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", "answer_count": 2, "answer_id": 1 }
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 }
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_...
{ "language": "en", "url": "https://math.stackexchange.com/questions/1235569", "timestamp": "2023-03-29T00:00:00", "source": "stackexchange", "question_score": "1", "answer_count": 4, "answer_id": 3 }
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", "url": "https://math.stackexchange.com/questions/1237953", "timestamp": "2023-03-29T00:00:00", "source": "stackexchange", "question_score": "3", "answer_count": 3, "answer_id": 0 }
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", "answer_count": 1, "answer_id": 0 }
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, $$\int_0^{\frac{\pi}{2}}\frac{1+\cos x}{2+\sin x}\mathrm{d}x=\int_0^{\frac{\pi}{2}}\frac{1}{2+\sin x}\mathrm{d}x+\int_0^{\frac{\pi}{2}}\frac{\...
{ "language": "en", "url": "https://math.stackexchange.com/questions/1238146", "timestamp": "2023-03-29T00:00:00", "source": "stackexchange", "question_score": "1", "answer_count": 1, "answer_id": 0 }
Use the generating function to solve a recurrence relation We have the recurrence relation $\displaystyle a_n = a_{n-1} + 2(n-1)$ for $n \geq 2$, with $a_1 = 2$. Now I have to show that $\displaystyle a_n = n^2 - n +2$, with $n \geq 1$ using the generating function. The theory in my book is scanty, so with the help o...
another observation is :$$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 $$a_n-a_1=2(1)+2(2)+2(3)+...2(n-1)=\\2(1+2+3+4+...+(n-1))=2 \...
{ "language": "en", "url": "https://math.stackexchange.com/questions/1242244", "timestamp": "2023-03-29T00:00:00", "source": "stackexchange", "question_score": "3", "answer_count": 3, "answer_id": 1 }
Magic square with not distinct numbers There's a 4x4 magic square: 4 0 1 0 3 0 2 0 0 3 0 2 0 4 0 1 Where 0s are different numbers, 1=1, 2=2, 3=3, 4=4. Only the rows and the columns have the same sum, the diagonals don't. Question: If rotating or reflecting the magic square counts as a different solution, how many dif...
Let's start by giving names to the unknown quantities $$ \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", "answer_count": 1, "answer_id": 0 }
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", "url": "https://math.stackexchange.com/questions/1248454", "timestamp": "2023-03-29T00:00:00", "source": "stackexchange", "question_score": "1", "answer_count": 2, "answer_id": 1 }
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", "url": "https://math.stackexchange.com/questions/1250756", "timestamp": "2023-03-29T00:00:00", "source": "stackexchange", "question_score": "1", "answer_count": 1, "answer_id": 0 }
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...
{ "language": "en", "url": "https://math.stackexchange.com/questions/1250880", "timestamp": "2023-03-29T00:00:00", "source": "stackexchange", "question_score": "1", "answer_count": 4, "answer_id": 2 }
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", "answer_count": 5, "answer_id": 3 }
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 $$|xyzw-6|\le$$ $$|xyzw-xyz\cdot 3|+|xyz\cdot 3-xy\cdo...
{ "language": "en", "url": "https://math.stackexchange.com/questions/1254547", "timestamp": "2023-03-29T00:00:00", "source": "stackexchange", "question_score": "2", "answer_count": 1, "answer_id": 0 }
Positivity of power function. Prove that $6^a-7^a+2\cdot 4^a-3^a-5^a\ge0$ for $-\frac{1}{2}\le a\le0$. I tried to do it by first derivative test but derivative become more complicated (same with 2nd derivative for convexity). Here is it's plotting.
For $a\in[-\frac12,0)$ and $x>0$ let $$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", "answer_count": 3, "answer_id": 0 }
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$$ We can prove by induction that $x_{n+1}\ge 2^n x_1$. Details of the induction, as requested: The base case is $n=1$,...
{ "language": "en", "url": "https://math.stackexchange.com/questions/1257835", "timestamp": "2023-03-29T00:00:00", "source": "stackexchange", "question_score": "2", "answer_count": 1, "answer_id": 0 }
Find the equation whose roots are each six more than the roots of $x^2 + 8x - 1 = 0$ Find the equation whose roots are each six more than the roots of $x^2 + 8x - 1 = 0$ I must use Vieta's formulas in my solution since that is the lesson we are covering with our teacher. My solution: Let p and q be the roots of the q...
Let $x_1,x_2$ be the roots of the equation: $x^2+8x-1 = 0$, and let $y_1 = x_1 +6, y_2 = x_2+6 \Rightarrow y_1+y_2 = x_1+x_2+12=-8+12 = 4, y_1y_2 = (x_1+6)(x_2+6)=x_1x_2+6(x_1+x_2)+36=-1+6\cdot (-8) + 36=-13\Rightarrow y^2-4y-13=0$ is the sought after equation.
{ "language": "en", "url": "https://math.stackexchange.com/questions/1259949", "timestamp": "2023-03-29T00:00:00", "source": "stackexchange", "question_score": "2", "answer_count": 3, "answer_id": 1 }
Compute the largest root of $x^4-x^3-5x^2+2x+6$ I want to calculate the largest root of $p(x)=x^4-x^3-5x^2+2x+6$. I note that $p(2) = -6$ and $p(3)=21$. So we must have a zero between two and three. Then I can go on calculating $p(\tfrac52)$ and see that the zero must lie in the interval $]2,\tfrac52[$. The answer is $...
You've : $p(x) = x^4 - x^3 - 5x^2 + 2x + 6 $ . Rearrange the given polynomial as : $$ 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}$$ Roots of $f(x)$ are : $t = 3 \ \te...
{ "language": "en", "url": "https://math.stackexchange.com/questions/1260906", "timestamp": "2023-03-29T00:00:00", "source": "stackexchange", "question_score": "3", "answer_count": 4, "answer_id": 3 }
I cannot solve this limit $$ \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 }
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", "answer_count": 2, "answer_id": 0 }
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", "source": "stackexchange", "question_score": "5", "answer_count": 2, "answer_id": 0 }
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 =...
{ "language": "en", "url": "https://math.stackexchange.com/questions/1271570", "timestamp": "2023-03-29T00:00:00", "source": "stackexchange", "question_score": "3", "answer_count": 4, "answer_id": 1 }
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", "question_score": "3", "answer_count": 4, "answer_id": 1 }
$f\left( x-1 \right) +f\left( x+1 \right) =\sqrt { 3 } f\left( x \right)$ 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.
{ "language": "en", "url": "https://math.stackexchange.com/questions/1274824", "timestamp": "2023-03-29T00:00:00", "source": "stackexchange", "question_score": "4", "answer_count": 2, "answer_id": 1 }
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 }
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", "answer_count": 1, "answer_id": 0 }
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)!}}$$
{ "language": "en", "url": "https://math.stackexchange.com/questions/1275769", "timestamp": "2023-03-29T00:00:00", "source": "stackexchange", "question_score": "3", "answer_count": 5, "answer_id": 3 }
Find the domain of $x$ in $4\sqrt{x+1}+2\sqrt{2x+3}\leq(x-1)(x^2-2)$ Solve this equation for $x$: $4\sqrt{x+1}+2\sqrt{2x+3}\leq(x-1)(x^2-2)$ I have no idea to solve that, but I know solutions are $x=-1$ or $x\ge 3$.
It seems the following. We can straightforwardly verify your answer with Mathcad as follows: Since $\sqrt{x+1}$ exists, $x\ge -1$. Since the left-hand side of the inequality is non-negative, its right-hand side is non-negative too, that is $x\ge\sqrt{2}$ or $-1\le x\le 1$. Squaring both sides of the inequality, we obta...
{ "language": "en", "url": "https://math.stackexchange.com/questions/1276510", "timestamp": "2023-03-29T00:00:00", "source": "stackexchange", "question_score": "3", "answer_count": 2, "answer_id": 0 }
Solve $\lim \limits_{x \to 0} (\frac{\sin x}{x})^\frac{1}{1-\cos x}$ without L'Hospital's Rule or Taylor expansion. I solved it with L'Hospital's Rule, but I had to use it 4 times! Is there any way to avoid it?
Can we use some basics: $\lim \limits_{x \to 0} (1+x)^\frac{1}{x} = e$ and $\lim \limits_{x \to 0} \frac{\sin x}{x} = 1$? So, we separate $1$: $$\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", "url": "https://math.stackexchange.com/questions/1277854", "timestamp": "2023-03-29T00:00:00", "source": "stackexchange", "question_score": "1", "answer_count": 1, "answer_id": 0 }
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", "answer_count": 2, "answer_id": 0 }
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
{ "language": "en", "url": "https://math.stackexchange.com/questions/1278706", "timestamp": "2023-03-29T00:00:00", "source": "stackexchange", "question_score": "5", "answer_count": 3, "answer_id": 1 }
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", "url": "https://math.stackexchange.com/questions/1279206", "timestamp": "2023-03-29T00:00:00", "source": "stackexchange", "question_score": "3", "answer_count": 3, "answer_id": 0 }
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 $$\left( 1 - \frac{1}{n} \ri...
{ "language": "en", "url": "https://math.stackexchange.com/questions/1282610", "timestamp": "2023-03-29T00:00:00", "source": "stackexchange", "question_score": "5", "answer_count": 4, "answer_id": 3 }
Integer solutions of $x^3 = 7y^3 + 6 y^2+2 y$? Does the equation $$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", "answer_count": 2, "answer_id": 0 }
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", "answer_count": 1, "answer_id": 0 }
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", "answer_count": 2, "answer_id": 0 }
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", "answer_count": 3, "answer_id": 0 }
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: $$\int _{ \pi /6 }^{ \pi /3 }{ \frac { dx }{ \sec x+\csc x } } $$ $$=\int _{\frac{\pi}{6}}^{ \frac{\pi}{3}} \dfrac{\sin x \cos x }{ \sin x+\cos ...
Integrate directly as follows \begin{align} \int _{ \pi /6 }^{ \pi /3 }{ \frac { dx }{ \sec x+\csc x } } & =\frac12 \int _{ \pi /6 }^{ \pi /3 } \frac{(\sin x +\cos x)^2-1 }{ \sin x+\cos x }dx \\ &= \frac{1}2\int _{ \pi /6 }^{ \pi /3 } \left(\sqrt2\cos(\frac\pi4-x) -\frac1{\sqrt2\cos(\frac\pi4-x)}\right)dx\\ &= \frac12...
{ "language": "en", "url": "https://math.stackexchange.com/questions/1288034", "timestamp": "2023-03-29T00:00:00", "source": "stackexchange", "question_score": "2", "answer_count": 4, "answer_id": 3 }
Find the diameter of iron ball. Iron weights $8$ times the weight of oak, Find the diameter of an iron ball whose weight is equal to that of a ball of oak $18$ cm in diameter. $a)\quad 4.5\quad cm \\\color{green}{b)\quad 9 \quad cm}\\c)\quad 12\quad cm\\d) \quad 15 \quad cm$ I tried $\dfrac43\times \pi\...
Since iron weighs $8$ times what oak weighs, you need the iron ball to have $\frac{1}{8}$ the volume of the oak ball. $$\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", "answer_count": 2, "answer_id": 0 }
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. $$ \begin{array}{cccccccccc} v & = & \frac{\langle v,u_{1} \rangle}{\langle u_{1},u_{1} \rangle} & u_{1} & + & \frac{\langle v,u_{2} \rangle}{\langle ...
{ "language": "en", "url": "https://math.stackexchange.com/questions/1288920", "timestamp": "2023-03-29T00:00:00", "source": "stackexchange", "question_score": "2", "answer_count": 4, "answer_id": 1 }
Derivative of $f(x) = \frac{\cos{(x^2 - 1)}}{2x}$ Find the derivative of the function $$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", "answer_count": 4, "answer_id": 1 }
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, $$\left(\frac{5}{n}\right)=\left(\frac{n}{5}\right)(-1)^{\frac{n-1}{2}\cdot\frac{5-1}{2}}=\left(\frac{n}{5}\right)=\begin{cases} \hphantom{-}1&\text{if }n\equiv 1,4\bmod 5\\ \hphantom{-}0&\text{if }n\equiv 0\bmod 5\\ -1 & \t...
{ "language": "en", "url": "https://math.stackexchange.com/questions/1294271", "timestamp": "2023-03-29T00:00:00", "source": "stackexchange", "question_score": "2", "answer_count": 1, "answer_id": 0 }
Show that there exists a vector $v$ such that $Av\neq 0$ but $A^2v=0$ Let $A$ be a $4\times 4$ matrix over $\mathbb C$ such that $rank A=2$ and $A^3=A^2\neq 0$.Suppose that $A$ is not diagonalisable. Then Show that there exists a vector $v$ such that $Av\neq 0$ but $A^2v=0$ My try:$\dim \operatorname{Im}(A)+\dim \ker ...
There are only two possible Jordan forms: $$\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}$$ only the first first one satisfies $A^2 = A^3.$ $\bf p.s.$ the nonzero vector $v$ you are looking for is in $\ker(A^2)\setminus \ker(A)$ we have $dim(\ker(A^2) = 3,dim(\ker(A) ...
{ "language": "en", "url": "https://math.stackexchange.com/questions/1295229", "timestamp": "2023-03-29T00:00:00", "source": "stackexchange", "question_score": "5", "answer_count": 1, "answer_id": 0 }
Find the number of possible values of $a$ Positive integers $a, b, c$, and $d$ satisfy $a > b > c > d, a + b + c + d = 2010$, and $a^2 − b^2 + c^2 − d^2 = 2010$. Find the number of possible values of $a.$ Obviously, factoring, $$(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", "question_score": "1", "answer_count": 2, "answer_id": 1 }
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 }
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 $$ to ...
{ "language": "en", "url": "https://math.stackexchange.com/questions/1304346", "timestamp": "2023-03-29T00:00:00", "source": "stackexchange", "question_score": "1", "answer_count": 2, "answer_id": 0 }
Yamabe's equation This is PDE Evans, 2nd edition: Chapter 9, Exericse 8(a). *(a) Assume $n \ge 3$. Find a constant $c$ such that $$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", "url": "https://math.stackexchange.com/questions/1304906", "timestamp": "2023-03-29T00:00:00", "source": "stackexchange", "question_score": "3", "answer_count": 2, "answer_id": 0 }
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 $$\frac{x^2}{x^2+4}...
{ "language": "en", "url": "https://math.stackexchange.com/questions/1307398", "timestamp": "2023-03-29T00:00:00", "source": "stackexchange", "question_score": "1", "answer_count": 6, "answer_id": 2 }
$5x\big(1+\frac{1}{x^2 +y^2}\big)=12$ ; $5y\big(1-\frac{1}{x^2 +y^2}\big)=4$ find $x$ and $y$ I already tried to solve using substitution and cross multiplication method . I got the first simplified (1)$$\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}$$ Now, $\binom{n}{r} \le n^r$ for all $r=0,\ldots,n$ (easily proven by induction; or just look a...
{ "language": "en", "url": "https://math.stackexchange.com/questions/1310555", "timestamp": "2023-03-29T00:00:00", "source": "stackexchange", "question_score": "6", "answer_count": 6, "answer_id": 3 }
What is the most unusual proof you know that $\sqrt{2}$ is irrational? What is the most unusual proof you know that $\sqrt{2}$ is irrational? Here is my favorite: Theorem: $\sqrt{2}$ is irrational. Proof: $3^2-2\cdot 2^2 = 1$. (That's it) That is a corollary of this result: Theorem: If $n$ is a positive integer...
This proof will look much better if someone can add a diagram to it. Suppose $\sqrt2=a/b$ with $a$ and $b$ positive integers chosen as small as possible. Draw a square of side $a$. In the upper left corner, place a square of side $b$, and another one in the lower right corner. The two $b$-squares have total area $2b^2...
{ "language": "en", "url": "https://math.stackexchange.com/questions/1311228", "timestamp": "2023-03-29T00:00:00", "source": "stackexchange", "question_score": "114", "answer_count": 19, "answer_id": 11 }
Evaluate the directional derivative of $f$ for the points and directions specified Evaluate the directional derivative of $f$ for the points and directions specified: (a) $f(x,y,z)=3x-5y+2z$ at $(2,2,1)$ in the direction of the outward normal to the sphere $x^2+y^2+z^2=9.$ (b) $f(x,y,z)=x^2-y^2$ at a general point of t...
(a) The outward normal of the sphere given is $\left(\frac{x}{3},\frac{y}{3}, \frac{1}{3}\sqrt{9-x^2-y^2}\right)$. The same at the given point is $\overline n=\left(\frac{2}{3},\frac{2}{3}, \frac{1}{3}\right).$ The vector of the partial derivatives is $\nabla f =(3,-5,2).$ The directional derivative is then $$\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 }
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}$$ Which conveniently sums to $3B = 9$, ...
{ "language": "en", "url": "https://math.stackexchange.com/questions/1313454", "timestamp": "2023-03-29T00:00:00", "source": "stackexchange", "question_score": "5", "answer_count": 2, "answer_id": 1 }
Prove $(a^2+b^2+c^2+d^2)^2≥(a+b)(b+c)(c+d)(d+a)$ I've been unsuccessfully trying to solve this contest-style problem for a while. Tried different substitutions and the such, but nothing helped. I presume the solution is related to Cauchy-Schwarz? Anyway, any hints would be appreciated! Given $a, b, c, d \in \mathbb{R...
By the AM-GM inequality: $$ (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).$$ The claim easily follows.
{ "language": "en", "url": "https://math.stackexchange.com/questions/1314431", "timestamp": "2023-03-29T00:00:00", "source": "stackexchange", "question_score": "3", "answer_count": 1, "answer_id": 0 }
Simple mod problem It’s kind of a silly question but I can't find a simple way for finding the value of variable $d$ . $(5*d) \mod 8 = 1$ I normally just do this recursively by saying $d=d+1$ until I get the right answer. Is there a better way of solving this?
To find the multiplicative inverse of $5$ modulo $8$, use the Extended Euclidean Algorithm. First, solve for the greatest common divisor of $5$ and $8$, which is $1$ since they are relatively prime. \begin{align*} 8 & = 1 \cdot 5 + 3\\ 5 & = 1 \cdot 3 + 2\\ 3 & = 1 \cdot 2 + 1\\ 2 & = 2 \cdot 1 \end{align*} Now, work...
{ "language": "en", "url": "https://math.stackexchange.com/questions/1315755", "timestamp": "2023-03-29T00:00:00", "source": "stackexchange", "question_score": "1", "answer_count": 4, "answer_id": 0 }
When is the series converges? Let the series $$\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}$$ $L < 1 \iff |2\s...
{ "language": "en", "url": "https://math.stackexchange.com/questions/1315805", "timestamp": "2023-03-29T00:00:00", "source": "stackexchange", "question_score": "2", "answer_count": 2, "answer_id": 0 }
What two numbers when multiplied gives $-25$ and when added, $-10$? Factors of $25$ are; $1, 5, 25$ $$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 }
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: $$38 = 45 + 23 - 2 \...
{ "language": "en", "url": "https://math.stackexchange.com/questions/1323118", "timestamp": "2023-03-29T00:00:00", "source": "stackexchange", "question_score": "2", "answer_count": 3, "answer_id": 1 }
High School Trigonometry ( Law of cosine and sine) I am preparing for faculty entrance exam and this was the question for which I couldn't find the way to solve (answer is 0). I guess they ask me to solve this by using the rule of sine and cosine: Let $\alpha$, $\beta$ and $\gamma$ be the angles of arbitrary triangle...
Consider the first summand, to begin with: $$ 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 }
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 }
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 $Q(x)$ $x^2 + 4x + 6=0$ $x^2=-(4x + 6)$ $$x^4 + ax^2 + b=(x^2 + 4x + 6)Q(x)$$ $$(-(4x + 6))^2 - a(4x + 6)+ b=0$$ $$16x^2+48x+36 - 4ax -6a+ b...
{ "language": "en", "url": "https://math.stackexchange.com/questions/1328210", "timestamp": "2023-03-29T00:00:00", "source": "stackexchange", "question_score": "4", "answer_count": 6, "answer_id": 5 }
Comparing two large numbers Can you compare two large exponential numbers, like $5^{44}$ and $4^{53}$ without taking their logs?
One approach is to figure that, roughly, $2^{10} \approx 10^3$, and $5^9 \approx 2,000,000 = 2 \cdot 10^6$. Then, \begin{align} 4^{53} &= (2^2)^{53} \\ &= 2^{106} \\ &= 2^{100} \cdot 2^6 \\ &= (2^{10})^{10} \cdot 2^6 \\ &\approx (10^3)^{10} \cdot 2^6 \\ &= 10^{30} \cdot 2^6 \end{align} By way of comparison, \begin{al...
{ "language": "en", "url": "https://math.stackexchange.com/questions/1329260", "timestamp": "2023-03-29T00:00:00", "source": "stackexchange", "question_score": "1", "answer_count": 4, "answer_id": 2 }
Integration of $\int \frac {x\arctan x}{(\sqrt{1+x^2})^3}$ How can I evaluate $$\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", "answer_count": 2, "answer_id": 1 }
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 }
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 $$ Notice the $N$ depe...
{ "language": "en", "url": "https://math.stackexchange.com/questions/1334870", "timestamp": "2023-03-29T00:00:00", "source": "stackexchange", "question_score": "9", "answer_count": 6, "answer_id": 0 }
Simplify P(n), where n is a positive integer : $ P(x)=\sum \limits_{k=1}^\infty \arctan\left(\frac{x-1}{(k+x+1)\sqrt{k+1}+(k+2)\sqrt{k+x}}\right). $ This is what I have tried, but I don't know what to do next, so I need help : $ P(x)=\sum \limits_{k=1}^\infty \arctan\left(\frac{x-1}{(n+x+1)\sqrt{n+1}+(n+2)\sqrt{n+x}}\r...
I'll assume $x$ is an positive integer. Define $$\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 }
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 }
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}$$
{ "language": "en", "url": "https://math.stackexchange.com/questions/1338957", "timestamp": "2023-03-29T00:00:00", "source": "stackexchange", "question_score": "2", "answer_count": 1, "answer_id": 0 }
Integral of binomial coefficients Let the integral in question be given by \begin{align} f_{n}(x) = \int_{1}^{x} \binom{t-1}{n} \, dt. \end{align} The integral can also be seen in the form \begin{align} f_{n}(x) = \frac{1}{n!} \, \int_{1}^{x} \frac{\Gamma(t) \, dt}{\Gamma(t-n)} = \frac{(-1)^{n+1}}{n!} \int_{1}^{1-x} (u...
Here's another variation of the theme based upon Stirling Numbers. Starting from \begin{align*} f_n(x)&=\int_1^x\binom{t-1}{n}dt =\frac{1}{n!}\int_1^x{(t-1)}_ndt =\frac{1}{n!}\int_0^{x-1}{(u)}_ndu \end{align*} we can use the Stirling Numbers of the first kind $s(n,k)$ which can be defined for $n\geq 0$ and...
{ "language": "en", "url": "https://math.stackexchange.com/questions/1339224", "timestamp": "2023-03-29T00:00:00", "source": "stackexchange", "question_score": "14", "answer_count": 2, "answer_id": 1 }
Compute definite integral Question: Compute $$\int_0^1 \frac{\sqrt{x-x^2}}{x+2}dx.$$ Attempt: I've tried various substitutions with no success. Looked for a possible contour integration by converting this into a rational function of $\sin\theta$ and $\cos \theta$.
$\displaystyle\int_0^1\frac{x^{1/2}(1-x)^{1/2}}{x+2}dx=\int_0^1\sum_{n=0}^{\infty}(-1)^n\frac{x^n}{2^{n+1}}x^{1/2}(1-x)^{1/2}dx=\sum_{n=0}^{\infty}\frac{(-1)^n}{2^{n+1}}\int_0^1x^{n+\frac{1}{2}}(1-x)^{\frac{1}{2}}dx$ $\displaystyle=\sum_{n=0}^{\infty}\frac{(-1)^n}{2^{n+1}}\beta(n+\frac{3}{2},\frac{3}{2})=\sum_{n=0}^{\i...
{ "language": "en", "url": "https://math.stackexchange.com/questions/1340612", "timestamp": "2023-03-29T00:00:00", "source": "stackexchange", "question_score": "6", "answer_count": 6, "answer_id": 5 }
Concluding three statements regarding $3$ real numbers. $\{a,b,c\}\in \mathbb{R},\ a<b<c,\ a+b+c=6 ,\ ab+bc+ac=9$ Conclusion $I.)\ 1<b<3$ Conclusion $II.)\ 2<a<3$ Conclusion $III.)\ 0<c<1$ Options By the given statements $\color{green}{a.)\ \text{Only conclusion $I$ can be derived}}$. $b.)\ $ Only conclusion $II$ can...
I shall show that only Conclusion I is correct. For a fixed $b$, we have $a+c=6-b$ and $ac=9-(a+c)b=9-(6-b)b=(b-3)^2$. Hence, the quadratic polynomial $x^2-(6-b)x+(b-3)^2$ has two distinct real roots $x=a$ and $x=c$. Therefore, the discriminant $(6-b)^2-4(b-3)^2=3b(4-b)$ of this quadratic is strictly positive (he...
{ "language": "en", "url": "https://math.stackexchange.com/questions/1341548", "timestamp": "2023-03-29T00:00:00", "source": "stackexchange", "question_score": "2", "answer_count": 2, "answer_id": 0 }
What is the last digit of $2003^{2003}$? What is the last digit of this number? $$2003^{2003}$$ Thanks in advance. I don't have any kind of idea how to solve this. Except that $3^3$ is $27$.
When calculating the last digit of $a\cdot b$ you only need to find the product of the last digit of $a$ and the last digit of $b$. so $2003^{2003}$ ends in the same digit as $3^{2003}$ $3^1$ ends in $3$ $3^2$ ends in $9$ $3^3$ ends in $7$ $3^4$ ends in $1$ $3^5$ ends in $3$ $3^6$ ends in $9$ $3^7$ ends in $7$ $3^8$ en...
{ "language": "en", "url": "https://math.stackexchange.com/questions/1342323", "timestamp": "2023-03-29T00:00:00", "source": "stackexchange", "question_score": "1", "answer_count": 7, "answer_id": 1 }
How to derive $\sum_{n=0}^\infty 1 = -\frac{1}{2}$ without zeta regularization On Wikipedia we find $$\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 }
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", "answer_count": 3, "answer_id": 0 }
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}$$ The actual solution to this problem uses a different method then mine. Since this function is really seams nice for integration, i t...
Intelligent question. The problem is that you first differentiate with respect to $t$, then substitute $t$ back for $x^3$. The correct approach is the reverse order: replace $t$ by $x^3$ and derive with respect to $x$. Otherwise, your solution is nice. Below you will find a slightly more direct one. Assume $x \ne 0$. T...
{ "language": "en", "url": "https://math.stackexchange.com/questions/1350095", "timestamp": "2023-03-29T00:00:00", "source": "stackexchange", "question_score": "3", "answer_count": 3, "answer_id": 2 }
Solving for y' in a fraction Given the equation $x+xy^2 = \tan^{-1}(x^2y)$ find $y'$. I have tried doing this but solving for $y'$ I need some help and would like your advice. Work so far... $$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}} $$ T...
{ "language": "en", "url": "https://math.stackexchange.com/questions/1351689", "timestamp": "2023-03-29T00:00:00", "source": "stackexchange", "question_score": "2", "answer_count": 2, "answer_id": 0 }
Stability analysis of Ralston's method Ralston's method is given by: $$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 }
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 $$ 0=3x^4-3y^4+28x^2z^2-28y^2z^2=(x...
{ "language": "en", "url": "https://math.stackexchange.com/questions/1354655", "timestamp": "2023-03-29T00:00:00", "source": "stackexchange", "question_score": "1", "answer_count": 1, "answer_id": 0 }
$ \lim_{x \to \infty} x(\sqrt{2x^2+1}-x\sqrt{2})$ Any ideas fot evaluating: $$ \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 }