negative integer.
Period values for some $a,b$ values:
$a=3$, $b=17$ ,$x=\frac{3 \pi}{17} \Rightarrow T=68$ (this example is given in the link) and $ \sum\limits_{k = 1 }^{68} f(k)=0 $
$a=1$, $b=2$ ,$x=\frac{ \pi}{2} \Rightarrow T=8$ and $ \sum\limits_{k = 1 }^8 f(k)=0 $
$a=1$, $b=3$ ,$x=\frac{ \pi}{3} \Rightarrow T=36$ and $ \sum\limits_{k = 1 }^{36} f(k)=0 $
$a=1$, $b=4$ ,$x=\frac{ \pi}{4} \Rightarrow T=16$ and $ \sum\limits_{k = 1 }^{16} f(k)=0 $
$a=1$, $b=5$ ,$x=\frac{ \pi}{5} \Rightarrow T=20$ and $ \sum\limits_{k = 1 }^{20} f(k)=0 $
$a=1$, $b=6$ ,$x=\frac{ \pi}{6} \Rightarrow T=72$ and $ \sum\limits_{k = 1 }^{72} f(k)=0 $
$a=1$, $b=7$ ,$x=\frac{ \pi}{7} \Rightarrow T=28$ and $ \sum\limits_{k = 1 }^{28} f(k)=0 $
$a=2$, $b=7$ ,$x=\frac{ 2\pi}{7} \Rightarrow T=14$ and $ \sum\limits_{k = 1 }^{14} f(k)=0 $
$a=3$, $b=7$ ,$x=\frac{ 3\pi}{7} \Rightarrow T=56$ and $ \sum\limits_{k = 1 }^{56} f(k)=0 $
$a=4$, $b=7$ ,$x=\frac{ 4\pi}{7} \Rightarrow T=14$ and $ \sum\limits_{k = 1 }^{14} f(k)=0 $
$a=5$, $b=7$ ,$x=\frac{ 5\pi}{7} \Rightarrow T=28$ and $ \sum\limits_{k = 1 }^{28} f(k)=0 $
Thanks a lot for answers.
Please note that: I have posted a new question to generalize the problem. the link to the question
|
1. Settings and main results
Let $a$ and $b$ be relatively prime integers. Let $\theta, e, F $ be defined by
\begin{align*}
\theta_n = \frac{a}{b}\left(\sum_{k=1}^{n} k^2 \right) + n, \qquad
e_n = \exp\{i\pi\theta_n\}, \qquad
F_m = \sum_{n=1}^{m} e_n.
\end{align*}
(Here, we extend $\sum$ by additivity to allow non-positive arguments for $\theta$ and $F$.) This definition is related to OP's question by $f(m) = \operatorname{Im}\left( F_m \right)$. In view of this, we will prove the following result.
Proposition. The smallest positive period $T_{\min}$ of $\{e_n\}$ is given by
$$ T_{\min} = \frac{4\gcd(b, 3)}{\gcd(a, 2)}b. \tag{1} $$
Moreover, $F$ has period $T_{\min}$ and satisfies
$$ \operatorname{Im} \left( \sum_{m=1}^{T_{\min}} F_m \right) = 0. $$
To establish this result, we aim to prove the following lemmas.
*
*
Lemma 1. An integer $T$ is a period of $\{e_n\}$ if and only if the following conditions hold
$ $
*
*$\text{(P1)} \ $ $T = 2bp$ for some integer $p$, and
*$\text{(P2)} \ $ $2 \mid ap$ and $3 \mid ap(2bp+1)(4bp+1)$.
*
Lemma 2. Let $T$ be a period of $\{e_n\}$ and write $U = \frac{T}{2}$. Then
*
*$e_{n+U} = e_U e_n$ and $e_{U-1-n} = -e_U \overline{e_n}$.
*$e_U = \pm 1$ and $e_{U-1} = -e_U$.
*If $e_U = 1$, then $U$ is also a period of $\{e_n\}$.
Let us see how this leads to the desired main result.
Proof of Proposition using Lemmas. It is easy to check that $\text{(1)}$ is the smallest positive $T$ satisfying both $\text{(P1)}$ and $\text{(P2)}$. Writing $U = T_{\min}/2$ for simplicity, it follows from the minimality of $T_{\min}$ and Lemma 2 that $U$ is not a period of $\{e_n\}$. In particular, we have $e_U = -1$. Then
$$ F_{T_{\min}}
= \sum_{n=1}^{U} (e_n + e_{U+n})
= \sum_{n=1}^{U} (e_n - e_n)
= 0. $$
Moreover, since $e_{U-1} + e_U = 0$ and $e_{-1} + e_0 = 0$, we have
$$ F_U
= \sum_{n=-1}^{U-2} e_n
= \sum_{n=1}^{U} e_{U-1-n}
= \sum_{n=1}^{U} \overline{e_n}
= \overline{F_U}. $$
This implies that $\operatorname{Im}(F_U) = 0$. Finally, it follows that
$$ \sum_{m=1}^{T} F_m
= \sum_{m=1}^{U} (F_m + F_{U+m})
= \sum_{m=1}^{U} (F_m + F_U - F_m)
= UF_U $$
and therefore $ \operatorname{Im}\left(\sum_{m=1}^{T} F_m \right) = 0$ as required.
2. Proofs of lemmas
Before proving these claims, we introduce an auxlilary quantity which will be useful throughout the solution. Set
$$ \Delta_{m,n} = \theta_{m+n} - \theta_m - \theta_n = \frac{a}{b}mn(m+n+1). $$
It is obvious that $e_{m+n} = e_m e_n \exp\{i\pi\Delta_{m,n}\}$ holds for any $m, n$. In particular, this implies that
$$ \text{$T$ is a period of $\{e_n\}$}
\quad \Leftrightarrow \quad
\begin{cases}
\theta_T \equiv 0 \pmod{2}, \\
\Delta_{T,n} \equiv 0 \pmod{2} \ \forall n \in \mathbb{Z}
\end{cases} \tag{2}$$
Now we proceed to prove Lemma 1 first.
Proof of Lemma 1. One direction is almost immediate. Indeed, assume that both $\text{(P1)}$ and $\text{(P2)}$ hold. Then we easily check that both $\Delta_{n,T}$ and $\theta_T$ are even integers, hence $T$ is a period in view of $\text{(2)}$. So we focus on proving the other direction.
Assume that $T$ is a period of $\{e_n\}$. Using $\text{(2)}$, we know that both
$$ \Delta_{T,2} - 2\Delta_{T,1} = \frac{2aT}{b} \quad \text{and} \quad \Delta_{T,2} - 3\Delta_{T,1} = -\frac{aT^2}{b} $$
are all even integers. Since $a$ and $b$ are relatively prime, the first identity implies that $q = T/b$ is an integer and hence the same is true for $S = qa = aT/b$. Then the second identity tells that $2 \mid ST$. Now let us expand $\theta_T$ as
$$ \theta_T = \frac{S(T+1)(2T+1)}{6} + T = \frac{S(2T^2 + 1)}{6} + \frac{ST}{2} + T. $$
Since $\frac{ST}{2} + T$ is integer and $2 \nmid 2T^2 + 1$, we obtain $2 \mid S$. Then we find that $2 \mid T$ as well, for otherwise $6(\theta_T - T)$ is not a multiple of $4$ while $6(\theta_T - T) = S(T+1)(2T+1) $ is a multiple of $4$, which is a contradiction.
So far we have proved that $b \mid T$ and $2 \mid S, T$. Since $q = \gcd(S, T)$, we may write $q = 2p$, proving $\text{(P1)}$. Plugging this back to $\theta_T$,
$$ 0
\equiv \theta_T
\equiv \frac{S(T+1)(2T+1)}{6}
\equiv \frac{ap(2bp+1)(4bp+1)}{3} \pmod {2}, $$
from which $\text{(P2)}$ follows. ////
Proof of Lemma 2. Let $T$ be a period of $\{e_n\}$ and let $p$ be as in Lemma 1. Write $U = \frac{T}{2}$. Then
$$ \Delta_{n,U} = apn(bpn+n+1)
\quad \text{and} \quad
\Delta_{n,U-1-n} = apn(bp-n-1) $$
are multiples of $ap$, which is even. So
$$e_{n+U} = e_n e_U \exp\{i\pi\Delta_{n,U}\} = e_n e_{U}.$$
Then plugging $n = U$ yields $1 = e_T = e_U^2$ and hence $e_U = \pm 1$. Similarly,
$$e_{U-1} = e_n e_{U-1-n} \exp\{i\pi\Delta_{n,U-1-n}\} = e_n e_{U-1-n}. $$
Then plugging $n = -1$ yields $e_{U-1} = e_{-1}e_U = -e_U$ and $e_{U-1-n} = -e_U \overline{e_n}$ as required. Finally, if $e_U = 1$, then we have $e_{n+U} = e_n$ and therefore $U$ is also a period. ////
|
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"timestamp": "2023-03-29T00:00:00",
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|
Solving $a! + b! = 2^n$
$a! + b! = 2^n$, find $a,b,n$
My obseravtions thus far:
*
*If $a>4$, $b<4$, otherwise $a! + b! = 0 \mod 10$
*If $a=1$, $b=1$
If we try all combination with the numbers 2 to 4, we get following solutions:
(1,1,1), (2,2,2), (2,3,3),(3,2,3)
Are there more solutions and how to find them?
|
There are no other solutions in $\mathbb N = \{ 1, 2, \dots \}$. There are a few more if you allow $0 \in \mathbb N$.
Indeed, we may assume that $a\le b$. Then $2^n = a! + b! = (a!)(1+c)$ and so $a!$ divides $2^n$. Therefore, $a!=1$ or $a!=2$, and so $a=1$ or $a=2$.
If $a=1$, then $b! = 2^n-1$ is odd and so $b!=1$, that is $b=1$. Then $n=1$.
If $a=2$, then $b! = 2^n-2 = 2(2^{n-1}-1)$ is even but not a multiple of $4$. Therefore, $b\le 3$.
If $a=2$ and $b=1$, then $a!+b!=3$, not a power of $2$.
If $a=2$ and $b=2$, then $a!+b!=4=2^2$, and so $n=2$.
If $a=2$ and $b=3$, then $a!+b!=8=2^3$, and so $n=3$.
If you allow $0 \in \mathbb N$, then you also get $(a,b,c)=(0,0,1), (0,1,1), (1,0,1)$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Find the minimum value of $\frac{a+b+c}{b-a}$ Let $f(x)=ax^2+bx+c$ where $(a<b)$ and $f(x)\geq 0$ $\forall x\in R$.
Find the minimum value of $$\frac{a+b+c}{b-a}$$
If $f(x)\geq 0$ $\forall x\in R$ then $b>a>0$ and $b^2-4ac\leq 0$ implying that $c>0$. After this not able to find way out.
|
If $f(x) \ge 0\to b^2-4ac \le 0$ then forming the lagrangian
$$
L(a,b,c,\lambda,\epsilon) =\phi(a,b,c) +\lambda(b^2-4ac+\epsilon^2)
$$
with
$$
\phi(a,b,c) = \frac{a+b+c}{b-a}
$$
The stationary points are computed by solving
$$
\nabla L = \left\{
\begin{array}{rcl}
\frac{a+b+c}{(b-a)^2}-4 c \lambda +\frac{1}{b-a}=0 \\
-\frac{a+b+c}{(b-a)^2}+2 b \lambda +\frac{1}{b-a}=0 \\
\frac{1}{b-a}-4 a \lambda =0 \\
b^2+\epsilon ^2-4 a c=0 \\
2 \epsilon \lambda =0 \\
\end{array}
\right.
$$
we obtain
$$
\left[
\begin{array}{cccccc}
a & b & c & \lambda & \epsilon & \phi\\
-\frac{b}{2} & b & -\frac{b}{2} & -\frac{1}{3 b^2} & 0 & 0 \\
\frac{b}{4} & b & b & \frac{4}{3 b^2} & 0 & 3 \\
\end{array}
\right]
$$
hence the feasible solution is
$$
a = \frac b4, c = b
$$
giving a minimum value of $3$
|
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|
For $x\geq 0$, what is the smallest value of $\frac{4x^2+8x+13}{6(x+1)}$? I know that I have to use the AM-GM inequality.
I tried separating the fraction:
$$\frac{4x^2+2x+7}{6(x+1)} + \frac{6(x+1)}{6(x+1)}$$
However, it doesn't seem to make either side of the inequality into a number.
I would appreciate some help, thanks!
|
$$\frac {4x^2+8x+13}{6(x+1)}$$
$$= \frac {2(x+1)}{3} + \frac {3}{2(x+1)}$$
$$ = y+\frac {1}{y} \ge 2$$
|
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|
Is there a way to show that $e^x=\lim_{n\to \infty }\left(1+\frac{x}{n}\right)^n$? I know that $$e:=\lim_{n\to \infty }\left(1+\frac{1}{n}\right)^n,$$
by definition. Knowing that, I proved successively that $$e^{k}=\lim_{n\to \infty }\left(1+\frac{k}{n}\right)^n,$$
when $k\in \mathbb N$, $k\in \mathbb Z$ and $k\in\mathbb Q$. Now, I was wondering : how can I extend this result over $\mathbb R$ ? I tried to prove that $f_n(x):=(1+\frac{x}{n})^n$ converge uniformly on $\mathbb R$ but unfortunately it failed (I'm not sure that it's even true). Any idea ?
My idea was to define the function $x\longmapsto e^x$ as $$e^x=\begin{cases}e^x& x\in \mathbb Q\\ \lim_{n\to \infty }e^{k_n}&\text{if }k_n\to x \text{ and }(k_n)\subset \mathbb Q\end{cases}.$$
But to conclude that $$e^x=\lim_{n\to \infty }\left(1+\frac{x}{n}\right)^n,$$
I need to prove that $f_n(x)=\left(1+\frac{x}{n}\right)^n$ converge uniformly on a neighborhood of $x$, but I can't do it. I set $$g_n(x)=f_n(x)-e^x,$$
but I can't find the maximum on a compact that contain $x$, and thus can't conclude.
|
Left side:
The exponential function may be written as a Taylor series:
$e^x=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+...$
Right side:
$(1+\frac{x}{n})^n$ is a binomial expansion like:
$(1+y)^n=\binom{n}{0}y^0+\binom{n}{1}y^1+\binom{n}{2}y^2+...+\binom{n}{n-1}y^{n-1}+\binom{n}{n}y^n$
Where $\binom{n}{k}$ is the Binomial coefficient given by the formula :
$\binom{n}{k}=\frac{n!}{k!(n-k)!}$
Some basic properties of $\binom{n}{k}$:
a)$\binom{n}{0}=1$ because $\frac{n!}{0!(n-0)!}=\frac{n!}{1*n!}$
b)$\binom{n}{1}=n$ because $\frac{n!}{1!(n-1)!}=\frac{(n-1)!*n}{(n-1)!}$
c)$\binom{n}{n-1}=n$ because $\frac{n!}{(n-1)!(n-(n-1))!}=\frac{(n-1)!*n}{(n-1)!*1!}$
d)$\binom{n}{n}=1$ because $\frac{n!}{n!(n-n)!}=\frac{1}{1!}$
e) The formula does exhibit a symmetry that is less evident from the multiplicative formula:
$\binom{n}{k}=\binom{n}{n-k}$
Returning:
$(1+\frac{x}{n})^n=1+n*\frac{x}{n}+\frac{n!}{2!(n-2)!}\frac{x^2}{n^2}+\frac{n!}{3!(n-3)!}\frac{x^3}{n^3}+...+\frac{n!}{3!(n-3)!}\frac{x^{n-3}}{n^{n-3}}+\frac{n!}{2!(n-2)!}\frac{x^{n-2}}{n^{n-2}}+n*\frac{x^{n-1}}{n^{n-1}}+\frac{x^n}{n^n}$
$(1+\frac{x}{n})^n=1+x+\frac{(n-1)n}{n^2}\frac{x^2}{2!}+\frac{(n-2)(n-1)n}{n^3}\frac{x^3}{3!}+...+\frac{(n-2)(n-1)n}{3!}\frac{x^{n-3}}{n^{n-3}}+\frac{(n-1)n}{2!}\frac{x^{n-2}}{n^{n-2}}+\frac{x^{n-1}}{n^{n-2}}+\frac{x^n}{n^n}$
$(1+\frac{x}{n})^n=1+x+\frac{n-1}{n}\frac{x^2}{2!}+\frac{(n-2)(n-1)}{n^2}\frac{x^3}{3!}+...+\frac{(n-2)(n-1)}{n^{n-4}}\frac{x^{n-3}}{3!}+\frac{n-1}{n^{n-3}}\frac{x^{n-2}}{2!}+\frac{x^{n-1}}{n^{n-2}}+\frac{x^n}{n^n}$
Let's analyze what happens for $n\rightarrow\infty$-here we have three types of limits:
-First type:
$\displaystyle\lim_{n \to \infty}\frac{n-1}{n}=\displaystyle\lim_{n \to \infty}[1+\frac{1}{n}]=1+0=1$
$\displaystyle\lim_{n \to \infty}\frac{(n-2)(n-1)}{n^2}=\displaystyle\lim_{n \to \infty}\frac{n^2-3n+2}{n^2}=\displaystyle\lim_{n \to \infty}[1-\frac{3}{n}+\frac{2}{n^2}]=1-0+0=1$
Hense $\displaystyle\lim_{n \to \infty}\Bigg(\frac{\displaystyle\prod_{i=1}^{k} (n-i)}{n^k}\Bigg)=1$
-Second type is
$\displaystyle\lim_{n \to \infty} \frac{x^{n-\alpha}}{n^{n-\beta}}$-Because ${n^{n-\beta}}$ grows much fasten than $x^{n-\alpha}$ hense: $\displaystyle\lim_{n \to \infty} \frac{x^{n-\alpha}}{n^{n-\beta}}=0$
-Third type:
$\displaystyle\lim_{n \to \infty}\Bigg(\frac{\displaystyle\prod_{i=1}^{k} (n-i)}{n^{n-k-1}}\frac{x^{n-k}}{k!}\Bigg)$
We have to show on the biggest power (similar to the first type) as the most relevant:
$\frac{\displaystyle\prod_{i=1}^{k} (n-i)}{n^{n-k-1}}\frac{x^{n-k}}{k!}\sim\frac{n^{k-1} }{n^{n-k-1}}\frac{x^{n-k}}{k!}=n^{k-1-(n-k-1)}\frac{x^{n-k}}{k!}=n^{2k-n}*\frac{x^{n-k}}{k!}=\frac{1}{k!}*\frac{x^{n-k}}{n^{n-2k}}$
Again: ${n^{n-\beta}}$ grows much faster than $x^{n-\alpha}$
Hense: $\displaystyle\lim_{n \to \infty}\Bigg(\frac{\displaystyle\prod_{i=1}^{k} (n-i)}{n^{n-k-1}}\frac{x^{n-k}}{k!}\Bigg)=0$
Our right side equals:
$\displaystyle\lim_{n \to \infty}(1+\frac{x}{n})^n=1+x+1*\frac{x^2}{2!}+1*\frac{x^3}{3!}+...+0+0+0+0$
$\displaystyle\lim_{n \to \infty}(1+\frac{x}{n})^n=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+...$
We got the same elements like in the Taylor series of $e^x$. Q.E.D.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Applying the quadratic Tschirnhausen transformation As per my previous question, I attempted to take dxiv's approach, though I can't seem to make much headway. Considering the simpler problem $x^3=x+a$ and the substitution $y=x^2+mx+n$, I got the following:
\begin{align}y^2&=x^4+2mx^3+(m^2+2n)x^2+2mnx+n^2\\&=(m^2+2n+1)x^2+(2mn+2m+a)x+(n^2+2am)\\&=\dot\ell x^2+\dot mx+\dot n\end{align}
\begin{align}y^3&=(m^2+2n+1)x^4+(2mn+2m+a)x^3+(n^2+2am)x^2\\&~~~~~+(m^3+2mn+m)x^3+(2m^2n+2m^2+am)x^2+(mn^2+2am^2)x\\&~~~~~+(m^2n+2n^2+n)x^2+(2mn^2+2mn+an)x+(n^3+2amn)\\&=(m^2+2n+1)x^4+(m^3+4mn+3m+a)x^3\\&~~~~~+(3m^2n+2m^2+3n^2+3am+n)x^2\\&~~~~~+(3mn^2+2am^2+2mn+an)x\\&~~~~~+(n^3+2amn)\\&=(3m^2n+3m^2+3n^2+3am+3n+1)x^2\\&~~~~~+(3mn^2+m^3+3am^2+6mn+3m+3an+2a)x\\&~~~~~+(am^3+n^3+6amn+3am+a^2)\\&=\ddot\ell x^2+\ddot mx+\ddot n\end{align}
Finding suitable coefficients for a cubic in $y$ leads me to
$$(2m-m^3+a-1)y^3+uy^2+vy=w\implies$$
\begin{cases}u=\ddot\ell m-\ddot m=3m^3n+3am^2-3mn-2m-3an-2a\\v=\dot\ell^2m-\ddot\ell\dot m-\dot\ell\ddot\ell m-\dot\ell\ddot m=\dots\end{cases}
which seems to leave me with quartics and possibly worse, when trying to set $u=v=0$.
Do I really need to expand that $v$? Have I made a mistake? Or is this simply a terrible path to go down and I should try using Newton's identities, as explained here?
|
I can follow the calculations all the way down to the last cubic. But when equating the following determinant to $\,0\,$ ...
$$
\scriptsize{
\left|
\begin{matrix}
1 & m & n - y\\
m^2+2n+1 & 2mn+2m+a & n^2+2am-y^2 \\
3m^2n+3m^2+3n^2+3am+3n+1 & 3mn^2+m^3+3am^2+6mn+3m+3an+2a & am^3+n^3+6amn+3am+a^2 - y^3
\end{matrix}
\right|
}
$$
... the coefficient I get for $\,y^2\,$ is $\,\color{blue}{u} = - (3 m^3 n \color{red}{+ 2 m^3} - 3 m n - 2 m - 3 a n - 2 a)\,$, instead of the posted $\,3m^3n\color{red}{+3am^2}-3mn-2m-3an-2a\,$. This $\,\color{blue}{u}\,$ factors as $\,(3 n + 2) ( - m^3 + a + m)\,$, which gives the immediate choice $\,n = -2/3\,$ to drop the $\,y^2\,$ term, then what's left is a quadratic in $\,m\,$ to drop the linear term next.
This is consistent with WA's collect[ resultant[ x^3 - x - a, y - x^2 - m x - n, x ], y ] $\small= -a^2 - a m^3 + y (-3 a m - m^2 + 3 n^2 + 4 n + 1) + 3 a m n + a m + m^2 n - n^3 - 2 n^2 + \color{red}{(-3 n - 2)} y^2 - n + y^3$.
[ EDIT ] To simplify the calculations somewhat, one can multiply the first column of the determinant above by $\,m\,$ and subtract it from the second column, which gives:
$$
\scriptsize{
\left|
\begin{matrix}
1 & 0 & n - y\\
m^2+2n+1 & -m^3+m+a & n^2+2am-y^2 \\
3m^2n+3m^2+3n^2+3am+3n+1 & - 3m^3n-2m^3+3mn+2m+3na+2a & am^3+n^3+6amn+3am+a^2 - y^3
\end{matrix}
\right| \\
= \left|
\begin{matrix}
1 & 0 & n - y\\
m^2+2n+1 & -m^3+m+a & n^2+2am-y^2 \\
3m^2n+3m^2+3n^2+3am+3n+1 & (- m^3 + m + a)(3 n + 2) & am^3+n^3+6amn+3am+a^2 - y^3
\end{matrix}
\right| \\
= -(m^3-m-a) \left|
\begin{matrix}
1 & 0 & n - y\\
m^2+2n+1 & 1 & n^2+2am-y^2 \\
3m^2n+3m^2+3n^2+3am+3n+1 & 3 n + 2 & am^3+n^3+6amn+3am+a^2 - y^3
\end{matrix}
\right|
}
$$
The latter determinant is a monic cubic in $y$ which (of course) matches WA's resultant.
|
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|
Is $\lim_{x\to -\infty} \frac{5x+9}{3x+2-\sqrt{4x^2-7}}=1$?
Evaluate $$\lim_{x\to-\infty} \frac{5x+9}{3x+2-\sqrt{4x^2-7}}$$
My attempt:
$$\lim_{x\to -\infty} \frac{5x+9}{3x+2-\sqrt{4x^2-7}}=\lim_{x\to-\infty}\frac{5x}{-\sqrt{4x^2}}=\frac{-5}{2}$$ According to the answer key, it actually equals $1$.
Thanks in advance.
|
Let $x<0.$ Then
$$\lim_{x\to-\infty} \frac{5x+9}{3x+2-\sqrt{4x^2-7}}=\lim_{x\to-\infty} \dfrac{5+\dfrac{9}{x}}{3+\dfrac{2}{x}+\sqrt{4-\dfrac{7}{x^2}}}=\frac{5+0}{3+0+\sqrt{4-0}}=1.$$
|
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|
Which Sign do I choose using the Half Angle Formula for sin for this? I'm evaluating $\sin\left(\frac{1}{2}\sin^{-1}\left(-\frac{7}{25}\right)\right).$
The first thing I did was rewrite it as $\sin\left(\frac{\beta }{2}\right)$
Then I said that
$\sin\left(\beta \right)=-\frac{7}{25}$
Using the Pythagorean Identity I found $cos\left(\beta \right)$
$\cos\left(\beta \right)=\pm\sqrt{1-\left(-\frac{7}{25}\right)^2}$
So $\cos\left(\beta \right)=\pm\frac{24}{25}$
Then to choose the sign of $\cos\left(\beta \right)$, I did this:
1) $\sin^{-1}\left(...\right)$: QI or QIV
2) $\sin\left(\beta \right)>0$ QI or QII
3) $\rightarrow \cos\left(\beta \right)$ is in QI
$\cos\left(\beta \right)=+\frac{24}{25}$
Then I applied this to the Half Angle Formula for Sine:
$\pm\sqrt{\frac{1}{25}\left(\frac{1}{2}\right)}$
$=\pm\frac{1}{5}\left(\frac{\sqrt{2}}{2}\right)$
$=\pm\frac{\sqrt{2}}{10}$
But which sign do I choose?
|
Let $\sin^{-1}\left(-\frac{7}{25}\right)=t$
As for real $x,-\dfrac\pi2\le\sin^{-1}x\le\dfrac\pi2,-\dfrac\pi2\le t<0$ as $t<0$
$\implies\sin\left(\dfrac t2\right)<0$
and $\cos t=+\sqrt{1-\sin^2t}=?$
$\sin\left(\dfrac t2\right)=-\sqrt{\dfrac{1-\cos t}2}=?$
|
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|
Prove there is no $x, y \in \mathbb Z^+ \text{ satisfying } \frac{x}{y} +\frac{y+1}{x}=4$ Prove that there is no $x, y \in \mathbb Z^+$ satisfying
$$\frac{x}{y} +\frac{y+1}{x}=4$$
I solved it as follows but I seek better or quicker way:
$\text{ Assume }x, y \in \mathbb Z^+\\ 1+\frac{y+1}{y}+\frac{x}{y} +\frac{y+1}{x}=1+\frac{y+1}{y}+4 \\
\Rightarrow \left(1+\frac{x}{y}\right)\left(1+\frac{y+1}{x}\right)=6+\frac{1}{y}\\
\Rightarrow (x+y)(x+y+1)=x(6y+1)\\
\Rightarrow x\mid (x+y) \;\text{ or }\; x\mid (x+y+1)\\
\Rightarrow x\mid y \;\text{ or }\; x\mid (y+1)\\
\text{Put}\; y=nx ,n \in \mathbb Z^+\;\Rightarrow\; \frac{x}{nx} +\frac{nx+1}{x}=4 \;\Rightarrow\; \frac{1}{n} +\frac{1}{x}=4-n \;\rightarrow\;(1)\\
\text{But}\; \frac{1}{n} +\frac{1}{x} \gt 0 \;\Rightarrow\; 4-n \gt 0 \;\Rightarrow\; n \lt 4\\
\text{Also}\; \frac{1}{n},\frac{1}{x}\le 1 \;\Rightarrow\; \frac{1}{n} +\frac{1}{x} \le 2 \;\Rightarrow\; 4-n \le 2 \;\Rightarrow\; n \ge 2\\
\;\Rightarrow\; n=2 \;\text{ or }\; 3, \;\text{substituting in eq. (1), we find no integral values for } x.\\ \text{The same for the other case.}\\
$
So is there any other better or intelligent way to get this result?
|
Hint
$$\frac{x}{y} +\frac{y+1}{x}=4\to \frac{x}{y}+\frac{y}{x}+\frac{1}{x}=4$$
Call $x/y=t\in \Bbb Q$.
$$t+\frac{1}{t}+\frac{1}{x}=4\to xt^2+t(1-4x)+x=0$$
By Rational Roots Theorem the candidates to be a rational root, $t$, are $\{\pm1,\pm x,\pm 1/x\}$.
Now, test every root and check the value of $x$ you get.
Can you finish?
|
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|
How is discriminant related to real $x$? Question in my text book
Solve for range of the function, $$y=\frac{x^2+4x-1}{3x^2+12x +20}$$
Text book says, cross multiply and express the obtained equation as a quadratic equation in $x$
So I get $ (3y-1)x^2 + (12y -4)x + (20y-1) = 0$
Now it says find discriminant $D$, so we have,
$$ D = -4 ( 3y-1)( 8y+3)$$
Now it says, set $D≥0$ as $x$ is real.
Wait what?
Isn't $x \in \mathbb{R}$ the domain for a quadratic function? Meaning "$x$" is always real? What does a discriminant got anything to do with $x$ being real, when all discriminant tells us is whether or not the ROOTS are real? Help please.
To be more specific about my doubt, here's an edit.
EDIT : I'm confused, setting discriminant $≥0$ would tell whether or not roots are real, meaning whether the graph of the quadratic function cuts/touches X axis. Now tell me what does this got anything to do with range? As far as I know, quadratic fucntions that don't have real roots are also continous throughout the X axis, meaning there SHOULD always be a corresponding $y-value$
|
$$\frac{x^2+4x-1}{3x^2+12x +20} = \frac{x^2+4x + \frac{20}{3}}{3x^2+12x +20} - \frac{ \frac{23}{3}}{3x^2+12x +20} = \frac{1}{3}- \frac{ \frac{23}{3}}{3x^2+12x +20}$$
Meanwhile, $$ 3x^2 + 12 x + 20 = 3 (x+2)^2 + 8 $$ is always positive
with a minimum value of $8$ when $x=-2,$ so the minimum value of the whole thing is
$$ \frac{1}{3}- \frac{ \frac{23}{3}}{8} = \frac{1}{3}-\frac{23}{24}= -\frac{5}{8} $$
It has an upper bound of $\frac{1}{3}$ but never achieves the bound, it just gets very close when $|x|$ is large.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Prove that $\sum_{j=0}^{\infty}(-1)^j\left(\frac{1}{5+6j}+\frac{1}{1+6j} \right) = \frac{\pi}{3}$ I was trying to solve the integral $$\int_{0}^{\infty}\frac{x^4}{1+x^6}dx$$
using series. Now I'm stuck at the series below.
How to prove that
$$\sum_{j=0}^{\infty}(-1)^j\left(\frac{1}{5+6j}+\frac{1}{1+6j} \right) = \frac{\pi}{3}?$$
|
$\begin{align}J&=\int_0^{\infty}\frac{x^4}{1+x^6}\,dx\\
&=\int_0^{\infty}\frac{2x^2-1}{3(x^4-x^2+1)}\,dx+\frac{1}{3}\int_0^{\infty}\frac{1}{1+x^2}\,dx\\
&=\int_0^{\infty}\frac{2x^2-1}{3(x^4-x^2+1)}\,dx+\frac{1}{3}\Big[\arctan x\Big]_0^{\infty}\\
&=\frac{1}{3}\int_0^{\infty}\frac{2x^2-1}{x^4-x^2+1}\,dx+\frac{1}{6}\pi
\end{align}$
Perform the change of variable $y=\dfrac{1}{x}$,
$\begin{align}K&=\int_0^{\infty}\frac{2x^2-1}{x^4-x^2+1}\,dx\\
&=\int_0^{\infty}\frac{2-x^2}{x^4-x^2+1}\,dx
\end{align}$
Therefore,
$\begin{align}2K&=\int_0^{\infty}\frac{(2x^2-1)+(2-x^2)}{x^4-x^2+1}\,dx\\
&=\int_0^{\infty}\frac{1+x^2}{x^4-x^2+1}\,dx\\
&=\int_{0}^{+\infty}\frac{\frac{1}{x^2}+1}{x^2-1+\frac{1}{x^2}}\,dx\\
&=\int_{0}^{+\infty}\frac{\frac{1}{x^2}+1}{\left(x-\frac{1}{x}\right)^2+1}\,dx
\end{align}$
Perform the change of variable $y=x-\dfrac{1}{x}$,
$\begin{align}2K&=\int_{-\infty}^{+\infty}\frac{1}{1+x^2}\,dx\\
&=\Big[\arctan x\Big]_{-\infty}^{+\infty}\\
&=\pi
\end{align}$
Therefore,
$\begin{align}K&=\frac{1}{2}\pi\end{align}$
Therefore,
$\begin{align}J&=\frac{1}{3}K+\frac{1}{6}\pi\\
&=\frac{1}{3}\times \frac{1}{2}\pi++\frac{1}{6}\pi\\
&=\boxed{\frac{1}{3}\pi}
\end{align}$
|
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|
Solve $\tan {(x-\frac{\pi}{4}})=-\tan{(x+\frac{\pi}{2})}$ The question:
Without the use of a calculator, solve for all values of $x$ if $\tan {(x-\frac{\pi}{4}})=-\tan{(x+\frac{\pi}{2})}.$
Using the compound angle formula for solving equations is normally easy, but I stumbled across this problem.
The $LHS$ is easy to expand, but when you apply the compound formula for the $RHS$,
\begin{align}
\tan{(x+\frac{\pi}{2})} & = \frac{\tan{(x)} + \tan{(\frac{\pi}{2})}}{1-\tan{(x)}\cdot\tan{(\frac{\pi}{2})}} \\
\end{align}
You might notice that this is a problem because I cannot evaluate $\tan{(\frac{\pi}{2})}$. So this is what I tried. First I tried writing
\begin{align}
\tan{(x+\frac{\pi}{2})} & = \frac{\sin{(x+\frac{\pi}{2})}}{\cos{(x+\frac{\pi}{2})}} \\
& = \frac{\cos (x)}{\sin (x)}
\end{align}
which I knew was wrong. Anyone know how to get around this?
|
Hint: $\tan(x)$ is an odd function so $\tan(-x)=-\tan(x)$
Solution:
By above we get $\tan{(x-\pi/4)}=\tan{(-x-\pi/2)}$ so
$$x-\pi/4 \equiv -x-\pi/2 \pmod{\pi}$$
(since the tangent function has a period of $\pi$)
$$2x \equiv -\frac{\pi}{4} \equiv \frac{3\pi}{4} \equiv \frac{7\pi}{4}\pmod{\pi}$$
Solving this yields solutions $x=\pi n +\frac{7\pi}{8}$ and $x=\pi n +\frac{3\pi}{8}$
|
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|
What is the next limit equal to? $$\lim_{n\to\infty}(\sqrt{n^2+n}-\sqrt[3]{n^3+1})=?$$
I tried amplifying the hole substraction to form the formula $$a^3-b^3$$ but didn't worked out. Can you help me figure it out?
|
The key here is the (not so) popular standard limit $$\lim_{x\to a} \frac{x^r-a^r} {x-a} =ra^{r-1}\tag{1}$$ The given expression (the one whose limit is to be evaluated) can be written as $$\dfrac{\sqrt{1+\dfrac{1}{n}}-\sqrt[3]{1+\dfrac{1}{n^3}}}{\dfrac{1}{n}}$$ Now we add and subtract $1 $ in the numerator and see that the expression can be written as $$\dfrac{\sqrt{1+\dfrac{1}{n}}-1}{\dfrac{1}{n}}-\frac{1}{n^2}\cdot\dfrac{\sqrt[3]{1+\dfrac{1}{n^3}}-1}{\dfrac{1}{n^3}}$$ Next we put $x=1+(1/n),y=1+(1/n^3)$ so that $x\to 1,y\to 1$ and the expression can be written as $$\frac{x^{1/2}-1}{x-1}-\frac{1}{n^2}\cdot\frac{y^{1/3}-1}{y-1}$$ and using $(1) $ the desired limit is easily seen to be $$\frac{1}{2}-0\cdot\frac{1}{3}=\frac{1}{2}$$
|
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|
Using inverse Laplace transform to solve differential equation The differential equation is as follows-
$$\frac{d^2 x}{dt^2} + 5 \frac{dx}{dt} + 6x = e^t $$
I use laplace transform to make it to become - $$X(s) = \frac{1}{(s-1)(s+3)(s+2)}$$
where $X(s)$ is the Laplace transform of $X(t)$
So now I am trying to find $X(t)$ using inverse transform.
From partial fractions-
$X(s) = \frac{1}{(s-1)(s+3)(s+2)} = \frac{A}{s-1} + \frac{B}{s+3} + \frac{C}{s+2} $
Numerator - $ 1 = A(s+3)(s+2) + B(s-1)(s+2) + C(s-1)(s+3) $
I am stuck from here on how to carry on this partial fraction
Can I sub all s values to be 0 ?
For example
$1 = A(0+3)(0+2)$
$1= B(0-1)(0+2) $
$1 = C (0-1)(0+3) $
|
$ s=0$ is not the best choice
There are three values to assign to $s$ which makes our life very easy.
$$1 = A(s+3)(s+2) + B(s-1)(s+2) + C(s-1)(s+3)$$
$$ s=1 \implies 12A=1 \implies A=\frac {1}{12}$$
$$ s=-3 \implies 4B=1 \implies B=\frac {1}{4}$$
$$ s=-2 \implies -3C=1 \implies C=\frac {-1}{3}$$
Now you proceed with the inverse Laplace Transform.
|
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|
Determine all real $x$ that satisfy $\det A=0$ I want to find all real $x$ that satisfy
$$
\textrm{det } X=
\begin{vmatrix}
x &2 &2 &2\\
2 &x &2 &2\\
2 &2 &x &2\\
2 &2 &2 &x
\end{vmatrix}\\
$$
My teacher does this by adding the three bottom rows to the top row
$$
\textrm{det } X=
\begin{vmatrix}
x+6 &x+6 &x+6 &x+6\\
2 &x &2 &2\\
2 &2 &x &2\\
2 &2 &2 &x
\end{vmatrix}\\
$$
and then subtracting a row of $2$'s from the bottom three rows
$$
\textrm{det } X=
(x+6)
\begin{vmatrix}
1 &1 &1 &1\\
0 &x-2 &0 &0\\
0 &0 &x-2 &0\\
0 &0 &0 &x-2
\end{vmatrix}.
$$
The answer is
$$
x\in \{-6,2\}.
$$
I think I understand the operations (although subtracting an arbitrary row of numbers from a matrix/determinant row is something I've never seen before, but I don't see why that wouldn't be allowed. Just like you can subtract arbitrary coefficients on both sides of an equation, right?), my main issue is why they are performed.
*
*Why can't I just in the same way subtract a row of $2$'s from the three bottom rows in the first determinant? If I do that I get a different answer.
*I know I want a column of all zeroes except one column-element, but why do I need to perform the first operation beforehand? Is it somehow necessary that all the top row elements to be the same, $(x+6)$?
|
Another way: the determinant is a fourth-degree polynomial $p(x)$ in variable $x$.
It is easy to see that $p(2) = p(-6) = 0$, like you did.
Then take the derivative:
\begin{align}
p'(x) &= \begin{vmatrix}
1 &2 &2 &2\\
0 &x &2 &2\\
0 &2 &x &2\\
0 &2 &2 &x
\end{vmatrix} + \begin{vmatrix}
x &0 &2 &2\\
2 &1 &2 &2\\
2 &0 &x &2\\
2 &0 &2 &x
\end{vmatrix} + \begin{vmatrix}
x &2 &0 &2\\
2 &x &0 &2\\
2 &2 &1 &2\\
2 &2 &0 &x
\end{vmatrix} + \begin{vmatrix}
x &2 &2 &0\\
2 &x &2 &0\\
2 &2 &x &0\\
2 &2 &2 &1
\end{vmatrix} \\
&=4 \begin{vmatrix}
x &2 &2\\
2 &x &2\\
2 &2 &x\\
\end{vmatrix} \\
\end{align}
Taking derivatives again we get $p''(x) = 4\cdot 3 \begin{vmatrix}
x &2 \\
2 &x \\
\end{vmatrix}$ so we can conclude that $p'(2) = p''(2) = 0$.
Therefore $2$ is a root of $p$ with muliplicity $3$ so $$p(x) = (x-2)^3(x+6)$$
Therefore the zeros are indeed only $-6$ and $2$.
|
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|
Prove, using first principles, that $\lim\limits_{(x,y)\to(0,0)}{x.y^2=0}$ My try is:
$\left|x.y^2-0\right|<\left|\sqrt{x^2+y^2}.(x^2+y^2)\right|< \delta.\delta^2=\epsilon$
where $0<\sqrt{x^2+y^2}< \delta$ what did I miss here?
|
Consider $x^2+y^2 <1$,
then $|y| \le 1$, and $y^2 \le |y|$.
Hence:
$|xy^2| \le |xy| \le (1/2)(x^2+y^2)$ .
Let $\epsilon >0$ be given.
Choose $\delta = \min (1,2\epsilon)$
Then $|x^2+y^2| \lt \delta$ implies
$|xy^2| \le$
$(1/2)(x^2+y^2) \lt (1/2)\delta =\epsilon$.
Used: $a^2+b^2 \ge 2|ab|$
|
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|
$\alpha$-limit set in $A=\{x\in \mathbb{R}^2|1<|x|<2\}$ Consider the system
$$
\dot{x}=-y+x(r^4-3r^2+1)\\
\dot{y}=x+y(r^4-3r^2+1)
$$
where $r^2=x^2+y^2$.
Let $A=\{x\in \mathbb{R}^2|1<|x|<2\}$.
Affirmation (fait accompli): $\dot{r}<0$ on the circle $r=1$ and $\dot{r}>0$ on the circle $r=2$.
My question (doubt) is: Why the affirmation above implies that the $\alpha$-limit set any trajectory that starts in $A$ is in $A$?
Thank You!
|
Considering
$$
\left\{
\begin{array}{rcl}
\dot{x}&=&-y+x(r^4-3r^2+1)\\
\dot{y}&=&x+y(r^4-3r^2+1)
\end{array}
\right.
\Rightarrow
\left\{
\begin{array}{rcl}
x\dot{x}&=&-x y+x^2(r^4-3r^2+1)\\
y\dot{y}&=&xy+y^2(r^4-3r^2+1)
\end{array}
\right.
\Rightarrow\frac 12\left(r^2\right)' = r^2(r^4-3r^2+1)
$$
Now analyzing
$$
\frac 12\left(r^2\right)' = r^2(r^4-3r^2+1)
$$
we conclude that for $r^2 = \frac 12\left(3\pm \sqrt 5\right)\;$ we have $\frac 12\left(r^2\right)' = 0\Rightarrow r_1 = \sqrt{\frac 12\left(3-\sqrt 5\right)}$ and $r_2 = \sqrt{\frac 12\left(3+ \sqrt 5\right)}\;$ are orbits.
We know that $r_1 < 1 < r_2 < 2\;$ and also we know that orbits in the phase plane do not cross. Those arguments suffice. Attached a stream plot for the system.
NOTE
We can also conclude from the signal of $(r^4-3r^2+1)$ the kind of attractiveness involved.
|
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|
Simplify a summation with indicator function I have this double summation that I want to simplify:
$$\sum^{q} _{j=-q} \sum^{q} _{i=-q} \mathbb{1}_{ \{h+j-i=0 \}}$$
A given solution says the answer is $2q+1-h$ for $|h| \leq 2q$ and $0$ otherwise.
I don't see this. For example, when I take $q=1$ and $h = 2*q = 2$ the summation sums up to $0$ which is unequal to $2q+1-h = 1$.
Anyone can help me out?
|
We have
$$\sum_{a=-q}^q \sum_{b=-q}^q \mathrm{1}_{h+a-b=0}
= \sum_{a=-q}^q \sum_{b=-q}^q \mathrm{Res}_{z=0} \frac{1}{z^{h+1+a-b}}
\\ = \mathrm{Res}_{z=0} \frac{1}{z^{h+1}}
\sum_{a=-q}^q \frac{1}{z^a} \sum_{b=-q}^q z^b
= \mathrm{Res}_{z=0} \frac{1}{z^{h+1}}
\left(\sum_{b=-q}^q z^b\right)^2
\\ = \mathrm{Res}_{z=0} \frac{1}{z^{h+1}}
\left(\frac{1}{z^q} \sum_{b=0}^{2q} z^b\right)^2
= \mathrm{Res}_{z=0} \frac{1}{z^{h+2q+1}}
\frac{(1-z^{2q+1})^2}{(1-z)^2}
\\ = \mathrm{Res}_{z=0} \frac{1}{z^{h+2q+1}}
\frac{1-2z^{2q+1}+z^{4q+2}}{(1-z)^2}.$$
The three pieces here are: first,
$$[z^{h+2q}] \frac{1}{(1-z)^2} = h+2q+1$$
for $h+2q\ge 0$ or $h\ge -2q,$ second
$$-2 [z^{h+2q}] \frac{z^{2q+1}}{(1-z)^2}
= -2 [z^{h-1}] \frac{1}{(1-z)^2}
= -2h$$
for $h+2q\ge 2q+1$ or $h\ge 1,$ third
$$[z^{h+2q}] \frac{z^{4q+2}}{(1-z)^2}
= [z^{h-2q-2}] \frac{1}{(1-z)^2}
= h-2q-1$$
for $h+2q\ge 4q+2$ or $h\ge 2q+2.$
Collecting everything we have that when $q=0$ the sum becomes
$\mathrm{1}_{h=0}$ so we may suppose that $q\ge 1.$ We then obtain
zero when $h\lt -2q.$ We get when $-2q\le h\lt 1$ the value $h+2q+1.$
Next for $1\le h\lt 2q+1$ we find $-h+2q+1.$ For $h=2q+1$ we get
$h+2q+1-2h = 0.$ Finally for $h\ge 2q+2$ we get zero, adding all three
pieces. We conclude with the closed form
$$\bbox[5px,border:2px solid #00A000]{
[[\; |h| \le 2q \;]] \times (2q+1-|h|).}$$
|
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|
How to prove $1 + \frac{1}{\sqrt{2}} +... +\frac{1}{\sqrt{n}} <\sqrt{n}\ .\left(2n-1\right)^{1/4} $
Prove that $$1 + \frac{1}{\sqrt{2}} +... +\frac{1}{\sqrt{n}} <\sqrt{n}\ .\Bigl(2n-1\Bigr)^{\frac{1}{4}} $$
My Approach :
I tried by applying Tchebychev's Inequality for two same sets of numbers;
$$1 , \frac{1}{\sqrt{2}} ,... ,\frac{1}{\sqrt{n}}$$
And got , $$\Bigl(1 + \frac{1}{\sqrt{2}} +... +\frac{1}{\sqrt{n}}\Bigr)^2 <n\Bigl(1 + \frac{1}{2} +... +\frac{1}{n}\Bigr) $$
Again I tried by applying Tchebychev's Inequality for another two same sets of numbers;
$$1,\frac{1}{2},...,\frac{1}{n}$$ And got, $$\Bigl(1 + \frac{1}{2} +... +\frac{1}{n}\Bigr)^2 <n\Bigl(1 + \frac{1}{2^2} +... +\frac{1}{n^2}\Bigr)$$
With these two inequities i tried solving further more, but i couldn't. So can you please help me solving this further. And if there is some other approach for this question then please answer that way too.
Thank you.
|
Using Cauchy Schwartz inequality with
*
*$u=(1,1,...,1)$
*$v=\left(1 , \frac{1}{\sqrt{2}} ,... ,\frac{1}{\sqrt{n}}\right)$
we have
$$\vec u \cdot \vec v\le |\vec u||\vec v| \implies 1 + \frac{1}{\sqrt{2}} +... +\frac{1}{\sqrt{n}} \le\sqrt{n}\ \cdot\left(\sqrt{1 + \frac{1}{2} +... +\frac{1}{n}}\right)$$
and thus it suffices to prove that
$$\sqrt{n}\ \cdot\left(\sqrt{1 + \frac{1}{2} +... +\frac{1}{n}}\right)<\sqrt{n}\ .\left(2n-1\right)^{\frac{1}{4}}$$
that is
$$1 + \frac{1}{2} +... +\frac{1}{n}<\sqrt{2n-1}$$
which is true for $n=2$ and it reduces to prove by induction
$$\sqrt{2n-1}+\frac{1}{n+1}<\sqrt{2n+1}$$
which is true indeed
$$\frac{1}{n+1}<\sqrt{2n+1}-\sqrt{2n-1}$$
$$\frac{1}{(n+1)^2}<2n+1+2n-1-2\sqrt{4n^2-1}$$
$$4n-\frac{1}{(n+1)^2}>2\sqrt{4n^2-1}$$
$$16n^2+\frac{1}{(n+1)^4}-\frac{8n}{(n+1)^2}>16n^2-4$$
$$\frac{1}{(n+1)^4}-\frac{8n}{(n+1)^2}+4>0$$
$$4(n+1)^4-8n(n+1)^2+1>0$$
$$4(n^4+4n^3+6n^2+4n+1)-8n(n^2+2n+1)+1>0$$
$$4n^4+8n^3+8n^2+8n+5>0$$
Therefore the given inequality holds for $n\ge 2$, for $n=1$ we can check directly that equality holds.
|
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|
Prove $\lim_{(x,y) \to (0,0)} \frac{\exp(xy)\cdot xy\cdot(x^2-y^2)}{x^2+y^2} =0$. I want to show that
$$\lim_{(x,y) \to (0,0)} \frac{\exp(xy)\cdot xy\cdot(x^2-y^2)}{x^2+y^2} =0.$$
Is it valid to do it like this:
$$\lim_{(x,y) \to (0,0)} \left|\frac{\exp(xy)\cdot xy\cdot(x^2-y^2)}{x^2+y^2}\right| \leq \lim_{(x,y) \to (0,0)} |\exp(xy)\cdot xy\cdot (x^2-y^2)|=0$$
|
We may restrict the neibourhood to $x^2+y^2<1$, then $e^{xy}<2$ and
$$\left|\dfrac{e^{xy}xy(x^2-y^2)}{x^2+y^2}\right|\leqslant\dfrac{2|xy|(|x|^2+|y|^2)}{x^2+y^2}\leqslant x^2+y^2$$
|
{
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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|
Compare $\arcsin (1)$ and $\tan (1)$
Which one is greater: $\arcsin (1)$ or $\tan (1)$?
How to find without using graph or calculator?
I tried using $\sin(\tan1)\leq1$, but how do I eliminate the possibility of an equality without using the calculator?
|
One can write:
$$
\frac{1}{\sqrt{1 + \tan^21}} = \cos1 = 1 - 2\sin^2\frac{1}{2} > \frac{17639}{32768},
$$
because
$$
\sin\frac{1}{2} < \frac{1}{2} - \frac{1}{2^3\cdot3!} + \frac{1}{2^5\cdot5!} = \frac{1920 - 80 + 1}{3840} < \frac{1845}{3840} = \frac{123}{256}.
$$
On the other hand, using Archimedes's lower bound, $\pi > 3\tfrac{10}{71}$:
$$
\frac{1}{\sqrt{1 + \left(\sin^{-1}1\right)^2}} = \frac{1}{\sqrt{1 + \left(\frac{\pi}{2}\right)^2}} < \frac{1}{\sqrt{1 + \left(\frac{223}{142}\right)^2}} = \frac{142}{\sqrt{69893}} < \frac{142}{\sqrt{69696}} = \frac{142}{264} = \frac{71}{132}.
$$
So, one can prove that $\tan1 < \sin^{-1}1$ by proving that:
$$
\frac{17639}{32768} > \frac{71}{132},
$$
which simplifies to $33 \times 17639 > 71 \times 8192$, that is, $582087 > 581632$ - which is true.
|
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"timestamp": "2023-03-29T00:00:00",
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|
if $a^2+b^2+c^2+2\rho(ab+bc+ca)\ge0$ then $\rho\ge-1/2$ This may be trivial but I am not able to prove that if $a^2+b^2+c^2+2\rho(ab+bc+ca)\ge0$ for $a,b,c\in\mathbb{R}$ then $\rho\ge-1/2$. Can anybody help me please?
Thanks!
|
If $a=b=1$ and $c=x> -1/2$ then we have $$-2\rho\leq {x^2+2\over 2x+1}=:f(x)$$
Since $f$ achieve minimum $1$ at $x=1$ we have $\boxed{\rho\geq-{1\over 2}}$.
And if $x<-{1\over 2}$ we get $$-2\rho\geq {x^2+2\over 2x+1}$$
Since $f$ achieve maximum $-2$ for $x=-3$ we have also $-2\rho \geq -2$ so $\boxed{\rho \leq 1}$.
|
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"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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|
Find the number of trailing zeros in 50! My attempt:
50! = 50 * 49 *48 ....
Even * even = even number
Even * odd = even number
odd * odd = odd number
25 evens and 25 odds
Atleast 26 of the numbers will lead to an even multiple (24 evens + 1 even * 1 odd) so at most 26 trailing zeros.
50 is divisible by 5: 10 times. Atleast 10 trailing zeros.
What is the answer?
|
There are $\lfloor \frac{50}{5}\rfloor$ numbers between $1$ and $50$ that are divisible by $5$. Similiarily $\lfloor \frac{50}{5^2}\rfloor$ and $\lfloor \frac{50}{5^3}\rfloor$ numbers divisible by $5^2$ and $5^3$ respectively.
Thus, the highest power of $5$ dividing $50!$ is
$$\lfloor \frac{50}{5}\rfloor + \lfloor \frac{50}{5^2}\rfloor + \lfloor \frac{50}{5^3}\rfloor = 10 + 2 + 0 = 12$$
With similiar arguments, one can show the greatest power of $2$ dividing $50!$ is greater then $12$. Thus, the greatest power of $10 = 2 \cdot 5$ dividing $50!$ is $12$.
|
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|
Evaluate $\lim_{x\rightarrow0}\left[x\left(\frac{1+3x}{1+2x}\right)^{\frac {1}{x}}\right]$ I want to evaluate the following limit:
$$\lim_{x\rightarrow0}\left[x\left(\frac{1+3x}{1+2x}\right)^{\frac {1}{x}}\right]$$
We have
$$\left(\frac{1+3x}{1+2x}\right)^{\frac 1 x} = e^{\frac{\log\left(\frac{1+3x}{1+2x}\right)}{x}}\equiv e^{h(x)}$$
$$h(x)= {\frac{\log\left(\frac{1+3x}{1+2x}\right)}{x}}$$
The argument of $\log\left(\frac{1+3x}{1+2x}\right)$ tends to 1, therefore:
$$\log\left(\frac{1+3x}{1+2x}\right)\sim\frac{1+3x}{1+2x}-1=\frac{x}{1+2x}\sim\frac {x}{2x}=\frac 1 2$$
This means that
$$h(x)\sim\frac 1 2 \cdot \frac 1 x = \frac 1 {2x}$$
We finally have for $x\rightarrow0$:
$$e^{h(x)}=e^{\frac {1}{2x}}\rightarrow+\infty$$
This should lead to an indeterminate form $[0\cdot\infty]$. Any hints on how to evaluate that limit?
|
$$\lim_{x\rightarrow0}\left[x\left(\frac{1+3x}{1+2x}\right)^{\frac {1}{x}}\right]=\lim_{x\to0}x \cdot\dfrac{\left(\lim_{x\to0}\left(1+3x\right)^{1/3x}\right)^3}{\left(\lim_{x\to0}\left(1+2x\right)^{1/2x}\right)^2} =\lim_{x\to0}x \cdot\dfrac{e^3}{e^2}=?$$
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
Double series convergent to $2\zeta(4)$? Using a computer I found the double sum
$$S(n)= \sum_{j=1}^n\sum_{k=1}^n \frac{j^2 + jk + k^2}{j^2(j+k)^2k^2}$$
has values
$$S(10) \quad\quad= 1.881427206538142 \\ S(1000) \quad= 2.161366028875634 \\S(100000) = 2.164613524212465\\$$
As a guess I compared with fractions $\pi^p/q$ where $p,q$ are positive integers and it appears
$$\lim_{n \to \infty} S(n) = \frac{\pi^4}{45} = 2\zeta(4) \approx 2.164646467422276 $$
I'd be interested in seeing a proof if true.
|
$$
\begin{align}
\frac{m^2+m\,n+n^2}{m^2\,(m+n)^2\,n^2}\, &=\frac{m^2+m\,n+n^2\,\color{red}{+m\,n-m\,n}}{m^2\,(m+n)^2\,n^2} \\[2mm]
&=\,\frac{(m+n)^2-m\,n}{(m+n)^2\,m^2\,n^2} \\[2mm]
&=\,\frac{1}{m^2\,n^2}-\frac{1}{m\,n\,(m+n)^2} \\[2mm]
&=\,\frac{1}{m^2\,n^2}-\frac{1}{m^3}\left(\frac{1}{n}-\frac{1}{m+n}-\frac{m}{(m+n)^2}\right) \\[2mm]
&=\,\color{brown}{\frac{1}{m^2\,n^2}}\color{green}{-\frac{1}{m^3}\left(\frac{1}{n}-\frac{1}{m+n}\right)}\color{blue}{+\frac{1}{m^2}\frac{1}{(m+n)^2}}
\end{align}
$$
$$
\begin{align}
\color{brown}{\large S_{\small 1}\,} &=\sum_{m=1}^{\infty}\sum_{n=1}^{\infty}\frac{1}{m^2\,n^2}=\sum_{m=1}^{\infty}\frac{1}{m^2}\sum_{n=1}^{\infty}\frac{1}{n^2}=\left(\zeta(2)\right)^2=\color{brown}{\frac{\pi^4}{36}} \\[4mm]
\color{green}{\large S_{\small 2}\,} &=\sum_{m=1}^{\infty}\sum_{n=1}^{\infty}\frac{1}{m^3}\left(\frac{1}{n}-\frac{1}{m+n}\right)=\sum_{m=1}^{\infty}\frac{1}{m^3}\sum_{n=1}^{\infty}\left(\frac{1}{n}-\frac{1}{m+n}\right) \\[2mm]
&=\sum_{m=1}^{\infty}\frac{1}{m^3}\sum_{n=1}^{\color{red}{m}}\frac{1}{n}=\sum_{m=1}^{\infty}\frac{H_{m}}{m^3}=\frac{5}{4}\zeta(4)=\color{green}{\frac{\pi^4}{72}}\tag{1} \\[4mm]
\color{blue}{\large S_{\small 3}\,} &=\sum_{m=1}^{\infty}\sum_{n=1}^{\infty}\frac{1}{m^2}\frac{1}{(m+n)^2}=\sum_{m=1}^{\infty}\frac{1}{m^2}\sum_{n=1}^{\infty}\frac{1}{(m+n)^2} \\[2mm]
&=\sum_{m=1}^{\infty}\frac{1}{m^2}\,\sum_{\color{red}{n=m+1}}^{\infty}\,\frac{1}{n^2}=\sum_{m=1}^{\infty}\frac{\psi^{\small (1)}(m+1)}{m^2} \\[2mm]
&=\sum_{m=1}^{\infty}\frac{1}{m^2}\left[\zeta(2)-\sum_{k=1}^{m}\frac{1}{k^2}\right]=\left(\zeta(2)\right)^2-\sum_{m=1}^{\infty}\frac{H_{m,2}}{m^2} \\[2mm]
&=\left(\zeta(2)\right)^2-\frac{1}{2}\left[\left(\zeta(2)\right)^2+\zeta(4)\right]=\frac{1}{2}\left[\left(\zeta(2)\right)^2-\zeta(4)\right]=\color{blue}{\frac{\pi^4}{120}}\tag{2}
\end{align}
$$
$$ \color{red}{\Longrightarrow\quad S}\,=S_1-S_2+S_3=\,\color{red}{\frac{\pi^4}{45}} $$
$\,H_m\,\,\,$ : Harmonic Number , $\,\{1\}\,$ : Equation (19) ,$\,\{2\}\,$ : Equation (43)
$\,{\large\psi}^{\small (1)}\,\,$ : Polygamma Function
|
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"timestamp": "2023-03-29T00:00:00",
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|
Positive Integers to make a square How many integers $n$ make the expression $7^n + 7^3 + 2\cdot7^2$ a perfect square?
Factoring $7^2$ we have that $7^n + 7^3 + 2\cdot7^2 = 7^2\cdot(7^{n-2} + 9)$.
How do we prove that the 2nd factor only has $1$ solution when $n = 3$?
|
An "equivalent" way of doing things, is to note that $7^3 + 2 \times 7^2 = 343+98=441 = 21^2$.
Now, if $7^n + 21^2 = y^2$ for some $y$, then $7^n = (y-21)(y+21)$.
So the prime factorization of $(y-21)$ and $(y+21)$ can both contain only the prime $7$, by unique factorization theorem. Therefore, $y - 21$ and $y+21$ are both powers of $7$. The difference between these powers of $7$ is $y+21 - (y-21) = 42$.
Only two such powers have a difference of $42$, which are $49$ and $7$.This is not very difficult to see from the fact that $7^n - 7^m = 7^{n-m}(7^{m} - 1) = 42$ forces $n-m = 1$ and $m = 1$. Therefore, $(y-21)(y+21) = 7 \times 49 = 343$, so $n = 3$ is forced from here.
Thinking of it, you could also have done $7^{n-2} + 9 = x^2$ since you factorized out $7^2$, but the same logic will apply, since $9 = 3^2$.
|
{
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|
Finding n-th term of a matrix I have to find the n-th power of the following matrix $$A=\begin{pmatrix} 2 & 0&1\\ 0 & -3& 0 \\ 1 & 0&2 \end{pmatrix}
$$ I computed a few powers $$A^2=\begin{pmatrix} 4+1& 0&4\\ 0 & 9& 0 \\ 4 & 0&1+4\end{pmatrix}$$
$$A^3 =\begin{pmatrix} 14 & 0&13\\ 0 & -27& 0 \\ 13& 0&14\end{pmatrix}$$
$$A^n=\begin{pmatrix} \frac12(3^n+1) & 0&\frac12(3^n-1)\\ 0 &( -3)^n& 0 \\ \frac12(3^n-1) & 0&\frac12(3^n+1)\end{pmatrix}$$ And is just induction left which is easy, but I found by check and try the n-th term, is it possible to find it in a proper way?
|
HINT
$A$ is a real and symmetric matrix therefore we can diagonalize that is find $P$ such that
$$A=PDP^{-1} \implies A^n=\overbrace{PDP^{-1}PDP^{-1}\ldots PDP^{-1}}^{n\,terms}=PD^nP^{-1}$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
$\lim_{x \rightarrow \infty}\frac{5^{x+1}+7^{x+1}}{5^x +7^x}$ $$\lim_{x \rightarrow \infty}\frac{5^{x+1}+7^{x+1}}{5^x +7^x}$$
I tried using L'Hospital rule, which yielded :
$$\lim_{x \rightarrow \infty}\frac{5^{x+1} \ln 5+7^{x+1} \ln 7}{5^x \ln 5 }$$
But I'm at the dead end... If I divide numerator and denominator by $5^x$ , I get a term $\frac{7^x}{5^x}$ ... which is unsolvable for the limit $x \rightarrow \infty $
However , the answer provided by book is $-7$ . I doubt there is mistake in the question.
|
The standard method is to factor out and cancel the highest power, which is here $7^x$:
We get:
$\lim_{x\to\infty} \frac{5^{x+1}+7^{x+1}}{5^x+7^x}=\lim_{x\to\infty} \frac{7^{x+1}\left(\left(\frac57\right)^{x+1}+1\right)}{7^x\left(\left(\frac57\right)^x+1\right)}=\lim_{x\to\infty} \frac{7\left(\left(\frac57\right)^{x+1}+1\right)}{\left(\frac57\right)^x+1}$
$=7\cdot\lim_{x\to\infty} \frac{\left(\frac57\right)^{x+1}+1}{\left(\frac57\right)^x+1}$
It is $\lim_{x\to\infty} \left(\frac{5}{7}\right)^x=0$, since $\frac57<1$
We get:
$=7\cdot\frac{0+1}{0+1}=7$
|
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|
What is the best algorithm for finding the last digit of an enormous exponent? I found most answers here not clear enough for my case such as
$$
123155131514315^{4515131323164343214547}
$$
I wrote the $n\bmod10$ in Python and execution time ran out. So I need a faster algorithm or method. Sometimes, the result is incorrect as it failed the test case.
|
If $a=10q+b$, then $a^n \equiv b^n \bmod 10$, and so it suffices to consider $b^n$, for $b = 0, 1, \dots, 9$.
These powers repeat as follows:
$0^n = 0, 0, \dots $
$1^n = 1, 1, \dots $
$2^n = 2, 4, 8, 6, 2, 4, 8, 6, \dots $
$3^n = 3, 9, 7, 1, 3, 9, 7, 1, \dots $
$4^n = 4, 6, 4, 6, \dots $
$5^n = 5, 5, \dots $
$6^n = 6, 6, \dots $
$7^n = 7, 9, 3, 1, 7, 9, 3, 1,\dots $
$8^n = 8, 4, 2, 6, 8, 4, 2, 6, \dots $
$9^n = 9, 1, 9, 1, \dots $
So, you only have to compute $n \bmod m$, where $m$ is the period corresponding to $b$, and look up the answer in the list above.
|
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|
Proving $\frac{\sin 30^\circ}{\sin 50^\circ}= \frac{\sin 40^\circ}{\sin 80^\circ}$
Prove that
$$\frac{\sin 30^\circ}{\sin 50^\circ}= \frac{\sin 40^\circ}{\sin 80^\circ}$$
I tried making $\sin 80^\circ=\sin(50^\circ+30^\circ)$, but it didn't go well. I also tried using, maybe, trig periodicities, but I still can't get it.
|
Let us first prove that $\sin(30^\circ) \cdot \sin(80^\circ) = \sin(40^\circ) \cdot \sin(50^\circ)$.
\begin{align}
LHS & = \sin(30^\circ) \cdot \sin(80^\circ) \\
&= \cos(60^\circ) \cdot \cos(10^\circ) \\
& = \frac{1}{2} \cos(10^\circ) \\
\end{align}
\begin{align}
RHS & = \sin(40^\circ) \cdot\sin(50^\circ) \\
& = \frac{1}{2}[\cos(40^\circ-50^\circ)-\cos(40^\circ+50^\circ)] \\
& = \frac{1}{2}[\cos(10^\circ) - \cos(90^\circ)] \\
& = \frac{1}{2} \cos(10^\circ) \\
\therefore LHS & = RHS
\end{align}
So we have $$\sin(30^\circ) \cdot \sin(80^\circ) = \sin(40^\circ) \cdot \sin(50^\circ)$$
And hence it follows that $$\frac{\sin(30^\circ)}{\sin(50^\circ)} = \frac{\sin(40^\circ)}{\sin(80^\circ)} $$
|
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|
Evaluate $f(x)=x\log{\left|\frac{x+2}{3-x}\right|}$ for $x\rightarrow\infty$ I have the following function:
$$f(x)=x\log{\left|\frac{x+2}{3-x}\right|}$$
I want to find the limit for $x\rightarrow+\infty$.
This is what I do. Since $x>=0$, I can remove the absolute value:
$$f(x)=x\log\left({\frac{x+2}{3-x}}\right)\sim x\left( \frac{x+2}{3-x}-1\right)=x\left(\frac{2x-1}{3-x}\right)=x\left(\frac{2x}{-x}\right)=-2x\rightarrow-\infty$$
The textbook reports that the limit is actually $5$. Why is my solution wrong?
|
We have for big $x > 0$
$$
\ln \left|\frac{x+2}{3-x}\right|^x = \ln \left(\frac{x+2}{x-3}\right)^x
$$
and also
$$
\lim_{x\to \infty}\left(\frac{x+2}{x-3}\right)^x = e^5
$$
hence
$$
\lim_{x\to\infty}x\ln\left|\frac{x+2}{3-x}\right| = 5
$$
|
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|
Prove that $1 + \frac{2}{3!} + \frac{3}{5!} + \frac{4}{7!} + \dotsb = \frac{e}{2}$
Prove that $1 + \frac{2}{3!} + \frac{3}{5!} + \frac{4}{7!} + \dotsb = \frac{e}{2}$.
This is problem 4 from page 303 of S.L.Loney's 'Plane Trigonometry'.
It seems fairly obvious that the series expansion $e^x$ will be used. However, I am unsure where to start. Should I consider other series?
|
Your series is $$\sum_{n=1}^{\infty}\frac{n}{(2n-1)!}.$$
Let $a_n=\frac{n}{(2n-1)!}$, then
$$a_n=\frac{1}{2}\frac{2n}{(2n-1)!}=\frac{1}{2}\frac{2n-1+1}{(2n-1)!}
=\frac{1}{2}\left(\frac{1}{(2n-2)!}+\frac{1}{(2n-1)!}\right).$$
So $$\sum_{n=1}^{\infty}\frac{n}{(2n-1)!}=\frac{1}{2}\sum_{n=1}^{\infty}\left(\frac{1}{(2n-2)!}+\frac{1}{(2n-1)!}\right)$$
$$=\frac{1}{2}\sum_{n=0}\frac{1}{n!}=\frac{1}{2}e.$$
|
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|
Any idea on how to find a upper bound for a limit in R^2? I'm working on this problem of continuity in $\mathbb{R}^2$ the statement is the following:
Prove by definition that
$$ \lim_{ (x,y) \to (0,0) } \frac{y^2}{3+\sin^2(x^2+y^2)+y^4}=0 $$
Given $\epsilon >0$, we have that
$$| \frac{y^2}{3+\sin^2(x^2+y^2)+y^4} | = y^2 \frac{1}{3+\sin^2(x^2+y^2)+y^4}$$
If we know that
$ \frac{1}{3+\sin^2(x^2+y^2)+y^4} \leq \frac{1}{\sin^2(x^2+y^2)} $ then we end with
$$ y^2 \frac{1}{3+\sin^2(x^2+y^2)+y^4} \leq y^2
\frac{1}{\sin^2(x^2+y^2)}$$
and even more,
$$y^2
\frac{1}{\sin^2(x^2+y^2)} \leq \frac{x^2+y^2}{\sin^2(x^2+y^2)} $$
I don't see any way to get rid out the $\frac{1}{\sin^2(x^2+y^2)}$ to conclude the $\delta$ since I can't say that $\frac{1}{\sin^2(x^2+y^2)} \leq 1$ because is the contrary, in fact $\frac{1}{\sin^2(x^2+y^2)} \geq 1$. Any ideas?
Thanks in advance.
|
$\frac{y^2}{3+\sin^2(x^2+y^2)+y^4}\leq \frac{y^2}{3}\rightarrow 0$
|
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|
If $\frac{\log x}{y - z} = \frac{\log y}{z - x} = \frac{\log z}{x - y}$, prove that $xyz = 1$ without using the following method If
$$\frac{\log x}{y-z} = \frac{\log y}{z-x} = \frac{\log z}{x-y}$$
prove that $$xyz = 1
$$.
without using this method :-
$$
\mbox{Let }\frac{\log x}{y-z} = \frac{\log y}{z-x} = \frac{\log z}{x-y} = k\\
\mbox{this gives three equations :-}\\
\mbox{1.}\log x = k(y -z)\\
\mbox{2.}\log y = k(z - x)\\
\mbox{3.}\log z = k(x - y)\\
\mbox{Adding eq. 1, 2 and 3, we get}\\
\log x + \log y + \log z = k(y - z) + k(z-x)+k(x-y)\\
\implies \log xyz = ky - kz + kz - kx + kx - ky\\
\implies \log xyz = 0\\
\implies \log xyz = \log 1\\
\implies xyz = 1\\
\mbox{(which is the required proof)}
$$
I am trying my best but not able to do it.
|
$$\frac{\log x}{y-z} = \frac{\log y}{z-x}$$
Implies
$$\frac{\log x}{y-z} = \frac{\log y}{z-x}=\frac{\log x+\log y}{(y-z)+(z-x)}$$
Therefore
$$\frac{\log xy}{y-x)}= \frac{\log z}{x-y}= \frac{-\log z}{y-x}
\Rightarrow \\
\log xy = -\log z \Rightarrow \\
\log xyz =0$$
|
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|
How to calculate $\int\frac{x}{x^2-x+1}\, dx$? $$\int \frac{x}{x^2-x+1}\, dx = \int \frac{x}{(x-\frac 1 2)^{2} + \frac 3 4}\, dx = \int \frac{x}{(x-\frac 1 2)^2 + (\frac {\sqrt{3}} {2})^2}$$
Substitute $u= \frac{2x-1}{\sqrt{3}}, du=\frac{2}{\sqrt{3}}dx$:
$$\frac {\sqrt{3}} 2 \int \frac{\frac{\sqrt{3}} {2}u + \frac 1 2}{(\frac{\sqrt{3}}{2}u)^2+(\frac {\sqrt{3}}{2})^2} = \int \frac{u}{u^2+1}du + \frac{1}{\sqrt{3}}\int\frac 1 {u^2+1}du.$$
This gives $$\frac 1 2\log({u^2+1})+\frac{1}{\sqrt{3}}\arctan{u}.$$
Substituting back in x yields $$\frac 1 2\log(\frac 4 3x^2-\frac 4 3 x+ \frac 4 3)+\frac 1 {\sqrt{3}}\arctan(\frac{2x-1}{\sqrt{3}})$$
However, according to Wolfram Alpha, the integral should evaluate to $$\frac 1 2 \log(x^2-x+1)+\frac{1}{\sqrt{3}}\arctan(\frac{2x-1}{\sqrt{3}})$$After working some time on the integral, I know how to reach this solution, but I don't understand why my first attempt didn't arrive at the correct answer. Do you see where I went wrong?
Thank you for any help!
|
Hello and welcome to math.stackexchange. The solution that you developed and the one given by Wolfram Alpha only differ by a constant (of integration), since
$$
\frac 1 2\log(\frac 4 3x^2-\frac 4 3 x+ \frac 4 3) =
\frac 1 2\log(x^2- x+ 1) + \frac 1 2 \log \frac 4 3 \, .
$$
|
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|
Integrate $\frac{x^2-4x+10}{x^2\sqrt x}$
Find the indefinate integral with respect to $x$ of $$\frac{x^2-4x+10}{x^2\sqrt x}$$
For this problem I first made each individual number in the numerator separate from each other for easier integration and then simplified
$$=\int \left(\frac{x^2}{x^2\sqrt x}-\frac{4x}{x^2\sqrt x}+\frac{10}{x^2\sqrt x}\right)dx$$
$$=\int \left(x^2x^{-\frac{3}{2}}-4x^1x^{-\frac{3}{2}}+10x^{-\frac{3}{2}}\right)dx$$
$$=\int \left(x^{\frac{1}{2}}-4x^{-\frac{1}{2}}+10x^{-\frac{3}{2}}\right)dx$$
I then integrated the expression to get
$$\frac{2x^{\frac{3}{2}}}{3}-8x^{\frac{1}{2}}-20x^{-\frac{1}{2}}$$
This is, however, wrong. Any ideas as to why?
Thanks in advance $:)$
|
HINT
Place $t=\sqrt{x}$, after you solve the integral of a rational function.
|
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|
Prove that $\frac{\sin x}{x}=(\cos\frac{x}{2}) (\cos\frac{x}{4}) (\cos \frac{x}{8})...$ How do I prove this identity:
$$\frac{\sin x}{x}=\left(\cos\frac{x}{2}\right) \left(\cos\frac{x}{4}\right) \left(\cos \frac{x}{8}\right)...$$
My idea is to let
$$y=\frac{\sin x}{x}$$
and
$$xy=\sin x$$
Then use the double angle identity $\sin 2x=2\sin x \cos x$ and its half angle counterparts repeatedly. I see some kind of pattern, but I can't seem to make out the pattern and complete the proof.
|
Partial answer:
The Taylor series of $\sin x$ is $x - \frac{x^3}{3!} + \frac{x^5}{5!} \cdots$, so:
$$\frac{\sin x}{x} = 1 - \frac{x^2}{3!}+\frac{x^4}{5!} \cdots.$$
Meanwhile, $\cos x = 1 - \frac{x^2}{2!} + \frac{x^4}{4!}$, and therefore:
$$\cos(\frac{x}{2^1}) = 1 - \frac{x^2}{2^2 \cdot 2!} + \frac{x^4}{2^4 \cdot 4!} - \frac{x^6}{2^6 \cdot 6!} \cdots$$
$$\cos(\frac{x}{2^2}) = 1 - \frac{x^2}{2^4 \cdot 2!} + \frac{x^4}{2^8 \cdot 4!} - \frac{x^6}{2^{12} \cdot 6!} \cdots$$
$$\cos(\frac{x}{2^3}) = 1 - \frac{x^2}{2^6 \cdot 2!} + \frac{x^4}{2^{12}\cdot 4!} - \frac{x^6}{2^{18} \cdot 6!} \cdots$$
and so on.
The constant term of the infinite product is $1$.
There is only one way to make an $x^2$ term: by multiplying the constant and the $x^2$ term. The coefficient of the $x^2$ term is:
$$-\frac{1}{2^2 \cdot 2!} - \frac{1}{2^4 \cdot 2!} - \frac{1}{2^6 \cdot 2!},$$
which is a geometric series with $a = \frac{1}{8}$, and $r = \frac{1}{4}$, which the formula gives as $\frac{\frac{1}{8}}{1- \frac{1}{4}} = \frac{1}{6}$.
|
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|
Expressing a linear map $\phi: \>{\mathbb R}^4\to{\mathbb R}^4$ in terms of a new base In terms of the standard coordinates $x$, $y$, $z$, $t$ a linear map $\phi: \>{\mathbb R}^4\to{\mathbb R}^4$ appears as
$$\phi{(x,y,z,t)} = (t,x,y,z)\ .$$ Now we are given a new base $$R = \{ (1,0,1,0),(0,1,0,1),(0,1,1,0),(0,1,1,1) \}$$ of $\mathbb{R}^4$ and should find the matrix $M_R(\phi)$.
Is my thinking correct? As I understand I need to find the matrix created by the linear transormation:
$$
\left( \begin{matrix}
0 & 0 & 0 & 1\\
1 & 0 & 0 & 0\\
0& 1& 0 & 0\\
0& 0& 1 & 0\\
\end{matrix} \right)
$$
then, solve
$$ \left[
\begin{array}{cccc|c}
1&0&0&0&a\\
0&1&1&1&b\\
1&0&1&1&c\\
0&1&0&1&d\\
\end{array}
\right] $$
by turning it to row echelon form to find out how the vector will look in the R basis. Is it the correct way to do a question like this?
|
The matrix $P = \begin{bmatrix}
1 & 0 & 0 & 0 \\
0 & 1 & 1 & 1 \\
1 & 0 & 1 & 1 \\
0 & 1 & 0 & 1 \\
\end{bmatrix}$ transforms the basis $R$ to the canonical basis $E$.
Therefore
$$M_R(\phi) = P^{-1}M_E(\phi) P = \begin{bmatrix}
1 & 0 & 0 & 0 \\
0 & 1 & 1 & 1 \\
1 & 0 & 1 & 1 \\
0 & 1 & 0 & 1 \\
\end{bmatrix}^{-1}
\begin{bmatrix}
0 & 0 & 0 & 1 \\
1 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 \\
0 & 0 & 1 & 0 \\
\end{bmatrix}\begin{bmatrix}
1 & 0 & 0 & 0 \\
0 & 1 & 1 & 1 \\
1 & 0 & 1 & 1 \\
0 & 1 & 0 & 1 \\
\end{bmatrix} = \begin{bmatrix}
0 & 1 & 0 & 1 \\
1 & 0 & -1 & 0 \\
0 & 0 & -1 & -1 \\
0 & 0 & 2 & 1 \\
\end{bmatrix}$$
|
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|
Prove $\frac{b-a+1}{ab} \leq \sum_{i=a}^b \frac{1}{i^2} \leq \frac{b-a+1}{(a-1)b}$ Given $b>a>1; \ a,b \in Z^+$. Prove $\frac{b-a+1}{ab} \leq \sum_{i=a}^b \frac{1}{i^2} \leq \frac{b-a+1}{(a-1)b}. \ $
I mostly care about the second inequality. Thank you!
|
For the second inequality just observe that
$$
\frac{1}{i^2} \leq \frac{1}{i(i-1)} = \frac{1}{i-1} - \frac{1}{i},
$$
hence
$$
\sum_{i=a}^b \frac{1}{i^2} \leq
\sum_{i=a}^b\left(\frac{1}{i-1} - \frac{1}{i}\right)
= \frac{1}{a-1} - \frac{1}{b} = \frac{b-a+1}{(a-1)b}\,.
$$
The first inequality can be proved by induction on $b$.
Let $a\in\mathbb{Z}^+$ be fixed, and let us prove that
$$
S(b) := \sum_{i=a}^{b} \frac{1}{i^2} - \frac{b-a+1}{ab}\geq 0
\qquad
\forall b\geq a.
$$
If $b = a$ then $S(b) = 0$.
Assume now that the inequality holds for some $b\geq a$, and let us prove that it holds for $b+1$.
Using the induction assumption $S(b) \geq 0$, we have that:
$$
\begin{split}
S(b+1) & = S(b) + \frac{1}{(b+1)^2} -\frac{b-a+2}{a(b+1)} + \frac{b-a+1}{ab}
\\ & \geq \frac{1}{(b+1)^2} -\frac{b-a+2}{a(b+1)} + \frac{b-a+1}{ab}
\\ & =
\frac{b-a+1}{ab(b+1)^2} > 0.
\end{split}
$$
|
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|
Sum the series $\frac{3}{1⋅2⋅4}+\frac{4}{2⋅3⋅5}+ \frac{5}{3⋅4⋅6}+...\text{(upto n terms)}$ $\frac{3}{1⋅2⋅4}+\frac{4}{2⋅3⋅5}+ \frac{5}{3⋅4⋅6}+...\text{(upto n terms)}$
The general term seems to be
$$T_r= \frac{r+2}{r(r+1)(r+3)}.$$
I see no way to telescope this because the factors of the denominator of the general term are not in arithmetic progression.
Do I have to use something else? Or am I missing some easy manipulation?
|
\begin{align*}
T_r
&= \frac{1}{r(r+1)} - \frac{1}{r(r+1)(r+3)} \\
&= \frac{1}{r(r+1)}- \left(\frac{1}{2}\cdot \frac{(r+3)-(r+1)}{r(r+1)(r+3)}\right) \\
&= \frac{1}{r}-\frac{1}{r+1}- \frac{1}{2}\left(\frac{1}{r}-\frac{1}{r+1}\right)+ \frac{1}{2}\cdot \frac{1}{r(r+3)} \\
&= \frac{1}{2}\left(\frac{1}{r}-\frac{1}{r+1}\right)+ \frac{1}{2}\cdot\frac{1}{3}\left(\frac{1}{r}-\frac{1}{r+3}\right) \\
&= \frac{2}{3}\left(\frac{1}{r}-\frac{1}{r+1}\right)+ \frac{1}{6}\left(\frac{1}{r+1}-\frac{1}{r+2}\right)+ \frac{1}{6}\left(\frac{1}{r+2}-\frac{1}{r+3}\right)
\end{align*}
You can now see telescope and quicky add the terms.
|
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|
Integration work: $\int\sqrt{\frac{2-x}{x-3}} \ \mathrm dx$
$$\int\sqrt{\dfrac{2-x}{x-3}}\mathrm dx$$
My approach
I=$$\int\sqrt{\dfrac{2-x}{x-3}}\mathrm dx$$
I= $$\int\frac{2-x}{\sqrt{-x^2+5x-6}}\mathrm dx$$
Next I substituted 2-x =t and processed but I am not getting the answer. Can you guys help me with this
|
$$\frac{2-x}{x-3}=t^2$$
$$\frac{1}{(x-3)^2}dx=2tdt$$
$$x=\frac{3t^2+2}{t^2+1}$$
$$x-3=\frac{3t^2+2}{t^2+1}-3=-\frac{1}{t^2+1}$$
$$\int t\cdot2tdt\cdot(\frac{-1}{t^2+1})^2=\int\frac{2t^2}{(t^2+1)^2}dt$$
Now solving
$${\displaystyle\int}\dfrac{t^2}{\left(t^2+1\right)^2}\,\mathrm{d}t$$
Write it as
$${\displaystyle\int}\dfrac{t^2+1-1}{\left(t^2+1\right)^2}\,\mathrm{d}t={\displaystyle\int}\dfrac{1}{\left(t^2+1\right)}\,\mathrm{d}t-{\displaystyle\int}\dfrac{1}{\left(t^2+1\right)^2}\,\mathrm{d}t$$
|
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|
Which is the partial solution of the ode 4th order I want to solve the ode $$y''''(t)+3y''(t)+y(t)=e^{-t}+t^2, \ \forall t>0$$ Forst we have to find the solution for the homogeneous problem using the characteristic polynomial, right?
We have the characteristic polynomial $$\lambda^4+3\lambda^2+1=0$$ which has the solutions $$\lambda=\pm i\sqrt{\frac{3-\sqrt{5}}{2}} \ \text{ and } \ \lambda=\pm i\sqrt{\frac{3+\sqrt{5}}{2}}$$ So we get the solutio of the homogenous part: $$y_h(t)=A\cos \left (\sqrt{\frac{3-\sqrt{5}}{2}} t\right )+B\sin \left (\sqrt{\frac{3-\sqrt{5}}{2}} t\right )+A\cos \left (\sqrt{\frac{3+\sqrt{5}}{2}} t\right )+B\sin \left (\sqrt{\frac{3+\sqrt{5}}{2}} t\right )$$ Is everything correct so far? How can we find the partial solution?
|
With $$\lambda=\pm i\sqrt{\frac{3-\sqrt{5}}{2}} \ \text{ and } \ \lambda=\pm i\sqrt{\frac{3+\sqrt{5}}{2}}$$
You have your $$y_c=A\cos \left (\sqrt{\frac{3-\sqrt{5}}{2}} t\right )+B\sin \left (\sqrt{\frac{3-\sqrt{5}}{2}} t\right )+A\cos \left (\sqrt{\frac{3+\sqrt{5}}{2}} t\right )+B\sin \left (\sqrt{\frac{3+\sqrt{5}}{2}} t\right )$$
correct.
For the particular solution you consider $$y_p = Ae^{-t} + Bt^2+Ct+D$$
Plug in your equation and find the cofficients.
|
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|
Does this matrix sequence always converge? Suppose $a_0, a_1, ... , a_{n-1}$ are real numbers from $(0; 1)$, such that $\sum_{k=0}^{n-1} a_k=1$. Suppose $A = (c_{ij})$ is a $n \times n$ matrix with entries $c_{ij} = a_{(i-j)\%n}$, where $\%$ is modulo operation. Is it always true that $\lim_{m \to \infty} A^m = \frac{1}{n} \begin{pmatrix} 1 & 1 & \cdots & 1 \\1 & 1 & \cdots & 1 \\ \vdots & \vdots & \ddots & \vdots \\ 1 & 1 & \cdots & 1 \end{pmatrix}$?
This statement is true for $n = 2$:
Suppose $A = \begin{pmatrix} a_0 & {1 - a_0} \\ {1 - a_0} & a_0 \end{pmatrix} = \begin{pmatrix} 1 & -1 \\ 1 & 1\end{pmatrix}^{-1}\begin{pmatrix} 1 & 0 \\ 0 & {1-2a_0} \end{pmatrix}\begin{pmatrix} 1 & -1 \\ 1 & 1 \end{pmatrix}$. Because $a_0$ is in $(0; 1)$, $1-2a_0$ is in $(-1;1)$. Thus $$\lim_{m \to \infty} A^m = \begin{pmatrix} 1 & -1 \\ 1 & 1\end{pmatrix}^{-1}\begin{pmatrix} 1 & 0 \\ 0 & 0\end{pmatrix}\begin{pmatrix} 1 & -1 \\ 1 & 1 \end{pmatrix} = \frac{1}{2}\begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix}$$
However, I do not know how to prove this statement for arbitrary $n$.
Any help will be appreciated.
|
Your matrix $A$ is a circulant matrix:
$$A = \begin{bmatrix}
a_0 & a_1 & \cdots & a_{n-1}\\
a_{n-1} & a_0 & \cdots & a_{n-2}\\
\vdots & \vdots & \cdots & \vdots \\
a_1 & a_2 & \cdots & a_0\\
\end{bmatrix}$$
$A$ is known to have eigenvalues equal to $\lambda_j = \sum_{k=0}^{n-1} a_k\omega_j^k$ with eigenvectors $\begin{bmatrix} 1 \\ \omega_j \\ \omega_j^2 \\ \vdots \\ \omega_j^{n-1}\end{bmatrix}$, where $\omega_j = e^{\frac{2\pi ij}{n}}$ for $j = 0, \ldots n-1$.
Therefore we can diagonalize $A$ as follows:
$$A = P^{-1}DP = \begin{bmatrix}
1 & 1 & \cdots & 1\\
\omega_0 & \omega_1 & \cdots & \omega_{n-1}\\
\vdots & \vdots & \cdots & \vdots \\
\omega_0^{n-1} & \omega_1^{n-1} & \cdots & \omega_{n-1}^{n-1}
\end{bmatrix}^{-1}
\begin{bmatrix}
\lambda_0 & 0 & \cdots & 0 \\
0 & \lambda_1 & \cdots & 0\\
\vdots & \vdots & \cdots & \vdots \\
0 & 0 & \cdots & \lambda_{n-1}\\
\end{bmatrix}
\begin{bmatrix}
1 & 1 & \cdots & 1\\
\omega_0 & \omega_1 & \cdots & \omega_{n-1}\\
\vdots & \vdots & \cdots & \vdots \\
\omega_0^{n-1} & \omega_1^{n-1} & \cdots & \omega_{n-1}^{n-1}
\end{bmatrix}$$
The triangle inequality for the eigenvalues gives
$$|\lambda_j| = \left|\sum_{k=0}^{n-1} a_k\omega_j^k\right| \le \sum_{k=0}^{n-1} a_k |\omega_j|^k = \sum_{k=0}^{n-1} a_k = 1$$
with equality holding if and only if $\{a_0, a_1\omega_j, \ldots, a_{n-1}\omega_j^{n-1}\}$ lie on the same side of a single line through the origin. Clearly this is true iff $j = 0$ so the eigenvalues satisfy $\lambda_0 = 1$ and $|\lambda_j| < 1$ for $j = 1, \ldots, n-1$.
Hence letting $m\to\infty$ in $A^m = P^{-1}D^mP$ gives
\begin{align}
\lim_{m\to\infty} A^m &= \begin{bmatrix}
1 & 1 & \cdots & 1\\
\omega_0 & \omega_1 & \cdots & \omega_{n-1}\\
\vdots & \vdots & \cdots & \vdots \\
\omega_0^{n-1} & \omega_1^{n-1} & \cdots & \omega_{n-1}^{n-1}
\end{bmatrix}^{-1}
\begin{bmatrix}
1 & 0 & \cdots & 0 \\
0 & 0 & \cdots & 0\\
\vdots & \vdots & \cdots & \vdots \\
0 & 0 & \cdots & 0\\
\end{bmatrix}
\begin{bmatrix}
1 & 1 & \cdots & 1\\
\omega_0 & \omega_1 & \cdots & \omega_{n-1}\\
\vdots & \vdots & \cdots & \vdots \\
\omega_0^{n-1} & \omega_1^{n-1} & \cdots & \omega_{n-1}^{n-1}
\end{bmatrix}\\
&= \begin{bmatrix}
1 & 1 & \cdots & 1\\
\omega_0 & \omega_1 & \cdots & \omega_{n-1}\\
\vdots & \vdots & \cdots & \vdots \\
\omega_0^{n-1} & \omega_1^{n-1} & \cdots & \omega_{n-1}^{n-1}
\end{bmatrix}^{-1} \begin{bmatrix} 1 \\ 1 \\ \vdots \\ 1\end{bmatrix}
\end{align}
You can calculate this by hand, or notice that the columns of $\lim_{m\to\infty} A^m$ satisfy the system $Px = \begin{bmatrix} 1 \\ 1 \\ \vdots \\ 1\end{bmatrix}$, which you can solve.
The sum $\sum_{k=0}^{n-1} \omega_k^j$ is equal to $0$ here, which would suggest that the limit doesn't have to be of the form you specified.
|
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|
How to prove $(a+1)(ab^2+1) \geq ((ab+1)^\frac{2}{k}+1)^k$? In the paper linear forms in the logarithms of real algebraic numbers close to 1, it is
written on page 8 that-
$$(a+1)(ab^2+1) \geq ((ab+1)^\frac{2}{k}+1)^k=(ab+1)^2+ \sum_{n=0}^{k-1} \binom{k-1}{n} (ab+1)^\frac{2n}{k} \cdots (1)$$
Clearly, $(a+1)(ab^2+1) > (ab+1)^2$, but right hand side of equation (1) has a sum $\sum_{n=0}^{k-1} \binom{k-1}{n} (ab+1)^\frac{2n}{k}$, so how we prove that-
$$(a+1)(ab^2+1) \geq (ab+1)^2+ \sum_{n=0}^{k-1} \binom{k-1}{n} (ab+1)^\frac{2n}{k} \cdots (2) ?$$
|
The inequality is obtained in two steps.
At first, for example by Rearrangement inequality, we have that
$$(a+1)(ab^2+1)=a^2b^2+ab\cdot b+a\cdot 1+1\ge a^2b^2+ab\cdot 1+a\cdot b+1$$
$$\ge a^2b^2+2ab+1= (ab+1)^2$$
but inequality holds $\iff$ $b=1$ and since $b>1$ we have that
$$(a+1)(ab^2+1)\color{red}> (ab+1)^2$$
then, second step, since the numbers are perfect $k$ powers we have that
$$(a+1)^{1/k}(ab^2+1)^{1/k}>(ab+1)^{2/k} \implies (a+1)^{1/k}(ab^2+1)^{1/k}\ge(ab+1)^{2/k}+1$$
that is
$$(a+1)(ab^2+1) \geq ((ab+1)^\frac{2}{k}+1)^k$$
|
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|
Finding the value of $\prod_{n=0}^\infty a_n$ with $a_0=1/2$ and $a_{n}=1+(a_{n-1}-1)^2$
Putting the value of the first term, we can see that the series goes like
$$1/2, 5/4, 17/16,...$$
I am unable to calculate the general term, and so not sum of the series. Please help me to find the general term or directly the infinite sum if possible.
|
Let $b_n=a_n-1$ so that $b_0=-\frac 12$ and $b_{n}=b_{n-1}^2$.
Consequently, $b_n=b_0^{2^n}$ hence
$$a_n=1+(-2)^{-2^n}$$
Let $x=\frac 12$, then
\begin{align}
(1+x)\prod_{n=0}^Na_n
&=(1+x)\prod_{n=0}^N(1+(-x)^{2^n})\\
&=(1+x)(1-x)(1+x^2)\cdots(1+x^{2^N})\\
&=(1-x^2)(1+x^2)\cdots(1+x^{2^N})\\
&=(1-x^4)(1+x^4)\cdots(1+x^{2^N})\\
&=(1-x^8)(1+x^8)\cdots(1+x^{2^N})\\
&=\cdots\\
&=(1-x^{2^N})(1+x^{2^N})\\
&=1-x^{2^{N+1}}\\
&\xrightarrow{N\to\infty}1
\end{align}
so that
$$\prod_{n=0}^\infty a_n=\frac 23$$
|
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|
Sum of three uniformly distributed random variables. Let $X,Y,Z$ be uniformly distributed $U(0,1)$. Then I know that the density function for the random variable $A= X+Y$ is
$$f(a) = \begin{cases}
a & a\in (0,1)\\
2-a & a\in[1,2)\\
0 & \text{ otherwise}
\end{cases}.
$$
and
$$
g(z)=\begin{cases}
1 & z\in (0,1)\\
0 & \text{otherwise}
\end{cases}
$$
My goal is to find the $B = A+Z.$ By the convolution theorem:
$$h(b) = \int_{-\infty}^{\infty}f(b-z)g(z)dz=\int_{0}^{1}f(b-z)dz.$$
After this I am not sure how to proceed. Any help will be much appreciated.
|
So we get
\begin{equation}
f(b-z) = \begin{cases}
b-z & b-z\in (0,1)\\
2-(b-z) & b-z\in[1,2)\\
0 & \text{ otherwise}
\end{cases}.
\end{equation}
We have that
\begin{equation}
0<z<1
\end{equation}
or
\begin{equation}
b-1<b-z<b
\end{equation}
If $b < 0$, then $b - z < 0$, so $h(b) = 0$ according to the boundaries we have. On the other hand if $b - 1> 2$ (or $b>3$), that is is $b-z>2$, we also get $f(b) = 0$. Other than that, we can distinguish three cases:
Case 1:
If $0<b<1$
\begin{equation}
h(b)
=
\int_{0}^{1}f(b-z)dz
=
\int_0^{b} b-z \ dz
=
\frac{b^2}{2}
\end{equation}
Case 2: If $1<b<2$
\begin{equation}
h(b)
=
\int_{0}^{1}f(b-z)dz
=
\int_{b-1}^{1} b-z \ dz
+
\int_{0}^{b-1} 2-(b-z) \ dz
= \frac{b(2-b)}{2}-\dfrac{b^2-4b+3}{2}
\end{equation}
Case 3: If $2<b<3$, we have
\begin{equation}
h(b)
=
\int_{0}^{1}f(b-z)dz
=
\int_{b-2}^{1} 2-(b-z) \ dz
=
\frac{b^2-6b+9}{2}
\end{equation}
Distribution of B
\begin{equation}
h(b) = \begin{cases}
\frac{b^2}{2} & b\in (0,1)\\
\frac{b(2-b)}{2}-\frac{b^2-4b+3}{2} & b\in(1,2)\\
\frac{b^2-6b+9}{2} & b\in(2,3)\\
0 & \text{ otherwise}
\end{cases}.
\end{equation}
|
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|
Evaluate $ \lim_{x \to 0} \left( {\frac{1}{x^2}} - {\frac{1} {\sin^2 x} }\right) $ $$\lim_{x\to0}\left({\frac{1}{x^2}}-{\frac{1}{\sin^2x}}\right)$$
Using the L'Hospital Rule I obtained the value $-1/4$, but the answer is given to be $-1/3$. I can't find the mistake. Here's what I did; please point out the mistake.
\begin{align}
\lim_{x\to0}\left({\frac{1}{x^2}}-{\frac{1}{\sin^2x}}\right)&=\lim_{x\to0}\frac{(\sin x+x)(\sin x-x)}{(x\sin x)(x\sin x)} \\[1ex]
&=\lim_{x\to0}\left(\frac{\sin x+x}{x\sin x}\right)\lim_{x\to0}\left(\frac{\sin x-x}{x\sin x}\right) \\[1ex]
&=\lim_{x\to0}\left(\frac{\cos x+1}{\sin x+x\cos x}\right)\lim_{x\to0}\left(\frac{\cos x-1}{\sin x+x\cos x}\right) \\[1ex]
&=\lim_{x\to0}\:(\cos x+1)\,\lim_{x\to0}\left(\frac{\cos x-1}{(\sin x+x\cos x)^2}\right) \\[1ex]
&=\lim_{x\to0}\frac{-\sin x}{(\sin x+x\cos x)(2\cos x-x\sin x)} \\[1ex]
&=-\lim_{x\to0}\left[\frac{1}{1+\cos x\left(\frac{x}{\sin x}\right)}\right]\left(\frac{1}{2\cos x-x\sin x}\right) \\[1ex]
&=-\frac{1}{2}\left[\lim_{x\to0}\,\frac{1}{1+\cos x}\right] \\[1ex]
&=-\frac{1}{4}
\end{align}
|
$$\lim_{x \to 0} \left( {\frac{1}{x^2}} - {\frac{1} {\sin^2 x} }\right)=\lim_{x \to 0}\frac{\sin^2 x-x^2}{x^2\sin^2 x}=\lim_{x \to 0}\frac{(\sin x-x)(\sin x+x)}{x^4}$$
$$=\lim_{x \to 0}\frac{(\sin x+x)}{x}\lim_{x \to 0}\frac{x(\sin x-x)}{x^4}=\lim_{x \to 0}\frac{2x(\sin x-x)}{x^4}=\lim_{x \to 0}\frac{2(\sin x-x)}{x^3}$$
$$=\lim_{x \to 0}\frac{2(\cos x-1)}{3x^2}=\lim_{x \to 0}\frac{-2\sin x}{6x}=\frac{-1}{3}.$$
|
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|
Probability of 0 in Binary Sequence A random binary sequence is produced as follows. The first coordinate equals $0$ with probability $0.6$ and equals $1$ with probability $0.4$. For any positive integer $n$, the $(n+1)^\text{th}$ coordinate is the same as the $n^\text{th}$ coordinate of the sequence, with probability $0.7$, and equals the opposite of the $n^\text{th}$ coordinate with probability $0.3$.
Calculate the probability that the $n^\text{th}$ coordinate is $0$ and also the limit of this probability as $n \to \infty$.
Progress: I rewrote this as a recurrence relation using the law of total probability and then tried to solve it using a characteristic polynomial of a first-order difference equation. This resulted in me getting a probability of $0.5$ as $n \to \infty$. I don't know if this is right, but I wanted to get the input of the MSE community.
|
Starting from an initial $0$ (prob. $=0.6$) then in the following $n-1$ you must have an even number ($2k$ ) of changes to end with a $0$.
Starting with $1$ instead (prob. $=0.4$), you must have $2j+1$ changes.
That means
$$
\eqalign{
& P_{\,0} (n) = 0.6\sum\limits_{0\, \le \,k\, \le \,\left\lfloor {\left( {n - 1} \right)/2} \right\rfloor } {\left( \matrix{
n - 1 \cr
2k \cr} \right)p^{\,2k} q^{\,n - 1 - 2k} } + \cr
& + 0.4\sum\limits_{0\, \le \,j\, \le \,\left\lfloor {\left( {n - 2} \right)/2} \right\rfloor } {\left( \matrix{
n - 1 \cr
2j + 1 \cr} \right)p^{\,2j + 1} q^{\,n - 2 - 2j} } = \cr
& = 0.5\left( {\sum\limits_{0\, \le \,k\, \le \,\left\lfloor {\left( {n - 1} \right)/2} \right\rfloor } {\left( \matrix{
n - 1 \cr
2k \cr} \right)p^{\,2k} q^{\,n - 1 - 2k} } + \sum\limits_{0\, \le \,j\, \le \,\left\lfloor {\left( {n - 2} \right)/2} \right\rfloor } {\left( \matrix{
n - 1 \cr
2j + 1 \cr} \right)p^{\,2j + 1} q^{\,n - 2 - 2j} } } \right) + \cr
& + 0.1\left( {\sum\limits_{0\, \le \,k\, \le \,\left\lfloor {\left( {n - 1} \right)/2} \right\rfloor } {\left( \matrix{
n - 1 \cr
2k \cr} \right)p^{\,2k} q^{\,n - 1 - 2k} } - \sum\limits_{0\, \le \,j\, \le \,\left\lfloor {\left( {n - 2} \right)/2} \right\rfloor } {\left( \matrix{
n - 1 \cr
2j + 1 \cr} \right)p^{\,2j + 1} q^{\,n - 2 - 2j} } } \right) = \cr
& = 0.5 + 0.1\left( {\sum\limits_{0\, \le \,k\, \le \,n - 1} {\left( \matrix{
n - 1 \cr
k \cr} \right)\left( { - 1} \right)^{\,k} p^{\,k} q^{\,n - 1 - k} } } \right) = \cr
& = 0.5 + 0.1\left( { - p + q} \right)^{\,n - 1} = 0.5 + 0.1\left( {0.4} \right)^{\,n - 1} \cr}
$$
A check using Markov matrices confirms that.
So, for $n \to \infty$ the probability tends to $1/2$.
|
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|
Range of $y = \frac{x^2-2x+5}{x^2+2x+5}$? How do I approach this problem? My book gives answer as $[\frac{3-\sqrt{5}}{2},\frac{3+\sqrt{5}}{2}]$. I tried forming an equation in $y$ and putting discriminant greater than or equal to zero but it didn't work. Would someone please help me?
I get $x^2 (y-1) + 2x (y+1) + (5y-5) =0$ and discriminant gives $2y^2 - y + 2 \leq 0$, which has complex roots.
|
Since
$$y = \frac{x^2 - 2x + 5}{x^2 + 2x + 5}$$
we obtain
\begin{align*}
y(x^2 + 2x + 5) & = x^2 - 2x + 5\\
yx^2 + 2yx + 5y & = x^2 - 2x + 5\\
(y - 1)x^2 + (2y + 2)x + 5(y - 1) & = 0\\
(y - 1)x^2 + 2(y + 1)x + 5(y - 1) & = 0
\end{align*}
which is equivalent to your equation $x^2(y - 1) + 2x(y + 1) + (5y - 5) = 0$. However, you should have obtained the discriminant
\begin{align*}
\Delta & = b^2 - 4ac\\
& = [2(y + 1)]^2 - 4(y - 1) \cdot 5(y - 1)\\
& = 4(y^2 + 2y + 1) - 20(y - 1)^2\\
& = 4y^2 + 8y + 4 - 20(y^2 - 2y + 1)\\
& = 4y^2 + 8y + 4 - 20y^2 + 40y - 20\\
& = -16y^2 + 48y - 16
\end{align*}
We want $\delta \geq 0$, so
\begin{align*}
0 & \leq -16y^2 + 48y - 16\\
16y^2 - 48y + 16 & \leq 0\\
y^2 - 3y + 1 & \leq 0\\
y^2 - 3y & \leq -1\\
y^2 - 3y + \frac{9}{4} & \leq \frac{5}{4}\\
\left(y - \frac{3}{2}\right)^2 & \leq \frac{5}{4}\\
\left|y - \frac{3}{2}\right| & \leq \frac{\sqrt{5}}{2}
\end{align*}
Thus,
$$-\frac{\sqrt{5}}{2} \leq y - \frac{3}{2} \leq \frac{\sqrt{5}}{2} \implies \frac{3 - \sqrt{5}}{2} \leq y \leq \frac{3 + \sqrt{5}}{2}$$
so the range of the function is
$$\left[\frac{3 - \sqrt{5}}{2}, \frac{3 + \sqrt{5}}{2}\right]$$
|
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|
Find this integral $I=\int\frac{x}{(1-3x)^{3/2}(x+1)^{3/2}}dx$ Find the integral
$$I=\int\dfrac{x}{(1-3x)^{3/2}(x+1)^{3/2}}dx$$
My try: $$(1-3x)(x+1)=-3x^2-2x+1=-3(x+1/3)^2+\dfrac{4}{3}$$
Thus
$$I=\int\dfrac{x}{(\frac{4}{3}-3(x+\frac{1}{3})^2)^{3/2}}dx$$
|
Hint:
For real calculus, we need $$-1<x<\dfrac13$$
$$\iff-\dfrac23<x+\dfrac13<\dfrac23$$
WLOG $x+\dfrac13=\dfrac23\cos2t$ where $0<2t<\pi,\cos t,\sin t,\sin2t>0$
$\iff3x+1=2\cos2t,3x=2\cos2t-1=4\cos^2t-3$
$3dx=-8\sin t\cos t\ dt$
$1-3x=1-(2\cos2t-1)=2(1-\cos2t)=4\sin^2t\implies(1-3x)^{3/2}=(2\sin t)^3$
$x+1=\dfrac{3x+3}3=\dfrac{2\cos2t-1+3}3=\dfrac{4\cos^2t}3\implies(1+x)^{3/2}=\dfrac{(2\cos t)^3}{3\sqrt3}$
$$I=\int\dfrac{-8\cdot3\sqrt3(4\cos^2t-3)\sin t\cos t\ dt}{3^28^2\sin^3t\cos^3t}dt$$
$$\implies-8\sqrt3I=\int\dfrac{4\cos^2t-3}{\sin^2t\cos^2t}dt=4\int\sec^2t\ dt-12\int\csc^22t\ dt$$
Can you take it from here?
|
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|
How to get the smallest positive integer solutions to $2\sqrt [3] {x} =3\sqrt [5] {y}$? How do I find the smallest positive integer solutions for $x$ and $y$ in the following equation? $$2\sqrt [3] {x} =3\sqrt [5] {y}$$
|
we have $$ 2\sqrt[3]{x} = 3\sqrt[5]{ y}$$
so cubing both side we have
$$2^3 x=3^3y^{\frac 35}$$
now
$$2^{ 3 \times 5}x^5=3^{ 3\times 5} y^3$$
so now we can write this
$$\frac{2^{ 3 \times 5}}{3^{ 3\times 5}}= \frac{y^3}{x^5}$$
so $$ y^3= (2^5)^3 $$
and $$ x^5= (3^3)^5 $$
so$$ y=32$$ and $$x=27$$
|
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|
What does $\sum\limits_{n=1}^{\infty}\left( {\frac{1}{n}-\frac{1}{\sqrt{n^2+x^2}}} \right)$ equal? I tried calculating it in Mathematica but it can't seem to figure out a formula for this. If you define x as a specific real number then the sum always converges and gives you a number. The output seems to be logarithmic because when I do $e^{\sum_{n=1}^{\infty}{\frac{1}{n}-\frac{1}{\sqrt{n^2+x^2}}}}$ the output becomes linear as the function goes to $\infty$. It isn't linear as x approaches 0 though. Does anyone know the answer to this or how to find it?
Here is a plot of $e^{\sum_{n=1}^{\infty}{\frac{1}{n}-\frac{1}{\sqrt{n^2+x^2}}}}$
|
For $x$ sufficiently small we have that
$$\frac{1}{\sqrt{n^2+x^2}}=\frac 1 n\frac{1}{\sqrt{1+(x/n)^2}}\sim\frac1n\left(1-\frac12\frac{x^2}{n^2}\right)=\frac 1n-\frac12\frac{x^2}{n^3}$$
thus
$$\frac{1}{n}-\frac{1}{\sqrt{n^2+x^2}} \sim \frac12\frac{x^2}{n^3}$$
and therefore
$$S(x)=\sum_{n=1}^{\infty}\left({\frac{1}{n}-\frac{1}{\sqrt{n^2+x^2}}}\right)\sim \frac12x^2\sum_{n=1}^{\infty} \frac1{n^3}=cx^2$$
which explains qualitatively the nonlinear beaviour for $x$ sufficiently small.
For $x$ large let consider an integral estimation that is
$$S(x,k)=\sum_{n=1}^{k}\left({\frac{1}{n}-\frac{1}{\sqrt{n^2+x^2}}}\right)\approx \int_1^k \left({\frac{1}{n}-\frac{1}{\sqrt{n^2+x^2}}}\right)dn=\left[\log \frac{n}{\sqrt{n^2+x^2}+n}\right]_1^k=\log \frac{k}{\sqrt{k^2+x^2}+k}+\log (\sqrt{1+x^2}+1)\to -\log 2+\log (\sqrt{1+x^2}+1)\sim \log x$$
|
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|
Solving homogeneous differential equation $\frac{dy}{dx}=\frac{x^2+8y^2}{3xy}$.
Solve the differential equation. Use the fact that the given equation is homogeneous
$$
\frac{dy}{dx}=\frac{x^2+8y^2}{3xy}
$$
First I multiply the right side by
$$
\frac{\frac{1}{x^2}}{\frac{1}{x^2}}
$$
Then
$$\frac{dy}{dx}=\frac{1+\frac{8y^2}{x^2}}{\frac{3y}{x}}
$$
$$
Let: v=\frac{y}{x}
$$ $$
\frac{dy}{dx}=v+x\frac{dv}{dx}
$$
Then substitute
$$
\frac{dy}{dx}=\frac{1+v^2}{3v}$$
$$
\frac{1+v^2}{3v}=v+x\frac{dv}{dx}$$
How do I find the solution from here?
|
Hint:
First a slight miscalculation (you missed the $8$) on your part, it should be:
$$
\frac{1+8v^2}{3v}=v+x\frac{\mathrm{d}v}{\mathrm{d}x}$$
After this just separate the variables and integrate:
$$\int \frac{\mathrm{d}x}{x}=\int \frac{3v\ \mathrm{d}v}{(1+5v^2)}$$
Can you proceed?
Update:
To help you remove your confusion I have written down the remaining steps:
$$\frac{10}{3}\ln(xc)=\ln\left(1+5\left(\frac{y}{x}\right)^2 \right)$$
$$\ln\left((xc)^{\frac{10}{3}}\right)=\ln\left(1+5\left(\frac{y}{x}\right)^2 \right)$$
$$\frac 15 cx^{\frac{10}{3}}-\frac 15=\frac{y^2}{x^2}$$
$$\frac 15 cx^{\frac{10}{3}} \cdot x^2-\frac {x^2}{5}=y^2$$
$$\implies y=\pm \sqrt{\frac{cx^{\frac{16}{3}}}{5}-\frac {x^2}{5}}$$
$$\implies y=\pm \sqrt{c_1x^{\frac{16}{3}}-\frac {x^2}{5}}$$
|
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|
Number of sequences formed from $1, 1, 1, 2,3,4,5,6$ in which all three $1$s appear before the $6$ I want to find number of sequences which contains $2,3,4,5,6$ exactly once and $1$ exactly three times. Also all three $1$s should be placed before $6$.
Example $1)1,1,1,2,3,4,5,6\\2)1,2,3,4,1,1,6,5$
I think that $6$ can not be at $1$st, $2$nd, and $3$rd place.
If we fix $6$ at fourth place then all three $1$'s are fixed at first three places and there are $4!$ such sequences possible.
How to calculate for $6$ be fixed at $5$th to $8$th place?
|
HINT: I will explain it for sixth place and leave the rest to you:
Since $6$ is on sixth place, there are five places left before it and three of them must be filled by $1$'s. For the rest of two, we can choose from $2,3,4,5$ with $\binom{4}{2}$ and arrange them in those five places with $\frac{5!}{3!}$ (since $1$'s are identical). For the places after sixth place, we can rearrange them with $2!$ so if $6$ is on sixth place, there are
$$\binom{4}{2}\cdot\frac{5!}{3!}\cdot2!$$
ways.
|
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|
Which pairs of positive integers (,) satisfy $^2−2^=153$? My attempt: Rearrange to $x^2=2^n + 153$ and with $2^n\geq 2\ $ it follows $x^2 \geq 155\ $.
The next square number is 169, so $x = 13$ and $n = 4$. A first solution. Since $2^n$ is even and 153 is odd, $x^2$ will be odd. So any candidate solution will have an even distance of $2m$ from a previous solution and the difference between these solutions is $(x+2m)^2 - x^2 = 4mx + 4m^2$. This difference can be expressed as a difference between two powers of 2, $4mx + 4m^2 = 2^p - 2^n$. My idea was to show that this doesn´t work so that the first solution is the only one.
|
Suppose $x \geq 0$. Studying the equation modulo $2$ reveals that $x$ must be odd. Studying it modulo $3$ reveals that $n$ must be even $n=2k$.
$$x^2-2^n=(x-2^k)(x+2^k)=153$$
So $x-2^k$ and $x+2^k$ are two integers that average to $x$ but multiply to $153$. But $153=51(3)=1(153)=17(9)$, so either $x=26$ or $x=77$ or $x=13$. But $x$ is odd, so we must have $x=13$ or $x=77$. But $x=77$ doesn’t work.
Modulo 2: $x^2 \equiv 1$ so $x \equiv 1$.
Modulo $3$: $x^2-(-1)^n \equiv 0$ so $x^2 \equiv (-1)^n$. But since any square is equivalent to $0$ or $1$ modulo $3$. Either $(-1)^n \equiv 0$ (which is obviously impossible) or $(-1)^n \equiv 1$ (which is only possible if $n$ is even).
|
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|
solve this 1st order linear equation $$(x+3)^2\frac{dy}{dx}=6-12y-4xy=6-y(12+4x)$$
a. Write it in standard form.
$$\frac{dy}{dx}+\frac{12+4x}{(x+3)^2}y=\frac6{(x+3)^2}$$
b. What is the integrating factor?
$$\frac{12+4x}{(x+3)^2} = \frac14x+\frac34 \implies$$
$$IF=e^{\int{\frac14x+\frac34}}=e^{\frac18x^2+\frac34x}$$
c. Integrate the DE subject to $y(0)=1$.
$$e^{\frac18x^2+\frac34x}\frac{dy}{dx}+ye^{\frac18x^2+\frac34x}(\frac14x+\frac34)=\frac{6e^{\frac18x^2+\frac34x}}{(x+3)^2} \implies$$
$$ye^{\frac18x^2+\frac34x}=\int{\frac{6e^{\frac18x^2+\frac34x}}{(x+3)^2}}+c$$
I have done this problem a few times and now getting kind of stuck. Please check my standard form and integrating factor.
Right now if everything else is correct I cannot solve the integral. Any help appreciated.
|
$$(x+3)^2\frac{dy}{dx}=6-4y(3+x)$$
$$(x+3)^2\frac{dy}{dx}+4y(3+x)=6$$
Multiply by $(x+3)^2$
$$(x+3)^4y'+4y(3+x)^3=6(x+3)^2$$
$$((x+3)^4y)'=6(x+3)^2$$
Integrate
$$(x+3)^4y=2(x+3)^3+C$$
$$y(x)=\frac 2 {(x+3)}+\frac C {(x+3)^4}$$
As pointed out in the comment this line is not correct
$$\frac{12+4x}{(x+3)^2} = \frac14x+\frac34 $$
This is correct
$$\frac{12+4x}{(x+3)^2} = \frac 4 {x+3}$$
|
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|
Any idea how to find $\lim_{x\to 0} \frac{\sqrt{1-\cos(x^2)}}{1-\cos(x)}$? $$\lim_{x\to 0} \frac{\sqrt{1-\cos(x^2)}}{1-\cos(x)}$$
I am trying to solve this limit for 2 days, but still cant find the solution which is $\sqrt{2}$ (that's what is written in the solution sheet)
I tried multiplying with the conjugate, tried with some identities but nothing much because of that $x^2$ in the $\cos$.
Then i tried L'Hopital because it is $\frac{0}{0}$ and still that $\cos$ in the square root is doing problems. I tried on symbolab calculator but it can't solve it.
So can someone help me how to solve this?
Thank you.
|
By L'Hopital's rule, we have
$$\begin{aligned}
\lim_{x \to 0}\frac{1 - \cos(x^2)}{(1 - \cos(x))^2}
&= \lim_{x \to 0}\frac{2x\sin(x^2)}{2(1 - \cos(x))\sin(x)} \\
&= \lim_{x \to 0}\frac{x}{\sin(x)}\frac{\sin(x^2)}{1 - \cos(x)} \\
&= \lim_{x \to 0}\frac{\sin(x^2)}{1 - \cos(x)} \\
\end{aligned}$$
where we have used that
$$\lim_{x \to 0}\frac{x}{\sin(x)} = 1$$
Now we can apply L'Hopital's rule again to obtain
$$\begin{aligned}
\lim_{x \to 0}\frac{\sin(x^2)}{1 - \cos(x)}
&= \lim_{x \to 0} \frac{2x\cos(x^2)}{\sin(x)} \\
&= \lim_{x \to 0} \frac{x}{\sin(x)} 2\cos(x^2) \\
&= \lim_{x \to 0} 2\cos(x^2) \\
&= 2
\end{aligned}$$
Summarizing what we have so far,
$$\lim_{x \to 0} \frac{1 - \cos(x^2)}{(1 - \cos(x))^2} = 2$$
Now take the square root of both sides and use the fact that $\sqrt{(\cdot)}$ is continuous at zero (note that we approach only from the right) to conclude that
$$
\begin{aligned}
\lim_{x \to 0}\frac{\sqrt{1 - \cos(x^2)}}{1 - \cos(x)}
&= \lim_{x \to 0}\sqrt{\frac{1 - \cos(x^2)}{(1 - \cos(x))^2}} \\
&= \sqrt{\lim_{x \to 0} \frac{1 - \cos(x^2)}{(1 - \cos(x))^2} } \\
&= \sqrt{2}
\end{aligned}$$
as desired.
|
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|
If $a^2+b^2 \gt a+b$ and $a,b \gt 0$ Prove that $a^3+b^3 \gt a^2+b^2$ I'm not too sure about this, I have been working on for some time and I reached a solution (not really too sure about)
Question: If $a^2+b^2 \gt a+b$ and $a,b \gt 0$ Prove that $a^3+b^3 \gt a^2+b^2$
My solution: Let $a \geq b$
From $a^2+b^2 \gt a+b$ we get $a^2-a \gt b-b^2$
Since $a \geq b$ we can get $a^3-a^2 \gt b^2-b^3$ $\Rightarrow$ $a^3+b^3 \gt a^2+b^2$
If this solution is incorrect, please explain why and attach the correct solution. Thank you.
|
Your proof is correct. Indeed:
$$\begin{cases}a^2-a \gt b-b^2\\
a\ge b>0\end{cases} \Rightarrow \\
a(a^2-a)>b(b-b^2) \Rightarrow \\
a^3-a^2>b^2-b^3 \Rightarrow \\
a^3+b^3>a^2+b^2.$$
Alternative proof. Consider $b=ax, x\ge 1$. Then:
$$a^2+b^2 \gt a+b \Rightarrow \\
a^2+a^2x^2>a+ax \stackrel{\text{divide by} \ a}{\Rightarrow} \\
a+ax^2>1+x \Rightarrow \\
a>\frac{1+x}{1+x^2} \ \ (1)$$
Hence:
$$a^3+b^3>a^2+b^2 \iff \\
a^3+a^3x^3>a^2+a^2x^2 \Rightarrow \\
a+ax^3>1+x^2 \iff \\
a(1+x^3)\stackrel{(1)}{>}\frac{1+x}{1+x^2}(1+x^3)\ge 1+x^2 \iff \\
(1+x)(1+x^3)\ge(1+x^2)^2 \iff \\
1+x+x^3+x^4\ge1+2x^2+x^4 \iff \\
x(1+x^2)\ge 2x^2 \iff \\
(x-1)^2\ge 0.$$
It is the Cauchy-Swarz inequality (given by Michael Rozenberg).
|
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|
Finding the result of the polynomial The question is that: Given that $x^2 -5x -1991 = 0$, what is the solution of $\frac{(x-2)^4 + (x-1)^2 - 1}{(x-1)(x-2)}$
I've tried to factorize the second polynomial like this:
$\frac{(x-2)^4 + (x-1)^2 - 1}{(x-1)(x-2)}
=\frac{(x-2)^4+x(x-2)}{(x-1)(x-2)}
=\frac{(x-2)((x-2)^3+x)}{(x-1)(x-2)}=\frac{(x-2)^3+x}{(x-1)}$
However I could not solve it with the given equation $x^2 -5x -1991 = 0$. I know that direct substitution may work, but I think that there is a neat solution to this polynomial. Thanks in advance!
Edited:
I've found the way to solve it with the help of lhf
$\frac{(x-2)^4+(x-1)^2-1}{(x-1)(x-2)}=\frac{((x-2)^2+1)((x-2)^2-1)+(x-1)^2}{(x-1)(x-2)}=\frac{(x^2-4x+5)(x-1)(x-3)+(x-1)^2}{(x-1)(x-2)}=\frac{(x^2-4x+5)(x-3)+x-1}{(x-2)}=\frac{(x^2-4x+5)(x-2)-x^2+4x-5+x-1}{(x-2)}=\frac{(x^2-4x+5)(x-2)-(x-2)(x-3)}{(x-2)}=x^2-5x+8$
|
Try making the question simpler.
Let $t=x-2$ then $x-1$ will become $t+1$. So the question
$$\dfrac{(x-2)^4 + (x-1)^2 - 1}{(x-1)(x-2)}$$
changes to
$$\dfrac{t^4 + (t+1)^2 - 1}{t(t+1)}$$
$$\implies\dfrac{t^4 + t²+1+2t - 1}{t(t+1)}$$
$$\implies\dfrac{t^4 + t²+2t}{t(t+1)}$$
$$\implies\dfrac{t(t²-t+2)(t+1)}{t(t+1)}$$
$$\implies t²-t+2$$
$$\implies (x-2)²-(x-2)+2$$
$$\implies x²+8-5x$$
Now,
$$x²-1991-5x=0$$
Or
$$x²-5x=1991$$
So
$$x²+8-5x=1991+8$$
$$=1999$$
|
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Simplifying $\frac{^nC_{1}}{1 \times 2}-\frac{^nC_{2}}{2 \times 3}+..+(-1)^{n+1}\frac{^nC_{n}}{n \times (n+1)}$ I was thinking about the following sum -
$\frac{^nC_{1}}{1 \times 2}-\frac{^nC_{2}}{2 \times 3}+\frac{^nC_{3}}{3 \times 4}+..+(-1)^{n+1}\frac{^nC_{n}}{n \times (n+1)} = 1 + \frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+..+\frac{1}{n+1}$
How to approach to prove this summation equality?, I could owrite the LHS as $2n - 3n(n-1) + 2n(n-1)(n-2) +...$ after simplification.
|
Firstly,
$$\frac{^nC_{1}}{1 \times 2}+\frac{^nC_{2}}{2 \times 3}+\frac{^nC_{3}}{3 \times 4}+..+\frac{^nC_{n}}{n \times (n+1)} =\left[\frac{^nC_{1}}{1}+\frac{^nC_{2}}{2}+\frac{^nC_{3}}{3}+..+\frac{^nC_{n}}{n}\right]-\left[\frac{^nC_{1}}{2}+\frac{^nC_{2}}{3}+\frac{^nC_{3}}{4}+..+\frac{^nC_{n}}{(n+1)}\right] $$
Now consider,
\begin{align*}
(1+x)^n-1 & = \binom{n}{1}x+\binom{n}{2}x^2+\dotsb+\binom{n}{n}x^n\\
\int_0^1((1+x)^n-1)\,dx & = \binom{n}{1}\frac{1}{2}+\binom{n}{2}\frac{1}{3}+\dotsb+\binom{n}{n}\frac{1}{n+1}\\
\frac{(1+x)^n-1}{x} & = \binom{n}{1}1+\binom{n}{2}x+\dotsb+\binom{n}{n}x^{n-1}\\
\int_0^1\frac{(1+x)^n-1}{x}\,dx & = \binom{n}{1}\frac{1}{1}+\binom{n}{2}\frac{1}{2}+\dotsb+\binom{n}{n}\frac{1}{n}.
\end{align*}
So your given series is
$$\int_0^1\frac{(1+x)^n-1}{x}\,dx-\int_0^1((1+x)^n-1)\,dx$$
Since the question was edited later:
Since you made the changes in the sign of the actual expression, so you need to redo what I have proposed with $(1-x)^n$.
|
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Using induction to prove $\sum_{i = 1}^{n} i\cdot 2^i = (n - 1) \cdot 2^{n + 1} + 2$ I have to prove the following proposition:
$$\sum_{i = 1}^{n} i\cdot 2^i = (n - 1) \cdot 2^{n + 1} + 2$$
I have the following proof thus far:
Proof by Induction
Basis Step $(n = 1)$
LHS: $\sum_{i = 1}^{n} i2^i = 1 \cdot 2^1 = 2$
RHS: $2^2(0) + 2 = 2$
LHS = RHS
Inductive Hypothesis
Assume that the proposition is true for $i = 1, 2, 3, \ldots, k$
Inductive Step $(n = k + 1)$
LHS: $\sum_{i = 1}^{k} i2^i + (k + 1) \cdot 2^{k + 1} = 1 \cdot 2^1 + 2 \cdot 2^2 + \ldots + k \cdot 2^k + (k + 1) \cdot 2^{k + 1}$
RHS: $(k - 1) \cdot 2^{k + 1} + 2 + (k + 1) \cdot 2^{k + 1}\\ = (k - 1) \cdot 2^{k + 1} + (k + 1) \cdot 2^{k + 1} + 2\\ = ((k - 1) + (k + 1)) \cdot 2^{k + 1} + 2\\ = (2k) \cdot 2^{k + 1} + 2$
LHS = RHS
QED
Am I correct? If not, am I headed in the right direction? I understand the basics of induction, but am continuing to apply it. Thank you!
|
Your base case is correct, but you need more information in the inductive hypothesis.
Base Case: $(n = 1)$
$$\sum_{i=1}^{1}i2^i = 1\cdot 2^1 = 2 = (1-1) \cdot 2^{1+1} + 2 $$
Inductive Hypothesis:
Assume that $$\sum_{i=1}^{k} i2^i = (k-1) \cdot 2^{k+1} + 2 \hspace{1in} \text{IH}$$ for some integer $k >1$. You must show that
$$\sum_{i=1}^{k+1} i2^i = k2^{k+2} + 2$$
Inductive Step:
$$\sum_{i=1}^{k+1} i2^i = \sum_{i=1}^{k} i2^i+ \sum_{i=k+1}^{k+1} i2^i = (k-1) \cdot 2^{k+1} + 2 + (k+1)\cdot 2^{k+1} \hspace{.5in} \text{By IH}$$
$$= (k-1)\cdot 2^{k+1} + 2$$
$$= 2^{k+1} [ (k-1) + (k+1)] + 2$$
$$= 2^{k+1}\cdot 2k + 2$$
$$= k2^{k+1}\cdot 2^1 + 2$$
$$= k2^{k+2} + 2$$
Thus,
$\sum_{i=1}^{n} i2^i = (n-1) \cdot 2^{n+1} + 2 \ \text{for any } n\geq 1$
QED
|
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Can someone help me finish this: evaluate $S_n = \frac{x}{1-x^2}+\frac{x^2}{1-x^4}+ ... + \frac{x^{2^{n-1}}}{1-x^{2^{n}}}$ I am asked to find the closed form solution for the below.
$$S_n = \frac{x}{1-x^2}+\frac{x^2}{1-x^4}+ ... + \frac{x^{2^{n-1}}}{1-x^{2^{n}}}$$
Just writing out the $S_1, S_2, S_3$, I have managed to find a pattern, which is:
$$S_n = \frac{S_{n-1}}{1-x^{2^n}} + \frac{x^{2^n}}{1-x^{2^{n+1}}}$$
I am not sure how to proceed onwards to solve this recurrence relation. Is there a clever trick I can do to solve it?
|
Recognize the geometric series and replace it:
$$\frac{x}{1-x^2}=x+x^3+x^5+x^7+\cdots$$
For all the rest, it's the same, just replace $x\to x^2,x^4,\ldots x^{2^n}$.
Now notice that the new terms are just filling in the missing terms:
$$S_2=x+x^2+x^3+\Box+x^5+x^6+x^7+\Box+x^9+\cdots$$
$$S_3=x+x^2+x^3+x^4+x^5+x^6+x^7+\Box+x^9+\cdots+x^{15}+\Box+x^{17}\cdots$$
From this, it's easy to see that the limit of this sequence is just the full geometric series without missing terms: $S_{\infty}=\frac{x}{1-x}$. If you stop at term $S_n$, you are missing all powers of $x^{2^n}$ (for example, $S_3$ is missing $x^8$, $x^{16}$ and so on), and the missing terms are simply the geometric series with $x^8$ ($x^{2^n}$ in general) instead of $x$. This makes it easy to write like this:
$$S_{n}=S_{\infty}(x)-S_{\infty}(x^{2^n})=\frac{x}{1-x}-\frac{x^{2^n}}{1-x^{2^n}}$$
|
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How do I complete the square of $y= -4x^2-2x-4$? $y = -4x^2 - 2x - 4$
I just can't figure this out, do I divide the second number and the third number by $4$ then by $2$ and then add the product to the second one and subtract it from the third one?
|
First, you need to isolate x^2 from the equation by factor the -4 out, which leave the equation like this:
$$y= -4(x^2+ \frac{1}{2} x ) -4 $$
We know that the equation of square binomials is:
$(x+b)^2= x^2+2xb+b^2$
To write the binomial $x^2 + \frac{1}{2}x$ to a square binomials, we need to find b^2
We got :
$2xb = \frac{1}{2} x $
=> $2b= \frac{1}{2}$
=> $b= \frac{1}{4}$
=> $b^2= \frac{1}{16}$
To balance the equation we need to add and subtract b^2 in the parentheses
$$y= -4(x^2+ \frac{1}{2} x +\frac{1}{16} -\frac{1}{16}) -4$$
Distribute the $-\frac{1}{16}$ out and combine like term
$$y= -4(x^2+ \frac{1}{2} x +\frac{1}{16}) +\frac{1}{4} -4$$
$$y= -4(x^2+ \frac{1}{2} x +\frac{1}{16}) -\frac{15}{4}$$
Now convert the trinominal inside of the parentheses to square binomial
$$y= -4(x^2+ 2\frac{1}{4} x +(\frac{1}{4})^2) -\frac{15}{4}$$
$$y= -4(x + \frac{1}{4})^2 -\frac{15}{4}$$
|
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Complex number equations ( process) Among the exercises I was solving these are the ones I don't understand and I don't know if the process or solutions are ok, so please correct whenever I went wrong. Thanks everyone.
(I'm supposed to solve by converting to trigonometric form when it's convenient and then finding roots.)
a) $z^3-z^{-3}=i$
$\frac{z^6-1}{z^{3}}=i$
$z^6-1=iz^3$
$z^3(z^3-i)=1$
1. $z^3=1$
$z=cis(\frac{2π+2kπ}{3})$, k=0,1,2
2. $z^3-i=1$
$z^3=\sqrt{2}cis(\frac{π}{4})$
$z=2^{\frac{1}{6}}cis(\frac{\frac{π}{4}+2kπ}{3})$, k=0,1,2
b) $(z-i)^6=-64$
$\sqrt{((z-i)^3)^2}=\sqrt{-64}$ square root to both sides
$(z-i)^3=\sqrt{-1*64}$
$(z-i)^3=8i$
$z-i=2cis(\frac{\frac{π}{2}+2kπ}{3})$
$z=2cis(\frac{5kπ}{6})+i=2cis(\frac{5kπ}{6})+cis(\frac{π}{2})$, k=0,1,2 ?
c) $z^5(1+i)=\bar z $ multiplying both sides by z
*$z^6(1+i)=\mid{z}\mid^2$ using absolute value on both sides
$\mid{z}\mid^6\mid{1+i}\mid=\mid{z}\mid^2$ dividing both sides by $\mid{z}\mid^2$
$\mid{z}\mid^4=\frac{1}{\mid{1+i}\mid}=\frac{1}{\sqrt{2}}$
$\mid{z}\mid=\frac{1}{2^\frac{1}{8}}$ then using * again
$z^6(1+i)=\frac{1}{2^\frac{1}{4}}$
$z^6=\frac{1}{2^\frac{1}{4}(1+i)}$
Don't know how to proceed.
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Hint: For $c$, let $z=re^{i\theta}$
\begin{align}
z^5(1+i) &= \bar z\\
r^5e^{5i\theta}(\sqrt{2}e^{i\pi/4}) &= re^{-i\theta} \\
r^4\sqrt{2} &= e^{-6i\theta-i\pi/4}
\end{align}
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"url": "https://math.stackexchange.com/questions/2945487",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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|
Prove that the quantity is an integer I want to prove that $\frac{n^3}{3}-\frac{n^2}{2}+\frac{n}{6} \in \mathbb{Z}, \forall n \geq 1$.
I have thought to use induction.
Base Case: For $n=1$, $\frac{n^3}{3}-\frac{n^2}{2}+\frac{n}{6}=\frac{1}{3}-\frac{1}{2}+\frac{1}{6}=0 \in \mathbb{Z}$.
Induction hypothesis: We suppose that it holds for $n=k$, i.e. that $\frac{k^3}{3}-\frac{k^2}{2}+\frac{k}{6} \in \mathbb{Z}$.
Induction step: We want to show that it holds for $n=k+1$.
$$\frac{(k+1)^3}{3}-\frac{(k+1)^2}{2}+\frac{k+1}{6}=\frac{k^3}{3}+\frac{k^2}{2}+\frac{k}{6}$$
Is everything right? If so, then we cannot use at the induction step the induction hypothesis, can we?
Or can we not get the desired result using induction?
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If you put everything over a common denominator you get $$f(n)=\frac {2n^3-3n^2+n}6=\frac {n(n-1)(2n-1)}6$$
Now it is easier, I think, to see what is going on for a straightforward induction, or you can write $2n-1=2n-4+3$ and split up into different fractions viz $$f(n)=\frac {n(n-1)(2n-4+3)}6=\frac {n(n-1)(n-2)}3+\frac {n(n-1)}2$$ and this is easily the sum of two integers since the product of $r$ successive integers is divisible by $r$ (easily proved by induction), or you can do the induction on $n$ based on this form of $f(n)$, which will should work easily.
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"url": "https://math.stackexchange.com/questions/2946269",
"timestamp": "2023-03-29T00:00:00",
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|
Maximum and minimum absolute value of a complex number
Let, $z \in \mathbb C$ and $|z|=2$. What's the maximum and minimum value of $$\left| z - \frac{1}{z} \right|? $$
I only have a vague idea to attack this problem.
Here's my thinking :
Let $z=a+bi$
Exploiting the fact that, $a^2+b^2=4$
We get $z-\dfrac{1}{z}=a-\dfrac{a}{4}+i\left(b+\dfrac{b}{4}\right)$
So
$$
\begin{split}
\left|z-\frac{1}{z}\right|
&=\sqrt{\left(a-\dfrac{a}{4}\right)^2+\left(b+\dfrac{b}{4}\right)^2}\\
&=\sqrt{4+\dfrac{1}{4}-\dfrac{a^2}{2}+\dfrac{b^2}{2}}\\
&=\sqrt{4+\dfrac{1}{4}+\dfrac{1}{2}(b^2-a^2)}
\end{split}
$$
The minimum value can be obtained if we can minimize $b^2-a^2$.
Setting $b=0$ gives
the minimum value $\sqrt{2+\dfrac{1}{4}}=\dfrac{3}{2}$
Now, comes the maximum value.
We can write $$\sqrt{4+\dfrac{1}{4}+\dfrac{1}{2}(b^2-a^2)}$$
$$=\sqrt{4+\dfrac{1}{4}+\dfrac{1}{2}(4-2a^2)}$$
$$=\sqrt{4+\dfrac{1}{4}+2-a^2}$$
$$=\sqrt{6+\dfrac{1}{4}-a^2}$$
Setting $a=0$ gives the maximum value $\sqrt{6+\dfrac{1}{4}}=\dfrac{5}{2}$.
I don't know if it's okay to set $b=0$ since $z$ would become a real number then.
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A (very) faster way:
We know that $z$ is in the circle centered at $0$ and radius $2$ and $1/z$ is in the circle of center $0$ and radius $1/2$. The maximum distance between a point of the former and a point of the latter is
$$2+\frac12=\frac52$$
Now, we need to show that there exists some $z$ such that this distance is reached. Take $z=2i$.
Can you deal with the minimum now?
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"timestamp": "2023-03-29T00:00:00",
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Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.