Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Inequality $\max\left(\frac{1-a^2}{3-2a},\frac{1-b^2}{3-2b}\right)\geq\frac{12}{19}\cdot\frac{3-a^2-2b^2}{6-a-3b}$ Let $0\leq a,b\leq 1$. Prove that
$$\max\left(\frac{1-a^2}{3-2a},\frac{1-b^2}{3-2b}\right)\geq\frac{12}{19}\cdot\frac{3-a^2-2b^2}{6-a-3b}.$$
Equality holds when $a=0$ and $b=2/3$. This seems to be the only... | hint: $$\frac{1-a^2}{3-2a}-\frac{1-b^2}{3-2b}=\frac{(a-b)(2-3a-3b+2ab)}{(2a-3)(2b-3)}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1892819",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
By using a geometric series and a factorisation, compute the first three terms of this given Taylor expansion By using the geometric series
$$\frac{1}{1-x} = \sum_{n=0}^\infty x^n$$
and the factorisation
$$\frac{1}{1-3x+2x^2}= \left(\frac{1}{1-2x}\right) \left(\frac{1}{1-x}\right)$$
compute the first three terms of the... | That could work, but an easier way is the following. Note that since
$$
\frac{1}{1-x} = \sum_{n=0}^\infty x^n
$$
it follows that
$$
\frac{1}{1-2x} = \sum_{n=0}^\infty (2x)^n = \sum_{n=0}^\infty 2^nx^n = 1 + 2x + 4x^2 +8x^3 + \cdots
$$
Now, since
$$
\frac{1}{1 - 3x + 2x^2} = \frac{1}{1 - 2x}\frac{1}{1-x}
$$
we can expa... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1893913",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
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Let $z$ be a complex number $\ne 0$. What is the absolute value of $z\sqrt{z}$? $\color{red}{\mathbf{EDIT}}$ The question was misinterpreted - it was actually: 'what is the absolute value of $z/\bar{z}$?'; I'am grateful for the answers given on the original problem though and will keep this up as is in case someone els... | Using polar coordinates, you write $z=re^{i\theta}$, from which $\bar z=re^{-i\theta}$, thus
$$
z/\bar z=\frac{re^{i\theta}}{re^{-i\theta}}=e^{2i\theta}
$$
whose absolute value is $1$ as requested.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1894307",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
Minimum and maximum with $1+\frac{a+x}{b+x}+\frac{a+x}{c+x}$ Let $a,b,c\geq 1$ and $x,y,z\geq 0$. What are the minimum and maximum of
$$f(a,b,c,x,y,z)=\frac{1}{1+\frac{a+x}{b+x}+\frac{a+x}{c+x}}+\frac{1}{1+\frac{b+y}{a+y}+\frac{b+y}{c+y}}+\frac{1}{1+\frac{c+z}{a+z}+\frac{c+z}{b+z}}?$$
Since each term is no more than $1... | Without loss of generality, take $a\ge b\ge c$.
By mediant inquality,
$$1 \le \frac{a+x}{b+x} \le \frac{a}{b}$$
$$1 \le \frac{a+x}{c+x} \le \frac{a}{c}$$
$$3 \le 1+\frac{a+x}{c+x}+\frac{a+x}{c+x} \le
1+\frac{a}{b}+\frac{a}{c}$$
$$\frac{b}{a} \le \frac{b+y}{a+y} \le 1$$
$$1 \le \frac{b+y}{c+y} \le \frac{b}{c}$$
$$2+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1895187",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
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Is the series $\sqrt{1} - \sqrt{2} + \sqrt{3} - \sqrt{4} + \dots$ summable? Is the series $\sqrt{1} - \sqrt{2} + \sqrt{3} - \sqrt{4} + \dots$ summable? I think it diverges although:
$$ \sqrt{n+1} - \sqrt{n} \approx \frac{1}{2\sqrt{n}}$$
for example by the Mean Value Theorem $f(x+1)-f(x) \approx f'(x)$ and then I might... | How about an Abel sum?
$$
\sum_{n=1}^\infty (-1)^{n+1}\sqrt{n}\;x^n = -\mathrm{Li}_{-1/2}(-x)
$$
for $|x|<1$ and converges as $x \to 1^-$ to the value
$$
-\mathrm{Li}_{-1/2}(-1) \approx 0.3801048
$$
So we call that value the Abel sum of the divergent series $\sum_{n=1}^\infty (-1)^{n+1}\sqrt{n}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1896464",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
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Find the second smallest integer such that its square's last two digits are $ 44 $ Given that the last two digits of $ 12^2 = 144 $ are $ 44, $ find the next integer that have this property.
My approach is two solve the equation $ n^2 \equiv 44 \pmod{100}, $ but I do not know how to proceed to solve that equation.
I t... | $n^2 \equiv 44 \pmod {100}$
By Chinese remainder theorem, $n^2 \equiv 0 \pmod 4$ and $n^2 \equiv 19 \pmod {25}$ is an solution for the formula above.
$$n^2 \equiv 0 \pmod 4 \tag{1}$$
$$n^2 \equiv 19 \pmod {25} \tag{2}$$
By (1), $n = 2k$. Substitude into (2),
$$k^2 \equiv 19 \times 4^{-1} \equiv 19 \times 19 \equiv 11 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1898485",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 3
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Limit of $\mathbf{tan}^2(\theta)[1-\mathbf{sin}(\theta)]$ Evaulate the limit of $\mathbf{tan}^2(\theta)[1-\mathbf{sin}(\theta)]$.
$$\lim \limits_{\theta \to \frac{\pi}{2}} \mathbf{tan}^2(\theta)[1-\mathbf{sin}(\theta)] $$
I have attempted the problem by direct substitution and get:
$$\mathbf{tan}^2(\frac{\pi}2)[1-\ma... | $$\lim _{ \theta \to \frac { \pi }{ 2 } }{ \tan } ^{ 2 }(\theta )[1-{ \sin }(\theta )]=\lim _{ \theta \to \frac { \pi }{ 2 } }{ \tan } ^{ 2 }(\theta )\frac { 1-\sin ^{ 2 }{ (\theta ) } }{ 1+{ \sin }(\theta ) } =\lim _{ \theta \to \frac { \pi }{ 2 } }{ \tan } ^{ 2 }(\theta )\frac { \cos ^{ 2 }{ (\theta ) } }{ 1+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1899409",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 0
} |
Show $\lim\left ( 1+ \frac{1}{n} \right )^n = e$ if $e$ is defined by $\int_1^e \frac{1}{x} dx = 1$ I have managed to construct the following bound for $e$, which is defined as the unique positive number such that $\int_1^e \frac{dx}x = 1$.
$$\left ( 1+\frac{1}{n} \right )^n \leq e \leq \left (\frac{n}{n-1} \right )^n$... | "$ ( 1+\frac{1}{n} )^n \leq e \leq (\frac{n}{n-1})^n$"$
I'm assuming you've shown $( 1+\frac{1}{n} )^n < ( 1+\frac{1}{n+1} )^{n+1}$ so we can say $\lim_{n\rightarrow \infty}( 1+\frac{1}{n} )^n = c \le e$ and likewise $\lim_{n\rightarrow \infty}(\frac{n}{n-1} )^n = d \ge e$.
So $\frac cd = \lim_{n\rightarrow \inft... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1900042",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 6,
"answer_id": 5
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Help proving the product of any four consecutive integers is one less than a perfect square Apparently this is a true statement, but I cannot figure out how to prove this. I have tried setting
$$(m)(m + 1)(m + 2)(m + 3) = (m + 4)^2 - 1 $$
but to no avail. Could someone point me in the right direction?
| I will show that
$16m(m+1)(m+2)(m+3) = 16(m^2 + 3m + 1)^2 - 16$
It will follow that
$$m(m+1)(m+2)(m+3) = (m^2 + 3m + 1)^2 - 1$$
Proof:
\begin{align}
16 m(m+1)(m+2)(m+3)
&=(2m)(2m+2)(2m+4)(2m+6)\\
& ---\text{let $2m = n-3$}---\\
&=(n-3)(n-1)(n+1)(n+3)\\
&=(n-3)(n+3) \cdot (n-1)(n+1) \\
&=(n^2 - 9)(n^2 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1900365",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 8,
"answer_id": 7
} |
Find the minimnum of the $\max{(2x_{1}+x_{2},2x_{2}+x_{3},\cdots,2x_{n-1}+x_{n})}$ Let $n$ is give postive integers,and for $x_{i}\ge 0$,such
$$x_{1}+x_{2}+\cdots+x_{n}=1$$
find $I$ minimum of the value
$$I=\max{(2x_{1}+x_{2},2x_{2}+x_{3},\cdots,2x_{n-1}+x_{n})}$$
I try to find the
$$2x_{1}+x_{2}\le I$$
$$2x_{2}+x_{3}... | Assume that $n\geq 2$. Observe that, as $x_1,x_2,\ldots,x_n\geq0$ (in particular, $x_{n-1}\geq 0$), we have
$$\begin{align}
S&:=\sum_{i=1}^{n-2}\,\left(\frac{2^i-(-1)^i}{3}\right)\,2^{n-2-i}\,\left(2x_{i}+x_{i+1}\right)+2^{n-2}\left(2x_{n-1}+x_n\right)
\\
&=\left(\frac{2^n-(-1)^n}{3}\right)\,x_{n-1}+2^{n-2}\,\sum_{i=1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1902192",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Prove that $\left(\frac{n}{2}\right)^n \gt n! \gt \left(\frac{n}{3}\right)^n \qquad n\ge 6 $
$$\left(\frac{n}{2}\right)^n \gt n! \gt \left(\frac{n}{3}\right)^n \qquad n\ge 6 $$
I tried it prove it by mathematical induction but failed .
For $n=6$
$$(3)^6 \gt 6!$$
Now for $n=k$
$$\left(\frac{k}{2}\right)^k \gt k!$$
No... | METHODOLOGY $1$: Proof by Induction
PRIMER:
In THIS ANSWER, I showed using Bernoulli's Inequality that $\left(1+\frac1n\right)^n$ is a monotonically increasing sequence. Therefore, its minimum is $2$ when $n=1$. It is clearly bounded by $3$ as can be shown applying the binomial theorem and applying simple estimate... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1902324",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Calculate the sum of $\sum_{n=1}^\infty (-1)^{n-1} \frac{n+1}{2^\frac{n}{2}}$ So, we have $$\sum_{n=1}^\infty (-1)^{n-1} \frac{n+1}{2^\frac{n}{2}}$$
I immediately noticed that $ln(1+x) = \sum_{n=1}^\infty (-1)^{n-1} \frac{x^n}{n}$ , for $|x|\lt 1$
It reminds of this sum, if we switch nominator and denominator and multi... | One may start with the standard finite evaluation:
$$
1+x+x^2+...+x^n=\frac{1-x^{n+1}}{1-x}, \quad |x|<1. \tag1
$$ Then by differentiating $(1)$ we get
$$
1+2x+3x^2+...+nx^{n-1}=\frac{1-x^{n+1}}{(1-x)^2}+\frac{-(n+1)x^{n}}{1-x}, \quad |x|<1, \tag2
$$ by making $n \to +\infty$ in $(2)$, using $|x|<1$, gives
$$
\sum_{n=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1902487",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Finding the value of theta using the $6$ trigonometric functions on an $xy$ plane. Here's a link of what I am trying to learn about.
http://www.webpages.uidaho.edu/learn/math/lessons/lesson03/3_05.htm
Now I have one question. How will I find the value for theta in each function if the hypotenuse has no exact square ro... | Based on a now deleted comment, it is my understanding that the terminal side of the angle passes through the point $(14, 5)$. You used the Pythagorean Theorem to conclude correctly that $r = \sqrt{221}$.
Using the trigonometric formulas
\begin{align*}
\sin\theta & = \frac{y}{r} & \csc\theta & = \frac{r}{y}\\
\cos\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1903325",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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If $a$, $b$, and $c$ are sides of a triangle, then $\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}<2$.
Let $a,b,c$ be the lengths of the sides of a triangle. Prove that
$$\sum_{\text{cyc}}\frac{a}{b+c}=\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}<2\,.$$
Attempt. By clearing the denominators, the required inequality is equiv... | Hint: Without loss of generality, suppose $c$ is the largest side. Hence, $c<a+b$. Also,
$$\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\leq\frac{a}{a+b}+\frac{b}{a+b}+\frac{c}{a+b}\,.$$
Note that the bound is sharp. In the limit $a\to 0$ and $b\to c$, we have the sum goes to $2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1903775",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 3
} |
Finding the matrix representation of a linear mapping
Let $V=\left \{ A\in M_2\left(\mathbb{C} \right )\mid \text{tr}(A)=0 \right \}$ and $T:V\rightarrow V$ defined by $T\left(A\right)=\left(3\,\mathrm{i}-2\right)A+\left(4-6\,\mathrm{i}\right)A^t$
Find the matrix representation of $T$ with regard to the basis $B=\le... | No the matrix is the matrix whose columns are the images. You need to find the images, and rewrite them as coordinate vectors with respect to $B$.
$$T(B_1) = (2-3i)B_1 + 0B_2 + 0B_3$$
$$T(B_2) = 0B_1 + (-2+3i)B_2 + (4-6i)B_3$$
$$T(B_3) = 0B_1 + (4-6i)B_2 + (-2+3i)B_3$$
Therefore
$$[T]_B = \begin{pmatrix} 2-3i & 0 & 0\\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1904341",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Prove inequality $\ln \left( \frac{e-e^x}{1-x} \right) \leq \sqrt{\frac{1+x+x^2}{3}}$ for $0The first function could be called 'exponential mean' of $y$ and $x$:
$$f(y,x)=\ln \left( \frac{e^y-e^x}{y-x} \right)$$
We can obtain it by Cauchy mean value theorem.
What is interesting, it appears numerically that:
$$\ln \lef... | A little demonstration (not a full solution, but I'll try to expand it later):
If we were allowed to use CAS, such as Mathematica (or spent a little time differentiating), we could just square both sides and expand the left side into series aroud $x=1$:
$$\ln^2 \left( \frac{e-e^x}{1-x} \right) =1+(x-1)+\frac{1}{3} (x-1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1904473",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Proving $\sum_{x=0}^{n-1} \cos\left(k +x{2\pi\over n}\right) =\sum_{x=0}^{n-1} \sin\left(k +x{2\pi\over n}\right) =0. $ Is there anybody who can help me show the following?
$$
\sum_{x=0}^{n-1} \cos\left(k +x{2\pi\over n}\right) =0
\qquad\hbox{and}\qquad
\sum_{x=0}^{n-1} \sin\left(k +x{2\pi\over n}\right) =0
$$
I c... | Let $k$ be an arbitrary real number. Consider the trigonometric identity
$$2 \cos A\, \sin B = \sin(A + B) - \sin(A - B)$$
Setting $A = k + x(\frac{2\pi}{n})$ and $B = \frac{\pi}{n}$, we get
$$2\cos\left(k + x\frac{2\pi}{n}\right) \sin\left(\frac{\pi}{n}\right) = \sin\left[k + \left(x + \frac{1}{2}\right)\frac{2\pi}{n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1905189",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
Trouble understanding proof of the inequality - $(\frac{1}{a}+1)(\frac{1}{b}+1)(\frac{1}{c}+1) \ge 64 $, for $a,b,c > 0$ and $a+b+c = 1$ I was looking into this problem in a book discussing inequalities, However I found the proof quite hard to understand.The problem is as follows:
Let $a,b,c$ be positive numbers with ... | I have any solution.
Note that
$a+1=a+a+b+c\geq 4\sqrt[3]{a^2bc} \tag{1}$
by AM-GM
Analog we get that
$b+1=b+a+b+c\geq 4\sqrt[3]{ab^2c} \tag{2}$
and
$c+1=c+a+b+c\geq 4\sqrt[3]{abc^2} \tag{3}$
So
$$(a+1)(b+1)(c+1)\geq 64abc\Longleftrightarrow (\dfrac{a+1}{a})(\dfrac{b+1}{b})(\dfrac{c+1}{c})\geq 64$$
The equality will ho... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1905278",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
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Discontinuous at infinitely many points While doing a worksheet on real analysis I came across the following problem.
$Q$. Let $f$ be a function defined on $[0,1]$ with the following property.
For every $y \in R$, either there is no $x$ in $[0,1]$ for which $f(x)=y$ or there are exactly two values of $x$ in $[0,1]$ for... | This example shows that the image of $f$ can be the whole of $\mathbb{R}$:
\begin{align*}
f(0) & = 0, \\
f(x) & = \begin{cases}
\frac{3}{2^n} - 8x &
\left( \frac{1}{2^{n + 1}} \leqslant x < \frac{5}{2^{n + 3}} \right) \\
\frac{1}{2^{n - 2}} - 8x &
\left( \frac{5}{2^{n + 3}} \leqslant x < \frac{3}{2^{n + 2}} \righ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1905560",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
"answer_id": 4
} |
prove $abc \ge 8$ for $\frac{1}{1+a}+\frac{1}{1+b}+\frac{1}{1+c}=1$ I am reading this book about inequalities and the chapter about AM-GM inequalities includes this problem:
Let $a,b,c$ be positive numbers for which $\frac{1}{1+a}+\frac{1}{1+b}+\frac{1}{1+c}=1$, prove that
$$abc \ge 8$$
The book does not provide full ... | If the problem states that a, b, and c are positive integers, this is pretty easy. Find a common denominator on the LHS, multiply both sides by the denominator and simplify. Then you get: $$3+2(a+b+c)+ab+ac+bc=1+(a+b+c)+ab+ac+bc+abc$$ Simplify this and you get $$2+a+b+c=abc$$ The lowest possible non-zero integers a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1906492",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
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Prove the inequality $\left(\frac1a+\frac1b+\frac1c\right)\left(\frac1{1+a}+\frac1{1+b}+\frac1{1+c}\right)\ge\frac9{1+abc}$
Let $a,b,c>0$. Prove the inequality
$$\left(\frac1a+\frac1b+\frac1c\right)\left(\frac1{1+a}+\frac1{1+b}+\frac1{1+c}\right)\ge\frac9{1+abc}$$
My work so far:
Use AM-GM:
$$\frac1{1+a}+\frac1{1+b}+... | By Rearrangement
$\sum\limits_{cyc}\frac{1}{a}\sum\limits_{cyc}\frac{1}{1+a}=\sum\limits_{cyc}\frac{1}{a(1+a)}+\sum\limits_{cyc}\frac{1}{b(1+a)}+\sum\limits_{cyc}\frac{1}{c(1+a)}\geq\sum\limits_{cyc}\frac{2}{b(1+a)}+\sum\limits_{cyc}\frac{1}{c(1+a)}$.
Now by AM-GM
$(1+abc)\sum\limits_{cyc}\frac{1}{b(1+a)}=\sum\limits... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1908753",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
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How can it be proven that $\frac{x}{y}+\frac{y}{x}\geq2$, with $x$ and $y$ positive? So I realized that I have to prove it with the fact that $(x-y)^2+2xy=x^2+y^2$
So $\frac{(x+y)^2}{xy}+2=\frac{x}{y}+\frac{y}{x}$ $\Leftrightarrow$ $\frac{(x+y)^2}{xy}=\frac{x}{y}+\frac{y}{x}-2$
Due to the fact that $(x+y)^2$ is a sq... | You made a few mistakes here.
First, actually $(x+y)^2=x^2+2xy+y^2$, what you need is $(x-y)^2+2xy=x^2+y^2$. Now, you get $xy>0$ but $(x-y)^2 \geq 0$, so $\frac{(x-y)^2}{xy} \geq 0$, and this gives you $$\frac{x^2-2xy+y^2}{xy} \geq 0$$
which gives $\frac{x}{y}+\frac{y}{x} \geq 2$, which is the desired inequality.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1911037",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Why does $f_A(X,X)>0$ for every $X$ (Question 8 of section 8.2 of Hoffman and Kunze's Linear Algebra) I'm studying Hoffman and Kunze's linear algebra book and I'm having troubles how to prove this exercise on page 276:
The only part of this exercise I couldn't prove was $f_A(X,X)>0$ for every real column matrix $X$.
| Let's compute:
$$
X^t A X = \begin{pmatrix} x & y \end{pmatrix}
\begin{pmatrix} a & c \\ c & b \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = x(ax+cy)+y(cx+by) = ax^2+2cxy + by^2
$$
Now use the additional hypothesis that $a, b>0$ and $ab > c^2$ (which comes from $\det A > 0$). So we get
$$
ax^2+2cxy+by^2 \geq ax... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1916249",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Integral of exponential of trigonometric functions I'm looking for a proof of the result
$$
J_0 \left( x \sqrt{a^2 + b^2}\right) = \frac{1}{2\pi} \int_0^{2\pi} e^{ix \, (a \cos \theta \, + \, b \sin \theta)} \, \text{d} \theta
$$
$J_0$ is the Bessel function of order $0$.
| We have that
$$a\cos\theta + b\sin\theta = \sqrt{a^2+b^2}\cos(\theta-\psi),\qquad \psi=\arctan\frac{b}{a}\tag{1}$$
hence
$$\frac{1}{2\pi}\int_{0}^{2\pi}e^{ix(a\cos\theta+b\sin\theta)}\,d\theta = \frac{1}{2\pi}\int_{0}^{2\pi}e^{ix\sqrt{a^2+b^2}\cos\theta}\,d\theta \tag{2}$$
and the claim follows by expanding $e^{ix\sqr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1917245",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Unable to justify solution for a problem with exponential constant $e$ involved. For the question,
$e^{2x} + e^x - 2 = 0.$
I was asked to solve for $x$.
What I performed,
$e^{2x} + e^x = 2.$
$e^x(e^x + 1) = 2.$
For the 2 solutions involved,
\begin{align}
e^x &= 2\\
\ln2 &= x\\
.69 &= x
\end{align}
OR
\begin{align}
e^x... | Let $y=e^{x}$. Then, your equation becomes
$$
y^{2}+y-2=0\implies (y+2)(y-1)=0.
$$
Thus, $y=-2$ or $y=1$. Hence, $e^{x}=-2$ or $e^{x}=1$. Since the first is impossible, it follows that $e^{x}=1\implies x=0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1918064",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
Finding values to make a set a basis Question:
For which values of $a\in \Bbb R$ is the set below a basis for $M_{2x2}(\Bbb R)$ as a vector space over $\Bbb R$?
$${(a,2a,2,3a), (1,2,2a,3), (1,2a,a+1,a+2), (1,a+1,2,2a+1)}$$ (these vectors should be matrices but I do not know how to do them in MathJax)
My thinking is tha... | In order for the matrices presented to form a basis they need to be linearly independent. This means that for scalars $x_i$, the equation:
$$
x_1 \begin{pmatrix} a & 2a \\ 2 & 3a \end{pmatrix}
+ x_2 \begin{pmatrix} 1 & 2 \\ 2a & 3 \end{pmatrix}
+ x_3 \begin{pmatrix} 1 & 2a \\ a+1 & a+2 \end{pmatrix}
+ x_4 \begin{pmat... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1918599",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Prove inequality $ (a+b+c)(a^7+b^7+c^7)\ge(a^5+b^5+c^5)(a^3+b^3+c^3)$ for $a,b,c\ge 0$ Prove that:
$$ (a+b+c)(a^7+b^7+c^7)\ge(a^5+b^5+c^5)(a^3+b^3+c^3)$$
I already know that this can be proven using Cauchy Schwarz, but I don't really see how to apply it here. I'm looking for hints.
| Using Holder's inequality, we have that
\begin{align*}
a^5+b^5+c^5 &= a^{\frac{1}{3}}a^{\frac{14}{3}} + b^{\frac{1}{3}}b^{\frac{14}{3}}+c^{\frac{1}{3}}c^{\frac{14}{3}}\\
&\le (a+b+c)^{\frac{1}{3}}(a^7+b^7+c^7)^{\frac{2}{3}},
\end{align*}
and
\begin{align*}
a^3+b^3+c^3 &= a^{\frac{2}{3}}a^{\frac{7}{3}} + b^{\frac{2}{3}}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1919468",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 2
} |
Mistake in basic algebra, I think? Problem prove $(2n+1)+(2n+3)+(2n+5)+...+(4n-1) =3n^2$
Induction proof: base case $n=1$ assume true for all $n$ prove for $n+1$.
The $n$th or last term becomes $(4(n+1)-1)=4n+3$.
We also sub $n+1$ in for all $n$ the $n-1$ term is $(4n-1)$ and the first term is $2(n+1)+1=2n+3$
The right... | $\sum_\limits{k=1}^{n} (2(n+k)-1) = 3n^2$
Base case: $n = 1$
$(2(1+1) - 1) = 3$
Suppose:
$\sum_\limits{k=1}^{n} (2(n+k)-1) = 3n^2$
We will show that
$\sum_\limits{k=1}^{n+1} (2((n+1)+k)-1) = 3(n+1)^2$
based on the inductive hypothesis
$\sum_\limits{k=1}^{n+1} (2((n+1)+k)-1)\\
\sum_\limits{k=1}^{n} (2(n+1)+k)-1) + 4n+3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1919985",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Prove: $\cos^3{A} + \cos^3{(120°+A)} + \cos^3{(240°+A)}=\frac {3}{4} \cos{3A}$ Prove that:
$$\cos^3{A} + \cos^3{(120°+A)} + \cos^3{(240°+A)}=\frac {3}{4} \cos{3A}$$
My Approach:
$$\mathrm{R.H.S.}=\frac {3}{4} \cos{3A}$$
$$=\frac {3}{4} (4 \cos^3{A}-3\cos{A})$$
$$=\frac {12\cos^3{A} - 9\cos{A}}{4}$$
Now, please help m... | If $\cos3x=\cos3A$
$3x=n360^\circ\pm3A$ where $n$ is any integer
$x=n120^\circ+ A$ where $n=0,1,2$
As $\cos3x=4\cos^3x-3\cos x,$ the roots of $4c^3-3c-\cos3A=0$ are
$c_{n+1}=\cos(n120^\circ+ A)$ where $n=0,1,2$
Using Vieta's formula, $$c_1+c_2+c_3=0, c_1c_2c_3=\dfrac{\cos3A}4$$ $$\text{ and }c_1c_2+c_2c_3+c_3c_1=-\dfr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1921191",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 8,
"answer_id": 7
} |
Stars and bars to find "how many $x$ digit numbers are there with sum of digits $y$"? This question poses a seemingly very simple method to solve problems of the sort "how many $x$ digit numbers are there with sum of digits $y$?", but I don't understand it. Why are the "bad" solutions in correspondence to the solutions... | The referred question asks for the number of non-negative integer solutions of
\begin{align*}
x_1+x_2+x_3+x_4+x_5=23
\end{align*}
with only one additional restriction $0\leq x_1\leq 9$. In this case we do not need the inclusion-exclusion principle.
In order to determine all integer solutions with
\begin{align*}
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1922623",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
} |
Proving this identity: $\frac{\sin^2\theta}{\cos^2\theta} + \frac{\cos^2\theta}{\sin^2\theta} = \sec^2\theta \csc^2\theta - 2$ I have tried solving this trig. identity, but I get stuck when it comes to the $-2$ part. Any suggestions?
$$\frac{\sin^2\theta}{\cos^2\theta} + \frac{\cos^2\theta}{\sin^2\theta} = \sec^2\thet... | hint: $\dfrac{a^2}{b^2} + \dfrac{b^2}{a^2} = \dfrac{a^4+b^4}{a^2b^2}= \dfrac{(a^2+b^2)^2}{a^2b^2}-2$. Apply this identity for $a = \cos \theta, b = \sin \theta$ to get the desired identity.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1923555",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 1
} |
Prove that $\binom{k}{p} \equiv \left\lfloor \frac{k}{p} \right\rfloor \pmod p$ for all odd prime number $p$ and $k\ge 3$ I don't really know how to start this exercise. Do I have to use p-adic valuation ?
If it's the case it will give $\nu_p(\binom{k}{p})= \sum \limits_{r=1}^{\infty}\left(\left\lfloor \frac{k}{p^r} \r... | Let us define $jp$, with $j$ positive integer, the greatest multiple of $p$ lower than $k$. Now let us consider the definition of $\binom{k}{p}$ as $$\frac {k (k-1)(k-2)(k-3)...[k-(p-1)]}{p!}$$ The numerator is formed by $p$ consecutive numbers and can be expressed as a product of factors that are $\equiv i \pmod p$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1924309",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
How do I prove that for $ x,y,z > 0$ $\frac{x}{y+z} + \frac{y}{x+z} + \frac{z}{x+y} > 3/2$? I'm solving a graduate entrance examination problem.
We are required to establish the inequality using the following result:
for $x,y > 0$, $\frac{x}{y} + \frac{y}{x} > 2$ (1), which is easy to prove as it is equivalent to $(x ... | Apply your formula three times:
$$\frac{x+y}{x+z}+\frac{x+z}{x+y}>2$$
$$\frac{y+x}{y+z}+\frac{y+z}{y+x}>2$$
$$\frac{z+x}{z+y}+\frac{z+y}{z+x}>2$$
Combining gives:
$$\frac{x+y}{x+z}+\frac{x+z}{x+y}+\frac{y+x}{y+z}+\frac{y+z}{y+x}+\frac{z+x}{z+y}+\frac{z+y}{z+x}>6$$
$$\frac{x+y+2z}{x+y}+\frac{x+2y+z}{x+z}+\frac{2x+y+z}{y... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1928892",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 0
} |
Approximation of square root I have a square root in a problem which needs to be approximated. I'm not entirely sure how to do this algebraically.
$$ \sqrt{10^2-(6.9\times 10^{-2})^2}$$
The answer the problem is proposing as the approximation is $$ 10[1-\frac 1 2(6.9)^2\times10^{-6}]$$
This hasn't exactly been the most... | Use the Taylor series for $(1+x)^{1/2}$.
First we rewrite $(A+B)^{\alpha}$ to be in a form similar to this :
$(A+B)^{\alpha} = A^{\alpha}(1 + \frac{B}{A})^{\alpha}$.
With $A \gg B$ the ratio $\frac{B}{A}$ is "small" and we do a Taylor series for $(1+x)^{\alpha}$
$$(1+x)^{\alpha} = 1 + \frac{\alpha}{1!}x + \frac{\alp... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1929331",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Show that $\frac 1x \ge 3 - 2\sqrt{x}$ for all positive real numbers $x$. Describe when we have equality. Show that $\frac 1x \ge 3 - 2\sqrt{x}$ for all positive real numbers $x$. Describe when we have equality.
I tried manipulating the equation but nothing helps. Any answer is greatly appreciated.
| $$
\frac{1}{x} \geq 3 - 2\sqrt{x} \iff \frac{1+2x\sqrt{x} -3x}{x} \geq 0 \iff \frac{(\sqrt{x}-1)^2(\sqrt{x} + \frac{1}{2})}{x} \geq 0
$$
We can cancel off the $(\sqrt{x}-1)^2$, and the $x$ on the bottom, since $x$ must be positive, and squares are always positive. This leaves us with $\sqrt{x} + \frac{1}{2} \geq 0$, wh... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1929829",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
$ax^3+8x^2+bx+6$ is exactly divisible by $x^2-2x-3$, find the values of $a$ and $b$ Find the values of $a$ and $b$ for which the polynomial $ax^3+8x^2+bx+6$ is divisible by $x^2-2x-3$.
| by using the long division we get
$$\frac{ax^3+8x^2+bx+6}{x^2-2x-3}=ax+(8+2a)+\frac{(7a+b+16)x+6a+30}{x^2-2x-3}$$
now the reminder should be zero
$$7a+b+16=0\tag 1$$
$$6a+30=0\tag2$$
$$a=-5, b=19$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1930266",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 9,
"answer_id": 0
} |
Proof of the inequality $2^{n} < \binom{2n}{n} < 2^{2n}$? As review for a midterm I am asked to prove the inequality:
$2^{n} < \binom{2n}{n} < 2^{2n}, n > 1.$
What I have is a two-part inductive proof. It is not hard to show for $2^{n} < \binom{2n}{n}$:
Base step:
Let $n=2$:
$2^{2} < \frac{(2n)!}{2!2!} < 2^{2*2}$
$4 < ... | You may avoid induction. We have
$$ \binom{2n}{n}=\frac{2n}{n}\cdot\frac{2n-1}{n-1}\cdot\ldots\cdot\frac{n+1}{2}>2^n $$
since every factor of the middle term, except the very first one, is greater than $2$.
On the other hand,
$$ \binom{2n}{n}<\sum_{k=0}^{2n}\binom{2n}{k} = 2^{2n} = 4^n $$
by the binomial theorem.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1932085",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
If $\cos A=\tan B$, $\cos B=\tan C$ … If $\cos A=\tan B$, $\cos B=\tan C$ and $\cos C=\tan A$, prove that $\sin A=\sin B=\sin C$.
My Attempt.
Let us consider $x$, $y$ and $z$ as:.
$$x = \tan^2A$$
$$y = \tan^2B$$
$$z = \tan^2C$$
$$\cos^2A = \tan^2B$$
$$\frac {1}{\sec^2A}= \tan^2B$$
$$\frac {1}{1 + \tan^2A} = \tan^2B$$
$... | $$\tan^2C=\cos^2B=\dfrac1{1+\tan^2B}=\dfrac1{1+\cos^2A}=\dfrac1{2-\sin^2A}$$
$$\cos^2C=\tan^2A$$
$$\implies\tan^2C=\sec^2C-1=\dfrac{1-\cos^2C}{\cos^2C}=\dfrac{1-\tan^2A}{\tan^2A}=\dfrac{1-2\sin^2A}{\sin^2A}$$
Equating the values of $\tan^2C$ and writing $\sin^2A=x$
$$\dfrac1{2-x}=\dfrac{1-2x}x\implies x^2-3x+1=0\implie... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1935885",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
$\lim_{z\to 0} \frac{z^2-\overline{z}}{|z|}$ $$\lim_{z\to 0} \frac{z^2-\overline{z}^2}{|z|}$$
If we suppose $z = x+iy$, then the limit becomes:
$$\lim_{(x,y)\to (0,0)}\frac{x^2-y^2+2xyi-(x-yi)^2}{\sqrt{x^2+y^2}} = \lim_{(x,y)\to(0,0)}\frac{x^2-y^2+2xyi-x^2+2xyi+y^2}{\sqrt{x^2+y^2}} = \lim_{(x,y)\to(0,0)}\frac{4ixy}{\sq... | HINT:
$$\left|\frac{xy}{\sqrt{x^2+y^2}}\right|\le \frac12\sqrt{x^2+y^2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1936086",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
0.1999... and 0.2000 are the same? From Sipser's Introduction to the Theory of Computation:
A slight problem arises because certain numbers, such as 0.1999... and 0.2000..., are equal even though their decimal representations are different.
Can someone explain? Because that's news to me.
| We can write
$$\begin{align}
0.1999 \cdots &= 0.1+0.09+0.099+ 0.999\cdots \\
&= \dfrac{1}{10}+\dfrac{9}{100}+\dfrac{9}{1000}+\dfrac{9}{10000} + \cdots\\
&= \dfrac{1}{10}+\dfrac{9}{10^2}+\dfrac{9}{10^3}+\cdots \\
&= \dfrac{1}{10}+9\left(\dfrac{1}{10^2}+\dfrac{1}{10^3}+\cdots\right) \\
&= \dfrac{1}{10}+\dfrac{9}{10}\left... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1936159",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Series and probability The task at hand:
Suppose there are 3 players (A,B,C), who have 3 different dices with the following chances (per roll) to win:
Player A has a dice with a $\frac{1}{3}$ chance to win
Player B has a dice with a $\frac{2}{5}$ chance to win
Player C has a dice with a $\frac{4}{7}$ chance to win
P... | Let $x$ be the probability that A wins. So $x$ equals P(A rolls a win) + P(B does not roll a win and C does not roll a win)*$x$. In other words:$$x = \frac{1}{3} + \frac{2}{3}\frac{3}{5}\frac{3}{7}x$$
and so
$$x = \frac{35}{3*29} \approx 0.4023.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1937593",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
How to prove $\sin 10^\circ = \frac{-1+\sqrt{9-8\sin 50^\circ}}{4}$? How to prove this identity?
$$\sin 10^\circ = \frac{-1+\sqrt{9-8\sin 50^\circ}}{4}$$
Is this a particular case of a more general identity? Also, is it possible to give a geometric proof of this equality?
| From the expression given for $\sin 10^{\circ}$ it is obvious that it is a root of a quadratic equation. Comparing $$\frac{-1 + \sqrt{9 - 8\sin 50^{\circ}}}{4}$$ with $$\frac{-b + \sqrt{b^{2} - 4ac}}{2a}$$ we can see that it almost fits with $a = 2, b = 3, c = \sin 50^{\circ}$ and we have then $$\frac{-1 + \sqrt{9 - 8 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1939066",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Calculate $\int{\frac{x^\alpha}{(x^2+a^2)^{\frac{3}{2}}}} \ \mathrm{d}x$ for $\alpha = 1,2,4,6$? What is the most straightforward way to calculate the following indefinite integral? $$\int{\frac{x^\alpha}{(x^2+a^2)^{\frac{3}{2}}}}\ \mathrm{d}x,$$
for $\alpha = 1, 2, 4,5,6$?
| HINTS:
*
*When $\alpha=1$, substitute $u=x^2+\text{a}^2$ and $\text{d}u=2x\space\text{d}x$:
$$\mathcal{I}_1(\text{a},x)=\int\frac{x}{\left(x^2+\text{a}^2\right)^{\frac{3}{2}}}\space\text{d}x=\frac{1}{2}\int\frac{1}{u^{\frac{3}{2}}}\space\text{d}u=-\frac{1}{\sqrt{u}}+\text{C}=\text{C}-\frac{1}{\sqrt{x^2+\text{a}^2}}$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1944931",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Finding dy/dx by implicit differentiation with the quotient rule I've been stuck on a certain implicit differentiation problem that I've tried several times now.
$$
\frac{x^2}{x+y} = y^2+6
$$
I know to take the derivatives of both sides and got:
$$
\frac{(x+y)2x-\left(1-\frac{dy}{dx}\right)x^2}{(x+y)^2} = 2y\frac{dy}{d... | An idea to avoid the cumbersome and annoying quotient rule: multiply by the common denominator
$$\frac{x^2}{x+y}=y^2+6\implies xy^2+6x+y^3+6y-x^2=0\implies$$
$$y^2+2xyy'+6+3y^2y'+6y'-2x=0\implies(2xy+3y^2+6)y'=2x-y^2-6\implies$$
$$y'=\frac{2x-y^2-6}{2xy+3y^2+6}$$
If nevertheless you want to use the quotient rule:
$$\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1945594",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Solving this $4$ variable diophantine equation Is there a way to solve this Diophantine equation in $a,b,c,d$? $$19a^3-33a^2b+3a^2c+30a^2d+21ab^2+24abc-12abd-15ac^2-54acd-30ad^2+ $$ $$2b^3-12b^2c-6b^2d+42bc^2+108bcd+60bd^2-7c^3-51c^2d-99cd^2-56d^3=0$$
Wolfram Alpha unfortunately cannot understand my input whenever I in... | Your equation is homogeneous of degree $3$, so if $(a,b,c,d)$ is a solution so is $(ta,tb,tc,td)$ for any $t$. Thus it makes sense to look for primitive solutions, which are solutions with greatest common divisor $1$. I found some $288$ primitive solutions with $a,b,c \in [-20\ldots -1, 1\ldots 20]$ and $d \in [1 \ld... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1945775",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Find limit of sequence sum Let us have sum of sequence (I'm not sure how this properly called in English): $$X(n) = \frac{1}{2} + \frac{3}{4}+\frac{5}{8}+...+\frac{2n-1}{2^n}$$
We need
$$\lim_{n \to\infty }X(n)$$
I have a solution, but was unable to find right answer or solution on the internet.
My idea:
This can be re... | Let
\begin{align*}
f(x) &= \sum_{n=1}^{\infty} \frac{x^{2n-1}}{2^n}\\
&=\frac{x}{2}\sum_{n=0}^{\infty}\left(\frac{x^2}{2}\right)^n\\
&=\frac{x}{2-x^2}.
\end{align*}
Then
\begin{align*}
f'(x) = \frac{2+x^2}{(2-x^2)^2}.
\end{align*}
Therefore,
\begin{align*}
\lim_{n\rightarrow \infty}X(n) = f'(1)=3.
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1947082",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
The system of equations: $\begin{cases} 2x^2=4y^2+3z^2+2; \\ 13x=4y+3z+29 \end{cases}$ Solve in positive integers the system of equations:
$$\begin{cases}
2x^2=4y^2+3z^2+2;
\\
13x=4y+3z+29
\end{cases}$$
My work so far:
I used wolframalpha: $x=3,y=1,z=2$.
| $$2x^2=4y^2+3z^2+2\tag1$$
$$13x=4y+3z+29\tag2$$
From $(1)(2)$,
$$13^2(4y^2+3z^2+2)=2(13x)^2=2(4y+3z+29)^2,$$
i.e.
$$644y^2+(-464-48z)y+489z^2-348z-1344=0$$
See this as a quadratic equation on $y$.
Since the discriminant has to be larger than or equal to $0$, we have to have
$$(-464-48z)^2-4\cdot 644(489z^2-348z-1344)\g... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1953480",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Find all functions $f:\mathbb Z \rightarrow \mathbb Z$ such that $f(0)=2$ and $f\left(x+f(x+2y)\right)=f(2x)+f(2y)$
Find all functions $f:\mathbb Z \rightarrow \mathbb Z$ such that $f(0)=2$ and
$$f\left(x+f(x+2y)\right)=f(2x)+f(2y)$$
for all $x \in \mathbb Z$ and $y \in \mathbb Z$
My work so far:
1) $x=0$ $$f\lef... | Supposing that we already know that $f(2n) = 2n + 2$ for $n \geq 0$, we can proceed as follows.
For some positive integer $k$, let $x = 2k$ and $y = -k$ in the functional equation. We get that
$$
f(2k + 2) = f( 2k + f(2k - 2k) ) = f(4k) + f(-2k).
$$
Now $2k + 2$ and $4k$ are positive even integers, and so this tells ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1954840",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 0
} |
Sum of series $1+\frac{1}{4}+\frac{1\times3}{4\times8}+\frac{1\times3\times5}{4\times8 \times12}+\cdots$
Sum of series :
$$1+\frac{1}{4}+\frac{1\times3}{4\times8}+\frac{1\times3\times5}{4\times8 \times12}+\cdots$$
$n$-th term of series is
$$a_{n} =\frac{1\times3 \times5\times\cdots \times(2n-1)}{4\times8\times12\... | Note, the general term $a_n, n\geq 1$ is
\begin{align*}
a_n=\frac{1\cdot 3\cdot 5\cdots (2n-1)}{4\cdot8\cdot 12\cdots (4n)}
=\frac{1\cdot 3\cdot 5\cdots (2n-1)}{(1\cdot 2\cdot 3\cdots n)\cdot 4^n}
=\frac{(2n-1)!!}{n!}\cdot \frac{1}{4^n}\\
\end{align*}
with $$(2n-1)!!=(2n-1)\cdot(2n-3)\cdots 5\cdot 3\cdot 1$$ double fac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1959377",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Tricky Limits Problem () Problem:
If $\lim_{x \to 0}{\sin2x\over x^3}+a+{b\over x^2}=0$ then find the value of $3a+b$.
My attempt: $\lim_{x \to 0}{\sin2x\over x^3}+a+{b\over x^2}=\lim_{x \to 0}{\sin2x\over 2x}({2\over x^2})+a+{b\over x^2}={2+b+ax^2\over x^2}$.From this we can conclude that $a=0$ and $b=-2$, hence $3a+b... | we have
$sin(X)=X-\frac{X^3}{6}+X^3\epsilon(X)$
with $\displaystyle{ \lim_{X\to 0} \epsilon(X)=0}$.
thus
$sin(2x)=2x-\frac{4x^3}{3}+x^3\epsilon(x)$.
so
$\frac{sin(2x)}{x^3}+a+\frac{b}{x^2}$
$=\frac{2}{x^2}-\frac{4}{3}+a+\frac{b}{x^2}+\epsilon(x)$.
the limit is $0$ if
$a=\frac{4}{3}$ and $b=-2$.
finally
we will have
$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1962646",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Finding other vectors that form a basis The question I'm working on is:
For the given vectors $v, w$ in $\mathbb R^4$, find other vectors $u_1, ..., u_n$ such that {${v, w, u_1,...,u_n}$} form a basis of $\mathbb R^4$.
$v = \begin{bmatrix} 1 \\ 2 \\ 3 \\ 4 \end{bmatrix}$ and $w = \begin{bmatrix} 4 \\ 3 \\ 2 \\ 1 \end{... | You want the determinant
$$
\begin{vmatrix}
1&4&a&w\\
2&3&b&x\\
3&2&c&y\\
4&1&d&z
\end{vmatrix}\ne0.
$$
Row reduction: If we subtract the previous row from each row, we get
$$
\begin{vmatrix}
1&4&a&w\\
1&-1&b-a&x-w\\
1&-1&c-b&y-x\\
1&-1&d-c&z-y
\end{vmatrix}.
$$
Repeat, and we get
$$
\begin{vmatrix}
1&4&a&w\\
0&-5&b-2a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1963125",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Find the derivative of $\sin { \left( 3{ x }^{ 2 }+x \right) } $ How do you find the derivative of
$$\sin { \left( 3{ x }^{ 2 }+x \right) } $$
using the derivative definition and not the chain rule.
This is how far I was able to get
Any help would be appreciated. Thanks!
| \begin{eqnarray*}\lim _{ h\rightarrow0 }&{ \frac { \sin { \left( 3{ \left( x+h \right) }^{ 2 }+x+h \right) } -\sin { \left( 3{ x }^{ 2 }+x \right) } }{ h } } \\=&\lim _{ h\rightarrow0 }{ \frac { 2\sin { \frac { \left( 3{ \left( x+h \right) }^{ 2 }+x+h \right) -\left( 3{ x }^{ 2 }+x \right) }{ 2 } \cos { \frac { ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1963332",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
How prove this equation has only one solution $\cos{(2x)}+\cos{x}\cdot\cos{(\sqrt{(\pi-3x)(\pi+x)}})=0$
Let $x\in (0,\dfrac{\pi}{3}]$.
Show that this equation
$$\cos{(2x)}+\cos{x}\cdot\cos{(\sqrt{(\pi-3x)(\pi+x)}})=0$$
has a unique solution $x=\dfrac{\pi}{3}$
I try to the constructor $$f(x)=\cos{(2x)}+\cos{x}\... | Not a solution, only a change of the question.
$\displaystyle a:=\sqrt{\frac{\pi-x}{2}+\sqrt{(\frac{\pi-x}{2})^2-x^2}}$ , $\enspace \displaystyle b:=\sqrt{\frac{\pi-x}{2}-\sqrt{(\frac{\pi-x}{2})^2-x^2}}$
=> $\enspace ab=x$ , $a^2+b^2=\pi-x$ , $a^2-b^2=\sqrt{(\pi-3x)(\pi+x)}$
The problem $f(x)=0$ with $\displaystyle 0... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1963425",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 2,
"answer_id": 0
} |
How do you find the area of a parallelogram with the vertices? How do you find the area of a parallelogram with the following vertices; $A(4,2)$, $B(8,4)$, $C(9,6)$ and $D(13,8)$.
| The absolute value of the cross product of two vectors $\vec{a}, \vec{b} \in \mathbb{R}^3$ spanning the parallelogram is its area:
$$A_\text{parallelogram}= \left|\vec{a}\times\vec{b}\right|$$
So in your case we have to write the points in $\mathbb{R}^2$ as vectors in $\mathbb{R}^3$ and apply the formula:
$\vec{AB} = ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1966062",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 6,
"answer_id": 1
} |
If 'z' and 'w' are two complex numbers, prove that: |z+w|<_|z|+|w|. If 'z' and 'w' are two complex numbers, prove that: |z+w|<_|z|+|w|.
My Attempt;
If z is given as x + iy, and w = u + iv,
then |z| = √(x^2 + y^2), = r, say;
and |w| = √(u^2 + v^2), = p, say
Then we can also view z as r cis θ,
and w as p cis φ (i.e., p ... | Perhaps writing the expression explicitly it can be clearer:
$$\begin{cases}z=a+bi\\{}\\w=x+iy\end{cases}\implies |z+w|=\sqrt{(a+x)^2+(b+y)^2}\stackrel?\le\sqrt{a^2+b^2}+\sqrt{x^2+y^2}$$
Well, now square both sides:
$$a^2+2ax+x^2+b^2+2by+y^2\stackrel?\le a^2+b^2+x^2+y^2+2\sqrt{(a^2+b^2)(x^2+y^2)}\iff$$
$$\iff2(ax+by)\s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1968175",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Prove that $\log_{2}(n) = O(n^{1-\epsilon})$ for any $0 \leq \epsilon < 1$ We assume that $n>0$. I would like to get some feedback on my proof of this statement.
When we want to show that $f(n) = O(g(n))$, we need to find some positive constants $c, n_{0}$ such that for every $n \geq n_{0}$ the following holds: $f(n) \... | Note: Here we analyse at some length OPs proof, identify some problems and propose an alternative while following OPs approach.
OPs proposed solution: $c=2^{\varepsilon}, n_0=4$
At first we check the proposed solution. OPs result $c=2^\varepsilon$ and $n_0=4$ gives
\begin{align*}
\log_2(n)\leq 2^{\varepsilon}n^{1-\v... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1968257",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
How do I prove that $x^n-y^n=(x-y)(x^{n-1}+x^{n-2}y+\dotsb+xy^{n-2}+y^{n-1})$ I am unsuccessfully attempting a problem from Spivak's popular book 'Calculus' 3rd edition. The problem requires proof for the following equation:
$$x^n-y^n=(x-y)(x^{n-1}+x^{n-2}+\dotsb+xy^{n-2}+y^{n-1})$$
The solution to the problem, contai... | I like using $\sum$ like this:
$\begin{array}\\
(x-y)(x^{n-1}+x^{n-2}y+\dotsb+xy^{n-2}+y^{n-1})
&=(x-y)\sum_{k=0}^{n-1} x^{n-1-k} y^k\\
&=x\sum_{k=0}^{n-1} x^{n-1-k} y^k-y\sum_{k=0}^{n-1} x^{n-1-k} y^k\\
&=\sum_{k=0}^{n-1} x^{n-k} y^k-\sum_{k=0}^{n-1} x^{n-1-k} y^{k+1}\\
&=\sum_{k=0}^{n-1} x^{n-k} y^k-\sum_{k=1}^{n} x^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1968874",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 0
} |
Calculate limit involving $\sin$ function Calculate the following limit:
$$\lim_{x \rightarrow 0} \frac{x-\overbrace{\sin (\sin (...(\sin x)...))}^{150\ \text{times}\ \sin}}{x^3}$$
I tried applying L'Hospital's rule, but it got too messy.
Thank you in advance!
| Use short form of series expansion:
$$
\sin(x)=x-\frac{x^3}{6}+O(x^5)\\
\sin(\sin(x))=\sin(x)-\frac{\sin^3(x)}{6}+O(\sin^5(x))\to\\
\sin(\sin(x))=x-\frac{x^3}{6}+O(x^5)-\frac{(x-\frac{x^3}{6}+O(x^5))^3}{6}+O(x^5)\to\\
\sin(\sin(x))=x-\frac{x^3}{3}+O(x^5)
$$
Important part here is to notice that each additional $\sin(..... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1969399",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 0
} |
How can we solve $\lim\limits_{n \to \infty} \left (1 + \frac{1}{n} \right)^n$? Taken from Wikipedia:
The number $e$ is the limit $$e = \lim_{n \to \infty} \left (1 + \frac{1}{n} \right)^n$$
Graph of $f(x) = \left (1 + \dfrac{1}{x} \right)^x$ taken from here.
Its evident from the graph that the limit actually appro... | An algebraic way is Binomial Expansion, which is given by
$$\begin{eqnarray*}
\left(1+\frac{1}{n}\right)^{\!n} &=& 1+n\left(\frac{1}{n}\right)+\frac{n(n-1)}{2!}\left(\frac{1}{n}\right)^{\!2}+\frac{n(n-1)(n-2)}{3!}\left(\frac{1}{n}\right)^{\!3}+\cdots \\ \\
&=& 1+1+\frac{n(n-1)}{n^2}\cdot\frac{1}{2!}+\frac{n(n-1)(n-2)}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1969934",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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How to evaluate $\lim\limits_{x\to 0} \frac{\arctan x - \arcsin x}{\tan x - \sin x}$ I have a stuck on the problem of L'Hospital's Rule,
$\lim\limits_{x\to 0} \frac{\arctan x - \arcsin x}{\tan x - \sin x}$ which is in I.F. $\frac{0}{0}$
If we use the rule, we will have
$\lim\limits_{x\to 0} \frac{\frac{1}{1+x^2}-\fra... | The first step is OK, but you should try and simplify things before going on.
The numerator is
$$
\frac{1}{1+x^2}-\frac{1}{\sqrt{1-x^2}}=
\frac{\sqrt{1-x^2}-1-x^2}{(1+x^2)\sqrt{1-x^2}}=
\frac{-x^2(3+x^2)}{(1+x^2)\sqrt{1-x^2}(\sqrt{1-x^2}+1+x^2)}
$$
The denominator is
$$
\frac{1}{\cos^2x}-\cos x=\frac{1-\cos^3x}{\cos^2x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1971171",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 3
} |
functional equation of type $f(x)-f(y) = f\left(\frac{x-y}{1-xy}\right)$ If $f(x)-f(y) = f\left(\frac{x-y}{1-xy}\right)$ and $f$ has domain equal to $(-1,1)$, then which of the following is the function satisfying the given functional equation?
Options:
(a) $2-\ln\left(\frac{1+x}{1-x}\right)$
(b) $\ln\left(\frac{1-x}{... | Taking $y=0$ gives $f(0) = 0$. Using this we can rewrite the functional equation as
$$\frac{f(x)-f(y)}{x-y} = \frac{1}{1-xy}\frac{f\left(\frac{x-y}{1-xy}\right) - f(0)}{\left(\frac{x-y}{1-xy}\right)}$$
Now assuming $f$ is differentiable at $x=0$ we can take the limit $y\to x$ above to obtain the ODE $$f'(x) = \frac{1}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1974085",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
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Equation which has to be solved with logarithms I need some help how to solve these equations for $x$. I think I have to use logarithms but still not sure how to do it and would be really grateful if someone could explain me.
$x^2 \cdot 2^{x + 1} +2 ^{\lvert x - 3\rvert + 2}
= x^2 \cdot 2^{\lvert x - 3\rvert + 4} + 2^... | $$x^2 \cdot 2^{x + 1} +2 ^{\lvert x - 3\rvert + 2}
= x^2 \cdot 2^{\lvert x - 3\rvert + 4} + 2^{x - 1}$$
Let us distinguish two cases, $x\ge3$ and $x\le3$, to get rid of the absolute value.
*
*$x\ge3$:
$$x^2 \cdot 2^{x + 1} +2 ^{x-1}
= x^2 \cdot 2^{x+1} + 2^{x - 1},$$ which is an identity !
*$x\le3$:
$$x^2 \c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1976125",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 2
} |
$x+\dfrac{1}{y}=y+\dfrac{1}{z}=z+\dfrac{1}{x}$, prove that $xyz=1$ Let $x>0$, $y>0$ and $z>0$ such that $x\neq y $ , $y\neq z$ and $x\neq z$. If $x+\dfrac{1}{y}=y+\dfrac{1}{z}=z+\dfrac{1}{x}$, prove that $xyz=1$
| Hint: Notice that
\begin{align} x-y &= \frac{y-z}{yz} \\
y-z &= \frac{z-x}{xz} \\
z-x &= \frac{x-y}{xy}.
\end{align}
Now multiply.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1976348",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Finding characteristic features of a parabola known by two points and their tangents This is from a math competition so it must not be something really long
If a parabola touches the lines $y=x$ and $y=-x$ at $A(3,3) $ and $b(1,-1)$ respectively, then
(A) equation of axis of parabola is $2x+y=0$
(B)slope of tangent... | Conditions (A),(B),(C),(D) are not needed to determine the tilted parabola ( $xy$ term non-zero). Because out of five constants needed to determine a conic, one can be reduced as zero determinant for parabola.You have given two points and two slopes which are quite sufficient.
$$ (x, y, y^{ \prime} ) = (3,3,1), (1,-1,... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1977306",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
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Given $P(x)=x^{4}-4x^{3}+12x^{2}-24x+24,$ then $P(x)=|P(x)|$ for all real $x$ $$P(x)=x^{4}-4x^{3}+12x^{2}-24x+24$$
I'd like to prove that $P(x)=|P(x)|$. I don't know where to begin. What would be the first step?
| By completing the square for $x^4+12x^2\cdots$,
$$P(x)=(x^2+6)^2-4x^3-24x-12=(x^2+6)^2-4x(x^2+6)-12\\
=(x^2+6)(x^2-4x+6)-12.$$
The minimum value of the first factor is $6$ and that of the second is $2$.
Or completing the square for $x^2(x^2-4x\cdots)$,
$$P(x)=x^2(x-2)^2+8x^2-24x+24=x^2(x-2)^2+8(x^2-3x+3).$$
As you can... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1981110",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 10,
"answer_id": 0
} |
Transform recursive sequence to direct. I am taking the GRE General Exam in a few weeks and there are some problems about sequences that I have found a bit difficult, e.g given a sequence in recursive form like $S_{n} = S_{n-1} - 10$ and some value for this sequence $S_{3}=0$ what is the value $S_{25}$?
I know that t... | This is easy - given a recursively defined sequence $S_n$ you:
(1) transform it to a form s.t. there are no "exceptions" assuming that $S_n$ is zero for negative $n$. E.g. if $S_n=2S_{n-1} - 4$ and $S_1=6$, the equation $S_n=2S_{n-1} - 4$ (assuming that $S_{-1}$=0) would output $S_0=2*0-4=-4$, so we have to add +4*[n=0... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1984274",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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If $A \in M_3(\mathbb C)$ and $\operatorname{tr}(A)=\operatorname{tr}(A^{-1})=0$ and $\det(A)=1$, then $A^3=I_3$ Assume that we have a $3\times 3$ matrix like $A$ in which the coordinates come from the set of complex numbers ($\mathbb C$).
We know that $\operatorname{tr}(A)=\operatorname{tr}(A^{-1})=0$ and $\det(A)=1$ ... | Solving the set of equations:
$$\begin{cases}
\lambda_1+\lambda_2+\lambda_3=0\ \\
\lambda_1^{-1}+\lambda_2^{-1}+\lambda_3^{-1}=0 \\
\lambda_1\lambda_2\lambda_3=1
\end{cases}
$$
you get $6$ solutions, but we can work just on one of these WLOG, that is:
$$\lambda_1 = 1, \lambda_2 = e^{-i\frac{2\pi}{3}} ~ \text{and} ~ \la... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1986427",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 2
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Trigonometric limit I couldn't find this limit , can someone help me?
$$\lim_{x \to \infty} \frac{\arctan(x+1) - \arctan(x)}{\sin\left(\frac{1}{x+1}\right) - \sin\left( \frac 1x\right)}$$
| Hint. By using Taylor series expansions, as $u \to 0$, one has
$$
\begin{align}
\arctan u&=u-\frac{u^3}{3}+o(u^3)
\\\sin u&=u-\frac{u^3}{6}+o(u^3)
\end{align}
$$ giving, as $x \to \infty$ $\big($with $u=\frac1{x}$ or $u=\frac1{x+1}$$\big)$,
$$
\begin{align}
\arctan x&=\frac{\pi}2-\arctan \frac1x =\frac{\pi}2-\frac1x+O\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1991898",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
$n$-derivative of function $f(x)=e^x \sin x$ at $x=0$ I have function $f(x) = e^x \sin{x}$ and must found $f^{(n)}(0)$
$f'(x) = e^x(\sin{x} + \cos{x}) $
$f''(x) = 2 e^x \cos{x}$
$f'''(x) = 2 e^x (\cos{x} - \sin{x})$
$f''''(x) = -4 e^x \sin{x}$
$f'''''(x) = -4 e^x (\sin{x} + \cos{x})$
I think $f^{(n)}(0) = \alpha (-1)^n... | I think the most appropriate approach is already given by @JackyChong.
Slightly more elaborated we have
\begin{align*}
\color{blue}{\left.\frac{d^n}{dx^n}\left(e^x\sin x\right)\right|_{x=0}}
&=\left.\frac{d^n}{dx^n}\left(\Im e^{(1+i)x}\right)\right|_{x=0}\\
&=\Im \left.\left(\frac{d^n}{dx^n}e^{(1+i)x}\right)\r... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1992070",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
"answer_id": 2
} |
Does ($\frac{p-1}{2})$! satisty $x^2$ = 1modp? I'm trying to show that for p prime = 3 mod4, ($\frac{p-1}{2})!^2$ = 1 modp.
I assume I use Wilson's for the factorial and Euler's Totient Function for the (p-1), but I'm not sure how to string them together.
| It is the case that, for prime $p \equiv 3 \pmod{4}$
$$
\frac{p-1}{2}!^2 \equiv 1 \pmod{p}
$$
To see why, first notice $k(p-k) \equiv -k^2 \equiv -(p-k)^2 \pmod{p}$. Begin with Wilson's Theorem
$$
(p-1)! = 1 \cdot 2 \cdot 3 \cdot 4\cdots (p-4)(p-3)(p-2)(p-1) \equiv -1 \pmod{p}
$$
Now rearrange this product in a clever ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1992547",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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If $A$ is a real $n \times n$ matrix satisfying $A^3 = I$ then Trace of $A$ is always If $A$ is a real $3 \times 3$ matrix satisfying $A^3$ = I such that $ A \neq I $ .Then, Trace of A is always
*
*$0$
*$1$
*$-1$
*$3$
I proceed as follows: from given,
$\min(x)=x-1$ or $x^2+x+1$ or $(x-1)(x^2+x+1)$
$\min(x)\... | Here are two matrices $A$ with $A^3 = A$:
$$
A = \begin{pmatrix}1 & 0 & 0\\ 0 & 1 & 0 \\ 0 & 0 & -1\end{pmatrix}\qquad A = \begin{pmatrix}-1 & 0 & 0\\ 0 & -1 & 0 \\ 0 & 0 & -1\end{pmatrix}
$$
Since these have diffferent traces, the question does not have a definite answer.
Edit: The matrix
$$
A = \begin{pmatrix} 0 & 1 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1994406",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
If $f(x)= \lim_ {n \to \infty} \sum_{r=1}^{n} \frac{n}{n^2+x^2r^2}$, find the required value We have
$$f(x)= \lim_ {n \to \infty} \sum_{r=1}^{n} \frac{n}{n^2+x^2r^2}$$, then find the value of $\sum_{k=1}^{3} k f(k)$.
Could someone give me slight hint as how to find $f(x)$ here.
| Since the terms of the sum are non-negative and monotonically decreasing in $r$, we can write
$$\int_1^{n+1} \frac{n}{n^2+x^2y^2}\,dy \le \sum_{r=1}^n\frac{n}{n^2+x^2r^2}\le \int_0^n \frac{n}{n^2+x^2y^2}\,dy \tag 1$$
Evaluation of the integrals in $(1)$ is straightforward and reveals
$$\frac{\arctan\left(\frac{x}{1+(n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1999572",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Find all $\alpha$ such that the series converges Find all values of $\alpha$ such that series $$\sum^\infty_{n=1} \left( \frac{1}{n \cdot \sin(1/n)} - \cos\left(\frac{1}{n}\right) \right)^\alpha$$ converges.
I used Maclaurin for $\sin$ and $\cos$ and got:
$$a_n = \left( \frac{1}{1 - \dfrac{1}{3!n^2} + \ldots} - 1 + \f... | HINT:
You are on the right track. Note that
$$\frac{1}{1-\frac1{6n^2}+O\left(\frac1{n^4}\right)}=1+\frac1{6n^2}+O\left(\frac1{n^4}\right)$$
Hence, we see that
$$\frac{1}{n\sin(1/n)}-\cos(1/n)=\frac2{3n^2}+O\left(\frac1{n^4}\right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1999706",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Compute the sum of a series given by non constant coefficient recurrence relation $a_k=(\frac{1}{2})^k-\frac{k}{n-k+1}a_{k-1}$ and $a_1=\frac{1}{2}+\frac{1}{n}(\frac{1}{2})^n-\frac{1}{n}$.($n\ge2$).I want to compute $\lim_{n\to\infty}\sum_{k=1}^{n-1}a_k$.
My first attempt is to find out a general form for $a_k$, and I ... | Preliminary Result
Using the Beta Function Integral, we get
$$
\begin{align}
\sum_{k=0}^m\frac{(-1)^k}{\binom{n}{k}}
&=(n+1)\sum_{k=0}^m\int_0^1(-1)^kt^k(1-t)^{n-k}\,\mathrm{d}t\\
&=(n+1)\sum_{k=0}^m\int_0^1(1-t)^n\left(\frac t{t-1}\right)^k\,\mathrm{d}t\\
&=(n+1)\int_0^1(1-t)^{n+1}\left(1-\left(\frac t{t-1}\right)^{m+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1999943",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
An Identity for Pell-numbers The Pell-numbers are defined recursively by:
$P_0 = 0,
P_1 = 1$ and $
P_{n+2} = 2P_{n+1} + P_n$
I am stuck trying to prove the identity:
$P_{2n+1}^2 - 1 = 4P_n P_{n+1}(P_{n+1}^2 - P_n^2)$
A proof would be great, otherwise a list of known Pell-identies would also help
| There is a well-know Pell-identity: $P_{2n+1} = P_n^2 + P_{n+1}^2$ which can be found in this paper. It can also be shown directly.
Using this, we just need to show: $(P_n^2 + P_{n+1}^2)^2 - 1 = 4P_n P_{n+1}(P_{n+1}^2 - P_n^2)$
This is done by induction: The statement clearly holds for $n = 0$, therefore suppose it ho... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2000382",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Is there anything like upper tridiagonal matrix? How to find the determinant of such a matrix? I want to find the determinant of the following matrix.
$$\left[\begin{matrix} -\alpha_1 & \beta_2 & -\gamma_3 & 0 & 0 & 0 & \cdots & 0&0
\\ 0 & -\alpha_2 & \beta_3 & -\gamma_4 & 0 & 0 & \cdots & 0 & 0
\\0 & 0 & -\alpha_3 & \... | The matrix that you put in the question is not square since it has size $(N-2)\times (N-1)$ (you can check by counting the $\alpha's$ in each row and column).
If your matrix is supposed to have the form
$$T_n=\begin{bmatrix}a_1 & b_1 & c_1 & 0 & \cdots &0\\ 0 & a_2 & b_2 & 0 & \cdots&0\\ 0 & 0 & a_3 & b_3 &\cdots & 0\\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2003709",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Evaluate $\lim_\limits{x \to +\infty}\left(x\ln (1+x)-x\ln x + \arctan\frac{1}{2x}\right)^{x^2\arctan x}$ $$\lim_{x \to +\infty}\left(x\ln (1+x)-x\ln x + \arctan\frac{1}{2x}\right)^{x^2\arctan x}$$
My attempt
\begin{align*}
&=\exp \lim_\limits{x \to +\infty} x^2\arctan x \cdot\ln\left[x\ln (1+x)-x\ln x + \arctan\frac{1... | $"=^{*}"$ denotes equality by L'Hospital's Rule.
Define $$\displaystyle f(x) = x\ln\Big(\frac{x+1}{x}\Big)\text{ and }g(x) = f(x)+\arctan\Big(\displaystyle \frac{1}{2x}\Big),\text { for } x > 0.$$ $$\displaystyle 1 = \lim_{x \to \infty}\frac{x}{x+1}=^{*}\lim_{x \to \infty}\frac{\ln\Big(\displaystyle \frac{x+1}{x}\Big)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2009310",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Showing $\sum\limits_{cyc}ab\le\frac14\left(\sum\limits_{cyc}a\right)^2$
$a,b,c,d>0$ with $a+b+c+d=4$ then $\sum\limits_{cyc}ab\le\frac14\left(\sum\limits_{cyc}a\right)^2$
How can I apply AM-GM here, is there a generalization, say we have $n$ variables summing up to $n$ then,
$\sum\limits_{cyc}a_1\cdot a_2\dots a_k\l... | Do we need that $a+b+c+d=4$ beacuse if we full expand we get that
$$a^2+b^2+c^2+d^2+2ac+2bd\geq 2ab+2bc+2cd+2da=2(a+c)(b+d)$$
or
$$(a+c)^2+(b+d)^2\geq 2(a+c)(b+d)$$
Which is true always, or maybe I did a mistake?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2009517",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Solve the integral $\int{\frac{x^2dx}{\sqrt[n]{x^2(3-x)}}} $ My attempt: $$\int{\frac{x^2dx}{\sqrt[n]{x^2(3-x)}}}=\int{\frac{x^2dx}{\sqrt[n]{\left( \frac{3-x}{x} \right)x^3 }}}$$
Let
$$ t^n = \frac{3-x}{x} ~~\rightarrow~~ nt^{n-1}dt=-3\frac{dx}{x^2} $$
or
$$ x = \frac{3}{t^n+1} ~~ \rightarrow~~dx=-\frac{3nt^{n-1}}{(t... | Let $x=3z$
\begin{align}
\int \frac{x^{2-2/n}}{(3-x)^{1/n}} dx &= 3^{-1/n} \int \frac{x^{2-2/n}}{(1-x/3)^{1/n}} dx \\
&= 3^{3-3/n} \int z^{2-2/n} (1-z)^{-1/n} dz \\
&= 3^{3-3/n} \mathrm{B}_{z} \left( 3-\frac{2}{n}, 1-\frac{1}{n} \right) \\
&= 3^{3-3/n} \frac{n}{3n-2} z^{3-2/n} {}_{2}\mathrm{F}_{1}\left(3-\frac{2}{n},\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2011861",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Prime factor of $A=14^7+14^2+1$ Find a prime factor of $A=14^7+14^2+1$. Obviously without just computing it.
| Same solution as Jyrki's, written differently
$$A=14^7+14^2+1=14^7-14+(14^2+14+1)\\
=14(14^6-1)+(14^2+14+1)=14(14-1)(14^2+14+1)(14^3+1)+(14^2+14+1)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2012344",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
If the matrices $A^3 = 0$, $B^3=0$ and $AB=BA$ then show this: The question: If $A$ and $B$ are square matrices of the same type such that $A^3=0$, $B^3=0$ and $AB=BA$. Show that
$$\left(I+A+\frac{A^2}{2!}\right)\left(I+B+\frac{B^2}{2!}\right)=I+(A+B)+\frac{(A+B)^2}{2!}+\frac{(A+B)^3}{3!}+\frac{(A+B)^4}{4!}$$
This is h... | Try to expand the right-hand side instead. You have
\begin{align}
\frac{1}{2}(A + B)^2 &= \frac{1}{2}A^2 + AB + \frac{1}{2}B^2\\[1mm]
\frac{1}{6}(A + B)^3 &= \frac{1}{2}A^2B + \frac{1}{2}AB^2\\[1mm]
\frac{1}{24}(A + B)^4 &= \frac{1}{4}A^2B^2,
\end{align}
where I have used that $A$ and $B$ commute as well as $A^k = B^k ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2012593",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
find upper bound M taylor inequality Let $f(x) = \frac{1}{1-x}$
Find an upper bound M for $|f^{(n+1)}(x)|$ on the interval $[-1/2,1/2]$
I found derivative of the function and plugged in the points to see which one gives the largest value.
$f^{'}(x)=\frac{1}{(1-x)^2}$
$f^{'}(1/2)=4$
The solution was: $2^{n+2}(n+1)!$
I ... | That is because you need to find the largest value of the $n$th derivative of $f$ over the interval [-0.5,0.5]. Given $f(x) = \dfrac{1}{1-x}$,
\begin{align*}
f'(x) & = \frac{1}{(1-x)^2}.\\
f''(x) & = \frac{1\cdot 2}{(1-x)^3}.\\
f'''(x) & = \frac{1\cdot 2\cdot 3}{(1-x)^4}.\\
\vdots & \qquad \vdots \\
f^{(n)}(x) & = \fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2013844",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find all solutions to $m^2 − n^2 = 105$, for which both m and n are integers Can I get a little help for this question?
Find all the integer solutions to $m^2 − n^2 = 105$.
| $(m+n)(m-n) = 105$
$(m+n)(m-n) = 3\cdot 5\cdot 7$
4 Solutions :
Solve them by yourself.
$m-n = 1$, $m+n = 3\cdot 5\cdot 7$.
$m-n = 3$, $m+n = 5\cdot 7$-
$m-n = 5$, $m+n = 3\cdot 7$.
$m-n = 7$, $m+n = 3\cdot 5$.
But notice those $m^2 - n^2$, it's clear that $m$ and/or $n$ could be negative - it would not change a thing... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2013934",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove using induction that $2^{4^n}+5$ is divisible by 21 I have to show, using induction, that $2^{4^n}+5$ is divisible by $21$. It is supposed to be a standard exercise, but no matter what I try, I get to a point where I have to use two more inductions.
For example, here is one of the things I tried:
Assuming that $2... | $2^{4^k}+5 \equiv (-1)^{4^k}+5 \equiv 1+5 \equiv 0 \pmod 3$
Proof by induction that $2^{4^k}+5 \equiv 0 \pmod 7$ for all non negative ingeters $k$.
If $k=0$, then $2^{4^k}+5 \equiv 2+5 \equiv 0 \pmod 7$
Suppose that $2^{4^m}+5 \equiv 0$ for some non negative integer $m$.
Then
$2^{4^{m+1}}+5 \equiv
2^{4\cdot 4^m}+5 \e... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2018239",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Let $p$ be a prime of the form $3k+2$ that divides $a^2+ab+b^2$. Prove that $a,b$ are both divisible by $p$. [Solution verification] Problem:Let $p$ be a prime of the form $3k+2$ that divides $a^2+ab+b^2$ for some $a,b\in \mathbb{Z}$. Prove that $a,b$ are both divisible by $p$.
My Attempt: $a^2+ab+b^2\equiv 0 \pmod p\R... | Start with $a^3\equiv b^3\bmod p$.
If $p$ divides $b$, then $p$ divides $a$.
If $p$ does not divide $b$, there is $c$ such that $bc \equiv 1 \bmod p$. Then $(ac)^3 \equiv 1 \bmod p$.
If $ac \not\equiv 1 \bmod p$, then the order of $ac \bmod p$ is $3$ and so $3$ divides $p-1$, which contradicts $p=3k+2$.
If $ac \equiv ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2018745",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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A Lagrange Mulipliers Problem My problem is this: Find the min and max values of $f(x,y)=x^2+3xy+y^2$ on the domain $(x-1)^2+y^2=1$.
I used lagrange multipliers to find that the $y$ coordinate satisfies $f\left(x\right)=8y+(46y^2)/3-16y^3-24y^4=0$ for any critical point $(x,y)$, and $x=2y^2+(2/3)y$
Using Wolfram Alph... | $f(x,y,\lambda) = x^2 + y^2 + xy + \lambda (x^2 - 2x + y^2)\\
\frac {\partial f}{\partial x} = 2(1+\lambda) x + y - 2\lambda = 0\\
\frac {\partial f}{\partial y} = x + 2(1+\lambda) y = 0\\
\frac {\partial f}{\partial \lambda} = x^2 - 2x + y^2 = 0\\
$
$x,y,\lambda = 0$ is one solution (and is the min)
$2\lambda = -\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2020489",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
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if a,b,c,d are integers such that 5 divides ($a^4+b^4+c^4+d^4$) then 5 also divides (a+b+c+d) i want to show that
if a,b,c,d are integers such that 5 divides ($a^4+b^4+c^4+d^4$) then 5 also divides (a+b+c+d) but im not quite sure how to aproach this problem.
| Hint
$$a^4-1=(a^2-1)(a^2+1)=(a-1)(a+1)(a^2-4+5)=(a-2)(a-1)(a+1)(a+2)+5(a^2-1)$$
Now, if $5 \nmid a$ one of $(a-2),(a-1),(a+1),(a+2)$ is divisible by $5$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2020821",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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Construction of multiplication and addition tables for GF(4) with Modulo ($x^2 + x + 1$) I am still trying to understand polynomial arithmetic with Galois Field.
Can someone explain to me what this question is asking to be constructed. I have searched all over google for GF(4) with modulo and I am not getting anywhere.... | What you're asking saying $GF(4)$ modulo $x^2+x+1$ is equivalent to do:
$$GF(2)\cong \frac{\mathbb{F}_2[x]}{x^2+x+1}$$
So the sum of $GF(2)$ is the same as in $\mathbb{F}_2[x]$ and then doing the modulo $x^2+x+1$: Since all your elements of degree $2$ or greater can be reduced considering $x^n=x^2x^{n-2}=(x+1)x^{n-2}$,... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2020956",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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If $ \sin \alpha + \sin \beta = a $ and $ \cos \alpha + \cos \beta = b $ , then show that $\sin(\alpha + \beta) = \frac {2ab } { a^2 + b^2} $ I've been able to do this, but I had to calculate $ \cos (\alpha + \beta) $ first. Is there a way to do this WITHOUT calculating $\cos(\alpha+\beta)$ first ?
Here's how I did it ... | $\sin \alpha+\sin \beta=2\sin(\frac{\alpha+\beta}{2})\cos(\frac{\alpha-\beta}{2})=a$ (1)
$\cos \alpha+\cos \beta=2\cos(\frac{\alpha+\beta}2)\cos(\frac{\alpha-\beta}2)=b$ (2)
Divide (1) and (2)
We get
$$\tan(\frac{\alpha+\beta}2) =\frac ab$$
We have the formula
$$\sin(\alpha+\beta) =\frac{2\tan(\frac{\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2021356",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
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Compute $\int_{-\infty}^{+\infty}(1+\frac{1}{v^2})\exp(-\frac{u^2}{2\sigma_1^2\sigma_2^2}(\sigma_2v+\frac{\sigma_1}{v})^2)\,dv$
Compute $$\int_{-\infty}^{+\infty}(1+\frac{1}{v^2})e^{-\frac{u^2}{2\sigma_1^2\sigma_2^2}(\sigma_2v+\frac{\sigma_1}{v})^2}\,dv$$
Does this integral has a close form solution? What if $\sigma_... | First we analyze the special case in which $u=\sqrt2$, and $\sigma_1=\sigma_2=1$. Noting that $\left(x+\frac 1x\right)^2=\left(x-\frac 1x\right)^2 +4$, we have
$$\begin{align}
I&=\int_{-\infty}^\infty\left(1+\frac1{x^2}\right)e^{-\left(x+\frac 1x\right)^2}\,dx\\\\
&=e^{-4}\int_{-\infty}^\infty\left(1+\frac1{x^2}\right... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2023685",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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Real values of $a$ for $x^4+(a-1)x^3+x^2+(a-1)x+1=0$ to have at least two negative roots
Problem Statement:-
Find all the values of $a$ for which the equation
$$x^4+(a-1)x^3+x^2+(a-1)x+1=0$$
possess at least two negative roots.
I know that there has been a post regarding this same problem here, but I have a dif... | From the place where you have $x+1/x\; (=z)$ in terms of $a,$ write $x+1/x=r$ where $$r=(-(a-1)-\sqrt {(a-1)^2+4})/2 <0.$$ So $x^2-rx+1=0,$ giving $$x=(r\pm \sqrt {r^2-4}\;)/2<0.$$ This yields two negative roots iff $r^2-4>0,$ that is, iff $|r|>2$ because $r^2-4<0$ yields no real negative $x$, and $r^2-4=0$ yields exac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2023856",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
How can $\frac{x^3-4x^2+4x}{x^2-4}$ be both $0$ and "undefined" when $x = 2$? Suppose I have a function defined as $$F(x)= \frac{x^3-4x^2+4x}{x^2-4}$$
Now I want to find the value of $F(2)$. I can do it in 2 ways:
*
*Put $x=2$ and solve the function. It will give:
$$F(2)=\frac{0}{0}$$ which is not defined.
*Solve $... | The reason is simply that
$$ \frac{x^3-4x^2+4x}{x^2-4}$$
and
$$\frac{x(x-2)}{x+2}$$
are two different expressions.
Their values indeed coincide for $x\ne2$ (and they are both undefined for $x=-2$), but they are not "mandated" to be equal at $x=2$.
This symptom reflects the difference between
$$\frac{x-2}{x-2}$$ and $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2027810",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 2
} |
Find the locus. let $A$ & $B$ be two points on the ellipse $\displaystyle \frac{x^2}{a^2}+\frac{y^2}{b^2} = 1$ such that they subtend $90^{0}$
on the centre of the ellipse. what is the locus of points of intersection of tangents at $A$ & $B$
| Let $\mathcal{E}$ be the ellipse $\left\{\; (u,v) \in \mathbb{R}^2 \;: \;\frac{u^2}{a^2} + \frac{v^2}{b^2} = 1\; \right\}$.
For any point $P = (x,y)$ outside $\mathcal{E}$, there are two tangents of $\mathcal{E}$ passing through $P$.
Let $A = (u_1,v_1)$ and $B = (u_2,v_2)$ be the intersections of these two tangents wit... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2028467",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
On the asymptotic behaviour of the Cauchy product of harmonic numbers $\sum_{k=1}^n H_k H_{n-k+1}$ In this post we take in our hands two simple tools. The first is the generating function of the harmonic numbers, you can see it in this Wikipedia, section 3, that holds for $|z|<1$. The second is the Cauchy product formu... |
Lemma For $n \ge 2$,
$$\begin{align}
\rho_n \stackrel{def}{=}\sum_{k=1}^{n-1} H_k H_{n-k}
&= (n+1)\left[\psi'(n+2)-\psi'(2) + (\psi(n+2)-\psi(2))^2 \right]\\
&= (n+1)\left[\psi'(n+2)-\left(\frac{\pi^2}{6}-1\right) + (H_{n+1} - 1)^2\right]
\end{align}
$$
where $\psi(x)$ is the digamma function.
In following expan... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2029378",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Find $\lim \limits_{n \to \infty}\left( 1 + \sqrt{2} + \sqrt[3]{3} + \dots \sqrt[n]{n} \right) \ln{2n+1 \over n}$ I am looking for
$$\lim \limits_{n \to \infty}\left( 1 + \sqrt{2} + \sqrt[3]{3} + \dots \sqrt[n]{n} \right) \ln{2n+1 \over n}$$
We notice $\ln{2n+1 \over n} = \ln\left({1 + {n+1 \over n}}\right)$. We also k... | Your reasoning is correct. But you could say that all terms in bracket (the first part) are greater than 1, and second part goes to $ln(2)$, which is non zero. So the total limit goes to Infinity.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2029922",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Finding height and area of non-right triangle - Heron's Formula? I would like to calculate the area for a triangle such that $a^2+b^2-c^2=1$ (an almost Pythagorean triple).
I know that the triangle is non-right, so I would like to use $\text{Area}=\frac{1}{2}ab\sin C$... but I do not know how to represent $\sin C$ sinc... | Your equation can be rewritten as $c^2=a^2+b^2-1$. Comparing it with the Law of Cosines $c^2=a^2+b^2-2ab\cos C$, we can see that $2ab\cos C=1$ or $\cos C=\frac{1}{2ab}$. Then
$$\sin C=\sqrt{1-\cos^2C}=\sqrt{1-\left(\frac{1}{2ab}\right)^2}=\sqrt{1-\frac{1}{4a^2b^2}},$$
and the area is
$$\text{Area}=\frac{1}{2}ab\sin C=\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2031032",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Solve system of equation using matrices (4 variables) The Question:
Solve using matrices.
$$2w-2x-2y+2z=10\\w+x+y+z=-5\\3w+x-y+4z=-2\\w+3x-2y+2z=-6$$
My work:
$$
\begin{bmatrix}
2&-2&-2&2&10\\
1&1&1&1&-5\\
3&1&-1&4&-2\\
1&3&-2&2&-6\\
\end{bmatrix}
\rightarrow
\begin{bmatrix}
1&-1&-1&1&5\\
0&2&2&0&-10\\
0&4&-1&1&-1... | The second matrix should be:
$$\begin{bmatrix}1&-1&-1&1&5\\0&2&2&0&-10\\0&4&2&1&-17\\0&4&-1&1&-11\end{bmatrix}$$
Note the row operations were:
$1)\; -R_1+R_2\to R_2\\
2) \;-3R_1+R_3\to R_3\\
3) \;-R_1+R_4\to R_4$
Also, in your "correct" answer, $y$ should be $-2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2036797",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
How many three-digit numbers can be generated from 1, 2, 3, 4, 5, 6, 7, 8, 9, such that the digits are in ascending order? Solution for the same is given below
Numbers starting with 12 – 7 numbers
Numbers starting with 13 – 6 numbers;
14 – 5, 15 – 4, 16 – 3, 17 – 2, 18 – 1.
Thus total number of
numbers starting from ... | You need to choose three digits from nine. Once the digits have been selected, the order in which they should be placed is determined. Hence $\binom{9}{3}$ numbers are possible.
You are seeing binomial coefficients in your method too: once the first digit is fixed, there are $\binom{n}{2}$ ways to choose the remainin... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2037138",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
How do I calculate surface integral? For the vector field $a = [-z^2–2z,-2xz+2y^2,-2xz-2z^2]^T$ and the area $F$ on the cylinder $x^2 + z^2 = 4$ , which is above the ground plane $z = 0$ , in front of the plane $x = 0$ and between the cross plane $y = 0$ and lies to the their parallel plane $y = 2$ , calculate the foll... | If you evaluate the dot product correctly, you get
\begin{align*}
\int_0^2\int_0^{\pi/2} &\left(-8\sin^3 u -8 \sin^2 u - 16 \cos u \sin^2 u - 16 \sin^2 u\right) \,du\,dv\\
&=(-8)\int_0^2 1 \,dv \cdot \int_0^{\pi/2} \left(\sin^3 u + \sin^2u + 2 \cos u \sin^2 u + 2 \sin^2 u\right)\,du \\
&=(-16)\int_0^{\pi/2}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2045013",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
How to find $ker(A)$ How can I find $ker(A)$ for
$A=
\begin{pmatrix}
2 & 1 & 1 \\
4 & 2 & 2 \\
3 & 0 & 1
\end{pmatrix}
$?
What does this represent?
Edit: OK, using the Hints i get
$2x+y+z=0$
$3x+z=0$
$y-x=0$
$y=x$
And now?
| Straight from wikipedia:
$$\ker(A) = \left\{\mathbf{v}\in V : A\cdot \mathbf{x} = \mathbf{0}\right\}$$
In your case $V=\mathbb{R}^3$ so you want to find all vectors in the usual 3 space which make this true:
$$A\bf{x}=
\begin{pmatrix}
2 & 1 & 1 \\
4 & 2 & 2 \\
3 & 0 & 1
\end{pmatrix}\begin{pmatrix}x\\y\\z\end{pmatri... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2045504",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"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.