Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Proof $13 \mid (k\cdot 2^n+1)$ if $n\equiv2 \pmod{12}$ and $k\equiv3 \pmod{13}$?
Proof $13 \mid (k \cdot 2^n + 1)$ if $n\equiv2 \pmod{12}$ and $k\equiv3 \pmod{13}$
Hint: for $k$ odd: $2^n \equiv-k' \pmod p$ and $kk' \equiv1 \pmod p$
My thoughts:
$13\mid(k-3) \Rightarrow k=13a+3$
and
$12|(n-2) \Rightarrow n=12b+2$
s... | If $n=12N+2$ and $k=13K+3$ then
$$
k2^n+1=13K2^n+3\cdot 4\cdot \left(\underbrace{2^{12}}_{\equiv 1\bmod{13}}\right)^{N}+1.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2337268",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Expand $\frac{1}{z-2}$ on $|z|>2$
Expand $\frac{1}{z-2}$ around $|z|>2$
So there is no singularity points
$$\frac{1}{(z-2)}=-\frac{1}{2-z}=-\frac{1}{2}\frac{1}{1-\frac{z}{2}}=-\frac{1}{2}\sum_{n=0}^{\infty} \frac{z^n}{2^n}=\sum_{n=0}^{\infty}-\frac{z^n}{2^{n+1}}$$
but we can also write:
$$\frac{1}{(z-2)}=\frac{1}{z}... | You already answered your own question:
To expand
$$\frac{1}{1-q}$$
as you did in both cases, you need $|q|<1$ for the series to converge. Thus, if $|z|<2$, the first one is right, whereas for $|z|>2$, the second one is right.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2339376",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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If the roots of an equation are $a,b,c$ then find the equation having roots $\frac{1-a}{a},\frac{1-b}{b},\frac{1-c}{c}$. Actually I have come to know a technique of solving this kind of problem but it's not exactly producing when used in a certain problem.
Say we have an equation
$$2x^3+3x^2-x+1=0$$
the roots of this ... | The equation $x^3+3x+1=0$ has roots $a,b,c$. Find the equation whose roots are $\frac{1-a}{a},\frac{1-b}{b},\frac{1-c}{c}$.
So let $y=\frac{1-x}{x}$ ... invert this equation, we have $x=\frac{1}{1+y}$, now substitute this into the original equation & you get $\color{red}{y^3+6y^2+9y+5=0}$.
Note that in your previous ex... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2339882",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
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A limit involving square roots Transcribed from photo
$$\require{cancel}
\lim_{n\to\infty}\sqrt{n+\tfrac12}\left(\sqrt{2n+1}-\sqrt{2n+3}\right)\\[12pt]
$$
$$
\lim_{n\to\infty}\sqrt{n+\tfrac12}\lim_{n\to\infty}\left(\sqrt{2n+1}-\sqrt{2n+3}\right)\frac{\sqrt{2n+1}+\sqrt{2n+3}}{\sqrt{2n+1}+\sqrt{2n+3}}\\[12pt]
$$
$$
\lim_... | $$\lim _{x\to \infty }\left(\sqrt{x+\frac{1}{2}}\right)\left(\sqrt{2x+1}-\sqrt{2x+3}\right)$$
As you said, we can rewrite:
$$\left(\sqrt{2x+1}-\sqrt{2x+3}\right)=\frac{-2}{\sqrt{2x+1}+\sqrt{2x+3}}$$
Try factoring the $x$
$$\frac{-2}{\sqrt{2x+1}+\sqrt{2x+3}}=\frac{-2}{\sqrt{\color{red}{x}(2+1/x)}+\sqrt{\color{red}{x}(2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2341318",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
} |
Eliminating parameter from 2 parametric equations I am given two parametric equations as-
$ x= g\sin^2(t) - f\sin(t)\cos(t) $
$ y = f\cos^2(t) - g\sin(t)\cos(t) $
and I have to eliminate the parameter $t$.
I tried taking $\sin(t)$ common from the first equation and $\cos(t)$ from the second, and divided the two to get ... | \begin{align}
\cos^2 t & = \frac 1 2 + \frac 1 2 \cos(2t) \\[10pt]
\sin^2 t & = \frac 1 2 - \frac 1 2 \cos(2t) \\[10pt]
\sin(t) \cos(t) & = \frac 1 2 \sin(2t) \\[10pt]
x & = g\sin^2 t - f\sin(t) \cos(t) = g\left( \frac 1 2 - \frac 1 2\cos(2t) \right) - f \cdot \frac 1 2 \sin(2t) \\[10pt]
y & = f \cos^2 t - g \sin(t)\co... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2342519",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Simplify $x^{\frac{2}{3}}+3x^{\frac{1}{3}}-2=0$ to $x^2 +45x-8=0$ Simplify ($x^{\frac{2}{3}}+3x^{\frac{1}{3}}-2=0$) to ($x^2 +45x-8=0$)
These two equations have the same solutions.
If given the first equation how would you go about simplifying such that you end up with the quadratic form ?
I tried cubing the first expr... | Let $\alpha$ and $\beta$ be the roots of $y^2+3y-2=0$. So
\begin{eqnarray*}
\alpha+\beta=-3 \\
\alpha \beta =-2.
\end{eqnarray*}
Now calculate the equation whose roots are $\alpha^3$ and $\beta^3$
\begin{eqnarray*}
\alpha^3+\beta^3&=&(\alpha+\beta)^3-3\alpha\beta(\alpha+\beta)&=&-45 \\
\alpha^3 \beta^3 &=&(\alpha\beta)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2342718",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Characteristic polynomial of a $7 \times 7$ matrix whose entries are $5$ Avoiding too many steps, what is the characteristic polynomial of the following $7 \times 7$ matrix? And why?
\begin{pmatrix}5&5&5&5&5&5&5\\5&5&5&5&5&5&5\\5&5&5&5&5&5&5\\5&5&5&5&5&5&5\\5&5&5&5&5&5&5\\5&5&5&5&5&5&5\\5&5&5&5&5&5&5\end{pmatrix}
| Here is a matrix $P,$ the columns are eigenvectors of your matrix. Note that $P$ is not orthogonal, although the columns are pairwise orthogonal.
$$
P =
\left( \begin{array}{rrrrrrr}
1 & -1 & -1 & -1 & -1 & -1 & -1 \\
1 & 1 & -1 & -1 & -1 & -1 & -1 \\
1 & 0 & 2 & -1 &... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2343155",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
If $f(x)=\frac{1}{x^2+x+1}$, how to find $f^{(36)} (0)$?
If $f(x)=\frac{1}{x^2+x+1}$, find $f^{(36)} (0)$.
So far I have tried letting $a=x^2+x+1$ and then finding the first several derivatives to see if some terms would disappear because the third derivative of $a$ is $0$, but the derivatives keep getting longer an... | Let $\omega$ be a complex cube root of $1$. Then Partial Fraction representation of $f(x)$ is given by
$f(x) = \dfrac{1}{x^2+x+1} = \dfrac{1}{(x-\omega)(x-\omega^2)} = \dfrac{1}{\omega - \omega^2}\Big(\dfrac{1}{x-\omega} - \dfrac{1}{x - \omega^2}\Big)$.
Find successive derivatives to show that
$f^{(36)}(x) = \dfrac{1}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2343565",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "42",
"answer_count": 6,
"answer_id": 1
} |
In $\triangle ABC$, we have $AB=3$ and $AC=4$. Side $\overline{BC}$ and the median from $A$ to $\overline{BC}$ have the same length. What is $BC$? In $\triangle ABC$, we have $AB=3$ and $AC=4$. Side $\overline{BC}$ and the median from $A$ to $\overline{BC}$ have the same length. What is $BC$?
Alright, so if d is the ... | You can also practice another method:
Note the areas of $\Delta ABD$ and $\Delta ACD$ are equal.
Labeling $BD=CD=x$ and $AD=2x$, we will use the Heron's formula:
$$S_{\Delta ABD}= \sqrt{\frac{3+3x}{2}\cdot \left(\frac{3+3x}{2}-3\right)\cdot \left(\frac{3+3x}{2}-2x\right)\cdot \left(\frac{3+3x}{2}-x\right)}=$$
$$S_{\De... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2345586",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
Algebraic identities Given that $$a+b+c=2$$ and $$ab+bc+ca=1$$
Then the value of
$$(a+b)^2+(b+c)^2+(c+a)^2$$ is how much?
Attempt:
Tried expanding the expression. Thought the expanded version would contain a term from the expression of $a^3+b^3+c^3-3abc$, but its not the case.
| Hint:
Express
$S_2=a^2+b^2+c^2$ in function of $s=a+b+c$ and $\sigma=ab+bc+ca$ from the algebraic identity for $(a+b+c)^2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2346162",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 8,
"answer_id": 2
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Solving Pinter 7.B.4 with a program Here is exercise 7.B.4 from 'A Book of Abstract Algebra' by Charles C. Pinter.
A solution to this using a C# program is posted.
Is there another good approach using a computer program?
Any language is welcome.
The subgroup of $S_5$ generated by
$
f = \begin{pmatrix}1 & 2 & 3 & 4 & 5... | This should be easy to do without a computer, as mentioned.
With a computer, you should be able to do this with any language. Using something designed for algebraic computation is easiest.
In MAGMA, I would type
{x : x in sub<Sym(5)|(1, 2), (3, 4, 5)>};
and it would list out all of the elements (I have them is cycle n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2346736",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
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for the cauchy problem , determine unique solution ,or no solution , or infinitely many solution for the cauchy problem , determine unique solution ,or no solution , or infinitely many solution
$u_x-6u_y=y$ with the date $u(x,y)=e^x$ on the line $y=-6x+2$
My attempt:
given $u_x-6u_y=y$ then $\frac{dx}{1}=\frac{dy}{-6... | given $u_x-6u_y=y$ then $\frac{dx}{1}=\frac{dy}{-6}=\frac{du}{y}\quad$ not $=\frac{du}{u}\quad$(typo).
$y^2+12u=F(6x+y)\quad$ is OK.
Condition $u(x,y)=e^x$ on the line $y=-6x+2\quad\to\quad y^2+12e^x=F(2)=$constant is impossible. Thus, there is no solution.
Condition $u(x,y)=1$ on the line $y=-x^2\quad\to\quad x^4+12=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2347887",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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How to obtain the sum of the following series? $\sum_{n=1}^\infty{\frac{n^2}{2^n}}$ It seems that I'm missing something about this.
First of all, the series is convergent: $\lim_{n\rightarrow\infty}\frac{2^{-n-1} (n+1)^2}{2^{-n} n^2}=\frac{1}{2}$ (ratio test)
What I tried to do is to find a limit of a partial sum $\li... | Suppose $|x|<1$ at the last you put $x=\frac12$
$$s_1=1+x+x^2+x^3+x^4+...=\frac{1}{1-x}$$nw find $s'_1$
$$s'_1=0+1+2x+3x^3+4x^3+5x^4+...+nx^{n-1}+...=(\frac{1}{1-x})'\\1+2x+3x^2+4x^3+...=\frac{1}{(1-x)^2}$$now multiply by $x$
$$S_2=xS'_1=x^1+2x^2+3x^3+4x^4+5x^5+...=\frac{x}{(1-x)^2}$$
now find $S_2'$
$$S'_2=1+2^2x^1+3^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2348438",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 6,
"answer_id": 3
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Find $\lim\limits_{x\to 1}\frac9{1-x^9}-\frac7{1-x^7}$ without using L'Hopital $$\lim_{x\to 1}\frac9{1-x^9}-\frac7{1-x^7}$$
I did this with l'Hôpital's rule, but how can we do this problem other than that? Any hints will be good.
| The difference of fractions equals $\frac{9(1-x^7) - 7(1-x^9)}{(1-x^7)(1-x^9)}$.
Try to divide out the common $(1-x)^2$ terms in both the bottom and the top.
In the bottom we get $(1+x+x^2+\ldots+x^8)(1+x + x^2 + \ldots + x^6)$
In the top $7x^7 + 14x^6 + 12x^5 + 10x^4 + 8x^3 + 6x^2 + 4x + 2$ (multiply by $(1-x)^2$ to c... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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If $a,b,c$ are distinct positive numbers , show that $\frac{a^8 + b^8 + c^8}{a^3 b^3 c^3} > \frac{1}{a} + \frac{1}{b} + \frac{1}{c}$ If $a,b,c$ are distinct positive numbers , show that $$\frac{a^8 + b^8 + c^8}{a^3 b^3 c^3} > \frac{1}{a} + \frac{1}{b} + \frac{1}{c}.$$ I am thinking of Tchebycheff's inequality for this ... | The given inequality can be written as
$$a^8+b^8+c^8 > a^2b^3c^3+b^2a^3c^3+c^2a^3b^3.$$
Since $(8,0,0)$ majorizes $(2,3,3)$, therefore we can apply Muirhead's inequality to get the result.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2355748",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 2
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The number of triangles with each side having integral length... The number of triangles with each side having integral length and the longest side is of 11 units is:-? MY ATTEMPT:- I applied following constraints: $12 \leq {a+b} \leq 22$ $a,b \geq 1$ I made different cases that a+b =12, a+b=13 and so on. My answer... | Your conditions are not quite correct. There is no stipulation on what $a + b$ is the only stipulation is that $11$ is the longest sides.
We need $a \le b \le 11$ so we don't count multiple instances the same and so that the maximum side(s) is $11$.
We need $a + b > 11$ or $a + b \ge 12$ to satisfy the (non-trivial) t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2356970",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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$\sqrt{9x^2-16}>3x+1$ I'm trying to solve the following inequality:
$$\sqrt{9x^2-16}>3x+1$$
Here's my attempt:
$\sqrt{9x^2-16}>3x+1$
$\longrightarrow 9x^2-16>9x^2+6x+1$
$\longrightarrow -16>6x+1$
$\longrightarrow x<-\frac{17}{6}$
Now, I need to check the constraints:
$9x^2-16 > 0$
$\longrightarrow (3x)^2 > 4^2$
$\lo... | $a > b \implies a^2 > b$ is not true if $b < 0$.
We have two cases:
1) if $3x + 1 \ge 0$ or $x \ge -\frac 13$ you correctly got $x < \frac {-17}6$. But that contradicts $x \ge -\frac 13$
So we must have case 2:
2) $3x + 1 < 0$ and $x < -\frac 13$.
From that we can't square both sides as $a \ge 0 > b$ does/can not ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2359030",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
Ratio of Height to the Radius of its base A conical tent is of given capacity.For the least amount of canvas required for it,find the ratio of its height to the radius of its base.
(capacity usually means its volume.Only units are different)
| Let $h$ be an altitude of the tent, $r$ be a radius of the base and $V$ be a volume of the tent.
Also, let $h=rx$.
Thus, $V=\frac{\pi r^2h}{3}$ or
$$V=\frac{\pi r^3x}{3},$$
which gives
$$r=\sqrt[3]{\frac{3V}{\pi{x}}}.$$
Thus, for an area of the canvas by AM-GM we obtain:
$$f(x)=\pi{r}\sqrt{h^2+r^2}=\sqrt[3]{9\pi V^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2359345",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Does $\lim_{n\rightarrow\infty} (\frac{1}{n+a} + \frac{1}{n+2a} + \cdots + \frac{1}{n+na}) $ converge? I want to check whether
$\lim_{n\rightarrow\infty} (\frac{1}{n+a} + \frac{1}{n+2a} + \cdots + \frac{1}{n+na}) $
converges or not. (a is a positive constant number.)
If it converges, how to find the value it converge... | $\frac{1}{n+a}+\dots+\frac{1}{n+an} = (\frac{1}{1}+\frac{1}{2}+\dots+\frac{1}{n+na})-(\frac{1}{1}+\frac{1}{2}+\dots+\frac{1}{n+a-1})$
$= \log(n+na)-\log(n+a-1)+o(1) = \log(\frac{n(a+1)}{n+a-1})+o(1)$, so the limit as $n \to \infty$ is $\log(a+1)$, since $\log$ is continuous.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2359410",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 3
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A sequence defined through an arc tangent and arc cotangent Let $$f:\mathbb{R^*}\to \mathbb{R}, f(x) = \arctan{\dfrac{1}{x}}-\operatorname{arccot}{\dfrac{1}{x}}$$
I found that this function is decreasing on $\mathbb{R^*}$ and a bijection for the same domain and for the range $(-\dfrac{3
\pi}{2};\dfrac{\pi}{2})$.
If we ... | $f(x) = \arctan{\frac{1}{x}}-arccot{\frac{1}{x}}$
Since
$\arctan(x)+arccot(x)
=\frac{\pi}{2}
$,
this becomes
$f(x)
= 2\arctan{\frac{1}{x}}-\frac{\pi}{2}
$.
If
$f(x_n) = \frac1{n}$.
then
$2\arctan{\frac{1}{x_n}}
=\frac{\pi}{2}+\frac1{n}
$
or
$\begin{array}\\
\frac{1}{x_n}
&=\tan(\frac{\pi}{4}+\frac1{2n})\\
&=\frac{\tan... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2359692",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Show that $\frac {1}{\sqrt{5}}[(\frac {1}{x+r_+}) - (\frac {1}{x+r_-}) = \frac {1}{\sqrt{5}x}[(\frac {1}{1-r_{+}x}) - (\frac {1}{1-r_{-}x})] $ I need to manipulate this equation:
$$
\frac {1}{\sqrt{5}}\left(\frac {1}{x+r_+} - \frac {1}{x+r_-}\right)
$$
to show that
$$ \frac {1}{\sqrt{5}}\left(\frac {1}{x+r_+} - \frac ... | \begin{align*}
\frac {1}{\sqrt{5}}\left(\frac {1}{x+r_+} - \frac {1}{x+r_-}\right)
&= \frac {1}{\sqrt{5}} \frac{x}{x} \left(\frac {1}{x+r_+} - \frac {1}{x+r_-}\right) &&,x \neq 0 \\
&= \frac {1}{\sqrt{5}\, x} \left(\frac {x}{x+r_+} - \frac {x}{x+r_-}\right) &&,x \neq 0 \\
&= \frac {1}{\sqrt{5}\, x} \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2359755",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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Compute $\int\limits_{-\infty}^{\infty}\frac{1}{(1+x^2)^{n+1}}dx$ via residue calculus. Let $\Gamma_R$ be the semicircle of radius $R$ in the upper half plane. Then,
\begin{align}
\int\limits_{-\infty}^{\infty}\frac{1}{(1+x^2)^{n+1}}dx
&= \lim_{R\to \infty}\int_{\Gamma_R}\frac{1}{(1+z^2)^{n+1}}dz \\
&= 2\pi i \operat... | Note that by residuals taking the upper semicircle as trajectory:
\begin{eqnarray*}
\int_{-\infty}^{+\infty} \frac{1}{(1+x^{2})^{n+1}}
& = & 2\pi i\text{Res}(f(z),i)\ i\text{ is a pole of order }n+1\\
% & = & 2\pi i\left(\frac{1}{n!}\frac{d^{n}}{dz^{n}}(z-i)^{n+1}f(z)\right)\\
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2362054",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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How to prove that $2^0 + 2^1 + 2^2 + 2^3 + \cdots + 2^n = 2^{n+1} - 1$ When I'm reading Computer Systems: A Programmer's Perspective, I met the sum of binary numbers and failed to prove it:
$$ 2^0 + 2^1 + 2^2 + 2^3 + \cdots + 2^n = 2^{n+1} - 1 $$
This might be preliminary knowledge, I'm not good at mathematics, any bod... | You are going to walk a distance of one mile. You walk half the distance and rest. You then walk half the remaining distance and rest. Continue doing this.
So now we can write down everyone's favorite infinite series,
$\frac{1}{2} +\frac{1}{4} +\frac{1}{8} +\frac{1}{16} + \dots = 1$
Multiply both sides by by $2^{n+1}$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2362116",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 9,
"answer_id": 7
} |
Evaluate the given limit: $\lim_{x\to \infty} (\sqrt {x-a} - \sqrt {bx})$
Evaluate the following limit.
$$\lim_{x\to \infty} (\sqrt {x-a} - \sqrt {bx})$$
My Attempt:
$$=\lim_{x\to \infty} (\sqrt {x-a} - \sqrt {bx})$$
$$=\lim_{x\to \infty} (\sqrt {x-a} - \sqrt {bx}) \times \dfrac {\sqrt {x-a} + \sqrt {bx}}{\sqrt {x-... | That is a good idea! Now, intuitively speaking, a square root grows slower than a linear function so you should get "some" infinity. However, it depends on the constants $a$ and $b$ what the result will look like.
E.g. if $a = 0$ and $b=2$, you would get
\begin{align*}
\lim_{x \rightarrow \infty} \frac{x - 2x}{\sqrt{x... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 2
} |
Radical problem: $\sqrt{7+2\sqrt{12}}+\sqrt{7-2\sqrt{12}}$ What is the value of
$$\sqrt{7+2\sqrt{12}}+\sqrt{7-2\sqrt{12}}$$
| Since $(2\pm\sqrt3)^2=7\pm4\sqrt3=7\pm2\sqrt{12}$,$$\sqrt{7+2\sqrt{12}}+\sqrt{7-2\sqrt{12}}=2+\sqrt3+2-\sqrt3=4.$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Why does $a^2x^2+(a^2+b^2-c^2)xy+b^2y^2>0,$ imply $(a^2+b^2-c^2)^2-4a^2b^2<0$? Why does $a^2x^2+(a^2+b^2-c^2)xy+b^2y^2>0,$ imply $(a^2+b^2-c^2)^2-4a^2b^2<0$, if $x, y$ are reals greater than $1$, and $a, b, c$ are positive reals?
A proof with all the math to go from one to the other would be nice.
| Below is old incorrect answer ignoring $x>1$ condition. Explanation:
The discriminant is less than $0$ iff for all $x$ the quadratic has no roots.
However, we only know the LHS has no roots for all $x>1$ so this cannot be used. In fact, the counterexample pointed out by @WillJagy shows this.
$$a^2x^2+(a^2+b^2-c^2)xy+b... | {
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How to prove $\ln{\frac{n+1}{n}}\le\frac{2}{n+1},\forall n\in\mathbb{N}^+$? I found an inequality: $$\ln\left(\frac{n+1}{n}\right)\le\frac{2}{n+1},\forall n\in\mathbb{N}^+.$$
I tried induction. It is obvious if $k=1$, when $n=k$, $\ln\sqrt{\left(\frac{k+1}{k}\right)^{k+1}}\le 1$, but bogged down for $n=k+1$:
$$\ln\sqrt... | Your question is equivalent to prove the following form
$$
(2-\frac{n-1}{n})\leq 2^{(1-\frac{n-1}{n+1})} \tag{1}
$$
The relation $(1)$ is true becuse $\frac{n-1}{n} \geq \frac{n-1}{n+1}$.
Edit:
If the relation $(1)$ be true then we can conclude that:
$$
\left\{
\begin{array}{c}
(2-\frac{n-1}{n})\leq 2^{(1-\frac{n-1}{n... | {
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"source": "stackexchange",
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Quadratic equations having a common root. I have a cubic polynomial $f(x)=x^3+px^2+qx+72$ which is divisible by both $x^2+ax+b$ and $x^2+bx+a$ (where a,b,p,q are constants and a$ \neq $b).I have to find the sum of the squares of the the roots of the cubic polynomial.
I tried to attempt it like this.
Since the quadratic... | Suppose $\alpha$ is the common root for the two quadratics $g(x)$ and $h(x)$. Then
\begin{align*}
g(\alpha)=\alpha^2+a\alpha+b &=0\\
h(\alpha)=\alpha^2+b\alpha+a &=0
\end{align*}
Then solving for $\alpha$ gives $\alpha=1$ (assuming $a\neq b$). This also tells us
$$a+b=-1.$$
This means the quadratic equation $g(x)=0$ h... | {
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In how many ways the sum of 5 thrown dice is 25? What I thought about is looking for the number of solutions to
$$x_{1}+x_{2}+x_{3}+x_{4}+x_{5}=25$$ such that $1\leq x_{i}\leq6$
for every $i$.
Now I know that the number of solutions to this such that
$0\leq x_{i}$ for every $i$ is $${5+25-1 \choose 5-1}={29 \choose 4}$... | When we multiply polynomials the formula is
$$ \left( \sum_i a_ix^i \right)\left( \sum_j b_jx^j \right) = \sum_k \left( \sum_{i + j = k} a_ib_j \right) x^k. $$
For a product of $5$ polynomials, we have:
$$ \left( \sum_{i_1} a_{1, i_1}x^{i_1} \right)\left( \sum_{i_2} a_{2, i_2}x^{i_2} \right)\left( \sum_{i_3} a_{3, i_3}... | {
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"source": "stackexchange",
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Finding solutions for expression to be a perfect square Determine all integers such that $$ n^4- n^2 +64$$ is the square of an integer.
The first two lines of the solution given in the textbook is as below:
Since $n^4-n^2+64>(n^2-1)^2. $ For some non negative integer $k$,
$n^4-n^2+64=(n^2+k)^2$.
I fail to understa... | $$(n^2-x)^2=n^4-n^2+64=n^4-2xn^2+x^2$$
where $x$ is an integer.Then,
$$n^2=\dfrac{x^2-64 }{2x-1 }\geq 0$$ hence by multiplying both sides by 4
$$4n^2=\dfrac{4x^2-1+1-256 }{2x-1 }=2x+1-\dfrac{255}{2x-1}$$
Since $255=1\cdot 3\cdot 5\cdot 17 $ then the ratio is an integer only if
$$2x-1=\pm 3^a5^b17^c$$ where $a,b,c \i... | {
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"source": "stackexchange",
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Find $ \ \ 8^{504} \equiv \pmod 5$ Find $ \ \ 8^{504} \equiv \pmod 5 $
Answer:
$ 8^{504} \equiv 2^{1512} \pmod 5 $
Now ,
$$ \begin{align}2^4 &\equiv 1 \pmod 5 \\\text{or, } \left(2^{4}\right)^{378} &\equiv 1^{378} \pmod 5 \\ \text{or, } 2^{1512} &\equiv 1 \pmod 5\\\text{or, } 8^{504} &\equiv \ 1 \pmod 5 \end{align}$$
... | As $gcd(8,5)=1$ so by Euler's theorem $8^4\equiv 1\pmod5$ as $\phi (5)=4$ Now 504 divide by 4 so $8^{504}\equiv 1\pmod 5$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Series Solution to the ODE $(x-1)y'' - xy' + y = 0$ with I.C. $y(0) = -3$ and $y'(0)=4$ \begin{align*}
(x-1)y'' - xy' + y = 0 &\iff (x-1)\sum_{n=2}^{\infty} n(n-1)c_n x^{n-2} - x\sum_{n=1}^{\infty} nc_n x^{n-1} + \sum_{n=0}^{\infty} c_n x^n = 0 \\
&\iff \sum_{n=2}^{\infty} n(n-1)c_n x^{n-1} - \sum_{n=2}^{\infty} n(n-1)... | other approach
The equation can be written as
$$(x-1)(y''-y')-(y'-y)=0$$
or
$$(x-1)z'-z=0$$
with $$z=y'-y $$
the solution is $$z=\lambda (x-1)=-7 (x-1) $$
Now look for series solution of
$$y'-y=-7 (x-1) $$
| {
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The minimum value of the integral a) Find $a$ and $b$ such that $\displaystyle \int_0^1 |x^2 - ax - b| \,dx$ is minimum.
b) How about $\displaystyle \int_0^1 |x^2 - ax - b|^2 \,dx$?
This is what I did for part a):
If $D=a^2+4b\leq 0$ then $|x^2-ax-b| =x^2-ax-b$ and we have:
$ I= \int_0^1 |x^2-ax-b|\,dx = \int_0^1 (x... | Part (b) is more immediately clear to me:
\begin{align*}
\int_0^1 |x^2+ax+b|^2 \, dx &= \int_0^1 (x^2+ax+b)^2 \, dx\\[0.3cm] &= \int_0^1 (x^4+2ax^3+(a^2+2b)x^2 + 2abx+b^2) \, dx\\[0.3cm] &= 1/5 + a^2/3 + 2 b/3 + b^2 + a/2 + ab
\end{align*}
Let $S=1/5 + a^2/3 + 2 b/3 + b^2 + a/2 + ab.$ This is extremised when $\partial ... | {
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Let $\ a,b,c$ be integers such that $ \ a^{2} + b^{2} = c^{2}$. If $c$ is divisible by $3$, prove that $a$ and $b$ are both divisible by $3$. Question:
Let $\ a,b,c$ be integers such that $ \ a^{2} + b^{2} = c^{2}$. If $c$ is divisible by $3$, prove that $a$ and $b$ are both divisible by $3$.
My attempt:
Proof by cont... | Note that $(a,b,c)$ is Pythagorean Triples, then we can choose non-zero integers $k;p,q$ and $\gcd (p,q)=1$ such that $$a=2kpq\\b=k(p^2-q^2)\\c=k(p^2+q^2)$$
Now since $3\mid c,$ we will have different cases:
Case 1:
If $3\mid k$, then $3\mid a$ and $3\mid b$
Case 2:
If $3\nmid k$ and $3\mid (p^2+q^2)$. Note that if $3\... | {
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Find $\int e^{x^4}(x+x^3+2x^5)e^{x^2}dx$ Find $\int e^{x^4}(x+x^3+2x^5)e^{x^2}dx$
$$I=\int e^{x^4}(x+x^3+2x^5)e^{x^2}dx$$
$$I=\int e^{x^4}(1+x^2+2x^4)e^{x^2}xdx$$Let $x^2=t$
$I=\int e^{t^2}(1+t+2t^2)e^t\frac{dt}{2}=\frac{1}{2}\int e^{t^2+t}(1+t+2t^2)dt$
I am stuck here.
| Assuming the answer is given by an elementary function, it has to be a function of the form $p(x) e^{x^2+x^4}$ with $p(x)$ being a quadratic polynomial. We have
$$\frac{d}{dx} x^2 e^{x^2+x^4} = 2(x+x^3+2x^5)e^{x^2+x^4} $$
hence such inspired guess immediately leads to
$$ \int (x+x^3+2x^5) e^{x^2+x^4}\,dx = C+\frac{x^2}... | {
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Find probability that confidence intervals are disjoint Given $X_1, \ldots, X_{100}$ and $Y_1,\ldots,Y_{50}$ which are independent random samples from identical distribution $N(\mu,1)$. Each of two statisticians is trying to build confidence interval for $\mu$ on a confidence level of $0.8$ ($1-\alpha=0.8$). You have t... | One interval has endpoints $\bar X \pm 1.28 \dfrac 1 {\sqrt{100}}$ and the other $\bar Y \pm 1.28 \dfrac1 {\sqrt{50}}.$
The intervals are disjoint if $\bar X + 1.28 \dfrac 1 {\sqrt{100}} < \bar Y - 1.28 \dfrac 1 {\sqrt{50}}$ or $\bar Y + 1.28 \dfrac 1 {\sqrt{50}} < \bar X - 1.28 \dfrac 1 {\sqrt{100}}.$
That happens if ... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Prove the inequality $\sum_{cyc}\frac{a}{1+\left(b+c\right)^2}\le \frac{3\left(a^2+b^2+c^2\right)}{a^2+b^2+c^2+12abc}$
Let $a>0$, $b>0$ and $c>0$ such that $a+b+c=3$. Prove the inequality
$$\frac{a}{1+\left(b+c\right)^2}+\frac{b}{1+\left(c+a\right)^2}+\frac{c}{1+\left(a+b\right)^2}\le \frac{3\left(a^2+b^2+c^2\right... | We need to prove that
$$\sum_{cyc}\left(\frac{a}{1+(b+c)^2}-a\right)\leq\frac{3(a^2+b^2+c^2)}{a^2+b^2+c^2+12abc}-3$$ or
$$\sum_{cyc}\frac{a(b+c)^2}{1+(b+c)^2}\geq\frac{36abc}{a^2+b^2+c^2+12abc}.$$
Now, by C-S
$$\sum_{cyc}\frac{a(b+c)^2}{1+(b+c)^2}=\sum_{cyc}\frac{a}{1+\frac{1}{(b+c)^2}}=$$
$$=\sum_{cyc}\frac{a^2}{a+\fr... | {
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"source": "stackexchange",
"question_score": "2",
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Prove $\frac{1}{\log_2\pi }+\frac{1}{\log_5\pi}>2$
Knowing that $\pi^2 < 10$. Prove that:
$$\frac{1}{\log_2\pi}+\frac{1}{\log_5\pi}>2.$$
I have tried to do this the following way:
$\log_2\pi+\log_5\pi>\frac{1}{2} \Leftrightarrow \log_2\pi+\frac{\log_2\pi}{\log_25}>\frac{1}{2}\Leftrightarrow 2\log_2\pi-\log_25>\frac... | Notice that:
\begin{align*}
\frac{1}{\log_2 \pi} + \frac{1}{\log_5 \pi}
&= \frac{1}{\left(\frac{\log_\pi \pi}{\log_\pi 2} \right)} + \frac{1}{\left(\frac{\log_\pi \pi}{\log_\pi 5} \right)} & \text{change of base} \\
&= \frac{\log_\pi 2}{\log_\pi \pi} + \frac{\log_\pi 5}{\log_\pi \pi} \\
&= \log_\pi 2 + \log_\pi 5 \\
&=... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Equation solving involving terms inside roots How do I solve
$$\left(x-\frac{1}{x}\right)^{1/2}+\left(1-\frac{1}{x}\right)^{1/2} = x$$
My Try I tried to take the first term as $t$ but then I had to square both sides twice and that led to a complex bi quadratic. I'm not sure even that'll solve the problem.
guys pleas... | $$\left(x-\frac{1}{x}\right)^{1/2}+\left(1-\frac{1}{x}\right)^{1/2} = x\iff \left(x-\frac{1}{x}\right)^{1/2} = x-\left(1-\frac{1}{x}\right)^{1/2}.$$ Squaring
$$x-\frac{1}{x} = x^2+1-\frac{1}{x}-2x\left(1-\frac{1}{x}\right)^{1/2}.$$ Simplifying
$$x = x^2+1-2x\left(1-\frac{1}{x}\right)^{1/2}.$$ That is
$$2x\left(1-\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2394163",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find the $[\Bbb{Q}(\sqrt2+\sqrt3)(\sqrt5):\Bbb{Q}(\sqrt2+\sqrt3)]$ We want to find the $[\Bbb{Q}(\sqrt2+\sqrt3)(\sqrt5):\Bbb{Q}(\sqrt2+\sqrt3)]$.
My first thought is to find the minimal polynomial of $\sqrt5$ over $\Bbb{Q}(\sqrt2+\sqrt3)$. And from this, to say that $[\Bbb{Q}(\sqrt2+\sqrt3)(\sqrt5):\Bbb{Q}(\sqrt2+\sqrt... | Here is an idea that avoids most of the theory.
The correspondence:
$$a+ b \sqrt{2} + c\sqrt{3} + d \sqrt{6} \mapsto a+ b \sqrt{2} - c\sqrt{3} - d \sqrt{6}$$ preserves sums and products. So if
$$\sqrt{5} = a+ b \sqrt{2} + c\sqrt{3} + d \sqrt{6}$$ then
$$(a+ b \sqrt{2} + c\sqrt{3} + d \sqrt{6})^2= 5$$ so
$$(a+ b \sqrt... | {
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"url": "https://math.stackexchange.com/questions/2396207",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Is there an analytic solution for the equation $\log_{2}{x}+\log_{3}{x}+\log_{4}{x}=1$?
I am looking for a close form solution for below equation.
$$\log_{2}{x}+\log_{3}{x}+\log_{4}{x}=1.$$
I solve it by graphing, but I don't know is there a way to find $x$ analytically ?
| $$\tag1
\log_{2}{x}+\log_{3}{x}+\log_{4}{x}=1$$
$$\tag2
\frac{\log{x}}{\log{2}}+\frac{\log{x}}{\log{3}}+\frac{\log{x}}{\log{4}} = 1$$
$$\tag3
\left(\frac{1}{\log{2}}+\frac{1}{\log{3}}+\frac{1}{\log{4}}\right)\log{x} = 1$$
$$\tag4
\left(\frac{1}{\log{2}}+\frac{1}{\log{3}}+\frac{1}{\log{4}}\right) = \frac{1}{\log{x}}$$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2397733",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 2
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Show that the integral is divergent $\int_0^\infty\frac{\ dx}{1+x^2\sin^2x}$
Show that the integral is divergent
$$\int_0^\infty\frac{\ dx}{1+x^2\sin^2x}$$
It has no point of discontinuity in range of integration. Also, I have found $\int_0^\infty\frac{\ dx}{1+x^2\sin^2x}>\frac{\pi}{2}$, but that seems to be of no... | We have that
\begin{align*}\int_0^\infty\frac{dx}{1+x^2\sin^2x}
&=\sum_{n=0}^{\infty}
\int_0^{\pi}\frac{dx}{1+(x+n\pi)^2\sin^2(x+n\pi)}
\\&\geq \sum_{n=0}^{\infty}
\int_0^{\pi/2}\frac{2dx}{1+\pi^2(1+n)^2\sin^2(x)}\\
&=\sum_{n=0}^{\infty}\frac{\pi}{\sqrt{1+\pi^2(n+1)^2}}\\
&\geq \frac{\pi}{\sqrt{1+\pi^2}}\sum_{n=0}^{\i... | {
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"url": "https://math.stackexchange.com/questions/2399407",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Show that if $a+b+c=0$, $2(a^4 + b^4+ c^4)$ is a perfect square
Show that for $\{a,b,c\}\subset\Bbb Z$ if $a+b+c=0$ then $2(a^4 + b^4+ c^4)$ is a perfect square.
This question is from a math olympiad contest.
I started developing the expression $(a^2+b^2+c^2)^2=a^4+b^4+c^4+2a^2b^2+2a^2c^2+2b^2c^2$ but was not able ... | $$a^2+b^2+c^2=(a+b+c)^2-2(ab+bc+ca)=?$$
Now $$(a^2)^2+(b^2)^2+(c^2)^2=(a^2+b^2+c^2)^2-2(a^2b^2+b^2c^2+c^2a^2)$$
$$a^2b^2+b^2c^2+c^2a^2=(ab+bc+ca)^2-abc(a+b+c)=?$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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find a and b such that the limit will exist and find the limit I am given this question $$\lim_{x\rightarrow 0} \frac{e^{\sqrt{1+x^2}}-a-bx^2}{x^4}$$
And asked to find the $a$ and $b$ such that the limit exists and also to compute the limit
My solution:
Let $t$=$\sqrt{1+x}$.
Then the Maclaurin polynomial is : $$\sqrt{... | Consider $f(x)=e^{1+x}$. By your reasoning, since $1+x=1+x+O(x^2)$, you get that:
$$f(x)=1+(1+x)+O(x^2)$$
But that isn't correct. If it was correct, then $f(0)=2$, but we know $f(0)=e$.
Now, if you did the full substitution you'd get:
$$f(x)=1+(1+x)+\frac{(1+x)^2}{2!}+\frac{(1+x)^3}{3!}+\cdots$$
The problem is that th... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Find polynomial : $P(x)=x^3+ax^2+bx+c$ Find all polynomials in $\mathbb{Q}[x]$ is of the form $P(x)=x^3+ax^2+bx+c$ which has $a, b, c$ as its roots.
Is my answer below correct?
| $x^3+ax^2+bx+c = (x-a)(x-b)(x-c)=x^3-(a+b+c)x^2+(ab+bc+ca)x-abc$
so $a=-(a+b+c)$, $b=ab+bc+ca$, $c=-abc$
Case 1 : $c=0$ so $b=ab$ then $a=1$ or $b=0$
if $a=1$, we get $(a,b,c)=(1,-2,0)$
if $b=0$, we get $(a,b,c)=(0,0,0)$
Case 2 : $c\not=0$ so $ab=-1$ then $-1+c(a+b)=b$ ---[1]
Since $a=-(a+b+c)$, so $c=-2a+\frac{1}{a}$ ... | {
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Proving that for $\{ a,b\} \subset \Bbb{R^{+}}$; $a+b=1 \implies a^2+b^2 \ge \frac{1}{2}$
For $\{ a,b\} \subset \Bbb{R^{+}}$; $a+b=1 \implies a^2+b^2 \ge
\frac{1}{2}$
I'm trying to prove this in the following way, but I'm not sure if it's correct. Could anyone please check it and see if it's okay?
$a+b=1 \implies (a... | For fun:
$a + b = 1$;
$a := \cos^2t;$ $b := \sin^2t$ , $ 0 \le t \lt 2π$.
$(a+b)^2 = a^2 +2ab +b^2 = $
$\cos^4t +2\cos^2t \sin^2t + \sin^4t =1$.
$\cos^4t + \sin^4t = 1 - (1/2)(\sin2t)^2$.
$\Rightarrow \cos^4t + \sin^4t \ge 1 - 1/2$ ,
since $ (\sin2t)^2 \le 1$.
Finally substituting back:
$a^2 + b^2 \ge 1/2$.
Recal... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2406268",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 7,
"answer_id": 6
} |
Let $h(4x-1)=2x+7$. For what value of $x$ is $h(x)=x$? Let $h(4x-1)=2x+7$. For what value of $x$ is $h(x)=x$?
If $h(a)=a$, then $4x-1=2x+7$ which implies $x=4$. So $a=15$ when I substitute $x=4$ into both linear equations. Is the value of $x$ $15$?
| The posted proof looks good. For an alternative one, let $4x-1=y \iff x = \frac{y+1}{4}\,$, then:
$$\,h(y) = h(4x-1)=2x+7=2\,\frac{y+1}{4}+7=\frac{y}{2}+\frac{15}{2}\,$$
Therefore $\,h(x)=\frac{x}{2}+\frac{15}{2}\,$, and $h(x)=x \iff x = \frac{x}{2}+\frac{15}{2} \iff x=15\,$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2407154",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
Solve $\frac{1}{x}<\frac{x}{2}<\frac{2}{x}$. First I consider the case $\frac{1}{x}<\frac{x}{2}$, manipulating to $\frac{(x-\sqrt{2})(x+\sqrt{2})}{2x}<0\Leftrightarrow x < -\sqrt{2}.$ Did this by subtracting $\frac{x}{2}$ from both sides and writing everything with a common denominator.
Second case to consider is $\fra... | Hint :$$\frac{1}{x}<\frac{x}{2} \to x\in (-\sqrt 2,0) \cup(\sqrt 2,+\infty)\tag{1}$$
$$\frac{x}{2}<\frac{2}{x}\to x\in(-\infty,-2)\cup (0,2) \tag{2}$$ then find $(1) \cap (2)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2407630",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 2
} |
Simple logarithmic differentiation Given the equation $\frac{dy}{dx} = \frac{-x}{y}$
How can you solve for $\frac{d^2y}{d^2x}$ using logarithmic differentiation?
Here is my work:
$ln(\frac{dy}{dx}) = ln(\frac{-x}{y})$
$ln(\frac{dy}{dx}) = ln(-x) - ln(y)$
(Take derivative of both sides)
$(\frac{d^2y}{d^2x})/(\frac{dy}{d... | You made a mistake. In particular, differentiating both sides of $$\log(\frac{dy}{dx}) = \log(-x)-\log(y)$$ gives
$$\frac{\frac{d^2y}{dx^2}}{\frac{dy}{dx}} = \frac{1}{x} - \frac{1}{y} \color{red}{\frac{dy}{dx}}.$$
Here $\frac{d}{dx} \log(y) = \frac{1}{y} \color{red}{\frac{dy}{dx}}$ comes from the chain rule.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2408228",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Show that $11^{10} \equiv 1 (\mod 100)$ $11^{10} \equiv 1 \pmod{100}$
I tried to solve by using euler's theorem, But I got stuck.
$\gcd(11, 100) = 1$
$11^{φ(100)} \equiv 1 \pmod{100}$
$11^{40} \equiv 1 \pmod{100}$
I don't know how to go on as $11^{40}$ is bigger than $11^{10}$
| $$\qquad{11^2\equiv 121\equiv 21\mod 100\\\\
(11^2)\times 11\equiv 231\equiv 31\mod 100\\
(11^4)\equiv 31\times 11\equiv 341\equiv 41\mod 100\\
(11^5)\equiv 41\times 11\equiv 451\equiv 51\mod 100\\
(11^6)\equiv 51\times 11\equiv 61\mod 100\\
(11^7)\equiv 61\times 11\equiv 71\mod 100\\
(11^8)\equiv 71\times 11\equiv 81\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2409057",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 7,
"answer_id": 6
} |
Finding the $nth$ term of a sequence Okay so I'm asking this quesion knowing a thing or two about sequences and general terms
What is the sum of the series :
$$1+\frac{1\cdot3}{6}+\frac{1\cdot3\cdot5}{6\cdot8}+\cdots$$
My Try: I tried calculating the general term $T_{n}$ for the sequence but I'm not able to understand... | $$1+\frac{1\cdot3}{6}+\frac{1\cdot3\cdot5}{6\cdot8}+\cdots$$
Numerator is the product of odd positive integers which can be written as
$1 \cdot 3\cdot 5=\dfrac{1 \cdot 2\cdot 3\cdot 4 \cdot 5 }{2 \cdot 4}=\dfrac{5! }{2^2\left( 1\cdot 2\right)}=\dfrac{(2\cdot 2 +1)!}{2^2\cdot 2!}$
$n-$th numerator will be $\dfrac{(2n+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2410655",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
$|x^2-5x+2|\leq 4$
Solve the following inequality.
$$|x^2-5x+2|\leq 4.$$
I know how to interpret $|x-a|$ as the distance between $x$ and $a$ along the $x$-axis, but how does one interpret an absolute value of $|ax^2+bx+c|$?
For the inequality $|x^2-5x+2|\leq 4$ I factored the LHS and got $$\left|\left(x+\frac{\sqrt... | It's just $$-4\leq x^2-5x+2\leq4,$$
which is $$x^2-5x-2\leq0$$ (which gives
$\frac{5-\sqrt{33}}{2}\leq x\leq\frac{5+\sqrt{33}}{2}$)
and $$x^2-5x+6\geq0$$ (which is
$x\geq3$ or $x\leq2$).
Finally, we obtain:
$$\left[\frac{5-\sqrt{33}}{2},2\right]\cup\left[3,\frac{5+\sqrt{33}}{2}\right]$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2412622",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
If the quadratic equation $x^2+px+q=0$ and $x^2+qx+p=0$ have a common root
If quadratic equations $x^2+px+q=0$ and $x^2+qx+p=0$ have a common root, prove that: either $p=q$ or $p+q+1=0$.
My attempt:
Let $\alpha $ be the common root of these equations. Since one root is common, we know:
$$(q-p)(p^2-q^2)=(q-p)^2.$$
Ho... | Let $\alpha$ be the common root.
\begin{align}
\alpha^2 + p \alpha + q &= 0 \\
\alpha^2 + q \alpha + p &= 0 \\
\hline
(p-q)\alpha + (q-p) &= 0 \\
(p-q)(\alpha-1) &= 0
\end{align}
$\alpha = 1$ or $p=q$.
If $\alpha = 1$, then $\alpha^2 + p \alpha + q = 0$ becomes
$1+p+q = 0$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2412731",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Finding the minimum value of $a^2+b^2+c^2$
Let $a$, $b$ and $c$ be $3$ real numbers satisfying $2 \leq ab+bc+ca$. Find the minimum value of $a^2+b^2+c^2$.
I've been trying to solve this, but I don't really know how to approach this. I thought of $(a+b+c)^2 = a^2+b^2+c^2 + 2(ab+bc+ca)$, but that gives me $a+b+c$, wh... | Again from Cauchy-Schwarz from a different perspective
$\sqrt{a^2c^2}+\sqrt{b^2c^2}+\sqrt{a^2c^2} \leq \sqrt{a^2+b^2+c^2}\sqrt{a^2+b^2+c^2}$
$a^2+b^2+c^2 \ge ab+bc+ac \ge 2$
$a^2+b^2+c^2 \ge 2$
İnequality holds for $a=b=c=\sqrt{\dfrac{2}{3}}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2413134",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 4
} |
Determinant of a symmetric matrix of order $2018$
Find out the determinant of the following matrix :
$$\begin{bmatrix}1^{2016} & 2^{2016} & ... & 2018^{2016}\\2^{2016} & 3^{2016} & ... & 2019^{2016}\\.. & .. & ... & ..\\2018^{2016} & 2019^{2016} & ... & 4035^{2016}\end{bmatrix}$$
Through examples of order $2\times 2$... | Suppose $p_1, p_2...,p_n $ are polynomials of degree $m-1 $ with atleast one has degree exactly $m-1$ . Let us consider a matrix $A$ whose $(i,j)-$th matrix has entry $p_i(u_j)$ , where $u_1,u_2,.......,u_n$ are $ n$ distinct number
let $p_i(x) = a_{i,1} + a_{i,2} x +.....+a_{i,m}x^{m-1} $
Now we can easily... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2413871",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Equation of one variable
Solve the following equation.
$$(8-x)\cdot(x^2-2x+16)^2+2x^4\cdot(x^2-2x+16)=16x^7,$$
where $x\in \mathbb{R}$
I already know that we need to prove $x=2$, but don't know how to show it...
I'd be grateful for some hints or solutions ;)
| It's $$(x-2)(16x^6+30x^5+65x^4+86x^3+240x^2+128x+1024)=0$$
and since
$$16x^6+30x^5+65x^4+86x^3+240x^2+128x+1024=$$
$$=(16x^6+30x^5+15x^4)+(50x^4+86x^3+37x^2)+(203x^2+128x+1024)>0,$$
we get the answer: $\{2\}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2414503",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Rules of Double Sums What are the (most important) rules of double sums? Below are some rules I encountered - are they all correct and complete? Offerings of clear intuition or proofs (or other additions) are most welcome to remove confusion.
*
*General case: $$\sum_{i=1}^m[x_i] \cdot \sum_{j=1}^n[y_j] = \sum_{i=1}^m... | Explanation to (3): Given relation is
\begin{align}
\left(\sum_{i = 1}^{n} x_{i} \right)^{2} & = \sum_{i = 1}^{n} x_{i} \sum_{j = 1}^{j} x_{j} \tag{1} \\
& = \sum_{i = 1}^{n} \sum_{j = 1}^{n}x_{i}x_{j} \tag{2} \\
& = \sum_{i = 1}^{n}x_{i}^{2} + \sum_{i = 1}^{n}\sum_{\substack{j = 1 \\ j \neq i}}^{n} x_{i}x_{j} \tag{3}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2416833",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "16",
"answer_count": 2,
"answer_id": 0
} |
Induction: Bounds on sum of inverses of first $n$ square roots I am trying to show by induction that $2(\sqrt{n}-1)<1+\dfrac{1}{\sqrt{2}}+\cdots+\dfrac{1}{\sqrt{n}}<2\sqrt{n}$.
For the upper bound, I have that if $1+\dfrac{1}{\sqrt{2}}+\cdots+\dfrac{1}{\sqrt{n}}<2\sqrt{n}$, then $1+\dfrac{1}{\sqrt{2}}+\cdots+\dfrac{1}{... | Try to use the induction method for $2(\sqrt{n}-1+\frac{1}{\sqrt{n}}) < \frac{1}{\sqrt{1}} + \frac{1}{\sqrt{2}} + \cdots + \frac{1}{\sqrt{n}}$
Check it for base cases like $n=1,2,3,4,5,6,7,8,9,10$ and it will fail for all $n=1,2,3,4,5,6,7$ and will hold for $n=8,9,10$.
Induction step : assume that $2(\sqrt{n}-1+\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2416918",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
$\lim\limits_ {n \to \infty} \left( \cos ^{2n} \left(\frac{k \pi}{3}\right)-\cos ^{2n}\left(\frac{k \pi}{5}\right)\right )=0$ For how many values of $k=1,2,3,....200$ $$\lim\limits_ {n \to \infty} \left( \cos ^{2n} \left(\frac{k \pi}{3}\right)-\cos ^{2n}\left(\frac{k \pi}{5}\right)\right )=0$$
My Try:
Let $A=\cos\left... | A different approach. It is worth mentioning that
$$0\leq \cos^2\left(\frac{k\pi}{3}\right)\leq1 \text{ and } 0\leq \cos^2\left(\frac{k\pi}{5}\right)\leq1$$
for $\forall k \in \mathbb{Z}$. As a result ($0\leq a \leq 1 \Rightarrow 0\leq a^n \leq 1, n\in \mathbb{N}$)
$$0\leq \cos^{2n}\left(\frac{k\pi}{3}\right)\leq1 \te... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2419227",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Discrete math $A\,\triangle\, B = C$ implies that $A\,\triangle\, C = B$ $A\,\triangle\, B = C$ implies that $A\,\triangle\, C = B$
I understand that the delta is the symmetric difference and that the symmetric difference of $A$ and $B$ is the set of elements that belong to exactly one of $A$ and $B$. How do I prove th... | We need to prove that given $C = A \ \Delta \ B$, then $B = A \ \Delta \ C$
A simple and elementary proof is to just use the definition of the symmetric difference, or $XOR$ as I like to call it, and setup some truth tables:
$\begin{array}{|c|c|c|} \hline
A & B & \mathbf{C} \\ \hline
0 & 0 & \mathbf{0} \\ \hli... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2419581",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
Find maximum of $f(x)=3\cos{2x}+4\sin{2x}.$ Find maximum of $f(x)=3\cos{2x}+4\sin{2x}.$ The shortest way to do this is to only consider $2x\in[0,2\pi)$, set $t=2x$ and note that
*
*Max$(\cos{t})=1$ for $t=0.$
*Max$(\sin{t})=1$ for $t=\frac{\pi}{2}.$
Max of these two functions added is when $t$ equals the angle ex... | This is a well-known problem: the maximum of $a\cos x+b\sin x$ is
$\sqrt{a^2+b^2}$. One way to see this is to use Cauchy-Schwarz:
$$a\cos x+b\sin x\le\sqrt{a^2+b^2}\sqrt{\cos^2x+\sin^2x}=\sqrt{a^2+b^2}$$
and then get equality by finding an $x$ with $\cos x=a/\sqrt{a^2+b^2}$
and $\sin x=b/\sqrt{a^2+b^2}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2420674",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
Integration: $\int_{2}^{\infty} \frac{\sqrt x}{x^2-1} dx$ I am having a problem integrating this term, I am not able to solve it by substitution either.
|
$$\mathscr{\text{Let } t= \sqrt x}$$
$$\mathscr{\text{.:} dt=\frac {1}{2\sqrt x} dx}$$
$$\mathscr{\text{.:} dx= 2t dt}$$
$$\int_0^\infty \frac{t}{t^4-1} 2t dt$$
$$=\int_0^\infty \frac {2t^2}{t^4-1} dt$$
$$= \int_0^\infty \frac{2t^2}{(t^2-1)(t^2+1)} dt$$
$$= \int_0^\infty \frac{(t^2-1)+(t^2+1)}{(t^2-1)(t^2+1)} dt$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2422300",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Finding $\int\frac{x}{\sqrt{3-2x-x^2}} dx$. I was looking for the integral of
$$\int\frac{x}{\sqrt{3-2x-x^2}} dx$$
My work:
$$\int \frac{x}{\sqrt{3-2x-x^2}} dx = \int \frac{x}{\sqrt{(3)+(-2x-x^2)}} dx $$
$$ = \int \frac{x}{\sqrt{(3)-(2x+x^2)}} dx $$
$$ = \int \frac{x}{\sqrt{(3)-(1+2x+x^2) +1}} dx $$
$$\int \frac{x}{... | use the Eulersubstution and set $$\sqrt{3-2x-x^2}=xt\pm \sqrt{3}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2422402",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 2
} |
Why $\sqrt{x^2}$ is not equal to $\big(\sqrt{x}\big)^2$? I'm reading Precalculus from James Stewart. In the book, the author says that $\sqrt{x^2}$ is not equal to $\big(\sqrt{x}\big)^2$. I was performing a couple of proofs and I ended up here:
If you have, for example:
$\sqrt{5^2} ≟ \big(\sqrt{5}\big)^2 \implies (\sqr... | You are not wrong about $\sqrt{(5)^2} = \left( \sqrt{5}\right) ^2$.
But if you think that this proves that $\sqrt{(-5)^2}$ is equal to $\left( \sqrt{-5}\right) ^2$, then you are mistaken.
$\sqrt{(-5)^2} = \sqrt{25} = 5$
While $\left( \sqrt{-5}\right) ^2 = \left( \sqrt{-5}\right)\left( \sqrt{-5}\right) = -5$ (or it doe... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2422719",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
If the partial sums $s_n $of $\sum a_n $ are bounded, show that the series $\sum \frac{a_n}{n} $ converges to $\sum \frac{s_n}{n (n+1)} $ My try: $\left(\frac{1}{n}\right)_{n=1}^\infty $is decreasing and $\lim_{n\to\infty} \frac{1}{n}=0$ with $\sum a_n $is bounded so by Dirichlet's Test $\sum \frac{a_n}{n}$ is conver... | Try the usual summation by parts as suggested in the comments.
Another approach:
$$\begin{align}\sum_{n=1}^N \frac{s_n}{n(n+1)} &= \sum_{n=1}^N \sum_{k=1}^n\frac{a_k}{n(n+1)} \\&= \sum_{n=1}^N \sum_{k=1}^N\frac{a_k\, 1_{k \leqslant n}}{n(n+1)} \\&= \sum_{k=1}^N \sum_{n=1}^N\frac{a_k\, 1_{k \leqslant n}}{n(n+1)} \\ &= ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2422876",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
How to calculate the determinant of this n by n matrix? Find the determinant of this n by n matrix.
$$
\begin{pmatrix}
0 & x_1 & x_2 & \cdots& x_k \\
x_1 & 1 & 0 & \cdots & 0 \\
x_2 & 0 & 1& \cdots & 0 \\
\vdots& \vdots& \vdots& \ddots & \vdots\\
x_k & 0 & 0 & \cdots& 1 \\
\end{pmatrix}
$$
where, $$ k=n-1 $$.
I am new... |
You can do one thing
$$
\begin{pmatrix}
0 & x_1 & x_2 & \cdots& x_k \\
x_1 & 1 & 0 & \cdots & 0 \\
x_2 & 0 & 1& \cdots & 0 \\
\vdots& \vdots& \vdots& \ddots & \vdots\\
x_k & 0 & 0 & \cdots& 1 \\
\end{pmatrix}
$$
Taking $x_r$ common from $(r+1)^{th}$ row for r=1 to k
$$
(\prod_{i=1}^k x_k)
\begin{pmatrix}
0 & x_1 &... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2423062",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 4
} |
Greatest Integer function limit problem I need help solving this problem without redefining the greatest interger function, is that possible?
$$\lim_{x \, \to \,\frac {1}{2}^-}[3x-\frac {1}{2}]$$
What about this rule
| Since we are interested in the behavior of $\left\lfloor 3x-\frac{1}{2}\right\rfloor$ as $x$ tends to $\frac{1}{2}$ from the left, we may assume that $\frac{1}{6}<x<\frac{1}{2}$. Then
$$\frac{3}{6}<3x<\frac{3}{2}$$
so
$$\frac{3}{6}-\frac{1}{2}<3x-\frac{1}{2}<\frac{3}{2}-\frac{1}{2}$$
which reduces to $0<3x-\frac{1}{2}<... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2423722",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Find maximize and minimize of $P=x+y$
For $\{x,y\}\subset\mathbb R$ such that $\sqrt{x+1}+\sqrt{y+1}=\sqrt{2}\left(x+y\right)$ find maximize and minimize of $P=x+y$
I found the maximize but minimize I have no idea. Help me.
| By AM-GM
$$2(x+y)^2=x+y+2+2\sqrt{(x+1)(y+1)}\leq x+y+2+(x+1+y+1)=2(x+y+2).$$
Thus, $$(x+y)^2-(x+y)-2\leq0$$ or
$$-1\leq x+y\leq2.$$
In the right inequality the equality occurs for $x=y=1$,
which says that $2$ is a maximal value of $P$.
Now, $\sqrt{.}$ is a concave function.
Thus, since $(x+1+y+1,0)\succ(x+1,y+1)$, by... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2423824",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Partial Differentiation Ex 2.2 Eng Mathematics by H.K. Das Test for continuity when
$f(x,y) = \frac{x^3\,y^3}{x^3+y^3}$ when $x\neq0, y\neq0$ and f(x,y)=0 when $x=0,y=0$.
| The domain of $f$ is $D=\mathbb{R^2}\setminus \{(x,-x):x\in\mathbb{R}\}$.
In $D$, let $y=x$ then, as $x\to 0$, also $y\to 0$ and
$$f(x,y)=\frac{x^3y^3}{x^3+y^3}=\frac{x^3}{2}\to 0.$$
In $D$, let $y=-x+x^4$ then, as $x\to 0$, also $y\to 0$ and
$$f(x,y)=\frac{x^3y^3}{x^3+y^3}=\frac{x^3(-1+x^3)^3}{1+(-1+x^3)^3}=\frac{-x^3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2425121",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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If $a,b$ are positive integers and $x^2+y^2\leq 1$ then find the maximum of $ax+by$ without differentiation. If $x^2+y^2\leq 1$ then maximum of $ax+by$
Here what I have done so far.
Let $ax+by=k$ . Thus $by=k-ax$.
So we can have that $$b^2x^2+(k-ax)^2 \leq b^2$$
$$b^2x^2+k^2-2akx +a^2x^2-b^2\leq 0 $$
By re-writing as a... | Use Cauchy Буняко́вський Schwarz, $|\langle (a,b), (x,y) \rangle| \le \|(a,b)\| \|x,y\|$. Choosing $(x,y)= {1 \over \|(a,b)\|} (a,b)$ shows equality.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2426188",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 0
} |
How to integrate $\int \frac {e^y}{y} dy$? The question is to evaluate $$\iint_R \frac {x}{y} e^y dx dy$$ where R is the region bounded by $0 \leq x \leq 1$ and $x^2 \leq y \leq x$.
So i write it as $$\int_0^1 \int_{x^2}^{x} \frac{x}{y} e^y dy dx$$.
The thing is, how do i evaluate $I=\int_{x^2}^{x} \frac{1}{y} e^y dy$?... | This is the Exponential integral, and it can't be expressed in terms of standard functions. However, one can first integrate over $x$ (with slightly more complicated bounds), and then the integration in $y$ becomes much more straightforward.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2427295",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
A floor function equation I want to solve below equation analytically $$\lfloor \frac{2x+1}{3}\rfloor +\lfloor \frac{\lfloor4x\rfloor+2}{3}\rfloor=1$$ but I have no idea to start .
Implicit: I solve it by graphing and the solution was $x\in[\frac {1}{4},1)$
I am thankful for any idea in advance .
I try to use $x=n+p... | If $x \geqslant 1,$ then $\lfloor{\frac{2 x+1}{3}} \rfloor + \lfloor{\frac{\lfloor{4 x \rfloor}+2}{3}} \rfloor \geqslant 1+2>1.$ If $x<.25,$ then $\lfloor{\frac{2 x+1}{3}} \rfloor + \lfloor{\frac{\lfloor{4 x \rfloor}+2}{3}} \rfloor < 0+1,$ so solutions will not be found in these intervals. Conversely, if $.25 \leqslant... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2427769",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 1
} |
Prove that $\mathbb{Z}[\omega]$ is an integral domain
Prove that $\mathbb{Z}[\omega]=\{ a+b \omega | a,b \in \mathbb{Z} \}$ is an integral domain, where $\omega^3=1$.
My solution:
Let $(a+b\omega)(c+d\omega)= (ac-bd)+(ad+bc-bd)\omega=0$
$\Rightarrow ac=bd \; \& \; ad+bc=bd. $
If $a\neq 0$, then put $c= bd/a$ to solv... | Since $ab = a^2 + b^2 \geq a^2 > 0$, we must have $b\neq 0$ and $a$ and $b$ must have the same sign. This gives
$$
a^2 + b^2 = ab\\
\frac{a}{b} + \frac{b}{a} = 1
$$
$a/b$ is positive, and therefore has a (possibly irrational) square root. We get
$$
0\leq\left(\sqrt\frac ab - \sqrt\frac ba\right)^2 = \frac{a}{b} + \fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2430461",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
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Let $A$ be a $4\times2$ matrix and $B$ be a $2\times4$ matrix. Find $BA$ when $AB$ is given.
Let $A$ be a $4\times2$ matrix and $B$ be a $2\times4$ matrix so that
$$AB=
\begin{pmatrix}
1 & 0 & -1 & 0 \\
0 & 1 & 0 & -1 \\
-1 & 0 & 1 & 0 \\
0 & -1 & 0 & 1
\end{pmatrix}.
$$
Find $BA$.
I have the ... | Let $C = (c_{ij}) = AB$, so $c_{ij} = a_{i1}b_{1j}+a_{i2}b_{2j}$, that means each entry $c_{ij}$ can be interpreted as the scalar product of row $i$ of $A$ and column $j$ of $B$. Let $a_{i:}$ be the $i$th row of $A$ and $b_{:j}$ the $j$-th column of $B$. Now we see that
$$a_{1:} \perp b_{:2} \text{ and } a_{1:} \perp ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2431735",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 4,
"answer_id": 2
} |
Prove $ \forall n \ge 4$, $n^{3} + n < 3^{n}$
Prove that $\forall n \ge 4$, $n^{3} + n < 3^{n}$
My attempt: Base case is trivial.
Suppose $ \ n \ge 4$, $n^{3} + n < 3^{n}$. Then,
$$ (n+1)^{3} + (n+1) = n^3 + n + 3n^2 + 3n + 1 +1 < 3^{n} + 3n^2 + 3n + 2 \\< 3\cdot 3^{n} + 3n^2 + 3n +2.$$
Not sure how to get rid of $3... | Because for $n\geq4$ we have:
$$3^n=(1+2)^n\geq1+2n+\frac{n(n-1)}{2}\cdot2^2+\frac{n(n-1)(n-2)}{6}\cdot2^3+$$
$$+\frac{n(n-1)(n-2)(n-3)}{24}\cdot2^4>n^3+n.$$
The last inequality is true because it's
$$3+6n+6n^2-6n+4n^3-12n^2+8n+2n^4-12n^3+22n^2-12n>3n^3+3n$$ or
$$2n^4-11n^3+16n^2-7n+3>0$$ or
$$2n^4-8n^3-3n^3+12n^2+4n^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2432292",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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If $ (x+\sqrt{x^2+1})(y+\sqrt{y^2+1})=1$, show that $x+y=0$
For $\{x,y\}\subset \Bbb R$, $(x+\sqrt{x^2+1})(y+\sqrt{y^2+1})=1.$
Prove that $x+y=0.$
Problem presented in a book, as being from Norway Math Olympiad 1985. No answer was presented. My developments are not leading to a productive direction. Sorry if this is... | Multiplying by $(x-\sqrt{x^2 +1})(y-\sqrt{y^2 +1})$ we get
$$(x-\sqrt{x^2 +1})(y-\sqrt{y^2 +1})=1$$
and thus $$(x+\sqrt{x^2 +1})(y+\sqrt{y^2 +1})=(x-\sqrt{x^2 +1})(y-\sqrt{y^2 +1})$$
hence
$$x(\sqrt{x^2 +1} +\sqrt{y^2 +1} )=-y(\sqrt{x^2 +1} +\sqrt{y^2 +1} )$$
therefore $$x=-y$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2432911",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 0
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$\arctan{x}=\arccos{2x}.$ First, let's determine some domains. By definition, it follows that $\arccos:[-1,1]\rightarrow[0,\pi]$ and $\arctan:\mathbb{R}\rightarrow\left[-\frac{\pi}{2},\frac{\pi}{2}\right]$. Since the domain of $\arctan$ is the entire reals, the restricting factor here should be the domain of arccos. Th... | Let $\arctan x=\arccos(2x)=u$
Using the definition of Principal values of Inverse trigonometric functions,
$-\dfrac\pi2\le u\le\dfrac\pi2,0\le u\le\pi\implies0\le u\le\dfrac\pi2\implies x\ge0$
We have $x=\tan u,2x=\cos u$
$$\dfrac1{(2x)^2}-x^2=1\implies4x^4+4x^2-1=0$$
$$\implies x^2=\dfrac{-1\pm\sqrt2}2$$
For real $x,... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2434627",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Calculating the limit $\lim\limits_{x \to 0^+} \frac{\sqrt{\sin x}-\sin\sqrt{ x}}{x\sqrt{x}}$
Calculate $$\lim\limits_{x \to 0^+} \dfrac{\sqrt{\sin x}-\sin \sqrt{x}}{x\sqrt{x}}$$
without use Taylor serie and L'Hôpital.
$$\lim\limits_{x \to 0^+} \dfrac{\sqrt{\sin x}-\sin \sqrt{x}}{x\sqrt{x}}\cdot\dfrac{\sqrt{\sin x... | The hint.
Prove that for all $x>0$ we have:
$$-\frac{1}{6}<\frac{\sin{x}-x}{x^3}<-\frac{1}{6}+\frac{x^2}{120}$$ and from this we obtain:
$$\lim_{x\rightarrow0^+}\frac{\sin{x}-x}{x^3}=-\frac{1}{6}.$$
Let $f(x)=\sin{x}-x+\frac{1}{6}x^3$, where $x>0$.
Thus, $f'''(x)=-\cos{x}+1\geq0$, which gives
$f''(x)>f''(0)=0$, $f'(x)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2435771",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 1
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Number of $6$ digit natural numbers with $3$ odd and $3$ even digits I know this has been asked before but I am stuck and need help.
For arranging $6$ digits there are $6! = 720$ ways.
One way is $3$ odd first then $3$ even, which is
$$5 \times 4 \times 3 \times 5 \times 4 \times 3.$$
This is one way, so we multiply... |
For arranging $6$ digits, there are $6! = 720$ ways.
This claim is true only if all six digits are distinct. However, $434152$ is a six-digit number with three even and three odd digits.
Find the number of six-digit natural numbers with three even and three odd digits.
Method 1: We add the number of six-digit num... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2435850",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
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Prove that for $n \ge 2$ the follow inequality holds $\frac {4^n}{n+1} \lt \frac{(2n)!}{(n!)^2}$. As already shown above I need to prove that $$\frac {4^n}{n+1} \lt \frac{(2n)!}{(n!)^2}\qquad \forall n \ge 2.$$
What I've come up with is the following:
$\underline{n=2:}\qquad$ $$\frac {16}{3} \lt 6$$
$\underline{n=k:}\... | For $k+1$ it is equivalent to
$$
\frac{4^k}{k+1}<\frac{(2k+1)(k+2)(2k)!}{2(k+1)^2(k!)^2}$$
It is enough to prove that
$$\frac{(2k+1)(k+2)}{2(k+1)^2}>1$$
Notice that the above fraction is monotonically decreasing and its limit at $k\to\infty$ is $1$, therefore it is always greater than $1$.
And also, you have a mistake,... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2437448",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
help in solving $z^2+|z|^2=1+2i$ I can't solve this:
$$z^2+|z|^2=1+2i$$
I've found $\sqrt{ \frac{2}{5}}$ for the modulus of z but I can't explicit about the angle of z .
The result of the book is, $ \pm (\frac{1}{\sqrt{2}}+i*\sqrt{2})$ but there are so many mistakes that I'm afraid is not very reliable
| $$z^2+|z|^2=(a+bi)^2+a^2+b^2=a^2+2abi-b^2+a^2+b^2=2a^2+2abi$$
So $$2a^2+2abi=1+2i$$
Equating real and imaginary parts:
$$\Re:2a^2=1\Rightarrow a=\pm\frac{1}{\sqrt{2}}$$
$$\Im:2(\pm\frac{1}{\sqrt{2}})bi=2i\Rightarrow\pm\frac{1}{\sqrt{2}}b=1\Rightarrow b=\pm\sqrt{2}$$
Thus,$z=\pm\frac{1}{\sqrt{2}}\pm\sqrt{2}i$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2442253",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Finding all values such that column vector is a linear combination Question:
For which values(s) of $ \ a$ is the column $ \ c = \begin{bmatrix} a\\a^2\\0\\a+1\end{bmatrix}$ a linear combination of the columns of,
$ \ x = \begin{bmatrix} 1\\1\\1\\1\end{bmatrix}$, $\ y = \begin{bmatrix} 1\\0\\-1\\0\end{bmatrix}$, $ \ z=... | A necessary condition is:
$$a^2 = r+t = a+1 \implies a = \frac{1\pm\sqrt{5}}{2}$$
Then you have $2r = a \implies r = s = \frac{1\pm\sqrt{5}}{4}$ so in that case the rows are indeed linearly dependent.
Thus, the rows are linearly dependent iff $a = \frac{1\pm\sqrt{5}}{2}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2442355",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
How would I solve the following? $$M = \begin{pmatrix}1&-4\\ -6&-6\end{pmatrix}$$
Find $c_1$ and $c_2$ such that $M^2 + c_1M +c_2I_2 = 0$, where $I_2$ is the identity $2 \times 2$ matrix.
The matrix above is $M$.
I got stuck after I merged the $c_1$ and $c_2$ into their respective matrices and calculated the square of ... | $M^2=\left(
\begin{array}{cc}
25 & 20 \\
30 & 60 \\
\end{array}
\right)$
$M^2 + c_1M +c_2I_2 = 0$ becomes
$\left(
\begin{array}{cc}
25 & 20 \\
30 & 60 \\
\end{array}
\right)+\left(
\begin{array}{cc}
c_1 & -4 c_1 \\
-6 c_1 & -6 c_1 \\
\end{array}
\right)+\left(
\begin{array}{cc}
c_2 & 0 \\
0 & c_2 \\
\end{array}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2442776",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Determine the derivative of $\arctan$ function Find $f'(x)$ for the function:
$f(x)= \arctan(\frac{a+x}{1-ax}))$ , $a\in R$
So this is what I've done:
$f(x) = \arctan x$
$f'(x) = \frac{1}{1+x^2}$
$x= \frac{a+x}{1-ax}$
$f'(x) = \frac{1}{1+\frac{(a+x)^2}{(1-ax)^2}}$
$f'(x) = \frac{(1-ax)^2}{(1-ax)^2+(a+x)^2}$
Is this co... | you must use the chain rule: $${1 \left( \left( -ax+1 \right) ^{-1}+{\frac { \left( a+x \right) a}{
\left( -ax+1 \right) ^{2}}} \right) \left( {\frac { \left( a+x
\right) ^{2}}{ \left( -ax+1 \right) ^{2}}}+1 \right) ^{-1}}
$$
simplified to
$$\frac{1}{1+x^2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2442892",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
No. of isosceles triangles possible of integer sides with sides $\leq n$ Prove that the no. of isosceles triangles with integer sides, no sides exceeding $n$ is $\frac{1}{4}(3n^2+1)$ or $\frac{3}{4}(n^2)$ according as n is odd or even, n is any integer.
How to do it? I found that under these conditions no. of triangles... | Fix the length of the two equal sides, say $k$. In how many ways can you choose the length of the base $b(k)$? Obviously $b(k) \ge 1$ and, for the triangle inequality, $b(k) < 2k$. But, since no side can exceed $n$, $b(k) \le n$. Putting these things together, we conclude that:
*
*when $2k - 1 \le n$, that is, when ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2443341",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 0
} |
$\frac{a^3}{b^2} + \frac{b^3}{c^2} + \frac{c^3}{a^2} \geq 3 \, \frac{a^2 + b^2 + c^2}{a + b + c}$ Proposition
For any positive numbers $a$, $b$, and $c$,
\begin{equation*}
\frac{a^3}{b^2} + \frac{b^3}{c^2} + \frac{c^3}{a^2}
\geq 3 \, \frac{a^2 + b^2 + c^2}{a + b + c} .
\end{equation*}
I am requesting an elementary, alg... | Also, SOS helps!
$$\text{LHS-RHS} = \sum\limits_{cyc}\left\{\frac{a^3}{b^2}-a-2(a-b)\right\}-\frac{3(a^2+b^2+c^2)-(a+b+c)^2}{a+b+c}$$
$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=\sum\limits_{cyc} {\frac { \left( a+2\,b \right) \left( a-b \right) ^{2}}{{b}^{2}}} - \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2446312",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 1
} |
In a circle, parallel chords of length $2$, $3$, and $4$ I was helping my comrade answering some questions taken from review classes when I stumbled upon this question. It looks like this:
In a circle, parallel chords of length $2$, $3$, and $4$ determine central angles of $A$, $B$, and $A+B$ radians, respectively, wh... | Note: this question is basically identical to the 1985 AIME Problem 9, so I'll summarise the solution below.
Because all chords of a given length in a given circle subtend the same arc and therefore the same central angle, we can rearrange our chords into a triangle with the circle as its circumcircle.
This triang... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2447262",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
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Prove that $\frac{ab}{a+b} + \frac{cd}{c+d} \leq \frac{(a+c)(b+d)}{a+b+c+d}$ $$\frac{ab}{a+b} + \frac{cd}{c+d} \leq \frac{(a+c)(b+d)}{a+b+c+d}$$
I tried applying a.m. g.m inequality to l.h.s and tried to find upper bound for l.h.s and lower bound for r.h.s but i am not getting answer .
| Not a beautiful proof I have to admit, but it is a proof:
Multiply 2 sides of equation with $(a+b)(c+d)(a+b+c+d)$, we get to the equivalent form:
$$
ab(c+d)(a+b+c+d) + cd(a+b)(a+b+c+d) \leq (a+c)(b+d)(a+b)(c+d).
$$
Work out both sides of the equation above to reduce it to the following equivalent form:
$$
2abcd \leq a^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2448601",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 0
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$f(x)=?$ if we have $f\left(\frac{x}{x^2+x+1}\right)=\frac{x}{x^2-x+1}$ $f(x)=?$
If we have $$f\left(\frac{x}{x^2+x+1}\right)=\frac{x}{x^2-x+1}$$ to fractions are very similar. I don't have an idea to find $f(x)$. Can someone show me a clue ?
| $$f\left( \dfrac{x}{x^{2}+x+1}\right)=\dfrac{x}{x^{2}-x+1}\rightarrow f\left( \dfrac{1}{x+\frac{1}{x}+1}\right)=\dfrac{1}{x+\frac{1}{x}-1}\rightarrow$$
$$f\left( \dfrac{1}{z+1}\right)=\dfrac{1}{z-1}$$
$$\dfrac{1}{z+1}=t\rightarrow \dfrac{1}{t}=z+1\rightarrow z=\dfrac{1}{t}-1\rightarrow z=\dfrac{1-t}{t}\rightarrow$$
$$f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2450683",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
Distributing 10 different folders between 7 people such that John gets at least 1 folder I did this by choosing 1 folder to give John ${10 \choose 1}=10$ and then saying that each of the remaining 9 folders can have 7 "states": They have to go to one of the 7 people. So that's $7^9$
So my answer is $10 \times 7^9$
My f... | Your friend is correct. You are counting cases in which John receives more than one folder multiple times.
If John receives exactly $k$ folders, then there are $6^{10 - k}$ ways to distribute the remaining $10 - k$ folders. Since John can receive exactly $k$ folders in $\binom{10}{k}$ ways, the number of ways to di... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2450910",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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If $\alpha,a,b$ are integers and $b\neq-1$, then prove that, if $\alpha$ satisfies the equation $x^2+ax+b+1=0$, $a^2+b^2$ must be composite.
Let $\alpha,a,b$ be integers such that $b\neq-1$. Assume that $\alpha$ satisfies the equation $x^2+ax+b+1=0$. Prove that the integer $a^2+b^2$ must be composite.
$\alpha=\frac{-... | We have
\begin{eqnarray*}
\alpha^2 + a \alpha +b+1=0.
\end{eqnarray*}
Multiply by $4$ and add $a^2$
\begin{eqnarray*}
(2 \alpha + a)^2 =a^2 -4(b+1).
\end{eqnarray*}
Now add & subtract $b^2$ & rearrange to
\begin{eqnarray*}
a^2+b^2=(2 \alpha + a+b+2)(2 \alpha + a-b-2).
\end{eqnarray*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2451572",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 3
} |
Basic arithmetic with matrices We have matrices.
$$A=\begin{bmatrix}
-2 & 0 \\
-5 & 6 \\
\end{bmatrix}
B^{-1}=\begin{bmatrix}
-7 & 8 \\
2 & -8 \\
\end{bmatrix}
C=\begin{bmatrix}
-15 & -2 \\
-8 & -14 \\
\end{bmatrix}
$$
We need to solve matrix $X$ from equation:
$$A^{-1}XB-C=0$$
$$X=AB^{-1}+C$$
$$X=\begin{bmatrix}
-2 &... | As point out by you that the order matters:
Pre-multiply and post-multiply are different.
Starting from
$${A^{ - 1}}XB - C = 0$$
Add $C$ on both sides
$${A^{ - 1}}XB = C$$
Now post-multiply by $B^{-1}$ on both sides
$${A^{ - 1}}X{B}{B^{ - 1}} = C{B^{ - 1}}$$
$${A^{ - 1}}X{I} = C{B^{ - 1}}$$
$${A^{ - 1}}X = C{B^{ - 1}}$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2451855",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
How can find domain and range of $f(x)=\sqrt{\sqrt 2 \lfloor x \rfloor - \lfloor\sqrt 2x \rfloor}$? How can find $D_{f(x)},R_{f(x)}$ when $$f(x)=\sqrt{\sqrt 2 \lfloor x \rfloor - \lfloor\sqrt 2x \rfloor} \ ?$$
I help some problem like this for a question here ... (Domain and range of a floor function)
But I get stuck o... | So, with the standard notation for the floor function, we are to study
$$ \bbox[lightyellow] {
y = \sqrt {\sqrt 2 \left\lfloor x \right\rfloor - \left\lfloor {\sqrt 2 x} \right\rfloor }
} \tag {1}$$
Let's first introduce some additional notation concerning the fractional part $\{x\}$
$$
x = \left\lfloor x \right\r... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2453266",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
how to $\int_0^{ +\infty} \frac{\sin(x)}{x+1}\, dx \leq \frac e5 \ln(\pi)$? I need to prove irreproachably that $$\int_0^{ +\infty} \frac{\sin(x)}{x+1}\, dx \leq \frac e5 \ln(\pi)$$ .
With an approximate calculation $\int_0^{ +\infty} \frac{\sin(x)}{x+1}\, dx\approx 0.62145$ and $\frac e5 \ln(\pi)\approx 0.62233$
We c... | The given integral equals
$$\begin{eqnarray*} \int_{0}^{\pi}\sin(x)\sum_{n\geq 0}\frac{(-1)^n}{x+n\pi+1}\,dx &=& \frac{1}{2\pi}\int_{0}^{\pi}\sin(x)\left[\,\psi\left(\tfrac{x+1}{2\pi}+\tfrac{1}{2}\right)-\psi\left(\tfrac{x+1}{2\pi}\right)\right]\,dx\\&=&\int_{\frac{1}{2\pi}}^{\frac{\pi+1}{2\pi}}\sin(2\pi z-1)\left[\,\p... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2460435",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 1
} |
The sum of series $\frac14+\frac{1\cdot3}{4\cdot6}+\cdots$ The problem I have here today is the following;
$$\frac{1}{4}+\frac{1\cdot3}{4\cdot6}+\frac{1\cdot3\cdot5}{4\cdot6\cdot8}+\cdots$$ the problem is exactly phrased like this (I can't say that the $\infty$ sign is a bit unnecessary at the end),
My Attempts
We can ... | The solution below is inspired by this other solution here.
Note that
$$\begin{align}
f(r)&=\frac {1\cdot 3\cdot 5\cdot\cdots \cdot(2r+1)}{4\cdot 6\cdot 8\cdot \cdots \cdot(2r+4)}\\
&=2\cdot \underbrace{\boxed{\frac {1\cdot 3\cdot 5\cdot\cdots \cdot(2r+1)}{2\cdot 4\cdot 6\cdot\cdots \cdot(2r+2)}}}_{A_r}\cdot \frac 1{2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2463183",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 1
} |
Find the volume between $x^2+y^2=9 , z=9-x^2-y^2, x^2+y^2+(z-16)^2=9$ So we have the surfaces$$x^2+y^2=9 $$$$ z=9-x^2-y^2$$$$ x^2+y^2+(z-16)^2=9$$
I was thinking this integral may be easier with cylindrical coordinates with the limits $0\le r\le 3$; $0\le\theta\le 2\pi$; and $9-r^2\le z\le S$.
Not sure how to define th... | Your approach by polar coordinates is correct.
Let $x=r\cos \theta, y=r\sin \theta$, then the first equation gives integrating area, while the second and third equation give the lower and upper bound:
$$
\begin{align}
0&\leq r\leq3 \\
9-r^2&\leq z\\
z&\leq 16-\sqrt {9-r^2}
\end{align}
$$
so we can write the double int... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2463772",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Equation of the tangent to a graph where the point is not on the graph. If $f(x)=(x+1)^{3/2}$, provided $x\geq -1$, I am asked to find the equation of all tangent lines to $f(x)$ at the point $(\frac{4}{3},3)$.
Simple enough. I first took the derivative which is:
$$f'(x)= \frac {3\sqrt{x+1}}{2}$$
Since $(\frac{4}{3},3)... | HINT
Use the fact that your line passes through $(k,f(k))$ and $(4/3,3)$, so the slope must also be
$$
m = \frac{\Delta y}{\Delta x} = \frac{3 - f(k)}{4/3 - k}
$$
and an alternate equation would be
$$
y - 3 = m(x - 4/3)
$$
but the equations have to match exactly...
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2465159",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Calculate $\iint_D x dxdy$ using polar coordinates Using polar coordinates, I want to calculate $\iint_D x dxdy$, where $D$ is the disk with center $(2,3)$ and radius $2$.
$$$$
I have done the following:
We have $D=\{(x,y)\mid (x-2)^2+(y-3)^2\leq 4\}$.
We use $(x,y)=(r\cos \theta, r\sin \theta)$.
From the inequali... | OFC your area of $\theta$ has to be restricted to the area where it "hits" the circle and as you already figured out your ansatz leads you to a very complicated calculation.
The reason is that your circle is not centered so first shift your integration area that you have a centered circle by substituting $\overline{x} ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2465787",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
How to find $\lim_{x\to1}\left(\tan\frac{\pi x}{4}\right)^{\tan\frac{\pi x}{2}}$?
Find
$$\lim_{x\to1}\left(\tan\frac{\pi x}{4}\right)^{\tan\frac{\pi x}{2}}$$
My attempt: ON THE basis of This post
$$\lim_{x\to1}\tan\frac{\pi x}{4} =1,\quad \lim_{x\to1}\tan\frac{\pi x}{2}=\infty$$
$$\implies\lim_{x\to1}\left(\tan\fr... | $$\lim_{x\rightarrow1}\left(\tan\frac{\pi x}{4}\right)^{\tan\frac{\pi x}{2}}=\lim_{x\rightarrow1}\left(1+\tan\frac{\pi x}{4}-1\right)^{\frac{1}{\tan\frac{\pi x}{4}-1}\cdot\tan\frac{\pi x}{2}\left(\tan\frac{\pi x}{4}-1\right)}=$$
$$=e^{-\lim\limits_{x\rightarrow}\frac{2\tan\frac{\pi x}{4}}{1+\tan\frac{\pi x}{4}}}=e^{-1}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2469040",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
Length of the chord of an ellipse whose midpoint is given Find the length of the chord of the ellipse $$\frac{x^2}{25}+\frac{y^2}{16}=1\tag1$$where mid point is $(\frac{1}{2},\frac{2}{3})$
My attempt: I know the equation of the chord of an ellipse whose mid-point is given is $T=S_1$,where T is $\left(\frac{x.x_1}{25}+\... | Line passing through midpoint $(\tfrac{1}{2}, \tfrac{2}{3})$ is
$$p(r)=\langle\tfrac{1}{2} + r \cos t, \tfrac{2}{3} + r \sin t\rangle \tag 1$$
where $r$ is the distance along the line and $t$ is the angle of slope of line.
Using your equation of chord, we can calculate $\cos t$ and $\sin t$:
$$\frac{x}{50}+\frac{y}{24... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2469193",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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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.