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Condition for the generators through any one of the ends of an equi-conjugate diameter of the principal elliptic section of the hyperboloid
Question: Show that the generators through any one of the ends of an equi-conjugate diameter of the principal elliptic section of the hyperboloid $\frac{x^2}{a^2}+\frac{y^2}{b^2}-\frac{z^2}{c^2} = 1$ are inclined to each other at an angle of $60 ^\circ$ if $a^2+b^2=6c^2$.
Solution: I took the point on the diameter as $(a\cos\theta,b\sin\theta,0)$ . So the two generators through this point are
$$\begin{align}
\frac{x-a\cos\theta}{a\sin\theta}=\frac{y-b\sin\theta}{-b\cos\theta}=\frac{z}{\pm c}
\end{align}$$
So the direction ratios of two generators are $(a\sin\theta,-b\cos\theta,c)$ and $(a\sin\theta, -b\cos\theta,-c)$ respectively.
$$\begin{align}
\cos\theta &=\frac{l_1l_2+m_1m_2+n_1n_2}{\sqrt{{l_1}^2+{m_1}^2+{n_1}^2}\sqrt{{l_2}^2+{m_2}^2+{n_3}^2}} \\
\implies \cos\theta &=\frac{a^2\sin^2\theta+b^2\cos^2\theta-c^2}{a^2\sin^2\theta+b^2\cos^2\theta+c^2}
\end{align}$$
Putting $\theta = 60^\circ$ and solving I got $3a^2+b^2 = 12c^2$
I am wondering whether the question is wrong or I missed anything or my approach itself is wrong. Appreciate any hint.
| $\alpha = 60^\circ$ is the angle between two generators not the parameter of end points of conjugal diameters.
$$\begin{align}
\cos\alpha &=\frac{l_1l_2+m_1m_2+n_1n_2}{\sqrt{{l_1}^2+{m_1}^2+{n_1}^2}\sqrt{{l_2}^2+{m_2}^2+{n_3}^2}} \\
\implies \cos\alpha &=\frac{a^2\sin^2\theta+b^2\cos^2\theta-c^2}{a^2\sin^2\theta+b^2\cos^2\theta+c^2}
\end{align}$$
Putting $\alpha = 60^\circ$ and $\theta = 45^\circ (\because $ equi-conjugal diameters means equal length of conjugal diameters i.e. $\sqrt {a^2cos^2\theta+b^2sin^2\theta} = \sqrt {a^2sin^2\theta+b^2cos^2\theta}$ which is possible for $\theta = 45^\circ$)
Putting values of $\theta$ and $\alpha$ you get the answer $a^2+b^2 = 6c^2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2533581",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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If $a,b,c$ are three complex numbers Find possible values of $\lvert a+b+c \rvert$ Given three complex numbers $a,b,c$ such that $\lvert a \rvert=\lvert b \rvert=\lvert c \rvert=1$ and $$\frac{a^2}{bc}+\frac{b^2}{ac}+\frac{c^2}{ab}=-1$$
Find which of the following are possible values of $\lvert a+b+c \rvert$
A)0
B)2
C)1.5
D)3
My try:
I assumed $a=e^{ix}$,$b=e^{iy}$, $c=e^{iz}$
Then we have
$$\cos (2x-y-z)+\cos (2y-x-z)+\cos (2z-x-y)=-1$$
$$\sin(2x-y-z)+\sin (2y-x-z)+\sin (2z-x-y)=0$$
Squaring and adding we get
$$3+2(\cos(3x-3y)+\cos(3y-3z)+\cos (3z-3x)=1$$
So
$$\cos(3x-3y)+\cos(3y-3z)+\cos (3z-3x)=-1$$
any clue here?
| We will use some facts such as $|a|=1$ implies $a\bar{a}=1$. So $\frac{1}{a}=\bar{a}$. Also let $w=a+b+c$, then $|w|^2=(a+b+c)(\bar{a}+\bar{b}+\bar{c})$.
\begin{align*}
\frac{a^2}{bc}+\frac{b^2}{ac}+\frac{c^2}{ab}&=-1\\
\frac{a^3}{abc}+\frac{b^3}{abc}+\frac{c^3}{abc}&=-1\\
a^3+b^3+c^3+abc & =0\\
a^3+b^3+c^3-3abc & =-4abc\\
(a+b+c)(a^2+b^2+c^2-ab-bc-ca)&=-4abc\\
(a+b+c)((a+b+c)^2-3ab-3bc-3ca)&=-4abc.
\end{align*}
we get
\begin{align*}
w^3-3w(ab+bc+ca)&=-4abc\\
w^3-3wabc\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)&=-4abc\\
w^3&=abc\left[3w\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)-4\right]\\
w^3&=abc\left[3w\left(\bar{a}+\bar{b}+\bar{c}\right)-4\right]\\
w^3&=abc\left[3|w|^2-4\right]\\
|w^3|&=\left|3|w|^2-4\right|\\
\tag{1}
\label{eq1}
|w|^3&=\left|3|w|^2-4\right|\\
\end{align*}
*
*Case 1: If $3|w|^2-4 \geq 0$, then \eqref{eq1} becomes $|w|^3-3|w|^2+4=0$ and this implies $|w|=2$.
*Case 2: If $3|w|^2-4 < 0$, then \eqref{eq1} becomes $|w|^3+3|w|^2-4=0$ and this implies $|w|=1$.
Thus $|w| \in \{1,2\}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2533698",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Only certain values seem to make an expression into perfect square, can you help me prove or disprove it? All the variables in this question are in integers. I am trying to prove that
$$\frac{4k-\Delta^2}3$$
is a perfect square only if $\Delta \in \{\pm(2a+b),\pm(a+2b),\pm(a-b)\}$ where $a,b$ are such that $k=a^2+ab+b^2$.
Example: $k=49$. The above expression becomes perfect square only for $\Delta \in \{2,7,11,13,14\}$ and their negatives.
$7$ and $14$ can be expressed as $a-b$ and $2a+b$ for $a=7,b=0$ which satisfy $7^2+7\cdot 0 + 0^2=49$.
$2, 11$ and $13$ can be expressed as $a-b, a+2b$ and $2a+b$ for $a=5, b=3$ which satisfy $5^2+5\cdot 3 + 3^2=49$.
Could you give me some hints on how to approach this and prove that any valid $\Delta$ will be expressable in those forms using $a,b$ that satisfy $k=a^2+ab+b^2$?
The background
To give some depth to my question - I was investigating the following equation in integers:
$$z^2=\frac{4k-\Delta^2}3$$
I discovered that it is solvable iff $k$ is a Löschian number (exists $a,b$ such that $k=a^2+ab+b^2$).
The proof:
For a Löschian $k=a^2+ab+b^2$ take any $\Delta \in \{\pm(2a+b),\pm(a+2b),\pm(a-b)\}$, insert $k,\Delta$ in RHS the and get a square. For the opposite direction - if $z, k, \Delta$ satisfy the equation, take $a=z$, $b=\frac{\Delta-z}2$ (can be shown to be integer) and see that $a^2+ab+b^2=\frac{3z^2+\Delta^2}4=k$ therefore $k$ is Löschian.
However, I also strongly suspect (based on numerical experiments) that the listed $\Delta$ values are the only ones that satisfy the equation. But I struggle to prove that.
| Suppose $\frac{4k - \Delta^2}{3}$ is a square, in particular say $$\frac{4k - \Delta^2}{3}=M^2.$$ Then $4k = \Delta^2 + 3 M^2$.
Reducing modulo 4, we see that $\Delta^2 + 3M^2 = 0 \pmod{4}$, so $\Delta^2 = M^2 \pmod{4}$. Hence, $\Delta=M \pmod{2}$, so they are both even or both odd.
If they are both odd:
Set $$a = \frac{\Delta + M}{2}, \quad b = \frac{M - \Delta}{2}$$ which are certainly both integers.
It can be checked that $a^2 + ab + b^2 = \frac{1}{4}(\Delta^2 + 3M^2) = k$.
Also, $\Delta = a-b$, so $\Delta$ can be written as $a-b$, where $a,b$ satisfy $k=a^2 + ab+b^2$
If they are both even:
Similarly, set $$a = \frac{\Delta-M}{2}, \quad b=M$$ both integers.
Again, we have $a^2 + ab+b^2 = k$, and $\Delta = 2a+b$, as required.
Ultimately, we have shown that for any $\Delta$ which makes $\frac{4k-\Delta^2}{3}$ a square number, there are $a,b$ satisfying $a^2 + ab+b^2=k$ and $\Delta = 2a+b$ or $\Delta = a-b$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2537381",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Closed-form expression $(x_{2^l})_{l\in \mathbb{N}_0}$ I have this problem I'm not sure how to solve.
Given $x \in \mathbb{R}^{\mathbb{N}_0}_{\geq 0}$ and
$x_k=\left\{\begin{matrix} 0 & k=0\\ 4x_{\left \lfloor \frac{k}{2}
> \right \rfloor} + k^2 & k \in \mathbb{N} \end{matrix}\right.$
Determine a closed-form expression for $(x_{2^l})_{l\in
> \mathbb{N}_0}$.
My solution attempt is as follows.
For $k \in \mathbb{N}$ we have $x_k=4x_{\left \lfloor \frac{k}{2} \right \rfloor} + k^2$. It follows
$$x_{2k}=4x_{\left \lfloor \frac{2k}{2} \right \rfloor} + 2k^2$$
$$=4x_{k} + 2k^2$$
For $l \in \mathbb{N}_0$ it follows
$$x_{2^{l+1}}=x_{2*2^l}=4x_{2^l} + 2 * 2^{2l}=4x_{2^l} + 2^{2l+1}$$
Is this correct? And how to get $x_{2^l}$ from here?
| Your calculation should be revised. We obtain for $l\in\mathbb{N}_0$:
\begin{align*}
x_{2^{l+1}}=4x_{2^l}+\color{blue}{\left(2^{l+1}\right)^2}=4x_{2^l}+4^{l+1}
\end{align*}
We start with $k=2^l$, calculate a few iterations and look if we can make a good guess of a closed formula. The validity of the formula could then be shown for instance with a proof by induction.
We obtain by successively using the recursion formula
\begin{align*}
\color{blue}{x_{2^l}}&=4x_{2^{l-1}}+\left(2^l\right)^2\\
&=4x_{2^{l-1}}+4^l=4\left(4x_{2^{l-2}}+\left(2^{l-1}\right)^2\right)+4^l=4^2x_{2^{l-2}}+4\cdot 4^{l-1}+4^l\tag{1}\\
&=4^2x_{2^{l-2}}+2\cdot 4^l=4^2\left(4x_{2^{l-3}}+\left(2^{l-2}\right)^2\right)+2\cdot4^l
=4^3x_{2^{l-3}}+4^2\cdot4^{l-2}+2\cdot 4^l\\
&=4^3x_{2^{l-3}}+3\cdot 4^l\\
&\qquad\vdots\\
&\color{blue}{=4^lx_{2^0}+l\cdot 4^l}\tag{2}
\end{align*}
Comment:
*
*In (1) we use the recursion formula once and do the same in the next line.
*In (2) we see the relationship between exponents and indices expressed in $l-1,l-2,l-3$ in the first lines and obtain the representation in the last line by setting $l-l$.
Since $x_0=0$ and $$x_1=4x_0+1=4\cdot0+1=1$$ we obtain
\begin{align*}
\color{blue}{x_{2^l}}=4^lx_{2^0}+l\cdot 4^l=4^l+l\cdot 4^l=\color{blue}{(l+1)4^l}
\end{align*}
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
} |
Find Jordan basis for the following nilpotent matrix I need to find the Jordan basis and Jordan form for this nilpotent matrix:
$$\begin{pmatrix}
-2i & 1 & 10+5i \\
4 & 2i & -10+19i \\
0 & 0 & 0 \\
\end{pmatrix}$$
The matrix is nilpotent so all eigenvalues are 0, therefore $Av_1=0$ and from that I got that $v_1=\begin{pmatrix}
1 \\ 2i \\ 0 \\ \end{pmatrix}$
Now $Av_2=v_1$ so $v_2=\begin{pmatrix} 0 \\ 1 \\ 0 \\ \end{pmatrix}$. This is where I'm stuck there seems to be no $v_3$ such that $Av_3=v_2$. How do I continue or what am I doing wrong?
| You have it a bit backwards. We have $A^3 = 0$ but $A^2 \neq 0.$ It appears you want the columns of the change of basis matrix, call that $V,$ as you want column vectors $v_1, v_2,v_3.$
Well,
$$
A^2 =
\left(
\begin{array}{rrr}
0 & 0 & -i \\
0 & 0 & 2 \\
0 & 0 & 0
\end{array}
\right)
$$
In order to arrange $A^2 v_3 \neq 0,$ take
$$
v_3 =
\left(
\begin{array}{rr}
0 \\
0 \\
1
\end{array}
\right)
$$
Next
$$
v_2 =A v_3 =
\left(
\begin{array}{rr}
10 + 5i \\
-10 + 19i \\
0
\end{array}
\right)
$$
$$
v_1 = Av_2 =A^2 v_3 =
\left(
\begin{array}{rr}
-i \\
2 \\
0
\end{array}
\right)
$$
we have set up $A v_1 = A^3 v_3 = 0 v_3 = 0.$
Put the columns in order, we have
$$
V =
\left(
\begin{array}{rrr}
-i & 10+5i & 0 \\
2 & -10 + 19i & 0 \\
0 & 0 & 1
\end{array}
\right)
$$
Note $\det V = -1,$ so finding $V^{-1}$ is not that bad.
Then
$$
J = V^{-1} AV =
\left(
\begin{array}{rrr}
0 & 1 & 0 \\
0 & 0 & 1 \\
0 & 0 & 0
\end{array}
\right)
$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "1",
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Find the greatest n for which the sum $\sum_{k=1}^n \lfloor\sqrt{k}\rfloor$ is a prime number. I've come across the sum:
$$\sum_{k=0}^n \lfloor\sqrt{k}\rfloor$$
According to Find a formula for $\sum\limits_{k=1}^n \lfloor \sqrt{k} \rfloor$
I approached the formula:
$$\sum_{k=0}^n \lfloor\sqrt{k}\rfloor = n\lfloor\sqrt{n}\rfloor-\frac{1}{3}\lfloor\sqrt{n}\rfloor^3-\frac{1}{2}\lfloor\sqrt{n}\rfloor^2+\frac{5}{6}\lfloor\sqrt{n}\rfloor$$
Now, programmaticaly I attempted this formula for first $100$ terms. The biggest possible $n$ for which this sum was a prime number should be $47$. Now, is there a precise mathematical approach to this? I can't seem to find a solution.
| Note that for $n\geq 1$, the sum $S_n$ is a positive integer which can be written as
$$S_n:=\sum_{k=0}^n \lfloor\sqrt{k}\rfloor =\frac{1}{6}\cdot\lfloor\sqrt{n}\rfloor\cdot\left(6n-(2\lfloor\sqrt{n}\rfloor^2+3\lfloor\sqrt{n}\rfloor-5)\right).$$
where $f_1:=\lfloor\sqrt{n}\rfloor$ and $f_2:=\left(6n-(2\lfloor\sqrt{n}\rfloor^2+3\lfloor\sqrt{n}\rfloor-5)\right)$ are positive integers.
Therefore if $f_1:=\lfloor\sqrt{n}\rfloor>6$ (i.e. $n\geq 49$) and
$$f_2=\left(6n-(2\lfloor\sqrt{n}\rfloor^2+3\lfloor\sqrt{n}\rfloor-5)\right)\geq \left(6n-(2n+3n-5)\right)=n+5>6$$
(i.e. $n\geq 2$) then $S_n=\frac{f_1\cdot f_2}{6}$ is not a prime (because $S_n$ is the product of two integer factors both greater than $1$).
Hence the largest $n$ such that $S_n$ is a prime, exists, it is less than $49$ and, according to your computations, we may conclude that it is equal to $47$ with $S_{47}=197$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Finding value of $c+d$
At a point $A(1,1)$ on the ellipse , equation of tangent is $y=x.$ If one of the
foci of the ellipse is $(0,-2)$ and coordinate of center of ellipse is $(c,d)$.
Then find value of $c+d$ (given length of major axis is $4\sqrt{10} unit$)
Attempt : assuming one foci is at $S_{1}(0,-2)$ and other is at $S_{2}(\alpha,\beta)$ and $A(1,1)$ be a point on ellipse. then $AS_{1}+AS_{2} = 4\sqrt{2}$
$\sqrt{10}+\sqrt{(1-\alpha)^2+(1-\beta)^2} = 4\sqrt{2}$
could some help me to solve it , thanks
| We have
*
*Sum of distances from focii $=2a = 4\sqrt{10}$ from which we obtain
$$(\alpha-1)^2+(\beta-1)^2 = 90$$
*Product of distances from the tangent (which is helpfully $x=y$) is $b^2$ ($b$ is the semi-minor axis) and hence $$4(\alpha-\beta)=4b^2$$ (Noting that $\alpha>\beta$)
*Distance between foci $=\sqrt{4(a^2-b^2)}$
This yields $$\alpha^2+(\beta+2)^2 = 160-4b^2$$
From these we obtain that $3 \alpha-\beta = 34$. Substituting in the first equation we get $\alpha=10$ and hence $\beta = 4$.
Thus $c=5, d=1$ and hence $c+d=6$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Finding roots of a specific Function How can I calculate the roots of this equation? What about in MATLAB?
x2=N(N(x2))
The plot of the function is in below:
| If I understand correctly, $N$ is the function defined piecewise by
$$
N(x) = \begin{cases}
5 & x \leq 2, \\
9 - 2x & 2 < x < 4, \\
1 & x \geq 4.
\end{cases}
$$
We can then calculate what $N(N(x))$ is.
*
*If $x \leq 2$ then $N(x) = 5$, so $N(N(x)) = N(5) = 1$.
*If $x \geq 4$ then $N(x) = 1$, so $N(N(x)) = N(1) = 5$.
*If $2 < x < 4$ then $N(x) = 9-2x$. Since we're plugging this back into $N$, we want to know when $2 < 9-2x < 4$. Subtracting $9$ from this gives $-7 < -2x < -5$, and dividing by $-2$ gives $5/2 < x < 7/2$.
In the range $5/2 < x < 7/2$ we thus have
$$
N(9-2x) = 9 - 2(9-2x) = 4x - 9.
$$
In the range $x \leq 5/2$ we have $9-2x \geq 4$, so that here $N(9-2x) = 1$.
In the range $x \geq 7/2$ we have $9-2x \leq 2$, so that here $N(9-2x) = 5$.
In summary,
$$
N(9-2x) = \begin{cases}
1 & x \leq 5/2, \\
4x - 9 & 5/2 < x < 7/2, \\
5 & x \geq 7/2.
\end{cases}
$$
Combining these three bullet points, we conclude that
$$
N(N(x)) = \begin{cases}
1 & x \leq 5/2, \\
4x - 9 & 5/2 < x < 7/2, \\
5 & x \geq 7/2.
\end{cases}
$$
Now, you are interested in determining which values of $x$ satisfy
$$
x = N(N(x)). \tag{1}
$$
Since $N(N(x)) = 1$ for $x \leq 5/2$, the only solution to equation $(1)$ with $x \leq 5/2$ is $x = 1$.
If $5/2 < x < 7/2$ then $N(N(x)) = 4x - 9$, and if we set
$$
x = 4x-9
$$
then we get $x = 3$. This falls in the range $5/2 < x < 7/2$, so this is a true solution to equation $(1)$.
If $x \geq 7/2$ then $N(N(x)) = 5$, so the only solution to equation $(1)$ in this interval is $x=5$.
In total, the solutions to equation $(1)$ are $x=1$, $x=3$, and $x=5$.
Here is a plot of $N(N(x))$ in blue versus $x$ in orange.
This was created in Mathematica using the code
n[x_] := Piecewise[{{5, x < 2}, {9 - 2 x, 2 < x < 4}, {1, x > 4}}];
Plot[{n[n[x]], x}, {x, 0, 6}, PlotPoints -> 70]
| {
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Prove that any field $F$ containing $\sqrt{a}+\sqrt{b}$ also contains $\sqrt{a}$ and $\sqrt{b}$.
Prove that any field $F$ containing $\sqrt{a}+\sqrt{b}$ also contains $\sqrt{a}$ and $\sqrt{b}$.
I started by taking $(\sqrt{a}+\sqrt{b})^2=2\sqrt{ab}+a+b$ and want to conclude that $\sqrt{ab}\in F$ but am unsure of this. Then I took $\sqrt{ab}(\sqrt{a}+\sqrt{b})=a\sqrt{b}+\sqrt{a}b\rightarrow b\sqrt{a}+a\sqrt{b}\in F$ and I know this one is because it is of the form $x\sqrt{a}+y\sqrt{b}\in F$. Thus $1\cdot \sqrt{a}+0\cdot \sqrt{b}=\sqrt{a}\in F$ and$ 0\cdot \sqrt{a}+1\cdot \sqrt{b}=\sqrt{b}\in F$. I think I have shown what is needed.
| You have a good start by showing that $a \sqrt{b} + b\sqrt{a} \in F$, but this does not show that $x\sqrt{a} + y\sqrt{b} \in F$ for an arbitrary choice of $x$ and $y$ ($a$ and $b$ are fixed constants).
What you want to do is add something to $a \sqrt{b} + b\sqrt{a}$ to cancel out one of the roots. Like maybe a multiple $x(\sqrt{a}+\sqrt{b})$ (here you can choose $x$ to be what you need).
| {
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"timestamp": "2023-03-29T00:00:00",
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Find the range of eccentricity of an ellipse such that the distance between its foci doesn't subtend any right angle on its circumference.
What is the range of eccentricity of ellipse such that its foci don't subtend any right angle on its circumference?
I thought that the eccentricity would definitely be more than $0$ and less than $\frac{1}{\sqrt2}$ The latter value is for an ellipse with $ae=b$, in which a right angle is subtended on an endpoint of the minor axis.
| Your answer is correct, except that $0$ should be included (there are no right angles subtended in a circle).
Here's a complete solution:
Using the parameterization $P=(a \cos\theta, b\sin\theta)$ for an origin-centered ellipse with major radius $a$ (in the $x$ direction) and minor radius $b$ (in the $y$ direction), consider the foci at points $F_{\pm}=(\pm c, 0)$, where $a^2 = b^2 + c^2$.
$\angle F_{+}PF_{-}$ will be a right angle if and only if
$$(F_{+}-P)\cdot(F_{-}-P) = 0 \tag{$\star$}$$
That is,
$$\begin{align}
0 &= (c - a \cos\theta )(-c-a\cos\theta) + (0 - b \sin\theta)(0-b\sin\theta) \\[4pt]
&= -c^2 + a^2 \cos^2\theta + b^2\sin^2\theta \\[4pt]
&= -c^2+a^2\cos^2\theta + ( a^2-c^2)(1-\cos^2\theta) \\[4pt]
&= a^2 - 2 c^2 + c^2 \cos^2\theta \tag{1}
\end{align}$$
Writing $c = ae$, where $e$ is the eccentricity, we can factor-out $a^2$ to get
$$e^2\cos^2\theta = 2 e^2 - 1 \tag{2}$$
In order for $(2)$ to be solvable for $\theta$, we obviously must have $e\neq 0$ (so that $\theta$ appears in the equation at all); then, for non-zero $e$, since $0\leq \cos^2\theta \leq 1$, the solvability of $(2)$ requires
$$
0 \leq 2-\frac{1}{e^2}\leq 1 \quad\to\quad
2 \geq \frac{1}{e^2}\geq 1 \quad\to\quad
\sqrt{\frac{1}{2}} \leq e \leq 1 \tag{3}$$
In other words, the equation is not solvable for $\theta$ ---that is, there are no subtended right angles--- for $e < {1\over\sqrt{2}}$ or $e > 1$ (although we dismiss the latter possibility, as such eccentricities belong to hyperbolas). Therefore, the desired range of eccentricities is
$$0 \leq e < \frac{1}{\sqrt{2}} \tag{$\star\star$}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2543194",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Proving that $a_n=\sqrt{n^2+n}-n$ converges, finding its limit and showing that its sequence ${(a_n)^{\infty}_{n=1}}$ is monotinic. As the title says, below I've proved the statement. Just posted here for verification and correction! And of course for other people to be inspired.
Let $a_n=\sqrt{n^2+n}-n$ with $n\in\mathbb{N}.$ Show that $a_n$ converges as $n\to\infty$ find its limit, and show that the sequence ${(a_n)^{\infty}_{n=1}}$ is monotinic.
Proof:
For convergence, we look for:
$\forall_{\epsilon\gt0}\exists_{n\in\mathbb{N}}\forall_{n\gt N}(|a_n-l|<\epsilon) \iff l-\epsilon<a_n<l+\epsilon.$ If this property holds for one and only one $l$, then $a_n$ convergence with its limit being unique.
$l-\epsilon<\sqrt{n^2+n}-n<l+\epsilon \iff (l+n-\epsilon)^2<n^2+n<(l+n+\epsilon)^2 \iff $
$n(n+l-\epsilon)+l(n+l-\epsilon)-\epsilon(n+l-\epsilon) <n^2+n< $
$n(n+l+\epsilon)+l(n+l+\epsilon)+\epsilon(n+l+\epsilon)$
$\epsilon$ can get randomly small, so it can be ommited for large $n$. Then;
$n(n+l)+l(n+l)<n^2+n< n(n+l)+l(n+l) \iff$
$n^2+2ln+l^2<n^2+n<n^2+2ln+l^2.$
As $n\to\infty$, all terms not containing $n$ can be omitted as they are unimpactfully small. Then;
$n^2+2ln<n^2+n<n^2+2ln \iff 2ln<n<2ln \iff n=2ln \iff l=\frac{1}{2}.$
This property only holds for one $l$, so $a_n$ is convergent, and with this, its limit $l=\frac{1}{2}$ has also been found.
For monotonity we look for:
$\forall_n(\sqrt{(n+1)^2+(n+1)}-(n+1)>\sqrt{n^2+n}-n)$.
For all $n$:
$1>0 \iff 4n^2+4n+1>4n^2+4n \iff (2n+1)^2>4(n^2+n) \iff$
$2n+1>2\sqrt{n^2+n} \iff n^2+2n+1>n^2+2\sqrt{n^2+n} \iff$
$(n+1)^2>n^2+2\sqrt{n^2+n} \iff (n+1)^2+n+1>n^2+n+2\sqrt{n^2+n}+1 \iff$
$\sqrt{(n+1)^2+n+1}>\sqrt{n^2+n}+1 \iff\sqrt{(n+1)^2+(n+1)}-(n+1)>\sqrt{n^2+n}-n$
So the series ${(a_n)^{\infty}_{n=1}}$ is monotonic.
Conclusion:
$a_n$ converges with its unique limit being $l=\frac{1}{2}$, and the series ${(a_n)^{\infty}_{n=1}}$ is monotinic. $\tag*{$\Box$}$
| $$a_n={n\over \sqrt{n^2+n}+n}= {1\over \sqrt{1+{1\over n}}+1}$$
so $$ 0<a_n\leq {1\over 2}$$
and since $\sqrt{1+{1\over n}}+1$ is decreasing so is $a_n$ increasing. So $a_n$ is convergent with limit ${1\over 2}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2543403",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Use Taylor’s Theorem to approximate $\sqrt{e}$ to within an error of magnitude at most $10^{-6}$ Could you please let me know if the following considered as a correct answer to this question.
Taylor Series
$f(x) = \sum_{k=0}^{n} \frac{f^{(k)}x_{0}}{k!}(x - x_{0})^{k}+ \frac{f^{(n+1)}(c)}{(n+1)!}(x - x_{0})^{(n+1)}$.
Take $f(x) = e^x$ and $x_{0} = 0$.
$f^{(0)}(x) = e^x \Rightarrow f^{(0)}(0) = 1$
$f^{(1)}(x) = e^x \Rightarrow f^{(1)}(0) = 1$
$f(x) = e^{x} = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + .. + \frac{e ^{c} x^{(n+1)}}{(n+1)!}$
Now, we take $x = \frac{1}{2}$
$e^{\frac{1}{2}} = 1 + \frac{1}{2} + \frac{1}{(2^2)2!} + \frac{1}{(2^3)3!} + .. + \frac{e ^{c}}{(2^{(n+1)})(n+1)!}$
For any n, the error term equals $E_{n} = \frac{e ^{c}}{(2^{(n+1)})(n+1)!}$ but $c$ is between $x$ and $x_{0}$, i.e., $0$ and $\frac{1}{2}$ which means $e^c \le e^{\frac{1}{2}} \le \sqrt{3}$
$E_{n} \le \frac{\sqrt{3}}{(2^{(n+1)})(n+1)!}$
For $n = 6$, $E_{6} \le \frac{\sqrt{3}}{46,080} = 3.76 \left( 10^{-5} \right)$. Then to have an error of magnitude at most $10^{-6}$, we should use $n = 7$?
| $$E_7 \leq \frac{\sqrt{3}}{2^8 \cdot 8!} \approx 1.67803161 \times 10^{-7} < 10^{-6}$$
$n=7$ suffices.
Note that Showing $n =6$ is not sufficient doesn't imply that $n=7 $ is sufficient, we still have to verify that it works.
Code to check that $n=7$ is indeed the smallest number of terms required.
| {
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"url": "https://math.stackexchange.com/questions/2551888",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Let $a^{4} + b^{4} + a^{2} b^{2} = 60 \ \ a,b \in \mathbb{R}$ Then prove that : Let $a^{4} + b^{4} + a^{2} b^{2} = 60 \ \ a,b \in \mathbb{R}$ Then prove that :
$$4a^{2} + 4b^{2} - ab \geq 30$$
My attempt: :
$$4a^{2} + 4b^{2} - ab \geq 30 \\ 4(a^2+b^2)-ab \geq30 \\4(60-a^2b^2)-ab\geq30\\ 240-30\geq4(ab)^2+ab\\ 4(ab)^2+ab \leq210$$
Now what ?
| We have $$[4(a^2+b^2)]^2=16(a^4+b^4+2a^2b^2) =16(60+a^2b^2)$$
Then, $$ [4(a^2+b^2)]^2 - (30 +ab)^2 =16(60+a^2b^2) - (30 +ab)^2 \\=15a^2b^2 -60 ab - 60 =\color{red}{15(ab-2)^2\ge0} $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2552804",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 2
} |
Finding: $\lim\limits_{x\to -\infty} \frac{6x^2+5\cos{\pi x}}{\sqrt{x^4+\sin{5\pi x}}}$
I'm running into problems with this limit:
$$\lim_{x\to -\infty} \frac{6x^2+5\cos{\pi x}}{\sqrt{x^4+\sin{5\pi x}}}$$
I've tried using l'Hospitals rule, however we will alway keep the $\cos(\pi x)$ expression, as well for $\sin(5\pi x)$. Also, terms do not cancel out with $\sin$ and $\pi$ since the product with $5$.
Can anyone give me a hint solving this?
Thanks in advance
Kind regards,
| Writing out like
$$\frac{6x^2+5\cos\pi x}{\sqrt{x^4+5\sin 5\pi x}} = \frac{x^2\left(6 + \frac{5\cos \pi x}{x^2}\right)}{x^2\sqrt{1 + \frac{5\sin 5\pi x}{x^4}}}$$
should do the trick.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2555213",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Calculate the derivative using limit definition. This is the function $f(x)$$=\frac{1}{\sqrt{3x-2}}$ .
I wrote that $$\lim_{h\to 0}\frac{\frac{\sqrt{3x+3h-2}}{3x+3h-2}-\frac{\sqrt{3x-2}}{3x-2}}{h}.$$
I am not able to continue further.
| Let $a = \sqrt{3x+h-2}$ and $b = \sqrt{3x-2}$, then $a^2 - b^2 = h$ which means $h \to 0$ as $a \to b$. So we've:
$$ \lim_{a \to b} \frac{1/a-1/b}{a^2-b^2} = \lim_{a \to b} \frac{b-a}{(ab)(a+b)(a-b)} = - \lim_{a \to b} \frac{1}{ab(a+b)} = -\frac{1}{2 b^3} = -\frac{1}{2\sqrt{(3x-2)^3}.} $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2556339",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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If $x,y,z\in {\mathbb R}$, Solve this system equation: If $x,y,z\in {\mathbb R}$, Solve this system equation:
$$
\left\lbrace\begin{array}{ccccccl}
x^4 & + & y^2 & + & 4 & = & 5yz
\\[1mm]
y^{4} & + & z^{2} & + & 4 & = &5zx
\\[1mm]
z^{4} & + & x^{2} & + & 4 & = & 5xy
\end{array}\right.
$$
This is an olympiad question in Turkey (not international), which that, I could not solve it.
My idea:
$xy=a \\ yz=b \\ xz=c$
$$
\left\lbrace\begin{array}{ccccccl}
a^2c^3 & + & b^3a & + & 4b^2c & = & 5b^3c
\\[1mm]
a^{3}b^2 & + & bc^3 & + & 4c^2a & = &5c^3a
\\[1mm]
b^{3}c^2 & + & a^3c & + & 4a^2b & = & 5a^3b
\end{array}\right.
$$
Yes, I know, this is a stupid idea, because it did not work at all ( last system equation is more difficult).
| Summing up all the equtions you find:
$$\sum_{cyc} x_i^4 + \sum_{cyc} x_i^2 - 5\sum_{cyc} x_ix_j +12= 0$$
Now manipulate in such way to have the equality as sum of squares!
EG
$$-5xy =\frac52 \left( x-y\right)^2-\frac52x^2-\frac52y^2$$
That is:
$$x^4+y^4+z^4-4x^2-4y^2-4z^2+\frac52 \left( x-y\right)^2+\frac52 \left(y-z\right)^2+\frac52 \left( z-x\right)^2+12=0$$
$$(x^2-2)^2+(y^2-2)^2+(z^2-2)^2+\frac52 \left( x-y\right)^2+\frac52 \left(y-z\right)^2+\frac52 \left( z-x\right)^2=0$$
The system has 2 different solutions:
$$x=y=z=\sqrt 2$$
and
$$x=y=z=-\sqrt 2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2556960",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 1
} |
Solution of the ordinary differential equation $y'(t)=-y^3+y^2+2y$ Consider the solution of the ordinary differential equation $y'(t)=-y^3+y^2+2y$ subject to $y(0)=y_0 \in (0,2). $ then $\lim \limits_{t \to \infty}y(t)$ belongs to
*
*{-1,0}
*{-1,2}
*{0,2}
*{0, $+\infty $}
My Attempt: $y'(t)=-y^3+y^2+2y \\ \Rightarrow y'(t)=-y(y^2-y-2) \\ \Rightarrow y'(t)=-y(y-2)(y+1) \\ \Rightarrow \frac{dy}{dt}=-y(y-2)(y+1) \\ \Rightarrow \frac{dy}{y(y-2)(y+1)}=-dt \\ \Rightarrow \int\frac{dy}{y(y-2)(y+1)}=\int-dt \\by \;\;partial \;fraction \; we \; get \\ \int [\frac{-1}{2y}+\frac{1}{6(y-2)}+\frac{1}{3(y+1)}]dy =-\int dt \\ \Rightarrow -\frac{1}{2} log|y| +\frac{1}{6} log|y-2| + \frac{1}{3} log|y+1|=-t + logc_1 \\ \Rightarrow \frac{-3log|y|+log|y-2|+2log|y+1|}{6}=-t + logc_1 \\ \Rightarrow log\frac{(y-2)(y+1)^2}{y^3}=-6t+6logc_1 \\ \Rightarrow log\frac{(y-2)(y+1)^2}{y^3}-logc_1^6=-6t \\ \Rightarrow log\frac{(y-2)(y+1)^2}{cy^3}=-6t \;\; where \;c_1^6=c \\ \Rightarrow \frac{(y-2)(y+1)^2}{cy^3}=e^{-6t}..........(1) \\ when \;\;t=0 \;\; \frac{(y_0-2)(y_0+1)^2}{y_0^3}=c \\ (1) \Rightarrow \frac{(y-2)(y+1)^2}{y^3}=\frac{(y_0-2)(y_0+1)^2}{y_0^3} e^{-6t}$
Now I am not getting how to apply the limit for this and how to find $\lim \limits_{t \to \infty} y(t)$
| Given that $$y'(t)=-y^3+y^2+2y=-y\left(y^2-y-2\right)=-y(y-2)(y+1)$$
Clearly at $~y=-1,~0,~2~$ we have $~y'=0~.$
So the direction field will be something like:
When $~y<-1~,$ slope of $~y~$ is increasing.
When $~-1<y<0~,$ slope of $~y~$ is decreasing.
When $~0<y<2~,$ slope of $~y~$ is increasing.
When $~y>2~,$ slope of $~y~$ is decreasing.
Given that $~y(0)=y_0\in(0,~2)~,$ which implies that $~\lim_{t\to\infty}y(t)=2~.$
Hence option $\bf{(2)}$ and option $\bf{(3)}$ are the only correct options.
Note: I know that there is an accepted answer of this old question. But One can check that this is an totally new idea. So I don't think the community should have any objection about it. Thank you.
| {
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"url": "https://math.stackexchange.com/questions/2563252",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove the inequality $\frac{b+c}{a(y+z)}+\frac{c+a}{b(z+x)}+\frac{a+b}{c(x+y)}\geq \frac{3(a+b+c)}{ax+by+cz}$
Suppose that $a,b,c,x,y,z$ are all positive real numbers. Show that
$$\frac{b+c}{a(y+z)}+\frac{c+a}{b(z+x)}+\frac{a+b}{c(x+y)}\geq \frac{3(a+b+c)}{ax+by+cz}$$
Below are what I've done, which may be misleading.
*
*I've tried to analyze when the equality holds：
1.1 Under the condition that $a=b=c$, it reduces to
$$\frac{2}{y+z}+\frac{2}{z+x}+\frac{2}{x+y}\geq \frac{9}{x+y+z}$$
1.2 Under the condition that $x=y=z$, it reduces to
$$\frac{b+c}{2a}+\frac{c+a}{2b}+\frac{a+b}{2c}\geq 3$$
Both are easy to verify. However, $ax+by+cz$ is not easy to deal with. Is there any famous inequality that I can use here?
1.3 Under the condition that $(x,y,z)$ and $(a,b,c)$ are in proportion, i.e. $x=at, y=bt, z=ct$ for some $t>0$, the inequality reduces to
$$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\geq \frac{3(a+b+c)}{a^2+b^2+c^2}$$
which can be proved by using Newton's inequality:
$$(ab+bc+ca)^2 \geq 3abc(a+b+c)$$
*I’ve also tried to construct a function
$$f(x,y,z)=\frac{b+c}{a(y+z)}+\frac{c+a}{b(z+x)}+\frac{a+b}{c(x+y)}- \frac{3(a+b+c)}{ax+by+cz}$$
and analyze its global minimum. But the first-order condition is complicated
$$\frac{3(a+b+c)}{(ax+by+cz)^2}=\frac{c+a}{ab(z+x)^2}+\frac{a+b}{ca(x+y)^2}$$
$$\frac{3(a+b+c)}{(ax+by+cz)^2}=\frac{b+c}{ab(y+z)^2}+\frac{a+b}{bc(x+y)^2}$$
$$\frac{3(a+b+c)}{(ax+by+cz)^2}=\frac{b+c}{ca(y+z)^2}+\frac{c+a}{bc(z+x)^2}$$
Maybe some convexity can be used here?
*Substitution has been considered. Let
$$u=\frac{a}{a+b+c},v=\frac{b}{a+b+c}, w=\frac{c}{a+b+c}$$
Then $u,v,w>0$ and $u+v+w=1$, which are just weights we assign to $x,y,z$. The inequality becomes
$$\frac{1}{u(y+z)}+\frac{1}{v(z+x)}+\frac{1}{w(x+y)}-\frac{3}{ux+vy+wz}\geq \frac{1}{y+z}+\frac{1}{z+x}+\frac{1}{x+y}$$
Define another function in variables $u,v,w$
$$g(u,v,w)=\frac{1}{u(y+z)}+\frac{1}{v(z+x)}+\frac{1}{w(x+y)}-\frac{3}{ux+vy+wz}$$
and consider the constrained optimization problem:
$$\min g(u,v,w)\\ \text{s.t. } u>0\\ v>0 \\w>0\\ u+v+w=1$$
The corresponding Lagrangian function can be
$$L_{\lambda}(u,v,w)=\frac{1}{u(y+z)}+\frac{1}{v(z+x)}+\frac{1}{w(x+y)}-\frac{3}{ux+vy+wz}+\lambda(u+v+w-1)$$
And the first-order conditions give
$$\frac{1}{u^2(y+z)}-\frac{3x}{(ux+vy+wz)^2}=\lambda\Rightarrow \frac{1}{u(y+z)}-\frac{3ux}{(ux+vy+wz)^2}=\lambda u\\ \frac{1}{v^2(z+x)}-\frac{3y}{(ux+vy+wz)^2}=\lambda\Rightarrow \frac{1}{v(z+x)}-\frac{3vy}{(ux+vy+wz)^2}=\lambda v\\ \frac{1}{w^2(x+y)}-\frac{3z}{(ux+vy+wz)^2}=\lambda\Rightarrow \frac{1}{w(x+y)}-\frac{3wz}{(ux+vy+wz)^2}=\lambda w\\ u+v+w=1$$
Summing up yields
$$\lambda=\frac{1}{u(y+z)}+\frac{1}{v(z+x)}+\frac{1}{w(x+y)}-\frac{3}{ux+vy+wz}$$
None of the methods I tried seems to work, and now I even doubt the truth of the inequality.
| Partial Proof:
If we start with your substitution at 3) and remark that we have with your condition :
$$ux+yv+zw=(x+y+z)(u+v+w)-u(y+z)-v(z+x)-w(x+y)=(x+y+z)-u(y+z)-v(z+x)-w(x+y)$$
So if we put :
$p=y+x$$\quad$$i=w(x+y)$
$q=z+x$$\quad$$k=v(z+x)$
$r=y+z$$\quad$$j=u(y+z)$
We get :
$$\frac{1}{i}+\frac{1}{j}+\frac{1}{k}+\frac{-3}{\frac{p+r+q}{2}-i-j-k}\geq \frac{1}{p}+\frac{1}{q}+\frac{1}{r} $$
With the condition :
$$\frac{i}{p}+\frac{k}{q}+\frac{j}{r}=1. $$
Now the idea is to use the relation between the incircle of center I and the side of a triangle ABC :
$$\frac{IA^2}{CA.AB}+\frac{IB^2}{BC.AB}+\frac{IC^2}{CA.BC}=1$$
So we put :
$p=CA.AB$$\quad$$i=IA^2$
$q=BC.AB$$\quad$$k=IB^2$
$r=CA.BC$$\quad$$j=IC^2$
We get :
$$\frac{1}{IA^2}+\frac{1}{IB^2}+\frac{1}{IC^2}+\frac{-3}{\frac{CA.AB+BC.AB+CA.BC}{2}-IA^2-IB^2-IC^2}\geq \frac{1}{CA.AB}+\frac{1}{BC.AB}+\frac{1}{CA.BC} $$
Furthermore we have the following relation :
$$\frac{1}{IA^2}+\frac{1}{IB^2}+\frac{1}{IC^2}=\frac{1}{r^2}-\frac{1}{2rR}$$
$$IA^2+IB^2+IC^2=s^2+r^2+8rR$$
$$CA.AB+BC.AB+CA.BC=s^2+(4R+r)r$$
$$\frac{1}{CA.AB}+\frac{1}{BC.AB}+\frac{1}{CA.BC}=\frac{1}{2rR} $$
Where $s$ denotes the semi-perimeter , $r$ the radius of the incircle and R the radius of the excircle
Finally we have :
$$\frac{1}{r^2}-\frac{1}{2rR}+\frac{-3}{-(s^2+r^2+8rR)+0.5(s^2+(4R+r)r)}\geq \frac{1}{2rR} $$
I finally found a theorem of Blundon wich characterizes the existence of a scalene triangle with a necessary and sufficient condition on $r$,$R$,$s$ .See here.But I don't know how to use it to have a condition on the inequality . Maybe somebody could do that .
| {
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"url": "https://math.stackexchange.com/questions/2564669",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "23",
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Quadratic equation with different indices My maths teacher gave me a worksheet to work through as I was getting slightly bored in lessons. However, there was one question which I cannot do. The worksheet gives the answer, but you are supposed to show how you did it. Here is the question:
$729 + 3^{2x+1} = 4\times3^{x+2}$
The sheet said that the answer was $x=2$ or $x=3$
Here is my working - I 'solved' by completing the square, as you will see:
\begin{align}
3^{2x+1}-4\times3^{x+2}+729& = 0 \\
3(3^{2x})-4\times(3^2)(3^x)+729& = 0\\
3(3^x)^2-36(3^x)+729& = 0\\
y &= 3^x\\
3y^2-36y+729 & = 0\\
3(y^2-12y+243) &= 0\\
(y-6)^2-36+243 &= 0\\
(y-6)^2+207 &= 0\\
(y-6)^2&=-207\\
y-6 &= \sqrt{207}i\\
y &= 6+\sqrt{207}i\\
3^x &= 6+\sqrt{207}i\\
x &= \log_3(6+\sqrt{207}i)\\
\end{align}
Clearly, this is not the answer as stated on the sheet. Have I answered this question wrong, or is the answer on the sheet wrong? Thank you in advance.
| Hint:
This seems to be a problem in arithmetic. Note that $729=3^6$, and rewrite the equation as
$$3^{2x+1}+3^6=4\cdot 3^{x+3}.$$
Some details
If $x$ is a natural number this implies $4$ divides the left-hand side. Depending on the values of $x$, factor out $3^6$ or $3^{2x+1}$. $4$ must divide the other factor. Now
*
*if $2x+1>6$, we factor out $3^6$, obtaining
$$3^6(3^{2x-5}+1)=4\cdot 3^{x+3},$$
so necessarily $\;x+3=6$ and $3^{2x-5}+1=4\iff 2x-5=1$. The first equality yields $\color{red}{x=3}$ and so does the second.
*if $2x+1<6$, factoring out $3^{2x+1}$, we get
$$3^{2x+1}(3^{5-2x}+1)=4\cdot 3^{x+3},$$
so $\,2x+1=x+3\,$ and $\,5-2x=1$. Both equations yield $\,\color{red}{x=2}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2565365",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Sequence : $a_{n+1}=2a_n-a_{n-1}+2$ Let $c$ be a positive integer. The sequence $a_1, a_2, \ldots$ is defined by $a_1=1, a_2=c$ and
$a_{n+1}=2a_n-a_{n-1}+2$ for all $n \geq 2$.
Prove that for each $n \in \mathbb{N}$ there exists $k \in \mathbb{N}$ such that $a_na_{n+1} = a_k$.
My attempt :
Trying with small numbers, I see that $a_n=(n-1)c+(n-2)^2$ and will prove by induction.
$a_1 = 1 , a_2 = c+0 = c$ is true.
Suppose that $a_k=(k-1)c+(k-2)^2$ and $a_{k+1}=kc+(k-1)^2$ are true.
$a_{k+2} = 2a_{k+1}+a_k+2 = 2[kc+(k-1)^2]-[(k-1)c+(k-2)^2]+2=(k+1)c+k^2$
so $a_n=(n-1)c+(n-2)^2$ is true.
$a_na_{n+1} = [(n-1)c+(n-2)^2][nc+(n-1)^2] $
$= n(n-1)c^2 + (n-1)^3c + n(n-2)^2c + (n^2-3n+2)^2 = (k-1)c+(k-2)^2$
Please suggest how to solve this equation.
| For $n=1$, $a_1a_2=c=a_2$.
In the following, $n\ge 2$.
You already have $a_n=(n-1)c+(n-2)^2$.
Then,
$$\begin{align}&a_na_{n+1}=a_k\\\\&\iff ((n-1)c+(n-2)^2)(nc+(n-1)^2)=(k-1)c+(k-2)^2\\\\&\iff k^2+(c-4)k-(c^2n^2-c^2n+2cn^3-7cn^2+7cn+n^4-6n^3+13n^2-12n)=0\\\\&\iff k=\frac{-c+4\pm\sqrt{\Delta}}{2}\end{align}$$
where
$$\begin{align}\Delta&=(c-4)^2+4(c^2n^2-c^2n+2cn^3-7cn^2+7cn+n^4-6n^3+13n^2-12n)\\\\&=4n^4+(8c-24)n^3+(4c^2-28c+52)n^2+(-4c^2+28c-48)n+(c-4)^2\\\\&=(2n^2+(2c-6)n-c+4)^2\end{align}$$
It follows from this that
$$a_na_{n+1}=a_k\iff k=n(n-3)+c(n-1)+4,-n(n+c-3)$$
Since the latter is non-positive for $n\ge 2$,
$$a_na_{n+1}=a_k\iff k=n(n-3)+c(n-1)+4$$
where $n(n-3)+c(n-1)+4$ is a positive integer.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2568850",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
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Transformation matrix - Are the matrices correct? I want to describe for the following $\mathbb{R}$-vector spaces $V$with basis $B$ the matrix $A_{f, B, B}$ for $f\in \text{End}_K(V)$:
*
*$V=\mathbb{R}^2, B=((1,0)^T, (0,1)^T)$ and $f$ the rotation by 45 degrees clockwise.
*$V=\mathbb{C}, B=(1,i)$ and $f(z)=a^2\overline{z}$ with $a=\cos \alpha+i\sin\alpha, \alpha \in [0,2\pi)$.
$$$$
The matrix $A_{f, B, B}$ contains the coefficients $a_{ij}$ of $f(b_j)=\sum a_{ij}b_i$, right?
We have the following:
*
*We have that $f(x)=\begin{pmatrix}\cos\frac{\pi}{4} & -\sin\frac{\pi}{4} \\ \sin\frac{\pi}{4} & \cos\frac{\pi}{4}\end{pmatrix}\cdot x=\frac{1}{\sqrt{2}}\begin{pmatrix}1 & -1 \\ 1 & 1\end{pmatrix}\cdot x$, right?
Therefore we have the following:
$$f(b_1)=f\begin{pmatrix}1\\0\end{pmatrix}=\frac{1}{\sqrt{2}}\begin{pmatrix}1 & -1 \\ 1 & 1\end{pmatrix}\cdot\begin{pmatrix}1\\0\end{pmatrix}=\frac{1}{\sqrt{2}}\begin{pmatrix}1 \\ 1 \end{pmatrix}=\frac{1}{\sqrt{2}}\begin{pmatrix}1 \\ 0 \end{pmatrix}+\frac{1}{\sqrt{2}}\begin{pmatrix}0 \\ 1 \end{pmatrix}=\frac{1}{\sqrt{2}}\cdot b_1+\frac{1}{\sqrt{2}}\cdot b_2 \\ f(b_2)=f\begin{pmatrix}1\\0\end{pmatrix}=\frac{1}{\sqrt{2}}\begin{pmatrix}1 & -1 \\ 1 & 1\end{pmatrix}\cdot\begin{pmatrix}0\\1\end{pmatrix}=\frac{1}{\sqrt{2}}\begin{pmatrix}-1 \\ 1 \end{pmatrix}=-\frac{1}{\sqrt{2}}\begin{pmatrix}1 \\ 0 \end{pmatrix}+\frac{1}{\sqrt{2}}\begin{pmatrix}0 \\ 1 \end{pmatrix}=-\frac{1}{\sqrt{2}}\cdot b_1+\frac{1}{\sqrt{2}}\cdot b_2$$
So, do we get $$A_{f, B, B}=\begin{pmatrix}\frac{1}{\sqrt{2}}& -\frac{1}{\sqrt{2}}\\ \frac{1}{\sqrt{2}}&\frac{1}{\sqrt{2}}\end{pmatrix}$$ ?
*We have the following: $$f(1)=a^2=(\cos\alpha+i\sin\alpha)^2=\cos^2\alpha-\sin^2\alpha-2i\cos\alpha\sin\alpha= (\cos^2\alpha-\sin^2\alpha)\cdot 1+(-2\cos\alpha\sin\alpha) \cdot i\\ f(i)=-a^2=-(\cos\alpha+i\sin\alpha)^2=-\cos^2\alpha+\sin^2\alpha+2i\cos\alpha\sin\alpha=-(\cos^2\alpha-\sin^2\alpha)\cdot 1+(2\cos\alpha\sin\alpha) \cdot i$$
So, do we get $$A_{f, B, B}=\begin{pmatrix}\cos^2\alpha-\sin^2\alpha& -(\cos^2\alpha-\sin^2\alpha)\\ -2\cos\alpha\sin\alpha&2\cos\alpha\sin\alpha\end{pmatrix}$$ ?
How could we justify that $f$ is a reflection at the line $\{at: t\in \mathbb{R}\}$ ?
| For the first, note that rotation matrix by $\theta$ clockwise is:
$$f(x)=\begin{pmatrix}\cos\ \theta & \sin\theta \\ -\sin\theta & \cos\theta \end{pmatrix}$$
Thus:
$$f(x)=\begin{pmatrix}\cos\frac{\pi}{4} & \sin\frac{\pi}{4} \\ -\sin\frac{\pi}{4} & \cos\frac{\pi}{4}\end{pmatrix}\cdot x=\frac{1}{\sqrt{2}}\begin{pmatrix}1 & 1 \\ -1 & 1\end{pmatrix}\cdot x$$
For the second we have:
$$a=\cos \alpha+i\sin\alpha\implies a^2= \cos^2 \alpha -\sin^2\alpha+2i\cos \alpha\sin \alpha $$
$$z=1\implies a^2 \bar z= a^2= \cos^2 \alpha -\sin^2\alpha+2i(\cos \alpha\sin \alpha)$$
$$z=i\implies a^2 \bar z= -ia^2= -i(\cos^2 \alpha -\sin^2\alpha)+2(\cos \alpha\sin \alpha) $$
Thus:
$$A_{f, B, B}=\begin{pmatrix}\cos^2\alpha-\sin^2\alpha&2(\cos\alpha\sin\alpha)\\ 2(\cos\alpha\sin\alpha)&-(\cos^2\alpha-\sin^2\alpha)\end{pmatrix}$$
To justify that $f$ is a reflection at the line $\{at: t\in \mathbb{R}\}$ note that:
$z\to f(z)=a^2 \bar z\to f(a^2 \bar z)=a^2 \bar a^2 z=z$
We can also verify this by matrix $A_{f, B, B}$, indeed:
$$A_{f, B, B}\cdot a=
\cos^3\alpha-\sin^2\alpha \cos\alpha+2(\cos\alpha\sin^2\alpha)+2i(\cos^2\alpha\sin\alpha)-i(\cos^2\alpha \sin \alpha-\sin^3\alpha)=
\cos^3\alpha+\sin^2\alpha\cos\alpha+i(\cos^2\alpha\sin\alpha+\sin^3\alpha)=\cos\alpha+i\sin\alpha=a$$
| {
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Finding the maximum of $p^3 + q^3 +r^3 + 4pqr$
$p$,$q$,$r$ are $3$ non-negative real numbers less than or equal to $1.5$ such that $p+q+r = 3$, what will be the maximum of $p^3 + q^3 + r^3 + 4pqr$ ?
I tried AM-GM on $p,q,r$ to get the maximum of $pqr$ as $1$, but on doing it for $p^3 + q^3 + r^3 $, I get the minimum, so I won't be able to combine them.
I tried to assume symmetry $p=q=r=1$, and the maximum value to be 7, but putting $p=1.1$, $q=1.1$ and $r=0.8$, I get the value to be $7.046 (> 7) $
How do I solve this with concepts like AM-GM?
PS: I do not know multivariable calculus concepts, however if it's not possible to solve it without using it, please show how to solve it with these concepts.
| Let $p+q+r=3u$, $pq+pr+qr=3v^2$ and $pqr=w^3$.
Hence, the condition does not depend on $w^3$ and we need to find a maximal value of $f,$ where
$$f(w^3)=27u^3-27uv^2+7w^3.$$
We see that $f$ increases,
which says that it's enough to solve our problem for the maximal value of $w^3$.
Now, $p$, $q$ and $r$ are three non-negative roots of the following equation.
$$(x-p)(x-q)(x-r)=0$$ or
$$x^3-3ux^2+3v^2x-w^3=0$$ or
$$x^3-3ux^2+3v^2x=w^3.$$
Let $u$ and $v^2$ will be constants and $w^3$ is changing.
Since the line $y=w^3$ and the graph of $y=x^3-3ux^2+3v^2x$ have three common points,
we see that $w^3$ will get a maximal value,
when a line $y=w^3$ is a tangent line to the graph of $y=x^3-3ux^2+3v^2x$,
which happens for equality case of two variables.
Also, we need to check the case, when one of our variables is equal to $\frac{3}{2}.$
*
*Two variables are equal.
Since our inequality is symmetric, we can assume $b=a$.
Thus, $c=3-2a\leq\frac{3}{2},$ which gives $\frac{3}{4}\leq a\leq\frac{3}{2}$ and we need to find a maximum of $g$, where
$$g(a)=2a^3+(3-2a)^3+4a^2(3-2a)$$ and easy to see that
$$\max_{\frac{3}{4}\leq a\leq \frac{3}{2}}g=g\left(\frac{3}{4}\right)=\frac{243}{32}.$$
2. One of variables is equal to $\frac{3}{2}.$
Let $c=\frac{3}{2}.$
Hence, $b=\frac{3}{2}-a$ and we need to find a maximal value of $h$, where
$$h(a)=a^3+\left(\frac{3}{2}-a\right)^3+\frac{27}{8}+3a(3-2a)$$ and easy to see that $$\max_{0\leq a\leq\frac{3}{2}}h=h\left(\frac{3}{4}\right)=\frac{243}{32},$$
which gives the answer: $\frac{243}{32}.$
| {
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"timestamp": "2023-03-29T00:00:00",
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Simplify $\frac{7^{ \log_{5} 15 }+3^{2+\log_{5}7}}{7^{\log_{5}3}}$ I know that the result of this expression is 16 but how do I get to that result?
$$\frac{7^{ \log_{5} 15 }+3^{2+\log_{5}7}}{7^{\log_{5}3}}$$
| $$\frac{7^{ \log_{5} 15 }+3^{2+\log_{5}7}}{7^{\log_{5}3}}=\frac{7^{ 1+\log_53 }+7^{\left(2+\log_{5}7\right)\log_73}}{7^{\log_{5}3}}=7+\frac{7^{\log_79+\log_53}}{7^{\log_53}}=7+9=16.$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How to prove $\{n!e\} = \frac{1}{n +1} + \frac{1}{(n+1)(n+2)} + \frac{1}{(n+1)(n+2)(n+3)} + \dots$ I have the definition $a_n = \frac{1}{n +1} + \frac{1}{(n+1)(n+2)} + \frac{1}{(n+1)(n+2)(n+3)} + \dots$
I need to show that:
$a)$ $0 < a_n < \frac{1}{n}$
$b)$ $a_n = n!e - \lfloor{n!e}\rfloor$
So I know that
$\frac{1}{n} = \frac{1}{n+1}+\frac{1}{(n+1)^2}+\frac{1}{(n+1)^3}+\dots$
$e = \sum_{n=0}^{+\infty}\frac{1}{n!}$
It's clear that $\{n!e\}=a_n<\frac{1}{n}$, but how to show it?
| Because $$n!e=n!\left(2+\frac{1}{2!}+...+\frac{1}{n!}\right)+\frac{1}{n+1}+\frac{1}{(n+1)(n+2)}+...$$ and
$$\frac{1}{n+1}+\frac{1}{(n+1)(n+2)}+...<\frac{1}{n+1}+\frac{1}{(n+1)^2}+...=\frac{\frac{1}{n+1}}{1-\frac{1}{n+1}}=\frac{1}{n}<1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2573192",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How I do evaluate this polynomial problem? Let $f(x)=x^2+ax+b$.
Suppose $f(f(x))=0$ equation has $4$ different real solutions $x_1,x_2,x_3,x_4$
and that two of them sum up to $-1$ (i.e. $\exists\, i\neq j$ such that $x_i+x_j=-1$)
Prove that $b\lt-\frac14$
| Let $y = f(x)$ then we have the following system
$$ \begin{cases} x^2 + ax + b = y & (1) \\ y^2 + ay + b = 0 & (2) \end{cases} $$
In order to have 4 distinct real solutions in $x$, we require two distinct real solutions in $y$, therefore $a^2-4b > 0$. Denote the two solutions as $y_1$ and $y_2$, then
$$ \begin{cases} x^2 + ax + b = y_1 & (3) \\ x^2 + ax + b = y_2 & (4) \end{cases} $$
We now need two dinstinct solutions for each value of $y$, therefore another necessary condition is $a^2 - 4(b-y_i) > 0$ or $a^2 - 4b + 4y_i > 0$, where $i=1,2$. Then $$ 2(a^2-4b) + 4(y_1+y_2) = 2(a^2-4b) - 4a > 0 $$
or $$4b < a^2 -2a \tag{5}$$
Denote $x_1,x_2$ as solutions to $(3)$ and $x_3,x_4$ as solutions to $(4)$. Then it follows that $$x_1 + x_2 = x_3 + x_4 = -a \tag{6} $$
If $a=1$, then $(6)$ satisfies the condition and $(5)$ gives $b < -\frac{1}{4}$.
If $a\ne 1$, it must be true that one root from $(3)$ and one root from $(4)$ add up to $-1$. Suppose $x_1 + x_3 = -1$, then adding $(3)$ and $(4)$ gives
$$ {x_1}^2 + {x_3}^2 + a(x_1+x_3) + 2b = y_1 + y_2 $$
$$ {x_1}^2 + {x_3}^2 = - 2b $$
Then
$$ 2x_1x_3 = (x_1+x_3)^2 - ({x_1}^2 + {x_3}^2) = 1 + 2b $$
It follows that $x_1$ and $x_3$ must be roots of the equation
$$ x^2 + x + \frac{1+2b}{2} = 0 $$
Since $x_1 \ne x_3$, we require
$$ 1 - 2(1+2b) > 0 $$
or $b < -\frac{1}{4}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2574097",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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$(x+\sqrt{x^2+1})(y+\sqrt{y^2+1})=1$ is true if and only if $x+y=0$ Prove that $(x+\sqrt{x^2+1})(y+\sqrt{y^2+1})=1$ is true if and only if $x+y=0$
I believe x and y could both be 0 as that satisfies the equations. Beyond that, I do not know how to prove this.
| You have $$(x+\sqrt{x^2+1})(y+\sqrt{y^2+1}) = 1$$
$$x+\sqrt{x^2+1} = \sqrt{y^2+1} - y$$
$$x+y = -\sqrt{x^2+1} + \sqrt{y^2+1}$$
$$(x+y)(\sqrt{x^2+1} + \sqrt{y^2+1}) = -x^2+y^2$$
$$(x+y)(\sqrt{x^2+1} + \sqrt{y^2+1} +x-y) = 0$$
Now, you get $x+y=0$ or $\sqrt{x^2+1} + \sqrt{y^2+1} +x-y = 0$. Can you go further?
| {
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More of this type $T_n^2+T_{n+1}^2+T_{n+2}^2=X^2?$ Given triangular numbers,
$$T_n:= {n(n+1)\over 2} = 1,3,6,10,15,...$$
Let $T_n, T_{n+1}$ and $T_{n+2}$ be three consecutive triangular numbers.
My question is: Are there more of $T_n^2+T_{n+1}^2+T_{n+2}^2$ that only yield a square number? Or this is $$T_3^2+T_4^2+T_5^2=6^2+10^2+15^2=19^2$$ the only one? I can't seem to find anymore.
| I've checked this for all $X$ up to $10^{10^5}$, here's how.
We are searching for integer solutions to $T_{n-1}^2 + T_n^2 + T_{n+1}^2 = X^2$ (note I've reparametrised, compared to the OP), which expands to
$$3 n^4 + 6 n^3 + 15 n^2 + 12 n + 4 =4X^2. $$
Curiously enough, this is an invertible expression, and we can instead say
$$n = \frac{1}{6} \left(\sqrt{24 \sqrt{3X^2 + 6} - 63}- 3\right)$$
assuming $n>0$.
For $n$ to be an integer, $\sqrt{24 \sqrt{3X^2 + 6} - 63}$ must be an integer, so $24 \sqrt{3X^2 + 6} - 63$ must be a square. Furthermore, it must be an integer, hence $ \sqrt{3X^2 + 6}$ must be an integer. (Note that $3X^2 + 6$ is an integer, so $ \sqrt{3X^2 + 6}$ is either an integer or irrational - being $\frac{1}{24}$ of an integer cannot work.) So, it suffices to search for integer solutions $(X,Y)$ to the equation $Y^2 - 3X^2 = 6$ such that $24Y-63$ is a square.
But, these are easy to generate, with some knowledge of Pell equations. I'll switch to talking about $x^2 - 3y^2 = 6$ for now, for consistency with literature. First observe that the equation $a^2 - 3b^2 = 1$ has a solution given by $(a,b) = (2, 1)$, and that this is the positive non-trivial solution with smallest $b$. By Theorem 5.3 (and Example 5.6) in Keith Conrad's first blurb on Pell equations, all solutions to this equation are generated by this one. Consulting Keith Conrad's second blurb on Pell equations, we can see that every other solution is a Pell multiple of this one. Considering the original equation $x^2 - 3y^2 = 6$ now, Theorem 3.3 in the second blurb tells us that every solution of this is given by a Pell multiple of a solution $(x_1, y_1)$ where $|x_1| \leq 22$ and $|y_1| \leq 7$. A quick computer check gives only $(3, 1)$ and $(9, 5)$ as the positive solutions here, and $(9, 5)$ is a Pell multiple of $(3, 1)$.
Hence, all solutions of $x^2 - 3y^2 = 6$ are Pell multiples of $(3, 1)$, where the next Pell multiple of pair $(x, y)$ is given by $(2x + 3y, x + 2y)$, so we can easily generate all the positive integer solutions in Haskell! (using the arithmoi library)
import Math.NumberTheory.Powers.Squares
recurse :: (Integer, Integer) -> (Integer, Integer)
recurse (a, b) = (2*a + 3*b, a + 2*b)
vals :: [(Integer, Integer)]
vals = iterate recurse (3, 1)
works :: (Integer, Integer) -> Bool
works (n, _) = isSquare' (24 * n - 63)
solutions :: Integer -> [(Integer, Integer)]
solutions limit = filter works bounded
where bounded = takeWhile (\(y, x) -> x < limit) vals
vals is a list of the solutions to the Pell equation, generated using the recurrence explained above. A pair $(Y, X) = (x, y)$ works if $24Y - 63$ is a square, checked using the arithmoi package for number theory. Finally, solutions limit checks every possible $(Y, X)$ that satisfies X < limit to see if it works or not.
solutions (10^(10^5)) gives [(3, 1), (33, 19)] and the pair $(3, 1)$ corresponds to the trivial solution $T_{-1}^2 + T_0^2 + T_1^2 = 1$, while $(33, 19)$ corresponds to the solution in the OP. In particular, $24 \times 33 - 63 = 729$, so $n=4$, and $T_3^2 + T_4^2 + T_5^2 = 19$.
My computer checks this in under two minutes, but fails for $10^{10^6}$, seemingly due to memory errors. Hopefully someone can grow this into a full proof using cleverer ideas about Pell equations, or check larger values.
EDIT: Now checked up to $10^{2\times {10^5}}$.
| {
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Is this a valid method of solving $3^x + 2^x = 35$? I want to know if this is a valid method of solving this equation:
$3^x + 2^x = 35$
$3^x+2^x = (7)(5)$
$3^x+2^x = (3+2^2)(2+3)$
$3^x+2^x = 3^2 + 2^3 + 18$
$3^x+2^x = 9 + 18 + 2^3 $
$3^x+2^x = 27 + 2^3 $
$3^x+2^x = 3^3 + 2^3 $
And now comes my problem. Is it correct to say that this last equation implies $x=3$? I don't think I can just compare terms just like it was a polynomial...
If this is not correct, is there a way of solving this equation algebraically?
| $3$ it's an unique root because $f(x)=3^x+2^x$ is an increasing function and $f(3)=35$.
| {
"language": "en",
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using a base 3 decimal to express as a base 10 fraction using geometric series
Express $0.\overline{21}_3$ as a base 10 fraction in reduced form.
So I was able to solve it by setting $x=\overline{.21}$, but the solution also briefly mentioned another way using the geometric series:
A quick way to get the answer by using the geometric series is: $(0.212121 \ldots)_3 = \frac{7}{9} + \frac{7}{81} + \frac{7}{729} + \dots = \frac{7}{8}.$
However, I'm having a hard time understanding how to actually use the geometric series (the above answer is not clear to me).
| You have
$(0.212121 \ldots)_3 = \frac{7}{9} + \frac{7}{81} + \frac{7}{729} + \dots$
and for $0<a<1$
$1+a+a^2+\ldots = \dfrac{1}{1-a}$
from which
$\begin {align}
(0.212121 \ldots)_3 &= 7\left(\frac{1}{9} + \frac{1}{81} + \frac{1}{729} + \dots
\right)\\
&= 7\left(\frac{1}{1-\frac{1}{9}} -1 \right)\\
&= 7\left(\frac{9}{8} -1 \right)\\
&= \frac{7}{8} \\
\end{align}$
You can also (perhaps this is what you actually did) multiply by $100_3 = 9_{\text{dec}}$ to see that for $x=0.\overline{21}_3$, $9x-x = 21_3 = 7_{\text{dec}}$ and solve from there.
| {
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Evaluate $\int \frac{a^2\sin^2 x+b^2\cos^2 x}{a^4\sin^2 x+b^4\cos^2 x}dx$ Evaluate
$$\int \frac{a^2\sin^2 x+b^2\cos^2 x}{a^4\sin^2 x+b^4\cos^2 x}\text dx$$
I would have given my attempt to this question but honestly, I think my attempts to solve this did nothing but only complicated it further.
Any hints or suggestions are welcome.Please verify that your method gives answer:$$\frac 1{a^2+b^2}(x+\tan^{-1} ({\frac {a^2\tan x}{b^2}}))+C$$
| Hint:
We have $$I = \int \frac{a^2\sin^2 x + b^2\cos^2 x}{a^4\sin^2 x + b^4\cos^2 x}\, dx = \int \frac{[a^2-b^2]\sin^2 x + b^2}{[a^4-b^4]\sin^2 x + b^4}\, dx $$ $$= (b^2-\frac{(a^2-b^2)b^4}{a^4-b^4})\int \frac{1}{(a^4-b^4)\sin^2 x + b^4}\, dx + \int \frac{a^2-b^2}{a^4-b^4} \, dx$$
The first integral can be easily calculated by using a substitution $u = \tan x$.
| {
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Why is my alternate method of calculating scalar products not working? The exercise is such: Given that $|\vec{a}| = 3$, $|\vec{b}| = 2$ and $\varphi = 60^{\circ}$ (the angle between vectors $\vec{a}$ and $\vec{b}$), calcluate scalar product $(\vec{a}+2\vec{b}) \cdot (2\vec{a} - \vec{b})$.
My initial thought was to solve the requested product by "sticking" separately calculated fragments together and then follow the definition of scalar product, which is
$$\vec{a} \cdot \vec{b} = |\vec{a}| \cdot |\vec{b}| \cdot \cos\varphi.$$
Via observation
$$\vec{a} \cdot \vec{a} = |\vec{a}| \cdot |\vec{a}| \cdot \cos0^{\circ} = |\vec{a}|^2 \Longrightarrow |\vec{a}| = \sqrt{\vec{a} \cdot \vec{a}}$$
we can get values of $|\vec{a}+2\vec{b}|$ and $|2\vec{a} - \vec{b}|$ (mind the notation: $a^2$ in this case represents $a^2 = |\vec{a}|^2$, and $b^2 = |\vec{b}|^2$):
\begin{align*}
|\vec{a}+2\vec{b}| &= \sqrt{(\vec{a}+2\vec{b}) \cdot (\vec{a}+2\vec{b})} =\sqrt{a^2 + 4|\vec{a}| \cdot |\vec{b}| \cdot \cos 60^{\circ} + b^2} = \\
&= \sqrt{3^2 + 4 \cdot 3 \cdot 2 \cdot \frac{1}{2}+2^2} = 5.
\end{align*}
Similarly,
\begin{align*}
|2\vec{a} - \vec{b}| &= \sqrt{(2\vec{a} - \vec{b}) \cdot (2\vec{a} - \vec{b})} = \sqrt{4a^2 - 4|\vec{a}| \cdot |\vec{b}| \cdot \cos 60^{\circ} + b^2} = \\
&=\sqrt{4 \cdot 3^2 - 4 \cdot 3 \cdot 2 \cdot \frac{1}{2} + 2^2} = 2\sqrt{7}.
\end{align*}
Now we plug both results in:
$$(\vec{a} + 2\vec{b}) \cdot (2\vec{a} - \vec{b}) = |\vec{a} + 2\vec{b}| \cdot |2\vec{a} - \vec{b}| \cdot \cos60^{\circ} = 5 \cdot 2\sqrt{7} \cdot \frac{1}{2} = 5\sqrt{7}.$$
However, this is not the right solution according to my textbook. The correct result is $19$. I thought about it for a bit and took a different route:
\begin{align*}
(\vec{a} + 2\vec{b}) \cdot (2\vec{a} - \vec{b}) &= \vec{a} \cdot 2\vec{a} - \vec{a} \cdot \vec{b} + 4 \vec{a} \cdot \vec{b} - 2\vec{b} \cdot \vec{b} = \\
&=2a^2 - |\vec{a}| \cdot |\vec{b}| \cdot \cos 60^{\circ} + 4 |\vec{a}| \cdot |\vec{b}| \cdot \cos 60^{\circ} - 2b^2 = \\
&= 2 \cdot 3^2 - 3 \cdot 2 \cdot \frac{1}{2} + 4 \cdot 3 \cdot 2 \cdot \frac{1}{2} - 2 \cdot 2^2 = 19.
\end{align*}
The second method clearly worked, while the first one failed miserably. But my question is why did my first approach fail? Did I get the wrong results when calculating $|\vec{a}+2\vec{b}|$ and $|2\vec{a} - \vec{b}|$? I have no idea. Please, help me understand my mistakes. Thank you in advance.
| I think it is failed because here
$(\vec{a} + 2\vec{b}) \cdot (2\vec{a} - \vec{b}) = |\vec{a} + 2\vec{b}| \cdot |2\vec{a} - \vec{b}| \cdot \cos60^{\circ} = 5 \cdot 2\sqrt{7} \cdot \frac{1}{2} = 5\sqrt{7}$
you are assuming that the angle between vectors $(\vec{a} + 2\vec{b})$ and $(2\vec{a} - \vec{b})$ is $60^\circ$ but it may not be (and apparently it is not).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2578792",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 2
} |
$\lim_{n\to \infty}[\frac{1}{a_{1}a_{2}}+\frac{1}{a_{2}a_{3}}\cdot\cdot\cdot\frac{1}{a_{n-1}a_{n}}].$ Let $a_{1}=1$ and $a_{n}=a_{n-1}+4$ for $n\geq 2.$ I have to find $\lim_{n\to \infty}[\frac{1}{a_{1}a_{2}}+\frac{1}{a_{2}a_{3}}\cdot\cdot\cdot\frac{1}{a_{n}a_{n-1}}].$ I tried it as
$\lim_{n\to \infty}[\frac{1}{a_{1}a_{2}}+\frac{1}{a_{2}a_{3}}\cdot\cdot\cdot\frac{1}{a_{n}a_{n}}]=\lim_{n\to \infty}[\frac{1}{a_{1}[a_{1}+4]}+\frac{1}{a_{2}[a_{2}+4]}\cdot\cdot\cdot\frac{1}{a_{n-1}[a_{n-1}+4]}]=\frac{1}{4}\lim_{n\to\infty}[\sum \frac{1}{a_{n}}-\sum \frac{1}{a_{n}+4}].$
Now please help me to solve the problem. Thanks .
| \begin{eqnarray}E(n)=\frac{1}{a_{1}a_{2}}+\frac{1}{a_{2}a_{3}}\cdot\cdot\cdot\frac{1}{a_{n}a_{n-1}}& =&{1\over 4}\Big(\frac{a_2-a_1}{a_1a_2}+\frac{a_3-a_2}{a_{2}a_{3}}\cdot\cdot\cdot\frac{a_{n+1-a_n}}{a_{n}a_{n-1}}\Big)\\
& =&{1\over 4}\Big(\frac{1}{a_1}-\frac{1}{a_2}+\frac{1}{a_2}-\frac{1}{a_3}\cdot\cdot\cdot\frac{1}{a_n}-\frac{1}{a_{n+1}}\Big)\\
& =&{1\over 4}\Big(\frac{1}{a_1}-\frac{1}{a_{n+1}}\Big)\\
\end{eqnarray}
since $a_n\to \infty$ we have $$\lim_{n\to \infty} E(n) = {1\over 4a_1} ={1\over 4}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2581602",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Why does multiplying by $\textbf{A}^T$ make a previously unsolvable linear system solvable Consider for instance the linear system:
$$\left(
\begin{array}{cc}
1 & 2 \\
3 & 4 \\
5 & 6 \\
\end{array}
\right).\left(
\begin{array}{c}
x \\
y \\
\end{array}
\right)=\left(
\begin{array}{c}
1 \\
2 \\
4 \\
\end{array}
\right)$$
This is over determined and thus has no solution. Yet, by simply multiplying both sides by $\textbf{A}^T$:
$$\left(
\begin{array}{ccc}
1 & 3 & 5 \\
2 & 4 & 6 \\
\end{array}
\right).\left(
\begin{array}{cc}
1 & 2 \\
3 & 4 \\
5 & 6 \\
\end{array}
\right).\left(
\begin{array}{c}
x \\
y \\
\end{array}
\right)=\left(
\begin{array}{ccc}
1 & 3 & 5 \\
2 & 4 & 6 \\
\end{array}
\right).\left(
\begin{array}{c}
1 \\
2 \\
4 \\
\end{array}
\right)$$
We find that the system now has a unique solution, which is the (x,y) that minimizes the squared error.
Now I understand the derivation of why multiplying by the transpose helps to find the pseudoinverse which then helps to perform OLS regression, but my question is perhaps a bit more fundamental.
How can multiplying both sides of an equation by a matrix change a system which previously had no solutions into one that has a unique solution? This seems to against what I assumed that the solutions to $\textbf{A}x = \textbf{B}$ were the same as the solutions to $\textbf{P}\textbf{A}x = \textbf{P}\textbf{B}$.
| The solutions to $Ax=b$ are the same as those of $PAx=Pb$ if $P$ is one-to-one, i.e. $\ker(P) = \{0\}$. If $P$ is not one-to-one, so that $P y = 0$ for some $y \ne 0$, then any $x$ such that $Ax = b + y$ is a solution of $PAx = Pb$ but not a solution of $Ax = b$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2583454",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
"answer_id": 0
} |
How can I prove that $\prod\limits_{n=1}^{\infty}\left(1-\frac{1}{x_{n}^{3/2}}\right)^{-1}=\frac{3}{2}$ if $x_{n}=x_{n-1}+\sqrt{x_{n-1}}+1, x_{0}=1$? If
$$x_{n}=x_{n-1}+\sqrt{x_{n-1}}+1, x_{0}=1$$
so we can say, that
$$\prod\limits_{n=1}^{\infty}\left(1-\frac{1}{x_{n}^{3/2}}\right)^{-1}=\left[\left(1-\frac{1}{3^{3/2}}\right)\left(1-\frac{1}{(4+\sqrt{3})^{3/2}}\right)\left(1-\frac{1}{(5+\sqrt{3}+\sqrt{4+\sqrt{3}})^{3/2}}\right)\cdots\right]^{-1}=\frac{3}{2}$$
All square roots are positive.
How can I prove it?
| Let $u_n = \sqrt{x_n}$, we have
$$u_n^2 = u_{n-1}^2 + u_{n-1} + 1 = \frac{u_{n-1}^3 - 1}{u_{n-1} - 1}
\implies u_{n-1}^3-1 = u_n^2(u_{n-1}-1)
$$
This leads to
$$1 - \frac{1}{x_n^{3/2}}
= \frac{u_n^3-1}{u_n^3}
= \frac{u_{n+1}^2(u_n-1)}{u_n^3}
= \frac{u_{n+1}^2(u_n^2-1)}{u_n^3(u_n+1)}
= \frac{u_{n+1}^2(u_n^2-1)}{u_n^2(u_{n+1}^2-1)}\\
= \left.\frac{x_n - 1}{x_n}\right/\frac{x_{n+1}-1}{x_{n+1}}
$$
The product at hand is a telescoping one with partial products
$$\prod_{n=1}^p \left(1 - \frac{1}{x_n^{3/2}}\right)^{-1}
=\left.\frac{x_1}{x_1 - 1}\right/\frac{x_{p+1}}{x_{p+1}-1}$$
Since $x_n = x_{n-1}+\sqrt{x_{n-1}} + 1 \ge x_{n-1} + 1$ and $x_0 = 1$, we can show $x_n \ge n + 1$ by induction. This means $$\lim\limits_{p\to\infty} x_p = \infty \implies \lim\limits_{p\to\infty}\frac{x_{p+1}}{x_{p+1}-1} = 1$$
As a result,
$$\prod_{n=1}^\infty \left(1 - \frac{1}{x_n^{3/2}}\right)^{-1} = \frac{x_1}{x_1 - 1} = \frac{3}{3-1} = \frac32$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2589746",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
How to calculate a combinatorial sum by hand My question is if there is an "easy" way to calculate
$$
\displaystyle \sum_{k=0}^{6}(2k)\binom{6}{k}(2)^k(3)^{6-k}
$$
I know that if it were just $\displaystyle \sum_{k=0}^{6}\binom{6}{k}(2)^k(3)^{6-k}$, this is the same as $(2+3)^6 = 5^6 = 15625$. But with that $2n$ term in there, I'm not sure if there is some clever trick to still do something like $(2+3)^6$. Thanks for the help!
| You can write this as
\begin{align}
\sum_{k=0}^{6}(2k)\binom{6}{k}(2)^k(3)^{6-k} &= 2\cdot3^6 \sum_{k=0}^{6}k\binom{6}{k}(2)^k(3)^{-k} \\
&= 2\cdot3^6 \sum_{k=0}^{6}k\binom{6}{k} \left(\frac{2}{3}\right)^k, \\
\end{align}
and use the identity
$$\sum_{k = 0}^n k \binom{n}{k} x^k = x \frac{d}{dx} (1 + x)^n = nx(1+x)^{n-1}. $$
This gives you the answer
$$ \sum_{k=0}^{6}(2k)\binom{6}{k}(2)^k(3)^{6-k} = 2 \cdot 3^6 \cdot 6 \cdot \frac{2}{3} \left( 1 + \frac23 \right)^5 = 75000.$$
The general formula, which you can prove in the same way, is
$$ \sum_{k = 0}^n k \binom{n}{k}x^ky^{n - k} = nx(x + y)^{n - 1}. $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2590615",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Distribution of $X+\frac{2}{X}$ when $X\sim\mathcal U(1,2)$ I have the following question:
If $X$ is a continuous random variable that is uniformly distributed on the interval $(1,2)$ what is the distribution function of $Y=X+\frac{2}{X}?$\
I have tried to calculate the inverse of the function $f(x)=x+\frac{2}{x}$ but didn't manage to complete the calculation. Any ideas?
| The function $x\mapsto x + \frac 2 x$ has a minimum point at $x=\sqrt 2,$ which is within the interval $(1,2),$ so it doesn't have an inverse on that interval. That makes things a bit more complicated. One thing that will make things simpler is that when $x={}$either $1$ or $2$ then $x+ \frac 2 x=3,$ i.e. it's the same at both endpoints.
Let $y = x+ \frac 2 x.$ Then $xy = x^2 + 2,$ or $x^2 - yx + 2 = 0,$ and that is a quadratic equation that can be solved for $x{:}$
$$
x = \frac {y \pm \sqrt{y^2 - 8}} 2.
$$
Since $x\mapsto x+ \frac 2 x$ is a curve that opens upward, you have $y\le{}$some specified number if $x$ is between two numbers.
Thus $Y\le y$ precisely if $\dfrac {y - \sqrt{y^2-8}} 2\le X\le\dfrac{y+\sqrt{y^2-8}} 2.$
The probability assoicated with that interval is the difference -- the larger one minus the smaller one. Thus the probability is $\sqrt{y^2-8}.$
This works for $2\sqrt 2 \le y\le 3,$ since those are the smallest and biggest values of $y$ when $1<x<2.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2593442",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Find the range of $f(x)=11\cos^2x+3\sin^2x+6\sin x\cos x+5$ I'm trying to solve this problem.
Find the range of $f(x)=11\cos^2x+3\sin^2x+6\sin x\cos x+5$
I have simplified this problem to $$f(x)= 8\cos^2x+6\sin x\cos x+8$$ and tried working with $g(x)= 8\cos^2x+6\sin x\cos x$.
I factored out the $2\cos x$ and rewrote the other factor as a linear combination of cosine.
It reduces down to $$g(x)=10\cos x\cos\left(x-\tan^{-1}\frac{3}{4}\right)$$
But then I'm stuck here. Please help me. Perhaps there's a different way to approach this?
| $$f(x)=8\left(\frac{1+\cos {2x}}{2}\right) +3\sin {2x}+8$$
$$f(x)=4\cos {2x}+3\sin {2x} +12$$
The range of $$4\cos {2x}+3\sin {2x}$$ is $[-5,5]$
Hence maximum and minimum values of expression are $17$ and $7$ respectively.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2594442",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 2
} |
Prove that $\sum^m_{k=0}\binom{m}{k}(-1)^k(\frac{1}{n+k+1}) = \frac{m!n!}{(m+n+1)!}$ I am looking for a more direct proof of the following identity: $$\sum^m_{k=0}\frac{\binom{m}{k}(-1)^k}{n+k+1} = \frac{m!n!}{(m+n+1)!}$$
My original proof comes from evaluating $\int^1_0{x^n(1-x)^m}{dx}$, $n, m \in \mathbb{N}_0$ in two ways:
1) Let $u = x^n$, $u' = nx^{n-1}$, $v' = (1-x)^m$, $v = -\frac{(1-x)^{m+1}}{m+1}$:
$$\int^1_0{x^n(1-x)^m}{dx} = \frac{x^n(1-x)^{m+1}}{m+1} + \frac{n}{m+1}\int^1_0{x^{n-1}(1-x)^{m+1}}{dx}$$
$x=1\Rightarrow \frac{x^n(1-x)^{m+1}}{m+1} = 0$ and $x=0\Rightarrow \frac{x^n(1-x)^{m+1}}{m+1} = 0$, therefore
$$\int^1_0{x^n(1-x)^m}{dx} = \frac{n}{m+1}\int^1_0{x^{n-1}(1-x)^{m+1}}{dx}$$
Repeating the process $n$ times gives:
$$\int^1_0{x^n(1-x)^m}{dx} = \frac{m!n!}{(m+n)!}\int^1_0{x^0(1-x)^{m+n}}{dx} = $$$$=\frac{m!n!}{(m+n)!}\cdot\frac{(1-1)^{m+n+1}}{m+n+1} - \frac{m!n!}{(m+n)!}\cdot\left(-\frac{(1-0)^{m+n+1}}{m+n+1}\right) = \frac{m!n!}{(m+n+1)!}$$
2)$$\int{x^n(1-x)^m}{dx} = \int{x^n\cdot\left(\sum^m_{k=0}\binom{m}{k}(-1)^kx^k\right)}{dx} = \int{\left(\sum^m_{k=0}\binom{m}{k}(-1)^kx^kx^n\right)}{dx} = $$
$$ = \sum^m_{k=0}\left(\binom{m}{k}(-1)^k\cdot\int{x^{n+k}}{dx}\right) = \sum^m_{k=0}\binom{m}{k}(-1)^k\frac{x^{n+k+1}}{n+k+1} =: F(x).$$
$$F(1) - F(0) = \sum^m_{k=0}\frac{\binom{m}{k}(-1)^k1^{n+k}}{n+k+1} - \sum^m_{k=0}\frac{\binom{m}{k}(-1)^k0^{n+k}}{n+k+1} = \sum^m_{k=0}\frac{\binom{m}{k}(-1)^k}{n+k+1}$$
| Induction on $m$.
The $m=0$ case is trivial.
For $m>0$,
$$
\begin{align}
\sum_{k=0}^m \frac{\binom{m}{k}(-1)^k}{n+k+1} &=
\sum_{k=1}^m \frac{\binom{m-1}{k-1}(-1)^k}{n+k+1} + \sum_{k=0}^{m-1} \frac{\binom{m-1}{k}(-1)^k}{n+k+1} \\ &=
\sum_{j=0}^{m-1} \frac{\binom{m-1}{j}(-1)^{j+1}}{(n+1)+j+1} + \sum_{k=0}^{m-1} \frac{\binom{m-1}{k}(-1)^k}{n+k+1} \\ &=
\frac{-(m-1)!(n+1)!}{(m+n+1)!} + \frac{(m-1)!n!}{(m+n)!} \\ &=
(-(n+1)+(m+n+1))\frac{(m-1)!n!}{(m+n+1)!} \\ &=
\frac{m!n!}{(m+n+1)!}
\end{align}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2596013",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Geometric proof for $\tan{(3x)}=\tan{(x)}\tan{\left(\frac{\pi}{3}-x\right)}\tan{\left(\frac{\pi}{3}+x\right)}$ Are there geometric proofs for the identitity
$$\tan{(3x)}=\tan{(x)}\tan{\left(\frac{\pi}{3}-x\right)}\tan{\left(\frac{\pi}{3}+x\right)}$$
My try:
Thank in advances.
|
$A,B,C,D,E$ are the same points in the problem.
$F= AB \cap CD$.
$G$ and $C$ are symmetric with respect to $E$. $H$ and $B$ are symmetric with respect to $E$.
$P = GD \cap AF$, $Q = AH \cap DF$. $R = GD \cap AH$.
(In my figure, I forgot to write $R$. Sorry.)
We want to show that $R$ is the incenter of $\triangle ADF$.
because $\angle AFD = \angle AED - \angle FAC - \angle FDB$,
$\angle AFD = \frac{\pi}{3}$.
$\angle EBC = \angle EBG = \angle EHC = \frac{\pi}{3} - x$.
and $\angle ECB = \angle EGB = \angle ECH = \frac{\pi}{6} + x$.
Because $\angle ABG = \angle EGB - \angle GAB$,
$\angle ABG = \frac{\pi}{6}$.
Also, because $\angle HDC = \frac{\pi}{6} -x$, and $\angle DCH = \angle EHC - \angle HDC$,
$\angle DCH = \frac{\pi}{6}$.
$\triangle DCH \equiv \triangle DGH$, and $\triangle ABG \equiv \triangle AHG$.
Thus, $\angle AHG = \angle DGH = \frac{\pi}{6}$. Thus, $RG=RH$.
Additionally, $\angle PRQ = \pi - \angle RGH - \angle RHG = \frac{2\pi}{3}$,
and $\angle PRQ + \angle PFQ = \pi$. Hence, four points $R, P, F, Q$ are on the same circle.
Because $\square BCHG$ is a rhombus, let $BG=CH=l$.
$\angle BPG = \pi - \angle AFD - \angle FDP = \frac{\pi}{3}+2x$.
In $\triangle BPG$,
$$\frac{GP}{\sin\frac{\pi}{6}} = \frac{l}{\sin(\frac{\pi}{3}+2x)}$$
Similarly, $\angle CQH = \pi - \angle AFD - \angle FAQ = \frac{2\pi}{3}-2x$.
In $\triangle CQH$,
$$\frac{HQ}{\sin\frac{\pi}{6}} = \frac{l}{\sin(\frac{2\pi}{3}-2x)}$$
$\sin(\frac{\pi}{3}+2x) = \sin(\frac{2\pi}{3}-2x)$, thus $GP=HQ$.
Hence, $RP = RG + GP = RH + HQ = RQ$.
Because $RP=RQ$ and $R,P,F,Q$ are on the same circle, $\angle RFP = \angle RFQ$.
Thus, we know $\angle RAF = 2x$, $\angle RFA = \angle RFD = \frac{\pi}{6}$,
$\angle RDF = \frac{\pi}{3}-2x$, and $\angle RAD + \angle RDA = \frac{\pi}{3}$.
Finally, we use the trigonometric form of Ceva's theorem.
$$\frac{\sin{\angle RAD}}{\sin{\angle RAF}} \times \frac{\sin{\angle RDF}}{\sin{\angle RDA}} \times \frac{\sin{\angle RFA}}{\sin{\angle RFD}} = 1$$
It is equal to
$$ \frac{\sin{\angle RAD}}{\sin 2x} \times \frac{\sin{\frac{\pi}{3}-2x}}
{\sin{\frac{\pi}{3}- \angle RAD}} = 1 $$
$\angle RAD$ should be $2x$. Because if $\angle RAD < 2x$, then the left-hand side of the above equation is less than $1$,
and if $\angle RAD > 2x$, then the left-hand side of the above equation is greater than $1$.
Thus $R$ is the incenter of $\triangle FAD$, and $\angle EAD = 3x$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2596954",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
Difference of values of continuous function $f:\left[\frac{1}{2 \sqrt{2}},2\sqrt{2}\right]\to\mathbf{R}$ is continuous and $f\left(2\sqrt{2}\right)-f\left(\frac{1}{2 \sqrt{2}}\right)=3$. How do I show that for some $x$ in the domain $f(2x)-f(x)=1$?
| Define $g(x) = f(2x) - f(x)$ for $\displaystyle x \in \left[\frac{1}{2\sqrt{2}}, \sqrt{2}\right]$. Note that $\displaystyle \frac{1}{2\sqrt{2}} = \frac{\sqrt{2}}{4}$, thus$$
g\left(\frac{\sqrt{2}}{4}\right) + g\left(\frac{\sqrt{2}}{2}\right) + g(\sqrt{2}) = f(2\sqrt{2}) - f\left(\frac{\sqrt{2}}{4}\right) = 3.
$$
If any of $\displaystyle g\left(\frac{\sqrt{2}}{4}\right)$, $\displaystyle g\left(\frac{\sqrt{2}}{2}\right)$ and $g(\sqrt{2})$ equals $1$ then it is already done. Otherwise, since their sum is $3$, they cannot all be less than $1$ or all be greater than $1$. Enumerate each possible scenario with respect to the relations of $\displaystyle g\left(\frac{\sqrt{2}}{4}\right)$, $\displaystyle g\left(\frac{\sqrt{2}}{2}\right)$, $g(\sqrt{2})$ and $1$, and apply the intermediate value theorem to prove the existence of the required $x$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2598043",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Average value of complex valued function property If $f:[a,b] \to \mathbb{C}$ is a continuous function define the average of $f$
$$A = \frac{1}{b-a} \int_a^b f(x)dx$$
where $\int f= \text{Re} (f) + i \int \text{Im}(f)$
Then I need to show that if $|f| \le |A|$ on $[a,b]$ then $f=A$. How to approach this?
| Assume that $A\ne 0$. We have
\begin{align*}
|A|&=\dfrac{1}{b-a}\left|\int_{a}^{b}f(x)dx\right|\leq\dfrac{1}{b-a}\int_{a}^{b}|f(x)|dx\leq\dfrac{1}{b-a}\int_{a}^{b}|A|dx=|A|,
\end{align*}
so
\begin{align*}
\dfrac{1}{b-a}\int_{a}^{b}(|A|-|f(x)|)dx=0,
\end{align*}
but $|A|-|f(\cdot)|\geq 0$ and $|A|-|f|$ is continuous, one has $|f|=A$.
Now we let $z$ be a complex number such that
\begin{align*}
zA=z\cdot\dfrac{1}{b-a}\int_{a}^{b}f(x)dx\geq 0,
\end{align*}
by writing $zf=u+iv$ for real $u,v$, then
\begin{align*}
|A|=\dfrac{1}{b-a}\left|\int_{a}^{b}f(x)dx\right|=\dfrac{1}{b-a}\int_{a}^{b}zf(x)dx=\dfrac{1}{b-a}\int_{a}^{b}u(x)dx+i\dfrac{1}{b-a}\int_{a}^{b}v(x)dx,
\end{align*}
so
\begin{align*}
\int_{a}^{b}v(x)dx=0,
\end{align*}
and hence
\begin{align*}
|A|\leq\dfrac{1}{b-a}\int_{a}^{b}u^{+}(x)dx\leq\dfrac{1}{b-a}\int_{a}^{b}|f(x)|dx=|A|,
\end{align*}
hence
\begin{align*}
\int_{a}^{b}(|f(x)|-u^{+}(x))dx=0,
\end{align*}
once again $|f(\cdot)|-u^{+}(\cdot)\geq 0$, so $|f|=u^{+}$. Then $u^{+}+u^{-}=|u|\leq|f|=u^{+}$ implies $u^{-}=0$. Also, $|f|^{2}=(u^{+})^{2}+|v|^{2}=|f|^{2}+|v|^{2}$ implies $v=0$, so $zf=u^{+}=|f|=|A|$.
Substitute $f=z^{-1}|A|$ into
\begin{align*}
A=\dfrac{1}{b-a}\int_{a}^{b}f(x)dx,
\end{align*}
one gets $z^{-1}=\dfrac{A}{|A|}$, so $f=A$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Prove that $\frac{1}{\sqrt{ab+a+2}}+ \frac{1}{\sqrt{bc+b+2}}+ \frac{1}{\sqrt{ac+c+2}} \leq \frac{3}{2}$ Let $abc=1$ and $a,b,c>0$.
Prove that
$$\frac {1}{\sqrt {ab+a+2}}+ \frac {1}{\sqrt {bc+b+2}}+ \frac {1}{\sqrt {ac+c+2}} \leq \frac {3}{2}.$$
I guess that $$\frac {1}{\sqrt {ab+a+2}}+ \frac {1}{\sqrt {bc+b+2}}+ \frac {1}{\sqrt {ac+c+2}} = \frac {3}{2}$$ if and only if $a=b=c=1$. I have spent two nights on this but it's just a mess and I still don't know how to apply Cauchy Schwarz or something to write a rigorous proof for this exercise.
| $\mathbf {Hint: }$
Let $$\frac{1}{\sqrt{ab+a+2}}+ \frac{1}{\sqrt{bc+b+2}}+ \frac{1}{\sqrt{ac+c+2}}=A$$
Applying C-S:
$$\left(\frac {1}{ab+a+2}+\frac {1}{bc+b+2}+\frac {1}{ac+c+2}\right)(3) \ge (A)^2$$
Hence $$A\le \sqrt{\left(\frac {1}{ab+a+2}+\frac {1}{bc+b+2}+\frac {1}{ac+c+2}\right)(3)}$$
Now consider by Titu's lemma we have
$$\frac {1}{ab+1} + \frac {1}{a+1} \ge \frac {4}{ab+a+2}$$
Hence our obtained inequality now has
$$A\le \sqrt{
\left(
\frac {1}{ab+a+2}+\frac {1}{bc+b+2}+\frac {1}{ac+c+2}
\right)
(3)
}\le
\sqrt {
\frac{3}{4}
\sum_{cyc} \left(\frac {1}{ab+1} + \frac {1}{a+1}
\right) }$$
But
$$\sum_{cyc} \left(\frac {1}{ab+1} + \frac {1}{a+1}\right) =\sum_{cyc} \left(\frac {c}{c+1} + \frac {1}{a+1}\right)=3 $$
Because $abc=1$
Hence
$$A \le \sqrt {\frac{3}{4} \sum_{cyc} \frac {c}{c+1} + \frac {1}{a+1}}=\sqrt {\frac{9}{4}}=\frac {3}{2}$$
Hope it helped now.
| {
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"timestamp": "2023-03-29T00:00:00",
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Generalising a question about divisibility of two numbers raised to a power Let $x$ and $y$ be integers greater than or equal to two. It is fairly straightforward to prove that if $x^3|y^2$ then $x|y$. I am interested in a more general question: when does $x^n| y^m$ imply that $x|y$ for $n,m\in\mathbb{N}$ with $n\geq m$? If we choose $n=4$ and $m=5$ then the statement is clearly false. But I can't figure out how to prove such a statement.
| Simple enough.
Let $x = \prod p_i^{k_i}$ be the prime factorisation of $x$.
Then $x^m|y^n$ means $p_i^{m\cdot k_i}|y^n$ so $p_i|y$ for each prime factor $p_i$ of $x$..
Let $l_i$ be "the power of $p_i$ in $y$". i.e. Let $p_i^{l_i}|y$ so that $p_i^{l_i + 1} \not \mid y$.
Then $p_i^{m\cdot k_i}|p_i^{n\cdot l_i}$ so $m\cdot k_i \le n\cdot l_i$ so $k_i \le \frac nm l_i$.
So $\ldots$ t
If $x^m|y^n$ then the prime factors of $x$ are also prime factors of $y$ and the powers of those prime factors are equal or less than $\frac nm$ the powers of the prime factors in $y$. Or the powers of the prime factors of $y$ are at leas $\frac mn$ of those in $x$.
In particular if $n \le m$ then $x|y$. But if $n > m$ we have limits of the powers of the prime factors.
Example if $54^5|y^7$ then as $54 = 3^3\cdot2$ then $3^{\lceil \frac 57\cdot 3\rceil} =3^3 |y$ and $2^{\lceil \frac 57\cdot 1\rceil} =2 |y$ so $54|y$ but if $54^5|w^8$ then $3^{\lceil \frac 58\cdot 3\rceil} =3^2 |w$ and $2|w$ so $18|w$ and indeed if $w = 18$ we have $18^8 = 3^{16\cdot2^8} = (3^3\cdot 2)^5\cdot(3\cdot2^3)= 54^5\cdot24$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How can I prove that $7+7^2+7^3+...+7^{4n} = 100*a$ (while a is natural number)? How can I prove that $7+7^2+7^3+...+7^{4n} = 100*a$ (while a is entire number) ?
I thought to calculate $S_{4n}$ according to:
$$ S_{4n} = \frac{7(7^{4n}-1)}{7-1} = \frac{7(7^{4n}-1)}{6} $$
But know, I don't know how to continue for get what that rquired.
I will be happy for help or hint.
After beautiful ideas for solving this question, someone know how to do it with induction too?
| Using the Formula for a Geometric Series
We need to show more than $7^4\equiv1\pmod{100}$. If that were all we knew, then because $2\mid6$, all we would know would be
$$
\frac{7\left(7^{4n}-1\right)}{6}\equiv0\pmod{50}
$$
However, since $7^4=2401\equiv1\pmod{800}$, we know that
$$
\frac{7\left(7^{4n}-1\right)}{6}\equiv0\pmod{400}
$$
Using Induction
Note that
$$
7^1+7^2+7^3+7^4=2800\equiv0\pmod{400}
$$
Therefore, if
$$
\sum_{k=0}^{n-1}\left(7^{4k+1}+7^{4k+2}+7^{4k+3}+7^{4k+4}\right)\equiv0\pmod{400}
$$
then
$$
\begin{align}
&\sum_{k=0}^n\left(7^{4k+1}+7^{4k+2}+7^{4k+3}+7^{4k+4}\right)\\
&=\underbrace{\sum_{k=0}^{n-1}\left(7^{4k+1}+7^{4k+2}+7^{4k+3}+7^{4k+4}\right)}_{0\bmod{400}\text{ by inductive hypothesis}}
+7^{4n}\underbrace{\vphantom{\sum_{k=0}^{n-1}}\left(7^1+7^2+7^3+7^4\right)}_{0\bmod{400}}\\[6pt]
&\equiv0\pmod{400}
\end{align}
$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 4
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Gradient descent proof: justify $\left(\dfrac{\kappa - 1}{\kappa + 1}\right)^2 \leq \exp(-\dfrac{4t}{\kappa+1})$ A claim on pg 279 of the notes states that:
How can the encircled be justified? Note that $\kappa := \beta /\alpha$ is a non-negative constant.
I tried using the definition of the exponential:
$$e^z = \lim\limits_{n \to \infty} (1+z/n)^n$$ to no avail, because I cannot produce such an $n$ for the term $\left(\dfrac{\kappa - 1}{\kappa + 1}\right)^2$
Can someone please show how this inequality was attained?
| You overlooked that in the last inequality the iterate changes to $1$ from $t$. So the inequality you need is just
\begin{align*}
\left(\left(1 - \frac {2} {\kappa + 1} \right)^2 \right)^t \le \left(e^{ \frac{-4} {\kappa +1} } \right)^t.
\end{align*}
For the remaining part you only need to note
\begin{align*}
\left(1 - \frac {2} {\kappa + 1} \right)^2 = 1 - \frac{4} {\kappa + 1} + \frac{4}{(\kappa +1 )^2} \\
e^{ \frac{-4} {\kappa +1} } = 1 - \frac{4}{\kappa + 1} + \frac{8}{(\kappa +1)^2} - \dots \ge 1 - \frac{4} {\kappa + 1} + \frac{4}{(\kappa +1 )^2}.
\end{align*}
| {
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"source": "stackexchange",
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Does $x+\sqrt{x}$ ever round to a perfect square, given $x\in \mathbb{N}$? I'll define rounding as $$R(x)=\begin{cases} \lfloor x \rfloor, & x-\lfloor x \rfloor <0.5 \\ \lceil x \rceil, & else\end{cases}$$
Does $x+\sqrt{x}$ ever round (to the nearest integer) to a perfect square, given $x\in \mathbb{N}$?
For example, $7+\sqrt{7}=9.646...$ which rounds up, and $57+\sqrt{57}=64.549...$ which also round up. Also, $6+\sqrt{6}$ and $57+\sqrt{57}$ both round down.
I think the positive integers $x$ such that $\lfloor x+\sqrt{x} \rfloor =k^2, k\in \mathbb{Z}$ are all of the form $n^2+n+1, n\in \mathbb{Z}^+$. The set of all $x$ begins as: $\{3, 7, 13, 21, 31, 43, 57, \dots \}$ and all those numbers are of the form $n^2+n+1$
I tried to find a pattern for whether the decimal part of $\sqrt{n^2+n+1}$ is less than $0.5$ or not, and I tried to modify $\sqrt{n^2+n+1}$ to $\sqrt{n^2+2n+1}=(n+1)^2$ but that didn't lead anywhere.
Is there an algebraic proof/disproof of my above claim?
Thanks.
| $$\sqrt{n^2+n+1}= \sqrt{\left(n+\frac12\right)^2+\frac 34} > n+\frac 12$$
And $$\sqrt{n^2+n+1} < \sqrt{n^2+2n+1} =n+1$$
$$\implies n+0.5 <\sqrt{n^2+n+1} <n+1$$
Thus $\rm{fractional part}{(n^2+n+1)}>0.5$
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "8",
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Evaluating: $\int \frac {1+\sin (x)}{1+\cos (x)} dx$
Evaluate: $\int \dfrac {1+\sin (x)}{1+\cos (x)} dx$
My Attempt:
$$=\int \dfrac {1+\sin (x)}{1+\cos (x)} dx$$
$$=\int \dfrac {(\sin (\dfrac {x}{2}) + \cos (\dfrac {x}{2}))^2}{2\cos^2 (\dfrac {x}{2})} dx$$
$$=\dfrac {1}{2} \int (\dfrac {\sin (\dfrac {x}{2}) + \cos (\dfrac {x}{2})}{\cos (\dfrac {x}{2})})^2 dx$$
$$=\dfrac {1}{2} \int (\tan (\dfrac {x}{2}) +1)^2 dx$$
How do I continue?
| *
*OP's question: how to continue?
*existing answers: start over with simpler steps
So I feel the need to respond to the original question.
\begin{align}
\int \frac {1+\sin (x)}{1+\cos (x)} dx
&=\frac {1}{2} \int (\tan (\frac {x}{2}) +1)^2 dx \\
&= \frac12 \int (\tan^2(\frac x2)+1) dx + \int \tan(\frac x2) dx \\
&= \frac12 \int \sec^2(\frac x2) dx + \int \tan(\frac x2) dx \\
&= \tan(\frac x2) + 2\ln|\sec \frac x2| + C
\end{align}
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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"answer_id": 3
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Related to matrix Find the values of $a$ and $b$ for which the equations are consistent
$$\begin{cases}
X+aY+Z=3 \\
X+2Y+2Z =b \\
X+5Y+3Z=9 \\
\end{cases}$$
I tried solving it although my phone is not uploading the snap of attempt I made. But I get to a point where I have to apply
$R_3-(R_2+\lambda)$
Is this valid?
| Consider equations$$(1)\qquad x+ay+z=3\\(2)\qquad x+2y+2z=b\\(3)\qquad x+5y+3z=9$$$(2)$ and $(3)$ are obviously linearly independent. Here we have two different cases:
Case 1: $(1)$ is independent from $(2)$ and $(3)$
At this case the determinant of $\begin{pmatrix} 1 & a &1 \\ 1 & 2 &2 \\1 & 5 &3 \\\end{pmatrix}$ is nonzero and the equations are consistent with unique solution for any $b$. The determinant is $-a-1$ so for $a\ne -1$ and $b$ free Case 1 would be achieved.
Case 2: $(1)$ is dependent to $(2)$ and $(3)$.
This happens when $a=-1$ therefore we have $$(1)=2*(2)-(3)$$ and it is consistent iff $$3=2b-9$$ or $b=6$.
In other words for $a\ne -1$, $b$ is free and for $a=-1$, $b=6$
Alternative way (using elementary row transformations):
Consider the matrix $\begin{pmatrix} 1 & a &1&3 \\ 1 & 2 &2&b \\1 & 5 &3&9 \\\end{pmatrix}$ and the following steps:$$\begin{pmatrix} 1 & a &1&3 \\ 1 & 2 &2&b \\1 & 5 &3&9 \\\end{pmatrix}$$$R_2-R_1\to R_2\\R_3-R_1\to R_3$$$\begin{pmatrix} 1 & a &1&3 \\ 0 & 2-a &1&b-3 \\0 & 5-a &2&6 \\\end{pmatrix}$$$R_1-R_2\to R_1\\R_3-2R_2\to R_3$ $$\begin{pmatrix} 1 & 2a-2 &0&6-b \\ 0 & 2-a &1&b-3 \\0 & 1+a &0&12-2b \\\end{pmatrix}$$Look at $R_3$. If $a+1\ne 0$ then there exists an answer and if $a+1=0$ there should forcibly be that $12-2b=0$ or $b=6$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Partial Fraction problem solution deviates from the Rule Question:
Compute $\displaystyle \int\frac{x^2+1}{(x^2+2)(x+1)} \, dx$
My Approach:
As per my knowledge this integral can be divided in partial Fraction of form $\dfrac{Ax+B}{x^2+px+q}$ and then do the following as per to integrate it.
Solution:
Taking $\dfrac{x^2+1}{(x^2+2)(x+1)}=\dfrac{Ax^2+Bx+C}{x^2+2}+\dfrac{D}{x+1}$
Rule given: Denominator g(x) contains quadratic tractor (may not be factorisable).
To each non-repeated quadratic factor of the form $x^2+px+q$(or $x^2+q$, $q$ not equal to $0$), there
Should be a partial Fraction of the form $Ax+B/(x^2+px+q)$.
My problem:
I can't understand why the solution provided deviates from the rule that I have studied to solve these kind of problems.
Book:
ISC MATHEMATICS XII
Publishers:
Kalyani
| Dividing $x^2+1$ by $x^2+2$ yields $1$ as the quotient and $-1$ as the remainder, so we have
$$
\frac{x^2+1}{x^2+2} = 1 - \frac 1 {x^2+2}.
$$
So
\begin{align}
& \frac{x^2+1}{(x^2+2)(x+1)} = \frac 1 {x+1} - \frac 1 {(x^2+2)(x+1)} \\[15pt]
= {} & \frac 1 {x+1} + \frac{Ax+B}{x^2+2} + \frac{\text{some constant}}{x+1} \\[15pt]
= {} & \frac{Ax+B}{x^2+2} + \frac C {x+1}
\end{align}
Thus I would decompose this into partial fractions in a way consistent with your approach.
| {
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"timestamp": "2023-03-29T00:00:00",
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System of ODE with non constant coefficients I have this system
$$\begin{bmatrix}\frac{dy_1(x)}{dx} \\ \frac{dy_2(x)}{dx} \end{bmatrix}=\begin{bmatrix}
\frac{1}{x} & \frac{1}{x} \\
\frac{4}{x} & \frac{1}{x}
\end{bmatrix}\begin{bmatrix}y_1(x) \\ y_2(x) \end{bmatrix}$$
I can solve system of ODE with constant coefficients and write the general solution but this is not the case. Does this system have a solution you can write with elementary functions? If this is the case how can I do? Is there a systematic procedure? Could you explain me?
Any help will be appreciated.
| Let $A=\begin{bmatrix}1 & 1 \\ 4 & 1\end{bmatrix}$, $y=\begin{bmatrix}y_1 \\ y_2\end{bmatrix}$ and write the equation as
$$
\frac{dy}{dx}-\frac{1}{x}Ay = 0.
$$
The characteristic polynomial of $A$ is $(\lambda-1)^2-4=(\lambda-3)(\lambda+1)$. So $A$ diagonalizable. It has a basis of eigenvectors
$$
X_1=\begin{bmatrix}1 \\ 2\end{bmatrix}, \; X_2 =\begin{bmatrix}-1 \\ 2\end{bmatrix}
$$
$AX_1 = 3X_1$ and $AX_2 = -X_2$.Then $y(x)=a(x)X_1+b(x)X_2$ can be substituted into the original equation to obtain
$$
a'(x)X_1+b'(x)X_2-\frac{3}{x}a(x)X_1+\frac{1}{x}b(x)X_2=0.
$$
This gives component equations
$$
a'(x)-\frac{3}{x}a(x)=0,\;\; b'(x)+\frac{1}{x}b(x)=0 \\
\frac{a'}{a}=\frac{3}{x},\;\; \frac{b'}{b}=-\frac{1}{x} \\
a = Ax^3,\;\; b=B\frac{1}{x}.
$$
The general solution is
$$
y(x) = Ax^3\begin{bmatrix}1 \\ 2\end{bmatrix}+B\frac{1}{x}\begin{bmatrix}-1 \\ 2\end{bmatrix},
$$
where $A$ and $B$ are arbitrary constants. So $y_1(x)=Ax^3-B/x$ and $y_2(x)=2Ax^3+2B/x$ are your solutions.
| {
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Solve the equation $(1-x)\sqrt{1-x^4}=x$
Solve the equation
$$(1-x)\sqrt{1-x^4}=x$$
My work so far:
$$(1-x)^2(1-x^4)=x^2$$
$$(1-2x+x^2)(1-x^4)=x^2$$
$$1-x^4-2x+2x^5+x^2-x^6=x^2$$
| You are completely correct, we obtain the polynomial equation
$$
x^6 - 2x^5 + x^4 + 2x - 1=0.
$$
The polynomial is irreducible over $\mathbb{Q}$, i.e., it does not factor into factors of smaller degree. We have exactly two real roots of the degree $6$ polynomial equation, namely
$x=0.492425875905$ and $x=- 0.935635630515$. The second one is not a solution of the original equation.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2625156",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Why $ (n+1)(1-x)c^n \rightarrow_{n \rightarrow +\infty} 0$ Simple question.
Why the function $f_n(x) = (n+1)(1-x)c^n \rightarrow_{n \rightarrow +\infty} 0$ if $|c| < 1$.
I know that $(1-x)c^n \rightarrow 0$ but $(n+1) \rightarrow + \infty$. Why $f_n(x) \rightarrow 0$ ?
| $a_n =( n+1)c^n$, where $|c| < 1.$
$|a_n| = (n+1)(|c|^n)$ .
Set $b:=1/|c|$, where $b >1$.
$b = 1+a$, $a>0$.
$b^n =(1+a)^n =$
$1+ na +n(n-1)/2! + ...$
Hence:
$|a_n| = \dfrac{n+1}{b^n}=$
$\dfrac{n+1}{1+na+(n(n-1)/2!)a^2 +...} \lt$
$\dfrac{2(n+1)}{n(n-1)a^2}=$
$\dfrac{2}{(n-1)a^2} +\dfrac{2}{n(n-1)a^2}.$
The limit $n \rightarrow \infty $ is?
| {
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"timestamp": "2023-03-29T00:00:00",
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Minimizing expression under constraint I was asked to minimize the expression $x^2+2y^2+3z^2$ under the constraint $xy+yz+zx=1$.
Using the Lagrange-multipliers method, the system to solve, to get eventual extrema, is
$$ \begin{cases} 2x = \lambda(y+z) \\ 4y=\lambda(x+z) \\ 6z=\lambda(x+y) \\ xy + yz + xz = 1\end{cases} $$
I'm having trouble in solving this system.
As for the minimization by other means, I tried the method suggested by Macavity in this post that uses only AM-GM but failed to find suitable coefficients to get it to work
Thanks for any advice, suggestion.
| We need to find a maximal value of $k$, for which the inequality
$$x^2+2y^2+3z^2\geq k(xy+xz+yz)$$ is true for all reals $x$, $y$ and $z$ or
$$3z^2-k(x+y)z+x^2+2y^2-kxy\geq0,$$ for which we need
$$k^2(x+y)^2-12(x^2+2y^2-kxy)\leq0$$ or
$$(12-k^2)x^2-2(k^2+6k)xy+(24-k^2)y^2\geq0,$$ for which we need $12-k^2>0$ and
$$(x^2+6k)^2-(12-k^2)(24-k^2)\leq0$$ or
$$k^3+6k^2-24\leq0.$$
Id est, it remains to find a maximal root of the equation
$$k^3+6k^2-24=0.$$
Indeed, let $k=4\cos\alpha-2.$
Thus, we get $$\cos3\alpha=\frac{1}{2},$$ which gives $$\alpha=\pm20^{\circ}+120^{\circ}n,$$ where $n\in\mathbb Z$ and we see that $\alpha=20^{\circ}$ is valid, which gives $k_{max}=4\cos20^{\circ}-2$ and we obtain:
$$\min_{xy+xz+yz=1}(x^2+2y^2+3z^2)=4\cos20^{\circ}-2.$$
| {
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Beckenbach Introduction to Inequalities Chapter 2: Show $(a+b)/2 \le ( (a^2 + b^2 )/2)^{1/2}$ I'm having trouble understanding the following problem
Problem
Beckenbach, Chapter 2 Pg 24 Ex 1
$$
\text{Show the following for all a, b}\quad
\frac{(a+b)}{2} \le \left(\frac{a^2 + b^2}{2}\right)^\frac{1}{2}
$$
The book provides answers in the back, the answer is shown as
$$
\text{equivalent to}\quad(a - b)^2 \ge 0
$$
My attempt
Thanks to this answer: Proving the inequality $\frac{a+b}{2} - \sqrt{ab} \geq \sqrt{\frac{a^2+b^2}{2}} - \frac{a+b}{2}$ I was able to show the following
$$
\begin{align}
\frac{a+b}{2} &\le \sqrt{\frac{a^2 + b^2}{2}} \\
\frac{a+b}{2} &\le \sqrt{\frac{(a+b)^2 + (a-b)^2}{4}}\\
\frac{a+b}{2} &\le \sqrt{\left(\frac{a+b}{2}\right)^2\left(1 + \left(\frac{a-b}{a+b}\right)^2\right)}\\
\frac{a+b}{2} &\le \frac{a+b}{2}\sqrt{1 + \left(\frac{a-b}{a+b}\right)^2}\\
0 &\le \sqrt{1 + \left(\frac{a-b}{a+b}\right)^2} - 1 \\
\text{Since}\quad\left(\frac{a-b}{a+b}\right)^2 \ge 0\quad\text{we're done}
\end{align}
$$
This feels ugly and I don't think is the way Beckenbach intended us to solve, considering this is in Chapter 2.
Question
*
*Is my attempt valid?
*Is there a more elegant way to show this, using $(a-b)^2 \ge 0$?
Thanks
| *
*Your proof has a tiny issue you need to address (and then it'll be correct): you forgot absolute values around $\frac{a+b}{2}$ once you factor $\left(\frac{a+b}{2}\right)^2$ out of the square root.
*Note that
$$
\frac{a+b}{2} \leq \frac{\lvert a\rvert +\lvert b\rvert}{2}\tag{1}
$$
so it suffices to prove
$$
\frac{\lvert a\rvert +\lvert b\rvert}{2} \leq\left(\frac{a^2+b^2}{2}\right)^{1/2} \tag{2}
$$
Square both sides of (2): this is equivalent, this everything is non-negative: so (2) is equivalent to
$$
\frac{a^2+b^2+2\lvert a\rvert\lvert b\rvert}{4} \leq \frac{a^2+b^2}{2}
$$
in turn equivalent to
$$
a^2+b^2+2\lvert a\rvert\lvert b\rvert \leq 2a^2+2b^2
$$
in turn equivalent to
$$
0 \leq a^2+b^2+2\lvert a\rvert\lvert b\rvert \tag{3}
$$
and the RHS is equal to $(\lvert a\rvert -\lvert b\rvert )^2$.
So (3) holds.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2630420",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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Solution to first order partial Cauchy problem? Solve the Cauchy Problem :
$$(x+y)u_{x} + (x-y)u_y = 1$$ with $ \; \; u(1,y)=\frac{1}{\sqrt{2}}$
My attempt:
We have the following characteristic equations :
$$\frac{dx}{x+y} = \frac{dy}{x-y} = \frac{du}{1}$$
Solving
$$\frac{dx}{x+y} = \frac{dy}{x-y} \Rightarrow xdx - ydy = xdy + ydy \Rightarrow x^2 - y^2 -2xy = c_1 $$ where $c_1$ is a constant
I am unable to find another pair of characteristic equations to solve.
In order to reduce it to canonical form, if I use transformation $\phi(x,y)= x^2 - y^2 -2xy $ and $\gamma(x,y)= y$
I am getting $(x+y)u_{\gamma} = 1$ which is again of no help.
How do I proceed?
| Your first characteristic equation is correct :
$$x^2-y^2-2xy=c_1$$
From this : $\quad x-y=\pm\sqrt{c_1+2y^2}$
Second characteristic, from $\quad\frac{dy}{x-y}=\frac{du}{1}=\frac{dy}{\pm\sqrt{c_1+2y^2}}$
$u-\int\frac{dy}{\pm\sqrt{c_1+2y^2}}=c_2$
$u\pm\frac{1}{\sqrt{2}}\ln\bigg|\pm\sqrt{2(c_1+2y^2)}+2y\bigg|=c_2$
$u\pm\frac{1}{\sqrt{2}}\ln\bigg|\pm\sqrt{2(x^2-y^2-2xy+2y^2)}+2y\bigg|=c_2$
$$u\pm\frac{1}{\sqrt{2}}\ln\bigg|\pm\sqrt{2}(x-y)+2y\bigg|=c_2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2631374",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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How to prove $\tan\Big[\frac{1}{2}\sin^{-1}\frac{3}{4}\Big]=\frac{4-\sqrt{7}}{3}$ Prove
$$
\tan\Big[\frac{1}{2}\sin^{-1}\frac{3}{4}\Big]=\frac{4-\sqrt{7}}{3}
$$
and justify why $\frac{4+\sqrt{7}}{3}$ is ignored.
My Attempt:
$$
\tan x=\frac{2\tan\frac{x}{2}}{1-\tan^2\frac{x}{2}}\implies\tan x-\tan x\tan^2\frac{x}{2}=2\tan\frac{x}{2}\\
\implies \tan^2\frac{x}{2}(\tan x)+2\tan\frac{x}{2}-\tan x=0\\
\implies \tan\frac{x}{2}=\frac{-2\pm2\sec x}{2\tan x}=\frac{-1\pm\sec x}{\tan x}
$$
Using this,
$$
\tan\bigg[\frac{\sin^{-1}\frac{3}{4}}{2}\bigg]=\frac{-1\pm\frac{4}{\sqrt{7}}}{\frac{3}{\sqrt{7}}}=\frac{-\sqrt{7}\pm4}{3}
$$
As $0\leq\frac{\sin^{-1}3/4}{2}\leq\frac{\pi}{4}\implies0\leq\tan(\frac{\sin^{-1}3/4}{2})\leq1$
$$
\implies \tan\bigg[\frac{\sin^{-1}\frac{3}{4}}{2}\bigg]=\frac{-\sqrt{7}+4}{3}
$$
Is my attempt correct and where is the value $\frac{4+\sqrt{7}}{3}$ to be excluded ?
| $$\tan\left[\frac{1}{2}\sin^{-1} \left(\frac{3}{4}\right)\right]=\frac{4-\sqrt{7}}{3}$$
$\sin^{-1}(x) = \arcsin(x) = 2\arctan \left(\frac x{1+\sqrt{1-x^2}}\right)$
See Inverse trigonometric functions
\begin{align}
\tan \left[\frac{1}{2}\sin^{-1} \left(\frac{3}{4}\right)\right] &=\tan \left[\frac{1}{2}2\arctan \left(\frac {\frac{3}{4}}{1+\sqrt{1-\left(\frac{3}{4}\right)^2}}\right)\right]\\
&= \frac {3}{4\left(1+\sqrt{1-\left(\frac{3}{4}\right)^2}\right)}
\end{align}
$\sqrt{1-\left(\frac{3}{4}\right)^2}=\frac{\sqrt{7}}{4}$
$$=\frac{3}{4\left(\frac{\sqrt{7}}{4}+1\right)} =\frac{3}{4\cdot \frac{4+\sqrt{7}}{4}}=\frac{3}{4+\sqrt{7}} = \frac{4-\sqrt{7}}{3}$$
$\tag*{$\Box$}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2632633",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 1
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Some implicit differentiation questions. Please check them [see desc.] I have a few implicit diffentiatial questions that I wanted to check. general question... how do you know that $y$ is a function of $x$? I assume the whole reason why these are called implicit differentiation questions is because $y$ is defined implicitly and it's hard or impossible to define $y$ as a function of $x$. Is that right?
1.
$$ \cos{xy} = 1+ \sin{y} $$
$$-\sin{xy} \cdot (\frac{dy}{dx} + y) = \cos{y} \cdot \frac{dy}{dx}$$
$$-\sin{xy} \cdot \frac{dy}{dx} - y(\sin{xy}) = \cos{y} \cdot \frac{dy}{dx}$$
$$ - y(\sin{xy}) = \cos{y} \cdot \frac{dy}{dx} + \sin{xy} \cdot \frac{dy}{dx}$$
$$ - y(\sin{xy}) = \frac{dy}{dx} ( \sin{xy} + \cos{y})$$
$$ \frac{- y(\sin{xy})}{( \sin{xy} + \cos{y})} = \frac{dy}{dx}$$
*$$x - y = x \cdot e^y$$
$$1 - \frac{dy}{dx} = x \cdot e^y \cdot \frac{dy}{dx} + e^y$$
$$ 1 - e^y = x \cdot e^y \frac{dy}{dx} + \frac{dy}{dx}$$
$$1 - e^y = \frac{dy}{dx}(x \cdot e^y + 1)$$
$$\frac{1-e^y}{x \cdot e^y + 1} = \frac{dy}{dx}$$
*$$y \cdot \cos{x} = x^2 + y^2$$
$$y(-\sin{x}) \cdot \cos{x} \cdot \frac{dy}{dx} = 2x + 2y \frac{dy}{dx}$$
$$ 2y \frac{dy}{dx} - \cos{x} \frac{dy}{dx} = y - \sin{x} - 2x$$
$$\frac{dy}{dx} (2y-cosx) = y - sinx - 2x$$
$$\frac{dy}{dx} = \frac{y - sinx - 2x}{2y - cosx}$$
Thank you.
| Note that in the case of implicit differentiation it is not necessarily the case that one variable is a function of the other in the strict sense, but the two functions are related. One can symbolize this as $x$ and $y$ being parametric functions of a variable $t$. An alternate approach to implicit differentiation is based on this idea.
Taking your first example:
\begin{eqnarray}
\cos{xy} &=& 1+ \sin{y}\\
\frac{d}{dt}\left(\cos{xy} \right)&=&\frac{d}{dt}\left(1+ \sin{y}\right)\\
-\left(x\frac{dy}{dt}+y\frac{dx}{dt}\right)\sin xy&=&(\cos y)\frac{dy}{dt}\\
-(x\,dy+y\,dx)\sin xy&=&\cos y\,dy\\
-y\sin xy\,dx&=&(x\sin xy+\cos y)\,dy\\
\frac{dy}{dx}&=&-\frac{y\sin xy}{x\sin xy+\cos y}
\end{eqnarray}
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Finding A Matrix Representation
Let $T=\begin{pmatrix}
3x+3y+3z \\
4x+7y+4z\\
6x+8y+6z
\end{pmatrix}$ And basis $B=\left\{\begin{pmatrix}
3 \\
1\\
2
\end{pmatrix},\begin{pmatrix}
4 \\
2\\
3
\end{pmatrix},\begin{pmatrix}
2 \\
1\\
2
\end{pmatrix}\right\}$ Find $[T]_B$
Now to find it directly I applied T on the elements of $B$ and wrote the coordinates vectors of the images with respect to basis $B$ and got
$$\begin{pmatrix}
-3 & -5 & -3 \\
10 & 15 & 8\\
-4 & -6 & -3
\end{pmatrix}$$
But how can I get the answer using basis change matrices?
I first need to find $[T]_B^{E}$ the transformation form the standard basis to $B$ which is
$$\begin{pmatrix}
-5 & -1 & -1 \\
2 & 2 & 1\\
5 & -1 & 0
\end{pmatrix}$$
And multiply it by the matrix of basis change $[I]_B^E$ and $[I]_E^B$?
Meaning $[I]_E^B\cdot[T]^E_B\cdot [I]_B^E$?
| Note that
$$[T]_E=
\begin{pmatrix}
3 & 3 & 3 \\
4 & 7 & 4\\
6 & 8& 6
\end{pmatrix}$$
and
$$[M]_{EB}=
\begin{pmatrix}
3 & 4 & 2 \\
1 & 2 & 1\\
2 & 3& 2
\end{pmatrix}$$
$$v_E=[M]_{EB} \cdot v_B\implies v_B=[M]_{EB}^{-1}\cdot v_E \implies v_B= [M]_{BE}\cdot v_E$$
thus for $w=T(v)$ we have
$$w_E=[T]_{E}\cdot v_E\implies [M]_{EB}\cdot w_B=[T]_{E}\cdot [M]_{EB}\cdot v_B \implies w_B=[M]_{BE}\cdot [T]_{E}\cdot[M]_{EB}\cdot v_B $$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Prove the positivity of a sequence Let $a>0$ a real number and $(u_n)$ the sequence defined by
$$
u_{n+1} = a - \frac{1}{u_n}\text{ and } u_0 = a.
$$
Question: Determine condition on the value of $a>0$ such that the sequence $(u_n)$ is always positive.
Attempt: I tried to establish a general formula of $u_n$ in order to set up a condition on $a$, by calculating $u_n$ with some $n$:
$$
u_1 = a - \frac{1}{u_0} = a - \frac 1a = \frac{a^2-1}{a},\\
u_2 = a - \frac{1}{u_1}= a - \frac{a}{a^2-1}=\frac{a^3-2a}{a^2-1},\\
u_3 = a - \frac{1}{u_2} = a - \frac{a^2-1}{a^3-2a}=\frac{a^4-3a^2+1}{a^3-2a},\\
u_4 = a - \frac{1}{u_3} = a - \frac{a^3-2a}{a^4-3a^2+1}=\frac{a^5-4a^3+2a-1}{a^4-3a^2+1}.
$$
But I wasn't successful, it seems that there is no general formula of $u_n$. So I would be appreciate for any suggestion of solution. Thank you!
| hint
Let $f (x)=a-1/x $.
$f $ is increasing at $(0,+\infty) $.
$(u_n) $ is monotonic.
$u_1 <u_0$ thus it is strictly decreasing.
the sequences terms are $>0$ if the limit (if it exists) is $\ge 0$.
the limit $l $ satisfies $l=a-1/l .$
if $a\ge 2$ then
$l=(a\pm\sqrt {a^2-4})/2>0$.
If $a<2$ , $u_2<0$.
So, the condition is $a\ge 2$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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PDF and CDF of Sum of 2 dice i'm trying to obtain the PDF and CDF of the sum of 2 dice toss.
There are tons of elementary exercise where is asked to find the exact probability,
but what about the PDF and CDF?
i thought this is a convolution of discrete uniform PDF. but i don't know where to start to find it.
it's like $P(X_1+X_2<y)=$ but from here i don't know how to continue... i can't substitute $1/n=X_1$ and $1/n=X_1$
this would make nosense.
Thank You
| How many ways are there to get a sum of $2$? We must get $1$ and then $1$ again with probability $\frac{1}{36}$.
How many ways are there to get a sum of $3$? We must either get a $1$ and then $2$ or vice versa with probability $\frac{2}{36}$.
How many ways are there to get a sum of $4$? We must either get a $1$ and then $3$, a $3$ and then $1$, or a $2$ and a $2$, giving a probability of $\frac{3}{36}$
Carrying on in this manner, we find that the probabilities appear to follow triangular shape. As commented, letting $X$ be the sum of the the two die, $X$ is a discrete random variable so we need a pmf.
$$ p_{X}(x)=
\begin{cases}
\frac{1}{36} & x = 12 \\
\frac{2}{36} & x=11 \\
\frac{3}{36} & x=10 \\
\frac{4}{36} & x=9 \\
\frac{5}{36} & x=8 \\
\frac{6}{36} & x=7 \\
\frac{5}{36} & x=6 \\
\frac{4}{36} & x=5 \\
\frac{3}{36} & x=4 \\
\frac{2}{36} & x=3 \\
\frac{1}{36} & x=2 \\
0 & \text{otherwise} \\
\end{cases} $$
If you wish to construct a probability density function, despite it not making much sense in this situation, you could so something like
$$ f_{X}(x)=
\begin{cases}
\frac{1}{36} & x \geq 12 \\
\frac{2}{36} & 11 \leq x \lt 12 \\
\frac{3}{36} & 10 \leq x \lt 11 \\
\frac{4}{36} & 9 \leq x \lt 10 \\
\frac{5}{36} & 8 \leq x \lt 9 \\
\frac{6}{36} & 7 \leq x \lt 8 \\
\frac{5}{36} & 6 \leq x \lt 7 \\
\frac{4}{36} & 5 \leq x \lt 6 \\
\frac{3}{36} & 4 \leq x \lt 5 \\
\frac{2}{36} & 3 \leq x \lt 4 \\
\frac{1}{36} & 2 \leq x \lt 3 \\
0 & x \lt 2 \\
\end{cases} $$
And the cumulative density function would be something like
$$ F_{X}(x)=
\begin{cases}
1 & x \geq 12 \\
\frac{35}{36} & 11 \leq x \lt 12 \\
\frac{33}{36} & 10 \leq x \lt 11 \\
\frac{30}{36} & 9 \leq x \lt 10 \\
\frac{26}{36} & 8 \leq x \lt 9 \\
\frac{21}{36} & 7 \leq x \lt 8 \\
\frac{15}{36} & 6 \leq x \lt 7 \\
\frac{10}{36} & 5 \leq x \lt 6 \\
\frac{6}{36} & 4 \leq x \lt 5 \\
\frac{3}{36} & 3 \leq x \lt 4 \\
\frac{1}{36} & 2 \leq x \lt 3 \\
0 & x \lt 2 \\
\end{cases} $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2639380",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
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} |
Sum of the series $\csc^{-1} \sqrt{10}+ \csc^{-1} \sqrt{50}+\csc^{-1}\sqrt{170}...$
Find the sum of the series:
$\csc^{-1}\sqrt{10}+ \csc^{-1}\sqrt{50}+\csc^{-1}\sqrt{170}...\csc^{-1}\sqrt{(n^2+1)(n^2+2n+2)}$
I converted the series to $\sum^{n} _{i=0}\arcsin \dfrac{1}{\sqrt{(i^2+1)(i^2+2i+2)}}$ and then tried to use $\arcsin(x)+\arcsin(y)= \arcsin({x\sqrt{1+y^2}+y\sqrt{1+x^2}}) \forall x,y \ge 0 $ and $x^2+y^2\le1$.
There was no clear pattern in the terms obtained. How to solve this question? And what is the general trick to solve inverse trigonometric summations?
| Use $$\arcsin\frac{1}{\sqrt{n^2+1}}-\arcsin\frac{1}{\sqrt{n^2+2n+2}}=\arcsin\frac{1}{\sqrt{(n^2+1)(n^2+2n+2)}}$$ and the telescopic sum.
Indeed,
$$\sin\left(\arcsin\frac{1}{\sqrt{n^2+1}}-\arcsin\frac{1}{\sqrt{n^2+2n+2}}\right)=$$
$$=\frac{1}{\sqrt{n^2+1}}\cdot\frac{n+1}{\sqrt{n^2+2n+2}}-\frac{n}{\sqrt{n^2+1}}\cdot\frac{1}{\sqrt{n^2+2n+2}}=\frac{1}{\sqrt{(n^2+1)(n^2+2n+2)}}.$$
Id est, $$\sum_{i=1}^n\arcsin\frac{1}{\sqrt{(i^2+1)(i^2+2i+2)}}=\sum_{i=1}^n\left(\arcsin\frac{1}{\sqrt{i^2+1}}-\arcsin\frac{1}{\sqrt{i^2+2i+2}}\right)=$$
$$=\arcsin\frac{1}{\sqrt{2}}-\arcsin\frac{1}{\sqrt{n^2+2n+2}}=\frac{\pi}{4}-\arcsin\frac{1}{\sqrt{n^2+2n+2}}.$$
The Manthanein's answer and mine they are the same.
Indeed, we need to prove that
$$\arcsin\frac{1}{\sqrt{n^2+2n+2}}+\arctan(n+1)=\frac{\pi}{2}$$ or
$$\cos\left(\arcsin\frac{1}{\sqrt{n^2+2n+2}}+\arctan(n+1)\right)=0$$ or
$$\sqrt{1-\frac{1}{n^2+2n+2}}\cdot\frac{1}{\sqrt{1+(n+1)^2}}-\frac{1}{\sqrt{n^2+2n+2}}\cdot\sqrt{1-\frac{1}{1+(n+1)^2}}=0,$$
which is obvious.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2644191",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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How to find the value of a and b in limit if the denominator = 0? Given that
$$
\lim_{x \rightarrow 3}\frac{4x^3-bx^2+2x+30}{x^3-ax+3a-27} = \frac{1}{2}
$$
Find the value of a and b.
I have using
$$
\lim_{x \rightarrow 3}4x^3-bx^2+2x+30 = 0
$$
To found that b = 16
Does this mean that
$$
\lim_{x \rightarrow 3}\frac{4x^3-16x^2+2x+30}{x^3-ax+3a-27} = \frac{1}{2}
$$
| You correctly realise that it must be a $0/0$ form since denominator goes zero for the limit, so numerator and denominator must have a common root. Since denominator has a root $x = 3$, numerator must also be divisible by $x-3$.
But as Hagan says, you don't know the multiplicity of the root $x=3$ in numerator or denominator.
*
*If there is an extra $(x-3)$ in numerator, limit will be zero
*If there is an extra $(x-3)$ in denominator, limit will not exist ($\pm \infty$ from different sides)
So we seek to factorise in such a way that multiplicity of root $x=3$ is same on numerator and denominator.
$$\frac{(x-3)(4x^2+(12-b)x+(2+3(12-b)))}{(x-3)(x^2+3x+9-a)}$$
Now in the numerator, the constant term $-3(2+3(12-b))$ must be equal to $30$ so that we get $b=16$. Substitute, we get numerator as
$$(x-3)(4x^2-4x-10)$$
From here, note that $4x^2-4x-10$ has no real roots, so we are good. Also we need $x=3$ to not be repeated root of denominator so we can say $a\neq 27$.
From the given limit, you can deduce that
$$\lim_{x\to 3} \frac{4x^2-4x-10}{x^2+3x+9-a} = \frac{14}{27-a} = \frac{1}{2}$$
which gives $a = -1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2645935",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
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Calculate the determinant $\left|\begin{smallmatrix} a&b&c&d\\ b&a&d&c\\ c&d&a&b\\d&c&b&a\end{smallmatrix}\right|$ Question: Calculate the following determinant
$$A=\begin{vmatrix} a&b&c&d\\ b&a&d&c\\ c&d&a&b\\d&c&b&a\end{vmatrix}$$
Progress: So I apply $R1'=R1+R2+R3+R4$ and get
$$A=(a+b+c+d)\begin{vmatrix} 1&1&1&1\\ b&a&d&c\\ c&d&a&b\\d&c&b&a\end{vmatrix}$$
Then, I apply $C2'=C2-C1,\, C3'=C3-C1$, etc to get
$$A=(a+b+c+d)\begin{vmatrix}1&0&0&0\\b&a-b&d-b&c-b\\c&d-c&a-c&b-c\\d&c-d&b-d&a-d \end{vmatrix}$$
Thus, $$A=(a+b+c+d)\begin{vmatrix}a-b&d-b&c-b\\d-c&a-c&b-c\\c-d&b-d&a-d \end{vmatrix}$$
But now I'm stuck here. I don't really want to expand the whole thing, because it is really messy. Is there any approach that don't require much calculation?
| This is a block matrix, you can calculate a determinant of a $2 \times 2$ matrix made of $2 \times 2$ matrices, and then expand that: $$A=\begin{vmatrix} M&N\\ N&M\end{vmatrix}$$ where $$M=\begin{bmatrix} a&b\\ b&a\end{bmatrix}$$ and $$N=\begin{bmatrix} c&d\\ d&c\end{bmatrix}$$
All matrices are symmetric, this is really simple (you don't have to think about commutativity of the matrix products).
$$A=\begin{vmatrix} M&N\\ N&M\end{vmatrix}=|M^2-N^2|$$
$$M^2=\begin{bmatrix} a^2+b^2&2ab\\ 2ab&a^2+b^2\end{bmatrix}$$
$$N^2=\begin{bmatrix} c^2+d^2&2cd\\ 2cd&c^2+d^2\end{bmatrix}$$
$$A=(a^2+b^2-c^2-d^2)^2-4(ab-cd)^2$$
The formula is correct, as can be checked by substituting some numbers.
Edit
A more general case of the determinant of a block matrix. (See here for example).
If $C$ and $D$ commute.
$$|M|= \begin{vmatrix} A & B \\C & D \end{vmatrix}=|AD-BC|$$
Edit 2
Just in case there are any more people claiming this is wrong, here's the check:
Simplify[Det[{{a, b, c, d}, {b, a, d, c}, {c, d, a, b}, {d, c, b, a}}] - (a^2 + b^2 - c^2 - d^2)^2 + 4 (a b - c d)^2]
| {
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"url": "https://math.stackexchange.com/questions/2647193",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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A sum of series problem: $\frac{3}{1!+2!+3!} + \frac{4}{2!+3!+4!} + \cdots + \frac{2008}{2006!+2007!+2008!}$ I have a question regarding the sum of this series:
$$\frac{3}{1!+2!+3!} + \frac{4}{2!+3!+4!} + \cdots + \frac{2008}{2006!+2007!+2008!}$$
My approach:
I found that this sum is equal to:
$$\sum_{n=3}^{2008}\frac{n}{(n-2)!+(n-1)!+(n)!}$$
I reduced it to :
$$\sum_{n=3}^{2008}\frac{1}{n(n-2)!}$$
Please suggest how to proceed further.
| $\sum_{n=3}^{2008}\frac{1}{n(n-2)!}=\sum_{n=3}^{2008}\frac{1}{n(n-2)!}.\frac{(n-1)}{(n-1)}=\sum_{n=3}^{2008}\frac{n-1}{n!}=\sum_{n=3}^{2008}(\frac{n}{n!}-\frac{1}{n!})=\sum_{n=3}^{2008}(\frac{1}{(n-1)!}-\frac{1}{n!})=\frac{1}{2!}-\frac{1}{3!}+\frac{1}{3!}-\frac{1}{4!}+...-\frac{1}{2007!}+\frac{1}{2007!}-\frac{1}{2008!}=\frac{1}{2!}-\frac{1}{2008!}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2647385",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 3
} |
number of distinct terms in binomial expansion of $\left(x+x^{-1}+x^2+x^{-2}\right)^{15}$ Question -> How do I find the number of distinct terms in the binomial expansion of $$\left(x+\frac{1}{x}+x^2+\frac{1}{x^2}\right)^{15}$$
Solution I tried ->
$$\left(\frac{x^3+x+x^4+1}{x^{2}}\right)^{15}=x^{-30}\left(1+x+x^3+x^4\right)^{15}$$ is in the form of
$$x^{-30}\left(a_{0}+a_{1}x+....a_{60}x^{60}\right)$$
I did not understand which powers of $x$ are present and which are absent.
How do I find the number of distinct terms?
| We can get any exponent from $0$ to $60$. Thus, there are $61$ distinct terms.
Consider
$$
\left(1+x+x^3+x^4\right)^{15}
$$
For $0\le k\le30$, there is a way to get $x^k$ by
$$
\color{#C00}{1}^{15-k+2\lfloor k/3\rfloor}\color{#C00}{x}^{k-3\lfloor k/3\rfloor}\color{#C00}{x}^{\color{#C00}{3}\lfloor k/3\rfloor}
$$
since $k-2\lfloor k/3\rfloor\le11$.
For $0\le k\le30$, there is a way to get $x^{60-k}$ by
$$
\color{#C00}{x}^{\color{#C00}{4}(15-k+2\lfloor k/3\rfloor)}\color{#C00}{x}^{\color{#C00}{3}(k-3\lfloor k/3\rfloor)}\color{#C00}{x}^{\lfloor k/3\rfloor}
$$
since $k-2\lfloor k/3\rfloor\le11$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2648294",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
How to prove that $x^2yz+xy^2z+xyz^2 \leq \frac{1}{3}$ where $x^2+y^2+z^2 = 1$ and $x,y,z >0$? How to prove that $x^2yz+xy^2z+xyz^2 \leq \frac{1}{3}$ where $x^2+y^2+z^2 = 1$ and $x,y,z >0$?
| By Cauchy–Schwarz inequality with
*
*$u=(x,y,z)$
*$v=(xyz,xyz,xyz)$
we have
$$u\cdot v\le |u|\cdot |v|$$
that is
$$x^2yz+xy^2z+xyz^2 \leq\sqrt3\,xyz \le \frac{1}{3}$$
indeed by AM-GM
$$\frac13=\frac{x^2+y^2+z^2}{3}\ge \sqrt[3]{x^2y^2z^2}\implies xyz \le\frac{1}{3\sqrt3}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2648962",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Powers of ten addition with negative power I have been stumped by my nephew's algebra homework.
He has a question:
$10^5 + 10^{-3}=$?
The multi-choice answers were:
$10^{-5},10^{15},10^{3},10^{2},10^{5},10^{8}$
(to 1 s.f.)
I thought it was as simple as adding the powers - therefore being $10^2$. Apparently not. Can someone help so I can assist?
| I will assume that by $10^5+10^{-3}$ you meant $10^5\times 10^{-3}$.
Say we have a value $x$ and we want to raise it to a power $n$. This means that we multiply $x$ by itself $n$ times. $$x^n = \underbrace{x\cdot x\cdot x\cdot\ldots\cdot x}_{n\text{ times.}}\tag1$$ It is confusing to most people when we say that $x^0 =1$, because how can we multiply $x$ by itself $0$ times? Here, we have to look at one of the Power Rules.
The first power rule is as follows: $$x^a\cdot x^b = x^{a + b}.\tag2$$ This is provable from $(1)$. Since $x^a$ and $x^b$ are all products of $x$, then when we multiply them together, the number of times $x$ is being multiplied by itself in total is of course $a + b$ times. Therefore, the product of $x^a$ and $x^b$ is always $x^{a+b}$.
So this means that since $n = n + 0$, we get that $x^n = x^n\cdot x^0$. Therefore, $x^0$ must be equal to $1$. $$x^0 = 1.$$
Now, substitute $x = 10$, $a = 5$, and $b = -3$ in $(2)$. You should have, $$\begin{align} 10^5\times 10^{-3} &= 10^{5+(-3)} \\ &= 10^{5-3} \\ &= 10^2.\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2650477",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
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Let $(x,y) \in \Bbb R^+$ Prove that $\Bigr(1+\frac{1}{x}\Bigl)\Bigr(1+\frac{1}{y}\Bigl)\ge \Bigr(1+\frac{2}{x+y}\Bigl)^2$ Let $(x,y) \in \Bbb R^+$ Prove that $$\Bigr(1+\frac{1}{x}\Bigl)\Bigr(1+\frac{1}{y}\Bigl)\ge \Bigr(1+\frac{2}{x+y}\Bigl)^2$$
My try
Well, i didn't see a way to factorize this, so i put it in WolframAlpha and i got that we can rewrite it like this, which is obviously true for $(x,y) \in \Bbb R^+$:
$$\frac{(y-x)^2(x+y+1)}{xy(x+y)^2}\ge 0$$
But i don't see the way to get there manually, im stuck here:
$$\frac{(x+y)(y+1)}{xy}\ge \frac{(x+y+2)^2}{(x+y)^2}$$
| it is just $AM-GM$
multiplying out we get
$$(x+y)^2+\frac{(x+y)^2}{x}+\frac{(x+y)^2}{y}+\frac{(x+y)^2}{xy}\geq (x+y)^2+4+4(x+y)$$ and this is
$$xy(x+y)^2+y(x+y)^2+x(x+y)^2+(x+y)^2\geq xy(x+y)^2+4xy+4xy(x+y)$$
$$(x+y)^3+(x+y)^2\geq 4xy+4xy(x+y)$$
and this is $$(x+y)^2(x+y+1)\geq 4xy(x+y+1)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2650818",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 8,
"answer_id": 3
} |
Find the equation of the plane which bisects the pair of planes $2x-3y+6z+2=0$ and $2x+y-2z=4$ at acute angles.
Find the equation of the plane which bisects the pair of planes $2x-3y+6z+2=0$ and $2x+y-2z=4$ at acute angles.
My try:
The normal to the plane $2x-3y+6z+2=0$ has direction ratios $2,-3,6$ and the normal to the plane $2x+y-2z=4$ has direction ratios $2,1,-2$.
Hence the angle between them is $\cos \theta=\vert\dfrac{n_1.n_2}{|n_1||n_2|}\vert=\dfrac{11}{45}$.
How to find the equation of the required plane from here?
Please help.
| Let $(x,y,z)$ be a point in the needed plane.
Thus, $$\frac{|2x-3y+6z+2|}{\sqrt{2^2+(-3)^2+6^2}}=\frac{|2x+y-2z-4|}{\sqrt{2^2+1^2+(-2)^2}},$$
which gives equations of two our planes.
I got
$$x+2y-4z-\frac{17}{4}=0;$$
$$10x-y+2z-11=0.$$
Now, take $A(1,1,1)$, which is placed on the second plane.
Let $AB$ be a perpendicular from $A$ to the plane $2x+y-2z-4=0$ (easy to see that $AB=1$) and $AC$ be a perpendicular from $A$ to the intersection line of given planes.
If $\measuredangle ACB<45^{\circ}$ then $10x-y+2z-11=0$ is the answer.
While if $\measuredangle ACB>45^{\circ}$ then $x+2y-4z-\frac{17}{4}=0$ is the answer.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2653872",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Collinearity when $\mathbf{a} \times \mathbf{b} + \mathbf{b} \times \mathbf{c} + \mathbf{c} \times \mathbf{a} = \mathbf{0}$
Let $\mathbf{a} = \begin{pmatrix}x_a\\y_a\\z_a\end{pmatrix}$, $\mathbf{b} = \begin{pmatrix}x_b\\y_b\\z_b\end{pmatrix}$, and $\mathbf{c} = \begin{pmatrix}x_c\\y_c\\z_c\end{pmatrix}$.
Show that $(x_a,y_a,z_a)$, $(x_b,y_b,z_b)$, and $(x_c,y_c,z_c)$ are collinear if and only if
$\mathbf{a} \times \mathbf{b} + \mathbf{b} \times \mathbf{c} + \mathbf{c} \times \mathbf{a} = \mathbf{0}.$
Since the cross product of two vectors gives an area, and for two vectors to give an area of $0$ they need to be on the same line (or they can be a point, but I'm assuming both are not $\mathbf 0$). However, in this problem, each of the cross products need not necessarily be $0$ since it's their sum that is $0$, and now I'm not sure what to do.
| Using that $\,\color{blue}{\mathbf{b} \times \mathbf{c} = -\,\mathbf{c} \times \mathbf{b}}\,$ and $\,\color{red}{\mathbf{a} \times \mathbf{a} = 0}\,$:
$$
\begin{align}
\mathbf{a} \times \mathbf{b} + \mathbf{b} \times \mathbf{c} + \mathbf{c} \times \mathbf{a} = \mathbf{0} \;\;&\iff\;\; \mathbf{a} \times \mathbf{b} \color{blue}{- \mathbf{c} \times \mathbf{b}} + \mathbf{c} \times \mathbf{a} \color{red}{- \mathbf{a} \times \mathbf{a}} = \mathbf{0} \\
& \iff\;\; (\mathbf{a} - \mathbf{c}) \times \mathbf{b} + (\mathbf{c} - \mathbf{a}) \times \mathbf{a} = 0 \\
& \iff\;\; (\mathbf{a} - \mathbf{c}) \times \mathbf{b} - (\mathbf{a} - \mathbf{c}) \times \mathbf{a} = 0 \\
& \iff\;\; (\mathbf{a} - \mathbf{c}) \times (\mathbf{b} - \mathbf{a}) = 0
\end{align}
$$
The latter equality is equivalent to $\,\mathbf{a} - \mathbf{c}\,$ and $\, \mathbf{b} - \mathbf{a}\,$ being collinear.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2654228",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Simplify the equation $16\cos^6(t) + 16\sin^6(t) + 48\sin^2(t)\cos^2(t)$ I am trying to simplify the equation
$$16\cos^6(t) + 16\sin^6(t) + 48\sin^2(t)\cos^2(t)$$
The goal here is to prove that this can be simplified in a scalar value.
Thanks!
| $(a^2+b^2)^3 = a^6 + 3a^4b^2 + 3a^2b^2 + b^6 \Rightarrow$
$ a^6+b^6 = (a^2+b^2)^3 - 3a^2b^2(a^2+b^2)$
Substituting $a=\sin(x)$ and $b=\cos(x)$:
$ \sin^6(x)+\cos^6(x) = (\sin^2(x)+\cos^2(x))^3 - 3\sin^2(x)\cos^2(x)(\sin^2(x)+\cos^2(x)) = \\
1 - 3\sin^2(x)\cos^2(x) \Rightarrow$
$ 16\sin^6(x)+16\cos^6(x) = 16 - 48\sin^2(x)\cos^2(x) \Rightarrow$
$ 16\sin^6(x)+16\cos^6(x) + 48\sin^2(x)\cos^2(x) = 16$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2654480",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Determine what condition the $n\times n$ matrix $A$ must fulfill so that the matrix $I_n - A$ has an inverse and this equals $I_n + A$.
Determine what condition the $n\times n$ matrix $A$ must fulfill so that the matrix $I_n - A$ has an inverse and this equals $I_n + A$.
I tried to apply it to a $2\times 2$ matrix obtaining the following result:
$A =
\begin{pmatrix}
a & b\\
c & d
\end{pmatrix}; $
$(I_n-A)^{-1} = \frac{1}{(1-a)(1-d)-bc} *
\begin{pmatrix}
1-d & b\\
c & 1-a
\end{pmatrix} =
\begin{pmatrix}
1+a & b\\
c & 1+d
\end{pmatrix} $
| $(I - A)^{-1} = I + A \Longleftrightarrow A^2 = 0; \tag 1$
for if
$A^2 = 0, \tag 2$
then
$(I - A)(I + A) = I + A - A - A^2 = I - A^2 = I; \tag 3$
likewise if
$(I - A)(I + A) = I, \tag 4$
then
$I - A^2 = (I - A)(I + A) = I, \tag 5$
then
$I = A^2 + I, \tag 6$
or
$A^2 = 0. \tag 7$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2654554",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
Block Matrix Inversion in Wikipedia Wikipedia provides two formulas for block-matrix inversion:
$$ {\begin{bmatrix}\mathbf {A} &\mathbf {B} \\\mathbf {C} &\mathbf {D} \end{bmatrix}}^{-1}={\begin{bmatrix}\mathbf {A} ^{-1}+\mathbf {A} ^{-1}\mathbf {B} (\mathbf {D} -\mathbf {CA} ^{-1}\mathbf {B} )^{-1}\mathbf {CA} ^{-1}&-\mathbf {A} ^{-1}\mathbf {B} (\mathbf {D} -\mathbf {CA} ^{-1}\mathbf {B} )^{-1}\\-(\mathbf {D} -\mathbf {CA} ^{-1}\mathbf {B} )^{-1}\mathbf {CA} ^{-1}&(\mathbf {D} -\mathbf {CA} ^{-1}\mathbf {B} )^{-1}\end{bmatrix}},$$
and
$${\begin{bmatrix}\mathbf {A} &\mathbf {B} \\\mathbf {C} &\mathbf {D} \end{bmatrix}}^{-1}={\begin{bmatrix}(\mathbf {A} -\mathbf {BD} ^{-1}\mathbf {C} )^{-1}&-(\mathbf {A} -\mathbf {BD} ^{-1}\mathbf {C} )^{-1}\mathbf {BD} ^{-1}\\-\mathbf {D} ^{-1}\mathbf {C} (\mathbf {A} -\mathbf {BD} ^{-1}\mathbf {C} )^{-1}&\quad \mathbf {D} ^{-1}+\mathbf {D} ^{-1}\mathbf {C} (\mathbf {A} -\mathbf {BD} ^{-1}\mathbf {C} )^{-1}\mathbf {BD} ^{-1}\end{bmatrix}}.$$
Is it true then that all of the following equalities are true?
\begin{align}\mathbf {A} ^{-1}+\mathbf {A} ^{-1}\mathbf {B} (\mathbf {D} -\mathbf {CA} ^{-1}\mathbf {B} )^{-1}\mathbf {CA} ^{-1}=&\;(\mathbf {A} -\mathbf {BD} ^{-1}\mathbf {C} )^{-1}\\
-\mathbf {A} ^{-1}\mathbf {B} (\mathbf {D} -\mathbf {CA} ^{-1}\mathbf {B} )^{-1}=&\; -(\mathbf {A} -\mathbf {BD} ^{-1}\mathbf {C} )^{-1}\mathbf {BD} ^{-1} \\
-(\mathbf {D} -\mathbf {CA} ^{-1}\mathbf {B} )^{-1}\mathbf {CA} ^{-1}=&\;-\mathbf {D} ^{-1}\mathbf {C} (\mathbf {A} -\mathbf {BD} ^{-1}\mathbf {C} )^{-1} \\
(\mathbf {D} -\mathbf {CA} ^{-1}\mathbf {B} )^{-1}=&\; \quad \mathbf {D} ^{-1}+\mathbf {D} ^{-1}\mathbf {C} (\mathbf {A} -\mathbf {BD} ^{-1}\mathbf {C} )^{-1}\mathbf {BD} ^{-1}\end{align}
| Assuming $A,B, C, D$ are fixed.
By uniqueness of the inverse, yes, those expression are equal provided those terms involved indeed exists.
| {
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"url": "https://math.stackexchange.com/questions/2659082",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How to extract $(x+y+z)$ or $xyz$ from the determinant Prove
$$\color{blue}{
\Delta=\begin{vmatrix}
(y+z)^2&xy&zx\\
xy&(x+z)^2&yz\\
xz&yz&(x+y)^2
\end{vmatrix}=2xyz(x+y+z)^3}
$$
using elementary operations and the properties of the determinants without expanding.
My Attempt
$$
\Delta\stackrel{C_1\rightarrow C_1+C_2+C_3}{=}\begin{vmatrix}
xy+y^2+yz+zx+yz+z^2&xy&zx\\
x^2+xy+xz+xz+yz+z^2&(x+z)^2&yz\\
x^2+xy+xz+xy+y^2+yz&yz&(x+y)^2\\
\end{vmatrix}\\
=\begin{vmatrix}
y(x+y+z)+z(x+y+z)&xy&zx\\
x(x+y+z)+z(x+y+z)&(x+z)^2&yz\\
x(x+y+z)+y(x+y+z)&yz&(x+y)^2\\
\end{vmatrix}\\
=(x+y+z)\begin{vmatrix}
y+z&xy&zx\\
x+z&(x+z)^2&yz\\
x+y&yz&(x+y)^2\\
\end{vmatrix}\\
\stackrel{R_1\rightarrow R_1+R_2+R_3}{=} \\(x+y+z)\begin{vmatrix}
2(x+y+z)&x^2+xy+xz+xz+yz+z^2&x^2+xy+zx+xy+y^2+yz\\
x+z&(x+z)^2&yz\\
x+y&yz&(x+y)^2\\
\end{vmatrix}\\
=(x+y+z)^2\begin{vmatrix}
2&x+z&x+y\\
x+z&(x+z)^2&yz\\
x+y&yz&(x+y)^2\\
\end{vmatrix}\\
=2(x+y+z)^2\begin{vmatrix}
1&\frac{x+z}{2}&\frac{x+y}{2}\\
x+z&(x+z)^2&yz\\
x+y&yz&(x+y)^2
\end{vmatrix}
\stackrel{R_2\rightarrow R_2-(x+z)R_1|R_2\rightarrow R_3-(x+y)R_1}{=}
2(x+y+z)^2\begin{vmatrix}
1&\frac{x+z}{2}&\frac{x+y}{2}\\
0&\frac{(x+z)^2}{2}&\frac{yz-x^2-xy-xz}{2}\\
0&\frac{yz-x^2-xy-xz}{2}&\frac{(x+y)^2}{2}
\end{vmatrix}
$$
How do I extract $x+y+z$ or $xyz$ from the determinant to find the solution ?
Note: In a similar problem How to solve this determinant there seems to be solutions talking about factor theorem and polynomials. I am basically looking for extracting the terms $(x+y+z)$ or $xyz$ from the given determinant to solve it only using the basic properties of determinants.
| $$
\Delta=\begin{vmatrix}
(y+z)^2&xy&zx\\
xy&(x+z)^2&yz\\
xz&yz&(x+y)^2
\end{vmatrix}=xyz\begin{vmatrix}
\frac{(y+z)^2}{x}&x&x\\
y&\frac{(x+z)^2}{y}&y\\
z&z&\frac{(x+y)^2}{z}
\end{vmatrix}=
xyz\begin{vmatrix}
\frac{(y+z)^2}{x}-x&x&0\\
y-\frac{(x+z)^2}{y}&\frac{(x+z)^2}{y}&y-\frac{(x+z)^2}{y}\\
0&z&\frac{(x+y)^2}{z}-z
\end{vmatrix}=
xyz\begin{vmatrix}
\frac{(y+z)^2-x^2}{x}&x&0\\
\frac{y^2-(x+z)^2}{y}&\frac{(x+z)^2}{y}&\frac{y^2-(x+z)^2}{y}\\
0&z&\frac{(x+y)^2}{z}
\end{vmatrix}=
xyz\begin{vmatrix}
\frac{[(y+z)-x][(y+z)+x]}{x}&x&0\\
\frac{[y-(x+z)][y+(x+z)]}{y}&\frac{(x+z)^2}{y}&\frac{[y-(x+z)][y+(x+z)]}{y}\\
0&z&\frac{[(x+y)-z][(x+y)+z]}{z}
\end{vmatrix}=
xyz(x+y+z)^2\begin{vmatrix}
\frac{[(y+z)-x]}{x}&x&0\\
\frac{[y-(x+z)]}{y}&\frac{(x+z)^2}{y}&\frac{[y-(x+z)]}{y}\\
0&z&\frac{[(x+y)-z]}{z}
\end{vmatrix}=
(x+y+z)^2\begin{vmatrix}
{[(y+z)-x]}&x^2&0\\
{[y-(x+z)]}&{(x+z)^2}&{[y-(x+z)]}\\
0&z^2&{[(x+y)-z]}
\end{vmatrix}
=(x+y+z)^2\begin{vmatrix}
{[(y+z)-x]}&x^2&0\\
{[y-(x+z)]}&{(x+z)^2}&{[y-(x+z)]}\\
0&z^2&{[(x+y)-z]}
\end{vmatrix}=
(x+y+z)^2\begin{vmatrix}
{[(y+z)-x]}&x^2&0\\
-2z&2xz&-2x\\
0&z^2&{[(x+y)-z]}
\end{vmatrix}=
(x+y+z)^2\begin{vmatrix}
{(y+z)}&x^2&\frac{x^2}{z}\\
0&2xz&0\\
\frac{z^2}{x}&z^2&{(x+y)}
\end{vmatrix}\\
=(x+y+z)^2.2xz.\Big[(x+y)(y+z)-\frac{x^2.z^2}{xz}\Big]\\
=(x+y+z)^2(2xz)(xy+xz+y^2+xz-xz)\\
=(x+y+z)^2(2xz)(xy+y^2+xz)\\=(x+y+z)^2.2xz.y(x+y+z)=2xyz(x+y+z)^2
$$
| {
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"url": "https://math.stackexchange.com/questions/2659331",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How many ways are there to get a sum of $25$ when $10$ distinct dice are rolled? I have come across the following problem and have given it a good attempt below. I am wondering if I have proceeded correctly, and if not if someone could show me the correct answer or maybe a more efficient solution, thanks!
How many ways are there to get a sum of $25$ when $10$ distinct dice are rolled?
Each die can be a number from $1$ to $6$. With $10$ dice, our generating function becomes:
$$g(x) = (x+x^2+x^3+x^4+x^5+x^6)^{10}$$
We want to find the coefficient of $x^{25}$. Observe that the expression inside the brackets is a finite geometric series with $a=x, r=x, n=6$. Thus we have
$$(x+x^2+x^3+x^4+x^5+x^6)^{10}$$
$$=\left(\frac{x(1-x^6)}{(1-x)}\right)^{10}$$
$$=x^{10}(1-x^6)^{10}(1-x)^{-10}$$
Then,
$$(1-x^6)^{10}(1-x)^{-10}$$
$$=\sum \binom{10}{i}(-x^6)^i \cdot\sum\binom{-10}{j}(-x)^j$$
In order to get terms that involve $x^{15}$ there are $3$ combinations of $i$ and $j$ to consider so that $6i+j = 15$
Thus we have:
$$\left[ \binom{10}{0}(-1)^0\binom{-10}{15}(-1)^{15}\right] + \left[ \binom{10}{1}(-1)^1 \binom{-10}{9}(-1)^9\right] + \left[\binom{10}{2}(-1)^2\binom{-10}{3}(-1)^3\right]$$
$$ = \left[\binom{10}{0}\binom{-10}{15}(-1)\right]+\left[ \binom{10}{1} \binom{-10}{9} \right] + \left[ \binom{10}{2} \binom{-10}{3}(-1)\right] $$
$$=\left[ \binom{10}{0}\binom{10+15-1}{15}(-1)^{15}(-1)\right] + \left[ \binom{10}{1} \binom{10+9-1}{9}(-1)^9\right] + \left[ \binom{10}{2}\binom{10+3-1}{3}(-1)^3(-1)\right]$$
$$= \left[ \binom{10}{0}\binom{24}{15}\right] - \left[ \binom{10}{1}\binom{18}{9}\right] + \left[ \binom{10}{2}\binom{12}{3}\right]$$
$$= 831204$$
| Can't comment yet so adding here: I haven't checked the math, but just for fun https://ideone.com/IKt8OR
| {
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Why does this innovative method of subtraction from a third grader always work? My daughter is in year $3$ and she is now working on subtraction up to $1000.$ She came up with a way of solving her simple sums that we (her parents) and her teachers can't understand.
Here is an example: $61-17$
Instead of borrowing, making it $50+11-17,$ and then doing what she was told in school $11-7=4,$ $50-10=40 \Longrightarrow 40+4=44,$ she does the following:
Units of the subtrahend minus units of the minuend $=7-1=6$
Then tens of the minuend minus tens of the subtrahend $=60-10=50$
Finally she subtracts the first result from the second $=50-6=44$
As it is against the first rule children learn in school regarding subtraction (subtrahend minus minuend, as they cannot invert the numbers in subtraction as they can in addition), how is it possible that this method always works? I have a medical background and am baffled with this…
Could someone explain it to me please? Her teachers are not keen on accepting this way when it comes to marking her exams.
| My way of explaining it:
\begin{align*}
61−17 &= (6 \cdot 10 + 1 \cdot 1) - (1 \cdot 10 + 7 \cdot 1) \\
&= (6 - 1) \cdot 10 + (1-7) \cdot 1 \\
&= (6 - 1) \cdot 10 - (7-1) \cdot 1 \\
&= 5 \cdot 10 - 6 \cdot 1 \\
&= 50 - 6 = 44 .
\end{align*}
On the one hand, you could generalise:
\begin{align*}
ABCD \ldots FG - abcd \ldots fg &= (A \cdot 10^\alpha + B \cdot 10^\beta + \ldots + F \cdot 10 + G) - (a \cdot 10^\alpha + b \cdot 10^\beta + \ldots + f \cdot 10 + g) \\
&= (A - a) \cdot 10^\alpha + (B - b) \cdot 10^\beta + \ldots + (F - f) \cdot 10 + (G-g) ,
\end{align*}
and then rearrange to adjust signs.
On the other hand, you can be flexible :-) :
\begin{align*}
61−17 &= (50 + 5 + 6) - (10 + 6 + 1) = 50 - 10 + 5 + 6 - 6 - 1 = 40 + 5 - 1 = 44\\
61−17 &= (17+3+40+1) - 17 = 3 + 40 + 1 = 44 \\
61−17 &= (99-38) - (34-17) = 99-38-34+17 = 99-72+17 = 27+17=44 \\
\ldots&
\end{align*}
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Proving $\frac{a}{a^2-a+1}+\frac{b}{b^2-b+1}+\frac{c}{c^2-c+1}+\frac{d}{d^2-d+1}\le \frac{8}{3}$ given $a+b+c+d=2$
Let $a,b,c,d\in \mathbb{R}$ and $a+b+c+d=2$. Prove that $$\frac{a}{a^2-a+1}+\frac{b}{b^2-b+1}+\frac{c}{c^2-c+1}+\frac{d}{d^2-d+1}\le \frac{8}{3}.$$
We have $$\frac{a}{a^2-a+1}\le \frac{4}{3}a\Longleftrightarrow \frac{-a\left(2a-1\right)^2}{3\left(a^2-a+1\right)}\le 0,$$ (Right)
So $$LHS\le \frac{4}{3}\left(a+b+c+d\right)=\frac{8}{3}=RHS$$
When $a=b=c=d=\frac{1}{2}$
But also $a=b=c=1;d=-1$ i don't know how to solve it.
| First, we simplify a bit by transforming $x = a-\frac12, y = b - \frac12, z = c - \frac12, w = d - \frac12$, so that we need to prove equivalently with $x+y+z+w=0$, the inequality
$$\sum \frac{2x+1}{4x^2+3} \leqslant \frac43$$
Here $\sum$ represents cyclic sums. Also as $\displaystyle 1 - 2\cdot\frac{2x+1}{4x^2+3} = \frac{(2x-1)^2}{4x^2+3}$, we may equivalently show
$$\sum \frac{(2x-1)^2}{4x^2+3} \geqslant \frac43$$
Now we need an upper bound on the denominator; for this, note
$$4x^2+3=3x^2+(y+z+w)^2+3\leqslant 3x^2+3(y^2+z^2+w^2)+3 = 3\sum x^2+3$$
Thus we have a common denominator now, therefore,
$$\sum \frac{(2x-1)^2}{4x^2+3} \geqslant \frac{\sum (2x-1)^2}{3\sum x^2+3} = \frac{4\sum x^2-4\sum x+4}{3(\sum x^2+1)}=\frac43$$
| {
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Proof Verification: Prove $f(n)<2$, when $f(1)=1$ and $f(n+1) = \sqrt{(2+f(n))}$ for all $n$ positive integers Proceed by induction:
Suppose $P(n)$ is the statement $f(n)<2$, when $n$ $f(1)=1$ and $f(n+1) = \sqrt{(2+f(n))}$ for all $n$ positive integers.
Base case: $P(1)=f(1)=1<2$. True.
Assume $P(n)$, i.e. $f(n)<2$ when $n$ $f(1)=1$ and $f(n+1) = \sqrt{(2+f(n))}$ for all $n$ positive integers, is true.Then $P(n+1)=\sqrt{(2+f(n))}< \sqrt{(2+2)}=2$. QED
If this is a correct proof, can you please explain how the step:$P(n+1)=\sqrt{(2+f(n))}<\sqrt{(2+2)}=2$ works. Namely, how can we substitute $f(n)$ with $2$.
Thank you!
| Alt. hint: obviously $f(n) \ge 0$ for all $\,n\,$, then:
$$
\begin{align}
f(n+1) = \sqrt{f(n)+2} \quad&\iff\quad \left(f(n+1)\right)^2 = f(n)+2 \\
&\iff\quad \left(f(n+1)\right)^2 -4 = f(n) - 2 \\
&\iff\quad \big(f(n+1)-2\big)\underbrace{\big(f(n+1)+2\big)}_{\gt 0}=f(n)-2
\end{align}
$$
It follows that all differences $f(n)-2$ have the same sign, and since $f(1) \lt 2$ they are all negative.
| {
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Let $a^2<2, b=2(a+1)/(a+2)$. Show $b^2<2$ (assignment) It is a part of my assignment.
$$ \text {Let }a^2<2, \quad b=2\frac {(a+1)}{(a+2)}\quad \text{ Show } b^2<2$$
I already proved that a
But, I am struggling to prove $b^2<2$.
My lecturer said that I need to manipulate $b^2$, which is larger than $b^2$ but less than 2. That is $4(a+1)^2/(a+2)^2$ < some number < 2.
I was trying this for hours, but I couldn't find the way to solve. Thanks for helping in advance.
| Hint: $\require{cancel}\;b^2 - 2 = \dfrac{4(a+1)^2}{(a+2)^2}-2=\dfrac{4a^2+\cancel{8a}+4-2a^2-\cancel{8a}-8}{(a+2)^2}=\dfrac{2a^2-4}{(a+2)^2}=\dfrac{2(a^2-2)}{(a+2)^2}\,$
| {
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Find the base of the kernel of $L(y)=x^2y''-3xy'+3y.$
Let $L:C^2(I)\rightarrow C(I), L(y)=x^2y''-3xy'+3y.$ Find the kernel of the linear transformation $L$. Can the solution of $L(y)=6$ be expressed in the form $y_H$+$y_L$, where $y_H$ is an arbitrary linear combination of the elements of ker L.
What I have tried:
Since ker L is subspace of $C^2(I)$ and dim $C^2(I)=2$, dim ker $L\leq 2$.
Let $y(x)=x^r$. Then substituting gives $L(x^r)=x^rr^2-3rx^r+3x^r$, hence $L(x^r)=0$ iff $x^rr^2-3rx^r+3x^r=0.$ $r$ can be solved using the quadratic formula.
$$r=\frac{3+i\sqrt3}{2} \vee r=\frac{3-i\sqrt3}{2}$$
$$y_1(x)=x^\frac{3+i\sqrt3}{2}, y_1(x)=x^\frac{3-i\sqrt3}{2} $$which are linearly independent(?).
How would one show that $y_1$ and $y_2$ are LI?
| Another way
$$x^2y''-3xy'+3y=0$$
For $x \neq 0$
$$y''-3 \left (\frac {xy'-y}{x^2} \right )=0$$
$$y''-3\left (\frac {y}{x}\right )'=0$$
$$y'-3 \left (\frac {y}{x} \right )=K_1$$
$$x^3y'-3{y}{x^2}=K_1x^3$$
$$\frac {x^3y'-3{y}{x^2}}{x^6}=\frac {K_1}{x^3}$$
$$\left (\frac {y}{x^3} \right )'=\frac {K_1}{x^3}$$
$$\frac {y}{x^3}=K_1\int\frac {dx}{x^3}=\frac {K_1}{x^2}+K_2$$
$$\boxed {y=K_1x+K_2x^3}$$
Both functions $(x,x^3)$ are linearly independant.
Show that $ \forall x \,\, c_1x+c_2x^3=0 \implies c_1=c_2=0$
| {
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Solve the reccurence relation $a_n = 3a_{n-1}+4_{n-2}, a_0=a_1 = 1$ This is my first time working through this type of problem and I am looking to see if I have worked it out correctly, thanks!
Solve the reccurence relation $$a_n = 3a_{n-1}+4a_{n-2}, a_0=a_1 = 1$$
First we get the characteristic equation:
$$x^n = 3x^{n-1}+4x^{n-2}$$
Dividing by the smallest we get,
$$x^2=3x+4$$
$$x^2-3x-4 = 0 $$
$$x = 4,-1$$
Using the auxiliary equation we get the general solution:
$$a_n = A_1x^{n+1} + A_2x^{n+1}$$
Using conditions in the equation and our solution for $x$ gives,
$$a_0 = 1 = A_1(4) + A_2(-1) = 4A_1 - A_2$$
$$a_1 = 1 = A_1(4)^2 + A_2(-1)^2 = 16A_1 + A_2$$
Solving for $A_1, A_2$ I got $A_1 = \frac{1}{10}, A_2 = -\frac{3}{5}$
Therefore plugging into the general solution I got,
$$a_n =\left(\frac{1}{10}\right)(4)^{n+1}+\left(-\frac{3}{5}\right)(-1)^{n+1}$$
Did my workings come out correct?
| Yes it is correct. We can check it with the use of generating functions.
Transform $a_n = 3a_{n-1} + 4a_{n-2}$ to $$a_{n+2} = 3a_{n+1} + 4a_n \tag{1}$$.
Let $A(x) := \sum_{n \geq 0} a_n x^n$ then we get from (1)
$$
\begin{eqnarray*}
\sum_{n \geq 0} a_{n+2} x^n &=& 3 \sum_{n \geq 0} a_{n+1} x^n + 4 \sum_{n \geq 0} a_n x^n \\
\Leftrightarrow \frac{A(x)}{x^2} - \frac{1}{x^2} - \frac{1}{x} &=&
3\Big(\frac{A(x)}{x} - \frac{1}{x}\Big) + 4 A(x) \\
\Leftrightarrow A(x) -1 -x &=& 3A(x)x-3x + 4A(x)x^2 \\
\Leftrightarrow A(x) &=& \frac{2x-1}{4x^2+3x-1}
\end{eqnarray*}
$$
With partial fractions we further obtain that $A(x) = \frac{3}{5(x+1)} + \frac{2}{5(1-4x)}$
and in series representation this is equivalent to
$$
A(x) = \sum_{n \geq 0} x^n \Big(\frac{1}{5} 3(-1)^n + \frac{2}{5} 4^n \Big)
$$
If we look now at the coefficient of $[x^n]$ in $A(x)$ we see that
$$
a_n = \Big(\frac{1}{5} 3(-1)^n + \frac{2}{5} 4^n \Big)
$$
| {
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Solving a system of equations: I am having problems solving a system of equations I haven't run into a system quite like this before and I am looking for some help, thanks!
$$a(1+\sqrt3)^2+b(1-\sqrt3)^2=3$$ $$a(1+\sqrt3)^3+b(1-\sqrt3)^3=8$$
How can I solve for $a$ and $b$?
| Write it as a system $Ax=v$ with $v=(3,8)^T$, $x=(a,b)^T$ and
$$
A=\begin{pmatrix} (1+\sqrt{3})^2 & (1-\sqrt{3})^2 \cr
(1+\sqrt{3})^3 & (1-\sqrt{3})^3\end{pmatrix}
$$
Since $\det(A)=-8\sqrt{3}\neq 0$, the inverse $A^{-1}$ exists, so the unique solution is
$$
\begin{pmatrix} a \cr b \end{pmatrix}=x=A^{-1}v=\frac{1}{4\sqrt{3}}\begin{pmatrix} 9-5\sqrt{3} \cr 15+9\sqrt{3} \end{pmatrix}
$$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Symmetric Olympiad inequality Prove that for all positive reals $a, b, c$ this inequality holds:
$$\sum_{cyc}\frac{a^3}{b^2 - bc + c^2} \ge \sum_{cyc}a$$
I have proved this in a very ugly way:
I multiplied with $(a^2 - ab + b^2)(b^2 - bc + c^2)(c^2 - ca + a^2)$ and with some work i managed to prove this with Schur. I find this way boring and time-consuming. My question is - Is there a nicer solution to this?
Source: Mildorf Inequalities
| Note by CS (with $\sum$ denoting cyclic sums):
$$ \sum \frac{a^3}{b^2-bc+c^2} \geqslant \frac{\left(\sum a^2 \right)^2}{\sum a(b^2-bc+c^2)}$$
Hence it is enough to show
$$\left(\sum a^2 \right)^2 \geqslant \left(\sum a\right) \cdot \sum a(b^2-bc+c^2)$$
$$\iff \sum a^4 + 2\sum a^2b^2 \geqslant \sum (2a^2b^2-a^2bc + b^3a+ ac^3)$$
which is just the fourth degree Schur inequality:
$$\sum a^4 + abc\sum a \geqslant \sum ab(a^2+b^2) $$
Equality is when $a=b=c$, or when any one variable is $0$ and the other two equal.
| {
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"timestamp": "2023-03-29T00:00:00",
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Integral of arctan form I am trying to solve this integral without techniques ($u$-sub/parts), just simplifying and inspection:
$$
\int \dfrac{1}{4+x^{2}}dx
$$
I notice an $\arctan$ form, but that $4$ in the denominator is confusing me, if there was a multiplying factor in front of the $x$ I would know what to do, but stumped on this case.
| One can take a different route with the following. Let $x = 2 t$ to obtain
\begin{align}
I &= \int \frac{dx}{4 + x^2} \\
&= \frac{1}{2} \, \int\frac{dt}{1 + t^2} = \frac{1}{4} \, \int\left(\frac{1}{1 + i t} + \frac{1}{1 - i t} \right) \, dt \\
&= \frac{1}{4} \, \left[ \frac{1}{i} \, \ln(1 + i t) - \frac{1}{i} \, \ln(1-i t) \right] + c_{1} \\
& = \frac{1}{4i} \, \ln\left(\frac{1+i t}{1- i t}\right) + c_{1}.
\end{align}
Now using
$$\tan^{-1}(y) = \frac{1}{2i} \, \ln\left(\frac{1+ i y}{1 - i y}\right)$$
then the integral becomes, after back substitution,
$$\int \frac{dx}{4 + x^2} = \frac{1}{2} \, \tan^{-1}\left(\frac{x}{2}\right) + c_{1} = \frac{1}{4i} \, \ln\left(\frac{2 +i x}{2 - i x}\right) + c_{0}.$$
| {
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Proving $\frac{\sqrt3\cos x-\sin x}{\sin 3x}> \frac{\sqrt3}{3x}-\frac13$ for small $x>0$
Prove that$$\frac{\sqrt3\cos x-\sin x}{\sin 3x}> \frac{\sqrt3}{3x}-\frac13$$
for small $x$ near $0$.
From Taylor expansion I can see that $$\frac{\cos x}{\sin3x}>\frac13x,\quad \frac{\sin x}{\sin3x}>\frac13,$$ but combining these two does not give actually what I want. I kindly thank you anyone that can provide at least an idea. I know this is true graphically but I want to see rigorously.
| Using the identity from @labbhattacharjee, for $0 < x < \dfrac{π}{3}$ there is\begin{align*}
0 < \frac{\sin 3x}{\sqrt{3}\cos x - \sin x} &= \frac{1}{2} \frac{\sin\left( 3\left( \dfrac{π}{3} - x \right) \right)}{\sin\left( \dfrac{π}{3} - x \right)} = \frac{1}{2} \left( 3 - 4\sin^2\left( \dfrac{π}{3} - x \right) \right)\\
&= \frac{1}{2} \left( 1 + 2\cos\left( \dfrac{2π}{3} - 2x \right) \right) = \frac{1}{2} (1 - \cos 2x + \sqrt{3} \sin 2x)\\
&= \sin^2 x + \sqrt{3} \sin x \cos x = \sin x (\sin x + \sqrt{3} \cos x)\\
&< x(x + \sqrt{3}) = x \cdot \frac{3 - x^2}{\sqrt{3} - x} < \frac{3x}{\sqrt{3} - x},
\end{align*}
thus$$
\frac{\sqrt{3}\cos x - \sin x}{\sin 3x} > \frac{\sqrt{3} - x}{3x} = \frac{1}{\sqrt{3} x} - \frac{1}{3}.
$$
| {
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"timestamp": "2023-03-29T00:00:00",
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$x^2+y^2+z^2 = 4, z \geq 4.$
Let $\mathbf{F}=3y \ \mathbf{i} -3xz \ \mathbf{j} + (x^2-y^2) \
\mathbf{k}.$ Compute the flux of the vectorfield
$\text{curl}(\mathbf{F})$ through the semi-sphere $x^2+y^2+z^2=4, \
z\geq 0$, by using direct parameterization of the surface and computation of $\text{curl}(\mathbf{F}).$
Denote the semisphere by $S$. In the book they start off by noting that
$$\iint_S xy \ dS = \iint_S xz \ dS = \iint_S yz \ dS = 0 \quad (1)\\
\iint_S x^2 \ dS = \iint_S y^2 \ dS = \iint_S z^2 \ dS \quad \quad \quad (2)\\
$$
They then proceed computing the curl of the field, which is $(2x-2y,-2x,-2z-3)$ and the normal of the surface, out from the surface is $\mathbf{N}=\frac{1}{2}(x,y,z).$ So
$$\text{curl}(\mathbf{F})\cdot \mathbf{N} = \frac{1}{2}(2x^2-4xy-2z^2-3z).$$
Using the symmetries (1) and (2) we get the integral
$$\frac{1}{2}\iint_S-3z \ dS.$$
They then proceed to compute $dS$ by arguing that: $S$ is a part of the implicitly defined surface $F(x,y,z)=4$, where $F(x,y,z)=x^2+y^2+z^2,$ so
$$dS=\left|\frac{\nabla F}{F_z}\right|dxdy = \left|\frac{(2x,2y,2z)}{2z}\right| \ dxdy= \left|\frac{(x,y,z)}{z}\right| \ dxdy=\frac{2}{z} \ dxdy \quad (3).$$
Question: Can someone intuitively, and with other examples, explain the symmetry properties (1), (2) and why equation (3) holds and what it says?
| Note that
*
*$(1)$ hold since $xy,xz,yz$ are antisymmetric function on the domain, indeed, for example, at each point xy in the I quadrant correspond -xy in th II and so on
*$(2)$ hold since $x^2,y^2,z^2$ are symmetric function on the domain and each single integral for $x^2,y^2,z^2$ must assume the same value for every semisphere
| {
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If $a+b+c<0$ and $ax^2+bx+c=0$ has no real roots, is it true that $c$ must be less than $0$? I decided to look at the graphs of the parabola to solve this problem. These are only two types that will fit this problem: parabolas that "open" upwards and parabolas that "open" downwards. These parabolas would never touch the $x$-axis, and this rules out the upwards opening parabola because $a+b+c>0$. However, I am not sure how to continue from here.
| By the quadratic formula a parabola has roots at $x = \frac {-b \pm \sqrt{b^2 - 4ac}}{2a}$ if such is a real number.
The only way for there not to be roots is if $b^2 - 4ac < 0$. Or if $b^2 < 4ac$.
If $c > 0$ then for this to happen we must have $0 < b^2 <4ac$ so $a > 0$ as well.
But $a+b+c < 0$ so $b$ must be negative. "More negative than $a,c$ are positive". In other words $b < -(a+c)<0$ or $|b| > |a+c|$.
So $b^2 > (a+c)^2 = a^2 + 2ac + c^2$
Now by AM-GM $a^2 + c^2 \ge 2 \sqrt{a^2c^2} = 2ac$
So $b^2 > a^2 + 2ac + c^2 > 4ac$ and thus real roots exist.
...
To get a grasp geometrically.
$ax^2 + bx + c = a(x^2 + \frac ba x) + c= a(x^2 + \frac ba x + \frac {b^2}{4a^2}) + c - \frac {b^2}{4a^2}) = a(x + \frac b{2a})^2 + c - \frac {b^2}{4a})$.
$a$ tells whether the parabola "goes up" or "goes down". $c - \frac {b^2}{4a})$ is the y value of the tip of the parabola.
to Not have roots either the tip is above the $x$-axis and it points up ($c > \frac {b^2}{4a}; a > 0$. Or the tip is below the $x$-axis and it points down ($c < \frac {b^2}{4a}; a < 0$).
In second case if $a < 0$ then $c< \frac {b^2}{4a} < 0$.
If the first case $c >0$ and $4ac > b^2$. ... which puts us where we were above.
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Finding $ a $ where $ P(x)-a $ divides with $ (x+1)^4 $ and $ P(x)+a $ divides with $ (x-1)^4 $ Let $ P(x) $ be a $ 7 $ degree polynomial with the coefficient of $ x^7 $ equal to $ 1 $. Let $ a \in\mathbb{R} $ such that $ P(x)-a $ divides with $ (x+1)^4 $ and $ P(x)+a $ divides with $ (x-1)^4 $
$ 1 ) $ Find the coefficient of $ x^5 $
The answer should be $ -\frac{21}{5} $
$ 2 ) $ Find $ a $
The answer should be $ \frac{16}{5} $
| $(x+1)^4 \mid (P(x)-a) \implies P(x) = f(x)(x+1)^4+a \implies (x+1)^3 \mid P'(x)$.
Similarly we have $(x-1)^3 \mid P'(x)$, so putting these together with $P(x)$ being monic of seventh degree, $P'(x) = 7(x^2-1)^3$.
$\implies P(x) = x^7 -\frac{21}5x^5+7x^3-7x + C$,
Now $P(1) = -a, P(-1) = a \implies C-\frac{16}5=-a, C + \frac{16}5=a \implies C=0, a = \frac{16}5$.
| {
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"url": "https://math.stackexchange.com/questions/2685002",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Find the largest cylinder inscribed inside a sphere. Is this calculation correct so far?
A right circular cylinder is inscribed in a sphere of radius $r$. Find the largest possible volume of such a cylinder.
I have that the radius of the cylinder is $r$, the radius of the sphere is $R$, and the height of the inscribed cylinder is $h$. So
$$(2r)^2+h^2=(2R)^2$$
$$4r^2 + h^2 = 4R^2$$
$$h=2\sqrt{R^2-r^2}$$
So volume of the cylinder is:
$$V=2\pi r^2\sqrt{R^2-r^2}$$
$$V'=4\pi r\sqrt{R^2-r^2}+2\pi r^2\frac d{dr}\sqrt{R^2-r^2}$$
And I think:
$$\frac d{dr}\sqrt{R^2-r^2}=\frac r{\sqrt{R^2-r^2}}$$
so
$$V'=4\pi r\sqrt{R^2-r^2}+\frac{2\pi r^3}{\sqrt{R^2-r^2}}$$
$$=4\pi r(R^2-r^2)+2\pi r^3$$
$$4\pi rR^2-4\pi r^3+2\pi r^3=4\pi rR^2-2\pi r^3$$
So finding critical values:
$$4\pi r R^2-2\pi r^3=0$$
$$4\pi r R^2=2\pi r^3$$
$$4\pi R^2=2\pi r^2$$
$$\frac{4\pi R^2}{2\pi}=r^2$$
$$r=\sqrt2R$$
Someone else who is better at math is getting $r=\sqrt{\frac23}R$. Where did I go wrong?
| You might enjoy the fact that you actually do not need derivatives. By the AM-GM inequality
$$V^2=16\pi^2\cdot \frac{r^2}{2}\cdot \frac{r^2}{2}\cdot(R^2-r^2)\leq 16\pi^2\left(\frac{R^2}{3}\right)^3 $$
i.e. $V\leq \frac{4\pi R^3}{3\sqrt{3}}$, with equality attained at $\frac{r^2}{2}=R^2-r^2$, i.e. at $r=R\sqrt{\frac{2}{3}}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2685819",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Yet another matrix equation
Solve the following matrix equation for $X$.
$$\left[\begin{array}{cc}
5 &-8\cr
8 &1
\end{array}\right] X + \left[\begin{array}{cc}
6 &6\cr
3 &5
\end{array}\right] = \left[\begin{array}{cc}
-1 &4\cr
-3 &-1
\end{array}\right] X$$
Please give me some hint to do this question. Thanks.
| This is a linear equation in four variables.
Let $X = \begin{bmatrix} x_{11} & x_{12} \\ x_{21} & x_{22} \end{bmatrix}$.
Consider each entry of the RHS and LHS as a seperate equation:
$$\begin{align}
5x_{11} - 8x_{21} + 6 &= -1x_{11}+4x_{12} & \text{(top left entry)}\\
5x_{12} - 8x_{22} + 6 &= -1x_{12}+4x_{22} & \text{(top right entry)}\\
8x_{11} - 1x_{21} + 3 &= -3x_{11}-1x_{12} & \text{(bottom left entry)}\\
8x_{12} - 1x_{22} + 5 &= -3x_{12}-1x_{22} & \text{(bottom right entry)}\\
\end{align}$$
Now you get a linear system of four equations wich can be solved using the usual methods.
As an alternative you can directly use matrix operations and solve
$$\left[\begin{array}{cc}
5 &-8\cr
8 &1
\end{array}\right] X + \left[\begin{array}{cc}
6 &6\cr
3 &5
\end{array}\right] = \left[\begin{array}{cc}
-1 &4\cr
-3 &-1
\end{array}\right] X$$
iff
$$\left[\begin{array}{cc}
6 &6\cr
3 &5
\end{array}\right] = \left[\begin{array}{cc}
-1 &4\cr
-3 &-1
\end{array}\right] X - \left[\begin{array}{cc}
5 &-8\cr
8 &1
\end{array}\right] X =\left[\begin{array}{cc}
-6 &12\cr
-11 &-2
\end{array}\right]X $$
iff
$$\left[\begin{array}{cc}
-6 &12\cr
-11 &-2
\end{array}\right]^{-1}\left[\begin{array}{cc}
6 &6\cr
3 &5
\end{array}\right] =X $$
(Note that the matrix that we are trying to invert is actually invertible since its determinant is $(-6)(-2)-(-11)12 = 144 \neq 0$.)
iff
$$X = \frac{1}{12}\left[\begin{array}{cc}
-4 & -6\cr
4 &3
\end{array}\right] $$
| {
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"timestamp": "2023-03-29T00:00:00",
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What are the maximum and minimum values of $4x + y^2$ subject to $2x^2 + y^2 = 4$? $$ 2x^2 + y^2 = 4 $$
$$ Y = \sqrt{4-2x^2} $$
$$4x + y = 2x^2 + \sqrt{4-2x^2}$$
Find the derivative of $$ 2x^2 + \sqrt{4-2x^2} $$ set as = 0
$$X^2 = 64/33$$
$$ F(64/33) = 34\sqrt{33}/33 $$
How to solve it the right way?
| Note that $y^2=-2x^2+4$
When performing substitution of $4x+y^2 \rightarrow 4x-2x^2+4$
Finding the vertex (maxima):
$$-2x^2+4x+4=-2(x^2-2x)+4=-2[(x-1)^2-1]+4=-2(x-1)^2+6$$
Which presents the relative(and only) maxima.
Note that there is no relative minima, as the parabola goes downwards.
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "2",
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Primes congruent to $1$ modulo $6$ are of the form $3a^2+b^2$ I am asked to prove that primes of the form $6k+1$, where $k$ is an integer can be written as $3a^2 + b^2$, where $a$ and $b$ are integers by using the fact that $\textbf Z[\omega]$ is a UFD, where $\omega^2 + \omega + 1 = 0$. However, I cannot see any connection. How can one connect these two seemingly unrelated facts? Thanks in advance.
| If $p \equiv 1 \pmod 6$ then $\left(\frac{-3}{p}\right) = \left(\frac{-1}{p}\right)\left(\frac{3}{p}\right) = (-1)^{\frac{p-1}{2}}\left(\frac{p}{3}\right)(-1)^{\frac{p-1}{2}\frac{3-1}{2}} = \left(\frac{p}{3}\right)=1$ so $-3$ is a perfect square mod $p$.
Therefore there exists a solution to the equation $x^2 + 3 = 0$ in $\mathbb{F}_p$.
Since $\omega = \frac{1 + \sqrt{-3}}{2}$, the polynomial $x^2 + x + 1$ will also have a root in $\mathbb{F}_p$.
By the third isomorphism theorem we know $\mathbb{F}_p[x]/\langle x^2 + x + 1 \rangle \simeq \mathbb{Z}[x]/\langle p, x^2 + x + 1 \rangle \simeq \mathbb{Z}[\omega]/(p)$.
Since $x^2 + x + 1$ is not irreducible in $\mathbb{F}_p$, this means $\mathbb{F}_p[x]/\langle x^2 + x + 1 \rangle$ is not an integral domain, and therefore $\mathbb{Z}[\omega]/(p)$ is not an integral domain either.
Therefore $(p)$ is not a prime ideal in $\mathbb{Z}[\omega]$ and hence not maximal either, and since $\mathbb{Z}[\omega]$ is a Dedekind domain and a UFD (and hence a PID), this means $(p)$ is properly contained in some ideal $(a + b\omega) \neq (1)$ and so $p$ is divisible by the non-unit $a + b\omega$ and $p$ is not an irreducible element in $\mathbb{Z}[\omega]$.
Therefore there exists a nontrivial factorization $p = u(a + b\omega)(c + d\omega)$, where $u$ is a unit in $\mathbb{Z}[\omega]^*$ and $p^2 = N(p) = N(u)N(a+b\omega)N(c+d\omega)=\pm(a^2-ab+b^2)(c^2-cd+d^2)$. The nontrivial factorization means that neither factor is a unit and so $p = a^2-ab+b^2$ (note that the quadratic form is positive-definite so it cannot be $-p$).
If $a$ is even and $b$ is odd, then by symmetry we can permute $a$ and $b$. If $a$ is odd and $b$ is odd then let $a' = b$ and $b' = b - a$. We see that $a'^2 - a'b' + b'^2 = b^2 - b(b-a) + (b-a)^2 = b^2 - b^2 + ab + b^2 - 2ab + a^2 = a^2 - ab + b^2$ so $p = a'^2 - a'b' + b'^2$ where $b'$ is now even.
Hence w.l.o.g. we can assume $b$ is even and we can rewrite the expression as $p = \left(a-\frac{b}{2}\right)^2 + 3\left(\frac{b}{2}\right)^2$.
| {
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$a+b+c=?$ by having $a^2+160=b^2+5$ and $a^2+320=c^2+5$ My question:
$a+b+c=?$ by having $a^2+160=b^2+5$ and $a^2+320=c^2+5$.
My work so far:
$a^2+160=b^2+5\Rightarrow (b-a)(a+b)=155=31\times 5$
$a^2+320=c^2+5\Rightarrow (c-a)(c+a)=315=5\times3^2\times 7$
And now, I'm stuck.
($a,b,c$ are a members of $\mathbb Z$ and are positive)
| Hint:
$$(b+a)(b-a)=31\times 5$$
Since $a,b$ are integers, and $5,31$ are prime numbers, what is the value of $(b+a)$ and $(b-a)$?
| {
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Solve $\frac{x^2}{\left(a+\sqrt{a^2+x^2}\right)^2}+\frac{x^2(1+x^2)}{\left(a+\sqrt{a^2+x^2(1+x^2)} \right)^2}=1$ I have been trying to solve the following equation for $x$:
\begin{align}
\frac{x^2}{\left(a+\sqrt{a^2+x^2}\right)^2}+\frac{x^2(1+x^2)}{\left(a+\sqrt{a^2+x^2(1+x^2)} \right)^2}=1,
\end{align}
for some fixed $a \in (0,1/2]$.
This equation came from an analysis of an electrical circuit. The solution is the current.
However, after trying to solve it by hand using the software it doesn't appear that there is a good way of finding a solution.
My question: Can we at least give a good estimate of the position of the positive zero? For example, can show that the zero belong to the specific interval?
Here is an equaivalent polynomial:
\begin{align}
x^4 - 4a^4 - 4a^3(a^2 + x^2)^{1/2} + x^6 - 4a^3(x^2(x^2 + 1) + a^2)^{1/2} - 4a^2(x^2(x^2 + 1) + a^2)^{1/2}(a^2 + x^2)^{1/2}=0.
\end{align}
| Your equation is simplified to:
$$\dfrac{t+\sqrt{1+t^2+\frac{a^4}{t^4}}}{2t} = (t+\sqrt{1+t^2})^2$$
with the substitution $a=xt.$ Conceivably, now you can explicitly find $a$ in terms of $t$ and then might succeed doing an asymptotic analysis, but I really doubt anything close to a closed form solution exists, despite the equation itself look like it should have a closed form solution.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Evaluate $ \lim_{n\to \infty} ( \lim_{x\to0} (1+\tan^2(x)+\tan^2(2x)+ \cdots + \tan^2(nx)))^{\frac{1}{n^3x^2}} $ $ \lim_{n\to \infty} ( \lim_{x\to0} (1+\tan^2(x)+\tan^2(2x)+ \cdots + \tan^2(nx)))^{\frac{1}{n^3x^2}} $
The answer should be $ {e}^\frac{1}{3} $
I haven't encountered problems like this before and I'm pretty confused, thank you.
I guess we must use the remarkable limit of $ \frac{\tan(x)}{x} $ when $ x $ approaches $ 0 $ by dividing and then multiplaying with $x$.
| Note that for $x\to 0$
*
*$\tan x= x+o(x^2)$
then
$$(1+\tan^2(x)+\tan^2(2x)+ \cdots + \tan^2(nx)))^{\frac{1}{n^3x^2}} = (1+x^2+4x^2+...+n^2x^2+o(x^2)) ^ {\frac{1}{n^3x^2}}=\large{e^{\frac{\log(1+x^2+4x^2+...+n^2x^2+o(x^2))}{n^3x^2}}=e^{\frac{x^2+4x^2+...+n^2x^2+o(x^2)}{n^3x^2}}=e^{\frac{n(2n+1)(n+1)}{6n^3}+o(1)} \stackrel{x\to0}\to e^{\frac{n(2n+1)(n+1)}{6n^3}}}$$
and for $n\to \infty$
$$\Large{e^{\frac{n(2n+1)(n+1)}{6n^3}}}\to e^\frac13$$
| {
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Find all the rational values of $x$ at which $y=\sqrt{x^2+x+3}$ is a rational number Question
Find all the rational values of $x$ at which $y=\sqrt{x^2+x+3}$
My attempt
Since we only have to find the rational values of $x$ and $y$, we can assume that
$$ x \in Q$$
$$ y \in Q$$
$$ y-x \in Q $$
Let$$ d = y-x$$
$$d=\sqrt{x^2+x+3}-x$$
$$d+x=\sqrt{x^2+x+3}$$
$$(d+x)^2=(\sqrt{x^2+x+3})^2$$
$$d^2 + x^2 + 2dx =x^2+x+3$$
$$d^2 +2dx = x +3$$
$$x = \frac{3-d^2}{2d-1}$$
$$d \neq \frac{1}{2}$$
So $x$ will be rational as long as $d \neq \frac{1}{2}$.
Now
$$ y = \sqrt{x^2+x+3}$$
$$ y = \sqrt{(\frac{3-d^2}{2d-1})^2 + \frac{3-d^2}{2d-1} + 3}$$
$$ y = \sqrt{\frac{(3-d^2)^2}{(2d-1)^2} + \frac{(3-d^2)(2d-1)}{(2d-1)^2} + 3\frac{(2d-1)^2}{(2d-1)^2}}$$
$$ y = \sqrt{\frac{(3-d^2)^2 + (3-d^2)(2d-1) + 3(2d-1)^2}{(2d-1)^2}} $$
$$ y = \frac{\sqrt{(3-d^2)^2 + (3-d^2)(2d-1) + 3(2d-1)^2}}{(2d-1)}$$
$$ y = \frac{\sqrt{d^4-2d^3+7d^2-6d+9}}{(2d-1)}$$
I know that again $d \neq \frac{1}{2}$ but I don't know what to do with the numerator. Help
| Hint:
The expression is of the following form
$$(a+b-c)^2 = a^2+b^2+c^2+2ab-2bc-2ac$$
| {
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Show that $\frac{1}{3}<\int_0^1\frac{1}{1+x+x^2} \, dx <\frac{\pi}{4}$ Show that $$\frac{1}{3}<\int_0^1\frac{1}{1+x+x^2}\,dx <\frac{\pi}{4}$$
I want to use if $f<g<h$ then $\int f<\int g<\int h$ formula for Riemann integration.
$1+x^2<1+x+x^2$ and it will give RHS as $$\frac{1}{1+x+x^2}<\frac{1}{1+x^2}$$
How to choose function $f$ and $h.$
| Alternative approach: for any $x\in(0,1)$ we have
$$ \frac{1}{1+x+x^2}=\frac{1-x}{1-x^3}=(1-x)\sum_{n\geq 0}x^{3n}=\sum_{n\geq 0}x^{3n}-x^{3n+1} \tag{1}$$
hence
$$ \mathcal{J}=\int_{0}^{1}\frac{dx}{1+x+x^2} = \sum_{n\geq 0}\frac{1}{(3n+1)(3n+2)} \tag{2}$$
and the series in the RHS of $(2)$ can be approximated through the Hermite-Hadamard inequality, since $g(x)=\frac{1}{(3x+1)(3x+2)}$ is convex on $\mathbb{R}^+$. We get
$$ \frac{1}{2}+\frac{1}{20}+\int_{2}^{+\infty}g(x)\,dx\leq \mathcal{J}\leq \frac{1}{2}+\frac{1}{20}+\int_{3/2}^{+\infty}g(x)\,dx \tag{3}$$
$$\frac{11}{20}+\frac{1}{3}\log\left(1+\frac{1}{7}\right)\leq \mathcal{J}\leq \frac{11}{20}+\frac{1}{3}\log\left(1+\frac{2}{11}\right)\tag{4}$$
so the distance between $\mathcal{J}$ and $\frac{3}{5}$ is at most $\frac{1}{100}$.
(2) and the reflection formula for the $\psi$ function imply $\mathcal{J}=\frac{\pi}{3\sqrt{3}}$, which can be proved by just completing the square, too. Another interesting approach is creative telescoping:
$$\begin{eqnarray*}\sum_{n\geq 0}\frac{1}{(3n+1)(3n+2)}&=&\frac{1}{2}+\sum_{n\geq 0}\left(\frac{1}{9(n+1)}-\frac{1}{9(n+2)}-\frac{2}{(3n+3)(3n+4)(3n+5)(3n+6)}\right)\\&=&\frac{1}{2}+\frac{1}{9}-\frac{1}{180}-2\sum_{n\geq 0}\frac{1}{(3n+6)(3n+7)(3n+8)(3n+9)}\end{eqnarray*} $$
also leading to the accelerated series $\mathcal{J}=\sum_{n\geq 1}\frac{1}{n\binom{2n}{n}}$.
| {
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What kind of matrix is this and why does this happen? So I was studying Markov chains and I came across this matrix \begin{align*}P=\left( \begin{array}{ccccc}
0 & \frac{1}{4} & \frac{3}{4} & 0 & 0\\
\frac{1}{4} & 0 & 0 & \frac{1}{4} & \frac{1}{2}\\
\frac{1}{2} & 0 & 0 & \frac{1}{4}& \frac{1}{4}\\
0 & \frac{1}{4} & \frac{3}{4} & 0& 0\\
0 & \frac{1}{4} & \frac{3}{4} & 0& 0\\
\end{array} \right).\end{align*}
I noticed (by brute force) that \begin{align*}P^2=\left( \begin{array}{ccccc}
\frac{7}{16} & 0 & 0 & \frac{1}{4} & \frac{5}{16}\\
0 & \frac{1}{4} & \frac{3}{4} & 0& 0\\
0 & \frac{1}{4} & \frac{3}{4} & 0& 0\\
\frac{3}{8} & 0 & 0 & \frac{1}{4}& \frac{1}{2}\\
\frac{7}{16} & 0 & 0 & \frac{1}{4} & \frac{5}{16}\\
\end{array} \right),\end{align*}
and \begin{align*}P^3=\left( \begin{array}{ccccc}
0 & \frac{1}{4} & \frac{3}{4} & 0& 0\\
\frac{7}{16} & 0 & 0 & \frac{1}{4} & \frac{5}{16}\\
\frac{7}{16} & 0 & 0 & \frac{1}{4} & \frac{5}{16}\\
0 & \frac{1}{4} & \frac{3}{4} & 0& 0\\
0 & \frac{1}{4} & \frac{3}{4} & 0& 0\\
\end{array} \right).\end{align*}
In fact; using a computer I found that every even power takes the form of the $P^2$ matrix and every odd power takes the form of the $P^3$ matrix.
I just wanted to know why that oscillation occurs? Is there a special name for the kind of matrix that $P$ is for it to exhibit that kind of behaviour?
| You can draw a graph for which your matrix is the matrix of probabilities of state change (some weights on the edges).
Paint states A, D and E as white, and B and C as black. Then any possible move changes the color.
Since P^n is the probablility matrix of moving from X to Y in exacly n steps (or sum of products of weights spotted by any n-step path) then the oscillation occurs because your graph is bipartite.
You can produce oscillation with other period if you wish this way.
| {
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"question_score": "24",
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Finding an isomorphism $\mathfrak{sp}(2) \to \mathbb{R}^3$ I am studying Lie groups and Lie algebras and I am trying to find an isomorphism between the groups/algebras and $\mathbb{R}^n$.
For $Sp(2,\mathbb{R})$, using the standard skew-symmetric matrix, and the condition
$$\begin{pmatrix}a & c \\ b & d \end{pmatrix} \begin{pmatrix}0 & 1 \\ -1 & 0\end{pmatrix} \begin{pmatrix}a & b \\ c & d\end{pmatrix} = \begin{pmatrix}0 & 1 \\ -1 & 0\end{pmatrix}$$
I got that the matrices of $Sp(2,\mathbb{R})$ are of the form
$$\begin{pmatrix}a & b \\ c & \frac{1+bc}{a}\end{pmatrix}$$
I am trying to come up with a similar thing for the Lie algebra $\mathfrak{sp}(2,\mathbb{R})$ but I end up with
$$\begin{pmatrix}0 & 1 \\ -1 & 0\end{pmatrix} \begin{pmatrix}a & b \\ c & d \end{pmatrix} + \begin{pmatrix}a & c \\ b & d\end{pmatrix} \begin{pmatrix}0 & 1 \\ -1 & 0\end{pmatrix}$$
$$= \begin{pmatrix}0 & a + d \\ -a-d & 0\end{pmatrix}$$ which does not have dimension $0$. According, to https://en.wikipedia.org/wiki/Table_of_Lie_groups this Lie algebra has dimension $n(2n+1)$ which is $3$ in this case, but I am not obtaining matrices isomorphic to $\mathbb{R}^3$. What am I doing wrong?
| I just noticed something really dumb. I forgot to use the constraint
$$\begin{pmatrix}0 & a + d \\ -a -d & 0 \end{pmatrix} = \begin{pmatrix} 0 & 0 \\ 0 & 0\end{pmatrix}$$
Which gives me $a = -d$.
Therefore, the matrices for $\mathfrak{sp}(2)$ are those of the form
$$\begin{pmatrix}a & b \\ c & -a\end{pmatrix}$$
Correct me if I'm wrong please.
| {
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Prob. 10, Sec. 3.3, in Bartle & Sherbert's INTRO TO REAL ANALYSIS, 4th ed: Is this sequence convergent? Here is Prob. 10, Sec. 3.3, in the book Introduction to Real Analysis by Robert G. Bartle & Donald R. Sherbert, 4th edition:
Establish the convergence or the divergence of the sequence $\left( y_n \right)$, where
$$ y_n \colon= \frac{1}{ n+1} + \frac{1}{n+2} + \cdots + \frac{1}{2n} \ \mbox{ for } n \in \mathbb{N}. $$
My Attempt:
For each $n \in \mathbb{N}$, we have
$$
\begin{align}
y_{n+1} - y_n &= \left( \frac{1}{ (n + 1) +1} + \frac{1}{ (n+1) +2 } + \cdots + \frac{1}{2(n+1)} \right) - \left( \frac{1}{ n+1} + \frac{1}{n+2} + \cdots + \frac{1}{2n} \right) \\
&= \left( \frac{1}{ n + 2} + \cdots + \frac{1}{2n+ 2 } \right) - \left( \frac{1}{ n+1} + \cdots + \frac{1}{2n} \right) \\
&= \frac{ 1 }{2n+1} + \frac{ 1}{2n+2} - \frac{1}{n+1} \\
&= \frac{1}{2n+1} - \frac{1}{2n+2} \\
&= \frac{1}{ (2n+1) (2n+2) } > 0,
\end{align}
$$
which shows that our sequence is monotonically increasing.
Also, for each $n \in \mathbb{N}$, we have
$$ y_n = \frac{1}{ n+1} + \frac{1}{n+2} + \cdots + \frac{1}{2n} < \underbrace{\frac{1}{n} + \frac{1}{n} + \cdots + \frac{1}{n} }_{ \mbox{ $n$ times } } = 1. $$
Thus our sequence, being monotonically increasing and bounded, is convergent.
Is what I've done so far all correct?
How to determine the limit of this sequence?
| Yes, you are all right. Because an increasing and bounded real number sequence is convergent, it's the axiom of Real number: A bounded set on $R$ has supremum.
For calculate the limit, consider function $$f(x)=\frac{1}{x+1}\ \ \text{on}\ \ [0,1]$$it's integrable, and we let $0<\frac{1}{n}<\frac{2}{n}<……<\frac{n-1}{n}<1$, hence
$$\int_0^1\frac{1}{x+1}=\lim_{n\rightarrow\infty}\sum_{i=1}^n\frac{1}{n}f(\frac{i}{n})=\sum_{i=1}^n\frac{1}{n+i}=\log(x+1)|_{x=0}^{x=1}=\log 2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2704730",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
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