Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Finding the remainder when $5^{55}+3^{55}$ is divided by $16$
Find the remainder when $5^{55}+3^{55}$ is divided by $16$.
What I try
$a^{n}+b^{n}$ is divided by $a+b$ when $n\in $ set of odd natural number.
So $5^{55}+3^{55}$ is divided by $5+3=8$
But did not know how to solve original problem
Help me please
| $3^4 = 81 = 16\cdot 5 + 1 \equiv 1 \mod(16)$
Then
$3^{52} = (3^{4})^{13} \equiv 1^{13} \mod(16) $
Then
$3^{55} = 3^3 \cdot 3^{52} \equiv 27 \cdot 1 \mod(16)$
And $27 \equiv 11 \mod(16)$
Thus
$3^{55} \equiv 11 \mod(16)$
Now,
$5^2 \equiv 9 \mod(16)$
So $(5^2)^{26} \equiv (3^{2})^{26} \mod(16)$
Then
$5^{52} \equiv 3^{52}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3490156",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 8,
"answer_id": 6
} |
Relation between two rational sequences approximating square root 2 We define recursive sequences $a_{n+1}=1+\frac 1{1+a_n}$, $a_1=1$ and $b_{n+1}=\frac{b_n^2+2}{2b_n}$, $b_1=1$. I wish to show that $b_{n+1}=a_{2^n}$.
This can be proven using closed forms expressions related to continued fractions. I know that $a_n$ ca... | Let $\left(\frac{1-\sqrt 2}{1+\sqrt 2}\right)^{2^n} = t$ for simplicity. Then,
$$\frac{b_{n+1}-\sqrt 2}{b_{n+1}+\sqrt 2} = t$$ implies
\begin{align}
b_{n+1} &= \sqrt 2 \frac{1+t}{1-t} \\
&=\sqrt2 \frac{(1+\sqrt 2)^{2^n} +(1-\sqrt 2)^{2^n}}{(1+\sqrt 2)^{2^n} - (1-\sqrt 2)^{2^n}}\\
&= a_{2^n}
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3491183",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
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"answer_id": 2
} |
If $f(x)=f(1-x)$, then $f'(x)=-f'(1-x)$ If $f(x)=f(1-x)$, then
$$\frac{df(x)}{dx}=\frac{df(1-x)}{dx}=\frac{df(1-x)}{d(1-x)}\frac{d(1-x)}{dx}$$
and since $\frac{d(1-x)}{dx}=-1\implies d(1-x)=-dx$,
$$\frac{df(x)}{dx}=\frac{df(1-x)}{-dx}\cdot (-1)=\frac{df(1-x)}{dx}\ne -\frac{df(1-x)}{dx},$$
where's the mistake?
| $f'(1-x)$ is $\left.\dfrac{df(t)}{dt}\right|_{t = 1-x} = \dfrac{df(1-x)}{d(1-x)}$, not $\dfrac{df(1-x)}{dx}$.
e.g. for $f(x) = x(x-1) = x^2 -x$,
$$\begin{align*}
f(1-x) &= (1-x)^2 - (1-x)\\
&= x^2 - 2x + 1 - 1 + x\\
&= f(x)\\
f'(x) &= 2x - 1\\
f'(1-x) &= 2(1-x) - 1\\
&= 1 - 2x\\
&= -f'(x)\\
\frac{df(1-x)}{dx} &= \fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3492656",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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If $\alpha, \beta, \gamma, \delta$ are distinct roots of equation $x^4 + x^2 + 1 = 0$ then $\alpha^6 + \beta^6 + \gamma^6 + \delta^6$ is I tried to find discriminant first and got two roots $$\frac{-1 + \sqrt{3}i}{2}$$ and $$\frac{-1 - \sqrt{3}i}{2}$$
I tried taking $x^2 = t$ and solving equation for root $w$ and $w^2$... | An alternative approach. Your polynomial is
$$ x^4+x^2+1 = (x^4+2x^2+1)-x^2 = (x^2+1)^2-x^2 = (x^2-x+1)(x^2+x+1) = \Phi_3(x)\Phi_6(x) $$
so for any root $\xi$ of the LHS, $\xi^6=1$. This implies $p_6=4$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3494035",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 6
} |
$I_{n}=\int_{-n}^{n}\{x+1\}\{x^2+2\}+\{x^2+2\}\{x^3+4\}dx$. Find $I_1$ where $\{\}$ denotes fractional part. $$I_{n}=\int_{-n}^{n}\{x+1\}\{x^2+2\}+\{x^2+2\}\{x^3+4\}dx$$. Find $I_1$ where $\{\}$ denotes fractional part.
$$I_1=\int_{-1}^{1}\{x+1\}\{x^2+2\}+\{x^2+2\}\{x^3+4\}dx$$
As $\{x+1\}=\{x\}$
$$I_1=\int_{-1}^{1}\{x... | Your calculations are correct. By employing a slightly different technique,
$$\int_0^1 \{x+1\}\{x^2+2\}dx= \int_0^1 \{x\}\{x^2\}dx= \int_0^1x^3dx = \frac 14;$$
$$\int_{-1}^0\{x+1\}\{x^2+2\}dx=\int_{-1}^0\{x+1\}\{(x+1)^2+1-2x\}dx=$$
$$\int_0^1\{x\}\{x^2+3-2x\}dx=\int_0^1\{x\}\{x^2+1-2x\}dx=\int_0^1\{x\}\{(1-x)^2\}dx=$$
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3495953",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Trouble finding the solution of $\int_0^{\pi/2}\sin^mx\cdot\cos^nx\,dx$ from the recurrence relation
Prove
\begin{align}
& \int_0^{\pi/2}\sin^mx \cos^nx\,dx \\[12pt]
= {} & \begin{cases}\dfrac{[(m-1)(m-3)\cdots][(n-1)(n-3)\cdots]}{(m+n)(m+n-2)\cdots}\cdot\dfrac{\pi}{2} & m,n\text{ even integers}\\[10pt]
\dfrac{[(m-1... | Your error is here:
$$I_{m,n}=\frac{(m-1)(m-3)\cdots 1}{(m+n)(m+n-2)\cdots\color {red}{2}}\cdot I_{0,n}.$$
The correct expression is:
$$I_{m,n}=\frac{(m-1)(m-3)\cdots 1}{(m+n)(m+n-2)\cdots\color {red}{(n+2)}}\cdot I_{0,n}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3497590",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Given $a, b, c > 0$ such that $a^2 + b^2 + c^2 + abc = 4$, prove that $\sum_{cyc}\frac{b}{\sqrt{(c^2 + 2)(a^2 + 2)}} \ge 1$.
Given positives $a, b, c$ such that $a^2 + b^2 + c^2 + abc = 4$, prove that $$\large \frac{a}{\sqrt{(b^2 + 2)(c^2 + 2)}} + \frac{b}{\sqrt{(c^2 + 2)(a^2 + 2)}} + \frac{c}{\sqrt{(a^2 + 2)(b^2 + 2)... | I like the following way.
Let $a=\frac{2x}{\sqrt{(x+y)(x+z)}}$ and $b=\frac{2y}{\sqrt{(x+y)(y+z)}},$ where $x$, $y$ and $z$ are positives.
Thus, $c=\frac{2z}{\sqrt{(x+z)(y+z)}}$ and we need to prove that:
$$\sum_{cyc}\frac{\frac{2x}{\sqrt{(x+y)(x+z)}}}{\sqrt{\left(\frac{4y^2}{(x+y)(y+z)}+2\right)\left(\frac{4z^2}{(x+z)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3501431",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
No positive integer solution for $x^2 + y^3 = z^6$? I am thinking about the following problem:
Is there any solution to $x^2 + y^3 = z^6$, where $x, y, z$ are positive integers?
I searched all possible solutions for $1 \leq z \leq 1000$ and there were no solution. So I think there is no solution, but I cannot prove i... | This is the proof referenced in the comments above.
The only solutions of this equation in nonzero integers are $(x,y,z)=(\pm 3k^3,-2k^2,\pm k)$, where $k$ is nonzero.
Let $(x,y,z)$ be a solution with minimal $|z|$. from equation we derive that $(z^3-x)(z^3+x)=y^3$. Obviously $\gcd (x,y)=\gcd(y,z)=\gcd(z,x)=1$. Thus th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3502645",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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A ring with $8$ elements
Let $(A,+,\cdot)$ be a unitary ring with 8 elements. Prove that :
$a)8=0$ and $k\neq 0$, for any odd $k$.
b)If $\exists a\in A$ such that $a^3+a+1=0$, then $a\neq 0$, $a\neq 1$, $2=0$, $a^7=1$ and $A$ is a field.
a) is pretty straightforward, it is obvious that $8=0$ from Lagrange's theorem ... | Suppose that $a = n \cdot 1$ for some $n \in \mathbb{Z}$. Then $\left( n^{3} +n +1 \right) \cdot 1 = 0$, and hence $n^{3} +n +1$ is even since it is a multiple of $\text{car}(A) \in \lbrace 2, 4, 8 \rbrace$. As this is not possible, $a \notin \mathbb{Z} \cdot 1$. In particular, $\mathbb{Z} \cdot 1 \neq A$, and so $\tex... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3504478",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
} |
We roll a fair die until a $5$ appears. What is the expected value of the minimum value rolled?
Question: Given a fair dice, we roll until we get a $5.$ What is the expected value of the minimum value rolled?
Answer is $\frac{137}{60}.$
There is a similar question asked in MSE but I do not understand the method used ... | The probability that the minimum is $X\in\{1,2,3,4,5\}$ can be found as follows:
The probability of rolling a sequence of length $k$ using numbers from the set $\{X,X+1,\dots,6\}-\{5\}$ is $\left(\frac{6-X}{6}\right)^k$. Similarly, the probability of getting a sequence with values in $\{X+1,\dots,6\}-\{5\}$ is $\left(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3505861",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 3
} |
Proving the inequality $\prod_{n=1}^\infty \left( 1+\frac1{n^2+\ln n} \right) < \frac72$ I am struggling to prove
$$\prod_{n=1}^k \left( 1+\frac1{n^2+\ln n} \right) < \frac72$$
For all $k \geq 1$. Clearly, this is true for $k=1$, so it suffices to show
$$\prod_{n=1}^\infty \left( 1+\frac1{n^2+\ln n} \right) < \frac72$... | Proof: We will use the following bound:
$$\frac{1}{x^2 + \ln x} < \frac{1}{x^2} - \frac{3}{2x^4}, \quad \forall x\ge 5.$$
proof: It suffices to prove that $f(x) = x^2 + \ln x - \frac{2x^4}{2x^2-3} > 0$ for $x\ge 5$.
We have $f'(x) = \frac{4x^4+6x^2+9}{x(2x^2-3)^2} > 0$ for $x\ge 5$. Note also that
$f(5) > 0$. The desir... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3506406",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 2
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Integration by substitution, why is the $u$ this value? $$\int\frac{1}{(x+3)\sqrt{x}}~dx$$
I was wondering how you integrate this. I know you use substitution however I think of using $\sqrt x~$ for $u$, however on the integration calculator it says to use $u = \dfrac{\sqrt x}{\sqrt 3}~.$ I don't understand why and wha... | Let $x=u^2$ so that $dx=2u\cdot du$. Then your integral is transformed into $$=\int\frac{2u\cdot du}{(u^2+3)\sqrt{u^2}}=2\int\frac{du}{u^2+3}$$
where the latter can be handled as an arctangent integral.
To flesh it out completely, $$=2\int\frac{du}{u^2+3}=2\int\frac{du}{3\left(\frac{u^2}{3}+1\right)}=\frac{2}{3}\int\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3506506",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
Find the limit of $a_n = n \sqrt{n}(\sqrt{n + 1} - a\sqrt{n} + \sqrt{n - 1})$ given that the sequence $(a_n)$ is convergent. I am given the sequence:
$$a_n = n\sqrt{n}(\sqrt{n + 1} - a\sqrt{n} + \sqrt{n - 1})$$
with $n \in \mathbb{N}^*$ and $a \in \mathbb{R}$. I have to find the limit of $(a_n)$ given that the sequence... | Your initial working leads well into using the generalized binomial expansion which gives the asymptotic behaviour as $n\to\infty$
\begin{align}
a_n
&=n^2\left(\sqrt{1+\frac1n}+\sqrt{1-\frac1n}-2\right)\\
&\sim n^2\left(1+\frac1{2n}-\frac1{8n^2}+\dots+1-\frac1{2n}-\frac1{8n^2}-\dots-2\right)\\
&=-\frac14\\
\end{align}
... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Why is $(1+\frac{1}{n})^n < (1+\frac{1}{m})^{m+1}$? Why is it that
$$
\left(1+\frac{1}{n}\right)^n < \left(1+\frac{1}{m}\right)^{m+1},
$$
for any natural numbers $m, n$?
I have tried expanding using the binomial series and splitting into cases. I understand why it is trivially true when $m=n$ but I am not sure if the... | For a simple answer to why is $\left(1+\frac{1}{n}\right)^n < \left(1+\frac{1}{m}\right)^{m+1}$? -- it is because of the inequality of geometric and arithmetic means. In particular
$a_n = \big(1 + \frac{1}{n}\big)^n \lt \big(1 + \frac{1}{n+1}\big)^{n+1} = a_{n+1}$
or equivalently, taking n+1 th roots
$a_{n}^\frac{1}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3512177",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 2
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Find locus of $\Delta ABC$ centroid with orthocentre at origin and side slopes 2, 3 and 5 Let $ABC$ be a triangle with slopes of the sides $AB$, $BC$, $CA$ are $2,3,5$ respectively. Given origin is the orthocentre of the triangle $ABC$. Then find the locus of the centroid of the triangle $ABC$.
Since sides have slo... | Hint: Fix a triangle $ABC$ with the required property. Show that any other triangle $A'B'C'$ with the required property is obtained from $\triangle ABC$ by a dilation about the origin. This proves that the locus of the centroid is a line passing through the origin (and the centroid $G$ of the fixed triangle $ABC$). ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3512736",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
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} |
Determine the conditions for $n$ and $\theta\neq0+2k\pi$ such that $(\frac{1+\cos\theta+i\sin\theta}{1-\cos\theta+i\sin\theta})^n \in\mathbb{R}$ Let $z=1+\cos\theta+i\sin\theta=|z|(\cos\alpha+i\sin\alpha)$ and $z'=1-\cos\theta+i\sin\theta=|z'|(\cos\alpha'+i\sin\alpha')$.
My line of reasoning is to convert the numerator... | $$\frac{1+\cos\theta+i\sin\theta}{1-\cos\theta+i\sin\theta}=\frac{2\cos^2\frac{\theta}{2}+2i\sin\frac{\theta}{2}\cos\frac{\theta}{2}}{2\sin^2\frac{\theta}{2}+2i\sin\frac{\theta}{2}\cos\frac{\theta}{2}}=-i\cot\frac{\theta}{2}(\cos\frac{\theta}{2}+i\sin\frac{\theta}{2})^2=$$
$$=\cot\frac{\theta}{2}(-\sin\theta+i\cos\thet... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3515051",
"timestamp": "2023-03-29T00:00:00",
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How to evaluate the following surface integral?
Evaluate $$\iint_S \left (x^4+y^4+z^4 \right )\ dS,$$ where $S=\left \{(x,y,z)\ :\ x^2+y^2+z^2 = a^2 \right \},\ a > 0.$
My attempt $:$ I have tried to find the integral using Gauss' divergence theorem which states that
Suppose $V$ is a subset of $\Bbb R^n$ (in this ... | Recognize the symmetry to simplify the integral first,
$$I=\int_S \left (x^4+y^4+z^4 \right )\ dS,=3\int_S z^4 \ dS$$
Then, integrate with spherical coordinates $z=a\cos\theta$, $dS = a^2\sin\theta d\theta d\phi$,
$$I=3 \int_0^{2\pi}\int_0^{\pi}(a\cos\theta)^4 a^2\sin\theta d\theta d\phi
=-6\pi a^6 \int_0^{\pi}\cos^4\t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3515605",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Can the quadratic formula be explained intuitively? Most people know what the quadratic formula is so I won’t post it here (Besides, I don’t know how to properly post formulas in general).
I was wondering if there is an intuitive explanation as to why the quadratic formula is structured the way it is.
| Using the factor theorem, suppose that you have an equation of the form
$$x^2 + Bx + C = 0.$$
If it has roots $r$ and $s$, then we have
$$x^2+Bx+C = (x-r)(x-s).$$
(This comes because you can write $x^2+Bx+C = (x-r)q(x)+t$, where $t$ is a remainder that must be constant by using Long Division, and then when you plug in ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3516253",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
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Asymmetric inequality in three variables $\frac{3(a+b)^2(b+c)^2}{4ab^2c} \geq 7+\frac{5(a^2+2b^2+c^2)}{(a+b)(b+c)}$ Consider three real positive variables $a,\ b$ and $c$. Prove that the following inequality holds:
$$\frac{3(a+b)^2(b+c)^2}{4ab^2c} \geq 7+\frac{5(a^2+2b^2+c^2)}{(a+b)(b+c)}$$
My progress: We can prove th... | Partial solution. I hope, it can help.
Let $a+c=2p$ and $ac=q^2,$ where $q>0.$
Thus, by AM-GM $p\geq q$ and we need to prove that
$$\frac{3(b^2+2bp+q^2)^2}{4q^2b^2}\geq7+\frac{5(b^2+4p^2-2q^2)}{b^2+2pb+q^2}.$$
Now, consider two cases:
*
*$b\geq q$, $p=q+u$ and $b=q+v$.
Thus, we need to prove that:
$$72uq^5+4(16u^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3519062",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
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Solving a limit by two methods with different results I'm considering this limit
$$\lim_{x\to 0}\frac{(1-\cos x^2)\arctan x}{e^{x^2}\sin 2x - 2x + \frac{2}{3}x^3}.$$
My first attempt was using the following equivalent infinitesimals
$$1-\cos x^2 \sim \frac{x^4}{2},\quad \arctan x \sim x, \quad \sin 2x \sim 2x, \quad e^... | Your first method is correct and the answer is indeed $0$. The problem with your second approach is that the Taylor series of $e^{x^2}\sin(2x)$ centered at $0$ begins with $2x\color{red}+\frac23x^3$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3519645",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
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Maximize product $(a^3-a^2+2)(b^3-b^2+2)(2c^3+5c^2+9)$ Consider three real variables such that $a^2+5b^2+5c^2 = 21$. Maximize the product:
$$P = (a^3-a^2+2)(b^3-b^2+2)(2c^3+5c^2+9)$$
My attempt (ideas): I suppose the first step is to give an argument that the maximum is reached when the variables are positive (or at le... | Incomplete solution: (Updated 2020-01-25)
Step 1 (incomplete): Establish that the maximum of $P$ occurs when all three factors of $P$ are non-negative.
First, let's observe that from the condition $a$ lies in $[-\sqrt{21},\sqrt{21}]$, while $b$ and $c$ lie in $\left[-\sqrt{\dfrac{21}{5}}, \sqrt{\dfrac{21}{5}}\right]$,... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3521250",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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Solution to two non polynomial equation If
$$x+\frac1x=-1$$
Find the value of
$$x^{99}+\frac{1}{x^{99}}$$
Is there any formal (traditional) way to solve these problems using only elementary algebra or high school math
| $$(x+\frac{1}{x})(x^2+\frac{1}{x^2})\Rightarrow x^3+\frac{1}{x^3}=2$$
$$(x+\frac{1}{x})(x^3+\frac{1}{x^3})\Rightarrow x^4+\frac{1}{x^4}=-1$$
$$(x+\frac{1}{x})(x^4+\frac{1}{x^4})\Rightarrow x^5+\frac{1}{x^5}=-1$$
$$(x+\frac{1}{x})(x^5+\frac{1}{x^5})\Rightarrow x^6+\frac{1}{x^6}=2
\\ \vdots$$
$$x^{99}+\frac{1}{x^{99}}=2$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3524359",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Find the function $f(x)$ if $f(x+y)=f(x-y)+2f(y)+xy$ I'm suffering from the procedure I found, which is contradictory.
The original question is:
Let $f$ be differentiable. For all $x, y \in \mathbb R$,
$$f(x+y)=f(x-y)+2f(y)+xy$$
Suppose $f'(0)=1$ and find $f(x)$.
My procedure was:
Put $x=y=0$:
$$f(0)=f(0)+2f... | In addition to the accepted answer, there is just no function $f$ whatsoever that satisfies the given functional equation, differentiable or otherwise. Consider the following special cases:
\begin{align}
f(1) &= f(1) + 2f(0) & x=1 \;& y=0 \\
f(2) &= f(0) + 2f(1) + 1 & x=1\;&y=1 \\
f(4) &= f(2) + 2f(1) + 3 &... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3524457",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
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Product of multiple sines Is there an identity for the following equation:
$\ f(x) = \sin(x.\pi/2)\cdot \sin(x.\pi/3)\cdot \sin(x.\pi/4) $
I am looking for an equation similar to this equation
| Express everything in terms of the trigonometric functions of $\frac {\pi x}{12}$ and working a little
$$4\sin \left(\frac{\pi x}{2}\right) \sin \left(\frac{\pi x}{3}\right) \sin
\left(\frac{\pi x}{4}\right)=\sin \left(\frac{\pi x}{12}\right)+\sin \left(\frac{5 \pi x}{12}\right)+\sin \left(\frac{7 \pi x}{12}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3526277",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Given that $z = i + i^{2016} + i^{2017}$, find $|z^{10}|$. I am told that
$$z = i + i ^ {2016} + i ^ {2017}$$
and I have to find $|z^{10}|$. This is what I tried:
$$i ^ {2016} = (i^4)^{504} = 1 ^ {504} = 1$$
$$i ^ {2017} = i \cdot (i^4)^{504} = i \cdot 1 = i$$
So we have that:
$$z = 1 + 2i$$
And then I worked towards ... | $z=1+2i$; $\overline{z}=1-2i$;
$|z|^2=z\cdot \overline{z}=1-4i^2=5$;
$|z|^{10}=(|z|^2)^5=5^5$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3527365",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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How many different solutions for this $\sqrt{3x^2 + 2x + 5} + \sqrt{x^2-4x+5} = \sqrt{2x^2 - 2x + 2} + \sqrt{2x^2+8}$
My attempt :
All the equation inside sqrt has $D<0$.
Let it be
$\sqrt A + \sqrt B = \sqrt C + \sqrt D$
2 possibilities
$$\sqrt A = \sqrt C \cap \sqrt B = \sqrt D$$
Find $x$ that match this condition.
$... | Let $a=\sqrt{3x^2+2x+5}, b=\sqrt{2x^2+8}, c=\sqrt{2x^2-2x+2}, d=\sqrt{x^2-4x+5}$
$\implies a-b=c-d$
Observe that $$3x^2+2x+5-(2x^2+8)=x^2+2x-3=2x^2-2x+2-(x^2-4x+5)$$
$\implies a^2-b^2=c^2-d^2$
$\implies a+b=\dfrac{c-d}{a-b}\cdot c+d=c+d$
Now add and subtract to find $$a=c$$
$$\implies\sqrt{3x^2+2x+5}=\sqrt{2x^2-2x+2}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3527924",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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integration of. $\int \frac{x}{x^3-3x+2}$ I am trying to integrate :
$\Large \int \frac{x}{x^3-3x+2}dx$
I decomposed the fraction and got :
$\Large \frac{x}{x^3-3x+2} = \frac {x}{(x-1)^2(x+2)}$
Then I tried to get two different fractions:
$ \Large \frac {x}{(x-1)^2(x+2)} = \frac {Ax}{(x-1)^2} + \frac {B}{(x+2)}$
W... | $$ I=\int \frac{x}{x^3-3x+2}dx$$
$$I= \int \frac {xdx}{(x-1)^2(x+2)}$$
Since $\frac {-1}{(x-1)^2}$ is an obvious derivative substitute $$u=\dfrac {1}{x-1} \implies -du=\dfrac {dx}{(x-1)^2}$$
The integral becomes:
$$I=-\int \dfrac {u+1}{3u+1}du$$
Which is easier to integrate.
$$u+1=u+\frac 1 3 +\frac 2 3$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3529446",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find $P$ such that $P^{-1}AP$ is a given matrix
Let
$$A = \begin{pmatrix} 1 & -3 & 0 \\ 3 & 4 & -3 \\ 3 & 3 & -2\end{pmatrix}$$
Find an invertible $P \in M_3(\mathbb{R})$ such that
$$B = P^{-1}AP = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 3 \\ 0 & -3 & 1\end{pmatrix}$$
The problem is that the characteristic polynom... | Let $P=\begin{pmatrix} 1 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 1 & 0 \end{pmatrix}$. $\;$ You can check that $P^{-1}AP=\begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 3 \\ 0 & -3 & 1 \end{pmatrix}=B$.
The way I found this matrix was as follows: in order to have $P^{-1}AP=B$, $P$ must send the standard basis vector $e_1$ to an eigenvector o... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3530589",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
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Find the remainder when $x^{100}$ is divided by $x^8 - x^6 + x^4 - x^2 + 1.$ I need help in the problem:
Find the remainder when $x^{100}$ is divided by $x^8 - x^6 + x^4 - x^2 + 1.$
I have tried factoring $x^{100}-1$ and adding 1 to that, but that hasn't helped. Could someone please help with this?
| $x^{100}-1=(x^{10}+1)(x^{90}-x^{80}+x^{70}-x^{60}+x^{50}-x^{40}+x^{30}-x^{20}+x^{10}-1)$
$=(x^8-x^6+x^4-x^2+1)(x^2+1)(x^{90}-x^{80}+x^{70}-x^{60}+x^{50}-x^{40}+x^{30}-x^{20}+x^{10}-1)$,
so $x^{100}=$
$(x^8-x^6+x^4-x^2+1)(x^2+1)(x^{90}-x^{80}+x^{70}-x^{60}+x^{50}-x^{40}+x^{30}-x^{20}+x^{10}-1) + 1.$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Proof by induction:$\frac{3}{5}\cdot\frac{7}{9}\cdot\frac{11}{13}\cdots\frac{4n-1}{4n+1}<\sqrt{\frac{3}{4n+3}}$ In the very beginning I'm going to refer to similar posts with provided answers:
Induction Inequality Proof with Product Operator $\prod_{k=1}^{n} \frac{(2k-1)}{2k} \leq \frac{1}{\sqrt{3k+1}}$ (answered by Öz... | Base case:
$n = 1$
$\frac {3}{5} < \sqrt {\frac {3}{7}}$
Since both are positive, we can square both sides.
$\frac {9}{25} < \frac {3}{7}$
Cross multiply
$63 < 75$
Inductive hypothesis
Suppose, $\prod_\limits{i=1}^{n} \frac {4i-1}{4i+1} < \sqrt {\frac {3}{4n+3}}$
We must show that when the hypothesis holds, $\prod_\li... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Obtain an expression for the nth term of the Geometric Progression. The 2nd, 6th and 8th terms of an Arithmetic progression are three successive terms of a Geometric progression.
Find the common ratio of the Geometric progression and obtain an expression for the $n$th term of the geometric progression.
Working so far:
... | Working backwards from the answer,
$\begin{align}\frac {16}{9}a(\frac{1}{2})^n&=xr^{n-1}\\ \frac {16}{9}a(\frac{1}{2})^n&=\frac{x(\frac{1}{2})^n}{\frac{1}{2}}\\\frac{16}{9}a&=2x\\\frac{8}{9}a&=x\end{align}$
So, that would mean $x=\frac{8}{9}a$, where $x$ is the first term of the G.P..
However, taking $\frac{(2)}{(1)}$:... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3541404",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Smarter way to solve $ \int_0^1\arctan(x^2)\,dx$ I'm trying to solve the following definite integral:
\begin{equation}
\int_0^1 \arctan(x^2)dx
\end{equation}
I tryed first by parial integration, finding:
\begin{equation}
x\arctan(x^2)\Bigl|_0^1-\int_0^1 \dfrac{2x^2}{1+x^4}dx
\end{equation}
Then:
\begin{equation}
\int_0... | I wouldn't necessarily say it is simpler, but it is more systematic and the same method can be applied to integrating general rational functions of $x$.
You already got to the point of calculating $$x\arctan(x^2)\bigg\vert_0^1 - \int\limits_0^1\frac{2x^2}{1+x^4}dx = \frac{\pi}{4} - \int\limits_0^1\frac{2x^2}{1+x^4}dx$$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3544203",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 0
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Sum of the series: $\sum_{n=2}^\infty\frac{(-1)^n}{n^2+n-2}$ is? This was a question I confronted in JAM 2016. I tried the following steps:
$$\displaystyle\sum_{n=2}^\infty\frac{(-1)^n}{n^2+n-2}
\implies\sum_{n=2}^\infty\frac{(-1)^n}{(n-1)(n+2)}\implies\sum_{n=2}^\infty(-1)^n\frac{1}{3}\cdot[\frac{1}{(n-1)}-\frac{1}{... | $$\frac{1}{3}\sum_{n=2}^{+\infty}(-1)^n\left(\frac{1}{n-1}-\frac{1}{n+2}\right)=$$
$$=\frac{1}{3}\sum_{n=1}^{+\infty}\frac{(-1)^{n+1}}{n}+\frac{1}{3}\sum_{n=1}^{+\infty}\frac{(-1)^{n+1}}{n}-\frac{1}{3}\left(1-\frac{1}{2}+\frac{1}{3}\right)=$$
$$=\frac{1}{3}\ln2+\frac{1}{3}\ln2-\frac{1}{3}\left(1-\frac{1}{2}+\frac{1}{3}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3544400",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Converge / Diverge $\sum_{n=1}^{\infty}\big( \frac{1}{\sqrt[n]{n!}}\big)^2$ $$\sum_{n=1}^{\infty}\big( \frac{1}{\sqrt[n]{n!}}\big)^2$$
$$2^n\leq n!$$
As $2\cdot 2 \cdots2\leq 1 \cdot 2 \cdots n!$
Therefore:
$$2\leq \sqrt[n]{n!}$$
$$2^2\leq \sqrt[n]{n!}^2$$
So:
$$\frac{1}{\sqrt[n]{n!}^2}\leq \frac{1}{4}$$
$$\sum_{n=1}^{... | Since
$$
e^n = 1 + n + \cdots + \frac{{n^n }}{{n!}} + \cdots > \frac{{n^n }}{{n!}},
$$
it holds that
$$
\sum\limits_{n = 1}^\infty {\left( {\frac{1}{{\sqrt[n]{{n!}}}}} \right)^2 } < \sum\limits_{n = 1}^\infty {\frac{{e^2 }}{{n^2 }}} = \frac{{(\pi e)^2 }}{6}.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3545317",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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If $h(x) = \dfrac{f(x)}{g(x)}$, then find the local minimum value of $h(x)$ Let $f(x) = x^2 + \dfrac{1}{x^2}$ and $g(x) = x – \dfrac{1}{x}$
, $x\in R – \{–1, 0, 1\}$. If $h(x) = \dfrac{f(x)}{g(x)}$, then the local minimum value
of $h(x)$ is :
My attempt is as follows:-
$$h(x)=\dfrac{x^2+\dfrac{1}{x^2}}{x-\dfrac{1}{x}}$... | Assuming that
$$h(x)=\frac{f(x)}{g(x)}\implies h'(x)=\frac{\left(x^2+1\right) \left(x^4-4 x^2+1\right)}{x^2 \left(x^2-1\right)^2}$$ Assuming $x\neq 0$ and $x\neq 1$ you are left with
$$x^4-4x^2+1=0 \implies (x^2)^2-4(x^2)+1=0$$ which is a quadratic in $x^2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3546598",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 5
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Finding $ \lim_{x\to1/2}\frac{f(x)}{x-1/2}$ given that $ \lim_{x\to1}\frac{f(x)}{x-1}=2$ and $\lim_{x\to-1}\frac{f(x)}{x+1}=6$ Consider the following question:
Find $\displaystyle \lim_{x\to1/2}\frac{f(x)}{x-1/2}$
given that $\displaystyle \lim_{x\to1}\frac{f(x)}{x-1}=2$ and
$\displaystyle \lim_{x\to-1}\frac{f(x)}{x+1... | Not enough information given.
For example, the polynomial $f(x)=ax^4+(1-2a)x^2+a-1$ with $a \in \mathbb{R}$ checks the hypothesis. The limit though, is finite only when $f\left(\frac{1}{2}\right) = 0$, which gives $a = \frac{4}{3}$ and the limit is $-1$,otherwise the limit can be $\pm \infty$. If we go for a polynomia... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3547336",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
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Summation of finite series: Let $f(r)$ be what? Find the sum of the first $n$ terms of
$\displaystyle \frac{1}{1\times4\times7}+\frac{1}{4\times7\times10}+\frac{1}{7\times10\times13}+...$
My working:
Let $\begin{align}\displaystyle\frac{1}{(3x-2)(3x+1)(3x+4)} &\equiv \frac{A}{3x-2}+\frac{B}{3x+1}+\frac{C}{3x+4}\\1&\equ... | In your final line, the $r$ should be replaced by $x$ to get
$$\sum_{x=1}^n \Bigl(\frac{1}{18(3x-2)} - \frac{1}{18(3x+1)}\Bigr) -\sum_{x=1}^n\Bigl(\frac{1}{18(3x+1)} - \frac{1}{18(3x+4)}\Bigr) \tag{1}\label{eq1A}$$
In each summation, the term being subtracted (i.e., $\frac{1}{18(3x+1)}$ and $\frac{1}{18(3x+4)}$) is the... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3550816",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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$abcd=6(1-a)(1-b)(1-c)(1-d)$ Positive integers $a \geqslant b \geqslant c \geqslant d$ satisfy:
$$abcd=6(1-a)(1-b)(1-c)(1-d)$$
$1.$ Prove that $d=2$
$2.$ Find all possible values of $a,b,c$.
What I know:
Since $x$ and $x-1$ are relatively prime for all positive integers $x$ :
$$\gcd(a,1-a)=\gcd(b,1-b)=\gcd(c,1-c)=\gcd... | We can rewrite this as:
$$\prod_{cyc} \frac{a}{a-1} =6$$
We clearly have $a \geqslant b \geqslant c \geqslant d > 1$. Also, we can see that $\frac{x}{x-1}$ is a decreasing function. If $d>2$ then,
$$\prod_{cyc} \frac{a}{a-1} < (1.5)^4<6$$
This forces $d=2$. We now have:
$$\frac{a}{a-1} \cdot \frac{b}{b-1} \cdot \frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3551012",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Let $\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}.... =A$. What is $\frac {1}{2^2}+\frac{1}{4^2}+\frac{1}{6^2}....$ in terms of A? Find the value of $\frac {1}{2^2}+\frac{1}{4^2}+\frac{1}{6^2}....\infty$ in terms of A
For the first expression, the $T_n$ term will be
$$T_n=\frac{1}{\frac{((n)(n+1))^2}{4}}$$
Now $$T_n=4\le... | Given expression$$\frac{1}{2^2}+\frac{1}{4^2}+\frac{1}{6^2}…$$
Which is equal to
$$\frac1{2^2}\left[\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}…\right]$$
Hence the given Series sums up to $$\frac{A}4$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3551477",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Prove that $n^4 \mod 8$ is identically equal to either 0 or 1, $\forall \ n\in \mathbb{N}$. I feel pretty confident about the first half of my solution for this, however I don't like how I used the induction hypothesis on the case for even integers, it feels like it isn't doing anything useful since it is really easy t... | The proof looks good, although as Doug M suggests, there are much easier ways to reach the same result. In particular, you're right to observe that you don't actually need the inductive hypothesis.
If you do want to apply induction, my advice is to start with a summary of your argument. Something like: "Let $P(n)$ be t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3552154",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Given a sequence $a_n =\frac{n^2 - 3n + (-1)^n}{3n^2 - 7n + 5}$, prove that $\lim_{n \to \infty} a_{n} = g$. The given Sequence is $a_n = \frac{n^2 - 3n + (-1)^n}{3n^2 - 7n + 5}$.
I showed that the Sequence $a_n$ converges towards a Value $g = \frac{1}{3}$.
How do I determine for each $\epsilon > 0$ an $n_0$ so that: $... | Your work is correct. You can continue with more inequalities. For $n>8$, we have $2n+8<3n$ and $9n^2-21n+15>n^2$ since $8x^2-21x+15=8\left(x-\frac{21}{16}\right)^2+\frac{39}{32}>0$ for all $x\in\mathbb{R}$. Hence,
$$\frac{2n+8}{9n^2-21n+15}<\frac{3n}{n^2}=\frac{3}{n} $$
Now you just need to find $n_1$ such that when $... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "2",
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Integrating $\frac{1}{\sqrt{2-\frac{1}{x^2}}}dx$ using two substitutions $z = x^2$ and $u = 2z-1$ An exercise asks to calculate $$\int_1^y\frac{1}{\sqrt{2-\frac{1}{x^2}}}dx$$
The solution manual states that with the two substitutions $z = x^2$ and then $u = 2z -1$ we get
$$\int_1^y\frac{1}{\sqrt{2-\frac{1}{x^2}}}dx = ... | $$z = x² \rightarrow dz = 2 x dx \rightarrow \frac{dz}{dx} = 2x$$
$$dx = \frac{dz}{2x} \rightarrow \int_1^y\frac{1}{\sqrt{2-\frac{1}{x^2}}}dx = \int_1^{y²}\frac{1}{\sqrt{2-\frac{1}{x^2}}} \frac{dz}{2x} = \int_1^{y²}\frac{1}{\sqrt{8x²-4}} dz = \int_1^{y²}\frac{1}{\sqrt{8z-4}} dz = \int_1^{y^2}\frac{1}{2\sqrt{2z-1}}dz$$
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3558485",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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Find the relative maximum and relative minimum of the function $f(x) = 49x + \frac{4}{x}$ Find the relative maximum and relative minimum of the function $f(x) = 49x + \frac{4}{x}=49x+4x^{-1}$
Solution:
Step 1: Find the values of $x$ where $f'(x)=0$ and $f'(x)$ DNE.
$f'(x)=49-4x^{-2}=49-\frac{4}{x^2}$
We can see that $... | First note that $f(-x)=-f(x)$, so the problem is essentially reduced to $x\ge0$.
That can be solved without calculus: by the AM-GM inequality, $49x+\dfrac4x\ge2\sqrt{49\times4}=28,$
and equality holds when $x=\dfrac27$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3558946",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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What can be inferred from this determinant regarding a triangle?
If
$$\Delta =
\begin{array}{|ccc|}
\sin A & \sin B & \sin C \\
\cos A & \cos B & \cos C \\
\cos^3 A & \cos^3 B & \cos^3 C \\
\end{array} = 0$$
Where A, B, C are the angles of a triangle. What can we say about the triangle? (Is the triangle equilate... | \begin{align}
\Delta&=
\sin\alpha \, \cos\beta \, \cos^3\gamma -\sin\alpha \, \cos\gamma \, \cos^3\beta
\\
&
-\cos\alpha \, \sin\beta \, \cos^3\gamma
+\cos\alpha \, \sin\gamma \, \cos^3\beta
\\
&+\cos^3\alpha \, \sin\beta \, \cos\gamma
-\cos^3\alpha \, \sin\gamma \, \cos\beta
=0
\tag{1}\label{1}
.
\end{align}
S... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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How to solve matrix polynomial, that is $A^{15}$ $A$ is a matrix that suffice $A^2 = A - I$ ($I$ is identity matrix.)
Of course, $A^{15} = {(A^2)}^7A$
I think i have to find the elements of matrix A. I tried using completing the square, $$A^2 - A + I = 0$$
$${(A-I/2)}^2 + (3/4) I = 0$$
Which leads to imaginary number.... | $A$ is a zero of $x^2 -(x - 1)=x^2-x+1$, which you may recognize as the $6$-th cyclotomic polynomial. Therefore, $A^6 = I$ and so $A^{15} = A^{12} A^3 = A^3$. Now, $A^3 = A^2-A = -I$.
A systematic way that does not require bright ideas is to use polynomial division. The quotient is irrelevant; only the remainder matter... | {
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Does $\int_{0}^{\infty} \frac{2x +3}{\sqrt {x^3 + 2x + 5}} \,dx $ converge? Does the following integral converge? I will post my solution, but I am unsure if it is true.
$$\int_{0}^{\infty} \frac{2x +3}{\sqrt {x^3 + 2x + 5}} \,dx $$
My solution
Let
$$ g(x) = \frac{2x}{\sqrt {x^3}}$$
$$ f(x) = \frac{2x +3}{\sqrt {x^3... | $$\int_{3}^{\infty} \frac{2x +3}{\sqrt {x^3 + 2x + 5}} \,dx > \int_{3}^{\infty} \frac{x}{\sqrt {2x^3 }} \,dx= \bigg| \sqrt {2x} \bigg|_{3}^{\infty}=\infty$$
You're correct. The integral diverges.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3569039",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Area of triangle in a circle. [Edexcel Specimen Papers Set 2, Paper 1H Q22]
The line $l$ is a tangent to the circle
$^2 + ^2 = 40$ at the point $A$. $A$ is the point $(2,6)$.
The line $l$ crosses the $x$-axis at the point $P$.
Work out the area of triangle $OAP$.
Any help is appreciated.Thank you.
I have attempted th... | The slope of the radius from $O(0,0)$ to $A(2,6)$ is $\frac{6}{2}=3$. The perpendicular slope of the tangent line at $(2,6)$ is the opposite reciprocal of the slope of $\overline{OA}$. Therefore the slope of $\overline{AP}=-\frac{1}{3}$.
The equation of the tangent line is found by using the slope and the coordinate ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3569778",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
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how to prove $x^2\ln{x}+x+e^x-3x^2>0$ Let $x>0$; show that:
$$f(x)=x^2\ln{x}+x+e^x-3x^2>0$$
It seem this inequality
$$f'(x)=2x\ln{x}+1+e^x-5x$$
$$f''(x)=2\ln{x}+e^x-3$$
$$f'''(x)=\dfrac{2}{x}+e^x>0$$
| We need to prove that $$\ln{x}+\frac{1}{x}+\frac{e^x}{x^2}-3>0$$ for $x>0$.
But $$\left(\ln{x}+\frac{1}{x}+\frac{e^x}{x^2}-3\right)'=\frac{x^2-x+e^x(x-2)}{x^3}$$ increases for $x>0$ because easy to show that
$$\left(\frac{x^2-x+e^x(x-2)}{x^3}\right)'=\frac{(x^2-4x+6)e^x-(x^2-2x)}{x^4}>0,$$
which says that the minimum ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3582813",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
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The circumradius of an isosceles triangle ABC is four times as that of inradius and A=B condition The circumradius of an isosceles triangle ABC is four times as that of inradius of the triangle, if A = B. Then
(1) $8 cos^2A – 8cosA + 1 = 0$
(2) $4 cos^2A – 10cosA + 1 = 0$
(3) $cos^2A – cosA – 3 = 0$
(4) $cos^2A – cosA... | Hint:
Using this
$$r=R(\cos A+\cos B+\cos C-1)$$
We have
$A=B\implies \cos B=\cos A,\cos C=\cos(\pi-A-A)=-\cos2A$
and $R=4r$
$\implies r=(4r)(\cos A+\cos A-\cos2A-1)$
Hope you can take it home from here?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3583794",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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If $A, B, C$ are angles of $\Delta ABC$ and $\sin (A-\pi /4) \sin (B-\pi/4) \sin (C-\pi/4)=\frac{1}{2\sqrt 2}$...
If $A, B, C$ are angles of $\Delta ABC$ and $\sin (A-\pi /4) \sin (B-\pi/4) \sin (C-\pi/4)=\frac{1}{2\sqrt 2}$, then prove that $\sum \tan A \tan B=\sum \tan A$
Solving the given equation, we get$$(\sin A... | From the given
\begin{align}0=&\frac{1}{2\sqrt 2}- \sin (A-\pi /4) \sin (B-\pi/4) \sin (C-\pi/4)\\
=&\frac{1}{2\sqrt 2}-\frac12\left[ \cos(A-B)-\cos(A+B-\frac\pi2)\right] \sin (C-\pi/4)\\
=&\frac{1}{2\sqrt 2}[ 1- (\cos(A-B)-\sin C) (\sin C - \cos C)]
\end{align}
Note that $\cos(A-B)\le 1$ and
\begin{align}
0 \ge & \fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3584153",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Trig substitution for $\sqrt{9-x^2}$ I have an integral that trig substitution could be used to simplify.
$$ \int\frac{x^3dx}{\sqrt{9-x^2}} $$
The first step is where I'm not certain I have it correct. I know that, say, $\sin \theta = \sqrt{1-cos^2 \theta}$, but is it correct in this case $3\sin \theta = \sqrt{9 - (3\c... | Maybe the following hint, based on Differential binomial, seems to be the same as @Toby's. However, you focused just on a trig substitution. You can take $$(9-x^2)=t^2,$$ and simplify the integral...
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3586503",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 1
} |
Proving $a^2 + b^2 + c^2 \geqslant ab + bc + ca$
Let $a, b, c \in \mathbb{R}$. Show that $a^2 + b^2 + c^2 \geqslant ab + bc + ca$.
My reasoning went as follows and I would like to know if it's correct.
$a^2 + b^2 + c^2 \geqslant ab + bc + ca$
$\Leftrightarrow a^2 + b^2 + c^2 -ab - bc - ca \geqslant 0$
$\Leftrightarro... | Your proof is correct and so is the identity $2a^2=a^2+a^2$. Alternatively, if you want to divide by $2$, then you can write
$a^2 + b^2 + c^2 \geqslant ab + bc + ca$
$\Leftrightarrow\frac {a^2 + b^2}{2} + \frac {b^2 + c^2}{2} + \frac {a^2 + c^2}{2}\geqslant ab + bc + ca$
$\Leftrightarrow\frac {a^2 + b^2}{2} + \frac {b^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3587457",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
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If $(x^2-5x+4)(y^2+y+1)<2y$ for all real $y$, then $x$ belongs to the interval $(2,b)$, then $b$ can be? If $(x^2-5x+4)(y^2+y+1)<2y$ for all real $y$, then $x$ belongs to the interval $(2,b)$, then $b$ can be?
$$y^2(x^2-5x+4)+y(x^2-5x+2)+(x^2-5x+4)<0$$
As it is true for all real y, hence $D<0$
$$(x^2-5x+2)^2-4(x^2-5x+4... | From $$y^2+y+1=(y+\tfrac12)^2+\tfrac34>0,$$
we know that we may divide by $y^2+y+1$.
What can we say about $\frac{2y}{y^2+y+1} $?
For $y\ge0$, this will be positive, but what about negative $y$?
We have $$y^2+y+1=(y-1)^2-y\ge-y$$
with equality iff $y=-1$. Hence $$\frac{2y}{y^2+y+1}\ge-2 $$with equality iff $y=-1$.
So... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3588892",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Compute $\left[\begin{smallmatrix}1-a & a \\ b & 1-b\end{smallmatrix}\right]^n$ Compute $\begin{bmatrix}1-a & a \\ b & 1-b\end{bmatrix}^n$, where the power of $n\in\mathbb N$ denotes multiplying the matrix by itself $n$ times; $a,b\in[0,1]$.
Edit:
I considered using induction, computed the desired matrix:
$$\begin{bm... | Diagonalize the matrix first.
Let $A$ be your matrix, and you get $A = S\cdot J \cdot S^{-1}$, where
$S = \begin{bmatrix}
1 & -a/b \\
1 & 1 \\
\end{bmatrix}$
and
$J = \begin{bmatrix}
1 & 0 \\
0 & 1-a-b \\
\end{bmatrix}$.
Then calculate $A^n = (S\cdot J \cdot S^{-1})^n = S \cdot J^n \cdot S^{-1}$ (just write things d... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3590279",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
One of my old inequality (very sharp) I'm proud to present one of my old inequality that I can't solve :
Let $a,b,c>0$ such that $a+b+c=1$ and $a\ge b \geq c $ then we have :$$\sqrt{\frac{a}{a^a+b^b}}+\sqrt{\frac{b}{b^b+c^c}}+\sqrt{\frac{c}{c^c+a^a}}\geq \sqrt{\frac{a}{a^a+c^c}}+\sqrt{\frac{c}{c^c+b^b}}+\sqrt{\frac{b... | We will apply Ji Chen's Symmetric Function Theorem for $n=3$
(see https://artofproblemsolving.com/community/c6h194103p1065812):
Symmetric Function Theorem: Let $d\in (0,1)$. Let $x, y, z, u, v, w$ be non-negative real numbers satisfying
$$x+y+z \ge u+v+w, \quad
xy+yz+zx \ge uv+vw+wu, \quad
xyz \ge uvw.$$
Then $x^d + y^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3590423",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Maxima and minima of $f(x) = \frac{\sqrt{x}(x-5)^2}{4}.$
Determine the local minima and maxima of the function $f:[0, \infty) \to \mathbb{R}$ $$f(x) = \frac{\sqrt{x}(x-5)^2}{4}.$$
Does $f$ have a maximum or minimum?
Computing the derivative gets me here:
$f'(x) = \frac14(2(x-5)\sqrt{x}+\frac{(x-5)^2}{2\sqrt{x}}),$
bu... | Hint:
By factoring, using a common denominator and combining terms, you get
$$\begin{equation}\begin{aligned}
f'(x) & = \frac{1}{4}\left(2(x-5)\sqrt{x}+\frac{(x-5)^2}{2\sqrt{x}}\right) \\
& = \frac{x-5}{4}\left(2\sqrt{x}+\frac{x-5}{2\sqrt{x}}\right) \\
& = \frac{x-5}{4}\left(\frac{4x}{2\sqrt{x}}+\frac{x-5}{2\sqrt{x}}\r... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3590967",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
On the integral $\int_0^{\sqrt{2}/2} \frac{\arctan \sqrt{1-2t^2}}{1+t^2} \, \mathrm{d}t$ I'm having a difficult time evaluating the integral
$$\mathcal{J} = \int_0^{\sqrt{2}/2} \frac{\arctan \sqrt{1-2t^2}}{1+t^2} \, \mathrm{d}t$$
This is integral arose after simplifying the integral $\displaystyle \int_{0}^{\pi/4 } \a... | Continue with\begin{align}
& \int_{0}^{\pi/4 } \tan^{-1} \sqrt{\frac{1-\tan^2 x}{2}}dx
\>\>\>\>\>\>\> \tan x =\sin t\\
=& \int_0^{\pi/2}\frac{\cos t}{2-\cos^2t}\ \tan^{-1}\frac{\cos t}{\sqrt2} \ dt\\
=& \int_0^{\pi/2}\int_0^{\pi/4} \frac{\sqrt2 \cos^2t}{(2-\cos^2t)(2\cos^2y+\sin^2 y\cos^2t)}dy\ dt\\
=& \ \frac\pi2\int_... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3592972",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
In triangle $ABC$, $AA_1$, $BB_1$, $CC_1$ divide sides in ratio of $1: 2$ and meet at $M$, $K$, $L$. Find area relation of $KLM$ and $ABC$
Points $A_1$, $B_1$, $C_1$ divide sides $BC$, $CA$, $AB$ equilateral triangle $ABC$ in a ratio of $1: 2$.
The line segments $AA_1$, $BB_1$, $CC_1$ determine the triangle $KLM$.
Is ... | According to Routh’s theorem
the ratio of the areas in general case is
\begin{align}
\frac{S_{KLM}}{S_{ABC}}
&=
\frac{(xyz-1)^2}{(xy+y+1)(yz+z+1)(zx+x+1)}
.
\end{align}
The question is a special case, for which
$x=y=z=\tfrac12$, so
\begin{align}
\frac{S_{KLM}}{S_{ABC}}
&=
\frac{(\tfrac18-1)^2}{(\tfrac14+\tfrac12+1)^3}
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3594577",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Evaluate discriminant of $x^3 +px+q$ given its roots? Consider the cubic polynomial $x^3 +px+q$ with $p, q ∈ \mathbb{Q}$. Suppose $α_1, α_2, α_3$ are the roots and thus $$x^3 + px + q = (x − α_1)(x − α_2)(x − α_3).$$
Let $D = (α_1 − α_2)^2(α_1 − α_3)^2(α_2 − α_3)^2$.
(a) Express $p$ and $q$ in terms of $α_1, α_2$ and $... | Per $α_1 + α_2 +α_3=0$ and $α_1 α_2 α_3=-q$,
$$(α_1 − α_2)^2 = (α_1 + α_2)^2 - 4α_1 α_2=α_3^2+\frac{4q}{α_3}
=\frac p{α_3}\left(\frac{3q}p-α_3\right)$$
Likewise, $(α_2 − α_3)^2 = \frac p{α_1}\left(\frac{3q}p-α_1\right)$ and $
(α_3 − α_1)^2 = \frac p{α_2}\left(\frac{3q}p-α_2\right)$. Then,
$$\begin{align}
D
& = (α_1 − ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3596718",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
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Tokyo university entrance exam - mathematics No.5 - (2)
Circle $C$ which radious is 1 and center is origin on $xy$-plane, We could imagine a cone $S$ that bottom is $C$ and top is $(0,0,2)$. Point $A$ is $(1,0,2)$ and point $P$ is on $C$, $T$ is a cone that $\overline{AP}$'s set
Calculate the volume of $R$ which share... | I am going to assume that $$S = \{ (x,y,z) \in \mathbb R^3 : x^2 + y^2 \le (z/2 - 1)^2 \cap 0 \le z \le 2 \}$$ and $$T = \{ (x,y,z) \in \mathbb R^3 : (x-z/2)^2 + y^2 \le (z/2-1)^2 \cap 0 \le z \le 2 \}$$ and you want to find the common volume of intersection of $S$ and $T$. It is immediately obvious that for a given $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3598447",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
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Find the subgroup of $S_4$ generated by $(1,2,4)$. What is the order of $(1,2,4)$? I am literally just being introduced to group theory this is all new. I know that the permutations that arise are $(4,1),(1,2),(2,4),(1,2,4)$ This is the same as saying that $4$ goes to $1$ then $1$ goes to $2$ and then $2$ goes to $4$ a... | I would like to explain very simply.
$$\sigma=(1,2,4)=\left( \begin{array}{cccc} 1 & 2 & 3 &4 \\
2 &4&3&1 \end{array} \right)$$$$$$then we must calculate $\sigma, \sigma^2,...$ until we reach $$e=\left( \begin{array}{cccc} 1 & 2 & 3 &4 \\
1 &2&3&4 \end{array} \right)=(1) $$$$not $$$$(1,2,3,4)=\left( \begin{array}{cccc}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3602851",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Lemniscate - two parametrizations I have a Lemniscate defined as $$ \|{F_{1}-Z}||^{2} ||F_{2}-Z||^{2}=1 $$ for $ F_1 = (-1,0), F_2=(1,0)$ in $\mathbb{R}^2$
And I should find the parametric equations using these two parametrizations
a) $x=r \cos(\phi)$, $y=r\sin(\phi)$
b) $x=x, y=x \sin(t)$
I understand that the result ... | A bit of algebra gives
$$
\left(y^2+x^2+1\right)^2=4x^2+1\tag1
$$
Applying $x=r\cos(\theta)$ and $y=r\sin(\theta)$, we have
$$
\left(r^2+1\right)^2=4r^2\cos^2(\theta)+1\tag2
$$
which has the solutions $r=0$ and
$$
\bbox[5px,border:2px solid #C0A000]{r^2=2\cos(2\theta)}\tag3
$$
Setting $y=x\sin(t)$ in $(1)$ gives
$$
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3603939",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Matrix rotation in space Determine the matrix of the linear transformation $F$ in the space defined by $v$ being projected onto the plane through the origin with the normal $(1,2,5)$ and then rotating $180^{\circ}$ around the vector
$e = (1,1,1)$.
Not showing all the steps, but the projected matrix should be
\begin{al... | One quick answer is as follows:
The matrix of a projection onto the plane with normal $v$ is given by $I - \frac{vv^T}{v^Tv}$. The matrix of a $180^\circ$ degree rotation in the plane with normal vector $v$ is given by $2 \frac{vv^T}{v^Tv} - I$. So, the matrix $P$ of the projection, $R$ of the rotation, and $F$ of the... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3605383",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Equation of the family of circle which touch the pair of lines $x^2-y^2+2y-1=0$ The question was to find the equation of the family of circles which touches the pair of lines, $x^2-y^2+2y-1 = 0$
So I tried as follows:-
The pair of lines is, by factoring the given equation, $x^2-y^2+2y-1 = 0$ is $x+y-1=0$, and $x-y+1=0$... | As yuou noted, the centre of the circle must lie on $y=1 \vee x=0$.
In the first case, we consider the centre to be $C(x_c,1)$ and we have:
$$(x-x_c)^2+(y-1)^2=r^2$$
The radius is the minimal distance from $C$ to the line $y=x+1$ or $y=-x+1$. So we have:
$$r=\frac{|x_c|}{\sqrt{2}}$$
Substituing, we have:
$$(x-x_c)^2+(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3609144",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 2
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Volume calculated by different orders of integration How to write volume of region
$$S=\{(x,y,z): (x-2)^2+y^2<4, z<3-x^2-y^2,z=-1\}$$ in orders x-y-z and y-z-x of integration.
I tried something but I don't get same result.
First, for x-y-z order I found proection to x-y plane
so I think integral is: $$\int\limits_{x... |
$$\int\limits_{x=0}^2 \int\limits_{y=-\sqrt{4 - (x - 2)^2}}^{\sqrt{4 - (x - 2)^2}} \int\limits_{z=-1}^{3-x^2-y^2} 1\ dx\ dy\ dz = \frac{64}{3} - 4 \pi$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3611569",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Why is there a reduction to REF to find the characteristic polynomial of a 4x4 matrix? I'm asked to find the characteristic polynomial of the matrix: $A = \small\begin{pmatrix}
2 & 2 & 0 & 0\\
2 & 2 & 0 & 0\\
0 & 0 & 2 & 2\\
0 & 0 & 0 & 4\\
\end{pmatrix}$. I've gotten $$\begin{equation}
\begin{split}
p_A(t) &= \det (A... | For an $n \times n$ matrix $A$ the best formula to use is $$\det(\lambda I-A))=\lambda^n+\sum_{i=1}^n \beta_i\lambda^{n-i} $$ where $\beta_i=(-1)^i$ sum of principal minors of order $i$. Note that since $n=4$, $$\det(A-\lambda I)=(-1)^4 \det(\lambda I-A)=\det(\lambda I-A)$$ $$\beta_1=-\text {trace} A=-(2+2+2+4)=-10,$$ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3611839",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to use prove this $p^4\equiv p\pmod {13}$
Let a prime number $p$, and $n$ a positive integer such $$p\mid n^4+n^3+2n^2-4n+3.$$
Show that $$p^4\equiv p\pmod {13}.$$
A friend of mine suggested that I might be able to use the results problem.
| Proof for the case when $n\bmod 13 \neq 3$
Let $n\bmod 13 = k$. Then, by substituting $k=0,1,\ldots,12$ and $k\neq 3$, we have
$$\left[n^4+n^3+2n^2-4n+3\right]\bmod 13= \left[k^4+k^3+2k^2-4k+3\right]\bmod 13\in\{1,3,9\},$$
which are all powers of $3$. Now, notice that $$n^4+n^3+2n^2-4n+3\bmod p=0\implies p\bmod 13 \in... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3612556",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 6,
"answer_id": 3
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Show that $1+2^n+2^{2n}$ is divisible by $7$, when $n$ is not a multiple of $3$ Problem taken from a paper on mathematical induction by Gerardo Con Diaz. Although it doesn't look like anything special, I have spent a considerable amount of time trying to crack this, with no luck.
Show that $1+2^n+2^{2n}$ is divisible b... | Whenever a number is not divisible by 3, it leaves remainder as 1 or 2. So, any number which is not divisible by 3 is of the form $n=3l+1$ or $n=3l+2$.
Now, when $n=3l+1$, we have $1+2^{3l+1}+2^{2(3l+1)}=1+2+4$ mod $7$.
and when
$n=3l+2$, we have $1+2^{3l+2}+2^{2(3l+2)}=1+4+16$ mod $7$ and $1+4+16=0$ mod $7$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3613766",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 3
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Application of Hensel's lemma $x^2 \equiv a (\mod 2^L)$ Let $a$ be an odd integer. And $L \geq 1$. I would like to know the number of solutions modulo $2^L$
to the congruence
$$
x^2 \equiv a \pmod {2^L}.
$$
Is it possible to conclude that there is number $C> 0$ such that the number of solutions modulo $2^L$ to the co... | Let us trace along an argument inspired by the proof that $a \equiv 1 \pmod{8}$ has square roots in $\mathbb{Z}_2$, in order to find out the number of square roots $\pmod{2^L}$.
First, since $a$ is odd and is a square, we know that if $L \ge 3$, then $a \equiv 1 \pmod{8}$; say $a = 8b + 1$. Then for $n^2 \equiv a \pmo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3614073",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
An A level question about partial differentiation
The equation of a curve is $2x^4+xy^3+y^4=10$. Show that $$\frac{dy}{dx}=-\frac{8x^3+y^3}{3xy^2+4y^3}.$$
I understand that you are to work out:
\begin{align}
\frac{dz}{dx} = 8x^3 + y^3\\
\frac{dz}{dy} = 3xy^2 + 4y^3
\end{align}
and therefore, $\dfrac{dy}{dx}$.
My ans... | You don't need to solve for $\frac{dz}{dx}$ or $\frac{dz}{dx}$.
We know $2x^4+xy^3+y^4=10$. I will take the derivative with respect to $x$ on both sides of the equation. So $\frac{d}{dx}(2x^4+xy^3+y^4)=\frac{d}{dx}10 \implies 8x^3+y^3+3xy^2\frac{dy}{dx}+4y^3\frac{dy}{dx}=0$. Trying to solving for $\frac{dy}{dx}$, we h... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3614462",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 1
} |
Prove that $\mathbb{Q}[\sqrt{3}, \sqrt{5}] = \mathbb{Q}[\sqrt{3}+ \sqrt{5}]$ Prove that $\mathbb{Q}[\sqrt{3}, \sqrt{5}] = \mathbb{Q}[\sqrt{3}+ \sqrt{5}]$
Take $x \in \mathbb{Q}[\sqrt{3}+ \sqrt{5}]. x = a_x + b_x( \sqrt{3} + \sqrt{5}) = a_x + b_x\sqrt{3} + b_x\sqrt{5} \in \mathbb{Q}[\sqrt{3}, \sqrt{5}] \Rightarrow \ma... | $(\sqrt 3 + \sqrt 5)(\sqrt 3-\sqrt 5) = -2$ so $\sqrt 3-\sqrt 5=\frac{-2}{\sqrt 3 + \sqrt 5}\in \mathbb Q[\sqrt{3} +\sqrt 5]$
$\sqrt 3 =\frac{(\sqrt 3-\sqrt 5)+(\sqrt 3+\sqrt 5)}2\in \mathbb Q[\sqrt{3}+\sqrt 5]$
$\sqrt 5 = (\sqrt 3 + \sqrt 5)-\sqrt 3\in \mathbb Q[\sqrt{3} + \sqrt{5}]$
So $\mathbb Q[\sqrt 3, \sqrt 5]\su... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3617442",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 5
} |
Continued fraction of $π$ using sums of cubes Recently I came across this identity:
$$\pi=3+\cfrac1{6+\cfrac{1^3+2^3}{6+\cfrac{1^3+2^3+3^3+4^3}{6+\cfrac{1^3+2^3+3^3+4^3+5^3+6^3}{6+\ddots}}}},$$
thus
$$\pi=3+\cfrac{1}{6+\cfrac{(1\cdot3)^2}{6+\cfrac{(2\cdot5)^2}{6+\cfrac{(3\cdot7)^2}{6+\ddots}}}}.$$
We also know that
$$... | The (first) identity is wrong; the continued fraction with "sums of cubes" doesn't converge. This can be shown using the known criterion: for a sequence $\{a_n\}$ of positive real numbers, the continued fraction $$a_0+\cfrac{1}{a_1+\cfrac{1}{a_2+\cfrac{1}{\ldots}}}$$ converges if and only if the series $\sum_{n=0}^{\in... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3618746",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Asymptotes of $r \theta \cos \theta = a \cos 2\theta$ Asymptotes are the lines which touch the curve at infinity.
Putting $u=\frac{1}{r}$, then
\begin{equation*}
\begin{split}
u &= \dfrac{\theta \cos \theta}{a \cos 2\theta} = F(\theta)\\
\end{split}
\end{equation*}
When $r \rightarrow \infty$, $u \rightarrow 0$, or
\be... | You miscalculated $F'\left(-\dfrac \pi 2\right)$. It is actually $\dfrac{-\pi}{2a}$. You will find it is the same line as for $\theta = \dfrac \pi 2$.
These asymptotes are actually rays, not lines. $\dfrac \pi 2$ is the asymptote for the top side of the graph and $-\dfrac \pi 2$ is the asymptote for the bottom side of ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3626248",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Shortest distance from circle to a line
Let $C$ be a circle with center $(2, 1)$ and radius $2$. Find the shortest distance from the line $3y=4x+20$.
This should be very simple, but I seem to end up with no real solutions.
The shortest distance would be from the center of the circle perpendicular to the line right?
S... | By substituting the equation of the line into the equation of the circle you are looking for points where the line intersects the circle. The fact that the resulting equation for $x$ has no real roots means that the line does not intersect the circle i.e. the shortest distance from the line to the circle will be greate... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3626410",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 7,
"answer_id": 1
} |
Prove $x^4-18x^2+36x-27$ can never be a nonzero square rational when $x$ is rational Prove $x^4-18x^2+36x-27$ can never be a square rational (excluding 0), when x is rational
I have tried to use modulus, but didn't get anywhere, any help would be greatly appreciated.
| Say $x^4-18*x^2+36*x-27 = n^2$
The polynomial factors up
$(x-3)*(x^3+3*x^2-9*x+9) = n^2$
It means that if $x-3$ is a perfect square, then $x^3+3*x^2-9*x+9
$ is also a perfect square or both aren't perfect square
Just like $36=9×4=12×3$
If $x-3$ is not a perfect square, then polynomial $x^3+3*x^2-9*x+9$ is not also a p... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3627258",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Prove that $1=\sqrt{2\cdot \color{red}{-1} + \sqrt{2\cdot \color{red}2 + \sqrt{2\cdot \color{red}{9}+\sqrt{2\cdot\color{red}{20}+\cdot \cdots}}}}$ I was playing around with square roots and I noticed that the number $1$ can be seemingly expressed as an infinite nested radical with an easy pattern. I then noticed that i... | This is the general identity:
$$2n+1=\sqrt{2(n+1)(2n-1)+\sqrt{2(n+2)(2n+1)+\sqrt{2(n+3)(2n+3)+\cdots}}}$$
Secondary (albeit similar) Proof:
This proof is my original one. @TheSimpliFire's is way more elegant and concise, but since that first proof is in the post, I will post mine here, just to contribute in benefit o... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3627916",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Continuous functions satisfying $f(f(x))=x$, for all $x \in \mathbb{R}$, and $\int_{-x}^{0} f(t)dt - \int_{0}^{x^2}f(t)dt=x^3$ for $x>0$
Find all continuous functions $f: \mathbb{R} \to \mathbb{R}$ satisfying:
I. For all $x\in \mathbb{R}$, $f(f(x))=x$
II. For all $x>0$, $\displaystyle\int_{-x}^{0} f(t)dt - \displaysty... | This is not an answer, but is too long for a comment. This is as far as I got.
Write the anti-derivative of $F$ as $f$.
Then by (2), for any $x>0$:
$$x^3 = F(0) - F(-x) - F(x^2) + F(0) = 2F(0) - F(-x) - F(x^2).$$
Taking a derivative:
$$3x^2 = f(-x) - 2xf(x^2) \iff f(-x) = 3x( x + \frac{2}{3}f(x^2)).$$
So if we know $f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3628256",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 1,
"answer_id": 0
} |
If $p=\frac {dy}{dx}$ then solve the equation $p^3-p(x^2+xy+y^2)+x^2y+xy^2=0$ If $p=\frac {dy}{dx}$ then solve the equation $p^3-p(x^2+xy+y^2)+x^2y+xy^2=0$
My Attempt:
$$p^3-p(x^2+xy+y^3)+x^2y+xy^2=0$$
$$p^3-p(x+y)^2+pxy+xy(x+y)=0$$
$$p(p^2-(x+y)^2)+xy(p+(x+y))=0$$
$$p\cdot (p+(x+y))\cdot (p-(x+y))+xy(p+(x+y))=0$$
$$(p... | We get three equations: $p=x,y-x-y$
We get three solutions: $$y=x^2/2+C, y=Ce^x, y=1-x+Ce^{-x}$$
The total solution of this first order ODE is
$$(x^2/2-C)(ye^{-x}-C)[(y-1+x)e^{x}-C]=0,$$
with only one contant of integration: $C$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3631226",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Calculate $\sum_{x=1}^{\infty} \frac{(x-1)}{2^{x}}$ I can prove it converges but I don't know at what value it converges.
$\sum_{x=1}^{\infty} \frac{(x-1)}{2^{x}}$
| Start with (for $\lvert u \rvert < 1$ the geometric series converges absolutely, we'll use $u = 1/2$ in the end, so all the manhandling is justified):
$\begin{align*}
\sum_{x \ge 1} u^x
&= \frac{u}{1 - u} \\
\frac{d}{d u} \sum_{x \ge 1} u^x
&= \frac{1}{(1 - u)^2} \\
\sum_{x \ge 1} x u^{x - 1}
&= \\
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3632074",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Showing there isn't such an integer $a \not\equiv 1$ that $a^7 \equiv 1 \pmod{21}$
Establish whether there exists an integer $a \in \mathbb{Z}, a \not\equiv 1$ such that $a^7 \equiv 1 \pmod{21}$
The first thing I tried was trying to compute the result of $a^7$ for some small $a$'s.
This is what I found:
By breaking u... | Well, you could just test them all. For any integer $a$ there must be an $i = 0, 1,2,....,20$ so that $a\equiv i\pmod{21}$ and $a^7\equiv i^7\pmod {21}$ and none of $0^7,2^7,....., 20^7$ are congruent to $1\pmod{21}$ and only $1^7\equiv 1 \pmod {21}$..
That's not satisfying as it's unlikely anyone want to take the ef... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3633407",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Let $s_{n}$ be a sequence such that $|s_{n+1}-s_{n}|<2^{-n}$. Prove that $s_{n}$ is Cauchy. Is this proof sound?
Proof: Let $\epsilon >0$. Since the sequence $\frac{1}{2^{n}}$ converges to $0$, there exists $N$ such that $n>N$ implies $\left|\frac{1}{2^{n}}-0\right|=\frac{1}{2^{n}}<\frac{\epsilon}{2}$. Assume without l... | It's right, and the condition can be more looser. We just need $ |s_{n+1}-s_{n}|\leqslant n^{-2} $. The key steps of proof are
\begin{align*}
|s_{n+p}-s_{n}| & \leqslant |s_{n+p}-s_{n+p-1}|+\dots+|s_{n+1}-s_{n}|\\
& \leqslant\frac{1}{n^{2}}+\frac{1}{(n+1)^{2}}+\dots+\frac{1}{(n+p-1)^{2}}\\
& \leqslant\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3634911",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Evaluate $\lim_{n\to\infty}\frac{1}{n^{p+1}}\cdot \sum_ \limits{i=1}^{n} \frac{(p+i)!}{i!} $, $p \in N$
Evaluate
$$\lim_{n\to\infty}\frac{1}{n^{p+1}}\cdot \sum_ \limits{i=1}^{n} \frac{(p+i)!}{i!} , p \in N$$
Now, I found this problem while doing some practice and I am curious on how to solve it . I have no good i... | A lower bound is given by
$$
\mathop {\lim }\limits_{n \to + \infty } \frac{1}{{n^{p + 1} }}\sum\limits_{i = 1}^n {\frac{{(p + i)!}}{{i!}}} \ge \mathop {\lim }\limits_{n \to + \infty } \frac{1}{{n^{p + 1} }}\sum\limits_{i = 1}^n {i^p } \\ = \mathop {\lim }\limits_{n \to + \infty } \frac{1}{n}\sum\limits_{i = 1}^n {... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3636667",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
Alternative methods of solving an angle in an equilateral triangle This question has already been answered here, but the OP said trigonometry was forbidden. I was thinking of different approaches that allow trigonometry, so I decided to post a new question.
An equilateral triangle is given with edges of the length $a$... | Not a trigonometric solution, but I decided to post it anyway.
It is clear that triangles $CXA$ and $AYB$ are congruent, hence $\angle CXA=\angle AYB$. Hence quadrilateral $BXTY$ is cyclic. The center of this circle is the midpoint $M$ of $XB$ as $MB=MX=\frac 13 a=MY$. Hence $\angle BTY =\frac 12 \angle BMY = 30^\circ$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3637969",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
What is this mystery function that wolfram alpha says my exponential generating function is equal too? The $E_n(x^n)$ is the mystery function
$$\sum_{d=0}^{\infty}\frac{x^{dn}}{\Gamma(dn+1)}=E_n(x^n)$$
Here are the first 3 values of the function
| For $$E_n=\sum_{d=0}^{\infty}\frac{x^{dn}}{\Gamma(dn+1)}$$ the first terms are simple
$$\left(
\begin{array}{cc}
n & E_n \\
1 & e^x \\
2 & \cosh (x) \\
3 & \frac{2}{3} e^{-x/2} \cos \left(\frac{\sqrt{3} x}{2}\right)+\frac{e^x}{3} \\
4 & \frac{\cos (x)}{2}+\frac{\cosh (x)}{2}
\end{array}
\right)$$ For $n>4$, they a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3642591",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Prove that $7^x=1+y^2+z^2$ has no solutions in positive integers $$7^x=1+y^2+z^2$$
So far I have not gotten any remarkable results. Analyzing $mod$ $3$ I get that $y$ and $z$ must be divisible by $3$. Further analyzing $mod$ $4$ I got that $x$, $y$, $z$ must all have the same parity. Looking at the equation $mod$ $8$ I... | $(7^{2k+1}-1)/2$ is an odd integer congruent to $3\bmod 4$ and hence has a prime of the form $4k+3$ appearing an odd number of times.
This implies there are no solutions for odd values of $x$, and we can induct over $v_2(x)$ and use the factorization $(7^{k}-1)(7^k+1)$ to get there are no solutions for any $x$ (you nee... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3643319",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
Does this inequality hold with some constant factor $c>0$? Does there exist a real number $c>0$ such that
$$ (x-1)^2+(y-1)^2-2(\sqrt{xy}-1)^2\ge c\big( (x-\sqrt{xy})^2+(y-\sqrt{xy})^2 \big) \tag{*}$$
holds for every positive real numbers $x,y$ such that $xy \ge \frac{1}{4}$.
Note that the LHS vanishes exactly when
$$ ... | Substitute
$$
r = 2\sqrt{xy} - 1
\qquad
q = x+y-2\sqrt{xy}.
$$
Thanks to AM-GM they are independent and can be any positive number.
It boils down to
$$
\frac{q+2r}{q+r+1}\ge c
$$
so $c$ cannot be positive, since for $q,r$ small enough, the expression converges to zero.
For example, if $x=1/2$ and $y=1/2+\varepsilon$,... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3643592",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Lagrange Method of Solving Cubic Equations Let $K$ be a field (for simplicity, one may assume $K\subseteq\mathbb{C}$), and let $d\in K$. Denote $w=\sqrt[3]{d}$, and $\zeta=(-1+\sqrt{-3})/2$. If $f(x)=x^3+ax^2+bx+c$ is the minimal polynomial of $x_{1}=\alpha+\beta w+\gamma w^2$, show that:
*
*The other two roots of $... | For the second part, it can be shown merely by calculation:
$$\begin{align}x_1+x_2+x_3&=(α+βw+γw^2 )+(α+ζβw+ζ^2γw^2 )+(α+ζ^2βw+ζγw^2)\\& =3α+βw(1+ζ+ζ^2 )+γw^2 (1+ζ+ζ^2 ),\end{align}$$
and by using $1+ζ+ζ^2=0$, as it is a root of $\Phi_3(x)=1+x+x^2=\frac{x^3-1}{x-1}$:
$$α=\frac{1}{3} (x_1 + x_2 + x_3).$$
In a similar fa... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3644877",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Median of the set of numbers which consists of all positive integers whose digits strictly increase from left to right Consider the set
$$S=\{1,2,3,4,5,6,7,8,9,12,13,14,15,16,17,18,19,23,24,\ldots,123456789\},$$
which consists of all positive integers whose digits strictly increase from left to right. This set is finit... | There are 9 of these numbers having 1 digit, $\binom{9}{2}$ having 2 digits, and in general $\binom{9}{k}$ having $k$ digits. The total number of elements in this set is thus $\binom{9}{1} + \binom{9}{2} + \cdots + \binom{9}{9} = 2^9-1$, by the binomial theorem, so the median is the $2^8$-th element.
The binomial th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3645829",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 1,
"answer_id": 0
} |
If $x+y+z=1$ Find the maximum of $\frac{x-y}{\sqrt{x+y}}+\frac{y-z}{\sqrt{y+z}}+\frac{z-x}{\sqrt{z+x}}$ Question -
Let $x, y, z$ be non-negative real numbers such that $x+y+z=1 .$ Find the maximum of
$$
\frac{x-y}{\sqrt{x+y}}+\frac{y-z}{\sqrt{y+z}}+\frac{z-x}{\sqrt{z+x}}
$$
My work -
first WLOG $x\ge y\ge z$
I rewrite ... | Your answer is not correct because your value $\sqrt6$ does not occur.
Also, you need to prove that we can assume $x\geq y\geq z$.
My solution:
Let $z=\min\{x,y,z\}$.
Now, we'll prove that if $y\geq x\geq z$ so
$$\frac{x-y}{\sqrt{x+y}}+\frac{y-z}{\sqrt{y+z}}+\frac{z-x}{\sqrt{z+x}}\leq\frac{y-x}{\sqrt{x+y}}+\frac{x-z}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3655009",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Second order ODE solution The task is to find a solution of second order ODE:
$$
2xx''-3(x')^2=4x^2; x(0) = 1; x'(0) = 0
$$
Now, I tried:
$$
x'' = 2x + \frac{3(x')^2}{2x}
$$
here we substitute $u(x) = x'$ to obtain:
$$
uu'=2x+\frac{3}{2}\frac{u^2}{x} \\
u' = 2\frac{x}{u} + \frac{3u}{2x}
$$
here I substitute $t = \frac{... | Let me work with $y(x)$, then the ODE is
$$yy''-(4y^2+3y'^2)=0~~~~(1)$$
Let $y'=p, y''=\frac{dp}{dx}=\frac{dp}{dy} \frac{dy}{dx}=p\frac{dp}{dy}$
Then (1) becomes
$$yp\frac{dp}{dy}-(4y^2+3p^2)=0 \implies ypdp-(4y^2+3p^2)dy=0~~~~(2)$$
The integrating factor for this in-exact ODE is
$$\mu=\exp[\int \frac{-6p-p}{py}dy=y^{-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3655192",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
How can I approach this inequality? Let $a, b$ and $c$ be three non-zero positive numbers. Show that:
$$\sqrt{\frac{2a}{a + b}} + \sqrt{\frac{2b}{b + c}} + \sqrt{\frac{2c}{a + c}} \leq 3$$
I know the triangular inequality would help here, but I don't know how to approach it.
I started by $a+b≥a$ then that gives $\frac{... | This is not the most elegant approach, but since the inequality is homogeneous we may as well assume $a=x>0, b=1, c=y>0$ and study the behaviour of
$$ f(x,y) = \sqrt{\frac{2x}{x+1}}+\sqrt{\frac{2}{1+y}}+\sqrt{\frac{2y}{x+y}} $$
over $(0,+\infty)^2$. If $x\to 0$ or $y\to 0$ we have $f(x,y)\leq 2\sqrt{2}<3$. By solving
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3659032",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
Prove that $\sqrt{\frac{2 a^{2}+b c}{a^{2}+2 b c}}+\sqrt{\frac{2 b^{2}+c a}{b^{2}+2 c a}}+\sqrt{\frac{2 c^{2}+a b}{c^{2}+2 a b}} \geq 2 \sqrt{2}$ Question -
Prove that for all non-negative real numbers a,b, c, we have
$$
\sqrt{\frac{2 a^{2}+b c}{a^{2}+2 b c}}+\sqrt{\frac{2 b^{2}+c a}{b^{2}+2 c a}}+\sqrt{\frac{2 c^{2}+a... | Take
$$
\sqrt{\frac{2 x+1}{x+2}}+\sqrt{\frac{2 y+1}{y+2}}+\sqrt{\frac{2 z+1}{z+2}}
$$
where $x=\frac{a^2}{bc}>0, y=\frac{b^2}{ac}>0, z=\frac{c^2}{ab}>0$.
Note that functions
$$f(w)=\sqrt{\frac{2 w+1}{w+2}}$$
are strictly increasing for $w\in[0,\infty)$.
Without the loss of generality assume that $x\geq y\geq z$. Hence... | {
"language": "en",
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"source": "stackexchange",
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Nonlinear Differential Equation of High Degree Any help, please? How can I start to solve them? I tried to use $y'=p$. Also I tried $x=e^x$ and so many methods, but I couldn't reach them to the end. I always got blocked in the middle. Thanks in advance.
First equation: $$6x^2y-6y'^2+(12x^2-3x^3)y'+x^5-6x^4=0.$$
Second ... | FIRST EQUATION :
$$6x^2y-6y'^2+(12x^2-3x^3)y'+x^5-6x^4=0.$$
By inspection one can see a particular solution :
$$y=\frac13 x^3\quad\implies\quad y'=x^2$$
$6x^2(\frac13 x^3)-6(x^2)^2+(12x^2-3x^3)(x^2)+x^5-6x^4=$
$=2x^5-6x^4+12x^4-3x^5+x^5-6x^4=0.$
This draw us to the change of function :
$$y(x)=\frac13 x^3+u(x)\quad\impl... | {
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"source": "stackexchange",
"question_score": "1",
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Finding $\frac{c\sin(A-B)}{a^2-b^2}-\frac{b\sin(C-A)}{c^2-a^2}$
If $a, b$ and $c$ (all distinct) are the sides of a triangle ABC opposite to the angles $A, B$ and $C$, respectively, then $\frac{c\sin(A-B)}{a^2-b^2}-\frac{b\sin(C-A)}{c^2-a^2}$ is equal to $?$
By opening, $\sin(A-B)$ as $\sin A\cos B-\cos A\sin B$ and ... | Use sine and cosine rules to evaluate
\begin{align}
& \frac{\sin(A-B)}{\sin(A-C)}
= \frac{\sin A\cos B - \cos A \sin B}{\sin A\cos C - \cos A \sin C}\\
=& \frac{a\frac{a^2+c^2-b^2}{2ac} - b\frac{b^2+c^2-a^2}{2bc} }
{a\frac{a^2+b^2-c^2}{2ab} - c\frac{b^2+c^2-a^2}{2bc} }
=\frac{\frac1c [(a^2+c^2-b^2 )- (b^2+c^2-a^2)] ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3663019",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
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My first linear algebra proof - is it accurate and written correctly? This is the first proof exercise from "Linear Algebra as an Introduction to Abstract Mathematics." It also happens to be my first proof (outside of proving things like properties of integers and trigonometric identities). I'm hoping someone can criti... | You can apply the following row operations in order to get that
\begin{align*}
\begin{cases}
ax_{1} + bx_{2} = 0\\\\
cx_{1} + dx_{2} = 0
\end{cases} & \Longrightarrow
\begin{cases}
acx_{1} + bcx_{2} = 0\\\\
cx_{1} + dx_{2} = 0
\end{cases}\\\\
& \Longrightarrow
\begin{cases}
(acx_{1} + bcx_{2}) - (acx_{1} + adx_{2}) = ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3667415",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Integer solutions to multivariate polynomial I'm looking for a way to show that an equation like $$15(1+x+x^{2})(1+y+y^{2})=16(xy)^{2}-1$$ has no solutions with x and y odd positive integers. The 16 and 15 can be changed but I'm trying to generalize it. Any help is appreciated.
| Factorization should be useful, but another idea is to note that
if the equation is $a(x^2+x+1)(y^2+y+1)=b(xy)^2-1$ and $a<b$, then $min(x,y)$ is upper-bounded by a function of $a,b$ and it is sufficient to examine those values (and solve for the other).
Specifically, here is one (not necessarily sharp) bound: $min(x,y... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Polynomial transformations and Vieta's formulas
Let $f(x)$ be a monic, cubic polynomial with $f(0)=-2$ and $f(1)=−5$. If the sum of all solutions to $f(x+1)=0$ and to $f\big(\frac1x\big)=0$ are the same, what is $f(2)$?
From $f(0)$ I got that $f(x)=x^3+ax^2+bx-2$ and from $f(1)=-5$ that $a+b = -4$ however I'm not sur... | If $\alpha$ is a root of $f(x)$ then $\alpha - 1$ is a root of $f(x+1)$ (and vice-versa). So the sum of roots of $f(x+1)$ is the sum of roots of $f(x)$, minus $3$ which by Vieta' formula is $-a$ minus $3 =-a-3$.
Now, the roots of $f(\frac 1x) = \frac{-2x^3+bx^2+ax+1}{x^3}$ are also the roots of $-2x^3+bx^2+ax+1$, the ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Prove that $x^2+y^2-\sqrt{2}xy < 4$ provided $1\le x,y \le \frac{5}{2}$. If possible I would like an elegant solution to the following problem:
Let $x,y\in\mathbb{R}$ such that $1\le x \le \frac{5}{2}$ and $1\le y \le \frac{5}{2}$. Prove that
$x^2+y^2-\sqrt{2}xy < 4$
I'm aware that you can use Lagrange multipliers bu... | The hessian of the function $f(x,y)=x^2+y^2-\sqrt2xy$ is
$$\begin{vmatrix}
2&-\sqrt2\\
-\sqrt2 &2
\end{vmatrix}=2>0,
$$
hence $f(x,y)$ is convex and therefore takes its maximal value on an extreme point of the domain. A simple check (which you have already done) shows that the maximal value is
$$f\left(1,\frac52\right... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3674487",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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Generator of a solution space In $\mathbb{F}_3$, consider:
$a+b+2c+d=0$
$2a+c+d=0$
$2a+b+c=0$
I got $a=0$, $c=-d$ and $b=d$, so we have $\mathbb{L}=\left \{ t \begin{pmatrix} 0\\1\\-1 \\ 1 \end{pmatrix} : t \in \mathbb{F}_3
\right \}$. Does this mean that $(0,1,-1,1)^T$ is the generator of the solution space?
| And easy way: form and then bring to echelon form the coefficients' matrix:
$$\begin{pmatrix}1&1&1&1\\
2&0&1&1\\
1&1&1&0\end{pmatrix}\longrightarrow\begin{pmatrix}1&1&1&1\\
0&2&0&0\\
0&0&0&2\end{pmatrix}\implies\begin{cases}d=0=b\\{}\\a=-c=2c\end{cases}$$
so the general solution is
$$\left\{\;\begin{pmatrix}2c\\0\\c\\0... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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Suppose $f \circ g(x) = \frac{x^2-6x+2}{x+1}$ and $g(x) = 1 - x$. Then $f(-1)$ is equal to... I tried to substitute the value of $g(x)$ to every $x$ in $f \circ g(x) = \frac{x^2-6x+2}{x+1}$ and ended up with $\frac{x^2+4x-3}{-x+2}$.
Although honestly I do not know if that helps, or what I could do next.
I thought mayb... | We have \begin{align*}f(-1) &= (f \circ g)(g^{-1}(-1)) \\ &= (f \circ g)(1-(-1)) \qquad \text{(as $g^{-1}(x)=1-x$)} \\ &= \frac{2^2-6\cdot 2 + 2}{2+1} \\ &= -2\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3681089",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
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Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.