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Max and Min$ f(x,y)=x^3-12xy+8y^3$ First I solved $f_y=0$ then plugged in my variable into $f_x$ to get an output and then plugged that output back into $f_y$ to get a point and did this again for $f_x$ I think I screwed up on the $(-7,\sqrt{3})$
$f_y=-12x+24y^2$
$-12x+24y^2=0$
$-12x=24y^2$
$x=2y^2$
$f_x(2y^2,y)=3(2y^2)^2-12y$
$3(4y^4)-12y=0$
$12y^4 - 36y=0$
$12y(y^3-3=0) \to y=0$ or $y=\sqrt{3}$
$f_y(x,0)=-12x=0 \to (0,0)$
$f_y(x,\sqrt{3}) = -12x+24(3)$
$f_y(x.sqrt{3})= -12x-84=0$
$f_y(x,\sqrt{3})=-12x=84$
$x=-7 \to (-7,\sqrt{3})$
$f_y=-12x+24y^2=0$
$-12x=24y^2$
$x=2y^2$
$f_x(2y^2,y)=3(2y^2)^2-12y=0$
$12y^4-12y=0$
$12y(y^3-1)=0$
$y=0$ or $y=1$
$f_y(x,o)= -12x=0 \to x=0$
$f_y(x,1)=-12x+24=0$
$f_y(x,1)=x=2$
Critical points: $(2,-1),(0,0),(-7,\sqrt{3})$
$D=(6x)(48y)-(-12)^2$
$D(0,0)=-144 so (0,0) is a saddle point
$D(,2,-1)= Min
D(-7,\sqrt{3})= Saddle point
| $$f_x=3x^2-12y$$
$$f_y=24y^2-12x$$
Now we let $\nabla f(x,y)=(f_x,f_y)=0$ which gives us the following:
$$
\left\{
\begin{array}{c}
f_x=3x^2-12y=0 \\
f_y=24y^2-12x=0
\end{array}
\right.
$$
Solving it we have
$$
\left\{
\begin{array}{c}
x=2y^2 \\
12y^4-12y=12y(y^3-1)=0
\end{array}
\right.
$$
Where $y=0\lor y=1$ and respectively $x=0\lor x=2$. Can you continue now?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2158562",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Find the number of roots of the equation $3^{|x|}-|2-|x||=1$
Find the number of roots of the equation $3^{|x|}-|2-|x||=1$
My working:
Let $t$ be any positive real number.
$3^{t}-|2-t|=1$
Case 1:
$t<2$
$3^{t}-2+t=1$
$3^{t}+t=3$
Case 2:
$t>2$
$3^{t}+2-t=1$
$3^t+1=t$
Now I don't know how to proceed further to solve these equations. I would require a hint for that.
| $f(x)=3^{|x|}-|2-|x||$ is an even function ,so
$f(x)=1$ has symmetric roots (like $x=\pm a$)
so ,solve the equation for $x \geq 0 $ and then add negative of root(s)
$$3^{|x|}-|2-|x||=1, x \geq 0 \\ 3^x-|2-x|=1 \to
\begin{cases}3^x-(2-x)=1 & x <2\\3^x+(2-x)=1 & x >2\end{cases}\\
\begin{cases}3^x=3-x & x <2\\3^x=-1-x & x >2\end{cases}\\
\begin{cases}3^x=3-x & x <2\\3^x=-1-x & x >2 \end{cases}$$
first one has a solution , but second one has not root because $x>2 \to -1-x<-3\\3^x>0$
finally the original equation has 2 roots
another observation is to take $3^{|x|}=1+|2-|x||$ there is $f(x)=g(x)$ plot them together ,so must be there two cross section between $f(x)$ and $g(x)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2158816",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
If $f_k(x)=\frac{1}{k}\left (\sin^kx +\cos^kx\right)$, then $f_4(x)-f_6(x)=\;?$ I arrived to this question while solving a question paper. The question is as follows:
If $f_k(x)=\frac{1}{k}\left(\sin^kx + \cos^kx\right)$, where $x$ belongs to $\mathbb{R}$ and $k>1$, then $f_4(x)-f_6(x)=?$
I started as
$$\begin{align}
f_4(x)-f_6(x)&=\frac{1}{4}(\sin^4x + \cos^4x) - \frac{1}{6}(\sin^6x + \cos^6x) \tag{1}\\[4pt]
&=\frac{3}{12}\sin^4x + \frac{3}{12}\cos^4x - \frac{2}{12}\sin^6x - \frac{2}{12}\cos^6x \tag{2}\\[4pt]
&=\frac{1}{12}\left(3\sin^4x + 3\cos^4x - 2\sin^6x - 2\cos^6x\right) \tag{3}\\[4pt]
&=\frac{1}{12}\left[\sin^4x\left(3-2\sin^2x\right) + \cos^4x\left(3-2\cos^2x\right)\right] \tag{4}\\[4pt]
&=\frac{1}{12}\left[\sin^4x\left(1-2\cos^2x\right) + \cos^4x\left(1-2\sin^2x\right)\right] \tag{5} \\[4pt]
&\qquad\quad \text{(substituting $\sin^2x=1-\cos^2x$ and $\cos^2x=1-\sin^2x$)} \\[4pt]
&=\frac{1}{12}\left(\sin^4x-2\cos^2x\sin^4x+\cos^4x-2\sin^2x\cos^4x\right) \tag{6} \\[4pt]
&=\frac{1}{12}\left[\sin^4x+\cos^4x-2\cos^2x\sin^2x\left(\sin^2x+\cos^2x\right)\right] \tag{7} \\[4pt]
&=\frac{1}{12}\left(\sin^4x+\cos^4x-2\cos^2x\sin^2x\right) \tag{8} \\[4pt]
&\qquad\quad\text{(because $\sin^2x+\cos^2x=1$)} \\[4pt]
&=\frac{1}{12}\left(\cos^2x-\sin^2x\right)^2 \tag{9} \\[4pt]
&=\frac{1}{12}\cos^2(2x) \tag{10}\\[4pt]
&\qquad\quad\text{(because $\cos^2x-\sin^2x=\cos2x$)}
\end{align}$$
Hence the answer should be ...
$$f_4(x)-f_6(x)=\frac{1}{12}\cos^2(2x)$$
... but the answer given was $\frac{1}{12}$.
I know this might be a very simple question but trying many a times also didn't gave me the right answer. Please tell me where I am doing wrong.
| ‘Tis I have a better solution to solve this math problem...
Let $a = \cos^2 x$ and $b = \sin^2 x,$ so $a + b = 1.$ Then
[(a + b)^2 = a^2 + 2ab + b^2 = 1,]so $a^2 + b^2 = 1 - 2ab.$ Also,
[(a + b)^3 = a^3 + 3a^2 b + 3ab^2 + b^3 = 1,]so
\begin{align*}
a^3 + b^3 &= 1 - (3a^2 b + 3ab^2) \\
&= 1 - 3ab(a + b) \\
&= 1 - 3ab.
\end{align*}Therefore,
\begin{align*}
f_4(x) - f_6(x) &= \frac{\sin^4 x + \cos^4 x}{4} - \frac{\sin^6 x + \cos^6 x}{6} \\
&= \frac{a^2 + b^2}{4} - \frac{a^3 + b^3}{6} \\
&= \frac{1 - 2ab}{4} - \frac{1 - 3ab}{6} \\
&= \boxed{\frac{1}{12}}.
\end{align*}
BTW, I solved this
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2159136",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
How do I find complex function knowing its real part and value in zero? For example I have a complex function $f(z)$, $z = x + iy$.
I know that $\operatorname{Re} f(z) = x^3 + 6 x^2 y + A x y^2 - 2 y^3$
and that $f(0) = 0$.
What should I do to find $f(z)$?
| First method: Cauchy-Riemann equations. The function $f$ is assumed analytic. Therefore, there is an open set around zero where it is holomorphic. The Cauchy-Riemann equations write
\begin{equation}
\frac{\partial\, \text{Im} f}{\partial y}(x+\text{i}y) = \frac{\partial\, \text{Re} f}{\partial x}(x+\text{i}y) = 3x^2 + 12xy + Ay^2
\end{equation}
and
\begin{equation}
\frac{\partial\, \text{Im} f}{\partial x}(x+\text{i}y) = -\frac{\partial\, \text{Re} f}{\partial y}(x+\text{i}y) = -6x^2 -2Axy + 6y^2 \, .
\end{equation}
Due to the equality of mixed partials, one has
\begin{equation}
\frac{\partial^2\, \text{Im} f}{\partial y\,\partial x}(x+\text{i}y) = \frac{\partial^2\, \text{Im} f}{\partial x\,\partial y}(x+\text{i}y) \, , \qquad\text{viz.}\qquad A = -3\, .
\end{equation}
Integrating $\partial\text{Im}/\partial y$ with respect to $y$, one obtains
\begin{equation}
\text{Im} f(x+\text{i}y) = 3 x^2 y + 6 x y^2 - y^3 + B(x)\, .
\end{equation}
Then, the expression of $\partial\text{Im}/\partial x$ imposes
\begin{equation}
B'(x) = -6x^2\, ,\qquad\text{i.e.}\qquad B(x) = -2x^3 + C\, ,
\end{equation}
where $C$ is in $\mathbb{R}$. Finally, the condition $f(0) = 0$ yields $C=0$, and
\begin{equation}
\text{Im} f(x+\text{i}y) = -2 x^3 + 3 x^2 y + 6 x y^2 - y^3 \, .
\end{equation}
Second method: Power series. The function $f$ is assumed analytic. Thus, let us look for an expression as a convergent power series in the vicinity of $z=0$:
\begin{equation}
f(z) = \sum_{n=0}^{+\infty} a_n\, z^n \, , \quad\text{for}\quad |z|<R \, ,
\end{equation}
where $(a_n)_{n\in\mathbb{N}}$ is a sequence of complex numbers. If $z=0$, the latter expression yields $f(0)=a_0=0$. If $z\neq 0$, we introduce the polar form $z = r e^{\text{i}\theta}$ with $r$ in $]0,R[$ and $\theta$ in $[0,2\pi[$. On the one hand, the real part of $f(z)$ satisfies
\begin{equation}
\text{Re} f(z) = \sum_{n=1}^\infty \text{Re}(a_n\, z^n) = \sum_{n=1}^\infty \left(\text{Re}(a_n)\cos(n\theta) - \text{Im}(a_n)\sin(n\theta)\right) r^n \, .
\end{equation}
On the other hand, the real part of $f(z)$ satisfies
\begin{equation}
\text{Re} f(z) = \left(\cos^3\!\theta + 6\cos^2\!\theta\sin\theta + A\cos\theta\sin^2\!\theta - 2\sin^3\!\theta\right) r^3 \, .
\end{equation}
Computing the Fourier series of the $\theta$-dependent factor, it rewrites as
\begin{equation}
\text{Re} f(z) = \left(\frac{A+3}{4}\cos\theta + \frac{1-A}{4}\cos(3\theta) + 2\sin(3\theta)\right) r^3 \, .
\end{equation}
Due to the uniqueness of the expression of $f$, one has
\begin{equation}
\left\lbrace
\begin{array}{l}
A = -3\, ,\\
\text{Re}(a_3) = (1-A)/4 = 1\, ,\\
\text{Im}(a_3) = -2\, ,\\
\text{Re}(a_{n}) = 0 \quad\text{for}\; n\neq 3\, ,\\
\text{Im}(b_{n}) = 0 \quad\text{for all}\; n\, .
\end{array}
\right.
\end{equation}
Finally, the following expression for $f$ is obtained:
\begin{equation}
f(z) = (1-2\text{i}) z^3 \, ,
\end{equation}
with imaginary part
\begin{equation}
\text{Im} f(x+\text{i}y) = -2 x^3 + 3 x^2 y + 6 x y^2 - y^3 \, .
\end{equation}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2159317",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Landing on Consecutive Spaces with Dice Rolls Imagine you are continuously rolling one six-sided fair die, and moving a token the amount on the die on a board with 1000 spaces. Assuming the token starts at zero, is there a general formula that describes the probability you will land on any three consecutive spaces? For instance, what is the probability you will land on either the 4th, 5th, or 6th spaces. What is the probability you will land on the 998th, 999th, 1000th spaces?
I know it converges to a probability by running simulations, but I don't know how to mathematically describe it.
| Let's start by defining $r(n)$ to be the probability that the sum is ever exactly $n$.
Let's express the probability that the sum will ever be $a$, $a+1$, or $a+2$ using $r$. Let's call that probability $P$.
For a set of positive integers $S$, define $p_s$ to be the probability that the sum will ever be in $S$.
Using the inclusion exclusion principal, we have
$$
P = p_{\{a\}}+p_{\{a+1\}}+p_{\{a+2\}}-p_{\{a,a+1\}}-p_{\{a,a+2\}}-p_{\{a+1,a+2\}}+p_{\{a,a+1,a_2\}}.$$
We can calculate all of these probabilities using the function $r$.
We have
$$
p_{\{a\}} = r(a), p_{\{a+1\}} = r(a+1), \text{ and } p_{\{a+2\}} = r(a+2).$$
Further,
$$p_{\{a,a+1\}} = r(a) \frac{1}{6}$$
and
$$p_{\{a+1,a+2\}} = r(a+1) \frac{1}{6}.$$
Since we can get to $a+2$ from $a$ in two ways, we have
$$
p_{\{a,a+2\}} = r(a)\frac{1}{6}+ r(a)\frac{1}{36} = \frac{7}{36}r(a).
$$
Finally,
$$p_{\{a,a+1,a_2\}} = r(a) \frac{1}{36}.$$
Putting this all together, we have
$$
P = \frac{2}{3}r(a)+\frac{5}{6}r(a+1)+r(a+2).
$$
Now, here's the complicated part. The closed form for $r$ is
$$
r(n) = \frac{2}{7} + \frac{1}{7} \sum_{i=2}^{7} A_i^n
$$
where the $A_i$ are complex numbers inside the unit circle. From this, we know that as $n$ goes to infinity, $r(n)$ goes to $\frac{2}{7}$, and so, as $a$ goes to infinity, $P$ approaches $\frac{5}{7}$.
If we really want to calculate $P$ exactly for some specific $a$, we can use the relation $\frac{2}{3}r(a)+\frac{5}{6}r(a+1)+r(a+2)$ given earlier and then go about calculating these $r$ values using the recurrence
$$
r(n) = \frac{1}{6}\sum_{i=1}^6 r(n-i)
$$
for $n>6$
with
$$r(1)= \frac{1}{6}
$$
$$
r(2)=\frac{1}{6}+\frac{1}{6} r(1) = \frac{7}{36}
$$
$$
r(3) = \frac{1}{6} + \frac{1}{6} r(2) + \frac{1}{6} r(1) = \frac{49}{216}
$$
$$
r(4) = \frac{1}{6} + \frac{1}{6} r(3) + \frac{1}{6} r(2) + \frac{1}{6} r(1) = \frac{343}{1296}
$$
$$
r(5) = \frac{1}{6} + \frac{1}{6} r(4) + \frac{1}{6} r(3) + \frac{1}{6} r(2) + \frac{1}{6} r(1) = \frac{2401}{7776}
$$
$$
r(6) = \frac{1}{6} + \frac{1}{6} r(5) + \frac{1}{6} r(4) + \frac{1}{6} r(3) + \frac{1}{6} r(2) + \frac{1}{6} r(1)
= \frac{16807}{46656}
$$
This way, we can calculate, for instance, that the probability of ever having a sum of $4$, $5$ or $6$ is exactly
$$
\frac{343}{432}
$$
and of ever having a sum of $1000$, $1001$ or $1002$ is
approximately $\frac{5}{7}+4.139\times10^{-138}$ (the exact fraction has
too many digits in its numerator and denominator to display here).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2161678",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
diophantine equation $x^2+y^2=3$ what is the solution of this diophantine equation $x^2+y^2=3$ ? with x and y are the rationnals number I have one solution : $x=\frac{a}{c}$ and $y=\frac{b}{c}$ a, b and c are the positive integer
I found a solution by doing a reasoning modulo 3
| Transform your search for rational solutions into a search for positive integer solutions by setting $x=a/c$, $y=b/c$, so the equation becomes
$$
a^2+b^2=3c^2
$$
Looking at this modulo $3$, we conclude that $a\equiv0\pmod{3}$ and $b\equiv{0}\pmod{3}$ (prove it). Therefore $a=3a_1$, $b=3b_1$ and so
$$
3a_1^2+3b_1^2=c^2
$$
which implies $c=3c_1$, giving $a_1^2+b_1^2=3c_1^2$.
Now we are done: assume a solution with positive $c$ exists and take one with minimal such $c$. We immediately get a contradiction.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2163153",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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Simplify this trigonometric equation (identities) Please simplify these trigonometric identities and describe step by step (wording).
$\frac{\sec \theta-\csc \theta}{1-\cot\theta}$
| Let $A = \frac{\sec\theta - \csc\theta}{1-\cot\theta}$
$A = \frac{\frac{1}{\cos\theta}-\frac{1}{\sin\theta}}{1-\frac{\cos\theta}{\sin\theta}}$
Now we make common denominators like so:
$A = \frac{\frac{\sin\theta}{\sin\theta \cdot \cos\theta}-\frac{\cos\theta}{sin\theta\cdot \cos\theta}}{\frac{\sin\theta}{\sin\theta}-\frac{\cos\theta}{\sin\theta}}$
$A = \frac{\sin\theta - \cos\theta}{\sin\theta\cdot\cos\theta} \cdot \frac{\sin\theta}{\sin\theta-\cos\theta}$
Now we can cancel out leaving $A = \frac{1}{\cos\theta} = \sec\theta$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2166247",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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Infinitely many $n$ such that $n, n+1, n+2$ are each the sum of two perfect squares. Prove that there exist infinitely many integers $n$ such that $n$, $n+1$, $n+2$ are all the sum of two perfect squares. Induction does not seem to be yielding any results.
| n = 0 is a trivial solution as 0 = $0^2 + 0^2, 1 = 1^2 + 0^2, 2 = 1^2 + 1^2$. Consider the pell equation $x^2 - 2y^2 = 1$, which has infinitely many solutions. Take $n = x^2 -1$. This implies $ n = y^2 + y^2$, $n+1 = x^2 + 0^2$ and $n+2 = x^2 + 1^2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2170494",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 4,
"answer_id": 2
} |
Does this system of equations has a solution? Let $R$ be an integral domain with identity. Let $A$ be an $n\times n$ matrix with entries from $R$. Assume that $A$ has the following properties:
(1) Every row of $A$ has exactly one $1$ and one $-1$.
(2) Every column of $A$ has exactly one $1$ and one $-1$.
(3) All other elements are $0$.
(4) $A_{ii}=1$ (all the diagonal entries are 1)
Let $B$ be an $n\times 1$. The $i$-th element of $B$ is $b_i$. Assume that $\sum_{i=1}^{n}b_i=0$.
Does $AX=B$ has a solution (where $X$ is an $n\times 1$ matrix) ?
| No: Consider the following system of equations (we work in $R = \mathbb{Z}$):
$$\begin{pmatrix}
1 & -1 & 0 & 0\\
-1 & 1 & 0 & 0\\
0 & 0 & 1& -1\\
0 & 0 & -1 & 1
\end{pmatrix}\cdot
\begin{pmatrix}
x_1 \\
x_2\\
x_3\\
x_4
\end{pmatrix}
=
\begin{pmatrix}
1\\
2\\
-1\\
-2
\end{pmatrix}$$
Then we would have that $1 = x_1 - x_2 = -(x_1 - x_2) = - 2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2171660",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Determine whether $\int^{\infty}_{2} \frac{1}{x\sqrt{x^2-4}} dx$ converges. Determine whether $$\int^{\infty}_{2} \frac{1}{x\sqrt{x^2-4}} dx$$ converges.
Doing some rough work, I realize that this function near $\infty$, behaves like $$\frac{1}{x\sqrt{x^2}} = \frac{1}{x^2}$$
I know this function converges, but I am having a hard time finding a $larger$ function that converges too.
I know that:
$$\frac{1}{x\sqrt{x^2-4}} \leq \frac{1}{\sqrt{x^2-4}}$$
but the bigger function diverges, so it doesn't work.
| Let $x-2=u$. Then $dx=du$. The integral becomes \begin{align*} \int_0^\infty \frac{du}{(u+2)\sqrt{(u+2)^2-4}}&=\int_0^\infty \frac{du}{(u+2)\sqrt{u^2+4u}}=\int_0^\infty \frac{du}{\sqrt{u^3+6u^2+8u}} \\ \\ &= \underbrace{\int_0^1 \frac{du}{\sqrt{u^3+6u^2+8u}}}_{=I_1} +\underbrace{\int_1^\infty \frac{du}{\sqrt{u^3+6u^2+8u}}}_{=I_2} \end{align*}
$\bullet$ Now, since $\displaystyle \frac{1}{\sqrt{u^3+6u^2+8u}} \geq \frac{1}{\sqrt{u^3}} = \frac{1}{u^{3/2}}$, $\ I_2$ converges by $p-$test ($3/2>1)$.
$\bullet$ Since, $\displaystyle \frac{1}{\sqrt{u^3+6u^2+8u}} \geq \frac{1}{\sqrt{8u}} = \frac{1}{\sqrt 8 u^{1/2}}$, $ \ \displaystyle \frac{I_1}{\sqrt 8}$ converges and hence $I_1$ converges by
$p-$test ($1/2<1)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2171854",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Proving: $\left(a+\frac1a\right)^2+\left(b+\frac1b\right)^2\ge\frac{25}2$
For $a>0,b>0$, $a+b=1$ prove:$$\bigg(a+\dfrac{1}{a}\bigg)^2+\bigg(b+\dfrac{1}{b}\bigg)^2\ge\dfrac{25}{2}$$
My try don't do much, tough
$a+b=1\implies\dfrac{1}{a}+\dfrac{1}{b}=\dfrac{1}{ab}$
Expanding: $\bigg(a+\dfrac{1}{a}\bigg)^2+\bigg(b+\dfrac{1}{b}\bigg)^2\\=a^2+b^2+\dfrac{1}{a^2}+\dfrac{1}{b^2}+4\\=(a+b)^2-2ab+\bigg(\dfrac{1}{a}+\dfrac{1}{b}\bigg)^2-\dfrac{2}{ab}+4\\=4-2ab-\dfrac{2}{ab}+1+\dfrac{1}{a^2b^2}\\=4-2\bigg(\dfrac{a^2b^2+1}{ab}\bigg)+\dfrac{a^2b^2+1}{a^2b^2}\\=4-\bigg(\dfrac{a^2b^2+1}{ab}\bigg)\bigg(2-\dfrac{1}{ab}\bigg)\\=4-\bigg(ab+\dfrac{1}{ab}\bigg)\bigg(2-\dfrac{1}{ab}\bigg)$
Seems to using Cauchy-Schwartz. Please help.
| Calculus will also work. Let $f(a)=(a+\frac {1}{a})^2 +(b+\frac {1}{b})^2$ with $b=1-a$ and, WLOG, $a\geq b.$ So $1/2\leq a<1.$
Since $db/da=-1$ we have $$f'(a)=2\left(a+\frac {1}{a}\right) \left(1-\frac {1}{a^2}\right)+2\left(b+\frac {1}{b}\right) \left(-1+\frac {1}{b^2}\right)=$$ $$=2(a-b)+2\left(\frac {1}{b^3}-\frac {1
}{a^3}\right)=2(a-b)+2\frac {(a^3-b^3)}{ab}$$ which is positive when $a>b>0$, that is, when $1/2<a<1$.
So $f(a)\geq f(1/2)=25/2.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2172634",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Prove by induction that $n^4+2n^3+n^2$ is divisible by 4
I'm trying to prove by induction that $n^4+2n^3+n^2$ is divisible by 4.
I know that P(1) it's true. Then $ n=k, P(k):k^4+2k^3+k^2=4w$ it's true by the hypothesis of induction.
When I tried to prove $n=k+1, P(k+1):(k+1)^4+(k+1)^3+(k+1)^2 = 4t$,
$$k^4+4k^3+6k^2+4k+1+2k^3+6k^2+6k+2+k^2+2k+1 = 4t.$$
$$(k^4+2k^3+k^2)+4k^3+6k^2+4k+1+6k^2+6k+2+2k+1 = 4t.$$
Can I replace $(k^4+2k^3+k^2)$ of the previous expression by $4w$ and then continue the proof? Is this correct?
Thanks.
| As an alternative to proving via induction, consider the following hints:
Hint #1: $n^4 + 2n^3 + n^2 = n^2(n^2 + 2n + 1) = \bigg(n(n+1)\bigg)^2$
Hint #2: The parity of $n$ is different from that of $n+1$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
Evaluating the following summation $\require{cancel}$
Evaluate $$\sum_{n=2}^{\infty}\ln\left(\frac{n^2-1}{n^2}\right)$$
My procedure:
$$\sum_{n=2}^{\infty}\ln\left(\frac{n^2-1}{n^2}\right)=\sum_{n=2}^{\infty}(\ln\left(n-1\right)-\ln(n)+\ln(n+1)-\ln(n))$$
If we evaluate a few terms:
$$\sum_{n=2}^{m}(\ln\left(n-1\right)-\ln(n)+\ln(n+1)-\ln(n))= $$
$$=0-\ln(2)\cancel{+\ln(3)}\cancel{-\ln(2)+\ln(2)}\cancel{-\ln(3)}\cancel{+\ln(4)} \cancel{-\ln(3)+\ln(3)}\cancel{-\ln(4)}+\ln(5)-\ln(4)+...+\ln(m+1)-\ln(m)$$
Thus: $$\lim_{m\to\infty}\sum_{n=2}^{m}\ln\left(\frac{n^2-1}{n^2}\right)=\lim_{m\to\infty}\left(-\ln(2)+\ln(m+1)-\ln(m)\right)=\boxed{-\ln(2)}$$
However, I am not really happy with this method. Is there any other way? A more elegant one?
| Another way would be to consider the exponentiation. Observe
\begin{align}
\exp\left(\sum^N_{n=2}\ln\frac{n^2-1}{n^2} \right)=&\ \prod^N_{n=2}\frac{n^2-1}{n^2} = \prod^N_{n=2}\frac{(n-1)(n+1)}{n \cdot n}\\
=&\ \frac{(2-1)(2+1)}{2\cdot 2}\cdot \frac{(3-1)(3+1)}{3\cdot 3}\cdots\frac{(N-2)N}{(N-1)\cdot (N-1)}\frac{(N-1)(N+1)}{N\cdot N}\\
=&\ \frac{1}{2}\frac{N+1}{N}
\end{align}
which means
\begin{align}
\sum^N_{n=2}\ln \frac{n^2-1}{n^2} = \ln\frac{N+1}{2N}.
\end{align}
Take the limit as $N\rightarrow \infty$ yields $-\ln 2$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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Limit of an indeterminate form with a quadratic expression under square root The problem is:
$$ \lim_{x\to0}\frac{\sqrt{1+x+x^2}-1}{x} $$
So far, I've tried substituting $\sqrt{1+x+x^2}-1$ with some variable $t$, but when $x\to0$, $t\to\sqrt{1}$.
I have also tried to rationalize the numerator, and applied l'hospital.
I simply can't figure out this limit.
Any help would be greatly appreciated! Thanks in advance.
| $$\lim _{ x\to 0 } \frac { \sqrt { 1+x+x^{ 2 } } -1 }{ x } \cdot \frac { \sqrt { 1+x+x^{ 2 } } +1 }{ \sqrt { 1+x+x^{ 2 } } +1 } =\lim _{ x\to 0 } \frac { x+{ x }^{ 2 } }{ x\sqrt { 1+x+x^{ 2 } } +1 } =\lim _{ x\to 0 } \frac { 1+x }{ \sqrt { 1+x+x^{ 2 } } +1 } =\frac { 1 }{ 2 } $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2176506",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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How to find the numbers of sets of positive numbers $\{a,b,c\}$ such that $(a)(b)(c) =2^{4}3^{5}5^{2}7^{3}$? How to find the numbers of sets of positive numbers $\{a,b,c\}$ such that $(a)(b)(c) =2^{4}3^{5}5^{2}7^{3}$ ?
$$\binom{4+2}{2} \binom{5+2}{2} \binom{2+2}{2} \binom{3+2}{2}$$
This would give the number of tuples $(a,b,c)$. But isolating the individual sets $\{a,b,c\}$ is harder as there would be duplicates. Any ideas how to do it?
| Let $S = \{(a,b,c) \colon abc = 2^4 3^5 5^2 7^3\}$. Let's split it into few smaller sets. $S_1 = \{(a,b,c) \in S \colon a \neq b;\, b \neq c;\, a \neq c\}$, $S_2 = \{(a,b,c) \in S \colon a=b;\, b \neq c\}$ and let $S_3, S_4$ be similar to $S_2$ only with equal pairs $a=c$ and $b=c$. It's easy to notice that $S = S_1 \cup S_2 \cup S_3 \cup S_4$, because the case $a=b=c$ is impossible for our condition (otherwise all powers should be divisible by 3). Also $|S_2| = |S_3| = |S_4|$.
The answer for the problem is ${|S_1| \over 6}$, because each possible set of three elements contributes to $S_1$ six times. We know $|S|$ hence all we have to do is find $|S_2|$. It's quite easy, let's consider some element of $S_2$. $(a,a,b) = (2^{a_1} 3^{a_2} 5^{a_3} 7^{a_4}, a, 2^{b_1} 3^{b_2} 5^{b_3} 7^{b_4})$ and from our condition
$$\left\{ \begin{array}{l} 2a_1 + b_1 = 4\\
2a_2 + b_2 = 5 \\
2a_3 + b_3 = 2\\
2a_4 + b_4 = 3
\end{array} \right.$$
Each condition is independent from the others and there are $3$ possibilities to satisfy the first condition, $3$ for the second, $2$ for the third and $2$ for the fourth. So $|S_2| = 3 \cdot 3 \cdot 2 \cdot 2$. $|S_1| = |S| - 3 |S_2|$. We're done.
| {
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"timestamp": "2023-03-29T00:00:00",
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Determine the eigenvector and eigenspace and the basis of the eigenspace The yellow marked area is correct, so don't check for accuracy :)
$A=\begin{pmatrix} 0 & -1 & 0\\ 4 & 4 & 0\\ 2 & 1 & 2
\end{pmatrix}$ is the matrix.
Characteristic polynomial is $-\lambda^{3}+6\lambda^{2}-12\lambda+8=0$
The (tripple) eigenvalue is $\lambda=2$.
Calculate the eigenvectors now:
$\begin{pmatrix}
-2 & -1 & 0\\ 4 & 2 & 0\\ 2 & 1 & 0 \end{pmatrix} \begin{pmatrix} x\\ y\\ z \end{pmatrix}= \begin{pmatrix} 0\\ 0\\ 0
\end{pmatrix}$
We get the equations:
$I: -2x-y=0 \Leftrightarrow y = -2x$
$II: 4x+2y=0$
$III: 2x+y=0 \Leftrightarrow 2x-2x=0 \Leftrightarrow 0=0$
We see that in every eequation $z$ is unknown, so we can choose an arbitrary $z$.
$x\begin{pmatrix}
1\\
-2\\
z
\end{pmatrix}$ and this is the eigenspace...?
And what is the basis of this eigenspace? Can I just set $x=1$ and some value for $z$? So this would be a correct basis of the eigenspace: $\begin{pmatrix}
1\\
-2\\
3
\end{pmatrix}$?
Now we need three linearly independent eigenvectors but I couldn't find them as I always got linearly dependent vectors...
I need a detailled, not too complicated answer that explains it well and I will give that answer a nice bounty (up to 200 rep) because I couldn't find another site explaining this correctly to me and I'm really in need of it.
| From your calculation you have that all eigenvectors are of the form $\begin{pmatrix}
x\\
-2x\\
z
\end{pmatrix} = x\begin{pmatrix}
1\\
-2\\
0
\end{pmatrix} + z\begin{pmatrix}
0\\
0\\
1
\end{pmatrix}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2177817",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 2
} |
$\sum_{n=1}^{N}\sin(x)\cos((2n-1)x) = \frac{1}{2}\sin(2Nx)$ I want to prove the following by induction $$\sum_{n=1}^{N}\sin(x)\cos((2n-1)x) = \frac{1}{2}\sin(2Nx)$$
here's my trial:
Base Step
N = 1 we have that $LHS = \sin(x)\cos(x)$ and $RHS = \frac{1}{2}\sin(2x) = \sin(x)\cos(x)$. Hence it works for N=1.
Inductive Step
Suppose it works for some $N$, i.e. $$\sum_{n=1}^{N}\sin(x)\cos((2n-1)x) = \frac{1}{2}\sin(2Nx)$$ I want to show that this implies that $$\sum_{n=1}^{N+1}\sin(x)\cos((2n-1)x) = \frac{1}{2}\sin(2(N+1)x)$$
My attempt
$$\sum_{n=1}^{N+1}\sin(x)\cos((2n-1)x) =\sum_{n=1}^{N}\sin(x)\cos((2n-1)x) + \sin(x)\cos((2(N+1)-1)x)$$ which gives $$ \sum_{n=1}^{N+1}\sin(x)\cos((2n-1)x) = \frac{1}{2}\sin(2Nx)+\sin(x)\cos((2N+1)x)$$
Now taking the last term I can see that $$\sin(x)\cos((2N-1)x) = \sin(x)(\cos(2Nx)\cos(x)-\sin(2Nx)\sin(x))$$ which gives $$\sin(x)\cos((2N-1)x) = \frac{1}{2}\cos(2Nx)\sin(2x)-\sin(2Nx)\sin^2(x)$$
But I really can't go any further.. any suggestions?
How do I prove it by induction?
EDIT
So in total we have $$\sum_{n=1}^{N+1}\sin(x)\cos((2n-1)x) =\frac{1}{2}\sin(2Nx)+\frac{1}{2}\cos(2Nx)\sin(2x)-\sin(2Nx)\sin^2(x)$$ and rearranging we have $$\sum_{n=1}^{N+1}\sin(x)\cos((2n-1)x) =\sin(2Nx)\left(\frac{1}{2}-\sin^2(x)\right)+\frac{1}{2}\cos(2Nx)\sin(2x) $$ and again $$\sum_{n=1}^{N+1}\sin(x)\cos((2n-1)x) =\frac{1}{2}\sin(2Nx)\left(1-2\sin^2(x)\right)+\frac{1}{2}\cos(2Nx)\sin(2x) $$ so finally $$\sum_{n=1}^{N+1}\sin(x)\cos((2n-1)x) =\frac{1}{2}\sin(2Nx)\cos(2x)+\frac{1}{2}\cos(2Nx)\sin(2x) $$ so then it's trivial $$\sum_{n=1}^{N+1}\sin(x)\cos((2n-1)x) =\frac{1}{2}\sin(2Nx+2x) = \frac{1}{2}\sin(2(N+1)x)$$
| So in total we have $$\sum_{n=1}^{N+1}\sin(x)\cos((2n-1)x) =\frac{1}{2}\sin(2Nx)+\frac{1}{2}\cos(2Nx)\sin(2x)-\sin(2Nx)\sin^2(x)$$ and rearranging we have $$\sum_{n=1}^{N+1}\sin(x)\cos((2n-1)x) =\sin(2Nx)\left(\frac{1}{2}-\sin^2(x)\right)+\frac{1}{2}\cos(2Nx)\sin(2x) $$ and again $$\sum_{n=1}^{N+1}\sin(x)\cos((2n-1)x) =\frac{1}{2}\sin(2Nx)\left(1-2\sin^2(x)\right)+\frac{1}{2}\cos(2Nx)\sin(2x) $$ so finally $$\sum_{n=1}^{N+1}\sin(x)\cos((2n-1)x) =\frac{1}{2}\sin(2Nx)\cos(2x)+\frac{1}{2}\cos(2Nx)\sin(2x) $$ so then it's trivial $$\sum_{n=1}^{N+1}\sin(x)\cos((2n-1)x) =\frac{1}{2}\sin(2Nx+2x) = \frac{1}{2}\sin(2(N+1)x)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2179633",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Does this radical ${a-1\over 2}\left({\sqrt[3]a+1\over \sqrt[3]a-1}-1\right)=...$ hold? Consider the nested radical $(1)$
$${a-1\over 2}\left({\sqrt[3]a+1\over \sqrt[3]a-1}-1\right)=2\sqrt[3]a+\left[(a^2-7a+1)+(6a-3)\sqrt[3]a+(6-3a)\sqrt[3]{a^2}\right]^{1/3}\tag1$$
An attempt:
$x=\sqrt[3]a$
$$\left({a-1\over 2}\left({x+1\over x-1}-1\right)-2x\right)^3=...\tag2$$
$(2)$ on the LHS becomes
$$\left({a-1\over x-1}-2x\right)^3=\left({a-1\over x-1}\right)^3-6x\left({a-1\over x-1}\right)^2+12x^2\left({a-1\over x-1}\right)-8x^3$$
We can't and don't how to simplify it any further.
Can anyone help us to prove that $(1)$ does hold?
| $y^3-1=(y-1)(y^2+y+1)$ ... let $y=\sqrt[3] a$. So I guess your LHS is
\begin{eqnarray*}
\frac{a-1}{2} \left( \frac{\sqrt[3] a+1 }{\sqrt[3] a-1}-1 \right) = (\sqrt[3] a)^2+\sqrt[3] a+1
\end{eqnarray*}
Now subtract $2 \sqrt[3] a$ from both sides and then cube both sides ... you will get the second term of the RHS ... but the algebra is pretty heavy.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2180701",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Noob question about $\int \frac{1}{\sin(x)}dx$ I manually integrate $\int \frac{1}{\sin(x)}dx$ as $$\int \frac{\sin(x)}{\sin^{2}(x)}dx = -\int \frac{1}{\sin^{2}(x)}d\cos(x) = \int \frac{1}{\cos^{2}(x) - 1}d\cos(x).$$
After replacing $u = \cos(x)$,
$$\int \frac{1}{u^{2} - 1}du = \int \frac{1}{u^{2} - 1}du = \frac {1} {2} \int \left(\frac {1} {u - 1} - \frac {1} {u+1}\right) du = \frac {1} {2} \ln\left(\frac {u-1} {u+1}\right) + C.$$
Substitute back to obtain
$$\frac {1} {2} \ln\left(\frac {\cos(x)-1}{\cos(x)+1}\right) + C.$$
The problem is that this solution is incorrect (I guess) because for example http://www.integral-calculator.com/ gives another solution
$$\frac {1} {2} \ln\left(\frac {1 - \cos(x)}{1 + \cos(x)}\right) + C.$$
And all other online solvers gives equivalent solution to
$$\frac {1} {2} \ln\left(\frac {1 - \cos(x)}{1 + \cos(x)}\right) + C.$$
The question is there I made a mistake?
Update: some of you may say that in complex space my answer is right but not so fast:
Take wolfram solver: integrate 1/sinx
The we get: $-ln(cot(x) + csc(x)) + C$ It is easy to see that it is equvalent to $$\frac {1} {2} \ln\left(\frac {1 - \cos(x)}{1 + \cos(x)}\right) + C.$$
$-\ln(\cot(x) + \csc(x)) + C = -\ln(\frac {\cos(x)} {\sin(x)} + \frac {1} {\sin(x)})$
then
$-\ln(\frac {\cos(x)} {\sin(x)} + \frac {1} {\sin(x)}) = -\frac {1}{2} \ln(\frac {(1+\cos(x))^{2}} {\sin^{2}x}) = -\frac {1}{2} \ln(\frac{1+\cos(x)+\cos(x)+\cos^{2}(x)} {1-\cos^{2}x}) = -\frac {1}{2} \ln(\frac {(1+\cos(x))(\cos(x)+\cos^{2})(x)} {1-\cos^{2}x}) = -\frac {1}{2} \ln(\frac {1+\cos(x)} {1-\cos(x)}) = \frac {1}{2} \ln(\frac {1-\cos(x)} {1+\cos(x)})$
| If you are considering antiderivatives as functions on connected domains $D$ that are open in $\mathbb{C}$ -- suitably chosen so that we don't have to worry about multivaluedness of the logarithm -- then you should still be okay. It's still true that the functions $z \mapsto \frac1{2} \ln \left(\frac{1 - \cos(z)}{1 + \cos(z)}\right)$ and $z \mapsto \frac1{2} \ln \left(\frac{\cos(z) - 1}{\cos(z) + 1}\right)$ are antiderivatives of $z \mapsto 1/\sin(z)$. Their difference is the constant $C = \ln(-1)/2$ for whatever branch of the logarithm you are considering. So you still get the same family of functions, up to a complex constant.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2181582",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Finding a basis for a set of $2\times2$ matrices
Find a basis for $M_{2\times2}$ containing the matrices $$\begin{pmatrix} 1 & 1 \\ 2 & 3 \end{pmatrix}$$ and $$\begin{pmatrix} 1 & 1 \\ 3 & 2 \end{pmatrix}$$
I know that every $2\times2$ matrix $\begin{pmatrix} a & b \\ c & d \end{pmatrix}$ can be written as:
$$a\begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix} + b \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix} + c \begin{pmatrix} 0 & 0 \\ 1 & 0 \end{pmatrix} + d \begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix}$$
so those matrices are a basis for the vector space of $2\times2$ matrices, but how do I apply this to specific matrices? I know how to find a basis for a set of vectors, but matrices confuse me.
| Here's a systematic approach: to begin, find a basis for $\Bbb R^4$ containing the vectors $(1,1,2,3)$ and $(1,1,3,2)$. Following the method that I outline here, we find that $\{(1,1,2,3),(1,1,3,2),(1,0,0,0),(0,1,0,0)\}$ is a basis for $\Bbb R^4$. Correspondingly,
$$
\left\{\pmatrix{1&1\\2&3},
\pmatrix{1&1\\3&2},
\pmatrix{1&0\\0&0},
\pmatrix{0&1\\0&0}
\right\}
$$
is a basis for $M_{2 \times 2}$
| {
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"url": "https://math.stackexchange.com/questions/2184690",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "2",
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Proof involving integer squares and parity Original proposition:
$a$, $b$ and $c$ are integers. Show that if $a^2+b^2=c^2$, at least
one of $a$ and $b$ is an even number.
My attempt:
Attemplting proof by contradiction:
Show that there exist a contradiction between the claim that $a^2+b^2=c^2$
...and $a=2k+1, b=2k+1$ where $a\in\mathbb{Z}$ and $a\in\mathbb{Z}$:
Or show:
$$(2k+1)^2+(2l+1)^2\neq c^2$$
I converted the above to:
$$4(k^2+k+l^2+l)+2\neq c^2$$
Perhaps it is possible to show that $4(k^2+k+l^2+l)$ is a perfect square and hence $4(k^2+k+l^2+l)+2$ cannot be a perfect square?
| Assume by way of contradiction $a$ and $b$ are odd, as you say we get that $$4(k^2+k+l^2+l)+2=c^2.$$ We have that $c^2$ is even thus $c$ is even, so $c=2m$ for some $m$. We get $$4(k^2+k+l^2+l)+2=4m^2.$$
divide both sides by two:
$$2(k^2+k+l^2+l)+1=2m^2$$
This is a contradiction since LHS is odd and RHS is even.
| {
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"timestamp": "2023-03-29T00:00:00",
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Using L'Hopital Rule, evaluate $\lim_{x \to 0} {\left( \frac {1} {x^2}-\frac {\cot x} {x} \right)}$ Using L'Hopital Rule, evaluate $$ \lim_{x \to 0} {\left( \frac {1} {x^2}-\frac {\cot x} {x} \right)}$$
I find this question weired.
If we just combine the two terms into one single fraction, we get$$\lim_{x \to 0} {\frac {1-x\cot x} {x^2}}=\frac10=\infty$$
If we follow L'Hopital Rule, this is $\infty-\infty$ form. We follow the following process to convert it into $\frac00$form.$$\infty_1 -\infty_2=\frac 1{\frac 1{\infty_1}}-\frac 1{\frac 1{\infty_2}}=\frac {\frac 1{\infty_2}-\frac 1{\infty_1}}{{\frac 1{\infty_1}}{\frac 1{\infty_2}}}$$
So we will get $$\lim_{x \to 0} {\left( \frac {1} {x^2}-\frac {\cot x} {x} \right)}=\lim_{x \to 0} {\frac {x\tan x-x^2}{x^3\tan x}}$$
If you keep differentiating using the rule you will get rid of the form of $\frac00$ in the third step of differentiation, which give you the answer $1 \over 3$. This method is very tedious. Trust me, you don't want to try.
I am wondering is there a smarter way of solving this question?
Thanks.
| first :
$$\lim_{x\to 0} \frac{\sin x-x}{x^3}=\frac{-1}{6} \tag{1}$$
$$\lim_{x\to 0} \frac{1-\cos x}{x^2}=\frac{3}{6} \tag{2}$$
$$\lim_{x\to 0} \frac{x}{\sin x}=1 \tag{3} $$
$$\frac{1}{x^2}-\frac{\cot x}{x}=\frac{\sin x-x}{x^3}\times \frac{x}{\sin x}+\frac{1-\cos x}{x^2}\times \frac{x}{\sin x}$$
$$L=\lim_{x\to 0} \frac{1}{x^2}-\frac{\cot x}{x}=\lim_{x\to 0}(\frac{\sin x-x}{x^3}\times \frac{x}{\sin x}+\frac{1-\cos x}{x^2}\times \frac{x}{\sin x})$$
Now using $(1) ,(2),(3)$:
$$L=(\frac{-1}{6}\times 1+\frac{3}{6}\times 1)=\frac{2}{6}$$
| {
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"url": "https://math.stackexchange.com/questions/2186061",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Find a prime number p that is simultaneously expressible in the forms $x^2 + y^2, u^2 + 2v^2$, and $r^2 + 3s^2$. background: From Burton, Number Theory 9.3#8 with a hint: (−1/p) = (−2/p) = (−3/p) = 1.]
Based on the previous problem 6:
$n^2+1, n^2+2$, or $n^2+3$ have solutions when $n^2\equiv-i\mod p$ $(\frac{-i}{p})=1$ This occurs when p = 2 or p ≡ 1 (mod 4); p = 2 or p ≡ 1 or 3 (mod 8);
p = 2, p = 3 or p ≡ 1 (mod 6) respectively.
Now to the original question, the hint is satisfied when $\require{enclose} \enclose{horizontalstrike}{p\equiv 1 \mod 12}\quad p\equiv 1 \mod 24$ but I don't see how to use this to find the prime satisfying the equations $x^2 + y^2, u^2 + 2v^2$, and $r^2 + 3s^2$.
| Let $p$ be an odd prime. Using Legendre symbol rules, one can deduce that
\begin{equation*}
\begin{aligned}
p &\equiv 1 \pmod 4, \\
p &\equiv 1,7 \pmod 8, \\
p &\equiv 1,11 \pmod{12}. \\
\end{aligned}
\end{equation*}
The above system has no solution unless $p$ is equivalent to $1$ modulo $4,8,12$. In this case, $$p \equiv 1 \pmod{24}.$$
The smallest such prime is $p = 73$, and indeed we find that
$$73 = 3^2 + 8^2 = 1^2 + 2\cdot6^2 = 5^2 + 3\cdot 4^2.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2187277",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
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prove that $\lim_{x \to 1}\frac{x^{2}-x+1}{x+1}=\frac{1}{2}$ Please check my proof
Let $\epsilon >0$ and $\delta >0$
$$0<x<\delta \rightarrow \frac{x^{2}-x+1}{x+1}-\frac{1}{2}<\epsilon $$
$$\frac{2x^{2}-2x+2-x+1}{2x+2}<\epsilon $$
$$\frac{2x^{2}-3x+1}{2x+2}<\epsilon $$
since $\frac{2x^2-3x+1}{2x+2} <\frac{2x^{2}-3x+2}{2}$
then $\frac{2x^{2}-3x+1}{2}< \epsilon $
$2x^{2}-3x+1<2\epsilon $
choose $2\epsilon =\delta $
then $\frac{2x^{2}-3x+1}{2}<\frac{2\epsilon }{2}=\epsilon $
by transitivity of inequality
$\frac{2x^{2}-3x+1}{2x+2}<\epsilon $
| This function is defined at continuous at x = 1.
(It would be a different problem is we were looking for the limit as x approaches -1)
Unless your professor is demanding epsilon-delta proofs for the practice, you know that the limit of a defined continuous f as x approaches 1 is
$f(1) = \frac{1^2 - 1 + 1}{1 + 1} = \frac{1}{2}$
(It is OK to do epsilon-delta proofs for practice if the professor thinks you need more practice; just not necessary here.)
| {
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"source": "stackexchange",
"question_score": "3",
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Showing that a function from $\mathbb{R}^2 \to\mathbb{R}$ is bounded
Show that
$$ f(x,y)= \begin{cases} \dfrac{xy^2}{x^2+y^4} & (x,y) ≠ (0,0) \\ 0 & (x,y) = (0,0) \end{cases}$$
is bounded.
I thought about splitting it up into different cases like $x<y$ but it turned out to be too many and I could not cover all of them.
As a hint I got the idea to use the arithmetic geometric inequality.
I hope someone can help me.
| The insight here is that the fraction is of the form $\frac{ab}{a^2+b^2}$, with $a=x$ and $b=y^2$.
\begin{align*}
(x-y^2)^2 &\geq 0 \\
\implies x^2 - 2xy^2 + y^4 &\geq 0 \\
\implies x^2 + y^4 &\geq 2xy^2 \\
\implies\frac{xy^2}{x^2 + y^4} &\leq \frac{1}{2}
\end{align*}
You can also apply the arithmetic mean-geometric mean inequality with $a = x^2$ and $b=y^4$. Then
\begin{align*}
\sqrt{ab} &\leq \frac{1}{2}(a + b) \\
\implies xy^2 &\leq \frac{1}{2}(x^2+y^4) \\
\implies \frac{xy^2}{x^2+y^4} &\leq \frac{1}{2}
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2191360",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Plotting complex arcs on an Argand diagram What would $arg(\frac{z-4i}{z+2i}) = \frac{\pi}{4}$ look like on an Argand diagram and why?
I don't know how to check my answer because Desmos does not support complex numbers.
| We have a Möbius transformation
$$z'=\frac{z-4i}{z+2i}$$
which maps circles to circles or straight lines, and straight lines to circles or straight lines. We are looking for the shape of the set of complex numbers which will be mapped to a straight line of slope $1$. We suspect that the shape is that of a circle.
Let $z=a+ib.$ Then
$$\frac{a+ib-4i}{a+ib+2i}=\frac{(a+ib-4i)(a-ib-2i)}{(a+i(b+2))(a-i(b+2))}=$$
$$=\frac{a^2+b^2-2b-8-i6a}{a^2+(b+2)^2}.$$
The argument is $\frac{\pi}4$. So,
$$\operatorname {atan}\left(-\frac{6a}{a^2+b^2-2b-8}\right)=\frac{\pi}4.$$
Taking the tangent of both sides we get
$$-\frac{6a}{a^2+b^2-2b-8}=1.$$
Thus
$$a^2+b^2-2b+6a-8=(a+3)^2+(b-1)^2=18.$$
This is a circle centered at $-3+i$ and of radius $\sqrt{18}.$
EDIT
Here is the plot of the locus of the $z$'s sought:
So, negative $a$'s are considered.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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} |
Find the co-efficient of $x^{18}$ in the expansion of $(x+1)(x+2)...(x+10)(2x+1)(2x+3)...(2x+19)$.
Find the co-efficient of $x^{18}$ in the expansion of
$$(x+1)(x+2)...(x+10)(2x+1)(2x+3)...(2x+19)$$
What I've done :
$$
(x+1)(x+2)...(x+10)(2x+1)(2x+3)...(2x+19)
\\
=\frac{(2x+1)(2x+2)(2x+3)...(2x+20)}{2^{10}}
$$
I can't think of any way to find the coefficient of $x^{18}$ in this expansion. What am I doing wrong ?
| Now to get a term with $x^{18}$ you have to choose $18$ of the $2x$ terms and two of the constant terms, so your term in the numerator is $$(2x)^{18}\sum_{i=1}^{19}\sum_{j=i+1}^{20} ij$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Simplification of $\binom{50}{0}\binom{50}{1} + \binom{50}{1}\binom{50}{2}+⋯+\binom{50}{49}\binom{50}{50}$ There was a post on this web site an hour ago asking for the sum of
\begin{equation*}
\binom{50}{0}\binom{50}{1} + \binom{50}{1}\binom{50}{2}+⋯+\binom{50}{49}\binom{50}{50}
\end{equation*}
expressed as a single binomial coefficient. (Four choices were provided in the post.) The post seems to have been deleted. I think it is worth keeping on this web site.
| There is another proof for this problem that I am mentioning here :
Since :
$$
\binom{50}{0}\binom{50}{1} + \binom{50}{1}\binom{50}{2}+⋯+\binom{50}{49}\binom{50}{50} \\
\qquad = \binom{50}{0}\binom{50}{49} + \binom{50}{1}\binom{50}{48}+⋯+\binom{50}{49}\binom{50}{0} \\$$
Consider the product :
$$(1+x)^{50} \times (1+x)^{50}$$
Look for the coefficient of $x^{49}$ in this product, it will be calculated as :
$$ x^0 ~~\text{from first bracket} , x^{49} ~\text{from second bracket} = \binom{50}{0} \times \binom{50}{49}$$
$$ x^1 ~~\text{from first bracket} , x^{48} ~\text{from second bracket} = \binom{50}{1} \times \binom{50}{48}$$
$$ x^2 ~~\text{from first bracket} , x^{47} ~\text{from second bracket} = \binom{50}{2} \times \binom{50}{47}
\\.
\\.
\\.
\\.$$
$$ x^{49} ~~\text{from first bracket} , x^{0} \text{from second bracket} = \binom{50}{49} \times \binom{50}{0}$$
So, coefficient of $x^{49}$ in expansion of $(1+x)^{50}(1+x)^{50} =$
$$\binom{50}{0}\binom{50}{49} + \binom{50}{1}\binom{50}{48}+⋯+\binom{50}{49}\binom{50}{0} \\$$
Which is coefficient of $x^{49}$ in expansion of $(1+x)^{100} =$
$$\binom{100}{49}$$
Thus :
$$\binom{50}{0}\binom{50}{49} + \binom{50}{1}\binom{50}{48}+⋯+\binom{50}{49}\binom{50}{0}=\binom{100}{49}$$
| {
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"url": "https://math.stackexchange.com/questions/2195269",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Vectors $v_1,v_2,v_3$ belong to vector space $\mathbb{Z}^{3}_{7}$ What's dimension of the subspace $U_7$?
$v_1=\begin{pmatrix} 1\\ 3\\ 5 \end{pmatrix} , v_2=\begin{pmatrix}
4\\ 5\\ 6 \end{pmatrix}, v_3 = \begin{pmatrix} 6\\ 4\\ 2
\end{pmatrix}$ are vectors of the vector space $\mathbb{Z}^{3}_{7}$,
over the field $\mathbb{Z}_{7}$. What's the dimension of $U_{7}=
\text{span}\left\{v_1,v_2,v_3\right\}$?
I think the elements of all vectors are alright because they are in $\mathbb{Z}_{7}= \left\{0,1,2,3,4,5,6\right\}$, if that is understood correctly at all by me (?)
Now I form the vectors to a matrix: $\begin{pmatrix}
1 & 4 & 6\\
3 & 5 & 4\\
5 & 6 & 2
\end{pmatrix}$ and transpose it: $\begin{pmatrix}
1 & 3 & 5\\
4 & 5 & 6\\
6 & 4 & 2
\end{pmatrix}$
Multiply first line with $4$:
$\begin{pmatrix}
4 & 12 & 20\\
4 & 5 & 6\\
6 & 4 & 2
\end{pmatrix}$
But $12$ and $20 \notin \mathbb{Z}_{7}$. So we have to do $12 \text{ mod } 7= 5$ and $20 \text{ mod } 7= 6$
Insert that in the matrix: $\begin{pmatrix}
4 & 5 & 6\\
4 & 5 & 6\\
6 & 4 & 2
\end{pmatrix}$
Now subtract first line second line, we get: $\begin{pmatrix}
4 & 5 & 6\\
0 & 0 & 0\\
6 & 4 & 2
\end{pmatrix}$
And thus the dimension of $U_{7}=
\text{span}\left\{v_1,v_2,v_3\right\}$ is $2$?
Another question, I had to check at the beginning if the vectors are linearly independent ?
| We could have simply reduced the matrix by rows (with rows = the given vectors):
$$\begin{pmatrix}
1&3&5\\
4&5&6\\
6&4&2\end{pmatrix}\stackrel{R_2-4R_1,\,R_3-6R_1}\longrightarrow\begin{pmatrix}
1&3&5\\
0&0&0\\
0&0&0\end{pmatrix}\implies$$
The second and third vectors are linearly dependent on the first one , and in fact:
$$\begin{pmatrix}4\\5\\6\end{pmatrix}=4\begin{pmatrix}1\\3\\5\end{pmatrix}\;,\;\;\begin{pmatrix}6\\4\\2\end{pmatrix}=6\begin{pmatrix}1\\3\\5\end{pmatrix}$$
and thus the dimension of the span of these three vectors over $\;\Bbb F_7\cong\Bbb Z/7\Bbb Z\;$ is one .
Keeping track of which vector is represented by what row (in case we interchange some of them) we can say, at the end of the row reduction, exactly which ones are lin. dependent on which ones, and we can even write down exactly what is the linear dependency. I think that by far this is the best method.
| {
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If $a+b+c=abc$ then $\sum\limits_{cyc}\frac{1}{7a+b}\leq\frac{\sqrt3}{8}$ Let $a$, $b$ and $c$ be positive numbers such that $a+b+c=abc$. Prove that:
$$\frac{1}{7a+b}+\frac{1}{7b+c}+\frac{1}{7c+a}\leq\frac{\sqrt3}{8}$$
I tried C-S:
$$\left(\sum_{cyc}\frac{1}{7a+b}\right)^2\leq\sum_{cyc}\frac{1}{(ka+mb+c)(7a+b)^2}\sum_{cyc}(ka+mb+c)=$$
$$=\sum_{cyc}\frac{(k+m+1)(a+b+c)}{(ka+mb+c)(7a+b)^2}.$$
Thus, it remains to prove that
$$\sum_{cyc}\frac{k+m+1}{(ka+mb+c)(7a+b)^2}\leq\frac{3}{64abc},$$
but I did not find non-negative values of $k$ and $m$, for which the last inequality is true.
If we replace $7$ with $8$ so for $(a,b,c)||(28,1,5)$ this inequality would be wrong. Around this point the starting inequality is true, but we see that we can'not free use AM-GM because in AM-GM the equality occurs, when all variables are equal.
Thank you!
| Not a proof just a simplification for the proof wich already exists :
We prove the following inequality :
Let $a,b,c>0$ such that $abc=a+b+c$ and $a\geq b \geq c$ then we have :
$$\sum_{cyc}\frac{1}{7a+b}\leq \sum_{cyc}\frac{1}{7b+a}$$
Proof :
We work with the following equivalent :
Let $a,b,c>0$ and $a\geq b \geq c$ then we have :
$$\sqrt{\frac{abc}{a+b+c}}\sum_{cyc}\frac{1}{7a+b}\leq \sqrt{\frac{abc}{a+b+c}}\sum_{cyc}\frac{1}{7b+a}$$
Remains to show :
Let $a,b,c>0$ and $a\geq b \geq c$ then we have :
$$\sum_{cyc}\frac{1}{7a+b}\leq \sum_{cyc}\frac{1}{7b+a}$$
We make the difference we get :
$$\sum_{cyc}\frac{1}{7a+b}-\sum_{cyc}\frac{1}{7b+a}=-\frac{(42 (a - b) (a - c) (b - c) (7 a^2 + 57 a b + 57 a c + 7 b^2 + 57 b c + 7 c^2))}{((7 a + b) (a + 7 b) (7 a + c) (a + 7 c) (7 b + c) (b + 7 c))}\leq 0$$
I think it simplify a part of the proof of River Li and maybe create a new approach to solve the initial inequality.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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proof by induction (summations) I'm trying to use induction to prove that
$$\frac{1}{(1)(2)}+\frac{1}{(2)(3)}+...+\frac{1}{(n)(n+1)}=1-\frac{1}{n+1}.$$
This is what I have so far:
Base Case: $n=1$. We get $$\sum_{i=1}^{k} \frac{1}{(i)(i+1)} = 1-\frac{1}{(k+1)} \implies \frac{1}{2} = \frac{1}{2}.$$
End Goal (what we want to show): $$\sum_{i=1}^{k+1} \frac{1}{(i)(i+1)} = 1-\frac{1}{k+2}.$$
So,
$$\begin{align}
&\sum_{i=1}^{k} \frac{1}{(i)(i+1)} + \frac{1}{(k+1)(k+2)}\\
&=1-\frac{1}{k+1}+\frac{1}{(k+1)(k+2)}\\
&= 1-\frac{k+2}{(k+1)(k+2)}+\frac{1}{(k+1)(k+2)}\\
&= 1-\frac{(k+2)+1}{(k+1)(k+2)}
\end{align}$$
But I don't know where to go from here. I can't cancel the $k+2$'s because I need one in the denominator so the LHS = RHS, but if I expand the numerator to $(k+1)+2$ and then cancel the $k+1$'s I'm left with a $2$ in the numerator when I need a $1$. Any suggestions?
| You are almost there. The trick goes as by writing
$$\frac{1}{(k+1)(k+2)}=\frac{1}{k+1}-\frac{1}{k+2}$$ so that
$$1-\frac{1}{k+1}+\frac{1}{(k+1)(k+2)}=1-\frac{1}{k+1}+\frac{1}{k+1}-\frac{1}{k+2}=1-\frac{1}{k+2}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2197679",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Explain why twice the sum $\binom{12}{0} + \binom{12}{1} + \binom{12}{2} + \binom{12}{3} + \binom{12}{4} + \binom{12}{5}$ is $2^{12}-\binom{12}6$ Can someone explain how is the RHS concluded? I did with sample numbers and it is all correct. but I can't figure out how C(12,6) comes to play.
$$
\binom{12}{0} + \binom{12}{1} + \binom{12}{2} + \binom{12}{3} + \binom{12}{4} + \binom{12}{5} = (2^{12} - \binom{12}{6}) / 2
$$
| Hint: We have
$$
\sum_{i=0}^n \binom{n}{i} = 2^n
$$
and
$$
\binom{n}{i} = \binom{n}{n-i}
$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Fibonacci sequence, proof verification. At the end of chapter $10$ of his "Book of Proof", Hammack calls for a proof of the Fibonacci sequence in a form that I did not encounter before, I would like to know if the proof that I give for it is correct.
*
*$30)$ Here $F_n$ is the $n$th Fibonacci number. Prove that
$$F_n=\frac{\left(\frac {1+\sqrt 5}2\right)^n-\left(\frac {1-\sqrt 5}2\right)^n}{\sqrt 5}\,.$$
First, I can see the golden ratio in the equation,
$$\frac {1+\sqrt 5}2 = \lim_{n\to \infty} \frac {F_{n+1}}{F_{n}} = \phi\,.$$
Then, I look at how $\frac {1-\sqrt 5}2$ relates to $\phi$,
$$\frac 1\phi = \frac 2{1+\sqrt 5} = \frac {2(1-\sqrt 5)}{1-5} = -\frac{1-\sqrt5}2\,.$$
Using this in the given formula for $F_n$ gives,
$$F_n = \frac 1{\sqrt 5}\left(\phi^n - \left(-\frac 1\phi\right)^n\right)\,.$$
So I will use that as a simplification for my induction proof.
I start by making sure that $F_n$ holds for $n=1$ and $n=2$ such that the inductive step can be taken,
$$F_1 = \frac 1{2 \sqrt 5}(1+\sqrt 5 -1 +\sqrt5) = \frac{2\sqrt 5}{2\sqrt5} = 1$$
$$F_2 = \frac 1{4 \sqrt 5}(1+2\sqrt 5+5 -1 +2\sqrt5 -5) = \frac{4\sqrt 5}{4\sqrt5} = 1$$
and I assume $F_{n} \,:\, n\in \Bbb Z^+$ is true.
We know that $F_n = F_{n-1}+F_{n-2}$ ergo,
$$\begin{align}
F_{n}+F_{n-1} & = \frac 1{\sqrt 5}\left(\phi^{n} - \left(-\frac 1\phi\right)^{n}\right) + \frac 1{\sqrt 5}\left(\phi^{n-1} - \left(-\frac 1\phi\right)^{n-1}\right)\\
& = \frac 1{\sqrt 5}\left[\left(\phi^{n} + \phi^{n-1}\right) - \left(\left( -\frac 1\phi\right)^{n} + \left(-\frac 1\phi\right)^{n-1}\right)\right]\\
& \require{cancel}\cancel{= \frac 1{\sqrt 5}\left[\phi^{n}(1 + \frac 1\phi) - \left(1 - \phi\right)\left( -\frac 1\phi\right)^{n}\right]} \\
\end{align}$$
$$\require{cancel}\cancel{1+\frac 1\phi = 1-\phi = \phi}$$
$$\require{cancel}\cancel{\therefore\quad F_{n}+F_{n-1} = \frac 1{\sqrt 5}\left(\phi^n - \left(-\frac 1\phi\right)^n\right) = F_{n+1}}$$
Edit
Following the suggestions in the comments I edited the question and as suggested, I tried using the fact that $\phi$ and $-\frac{1}{\phi}$ are the roots of $x^2=x+1$ which I overlooked the first time.
So with that information in hand, I multiply $\phi^2 =\phi + 1$ by $\phi^{n-1}$ and $\left(-\frac 1\phi\right)^2 = \left(-\frac 1\phi\right) + 1$ by $\left(-\frac {1}\phi\right)^{n-1}$ as suggested to see,
$$\phi^{n+1} = \phi^n + \phi^{n-1}$$
$$\left(-\frac 1\phi\right)^{n+1} = \left(-\frac 1\phi\right)^n + \left(-\frac {1}\phi\right)^{n-1}$$
Then, knowing $F_1 =F_2 =1$ and assuming $F_{n} \,:\, n\in \Bbb Z^+$ is true, I use what I already found for $F_n+F_{n-1}$ and get,
$$\begin{align}
F_{n}+F_{n-1} & = \frac 1{\sqrt 5}\left(\phi^{n} - \left(-\frac 1\phi\right)^{n}\right) + \frac 1{\sqrt 5}\left(\phi^{n-1} - \left(-\frac 1\phi\right)^{n-1}\right) \\
& = \frac 1{\sqrt 5}\left[\left(\phi^{n} + \phi^{n-1}\right) - \left(\left( -\frac 1\phi\right)^{n} + \left(-\frac 1\phi\right)^{n-1}\right)\right]\\
& = \frac 1{\sqrt 5}\left[\phi^{n+1} - \left(-\frac 1{\phi}\right)^{n+1}\right] = F_{n+1} \\
\end{align}$$
$$\frac{F_{n}\implies F_{n+1}\;,\;F_{n}}{\therefore\;F_{n+1}}\qquad \text{modus ponens} \tag*{$\blacksquare$}$$
Is this proof correct and acceptable?
| Yes, your proof seems correct and acceptable (given the edit).
| {
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Show that for $p$ to be odd prime and $p \equiv 3$ mod $4$, then $x^2+y^2 = p$ has no integer solution Show that for $p$ to be odd prime and $p \equiv 3$ mod $4$, then $x^2+y^2 = p$ has no integer solution. I have no idea how can i apply quadratic reciprocity to the equation $x^2+y^2 = p$ or should use other method.
| $p$ is odd then $x$ and $y$ have different parity. We can assume without loss of generality that $x$ is odd and $y$ is even.
We set $$x=2X+1$$
$$y=2Y$$
Then $$x^2+y^2=4X^2+4X+1+4Y^2=4(X^2+X+Y^2)+1=p$$
Hence, We only have solutions if only:
$$p\equiv 1\pmod 4$$
| {
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"timestamp": "2023-03-29T00:00:00",
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$\sqrt{x^{2}+\frac{1}{n}}$ converges uniformly to $\left|x\right|$ Im trying to prove that $f_{n}\left(x\right)=\sqrt{x^{2}+\frac{1}{n}}$ converges uniformly to $f(x) = \left|x\right|$ in $[-1,1]$.
So for evary $\varepsilon$ exists $N \in \mathbb{N}$ s.t for all $n>N$ and for all $x \in [-1,1]$ $\left|f_{n}\left(x\right)-f\left(x\right)\right| \le \varepsilon$
And this how i tried to continue :
$\left|f_{n}\left(x\right)-f\left(x\right)\right|=\left|\sqrt{x^{2}+\frac{1}{n}}-\left|x\right|\right|=\left|\sqrt{x^{2}+\frac{1}{n}}-\sqrt{x^{2}}\right|=\left|\frac{\left(\sqrt{x^{2}+\frac{1}{n}}-\sqrt{x^{2}}\right)\left(\sqrt{x^{2}+\frac{1}{n}}+\sqrt{x^{2}}\right)}{\sqrt{x^{2}+\frac{1}{n}}+\sqrt{x^{2}}}\right|=\left|\frac{x^{2}+\frac{1}{n}-x^{2}}{\sqrt{x^{2}+\frac{1}{n}}+\sqrt{x^{2}}}\right|=\left|\frac{\frac{1}{n}}{\sqrt{x^{2}+\frac{1}{n}}+\sqrt{x^{2}}}\right|$
Which lead me to the same problem of $\left|\sqrt{x^{2}+\frac{1}{n}}-\sqrt{x^{2}}\right|$
any advice ?
| We have
$$\left\vert \sqrt{x^2 + \frac{1}{n}} + \sqrt{x^2}\right\vert \geq \frac{1}{\sqrt{n}},$$
thus by your work thus far
$$\vert f_n(x) - \sqrt{x} \vert \leq \frac{\frac{1}{n}}{\frac{1}{\sqrt{n}}} = \frac{1}{\sqrt{n}}.$$
| {
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If $\frac{a\sqrt{2}+b}{b\sqrt{2}+c}$ is rational, then $a+b+c|ab+bc+ca$ Let $a,b,c\in\mathbb{N_{>0}}$. Prove that if $$\frac{a\sqrt{2}+b}{b\sqrt{2}+c}$$ is a rational number, then $a+b+c|ab+bc+ca$.
Hints?
My approach was to assume that
$$ \frac{a\sqrt{2}+b}{b\sqrt{2}+c}=\frac{p}{q}$$
then
$$ pb\sqrt{2}+pc=qa\sqrt{2}+qb$$
or
$$ \sqrt{2}(pb-qa)=qb-pc$$
squaring:
$$ 2(pb-qa)^2=(qb-pc)^2$$
but that doesn't seems to do much.
| Let $\frac{a\sqrt2+b}{b\sqrt2+c}=r\in\mathbb Q$.
Hence, $(a-rb)\sqrt2=rc-b$.
If $a-rb\neq0$ so we get a contradiction.
Thus, $a-rb=0$ and $rc-b=0$ or $a=rb$ and $c=\frac{b}{r}$.
Id est, $$\frac{ab+ac+bc}{a+b+c}=\frac{b^2\left(1+r+\frac{1}{r}\right)}{b\left(1+r+\frac{1}{r}\right)}=b$$
and we are done!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2202753",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How to prove that $\frac{1}{n+1}+\frac{1}{n+2}+...+\frac{1}{2n} \le \frac{n}{n+1}$? I have a series $$a_n = \frac{1}{n+1}+\frac{1}{n+2}+...+\frac{1}{2n} \quad n \ge 1$$
For example, $a_3 = \frac{1}{4}+\frac{1}{5}+\frac{1}{6}$.
I need to prove that for $n \ge 1$:
$$a_n = \frac{1}{n+1}+\frac{1}{n+2}+...+\frac{1}{2n} \le \frac{n}{n+1}$$
I guess one could say that:
$$
\sum_{i=1}^n\frac{1}{n+i} \le \sum_{i=1}^n\frac{n}{n+1}
$$
However, I'm not sure this is rigorous enough (for example, in $\sum_{i=1}^n\frac{1}{n+i}$ how do we really know that the index goes from $1$ to $n$) and I think this needs to be proven via induction.
So the base case is:
$$a_1 = \frac{1}{2} \le \frac{1}{2} = \frac{n}{n+1}$$
The step: suppose $a_n \le \frac{n}{n+1}$ then let's prove that $$a_{n+1} \le \frac{n+1}{n+2}$$
The above can be developed as:
$$
\frac{1}{n+1}+\frac{1}{n+3}+...+\frac{1}{2(n+1)} \le \frac{n}{n+1}+\frac{1}{n+2}
$$
This is where I get stuck. If I could somehow prove that the number of terms to the left $\le$ the terms to the left I would be golden. Or maybe there's another way.
| This is wrong:
$$a_{n+1}=\frac{1}{n+1}+\frac{1}{n+3}+...+\frac{1}{2(n+1)}$$
You should have:
$$a_{n+1}=\frac{1}{n+2}+\frac{1}{n+3}+...+\frac{1}{2(n+1)}$$
Note that this has $1$ extra term than $a_n$ because the series is defined using $n$ twice - once for $n+1$ and again at $2n$.
From here subtract $a_n$ and rearrange to get:
$$a_{n+1}=a_{n}-\dfrac1{n+1}
+\dfrac1{2n+1}+\dfrac1{2n+2}$$
We have:
$$\dfrac1{n+1}=\dfrac2{2(n+1)}=\dfrac1{2n+2}+\dfrac1{2n+2}$$
So the extra fractions become:
$$
\begin{align}
&-\dfrac1{n+1}+\dfrac1{2n+1}+\dfrac1{2n+2}\\
&=-\dfrac1{2n+2}-\dfrac1{2n+2}+\dfrac1{2n+1}+\dfrac1{2n+2}\\
&=-\dfrac1{2n+2}+\dfrac1{2n+1}\\
&=\dfrac1{2n+1}-\dfrac1{2n+2}\\
&=\dfrac1{(2n+1)(2n+2)}
\end{align}
$$
As $a_n \le \dfrac{n}{n+1}$ we need to prove:
$$\dfrac{n}{n+1}+\dfrac1{(2n+1)(2n+2)}\le\dfrac{n+1}{n+2}$$
Or equivalently that:
$$\dfrac1{(2n+1)(2n+2)}\le\dfrac{n+1}{n+2}-\dfrac{n}{n+1}$$
The right hand side is:
$$\dfrac{n+1}{n+2}-\dfrac{n}{n+1}=\dfrac{n^2+2n+1-n^2-2n}{(n+1)(n+2)}=\dfrac1{(n+1)(n+2)}$$
and we are done ($(n+1)(n+2)\le(2n+1)(2n+2)$ for $n\in\mathbb{N}$).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2204141",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How do you derive the following two summations?
I tried using the geometric series formula, $$ \sum_{n=0}^{k-1} ar^{n}=a\left (\frac{1-r^{n}}{1-r} \right ) $$, but I got $1-2\left ( \frac{1}{2}^{n-1} \right )$ for the first one and $2-2\left ( \frac{1}{2}^{n-1} \right )$ for the second.
Edit: I derived the first answer by doing $1\cdot \frac{1-\left (\frac{1}{2}\right)^{n-1}}{1-\frac{1}{2}}$, which results in $2\cdot\left(1-\left (\frac{1}{2}\right)^{n-1} \right)$ = $2-2\left (\frac{1}{2}\right)^{n-1}$. Since it starts out as n=1, I did $2-2\left (\frac{1}{2}\right)^{n-1}-1$, so the final answer I got is $1-2\left ( \frac{1}{2}^{n-1} \right )$
| $\sum_{n=0}^{n-1}a\left (\frac{1-r^{n}}{1-r} \right ) =
\frac {a(n-1)}{1-r}- \frac{1}{1-r}\sum_{n=0}^{n-1} r^n $
$=\frac {a(n-1)}{1-r}- a\frac{(1-r^n)}{(1-r)^2}$ [by multiplying by $\frac{1-r}{1-r}$]
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2205264",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Blockwise cofactor matrix identity Wikipedia gives an identity for blockwise inversion, assuming the appropriate inverses exist:
$$\begin{bmatrix}\mathbf{A} & \mathbf{B} \\ \mathbf{C} & \mathbf{D}\end{bmatrix}^{-1} = \begin{bmatrix} \mathbf{A}^{-1}+\mathbf{A}^{-1}\mathbf{B}(\mathbf{D}-\mathbf{CA}^{-1}\mathbf{B})^{-1}\mathbf{CA}^{-1} & -\mathbf{A}^{-1}\mathbf{B}(\mathbf{D}-\mathbf{CA}^{-1}\mathbf{B})^{-1} \\ -(\mathbf{D}-\mathbf{CA}^{-1}\mathbf{B})^{-1}\mathbf{CA}^{-1} & (\mathbf{D}-\mathbf{CA}^{-1}\mathbf{B})^{-1} \end{bmatrix}$$
Is there a corresponding general formula for the matrix of cofactors:
$$\operatorname{adj}^T\begin{bmatrix}\mathbf{A} & \mathbf{B} \\ \mathbf{C} & \mathbf{D}\end{bmatrix}$$
for when $\mathbf{A}^{-1}$ or $(\mathbf{D}-\mathbf{C}\mathbf{A}^{-1}\mathbf{B})^{-1}$ do not necessarily exist?
Is there at least one for bordered matrices, i.e. $\mathbf{D} \in \mathbb{C}^{1\times 1}$?
| $\newcommand{\inv}[1]{{#1}^{-1}}
\newcommand{\adj}{\operatorname{adj}}
$
Let $M = \begin{pmatrix} A_{m\times m} & B_{m \times n} \\ C_{n \times m} & D_{n \times n}\end{pmatrix}$ then the following always holds
$$
\begin{pmatrix} (\det{D})^{m-1} I_m & 0 \\ 0 & (\det(A))^{n-1}I_n \end{pmatrix} \adj M = \begin{pmatrix}\adj(A\det(D)-B\adj(D)C) & 0 \\ 0 & \adj(D\det(A) - C\adj(A)B)\end{pmatrix}\begin{pmatrix} \det(D)I_m & -B\adj D\\
-C\adj A & \det(A)I_n\end{pmatrix}.\tag{0}$$
Proof: Assume, initially , that $M,A,D$ are invertible. This also implies the invertibility of $F = D - C\inv{A}B$ and $G=A-B\inv{D}C.$
We have $\det(M) = \det(A) \det(F) = \det(D)\det(G).$
Under these assumptions $$M^{-1} = \begin{pmatrix} \inv{G} & 0 \\ 0 & \inv{F} \end{pmatrix}\begin{pmatrix} I_m & -B\inv{D}\\ -C\inv{A} & I_n\end{pmatrix}$$
and $\adj M = \det(M) \inv{M} = \begin{pmatrix} \det(G)\inv{G} & 0 \\ 0 & \det(F) \inv{F} \end{pmatrix}\begin{pmatrix} \det(D)I_m & -B\det(D)\inv{D}\\
-C\det(A)\inv{A} & \det(A)I_n\end{pmatrix}.$
So, $\adj M = \begin{pmatrix} \adj{G} & 0 \\ 0 & \adj(F) \end{pmatrix}\begin{pmatrix} \det(D)I_m & -B\adj D\\
-C\adj A & \det(A)I_n\end{pmatrix}. \tag{1}$
Now $G = \dfrac{A\det(D) - B\adj(D)C}{\det(D)},$ so
$ \adj(G) = \left(\dfrac{1}{\det(D)}\right)^{m-1}\adj(A\det(D)-B\adj(D)C)$ and similarly $\adj(F) = \left(\dfrac{1}{\det(A)}\right)^{n-1}\adj(D\det(A) - C\adj(A)B). \tag{2}$
Plugging (2) into (1) we get
$$
\begin{pmatrix} (\det{D})^{m-1} I_m & 0 \\ 0 & (\det(A))^{n-1}I_n \end{pmatrix} \adj M = \begin{pmatrix}\adj(A\det(D)-B\adj(D)C) & 0 \\ 0 & \adj(D\det(A) - C\adj(A)B)\end{pmatrix}\begin{pmatrix} \det(D)I_m & -B\adj D\\
-C\adj A & \det(A)I_n\end{pmatrix}.$$
In the space of $n \times n$ real (or complex) matrices, the invertible matrices are dense. Given any $M = \begin{pmatrix} A & B \\ C & D \end{pmatrix}$ we will find a sequence of matrices $M_n = \begin{pmatrix} A_n & B_n \\ C_n & D_n \end{pmatrix}$ such that $M_n,A_n,D_n$ are invertible and $\lim_{n\to\infty}M_n = M.$
Since (0) holds for $M_n,A_n,B_n,C_n$ and since the LHS and RHS are continuous functions we can let $n \to \infty$ and get the result for all $M$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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A natural number $a$ has four digits and $a^2$ ends with the same four digits as that of $a$. Find the value of $a$?
A natural number $a$ has four digits and $a^2$ ends with the same four digits as that of $a$. Find the value of $a$?
This question appeared in a math contest. I tried for a while and came up with this condition: $10^4|(a-1)a$. But I'm not able to take it further. It is clear that out of the two ($a-1$ and $a$) the one which is odd will contain no powers of 2 and hence the other one would contain the $2^4$. But there are to many possibilities for one to see by trial and error.
| The equation $a^2 = a$ has exactly two solutions ($0$ and $1$) modulo any prime power, because:
*
*Mod $p$, it has only two roots $0$ and $1$, because it's quadratic, and $\mathbb F_p$ is a field;
*By Hensel's lemma, since the derivative of $a^2-a$ is $2a-1$ and is nonzero when $a=0$ or $a=1$ we can lift those solutions.
So we know that:
*
*$a^2 \equiv a \pmod{2^4}$ only if $a \equiv 0 \pmod{2^4}$ or $a \equiv 1 \pmod{2^4}$,
*$a^2 \equiv a \pmod{5^4}$ only if $a \equiv 0 \pmod{5^4}$ or $a \equiv 1 \pmod{5^4}$.
However, modulo $10^4$, we have more freedom, because we can glue these solutions any way we like by the Chinese remainder theorem. Our options are
\begin{align}
a\equiv 0 \pmod{2^4}\text{ and }a\equiv 0 \pmod{5^4} &\implies a \equiv 0 \pmod{10^4} \\
a\equiv 0 \pmod{2^4}\text{ and }a\equiv 1 \pmod{5^4} &\implies a \equiv 9376 \pmod{10^4} \\
a\equiv 1 \pmod{2^4}\text{ and }a\equiv 0 \pmod{5^4} &\implies a \equiv 625 \pmod{10^4} \\
a\equiv 1 \pmod{2^4}\text{ and }a\equiv 1 \pmod{5^4} &\implies a \equiv 1 \pmod{10^4}
\end{align}
(To compute the two middle values, take $2^4$ times the inverse of $2^4$ modulo $5^4$, and $5^4$ times the inverse of $5^4$ modulo $2^4$, respectively.)
So in general (modulo $10^n$ for other $n$) there's two solutions other than $0$ and $1$, but because $625$ is not actually a four-digit number, the correct one here is just $9376$.
I guess a simpler way to get to the Chinese remainder step is to observe that if $10^4 \mid a(a-1)$, one of $a$ and $a-1$ is odd, and one of $a$ and $a-1$ is not divisible by $5$, so we must have $2^4 \mid a$ and $5^4 \mid a-1$ or the reverse to get a nontrivial solution. But I don't see a way to avoid the Chinese remainder theorem.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Solution of non exact differential equations with integration factor depend both $x$ and $y$ I'm not finding any general description to solve a non exact equation which's integrating factor depend both on $x$ and $y$.
I'm on this problem $$(2x^{2}-y)dx+(x+y^2)dy=0 $$
I am trying to solve and kind of stuck now which is given below.
| $(2x^2-y)~dx+(x+y^2)~dy=0$
$(y^2+x)\dfrac{dy}{dx}=y-2x^2$
Which relates to an ODE of the form http://science.fire.ustc.edu.cn/download/download1/book%5Cmathematics%5CHandbook%20of%20Exact%20Solutions%20for%20Ordinary%20Differential%20EquationsSecond%20Edition%5Cc2972_fm.pdf#page=166.
Let $u=\dfrac{1}{x}$ ,
Then $\dfrac{dy}{dx}=\dfrac{dy}{du}\dfrac{du}{dx}=-\dfrac{1}{x^2}\dfrac{dy}{du}=-u^2\dfrac{dy}{du}$
$\therefore-\left(y^2+\dfrac{1}{u}\right)u^2\dfrac{dy}{du}=y-\dfrac{2}{u^2}$
$(u^2y^2+u)\dfrac{dy}{du}=\dfrac{2}{u^2}-y$
Let $t=uy$ ,
Then $y=\dfrac{t}{u}$
$\dfrac{dy}{du}=\dfrac{1}{u}\dfrac{dt}{du}-\dfrac{t}{u^2}$
$\therefore(t^2+u)\left(\dfrac{1}{u}\dfrac{dt}{du}-\dfrac{t}{u^2}\right)=\dfrac{2}{u^2}-\dfrac{t}{u}$
$\left(\dfrac{t^2}{u}+1\right)\dfrac{dt}{du}-\dfrac{t^3}{u^2}-\dfrac{t}{u}=\dfrac{2}{u^2}-\dfrac{t}{u}$
$\left(\dfrac{t^2}{u}+1\right)\dfrac{dt}{du}=\dfrac{t^3+2}{u^2}$
$(t^3+2)\dfrac{du}{dt}=t^2u+u^2$
$\dfrac{du}{dt}-\dfrac{t^2u}{t^3+2}=\dfrac{u^2}{t^3+2}$
Luckily this becomes a Bernoulli ODE.
If we go back to the original problem,
and if we let $t=\dfrac{y}{x}$ ,
Then $y=xt$
$\dfrac{dy}{dx}=x\dfrac{dt}{dx}+t$
$\therefore(x^2t^2+x)\left(x\dfrac{dt}{dx}+t\right)=xt-2x^2$
$(xt^2+1)x^2\dfrac{dt}{dx}+x^2t^3+xt=xt-2x^2$
$(xt^2+1)x^2\dfrac{dt}{dx}=-(t^3+2)x^2$
$(t^3+2)\dfrac{dx}{dt}=-t^2x-1$
$\dfrac{dx}{dt}+\dfrac{t^2x}{t^3+2}=-\dfrac{1}{t^3+2}$
Which is also luckily this becomes a linear ODE.
I.F.$=e^{\int\frac{t^2}{t^3+2}dt}=e^{\int\frac{d(t^3+2)}{3(t^3+2)}}=e^\frac{\ln(t^3+2)}{3}=\sqrt[3]{t^3+2}$
$\therefore\dfrac{d\left(x\sqrt[3]{t^3+2}\right)}{dt}=-\dfrac{1}{(t^3+2)^\frac{2}{3}}$
$x\sqrt[3]{t^3+2}=C-\int^t\dfrac{d\tau}{(\tau^3+2)^\frac{2}{3}}$
$x\sqrt[3]{\dfrac{y^3}{x^3}+2}=C-\int^\frac{y}{x}\dfrac{d\tau}{(\tau^3+2)^\frac{2}{3}}$
$\sqrt[3]{2x^3+y^3}+\int^\frac{y}{x}\dfrac{d\tau}{(\tau^3+2)^\frac{2}{3}}=C$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2208786",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Evaluation of the series $\sum_{n=1}^{\infty} \frac{n^2+2n+3}{2^n}$. Find the following sum
$$\sum_{n=1}^{\infty} \frac{n^2+2n+3}{2^n}$$
Could someone give some hint to proceed in this question?
| The given series is absolutely convergent, and by setting $S=\sum_{n\geq 1}\frac{n^2+2n+3}{2^n}$ we also get:
$$ 2S = \sum_{n\geq 1}\frac{n^2+2n+3}{2^{n-1}} = 6+\sum_{n\geq 1}\frac{n^2+4n+6}{2^n}\tag{1}$$
from which:
$$ S = 2S-S = 6+\sum_{n\geq 1}\frac{2n+3}{2^n}\tag{2} $$
and by setting $T=\sum_{n\geq 1}\frac{2n+3}{2^n}$ we also get:
$$ 2T = \sum_{n\geq 1}\frac{2n+3}{2^{n-1}} = 5+\sum_{n\geq 1}\frac{2n+5}{2^n}\tag{3} $$
$$ T = 2T-T = 5+\sum_{n\geq 1}\frac{2}{2^n} = 7 \tag{4} $$
so $S=6+7=\color{red}{13}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2209638",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Solving $\sqrt[3]{{x\sqrt {x\sqrt[3]{{x \ldots \sqrt x }}} }} = 2$ with 100 nested radicals I have seen a book that offers to solve the following equation:
$$\underbrace {\sqrt[3]{{x\sqrt {x\sqrt[3]{{x \ldots \sqrt x }}} }}}_{{\text{100 radicals}}} = 2$$
The book also contains the answer:
$$x = {2^{\left( {\frac{{5 \times {6^{50}}}}{{3 \times ({6^{50}} - 1)}}} \right)}}$$
How did they get the answer for such equation? I tried to obtain the recurrence relation, but could not find the way to get the above answer.
EDIT
$${u_{100}} = 2,$$
$$\sqrt[3]{{x{u_{99}}}} = 2,$$
$$x{u_{99}} = {2^3},$$
$${u_{99}} = \sqrt {x{u_{98}}} ,$$
$$x\sqrt {x{u_{98}}} = {2^3},$$
$${x^2}x{u_{98}} = {({2^3})^2},$$
$${x^3}{u_{98}} = {2^6},$$
$${u_{98}} = \sqrt[3]{{x{u_{97}}}},$$
$${x^3} \times \sqrt[3]{{x{u_{97}}}} = {2^6},$$
$${x^9}x{u_{97}} = {({2^6})^3},$$
$${x^{10}}{u_{97}} = {2^{18}},$$
$${u_{97}} = \sqrt {x{u_{96}}} ,$$
$${x^{10}}\sqrt {x{u_{96}}} = {2^{18}},$$
$${x^{20}}x{u_{96}} = {({2^{18}})^2},$$
$${x^{21}}{u_{96}} = {2^{36}},$$
$${u_{96}} = \sqrt[3]{{x{u_{95}}}},$$
$${x^{21}} \times \sqrt[3]{{x{u_{95}}}} = {2^{36}},$$
$${x^{63}}x{u_{95}} = {2^{108}}$$
$${x^{64}}{u_{95}} = {2^{108}},$$
$${u_{95}} = \sqrt {x{u_{94}}} ,$$
$${x^{64}}\sqrt {x{u_{94}}} = {2^{108}},$$
$${x^{128}}x{u_{94}} = {2^{216}},$$
$${x^{129}}{u_{94}} = {2^{216}},$$
$$ \ldots $$
but I still have no idea how to find a generalized formula which allows to obtain the answer.
| Let $u_n$ be the expression with $n$ radicals. Then $u_{100} = 2$. Then $u^3_{100} = xu_{99} = 2^3$. Next $(xu_{99})^2 = x^3xu_{98} = 2^9$. Keep going and try to find a pattern.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2211193",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 1
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Solve $\frac{2}{\sin x \cos x}=1+3\tan x$ Solve this trigonometric equation given that $0\leq x\leq180$
$\frac{2}{\sin x \cos x}=1+3\tan x$
My attempt,
I've tried by changing to $\frac{4}{\sin 2x}=1+3\tan x$, but it gets complicated and I'm stuck. Hope someone can help me out.
| Multiply both sides by $sin(x)cos(x)$ and rearrange to get $2-3sin^2(x)=sin(x)cos(x)$.
Square both sides: $4-12sin^2(x)+9sin^4(x)=sin^2(x)cos^2(x)$.
Substitute $1-sin^2(x)$ for $cos^2(x)$ to get $$
4-12sin^2(x)+9sin^4(x) = sin^2(x) - sin^4(x)$$
$$10sin^4(x)-13sin^2(x)+4=0$$ Substitute $u=sin^2(x)$ to get a quadratic equation, with solution $u=\frac{13\pm\sqrt{169-160}}{20}=\frac{4}{5},\frac{1}{2}$
Then solve for $x$ given $u$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2211823",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 4
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Finding the general formula of the sequence $ a_n = 7a_{n-2} + 6a_{n-3} $ Given the equation $ a_n = 7a_{n-2} + 6a_{n-3} $, and $ a_0 = a_1 = a_2 = 1 $, how do I find the general equation?
I have tried to express this sequence in matrix form, $ Q = Xb $ as follows
$$ Q = \begin{bmatrix} a_{n+1} \\ a_{n-2} \\ a_{n-3} \\ \end{bmatrix}
\hspace{1cm}
X =\begin{bmatrix}0 & 7 & 6 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{bmatrix}
\hspace{1cm}
b = \begin{bmatrix} a_{n} \\ a_{n-1} \\ a_{n-2} \\ \end{bmatrix} $$
I attempt diagonalize $X$ in order to find $X^n$, but the eigenvalues of $X$ seem to be not real
Can anyone see if my method is correct?
| We have $a_n = 7a_{n-2}+6a_{n-3}$ and we can write:
\begin{align}
a_n &= 0a_{n-1}+7a_{n-2}+6a_{n-3}\\
a_{n-1} &= 1a_{n-1} + 0a_{n-2} +0a_{n-3}\\
a_{n-2} &= 0a_{n-1}+1a_{n-2}+0a_{n-3}
\end{align}
Setting
$$A = \begin{pmatrix}
0 & 7 & 6\\
1 & 0 & 0\\
0 & 1 & 0
\end{pmatrix}\ \text{and}\ \ x_n = \begin{pmatrix} a_n\\ a_{n-1}\\ a_{n-2}\end{pmatrix}$$
we get that $$x_n = Ax_{n-1} = A^2x_{n-2} =\ldots = A^{n-2}x_2$$
Eigenvalues of $A$ are $\{-1,-2,3\}$ with corresponding eigenvectors $(1,-1,1), (4,-2,1), (9,3,1)$. Can you finish from here?
Another way to do this is using generating functions. Let $f(x) = \sum_{n=0}^\infty a_nx^n$. We have that $a_n = 7a_{n-2}+6a_{n-3}$ and we can multiply it by $x^n$ and sum it for all $n\geq 3$ to get
\begin{align}&\sum_{n=3}^\infty a_nx^n = 7\sum_{n=3}^\infty a_{n-2}x^n + 6\sum_{n=3}^\infty a_{n-3}x^n\\
\implies &\sum_{n=3}^\infty a_nx^n = 7x^2\sum_{n=1}^\infty a_nx^n + 6x^3\sum_{n=0}^\infty a_nx^n\\
\implies &f(x) -a_2x^2-a_1x-a_0 = 7x^2(f(x)-a_0)+6x^3f(x)\\
\implies &f(x)=\frac{1+x-6x^2}{1-7x^2-6x^3}\\
\implies &f(x)=\frac 32\cdot\frac 1{1+x} -\frac 45\cdot\frac 1{1+2x}+\frac 3{10}\frac 1{1-3x}\\
\implies &f(x) = \frac 32\sum_{n=0}^\infty (-1)^nx^n-\frac 45\sum_{n=0}^\infty (-2)^nx^n+\frac 3{10}\sum_{n=0}^\infty 3^nx^n\\
\implies &f(x) = \sum_{n=0}^\infty \left(\frac 32(-1)^n-\frac 45(-2)^n + \frac 3{10}\cdot 3^n\right)x^n\\
\implies &a_n = (-1)^n\left(\frac 32 -\frac{2^{n+2}}5\right)+\frac{3^{n+1}}{10}
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2213005",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
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Compositions of 50 with unique restrictions solve by generating function. Consider the compositions of 50 that have exactly 5 summands, such that the first and last summands are odd, all other summands are even, and no summand is greater than 20.
Use a generating function to count the number of such compositions.
Simplify your answer to an integer.
$(x^2+x^4+...+x^{20})^3=x^6 (1+x^2+...+x^{18})^3$
i think this counts all the even summands (there is 3 of them so i cubed it.)
the odd summands should be $(x+x^3+....+x^{19})^2$
so we have $g(x)= x^8(1+x^2+....+x^{18})^2 (1+x^2+...+x^{18})^3= x^8(1+x^2+...+x^{18})^5 = \frac{1-y^{9+1}}{1-y}=(\frac{1-x^{20}}{1-x^2})^5x^8$
theres about half a dozen things im lost on.
The fact that its a composition means the odd sum should be $(x+x^3+...+x^{19})$ right?
$ \frac{1-y^{9+1}}{1-y}=(\frac{1-x^{20}}{1-x^2})^5x^8$ im not sure this substitution is right could easily be $x^{19}$ instead of 20
Lastly from $(\frac{1-x^{20}}{1-x^2})^5x^8$ how do i push this into a suming serie so i can talk about the coefficient of the $x^{50}$ term?
| You have it all correct so far.
To alay some of your concerns I have personally always found it far easier to remember the formula for the sum of a finite geometric sequence as
$$\text{sum of geometric sequence}=\frac{(\text{first term})\: -\: (\text{term after last term})}{1\: -\: \text{ratio}}$$
In this case you have $1+x^2+x^4+\ldots +x^{18}$ with a ratio of $x^2$, so the term after last would be $x^{20}$ and obviously the first term is $1$, hence
$$1+x^2+x^4+\ldots +x^{18}=\frac{1-x^{20}}{1-x^2}$$
So you have correctly arrived at the generating function
$$x^8\left(\frac{1-x^{20}}{1-x^2}\right)^5$$
$$\implies x^8\frac{(1-x^{20})^5}{(1-x^2)^5}$$
$$\implies x^8(1-x^{20})^5(1-x^2)^{-5}$$
$$\implies x^8\left(\sum_{k=0}^{5}(-1)^k\binom{5}{k}x^{20k}\right)(1-x^2)^{-5}$$
The binomial with negative index has a Taylor expansion, or we could substitute $y=x^2$ and realise that the combinatorial interpretation for the $r^{\text{th}}$ coefficient of this is just the number of ways of placing $r$ balls into $5$ bins, there are $\binom{4+r}{4}$ ways to do this using stars and bars, hence
$$\begin{align}(1-x^2)^{-5}=(1-y)^{-5}&=\sum_{r\ge 0}\binom{4+r}{4}y^r\\&=\sum_{r\ge 0}\binom{4+r}{4}x^{2r}\end{align}$$
we can then multiply out
$$\left(\sum_{k=0}^{5}(-1)^k\binom{5}{k}x^{20k+8}\right)\left(\sum_{r\ge 0}\binom{4+r}{4}x^{2r}\right)$$
$$\implies \sum_{r\ge 0}\left(\sum_{k=0}^{5}\left((-1)^k\binom{5}{k}\binom{4+r}{4}x^{20k+2r+8}\right)\right)$$
If we re-label $p=20k+2r+8$ and sum over $p$ instead of $r$ then $r=(p-20k-8)/2$ so our generating function is
$$\sum_{p\ge 8}\left(\sum_{k=0}^{5}(-1)^k\binom{5}{k}\binom{\frac{p}{2}-10k}{4}\right)x^{p}$$
the coefficient of $x^{50}$ is
$$\sum_{k=0}^{5}(-1)^k\binom{5}{k}\binom{25-10k}{4}$$
or, defining $\binom{a}{b}=0$ for $a<0$ we have
$$\binom{5}{0}\binom{25}{4}-\binom{5}{1}\binom{15}{4}+\binom{5}{2}\binom{5}{4}=5\,875$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2213614",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Calculating inverse of $(7+9i)$ How can I calculate $(7+9i)^{-1}$?
So I have:
$(7+9i)^{-1}$
$(a+bi) \cdot (7+9i)$
$7a + 9ai + 7 bi + 9bi^2 = 1 + 0i$
$7a + (9a + 7b)i - 9b = 1$
So there are two equations:
1) $7a - 9b = 1$
2) $9a + 7b = 0$
So getting a from the first equation:
$a = \frac{9}{7}b$
Inserting it in the second one:
$9 \cdot \frac{9}{7}b + 7b = 0$
$\frac{81}{7}b + 7b = 0$
$b = \frac{130}{7}$
The correct solution should be: $\frac{7}{130}- \frac{9i}{130}$
Question: My solution looks close but wrong is wrong. Where is my mistake here?
| $$(7+9i)^{-1}=\frac{1}{7+9i}=\frac{1}{7+9i} \times \frac{7-9i}{7-9i}=\frac{7-9i}{7^2-(9i)^2}=\frac{7-9i}{130}= \frac{7}{130}-\frac{9i}{130}$$
Also in your solution $7a-9b=1 \implies a= \dfrac{1+9b}{7} \neq \dfrac{9b}{7}$
| {
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"url": "https://math.stackexchange.com/questions/2214667",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 2
} |
Show $\lim_{(x,y)\to(0,0)}\frac{x^3-xy^3}{x^2+y^2}=0$ I need to show if the following limit is exists (and exists $0$),
$$
f(x,y)=\frac{x^3-xy^3}{x^2+y^2},
$$
for $\vec x\to 0$. I tried out the following:
$$
\frac{x^3-xy^3}{x^2+y^2}\leq\frac{x^3-xy^3}{x^2}=1-\frac{y^3}{x^2},
$$
but obviously I'm stuk here. A similar approach by omitting $y^2$ gives the same problem. What could I do next?
| Note that
$$\left|x^3-xy^3\over x^2+y^2\right|\le|x|\left|x^2\over x^2+y^2\right|+|xy|\left|y^2\over x^2+y^2\right|\le|x|+|xy|$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2216579",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Can anything interesting be said about this fake proof? The Facebook account called BestTheorems has posted the following. Can anything of interest be said about it that a casual reader might miss?
Note that \begin{align}
\small 2 & = \frac 2{3-2} = \cfrac 2 {3-\cfrac2 {3-2}} = \cfrac 2 {3 - \cfrac 2 {3 - \cfrac 2 {3-2}}} = \cfrac 2 {3 - \cfrac 2 {3 - \cfrac 2 {3 - \cfrac 2 {3-2}}}} = \cfrac 2 {3 - \cfrac 2 {3 - \cfrac 2 {3 - \cfrac 2 {3 - \cfrac 2 {\ddots}}}}} \\[10pt]
\text{and } \\
\small 1 & = \frac 2 {3-1} = \cfrac 2 {3 - \cfrac 2 {3-1}} = \cfrac 2 {3 - \cfrac 2 {3 - \cfrac 2 {3-1}}} = \cfrac 2 {3 - \cfrac 2 {3 - \cfrac 2 {3 - \cfrac 2 {3-1}}}} = \cfrac 2 {3 - \cfrac 2 {3 - \cfrac 2 {3 - \cfrac 2 {3 - \cfrac 2 {\ddots}}}}} \\
\text{So } & 2=1.
\end{align}
| $$x = \cfrac 2 {3 - \cfrac 2 {3 - \cfrac 2 {3 - \cfrac 2 {3 - \cfrac 2 {\ddots}}}}}$$
$$x = \frac 2 {3 - x}\\
x^2 - 3x + 2 = 0\\
(x-1)(x-2) = 0$$
$1,2$ are both solutions. However, if we consider this recurrence relation:
$$x_n = \frac 2 {3 - x_{n-1}}$$
when $x_{n-1} <1 \implies x_{n-1}<x_n<1$
And the squence converges to 1.
And
$$1<x_{n-1} < 2 \implies 1<x_n <x_{n-1}$$
and the sequence again converges.
but,
$$2<x_{n-1} < 3 \implies x_{n-1}<x_n$$
and
$$x_{n-1} >3 \implies x_n< 0$$
The sequence isn't stable in a neighborhood of $2.$
and it converges to $1$ for nearly all starting conditions.
| {
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"url": "https://math.stackexchange.com/questions/2216964",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "30",
"answer_count": 4,
"answer_id": 1
} |
Integration by parts $\int xe^{x^2} dx$.
Find with integration by parts $$\int xe^{x^2} dx$$.
$$\int xe^{x^2} dx$$
Let $u(x) = x \ \ \ v^{'}(x) = e^{x^2}$
Now I can write the integral as $$\int xe^{x^2} dx = x \int e^{x^2} dx - \int\left(\int e^{x^2} dx \right) dx$$
Even after trying everything I am unable to solve $v^{'}(x) = e^{x^2}$ for $v(x)$.
I can do this question substitution $u = x^2$ but I told use integration by parts to do this question. What should I do ?
| Take $u=e^{x^2}$ then $du=2xe^{x^2}$ and $v=\frac{x^2}{2}$ so
$$\int xe^{x^2}dx=\frac{x^2e^{x^2}}{2}-\int x^3e^{x^2}dx$$
Take $u=e^{x^2}$ and $dv=x^3$ then
$$\int x^3e^{x^2}dx=\frac{x^4}{4}e^{x^2}-\frac{1}{2}\int x^5e^{x^2}dx$$
Similary
$$\int x^5e^{x^2}dx=\frac{x^6}{6}e^{x^2}-\frac{1}{3}\int x^7e^{x^2}dx$$
In general $$\int x^{2n+1}e^{x^2}dx=\frac{x^{2n+2}}{2n+2}e^{x^2}-\frac{1}{n+1}\int x^{2n+3}e^{x^2}dx$$
From that we can take a few terms and notice that $$\int xe^{x^2}dx=\frac{x^2}{2}e^{x^2}-\frac{x^4}{4}e^{x^2}+\frac{x^6}{12}e^{x^2}-\frac{x^8}{48}e^{x^2}+\frac{x^{10}}{240}e^{x^2}+\cdots\\\int xe^{x^2}dx=\frac{e^{x^2}}{2}\sum_{k=1}^{\infty}\frac{(-1)^{k+1}x^{2k}}{k!}$$Now with few adjustments you'll get$$e^{-x^2}=\sum_{k=0}^\infty\frac{(-1)^kx^{2k}}{k!}=1+\sum_{k=1}^\infty\frac{(-1)^kx^{2k}}{k!}=1-\sum_{k=1}^\infty\frac{(-1)^{k+1}x^{2k}}{k!}\\\int xe^{x^2}dx=\frac{e^{x^2}}{2}\left(1-e^{-x^2}\right)=\frac{e^{x^2}}{2}+C$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2217850",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
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Expand binomially to prove trigonometric identity Prompt: By expanding $\left(z+\frac{1}{z}\right)^4$ show that $\cos^4\theta = \frac{1}{8}(\cos4\theta + 4\cos2\theta + 3).$
I did the expansion using binomial equation as follows
$$\begin{align*}
\left(z+\frac{1}{z}\right)^4 &= z^4 + \binom{4}{1}z^3.\frac{1}{z} + \binom{4}{2}z^2.\frac{1}{z^2} + \binom{4}{3}z^3.\frac{1}{z}+\frac{1}{z^4}\\
&=z^4+4z^2+6+\frac{4}{z^2}+\frac{1}{z^4}\\
&=z^4+\frac{1}{z^4}+4\left(z^2+\frac{1}{z^2}\right) + 6. -(eqn 1)
\end{align*}
$$
I'm not sure how to go on about rest of the problem.
[update]
Reading comments, I tried assuming $z = e^{i\theta}$
$2\cos\theta = e^{i\theta} + e^{-i\theta}$
$(2\cos\theta)^4 = (e^{i\theta} + e^{-i\theta})^4$
$=e^{4i\theta} + e^{-4i\theta} + 8(e^{2i\theta}+e^{-2i\theta})+6$ (from eqn 1)
| Let $\cos \theta=z+\dfrac1z.$
Then $\cos2\theta=2\cos^2\theta-1=2z^2+3+\dfrac2{z^2}$ and similarly you can find $\cos 4\theta.$
Compare them with your binomial identity.
Good luck.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Can someone please check if my reasoning for this proof is valid or not? I have already seen the other questions about this proof. I'm just trying a different sort of method, though I'm not sure if it's valid or not.
Context for the main question: Prove by induction $2^n\gt n^3$ for $n\ge10$
Obviously, the base case works for n=10
$1024=2^{10}\gt1000=10^3$
The induction hypothesis: Assume $P_n$ is true $\rightarrow$ $2^n\gt n^3$
I want to then prove that $2^{n+1}\gt (n+1)^3$
Now, using the induction hypothesis:
$2^n\gt n^3$
multiply both sides by 2
$2^{n+1}\gt 2n^3$
Using the fact that $n\ge10$ this implies that $n^3\ge10n^2$
$2n^3=n^3 +n^3\gt n^3 +10n^2=n^3 +3n^2 +7n^2$
Using the fact that $7\gt 1$ this implies that $7n\gt n$ since n is positive.
$n^3 +3n^2 +7n^2\gt n^3 +3n^2 +n^2$
Once again, using $n\ge 10$ this implies $n^2\ge 10n$
$n^3 +3n^2 +n^2\gt n^3 +3n^2 +10n=n^3 +3n^2 +3n+7n$
Again $7\gt 1$
$n^3 +3n^2 +3n+7n\gt n^3 +3n^2 +3n +n$
Using $n\ge 10$ one last time
$n^3 +3n^2 +3n +n\gt n^3 +3n^2 +3n+10$
Since $10\gt 1$
$n^3 +3n^2 +3n+10\gt n^3 +3n^2 +3n+1=(n+1)^3$
Thus, through the chain of inequalities, I have proved that $2^{n+1}\gt (n+1)^3$.
QED
Sorry if there are any errors in my reasoning. Thank you for reading and feedback.
| Here's an alternate method: we have in our assumption that $$2^n>n^3,\quad n\geq 10$$
Then consider $f(n) = 2^{n+1}-(n+1)^3$
$f(n) = 2\cdot 2^n-n^3-3n^2-3n-1$
$f(n)= (2^n-n^3)+(2^n-3n^2-3n-1)$
Let $g(n) = n^3-3n^2-3n-1\implies g(n) <2^n-3n^2-3n-1$ using $2^n>n^3$
$g'(n) = 3n^2-6n-3 = 3(n^2-2n-1) =3[(n-1)^2-2] >0$ for $n\geq 10$
$g(10) = 693\implies g(n) >0,\quad n\geq 10$
Therefore $f(n)$ is the sum of two positive functions of $n$ $\implies f(n)>0,\quad n\geq 10$ as required
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2221189",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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} |
Suppose $x^2+y^2=4$ and $z^2+t^2=9$ and $xt+yz=6$. What are $x,y,z,t$? Let $x^2+y^2=4$ and $z^2+t^2=9$ and $xt+yz=6$.
What are $x,y,z,t$?
| Presumably, this should be solved over $\Bbb R$.
In this case the easiest way to think about this is that you have two (real) vectors $v = (x,y)$ and $w = (t,z)$.
Then,
$$\|v\| = \sqrt{x^2+y^2} = \sqrt 4 = 2,$$
$$\|w\| = \sqrt{t^2+z^2} = \sqrt 9 = 3,$$
$$v\cdot w = xt + yz = 6.$$
If we denote with $\varphi$ the angle between $v$ and $w$, we have
$$\cos\varphi =\frac{v\cdot w}{\|v\|\|w\|} = 1\implies \varphi = 0\implies w =\lambda v,\ \lambda>0.$$
Since $\|w\| = \lambda\| v\|$, we conclude that $\lambda = \frac 32$.
Considering polar coordinates, we can write $v = r(\cos\alpha,\sin\alpha)$, $w = \frac 32 r(\cos\alpha,\sin\alpha)$, where $\alpha\in[0,2\pi)$. Note that $$\|v\| =r\sqrt{\cos^2\alpha + \sin^2\alpha} = r\implies r = 2,$$ meaning that $v = 2(\cos\alpha,\sin\alpha)$, $w = 3(\cos\alpha,\sin\alpha)$, or
\begin{array}{cc}
\begin{align}
x &= 2\cos\alpha,\\
y &= 2\sin\alpha,\\
t &= 3\cos\alpha,\\
z &= 3\sin\alpha.
\end{align}&\tag{1}\end{array}
Returning to the original system we get
$$x^2+y^2 = 4\cos^2\alpha+4\sin^2\alpha = 4,$$
$$t^2+z^2 = 9\cos^2\alpha+9\sin^2\alpha = 9,$$
$$xt+yz = 6\cos^2\alpha + 6\sin^2\alpha = 6,$$
and thus, all (real) solutions are given by $(1)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2222245",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Limit of a 2-variable function Given problem:
$$
\lim_{x\to0,y\to0}(1+x^2y^2)^{-1/(x^2+y^2)}
.$$
I tried to do it with assigning y to $y = kx$, but that didn't help me at all. Also one point, I can't use L'Hospital's rule.
| Using Polar coordinates, $x = r\cos\theta$ and $y = r\sin\theta$, the problem boils down to $$ \lim_{r\to0}\space(1 + r^4\cos^2\theta\sin^2\theta)^{\frac{-1}{r^2}}$$
Now prove that independent of the value of $\theta$, the limit exists and solve for it then.
EDIT:
Now let $$L = \lim_{r\to0}\space(1 + r^4\cos^2\theta\sin^2\theta)^{\frac{-1}{r^2}}$$
Taking $\log$ on both sides,
$$\log L = \lim_{r\to0}\space -\dfrac{\log(1 + r^4\cos^2\theta\sin^2\theta)}{r^2}$$As $\log (1+x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \cdots$,
$$\log L = \lim_{r\to0}\space - \dfrac{1}{r^2}\left (r^4\cos^2\theta\sin^2\theta - \dfrac{\left (r^4\cos^2\theta\sin^2\theta\right )^2}{2}) + \cdots \right ) = 0$$
$$\implies \log L =0 \implies L = 1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2223838",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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If $a$ , $b$, $c$ are positive integers satisfying the following $(a^2 +2)(b^2+3)(c^2+4)=2014$ What is the value of $ a+b+c $ If $a$ , $b$, $c$ are positive integers satisfying the following
$(a^2 +2)(b^2+3)(c^2+4)=2014$
What is the value of $ a+b+c $
I need details because i don't know how to solve simlar problems ?
Thank you for your help.
| let us consider following program
z=input('enter your number : ');
string='';
for ii=2:z
s=0;
while z/ii==floor(z/ii) % check if z is divisible by ii
z=z/ii;
s=s+1;
end
if s>0
str =[num2str(ii) '^' num2str(s) ];
string=strcat(string,str);
string= strcat(string,'*');
% If z = 1, no more divisions are necessary,
% thus breaks the loop and quits
if z == 1
break
end
end
end
string=string(1:end-1);% remove last sign of multiplicaiton
fprintf('prime factorization is %s\n',string);
we would have
>> integer_factorization
enter your number : 2014
prime factorization is 2^1*19^1*53^1
now
x=2
y=19
z=53
$a^2+2=2$
$a^2=0$
which means $a=0$
$c^2+4=53$
$c^2=53-4=49$
which means that $c=7$ or $c=-7$ depend we have positive integers or integers and also
$b^2+3=19$
we have $ b=4 $ or $b-4$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2225235",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Uniform convergence of $\sum_{n = 1}^\infty {\arctan{\frac{2x}{x^2+n^2}}}$ on $\mathbb R$
Check if $\sum_\limits{n = 1}^\infty{\arctan{\dfrac{2x}{x^2+n^2}}}$ is uniformly convergent series of function on $\mathbb{R}$.
"Uniform convergence" really messes up my mind.
Since $\dfrac{2x}{x^2+n^2} \approx \arctan{\dfrac{2x}{x^2+n^2}}$ when $n\to\infty$, could I use the uniform convergence of $\sum_\limits{n = 1}^\infty{\dfrac{2x}{x^2+n^2}}$ to check the uniform convergence of $\sum_\limits{n = 1}^\infty{\arctan{\dfrac{2x}{x^2+n^2}}}$?
Thank you in advance.
| Your heuristics may give some idea, but any such heuristics should be justified in order to yield a valid proof.
Let me demonstrate a possible way. It follows from AM-GM inequality that
$$ \left|\frac{2x}{x^2 + n^2}\right| \leq \frac{1}{n} $$
uniformly in $x \in \Bbb{R}$. Together with the inequality $\frac{1}{2}|x| \leq |\arctan x| \leq |x|$ which holds for $|x| \leq 1$, we find that
$$ \frac{1}{2} \left\| \sum_{n=M}^{N} \frac{2x}{x^2 + n^2} \right\|_{\sup}
\leq \left\| \sum_{n=M}^{N} \arctan\left( \frac{2x}{x^2 + n^2} \right) \right\|_{\sup}
\leq \left\| \sum_{n=M}^{N} \frac{2x}{x^2 + n^2} \right\|_{\sup}. $$
That is, $\sum_{n=1}^{\infty} \arctan( \frac{2x}{x^2 + n^2} )$ is uniformly Cauchy if and only if $\sum_{n=1}^{\infty} \frac{2x}{x^2 + n^2}$ is uniformly Cauchy. This is enough to conclude that we can investigate the uniform convergence of $\sum_{n=1}^{\infty} \frac{2x}{x^2 + n^2}$ .
Now let us prove that the series does not converge uniformly. @Did already provided a short and elegant solution for this, but let me articulate a little bit by showing that, loosely speaking, 'positive mass' concentrates near the infinity.
Let $S(x) = \sum_{n=1}^{\infty} \frac{2x}{x^2 + n^2}$ and we write $S(x)$ as
$$ S(x) = \sum_{n = 1}^{\infty} \frac{2}{1 + (\frac{n}{x})^2} \frac{1}{x}. $$
We can recognize $S(x)$ as a Riemann sum of the function $f(t) = \frac{2}{1+t^2}$. To prove that this is indeed true, let $x > 0$ and notice that
$$ \int_{\frac{n}{x}}^{\frac{n+1}{x}} \frac{2}{1+t^2} \, dt
\leq \frac{2}{1 + (\frac{n}{x})^2} \frac{1}{x}
\leq \int_{\frac{n-1}{x}}^{\frac{n}{x}} \frac{2}{1+t^2} \, dt $$
for all $n \geq 1$. Summing over $n$ and taking limit as $x \to \infty$, this proves that
$$ \lim_{x\to\infty} S(x) = \int_{0}^{\infty} \frac{2}{1+t^2} \, dt = \pi. $$
This is enough to conclude that $S(x)$ does not converge uniformly, since any partial sum $S_N(x) = \sum_{n=1}^{N} \frac{2x}{x^2 + n^2}$ satisfies
$$ \lim_{x\to\infty} S_N(x) = \sum_{n=1}^{N} \lim_{x\to\infty} \frac{2x}{x^2 + n^2} = 0 $$
and hence there is no way we have $\| S_N - S \|_{\sup} \to 0$ as $N \to \infty$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2226502",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Which of the following is true for the definite integrals shown above? $$J = \int_{0}^{1}\sqrt{1-x^4}\,dx$$
$$K = \int_{0}^{1}\sqrt{1+x^4}\,dx$$
$$L = \int_{0}^{1}\sqrt{1-x^8}\,dx$$
Which of the following is true for the definite integrals shown above?
(A) $J<L<1<K$
(B) $J<L<K<1$
(C) $L<J<1<K$
(D) $L<J<K<1$
(E) $L<1<J<K$
What is the smartest way of solving this question other than solving the integrals as this question must take from me at most 2.5 minutes during the examination?
| $\begin{align}L-J&=\int_0^1 \sqrt{1-x^8}dx-\int_0^1 \sqrt{1-x^4}dx\\
&=\int_0^1 \left(\sqrt{1-x^8}-\sqrt{1-x^4}\right)dx\\
&=\int_0^1 \dfrac{\left(\sqrt{1-x^8}\right)^2-\left(\sqrt{1-x^4}\right)^2}{\sqrt{1-x^8}+\sqrt{1-x^4}}dx\\
&=\int_0^1 \dfrac{x^4(1-x^4)}{\sqrt{1-x^8}+\sqrt{1-x^4}}dx>0
\end{align}$
Therefore $L>J$ and $(C),(D),(E)$ are false.
$\begin{align} K-1&=\int_0^1 \sqrt{1+x^4}dx-\int_0^1 1 dx\\
&=\int_0^1 \left(\sqrt{1+x^4}-1\right)dx\\
&=\int_0^1 \dfrac{\left(\sqrt{1+x^4}\right)^2-1^2}{\sqrt{1+x^4}+1}dx\\
&=\int_0^1 \dfrac{x^4}{\sqrt{1+x^4}+1}dx>0\\
\end{align}$
Therefore $K>1$ and $(B)$ is false.
Therefore $(A)$ is true
I have supposed the question has an unique answer.
Otherwise to be sure $(A)$ is the good answer it remains to prove that $L<1$. Same method can be used to prove this.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Finding orthogonal surface Find the surface which is orthogonal to the system $z=cxy(x^2+y^2)$ and which passes through the hyperbola $(x^2-y^2)=a^2$.
| We can consider $c$ as function of $x,y,z$.
$G(x,y,z)=c=\dfrac{z}{xy(x^2+y^2)}$ Its gradient is a vector orthogonal to each of the surfaces $G(x,y,z)=c$ constant.
$\nabla G=-\dfrac{z(y^2+3x^2)}{yx^2(x^2+y^2)^2}\hat i-\dfrac{z(x^2+3y^2)}{y^2x(x^2+y^2)^2}\hat j+\dfrac{1}{xy(x^2+y^2)}\hat k$
We have to find a function $z(x,y)$ orthogonal in each of its points to one of the surfaces with $G(x,y,z)=c$ constant. In each point of $z(x,y)$, the vectors, the orthogonal to $z\to(z_x,z_y,-1)$ and the orthogonal to $G\to\nabla G$, have to be orthogonals, so their dot product is zero.
For convenience we can use some other vector proportional to $\nabla G$
$W=\dfrac{(y^2+3x^2)}{x}\hat i+\dfrac{(x^2+3y^2)}{y}\hat j-\dfrac{(x^2+y^2)}{z}\hat k=a\hat i+b\hat j+c\hat k$
But still is valid: $(a,b,c)·(z_x,z_y,-1)=0$ Or, making $u=z$,
$\dfrac{y^2+3x^2}{x}u_x+\dfrac{x^2+3y^2}{y}u_y=\dfrac{-(x^2+y^2)}{u}$
It's a first order semilinear pde. To solve it, we use the method of characteristics. We write,
$\dfrac{xdx}{y^2+3x^2}=\dfrac{ydy}{x^2+3y^2}=\dfrac{-udu}{y^2+x^2}$
First surface for the characteristics: Adding the first and second ratios and equating to the third $\dfrac{xdx+ydy}{4(x^2+y^2)}=\dfrac{-udu}{y^2+x^2}\implies 2xdx+2ydy=-8udu\implies x^2+y^2=-4u^2+c_1$
Second surface: Substracting the second ratio to the first and equating to the third, $\dfrac{xdx-ydy}{2(x^2-y^2)}=\dfrac{-udu}{x^2+y^2}\implies\dfrac{2xdx-2ydy}{x^2-y^2}=\dfrac{-4udu}{-4u^2+c_1}\implies$
$\implies\ln\vert x^2-y^2\vert=(1/2)\ln\vert -4u^2+c_1\vert+k\implies (x^2-y^2)^{2}=(x^2+y^2)c_2$
Or $\dfrac{(x^2-y^2)^2}{x^2+y^2}=c_2$
For the general solution we have to set $c_1=f(c_2)$ for some $f$ single variable, differentiable function.
$$4u^2=f\left(\dfrac{(x^2-y^2)^2}{x^2+y^2}\right)-(x^2+y^2)$$
Now we have to impose the boundary conditions, so is, $u(x,y)=0$ for the hyperbola $x^2-y^2=a^2$ ($y^2=x^2-a^2$)
$$0=f\left(\frac{a^4}{2x^2-a^2}\right)-(2x^2-a^2)$$
Let $v=\dfrac{a^4}{2x^2-a^2}$, then $2x^2-a^2=\dfrac{a^4}{v}$, so is,
$f(v)=\dfrac{a^4}{v}$ Leading to the two particular solutions,
$$u(x,y)=\pm\sqrt{\dfrac{a^4(x^2+y^2)}{4(x^2-y^2)^2}-\dfrac{x^2+y^2}{4}}$$
The solution is not unique, but both solutions, understood as surfaces, form the surface solution.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Find the number of distinct arrangements of 5 unequal positive integers such that their sum is 20 Let the integers be $n_1, n_2, n_3, n_4$ and $n_5$.
It's given in the question that $n_1 + n_2 + n_3 + n_4 + n_5 = 20$.
I thought of taking $n_2$ as $n_1 + a$, $n_3$ as $n_1 + a + b$ and so on... where a, b,... are not equal to 0.
So I got this expression:
$5n_1 + 4a + 3b + 2c + d = 20$
After this, I'm not able to continue. How do I proceed?
Thanks in advance.
| Integer partitions of $n$ into $k$ parts is given by the recurrence
$$p(n,k)=p(n-1,k-1)+p(n-k,k)\tag{1}\label{1}$$
Since a partition of $n$ into $k$ parts either has $1$ as it's smallest part in $p(n-1,k-1)$ ways or it has it's smallest part greater than $1$ in $p(n-k,k)$ ways. This recurrence has $p(n,1)=1$ and $p(1,1)=1$ and $p(1,k)=0$ for $k\gt 1$.
If, say, the integer partition is
$$\sum_{r=1}^{k}n_r=n$$
such that $n_1\le n_2\le \cdots\le n_k$ we make the substitution $n_r'=n_r+(r-1)$ then we have
$$\sum_{r=1}^{k}n_r'-\sum_{r=1}^{k}(r-1)=n$$
$$\implies\sum_{r=1}^{k}n_r'=n+\binom{k}{2}=n'$$
for $n_1'\lt n_2'\lt\cdots\lt n_k'$ this is a bijection between integer partitions of $n$ into $k$ parts and integer partitons of $n'$ into $k$ distinct parts.
In other words the number of partitions of $n'$ into $k$ distinct parts is equal to the number of partitions of $n=n'-\binom{k}{2}$ into $k$ parts.
$$p_d(n',k)=p(n'-\text{C}(k,2),k)$$
In your case $n'=20$ and $k=5$ and $\binom{5}{2}=10$ so $n=20-10=10$.
Reading off $p(10,5)$ in the table formed by the recurrence $\eqref{1}$
$$\begin{array}{cc}
&k\\
n&\begin{array}{c|cccccccccccc}
p(n,k)&1 &2 &3 &4 &5 &6 &7 &8 &9 &10 &11 &12\\\hline
1 &1&&&&&&&&&&&\\
2 &1 &1&&&&&&&&&&\\
3 &1 &1 &1&&&&&&&&&\\
4 &1 &2 &1 &1&&&&&&&&\\
5 &1 &2 &2 &1 &1&&&&&&&\\
6 &1 &3 &3 &2 &1 &1&&&&&&\\
7 &1 &3 &4 &3 &2 &1 &1&&&&&\\
8 &1 &4 &5 &5 &3 &2 &1 &1&&&&\\
9 &1 &4 &7 &6 &5 &3 &2 &1 &1&&&\\
10 &1 &5 &8 &9 &\bbox[#FFA,10px]{7} &5 &3 &2 &1 &1&&\\
11 &1 &5 &10 &11 &10 &7 &5 &3 &2 &1 &1&\\
12 &1 &6 &12 &15 &13 &11 &7 &5 &3 &2 &1 &1\\
\end{array}\end{array}$$
Thus, we have
$$p_d(20,5)=p(10,5)=7\tag{Answer 1}$$
If you want the order of the parts to count then multiply that by $5!$ to give
$$7\cdot 5!=840\tag{Answer 2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2232434",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Largest interval of definition furnished by Picard's theorem Consider the IVP $$y'=\frac{1}{(3-x^2)(9-y^2)},\;y(0)=0.$$
I need to find the largest interval of definition furnished by Picard's theorem. I know the positive endpoint of the interval is given by $\min (a,\frac bM)$ where $|x|<a,|y|<b$ and $M=\sup_B |f(x,y)|$ with $B$ being the aforementioned rectangle.
For any $a,b$, which necessarily satisfy $a<\sqrt 3,b<3$ we have $\frac bM=b(3-a^2)(9-b^2)$. So I think I need to maximize $$\min(a,b(3-a^2)(9-b^2)).$$ How to do this? Alternatively, what am I doing wrong?
| Picard's theorem by itself doesn't furnish a particular interval of definition, it just says there is one. There are many different ways to find an explicit interval. In this case the d.e. is separable, leading to the implicit solution
$$ - \frac{\text{arctanh}(x/\sqrt{3})}{\sqrt{3}}
- \frac{y^3}{3} + 9 y = 0 $$
This has $y=\pm 3$ when $x = \pm \sqrt{3} \tanh(18 \sqrt{3})$. So the actual interval is $(-\sqrt{3} \tanh(18 \sqrt{3}), \sqrt{3} \tanh(18 \sqrt{3}))$.
EDIT: I would say it's not Picard's theorem that furnishes the interval, but it may be a particular proof of Picard's theorem. Anyway, you want to maximize
$t$ such that $t \le a$, $t \le b (3-a^2)(9-b^2)$, $a \le \sqrt{3}$, $b \le 3$.
Since $a$ is an increasing function of $a$ while $b(3-a^2)(9-b^2)$ is a decreasing function of $a$ in the given region, the maximum should occur at a point where
$a = b(3-a^2)(9-b^2)$. Thus we're looking for a point on the curve $a = b(3-a^2)(9-b^2)$ with $a$ as large as possible. That curve looks like this:
On this curve we want a point where the tangent is vertical, so
with
$$ g(a,b) = a - b(3-a^2)(9-b^2)$$
we take
$$ \dfrac{\partial g}{\partial b} = -3 (a^2 - 3)(b^2-3) = 0$$
Since we can't have $g=0$ with $a^2=3$, we must have $b^2 = 3$, and then
$t = a = \sqrt{3}(\sqrt{1297}-1)/36 \approx 1.684606387$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2232651",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Find coefficients of $x^{2012}$ in $(x+1)(x^2+2)(x^4+4)\cdots (x^{1024}+1024)$ Find coefficients of $x^{2012}$ in $(x+1)(x^2+2)(x^4+4)\cdots (x^{1024}+1024)$
Attempt: i have break $2012$ in to sum of power of $2$
as $2012 = 2^{10}+2^{9}+2^{8}+2^7+2^6+2^4+2^3+2^2$
but wan,t be able to go further, could some help me , thanks
| $2012 = 2047 - 35 = (\sum_{k=0}^{10}{2^k}) - 35$
$35 = 2^5+2^1+2^0 => 2012 = 2^{10}+2^9+2^8+2^7+2^6+2^4+2^3+2^2$
So this means that the for the parenthesis with the power of $x$ in this set: ${10, 9, 8, 7, 6, 4, 3, 2}$, the part with $x$ is multiplied. So for the rest of the parenthesis, the number is chosen. So the coefficient of $x^{2012}$ is:
$2^5*2^1*2^0 = 2^6 = 64$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2232769",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
transition matrix and coordinate vector Could you please help me solve these quesitons??
Consider the bases S={$u_1, u_2, u_3$}, and T={$v_1, v_2, v_3$}, with
$u_1$=[-3, 0, -3], $u_2$=[-3, 2, -1], $u_3$=[1, 6, -1],
$v_1$=[-6, -6, 0],
$v_2$=[-2, -6, 4], $v_3$=[-2, -3, 7]
(a) Find the transition matrix from $S$ to $T$
(b) Using the result in (a), compute the coordinate vector $[w]_T$ where $w$=[-5, 8, -5]
| As noted by @Emilio Novati, the key idea is to connect to the standard basis.
Vectors will be colored according to the basis membership, and named chromatically:
$$
\color{blue}{\mathbf{B}\ (standard)}, \qquad
\color{red}{\mathbf{R}\ (u)}, \qquad
\color{green}{\mathbf{G}\ (v)}.
$$
$$
\mathbf{R}_{\color{red}{R}\to \color{blue}{A}}=
\color{black}{\left[
\begin{array}{rrr}
-3 & -3 & 1 \\
0 & 2 & 6 \\
-3 & -1 & -1 \\
\end{array}
\right]}
$$
Example: the second vector in the $\color{red}{\mathbf{S}}$ basis has the following coordinates in the $\color{blue}{standard}$ basis
$$
\mathbf{R}_{\color{red}{S}\to \color{blue}{A}}
\color{red}{\left[
\begin{array}{c}
0 \\
1 \\
0
\end{array}
\right]}
=
\color{blue}{\left[
\begin{array}{r}
-3 \\
2 \\
-1
\end{array}
\right]}
$$
You may think of the matrix $\mathbf{R}_{\color{red}{S}\to \color{blue}{A}}$ as an operator which takes a $\color{blue}{blue}$ vector and returns a $\color{red}{red}$ vector.
The inverse matrix is a map which connects vectors in the $\color{blue}{standard}$ basis to vectors in the $\color{red}{\mathbf{R}}$ basis:
$$
\mathbf{R}^{-1}_{\color{blue}{B}\to \color{red}{U}}
%
\color{blue}{\left[
\begin{array}{c}
0 \\
0 \\
1
\end{array}
\right]}
=
%
\frac{1}{24}
\left[
\begin{array}{rrr}
2 & -2 & -10 \\
-9 & 3 & 9 \\
3 & 3 & -3 \\
\end{array}
\right]
%
\color{blue}{\left[
\begin{array}{c}
0 \\
0 \\
1
\end{array}
\right]}
=
%
\color{red}{\frac{1}{24}
\left[
\begin{array}{r}
-10 \\
9 \\
-3
\end{array}
\right]}
$$
For the $\color{green}{\mathbf{G}}$ basis, the maps are
$$
\mathbf{G}_{\color{green}{G}\to \color{blue}{A}}=
\color{black}{\left[
\begin{array}{rrr}
-6 & -6 & 0 \\
-2 & -6 & 4 \\
-2 & -3 & 7 \\
\end{array}
\right]}, \qquad
%
\mathbf{G}^{-1}_{\color{blue}{A}\to \color{green}{G}}=
\color{black}{\left[
\begin{array}{rrr}
-5 & 7 & -4 \\
1 & -7 & 4 \\
-1 & -1 & 4 \\
\end{array}
\right]}
$$
Transition from $S$ to $T$
We connect all bases through the hub of the $\color{blue}{standard}$ basis. Start with a vector in the $\color{red}{\mathbf{R}}$ basis, map that to a vector in the $\color{blue}{standard}$ basis, then map that to a vector in the $\color{green}{\mathbf{G}}$ basis:
$$
\color{red}{
\left[
\begin{array}{c}
x_{1} \\
y_{1} \\
z_{1}
\end{array}
\right]}
%
\quad \Longrightarrow \quad
%
\color{blue}{
\left[
\begin{array}{c}
x_{2} \\
y_{2} \\
z_{2}
\end{array}
\right]}
%
\quad \Longrightarrow \quad
%
\color{green}{
\left[
\begin{array}{c}
x_{3} \\
y_{3} \\
z_{3}
\end{array}
\right]}
%
$$
The formal steps are
$$
\begin{align}
%
\mathbf{R}_{\color{red}{R}\to \color{blue}{B}}
\color{red}{
\left[
\begin{array}{c}
x_{1} \\
y_{1} \\
z_{1}
\end{array}
\right]}
%
&=
%
\color{blue}{
\left[
\begin{array}{c}
x_{2} \\
y_{2} \\
z_{2}
\end{array}
\right]} \\[3pt]
%
\mathbf{G}^{-1}_{\color{blue}{B}\to \color{green}{G}}
\left(
\mathbf{S}_{\color{red}{R}\to \color{blue}{B}}
\color{red}{
\left[
\begin{array}{c}
x_{1} \\
y_{1} \\
z_{1}
\end{array}
\right]} \right)
%
&=
%
\mathbf{G}^{-1}_{\color{blue}{B}\to \color{green}{G}}
\color{blue}{
\left[
\begin{array}{c}
x_{2} \\
y_{2} \\
z_{2}
\end{array}
\right]}
%
=
%
\color{green}{
\left[
\begin{array}{c}
x_{3} \\
y_{3} \\
z_{3}
\end{array}
\right]}
%
\end{align}
$$
The operator which maps $\color{red}{red}$ vectors to $\color{green}{green}$ is
$$
\mathbf{X}_{\color{red}{R}\to \color{green}{G}} =
\mathbf{G}^{-1}_{\color{blue}{B}\to \color{green}{G}}
\mathbf{R}_{\color{red}{R}\to \color{blue}{B}} =
%
\frac{1}{24}
\left(
\begin{array}{rr}
27 & 33 & 41 \\
-15 & -21 & -45 \\
-9 & -3 & -11 \\
\end{array}
\right)%
$$
Computation
Turn a $\color{red}{red}$ vector into a $\color{green}{green}$ vector:
$$
\mathbf{X}_{\color{red}{R}\to \color{green}{G}}
\color{red}{
\left[
\begin{array}{r}
-5 \\
8 \\
-5
\end{array}
\right]}
=
\color{green}{
\frac{1}{18}
\left[
\begin{array}{r}
-57 \\
99 \\
57
\end{array}
\right]}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2232952",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
The function $(1+x)(1+x^4)(1+x^{16})(1+x^{64})\cdots =\prod_{n\geq 0} (1+x^{4^n})$ Let $$f(x)=(1+x)(1+x^4)(1+x^{16})(1+x^{64})\cdots=\prod_{n\geq 0} (1+x^{4^n})$$
Then what is $f^{-1}(\frac{8}{5f(3/8)})?$
The answer should be a rational number.
My attempt: I tried to take a log of the expression to turn it into a sum, but that did not simplify. So I am clueless on how to proceed. Any suggestions?
| It's hard to say things about $f(x)$ for general $x$. But we have
\begin{align}
(1-x) f(x) f(x^2) &= (1-x)(1+x)(1+x^2)(1+x^4)(1+x^8)\dotsb \\
&= (1-x^2)(1+x^2)(1+x^4)(1+x^8)\dotsb \\
&= (1-x^4)(1+x^4)(1+x^8) \dotsb \\
&= (1-x^8)(1+x^8) \dotsb \\
&= \dots = 1.
\end{align}
(Assuming convergence.)
If $x = f^{-1}\left(\frac{8}{5f(3/8)}\right)$, then $\frac58 \cdot f(\frac38) \cdot f(x)=1$, and the above identity tells us that $\frac58 \cdot f(\frac38) \cdot f(\frac{9}{64}) = 1$. So one possible answer is $\frac{9}{64}$.
To verify that this solution is valid (i.e., to verify convergence) and that it is the only solution, note that $$\log f(x) = \sum_{n\ge 1} \log\left(1 + x^{4^n}\right) \le \sum_{n \ge 1} x^{4^n}$$ which converges for $|x| < 1$, and $f$ is increasing (because each factor is increasing) and therefore injective.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2236115",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
Find the value of $(\log_{a}b + 1)(\log_{b}c + 1)(\log_{c}a+1)$ if $\log_{b}a+\log_{c}b+\log_{a}c=13$ and $\log_{a}b+\log_{b}c+\log_{c}a=8$ Let $a,b$, and $c$ be positive real numbers such that
$$\log_{a}b + \log_{b}c + \log_{c}a = 8$$
and
$$\log_{b}a + \log_{c}b + \log_{a}c = 13.$$
What is the value of
$$(\log_{a}b + 1)(\log_{b}c + 1)(\log_{c}a + 1) ?$$
I tried to convert the entire thing to fractional logs and multiply the expression and add the two equations but it did not help.
| Here's how I solved this:
First, expand the expression. For conciseness, express it as $(x+1)(y+1)(z+1)$:
$$(x+1)(y+1)(z+1) = xyz + \Big(xy + yz + xz\Big) + \Big(x + y + z\Big) + 1$$
We recognize immediately that $x+y+z$ is simply the first equation we're given, with value 8.
It must be the case that the remaining terms are related (at least partially) to the second equation, so let's play with that a little bit. Is there a relationship between the second equation and the objects in the first?
If you don't know the relationship off the top of your head (I had forgotten) $\log_b{a}$ certainly looks close to $\log_a{b}$; expressing
$$\log_b{a} = x$$
$$\Leftrightarrow b = a^x$$
$$\Leftrightarrow 1 = x \log_b{a}$$
We quickly see that $\log_b{a} = \frac1{\log_a{b}} = \frac1x$, so the second equation is just
$$\frac1x + \frac1y + \frac1z = 13$$
The natural thing to do is to combine the terms on the LHS to get
$$\frac{xy + xz + yz}{xyz} = 13$$
This is excellent, since it directly relates to the unaccounted terms from our above expression and tells us that $xy+xz+yz=13xyz$, so
$$(x+1)(y+1)(z+1) = 14xyz + 9$$
So we're left with needing to know the value for $xyz = \log_a{b} \log_b{c} \log_c{a}$, which looks cyclical. Recalling that $\log_b{a}$ and $\log_a{b}$ have an inverse/cyclical relationship, that $\log_b{a} \log_a{b} = 1$, we see if a similar identity might hold for this extended case:
$$\log_a{b} \log_b{c} \log_c{a} = 1$$
We can quickly verify this is indeed the case, so that $xyz = 1$ and the result that $(x+1)(y+1)(z+1) = 23$ follows immediately.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2236306",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 5,
"answer_id": 4
} |
how to calculate the multivariable limit? I need to show that the function $f(x,y)=\frac{x^5+y^6}{|x|^3+|y|^3}$ is continuous at (0,0), so I want to show that
$ \lim_{(x,y)\to(0,0)}\frac{x^5+y^6}{|x|^3+|y|^3} =0$
I tried switching to polar coordinates but did not see how that helps. what would be a good way to calculate the limit?
| It's enough to show $\;\dfrac{\lvert x^5+y^6\rvert}{\lvert x\rvert^3+\lvert y\rvert^3}$ tend to $0$. Now, setting $x=r\cos\theta$, $y=r\sin\theta$, we have
$$\dfrac{\lvert x^5+y^6\rvert}{\lvert x\rvert^3+\lvert y\rvert^3}\le\dfrac{\lvert x\rvert^5+\lvert y^6\rvert}{\lvert x\rvert^3+\lvert y\rvert^3}=\frac{r^5\bigl(\lvert\cos\theta\rvert^5+r\lvert\sin\theta\rvert^6\bigr)}{r^3\bigl(\lvert\cos\theta \rvert^3+\lvert\sin\theta \rvert^3}\le\frac{r^2(1+r)}{\lvert\cos\theta \rvert^3+\lvert\sin\theta \rvert^3}.$$
There remains to show $\lvert\cos\theta \rvert^3+\lvert\sin\theta \rvert^3$ has a positive lower bound. For this, we may suppose $0\le \theta\le \pi/2$, so that
\begin{align}
\lvert\cos\theta \rvert^3+\lvert\sin\theta \rvert^3&=\cos\theta^3+\sin^3\theta=(\sin\theta+\cos\theta)(1+\sin\theta\cos\theta)\\&\ge\sin\theta+\cos\theta=\sqrt2\sin\Bigl(\theta+\frac\pi4\Bigr)\ge1\qquad\text{if }\;0\le \theta\le \frac\pi2.
\end{align}
Thus $\;\dfrac{\lvert x^5+y^6\rvert}{\lvert x\rvert^3+\lvert y\rvert^3}\le r^2(1+r)$, which tends to $0$ when $r$ tends to $0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2240041",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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How to show that $\mathbb{Z}\left[\frac{1 + \sqrt{5}}{2}\right]$ is finitely generated? We can say that $\mathbb{Z}\left[\cfrac{1 + \sqrt{5}}{2}\right]$ is finitely generated if minimal polynomial of $\cfrac{1 + \sqrt{5}}{2}$ is in $\mathbb{Z}[X]$. After some calculations it can be shown that $f(X) = X^2 - X - 1 \in \mathbb{Z}[X]$ is the minimal polynomial of $\cfrac{1 + \sqrt{5}}{2}$. I think that $\mathbb{Z}\left[\cfrac{1 + \sqrt{5}}{2}\right]$ is generated by $\left \{1, \cfrac{1 + \sqrt{5}}{2}\ \right \}$ so that any element of it can be written in the form $a + b \cfrac{1 + \sqrt 5}{2}$ or simply $\cfrac{c}{2} + \cfrac{d \sqrt{5}}{2}$ with $c, d \in \mathbb{Z}$.
By this claim, if I take some $f\in \mathbb{Z}\left[\cfrac{1 + \sqrt{5}}{2}\right]$ then I should be able to write it in the form above. Now let $f = a_n \left(\cfrac{1 + \sqrt{5}}{2} \right)^n + a_{n - 1} \left(\cfrac{1 + \sqrt{5}}{2} \right)^{n - 1} + \dots + a_1 \left(\cfrac{1 + \sqrt{5}}{2} \right) + a_0$, each $a_i \in \mathbb{Z}$. The $k^{th}$ term in the partial sum above is equal to $a_k.\cfrac{n!.5^{\frac{k}{2}}}{k!(n-k)!2^k}$ . I can not see that how these terms cancel out each other and in the end we have something like $\cfrac{c}{2}+\cfrac{d\sqrt{5}}{2}$. How do we show this?
| Elements of $\mathbb Z[(1 + \sqrt{5})/2]$ take the form $g((1 + \sqrt{5})/2)$ for some polynomial $g(X) \in \mathbb Z[X]$. By the division theorem, you can write $g(X) = q(X)f(X) + r(X)$ where $r = 0$ or $\deg r < \deg f = 2$ and $q, r \in \mathbb{Z}[X]$. Then
$$ g\left( \frac{1 + \sqrt{5}}{2} \right) = q\left( \frac{1 + \sqrt{5}}{2} \right)f\left( \frac{1 + \sqrt{5}}{2} \right) + r\left( \frac{1 + \sqrt{5}}{2} \right) = r\left( \frac{1 + \sqrt{5}}{2} \right). $$
Since $f((1 + \sqrt{5})/2) = 0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2240560",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Find all $n$ for which $504$ doesn't divide $n^8-n^2$ I have seen a similar problem where someone showed that it does divide $n^9-n^3$ but here one has to show that it isn't the case for $n^8-n^2$ for all $n$.
| By the Chinese remainder theorem, $n^8-n^2$ is divisible by $504$ if and only if it is divisible by $8$, $7$ and $9$.
*
*Mod $8$, $n^2\equiv 0$ or $1$, except if $n\equiv \pm2\mod8$, hence $n^8-n^2=n^2(n^6-1)\equiv 0$, except if $n\equiv \pm2\mod8$, in which case $n^8-n^2\equiv 4$.
*Mod $7$, Lil' Fermat says $n^7\equiv n\mod 7$, hence $n^8-n^2\equiv n^2-n^2=0\mod7$.
*Mod $9$, $n^2\equiv 0$ if $n\equiv 0,\pm3$, and $n^6\equiv 1$ if $n\equiv \pm1,\pm 2, 4$, so $n^8-n^2\equiv 0\mod9$ for all $n$.
As a conclusion, $n^8-n^2\equiv 0\mod 504$, except if $n\equiv\pm2$ (i.e. $n\equiv 2,6$) $\bmod8$. As observed by @lhf, the simplifies to $n\equiv 2\mod4$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2241697",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Sum of all possible angles. If $$\tan\left(\frac{\pi}{12}-x\right) , \tan\left(\frac{\pi}{12}\right) , \tan \left(\frac{\pi}{12} +x\right)$$ in order are three consecutive terms of a GP then what is sum of all possible values of $x$.
I am not getting any start, can anybody provide me a hint?
| Use :$$\Big(\tan\frac{\pi}{2}\Big)^2=\tan\Big(\frac{\pi}{12}-x\Big)\tan\Big(\frac{\pi}{12}+x\Big) \tag1$$ (Condition for G.P)
And :
$$\tan (A\pm B) = \frac{\tan A \pm \tan B}{1\mp \tan A\tan B}$$
Let $\tan \dfrac{\pi}{12} =c$
Put in $(1)$
$$c^2=\frac {c+\tan x}{1-c\tan x} \cdot \frac{c-\tan x}{1+c\tan x} \implies c^2-c^4\tan^2 x =c^2-\tan^2 x \implies \tan x =0$$
Sum of solutions in a bounded interval will be sum all angles in the form $k\pi ~;~ k \in \mathbb Z$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2242177",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Residue of a function f
$\int \dfrac{w^2}{(w^2+1)^2(w^2+2w+2)}=\dfrac{p \pi}{q}$, limits are from minus infinity to infinity.
| After more comments see
\begin{eqnarray}
residue(f,-1+ i) &=& \dfrac{3}{25}-\dfrac{4}{25}i\\
residue(f,-1- i) &=& \dfrac{3}{25}+\dfrac{4}{25}i\\
residue(f, i) &=& -\dfrac{3}{25}+\frac{9 i}{100}\\
residue(f, -i) &=& -\dfrac{3}{25}-\frac{9 i}{100}
\end{eqnarray}
for integral $\displaystyle\int_{-\infty}^\infty \dfrac{x^2}{(x^2+1)^2(x^2+2x+2)}dx$ which lies in upper half plane, only residues in $i$ and $-1+i$ are important that concludes
$$2\pi i\left(\dfrac{3}{25}-\dfrac{4}{25}i-\dfrac{3}{25}+\frac{9 i}{100}\right)=2\pi i\left(-\frac{7 i}{100}\right)=\dfrac{7}{50}\pi$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2244901",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How to solve this using generating functions? I have an equation:
$t_n =2t_{n-1}+t_{n-2}; t_0=0, t_1=1$
So, I rewrote this using generating functions as:
$\sum_{n=0}^{\infty} t_nz^n = 2\sum_{n=0}^{\infty} t_{n-1}z^n+\sum_{n=0}^{\infty} t_{n-2}z^n$
$g(z)=2zg(z)+z^2g(z)$
$g(z)={1\over{1-2z-z^2}}$
I don't know how to proceed after this? Can you help me please?
Thanks.
| The correct and most effective approach (re. to "Concrete Mathematics") starts from writing the recurrence
in such a way that it incorporates the initial conditions and is valid for all the values of the index $n$.
Since the $t_n$ are assumed to be null for $n < 0$, we shall rewrite your recurrence as:
$$
t_{\,n} = 2\,t_{\,n - 1} + t_{\,n - 2} + \left[ {1 = n} \right]\quad \left| {\;\forall n} \right.
$$
where we make profitable use of the Iverson bracket
$$
\left[ P \right] = \left\{ {\begin{array}{*{20}c}
1 & {P = TRUE} \\
0 & {P = FALSE} \\
\end{array} } \right.
$$
Now we multiply by $z^n$ and sum
$$
\eqalign{
& g(z) = \sum\limits_{0\, \le \,n} {t_{\,n} \,z^{\,n} } = 2\,\sum\limits_{0\, \le \,n} {t_{\,n - 1} \,z^{\,n} } + \sum\limits_{0\, \le \,n} {t_{\,n - 2} \,z^{\,n} } + \sum\limits_{0\, \le \,n} {\left[ {1 = n} \right]\,z^{\,n} } = \cr
& = 2\,z\,g(z) + z^2 g(z) + z \cr}
$$
so
$$
g(z) = {z \over {1 - 2z - z^2 }}
$$
To this expression we can apply partial fraction decomposition to obtain
$$
\eqalign{
& g(z) = {z \over {1 - 2z - z^2 }} = - {z \over {\left( {z - \left( { - 1 - \sqrt 2 } \right)} \right)\left( {z - \left( { - 1 + \sqrt 2 } \right)} \right)}} = \cr
& = - {z \over {\left( {z - a} \right)\left( {z - b} \right)}} = - {1 \over {a - b}}\left( {{a \over {\left( {z - a} \right)}} - {b \over {\left( {z - b} \right)}}} \right) = \cr
& = {1 \over {a - b}}\left( {{1 \over {\left( {1 - z/a} \right)}} - {1 \over {\left( {1 - z/b} \right)}}} \right) = \cr
& = {1 \over {a - b}}\sum\limits_{0\, \le \,n} {\left( {{1 \over {a^{\,n} }} - {1 \over {b^{\,n} }}} \right)\,z^{\,n} } \cr}
$$
Therefore we get for $t_n$
$$
\eqalign{
& t_{\,n} = {1 \over {a - b}}\left( {{1 \over {a^{\,n} }} - {1 \over {b^{\,n} }}} \right) = {1 \over {2\sqrt 2 }}\left( {{1 \over {\left( {\sqrt 2 - 1} \right)^{\,n} }} - {1 \over {\left( { - 1 - \sqrt 2 } \right)^{\,n} }}} \right) = \cr
& = {1 \over {2\sqrt 2 }}\left( {{1 \over {\left( {\sqrt 2 - 1} \right)^{\,n} }} - {{\left( { - 1} \right)^{\,n} } \over {\left( {\sqrt 2 + 1} \right)^{\,n} }}} \right) = \cr
& = {{\left( {\sqrt 2 + 1} \right)^{\,n} - \left( { - 1} \right)^{\,n} \left( {\sqrt 2 - 1} \right)^{\,n} } \over {2\sqrt 2 }} = \cr
& = {1 \over {2\sqrt 2 }}\sum\limits_{\left( {0\, \le } \right)\,k\,\left( { \le \,n} \right)} {\left( {\left( \matrix{
n \cr
k \cr} \right)\,\sqrt 2 ^{\,k} - \left( { - 1} \right)^{\,n} \left( \matrix{
n \cr
k \cr} \right)\,\sqrt 2 ^{\,k} \left( { - 1} \right)^{\,n - k} } \right)} = \cr
& = {1 \over {2\sqrt 2 }}\sum\limits_{\left( {0\, \le } \right)\,k\,\left( { \le \,n} \right)} {\left( {\left( \matrix{
n \cr
k \cr} \right)\, - \left( { - 1} \right)^{\,k} \left( \matrix{
n \cr
k \cr} \right)\,} \right)\sqrt 2 ^{\,k} } = \cr
& = \sum\limits_{\left( {0\, \le } \right)\,j\,\left( { \le \,n} \right)} {\left( \matrix{
n \cr
2j + 1 \cr} \right)2^{\,j} } \cr}
$$
whose first values are:
$$
t = \left\{ {0,1,2,5,12,29,70,169, \cdots } \right\}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2246960",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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Detailing two integration equalities Could you please detail the follwing two equalities (I can not understand why they are valid):
\begin{align*}
\int_{0}^{\infty}\frac{\sin(2 \pi x)}{x+1}\, dx&=\int_{0}^{\infty}\frac{\sin( \pi x)}{x+2}\,dx\\
&=\int_{0}^{1}\sin( \pi x) \left[ \frac{1} {x+2} -\frac{1} {x+3}+\frac{1} {x+4}-\frac{1} {x+5}+\dotsb\right] \,dx
\end{align*}
[EDIT] the first one is a simple variable change. But the second I did not get.
| The second one in the OP is not quite correct. The argument of the sine function should be $\pi x $, not $2\pi x$.
Note that we can write
$$\begin{align}
\int_0^\infty \frac{\sin(\pi x)}{x+2}\,dx&=\sum_{k=1}^\infty \int_{k-1}^k \frac{\sin(\pi x)}{x+2}\,dx\\\\
&=\sum_{k=1}^\infty \int_0^1 \frac{\sin(\pi (x+k-1))}{x+k+1}\,dx\\\\
&=\sum_{k=1}^\infty \int_0^1 \frac{(-1)^{k-1}\sin(\pi x)}{x+k+1}\,dx\\\\
&=\int_0^1 \sin(\pi x)\left(\frac{1}{x+2}-\frac{1}{x-3}+\frac{1}{x+4}-\frac{1}{x+5}+-...\right)\,dx
\end{align}$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Knowing that for any set of real numbers $x,y,z$, such that $x+y+z = 1$ the inequality $x^2+y^2+z^2 \ge \frac{1}{3}$ holds.
Knowing that for any set of real numbers $x,y,z$, such that $x+y+z = 1$ the inequality $x^2+y^2+z^2 \ge \frac{1}{3}$ holds.
I spent a lot of time trying to solve this and, having consulted some books, I came to this:
$$2x^2+2y^2+2z^2 \ge 2xy + 2xz + 2yz$$ $$2xy+2yz+2xz = 1-(x^2+y^2+z^2) $$ $$2x^2+2y^2+2z^2 \ge 1 - x^2 -y^2 - z^2 $$ $$x^2+y^2+z^2 \ge \frac{1}{3}$$
But this method is very unintuitive to me and I don't think this is the best way to solve this. Any remarks and hints will be most appreciated.
| This is not a proof in itself, but if you've studied statistics, then you've seen a proof that
$$0\le V(X)=E(X^2)-E(X)^2$$
If we now consider a random variable $X$ with three equally likely values, $X=x,y$, and $z$, then we have
$$E(X)={x+y+z\over3}\qquad\text{and}\qquad E(X^2)={x^2+y^2+z^2\over3}$$
If, in addition, we assume $x+y+z=1$, then we have $E(X)={1\over3}$, which implies $E(X^2)\ge\left(1\over3\right)^2={1\over9}$, or $x^2+y^2+z^2\ge{1\over3}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2247973",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "21",
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"answer_id": 7
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Prove that $\lim_{(x,y)\to (0.0)} \frac{4xy^2 - 3x^3}{x^2 - y^2}$ does not exist I'm struggling with this limit. I have to approach using different curves and show that there is one curve wich prove that this limit does not exist, despite the fact that when trying with a lot of curves shows that the limit is 0.
If someone helps me to find that curve I will be very pleased. Thanks!
$$\lim_{(x,y)\to (0.0)} \frac{4xy^2 - 3x^3}{x^2 - y^2}$$
| By writing $x,y$ in polar coordinates,
$$\lim_{(x,y)\to (0,0)} \frac{4xy^2 - 3x^3}{x^2 - y^2}=\lim_{r\to 0}\ r\cdot\frac{4\cos\theta\sin^2\theta - 3\cos^3\theta}{\cos^2\theta - \sin^2\theta}$$
indicating that the limit may not be zero only when $\tan^2\theta\to 1$, that is $\theta\to \frac{\pi}{4}$ or $\theta\to\frac{7\pi}{4}$.
Therefore, we shall choose a curve which the slope at origin is $\tan\frac{\pi}{4}=1$ or $\tan\frac{7\pi}{4}=-1$. For example, $y=x^2+x$ would be a choice.
$$\lim_{x\to 0} \frac{4x(x^2+x)^2 - 3x^3}{x^2 - (x^2+x)^2}=\lim_{x\to 0}\frac{4x^5+8x^4+4x^3-3x^3}{x^2 - x^4-2x^3-x^2}=\lim_{x\to 0}\frac{4x^2+8x+1}{-x-2}=-\frac{1}{2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2249619",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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$2x^2 + 3x +4$is not divisible by $5$ I tried by $x^2 \equiv 0, 1, 4 \pmod 5$ but how can I deal with $3x$?
I feel this method does not work here.
| $5$ is small. You could just consider all possibilities $\mod{5}$. There are only $0,1,2,3,4$. If any of these get you $0$, then, it can be possibly divisible by $5$. Otherwise, not. Let $f(x)=2x^2+3x+4$. So, $x\equiv0\implies f(0)\equiv4$, $x\equiv1\implies f(1)\equiv 2+3+4\equiv4\pmod{5}$, $x\equiv2\implies f(2)\equiv8+6+4\equiv3$, $x\equiv3\implies f(3)\equiv 18+9+4\equiv 1\pmod{5}$, and finally $x\equiv4\implies f(4)\equiv 32+12+4\equiv3$. An even faster way to do the calculations would just be to see that $3\equiv-2$ and $4\equiv-1$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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If $ab$ is a square number and $\gcd(a,b) = 1$, then $a$ and $b$ are square numbers.
Let $n, a$ and $b$ be positive integers such that $ab = n^2$. If $\gcd(a, b) = 1$, prove that there exist positive integers $c$ and $d$ such that $a = c^2$ and $b = d^2$
So far I have tried this:
Since $n^2 = ab$ we have that $n = \sqrt{ab}$.
Because $\gcd(a,b) = 1$, there exists integers $k$ and $l$ such that $ak + bl = 1$. This means that $\sqrt{a}(k\sqrt{}) + \sqrt{b}(l\sqrt{b}) = 1$.
Hence $\sqrt{a}$ and $\sqrt{b}$ are both positive integers and we can set $\sqrt{a} = c$ for some arbitrary integer $c$ and $\sqrt{b} = d$ for some arbitrary integer $d$. Therefore, $a = c^2$ and $b = d^2$.
| This:
$\sqrt{a}$ and $\sqrt{b}$ are both positive integers
does not follow from
$\sqrt{a}(k\sqrt{a}) + \sqrt{b}(l\sqrt{b}) = 1$.
For instance,
$$ \sqrt{2}((-1)\sqrt{2}) + \sqrt{3}((1)\sqrt{3}) = 1$$
($a = 2, b = 3, k = -1, l = 1$).
What you need to do instead is use prime factorization.
| {
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"timestamp": "2023-03-29T00:00:00",
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Find $\lim\limits_{x \to 0}{\frac{1}{x}(\frac{1}{\sin(x)}-\frac{1}{\sinh(x)})}$.
$$\lim\limits_{x \to 0}{\frac{1}{x}\left(\frac{1}{\sin(x)}-\frac{1}{\sinh(x)}\right)}$$
I replaced $\sinh(x)$ by $\displaystyle\frac{e^{x}-e^{-x}}{2}$ and used L'Hopital's rule twice but this expression becomes very large.
Could you please help me
| Disclaimer: I am going to write the mother of overkills, just for fun.
From the Weierstrass product for the sine function we have
$$ \frac{x}{\sin(x)} = \prod_{n\geq 1}\left(1-\frac{x^2}{\pi^2 n^2}\right)^{-1} = \prod_{n\geq 1}\left(1+\frac{x^2}{\pi^2 n^2}+\frac{x^4}{\pi^4 n^4}+\ldots\right) \tag{1}$$
and since $\frac{x}{\sin x}$ is an even analytic function in a neighbourhood of the origin we have
$$ \frac{x}{\sin(x)}=1+a_2 x^2+o(x^3),\qquad a_2 = \frac{1}{\pi^2}\sum_{n\geq 1}\frac{1}{n^2} = \frac{\zeta(2)}{\pi^2}\tag{2} $$
and by replacing $x$ with $ix$ we also get
$$ \frac{x}{\sinh(x)}=1-a_2 x^2+o(x^3),\qquad a_2 = \frac{1}{\pi^2}\sum_{n\geq 1}\frac{1}{n^2} = \frac{\zeta(2)}{\pi^2}\tag{3} $$
so:
$$ \lim_{x\to 0}\frac{1}{x}\left(\frac{1}{\sin(x)}-\frac{1}{\sinh x}\right) = \lim_{x\to 0}\frac{1}{x^2}\left(\frac{x}{\sin(x)}-\frac{x}{\sinh x}\right) = 2a_2 = \color{red}{\frac{1}{3}}.\tag{4}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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How can all the options be correct in this question? Question: The equations of two sides $AB$ and $AC$ of an isosceles triangle $ABC$ are $x+y=5$ and $7x-y=3$ respectively. If the area of the Triangle $ABC$ is 5 sq. units , then the possible equations of the side $BC$ is (are): ?
$A). x-3y+1=0$
$B). 3x+y+2=0$
$C). x-3y-21=0$
$D). 3x+y-12=0$
Answer given :$A,B,C,D$
Making an attempt: Obviously, $A$ must be $(1,4)$. Now after that, if I consider a point $(x_1,y_1)$ on $AB$ and $(x_2,y_2)$ on $BC$, I could see that the area should be :
$$\frac{1}{2} \begin{bmatrix}1&4&1\\x_1&y_1&1\\x_2&y_2&1 \end{bmatrix}=5$$
(treat it as a determinant please)
And, I assumed that $AB=BC$ which gave me:
$$(2-x_1-x_2)(x_2-x_1)=(8-y_1-y_2)(y_1-y_2)$$
But sadly none of the equations could satisfy these conditions. True enough, my assumption can be wrong but I can't seem to make any headway in this question. Please Guide
| WLOG, $B(a,5-a);C(b,7b-3)$
We have $$(a-1)^2+(5-a-4)^2=(b-1)^2+(7b-3-4)^2$$
$$\iff2(a-1)^2=50(b-1)^2\implies a-1=\pm5(b-1)$$
For $a-1=5(b-1), a=5b-4$
$\implies|BC|^2=(a-b)^2+(5-a-7b+3)^2=(5b-4-b)^2+\{8-7b-(5b-4)\}^2=(b-1)^2(4^2+12^2)$
Now if $M$ is the midpoint of $BC,M:\left(\dfrac{a+b}2,\dfrac{5-a+7b-3}2\right)$
Replace $a$ with $5b-4$
Now $|AM|$ is also the perpendicular distance of $BC$ from $A$
We have $$\dfrac{|AM|\cdot|BC|}2=5$$
Similarly for $a-1=-5(b-1)$
| {
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Summation splitting: $\sum_{n=1}^{\infty} \frac{n}{3\cdot 5\cdot 7 \dots (2n+1)}$ $$\sum_{n=1}^{\infty} \frac{n}{3\cdot 5\cdot 7 \dots (2n+1)}$$
Can someone please help me in solving this problem, I tried to take the summation of the numerator and denominator individually but my teacher said that it is wrong to do the summation individually, can somebody please explain why is it wrong to take individual summation and please recommend the correct way of solving this problem
| Hint. One may observe that, for $n\ge1$,
$$
\begin{align}
\frac {n}{1\cdot 3\cdots (2n+1)} &=\color{red}{\frac12} \cdot \frac {(2n+1)-1}{1\cdot 3\cdots (2n+1)}
\\\\&=\color{red}{\frac12} \cdot \frac {1}{1\cdot 3\cdots (2n-1)}-\color{red}{\frac12} \cdot \frac {1}{1\cdot 3\cdots (2n+1)},
\end{align}
$$ then, by telescoping terms, one obtains
$$
\sum_{n=1}^\infty\frac {n}{1\cdot 3\cdots (2n+1)}=\color{red}{\frac12}.
$$
| {
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} |
Write the function $\frac{1}{(z+1)(3-z)}$ as a Laurent series. $$f(z)=\frac{1}{(z+1)(3-z)}=\frac{1}{4z+4} + \frac{1}{12-4z}$$
$$\frac{1}{4z+4}=\frac{1}{4z}\frac{1}{1-\frac{-1}{z}}=\frac{1}{4z}\sum_{k=0}^{\infty} \left(\frac{-1}{z}\right)^k$$
$$\frac{1}{12-4z}=\frac{1}{12}\frac{1}{1-\frac{z}{3}}=\frac{1}{12}\sum_{k=0}^{\infty} \left(\frac{z}{3}\right)^k$$
$$f(z)=\frac{1}{4z}\sum_{k=0}^{\infty} \left(\frac{-1}{z}\right)^k+\frac{1}{12}\sum_{k=0}^{\infty} \left(\frac{z}{3}\right)^k$$
I can rewrite that as $$f(z)=\frac{1}{4z}\sum_{k=-\infty}^{0} (-1)^k z^k+\frac{1}{12}\sum_{k=0}^{\infty} \left(\frac{1}{3}\right)^k z^k$$.
I need to move the $\frac{1}{4z}$ and $\frac{1}{12}$ into the sums but finding a series that will converge to each, but I have no idea what to use for either. Any suggestions? Am I taking a wrong approach or is there an obvious series to use for this?
Edit: The center is $0$ and the region in $1 \le |z| \le 3$.
I think I can use a geometric sequence to say $\frac{1}{4z}=\sum_{k=0}^{\infty}\frac{1}{2}(\frac{1}{2})^{k-1}\frac{z}{k}$ and $\frac{1}{12}=\sum_{k=0}^{\infty}\frac{1}{36}(\frac{2}{9})^{k-1}\frac{z}{k}$. I'm pretty sure that's true, but it seems like it makes the whole thing a complicated mess.
|
The function
\begin{align*}
f(z)&=\frac{1}{(z+1)(3-z)}\\
&=\frac{1}{4(z+1)}-\frac{1}{4(z-3)}
\end{align*}
has two simple poles at $-1$ and $3$.
Since we want to find a Laurent expansion with center $0$, we look at the poles $-1$ and $3$ and see they determine three regions.
\begin{align*}
|z|<1,\qquad\quad
1<|z|<3,\qquad\quad
3<|z|
\end{align*}
*
*The first region $ |z|<1$ is a disc with center $0$, radius $1$ and the pole $-1$ at the boundary of the disc. In the interior of this disc all two fractions with poles $-1$ and $3$ admit a representation as power series at $z=0$.
*The second region $1<|z|<3$ is the annulus with center $0$, inner radius $1$ and outer radius $3$. Here we have a representation of the fraction with poles $-1$ as principal part of a Laurent series at $z=0$, while the fraction with pole at $3$ admits a representation as power series.
*The third region $|z|>3$ containing all points outside the disc with center $0$ and radius $3$ admits for all fractions a representation as principal part of a Laurent series at $z=0$.
A power series expansion of $\frac{1}{z+a}$ at $z=0$ is
\begin{align*}
\frac{1}{z+a}&=\frac{1}{a}\cdot\frac{1}{1+\frac{z}{a}}\\
&=\sum_{n=0}^{\infty}\frac{1}{a^{n+1}}(-z)^n
\end{align*}
The principal part of $\frac{1}{z+a}$ at $z=0$ is
\begin{align*}
\frac{1}{z+a}&=\frac{1}{z}\cdot\frac{1}{1+\frac{a}{z}}=\frac{1}{z}\sum_{n=0}^{\infty}\frac{a^n}{(-z)^n}
=-\sum_{n=0}^{\infty}\frac{a^n}{(-z)^{n+1}}\\
&=-\sum_{n=1}^{\infty}\frac{a^{n-1}}{(-z)^n}
\end{align*}
We can now obtain the Laurent expansion of $f(z)$ at $z=0$ for all three regions.
*
*Region 2: $1<|z|<3$
\begin{align*}
f(z)&=\frac{1}{4(z+1)}-\frac{1}{4(z-3)}\\
&=-\frac{1}{4}\sum_{n=1}^\infty\frac{1}{(-z)^n}-\frac{1}{4}\sum_{n=0}^\infty \frac{1}{(-3)^{n+1}}(-z)^n\\
&=\frac{1}{4}\sum_{n=1}^\infty(-1)^{n+1}\frac{1}{z^n}+\frac{1}{4}\sum_{n=0}^\infty \frac{1}{3^{n+1}}z^n\\
\end{align*}
The Laurent expansion for the other regions can be calculated similarly.
| {
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Proving that $3^n+7^n+2$ is divisible by 12 for all $n\in\mathbb{N}$. Can someone help me prove this? :( I have tried it multiple times but still cannot get to the answer.
Prove by mathematical induction for $n$ an element of all positive integers that $3^n+7^n+2$ is divisible by 12.
| With induction:
When $n=1$: $3^1+7^1+2=12$ is divisible by $12$. Therefore, we assume that $(3^k+7^k+2)$ is divisible by $12$ for some $k≥1$ (thus, $3^k+7^k+2=12a$ where $a$ is a function of $3$ and $7$).
Then, $$3^{k+1}+7^{k+1}+2=3×3^k+7×7^k+2$$
$$=3×3^k+7^k(3+4)+6-6+2$$
$$=3×3^k+3×7^k+6+4×7^k-4$$
$$=3(3^k+7^k+2)+4(7^k-1)$$
$$=3×(12a)+4(7-1)(7^{k-1}+7^{k-2}+⋯+2+1)$$
$$=36a+24(7^{k-1}+7^{k-2}+⋯+2+1)$$
$$=12b$$ where $b$ is also a function of $3$ and $7$.
Thus, $(3^{k+1}+7^{k+1}+2)$ is also divisible by $12$.
| {
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"question_score": "3",
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Which is greater $x_1$ or $x_2$? $$x_1=\arccos\left(\frac{3}{5}\right)+\arccos\left(\frac{2\sqrt{2}}{3}\right)$$
$$x_2=\arcsin\left(\frac{3}{5}\right)+\arcsin\left(\frac{2\sqrt{2}}{3}\right)$$ We have to find which is greater among $x_1$ and $x_2$
If we add both we get $$x_1+x_2=\pi$$ If we use formulas we get
$$x_1=\arccos\left(\frac{6\sqrt{2}-4}{15}\right)$$ and
$$x_2=\arcsin\left(\frac{3+8\sqrt{2}}{15}\right)$$
but how to compare now?
| Ok fine similarly we have $$x_1=arccos\left(\frac{3}{5}\right)+arccos\left(\frac{2\sqrt{2}}{3}\right)=arcsin\left(\frac{4}{5}\right)+arcsin\left(\frac{1}{3}\right)$$
Now $$\frac{4}{5} \lt \frac{\sqrt{3}}{2}$$
hence
$$arcsin\left(\frac{4}{5}\right) \lt \frac{\pi}{3}$$ and
$$\frac{1}{3} \lt \frac{1}{2}$$ so
$$arcsin\left(\frac{1}{3}\right) \lt \frac{\pi}{6}$$
hence
$$x_1 \lt \frac{\pi}{2}$$
Finally $$x_2 \gt \frac{\pi}{2} \gt x_1$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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"answer_count": 4,
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Proving that a sequence is monotone Let $ \{s_m\}_{m=1}^\infty$ be the sequence defined:
$s_m = \sum_{k=1}^{m} \frac{1}{\sqrt{m^2 + k}}$
I have already proven that $s_m\to 1$ as $m\to \infty $ but i'm having trouble to show that $s_m$ is a monotonic sequence.
Would appreciate any help or advice, thanks in advance.
| The power series for
$(1+x)^{-1/2}$
with $0 < x < 1$
is an enveloping series,
in that its sum is between
any two consecutive finite sums.
Since
$(1+x)^{-1/2}
=1-\frac{x}{2}+\frac{3x^2}{8}-\frac{5x^3}{16}+...
$
we have
$1-\frac{x}{2}
\lt (1+x)^{-1/2}
\lt 1-\frac{x}{2}+\frac{3x^2}{8}
$
or
$(1+x)^{-1/2}
=1-\frac{x}{2}+ax^2
$
where
$0 < a < \frac38
$.
This is our basic inequality.
Note:
All the $a$ and $a_m$
in what follows are
values satisfying
$0 < a < \frac38$.
$\begin{array}\\
s_m
&= \sum_{k=1}^{m} \frac{1}{\sqrt{m^2 + k}}\\
&= \frac1{m}\sum_{k=1}^{m} \frac{1}{\sqrt{1 + k/m^2}}\\
&= \frac1{m}\sum_{k=1}^{m} (1 + k/m^2)^{-1/2}\\
&= 1-\frac1{2m}\sum_{k=1}^{m}
(k/m^2)+\frac1{m}\sum_{k=1}^{m}a_k(k/m^2)^2\\
&= 1-\frac{m(m+1)}{4m^3}+a_m\frac{m(m+1)(2m+1)}{6m^5}\\
&= 1-\frac{m+1}{4m^2}+a_m\frac{(m+1)(2m+1)}{6m^4}\\
\end{array}
$
In what follows,
"(W)" means that
Wolfy helped.
$\begin{array}\\
s_{m+1}-s_m
&=(1-\frac{m+2}{4(m+1)^2}+a_{m+1}\frac{(m+2)(2m+3)}{6(m+1)^4})-(1-\frac{m+1}{4m^2}+a_m\frac{(m+1)(2m+1)}{6m^4})\\
&=\frac{m+1}{4m^2}-\frac{m+2}{4(m+1)^2}+a_{m+1}\frac{(m+2)(2m+3)}{6(m+1)^4}-a_m\frac{(m+1)(2m+1)}{6m^4}\\
&=\frac{m^2 + 3 m + 1}{4 m^2 (m + 1)^2}+a_{m+1}\frac{(m+2)(2m+3)}{6(m+1)^4}-a_m\frac{(m+1)(2m+1)}{6m^4}
\qquad\text{(W)}\\
&\gt\frac{m^2 + 3 m + 1}{4 m^2 (m + 1)^2}-\frac38\frac{(m+1)(2m+1)}{6m^4}\\
&=\frac{2 m^4 + 5 m^3 - 5 m^2 - 5 m - 1}{16 m^4 (m + 1)^2}
\qquad\text{(W)}\\
&\gt 0\\
\end{array}
$
| {
"language": "en",
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"source": "stackexchange",
"question_score": "2",
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Determining the number of solutions using Generating functions I have the equation
$u_1 + u_2 + ... + u_5 = 24$
with the restrictions
$1 \le u_i \le 7, i = 1,...5$
I've managed to work up to the point of finding the coefficient of $x^{24}$ in
$(X+X^2+...X^7)^5 = X^5(1+X+...X^6)^5$
I know my final answer will need to be in binomial form similar to this $\binom{10}{5}$
However I am unsure where to go, or if this is even correct. Any help would be greatly appreciated.
| Everything is fine with your generating function. It is convenient to use the coefficient of operator $[x^n]$ to denote the coefficient of $x^n$ in a series.
We obtain
\begin{align*}
[x^{24}]&(x+x^2+\cdots+x^7)^5\\
&=[x^{24}]x^5(1+x+\cdots+x^6)^5\\
&=[x^{19}]\left(\frac{1-x^{7}}{1-x}\right)^5\tag{1}\\
&=[x^{19}](1-x^7)^5\cdot\frac{1}{(1-x)^5}\\
&=[x^{19}](1-5x^{7}+10x^{14})\sum_{n=0}^\infty\binom{-5}{n}(-x)^n\tag{2}\\
&=\left([x^{19}]-5[x^{12}]+10[x^5]\right)\sum_{n=0}^\infty\binom{n+4}{4}x^n\tag{3}\\
&=\binom{23}{4}-5\binom{16}{4}+10\binom{9}{4}\tag{4}\\
&=1015
\end{align*}
Comment:
*
*In (1) we use the geometric series formula and apply the rule
\begin{align*}
[x^{p-q}]A(x)=[x^p]x^qA(x)
\end{align*}
*In (2) we use the binomial series expansion and expand $(1-x^7)^5$. We skip terms with exponent $21$ and greater since they do not contribute to $[x^{19}]$.
*In (3) we use the linearity of the coefficient of operator, apply again the rule as in (1) and we use the binomial identity $$\binom{-p}{q}=\binom{p+q-1}{p-1}(-1)^q$$
*In (4) we select the coefficients accordingly.
| {
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Laurent series of $\frac{1}{z^4+z^2}$ about $z=0$ Okay, I get $\frac{1}{z^4+z^2}=\frac{1}{z^2}-\frac{1}{z^2+1}$
And I know I need to use geometric series sum but couldn't quite find it. Can someone give me a hint?
| One of the two terms is easy $$\frac{1}{1+z^2} = \frac{1}{1-(-z^2)}=\sum_{m=0}^{+\infty} (-1)^mz^{2m}.$$ For the other, notice that $\frac{1}{z^2}$ is already in a Laurent form at $z=0.$ So the developpement is $$\frac{1}{z^4+z^2} = \frac{1}{z^2} - \sum_{m=0}^{+\infty} (-1)^mz^{2m}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2260613",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Volume of the revolution solid Let $K \subset \mathbb R^2$ be area bounded by curves $x=2$, $y=3$, $xy=2$ and $xy=4$.
What is the volume of the solid we get after rotating $K$ around $y$-axis?
I tried to express the curves in terms of $y$ and solved the integral $\displaystyle \pi\left(\int_2^3 \frac{16}{y^2}dy - \int_1^3\frac{4}{y^2}dy \right)$ but it came out zero. I got the integral by subtracting the volume of the inner shape from the volume of the outer shape, if that makes any sense.
| $y = 3$ intersects $xy = 2$ at $x = \frac 23$ and $xy = 4$ at $x = \frac 43$
By shells I get.
$2\pi\int_\frac {2}{3}^{\frac 43} (3-\frac 2x)x \ dx + 2\pi\int_{\frac 43}^2 (\frac 4x -\frac 2x)x \ dx\\
2\pi\int_\frac {2}{3}^{\frac 43} 3x-2 \ dx + 2\pi\int_{\frac 43}^2 2\ dx\\
2\pi(\frac{3}{2}x^2 -2x|_\frac {2}{3}^{\frac 43} + 2x| _{\frac 43}^2)\\
2\pi(\frac 83 - \frac 23 - \frac 83 + \frac 43 + 4 - \frac 83) \\
4\pi$
| {
"language": "en",
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"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Monotonicity of $f(x,y)=\frac{x^2+y^2}{x+y}$ $f(x,y)=\frac{x^2+y^2}{x+y} $ where $x,y \in [0,1]$
let $x\leq r$ and $y\leq s$ I have to prove\disprove $f(x,y)\leq f(r,s)$.
Clearly $x^2+y^2\leq r^2+s^2$ and $x+y\leq r+s$ but then $\frac{1}{x+y}\geq \frac{1}{r+s}$ ??
| It is not true.
As one of many counterexamples, consider $x=r=1$ and $y=0, s=\frac12$
Then $\dfrac{1^2+0^2}{1+0}=1 \gt \dfrac56 = \dfrac{1^2+\left(\frac12\right)^2}{1+\frac12}$ so $f(x,y) \not \le f(r,s)$ even though $x \le r$ and $y \le s$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2263952",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Is it true for any associative groupoid which elements are $a,b,c$, that $(a\ \circ \ b) \circ c = (b \ \circ c) \circ a $? As I know an associative groupoid is a semigroup. I'm assuming that it can be true, but I'm not sure, how can I give a proof for it?
| Let $G = (\{a,b,c\},\circ)$ be the groupoid where $\circ$ obeys the following table:
$$
\begin{array}{c|ccc} \circ & a & b & c \\ \hline a & a & a & a \\ b & b & b & b \\ c & a & a & c \end{array}
$$
One can check that $\circ$ is an associative operation. However
$$(a\circ b)\circ c = a \circ c = a \neq b = b\circ a = (b\circ c)\circ a.$$
My code found $7$ such groupoids (that is, up to isomorphism). Other than the one above, these are:
$$
\begin{array}{lcccr}
\begin{array}{c|ccc} \circ & a & b & c \\ \hline a & a & a & a \\ b & b & b & b \\ c & b & b & b \end{array} & &
\begin{array}{c|ccc} \circ & a & b & c \\ \hline a & a & a & a \\ b & b & b & b \\ c & c & c & c \end{array} & &
\begin{array}{c|ccc} \circ & a & b & c \\ \hline a & a & a & c \\ b & a & b & c \\ c & a & a & c \end{array} \\ \\
\begin{array}{c|ccc} \circ & a & b & c \\ \hline a & a & b & a \\ b & a & b & b \\ c & a & b & c \end{array} & &
\begin{array}{c|ccc} \circ & a & b & c \\ \hline a & a & c & c \\ b & a & b & c \\ c & a & c & c \end{array} & &
\begin{array}{c|ccc} \circ & a & b & c \\ \hline a & b & b & c \\ b & b & b & c \\ c & b & b & c \end{array}
\end{array}
$$
If my code is correct this are the only $3$-element associative groupoids (up to isomorphism) that don't satisfy the identity in the stated problem.
| {
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"source": "stackexchange",
"question_score": "1",
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Probability problem with tissues Girl A has 10 new tissues and 5 used tissues in her drawer. Girl B comes to visit girl A. Girl A makes her friend a coffee and randomly takes out 2 tissues out of the drawer. Later she puts them back into the drawer. The next day she takes out 2 tissues again. What is the probability that both of the tissues are used?
The answer is 359/2205 . I've been trying to solve this without any luck.
| Initially we have ten unused tissues and five used ones. There are three different use cases:
*
*The girl takes out two used tissues on the first day. Both the first and the second drawn tissue must be used, so this probability equals $\frac{5}{15} \cdot \frac{4}{14}$;
*The girl takes out one used and one unused tissue on the first day, which comes down to either taking first one used and then one unused tissue, or taking first one unused and then one used tissue. As such, the probability of this happening equals $\frac{10}{15} \cdot \frac{5}{14} + \frac{5}{15} \cdot \frac{10}{14}$;
*The girl takes out two unused tissues on the first day, which has a probability of $\frac{10}{15} \cdot \frac{9}{14}$.
The probability of taking out two used tissues on the second day equals $\frac{5}{15} \cdot \frac{4}{14}$ in the first case, $\frac{6}{15} \cdot \frac{5}{14}$ in the second and $\frac{7}{15} \cdot \frac{6}{14}$ in the third. Overall, we get:
$$P[2\,used\,tissues] = \Big(\frac{5}{15} \cdot \frac{4}{14}\Big) \Big(\frac{5}{15} \cdot \frac{4}{14}\Big) + \Big(\frac{10}{15} \cdot \frac{5}{14} + \frac{5}{15} \cdot \frac{10}{14}\Big) \Big(\frac{6}{15} \cdot \frac{5}{14}\Big) + \Big(\frac{10}{15} \cdot \frac{9}{14}\Big) \Big(\frac{7}{15} \cdot \frac{6}{14}\Big) = \frac{7180}{44100} = \frac{359}{2205}$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
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How many four digit numbers divisible by 29 have the sum of their digits 29?
How many four digit numbers divisible by $29$ have the sum of their digits $29$?
A way to do it would be to write $1000a+100b+10c+d=29m$ and $a+b+c+d=29$ and then form equations like $14a+13b+10c+d=29m'$ and eventually $4a + 3b – 9d = 29 (m'' – 9)$. Analysing this equation for integer solutions using the advantage we have $\to$ $29$ is a prime; will give the solutions, but is tedious work.
Are there better solutions?
| $(a=4\land b=9\land c=8\land d=8\land m=172)\lor (a=7\land b=5\land c=9\land
d=8\land m=262)\lor (a=7\land b=8\land c=5\land d=9\land m=271)\lor (a=9\land
b=6\land c=8\land d=6\land m=334)\lor (a=9\land b=9\land c=4\land d=7\land
m=343)$
So 5 numbers which are $4988,7598,7859,9686,9947$.
| {
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"source": "stackexchange",
"question_score": "6",
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A number that is a triangle number and also a square number. This is a problem from standupmaths, from Youtube.
Question:
Think of a number which is a triangle number and a square number.
This can be expressed using a single equation:
$$x^2=\frac{y(y+1)}{2}$$
If you use trial and improvement, you have the square numbers:
$[1,4,9,16,25,36,49,64,81,100,121,144,169,196,225,256,289...]$
And triangle numbers:
$[1, 3, 6, 10, 15, 21, 28, 36, 45, 55, 66, 78, 91, 105, 120, 136, 153, 171, 190, 210...]$
The only numbers that appears in both of these lists are $$36, 6^2 = 36, \frac{8*9}{2} = 36, 1, 1^2=1, \frac{1*2}{2}=1$$
Are there any beyond this?
Well, $2x^2=y^2+y$. How do I go on from this? What do I have to use?
What are the numbers where the number you substitute in are equal, and their results are equal?
Lets say:
What if $x=y$?
$$x^2=\frac{x(x+1)}{2}$$
$2x^2=x(x+1)$
$2x^2=x^2+x$
$x^2-x=0$
$x(x-1)=0$
$x$ or $(x-1)$ has to be $0$.
Therefore, $x = 0.5\pm 0.5$
Second question:
But is $0$ technically a square or triangle number?
| Following Shark, here is how to solve the Pell equation $z^2 - 8 x^2 = 1$ by hand, although one can easily guess the first as $9-8=1:$
Method described by Prof. Lubin at Continued fraction of $\sqrt{67} - 4$
$$ \sqrt { 8} = 2 + \frac{ \sqrt {8} - 2 }{ 1 } $$
$$ \frac{ 1 }{ \sqrt {8} - 2 } = \frac{ \sqrt {8} + 2 }{4 } = 1 + \frac{ \sqrt {8} - 2 }{4 } $$
$$ \frac{ 4 }{ \sqrt {8} - 2 } = \frac{ \sqrt {8} + 2 }{1 } = 4 + \frac{ \sqrt {8} - 2 }{1 } $$
Simple continued fraction tableau:
$$
\begin{array}{cccccccccc}
& & 2 & & 1 & & 4 & \\
\\
\frac{ 0 }{ 1 } & \frac{ 1 }{ 0 } & & \frac{ 2 }{ 1 } & & \frac{ 3 }{ 1 } \\
\\
& 1 & & -4 & & 1
\end{array}
$$
$$
\begin{array}{cccc}
\frac{ 1 }{ 0 } & 1^2 - 8 \cdot 0^2 = 1 & \mbox{digit} & 2 \\
\frac{ 2 }{ 1 } & 2^2 - 8 \cdot 1^2 = -4 & \mbox{digit} & 1 \\
\frac{ 3 }{ 1 } & 3^2 - 8 \cdot 1^2 = 1 & \mbox{digit} & 4 \\
\end{array}
$$
Anyway, given a solution $(z,x)$ in positive integers to $z^2 - 8 x^2 = 1,$ we get the next in an infinite sequence by
$$ (z,x) \mapsto (3z + 8x, z + 3x),$$ so
$$ ( 1,0 ), $$
$$ (3,1), $$
$$ ( 17,6),$$
$$ (99 ,35 ), $$
$$ ( 577, 204 ), $$
$$ (3363 , 1189 ), $$
By Cayley -Hamilton, the coordinates $z_n, x_n$ obey
$$ z_{n+2} = 6 z_{n+1} - z_n, $$
$$ x_{n+2} = 6 x_{n+1} - x_n. $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2267316",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
If $\sin x + \sin^2 x =1$ then find the value of $\cos^8 x + 2\cos^6 x + \cos^4 x$ If $\sin x + \sin^2 x =1$ then find the value of $\cos^8 x + 2\cos^6 x + \cos^4 x$
My Attempt:
$$\sin x + \sin^2 x=1$$
$$\sin x = 1-\sin^2 x$$
$$\sin x = \cos^2 x$$
Now,
$$\cos^8 x + 2\cos^6 x + \cos^4 x$$
$$=\sin^4 x + 2\sin^3 x +\sin^2 x$$
$$=\sin^4 x + \sin^3 x + \sin^3 x + \sin^2 x$$
$$=\sin^3 x(\sin x +1) +\sin^2 x(\sin x +1)$$
$$=(\sin x +1) (\sin^3 x +\sin^2 x)$$
How do I proceed further?
| Here is a shorter way to do it. Your condition can be rewritten in the forms
$$\sin x = \cos^2(x) \implies \sin^n(x) = \cos^{2n}(x) \implies \sin^n(x)+\sin^{n-1}(x) = \sin^{n-2}(x)$$
We thus rewrite your cosines as
$$\begin{align}
\cos^8 x + 2\cos^6 x + \cos^4 x &= \sin^4(x)+2\sin^3(x)+\sin^2(x)\\ &= [\sin^4(x)+\sin^3(x)]+[\sin^3(x)+\sin^2(x)]\\ &= \sin^2(x)+\sin(x)\\&=1
\end{align}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2269298",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 1
} |
For any value of $n\ge5$, the value of $1+ \frac{1}{2}+ \frac{1}{3}+...+\frac{1}{2^n+1}$ lies between
For any value of $n\ge5$, the value of $$1+ \frac{1}{2}+ \frac{1}{3}+...+\frac{1}{2^n+1}$$ lies between
A) $0$ and $\frac{n}{2}$;
B) $\frac{n}{2}$ and $n$;
C) $n$ and $2n$;
D) none of them
I know this is Harmonic progression, but how exactly am I supposed to know the interval?
| Note that
$$\sum\limits_{k=1}^{2^n+1} \dfrac{1}{k} > \sum\limits_{k=2}^{2^n}\dfrac{1}{k} = \dfrac{1}{2} + \dfrac{1}{3} + \dfrac{1}{4} + \dfrac{1}{5} + \dots \dfrac{1}{2^n} > \dfrac{1}{2} + \dfrac{1}{4} + \dfrac{1}{4} + \dfrac{1}{8} + \dots + \dfrac{1}{2^n} = \sum\limits_{k=1}^n \dfrac{2^{k-1}}{2^k} = \dfrac{n}{2}$$
On the other hand,
$$\sum\limits_{k=1}^{2^n+1} \dfrac{1}{k} \le \sum\limits_{k=1}^{2^{n+1} - 1} \dfrac{1}{k} = 1 + \dfrac{1}{2} + \dfrac{1}{3} + \dfrac{1}{4} + \dfrac{1}{5} + \dots \dfrac{1}{2^{n+1}-1} < 1+ \dfrac{1}{2} + \dfrac{1}{2} + \dfrac{1}{4} + \dfrac{1}{4} + \dots + \dfrac{1}{2^n} = \sum\limits_{k=0}^{n} \dfrac{2^k}{2^k} = n + 1$$
You can decrease the upper bound to $n$ because $\dfrac{1}{2} - \dfrac{1}{3} + \dfrac{1}{4}-\dfrac{1}{5} + \dfrac{1}{4}-\dfrac{1}{6} \dots + \dfrac{1}{2^{n+1}-1}-\dfrac{1}{2^n} > 1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2272741",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
extrema of functions of two variables An experiment was conducted by a group of students to analyze the performance of a subject if stimulus A and stimulus B are used. It was found that if $x$ units of stimulus A and $y$ units of stimulus B were applied, the performance of the subject can be measured using the following equation:
$$
f(x,y) = C + xy e^{1-x^2 -y^2}
$$
where C is a positive constant. How many units of each stimuli yield the maximum performance?
| Since you haven't mentioned it, I am assuming that stimuli A and B both take values in ${\rm I\!R}$.
Let's compute the gradient of the function we want to maximize:
$$
\nabla f (x, y) = \begin{pmatrix}
\frac{\partial f}{\partial x} (x, y)
\\
\frac{\partial f}{\partial y} (x, y)
\end{pmatrix}
= \begin{pmatrix}
(1-2x^2)ye^{1-x^2-y^2}
\\
(1-2y^2)xe^{1-x^2-y^2}
\end{pmatrix}
$$
We need to find critical points, i.e. points where the gradient is equal to the zero vector. In this case:
$$
1-2x^2 = 0 \implies x^2 = \frac{1}{2} \implies x = \pm \frac{\sqrt{2}}{2}
$$
$$
1-2y^2 = 0 \implies y^2 = \frac{1}{2} \implies y = \pm \frac{\sqrt{2}}{2}
$$
So we have four critical points:
$(\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2})$,
$(-\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2})$,
$(\frac{\sqrt{2}}{2}, -\frac{\sqrt{2}}{2})$ and
$(-\frac{\sqrt{2}}{2}, -\frac{\sqrt{2}}{2})$. Out of these, we need to identify which are the maxima (if any), so we compute the Hessian matrix:
$$
H(f(x, y)) = \begin{pmatrix}
\frac{\partial^2 f}{\partial x ^2} (x, y)
&
\frac{\partial^2 f}{\partial x \partial y} (x, y)
\\
\frac{\partial^2 f}{\partial y \partial x} (x, y)
&
\frac{\partial^2 f}{\partial y ^2} (x, y)
\end{pmatrix}
= \begin{pmatrix}
(4x^2-6)xye^{1-x^2-y^2}
&
(4x^2y^2-2x^2-2y^2+1)e^{1-x^2-y^2}
\\
(4x^2y^2-2x^2-2y^2+1)e^{1-x^2-y^2}
&
(4y^2-6)xye^{1-x^2-y^2}
\end{pmatrix}
$$
Note that for $x = \pm \frac{\sqrt{2}}{2}$ and $y = \pm \frac{\sqrt{2}}{2}$:
$$
x^2 = y^2 = \frac{1}{2}
$$
$$
e^{1-x^2-y^2} = e^{1-\frac{1}{2}-\frac{1}{2}} = 1
$$
Therefore, only for working with these four points, the Hessian matrix can be reduced to the following:
$$
H(f(x, y)) = \begin{pmatrix}
-4xy & 0
\\
0 & -4xy
\end{pmatrix}
$$
and the determinant is:
$$
det(H(f(x, y))) = 16x^2y^2
$$
which means that all of these four points are extrema (not saddle points), and only those which satisfy $-4xy < 0$ are maxima. This last condition is satisfied only when $x$ and $y$ have the same sign.
Therefore, our function is maximized for $(x, y) = (-\frac{\sqrt{2}}{2}, -\frac{\sqrt{2}}{2})$ and $(x, y) = (\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2})$.
Plotting the function, we can see the two maxima:
Note that in this case, I plotted the function for $C = 1$, but the value of $C$ does not affect the result in any way as its effect can be seen as a simple displacement of the image of the function in the codomain.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2272947",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Calculating general $n\times n$ determinant I'm given determinant $\begin{vmatrix}
1 & 2 &3 & \cdots & n -1 & n \\
2 & 3 &4 & \cdots & n & 1 \\
3 & 4 &5 & \cdots & 1 & 2 \\
\vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\
n & 1 & 2 & \cdots & n-2 & n-1
\end{vmatrix}$ and I have to calculate it's value only using Laplace formula and properties of antilinear $n$-forms.
First, I subtract $i-1$-th column from $i$-th column, which gives me
$\begin{vmatrix}
1 & 1 &1 & \cdots & 1 & 1 \\
2 & 1 &1 & \cdots & 1 & 1-n \\
3 & 1 &1 & \cdots & 1 - n & 1 \\
\vdots & \vdots & \vdots & \ddots & \vdots & \vdots\\
n & 1-n & 1 & \cdots & 1 & 1
\end{vmatrix}$.
Then, subtracting $i-1$-th row from $i$-th row yields us
$\begin{vmatrix}
1 & 1 &1 & \cdots & 1 & 1 \\
1 & 0 &0 & \cdots & 0 & -n \\
1 & 0 &0 & \cdots & - n & 0 \\
\vdots & \vdots & \vdots & \ddots & \vdots & \vdots\\
1 & -n & 0 & \cdots & 0 & 0
\end{vmatrix}$.
Again, subtracting $i-1$-th column from $i$-th column
$\begin{vmatrix}
1 & 1 &0 & \cdots & 0 & 0 \\
1 & 0 &0 & \cdots & 0 & -n \\
1 & 0 &0 & \cdots & - n & n \\
\vdots & \vdots & \vdots & \ddots & \vdots & \vdots\\
1 & -n & n & \cdots & 0 & 0
\end{vmatrix}$.
Two last steps before using Laplace formula are adding $i-1$ to $i$-th row
$\begin{vmatrix}
1 & 1 &0 & \cdots & 0 & 0 \\
2 & 1 &0 & \cdots & 0 & -n \\
2 & 0 &0 & \cdots & - n & 0 \\
\vdots & \vdots & \vdots & \ddots & \vdots & \vdots\\
2 & -n & 0 & \cdots & 0 & 0
\end{vmatrix}$
and using properties of antilinear $n$-form flipping determinant
$(-1)^\frac{n(n-1)}{2}\begin{vmatrix}
0 & 0 &0 & \cdots & 1 & 1 \\
-n & 0 &0 & \cdots & 1 & 2 \\
0 & -n &0 & \cdots & 0 & 2 \\
\vdots & \vdots & \vdots & \ddots & \vdots & \vdots\\
0 & 0 & 0 & \cdots & -n & 2
\end{vmatrix}$.
Final step is using Laplace formula on 1st row :
$$(-1)^\frac{n(n-1)}{2}\begin{vmatrix}
0 & 0 &0 & \cdots & 1 & 1 \\
-n & 0 &0 & \cdots & 1 & 2 \\
0 & -n &0 & \cdots & 0 & 2 \\
\vdots & \vdots & \vdots & \ddots & \vdots & \vdots\\
0 & 0 & 0 & \cdots & -n & 2
\end{vmatrix} = (-1)^\frac{n(n-1)}{2}((-n)^{n-1} - 2(-n)^{n-2}) = (-1)^{\frac{n(n-1)^2}{2}}(n^{n-1} + 2n^{n - 2}).$$
But my textbook states that result should be $(-1)^{\frac{n(n-1)}{2}}\cdot\frac{1}{2}(n^n + n^{n-1})$. What I am doing wrong in my calculations?
| In your second step, when you subtract row $i-1$ from row $i$, there should be a diagonal of $n$'s under your diagonal of $-n$'s, because $1-(1-n)=n$. For example, the third entry in the last column of your third determinant should be $n$, not $0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2273869",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Find all natural numbers $n$ such that $3^n+81$ is a perfect square. Find all natural numbers $n$ such that $3^n+81$ is a perfect square.
My solution:
$3^n=a^2-81 \Rightarrow 3^n=(a-9)(a+9)$
$gcd(a-9,a+9) \mid 18$
because $a+9>a-9$ then $a-9$ can be $1,3,9$that only $a=9$ is the answer so we have only one solution.I just wanted to verify my answer is it right?
| You get $3^n = (a-9)(a+9)$ in the question. But the only factors of $3^n$ are other powers of $3$, so you need $a-9$ and $a+9$ to both be powers of $3$. The only powers of three differing by $18$ are $9$ and $27$. So $a = 18$, and the only solution is $18^2 = 3^5 + 81$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2276153",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Given that $x+\frac{1}{x}=\sqrt{3}$, find $x^{18}+x^{24}$ Given that $x+\frac{1}{x}=\sqrt{3}$, find $x^{18}+x^{24}$
Hints are appreciated. Thanks in advance.
| As $x\ne0,$ on simplification we find $$x^2+1=\sqrt3x$$
Squaring we get $$x^4+2x^2+1=3x^2\iff x^4-x^2+1=0$$
Now
$$ x^6+1=(x^2+1)(x^4-x^2+1)$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 2
} |
For integer m greather than 2, $\frac{1}{m} + \frac{1}{m+2}$, the numerators and denomitors are primitive pythagorean triples $a$ and $b$ For $m = 2$, the fraction is $\frac{3}{4}$. for $m=3$, the fraction is$\frac{8}{15}$. I was wondering why numerators and demoninators of $\frac{1}{m} + \frac{1}{m+2}$ show primitive pythagorean triples a and b.
| There is a simple form to proof:
$(m^2+2m)^2 + (2(m+1))^2 = ((m+1)^2 - 1)^2 + 4 (m+1)^2 = (m+1)^4 + 1 -2(m+1)^2 + 4(m+1)^2 = (m+1)^4 -2(m+1)^2 + 1 = ((m+1)^2 - 1)^2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2277117",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Proving Ramanujan's Nested Cube Root Ramanujan's Nested Cube:
If $\alpha,\beta$ and $\gamma$ are the roots of the cubic equation$$x^3-ax^2+bx-1=0\tag{1}$$then, they satisfy$$\alpha^{1/3}+\beta^{1/3}+\gamma^{1/3}=(a+6+3t)^{1/3}\tag{2.1}$$
$$(\alpha\beta)^{1/3}+(\beta\gamma)^{1/3}+(\gamma\alpha)^{1/3}=(b+6+3t)^{1/3}\tag{2.2}$$where$$t^3-3(a+b+3)t-(ab+6(a+b)+9)=0\tag3$$
The formula (2.1) is what Ramanujan used to get
$$\sqrt[3]{\cos\tfrac {2\pi}7}+\sqrt[3]{\cos\tfrac {4\pi}7}+\sqrt[3]{\cos\tfrac {8\pi}7}=\sqrt[3]{\tfrac 12\left(5-3\sqrt[3]7\right)}$$
by starting with $x^3+x^2-2x-1=0$ along with its trigonometric roots $\cos\frac {2\pi}7,\>\cos\frac {4\pi}7, \>\cos\frac {8\pi}7$ on LHS, and then getting RHS with $a=-1,b=-2$.
Question:
*
*How to prove the formulas (2.1) and (2.2)?
*Is there a standard procedure to find trigonometric roots of a polynomial?
I first started off with a function $x^3-px^2+qx-1=0$ and assumed that the roots were $\alpha^{1/3},\beta^{1/3},\gamma^{1/3}$. That way, by Vieta's formula, we have$$\alpha^{1/3}+\beta^{1/3}+\gamma^{1/3}=p$$$$(\alpha\beta)^{1/3}+(\beta\gamma)^{1/3}+(\gamma\alpha)^{1/3}=-q$$However, I'm not sure how to represent the RHS in $(2.1)$ or $(2.2)$
EDIT: I found a proof, but something doesn't match up. I have posted another question here
| Let $$\sqrt[3]{\cos\tfrac {2\pi}7}+\sqrt[3]{\cos\tfrac {4\pi}7}+\sqrt[3]{\cos\tfrac {8\pi}7}=x$$ and
$$\sqrt[3]{\cos\tfrac {2\pi}7\cos\tfrac {4\pi}7}+\sqrt[3]{\cos\tfrac {2\pi}7\cos\tfrac {8\pi}7}+\sqrt[3]{\cos\tfrac {4\pi}7\cos\tfrac {8\pi}7}=y.$$
Hence, since $$\cos\tfrac {2\pi}7+\cos\tfrac {4\pi}7+\cos\tfrac {8\pi}7=-\frac{1}{2},$$
$$\cos\tfrac {2\pi}7\cos\tfrac {4\pi}7+\cos\tfrac {2\pi}7\cos\tfrac {8\pi}7+\cos\tfrac {4\pi}7\cos\tfrac {8\pi}7=-\frac{1}{2}$$ and
$$\cos\tfrac {2\pi}7\cos\tfrac {4\pi}7\cos\tfrac {8\pi}7=\frac{1}{8},$$
we obtain:
$$x^3=\cos\tfrac {2\pi}7+\cos\tfrac {4\pi}7+\cos\tfrac {8\pi}7+3xy-3\sqrt[3]{\cos\tfrac {2\pi}7\cos\tfrac {4\pi}7\cos\tfrac {8\pi}7}=-\frac{1}{2}+3xy-\frac{3}{2}=3xy-2$$ and
$$y^3=\cos\tfrac {2\pi}7\cos\tfrac {4\pi}7+\cos\tfrac {2\pi}7\cos\tfrac {8\pi}7+\cos\tfrac {4\pi}7\cos\tfrac {8\pi}7+$$
$$+3xy\sqrt[3]{\cos\tfrac {2\pi}7\cos\tfrac {4\pi}7\cos\tfrac {8\pi}7}-3\sqrt[3]{\cos^2\tfrac {2\pi}7\cos^2\tfrac {4\pi}7\cos^2\tfrac {8\pi}7}=$$
$$=-\frac{1}{2}+\frac{3}{2}xy-\frac{3}{4}=\frac{3}{2}xy-\frac{5}{4}.$$
Thus,
$$x^3y^3=(3xy-2)\left(\frac{3}{2}xy-\frac{5}{4}\right)$$ or
$$4x^3y^3=18x^2y^2-27xy+10$$ or
$$8x^3y^3-36x^2y^2+54xy-20=0$$ or
$$(2xy-3)^3+7=0$$ or
$$xy=\frac{1}{2}\left(3-\sqrt[3]7\right),$$
which gives
$$x^3=\frac{3}{2}\left(3-\sqrt[3]7\right)-2$$ or
$$x=\sqrt[3]{\frac{1}{2}\left(5-3\sqrt[3]7\right)}$$
and we are done!
In the general we just need to solve a cubic equation of $xy$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2278338",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 2,
"answer_id": 1
} |
Binomial coefficients equality or maybe probability Let $m,n$ be positive integers.
Evaluate the following expression:
$$
F(m,n) = \sum\limits_{i=0}^n\frac{\binom{m+i}{i}}{2^{m+i+1}}+
\sum\limits_{i=0}^m\frac{\binom{n+i}{i}}{2^{n+i+1}}.
$$
Calclulations give the hypothesis that $$F(m,n)=1,$$ for all positive integers $m,n$.
Also if $m=n$, then
$$
F(m,m) = \sum\limits_{i=0}^m\frac{\binom{m+i}{i}}{2^{m+i}} = \sum\limits_{i=0}^m\frac{\binom{m+i}{m}}{2^{m+i}}.
$$
The numerator of every summand is equal to the number of $m$-subsets in $m+i$-set and denominator is equal to the number of subsets in $m+i$-set. So, I think it maybe the key to solution.
| Here is an answer in two steps based upon generating functions
First step: The following identity holds true for $m,n\geq 0$
\begin{align*}
\sum_{i=0}^n\binom{m+i}{i}\frac{1}{2^{m+i+1}}=\frac{1}{2^{m+n+1}}\sum_{i=0}^n\binom{m+n+1}{i}\tag{1}
\end{align*}
In order to show (1) it is convenient to use the coefficient of operator $[z^i]$ to denote the coefficient of $z^i$. This way we can write e.g.
\begin{align*}
[z^i](1+z)^n=\binom{n}{i}
\end{align*}
We obtain
\begin{align*}
\color{blue}{\sum_{i=0}^{n}\binom{m+i}{i}\frac{1}{2^{m+i+1}}}
&=\frac{1}{2^{m+n+1}}\sum_{i=0}^n\binom{m+n-i}{n-i}2^i\tag{2}\\
&=\frac{1}{2^{m+n+1}}\sum_{i=0}^\infty[z^{n-i}](1+z)^{m+n-i}2^i\tag{3}\\
&=\frac{1}{2^{m+n+1}}[z^n](1+z)^{m+n}\sum_{i=0}^\infty\left(\frac{2z}{1+z}\right)^i\tag{4}\\
&=\frac{1}{2^{m+n+1}}[z^n](1+z)^{m+n}\frac{1}{1-\frac{2z}{1+z}}\tag{5}\\
&=\frac{1}{2^{m+n+1}}[z^n]\frac{(1+z)^{m+n+1}}{1-z}\\
&=\frac{1}{2^{m+n+1}}\sum_{i=0}^n\binom{m+n+1}{i}[z^{n-i}]\frac{1}{1-z}\tag{6}\\
&\color{blue}{=\frac{1}{2^{m+n+1}}\sum_{i=0}^n\binom{m+n+1}{i}}\tag{7}
\end{align*}
and (1) follows.
Comment:
*
*In (2) we exchange the order of summation by letting $i\rightarrow n-i$.
*In (3) we apply the coefficient of operator and set the upper limit of the sum to infty without changing anything since we are adding zeros only.
*In (4) we use the linearity of the coefficient of operator and apply the rule $$[z^{p-q}]A(z)=[z^p]z^qA(z)$$
*In (5) we use the geometric series expansion.
*In (6) we select the coefficient of $[z^{n-i}]$ in $(1+z)^{m+n+1}$ and restrict the upper limit of the sum to $n$ since the exponent of $z^{n-i}$ is non-negative.
*In (7) we note the coefficient of $[z^{n-i}]$ in $\frac{1}{1-z}=1+z+z^2+\cdots$ is $1$.
Second step: $F(m,n)$
Now it's time to harvest. We obtain from (7)
\begin{align*}
\color{blue}{\frac{1}{2^{m+n+1}}\sum_{i=0}^m\binom{m+n+1}{i}}
&=1-\frac{1}{2^{m+n+1}}\sum_{i=m+1}^{m+n+1}\binom{m+n+1}{i}\tag{8}\\
&=1-\frac{1}{2^{m+n+1}}\sum_{i=0}^{n}\binom{m+n+1}{n-i}\tag{9}\\
&\color{blue}{=1-\frac{1}{2^{m+n+1}}\sum_{i=0}^{n}\binom{m+n+1}{i}}\tag{10}\\
\end{align*}
Comment:
*
*In (8) we use $\sum_{i=0}^{p}\binom{p}{i}=2^p$.
*In (9) we shift the index to start with $i=0$ and we use the binomial identity $\binom{p}{q}=\binom{p}{p-q}$.
*In (10) we change the order of summation by letting $i\rightarrow n-i$.
We conclude from (7) and (10)
\begin{align*}
\color{blue}{F(m,n)}&=\sum_{i=0}^{n}\binom{m+i}{i}\frac{1}{2^{m+i+1}}
+\sum_{i=0}^{m}\binom{n+i}{i}\frac{1}{2^{n+i+1}}\\
&=\frac{1}{2^{m+n+1}}\sum_{i=0}^n\binom{m+n+1}{i}
+\frac{1}{2^{m+n+1}}\sum_{i=0}^m\binom{m+n+1}{i}\\
&=\frac{1}{2^{m+n+1}}\sum_{i=0}^n\binom{m+n+1}{i}+\left(1-\frac{1}{2^{m+n+1}}\sum_{i=0}^{n}\binom{m+n+1}{i}\right)\\
&\color{blue}{=1}
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2278964",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
How to calculate the volume of an ellipsoid with triple integral I'm having some troubles since this morning on an exercise.
I need to find the volume of the ellipsoid defined by $\frac{x^2}{a^2} + \frac{y^2}{a^2} + \frac{z^2}{a^2} \leq 1$.
So at the beginning I wrote
$\left\{\begin{matrix} -a\leq x\leq a
\\ -b\leq y\leq b
\\ -c\leq z\leq c
\end{matrix}\right.$
Then I wrote this as integral : $\int_{-c}^{c}\int_{-b}^{b}\int_{-a}^{a}1 dxdydz $.
I found as a result : $8abc$
But I knew it was incorrect. I browsed this website and many others for 2 hours and yeah I found that we need to express b and c in terms of x. But I don't know why.. Is there an intuitive way to understand that ? I have already done some double integral with b expressed in terms of a but I never figured out why..
| If you are sticking with cartesian coordinates, then you will consider the points inside of the ellipsoid. The limits you wrote in your OP consider all points within $\pm a$,$\pm b$,$\pm c$ as @u8y7451 mentioned.
First, let's find our range of z.
$\frac{x^2}{a^2} + \frac{y^2}{b^2} + \frac{z^2}{c^2} \le 1$
$\frac{z^2}{c^2} \le 1- \frac{x^2}{a^2} - \frac{y^2}{b^2} $
$-c\sqrt{(1- \frac{x^2}{a^2} - \frac{y^2}{b^2})} \le z \le c\sqrt{(1- \frac{x^2}{a^2} - \frac{y^2}{b^2})}$
Next, we look at the projection of the ellipsoid onto the x-y plane.
$\frac{x^2}{a^2} + \frac{y^2}{b^2} \le 1$
This gives our value of y as a function of x.
$\frac{y^2}{b^2} \le 1 - \frac{x^2}{a^2}$
$-b\sqrt{(1 - \frac{x^2}{a^2})} \le y \le b\sqrt{(1 - \frac{x^2}{a^2})}$
$-a \le x \le a$
This should give you the proper ranges for your integrals as:
$\int\limits_{-a}^{a} \int_{-b\sqrt{1-x^2/a^2}}^{b\sqrt{1-x^2/a^2}}\int_{-c\sqrt{1-x^2/a^2 -y^2/b^2}}^{c\sqrt{1-x^2/a^2 -y^2/b^2}}1 dzdydx$
Apologies for the ugliness of that final formula.
EDIT: some further explanation.
Suppose you would like to calculate the area inside of an ellipse centered at the origin, with vertices at $(\pm a,0)$ and $(0,\pm b)$ (see image). To do so, you need to decide what values of $(x,y)$ are inside. Clearly, the limits along the $x$-axis are $[-a,+a]$. So you could start at $x=-a$ and move to the right. Now, at each $x$, what values of $y$ are inside the ellipse? From the formula for an ellipse:
$\frac{x^2}{a^2} + \frac{y^2}{b^2} \le 1 \implies $
$-b\sqrt{(1 - \frac{x^2}{a^2})} \le y \le b\sqrt{(1 - \frac{x^2}{a^2})}$
Now, we could stop here, and calculate the area with a single integral:
$2b\int\limits_{-a}^{a}\sqrt{(1 - \frac{x^2}{a^2})}dx$. [eqn 1]
Effectively we are integrating the height of the ellipse by the width $dx$ (see Fig 1A). This is what most students are more familiar with.
Alternatively, we can go one more step, and instead of integrating columns, we integrate over tiny areas of width $dx$ and height $dy$, centered around each point within the ellipse (this is a little hand-wavy but hopefully Fig 1B makes it clearer). So as before, we start at $x=-a$ moving to the right, and for each x we start at the lower limit of $y$ moving up.
$\int\limits_{-a}^{a} \int_{-b\sqrt{1-x^2/a^2}}^{b\sqrt{1-x^2/a^2}}dydx$ [eqn 2]
The original problem just takes this one step further into the z-dimension.. Hope this helps!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2279468",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
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