Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Proof of rational numbers I am recently starting in the subject of pure mathematics and this problem has eluded me.
Any rational number $\frac{p}{q}$ can be expressed in a simple series form:
$$\frac{1}{1},\frac{2}{1},\frac{1}{2},\frac{3}{1},\frac{2}{2},\frac{1}{3},\frac{4}{1},\frac{3}{2},\frac{2}{3},\frac{1}{4},...$$
Show that $\frac{p}{q}$ is the $[\frac{1}{2}(p+q-1)(p+q-2)+q]th$ of the series (where $p$ and $q$ are coprime and $p \neq 0$).
| Hint. Divide the numbers in rows:
$$\frac{1}{1}$$
$$\frac{2}{1},\frac{1}{2}$$
$$\frac{3}{1},\frac{2}{2},\frac{1}{3}$$
$$\frac{4}{1},\frac{3}{2},\frac{2}{3},\frac{1}{4}.$$
In the first $n$ rows there are $1+2+\dots+n=n(n+1)/2$ fractions.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1935563",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Prove that $\sin^4\theta+\cos^4\theta=\frac{3+\cos4\theta}{4}$ I have a question that goes exactly like this:
By considering $(\sin^2\theta+\cos^2\theta)^2$ and $(\sin^2\theta+\cos^2\theta)^3$ prove that
$$
\text{a) }\sin^4\theta+\cos^4\theta=\frac{3+\cos4\theta}{4},\qquad\text{b) }\sin^6\theta+\cos^6\theta=\frac{5+3\cos4\theta}{8}
$$
I have not idea how to do this. Please help.
| From Euler's Identity and subsequent use of the binomial theorem, we have
$$\begin{align}
\cos(4\theta)&=\text{Re}(e^{i4\theta})\\\\
&=\text{Re}\left(\cos(\theta)+i\sin(\theta)\right)^4\\\\
&=\text{Re}\left(\sum_{k=0}^4\binom{4}{k}\cos^{4-k}(\theta)i^k\sin^k(\theta)\right)\\\\
&=\cos^4(\theta)-6\cos^2(\theta)\sin^2(\theta)+\sin^4(\theta)
\end{align}$$
Next, noting that $\sin^2(\theta)\cos^2(\theta)=\frac12-\frac12(\cos^4(\theta)+\sin^4(\theta))$ (this is a trivial consequence of $(\cos^2(\theta)+\sin^2(\theta))^2=1$), we obtain the coveted equality
$$\cos^4(\theta)+\sin^4(\theta)=\frac{3+\cos(4\theta)}{4}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1936121",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 8,
"answer_id": 6
} |
Compute $\int^{\pi/2}_0 \frac{dx}{(a^2\cos^2 x + b^2 \sin ^2 x)^2}$ I have tried solving this for about an hour and will probably resort to head banging in some time:
$$\int ^{\frac{\pi}{2}}_{0} \dfrac{dx}{(a^2\cos^2 x + b^2 \sin ^2 x)^2}$$
I first divided by $\cos^4 x$ and then subsequently put $\tan x = t$, to get:
$$\int ^{\infty}_{0} \dfrac{1+t^2}{(a^2 + b^2 t^2)^2}dt$$
This has become unmanageable. Neither splitting the numerator, nor Partial fraction (taking t^2 = z and applying partial fraction) seems to work.
| Say that $ a, b >0 .$ From the substitution $t \mapsto \dfrac{a}{b}t$ followed by $ t \mapsto \dfrac{1}{t}$ we get
$$
I = \int_{0}^{\infty}\dfrac{1+t^{2}}{(a^{2}+b^{2}t^{2})^{2}}\, dt = \dfrac{1}{a^{3}b^{3}} \int_{0}^{\infty}\dfrac{b^2+a^2t^{2}}{(1+ t^{2})^{2}}\, dt
$$
and
$$
I = \dfrac{1}{a^{3}b^{3}} \int_{0}^{\infty}\dfrac{b^2t^2+a^2}{(1+ t^{2})^{2}}\, dt .
$$
Consequently
$$
2I = \dfrac{a^2+b^2}{a^3b^3}\int_{0}^{\infty}\dfrac{1}{1+t^2}\, dt = \dfrac{\pi}{2}\dfrac{a^2+b^2}{a^3b^3}
$$
and
$$
I = \dfrac{\pi}{4}\dfrac{a^2+b^2}{a^3b^3}.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1939026",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
"answer_id": 0
} |
How to show this conjecture：$\frac{1}{\sqrt{a_{1}}}+\frac{1}{\sqrt{a_{2}}}+\cdots+\frac{1}{\sqrt{a_{n}}}\ge\frac{2n}{\sqrt{a_{1}}+\sqrt{a_{n}}}$ I have prove this inequality
$$\dfrac{1}{\sqrt{1}}+\dfrac{1}{\sqrt{2}}+\cdots+\dfrac{1}{\sqrt{n}}>2\sqrt{n+1}-2$$
because
$$\sum_{k=1}^{n}\dfrac{1}{\sqrt{k}}>\sum_{k=1}^{n}\dfrac{2}{\sqrt{k}+\sqrt{k+1}}=2\sum_{k=1}^{n}(\sqrt{k+1}-\sqrt{k})=2\sqrt{n+1}-2$$
so
$$\sum_{k=1}^{n}\dfrac{1}{\sqrt{k}}>2\sqrt{n+1}-2$$
Now I conjecture：
if postive arithmetic progression $\{a_{n}\}$ ,and the common difference of successive members is $d>0$.have
$$\dfrac{1}{\sqrt{a_{1}}}+\dfrac{1}{\sqrt{a_{2}}}+\cdots+\dfrac{1}{\sqrt{a_{n}}}\ge\dfrac{2n}{\sqrt{a_{1}}+\sqrt{a_{n}}}$$
| Using the AM-HM inequality states: if $x_1\leq\ldots\leq x_n$ are nonnegative numbers, then
\begin{align}
\frac{x_1+\ldots +x_n}{n} \geq \frac{n}{\frac{1}{x_1}+\ldots+\frac{1}{x_n}}
\end{align}
which implies
\begin{align}
\frac{1}{x_1} + \ldots + \frac{1}{x_n} \geq \frac{n^2}{x_1+ \ldots + x_n} \geq \frac{n}{x_1 + x_n}.
\end{align}
However, we might be able to do better.
Special Case: Let us consider arithmetic progression of the form $a_k = dk$ for $k = 1, \dots, n$, then the desired inequality becomes
\begin{align}
\sum^n_{k=1} \frac{1}{\sqrt{k}} \geq \frac{2n}{1+\sqrt{n}} \ \ \ (*)
\end{align}
To establish $(\ast)$, we will use Euler Maclaurin Series
\begin{align}
\sum^n_{k=1} \frac{1}{\sqrt{k}} = 1 + \int^n_1\frac{dx}{\sqrt{x}} + \frac{1}{2}\left(\frac{1-\sqrt{n}}{\sqrt{n}} \right)+\frac{1}{24}\left( 1-n^{-3/2}\right)+R
\end{align}
where
\begin{align}
|R| \leq \frac{3\zeta(2)}{2(2\pi)^2}\int^n_1 \frac{dx}{x^{5/2}} = \frac{1}{24}(1-n^{-3/2}).
\end{align}
Then it follows
\begin{align}
\sum^n_{k=1} \frac{1}{\sqrt{k}}-\frac{2n}{1+\sqrt{n}} =&\ 1+ \frac{2(n-1)}{1+\sqrt{n}}+\frac{1}{2}\left(\frac{1-\sqrt{n}}{\sqrt{n}} \right)-\frac{2n}{1+\sqrt{n}}+\text{ something positive}\\
\geq&\ \frac{\sqrt{n}-1}{\sqrt{n}+1}+\frac{1}{2}\left(\frac{1-\sqrt{n}}{\sqrt{n}} \right) = \frac{1}{2}\frac{(\sqrt{n}-1)^2}{\sqrt{n}(\sqrt{n}+1)}.\\
\end{align}
Thus, it follows
\begin{align}
\sum^n_{k=1} \frac{1}{\sqrt{k}} \geq \frac{2n}{1+\sqrt{n}}.
\end{align}
Comment: Using the same idea, we might be able to work with arithmetic progression of the form $a_k = d(m+k)$ for $k=1, \ldots, n$. However, I have no clue how to get the most general arithmetic progression $a_k = d(m+k)+\ell$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1939123",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Evaluation of Trigonometric Integration $\int \sin x\sqrt{\sin 2x}dx$.
Evaluation of $$\int \sin x\sqrt{\sin 2x}dx$$
$\bf{My\; Try::}$ Let $$I = \int \sin x\sqrt{\sin 2x}dx = \sqrt{2}\int \frac{\tan x\sqrt{\tan x}}{\sec x}dx = \sqrt{2}\int \frac{(\tan x)^{\frac{3}{2}}}{\sec^3 x}\cdot \sec^2 xdx$$
So $$I = \sqrt{2}\int \frac{(\tan x)^{\frac{3}{2}}}{\sqrt{(1+\tan^2 x)^3}}\cdot \sec^2 xdx$$
Now Put $\tan x= t^2\;,$ Then $\sec^2 xdx = 2tdt$
So $$I = 2\sqrt{2}\int \frac{t^4}{\sqrt{(1+t^8)^3}}dt$$
Now how can i solve it after that, Help required, Thanks
| Note that $\sin{2x} \equiv 1-(\sin{x}-\cos{x})^2 \equiv (\sin{x}+\cos{x})^2 -1$
Also note that $\text{d}(\sin{x}-\cos{x}) - \text{d}(\sin{x}+\cos{x}) = 2\sin{x}\,\text{d}x$
Substitute and separate the integral into two different integrals which can be evaluated independently.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1939243",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
In the expansion of $(1+x+x^2)^n$, find the required value If $n \in $ odd integer, such that:
$$(1+x+x^2)^n=a_0+a_1x+a_2x^2+.....+a_{2n}x^{2n}$$ and
$$(a_1)^2-(2a_2)^2+(3a_3)^2-(4a_4)^2+......-(2na_{an})^2=- \lambda n^2a_{n-1}$$ then find the value of $\lambda$.
Is there any better approach to this question rather than binomial expansion and then collection of required coefficient?
| $$y(x):=(1+x+x^2)^n$$
Suppose ' represents differentiation w.r.t $x$.
$$
y'(x)=n(1+x+x^2)^{n-1}(1+2x)=\sum_{k=1}^{2n}ka_kx^{k-1}
$$
Now the constant term in $y'(-x)y'(x^{-1})$ is
$$
a:=a_1-(2a_2)^2+(3a_3)^2-\cdots-(2na_{2n})^2
$$
Now
$$
y'(-x)y'(x^{-1})={n^2(1+x^2+x^4)^{n-1}(2-3x-2x^2)\over x^{2n-1}}
$$
Let $g(x)=(1+x+x^2)^{n-1}=\sum_{k=0}^{2n-2}b_{k}x^k$. Clearly $g$ is a polynomial and $g(x^2)$ does not contain odd powers of $x$. Therefore coefficients of $x^{2n-1}$ in $g(x^2)$ and $x^2g(x^2)$ are $0$. Hence
$$
\begin{align}a&=-3n^2\times(\text{Coefficient of }x^{2n-2}\text{ in }g(x^2))\\
&=-3n^2\times(\text{Coefficient of }x^{n-1}\text{ in }g(x))\\
&=-3n^2b_{n-1}\end{align}
$$
Now
$$g(x)=\sum_{j=0}^{n-1}\binom{n-1}{j}x^j(1+x)^j=\sum_{j=0}^{n-1}\sum_{r=0}^j\binom{n-1}{j}\binom{j}{r}x^{j+r}\\
\begin{align}\therefore \;b_{n-1}&=\sum_{r\le j,\;j+r=n-1}\binom{n-1}{j}\binom{j}{r}\\
&=\sum_{r=0}^{\left\lfloor\frac{n-1}{2}\right\rfloor}\binom{n-1}{n-1-r}\binom{n-1-r}{r}\\
\end{align}$$
Similar calculation shows
$$
a_{n-1}=\sum_{r=0}^{\left\lfloor\frac{n-1}{2}\right\rfloor}\binom{n}{n-1-r}\binom{n-1-r}{r}
$$
Finally,
$$\lambda=3\frac{b_{n-1}}{a_{n-1}}=3\frac{\sum_{r=0}^{\left\lfloor\frac{n-1}{2}\right\rfloor}\binom{n-1}{r}\binom{n-1-r}{r}}{\sum_{r=0}^{\left\lfloor\frac{n-1}{2}\right\rfloor}\binom{n}{r+1}\binom{n-1-r}{r}}$$
In case $n$ is odd we may remove the floor brackets.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1939679",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
$120^{1492}\pmod{100}$ How can I solve it?
I can solve it using the fact: if $b\mid a$ then $b^{n}\mid a^{n}$ (I have proved it using induction)
$\frac{120}{10}=12$ ,then $\frac{120^{2}}{100}$
this mean $120^{2}\equiv0\pmod{10}$
so, $120^{1492}\equiv (120^{2})^{746}\equiv 0^{746}\equiv 0\pmod{100}$
I'm asking for another way.
I tried this: $120=2^{3}\cdot3\cdot5$,so:$$[2^{4467}\pmod{100}\cdot3^{1492}\pmod{100}\cdot5^{1492}\pmod{100}]\pmod{100}$$
$3^{1492}\equiv41\pmod{100}$ using Euler's theorem
but I stuck with $$[2^{4467}\pmod{100}\cdot41\cdot5^{1492}\pmod{100]\pmod{100}}$$
| Use this teorem: If $a \equiv b \pmod{m} \implies a^n \equiv b^n \pmod{m}$, then: $$120 \equiv 20 \pmod{100} \implies 120^{1492} \equiv 20^{1492} \pmod{100}$$
If you look the values of $20^n$ you gonna see this:
$$\begin{align}
20^1 \equiv 20 \pmod {100} \\
20^2 \equiv 0 \pmod {100} \\
20^3 \equiv 0 \pmod{100} \\
... \\
20^{1492} \equiv 0 \pmod {100}
\end{align}$$
Then, for transitive property $120^{1492} \equiv 0 \pmod{100}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1939929",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Solve the equation $\cos^n(x) - \sin^n(x)=1$
Solve the equation $\cos^n(x) - \sin^n(x)=1,n \in \mathbb{N}-\{0\}$
If $n$ is even then $\cos^n(x) = \sin^n(x)+1$ is only possible if $\sin(x)=0$ therefore the solution is $x=k\pi, k \in \mathbb{Z}$.
I'm having problems with $n$ odd case.
UPDATE
For $n=1$ we have $\cos(x) - \sin(x)=1$ and by squaring we get $sin(x)cos(x)=0$ which leads to $x=k\pi$ or $x=\pm \frac \pi 2 + k\pi$. From these solutions, only $x=2k\pi$ and $x=- \frac \pi 2 + 2k\pi$ are valid.
Also $x=2k\pi$ and $x=- \frac \pi 2 + 2k\pi$ are solutions for all odd $n$
| I found a simple proof. For $n \ge 3$ odd we have $\cos^n(x) + (-\sin(x))^n=1 \tag1$ First, we know $\sin^2(x) + \cos^2(x)=1$
Also $|\cos^n(x)|\lt \cos^2(x)$ and $|\sin^n(x)|\lt \sin^2(x)$ for $x \not = k\pi$ and $x \not = \pm \frac \pi 2 + k\pi$ therefore the equality (1) cannot hold for $x \not = k\pi$ and $x \not = \pm \frac \pi 2 + k\pi$
It's easy to conclude from here.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1940444",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Simplifcation of a sum I’m unsure if I’m allowed to simplify like this:
$$\begin{align*}
\sum_{n=2}^\infty \frac{2^{n+3}+7^{n-1} \cdot(n-1)!}{10^n(n-1)!} &= \sum_{n=2}^\infty \frac{2^{n+3}}{10^n(n-1)!} + \sum_{n=2}^\infty \frac{7^{n-1} \cdot(n-1)!}{10^n(n-1)!}\\
&= \sum_{n=2}^\infty \frac{2^{n+3}}{10^n(n-1)!} + \frac{1}{7}\sum_{n=2}^\infty \frac{7^{n} \cdot(n-1)!}{10^n(n-1)!}\\
&= \sum_{n=2}^\infty \frac{2^{n+3}}{10^n(n-1)!} + \frac{1}{7}\sum_{n=2}^\infty \frac{7}{10}\\
&= \sum_{n=2}^\infty \frac{2^{n+3}}{10^n(n-1)!} + \frac{1}{10}
\end{align*}$$
Please tell me whether this is right or wrong and if possible how else the manipulated part could be simplified. Thanks in advance!
| Your first two steps are correct, but then you went astray.
$$\sum_{n=2}^\infty\frac{7^n(n-1)!}{10^n(n-1)!}=\sum_{n=2}^\infty\frac{7^n}{10^n}=\sum_{n=2}^\infty\left(\frac7{10}\right)^n\;,$$
not $\sum_{n=2}^\infty\frac7{10}$. Now
$$\frac17\sum_{n=2}^\infty\left(\frac7{10}\right)^n=\frac17\cdot\frac{\left(\frac7{10}\right)^2}{1-\frac7{10}}=\frac17\cdot\frac{10}3\cdot\frac7{10}\cdot\frac7{10}=\frac7{30}\;,$$
since the sum is just a geometric series. And you should note that $\sum_{n=2}^\infty\frac7{10}$ diverges, so $\frac17\sum_{n=2}^\infty\frac7{10}$ definitely isn’t $\frac1{10}$ even if this were the right expression to be evaluating.
The first summation can also be simplified. For starters,
$$\sum_{n=2}^\infty\frac{2^{n+3}}{10^n(n-1)!}=8\sum_{n=2}^\infty\frac{2^n}{10^n(n-1)!}=8\sum_{n=2}^\infty\left(\frac15\right)^n\frac1{(n-1)!}\;;$$
this at least gives the two exponentials the same exponent, so that they can be combined into a single exponential. But we can do better if we remember that
$$e^x=\sum_{n=0}^\infty\frac{x^n}{n!}\;.\tag{1}$$
We first pull out a factor of $\frac15$ to make the exponent on $\frac15$ match the factorial:
$$8\sum_{n=2}^\infty\frac{2^n}{10^n(n-1)!}=8\sum_{n=2}^\infty\left(\frac15\right)^n\frac1{(n-1)!}=\frac85\sum_{n=2}^\infty\left(\frac15\right)^{n-1}\frac1{(n-1)!}\;.$$
Now shift the index of summation one place: as $n$ runs up from $2$, $n-1$ runs up from $1$, and
$$\frac85\sum_{n=2}^\infty\left(\frac15\right)^{n-1}\frac1{(n-1)!}=\frac85\sum_{n=1}^\infty\frac{(1/5)^n}{n!}\;.$$
Now use $(1)$ to express this in closed form, i.e., without a summation.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1942281",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Find $\int_0^{2\pi}\frac1{5-4\cos x}\ dx$ $$\int_0^{2\pi}\frac1{5-4\cos x}\ dx$$
How do I compute this integral? An online integral calculator gives an antiderivative as
$$\frac{2\arctan(3\tan\frac x2)}3$$
but then gives the definite integral as $\frac{2\pi}3$. Obviously this doesn't make sense as the antiderivative vanishes at $x=0$ and $x=2\pi$.
| Using complex analysis
your integral can be expressed as
$$ \int \limits^{2\pi }_{0}\frac{1}{5-4\cos \left( x\right) } dx=\oint \frac{1}{5-2\left( z+\frac{1}{z} \right) iz} dz$$
Now we have two singularites at $z=2$and at $z=\frac {1}{2} $
Now apply Residue theorem
$$\oint \frac{1}{5-2\left( z+\frac{1}{z} \right) iz} dz = 2\pi i \ \text {Res}\sum \left( f\left( z\right) , z\right) $$
We have only one singularity in the contour.
So the Residue will be
$$ 2 \pi i \ \text {Res} \sum(f (z), z)=2 \pi i \left[\lim \limits_{z\to \frac{1}{2} }\left( 2z-1\right) \frac{1}{2i\left( -z+2\right) \left( 2z-1\right) } =\frac{1}{3i} \right]=\frac {2 \pi}{3} $$
So
$$\color{Brown }{\int \limits^{2\pi }_{0}\frac{1}{5-4\cos \left( x\right) } dx=\boxed {\frac{2\pi }{3}} }$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1942983",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 2
} |
Properties of the rotation matrix $$R(\theta) = \begin{pmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{pmatrix}$$
Consider the above rotation matrix.
$a)$ Show that $R(\theta)$ is non-singular with: $R^{-1}(\theta) = R(-\theta)$
$b)$ Show that $R(\theta)^T = R(-\theta)$
$c)$ For what angles $\theta \in \mathbb{R}$ is $R(\theta)$ symmetric?
I only somewhat understand part $b$. I know that a skew-matrix is one in which its transpose is also its negative. Apart from that I'm not sure how to implement that into the question and I have no idea how to do part $a$ and $c$. I was wondering if someone could help me get started.
| a
A quick singularity check:
$$
\det \mathbf{R} \left( \theta \right)
= \cos^{2} \left( \theta \right) + \sin^{2} \left( \theta \right) = 1
$$
Because the determinant is not $0$, the rotation matrix is not singular.
Is
$$
\mathbf{R} \left( \theta \right)^{-1} = \mathbf{R} \left( -\theta \right)?
$$
Check via multiplication
$$
\begin{align}
\mathbf{R} \left( \theta \right) \mathbf{R} \left( -\theta \right) &=
%
\left[
\begin{array}{rr}
\cos ( \theta ) & -\sin (\theta ) \\
\sin (\theta ) & \cos (\theta ) \\
\end{array}
\right]
%
\left[
\begin{array}{rr}
\cos (\theta ) & \sin (\theta ) \\
-\sin (\theta ) & \cos (\theta ) \\
\end{array}
\right] \\[5pt]
%
&=
%
\left[
\begin{array}{cc}
\cos ^2(\theta )+\sin ^2(\theta ) & 0 \\
0 & \cos ^2(\theta )+\sin ^2(\theta ) \\
\end{array}
\right]
\\[5pt]
%
&=
%
\left[
\begin{array}{cc}
1 & 0 \\
0 & 1 \\
\end{array}
\right] \\[2pt]
%
&= \mathbf{I}_{2}
%
\end{align}
$$
b
Show that the transpose matrix is $\mathbf{R}\left( -\theta \right)$.
$$
\begin{align}
\mathbf{R}\left( -\theta \right) &=
\left[
\begin{array}{rr}
\cos (\theta ) & \sin (\theta ) \\
-\sin (\theta ) & \cos (\theta ) \\
\end{array}
\right] \\[3pt]
%
\mathbf{R}^{T}\left( \theta \right) &=
\left[
\begin{array}{rr}
\cos (\theta ) & \sin (\theta ) \\
-\sin (\theta ) & \cos (\theta ) \\
\end{array}
\right] \\
\end{align}
%
$$
c
When does $\mathbf{R} \left( \theta \right)$ have a symmetric form like
$$
\left[
\begin{array}{rr}
a & b \\
b & b \\
\end{array}
\right] ?
$$
The diagonal entries are the same, $\cos \left( \theta \right)$. They pose no constraint. For the off-diagonal entries, the question is when does
$$
\sin \left( \theta \right) = - \sin \left( \theta \right)?
$$
The answer is when $\sin \left( \theta \right) = 0?$ which occurs for
$$
\theta = 2k\pi, \quad k\in\mathbb{Z}.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1943175",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Double Integral Math Olympiad Problem I was taking a Math Olympiad test and one of the questions was to calculate the following double integral:
$$\int_0^\infty\int_0^\infty\frac{\log|(x+y)(1-xy)|}{(1+x^2)(1+y^2)}\ \mathrm{d}x\ \mathrm{d}y$$
Here, as usual, $\log a$ and $|a|$ are the natural logarithm and absolute value of $a$ respectively.
I'm guessing that you're not supposed to solve it analytically, but rather find some symmetry argument or clever simplification that would make it straightforward. Since I don't even know where to start, any help is welcome.
In case you want to know, this was taken from the 2016 Rio de Janeiro State Math Olympiad, known in Portuguese as OMERJ.
| Here is a solution following @Sirzh's hint. Let
$$ I = \int_{0}^{\infty}\int_{0}^{\infty} \frac{\log|(x+y)(1-xy)|}{(1+x^2)(1+y^2)} \, dxdy $$
and
$$ J = \int_{0}^{\frac{\pi}{2}} \log \cos \theta \, d\theta = -\frac{\pi}{2}\log 2. \tag{1}$$
Applying the substitution $x = \tan u$ and $y = \tan v$ to $I$, we get
\begin{align*}
I
&= \int_{0}^{\frac{\pi}{2}}\int_{0}^{\frac{\pi}{2}} \log|(\tan u + \tan v)(1 - \tan u \tan v)| \, dudv \\
&= \int_{0}^{\frac{\pi}{2}}\int_{0}^{\frac{\pi}{2}} \log|\sin(u+v)\cos(u+v)| \, dudv - \int_{0}^{\frac{\pi}{2}}\int_{0}^{\frac{\pi}{2}} \log(\cos^2 u \cos ^2 v) \, dudv \tag{2} \\
&= \int_{0}^{\frac{\pi}{2}}\int_{0}^{\frac{\pi}{2}} \log\left|\frac{\sin(2u+2v)}{2}\right| \, dudv - 2\pi J. \\
&= \int_{0}^{\frac{\pi}{2}} \left( \frac{1}{2} \int_{v}^{\pi+v} \log \left| \frac{\sin \theta}{2} \right| \, d\theta \right) dv - 2\pi J
\qquad (\theta = 2u+2v)
\end{align*}
Now since $\theta \mapsto \log \left| \frac{\sin \theta}{2} \right|$ is $\pi$-periodic, for any $v$ we have
\begin{align*}
\frac{1}{2} \int_{v}^{\pi+v} \log \left| \frac{\sin \theta}{2} \right| \, d\theta
&= \frac{1}{2} \int_{0}^{\pi} \log \left| \frac{\sin \theta}{2} \right| \, d\theta \\
&= \int_{0}^{\frac{\pi}{2}} \log \left| \frac{\sin \theta}{2} \right| \, d\theta
= J - \frac{\pi}{2}\log 2.
\end{align*}
Therefore
$$I = \frac{\pi}{2} \left( J - \frac{\pi}{2}\log 2 \right) - 2\pi J = \frac{\pi^2}{2}\log 2.$$
$\text{(1)}$ : You may use the Fourier expansion
$$ \log |\cos\theta| = \Re \log\left(\frac{1+e^{2i\theta}}{2} \right) = -\log 2 + \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n} \cos (2n\theta) $$
to compute $J$. Alternatively, one can compute $J$ from the following equation
$$J
= \int_{0}^{\frac{\pi}{2}} \log \sin\theta \, d\theta
= \int_{0}^{\frac{\pi}{2}} \log \sin(2\theta) \, d\theta
= \int_{0}^{\frac{\pi}{2}} \log (2\sin \theta \cos \theta) \, d\theta
= \frac{\pi}{2}\log 2 + 2J. $$
$\text{(2)}$ : Apply the addition formula to
$$ (\tan u + \tan v)(1 - \tan u \tan v) = \frac{(\sin u \cos v + \cos u \sin v)(\cos u \cos v - \sin u \sin v)}{\cos^2 u \cos^2 v}. $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1945637",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 2,
"answer_id": 1
} |
Find one side of a triangle using trigonometry impossible question In the isosceles triangle $ABC$ where $a=|BC|,\ b=|CA|,\ c=|AB|$ and angle at corner $A = x$
and angle at corner $B = y$ and angle at corner $C = z$ ($z>90$ degrees)
$a=b=4$ ; $\sin z = 3/8$
Find value of $c$ (which is side $|AB|$ as stated above)
My friends father is math teacher in high school and he couldn't even solve it,
can anyone help?
| @grixor's answer, using the law of cosines is (currently, almost) correct, but perhaps uses a tool that is not yet in the student's toolbox. We can derive this result from the problem as given.
Extend $\overline{AC}$ until it intersects the perpendicular to $\overline{AC}$ through $B$ and call this intersection $D$. Call the measure of angle $BCD$ $\theta$ and note that $\theta = 180^\circ - z$. Then the right triangle $BCD$ supplies \begin{align}
\sin \theta &= \frac{|BD|}{a} \\
\cos \theta &= \frac{|CD|}{a} \text{.} \\
\end{align}
Using supplementary angle identities, we have $\sin \theta = \sin z = |BD|/a$ and $\cos \theta = -\cos z = |CD|/a$. Rewriting in terms of the quantities we want, $|BD| = a \sin z$ and $|CD| = -a \cos z$.
Now shift attention to the right triangle $ABD$. From this triangle and the Pythagorean theorem, we see \begin{align}
c^2 &= |BD|^2 + (b+|CD|)^2 \\
&= (a \sin z)^2 + (b - a \cos z)^2 \\
&= a^2 \sin^2 z + b^2 - 2 a b \cos z + a^2 \cos^2 z \\
&= a^2( \sin^2 z + \cos^2 z) +b^2 - 2 a b \cos z \\
&= a^2 + b^2 - 2 a b \cos z \text{.}
\end{align} We have used the identity $\sin^2 z + \cos^2 z = 1$. We use it again to get $\cos z = -\sqrt{1 - \sin^2 z}$. (We know we want the negative root because we know $z$ is in the second quadrant, so cosine is negative.) Collecting our algebra together, we have $$ c^2 = a^2 + b^2 + 2 a b \sqrt{1 - \sin^2 z} \text{.} $$
We're given $a = b = 4$ and $\sin z = 3/8$, so we can evaluate:\begin{align}
c^2 &= 4^2 + 4^2 + 2 \cdot 4^2 \sqrt{1 - (3/8)^2} \\
&= 32 + 32\sqrt{\frac{64-9}{64}} \\
&= 32 + 32\sqrt{55/64} \text{.}
\end{align} Now we know $c = \sqrt{32 + 32\sqrt{55/64}}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1946478",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Solving $\lim\limits_{x \to 0} \frac{\ln(1+x)}{\ln(1+4x+x^2)}$ without L'Hospital or Sandwiching Using the L'Hospital I got: $\lim\limits_{x \to 0} \frac{\ln(x+1)}{\ln(1+4x+x^2)} = \lim\limits_{x \to 0}\frac{\frac{1}{1+x}}{\frac{4+2x}{1+4x+x^2}}= \frac{1}{4}$,
I then wondered, if I could sandwich it andobserved that for: $\frac{x-1}{x} \le \ln(x) \le x-1 $, that:$ \frac{1}{4} \xleftarrow{x \to 0} \frac{\frac{x+1-1}{x+1 }}{1+4x+x^2-1} \le \frac{\ln(x+1)}{\ln(1+4x+x^2)} \le \frac{x+1-1}{\frac{1+4x+x^2-1}{1+4x+x^2}} \xrightarrow{x \to 0}\frac{1}{4} $
I then however wondered, if it were directly possible to get the result by cleverly manipulating the variables, so that the term simplifies itself to $\frac{1}{4}$ in the limit. Simply substituting $x+1$ or $1+4x+x^2$ with $e^u$ seems rather cumbersone, if (!) it is constructive at all. I am afraid, that there is just no real nice simplification of above. But maybe someone knows another nice, elegant way to get this limit :)
I am always happy to expand and practice my toolkit.
As always thanks in advance.
| Note that we can write $\log(1+4x+x^2)=\log(1+4x)+\log\left(1+\frac{x^2}{1+4x}\right)$.
Therefore, we have
$$\begin{align}
\frac{\log(1+x)}{\log(1+4x+x^2)}&=\frac{\log(1+x)}{\log(1+4x)+\log\left(1+\frac{x^2}{1+4x}\right)}\\\\&=\frac{\log(1+x)}{\log(1+4x)}\left(1+\frac{\log\left(1+\frac{x^2}{1+4x}\right)}{\log(1+4x)}\right)^{-1}
\end{align}$$
Inasmuch as $\displaystyle \lim_{x\to0}\left(1+\frac{\log\left(1+\frac{x^2}{1+4x}\right)}{\log(1+4x)}\right)=1$, the problem boils down to evaluating the limit
$$\begin{align}
\lim_{x\to 0}\frac{\log(1+x)}{\log(1+4x)}&=\frac14\lim_{x\to 0}\left(\frac{\log(1+x)}{x}\,\,\frac{4x}{\log(1+4x)}\right)\\\\
&=\frac14
\end{align}$$
where we used
$$\begin{align}
\lim_{h\to 0}\frac{\log(1+h)}{h}&=\lim_{h\to 0}\frac{\log(1+h)-\log(1)}{h}\\\\
&=\left. \left(\frac{d\log(x)}{dx}\right)\right|_{x=1}\\\\
&=\left. \left(\frac{1}{x}\right)\right|_{x=1}\\\\
&=1
\end{align}$$
with $h=x$ and $h=4x$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1946768",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Is my proof that $x^2+y^2+z^2 ≥ xy+yz+xz$ correct? The question:
Prove that $x^2+y^2+z^2 ≥ xy+yz+xz$ for all real numbers $x$, $y$ and $z$.
This problem has been posed before, but my question is whether my proof below is correct, since it seems the other answers to this problem are different.
If $x$, $y$ and $z$ are real numbers then $(x-y-z)^2 \geqslant 0$. That is $x^2 + y^2 +z^2 -2xy-2xz-2yz \geqslant 0$. But this implies that $\frac{x^2 + y^2 +z^2}{2} \geqslant xy + xz + yz$, so because $x^2 + y^2 +z^2 \geqslant \frac{x^2 + y^2 +z^2}{2} \geqslant xy + xz + yz$ we get the desired result.
| another way $$xy+yz+zx\le \sqrt { \left( x^{ 2 }+{ y }^{ 2 }+{ z }^{ 2 } \right) \left( x^{ 2 }+{ y }^{ 2 }+{ z }^{ 2 } \right) } =x^{ 2 }+{ y }^{ 2 }+{ z }^{ 2 }\\ \\ \\ $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1948633",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Find the equation of the circle passing ... Find the equation of the circle passing through the points $P(5,7)$, $Q(6,6)$ and $R(2,-2)$.
My Attempt:
Let the equation of the circle be:
$$x^2+y^2+2gx+2fy+c=0$$
The point $P(5,7)$ lies on the circle then,
$$5^2+7^2+10g+14f+c=0$$
$$10g+14f+c=-74$$-----(1)
The point $Q(6,6)$ lies on the circle then,
$$6^2+6^2+12g+12f+c=0$$
$$12g+12f+c=-72$$-------(2)
The point $R(2,-2)$ lies on the circle the,
$$2^2+(-2)^2+4g-4f+c=0$$
$$4g-4f+c=0$$-----(3).
Now, please help me from here.
| Another way to do this is to use the geometric fact that the perpendicular bisector of a chord passes through the center of a circle. The line through P(5,7) and Q(6,6)I has slope (7- 6)/(5- 6)= -1 and midpoint (11/2, 13/2). The perpendicular bisector is y= (x- 11/2)+ 13/2)= x+ 1. The line through Q(6, 6) and R(2,−2) has slope (6-(-2))/(6- 2)= 8/4= 2 and midpoint (4, 2). The perpendicular bisector is y= -(1/2)(x- 4)+ 2= -(1/2)x+ 4.
Those two lines intersect when y= x+ 1= -(1/2)x+ 4 so (3/2)x= 3, x= 2. Then y= 2+ 1= 3. The center of this circle is (2, 3) and the radius is $\sqrt{(2- 6)^2+ (6- 3)^2}= \sqrt{25}= 5$. The equation of that circle is $(x- 2)^2+ (y- 3)^2= 25$.
As a check $(5- 2)^2+ (7- 3)^2= 9+ 16= 25$, $(6- 2)^2+ (6- 3)^2= 16+ 9= 25$, and $(2- 2)^2+ (-2- 3)^2= 0+ 25= 25$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1948723",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 2
} |
Find $a$ such that $(x+a)(x+1991)+1=(x+b)(x+c)$ with $a,b,c\in\Bbb Z$
Find all integer values of $a$ such that the quadratic expression $(x+a)(x+1991)+1$ can be factored as a product $(x+b)(x+c)$ where b and c are integers.
I tried to do it by comparing the two expressions but I can't proceed.
| We have that $(x+a)(x+1991) + 1 = x^2 + (1991 + a)x + (1991a + 1)$. So if such factorisation exists we must have that the discriminant of the quadratic equation is a square. Now:
$$D = (1991 + a)^2 - 4(1991a + 1) = 1991^2 + 2 \cdot 1991a + a^2 - 4 \cdot 1991a - 4 = (1991 - a)^2 - 4$$
But then $\sqrt{D}, 2, (1991 - a)$ make a Pythagorean triplet, but we know that the only squares whose difference is $4$ are $4$ and $0$, therefore we must have $1991-a = \pm 2 \implies a_1=1989, a_2 = 1993$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1949062",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
How to find the inverse of $f(x)=\dfrac{-x|x|}{1+x^2}$? I can see that
$f(x)= \dfrac{-(x^2)}{1+x^2}$ for $x \geq 0$
and $f(x)= \dfrac{(x^2)}{1+x^2}$ for $x<0$.
Help me proceed to find the inverse.
| You can usted the fact that f(x) is a Odd function, so if $x>0$ then $y=f(x)=\frac{-x^2}{1+x^2}$, so $y+yx^2=-x^2$ and $x^2=\frac{-y}{1+y}$ finally $f(x)^{-1}=\sqrt{\frac{-x}{1+x}}$ for $x>0$, and $f(x)^{-1}=\sqrt{\frac{x}{1-x}}$ for $x\leq0$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1953547",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
What will be the value of the following determinant without expanding it? $$\begin{vmatrix}a^2 & (a+1)^2 & (a+2)^2 & (a+3)^2 \\ b^2 & (b+1)^2 & (b+2)^2 & (b+3)^2 \\ c^2 & (c+1)^2 & (c+2)^2 & (c+3)^2 \\ d^2 & (d+1)^2 & (d+2)^2 & (d+3)^2\end{vmatrix} $$
I tried many column operations, mainly subtractions without any success.
| If you expand the binomials and you subtract the $1^{\rm st}$ column from the 2nd, 3rd, and 4th you get
$$
\begin{vmatrix}a^2 & 2a+1 & 4a+4 & 6a+9 \\ b^2 & 2b+1 & 4b+4 & 6b+9 \\ c^2 & 2c+1 & 4c+4 & 6c+9 \\ d^2 & 2d+1 & 4d+4 & 6d+9\end{vmatrix}.
$$
Next subtract 2 times the 2$^{\rm nd}$ column from the 3$^{\rm rd}$ one, and 3 times the 2$^{\rm nd}$ column from the 4$^{\rm th}$ one:
$$
\begin{vmatrix}a^2 & 2a+1 & 2 & 6 \\ b^2 & 2b+1 & 2 & 6 \\ c^2 & 2c+1 & 2 & 6 \\ d^2 & 2d+1 & 2 & 6\end{vmatrix}=0.
$$
Now it is clear that the determinant is zero, as the fourth column is a multiple of the third one (or, factor 3 from the fourth column to get two equal columns).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1953843",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "27",
"answer_count": 7,
"answer_id": 3
} |
If $x*y$ is the operation $x*y = xy/(x+y+1)$, then how does $x*y*z = xyz(x+y+1)/(x+y+z+xy+xz+1)$? $$x*y = \frac{xy}{x+y+1}$$
What I got was,
$$x*(y*z) = x* \frac{yz}{y+z+1} =x\frac{\frac{yz}{y+z+1}}{x+\frac{yz}{y+z+1}+1}$$
The book's answer is
$$x*(y*z) = x*\frac{yz}{y+z+1} = \frac{xyz(y+z+1)}{x+y+z+xy+xz+1}$$
| $x*y=\frac{xy}{x+y+1}$
We put $ b=x*y=\frac{xy}{x+y+1}. $
$b*z=\frac{bz}{b+z+1}=\frac{\frac{xy}{x+y+1}z}{\frac{xy}{x+y+1}+z+1}$
$x*y*z=\frac{zxy}{xy+xz+yz+z+x+y+1}$
So: the book is wrong
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1954170",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
If the value of $xy$ , If $ x^2 + y^2 = 34$ and $x + y =8$ If $ x^2 + y^2 = 34$ and $x + y =8$ find the value of xy
So what should I be considering here the sum of two squares or..
I'm struggling here & any help to figure this out will be kindly appreciated
| $x^2$ + $y^2$ = $34$ $-->$ Equation 1
$x + y = 8$ $-->$ Equation 2
Now, we can rewrite Equation 2 with respect to x. This value will be substituted into Equation 1.
$x = 8 - y$
$x^2 + y^2 = 34$
$(8 - y)^2 + y^2 = 34$
$64 - 16y + y^2 + y^2 = 34$
$64 - 16y + 2y^2 = 34$
$2y^2 - 16y + 30 = 0$
Solving the quadratic above, the roots are: $$y1 = 5$$ and $$y2 = 3$$
Now plug both of these values into Equation 2. This will help you find the corresponding x-value. Once this is obtained, multiply the two numbers.
$x + y = 8$
$x + 5 = 8$
$x = 3$
$$xy = 3(5) = 15$$
This is the first solution.
$x + y = 8$
$x + 3 = 8$
$x = 5$
$$xy = 5(3) = 15$$
In both cases, the solution is 15, and this is due to the symmetry of the two points. Therefore, xy must be equal to 15.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1962121",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Upper bound on $\displaystyle\sum_{\text{cyc}}\dfrac{a}{a^3+b^2+c}$ Let $a,b,c$ be positive reals such that $a+b+c=3$. Determine the largest possible value of $$\dfrac{a}{a^3+b^2+c}+\dfrac{b}{b^3+c^2+a}+\dfrac{c}{c^3+a^2+b}.$$
Experimenting for some values of $(a,b,c)$ I conjectured that the maximum value attained is $1$. But I am unable to prove this upper bound. Any hints or solutions are welcome.
| Hint:
Using Holder: $(a^3+b^2+c)(1+b+c)(1+1+c) \ge (a+b+c)^3$ to yield $$\frac{a}{a^3+b^2+c} \le \frac{a(1+b+c)(2+c)}{(a+b+c)^3}$$ and reduce the inequality to $5\sum ab + \sum a^2b + 3abc \le 21$ which is easy.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1962291",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
} |
Expressing $(\alpha-1)^{-1}$ where $\alpha^3-\alpha^2+\alpha+2=0$ Let $E=\Bbb{Q}[\alpha]$ where $\alpha^3-\alpha^2+\alpha+2=0$
I would like to express $(\alpha^2+\alpha+1)(\alpha^2-1)$ and $(\alpha-1)^{-1}$ on the form $a+b\alpha+c\alpha^2$ where $(a,b,c)\in\Bbb{Q}^3.$
For the first one I get $-4\alpha-2$, but for the inverser I am not sure how can I do that, I would like to say that it's $\alpha-1$ but it seems 'silly'.
| Let's try to represent $(\alpha - 1)^{-1}$ as $a + b\alpha + c\alpha^2$. We must have
$$ (\alpha - 1)(a + b\alpha + c\alpha^2) = a\alpha + b\alpha^2 + c\alpha^3 - a - b\alpha - c\alpha^2 \\
= c\alpha^3 + (b - c)\alpha^2 + (a-b)\alpha -a = \\
c(\alpha^3 - \alpha^2+\alpha +2) +b\alpha^2+(a-b-c)\alpha - (2c + a) = 1. $$
Thus, by comparing coefficients, we see that we want
$$ b = 0, a - b - c = 0, 2c + a = -1 $$
which is a linear system of equations whose solution is $b = 0$ and $a = c = -\frac{1}{3}$. That is, $(\alpha - 1)^{-1} = -\frac{1}{3} (1 + \alpha^2)$. Checking ourselves, we indeed have
$$ -\frac{1}{3}(\alpha - 1)(1 + \alpha^2) = -\frac{1}{3} \left( \alpha^3 - \alpha^2 + \alpha - 1 \right) = -\frac{1}{3} \left( \alpha^3 - \alpha^2 + \alpha + 2 - 3 \right) \\
= -\frac{1}{3}( -3) = 1.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1963526",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Prove the inequality $\frac{1}{1+a+b}+\frac{1}{1+c+b}+\frac{1}{1+a+c}+a+b+c \le 3 + \frac 13 (ab+bc+ac)$ Let $a,b,c \in [0;1]$. Prove that
$$\frac{1}{1+a+b}+\frac{1}{1+c+b}+\frac{1}{1+a+c}+a+b+c \le 3 + \frac 13 (ab+bc+ac).$$
I used Buffalo way. Can anyone offer an easier way
| Let $f(a,b,c)=\sum\limits_{cyc}\left(\frac{1}{1+a+b}+a-1-\frac{ab}{3}\right)$.
Hence, $f$ is a convex function of $a$, of $b$ and of $c$.
Thus, $\max\limits_{\{a,b,c\}\subset[0,1]}f=\max\limits_{\{a,b,c\}\subset\{0,1\}}f=f(0,0,0)=0$.
Done!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1963921",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Trying to prove, by induction, that $2^{4n} + 5 $ is divible by $21.$ I want show by induction
$$ 21 \mid (2^{4n}+5) $$
So I assume:
$ 2^{4k}+5= 21p$
to prove that $ 21 \mid 2^{4(k+1)}+5 $
So I get it:
$2^{4(k+1)}+5 = 2^{4k+4}+5 = 2^{4k}2^{4}+
2^{4}2^{4k}+5 = 2^{4k} 16 +5 $ =
$16(2^{4k} +5 -5 )+5 = 16(21p-5)+5 = 16 \cdot 21p - 80+5 = 16 \cdot 21p - 75 $
But its not divisible by 21. Whats I doing wrong ?
| for $n=2$
$2^{4n}+5=256+5=261$
and
$261=21\times12+9$
so your identity is not true for $n=2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1966026",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
conditional probability that a freshman will graduate
A university finds that 75% of its graduating seniors scored above 80 on the entrance
exam, while only 50% of those who fail to graduate score above 80. They also find that half
of the entering freshmen graduate. What is the conditional probability that a freshman will
graduate given that:
(a) he or she scored above 80 on the entrance exam;
(b) he or she scored 80 or below on the entrance exam?
My solution:
P(graduating seniors with score above 80)=$\frac{3}{4}$
P(non-graduating seniors with score above 80)=$\frac{1}{2}$
P(freshman will graduate)=$\frac{1}{2}$
(a)Using Bayes, $\frac{P(freshman will graduate, scored above 80)}{P(scored above 80)}$=
$\frac{1}{2}$ * $\frac{3}{4}$/$\frac{1}{2}$+$\frac{3}{4}$=$\frac{3}{10}$
(b)$\frac{1}{2}$
As given condition is either he scores above or below, it doesn't make any difference to the result.
Am I correct?
| We introduce the following notation.
*
*$\Omega$ - the set of admitted students, which we assume to be finite and non-empty.
*$A$ - the set of graduating students.
*$B$ - the set of students who scored above $80$ on the entrance exam.
*$P$ - the uniform probability on $\Omega$.
It is given that
$$
\begin{align}
P(B|A) &= \frac{3}{4} \\
P(B|A^c) &= \frac{1}{2} \\
P(A) &= \frac{1}{2}
\end{align}
$$
a. We are to calculate $P(A|B)$.
$$
\begin{align}
P(B) &\overset{\text{total prob.}}{=} P(B|A)P(A)+P(B|A^c)P(A^c) \\
&= \frac{3}{4}\frac{1}{2}+\frac{1}{2}\left(1-\frac{1}{2}\right) \\
&= \frac{5}{8} \\
P(A|B) &\overset{\text{Bayes}}{=} \frac{P(B|A)P(A)}{P(B)} \\
&= \frac{\frac{3}{4}\frac{1}{2}}{\frac{5}{8}} \\
&= \frac{3}{5}
\end{align}
$$
b. We are to calculate $P(A|B^c)$.
$$
\begin{align}
P(A|B^c) &\overset{\text{Bayes}}{=} \frac{P(B^c|A)P(A)}{P(B^c)} \\
&= \frac{\left(1-P(B|A)\right)P(A)}{1-P(B)} \\
&= \frac{\left(1-\frac{3}{4}\right)\frac{1}{2}}{1-\frac{5}{8}} \\
&= \frac{1/8}{3/8} \\
&= \frac{1}{3}
\end{align}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1967945",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Is this the shortest proof of Fermat's last theorem for exponent 3? To begin with, here we only consider the case of $n^3 + (n+1)^3 = (n+2)^3$, that is the case of consecutive integers $(x,y,z)$ in the well known equation. The general case $x^3 + y^3 = z^3$ is much more difficult to prove (and the work is yet to be done with the method outlined below).
Here we use the integer representation of a cube $n^3$ as a sequence of consecutive odd numbers. We get the sequence by writing $n^3=n\cdot n^2$. This means that we need $n$ terms with an average value of $n^2$ to represent the cube $n^3$. This representation has a big advantage over other possible representations and it is that consecutive cubes $n^3$, $(n+1)^3$, and $(n+2)^3$ have a representation that ends when the next one starts (that is they do not share any term).
For example $2^3=2\cdot 2^2= 3+5$ and $3^3=3\cdot 3^2=7+9+11$. This is a well known property of cubes when represented as a sequence of consecutive odd integers (https://en.wikipedia.org/wiki/Cube_(algebra)). This means that we can concatenate the left hand side of the equation to form a new sequence. And we need to figure out what integer this concatenated sequence represents.
It was not really difficult to do that.
So we gave an example and show why the left hand side cannot represent a cube whose value is the right hand side of the equation.
$$6^3 + 7^3 = 8^3.$$
We write down the representation of each sequence:
$$6^3 = 6\cdot 6^2 = 31 + 33 + 35 + 37 + 39 + 41,$$ that is 6 terms with average value of $6^2=36$.
We do the same for $7^3$.
$$7^3 = 7\cdot7^2 = 43 + 45 + 47 + 49 + 51 + 53 + 55,$$ that is 7 terms with average value of $7^2=49$.
When we concatenate the two sequences into one we will call lhs, we get:
$$\text{lhs} = 31 + 33 + 35 + 37 + 39 + 41 + 43 + 45 + 47 + 49 + 51 + 53 + 55.$$
This sequence has $(6+7)=13$ terms with average $(n^2 + (n+1)^2 +1)/2$. In our case, the lhs is therefore a sequence with 13 terms with average 43 which represents the integer $m=13\cdot 43=559=13\cdot43$.
The right hand side is $\text{rhs}=8^3=8\cdot8^2$, so it will have 8 terms with an average of $64$. $$\text{rhs} = 57 + 59 + 61 + 63 + 65 + 67 + 69 + 71.$$ So looking at the result, we see that $\text{lhs}=13\cdot 43$ cannot be equal to $\text{rhs}=8^3$. That is $(2n+1)\cdot (n^2 + (n+1)^2 +1)/2$ cannot be equal to $(n+2)^3$.
I am not sure if this method can be extended to the general case where $x,y$ and $z$ can take any value (not necessarily consecutive). But I think it's worth looking into it because of the simplicity of the method.
| No.
The presumably shortest proof for the case you consider is this
Assume $$n^3+(n+1)^3=(n+2)^3. $$
Then
$$ 0=n^3+(n+1)^3-(n+2)^3= n^3-3n^2-9n-7=(n^2-3n-9)\cdot n-7.$$
We conclude that $n\mid 7$, but neither $1^3+2^3=3^3$, nor $7^3+8^3=9^3$. $\square$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1969973",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Prove $30|(a^3b-ab^3) $ Prove that if three distinct integers are chosen at random then there will exists two among them, say $a$ and $b$ such that $30|(a^3 b-ab^3)$
| By Fermat's little theorem (or simply by direct calculation), $x^2\equiv x$ mod $2$ and $x^3\equiv x$ mod $3$ for any integer $x$. It follows that
$$a^3b\equiv ab\equiv ab^3\mod 6$$
for any pair of integers $a,b$, hence $6$ divides $a^3b-ab^3$ for any pair. It remains to show that $5$ divides $a^3b-ab^3=ab(a^2-b^2)$ for at least one of the three pairs. But there are only three squares mod $5$, namely $0$, $1$, and $-1$. So given three integers, either two of their squares, say $a^2$ and $b^2$ belong to the same residue class, in which case $5$ divides $a^2-b^2$, or one of the three squares, say $a^2$ is $0$ mod $5$, in which case $5$ divides $a$ (and you can let $b$ be either of the other two numbers). In either case $5$ divides $ab(a^2-b^2)=a^3b-ab^3$ for the chosen pair, and we are done.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1972373",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
} |
Develop $(\frac{2}{x}-x)^8$ $(\frac{2}{x}-x)^8$
How do I develop this?
My book gives me four options to what that one of the terms of the expression is supposed to look like after the expression is developed. The right option is:$$^8C_3*(\frac{2}{x})^3*(-x)^5$$
I even tried graphing both expressions on my graphical calculator and the graphs are different. Could this be a mistake from my book? If not, how do I solve this?
| The function is:
$$ f(x) = \left( \frac{2}{x} - x \right)^8 = \frac{(2-x^2)^8}{x^8} = x^8 \, \left(\frac{2}{x^2} - 1 \right)^8. $$
These have the expansions:
\begin{align}
f(x) &= \sum_{k=0}^{8} \binom{8}{k} \, \left(- \frac{2}{x}\right)^{8-k} \, (-x)^k \\
&= \frac{1}{x^8} \, \sum_{k=0}^{8} \binom{8}{k} \, 2^{8-k} \, (-1)^k \, x^{2k}
\end{align}
other expansions follow in a similar manor.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1975013",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Prove that any odd number can be a leg in a Pythagorean Triple $(a,b,c)$ where $a^2+b^2=c^2$ is a Pythagorean triple.
My first thought was to do a proof by cases. I have three cases :
*
*a is odd, b is odd.
*a is odd, b is even.
*a is even, b is odd.
I'll just show case one :
$(2k+1)^2 + (2k)^2 = c^2$
I am kind of stuck relating how this answers the question. Thank you for any guidance in the right direction.
| Hint:
Pythagorean Triples comes in the form: $(x^2-y^2,2xy,x^2+y^2)$ where $x$ and $y$ are any positive integers with $x>y$.
Can you use this to show that any odd number is possible?
Edit:
$(x^2-y^2,2xy,x^2+y^2)$ is equivalent to $(a,b,c)$ because:
$$\big(x^2-y^2\big)^2\ +\ \big(2xy\big)^2\ =\ \big(x^2+y^2\big)^2$$
Answer
A Pythagorean Triple can be made from the values $(x^2-y^2,2xy,x^2+y^2)$
We require one of the numbers to be odd and be any possible odd number.
We will show that $x^2-y^2$ can represent any odd number with suitable choices of $x$ and $y$.
So let $2n-1$ be the target odd number.
If $x=n$ and $y=n-1$ then $x^2-y^2=n^2-(n-1)^2=2n-1$
So we can create the Pythagorean Triple.
To expand upon that, given an odd number $k$ we can construct a Pythagorean Triple as follows:
We have $k=2n-1$ so $x=n=\frac{k+1}{2}$ and $y=n-1=\frac{k-1}{2}$.
Hence the Pythagorean Triple is:
$$\bigg\{\left(\frac{k+1}{2}\right)^2-\left(\frac{k-1}{2}\right)^2,2\cdot\left(\frac{k+1}{2}\right)\cdot\left(\frac{k-1}{2}\right),\left(\frac{k+1}{2}\right)^2+\left(\frac{k-1}{2}\right)^2\bigg\}$$
$$=\bigg\{\frac{k^2+2k+1-k^2+2k-1}{4},\frac{k^2-1}{2},\frac{k^2+2k+1+k^2-2k+1}{4}\bigg\}$$
$$=\bigg\{k,\frac{k^2-1}{2},\frac{k^2+1}{2}\bigg\}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1975146",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
If $n > 1$ is an integer not of the form $6k + 3$, prove that $n^{2} + 2^{n}$ is composite. I try to divide in to $5$ case $6k, 6k+1, 6k+2, 6k+4, 6k+5$
and put them in $n^{2}$ and consider the possible last digit 0f $2^n$
and sum with those form
I can only prove $6k, 6k+2, 6k+4$ case .the last digit is even number and can divide by $2$.
But I can't prove $6k+1, 6k+5$ case because might be odd and not guarantee Ican divide by some numbers
Anyone can help me by give a hint ?
| $n = 6k + i$. If $i$ is even then $n$ is even and $n^2 + 2^n$ is even and as $n^2 + 2^n > 2$ it must be composite.
So it remains to check for $n = 6k \pm 1$.
$(6k \pm 1)^2 + 2^n = 36k^2 \pm 12k + (1 + 2^n)$.
$3|36k^2 \pm 12k$ so if we can show $3|1+2^n$ we are done as $3|n^2 + 2^n$ and as $n^2 + 2^n > 3$ if $n > 1$.
We see $3|1+2^1, 1+2^3, 1+2^5...$ but $3\not \mid 1+2^0, 1+2^2, 1+2^4,...$ so it really looks like $3|1 + 2^{odd}$ which is the case for $n= 6k \pm 1$.
That really seems familiar... we can maybe search it. ... Or we can just do it ourselves.
$2^{2m + 1} +1 = 2^{2m}2 + 1 = 4^m*2 + 1$ and $4^m*2 + 1 \equiv 1^m*2 + 1 \equiv 0 \mod 3$.
So... that's there... $3|2^{odd} + 1$ so...
Case 1-3: $n = 6k + 0,2,4$ then $2|n^2 + 2^n$ and as $n> 0$, $n^2 + 2^n \ne 2$.
Case 4,5: $n = 6k +1,5$ then $3|n^2 + 2^n$ as $n$ is odd. And as $n > 1$, $n^2 + 2^n \ne 3$.
Case 6: $n = 6k + 3$, then $n^2 + 2^n$ may or may not be prime. We weren't asked to check.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1975284",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Looking for $\tan\frac{1}{2}\theta$ Given that $$\frac{\cos \theta}{1-\sin\theta}=a,$$
how can I find the value of $\tan\frac{1}{2}\theta$?
I went in this way,
but the matter is in the box as shown (the fallacies).
| $$\frac{\cos\theta}{1-\sin\theta}=a$$
$$\frac{\frac{1-\tan^2\frac{\theta}{2}}{1+\tan^2\frac{\theta}{2}}}{1-\frac{2\tan\frac{\theta}{2}}{1+\tan^2\frac{\theta}{2}}}=a$$
$$\frac{1-\tan^2\frac{\theta}{2}}{1+\tan^2\frac{\theta}{2}-2\tan\frac{\theta}{2}}=a$$
$$(a+1)\tan^2\frac{\theta}{2}-2a\tan\frac{\theta}{2}+(a-1)=0$$
now solve quadratic for $\tan\frac{\theta}{2}$,
$$\tan\frac{\theta}{2}=\frac{2a\pm\sqrt{4a^2-4(a+1)(a-1)}}{2(a+1)}=\frac{a\pm1}{a+1}$$
hope it answers
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1976666",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
A curious case of $1729$ Ramanujan's Taxicab number 1729 is famous for being the smallest positive integer which can be written as the sum of two positive cubes in two different ways. On a different note, I observed that
$12^3 + 1^3 = 1729$
$12^2 + 1^2 = 1\cdot7\cdot2\cdot9 + 1 + 7 + 2 + 9$
$12^1 + 1^1 = -1+7-2+9$
Question 1: Is there any other $n$ number with the property that if $n = a^3 + b^3$ for some positive $a$ and $b$ then
$$
a^2 + b^2 = \text{Products of the digits of $n$} \ + \ \text{Sum of the digits of $n$}
$$
Question 2: While searching for a solution of Question 1, the program run by Peter has found only two solution, $(6,11)$ and $(1,12)$ for $1 \le a,b \le 20000$. Looks like there are no more solution. Can this be proven or disproven?
| If I understood correctly the problem is
Find numbers $n$ such that there exist a pair $(a,b)$ with the property that
$a^3+b^3=n$
$a^2+b^2$=products of the digits of n + Sum of the digits of n
Then, the number $1547$ is a solution, for the pair $(a,b)=(11,6)$:
$11^3+6^3=1547\\11^2+6^2=1\cdot 5 \cdot 4 \cdot 7 + 1 + 5 + 4 + 7\\11^1+6^1 = 1+5+4+7$
And of course also the number $0$ is a solution. I also feel that there are no more solutions.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1977686",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 3,
"answer_id": 0
} |
prove that if x is a real number, then $\lfloor2x\rfloor=\lfloor x\rfloor+\lfloor x + \frac{1}{2}\rfloor$ prove that if $x$ is a real number, then $\lfloor2x\rfloor=\lfloor x\rfloor + \lfloor x + \frac{1}{2}\rfloor$
I've seen the proof in my discrete math book, but it was very confusing. I don't understand why they considered those cases and why they chose $\frac{1}{2}$ specifically. It would be great if someone can explain better or provide a much clearer proof.
Proof from the book:
Let x = n + $\epsilon$ where n is an integer and 0 $\le\epsilon \lt 1$
case 1: Consider 0 $\le\epsilon \lt \frac{1}{2}$
2x = 2n + 2$\epsilon$ and $\lfloor$2x$\rfloor$ = 2n because 0 $\le 2\epsilon \lt 1$
Similarly, x + $\frac{1}{2}$= n + ( $\frac{1}{2}$+ $\epsilon$), so $\lfloor$x + $\frac{1}{2}$$\rfloor$ = n,
because 0 < $\frac{1}{2}$ + $\epsilon$ < 1.
Consequently,$\lfloor$2x$\rfloor$ = 2n and $\lfloor$x$\rfloor$ + $\lfloor$x + $\frac{1}{2}$$\rfloor$ = n + n = 2n.
case 2: $\frac{1}{2}$ ≤ $\epsilon$< 1.
2x = 2n + 2$\epsilon$ =(2n + 1) + (2$\epsilon$ − 1).
Because 0 ≤ 2$\epsilon$ − 1 < 1, it follows that $\lfloor$2x$\rfloor$ = 2n + 1. Because$\lfloor$x + $\frac{1}{2}$$\rfloor$ = $\lfloor$n + ( $\frac{1}{2}$+ $\epsilon$)$\rfloor$ = $\lfloor$n + 1 + ($\epsilon$ − $\frac{1}{2}$ )$\rfloor$ and 0 ≤ $\epsilon$ − $\frac{1}{2}$ < 1,
it follows that $\lfloor$x + $\frac{1}{2}$$\rfloor$ =n + 1.
Consequently, $\lfloor$2x$\rfloor$ = 2n + 1 and $\lfloor$x$\rfloor$ + $\lfloor$x + $\frac{1}{2}$$\rfloor$= n + (n + 1) = 2n + 1.
This concludes the proof.
| Let $x=n+\varepsilon$ where $n$ is an integer and $0\le\varepsilon\lt1.$ Then $$\lfloor2x\rfloor=\lfloor2n+2\varepsilon\rfloor=2n+\lfloor2\varepsilon\rfloor,$$
and
$$\lfloor x\rfloor+\lfloor x+\frac12\rfloor=\lfloor n+\varepsilon\rfloor+\lfloor n+\varepsilon+\frac12\rfloor=n+\lfloor\varepsilon\rfloor+n+\lfloor\varepsilon+\frac12\rfloor=2n+\lfloor\varepsilon+\frac12\rfloor,$$
so the identity
$$\lfloor2x\rfloor=\lfloor x\rfloor+\lfloor x+\frac12\rfloor$$
is equivalent to the identity (for $0\le\varepsilon\lt1$)
$$\lfloor2\varepsilon\rfloor=\lfloor\varepsilon+\frac12\rfloor.$$
Since $0\le2\varepsilon\lt2,$ the value of $\lfloor2\varepsilon\rfloor$ is either $0$ or $1,$ namely,
$$\lfloor2\varepsilon\rfloor=\begin{cases}
1\text{ if }2\varepsilon\ge1,\\
0\text{ otherwise.}
\end{cases}$$
Likewise, since $\frac12\le\varepsilon+\frac12\lt\frac32,$ the value of $\lfloor\varepsilon+\frac12\rfloor$ is either $0$ or $1,$ namely,
$$\lfloor\varepsilon+\frac12\rfloor=\begin{cases}
1\text{ if }\varepsilon+\frac12\ge1,\\
0\text{ otherwise.}
\end{cases}$$
Finally, to see that $\lfloor2\varepsilon\rfloor=\lfloor\varepsilon+\frac12\rfloor,$ note that the inequality $2\varepsilon\ge1$ and the inequality $\varepsilon+\frac12\ge1$ both have the same solution set, namely, $\varepsilon\ge\frac12.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1979483",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Find matrix $A$. How many solutions are there? The matrices below is matrix multiplcation of $AB=C$ but someone has changed the order of the coloumns in $C$ and erased what was in $A$. What is matrix $A$? Is it the only one? I have no idea how to start. I appreciate all the help I can get.
$$\begin{bmatrix} & & & \\ & & & \\ & & &\\& & & & \end{bmatrix}
\cdot\begin{bmatrix}2 & 0 &-1 & -2&-4&-4\\-1 & 1 & 0 & -1&2&-1\\ -4& 0 & 1 & 3&8&5\\ -1&0&0&1&2&1\end{bmatrix}=\begin{bmatrix}1 & 1 &3 & -1&0&2\\0 & 1 & 1 & 0&0&0\\0 & 0 & 0&-1&0& 2\\ 0&0&0&0&1&0\end{bmatrix}$$
| No solution (In case you don't shuffle $C$)
P.S: Thanks @JeanMarie
Denote the $(i,j)^{th}$ entry of $A$ by $a_{ij}$.
$$\begin{bmatrix} a_{11} &a_{12} &a_{13} &a_{14} \\ a_{21} &a_{22} &a_{23} &a_{24} \\ a_{31} &a_{33} &a_{33} &a_{34} \\a_{41} &a_{42} &a_{43} &a_{44} \end{bmatrix}
\cdot\begin{bmatrix}2 & 0 &-1 & -2&-4&-4\\-1 & 1 & 0 & -1&2&-1\\ -4& 0 & 1 & 3&8&5\\ -1&0&0&1&2&1\end{bmatrix}=\begin{bmatrix}1 & 1 &3 & -1&0&2\\0 & 1 & 1 & 0&0&0\\0 & 0 & 0&-1&0& 2\\ 0&0&0&0&1&0\end{bmatrix}$$
Multiply the first row of $A$ by the first column of $B$ (inner product), you get
$$T = 2a_{11} - a_{21} - 4a_{31} - a_{41} = 1$$
Multiply the first row of $A$ by the fifth column of $B$ (inner product), you get
$$-4a_{11} + 2a_{21} + 8a_{31} + 2a_{41} = 0$$
CONTRADICTION!! Since
$$-4a_{11} + 2a_{21} + 8a_{31} + 2a_{41} = 0 = -2 (2a_{11} - a_{21} - 4a_{31} - a_{41})= - 2T = -2 = 0 !!$$
Therefore NO SOLUTION
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1980645",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Find the sum of $\sum\limits_{n=1}^{\infty} \frac{x^{n}}{n(n+1)}$
Find the sum of $\sum\limits_{n=1}^{\infty} \frac{x^{n}}{n(n+1)}$ on its domain of convergence.
This is my idea.
We have the radius of convergence is $R=1$. And
$\sum_{n=1}^{\infty} \dfrac{x^{n}}{n(n+1)}=\dfrac{1}{x}\cdot \sum_{n=1}^{\infty} \dfrac{x^{n+1}}{n(n+1)}$.
Then let $f(x)=\sum_{n=1}^{\infty} \dfrac{x^{n+1}}{n(n+1)}$.
We have $f'(x)=\sum_{n=1}^{\infty} \dfrac{x^{n}}{n}$ and $f''(x)=\sum_{n=1}^{\infty} x^{n-1}=\dfrac{1-x^n}{1-x}$
Then if I take the primity of $f''(x)$, I get:
$f'(x)=\displaystyle\int\limits_{0}^{x}\dfrac{1-t^n}{1-t}\mathrm{d}t=-\ln(1-x)-\displaystyle\int\limits_{0}^{x}\dfrac{t^n}{1-t}\mathrm{d}t$
Then I will get a sum again, and that's not the goal.
What can I do then? Thank you so much.
| If you know (or can assume) the logarithm series for $|x|<1$:
$$-\log(1-x) = \sum_{n=1}^{\infty} \frac{x^n}{n}$$
then you can forego the derivation/integration steps by noticing that:
$$
\begin{align}
\sum_{n=1}^{\infty} \frac{x^{n}}{n(n+1)} & = \sum_{n=1}^{\infty} x^n \left(\frac{1}{n} - \frac{1}{n+1} \right) \\
& = \sum_{n=1}^{\infty} \frac{x^n}{n} - \frac{1}{x} \sum_{n=2}^{\infty} \frac{x^n}{n} \\
& = -\log(1-x) - \frac{1}{x}(-\log(1-x) - x) \\
& = \left(\frac{1}{x}-1\right)\log(1-x) \;+\; 1
\end{align}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1980875",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
find $b$ if function $f(x)$ is strictly decreasing function if $\displaystyle f(x) = \frac{2}{\sqrt{3}}\tan^{-1}\left(\frac{2x+1}{\sqrt{3}}\right)-\ln(x^2+x+1)+(b^2-5b+3)x+c$ is strictly decreasing
function forall real values of $x.$ then $b$
$\displaystyle f'(x) = 2\sqrt{3} \; \frac{1}{3+(2x+1)^2}-\frac{2x+1}{x^2+x+1}+(b^2-5b+3)$
function $f(x)$ is strictly decreasing if $f'(x)<0$ for all real $x$
$\displaystyle 2\sqrt{3} \; \frac{1}{3+(2x+1)^2}-\frac{2x+1}{x^2+x+1}+(b^2-5b+3)<0$
could some help me with this, thanks
| I think that you made a mistake in the derivative $$f(x)=\frac{2}{\sqrt{3}}\tan^{-1}\left(\frac{2x+1}{\sqrt{3}}\right)-\ln(x^2+x+1)+(b^2-5b+3)x+c$$ should lead, after basic simplifications, to $$f'(x)=(b^2-5b+3)-\frac{2 x}{x^2+x+1}$$ Since we are in the real domain, $(x^2+x+1)$ does not show any real root, then $f'x)<0$ reduces to $$(b^2-5b+3)(x^2+x+1)-2x <0$$
I am sure that you can take it from here.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1982621",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
When does $\int_0^{\infty}\frac{7x}{x^2+1}-\frac{7C}{3x+1}dx$ converge? I am required to find the values of $C$ for which the integral
$$\int_0^{\infty}\frac{7x}{x^2+1}-\frac{7C}{3x+1}dx$$ converges.
I know by experimentation that it converges when $C=3$. I am, however, unable to show this in a rigorous way.
I get stuck when I need to evaluate $$\lim_{x\rightarrow\infty}(\frac{7}{2}\ln(x^2+1)-\frac{7C}{3}ln(3x+1))$$
Any help would be appreciated.
Taking the hint from DonAntonio, I wrote the following solution. Let me know if there is anything wrong with the way I expanded DonAntonio's solution.
$$\frac{7x}{x^2+1}-\frac{7C}{3x+1}=7\cdot\,\frac{(3-C)x^2+x-C}{(x^2+1)(3x+1)}$$
When $C=3$
$$7\cdot\,\int_0^{\infty}\frac{(3-3)x^2+x-3}{(x^2+1)(3x+1)}=7\cdot\,\int_0^{\infty}\frac{x-3}{(x^2+1)(3x+1)}dx<\int_0^{\infty}\frac{x}{x^3}dx = \int_0^{\infty}\frac{1}{x^2} dx$$
Which is convergent.
When $C\neq3$
$$7\cdot\,\int_0^{\infty}\frac{(3-C)x^2+x-C}{(x^2+1)(3x+1)}=7\cdot\,\int_0^{\infty}\frac{(3-C)x^2}{(x^2+1)(3x+1)}+7\cdot\,\int_0^{\infty}\frac{x-C}{(x^2+1)(3x+1)}dx$$
$$=|7\cdot\,\int_0^{\infty}\frac{(3-C)x^2}{(x^2+1)(3x+1)}dx+K| > \int_0^{\infty}\frac{x^2}{x^4} dx = \int_0^{\infty}\frac{1}{x^2} $$
For some constant $K$
Which is divergent.
| Hints:
$$\frac{7x}{x^2+1}-\frac{7C}{3x+1}=7\cdot\,\frac{(3-C)x^2+x-C}{(x^2+1)(3x+1)}$$
Now, if $\;C\neq3\;$ then comparison with some multiple of $\;\frac1x\;$ will give you divergence...
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1982852",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
How to factorise quadratic equation in the form of $x^2+6xy+y^2$? My Text book has some questions of quadratic equations in the form of
$x^2+xy+y^2$
1)Is there any strategy or proper way to factorise quadratic equations in such way?
2)Is that a type of quadratic trinomials?if yes why?
| Hint: let $z = \frac{x}{y}$ then $a x^2+b xy+c y^2=y^2(a z^2+b z+c)$. The latter is a quadratic in $z$. Factoring it, then substituting $\frac{x}{y}$ back gives the factorization in terms of $x,y$.
[ EDIT ] Here is the fully worked out example for factoring $8x^2+6xy+y^2$.
*
*Let $z=\frac{x}{y}$ then write $8x^2+6xy+y^2 = y^2(8 z^2 + 6 z + 1)$.
*Factor $8 z^2 + 6 z + 1$ (either by inspection, or) by actually calculating its roots using the quadratic formula:
$$z_{1,2} = \frac{-6 \pm \sqrt{6^2 - 4 \cdot 8}}{16}= \frac{-6 \pm 2}{16} = \frac{-3 \pm 1}{8}$$
so $z_1 = -\frac{1}{2}$, $z_2=-\frac{1}{4}$
*It follows that $8 z^2 + 6 z + 1 = 8(z-z_1)(z-z_2) = 8(z+\frac{1}{2})(z+\frac{1}{4})=(2 z+1)(4 z+1)$
*Substituting back $z=\frac{x}{y}$:
$$y^2(8 z^2 + 6 z + 1)=y^2(2 \frac{x}{y}+1)(4 \frac{x}{y}+1)=(2x+y)(4x+y)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1983318",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Geometry with Trigonometry From the given figure, prove that
$$\cot\ \theta=\cot\ A+\cot\ B+\cot\ C$$
I could not even start to the first point. Please help me to solve this.
Thanks in advance
| If $O$ is the intersection of the inner lines,
$\angle ABO=B-\theta\implies\angle AOB=\pi-B,$
$\angle BCO=C-\theta\implies\angle COB=\pi-C$
Writing $AB=c$, $BC=a$, $CA=b$ (conventionally)
Using Sine Law for $\triangle AOB,$ $$\dfrac{AB}{\sin(\pi-B)}=\dfrac{OB}{\sin\theta}\implies\dfrac c{\sin B}=\dfrac{OB}{\sin\theta}\ \ \ \ (1)$$
Similarly from $\triangle COB,$ $$\dfrac{OB}{\sin(C-\theta)}=\dfrac a{\sin C}\ \ \ \ (2)$$
Equating the values of $OB,$
$$\sin C\sin\theta=\dfrac ac\sin B\sin(C-\theta)\ \ \ \ (3)$$
Now using Sine Law for $\triangle ABC,$ $$\dfrac a{\sin A}=\dfrac c{\sin C}\iff\dfrac ac=\dfrac{\sin A}{\sin C}\ \ \ \ (4)$$
By $(3),(4)$
$$\sin\theta\cdot\sin^2C=\sin A\sin B\sin(C-\theta)=\sin A\sin B\{\sin C\cos\theta-\cos C\sin\theta\}$$
$$\implies\dfrac{\cos\theta}{\sin\theta}=\dfrac{\sin^2C+\sin A\sin B\cos C}{\sin A\sin B\sin C}\ \ \ \ (5)$$
Again, $$\cot A+\cot B+\cot C=\dfrac{\cos A\sin B\sin C+\sin A\cos B\sin C+\sin A\sin B\cos C}{\sin A\sin B\sin C}$$
$$=\dfrac{(\cos A\sin B+\sin A\cos B)\sin C+\sin A\sin B\cos C}{\sin A\sin B\sin C}\ \ \ \ (6)$$
Finally use $(\cos A\sin B+\sin A\cos B)\sin C$
$=\sin(A+B)\sin C=\sin(\pi-C)\cdot\sin C=\sin^2C$ on $(6)$
and compare $(5),(6)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1984128",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
The sum of two integers is a square. The sum of their squares is also a square. The sum of two integers is a square. The sum of their squares is also a square. E.g.
$$
-3 + 4 = 1^2, \ (-3)^2 + 4^2 = 5^2
$$
$$
9 + 40 = 7^2, \ 9^2 + 40^2 = 41^2
$$
Question: General solution/form of all such pairs.
Clarification: By general form we mean a parametric form like Euclid's solution for Pythagorean triplets $a = r^2 - s^2, \ b = 2rs$ and $c = r^2 + s^2$ which covers all Pythagorean triplets.
| WLOG for $d^2=a^2+b^2, a=2mn, b=m^2-n^2$
$a+b=m^2-n^2+2mn$
Let $m^2-n^2+2mn=(m+pn)^2\iff n(p^2+1)=2m(1-p)$
$\dfrac n{2(1-p)}=\dfrac m{1+p^2}=r$(say)
Now $(1+p^2,1-p)|(1+p^2,1-p^2)$ which must divide $1+p^2+1-p^2=2$
Case $\#1:$ $(1+p^2,2(1-p))=1$ if $p$ is even
$n=2r(1-p^2),m=r(1+p^2)$
Case $\#2:$
For odd $p, p^2\equiv1\pmod8, 1+p^2\equiv2$
In that case, $(1+p^2,2(1-p))=2$
$n=r(1-p)$ and $m=\dfrac{r(1+p^2)}2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1985697",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Divisibility of $8^{2000}+12^{2000}-5^{2000}-1$ Choose the correct option:
$8^{2000}+12^{2000}-5^{2000}-1$ is divisible by:
$(1)$ $323$
$(2)$ $221$
$(3)$ $299$
$(4)$ $237$
Could someone give me slight hint to solve this problem?
| The expression $E := 8^{2000}+12^{2000}-5^{2000}-1$ will obviously be divisible by $K$ if each term $ \equiv 1 \bmod K$, but may also be divisible by $K$ even if this is not true.
The first thing to do is factor the four numbers given : $323= 17\times 19$, $221 = 13 \times 17$, $299 = 13 \times 23$, $237=3 \times 79$.
Because $16$ divides $2000$, we know from Fermat's Little Theorem that $17$ divides $E$ -- for any $A$ not divisible by $17$, $A^{2000} \equiv (A^{16})^{125} \equiv 1^{125} \equiv 1 \bmod 17$, and then $E \equiv 1+1-1-1 \equiv 0 \bmod 17$
Similarly we know that $3 \not\mid E$, since $8^{2000} \equiv 5^{2000} \equiv 1 \bmod 3$ but $12^{2000} \equiv 0 \bmod 3$, so $E \equiv 1+0-1-1 \equiv -1 \bmod 3$ (eliminating option (4)).
Because $8 = 2^3$, we know that $8^4 \equiv 1 \bmod 13$, and since $5^2 =25 \equiv -1 \bmod 13$, $5^4 \equiv 1 \bmod 13$ also. Clearly $12^4 \equiv-1^4 \equiv 1 \bmod 13$ also and since $4 \mid 2000$ we know that $13 \mid E$.
This gives us answer (2), $221$ as a correct choice since we know $E$ is divisible by both $13$ and $17$.
As it happens, $E$ is also divisible by $23$ but in that case the components of $E$ do not come out to $1 \bmod 23$. So there is another valid answer on the list, $299$, but much more difficult to demonstrate.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1986502",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
Show that $\{a_n\}$ is an increasing sequence and bounded from above Question
Let $a_1$ = 1 and $a_{n+1} = \sqrt{1+2a_n}$. Show that $\{a_n\}$ is increasing and bounded from above.
Attempted solution
showing that the sequence is increasing:
\begin{align*}
a_n &\geq 1 \quad \forall n = {1,2,3,...} \\
\Rightarrow 1+2a_n &\geq 3 \\
\Rightarrow \sqrt{1+2a_n} &\geq \sqrt{3} \\
\Rightarrow a_{n+1} &\geq \sqrt{3} \quad \forall n = {1,2,3,...} \\
\Rightarrow a_{n+1} &\geq a_n
\end{align*} Is this a valid approach?
Assuming that I have shown that the sequence is increasing, I show that it is bounded by doing the following
\begin{align*}
a_{n+1} &\geq a_n \\
\Rightarrow \sqrt{1+2a_n} &\geq a_n \\
\Rightarrow 1+2a_n &\geq a^2_n \\
\Rightarrow (a_n+1)^2 - a^2_n &\geq a^2_n \\
\Rightarrow \frac{a_n + 1}{a_n} &\geq \sqrt{2} \\
\Rightarrow a_n &\leq 1+\sqrt{2} \quad \forall n = 1,2,3,...
\end{align*}
Is there another more straightforward way to show that this sequence is bounded?
| You have taken $a_1 = 1$. For this value $ { a_{n+1}} $ keeps on increasing. But had you taken $ a_1$ greater than $ 1\, + \, √2, a_{n+1}$ decreases. Now, for all values of $ a_1$ less than $1\, +√2, \, a_{n+1} $ increases. And as long as $a_n$ is less than $ 1+√2,$ the $a_{n+1}$ is also less than $1+√2.$ Thus for $a_1$ less than $1+√2,$ sequence is increasing and bounded by $ 1+√2.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1987861",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Integrate $I=\int_0^1\frac{\arcsin{(x)}\arcsin{(x\sqrt\frac{1}{2})}}{\sqrt{2-x^2}}dx$ How to prove
\begin{align}
I &= \int_0^1\frac{\arcsin{(x)}\arcsin{(x\sqrt\frac{1}{2})}}{\sqrt{2-x^2}}dx \\
&= \frac{\pi}{256}\left[ \frac{11\pi^4}{120}+2{\pi^2}\ln^2{2}-2\ln^4{2}-12\zeta{(3)}\ln{2} \right]
\end{align}
By asking $$x=\sqrt{2}y$$ then using integration by parts, we have
$$I=\frac{\pi^5}{2048}-\frac{1}{4}\int_0^1{\arcsin^4\left( \frac{z}{\sqrt{2}}\right) }\frac{dz}{\sqrt{1-x^2}}$$
But how to calculate this integral? I would appreciate your help
| This integral can be done by recognizing that
$$\frac{\arcsin{\frac{x}{\sqrt{2}}}}{\sqrt{2-x^2}} = \sum_{n=0}^{\infty} \frac{2^n x^{2 n+1}}{(2 n+1) \binom{2 n}{n}}$$
and that
$$\int_0^1 dx \, x^{2 n+1} \arcsin{x} = \frac{\pi}{4 (n+1)} \left [1- \frac1{2^{2 n+2}} \binom{2 n+2}{n+1}\right ] $$
To see this, integrate by parts and see this answer.
With a bit of algebra, we find that the integral is equal to
$$\frac{\pi}{2} \sum_{n=0}^{\infty} \frac{2^n}{(2 n+1)(2 n+2) \binom{2 n}{n}} - \frac{\pi}{16} \sum_{n=0}^{\infty} \frac1{2^n (n+1)^2}$$
The first sum may be evaluated by recognizing that it is
$$\int_0^1 dx \frac{\arcsin{\frac{x}{\sqrt{2}}}}{\sqrt{2-x^2}} = \frac{\pi^2}{32}$$
The second sum is recognized as $\operatorname{Li_2}{\left ( \frac12 \right )} = \frac{\pi^2}{6}-\log^2{2} $
Putting all of this together, we find that the integral is equal to
$$\int_0^1 dx \, \frac{\arcsin{x} \arcsin{\frac{x}{\sqrt{2}}}}{\sqrt{2-x^2}} = \frac{\pi^3}{192} + \frac{\pi}{16} \log^2{2} $$
Numerical evaluation in Mathematica confirms the result, which differs from that asserted by the OP.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1990442",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 0
} |
Derivative of sum over sum? Consider the following expression:
$$f(r)=\sum_{t=1}^T\frac{c}{(1+r)^t\sum_{k=1}^T\frac{c}{(1+r)^k}}\cdot t$$
with $c,r,T>0$.
Is it true, that $f(r)$ is decreasing in $r$?
I.e. $f(0)\geq f(r),\quad \forall r>0$?
| Hint: The function can be considerably simplified, which should ease the analysis of monotonicity.
In the following I use for convenience only $f_T$ instead of $f$ to indicate the dependency of the parameter $T>0$.
Observe that the inner sum does not depend on the index $t$ of the outer sum and can so be factored out. We obtain for $c,r,T>0$
\begin{align*}
f_T(r)&=\sum_{t=1}^T\frac{c}{(1+r)^t\sum_{k=1}^T\frac{c}{(1+r)^k}}\cdot t\\
&=\left(\sum_{k=1}^T\frac{1}{(1+r)^k}\right)^{-1}\sum_{t=1}^T\frac{t}{(1+r)^t}\tag{1}\\
&=\left(\frac{\frac{1}{1+r}-\left(\frac{1}{1+r}\right)^{T+1}}{1-\frac{1}{1+r}}\right)^{-1}\sum_{t=1}^T\frac{t}{(1+r)^t}\tag{2}\\
&=\frac{r}{1-\left(\frac{1}{1+r}\right)^{T}}\sum_{t=1}^T\frac{t}{(1+r)^t}\tag{3}\\
&=\frac{r(1+r)^{T}}{(1+r)^T-1}\sum_{t=1}^T\frac{t}{(1+r)^t}\tag{4}
\end{align*}
Comment:
*
*In (1) we cancel $c$ and factor out the inner sum.
*In (2) we apply the formula for the finite geometric series.
*In (3) and (4) we do some simplifications.
Now we derive a closed formula for the series in (4).
We obtain from (4) with $q:=\frac{1}{1+r}$
\begin{align*}
\sum_{t=1}^T\frac{t}{(1+r)^t}&=\sum_{t=1}^Ttq^t
=q\cdot\sum_{t=1}^Ttq^{t-1}
=q\cdot\frac{d}{dq}\left(\sum_{t=1}^Tq^t\right)\\
&=q\cdot\frac{d}{dq}\left(\frac{1-q^{T+1}}{1-q}\right)\\
&=q\cdot\frac{Tq^{T+1}-(T+1)q^{T}+1}{\left(1-q\right)^2}\\
&=\frac{1}{1+r}\cdot\frac{T\left(\frac{1}{1+r}\right)^{T+1}-(T+1)\left(\frac{1}{1+r}\right)^T+1}{\left(1-\frac{1}{1+r}\right)^2}\\
&=\frac{1+r}{r^2}\cdot\left(T\left(\frac{1}{1+r}\right)^{T+1}-(T+1)\left(\frac{1}{1+r}\right)^T+1\right)\\
\end{align*}
$$ $$
Putting this in (4) we obtain
\begin{align*}
f_T(r)&=\frac{1}{r}\cdot\frac{(1+r)^{(T+1)}}{(1+r)^T-1}\cdot\left(T\left(\frac{1}{1+r}\right)^{T+1}-(T+1)\left(\frac{1}{1+r}\right)^T+1\right)\\
&=\frac{1}{r}\cdot\frac{(1+r)^{T+1}-(1+T)r-1}{(1+r)^{T}-1}\\
&=1+\frac{1}{r}-\frac{T}{(1+r)^T-1}
\end{align*}
Note: The derivation of $f_T$ assumes $r>0$ according to the stated problem. We get
\begin{align*}
\lim_{r\rightarrow 0}\left(1+\frac{1}{r}-\frac{T}{(1+r)^T-1}\right)=\frac{1}{2}\left(T+1\right)
\end{align*}
which coincides with $f_T$ evaluated at $r=0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1997527",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Show that $a^{bc}+b^{ca}+c^{ab}>2$ for $a,b,c\in(0,1)$
Show that $a^{bc}+b^{ca}+c^{ab}>2$ for $a,b,c\in(0,1)$.
I have no idea how to attempt this exponential inequality and couldn't find the condition for equality. Any hints will be appreciated.
| Let $\{x,y\}\subset(0,1)$.
Hence, by Bernoulli $\left(\frac{1}{x}\right)^y=\left(1+\frac{1}{x}-1\right)^y\leq1+y\left(\frac{1}{x}-1\right)=\frac{x+y-xy}{x}$.
Id est, $\sum\limits_{cyc}a^{bc}\geq\sum\limits_{cyc}\frac{a}{a+bc-abc}$.
Thus, it remains to prove that $\sum\limits_{cyc}\frac{a}{a+bc-abc}\geq2$, which is
$2a^2b^2c^2-2abc+1+\sum\limits_{cyc}(-2a^2b^2c+a^2bc+a^2b+a^2c-a^2)\geq0$ 0r
$$(1-a^2)(1-b^2)(1-c^2)+\sum\limits_{cyc}(a^2b^2c^2-2a^2b^2c+a^2bc)+$$
$$+\frac{1}{2}\sum\limits_{cyc}(a^2b+a^2c-2a^2b^2)+\frac{1}{2}\sum\limits_{cyc}\left(a^2b+a^2c-\frac{4}{3}abc\right)\geq0,$$
which is obvious because
$$\sum\limits_{cyc}(a^2b+a^2c-2a^2b^2)=\sum\limits_{cyc}(a^2(b-b^2)+a^2(c-c^2))\geq0,$$
$$\sum\limits_{cyc}(a^2b^2c^2-2a^2b^2c+a^2bc)=abc\sum\limits_{cyc}a(1-b)(1-c)\geq0$$ and
$$\sum\limits_{cyc}\left(a^2b+a^2c-\frac{4}{3}abc\right)\geq\sum\limits_{cyc}\left(a^2b+a^2c-2abc\right)=\sum\limits_{cyc}c(a-b)^2\geq0$$
Done!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1999909",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
} |
Find $w+\frac{1}{w} + \frac{1}{w^2}$ where $w^4=1$, $w\neq 1$ Let $w$ be a root of the equation $z^4 = 1$ and $w$ is not equal to $1$.
I would like to find the value of $w+\frac{1}{w} + \frac{1}{w^2}$.
So for this question since w is a root,
I first substituted w and got $w^4=1$
Which is $w * w^3 = 1$.
Since $w$ is not equal to $1$,
$w^3=1$
Which can be rewritten as $(w-1) (1 + w + w^2) = 0$
And since $w \neq 1,$ $(1 + w + w^2) = 0$
On simplifying the given equation,
I get
$w^3 + w + 1
= 2+w$ (Is this right approach to go about it?)
| Solution one:
The four roots are $1,-1,i,-i$. Simply plugging in the last three we get $-1$.
Solution two:
$z^4-1=0$ implies $(z^2-1)(z^2+1)=0$ implies $(z-1)(z+1)(z^2+1)=0$. Since we don't consider $1$, we may divide by $(z-1)$ to get: $(z+1)(z^2+1)=0$ or $z^3+z^2+z+1=0$.
Note $$\omega+\frac{1}{\omega}+\frac{1}{\omega^2}=\frac{\omega^3+\omega+1}{\omega^2}=\frac{-\omega^2}{\omega^2}=-1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2000113",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
functional equation of type $f(x+f(y)+xf(y)) = y+f(x)+yf(x)$
If $f:\mathbb{R}-\{-1\}\rightarrow \mathbb{R}$ and $f$ is a differentiable function that satisfies $$f(x+f(y)+xf(y)) = y+f(x)+yf(x)\forall x,y \in \mathbb{R}-\{-1\}\;,$$ Then value of $\displaystyle 2016(1+f(2015)) = $
$\bf{My\; Try::}$ Using partial Differentiation, Differentiate w r to $x$ and $y$ as a constant
$$f'(x+f(y)+xf(y)) \cdot (1+f(y)) = f'(x)+yf'(x)$$
Similarly Differentiate w r to $y$ and $x$ as a constant
$$f'(x+f(y)+xf(y))\cdot (f'(y)+xf'(y)) = 1+f(x)$$
Now Divide these two equation, We get $$\frac{1+f(y)}{(1+x)f'(y)} = \frac{(1+y)f'(x)}{1+f(x)}$$
Now How can i solve it after that, Help required, Thanks
| Your last equation tells us that
$$
\frac{(1+x)f'(x)}{1+f(x)}
$$
has to be constant, and that this constant has to be equal to its reciprocal, i.e., it is $\pm1$.
Integration gives
$$
\ln|1+f(x)|=\pm \ln|1+x|+C
$$
and thus
$$
f(x)=C(1+x)-1\text{ or } f(x)=\frac{C}{1+x}-1
$$
This now needs to be tested against the original functional equation.
The equation can be rewritten as $1+f((1+x)(1+f(y))-1)=(1+y)(1+f(x))$.
\begin{array}{rl|l}
f(x)&=C(1+x)-1&=\dfrac{C}{1+x}-1\\ \hline
(1+x)(1+f(y))&=C(1+x)(1+y) & =C\dfrac{1+x}{1+y}\\
1+f((1+x)(1+f(y))-1)&=C^2(1+x)(1+y) & =\dfrac{1+y}{1+x}\\
(1+y)(1+f(x))&=C(1+y)(1+x) & = C\dfrac{1+y}{1+x}
\end{array}
which gives $f(x)=-1$, $f(x)=x$ and $f(x)=\dfrac{1}{1+x}-1=-\dfrac{x}{1+x}$ as valid solutions.
Thus the possible values for $(1+x)(1+f(x))$, for $x=2015$ or elsewhere, are $0$, $(1+x)^2$ or $-1$.
PS: As discussed with Patrick Stevens and egreg, for the function $f\equiv -1$ the functional equation is nowhere defined, as the argument of $f$ on the left would be $x(1+f(y))+f(y)=-1$ and thus outside the domain of $f$.
If one wants to simplify the functional equation first, set $g(x)=1+f(x-1)$, $u=1+x$, $v=1+x$ then the reduced form is $$g(ug(v))=vg(u).$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2001644",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 0
} |
Find the inverse of 17 mod 41 Questions
(1) Find the inverse of $17 \mod 41$.
(2) Find the smallest positive number n so that $17n \equiv 14 \pmod{41}$
For the first question, my attempt is as follows:
$$41-17\cdot2=7$$
$$17-7\cdot2=3$$
$$7-3\cdot2=1$$
$$7-2(17-7\cdot2)=1$$
$$7-2\cdot17=1$$
$$41-17\cdot2-2\cdot17=1$$
$$41-4\cdot17=1$$
So the inverse of $17$ is $-4$.
That is, the inverse of $17$ is $37$
Am I right?
| \begin{align*}
41 &= 2 \cdot 17 + 7 & 7 &= 1 \cdot 41 - 2 \cdot 17 \\
17 &= 2 \cdot 7 + 3 & 3 &= 1 \cdot 17 - 2 \cdot (1 \cdot 41 - 2 \cdot 17) \\
&& &= 5 \cdot 17 - 2 \cdot 41 \\
7 &= 2 \cdot 3 + 1 & 1 &= 1 \cdot 7 - 2 \cdot 3 \\
&& &= 1\cdot(1 \cdot 41 - 2 \cdot 17) - 2 \cdot (5 \cdot 17 - 2 \cdot 41) \\
&& &= 5 \cdot 41 - 12 \cdot 17
\end{align*}
Therefore $17^{-1} \cong -12 \cong 29 \pmod{41}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2003116",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 8,
"answer_id": 2
} |
Limit of product with cubes $\lim\limits_{n \to \infty}\frac{2^3-1}{2^3+1}\frac{3^3-1}{3^3+1}\dots\frac{n^3-1}{n^3+1}$ I am trying to evaluate
$$\lim\limits_{n \to \infty}{2^3-1 \over 2^3+1}{3^3-1 \over 3^3+1}\dots{n^3-1 \over n^3+1}$$
It seems to be the perfect candidate for a factoring formula, however, I get stuck rewriting the fractions. Do you have any suggestions?
| $$
\begin{align}
\prod_{k=2}^\infty\frac{k^3-1}{k^3+1}
&=\lim_{n\to\infty}\prod_{k=2}^n\frac{(k-1)\left(k^2+k+1\right)}{(k+1)\left(k^2-k+1\right)}\\
&=\lim_{n\to\infty}\underbrace{\prod_{k=2}^n\frac{k-1}{k+1}}\underbrace{\prod_{k=2}^n\frac{k(k+1)+1}{(k-1)k+1}}\\
&=\lim_{n\to\infty}\,\frac{1\cdot2}{n(n+1)}\quad\frac{n(n+1)+1}{3}\\[3pt]
&=\frac23
\end{align}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2004071",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
How to symbolically solve a system of linear equations A fellow user had an interesting question:
Less-tedious way of solving this system of linear equations?
Is there a shortcut to solving for $w_1, w_2, w_3$ in $$
\left[ \begin{array}{ccc} 1 & 1 & 1\\ \frac{3}{4}a+\frac{1}{4}b &
\frac{1}{2}a+\frac{1}{2}b & \frac{1}{4}a+\frac{3}{4}b\\
(\frac{3}{4}a+\frac{1}{4}b)^2 & (\frac{1}{2}a+\frac{1}{2}b)^2 &
(\frac{1}{4}a+\frac{3}{4}b)^2\\ \end{array} \right] \cdot
\left[ \begin{array}{c} w_1\\ w_2\\ w_3 \end{array} \right] =
\left[ \begin{array}{c} b-a\\ \frac{b^2-a^2}{2}\\
\frac{b^3-a^3}{3} \end{array} \right] $$ where $a, b$ are
constants? (I'm trying to derive the formula for a 3-point open
Newton-Cotes quadrature rule.) Thanks!
Unfortunately the question was deleted (link), but I think it might be interesting in general.
| One way is to use the free Maxima computer algebra system.
You can enter your matrix like this: | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2004873",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
Periodic Continued Fraction Formula for $\sqrt{a^2+4}$ for any non-zero integer $a$ If $$x = \sqrt{ a^2 + 4 } $$ then I have conjectured that:
$$ x = [ a ; \overline{ \frac{a}{2} , 2a }] \ \ \ \text{ if a is even}$$
and $$ x = [ a ; \overline{ \frac{a}{2}, 1, 1, \frac{a}{2}, 2a } ]\ \ \ \text{ if a is odd} $$ where we take the floor of $\frac{a}{2} $
The even version is easy enough to prove. But for the odd version, we note that $$\frac{a}{2} = [\frac{a}{2}] + \frac{1}{2} = [\frac{a}{2}] + \frac{1}{1+\frac{1}{1}}$$
and I suspect that this is why the pair of ones emerges, but still a bit stuck!
| Follow the usual development of a continued fraction. It is important to express the first term as $\frac {a-1}2$ as that can be used in algebra instead of $\frac a2$ rounded down.
$$\frac 1{\sqrt{a^2+4}-a}=\frac {a-1}2+\frac {\sqrt{a^2+4}-a+2}4\\
\frac 4{\sqrt{a^2+4}-a+2}=1+\frac {\sqrt{a^2+4}-2}a\\
\frac a{\sqrt{a^2+4}-2}=1+\frac{\sqrt{a^2+4}+2-a}a\\\frac a{\sqrt{a^2+4}-(a-2)}=\frac{a-1}2+\frac {\sqrt{a^2+4}-a}4\\
\frac a{\sqrt{a^2+4}-a}=2a+(\sqrt{a^2+4}-a)$$
and you can read off the desired repeat.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2008944",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
How to find the coefficient of $x^{98}$ in $(x-1)(x-2)(x-3).....(x-100)$ Anyone please teach me how to solve such questions... I am thinking of solving it using binomial theory but still no idea how to do it..
Plz help
| $$(x-1)(x-2)(x-3)…(x-100)=x^{100} (1-\frac{1}{x})(1-\frac{2}{x})...(1-\frac{100}{x})$$
Denote $y=\frac{1}{x}$. Then
$$x^{100} (1-\frac{1}{x})(1-\frac{2}{x})...(1-\frac{100}{x}) =x^{100} (1-y)(1-2y)...(1-100y)$$
and the problem becomes: Find the coefficient of $y^2$ in
$$P(y)=(1-y)(1-2y)...(1-100y)$$
This coefficient is
$$\frac{P''(0)}{2!}$$
Derivating twice and plugging $x=0$ we get
$$P''(0)=\sum_{k=1}^{100}\sum_{j=1, j\neq k}^{100} kj$$
thus your coefficient is
$$\frac{1}{2} \sum_{k,j =1 , k\neq j}^{100} kj= \frac{1}{2} \left( \left( \sum_{k=1}^{100} k \right)^2 - \left( \sum_{k=1}^{100} k^2 \right) \right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2010183",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Inequality $9^x-k\cdot 3^x-k+3\leq 0$ is satisfied for at least one real $x$
If the Inequality $9^x-k\cdot 3^x-k+3\leq 0$ is satisfied for at least one real $x$ in $(0,1)\;,$ Then $k\in $
$\bf{My\; Try::}$ Let $3^x=y>0\;,$ Then inequality is
$$y^2-ky-k+3\leq 0\Rightarrow y^2-ky-k+3\leq 0\forall y\in (1,3)$$
So $$y^2-ky+\frac{k^2}{4}-\frac{k^2}{4}-k+3\leq 0$$
So $$\bigg(y-\frac{k}{2}\bigg)^2-\bigg(\frac{k^2+4k-12}{4}\bigg)\leq 0$$
Now above $\displaystyle 1<y<3\Rightarrow 1-\frac{k}{2}<y-\frac{k}{2}<3-\frac{k}{2}\Rightarrow \left(3-\frac{k}{2}\right)^2<\left(y-\frac{k}{2}\right)^2<\left(3-\frac{k}{2}\right)^2$
Now how can i solve it,Help required, Thanks
| Given: the inequality $9^x-k3^x -k + 3 \leq 0$ has at least one real x in $(0,1)$.
Transform the above inequality by replacing $3^x$ by $y$ to get
$y^2 - ky -k + 3 \leq 0$ which has atleast one real $y$ in $(3^0, 3^1) = (1,3)$. Now consider the equality $y^2 - ky -k + 3 = 0$. Note that the coefficient of $y^2$ is greater than $0$ which implies that the curve traced by the inequality will be convex. In general, one should remember the following rules for a curve traced by general quadratic equation $f(x) = ax^2 + bx + c = 0, \, a\neq 0$ which has a discriminant of $D = b^2 - 4ac$,
*
*$D > 0$
*
*$a>0 \Rightarrow$ convex with two distinct real roots $x_1$ and $x_2$ (say $x_1<x_2$) and $f(x) \leq 0 \, \forall \, x\in(x_1,x_2)$ and $f(x)\geq0$ elsewhere.
*$a<0 \Rightarrow$ concave with two distinct real roots $x_1$ and $x_2$ (say $x_1<x_2$) and $f(x) \geq 0 \, \forall \, x\in(x_1,x_2)$ and $f(x)\leq0$ elsewhere.
*$D = 0$
*
*$a>0 \Rightarrow$ convex with two equal real roots $x_1$ and $f(x_1) = 0$ and $f(x)>0$ elsewhere.
*$a<0 \Rightarrow$ concave with two equal real roots $x_1$ and $f(x_1) = 0$ and $f(x)<0$ elsewhere.
*$D < 0$
*
*$a>0 \Rightarrow$ convex with no real root and $f(x) > 0 \, \forall \, x$.
*$a<0 \Rightarrow$ concave with no real root and $f(x) < 0 \, \forall \, x$.
Note that convex means curve opening up and concave means curve opening down.
Now if the inequality $y^2 - ky -k + 3 \leq 0$ with $a=1>0$ has to have atleast one real y in $(1,3)$, then the equality $y^2 - ky -k + 3 = 0$ should have a discriminant $D \geq 0$ (by eliminating the wrong choices from above).
$D\geq0 \\ \Rightarrow k^2+4k-12\geq0 \\ \Rightarrow (k+6)(k-2) \geq0 \\ \Rightarrow k \in (-\infty,-6)\cup(2,\infty)$.
And the roots of the equation will be $y_1 = \frac{k - \sqrt{(k+6)(k-2)}}{2}$ and $y_2 = \frac{k + \sqrt{(k+6)(k-2)}}{2}$.
Now one of them should lie in $(1,3)$ and other should be greater than $0$ $(y=3^x>0 \, \forall \, x)$. So we need to check two cases:
CASE 1
$y_1 \in (1,3)$ and $y_2 > 0$
$y_1 > 1 \\ \Rightarrow \frac{k - \sqrt{(k+6)(k-2)}}{2} > 1 \\ \Rightarrow (k-2) > \sqrt{(k+6)(k-2)}$
which is never satisfied for $k \in (-\infty,-6)\cup(2,\infty)$. Now, switch to case 2.
CASE 2
$y_1 > 0$ and $y_2 \in (1,3)$
$y_1 > 0 \\ \Rightarrow \frac{k - \sqrt{(k+6)(k-2)}}{2} > 0 \\ \Rightarrow k > \sqrt{(k+6)(k-2)}$.
For $k>2$, both sides will be positive and can be squared without changing inequality which will give,
$k^2 > (k+6)(k-2) \\ \Rightarrow 4k-12 < 0 \\ \Rightarrow k<3$.
For $k<-6$, the inequality is not satisfied.
This further reduces the possibilities of $k$ to $(2,3)$.
Finally, $y_2 \in (1,3)$
$\frac{k + \sqrt{(k+6)(k-2)}}{2} > 1 \\ \Rightarrow (k-2) > -\sqrt{(k+6)(k-2)}$
which is always true for $k>2$.
$\frac{k + \sqrt{(k+6)(k-2)}}{2} < 3 \\ \Rightarrow (k-6) > -\sqrt{(k+6)(k-2)}$.
Both sides will be negative for $k \in (2,3)$ therefore can be squared with change of inequality
$\Rightarrow (k-6) > -\sqrt{(k+6)(k-2)} \\ \Rightarrow (k-6)^2 > (k+6)(k-2) \\ \Rightarrow -12k + 36 > 4k -12 \\ \Rightarrow 16k - 48 < 0 \\ \Rightarrow k<3$.
Therefore, the final solution is $k \in (2,3)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2015281",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
To show: $\det\left[\begin{smallmatrix} -bc & b^2+bc & c^2+bc\\ a^2+ac & -ac & c^2+ac \\ a^2+ab & b^2+ab & -ab \end{smallmatrix}\right]=(ab+bc+ca)^3$ I've been having quite some trouble with this question.
I'm required to show that the below equation holds, by using properties of determinants (i.e. not allowed to directly expand the determinant before greatly simplifying it).
$$\begin{vmatrix}
-bc & b^2+bc & c^2+bc\\
a^2+ac & -ac & c^2+ac \\
a^2+ab & b^2+ab & -ab
\end{vmatrix}= (ab+bc+ca)^3$$
I tried everything:
$R_1\to R_1+R_2+R_3$ and similar transformations to extract that $ab+bc+ca$ term, but to no avail. $C_2\to C_1+C_2$ and $C_3\to C_3+C_1$ seemed to be a good lead, but I couldn't follow up.
How can I solve this question?
| If $a=b=c=0$ then the equality holds.
WLG we assume $a\neq 0$ then
$$\begin{vmatrix}
-bc & b^2+bc & c^2+bc\\
a^2+ac & -ac & c^2+ac \\
a^2+ab & b^2+ab & -ab
\end{vmatrix}\\=\frac{1}{a}\begin{vmatrix}
a(-bc)+b(a^2+ac)+c(a^2+ab) & b(ac+bc+ca) & c(ac+bc+ca)\\
a^2+ac & -ac & c^2+ac \\
a^2+ab & b^2+ab & -ab
\end{vmatrix}\\(applying R_1 \to \frac1a(aR_1+bR_2+cR_3))\\
=\frac{1}{a}\begin{vmatrix}
a(ab+bc+ca) & b(ac+bc+ca) & c(ac+bc+ca)\\
a^2+ac & -ac & c^2+ac \\
a^2+ab & b^2+ab & -ab
\end{vmatrix}\\
=\frac{1}{a}(ab+bc+ca)\begin{vmatrix}
a & b & c\\
a^2+ac & -ac & c^2+ac \\
a^2+ab & b^2+ab & -ab
\end{vmatrix}\\
=\frac{1}{a}(ab+bc+ca)\begin{vmatrix}
a & b & c\\
0 & -(ab+bc+ac) & 0 \\
0 & 0 & -(ab+bc+ca)
\end{vmatrix}\\
(applying R_2 \to R_2-(a+c)R_1\, and\, R_3 \to R_3-(a+b)R_1)\\
=(ab+bc+ca)^3$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2016733",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 0
} |
Prove the inequality $xyz \geq xy+yz+xz \implies \sqrt{x y z } \geq \sqrt{x}+\sqrt{y}+\sqrt{z}$ Let $x,y,x>0$ and $xyz \geq xy+yz+xz.$ Prove that
$$\sqrt{x y z } \geq \sqrt{x}+\sqrt{y}+\sqrt{z}. $$
Solution.
Using AM GM inequality we have
\begin{gather*}
x y+xz \geq 2 \sqrt{x y x z}=2 x \sqrt{yz},\\
xy+yz \geq 2 y \sqrt{x z},\\
x z +y z \geq 2 z \sqrt{xz}.
\end{gather*}
Add and get
$$
xy+xz+yz \geq x \sqrt{yz}+y \sqrt{x z}+z \sqrt{xz}.
$$
By condition
$$
xyz \geq x \sqrt{yz}+y \sqrt{x z}+z \sqrt{xz}
$$
Dividing by $\sqrt{xyz}$ we obtain the inequality.
Question. Are there another ways to prove it?
| You could use this famous inequality: $$xy+yz+xz\geq \sqrt{3xyz(x+y+z)}$$ (which can be easily proved by AM-GM).
Using that inequality we have $$xyz\geq xy+yz+zx\geq \sqrt{3xyz(x+y+z)},$$
from where is $$\sqrt{xyz}\geq \sqrt{3(x+y+z)}=\sqrt{(1+1+1)(x+y+z)}$$
By Cauchy-Schwartz inequality, we have
$$\sqrt{(1+1+1)(x+y+z)} \geq (\sqrt{x}+\sqrt{y}+\sqrt{z})$$
and finally we can conclude $$\sqrt{xyz}\geq\sqrt{x}+\sqrt{y}+\sqrt{z}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2017063",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
Show the coefficient of $x^3$ in the expansion of $(1+x+x^2+x^3)(1+x^2)(1+x^3)$ is the same as the number of partitions of $3$ (let a partition of $n \in \Bbb N, n>0 $ be a sum of positive integers)
How would you show that the number of different partitions of $3$ is the coefficient of
$x^3$
in the
expansion of:
$(1+x+x^2+x^3)(1+x^2)(1+x^3)?$
My attempt:
since there are $3$ brackets to choose factors of $x^3$ from, we can say the following:
$ x^3 = x^a x^b x^c, a \in \{0,1,2,3\}, b \in \{0,2\}, c \in \{0,3\}$
The coefficient of $x^3$ in the expansion will equal the number of solutions satisfying $ x^3 = x^a x^b x^c$.
$a=0 \Rightarrow b=0 $ and $ c=3$,
$a=1 \Rightarrow b=2 $ and $ c=0$,
$a=2 \Rightarrow$ no solution.
$a=3 \Rightarrow b=0 $ and $ c=0$,
$\Rightarrow$ the coefficient of $x^3$ is $3$ which equals the number of partitions of $3$.
Do I understand what the question is asking? Because I'm wondering if just multiplying the brackets and obtaining the coefficient is an equivalent solution to the question.
Thanks.
| Here are two variations, both being equally valid with respect to the stated problem. Maybe the second variant is preferable since it provides some additional insight to the representation of the polynomial.
Algebraic variant:
We calculate the coefficient of $x^3$ of the polynomial and compare it with $p(3)$, the number of partitions.
Since $3$ admits a representation as sum of positive integers without respecting order of summands by
\begin{align*}
3&=3\\
&=2+1\\
&=1+1+1
\end{align*}
we see $p(3)=3$.
On the other hand using the coefficient of operator $[x^n]$ to denote the coefficient of $x^n$ in a polynomial, we obtain
\begin{align*}
[x^3]&(1+x^3)(1+x^2)(1+x+x^2+x^3)\\
&=\left([x^3]+[x^0]\right)(1+x^2)(1+x+x^2+x^3)\tag{1}\\
&=\left([x^3]+[x^1]+[x^0]\right)(1+x+x^2+x^3)\tag{2}\\
&=1+1+1\tag{3}\\
&=3
\end{align*}
Comment:
*
*In (1) we consider the coefficient of $[x^3]$ in $1+x^3$ and use the rule
\begin{align*}
[x^p]x^qA(x)=[x^{p-q}]A(x)
\end{align*}
*In (2) we do the same with $1+x^2$ noting that $[x^0]x^2=0$.
*In (3) we select the coefficients of $1+x+x^2+x^3$ accordingly.
Variant based upon generating functions:
If we want to count the number of partitions of a number $n$, we have to consider summands $k$ with $1\leq k\leq n$. The summand $k$ may occur zero, one, two or more times, but at most $\left\lfloor\frac{n}{k}\right\rfloor$ times. The contribution of $k$ encoded as powers of $x$ in a generating function is therefore
\begin{align*}
1+x^k+x^{2k}+\cdots+x^{\left\lfloor\frac{n}{k}\right\rfloor k}
\end{align*}
In case of $p(n)$ with $n=3$ we have to consider the summands $3,2$ and $1$.
*
*Summand: $3$ may occur zero or one times giving $\longrightarrow 1+x^3$
*Summand: $2$ may occur zero or one times giving $\longrightarrow 1+x^2$
*Summand: $1$ may occur zero, one, two or three times $\longrightarrow 1+x+x^2+x^3$
We conclude: $p(3)$ is the coefficient of $x^3$ in
\begin{align*}
(1+x^3)(1+x^2)(1+x+x^2+x^3)
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2019205",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Showing an infimum of an expression involving alternating series is positive in a given variable as another variable tends to infinity Let $\Bbb D$ be the unit disc in the complex plane. For the last month or so, I have been trying to prove that, if $\lambda \in \Bbb D$, and if $n \geq 4$ is an integer,
$$\inf_{\lambda \in \Bbb D}\Big[\frac{1+|\lambda|^2}{2} + Re(\lambda)\Big[1 - \frac{(2\pi)^{2}}{(n+2)(n+3)}+\frac{(2\pi)^{4}}{(n+2)(n+3)(n+4)(n+5)}-...\Big]-Im(\lambda)\Big[\frac{(2\pi)}{(n+2)} - \frac{(2\pi)^{3}}{(n+2)(n+3)(n+4)}+...\Big]\Big]$$
$\textit{decreases}$ to $0$ as $n \to \infty$, but that for any fixed $n \geq 4$, this infimum is always positive. I have been able to prove it along the coordinate axes within the disc, and in quadrants 1,3,4 of the disc. What remains is quadrant 2, which for obvious reasons makes the estimate more difficult.
I know that the first series converges to $1$ as $n \to \infty$, and the second series converges to $0$ as $n \to \infty$. I also have used wolfram alpha extensively to work on this, which has shown that for $n$ as large as I have entered, the statement holds.
Empirically, all the evidence stacks up in my favor, but I can not rigorously show what I need to show here.
| The following partial answer only shows that the infimum above is positive, and does so in a bit of a hack-y manner. An elegant answer would be interesting.
Let $\alpha_n$ and $\beta_n$ be the expressions multiplied respectively by $x$ and $y$ in the question. We first show that $\alpha_n^2 + \beta_n^2 < 1$ for $ n \ge 4$.
Note that $\alpha_n$ and $\beta_n$ are defined by alternating sums of monotonically decreasing terms (since $2\pi \le 7 \le n+3$), ergo, they're alternating series. This lets us use the upper bounds by truncation. For instance, $$\alpha_n < 1 -\frac{(2\pi)^{2}}{(n+2)(n+3)}+\frac{(2\pi)^{4}}{\prod_{j = 0}^3(n+2 + j)} $$ $$ \beta_n < \frac{2\pi}{n+2} - \frac{(2\pi)^3}{\prod_{j = 0}^2(n+2 + j)} + \frac{(2\pi)^5}{\prod_{j = 0}^4(n+2 + j)} $$
This form is messy but possible to attack computationally - it merely requires calculating the roots of some $8^{th}$ degree polynomial. Define $$f(n) := \left(1 -\frac{(2\pi)^{2}}{(n+2)(n+3)}+\frac{(2\pi)^{4}}{\prod_{j = 0}^3(n+2 + j)} \right) ^2 + \left(\frac{2\pi}{n+2} - \frac{(2\pi)^3}{\prod_{j = 0}^2(n+2 + j)} + \frac{(2\pi)^5}{\prod_{j = 0}^4(n+2 + j)} \right)^2,$$ and use your favourite package to determine when $f(n) < 1$. For instance, Wolfram Alpha tells me that $f(n) <1$ for $n \ge 3$, and since the bounds above hold for $n \ge 4$, we have that $\alpha_n^2 + \beta_n^2 < f(n) < 1$ for $n \ge 4$.
Next, we calculate the infimum. Note that
$$ \frac{1 + x^2 + y^2}{2} + \alpha x - \beta y = \frac{1}{2} + \frac{x^2 + 2\alpha x + y^2 - 2\beta y}{2}$$ $$= \frac{1 + (x + \alpha)^2 + (y - \beta)^2}2 - \frac{\alpha^2 + \beta^2}{2}$$ If $\alpha^2 + \beta^2 < 1 $, then the above is minimised at the point $(-\alpha, \beta) \in \mathbb{D}$, since otherwise at least one of $(x + \alpha)^2$ or $(y -\beta)^2$ is positive. Since we know that $\alpha_n^2 + \beta_n^2 <1$, $$ c_n := \inf_\mathbb{D} \frac{1 + x^2 + y^2}{2} + \alpha_n x - \beta_n y = \frac{1 - (\alpha_n^2 + \beta_n^2)}{2} > 0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2019397",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Factorise $f(x) = x^3+4x^2 + 3x$ Not sure if this belongs here, but I'm slowly trudging through my studies for Math 1 and wondered if y'all could give feedback and/or corrections on the following factorisation question:
$$ \text{Factorise}: f(x) = x^3+4x^2+3x $$
Firstly, the GCD of the above is $x$:
$$x(x^2+4x+3)$$
Now take $x^2+4x+3$ and factorise that:
$$ x^2+4x+3 $$
Using the box method, enter the first term $x^2$ into the upper left corner, and the last term $3$ into the lower right corner.
\begin{array}{|c|c|}
\hline
x^2 & \\
\hline
& 3 \\
\hline
\end{array}
Then find HCF of 3:
$$3\\
1 | 3
$$
Enter the values $1x$ and $3x$ into the other two boxes:
\begin{array}{|c|c|}
\hline
x^2 & 1x \\
\hline
3x& 3 \\
\hline
\end{array}
Now factorise the rows and columns:
$$ x^2 + 1x = x(x+1)\\
x^2 + 3x = x(x+3)\\
1x + 3=1(x+3)\\
3x +3=3(x+3)
$$
Therefore:
$$x^2+4x+3=(x+1)(x+3)$$
It follows that:
$$f(x) = x^3+4x^2+3x=x(x+1)(x+3)$$
Any feedback on method and/or corrections are gladly accepted! Be gentle, I'm a struggling student you know...
| Everything is more or less correct about the way you approach the problem. You could have also opted for the middle term factorisation method or the method of vanishing method. However you have to correct one thing ...
Now factorise the rows and columns:
$$ x^2 + 1x = x(x+1)\\
x^2 + 3x = x(x+3)\\
1x + 3=1(x+3)\\
3x +3=1(x+3)
$$
The last line ought to be $3x +3=3(x+1)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2020321",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 7,
"answer_id": 2
} |
Find all positive three - digit integers $\overline{abc}$ such that $\overline{abc}=5abc$ Number is divisible by $5$ if the last digit is $0$ or $5$. Since $c\neq 0$, $c=5$.
Using brute force (checking each combination with three digits), I get that the only solution is $175$ : $a=1,b=7,c=5$.
How to solve a problem efficiently?
| Once you have $c=5$, it becomes clear that $100a+10b+5=25ab$. Manipulating this, we have:
$$
0=5ab-20a-2b-1=5a(b-4)-2(b-4)-8-1\implies \color{blue}{\boxed{9=(5a-2)(b-4)}}.
$$
Note that $5a-2$ can only be $-2$, $3$, $8$ or bigger than $9$. We conclude that $5a-2=3=b-4$, yielding the answer you have found.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2020594",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Solve for $x$ in $3\cdot 5^{x^2}+3^{x^2+1}-8\cdot 3^{x^2}=0$ $3\cdot 5^{x^2}+3^{x^2+1}-8\cdot 3^{x^2}=0$
Note that I didn't forget any parenthesis. I copied it as it is in my book.
I tried:
$$3\cdot 5^{x^2}+3^{x^2+1}-8\cdot 3^{x^2}=0 \Leftrightarrow 3\cdot 5^{x^2} - 5 \cdot 3^{2x} = 0 \Leftrightarrow 5 \cdot 3^{x^2} = 3 \cdot 5^{x^2} \Leftrightarrow ???$$
What do I do next?
Also, this $3^{x^2+1}$ confuses me. Does it expand to $3^{x^2}\cdot 3$ or $3^{x^2}\cdot 3^x$?
| $3^{x^2+1}$ indeed equals $3\cdot 3^{x^2}$. Proceeding from your last equality:
$$
5\cdot 3^{x^2}=3\cdot 5^{x^2}\iff 3^{x^2-1}=5^{x^2-1}\iff(x^2-1)\log(3/5)=0.
$$
The last $\iff$ above is obtained by taking $\log$ and simplifying. Can you continue from here?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2020775",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Prove that $\triangle ABC$ is equilateral, If $(a+b-c)^{a}\cdot (b+c-a)^{b}\cdot (c+a-b)^{c}\geq a^a\cdot b^b\cdot c^c$
If $a,b,c$ are the sides of a $\triangle ABC\;,$ such that $$\left(1+\frac{b-c}{a}\right)^{a}\left(1+\frac{c-a}{b}\right)^{b}\left(1+\frac{a-b}{c}\right)^{c}\geq 1,$$
then proving $\triangle$ is equilateral.
$\bf{My\; Try::}$ We can write it as $$\left(1+\frac{b-c}{a}\right)^{a}\left(1+\frac{c-a}{b}\right)^{b}\left(1+\frac{a-b}{c}\right)^{c}\geq 1$$
So $$\left(\frac{a+b-c}{a}\right)^{a}\cdot \left(\frac{b+c-a}{b}\right)^{b}\cdot \left(\frac{c+a-b}{c}\right)^{c}\geq 1$$
So $$(a+b-c)^{a}\cdot (b+c-a)^{b}\cdot (c+a-b)^{c}\geq a^a\cdot b^b\cdot c^c$$
Now how can i solve after that, Help required, Thanks
| The title's inequality is wrong. I'll prove a reversed inequality.
Let $a=y+z$, $b=x+z$ and $c=x+y$.
Since our inequality does not changed after substitution $a\rightarrow ka$...,
we can assume that $x+y+z=3$.
And we need to prove that $\sum\limits_{cyc}(3-x)\ln\frac{2x}{3-x}\leq0$ or
$\sum\limits_{cyc}\left(3(x-1)-(3-x)\ln\frac{2x}{3-x}\right)\geq0$, which is true because
$3(x-1)-(3-x)\ln\frac{2x}{3-x}\geq0$ is true for all $x\in(0,3)$.
The equality occurs for $x=y=z=1$.
Thus, $a=b=c$ and we are done!
Maybe you mean the following problem.
If $a,b,c$ are the sides of a $\triangle ABC\;,$ such that $$\left(1+\frac{b-c}{a}\right)^{a}\left(1+\frac{c-a}{b}\right)^{b}\left(1+\frac{a-b}{c}\right)^{c}\geq 1,$$
then proving $\triangle$ is equilateral.
Let $a+b+c=1$.
Hence, by Jensen $a\ln\left(1+\frac{b-c}{a}\right)\leq\ln\sum\limits_{cyc}(a+b-c)=1$.
Thus, $\left(1+\frac{b-c}{a}\right)^{a}\left(1+\frac{c-a}{b}\right)^{b}\left(1+\frac{a-b}{c}\right)^{c}\leq 1$ and with the given we obtain $a=b=c$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2021321",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
For 1/A = 1/B + 1/C, why is A $\frac{1}{a}$ =$\frac{1}{b}$ +$\frac{1}{c}$
I read that $a$ is always < the smaller of $b$ and $c$
In the case of $0$ < $B$ $<1$ and $0$ < $C$ $<1$,
I can understand the rule as:
Firstly,
$$\frac{1}{A}=\frac{1}{B} +\frac{1}{C}=\frac{C+B}{B*C}$$
Now looking at 3 cases:
Case 1
$$\frac{1}{A}=\frac{1}{0.5} +\frac{1}{0.5}=\frac{0.5+0.5}{0.5*0.5}=\frac{1}{0.25}$$
Case 2
$$\frac{1}{A}=\frac{1}{0.2} +\frac{1}{0.9}=\frac{0.2+0.8}{0.2*0.8}=\frac{1}{0.04}$$
Case 3
$$\frac{1}{A}=\frac{1}{0.001} +\frac{1}{0.999}=\frac{0.001+0.999}{0.001*0.999}=\frac{1}{0.0009}$$
So if we look at the $B*C$ term,
$$D=B*C$$
When $0$ < $B$ $<1$ and $0$ < $C$ $<1$
The smaller value (of $B$ or $C$) brings the product $B*C$ down to its level.
But how does the same hold true for when $B >1$ and $C>1$
Case 4
$$\frac{1}{A}=\frac{1}{1} +\frac{1}{1}=\frac{1+1}{1*1}=\frac{2}{1}$$
Hence $$A=1/2$$
Case 5
$$\frac{1}{A}=\frac{1}{5} +\frac{1}{5}=\frac{5+5}{5*5}=\frac{10}{25}$$
Hence $$A=25/10=2.5$$
I can see in both of these cases $A$ is still $<B$ and $A<C$. I know there is something going on here in relation to the A value and the result on the RHS of the equation but I can't quite put my finger on what causes $A$ to be less than $B$ and less than $C$ where $B$ and $C$ $>1$
| Since everything is strictly positive, it's simply the fact that $$\frac1A=\frac1B+\frac1C>\frac1B$$ And $$\frac1A>\frac1B\implies B>A$$ Same for $C$. Since $A$ is smaller than boh, it is smaller than the minimum.
I don't really understand why in you decided to consider only three numbers.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2022100",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Simplify $\frac {2\sqrt{x^2+x+1}-2-x} {x^2} $ In order to find the limit of the function at $0$ I need to simplify the fraction $$\frac {2\sqrt{x^2+x+1}-(2+x)} {x^2} $$
so that at $0$ it didn't look like $\frac 0 0$. I'm almost certain that it can be solved with a clever substitution of the variable but I didn't manage to find one.
I've also tried multiplying numerator and denominator by $2\sqrt{x^2+x+1}+2+x$, but without any success.
| Hint. By Taylor expansion, it is easy to see that for $t\to 0$,
$$\sqrt{1+t}=1+\frac{t}{2}-\frac{t^2}{8}+o(t^2).$$
Hence
$$2\sqrt{x^2+x+1}=2\sqrt{1+(x+x^2)}=2+x+x^2-\frac{(x+x^2)^2}{4}+o((x+x^2)^2)\\=2+x+\frac{3}{4}x^2+o(x^2).
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2022345",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Let $\frac{{{x^5}}}{{1 - {x^5} + {x^{10}}}} = k$. What is $\frac{{{x^{15}}}}{{1 - {x^{15}} + {x^{30}}}}$? Let $\frac{{{x^5}}}{{1 - {x^5} + {x^{10}}}} = k$ and $k\in \mathbb{R}$.
What is $\frac{{{x^{15}}}}{{1 - {x^{15}} + {x^{30}}}}$?
| Note that
$$
k=\frac{x^5}{1-x^5+x^{10}}=\frac{1}{x^{-5}-1+x^{5}}\implies x^{-5}-1+x^{-5}=k^{-1}.
$$
It follows that
\begin{aligned}
k^{-2}&=1+(x^{-10}+x^{10})-2(x^{-5}-1+x^5)\implies(x^{-10}+x^{10})=k^{-2}-1+2k^{-1},\\
k^{-3}&=(x^{-5}-1+x^{-5})^3=(x^{-15}-1+x^{15})+6(x^{-5}-1+x^5)-3(x^{-10}+x^{10}).
\end{aligned}
We conclude
$$
x^{-15}-1+x^{-15}=k^{-3}-6k^{-1}+3(k^{-2}-1+2k^{-1})=k^{-3}+3k^{-2}-3
$$
and the desired answer the reciprocal of the rightmost expression above:
$$
\frac{k^3}{1+3k-3k^3}.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2022747",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Is $\frac{a^4}{(b-a)(c-a)}+\frac{b^4}{(c-b)(a-b)}+\frac{c^4}{(a-c)(b-c)} $ always an integer? In a textbook I found the rather strange identity:
$$ \frac{2^4}{(5-2)(3-2)}+\frac{3^4}{(5-3)(3-2)}+\frac{5^4}{(5-3)(5-2)}= \frac{414}{6}=69 $$
just kind if out of nowhere and I wonder if it generalizes and why. Perhaps it is that:
$$ \frac{a^4}{(b-a)(c-a)}+\frac{b^4}{(c-b)(a-b)}+\frac{c^4}{(a-c)(b-c)} \tag{$\ast$} $$
is always an integer? On the same page I found the formula:
$$ \left|\begin{array}{cccc} 1 & a & a^2 & a^4 \\
1 & b & b^2 & b^4 \\
1 & c & c^2 & c^4 \\
1 & d & d^2 & d^4 \\ \end{array} \right| = P \times (a+b+c+d)$$
My guess is that $P$ is short-hand for the product of all the different pairs of numbers:
$$ P = (a-b)(a-c)(a-d)(b-c)(b-d)(c-d) $$
This is not the right formula I'm afraid. However, I do see:
$$ \left|\begin{array}{cccc} 1 & 2 & 16 \\
1 & 3 & 81 \\
1 & 5 & 625 \end{array} \right| = P \times \stackrel{2}{C}(2,3,5)$$
which could mean anything. The book is in German - which I do not understand, and the man has invented this wonderful symbol $\stackrel{2}{C}$.
Mainly I am wondering if other integer identities can be found this way. Or is it only the case for $(a,b,c) = (2,3,5)$.
| The statement is true even when you replace the numerators by any polynomial with integer coefficients or increase the number of variables. More precisely,
For any $f(x) \in \mathbb{Z}[x]$ and distinct $a_1, \ldots, a_n \in \mathbb{Z}$,
$\displaystyle\;\sum\limits_{k=1}^n \frac{f(a_k)}{\prod\limits_{j=1,\ne k}^n (a_k - a_j)}\;$ is an integer.
I will only prove the statement for $3$ distinct integers $a,b,c$. The proof for other $n$ is similar.
For any $f(x) \in \mathbb{Z}[x]$, long divide it by $(x-a)(x-b)(x-c)$.
This gives us two polynomials $g(x), h(x) \in \mathbb{Z}[x]$ such that
$$f(x) = (x-a)(x-b)(x-c)g(x) + h(x)$$
and $h(x) = Ax^2 + Bx + C$ is at most quadratic.
Apply partial fraction decomposition to $\displaystyle\;\frac{h(x)}{(x-a)(x-b)(x-c)}\;$ and notice
$$f(a) = h(a),\quad f(b) = h(b)\quad\text{ and }\quad f(c) = h(c)$$
We obtain
$$\begin{align}
\frac{h(x)}{(x-a)(x-b)(x-c)} &=
\frac{1}{x-a}\frac{h(a)}{(a-b)(a-c)} + \frac{1}{x-b}\frac{h(b)}{(b-a)(b-c)}
+ \frac{1}{x-c}\frac{h(c)}{(c-a)(c-b)}\\
&= \frac{1}{x-a}\frac{f(a)}{(a-b)(a-c)} + \frac{1}{x-b}\frac{f(b)}{(b-a)(b-c)}
+ \frac{1}{x-c}\frac{f(c)}{(c-a)(c-b)}
\end{align}
$$
This implies
$$\frac{f(a)}{(a-b)(a-c)} + \frac{f(b)}{(b-a)(b-c)} + \frac{f(c)}{(c-a)(c-b)}
= \lim_{x\to\infty} \frac{xh(x)}{(x-a)(x-b)(x-c)} = A$$
The coefficients of $x^2$ in $h(x)$ which is an integer.
For the original problem, we can take $f(x) = x^4$ and conclude
$$\frac{a^4}{(a-b)(a-c)} + \frac{b^4}{(b-a)(b-c)} + \frac{c^4}{(c-a)(c-b)} \in \mathbb{Z}
\quad\text{ for distinct }\; a,b,c \in \mathbb{Z}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2023628",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 1
} |
Solving $a^2-b^2-a=0$ ∧ $-2ab+b={1\over 2}$ I want to solve
$$a^2-b^2-a=0 ∧ -2ab+b={1\over 2}$$
I'm a bit stuck in the first part. I thought of the solutions, $a={1±\sqrt {1+4b^2} \over 2}$. But can't figure out solutions to the second part. Am I approaching it the right way?
| I would solve it as $b=\pm \sqrt{a^2-a}$ and $b=\frac{1}{2(1-2a)}$.
So $$a^2-a = \frac{1}{4(1-2a)^2}$$
Or:
$$\left(2a-1\right)^2 - 1 = \frac{1}{(1-2a)^2}$$
Letting $u=(2a-1)^2$, you have $u-1=\frac{1}{u}$ or $u^2-u-1=0$ or $u=\frac{1\pm\sqrt{5}}{2}$. Then:
$$a=\frac{1\pm\sqrt{\frac{1\pm\sqrt{5}}{2}}}{2}$$
Not that $a$ is complex if $u=\frac{1-\sqrt{5}}{2}$.
So for real solutions, we get $a=\frac{1\pm\sqrt{\frac{1+\sqrt{5}}{2}}}{2}$ and $$b=\frac{1}{2(1-2a)}=\frac{1}{\mp 2\sqrt{\frac{1+\sqrt{5}}{2}}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2030374",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Finding the length of the shortest ladder
A wall 8 feet high is 1 foot from a house. Find the length $L$ of the
shortest ladder over the wall to the house. Draw a triangle with
height $y$, base $1 + x$, and hypotenuse $L$.
The shortest ladder is placed straight up close to the wall. It has length $8+8=16$ for going up and down. Does this even need calculus? Or have I misunderstood the problem?
|
The triangles $ABC$ and $AB'C'$ satisfy the relation
$$
\frac{y}{1 + x} = \frac{8}{x}
$$
So that
$$
y = \frac{8}{x}(1 +x) \tag{1}
$$
The length of the ladder (red line) follows from
$$
L^2 = (1 + x)^2 + y^2 \stackrel{(1)}{=} (1 + x)^2 + \frac{64}{x^2}(1 + x)^2 = (1 + x)^2\frac{x^2 + 64}{x^2}
$$
Call $f(x) = L^2$, and note that minimizing $f(x)$ is equivalent to minimize $L$, therefore we want to find the minimum of
$$
f(x) = (1 + x)^2\frac{x^2 + 64}{x^2}
$$
Which can be done by solving the problem
$$
\frac{df}{dx} = 2\frac{-64 - 64 x + x^3 + x^4}{x^3} = 0
$$
whose solution is $x = 4$. It is easy to see that for this value $f$ has a minimum since
$$
\left.\frac{d^2f}{dx^2}\right|_{x = 4} = \frac{15}{2} > 0
$$
The length of the ladder is then
$$
L = \sqrt{f(4)} = 5\sqrt{5}\;{\rm ft}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2030973",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Evaluation of $\sum_{n=1}^{\infty} (\frac{1}{4n-3}-\frac{1}{4n-1}) $ Prove that
$$\sum_{n=1}^{\infty} (\frac{1}{4n-3}-\frac{1}{4n-1})=\frac{\pi}{4}$$
Could someone give me slight hint as how to convert above expression to $\int_{0}^{1} \tan^{-1}x.dx$
| By IBP,
\begin{eqnarray*}\int \tan^{-1}x.dx&=&x\times tan^{-1}x|-\int x\times \frac{1}{1+x^2}.dx \\
&=&x\times tan^{-1}x|-\int \frac{d(x^2+1)}{2dx}\times \frac{1}{1+x^2}.dx\\
&=&x\times tan^{-1}x|-\ln|x^2+1| |
\end{eqnarray*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2032479",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Finding and proving linear independence of solutions of second order differential equation Show that the solutions of the differential equation $2x^2y'' + 10xy' + 8y = 0$ are linearly
independent in two ways and find its particular solution satisfying the initial conditions
$y(1) = 3$ and $y'(1) = 5$
I have tried this problem by using power series method but ended up getting only the trivial solution as my answer. Any help would be appreciated.
| Your equation is a Cauchy-Euler equation which can be solved by assuming that the solution is of the form $y(x) = x^k$ for some $k \in\mathbb{R}$.
Plugging this in gives you:
$$2x^2(k-1)kx^{k-2} + 10xkx^{k-1} + 8x^k = 0$$
Factoring out $x^k$, we obtain:
$$x^k(2k^2 - 2k + 10k + 8) = x^k(2k^2+8k+8) = 0$$
However, this must hold for any value of $x$, so we know that $2k^2 + 8k + 8 = 0$ must hold, which is equivalent to $k^2 + 4k + 4 = 0$. We can easily solve this with the quadratic formula and obtain $k = -2$ as the only solution.
Because this is a repeated solution to our quadratic, the general solution of the initial ODE is given by
$$y(x) = C_1 y_1 (x) + C_2 y_2(x)=\frac{C_1}{x^2} + \frac{C_2\log(x)}{x^2}$$
Using our first initial condition $y(1) = 3$, we have
$$3 = y(1) = \frac{C_1}{1} + \frac{C_2\log(1)}{1} = C_1$$
Plugging this in and calculating $y'(x)$, we have
$$y'(x) = -\frac{6}{x^3} + C_2\frac{1-2\log(x)}{x^3}$$
and thus, with our second initial condition that $y'(1) = 5$, we obtain
$$5 = y'(1) = -6 + C_2 \Rightarrow C_2 = 11$$
Our particular solution is therefore
$$y(x) = \frac{6}{x^2} + \frac{11\log(x)}{x^2}$$
The linear independence of the two solutions of the general equation is pretty obvious and can be shown directly (by showing that no $C_1, C_2 \in \mathbb{R}$ with at least one of them non-zero exist such that $y(x) = 0$ for all $x$ - this is fairly straightforward), or by calculating the Wronskian determinant.
\begin{align*} W(x) &= \det\begin{pmatrix}
y_1(x) & y_2(x)\\
y_1 '(x) & y_2 '(x)
\end{pmatrix} = \det\begin{pmatrix}
\frac{1}{x^2} & \frac{\log(x)}{x^2}\\
-\frac{2}{x^3} & \frac{1-2\log(x)}{x^3}
\end{pmatrix}\\
&= \frac{1 - 2\log(x) + 2\log(x)}{x^5} = \frac{1}{x^5}\neq 0 \end{align*}
Since the Wronskian determinant is not identically $0$, $y_1(x)$ and $y_2(x)$ form a fundamental system and are thus linearly independent.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2033118",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Convergence and evaluation of $\sum_{n=1}^{\infty}\ln {(2n+1)n \over (n+1)(2n-1)}$ I am trying to evaluate
$$\sum_{n=1}^{\infty}\ln{(2n+1)n \over (n+1)(2n-1)}$$
First, we notice
$${(2n+1)n \over (n+1)(2n-1)} = {2n^2+n -1 + 1 \over 2n^2+n-1} = 1 + {1 \over 2n^2+n-1}= 1 + {1 \over (n+1)(2n-1)}$$
Then, we use ${x \over 1+x} \le \ln(1+x) \le x$:
$${1 \over (2n+1)n} \le \ln\left(1 + {1 \over (n+1)(2n-1)} \right) \le {1 \over (n+1)(2n-1)}$$
Now, I am going for the squeeze theorem with the partial sum
$${1 \over 3} + {1 \over 10} + \dots + {1 \over (2p+1)p} \le \ln\left( 1 + {1 \over 2} \right) + \ln\left( 1 + {1 \over 9} \right) + \dots +\ln\left( 1 + {1 \over (p+1)(2p-1)} \right) \\ \le {1 \over 2} + {1 \over 9} + \dots + {1 \over (p+1)(2p-1)}$$
I get stuck on reorganizing the terms.
| Hint. This may be seen a limit of two telescoping sums:
$$
\sum_{n=1}^N\ln{(2n+1)n \over (n+1)(2n-1)}=\sum_{n=1}^N\left(\ln(2n+1)-\ln(2n-1)\right)-\sum_{n=1}^N\left(\ln(n+1)-\ln(n)\right).
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2034917",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Calculating sign of a permutation of unknown size, but with a pattern I'm trying to calculate the determinant of the following matrix, using the Leibniz formula:
$$
\begin{pmatrix}
0 & 1 & 0 & \cdots & 0 & 1 \\
1 & 0 & 1 & \ddots & \ddots & 0\\
0 & 1 & \ddots & \ddots & \ddots & \vdots\\
\vdots & \ddots & \ddots & \ddots & 1 & 0\\
0 & \ddots & \ddots & 1 & 0 & 1\\
1 & 0 & 0 & 0& 1 & 0
\end{pmatrix}.
$$
So, the only permutation that are relevant (non-zero product) are:
$\sigma_1 =
\begin{pmatrix}
1 & 2 & 3 & 4 & 5 & 6 & ... & n \\
2 & 3 & 4 & 5 & 6 & 7 & ... & 1
\end{pmatrix}$
$\sigma_2 =
\begin{pmatrix}
1 & 2 & 3 & 4 & 5 & 6 & ... & n \\
n & 1 & 2 & 3 & 4 & 5 & ... & 1
\end{pmatrix} $
$\sigma_3 =
\begin{pmatrix}
1 & 2 & 3 & 4 & 5 & 6 & ... & n-1 & n \\
2 & 1 & 4 & 3 & 6 & 5 & ... & n & n-1
\end{pmatrix} $
$\sigma_4 =
\begin{pmatrix}
1 & 2 & 3 & 4 & 5 & 6 & ... & n-2 & n-1 & n \\
n & 3 & 2 & 5 & 4 & 7 & ... & n-1 & n-2 & 1
\end{pmatrix}$
How would I proceed to determine $\operatorname{Sgn}(\sigma)$ based on the value of $n$?
Thanks!
| $\mathrm{sgn}(\sigma)=(-1)^{\mathrm{inv}(\sigma)}$, where $\mathrm{inv}(\sigma)$ is the number of inversions of $\sigma$, that is to say, the number of pairs of positions $(i,j)$ where $i<j$ but $\sigma(i)>\sigma(j)$ (a larger value is to the left of a smaller value).
For the permutations you listed, the numbers of inversions are
$$
\begin{split}
\mathrm{inv}(\sigma_1)&=n-1,\\
\mathrm{inv}(\sigma_2)&=n-1,\\
\mathrm{inv}(\sigma_3)&=\left\lfloor\frac{n}{2}\right\rfloor,\\
\mathrm{inv}(\sigma_4)&=\left(\left\lfloor\frac{n}{2}\right\rfloor-1\right)+(n-2)+(n-2)+1=\left\lfloor\frac{n}{2}\right\rfloor+2n-4.
\end{split}
$$
In particular, the parity of the number of inversions is the same for $\sigma_1$ and $\sigma_2$ as well as for $\sigma_3$ and $\sigma_4$, so $\mathrm{sgn}(\sigma_1)=\mathrm{sgn}(\sigma_2)$ and $\mathrm{sgn}(\sigma_3)=\mathrm{sgn}(\sigma_4)$. Thus, the determinant can only take values $-4$, $0$, $4$.
I also see that you implicitly assume that $n$ is even. In that case, $n-1$ is odd, so $\mathrm{sgn}(\sigma_1)=\mathrm{sgn}(\sigma_2)=-1$, so the determinant is $-4$ if $\frac{n}{2}$ is odd, and $0$ if $\frac{n}{2}$ is even.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2040819",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
An interesting definite integral $\int_0^1(1+x+x^2+x^3+\cdot\cdot\cdot+x^{n-1})^2 (1+4x+7x^2+\cdot\cdot\cdot+(3n-2)x^{n-1})~dx=n^3$ How to prove
$~~ \forall n\in\mathbb{N}^+$,
\begin{align}I_n=\int_0^1(1+x+x^2+x^3+\cdot\cdot\cdot+x^{n-1})^2 (1+4x+7x^2+\cdot\cdot\cdot+(3n-2)x^{n-1})~dx=n^3.\end{align}
My Try:
Define $\displaystyle S(n)=\sum_{k=0}^{n-1}x^k=1+x+x^2+x^3+\cdot\cdot\cdot+x^{n-1}=\frac{x^n-1}{x-1}$. Then,
\begin{align}\frac{d}{dx}S(n)=S'(n)=1+2x+3x^2+\cdot\cdot\cdot(n-1)x^{n-2}=\sum_{k=0}^{n-1}kx^{k-1}.\end{align}
Therefore,
\begin{align}
I_n&=\int_0^1 S^2(n)\left(3S'(n+1)-2S(n)\right)~dx\\
&=3\int_0^1 S^2(n)S'(n+1)~dx-2\int_0^1 S^3(n)~dx\\
&=3\int_0^1 S^2(n)(S'(n)+nx^{n-1})~dx-2\int_0^1 S^3(n)~dx\\
&=3\int_0^1 S^2(n)~d(S(n))+3\int_0^1 S^2(n)(nx^{n-1})~dx-2\int_0^1 S^3(n)~dx\\ &=n^3-1+\int_0^1 S^2(n)(3nx^{n-1}-2S(n))~dx\\
&=n^3-1+\int_0^1 \left(\frac{x^n-1}{x-1}\right)^2\left(3nx^{n-1}-2\cdot\frac{x^n-1}{x-1}\right)~dx
\end{align}
So the question becomes:
Prove \begin{align}I'=\int_0^1 \left(\frac{x^n-1}{x-1}\right)^2\left(3nx^{n-1}-2\cdot\frac{x^n-1}{x-1}\right)~dx=1.\end{align}
\begin{align}I'&=\int_0^1 \frac{3nx^{n-1}(x^n-1)^2}{(x-1)^2}-\frac{2(x^n-1)^3}{(x-1)^3}~dx\\
&=\int_0^1 \frac{(x-1)^2\left(\frac d {dx} (x^n-1)^3\right)-2(x^n-1)^3(x-1)}{(x-1)^4}~dx\\
&=\int_0^1 \frac d {dx} \left(\frac{(x^n-1)^3}{(x-1)^2}\right)~dx\\
&=\lim_{x \to 1} \frac{(x^n-1)^3}{(x-1)^2}-\frac{(0^n-1)^3}{(0-1)^2}\\
\end{align}
$$\therefore I'=1.$$
\begin{align}\therefore I_n=n^3.\end{align}
There MUST be other BETTER ways evaluating $I_n$.
Could anyone give me some better solutions? Thanks.
| First apply the substitution $x = t^3$. Then
\begin{align*}
I_n
&= \int_{0}^{1} (1 + t^3 + \cdots + t^{3n-3})^2 (1 + 4t^3 + \cdots + (3n-2)t^{3n-3}) \cdot 3t^2 \, dt \\
&= \int_{0}^{1} 3 (t + t^4 + \cdots + t^{3n-2})^2 (1 + 4t^3 + \cdots + (3n-2)t^{3n-3}) \, dt.
\end{align*}
Now let $u = u(t) = t + t^4 + \cdots + t^{3n-2}$. Then
$$ 3 (t + t^4 + \cdots + t^{3n-2})^2 (1 + 4t^3 + \cdots + (3n-2)t^{3n-3}) = 3u^2 \frac{du}{dt}.$$
Therefore
$$ I_n = \left[ u(t)^3 \right]_{t=0}^{t=1} = u(1)^3 - u(0)^3 = n^3. $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2042986",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "21",
"answer_count": 1,
"answer_id": 0
} |
Prove that $\vert\sin(x)\sin(2x)\sin(2^2x)\cdots\sin(2^nx)\vert < \left(\frac{\sqrt{3}}{2}\right)^n$ For $x$ in $\mathbb R^*$ and $n$ in $\mathbb N$, define
$$U_n = \sin(x)\sin(2x)\sin(2^2x)\cdots\sin(2^nx) = \prod_{k=0}^n\sin(2^k x)$$
Prove that $$\vert U_n\vert \leq \left(\frac{\sqrt{3}}{2}\right)^n$$
Can anyone help please.
(Edited by @River Li) I found it was B6(c) in the 34th Putnam (1973):
B-6: On the domain $0 \le \theta \le 2\pi$:
(a) Prove that $\sin^2\theta \sin 2\theta$
takes its maximum at $\pi/3$ and $4\pi/3$
(and hence its maximum at $2\pi/3$ and $5\pi/3$).
(b) Show that
$$\left|\sin^2\theta \, \Big[\sin^3(2\theta)
\cdot \sin^3(4\theta) \cdots \sin^3(2^{n - 1}\theta)\Big]\, \sin (2^n\theta)\right|$$
takes its maximum at $\theta = \pi/3$.
(The maximum may also be attained at other points.)
(c) Derive the inequality:
$$\sin^2\theta \cdot \sin^2(2\theta) \cdot \sin^2(4\theta) \cdots \sin^2(2^n\theta)\le (3/4)^n.$$
See: The American Mathematical Monthly, Vol. 81, No. 10, Dec., 1974, (Page 1086-1095)
or https://prase.cz/kalva/putnam/putn73.html
| Easy to see that $|U_1(x)|\leq\frac{\sqrt3}{2}$ and $|U_2(x)|\leq\frac{3}{4}$ for every $x$.
Assume that $|U_k(x)|\leq\left(\frac{\sqrt3}{2}\right)^k$ for every real $x$ and every $k\leq n$, for some $n\ge2$.
*
*If $|\sin{x}|\leq\frac{\sqrt3}{2}$, then $$|U_{n+1}(x)|=|\sin x|\cdot|U_n(2x)|\leq\frac{\sqrt3}{2}\cdot\left(\frac{\sqrt3}{2}\right)^n=\left(\frac{\sqrt3}{2}\right)^{n+1}$$
*If $|\sin{x}|\geq\frac{\sqrt3}{2}$, then $|\cos{x}|\leq\frac{1}{2}$ and $|\sin{x}\sin2x|=|2(1-\cos^2x)\cos{x}|\leq\frac{3}{4}$.
Thus, $$|U_{n+1}(x)|=|\sin{x}\sin2x|\cdot|U_{n-1}(4x)|\leq\frac{3}{4}\cdot\left(\frac{\sqrt3}{2}\right)^{n-1}=\left(\frac{\sqrt3}{2}\right)^{n+1}$$
By induction on $n\ge1$, we are done.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2043127",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 0
} |
Rate of convergence of ${x_n}=\arctan ({x_{n-1}})$ Find the rate of convergence of :
${x_n}=\arctan ({x_{n-1}})$. Well,
$$\lim\limits_{n\to\infty}\frac{{x_{n+1}}}{{x_n}}=\lim\limits_{n\to\infty}\frac{\arctan({x_n})}{\arctan({x_{n-1}})}=l,$$
where $l$ is the point where the series converges to. I can't calculate this limit so I can not find $l$.
The next step would be to calculate: $$\lim\limits_{n\to\infty}\frac{{x_{n+1}}-l}{({x_n}-l)^p}$$
Basically this question has to do with the $l$ calculation which will help me define the rate of converge for ${x_n}$. Any hint about the limit would be appreciated. Thanks in advance!
| As pointed out in comments, both the limits $\lim_{n\to\infty} \frac{x_{n+1}}{x_n}$ and $\lim_{n\to\infty} \frac{x_{n+1}}{x_n^p}$ are easy to compute given the observation that $x_n \to 0$. So let me demonstrate a method of finding an asymptotic expansion of $(x_n)$ instead.
A hint is that you can estimate the growth of $x_n^{-2}$ more easily. For instance, assuming that $x_1 > 0$,
Step 1. Since $(x_n)$ is strictly decreasing and bounded, $(x_n)$ converges. The limit must be a fixed point of $\arctan$, which is exactly $0$.
Step 2. Notice that $\frac{1}{\arctan^2 x} = \frac{1}{x^2} + \frac{2}{3} + \mathcal{O}(x^2)$. From this,
$$ \frac{1}{x_{n+1}^2} - \frac{1}{x_n^2} = \frac{2}{3} + \mathcal{O}(x_n^2). $$
In view of Stolz–Cesàro theorem, we have
$$ \lim_{n\to\infty} \frac{x_n^{-2}}{n} = \lim_{n\to\infty} \frac{\frac{2}{3} + \mathcal{O}(x_n^2)}{(n+1) - n} = \frac{2}{3} $$
and hence $x_n^{-2} \sim \frac{2}{3}n$.
Step 3. Now using $\frac{1}{\arctan^2 x} = \frac{1}{x^2} + \frac{2}{3} - \frac{1}{15}x^2 + \mathcal{O}(x^4)$ and the previous step,
$$ \lim_{n\to\infty} \frac{x_n^{-2} - \frac{2}{3}n}{\log n} = \lim_{n\to\infty}\frac{-\frac{1}{15}x_{n+1}^2 + \mathcal{O}(x_n^4)}{\log(n+1) - \log n} = -\frac{1}{10} $$
and hence $x_n^{-2} = \frac{2}{3}n - \frac{1}{10}\log n + o(\log n)$.
Step 4. Now write
$$ \frac{1}{x_n^2} - \left( \frac{2}{3}n - \frac{1}{10}\log n \right)
= \frac{1}{x_1^2} - \frac{2}{3} + \sum_{k=1}^{n-1} \left( \frac{1}{x_{k+1}^2} - \frac{1}{x_k^2} - \frac{2}{3} + \frac{1}{10}\log\left(1+\frac{1}{k}\right) \right). $$
Using the previous step, it is easy to check that the right-hand side converges as $n\to\infty$ with the error term decaying at most as fast as $\mathcal{O}(\frac{\log n}{n})$. So there exists a constant $C$ such that
$$ x_n^{-2} = \frac{2}{3}n - \frac{1}{10}\log n + C + \mathcal{O}\left(\frac{\log n}{n}\right). $$
Using this estimation above, we can easily check that
$$ x_n = \sqrt{\frac{3}{2n}} \left( 1 + \frac{3}{40}\frac{\log n}{n}\right) + \mathcal{O}\left(\frac{1}{n^{3/2}}\right). $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2043709",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Finding the joint Probability distribution of $X$ and $Y$? If the joint probability distribution of X and Y is given by
$$f(x,y)= \frac{(x-y)^2}{7}, \text{for }x=1,2,3;y=1,2 $$
$(1)$ Find the probability distribution of $U = X + Y; $
$(2)$ Find the conditional probability distribution of $X$ given $U =4.$
In order to solve this problem one must draw a chart.
$$\begin{array}{|c|c|c|c|}
\hline
(x,y)& (1,1) & (1,2) & (2,1) & (2,2) & (3,1) & (3,2) \\ \hline
f(x,y)& 0 & \frac{1}{7} & \frac{1}{7} & 0 & \frac{4}{7} & \frac{1}{7}\\ \hline
U=x+y& 2 & 3 &3 & 4 & 4 & 5\\ \hline
x & & &\\ \hline
\end {array}$$
How does one fill up the rest of the table and answer questions one and two.
EDIT
In order to find solve $(1)$ one must add all the related $f(x,y)$ relations. Thus
$(1)$ $$ \quad P(U=2) =0, \\ P(U=3) = \frac{1}{7} + \frac{1}{7} = \frac27, \\ P(U=4)= 0+\frac{4}{7} = \frac47, \\ P(U=5)=\frac{1}{7}$$
One must use this notation to solve.
$$P(x=1|U=4)= \frac{P(x=1,U=4)}{P(U=4)} = ? \\P(x=2|U=4) = ? \\ P(x=3|U=4) = ? $$
Knowing this does anyone know how to solve $(2)$ using this notation? What does one substitute for this question to derive the answer?
| Ok so the first thing you notice is that so far your attempt has
$$
\begin{align}
P(U = 2) &= 0, \\
P(U = 3) &= \frac{1}{7}, \\
P(U = 4) &= \frac{4}{7}, \\
P(U = 5) &= \frac{1}{7}
\end{align}
$$
and zero elsewhere, but summing over all possible situations only takes us to $\frac{6}{7}$ so something has clearly gone wrong! So what you have missed is that
$$
P(U=3) = P(x=1,y=2) + P(x=2,y=1) = \frac{2}{7}.
$$
For the second part of your question look at your table and study the different combinations of $x,y$ that will make $U=4$ and then look at the joint probability of these combinations, and you should see clearly what the distribution of $x$ must be.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2043984",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Finding the area inside the plot $x^4+y^4=x^2+y^2$ Find the area inside the plot $x^4+y^4=x^2+y^2$.
| Using polar coordinates, we have $x = r\cos \theta, y = r\sin \theta\implies x^4+y^4 = r^4(\cos ^4 \theta + \sin^4 \theta)= r^4(1 - 2\sin^2\theta\cos^2\theta)= x^2+y^2 = r^2\implies r^2 = \dfrac{1}{1-2\sin^2\theta\cos^2\theta}\implies r = \dfrac{1}{\sqrt{1-2\sin^2\theta\cos^2\theta}}\implies \text{Area} = 4\displaystyle \int_{0}^{\frac{\pi}{2}}\int_{0}^{\frac{1}{\sqrt{1-2\sin^2\theta\cos^2\theta}}}rdrd\theta$. Can you continue?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2044460",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 0
} |
How to find the coordinates under a basis when variables are involved I understand how to use row operations when trying to find coordinates of a vector under a basis when both are involving numbers. Simply create an augmented matrix and row reduce. But how do I go about finding the coordinates of the vector ($2x+1+ \frac{x^3}{3} + \frac{x^2}{2}$) under the basis B = ($x,x^2+x, 3-2x,x^3$)
I already have the solution, can someone please explain the steps to arrive there?
| Writing the vector as a combination of the base we have:
$$2x+1+ \frac{x^3}{3} + \frac{x^2}{2}=a(x)+b(x^2+x)+c(3-2x)+d(x^3)$$
$$2x+1+ \frac{x^3}{3} + \frac{x^2}{2}=dx^3+bx^2+(a+b-2c)x+3c$$
and using polynomial identity we get:
$$d=1/3$$
$$b=1/2$$
$$a+b-2c=2$$
$$3c=1$$
so
$$a=13/6$$
$$b=1/2$$
$$c=1/3$$
$$d=1/3$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2045263",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Solve $\frac{5}{x+3}+\frac{2}{2x+6}=4$ I want to solve $\frac{5}{x+3}+\frac{2}{2x+6}=4$ for $x$
I try this;
$5(2x+6)+2(x+3)=4(x+3)(2x+6)$
$12x+36 = 4(2x^2+12x+18)$
$8x^2+36x+36=0$
Where I would then go to the formula, however the answer says -1.5, what am I doing wrong.
| The quadratic equation you have obtained has two solutions: $-3$ and $-\frac{3}{2}$. The latter one is the correct answer. The former one is incorrect because it would lead to division by $0$ when we plug it into the original equation.
However, the problem need not be solved via quadratic equation. Observe that
$\frac{5}{x + 3} + \color{red}{\frac{2}{2x + 6}} = \frac{5}{x + 3} + \color{red}{\frac{1}{x + 3}} = \frac{6}{x + 3}$. Then the equation could be written as $4(x + 3) = 6$, so $x = \frac{6}{4} - 3 = -\frac{3}{2}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2046493",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 7,
"answer_id": 1
} |
Evaluate the limit (if it exists) $\{(\frac{n^3+4n-5}{n^6+2n^2-3},n-7[\frac{n}{7}])\}_{n=1}^{\infty}$
$$\left\{\frac{n^3+4n-5}{n^6+2n^2-3},n-7\left[\frac{n}{7}\right]\right\}_{n=1}^{\infty}$$
where $\left[\frac{n}{7}\right]$ is the floor and ceiling function
The sequence converges if each element convarges so:
$$\lim_{n\to \infty} \frac{n^3+4n-5}{n^6+2n^2-3}=\lim_{n\to \infty} \frac{\frac{n^3}{n^6}+\frac{4n}{n^6}+\frac{-5}{n^6}}{\frac{n^6}{n^6}+\frac{2n^2}{n^6}+\frac{-3}{n^6}}=0$$
$$\lim_{n\to \infty}n-7\left[\frac{n}{7}\right]=\infty\cdot \infty=\infty$$
So $$\left\{\left(\frac{n^3+4n-5}{n^6+2n^2-3},n-7\left[\frac{n}{7}\right]\right)\right\}_{n=1}^{\infty}=\{0,\infty\}$$
And therefore there is not limit?
| The remainder-like function
$$
n-7\left[\frac n7\right]
$$
cycles among the values $\{0,1,2,3,-3,-2,-1\}$. Since one of the components does not converge, the sequence of pairs does not converge.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2050851",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Prove that $\sum\limits_{cyc}\frac{a}{\sqrt{a+3b}}\geq\sqrt{a+b+c+d}$ Let $a$, $b$, $c$ and $d$ be positive numbers. Prove that:
$$\frac{a}{\sqrt{a+3b}}+\frac{b}{\sqrt{b+3c}}+\frac{c}{\sqrt{c+3d}}+\frac{d}{\sqrt{d+3a}}\geq\sqrt{a+b+c+d}$$
I tried Holder, AM-GM and more, but without success.
| WLOG, assume that $d = \min(a,b,c,d)$.
Squaring both sides and using AM-GM, it suffices to prove that
\begin{align}
&\sum_{\mathrm{cyc}} \frac{a^2}{a+3b}
+ \sum_{\mathrm{cyc}} \frac{4ab}{a+3b + b+3c}
+ \frac{4ac}{a+3b + c+3d} + \frac{4bd}{b+3c + d+3a} \\
\ge\ & a + b + c + d.
\end{align}
After clearing the denominators, it suffices to prove that
$f(a, b, c, d)\ge 0$ where $f(a,b,c,d)$ is a homogeneous polynomial of degree $11$.
The Buffalo Way works. Let $c=d+s, \ b = d+t, \ a = d+r; \ s, t, r\ge 0$.
We have $$f(d+r, d+t, d+s, d) = q_9d^9 + q_8d^8 + \cdots + q_1d + q_0.$$
Here $q_9 = 12582912r^2+16777216rs-20971520rt+12582912s^2-20971520st+12582912t^2$, etc.
It suffices to prove that $q_9, q_8, \cdots, q_0 \ge 0$.
I used Mathematica Resolve and the Buffalo Way to verify it. We are done.
However, nice or simple proof for $q_9, q_8, \cdots, q_0 \ge 0$ is expected.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2051278",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "22",
"answer_count": 2,
"answer_id": 1
} |
Evaluate $\displaystyle \int_C \frac{1}{4z^2+4z-3}$ where $C=C^{+}_{1}(0)$.
Evaluate $\displaystyle \int_C \frac{1}{4z^2+4z-3}$ where $C=C^{+}_{1}(0)$.
I am not sure whether I understand how to use Cauchy's theorem which states that:
If $f$ is a analytic in a simply connected domain $D$, and $C$ is a simple closed contour lying in $D$, then $\displaystyle \int_C f(z)dz=0$. (From this also follows the deformation of contours).
Anyways, what I have done so far is in decomposing $f(z)=\frac{1}{4z^2+4z-3}=\frac{1}{4}\left(\frac{1}{2z-1}-\frac{1}{2z+3}\right)$.
Now, I look at the singularities of $f$ which are $z=\frac{1}{2},-\frac{3}{2}$. My contour is the circle (with positive orientation) centered at $0$ with radius $1$ (so the integral part of $\frac{1}{4}\frac{1}{2z+3}$ just goes to $0$ by Cauchy). I then make a contour $C_{1/2}$ around the singularity $\frac{1}{2}$ so I can apply the deformation theorem. I'm then left with $\frac{1}{4}{2\pi i}=\frac{\pi}{2}i$. But apparently the answer is $\frac{\pi}{4}i$. Where did I go wrong here?
Thanks.
| $\displaystyle\frac{1}{4}\frac{1}{2z-1}=\frac{1}{8}\frac{1}{z-\frac{1}{2}}$, so by Cauchy Integral Formula for discs,
\begin{equation*}
\frac{1}{8}\int_{B_1(0)}\frac{dz}{z-\frac{1}{2}}=\frac{1}{8}2\pi i=\frac{1}{4}\pi i
\end{equation*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2052220",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Simple partial fraction expansion I would like to ask for help solving a simple Partial fraction example:
(It contains complex roots)
$\frac{1}{(x^3 - 3x^2 + 4x - 2)}$
According to my book we can factor it as a product of the real roots and quadratic factors:
$\frac{1}{(x-1)(x^2-2x+2)}$
Quadratic form should adopt the following scheme:
$\frac{ax+b}{x^2+px+q}$
Using the same system as in the case that all the roots are real:
$\frac{a}{x-1}+\frac{ax+b}{x^2-2x+2}==>\frac{a(x^2-2x+2)+(ax+b)(x-1)}{(x-1)(x^2-2x+2)}$
And then we solve the equation:
$a(x^2-2x+2)+(ax+b)(x-1) = 1$
The book expose the solution:
$\frac{1}{x-1}+\frac{1-x}{x^2-2x+2}$
I just can't figure out how they do the transition.
I am learning to use "Inverse Laplace Partial Fractions with Complex Roots" by converting the quadratic form to a product of complex roots:
$(x^3 - 3x^2 + 4x - 2)=(x-1)(x-1-i)(x-1+i)$
But it is just way out of the current scope (is not explained in the book, I found it during research). There must be some simple method/analysis but I just can't see it.
| You need to start from
$$\frac{a}{x-1}+\frac{bx+c}{x^2-2x+2}=\frac{1}{(x-1)(x^2-2x+2)}$$
$$a(x^2-2x+2)+(bx+c)(x+1)=1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2052705",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Proving one of the Binomial Identities The binomial identity is
$$\sum^{\left \lfloor{n/3}\right \rfloor }_{k=-\left \lfloor{n/3}\right \rfloor} (-1)^k \binom{2n}{n+3k} = 2 \cdot 3^{n-1} $$
and this is valid for all positive integers $n$. What would be some proofs to show that this identity is true?
| This can be proved by induction on $n$. Instead of writing it up as efficiently as possible, I’ll show you how I got there.
First note that there’s no need to specify limits on the index $k$: the sum is taken over all $k$ for which the binomial coefficient is non-zero, with the usual convention that $\binom{n}k=0$ if $k>n$ or $k<0$. With the induction hypothesis
$$\sum_k(-1)^k\binom{2n}{n+3k}=2\cdot3^{n-1}\;,\tag{1}$$
the natural approach to the induction step looks like this:
$$\begin{align*}
\sum_k(-1)^k\binom{2n+2}{n+1+3k}&=\sum_k(-1)^k\left(\binom{2n}{n-1+3k}+2\binom{2n}{n+3k}+\binom{2n}{n+1+3k}\right)\\
&=4\cdot3^{n-1}+\sum_k(-1)^k\binom{2n}{n-1+3k}+\sum_k(-1)^k\binom{2n}{n+1+3k}\\
&=4\cdot3^{n-1}+\sum_k(-1)^k\binom{2n}{n+1-3k}+\sum_k(-1)^k\binom{2n}{n+1+3k}\\
&=4\cdot3^{n-1}+2\sum_k(-1)^k\binom{2n}{n+1+3k}\;,
\end{align*}$$
and we’d like to know that
$$\sum_k(-1)^k\binom{2n}{n+1+3k}=3^{n-1}\;.\tag{2}$$
Fine: we can try proving both identities by induction simultaneously. In that case $(2)$ is also part of our induction hypothesis, alongside $(1)$, so we can finish off the calculation with
$$4\cdot3^{n-1}+2\sum_k(-1)^k\binom{2n}{n+1+3k}=4\cdot3^{n-1}+2\cdot3^{n-1}=2\cdot3^n\;,$$
as desired, and the rest of our induction step looks like this:
$$\begin{align*}
\sum_k(-1)^k\binom{2n+2}{n+2+3k}&=\sum_k(-1)^k\left(\binom{2n}{n+3k}+2\binom{2n}{n+1+3k}+\binom{2n}{n+2+3k}\right)\\
&=2\cdot3^{n-1}+2\cdot 3^{n-1}+\sum_k(-1)^k\binom{2n}{n+2+3k}\\
&=4\cdot3^{n-1}+\sum_k(-1)^k\binom{2n}{n+2+3k}\\
&=4\cdot3^{n-1}-\sum_k(-1)^k\binom{2n}{n-1+3k}\\
&=4\cdot3^{n-1}-\sum_k(-1)^k\binom{2n}{n+1-3k}\\
&=4\cdot3^{n-1}-\sum_k(-1)^k\binom{2n}{n+1+3k}\\
&=4\cdot3^{n-1}-3^{n-1}\\
&=3^n\;,
\end{align*}$$
again as desired.
This is yet another induction in which it seems to be easier to prove a stronger result.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2054777",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Four Equations in Four Unknowns
Four Equations in Four Unknowns
Completely solve the following equation!
$$\begin{eqnarray}
x&+&y&+&z&+&w&=&10 \\
x^2&+&y^2&+&z^2&+&w^2&=&30 \\
x^3&+&y^3&+&z^3&+&w^3&=&100\\
&&&&&&xyzw&=&24
\end{eqnarray}$$
This is problem 3 on page 3 of
Mathematical Quickies
270 Stimulating Problems with Solutions
by Charles W. Trigg
Dover Publications, Inc., New York
ISBN 0-486-24949-2
Here is the solution from page 78 of the book:
By inspection $(1,2,3,4)$ is a solution if the first and fourth equation and satisfies the second and third equations. Since the equations are symmetrical in $x$, $y$, $z$, $w$ the other $23$ permutations of $1,2,3,4$ are solutions also.
But these are all the solutions, since the product of the degrees of the equations is 4!
I assume that the exclamation mark at the end of the last sentence is a factorial symbol because the product of the degree of the equation is 24.
This seems to be a property that is similar to the fact that a univariate polynomial of degree $n$ has at most $n$ zeroes. But I can't see how to generalize this to multivariate equations.
Why does this system of equations where the product of the degree of the equations is 24 has at most 24 solutions?
| I. Yes, you are correct that a univariate polynomial of degree $n$ has $n$ zeros (counting multiplicity). However, your system are just the roots of a univariate in disguise.
The clue is the elementary symmetric polynomials $x+y+z+w$ and $xyzw$. If these unknowns are indeed the roots of the quartic,
$$F(u)=u^4+au^3+bu^2+cu+d=0$$ then,
$$\begin{aligned}
x+y+z+w &=10 = -a\\
x^2+y^2+z^2+w^2 &=30 = a^2-2b\\
x^3+y^3+z^3+w^3 &=100=-a^3 + 3 a b - 3 c\\
xyzw &=24=d
\end{aligned}\tag1$$
It is easy to solve for $a,b,c,d\,$ above and we get,
$$F(u)=u^4 - 10u^3 + 35u^2 - 50u + 24 = 0$$
$$F(u)=(u - 1)(u - 2)(u - 3)(u - 4)=0$$
hence these, including their permutations, are all the solutions.
II. Let $n_r$ be the number of roots and $n_d$ be the product of the degrees. By Bezout's theorem, then $n_r$ is at most equal to $n_d$. The observation that your example has $n_r=n_d=24$ is just a peculiarity of the system. If we tweak it slightly,
$$\begin{aligned}
x+y+z+w &=10 = -a\\
x^2+y^2+z^2+w^2 &=30 = a^2-2b\\
\color{blue}{x^5+y^5+z^5+w^5} &=100=-a^5 + 5 a^3 b - 5 a b^2 - 5 a^2 c + 5 b c + 5 a d\\
xyzw &=24=d
\end{aligned}\tag2$$
Solving for $a,b,c,d,\,$ they turn out to be rational so,
$$F(u)=u^4 - 10u^3 + 35u^2 - \color{blue}{\tfrac{602}{13}}u + 24 = 0$$
though $F(u)$ is no longer rationally factorable. Like the previous, by including the permutations of the $u_i$, there are again $n_r=24$ roots, but the product of the degrees is different now as $n_d=1\times2\times5\times4 = 40$ so $n_r\neq n_d$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2055106",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
} |
Inverse laplace transform $1/(s^2+9)^2$
Find inverse laplace transform of $$\frac{1}{(s^2+9)^2}$$
I've tried to decompose the fraction using
$$\frac{As+B}{s^2+9}+\frac{Cs+D}{(s^2+9)^2}$$
$$1=(As+B)(s^2+9)+Cs+D$$
yet D=1, still giving me the same exact equation
$$\frac{1}{(s^2+9)^2}$$
any help?
| One could make an ansatz with "known" expressions from a simple Laplace Transform Table:
$$\frac1{(s^2+a^2)^2}=\frac{Aa}{s^2+a^2}+\frac{Bs}{s^2+a^2}+\frac{2Cs}{(s^2+a^2)^2}+\frac{D(s^2-a^2)}{(s^2+a^2)^2},$$
given the Laplace transforms of $\sin at$, $\cos at$, $t\sin at$, and $t\cos at$, respectively. Hence, we get $A=1/(2a^3)$, $B=C=0$, and $D=-1/(2a^2)$. Therefore
$$\frac1{(s^2+a^2)^2}=\frac1{2a^3}\cdot\frac a{s^2+a^2}-\frac1{2a^2}\cdot\frac{s^2-a^2}{(s^2+a^2)^2}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2056955",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 2
} |
Is there a "clean" way of parametrizing the intersection of the sphere and saddle? I was playing around, trying to find a parametrization for the intersection of the two surfaces,
$$x^2+y^2+z^2=1$$
$$z=x^2-y^2,$$
but I wasn't able to get anything nice-looking. Any suggestions, or is this a hopeless endeavor?
EDIT: Thanks, Thomas Andrews. With the corrected formulas,
$$ a(t) = \sqrt{\frac{2}{1+\sqrt{1+4\cos^2 2t}}} $$
$$ (x(t),y(t),z(t)) = (a(t)\cos t, a(t)\sin t, a(t)^2 \cos2t), $$
we get a very nice picture!
| $$x^2+y^2=1-z^2\\
x^2-y^2=z$$
So $x^2=\frac{1+z-z^2}{2}$ and $y^2=\frac{1-z-z^2}{2}$.
Now you just need to pick the range of values for $z$ where both $1-z-z^2$ and $1+z-z^2$ are positive. This is when $1-z^2\geq |z|$ or $1-3z^2+z^4\geq 0$ and $|z|\leq 1$. But $1-3z^2+z^4=\left(z^2-\frac{3}{2}\right)^2 -\frac{5}{4}$, so you need $$z^2\leq \frac{3-\sqrt{5}}{2} \text{ or } z^2\geq\frac{3+\sqrt{5}}{2}$$
The right side is irrelevant, since it is already greater than the known bound $1$. Also: $\frac{3-\sqrt{5}}{2}=\left(\frac{\sqrt{5}-1}{2}\right)^2$. So you get that $|z|\leq\frac{\sqrt{5}-1}{2}$ and:
$$(x,y,z)=\left(\pm\sqrt{\frac{1+z-z^2}{2}},\pm\sqrt{\frac{1-z-z^2}{2}},z\right)$$
To correct the nice parameterization by Servaes in his wrong but close answer, if $(x,y,z)=(a\cos t,a\sin t, z)$ then $z=a^2(\cos^2 t -\sin^2 t)=a^2\cos 2t$.
Then $1=x^2+y^2+z^2=a^2+a^4\cos^2 2t$ or $$a^2=\frac{-1\pm\sqrt{1+4\cos^2 2t}}{2\cos^2 2t}=\frac{2}{1\pm\sqrt{1+ 4\cos^2 2t}}$$
Since you need $a^2\geq 0$, you can eliminate the $-$ case from $\pm$ and you get:
$$a(t)=\sqrt{\frac{2}{1+\sqrt{1+4\cos^2 2t}}}$$
and $$(x,y,z)=(a(t)\cos(t),a(t)\sin(t),a(t)^2\cos(2t))$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2057231",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
How to work with triangle inequality using rational coordinates and squared distance? Given three points in some Euclidean space, $\mathbf{r}, \mathbf{g}, \mathbf{b} \in \mathbb{R}^d$, I know via the triangle inequality that:
$$\|\mathbf{r} - \mathbf{b}\| \leq \|\mathbf{r} - \mathbf{g}\| +\|\mathbf{g} - \mathbf{b}\|.$$
In the figure below, suppose we only know how to measure distance between the red and green points and between the green and blue points. Using the triangle inequality, we can put an upper bound on distance the between the red and blue points:
So far, just elementary geometry. However, now I'd like to do some computation with my points using "rational coordinates:" $\mathbf{r}, \mathbf{g}, \mathbf{b} \in \mathbb{Q}^d$, for example using gmp or cgal exact arithmetic types. It is easy to verify that the squared distance is also a rational: $\|\mathbf{r} - \mathbf{b}\|^2 = (\mathbf{r}-\mathbf{b})\cdot(\mathbf{r}-\mathbf{b}) \in \mathbb{Q}$. But the distance, $\|\mathbf{r}-\mathbf{b}\| = \sqrt{(\mathbf{r}-\mathbf{b})\cdot(\mathbf{r}-\mathbf{b})}$ may be irrational.
Clearly, the triangle inequality does not hold for squared Euclidean distances. The best upper bound I can come up with is rather loose:
$$\|\mathbf{r} - \mathbf{b}\|^2 \leq \left(\|\mathbf{r} - \mathbf{g}\| +\|\mathbf{g} - \mathbf{b}\|\right)^2,\\
\|\mathbf{r} - \mathbf{b}\|^2 \leq \|\mathbf{r} - \mathbf{g}\|^2 +\|\mathbf{g} - \mathbf{b}\|^2 + 2 \|\mathbf{r} - \mathbf{g}\| \|\mathbf{g} - \mathbf{b}\|,\\
\|\mathbf{r} - \mathbf{b}\|^2 \leq \|\mathbf{r} - \mathbf{g}\|^2 +\|\mathbf{g} - \mathbf{b}\|^2 + 2\max\left( \|\mathbf{r} - \mathbf{g}\|, \|\mathbf{g} - \mathbf{b}\|\right),\\\|\mathbf{r} - \mathbf{b}\|^2 \leq \min\left( \|\mathbf{r} - \mathbf{g}\|, \|\mathbf{g} - \mathbf{b}\|\right) + 3\max\left( \|\mathbf{r} - \mathbf{g}\|, \|\mathbf{g} - \mathbf{b}\|\right).\\
$$
Concretely, is there a tighter upper bound on the squared distance between two points using squared distances to an third, intermediary point?
More generally, is there a standard way/trick to work with the triangle inequality using just squared distances? Or using some other feature of rational coordinates?
| The bound for the triangle inequality is, and always is, the sum of the two euclidean distances. However, the question is how to represent this sum, and how to do calculations with it.
Let's say you have $A = |r - g|^2$ and $B = |g - b|^2$, which are rational numbers. Then the bound you want is simply $x = \sqrt{A} + \sqrt{B}$.
In general, the number $x$ generates a quadric extension of $\mathbb{Q}$, hence one has to work in this big number field.
The usual trick is then to write down a minimal polynomial for $x$, which is:
$$F(X)=(X - \sqrt{A} - \sqrt{B})(X - \sqrt{A} + \sqrt{B})(X + \sqrt{A} - \sqrt{B})(X + \sqrt{A} + \sqrt{B})\\
= X^4 - 2(A + B) X^2 + (A - B)^2$$
Since the polynomial has four real roots, the inequality $X > x$ is not equivalent to $F(X) > 0$. But if $X$ is bigger than $\max(\sqrt{A}, \sqrt{B})$, then it is clear that $X > x$ if and only if $F(X) > 0$.
I summarize all these in the following algorithm.
Input: positive rational numbers $A$, $B$ and $X$.
Output: true or false, indicating whether $X$ is bigger than $\sqrt{A} + \sqrt{B}$.
*
*if $X^2 \leq A$ or $X^2 \leq B$, return false;
*if $X^4 - 2(A + B) X^2 + (A - B)^2 > 0$, return true; otherwise return false.
Note that the input $X$ need not be rational, it can be any real number (e.g. float types). But in view of your application, you would probably only use the case where $X$ is rational.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2057364",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Find the Sum of an Infinite Series Compute
$${1 \cdot \frac {1}{2} + 2 \cdot \frac {1}{4} + 3 \cdot \frac {1}{8} + \dots + n \cdot \frac {1}{2^n} + \dotsb.}$$
I tried creating a partition for this but no such luck.
What would the the equation to generate the sum for the nth term
| Let $S=\sum_{n\geq 1}\frac{n}{2^n}.$ Then
$$2S=\sum_{n\geq 1}\frac{2n}{2^n}=\sum_{n\geq 1}\frac{n}{2^{n-1}}=\sum_{n\geq 1}\frac{(n-1)+1}{2^{n-1}} = 2+\sum_{n\geq 1}\frac{n-1}{2^{n-1}} = 2+S $$
hence $S=2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2059172",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
How many principal minors a 3x3 matrix can have? I'm reading the book Polarimetric Radar Imaging: From basics to applications
On page 75, there's this claim:
$$T_3 = \begin{bmatrix}
2A_0 & C-jD & H+jG\\
C+jD & B_0+B & E+jF\\
H-jG & E-jF & B_0-B
\end{bmatrix}$$
As the coherency $T_3$ matrix in such a case is a rank 1 Hermitian matrix, it follows
that its nine principal minors are zero, with:
$$\bbox[yellow]{2A_0(B_0+B)-C^2-D^2=0}\\
\bbox[yellow]{2A_0(B_0-B)-G^2-H^2=0}\\
-2A_0E+CH-DG=0\\
\bbox[yellow]{B_0^2-B^2-E^2-F^2=0}\\
C(B_0-B)-EH-GF=0\\
-D(B_0-B)+FH-GE=0\\
2A_0F-CG-DH=0\\
-G(B_0+B)+FC-ED=0\\
H(B_0+B)-CE-DF=0$$
I am wondering because I know that a $3\times 3$ matrix has nine $2\times 2$ submatrices and so nine minors, but only 3 of them (the highlighted ones are principal minors)?!!
The others are only submatrices (not principals) and so the other six equations are resulted from $\Re{|A_{({1,3},{1,2})}|}=0$,$\Im{|A_{({1,3},{1,2})}|}=0$,$\Re{|A_{({2,3},{1,2})}|}=0$,$\Im{|A_{({2,3},{1,2})}|}=0$,$\Re{|A_{({2,3},{1,3})}|}=0$ , $\Im{|A_{({2,3},{1,3})}|}=0$
Knowing that when a $3\times 3$ matrix is of rank 1, the determinant of all of its $2\times 2$ submatrices ($2\times 2$ minors) should be zero not just the principal ones.
| For a general $3\times 3$ matrix,
$$A =\begin{bmatrix}
a_{11} & a_{12} & a_{13}\\
a_{21} & a_{22} & a_{23}\\
a_{31} & a_{32} & a_{33}
\end{bmatrix}
$$ there is one third order principal minor namely $|A|$.
There are three second order principal minors:
$\begin{vmatrix}
a_{11} & a_{12}\\
a_{21} & a_{22}
\end{vmatrix}$ formed by deleting column $3$ and row $3$.
$\begin{vmatrix}
a_{11} & a_{13}\\
a_{31} & a_{33}
\end{vmatrix}$ formed by deleting column $2$ and row $2$.
$\begin{vmatrix}
a_{22} & a_{23}\\
a_{32} & a_{33}
\end{vmatrix}$ formed by deleting column $1$ and row $1$.
There are three first order principal minors: $|a_{11}|, |a_{22}|, a_{33}|$ .
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2059590",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
} |
Prove that $\sum\limits_{cyc}\frac{a^2}{a^3+2}\leq\frac{4}{3}$ if $a, b, c, d > 0$ and $abcd=1$
Let $a$, $b$, $c$ and $d$ be positive numbers such that $abcd=1$. Prove that:
$$\frac{a^2}{a^3+2}+\frac{b^2}{b^3+2}+\frac{c^2}{c^3+2}+\frac{d^2}{d^3+2}\leq\frac{4}{3}.$$
Vasc's LCF Theorem does not help here. Also I tried MV method, but without success.
| Let : $a= e^{x_1}\ , \ b=e^{x_2}\ , \ c=e^{x_3} \ , \ d =e^{x_4}$
So we must to prove:
$$f(x_1)+f(x_2)+f(x_3)+f(x_4) \le \dfrac{4}{3}$$
for $x_1+x_2+x_3+x_4=0 $
$$f(x)= \dfrac{e^{2x}}{e^{3x}+2}$$
Since : $f'(x)= \dfrac{e^{2x}(4-e^{3x})}{(2+e^{3x})^2}$ and $f''(x)=\dfrac{e^{2x}(e^{6x}-26e^{3x}+16)}{(2+e^{3x})^3}$
We only need to consider the inequality in case : $x_1=x_2=x_3=t\ , \ x_4= -4t$
$\Leftrightarrow a=b=c=x \ ,\ d=\dfrac{1}{x^3}$
$g(x)=\dfrac{3x^2}{x^3+2}+\dfrac{x^3}{1+2x^9} \ , x>0$
$g'(x)=\dfrac{3x(1-x^4)(4x^{17}-16x^{14}+4x^{13}+4x^{12}-16x^{10}+20x^9+8x^8+4x^5+8x^4-x^3+4x+4)}{(x^3+2)^2(2x^9+1)^2}$
Maximum is attained at $x=1\ ,\ g(1)=\dfrac{4}{3}$
Equality holdes for : $(a=b=c=d=1)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2061972",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 3,
"answer_id": 2
} |
Find the value of csc$\frac{16\pi}{3}$. Q. Find the value of csc$\frac{16\pi}{3}$.
A. This angle is equal to two full revolutions plus $\frac{4\pi}{3}$. I got this by subtracting, $\frac{16\pi}{3} - 4\pi = \frac{16\pi}{3} - \frac{12\pi}{3} = \frac{4\pi}{3}$. The terminal side is in Quadrant III. The reference angle is $\frac{4\pi}{3} - \pi = \frac{\pi}{3}$. The cosecant of $\frac{\pi}{3}$ is $\frac{2\sqrt{3}}{3}$. Using the negative, the answer is $-\frac{2\sqrt{3}}{3}$.
My questions:
is how do one tell how many revolutions it takes, based off the fraction they give?
And how does one find the cosecant of $\frac{\pi}{3}$? On the calculator I use sine and then take the reciprocal of sine, but sine come out to $.018276028$. And I'm not sure on how to get the reciprocal from that or even find the cosecant of that number.
| Just observe that $16=5\cdot 3+1$.
The cosecant of $16\pi/3$ is the reciprocal of the sine; from $16=5\cdot3+1$, we get
$$
\frac{16\pi}{3}=5\pi+\frac{\pi}{3}=\pi+\frac{\pi}{3}+4\pi
$$
you have
$$
\sin\frac{16\pi}{3}=\sin\left(\pi+\frac{\pi}{3}\right)=
-\sin\frac{\pi}{3}=-\frac{\sqrt{3}}{2}
$$
Therefore
$$
\csc\frac{16\pi}{3}=-\frac{2}{\sqrt{3}}=-\frac{2\sqrt{3}}{3}\approx-1.1547
$$
*
*You probably had the calculator set to angles in degrees, rather than radians.
*In a right triangle having one angle of $\pi/3$, the other acute angle is $\pi/6$. Thus the triangle is half of an equilateral triangle, and so $\cos\frac{\pi}{3}=\frac{1}{2}$. With Pythagoras' theorem, $\sin\frac{\pi}{3}=\frac{\sqrt{3}}{2}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2063947",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
Logic: Proving tautological consequence I'm having trouble proving this tautological consequence. I'd hope that you guys can maybe oversee my process and identify errors, because I went over this couple of times and I arrive at the same conclusion.
The question goes like this:
$A \Rightarrow B$
$C \Rightarrow B$
Therefore: $(A \lor C) \Rightarrow B$
Q: Show that the conclusion of the arguments is a tautological
consequence of the premises using truth tables.
This is how I tried to solve it:
My solution to the problem. It seems like I'm missing something very fundamental here.
| The problem is really:
A implies B
C implies B is
(A implies B) and (C implies B)
And then it should be easily provable
edit:
\begin{array}{ccc|c@{}ccc@{}ccccc@{}ccc@{}ccc@{}ccc@{}c}
a&b&c&(&a&\lor&c&)&\rightarrow&b&\leftrightarrow&(&a&\rightarrow&b&)&\land&(&c&\rightarrow&b&)\\\hline
1&1&1&&1&1&1&&1&1&\mathbf{1}&&1&1&1&&1&&1&1&1&\\
1&1&0&&1&1&0&&1&1&\mathbf{1}&&1&1&1&&1&&0&1&1&\\
1&0&1&&1&1&1&&0&0&\mathbf{1}&&1&0&0&&0&&1&0&0&\\
1&0&0&&1&1&0&&0&0&\mathbf{1}&&1&0&0&&0&&0&1&0&\\
0&1&1&&0&1&1&&1&1&\mathbf{1}&&0&1&1&&1&&1&1&1&\\
0&1&0&&0&0&0&&1&1&\mathbf{1}&&0&1&1&&1&&0&1&1&\\
0&0&1&&0&1&1&&0&0&\mathbf{1}&&0&1&0&&0&&1&0&0&\\
0&0&0&&0&0&0&&1&0&\mathbf{1}&&0&1&0&&1&&0&1&0&
\end{array}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2066844",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Number of positive integers $n$ The question is to find out the number of positive integers $n$ such that $nx^4+4x+3 \leq 0 $ for some real $x$ (without using a graphic calculator).
My attempt at the solution:
We have $nx^4+4x+3 \leq 0 $ which is $nx^4 \leq -3-4x$ or $n \leq -\frac{(3+4x)}{x^4}$. It can easily be observed that for $x=-1$ the condition of $n$ being a positive integer is satisfied. But I could not get others. Moreover to satisfy that $n$ is positive we must have $x<-3/4$.
Please help me in this regard. Thanks.
| Solution WITHOUT calculus.
Note that
$$nx^4+4x+3=(n-1)x^4+x^4+4x+3=(n-1)x^4+((x-1)^2+2)(x+1)^2.$$
Hence for $n=1$, we have that the above polynomial is $0$ at $x=-1$.
If $n\geq 2$ then, for any $x\in\mathbb{R}$,
$$nx^4+4x+3\geq x^4+2(x+1)^2>0$$
because the squares $x^4$ and $(x+1)^2$ are zero for different values of $x$ (namely $x=0$ and $x=-1$).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2067147",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
prove $\frac{a}{b}+\frac{b}{a}+\frac{c}{d}+\frac{d}{c}\le\frac{1}{64abcd}$ Let $a,b,c,d$ be positive real no such that $a+b+c+d\le1$
prove that$$\frac{a}{b}+\frac{b}{a}+\frac{c}{d}+\frac{d}{c}\le \frac{1}{64abcd}$$
I did this way$$\frac{a^2cd+b^2cd+c^2ab+d^2ab}{abcd}\le\frac{1}{64abcd}$$
$\implies$ $a^2cd+b^2cd+c^2ab+d^2ab\le\frac{1}{64}$
$\implies$ $(ac+bd)(ad+bc)\le\frac{1}{64}$
now I don't know how to proceed.
| By AM-GM $(ac+bd)(ad+bc)\leq\left(\frac{ac+bd+ad+bc}{2}\right)^2=\left(\frac{(a+b)(c+d)}{2}\right)^2\leq\left(\frac{\left(\frac{a+b+c+d}{2}\right)^2}{2}\right)^2\leq\frac{1}{64}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2067312",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Evaluation of limit $\lim_{n\rightarrow \infty}\sum^{n}_{k=1}\left(\sqrt{1+\frac{k}{n^2}}-1\right)$
Evaluation of the limit $$\lim_{n\rightarrow \infty}\sum^{n}_{k=1}\left(\sqrt{1+\frac{k}{n^2}}-1\right).$$
I don't understand how can I start the above problem because Riemann sums and integrals are not applicable here.
It seems that we can solve it using the Squeeze theorem, but I don't understand
which inequality I have to use. Help required, thanks.
| Note that $\sqrt{1+k/n^2} -1 = \frac{k}{2n^2} + O\left(\left( \frac{k}{n^2} \right)^2\right)$ by the Maclaurin series of $\sqrt{1+x}$. We use the notation $O(f(n))$ for some term thats bounded for sufficiently large $n$ by $C |f(n)|$ for some constant $C>0$ .
So, $\sum_{k=1}^n \left( \sqrt{1+k/n^2} -1 \right) = \sum_{k=1}^n \left( \frac{k}{2n^2} + O\left(\left( \frac{k}{n^2} \right)^2\right) \right) = \frac{n(n+1)}{2} \frac{1}{2n^2} + O\left(n\left( \frac{k}{n^2} \right)^2 \right) = \frac{n(n+1)}{2} \frac{1}{2n^2} + O(1/n)$ where the first equalityfollows from maclaurin series, second equality follows from evaluating the sum of the first term, and the sum containing $n$ terms, and the third equality follows from $k \leq n$.
Now, take the limit as $n \to \infty$ to get $\frac{1}{4}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2069244",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Show that $\arctan\left({1\over A}\right)=\arctan\left({1\over B}\right)+\arctan\left({1\over C}\right)$ $1,1,2,3,5,8,...$ for $n=1,2,3,4,5,...$ ;It is the n-th Fibonacci numbers.
Show that,
$\arctan\left({1\over xF_{2n-1}+F_{2n}}\right)=\arctan\left({1\over xF_{2n-1}+F_{2n+1}}\right)+\arctan\left({1\over x^2F_{2n-1}+xF_{2n+2}+F_{2n+2}}\right)$
I try:
Using
$$\arctan\left({1\over a}\right)-\arctan\left({1\over b}\right)=\arctan\left({b-a\over ab+1}\right)$$
Just ignore the arctan for the sake of space.
$$\arctan{1\over xF_{2n-1}+F_{2n}}-\arctan{1\over xF_{2n-1}+F_{2n+1}}=\arctan{1\over x^2F_{2n-1}+xF_{2n+2}+F_{2n+2}}=T$$
$$\arctan{F_{2n+1}-F_{2n}\over x^2F_{2n-1}^2+xF_{2n-1}(F_{2n}+F_{2n+1})+F_{2n}F_{2n+1}+1}=T$$
$$\arctan{F_{2n+1}-F_{2n}\over x^2F_{2n-1}^2+xF_{2n-1}F_{2n+2}+F_{2n}F_{2n+1}+1}=T$$
$$\arctan{F_{2n-1}\over x^2F_{2n-1}^2+xF_{2n-1}F_{2n+2}+F_{2n}F_{2n+1}+1}=T$$
Any hints on how to simplify this further?
| We make use of the identity:$$F_{2n-1}^2 -F_{2n}^2 + F_{2n}F_{2n-1} = 1...(1)$$ We can rewrite this as: $$2F_{2n}F_{2n-1} + F_{2n-1}^2 = F_{2n}^2 + F_{2n}F_{2n-1} + 1$$ $$\implies F_{2n-1}[2F_{2n} + F_{2n-1}] = F_{2n}[F_{2n} + F_{2n-1}] + 1...(2)$$ Now, we use the identity: $$F_{n+m} = F_nF_{m+1} + F_{n-1}F_m$$ to change the bracketed terms of $(2)$ into: $$F_{2n-1}F_{2n+2} = F_{2n}F_{2n+1} + 1$$ This is the same condition that is obtained by equating the $\arctan \frac{F_{2n-1}}{x^2F_{2n-1}^2 + xF_{2n-1}F_{2n+2} + F_{2n}F_{2n+1} + 1}$ and $\arctan \frac{1}{x^2F_{2n-1} + xF_{2n+2} +F_{2n+2}}$.
Hope it helps.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2069332",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
How to solve $x+x^2+x^3+x^4+x^5=62$? $$x+x^2+x^3+x^4+x^5=62$$
The roots in the set of Real numbers is $x=2$, but how solve it?
| Apply the rational roots theorem:
$$62\implies1,2,31,62$$
Thus, it is easy enough to check if $x=\pm1,\pm2,\pm31,\pm62$ are roots. Indeed, if we check, we find that $x=2$ is the only rational root. The remaining roots can be deduced to be complex since after factoring out, we find it has only one real relative minimum at $x=-1.55$ such that $\left(x^4+3 x^3+7 x^2+15 x+31\right)>0$ for all $x$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2069590",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Proving $\sqrt{x+y-x^2}+\sqrt{y+z-y^2}+\sqrt{z+w-z^2}+\sqrt{w+x-w^2}\le4\sqrt2-3$ I found this inequality using unusual calculations in maths Olympics
and I wonder if some clever teenager could prove it using their elementary knowledge of mathematics.
Let $x,y,z,w$ be non-negative numbers such that $$x+y+z+w=1$$ Prove that
$$\sqrt{x+y-x^2}+\sqrt{y+z-y^2}+\sqrt{z+w-z^2}+\sqrt{w+x-w^2}\le4\sqrt2-3$$
| It depends on what you consider to be elementary.
By cauchy-schwarz:
$$ \sum\limits_{cyc} \sqrt{x+y-x^2} \leq2\sqrt{\sum\limits_{cyc}x+y-x^2}=2\sqrt{\frac{7}{4}+\frac{1}{4}(x+y+w+z)^2-(x^2+y^2+w^2+z^2)}.$$
Now, by jenson we have that $$(x+y+w+z)^2\leq 4(x^2+y^2+z^2+w^2). $$
Hence we conclude that $$ \sum\limits_{cyc} \sqrt{x+y-x^2} \leq \sqrt{7}.$$
Edit: I would consider this elementary since jensen and cauchy-schwarz are some of the first inequalities one learns after AM-GM. Moreover, we have equality in the above if and only if$x=y=w=z=1/4.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2069671",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
$t^3$ is never equal to
Let $t$ be a real number such that $t^2=at+b$ for some positive integers $a$ and $b$. Then for any choice of positive integers $a$ and $b$, $t^3$ is never equal to:
(A). $4t+3$
(B). $8t+5$
(C). $10t+3$
(D). $6t+5$
We have $t^3 = at^2+bt$ and so for (A) we have $t^3 = 4t+3 = 4(at^2+bt)+3$. I don't see how to continue since this is a cubic equation. Is there an easier way of finding a contradiction?
| $$t^2=at+b\\ t^3=at^2+bt \to t^3=a(t^2)+bt\\t^3=a(at+b)+bt \\t^3=(a^2+b)t+ab$$
$$t^3=(a^2+b)t+ab=4t+3 \to ab=3 \to \\a=1 ,b=3 \to t^3=(1^2+3)t+1.3 \\$$
$$t^3=(a^2+b)t+ab=10t+3 \to ab=3 \to \\a=3 ,b=1 \to t^3=(3^2+1)t+1.3 \\$$
$$t^3=(a^2+b)t+ab=6t+5 \to ab=5 \to \\a=1 ,b=5 \to t^3=(1^2+5)t+1.5 \\$$
so answer is :"B"
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2071932",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.