Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Prove that $\int J_5(x) dx= -J_4(x) - \frac {4}{x} J_3(x) - \frac {8}{x^2}J_2(x) +c$
Prove that $$\int J_5(x) dx= -J_4(x) - \frac {4}{x} J_3(x) - \frac
{8}{x^2}J_2(x) +c$$ where $J_n (x)$ is the Bessel function of first kind
and order $n $.
My attempt:
I know the following recurrence relations:
$$\frac {d}{dx}\le... | I will use the second of your recurrence relations:
$$
\frac{d}{dx}[x^{-n}J_n(x)] = -x^{-n}J_{n + 1}(x)
$$
And consider the next cases
*
*$n = 2$
$$
\frac{d}{dx}[x^{-2}J_2(x)] = -x^{-2}J_{3}(x) ~~~\Rightarrow~~~ \int dx\; x^{-2}J_3(x) = -x^{-2}J_2(x) \tag{1}
$$
*$n = 3$
$$
\frac{d}{dx}[x^{-3}J_3(x)] = -x^{-3}J_{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2046887",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Finding the determinant using row operations. So I have to find the determinant of $\begin{bmatrix}3&2&2\\2&2&1\\1&1&1\end{bmatrix}$ using row operations. From what I've learned, the row operations that change the determinate are things like swaping rows makes the determinant negative and dividing a row by a value mean... | Your error is when you multiply the third row by $2$ and subtract the second row; this introduces a factor $2$ that you have to remove.
Use a more systematic way:
\begin{align}
\begin{bmatrix}
3&2&2\\
2&2&1\\
1&1&1
\end{bmatrix}
&\to
\begin{bmatrix}
1&2/3&2/3\\
2&2&1\\
1&1&1
\end{bmatrix}
&&R_1\gets \tfrac{1}{3}R_1 &&\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2046993",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Find $\int \frac{\sin(x)}{1+\sin(x)\cos(x)} dx$
Find $$\int \frac{\sin(x)}{1+\sin(x)\cos(x)} dx$$
What I have tried
First method was to try $u$ substitution
Let $u=\cos(x)$ then $-du=\sin(x)dx$ then $\sin(x)=\sqrt{1-u^2} $ which transforms our integral into
$$ \int \frac{\sin(x)}{1+\sin(x)\cos(x)} dx=-\int \frac{... | \begin{align*} &~~~~~\int\frac{\sin x}{1+\sin x\cos x}{\rm d}x\\ &=\int\frac{2\sin x}{2+2\sin x\cos x}{\rm d}x\\ &=\int\frac{(\sin x+\cos x)+(\sin x-\cos x)}{2+2\sin x\cos x}{\rm d}x\\ &=\int\frac{\sin x+\cos x}{2+2\sin x\cos x}{\rm d}x+\int\frac{\sin x-\cos x}{2+2\sin x\cos x}{\rm d}x\\ &=\int\frac{\sin x+\cos x}{3-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2048910",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Find the greatest value of $a^2\cdot b^3$ where $a+b=10$ Find the greatest value of $a^2\cdot b^3$ where $a,b$ are positive real numbers satisfying $a+b=10$. Determine the values of $a,b$ for which the greatest value is attained.
At first I tried to find the value by putting $a=10-b$, I found a expression, but I got no... | The Am-GM inequality says that $\{a_n\}$ are given then $\frac {\sum a_n}{n} \geq \sqrt[n]{\prod a_n}$.
Rewrite $a+b = 10$ as $2 \times \frac a2 + 3 \times \frac b3 = 10$. Now, apply AM-GM and see what happens, noting that equality occurs when $\frac a2 = \frac b3$.
Applying AM-GM with the set $\{\frac a2,\frac a2,\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2049448",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
Limit of $\left(1+\frac{2}{n^2}\right)^n $
Compute$$\lim_{n\to\infty}\left(1+\frac{2}{n^2}\right)^n $$
I dont know how to do it without using continuity of exponential function
I mean I cannot do this:
$$\lim_{x\to\infty}a_n =a \wedge \lim_{x\to\infty}b_n =b\Rightarrow \lim_{x\to\infty}{a_n}^{b_n} =a^b$$
| Use Bernoulli:
$$\left(1+\frac{2}{n^2}\right)^n= \frac{1}{\left(\frac{n^2}{n^2+2}\right)^n}= \frac{1}{\left(1-\frac{2}{n^2+2}\right)^n}$$
And by Bernoulli
$$\left(1-\frac{2}{n^2+2}\right)^n \geq 1-\frac{2n}{n^2+2}=\frac{n^2-2n+2}{n^2+2}$$
Therefore
$$1 \leq \left(1+\frac{2}{n^2}\right)^n \leq \frac{n^2+2}{n^2-2n+2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2050128",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
Evaluating $\int_{ - \infty }^\infty {\frac{{{e^{7\pi x}}}}{{{{\left( {{e^{3\pi x}} + {e^{ - 3\pi x}}} \right)}^3}\left( {1 + {x^2}} \right)}}dx}$
How to evaluate this integral?
$$\int_{ - \infty }^\infty {\frac{{{e^{7\pi x}}}}{{{{\left( {{e^{3\pi x}} + {e^{ - 3\pi x}}} \right)}^3}\left( {1 + {x^2}} \right)}}dx}$$
M... | The given integral equals
$$ I=\frac{1}{8}\int_{0}^{+\infty}\frac{2\cosh(7\pi x)}{(1+x^2)\cosh^3(3\pi x)}\,dx = \frac{1}{8}\int_{\mathbb{R}}\frac{\cosh(7\pi x)}{(1+x^2)\cosh^3(3\pi x)}\,dx $$
and by computing the residues of the integrand function at $x=i$ and at $(2k+1)\frac{i}{2}$ we get:
$$ I = \frac{\pi}{8}+\frac{2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2051160",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
$f(0)=0$ ; $f(x):=\int_0^x \cos \frac 1t \cos \frac 3t \cos \frac 5t \cos \frac 7t\,dt$, $\forall x \ne 0$. Is $f$ differentiable at $0$? Let $f:\mathbb R \to \mathbb R$ be defined by $f(0)=0$ and
$$
f(x):=\int_0^x \cos \dfrac 1t \cos \dfrac 3t \cos \dfrac 5t \cos \dfrac 7t \,dt ,\quad\forall x \ne 0.
$$
Then is $f$ di... | First write the integrand as
$$\begin{aligned}
\cos\frac{1}{t}&\cos\frac{3}{t}\cos\frac{5}{t}\cos\frac{7}{t}\\
&=\frac{1}{8}\Bigl(1+\cos\frac{2}{t}+\cos\frac{4}{t}+\cos\frac{6}{t}+\cos\frac{8}{t}+\cos\frac{10}{t}+\cos\frac{14}{t}+\cos\frac{16}{t}\Bigr).
\end{aligned}
$$
The, let $f_k(x)=\int_0^x\cos\frac{k}{t}\,dt$ and... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2051257",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 0
} |
What is wrong with the "proof" for $\ln(2) =\frac{1}{2}\ln(2)$? I have got a question which is as follows:
Is $\ln(2)=\frac{1}{2}\ln(2)$??
The following argument seems suggesting that the answer is yes:
We have the series $1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\cdots$
which has a mathematically determined value $... | Such regrouping of a series is just not guaranteed not to alter the limit or even preserve convergence.
It would be admissible if the series were absolutely convergent, that is the series of absolute values would converges. But, the current series does not converge absolutely.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2051691",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 5,
"answer_id": 0
} |
Functional equation: $f: \mathbb{R} \rightarrow \mathbb{R}$, $f((x + 1) f(y)) = y (f(x) + 1)$ Determine all $f: \mathbb{R} \rightarrow \mathbb{R}$ such that, for all $x, y \in \mathbb{R}$, $$f((x + 1) f(y)) = y (f(x) + 1).$$
Source: Vuong Lam Huy, on a Facebook group.
| Let $f : \mathbb{R} \to \mathbb{R}$ satisfy
$$ f((x+1)f(y)) = (f(x)+1)y, \quad : \forall x,y \in \mathbb{R} \tag{1} $$
Lemma 1. $f$ is bijective, $f(0) = 0$ and $f(f(x)) = x$.
Proof. Plug $y = 0$ to $\text{(1)}$. Then $f((x+1)f(0)) = 0$ In particular, $f(f(0)) = 0$.
Now plugging $x = f(0)$ to $\text{(1)}$, we have $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2051991",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
The Most general solution satisfying equations $\tan x=-1$ and $\cos x=1/\sqrt{2}$ The most general value of $x$ satisfying the equations $\tan x=-1$ and $\cos x=1/\sqrt{2}$, is found to be $x=2n\pi+\frac{7\pi}{4}$.
My approach:
$$
\frac{\sin x}{\cos x}=-1\implies \sqrt{2}\sin x=-1\implies\sin x=\frac{-1}{\sqrt{2}}=\si... | $\tan \theta = -1 =\tan (\frac{3\pi}{4}) =\tan(\frac{7\pi}{4})$.
$\tan (\pi -\frac{3\pi}{4}) = \tan(\pi +\frac{3\pi}{4})$.
$\cos \theta =\frac{1}{\sqrt{2}}$.
$\cos \frac{\pi}{4} = \frac{1}{\sqrt{2}}$.
$\cos (2\pi -\frac{7\pi}{4}) =\frac{1}{\sqrt{2}} = \cos \frac{7\pi}{4}$.
Hence, the principal value for both $\tan \the... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2052250",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
General solution to $(\sqrt{3}-1)\cos x+(\sqrt{3}+1)\sin x=2$ $(\sqrt{3}-1)\cos x+(\sqrt{3}+1)\sin x=2$ is said to have a general solution of $x=2n\pi\pm\frac{\pi}{4}+\frac{\pi}{12}$.
My Approach:
Considering the equation as
$$
a\cos x+b\sin x=\sqrt{a^2+b^2}\Big(\frac{a}{\sqrt{a^2+b^2}}\cos x+\frac{b}{\sqrt{a^2+b^2}}\... | $$(\sqrt{3}-1)\cos x+(\sqrt{3}+1)\sin x=2$$
divide by $2\sqrt{2}$:
$$\frac{\sqrt{3}-1}{2\sqrt{2}}\cos x+\frac{\sqrt{3}+1}{2\sqrt{2}}\sin x=\frac{1}{\sqrt{2}}$$
now use
$$\cos a \cos b+\sin a \sin b=\cos (a-b)$$
$$
\cos^2 a+\sin^2 a=\bigg(\frac{\sqrt{3}-1}{2\sqrt{2}}\bigg)^2+\bigg(\frac{\sqrt{3}+1}{2\sqrt{2}}\bigg)^2=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2053720",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 3
} |
Use Mathematical Induction to prove equation? Use mathematical induction to prove each of the following statements.
let $$g(n) = 1^3 + 2^3 + 3^3 + ... + n^3$$
Show that the function $$g(n)= \frac{n^2(n+1)^2}{4}$$ for all n in N
so the base case is just g(1) right? so the answer for the base case is 1, because 4/4 = 1
t... | Show that: $$1^3 + 2^3 + 3^3 + ... + n^3= \frac{n^2(n+1)^2}{4}$$
$(1)$ First step is to evaluate the expression for $n=1$, $$1^3=\frac{1\times(1+1)^2}{4}$$
Which is $1=1$, which means we can proceed.
$(2)$ Now assume it is true for $n$, $$1^3 + 2^3 + 3^3 + ... + n^3= \frac{n^2(n+1)^2}{4}$$
$(3)$ Lastly prove for $n+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2055695",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 3
} |
Find all integer solutions $(x,y)$ such that $2x^2 + y^2 = 3x^2y$
Find all integer solutions $(x,y)$ such that $2x^2 + y^2 = 3x^2y$.
We can rearrange the given equation to $$y^2 = x^2(3y-2)\tag1$$ Thus $3y-2$ must be a perfect square and so $3y-2 = k^2$.
How can we continue?
| I. The method described by the OP is a bit round-about. To continue, solve for $y$ in $3y-2=k^2$ then substitute into $y^2-x^2(3y-2)=0$. We get,
$$(2 + k^2 - 3 k x) (2 + k^2 + 3 k x)=0$$
Solve for $k$,
$$k =\frac{3x\pm\color{blue}{\sqrt{-8+9x^2}}}{2}$$
Thus one needs to solve $-8=z^2-9x^2$ in the integers. It is a Pell... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2056506",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 1
} |
Evaluating $\int_0^{\infty} \frac{x^2}{x^5 + 1} \ dx$ The following is a problem from an older exam which the instructor didn't provide solutions to.
Evaluate $$\int_0^{\infty} \frac{x^2}{x^5 + 1} \ dx$$
using only real-analytic techniques.
My work: this boils down to factoring $x^4 - x^3 + x^2 - x+ 1$ which I am... |
My work: this boils down to factoring $x^4 - x^3 + x^2 - x+ 1$ which I am unable to do by hand.
Because you seem to be looking for an approach without complex numbers; here is a not so neat, 'brute force' but real-valued approach.
Since $x^4 - x^3 + x^2 - x+ 1$ has no real roots, it has a factorization of the form:
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2061215",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 0
} |
How to factorize $a^2-b^2-a+b+(a+b-1)^2$? The answer is $(a+b-1)(2a-1)$ but I have no idea how to get this answer.
| Expand it and treat it as a quadratic expression in $a.$ That is if $C\ne 0$ then $Ca^2+Da+E=\;[a-(D- F)/2C]\;[a-(D+F)/2C)]\;$ where $F=\sqrt {D^2-4CE}.$
In this Q, we get $D^2-4CE=4b^2-4b+1=(2b-1)^2$ so it will simplify.
That is $$a^2-b^2-a+b+(a+b-1)^2=$$ $$= a^2-b^2-a+b +(a^2+2ab+b^2)-2(a+b)+1=$$ $$=2a^2+2ab -3a+b-1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2063343",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
$\sin{x}+\sin{2x}+\sin{3x}=1+\cos{x}+\cos{2x}$ - solving a trigonometric equation I'm looking for hints regarding such equation:
$$ \sin{x}+\sin{2x}+\sin{3x}=1+\cos{x}+\cos{2x} $$
I'd be particularly interested in any clever method for dealing with such problem.
| $$\sin x +\sin 3x=2\cdot\sin 2x \cdot\cos x$$
$$\cos 2x=2\cos^2x-1$$
Then
$$2\cdot\sin{2x}\cdot\cos x+\sin{2x}=1+\cos{x}+2\cos^2x-1$$
$$\sin{2x}\cdot(2\cos x+1)=\cos{x}\cdot(2\cos x+1) $$
$$(2\cos x +1)\cdot(\sin 2x -\cos x)=0$$
$$(2\cos x +1)\cdot(\cos x)\cdot(2\sin x -1)=0$$
$$\cos x=-\frac{1}{2} \Rightarrow x=\pm\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2065171",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Find the last Digit of $237^{1002}$? I looked at alot of examples online and alot of videos on how to find the last digit But the thing with their videos/examples was that the base wasn't a huge number. What I mean by that is you can actually do the calculations in your head. But let's say we are dealing with a $3$ di... | Simple version without the notation:
$7 \times 1 = 7$
$7 \times 7 = 49$
$7 \times 9 = 63$
$7 \times 3 = 21$
Just look at the last digit in each case.
So the last digit of $7^1$ is $7$. The last digit of $7^2$ is $9$. The last digit of $7^3$ is $3$. And, the last digit of $7^4$ is $1$.
Thus the last digit of $7^5$ is... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2065373",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 11,
"answer_id": 3
} |
How to integrate $\int_a^b (x-a)(x-b)\,dx=-\frac{1}{6}(b-a)^3$ in a faster way? $\displaystyle \int_a^b (x-a)(x-b)\,dx=-\frac{1}{6}(b-a)^3$
$\displaystyle \int_a^{(a+b)/2} (x-a)(x-\frac{a+b}{2})(x-b)\, dx=\frac{1}{64}(b-a)^4$
Instead of expanding the integrand, or doing integration by part, is there any faster way to ... | One way is to use Simpson's rule.
Without it, one could argue what is faster, but:
If $A=(a+b)/2$ and $D=(b-a)/2$, then your first integrand is
$$
(x-A+D)(x-A-D)=(x-A)^2-D^2.
$$
Thus
$$
\begin{aligned}
\int_a^b (x-a)(x-b)\,dx &=\frac{1}{3}\bigl((b-A)^3-(a-A)^3\bigr)-2D^3\\
&=\frac{1}{3}\bigl(D^3+D^3)-2D^3=-\frac{1}{6}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2065639",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 6,
"answer_id": 5
} |
Help verify the proof that $\min \left({\lfloor{N/2}\rfloor, \lfloor{(N + 2)/p}\rfloor}\right) = \lfloor{(N + 2)/p}\rfloor$ for $N \ge p$ Well, I think that I have this correct, however my proof seams cumbersome, so can someone check these results or suggest a simplified proof.
Prove that ($N \ge p$)
\begin{equation*}
... | You can prove this a bit easier using for example using proof by contradiction:
Let's assume $\left\lfloor{\frac{N}{2}}\right\rfloor < \left\lfloor{\frac{N + 2}{p}}\right\rfloor$. This implies $\frac{N}{2} < \frac{N+2}{p}$ or after simplification $p < 2+\frac{4}{N}$.
Now cases where $N\leq 4$ can be easily checked by ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2066562",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Common proof for $(1+x)(1+x^2)(1+x^4)...(1+x^{2^n})=\dfrac{1-x^{2^{n+1}}}{1-x} $ I'm asking for an alternative (more common?) proof of the following equality, more specifically an alternative proof for the inductive step:
$$(1+x)(1+x^2)(1+x^4)...(1+x^{2^n})=\dfrac{1-x^{2^{n+1}}}{1-x} (x\neq 1)$$
This is how I proved it... | On the left-hand side, the ratio of $\Pi_n$ over $\Pi_{n-1}$ is
$$1+x^{2^n},$$
and on the right-hand side,
$$\frac{\dfrac{1-x^{2^{n+1}}}{1-x}}{\dfrac{1-x^{2^n}}{1-x}}=\frac{1-(x^{2^n})^2}{1-x^{2^n}}.$$
Alternatively,
$$\frac{1-x^{2^{n+1}}}{1-x}=\sum_{k=0}^{2^n}x^k.$$
Then
$$(1+x^{2^n})\sum_{k=0}^{2^n}x^k=\sum_{k=0}^{2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2067075",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 2
} |
Why is $x^x-(\sin x)^x\sim\frac{1}{6}x^3 $ when $x\to 0$? I'm learning Taylor's expansion.
The given solution of the problem is:
when $x\to 0$, $x^x-(\sin x)^x=x^x(1-(\frac{\sin x}{x})^x)\sim\frac{1}{6}x^3 $
But I don't know how to use Taylor's formula to get:
$x^x(1-(\frac{\sin x}{x})^x)\sim\frac{1}{6}x^3 $
| Just use
$$\sin(x) \approx x - \frac{x^3}{6}$$
$$\frac{\sin x}{x} \approx 1 - \frac{x^2}{6}$$
$$x^x\left(1 - \left(1 - \frac{x^2}{6}\right)^x\right)$$
Now the bracket factor
$$\left(1 - \frac{x^2}{6}\right)^x$$
is small, so you can use Binomial expansion: $(1+X)^a = 1 + aX + \ldots$
That is
$$x^x\left(1 - \left( 1 - x\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2067211",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Prove that if {a;b} $\in \mathbb R^+$ then $a^2+b^2>ab$ I have tried factoring it already, but it doesn't seem to evolve much:
First I multiply each side by $2$:
$ 2(a^2+b^2)>2ab$
Then I substitute using the relation $(a+b)^2=a^2+2ab+b^2$ and it becomes:
$2(a^2+b^2)>(a+b)^2 - (a^2+b^2)$
and then:
$3(a^2+b^2)>(a+b)^2$
A... | $$
(a-b)^2 \geq 0 \\
(a-b)^2 = a^2 - 2ab + b^2 \\
a^2 + b^2 \geq 2ab \geq ab
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2069625",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 6,
"answer_id": 2
} |
Is there a matrix $A$ such that: $A^4=\begin{pmatrix}0 & 2 & -1 & 1\\ 0 & 0 & 3 &1\\ 0 & 0& 0 & 4\\ 0 & 0 & 0 & 0 \\ \end{pmatrix} ~?$
Is there a matrix $A\in \mathbb{C}^{4\times 4}$ such that: $$A^4=\left(\begin{array}{cccc}
0 & 2 & -1 & 1\\
0 & 0 & 3 &1\\
0 & 0& 0 & 4\\
0 & 0 & 0 & 0 \\
\end{array} \right) ~?$$
An... | If an $n\times n$ matrix satisfies $A^m=0$ for some $m$ then it satisfies $A^n=0$.
notice that if $A^4$ was equal to that matrix then $(A^4)^4=0$, but then $A^4=0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2070036",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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"answer_id": 0
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Calculate limit $\lim_{x \to 2} \frac{(2^x)-4}{\sin(\pi x)}$ without L'Hopital's rule How to calculate limit: $\lim_{x \to 2} \frac{(2^x)-4}{\sin(\pi x)}$ without L'Hopital's rule?
If $x = 2$, I get uncertainty $\frac{0}{0}$
| $\lim_{x \to 2} \frac{(2^x)-4}{\sin(\pi x)}
$
Putting $x = y+2$,
$\begin{array}\\
\dfrac{(2^x)-4}{\sin(\pi x)}
&=\dfrac{(2^{y+2})-4}{\sin(\pi (y+2))}\\
&=4\dfrac{(2^{y})-1}{\sin(\pi y)}
\qquad\text{since } 2^{y+2} = 4\cdot 2^y
\text{ and }\sin(\pi (y+2))=\sin(\pi y + 2\pi)=\sin(\pi y)\\
&=4\dfrac{e^{y\ln 2}-1}{\sin(\p... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2072811",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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Find coefficient of $x^n$ in $(1+x+2x^2+3x^3+.....+nx^n)^2$ Find coefficient of $x^n$ in
$(1+x+2x^2+3x^3+.....+nx^n)^2$
My attempt:Let $S=1+x+2x^2+3x^3+...+nx^n$
$xS=x+x^2+2x^3+3x^4+...+nx^{n+1}$
$(1-x)S=1+x+x^2+x^3+....+x^n-nx^{n+1}-x=\frac{1-x^{n+1}}{1-x}-nx^{n+1}-x$
$S=\frac{1}{(1-x)^2}-\frac{x}{1-x}=\frac{1-x+x^2... | Hint: use
$$ (f(x)g(x))^{(n)}=\sum_{k=0}^n\binom{n}{k}f^{(n-k)}(x)g^k(x).$$
The coefficient of $x^n$ is $(f(x)g(x))^{(n)}(0)/n!$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2075434",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
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Why is the solution to $x-\sqrt 4=0$ not $x=\pm 2$? If the equation is $x-\sqrt 4=0$, then $x=2$.
If the equation is $x^2-4=0$, then $x=\pm 2$.
Why is it not $x=\pm 2$ in the first equation?
| The root function $f(x) = \sqrt{x}$ is defined to be nonnegative.
So $\sqrt{2} = 1.4...$, however the equation $x^2 - 4=0$ has to solutions:
If $x$ is such a solution, then so is $-x$ since $(-x)^2-4 = (-1)^2x^2-4 = x^4-4 = 0$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2075745",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 13,
"answer_id": 7
} |
Show that:$\sum\limits_{n=1}^{\infty}{n\over (4n^2-1)(16n^2-1)}={1\over 12}(1-\ln{2})$ Show that
$$\sum_{n=1}^{\infty}{n\over (4n^2-1)(16n^2-1)}={1\over 12}(1-\ln{2})$$
My try:
We split into partial decomposition
$$n={A\over 2n-1}+{B\over 2n+1}+{C\over 4n-1}+{D\over 4n+1}$$
Setting $n={1\over 2}$, ${-1\over2}$ we have... | Note that
$$S = \sum_{n=1}^{\infty}{n\over (4n^2-1)(16n^2-1)}= \sum_{n=1}^\infty \frac{1}{12n}\left\{\frac{1}{4n^2-1}-\frac{1}{16n^2-1} \right\}$$
Hence
$$ \frac{1}{12n}\left\{\frac{1}{4n^2-1}-\frac{1}{16n^2-1} \right\}= \frac{1}{24n}\left\{\frac{1}{2n-1}-\frac{1}{2n+1}+\frac{1}{4n+1}-\frac{1}{4n-1} \right\}$$
Using... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2075828",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 3,
"answer_id": 0
} |
Show that $\sqrt{4 + 2\sqrt{3}} - \sqrt{3}$ is rational.
Show that $\sqrt{4 + 2\sqrt{3}} - \sqrt{3}$ is rational.
I've tried to attempt algebra on this problem. I noticed that there is some kind of nesting effect when trying to solve this. Please help me to understand how to attempt to denest this number.
Any help wo... | Personally I like this method which is the same thing as the other answers.$\sqrt{4+2\sqrt{3}}=\sqrt{a}+\sqrt b$ squaring both sides $4+2\sqrt 3=a+2\sqrt{ab}+b$ for this to be true we must equate the parts with and without the radicals so $a+b=4$ and $2\sqrt{ab}=2\sqrt{3}$ so $ab=3$ now we have a system of equations th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2076737",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 6,
"answer_id": 2
} |
proving $ \binom{n}{0}-\binom{n}{1}+\binom{n}{2}+\cdots \cdots +(-1)^{n-1}\binom{n}{m-1}=(-1)^{m-1}\binom{n-1}{m-1}$ proving $\displaystyle \binom{n}{0}-\binom{n}{1}+\binom{n}{2}+\cdots \cdots +(-1)^{\color{red}{m}-1}\binom{n}{m-1}=(-1)^{m-1}\binom{n-1}{m-1}.$
$\displaystyle \Rightarrow 1-n+\frac{n(n-1)}{2}+\cdots \cdo... | HINT: Fix a random $n$ and do induction on $m$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Can I express $x^7y+xy+x+1$ by a repeated use of operation $xy+x+y+1$? A certain calculator can only give the result of $xy+x+y+1$ for any two real numbers $x$ and $y$.
How to use this calculator to calculate $x^7y+xy+x+1$ for any given $x$ and $y$?
When $x$ and $y$ are equal, it will give $(x+1)^2$. But I cannot pr... | Give $x^7$ and $y$ to the calculator, to get $R_1 = x^7y + x^7+y +1$.
Give $x$ and $y$ to the calculator, to get $R_2 = xy+x+y+1$.
Add $R_1$ and $R_2$ to get, $R = R_1+R_2 = x^7y + x^7 + x+xy + 2y+1$.
Subtract $x^7$ and $2y$ from $R$ to get the required.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2078290",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Find $AC: CB$ in $\triangle XYZ$ Problem:
In $\triangle XYZ$, $XY = 4$, $YZ = 7$, and $XZ = 9$. Let $M$ be the midpoint of $\overline{XZ}$, and let $A$ be the point on $\overline{XZ}$ such that $\overline{YA}$ bisects angle $XYZ$. Let $B$ be the point on $\overline{YZ}$ such that $\overline{YA} \perp \overline{AB}$. Le... | You have already got
$$\dfrac{11}{3} \times \dfrac{AC}{CB} \times \dfrac{YB}{7} = 1\tag1$$
Also, we have
$$YB=\frac{AY}{\cos \frac Y2}\tag2$$
So, we want to find $\cos\frac Y2$ and $AY$.
By the law of cosines,
$$9^2=4^2+7^2-2\cdot 4\cdot 7\cos Y\implies \cos Y=-\frac 27$$
from which we have
$$\cos\frac Y2=\sqrt{\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2079080",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Evaluate the triple integral problem Evaluate $\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}(x^2-2xy)e^{-Q}dxdydz$
, where $Q=3x^2+2y^2+z^2+2xy$.
| Let $\Sigma$ be the symmetric $3 \times 3$ matrix which is specified by $\frac{1}{2}\mathrm{x}^{\mathsf{T}}\Sigma^{-1} \mathrm{x} = Q(\mathrm{x})$ for $\mathrm{x}\in \Bbb{R}^3$. Then $\Sigma$ satisfies
$$ \Sigma^{-1} = \begin{pmatrix} 6 & 2 & 0 \\ 2 & 4 & 0 \\ 0 & 0 & 2 \end{pmatrix}, \qquad
\Sigma = \frac{1}{10} \begi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2079205",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Help find closed form for:$\sum_{n=0}^{\infty}{(-1)^n\left({{\pi\over 2}}\right)^{2n}\over (2n+k)!}=F(k)$ What is the closed form for
$$\sum_{n=0}^{\infty}{(-1)^n\left({{\pi\over 2}}\right)^{2n}\over (2n+k)!}=F(k)?$$
My try:
I have found a few values of $F(k)$, but was unable to find a closed form for it.
$F(0)=0$
$F(1... | Note that
$$\sum_{n=0}^\infty \frac{(-1)^nx^{2n}}{(2n)! } = \cos(x)$$
You can reach the result by integrating $k$-times.
$$\sum_{n=0}^\infty \frac{(-1)^nx^{2n}}{(2n+k)! } = \frac{1}{x^{k}}\int^{x}_0 \mathrm{d}t_{k-1}\int^{t_{k-1}}_0 \mathrm{d}t_{k-2} \cdots\int^{t_1}_0\mathrm{d}t_0\cos(t_0) $$
For example when $k=1$
$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2080167",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 0
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If $a+b+c+d+e=0$ so $180(a^6+b^6+c^6+d^6+e^6)\geq11(a^2+b^2+c^2+d^2+e^2)^3$ Let $a$, $b$, $c$, $d$ and $e$ be real numbers such that $a+b+c+d+e=0$. Prove that:
$$180(a^6+b^6+c^6+d^6+e^6)\geq11(a^2+b^2+c^2+d^2+e^2)^3$$
I tried Holder, uvw and more, but without success.
| Let $x = (a,b,c,d,e)$.
By homogeneity, let $g(x) = a^2+b^2+c^2+d^2+e^2 = 1$. Further, as stated in the question, $h(x) =a+b+c+d+e=0$.
Let $$f(x) = a^6+b^6+c^6+d^6+e^6 $$ and we have to show, under the conditions above, that $f(x) \geq 11/180$. For $x_0 = \frac{1}{\sqrt{30}}(-3,-3,2,2,2)$ we have equality, $f(x_0)=11/18... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2081528",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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Relation between inverse tangent and inverse secant I've been working on the following integral
$$\int\frac{\sqrt{x^2-9}}{x^3}\,dx,$$
where the assumption is that $x\ge3$. I used the trigonometric substitution $x=3\sec\theta$,which means that $0\le\theta<\pi/2$. Then, $dx=3\sec\theta\tan\theta\,dx$, and after a large n... |
Pulled straight off wikipedia.
You seem to have used ** intersection of tan($\theta$) and arcsec(x) **
Also, that "constant" you were referring to is probably $\frac{\pi}{12}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2082980",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
Quadratic equation system $A^2 + B^2 = 5$ and $AB = 2$ Given a system of equations
$A^2 + B^2 = 5$
$AB = 2$
what is the correct way to solve it?
I see immediately that the answers are
*
*$A=1, B=2$
*$A=2, B=1$
*$A=-1, B=-2$
*$A=-2, B=-1$
but I don't understand the correct way of getting there. I have tried to is... | If
$a^2+b^2 = c$
and
$ab = d$
then
$(a+b)^2
=a^2+2ab+b^2
= c+2d
$
and
$(a-b)^2
=a^2-2ab+b^2
= c-2d
$.
Therefore
$a+b = \sqrt{c+2d}$
and
$a-b = \sqrt{c-2d}$
so
$a = \frac12(\sqrt{c+2d}+\sqrt{c-2d})
$
and
$b = \frac12(\sqrt{c+2d}-\sqrt{c-2d})
$.
If $c=5, d=2$,
then
$\sqrt{c+2d} = 3$
and
$\sqrt{c-2d} = 1$
so
$a= 2$
and
$b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2084539",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 10,
"answer_id": 8
} |
Question on completing the square obtaining the form $a(x+p)^2+q$ From the last term in completing the square from the quadratic $ax^2+bx+c$,
I was just wondering how $$-\left(\frac{b}{2a}\right)^2+c = \left(c-\frac{b^2}{4a}\right)$$
I would have gotten $$-\left(\frac{b}{2a}\right)^2+c=\left(c-\frac{b^2}{(2a)^2}\right... | You forgot to multiply the $\left(\frac{b}{2a}\right)^2$ by $a$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2085699",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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show that $a_{n+874}=a_{n}$,if such $a_{n+2}=\left\lceil \frac{4}{3}a_{n+1}-a_{n}+0.5\right\rceil$
Let the sequence $\{a_{n}\}$ be such that $a_{1}=1, a_{2}=100$, and $$a_{n+2}=\left\lceil \dfrac{4}{3}a_{n+1}-a_{n}+0.5\right\rceil$$
Prove that the sequence $\{a_{n}\}$ is periodic.
I have used a computer and found t... | This is not an answer, but rather an observation.
The following graph shows the set of points $P = \{(a_n, a_{n+1}) : n \geq 1\}$.
$\hspace{8em}$
Notice that they are confined in a very narrow region and are clustered near an ellipse. This ellipse is not hard to identify. Indeed, if a sequence $(b_n)$ satisfies
$$ b_{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2086289",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "17",
"answer_count": 2,
"answer_id": 0
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Finding a basis for this subspace of $n\times n$ matrices $M(n,\mathbb{F})$ denotes the set of all $n\times n$ matrices with entries in $\mathbb{F}$.
Assume that $M(n,\mathbb{F})$ is a vector space over $\mathbb{F}$. Suppose $(\mu_{ij}) \in M(n,\mathbb{F})$ where $\mu_{ij} \mu_{ji}=1$ for all $i,j$. Let $U = \{(m_{ij... | Hint:
If $D$ is the diagonal and $U$ is the upper triangular part you can write
$$M = D+U+ K_1 U^T K_2$$
Now can you find $K_1,K_2$ matrices in terms of the $\mu$s which accomplish what you want?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2087122",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Inequality with a+b+c=1 Let $a,b,c$ be positive reals such that $a+b+c=1$. Prove that
$$\dfrac{bc}{a+5}+\dfrac{ca}{b+5}+\dfrac{ab}{c+5}\le \dfrac{1}{4}.$$
Progress: This is equivalent to show that
$$\dfrac{1}{a^2+5a}+\dfrac{1}{b^2+5b}+\dfrac{1}{c^2+5c}\le \dfrac{1}{4abc}.$$
I'm not sure how to proceed further. Also e... | We'll prove a stronger inequality: $\sum\limits_{cyc}\frac{bc}{5+a}\leq\frac{1}{16}$.
Indeed, we need to prove that $$\sum\limits_{cyc}\frac{bc}{6a+5b+5c}\leq\frac{a+b+c}{16}$$ or
$$\sum\limits_{cyc}\left(\frac{a}{16}-\frac{bc}{6a+5b+5c}\right)\geq0$$ or
$$\sum\limits_{cyc}\frac{6a^2+5ab+5ac-16bc}{6a+5b+5c}\geq0$$ or
$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2088687",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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How to integrate $\int\sqrt{\frac{4-x}{4+x}}$? Let
$$g(x)=\sqrt{\dfrac{4-x}{4+x}}.$$
I would like to find the primitive of $g(x)$, say $G(x)$.
I did the following: first the domain of $g(x)$ is $D_g=(-4, 4]$. Second, we have
\begin{align}
G(x)=\int g(x)dx &=\int\sqrt{\dfrac{4-x}{4+x}}dx\\
&=\int\sqrt{... | Solving a more general problem, using integration by parts:
$$\mathcal{I}_\text{n}\left(x\right)=\int\sqrt{\frac{\text{n}-x}{\text{n}+x}}\space\text{d}x=x\sqrt{\frac{\text{n}-x}{\text{n}+x}}+\text{n}\int\frac{x}{\left(\text{n}+x\right)^2\sqrt{\frac{2\text{n}}{\text{n}+x}-1}}\space\text{d}x$$
Now, substitute $\text{u}=\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2089334",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
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Solving proportions with $3$ ratios,$ x:3:y = -2:3:-4$ Proportion seems simple enough for me. Example is $4:x = 2:5,$ and the answer is $x = 10.$
My problem is how do I solve for proportions with $3$ ratios like $x:3:y = -2:3:-4$ ?
Do I write it like
$$\frac{\frac{x}{3}}{y} = \frac{\frac{-2}{3}}{-4} $$
?
| We have: $x:3:y=-2:3:-4$
$\Rightarrow x:3=-2:3$
$\Rightarrow \dfrac{x}{3}=-\dfrac{2}{3}$
$\Rightarrow x=-2$
and
$\Rightarrow 3:y=3:-4$
$\Rightarrow \dfrac{3}{y}=-\dfrac{3}{4}$
$\Rightarrow \dfrac{y}{3}=-\dfrac{4}{3}$
$\Rightarrow y=-4$
$\therefore x=-2,y=-4$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2089636",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
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Let $f(x) = x^3-\frac{3}{2}x^2+x+\frac{1}{4}.$ Then the value of $ \int^{3/4}_{1/4}f(f(x))\mathrm dx$ If $\displaystyle f(x) = x^3-\frac{3}{2}x^2+x+\frac{1}{4}.$ then the value of $\displaystyle \int^{\frac{3}{4}}_{\frac{1}{4}}f(f(x))\mathrm dx$
$\displaystyle \int^{\frac{3}{4}}_{\frac{1}{4}}(f(x))^3-1.5(f(x))^2+f(x)+... | As @dxiv says in the comments, we can write $$f (x) = (x-\frac {1}{2})^3 +\frac {1}{4}x +\frac {3}{8} $$ Thus, substituting $u=x- \frac {1}{2}$ gives us $f (u)=u^3+\frac {1}{4}(u+\frac {1}{2}) +\frac {3}{8} = u^3+\frac {1}{4}u +\frac {1}{2} $. Thus, $$f (f (u)) =(u^3+\frac {1}{4}u +\frac {1}{2})^3 +\frac {1}{4}[u^3 +\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2089877",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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Show that $\sum\limits_{n=1}^{32}\frac1{n^2}=1 + \frac{1}{4} +\frac{1}{9} +\dots+ \frac{1}{1024}<2$ Show that
$$1 + \frac{1}{4} +\frac{1}{9} +\dots+ \frac{1}{1024} <2$$
I know that the denominators are perfect squares starting from that of $1$ till $32$. Also I know about this identity
$$\frac{1}{n(n+1)} > \frac{1}{(... | For $n>0$, we have
$$\frac{1}{(n+1)^2}\leq \frac{1}{n(n+1)}=\frac{1}{n}-\frac{1}{n+1}$$
and by sum,
$$1+\sum_{n=1}^{31}\frac{1}{(n+1)^2}\leq 2-\frac{1}{32}<2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2091928",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
discrete mathematics combinatorics - counting
I have $52$ cards from $4$ series ($13 \times 4=52 \text{ cards}$). Each series are
numbered from $1$ to $13$. In how many possible ways you can draw from the
deck $6$ cards, so that all your six cards has two numbers exactly?
My answer:
First card - $\binom{13}{1} \c... |
The number of different ways to obtain $2$ out of $13$ different numbers is
\begin{align*}
\binom{13}{2}
\end{align*}
Next we consider all valid configurations which can be obtained with two numbers.
Since there are always four cards with the same number and we take six cards from the deck there are three typ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2092165",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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Solve the system of equations ... Solve the system of equations :
(EDIT : The problem does not say anything about the nature of $x$ and $y$ (integer, natural number ,..etc.) )
$4xy + 4(x^2 + y^2) + {\frac {3} { (x+y) ^ 2 } } = \frac {85} {3} $
$2x + {\frac {1} {x+y}} = \frac {13} {3}$
I do not know how to approach the... | Perhaps this can help: Let $x+y=v$ and $x-y=w$, then the equations can be written as
\begin{align*}
3\left(v^2+\frac{1}{v^2}\right)+w^2 & = \frac{85}{3}\\
\left(v+\frac{1}{v}\right)+w & = \frac{13}{3}.
\end{align*}
Now let $v+\frac{1}{v}=t$, then the system can be rewritten as
\begin{align*}
3t^2+w^2 & = \frac{85}{3}+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2094156",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Positive pairs of integral values satisfying $2xy − 4x^2 +12x − 5y = 11$ The number of positive pairs of integral values of $(x, y)$ that solves
$2xy − 4x^2 +12x − 5y = 11$ is?
I rearranged it to $(2x-5)(y+1-2x)=6$, which took quite a bit of time.
So it can be $2*3$ , $3*2$, $6*1$ or $1*6$ which gives us 2 possible pos... | First note that:
$$y=\frac{11 - 12 x + 4 x^2}{2 x-5}$$
Applying division:
$$y=\frac{(2x-1)(2x-5)+6}{2x-5}=(2x-1)+\frac{6}
{2x-5}$$
To be an integer $2x-5$ must divide $6$ so $2x-5$ must be $-1,-3,1,2$ or $3$ or $6$. Check then by the positive integral solutions.
| {
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Any solution for $\iiint\frac{x^2+2y^2}{x^2+4y^2+z^2}\,dv$ I tried to solve this triple integral but couldn't integrate the result.
$$\iiint\frac{x^2+2y^2}{x^2+4y^2+z^2}\,dv$$ and the surface to integrate in is $$x^2+y^2+z^2\le1$$
Is there any way to transform the integral into polar coordinates?
| Trivial by symmetry: the unit ball $B\subset\mathbb{R}^3$ is invariant under the transformation $(x,y,z)\mapsto(z,y,x)$, hence:
$$\iiint_B \frac{x^2+2y^2}{x^2+4y^2+z^2}\,d\mu = \frac{1}{2}\left(\iiint_B \frac{x^2+2y^2}{x^2+4y^2+z^2}\,d\mu+\iiint_B \frac{z^2+2y^2}{x^2+4y^2+z^2}\,d\mu\right) = \frac{1}{2}\mu(B) = \color{... | {
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Show the following equality Basically I want to show the following:
$$\sqrt{2}\ |z|\geq\ |\operatorname{Re}z| + |\operatorname{Im}z|$$
So what I did is the following:
Let $z = a + bi$
Consider the following:
$$2|z|^2 = 2a^2 + 2b^2 = a^2 + b^2 +a^2+b^2$$
Since $(a-b)^2\geq0$, hence $a^2+b^2\geq 2ab$
Thus $2|z|^2 \geq a^... | Let $z=a+bi$. Then $|z|^2=a^2+b^2$ and Re$z=a$ and Im$z=b$. Observe that
$$(\sqrt{2}|z|)^2=2(a^2+b^2)$$ and $$\Big(|\text{Re}z|+|\text{Im}z|\Big)^2=a^2+2|ab|+b^2.$$
Now,
$$
\begin{align}
(|a|-|b|)^2\geq 0&\iff a^2+b^2-2|ab|\geq 0\\
&\iff2(a^2+b^2)\geq a^2+2|ab|+b^2\\
&\iff (\sqrt{2}|z|)^2\geq\Big(|\text{Re}z|+|\text{Im... | {
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Sum of reciprocals of the triangle numbers
Consider the sum of $n$ terms :
$S_n = 1 + \frac{1}{1+2} + \frac {1}{1+2+3} + ... + \frac {1}{1+2+3+...+n}$
for $n \in N$.
Find the least rational number $r$ such that $S_n < r$, for all $n \in N$.
My attempt :
$S_n = 2(1-\frac{1}{2} + \frac {1}{2} - \frac{1}{3} + .... + \... | Let $r=2$, and we can see that
$$\frac{2n-2}n=2-\frac2n<2$$
Similarly, as $n\to\infty$, the limit is $2$, so this is the least rational number satisfying the inequality.
| {
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Prove that if $a$ and $b$ are integers with $a\not= 0$ and $x$ is a positive integer such that $ax^2 + bx + b − a = 0$, then $a|b$. Prove that if $a$ and $b$ are integers with $a\not= 0$ and $x$ is a positive integer such that $ax^2 + bx + b − a = 0$, then $a|b$.
I will use the backwards and forwards method to prove th... | $a, b$ are integers with $a \not= 0$ and $x$ is a positive integer such that $ax^2 + bx + b − a = 0$.
B: $a|b$ ($b$ is divisible by $a$)
B1:
How can I show that one integer (namely, $b$) is divisible by another (namely, $a$)?
Show that there is an integer $k$ such that $\dfrac{b}{a} = k$.
A1: $ax^2 + bx + b − a = 0$ (... | {
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Proof that $\frac{a}{a+3b+3c}+\frac{b}{b+3a+3c}+\frac{c}{c+3a+3b} \ge \frac{3}{7}$ for all $a,b,c > 0$ So I am trtying to proof that $\frac{a}{a+3b+3c}+\frac{b}{b+3a+3c}+\frac{c}{c+3a+3b} \ge \frac{3}{7}$ for all $a,b,c > 0$. First I tried with Cauchy–Schwarz inequality but got nowhere.
Now I am trying to find a conve... | Cauchy-Schwarz actually works pretty well.
We have with Cauchy-Schwarz if we put $x_1=\sqrt{\frac{a}{a+3b+3c}}$, $x_2=\sqrt{\frac{b}{b+3c+3a}}$ and $x_3=\sqrt{\frac{c}{c+3a+3b}}$ as well as $y_1=\sqrt{a(a+3b+3c)}$, $y_2=\sqrt{b(b+3c+3a)}$ and $y_3=\sqrt{c(c+3a+3b)}$
$$
\left(x_1^2+x_2^2+x_3^2\right)\left(y_1^2+y_2^2+y_... | {
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For which $a$, does $p^2+p+(a-1)$ have only rational solutions?
For which $a$, does $p^2+p+(a-1)$ have only rational solutions?
The roots of $p^2+p+(a-1)$ are $\frac{-1+\sqrt{(5-4a)}}{2}$ and $\frac{-1-\sqrt{(5-4a)}}{2}$, and they are rational iff $\sqrt{5-4a}$ is rational, then iff $5-4a$ is a square of some rationa... | It has only rational solutions if $x^2+x+a-1=(x-r)(x-s)=x^2-(r+s)x+rs$ with rational $r,s$. This says that $a=rs+1=r(-1-r)+1=1-r-r^2$. Hence the quadratic equation has only rational solutions if $a$ is representable by $1-r-r^2$ for some rational $r$. For example, $a=\sqrt{2}$ is not of this form.
| {
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If $q^k n^2$ is an odd perfect number with Euler prime $q$, does $I(n^2) \geq 5/3$ imply $k=1$? Let $\sigma(x)$ denote the sum of the divisors of $x \in \mathbb{N}$. Denote the abundancy index of $y \in \mathbb{N}$ by $I(y)=\sigma(y)/y$.
If $\sigma(N)=2N$, then $N$ is said to be perfect.
Euler proved that every odd pe... | In an answer to a related MSE question, we get that
$$I(n^2) = \dfrac{2{q^k}(q - 1)}{q^{k+1} - 1} = 2 - 2\dfrac{q^k - 1}{q^{k+1} - 1}.$$
Then $I(n^2) \geq 5/3$ implies that
$$2 - 2\dfrac{q^k - 1}{q^{k+1} - 1} \geq \dfrac{5}{3}$$
$$\dfrac{1}{6} \geq \dfrac{q^k - 1}{q^{k+1} - 1}$$
$$q^{k+1} - 6q^k + 5 \geq 0,$$
which doe... | {
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How do you compute the $\gcd(1+n+n^2,1+n+n^2+s+2ns+s^2)$ I would like to prove the following claim which I think is true:
Claim: Let $n,$ $m$ and $s$ be positive numbers. Fix $s$, then for every positive number $n$ the $\gcd(1+n+n^2,1+n+s+n^2+2ns+s^2)$ will be equal to a divisor of $1+5s^2+s^4.$
For example for every... | No way a statement like this can be right.
$$\gcd(1+n+n^2, 1+n+s+n^2+2ns+s^2) = \gcd(1+n+n^2, s(1+2n+n^2)) = \gcd(1+n+n^2, sn) = \gcd(1+n+n^2, s)$$. So simply take $s = n^2+n+1$, then it can never be $s |1+5s^2+s^4$, unless $s=1$.
| {
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Prove $n^5+n^4+1$ is not a prime I have to prove that for any $n>1$, the number $n^5+n^4+1$ is not a prime.With induction I have been able to show that it is true for base case $n=2$, since $n>1$.However, I cannot break down the given expression involving fifth and fourth power into simpler terms. Any help?
| $n^5 + n^4 + 1 = n^5 + n^4 + n^3 – n^3 – n^2 − n + n^2 + n + 1$
$\implies$$ n^3(n^2 + n + 1) − n(n^2 + n + 1) + (n^2 + n + 1)$
=$ (n^2 + n + 1)(n^3 − n + 1)$
Hence, for $n>1$, $n^5 + n^4 + 1$ is not a prime number.
| {
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If $f(1)$ and $f(i)$ real, then find minimum value of $|a|+|b|$ A function $f$ is defined by $$f(z)=(4+i)z^2+a z+b$$ $(i=\sqrt{-1})$for all complex number $z$ where $a$ and $b$ are complex numbers. If $f(1)$ and $f(i)$ are both purely real, then find minimum value of $|a|+|b|$
Now
$f(1)=4+i+a+b$ which means imaginary ... | Setting $a = a_1 + i a_2, b = b_1 + i b_2$ for four real unknowns $a_k, b_k$.
Real $f(1) = 4 + i + a + b$ and real $f(i) = -4 - i + a i + b$ give the equations
$$
\text{Im}(f(1)) = 1 + a_2 + b_2 = 0 \\
\text{Im}(f(i)) = -1 + a_1 + b_2 = 0
$$
Then we have to minimize
$$
g(a_1, a_2, b_1, b_2)
= \lvert a \rvert + \lvert ... | {
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Optimization and maximum geometry What is the side length of the largest square that will fit inside an equilateral triangle with sides of length 1.
I created two equations:
Square: Area=$x^2$
Triangle: Area= $\sqrt{3 /4}$
However, how can I find the maximum??
I did $\sqrt{3 /4}=x^2$ ang got fourth root $3 / 2$ which i... |
$1)$ Suppose that $DEFG$ is a square. Let's call $x$ the side of the square.
We also have that $\Delta ADE$ is an equilateral triangle, so $AD=x$.
Now, for the triangle $BDG$ we have:
$$\sin 60°=\frac{DG}{BD}=\frac{x}{BD} \rightarrow BD=\frac{2\sqrt{3}x}{3}$$
$$AB=1=AD+BD=x+\frac{2\sqrt{3}x}{3} \rightarrow x=\frac{3}... | {
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How do I divide a square root of a fraction by another square root of another fraction? My teacher gave this challenger in class today and I can't figure how to solve it:
$\frac{\sqrt{\frac23} - \sqrt{\frac32}}{\sqrt{\frac13} - \sqrt{\frac12}}$
| Given,
$\frac{\sqrt{\frac23} - \sqrt{\frac32}}{\sqrt{\frac13} - \sqrt{\frac12}}$
On rationalizing,
$\frac{\sqrt{\frac23} - \sqrt{\frac32}}{\sqrt{\frac13} - \sqrt{\frac12}} \cdot \frac{\sqrt{\frac13} + \sqrt{\frac12}}{\sqrt{\frac13} + \sqrt{\frac12}}$
= $\frac{\sqrt{\frac29} + \sqrt{\frac13} - \sqrt{\frac12} - \sqrt{\fr... | {
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Solve $y'=\frac{x+y-2}{x-y}$ ODE
$$y'=\frac{x+y-2}{x-y}$$
$$\begin{vmatrix}
1 & 1\\
1 & -1
\end{vmatrix}\neq0 $$
Substation: $x=u+\alpha$ , $y=v+\beta$
$\frac{dy}{dx}=\frac{dv}{du}$
$$\begin{pmatrix}
1 & 1\\
1 & -1
\end{pmatrix}\begin{pmatrix}
\alpha \\
\beta
\end{pmatrix} =\begin{pmatrix... | $u\frac{dz}{du}=\frac{1+z}{1-z}-z=\frac{z^2+1}{1-z}$ thus
$$\frac{1-z}{1+z^2}dz=\frac{1}{u}du$$
or
$$\left(\frac{1}{1+z^2}-\frac{z}{1+z^2}\right)dz=\frac{1}{u}du$$
as a result
$$\tan^{-1}(z)-\frac12\ln(1+z^2)=C+\ln(u)$$
| {
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Derivative of arcsin In my assignment I need to analyze the function
$f(x)=\arcsin \frac{1-x^2}{1+x^2}$
And so I need to do the first derivative and my result is:
$-\dfrac{4x}{\left(x^2+1\right)^2\sqrt{1-\frac{\left(1-x^2\right)^2}{\left(x^2+1\right)^2}}}$
But in the solution of this assignment it says
$f'(x)=-\frac... | Divide et impera.
Under the square root you have
$$
1-\frac{(1-x^2)^2}{(1+x^2)^2}=
\frac{(1+x^2)^2-(1-x^2)^2}{(1+x^2)^2}=\frac{4x^2}{(1+x^2)^2}
$$
so the big square root is actually
$$
\frac{2|x|}{1+x^2}
$$
Thus your formula becomes
$$
-\frac{4x}{(1+x^2)^2}\frac{1+x^2}{2|x|}=-\frac{2x}{|x|(1+x^2)}
$$
| {
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Find all possible positive integer $n$ such that $3^{n-1} + 5^{n-1} \mid 3^n + 5^n $
Find all possible positive integers $n$ such that $3^{n-1} + 5^{n-1} \mid 3^n + 5^n $.
I don't know how to start with. Any hint or full solution will be helpful.
| Let's suppose that $3^{n-1}+5^{n-1}\mid 3^n+5^n$, so there is some positive integer $k$ such that $3^n+5^n=k(3^{n-1}+5^{n-1})$. Now, if $k\ge 5$ we have $$k(3^{n-1}+5^{n-1})\ge 5(3^{n-1}+5^{n-1})=5\cdot 3^{n-1}+5\cdot 5^{n-1}>3^n+5^n.$$
This means that $k\le 4$. On the other hand, $3^n+5^n=3\cdot 3^{n-1}+5\cdot 5^{n-1}... | {
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How to find $\lim_{x\to1}\frac{\sqrt[3]{x+7}+2\sqrt{3x+1}-6}{\sqrt[4]{x}-1}$ without using L'Hospital Rule? Compute the following limit. I've tried using l'Hospital. And it work the result was $\dfrac{7}{3}$. But how can I do this without this rule? I am trying to help I friend who hasn't done derivatives yet.
$$\lim_{... | Letting $u=\sqrt[4]{x}$, you have that: $\frac{\sqrt[4]x -1}{x-1}=\frac{u-1}{u^4-1}=\frac{1}{1+u+u^2+u^3}\to \frac{1}{4}$ as $u\to 1$ and hence also as $x\to 1$.
Letting $v=\sqrt[3]{x+7}$ we get that $\frac{\sqrt[3]{x+7}-2}{x-1}=\frac{v-2}{v^3-8}=\frac{1}{v^2+2v+4}\to \frac{1}{12}$ as $v\to 2$ and hence as $x\to 1$.
Fi... | {
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Explain why $(a−b)^2 = a^2 −b^2$ if and only if $b = 0$ or $b = a$. This is a question out of "Precalculus: A Prelude to Calculus" second edition by Sheldon Axler. on page 19 problem number 54.
The problem is Explain why $(a−b)^2 = a^2 −b^2 $ if and only if $b = 0$ or $b = a$.
So I started by expanding $(a−b)^2$ to $(a... | This is a geometrical approach.
Let $a-b=d$. Then $d^2 + b^2 = a^2$ and the using Converse of Pythagoras theorem there must be a triangle having $a, b, a-b$ as sides, which is impossible because $b + (a - b) = a \not \gt a$.
| {
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How can I find $\sum_{cyc}\sin x\sin y$ $x$, $y$ & $z$ are real number such that
$$\frac{\sin{x}+\sin{y}+\sin{z}}{\sin{(x+y+z)}}=\frac{\cos{x}+\cos{y}+\cos{z}}{\cos{(x+y+z)}}=2$$
find the value
$$\sum_{cyc}\sin x\sin y$$
All help would be appreciated.
| Since
\begin{align*}
2 \cos(x+y+z) &= \cos x + \cos y + \cos z \\
2 \sin(x+y+z) &= \sin x + \sin y +\sin z
\end{align*}
\begin{align*}
2 e^{i(x+y+z)} & = e^{ix} + e^{iy} + e^{iz}
\end{align*}
Multiplying throughout by $e^{-ix}$, we get
\begin{align*}
2e^{i(y+z)} = 1 + e^{i(y-x)} + e^{i(z-x)}
\end{align*}
Equating the... | {
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Prove $a = 2 \cdot 10^{2016} + 16$ cannot be a perfect square
Prove the number $a = 2 \cdot 10^{2016} + 16$ cannot be a perfect square.
I tried studying $a \mod b$ for different integers $b$, without succes.
| Let $$k^2 = 2\cdot 10^{2016} + 16$$ Then we have $(k-4)(k+4)=2\cdot 10^{2016}$. Thus we have two factors of $2\cdot 10^{2016}$ with a difference of $8$.
Now if these factors, say $a$ and $b$, both contain $5$ in their prime factorization, then we have that $|a-b|$ is a multiple of $5$ which is not $8$. Thus let $a=2^... | {
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Show that $a^3+a^2b+ab^2+b^3$ is not a prime, if a and b are positive integers I approached it by attempting to factor it and then show that one factor can't be one and the other can't be prime. This gets nowhere, as you can't factor this expression. Is there another way to do it?
| $a^3+b^3+a^2b+ab^2\\=(a+b)(a^2-ab+b^2)+a^2b+ab^2\\=(a+b)(a^2-ab+b^2)+ab(a+b)\\=(a+b)(a^2-ab+b^2+ab)\\=(a+b)(a^2+b^2)$
| {
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Show that determinant of $\small\begin{pmatrix}2 & 2 & 8\\ 3& 2 & 3 \\ 4 & 5 & 6\end{pmatrix}$ is divisible by $19$ Using that the numbers
$228,
323$
and
$456$
are
divisible
by
$19$.
Show
that
the
determinant of matrix
$\begin{pmatrix}2 & 2 & 8\\ 3& 2 & 3 \\ 4 & 5 & 6\end{pmatrix}$
is
divisible
by
$19$.
| $$\det\begin{pmatrix}2 & 2 & 8\\ 3& 2 & 3 \\ 4 & 5 & 6\end{pmatrix}=\det\begin{pmatrix}22 & 2 & 8\\ 32& 2 & 3 \\ 45 & 5 & 6\end{pmatrix}=\det\begin{pmatrix}\color{red}{228} & 2 & 8\\ \color{red}{323}& 2 & 3 \\ \color{red}{456} & 5 & 6\end{pmatrix}=\color{red}{19}\cdot\det\begin{pmatrix}12 & 2 & 8\\ 17& 2 & 3 \\ 24 & 5 ... | {
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Prove that $a^{\frac {1}{4}}+b^{\frac {1}{4}}+c^{\frac {1}{4}}=0\Rightarrow a+b+c =2(\sqrt {ab}+\sqrt {bc}+\sqrt {ca})$ If $a^{\frac {1}{4}}+b^{\frac {1}{4}}+c^{\frac {1}{4}}=0$, prove that $a+b+c =2(\sqrt {ab}+\sqrt {bc}+\sqrt {ca})$.
I guess that the given expression is true if and only if $a=b=c=0$. Is it true? Or,i... | Maybe you mean that
$$2(\sqrt{ab}+\sqrt{ac}+\sqrt{bc})-a-b-c=$$
$$=(\sqrt[4]a+\sqrt[4]b+\sqrt[4]c)(\sqrt[4]a+\sqrt[4]b-\sqrt[4]c)(\sqrt[4]a-\sqrt[4]b+\sqrt[4]c)(-\sqrt[4]a+\sqrt[4]b+\sqrt[4]c),$$
which is not necessary.
| {
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"answer_id": 1
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Find all prime solutions of equation $5x^2-7x+1=y^2.$ Find all prime solutions of the equation $5x^2-7x+1=y^2.$
It is easy to see that
$y^2+2x^2=1 \mod 7.$ Since $\mod 7$-residues are $1,2,4$ it follows that $y^2=4 \mod 7$, $x^2=2 \mod 7$ or $y=2,5 \mod 7$ and $x=3,4 \mod 7.$
In the same way from $y^2+2x=1 \mod 5$ w... | If there is a solution it has more than 4000 digits, which makes me think there is no solution beyond the one two already mentioned.
In Mathematica,
i=1;
ans=Solve[5 x ^2-7x+1==y^2&&x>0&&y>0,{x,y},Integers];
ans2=ans/.C[1]->i//RootReduce
Dynamic@i
Dynamic[IntegerLength@ans2[[All,All,2]]]
and then run
While[FreeQ[P... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2117054",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 5,
"answer_id": 4
} |
Find the value of $a^4-a^3+a^2+2$ when $a^2+2=2a$ Find the value of $a^4-a^3+a^2+2$ when $a^2+2=2a$
My Attempt,
$$a^4-a^3+a^2+2=a^4-a^3+2a$$
$$=a(a^3-a^2+2)$$
What's next?
| Obviously $a \not = 0$. Then we have $a^4 = 2a^3 - 2a^2$ by multiplying the condition by $a^2$, so the equation becomes: $a^3 - a^2 + 2$. Similarly $a^3 = 2a^2 - 2a$, so the equation becomes: $a^2 - 2a + 2 = 0$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2117727",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
Coefficient of $x^2$ in $(x+\frac 2x)^6$ I did $6C4 x^2\times (\dfrac 2x)^4$ and got that the coefficient of $x^2$ is $15$, but the answer is $60$, why? Did I miss a step?
| We have $$\displaystyle\left(x+\dfrac{2}{x}\right)^6=\sum_{k=0}^{6}\binom{6}{k}x^{6-k}\left(\frac{2}{x}\right)^k=\sum_{k=0}^{6}2^k\binom{6}{k}x^{6-2k}.$$ Then, $6-2k=2\Leftrightarrow k=2,$ so $$\text{coef }x^2=2^2\binom{6}{2}=60.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2118037",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 3
} |
smallest positive integer $r$ so that $5^{33333}≡ r (mod 11)$ Title is pretty self explanatory. I tried using the division algorithm to get some a hint about the $5^{33333}$. This was not helpful.
| $$5\equiv 5\mod 11\\
5^2=25\equiv 3\mod 11\\
5^3= 5^2\cdot 5\equiv 3\cdot 5=15\equiv 4\mod 11\\
5^4= 5^3\cdot 5\equiv 4\cdot 5=20\equiv 9\mod 11\\
5^5= 5^4\cdot 5\equiv 9\cdot 5=45\equiv 1\mod 11\\
$$
So, you know that $5^5\equiv 1\mod 11$.
You also know that $33333=5\cdot 6666+3$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2123670",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
Does flipping the negative in a fraction flip all terms both sides? This is an expression that contains a negative in the denominator. If I were to take the negative and place it in the numerator, would this change all positive terms to negative and vice-versa?
Negative in denominator:
$\frac{a+3-y^2}{-2}$
After flipp... | Yes, but your language is a little vague and misleading. Things to notice:
$1/1 = 1$ (Duh)
$1/(-1) = -1$ (Not so duh; but pretty easy)
$1*1 = 1; (-1)*1= 1*(-1) = -1; (-1)(-1) = 1$ ("A negative times a negative is a positive". To some this is obvious; to some this makes no sense. But most know this as a rule [$*$])... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2124391",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
Max and min of a function with absolute values I have this function :
$f(x) =
\begin{cases}
\left\lvert\frac{x-3}{x+3}\right\rvert\log \left\lvert\frac{x+3}{x-3}\right\rvert -3, & \text{if $x\neq \pm3$} \\[2ex]
-3, & \text{if $x= \pm3$}
\end{cases}$
How do you find the derivative?
I usually split the functions when t... | I think what you have is correct. Namely, you can prove using your derivatives on different intervals that the function is strictly decreasing of $x<-3$. On $(-3,3)$ and on $(3,\infty)$ the function is probably concave and hence has one local maximum on each interval. Comparing values at those local maxima with $-3$, v... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2126046",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
$y=x^2-x^3$ Area under curve I have to use the limit process to fund the area of the region between the graph of the function, and the x-axis over the given interval.
$f(x)=x^2-x^3$ in the interval of $[-1,0]$
I started the problem by solving for $\Delta x$, and got that $\Delta x=\frac{1}{n}$. Then I got $c_i=a+i\Del... | To find the area under the curve, evaluate:
$\lim_\limits{n \to \infty}\sum_\limits{i=1}^n(f(c_i)\Delta x)$
$\Delta x = \frac{b-a}{n} = \frac{0 - (-1)}{n} = \frac{1}{n}$
$c_i = a + i\Delta x = -1 + \frac{i}{n}$
$f(c_i) = (c_i)^2 - (c_i)^3 = \frac{4i^2}{n^2} - \frac{5i}{n} - \frac{i^3}{n^3} + 2$
So we have,
$\lim_\limit... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2126692",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Derive $\cos(3\theta)=(4\cos\theta)^3 − 3\cos\theta$ I'm having trouble with the following derivation:
Q:
We can use Euler's Theorem ($e^{i\theta} = \cos\theta + i\sin\theta$), where $e$ is the base of the natural logarithms, and $i = \sqrt{-1}$, together with the binomial theorem as above, to derive a number of trigo... | Continue from your answer,
$\cos(3\theta) = (\cos \theta)^3 - 3(\sin \theta)^2\cos \theta$
$= (\cos \theta)^3 - 3(1-\cos^2 \theta)\cos \theta$
$= (\cos \theta)^3 - 3\cos \theta(1-\cos^2 \theta)$
$= \cos^3 \theta - 3\cos \theta + 3\cos^3\theta$
$= 4\cos^3 \theta - 3\cos \theta$
Similarly you can change $\cos^2\theta = 1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2126778",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
What is the greatest area difference between two "nice triangles"? We call a triangle "nice if all angles are between $45$ and $90$ degrees (including $90$ and $45$ itself) and all sides are between $1$ and $2$ (including $1$ and $2$ itself). What is the greatest area difference between two "nice triangles"?
My attempt... | The Maximal triangle has two sides of length $2.$
for any triangle that had only 2 sides of lenght 2, we could extend the second longest side until it was length 2, and create a larger triangle.
$Area = \frac 12 BC\sin a\\
B,C = 2$
maximize $2\sin a$
constrained by:
$4 \sin \frac {a}{2} \le 2$
$\sin \frac {a}{2}\le \fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2127099",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
What is $x$, if $3^x+3^{-x}=1$? I came across a really brain-racking problem.
Determine $x$, such that $3^x+3^{-x}=1$.
This is how I tried solving it:
$$3^x+\frac{1}{3^x}=1$$
$$3^{2x}+1=3^x$$
$$3^{2x}-3^x=-1$$
Let $A=3^x$.
$$A^2-A+1=0$$
$$\frac{-1±\sqrt{1^2-4\cdot1\cdot1}}{2\cdot1}=0$$
$$\frac{-1±\sqrt{-3}}{2}=0$$
I... | Just building upon previous comments, it doesn't have as you pointed out a Real result but the Imaginary solution can be analytically found as:
$3^x = \dfrac{1\pm\sqrt{3} i}{2}$
Reexpressing rhs in polar notation:
$3^x = e^{i \dfrac{\pi}{3}}$
And changing lhs basis to $e$
$3^x = e^{\ln{3^x}} = e^{x\ln{3}} $
Then:
$\box... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2128506",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
"answer_id": 3
} |
Continuity and integration using Dominated Convergence theorem I have to use the Dominated Convergence Theorem to show that $\lim \limits_{n \to \infty}$ $\int_0^1f_n(x)dx=0$ where $f_n(x)=\frac{n\sqrt{x}}{1+n^2x^2}$.
I did the following:
$$\frac{n\sqrt{x}}{1+n^2x^2} <\frac{n\sqrt{x}}{n^2x^2} = \frac{x^{-\frac{3}{2}}... | Let $f_n(x)=\frac{n\sqrt{x}}{1+n^2x^2}$. Then, differentiating with respect to $n$ reveals
$$\begin{align}
\frac{df_n(x)}{dn}&=\frac{\sqrt{x}}{1+n^2x^2}-\frac{2n^2x^{5/2}}{(1+n^2x^2)^2}\\\\
&=\frac{\sqrt{x}(1-n^2x^2)}{(1+n^2x^2)^2}\tag1
\end{align}$$
From $(1)$, it is easy to see that $\frac{df_n(x)}{dn}=0$ when $n=1/... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2130472",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Let $a,b\in \mathbb{N}$ where $\gcd(a,b)=1$. Describe the set $\mathcal{S}=\{ax+by\mid x,y\ge0 \text{ and } x,y\in \mathbb{Z}\}$.
Let $a,b\in \mathbb{N}$ where $\gcd(a,b)=1$. Describe the set $$\mathcal{S}=\{ax+by \mid x,y\geq0 \hspace{0.25cm} \text{ and } \hspace{0.25cm} x,y\in \mathbb{Z}\}.$$
Since $a,b\in \mathbb... | $S \subset \mathbb N$ (let's include $0$ in our natural numbers) clearly. $a\mathbb N \subset S$ and $b\mathbb N \subset S$ so if $a = 1$ or $b =1$, $S = \mathbb N$.
So it suffices to assume $a > 1$ and $b > 1$.
Okay some bold claims:
1) $n = a(b-1) + b(-1) = a(-1) + b(a-1)=ab-a-b \not \in S$.
2) For any $n > ab-a-b$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2131050",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Prove that $\sum\limits_{n=1}^{\infty} \frac{n}{3^n} = \frac{3}{4}$ How can you derive that
$$ \sum\limits_{n=1}^{\infty} \frac{n}{3^n} = \frac{3}{4} \, ?$$
I suspect some clever use of the geometric series will do, but I don't know how.
| Take a geometric series with term $r$ and take derivative with respect to $r$ both sides:
$\frac{d}{dr}\sum_{n=0}^{\infty}r^n=\frac{d}{dr}\frac{1}{1-r}.$
Derivate,
$\sum_{n=0}^{\infty}nr^{n-1}=\frac{1}{(1-r)^2}.$
At the left hand side, for $n=0$ the term is zero, so we can start the sum at $n=1$, and also multiply $r$ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2131479",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 5,
"answer_id": 2
} |
Antiderivative of a product of a rational function and a square root. Let $0<x<1$ and $p$ be a non-negative integer.
Consider a following definite integral:
\begin{equation}
{\mathbb I}_p(x):=\int\limits_1^x \frac{(1-v^2)^p}{v^{2p+2}} \cdot \sqrt{\frac{1-v}{1+v}} dv
\end{equation}
We have computed it in a following way... | $\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,}
\newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
\newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
\newcommand{\dd}{\mathrm{d}}
\newcommand{\ds}[1]{\displaystyle{#1}}
\newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
\newc... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2133670",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Evaluate the integral $\int \frac{x^2(x-2)}{(x-1)^2}dx$
Find $$\int \frac{x^2(x-2)}{(x-1)^2} dx .$$
My attempt:
$$\int \frac{x^2(x-2)}{(x-1)^2}dx = \int \frac{x^3-2x^2}{x^2-2x+1}dx $$
By applying polynomial division, it follows that
$$\frac{x^3-2x^2}{x^2-2x+1} = x + \frac{-x}{x^2-2x+1}$$
Hence $$\int \frac{x^3-2x^2}{... | I would let $u=x-1$ which makes the integrand $(u+1)^2(u-1)/u^2.$ After you multiply out the top you have a sum of powers of $u$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2135003",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 7,
"answer_id": 4
} |
$\lim_{z \to \exp(i\pi/3)} \frac{z^3+8}{z^4+4z+16}$ Find $$\lim_{z \to \exp(i \pi/3)} \dfrac{z^3+8}{z^4+4z+16}$$
Note that
$$z=\exp(\pi i/3)=\cos(\pi/3)+i\sin(\pi/3)=\dfrac{1}{2}+i\dfrac{\sqrt{3}}{2}$$
$$z^2=\exp(2\pi i/3)=\cos(2\pi/3)+i\sin(2\pi/3)=-\dfrac{1}{2}+i\dfrac{\sqrt{3}}{2}$$
$$z^3=\exp(3\pi i/3)=\cos(\pi)+i\... | You can make your life very easy by noticing that $(\mathrm e^{\mathrm i \pi/3})^3=\mathrm e^{\mathrm i \pi} = -1$.
This means that if $z=\mathrm e^{\mathrm i \pi/3}$, then $z^3=-1$ and hence $z^4=-z$.
$$\frac{z^3+8}{z^4+4z+16} = \frac{7}{3z+16}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2136937",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Find the $x$ such $|AB|=\sin{x},|BC|=\cos{x},|CD|=\tan{x},|DA|=\cot{x},|AC|=\sec{x},|BD|=\csc{x}$ Find $x$, such there exsit four point $A,B,C,D$ on the plane,such
$$|AB|=\sin{x},|BC|=\cos{x},|CD|=\tan{x},|DA|=\cot{x},|AC|=\sec{x},|BD|=\csc{x}$$
I think it's nice problem. I'm looking for an arbitrary quadrilateral six ... | There are no such $x\in\mathbb R$.
We may suppose that
$$0\lt x\lt \frac{\pi}{2},\quad A(0,0),\quad B(\sin x,0),\quad C(c_1,c_2),\quad D(d_1,d_2)$$
with $c_2\ge 0$.
Since $C(c_1,c_2)$ where $c_2\ge 0$ is both on the circle $X^2+Y^2=|AC|^2$ and on the circle $(X-|AB|)^2+Y^2=|BC|^2$, we get
$$c_1=\frac{\sec^2x+\sin^2x-\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2137636",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Can the quadratic formula be used with variable coefficients? Could we use the quadratic formula on an expression such as $z^2xy - zx^2y+y = 0$ to find $z$ in terms of $x$ and $y$?
| Write $$z^2xy - zx^2y + y = 0$$
as $$(xy)z^2 - (x^2y)z + (y) = 0$$
so that you can see $a = xy$, $b=-x^2y$ and $c=y$.
Then
\begin{align}
z &= \dfrac{-b \pm \sqrt{b^2-4ac}}{2a} \\
z &= \dfrac{x^2y \pm \sqrt{x^4y^2-4xy^2}}{2xy} \\
z &= \dfrac{x^2y \pm |y|\sqrt{x^4-4x}}{2xy} \\
z &= \dfrac{x^2 \pm \sqrt... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2139128",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 5,
"answer_id": 0
} |
Pseudoinverse of $2\times 2$ matrix How can I find the Moore-Penrose pseudoinverse of the $2 \times 2$ complex matrix
$$A=\begin{pmatrix}0&a\\0&b\end{pmatrix}$$
for $a \neq 0$ and $b \neq 0$?
Here I want to use the limit formula
$$A^+=\lim_{\epsilon \to 0} (\epsilon I+A^*A)^{-1}A^*$$
since $\mbox{rank}(A)=1$, which ... | Computing eigendecompositions using SymPy:
>>> from sympy import *
>>> a, b = symbols('a b')
>>> M = Matrix([[0,a],[0,b]])
>>> (M.T * M).eigenvects()
[(0, 1, [Matrix([
[1],
[0]])]), (a**2 + b**2, 1, [Matrix([
[0],
[1]])])]
>>> (M * M.T).eigenvects()
[(0, 1, [Matrix([
[-b/a],
[ 1]])]), (a**2 + b**2, 1, [Matrix([
[a/b]... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2141266",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
rational numbers of the form $\frac{a^3+b^3}{c^3+d^3}$ Show that all positive rational numbers can be written in the form
$$\frac{a^3+b^3}{c^3+d^3}$$
where $a,b,c,d$ are positive integers.
| Xam commented on 2017 Feb 13 that an answer was given here by zabelman on 2005 Dec 17. Also on 2017 Feb 13, timon92 mentioned the following answer found in a post by mathlove on 2015 Mar 11:
For every positive rational number $r$, there exists a set of four positive integers $(a,b,c,d)$ such that
$$r=\frac{a^\color... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2142429",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
$\lim_{x\to2}\frac{\sqrt{1+\sqrt{x+2}-\sqrt3}}{x-2}$ $\lim_{x\to2}\frac{\sqrt{1+\sqrt{x+2}-\sqrt3}}{x-2}$
My try:
$\lim_{x\to2}\frac{\sqrt{1+\sqrt{x+2}-\sqrt3}}{x-2}=$
$\lim_{x\to2}\frac{\sqrt{1+\sqrt{x+2}-\sqrt3}}{x-2}\times\frac{\sqrt{1+\sqrt{x+2}+\sqrt3}}{\sqrt{1+\sqrt{x+2}+\sqrt3}}=$
$\lim_{x\to2}\frac{\sqrt{x+2\sq... | Following my comment, the question would make "more sense" (seeing that you know the method of multiplying with a conjugate expression) if the numerator was $\sqrt{1+\sqrt{x+2}}-\sqrt3$.
If you did copy the question correctly, then it might have been a bit of a trick question since you don't need that method; it is no... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2142540",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
Given the positive numbers $a, b, c$. Prove that $\frac{a}{\sqrt{a^2+1}}+\frac{b}{\sqrt{b^2+1}}+\frac{c}{\sqrt{c^2+1}}\le \frac{3}{2}$ Given the positive numbers $a, b, c$ satisfy $a+b+c\le \sqrt{3}$. Prove that $\frac{a}{\sqrt{a^2+1}}+\frac{b}{\sqrt{b^2+1}}+\frac{c}{\sqrt{c^2+1}}\le \frac{3}{2}$
My Try (Edited from C... | By AM-GM and P-M we obtain:
$$\sum_{cyc}\frac{a}{\sqrt{a^2+1}}=\sum_{cyc}\frac{a}{\sqrt{a^2+3\cdot\frac{1}{3}}}\leq\sum_{cyc}\frac{a}{\sqrt{4\sqrt[4]{\frac{a^2}{27}}}}=
\frac{3\sqrt[8]{27}}{2}\frac{\sum\limits_{cyc}\sqrt[4]{a^3}}{3}\leq$$
$$\leq\frac{3\sqrt[8]{27}}{2}\left(\frac{a+b+c}{3}\right)^{\frac{3}{4}}\leq\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2144208",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
A problem with two dimensional vectors and scalar multiples
Let $\vec{x}$ and $\vec{y}$ be non-zero vectors. Show that any two-dimensional vector can be expressed in the form
$s \vec{x} + t \vec{y}$, where $a$ and $b$ are real numbers, if and only if of the vectors $\vec{x}$ and $\vec{y}$, one vector is not a scalar... | Suppose $\vec{x} = k \vec{y}$. Then, $s\vec{x} + t\vec{y} = (s+kt)\vec{x}$. The locus of such points as $s$ and $t$ range over the reals is a line. So, not every two-dimensional vector can be expressed as $$s\vec{x} + t\vec{y}.$$For the other direction, let $\vec{x} = \begin{pmatrix} a \\ b \end{pmatrix}$ and $\vec{y} ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2145233",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Solve the following system of homogeneous linear equation. $$2x-y+z=0, 3x+2y-z=0,x+4y+3z=0$$
$$
\begin{vmatrix}
2 & -1 & 1 \\
3 & 2 & -1 \\
1 & 4 & 3 \\
\end{vmatrix}
$$
By reducing row $$R_1=R_1-2R_3\\R_2=R_3-3R_3$$
we get
\begin{vmatrix}
0 & -9 & -5 \\
0 & -10 ... | Once all equations are equal to zero then:
$$\begin{vmatrix}
0 & -9 & -5 \\
0 & -10 & -10 \\
1 & 4 & 3 \\
\end{vmatrix}=\begin{vmatrix}
0 & -9 & -5 \\
0 & 1 & 1 \\
1 & 4 & 3 \\
\end{vmatrix}$$
And now $R_1=R_1+9R_2$ and get
$$\begin{vmatrix}
0 & ... | {
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"timestamp": "2023-03-29T00:00:00",
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How to validate the following inequality? $\sqrt{2(x+y+z)}\leq a\sqrt{x+y}+b\sqrt{z}.$ I want to find the least values of $a$ and $b$ for which the above inequality holds good for all nonnegative real values of $x, y, z.$
| $$\tag{1}\sqrt{2(x+y+z)}\leq a\sqrt{x+y}+b\sqrt{z}$$
Let us proceed by successive equivalent propositions, reaching at the end two unique values for $a$ and $b$.
First of all, you do not need three variables: you can amalgamate $x$ and $y$ into a single variable $t:=x+y$.
$$\tag{2}\sqrt{2(t+z)}\leq a\sqrt{t}+b\sqrt{z}$... | {
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"timestamp": "2023-03-29T00:00:00",
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To find value of $\sin 4x$ Given $\tan x = \frac { 1+ \sqrt{1+x}}{1+ \sqrt{1-x}}$. i have to find value of $\sin 4x$. i write $\sin 4x=4 \frac{ (1-\tan^2 x)(2 \tan x)}{(1+\tan^2 x)^2}$ but it seems very complicated to do this? Any other methods?
Thanks
| $a=\sqrt{1+y}\qquad b=\sqrt{1-y}\qquad t=\tan(x)=\frac{1+a}{1+b}\qquad \sin(4x)=\frac{4t(1-t^2)}{(1+t^2)^2}$
$a^2+b^2=2$
$a^2-b^2=2y$
Let's have fun:
$\displaystyle{\sin(4x)=4\left(\frac{1+a}{1+b}\right)\frac{(1+b)^2-(1+a)^2}{(1+b)^2}\frac{(1+b)^4}{((1+a)^2+(1+b)^2)^2}}$
$((1+a)^2+(1+b)^2)^2=(1+2a+a^2+1+2b+b^2)^2=4(2+a... | {
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"url": "https://math.stackexchange.com/questions/2148360",
"timestamp": "2023-03-29T00:00:00",
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Computing $7^{13} \mod 40$ I wanted to compute $7^{13} \mod 40$. I showed that
$$7^{13} \equiv 2^{13} \equiv 2 \mod 5$$
and
$$7^{13} \equiv (-1)^{13} \equiv -1 \mod 8$$.
Therefore, I have that $7^{13} - 2$ is a multiple of $5$, whereas $7^{13} +1$ is a multiple of $8$. I wanted to make both equal, so I solved $-2 + 5k... | You could note that $\varphi(5) = 5-1 = 4$ and $\varphi(8) = 8- 4 = 4$. Hence $7^4 \equiv 1 \pmod 5$ and $7^4 \equiv 1 \pmod 8$, in which case $7^4 \equiv 1 \pmod{40}$.
Hence $7^{13} \equiv (7^4)^3 \cdot 7 \equiv 7 \pmod{40}$.
| {
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"url": "https://math.stackexchange.com/questions/2149062",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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"answer_id": 4
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Solve for the general solution of the equation $(2y^2+3xy-2y+6x)dx + x(x+2y-1)dy=0$ I know that this equation is neither exact nor homogeneous but after I put it in the form \[\frac{dy}{dx} + \frac{2y^2+3xy-2y+6x}{x(x+2y-1)}=0\]
I don't know how to proceed with it. Maybe some substitutions I overlooked?
Thanks!
| Or you can calculate directly an integrating factor.
Integrating factor. If you write your equation as the line-field represented by the one-form $$(2y^2+3xy-2y+6x)dx + x(x+2y-1)dy=0$$ set $A(x,y) = 2y^2+3xy-2y+6x$ and $B(x,y) = x(x+2y-1)$ and look for a function $\mu(x,y)$ such that the form $\mu \, A \, dx + \mu \, ... | {
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Divide $1^2,2^2..,81^2$ numbers into 3 groups. Given $1^2,2^2,3^2,.....,81^2$ numbers. How can I divide them into $3$ groups with $27$ numbers in each so that they have the same sum.
Is there any algorithm to solve this task?
Thanks in Advance.
| We take a sequence of 9 consecutive squares, $n^2, (n+1)^2,...,(n+8)^2$. Then,
$$(n+0)^2 + (n+4)^2 + (n+8)^2 = 3n^2 + 24n + 80$$
$$(n+1)^2 + (n+5)^2 + (n+6)^2 = 3n^2 + 24n + 62$$
$$(n+2)^2 + (n+3)^2 + (n+7)^2 = 3n^2 + 24n + 62$$
So, we can divide 27 squares into 3 groups of equal sum, by rotating the dominant group ou... | {
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"url": "https://math.stackexchange.com/questions/2150253",
"timestamp": "2023-03-29T00:00:00",
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What is the remainder when $15^{40}$ divided by $1309$? I know that $$ 15 \equiv 1\pmod{7}, N \equiv 1\pmod{7},$$ but cannot proceed further.
| Note that $1309=7 \cdot 11 \cdot 17$. Let $N=15^{40}$.
Since $15 \equiv 1\pmod{7}$, $N \equiv 1\pmod{7}$.
Since $15^{10} \equiv 1\pmod{11}$, $N=\big(15^{10}\big)^4 \equiv 1\pmod{11}$.
Since $15 \equiv -2\pmod{17}$, $15^4 \equiv (-2)^4 \equiv -1\pmod{17}$ and $N=\big(15^4\big)^{10} \equiv (-1)^{10}=1\pmod{17}$.
Therefor... | {
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"timestamp": "2023-03-29T00:00:00",
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Limit as $x \to \infty$ of $\frac{x^5+x^3+4}{x^4-x^3+1}$ Suppose we have to find the following limit
$\lim_{x\to\infty}\frac{x^5+x^3+4}{x^4-x^3+1}$
Now, if we work with the De L'Hopital rule with successive differentiations we get $L=+\infty$
But if we work like this instead: $$L=\lim_{x\to\infty}\frac{x^5(1+\frac{1}{x... | First off, division by $0$ is undefined. Try this, $$\lim_{x\to\infty}\frac{x^5+x^3+4}{x^4-x^3+1}=\lim_{x\to\infty}\frac{x+\frac{1}{x}+\frac{4}{x^4}}{1-\frac{1}{x}+\frac{1}{x^4}}=\infty$$
This shows more clearly the limit without resulting in $\frac{1}{0}$
| {
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"url": "https://math.stackexchange.com/questions/2152654",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.