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Find the sum of series $\sum_{n=2}^\infty\frac{1}{n(n+1)^2(n+2)}$ How to find the sum of series $\sum_{n=2}^\infty\frac{1}{n(n+1)^2(n+2)}$ in the formal way? Numerically its value is $\approx 0.0217326$ and the partial sum formula contains the first derivative of the gamma function (by WolframAlpha).
| There is an alternate method and is as follows.
Notice that
$$ \frac{1}{n(n+1)(n+2)} = \frac{(n-1)!}{(n+2)!} = \frac{1}{2!} \, B(n,3) $$
where $B(x,y)$ is the Beta function. Using an integral form of the Beta function the summation becomes
\begin{align}
S &= \sum_{n=2}^{\infty} \frac{1}{n \, (n+1)^{2} \, (n+2)} \\
&= \frac{1}{2} \, \int_{0}^{1} \left( \sum_{n=2}^{\infty} \frac{x^{n-1}}{n+1} \right) \, (1-x)^{2} \, dx \\
&= - \frac{1}{2} \, \int_{0}^{1} \left(x + \frac{x^{2}}{2} + \ln(1-x) \right) \, \frac{(1-x)^{2}}{x^{2}} \, dx \\
&= - \frac{1}{4} \, \int_{0}^{1} (2+x) (1-x)^{2} \, \frac{dx}{x} - \frac{1}{2} \, \int_{0}^{1} \left( 1 - \frac{1}{x} \right)^{2} \, \ln(1-x) \, dx \\
&= - \frac{1}{4} \left[ \frac{x^{3}}{3} - 3 x + 2 \ln(x) \right]_{0}^{1} - \frac{1}{2} \left[ (x-1) \, \ln(1-x) - x \right]_{0}^{1} + \left[ - Li_{2}(x) \right]_{0}^{1} \\
& \hspace{20mm} - \frac{1}{2} \left[ \frac{(x-1) \ln(1-x) - x \ln(x)}{x} \right]_{0}^{1} \\
&= \frac{2}{3} + \frac{1}{2} - Li_{2}(1) - \frac{1}{2} \, \lim_{x \to 0} \left\{ - \ln(x) - \frac{\ln(1-x)}{x} + \ln(x) \right\} \\
&= \frac{5}{3} - \zeta(2).
\end{align}
This leads to the known result
\begin{align}
\sum_{n=2}^{\infty} \frac{1}{n \, (n+1)^{2} \, (n+2)} = \frac{5}{3} - \frac{\pi^{2}}{6}.
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1378465",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Greatest of the numbers given To find out the greatest among the number given below:
$3^{1/3}, 2^{1/2}, 6^{1/6}, 1, 7^{1/7}$
I have plotted the following graph using graph plotter which is shown below:
It can be concluded that $3^{1/3}$ is the greatest.
I want to know that is there any other method to find greatest among the such numbers.
| Take lcm$(3,2,6,7)=42$
We need to check for $3^{1/3},2^{1/2},6^{1/6},7^{1/7}$
equivalently taking $42$nd power of each $3^{14},2^{21},6^7,7^6$
Now $2^3<3^2\iff(2^3)^7<(3^2)^7$
Again, $3^{14}-6^7=3^7(3^7-2^7)>0$
and finally $3^7>3^6=729>343=7^3\implies(3^7)^2>(7^3)^2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1381115",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 0
} |
Badly formed question? $\|x\|=1=\|y\|$ and $\|x+y\|=\|x\|+\|y\|$, there is a line segment in the unit sphere Show that if a normed linear space $X$ contains linearly independent vectors $x$ and $y$ such that $\|x\|=1$ and $\|y\|=1$ with $\|x+y\|=\|x\|+\|y\|$, then there is a line segment contained in the unit sphere of $X$.
Now since the unit sphere in $X$ is defined as $X=\{a\in X: \|a\|=1\}$, and this means that $x,y$ both lie on the unit sphere of $X$. But $\|x+y\|=2$ and hence they must lie on opposite ends of the sphere, and hence form a straight line through the origin. But then these two vectors aren't linearly independent?
Is this question badly formed or what am I missing?
| HINT:
You have $\|\frac{1}{2}x + \frac{1}{2}y\| = 1$. Let's show for instance that
$\|\frac{1}{3}x + \frac{2}{3}y\| = 1$. Note that
$$\left\|\frac{1}{3}x + \frac{2}{3}y\right\| \le \frac{1}{3}\|x\| + \frac{2}{3}\|y\|=1$$
If we had $\|\frac{1}{3}x + \frac{2}{3}y\| <1$ then
$$\left\|\frac{1}{2}x + \frac{1}{2}y\,\right\| = \left\|\frac{1}{4} x + \frac{3}{4}\left( \frac{1}{3}x + \frac{2}{3}y\right)\, \right\| \le \frac{1}{4}\left\|x\right\| + \frac{3}{4}\cdot \left\|\frac{1}{3}x + \frac{2}{3}y\,\right\| < 1$$
contradiction.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1381671",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Area of a triangle with sides $\sqrt{x^2+y^2}$,$\sqrt{y^2+z^2}$,$\sqrt{z^2+x^2}$ Sides of a triangle ABC are $\sqrt{x^2+y^2}$,$\sqrt{y^2+z^2}$ and $\sqrt{z^2+x^2}$ where x,y,z are non-zero real numbers,then area of triangle ABC is
(A)$\frac{1}{2}\sqrt{x^2y^2+y^2z^2+z^2x^2}$
(B)$\frac{1}{2}(x^2+y^2+z^2)$
(C)$\frac{1}{2}(xy+yz+zx)$
(D)$\frac{1}{2}(x+y+z)\sqrt{x^2+y^2+z^2}$
I tried applying Heron's formula but calculations are very messy and simplification is difficult.I could not think of any other method to find this area.Can someone assist me in solving this problem.
| For such form of the side lengths, the most convenient
would be a variation of the
Heron's formula
for the area:
\begin{align}
S&=\tfrac14\sqrt{4a^2b^2-(a^2+b^2-c^2)^2}
\\
S&=\tfrac14\sqrt{
4(x^2+y^2)(y^2+z^2)-
(x^2+y^2+y^2+z^2-z^2-x^2)^2
}
\\
&=\tfrac14\sqrt{4(x^2+y^2)(y^2+z^2)-4y^4}
\\
&=\tfrac12\sqrt{x^2 y^2+y^2 z^2+z^2 x^2}.
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1381791",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 3
} |
Two numbers that cannot both be squares I was wondering where to start with the following question:
Show for $a,b \in \mathbb{N}$ that $a+b^2$ and $a^2+b$ cannot be both squares.
Here $\mathbb{N}$ is the positive integers ($0$ not included).
| Consider $a < b$:
Clearly, $b^2 < a+b^2$. Further, we see that
$$
a+b^2 < b + b^2 = b(b+1) < (b+1)^2
$$
Hence, $b^2 < a+b^2 < (b+1)^2$. Thus, $a+b^2$ is not a square. On the other hand, $a^2+b$ may be a square, depending on the choice of $a$ and $b$.
If we have $a=b$, then neither $a^2+b$ nor $a+b^2$ are squares, as they would both simplify to $a(a+1)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1384126",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
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"answer_id": 0
} |
Find $\lim\limits_{x\to 0}\frac{\sqrt{2(2-x)}(1-\sqrt{1-x^2})}{\sqrt{1-x}(2-\sqrt{4-x^2})}$ I use L'Hospitals rule, but can't get the correct limit.
Derivative of numerator in function is
$$\frac{-3x^2+4x-\sqrt{1-x^2}+1}{\sqrt{(4-2x)(1-x^2)}}$$
and derivative of denominator is
$$\frac{-3x^2+2x-2\sqrt{4-x^2}+4}{2\sqrt{(1-x)(4-x^2)}}$$
Now, L'Hospitals rule must be applied again. Is there some easier way to compute the limit?
Limit should be $L=4$
| First note that the factors $\sqrt {2 (2-x)}$ and $\sqrt {1-x}$ tend to $2$ and $1$ respectively, therefore they don't raise any problem. It remains to compute the limit of $\frac {1 - \sqrt {1-x^2}} {2 - \sqrt {4-x^2}}$. Note that this fraction can be rewritten as $\frac {1 - (1-x^2)} {1 + \sqrt {1-x^2}} \frac {2 + \sqrt {4 - x^2}} {4 - (4-x^2)} = \frac {2 + \sqrt {4 - x^2}} {1 + \sqrt {1-x^2}}$ which clearly tends to $2$. Putting all the pieces together, the limit is $4$.
(The core fact was that $a-b = \frac {a^2 - b^2} {a+b}$.)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1385320",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 3
} |
The range of the function $f:\mathbb{R}\to \mathbb{R}$ given by $f(x)=\frac{3+2 \sin x}{\sqrt{1+ \cos x}+\sqrt{1- \cos x}}$ The range of the function $f:\mathbb{R}\to \mathbb{R}$ given by $f(x)=\frac{3+2 \sin x}{\sqrt{1+ \cos x}+\sqrt{1- \cos x}}$ contains $N$ integers. Find the value of $10N$.
I tried to find the minimum and maximum value of the function.First i simplified the function.
$f(x)=\frac{3+2 \sin x}{\sqrt{1+ \cos x}+\sqrt{1- \cos x}}=\frac{1+4\sin^2\left(\frac{x}{2}+\frac{\pi}{4}\right)}{2\sin \left(\frac{x}{2}+\frac{\pi}{4}\right)}$
Then i differentiated the function and equate it to zero to get the critical points.
Critical point equations are $\cos\left(\frac{x}{2}+\frac{\pi}{4}\right)=0$
$\sin\left(\frac{x}{2}+\frac{\pi}{4}\right)=\frac{1}{2},\sin\left(\frac{x}{2}+\frac{\pi}{4}\right)=\frac{-1}{2}$
When i checked plotted the function on desmos.com graphing calculator,i found minimum value to be $0.5$ and maximum value to be $2.5$.
Which i cannot get by my critical points.Where have i gone wrong?Please help me.
| Your simplified equation is correct only for $0 \leq x \leq \pi$. During your derivation be sure to consider both positive and negative square roots in the denominator. The result is an alternate version of your simplified equation
$f(x)=\frac{3+2 \sin x}{\sqrt{1+ \cos x}+\sqrt{1- \cos x}}=\frac{5-4\sin^2\left(\frac{x}{2}-\frac{\pi}{4}\right)}{2\sin \left(\frac{x}{2}-\frac{\pi}{4}\right)}$
valid for $\pi < x < 2\pi$. When you work this through you will get another critical point at $\cos(x/2 - \pi/4)=0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1386474",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
find a horizontal speed $\frac{dx}{dt}$ A girl slides down a slide in the shape of the parabola $y=(x−1)^2$ for $0≤x≤1$. Her vertical speed is $\frac{dy}{dt}=−y(1−y)$. Find her horizontal speed $\frac{dx}{dt}$ when $y=\frac 12$.
I have found that $$\frac{dy}{dt}=2(x+1)\frac{dx}{dt}$$, since $\frac{dy}{dt}=−y(1−y)$ and $y=\frac 12$ I got $$\frac{dx}{dt}=\frac {-\frac 12(1-\frac 12)}{2(x+1)}$$, but what to do with x now?
| Use the chain rule:
$$
\frac { dx }{ dt } \quad =\quad \frac { dx }{ dy } \frac { dy }{ dt } \\ \qquad \quad =\quad \frac { 1 }{ \frac { dy }{ dx } } y(y-1)\\ \qquad \quad =\quad \frac { y(y-1) }{ 2(x-1) } \\ \frac { dx }{ dt } \quad =\quad \frac { y(y-1) }{ -2\sqrt { y } }
$$
And now all you have to do is substitute $y = \frac{1}{2}$ to get:
$$
\\ \frac { dx }{ dt } \quad =\quad \frac { \frac { 1 }{ 2 } (\frac { 1 }{ 2 } -1) }{ -2\sqrt { \frac { 1 }{ 2 } } } \quad =\quad \frac { 1 }{ 4\sqrt { 2 } }
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1388215",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Solve this trigonometric equation $\cos3x\cos x=\sin 3x$ Solve this equation $$ \cos 3x\cos x=\sin 3x$$
I tried converting product into sum but with no results. I think they forgot to add $\sin x$ after $\sin 3x$.
| We have$$\begin{align}\\&\cos 3x\cos x=\sin 3x\\&\iff (4\cos^3x-3\cos x)\cos x=3\sin x-4\sin^3x\\&\iff 4\cos^4x-3\cos^2x=3\sin x-4\sin^3x\\&\iff 4(1-\sin^2x)^2-3(1-\sin^2x)=3\sin x-4\sin^3x\end{align}$$
Here, let $t=\sin x$.
$$\begin{align}\\& 4(1-t^2)^2-3(1-t^2)=3t-4t^3\\&\iff 4(1-2t^2+t^4)-3+3t^2-3t+4t^3=0\\&\iff 4-8t^2+4t^4-3+3t^2-3t+4t^3=0\\&\iff 4t^4+4t^3-5t^2-3t+1=0\\&\iff 16t^4+16t^3-20t^2-12t+4=0\\&\iff (4t^2+2t-3)^2-5=0\\&\iff (4t^2+2t-3-\sqrt 5)(4t^2+2t-3+\sqrt 5)=0\\&\iff \sin x=t=\frac{-1\pm\sqrt{13+4\sqrt 5}}{4},\frac{-1\pm\sqrt{13-4\sqrt 5}}{4}\end{align}$$
Here, note that
$$\frac{-1-\sqrt{13+4\sqrt 5}}{4}\lt -1.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1390616",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Why does the elliptic curve for $a+b+c = abc = 6$ involve a solvable nonic? The curve discussed in this OP's post,
$$\color{brown}{-24a+36a^2-12a^3+a^4}=z^2\tag1$$
is birationally equivalent to an elliptic curve. Following E. Delanoy's post, let $G$ be the set of rational numbers that solve $(1)$. Courtesy of Aretino's comment, it is known that $G$ is invariant by the transformation,
$$f(a)=\frac{-54a(-6+12a-6a^2+a^3)^2}
{−216+1296a^2−2160a^3+1296a^4−108a^5−234a^6+108a^7−18a^8+a^9}$$
thus both $a$ and $f(a)$ are solutions to $(1)$. Equivalently,
$$f(a)= \frac{54a^2(-6+12a-6a^2+a^3)^2}{9a^2(-6+9a-6a^2+a^3)^2-(3-3a^2+a^3)^2\color{brown}{(-24a+36a^2-12a^3+a^4)}}$$
Q: Let $P(a)$ be the irreducible nonic denominator. Why does it have a solvable Galois group?
In other words, the eqn $P(a) = 0$ is solvable in radicals. There is an online Magma calculator that computes the Galois group and the command is,
Z := Integers(); P < x > := PolynomialRing(Z); f := -216 + 1296*x^2 - 2160*x^3 + 1296*x^4 - 108*x^5 - 234*x^6 + 108*x^7 - 18*x^8 + x^9; G, R := GaloisGroup(f); G;
It says that the order is $54$ and hence is solvable.
P.S. This is the second time within a short period that I've come across a polynomial identity that surprisingly has a solvable Galois group. (The explicit identity is given in the first link above.) See also the recent post, Why does $x^2+47y^2 = z^5$ involve solvable quintics?
| Тhat is a beautiful solvable nonic! Here is its smallest real root:
With[{α = Sin[ArcSin[(1 + 3 × 3^(2/3))/8] / 3],
β = 3 + 9 × 3^(1/3) + 7 × 3^(2/3),
γ = 18 + 9 × 3^(1/3) + 2 × 3^(2/3)},
With[{
ρ = -51 - 33 × 3^(1/3) + 121 × 3^(2/3) - 8 α β + 4 α^2 (β - 60 × 3^(2/3)),
σ = 12 - 9 × 3^(1/3) - 7 × 3^(2/3) + α γ + 2 α^2 β,
τ = 53 + 24 × 3^(1/3) + 27 × 3^(2/3) - 8 α (11 + 3 × 3^(1/3) - 3^(2/3)) -
16 α^2 (1 + 3 × 3^(1/3) + 4 × 3^(2/3))},
(90 (45 - 4 β) + 180 α (2 γ - 15) + 720 α^2 β -
5^(2/3) (135 × 6^(1/3) ρ^(1/3) + 36 σ τ^(1/3))) / 2025]]
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1391806",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
"answer_id": 1
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LU decomposition for cyclic tridiagonal matrices It is known that a tridiagonal matrix
$$
A = \begin{pmatrix}
b_1 & c_1 & 0 & 0 & \dots & 0\\
a_2 & b_2 & c_2 & 0 & \dots & 0\\
0 & a_3 & b_3 & c_3 & \dots & 0\\
0 & 0 & \ddots &\ddots &\ddots & 0\\
0 & \dots & 0 & a_{n-1} & b_{n-1} & c_{n-1}\\
0 & \dots & 0 & 0 & a_{n} & b_{n}\\
\end{pmatrix}
$$
has a very simple LU decomposition, that is
$$
A = \begin{pmatrix}
1 & 0 & 0 & 0 & \dots & 0\\
l_2 & 1 & 0 & 0 & \dots & 0\\
0 & l_3 & 1 & 0 & \dots & 0\\
0 & 0 & \ddots &\ddots &\ddots & 0\\
0 & \dots & 0 & l_{n-1} & 1 & 0\\
0 & \dots & 0 & 0 & l_{n} & 1\\
\end{pmatrix}
\begin{pmatrix}
u_{1} & v_{1} & 0 & 0 & \dots & 0\\
0 & u_{2} & v_{2} & 0 & \dots & 0\\
0 & 0 & u_{3} & v_{3} & \dots & 0\\
0 & 0 & \ddots &\ddots &\ddots & 0\\
0 & \dots & 0 & 0 & u_{n-1} & v_{n-1}\\
0 & \dots & 0 & 0 & 0 & u_{n}\\
\end{pmatrix}
$$
which can be computed easily in $O(n)$ operations via regular LU decomposition algorithm.
I'm interested in the LU decomposition of a cyclic tridiagonal matrix, that is
$$
A' = \begin{pmatrix}
b_1 & c_1 & 0 & 0 & \dots & a_1\\
a_2 & b_2 & c_2 & 0 & \dots & 0\\
0 & a_3 & b_3 & c_3 & \dots & 0\\
0 & 0 & \ddots &\ddots &\ddots & 0\\
0 & \dots & 0 & a_{n-1} & b_{n-1} & c_{n-1}\\
c_n & \dots & 0 & 0 & a_{n} & b_{n}\\
\end{pmatrix}
$$
with $a_1$ and $c_n$ wrapped. I know that there are efficient $O(n)$ algorithms to solve a system with that kind of matrix. But I'm interested in a LU-like decomposition. I found that this matrix can be represented as
$$
A' = \begin{pmatrix}
1 & 0 & 0 & 0 & \dots & l_1\\
l_2 & 1 & 0 & 0 & \dots & 0\\
0 & l_3 & 1 & 0 & \dots & 0\\
0 & 0 & \ddots &\ddots &\ddots & 0\\
0 & \dots & 0 & l_{n-1} & 1 & 0\\
0 & \dots & 0 & 0 & l_{n} & 1\\
\end{pmatrix}
\begin{pmatrix}
u_{1} & v_{1} & 0 & 0 & \dots & 0\\
0 & u_{2} & v_{2} & 0 & \dots & 0\\
0 & 0 & u_{3} & v_{3} & \dots & 0\\
0 & 0 & \ddots &\ddots &\ddots & 0\\
0 & \dots & 0 & 0 & u_{n-1} & v_{n-1}\\
v_{n} & \dots & 0 & 0 & 0 & u_{n}\\
\end{pmatrix}.
$$
If I denote $A' = A + a_1 Z + c_n Z^\top$ then
$$
(L + l_1 Z)(U + v_{n} Z^\top) =
\underbrace{LU + l_1 v_n ZZ^\top}_{\text{tridiagonal}} + l_1 u_n Z + v_n Z^\top
$$
Thus
$$
v_n = c_n\\
l_1 u_n = a_1\\
LU = A - l_1 v_n ZZ^\top = A - \frac{a_1 c_n}{u_n} ZZ^\top.
$$
But I would like to find an effective algorithm to perform such factorization. I tried performing symbolical decomposition of $A - \alpha ZZ^\top$ with $\alpha$ being unknown, but that uses $\Omega(n^2)$ memory, thus is not an $O(n)$ algorithm.
| Actually, I was mistaken. There is a way to perform a symbolical decomposition of $B = A - \frac{a_1 c_n}{u_n} ZZ^\top$ with $u_n$ being a variable effectively in $O(n)$ operations.
One can note, that the first diagonal element of the matrix $B$ that is $b_1 - \frac{a_1 c_n}{u_n}$ has the following form
$$\frac{b_1 u_n - a_1 c_n}{u_n} = p_1\frac{u_n - q_2}{u_n - q_1}$$
with $p_1 = b_1, q_1 = 0, q_2 = \frac{a_1 c_n}{b_1}$.
The elements of $L,U$ are rational functions of $u_n$.
Considering rank-1 update for the LU decomposition procedure
$$
\begin{pmatrix}
p_i\frac{u_n - q_{i+1}}{u_n - q_i} & c_i & \\
a_{i+1} & b_{i+1} & \ddots \\
& \ddots & \ddots
\end{pmatrix}
= \begin{pmatrix}
1 & 0 & \\
l_{i+1} & L_{i+1} & \ddots \\
& \ddots & \ddots
\end{pmatrix}
\begin{pmatrix}
p_i \frac{u_n - q_{i+1}}{u_n - q_i} & v_i & \\
0 & U_{i+1} & \ddots \\
& \ddots & \ddots
\end{pmatrix}
$$
one can obtain a recurrent formula to compute $p_i, q_i$:
$$
v_i = c_i\\
l_{i+1} = \frac{a_{i+1}}{p_i}\frac{u_n - q_i}{u_n - q_{i+1}}\\
u_{i+1} = p_{i+1} \frac{u_n - q_{i+2}}{u_n - q_{i+1}} = b_{i+1} - l_{i+1} v_i
= b_{i+1} - \frac{u_n - q_i}{u_n - q_{i+1}} \frac{a_{i+1} c_i}{p_i} =\\=\frac{u_n - q_{i+1}}{u_n - q_{i+1}}b_{i+1} - \frac{u_n - q_i}{u_n -q_{i+1}} \frac{a_{i+1} c_i}{p_i} =
\frac{\left(b_{i+1} - \frac{a_{i+1} c_i}{p_i}\right)u_n - \left(
q_{i+1}b_{i+1} - q_i\frac{a_{i+1} c_i}{p_i}
\right)}{u_n - q_{i+1}}\\
p_{i+1} = b_{i+1} - \frac{a_{i+1} c_i}{p_i}\\
q_{i+2} = b_{i+1}\frac{q_{i+1}}{p_{i+1}} - \frac{a_{i+1} c_i}{p_i} \frac{q_i}{p_{i+1}}
$$
Finally $u_n$ can be determined from
$$
u_n = p_n \frac{u_n - q_{n+1}}{u_n - q_{n}}\\
u_n^2 - \left(
q_n + p_n
\right) u_n + p_n q_{n+1} = 0\\
u_n = \frac{(q_n + p_n) \pm \sqrt{(q_n + p_n)^2 - 4 p_n q_{n+1}}}{2}.
$$
It seems that this type of decomposition is not unique, since the last equation has two roots. I'm curious which one of them gives a numerically stable decomposition and whether this algorithm is stable at all.
Applied to sample matrix
$$
A = \begin{pmatrix}
3 & 1 & & & & 1 \\
1 & 1 & 1 & & & \\
& 1 & 0 & 1 & & \\
& & 1 & 1 & 1 & \\
& & & 1 & 3 & 1 \\
1 & & & & 1 & 4
\end{pmatrix}
$$
this method yields two decompositions
$$
A = \begin{pmatrix}
1 & & & & & \frac{1}{3}
\left(4+\sqrt{10}\right) \\
1+\sqrt{\frac{2}{5}} & 1 &&&&\\
& -\sqrt{\frac{5}{2}} & 1 &&&\\
&& \sqrt{\frac{2}{5}} & 1 &&\\
&&& \frac{1}{3}\left(5+\sqrt{10}\right) & 1 &\\
&&&&2+\sqrt{\frac{5}{2}} & 1
\end{pmatrix} \cdot \qquad \\ \qquad \cdot
\begin{pmatrix}
\frac{1}{3}
\left(5-\sqrt{10}\right) & 1 &
&&& \\
& -\sqrt{\frac{2}{5}} & 1 &&
&\\
& & \sqrt{\frac{5}{2}} & 1 &
& \\
& & & 1-\sqrt{\frac{2}{5}} &
1 & \\
& & & & \frac{1}{3}
\left(4-\sqrt{10}\right) & 1 \\
1 & & & & &
2-\sqrt{\frac{5}{2}}
\end{pmatrix} = \\ =
\begin{pmatrix}
1 & & & & &
\frac{1}{2+\sqrt{\frac{5}{2}}}
\\
1-\sqrt{\frac{2}{5}} & 1 & & &
& \\
& \sqrt{\frac{5}{2}} & 1 & &
& \\
& & -\sqrt{\frac{2}{5}} & 1 &
& \\
& & & \frac{1}{3}
\left(5-\sqrt{10}\right) & 1 & \\
& & & &
2-\sqrt{\frac{5}{2}} & 1
\end{pmatrix} \cdot \qquad \\ \qquad \cdot
\begin{pmatrix}
\frac{1}{3}
\left(5+\sqrt{10}\right) & 1 &
& & & \\
& \sqrt{\frac{2}{5}} & 1 & &
& \\
& & -\sqrt{\frac{5}{2}} & 1 & & \\
& & & 1+\sqrt{\frac{2}{5}} &
1 & \\
& & & & \frac{1}{3}
\left(4+\sqrt{10}\right) & 1 \\
1 & & & & &
2+\sqrt{\frac{5}{2}}
\end{pmatrix}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1391970",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Evaluate the limit $\lim_{x \to 0} \left(\frac{1}{x^{2}}-\frac{1}{\tan^{2}x}\right)$
Evaluate the limit $$\lim_{x \to 0}\left( \frac{1}{x^{2}}-\frac{1}{\tan^{2}x}\right)$$
My attempt
So we have $$\frac{1}{x^{2}}-\frac{\cos^{2}x}{\sin^{2}x}$$
$$=\frac{\sin^2 x-x^2\cos^2 x}{x^2\sin^2 x}$$
$$=\frac{x^2}{\sin^2 x}\cdot\frac{\sin x+x\cos x}{x}\cdot\frac{\sin x-x\cos x}{x^3}$$
Then I have $3$ limits to evaluate
$$\lim_{x \to 0}\frac{x^2}{\sin^2 x}=\left(\lim_{x \to o}\frac{x}{\sin x}\right)^2=1^2=1$$
$$\lim_{x \to 0}\frac{\sin x+x\cos x}{x}=\lim_{x \to 0}\left(\frac{\sin x}{x} + \cos x\right)=1+1=2$$
Now I'm having trouble with the last one which is
$$\lim_{x \to 0}\frac{\sin x-x\cos x}{x^3}=?$$
Thanks for any help.
| It is done much simpler in the following manner with just one application of LHR.
\begin{align}
L &= \lim_{x \to 0}\left(\frac{1}{x^{2}} - \frac{1}{\tan^{2}x}\right)\notag\\
&= \lim_{x \to 0}\frac{\tan^{2}x - x^{2}}{x^{2}\tan^{2}x}\notag\\
&= \lim_{x \to 0}\frac{\tan^{2}x - x^{2}}{x^{4}}\cdot\frac{x^{2}}{\tan^{2}x}\notag\\
&= \lim_{x \to 0}\frac{\tan^{2}x - x^{2}}{x^{4}}\cdot 1\notag\\
&= \lim_{x \to 0}\frac{\tan x - x}{x^{3}}\cdot \frac{\tan x + x}{x}\notag\\
&= \lim_{x \to 0}\frac{\tan x - x}{x^{3}}\cdot \lim_{x \to 0}\left(\frac{\tan x}{x} + 1\right)\notag\\
&= 2\lim_{x \to 0}\frac{\tan x - x}{x^{3}}\notag\\
&= 2\lim_{x \to 0}\frac{\sec^{2}x - 1}{3x^{2}}\text{ (via LHR)}\notag\\
&= \frac{2}{3}\lim_{x \to 0}\frac{\tan^{2}x}{x^{2}}\notag\\
&= \frac{2}{3}\notag
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1392209",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 2
} |
real values of $x$ in $\sqrt{5-x} = 5-x^2$.
Calculate the real solutions $x\in\mathbb{R}$ to
$$
\sqrt{5-x} = 5-x^2
$$
My Attempt:
We know that $5-x\geq 0$ and thus $x\leq 5$ and
$$
\begin{align}
5-x^2&\geq 0\\
x^2-\left(\sqrt{5}\right)^2&\leq 0
\end{align}
$$
which implies that $-\sqrt{5}\leq x \leq \sqrt{5}$. Now let $y=\sqrt{5-x}$. Then
$$
\tag1 y^2=5-x
$$
and the equation converts into
$$
\begin{align}
y &= 5-x^2\\
x^2 &= 5-y\\
y^2-x^2 &= 5-x-(5-y)\\
y^2-x^2 &= y-x\\
(y-x)(y+x)-(y-x) &= 0\\
(y-x)(y+x-1) &= 0
\end{align}
$$
So either $y=x$ or $x+y=1$.
Case 1 ($y=x$):
We can plug this into $(1)$ to get
$$
\begin{align}
y^2 &= 5-x\\
x^2 &= 5-x\\
x^2+x-5 &= 0\\
x &= \frac{-1\pm \sqrt{1+20}}{2}
\end{align}
$$
Since $-\sqrt{5}\leq x\leq \sqrt{5}$, the only solution is
$$
x = \frac{-1+\sqrt{21}}{2}
$$
Case 2 ($y=1-x$):
We can plug this into $(1)$ to get
$$
\begin{align}
y^2 &= 5-x\\
(1-x)^2 &= 5-x\\
1+x^2-2x &= 5-x\\
x^2-x-4 &= 0\\
x &= \frac{1\pm\sqrt{17}}{2}
\end{align}
$$
Since $-\sqrt{5}\leq x\leq \sqrt{5}$, the only solution is
$$
x = \frac{1-\sqrt{17}}{2}
$$
So final solution is
$$
x \in \left\{\displaystyle \frac{1-\sqrt{17}}{2}, \frac{-1+\sqrt{21}}{2} \right\}
$$
Is it possible to solve this problem using geometry? For example, could we use the properties of a half-circle and a parabola?
| You can see the solutions as the abscissa of the intersection points of the two curves $y=5-x^2$ (a parabola) and $y=\sqrt{5-x}$ (upper half of a parabola). These twoo parabolas intersect at 4 points, but only two of them lie in the $y>0$ half-plane.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1393046",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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} |
maximum value of $a+b+c$ given $a^2+b^2+c^2=48$? How can i get maximum value of this $a+b+c$ given $a^2+b^2+c^2=48$ by not using AM,GM and lagrange multipliers .
| You can use the identity
$$(a+b+c)^2=3(a^2+b^2+c^2)-(a-b)^2-(b-c)^2-(c-a)^2$$
The right-hand side is clearly maximised when $a=b=c$ and then $a+b+c=12$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1393590",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 2
} |
$\lim _{ x\to 1 }{ \frac { \sqrt { x } +\sqrt [ 3 ]{ x } -2 }{ x-1 } } $ I have been trying to algebraically solve this limit problem without using L'Hospital's rule but was whatsoever unsuccessful:
$$\lim _{ x\to 1 }{ \frac { \sqrt { x } +\sqrt [ 3 ]{ x } -2 }{ x-1 } } $$
Thanks in advance!
| Let $x = u^6$.
$$\lim _{ x\to 1 }\dfrac { \sqrt { x } +\sqrt [ 3 ]{ x } -2 }{ x-1 } = \lim_{u\to 1}\dfrac{u^3 + u^2 - 1 - 1}{u^6-1}$$
$$
= \lim_{u\to 1}\dfrac{u^3 - 1}{(u^2)^3-1^3} + \lim_{u\to 1}\dfrac{u^2 - 1}{(u^2)^3-1^3}
$$
$$
= \lim_{u\to 1}\dfrac{(u - 1)(u^2+u+1)}{(u-1)(u+1)(u^4 + u^2 +1)} + \lim_{u\to 1}\dfrac{(u - 1)(u+1)}{(u-1)(u+1)(u^4 + u^2 +1)}
$$
$$
= \dfrac{3}{6} + \dfrac{2}{6} = \dfrac{5}{6}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1393768",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 3
} |
Prove that neither $A$ nor $B$ is divisible by $5$ Let the sum
$$ {1+ \frac12 + \frac13 + \frac 14+ \dots +\frac1{99} + \frac 1{100}}$$
be written as $\frac AB$, where $A$ and $B$ are positive integers with no common factors. Show that neither $A$ nor $B$ is divisible by $5$.
Using Mathematica, I found the sum is $$\frac AB=\frac {14466636279520351160221518043104131447711}{2788815009188499086581352357412492142272}$$
| Firstly $\frac{1}{25} + \frac{1}{50} + \frac{1}{75} + \frac{1}{100} = \frac{1}{25} \left( \frac{1}{1} + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} \right) = \frac{1}{12}$.
Also $\frac{1}{25k+5} + \frac{1}{25k+10} + \frac{1}{25k+15} + \frac{1}{25k+20} = \frac{1}{5} \left( \frac{(5k+4)+(5k+1)}{(5k+1)(5k+4)} + \frac{(5k+3)+(5k+2)}{(5k+2)(5k+3)} \right)$
$\ \ = \frac{2k+1}{(5k+1)(5k+4)} + \frac{2k+1}{(5k+2)(5k+3)}$ for any integer $k$.
Therefore $\sum_{k=1}^{100} \frac{1}{k}$ has no factor of $5$ in the reduced fraction form.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1394206",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
"answer_count": 1,
"answer_id": 0
} |
Integral of $\frac{1}{x^2+x+1}$ and$\frac{1}{x^2-x+1}$ How to integrate two very similar integrals. I am looking for the simplest approach to that, it cannot be sophisticated too much as level of the textbook this was taken from is not very high. $$\int \frac{1}{x^2+x+1} dx$$ and$$\int \frac{1}{x^2-x+1} dx$$
| $$\int { \frac { dx }{ { x }^{ 2 }+x+1 } =\int { \frac { dx }{ { \left( x+\frac { 1 }{ 2 } \right) }^{ 2 }+\frac { 3 }{ 4 } } } } =\int { \frac { d\left( x+\frac { 1 }{ 2 } \right) }{ { \left( x+\frac { 1 }{ 2 } \right) }^{ 2 }+\frac { 3 }{ 4 } } } =\frac { 2 }{ \sqrt { 3 } } \arctan { \left( \frac { 2 }{ \sqrt { 3 } } \left( x+\frac { 1 }{ 2 } \right) \right) \quad +C } $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1394731",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 5
} |
Solve the equation in the questions Solve the equation $$(x+3)(\sqrt{2x^2+6x+2}-2x)=\sqrt[3]{x^2+1}+(x^2-7)\sqrt{x+3}$$
It's little complicated, any help will be appreciated.
| $$ (x+3)\sqrt{2x^2+6x+2}-(2x^2+6x+2)=\sqrt[3]{x^2+1}-2+(x^2-7)\sqrt{x+3}$$
$$\Rightarrow \sqrt{2x^2+6x+2}\frac{7-x^2}{x+3+\sqrt{2x^2+6x+2}}=\frac{x^2-7}{\sqrt[3]{(x^2+1)^2}+2\sqrt[3]{x^2+1}+4}+(x^2-7)\sqrt{x+3}$$
And so the solutions $${x\in\{-\sqrt 7,+\sqrt 7\}}$$
And $$-\frac{\sqrt{2x^2+6x+2}}{x+3+\sqrt{2x^2+6x+2}}=\frac{1}{\sqrt[3]{(x^2+1)^2}+2\sqrt[3]{x^2+1}+4}+\sqrt{x+3}$$
Which clearly has no more solution since $LHS\le 0<RHS$
AND WE'RE DONE
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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} |
Show that $f$ is a decreasing function It's given that $f(x)=\frac{1}{x^3}-x^3$ for $x>0$ show that $f$ is a decreasing function.
My attempt $f'(x)=-3x^{-4}-3x^2$
$x^6=-1$
How to continue by my attempt ?
| Notice, we have $$f(x)=\frac{1}{x^3}-x^3$$
$$\frac{d}{dx}(f(x))=\frac{d}{dx}\left(\frac{1}{x^3}-x^3\right)$$
$$f'(x)=\frac{-3}{x^4}-3x^2$$
$$f'(x)=-3\left(\frac{1}{x^4}+x^2\right)$$
$$f'(x)=-3\left(\left(\frac{1}{x^2}\right)^2+x^2\right)\tag 1$$
We know that $$\left(\frac{1}{x^2}\right)^2> 0\ \ \ \ \forall \ x>0 \tag 2$$ & $$\left(x\right)^2> 0\ \ \ \ \forall \ x>0 \tag 3$$ From (2) & (3) we get $$\left(\frac{1}{x^2}\right)^2+x^2> 0$$$$\iff -3\left(\left(\frac{1}{x^2}\right)^2+x^2\right)<0\ \ \ \ \forall \ \ x>0\tag 4$$ Hence, from (1) & (4), we find that $$f'(x)<0\ \ \ \forall \ \ \ x>0$$
Hence, the function $f(x) $ is decreasing for all $x>0$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1395266",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
$\int\limits_{0}^{\pi/2}\frac{1+2\cos x}{(2+\cos x)^2}dx$ $\int\limits_{0}^{\pi/2}\frac{1+2\cos x}{(2+\cos x)^2}dx$
I tried to solve it.
$\int\limits_{0}^{\pi/2}\frac{1+2\cos x}{(2+\cos x)^2}dx=\int\limits_{0}^{\pi/2}\frac{4+2\cos x}{(2+\cos x)^2}-\frac{3}{(2+\cos x)^2}dx=\int\limits_{0}^{\pi/2}\frac{2}{2+\cos x}-\frac{3}{(2+\cos x)^2}dx$
But i could not solve further.Please help me in completing.
| Considering $$I=\int\frac{1+2\cos (x)}{(2+\cos (x))^2}\,dx$$ you can do several things.
First, considering the squared denominator, you could suppose that the result of integration would be something like $$\frac {a+b\sin(x)+c\cos(x)}{2+\cos(x)}$$ which, one differentiated, would give $$\frac{(a-2 c) \sin (x)+2 b \cos (x)+b}{(2+\cos (x))^2}$$ So, $$\frac{(a-2 c) \sin (x)+2 b \cos (x)+b}{(2+\cos (x))^2}=\frac{1+2\cos (x)}{(2+\cos (x))^2}$$ Removing the denominator let us with $$(a-2 c) \sin (x)+2 (b-1) \cos (x)+b-1=0$$ So, $b=1$, $a-2c=0$ makes $$I=\frac{\sin(x)}{2+\cos (x)}$$
The second way would use the tangent half-angle substitution $t=\tan(\frac x2)$ which would lead to $$I=\int\frac{1+2\cos (x)}{(2+\cos (x))^2}\,dx=-\int\frac{2 \left(t^2-3\right)}{\left(t^2+3\right)^2}\,dt$$ which is not very difficult to integrate.
I am sure that you can takle from here.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1395338",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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} |
Prove this inequality $(a_{1}a_{2}\cdots a_{n})\sqrt{1-a_{n+1}}+\sqrt{n-1}\cdot a_{n+1}<\sqrt{n}$ Assmue that $a_{i}\in (0,1),i=1,2,3,\cdots,n$,show that
$$(a_{1}a_{2}\cdots a_{n})\sqrt{1-a_{n+1}}+\sqrt{n-1}\cdot a_{n+1}<\sqrt{n},$$
I've tried many things but all have failed.
| Let $C_n:=a_1\cdots a_n$. Note, that for $n=1$, the inequality is true:
$$
a_1\cdot\sqrt{1-a_{n+1}}<1\cdot\sqrt{1-0}=1
$$
Now we have the following for $n>1$:
$$
0≤\sqrt{n-1}\left(\sqrt{1-a_{n+1}}-\frac{C_n}{2\sqrt{n-1}}\right)^2 \iff \\
0≤\sqrt{n-1}\cdot (1-a_{n+1})-C_n\sqrt{1-a_{n+1}}+\frac{C_n^2}{4\sqrt{n-1}}\iff\\ C_n\sqrt{1-a_{n+1}}+\sqrt{n-1}\cdot a_{n+1}≤\sqrt{n-1}+\frac{C_n^2}{4\sqrt{n-1}}<\sqrt{n-1}+\frac{1}{4\sqrt{n-1}}
$$
Where in the last inequality we used $C_n<1$. Therefore, we have to prove
$$
\sqrt{n-1}+\frac{1}{4\sqrt{n-1}}<\sqrt{n}\iff \frac{1}{4\sqrt{n-1}}<\sqrt{n}-\sqrt{n-1}=\frac{1}{\sqrt{n}+\sqrt{n-1}}\iff \\\sqrt{n}<3\sqrt{n-1}\iff n<9n-9 \iff \frac{9}{8}<n
$$
So we conclude that the original inequality is true.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1395624",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
Multivariable limit of $\sqrt{x^2+y^2} \sin (\frac{1}{x^2+y^2})$ Find Multivariable limit of $$\lim \limits_{(x,y) \rightarrow (0,0) }\sqrt{x^2+y^2} \sin \left(\frac{1}{x^2+y^2}\right)$$
Limit is obviously zero, but my question is how to simplify it a bit more to make it more obvious?
| Notice that
$$
0 \leq \sqrt{x^2+y^2} \left| \sin \left( \frac{1}{x^2+y^2} \right) \right| \leq \sqrt{x^2+y^2}
$$
and by squeezing...
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1396910",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find the number of real roots of $1+x/1!+x^2/2!+x^3/3! + \ldots + x^6/6! =0$.
Find the number of real roots of $1+x/1!+x^2/2!+x^3/3! + \ldots + x^6/6! =0$.
Attempts so far:
Used Descartes signs stuff so possible number of real roots is $6,4,2,0$
tried differentiating the equation $4$ times and got an equation with no roots hence proving that above polynomial has $4$ real roots.
But using online calculators I get zero real roots. Where am I wrong?
| Let $f(x)= 1+x+\frac{x^2}{2!}+\ldots +\frac{x^6}{6!}$
Differentiating $f(x)$:
$$f'(x)= 1+x+\frac{x^2}{2!}+\ldots+\frac{x^5}{5!}$$
Consider $a$ as one real root for $f'(x)=0$, i.e. $f'(a)=0$. Obviously $a\ne 0$.
Now, $f(a)= 0+ \frac{a^6}{6!}>0$. This means that all the potential local minima of $f(x)$ are already positive. Hence, there are no real roots for $f(x)=0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1397428",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 10,
"answer_id": 9
} |
Proving a ratio that has a relation with the Perpendicular bisectors and circumcircle $ABC$ is a triangle, $D$ is a point on the side $BC$ of $\triangle ABC$, $R_b$ is circumradius of $\triangle ABD$ , and $R_c$ is the circumradius of $\triangle ACD$.
Prove that
$$ {Rb\over Rc} ={AB\over AC}$$
Thanks for your help
| Applying sine rule in $\triangle ABD$ as follows $$\frac{\sin\angle ABD}{AD}=\frac{\sin\angle ADB}{AB} $$ $$\implies \color{red}{\sin \angle ABD=\frac{AD}{AB}\sin\angle ADB}$$ Now, the circumscribed radius $R_b$ of $\triangle ABD$ is given as $$R_b=\frac{\text{side of triangle }}{2\times\sin \text{(opposite angle)}}$$$$=\frac{AD}{2\sin \angle ABD}=\frac{AD}{2\times \frac{AD}{AB}\sin\angle ADB }$$$$R_b=\frac{AB}{2\sin \angle ADB}\tag 1$$
Similarly, applying sine rule in $\triangle ACD$ as follows
$$\frac{\sin\angle ACD}{AD}=\frac{\sin\angle ADC}{AC} $$ $$\implies \color{red}{\sin \angle ACD=\frac{AD}{AC}\sin\angle ADC}$$ Now, the circumscribed radius $R_c$ of $\triangle ACD$ is given as $$R_c=\frac{AD}{2\sin \angle ACD}=\frac{AD}{2\times \frac{AD}{AC}\sin\angle ADC} $$ $$R_c=\frac{AC}{2\sin \angle ADC}$$ But, we have $$\angle ADB+\angle ADC=180^\circ\iff \angle ADC=180^\circ-\angle ADB$$ Now, setting the value of $\angle ADC$, we get $$ R_c=\frac{AC}{2\sin(180^\circ- \angle ADB)}$$ $$R_c=\frac{AC}{2\sin \angle ADB}\tag 2$$
Now, dividing (1) by (2), we get $$\frac{R_b}{R_c}=\frac{\frac{AB}{2\sin \angle ADB}}{\frac{AC}{2\sin \angle ADB}}=\frac{AB}{AC}$$
Hence, proved that
$$\bbox[5px, border:2px solid #C0A000]{\color{red}{\frac{R_b}{R_c}=\frac{AB}{AC}}}$$
| {
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"url": "https://math.stackexchange.com/questions/1402579",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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$\int\frac{1+x^2}{x^4+3x^3+3x^2-3x+1}dx$ $$\int\frac{1+x^2}{x^4+3x^3+3x^2-3x+1}dx$$
I tried to solve it.
$$\int\frac{1+x^2}{x^4+3x^3+3x^2-3x+1}dx=\int\frac{1+x^2}{(x^2+1)^2+3x^3+x^2-3x+1}dx=\int\frac{1+x^2}{(x^2+1)^2+3x(x^2-1)+x^2+1}dx$$
But this does not seem to be solving.Please help.
| As already pointed out, the roots of the polynomial are complex.
The analytic solving is a boring task. Only the main intermediate results are reported below :
| {
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Upper and lower bounds for $S(n) = \sum_{i=1}^{2^{n}-1} \frac{1}{i} = 1+\frac{1}{2}+ \cdots +\frac{1}{2^n-1}.$
For a positive integer $n$ let $S(n) = \sum_{i=1}^{2^{n}-1} \frac 1i = 1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+ \cdots +\frac{1}{2^n-1}.$
Then which of the following are true.
*
*(a) $S(100)\leq 100$.
*(b) $S(100)>100$.
*(c) $S(200)\leq 100$.
*(d) $S(200)>100$.
My attempt
*
*For the upper bound
$$\begin{align}
S(n) &= 1 + \left( \frac 12 + \frac 13 \right) + \left( \frac 14 + \frac 15 + \frac 16 + \frac 17 \right) + \cdots + \left( \frac 1{2^{n-1}} + \frac 1{2^{n-1}+1} + \cdots + \frac 1{2^n-1} \right)
\\ &< 1 + \left( \frac 12 + \frac 12 \right) + \left( \frac 14 + \frac 14 + \frac 14 + \frac 14 \right) + \cdots + \left( \frac 1{2^{n-1}} + \frac 1{2^{n-1}} + \cdots + \frac 1{2^{n-1}} \right) \\
&= \underbrace{1 + 1 + \cdots + 1}_{n\text{ times}} \\
&= n.
\end{align}$$
So we get $S(n) < n$ (for $n > 1$), and in particular $S(100) < 100$.
Now I did not understand how to calculate a lower bound, or if there is any other method by which we can solve this.
| Group the terms
exactly as you have,
but get a lower bound
that can be easily manipulated.
I'll copy, paste, and edit
your equations to show what I mean.
$$\displaystyle S(n) = 1+\underbrace{\frac{1}{2}+\frac{1}{3}}+\underbrace{\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}}+....+\underbrace{\frac{1}{2^{n-1}}+..+\frac{1}{2^n-1}}\\
>1+\underbrace{\frac{1}{4}+\frac{1}{4}}+\underbrace{\frac{1}{8}+\frac{1}{8}+\frac{1}{8}+\frac{1}{8}}+....+\underbrace{\frac{1}{2^{n}}+..+\frac{1}{2^{n}}}\\
=1+\frac12+\frac12....(\bf{n-times}) = 1+\frac{n}{2}$$
So we get $$\displaystyle S(n)
= 1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+..........+\frac{1}{2^n-1}
>1+\frac{n}{2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1404602",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Maximize $xy^2$ on the ellipse $x^2+4y^2=4$ I was using Lagrange multiplier, any steps gone wrong?
$$f(x,y)=xy^2$$
$$c(x,y)=x^2+4y^2$$
Partial Derivatives
$$\frac {\partial f}{\partial x} = y^2 $$
$$\frac {\partial f}{\partial y} = 2xy $$
$$\frac {\partial c}{\partial x} = 2x $$
$$\frac {\partial c}{\partial y} = 8y $$
Find Lambda
$$y^2=2x\lambda$$
$$\lambda=\frac {y^2}{2x} $$
$$ 2xy=8y\lambda $$
$$\lambda= \frac{x}{4} $$
Let
$$ \frac {y^2}{2x}=\frac {x}{4} $$ yields $$y=\frac {x}{\sqrt 2}$$ and $$y=-\frac {x}{\sqrt 2}$$
Bringing $$y=\frac {x}{\sqrt 2}$$ into C
$$ x^2 + 2x^2 = 4 $$
$$ 3x^2=4 $$ yields $$ x=\frac {2}{\sqrt 3} and -\frac {2}{\sqrt 3} $$
Since $$ y=\frac {x}{\pm\sqrt 2} $$
plugging $x=\frac {2}{\pm\sqrt 3}$ into y I got 4 points:
$(\frac {2}{\sqrt 3},\frac {2}{\sqrt 6}),(-\frac {2}{\sqrt 3},\frac {2}{\sqrt 6}),(-\frac {2}{\sqrt 3},-\frac {2}{\sqrt 6}),(\frac {2}{\sqrt 3},-\frac {2}{\sqrt 6})$
is every step ok up to this point?
| You can suppose $x,y>0$ otherwise $xy^2\leq 0$. Then you can use that $\frac{x_1+x_2+x_3}{3}\geq \sqrt[3]{x_1x_2x_3}$ and the equality holds iff $x_1=x_2=x_3$. Note that $(x^2+2y^2+2y^2)^3\geq 27x^2y^4$. So $64\geq 27(xy^2)^2$ the eqaulity holds iff $x=\sqrt{2}y$ or $x=-\sqrt{2}y$. Therefore $xy^2\leq \frac{8}{3\sqrt{3}}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1406260",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Prove $\lim \frac{3^n + 2^n}{5\cdot3^n + 7\cdot2^n} = \frac{1}{5}$ Prove using limit definition.
$$\lim \frac{3^n + 2^n}{5\cdot 3^n + 7\cdot 2^n} = \frac{1}{5}
$$
My try:
$$\left| {\frac{3^n + 2^n}{5\cdot 3^n + 7\cdot 2^n} - \frac{1}{5}} \right| < \varepsilon
$$
$$ \Leftrightarrow \left| \frac{3^n + 2^n - 5\cdot 3^n - 7\cdot 2^n}{5(5\cdot3^n + 7\cdot2^n)} \right| < \varepsilon \
$$
$$ \Leftrightarrow \frac{4\cdot 3^n + 5\cdot 2^n}{5(5\cdot 3^n + 7\cdot 2^n)} < \varepsilon
$$
I can increase numerator next and simplify like this:
$$\frac{4\cdot 3^n + 5\cdot 2^n}{5(5\cdot 3^n + 7\cdot 2^n)} < \,\frac{5\cdot 3^n + 7\cdot 2^n}{5\cdot (5\cdot 3^n + 7\cdot 2^n)} = \frac{1}{5}
$$
But in this case I'm loosing the $n$ in the dominator.
| HINT
$$
\frac{ 3^n + 2^n }{ 5 \times 3^n + 7 \times 2^n }
= \frac{ \color{red}{1} + (2/3)^n }{ \color{red}{5} + 7 \times (2/3)^n }
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1408198",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Prove that $\frac{2^{x+1}+(x+1)^2}{2^x+x^2}\rightarrow 2$ as $x \rightarrow \infty$ Could someone please show me the proof that $\frac{2^{x+1}+(x+1)^2}{2^x+x^2}\rightarrow 2$ as $x \rightarrow \infty$
I have no idea where to begin with this one.
Thanks.
| Notice, we have $$\lim_{x\to \infty}\frac{2^{x+1}+(x+1)^2}{2^x+x^2}$$
$$=\lim_{x\to \infty}\frac{2\cdot 2^{x}+x^2+2x+1}{2^x+x^2}$$
$$=\lim_{x\to \infty}\frac{(2^{x}+x^2)+(2^x+2x+1)}{2^x+x^2}$$
$$=1+\lim_{x\to \infty}\frac{2^x+2x+1}{2^x+x^2}$$
Using L-Hospital's rule 3 times:
$$=1+\lim_{x\to \infty}\frac{2^x\ln 2+2}{2^x\ln 2+2x}$$
$$=1+\lim_{x\to \infty}\frac{2^x(\ln 2)^2}{2^x(\ln 2)^2+2}$$
$$=1+\lim_{x\to \infty}\frac{2^x(\ln 2)^3}{2^x(\ln 2)^3}$$
$$=1+\lim_{x\to \infty}1=1+1=2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1408900",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Calculate $\lim_{n\to\infty} (n - \sqrt {{n^2} - n} )$ Calculate limit:
$$\lim_{n\to\infty} (n - \sqrt {{n^2} - n})$$
My try:
$$\lim_{n\to\infty} (n - \sqrt {{n^2} - n} ) = \lim_{n\to\infty} \left(n - \sqrt {{n^2}(1 - \frac{1}{n}} )\right) = \lim_{n\to\infty} \left(n - n\sqrt {(1 - \frac{1}{n}})\right)$$
$$\sqrt {(1 - \frac{1}{n}} ) \to 1$$
$$\lim_{n\to\infty} \left(n - n\sqrt {(1 - \frac{1}{n}})\right) = \lim_{n\to\infty} (n - n) = 0$$
Ok, it's not correct. In articles i found method:
$$\lim_{n\to\infty} \frac{{(n - \sqrt {{n^2} - n} )(n + \sqrt {{n^2} - n} )}}{{n + \sqrt {{n^2} - n} }} = \lim_{n\to\infty} \frac{n}{{n + \sqrt {{n^2} - n} }} = \lim_{n\to\infty} \frac{1}{{1 + \sqrt {1 - \frac{1}{n}}}} = \frac{1}{2}$$
Why the second method is valid, but the first one not? Whats difference?
| Given $\displaystyle \lim_{n\rightarrow \infty}\left[n-\sqrt{n^2-n}\right] = \lim_{n\rightarrow \infty}\left[n-n\cdot \left(1-\frac{1}{n}\right)^{\frac{1}{2}}\right]$
Now Using $\displaystyle (1+x)^{n} = 1+nx+\frac{n(n-1)}{2}x^2+.....$
So we get $\displaystyle \lim_{n\rightarrow \infty}\left[n-n\left(1-\frac{n}{2n}+\frac{1}{2}\cdot \frac{1}{2}\cdot \frac{1}{2n^2}+.......\infty\right)\right] = \frac{1}{2}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1409200",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 3
} |
How can the trigonometric equation be proven? This question :
https://math.stackexchange.com/questions/1411700/whats-the-size-of-the-x-angle
has the answer $10°$. This follows from the equation
$$2\sin(80°)=\frac{\sin(60°)}{\sin(100°)}\times \frac{\sin(50°)}{\sin(20°)}$$
which is indeed true , which I checked with Wolfram.
*
*How can this equation be proven ?
| $$\frac{\sin 80^\circ \sin 20^\circ\sin 100^\circ }{\sin 50^\circ}=\frac{\sin 80^\circ \sin 20^\circ\cdot 2\sin 50^\circ \cos 50^\circ }{\sin 50^\circ}=2\sin 80^\circ\sin 20^\circ\cos 50^\circ=2\sin 80^\circ\sin 20^\circ\sin40^\circ$$
as $\cos 50^\circ=\sin40^\circ$
Now by Prove $ \sin(A+B)\sin(A-B)=\sin^2A-\sin^2B $, $$\sin\left(60^\circ-x\right)\sin\left(60^\circ+x\right)=\sin^260^\circ-\sin^2x$$
$$\implies\sin x\sin\left(60^\circ-x\right)\sin\left(60^\circ+x\right)=\dfrac{\sin3x}4$$
Here $x=20^\circ$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1412021",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Trying to solve $\sqrt{7-4\sqrt2 \sin x}=2\cos(x)-\sqrt2 \tan(x)$ The equation is
$$\sqrt{7-4\sqrt2 \sin x}=2\cos(x)-\sqrt2 \tan(x)$$
We get the system
$$
\begin{cases}
7-4\sqrt 2 \sin(x)=4\cos^2(x)-2\sqrt2\cos(x)\tan(x)+2\tan^2(x) \\
2\cos(x)-\sqrt2 \tan(x)\ge 0
\end{cases}
$$
I transformed the equation thus:
$$7(\sin^2(x)+cos^2(x))-4\sqrt 2 \sin(x)=4\cos^2(x)-2\sqrt2\sin(x)+2\tan^2(x)$$
$$7\sin^2(x)+3cos^2(x)-2\sqrt 2 \sin(x)-2\frac{\sin^2(x)}{1-sin^2(x)}=0$$
I multiply the whole equation by $(1-sin^2(x))$ and then substitute sin(x) with t:
$$4t^4-2\sqrt2 t^3+t^2+2\sqrt2 t - 3 =0$$
And here I'm stuck. The polynomial is seemingly non-factorizable.
A hint would be welcome. (0:
P.S. The problem as it is presented in the texbook:
| $$2\cos x-\sqrt2\tan x=\dfrac{2-2\sin^2x-\sqrt2\sin x}{\cos x}=-\sqrt2\cdot\dfrac{\sqrt2\sin^2x+\sin x-\sqrt2}{\cos x}$$
Now $\sqrt2\sin^2x+\sin x-\sqrt2=(\sqrt2\sin x-1)(\sin x+\sqrt2)$ and $\sin x+\sqrt2\ge\sqrt2-1>0$
So we need $\dfrac{\sqrt2\sin x-1}{\cos x}\le0$
If $\cos x>0, \sqrt2\sin x-1\le0\iff\sin x\le\dfrac1{\sqrt2}$
$2n\pi\le x\le 2n\pi+\dfrac\pi4\ \ \ \ (1A)\ \ $ OR $\ \ 2n\pi+\dfrac{3\pi}2<x\le2n\pi+2\pi\ \ \ \ (1B)$
$(1A),(1B)$ can be clubbed as $2n\pi-\dfrac\pi2< x\le 2n\pi+\dfrac\pi4\ \ \ \ (2)$
Else if $\cos x<0, \sqrt2\sin x-1\ge0\iff\sin x\ge\dfrac1{\sqrt2}$
$2n\pi+\dfrac\pi2\le x\le 2n\pi+\dfrac{3\pi}4\ \ \ \ (3)$
Now from the other answer, $$\cos2x=2\cos^2x-1=-\dfrac12=\cos\left(\pi-\dfrac\pi3\right)$$
$$2x=2m\pi\pm\dfrac{2\pi}3\iff x=m\pi\pm\dfrac\pi3$$
Can you check which values of $x$ conform to $(1A),(1B)$ or $(2)$ and $(3)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1412884",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Find the solution set of the equation $5.(\frac{1}{25})^{\sin^2x}+4.5^{\cos2x}=25^{\frac{\sin2x}{2}}$ Problem :
Find the solution set of the equation $5.(\frac{1}{25})^{\sin^2x}+4.5^{\cos2x}=25^{\frac{\sin2x}{2}}$ where $x \in [0,2\pi]$
My approach :
$5.(\frac{1}{25})^{\sin^2x}+4.5^{1-2\sin^2x}=25^{\frac{\sin2x}{2}}$
Unable to understand how to use $\sin2x$ in R.H.S. to solve further , can you please guide further thanks
| $$5\cdot (\frac { 1 }{ 25 } )^{ sin^{ 2 }x }+4\cdot { \left( 5 \right) }^{ cos2x }=\left( 25 \right) ^{ \frac { sin2x }{ 2 } }\\ { 5 }^{ -2\sin ^{ 2 }{ x+1 } }+4\cdot { 5 }^{ 1-2\sin ^{ 2 }{ x } }={ 5^{ sin2x } }\\ 5\cdot { 5 }^{ 1-2\sin ^{ 2 }{ x } }={ 5^{ sin2x } }\\ { 5 }^{ 2-2\sin ^{ 2 }{ x } }=5^{ sin2x }\\ 2-2\sin ^{ 2 }{ x } =\sin { 2x } \\ 1-\sin ^{ 2 }{ x } =\sin { x } \cos { x } \\ \cos ^{ 2 }{ x } =\sin { x } \cos { x } \\ \cos { x } \left( \cos { x } -\sin { x } \right) =0\\ \cos { x } =0,\cos { x } -\sin { x } =0 $$
Can you proceed from here?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1414941",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Prove that $1^2 + 2^2 + ..... + (n-1)^2 < \frac {n^3} { 3} < 1^2 + 2^2 + ...... + n^2$ I'm having trouble on starting this induction problem.
The question simply reads : prove the following using induction:
$$1^{2} + 2^{2} + ...... + (n-1)^{2} < \frac{n^3}{3} < 1^{2} + 2^{2} + ...... + n^{2}$$
| I would start by proving each side of the equation using induction.
First prove that $1^2+2^2+..+n^2 = \sum\limits_{i=1}^n i^2 \geq \frac{n^3}{3}$
Induction says that we must first check that the relationship holds for $n=1$.
$\sum\limits_{i=1}^1 i^2 = 1 \geq \frac{1^3}{3}=\frac{1}{3}$
So it holds for $n=1$. Now we assume that it is true that $\sum\limits_{i=1}^n i^2 \geq \frac{n^3}{3}$ for $n=k$, and prove that the result can be obtained from $n=k+1$.
$\sum\limits_{i=1}^{k+1} i^2 = \sum\limits_{i=1}^k i^2 + (k+1)^2$
$\sum\limits_{i=1}^{k+1} i^2 \geq \frac{k^3}{3} + k^2+2k+1$
$\sum\limits_{i=1}^{k+1} i^2 \geq \frac{k^3+3k^2+3k+1}{3}+k+\frac{2}{3}$
$\sum\limits_{i=1}^{k+1} i^2 \geq \frac{(k+1)^3}{3}+k+\frac{2}{3}$
$\sum\limits_{i=1}^{k+1} i^2 \geq \frac{(k+1)^3}{3}$
Since $k+\frac{2}{3}$ is a positive number, removing it from the right side still maintains the inequality, and the first part of your question has been proved by induction. Do a similar thing to the left hand side to prove it.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1416472",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 3
} |
How do I solve $x^4-3x^2+2=0$? How do I solve $x^4-3x^2+2=0$ ?
I would appreciate some kind of hint here. I have no clue how to start this problem.
| Consider the product $(y-2)(y-1)$ for which the expansion is $y^2 - 3 y +2$. Now let $y = x^2$ to obtain $x^4 - 3 x^2 + 2 = (x^2 -2)(x^2 -1) = 0$. This leads to $x^2 = 2$ or $x^2 =1$. These two equations yields $x \in \{ - 1, 1, - \sqrt{2}, \sqrt{2} \}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1419188",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Solve logarithmic equation $\log_{\frac{x}{5}}(x^2-8x+16)\geq 0$ Find $x$ from logarithmic equation:
$$\log_{\frac{x}{5}}(x^2-8x+16)\geq 0 $$
This is how I tried:
$$x^2-8x+16>0$$
$$ (x-4)^2>0 \implies x \not = 4$$
then
$$\log_{\frac{x}{5}}(x^2-8x+16)\geq \log_{\frac{x}{5}}(\frac{x}{5})^0 $$
because of base $\frac{x}{5}$, we assume $x \not\in (-5,5)$, then
$$x^2-8x+16 \geq 1$$
$$ (x-3)(x-5) \geq 0 \implies$$
$$ \implies x \in {(- \infty,-5) \cup (5, \infty)} \cap x\not = 4 $$
But this is wrong, because the right solution is $$x \in {(3,4) \cup (4,6)} $$
I'm sorry if I used the wrong terms, English is not my native language.
| $$\log_{\frac{x}{5}}(x^2-8x+16)\geq 0$$
$$=\frac{\log{(x^2-8x+16)}}{\log\frac{x}{5}}\geq0$$
So $x$ is non-positive. And now raising both sides on the base $10$.
$$x^2-8x+16\geq1$$
$$x^2-8x+15\geq0$$
$$(x-5)(x-3)\geq0$$
$$x\geq5; \ x\geq3$$
But $x$ cannot be $5$. So answer is $[3,5)\cup(5,\infty)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1422858",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Pythagorean identities. I was remembering about Pythagorean trigonometric identities so I grabbed some of my old notebooks and I was having so much fun, solving some equalities, until I found these four equalities that I just can't find the way to solve them, probably they are easier than I thought but I really need some help, otherwise I wont sleep tonight!
I'll just show you how I tried one, because the rest of them are all pretty the same, but I'll post all of them in case someone can help me with all.
$$\frac{\sin (\theta)}{\cot(\theta)\csc(\theta)-\cos(\theta)}= \frac{\cos(\theta)-\sec(\theta)}{\sin(\theta)-\csc(\theta)}$$
So, I started with
$$LHC= \frac{\sin(\theta)}{\Big(\frac{\cos(\theta)}{\sin(\theta)}\Big) \Big(\frac{1}{\sin(\theta)}\Big)-\cos(\theta)} = \frac{\sin(\theta)}{\Big(\frac{\cos(\theta)}{\sin^2(\theta)}\Big)-\cos(\theta)}=\frac{\sin(\theta)}{\frac{\cos(\theta)-\cos(\theta)\sin(\theta)}{\sin^2(\theta)}}=\frac{\sin^3(\theta)}{\cos(\theta)-\cos(\theta)\sin^2(\theta)}=\frac{\sin^3(\theta)}{\cos(\theta)(1-\sin^2(\theta))}=\frac{\sin^3(\theta)}{\cos(\theta)\cos^2(\theta)}= \tan^3 (\theta)$$
$$RHC = \frac{\cos (\theta) - \frac{1}{\cos(\theta)}}{\sin(\theta)-\frac{1}{\sin(\theta)}} = \frac{\frac{\cos^2(\theta)-1}{\cos(\theta)}}{\frac{\sin^2(\theta)-1}{\sin(\theta)}}= \frac{\sin(\theta)(\cos^2-1)}{\cos(\theta)(\sin^2(\theta)-1)}= \frac{\sin(\theta)(-\sin(\theta)}{\cos(\theta)(-\cos(\theta)} = \tan^3(\theta).$$
And I seriously don't know how to manage to get to $\frac{\cos(\theta)-\sec(\theta)}{\sin(\theta)-\csc(\theta)}.$
And I have the same issue with
$$\frac{1+\cot^3 (\epsilon)}{1+\cot (\epsilon)}= \csc^2 (\epsilon) - \cot (\epsilon),$$
and
$$\sin^2 (\lambda)+\tan(\lambda) = \frac{\tan^3(\lambda)-1}{\tan (\lambda)-1}.$$
Please guys, help me to manipulate these equations, I'll appreciate it!
| Lets start from the LHS, then multiply by $\frac{RHS}{RHS}$
$$\begin{array}{lll}
\displaystyle\frac{\sin\theta}{\cot\theta\csc\theta-\cos\theta}&=&\displaystyle\frac{\sin\theta}{\cot\theta\csc\theta-\cos\theta}\cdot\frac{\frac{\cos\theta-\sec\theta}{\sin\theta-\csc\theta}}{\frac{\cos\theta-\sec\theta}{\sin\theta-\csc\theta}}\\
\displaystyle&=&\displaystyle\bigg(\frac{\sin\theta}{\cot\theta\csc\theta-\cos\theta}\cdot \frac{\sin\theta-\csc\theta}{\cos\theta-\sec\theta}\bigg)\cdot\frac{\cos\theta-\sec\theta}{\sin\theta-\csc\theta}\\
&=&\displaystyle\frac{\sin^2\theta\cos\theta}{\sin^2\theta\cos\theta}\cdot\displaystyle\bigg(\frac{\sin\theta}{\cot\theta\csc\theta-\cos\theta}\cdot \frac{\sin\theta-\csc\theta}{\cos\theta-\sec\theta}\bigg)\cdot\frac{\cos\theta-\sec\theta}{\sin\theta-\csc\theta}\\
&=&\displaystyle\displaystyle\bigg(\frac{\sin^2\theta\cos\theta}{\cos\theta-\cos\theta\sin^2\theta}\cdot \frac{\sin^2\theta-1}{\cos^2\theta-1}\bigg)\cdot\frac{\cos\theta-\sec\theta}{\sin\theta-\csc\theta}\\
&=&\displaystyle\displaystyle\bigg(\frac{\sin^2\theta\cos\theta}{\cos\theta(1-\sin^2\theta)}\cdot \frac{-\cos^2\theta}{-\sin^2\theta}\bigg)\cdot\frac{\cos\theta-\sec\theta}{\sin\theta-\csc\theta}\\
&=&\displaystyle\displaystyle\bigg(\frac{\sin^2\theta\cos\theta}{\cos^3\theta}\cdot \frac{-\cos^2\theta}{-\sin^2\theta}\bigg)\cdot\frac{\cos\theta-\sec\theta}{\sin\theta-\csc\theta}\\
&=&\displaystyle1\cdot\frac{\cos\theta-\sec\theta}{\sin\theta-\csc\theta}=\displaystyle \frac{\cos\theta-\sec\theta}{\sin\theta-\csc\theta}\\
\end{array}$$
In general
$$\frac{LHS}{RHS}\cdot RHS = RHS$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1423609",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Double u-substitution Problem I have to find $$\int\limits_{0}^{1}x(x+a)^{1/p}\text{ d}x$$ and I'm told to use two substitutions, $u=(x+a)^{1/p}$ and $u=x+a$.
However, I can't find the derivative of $u$, $\text{d}u$, to replace once I've done the substitution - could someone offer a bit of insight into what I'm missing here? Thank you!
| As the others have suggested, we need to use the anti-power rule:
$$\int{x^a}dx=\frac{1}{a+1}x^{a+1}$$
Also as the comment has suggested, we may use $x=(x+a)-a$ to substitute the first $x$:
$$\int_{0}^{1}x(x+a)^{\frac{1}{p}}dx = \int_{0}^{1}[(x+a)-a](x+a)^{\frac{1}{p}}dx$$
This works because $(x+a)-a = x+a-a = x$. Let us continue with the expansion:
$$\int_{0}^{1}[(x+a)(x+a)^{\frac{1}{p}} - a(x+a)^{\frac{1}{p}}]dx$$
$$=\int_{0}^{1}(x+a)^{\frac{1}{p}+1}dx - a\int_{0}^{1}(x+a)^{\frac{1}{p}}dx$$
Now we use the anti-power rule to continue: ($\frac{1}{p}+1=\frac{1+p}{p}$)
$$\int_{0}^{1}(x+a)^{\frac{1+p}{p}}dx - a\int_{0}^{1}(x+a)^{\frac{1}{p}}dx$$
$$=[(1+\frac{1+p}{p})(x+a)^{1+\frac{1+p}{p}} - a((1+\frac{1}{p})(x+a)^{1+\frac{1}{p}}]|_{0}^{1}$$
$$=\frac{1+2p}{p}(1+a)^{\frac{1+2p}{p}}-\frac{1+2p}{p}(a)^{\frac{1+2p}{p}}-a[\frac{p+1}{p}(1+a)^{\frac{p+1}{p}}-\frac{p+1}{p}(a)^{\frac{p+1}{p}}]$$
$$=\left(\frac{1}{p}+2\right)(1+a)^{\frac{1}{p}}(1+a)^2 - \left(\frac{1}{p}+1\right)(1+a)^{\frac{1}{p}}(1+a)a - \left(\frac{1}{p}+2\right)a^{\frac{1}{p}}a^2 + \left(\frac{1}{p}+1\right)a^{\frac{1}{p}}a^2$$
$$=(1+a)^{\frac{1}{p}}(1+a)^2 \left( \left(\frac{1}{p}+2\right) - \left(\frac{a}{1+a}\right) \right) - a^{\frac{1}{p}}a^2$$
$$=(1+a)^{\frac{1}{p}}(1+a)\left( \frac{1+a+2p+ap}{p} \right) - a^{\frac{1}{p}}a^2$$
| {
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Find $\int_1^{\frac{5}{2}}\frac{|x|}{[2x-5]}dx$ Find $\int_1^{\frac{5}{2}}\frac{|x|}{[2x-5]}dx$, where $[x]$ is the greatest integer function.
First, $|x|=x$ in the interval $[1,5/2]$. Next we divide subintervals, so that
for $[1,3/2), [2x-5]=-3$, $[3/2,2), [2x-5]=-2,$ and for $[2,5/2) [2x-5]=-1$.
Hence $$\int_1^{\frac{5}{2}}\frac{|x|}{[2x-5]}dx=-\frac{1}{3}\int_1^{3/2}xdx-\frac{1}{2}\int_{3/2}^2xdx-\int_2^{5/2}xdx\\
=-\frac{1}{3}\cdot\frac{1}{2}(9/4-1)-\frac{1}{2}\cdot\frac{1}{2}(4-9/4)-\frac{1}{2}(25/4-4)=-85/48$$
However, the answer to this problem says it should be $-157/48$. I don't understand what's wrong with my solution. I would appreciate it if anyone can help me out.
| Your answer is correct.
Another way to perform the calculation is to first observe as you did that $|x| = x$ for $x > 0$. Then the function $$f(x) = \frac{|x|}{\lfloor 2x-5 \rfloor} = \begin{cases} -x/3, & 1 \le x < 3/2 \\ -x/2, & 3/2 \le x < 2 \\ -x, & 2 \le x < 5/2 \end{cases}$$ on the interval of interest, and more generally, for each $k \in \mathbb Z$, $$f(x) = \begin{cases} -x/k, & \left(\frac{k+5}{2} \le x < \frac{k+6}{2}\right) \cap (k \le -6) \\ x/k & \left(\frac{k+5}{2} \le x < \frac{k+6}{2}\right) \cap (k \ge -5, k \ne 0) \\ \text{undefined} & 5/2 \le x < 3. \end{cases}$$ demonstrates that the integral is the sum of three trapezoidal areas, each with equal heights $$h = 5/2 - 2 = 2 - 3/2 = 3/2 - 1 = 1/2.$$ Therefore, the area is expressed as sum of the mean base length times the height of each trapezoid, or even more simply, the total mean base lengths times the height: $$\int_{x=1}^{5/2} f(x) \, dx = \frac{1}{2}\left(f(5/4) + f(7/4) + f(9/4)\right) = -\frac{1}{2}\left(\frac{5}{12} + \frac{7}{8} + \frac{9}{4} \right) = -\frac{85}{48}.$$
| {
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A Floor Function based limit $\large{ \lim_{n \to \infty} \dfrac{1}{n^2} \sum_{k=1}^{n-1} \ k \cdot \left \lfloor x + \dfrac{n-k-1}{n} \right \rfloor = \ ? }$
Find the value of the above limit upto 3 decimal places when $(x = \sqrt{2015})$.
Any suggestions on how to handle the floor function?
| By my comment above, some of the floor functions will evaluate to $\lfloor x\rfloor$, and some to $\lfloor x + 1\rfloor = \lfloor x\rfloor +1$. If we move that $+1$ out to a separate sum, we get
$$
\frac{1}{n^2} \sum_{k=1}^{n-1} \ k \cdot \left \lfloor x + \dfrac{n-k-1}{n} \right \rfloor = \frac{1}{n^2}\left( \sum_{k=1}^{n-1} \ k\lfloor x\rfloor + \sum_{k = 1}^{i_n} k\right)
$$
for some $i_n$. More specifically, $i_n$ is the largest integer that satisfies
$$
\frac{n-i_n-1}{n} \geq1-\{x\}\\
\frac{i_n+1}{n}\leq \{x\}
$$
where $\{x\} = x-\lfloor x\rfloor$ is the fractional part of $x$. In any case, it is roughly proportional to $n$ as $n$ grows, with the proportionality factor closing in on $\{x\}$. Let $a_n = i_n/n$ be this proportionality factor.
These two sums are arithmetic, which means we can calculate them as follows:
$$
\sum_{k=1}^{n-1} \ k\lfloor x\rfloor + \sum_{k = 1}^{i_n} k = \frac{n(n-1)}{2}\lfloor x\rfloor + \frac{i_n(i_n+1)}{2}\\
= \frac{n^2 - n}{2}\lfloor x\rfloor + \frac{a_n^2n^2+ a_nn}{2}
$$
Now, this is supposed to be multiplied by $1/n^2$, and this yields:
$$
\frac{1-1/n}{2}\lfloor x\rfloor + \frac{a_n^2 + a_n/n}{2}
$$
In this expression it is safe to let $n$ go to $\infty$, and we get
$$
\frac{\lfloor x\rfloor + \{x\}^2}{2}
$$
(remember that $\lim_{n \to \infty} a_n = \{x\}$). Lastly, we can insert $x = \sqrt{2015}$, which gives
$$
\frac{44 + (\sqrt{2015} - 44)^2}{2} = \frac{44 + 2015 - 88\sqrt{2015} + 1936}{2}\\
= \frac{3995 - 88\sqrt{2015}}{2}
$$
| {
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Finding positive integral solutions of $3^a=b^2+2$ I am trying to find integral solutions of the equation $3^a=b^2+2$. I could get that $a$ and $b$ have to be odd. But I am unable to get any further. I believe that $(a,b)=(1,1);(3,5)$ are the only solutions but I am unable to show that. If any one has any hints, it would be great. Thanks
| The quadratic field $\mathbb Q(\sqrt {-2})$ we will use below in solving this question, is a very good field: its ring of integers is $\mathbb Z[\sqrt {-2}]$ (because $-2\equiv 2 (mod\space 4)) $; its only units are $\pm1$ (because $-2<0$ and distinct of $-1$ and$-3$); it is a registered unique factorization domain and, more, an Euclidean domain with Euclidean function defined by $|N(x)|$ where N is the norm . On the other hand $-2$ is clearly a quadratic residue modulo 3 (because $-2=1$ in $\mathbb F_3$) so, according to the theory, 3 is decomposed into two primes $p_1p_2$ in the ring $\mathbb Z[\sqrt {-2}]$. We have in fact $3=(1+\sqrt{-2})(1-\sqrt {-2})$ so we have
$$3^a=b^2+2\iff 3^a=(1+\sqrt{-2})^a(1-\sqrt {-2})^a=(b+\sqrt{-2})(b-\sqrt{-2})$$
Suppose $x+y\sqrt{-2}$ is a common factor of $(b+\sqrt{-2})$ and $(b-\sqrt{-2})$ then it divides their difference $2\sqrt{-2}$ then, taking norms in the Euclidean domain $\mathbb Z[\sqrt {-2}]$, one has $x^2+2y^2$ divides $8$; hence $x^2+2y^2=1,2,4,8$ which respectively gives $(x,y)=(1,0),(0,1),(2,0),(0,2)$.
None of these give proper factors neither of $(b+\sqrt{-2})$ nor $(b-\sqrt{-2})$ hence $(b+\sqrt{-2})$ and $(b-\sqrt{-2})$ are coprime.
Thus, by unique factorization one has $(1+\sqrt{-2})^a = b\pm\sqrt{-2}$ from which
$A_a+B_a\sqrt{-2}= b\pm\sqrt{-2} \Rightarrow B_a=\pm1$
Now calculating $B_a$ for $a\ge1$ we get the only values of the exponent $a$ for which we have $B_a=\pm1$ are $1$ and $3$; in fact,
$a=1$ gives $(1+\sqrt{-2})=b+\sqrt{-2}$ hence the obvious solution $\boxed{(a,b)=(1,1)}$
$a=3$ gives $(1+\sqrt{-2})^3=1+3\sqrt{-2}+3(\sqrt{-2})^2+(\sqrt{-2})^3=-5+\sqrt{-2}$ hence the solution $\boxed{(a,b)=(3,5)}$ deduced from $3^3=27=b^2+2$.
The other values of $a$ are such that $|B_a|\gt1$ as one can see taking the odd powers in
$(1+\alpha)^a= 1+a\alpha+\binom a2\alpha^2+\cdot\cdot \cdot +\binom ak\alpha^k + \cdot \cdot\cdot +a{\alpha}^{a-1}+ {\alpha}^{a}+ $ with $\alpha=\sqrt{-2}$ so one has
$$B_a= a+\binom a3(-2)+\binom a5(-2)^2+\cdot\cdot $$ in which it is verified $|B_a|\gt1$ when $a\neq 1,3$.
Finally the only solutions are the two above boxed ones.
| {
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In how many ways can a selection be done of $5$ letters?
In how many ways can a selection be done of $5$ letters out of $5 A's, 4B's, 3C's, 2D's $ and $1 E$.
$ a) 60 \\
b) 75 \\
\color{green}{c) 71} \\
d.) \text{none of these} $
Number of ways of selecting $5$ different letters = $\dbinom{5}{5} = 1$ way
Number of ways to select $2$ similar and $3$ different letter = $\dbinom{4}{1}\times \dbinom{4}{3}=16$.
Number of ways of selecting $2$ similar + $2$ more similar letter and $1$ different letter = $\dbinom{4}{2}\times \dbinom{3}{1}=18$.
Number of ways to select $3$ similar and $2$ different letter = $\dbinom{4}{2}\times \dbinom{3}{1}=18$.
Number of ways to select $3$ similar and another $2$ other similar = $\dbinom{3}{1}\times \dbinom{3}{1}=9$
Number of ways to select $4$ similar and $1$ different letter = $\dbinom{2}{1}\times \dbinom{4}{1}=8$
Ways of selecting
$5$ similar letters = $1$
Total ways = $1+16+18+18+9+8+1= 71$
Well I have the solution But I am not able to fully understand it.
Or if their could be an $\color{red}{\text{alternate way}}$ than it would be great.
I have studied maths up to $12$th grade.
| Second line: You can get 2 of a type from A,B,C or D so $4\choose 1$. Then ${4\choose3}$ to select 3 different letters from the 4 types not yet selected, and finally, apply multiplication principle.
Second last line: 4 of a type only from A or B, so ${2\choose1}$, 1 more selection from the remaining 4 types, ${4\choose 1}$, and multiply.
You should be able to understand the remaining along similar lines.
| {
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If $a,b,c$ and $d$ are positive reals such that $a+b+c+d=1$.Find the minimum value of $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}$ If $a,b,c$ and $d$ are positive reals such that $a+b+c+d=1$.
Find the minimum value of $\displaystyle \frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}$.
I have proved it by using Lagrange's multiplier method.
I also want to solve it by Cauchy Schwartz inequality but could not.Please help me.
Lagrange's multiplier method:
Let $f\displaystyle \equiv \frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}$ and $\displaystyle g\equiv a+b+c+d-1=0$
Let $\displaystyle h\equiv \frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}-\lambda(a+b+c+d-1)$
$\displaystyle \frac{\partial h}{\partial a}=\frac{-1}{a^2}-\lambda=0$
$\displaystyle \frac{\partial h}{\partial b}=\frac{-1}{b^2}-\lambda=0$
$\displaystyle \frac{\partial h}{\partial c}=\frac{-1}{c^2}-\lambda=0$
$\displaystyle \frac{\partial h}{\partial d}=\frac{-1}{d^2}-\lambda=0$
$\therefore a=b=c=d$
As $\displaystyle a+b+c+d=1\therefore a=b=c=d=\frac{1}{4}\therefore\displaystyle \frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}=16$
Is my way and answer correct?
Please help me with Cauchy Schwartz inequality method.
| Given $a+b+c+d = 1$ and We have to calculate $\bf{Minimum}$ of $\displaystyle \left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}\right)$
where $a,b,c,d>0$
Using $\bf{A.M\geq H.M}$ Inequality
$$\displaystyle \frac{a+b+c+d}{4}\geq \frac{4}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}}\Rightarrow \left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}\right)\geq 16$$
And equality hold when $$a=b=c=d$$
Using $\bf{Cauchy -Schwartz}$ Inequality
$$\displaystyle \left[\left(\sqrt{a}\right)^2+\left(\sqrt{b}\right)^2+\left(\sqrt{c}\right)^2+\left(\sqrt{d}\right)^2\right]\cdot \left[\left(\frac{1}{\sqrt{a}}\right)^2+\left(\frac{1}{\sqrt{b}}\right)^2+\left(\frac{1}{\sqrt{c}}\right)^2+\left(\frac{1}{\sqrt{d}}\right)^2\right]\geq (1+1+1+1)^2$$
So We get $$\displaystyle \left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}\right)\geq 16$$
And equality hold when $$\frac{a}{1}=\frac{b}{1}=\frac{c}{1}=\frac{d}{1}$$
| {
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$2\sqrt n-2<1+\frac{1}{\sqrt2}+\frac{1}{\sqrt3}+....+\frac{1}{\sqrt n}<2\sqrt n-1$ If $n\in N$,then prove that $2\sqrt n-2<1+\frac{1}{\sqrt2}+\frac{1}{\sqrt3}+....+\frac{1}{\sqrt n}<2\sqrt n-1$
How should i prove this inequality,neither AM-HM nor any other inequality theorem is working here.
| This inequality comes in hand: $$ \sqrt{n+1} - \sqrt{n} < \frac{1}{2\sqrt{n}} < \sqrt{n} - \sqrt{n-1} $$
Part 1: $ \sqrt{n+1} - \sqrt{n} < \frac{1}{2\sqrt{n}} $
$$ \sqrt{n+1} - \sqrt{n} = (\sqrt{n+1} - \sqrt{n}) \times \frac{\sqrt{n+1} + \sqrt{n}}{\sqrt{n+1} + \sqrt{n}} = \frac{1}{\sqrt{n+1} + \sqrt{n}}$$
Since $\sqrt{n+1} > \sqrt{n} $, then $\, \sqrt{n+1} + \sqrt{n} > 2 \sqrt{n} \implies \frac{1}{\sqrt{n+1} + \sqrt{n}} < \frac{1}{2\sqrt{n}}$
So: $$ \sqrt{n+1} - \sqrt{n} < \frac{1}{2\sqrt{n}}$$
Part 2: $ \sqrt{n} - \sqrt{n-1} > \frac{1}{2\sqrt{n}}$
$$ \sqrt{n} - \sqrt{n-1} = (\sqrt{n} - \sqrt{n-1}) \times \frac{\sqrt{n} + \sqrt{n-1}}{\sqrt{n} + \sqrt{n-1}} = \frac{1}{\sqrt{n} + \sqrt{n-1}}$$
Since $\sqrt{n-1} < \sqrt{n} \implies \sqrt{n-1} + \sqrt{n} < 2 \sqrt{n} \implies \frac{1}{\sqrt{n} + \sqrt{n-1}} > \frac{1}{2\sqrt{n}} $
So: $$\sqrt{n} - \sqrt{n-1} > \frac{1}{2\sqrt{n}} $$
Now you have the following inequalities:
$$\sqrt{2} - \sqrt{1} < \frac{1}{2\sqrt{1}} < \sqrt{1} - \sqrt{0} $$
$$\sqrt{3} - \sqrt{2} < \frac{1}{2\sqrt{2}} < \sqrt{2} - \sqrt{1} $$
$$\sqrt{4} - \sqrt{3} < \frac{1}{2\sqrt{3}} < \sqrt{3} - \sqrt{2} $$
$$ \ldots $$
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Can I show $\frac{1}{n+1} + \frac{1}{n+2} + \cdots + \frac{1}{2n} \leq 1$ by induction? It is known that $a_{n+1} > a_n$. I tried to prove
$$\frac{1}{n+1} + \frac{1}{n+2} + \cdots + \frac{1}{2n} \leq 1$$
via induction.
Base case: $a_0 = \frac{1}{0+1} = 1 \leq 1$
Inductive step: assume $a_n \leq 1$, prove $a_{n+1} \leq 1$.
$a_n = \frac{1}{n+1} + \frac{1}{n+2} + \cdots + \frac{1}{2n} \leq 1$
$\frac{1}{n+2} + \frac{1}{n+3}+ \cdots + \frac{1}{2n} + \frac{1}{2n+1} + \frac{1}{2n+2} \leq 1-\frac{1}{n+1} + \frac{1}{2n+1} + \frac{1}{2n+2}$
But $-\frac{1}{n+1} + \frac{1}{2n+1} + \frac{1}{2n+2}$ is positive, so I'm not sure where to go from here.
Normally, the inductive step would involve a change on the right hand side, but in this case, it remains the same. Makes me wonder if induction is even the correct approach?
| The result is "obvious." There are $n$ terms in the sum, each $\le \frac{1}{n+1}$. So the sum is $\le \frac{n}{n+1}$, which is $\lt 1$.
If we, unreasonably, want to prove the result by induction, we run into the problem mentioned in the OP. The solution is to strengthen the induction hypothesis. We prove by induction that our sum is $\le \frac{n}{n+1}$.
So suppose we know that for a particular $k$ we have
$$\frac{1}{k+1}+\frac{1}{k+2}+\cdots +\frac{1}{2k}\le \frac{k}{k+1}.\tag{1}$$
We want to show that
$$\frac{1}{k+2}+\frac{1}{k+3}+\cdots +\frac{1}{2k}+\frac{1}{2k+1}+\frac{1}{2k+2}\le \frac{k+1}{k+2}.\tag{2}$$
Now if we use the induction hypothesis (1) and play a bit with inequalities, we can prove (2).
| {
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Euler's constant greater than 0 for all values of n? If Euler's constant is described as the limit as n approaches infinity of the following:
$$t_n = 1 + \frac 12 + \frac 13 \cdots + \frac1n -\ln(n)$$
How can one prove that $t_n$ is greater than $0$ for all values of $n$?
Thanks!
| First, put $s_n = 1+\dfrac{1}{2}+\cdots + \dfrac{1}{n}$, then
$1+\dfrac{1}{2}+\cdots + \dfrac{1}{n} - \ln n > 0$ because:
$1 = \displaystyle \int_{1}^2 1dx \geq \displaystyle \int_{1}^2 \dfrac{1}{x}dx$,
$\dfrac{1}{2} = \displaystyle \int_{2}^3 \dfrac{1}{2}dx \geq \displaystyle \int_{2}^3 \dfrac{1}{x}dx$, and
$.....$
$\dfrac{1}{n-1} = \displaystyle \int_{n-1}^n \dfrac{1}{n-1}dx\geq \displaystyle \int_{n-1}^n \dfrac{1}{x}dx$, and add these integrals we have:
$s_n = s_{n-1} + \dfrac{1}{n} \geq \displaystyle \int_{1}^n \dfrac{1}{x}dx + \dfrac{1}{n}=\ln n + \dfrac{1}{n} > \ln n\Rightarrow t_n = s_n - \ln n > 0$
| {
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Which one is greater $\left(5/2\right)^{2/5}$ or $\left(7/2\right)^{2/7}$? I have a question here:
Which of $\left(5/2\right)^{2/5}$ and $\left(7/2\right)^{2/7}$ is greater?
I tried comparison by the function $y=x^{1/x}$ and found the derivative as follows
$$\frac{\partial y}{\partial x}=\frac{1}{x}(x^{\frac{1}{x}-1})+x^{1/x}\ln x=x^{1/x}(1/x^2+\ln x)$$
I got stuck here, what to do next?
My teacher told me the answer is $\left(5/2\right)^{2/5}$. I want to know how it is.
| It is enough to notice that:
$$ f(x) = \log\left(x^{\frac{1}{x}}\right) = \frac{\log x}{x} \tag{1}$$
fulfills:
$$ f'(x) = \frac{1-\log x}{x^2},\qquad f''(x) = \frac{2\log x-3}{x^3} \tag{2}$$
hence $f(x)$ is a concave function over the interval $I=\left[\frac{5}{2},\frac{7}{2}\right]$, increasing from $\frac{5}{2}$ to $e$ and decreasing from $e$ to $\frac{7}{2}$. Since the value of $f'''(x)=\frac{11-6\log x}{x^4}$ on $I$ is positive and quite small, the graph of $f(x)$ over $I$ is essentially the graph of a concave-down parabola with vertex at $x=e$. Since $\frac{5}{2}$ is the endpoint of $I$ closer to $e$,
$$ f\left(\frac{5}{2}\right) > f\left(\frac{7}{2}\right).\tag{3}$$
To prove it in a more formal way, compute the second-order Taylor expansion of $f(x)$ centered at $x=e$ and use Lagrange's remainder:
$$f(x) = f(e)+f'(e)(x-e)+\frac{f''(e)}{2}(x-e)^2+\frac{f'''(\xi)}{6}(x-e)^3,\qquad \xi\in I.\tag{4}$$
That gives $f(5/2)>\frac{73}{200}>f(7/2)$ and we are done.
An alternative to prove $5^7>4\cdot 7^5$ (that is equivalent to the previous claim) is to prove:
$$ \left(\frac{1+\frac{1}{6}}{1-\frac{1}{6}}\right)^6<\frac{35}{4}\tag{5}$$
by exploiting the Taylor series of the function $g(x)=\left(\frac{1+x}{1-x}\right)^6$:
$$ g(x) = 1+\sum_{n\geq 1}\frac{4n}{15}(23+20n^2+2n^4)\,x^n \tag{6} $$
and proving the inequality:
$$ \forall x\in(0,1),\qquad \left(\frac{1+\frac{x}{6}}{1-\frac{x}{6}}\right)^6\leq 1+2x+2x^2+3x^3\tag{7}$$
from which:
$$ \left(\frac{1+\frac{1}{6}}{1-\frac{1}{6}}\right)^6\leq 8.\tag{8}$$
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Unique Solution to a System of Equivalences
Show that $(2, 3, 7)$ is the only solution set to the following system of equivalences:
\begin{align}
ab = -1 & \mod c \\
ac = -1 & \mod b \\
bc = -1 & \mod a \\
\end{align}
I already attempted systems of equations like $ab + cm = -1$, $ac + bn = -1 \ldots$ but each time I was just running in circles. So far, I know that $a, b, c$ are pairwise relatively prime. Moreover, I must have at least two odd numbers. Otherwise, the two even numbers are not relatively prime. What other facts can I use in this question?
| I suppose you don't count solutions like $(1,1,1)$ although they are technically solutions to the system. So for now assume w.l.o.g. $1<a<b<c$.
The equations are equivalent to $c \mid ab+1,b \mid ac+1,a \mid bc+1$.
Since (as you stated) $a,b,c$ clearly must be pairwise relatively prime, we conclude that
$$abc \mid (ab+1)(ac+1)(bc+1)=a^2b^2c^2+(abc)(a+b+c)+ab+ac+bc+1$$ and hence $abc \mid ab+ac+bc+1$.
Now, let $$\frac{ab+ac+bc+1}{abc}=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{abc}=n \in \mathbb{N}$$
Then $n <\frac{4}{a} \le \frac{4}{2}=2$ and hence $n=1$ i.e. $ab+ac+bc+1=abc$.
Also $a<4$ and hence $a=2$ or $a=3$.
If $a=3$ you have $b \ge 4$ and $c \ge 5$ and hence $$1=n \le \frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{60}<1$$ Absurd!
Hence we must have $a=2$.
Then it follows that $bc+1+2b+2c=2bc$ i.e. $bc-2b-2c-1=0$ or $(b-2)(c-2)=5$.
Hence $b-2=1$ and $c-2=5$ which gives exactly the solution $(a,b,c)=(2,3,7)$ you stated above...
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1438530",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Solving the nonlinear Diophantine equation $x^2-3x=2y^2$ How can I solve (find all the solutions) the nonlinear Diophantine equation $x^2-3x=2y^2$?
I included here what I had done so far. Thanks for your help.
Note: The equation above can be rewritten into $x^2-3x-2y^2=0$ which is quadratic in $x$. By quadratic formula we have the following solutions for $x$.
\begin{equation}
x=\frac{3\pm\sqrt{9+8y^2}}{2}
\end{equation}
I want $x$ to be a positive integer so I will just consider:
\begin{equation}
x=\frac{3+\sqrt{9+8y^2}}{2}
\end{equation}
From here I don't know how to proceed but after trying out values for $1\leq y\leq 1000$ I only have the following $y$ that yields a positive integer $x$.
\begin{equation}
y=\{0,3,18,105,612\}.
\end{equation}
Again, thanks for any help.
| Multiply through by $4$ and complete the square, $(2x-3)^2 - 8 y^2 = 9,$ so that $(2x-3)^2 + y^2 \equiv 0 \pmod 3.$ It follows that both $(2x-3)$ and $y$ are divisible by $3,$ therefore $x$ as well. Write $x=3t, y=3v.$ So far $(2t-1)^2 - 8 v^2 = 1.$ Let $u=2t-1,$ so $u^2 - 8 v^2 = 1.$
...if we write $$ x = \frac{3(1+u)}{2} $$ and $$ y = 3v, $$
we get $$ u^2 - 8 v^2 = 1. $$ This gives a sequence of pairs, $u=1,3,17,99,...$ and $v=0,1,6,35...$
see https://oeis.org/A001541 and https://oeis.org/A001109
These obey
$$ u_{n+1} = 3 u_n + 8 v_n, $$
$$ v_{n+1} = u_n + 3 v_n. $$ By Cayley-Hamiton
$$ u_{n+2} = 6 u_{n+1} - u_n, $$
$$ v_{n+2} = 6 v_{n+1} - v_n. $$ Then
$$ x_{n+2} = 6 x_{n+1} - x_n - 6, $$
$$ y_{n+2} = 6 y_{n+1} - y_n. $$ The (non-negative) $(x,y)$ pairs begin
$$ (3,0) $$
$$ (6,3) $$
$$ (27,18) $$
$$ (150,105) $$
$$ (867,612) $$
$$ (5046,3567) $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1439519",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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} |
Troubles with solving $\sqrt{2x+3}-\sqrt{x-10}=4$ I have been trying to solve the problem $\sqrt{2x+3}-\sqrt{x-10}=4$ and I have had tons problems of with it and have been unable to solve it. Here is what I have tried-$$\sqrt{2x+3}-\sqrt{x-10}=4$$ is the same as $$\sqrt{2x+3}=4+\sqrt{x-10}$$ from here I would square both sides $$(\sqrt{2x+3})^2=(4+\sqrt{x-10})^2$$
which simplifies to $$2x+3=16+x-10+8\sqrt{x-10}$$ I would then isolate the radical $$x-3=8\sqrt{x-10}$$ then square both sides once again $$(x-3)^2=(8\sqrt{x-10})^2$$ which simplifies to $$x^2-6x+9=8(x-10)$$ simplified again $$x^2-6x+9=8x-80$$ simplified once again $$x^2-14x+89=0$$ this is where I know I have done something wrong because the solution would be $$14 \pm\sqrt{-163 \over2}$$ I am really confused and any help would be appreciated
| Alternatively, take the reciprocal to get:
$$\frac{1}{\sqrt{2x + 3} - \sqrt{x - 10}} = \frac{\sqrt{2x + 3} + \sqrt{x - 10}}{(2x+3)-(x-10)} = \frac{\sqrt{2x + 3} + \sqrt{x - 10}}{x+13}$$
which we know is equal to $\frac 1 4$.
Thus we have the following:
$$\sqrt{2x+3} - \sqrt{x+10} = 4 \tag{1}$$
$$\sqrt{2x + 3} + \sqrt{x - 10} = \frac{1}{4}(x+13) \tag{2}$$
Add the two together to get:
$$2 \sqrt{2x + 3} = 4 + \frac{1}{4}(x + 13)$$
$$8\sqrt{2x+3} = 16 + (x + 13)$$
$$64(2x + 3) = x^2 + 58x + 841$$
$$x^2-70x+649=0$$
$$(x - 11)(x - 59) = 0 \implies x = \boxed{11, 59}.$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
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Polynomial Division by product of two polynomials f(x) is a polynomial, when it is divided by (x-3) it leaves remainder 15. when f(x) is divided by square of(x-1) it leaves remainder 2x+1. Find the remainder when f(x) is divided by product of two above divisors.
| HINT:
Let $f(x)=g(x)(x-3)(x-1)^2+Ax^2+Bx+C$
$15=f(3)=0+3^2A+3B+C$
$2x+1=A\{x^2-(x-1)^2\}+Bx+C$
$\iff2x+1=x(B+2A)+C-A$
Compare the constants & the coefficients of $x$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1441782",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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$A$ is a unitary ring with $xy=1$ implies $yx=1$, prove that if $a^2=3b^2$ and $3ab=1+4ba$, then $ab=ba$ Let $(A,+, \cdot)$ be a unitary ring with the property that if $xy=1$ for $x,y \in A$, then $yx=1$. Let $a,b \in A$ with $a^2=3b^2$ and $3ab=1+4ba$. Prove that $ab=ba$.
We easily get that $(a-2b)(2a+3b)=1$, so, by hypothesis, we have that $(2a+3b)(a-2b)=1$. From here we get that $3ba=1+4ab$.
Then, $3ab-4ba=1=3ba-4ab$, and so $7ab=7ba$.
Now, $9ab=3+12ba$ and $12ba=4+16ab$, so $9ab=3+4+16ab$, thus $7ab=-7$.
Of course, we have $7ba=-7$ too.
Next, from $3ab=1+4ba$ and $3ba=1+4ab$ we get that $1+ba=3ab-3ba=-1-ab$, so $ab+ba=-2$.
We can go further and write that $aba+ba^2=-2a$ and $a^2b+aba=-2a$.
We get that $a^2b=ba^2$ and, analogously $ab^2=b^2a$.
Now, we can easily prove that $ab=6ba+5$, $ba=6ab+5$, $2ab=5ba+3$ and $2ba=5ab+3$. Here I'm stuck.
| Counterexample: $A = \mathbb{F}_7^{2\times 2}$ (the $2\times 2$ matrix ring over $\mathbb{F}_7$), $a = \left(\begin{matrix} 0 & 1 \\ 0 & 0 \end{matrix}\right)$, $b = \left(\begin{matrix} 0 & 0 \\ 5 & 0 \end{matrix}\right)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1442392",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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(Beginner) Intuition to solve a functional equation and steps for this particular- Please can someone please tell me the intuition behind solving a functional equation?
For example, $$f(x)+f(y)=2f\left(\frac{x+y}2\right)f'\left(\frac{x-y}2\right)$$
Now at the first look, it seems like $f(x)=\sin x$, but how to solve it rigorously?
P.S: As written in the title, I am a beginner and I have no experience at solving a functional equation. I know little bit about series expansions and I know calculus upto 12th grade. (Derivatives, Integrals (Impropers-a bit), Limits)
| Taking $x=y$ gives
$$ 2f(x) = 2f(x)f'(0), \tag{1} $$
so either $f(x) \equiv 0$ or $f'(0)=1$. Suppose it is the latter (else we have found $f$). Setting $x=-y$ gives
$$ f(x)+f(-x) = 2f(0)f'(x). \tag{2} $$
Differentiating the original equation with respect to $x$ gives
$$ f'(x) = f'\left( \frac{x+y}{2} \right)f'\left( \frac{x-y}{2} \right) + f\left( \frac{x+y}{2} \right)f''\left( \frac{x-y}{2} \right), \tag{3} $$
and setting $y=-x$ gives
$$ f'(x) = f'(0)f'(x)+f(0)f''(x) = f'(x)+f(0)f''(x). $$
This says that either $f''(x) \equiv 0$ or $f(0)=0$. It is easy to check that the only functions with $f''(x) \equiv 0$ which satisfy the equation are $f(x) =0$ and $f(x)=x$, so now assume $f(0)=0$. (2) then gives that
$$ f(-x) = -f(x), $$
so we also have $f''(0)=0$. Differentiating (3) with respect to $y$ gives
$$ 0 = f''\left( \frac{x+y}{2} \right)f'\left( \frac{x-y}{2} \right) - f'\left( \frac{x+y}{2} \right)f''\left( \frac{x-y}{2} \right) + f\left( \frac{x+y}{2} \right)f''\left( \frac{x-y}{2} \right) - f\left( \frac{x+y}{2} \right)f'''\left( \frac{x-y}{2} \right) \\
= f''\left( \frac{x+y}{2} \right)f'\left( \frac{x-y}{2} \right) - f\left( \frac{x+y}{2} \right)f'''\left( \frac{x-y}{2} \right). $$
Now, if we take $y=x$, this reduces to
$$ f'''(0) f(x) = f''(x) f'(0) = f''(x). $$
At this point, we can write down the general solution to this equation: setting $f'''(0)=A^2$ gives
$$ f(x) = B\sinh{Ax}+C\cosh{Ax}, $$
and imposing the boundary conditions $f(0)=0$, $f'(0)=1$ gives general solution
$$ f(x) = \frac{1}{A}\sinh{Ax}. $$
It is simple to check that this always satisfies the functional equation, and further, you can recover the $f(x)=x$ solution by taking the limit as $A \to 0$.
(Remark: this does include the sine solution you expected, since
$$ \sin{ax} = \frac{\sinh{iax}}{i}, $$
so taking $A=ia$ gives the sine solution.
Therefore we have shown
If $f(x)$ is thrice-differentiable and satisfies the functional equation
$$ f(x)+f(y) = 2f\left( \frac{x+y}{2} \right)f'\left( \frac{x-y}{2} \right), $$
then $f(x)$ is one of
*
*$0$
*$x$
*$A^{-1}\sinh{Ax}$, some $A \in \mathbb{C}$.
EDIT:
In fact, we only need twice-differentiability, because setting $x=0$,$y=2z$ in (3) gives
$$ 1 = (f'(z))^2-f(z)f''(z), $$
which is a thoroughly rotten-looking DE, but we can solve it by noticing that
$$ \left(\frac{f'}{f}\right)' = \frac{ff''-f'^2}{f^2}, $$
so an equivalent is
$$ -\frac{1}{f^2} = \left(\frac{f'}{f}\right)' $$
Multiply by $2f'/f$:
$$ -2\frac{f'}{f^3} = 2\left(\frac{f'}{f}\right)\left(\frac{f'}{f}\right)' $$
Integrating this once, we have
$$ a^2+\frac{1}{f^2} = \frac{f'^2}{f^2}, $$
and rearranging this gives
$$ 1 = \frac{f'^2}{a^2 f^2+1} $$
Taking the positive square root (since $f'(0)=1$) and integrating,
$$ z-0 = \int_0^{f(z)} \frac{dt}{\sqrt{1+a^2 t^2}} $$
Now, this is easy to integrate by setting $u=a^{-1}\sinh{t}$: then the integral just simplifies to $a^{-1}\int_0^{\arg\sinh{(af(z))}} du =\arg\sinh{(af(z))}, $
and hence $f(z)=\frac{1}{a}\sinh{az}$ as before. (And $a=0$ gives the $f(x)=x$ solution.)
| {
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"url": "https://math.stackexchange.com/questions/1444918",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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GCSE level trigonometry question In the UK we do GCSE exams and this is one of the GCSE questions I can't solve at this time of night lol. First part is easy but I can't see a way around part B. Any help would be much appreciated.
http://s17.postimg.org/8me1qdbkf/Screen_Shot_2015_09_21_at_21_28_51.png
http://postimg.org/image/9oo88wudn/
| $\angle DBC = 45^\circ$ and $\angle BCD = \arctan \dfrac{18} 6 $. Therefore $\angle BDC = 180^\circ - 45^\circ - \arctan\dfrac{18} 6$.
So then you have the three angles of $\triangle BDC$ and you have the length of one of the sides, so you can use the law of sines.
$$\sin x = \sin \angle BCD = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{18}{\sqrt{18^2+6^2}} = \frac{3}{\sqrt{3^2 + 1^2}}. $$
$$\sin \angle DBC = \frac{\sqrt 2} 2$$
\begin{align}
\sin \angle BDC & = \sin\left( 180^\circ - 45^\circ - \arcsin\frac 3 {\sqrt{10}} \right) \\[10pt]
& = \sin\left( 45^\circ + \arcsin\frac 3 {\sqrt{10}} \right) \\[10pt]
& = \sin 45^\circ \cos\arcsin\frac 3 {\sqrt{10}} + \cos 45^\circ \sin\arcsin\frac 3 {\sqrt{10}} \\[10pt]
& = \frac{\sqrt 2}2 \cdot \frac 1 {\sqrt{10}} + \frac{\sqrt 2} 2\cdot\frac 3 {\sqrt{10}} = \frac 1 {2\sqrt 5} + \frac 3 {2\sqrt 5} \\[10pt]
& = \frac 2 {\sqrt 5}.
\end{align}
The law of sines then tells us that
$$
\frac{6\text{ cm}}{2/\sqrt 5} = \frac{BD}{3/\sqrt{10}}.
$$
Dividing both sides by $\sqrt 5$, we get
$$
\frac{6\text{ cm}} 2 = \frac{BD\cdot\sqrt 2} 3.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1445581",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
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How many pairs of positive integers $(x,y)$ satisfy the equation $x^2 - 10! = y^2$? I have a question about number theory:
How many pairs of positive integers $(x,y)$ satisfy the following equation? $$x^2 - 10! = y^2$$
My attempt:
Move the $y^2$ from right to the left and 10! From left to the right such that:
$$x^2-y^2= 10!$$
$$(x-y)(x+y)=10!$$
$$(x-y)(x+y)= 10.9.8.7.6.5.4.3.2$$
Until this step, I think I will get many possibility to get the answer, such that:
$$(x-y)= 10.9$$ and $$(x+y)= 8!$$
Also
$$(x-y)= 10.9.8$$ and $$(x+y)= 7!$$
And another possibilities, like $$(x-y)=10.9.8.7$$ and $$(x+y)=6!$$
And so on until $$(x-y)=10.9.8.7.6.5.4.3.2$$ and $$(x+y)=1$$
And I think solving all the possibility is a bit tedious work. Can somebody help me to find a better solution to solve this kind of problem?
Thanks
| Write
$10!
= 10.9.8.7.6.5.4.3.2
=2^83^45^27
$.
You want
$(x+y)(x-y)=10!=ab$
with
$x+y=a$ and $x-y=b$.
Then
$x = (a+b)/2$ and
$y = (a-b)/2$.
So $a$ and $b$
have to have the same parity.
Then choose $2^a3^b5^c7^d$
with all possible choices
satisfying the above.
Since they have to
have the same parity,
$a$ must have
$2^n$ where
$1 \le n \le 6$.
This is a start.
| {
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"url": "https://math.stackexchange.com/questions/1446072",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
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Solving $\lim _{x\to 1}\left(\frac{1-\sqrt[3]{4-3x}}{x-1}\right)$ $$\lim _{x\to 1}\left(\frac{1-\sqrt[3]{4-3x}}{x-1}\right)$$
So
$$\frac{1-\sqrt[3]{4-3x}}{x-1} \cdot \frac{1+\sqrt[3]{4-3x}}{1+\sqrt[3]{4-3x}}$$
Then
$$\frac{1-(4-3x)}{(x-1)(1+\sqrt[3]{4-3x})}$$
That's
$$\frac{3\cdot \color{red}{(x-1)}}{\color{red}{(x-1)}(1+\sqrt[3]{4-3x})}$$
Finally
$$\frac{3}{(1+\sqrt[3]{4-3x})}$$
But this evaluates to
$$\frac{3}{2}$$
When the answer should be
$$1$$
Where did I fail?
| If you let $t=4-3x,\;$ so $x=\frac{1}{3}(4-t)$, you get
$\displaystyle\lim_{x\to 1}\frac{1-\sqrt[3]{4-3x}}{x-1}=\lim_{t\to1}\frac{1-\sqrt[3]{t}}{\frac{1}{3}(1-t)}=\lim_{t\to1}\frac{3(1-t^{1/3})}{(1-t^{1/3})(1+t^{1/3}+t^{2/3})}=\lim_{t\to1}\frac{3}{1+t^{1/3}+t^{2/3}}=1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1447217",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 5
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Find the determinant of $n\times n$ matrix Suppose, $ M=\begin{bmatrix}\begin{array}{ccccccc}
-x & a_2&a_3&a_4&\cdots &a_n\\
a_{1} & -x & a_3&a_4&\cdots &a_n\\
a_1&a_{2} & -x &a_4&\cdots &a_n\\
\vdots&\vdots&\vdots&\vdots &\ddots&\vdots\\
a_1&a_{2} & a_3&a_4&\cdots & -x\\
\end{array}\end{bmatrix}$, then how to find the $\det (M)$?
Proof: First I started by taking the $a_i$ from each $i$-th columns, then$ |M|=\prod_{i=1}^{n}{a_i} \begin{vmatrix}\begin{array}{ccccccc}
\frac{-x}{a_1} & 1&1&1&\cdots &1\\
{1} & \frac{-x}{a_2} & 1&1&\cdots &1\\
1&1 & \frac{-x}{a_3} &1&\cdots &1\\
\vdots&\vdots&\vdots&\vdots &\ddots&\vdots\\
1&1 & 1&1&\cdots & \frac{-x}{a_n}\\
\end{array}\end{vmatrix}$. After this is there any easiest way to find the determinant.
| Subtract the first row from each of the other rows. Most of the terms are now zero, and you can expand across the first row. Each product misses one of the $(x+a_i)$ factors, replaced by $a_i$. So the determinant is
$$(-1)^n\prod_i(x+a_i)\left[\frac x{x+a_1}-\frac {a_2}{x+a_2}-\frac {a_3}{x+a_3}...\right]\\
=(-1)^n\prod_i(x+a_i)\left[1-\sum_i\frac{a_i}{x+a_i}\right]$$
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "1",
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Applic. Of derivatives The straight line $x\cos \alpha + y \sin \alpha = p$ touches the curve $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$. Prove that $a^{2}\cos 2\alpha+b^{2} \sin 2\alpha=p^{2}$
| Notice, solving the given equations by setting $y=\frac{p-x\cos \alpha}{\sin\alpha}$ in the equation of the curve: $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$, we get $$\frac{x^2}{a^2}+\frac{\left(\frac{p-x\cos \alpha}{\sin\alpha}\right)^2}{b^2}=1$$ $$b^2x^2\sin^2\alpha+a^2(p-x\cos \alpha)^2=a^2b^2\sin^2\alpha $$
$$(a^2\cos^2\alpha+b^2\sin^2\alpha)x^2-2(a^2p\cos\alpha)x+(a^2p^2-a^2b^2\sin^2\alpha)=0$$
Since, the straight line touches the curve (ellipse), hence the roots of the above quadratic equation should be equal hence the discriminant $\Delta=B^2-4AC=0$
$$(-2a^2p\cos\alpha)^2-4(a^2\cos^2\alpha+b^2\sin^2\alpha)(a^2p^2-a^2b^2\sin^2\alpha)=0$$
$$a^4b^2\sin^2\alpha\cos^2\alpha+a^2b^4\sin^4\alpha-a^2b^2p^2\sin^2\alpha=0$$
Dividing the equation by $a^2b^2\sin^2\alpha$, we get
$$a^2\cos^2\alpha+b^2\sin^2\alpha-p^2=0$$ $$\color{red}{a^2\cos^2\alpha+b^2\sin^2\alpha=p^2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1449645",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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Is it possible to describe the Collatz function in one formula? This is related to Collatz sequence, which is that
$$C(n) = \begin{cases} n/2 &\text{if } n \equiv 0 \pmod{2}\\ 3n+1 & \text{if } n\equiv 1 \pmod{2} .\end{cases}$$
Is it possible to describe the Collatz function in one formula? (without modular conditions)
| Let f be the iteration function for every even number and g for every uneven number.
\begin{align}
f(n) = \frac{n}{2}
\end{align}
\begin{align}
g(n) = 3n + 1
\end{align}
It is possible to use the right equation for the right number by using a sine function with a period of 2. This way, even numbers get iterated by f and uneven numbers by g. Let c be the sine function.
\begin{align}
c(n) = a\cdot cos(\frac{2\pi}{d}\cdot n) + b
\end{align}
, where
\begin{align}
d = 2
\end{align}
and a is the difference between both functions divided by 2.
\begin{align}
a = \frac{f(n)-g(n)}{2}
\end{align}
'b' is the average of both functions.
\begin{align}
b = \frac{f(n)+g(n)}{2}
\end{align}
Now, time to rewrite c.
\begin{align}
c(n) = \frac{f(n)-g(n)}{2}\cdot cos(\pi n) + \frac{f(n)+g(n)}{2}
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1449874",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "39",
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"answer_id": 3
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Partial Fraction Decomposition trouble $\frac{1}{(1+x^2)(1+x^{2015})}$ I wish to decompose $\frac{1}{(1+x^2)(1+x^{2015})}$
I have $1 = (Ax+B)(1+x^{2015}) + (Cx+D)(1+x^2) = Ax^{2016} + Bx^{2015} + Cx^{3} + Dx^{2} + x(A+C) + B+D$
Doesn't this imply that $B+D = 1$, but we also have that $B=D = 0$ as well.
What did I do wrong?
I am examining the integral $\int_{0}^{\infty} \frac{dx}{(1+x^2)(1+x^{2015})}$
| There is a trick:
$$\begin{eqnarray*}\int_{0}^{+\infty}\frac{dx}{(1+x^2)(1+x^m)}&=&\int_{0}^{1}\frac{dx}{(1+x^2)(1+x^m)}+\int_{1}^{+\infty}\frac{dx}{(1+x^2)(1+x^m)}\\&=&\int_{0}^{1}\frac{dx}{1+x^2}\left(\frac{1}{1+x^m}+\frac{1}{1+x^{-m}}\right)\\&=&\int_{0}^{1}\frac{dx}{1+x^2}=\arctan(1)=\color{red}{\frac{\pi}{4}}\end{eqnarray*}$$
no matter what $m$ is.
Explanation: split the integration range into $(0,1)\cup (1,+\infty)$; use the change of variable $x\mapsto\frac{1}{x}$ on the unbounded piece and recombine the two pieces in a single integral over $(0,1)$; simplify, simplify more.
| {
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"url": "https://math.stackexchange.com/questions/1449951",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove that $2^n+(-1)^{n+1}$ is divisible by 3. Prove that $2^n+(-1)^{n+1}$ is divisible by 3 for $n\in\mathbb{N}$.
My attempt:
For $n=1$:
$2^1+(-1)^2 = 2 + 1 = 3, 3 |3$
We assume that $3|(2^n+(-1)^{n+1})$
Then for $n+1$:
$2^{n+1} + (-1)^{n+2} = 2\cdot 2^n -(-1)^{n+1}$
I am trying to show that for $n+1$, the expression is a constant multiple of the case for $n$ plus or minus a multiple of $3$. However, the minus sine I pulled out of the alternating term is tripping me up. Am I going about this the right way?
| If $n=1$, then $2^{n} + (-1)^{n+1} = 3$, which is divisible by $3$; if $n \geq 1$ such that $2^{n} + (-1)^{n+1} = 3q$ for some integer $q \geq 1$, then
$$
2^{n+1} + (-1)^{n+2} = 2[ 3q - (-1)^{n+1} ] + (-1)^{n+2} = 6q + 2(-1)^{n+2} + (-1)^{n+2} = 6q + 3(-1)^{n+2},
$$
which is again divislble by $3$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1450472",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Solving $\lim_{x\to0}\frac{1-\sqrt{\cos x}}{x^2}$ I've been trying to solve this over and over without L'Hopital but keep on failing:
$$\lim_{x\to0}\frac{1-\sqrt{\cos x}}{x^2}$$
My first attempt involved rationalizing:
$$\frac{1-\sqrt{\cos x}}{x^2} \cdot \frac{1+\sqrt{\cos x}}{1+\sqrt{\cos x}} = \frac{1-\cos x}{x\cdot x \cdot (1+\sqrt{\cos x})}$$
Using the rule $\frac{1-\cos x}{x} = 0$ for $x\to0$ is useless because we would end up with
$$\frac{0}{0\cdot x \cdot \sqrt{\cos x}} = \frac{0}{0}$$
But hey, perhaps we can rationalize again?
$$\frac{1-\cos x}{(x^2+x^2\sqrt{\cos x})}\cdot \frac{(x^2-x^2\sqrt{\cos x})}{(x^2-x^2\sqrt{\cos x})}$$
Resulting in
$$\frac{(1-\cos x)(x^2-x^2\sqrt{\cos x})}{x^4 - x^4\cdot\cos x} = \frac{(1-\cos x)(x^2-x^2\sqrt{\cos x})}{x^4 \cdot (1-\cos x)}$$
Cancelling
$$\frac{(x^2-x^2\sqrt{\cos x})}{x^4} = \frac{1-\sqrt{\cos x}}{x^2}$$
Well that was hilarious. I ended up at the beginning! Dammit.
My second attempt was to use the definition $\cos x = 1 - 2\sin \frac{x}{2}$:
$$\lim_{x\to0}\frac{1-\sqrt{\cos x}}{x^2} = \frac{1-\sqrt{1 - 2\sin \frac{x}{2}}}{x^2}$$
And then rationalize
$$\frac{1-\sqrt{1 - 2\sin \frac{x}{2}}}{x^2} \cdot \frac{1+\sqrt{1 - 2\sin \frac{x}{2}}}{1+\sqrt{1 - 2\sin \frac{x}{2}}}$$
$$\frac{1-(1 - 2\sin \frac{x}{2})}{x^2+x^2\sqrt{1 - \sin \frac{x}{2}}} = \frac{- 2\sin \frac{x}{2}}{x^2\left(1+\sqrt{1 - \sin \frac{x}{2}}\right)}$$
I want to make use of the fact that $\frac{\sin x}{x} = 1$ for $x\to0$, so I will multiply both the numerator and denominator with $\frac{1}{2}$:
$$\frac{-\sin \frac{x}{2}}{\frac{x}{2}\cdot x\left(1+\sqrt{1 - \sin \frac{x}{2}}\right)}$$
Then
$$\frac{-1}{x\left(1+\sqrt{1 - \sin \frac{x}{2}}\right)}$$
Well clearly that's not gonna work either. I will still get $0$ in the denominator.
The correct answer is $\frac{1}{4}$. I can kind of see why is the numerator $1$, but no idea where is that $4$ going to come out of.
I don't know how am I supposed to solve this without L'Hopital.
| HINT: $$\frac{1-\sqrt{\cos(x)}}{x^2}=\frac{1-\cos(x)}{x^2(1+\sqrt{\cos(x)})}=\frac{1-\cos(x)^2}{x^2(1+\sqrt{\cos(x)})(1+\cos(x))}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1453288",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Show that $\arctan\frac{1}{2}+\arctan\frac{1}{3}=\frac{\pi}{4}$
Show that $\arctan\frac{1}{2}+\arctan\frac{1}{3}=\frac{\pi}{4}$.
Attempt:
I've tried proving it but it's not equating to $\frac{\pi}{4}$. Please someone should help try to prove it. Is anything wrong with the equation? If there is, please let me know.
| Sine addition identity: $$\sin(\alpha + \beta) = \sin \alpha \cos \beta + \cos \alpha \sin \beta.$$
Cosine addition identity: $$\cos(\alpha + \beta) = \cos \alpha \cos \beta - \sin \alpha \sin \beta.$$
From the above, we obtain the tangent addition identity: $$\begin{align*} \tan (\alpha + \beta) &= \frac{\sin (\alpha + \beta)}{\cos (\alpha + \beta)} \\ &= \frac{\sin \alpha \cos \beta + \cos \alpha \sin \beta}{\cos \alpha \cos \beta - \sin \alpha \sin \beta} \\ &= \frac{\tan \alpha + \tan \beta}{1 - \tan \alpha \tan \beta}. \end{align*}$$
From this, we now let $x = \tan \alpha$, $y = \tan \beta$, or equivalently, $\alpha = \tan^{-1} x$, $\beta = \tan^{-1} y$, and substitute: $$\tan(\tan^{-1} x + \tan^{-1} y) = \frac{x+y}{1-xy}.$$ Now taking the inverse tangent of both sides, we obtain the inverse tangent identity: $$\tan^{-1} x + \tan^{-1} y = \tan^{-1} \frac{x+y}{1-xy}.$$ Now let $x = 1/2$, $y = 1/3$, and simplify the right hand side.
| {
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"timestamp": "2023-03-29T00:00:00",
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$\int_{c}^{c+2L} cos(\frac{2\pi nx}{L})$ Wolfram Alpha gives me this result but when doing I dont get it $\int_{c}^{c+2L} cos(\frac{2\pi nx}{L}) = \frac{L sin(2\pi n) cos(\frac{2 \pi n(c+L)}{L})}{\pi n}$ dx with $n = 0,1,2,...$
I put this integral in Wolfram Alpha and gives me this result, but when I try to do it at hand I arrive to other result. Is Wolfram Alpha wrong?
| Upon integrating, we have for $n\ne 0$, the term
$$\begin{align}
\sin \left(\frac{2\pi n(c+2L)}{L}\right)-\sin \left(\frac{2\pi nc}{L}\right)&=\sin \left(\frac{2\pi nc}{L}\right)\cos(4\pi n)\\\\
&+\cos \left(\frac{2\pi nc}{L}\right)\sin(4\pi n)\\\\
&-\sin \left(\frac{2\pi nc}{L}\right)\\\\
&=\sin \left(\frac{2\pi nc}{L}\right)\left(cos(4\pi n)-1\right)\\\\
&+\cos \left(\frac{2\pi nc}{L}\right)\sin(4\pi n)\\\\
&=\sin \left(\frac{2\pi nc}{L}\right)\left(-2\sin^2(2\pi n)\right)\\\\
&+\cos \left(\frac{2\pi nc}{L}\right)\left(2\sin(2\pi n)\cos(2\pi n)\right)\\\\
&=2\sin(2\pi n)\cos \left(\frac{2\pi nc}{L}+2\pi n\right)\\\\
&=2\sin(2\pi n)\cos \left(\frac{2\pi n(C+L)}{L}\right)\\\\
\end{align}$$
Therefore, the integral is for $n\ne 0$
$$\int_c^{c+2L}\cos (2\pi nx/L)\,dx=
\begin{cases}
\frac{L\sin(2\pi n)\cos \left(\frac{2\pi n(C+L)}{L}\right)}{\pi n}&,n\ne 0\\\\
2L&,n=0
\end{cases}$$
If $n$ is an integer, then we have
$$\int_c^{c+2L}\cos (2\pi nx/L)\,dx=
\begin{cases}
0&,n\ne 0\\\\
2L&,n=0
\end{cases}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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How to prove that $(1+x^2)(1+y^2)(1+z^2)$ is a square of an integer? $x,y,z \in \Bbb Z$, if $xy+yz+xz=1$ then prove that $(1+x^2)(1+y^2)(1+z^2)$ is a square of an integer
$(1+x^2)(1+y^2)(1+z^2)=1+x^2+y^2+z^2+(xy)^2+(yz)^2+(xz)^2+(xyz)^2$
$(xy)^2+(yz)^2+(xz)^2+2xyz(x+y+z)=1$
| Let me try. $$(x+y+z)^2 = x^2 + y^2 + z^2 + 2(xy+yz+zx) = x^2 + y^2 + z^2 + 2.$$
So, $$(1+x^2)(1+y^2)(1+z^2) = 1 + (x^2+y^2+z^2) + (x^2y^2+y^2z^2 + z^2x^2) +(xyz)^2 = 1 + (x+y+z)^2-2 + 1 - 2xyz(x+y+z) + (xyz)^2 = (x+y+z-xyz)^2.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1471345",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Where am I going wrong in solving this exponential inequality? $$(3-x)^{ 3+x }<(3-x)^{ 5x-6 }$$
Steps I took:
$$(3-x)^{ 3 }\cdot (3-x)^{ x }<(3-x)^{ -6 }\cdot (3-x)^{ 5x }$$
$$\frac { (3-x)^{ 3 }\cdot (3-x)^{ x } }{ (3-x)^{ 3 }\cdot (3-x)^{ x } } <\frac { \frac { 1 }{ (3-x)^{ 6 } } \cdot (3-x)^{ 5x } }{ (3-x)^{ 3 }\cdot (3-x)^{ x } } $$
$$1<\frac { 1 }{ (3-x)^{ 9 } } \cdot (3-x)^{ 4x }$$
$$(3-x)^{ 9 }<(3-x)^{ 4x }$$
$$4x>9$$
$$x>2.25$$
This answer seems to be wrong. I am not sure where I went wrong in the steps that I took. What did I do wrong?
| $$(3-x)^{ 3+x }<(3-x)^{ 5x-6 }$$
For $3>x$
$$(3+x)\log_{3-x}(3-x)<(5x-6 )\log_{3-x}(3-x)$$
$$3-x<5x-6$$
$$x>1.5$$
Thus, $$1.5<x<3$$
| {
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"timestamp": "2023-03-29T00:00:00",
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How to evaluate $ \int_0^\infty \frac{\log x}{(x^2+a^2)^2} dx $ Evaluate $$ \int_0^\infty \frac{\log x}{(x^2+a^2)^2} dx $$ $$(a>0) $$
How can I use contour appropriately?
What is the meaning of this integral?
(additionally posted)
I tried to solve this problem.
First, I take a branch $$ \Omega=\mathbb C - \{z|\text{Re}(z)=0\; \text{and} \; \text{Im}(z)\le0\} $$
Then ${\log_\Omega z}=\log r +i\theta (-\frac{\pi}{2}\lt\theta\lt\frac{3\pi}{2})$
Now, $\frac{\log z}{(z^2+a^2)^2}$ is holomorphic in $\Omega - \{ai\}$ with double poles at $ai$.
Now I'll take the contour which forms an indented semicircle.
For any $0\lt\epsilon\lt{a}$, where $\max (1,a)\lt R$, $\Gamma_{R,\epsilon}\subseteq\Omega - \{ai\}$ and in $\Omega$, $i=e^{i\pi/2}$.
Now using the residue formula, $$2\pi{i}\operatorname*{Res}_{z=ai}\frac{\log_\Omega{z}}{(z^2+a^2)^2}=2\pi{i}\operatorname*{lim}_{z\to ai}\frac{d}{dz}(z-ai)^2\frac{\log_\Omega{z}}{(z^2+a^2)^2}=\frac{\pi}{2a^3}(\log_\Omega{ai}-1)$$
Now, the last part, take $i=e^{i\pi/2}$, then is equal to $\frac{\pi}{2a^3}(\log{a}-1+i\pi/2)$
So, I can split integrals by four parts,
$$\int_{\epsilon}^R dz + \int_{\Gamma_R} dz + \int_{-R}^{-\epsilon} dz + \int_{\Gamma_\epsilon} dz$$
First, evaluate the second part,
$$\left|\int_{\Gamma_R} dz\right|\le\int_0^{\pi}\left|\frac{\log_\Omega{Re^{i\theta}}}{(R^2e^{2i\theta}+a^2)^2}iRe^{i\theta}\right|d\theta$$
Note that
$$\left|\log_\Omega{Re^{i\theta}}\right|=\left|\log R+i\theta\right|\le\left|\log R\right|+|\theta|$$
$$\left|R^2e^{2i\theta}+a^2\right|\ge R^2-a^2\quad (R\gt a)$$
Then, 2nd part $\le\frac{R(\pi R+\frac{\pi^2}{2})}{(R^2+a^2)^2}\to 0\; \text{as} \; R \to \infty\quad \left|\log R\right|\lt R\;\text{where}\;(R\gt 1)$
So, 4th part similarly, goes to $\;0$.
Then 3rd part, substitute for $\;t=-z$,
$$\int_\epsilon^{R}\frac{\log t}{(t^2+a^2)^2}dt + i\pi\int_\epsilon^{R}\frac{dt}{(t^2+a^2)^2}$$
And $\;i\pi\lim\limits_{{\epsilon \to 0},\;{R\to\infty}}\int_\epsilon^{R}\frac{dt}{(t^2+a^2)^2}=\frac{\pi}{4a^3}$
With tedious calculations, I got $\frac{\pi}{4a^3}(\log a -1)$.
| Real Method
We first evaluate the integral
$$
J=\int_0^{\infty} \frac{\ln x}{x^2+a^2} d x
$$
Putting $x=\tan \theta$ yields
$$
\begin{aligned}
J& =\int_0^{\infty} \frac{\ln x}{x^2+a^2} d x \\
& =\int_0^{\frac{\pi}{2}} \frac{\ln (a \tan \theta)}{a^2 \sec ^2 \theta} \cdot a \sec ^2 \theta d \theta \\
& =\frac{1}{a} \int_0^{\frac{\pi}{2}} \ln (a \tan \theta) d \theta \\
& =\frac{1}{a} \int_0^{\frac{\pi}{2}} \ln a d \theta+ \underbrace{\int_0^{\frac{\pi}{2}} \ln (\tan \theta) d \theta }_{=0} \\
& =\frac{\pi}{2} \frac{\ln a}{a}
\end{aligned}
$$
Differentiating both sides w.r.t. $a$ yields
$$
\begin{aligned}
& -2 a \int_0^{\infty} \frac{\ln x}{\left(x^2+a^2\right)^2} d x=\frac{\pi}{2} \frac{1-\ln a}{a^2} \\
\Rightarrow & \int_0^{\infty} \frac{\ln x}{\left(x^2+a^2\right)^2}=\frac{\pi(\ln a-1)}{4 a^3}
\end{aligned}
$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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} |
On solving the equation $\sin ^{2}\left ( z \right )= i\pi$ One obvious approach is to rewrite $\sin\left ( z \right )$ as $\frac{e^{iz}-e^{-iz}}{2i}$ and then proceed applying an appropriate substitution. However that leads to a quartic complex equation, which gets a bit(actually a lot) messy. Is there another(maybe obscure) approach suitable for this problem?
| $$
sin(z) = \pm \sqrt{i \pi} \\
e^{iz} - e^{-iz} = \pm 2i \sqrt{i \pi}
$$
Let $e^{iz} = x$, then $x^2 \pm 2i \sqrt{i \pi} x - 1 = 0$.
Hence:
$e^{iz} = x = \frac{\pm 2i \sqrt{i \pi} \pm \sqrt{-4i \pi +4}}{2} = \pm i \sqrt{i \pi} \pm \sqrt{1-i \pi}$
Thus, $z = -i \log(\pm i \sqrt{i \pi} \pm \sqrt{1-i \pi})$.
Or:
$$
z_1 = -i \log(i \sqrt{i \pi} + \sqrt{1-i \pi}) \\
z_2 = -i \log(i \sqrt{i \pi} - \sqrt{1-i \pi}) \\
z_3 = -i \log(-i \sqrt{i \pi} + \sqrt{1-i \pi}) \\
z_4 = -i \log(-i \sqrt{i \pi} - \sqrt{1-i \pi})
$$
As $\log(-1) = i (1 + 2k) \pi$, $z_3 = z_2 + i(1+2k) \pi$, and $z_4 = z_1 + i(1+2k) \pi$, for $k$ integer.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1481878",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Evaluating $\sum_{n=1}^\infty \frac{n^3}{e^{2\pi n}-1}$ using inverse Mellin transform inspiration on the post Evaluating $\sum_{n=1}^{\infty} \frac{n}{e^{2 \pi n}-1}$ using the inverse Mellin transform
it is possible to calculate in close form
$$\sum _{k=1}^{\infty } -\frac{k^3}{e^{2 \pi k}-1}=\frac{3840 \pi ^4 \psi _{e^{2 \pi }}^{(1)}(1)+480 \pi ^2 \psi _{e^{2 \pi }}^{(3)}(1)-704 \pi ^6-5760 \pi ^5+3 \Gamma \left(\frac{1}{4}\right)^8}{23040 \pi ^6}$$ using euler sum I appreciatte some comment.
I like to give another series it will interesting using elliptic theta function theory to prove it
$$\sum _{k=1}^{\infty } \frac{\left(k \left(-\log \left(\frac{\pi }{2}\right)\right)\right)^3}{e^{2 \pi \left(k \log \left(\frac{\pi }{2}\right)\right)}+1}$$
$$\frac{\log ^4\left(\frac{\pi }{2}\right) \psi _{e^{-\frac{\pi }{\log \left(\frac{\pi }{2}\right)}}}^{(3)}(1)-\log ^4\left(\frac{\pi }{2}\right) \psi _{e^{-\frac{\pi }{\log \left(\frac{\pi }{2}\right)}}}^{(3)}\left(-\frac{\left(i \pi -\frac{\pi }{\log \left(\frac{\pi }{2}\right)}\right) \log \left(\frac{\pi }{2}\right)}{\pi }\right)}{16 \pi ^4 \log \left(\frac{\pi }{2}\right)}-\frac{1}{240} \log ^3\left(\frac{\pi }{2}\right)-\frac{7}{1920 \log \left(\frac{\pi }{2}\right)}$$
sorry for the latex type i do not to improve
| This is an alternative approach which is too long for comment. If we put $q = e^{-\pi}$ then the desired sum is $$-\sum_{n = 1}^{\infty}\frac{n^{3}}{q^{-2n} - 1} = -\sum_{n = 1}^{\infty}\frac{n^{3}q^{2n}}{1 - q^{2n}} = \frac{1 - Q(q^{2})}{240}$$ and we know that $$Q(q^{2}) = \left(\frac{2K}{\pi}\right)^{4}(1 - k^{2} + k^{4})$$ For $q = e^{-\pi}$ we have $k = 1/\sqrt{2}$ so that $(1 - k^{2} + k^{4}) = 1 - 1/2 + 1/4 = 3/4$ and $$K(k) = \frac{1}{4\sqrt{\pi}}\Gamma\left(\frac{1}{4}\right)^{2}$$ so that $$Q(q^{2}) = \frac{3}{4}\left(\frac{2K}{\pi}\right)^{4} = \frac{3}{64\pi^{6}}\Gamma\left(\frac{1}{4}\right)^{8}$$ and therefore the desired sum is equal to $$\frac{64\pi^{6} - 3\Gamma(1/4)^{8}}{15360\pi^{6}}$$ The expression for $Q(q^{2})$ in terms of $K, k$ is derived here. Using the same technique and expression for $R(q^{2})$ we can get the surprisingly simple and beautiful result $$\sum_{n=1}^{\infty}\frac{n^{5}}{e^{2\pi n} - 1} = \frac{1}{504}$$ For the second sum mentioned in the question we let $$q = \exp(-2\pi\log(\pi/2)) = \left(\frac{2}{\pi}\right)^{2\pi}$$ and then the desired sum is equal to $$S = -\left(\log\left(\frac{\pi}{2}\right)\right)^{3}\sum_{n = 1}^{\infty}\frac{n^{3}q^{n}}{1 + q^{n}} = -\left(\log\left(\frac{\pi}{2}\right)\right)^{3}\cdot A$$ where the sum
\begin{align}
A &= \sum_{n = 1}^{\infty}\frac{n^{3}q^{n}}{1 + q^{n}}\notag\\
&= \sum_{n = 1}^{\infty}n^{3}q^{n}\left(\frac{1}{1 + q^{n}} - \frac{1}{1 - q^{n}} + \frac{1}{1 - q^{n}}\right)\notag\\
&= \sum_{n = 1}^{\infty}n^{3}q^{n}\left(\frac{1}{1 - q^{n}} - \frac{2q^{n}}{1 - q^{2n}}\right)\notag\\
&= \sum_{n = 1}^{\infty}\frac{n^{3}q^{n}}{1 - q^{n}} - 2\sum_{n = 1}^{\infty}\frac{n^{3}q^{2n}}{1 - q^{2n}}\notag\\
&= \frac{Q(q) - 1}{240} - \frac{Q(q^{2}) - 1}{120}\notag\\
&= \frac{1 + Q(q) - 2Q(q^{2})}{240}\notag\\
&= \frac{1}{240} + \frac{1}{240}\left(\frac{2K}{\pi}\right)^{4}(1 + 14k^{2} + k^{4} - 2 + 2k^{2} - 2k^{4})\notag\\
&= \frac{1}{240} - \frac{1}{240}\left(\frac{2K}{\pi}\right)^{4}(1 - 16k^{2} + k^{4})\notag\\
\end{align}
and hence $$S = \frac{1}{240}\left(\log\left(\frac{\pi}{2}\right)\right)^{3}\left(\frac{2K}{\pi}\right)^{4}(1 - 16k^{2} + k^{4}) - \frac{1}{240}\left(\log\left(\frac{\pi}{2}\right)\right)^{3}$$ I doubt if it can be put into a closed form which is as simple as that for the previous sum.
| {
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"timestamp": "2023-03-29T00:00:00",
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Find the number of integer solutions for $x_1+x_2+x_3 = 15$ under some constraints by IEP. For equation
$$
x_1+x_2+x_3 = 15
$$
Find number of positive integer solutions on conditions:
$$
x_1<6, x_2 > 6
$$
Let: $y_1 = x_1, y_2 = x_2 - 6, y_3 = x_3$
than, to solve the problem, equation $y_1+y_2 +y_3 = 9$ where $y_1 < 6,0<y_2, 0<y_3 $ has to be solved. Is this correct?
To solve this equation by inclusion-exclusion, number of solution without restriction have to be found $C_1 (3+9-1,9)$ and this value should be subtracted by $C_2 (3+9-7-1,2)$ , (as the negation of $y_1 < 6$ is $y_1 \geq 7$).
Thus:
$$
55-6=49
$$
Is this the correct answer ?
Problem must be solved using inclusion-exclusion...
| You solved the problem in the nonnegative integers rather than the positive integers.
You wish to determine the number of solutions of the equation
$$x_1 + x_2 + x_3 = 15 \tag{1}$$
in the positive integers subject to the constraints $x_1 < 6$ and $x_2 > 6$.
Let's deal with the constraint $x_2 > 6$ first. Let $y_2 = x_2 - 6$. Then $y_2$ is a positive integer. Substituting $y_2 + 6$ for $x_2$ in equation 1 yields
\begin{align*}
x_1 + y_2 + 6 + x_3 & = 15\\
x_1 + y_2 + x_3 & = 9 \tag{2}
\end{align*}
Equation 2 is an equation in the positive integers. A particular solution of equation 2 in the positive integers corresponds to a placement of addition signs in two of the eight spaces between successive ones in a row of nine ones. For instance,
$$1 1 + 1 1 1 1 1 + 11$$
corresponds to the solution $x_1 = 2$, $y_2 = 5$, and $x_3 = 2$ (or $x_1 = 2$, $x_2 = 11$ and $x_3 = 2$ in equation 1), while
$$1 1 1 1 + 1 1 + 1 1 1$$
corresponds to the solution $x_1 = 4$, $y_2 = 2$, and $x_3 = 3$ (or $x_1 = 4$, $x_2 = 8$, and $x_3 = 3$ in equation 1). Therefore, the number of solutions of equation 2 is
$$\binom{8}{2}$$
From these, we must exclude those solutions in which $x_1 \geq 6$. Assume $x_2 \geq 6$. Let $y_1 = x_1 - 5$. Then $y_1$ is a positive integer. Substituting $y_1 + 5$ for $x_1$ in equation 2 yields
\begin{align*}
y_1 + 5 + y_2 + x_3 & = 9\\
y_1 + y_2 + y_3 & = 4 \tag{3}
\end{align*}
Equation 3 is an equation in the positive integers. The number of solutions of equation 3 is the number of ways we can place two addition signs in the three spaces between successive ones in a row of four ones, which is
$$\binom{3}{2}$$
Hence, the number of solutions of equation 1 subject to the constraints $x_1 < 6$ and $x_2 > 6$ is
$$\binom{8}{2} - \binom{3}{2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1483343",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Olympiad inequality problem with $a+b+c+abc=4$ If $a,b,c \in \mathbb R_{> 0}$ and $a+b+c+abc=4$, prove that
$$({a\over {\sqrt {b+c}}}+{b\over {\sqrt {c+a}}}+{c\over {\sqrt {a+b}}})^2(ab+bc+ca) \ge {\frac 12}(4-abc)^3$$
This can be solved by AM-GM-HM or the Cauchy-Schwarz inequality. I'd tried for some hours but couldn't solve it. Can anyone help me? Thanks in advance:).
| Consider using the following Hölder's inequality:
If $a_1,\ldots, a_n$ and $b_1,\ldots, b_n$ are positive real numbers, then $$\sum_{k = 1}^n a_kb_k \le \left(\sum_{k = 1}^n a_k^p\right)^{1/p}\left(\sum_{k = 1}^n b_k^q\right)^{1/q}$$ where $p$ and $q$ are real numbers such that $p,q \ge 1$ and $\frac{1}{p} + \frac{1}{q} = 1$.
Write
$$ab + bc + ca = \frac{ab + bc}{2}+\frac{bc + ca}{2} + \frac{ca + ab}{2} = \frac{a(b + c)}{2}+\frac{b(c + a)}{2}+\frac{c(a + b)}{2}.$$
Then
\begin{align}&\left(\frac{a}{\sqrt{b+c}} + \frac{b}{\sqrt{c + a}} + \frac{c}{\sqrt{a+b}}\right)^{2/3}\left(ab + bc + ca\right)^{1/3}\\
&=\left(\frac{a}{\sqrt{b+c}} + \frac{b}{\sqrt{c + a}} + \frac{c}{\sqrt{a+b}}\right)^{2/3} \left[\frac{a(b + c)}{2} + \frac{b(c + a)}{2} + \frac{c(a + b)}{2}\right]^{1/3}\\
&\ge \frac{a^{2/3}}{(b+c)^{1/3}}\cdot \frac{a^{1/3}(b+c)^{1/3}}{2^{1/3}} + \frac{b^{2/3}}{(c + a)^{1/3}}\cdot \frac{b^{1/3}(c+a)^{1/3}}{2^{1/3}} + \frac{c^{2/3}}{(a + b)^{1/3}}\cdot \frac{c^{1/3}(a + b)^{1/3}}{2^{1/3}}\\
&= \frac{a+b+c}{2^{1/3}}.
\end{align}
The second to last step follows from Hölder's inequality with $p=3/2$ and $q=3$. Since $a + b + c = 4 - abc$, then
$$\left(\frac{a}{\sqrt{b+c}} + \frac{b}{\sqrt{c + a}} + \frac{c}{\sqrt{a+b}}\right)^{2/3}\left(ab + bc + ca\right)^{1/3} \ge \frac{4-abc}{2^{1/3}}.$$
The result is obtained by cubing both sides of this inequality.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1483425",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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} |
Find the partial sums of $4+44+444+\cdots$
Find the sum to $n$ terms of $4+44+444+\cdots $
My attempts:
*
*Used successive difference method
*Used $4+(4+40)+(4+40+400)\dots$ method
Failed to get a formula for partial sum in both ways. What am I missing?
| We can Write $\displaystyle 4 = 4(1) = \frac{4}{9}\left[9\right] = \frac{4}{9}\left[10-1\right]$
Similarly $\displaystyle 44 = 4(11)=\frac{4}{9}\left[99\right] = \frac{4}{9}\left[10^2-1\right]$
Similarly $\displaystyle 444 = 4(111)=\frac{4}{9}\left[999\right] = \frac{4}{9}\left[10^3-1\right]$
Now you can proceed from here.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Prove by induction: $\sum\limits_{k=1}^{n}(-1)^{k+1}{n\choose k}\frac{1}{k}=1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{n}$ $\sum\limits_{k=1}^{n}(-1)^{k+1}{n\choose k}\frac{1}{k}=1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{n}$
For $n=1$ equality is true.
For $n=m$
$m-{m\choose 2}\frac{1}{2}+...+(-1)^{m+1}\frac{1}{m}=1+\frac{1}{2}+...+\frac{1}{m}$
For $n=m+1$
$\left(\sum\limits_{k=1}^{m}(-1)^{k+1}{m\choose k}\frac{1}{k}\right)+(-1)^{m+2}\frac{1}{m+1}=1+\frac{1}{2}+...+\frac{1}{m+1}$
If $m$ is even, equality is true, but not if $m$ is odd.
Is this correct?
| Define
$$
f(n)=\sum_{k=1}^n(-1)^{k-1}\binom{n}{k}\frac1k\tag1
$$
Then, $f(0)=0$ and for $n\gt0$,
$$
\begin{align}
f(n)
&=\sum_{k=1}^n(-1)^{k-1}\binom{n}{k}\frac1k\tag2\\
&=\sum_{k=1}^n(-1)^{k-1}\left[\binom{n-1}{k}+\binom{n-1}{k-1}\right]\frac1k\tag3\\
&=\sum_{k=1}^{n-1}(-1)^{k-1}\binom{n-1}{k}\frac1k+\sum_{k=1}^n(-1)^{k-1}\binom{n-1}{k-1}\frac1k\tag4\\
&=f(n-1)+\frac1n\sum_{k=1}^n(-1)^{k-1}\binom{n}{k}\tag5\\
&=f(n-1)+\frac1n\tag6
\end{align}
$$
Explanation:
$(3)$: Pascal's Rule
$(4)$: distribute the summation
$(5)$: apply $(1)$ for $n-1$ and $\frac1k\binom{n-1}{k-1}=\frac1n\binom{n}{k}$
$(6)$: $\sum\limits_{k=1}^n(-1)^{k-1}\binom{n}{k}=1$
Therefore, $f(n)$ is the $n^\text{th}$ Harmonic Number for $n\ge0$.
| {
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"timestamp": "2023-03-29T00:00:00",
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If $x^2+4y^2=4.$ Then find range of $ x^2+y^2-xy$
If $x^2+4y^2=4.$ Then find range of $ x^2+y^2-xy$
$\bf{My\; Try::}$ Given $$x^2+4y^2 = 4\Rightarrow \frac{x^2}{4}+\frac{y^2}{1} = 1$$, So parametric Coordinate for Ellipse are
$x = 2\cos \phi$ and $y = \sin \phi$.
Now Let $$f\left(x,y\right) = x^2+y^2-xy = 4-4y^2+y^2-xy = 4-3y^2-xy = 4 - 3\sin^2 \phi - 2\sin \phi \cdot \cos \phi$$
So $$f\left(\phi \right) = 4-\frac{3}{2}\left(1-\cos 2\phi \right) - \sin 2\phi = \frac{5}{2} + \frac{1}{2}\left(3\cos 2 \phi -2 \sin 2\phi \right)$$
Now Range of $$-\sqrt{13}\leq \left(3\cos 2 \phi - 2\sin 2\phi \right)\leq \sqrt{13}$$
So $$\frac{1}{2}\cdot \left(5-\sqrt{13} \right) \leq f\left(\phi \right)\leq \frac{1}{2}\cdot \left(5+\sqrt{13} \right)$$
My question is can we solve it using Inequality or any other method, If yes Then plz explain here
Thanks
| Let's solve $\max_{(x,y)}x^2+y^2-xy$ s.t. $x^2+4y^2=4$.
*
*Let first $x=-\sqrt{4-4y^2}$ to have $$f(y)=-3 y^2+\sqrt{4-4 y^2} y+4$$ here we have $\frac{\sqrt{1-y^2}}{2}f'(y)=-2 y^2-3 \sqrt{1-y^2} y+1$. Solving this equation we get that $y_1=-\sqrt{\frac{1}{2}+\frac{3}{2 \sqrt{13}}}$ and $y_2=\sqrt{\frac12-\frac{3}{2 \sqrt{13}}}$. Now note that $$f''(y)=\frac{4 y^3+6 \sqrt{1-y^2} y^2-6 \sqrt{1-y^2}-6 y}{\left(1-y^2\right)^{3/2}}$$ With some "fun" we get $$f''(y_1)=\sqrt{26-6 \sqrt{13}}$$ $$f''(y_2)=-\sqrt{26+6 \sqrt{13}}$$
with some more fun we obtain $f(y_1)=\frac{1}{2} \left(5-\sqrt{13}\right)$ and $f(y_2)=\frac{1}{2} \left(5+\sqrt{13}\right)$.
*Letting $x=\sqrt{4-4y^2}$ we obtain the same max and min.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Find the area of the triangle There are two points $N$ and $M$ on the sides $AB$ and $BC$ of the triangle $ABC$ respectively. The lines $AM$ and $CN$ intersect at point $P$. Find the area of the triangle $ABC$, if areas of triangles $ANP, CMP, CPA$ are $6,8,7$ respectively.
| Simple Geometry Approach
Note that, with the same altitude, ratio of areas of two triangles is equal to the ratio of their bases.
Therefore, $\dfrac {x}{y} = \dfrac {NP}{PC} = \dfrac {u}{v + z}$ ….. (*)
Similarly, $\dfrac {z}{y} = \dfrac {v}{u + x}$ ….. (#)
After eliminating v from (*) and (#) and making u as subject, we have
$ u = \dfrac {xyz +x^2z}{y^2 - xz}$
Similarly, or by symmetry, we have $v = \dfrac {xyz +z^2x}{y^2 - xz}$
∴ $[\triangle ABC] = x + y + z + u + v = x + y + z + \dfrac {xyz +x^2z}{y^2 - xz} + \dfrac {xyz +z^2x}{y^2 - xz} = ... = 1365$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1490393",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Proof of the hockey stick/Zhu Shijie identity $\sum\limits_{t=0}^n \binom tk = \binom{n+1}{k+1}$ After reading this question, the most popular answer use the identity
$$\sum_{t=0}^n \binom{t}{k} = \binom{n+1}{k+1},$$
or, what is equivalent,
$$\sum_{t=k}^n \binom{t}{k} = \binom{n+1}{k+1}.$$
What's the name of this identity? Is it the identity of the Pascal's triangle modified.
How can we prove it? I tried by induction, but without success. Can we also prove it algebraically?
Thanks for your help.
EDIT 01 : This identity is known as the hockey-stick identity because, on Pascal's triangle, when the addends represented in the summation and the sum itself are highlighted, a hockey-stick shape is revealed.
| First proof
Using stars and bars, the number of ways to put $n$ identical objects to $k$ bins(empty bin allowed) is $\binom{n+k-1}{k-1}$.
If we reduce the number of bins by one, the number of ways to put $n$ identical objects to $k-1$ bins is $\binom{n+k-2}{k-2}$.
We can also count the number of ways to put $n$ identical objects to $k$ bins using the answer for $k-1$ bins. Split them into different cases based on how many objects you put in the first bin.
\begin{array} {|r|r|}\hline \text{Number of objects for first bin} & \text{Number of objects remaining for other bins} & \text{Number of ways to distribute} \\ \hline 0 & n & \binom{n+k-2}{k-2} \\ \hline 1 & n-1 & \binom{n+k-3}{k-2} \\ \hline 2 & n-2 & \binom{n+k-4}{k-2} \\ \hline \vdots & \vdots & \vdots \\ \hline n-2 & 2 & \binom{k}{k-2} \\ \hline n-1 & 1 & \binom{k-1}{k-2} \\ \hline n & 0 & \binom{k-2}{k-2} \\ \hline \end{array}
Therefore,
$$\binom{n+k-1}{k-1} = \binom{k-2}{k-2} + \binom{k-1}{k-2} + \binom{k}{k-2} +\dots+ \binom{n+k-4}{k-2} + \binom{n+k-3}{k-2} + \binom{n+k-2}{k-2}$$
Rename the variables: Let $m+1 = n+k-1$ and $r+1 = k-1$. We get:
$$\binom{m+1}{r+1} = \binom{r}{r} + \binom{r+1}{r} + \binom{r+2}{r} +\dots+ \binom{m-2}{r} + \binom{m-1}{r} + \binom{m}{r}$$
This is the identity when expanded.
Second proof
Suppose we have $m+1$ people where everyone has different age and we want to select $r+1$ people to be in some committee. There are $$\binom{m+1}{r+1} $$ ways to form the committee.
Alternatively, we can count this by splitting the committee based on who the oldest person in the committee is.
We can have the oldest person to be the oldest in the committee. Then there are $m$ people left and we still need to add $r$ people to be in the committee. So there are $\binom{m}{r}$ ways.
We can have the second oldest person to be the oldest in the committee. Then there are $m-1$ people left since we must exclude the first oldest person and second oldest(already selected) and we still need to add $r$ people to be in the committee. So there are $\binom{m-1}{r}$ ways.
In general, suppose we have the kth oldest person to be the oldest in the committee. Then there are $m-k+1$ possible people left to add(excluding kth oldest) since we must exclude the oldest members up to the $k$th oldest person. We still need to add $r$ people to be in the committee. So there are ${m-k+1 \choose r}$ ways.
In each case, there needs to be at least $r$ people left to choose so the last case is when we select the $(m+1-r)$th oldest person. This will give us ${r \choose r}$ ways.
So
$$\binom{m+1}{r+1} = \binom{r}{r} + \binom{r+1}{r} + \binom{r+2}{r} +\dots+ \binom{m-2}{r} + \binom{m-1}{r} + \binom{m}{r}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1490794",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "73",
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Finding a triangle ratio.
In the triangle ABC, the point P is found on the side AB. AC = 6 cm, AP = 4 cm, AB = 9 cm. Calculate BC:CP.
For some reason, I cannot get this even though I tried for half an hour.
I got that, $BC/CP < 17/10 = 1.7$ by the triangle inequality. $AP/PB = 4/5$
But that does not help one but, I'm very stuck!
| Notice, let $BC=x$ & $CP=y$ now, applying cosine rule in $\triangle ABC$ $$\cos\angle BAC=\frac{AB^2+AC^2-BC^2}{2(AB)(AC)}=\frac{9^2+6^2-y^2}{2(9)(6)}=\frac{117-y^2}{108}\tag 1$$
similarly, applying cosine rule in $\triangle APC$
$$\cos\angle BAC=\frac{AP^2+AC^2-PC^2}{2(AP)(AC)}=\frac{4^2+6^2-x^2}{2(4)(6)}=\frac{52-x^2}{48}\tag 2$$
Equating (1) & (2), we get
$$\frac{117-y^2}{108}=\frac{52-x^2}{48}$$
$$468-4y^2=468-9x^2$$
$$\frac{y^2}{x^2}=\frac{9}{4}\implies \frac{y}{x}=\frac{3}{2}$$
$$\color{red}{BC:CP=3:2}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Proof by induction $\frac{1}{1^2} + \frac{1}{2^2} + \cdots + \frac{1}{n^2} < 2$ Proof by induction $\frac{1}{1^2} + \frac{1}{2^2} + \cdots + \frac{1}{n^2} < 2 \ \ \ n \in \mathbb{N}$
So for $n=1$
$$ 1 < 2$$
For $n > 1$
Assumption: $$\frac{1}{1^2} + \frac{1}{2^2} + \cdots + \frac{1}{n^2} < 2$$
Hypothesis (inductive step):
$$\frac{1}{1^2} + \frac{1}{2^2} + \cdots + \frac{1}{n^2} + \frac{1}{(n+1)^2} < 2$$
So using assumption and hypothesis I have:
$$\frac{1}{1^2} + \frac{1}{2^2} + \cdots + \frac{1}{n^2} < 2 - \frac{1}{(n+1)^2} < 2 $$
So then: $$ \frac{1}{(n+1)^2} > 0 $$ which is always true
I was told it's relatively "hard" one. Thus I think I made sth stupid here.
| inequility： $\frac{2}{n^2}\leq \frac{2}{n^2-1}=\frac{1}{n-1}-\frac{1}{n+1}$
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "1",
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"answer_id": 1
} |
Finding the coordinate C. A triangle $A$, $B$, $C$ has the coordinates:
$A = (-1, 3)$
$B = (3, 1)$
$C = (x, y)$
$BC$ is perpendicular to $AB$. Find the coordinates of $C$
My attempt:
Grad of $AB$ =
$$\frac{3-1}{-1-3} = -0.5$$
Grad of $BC = 2$ ($-0.5 \times 2 = -1$ because AB and BC are perpendicular).
Equation of $BC$
$(y-1) = 2(x-3)$
$y = 2x - 5$
Equation of $AC$
$(y-3) = m(x--1)$
$y = mx+m+3$
I do not know how to proceed further. Please help me out.
| Yes, the gradient of line AB is $-1/2$ so the gradient of any line perpendicular to $AB$ is $2$. The line with gradient $2$ through $B = (3, 1)$ is $y = 2(x- 3)+ 1$ or $y= 2x- 5$. All points on that line, $(x, 2x- 5)$, satisfy the conditions for C.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Why does $\lim\limits_{n \to \infty } ((1 + x)(1 + {x^2})(1 + {x^4})\ldots(1 + {x^{{2^n}}})) = \frac{1}{{1 - x}}$? Let $\left| x \right| < 1$.
Why does $$\lim\limits_{n \to \infty } ((1 + x)\cdot(1 + {x^2})\cdot(1 + {x^4})\cdot\ldots\cdot(1 + {x^{{2^n}}})) = \frac{1}{{1 - x}}$$
| Using $$\displaystyle \lim_{n\rightarrow \infty}\frac{1}{1-x}\left[\overbrace{(1-x)(1+x)}^{(1-x^{2^{1}})}(1+x^2)(1+x^4).......(1+x^{2^{n}})\right]$$
So we get $$\displaystyle \lim_{n\rightarrow \infty}\frac{1}{1-x}\left[\overbrace{(1-x^2)(1+x^2)}^{(1-x^{2^2})}(1+x^4)....(1+x^{2^{n}})\right]$$
In a similar way we get the limit $$\displaystyle = \lim_{n\rightarrow \infty}\frac{1-x^{2^{n+1}}}{1-x}\;,$$ Now Given $-1 <x<1$
So we get limit $$\displaystyle = \frac{1}{1-x}\;,$$ Bcz $x^{2^{n+1}}\rightarrow 0\;,$ When $n\rightarrow \infty$
| {
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"timestamp": "2023-03-29T00:00:00",
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Multivariable Epsilon Delta Proof I am trying to prove that a function $\ f(x,y)=y^2x$ is continuous everywhere, using epsilon delta proofs.
I've set it up like:
$\ |x^2y-a^2b| < \epsilon $
and
$\ \sqrt{\left(x-a\right)^2+\left(y-b\right)^2} < \delta $
But other than that, have absolutely no clue about what operations I need to do to the top equation. If someone could please point me in the right direction that would be great.
| Fix $(a,b)\in \Bbb R^2$. For all $(x,y)\in \Bbb R^2$,
\begin{align}|f(x,y) - f(a,b)| &= |y^2x - b^2 a|\\
& = |(y^2x - y^2 a) + (y^2a - b^2 a)|\\
& \le |y^2x - y^2a| + |y^2a - b^2a|\\
& = y^2|x - a| + |y^2 - b^2||a|.
\end{align}
Now if $(x - a)^2 + (y - b)^2 < 1$, then in particular $|y - b| < 1$. So by the triangle inequality, $|y| \le |y - b| + |b| < 1 + |b|$ and $|y + b| \le |y - b| + 2|b| < 1 + 2|b|$. Thus
\begin{align}y^2|x - a| + |y^2 - b^2||a| &= y^2|x - a| + |a||y + b||y - b| \\
&<(1 + |b|)^2|x - a| + (1 + 2|b|)|a||y - b|.
\end{align}
The last expression can be made less than a positive number $\epsilon$ by having $|x - a|$ and $|y - b|$ less than
$$\frac{\epsilon}{(1 + |b|)^2 + |a|(1 + 2|b|)},$$
which can be done by having
$$\sqrt{(x - a)^2 + (y - b)^2} < \frac{\epsilon}{(1 + |b|)^2 + |a|(1 + 2|b|)}.$$
Hence, given $\epsilon > 0$, set
$$\delta = \min\left\{1, \frac{\epsilon}{(1 + |b|)^2 + |a|(1 + 2|b|)}
\right\}.$$
For all $(x,y)\in \Bbb R^2$, $(x - a)^2 + (y - b)^2 < \delta^2$ implies
$$|f(x,y) - f(a,b)| < (1 + |b|)^2|x - a| + (1 + 2|b|)|a||y - b| < \epsilon.$$
| {
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Indefinite integration of $\ln(1+x^2)\arctan x$ We need to evaluate $$\int \ln(1+x^2)\arctan x \, \mathrm{d}x$$
My thoughts were to set $u = \arctan x \implies x = \tan u$ so that our integral is transformed to $$\int 2u \sec^2 u \ln \sec u \, \mathrm{d}u$$
Am I on the right path? How do I continue from here? Are there alternative ways of doing this?
| We integrate by parts. First we find $\int \arctan x\, dx$.
$$\int \arctan x\, dx = (\arctan x)(x)-\int \frac{x}{x^2+1}\, dx$$
$$=(\arctan x)(x)-\frac{1}{2}\int \frac{d\left(x^2+1\right)}{x^2+1}=(\arctan x)(x)-\frac{1}{2}\ln(x^2+1)+C$$
$$\int \left(\ln\left(1+x^2\right)\right)\left(\arctan x\, dx\right)$$
$$=\ln\left(1+x^2\right)\left((\arctan x)(x)-\frac{1}{2}\ln(x^2+1)\right)-$$
$$-\int \left(\frac{2x}{1+x^2}\right)\left((\arctan x)(x)-\frac{1}{2}\ln(x^2+1)\right)\, dx$$
It's enough to find:
$$\int \left(\frac{2x}{1+x^2}\right)\left((\arctan x)(x)-\frac{1}{2}\ln(x^2+1)\right)\, dx=$$
$$=2\int \arctan x\, dx -2\int \frac{1}{1+x^2}\arctan x\, dx-\int \frac{x\ln\left(x^2+1\right)}{x^2+1}\, dx$$
We have $3$ integrals. We know the $1$st one. The $2$nd one is:
$$\int \frac{1}{1+x^2}\arctan x\, dx=\int \arctan x\, d(\arctan x)=\frac{(\arctan x)^2}{4}+C$$
The $3$rd one is:
$$\int \frac{x\ln\left(x^2+1\right)}{x^2+1}\, dx=\frac{1}{2}\int \frac{\ln(x^2+1)}{x^2+1}\, d\left(x^2+1\right)$$
$$=\frac{1}{2}\int \ln\left(x^2+1\right)\, d\left(\ln\left(x^2+1\right)\right)$$
$$=\frac{\left(\ln\left(x^2+1\right)\right)^2}{4}+C$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Write on my own my first mathematical induction proof I am trying to understand how to write mathematical induction proofs. This is my first attempt.
Prove that the sum of cubic positive integers is equal to the formula
$$\frac{n^2 (n+1)^2}{4}.$$ I think this means that the sum of cubic positive integers is equal to an odd number. However, let's go on proving...
1) I start by proving the base case $n=1$ and I show that the formula holds.
2) I assume than any number $k$ other than $1$, which appartains at $N$, holds for the formula and I write the same formula but with $k$ which replaces $n$.
3) For mathematical induction, I assume that the formula holds also for $k+1$ = $n$
So, the left side of the equation should be:
$$\sum^{k+1}_{i=1} i^3 = 1^3 + 2^3 + 3^3 + ... + (k+1)^3$$
I am wondering about which one of these 2 forms (equivalents, I think) should have the right side :
this one, with $k+1$ in place of the $n$ of the original formula / or $k$ in the second version: $\frac{(k+1)^2[(k+1)+1]^2}{4}$ or this one: $\frac{k^2(k+1)^2 }{4} + (k+1)^3$ ?
I think that, in order for the proof to be convincing, we should write an equivalent statement for the original form of the formula, namely $$\sum^{n}_{i=1} i^3= \frac{n^2(n+1)^2}{4}$$ and perhaps we do it by showing that after algebraic passages $\frac{k^2(k+1)^2 }{4} + (k+1)^3$ is equal to $\frac{(k+1)^2[(k+1)+1]^2}{4}$ ?
Sorry for my soliloquy but it helps to understand and I would appreciate confirmation from you!
| You should use the second one:
Suppose it holds for the first $k$ numbers. So their sum is equal to $\frac{k²(k+1)^2}{4}$. Then the first sum of the first $k+1$ is equal to $1^3 + 2^3 + 3^3 + ... + (k+1)^3=\frac{k²(k+1)^2}{4}+(k+1)^3=\frac{k²(k+1)^2}{4}+\frac{4(k+1)^3}{4}$ which is equal to $\frac{k²(k+1)^2+4(k+1)^3}{4}=\frac{(k+1)²(k²+4k+4)}{4}=\frac{(k+1)²(k+2)²}{4}$.
Which is precisely what you need.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Prove: If $|z|<\frac{1}{2}\Rightarrow |(1+i)z^3+iz|<\frac{3}{4},z\in\mathbb{C}$ Prove: If $|z|<\frac{1}{2}\Rightarrow |(1+i)z^3+iz|<\frac{3}{4},z\in\mathbb{C}$
$|z|=\sqrt{x^2+y^2}<\frac{1}{2}\Rightarrow x^2+y^2 <\frac{\sqrt{2}}{2}$
$$|(1+i)z^3+iz|=|(x^3-3xy-3x^2y-y-y^3)+i(x^3-3xy+3x^2y+x-y^3)|=\sqrt{(x^3-3xy-3x^2y-y-y^3)^2+(x^3-3xy+3x^2y+x-y^3)^2}$$
Is there another approach rather than expanding these expressions?
| HINT:
Assuming $z=ai$ with $a\in\mathbb{R}$:
$$|(1+i)(ai)^3+i(ai)|<\frac{3}{4}\Longleftrightarrow$$
$$|(1+i)(ai)^3-a|<\frac{3}{4}\Longleftrightarrow$$
$$|(1-i)a^3-a|<\frac{3}{4}\Longleftrightarrow$$
$$|a((1+i)a^2-i)|<\frac{3}{4}\Longleftrightarrow$$
$$|a||(1+i)a^2-i|<\frac{3}{4}\Longleftrightarrow$$
$$\sqrt{a^6+(a^3-a)^2}<\frac{3}{4}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Understanding telescoping series? The initial notation is:
$$\sum_{n=5}^\infty \frac{8}{n^2 -1}$$
I get to about here then I get confused.
$$\left(1-\frac{3}{2}\right)+\left(\frac{4}{5}-\frac{4}{7}\right)+...+\left(\frac{4}{n-3}-\frac{4}{n-1}\right)+...$$
How do you figure out how to get the $\frac{1}{n-3}-\frac{1}{n-1}$ and so on? Like where does the $n-3$ come from or the $n-1$.
| The correct way to analyze this is to write
$$\begin{align}
\sum_{n=5}^N\frac{2}{n^2-1}&=\sum_{n=5}^{N}\left(\frac{1}{n-1}-\frac{1}{n+1}\right)\\\\
&=\left(\frac14-\frac16\right)+\left(\frac15-\frac17\right)+\left(\frac16-\frac18\right)+\cdots \\\\
&+\left(\frac1{N-3}-\frac1{N-1}\right)+\left(\frac1{N-2}-\frac1N\right)+\left(\frac1{N-1}-\frac1{N+1}\right)\\\\
&=\frac14+\frac15+\frac{1}{N}+\frac{1}{N+1}
\end{align}$$
Therefore,
$$\sum_{n=5}\frac{8}{n^2-1}=4\lim_{N\to \infty}\left(\frac14+\frac15+\frac{1}{N}+\frac{1}{N+1}\right)=\frac95$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Evaluate the Integral $\int \sqrt{1-4x^2}\ dx$ $\int \sqrt{1-4x^2}\ dx$
I am confused as I get to the end. Why would I use a half angle formula? And why is it necessary to use inverses?
| This is called Trigonometric Substitution.
$$\int\sin^2 x\cos xdx, \color{green}{\text{let }u=\sin x}\implies du=\cos xdx$$
$$\int\sin^2 x(\cos xdx)=\int u^2du =\frac{u^3}{3}+C =\frac{\sin^3 x}{3}+C$$
Whereas this is called Inverse Trigonometric Substitution
$$\int\sqrt{1-4x^2}dx = \int\sqrt{1-(2x)^2}dx, \color{green}{\text{let }\sin u=2x}\implies \cos u du=2dx$$
$$\begin{array}{lll}
\int\sqrt{1-(2x)^2}dx&=&\frac{1}{2}\int 2\sqrt{1-(2x)^2}dx\\
&=&\frac{1}{2}\int \cos u\sqrt{1-\sin^2 u}du\\
&=&\frac{1}{2}\int \cos^2 udu\\
&=&\dots\\
\end{array}$$
Notice that the substitution $\sin u = 2x$ can be rewritten as
$$u = \sin^{-1} (2x)$$
Compare the differences between the substitutions in the two examples, and it should become clear why inverses are necessary.
It may be instructive to use implicit differentiation to derive
$$\frac{d}{dt}\sin^{-1}t = \frac{1}{\sqrt{1-t^2}}$$
and then explicitly make the substitution $u = \sin^{-1}(2x)$
$$\frac{du}{dx} = \frac{d}{d2x}\sin^{-1}(2x)\cdot\frac{d}{dx}2x = \frac{2}{\sqrt{1-(2x)^2}}$$
$$du = \frac{2dx}{\sqrt{1-(2x)^2}}$$
Getting back to our integral
$$\int\sqrt{1-(2x)^2}dx=\int\sqrt{1-(2x)^2}\cdot\frac{\color{green}{2}\sqrt{1-(2x)^2}}{2\color{green}{\sqrt{1-(2x)^2}}}dx = \int\frac{1}{2}(1-(2x)^2)\cdot\color{green}{\frac{2}{\sqrt{1-(2x)^2}}}dx$$
$$= \int\frac{1}{2}(1-\sin^2(\sin^{-1}(2x)))\cdot\frac{2}{\sqrt{1-(2x)^2}}dx$$
$$= \int\frac{1}{2}(1-\sin^2 u)du = \frac{1}{2}\int\cos^2 udu=\dots$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Associative operation on a set $S$ Let $S$ be a non-empty set and $n\in \mathbb{N}, n\geq 2$ a fixed integer. We consider an associative operation $"\cdot"$ on $S$ with the following properties:
*
*$x^{n+1}=x, \forall x\in S$
*$xy^nx=yx^ny, \forall x,y\in S$
Show that this operation is commutative. This problem has been given at a romanian contest. The proof uses very complicated substitutions and I am wondering if there exists an elegant proof of this. Thank you!
| We have
$$ \tag1x^2 = (x^2)^{n+1}=x(x^2)^nx=x^2x^nx^2=xx^{n+1}x^2=x^4$$
and consequently
$$\tag2 x = x^{n+1}=x^2x^{n-1}\stackrel{(1)}=x^4x^{n-1}=x^2x^{n+1}=x^3$$
and
$$ \tag3x^n=x^{n-1}x\stackrel{(2)}=x^{n-1}x^3=x^{n+1}x=x^2$$
for all $x$.
Now
$$\tag4 xy\cdot yx=xy^2x\stackrel{(3)}=xy^nx=yx^ny\stackrel{(3)}=yx^2y=yx\cdot xy$$
so that $xy$ and $yx$ commute for all $x,y$.
Next,
$$\tag5\begin{align} xy&\stackrel{(2)}=xy^3\stackrel{(2)}=(xy)^3y^2\stackrel{(2)}=(xy)^5y^2\\&=x(yx)^2\cdot y(xy)^2y\cdot y\\
&\stackrel{(4)}=x(yx)^2\cdot (xy)y^2(xy)\cdot y\\
&=x(yx)^2x\cdot y^3xy^2\stackrel{(3)}=x(yx)^2x\cdot yxy^2\\
&\stackrel{(4)}=(yx)x^2(yx)yxy^2\stackrel{(2)}=(yx)^3y^2\stackrel{(2)}=yxy^2\end{align}$$
and
$$\tag6 xy\stackrel{(5)}=yxyy\stackrel{(5)}=yyxyyy\stackrel{(2)}=y^2xy$$
so that $y$ commutes with $yxy$ as well as with $xy^2$.
The operation $x*y:=yx$ has the same properties as "$\cdot$". Thus we obtain the corresponding result
$$ yx = x*y\stackrel{(5^*)}=y*x*y*y = yyxy \stackrel{(6)}= xy.$$
If we are bored, we can as well translate the derivation of $(5)$ directly to one of $(5^*)$ and thus show:
$$\begin{align} yx&\stackrel{(2)}=y^3x\stackrel{(2)}=y^2(yx)^3\stackrel{(2)}=y^2(yx)^5\\&= y\cdot y(yx)^2y\cdot (xy)^2x\\
&\stackrel{(4)}=y\cdot (yx)y^2(yx)\cdot(xy)^2x\\
&=y^2xy^3\cdot x(xy)^2x\stackrel{(3)}=y^2xy\cdot x(xy)^2x\\
&\stackrel{(4)}=y^2xy(xy)x^2(xy)\stackrel{(2)}=y^2(xy)^3\stackrel{(2)}=y^2xy\\
&\stackrel{(6)}=xy.\end{align}$$
| {
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"url": "https://math.stackexchange.com/questions/1509184",
"timestamp": "2023-03-29T00:00:00",
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Epsilon and Delta proof for limit $\lim_{x\to-1} \frac{x+5}{2x+3}=4$ I have a function $f(x)=\frac{x+5}{2x+3}$ and want to show that the limit
$$\lim_{x\to -1} f(x) = 4$$
is true using $\epsilon$-$\delta$. I have a trouble finding $\delta$ so that $f(x)-4$ is less than every $\epsilon$.
Can anyone help?
| Given $\epsilon > 0$, for $x \in (-1.2, -0.8)$, $|2x + 3| > 0.6$. Take $\delta = \min(0.2, \epsilon/12)$, then for all $|x - (-1)| < \delta$, we have
\begin{align}
& \left|f(x) - 4\right| \\
= & \left|\frac{x + 5}{2x + 3} - 4\right| \\
= & \left|\frac{-7x - 7}{2x + 3}\right| \\
= & 7\left|\frac{x + 1}{2x + 3}\right| \\
\leq & 7|x + 1|/0.6 \\
< & 12 |x + 1| < \epsilon.
\end{align}
| {
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"timestamp": "2023-03-29T00:00:00",
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Solving $\sqrt\frac{x^2+2x-2}{x^2+3x+2} + \sqrt\frac{x^2+3x+2}{x^2+2x-2}=\frac52 $ Got these two problems on an exam recently and was unsure if I managed to answer them correctly:
$$\sin^4x + \cos^4x \ge \frac58, \ x \in \left\{-\frac\pi4,\frac\pi4\right\}$$
Second:
$$\sqrt\frac{x^2+2x-2}{x^2+3x+2} + \sqrt\frac{x^2+3x+2}{x^2+2x-2}=\frac52 $$
Anyone able to provide the correct answer or method of solving these?
| For the first, start from $\sin^2x+\cos^2x=1$ and square both sides.
For the second, note that the second part is the reciprocal of the first. So you need to solve $p+\frac 1p=\frac 52$ first.
But what have you tried?
| {
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"timestamp": "2023-03-29T00:00:00",
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Solve $ \left|\frac{x}{x+2}\right|\leq 2 $ I am experiencing a little confusion in answering a problem on Absolute Value inequalities which I just started learning. This is the problem: Solve:
$$
\left|\frac{x}{x+2}\right|\leq 2
$$
The answer is given to be $x\leq-4$ or $x \geq-1$
This is my attempt to solve the problem:
By dividing, $\left|\frac{x}{x+2}\right|\leq 2$ is equivalent to $\left |1-\frac{2}{x+2}\right|\leq2$ which is also equivalent to $\left |\frac{2}{x+2}-1\right|\leq2$
So, $-2\leq\frac{2}{x+2}-1\leq2$ which is equivalent to $-\frac{1}{2}\leq\frac{1}{x+2}\leq\frac{3}{2}$
Case 1: $x+2>0$. Solving $-\frac{1}{2}\leq\frac{1}{x+2}\leq\frac{3}{2}$, I get $x\geq-4$ and $x\geq-\frac{4}{3}$ which is essentially $x\geq-\frac{4}{3}$.
Case 2: $x+2<0$. Solving $-\frac{1}{2}\times(x+2)\geq{1}\geq\frac{3}{2}\times(x+2)$, I get $x\leq-4$ and $x\leq-\frac{4}{3}$ which is essentially $x\leq-4$.
So, the solutions are: $x\leq-4$ or $x\geq-\frac{4}{3}$.
I couldn't get $x \geq-1$ as a solution. Did I do anything wrong? The book I am using is Schaum's Outlines-Calculus. Another question I would like to ask is that am I using 'and' and 'or' correctly in the above attempt to solve the problem? I have had this problem many times.
| Too long for a comment:
As long as $x\in\Bbb{R}$, you can consider $|\frac{x}{x+2}|$ as $(\frac{x}{x+2})\text{sign}(\frac{x}{x+2})$
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "3",
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Solving $z^6=64$
$$z^6=64$$
My attempt:
$$\stackrel{\text{Euler}}{=}[r(\cos(\theta)+i\sin(\theta))]^{6}=64$$
$$\stackrel{\text{De moivre}}{=}r^6(\cos(6\theta)+i\sin(6\theta))=64$$
$$\stackrel{\text{Euler}}{=}r^6e^{i6\theta}$$
I know that I should get $6$ roots..
| $$z^6=64\Longleftrightarrow$$
$$z^6=\left|64\right|e^{\arg\left(64\right)i}\Longleftrightarrow$$
$$z^6=64e^{oi}\Longleftrightarrow$$
$$z=\left(64e^{2\pi ki}\right)^{\frac{1}{6}}\Longleftrightarrow$$
$$z=2e^{\frac{1}{6}\cdot 2\pi ki}\Longleftrightarrow$$
$$z=2e^{\frac{1}{3}\cdot \pi ki}$$
With $k\in\mathbb{Z}$ and $k:0-5$
So the solutions are:
$$z_0=2e^{\frac{1}{3}\cdot \pi \cdot 0 i}=2$$
$$z_1=2e^{\frac{1}{3}\cdot \pi \cdot 1 i}=2e^{\frac{\pi}{3}i}$$
$$z_2=2e^{\frac{1}{3}\cdot \pi \cdot 2 i}=2e^{\frac{2\pi}{3}i}$$
$$z_3=2e^{\frac{1}{3}\cdot \pi \cdot 3 i}=-2$$
$$z_4=2e^{\frac{1}{3}\cdot \pi \cdot 4 i}=2e^{-\frac{2\pi}{3}i}$$
$$z_5=2e^{\frac{1}{3}\cdot \pi \cdot 5 i}=2e^{-\frac{\pi}{3}i}$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How to compute this finite sum $\sum_{k=1}^n \frac{k}{2^k} + \frac{n}{2^n}$? I do not know how to find the value of this sum:
$$\sum_{k=1}^n \frac{k}{2^k} + \frac{n}{2^n}$$
(Yes, the last term is added twice).
Of course I've already plugged it to wolfram online, and the answer is $$2-\frac{1}{2^{n-1}}$$
But I do not know how to arrive at this answer.
I am not interested in proving the formula inductively :)
| A trick with generating functions:
$\begin{align}
\sum_{k \ge 0} 2^{-k} z^k
&= \frac{1}{1 - 2^{-1} z} \\
z \frac{\mathrm{d}}{\mathrm{d} z} \frac{1}{2 (1 - 2^{-1} z)}
&= \frac{z}{2 (1 - 2^{-1} z)^2} \\
&= \sum_{k \ge 0} k 2^{-k} z^k \\
\frac{1}{1 - z} \cdot \frac{z}{(1 - 2^{-1} z)^2}
&= \sum_{n \ge 0} \sum_{0 \le k \le n} k 2^{-k} z^n
\end{align}$
Thus you are interested in the coefficient of $z^n$ in the above:
$\begin{align}
[z^n] \frac{z}{(1 - z) (1 - 2^{-1} z)^2}
&= [z^n] \left(
\frac{4}{1 - z}
- \frac{2}{1 - 2^{-1} z}
- \frac{2}{(1 - 2^{-1} z)^2}
\right) \\
&= 4 - 2 \cdot 2^{-n} - 2 (n + 1) \cdot 2^{-n} \\
&= 4 - \frac{n + 2}{2^{n - 1}}
\end{align}$
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "4",
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"answer_id": 4
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Find the limit? How to calculate this limit ?
$$\lim _{x\to 1}\frac{\sqrt{x^2-1}}{\sqrt[3]{x-1}}$$
An image for clarification.
| $$\lim _{ x\rightarrow 1 }{ \left( \frac { \sqrt { { x }^{ 2 }-1 } }{ \sqrt [ 3 ]{ x-1 } } \right) = } \lim _{ x\rightarrow 1 }{ \left( \sqrt [ 6 ]{ \frac { { \left( { x }^{ 2 }-1 \right) }^{ 3 } }{ { \left( x-1 \right) }^{ 2 } } } \right) = } \lim _{ x\rightarrow 1 }{ \left( \sqrt [ 6 ]{ \frac { { \left( { x }-1 \right) ^{ 3 }\left( { x }+1 \right) }^{ 3 } }{ { \left( x-1 \right) }^{ 2 } } } \right) = } \lim _{ x\rightarrow 1 }{ \left( \sqrt [ 6 ]{ \left( { x }-1 \right) \left( { x }+1 \right) ^{ 3 } } \right) =0 } $$
| {
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"timestamp": "2023-03-29T00:00:00",
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Show that $\sqrt{x-1}+\sqrt{y-1}\leq \sqrt{xy}$ How can one show that $\sqrt{x-1}+\sqrt{y-1}\leq \sqrt{xy}$
Assuming that :
$\sqrt{x-1}+\sqrt{y-1}\leq \sqrt{xy}$
So
$(\sqrt{x-1}+\sqrt{y-1})^2\leq xy$
$\sqrt{(x-1)(y-1)} \leq xy-x-y+2$
$ (y-1)(x-1)+3 \leq \sqrt{(x-1)(y-1)}$
Here I'm stuck !
| WLOG $\sqrt x=\sec A,\sqrt y=\sec B$ where $0<A,B<\dfrac\pi2$
$\sqrt{x-1}+\sqrt{y-1}=\tan A+\tan B=\dfrac{\sin(A+B)}{\cos A\cos B}\le\sqrt{\sec A\sec B}=?$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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What are roots of $x^{3}+3x^{2}+4x+1$? There are no divisor of 1 in this polynomial for which would be satisfied $x^{3}+3x^{2}+4x+1=0$.
How to find roots here?
| This can be done by hand with a bunch of substitutions.
Let $y=x+1$, then
$$1+4(y-1)+3(y-1)^2+(y-1)^3=0$$
If we expand all the terms we get
$$1+4y-4+3y^2-6y+3+y^3-3y^2+3y-1=0$$
which boils down to
$$y^3+y-1=0$$
Now we can do a change in coordinates with another substitution.
Let $y=u+\frac{\lambda}{u}$. We will find $\lambda$ later. This yields.
$$-1+u+\frac{\lambda}{u}+\left ( u + \frac{\lambda}{u} \right )^3=0$$
We can expand this by multiplying by $u^3$ which gives
$$u^4(3\lambda + 1) + u^2\lambda(3\lambda+1)-u^3+u^6+\lambda^3=0$$
We want the coefficients on the $u^2$ and $u^4$ terms to be $0$. So $\lambda=-\frac{1}{3}$.
Substituting we now have
$$u^6-u^3-\frac{1}{27}=0$$
If we perform yet another substitution, we can finally get to a quadratic.
Let $z=u^3$.
$$z^2-z+\frac{1}{27}=0$$
Using the quadratic formula we get
$$z=\frac{1}{18}(9+\sqrt{93})$$.
Now we climb back up our ladder of subsitutions
$$u^3=\frac{1}{18}(9+\sqrt{93})$$
$$u=\sqrt[3]{\frac{1}{18}(9+\sqrt{93})}$$
We substitute for $u$ and get
$$y=\sqrt[3]{\frac{1}{18}(9+\sqrt{93})}-\frac{1}{3\sqrt[3]{\frac{1}{18}(9+\sqrt{93})}}$$
Now to get $x$ we just subtract 1.
$$x=\left( \frac{1}{18}(9+\sqrt{93})\right )^{1/3} - \frac{1}{3}\left( \frac{1}{18}(9+\sqrt{93})\right )^{-1/3}-1$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Modified Leibnitz integral: $\lim\limits_{a \to\infty}\frac1a\int _0^\infty\frac{(x^2+ax+1)\arctan(\frac{1}{x})}{1+x^4}dx=?$ $\lim\limits_{a \to \infty} \frac{1}{a} \int _0^\infty\frac{(x^2+ax+1)\arctan(\frac{1}{x})}{1+x^4}dx $ ,where $a$ is a parameter.
ATTEMPT:- Let $I(a)=\frac{1}{a} \int _0^\infty \frac{(x^2+ax+1)\arctan(\frac{1}{x})}{1+x^4}dx$
Now by Leibnitz theorem, $I'(a)= -\int _0^\infty\frac{(1+x^2)(arccot(x))}{(1+x^4)a^2}dx$
Substituting $x=cot\theta.$
$\implies$ $I'(a)= -\int _0^\frac{\pi}{2}\frac{(\theta)(cosec^4\theta)}{(1+cot^4\theta)a^2}d\theta$
Also $I'(a)= -\int _0^\frac{\pi}{2}\frac{(\frac{\pi}{2})(sec^4\theta)}{(1+tan^4\theta)a^2}d\theta +\int _0^\frac{\pi}{2}\frac{(\theta)(sec^4\theta)}{(1+tan^4\theta)a^2}d\theta$
$\implies $$2I'(a)= -\int _0^\frac{\pi}{2}\frac{(\frac{\pi}{2})(sec^4\theta)}{(1+tan^4\theta)a^2}d\theta+\int _0^\frac{\pi}{2}\frac{(\theta)(sec^4\theta)}{(1+tan^4\theta)a^2}d\theta -\int _0^\frac{\pi}{2}\frac{(\theta)(cosec^4\theta)}{(1+cot^4\theta)a^2}d\theta$
Note:The last two integrals add to ZERO.
$\implies$ $2I'(a)=-\int _0^\frac{\pi}{2}\frac{(\frac{\pi}{2})(sec^4\theta)}{(1+tan^4\theta)a^2}d\theta$
This integral can be easily evaluated by putting $tan\theta =t$
However I am getting $I(a)=\frac{\pi^2}{4\sqrt{2}a}.$ which is not the correct answer.
| Hint. One may recall that
$$
\arctan x +\arctan \frac1x =\frac{\pi}2,\qquad x>0.
$$
Then by the change of variable $x \to \dfrac1x$, one gets that
$$
aI(a):=\int _0^\infty\frac{(x^2+ax+1)\arctan(\frac1x)}{1+x^4}dx=\int _0^\infty\frac{(x^2+ax+1)\arctan x}{1+x^4}dx
$$ Thus we obtain
$$
aI(a)=\frac{\pi}2\int_0^\infty\frac{(x^2+ax+1)}{1+x^4}dx-aI(a)
$$ or
$$
aI(a)=\frac{\pi}4\int_0^\infty\frac{(x^2+ax+1)}{1+x^4}dx.
$$ Now the latter integral is classically evaluated, giving
$$
aI(a)=\frac{\pi}4 \times \frac{\pi}4 \left(a+2 \sqrt{2}\right)
$$ and
$$
I(a)=\frac{\pi^2}{16}\left(1+ \frac{2\sqrt{2}}a\right), \qquad a\neq0.
$$
One may deduce that, as $a \to \infty$,
$$
I(a)\to\frac{\pi^2}{16}.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1520120",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Express $-a^3 b + a^3 c+ab^3-a c^3-b^3c+bc^3$ as a product of linear factors.
Express
$$-a^3 b + a^3 c+ab^3-a c^3-b^3c+bc^3$$
as a product of linear factors.
I have tried rewriting the expression as:
$$ab^3-a^3b + a^3c-ac^3 +bc^3-b^3c$$
$$= ab(b^2-a^2)+ac(a^2-c^2)+bc(c^2-b^2)$$
$$= ab(b-a)(b+a) + ac(a-c)(a+c)+bc(c-b)(c+b)$$
Which is at least a product of linear factors.
However I am now stuck as to how to proceed.
Also I would prefer the $(b^2-a^2)$ and the $(c^2-b^2)$ factors of my second line of working to be in the form $(a^2-b^2)$ and $(b^2-c^2)$, for 'neatness' , if possible
| This is maybe not the most elegant answer, but I think it's relatively efficient. By trial and error we find that $a=b$, $a=c$ and $b=c$ make the expression vanish. Hence we write it as
$$\begin{align}
(a-b)(a-c)(b-c)f&=(a^2b-a^2c-ab^2+ac^2+b^2c-bc^2)f\\
&=-a^3 b + a^3 c+ab^3-a c^3-b^3c+bc^3
\end{align}$$
Now you could either divide one side by the other, or stare at it for a bit and note that $f=(a+b+c)$ makes it work.
This tactic only works when the things you're doing is a homework exercise, since you can be pretty sure that the resulting factorisation will be nice and easy, so trying all 'easy relations' will get you quite far generally.
Note that since all coefficients are $1$, and you expect the factorisation to be easy, it wouldn't even be that far fetched to also just try $a=-b-c$ and see what comes out. This would have given you the solution without any calculations
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1521015",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
Solve $\lim\limits_{x \to 1} \frac{2x-2}{(x^3-x)^2}$ without L'Hôpital's rule Find limit without L'Hôpital's rule: $\lim\limits_{x \to 1} \frac{2x-2}{(x^3-x)^2}$
I'm learning limits and I'm stuck with this example. So far, I tried modify denominator but I have no luck with it.
| $$\lim\limits_{x \to 1} \frac{2x-2}{(x^3-x)^2}=\lim\limits_{x \to 1} \frac{2(x-1)}{x^2(x^2-1)^2}=\lim\limits_{x \to 1} \frac{2(x-1)}{x^2(x-1)^2(x+1)^2}$$
Therefore $\lim\limits_{x \to 1^+} \frac{2}{x^2(x-1)(x+1)^2}=+\infty$
and $\lim\limits_{x \to 1^-} \frac{2}{x^2(x-1)(x+1)^2}=-\infty$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1522989",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Solve the limit without using L'Hopitals $\lim _{x\to 1}\left(\frac{\sqrt[3]{x+7}-2}{2x^2+3x-5\:}\right)$ $$\lim _{x\to 1}\left(\frac{\sqrt[3]{x+7}-2}{2x^2+3x-5\:}\right)$$
I think I need to do in the numerator by multiplying the difference of cubes.
$\lim _{x\to 1}\left(\frac{\left(\sqrt[3]{x+7}-2\right)\cdot \left(\sqrt[3]{\left(x+7\right)^2}+2\sqrt[3]{x+7}+4\right)}{\left(2x^2+3x-5\right)\:\cdot \left(\sqrt[3]{\left(x+7\right)^2}+2\sqrt[3]{x+7}+4\right)}\right)=\lim _{x\to 1}\left(\frac{x+7-8}{\left(2x^2+3x-5\right)\:\cdot \left(\sqrt[3]{\left(x+7\right)^2}+2\sqrt[3]{x+7}+4\right)}\right)=\lim _{x\to 1}\left(\frac{x-1}{\left(2x^2+3x-5\right)\:\cdot \left(\sqrt[3]{\left(x+7\right)^2}+2\sqrt[3]{x+7}+4\right)}\right)$
Here's what to do with the denominator?
and whether I chose the idea of a solution?
| $$
\begin{aligned}
\lim _{x\to 1}\left(\frac{\sqrt[3]{x+7}-2}{2x^2+3x-5\:}\right)
& = \lim _{t\to 0}\left(\frac{\sqrt[3]{\left(t+1\right)+7}-2}{2\left(t+1\right)^2+3\left(t+1\right)-5\:}\right)
\\& = \lim _{t\to 0}\left(\frac{\sqrt[3]{t+8}-2}{2t^2+7t}\right)
\\& = \lim _{t\to 0}\left(\frac{2+\frac{1}{12}t-\frac{1}{288}t^2+o\left(t^2\right)-2}{2t^2+7t}\right)
\\& = \color{red}{\frac{1}{84}}
\end{aligned}
$$
Solved with substitution $\color{red}{t = x-1}$ and Taylor expansion
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1526961",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Find the minimum of $(u-v)^2+ (\sqrt{2-u^2}-\frac{9}{v})^2$ for $00$.
Find the minimum of $\displaystyle (u-v)^2+ \left(\sqrt{2-u^2}-\frac{9}{v}\right)^2$ for
$0<u<\sqrt{2}$ and $v>0$.
I think I have to use the Arithmetic and Geometric Means Inequalities. Or $\displaystyle \frac{1}{2}(u-v)^2+ \left(\sqrt{2-u^2}-\frac{9}{v}\right)^2 \geq \frac{1}{4}\left(u-v+\sqrt{2-u^2}-\frac{9}{v}\right)^2$ by Chebychev inequality. Is anyone is able to give me a hint how to finish it?
I think I have to use the Arithmetic and Geometric Means Inequalities.
| $\bf{Using\; Cauchy\; Schwartz\; Inequality::}$
$$[1^2+1^2]\cdot \left[(u-v)^2+\left(\sqrt{2-u^2}-\frac{9}{v}\right)^2\right]\geq \left[v+\frac{9}{v}-\left(u+\sqrt{2-u^2}\right)\right]^2$$
Now Using $\bf{A.M\geq G.M}\;,$ We get
$\displaystyle v+\frac{9}{v}\geq 6$ and equality hold when $v=3$ and $\displaystyle \left(u+\sqrt{2-u^2}\right)^2\leq 2(u^2+2-u^2)=4$
So we get $\displaystyle \left(u+\sqrt{2-u^2}\right)\leq 2$ and equality hold when $u=1$
So we get $$2\cdot \left[(u-v)^2+\left(\sqrt{2-u^2}-\frac{9}{v}\right)^2\right]\geq \left[6-2\right]^2 = 16$$
So $$\left[(u-v)^2+\left(\sqrt{2-u^2}-\frac{9}{v}\right)^2\right]\geq 8$$
And equality hold ,When $u=3$ and $v=1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1530179",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
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