Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Solving 2nd Order ODE w/Laplace Transforms I am having difficulty with this problem:
*Note: The Delta3(t) is the delta dirac function, also the answer in the image is WRONG.
Attempt at solution :
*
*Let Laplace{y(t)}=Y
*Take Laplace of LHS and RHS.
*Solve for Y.
*Take inverse Laplace of Y, giving me a function o... | Define the Laplace transform as
\begin{align}
f(s) = \int_{0}^{\infty} e^{-s t} \, y(t) \, dt
\end{align}
then for the differential equation
\begin{align}
y'' - 3 y' -4 y = -4 t + \delta(t),
\end{align}
where $y(0) = -2$ and $y'(0) = -1$ then
\begin{align}
s^{2} f(s) - y'(0) - s y(s) - 3s f(s) + 3 y(0) - 4 f(s) = - \fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1233636",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Use Euclid's algorithm to find the multiplicative inverse $11$ modulo $59$ I was wondering if this answer would be correct
the multiplicative of $11$ modulo $59$ would be $5$ hence $5\cdot11 \equiv 4 \pmod{59}$.
Is this correct?
| No. A multiplicative inverse $a$ has $5 \cdot a \equiv 1 \pmod{59}$ and $11$ has $5 \cdot 11 \equiv -4 \not\equiv 1\pmod{59}$. If you want to find the inverse, use the Euclidean algorithm, as you said:
\begin{align*}
59 &= 5 \cdot 11 + 4\\
11 &= 2 \cdot 4 + 3\\
4 &= 1 \cdot 3 + 1\\
\end{align*}
Hence
\begin{al... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1234166",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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If the equation $|x^2+4x+3|-mx+2m=0$ has exactly three solutions then find value of m. Problem :
If the equation $|x^2+4x+3|-mx+2m=0$ has exactly three solutions then find value of $m$.
My Approach:
$|x^2+4x+3|-mx+2m=0$
Case I : $x^2+4x+3-mx+2m=0$
$\Rightarrow x^2+ x (4-m) + 3+2m=0 $
Discriminant of above qudratic ... | If you recognize that $x^2 + 4x + 3 = (x+2)^2 - 1,$
then it can quickly be seen that the graph of that function is an
upward-opening parabola with its minimum at $x=-2$, where the value of the
function is $-1$.
Clearly for values of $x$ far from $-2$, the function is positive,
those parts of the graph of the function a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1234482",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 1
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Prove by strong induction that $2^n$ divides $p_n$ for all integers n ≥ 1
Let $p_1 = 4$, $p_2 = 8$, and $p_n = 6p_{n−1} − 4p_{n−2}$ for each integer $n ≥ 3$.
Prove by strong induction that $2^n$ divides $p_n$ for all integers $n ≥ 1$
I got up to the base step where by you prove for $p_3$ but unsure about the strong... | If $2^n$ divides $p_n$ then we have that $p_n = \lambda_n 2^n$ for some integer $\lambda_n$. Therefore, the inductive hypothesis states that:
$$
p_n = \lambda_n2^n = 6p_{n - 1} - 4p_{n - 2}
$$
If we assume that it holds for both $n - 1$ and $n - 2$ (this is the strong induction) then we get that: $p_{n - 1} = \lambda_... | {
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"url": "https://math.stackexchange.com/questions/1235569",
"timestamp": "2023-03-29T00:00:00",
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Prove the set $\mathcal{O}_d := \left\{\frac{a + b\sqrt{d}}{2}:a,b \in \mathbb{Z}, a \equiv b \mod 2 \right\}$ is a ring.
Prove that if $d$ is a non-square integer with $d \equiv 1 \mod 4$ then the set $$\mathcal{O}_d := \left\{\frac{a + b\sqrt{d}}{2}:a,b \in \mathbb{Z}, a \equiv b \mod 2 \right\}$$ is a ring, and in ... | Hint $\ $ Show that $\,{\cal O}_d = \Bbb Z\left[w\right],\ w = (1\!+\!\sqrt d)/2,\,$ using $\ w^2\! -w + n = 0,\,\ n =(1\!-\!d)/4$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Consider the equation $|z + 3i|=3|z|$ for complex z and give a geometric description of the set S of all solutions. Writing $z$ in the form $a+ib$ and then rearranging gives $-8a^2-8b^2+6b+9=0$. The most promising form I could manage from this is $(b-\frac{3}{8})^2=(\frac{9}{8}-a)(\frac{9}{8}+a)$ but I still do not kno... | If $z=x+iy$ then
\begin{align*}
|z+3i|&=3|z|&\;\iff\;&&x^2+(y+3)^2&=9(x^2+y^2)\\
& &\;\iff\;&&8x^2+8y^2-6y-9&=0\\
& &\;\iff\;&& x^2+ y^2-\frac{3}{4}y-\frac{9}{8}&=0\\
& &\;\iff\;&& x^2+ \left(y-\frac{3}{8}\right)^2&=\frac{81}{64}\\
\end{align*}
Therefore the set $S$ is a cir... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1238025",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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prove Inequalities for integrals prove
$\frac{\pi}{6}+\frac{1}{3}\leq \int_0^\frac{\pi}{2}\frac{1+\cos(x)}{2+\sin(x)}dx \leq \frac{\pi}{4}+\frac{1}{2}$
I got to the point where $\frac{1}{3} \leq f(x) \leq 1$,
so $\frac{\pi}{6} \leq \int_0^\frac{\pi}{2}f(x)dx \leq \frac{\pi}{2}$ and where do I go from here , thanks
| The upper bound is easy:$$\int_0^{\frac{\pi}{2}}\frac{1+\cos x}{2+\sin x}\mathrm{d}x\leq \int_2^{\frac{\pi}{2}}\frac{1}{2}(1+\cos x)\mathrm{d}x=\pi/4+1/2.$$
For the lower bound, $$\int_0^{\frac{\pi}{2}}\frac{1+\cos x}{2+\sin x}\mathrm{d}x=\int_0^{\frac{\pi}{2}}\frac{1}{2+\sin x}\mathrm{d}x+\int_0^{\frac{\pi}{2}}\frac{\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1238146",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Use the generating function to solve a recurrence relation We have the recurrence relation $\displaystyle a_n = a_{n-1} + 2(n-1)$ for $n \geq 2$, with $a_1 = 2$.
Now I have to show that $\displaystyle a_n = n^2 - n +2$, with $n \geq 1$ using the generating function.
The theory in my book is scanty, so with the help o... | another observation is :$$a_n=a_{n-1}+2(n-1) ,\space \space \space a_1=2$$ $$\rightarrow a_n-a_{n-1}=2(n-1)\\$$ put $n=1,2,3,..(n-1)$ $$a_2-a_1=2(2-1
)=2(1)\\ a_3-a_2=2(3-1)=2(2)\\a_4-a_3=2(4-1)=2(3)\\...\\a_n-a_{n-1}=2(n-1)=2(n-1)\\$$no look at sum of them $$a_n-a_1=2(1)+2(2)+2(3)+...2(n-1)=\\2(1+2+3+4+...+(n-1))=2 \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1242244",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
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Magic square with not distinct numbers There's a 4x4 magic square:
4 0 1 0
3 0 2 0
0 3 0 2
0 4 0 1
Where 0s are different numbers, 1=1, 2=2, 3=3, 4=4.
Only the rows and the columns have the same sum, the diagonals don't.
Question: If rotating or reflecting the magic square counts as a different solution, how many dif... | Let's start by giving names to the unknown quantities
$$
\begin{bmatrix}
4 & a & 1 & e \\
3 & b & 2 & f \\
c & 3 & g & 2 \\
d & 4 & h & 1
\end{bmatrix}
$$
and call the total sum of the rows and columns $T$. We can now subtract the 1st row from the 2nd column as follows
$$(a+b+3+4)-(4+a+1+e)=T-T=0$$
$$|| \quad\quad\qua... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1243053",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Determining the Laurent Series I need to determine the Laurent series of this function:
$$\frac{1}{(z-1)(z+5)}$$ Inside the annulus: $$\left\{z|1<|z-2|<6\right\}$$ Any help appreciated.
| Here is just a set up: $\frac{1}{1-x}=1+x+x^2+x^3+x^4+.....$,then
$\frac{1}{1+x}=1-x+x^2-x^3+x^4-.....$ then $\frac{1}{1+x/5}=\frac{5}{5+x}=1-(x/5)+(x/5)^2-(x/5)^3+(x/5)^4-.....$ Now divide by 5 to get $\frac{1}{5+x}=1/5-(x/5)/5+(x/5)^2/5-(x/5)^3/5+...$ And then integrate. Can you now work out the details?
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Draw the line segment joining the centers of two circles. Where does it meet the circles? I'm trying to construct a line segment between two circles. Given each radius and $x$, $y$ center of each circle, how can I find the endpoints for the blue line segment?
| Let the centres of the two circles be $(x_1, y_1)$ and $(x_2, y_2)$, where either $x_1 < x_2$ or $x_1 = x_2$ and $y_1 < y_2$, and radii $r_1$, $r_2$.
Suppose $x_1 = x_2$. In this case, the blue line segment is vertical, and its endpoints can easily be seen to be $(x_1, y_1 + r_1)$ and $(x_2, y_2 - r_2) = (x_1, y_2 - r_... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Precalculus unit circle with imaginary axis. (a) Suppose $p$ and $q$ are points on the unit circle such that the line through $p$ and $q$ intersects the real axis. Show that if $z$ is the point where this line intersects the real axis, then $z = \dfrac{p+q}{pq+1}$.
(b) Let $P_1 P_2 \dotsb P_{18}$ be a regular 18-gon. S... | (b) Let $\omega = e^{2 \pi i/18}$, a primitive $18^{\text{th}}$ root of unity. Then $\omega^{18} = 1$. Also, $\omega^{18} - 1 = 0$, which factors as
$(\omega^9 - 1)(\omega^9 + 1) = 0.$Since $\omega^9 = e^{2 \pi i/2} = e^{\pi i} = -1 \neq 1$, $\omega$ must satisfy the equation $\omega^9 + 1 = 0$. We can factor this equa... | {
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Prove that $f : [-1, 1] \rightarrow \mathbb{R}$, $x \mapsto x^2 + 3x + 2$ is strictly increasing.
Prove that $f : [-1, 1] \rightarrow \mathbb{R}$, $x \mapsto x^2 + 3x + 2$ is strictly increasing.
I do not have use derivatives, so I decided to apply the definition of being a strictly increasing function, which should ... | Yes, to see more clearly, let $x=a+2, y =a+1, x+c = b+2, y+c = c+2$, compute $(x+c)(y+c)- xy$. Notice that $x, c$ are positive, $y$ is non-negative.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1253120",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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The implication $\sqrt{(x-2)^2+(y-1)^2+(z-1)^2+(w-3)^2}Given a real number $a>0$ find a $b>0$ such that
$\sqrt{(x-2)^2+(y-1)^2+(z-1)^2+(ω-3)^2}<b\Longrightarrow |xyzω-6|<a$
I tried the procedure followed in another one of my questions, but it doesn't work.
| It seems the following.
The question seems to be a numerical adjustment of the continuity of multiplication. Assume that $$\sqrt{(x-2)^2+(y-1)^2+(z-1)^2+(w-3)^2}<b.$$ Then $|x-2|<b$, $|y-1|<b$, $|z-1|<b$, and $|w-3|<b$.
We have $2\cdot 1\cdot 1\cdot 3=6$ and
$$|xyzw-6|\le$$ $$|xyzw-xyz\cdot 3|+|xyz\cdot 3-xy\cdo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1254547",
"timestamp": "2023-03-29T00:00:00",
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Positivity of power function. Prove that
$6^a-7^a+2\cdot 4^a-3^a-5^a\ge0$ for $-\frac{1}{2}\le a\le0$.
I tried to do it by first derivative test but derivative become more complicated (same with 2nd derivative for convexity).
Here is it's plotting.
| For $a\in[-\frac12,0)$ and $x>0$ let $$g_a(x)=x^a-(x+1)^a-(x+3)^a$$
Then $$f(a)=g_a(4)-g_a(3)$$
By the Mean Value Theorem this equals $g_a'(\theta)=ag_{a-1}(\theta)$ for some $\theta\in [3,4]$. We have
$$\begin{align}\frac{g_{a-1}(\theta)}{\theta^{a-1}}&=1-(1+\tfrac1\theta)^{a-1}-(1+\tfrac3\theta)^{a-1}\\&\le 1-(1+\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1257102",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Solution Invariant Explanation Trick
Suppose not all 4 integers, $a,b,c,d$ are equal. Start with $(a,b,c,d)$ and repeatedly replace $(a,b,c,d)$ by $(a−b,b−c,c−d,d−a)$. Then show that at least one number of the quadruple will become arbitrarily large.
Solution:
We have
$$(a_{n+1},b_{n+1},c_{n+1},d_{n+1}) = (a_n-b_n... | Set $x_n=a_n^2+b_n^2+c_n^2+d_n^2$. The relationship, to be used repeatedly, is $$\color{red}{ x_{n+1}\ge 2 x_n,\text{ for all }n\in\mathbb{N}}.$$ Hence $$x_4\ge 2x_3\ge 2(2x_2)\ge 2(2(2x_1))=2^3x_1$$
We can prove by induction that $x_{n+1}\ge 2^n x_1$.
Details of the induction, as requested:
The base case is $n=1$,... | {
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Find the equation whose roots are each six more than the roots of $x^2 + 8x - 1 = 0$
Find the equation whose roots are each six more than the roots of $x^2 + 8x - 1 = 0$
I must use Vieta's formulas in my solution since that is the lesson we are covering with our teacher.
My solution:
Let p and q be the roots of the q... | Let $x_1,x_2$ be the roots of the equation: $x^2+8x-1 = 0$, and let $y_1 = x_1 +6, y_2 = x_2+6 \Rightarrow y_1+y_2 = x_1+x_2+12=-8+12 = 4, y_1y_2 = (x_1+6)(x_2+6)=x_1x_2+6(x_1+x_2)+36=-1+6\cdot (-8) + 36=-13\Rightarrow y^2-4y-13=0$ is the sought after equation.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1259949",
"timestamp": "2023-03-29T00:00:00",
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Compute the largest root of $x^4-x^3-5x^2+2x+6$ I want to calculate the largest root of $p(x)=x^4-x^3-5x^2+2x+6$. I note that $p(2) = -6$ and $p(3)=21$. So we must have a zero between two and three. Then I can go on calculating $p(\tfrac52)$ and see that the zero must lie in the interval $]2,\tfrac52[$. The answer is $... | You've : $p(x) = x^4 - x^3 - 5x^2 + 2x + 6 $ .
Rearrange the given polynomial as : $$ p(x) =(x^4 - 5x^2 + 6) - x(x^2 - 2) $$
For the first term, let us factor it separately. Say, $x^2 = t$ . So, you've :
$$\begin{align} f(x) =&\ x^4 - 5x^2 + 6\\
f(x) =&\ t^2 - 5t + 6 \end{align}$$
Roots of $f(x)$ are : $t = 3 \ \te... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1260906",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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I cannot solve this limit $$
\lim_{n\to\infty}\frac{(\frac{1}{n}+1)^{bn+c+n^2}}{e^n}=e^{b-\frac{1}{2}}
$$
I am doing it like this, and I cannot find the mistake:
$$
\lim_{n\to\infty}\frac{1}{e^n}e^{n+b+c/n}=
\lim_{n\to\infty}e^{n+b-n+c/n}=e^b
$$
| I would say that $$\log_e\left(\frac{(\frac{1}{n}+1)^{bn+c+n^2}}{e^n}\right) $$ $$ = (bn+c+n^2)\log_e\left(\frac{1}{n}+1\right) - n$$ $$= (bn+c+n^2)\left(\frac{1}{n}-\frac{1}{2\,{n}^{2}}+\frac{1}{3\,{n}^{3}}-\frac{1}{4\,{n}^{4}}+\cdots\right) - n$$ $$= b - \frac{b}{2n} + o(1/n) +\frac{c}{n} - o(1/n) +n -\frac{1}{2} +... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1264971",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
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Integration of $\int\frac{x^{4}}{x+1}dx$ Could you help me with the following integral:
$$\int\frac{x^{4}}{x+1}dx$$
| HINT: $$x^4=x^4-1+1=(x-1)(x+1)(x^2+1)+1$$
Generalization : For integer $n\ge1,$
$$x^n-1=(x-1)(x^{n-1}+x^{n-2}+\cdots+x+1)$$
For $x=y^2,y^{2n}-1=(y^2-1)(y^{2(n-1)}+y^{2(n-2)}+\cdots+y^2+1)$
$\iff y^{2n}-1=(y-1)(y+1)(y^{2n-2}+y^{2n-4}+\cdots+y^2+1)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1269431",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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To find the smallest integer with $n$ distinct divisors For example, if $n=20$, how can I find the smallest integer which has exactly $20$ distinct divisors? Can someone give me some hints?
| If $m=p_1^{a_1}p_2^{a_2}\ldots p_k^{a_k}$, we have number of divisors to be
$$(1+a_1)(1+a_2)\ldots(1+a_k)$$
For this to be $20$, we want
$$(1+a_1)(1+a_2)\ldots(1+a_k) = 20 = 2 \cdot 2 \cdot 5$$
This means $m$ can have at most $3$ distinct prime divisors.
*
*$3$ prime divisors. The smallest possible $m$ is $2^4 \cdot... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1270202",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "5",
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Solve this logarithmic equation: $2^{2-\ln x}+2^{2+\ln x}=8$ Solve this logarithmic equation: $2^{2-\ln x}+2^{2+\ln x}=8$. I thought to write
$$\dfrac{2^2}{2^{\ln(x)}} + 2^2 \cdot 2^{\ln(x)} = 2^3 \implies \dfrac{2^2 + 2^2 \cdot \left(2^{\ln(x)}\right)^2}{2^{\ln(x)}} = 2^3$$
What should I do else?
| Since $2^2+2^2 = 4+4 = 8$, can you just let $\ln x = 0$ and find $x = 1$?
ETA: This can be shown to be unique by finding the derivative with respect to $x$:
$$
\frac{d}{dx} 2^{2-\ln x} + 2^{2+\ln x}
= \frac{\ln 2}{x} \left(2^{2+\ln x}-2^{2-\ln x}\right)
$$
and provided that $x > 0$, this only equals $0$ at $\ln x =... | {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "3",
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Prove that number of zeros at the right end of the integer $(5^{25}-1)!$ is $\frac{5^{25}-101}{4}.$
Prove that number of zeros at the right end of the integer $(5^{25}-1)!$ is $\frac{5^{25}-101}{4}.$
Attempt:
I want to use the following theorem:
The largest exponent of $e$ of a prime $p$ such that $p^e$ is a divisor... | As shown in this answer, the number of factors of a prime $p$ in $n!$ is
$$
\frac{n-\sigma_p(n)}{p-1}
$$
where $\sigma_p(n)$ is the sum of the digits in the base-$p$ expansion of $n$.
The base-$5$ expansion of $5^{25}-1$ consists of $25\,\,4$s, thus
$$
\sigma_{5}(5^{25}-1)=25\cdot4=100
$$
Thus, the number of factors o... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1272660",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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$f\left( x-1 \right) +f\left( x+1 \right) =\sqrt { 3 } f\left( x \right)$ Let f be defined from real to real
$f\left( x-1 \right) +f\left( x+1 \right) =\sqrt { 3 } f\left( x \right)$
Now how to find the period of this function f(x)?
Can someone provide me a purely algebraic method to solve this problem please?
Update:... | Algebraic (non-coanstant)? NO.
But, for example
$$
f(x) = \sin\frac{\pi x}{6}
$$
is a solution.
| {
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"timestamp": "2023-03-29T00:00:00",
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"answer_id": 1
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How does $-\frac{1}{x-2} + \frac{1}{x-3}$ become $\frac{1}{2-x} - \frac{1}{3-x}$ I'm following a solution that is using a partial fraction decomposition, and I get stuck at the point where $-\frac{1}{x-2} + \frac{1}{x-3}$ becomes $\frac{1}{2-x} - \frac{1}{3-x}$
The equations are obviously equal, but some algebraic mani... | We have $- \frac{1}{x-2} + \frac{1}{x-3} = \frac{1}{-1} \times \frac{1}{x-2} + \frac{-1}{-1} \times \frac{1}{x-3} = \frac{1}{(-1)\times (x-2)} + \frac{(-1)\times 1}{(-1)\times (x-3)} = \frac{1}{-x+2} + \frac{-1}{-x+3}= \frac{1}{-x+2} - \frac{1}{-x+3} $. And we have the result by just nature of your own question.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1275071",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 7,
"answer_id": 3
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Generating Series and Recurrence Relation and Closed Form We have the following recurrence relation:
$b_n=2b_{n-1}+b_{n-2}$
and initial conditions $b_0=0, b_1=2$
I use the generating series method to solve as following:
Let
$B(x)=b_0+b_1x+b_2x^2+...+b_nx^n+...$
$-xB(x)=-b_0x-b_1x^2-...-b_{n-1}x^n-...$
$-x^2B(x)=-b_0x... | Shift indices to get:
$$
b_{n + 2} = 2 b_{n + 1} + b_n
$$
Define the generating function:
$$
B(z) = \sum_{n \ge 0} b_n z^n
$$
Multiply the recurrence by $z^n$, sum over $n \ge 0$ and recognize the resulting sums:
$$
\frac{B(z) - b_0 - b_1 z}{z^2}
= 2 \frac{B(z) - b_0}{z} + B(z)
$$
As partial fractions:
$\begin{align}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1275451",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to compute $\sum_{n=0}^{\infty}{\frac{3^n(n + \frac{1}{2})}{n!}}$? I have to compute the series $\displaystyle\sum_{n=0}^{\infty}{\frac{3^n(n + \frac{1}{2})}{n!}}$.
$$\displaystyle\sum_{n=0}^{\infty}{\frac{3^n(n + \frac{1}{2})}{n!}} = \sum_{n=0}^{\infty}{\frac{3^n\frac{1}{2}}{n!}} + \sum_{n=0}^{\infty}{\frac{3^nn}... | Hint:
$$\sum_{n=0}^{\infty}{\frac{3^nn}{n!}}=0+3\sum_{n=1}^{\infty}{\frac{3^{n-1}}{(n-1)!}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1275769",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 3
} |
Find the domain of $x$ in $4\sqrt{x+1}+2\sqrt{2x+3}\leq(x-1)(x^2-2)$
Solve this equation for $x$: $4\sqrt{x+1}+2\sqrt{2x+3}\leq(x-1)(x^2-2)$
I have no idea to solve that, but I know solutions are $x=-1$ or $x\ge 3$.
| It seems the following.
We can straightforwardly verify your answer with Mathcad as follows:
Since $\sqrt{x+1}$ exists, $x\ge -1$. Since the left-hand side of the inequality is non-negative, its right-hand side is non-negative too, that is $x\ge\sqrt{2}$ or $-1\le x\le 1$. Squaring both sides of the inequality, we obta... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1276510",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 0
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Solve $\lim \limits_{x \to 0} (\frac{\sin x}{x})^\frac{1}{1-\cos x}$ without L'Hospital's Rule or Taylor expansion. I solved it with L'Hospital's Rule, but I had to use it 4 times! Is there any way to avoid it?
| Can we use some basics: $\lim \limits_{x \to 0} (1+x)^\frac{1}{x} = e$ and $\lim \limits_{x \to 0} \frac{\sin x}{x} = 1$?
So, we separate $1$: $$\large\lim \limits_{x \to 0} \left(\frac{\sin x}{x}\right)^\frac{1}{1-\cos x} = \lim \limits_{x \to 0} \left(1+\frac{\sin x}{x}-1\right)^\frac{1}{1-\cos x}
= \lim \limits_{x \... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Calculating all possible sums of the numbers $2^0, 2^1, \ldots, 2^{(n-1)}$ Using the simple equation $2^{n-1}$ you get values such as:
$1,2,4,8,16,32,64,128,256,$ etc.
How can I find all possible number combinations within this range? For example, the numbers $1,2,4,8$ give:
\begin{align}
1 &= 1 \\
2 &= 2 \\
1 + 2 &= ... | Computing the decomposition of a nonnegative integer into sums of powers of $2$ is essentially determining the binary (base-$2$)-representation of a number.
For example, we can decompose $42$ as
$$42 = 32 + 8 + 2 = 1 \cdot 2^5 + 0 \cdot 2^4 + 1 \cdot 2^3 + 0 \cdot 2^2 + 1 \cdot 2^1 = 0 \cdot 2^0,$$
so in binary we writ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1278579",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find $\int \frac{1}{x^4+x^2+1} \,\, dx$ Find $$\int \frac{1}{x^4+x^2+1} \,\, dx$$
I tried to find like that:
$\int \frac{1}{x^4+x^2+1} = \int \frac{\frac{1}{2}x + \frac{1}{2}}{x^2+x+1} \,\, dx + \int \frac{-\frac{1}{2}x + \frac{1}{2}}{x^2-x+1} \,\, dx = \frac{1}{2} \Big(\int \frac{2x + 1}{x^2+x+1} - \int \frac{x}{x^2+... | The above method is a good way to solve it, however I think you are trying to solve the wrong question because its integral is very complicated.Take a look at this: http://www.wolframalpha.com/input/?i=integral+1%2F%28x%5E4%2Bx%5E2%2B1%29
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Show that $f:= \frac{1 - 2x}{x^2 - 1}$ is monotonically increasing in the open interval $(-1, 1)$ Show that $f:= \frac{1 - 2x}{x^2 - 1}$ is monotonically increasing in the open interval $(-1, 1)$
To show a function is monotonically increasing, I started by saying that:
A function $f$ is monotonically increasing in an ... | A simpler solution:
$$ f(x)=\frac{1-2x}{x^2-1}=\frac{1}{2}\left(\frac{1}{1-x}-\frac{3}{x+1}\right)\tag{1}$$
and both $\frac{1}{1-x}$ and $\frac{-1}{x+1}$ are increasing over $(-1,1)$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Computing a limit similar to the exponential function I want to show the following limit:
$$
\lim_{n \to \infty}
n
\left[
\left( 1 - \frac{1}{n} \right)^{2n}
- \left( 1 - \frac{2}{n} \right)^{n}
\right]
= \frac{1}{e^{2}}.
$$
I got the answer using WolframAlpha, and it seems to be correct numerically, but I ... | In the same spirit as Lucian, you can consider the general expansion for large values of $n$ $$ \left( 1 - \frac{a}{n} \right)^{b\,n}= \left(1-\frac{a^2 b}{2 n}+\frac{a^3 b (3 a b-8)}{24 n^2}-\frac{a^4 b
\left(a^2 b^2-8 a b+12\right)}{48 n^3}+\cdots\right)e^{-a b} $$ Applied to your case $$\left( 1 - \frac{1}{n} \ri... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1282610",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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"answer_id": 3
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Integer solutions of $x^3 = 7y^3 + 6 y^2+2 y$? Does the equation $$x^3 = 7y^3 + 6 y^2+2 y\tag{1}$$ have any positive integer solutions? This is equivalent to a conjecture about OEIS sequence A245624.
Maple tells me this is a curve of genus $1$, and its Weierstrass form is $s^3 + t^2 + 20 = 0$, with $$ \eqalign{ s = \... | $$ \gcd(y, 7y^2 + 6 y + 2) = 1,2 $$
The first case is odd $y,$ so that $7y^2 + 6y+2$ is odd and $\gcd(y, 7y^2 + 6 y + 2) = 1.$ Both $y$ and $7y^2 + 6 y + 2$ must be cubes. Take $y = n^3.$ We want $7n^6 + 6 n^3 + 2$ to be a cube. Cubes are $1,0,-1 \pmod 9.$ If $n \equiv 0 \pmod 3,$ then $7n^6 + 6 n^3 + 2 \equiv 2 \pmod... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1283013",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
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Trigonometric root of a polynomial
If $4\cos^2 \left(\dfrac{k\pi}{j}\right)$ is the greatest root of the equation
$$x^3-7x^2+14x-7=0$$
where $\gcd(k,j)=1$
Evaluate $k+j$
I tried factorizing the equation but it wasn't of much help.
Btw, the answer given in my book is $k=1$ and $j=14$
Any help will be appreci... | Let $z=\frac{k\pi}{j}$. If $4\cos^2z$ is the root of $x^3−7x^2+14x−7=0$, then
$2\cos z$ is the root of $x^6−7x^4+14x^2−7=0$.
Let's plug $x=2\cos z=2\frac{e^{iz}+e^{-iz}}{2}$ in and see what happens.
The expression comes out to be $e^{-6 i z} \left(-e^{2 i z}+e^{4 i z}-e^{6 i z}+e^{8 i z}-e^{10 i z}+e^{12 i z}+1\right)=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1285202",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Roots of a polynomial whose coefficients are ratios of binomial coefficients
Prove that $\left\{\cot^2\left(\dfrac{k\pi}{2n+1}\right)\right\}_{k=1}^{n}$ are the roots of the equation
$$x^n-\dfrac{\dbinom{2n+1}{3}}{\dbinom{2n+1}{1}}x^{n-1} + \dfrac{\dbinom{2n+1}{5}}{\dbinom{2n+1}{1}}x^{n-2} - \ldots \ldots \ldots + \d... | We have
$$\frac{e^{m i \theta}}{\sin^m \theta} = \frac{(\cos \theta + i \sin \theta)^m}{\sin^m \theta} = (\cot \theta + i)^m $$
for $m \in \mathbb{N}$ and $\theta \in \mathbb{R}$. Take $m=2n+1$ and $\theta = k\pi/(2n+1)$ to get
$$\frac{(-1)^k}{\sin^{2n+1}\theta} = \sum_{r=0}^{2n+1} \binom{2n+1}{r}i^r \cot^{2n+1-r} \the... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1285406",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Determinant of symmetric matrix $(A-\lambda I)$ If we have a matrix $(A-\lambda I)$ which is:
$\left(
\begin{array}{ccc}
1-\lambda & -1 & 2 \\
-1 & 1-\lambda & 2 \\
2 & 2 & 2-\lambda \\
\end{array}
\right)
$
Then it's determinant can be written as : $(-1)^n(\lambda-\lambda_1)(\lambda-\lambda_2)(\lambda-\lambda_3)$. ... | $\left( \begin{array}{ccc}
1-\lambda & -1 & 2 \\
-1 & 1-\lambda & 2 \\
2 & 2 & 2-\lambda \\
\end{array} \right)
$ = $\left( \begin{array}{ccc}
2-\lambda & -2+\lambda & 0 \\
-1 & 1-\lambda & 2 \\
2 & 2 & 2-\lambda \\
\end{array} \right)
=(-2+\lambda)
\left( \begin{array}{ccc}
-1 & 1 & 0 \\
-1 & 1-\lambda & 2 \\
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1286219",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Cannot understand an Integral $$\displaystyle \int _{ \pi /6 }^{ \pi /3 }{ \frac { dx }{ \sec x+\csc x } } $$
I had to solve the integral and get it in this form.
My attempt:
$$\int _{ \pi /6 }^{ \pi /3 }{ \frac { dx }{ \sec x+\csc x } } $$
$$=\int _{\frac{\pi}{6}}^{ \frac{\pi}{3}} \dfrac{\sin x \cos x }{ \sin x+\cos ... | Integrate directly as follows
\begin{align}
\int _{ \pi /6 }^{ \pi /3 }{ \frac { dx }{ \sec x+\csc x } }
& =\frac12 \int _{ \pi /6 }^{ \pi /3 } \frac{(\sin x +\cos x)^2-1 }{ \sin x+\cos x }dx \\
&= \frac{1}2\int _{ \pi /6 }^{ \pi /3 } \left(\sqrt2\cos(\frac\pi4-x)
-\frac1{\sqrt2\cos(\frac\pi4-x)}\right)dx\\
&= \frac12... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1288034",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 3
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Find the diameter of iron ball.
Iron weights $8$ times the weight of oak, Find the diameter of an iron ball whose weight is equal to that of a ball of oak $18$ cm in diameter.
$a)\quad 4.5\quad cm \\\color{green}{b)\quad 9 \quad cm}\\c)\quad 12\quad cm\\d) \quad 15 \quad cm$
I tried
$\dfrac43\times \pi\... | Since iron weighs $8$ times what oak weighs, you need the iron ball to have $\frac{1}{8}$ the volume of the oak ball.
$$\begin{align}
\frac{4}{3} \times \pi \times \left(\frac{d}{2}\right)^3 &=
\color{red}{\frac{1}{8}} \times \frac{4}{3} \times \pi \times (9)^3 \\
\frac{d^3}{8} &= \frac{1}{8} \times (9)^3 \\
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1288210",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find the coordinates for $v$ in the subspace $W$ spanned by the following vectors:
I'm confused as to how I solve this. I was told to project each $u$ onto $v$ but how would that get me to the answer that's required?
| Resolve the data vector into its components in the $u$ coordinate system. because the $u$ vectors are orthogonal, we can solve for the projection terms independently.
$$
\begin{array}{cccccccccc}
v & = &
\frac{\langle v,u_{1} \rangle}{\langle u_{1},u_{1} \rangle} & u_{1} & + & \frac{\langle v,u_{2} \rangle}{\langle ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1288920",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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Derivative of $f(x) = \frac{\cos{(x^2 - 1)}}{2x}$ Find the derivative of the function $$f(x) = \frac{\cos{(x^2 - 1)}}{2x}$$
This is my step-by-step solution: $$f'(x) = \frac{-\sin{(x^2 - 1)}2x - 2\cos{(x^2 -1)}}{4x^2} = \frac{2x\sin{(1 - x^2)} - 2\cos{(1 - x^2)}}{4x^2} = \frac{x\sin{(1 - x^2)} - \cos{(1 - x^2)}}{2x^2} ... | In the first step, when you take the derivative of $\cos(x^2 - 1)$ in the quotient rule calculation, you're getting $-\sin(x^2-1)$, when you should be getting $-2x\sin(x^2 - 1)$ by the chain rule. This is then multiplied by the $2x$ from the denominator, giving $-4x^2\sin(x^2 - 1)$.
This is the only error--you'll notic... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1293272",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Properties of Jacobi symbol Let $\left(\frac{a}{n}\right) $ be Jacobi symbol .
It is well known that Jacobi symbol for $a=-1$ and $a=2$ satisfies the following:
$\left(\frac{-1}{n}\right) =
\begin{cases}
1, & \text{if } n \equiv 1 \pmod{4} \\
-1, & \text{if } n \equiv 3 \pmod{4}
\end{cases}$
and ,
$\left(\frac{2}{n}\r... | Yes, it does for all $a$ - that's arguably the point of quadratic reciprocity.
For one example,
$$\left(\frac{5}{n}\right)=\left(\frac{n}{5}\right)(-1)^{\frac{n-1}{2}\cdot\frac{5-1}{2}}=\left(\frac{n}{5}\right)=\begin{cases}
\hphantom{-}1&\text{if }n\equiv 1,4\bmod 5\\
\hphantom{-}0&\text{if }n\equiv 0\bmod 5\\
-1 & \t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1294271",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Show that there exists a vector $v$ such that $Av\neq 0$ but $A^2v=0$ Let $A$ be a $4\times 4$ matrix over $\mathbb C$ such that $rank A=2$ and $A^3=A^2\neq 0$.Suppose that $A$ is not diagonalisable. Then
Show that there exists a vector $v$ such that $Av\neq 0$ but $A^2v=0$
My try:$\dim \operatorname{Im}(A)+\dim \ker ... | There are only two possible Jordan forms:
$$\pmatrix{0&1&0&0\\0&0&0&0\\0&0&0&0\\0&0&0&1}, \quad \pmatrix{0&0&0&0\\0&0&0&0\\0&0&1&1\\0&0&0&1}$$ only the first first one satisfies $A^2 = A^3.$
$\bf p.s.$
the nonzero vector $v$ you are looking for is in $\ker(A^2)\setminus \ker(A)$ we have $dim(\ker(A^2) = 3,dim(\ker(A) ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1295229",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Find the number of possible values of $a$
Positive integers $a, b, c$, and $d$ satisfy $a > b > c > d, a + b + c + d = 2010$,
and $a^2 − b^2 + c^2 − d^2 = 2010$. Find the number of possible values of $a.$
Obviously, factoring,
$$(a-b)(a+b) + (c-d)(c+d) = 2010$$
$$a + b + c + d = 2010$$
Substituting you get:
$$(a-b... | Denote $$b=a-x, \quad d=c-y. \qquad(x,y\in\mathbb{N}).\tag{1}$$
Then
$$
a+b+c+d = 2a-x+2c-y = 2010,\\
a^2-b^2+c^2-d^2=(2a-x)x+(2c-y)y=2010.\tag{2}
$$
Since $x,y\in\mathbb{N}$, then unique solution of $(2)$ has $$x=y=1.\tag{3}$$
So, solution has form
$$
(a,\;a-1,\;c,\;c-1).\tag{4}
$$
$(2),(4)\Rightarrow$
$$a+c-1=1005,$$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1295944",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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What mistake have I made when trying to evaluate the limit $\lim \limits _ {n \to \infty}n - \sqrt{n+a} \sqrt{n+b}$? Suppose $a$ and $b$ are positive constants.
$$\lim \limits _ {n \to \infty}n - \sqrt{n+a} \sqrt{n+b} = ?$$
What I did first:
I rearranged $\sqrt{n+a} \sqrt{n+b} = n \sqrt{1+ \frac{a}{n}} \sqrt{1+ \frac{b... | Another approach to solving this is via L'Hopital's rule. For this, we first need to do a bit of rearranging...
$$
\lim_{n\to\infty} \frac{1-\sqrt{1+a/n}\sqrt{1+b/n}}{1/n}
$$
which, letting $m=1/n$, gives us
$$
\lim_{m\to0} \frac{1-\sqrt{1+am}\sqrt{1+bm}}{m}
$$
It is from here that L'Hopital's rule (in this case, it's ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1297035",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 7,
"answer_id": 6
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Solving differential equation $x'=\frac{x+2t}{x-t}$ I am trying to solve the following differential equation:
$$x'=\frac{x+2t}{x-t}$$
with initial value condition: $x(1)=2$
This is what I have so far:
Substitution: $u=\frac{x}{t}$
$$\implies u't+u=\frac{2t+tu}{-t+tu}$$
Separation of variables:
$$\implies \frac{u'(u-1)}... | here is another way to do this. we have $$\frac{dx}{dt} = \frac{x+2t}{x-t}. $$ we can rewrite this as an exact differential in the form $$0=(t-x) dx + (x+2t) dt = f_x\, dx + f_t \, dt $$ we have $$f_{xt} = f_{tx} =1 $$
we can integrate $$ f_x = t-x \to f = tx-\frac12x^2 + C(t) \to f_t = x + C'=x+2t\to C= t^2 $$ to ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1304346",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Yamabe's equation This is PDE Evans, 2nd edition: Chapter 9, Exericse 8(a).
*(a) Assume $n \ge 3$. Find a constant $c$ such that $$u(x) := (1+|x|^2)^{\frac{2-n}2}$$ solves Yamabe's equation $$-\Delta u = cu^{\frac{n+2}{n-2}} \quad \text{in }\mathbb{R}^n.$$ Note the appearance of the critical exponent $\frac{n+2}{n... | You should have $$u_{x_i x_i} = (2-n)\left[-n(1+|x|^2)^{-\frac{n+2}2} x_i^2 + (1+|x|^2)^{-\frac n2}\right].$$
That is, not $x_i$, but $x_i^2$ in the squred bracket. Therefore, we have
$$-\Delta u=-\sum_{i=1}^nu_{x_i x_i} = -(2-n)\left[-n(1+|x|^2)^{-\frac{n+2}2} |x|^2 + n(1+|x|^2)^{-\frac n2}\right]=n(n-2)(1+|x|^2)^{-\f... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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"answer_id": 0
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For every $n$ and $x$ there is an $m$ such that $3^n\mid 4^mx + (4^m-1)/3$ I am interested to know if for any $x\in 2\mathbb N +1$ and $n\in\mathbb N$ there exists a number $m\in \mathbb N\cup\{0\}$ such that
$3^n$ divides $$4^mx + \frac{4^m-1}{3}$$
I can easily show this if $x\equiv 1 \mod 3^n$, however I don't seem t... | Generalizing slightly: Let $p$ be an odd prime (in the question we have $p=3$) and $n$ be a positive integer. We want to show that for every integer $x$ there exists an $m\in\mathbb N$ such that
$$p^n\mid (p+1)^mx+\dfrac{(p+1)^m−1}p$$
We multiply this by $p$ on both sides to get
$$p^{n+1}\mid (p+1)^m(px+1)−1$$
or in ot... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1305667",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Remainder of $(1+x)^{2015}$ after division with $x^2+x+1$
Remainder of $(1+x)^{2015}$ after division with $x^2+x+1$
Is it correct if I consider the polynomial modulo $5$
$$(1+x)^{2015}=\sum\binom{2015}{n}x^n=1+2015x+2015\cdot1007x^2+\cdots+x^{2015}$$
RHS stays the same and then The remainder must be of the form $Ax+B... | Method $\#1:$
Observe that $$1+x\equiv-x^2\pmod{x^2+x+1}\text{ and }x^3\equiv1$$
$$\implies(1+x)^{3n+2}\equiv(-x^2)^{3n+2}\equiv(-1)^{3n+2}(x^3)^{2n+1}\cdot x\equiv(-1)^nx$$ as $3n+2\equiv n\pmod2$
Method $\#2:$
Observe that $$(1+x)^2\equiv x\pmod{x^2+x+1}\text{ and }x^3\equiv1$$
$$\implies(1+x)^{6n+5}=\{(1+x)^2\}^{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1306470",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 2
} |
Determine whether or not the limit exists: $\lim_{(x,y)\to(0,0)}\frac{(x+y)^2}{x^2+y^2}$
Determine whether or not the limit $$\lim_{(x,y)\to(0,0)}\frac{(x+y)^2}{x^2+y^2}$$
exists. If it does, then calculate its value.
My attempt:
$$\begin{align}\lim \frac{(x+y)^2}{x^2+y^2} &= \lim \frac{x^2+y^2}{x^2+y^2} + \lim \fr... | Here's another way
$$ \lim\limits_{(x,y)\to(0,0)} \frac{(x+y)^2}{x^2+y^2} $$
Using polar coordinates, we have
$$ \lim\limits_{r\to 0^+} \frac{\left(r\cos\phi+r\sin\phi\right)^2}{r^2\cos^2\phi + r^2\sin^2\phi} $$
$$ = \lim\limits_{r\to 0^+} \frac{r^2\left(\cos\phi+\sin\phi\right)^2}{r^2\left(\cos^2\phi + \sin^2\phi\righ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1307149",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 6,
"answer_id": 1
} |
Linear Transformation Matrix with polynomials A linear transformation $T : P_2 \to P_2$ has matrix with respect to $S$ given by:
$$[T]\,( S) = \begin{bmatrix}
1/2&-3&1/2\\
-1&4&-1\\
1/2&2&1/2\\
\end{bmatrix}
$$
How do you find $T(a+bx+cx^2)$?
Thank you!!
| Every polynomial $S$ of degree $2$ s.t. $S \in P_2$ can be represented as a vector in three dimensional space:
$$
S = a + bx + cx^2 \quad \iff \quad
\begin{bmatrix}
a \\ b \\ c \\
\end{bmatrix}
\cdot
\begin{bmatrix}
1 \\ x \\ x^2 \\
\end{bmatrix},
$$
therefore we can associate $S$ with a 3D vector
$$
S \longleftr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1307305",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
find the minimum value of $\sqrt{x^2+4} + \sqrt{y^2+9}$ Question:
Given $x + y = 12$, find the minimum value of $\sqrt{x^2+4} + \sqrt{y^2+9}$?
Key:
I use $y = 12 - x$ and substitute into the equation, and derivative it.
which I got this
$$\frac{x}{\sqrt{x^2+4}} + \frac{x-12}{\sqrt{x^2-24x+153}} = f'(x).$$
However, aft... | Setting $f'(x)=0$ gives
$$\frac{x}{\sqrt{x+4}}+\frac{x-12}{\sqrt{x^2-24x+153}}=0 $$
or
$$\frac{x}{\sqrt{x+4}}=-\frac{x-12}{\sqrt{x^2-24x+153}}\tag1$$
Squaring both sides of $(1)$ gives
$$\left(\frac{x}{\sqrt{x^2+4}}\right)^2=\left(\frac{x-12}{\sqrt{x^2-24x+153}}\right)^2$$
which can be rewritten as
$$\frac{x^2}{x^2+4}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1307398",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 2
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$5x\big(1+\frac{1}{x^2 +y^2}\big)=12$ ; $5y\big(1-\frac{1}{x^2 +y^2}\big)=4$ find $x$ and $y$ I already tried to solve using substitution and cross multiplication method .
I got the first simplified
(1)$$\frac{12}{5x}=1+ \frac{1}{x^2} +y^2$$
(2)
$$\frac{4}{5y}=1- \frac{1}{x^2+y^2}$$
Adding (1) and (2)
$$12y +4x= 50xy$... | Let $x=r\cos\varphi,\, y=r\sin\varphi$
$$\begin{cases}
\frac{12}{5r\cos\varphi}=\frac{r^2+1}{r^2}\\
\frac{4}{5r\sin\varphi}=\frac{r^2-1}{r^2}
\end{cases}$$
$$\begin{cases}
\frac{5\cos\varphi}{12}=\frac{r}{r^2+1}\\
\frac{5\sin\varphi}{4}=\frac{r}{r^2-1}
\end{cases}$$
$$\begin{cases}
\cos\varphi=\frac{12}{5}\cdot\frac{r}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1308201",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Showing that $\sqrt[105]{105}>\sqrt[106]{106}$ How can one prove that
$$\sqrt[105]{105}>\sqrt[106]{106}\text{ ?}$$
Induction on the statement $$\sqrt[n]{n}>\sqrt[n+1]{n+1} \text{ for } n \in \mathbb{N}| n>2$$ would yield $\sqrt[3]{3}>\sqrt[4]{4}$ at the base step $n=3$, which we cannot assume.
So, based on the proper... | All you need here is the binomial theorem for positive integer exponents:
$$(1+x)^n = \sum_{r=0}^n \binom{n}{r}x^r$$
Fix $n \ge 3$, and put $x = \frac{1}{n}$:
$$\left(1+\frac{1}{n}\right)^n = \sum_{r=0}^n \binom{n}{r}n^{-r}$$
Now, $\binom{n}{r} \le n^r$ for all $r=0,\ldots,n$ (easily proven by induction; or just look a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1310555",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 6,
"answer_id": 3
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What is the most unusual proof you know that $\sqrt{2}$ is irrational? What is the most unusual proof you know that $\sqrt{2}$ is irrational?
Here is my favorite:
Theorem: $\sqrt{2}$ is irrational.
Proof:
$3^2-2\cdot 2^2 = 1$.
(That's it)
That is a corollary of
this result:
Theorem:
If $n$ is a positive integer... | This proof will look much better if someone can add a diagram to it.
Suppose $\sqrt2=a/b$ with $a$ and $b$ positive integers chosen as small as possible. Draw a square of side $a$. In the upper left corner, place a square of side $b$, and another one in the lower right corner. The two $b$-squares have total area $2b^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1311228",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "114",
"answer_count": 19,
"answer_id": 11
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Evaluate the directional derivative of $f$ for the points and directions specified Evaluate the directional derivative of $f$ for the points and directions specified:
(a) $f(x,y,z)=3x-5y+2z$ at $(2,2,1)$ in the direction of the outward normal to the sphere $x^2+y^2+z^2=9.$
(b) $f(x,y,z)=x^2-y^2$ at a general point of t... | (a)
The outward normal of the sphere given is $\left(\frac{x}{3},\frac{y}{3}, \frac{1}{3}\sqrt{9-x^2-y^2}\right)$. The same at the given point is $\overline n=\left(\frac{2}{3},\frac{2}{3}, \frac{1}{3}\right).$ The vector of the partial derivatives is $\nabla f =(3,-5,2).$ The directional derivative is then
$$\nabla f ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1312172",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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Generating function for $l_{n+1} = 3l_n+1$. I have the sequence $l_{n+1} = 3l_n+1$ for $l_0 = 0$ or $1$ (This just shifts the sequence one index back or forward.) The first terms are $(0),1,4,13,40,121,364,\ldots$
So I am looking for an expression for $S(x) = \sum\limits_{n=0}^\infty l_n x^n$ I tried to standard approa... | Your mistake is that $\sum_{n=0}^{\infty} x^{n+1} = \frac{x}{1-x}$, not $\frac{1+x}{1-x}$.
Fixing that, the rest of the steps are:
$$S(x) = \frac{l_0}{1 - 3x} + \frac{x}{(1-x)(1-3x)}$$
With $l_0 = 0$, that's:
$$\begin{split}
S(x) &= \frac{-\frac12}{1-x} + \frac{\frac12}{1-3x} \\
&= -\frac12\sum_{n=0}^{\infty}x^n + \fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1312313",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Is this a valid partial fraction decomposition?
Write $\dfrac{4x+1}{x^2 - x - 2}$ using partial fractions.
$$ \frac{4x+1}{x^2 - x - 2} = \frac{4x+1}{(x+1)(x+2)} = \frac{A}{x+1} + \frac{B}{x-2} = \frac{A(x-2)+B(x+1)}{(x+1)(x-2)}$$
$$4x+1 = A(x-2)+B(x+1)$$
$$x=2 \Rightarrow 4 \cdot2 + 1 = A(0) + B(3) \Rightarrow B = 3$$... | $A(x-2)+B(x+1) = 4x+1$ gives us a system of two equations:
$$\begin{split}
A+B &= 4 \\
-2A + B &= 1
\end{split}$$
But really the first equation as really $(A+B)x = 4x$, so the substitution $x=2$ simply gives us the system:
$$\begin{split}
2A+2B &= 8 \\
-2A + B &= 1 \\
\end{split}$$
Which conveniently sums to $3B = 9$, ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1313454",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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"answer_id": 1
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Prove $(a^2+b^2+c^2+d^2)^2≥(a+b)(b+c)(c+d)(d+a)$ I've been unsuccessfully trying to solve this contest-style problem for a while.
Tried different substitutions and the such, but nothing helped. I presume the solution is related to Cauchy-Schwarz? Anyway, any hints would be appreciated!
Given $a, b, c, d \in \mathbb{R... | By the AM-GM inequality:
$$ (a+b)(c+d) \leq \frac{1}{4}(a+b+c+d)^2, $$
$$ (b+c)(d+a) \leq \frac{1}{4}(a+b+c+d)^2, $$
and by the AM-QM inequality:
$$ \frac{1}{4}(a+b+c+d)^2 \leq (a^2+b^2+c^2+d^2).$$
The claim easily follows.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1314431",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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Simple mod problem It’s kind of a silly question but I can't find a simple way for finding the value of variable $d$ .
$(5*d) \mod 8 = 1$
I normally just do this recursively by saying $d=d+1$ until I get the right answer. Is there a better way of solving this?
| To find the multiplicative inverse of $5$ modulo $8$, use the Extended Euclidean Algorithm.
First, solve for the greatest common divisor of $5$ and $8$, which is $1$ since they are relatively prime.
\begin{align*}
8 & = 1 \cdot 5 + 3\\
5 & = 1 \cdot 3 + 2\\
3 & = 1 \cdot 2 + 1\\
2 & = 2 \cdot 1
\end{align*}
Now, work... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1315755",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
When is the series converges?
Let the series $$\sum_{n=1}^\infty \frac{2^n \sin^n x}{n^2}$$. For $x\in (-\pi/2, \pi/2)$, when is the series converges?
By the root-test:
$$\sqrt[n]{a_n} = \sqrt[n]{\frac{2^n\sin^n x}{n^2}} = \frac{2\sin x}{n^{2/n}} \to 2\sin x$$
Thus, the series converges $\iff 2\sin x < 1 \iff \sin x ... | I like the ratio test here:
$$\begin{split}
L &= \lim_{n\to\infty} \left|\frac{2^{n+1} \sin^{n+1} x}{(n+1)^2} \frac{n^2}{2^n\sin^n{x}} \right| \\
&=\lim_{n\to\infty} \left|\frac{2n^2\sin{x}}{(n+1)^2}\right| \\
&=\lim_{n\to\infty} \left|\frac{2\sin{x}\cdot n^2}{n^2}\right| \\
&= |2\sin{x}|
\end{split}$$
$L < 1 \iff |2\s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1315805",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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What two numbers when multiplied gives $-25$ and when added, $-10$? Factors of $25$ are; $1, 5, 25$
$$5 \times 5 = 25$$
$$5 + 5 = 10$$
$$-5 \times 5 = -25$$
$$-5 + 5 = 0$$
How can I solve this?
Thanks
| Well one cannot solve this if only integers may be used.
The answer is:
$$(a,b) = (5(\sqrt{2} - 1), -5(1+\sqrt{2}))$$
We can see this as we have
$$a*b = -25,\qquad a+b =-10.$$
From this it follows that $a = -10-b$, thus
$$-10b -b^2 = -25, \Rightarrow b^2+10b-25 = 0.$$
This can be solved quite easily, as this gives
$$b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1316134",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
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Help with solving this Recurrence Relation I really need help with this question
Would anyone please give a simple step-by-step on how to solve this Recurrence Relation??
$a_n = 2a_{n-1} - 2a_{n-2}$ where $a_0 = 1$ and $a_1 = 3$
It would really be great if someone could explain how to solve this
So far I made it ... | A generatingfunctionological solution would be to define $A(z) = \sum_{n \ge 0} a_n z^n$, write the recurrence shifting indices as:
$$
a_{n + 2} = 2 a_{n + 1} - 2 a_n
$$
Multiply the recurrence by $z^n$, sum over $n \ge 0$ and recognize some sums:
$$
\frac{A(z) - a_0 - a_1 z}{z^2}
= 2 \frac{A(z) - a_0}{z} - 2 A(z)
$$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1316439",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
solving difficult complex number proving if $z= x+iy$ where $y \neq 0$ and $1+z^2 \neq 0$, show that the number $w= z/(1+z^2)$ is real only if $|z|=1$
solution :
$$1+z^2 = 1+ x^2 - y^2 +2xyi$$
$$(1+ x^2 - y^2 +2xyi)(1+ x^2 - y^2 -2xyi)=(1+ x^2 - y^2)^2 - (2xyi)^2$$
real component
$$(1+ x^2 - y^2)x - yi(2xyi) = x + x^3... | It's easier to work with $w=\dfrac{1+z^2}z=\dfrac1z+z$ (that doesn't change the claim).
Taking the imaginary part,
$$-\frac y{x^2+y^2}+y=y\left(1-\frac1{x^2+y^2}\right)=0,$$
i.e.
$$x^2+y^2=1.$$
Or, in polar coordinates,
$$-\frac1r\sin(\phi)+r\sin(\phi)=0\implies r=1.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1316782",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 3
} |
Solve $p_{n+1} + \frac 16 p_n = \frac 1 2 (\frac 5 6 ) ^{n-1}$ I'm trying to solve:
$$p_{n+1} + \frac 16 p_n = \frac 1 2 \left(\frac 5 6 \right) ^{n-1}$$
with initial condition: $p_1 = 1$.
First, I search particular solution of the form $p_n^* = \lambda (\frac 5 6 ) ^n$. I found $\lambda = \frac 3 5$.
Next, I know tha... | Your answer is correct! Here's a full solution to the recurrence.
We have the following:
$$
\begin{align*}
p_{n+1} + \dfrac{1}{6} p_n &= \dfrac{1}{2}\left(\dfrac{5}{6}\right)^{n-1} \\
p_{n} + \dfrac{1}{6} p_{n-1} &= \dfrac{1}{2}\left(\dfrac{5}{6}\right)^{n-2} \\
\end{align*}
$$
Dividing the two, we have:
$$
\begin{alig... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1320529",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
System of equations with complex numbers-circles The system of equations
\begin{align*}
|z - 2 - 2i| &= \sqrt{23}, \\
|z - 8 - 5i| &= \sqrt{38}
\end{align*}
has two solutions $z_1$ and $z_2$ in complex numbers. Find $(z_1 + z_2)/2$.
So far I have gotten the two original equations to equations of circles,
$(a-2)^2 +(b-2... | You're given the distances of the solution points to $A = 2+2i$ and $B = 8+5i.$
One of the solutions and these two complex numbers give you a triangle in the complex plane, and you know the lengths of all the sides.
We can use Law of Cosines to find the angle $\theta$ that has $2+2i$ as its vertex:
$$38 = 45 + 23 - 2 \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1323118",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
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High School Trigonometry ( Law of cosine and sine) I am preparing for faculty entrance exam and this was the question for which I couldn't find the way to solve (answer is 0). I guess they ask me to solve this by using the rule of sine and cosine:
Let $\alpha$, $\beta$ and $\gamma$ be the angles of arbitrary triangle... | Consider the first summand, to begin with:
$$
C=\frac{b-2a\cos\gamma}{a\sin\gamma}
$$
The law of cosines tells you that
$$
c^2=a^2+b^2-2ab\cos\gamma
$$
so we can write
$$
b-2a\cos\gamma=\frac{c^2-a^2}{b}
$$
and so we have
$$
C=\frac{c^2-a^2}{ab\sin\gamma}
$$
The law of sines is
$$
\frac{a}{\sin\alpha}=\frac{b}{\sin\bet... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1326863",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 8,
"answer_id": 2
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Solution to Fibonacci Recursion Equations Let the sequence $(a_n)_{n\geq0}$ of the fibonacci numbers: $a_0 = a_1 = 1, a_{n+2} = a_{n+1} + a_n, n \geq 0$
Show that:
i) $$a^2_n - a_{n+1}a_{n-1} = (-1)^n \text{ for }n\geq1$$
I try to show this with induction:
$n=1:$
\begin{align*}
a^2_1 - a_{2}a_{0}&=1 ... | Since the first term of the induction process for part one has been established the t remains to be seen for general $n$. Now,
\begin{align}
a_{n+1}^{2} - a_{n} \, a_{n+2} &= a_{n+1}^{2} - a_{n} \, ( a_{n+1} + a_{n} )\\
&= a_{n+1} ( a_{n+1} - a_{n} ) - a_{n}^{2} \\
&= a_{n-1} \, a_{n+1} - a_{n}^{2} = (-1)^{n+1}.
\end{a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1328131",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
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Given $x^2 + 4x + 6$ as factor of $x^4 + ax^2 + b$, then $a + b$ is I got this task two days ago, quite difficult for me, since I have not done applications of Vieta's formulas and Bezout's Theorem for a while. If can someone solve this and add exactly how I am supposed to use these two theorem's on this task, I would ... | If $x^2 + 4x + 6$ is factor of $x^4 + ax^2 + b$, then it must satify
$$x^4 + ax^2 + b=(x^2 + 4x + 6)Q(x)$$ where $Q(x)$ is a second degree polynominal. You do not need to find $Q(x)$
$x^2 + 4x + 6=0$
$x^2=-(4x + 6)$
$$x^4 + ax^2 + b=(x^2 + 4x + 6)Q(x)$$
$$(-(4x + 6))^2 - a(4x + 6)+ b=0$$
$$16x^2+48x+36 - 4ax -6a+ b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1328210",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 5
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Comparing two large numbers Can you compare two large exponential numbers, like $5^{44}$ and $4^{53}$ without taking their logs?
| One approach is to figure that, roughly, $2^{10} \approx 10^3$, and $5^9 \approx 2,000,000 = 2 \cdot 10^6$.
Then,
\begin{align}
4^{53} &= (2^2)^{53} \\
&= 2^{106} \\
&= 2^{100} \cdot 2^6 \\
&= (2^{10})^{10} \cdot 2^6 \\
&\approx (10^3)^{10} \cdot 2^6 \\
&= 10^{30} \cdot 2^6
\end{align}
By way of comparison,
\begin{al... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1329260",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
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Integration of $\int \frac {x\arctan x}{(\sqrt{1+x^2})^3}$ How can I evaluate
$$\int \frac{x \arctan x}{(\sqrt{1+x^2})^3}\,dx$$
I think it must be done by parts, but I can't get any appropriate substitution or such.
| There is no magic in this calculation, the key is familiarity, familiar with $\int\frac{dx}{1+x^2}=\arctan(x)+C$.
Let $\theta=\arctan(x)$, then we have $d\theta=\frac{dx}{1+x^2}$, $\frac{x}{\sqrt{1+x^2}}=\sin\theta$ and $\frac{1}{\sqrt{1+x^2}}=\cos\theta$, so
$$
I=\int\frac{x\theta}{\sqrt{1+x^2}}\centerdot\frac{dx}{1+x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1329604",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find the cubic equation of $x=\sqrt[3]{2-\sqrt{3}}+\sqrt[3]{2+\sqrt{3}}$ Find the cubic equation which has a root $$x=\sqrt[3]{2-\sqrt{3}}+\sqrt[3]{2+\sqrt{3}}$$
My attempt is
$$x^3=2-\sqrt{3}+3\left(\sqrt[3]{(2-\sqrt{3})^2}\right)\left(\sqrt[3]{(2+\sqrt{3})}\right)+3\left(\sqrt[3]{(2-\sqrt{3})}\right)\left(\sqrt[3]{(... | Probably it is better to notice, for first, that the product between $2-\sqrt{3}$ and $2+\sqrt{3}$ is one. After that, the minimal polynomial of $a=\sqrt[3]{2+\sqrt{3}}$ over $\mathbb{Q}$ is quite trivially:
$$ p(x)=(x^3-2)^2-3 = x^6-4x^3+1 $$
hence $a$ is a root of:
$$ x^3+x^{-3}-4 = \left(x+\frac{1}{x}\right)^3 - 3\l... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1331417",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 8,
"answer_id": 3
} |
Find the remainder when $f(x)$ is divided by $x^2+x-2$
When $f(x)$ is divided by $x-1$ and $x+2$, the remainders are $4$ and $-2$ respectively. Find the remainder when $f(x)$ is divided by $x^2+x-2$.
Please help. The answer is $2x+2$.
I tried to understand the link between the first sentence and the second sentence b... | Use the Chinese Remainder theorem (Wikipedia link) in the ring of polynomials.
You'll want to look at the polynomials $g=\frac{1}{3}(x+2)$ and $h=-\frac{1}{3}(x-1)$, which satisfy
$$\begin{align*}
g&\equiv 1\bmod (x-1) & h&\equiv 0\bmod(x-1)\\
g&\equiv 0\bmod (x+2) & h&\equiv 1\bmod (x+2)
\end{align*}$$
Thus, for any $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1333007",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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Triple fractions I've got this simple assignment, to find out the density for a give sphere with a radius = 2cm and the mass 296g. It seems straightforward, but it all got hairy when i've got to a fraction with three elements(more precisely a fraction divided by a number actually this was wrong, the whole point was tha... | $$\frac{a}{\frac{b}{c}}\ne\frac{\frac{a}{b}}{c} \tag 1$$
The left-hand side of $(1)$ can be written as
$$\frac{a}{\frac{b}{c}}=\frac{ac}{b}$$
whereas the right-hand side of $(1)$ can be written as
$$\frac{\frac{a}{b}}{c}=\frac{a}{bc}$$
Let's look at an example: Suppose $a=3$, $b=6$, and $c=2$. Then, we have
$$\frac{a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1333886",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 0
} |
Distance between orthocenter and circumcenter.
Let $O$ and $H$ be respectively the circumcenter and the orthocenter of triangle $ABC$. Let $a$, $b$ and $c$ denote the side lengths. We are given that $a^2+b^2+c^2=29$ and the circumradius is $R=9$. We need to find $OH^2$.
I know that there is formula $OH^2=9R^2-(a^2+b^2... |
since
$$\angle HAO=|B-C|.OA=R,AH=2R\cos{A}$$
use cosin theorem we have
\begin{align*}
OH^2&=AH^2+AO^2-2AH\cdot AO\cos{|B-C|}
=4R^2\cos^2{A}+R^2-4R^2\cos{A}\cos{(B-C)}\\
&=5R^2-4R^2\sin^2{A}+4R^2\cos{(B+C)}\cos{(B-C)}\\
&=9R^2-4R^2(\sin^2{A}+\sin^2{B}+\sin^2{C})\\
&=9R^2-(a^2+b^2+c^2)
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1334039",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
How to find sum of the infinite series $\sum_{n=1}^{\infty} \frac{1}{ n(2n+1)}$ $$\frac{1}{1 \times3} + \frac{1}{2\times5}+\frac{1}{3\times7} + \frac{1}{4\times9}+\cdots $$
How to find sum of this series?
I tried this: its $n$th term will be = $\frac{1}{n}-\frac{2}{2n+1}$; after that I am not able to solve this.
| First, we can rewrite the partial sum as an integral
$$\sum_{n=1}^N \frac{1}{n(2n+1)} = 2\sum_{n=1}^N \left(\frac{1}{2n} - \frac{1}{2n+1}\right)
= 2\sum_{n=1}^N \int_0^1 (z^{2n-1} - z^{2n}) dz\\
= 2 \int_0^1 z(1-z)\left(\sum_{n=0}^{N-1} z^{2n}\right) dz
= 2 \int_0^1 \frac{z}{1+z}( 1 - z^{2N} ) dz
$$
Notice the $N$ depe... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1334870",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 6,
"answer_id": 0
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Simplify P(n), where n is a positive integer : $ P(x)=\sum \limits_{k=1}^\infty \arctan\left(\frac{x-1}{(k+x+1)\sqrt{k+1}+(k+2)\sqrt{k+x}}\right). $ This is what I have tried, but I don't know what to do next, so I need help
:
$ P(x)=\sum \limits_{k=1}^\infty \arctan\left(\frac{x-1}{(n+x+1)\sqrt{n+1}+(n+2)\sqrt{n+x}}\r... | I'll assume $x$ is an positive integer. Define
$$\begin{cases}
a_n &= \frac{x-1}{(n+x+1)\sqrt{n+1}+(n+2)\sqrt{n+x}}\\
u_n &= \sqrt{n+x}\\
v_n &= \sqrt{n+1}
\end{cases}$$
We have
$$a_n = \frac{u_n^2-v_n^2}{(u_n^2+1)v_n + (v_n^2+1)u_n} = \frac{u_n-v_n}{1 + u_nv_n}
$$
This leads to
$$\tan^{-1}a_n
= \tan^{-1}\left(\frac{u... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1337393",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
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Let $z \in \Bbb Z_m$, when is $z^2 \equiv 1$? Let $z \in \Bbb Z_m$. When is $z^2=1, (z\neq1)$?
I know that for $m$ prime, $z=p-1$ is it's own inverse, but what about nonprime $m$?
Is $p-1$ the only self inverse element in $\Bbb Z_p$ ?
| Let $p$ be an odd prime throughout.
Claim 1: In $\mathbb{Z}/p\mathbb{Z}$, the only numbers $n$ such that $n^2 \equiv 1$ are $\pm 1$.
Proof. Suppose $n$ is such that $n^2 \equiv 1 \pmod p$. Then we must have that $p \mid n^2 - 1 = (n+1)(n-1)$. Since $p$ is prime, we know that we then have $p \mid n+1$ or $p \mid n-1$. I... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1338083",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
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Prove that $\sin{\frac{2\pi x}{x^2+x+1}}=\frac{1}{2}$ has no rational roots.
Show that the following equation has no rational roots.
$$\sin{\frac{2\pi x}{x^2+x+1}}=\frac{1}{2}$$
This is what I've tried:
$$\left ( \frac{2\pi x}{x^2+x+1}=\frac{\pi}{6}+2k\pi \right)\lor\left (\frac{2\pi x}{x^2+x+1}=\frac{5\pi}{6}+2l\p... | You are almost there. If the equation has rational roots, then $\Delta$ must be a perfect square (why?).
To show that $\Delta$ is not a perfect square just observe that
$$\Delta=96k^2-316k+11 \equiv 3 \pmod{4}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1338957",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Integral of binomial coefficients Let the integral in question be given by
\begin{align}
f_{n}(x) = \int_{1}^{x} \binom{t-1}{n} \, dt.
\end{align}
The integral can also be seen in the form
\begin{align}
f_{n}(x) = \frac{1}{n!} \, \int_{1}^{x} \frac{\Gamma(t) \, dt}{\Gamma(t-n)} = \frac{(-1)^{n+1}}{n!} \int_{1}^{1-x} (u... |
Here's another variation of the theme based upon Stirling Numbers.
Starting from
\begin{align*}
f_n(x)&=\int_1^x\binom{t-1}{n}dt
=\frac{1}{n!}\int_1^x{(t-1)}_ndt
=\frac{1}{n!}\int_0^{x-1}{(u)}_ndu
\end{align*}
we can use the Stirling Numbers of the first kind $s(n,k)$ which can be defined for $n\geq 0$ and... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1339224",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 2,
"answer_id": 1
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Compute definite integral Question: Compute
$$\int_0^1 \frac{\sqrt{x-x^2}}{x+2}dx.$$
Attempt: I've tried various substitutions with no success. Looked for a possible contour integration by converting this into a rational function of $\sin\theta$ and $\cos \theta$.
| $\displaystyle\int_0^1\frac{x^{1/2}(1-x)^{1/2}}{x+2}dx=\int_0^1\sum_{n=0}^{\infty}(-1)^n\frac{x^n}{2^{n+1}}x^{1/2}(1-x)^{1/2}dx=\sum_{n=0}^{\infty}\frac{(-1)^n}{2^{n+1}}\int_0^1x^{n+\frac{1}{2}}(1-x)^{\frac{1}{2}}dx$
$\displaystyle=\sum_{n=0}^{\infty}\frac{(-1)^n}{2^{n+1}}\beta(n+\frac{3}{2},\frac{3}{2})=\sum_{n=0}^{\i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1340612",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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"answer_id": 5
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Concluding three statements regarding $3$ real numbers.
$\{a,b,c\}\in \mathbb{R},\ a<b<c,\ a+b+c=6 ,\ ab+bc+ac=9$
Conclusion $I.)\ 1<b<3$
Conclusion $II.)\ 2<a<3$
Conclusion $III.)\ 0<c<1$
Options
By the given statements
$\color{green}{a.)\ \text{Only conclusion $I$ can be derived}}$.
$b.)\ $ Only conclusion $II$ can... | I shall show that only Conclusion I is correct. For a fixed $b$, we have
$a+c=6-b$ and $ac=9-(a+c)b=9-(6-b)b=(b-3)^2$. Hence, the quadratic polynomial
$x^2-(6-b)x+(b-3)^2$ has two distinct real roots $x=a$ and $x=c$. Therefore,
the discriminant $(6-b)^2-4(b-3)^2=3b(4-b)$ of this quadratic is strictly
positive (he... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1341548",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
What is the last digit of $2003^{2003}$?
What is the last digit of this number?
$$2003^{2003}$$
Thanks in advance.
I don't have any kind of idea how to solve this. Except that $3^3$ is $27$.
| When calculating the last digit of $a\cdot b$ you only need to find the product of the last digit of $a$ and the last digit of $b$.
so $2003^{2003}$ ends in the same digit as $3^{2003}$
$3^1$ ends in $3$
$3^2$ ends in $9$
$3^3$ ends in $7$
$3^4$ ends in $1$
$3^5$ ends in $3$
$3^6$ ends in $9$
$3^7$ ends in $7$
$3^8$ en... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1342323",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 7,
"answer_id": 1
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How to derive $\sum_{n=0}^\infty 1 = -\frac{1}{2}$ without zeta regularization On Wikipedia we find $$\displaystyle \bbox[5px,border:1px solid #F5A029]{1 + 1 + 1+\dots =\sum_{n=0}^\infty 1 = -\frac{1}{2}}$$ using (the rather complicated) zeta-function regularization. I asking for an elementary derivation possibly base... | For $f(x)=\sum_0^\infty a_n x^n$, a real number $R$, $R\neq 1$, and
$g(x)=f(x)-Rf(x^2)$, if $g(1)$ is Abel summable, that is,
$g(1)=\lim_{x\to 1^-} g(x)$, an elementary Ramanujan sum of $f(1)$ is
defined by
$$
f(1)=\frac{g(1)}{1-R}.
$$
Inversely, for an Abel summable series $g(1)=\sum_0^\infty g_n$ and
a real number $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1342394",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 7,
"answer_id": 4
} |
How to determine a basis and the dimension for this vectorspace? Determine a basis and the dimension for the following vectorspace:
\begin{align*} W = \left\{A \in \mathbb{R}^{3 \times 3} \mid A \ \text{is a diagonal matrix and} \ \sum_{i=1}^3 A_{ii} = 0\right\} \end{align*} I know the dimension of all $(3 \times 3)$-d... | Note that $A\in W$ if and only if
$$
A=\begin{bmatrix}a&0&0\\0&b&0\\0&0&c\end{bmatrix}
$$
where $a+b+c=0$. It follows that
$$
A=\begin{bmatrix}a&0&0\\0&b&0\\0&0&-a-b\end{bmatrix}
=a\begin{bmatrix}1&0&0\\0&0&0\\0&0&-1\end{bmatrix}+
b\begin{bmatrix}0&0&0\\0&1&0\\0&0&-1\end{bmatrix}
$$
This proves that
$$
\left\{
\begin{b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1342553",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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How do you multiply this How can you multiply these ordinal numbers:
$(\omega+1)(\omega+1)(\omega2+2)$
I tried and have gotten to this:
$(\omega^2+1)(\omega2+2)$
Is that the correct way, or did i made a mistake?
| Ordinal multiplication is distributive in the rightmost element. Therefore
$$(\omega+1)(\omega+1)=(\omega+1)\omega+(\omega+1)=\omega^2+\omega+1$$
Therefore:
\begin{align*}
(\omega+1)(\omega+1)(\omega\cdot2+2) &=(\omega+1)(\omega+1)\omega\cdot2+(\omega+1)(\omega+1)2 \\ &=(\omega+1)(\omega+1)\omega+(\omega+1)(\omega+1)\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1344329",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
This general inequality maybe is true? $\sum_{i=1}^{n}\frac{i}{1+a_{1}+\cdots+a_{i}}<\frac{n}{2}\sqrt{\sum_{i=1}^{n}\frac{1}{a_{i}}}$
Let $a_{1},a_{2},\ldots,a_{n}>0$ and prove or disprove
$$\dfrac{1}{1+a_{1}}+\dfrac{2}{1+a_{1}+a_{2}}+\cdots+\dfrac{n}{1+a_{1}+a_{2}+\cdots+a_{n}}\le\dfrac{n}{2}\sqrt{\dfrac{1}{a_{1}}+... | ONLY AN IDEA (and no solution, or at least a partial solution):
From this question (thx to the comments) we have
\begin{align*}
\sum_{k=1}^n\frac{k}{1+a_1+\ldots+a_k}\leq
\sum_{k=1}^n\frac{k}{(1/n+a_1)+\ldots+(1/n+a_k)}\leq2\left(\frac{1}{1/n+a_1}+\ldots+\frac{1}{1/n+a_n}\right)=2x,
\end{align*}
where $x:=\sum_{k=1}^n\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1344644",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
"answer_count": 2,
"answer_id": 1
} |
Solve integrals using residue theorem? $$\int_{0}^{\pi}\frac{d\theta }{2+\cos\theta}$$
$$\int_{0}^{\infty}\frac{x }{(1+x)^6} dx$$
My problem is that I don't know how to start solving these integrals, or how to convert them into usual types that can be solved.
| Hint for the First one First compute
$$
I=\int_0^{2\pi} \frac{d\theta}{2 + \cos\theta}=\int_0^{2\pi}R(\cos(\theta), \sin(\theta)) d\theta
$$
Where $R$ is the rational function given by $$R(x,y)=\frac{1}{2+x}$$
How to do this using the residue theorem? Put $z=e^{it}=\cos(t)+i\sin(t)$, thus
$$
\cos(t)=Re(z)=\frac{z+z^{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1344724",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Convert the equation to rectangular form $r = \frac {6}{1-\sinθ}$ Convert the equation to rectangular form $r = \frac {6}{1-\sinθ}$
The answer should be: $y = \frac{1}{12} x^2 -3$
But how to arrive at the answer?
I tried replacing r with $\sqrt{x^2 + y^2}$, then $\sinθ$ as $\frac{y}{r}$ but to no avail.
I also ended up... | Write
$$6=r(1-\sin\theta)=r-r\sin\theta=\sqrt{x^2+y^2}-y$$
Then
$$6+y=\sqrt{x^2+y^2}\implies12y+36=x^2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1345356",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Solving a system of non-linear equations Let
$$(\star)\begin{cases}
\begin{vmatrix}
x&y\\
z&x\\
\end{vmatrix}=1, \\
\begin{vmatrix}
y&z\\
x&y\\
\end{vmatrix}=2, \\
\begin{vmatrix}
z&x\\
y&z\\
\end{vmatrix}=3.
\end{cases}$$
Solving the above system of three non-linear equations with three unknowns.
... | The resultant of $x^2-yz-1$ and $y^2-xz-2$ with respect to $z$ is $x^3-y^3-x+2y$. The resultant of $x^2 - yz - 1$ and $z^2 - xy - 3$ with respect to $z$ is $x^4-x y^3-2 x^2-3 y^2+1$. The resultant of $x^3-y^3-x+2y$ and
$x^4-x y^3-2 x^2-3 y^2+1$ with respect to $x$ is $-y^4(18 y^2-1)$. So either
$y = 0$ or $y = \pm ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1347813",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 4,
"answer_id": 2
} |
value of an $\sum_3^\infty\frac{3n-4}{(n-2)(n-1)n}$ I ran into this sum $$\sum_{n=3}^{\infty} \frac{3n-4}{n(n-1)(n-2)}$$
I tried to derive it from a standard sequence using integration and derivatives, but couldn't find a proper function to describe it.
Any ideas?
| Setting
$$\frac{3n-4}{n(n-1)(n-2)}=\frac{A(n-1)-B}{(n-2)(n-1)}-\frac{An-B}{(n-1)n}$$
gives you $A=3,B=2$, i.e.
$$\frac{3n-4}{n(n-1)(n-2)}=\frac{3(n-1)-2}{(n-2)(n-1)}-\frac{3n-2}{(n-1)n}.$$
Hence, we have
$$\begin{align}\sum_{n=3}^{\infty}\frac{3n-4}{n(n-1)(n-2)}&=\lim_{m\to\infty}\sum_{n=3}^{m}\left(\frac{3(n-1)-2}{(n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1348046",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Expanding $\frac{2x^2}{1+x^3}$ to series So I was doing some series expansion problems and stumbled upon this one ( the problem is from Pauls Online Notes )
$$f(x) = \frac{2x^2}{1+x^3}$$
The actual solution to this problem uses a different method then mine. Since this function is really seams nice for integration, i t... | Intelligent question. The problem is that you first differentiate with respect to $t$, then substitute $t$ back for $x^3$. The correct approach is the reverse order: replace $t$ by $x^3$ and derive with respect to $x$. Otherwise, your solution is nice. Below you will find a slightly more direct one.
Assume $x \ne 0$. T... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1350095",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
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Solving for y' in a fraction Given the equation $x+xy^2 = \tan^{-1}(x^2y)$ find $y'$.
I have tried doing this but solving for $y'$ I need some help and would like your advice.
Work so far...
$$1+y^2+2xy\left(\frac{dy}{dx}\right)= \frac{2xy+x^2\left(\frac{dy}{dx}\right)}{1+x^4y^2}$$
What can be done now to solve for $y'... | So you have the expression
$$ 1 + y^{2} + 2xy(\dfrac{dy}{dx}) = \dfrac{2xy + x^{2}(\dfrac{dy}{dx})}{1 + x^{4}y^{2}}$$
And we can break up the term on the left hand side in this way: $$\dfrac{2xy + x^{2}(\dfrac{dy}{dx})}{1 + x^{4}y^{2}} = \dfrac{2xy }{1 + x^{4}y^{2}} + \dfrac{x^{2}(\dfrac{dy}{dx})}{1 + x^{4}y^{2}} $$
T... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1351689",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Stability analysis of Ralston's method Ralston's method is given by:
$$y_{n+1} = y_n + \frac{h}3(f(t_n,y_n)+2f(t_n+\frac34h, y_n + \frac34h f(t_n,y_n)))$$
carry out a stability analysis of this method to determine the condition for stability based on the test problem:
$$\frac{dy}{dt} -\lambda y$$
From what I understand... | Usually, stability of a method is studied on a test ODE
$$
y' = f(t,y) \quad\Leftrightarrow\quad y' - f(t,y) = 0
$$
with $f(t,y) = \lambda y,\; \lambda \in \mathbb{C}^{-}$.
So
$$
f(t_n, y_n) = \lambda y_n\\
f(t_n + \tfrac{3}{4}h, y_n + \tfrac{3}{4}h f(t_n, y_n)) =
f(t_n + \tfrac{3}{4}h, y_n + \tfrac{3}{4}h \lambda y_n)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1354505",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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System of Congruences with Special Symmetry Show that the following system of congruences
\begin{align}
\begin{cases}
3 x^4 - 7 x^2 y^2 - 7 x^2 z^2 - 35 y^2 z^2 \equiv 0 \pmod{p} \\
3 y^4 - 7 x^2 y^2 - 7 y^2 z^2 - 35 x^2 z^2 \equiv 0 \pmod{p} \\
3 z^4 - 7 y^2 z^2 - 7 x^2 z^2 - 35 x^2 y^2 \equiv 0 \pmod{p}
\end{cases}
\... | The symmetries are, indeed, the key. I let the variables range over the field $\Bbb{Z}_p$ to avoid having to write $\pmod p$ all the time.
Assume first that the squares $x^2,y^2,z^2$ are all pairwise non-congruent modulo $p$. Then subtracting your second equation from the first gives
$$
0=3x^4-3y^4+28x^2z^2-28y^2z^2=(x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1354655",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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$ \lim_{x \to \infty} x(\sqrt{2x^2+1}-x\sqrt{2})$ Any ideas fot evaluating:
$$ \lim_{x \to \infty} x(\sqrt{2x^2+1}-x\sqrt{2})$$
thanks.
| While one way to proceed here is to write the term
$\sqrt{2x^2+1}-\sqrt{2}x=\frac{1}{\sqrt{2x^2+1}+\sqrt{2}x}$
a second way is to write
$$\sqrt{2x^2+1}=\sqrt{2}x\left(1+\frac{1}{4x^2}+O\left(\frac{1}{x^4}\right)\right)$$
so that
$$\begin{align}
x\left(\sqrt{2x^2+1}-\sqrt{2}x\right)&=\sqrt{2}x^2\left(1+\frac{1}{4x^2}+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1355259",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Shortest distance of the point $(0,c)$ from the parabola $y=x^2$ Shortest distance of the point $(0,c)$ from the parabola $y=x^2$ ? (Where $0\le c \le5$)
My approach: I wrote the distance formula by taking parametric coordinates as $(t,t^2)$ and then differentiated the equation.I got the extremum as $x=\sqrt (c)$,but ... | Given is
$$
y = x^2
$$
So the normal for point $(x_o,x_o^2)$ can be written as
$$
(x,y) = (x_o - 2\alpha x_o, x_o^2 + \alpha )
$$
The length of the normal from point $(x_o,x_o^2)$ is given by
$$
\ell = \alpha \sqrt{4 x_o^2 + 1}
$$
We need to solve
$$
(0,c) = (x_o - 2\alpha x_o, x_o^2 + \alpha )
$$
or
$$
\left[
\begin{a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1356040",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Probability that team $A$ has more points than team $B$
Seven teams play a soccer tournament in which each team plays every other team exactly once. No ties occur, each team has a $50\%$ chance of winning each game it plays, and the outcomes of the games are independent. In each game, the winner is awarded a point and... | $A$ and $B$ have played one game against each pther, and $B$ lpst.
Bpth teams have each to play against the remaining five teams (independently). Let $X$ be the count of those games $A$ wins (ie: the points), and let $Y$ be the count of games $B$ wins. Since $A$ has at least one point from the game between the two ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1358776",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
Integer solutions for $x^2+y^2=208$ Which steps I can follow to find the integer solutions for the equation $x^2 + y^2 = 208$?
| A positive integer is the sum of two squares iff primes $\equiv 3\pmod 4$ (following the Fermat's theorem on sums of two squares) of its factorization are raised to even exponents.
$$208=2^4\cdot 13$$
So, $208$ meets the condition.
Now, $13=3^2+2^2$, so $208=4^2(3^2+2^2)=12^2+8^2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1359674",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Finding the area between : $\frac{\pi}{3}\leq \theta \leq \frac{2 \pi}{3}, 0 \leq r \leq 1$ using polar coordinates
I need to find the area between : $\frac{\pi}{3}\leq \theta \leq \frac{2 \pi}{3}, 0 \leq r \leq 1$ using polar coordinates $x=r\cos \theta , \, y=r\sin \theta$
$$x^2+y^2=r \Longrightarrow r=1$$
My attem... | Here's how it should look:
$$\int_{\pi/3}^{2 \pi / 3} \int_0^1 \color{blue}{r} \, dr \, d \theta = \int_{\pi/3}^{2 \pi/3} \left. \color{blue}{\frac{r^2}{2}} \right|_0^1 d \theta$$
$$= \int_{\pi/3}^{2 \pi/3} \frac{1}{2} \, d \theta = \left. \frac{1}{2} \theta \right|_{\pi/3}^{2\pi/3} = \color{red}{\frac{\pi}{6}}$$
You d... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1359921",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"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.