Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Sum of the series $1+\frac{1\cdot 3}{6}+\frac{1\cdot3\cdot5}{6\cdot8}+\cdots$ Decide if the sum of the series
$$1+\frac{1\cdot3}{6}+\frac{1\cdot3\cdot5}{6\cdot8}+ \cdots$$
is: (i) $\infty$, (ii) $1$, (iii) $2$, (iv) $4$.
| Well , imagination is the magic and you will see the best and most elegent method that follows as such:
$$1+\frac{1\cdot3}{6}+\frac{1\cdot3\cdot5}{6\cdot8}+ \cdots=1+\frac{1\cdot3\cdot(6-5)}{6}+\frac{1\cdot3\cdot5\cdot(8-7)}{6\cdot8}+\cdots$$
$$=1+\frac{1\cdot3\cdot6}{6}-\frac{1\cdot3\cdot5}{6}+\frac{1\cdot3\cdot5\cdot... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/479610",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 8,
"answer_id": 2
} |
relation between inscribed and circumsdribed circle Let T be the triangle with side lengths $b_1,b_2,b_3$, and $r_{insc}$ and $r_{cir}$
be the radii of the inscribed and circumscribed circles,respectively,
I need to show that
$$ \frac {r_{insc}}{r_{cir}}=\frac {(b_2+b_3-b_1)(b_3+b_1-b_2)(b_2+b_1-b_3)}{2b_1b_2b_3}$$
Ma... | Let's side of the triangle be $a,b,c$ and $R$ and $r$ are radii of circumscribed and inscribed circles, respectively.
The equaiton now looks like
$$\frac{r}{R} = \frac{(a+b-c)(a+c-b)(b+c-a)}{2abc}$$
We multiply both sides by $\frac{a+b+c}{8}$
$$\frac{r(a+b+c)}{8R} = \frac{a+b-c}{2} \times \frac{a+c-b}{2} \times \frac{b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/479895",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Matrices determinants problem Could you please give at least a hint to prove this? I really don't know how to start.
Let $A$ be the matrix
$A= \left( \begin{array}{ccc}
A_{11} & A_{12} \\
A_{21} & A_{22} \end{array} \right)$,
where $A_{11}$ and $A_{22}$ are square matrices of order $k$ and $m$ respectively. Prove that ... | Note that
\begin{align*}
\left(\begin{array}{cc}D\cdot A_{11}&D\cdot A_{12}\\A_{21}&A_{22}\end{array}\right)=\left(\begin{array}{cc}D&0\\0&I\end{array}\right)\cdot\left(\begin{array}{cc}A_{11}& A_{12}\\A_{21}&A_{22}\end{array}\right),
\end{align*}
where $I$ is the identity matrix of order $m$. The first matrix in the s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/481818",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Compute $ \int_{-\infty}^{\infty} \frac{x^2}{(1+x^2)^2} dx$ Compute $$ \int_{-\infty}^{\infty} \frac{x^2}{(1+x^2)^2} dx$$
Of course we have $$ \int_{-\infty}^{\infty} \frac{x^2}{(1+x^2)^2} dx = 2 \int_{0}^{\infty} \frac{x^2}{(1+x^2)^2} dx = 2 \int_{0}^{\infty} \left( \frac{x}{1+x^2} \right) ^2 dx = \lim_{ A \to \inft... | I know I'm a bit late for the party, but here is another solution:
$$\begin{align}
2\int_{0}^\infty \frac{x^2}{(1+x^2)^2}\mathrm dx &=
\frac22\int_{-\infty}^\infty \frac{y^{\frac{2-1}{2}}}{(1+y)^2}\mathrm dx \\
&= B\bigl(\tfrac12 + 1, 2-\tfrac12 -1\bigr) = B\bigl(\tfrac32,\tfrac12\bigr) \\
&= \frac{\Gamma\bigl(\tfrac32... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/483180",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 7,
"answer_id": 5
} |
How to show $\sqrt{4+2\sqrt{3}}-\sqrt{3} = 1$ I start with $x=\sqrt{4+2\sqrt{3}}-\sqrt{3}$, then
$\begin{align*}
x +\sqrt{3} &= \sqrt{4+2\sqrt{3}}\\
(x +\sqrt{3})^2 &= (\sqrt{4+2\sqrt{3}})^2\\
x^2 + (2\sqrt{3})x + 3 &= 4+ 2\sqrt{3}\\
x^2 + (2\sqrt{3})x - 1 - 2\sqrt{3} &= 0
\end{align*}$
So I have shown that there is so... | We have $4+2 \sqrt 3= 3+ 2 \sqrt 3+1= \left( \sqrt 3+1\right)^2$. Therefore $\sqrt{4+2 \sqrt 3}= \sqrt 3+1$.
Thus, $\sqrt{4+2 \sqrt 3}- \sqrt 3=1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/483398",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 4,
"answer_id": 3
} |
Solving a system of equations using modular arithmetic modulo 5 Give the solution to the following system of equations using modular arithmetic modulo 5:
$4x + 3y = 0 \pmod{5}$
$2x + y \equiv 3 \pmod{5}$
I multiplied $2x + y \equiv 3 \pmod 5$ by $-2$, getting $-4x - 2y \equiv -6 \pmod{5}$.
$-6 \pmod{5} \equiv 4 \... | You may consult Maple as follows to get that $(2,4)$ is the only solution for the system.
[> msolve({2*x+y = 3, 4*x+3*y = 0}, 5);
{x=2,y=4}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/484467",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 1
} |
Solving a differential equation with composite functions If $c=f(a+e^b)+g(a-e^b)$ where $f$ and $g$ are functions of $a+b^2$ and $a-b^2$ respectively, find $c$ such that when $b=0$, you find that $c=0$ and $\frac{\partial c}{\partial b}=1+a$.
| You say, that
$$
c(a,b) = f(a+e^b) + g(a-e^b)
$$
where $f, g : \mathbb R \to \mathbb R$, and you need to find some particular form of $c(a,b)$ so that
\begin{align}
c(a,0) &= 0\\
\left . \frac {\partial c}{\partial b} \right |_{b = 0} &= 1 + a
\end{align}
Use first restriction.
$$
c(a,0) = f(a+1) + g(a-1) = 0 \tag 1
$$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/485120",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
} |
Help with minimization problem help me, if $x$ and $y$ are real such that $3x-4y = 12$, determine the minimum value of $z = x ^ 2 + y ^ 2$?$$$$I thought of $$3x-4y = 12\Longrightarrow x=4\frac{y+3}{3}\\z = x ^ 2 + y ^ 2\Longrightarrow z = \left(4\frac{y+3}{3}\right) ^ 2 + y ^ 2$$ and then?
| The quantity $x^2+y^2$ is minimized when $\sqrt{x^2+y^2}$ is minimized, and $\sqrt{x^2+y^2}$ is just the distance from the point $\langle x,y\rangle$ to the origin. Thus, you’re looking for the point on the line $3x-4y=12$ that is closest to the origin. Call this line $\ell$; the point closest to the origin is the poin... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/485277",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 0
} |
Doubt in derivative problem I'm in trouble with the following problem:
Give $f(x) = x^3 - 3x + 1$ and $g(x) = x^3 + ax^2 + b$, determine the values of $a$ and $b$ in such a way that both $f(x)$ and $g(x)$ have the same relative maximum and minimum.
I already know that the relative maximum and minimum can be found thoug... | As the comments have pointed out, we must match up the function values at the relative min and max (which, in this case, do not have the same $x$-values).
Since $f'=0$ has $x=\pm1$ as roots, we substitute these critical points into $f$ to obtain $f(-1)=-1+3+1=3$ and $f(1)=1-3+1=-1$. You can verify for yourself that the... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/486796",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
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Proving statements by its contrapositive Prove the following statement by proving its contrapositive:
“If $n^3 + 2n + 1$ is odd then n is even”
Therefore: $\lnot q \rightarrow \lnot p =$ "if $n^3 + 2n + 1$ is even then $n$ is odd.
So for this I began assuming that: $n=2k+1$
$(2k+1)^3 +2(2k+1)+1 = 8k^3+12k^2 +10k+4 = ... | $\displaystyle n^3+2n+1=n^3-n^2+n^2+n+n+1$
$=n^2(n-1)+n(n+1)+n+1\equiv n+1\pmod2$ as the product of any two consecutive integer is divisible by $2$
$\implies n^3+2n+1$ and $n$ has opposite parity
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/486981",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 2
} |
Proving inequality $(a+\frac{1}{a})^2 + (b+\frac{1}{b})^2 \geq \frac{25}{2}$ for $a+b=1$ If $a, b$ are positive real numbers and $a+b = 1$, prove that :
$$\left(a+\frac{1}{a}\right)^2 + \left(b+\frac{1}{b}\right)^2 \geq \frac{25}{2}$$
I can see that the value $\frac{25}2$ is attained for $a=b=\frac12$. But I do not kno... | Generalization:
If $\sum_{1\le r\le n}a_r=S$ where $a_i$s are positive real numbers
$$\sum_{1\le r\le n}\left(a_r+\frac1{a_r}\right)^2=\sum_{1\le r\le n}a_r^2+\sum_{1\le r\le n}a_r^{-2}+2n$$
We know $$\frac{\sum_{1\le r\le n}a_r^m}n> \text{ or } <\left(\frac{\sum_{1\le r\le n}a_r}n\right)^m$$ according as $m$ lies or... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/487486",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 10,
"answer_id": 0
} |
Sum $ \sum\limits_{k=1}^{n} (-1)^k \frac{2k+3}{k(k+1)} $ I have the following Sum
$$ \sum\limits_{n=1}^{\infty} (-1)^n \frac{2n+3}{n(n+1)} $$
and I need to calculate the sums value by creating the partial sums.
I started by checking if $$\sum\limits_{n=1}^{\infty} \left| (-1)^n \frac{2n+3}{n(n+1)} \right|$$ converge... | $\dfrac{2n+3}{n(n+1)} = \color{red}{\dfrac{3}{n}} - \color{blue}{\dfrac{1}{n+1}}$,
so
$$
\sum_{n=1}^{\infty} (-1)^n \dfrac{2n+3}{n(n+1)}
=\color{red}{\sum_{n=1}^{\infty} (-1)^n \dfrac{3}{n}} - \color{blue}{\sum_{n=1}^{\infty} (-1)^n \dfrac{1}{n+1}}.
$$
(all series are convergent here).
A)
$$
\color{red}{\sum_{n=1}^{\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/487695",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
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Limit $\lim\limits_{x\to\infty} \frac{5x^2}{\sqrt{7x^2-3}}$ Evaluate the following limit: $$\lim_{x\to\infty} \frac{5x^2}{\sqrt{7x^2-3}}$$
I'm not really sure what to do when there is a square root for an infinity limit.
Please Help!
| You can divide numerator and denominator by $x$ to find that the limit $\to + \infty$.
$$ \large \frac{5x^2}{\sqrt{7x^2-3}}\cdot \frac{\frac {1}{x}}{\frac{1}{\sqrt{x^2}}} = \frac {\frac {5x^2}{x}}{\sqrt {\frac {7x^2}{x^2} - \frac 3{x^2}}} = \frac {5x}{\sqrt{7 - \frac 3{x^2}}} = \frac {5x}{7}$$
This gives us that $$\li... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/487752",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 0
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Making $R(x)=\frac{x^3+8x^2+18x+15}{x+5}$ continuous from $x = -5$ point The function $R(x)=\frac{x^3+8x^2+18x+15}{x+5}$ is not defined at the point $x = -5$. How should it be defined to make it continuous at this point.
I'm pretty sure you have to factor the top but for some reason I'm having a hard time with this pro... | This is where polynomial long division comes in handy.
We want to remove the removable singularity at $x = -5$, so we test to determine the quotient when dividing the numerator by the factor $(x + 5)$.
Doing so gives us: $$x^3 + 8x^2 + 18x + 15 = (x+5) (x^2 + 3x + 3)$$
This gives us $$\frac{x^3+8x^2+18x+15}{x+5} \impli... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/489711",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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find this limit $\lim_{x\to0^{+}}\frac{\tan{(\tan{x})}-\tan{(\sin{x})}}{\tan{x}-\sin{x}}$ find the limit.
$$\lim_{x\to0^{+}}\dfrac{\tan{(\tan{x})}-\tan{(\sin{x})}}{\tan{x}-\sin{x}}$$
my try:
$$\tan{x}=x+\dfrac{1}{3}x^3+o(x^3),\sin{x}=x-\dfrac{1}{6}x^3+o(x^3)$$
so
$$\tan{(\tan{x})}=\tan{x}+\dfrac{1}{3}(\tan{x})^3+o(\t... | Since all the functions involved are continuosly differentiable we can apply the Mean Value Theorem to get
$$
\frac{\tan(\tan x)-\tan(\sin x)}{\tan x-\sin x}=\frac1{\cos^2 z}
$$
for some $z(x)$ with $\sin x\leq z(x)\leq\tan x$. By the Squeeze Rule, $z\to0$ when $x\to0$, and so the limit is $1/\cos^20=1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/490470",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 2
} |
Compute $\int_0^n \left[\frac x {x+1} + \frac x {2x+4} + \frac x {3x+9} + \cdots\right] \, dx$ for $x>0$ I want to compute $\displaystyle \int_0^n \left[\frac x {x+1} + \frac x {2x+4} + \frac x {3x+9} + \cdots \right] \, dx$ for $x>0$
My attempt: The integrand can be written as a sum of $\displaystyle f_k(x)=\frac x{kx... | \begin{align}
\sum_{k = 1}^{n}{x \over kx + k^{2}}
&=
\sum_{k = 1}^{n}\left({1 \over k} - {1 \over k + x}\right)
=
\sum_{k = 0}^{n - 1}{1 \over k + 1} - \sum_{k = 0}^{n - 1}{1 \over k + x + 1}
\\[3mm]&=
\left\lbrack\Psi\left(1 + n\right) - \Psi\left(1\right)\right\rbrack
-
\left\lbrack\Psi\left(1 + n + x\right) - \Psi\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/493696",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 2
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Relationships between the elements $(a,b,c,d)$ of a solution to $A^2+B^2+4=C^2+D^2$ I have reduced a certain equation (in positive integers) to the equation
$$A^2 + B^2 + 4 = C^2 + D^2. \quad(\star)$$
Assume the positive integers $(a,b,c,d)$ are any solution to $(\star)$. Are there any algebraic restrictions on [i.e., ... | A subset of integer solutions to,
$$A^2+B^2+4 = C^2+D^2\tag{1}$$
can be reduced to solving the Pell equation $x^2-ny^2 = 1$ for any non-square $n$ (where $n=5$ will involve the Fibonacci numbers). Consider the more general,
$$x_1^2+x_2^2+x_3^2 = y_1^2+y_2^2+y_3^2\tag{2}$$
The complete solution is,
$$(a+b)^2+(c+d)^2+(e+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/494534",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 0
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Calculate $(2013)^4 - 4(2011)^4+6(2009)^4-4(2007)^4+(2005)^4 = $ Calculate $(2013)^4 - 4(2011)^4+6(2009)^4-4(2007)^4+(2005)^4 = $
Try:: Let $x = 2009$, Then expression convert into $(x+4)^4-4(x+2)^4+6x^4-4(x-2)^4+(x-4)^4$
$\left\{(x+4)^4+(x-4)^4\right\}-4\left\{(x+2)^4+(x-2)^4\right\}+6x^4$
But This is very Complicated... | Hint: This is a finite differences problem.
If $x(n)$ is any function, then $$(\Delta^4 x)(n)= x(n+4)-4x(n+3)+6x(n+2)-4x(n+1)+x(n)$$
Where $\Delta$ is the finite difference operator.
Now, since $x(n)=(2n+1)^4$, is a polynomial of degree $4$ with lead coefficient $2^4$ we know that $(\Delta^4x)(n)$ is constant, equal t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/494686",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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Different Ways of Integrating $3\sin x\cos x$ I am asking this question for my son who is in (equivalent) twelfth
grade and I failed to answer his query.
When he tries to integrate $3\sin x\cos x$, he finds that
this can be done in at least following three ways.
And these three ways do not produce equivalent results.
O... | Actually, we are to find the antiderivative of the given function, and all the functions you have found out to be potential answers are equally correct because any of them is just a constant shifted version of another.
And these are not the only candidate to be the answers, we can find infinitely many more. When we d... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/495159",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 3,
"answer_id": 2
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Finding the equation of a circle given two points and the radius Can't seem to figure this out - the question is:
There are exactly two circles of radius $r = \sqrt{5}$ through the points $(6,3)$ and $(7,2)$. Find the equations of both circles.
I was thinking that I would find the equation of the line passing through... | Here is a mostly 'square free' approach:
The two constraints are $\|x-(6,3)\|^2 = 5$ and $\|x-(7,2)\|^2 = 5$. Expanding gives $\|x\|^2-2 \langle (6,3), x \rangle +45 = 5$ and $\|x\|^2-2 \langle (7,2), x \rangle +53 = 5$,
subtracting and simplifying gives $\langle (1,-1), x \rangle = 4$. Note that this is a line at $45 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/496070",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 9,
"answer_id": 2
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LU decomposition by hand Can someone show me a step by step solution to calculate the $LU$ decompisition of the following matrix:
$A = \begin{bmatrix}
5 & 5 & 10 \\
2 & 8 & 6 \\
3 & 6 & -9
\end{bmatrix}$
I am getting the correct U by I am having difficult getting the correct L.
Here is the co... | $A = \begin{pmatrix}
5 & 5 & 10 \\
2 & 8 & 6 \\
3 & 6 & -9
\end{pmatrix}$
First step (working on the first column):
a) Multiply the first row with $l_{21}= \frac{2}{5}$ and subtract it from row 2
b) Multiply the first row with $l_{31}= \frac{3}{5}$ and subtract it from row 2
This leads to
$U^{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/496889",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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How do I solve this Partial Fractions question? $$\frac{x^4}{(x^2-3)(x^2+3)}\;$$
How would I do this?
My attempt started with this:
$$\frac{A+Bx}{(x^2-3)}\; + \frac{C+Dx}{(x^2+3)}$$
But when I start working it all out, A, B, C and D all go to 0. What am I doing wrong?
| Now that the $x$ in the numerator is really supposed to be $x^4$, we first need the degree in the numerator to be less than that of the denominator before employing partial fraction decomposition. Note that the denominator can be expressed as the difference of squares: $x^4 - 9$, which makes polynomial division very st... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/497443",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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Improving bound on $\sqrt{2 \sqrt{3 \sqrt{4 \ldots}}}$ An old challenge problem I saw asked to prove that $\sqrt{2 \sqrt{3 \sqrt{4 \ldots}}} < 3$. A simple calculation shows the actual value seems to be around $2.8$, which is pretty close to $3$ but leaves a gap. Can someone find $C < 3$ and prove $\sqrt{2 \sqrt{3 \sq... | Reusing my answer on Quora and showing how it can be used to improve bounds
Write the expression as
$$\displaystyle 2\sqrt[2]{\frac{3}{2}}\sqrt[4]{\frac{4}{3}}\sqrt[8]{\frac{5}{4}}\sqrt[16]{\frac{6}{5}}...$$
$$\displaystyle \prod\limits_{n=1}^{+\infty} \left ( 1 + \frac{1}{n} \right )^{\frac{1}{2^{n-1}}}$$
$$\displayst... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/498774",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 6,
"answer_id": 5
} |
Multiplication of summations The length of the following equation depends on the value of the variable n. How can I write this equation? How can I reformulate it?
$$f(n)=x+y\cdot\sum_{a=1}^{n}\left(x+y\cdot\sum_{b=1}^{n-1}\left(x+y\cdot\sum_{c=1}^{n-2}\left(x+y\cdot\sum_{d=1}^{n-3}....\right)\right)\right)$$
Moreover I... | Given that summing $m$ times the same thing amounts tom multiplying it by $m$, you get
$$
f(n)=x+yn(x+y(n-1)(x+y(n-2)(\cdots(x+y.1(x+0y))\ldots)))
$$
which resembles a polynomial in $y$ expressed by a Horner scheme; you get
$$
\begin{align}
f(n)&=x+xny+xn(n-1)y+n(n-1)(n-2)y^3x+\cdots+n(n-1)\ldots2xy^{n-1}+n!xy^n
\\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/499985",
"timestamp": "2023-03-29T00:00:00",
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steady state solution to differential equation - checking my work EDIT: fixed a stray negative sign.
The problem as given:
$y'' + 2y' + 5y = 10\cos t$
We want to find the general solution and the steady-state solution. We're using $\mu y'' + c y' + k y = F(t)$ as our general form.
OK, so I first want the general so... | Answering my own question and taking Artem's advice, I tried this....
$y'' + 2y' + 5y = 10\cos t$
We want to find the general solution and the steady-state solution. We're using $\mu y'' + c y' + k y = F(t)$ as our general form.
the characteristic equation is $\lambda^2 + 2\lambda + 5$ and it has imaginary roots at ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "2",
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Skew operator Squared Proof I have a linear algebra question that is a proof and I am unsure how to approach this problem.
For any vector $s$, show that $$(s\times)^2 = ss^T-s^TsI$$ where the skew operator $s\times$ is defined by $$s\times\triangleq\begin{bmatrix}0&-c_s&b_s\\c_s&0&-a_s\\-b_s&a_s&0\end{bmatrix}$$
| So you want to show
$$\begin{bmatrix} 0 & -c & b \\ c & 0 & -a \\ -b & a & 0\end{bmatrix}^2=\begin{bmatrix}a \\ b \\ c\end{bmatrix}\begin{bmatrix} a & b & c\end{bmatrix}-(a^2+b^2+c^2)\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{bmatrix}.$$
What's wrong with simplifying both sides and seeing that they're the ... | {
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How prove this $ \sqrt{\frac{a}{a+3b+5bc}}+\sqrt{\frac{b}{b+3c+5ca}}+\sqrt{\frac{c}{c+3a+5ab}}\geq 1.$ Let $a,b,c$ be nonnegative real numbers such that $a+b+c=3$, Prove that
$$ \sqrt{\frac{a}{a+3b+5bc}}+\sqrt{\frac{b}{b+3c+5ca}}+\sqrt{\frac{c}{c+3a+5ab}}\geq 1.$$
This problem is from http://www.artofproblemsolving.... | Let $\displaystyle A = \sqrt{\frac{a}{a+3b+5bc}}+\sqrt{\frac{b}{b+3c+5ca}}+\sqrt{\frac{c}{c+3a+5ab}}$ and $\displaystyle B = \sum_{cyc}a^2(a+3b+5bc)$.
Then by Hölder's inequality we have $A^2B \ge (a+b+c)^3 = 27$.
So it is sufficient to prove that $B \le 27$
$$B = \sum_{cyc}a^3 + 3 \sum_{cyc}a^2b+5\sum_{cyc}a^2bc$$
As... | {
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Common linear and Quadratic factors (i) If $ax^5+bx^2+c$ has a factor of the form $x^2+px+1$ prove that:
$(a^2-c^2)(a^2-c^2+bc)=(ab)^{2}$
(ii)In this case prove that:
$ax^5+bx^2+c$ and $cx^5+bx^3+a$ have a common quadratic factor
| Part (i) is a continuation of Part (ii).
m is a root of g(x) = 0 and hence is a root of f(x) = 0.
Therefore, $am^5 + bm^2 + c = 0$……(4)
Using n = 1/m as root, from part (ii), we have $a + bm^3 +cm^5 = 0$…….(5)
Eliminating a from (4) & (5) by performing $(5)*m^5 – (4)$, we have
$bm^2(m^6 – 1) + c[m^T – 1] = 0$……(6) [T =... | {
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integrating $\int \frac{dt}{(t+2)^2(t+1)}$ I'm practicing to solve a whole, and I am not able to solve this one, could you help me? $$\int \frac{dt}{(t+2)^2(t+1)}$$I tried $$\frac{1}{(t+2)^2(t+1)}=\frac{A}{(t+2)^2}+\frac{B}{(t+2)}+\frac{C}{(t+1)}\\1=A(t+1)+B(t+2)(t+1)+C(t+2)^2\\t=-2\Longrightarrow 1=-A\Longrightarrow \... | Differentiating $$\frac{1}{t+2} - \ln |t + 2| + \ln|t+1| + c$$ gives you $$-\frac{1}{(t+2)^2} - \frac{1}{t + 2} + \frac{1}{t+1}$$ which is certainly what you got when you split the fraction initially. To check that your A, B and C were correct, see that this is just $$\frac{-(t + 1) - (t + 2)(t+1) + (t+2)^2}{(t+2)^2(t+... | {
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"source": "stackexchange",
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minimal polynomial of a matrix with some unknown entries Question is to prove that :
characteristic and minimal polynomial of $ \left( \begin{array}{cccc}
0 & 0 & c \\
1 & 0 & b \\
0 & 1 & a \end{array} \right) $ is $x^3-ax^2-bx-c$.
what i have done so far is :
characteristic polynomial of a matrix $A$ is given by $\... | Here's how I would argue the point: clearly the minimal polynomial $m_A(x)$ of $A$,
where
$A = \begin{bmatrix} 0 & 0 & c \\ 1 & 0 & b \\ 0 & 1 & a \end{bmatrix}, \tag{1}$
must satisfy $\deg m_A(x) \le 3$, since the characteristic poynomial $p_A(x)$ of $A$ satisfies $\deg p_A(x) = 3$. Furthermore, we also have $m_A(... | {
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"question_score": "4",
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"answer_id": 2
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If $ k = \sqrt{n\cdot (n+1)\cdot(n+2)\cdot (n+3)}$. Then $\lfloor k \rfloor =$ If $ k = \sqrt{n\cdot (n+1)\cdot(n+2)\cdot (n+3)}$, where $n\in \mathbb{N}$. Then $\lfloor k \rfloor = $
$\underline{\bf{My\; Try}}::$ We can write the expression $n\cdot (n+1)\cdot(n+2)\cdot (n+3) = (n^2+3n).(n^2+3n+2)$
$ = (n^2+3n)^2+2\cdo... | You are using the right approach, we need to establish upper and lower bounds on k. The lower bound needs to be an integer and the upper bound must be no greater than 1 plus the lower bound. This will result in the lower bound being the floor of k. jim established the correct upper bound by adding 1 under the radical.
... | {
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"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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Finding the product of real and imaginary roots separately If I am given a polynomial of nth degree and asked to fond the product of real and imaginary roots what steps should I take?
I know how to calculate the sum or product of all roots of a polynomial of nth degree but how to separately find the product of real an... | This is too long for a comment. Just found a partial answer for $x^4+a_3x^3+a_2x^2+a_1x+a_0=0$. Let $y$ and $x$ denote the product of the real and complex roots and $b$ and $a$ the sums of the real and complex roots resp. Playing Vieta with the coefficients and those sums and product, you will find the following sy... | {
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"question_score": "2",
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How find this equation $\prod\left(x+\frac{1}{2x}-1\right)=\prod\left(1-\frac{zx}{y}\right)$ let $x,y,z\in(0,1)$, find the pairs of $(x,y,z)$ such
$$\left(x+\dfrac{1}{2x}-1\right)\left(y+\dfrac{1}{2y}-1\right)\left(z+\dfrac{1}{2z}-1\right)=\left(1-\dfrac{xy}{z}\right)\left(1-\dfrac{yz}{x}\right)\left(1-\dfrac{zx}{y}\ri... | It is fairly obvious that, since the expressions on both sides are symmetrical with regards to each of the three variables, they can easily be reduced to $$\left(t + \frac1{2t} - 1\right)^3 = \left(1 - \frac{t\ \cdot\ t}t\right)^3 \qquad\iff\qquad \left(t + \frac1{2t} - 1\right) = (1 - t) \quad | \cdot 2t$$ $$2t^2 + 1 ... | {
"language": "en",
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"source": "stackexchange",
"question_score": "9",
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Diophantus math Find two numbers such that their difference and also the difference of their cubes are given numbers; say, their difference is 6 and the difference of their cubes are 504. Call the numbers $x + 3$ and $x - 3$.
| Well, using the difference of cubes formula $a^3-b^3=(a-b)(a^2+ab+b^2),$ we get $$\begin{align}504 &= (x+3)^3-(x-3)^3\\ &= \bigl((x+3)-(x-3)\bigr)\left((x+3)^2+(x+3)(x-3)+(x-3)^2\right)\\ &= 6\bigl((x^2+6x+9)+(x^2-9)+(x^2-6x+9)\bigr)\\ &= 6(3x^2+9)\\ &= 18x^2+54.\end{align}$$ You can take it from there, yes?
| {
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How many even numbers of four digits can be formed with the digits 0,1,2,3,4,5 and 6 no digit being used more? My attempt to solve this problem is:
First digit cannot be zero, so the number of choices only $6 (1,2,3,4,5,6)$
The last digit can be pick from $0,2,4,6$, so the number of choices only 4
Second digit can be o... | Consider two cases:
Case 1: first digit even
There are three ways to find the first digit, if the first digit is even.
There are 2,4, and 6.
There are three ways to find the last digit, because one is taken for the first digit.
There are 5 ways to find the second digit and 4 ways to find the third digit.
So, the total ... | {
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"source": "stackexchange",
"question_score": "4",
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Inequality....(RMO $1994$...question $8$) If $a$, $b$, $c$ are positive real numbers such that $a+b+c = 1$, prove that
$$(1+a)(1+b)(1+c) ≥ 8(1-a)(1-b)(1-c)\text{.}$$
| One quite natural start is to replace $1-a$ by $b+c$ as there are more inequalities for nonnegative reals than reals. (In this case all terms are positive but minus signs are ugly in this problem so let's remove them.) So we have to prove that $$(1+a)(1+b)(1+c)\geq 8(b+c)(a+c)(a+b).$$ Also, there are number eight in th... | {
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"question_score": "8",
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Prove that if $\gcd(a,b)=1$, then $\gcd(a^2,b^2)=1$ So, if $\gcd(a,b)=1$, then $\gcd(a^2,b^2)=1$ means $1=ax+by$, and want to show $a^2x+b^2y=1$.
By squaring $1=ax+by$ both sides, I get, $1=(ax)^2+2(ax)(by)+(by)^2$, but this doesn't help my proof.
| First show
$\gcd(a, b^2) = \gcd(a^2, b) = 1$
$\gcd(a,b) = 1$ means that there are $x$, $y$ ($x$, $y$ are integers) such that $ax + by =1$
$ax + by(ax+by) =1$
$a(x+bxy) + b^2y^2 =1$
That means $\gcd(a,b^2)=1$
By a similar derivation, $\gcd(a^2,b)=1$.
$ar_1(ar_2+b^2s_2) +b^2s_1 = 1$
$a^2r_1^2+b^2(ar_1s_1+s_1)=1$
Therefo... | {
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"source": "stackexchange",
"question_score": "10",
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How to solve this system of equation? I need to solve the following system of $(x,y)$:
\begin{cases}
3y^3+3x\sqrt{1-x}=5\sqrt{1-x}-2y\\
x^2-y^2\sqrt{1-x}=\sqrt{2y+5}-\sqrt{1-x}
\end{cases}
| The first equation can be written as
$x=1-y^{2}$
because it is manipulated in this way.
We bring to the first member the terms in x and to the second member the terms in y:
$(5-3x)\sqrt{1-x}=y(3y^{2}+2)$.
We see this equation as the equality of two products, that is:
$\sqrt{1-x}=y$,
$-3x+5=3y^{2}+2$.
The solution of th... | {
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"question_score": "1",
"answer_count": 3,
"answer_id": 1
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How do I prove $2^{n+1} + 2n + 1 = 2^{n+2} - 1$ I am attempting to prove using induction:
$\sum_0^n 2i = 2^{n + 1} - 1$
I have gotten to the point where I need to show:
$2^{n+1} + 2n + 1 = 2^{n+2} - 1$
How do I prove this? Or should I be proving the initial question a different way
| You missed the power of $2$ in the Left hand side
and the range of $i$ should start from $0$ instead of $1$
We find $\displaystyle\sum_0^1 2^i =1+2=3 $ and $ 2^{1 + 1} - 1=4-1=3$
So, $\displaystyle\sum_0^n 2^i = 2^{n + 1} - 1$ is true for $n=1$
Let $\displaystyle\sum_0^n 2^i = 2^{n + 1} - 1$ is true for $n=m$
$$\impl... | {
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"source": "stackexchange",
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Determine $p$ for one to have $\displaystyle\frac{1}{5}+\frac{1}{6}=\frac{1+1}{5+6}=\frac{2}{11}$, in $\mathbb{Z_p}$. Determine $p$ for one to have $\displaystyle\frac{1}{5}+\frac{1}{6}=\frac{1+1}{5+6}=\frac{2}{11}$, in $\mathbb{Z_p}$.
I know that $p$ must, $13, 43, 61, 101,103$.
| In addition to $p=121-60=61$ we can use the extended Euclidean algorithm to obtain $5^{-1}=-12$, $6^{-1}=-10$ and $11^{-1}=-11$ in $\mathbb{Z}/61 \mathbb{Z}$. Hence
$$
\frac{1}{5}+\frac{1}{6}=-12-10=-22=2\cdot (-11)=\frac{2}{11}.
$$
| {
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"source": "stackexchange",
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"answer_id": 1
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How to solve this symmetric system of equations? How many solutions are there to this equation?
$$\begin{align*}
x^2-y^2&=z\\
y^2-z^2&=x\\
z^2-x^2&=y
\end{align*}$$
| We have
\begin{align}
x^2 - y^2 & = z\\
y^2 - z^2 & = x\\
z^2 - x^2 & = y
\end{align}
Adding the first two equations, we get that
$$x^2-z^2 = z+x \implies z=-x \text{ or }x-z=1$$
Similarly, we get that
$$x=-y \text{ or }y-x=1$$
$$y=-z \text{ or }z-y=1$$
Hence, there are $8$ possible choices of getting $3$ equations. Bu... | {
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"source": "stackexchange",
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The sum of three square roots bounded below by $\sqrt{82}$
Let $a,b,c >0$ and $a+b+c \le 1$. Prove: $$\sqrt{a^2+1/a^2}+\sqrt{b^2+1/b^2}+\sqrt{c^2+1/c^2} \ge \sqrt{82}$$
Progress: I tried to have 3 vectors $(a,1/a)$, $(b,1/b)$ and $(c,1/c)$, played around with the vector inequalities but no success.
| since $f（x)=x^2+\dfrac{1}{x^2}$ is mono decreasing function when $x \in (0,1] $, we can prove $a+b+c=1$ case first, for any $a'+b'+c' <1$, let $p(a'+b'+c')=1 \implies p>1$, then $a=pa'>a',b=pb'>b',c=pc'>c' \implies \sum \sqrt{a'^2+\dfrac{1}{a'^2}}>\sum\sqrt{a^2+\dfrac{1}{a^2}}$
to prove the case $a+b+c=1$, it is same a... | {
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"source": "stackexchange",
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"answer_id": 3
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$x,y,z$ positive real numbers , $x+y+z=3$ $\implies x^4y^4z^4(x^3+y^3+z^3)≤3$ If $x,y,z$ are positive real numbers with $x+y+z=3$ then how to prove (without using calculus) that $\space$ $x^4y^4z^4(x^3+y^3+z^3)≤3$ ?
| We have,
$$(x+y+z)^3=x^3+y^3+z^3+3(x+y)(y+z)(x+z) \ge x^3+y^3+z^3+\frac{8}{3}(x+y+z)(xy+yz+xz) = x^3+y^3+z^3+(xy+yz+xz)+(xy+yz+xz)..[8 \text{times}] \ge 9\sqrt[9]{(x^3+y^3+z^3)(xy+yz+xz)^8}$$
Since $$(xy+yz+xz)^2 = x^2y^2+y^2z^2+x^2z^2+2xyz(x+y+z) \ge 3xyz(x+y+z)$$ we have,
$$ 9\sqrt[9]{(x^3+y^3+z^3)(xy+yz+xz)^8} \ge... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 5,
"answer_id": 0
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How would one discover this finite product identity? I recently found the following finite product identity in a table of products:
\begin{align}
\prod_{k=0}^{n-1}\left[\sinh^2y+\sin^2\left(x+\frac{k\pi}{n}\right)\right]=2^{1-2n}(\cosh(2ny) -\cos(2nx))
\end{align}
I'd like to know how one would have gone about discov... | Discovering such identities often is just recognising things one has already seen elsewhere.
The factors of the product are easily recognised as
$$\left\lvert\sin \left(x+\frac{k\pi}{n} +iy\right) \right\rvert^2$$
when one has some experience with complex analysis. Euler's formulae make the connection between the trigo... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove that x is rational Let $x$ be a real number with the properties that $x^3+x$ and $x^5+x$ are rational.
Prove that $x$ is rational. Denote $a=x^3+x$; $b=x^5+x$. We can multiply and add them together until we get the desired result. I also know some non-elementary proofs of this, but have you some nice elementary p... | We may take $x>0$. $\frac{x^5+x}{x^3+x}$ is rational, so $bx^4+b = ax^2 +a$ for some integers $a,b$ and $x = \sqrt{k}$, where $k=c+\sqrt{d}$ is the root of a quadratic equation, with $c,d\in\mathbb{Q}$, $d=0$ or $d>0$ not a square.
If $k$ is rational ($d=0$) but not a square, $x^2-1$ is rational and $x^5+x = (x^3+x)(x^... | {
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"source": "stackexchange",
"question_score": "11",
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How prove this $\overrightarrow{r_{i}}\cdot\overrightarrow{r_{j}}\ge\frac{1}{2}$ let $a,b,c$ are real numbers,and such $a+b+c=0,a^2+b^2+c^2=1$, we define:
$\overrightarrow{r}=(x_{i},y_{i},z_{i})(i=1,2,3,4,5,6)$,where $\{x_{i},y_{i},z_{i}\}=\{a,b,c\}$,
show that: there are exst $\overrightarrow{r_{i}}\neq\overrightarro... | The points $r_i$ lie on a unit circle (the intersection of the plane $x+y+z=0$ and the sphere $x^2 + y^2 + z^2 = 1$). There isn't enough "room" on a circle for 6 points that are all more than 60 degrees apart. So, there must be two of them (say $r_i$ and $r_j$) whose angles differ by 60 degrees or less. Then $\vec{r_i}... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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On $n! + k^2 = m^2$ Brocard's problem is the case $k=1$ of,
$$n!+k^2 = m^2\tag{1}$$
However, for general $k$,
$$4!+1 = 5^2\tag{2}$$
$$6!+3^2 = 27^2\tag{3}$$
$$8!+9^2 = 201^2\tag{4}$$
and using the first two, we can observe two polynomial identities,
$$(x+1)(x+2)(x+3)(x+4) + 1 = (x^2+5x+5)^2\tag{5}$$
$$(2x+1)(2x+2)(3... | Let's pretend I haven't seen any of the identities above. I just know the fact that $4!+1=5^2$ and I want to find an identity of this form:
$$
(ax+1)(bx+2)(cx+3)(dx+4) + 1 = P(x)^2
$$
Where $a, b, c, d \in \mathbb{N}$ and $P\in \mathbb{N}[x]$. Well, since $4!+1=5^2$, we know that the constant term of $P$ is $5$. So now... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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Evaluating $\int_0^1 \frac{\log x \log \left(1-x^4 \right)}{1+x^2}dx$ I am trying to prove that
\begin{equation}
\int_{0}^{1}\frac{\log\left(x\right)
\log\left(\,{1 - x^{4}}\,\right)}{1 + x^{2}}
\,\mathrm{d}x = \frac{\pi^{3}}{16} - 3\mathrm{G}\log\left(2\right)
\tag{1}
\end{equation}
where $\mathrm{G}$ is Catalan's Con... | $$I=\int_0^1 \frac{\ln x \log \left(1-x^4 \right)}{1+x^2}dx$$
Let,
\begin{align*}
\displaystyle A&=\int_0^1 \dfrac{x\arctan x\ln x}{1+x^2}dx\\
\displaystyle B&=\int_0^1 \dfrac{\ln x \ln(1+x^2)}{1+x^2}dx\\
\displaystyle C&=\int_0^1 \dfrac{\arctan x\ln x}{1+x}dx\\
\end{align*}
From Evaluating $\int_0^{\pi/4} \ln(\tan x)\... | {
"language": "en",
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"question_score": "33",
"answer_count": 7,
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} |
Prove that $\log^25 + \log^27 > \log12$. Prove that $\log^25 + \log^27 > \log12$.
What I tried so far:
$\log^25 + \log^27 > \log3 + \log4$
$(\log5 + \log7)^2 - 2 \cdot \log5 \cdot\log7 > \log3 + \log4$
But it seems that I'm not even near the result.
Every suggestion / hint would be appreciated :)
Thanks in advance.
EDI... | Without actually computing exact logs,...
$$\log_{10}5 \cdot \log_{10}5 + \log_{10}7\cdot \log_{10}7 > \log_{10}12$$
$$\iff \frac{\log_{10}5}{\log_{10}7} + \frac{\log_{10}7}{\log_{10}5} > \frac{\log_{10}12}{\log_{10}5 \cdot \log_{10}7}$$
Now LHS $>2$ as it is the sum of a positive number ($\neq 1$) and its reciprocal... | {
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"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
How to find the determinant of this matrix I have the following matrix:
$
\begin{bmatrix}
a & 1 & 1 & 1 \\
1 & a & 1 & 1 \\
1 & 1 & a & 1 \\
1 & 1 & 1 & a \\
\end{bmatrix}
$
My approach is to rref the matrix so that i can find the determinant by multiplying along the diagonals.
I attempted to do an rref and ended u... | How about the LU decomposition (modulo a rearrangements of rows)?
\begin{align*}
&\left[\begin{array}{cccc}0&1&0&0\\0&0&0&1\\0&0&1&0\\1&0&0&0\end{array}\right]\times\left[\begin{array}{cccc}
a&1&1&1\\
1&a&1&1\\
1&1&a&1\\
1&1&1&a\end{array}\right]\\=&\left[\begin{array}{cccc}1&0&0&0\\1&1&0&0\\1&1&1&0\\a&1+a&-1&1\end{arr... | {
"language": "en",
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"question_score": "3",
"answer_count": 5,
"answer_id": 1
} |
Likelihood of sum of dice roll is exactly 1 million A board game is set up such that there is a number line with squares numbered 0..1 million. You roll a standard 6 sided die and move forward the number of spaces that you roll. Eventually you will either land on, or pass the 1 millionth spot. What is the probability y... | Let $S(\omega) \subset \mathbb{N}$ be the spaces that are landed on during trial $\omega$. That is, if $d_k(\omega) \in \{1,...,6\}$ are the die rolls, then $S(\omega) = \{ \sum_{k=1}^n d_k(\omega) \}_{n=1}^\infty $.
For $k \ge 1$, let $r_k(\omega) = \min \{ j-k | j \ge k, \ j \in S(\omega) \}$. Note that $r_k(\omega) ... | {
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"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
Did I answer this probability question correctly? Hello I have the following question to answer:
Imagine three boxes, each of which has three slips of paper in it each with a number
marked on it. The numbers for box A are 2, 4 and 9, for box B 1, 6 and 8, and for box C
3, 5 and 7. One slip is drawn, independently and w... | There are $3$ equally likely possibilities for A's number, $2$, $4$, and $9$.
If A got $2$, then the probability that A beat B is $\frac{1}{3}$, for B must get $1$.
If A got $4$, then again the probability that A beat B is $\frac{1}{3}$.
If A got $9$, then the probability that A beat B is $1$, or more prettily $\frac... | {
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"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
How to find the $f^{-1}(x)$ of $f(x)=x^{3}-12x+\frac{48}{x}-\frac{64}{x^{3}}$ It is a question from a quiz.
The following is the whole question.
Let
\begin{eqnarray}
\\f(x)=x^{3}-12x+\frac{48}{x}-\frac{64}{x^{3}} , \space x\in (-\infty ,0),
\end{eqnarray}
find $f^{-1}(x)$. Hint : $f(x)$ can be written in the form,... | First of all, $f$ does not have a inverse function because you always have two root of $f - y$. One positive root, one negative root.
If you add the constraints that $x$ is positive and $y$ is real, then there will be an inverse function.
$$(x - \frac4x)^3=y$$
Since $y$ is real:
$$x - \frac4x=\sqrt[3]y$$
Since $x$ is n... | {
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"source": "stackexchange",
"question_score": "3",
"answer_count": 7,
"answer_id": 5
} |
Prove $u_n$ has the limit given Let $u_n = \sqrt{u_n+1}~ and ~u_1=1$. Prove that $\displaystyle\lim_{n\to\infty} u_n = \frac{1}{2}(1+\sqrt5)$
Here is what I did.
First, to show the sequence converges:
$u_1 = 1, \space
u_2 = \sqrt 2 = 2^{(1/2)}, \space
~u_3 = \sqrt {1 + 2^{(1/2)}} < \sqrt{2*2^{(1/2)}} = 2^{(3/4)} $
t... | $$\left. \begin{align} u_{_{\infty\ +\ 1}} &= \sqrt{u_\infty + 1} \\ \\ \infty + 1 &= \infty \end{align}\ \right\} \iff u_\infty = \sqrt{u_\infty + 1} \iff u_\infty^2 = u_\infty + 1 \iff u_\infty^2 - u_\infty - 1 = 0$$ $$u_\infty = \frac{1\pm\sqrt5}2$$ Obviously, since a radical is always positive, the only possible so... | {
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"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Show that $\frac{(1-y^{-1})^2}{(1-y^{-1 -c})(1-y^{-1+c} )}$ is decreasing in $y > 1 $. I am interested in the function
$f(y) = \frac{(1-y^{-1})^2}{(1-y^{-1 -c})(1-y^{-1+c} )},$
for values of $c \in (0,1)$, and $y > 1$, and have been trying to show that the function is decreasing.
I have tried differentiating the functi... | Let $f_c(x)=\frac{(1-x^{-1})^2}{(1-x^{-1-c})(1-x^{-1+c})}$. Then $\frac{\partial}{\partial c}(\ln f_c(x))=-\ln x(\frac{1}{x^{c-1}-1}+\frac{1}{x^{c+1}-1}+1)$, so since $f_0(x)=1$, it suffices to show that $\frac{1}{x^{c-1}-1}+\frac{1}{x^{c+1}-1}$ is increasing with respect to $x > 1$. $\frac{1}{x^{c-1}-1}$ is negative a... | {
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"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 1
} |
Prove that there are no positive integers $a,b,c,d$ such that $a^2 + b^2 = c^2$ and $a^2 - b^2 = d^2$ Prove that there are no positive integers $a,b,c,d$ such that $a^2 + b^2 = c^2$ and $a^2 - b^2 = d^2$?
Could you show me the proof?
| Summing and subtracting, you get $$\begin{cases} 2a^2=c^2+d^2\\2b^2=c^2-d^2\end{cases}$$ Thus $c,d$ have the same parity. If they are both even, we obtain $2a^2=4c'^2+4d'^2$ and $2b^2=4c'^2-4d'^2$, so $$\begin{cases} a^2=2(c'^2+d'^2)\\b^2=2(c'^2-d'^2)\end{cases}$$
Since $2\mid a^2,b^2$, we must have $a^2=4a'^2$, $b^2=4... | {
"language": "en",
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"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 3
} |
None exact first order ODE i have to solve the following $1^{st}$ order differential equation
$(xy+1)dx+(2y-x)dy=0$
i am in the elementary differential class,and have not learned multivariate functions,
the equation below is none exact,since
$M_y=x\ne N_x=-1$
so i am looking for a substitution that can make it exact b... | Approach $1$:
$(xy+1)~dx+(2y-x)~dy=0$
$(xy+1)~dx=(x-2y)~dy$
$(x-2y)\dfrac{dy}{dx}=xy+1$
Let $u=\dfrac{x}{2}-y$ ,
Then $y=\dfrac{x}{2}-u$
$\dfrac{dy}{dx}=\dfrac{1}{2}-\dfrac{du}{dx}$
$\therefore2u\left(\dfrac{1}{2}-\dfrac{du}{dx}\right)=x\left(\dfrac{x}{2}-u\right)+1$
$u-2u\dfrac{du}{dx}=\dfrac{x^2}{2}-xu+1$
$2u\dfrac{d... | {
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"answer_count": 2,
"answer_id": 0
} |
Prove that $\sum_{k=0}^{m}\frac{\binom{m}{k}}{\binom{n}{k}}=\frac{n+1}{n+1-m}$ Prove that $\sum_{k=0}^{m}\dfrac{\binom{m}{k}}{\binom{n}{k}}=\dfrac{n+1}{n+1-m}$.
We know, n > m. From the right side. we have $\dfrac{n+1}{n+1-m}=\dfrac{1}{1-\dfrac{m}{n+1}}$. since n > m. $0<\dfrac{m}{n+1}<1$. Then
$\dfrac{1}{1-\dfrac{m}{n... | Partial Fractions
Using the Heaviside Method for Partial Fractions, we get
$$
\begin{align}
\sum_{k=0}^m\frac{\binom{m}{k}}{\binom{n}{k}}
&=1+\sum_{k=1}^m\frac{m(m-1)(m-2)\dots(m-k+1)}{n(n-1)(n-2)\dots(n-k+1)}\tag{1}\\
&=1+m\sum_{k=1}^m\sum_{j=0}^{k-1}\frac{(-1)^{k-j-1}\binom{m-1}{k-1}\binom{k-1}{j}}{n-j}\tag{2}\\
&=1+... | {
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"source": "stackexchange",
"question_score": "8",
"answer_count": 4,
"answer_id": 2
} |
How can I find the value of $\ln( |x|)$ without using the calculator? I want to know if there is a way to find for example $\ln(2)$, without using the calculator ?
Thanks
| We can represent the logarithm of positive rational numbers as follows.
First, consider the following null conditionally convergent series (cancelled harmonic series):
$$0=(1-1)+\left(\frac{1}{2}-\frac{1}{2}\right)+\left(\frac{1}{3}-\frac{1}{3}\right)+\left(\frac{1}{4}-\frac{1}{4}\right)+\left(\frac{1}{5}-\frac{1}{5}\r... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 9,
"answer_id": 7
} |
$123^{561}$ find last $2$ digits Modular Exponentiation mod $100$? $123^{561}$
Find the last $2$ digits
Can't I work this out using modular exponentiation working mod $100$?
$123^2 = 29\pmod{100}$
$123^4 = (123^2)^2 = 41\pmod{100}$
$123^8 = (123^4)^2 = 81\pmod{100}$
$123^{16} = (123^8)^2 = 61\pmod{100}$
$123^{32} = (12... | While dealing with modulo of composite number, Carmichael function is general more, if not equally efficient than Euler's Totient Theorem .
Here $\lambda(100)=$lcm$(\lambda(5^2),\lambda(2^2))=$lcm$(20,4)=20$
$\implies a^{20}\equiv1\pmod{100}\ \ \ \ (1)$ if $(a,100)=1\iff (a,10)=1$
whereas $\phi(10^2)=10\cdot\phi(10)=4... | {
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"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
How to prove $\sum_{k=1}^n \frac{2^k}{k}< 3\frac{2^n}{n}$? How to prove $$\sum_{k=1}^n \frac{2^k}{k}< 3\frac{2^n}{n}$$ and further $$\lim_{n\rightarrow \infty}\frac{n}{2^n}\sum_{k=1}^n
\frac{2^{k}}{k} = 2$$?
These results are verified by computer, yet I can't figure out a neat proof.
| I'll answer the second part of the question. It is helpful to notice that the sum
$$
S_n = \sum_{k=1}^n \frac{2^k}{k}
$$
is dominated by only a few terms near $k=n$. In a sense, the terms near the end constitute the principal contribution to the whole sum, while the terms near the beginning are negligible. We can th... | {
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"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 0
} |
Improper Integral $\int\limits_0^1\frac{\ln(x)}{x^2-1}\,dx$ How can I prove that?
$$\int_0^1\frac{\ln(x)}{x^2-1}\,dx=\frac{\pi^2}{8}$$
I know that
$$\int_0^1\frac{\ln(x)}{x^2-1}\,dx=\sum_{n=0}^{\infty}\int_0^1-x^{2n}\ln(x)\,dx=\sum_{n=0}^{\infty}\frac{1}{(2n+1)^2}=\frac{\pi^2}{8}$$
but I want another method.
| We have$$\int_0^1\frac{\ln(x)}{x^2-1}\,\mathrm dx=\int_0^1\frac{\ln(1-x)}{(1-x)^2-1}\,\mathrm dx=\int_0^1\frac{\ln(1-x)}{x(x-2)}\,\mathrm dx$$
We will generalize by introducing parameter $\alpha$ such that
$$I(\alpha )=\int_0^1\frac{\ln(1-\alpha x )}{x(x-2)}\,\mathrm dx$$
And we have $I(0)=0$
Then
$$I'(\alpha )=-\int_0... | {
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"source": "stackexchange",
"question_score": "20",
"answer_count": 8,
"answer_id": 1
} |
Calculating Value Of Trigonometric Formula when $~\tan\theta+\sin\theta=\dfrac{1}{2}$, evaluate $~(\sin^{2}\theta-\sin 2\theta)$
is it possible to get the exact value?
I got $~\sin^{2}\theta-\sin 2\theta=\dfrac{\tan^{2}\theta-2\tan\theta}{1+\tan^{2}\theta}$, but can't calculate it more.
| $\large \tan \theta + \sin \theta=\frac{1}{2} $; Squaring we get
$\large\tan^2 \theta+2.\sin \theta.\tan \theta+\sin^2 \theta=\frac{1}{4}$
$\large\frac{\sin^2 \theta}{\cos^2 \theta}+2.\frac{\sin^2 \theta}{\cos\theta}+\sin^2 \theta=\frac{1}{4}$
$\large\sin^2\theta[\frac{1}{\cos^2\theta}+2.\frac{1}{\cos\theta}+1]=\frac{1... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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If $a^3 + b^3 +3ab = 1$, find $a+b$
Given that the real numbers $a,b$ satisfy $a^3 + b^3 +3ab = 1$, find $a+b$.
I tried to factorize it but unable to do it.
| $$a^3+b^3-1+3ab=(a+b)^3-1-3a^2b-3ab^2+3ab=$$
$$=(a+b-1)((a+b)^2+a+b+1)-3ab(a+b-1)=$$
$$=(a+b-1)(a^2+b^2-ab+a+b+1).$$
Thus, $$a+b=1$$ or
$$a^2+b^2-ab+a+b+1=0$$ or
$$(a-b)^2+(a+1)^2+(b+1)^2=0,$$ which gives $a=b=-1$ and $$a+b=-2.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/540084",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 9,
"answer_id": 8
} |
How to derive this second derivative using the quotient rule? If a given first derivative is: $\ {dy \over dx} = {-48x \over (x^2+12)^2} $
What are the steps using the quotient rule to derive the second derivative: $\ {d^2y \over dx^2} = {-144(4-x^2) \over (x^2+12)^3} $
My Steps:
\begin{align*}
{d^2y \over dx^2}
&= {-4... | Recall the Quotient Rule:
$$(\frac{f}{g})' = \frac{f'\cdot g - g' \cdot f}{g^2}$$
So you have the first derivative, $\frac{dy}{dx}= \frac{-48x}{(x^2+12)^2}$. Just apply the quotient rule:
Let $f(x) = -48x$. So $f'(x) = -48$.
Let $g(x) = (x^2 + 12)^2$. So, by the chain rule, $g'(x) = 2(x^2 + 12) \cdot 2x = 4x(x^2+12)$.
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/541696",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
How to find the minimum $n$
Find the smallest $n$ such that the sequence
$1000n,1000n+1,1000n+2,\cdots,1000n+2013$ does not contain a perfect square.
My try:
$n=1$ does not work because
$$1000n=1000,1000n+1=1001,\cdots,1000n+4=1024,\cdots,1000n+2013=3013$$
contains the square numbers $32^2=1024,33^2=1089,\cd... | Let $x^2$ be the largest perfect square $\leq 1000n-1$. Then $(x+1)^2 \geq 1000n+2014$ so $(2x+1)=(x+1)^2-x^2 \geq 2015$ so $x \geq 1007$.
Put $x=1000+y, y \geq 0$, so that $x^2 \equiv y^2 \pmod{1000}$. If $y \leq 30$, then $x^2 \equiv 0, 1, 4, \ldots ,900 \pmod{1000}$ so $x^2 \leq 1000n-100$. Thus $(2x+1)=(x+1)^2-x^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/542340",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Convergence of $\frac{1}{\sqrt{2\pi i\varepsilon}}\exp\left( -\frac{x^2}{2i\varepsilon}\right)$ as $\varepsilon \to 0$ Is it true that $\frac{1}{\sqrt{2\pi i\varepsilon}}\exp\left( -\frac{x^2}{2i\varepsilon}\right)$ converges (in some sense) to $\delta_0$, Dirac delta distribution at point $0$, as $\varepsilon \to 0$ ... | So the $i$ is intentional. Then let's look how that function acts as a distribution, hence let $\varphi$ an arbitrary test function:
$$\begin{align}
\frac{1}{\sqrt{2\pi i\varepsilon}}\int_{-\infty}^\infty \exp \left(i\frac{x^2}{2\varepsilon}\right)\varphi(x)\,dx &= \frac{1}{\sqrt{2\pi i\varepsilon}}\int_{-\infty}^\inft... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/542662",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Calculate $a^3+b^3+c^3+d^3$ for the real roots of $x^4+2x^3-3x^2-3x+2$ $a,b,c,d \in \mathbb{R}$ are the real roots of $x^4+2x^3-3x^2-3x+2$.
Calculate $a^3+b^3+c^3+d^3$.
With approximation i found out, that $a^3+b^3+c^3+d^3 = -17$, but how can I proof that without calculating the roots exactly?
Cheers
| thanks for your help.
$P(x) = x^4 + 2x^3 -3x^2 -3x + 2:$
$\sigma_1 = -2$
$\sigma_2 = -3$
$\sigma_3 = 3$
$\sigma_4 = 2$
$a^3 + b^3 + c^3 + d^3 = \sigma_1^3 - 3*\sigma_1 * \sigma_2 + 3 * \sigma_3$
We get: $a^3 + b^3 + c^3 + d^3 = (-2)^3 - 3 * (-2) * (-3) + 3 * 3 = -17$
| {
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"url": "https://math.stackexchange.com/questions/544499",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 2
} |
Prove: Sum and Difference of two distinct positive integers are both perfect squares. I'm trying to prove that there exists two distinct positive integers whose sum and difference are both perfect squares. I cannot find any pattern or characteristic between the pairs of numbers that work i.e.
*
*4, 5
*6, 10
*8,... | Let those positive integers be $a$ and $b$.Then,
$a+b=x^2$
$a-b=y^2$
Adding the two we get $a=\frac{x^2+y^2}{2}$ and $b=\frac{x^2-y^2}{2}$
Since, a and b are integers we must have $x^2+y^2$ and $x^2-y^2$ must be even, for that we must have x and y both even or both odd.
Now for finding such pairs take any even $x,y$ fo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/545123",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Find asymptotics in a given form $n=(e+o(1))^{f(s)}$
Let $p\to\infty$, $s={\binom {p^4} p}$ and $n={\binom {p^4}{p^2}}$. Find a function $f(s)$ in the following form $$\large n=(e+o(1))^{f(s)}$$
I've tried to use the followinf asymptotics for binomial coeffs:
$$\binom n k = \frac{n^k \exp\left(k^2/2n\right)}{k!}(1+O... | For the beginning some preparatory computations:
$$
\begin{align}
{n\choose k}
&=\frac{n\ldots (n-k+1)}{k!}\\
&=\frac{n^k}{k!}\left(1-\frac{1}{n}\right)\ldots\left(1-\frac{k-1}{n}\right)\\
&=\frac{n^k}{k!}\exp\left(\sum_{i=1}^{k-1}\ln\left(1-\frac{i}{n}\right)\right)\\
&=\frac{n^k}{k!}\exp\left(\sum_{i=1}^{k-1}\left(-\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/545980",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Stochastics with induction prove that for all $n \in \mathbb{N}$:
$\sum_{r=0}^n \binom{n}{r}(-1)^{r} = 0$.
The base step is easy, i only get lots of problems when i try to mess with the sum boundries....
so far i've tried:
$\sum_{r=0}^{n+1} \binom{n+1}{r}(-1)^{r} = \sum_{r=0}^{n+1}(\binom{n}{r}+\binom{n}{r-1})(-1)^{r} ... | Perhaps this way:
$$
\begin{split}
\sum_{r=0}^{n+1} \binom{n+1}{r} (-1)^r
&= \binom{n+1}{0} + \binom{n+1}{n+1} (-1)^{n+1}
+ \sum_{r=1}^n \binom{n+1}{r} (-1)^r \\
&= 1 + (-1)^{n+1}
+ \sum_{r=0}^{n-1} \binom{n+1}{r+1} (-1)^{r+1} \\
&= 1 + (-1)^{n+1}
- \sum_{r=0}^{n-1} \binom{n+1}{r+1} ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/548059",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Find the Area of the ellipse Given $$\frac{x^2}{a^2} + \frac{y^2}{b^2}=1$$
where $a>0$, $b>0$
I tried to make $y$ the subject from the equation of the ellipse and integrate from $0$ to $a$. Then multiply by $4$ since there are $4$ quadrants.
$$Area=4\int^a_0\left(b^2-x^2\left(\frac{b^2}{a^2}\right)\right)^\frac{1}{2}dx... | Here is my proof if it is any use to anyone.
The equation of an ellipse is given by: $$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$$
Rearrange and write this ( the equation of an ellipse ) in terms of y: $$y=b\sqrt{1-\frac{x^2}{a^2}}$$
Lets find the area of one quarter of the ellipse and multiple that by 4 to get the area of the... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/548876",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 5,
"answer_id": 3
} |
Using Newton's method calculate $ \frac{1}{\sqrt{a}} $ without division Suggest algorithm for the numerical calculation $ \frac{1}{\sqrt{a}} \ a > 0 $ without division, use Newton's method.
My idea is:
$$ \frac{1}{\sqrt{a}} = (\sqrt{a})^{-1} = a^{-\frac{1}{2}}$$
$$ x = a^{-\frac{1}{2}} \Rightarrow f(x)=x^2 - a^{-1}$$... | After using dark magic to make an initial guess $y_0$, the fast inverse square root algoritm essentially approximates $y=1/\sqrt{x}$, $x > 0$, by using the iteration
$$
y_{n+1} = \frac{3}{2}y_n - \frac{1}{2}xy_n^3.
$$
There is no division here, unless you count the multiplication by the constants $3/2$ and $1/2$.
If th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/550177",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
Proving the form of a sequence's terms How do I go about attacking this problem and what is it asking?
Suppose that $\alpha^2 = \alpha + 1$ and suppose $F_n$ denotes the Fibonacci sequence.
Show that $\alpha^3 = 2\alpha +1, \alpha^4 = 3\alpha +2$
| More explicitly,
since
$x^2 = x+1$,
$x^3 = x(x^2)
=x(x+1)
=x^2+x
=(x+1)+x
=2x+1
$.
Do the same for
$x^4$.
Essentially,
any power of $x$,
say $x^n$ with $n \ge 3$,
can be replaced
by lower powers of $x$
via
$x^n
=x^{n-2}x^2
=x^{n-2}(x+1)
=x^{n-1}+x^{n-2}
$.
Keep on applying this
until only linear and constant terms are ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/550934",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Simplifying square root with fraction I'm not sure about this equality $$4(-3+\sqrt {15})/4)^2 = (9-6 \sqrt{15} +15)/4$$
Hope some one can enlighten me. I will be facing more of such fractions, please guide me on how to solve/simplify in easy method.
Thanks :)
| I’ll even finish the simplification:
$$\begin{align*}
4\left(\frac{-3+\sqrt{15}}4\right)^2&=4\cdot\frac{(-3+\sqrt{15})(-3+\sqrt{15}}{4\cdot4}\\
&=\frac{(-3)^2+2(-3)\sqrt{15}+(\sqrt{15})^2}4\\
&=\frac{9-6\sqrt{15}+15}4\\
&=\frac{24-6\sqrt{15}}4\\
&=6-\frac32\sqrt{15}\;.
\end{align*}$$
In short: cancel a factor of $4$, a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/555689",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Show that $\frac{(2n)!}{(n)!}=2^n(2n-1)!!$ Show that $\frac{(2n)!}{(n)!}=2^n(2n-1)!!$ is the question I am struggling with.
I started by saying: $(2n)!=2n(2n-1)(2n-2)(2n-3)...3*2*1$
But then I'm stuck.
| $$\frac{2n!}{n!} = \frac{1\cdot2\cdot\ldots\cdot2n}{1\cdot2\cdot\ldots \cdot n}=\frac{1\cdot3\cdot\ldots\cdot(2n-1)\cdot2\cdot4\cdot\ldots\cdot2n}{1\cdot2\cdot\ldots \cdot n} = \frac{1\cdot3\cdot\ldots(2n-1)\cdot2^n\cdot 1\cdot2\cdot\ldots \cdot n}{1\cdot2\cdot\ldots \cdot n} = 2^n\cdot1\cdot3\cdot\ldots \cdot (2n-1)=2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/558104",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Elementary proof that $\pi < \sqrt{5} + 1$ I wanted to show that
$$ \frac{\pi}{4\phi} < \frac{1}{2} $$
Where $\phi$ is the golden ratio. I have confirmed the results numerically, and by
simple algebra the inequality simplifies down to
$$
\pi < \sqrt{5} + 1
$$
This is a weaker relation than what was shown here. Prove t... | Below are two geometric ways to get better approximations than the one requested. Neither of the two involve the expression $\sqrt{5}+1$ directly so it would still take some work to show that these results are in fact stronger. However, a trick combining both results will show that $\pi < \frac{16}{5}$ which is quick... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/560315",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 6,
"answer_id": 3
} |
Find the inverse of a matrix with a variable $$X=
\begin{pmatrix}
2-n & 1 & 1 & 1 & \ldots & 1 & 1 \\
1 & 2-n & 1 & 1 & \ldots & 1 & 1 \\
\vdots & \vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\
1 & 1 & 1 & 1 & \ldots & 2-n & 1 \\
1 & 1 & 1 & 1 & \ldots & 1 & 2-n\end{pmatrix}_{n\times n}
$$
Which means that the ... | Hint: Use the Sherman–Morrison formula on $$X=\text{diag}(1-n)_{n\times n}+\begin{pmatrix} 1\\ \vdots \\1\end{pmatrix}_{n\times 1}\begin{pmatrix}1 & \ldots &1 \end{pmatrix}_{1\times n}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/561047",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Prove that $\sin(x + \alpha)$, $\sin(x + \beta)$ and $\sin(x + \gamma)$ are linearly dependent.
Functions $f$ and $g$ are independent on an interval $D$ if $af(x) + bg(x) = 0$ implies that $a = 0$ and $b = 0$ $\forall x \in D$
let $\alpha$, $\beta$, $\gamma$ be real constants. Prove that
$\sin(x + \alpha)$, $\sin(x +... | Explain how you got that expression for $W(f,g,h)$ equal to $0$. Otherwise the argument is good.
Another idea: consider rewriting the functions with the sum-angle formulas, then interpret the three functions as three vectors in a two-dimensional vector space spanned by $\sin$ and $\cos$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/562034",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 1
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Proving that a polynomial about the volume of a tetrahedron is irreducible We know that the volume of a tetrahedron $ABCD$ can be represented as
$$144V^2=(a^2b^2d^2+b^2c^2e^2+c^2a^2f^2+b^2a^2e^2+c^2b^2f^2+a^2c^2d^2+c^2e^2f^2+a^2f^2d^2+b^2d^2e^2+c^2d^2f^2+a^2e^2d^2+b^2f^2e^2)-(a^2b^2c^2+a^2e^2f^2+b^2f^2d^2+c^2d^2e^2)$$
... | Notice there are no linear terms in $a$. So it is a polynomial of the form $Aa^2+B$. So, $B/A$ must be a square.
$A=b^2(d^2+e^2+f^2e^2+c^2)+(e^2f^2+c^2f^2+c^2d^2+f^2d^2+c^2d^2f^2+e^2d^2)$
and
$B=b^2(f^2d^2+c^2e^2+c^2f^2+d^2e^2)+(c^2d^2e^2+c^2e^2f^2)$
if I didn't miss some term.
Check that $A$ and $B$ don't have a comm... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/563923",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Easy exponents question I have the GRE Friday... I got hung up on this easy exponents problem (I think it was these exponents, don't recall exactly)
$$\frac{6^{14}}{2^7 \times 3^5} = ? $$
The answer is $2^73^9$, but could anyone double check for me?
| Essentially yes. You need to use the rules $(xy)^a = x^a y^a$ and also $x^ax^b = x^{a+b}$ (the latter of which sometimes looks like $\frac{x^a}{x^b} = x^a x^{-b} = x^{a-b}$ for division). Here you first factor $6$ and then simplify. Explicitly:
$$
\frac{6^{14}}{2^7\cdot 3^5} = \frac{(2\cdot3)^{14}}{2^7\cdot 3^5} = \fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/566200",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 1
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Inequality: $ab^2+bc^2+ca^2 \le 4$, when $a+b+c=3$. Let $a,b,c $ are non-negative real numbers, and $a+b+c=3$.
How to prove inequality
$$
ab^2+bc^2+ca^2\le 4.\tag{*}
$$
In other words, if $a,b,c$ are non-negative real numbers, then how to prove inequality
$$
27(ab^2+bc^2+ca^2)\le 4(a+b+c)^3.\tag {**}
$$
$\color{gray}... | Let $\{a,b,c\}=\{x,y,z\}$, where $x\geq y\geq z$.
Hence, by Rearrangement and AM-GM we obtain:
$$ab^2+bc^2+ca^2=ab\cdot b+bc\cdot c+ca\cdot a\leq xy\cdot x+xz\cdot y+yz\cdot z=y(x^2+xz+z^2)\leq$$
$$\leq y(x+z)^2=4y\left(\frac{x+z}{2}\right)^2\leq4\left(\frac{y+\frac{x+z}{2}+\frac{x+z}{2}}{3}\right)^3=4.$$
Done!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/566768",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 2
} |
Probability of group I have seen a problem in Probability that I am stuck :
Imagine that we have $12$ students in class and among this students there are $3$ honor students. Say that a teacher wants to assign a group project and wants to balance the groups out by forming 3 groups of students with exactly one honor stud... | Imagine that we line up the students at random, and assign the first $4$ to one group, the next $4$ to another, and the last $4$ to another.
We can analyze as follows. We have $3$ copies of the letter $H$, and $9$ copies of the letter $M$. We make a $12$-letter word. What is the probability there will be an $H$ among... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/567299",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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On the equation $(1-x)^2/x + (1-y)^2/y + (1-z)^2/z + 4 = 0$ The problem is to solve the equation,
$$\frac{(1-x)^2}{x} + \frac{(1-y)^2}{y} + \frac{(1-z)^2}{z} + 4 = 0\tag{1}$$
in the rationals. Treating this as an equation in $z$, easy solutions would involve $z = \pm 1$, $z = \pm x, \pm y$. More complicated ones would ... | After some further thought and computation, if $f=(k^2-1)/2k$ then $u=2(k-1)^2$ gives
\begin{equation*}
v= \pm \frac{(k-1)^2(k^2-2k-1)(k^2-2k+3)}{2k^2}
\end{equation*}
This point gives the parametric form
\begin{equation*}
f=\frac{k^2-1}{2k} \hspace{2cm}
g=\frac{2k}{k^2-1} \hspace{2cm} h=\frac{k(1-k)}{k+1}
\end{equati... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/569536",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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Prove $\left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\right)^{2}\geq (a+b+c)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)$
Prove that $$\left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\right)^{2}\geq (a+b+c)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)$$ for $a,b,c>0$
Any hints/solutions?
| Expanding the terms, we want to show that
$$ \frac{a^2}{b^2} + \frac{b^2}{c^2} + \frac{c^2}{a^2} + \frac{a}{c} + \frac{b}{a} + \frac{c}{b} \geq 1 + 1 + 1 + \frac{a}{b} + \frac{b}{c} + \frac{c}{a} $$
This is true because
$ \frac{a^2}{b^2} + \frac{b}{a} + \frac{b}{a} \geq 3$ and $ \frac{a^2}{b^2} + 1 \geq 2 \frac{a}{b}$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/570527",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Evaluate the limit $\lim_{x\rightarrow 0} \frac{\sqrt{1-\sin(5x)}-\sqrt{1+\sin(5x)}}{x^2+x}$ Trying to find
$$\lim_{x\rightarrow 0} \dfrac{\sqrt{1-\sin(5x)}-\sqrt{1+\sin(5x)}}{x^2+x}=\lim_{x\rightarrow 0} \dfrac{(1-\sin(5x))-(1+\sin(5x))}{(x^2+x)(\sqrt{1-\sin(5x)}+\sqrt{1+\sin(5x)})}=\lim_{x\rightarrow 0} \dfrac{-2\s... | Recall
$$\sin^2(a) + \cos^2(a) = 1$$
Hence,
$$1 \pm \sin(2a) = ( \cos(a) \pm \sin(a))^2$$
For small $x$, we have
$$\sqrt{1 \pm \sin(5x)} = ( \cos(5x/2) \pm \sin(5x/2))$$
Use this along with the fact that $\lim_{x \to 0} \dfrac{\sin(\alpha x)}{x} = \alpha$ to conclude the limit.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/570665",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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How prove this $\sqrt[5]{1782+\sqrt[3]{35+15\sqrt{6}}+\cdots}$ is positive integer numbers. Prove that
$$\sqrt[5]{1782+405\sqrt[3]{35+15\sqrt{6}}+405\sqrt[3]{35-15\sqrt{6}}}-\sqrt[3]{35+15\sqrt{6}}-\sqrt[3]{35-15\sqrt{6}}\in N$$
This problem from this
My try: let
$$x=\sqrt[3]{35+15\sqrt{6}}+\sqrt[3]{35-15\sqrt{6}}$$
t... | This isn't a complete answer, but I think it might be enough for you to go on:
Note that $x=-3$ is a root of the polynomial. Therefore $x+3$ divides $x^5-405x-972$. Dividing yields $x^4-3x^3+9x^2-27x-324$. It turns out that $x=-3$ is a root of this quotient as well. Dividing again by $x+3$ yields $x^3-6x^2+27x-108$. Th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/571426",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 2,
"answer_id": 0
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Evaluate $ \sum\limits_{n=1}^{\infty}\frac{n}{n^{4}+n^{2}+1}$ The question was: Evaluate, ${\textstyle {\displaystyle \sum_{n=1}^{\infty}\frac{n}{n^{4}+n^{2}+1}}}.$
And I go, since $\frac{n}{n^{4}+n^{2}+1}\sim\frac{1}{n^{3}}$ and we know that ${\displaystyle \sum_{n=}^{\infty}\frac{1}{n^{3}}}$ converges. so ${\displays... | HINT:
As $n^4+n^2+1=(n^2+1)^2-n^2=(n^2+1-n)(n^2+1+n)$
and $(n^2+1+n)-(n^2+1-n)=2n$
$$\frac n{n^4+n^2+1}=\frac12\left(\frac{2n}{(n^2+1-n)(n^2+1+n)}\right)$$
$$=\frac12\left(\frac{(n^2+1+n)-(n^2+1-n)}{(n^2+1-n)(n^2+1+n)}\right)$$
$$=\frac12\left(\frac1{n^2-n+1}-\frac1{n^2+n+1}\right)$$
Also observe that $: (n+1)^2-(n+1)+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/571973",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 6,
"answer_id": 2
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Solution of differential equation $x^3dy -x\ln x dx = (1+x^2) \tan^{-1}x dx -x dy $.... Solution of differential equation
$x^3dy -x \ln x dx = (1+x^2) \tan^{-1}x dx -x dy $ ( options are )
(a) $y = \frac{\ln x}{\tan^{-1}x}+c$
(b) $y = \tan^{-1}x \ln x +c$
(c) $y =\frac{\tan^{-1}x}{\ln x }+c$
(d) none of these
... | It follows from integration by parts that $$\int\frac{\tan^{-1}x}{x}dx =\tan^{-1}x\ln x-\int \frac{\ln x}{1+x^2}dx$$ and hence we have $$\int\frac{\tan^{-1}x}{x}dx+\int \frac{\ln x}{1+x^2}dx=\tan^{-1}x\ln x+c.$$ Therefore, $y=\tan^{-1}x\ln x+c$ is the solution.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/572061",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Find a power series solution centered at 0 (Differential equations Here's the problem: $$(x-1)y''+y'=0$$ This is the work that I've already done: $$y=\sum_{n=0}^{\infty}a_{n}x^n$$
$$y'=\sum_{n=0}^{\infty}(a_{n+1})(n+1)x^n$$
$$y''=\sum_{n=0}^{\infty}(a_{n+2})(n+2)(n+1)x^n$$
I then plug those into the original equation:... | Setting
$$
y(x)=\sum_{n=0}^\infty a_nx^n,
$$
we have
\begin{eqnarray}
(x-1)y''(x)+y'(x)&=&(x-1)\sum_{n=2}^\infty n(n-1)a_nx^{n-2}+\sum_{n=1}^\infty na_nx^{n-1}\\
&=&\sum_{n=2}^\infty n(n-1)a_nx^{n-1}-\sum_{n=2}^\infty n(n-1)a_nx^{n-2}+\sum_{n=1}^\infty na_nx^{n-1}\\
&=&\sum_{n=1}^\infty n^2a_nx^{n-1}-\sum_{n=2}^\infty... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/572420",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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How prove this Polynomial $g(x)=\sum_{i=1}^{n}a^m_{i}x^i$have only real roots? Question 1:
let Polynomial $f(x)=\displaystyle\sum_{i=0}^{3}a_{i}x^i,$ have three real numbers roots,where $a_{i}>0,i=1,2,3$.
show that:
$$g(x)=\sum_{i=0}^{3}a^m_{i}x^i$$ have only real roots,where $m\in R,m\ge 1$
My try:
case (1):
suppose... | As I explained in my comment, it is enough to do case (2).
Let $x,y \ge 0$. Then $(X+x)(X+x)(X+y) = X^3+(2x+y)X^2 + (x^2+2xy)X + x^2y$.
The discriminant of $X^3+bX^2+cX+d$ is, according to wikipedia, $\Delta(b,c,d) = b^2c^2-4c^3-4b^3d-27d^2+18bcd$.
The derivative of $m \mapsto \Delta(b^m,c^m,d^m)$ at $m=1$ is
$$\delta... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/574220",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 3,
"answer_id": 1
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Finding a recursive definition and computing $B(10)$ For $n \geq 1$, let $B(n)$ be the number of ways to express $n$ as the sum of $1$s and $2$s, taking order into account. Thus $B(4) = 5$ because $4 = 1 + 1 + 1 + 1 = 1 + 1 + 2 = 1 + 2 + 1 = 2 + 1 + 1 = 2 + 2$.
(a) Compute $B(i)$ for $1 \leq i \leq 5$ by showing all th... | Hint: There are two questions for you to answer. Suppose that $n\ge 3.$ How many ways are there to obtain $n$ as such a sum where the first number is $1$? What if the first number is $2$? What can you then conclude?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/574288",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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How to prove this limit exists: $a_{n+2}=\sqrt{a_{n+1}}+\sqrt{a_{n}}$ Question:
Consider a sequence $\{a_{n}\}$ such that $a_{1},a_{2}>0$, and for all $n \in \mathbb{N}$ we have:
$$a_{n+2}=\sqrt{a_{n+1}}+\sqrt{a_{n}}$$
Prove that :
$\displaystyle\lim_{n\to\infty}a_{n}$ exists and find this limit.
My work: If this l... | and I have consider other solution
if $a_{0}>0,a_{1}>0,$,and $a_{n+2}=\sqrt{a_{n+1}}+\sqrt{a_{n}}$,then $a_{n}\to 4$
pf: if $0<a_{0}\le a_{1}\le 1$,then $$a_{2}=\sqrt{a_{1}}+\sqrt{a_{1}}\ge a_{1}$$
we note
$$a_{n+2}-a_{n+1}=(\sqrt{a_{n+1}}+\sqrt{a_{n}})-(\sqrt{a_{n}}+\sqrt{a_{n-1}})=(\sqrt{a_{n+1}}-\sqrt{a_{n}})+(\sqrt... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/574549",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 5,
"answer_id": 1
} |
Find a primitive of $x^2\sqrt{a^2 - x^2}$ I'm stuck on this problem for the last while now and any help would be appreciated. I need to find the indefinite integral of $$\int x^2\sqrt{a^2-x^2} dx$$
and show that it equals $$\frac x8(2x^2-a^2)\sqrt{a^2-x^2}+{a^4\over 8}\arcsin{x\over a}+c$$
I've tried using the substi... | $$ \int x^2 \sqrt{a^2 - x^2} ~dx ~ = ~ a^4 \int \sin^2 u \cos^2 u ~du ~ = ~ a^4 \int \sin^2 u - \sin^4 u ~ du \quad (x = a \sin u)$$
Using the double angle formulae $\cos 2A=2\cos^2A-1=1-2\sin^2A$
$$\begin{align}
\sin^2u &=\dfrac{1-\cos{2u}}{2}\\
\\
\sin^4u &=\left(\sin^2u\right)^2\\
\\
&=\left(\dfrac{1-\cos{2u}}{2} \r... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/575933",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 4
} |
Solving $(z+1)^5 = z^5$ The question says to solve this equation: $(z+1)^5 = z^5$
I did. Just want to find out if I did it properly and if my run-around logic makes sense.
First I begin my writing the equations as:
$$ (z+1)^5 = z^5$$
$$ \mathbf{e}^{5 \mathbf{Log}(z+1)} = \mathbf{e}^{5 \mathbf{Log}(z)} $$
So $$ \mathbf... | How about, let $ u = z-1/2 $. Then in terms of $ u $ you have:
$$
\left(u+\frac{1}{2}\right)^5 = \left(u - \frac{1}{2}\right)^5
$$
Upon expansion:
$$
5 u^4 + \frac{5}{2} u^2 + \frac{1}{16} = 0
$$
A quadratic equation in $ u^2 $. Can you get it from here?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/577847",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "17",
"answer_count": 5,
"answer_id": 0
} |
Proving for every odd number $x$, $x^2$ is always congruent to $1$ or $9$ modulo $24$ A problem I have been presented with asks the following:
Prove for every odd number $x$, $ x^2$ is always congruent to $1$ or $9$ modulo $24$.
This seems odd and non-intuitive to me. Of course, it must be true other wise they wouldn't... | $$x^2-1=(x-1)(x+1)$$
is the product of two consecutive even numbers. Thus, one is multiple of $4$ and the other is even. This shows that $x^2-1$is a multiple of $8$.
Moreover, one of $x-1,x,x+1$ is a multiple of three.
Case 1: $x$ is not a multiple of $3$. Then $x^2-1$ must be a multiple of three. Thus
$x^2-1$ is a m... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/578749",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 7,
"answer_id": 6
} |
question about the limit $\lim_{h\to0}\frac{\arcsin(x+h)-\arcsin(x)}{h}$ Because $\sin'(x)=\cos(x)$ we can prove that $\arcsin'(x)=\frac{1}{\sqrt{1-x^2}}$. but, by definition we have
$$\arcsin'(x)=\lim_{h\to0}\frac{\arcsin(x+h)-\arcsin(x)}{h}\tag{1}$$
therefore,
$$\lim_{h\to0}\frac{\arcsin(x+h)-\arcsin(x)}{h}=\frac{1}{... | $$\text{Using }\arcsin p-\arcsin q=\arcsin(p\sqrt{1-q^2}-q\sqrt{1-p^2}),$$
$$\arcsin(x+h)-\arcsin x=\arcsin((x+h)\sqrt{1-x^2}-x\sqrt{1-(x+h)^2})$$
$$\displaystyle\lim_{h\to0}\frac{\arcsin(x+h)-\arcsin x}h $$
$$=\lim_{h\to0}\frac{\arcsin((x+h)\sqrt{1-x^2}-x\sqrt{1-(x+h)^2})}h$$
$$=\lim_{h\to0}\frac{\arcsin((x+h)\sqrt{1-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/579170",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 4,
"answer_id": 1
} |
Is this matrix positive-semidefinite in general? for the matrix written below I was wondering if one can show that it is positive-semidefinite for $n>3$ and $0< \alpha<1$. (Or not. For $n=2, 3$ it works by showing that all principal minors are non-negative.)
$$
C_{n,n} =
\begin{pmatrix}
1 & \alpha^1& \alpha^2 & \cdo... | Taking the first column, and substracting to it $\alpha$ times the column 2, we get $\det C_{n,n}=(1-\alpha^2)\det C_{n-1,n-1}$, hence we can conclude by Sylvester's criterion.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/579242",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
} |
Evaluate $\tan\left(2\sin^{-1}\frac{\sqrt{5}}{5}\right)$ without using a calculator Evaluate without using a calculator:
$\displaystyle{\tan\left(2\sin^{-1}\left(\sqrt{5} \over 5\right)\right).}$
So I built my triangle hyp=$5$, adj=$2\sqrt{5}$, opp=$\sqrt{5}$.
$$
\tan\left(2\theta\right) = 2\sin\left(\theta\right)\cos\... | $\displaystyle \frac{\sqrt{5}}{5} = \frac{1}{\sqrt{5}}$. So, draw a triangle with
sides $1$, $2$, and $\sqrt{5}$ which should be a right triangle since $1^2 + 2^2 = \left(\sqrt{5}\right)^2$. The sine of one of the angles of this triangle, call
it $\theta$, is $\sin(\theta) = \frac{1}{\sqrt{5}}$, so its tangent $\tan... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/579839",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Find all solutions of the equality $y^2=x^3+23$ for integers $x,y$
Find all solutions of the equality $y^2=x^3+23$ for integers $x,y$
I guess, that x cannot be even. because we can apply mod 4 test
say $x=2k$ then $y^2=8k^3+23\equiv3\mod(4)$
but, this is not possible, since the square of an integer is congruent to $... | There are no integer solutions.
You noted already that $x$ cannot be even, because the equation would then
fail mod $4$; so we need only consider $x$ odd. Then $x^3+23$ is even,
so must be a multiple of $4$, whence $x \equiv 1 \bmod 4$.
We cannot obtain a contradiction from further congruence considerations.
But since... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/579928",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 1,
"answer_id": 0
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