Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Solving trigonometric equations: $\cos(3x)+ \cos(x)=0$ Recently I worked on a problem where I had to solve $$\cos(3x)+\cos(x)=0$$
When I tried calculating it by evaluating $x$, $x=2nπ+\dfrac{\pi}{4}$, $x=2n\pi-\dfrac{\pi}{4}$, or $x=2n\pi+\dfrac{\pi}{2}$ was the solution I reached.
Unfortunately, it was apparently wro... | Symmetry is always a powerful and nice instrument
$$
\eqalign{
& 0 = \cos (3x) + \cos x = \cr
& = \cos (2x + x) + \cos (2x - x) = \cr
& = \cos (2x)\cos x - \sin (2x)\sin x + \cos (2x)\cos x + \sin (2x)\sin x = \cr
& = 2\cos (2x)\cos x \cr}
$$
and then
$$
\eqalign{
& \cos (2x) = 0\quad \vee \quad... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3729544",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 5
} |
Pseudo determinant of product of two square matrices Let $A$ and $B$ be square symmetric matrices. $A$ is singular and $B$ is non-singular. Is there a way to decompose:
$Det(AB)$ in terms of $Det(A)$ and $det(B)$.
$Det(.)$ refers to the pseudo determinant and $det(.)$ refers to the usual determinant of a square non-s... | No. For example, consider
\begin{align*}
A_1 &= \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix} \\
A_2 &= \begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix} \\
B &= \begin{pmatrix} 1 & 0 \\ 0 & 2 \end{pmatrix}.
\end{align*}
Then, $\operatorname{Det} A_1 = \operatorname{Det} A_2 = 1$, but $\operatorname{Det}(A_1 B) = 1$ and $\op... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3732573",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
For what value of $m$ the is sum $\sum_{i = 0}^{m} {10 \choose i}{20 \choose m - i}$ where ${p\choose q} = 0$, if $pFor what value of $m$ the is sum
$$\sum_{i = 0}^{m} {10 \choose i}{20 \choose m - i}\text{where ${p\choose q}$} = 0\text{, if $p<q$, a maximum}$$
My approach
$$\sum_i^{m} {10 \choose i}{20 \choose m - i} ... | Vandermonde's Identity says that
$$
\sum_{i=0}^m\binom{10}{i}\binom{20}{m-i}=\binom{30}{m}\tag1
$$
The central binomial coefficient is the greatest. Therefore, the maximum of $(1)$ is when $m=15$; that is, $\binom{30}{15}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3734305",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 0
} |
Added angle formula to solve this indefinite integral $\int\frac{2\cos x-\sin x}{3\sin x+5\cos x }\,dx$ Starting from this very nice question Integrate $\int\frac{2\cos x-\sin x}{3\sin x+5\cos x }\,dx$ and the relative answers, I would to understand because this integral $$\int\frac{2\cos x-\sin x}{3\sin x+5\cos x }\,d... | Answer for @Koro: Yes with some calculus I obtain:
$$\int \left(\cos \left(a+b\right)-\tan \left(t\right)\sin \left(a+b\right)\right)dt=t\cos \left(a+b\right)+\sin \left(a+b\right)\ln \left|\cos \left(t\right)\right|+k, \quad k\in\mathbb R$$
Hence,
$$\int\frac{2\cos x-\sin x}{3\sin x+5\cos x }dx=$$
$$=\sqrt{\frac{5}{34... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3734680",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
How to simplify $\cos(\frac{1}{3}\cos^{-1}(x))$ and $\sin(\frac{1}{3}\cos^{-1}(x))$? How to simplify this trigonometric expression
$$
\cos \left(\frac{1}{3} \cos ^{-1}(b)\right)
$$
I want to write this equation as
$$
\cos \left(\frac{1}{3} \cos ^{-1}(b)\right) = \cos(\cos^{-1}(B)) = B
$$
Where $B$ is written in terms o... | Use trig. identity $$4\cos^3x-3\cos x=\cos 3x$$
substitute $x=\frac13\cos^{-1}b$
$$4\cos^3\left(\frac13\cos^{-1}b\right)-3\cos\left(\frac13\cos^{-1}b\right)=b$$
$$4B^3-3B-b=0$$
Solve above cubic equation for $B$ i.e. $\cos\left(\frac13\cos^{-1}b\right)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3739741",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Prove $\int_0^1dx\int_x^{1/x}\frac{y^2dy}{(x+y)^2\sqrt{1+y^2}}=\sqrt{2}-\frac{1}{2}$ by change of order of integration. I was trying to prove the following by changing the order of integration:
$$\int_0^1dx\int_x^{1/x}\frac{y^2dy}{(x+y)^2\sqrt{1+y^2}}=\sqrt{2}-\frac{1}{2}$$
Splitting the region $R=\big\{0\leq x\leq1, x... | Without seeing your computation, it is not possible to pinpoint your error. Let $$f(x,y) = \frac{y^2}{(x+y)^2\sqrt{1+y^2}} .$$
Then $$\int f(x,y) \, dx = -\frac{y^2}{(x+y)\sqrt{1+y^2}} + C$$ and in particular, for $g > 0$,
$$F(g,y) = \int_{x=0}^g f(x,y) \, dx = \frac{y^2}{\sqrt{1+y^2}}\left(\frac{1}{y} - \frac{1}{g+y}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3740661",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How to prove $(n + \frac{1}{2})\log(1+\frac{1}{n})$ is increasing in $n$? Is $(n + \frac{1}{2})\log(1+\frac{1}{n})$ increasing in $n$?
I attempted to differentiate $p_n=(n + \frac{1}{2})\log(1+\frac{1}{n})$ with respect to $n$.
$p_n^{'} = \log(1+\frac{1}{n}) + \frac{(n+\frac{1}{2})}{(1+\frac{1}{n})}(-\frac{1}{n^2}) = \... | When in doubt, plot the function. As you can see, the function is strictly decreasing.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3740793",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
How prove this $\sum_{i=n+2}^{+\infty}\frac{1}{i^2}>\frac{2n+5}{2(n+2)^2}$ let $n$ be postive integer,show that
$$\sum_{i=n+2}^{+\infty}\dfrac{1}{i^2}>\dfrac{2n+5}{2(n+2)^2}\tag{1}$$
I know
$$\sum_{i=n+2}^{+\infty}\dfrac{1}{i^2}>\int_{n+2}^{+\infty}\dfrac{1}{x^2}dx=\dfrac{1}{n+2}$$
But $$\dfrac{1}{n+2}-\dfrac{2n+5}{2(n... | By using $\frac{1}{q^2} = \int_0^\infty \mathrm{e}^{-x q} x \mathrm{d} x$ for $q > 0$, we have (the summation and the integral are interchangeable by Fubini/Tonelli theorems)
\begin{align}
\sum_{i=n+2}^\infty \frac{1}{i^2} &= \sum_{i=n+2}^\infty \int_0^\infty \mathrm{e}^{-x i} x \mathrm{d} x\\
&= \int_0^\infty \sum_{i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3743211",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
how can I integrate $\int \dfrac{2x+3}{x^3-9x}dx$? How can I integrate this $$\int \dfrac{2x+3}{x^3-9x}dx?$$
My work
$$\dfrac{2x+3}{x^3-9x}=\dfrac{2x+3}{x(x+3)(x-3)}$$
$$=\dfrac{A}{x-3}+\dfrac{B}{x}=\dfrac{C}{x+3}$$
after solving, I got $A=1/2$, $B=-1/3$, $C=-1/6$
partial fractions decompositions:
$$\dfrac{2x+3}{x^3-9x... | $$\left(\dfrac{1}{2}\ln (x-3)- \dfrac{1}{3}\ln (x)- \dfrac{1}{6}\ln (x+3)+C\right)'=\frac1{2(x-3)}-\frac1{3x}-\frac1{6(x+3)}
\\=\frac{18(x^2+3x)-12(x^2-9)-6(x^2-3x)}{12x(x^2-9)}
\\=\frac{72x+108}{36(x^3-9x)}.$$
It seems that you are right.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3744234",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Problematic solution related to finding the function range
The Question: Find the range of the following function
$$y=\dfrac{x}{2}+\dfrac{8}{x}$$
Solution $-1.$ (the solution given to me)
By Cauchy inequality,
$$\dfrac{x}{2}+\dfrac{8}{x}≥2\sqrt{ \dfrac{x}{2}× \dfrac{8}{x}}=4$$ where $x>0$ $$\dfrac{x}{2}+\dfrac{8}{x... | y' = 1/2 - 8/x$^2$
Solving y' = 0, x$^2$ = 16.
Thus for positive x,
the minimum of y is y(4) = 4
and the range of y is [4,$\infty$).
Simular for negative x.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3745060",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Line that passes through the Centroid of the ABC triangle and the sectioned areas. (ratio of the triangle areas: $2≤r≤2.25$) The line through Centroid $G$ divides the triangle $ABC$ into two figures. I need help with formal proof that the ratio between the areas of triangle ABC and the one that was sectioned is between... | If you are so inclined to fill out the details, then you can solve it by a simple Menelaus.
Let $ED$ intersect $AB$ at $F.$ If the two are parallel, then the ratio is precisely $\dfrac 94$, so we can assume they must intersect.
Then, draw the median $CM$, passing through $G$, and write the Menelaus' theorem for the tri... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3748043",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Prove The following inequality $(ax+by)^2 \le ax^2+by^2$ for $a+b=1$ Prove The following inequality $(ax+by)^2 \le ax^2+by^2$ for $a+b=1, 0 \le a,b \le 1$
I tried expanding the equation and substituting $b=1-a$
\begin{equation}
(ax+by)^2=a^2x^2+2abxy+b^2y^2=a^2x^2+2axy-2a^2xy+b^2y^2
\end{equation}
The middle member $2a... | We can use $\frac{x^2}{p}+\frac{y^2}{q}\geq\frac{(x+y)^2}{p+q}$,
$$ax^2+by^2=\frac{(ax)^2}{a}+\frac{(by)^2}{b}\geq\frac{(ax+by)^2}{a+b}$$
$$\Rightarrow ax^2+by^2\geq{(ax+by)^2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3749567",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Integer solutions of $2a+2b-ab\gt 0$ Let $a\in\mathbb{N}_{\ge 3}$ and $b\in\mathbb{N}_{\ge 3}$. What are the solutions of the Diophantine inequality
$$2a+2b-ab\gt 0?$$
By guessing, I found 5 solutions:
$$\text{1)}\, a=3,\, b=3$$
$$\text{2)}\, a=3,\, b=4$$
$$\text{3)}\, a=4,\, b=3$$
$$\text{4)}\, a=5,\, b=3$$
$$\text{5)... | My thinking is: in general $2a +2b$ is only two $a$s and two $b$s but $ab$ is $b$ number of $a$s which can be much more than $2$ if $b$ is large. Likewise $ab$ is $a$ number of $b$s and even if we used two "of those $a$s" to account for $2a$ we still have more than $2b$ is $a$ is large enough. So if $a,b$ are "large en... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3749995",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 3
} |
Proof that any number is equal to $1$ Before I embark on this bizzare proof, I will quickly evaluate the following infinite square root; this will aid us in future calculations and working:
Consider $$x=\sqrt{2+\sqrt{{2}+\sqrt{{2}+\sqrt{{2}...}}}}$$
$$x^2-2=\sqrt{2+\sqrt{{2}+\sqrt{{2}+\sqrt{{2}...}}}}=x \implies x^2-x-... | As you go along the last square root has $x^{2n}+\frac{1}{x^{2n}}$ which diverges, so it can't be ignored as $n\to \infty$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3750275",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
Remainder when $\prod_{n=1}^{100}(1- n^{2} +n^{4})$ is divided by $101$
What is the remainder when the expression
$$\prod_{n=1}^{100}(1- n^{2} +n^{4})$$
is divided by $101$?
If $\zeta=\dfrac{-1+\sqrt{-3}}{2}$, then
$$1-n^2+n^4=(1-n+n^2)(1+n+n^2)=(-\zeta-n)(-\bar{\zeta}-n)(\zeta-n)(\bar{\zeta}-n).$$
We then have
$$\pr... | Modulo $101$, the set of values $0^3, 1^3,\dots,100^3$ is a permutation of $0,1,2,\dots,100.$ This is because $101$ is prime and $3$ is not a divisor of $100.$
But $$n^4-n^2+1=\frac{n^6+1}{n^2+1}$$
Now, if $n=10,91$ then $n^2+1$ is divisible by $101.$ The other terms are a permutation, so:
$$\begin{align}\prod_{n=1}^{1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3750625",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Trigonometry (For what values of $(\cos A-1/5)(\cos B-1/5)(\cos C-1/5) \leq k$ hold in all triangles $ABC$? For what values of $k$ does
$$
\bigl(\cos A-\frac{1}{5}\bigr)\bigl(\cos B-\frac{1}{5}\bigr)\bigl(\cos C-\frac{1}{5}\bigr ) \leqslant k\:
$$
hold in all triangles $ABC$?
| \begin{align}
(\cos A-\tfrac15)(\cos B-\tfrac15)(\cos C-\tfrac15)
\le k
\tag{1}\label{1}
.
\end{align}
Given that for $\triangle ABC$ we know its semiperimeter $\rho$,
inradius $r$ and circumradius $R$,
let $u=\rho/R$, $v=r/R$. It is known that
for a valid triangle (including degenerate cases)
\begin{align}
v&\in[0,\tf... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3758766",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Solve Complex Equation $z^3 = 4\bar{z}$ I'm trying to solve for all z values where $z^3 = 4\bar{z}$.
I tried using $z^3 = |z|(\cos(3\theta)+i\sin(3\theta)$ and that $|z| = \sqrt{x^2+y^2}$ so:
$$z^3 = \sqrt{x^2+y^2}(\cos(3\theta)+\sin(3\theta))$$ and $$4\bar z = 4x-4iy = 4r\cos(\theta)-i4r\sin(\theta)$$
but I have no id... | Just for fun, let's do this by expanding $(x+iy)^3=4(x-iy)$ with $x,y\in\mathbb{R}$ and separate the real and imaginary parts. We wind up with
$$x(x^2-(3y^2+4))=0\quad\text{and}\quad y(y^2-(3x^2+4))=0$$
If $x=0$ then $y(y^2-4)=0$, so $y=0,\pm2$, hence $z=0$, $2i$ and $-2i$ are solutions.
If $x\not=0$, then we must have... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3759497",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
How many unordered pairs of positive integers $(a,b)$ are there such that $\operatorname{lcm}(a,b) = 126000$?
How many unordered pairs of positive integers $(a,b)$ are there such that $\operatorname{lcm}(a,b) = 126000$?
Attempt:
Let $h= \gcd(A,B)$ so $A=hr$ and $B=hp$, and $$phr=\operatorname{lcm}(A,B)=3^2\cdot 7\cdo... | $126000 = 2^4\cdot 3^2\cdot 5^3\cdot 7^1$
For two numbers to have an LCM of $126000$, either $a$ or $b$ has to have $2^4$ in its prime factorization, and similarly with $3^2, 5^3, $ and $7$.
We need either $a$ and/or $b$ to have $2^4$, so we assign $2^4$ to $a$ first. Then, $b$ can have a factor from $2^0$ to $2^4$, fo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3760396",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 3
} |
Solving $\sqrt{1+\sqrt{2x-x^2}} + \sqrt{1-\sqrt{2x-x^2}} = \sqrt{4-2x}$ Can we find the solutions for this equation?
$$\sqrt{1+\sqrt{2x-x^2}} + \sqrt{1-\sqrt{2x-x^2}} = \sqrt{4-2x}, \quad x \in \mathbb{R}$$
I tried to amplify the second square root in the $LHS$ with the conjugate and then use AM-GM in order to find whe... | Note
$$LHS^2= \left(\sqrt{1+\sqrt{2x-x^2}} + \sqrt{1-\sqrt{2x-x^2}} \right)^2\\
=2+2\sqrt{(1 -x)^2}= {4-2x}=RHS^2
$$
So, the solution is just the domain of the equation, which is jointly determined by
$$2x-x^2\ge 0,\>\>\>1-\sqrt{2x-x^2}\ge 0,\>\>\>4-2x\ge 0$$
which leads to $0\le x\le 1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3763208",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 1
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Computing $2 \binom{n}{0} + 2^2 \frac{\binom{n}{1}}{2} + 2^3 \frac{\binom{n}{2}}{3} + \cdots + 2^{n+1} \frac{\binom{n}{n}}{n+1}$ How can I compute the sum $2 \binom{n}{0} + 2^2 \frac{\binom{n}{1}}{2} +
2^3 \frac{\binom{n}{2}}{3} + \cdots + 2^{n+1} \frac{\binom{n}{n}}{n+1}$? I think I should expand $(1+ \sqrt{2})^n$ or... | The hint given by @ThomasAndrews indicates a purely algebraic approach.
We obtain
\begin{align*}
\color{blue}{\sum_{j=0}^n\frac{2^{j+1}}{j+1}\binom{n}{j}}
&=\frac{1}{n+1}\sum_{j=0}^n2^{j+1}\binom{n+1}{j+1}\tag{1}\\
&=\frac{1}{n+1}\sum_{j=1}^{n+1}2^j\binom{n+1}{j}\tag{2}\\
&\,\,\color{blue}{=\frac{1}{n+1}\left(3^{n+1}-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3764054",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
Simplifying trigonometry function by substitution I want to simplify the following expression: $\frac{{\left( {1 + \sqrt 3 \tan {1^{\circ}}} \right)\left( {1 + \sqrt 3 \tan {2^{\circ}}} \right)\left( {\tan {1^{\circ}} + \tan {{59}^{\circ}}} \right)\left( {\tan {2^{\circ}} + \tan {{58}^{\circ}}} \right)}}{{\left( {1 + {... | $$
\tan 59^{\circ} = \frac{\sqrt{3}-\tan 1^{\circ}}{1+\sqrt{3} \tan 1^{\circ}}
$$
$$
\tan 59^{\circ}+\tan 1^{\circ} = \frac{\sqrt{3}(1+\tan^2 1^{\circ})}{1+\sqrt{3} \tan 1^{\circ}},
$$so we have
$$
(1+\sqrt{3} \tan 1^{\circ})(\tan 59^{\circ}+\tan 1^{\circ})=\sqrt{3}(1+\tan^2 1^{\circ})
$$A similar argument works for $\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3764542",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
What should $n$ be equal to, so that $5^{2n+1}2^{n+2} + 3^{n+2}2^{2n+1}$ is completely divisible by $19$? What should $n$ be equal to, so that the number:
$$5^{2n+1}2^{n+2} + 3^{n+2}2^{2n+1}$$
is completely divisible by 19? I broke it into this:
$$20\cdot 2^{n}\cdot 25^{n}+18\cdot 3^{n}\cdot 4^{n}$$ But what should i d... | You can prove by induction that $20\cdot2^n\cdot25^n+18\cdot3^n\cdot4^n$
$=20\cdot50^{n}+18\cdot12^{n}$ is divisible by $19$.
It's obviously true for $n=0$, since $20+18=38=2\cdot19$.
Now assume $19$ divides $20\cdot50^{n}+18\cdot12^{n}$.
Then $20\cdot50^{n+1}+18\cdot12^{n+1}=50\cdot20\cdot50^{n+1}+12\cdot18\cdot12^{n+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3764765",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
Prove that $|a + b| = |a| + |b| \iff a\overline{b} \ge 0$ I'm reading a complex analysis book. In this book, the author establishes the following statement
If $a,b \in \mathbb{C}$, then $|a + b| = |a| + |b| \iff \left(a\overline{b}\in \mathbb{R}\right) \wedge \left(a\overline{b}\ge 0\right)$
This statement didn't s... | HINT:
Note that
$$|a+b|^2=|a|^2+|b|^2+2\text{Re}(a\bar b)$$
while $(|a|+|b|)^2=|a|^2+|b|^2+2|a||b|$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3765616",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
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Use Cauchy formula to solve $\int _0^{2\pi} \frac{dt}{a\cos t+ b\sin t +c} $ given $\sqrt{(a^2+b^2)}=1Use Cauchy formula to solve $$\int _0^{2\pi} \frac{dt}{a \cos t+ b \sin t +c} $$ Given $\sqrt{(a^2+b^2)}=1<c$.
I tried a variable substitution, but nothing elegant. Can anyone solve it using Cauchy formula?
Edit: there... | It may still be desirable to integrate as follows,
\begin{align}
& \int_0^{2\pi} \frac{dt}{a\cos t+ b\sin t +c}\\
=& \int_0^{2\pi} \frac{dt}{\sqrt{a^2+b^2}\cos (t-\theta) +c}
= \int_0^{\pi} \frac{2dt}{\sqrt{a^2+b^2}\cos t +c}\\
= & \int_0^{\pi} \frac{2dt}{2\sqrt{a^2+b^2}\cos^2\frac t2+c-\sqrt{a^2+b^2}}\\
= & \int_0^{\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3767829",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Positive integer solutions to $\frac{1}{a} + \frac{1}{b} = \frac{c}{d}$ I was looking at the equation $$\frac{1}{a}+\frac{1}{b} = \frac{c}{d}\,,$$ where $c$ and $d$ are positive integers such that $\gcd(c,d) = 1$.
I was trying to find positive integer solutions to this equation for $a, b$, given any $c$ and $d$ that s... | $\frac{1}{a}+\frac{1}{b} = \frac{c}{d}\,,$
Above has solution:
$a=(3k-2)$
$b=(k-2)(3k-2)$
$c=(k-1)(3k-2)$
$d=(k-2)(3k-2)^2$
For k=5 we get:
$(a,b,c,d)=(13,39,52,507)$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Show $C(t)=(2\cos^3(t),\sin(2t),2\sin(t))$ is on the intersection of $x^2+y^2+z^2=4$ and $x^2+y^2=2x$ Given a curve defined as: $$C(t)=(2\cos^2(t),\sin(2t),2\sin(t))$$ for $0\le t\le2\pi$
Show that the curve is on the intersection of the sphere $x^2+y^2+z^2=4$ and $x^2+y^2=2x$.
The intersection of the two equation sa... | Check if it belongs to the first surface
Replacing the curve in the first equation we get
$$x^2 + y^2 + z^2 =
4\cos^4(t )+ \sin^2(2t) + 4\sin^2(t) =4\cos^4(t )+ 4\sin^2(t)\cos^2(t) + 4\sin^2(t)$$
which is
$$4\cos^4(t )+ \sin^2(2t) + 4\sin^2(t) =4\cos^2(t)[\cos^2(t) + \sin^2(t)] + 4\sin^2(t) = 4\cos^2(t) + 4\sin^2(t) =... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3770695",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Given that $f(x)$ is a polynomial of degree $3$, and its remainders are $2x - 5$ and $-3x + 4$ when divided by $x^2 - 1$ and $x^2 - 4$ respectively. So here is the Question :-
Given that $f(x)$ is a polynomial of degree $3$, and its remainders are $2x - 5$ and $-3x + 4$ when divided by $x^2 - 1$ and $x^2 - 4$ respectiv... | Hint
You have
$$f(x)=(x^2-1)(ax+b)+(2x-5)$$
and
$$f(x)=(x^2-4)(cx+d)+(-3x+4)$$
From the second you get $f(1)=-3(c+d)+1$ and from the first you get $f(1)=-3$. Thus
$$\color{red}{c+d=\frac{4}{3}}$$
From the first you get $f(2)=3(2a+b)-1$ and from the second you get $f(2)=-2$. Thus
$$\color{red}{2a+b=\frac{-1}{3}}$$
You a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3770874",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 0
} |
Find the complete solution set of the equation ${\sin ^{ -1}}( {\frac{{x + \sqrt {1 - {x^2}} }}{{\sqrt 2 }}} ) = \frac{\pi }{4} + {\sin ^{ - 1}}x$ is The complete solution set of the equation ${\sin ^{ - 1}}\left( {\frac{{x + \sqrt {1 - {x^2}} }}{{\sqrt 2 }}} \right) = \frac{\pi }{4} + {\sin ^{ - 1}}x$ is
(A)[-1,0]
(B)... | First of all we need $-1\le x\le1$ for real $\arcsin(x)$
Using Proof for the formula of sum of arcsine functions $ \arcsin x + \arcsin y $
the equality will be valid
if $$x^2+\dfrac12\le1\iff x^2\le1-\dfrac12\iff?\le x\le?$$
Or if $x^2+\dfrac12>1$ and $(\sqrt2x<0\iff x<0)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3771166",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 3
} |
Find $ED$ in the triangle $ABC$
$AB = 6$, $AB \| EG$, $BC = 8$, $AC = 10$
I only have this:
From here, we know $BF =FG, AF=DA, GC=DC, OG=OF=OD$, also $$\frac{6}{GE}=\frac{8}{GC}=\frac{10}{EC}$$
And, we can see $$\frac{OG}{OC} = \frac{GC}{EC} \implies OG = \frac{4}{5}OE$$
From the same $$\frac{4}{5}EC=CG=DC=EC - ED \im... | HInt:$$AB = 6,BC = 8,AC = 10\\AC^2=BC^2+AB^2\\10^2=6^2+8^2$$ SO $$\triangle ABC$$ Is right angle, $\widehat{B}=90$
IF you draw a line $OF$ then $BGOF$ is a rectangle so,$ BF =FG$
then for $OG=OF=OD$ they are the radius of the circle .
after that
$GA,HA$ are tangent to the circle, so
$$OH=OG=R\\H=G=90\\OA=OA \\\to \tri... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3771780",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Convergence or divergence of $\sum^{\infty}_{n=1}\frac{2^n\cdot n^5}{n!}$ using integral test
Finding convergence or Divergence of series $$\sum^{\infty}_{n=1}\frac{2^n\cdot n^5}{n!}$$ using integral test.
What i try::
$$\frac{2}{1}+\frac{4\cdot 32}{2!}+\frac{8\cdot 3^5}{3!}+\sum^{\infty}_{n=4}\frac{2^n\cdot n^5}{n!}... | All solutions given are very good and effective, as a further alternative by root test we obtain
$$\sqrt[n]{\frac{2^n\cdot n^5}{n!}}=\frac{2\cdot \sqrt[n]{n^5}}{\sqrt[n]{n!}}\to 0$$
indeed by standard limits
*
*$\sqrt[n]{n^5}\to 1$
*$\sqrt[n]{n!}\to \infty$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3771866",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 3
} |
Write the polynomial of degree $4$ with $x$ intercepts of $(\frac{1}{2},0), (6,0)$ and $(-2,0)$ and $y$ intercept of $(0,18)$. Write the polynomial of degree $4$ with $x$ intercepts of $(\frac{1}{2},0), (6,0) $ and $ (-2,0)$ and $y$ intercept of $(0,18)$.
The root ($\frac{1}{2},0)$ has multiplicity $2$.
I am to write t... | Both forms are equivalent:
$$
-6(x-\frac{1}{2})^2(x+2)(x-6)\\
=-\frac{3}{2}\cdot 2^2(x-\frac{1}{2})^2(x+2)(x-6)\\
=-\frac{3}{2}(2(x-\frac{1}{2}))^2(x+2)(x-6)\\
=-\frac{3}{2}(2x-1)^2(x+2)(x-6)
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3772371",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Multiplication group for $\mathbb Z_n$ modulo $n$ By definition:
Let $\mathbb Z^+_n = \{[0],[1],[2],\ldots,[n−1]\}$
$\mathbb Z^+_4 = \{[0],[1],[2],[3]\},$
but
how $\mathbb Z^*_{12}$ is $\{[1],[5],[7],[11]\}$ ?
how $\mathbb Z^*_{7}$ is $\{[1],[2],[3],[4],[5],[6]\}$ ?
| $\require{cancel}$
\begin{align}
\frac 1 {12} & = \frac 1 {12} & \text{No cancellation occurs here.} \\[6pt]
\frac 2 {12} & = \frac {\cancel2\times1}{\cancel2\times6} = \frac 1 6 \\[6pt]
\frac 3 {12} & = \frac {\cancel3\times 1}{\cancel3\times 4} = \frac 1 4 \\[6pt]
\frac 4 {12} & = \frac{\cancel4\times1}{\cancel4\time... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3773712",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
For which values of $\eta\in\mathbf{N}$, $\frac{10^2\cdot\eta}{\varphi\cdot\eta-1}$ is an integer ($\varphi\in\mathbf{N}_0$)? Recently, I have found this problem:
Determine for which values of $\eta\in\mathbf{N}$ and $\varphi\in\mathbf{N}_0$, the quantity:
$$\gamma=\frac{10^2\cdot\eta}{\varphi\cdot\eta-1}$$
is a posit... | Observe that $\varphi\cdot\eta -1$ is prime with $\eta$ (use Euclidean Division). So $\gamma$ is an integer iff $\varphi\cdot\eta -1$ divides $10^2$, do you know how to proceed?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3774283",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
How to find the f-related vector fields So lets say you have a vector field $X = (2x+y) \partial x + x \partial y$ and a diffeomorphism $f(x,y) = (x-2y,2x+y)$, then I need to know how to find the f-related vector fields of X, lets say $\overline{X}$. That is, those that meet:
$\overline{X} \circ f = \partial f \circ X ... | I'll use the notation of An Introduction to Manifolds (second edition) written by Loring W.Tu (in particular, chapter one), so that $$X = (2x+y) \frac{\partial}{\partial x} + x \frac{\partial}{\partial y}$$
A vector field $\overline{X}$ is $f$-related to $X$ if and only if $f_{*}X=\overline{X}$. More precisely, $f_{*, ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3775610",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Prove by induction $1 + z + z^2 + .... +z^n = \frac{z^{n+1}-1}{z-1}$ for $z \neq$ 1 Prove $$1 + z + z^2 \ldots +z^n = \frac{z^{n+1}-1}{z-1} \quad \text{ where } z\ne 1 $$
So I started to prove this by induction.
What have I done so far:
Let $n= 1$
$$1+ z + z^2 \ldots+z^n = \frac{z^{n+1}-1}{z-1} \\
1+z \stackrel {?}{=}\... | This is provable using standard algebra; however, if you wish to do this by induction:
For ${n=1}$, we get ${1 + z = \frac{z^2-1}{z-1}=z+1}$, so it works. Now assume
$${1 + z + z^2 + ... + z^k = \frac{z^{k+1}-1}{z-1}}$$
This would imply that
$${1 + z + z^2 + ... + z^{k} + z^{k+1} = \frac{z^{k+1}-1}{z-1} + z^{k+1}}$$
No... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3776242",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
Graphing $ \lim _{n \rightarrow \infty} \frac{{|x+1 \mid}^{n}+x^{2}}{|x|+x^{2n}} $ Graph: $$
\lim _{n \rightarrow \infty} \frac{{|x+1 \mid}^{n}+x^{2}}{|x|+x^{2n}}
$$
Here I tried first to resolve the limit, dividing by ${x}^{2n}$ in both numerator and plugging in $\infty$ I obtain the limit to be zero for all values o... | Case by case:
If $x>\varphi = \frac{1+\sqrt{5}}{2}$, then $x+1<x^2$ and we can divide by $x^{2n}$ to se that:
$$
\lim_n \frac{(x+1)^n + x^2}{x+(x^2)^n}=\lim_n \frac{(\frac{x+1}{x^2})^n + x^{2-2n}}{x^{1-2n}+1}=\frac{0+0}{0+1}=0
$$
If $0<x<\varphi$, then $x+1>x^2$ , thus
$$
\lim_n \frac{(x+1)^n + x^2}{x+(x^2)^n}=\lim_n \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3777527",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Proving $(1+a^2)(1+b^2)(1+c^2)\geq8 $ I tried this question in two ways-
Suppose a, b, c are three positive real numbers verifying $ab+bc+ca = 3$. Prove that $$ (1+a^2)(1+b^2)(1+c^2)\geq8 $$
Approach 1:
$$\prod_{cyc} {(1+a^2)}= \left({a^2\over2}+1+{a^2\over2}\right)\left({b^2\over2}+{b^2\over2}+1\right)\left(1+{c^2\... | Another way.
Let $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.
Thus, $v^2=1$ and we need to prove that:
$$a^2b^2c^2+\sum_{cyc}(a^2b^2+a^2)\geq7$$ or $f(w^3)\geq0,$ where $$f(w^3)=w^6+(9v^4-6uw^3)v^2+(9u^2-6v^2)v^4-7v^6.$$
But, $$f'(w^3)=2w^3-6uv^2w^3<0,$$ which says that $f$ decreases and it's enough to prove our inequali... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3779630",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 7,
"answer_id": 4
} |
Solve $\sqrt[4]{x}+\sqrt[4]{x+1}=\sqrt[4]{2x+1}$ Solve $\sqrt[4]{x}+\sqrt[4]{x+1}=\sqrt[4]{2x+1}$
My attempt:
Square both sides three times
$$\begin{align*}
36(x^2+x)&=4(\sqrt{x^2+x})(2x+1+\sqrt{x^2+x})\\
(\sqrt{x^2+x})(35\sqrt{x^2+x}-4(2x+1))&=0
\end{align*}$$
This means $0,-1$ are solutions but I can't make sure that... | You have $x^{1/4}+(x+1)^{1/4}=(2x+1)^{1/4}$
Raise both sides to the power $4$ and you have:
$$x+(x+1)+4x^{3/4}(x+1)^{1/4}+6x^{1/4}(x+1)^{1/4}+4x^{1/4}(x+1)^{3/4}=2x+1$$
$$x^{1/4}(x+1)^{1/4}\Big(2x^{1/2}+3x^{1/4}(x+1)^{1/4}+2(x+1)^{1/4}\Big)=0$$
From here we have $x=0$ or $x=-1$ as solution.
Now if $x\neq 0$ and $x\neq ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3779710",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 2
} |
Counting the number of integers with given restrictions Question: Consider the numbers $1$ through $99,999$ in their ordinary decimal representations. How many contain exactly one of each of the digits $2, 3, 4, 5$?
Answer: $720$.
Attempt at deriving the answer:
We have two cases: four digit numbers and five digit num... | Four digit numbers are accounted for by the $4!=\underline{24}$ ways to arrange the four digits $2,3,4,5$.
Five digit numbers not containing $0$ are accounted for by $5\times5!=\underline{600}$ (arranging the four digits $2,3,4,5$ then one from $\{1,6,7,8,9\}$).
Five digit numbers containing $0$ are accounted for by th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3780987",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Find value of $AB$
Let
$$A=\dfrac{1}{\sin20^\circ\cos40^\circ}+\dfrac{1}{\sqrt{3}\cos20^\circ\cos40^\circ}$$
and
$$B=\cot40^\circ+\cot230^\circ-\tan185^\circ(\tan230^\circ-\cot50^\circ)$$
Find $AB$.
My attempt :
\begin{align*}
A&=\dfrac{1}{\cos40^\circ}(\dfrac{1}{\sin20^\circ}+\dfrac{1}{\sqrt{3}\cos20^\circ})\\
&=\df... | \begin{align*}
A &= \frac{4(\sin 60^{\circ} \cos 20^{\circ} + \cos 60^{\circ} \sin 20^{\circ})}{\sqrt{3} \sin 40^{\circ}\cos 40^{\circ}} = \frac{4\sin 80^{\circ}}{\sqrt{3} \sin 40^{\circ}\cos 40^{\circ}} \\&= \frac{8\sin 80^{\circ}}{\sqrt{3}\sin 80^{\circ}} = \frac{8\sqrt{3}}{3}\\
B &= \cot 40^{\circ} + \cot 230^{\circ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3783603",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Minimal polynomial of $\alpha + \beta$ over $\mathbb{Q}$ Let $\alpha,\beta \in \mathbb{C}$ such that $\alpha^3+\alpha+1=0$ and $\beta^2+\beta-3=0$ . Find the minimal polynomial of $\alpha+\beta$ over $\mathbb{Q}$.
I was trying the usual trick for this kind of problems, that is
Let $\gamma = \alpha+\beta$, and therefore... | Let $\alpha=\alpha_1$ be one of the roots, $\alpha_i,i=1,2,3,$ of $x^3+x+1$ and let $\beta=\beta_1$ be one of the roots, $\beta_j,j=1,2,$ for $x^2+x-3$. Consider the polynomial $$f(x)=\prod_{i,j}(x-(\alpha_i+\beta_j)).$$ Then by the relation between coefficients of a polynomial and its roots, one sees that $f(x)\in {\m... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3786857",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 3
} |
Limit of ${ \lim_{(x,y)\to(0,0)} {(\left| x \right| + \left| y \right|) \ln{(x^2 + y^4)} }}$ This question comes from Hubbard's textbook "Vector Calculus, Linear Algebra, and Differential Forms: A Unified Approach" (5th edition, exercise 1.5.16b)
I wrote a $\epsilon -\delta$ proof to show ${ \lim_{(x,y)\to(0,0)} {(\lef... | Given that the function is even with respect to $x$ and with respect to $y,$ it is enough to consider $x,y\geq0,$ and $(x,y)\neq(0,0).$ I would write
$$
\lim_{(x,y)\to(0,0)}\frac{x+y}{\sqrt[4]{x^2+y^4}}\cdot\sqrt[4]{x^2+y^4}\log(x^2+y^4).
$$
Now, it should be rather easy to see that
$$
\lim_{(x,y)\to(0,0)}\sqrt[4]{x^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3787095",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
An interesting limit: $\lim_\limits{n\to\infty}\frac{\sin 1\sin\sqrt{1}+\sin 2\sin\sqrt{2}+\sin 3\sin\sqrt{3}+\cdots+\sin n\sin\sqrt{n}}{n}$ I would like to prove that
$$\lim_\limits{n\to\infty}\frac{\sin 1\sin\sqrt{1}+\sin 2\sin\sqrt{2}+\sin 3\sin\sqrt{3}+\cdots+\sin n\sin\sqrt{n}}{n}=0$$
but I am stuck.
I tried to so... | Let $S_n$ be given by
$$S_n=\sum_{k=1}^n \sin(k)\sin\sqrt{k}\tag1$$
Applying summation by parts to the sum in $(1)$ reveals
$$S_n=\sin(\sqrt {n+1})\sum_{k=1}^{n}\sin(k)-\sum_{k=1}^n \left(\sum_{\ell=1}^k \sin(\ell)\right)\left(\sin(\sqrt {k+1})-\sin(\sqrt{k})\right)\tag 2$$
ESTIMATES:
The sum $\sum_{\ell=1}^k \sin(\el... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3788621",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "18",
"answer_count": 2,
"answer_id": 1
} |
Computing $\int_0^1 \frac{\arcsin \sqrt x}{x^2-x+1} dx$
Compute:
$$\int_0^1 \frac{\arcsin \sqrt x}{x^2-x+1} dx$$
Answer: $\frac{\pi^2}{6\sqrt 3}$
My Attempt:
The obvious substitution: $\arcsin \sqrt x=t$. This transforms my integral (say $I$) to:
$$I=\int_0^{\frac{\pi}{2}} \frac{t\cdot 2\sin t\cos t}{\sin^4-\sin^2+1}... | We could complete this integral with one integration-by-parts:
$$\int_0^1 \frac{\arcsin\sqrt{x}}{\left(x-\frac{1}{2}\right)^2+\frac{3}{4}}dx = \frac{2}{\sqrt{3}}\arctan\left(\frac{2x-1}{\sqrt{3}}\right)\arcsin\sqrt{x}\Biggr|_0^1 - \frac{1}{\sqrt{3}}\int_0^1\frac{\arctan\left(\frac{2x-1}{\sqrt{3}}\right)}{\sqrt{x}\sqrt{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3790387",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
} |
Find the minimum value of $x_1^2+x_2^2+x_3^2+x_4^2$ subject to $x_1+x_2+x_3+x_4=a$ and $x_1-x_2+x_3-x_4=b$. Question: Find the minimum value of $x_1^2+x_2^2+x_3^2+x_4^2$ subject to $x_1+x_2+x_3+x_4=a$ and $x_1-x_2+x_3-x_4=b$.
My attempt: It can be easily seen that $x_1+x_3=\frac{a+b}{2}$ and $x_2+x_4=\frac{a-b}{2}$. Fu... | By your work and by C-S
$$x_1^2+x_2^2+x_3^2+x_4^2\geq\frac{1}{2}\left(\frac{a+b}{2}\right)^2+\frac{1}{2}\left(\frac{a-b}{2}\right)^2=\frac{a^2+b^2}{4}.$$
The equality occurs for $x_1=x_3=\frac{a+b}{4}$ and $x_2=x_4=\frac{a-b}{4},$ which says that we got a minimal value.
We used the following C-S:
$$x^2+y^2=\frac{1}{2}(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3790537",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
Why does $-8^{\frac{1}{3}}$ have $2$, $e^{\frac{\pi}{3}}$ and $e^{\frac{5\pi}{3}}$? Use DeMoivre’s theorem to find $-8^{\frac{1}{3}}$. Express your answer in complex form.
Select one:
a. –2
b. – 2, 2 cis ($\pi$/3)
c. – 2, 2 cis ($\pi$/3), 2 cis (5$\pi$/3)
d. 2, 2 cis ($\pi$/3), 2 cis (5$\pi$/3)
e. None of these
The c... | We can rewrite,
$$-8^{\frac13}=2.(-1)^{\frac13}=2\cdot \text{CiS}\left(\frac{2k\pi + \pi}{3}\right)$$
And on substituting,
$k=0,1 $ and $2$ we get our desired answer
| {
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"url": "https://math.stackexchange.com/questions/3791784",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Calculate integral $\int^{+\infty}_0 \frac{e^{-x^2}}{(x^2+\frac{1}{2})^2} dx$? I've posted a similar integral earlier, in which the Goodwin-Staton Integral is involved, making the integral unsolvable. Now I make a little modification to make it solvable and give my answer below.
| $$\begin{array}\ \int^{+\infty}_0 \frac{e^{-x^2}}{(x^2+\frac{1}{2})^2} dx &= 2\bigg[\int^{+\infty}_0 \frac{e^{-x^2}}{x^2+\frac{1}{2}} dx - \int^{+\infty}_0 \frac{x^2e^{-x^2}}{(x^2+\frac{1}{2})^2} dx\bigg]\\ &= 2\bigg[\int^{+\infty}_0 \frac{e^{-x^2}}{x^2+\frac{1}{2}} dx + \frac{1}{2}\int^{+\infty}_0xe^{-x^2}d(\frac{1}{x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3792663",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
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find a value of $a$ so the equation below has some solutions. I need find a value of $a$ so the equation below has some solutions.
$\begin{pmatrix}
3 & 1 &-5&2\\
-1 & 0 &2&-1\\
2&1&-3&1\\
3&-2&4&-1
\end{pmatrix}
\left(
\begin{array}{c}
x_1 \\
x_2 \\
x_3 \\
x_4 \\
\end{array}
\right)=
\left(
\beg... | You would be right, but you made a mistake in your computations. Last row should be all zeroes. Actually, the RREF is
$$
\begin{bmatrix}
1&0&0&0&-a/6+1/3\\
0&1&0&-1/2&-5a/12-2/3\\
0&0&1&-1/2&-7a/12+2/3\\
0&0&0&0&0
\end{bmatrix}.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3793798",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Maximum of the function $F(x,y,z):=\frac{x(1-x)y(1-y)z(1-z)}{1-(1-xy)z}$ I am stuck with finding the maximum of the function
\begin{align}
F(x,y,z):=\frac{x(1-x)y(1-y)z(1-z)}{1-(1-xy)z}
\end{align}
on the compact interval $[0,1]\times[0,1]\times[0,1].$
I think the Lagrange multiplier may work, but I have no idea to set... | Taking partial derivatives with respect to $x,y,z$ we have
$$\begin{cases}
\dfrac{\partial F}{\partial x} = -\dfrac{(y - 1) y (z - 1) z (x^2 y z - 2 x (z - 1) + z - 1)}{(z (x y - 1) + 1)^2}\\
\dfrac{\partial F}{\partial y} =-\dfrac{(x - 1) x (z - 1) z (x y^2 z - 2 y z + 2 y + z - 1)}{(x y z - z + 1)^2}\\
\dfrac{\partia... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3797153",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
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Solving $\sqrt{9-x^2} > x^2 + 1$ without graphic calculator for the exact form
Is there any way to solve this inequality without using a graphic calculator to get the exact form?
$$\sqrt{9-x^2} > x^2 + 1$$
I've tried completing the square but I end up with
$$\frac{3 - \sqrt{41}}{2} < x^2 < \frac{3 + \sqrt{41}}{2}$$
w... | Hint:
*
*If $A>B>0$ then $A^2>B^2$. Apply this to $\sqrt{9-x^2}>x^2+1$.
*As $x^2+1$ is always positive then $9-x^2>0$.
Both must be satisfied. So you need $\cap$ to intersect the solution sets from both cases.
The hint is exapanded.
\begin{gather}
9-x^2>x^4+2x^2+1\\
x^4+3x^2-8<0\\
(x^2-\frac{-3+\sqrt{41}}{2})(x^2-\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3797646",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Find $S=2016^2 + 2015^2 +2014^2 -2013^2 -2012^2 -2011^2 .....+6^2 +5^2 +4^2-3^2-2^2-1^2$ Simplifying the expression as
$$2016^2+2015^2+2014^2+2013^2+2012^2+2011^2...6^2+5^2+4^2+3^2+2^2+1^2-2(2013^2+2012^2+2011^2 +2007^2+2006^2+ 2005^2....3^2+2^2+1^2)$$
Now the left part can be evaluated by using $\sum = \frac{(n)(n+1)(... | Note that $2016=6\cdot 336.$ Then rewrite the whole sum as:
$$S = \sum_{k=1}^{336}\left((6k)^2+(6k-1)^2+(6k-2)^2-(6k-3)^2-(6k-4)^2-(6k-5)^2\right) = $$
$$ = \sum_{k=1}^{336}(108k-45)=...$$
Or, if you insist on your own method, then the thing in the bracket you are struggling with is simply:
$$T = \sum_{k=1}^{335}\left(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3797752",
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"source": "stackexchange",
"question_score": "2",
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Given that $a,b,c$ satisfy the equation $x^3-2007 x +2002=0$, then find $\frac{a-1}{a+1}+\frac{b-1}{b+1} +\frac{c-1}{c+1}$
Given that $a,b,c$ satisfy the equation $x^3-2007 x +2002=0$, then find $\frac{a-1}{a+1}+\frac{b-1}{b+1} +\frac{c-1}{c+1}$
The concept of transformation of roots can be applied here.
So replace
$... | The equation is expanded to $$\frac{3abc+(ab+ac+bc)-(a+b+c)-3}{abc+(ab+ac+bc)+(a+b+c)+1}.$$
By Vieta's, $$abc = -2002$$ $$ab+ac+bc = -2007$$ $$a+b+c = 0$$
Therefore, the expression is $$\frac{3(-2002) + (-2007) - 0 - 3}{-2002 + (-2007) + 0 + 1}.$$
This simplifies to $$\boxed{+2.}$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to prove $\frac{a}{b}=\frac{c}{d} \implies \frac{a+c}{b+d}=\frac{a-c}{b-d}$? My question:
How do I prove the following property of ratio?
$$\frac{a}{b}=\frac{c}{d} \implies \frac{a+c}{b+d}=\frac{a-c}{b-d} \tag{1}$$
$a,b,c,d \in \mathbb{R} \backslash \{ 0 \}$
I want to use the result in argument below.
Use the s... | To prove your requested expression (note there are limitations as Gerry Myerson's question comment explains, e.g., $b \neq 0$, $d \neq 0$, $b + d \neq 0$, $b - d \neq 0$, etc.), cross-multiply and use other manipulations to get
$$\begin{equation}\begin{aligned}
\frac{a}{b} & = \frac{c}{d} \\
ad & = bc \\
ad + cd & = bc... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3802366",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Given four edge-lengths of a quadrangle $a,b,c,d$ so that $a\leq b\leq c\leq d$. Prove that $a^{2}+b^{2}+c^{2}+d^{2}<2\left(ab+ac+ad+bc+bd+cd\right).$
Given four edge-lengths of a quadrangle $a, b, c, d$ so that $a\leq b\leq c\leq d$. Prove that
$$a^{2}+ b^{2}+ c^{2}+ d^{2}< 2\left ( ab+ ac+ ad+ bc+ bd+ cd \right )$$
... | \begin{align}\Rightarrow 2\left ( ab+ ac+ ad+ bc+ bd+ cd \right ) & = 2\left ( a+ b+ c \right )d+ ab+bc+bc+ca+ca+ab\\
& \gt 2d^{2}+ \left (c+ a \right)b + \left (a+ b \right)c + \left (b+ c \right)a\\
& \gt 2d^{2} + b^{2} + \left (a + b \right)c + a^{2} \\
& \gt c^2 + d^{2} + b^{2} + \left (a + b \right)c + a^{2} \\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3802716",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
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Find the infinite sum $\sum_{n\geq0} \frac{2^n}{1+x^{2^n}}$ for $|x|>1$ I tried finding a pattern when expanding the sum to get
$$\frac{x^{n-1}+2x^{n-2}+\cdots+(n-1)x+n}{x^n+x^{n-1}+\cdots+x+1}$$
where nothing seems to cancel out, and the solution to the sum seems to be $\frac{1}{x-1}$, but how would one get to a simpl... | You are trying to solve $$\sum_{n = 0}^{\infty} \frac{2^n}{1+x^{2^n}}$$
Each term can be be split as $\frac{2^n}{1+x^{2^n}} = \frac{2^n}{-1+x^{2^n}} - \frac{2^{n+1}}{-1+x^{2^{n+1}}}$. Let $a_n = \frac{2^n}{-1+x^{2^n}}$. The sum is then $$\sum_{n = 0}^{\infty} \left( a_n - a_{n+1}\right)$$
which telescopes to $a_0$, or ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3803590",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find the minimum value of $f(x) = x + \frac{x}{x^2 + 1} + \frac{x(x + 4)}{x^2 + 2} + \frac{2(x + 2)}{x(x^2 + 2)}$ Find the minimum value of
$$f(x) = x + \frac{x}{x^2 + 1} + \frac{x(x + 4)}{x^2 + 2} + \frac{2(x + 2)}{x(x^2 + 2)}$$ for $x > 0.$
I'm not sure how to start this problem. I think AM-GM is the best way, but I'... | $$f(x)=x + \frac{x}{x^2 + 1} + \frac{x(x + 4)}{x^2 + 2} + \frac{2(x + 2)}{x(x^2 + 2)}$$
$$f(x)=\frac{x^6+x^5+8 x^4+3 x^3+12 x^2+2 x+4}{x \left(x^2+1\right) \left(x^2+2\right)}$$
$$f'(x)=\frac{ x^{10}+x^8-2 x^6-8 x^4-12 x^2-8}{\big[x \left(x^2+1\right) \left(x^2+2\right)\big]^2}$$
$$x^{10}+x^8-2 x^6-8 x^4-12 x^2-8=\left... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3804362",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove that$(1-\omega+\omega^2)(1-\omega^2+\omega^4)(1-\omega^4+\omega^8)…$ to 2n factors$=2^{2n}$ Prove that $(1-\omega+\omega^2)(1-\omega^2+\omega^4)(1-\omega^4+\omega^8)…$ to 2n factors$=2^{2n}$ where $\omega$ is the cube root of unity
My attempt:
$(1+\omega^n+\omega^{2n})=0$
$\Rightarrow (1-\omega^n+\omega^{2n})=-2\... | Since $\omega$ is the third root of unity, and also since $2^{2n}-1\equiv 0 \mod{3}$
$$
\begin{align}
1+\omega^{2^{i}}+\omega^{2^{i+1}}&=0\\
\\
1-\omega^{2^{i}}-\omega^{2^{i+1}}&=2\omega^{2^{i}}\\
\\
\prod_{i=0}^{2n-1}{\left(1-\omega^{2^{i}}+\omega^{2^{i+1}}\right)}&=\prod_{i=0}^{2n-1}{\left(2\omega^{2^{i}}\right)}\\
&... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3805820",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
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Values of the Diophantine equation in an interval Set $a$ and $b$ so that for each $k$, with $ a < k \leq a + b $,
there exist $ x $ and $ y $ with $ x \geq 2 $ and $ y \geq 1 $ such that $ k = 6xy + x -y $
Is it possible to proof that, set $a$, there is a value of $b$ for which this cannot happen?
For example by pro... | $k = 6xy + x - y = (6x - 1)y + x$
imagining this number as if $x$ was fixed, and $y$ runs through the positive integers. With this logic we see that incrementing/decrementing $y$ by 1, the next value of $k$ is $6x - 1$ away.
Now if $1 < k - a < 6x - 1$, then it means $a$ is strictly smaller than $k$ and decrementing $y... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3806412",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Show that if $3^x+4^y=12^z \Rightarrow z=\frac{xy}{x+y}$ Show that if $3^x+4^y=12^z \Rightarrow z=\frac{xy}{x+y}$
$\textbf{My Attempt}$
$$3^x+4^y=12^z \Rightarrow z=\frac{xy}{x+y}$$
$$\Rightarrow \log(3^x4^y)=\log(12^z)\Rightarrow \log \big(\frac{3^x4^y}{12^z}\big)=0\Rightarrow \frac{3^x4^y}{12^z}=1 \Rightarrow 3^x4^y=... | It's wrong.
Try $x=z=1$ and $y=\log_23.$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Proving $\sin\frac{a\pi}{2}\,\sin\frac{b\pi}{2}=\sin\frac{ab\pi}{2}$ for $a,b\in\mathbb{N}$? I was playing with the graphs of $f:\,\mathbb{R}\to\mathbb{R},\, f(x)=\sin axt\pi$ and $g:\,\mathbb{R}\to\mathbb{R},\, g(x)=\sin xt\pi\sin at\pi$.
What I conjectured is the following:
If $t=\frac{m}{2}$ where $m$ is $1$ above ... | We have that
*
*$a\lor b$ even
$$\implies \sin\frac{a\pi}{2}\,\sin\frac{b\pi}{2}=\sin \frac{ab\pi}{2}=\sin (k\pi)=0$$
*
*$a\land b$ odd
$$\implies \sin\frac{(2A+1)\pi}{2}\,\sin\frac{(2B+1)\pi}{2}=\sin \frac{(2A+1)(2B+1)\pi}{2}=$$
$$\sin \frac{(4AB+2A+2B+1)\pi}{2}=\sin \left(\frac \pi 2+k\pi\right)=\pm1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3807762",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
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Counting and Symmetry The digits $1,2,3,4$ and $5$ can be arranged to form many different $5$-digit positive integers with five distinct digits. In how many such integers is the digit $1$ to the left of the digit $2$? (The digits 1 and 2 do not have to be next to each other.)
Case 1: 2 is in the second spot
We get $3!=... | A different way is $\binom{5}{2}$ ways to pick 2 slots for 1 and 2, where 1 is always the left one, and $3!$ ways to arrange the rest of the numbers in their own 3 slots. So total number of cases is
$$
\binom{5}{2} \cdot 3! = 10 \cdot 6 = 60.
$$
This also changes the probability:
$$
\frac{60}{5!} = \frac{60}{120} = \f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3812479",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove $\sum_{m=i}^{n}2^{n-m}\binom{m}{i}=\binom{n+1}{i+1}+\ldots+\binom{n+1}{n+1}=\sum_{m=i}^{n}\binom{n+1}{m+1}$ without induction I would like to prove the following: $$\sum_{m=i}^{n}2^{n-m}\binom{m}{i} = \binom{n+1}{i+1}+\binom{n+1}{i+2}+\ldots+\binom{n+1}{n+1} = \sum_{m=i}^{n}\binom{n+1}{m+1}$$
which is quite easy ... | Here is an algebraic proof. We get for the RHS
$$\sum_{m=q}^n {n+1\choose m+1} =
\sum_{m=q}^n {n+1\choose n-m}
= [z^n] \sum_{m=q}^n z^m (1+z)^{n+1}.$$
Here the coefficient extractor enforces the range and we get
$$[z^n] (1+z)^{n+1} \sum_{m\ge q} z^m
= [z^n] (1+z)^{n+1} \frac{z^q}{1-z}.$$
This is
$$\mathrm{Res}_{z=0} \f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3812841",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 1
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prove $a^3+b^3+c^3+3abc\ge \sum_{cyc}ab(a+b)$
prove $$a^3+b^3+c^3+3abc\ge \sum_{cyc}ab(a+b),$$$a,b,c>0$
Obviously this is a direct consequence of the third degree schur's inequality.
I was wondering if this could be proved without this theorem ,or uvw but through basic methods like AM-GM,C-S etc.
| Another way.
Let $a\geq b\geq c$.
Thus, $$\sum_{cyc}(a^3-a^2b-a^2c+abc)=\sum_{cyc}(a^3-abc-a^2b-a^2c+2abc)=$$
$$=\sum_{cyc}(a-b)^2\left(\frac{a+b+c}{2}-c\right)=\frac{1}{2}\sum_{cyc}(a-b)^2(a+b-c)\geq$$
$$\geq\frac{1}{2}((a-c)^2(a+c-b)+(b-c)^2(b+c-a))\geq$$
$$\geq\frac{1}{2}((b-c)^2(a-b)+(b-c)^2(b-a))=0.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3815419",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 6,
"answer_id": 3
} |
Number of real roots $x^8-x^5+x^2-x+1=0$ Find the number of real roots of $x^8-x^5+x^2-x+1$.
My attempt: $f(x)$ has $4$ sign changes and $f(-x)$ has no sign changes, so the possibility of having real roots is $4+0=4$. Since this is a polynomial of degree $8$ it should have $8-4=4$ imaginary roots.
It is quite impossibl... | Another way: note that
$$
x^8-x^5+x^2-x+1=\left(x^4-\frac{1}{2}x\right)^2+\frac{1}{2}x^2+\left(\frac{1}{2}x-1\right)^2>0.
$$
Equality cannot occur since $x^4-\frac{1}{2}x$, $x$ and $\frac{1}{2}x-1$ cannot vanish together.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3816206",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 1
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System of generators for $\Bbb Z_9\times \Bbb Z_{18}$. Is $\{(4,3),(3,5)\}$ a system of generators for $\Bbb Z_9\times \Bbb Z_{18}$?
I tried to generate the elements using the straightforward method $(4,3), (8,6), (3, 9)$ ... but it takes too long.
Can I use a faster method?
| Let $v_1=(4,3)$ and $v_2=(3,5)$. Then
$$
\begin{pmatrix}v_1\\v_2\end{pmatrix}
=
\begin{pmatrix}4&3\\3&5\end{pmatrix}
\begin{pmatrix}e_1\\e_2\end{pmatrix}
$$
where $e_1=(1,0)$ and $e_2=(0,1)$.
The matrix has determinant $11$, which is invertible in the ring $\Bbb Z_9\times \Bbb Z_{18}$. Therefore, $v_1$ and $v_2$ genera... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3817747",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
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combinatorial identity another solution? I would appreciate if somebody could help me with the following problem:
I need to show that,
$$\sum^{33}_{k=0}\binom{100}{3k}=\sum^{49}_{k=0}4^k$$
My proof is the following:
$$(1+x)^{100} = a_0 + a_1x + a_2x^2 + \cdots + a_{100}x^{100}$$
Let
$A = a_0 + a_3 + a_6 + \cdots + a_{9... | $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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\newcommand{\dd}{\mathrm{d}}
\newcommand{\ds}[1]{\displaystyle{#1}}
\newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
\new... | {
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"timestamp": "2023-03-29T00:00:00",
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Book: Actuarial Mathematics for Life Contingent Risks Exercise 2.4 Exercise $2.4$ Let $$S_0(x) =\exp\left({-\left( Ax+ \frac{Bx^2}2 +\frac{C(D^x-1)}{\ln D}\right)}\right)$$
where $A, B, C , D>0$
$(b)$ Derive a formula for $S_x(t)$
I know the equation for $S_x(t)= \dfrac{S_0(x+t)}{S_0(x)}$
Right now I am at: $\dfrac{\e... | It is better to let $$f(x) = Ax + \frac{Bx^2}{2} + \frac{C(D^x - 1)}{\log D}$$ for fixed constants $A, B, C, D$, then observe $$S_x(t) = \frac{S_0(x+t)}{S_0(x)} = \frac{\exp(-f(x+t))}{\exp(-f(x))} = \exp(-f(x+t)+f(x)).$$ Then $$\begin{align}
f(x) - f(x+t)
&= \left(Ax - A(x+t)\right) + \left(\frac{Bx^2}{2} - \frac{B(x... | {
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"url": "https://math.stackexchange.com/questions/3819986",
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Finding $k$ such that $\frac{3x^2+kx-44}{x^2-121}$ simplifies to $\frac{3x-4}{x-11}$
If the rational expression $\frac{3x^2+kx-44}{x^2-121}$, where $k$ is an element of $\mathbb{W}$, simplifies to $\frac{3x-4}{x-11}$, then the value of $k$ must be?
my work:
$$\frac{3x^2+kx-44}{x^2-121}= \frac{3x-4}{x-11}$$
so the ans... | Hint:
It simply means that $x+11$ divides the numerator, i.e. $-11$ is a root of the numerator, the other root being $4/3$. Use Vieta's relations to determine $k$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3820408",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Find the following limit and prove its value by definition. $\lim _{n\to \infty} (n^2 + 3n + 7)/(2n^2 −5n − 4)$. Find the following limit and prove its value by definition. $\lim_{n\to \infty} \frac{n^2 + 3n + 7}{2n^2 −5n − 4}$. I am supposed to use$$ \left|\frac{n^2 + 3n + 7}{2n^2 −5n − 4} - \frac{1}{2}\right| < ε $$t... | Since
$$\frac{n^2 + 3n +7}{2n^2-5n+4} - \frac{1}{2} = \frac{2n^2 + 6n +14 - 2n^2+5n-4}{4n^2-10n-8} = \frac{11n-10}{4n^2-10n-8} $$
Then you could simply bound the right hand side,
$$|\frac{11n-10}{4n^2-10n-8} | \leq |\frac{11n}{4n^2-10n-8}| \leq |\frac{11n}{4n^2-11n}| = |\frac{11}{4n-11}| \leq |\frac{11}{4n-n}| \leq \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3820669",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Divisibility of higher power polynomials I am stuck with this problem suggested to me by a friend . The problem reads,
The power of $x^2+xy+y^2$ by which the polynomial $(x+y)^7-x^7-y^7$ is divisible are ?
I tried it by writing $x^7+y^7$ expansion and simplifying the polynomials in factors but that doesn't help as the ... | By the Binomial theorem we obtain:
$$(x+y)^7-x^7-y^7=7xy(x^5+3x^4y+5x^3y^2+5x^2y^3+3xy^4+y^5)=$$
$$=7xy(x+y)(x^4-x^3y+x^2y^2-xy^3+y^4+3xy(x^2-xy+y^2)+5x^2y^2)=$$
$$=7xy(x+y)(x^4+2x^3y+3x^2y^2+2xy^3+y^4)=7xy(x+y)(x^2+xy+y^2)^2.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3820979",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Maximizing $y=\tan(x+\frac{2\pi}{3})-\tan(x+\frac{\pi}{6})+\cos(x+\frac{\pi}{6})$ for $x\in[-\frac{5\pi}{12}, -\frac{\pi}{3}]$
Let $x \in [-\frac{5\pi}{12}, -\frac{\pi}{3}]$. Find the maximum value of $$y=\tan(x+\frac{2\pi}{3})-\tan(x+\frac{\pi}{6})+\cos(x+\frac{\pi}{6})$$
The above question is from the China Mathema... | The given proof is very clever and efficent.
As an alternative we have that
$$y=\frac{2}{\sin(2z)}+\cos(z)=\frac{1}{\sin(z)\cos(z)}+\cos(z)$$
which is equivalent to
$$y(t)=\frac{1}{t\sqrt{1-t^2}}+t \quad t=\left(\cos \left(\frac \pi 4\right),\cos \left(\frac \pi 6\right)\right)=\left(\frac{\sqrt 2}2,\frac{\sqrt 3}2\rig... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3823912",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Integrate $\ln(\sqrt{1-x}+\sqrt{1+x})$ Integrate $$\ln(\sqrt{1-x}+\sqrt{1+x})$$
My attempts:
i) Multiplying and dividing its conjugate, we get
$$I=\int \ln(-2x)dx-\int\ln(\sqrt{1-x}-\sqrt{1+x})dx$$
The second term is equally challenging as the original one, so its of no use.
ii) $\sqrt{1-x}+\sqrt{1+x}=t \implies\frac{... | I think IBP approach you did (as @AnginaSeng said) is on the right track. We have
$$
\frac{x}{\sqrt{1-x^{2}}}\frac{\sqrt{1-x} - \sqrt{1+x}}{\sqrt{1-x} +\sqrt{1+x}} = \frac{x}{\sqrt{1-x^{2}}} \frac{(\sqrt{1-x} - \sqrt{1+x})^{2}}{-2x} =- \frac{2 - 2\sqrt{1-x^{2}}}{2\sqrt{1-x^{2}}} = 1 - \frac{1}{\sqrt{1-x^{2}}}
$$
and I ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3824400",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Can we prove that $\operatorname{Tr}(ABC) = \operatorname{Tr}(CBA)$? I would like to verify the claim:
$$\operatorname{Tr}(ABC) = \operatorname{Tr}(CBA)$$
I tried verifying through an example:
Given the following $3$ different matrices:
\begin{align}
A & = \begin{pmatrix}
1 & 2 & 3 & 4 \\
5 & 6 & 7 & 8 \\
9 & 10 & 11... | As suggested in the comments, the statement $\text{Tr}(ABC)=\text{Tr}(CBA)$ is not true in general, so you should not be concerned that you found a counterexample. Actually, there is a simpler counterexample. Take
$$ A=
\begin{bmatrix}
5&2\\
0&0
\end{bmatrix},\:\:\:
B=
\begin{bmatrix}
2&7\\
5&3
\end{bmatrix},\:\:\:
C=
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3824594",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Proving that $\frac{1}{(2k+1)^4} - 2 \sum_{n=k+1}^\infty \left(\frac{1}{(2n)^4} - \frac{1}{(2n+1)^4}\right)$ is non negative Prove that the sequence
$$s_k = \frac{1}{(2k+1)^4} - 2 \sum_{n=k+1}^\infty \left(\frac{1}{(2n)^4} - \frac{1}{(2n+1)^4}\right)$$ is non-negative.
I would appreciate an elementary proof. I tried us... | $$\Delta_k=\frac{1}{(2k+1)^4} - 2 \sum_{n=k+1}^\infty \left(\frac{1}{(2n)^4} - \frac{1}{(2n+1)^4}\right)$$
$$\sum_{n=k+1}^\infty \frac{1}{(2n)^4}=\frac{1}{96} \psi ^{(3)}(k+1)\qquad \qquad\sum_{n=k+1}^\infty \frac{1}{(2n+1)^4}=\frac{1}{96} \psi ^{(3)}\left(k+\frac{3}{2}\right)$$ Using asymptotics
$$\Delta_k=\frac 1 {16... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3833665",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Concrete Mathematics: Clarifying expressing sum in terms of $H_n$ leading to equation 2.14 In Concrete Mathematics (Knuth, Patashnik, and Graham) we have an intermediate solution to the quick-sort recurrence of
$$
C_n = 2(n + 1)\sum^n_{k=1}\frac{1}{k + 1}
$$
Given the sum part of that resembles the Harmonic summation o... | $$\sum_{k=1}^n \frac{1}{k+1} = \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{n+1}.$$ So if we compare this to $$H_n = \sum_{k=1}^n \frac{1}{k} = 1 + \frac{1}{2} + \cdots + \frac{1}{n},$$ we can see that in the first sum we are missing the first term $1$, and we have an extra last term, $\frac{1}{n+1}$. Therefore, $$\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3834268",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Compute the limit $\lim\limits_{t \to + \infty} \int_0^{+ \infty} \frac{ \mathrm d x}{e^x+ \sin tx} $ I have been working on this limit for days, but I am not getting it. The question is
Compute the limit $$\lim_{t \to + \infty} \int_0^{+ \infty} \frac{ \mathrm d x}{e^x+ \sin (tx)}$$
Note that the integral is well de... | For each $t > 0$, we have
\begin{align}
\int_0^\infty \frac{1}{\mathrm{e}^x + \sin t x}\mathrm{d} x
&= \int_0^\infty \sum_{k=0}^\infty (-1)^k \mathrm{e}^{-(k+1)x} (\sin t x)^k\mathrm{d} x\\
&= \sum_{k=0}^\infty (-1)^k \int_0^\infty \mathrm{e}^{-(k+1)x} (\sin t x)^k\mathrm{d} x\\
&= \sum_{k=0}^\infty (-1)^k\frac{1}{k+1}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3836576",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "17",
"answer_count": 4,
"answer_id": 2
} |
Convergence of the series $ \frac{2x}{1+x^2}+\frac{4x^3}{1+x^4}+\frac{8x^7}{1+x^8}+...$ The series $ \frac{2x}{1+x^2}+\frac{4x^3}{1+x^4}+\frac{8x^7}{1+x^8}+...$
(A) is uniformly convergent for all x
(B) is convergent for all x but the convergence is not uniform
(C) is convergent only for $|x|\le \frac{1}{2}$ but the co... | The answer is D). For $|x| \leq \frac1 2 $ we have $|\frac {n x^{2n-1}} {1+x^{2n}}| \leq n|x|^{2n-1} $ and $ \sum n |\frac 1 2 |^{n}$ is convergent.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3837139",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
How many 5-digit sequences of $\{1, 2, 3, 4\}$ are there in which $1$, $2$, $3$ and $4$ all appear? I'm trying to work out how many possible 5 digit sequences there are using the numbers $\{1, 2, 3, 4\}$ in which each number appears at least once.
I thought it might be $4\times 3\times 2\times 1\times 4$ but it's not o... | Here's an alternative approach via inclusion-exclusion, where the four properties to be avoided are that $j\in\{1,2,3,4\}$ does not appear:
\begin{align}
\sum_{k=0}^4 (-1)^k \binom{4}{k} (4-k)^5
&= \binom{4}{0} 4^5 - \binom{4}{1} 3^5 + \binom{4}{2} 2^5 - \binom{4}{3} 1^5 + \binom{4}{4} 0^5 \\
&= 1024 - 972 + 192 - 4 +... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3837395",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Alternative approach for proving that for any $x\in\mathbb R^+$, $x^2+3x+\frac{1}{x} \ge \frac{15}{4}$. Let $x \in \mathbb{R^+}$. Prove that:
$$x^2+3x+\frac{1}{x} \ge \frac{15}{4}.$$
While this is indeed easily proven using derivatives, where the minimum is obtained when $x=\frac{1}{2}$, is it possible to prove it thro... | Here is a verification approach by factorization:
We want to show
$$x^2+3x+\frac1x \ge \frac{15}4$$
We know that $x > 0$,
Hence it is equivalent to showing:
$$4x^3 + 12x^2 + 4 \ge 15x$$
$$4x^3+12x^2-15x+4 \ge 0$$
Observe that $x=-4$ is a root.
$$(x+4)(4x^2-4x+1) \ge 0$$
$$(x+4)(2x-1)^2 \ge 0$$
Since $x>0$, clearly, $x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3838183",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 7,
"answer_id": 0
} |
How to prove $5^{2n+1}-3^{2n+1}-2$ is divisible by $30$ for even $n$? How to prove $5^{2n+1}-3^{2n+1}-2$ is divisible by $30$ for all positive even integer $n$?
Here are my efforts:
Let $f(n)=5^{2n+1}-3^{2n+1}-2$.
For $n=2$,
$$f(1)=5^{2\cdot 2+1}-3^{2\cdot2+1}-2=2880=30\times 96$$
So, the statement is true for $n=2$.
... | Suppose $f(k)=5^{2k+1}-3^{2k+1}-2=30t$
for $n=k+2$,
\begin{align}f(k+2)&=5^{2(k+2)+1}-3^{2(k+2)+1}-2\\
&=(5^4)\cdot (5^{2k+1})-(3^4)\cdot (3^{2k+1})-2\\
&=30(80t)+160+544\cdot (5^{2k+1}) \\
&= 30(80t) + 150 + 10 + 540(5^{2k+1}) + 4(5^{2k+1})\end{align}
It suffices to prove that $10+4(5^{2k+1})$ is divisible by $30$.
\b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3838423",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 2
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Starting with $1,9,9,3$ the sequence continues as the last digit of the sum of preceding 4 terms. Will subsequence $7, 3, 6, 7$ ever appear in it? Formally, $a_n = (a_{n-1} + a_{n-2} + a_{n-3} + a_{n-4})$ mod $10$.
Is there a mathematical way of knowing whether $7, 3, 6, 7$ will ever occur in the sequence?
One idea I w... | If $X_n = \pmatrix{a_n\cr a_{n+1}\cr a_{n+2}\cr a_{n+3}\cr}$, we have $X_{n+1} = M X_n \mod 10$ where $M = \pmatrix{0 & 1 & 0 & 0\cr 0 & 0 & 1 & 0\cr 0 & 0 & 0 & 1\cr 1 & 1 & 1 & 1\cr}$. That matrix has characteristic polynomial $P(\lambda) = \lambda^4 - \lambda^3 - \lambda^2 - \lambda - 1$, which is irreducible both ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3840038",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
how to show that $\csc x - \csc\left(\frac{\pi}{3} + x \right) + \csc\left(\frac{\pi}{3} - x\right) = 3 \csc 3x$?
How to show that $\csc x - \csc\left(\frac{\pi}{3} + x \right) + \csc\left(\frac{\pi}{3} - x\right) = 3 \csc 3x$?
My attempt:
\begin{align}
LHS &= \csc x - \csc\left(\frac{\pi}{3} + x\right) + \csc\left(\... | Using the identities
$$\sin A \sin B = \frac12 (\cos (A-B) - \cos (A+B))$$
$$\sin A \cos B = \frac12 (\sin (A+B) + \sin (A-B))$$
we have
\begin{align}
&\phantom{=}\sin x \sin\left(\frac{\pi}{3} + x\right) + \sin\left(\frac{\pi}{3} - x\right) \cos\left(x + \frac{\pi}{6}\right)\\&=\frac12\left(\cos \left(-\frac\pi3\right... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3840253",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
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Finding the magic number as following Let $s$ and $t$ be distinct positive integers with $s+t$ and $s-t$ are a square numbers.
A pair $(s,t)$ called magic if there is exist positive integer $u$, such that
$12s^2 + t^2 = 4t^2u^3$. Does it exist a magic number?
I try that $s+t = m^2$ and $s-t = n^2$ for some positive int... | Too long for a comment. As Ross Molikan points out, the problem boils down to solve $3k^2+1=4u^3$.
We work in the ring $R=\mathbb{Z}[j]$, where $j=e^{2i\pi/3}$. The ring $R$ is a PID (it is even euclidean), hence a UFD.
Set $z=1+k\sqrt{-3}=k+1+2k j$. The equation may be rewritten $zz^*=4u^3$, where $*$ denotes complex ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3841828",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Limit of multiple absolute values Let $f(x) = \frac{x^2+2x-3|1-x^2|+1}{2|x+1|-|x^2+x|}$. Find $\lim_{x\rightarrow -1}f(x)$.
I broke up each absolute value into parts:
$|1-x^2| = \begin{cases}
1-x^2 & -1 \leq x\leq 1 \\
-(1-x^2) & x>1,x<-1
\end{cases}
$ ,
$|x+1| = \begin{cases}
x+1 & x\geq -1 \\... | $f(x) = \frac{x^2+2x-3|1-x^2|+1}{2|x+1|-|x^2+x|}$
$\lim_{x\rightarrow -1^-}f(x)=\lim_{x\rightarrow -1^-}\frac{x^2+2x-3|1-x^2|+1}{2|x+1|-|x^2+x|}=\lim_{x\rightarrow -1^-} \frac{x^2+2x-3(x^2-1)+1}{2(-(x+1))-(x^2+x)}=\lim_{x\rightarrow -1^-}f(x)\frac{-2x^2+2x+4}{-x^2-3x-2}=\lim_{x\rightarrow -1^-} \frac{-2(x+1)(x-2)}{-(x+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3842370",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Evaluate $\lim_{x\to \infty} \frac{e^{\frac{1}{x^2}}-1}{2\tan^{-1} (x^2)-\pi}$ $$\lim_{x\to \infty} \frac{ e^{\frac{1}{x^2}} -1}{\frac{1}{x^2}} \frac{1}{(2\tan^{-1}(x^2)-\pi)x^2}$$
$$=\lim_{x\to \infty} \frac{1}{x^2(2\tan^{-1} x^2 -\pi)}$$
How do I proceed?
| From here
$$\frac{ e^{\frac{1}{x^2}} -1}{\frac{1}{x^2}} \frac{1}{\left(2\tan^{-1}(x^2)-\pi\right)x^2}$$
we have that by standard limit
$$\frac{ e^{\frac{1}{x^2}} -1}{\frac{1}{x^2}} \to 1$$
and by $\arctan x = \frac \pi 2 -\arctan \frac1x$
$$\frac{1}{\left(2\tan^{-1}(x^2)-\pi\right)x^2}=-\frac12\frac{\frac1{x^2}}{\tan^{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3843885",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
If $\frac1x+\frac1y+\frac1z=0, xyz \neq 0$, $\sqrt[3]{\tiny\frac{x^9+y^9+z^9-3xyz(x^6+y^6+z^6)+6x^3y^3z^3}{x^6+y^6+z^6-3x^2y^2z^2}} =?$ If $\frac1x+\frac1y+\frac1z=0,$ and $xyz \neq 0$, What is $$K =\sqrt[3]{\frac{x^9+y^9+z^9-3xyz(x^6+y^6+z^6)+6x^3y^3z^3}{x^6+y^6+z^6-3x^2y^2z^2}} ?$$
Source: Lumbreras Editors
I found t... | Use the identity:
$$a^3+b^3+c^3 - 3abc = (a+b+c)(a^2+b^2+c^2-ab-bc-ca).$$
Then the denominator, $A$, inside the square root is:
\begin{equation}A = (x^9+y^9+z^9-3x^3y^3z^3) -3xyz(x^6+y^6+z^6 - 3x^2y^2z^2)\end{equation}
The first summand is then further decomposed as:
\begin{align}
x^9+y^9+z^9-3x^3y^3z^3 = (x^3+y^3+z^3)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3844394",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Find the minimum of $P = (a - b)(b - c)(c - a)$
Given $a, b, c$ are real numbers such that $a^2 + b^2 + c^2 = ab + bc + ca + 6$. Find the minimum of
$$P = (a - b)(b - c)(c - a)$$
My solution:
*
*We have:
$$a^2 + b^2 + c^2 = ab + bc + ca + 6$$
$$\implies 2a^2 + 2b^2 + 2c^2 = 2ab + 2bc + 2ca + 12$$
$$\implies (a - b... | The inequality
$$(a - b)^2 + (b - c)^2 + (c - a)^2 \geq 3 \sqrt[3]{((a - b)(b - c)(c - a))^2}$$
become equality when $a-b=b-c=c-a,$ or $a=b=c,$ but for this value then
$$(a-b)(b-c)(c-a)=0 \ne -8.$$
This is my solution, we have
$$P^2 = \frac{4(a^2+b^2+c^2-ab-bc-ca)^3-(a+b-2c)^2(b+c-2a)^2(c+a-2b)^2}{27} \quad (1)$$
$$\le... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3844805",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 6,
"answer_id": 1
} |
Simplification of Combinatorial Expression I have been solving a problem and have been stuck on it for hours now.
I can't recall any combinatorial approach/formula for the below expression :
$$\sum_{i=a+1}^n {i-1 \choose a}{n-i \choose k-a}$$
Any sort of help would be appreciated. I know it is a norm to share my approa... | What follows is admittedly not the simplest proof. We post here by way
of enrichment and to showcase four techniques, coefficient extractors,
the Iverson bracket, residues and the Leibniz rule.
We seek to show that
$$\sum_{q=a+1}^n {q-1\choose a} {n-q\choose k-a}
= {n\choose k+1}$$
where $k\ge a$ for the binomial coef... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3845061",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
} |
If $x,y,z\in(1,3)$ and $xy+yz+zx=26$ then prove that $x+y+z>\frac{35}{4}$(Sweden 1971) If $x,y,z\in(1,3)$ and $xy+yz+zx=26$ then prove that $x+y+z>\frac{35}{4}$
I was trying to do the question above using homogenization. I was trying to do it as follows:
$\sqrt{xy+yz+xz}=\sqrt{26}$
Hence $x+y+z>\frac{35}{4*\sqrt{26}}*\... | We can use AM-GM inequality to get
$$26 = xy + yz + xz \leq \frac{x^2+y^2}{2} + \frac{y^2+z^2}{2} + \frac{x^2+z^2}{2} = x^2+y^2+z^2$$
Then add the two inequalities
$$x^2+y^2+z^2+2xy+2yz+2xz = (x+y+z)^2 > 78$$
$$\implies x+y+z > \sqrt{78} > \frac{35}{4}$$
Oddly enough, this only relies on $x+y+z$ being positive, and we ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3847360",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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If AM=a, NC=b, BL=? Source: "Problemas selectos" by Lumbreras Editors.
The bisectors harmonically divide the segment AC: $→ \dfrac{NC}{MN}=\dfrac{AM+MN+NC}{AM}$
By metric relation in triangle ABC: $ → (AM + MN + NC)(MN + NC) = (BL + LC)^2$
It is what I managed to relate, I did not advance further.
| Let $O$ be the center of the circle, $r$ be the radius, $h=MT$ and $x=MN$. We have
$$2r=AN=AM+MN=a+x$$
$$r^2=MT^2+OM^2=h^2+(ON-MN)^2=h^2+(r-x)^2$$
$$2rx=h^2+x^2$$
$$(r+b)^2=(ON+NC)^2=OC^2=OT^2+TC^2=r^2+MT^2+MC^2=r^2+h^2+(MN+NC)^2=r^2+h^2+(x+b)^2$$
$$2br=h^2+x^2+2bx=2rx-x^2+x^2+2bx=2xr+2bx$$
$$2r(b-x)=2bx$$
$$(a+x)(b-x)... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Is $(-2)^{\frac{7}{5}}=((-2)^7)^{\frac{1}{5}}$ or not? In Wolfram Alpha this statement is false. But how? Because $(a)^{\frac{b}{c}}=(a^b)^{\frac{1}{c}}$. Is there any condition. Please tell me.
| We have that $(a)^{\frac{b}{c}}=(a^b)^{\frac{1}{c}}=(a^\frac1c)^{b}$ holds for any $a\ge 0$.
For the case $(-a)^{\frac{b}{c}}$ with $b$ and $c$ coprime we need to consider the following cases:
*
*$b$ odd and $c$ even
$$\left[(-a)^b\right]^\frac1c=\left[(-1)^b\cdot(a^b)\right]^\frac1c=\sqrt[c] {(-1)^b}\cdot a^{\frac{... | {
"language": "en",
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"question_score": "1",
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Use the $\varepsilon - \delta$ definition of the limit to verify that $\lim_{(x,y)\to(2,5)} xy = 10$.
Question:
Use the $\varepsilon - \delta$ definition of the limit to verify that
$\lim\limits_{(x,y)\to(2,5)}xy = 10$.
Hint given:
$xy − 10 = (x − 2)(y − 5) + 5(x − 2) + 2(y − 5).$
My solution
$\forall \space\varepsil... | Note that since $\epsilon >0$, $\sqrt{ 49 + 4\epsilon } > \sqrt{49} = 7$. So your $\delta$ is negative.
To remedy this, you can choose
$$ \delta = \frac{-7+\sqrt {49+4\epsilon}}{4},$$
which is positive and $\delta^2 + 7\delta = \epsilon$.
Another method: note that you obtained
$$ |xy-10| \le \delta^2 + 7\delta$$
and yo... | {
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"source": "stackexchange",
"question_score": "2",
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Prove that if x,y,z are positive integers such that $x^3+y^3+z^3=3(x+y+z+xyz)$ then they must be consecutive numbers I was thikning going for contradiction, assuming they're not consecutive but I found it very hard to find a statement that contradicts that condition. Maybe stating that there exist some $m, n \in \Bbb{Z... | Use the well-known identity $$x^3+y^3+z^3-3xyz = \dfrac12(x+y+z)\left((x-y)^2+(y-z)^2+(z-x)^2\right)$$ to have, from your given equation (since the left hand side equals $3(x+y+z)$, $$\begin{aligned} 3(x+y+z) & = \dfrac12(x+y+z)\left((x-y)^2+(y-z)^2+(z-x)^2\right) \\ \implies 6 &= (x-y)^2+(y-z)^2+(z-x)^2 \ (\because x,... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3856174",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
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Let f(x) = $x^2+ax+b,a,b \in R$. If $f(1)+f(2)+f(3)=0$, then the nature of the roots of the equation $f(x) =0$ is ..... Let $f(x) = x^2+ax+b$ for $a,b \in \mathbb{R}$. If $f(1)+f(2)+f(3)=0$, then the nature of the roots of the equation $f(x) =0$ is
(A) real
(B) imaginary
(C) real and distinct
(D) equal roots
My attemp... | Brute force,
\begin{align}
\Delta &= a^2-4b \\
&=a ^2-4\left( -\frac{14+6a}{3} \right) \\
&= a^2+8a \color{red}{+16} +\frac{56}{3} \color{red}{-16} \\
&= (a+4)^2+\frac{8}{3} \\
&> 0
\end{align}
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
If $a+b+c=3$, then show $\sqrt{a^2 + ab +b^2}+ \sqrt{b^2 + bc +c^2}+\sqrt{c^2 + ac +a^2} \geq \sqrt{3}$ If $a+b+c=3$, and $a,b,c$ are positive real numbers, then show $\sqrt{a^2 + ab +b^2} + \sqrt{b^2 + bc +c^2} +\sqrt{c^2 + ac +a^2}\geq \sqrt{3}$
Normally when I do inequalities I try to first find where equality would... | It's true even for any reals $a$, $b$ and $c$ such that $a+b+c=3.$
Indeed, by C-S $$\sum_{cyc}\sqrt{a^2+ab+b^2}=\sqrt{\sum_{cyc}\left(2a^2+ab+2\sqrt{(a^2+ab+b^2)(a^2+ac+c^2)}\right)}=$$
$$=\sqrt{\sum_{cyc}\left(2a^2+ab+2\sqrt{\left(\left(a+\frac{b}{2}\right)^2+\frac{3}{4}b^2\right)\left(\left(a+\frac{c}{2}\right)^2+\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3861614",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
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Have I done my homogenization correctly for this question and if so how do I finish it off? We have $a,b,c>0$ with $a^2+b^2+c^2=1$ prove that:
$\frac{a}{1+bc}+\frac{b}{1+ac}+\frac{c}{1+ab}>1$
As soon as I saw this question, I immediately thought of using homogenization in the following manner:
$\frac{a}{1+bc}+\frac{b}{... | Observe that $$\frac{a}{1+bc}+\frac{b}{1+ca}+\frac{c}{1+ab} = \sum_{cic} \frac{a^2}{a+abc} \geqslant \sum_{cyc} \frac{a^2}{a+a \cdot \frac{b^2+c^2}{2}}$$ Using $a^2+b^2+c^2=1$, we obtain $$ \sum_{cyc} \frac{a^2}{a+a \cdot \frac{b^2+c^2}{2}}= \sum_{cyc} \frac{a^2}{a+a \cdot \frac{1-a^2}{2}} = \sum_{cyc} \frac{a^2}{1-\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3865582",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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Triple integration with $x^2+y^2-z^2$ I'm having this problem.
$$\iiint_{K}(x^2+y^2-z^2)dxdydz,\quad K: x^2+y^2+z^2\leq 1$$
My solution.
$$x=\rho \sin\phi \cos\Theta,\ y=\rho \sin\phi \sin\Theta, z=\rho \cos \phi\\
dxdydz=\rho^2 \sin\phi d\rho d\phi d\theta\\
0\leq \rho \leq 1\\
0 \leq \phi \leq \pi\\
0 \leq \theta \le... | The integrals over $y^2-z^2$ cancel due to symmetry. It remains to integrate over $x^2$ over the volume which gives $\frac{4\pi}{15}$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
System of equations from roots of polynomial I'm given the equation $3072x^4-2880x^3+840x^2-90x+3=0$ and told that its roots are $\alpha, \alpha r, \alpha r^2, \alpha r^3,$ for some $r\in \mathbb{R}$.
By considering the sum of the roots, the product, etc. I've found that \begin{gather}\alpha(1+r+r^2+r^3)=\frac{15}{16} ... | Since you see powers of $2$, you can use a factorization:
$$3072x^4-2880x^3+840x^2-90x+3=3(1024x^4-960x^3+280x^2-30x+1)=$$
$$=3(1024x^4-64x^3-896x^3+56x^2+224x^2-14x-16x+1)=$$
$$=3(16x-1)(64x^3-56x^2+14x-1)=$$
$$=3(16x-1)(64x^3-8x^2-48x^2+6x+8x-1)=$$
$$=3(16x-1)(8x-1)(8x^2-6x+1)=2(16x-1)(8x-1)(4x-1)(2x-1).$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 0
} |
$\sqrt{1-x^2}$ is not differentiable at $x = 1$. Please give me an easier proof if exists.
Prove that $\sqrt{1-x^2}$ is not differentiable at $x = 1$.
My proof is here:
Let $0 < h \leq 2$.
$\frac{\sqrt{1-(1-h)^2}-\sqrt{1-1^2}}{-h} = \frac{\sqrt{2h-h^2}}{-h} = \frac{2h-h^2}{-h\sqrt{2h-h^2}} = \frac{h-2}{\sqrt{2h-h^2}... | It all depends on how much you "want to prove".
The function $f(x)= \sqrt{1-x^2}$ has domain $x\in [-1,1]$. The derivative of $f$, if well-defined, can only be defined on $(-1,1)$. If possible, you can extend the definition to $x=\pm 1$ by taking the left- and right-limits.
The derivative of $f$ is
$$f'(x) = \lim_{h\to... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Variance of $X_n$ if we have convergence in probability for a random variable with $|X_n| >1$ Suppose $X_n$ is a sequence of random variables with $|X_n| \leq 1$. Assume $X_n \overset{p}{\to} X$. True or false: $Var(X_n)\to 0$. If false, give a counterexample. If true, give an argument.
Convergence of $\operatorn... | I added an edit that posted my answer after following some advice given! Thanks to Teresa!
Let $X \sim U[0,1]$. Since $X_n \overset{p}{\to} X$ we know $X_n \overset{d}{\to} X$. This implies that $X_n \sim U[0,1]$ $Var(X_n) = Var(X)$ and we will show this $\neq 0$ for at least the following case. Using the definitio... | {
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"source": "stackexchange",
"question_score": "1",
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