Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Proving Trigonometric Equality I have this trigonometric equality to prove:
$$\frac{\cos x}{1-\tan x}-\frac{\sin x}{1+\tan x}=\frac{\cos x}{2\cos^2x-1}$$
I started with the left hand side, reducing the fractions to common denominator and got this:
$$\frac{\cos x+\cos x\tan x-\sin x+\sin x\tan x}{1-\tan^2x}\\=\frac{\cos... | Here's a hint: $$\frac{\cos x + \frac{\sin x}{\cos x}^2}{1 - \left(\frac{\sin x}{\cos x}\right)^2} = \frac{\frac{\sin x^2 + \cos x^2}{\cos x}}{\frac{\cos x^2 - \sin x^2}{\cos x^2}}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2299015",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 7,
"answer_id": 3
} |
Number equal to the sum of digits + product of digits) Are every number equal to (sum of digits + product of digits) in a given base only two digits long ?
Thought about limiting like this :
$$b^{(n - 1)} \leq N = \text{Product digits} + \text{Sum digits} \leq (b - 1)*n + (b - 1)^{(n)}$$
Obviously the set is finite (w... | Bear with me:
For $k \ge 1; b\ge 2$ $(b - 1)^k \le b^k + (-1)^k$. Pf: induction: For $k =1$ then $(b-1)^1 = b^1 + (-1)^1$. For $k = 2$ then $(b-1)^2 = b^2 -2b +1 < b^2 + 1$. If $(b -1)^n \le b^n + (-1)^n$ then $$(b-1)^{n+1} = (b-1)^n(b-1) \le (b^n + (-1)^n)(b-1) = b^{n+1} + (-1)^nb - b^n +(-1)^{n+1} = b^{n+1} + (-1)^{n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2304977",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Galois correspondence for the splitting field of $x^3-3$ over $\mathbb{Q}$ I am struggling to find the fixed fields, am I missing a trick here or is it really this nitty gritty?
By the tower law $[\mathbb{Q}(\sqrt[3]{3}, \zeta_3):\mathbb{Q}] = [(\mathbb{Q}(\sqrt[3]{3}, \zeta_3): \mathbb{Q}(\sqrt[3]{3})][\mathbb{Q}(\sqr... | How can $\sigma_4$ and $\sigma_5$ have the same fixed field?
The fixed field of $\sigma_4$ is $\Bbb Q(\alpha_3)=\Bbb Q(\zeta_3^2
\sqrt[3]3)$.
The fixed field of $\sigma_5$ is $\Bbb Q(\alpha_2)=\Bbb Q(\zeta_3
\sqrt[3]3)$.
The fixed field of $\sigma_6$ is $\Bbb Q(\alpha_1)=\Bbb Q(\sqrt[3]3)$.
The fixed field of $\sigma_2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2309162",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Radius of converge of Laurent series for $\frac{1}{\sin z}$ I want to show that the radius of convergence of Laurent series for $\frac{1}{\sin z}$ is $\pi$
I showed that:
\begin{align}
\frac 1 {\sin z} & = \frac 1 {z - \frac{z^3}{3!} + \frac{z^5}{5!} + \cdots } = \frac 1 {z\big(1 - (\frac{z^2}{3!} - \frac{z^4}{... | Since $\text{Res}\left(\frac{1}{\sin z},z=k\pi\right) = (-1)^m $ for any $k\in\mathbb{Z}$, we have
$$ \frac{1}{\sin z} = \frac{1}{z}+\sum_{m\geq 1}(-1)^m\left(\frac{1}{z-m\pi}+\frac{1}{z+m\pi}\right) \tag{1}$$
over any compact subset of $\mathbb{C}\setminus\pi\mathbb{Z}$. In particular:
$$ \frac{1}{\sin z}-\frac{1}{z} ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2309549",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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AM-GM Inequality concept challenged! If $a,b,c$ are positive numbers and $abc=9$, then find the minimum value of $(1+a)(1+b)(1+c)=?$
Someone's approach is
but for this to be possible, $a=b=c=1$. which doesn't satisfies $abc=9$.
Where does the above method go wrong and what should be the correct answer?
| By AM-GM
\begin{align}
1+a &\geq 2 \sqrt{a} \\
1+b &\geq 2 \sqrt{b} \\
1+c &\geq 2 \sqrt{c}
\end{align}
Thus
\begin{align}
(1+a)(1+b)(1+c) &\geq 2^3 \sqrt{a} \sqrt{b}\sqrt{c} \\
&= 8.\sqrt{9} \ldots
\end{align}
Can you take it from here?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2311772",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Arclength, finding definite integral Find the length of the segment between $x=5$ and $x=2$ for the following curve.
$$f(x)= \frac{1}{2}x^2-\frac{1}{4}\log\left(x\right)$$
My working so far.
$$f'\left(x\right)=x-\frac{1}{4x}.$$
Then
$$L = \int _2^5\:\sqrt{1+\left(f'\left(x\right)\right)^2}\,dx
=\int _2^5\:\sqrt{x^2+\fr... | Hint. We have that
$$1+\left(x-\frac{1}{4x}\right)^2=\frac{(4x)^2+16x^4-8x^2+1}{(4x)^2}=\frac{16x^4+8x^2+1}{(4x)^2}=\left(\frac{4x^2+1}{4x}\right)^2.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2311852",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Prove that the product is never a perfect square
Prove that for nonnegative integers $x_1,\ldots,x_{2011}$ and $y_1,\ldots,y_{2011}$ the product $$(2x_1^2+3y_1^2)(2x_2^2+3y_2^2) \cdots (2x_{2011}^2+3y_{2011}^2)$$ is never a positive perfect square.
I thought about generalizing this question to any odd subscript $n$ i... | Case n=1
Let $z^2=2x^2+3y^2$
$z^2=2x^2 [3]$.
If x is not divisible by 3, then $z^2=2[3]$ which is impossible. Then $x$ is divisible by 3, and consequently $z$ is also a multiple of 3.
Let us look at the prime decomposition of $z$ such that $z^2=2x^2+3y^2$. 3 is a prime factor of $z$ and let us not $p$ its power: $z=3^p... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2312774",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
"answer_id": 0
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$1 + \frac{1}{3}\frac{1}{4} + \frac{1}{5}\frac{1}{4^2} + \frac{1}{7}\frac{1}{4^3} + ..........$ $1 + \frac{1}{3}\frac{1}{4} + \frac{1}{5}\frac{1}{4^2} + \frac{1}{7}\frac{1}{4^3} + ......$
Can anyone help me out how to solve this.
My try : I was thinking about the expansion of $tan^{-1}x$. But in that series positive ... | A different approach would be to use binomial theorem . Consider the series expansion of $\ln (\frac {1}{1-x}),\ln (\frac {1}{1+x})$ using binomial expression for negative index we have $\ln (\frac {1}{1-x})-\ln (\frac {1}{1+x})=2 (x+\frac {x^3}{3}+\frac {x^5}{5}+.. ) $ now pull out $x $ common from $RHS $ and put $x=\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2313339",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Show which of $6-2\sqrt{3}$ and $3\sqrt{2}-2$ is greater without using calculator How do you compare $6-2\sqrt{3}$ and $3\sqrt{2}-2$? (no calculator)
Look simple but I have tried many ways and fail miserably.
Both are positive, so we cannot find which one is bigger than $0$ and the other smaller than $0$.
Taking the fi... | Here's yet another way, for those who aren't comfortable with the $\gtrless$ or $\sim$ notation.
We can use crude rational approximations to $\sqrt 2$ and $\sqrt 3$.
$$\begin{align}
\left(\frac{3}{2}\right)^2 = \frac{9}{4} & \gt 2\\
\frac{3}{2} & \gt \sqrt 2\\
\frac{9}{2} & \gt 3\sqrt 2
\end{align}$$
And
$$\begin{align... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2317625",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 7,
"answer_id": 5
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If $A$ is a non-square matrix with orthonormal columns, what is $A^+$? If a matrix has orthonormal columns, they must be linearly independent, so $A^+ = (A^T A)^{−1} A^T$ . Also, the fact that its columns are orthonormal gives $A^T A = I$. Therefore,
$$A^+ = (A^T A)^{−1} A^T = (I)^{-1}A^T = A^T$$
Thus, $A^+ = A^T$. Am ... | Problem statement
Start with a matrix $$A\in\mathbb{C}^{m\times n}$$ where $m>n$, and
a valid statement for the pseudoinverse matrix
$$
\mathbf{A}^{+} = \left( \mathbf{A}^{*} \mathbf{A} \right)^{-1} \mathbf{A}^{*}
$$
We know (see links) that this matrix is a left inverse:
$$
\mathbf{A}^{+} \mathbf{A} = \mathbf{I}_{n}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2318428",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 0
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Jordan form dependency on choosing a specific kernel. Let $A=\begin{pmatrix}2 && 2 && 3 \\ 1 && 3 && 3\\ -1 && -2 && -2\end{pmatrix}$
Finding the eigenvalues produce that $\lambda_1=\lambda_2=\lambda_3=1$
The geometry of $A$ is $2$, therefore exists only two blocks, and therefore the only possibility is $J_1(1),J_2(1)... | The procedure for finding the basis is to start from a generalized eigenvector, that is,
a vector $v_2$ such that $(A-I)^2v_2=0$, but $(A-I)v_2\ne 0$. Since $(A-I)^2=0$, you can take any vector in $\Bbb{R}^3\setminus Ker(A-I)$.
Then $v_1=(A-I)v_2$ will be an eigenvector of $A$.
Note that $Im(A-I)$ is one-dimensional... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2318640",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Find all real coefficient polynomial such $f(x)|f(x^2-2)$
Find all real coefficient polynomial with the degree is $3$,and such
$$f(x)|f(x^2-2),\forall x\in Z $$
I try: let $f(x)=ax^3+bx^2+cx+d$,then $$f(x^2-2)=ax^6-6ax^4+12ax^2-8a+bx^4-4bx^2+4b+cx^2-2c+d$$,then such
$$(ax^3+bx^2+cx+d)|ax^6-6ax^4+12ax^2-8a+bx^4-4bx... | Suppose $f(x) \mid f(x^2 - 2)$. Ghartal's argument has already shown that if $x$ is a root of $f$, then $x^2 - 2$ is also a root of $f$. In fact, by the same argument, we have that for all $x \in \mathbb{C}$, the multiplicity of $x^2 - 2$ as a root of $f$ is at least the multiplicity of $x$ as a root of $f$.
I claim th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2319569",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Find Angle Between Two Curves at Point of Intersection Find angle between these two curves at point of intersection :
$$K_1: x^2y^2 + y^4 = 1$$ and
$$K_2 : x^2 + y^2 = 4 $$
Thanks!
| To find the points of intersection, we see that the equation of the curve $K_1$ can be written as:
\begin{align*}
x^2 y^2 + y^4 &= 1 \\
y^2 (x^2 + y^2) &= 1
\end{align*}
Since the points of intersection lies on both curve $K_1$ and $K_2$, it must satisfy both the above equation and the equation of $K_2$, meaning:
\begi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2320284",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
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To find the last digit of the number $4^{7^5}$? To find the last digit of the number $4^{7^5}$, I use the following method.
first, we know $4\equiv_{10}4,4^2\equiv_{10}6,4^3\equiv_{10}4,...,4^{2n}\equiv_{10}6,4^{2n+1}\equiv_{10}4$
Then$7^1\equiv_21,7^2\equiv_21,7^3\equiv_21,...,7^5\equiv_21$
Therefore, $4^{7^5}\equiv4... | Your method is correct. Another way is to consider the easy fact that
$$2^{4n+k}= \begin{cases}2\pmod {10}\text{ for}\space k=1\\4\pmod {10}\text{ for}\space k=2\\8\pmod {10}\text{ for}\space k=3\\6\pmod {10}\text{ for}\space k=0\end{cases}$$ and the calculation $$7^5=16807$$ Then $$4^{7^{5}}=2^{33614}=2^{33612+2}=4\pm... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2323684",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
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Find $P$ such that $P^TAP = D$ where $D$ is a diagonal matrix.
Let $$A = \left(\begin{array}{cc} 2&3 \\ 3&4 \end{array}\right) \in
M_n(\mathbb{C})$$
Find $P$ such that $P^TAP = D$ where $D$ is a diagonal matrix.
How can I find $P$? I am doing Gauss but it does not work?$$A = \left(\begin{array}{cc|cc} 2&3&1&0\\ 3&4&... | I think Jose comes closest to solving this problem in the way that I would, but his answer doesn't expand too much on the why of it all, so I'm writing up a separate answer.
If there is such a $D$ and $P$ such that $A = P^T D P$, then we say that $A$ is diagonalizable. One can prove that a valid diagonalization of $A$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2324359",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 1
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Infinitely many positive integers of the form $1998k+1$ such that all digits in their decimal representation are equal
Prove that there exist infinitely many positive integers of the form $1998k+1, k \in \mathbb{N},$ such that all digits in their decimal representation are equal.
We need to find integer solutions to... | Credit of the full answer to Mlazhinka Shung Gronzalez LeWy.
Let $N = 10^{3n} - 1 = 999\ldots 9$. By factorisation, $N$ is a multiple of $999$, and the other quotient is $$\frac{10^{3n}-1}{10^3-1} = 10^{3(n-1)}+ \cdots + 10^6 + 10^3+1.$$
I further would like the quotient be a multiple of $9$, so that $N$ is a multiple ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2324423",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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Let $a^4+b^4+c^4=3$. Prove that $a^2+b^2+c^2\geq a^2b+b^2c+c^2a$ Let $a^4+b^4+c^4=3$. Prove that $a^2+b^2+c^2\geq a^2b+b^2c+c^2a$
My proof :D
We have the inequality
$\sum_{cyc}^{ }a^2.\sum_{cyc}^{ }a\geq \sum_{cyc}^{ }a^2b $
which is equivalent to $\sum_{cyc}^{ }b(a-b)^2$ (true)(1)
In another side, by AM-GM, we have: $... | Let $0\le z \le 3$. Then, we have the inequality $z^2 - 2z - 3 \le 0$. We need this because we have $\frac{a^2+b^2+c^2}{3} \le \left(\frac{a^4 + b^4 + c^4}{3}\right)^{1/2} = 1$.
Now, $a^2b+b^2c+c^2a \le (a^2+b^2+c^2)^{1/2}(a^2b^2+b^2c^2+c^2a^2)^{1/2}$, thus we need to prove $a^2b^2+b^2c^2+c^2a^2 \le a^2+b^2+c^2$. We h... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2324574",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
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Forming a quadratic equation with the given roots Question: If $\frac{p^2}{q}$ and $\frac{q^2}{p}$ are the roots of the equation $2x^2+7x-4=0$, find the equation whose roots are $p$ and $q$($p+q$ is real).
My attempt: The required equation is $x^2-(p+q)x+pq$
$(\frac{p^2}{q})(\frac{q^2}{p})=\frac{-4}{2}$
$pq=-2$
$\frac{... | $$\frac{p^2}{q}+\frac{q^2}{p}=-\frac{7}{2}$$ and $pq=-2$.
Thus, $$p^3+q^3=7$$ or
$$(p+q)((p+q)^2+6)=7$$ or
$$(p+q)^3+6(p+q)-7=0,$$
which gives $p+q=1$ and we get the answer:
$$x^2-x-2=0.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2324912",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find this limit. Compute the value of the limit :
$$
\lim_{x\to\infty}{\frac{1-\cos x\cos2x\cos3x}{\sin^2x}}
$$
I've tried simplifying the expression to
$$
\lim_{x\to\infty}\frac{-8\cos^6x+10\cos^4x-3\cos^2x+1}{\sin^2x}
$$
But I don't know what to do after this.
| Generalization:
For $$\lim_{x\to0}\dfrac{1-\cos ax\cos bx\cos cx}{\sin^2x}=\lim_{x\to0}\dfrac1{1+\cos ax\cos bx\cos cx}\cdot\lim_{x\to0}\dfrac{1-\cos^2ax\cos^2bx\cos^2cx}{\sin^2x}$$
$$=\dfrac12\cdot\lim_{x\to0}\dfrac{1-(1-\sin^2ax)(1-\sin^2bx)(1-\sin^2cx)}{\sin^2x}$$
$$=\dfrac12\cdot\lim_{x\to0}\left(\left(\dfrac{\sin ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2325217",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 7,
"answer_id": 5
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Shortest distance between an ellipse and a hyperbola.
Suppose we have a hyperbola $xy=c$ $(c\neq0)$ and an ellipse $x^2/a^2+y^2/b^2=1$ $(a>b>0)$ that do not intersect. What is a quick way to calculate the shortest distance between these two curves?
I thought of setting a point $(a\cos\theta,b\sin\theta)$ on the ellip... | There doesn't seem to be a nice closed-form solution to this, but the following might be helpful..
For an ellipse $\frac {x^2}{a^2}+\frac {y^2}{a^2}=1$, the equation of the normal at point $P(a \cos\theta, b\sin\theta)$ on the ellipse is
$$\frac {a\sin\theta}{b\cos\theta}x-y=a\left(\frac{a^2-b^2}{ab}\right)\sin\theta\t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2325673",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
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Proof that $x^3+y^4=z^{31}$ has infinitely many solutions This is a question from RMO 2015.
Show that there are infinitely many triples (x,y,z) of integers such that $x^3+y^4=z^{31}.$
This is how I did my proof:
Suppose $z=0,$ which is possible because $0$ is an integer. Then $x^3+y^4=0 \Rightarrow y^4=-x^3.$ Now, supp... | A small variation on the 'official' solution: instead of proving that the equation $12r+1=31k$ has infinitely many positive integer solutions we find a particular one, let's say $k=7,r=18$; it follows that $x=2^{4\cdot 18},y=2^{3\cdot 18},z=2^7$ is a particular solution for our equation and from here it is easy to noti... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2326050",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 1
} |
Calculate the sum of the series $\sum_{1\leq aCalculate the sum of the series:
$$S = \sum_{1\leq a<b<c \\a,b,c\in \mathbb{N}} \frac{1}{2^a 3^b 5^c}$$
My attempt:
$$S = \sum_{1\leq a<b<c \\a,b,c\in \mathbb{N}} \frac{1}{2^a 3^b 5^c} = \sum _{c=3}^{\infty } \sum _{b=2}^{c-1} \sum _{a=1}^{b-1} \frac{1}{2^a 3^b 5^c}$$
Is it... | $$
S=
\sum _{c=3}^{\infty } \sum _{b=2}^{c-1} \sum _{a=1}^{b-1} \frac{1}{2^a 3^b 5^c}
=
\sum _{c=3}^{\infty } \sum _{b=2}^{c-1} \left(1-\frac{1}{2^{b-1}}\right) \frac{1}{3^b5^c}
=
\sum _{c=3}^{\infty } \sum _{b=2}^{c-1}\frac{1}{3^b5^c}-2\sum _{c=3}^{\infty } \sum _{b=2}^{c-1}\frac{1}{6^b5^c},
$$
let us say $S_1-2S_2$. ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2327827",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
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Help with polynomial problem, please I am stuck on the following:
If the equation
$$x^4-ax^3+2x^2-bx+1=0$$
has a real root, then show that
$$a^2 + b^2 \geq 8\,.$$
Can I have a hint to go ?
EDIT
My solution after considerable effort:
Since $\,x=0\,$ is not a root of the polynomial we can divide it by $\,x^2$. Th... | since $x=0$ isn't a root of the polynomial ; we can divide it by $x^2 $ yields
$$x^2-ax+2-b/x +1/x^2=0$$
$$\left(x-\frac{a}{2}\right)^2 + \left(\frac{1}{x} - \frac{b}{2}\right)^2= \frac{a^2}{4}+\frac{b^2}{4}-2 = \frac {a^2+b^2-8}{4} \;\;\gt\;\;0$$
So
$$\frac {a^2+b^2-8}{4} \ge 0 \;\;\implies\;\; a^2+b^2-8 \ge 0 \;\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2327928",
"timestamp": "2023-03-29T00:00:00",
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"answer_id": 0
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How to prove this inequality $\frac{a}{b+c} + \frac{b}{c+a} + \frac{c}{a+b} \ge 1.5$? Given $a$, $b$, and $c$ are three positive real numbers. How to prove that sum of $\frac{a}{b+c}$, $\frac{b}{c+a}$ and $\frac{c}{a+b}$ is greater than or equal to $1.5$?
| by Cauchy Schwarz we have
$$\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\geq \frac{a^2}{ab+ac}+\frac{b^2}{bc+ab}+\frac{c^2}{ac+bc}\geq \frac{(a+b+c)^2}{2(ab+bc+ca)}\geq \frac{3}{2}$$ if and only if $$a^2+b^2+c^2\geq ab+bc+ca$$ which is true.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2328829",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Which functions have a compositional inverse which is the antiderivative? Which functions $f$ of one variable have a compositional inverse which is the antiderivative of $f$?
| Following up Arthur's suggestion and assuming a form $a x^{b}$ for $f$:
Let $g(x) = f^{-1}(x)$ and let $g'(x) = f(x)$
$$
f = a x^b \Rightarrow g = \frac{a}{b+1} x^{b+1}
$$
I accept I've ignored a possible integration constant for $g$ above.
$$
f (g (x)) = x
\Rightarrow
f (\frac{a}{b+1} x^{b+1}) = x
$$
$$
a \left(
\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2328948",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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If $x^3 - 5x^2+ x=0$ then find the value of $\sqrt {x} + \dfrac {1}{\sqrt {x}}$ If $x^3 - 5x^2+ x=0$ then find the value of $\sqrt {x} + \dfrac {1}{\sqrt {x}}$
My Attempt:
$$x^3 - 5x^2 + x=0$$
$$x(x^2 - 5x + 1)=0$$
Either,
$x=0$
And,
$$x^2-5x+1=0$$
??
| $$\sqrt{x}+\frac 1{\sqrt{x}} = \frac{(x+1)}{\sqrt{x}}$$
$$\left(\frac{x+1}{\sqrt{x}}\right)^2 = \frac{(x^2+2x+1)}x = \frac{(x^2-5x+1)+7x}x = 7x/x = 7$$($x$ is not equal to $0$)
Therefore, $$\sqrt{x}+\frac 1{\sqrt{x}} = \sqrt{7}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2329371",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 4
} |
Solve $x^2 + (y-1)^2 + (x-y)^2 - \frac{1}{3} = 0.$ I came across an interesting equation with two variables $x,y\in\mathbb{R}$, $x^2 + (y-1)^2 + (x-y)^2 - \frac{1}{3} = 0.$
This, once expanded, can be simplified to $3x^2 - 3xy + 3y^2 - 3y + 1=0.$
How can one proceed to solve it algebraically? The solution according to ... | Solving the quadratic equation $x^2+(y-1)^2+(x-y)^2=\frac13$ (where the unknown is $x$) gives you$$x=\frac{\left(3 y-\sqrt{3} \sqrt{-9 y^2+12y-4}\right)}6.$$But $-9y^2+12y-4=-(3y-2)^2$. Therefore it is a real number greater than or equal to $0$ if and only if $y=\frac23$ and, when that happens, $x=\frac13$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2329534",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
} |
Prove that $\frac{1}{2\pi i}\int_\mathcal{C} |1+z+z^2+\cdots+z^{2n}|^2~dz =2n$ where $\mathcal{C}$ is the unit circle On the generalization of a recent question, I have shown, by analytic and numerical means, that
$$\frac{1}{2\pi i}\int_\mathcal{C} |1+z+z^2+\cdots+z^{2n}|^2~dz =2n$$
where $\mathcal{C}$ is the unit circ... | $$
\frac{1}{2\pi i}\int_0^{2\pi}\sum_{l=0}^{2n}e^{il\theta}\sum_{m=0}^{2n}e^{-im\theta}ie^{i\theta}d\theta=\frac{1}{2\pi}\sum_{l,m=0}^{2n}\int_0^{2\pi}\exp(i\theta(l-m+1))d\theta
$$
I think the last integral is zero unless $l-m+1=0$, which happens $2n$ times when $l$ and $m$ go between 0 and 2n.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2331657",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 3
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If $ax^2+\frac{b}{x}\ge c$ for all positive $x$ where $a>0$ and $b>0,$ then show that $27ab^2\ge4c^3$ If $ax^2+\frac{b}{x}\ge c$ for all positive $x$ where $a>0$ and $b>0,$ then show that $27ab^2\ge4c^3$
I considered the function $f(x)=ax^2+\frac{b}{x}$ and I put $f'(x)=0$ to find $x^3=\frac{b}{2a}$
I am stuck here.
| We have $x=(\frac {b}{2a})^{\frac {1}{3}} $. From second derivative we find that at this point we have a minima as $f''(x)>0$ . Thus rearranging original equation we have $ax^3+b\geq cx $ plugging in value we have $\frac {3b}{2}\geq c.(\frac {b}{2a})^{\frac {1}{3}} $ cubing and rearranging $27ab^2\geq 4c^3$. Hence the... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2333879",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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$\sin(2\arcsin(\frac{1}{3}))=? $ A broad one I have written it as $$\sin(2\arcsin(\frac{1}{3}))=2\sin(\arcsin(\frac{1}{3}))\cos(\arcsin(\frac{1}{3})).$$Then, solved for $$\cos(\arcsin(\frac{1}{3}))=\frac{\sqrt{10}}{3}$$ $$\arcsin(\frac{1}{3})=\theta$$$$\sin\theta=\frac{1}{3}\Rightarrow\cos\theta=\frac{\sqrt{10}}{3}.$$B... | $$\sin2\arcsin\frac{1}{3}=2\cdot\frac{1}{3}\cdot\sqrt{1-\frac{1}{9}}=\frac{4}{9}\sqrt2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2334116",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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Reduction of order of differential equation! Given that y=x is a solution to the following differential equation
$$(x^2-x+1)\frac{d^2y}{dx^2}-(x^2+x)\frac{dy}{dx}+(x+1)y=0$$
Find a linearly independent solution to the new differential equation by reducing the order. Write a general solution.
I am stuck somewhere here.... | We are given
$$\tag 1 (x^2-x+1) y'' - (x^2 + x) y' + (x+1) y = 0$$
You made a good Reduction of Order substitution using $y = v x \implies y' = v + v' x$ and we reduce $(1)$ to
$$\tag 2 x(x^2-x+1)\dfrac{d^2v}{dx^2}+(-x^3+x^2-2x+2)\dfrac{dv}{dx}=0$$
Next, you correctly let $\dfrac{dv}{dx} = w$ and reducing $(2)$ we get ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2335342",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Significance of the Triangle Inequality After working through a few problems regarding the Cauchy Integral Formula, I'm still a little confused on the significance of the triangle inequality. Why do we use it and what information does it tell us?
See the following example below.
Determine $\int_{0}^{\infty} \frac{x^... | A fix on this argument, motivated by Fightclub's observation, is this: We know that for every $R>1$:
$$ \underbrace{\int_{C_1} \frac{z^2}{(z^2+1)^2}dz}_{\int_{-R}^R \frac{x^2}{(x^2+1)^2}dx} + \int_{C_2} \frac{z^2}{(z^2+1)^2}dz = \pi/2 $$
Hence:
$$\lim_{R\rightarrow\infty} \int_{-R}^R \frac{x^2}{(x^2+1)^2}dx + \lim_{R... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2335550",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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How to properly simplify and trig substitute this integral? $\int\frac{{\sqrt {25-x^2}}}x\,dx$
I see that it is the form $a-x^2$.
Let $x = 5\sin\theta$
Then substitute this $x$ value in and
get $\int\frac {\sqrt {25-(5 \sin\theta)^2}}{5 \sin\theta}\, d\theta $
Take the root, simplified to $\int\frac { {5-(5 \sin\theta... | If $x = 5\sin\theta$ then $dx = 5\cos\theta\, d\theta.$ That was neglected.
You also replaced $\sqrt{25 - (5\sin\theta)^2}$ with $5 - 5\sin\theta.$ That is also not correct. Notice that, for example,
$$
\sqrt{5^2 - 3^3} = \sqrt{25-9} = \sqrt{16} = 4 \ne 5-3, \text{ so } \sqrt{5^2-3^2} \text{ differs from } 5-3.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2335930",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Stability of a fixed point in a nonlinear system with no linear part The following nonlinear system has a fixed point in the origin. I want to know if this fixed point is stable:
\begin{align*}
&\dot{\alpha}=\alpha^2-2\beta(\alpha+\beta)\\
&\dot{\beta}=\beta^2-2\alpha(\alpha+\beta)
\end{align*}
Note that the linearizat... | This is mostly a recap of the observations made in the comments, plus some more analysis, because I think it's a nice problem to analyse.
First, both the functional form of the system and the reflection symmetry (see also the phase plane) suggest it's a good idea to introduce $x = \alpha+\beta$, $y = \alpha-\beta$, to ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2336200",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
Proving trigonometric identity $\cos^6A+\sin^6A=1-3 \sin^2 A\cos^2A$ Show that
$$\cos^6A+\sin^6A=1-3 \sin^2 A\cos^2A$$
Starting from the left hand side (LHS)
\begin{align}
\text{LHS} &=(\cos^2A)^3+(\sin^2A)^3 \\
&=(\cos^2A+\sin^2A)(\cos^4A-\cos^2A\sin^2A+\sin^4A)\\
&=\cos^4A-\cos^2A\sin^2A+\sin^4A
\end{align}
Can a... |
(a+b)^3= a^3 + b^3 + 3ab(a + b)
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 2
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Solve three equations with three unknowns Solve the system:
$$\begin{cases}a+b+c=6\\ab+ac+bc=11\\abc=6\end{cases}$$
The solution is:
$a=1,b=2,c=3$
How can I solve it?
| Without Vieta's formulas, the polynomial can be derived the hard way by elimination. From the first and last equations:
$$
b+c=6-a \\
bc = 6 / a
$$
Substituting the above into the middle equation:
$$
11=a(b+c)+bc=a(6-a)+6/a \;\;\iff\;\; a^3-6 a^2+11a - 6 = 0
$$
By inspection, the latter equation has $a=1$ as a root, th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2339834",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 4
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Trig Identity Proof $\frac{1 + \sin\theta}{\cos\theta} + \frac{\cos\theta}{1 - \sin\theta} = 2\tan\left(\frac{\theta}{2} + \frac{\pi}{4}\right)$ I've been working on this for like half an hour now and don't seem to be getting anywhere. I've tried using the double angle identity to write the LHS with $\theta/2$ and have... | Let $\dfrac\theta2+\dfrac\pi4=y\implies\theta=2y-\dfrac\pi2$
$$\sin\theta=\cdots=-\cos2y,\cos\theta=\cdots=\sin2y$$
$$\dfrac{1+\sin\theta}{\cos\theta}=\dfrac{1-\cos2y}{\sin2y}=\dfrac{2\sin^2y}{2\sin y\cos y}=?$$
$$\dfrac{\cos\theta}{1-\sin\theta}=\dfrac{\sin2y}{1+\cos2y}=?$$
See also : Solve $\frac{\cos x}{1+\sin x} + ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2341368",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
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How to find this simplification integral involving products of roots? Consider the following integral
$$
f = \int_0^1 \frac{1}{\sqrt{-\frac{1}{2} \, t^{2} + 1} \sqrt{-t^{2} + 1}} \mathrm \,dt.
$$
If we change variable by letting $x^2=t^2/(2-t^2)$, then we have
$$
f = \int_0^1 \sqrt{2} \cdot\sqrt{\frac{1}{1-x^{4}}} \mat... | $$x^2=\frac{t^2}{2-t^2}$$
$$t=x \sqrt{\frac{2}{x^2+1}}$$
$$dt = \sqrt{2} \sqrt{\frac{1}{x^2+1}} \left(1-\frac{x^2}{x^2+1}\right) dx$$
So the integrand becomes is
$$\frac{\sqrt{2} \left(\sqrt{\frac{1}{x^2+1}} \left(1-\frac{x^2}{x^2+1}\right)\right)}{\sqrt{1-\frac{2 x^2}{x^2+1}} \sqrt{1-\frac{x^2}{x^2+1}}}$$
and finally... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2341785",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
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What is a simple (or not) way of finding the lengths of the diagonals of this rhombus? I recently taught a group of geometry students about properties of rhombuses, and gave them a set of homework exercises created by a previous instructor which included the following problem.
If a rhombus has sides of length $25$ mm ... | As I mentioned in the question, here are the two ways that I have found to solve this problem.
Method 1: Trigonometry
If we draw in the diagonals, we get the following image, and we know that each diagonal bisects the other as well as the angles of the rhombus:
That means that we can say
\begin{align*}
a &= 25\sin(22.5... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2342388",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 7,
"answer_id": 0
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Proving right angle triangle A triangle $A$$B$$C$ is integer sided and has inradius $1$.prove that it is right angled. Please help.give some hints please.
| My idea :
$r = \Delta/s = 1 \implies \Delta = s \implies \Delta^2 = s^2 \implies (s-a)(s-b)(s-c) = s \implies (b+c-a)(c+a-b)(a+b-c) = 4(a+b+c) $.
If all $a,b,c$ are even or odd , then the parity doesn't satisfy. Thus that means $a+b+c$ is even. So $s$ is an integer. Thus by the above argument, we have $ s-a|s,s-b|s,s-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2343094",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find coefficient of $x^2$ in a complicated expansion
Find the coefficient of $x^2$ in the expansion of $(4-x^2)[(1+2x+3x^2)^6+(1+4x^3)^5]$
I noticed that since we only care about the coefficient of $x^2$, only the coefficients of $x^2$ inside the square brackets, as well as the constant terms, will matter. From there... | Let $A=\big[(1+2x+3x^2)^6+(1+4x^3)^5\big]$.
$$\begin{align}
\color{green}{[x^2]}A
&= \color{green}{[x^2]}(1+2x+3x^2)^6+\color{green}{[x^2]}(\underbrace{1+4x^3}_{\color{blue}{\text{no $x,x^2$ terms}}})^5\\
&=\color{green}{[x^2]}\bigg[\binom 6{5,0,1}1^5(2x)^0(3x^2)^1+\binom 6{4,2,0}1^4(2x)^2(3x^2)^0\bigg]+\color{blue}0\\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2344095",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
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For which positive integer $k$ does the series $\sum_{n=1}^\infty\frac{\sin(n\pi/k)}{n}$ converge?
For which positive integer $k$ does the series $$\sum_{n=1}^\infty\frac{\sin(n\pi/k)}{n}$$ converge?
The cases when $k=1$ or $k=2$ are trivial. For $k>2$, I don't see how to approach it. If one looks at the partial sums... | $$\sum_{n=1}^\infty\frac{\sin (n\pi/k)}{n} = \sum_{n=1}^\infty\frac{\pi}{n\pi}\sin2\pi \frac{n}{2}\frac1{k}$$
The right hand side is a sawtooth wave of period $2$, at $t = 1/k$:
$$f(t) = \frac{\pi}{2}-\frac{\pi}2t, \quad 0<t<2\\
f(t+2) = f(t)\\
f(2n) = 0, \quad n\in\mathbb Z$$
So specifically,
$$\sum_{n=1}^\infty\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2346681",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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the integral of $1/(x\sqrt{4x+1})$ I integrated $\frac{1}{x\sqrt{4x + 1}}$ and got I $\ln\left|\frac{\sqrt{4x+1}-1}{2\sqrt{x}}\right| + K$, where K is a constant term. But my book gives the answer as $\ln\left|\frac{\sqrt{4x+1}-1}{\sqrt{4x+1} + 1}\right| + K$. Where did I make a mistake?
Here is how I solved it.
| Let $\sqrt{4x+1}=t$.
Hence, $\frac{4}{2t}dx=dt$ and $x=\frac{t^2-1}{4}$.
Thus,
$$\int\frac{1}{x\sqrt{4x+1}}dx=2\int\frac{1}{t^2-1}dt=\int\left(\frac{1}{t-1}-\frac{1}{t+1}\right)dt=\ln\left|\frac{t-1}{t+1}\right|+C$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2346771",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Continuity of $f(x,y) = \dfrac{\sin(x^2+y^2)}{x^2+y^2}$ at $(x,y) = (0,0)$ Continuity of $f(x,y) = \dfrac{\sin(x^2+y^2)}{x^2+y^2}$ at $(x,y) = (0,0)$
I need to specify the value of $f(x,y)$ which makes the given function continuous.
I had tried to give a arbitral relationship between $y$ and $x$, but as $(x, y)$ gets c... | For $\lim_{(x,y)\rightarrow (0,0)} \frac{sin(x^2 +y^2}{x^2+y^2}$, using $r^2=x^2+y^2$ we get $\lim_{r\rightarrow 0} \frac{sin(r^2)}{r^2}$=$\lim_{r \rightarrow 0} \frac{2 \cdot r \cdot \cos(r^2)}{2r}$=$\lim_{r \rightarrow 0} \cos(r^2)=1$, therefore $\lim_{(x,y)\rightarrow (0,0)} \frac{sin(x^2 +y^2}{x^2+y^2}=1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2348565",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Find the minimum value of $P=\sum _{cyc}\frac{\left(x+1\right)^2\left(y+1\right)^2}{z^2+1}$
For $x>0$, $y>0$, $z>0$ and $x+y+z=3$ find the minimize value of $$P=\frac{\left(x+1\right)^2\left(y+1\right)^2}{z^2+1}+\frac{\left(y+1\right)^2\left(z+1\right)^2}{x^2+1}+\frac{\left(z+1\right)^2\left(x+1\right)^2}{y^2+1}$$
W... | Let $x=y=z=1$.
Hence, $P=24$.
We'll prove that it's a minimal value.
Indeed, by C-S
$$\sum_{cyc}\frac{(x+1)^2(y+1)^2}{z^2+1}=\sum_{cyc}\frac{(x+1)^2(y+1)^2(x+y)^2}{(z^2+1)(x+y)^2}\geq\frac{\left(\sum\limits_{cyc}(x+1)(y+1)(x+y)\right)^2}{\sum\limits_{cyc}(z^2+1)(x+y)^2}.$$
Thus, it remains to prove that
$$\left(\sum\li... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2349388",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Common complex roots
If the equations $ax^2+bx+c=0$ and $x^3+3x^2+3x+2=0$ have two common roots then show that $a=b=c$.
My attempts:
Observing $-2$ is a root of $x^3+3x^2+3x+2=0\implies x^3+3x^2+3x+2=(x+2)(x^2+x+1)=0$
Hence $ax^2+bx+c=0$ can have complex roots in common, comming from $(x^2+x+1)=0$
Both the roots of $... | If $a$, $b$, and $c$ are not real, then the question as stated is false. The roots of $x^3 + 3x^2 + 3x + 2$ are $-2$ and $-\frac{1}{2} \pm \frac{\sqrt{3}}{2} i$; and the polynomial $x^2 + (\frac{5}{2} + \frac{\sqrt{3}}{2} i)x + (1 + \sqrt{3} i) = (x + 2)(x + \frac{1}{2} + \frac{\sqrt{3}}{2} i)$ shares two roots with ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2353367",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
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If a is a non real root of $x^7 = 1$, find the equation whose roots are $ a + a^6 , a^2 + a^5, a^3 + a^4$ If a is a non real root of $ x^7 = 1$, find the equation whose roots are $a + a^6 , a^2 + a^5, a^3 + a^4$. This is one of the questions I have encountered while preparing for pre rmo. I feel the question requires t... | Like factor $z^7-1$ into linear and quadratic factors and prove that $ \cos(\pi/7) \cdot\cos(2\pi/7) \cdot\cos(3\pi/7)=1/8$,
we can prove $a^r+a^{7-r}=a^r+a^{-r}=2\cos\dfrac{2r\pi}7$ for $r=1,2,3$
Now if $7x=2r\pi,$
$\cos4x=\cos(2r\pi-3x)=\cos3x\ \ \ \ (1) $
So, the equation whose roots are $\cos\dfrac{2r\pi}7$ for $r... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2353710",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Proving an inequality of trigonometric functions I am trying to prove that $$(\cos x)^{\cos x} > (\sin x)^{\sin x}$$ for $x \in \left(0, \frac{\pi}{4}\right)$.
I have plotted the graph of $y=(\cos x)^{\cos x} - (\sin x)^{\sin x}$ and see that my conjecture is supported.
Therefore, I tried to prove an equivalent stateme... | (sin x + cos x)^cos x = sin x^cos x + cos x^cos x + P1( sin x, cos x)
(sin x + sin x) ^sin x = sin x^sin x + cos x^sin x + P2( sin x, cos x)
Where P1 and P2 are polynomials with factors sin x and cos x.
But cos x > sin x in range 0 < x < π /4 , therefore:
(sin x + cos x)^cos x > (sin x + sin x) ^sin x;
⇒ sin x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2354624",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 3
} |
If $ a=\frac{\sqrt{x+2} + \sqrt {x-2}}{\sqrt{x+2} -\sqrt{x-2}}$ then what is the value of $a^2-ax$ is equal to
If $$ a=\frac{\sqrt{x+2} + \sqrt {x-2}}{\sqrt{x+2} -\sqrt{x-2}}$$ then
the value of $a^2-ax$ is equal to:
a)2 b)1 c)0 d)-1
Ans. (d)
My attempt:
Rationalizing $a$ we get,
$ x+ \sqrt {x^2-4}$
$a^2=(x+\sqrt{... | HINT: write $a$ in the form $$a=\frac{\sqrt{x+2}+\sqrt{x-2}}{\sqrt{x+2}-\sqrt{x-2}}=\frac{(\sqrt{x+2}+\sqrt{x-2})^2}{4}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2356126",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 7,
"answer_id": 4
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Calculating limits of areas of circle
If $r$ is the radius of circle then I calculated $f(t)$ and $g(t)$ to be
$$f(t)=\frac{1}{2}r^2t-\frac{1}{2}r^2\sin t=\frac{1}{2}r^2(t-\sin t)$$
$$AB^2=4r^2\sin^2 \frac{t}{2}= 2AC^2+2AC^2\cos(\pi-t)$$
$$g(t)=\frac{1}{2}AC^2\sin (\pi-t)-\frac{1}{2}r^2(t-\sin t )= \frac{r^2\sin^2 \f... | Alternatively:
$$\begin{align}
f(t)=&\frac{1}{2}r^2t-\frac{1}{2}r^2\sin{t}=\frac{1}{2}r^2(t-\sin t). \\
AC=&r\tan{\frac{t}{2}}. \\
S_{ABC}=&\frac{1}{2}AC^2\sin{(\pi-t)}=\frac{1}{2}r^2\tan^2{\frac{t}{2}}\sin{t}=\frac{1}{2}r^2\frac{1-\cos t}{1+\cos t}\sin t. \\
g(t)=&S_{ABC}-f(t)=\frac{1}{2}r^2(\frac{1-\cos t}{1+\cos t}\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2358334",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
inequality involving the product of lengths of edges of quadrilateral Is there a constant $C$ such that for every quadrilateral with sides $a,b,c,d$ and area $S$, the following inequality holds:
$$abcd\ge C\cdot S^2$$
My attempt:
It is known, that $$S^2\le (s-a)(s-b)(s-c)(s-d)=\frac{1}{16}(-a+b+c+d)(a-b+c+d)(a+b-c+d)(... | There is no positive constant $C$. Take a non degenerate convex quadrilateral $ABCD$ and then move $D$ to $A$ along the initial side $DA$. Then the product $abcd$ goes to zero (because $d=|DA|$ goes to zero and $c=|DC|$ remains bounded). Moreover $S$ is greater or equal to the area of the triangle $ABC$ which is posi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2359519",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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A right angle at the focus of a hyperbola
$P$ is a point on a hyperbola. The tangent at $P$ cuts a directrix at point $Q$. Prove that $PQ$ subtends a right angle to the focus $F$ corresponding to the directrix.
I have tried to use the general equation of the hyperbola and gradient method to show, but too many unknown... | Here is a solution for parabola. You can use the similar way for hyperbola.
Let $y^2=2px$ be an equation of our parabola, $P(x_1,y_1)$.
Hence, $F\left(\frac{p}{2},0\right)$ and $x=-\frac{p}{2}$ is an equation of the directrix.
If $x_1=\frac{p}{2}$ then since $yy_1=p(x+x_1)$ is an equation of the tangent,
$Q\left(-\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2360164",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 1
} |
Divergent series $\sum _{n=1}^{\infty } \frac{1}{\sqrt{k}} \cos (kx)$ Working in the series above and using maclaurin euler formula i get
$$\sum _{n=1}^{\infty } \frac{\sqrt{2 \pi } \sqrt{\frac{2 \pi n}{x \sqrt{\frac{4 \pi ^2 n^2}{x^2}-1}}+1}}{\sqrt[4]{x^2-4 \pi ^2 n^2}}+\frac{\sqrt{\frac{\pi }{2}}}{\sqrt{x}}=\sum _{n... | I am not sure that your approach will get you the result.
This is my hint. The given series does not converge if $x$ is a integer multiple of $2\pi$, i.e $x=2\pi m$ with $m\in\mathbb{Z}$ because
$$\sum _{k=1}^{\infty } \frac{\cos (k 2\pi m)}{\sqrt{k}} =\sum _{k=1}^{\infty } \frac{1}{\sqrt{k}}=+\infty$$
Otherwise, note... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2360352",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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Find minimum of $a+b+c+\frac1a+\frac1b+\frac1c$ given that: $a+b+c\le \frac32$
Find minimum of $a+b+c+\frac1a+\frac1b+\frac1c$ given that: $a+b+c\le \frac32$ ($a,b,c$ are positive real numbers).
There is a solution, which relies on guessing the minimum case happening at $a=b=c=\frac12$ and then applying AM-GM inequal... | Hint: by the AM-HM (arithmetic-harmonic mean) inequality:
$$
\frac{a+b+c}{3} \ge \frac{3}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}} \;\;\iff\;\; \frac{1}{a}+\frac{1}{b}+\frac{1}{c} \ge \frac{9}{a+b+c}
$$
Let $x=a+b+c \in (0,\frac{3}{2}]\,$, then the expression to be minimized can be written as:
$$
a+b+c+\frac1a+\frac1b+\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2360813",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 0
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Factorization of polynomial with prime coefficients I'm interested in the problem linked with this answer.
Let $ f(x) = a_n + a_1 x + \dots + a_{n-1} x^{n-1} $ be polynomial with distinct $a_i$ which are primes.
(Polynomials like that for $n= 4 \ \ \ \ f(x) = 7 + 11 x + 17 x^2 + 19 x^3 $)
*
*Is it for ... | This is certainly possible. The easiest way is to use pairs of twin primes and $n=4$. Such as
$$
f(x)=7+5x+11x^2+13x^3
$$
where $f(-1)=0$ and $x+1$ is a common factor of $f(x)$ and $x^4-1.$
Extending the same idea to third roots of unity. Consider
$$
f(x)=7+5x+17x^2+29x^3+31x^4+19x^5.
$$
Because $7+29=5+31=17+19=36$ w... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2362790",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Rotate the parabola $y=x^2$ clockwise $45^\circ$. I used the rotation matrix to do this and I ended up with the equation: $$x^2+y^2+2xy+\sqrt{2}x-\sqrt{2}y=0$$
I tried to plot this but none of the graphing softwares that I use would allow it.
Is the above the correct equation for a parabola with vertex (0,0) and axis o... | Let us start with general conic section
$$Ax^2+Bxy+Cy^2+Dx+Ey+F = 0$$
or equivalently, we can write it as
$$\begin{pmatrix} x & y & 1 \end{pmatrix}\begin{pmatrix} A & B/2 & D/2\\ B/2 & C & E/2\\ D/2 & E/2 & F \end{pmatrix}\begin{pmatrix} x\\ y\\ 1 \end{pmatrix}=0$$
(we will denote the above 3x3 matrix with $M$)
So, let... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2363075",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 5,
"answer_id": 0
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Inverse of a non-trivial exponential function I am asked to determine the inverse function of this function,
$$f(x)=2^{x}+3^{x}$$
The inverse function can not be found explicitly, since there is no way to explicitly clear x, but this does not mean that it has no inverse.
I could show a way to find the inverse of th... | You could, for example, write the solution to $2^x + 3^x = y$ as a series in powers of $y-2$:
$$\eqalign{x &= \frac{y-2}{\ln(6)} - \left(\ln(3)^2 + \ln(2)^2\right) \frac{(y-2)^2}{2 \ln(6)^3}\cr +& \left(2 \ln(3)^4 - \ln(3)^3\ln(2)+6\ln(3)^2\ln(2)^2-\ln(3)\ln(2)^3 + 2 \ln(2)^4\right)\frac{(y-2)^3}{6 \ln(6)^5}\cr - &\lef... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2363488",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Find n so that the following converges $\int_1^{+ \infty} \left( \frac{nx^2}{x^3 + 1} - \frac 1 {13x + 1} \right) dx$ Question
Determine $n$ such that the following improper integral is convergent
$$ \int_1^{+ \infty} \left(
\frac{nx^2}{x^3 + 1}
-\frac{1}{13x + 1}
\right) dx
$$
I'm not sure how to go abou... | A simpler approach:
Since we know that:
$$\int_1^\infty\frac1{x^a}~\mathrm dx<\infty\iff a>1$$
It follows that if we know
$$\lim_{x\to\infty}\frac{\frac{nx^2}{x^3+1}-\frac1{13x+1}}{1/x^a}=c\ne0$$
Then the integral converges iff $a>1$.
For $n=\frac1{13}$, we find that using $a=2$ satisfies the limit, so it converges fo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2365228",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 0
} |
Evaluate $\lim_{ x\to \infty} \left( \tan^{-1}\left(\frac{1+x}{4+x}\right)-\frac{\pi}{4}\right)x$
Evaluate
$$\lim_{ x\to \infty} \left( \tan^{-1}\left(\frac{1+x}{4+x}\right)-\frac{\pi}{4}\right)x$$
I assumed $x=\frac{1}{y}$ we get
$$L=\lim_{y \to 0}\frac{\left( \tan^{-1}\left(\frac{1+y}{1+4y}\right)-\frac{\pi}{4... | $$
\begin{aligned}
\lim _{x\to \infty }\left(\arctan\left(\frac{1+x}{4+x}\right)-\frac{\pi \:}{4}\right)x
& = \lim _{t\to 0}\left(\frac{\arctan \left(\frac{t+1}{4t+1}\right)-\frac{\pi }{4}}{t}\right) \\
& = \lim _{t\to 0}\left(\frac{\left(\frac{\pi \:}{4}-\frac{3}{2}t+o\left(t\right)\right)-\frac{\pi \:}{4}}{t}\right)\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2365914",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 8,
"answer_id": 7
} |
show that $xyz\in N$ if $x^n+y^n+z^n\in Z$ Let $x,y,z\in R$ and such for any postive integers $n$ have
$$a_{n}=x^n+y^n+z^n\in Z$$
show that $xyz\in Z$
I have use
$$a_{n+3}=(x+y+z)a_{n+2}-(xy+yz+xz)a_{n+1}+xyza_{n}$$
since $2(xy+yz+xz)=(x+y+z)^2-(x^2+y^2+z^2)\in Z$
but $xy+yz+xz$ can't integers,because $2(xy+yz+xz)$ is ... | $2(xy+xz+yz)=(x+y+z)^2-(x^2+y^2+z^2)\in\mathbb Z$ and since
$$(x+y+z)^3=x^3+y^3+z^3+3(x+y+z)(xy+xz+yz)-3xyz,$$ we see that $6xyz\in\mathbb Z$.
Now, $$x^4+y^4+z^4=$$
$$=(x+y+z)^4-4(x+y+z)^2(xy+xz+yz)+2(xy+xz+yz)^2+4(x+y+z)xyz$$ or
$$3(x^4+y^4+z^4)=$$
$$=3(x+y+z)^4-12(x+y+z)^2(xy+xz+yz)+6(xy+xz+yz)^2+12(x+y+z)xyz,$$
whic... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2368617",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Horizontal line(s) that intersect $f(x)=x-2+\frac{5}{x}$ in two points. Compute exactly the value(s) of $q$ for which the horizontal line $y = q$
intersects the graph of $f(x)$ in two points that are located on a distance $4$ from each other.
While I found the two lines as $y = 4$ and $y = -8$
, I do not know how to fi... | Intersection of $y$ and $f$:
$$
q = x - 2 + 5/x \iff \\
qx = x^2 -2x + 5 \wedge x \ne 0
$$
So we remember to discard a solution $x=0$ and solve the quadratic equation:
$$
0 = x^2 - (2 + q)x + 5 = (x - (1+q/2))^2 - (1+q/2)^2 + 5 \iff \\
(x-(1+q/2))^2 =(1+q/2)^2 - 5 \iff \\
x = (1+q/2) \pm \sqrt{(1+q/2)^2 -5}
$$
As usu... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2369199",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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Find the sum of all values of $a$ satisfying that there exist positive integers $a,b$ satisfying $(a-b) \sqrt{ab}=2016$
Find the sum of all values of $a$ satisfying that there exist positive integers $a,b$ satisfying $$(a-b) \sqrt{ab}=2016$$
my try
let $a= x^2$ , $b= y^2$ $\to$ $(a-b) \sqrt{ab}$=$(x^2-y^2)(xy)=2016... | A start of a solution.
The prime factorization of $2016$ is $2^5\cdot3^2\cdot7$.
In order for $\sqrt{ab}$ to be a integer, both $a$ and $b$ must be multiples of squares and the same factors, i.e. $a=zx^2,b=zy^2, x>y$. Thus:
$$(a-b)\sqrt{ab}=z^2(x^2-y^2)xy = 2016$$
Use the fact that if neither of two numbers is divisib... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2369922",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
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ratio between the area of square $wxyz$ and the area of square $ abcd$ equal?
$ABCD$ is a square and $H$ is an interior point, which divides it for four triangles. If $W$, $X,$ $Y$ and $Z$ are the centroids of triangles $AHD$, $AHB$, $BHC$ and $CHD$ respectively ,
then what is the ratio between the area of... | $XW=YZ=\frac{1}{3}BD$, $XY=WZ=\frac{1}{3}AC$ and $XYZW$ is square.
Thus, the ratio is $\frac{2}{9}$ because $S_{ABCD}=\frac{1}{2}AC\cdot BD$ and
$$\frac{S_{XYZW}}{S_{ABCD}}=\frac{\frac{1}{3}AC\cdot\frac{1}{3}BD}{\frac{1}{2}AC\cdot BD}=\frac{2}{9}$$
For example, let $M$ is a midpoint of $BH$.
Hence, $$\frac{XY}{AC}=\fr... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How to simplify an expression involving several square roots without a calculator? $$\frac{5 \sqrt{7}}{4\sqrt{3\sqrt{5}}-4\sqrt{2\sqrt{5}}}- \frac{4 \sqrt{5}}{\sqrt{3\sqrt{5}}-\sqrt{2\sqrt{5}}}$$
This type of questions are common in the university entrance examinations in our country but the calculators are not allowed... | Hint: apply $\frac{1}{\sqrt{a} - \sqrt{b}}=\frac{\sqrt{a} + \sqrt{b}}{a - b}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2377485",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
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How to find out $ k $ in this problem If $ k $ is an non negative integer, such that $ 24^k $ divides $ 13! $, what is $ k $ then?
Explanation needed, as I am new to factorials.
| The highest power of a prime $p$ in $n!$ is given by $$\sum_{k=1} \Biggl \lfloor \frac{n}{p^k}\Biggr \rfloor $$
$24^k = (2^3 \times 3)^k = 2^{3k} \times 3^k$
Highest Power of $2$ in $13!$
Let the highest power be $x$.
$x= \lfloor \frac{13}{2} \rfloor + \lfloor \frac{13}{2^2} \rfloor + \lfloor \frac{13}{2^3} \rfl... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2378520",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 3
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Deteminant of a matrix If $P= \left(\begin{array}{ll}
A & B \\
C & D \\
\end{array}\right)$, where A, B, C, D are tridiagonal matrices. Then how to define its determinant ?
| See Determinants of Block Matrices.
A block matrix is a matrix that is defined using smaller matrices, called blocks. For example,
$P= \left(\begin{array}{ll}
A & B \\
C & D \\
\end{array}\right)$
where $A$, $B$, $C$, and $D$ are themselves matrices, is a block matrix. In the specific example
$A= \left(\beg... | {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "3",
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If $\sum_{k=0}^n \frac{A_k}{x+k}=\frac{n!}{x(x+1)(x+2)(x+3)\cdots(x+n)}$ Then find $A_7$ If $$\sum_{k=0}^n \frac{A_k}{x+k}=\frac{n!}{x(x+1)(x+2)(x+3)\cdots(x+n)}$$ Then find $A_7$
My Try:
I have considered a function
$$f(x)=x^{A_0}(x+1)^{A_1}(x+2)^{A_2}\cdots(x+n)^{A_n}$$
taking natural log on both sides and then diff... | Multiplying by $x+7$ and for $x = -7$ you have $A_7 = -\frac{n!}{7!(n-7)!}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2381927",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 1
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How to solve $n$-th derivative of logarithm function? How to slove the $n$-th derivative of logarithm function
$$\frac{d^n}{dx^n} \ln \left(\frac{p}{ax+b}+\frac{1-p}{b}\right) $$
where $p \in [0,1]$, $a$ and $b$ are positive numbers
| These are the first derivatives. Try to find a general formula for the $n-$th
$$
\begin{array}{r|r}
\text{order} & \text{derivative}\\
\hline
1 & -\frac{a}{b+a x}+\frac{a-a p}{b-a (p-1) x}\\
2 & \frac{a^2}{(b+a x)^2}-\frac{(a-a p)^2}{(b-a (p-1) x)^2} \\
3 & 2\left(-\frac{ a^3}{(b+a x)^3} +\frac{ (a-a p)^3}{(b-a (p-1... | {
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"question_score": "1",
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Find $\lvert m\rvert$ given $\sqrt[3]{m+9}=3+\sqrt[3]{m-9}$ I've already found the solution, but I was wondering if there is a faster or alternative method. My solution is found below:
$$\sqrt[3]{m+9}=3+\sqrt[3]{m-9}$$
$$(m+9)^\frac{1}{3}-(m-9)^\frac{1}{3}=3$$
$$\bigl((m+9)^\frac{1}{3}-(m-9)^\frac{1}{3}\bigr)^3=27$$
$$... | Let $\sqrt[3]{m+9}=a$, $-3=b$ and $-\sqrt[3]{m-9}=c$.
Hence, we have
$$a+b+c=0.$$
But $$a^3+b^3+c^3-3abc=a^3+3a^2b+3ab^2+b^3+c^3-3a^2b-3ab^2-3abc=$$
$$=(a+b)^3+c^3-3ab(a+b+c)=(a+b+c)((a+b)^2-(a+b)c+c^2-3ab)=$$
$$=(a+b+c)(a^2+b^2+c^2-ab-ac-bc).$$
We see that $$ a^2+b^2+c^2-ab-ac-bc=\frac{1}{2}((a-b)^2+(a-c)^2+(b-c)^2),$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2386952",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Area between arc and line I have been working on a problem for a few days now. This was a challenge problem on a lecture for Trigonometry. I managed to find an equation for the radius, but wasn't able to solve it.
Problem: Line $AB$ is drawn such that $\overline{AB} = 20$. Minor arc $AB$ is drawn with endpoints $AB$ su... | hint
Let $t $ be the angle $\angle ACB .$
Then
$$21=rt $$
and
$$r\sin(t/2)=10 .$$
the area we want is the difference
$$S=\frac {r^2}{2}t-\frac{20.h}{2} $$
with
$$h=r\cos(t/2) $$
hence
$$S=\frac {21}{2}r-10\sqrt {r^2-100} $$
with $r $ satisfying
$$r\sin (\frac {21}{2}r)=10$$
| {
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"url": "https://math.stackexchange.com/questions/2387122",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Find the real and imaginary part of z let $z=$ $$ \left( \frac{1 + \sin\theta + i\cos\theta}{1 + \sin\theta - i\cos\theta} \right)^n$$
Rationalizing the denominator:
$$\frac{1 + \sin\theta + i\cos\theta}{1 + \sin\theta - i\cos\theta}\cdot\left( \frac{1 + \sin\theta + i\cos\theta}{1 + \sin\theta + i\cos\theta}\right) =... | we know that $a+ib=\sqrt{a^{2}+b^{2}}\cdot e^{i\arctan \frac{b}{a}}\Rightarrow \frac{a+ib}{a-ib}=e^{2i\arctan \frac{b}{a}}$ and $\cos \theta +i\sin \theta =e^{i\theta }$ then:
$\left( \frac{1+\sin \theta +i\cos \theta }{1+\sin \theta -i\cos \theta } \right)^{n}=e^{i\cdot 2n\arctan \frac{\cos \theta }{1+\sin \theta }}=\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2387172",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 8,
"answer_id": 4
} |
Solving $\frac{1}{(x-1)} - \frac{1}{(x-2)} = \frac{1}{(x-3)} - \frac{1}{(x-4)}$. Why is my solution wrong? I'm following all hitherto me known rules for solving equations, but the result is wrong. Please explain why my approach is not correct.
We want to solve:
$$\frac{1}{(x-1)} - \frac{1}{(x-2)} = \frac{1}{(x-3)} - \f... | But as I suggested in my comment, the better approach is to first simplify each side separately . . .
\begin{align*}
&\frac{1}{x-1} - \frac{1}{x-2} = \frac{1}{x-3} - \frac{1}{x-4}\\[4pt]
\implies\;
&\frac{(x-2)-(x-1)}{(x-1)(x-2)}=\frac{(x-4)-(x-3)}{(x-3)(x-4)}\\[4pt]
\implies\;
&\frac{-1}{(x-1)(x-2)}=\frac{-1}{(x-3)(x-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2388701",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 7,
"answer_id": 2
} |
If $x$, $y$ and $z$ are distinct positive integers and $x+y+z=11$ then what is the maximum value of $(xyz+xy+yz+zx)$?
If $x$, $y$ and $z$ are distinct positive integers and $x+y+z=11$ then what is the maximum value of $(xyz+xy+yz+zx)$?
We know that product is maximum when difference between $x$, $y$ and $z$ is minimu... | the way the question is worded, $x$ and $y$ and $z$ cannot be equal (they must instead be distinct) so $y$ and $z$ cannot be $4$.
so the correct answer is choosing $5$, $4$ and $2$ for $x$, $y$ and $z$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2391573",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 2
} |
Solution for the inequality $x \lt \frac 1 x$ I am unable to find correct solution for the inequality:
$x \lt \frac 1 x, x \in \Bbb R, x \ne 0$.
My solution first takes $x > 0$, then we have:
$x < \frac 1x$ (multiply both sides by $x$)
$\iff x^2 < 1$ (taking square root)
$\iff x < 1 \lor x < -1$
$\iff 0 < x < 1$ sin... | $$x<\frac { 1 }{ x } \\ x-\frac { 1 }{ x } <0\\ \frac { { x }^{ 2 }-1 }{ x } <0\\ \frac { x\left( x-1 \right) \left( x+1 \right) }{ { x }^{ 2 } } <0\\ x\left( x-1 \right) \left( x+1 \right) <0\\ x\in \left( -\infty ;-1 \right) \cup \left( 0;1 \right) $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2394313",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 7,
"answer_id": 0
} |
Is it possible to solve for $\theta$ in this multivariable equation? Question: Given the equation $$c^2=a^2+(\frac{pq}{\sqrt{q^2\sin^2{\theta}+p^2\cos^2{\theta}}})^2-2a\frac{pq}{\sqrt{q^2\sin^2{\theta}+p^2\cos^2{\theta}}}\cos{\theta},$$
is it possible to solve for $\theta$ in terms of $a,c,p,$ and $q$?
Original Problem... | For this problem, I should write two equations
$$c^2=a^2+x^2-2ax \cos(\theta) \tag 1$$
$$x=\frac{pq}{\sqrt{q^2 \sin^2(\theta)+p^2\cos^2(\theta)}}\tag 2$$ From $(1)$ we get $$\cos(\theta)=\frac{a^2-c^2+x^2}{2 a x}\tag 3$$ Replacing in $(2)$, squaring, simplifying and so on, we end with $$Ax^4+Bx^2+C=0\tag 4$$ (which is ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2394546",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Proving $x-\frac{x^3}{6} < \sin(x) < x - \frac{x^3}{6} + \frac{x^5}{120} ~~ \forall x \in \Bbb R^+$ using Taylor's expansion I'm trying to prove
$$x-\frac{x^3}{6} < \sin(x) < x - \frac{x^3}{6} + \frac{x^5}{120} ~~ \forall x \in \Bbb R^{+}$$
I wrote $\sin(x)=x-x^3/3!+x^5/5!-x^7/7!+\dots$ and then the expression for ... | For any $x>0$ we have $\sin(x)<x$, hence by applying $\int_{0}^{t}(\ldots)\,dx$ to both sides we get $1-\cos t < \frac{t^2}{2}$. By applying $\int_{0}^{x}(\ldots)\,dt$ to both sides we get $x-\sin x<\frac{x^3}{6}$, which can be rearranged as $\sin(x)>x-\frac{x^3}{6}$. By performing the same trick again we also get $\si... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2394867",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 0
} |
Partial fractions - integration $$\int \frac{4}{(x)(x^2+4)} $$
By comparing coefficients,
$ 4A = 4 $,
$A = 1$
$1 + B = 0 $,
$B= -1 $
$xC= 0 $,
$C= 0 $
where $\int \frac{4}{x(x^2+4)}dx =\int \left(\frac{A}{x} + \frac{Bx+C}{x^2 + 4}\right)dx$.
So we obtain $\int \frac{1}{x} - \frac{x}{x^2+4} dx$.
And my final an... | I'm not necessarily sure how you got a factor of $x$ in the second term, but here is my go at it:
$$\frac{4}{x(x^2+4)}= \frac{A}{x} + \frac{Bx+C}{x^2+4}=\frac{A(x^2+4)+x(Bx+C)}{x(x^2+4)}=\frac{Ax^2+Bx^2+Cx+4A}{x(x^2+4)}$$
Then, by equating the numerator of the first and last expressions we have:
$$4=x^2(A+B) +Cx +4A$$
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2396334",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
Obtain the value of $\int_0^1f(x) \ dx$, where $f(x) = \begin{cases} \frac {1}{n}, & \frac{1}{n+1}
Is the function Riemann integrable? If yes, obtain the value of $\int_0^1f(x) \ dx$
$f(x) =
\begin{cases}
\frac {1}{n}, & \frac{1}{n+1}<x\le\frac{1}{n}\\
0, & x=0
\end{cases}$
My attempt
$f$ is bounded and monotonically ... | since $(0,1) = \bigcup_{n=1}^{\infty} ( \frac1{n+1} , \frac1{n })$. (pairwise disjoint union)
$$\int_{0}^1f(x)dx = \sum_{n=1}^{ \infty} \int_{1/n+1}^{1/n}f(x)dx = \sum_{n=1}^{ \infty} \int_{1/n+1}^{1/n}\frac1ndx=\sum_{n=1}^{ \infty} \frac1{n^2 }- \frac1{(n+1)n}= \frac{\pi^2}{6}-1
$$
Given that $ \frac1{(n+1)n}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2396444",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Integration with logarithmic expression. I'm trying to solve an integration problem from the book which is the $$\int\frac{\sqrt{9-4x^2}}{x}dx$$ using trigonometric substitution. The answer from the book is $$3\ln\left|\frac{3-\sqrt{9-4x^2}}{x}\right|+\sqrt{9-4x^2}+C.$$
I have almost the same solution where there's a ... | It is easy to mess up with the constants in such integrals. But we can temporarily ignore them because a simple scaling of the variable and of the integrand can bring us to
$$I:=\int\frac{\sqrt{1-t^2}}tdt.$$
This calls for the substitution $1-t^2=u^2$, giving $dt/t=-u\,du/2t^2$ and
$$I\propto\int\frac{u^2}{u^2-1}du=\in... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2400006",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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How to find $\lim_{x \to \infty} \frac{ex^{x+1}-x(x+1)^x}{(x+1)^x}$ I came across this problem a few days ago and I have not been able to solve it. Wolfram Alpha says the answer is 1/2 but the answer I came up with is 0. Can anyone see what is wrong with my work and/or provide the correct way of solving this problem?
$... | Consider $$y=\frac{ex^{x+1}-x(x+1)^x}{(x+1)^x}=\frac{ex^{x+1}}{(x+1)^x}-x$$ Now
$$z=\frac{ex^{x+1}}{(x+1)^x}\implies \log(z)=1+(x+1)\log(x)-x\log(x+1)=1+\log(x)-x \log\left(1+\frac 1x\right)$$ Now, using Taylor $$\log\left(1+\frac 1x\right)=\frac{1}{x}-\frac{1}{2 x^2}+\frac{1}{3
x^3}+O\left(\frac{1}{x^4}\right)$$ ma... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2400449",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 3
} |
Infinite power summation where no common ratio can be seen Problem in book states to evaluate the sum:
$$\sum_{i=0}^\infty \frac{i^2}{4^i}$$
I can come up with the series S = $\frac{1^2}{4^1} + \frac{2^2}{4^2} + \frac{3^2}{4^3} + \frac{4^2}{4^4} + \frac{5^2}{4^5} + ...$
And 4S = $ 1 + \frac{2^2}{4^1} + \frac{3^2}{4^2} ... | Consider $$\sum_{i=0}^\infty {i^2}{x^i}=\sum_{i=0}^\infty {[i(i-1)+i]}{x^i}=\sum_{i=0}^\infty {i(i-1)}{x^i}+\sum_{i=0}^\infty {i}{x^i}=x^2\sum_{i=0}^\infty {i(i-1)}{x^{i-2}}+x\sum_{i=0}^\infty {i}{x^{i-1}}$$ that is to say
$$\sum_{i=0}^\infty {i^2}{x^i}=x^2\left(\sum_{i=0}^\infty {x^{i}} \right)''+x\left(\sum_{i=0}^\i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2405902",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Let $a = \frac{9+\sqrt{45}}{2}$. Find $\frac{1}{a}$ I've been wrapping my head around this question lately:
Let
$$a = \frac{9+\sqrt{45}}{2}$$
Find the value of
$$\frac{1}{a}$$
I've done it like this:
$$\frac{1}{a}=\frac{2}{9+\sqrt{45}}$$
I rationalize the denominator like this:
$$\frac{1}{a}=\frac{2}{9+\sqrt{45}} \ti... | $$\frac { 1 }{ a } =\frac { 9-\sqrt { 45 } }{ 18 } =\frac { 3\left( 3-\sqrt { 5 } \right) }{ 18 } =\frac { 3-\sqrt { 5 } }{ 6 } $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2407843",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
In expansion of $(1+x+x^{2}+x^{3}+....+x^{27})(1+x+x^{2}+x^{3}+....x^{14})^{2}$ . Find the coefficient of $x^{28}$ I am not able to apply binomial theorem here
$(1+x+x^{2}+x^{3}+....+x^{27})(1+x+x^{2}+x^{3}+....x^{14})^{2}$
Please help me to find the coefficient of$ x^{28}$
Any help will be appreciated.
| \begin{eqnarray*}
(1+x+x^{2}+x^{3}+\cdots+x^{27})(1+x+x^{2}+x^{3}+\cdots+x^{14})^{2} \\
= (1+x+x^{2}+\cdots+x^{27}) (1+2x+3x^{2}+\cdots +14x^{13}+15x^{14} +14x^{15} \cdots +2x^{27}+ x^{28}) \\
= \cdots+x^{28}( \underbrace{1+2 +\cdots + 15}_{120} + \underbrace{14+ \cdots +2}_{104})+ \cdots
\end{eqnarray*}
EDIT
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2408611",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 3
} |
Solve in positive integers, $ x^6+x^3y^3-y^6+3xy(x^2-y^2)^2=1$ Solve in positive integers, $$ x^6+x^3y^3-y^6+3xy(x^2-y^2)^2=1$$
My attempt :
$ x^3y^3+x^6-y^6+3xy(x^2-y^2)^2$
$=x^3y^3+3x^2y^2(x^2-y^2)+3xy(x^2-y^2)^2+(x^2-y^2)^3$
$=(xy+(x^2-y^2))^3$
$=(x^2+xy-y^2)^3 = 1$
so $x^2+xy-y^2 = 1$
Please suggest how to proceed.... | Cassini's identity for the Fibonacci numbers is
$$
F_{n-1}F_{n+1}-F_{n}^{2}=(-1)^{n}
$$
Therefore
$$
1
=F_{2n-1}F_{2n+1}-F_{2n}^{2}
=F_{2n-1}(F_{2n-1}+F_{2n})-F_{2n}^{2}
=F_{2n-1}^2+F_{2n-1}F_{2n}-F_{2n}^{2}
$$
and $x=F_{2n-1}$, $y=F_{2n}$ are solutions of $x^2+xy-y^2 = 1$.
Similarly,
$$
F_{2n-1}^2-F_{2n-1}F_{2n-2}-F_... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2412268",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Show that $7 |(n^6 + 6)$ if $7 ∤ n$, $∀ n ∈ ℤ$
Show that $7 |(n^6 + 6)$ if $7 ∤ n$, $∀ n ∈ ℤ$
I need to prove this by the little Fermat's theorem.
My attempt
$n^6 \equiv -6 \pmod 7$
To show $7 ∤ n$ I need to show that $N$ is not congruent to $0$ mod $7$.
as $-6 \equiv 1\pmod 7$
$n^6 \equiv 1\pmod7$
But now, How can ... | Because $n^6+6=n^6-1+7$ and $n^6-1$ divisible by $7$.
$$n^6-1=(n-1)(n+1)(n^2-n+1)(n^2+n+1)$$
and now easy to check $n\equiv\pm1,\pm2\pm3(\mod7)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2412519",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Constrained Optimisation Problem: Confusion with Algebra/Contradiction I completed the following constrained optimisation problem:
Maximum and minimum values of $f(x,y) = x^2 + 2xy + y^2$ on the ellipse $g(x,y) = 2x^2 + y^2 - xy - 4x = 0$.
$\nabla f = \lambda \nabla g$
$\therefore (2x + 2y, 2x + 2y) = \lambda(4x - y ... | You don't have to solve like that and exclude $x=2y$ or other limitations.
The system
$
\left\{ {\begin{array}{*{20}{l}}
{2x + 2y = \lambda \left( {4x - y - 4} \right)} \\
{2x + 2y = \lambda \left( {2y - x} \right)} \\
{2{x^2} + {y^2} - xy - 4x = 0}
\end{array}} \right.
$
can be solved looking at the first tw... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2413361",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Finding divisibility rules Find the divisibility test for 13.
Let's first find a test for small 3-digit numbers first and then find a general test. We'll be finding tests which involves separation of last digit. These tests will be similar to the tests of 19, 23, 29, etc.
Let $\overline { abc }$ be any number such tha... | Although you haven't asked a question, here is a quicker way to reach the same result. Using the division algorithm, write your number as $N=10m + d$ where $0\le d \le 9$.
Then $13 \mid N\iff 13 \mid 4N \iff 13\mid 40m+4d \iff 13\mid m+4d$
(The first equivalence in the chain follows because $4$ and $13$ are relatively... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2413452",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Proving/disproving $∀n ∈ \text{positive integers}$, $\left\lceil{\frac{4n^2+1}{n^2}}\right \rceil = 5$? Is my proof getting anywhere for proving/disproving $∀n ∈ \text{postive integers}$, $$\left\lceil{\frac{4n^2+1}{n^2}}\right \rceil = 5\text{?}$$
$$\left\lceil{\frac{4n^2}{n^2}}+\frac{1}{n^2}\right \rceil = 5$$
$$\lef... | Your proof is probably okay if you specify all statements are $\iff$ results. That way one can reverse direction. Otherwise you are assuming what you need to prove.
$[\frac {4n^2 + 1}{n^2}] = 5 \iff$
$[4 + \frac 1{n^2}] = 5 \iff$
$[\frac 1{n^2}] = 5 - 4$ (actually do you know if that is true? I don't feel comfortabl... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2414453",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 3
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Find all of the points of the form $(x, −1)$ which are $4$ units from the point $(3, 2)$
Find all points of the form $(x, −1)$, which are $4$ units from the point
$(3, 2)$.
I understand the distance formula, I think. Also, I don't know how to format for math on here yet so I apologize for that.
\begin{align}
&\t... | Suppose you have two points in a 2-D coordinate system such as $(x_{1}, y_{1}) = (3, 2)$ and $(x_{2}, y_{2}) = (x, -1)$ and distance between them is 4 units. Hence you can write as
\begin{align}
& d = \sqrt{(x_{1} - x_{2})^{2} + (y_{1} - y_{2})^{2}} \\
\implies & 4 = \sqrt{(3-x)^{2} + (2-(-1))^{2}} \\
\implies & 16 =... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2416760",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 2
} |
What is $\lim\limits_{n\to\infty}\sum\limits^n_{i=0}F_{2i}-\phi F_{2i-1}$? Firstly, the question is:
Evaluate
$$(1-0\phi)+(2-1\phi)+(5-3\phi)+(8-5\phi)+\cdots=\lim_{n\to\infty}\sum^n_{i=0}F_{2i}-\phi F_{2i-1}$$
where $F_0=1,$
$F_1=1$ and
$F_n=F_{n-1}+F_{n-2}$ for all integer n (even n negative).
I attempted the basic m... | Since $$\begin{pmatrix} 1 & 1 \\ 1 & 0 \end{pmatrix} \begin{pmatrix} F_i \\ F_{i-1} \end{pmatrix} = \begin{pmatrix} F_{i+1} \\ F_i \end{pmatrix},$$ you are looking for the limit $$\lim_{N \rightarrow \infty} \begin{pmatrix} 1 & -\phi \end{pmatrix} \begin{pmatrix} F_{2i} \\ F_{2i-1} \end{pmatrix} = \sum_{i=0}^N \begin{p... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2417071",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Solve this trigonometric equation ${\tan ^2}x + {\cot ^2}x - 3(\tan x - \cot x) = 0$
Solve the following equation. $${\tan ^2}x + {\cot ^2}x - 3(\tan x - \cot x) = 0.$$
I can't figure out the way to solve this equation. This was my attempt
\begin{array}{l}
{\tan ^2}x + {\cot ^2}x - 3(\tan x - \cot x) = 0\\
\Leftrigh... | As an alternative to Michael Rozenberg's answer, let $u=\tan x$. The equation becomes
$$u^2+{1\over u^2}-3\left(u-{1\over u}\right)=0$$
Multiplying through by $u^2$ to clear denominators, we wind up with
$$u^4-3u^3+3u+1=0$$
The quartic factors into two quadratics:
$$u^4-3u^3+3u+1=(u^2-u-1)(u^2-2u-1)$$
The rest follow... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2417589",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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How to prove that $\lim_{n \to \infty} \frac{\sqrt {n^2 +2}}{4n+1}=\frac14$?
How to prove that $\lim_{n \to \infty} \frac{\sqrt {n^2 +2}}{4n+1}=\frac14$?
I started my proof with Suppose $\epsilon > 0$ and $m>?$ because I plan to do scratch work and fill in.
I started with our conergence definition, i.e. $\lvert a_n -... | Update If you really need to use a complete $\epsilon$-definition type proof, then using this method will only require that you additionally prove $\lim_{n\to\infty} \frac{n}{4n+1} = \frac{1}{4}$, which is considerably easier. (Just note that $\left|\frac{n}{4n+1}-\frac{1}{4}\right| = \frac{1}{4(4n+1)}$)
We will bound ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2420408",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 5
} |
Finding a substitution that eliminates the squared term from a cubic equation So I am a little stumped here, and it could be simple. I'm just not exactly sure how to approach this.
The question reads:
Find $\omega_0$ in the set of Complex numbers, such that the substitution $z = \omega - \omega_0$ reduces the cubic e... | Letting $z=\omega - \omega_0$ and getting
$$ z^3 + Az^2 + Bz + C = \omega^3 + A'\omega^2 +B'\omega +C'$$
Can be accomplished by doing the following synthetic divisions
\begin{array}{r|ccccc}
& 1
& A
& B
& C
\\
-\omega_0 & 0
& -\omega_0
& \omega_0^2 - A\omeg... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2424827",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Finding derivative of $\frac{x}{x^2+1}$ using only the definition of derivative I think the title is quite self-explanatory. I'm only allowed to use the definition of a derivative to differentiate the above function. Sorry for the formatting though.
Let $f(x) = \frac{x}{x^2+1}$
$$
\begin{align}
f'(x)&= \lim_{h\to 0} \... |
Your last step is wrong, the actual steps will be,
$$=\lim_{h\to 0} \frac{\frac{-\color{red}{(hx^2-h+h^2x)}}{(1+x^2)(x^2+2hx+h^2+1)}}{\color{red}{h}}$$
$$=\lim_{h\to 0} \frac{\frac{-\color{blue}{(x^2-1+hx)}}{(1+x^2)(x^2+2hx+h^2+1)}}{\color{blue}{1}}$$
$$=\frac{1-x^2}{(1+x^2)^2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2426142",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Solve the following nonlinear system of equations:$ xy+xz+yz=12 , xyz=2+x+y+z$
Solve the following system of equations in $\Bbb R^+$:
$$
\left\{
\begin{array}{l}
xy+yz+xz=12 \\
xyz=2+x+y+z\\
\end{array}
\right.
$$
I did as follows.
First equation yields to: $(x+y+z)^2-x^2-y^2-z^2=24$. Now, using second equat... | [ EDIT ] The question has been changed to require $\,x,y,z \ge 0\,$ now, which renders the following not applicable any longer.
(Too long for a comment.) Let $x=a+2, y=b+2, z=c+2\,$, then the conditions rewrite as:
$$\require{cancel}
\begin{cases}
\begin{align}
a b + b c + a c + 4(a + b + c) + \cancel{12} &= \cancel{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2427470",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 6,
"answer_id": 2
} |
For what values of $a$ does the system have infinite solutions? Find the solutions. The system is
$$\left\{
\begin{array}{rcr}
x+ay+z & = & 1 \\
ax+y+z & = & 1+a \\
x-y+z & = & 2+a
\end{array}
\right.$$
After row reducing I got
$$\left\{
\begin{array}{rcr}
x+ay+z & = & 1 \\
-(1+a)y+0 & = & 1+a \\
(1-a)z& = & 2-a + (1... | Something has gone wrong with the row reduction. First, we eliminate $x,$ which gives us $$\left\{\begin{array}{rcl}x+ay+z & = & 1 \\(1-a^2)y+(1-a)z & = & 1 \\-(1+a)y & = & 1+a.\end{array}\right.$$ Next, we exchange the second and third equations, then use the (now) second equation to eliminate $y$ from the (now) thir... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2427722",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
Distinct integer solutions to $a^2+b^2=c^2+d^2$ I'm trying to find integer solutions to
$$a^2+b^2=c^2+d^2$$
with values $a> c > d > b>0$
Or in other words, two triangles with integer legs and equal hypotenuse lengths, not necessarily integer. Seems like a Diophantine equation to me, but I only learned how to solve Dio... | Strong Hint:
$$\begin{align} a^2 + b^2 &= c^2 + d^2 \\ \implies a^2 + b^2 - c^2 &= d^2 \\ \implies a^2 + (b + c)(b - c) &= d^2 \\ \implies (b + c)(b - c) &= d^2 - a^2 \\ \implies (b + c)(b - c) &= (d + a)(d - a) \end{align}$$.
$$\therefore a^2 + b^2 \neq \{p : p = \text{prime number}\} \iff a^2 + b^2 = c^2 + d^2$$
Now ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2429520",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 3
} |
Limit of infinite sum. Let $x$ be a real number such as $0 < x < 1$, compute the following series
$$\sum^\infty _{n=2} \frac{x^n}{n-1}$$ $$\sum^\infty _{n=2} \frac{x^n}{n+1}
$$
For instance, I made something like this : $$\frac{x^n}{n-1} = \frac{x^n}{n(1-\frac{1}{n})} \sim _{n \rightarrow \infty} \frac{x^n}{n}(1+\frac... | Let $n^\prime = n-1$:
$\sum\limits_{n=2}^\infty = \sum\limits_{n^\prime=1}^\infty {x^{n^\prime + 1} \over n^\prime} = x \sum\limits_{n^\prime}^\infty {x^{n^\prime} \over n^\prime} = -x \ln(1-x)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2429613",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Find the total number of coefficients in the expansion of $(x^2-x +1)^{22}$ that are divisible by $6$. Find the total number of coefficients in the expansion of $(x^2-x +1)^{22}$ that are divisible by $6$.
So I am really stuck on this one.. I immediately defeated when I saw that I need to compute $\binom {22} {a,b,c}$... | Using that $(a+b+c)^3\equiv a^3+b^3+c^3\pmod 3$ repeatedly, we have, in mod $3$,
$$\begin{align}&(x^2-x+1)^{22}\\\\&=(x^2-x+1)(x^2-x+1)^3((x^2-x+1)^9)^2\\\\&\equiv (x^2-x+1)(x^6-x^3+1)(x^{18}-x^9+1)^2\\\\&\equiv (x^2-x+1)(x^6-x^3+1)(x^{36}+x^{27}+x^9+1)\\\\&\equiv x^{44}-x^{43}+x^{42}-x^{41}+x^{40}-x^{39}+x^{38}-x^{37}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2429982",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 0
} |
Sum of terms $1^2-2^2+3^2-...$ Prove that: $$1^2-2^2+3^2-4^2+...+(-1)^{n-1}n^2=\frac{(-1)^{n-1}n(n+1)}{2}$$
I proved it by induction but is there any other way to solve it?
If it was not a proof but rather a question like find the term,how to solve it?
I realized that alternate terms were under same sign but can't unde... | Notice $$n^2 = \frac{(n-1)n + n(n+1)}{2}
\quad\implies\quad (-1)^{n-1} n^2 = (-1)^{n-1}\frac{(n-1)n}{2} - (-1)^{n}\frac{n(n+1)}{2}$$
The sum is a telescoping sum and
$$\require{cancel}\sum_{k=1}^n (-1)^{k-1} k^2 = \color{red}{\cancelto{0}{\color{grey}{(-1)^{1-1}\frac{(1-1)1}{2}}}}- (-1)^n \frac{n(n+1)}{2}
= (-1)^{n-1}\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2430887",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 8,
"answer_id": 0
} |
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.