Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Number of integer solutions of the equation $x + y + z = 30$ if $x \geq 2$, $y \geq 0$, $z \geq -3$ I am trying to solve a problem using permutation/combination but cannot figure out how to proceed.
Suppose the sum of three variables $x, y, z$ is $30$. If $x\ge2, y\ge0, z\ge-3$, how many integer solutions exist?
I understand that $2\le x\le33, 0\le y\le31, -3\le z\le28$. A simple simulation shows that there are $528$ solutions. However, I am unable to calculate this mathematically. I would like a hint so that I can try this on my own.
| We wish to find the number of solutions of the equation
$$x + y + z = 30 \tag{1}$$
in the integers subject to the restrictions that $x \geq 2$, $y \geq 0$, and $z \geq -3$.
Let $x' = x - 2$, $y' = y$, and $z' = z + 3$. Then $x', y', z'$ are nonnegative integers. Substituting $x' + 2$ for $x$, $y'$ for $y$, and $z' - 3$ for $z$ in equation 1 yields
\begin{align*}
x' - 2 + y' + z' + 3 & = 30\\
x' + y' + z' & = 31 \tag{2}
\end{align*}
Equation 2 is an equation in the nonnegative integers. A particular solution of equation 2 corresponds to the placement of two addition signs in a row of $31$ ones. For instance,
$$1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 + 1 1 1 1 1 1 1 1 1 1 +$$
corresponds to the solution $x_1 = 21$, $x_2 = 10$, $x_3 = 0$, while
$$1 1 1 1 1 1 1 + 1 1 1 1 1 1 1 1 1 1 1 + 1 1 1 1 1 1 1 1 1 1 1 1 1$$
corresponds to the solution $x_1 = 7$, $x_2 = 11$, $x_3 = 13$. The number of solutions of equation 2 is the number of ways we can place two addition signs in a row of $31$ ones, which is equal to the number of ways we can select which $2$ of the $33$ positions required for $31$ ones and $2$ addition signs will be filled with addition signs.
$$\binom{31 + 2}{2} = \binom{33}{2} = \frac{33!}{2!31!} = \frac{33 \cdot 32 \cdot 31!}{2 \cdot 1 \cdot 31!} = \frac{33 \cdot 32}{2} = 33 \cdot 16 = 528$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2909070",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
Finding joint CDF from joint PDF.
Suppose $X$ and $Y$ are random variables with joint density
$$
f(x,y) = \left\{
\begin{array}{ll}
1/\pi & \quad\text{if}\ x^{2} + y^{2} \leq 1 \\
0 & \quad \mathrm {otherwise}
\end{array}
\right.$$
Find the joint CDF of $X$ and $Y$.
How do I proceed? Please help me as I am a beginner of this topic.
| I think the joint CDF $F(x,y)= \mathbb P(X \le x,Y \le Y)$ may be fairly involved, and looking at this diagram suggests the following to me:
*
*If $x \le -1$, or if $y \le -1$, or if $x \le 0$ and $y \le 0$ and $x^2+y^2 \ge 1$:
$$F(x,y)=0$$
*
*If $x \ge 1$ and $y \ge 1$:
$$F(x,y)=1$$
*
*If $-1 \le x \le 1$ and $y \ge 1$, or if $-1 \le x \le 0$ and $0 \le y \le 1$ and $x^2+y^2 \ge 1$:
$$F(x,y)= \frac1\pi\left( \frac\pi 2 + \sin^{-1}(x) + x\sqrt{1-x^2} \right)$$
*
*If $x \ge 1$ and $-1 \le y \le 1$, or if $0 \le x \le 1$ and $-1 \le y \le 0$ and $x^2+y^2 \ge 1$:
$$F(x,y)= \frac1\pi\left( \frac\pi 2 + \sin^{-1}(y) + y\sqrt{1-y^2} \right)$$
*
*If $0 \le x \le 1$ and $0 \le y \le 1$ and $x^2+y^2 \ge 1$:
$$F(x,y)= \frac1\pi\left( \sin^{-1}(x)+ \sin^{-1}(y) + x\sqrt{1-x^2} + y\sqrt{1-y^2} \right)$$
*
*If $x^2+y^2 \le 1$:
$$F(x,y)= \frac1\pi\left( \frac\pi 4 + \frac1{2}\sin^{-1}(x) + \frac1{2} \sin^{-1}(y) + \frac1{2} x \sqrt{1-x^2} + \frac1{2}y \sqrt{1-y^2} + x y \right)$$
The interesting parts of the marginal CDFs for $X$ and $Y$ are in the third and fourth points above respectively
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2909296",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Can we prove this inequality $1+\sqrt{2}+\cdots+\sqrt{n}<\frac{4n+3}{6}\sqrt{n}$ using integrals by parts? Can we prove this inequality $1+\sqrt{2}+\cdots+\sqrt{n}<\dfrac{4n+3}{6}\sqrt{n}$ using integrals by parts?
I need to prove this inequality, and I tried this way:
$$\begin{split}
\int_0^n\sqrt{x} \,dx &=\sum\limits_{k=1}^n \int_{k-1}^k \sqrt{x}\, dx\\
&=\sum\limits_{k=1}^n \int_0^1 \sqrt{t+k-1} \, dt\\
&=\sum\limits_{k=1}^n \sqrt{k}-\sum\limits_{k=1}^n \int_0^1\dfrac{t}{2\sqrt{t+k-1}}\,dt
\end{split}$$
Thus
$$\sum\limits_{k=1}^n \sqrt{k}=\int_0^n \sqrt{x}\, dx+\sum\limits_{k=1}^n \int_0^1 \dfrac{t}{2\sqrt{t+k-1}}\, dt$$.
But maybe I didn't find some inequality trick, I cannot find a proper approxmation of the sum of the integrals.
| First, see that we have
$$
\int_0^n \sqrt{x} \;dx = \frac{4n}{6}\sqrt{n}
$$
which gives us a "part" of the demanded result. Moreover, it suggests that we should try to prove the following:
$$
\sum_{k=1}^n \int_0^1 \frac{t}{2\sqrt{t+k-1}} \;dt < \frac{3}{6}\sqrt{n} = \frac{1}{2}\sqrt{n}
$$
Evaluating the integral we get:
\begin{align}
\int_0^1 \frac{t}{2\sqrt{t+k-1}} \;dt &=
\int_{k-1}^{k} \frac{x-k+1}{2\sqrt{x}} \;dx =
\frac{1}{2} \int_{k-1}^{k} \sqrt{x} \;dx - (k-1)
\int_{k-1}^{k} \frac{1}{2\sqrt{x}} \;dx \\
&=
\frac{1}{3} \left(-2k\sqrt{k} + 2k\sqrt{k-1} + 3\sqrt{k} - 2\sqrt{k-1}\right)
\end{align}
However, for all $k \geq 1$, you can prove that:
$$
\frac{1}{3} \left(-2k\sqrt{k} + 2k\sqrt{k-1} + 3\sqrt{k} - 2\sqrt{k-1}\right) <
\frac{1}{2} \left(\sqrt{k}-\sqrt{k-1}\right)
$$
So we end up with a telescoping sum and the result we wanted:
$$
\sum_{k=1}^n \int_0^1 \frac{t}{2\sqrt{t+k-1}} \;dt <
\frac{1}{2} \sum_{k=1}^n \left(\sqrt{k}-\sqrt{k-1}\right) =
\frac{1}{2} \sqrt{n}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2912228",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
Finding a closed form for $\sum_{n = 1}^{\infty} \frac{1}{(1 + x^2)^{n}}$ So I got the sum
$$\sum_{n = 1}^{\infty} \frac{1}{(1 + x^2)^{n}}$$
and I want a closed form.
Can I just do
$$\frac{1}{1 - r} = \frac{1}{1 - \frac{1}{1 + x^2}} = \frac{1}{x^2} + 1 $$
for a closed form?
| The formula that you are using for the sum of a geometric series, namely that
$$
\sum_{n=0}^\infty ar^n = \frac{a}{1-r},
$$
is valid only when $|r|<1$. In your case, the common ratio $r$ is
$$
\frac{1}{1+x^2},
$$
which is less than $1$ for all $x \neq 0$. So, for $x \neq 0$, you can express your series as
$$
\frac{\frac{1}{1+x^2}}{1-\frac{1}{1+x^2}} = \frac{1}{x^2}.
$$
Note that you missed the initial term $a = \frac{1}{1+x^2}$ in your computation.
When $x = 0$, the series diverges because it is the infinite sum $1 + 1 + 1 + \cdots$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2913996",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Determinant of Symmetric Matrix $\mathbf{G}=a\mathbf{I}+b\boldsymbol{ee}^T$ To give the close-form of $\det(\mathbf{G})$, where $\mathbf{G}$ is
\begin{align}
\mathbf{G}=a\mathbf{I}+b\boldsymbol{ee}^T
\end{align}
in which $a$ and $b$ are constant, and $\boldsymbol{e}$ is a column vector with all elements being $1$. In addition, $(\cdot)^T$ is transposition operation. $\mathbf{G}$ is $u\times u$. We use $\mathbf{G}_u$ to underline the dimension of $\mathbf{G}$.
The question is to determine $\det(\mathbf{G})$.
As I know: We rewrite $\mathbf{G}_u$ as
\begin{align}
\mathbf{G}_u=\left[\begin{array}{ccc}
\mathbf{G}_{u-1} & b\\
b & a+b
\end{array}
\right]
\end{align}
Using the determinant of block matrix lemma
\begin{align}
\det\left[\begin{array}{ccc}
\mathbf{A} & \mathbf{B}\\
\mathbf{C} & \mathbf{D}
\end{array}
\right]=\det(\mathbf{A})\det(\mathbf{D}-\mathbf{C}\mathbf{A}^{-1}\mathbf{B})
\end{align}
We then have
\begin{align}
\det(\mathbf{G}_u)=\det(\mathbf{G}_{u-1})\det(a+b-b^2\boldsymbol{e}^T\mathbf{G}_{u-1}^{-1}\boldsymbol{e})
\end{align}
It still needs to get $\mathbf{G}_{u-1}^{-1}$ via matrix inverse lemma
\begin{align}
(\mathbf{A}+\mathbf{BC})^{-1}=\mathbf{A}^{-1}-\mathbf{A}^{-1}\mathbf{B}(\mathbf{I}+\mathbf{CA}^{-1}\mathbf{B})^{-1}\mathbf{CA}^{-1}
\end{align}
We then have
\begin{align}
\mathbf{G}_{u}
&=\frac{1}{a}\mathbf{I}-\frac{1}{a^2}b\boldsymbol{e}\left(1+\frac{b}{a}\boldsymbol{e}^T\boldsymbol{e}\right)^{-1}\boldsymbol{e}^T\\
&=\frac{1}{a}\mathbf{I}-\frac{b}{a(a+bu)}\boldsymbol{ee}^T
\end{align}
where $\boldsymbol{e}^T\boldsymbol{e}=u$ is used. Plugging it into
\begin{align}
&\det(a+b-b^2\boldsymbol{e}^T\mathbf{G}_{u-1}^{-1}\boldsymbol{e})\\
=&a+b-b^2\left({\frac{u-1}{a}-\frac{b(u-1)^2}{a[a+b(u-1)]}}\right)
\end{align}
Then
\begin{align}
\det(\mathbf{G}_u)=\det(\mathbf{G}_{u-1})\left[{a+b-b^2\left({\frac{u-1}{a}-\frac{b(u-1)^2}{a[a+b(u-1)]}}\right)}\right]
\end{align}
Although, the connection between $\mathbf{G}_u$ and $\mathbf{G}_{u-1}$ is found, I can't give the expression of $\mathbf{G}_u$. Please give me hand, thanks a lot!
| A More Basic Approach
Let us consider for illustration, $G_4$:
$$
G=aI_{4\times4}+bee^T \\
\text{where } e = \begin{bmatrix} 1\\ 1\\ 1\\ 1 \end{bmatrix}
\\
\text{Hence, } G =
\begin{bmatrix}
a & 0 & 0 & 0 \\
0 & a & 0 & 0 \\
0 & 0 & a & 0 \\
0 & 0 & 0 & a \\
\end{bmatrix} +
\begin{bmatrix}
b & b & b & b \\
b & b & b & b \\
b & b & b & b \\
b & b & b & b \\
\end{bmatrix} =
\begin{bmatrix}
a + b & b & b & b \\
b & a + b & b & b \\
b & b & a + b & b \\
b & b & b & a + b \\
\end{bmatrix}
$$
Taking the determinant, let us apply the transformation $\left(R_1 \rightarrow R_1 + R_2 + R_3 + R_4 \right)$,
$$
\det(G) = \begin{vmatrix}
a + 4b & a + 4b & a + 4b & a + 4b \\
b & a + b & b & b \\
b & b & a + b & b \\
b & b & b & a + b \\
\end{vmatrix}
$$
Applying now $\left( R_1 \rightarrow \frac{1}{a+4b} R_1 \right)$ and $\prod_{k=2}^{4}{\left(R_k \rightarrow R_k - bR_1\right)}$
$$
= (a+4b) \cdot \begin{vmatrix}
1 & 1 & 1 & 1 \\
0 & a & 0 & 0 \\
0 & 0 & a & 0 \\
0 & 0 & 0 & a \\
\end{vmatrix}
\\
\text{So } \det{(G)} = (a+4b)\cdot a^3
$$
Similarly, for $G_n$, upon applying the transformations $\left( R_1 \rightarrow \frac{1}{a+nb} \sum_{k=1}^{n}{R_k} \right)$ and $\left( \prod_{k=2}^{n} \left( R_k \rightarrow R_k - bR_1 \right) \right)$ and then simply expanding,
$$
\boxed {\det{(G_n)} = (a+nb) \cdot a^{n-1}}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2916569",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 3
} |
How to evaluate $1+\frac{2^2}{3!}+\frac{3^2}{5!}+\frac{4^2}{7!}+\cdots$? I learnt that $\displaystyle \sum_{n=0}^{\infty} \frac{n+1}{(2n+1)!} = \frac{e}{2}$.
I am wondering what the closed form for $\displaystyle \sum_{n=0}^{\infty} \frac{(n+1)^2}{(2n+1)!}$ is.
I tried using the fact that $ 1+3+5+\cdots+(2n-1) = n^2$, but it was not fruitful.
Could people give me some hints on how to approach this problem?
To view the general formula, please visit The value of $\sum_{n=0}^{\infty}\frac{(n+1)^k}{(2n+1)!}$, where $k\in\mathbb{W}$
| Note that
$$\sinh x = \frac{e^x - e^{-x}}{2} = x + \frac{x^3}{3!} + \frac{x^5}{5!} + \cdots$$
Then
$$x \sinh x = x^2 + \frac{x^4}{3!} + \frac{x^6}{5!} + \cdots$$
Now we replace $x$ with $\sqrt{x}$ getting
$$\sqrt{x} \sinh \sqrt{x} = x + \frac{x^2}{3!} + \frac{x^3}{5!} + \cdots$$
We take the derivative
$$(\sqrt{x} \sinh \sqrt{x})' = 1 + \frac{2x}{3!} + \frac{3x^2}{5!} + \cdots$$
Then we multiply by $x$
$$x (\sqrt{x} \sinh \sqrt{x})' = x + \frac{2x^2}{3!} + \frac{3x^3}{5!} + \cdots$$
and take the derivative again
$$\left( x (\sqrt{x} \sinh \sqrt{x})' \right)' = 1 + \frac{2^2 x}{3!} + \frac{3^2 x^2}{5!} + \cdots$$
Finally, we set $x=1$
$$\left. \left( x (\sqrt{x} \sinh \sqrt{x})' \right)' \right|_{x=1} = 1 + \frac{2^2}{3!} + \frac{3^2}{5!} + \cdots$$
Thus, you only need to expand the left hand side analytically to find the value of the series.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2917988",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 1
} |
Limit of quotient I recently learned that when you are solving for the limit of a quotient, you have to divide everything by the highest number in the denominator, like this
$$ \lim_{x \to \infty} \frac{\sqrt{4 x^2 - 4}}{x+5} = \lim_{x \to \infty} \frac{\sqrt{4 - \frac{4}{x^2}}}{1 + \frac{5}{x}} = 2.$$
But I don't quite understand how when $x \to - \infty$ , the answer changes to $-2$, since when you divide everything by the highest power, it gives
$$\frac{\sqrt{4-\frac{4}{x^2}}}{1+\frac{5}{x}},$$ meaning that even if you put negative infinity in the place of $x$, it still only gives $0$ and leaves $2$ as the final answer.
Why does the answer change to $-2$? I understand that it must be $-2$ when I look at the graph, I just don't understand the algebra part of it.
| $\dfrac{\sqrt{4 \cdot 5^2}}{5}
= \dfrac{\sqrt{4 \cdot 5^2}}{\sqrt{5^2}}
= \sqrt{\dfrac{4 \cdot 5^2}{5^2}}
= 2$
$\dfrac{\sqrt{4 \cdot (-5)^2}}{-5}
= \dfrac{\sqrt{4 \cdot (-5)^2}}{-\sqrt{(-5)^2}}
= -\sqrt{\dfrac{4 \cdot (-5)^2}{(-5)^2}}
= -2$
If $x < 0$, then $x = -\sqrt{x^2}$.
$\dfrac{\sqrt{4 \cdot x^2}}{x}
= \dfrac{\sqrt{4 \cdot x^2}}{-\sqrt{x^2}}
= -\sqrt{\dfrac{4 \cdot x^2}{x^2}}
= -2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2918223",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 2
} |
Getting a closed form from $\sum_{i=1}^{n-1} \sum_{j=i+1}^{n} \sum_{k=1}^{j} 1$ I need to get a closed form from
$$
\sum_{i=1}^{n-1} \sum_{j=i+1}^{n} \sum_{k=1}^{j} 1
$$
Starting from the most outer summation, I got
$$
\sum_{k=1}^{j} 1 = j
$$
But now I don't know how to proceed with:
$$
\sum_{i=1}^{n-1} \sum_{j=i+1}^{n} j
$$
Could you guys please help me?
Thanks in advance.
| Thanks @Winther for pointing out the previous mistake
\begin{equation}
\sum_{i=1}^{n-1} \sum_{j=i+1}^{n} \sum_{k=1}^{j} 1
\end{equation}
We know that
\begin{equation}
\sum_{k=1}^{j} 1 = j
\end{equation}
So
\begin{equation}
\sum_{i=1}^{n-1} \sum_{j=i+1}^{n} \sum_{k=1}^{j} 1 =
\sum_{i=1}^{n-1} \sum_{j=i+1}^{n} j
\end{equation}
\begin{equation}
\sum_{j=i+1}^{n} j
=
\sum_{j=1}^{n} j
-
\sum_{j=1}^{i} j
=
\frac{n(n+1)}{2}
-
\frac{i(i+1)}{2}
\end{equation}
\begin{equation}
\sum_{i=1}^{n-1} \sum_{j=i+1}^{n} \sum_{k=1}^{j} 1 =
\sum_{i=1}^{n-1} ( \frac{n(n+1)}{2}
-
\frac{i(i+1)}{2})
=
\frac{n(n+1)(n-1)}{2}
-
\frac{1}{2}
\sum_{i=1}^{n-1}
i+i^2
\end{equation}
But
\begin{align}
\sum_{i=1}^{n-1}
i &= \frac{(n-1)n}{2} \\
\sum_{i=1}^{n-1}
i^2 &= \frac{(n-1)n(2n-1)}{6}
\end{align}
So
\begin{equation}
\sum_{i=1}^{n-1} \sum_{j=i+1}^{n} \sum_{k=1}^{j} 1 =
\frac{n(n+1)(n-1)}{2}
-
\frac{1}{2}
(\frac{(n-1)n}{2} + \frac{(n-1)n(2n-1)}{6})
\end{equation}
Let's arrange
\begin{equation}
\sum_{i=1}^{n-1} \sum_{j=i+1}^{n} \sum_{k=1}^{j} 1 =
\frac{6n(n+1)(n-1) - 3n(n-1) - n(n-1)(2n-1)}{12}
\end{equation}
We arrive at the most compact form,
\begin{equation}
\sum_{i=1}^{n-1} \sum_{j=i+1}^{n} \sum_{k=1}^{j} 1 =
\frac{(n-1)n(6n + 6 - 3 - 2n + 1)}{12}
=
\frac{(n-1)n(n + 1)}{3}
\end{equation}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2918464",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
If $x>0$ real number and $n>1$ integer, then $(1+x)^n>\frac{1}{2}n(n-1)x^2$ If $x>0$ real number and $n>1$ integer, then $(1+x)^n>\frac{1}{2}n(n-1)x^2$.
Is there a way to prove it without using the Binomial Theorem? Is it possible to use the Bernoulli's Inequality to prove it? If yes, please show me.
I tried to prove it by induction on $n$, where $x>0$ is fixed. But it doesn't let me reach the conclusion. Here's what I did:
Fix $x>0$. We instead prove that
\begin{equation*}
\frac{2(1+x)^n}{n(n-1)}> x^2
\end{equation*}
for all $n>1$. Clearly, this is true for $n=2$. Assume this statement holds for $n=k$, for some $k\geq 2$. We have
\begin{equation*}
\frac{2(1+x)^{k+1}}{(k+1)k}=\frac{2(1+x)^k}{k(k-1)}\frac{(x+1)(k-1)}{k+1}>\frac{(x+1)(k-1)}{k+1}x^2.
\end{equation*}
Since $y\mapsto \frac{y-1}{y+1}$ is increasing on $\mathbb{R}\setminus \{-1\}$, we get $\frac{y-1}{y+1}\geq \frac{1}{3}$ for all $y\geq 2$. Therefore,
\begin{equation*}
\frac{(x+1)(k-1)}{k+1}x^2\geq \frac{1}{3}(x+1)x^2>\frac{1}{3}x^2.
\end{equation*}
As you can see, I can not reach $>x^2$ instead of $>\frac{1}{3}x^2$.
| Let, $f(x)=(1+x)^n-\frac{1}{2} n(n-1) x^2.$
So, $f'(x)=n(1+x)^{n-1}-n(n-1)x>0 \forall n>1$
Since,for $n>1$ and $x\ge 0,$ by A.M.-G.M. inequality we have,
$$\frac{1.(1+x)^{n-1}+(n-2).1}{n-1}>(1+x)$$
$$\implies (1+x)^{n-1}>(n-1)x$$
Therefore, $f$ is strictly increasing.
Hence we have,
$$f(x)>f(0)\implies (1+x)^n>\frac{1}{2}n(n-1) x^2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2919501",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Calculating the sum of the infinite series $\frac{1}{5} + \frac{1}{3}\frac{1}{5^3} + \frac{1}{5} \frac{1}{5^5} +......$ How do I calculate the sum of the infinite series?
$$\frac{1}{5} + \frac{1}{3}.\frac{1}{5^3} + \frac{1}{5}. \frac{1}{5^5} +......$$
My attempt :
I know that $$\log (\frac{1+x}{1-x}) = 2 \, \left(x + \frac{x^3}{3} + \frac{x^5}{5} +.....+\frac{x^{2r-1}} {2r-1}+..\right)$$
Now
\begin{align}\frac{1}{5} + (\frac{1}{3}.\frac{1}{5^3} + \frac{1}{5}. \frac{1}{5^5} +......) &= \frac{1}{5} + \log \frac{(1 +1/5)} {(1-1/5)} - \frac{5}{2} \\ &=\frac{1}{5} + \log (\frac{6}{4})-10= -\frac{49}{5} + \log \frac{3}{2}
\end{align}
Please verify my answer, thank you!
| From your own formula,
$$\frac15+\frac13\frac1{5^3}+\frac15\frac1{5^5}+\cdots=\frac12\log\frac{1+\dfrac15}{1-\dfrac15}=\frac12\log\frac32.$$
Check:
The series is fast converging. With the first five terms, $0.202732552127\cdots$, vs. the exact value $0.2027325540541\cdots$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2925053",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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No of matrices of rank 3 I was tackling this problem
How many 4 x 3 matrices can be formed of rank 3 where entries are coming from a field of 3 elements.
When I solved it as finding total no of linearly independent sets with cardinality 4 in F^4(F). Assuming that each column of the matrix is coming from F^4(F).
I am getting correct answer as follows
(3^4-1)(3^4-3)(3^4-3^2)
But if I am solving this problem as taking each row from F^3 (F).
Then our task is to find matrices with rank 3. So any ordered basis can serve to form any three rows of A and remaining row can be anything
So I am getting (3^3-1)(3^3-3)(3^3-3^2)3^3.
But it doesn't matches with that answer which I got using columns. Where I am doing mistake I am unable to identify.
| You're only counting matrices where the upper $3\times3$ part has full rank. You need to add three more cases: If dependence first occurs in the third row, we have $(3^3-1)(3^3-3)3^2(3^3-3^2)$ options; if it first occurs in the second row, we have $(3^3-1)3(3^3-3)(3^3-3^2)$ options, and if it first occurs in the first row, we have $1(3^3-1)(3^3-3)(3^3-3^2)$ options, for a total of
\begin{eqnarray*}
&&(3^3-1)(3^3-3)(3^3-3^2)3^3+(3^3-1)(3^3-3)3^2(3^3-3^2)+(3^3-1)3(3^3-3)(3^3-3^2)+1(3^3-1)(3^3-3)(3^3-3^2)
\\
&=&
(3^3-1)(3^3-3)(3^3-3^2)(3^3+3^2+3+1)
\\
&=&
(3^3-1)(3^3-3)(3^3-3^2)\frac{3^4-1}{3-1}
\\
&=&
(3^3-1)(3^3-3)\frac{3^3-3^2}{3-1}(3^4-1)
\\
&=&
(3^3-1)(3^3-3)3^2(3^4-1)
\\
&=&
(3^4-3)(3^4-3^2)(3^4-1)
\\
&=&
(3^4-1)(3^4-3)(3^4-3^2)\;.
\end{eqnarray*}
One more piece of advice: You would probably have found the error yourself if you'd tried the simpler case of a $2\times1$ matrix.
| {
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show that $ \ a_n \leq x \leq a_n+\frac{1}{b^n} \ $ for any base $b \neq 10 $ holds for any real number in $[0,1]$. Every infinite sequence $ \ 0. a_1a_2a_3 \cdots \cdots $ represents a real number $ \ x \in [0,1] $ and $ \ \sum\limits_{i=1}^{n} \frac{a_i}{10^i} \leq x \leq \sum\limits_{i=1}^{n} \frac{a_i}{10^i} +\frac{1}{10^n} $ in decimal system for all $ n \in \mathbb{N}$.
Prove the similar inequality $ \ \sum\limits_{i=1}^{n} \frac{a_i}{b^i} \leq x \leq \sum\limits_{i=1}^{\infty} \frac{a_i}{b^i}+\frac{1}{b^n} \ $ for any base $b \neq 10 $.
Answer:
For any base $ \ b $ other than $ 10$ , a real number $ \ x \in [0,1] $ can be expressed as
$ x= \sum\limits_{n=0}^{\infty} a_n b^{-n} =a_0+\frac{a_1}{b}+\frac{a_2}{b^2}+\cdots +\frac{a_n}{b^n}+\cdots \cdots , \ \text{where $a_0 < 1$}. $
Thus, $ \ x \in [0,1] $.
But how to show that
$ \sum\limits_{i=1}^{n} \frac{a_i}{b^i} \leq x \leq \sum\limits_{i=1}^{n} \frac{a_i}{b^i} +\frac{1}{b^n} $ holds good?
Help me showing this
| Take the difference
$$0\leq x - \sum\limits_{i=1}^{n} \frac{a_i}{b^i} = \sum\limits_{i=n+1}^{\infty} \frac{a_i}{b^i} \leq ...$$
and because $0\leq a_i \leq b-1$ and assuming an integer base $b>1$, we have
$$ ... \leq \sum\limits_{i=n+1}^{\infty} \frac{b-1}{b^i}=
\frac{b-1}{b^{n+1}} \left(\color{red}{\sum\limits_{i=0}^{\infty} \frac{1}{b^{i}}}\right)=
\frac{b-1}{b^{n+1}} \left(\frac{1}{1-\frac{1}{b}}\right)=\frac{1}{b^n}$$
the sum in red is an infinite geometric progression with ratio $\left|\frac{1}{b}\right|<1$.
| {
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Solving $\sqrt{8-x^2}-\sqrt{25-x^2}\geq x$
I would like to find the solution of
$$\sqrt{8-x^2}-\sqrt{25-x^2}\geq x.$$
My try:
First I used the hint of this answer.
$$ \frac{8-x^2-25+x^2}{\sqrt{8-x^2}+\sqrt{25-x^2}}\geq x \leftrightarrow \frac{-17}{\sqrt{8-x^2}+\sqrt{25-x^2}}\geq x.$$
Then the solution can be found by
$$\left(-17\right)^2\geq \left(x\sqrt{8-x^2}+x\sqrt{25-x^2}\right)^2.$$
But I think this is not the best approach.
| Dinesh's answer is alright, but there is another approach. Your question can be rephrased as finding the range of $x$ such that $f(x)\geq 0$, where
$$f(x) = \sqrt{8-x^2}-\sqrt{25-x^2}-x.$$
We find the derivative of $f$ then set it to be $0$ to find its extreme values:
$$ f'(x) = -\frac{x}{\sqrt{8-x^2}}+\frac{x}{\sqrt{25-x^2}}-1=0 $$
Which implies that
$$ x\left(\sqrt{8-x^2}-\sqrt{25-x^2}\right)=\sqrt{(8-x^2)(25-x^2)}. $$
Solve this (with some effort) to get that the critical point of $f$, which happens to be a maximum, occurs at about $x=-2.37$ where the $y$ value is $-0.49<0$. Thus $f$ is always less than $0$ and there is no solution.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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} |
Positive Definite Matrix (Block Matrix) Let $B$ be an $m\times n$ matrix. Is $$ A=\begin{pmatrix} I & B \\ B^T & I+B^TB \end{pmatrix} $$ positive definite?
Attempt:
Let $\mathbf{z}=\begin{pmatrix} \mathbf{x} \\ \mathbf{y} \end{pmatrix}$. To show that $A$ is positive definite, $\mathbf{z}^TA\mathbf{z}>0$. Expanding $\mathbf{z}^TA\mathbf{z}>0$ gives $\mathbf{x}^T\mathbf{x}+\mathbf{y}^T\mathbf{y}+(B\mathbf{y})^T(B\mathbf{y})+2\mathbf{x}^TB\mathbf{y}$. The first three terms are positive, but what can be concluded about the $2\mathbf{x}^TB\mathbf{y}$ term?
| Sylvester's Law of Inertia.
We have your symmetric $A.$ We are going to construct $P^T A P = D$ where $\det P \neq 0$ and $D$ is diagonal. Then the count of positive eigenvalues for $A$ is the same as the count of positive eigenvalues of $D$
$$
\left(
\begin{array}{cc}
I&0 \\
-B^T& I \\
\end{array}
\right)
\left(
\begin{array}{cc}
I&B \\
B^T&I+B^T B \\
\end{array}
\right)
\left(
\begin{array}{cc}
I&-B \\
0&I \\
\end{array}
\right) =
\left(
\begin{array}{cc}
I&0 \\
0&I \\
\end{array}
\right)
$$
Sylvester says that your $A$ is positive definite.
Or, given my invertible
$$
P=
\left(
\begin{array}{cc}
I&-B \\
0&I \\
\end{array}
\right)
$$
let us switch to $Q = P^{-1}$
$$
Q=
\left(
\begin{array}{cc}
I&B \\
0&I \\
\end{array}
\right)
$$
and write $Q^T D Q = A.$ Since $D$ turned out to be the identity matrix
$$
\left(
\begin{array}{cc}
I&0 \\
B^T&I \\
\end{array}
\right)
\left(
\begin{array}{cc}
I&B \\
0&I \\
\end{array}
\right) =
\left(
\begin{array}{cc}
I&B \\
B^T&I+B^T B \\
\end{array}
\right)
$$
As $Q$ is nonsingular this again says $Q^T Q = A$ is positive definite. We can write using dot products, you wrote a column vector
$$
z=
\left(
\begin{array}{c}
x \\
y \\
\end{array}
\right)
$$
after which
$$
Qz=
\left(
\begin{array}{c}
x + B y \\
y \\
\end{array}
\right)
$$
and
$$ (Qz)^T (Qz) = |x+By|^2 + |y|^2 $$
If $y\neq0$ this is positive. If $y=0,$ we are left with $|x|^2,$ and this is positive unless $x$ is also $0$
| {
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For positve $a$, $b$, $c$, $d$ with $a+b+c+d\leq 1$, prove that $\sqrt[4]{(1-a^4)(1-b^4)(1-c^4)(1-d^4)}\geq255\cdot a b c d .$
Let $a,b,c,d\in\mathbb R_+$ such that $a+b+c+d\leqslant1$. Prove that$$
\sqrt[4]{\smash[b]{(1-a^4)(1-b^4)(1-c^4)(1-d^4)}}\geqslant255·abcd.
$$
My observations:
I can see that all of $a,b,c,d$ are positive fractions which makes all the bracketed factors in LHS positive. Now if we raise both sides to the power $4$, and replace the $4$th powers of $a,b,c,d$ with $A,B,C,D$, our inequality gets reduced to
$$(1-A)(1-B)(1-C)(1-D)\ge 255^4 \cdot ABCD$$
Now applying AM $\ge$ GM we get two results,
$$A+B+C+D\ge 4\cdot\sqrt[4]{ABCD}$$
and
$$4-(A+B+C+D)\ge 4\cdot\sqrt[4]{\smash[b]{(1-A)(1-B)(1-C)(1-D)}}$$
Can I somehow use both these results to prove the inequality? Please help
| By AM-GM we obtain:
$$\sqrt[4]{\prod\limits_{cyc}(1-a^4)}\geq\sqrt[4]{\prod\limits_{cyc}((a+b+c+d)^4-a^4)}=$$
$$=\sqrt[4]{\prod_{cyc}\left((b+c+d)\left((a+b+c+d)^3+(a+b+c+d)^2a+(a+b+c+d)a^2+a^3\right)\right)}\geq$$
$$\geq\sqrt[4]{\prod_{cyc}\left(3\sqrt[3]{bcd}\left(64\sqrt[4]{a^3b^3c^3d^3}+16a\sqrt{abcd}+4\sqrt[4]{abcd}a^2+a^3\right)\right)}\geq$$
$$\geq\sqrt[4]{\prod_{cyc}\left(3\sqrt[3]{bcd}\cdot85\sqrt[85]{(abcd)^{48}\cdot a^{16}\cdot(abcd)^8\cdot(abcd)\cdot a^8\cdot a^3}\right)}=$$
$$=255\sqrt[4]{\prod_{cyc}a^{\frac{84}{85}}b^{\frac{256}{255}}c^{\frac{256}{255}}d^{\frac{256}{255}}}=255abcd.$$
| {
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Miscalculating the determinant I am learning linear algebra and am getting stuck when trying to calculate the determinant using elementary row operations. Consider the matrix A.
\begin{vmatrix}
0 & 1 & 2 & 3 \\
1 & 1 & 1 & 1 \\
-2 & -2 & 3 & 3 \\
1 & 2 & -2 & -3 \\
\end{vmatrix}
According to the solution in my textbook and Matlab the determinant should be 10. I however find -20. Here is what I did.
I first interchanged row 1 and row 3.
\begin{vmatrix}
1 & 2 & -2 & -3 \\
1 & 1 & 1 & 1 \\
-2 & -2 & 3 & 3 \\
0 & 1 & 2 & 3
\end{vmatrix}
I then substracted row 1 from row two. I also added the first row twice to the third row.
\begin{vmatrix}
1 & 2 & -2 & -3 \\
0 & -1 & 3 & -2 \\
0 & 2 & -1 & 9 \\
0 & 1 & 2 & 3
\end{vmatrix}
Then, I added the second row twice to the third row and once to the fourth row.
\begin{vmatrix}
1 & 2 & -2 & -3 \\
0 & -1 & 3 & -2 \\
0 & 0 & 5 & 5 \\
0 & 0 & 5 & 1
\end{vmatrix}
My final operation was to substract the third row from the fourth row, which gave:
\begin{vmatrix}
1 & 2 & -2 & -3 \\
0 & -1 & 3 & -2 \\
0 & 0 & 5 & 5 \\
0 & 0 & 0 & -4
\end{vmatrix}
Finally, I calculated the determinant: $(-1)^1 \cdot 1 \cdot -1 \cdot 5 \cdot -4 = -20$ The $(-1)^1$ is there since I did one operation in which I interchanged two rows.
I would really appreciate if you could tell me what I did wrong.
Martijn
| Actually, the first thing that you did was to exchange rows $1$ and $4$, not $1$ and $3$. After that, when you subtracted row $1$ from row $2$, you should have got a $4$ at the end of the row. And when you add twice the first row to the third one, the last entry should become $-3$, not $9$.
| {
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Solution for $\frac{3}{x-9} \gt \frac{2}{x+2}$ What I did:
$\frac{3}{x-9} \gt \frac{2}{x+2}$
$3(x+2) \gt 2(x-9)$
$3x+6 \gt 2x-18$
$x \gt -24$
When typing this in in symbolab, it showed me that the solution is $-24 \lt x \lt -2$ or $x \gt 9$
What did i do wrong ?
How come i didnt get the correct solution ?
| Consider the function $f(x) = \dfrac{3}{x-9} - \dfrac{2}{x+2}$.
The function $f(x)$ has a zero at $x=-24$ and poles at $x=-2$ and at $x=9$. The two poles and the one zero divide the $x-$axis into four intervals
$$(-\infty, -24), \ (-24, -2), \ (-2, 9), (9, \infty) $$
In each of those intervals, $f(x)$ is either entirely positive or entirely negative.
We note that
$\text{$-100 \in (-\infty, -24)$ and $f(-100) < 0$}$
$\text{$-13 \in (-24, -2)$ and $f(-13) = \dfrac{1}{22}$} > 0$
$\text{$0 \in (-2, 9)$ and $f(0) = -\dfrac 43 < 0$}$
$\text{$100 \in (9, \infty)$ and $f(100) > 0$}$
So the solution set is $(-24, -2) \cup (9, \infty)$.
| {
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Prove by induction that, for all positive integers n, the following inequality holds: (1-2-1)(1-2-2)...(1-2-n) ≥ 1/4 + 2-(n+1)
My solution so far:
After verifying the base case for n = 1, I began the inductive step:
(1-2-1)(1-2-2)...(1-2-(k+1)) ≥ 1/4 + 2-(k+2)
Subtracting 1/4 + 2-(k+1) on the RHS as it's an inequality:
(1-2-1)(1-2-2)...(1-2-(k+1)) ≥ 1/4 + 2-(k+2) - (1/4 + 2-(k+1))
1 - 2-(k+1) ≥ 2-(k+2) - 2-(k+1)
Taking common denominator on RHS (skipped a few algebraic steps):
(2k+1 - 1)/2k+1 ≥ 1/(2k+2) - 1/(2k+1)
Taking common denominator on LHS resulting in:
(2k+1 - 1)/2k+1 ≥ - 1/(2-(k+1))
It's my first time learning about induction so I'm wondering if the last step is sufficient for proving the proposition since the numerator on the LHS is greater than the numerator on the RHS, and they have the exact same denominator, or is this just the completely wrong approach? Any hints and/or tips would be helpful. Thanks.
| Assume the inequality holds for some $n = k$, $k \ge 1$,
$$\left(1-2^{-1}\right)\left(1-2^{-2}\right)\cdots\left(1-2^{-k}\right) \ge \frac14 + 2^{-(k+1)}$$
For $n = k+1$,
$$\begin{align*}
LHS &= \left(1-2^{-1}\right)\left(1-2^{-2}\right)\cdots\left(1-2^{-k}\right)\left(1-2^{-(k+1)}\right)\\
&\ge \left(\frac14 + 2^{-(k+1)}\right)\left(1-2^{-(k+1)}\right)\\
&= \frac14 - \frac14\cdot 2^{-(k+1)} + 2^{-(k+1)}-2^{-2(k+1)}\\
&= \frac14 + 2^{-(k+2)}\left[-\frac14\cdot2 + 2-2^{-k}\right]\\
&= \frac14 + 2^{-(k+2)}\left[\frac32-2^{-k}\right]\\
&\ge \frac14 + 2^{-(k+2)}\left[\frac32-2^{-1}\right]\\
&= \frac14 + 2^{-(k+2)}\cdot 1\\
&= RHS
\end{align*}$$
| {
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Showing that $\sum_{x \in \mathbb{F}_p} \left(\frac{x^2-1}{p}\right) = -1$, where $\left(\frac{x}{p}\right)$ is the Legendre symbol The question is,
Show that $$\sum_{x \in \mathbb{F}_p} \left(\frac{x^2-1}{p}\right) = -1$$ where the operation $\left(\frac{x}{p}\right) = \pm 1$ if $x$ is a quadratic residue/non-residue and $0$ otherwise.
I was able to follow the argument given here Prove that $\sum_{X=0}^{p-1} \left(\frac{X^{2}+A}{p}\right)=-1$ for the $\left(\frac{-1}{p}\right) = -1$ case, but I'm not sure how to make progress on the other case (i.e $p\equiv 1$ mod $4$).
One thought I had that I couldn't quite make work was to try and use the formula $\left(\frac{x}{p}\right) \equiv x^r$ mod $p$, where $r = (p-1)/2$ and then expanding using the binomial theorem.
Any thoughts or hints would be greatly appreciated!
| This is my attempt at what I believe is a full answer following the hints given in the comments.
We first show that there are exactly $p-1$ solutions to the equation $$y^2 = x^2 - 1$$ for pairs $(x,y)\in \mathbb{F}_p^2$. If $y^2 = x^2 -1$ then $(x-y)(x+y) = 1$. Letting $$u = x-y$$ $$v = x+y$$ we see that for any choice of $(u,v) \in \mathbb{F}_p^2$ we can uniquely solve the system of equations for $x$ and $y$. Thus the number of solutions to $1 = (x-y)(x+y)$ is just the number of solutions to $1 = uv$. There are $p-1$ choices for $u$, and once $u$ is chosen, there is only one $v$ for which $1 = uv$. We conclude that there are exactly $p-1$ solutions to $y^2 = x^2 -1$
We will use this information to count the number of distinct $x$ values for which $(\frac{x^2-1}{p}) = 1$. We have the two solutions $(\pm1,0)$, and then for each other value of $x$ for which $y^2 = x^2 -1$ has a solution, we get two solutions, $(x,\pm y)$. Thus we conclude that there should be $\frac{p-3}{2}$ distinct values of $x$ for which $(\frac{x^2-1}{p}) = 1$. Since there are two values of $x$ for which $(\frac{x^2-1}{p}) = 0$, we conclude that the remaining $p-\frac{p-3}{2} - 2$ values of $x$ evaluate $(\frac{x^2-1}{p}) $ as $-1$. Thus our sum is just $\frac{p-3}{2} - (p-\frac{p-3}{2}-2) = p-3+2-p = -1$.
| {
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Given two points that are joined by a line that is a tangent to a curve, find the missing constant in the equation for the curve How do I solve a question like this?
I have been given two points that make up a line tangent to $y=\frac{a}{(x+2)^2}$. I need to determine $a$.
First, I calculated the equation of the tangent line with $\frac{y2-y1}{x2-x1}$, and also found $\frac{dy}{dx}$. I had planned to set the equal derivative to the slope of the tangent and plug in an $x$ value from one of the points, however they both got a different result.
Is there a step I am missing?
| The tangent point and slope there are the same for both equations.
Therefore $mx + c = \frac{a}{(x+2)^2}$ and $m = -\frac{2a}{(x+2)^3}$ for a
certain $x$ value.
To find the x value, set $(mx + c)(x+2)^2 = a$
And $m(x+2)^3 = -2a$
So $-\frac{m}{2}(x+2)^3 = (mx + c)(x+2)^2$
$-\frac{m}{2}(x+2) = mx + c$
$-\frac{mx}{2} - m = mx + c$
$\frac{3mx}{2} = -m - c$
$3mx = -2m - 2c$
$x = -\frac{2m+2c}{3m}$
Once you have $x$, find $y$ from $y = mx + c$ and then find $a$ from $y = \frac{a}{(x+2)^2}$
| {
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Evaluating $\mathcal{P}\int_0^\infty \frac{dx}{(1 + a^2x^2)(x^2 - b^2)}$ I am trying to evaluate this integral
$$\mathcal{P}\int_0^\infty \frac{dx}{(1 + a^2x^2)(x^2 - b^2)}$$
with $\mathcal{P}$ the principal value and $a,b>0$.
I already know the answer to be $$ - \frac{a\pi}{2(1 + a^2b^2)}$$
after fiddling with Mathematica putting numbers for $a$ and $b$.
The poles of this function are at $x=\pm b, \pm \frac{i}{a^2}$ so I cannot find a proper contour.
Any help in getting the answer will be appreciated!
| Use that $$\frac{1}{(1+a^2x^2)(x^2-b^2)}=1/2\,{\frac {1}{ \left( 1+{a}^{2}{b}^{2} \right) b \left( -b+x
\right) }}-{\frac {{a}^{2}}{ \left( 1+{a}^{2}{b}^{2} \right) \left(
1+{a}^{2}{x}^{2} \right) }}-1/2\,{\frac {1}{ \left( 1+{a}^{2}{b}^{2}
\right) b \left( b+x \right) }}
$$
| {
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"source": "stackexchange",
"question_score": "3",
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Prove by the limit definition that $\lim_{x\rightarrow 2}\frac{x^2-1}{x-3}=-3$ I was reviewing $\mathbb{R}-$analisys with a friend and I'm thinking about one of the questions...
Prove by the $\epsilon-\delta$ limit definition that $\lim_{x\rightarrow 2}\frac{x^2-1}{x-3}=-3$.
My answer was very long, someone could do a better answer? Does is it OK in this way? Thanks very much.
My attempt
We have $|f(x)-(-3)|=|f(x)+3|=|\frac{x^2-1}{x-3}+3|=|\frac{x^2-1+3x-9}{x-3}|=|\frac{x^2+3x-10}{x-3}|=|\frac{(x-2)(x+5)}{x-3}| \qquad (i)$
and
$|x+5|=|x+7-2|<|x-2|+|7|=|x-2|+7 \qquad (ii)$
$ |x-3|=|x-2-1|>|x-2|+|-1|=|x-2|+1\Longrightarrow \dfrac{1}{|x-3|}<\dfrac{1}{|x-2|+1} \qquad(iii)$
So, by (ii) and (iii)
$|\frac{(x-2)(x+5)}{x-3}|=\dfrac{|x-2||x+5|}{|x-3|}<\dfrac{|x-2|(|x-2|+7)}{|x-2|+1} \qquad (iv)$
But also, if $|x-2|<\delta \qquad (v)$, we have
$|\frac{(x-2)(x+5)}{x-3}|<^{(iv)}\dfrac{|x-2|(|x-2|+7)}{|x-2|+1}<^{(v)}\dfrac{\delta(\delta+7)}{\delta+1} \qquad(vi)$
And more than that
$\dfrac{\delta(\delta+7)}{\delta+1}<\dfrac{\delta(\delta+1)}{\delta+1}=\delta \qquad (vii)$
So, if $\delta<\epsilon$, we have
$|f(x)-(-3)|=^{(i)}|\frac{(x-2)(x+5)}{x-3}|<^{(vi)}\dfrac{\delta(\delta+7)}{\delta+1}<^{(vii)}\delta<\epsilon$, Q.E.D.
| Hint: Recall that $0<a<b$ implies that $\frac{1}{a}>\frac{1}{b}>0$. So,
$$0<\vert x-2 \vert<\delta$$
$$\Longrightarrow \ 1<\vert x-2 \vert+1<\delta+1 $$
$$\Longrightarrow \ 1>\frac{1}{\vert x-2\vert+1}>\frac{1}{\delta+1} $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2942204",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Is there any other way to solve this difficult integral $ \int \frac{1}{ (a^2 \cos ^2 x + b ^ 2 \sin ^2x) ^2} \ dx $ $$
\int \frac{1}{ (a^2 \cos ^2 x + b ^ 2 \sin ^2x) ^2} \ dx
$$
So this is the question .
The solution given in book is to divide numerator and denominator by $\cos ^4x$ and then substitute $\tan x = t$ in the resulting integrand.
Other way of doing this was to substitute $b \tan x = a \tan t$.
So I was thinking is not there any other way to solve this as it seems to a complicated problem as the given methods are very lengthy while solving.
Any simpler\shorter method anyone could think of ?
| Note that
$$I(p,q)=\int \frac{dx}{ a^2\cos ^2 x + b^2 \sin ^2x}
=\int \frac{d(\tan x)}{ a^2+b^2\tan ^2 x }
=\frac1{ab}\tan^{-1}\bigg({\frac ba}\tan x \bigg)
$$
and evaluate
\begin{align}
&\int \frac{1}{ (a^2\cos ^2 x + b^2 \sin ^2x)^2}dx
= -\frac12\left(\frac{I’_a}a+\frac{I’_b}b\right)\\
=&\ \frac1{2ab}\left(\frac1{a^2}+\frac1{b^2}\right)\tan^{-1}\left(\frac ba\tan x \right)
+ \frac12\left(\frac1{a^2}-\frac1{b^2}\right)\frac{\tan x}{b^2\tan^2x+a^2}
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2943314",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Tournament probability question If we have 8 teams in the quarter final
if 3 teams are from the same country
what is the probability that 2 of the 3 will be competing against each other in the quarter final?
From my understanding, if we only had 2 team, probability is basically 1/7 because we are drawing 1 from the remaining 7 teams to pair up.
But what about 3 teams, 4 teams? If we have 5 teams then definitely the probability will be over 1 because we are guaranteed to have a pair that are from same country.
I want to know how to calculate this
| It's the number of $4$ match sets containing a pairing of $2$ of the $3$ teams from the same country divided by the total number of $4$ match sets.
We can number teams $1$ to $8$ with $1$ to $3$ being from the same country. For the number of $4$ match sets containing a pairing of $2$ of the $3$ teams from the same country, there are $3$ possible pairings $(1,2)(1,3)(2,3)$ and for every one of those, from the remaining $6$ teams there are $5$ ways to match the next team, and from the remaining $4$ there are $3$ ways to match the next team leaving a matched pair. This number is therefore $3\cdot 5\cdot 3 = 45$
For the total, there are $7$ ways to match the first team and from the remaining $6$, there are $5$ ways to match the second team and from the remaining $4$ there are $3$ ways to match the third team leaving a matched pair. The total is therefore $7\cdot 5\cdot 3 = 105$
$$P = \frac{3\cdot 5\cdot 3}{7\cdot 5\cdot 3} = \frac{3}{7}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2943541",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
product of two Jacobi matrices Let
$$
A = \begin{bmatrix}
a_1 & b_1 \\
b_1 & a_2 & b_2 \\
& b_2 & \ddots & \ddots \\
& & \ddots & \ddots & b_{n-1} \\
& & & b_{n-1} & a_n
\end{bmatrix},\ B = \begin{bmatrix}
1&0&0&\cdots\\
0&a_2 & b_2 \\
0&b_2& a_3 & \ddots & \\
\vdots&\ddots & \ddots & b_{n-1} \\
& & b_{n-1} & a_n
\end{bmatrix},
$$
where $b_i \neq 0$ and all entries are real. The following matrix is $A^{-1} \times B$
$$
C = \begin{bmatrix}
*&*&0&0&\cdots&0\\
*&*&0&0&\cdots&0 \\
*&*&1&0&\cdots&0 \\
*&*&0&1&\cdots&0 \\
\vdots&\vdots&&\ddots&\ddots&\\
*&*&0&0&\cdots&1 \\
\end{bmatrix},
$$
As you see an identity matrix appears in $C[2 \cdots n]$ and rows 1, 2 from the third column are zero. I need to prove that it always happen. I tried this using syms in Matlab and it verified this result. I would be appreciated for any help.
update
Fonseca 2001 in his paper Explicit inverses of some tridiagonal matrices gave the following relations for a symmetric tridiagonal matrix:
$$
A^{-1} =
\begin{bmatrix}
u_1v_1&u_1v_2&u_1v_3&\cdots&u_1v_n\\
u_1v_2&u_2v_2&u_2v_2&\cdots&u_2v_2\\
u_1v_3&u_2v_3&u_3v_3&\cdots&u_3v_3\\
\vdots\\
u_1v_n&u_2v_n&u_3v_n&\cdots&u_nv_n\\
\end{bmatrix}
$$
such that
\begin{align}
v_1&= \dfrac{1}{d_1}\\
v_k&= -\dfrac{b_{k-1}}{d_k}v_{k-1}, k=2, \cdots,n\\
d_n&=a_n\\
d_i&=a_i - \dfrac{b_ic_i}{d_{i+1}}, i=n-1, \cdots, 1\\
u_n &= \dfrac{1}{\delta_nv_n}\\
u_k&=-\dfrac{b_k}{\delta_k}u_{k+1}, k=n-1, \cdots, 1\\
\delta_1&=a_1\\
\delta_i&=a_i-\dfrac{b_{i-1}c_{i-1}}{\delta_{i-1}}, i=2, \cdots, n-1
\end{align}
I tried to replace these variables in their position to get the appropriate results, but I failed.
I would like to know is there any block version of computing inverse of a matrix? because two blocks of $A, B$ are the same, it may help.
thanks in advance.
| You have
$$
B=A-D,
$$
where
$$
D=\begin{bmatrix}
a_1-1&b_1&0&0&\cdots&0\\
b_1&0&0&0&\cdots&0 \\
0&0&0&0&\cdots&0 \\
0&0&0&0&\cdots&0 \\
\vdots&\vdots&&\ddots&\ddots&\\
0&0&0&0&\cdots&0 \\
\end{bmatrix}.
$$
Then
$$
C=A^{-1}B=I-A^{-1}D.
$$
Since $D$ has nonzero entries only at the first two columns, we have
$$
A^{-1}D= \begin{bmatrix}
*&*&0&0&\cdots&0\\
*&*&0&0&\cdots&0 \\
*&*&0&0&\cdots&0 \\
*&*&0&0&\cdots&0 \\
\vdots&\vdots&&\ddots&\ddots&\\
*&*&0&0&\cdots&0 \\
\end{bmatrix},
$$
and so
$$
C=\begin{bmatrix}
1&0&0&0&\cdots&0\\
0&1&0&0&\cdots&0 \\
0&0&1&0&\cdots&0 \\
0&0&0&1&\cdots&0 \\
\vdots&\vdots&&\ddots&\ddots&\\
0&0&0&0&\cdots&1 \\
\end{bmatrix}-
\begin{bmatrix}
*&*&0&0&\cdots&0\\
*&*&0&0&\cdots&0 \\
*&*&0&0&\cdots&0 \\
*&*&0&0&\cdots&0 \\
\vdots&\vdots&&\ddots&\ddots&\\
*&*&0&0&\cdots&0 \\
\end{bmatrix},
$$
which is of the desired form.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2944395",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Real Part of the Dilogarithm It is well known that
$$\frac{x-\pi}{2}=-\sum_{k\geq 1}\frac{\sin{kx}}{k}\forall x\in(0,\tau),$$
which gives
$$\frac{x^2}{4}-\frac{\pi x}{2}+\frac{\pi^2}{6}=\sum_{k\geq 1}\frac{\cos(kx)}{k^2}.$$
Note that
$$\textrm{Li}_2(e^{ix})=\sum_{k\geq 1}\frac{\cos(kx)+i\sin(kx)}{k^2}$$
This means that
$$\mathfrak{R}\textrm{Li}_2(e^{ix})=\frac{x^2}{4}-\frac{\pi x}{2}+\frac{\pi^2}{6}$$
unless I'm wrong on one of the above statements. Now, with the previous restrictions on $x$, we have that this formula is valid for all complex numbers lying on the complex unit circle as inputs. My question is: is there a way to define something along these lines (a finite degree polynomial, preferably) for complex inputs of the dilogarithm that don't necessarily lie on the complex unit circle? More specifically,
$$\mathfrak{R}\textrm{Li}_2(re^{ix})=?$$
| To simplify, you can use the Inverse Tangent Integral $\operatorname{Ti_{2}}$ and the Clausen Functions $\operatorname{C_{n}}$ and $\operatorname{S_{n}}$.
$$
\begin{align*}
\operatorname{Li_{2}}\left( e^{x \cdot i} \right) &= \sum_{k = 1}^{\infty} \left( \frac{\cos\left( k \cdot x \right) + \sin\left( k \cdot x \right) \cdot i}{k^{2}} \right)\\
\operatorname{Li_{2}}\left( e^{x \cdot i} \right) &= \sum_{k = 1}^{\infty} \left( \frac{\cos\left( k \cdot x \right)}{k^{2}} \right) + \sum_{k = 1}^{\infty} \left( \frac{ \sin\left( k \cdot x \right)}{k^{2}} \right) \cdot i\\
\operatorname{Li_{2}}\left( e^{x \cdot i} \right) &= \operatorname{C_{2}}\left( x \right) + \operatorname{S_{2}}\left( x \right) \cdot i\\
\operatorname{Li_{2}}\left( e^{x \cdot i} \right) &= \frac{1}{6} \cdot \pi^{2} - \frac{1}{2} \cdot \pi \cdot x + \frac{1}{4} \cdot x^{2} + \operatorname{S_{2}}\left( x \right) \cdot i\\
\end{align*}
$$
$$\fbox{$\Re\left( \operatorname{Li_{2}}\left( e^{x \cdot i} \right) \right) = \frac{1}{6} \cdot \pi^{2} - \frac{1}{2} \cdot \pi \cdot x + \frac{1}{4} \cdot x^{2}$}$$
and
$$
\begin{align*}
\operatorname{Li_{2}}\left( x \cdot i \right) &= \sum_{k = 1}^{\infty} \left( \frac{\cos\left( k \cdot x \cdot i \right) + \sin\left( k \cdot x \cdot i \right) \cdot i}{k^{2}} \right)\\
\operatorname{Li_{2}}\left( x \cdot i \right) &= \frac{1}{4} \cdot \operatorname{Li_{2}}\left( -x^{2} \right) + \operatorname{Ti_{2}}\left( x \right) \cdot i\\
\operatorname{Li_{2}}\left( x \cdot i \right) &= \frac{1}{4} \cdot \operatorname{Li_{2}}\left( -x^{2} \right) + \frac{1}{4} \cdot \phi\left( -x^{2};\, 2,\, \frac{1}{2} \right) \cdot i\\
\end{align*}
$$
$$\fbox{$\Re\left( \operatorname{Li_{2}}\left( x \cdot i \right) \right) = \frac{1}{4} \cdot \operatorname{Li_{2}}\left( -x^{2} \right)$}$$
You can write it even more generally:
To find the real part of the dilogarithm, you can use the Bloch-Wigner-Dilogarithm or Bloch–Wigner function $\operatorname{D_{2}}$, which is defined as $\operatorname{D_{2}}\left( z \right) \equiv \Im\left( \operatorname{Li_{2}}\left( z \right) \right) + \arg\left( 1 - z \right) \cdot \ln\left( \left| z \right| \right)$ for $z \in \mathbb{C} \backslash \left\{ 0,\, 1 \right\}$.
$$\fbox{$\begin{align*}
\Re\left( \operatorname{Li_{2}}\left( z \right) \right) &= \operatorname{Li_{2}}\left( z \right)- \left( \operatorname{D_{2}}\left( z \right) - \arg\left( 1 - z \right) \cdot \ln\left( \left| z \right| \right) \right) \cdot i\\
\end{align*}$}$$
Via using Ramakrishnan's equality we can get this relation too:
$$
\begin{align*}
\operatorname{D_{m}}\left( z \right) &\equiv \Re\left[ i^{m + 1} \cdot \left( \sum_{k = 1}^{m} \left( \frac{\left( -\ln\left( \left| z \right| \right) \right)^{m - k}}{\left( m - k \right)!} \cdot \operatorname{Li_{k}}\left( z \right) \right) - \frac{\left( -\ln\left( \left| z \right| \right) \right)^{m}}{2 \cdot m!} \right) \right]\\
\operatorname{D_{2}}\left( z \right) &= \Re\left[ i^{2 + 1} \cdot \left( \sum_{k = 1}^{2} \left( \frac{\left( -\ln\left( \left| z \right| \right) \right)^{2 - k}}{\left( 2 - k \right)!} \cdot \operatorname{Li_{k}}\left( z \right) \right) - \frac{\left( -\ln\left( \left| z \right| \right) \right)^{2}}{2 \cdot 2!} \right) \right]\\
\operatorname{D_{2}}\left( z \right) &= \Re\left[ -i \cdot \left( \sum_{k = 1}^{2} \left( \frac{\left( -\ln\left( \left| z \right| \right) \right)^{2 - k}}{\left( 2 - k \right)!} \cdot \operatorname{Li_{k}}\left( z \right) \right) - \frac{\left( -\ln\left( \left| z \right| \right) \right)^{2}}{2 \cdot 2!} \right) \right]\\
\operatorname{D_{2}}\left( z \right) &= \Re\left[ -i \cdot \left( \frac{\left( -\ln\left( \left| z \right| \right) \right)^{2 - 1}}{\left( 2 - 1 \right)!} \cdot \operatorname{Li_{1}}\left( z \right) + \frac{\left( -\ln\left( \left| z \right| \right) \right)^{2 - 2}}{\left( 2 - 2 \right)!} \cdot \operatorname{Li_{2}}\left( z \right) - \frac{\left( -\ln\left( \left| z \right| \right) \right)^{2}}{2 \cdot 2!} \right) \right]\\
\operatorname{D_{2}}\left( z \right) &= \Re\left[ -i \cdot \left( \ln\left( \left| z \right| \right) \cdot \ln\left( 1 - z \right) + \operatorname{Li_{2}}\left( z \right) - \frac{\ln\left( \left| z \right| \right)}{4} \right) \right]\\
\\
\Im\left( \operatorname{Li_{2}}\left( z \right) \right) &= \Re\left[ -i \cdot \left( \ln\left( \left| z \right| \right) \cdot \ln\left( 1 - z \right) + \operatorname{Li_{2}}\left( z \right) - \frac{\ln\left( \left| z \right| \right)}{4} \right) \right] - \arg\left( 1 - z \right) \cdot \ln\left( \left| z \right| \right)\\
\end{align*}
$$
$$\fbox{$\begin{align*}
\Re\left( \operatorname{Li_{2}}\left( z \right) \right) = ~&\operatorname{Li_{2}}\left( z \right) - (\Re\left[ -i \cdot \left( \ln\left( \left| z \right| \right) \cdot \ln\left( 1 - z \right) + \operatorname{Li_{2}}\left( z \right) - \frac{\ln\left( \left| z \right| \right)}{4} \right) \right]\\
&- \arg\left( 1 - z \right) \cdot \ln\left( \left| z \right| \right) + \arg\left( 1 - z \right) \cdot \ln\left( \left| z \right| \right) ) \cdot i\\
\end{align*}$}$$
But there are no further elementary simplifications. especially not with polynomials of finite degree.
| {
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"url": "https://math.stackexchange.com/questions/2947873",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Simplify the Dot Product in terms of $a$ and $b$ Where $a$ and $b$ are arbitrary vectors
$(a+2b) \cdot (2a-b)$
$$a\cdot(2a-b)+2b\cdot(2a-b) = 2(a\cdot a)-a\cdot b+1(b\cdot a)-2(b\cdot b)$$
$$=2(a)-ab+4ab-2(b)^2$$
$$=2a^2-2b^2$$
$$=2(a^2-b^2)$$
Where did i go wrong in simplifying this?
| $$(\mathbf a+2\mathbf b)\cdot(2\mathbf a-\mathbf b)=2\mathbf a\mathbf a-\mathbf a\mathbf b+4\mathbf b\mathbf a-2\mathbf b\mathbf b=2(|\mathbf a|^2-|\mathbf b|^2)+3\mathbf a\mathbf b$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2948870",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Proving $7|13\cdot 6^{n}+8\cdot 13^{n}$ for all natural numbers. I would like to verify whether my proof is correct. The answer sheet used a much more intuitive and logical approach but I think mine is correct also.
To prove: $7|13\cdot 6^{n}+8\cdot 13^{n}$ for all natural numbers
Proof: We proceed by induction and show the base case holds.
$13+8=21 $. Since 7 divides 21 the base case holds.
We assume $7|13\cdot 6^{m}+8\cdot 13^{m}$ and we need to show $7|13\cdot6^{m+1}+8\cdot 13^{m+1}$.
Since, $7|13\cdot 6^{m}+8\cdot 13^{m}$, we have, $13\cdot 6^{m}+8\cdot 13^{m} =7x, \space x \in \mathbb{N}$.
We rewrite $13\cdot 6^{m+1}+8\cdot 13^{m+1}$ as $6(13\cdot 6^m)+13(8\cdot 13^m)$ and notice, $13\cdot 6^m=7x-8\cdot13^m$.
We substitute and find,
$6(7x-8\cdot13^m)+13(8\cdot13^m)=42x-48\cdot13^m+104\cdot13^m = 42x+56\cdot13^m=7(6x+8\cdot13^m)$
Since $6x+8\cdot13^m \space \in \mathbb{N}$
$\\ \therefore $ By the principle of mathematical induction, $7|13\cdot 6^{n}+8\cdot 13^{n}$ for all natural numbers $n \space \blacksquare$.
| $$13\cdot 6^{n}+8\cdot 13^{n}\equiv (-1)(-1)^n+(1)(-1)^n =0 \text { mod (7) }$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2949124",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 2
} |
Putting the general equation for a circle in terms of x (or y) After about a year of inactivity, I'm back!!
I've been trying to solve a series of math equations by putting them in terms of x, but I've run into a problem with the famous circle equation:
$$a = \sqrt{ ( x - b ) ^ 2 + ( y - c ) ^ 2}$$
where:
a is the radius of the circle,
b is how far to the right the circle is
c is how far upwards the circle is.
After trying for about 3 weeks, i've come up with this much so far:
$$\begin{matrix}
\text{square both sides} & a ^ 2 = ( x - b ) ^ 2 + ( y - c ) ^ 2 \\
\text{isolate term with x} & ( x - b ) ^ 2 = a ^ 2 - ( y - c ) ^ 2\\
\text{square root all terms} & | x - b | = \sqrt{ a ^ 2 - ( y - c ) ^ 2 }\\
\text{absolute value} & \sqrt{ ( x - b ) ^ 2 } = \sqrt{ a ^ 2 - ( y - c ) ^ 2 }\\
\text{dead end, trying another path} & ( x - b ) ^ 2 = ( a + ( y - c ) ) \times (a - ( y - c ) )\\
\text{}
\end{matrix}$$
After that point, it just becomes a mess of trying stuff and not getting anywhere. The reason I need it in terms of x (or y) is so I can use it for a math equation I'm doing, where I find an equation to graph every letter. Could someone help? Or at least give me a suggestive nudge?
| Does this work:
$a = \sqrt{ ( x - b ) ^ 2 + ( y - c ) ^ 2}$
$a^2=(x-b)^2+(y-c)^2$
$(y-c)^2=a^2-(x-b)^2$
$y-c=\pm \sqrt {a^2-(x-b)^2}$
$y=c\pm \sqrt {a^2-(x-b)^2}$
For an equation of the form $x=f(y)$
$a = \sqrt{ ( x - b ) ^ 2 + ( y - c ) ^ 2}$
$a^2=(x-b)^2+(y-c)^2$
$(x-b)^2=a^2-(y-c)^2$
$x-b=\pm\sqrt {a^2-(y-c)^2}$
$x=b\pm \sqrt {a^2-(y-c)^2}$
It renders on Geogebra as:
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2952760",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
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Solve $9^x-2^{x+\frac{1}{2}}=2^{x+\frac{7}{2}}-3^{2x-1}$ $$9^x-2^{x+\frac{1}{2}}=2^{x+\frac{7}{2}}-3^{2x-1}.$$
The equation states solve for $x$.
What I first did was put like bases together.
$$3^{2x}+3^{2x-1}= 2^{x+\frac{7}{2}}+ 2^{x+\frac{1}{2}}.$$
Then I factored $3^{2x}$ and $2^x$
$$3^{2x}(1+\frac{1}{3})=2^x(2^{\frac{7}{2}}+2^{\frac{1}{2}}),$$
then I got
$$\frac{3^{2x}}{2^x}=9\sqrt{2}.$$
From here I took $\log$s, but the answer wasn't nice.
What to do?
| Your last line is wrong.
It should be $$3^{2x-3}=(\sqrt2)^{2x-3},$$ which gives $x=1.5$.
You got:
$$3^{2x}\left(1+\frac{1}{3}\right)=2^x(2^{\frac{7}{2}}+2^{\frac{1}{2}})$$ or
$$3^{2x-1}=2^{x-2}2^{\frac{1}{2}}(1+8)$$ or
$$3^{2x-3}=2^{x-\frac{3}{2}}$$
| {
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"url": "https://math.stackexchange.com/questions/2955218",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
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Is there an elegant way to determine $Av$ given $Au_1, Au_2$, and $Au_3$ for a $3\times3$ matrix $A$? Let A be a 3x3 matrix such that ${A} \begin{pmatrix} 3 \\ 4 \\ 5 \end{pmatrix} = \begin{pmatrix} 2 \\ 7 \\ -13 \end{pmatrix}, \quad \ {A} \begin{pmatrix} 4 \\ 5 \\ 6 \end{pmatrix} = \begin{pmatrix} -6 \\ 0 \\ 4 \end{pmatrix}, \quad \ {A} \begin{pmatrix} 5 \\ -9 \\ 1 \end{pmatrix} = \begin{pmatrix} 3 \\ 3 \\ -11 \end{pmatrix}.$
Find $\ {A} \begin{pmatrix} 3 \\ -11 \\ -1 \end{pmatrix}$
My original approach to this problem was to let $\ {A} $ be some 3x3 matrix with variable elements, and solve for each one, But i was just wondering if there was a more elegant way to solve it.
Thanks in advance!
| Note that the given conditions can be generalized:
$$\ {A} \begin{pmatrix} 3&4&5 \\ 4&5&-9 \\ 5&6&1 \end{pmatrix}=\begin{pmatrix} 2&-6&3 \\ 7&0&3 \\ -13&4&-11 \end{pmatrix}.$$
Find $A$:
$$A=\begin{pmatrix} 2&-6&3 \\ 7&0&3 \\ -13&4&-11 \end{pmatrix}\cdot \begin{pmatrix} 3&4&5 \\ 4&5&-9 \\ 5&6&1 \end{pmatrix}^{-1}=\begin{pmatrix} 2&-6&3 \\ 7&0&3 \\ -\frac{97}{4}&-\frac{19}{2}&\frac{99}{4} \end{pmatrix}.$$
At last:
$$A\cdot \begin{pmatrix}3\\-11\\-1\end{pmatrix}=\begin{pmatrix}69\\18\\7\end{pmatrix}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2958725",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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If $x>0, y>0,x+y=\frac{\pi}{3}$ then maximum value of $\tan x\tan y$
If $x>0, y>0$ and $x+y=\frac{\pi}{3}$, then find the maximum value of $\tan x\tan y$
My Attempt
$x>0, y>0, x+y=\frac{\pi}{3}\implies x, y$ in $1^\text{st}$ quadrant.
$\tan x, \tan y>0, \tan(x+y)=\sqrt{3}, \tan x\tan y>0$
$$
\tan x+\tan y\geq2\sqrt{\tan x.\tan y}\implies \tan^2x+\tan^2y+2\tan x\tan y\geq 4\tan x\tan y\\
2\tan x\tan y\leq\tan^2x+\tan^2y\implies \color{red}{?}
$$
or
$$
1-\tan x\tan y=\frac{\tan x+\tan y}{\tan(x+y)}\implies\tan x\tan y=1-\frac{\tan x+\frac{\sqrt{3}-\tan x}{1+\sqrt{3}\tan x}}{\sqrt{3}}\\
\tan x\tan y=1-\frac{\frac{\tan x+\sqrt{3}\tan^2x+\sqrt{3}-\tan x}{1+\sqrt{3}\tan x}}{\sqrt{3}}=1-\frac{1+\tan^2x}{1+\sqrt{3}\tan x}\leq\color{red}{?}\\
=\frac{1+\sqrt{3}\tan x-1-\tan^2x}{1+\sqrt{3}\tan x}=\frac{\tan x(\sqrt{3}-\tan x)}{1+\sqrt{3}\tan x}=\color{red}{?}
$$
Note: I prefer not to do differentiation
| For $-\dfrac{\pi}{6} < t < \dfrac{\pi}{6}$, let $x = \dfrac{\pi}{6} - t \ $ and $ \ y = \dfrac{\pi}{6} + t$.
Then $x > 0$, $y>0$ and $x + y = \dfrac{\pi}{3}$.
Let
\begin{align}
f(t)
&= \tan(x) \tan(y) \\
&= \tan\left(\dfrac{\pi}{6} - t \right) \tan \left(\dfrac{\pi}{6} + t \right) \\
&= \dfrac{1-\sqrt 3 \tan t}{\sqrt 3 + \tan t} \cdot
\dfrac{1+\sqrt 3 \tan t}{\sqrt 3 - \tan t} \\
&= \dfrac{1 - 3 \tan^2 t}{3 - \tan^2 t} \\
&= \dfrac{3 \tan^2 t - 1}{\tan^2 t - 3} \\
&= 3 - \dfrac{8}{3 - \tan^2 t}
\end{align}
Note that $f(t)$ is an even function on the interval $\left[-\dfrac{\pi}{6}, \dfrac{\pi}{6} \right]$.
\begin{align}
t \in \left[-\dfrac{\pi}{6}, 0 \right]
&\implies -\dfrac{1}{\sqrt 3} \le \tan t \le 0 \\
&\implies 0 \le \tan^2 t \le \dfrac 13 \\
&\implies \dfrac 83 \le 3 - \tan^2 t \le 3 \\
&\implies \dfrac 83 \le \dfrac{8}{3 - \tan^2 t} \le 3 \\
&\implies 0 \le f(t) \le \dfrac 13
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2958959",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Find exhaustive range of $k$ such that $f(x)=\frac{x-1}{k-x^2}$ never belongs to $\left[-1 \:\: \frac{-1}{3}\right]$ Find exhaustive range of $k$ such that $$f(x)=\frac{x-1}{k-x^2}$$ never belongs to $\left[-1 \:\: \frac{-1}{3}\right]$
My try:
Letting $$y=\frac{x-1}{k-x^2}$$ we get
$$yx^2+x-(1+ky)=0$$ and since $x\in \mathbb{R}$ we have $D >0$ we get
$$4ky^2+4y+1 \gt 0$$
Now the above inequality holds true when $y \in \left( -\infty \: -1\right) \cup \left(\frac{-1}{3} \: \infty\right)$
Any clue here?
| $f(x)=\frac{x-1}{k-x^2}$ never belongs to $[-1,-\frac{1}{3}]$
$\iff$ For every $y$ satisfying $-1\le y\le -\frac 13$, there are no $x$ such that $y=\frac{x-1}{k-x^2}$
$\iff$ For every $y$ satisfying $-1\le y\le -\frac 13$, there are no $x$ such that $y(k-x^2)=x-1$ and $k-x^2\not=0$
(If $k-x^2=0$, then $x=1$ implying $k=1$. But then $f(0)=-1$, so $k\not=1$.)
$\iff k\not=1$, and for every $y$ satisfying $-1\le y\le -\frac 13$, there are no $x$ such that $yx^2+x-yk-1=0$
$\iff k\not=1$, and if $-1\le y\le -\frac 13$, then $1^2-4y(-yk-1)\lt 0$
(if $k=0$, then $f(x)=\frac{x-1}{-x^2}$. Its range is $[-\frac 14,\infty).$ So, $k=0$ is sufficient.)
$\iff k\not=1$ and $k=0$ is sufficient, and if $-1\le y\le -\frac 13$, then $g(k):=4k(y+\frac{1}{2k})^2-\frac{1}{k}+1\lt 0$
$\iff k\not=1$ and $k=0$ is sufficient, and (($k\lt 0$ and $-\frac 13\lt -\frac{1}{2k}$ and $g(-\frac 13)\lt 0$) or ($k\lt 0$ and $-\frac{1}{2k}\lt -1$ and $g(-1)\lt 0$) or ($k\lt 0$ and $-1\le -\frac{1}{2k}\le -\frac 13$ and $g(-\frac{1}{2k})\lt 0$) or ($k\gt 0$ and $-\frac 13\lt -\frac{1}{2k}$ and $g(-1)\lt 0$) or ($k\gt 0$ and $-\frac{1}{2k}\lt -1$ and $g(-\frac 13)\lt 0$) or ($k\gt 0$ and $-1\le -\frac{1}{2k}\le -\frac 13$ and $g(-1)\lt 0$ and $g(-\frac 13)\lt 0$))
$\iff k\not=1$ and $k=0$, and ($k\lt 0$ or $0\lt k\lt\frac 12$ or $\frac 12\le k\lt \frac 34$)
$$\iff \color{red}{k\in\left(-\infty,\frac 34\right)}$$
Another way :
Let us separate it into cases :
*
*Case 1 : If $k=0$, then $f(x)=\frac{1-x}{x^2}$ and $f'(x)=\frac{x-2}{x^3}$, and its range is $[f(2),\infty)$, i.e. $[-\frac 14,\infty)$.
*Case 2 : If $k\in\mathbb [\frac 34,\infty)$, then $f\left(\frac{1-\sqrt{4k-3}}{2}\right)=-1$ with $k-\left(\frac{1-\sqrt{4k-3}}{2}\right)^2\not=0$.
*Case 3 : If $k\lt 0$, then $k-x^2\lt 0$ for every $x$. Then, we have $$f(x)\gt -\frac 13\iff x^2-3x+3-k\gt 0$$ which holds for every $x$ since $D=4k-3\lt 0$. So, $f(x)\gt -\frac 13$ for every $x$.
*Case 4 : If $k\in(0,\frac 34)$, then $f(x)$ is not defined when $x=\pm\sqrt k$. We have $$f'(x)=\frac{(x-(1+\sqrt{1-k}))(x-(1-\sqrt{1-k}))}{(k-x^2)^2}$$ with $-\sqrt k\lt 0\lt 1-\sqrt{1-k}\lt\sqrt k\lt 1\lt 1+\sqrt{1-k}$ and $$\lim_{x\to\pm\infty}f(x)=0,\quad \lim_{x\to {-\sqrt k}^{\pm}}f(x)=\mp\infty,\quad \lim_{x\to{\sqrt k}^{\pm}}f(x)=\pm\infty$$$$f(1\pm\sqrt{1-k})=\frac{-1\pm\sqrt{1-k}}{2k}\lt 0,\quad f(0)=-\frac 1k\lt 0,\quad f(1)=0$$So, considering the graph of $y=f(x)$, a necessary and sufficient is $$f(1-\sqrt{1-k})\lt -1\qquad\text{and}\qquad f(1+\sqrt{1-k})\gt -\frac 13,$$i.e.$$0\lt k\lt\dfrac 34$$
Therefore, from the four cases, the answer is
$$k\in\left(-\infty,\frac 34\right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2960218",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Prove that $\frac{1+\sin\theta + i\cos\theta}{1+\sin\theta-i\cos\theta}=\sin\theta+i\cos\theta.$ Prove that $$\frac{1+\sin\theta + i\cos\theta}{1+\sin\theta-i\cos\theta}=\sin\theta+i\cos\theta.$$
Hence, show that $$(1+\sin\frac{\pi}{5}+i\cos\frac{\pi}{5})^5+i(1+\sin\frac{\pi}{5}-i\cos\frac{\pi}{5})^5=0.$$
In the first part, I tried realising LHS so I got $$LHS=\frac{(1+\sin\theta+i\cos\theta)^2}{(1+\sin\theta)^2+\cos^2\theta}=\frac{1+2\sin\theta+2i\cos\theta+2i\cos\theta\sin\theta+\sin^2\theta+i^2\cos^2\theta}{2+2\sin\theta}.$$
but now I am stuck :( . Any help would be greatly appreciated, thanks!
| Hint:
$$\frac{1+sin\theta + icos\theta}{1+sin\theta-icos\theta}=sin\theta+icos\theta \iff 1+sin\theta + icos\theta =\underbrace{(1+sin\theta-icos\theta)(sin\theta+icos\theta)}_{R}$$
so calculate the right side and...
$$ R = sin\theta+icos\theta + \underbrace{sin^2\theta-i^2cos ^2\theta}_{=1}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2961689",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Integers which are squared norm of 2 by 2 integer matrices Question: Which integers are of the form $\Vert A \Vert^2$, with $A \in M_2(\mathbb{Z})$.
The code below provides the first such integers: $0, 1, 2, 4, 5, 8, 9, 10, 13, 16, 17, 18, 20, 25, 26$.
By searching this sequence on OEIS, we find: "Numbers that are the sum of 2 squares" A001481.
Are these integers exactly those which are the sum of two squares ?
Research
First, $\Vert A \Vert^2$ is the largest eigenvalue of $A^*A$, so for $A = \left( \begin{matrix} a & b \cr c & d \end{matrix} \right)$ and $a,b,c,d \in \mathbb{Z}$, so we get:
$$\Vert A \Vert^2 = \frac{1}{2} \left(a^2+b^2+c^2+d^2+\sqrt{(a^2+b^2+c^2+d^2)^2 - 4(ad-bc)^2}\right)$$
Obviously, every sum of two squares is of the expected form , because by taking $c=d=0$, we get $\Vert A \Vert^2=a^2+b^2$.
Then it remains to prove that there is no other integer (if true).
Now, recall that:
Sum of two square theorem
An integer greater than one can be
written as a sum of two squares if and only if its prime decomposition
contains no prime congruent to 3 (mod 4) raised to an odd power.
By taking $c=ra$ and $d=rb$, we get that $\Vert A \Vert^2 = (r^2+1)(a^2+b^2)$, which is also a sum of two square because the following equation occurs (proof here):
$$r^2 \not \equiv -1 \mod 4s+3$$
A necessary condition for $\Vert A \Vert^2$ to be an integer, is that $(a^2+b^2+c^2+d^2)^2 - 4(ad-bc)^2$ must be a square $X^2$, so that $(X,2(ad-bc),a^2+b^2+c^2+d^2)$ is a Pythagorean triple, so must be of the form $(k(m^2-n^2),2kmn,k(m^2+n^2)$, and then $\Vert A \Vert^2 = km^2$. So it remains to prove that $k$ must be a sum of two squares.
sage: L=[]
....: for a in range(-6,6):
....: for b in range(-6,6):
....: for c in range(-6,6):
....: for d in range(-6,6):
....: n=numerical_approx(matrix([[a,b],[c,d]]).norm()^2,digits=10)
....: if n.is_integer():
....: L.append(int(n))
....: l=list(set(L))
....: l.sort()
....: l[:20]
....:
[0, 1, 2, 4, 5, 8, 9, 10, 13, 16, 17, 18, 20, 25, 26, 29, 32, 34, 36, 37]
| Yes. If $A$ is a $2\times2$ integer matrix such that $n=\|A\|^2$ is an integer, $n$ must be the sum of two integer squares. Conversely, if $n$ is the sum of two integer squares, then $n=\|A\|^2$ for some $2\times2$ integer matrix $A$.
Proof. Suppose $A$ is a $2\times2$ integer matrix such that $n=\|A\|^2$ is an integer. We want to show that $n$ is the sum of two integer squares. This is clearly true if $n$ is $0$ or $1$. Suppose $n>1$. Then $A^TA-nI$ is a singular matrix with integer entries. Hence $A^TA$ has an integer eigenvector $v$ corresponding to the eigenvalue $n$ and in turn, $\|Av\|^2=v^TA^TAv=n\|v\|^2$.
Since both $\pmatrix{x\\ y}:=v$ and $\pmatrix{a\\ b}:=Av$ are integer vectors, the previous equality implies that $n(x^2+y^2)=a^2+b^2$. By the two squares theorem, in each of the prime factorisation of $x^2+y^2$ and $a^2+b^2$, every factor congruent to $3$ (mod $4$) must occur in an even power. Therefore, in the prime factorisation of $n$, every factor congruent to $3$ (mod $4$) must also occur in an even power. Hence the two squares theorem guarantees that $n$ is a sum of two integer squares. This proves one direction of our assertion.
For the other direction, suppose $n=a^2+b^2$ for some two integers $a$ and $b$. Then $\|A\|^2=n$ when $A=\pmatrix{a&-b\\ b&a}$ or $\pmatrix{a&0\\ b&0}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2963069",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 1,
"answer_id": 0
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Proving a binomial coefficient inequality per induction $\left(\begin{array}{c}2n\\ n+k\end{array}\right) < \left(\begin{array}{c}2n\\ n\end{array}\right), n,k \in \mathbb{N}, 1\leq k \leq n$
Proving this by induction on $k$ is no problem. However, I want to prove this by induction on $n$, but I am stuck because of a too "harsh" inequality estimation:
$\left(\begin{array}{c}2(n+1)\\ (n+1)+k\end{array}\right) = \frac{(2n+1)(2n+2)}{(n+1-k)(n+1+k)} \left(\begin{array}{c}2n\\ n+k\end{array}\right) < (induction \space hypothesis) \frac{(2n+1)(2n+2)}{(n+1-k)(n+1-k)} \left(\begin{array}{c}2n\\ n\end{array}\right)$
But $\frac{(2n+1)(2n+2)}{(n+1-k)(n+1+k)} \left(\begin{array}{c}2n\\ n\end{array}\right) \nleq \left(\begin{array}{c}2(n+1)\\ n+1\end{array}\right) $
which ruins the inductive step.
I don't know if I am getting something wrong or one has to handle it differently.
Thanks in advance.
| Choose $c$ s.t. $\left(\begin{array}{c}2n\\ n+k\end{array}\right) < c \times \left(\begin{array}{c}2n\\ n\end{array}\right) < \left(\begin{array}{c}2n\\ n\end{array}\right)$ . $c = \frac{(n+1-k)(n+1+k)}{(n+1)(n+1)} $. The inductive step (first inequality) works. The second inequality is true since $ \frac{(n+1-k)(n+1+k)}{(n+1)(n+1)} = 1 - \frac{k^2}{(n+1)^2} < 1 $
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2963266",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Using the sequential definition of a limit to show $\lim_{x\to 1} \frac{x^2 - 1}{\sqrt{x} - 1} = 4.$ Very recently, I posted this thread: Using the sequential definition of a limit to show $\lim_{x\to 0} \frac{x^2}{x} = 0.$ My solution for that proof was correct, but now I'm having trouble showing that $\lim_{x\to 1} \frac{x^2 - 1}{\sqrt{x} - 1} = 4$. For reference, here is my definition:
I have the following definition for a limit:
Definition: Given a function $f : D \rightarrow \mathbb{R}$ and a limit point $x_{0}$ of its domain $D$, for a number $\ell$, we write
$$ \lim_{x\to x_{0}} f(x) = \ell$$
provided that whenever $\{x_{n}\}$ is a sequence in $D \ - \{x_{0}\}$ that converges to $x_{0}$,
$$\lim_{n\to\infty} f(x_{n}) = \ell. $$
Using this definition, here is my attempt:
Let $\{x_{n}\}$ be a sequence in $\mathbb{R} - \{1\}$ such that $\{x_{n}\}$ converges to $1$. This means for all $\epsilon > 0$, there exists an index $N$ so that
$$|x_{n} - 1| < \epsilon$$
for all $n \geq N$. Now, we need to show for all $\epsilon > 0$, there exists an index $N_{2}$ so that
$$\left|\frac{x_{n}^2 - 1}{\sqrt{x_{n}} - 1} - 4\right| < \epsilon$$
for $n\geq N_{2}$.
So, I'm having trouble finding such an index $N_{2}$. I tried writing the expression as follows:
$$\left|\frac{x_{n}^{2} - 1}{\sqrt{x_{n}} - 1} - 4\right| \\$$
$$= \left|\frac{x_{n}^{2} - 1 - 4\sqrt{x_{n}} + 4}{\sqrt{x} - 1} \right| $$
$$\leq \left|\frac{x_{n}^{2} + 3}{\sqrt{x_{n}} - 1} \right|,$$
but I couldn't get anywhere after this. Can someone please help me finish this proof?
EDIT: An attempt based on current answers:
$$\begin{align*}
\left|\frac{x_{n}^{2} - 1}{\sqrt{x_{n}} - 1} - 4\right| = \left| \frac{(\sqrt{x_{n}} - 1)(1 + \sqrt{x_{n}} + x_{n} + \sqrt{x_{n}^{3}})}{\sqrt{x_{n} - 1}} - 4\right| \\[1em]
= \left|1 + \sqrt{x_{n}} + x_{n} + \sqrt{x_{n}^{3}} - 4\right|
\end{align*},$$
but I get nowhere from here.
| Hint
Use that
$$t^4-1=(t-1)(1+t+t^2+t^3)$$
with $t=\sqrt x.$
Edit
You got
$$\left|\frac{x_{n}^{2} - 1}{\sqrt{x_{n}} - 1} - 4\right|
= \left|1 + \sqrt{x_{n}} + x_{n} + \sqrt{x_{n}^{3}} - 4\right|.$$
Assume that $1-\delta<x_n\le 1.$ Then we have
$$1 + \sqrt{x_{n}} + x_{n} + \sqrt{x_{n}^{3}}>1+\sqrt{1-\delta}+1-\delta+\sqrt{(1-\delta)^3}>1+3\sqrt{(1-\delta)^3}.$$
Thus $$0<4-(1 + \sqrt{x_{n}} + x_{n} + \sqrt{x_{n}^{3}})<3(1-\sqrt{(1-\delta)^3}).$$ Now
$$3(1-\sqrt{(1-\delta)^3})<\epsilon\iff \delta <1-\sqrt[3]{\left(1-\frac{\epsilon}{3}\right)^2}.$$
Assume that $1\le x_n<1+\delta.$ Then we have
$$1 + \sqrt{x_{n}} + x_{n} + \sqrt{x_{n}^{3}}<1+\sqrt{1+\delta}+1+\delta+\sqrt{(1+\delta)^3}<1+3\sqrt{(1+\delta)^3}.$$
Thus $$0<1 + \sqrt{x_{n}} + x_{n} + \sqrt{x_{n}^{3}}-4<3(\sqrt{(1+\delta)^3}-1).$$ Now
$$3(\sqrt{(1+\delta)^3}-1)<\epsilon\iff \delta <\sqrt[3]{\left(1+\frac{\epsilon}{3}\right)^2}-1.$$
Finally, we have shown that $$\delta<\min\{1-\sqrt[3]{\left(1-\frac{\epsilon}{3}\right)^2}, \sqrt[3]{\left(1+\frac{\epsilon}{3}\right)^2}-1 \}\implies \left|1 + \sqrt{x_{n}} + x_{n} + \sqrt{x_{n}^{3}} - 4\right|<\epsilon.$$ But since $\lim_n x_n=1$ for all $\delta>0$ there exists $N\in\mathbb{N}$ such that $$n\ge N\implies 1-\delta<x_n<1+\delta.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2965202",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
Help with this proof by induction with inequalities. Show that mathematical induction can be used to
prove the stronger inequality $\frac{1}{2}\cdot ...\cdot \frac{2n-1}{2n} < \frac{1}{\sqrt{3n + 1}}$ for all integers greater than 1, which, together with a verification for the case where n = 1, establishes the weaker inequality we originally tried to prove using mathematical induction.
Base Case: P(2)
\begin{aligned}
\frac{1}{2}\cdot\frac{2(2)-1}{2(2)} &< \frac{1}{\sqrt{3(2) + 1}}\\
\frac{3}{8} &< \frac{1}{\sqrt{7}}\\
\frac{1}{8} &< \frac{1}{3\sqrt{7}}\\
\end{aligned}
This is true as $8 > 3\sqrt{7}$.
Inductive Hypothesis: $\frac{1}{2}\cdot ...\cdot \frac{2n-1}{2n} < \frac{1}{\sqrt{3n+1}}$
In the inductive step, we want to show that $\frac{1}{2}\cdot ...\cdot \frac{2n-1}{2n} \cdot \frac{2n+1}{2n+2} < \frac{1}{\sqrt{3n+4}}$.
Using the inductive hypothesis, we can get to the following:
\begin{aligned}
\frac{1}{2}\cdot ...\cdot \frac{2n-1}{2n} \cdot \frac{2n+1}{2n+2} &< \frac{1}{\sqrt{3n+1}}\cdot \frac{2n+1}{2n+2}\\
&< \frac{1}{\sqrt{3n+1}}\cdot 1\\
\end{aligned}
I am not sure how to get to $< \frac{1}{\sqrt{3n+4}}$ from here because i know that if the denominator would get bigger by adding 3 to it so the inequality wouldn't follow...
| Continuing where you left. You need to show $\large{\frac{1}{\sqrt{3n+1}}\cdot \frac{2n+1}{2n+2}<\frac{1}{\sqrt{3n+4}}}$, as square roots defined as positive numbers we can multiply both side by $\sqrt{3n+1},$ and this implies $$\frac{2n+1}{2n+2}< \frac{\sqrt{3n+1}}{\sqrt{3n+4}}\implies (2n+1)\sqrt{3n+4}<(2n+2)\sqrt{3n+1}\\ \implies (4n^2+4n+1)(3n+4)<(4n^2+8n+4)(3n+1)\\ \implies(4n^2+4n+1)\big(3n+4-3n-1)<(4n+3)(3n+1)\\ \implies 12n^2+12n+3<12n^2+13n+3 $$
which is true. Hence we are done.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2965485",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Finding minimum value of $ \frac{x^2 +y^2}{y} $
Finding the minimum value of $\displaystyle \frac{x^2 +y^2}{y}.$ where $x,y$ are
real
numbers satisfying $7x^2 + 3xy + 3y^2 = 1$
Try: Equation $7x^2+3xy+3y^2=1$ represent Ellipse
with center is at origin.
So substitute $x=r\cos \alpha $ and $y=r\sin \alpha$
in $7x^2+3xy+3y^2=1$
$$3r^2+4r^2\cos^2 \alpha+3r^2\sin \alpha \cos \alpha =1$$
$$3r^2+2r^2(1+\cos 2 \alpha)+\frac{3r^2}{2}\sin 2 \alpha =1$$
$$8r^2+r^2(4\cos 2 \alpha+3\sin \alpha)=2$$
So $$r^2=\frac{2}{8+(4\cos 2 \alpha+3\sin \alpha)}$$
$$\frac{2}{8+5}=\frac{2}{13}\leq r^2\leq \frac{2}{8-5}=\frac{2}{3}$$
we have to find minimum of $$\frac{x^2+y^2}{y}=\frac{r}{\sin \alpha}$$
How can i find it, could some help me
| The problem has no solution (assuming you are looking for a (global) minimum rather than just a local minimum). As we approach the point $(1/\sqrt{7},0)$ along the constraint curve from below the $x$-axis, the value of the function goes to $-\infty$:
The points
$$\left(\frac{\sqrt{28-75{y}^{2}}-3y}{14},y\right)$$
satisfy the constraint. We have
$$\begin{align*}f\left(\frac{\sqrt{28-75{y}^{2}}-3y}{14},y\right)&=
%\frac{65{y}^{2}+14-3y\sqrt{28-75{y}^{2}}}{98y}=
65y+\frac{1}{7y}-\frac{3}{98}\sqrt{28-75{y}^{2}}\to-\infty\end{align*}$$
as $y\to0^-$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2965865",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
What is the smallest positive integer $n$ such that $(a+1)^{7^n} = (a+1)$? In $\mathbb{F} = \mathbb{F}_7[X]/(X^2+1)$ let $a$ be the class of $X$. What is the smallest positive integer $n$ such that $(a+1)^{7^n} = (a+1)$ ?
I know that the answer is $2$ but I don't see how it works. Can someone explain the logic to me? Because if I get it correctly, $a = X$ in this case and we're reducing $mod$ $X^2 +1$ so how does $ a^{49} +1$ equal to $a+1$ ?
| The question can be rephrased as
What is the smallest positive integer $n$ such that $(a+1)^{7n-1} = 1$?
Let $m$ be the order of $a+1$ in the multiplicative group $\Bbb{F}^\times$, which has order $48$.
Then $m$ divides $\gcd(7n-1,48)$. Let's see the possibilities:
$$
\begin{array}{r|rrrrrrr}
n & 1 & 2 & 3 & 4 & 5 & 6 & 7 \\
7n-1 & 6 & 13 & 20 & 27 & 34 & 41 & 48 \\
\gcd(7n-1,48) & 6 & 1 & 4 & 3 & 2 & 1 & 48
\end{array}
$$
By checking that $(a+1)^k \ne 1$ for $k=1,2,3,4,6$, we arrive at $n=7$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2969252",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Evaluate $\int\limits_{|z|=1}\frac{\sin{1/z}}{(z-2)^2}dz.$ This is a question from an old exam. The two integrals are seemingly similar, but the first one seems quite tedious compared to the other one. It seems, according to the prof solution, that the hard part in the first integral is computing the residue at $z=0.$
Evaluate the intergals
a) $$\int\limits_{|z|=1}\frac{\sin{(1/z)}}{(z-2)^2} \ dz$$
b) $$\int\limits_{|z|=3}\frac{\sin{(1/z)}}{(z-2)^2} \ dz$$
In the first, we have $2$ poles, first one is $z_1=0$ and the second one is a pole of order $2$, which is $z_2=2.$ However, only $z_1$ is inside our unitcircle so we only need to compute $\text{Res}_{z_1}(f(z))$ and apply the residue theorem.
Since our function is of the form $f(z)=\frac{g(z)}{(z-z_1)^k}$ the residue is given by
$$\text{Res}_{z_1}(f(z))=\frac{g^{k-1}(z_1)}{(k-1)!},$$
and here $k=2$ and $(\sin(1/z))'=-\cos(1/z)/z^2$, so
$$\text{Res}_{z_1}(f(z))=-\frac{\cos\frac{1}{z_1}}{z_1^2}=...\text{this is the moment I realised I'm screwed.}$$
Howver, computing the residue at $z_2=2$ for the other integral I can use conventional methods, but not here.
Can someone break down the main difference between these integrals and show how to find the residue at $z_1=0?$
EDIT:
I need to solve $a)$ using Laurent series.
| Re $(a)$ and series, given
$$\sin{z}=z-\frac{z^3}{3!}+\frac{z^5}{5!}-\frac{z^7}{7!}+\frac{z^9}{9!}+...=
\sum\limits_{k=0}\frac{(-1)^k}{(2k+1)!}z^{2k+1} \tag{1}$$
we have
$$\sin{\frac{1}{z}}=\sum\limits_{k=0}\frac{(-1)^k}{(2k+1)!}\frac{1}{z^{2k+1}}$$
then
$$\int\limits_{|z|=1}\frac{\sin{(1/z)}}{(z-2)^2} dz=\sum\limits_{k=0}\frac{(-1)^k}{(2k+1)!}\int\limits_{|z|=1}\frac{1}{z^{2k+1}(z-2)^2}dz \tag{2}$$
Using Cauchy's integral formula:
$$f^{(n)}(a)=\frac{n!}{2\pi i} \int\limits_{\gamma}\frac{f(z)}{(z-a)^{n+1}}dz \tag{3}$$
where $f(z)=\frac{1}{(z-2)^2}$, because $2$ is outside $|z|=1$, we have
$$f^{2k}(0)=\frac{(2k)!}{2\pi i}\int\limits_{|z|=1}\frac{1}{z^{2k+1}(z-2)^2}dz$$
then $(2)$ becomes
$$\int\limits_{|z|=1}\frac{\sin{(1/z)}}{(z-2)^2} dz=2\pi i\left(\sum\limits_{k=0}\frac{(-1)^k}{(2k+1)!}\frac{f^{2k}(0)}{(2k)!}\right) \tag{4}$$
But
$$f^{'}(z)=-\frac{2}{(z-2)^3}$$
$$f^{''}(z)=\frac{2\cdot3}{(z-2)^4}$$
$$f^{'''}(z)=-\frac{2\cdot3\cdot4}{(z-2)^5}$$
$$f^{(4)}(z)=\frac{2\cdot3\cdot4\cdot5}{(z-2)^6}$$
$$...$$
$$f^{(2k)}(z)=\frac{(2k+1)!}{(z-2)^{2k+2}} \tag{5}$$
and $(4)$ becomes
$$\int\limits_{|z|=1}\frac{\sin{(1/z)}}{(z-2)^2} dz=
2\pi i\left(\sum\limits_{k=0}\frac{(-1)^k}{(2k)!}\frac{1}{(-2)^{2k+2}}\right)=
2\pi i \frac{1}{4}\left(\sum\limits_{k=0}\frac{(-1)^k}{(2k)!}\frac{1}{2^{2k}}\right) \tag{6}$$
but
$$\sum\limits_{k=0}\frac{(-1)^k}{(2k)!}\frac{1}{2^{2k}}=\cos{\frac{1}{2}}$$
from cosine series expansion, thus
$$\int\limits_{|z|=1}\frac{\sin{(1/z)}}{(z-2)^2} dz=
\frac{\pi i}{2} \cos{\frac{1}{2}} \tag{7}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2969634",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Find Derivative of $y=\tan^{-1}\left(\sqrt{\frac{x+1}{x-1}}\right)$ for $|x|>1$?
Question. If $y=\tan^{-1}\left(\sqrt{\frac{x+1}{x-1}}\right)$ for $|x|>1$ then $\frac{d}{dx}=?$
Answer: $\displaystyle\frac{dy}{dx}=-\frac{1}{2|x|\sqrt{x^2-1}}$
My 1st attempt- I followed the simple method and started by taking darivative of tan inverse and the following with chain rule and i got my answer as ($-\frac{1}{2x\sqrt{x^2-1}}$), which is not same as the above correct answer. 2nd method is that you can substitute $x=\sec\left(\theta\right)$ and while solving in last step we will get $\sec^{-1}\left(\theta\right)$ whose derivative contains $\left|x\right|$, but still i searched and don't know why its derivative has $\left|x\right|$
Here's my attempt stepwise
$\displaystyle\frac{dy}{dx}=\frac{1}{1+\left(\sqrt{\frac{x+1}{x-1}}\right)^2}\cdot\frac{1}{2\sqrt{\frac{x+1}{x-1}}}\cdot\frac{\left(x-1\right)-\left(x+1\right)}{\left(x-1\right)^2}$
$\displaystyle=\frac{\left(x-1\right)}{\left(x-1\right)+\left(x+1\right)}\cdot\frac{1\sqrt{x-1}}{2\sqrt{x+1}}\cdot-\frac{2}{\left(x-1\right)^2}$
$\displaystyle=-\frac{1}{2x}\cdot\frac{\left(x-1\right)\sqrt{x-1}}{\left(x-1\right)^2}\cdot\frac{1}{\sqrt{x+1}}$
$\displaystyle=-\frac{1}{2x\sqrt{x-1}\sqrt{x+1}}$
$\displaystyle=-\frac{1}{2x\sqrt{x^2-1}}$
Can you tell what i am doing wrong in my 1st attempt?
| You start right:
\begin{align}
\frac{dy}{dx}
&=\frac{1}{1+\left(\sqrt{\dfrac{x+1}{x-1}}\right)^2}
\frac{1}{2\sqrt{\dfrac{x+1}{x-1}}}
\frac{(x-1)-(x+1)}{(x-1)^2}\\
&=\frac{1}{2}\frac{x-1}{(x-1)+(x+1)}
\sqrt{\dfrac{x-1}{x+1}}
\frac{-2}{(x-1)^2}
\end{align}
but then make a decisive error in splitting the square root in the middle and then use the wrong fact that $t=\sqrt{t^2}$, which only holds for $t\ge0$.
You can go on with
$$
=-\frac{1}{2x}\sqrt{\dfrac{x-1}{x+1}}\frac{1}{x-1}
$$
and now you have to split into the cases $x>1$ and $x<-1$ or observe that $x(x-1)=|x|\,|x-1|$ (when $|x|>1$) so you can write
$$
=-\frac{1}{2|x|}\sqrt{\dfrac{x-1}{x+1}}\frac{1}{|x-1|}
=-\frac{1}{2|x|}\sqrt{\dfrac{x-1}{x+1}\frac{1}{(x-1)^2}}
$$
and finish up.
You might try simplifying the expression, before plunging in the computations.
You have, by definition,
$$
\sqrt{\frac{x+1}{x-1}}=\tan y
$$
so that
$$
x+1=x\tan^2y-\tan^2y
$$
that yields
$$
x=\frac{1+\tan^2y}{\tan^2y-1}=\frac{1}{\cos^2y}\frac{\cos^2y}{-\cos2y}=-\frac{1}{\cos2y}
$$
Hence $\cos2y=-1/x$ and $y=\frac{1}{2}\arccos(-1/x)$. Thus
$$
\frac{dy}{dx}=-\frac{1}{2}\frac{-1}{\sqrt{1-\dfrac{1}{x^2}}}\frac{-1}{x^2}=
=-\frac{1}{2}\frac{-\sqrt{x^2}}{\sqrt{x^2-1}}\frac{-1}{x^2}=
-\frac{1}{2|x|\sqrt{x^2-1}}
$$
because
$$
\frac{\sqrt{x^2}}{x^2}=\frac{|x|}{|x|^2}=\frac{1}{|x|}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2970280",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 1
} |
Verify the proof that $x_n = \ln^2(n+1) - \ln^2n$ is a bounded sequence.
Let $n\ \in \mathbb N$ and:
$$
x_n = \ln^2(n+1) - \ln^2n
$$
Prove that $x_n$ is a bounded sequence.
I've taken the following steps. Consider $x_n$
$$
\begin{align}
x_n &= \ln^2(n+1) - \ln^2n = \\
&= (\ln(n+1) + \ln n)(\ln (n+1) - \ln n) = \\
&= \ln \frac{n + 1}{n}\cdot \ln (n(n+1)) = \\
&= \ln\left({1 + {1\over n}}\right)\cdot \ln(n(n+1))
\end{align}
$$
Now multiply and divide by $n$:
$$
\begin{align}
x_n &= {n \over n} \ln\left({1 + {1\over n}}\right)\cdot \ln(n(n+1)) = \\
&= \ln\left({1 + {1\over n}}\right)^n \cdot\ln \sqrt[^n]{(n(n+1))}
\end{align}
$$
Now consider $\left({1 + {1\over n}}\right)^n$. There are plenty of proofs that it is bounded. In my case I've used expansion with binomial coefficients to prove that :
$$
2< \left({1 + {1\over n}}\right)^n < 3 \implies \\
\ln2 < \ln \left({1 + {1\over n}}\right)^n < \ln3
$$
So now we want to prove that $\ln \sqrt[^n]{(n(n+1))}$ is bounded. Start with the following:
$$
\ln \sqrt[^n]{n(n+1)} < \ln \sqrt[^n]{(n+1)^2}
$$
Consider the following equation:
$$
\begin{align}
\sqrt[^n]{(n+1)^2} &= 1+a_n \iff \\
\iff (n+1)^2 &= (1+a_n)^n = \sum_{k=0}^{n}\binom{n}{k}a_n^k
\end{align}
$$
Now:
$$
\sum_{k=0}^{n}\binom{n}{k}a_n^k \ge \frac{n(n+1)}{2}a_n^2 \implies \\
\implies (n+1)^2 \ge \frac{n(n+1)}{2}a_n^2 \implies \\
\implies a_n \le \sqrt{2 + {2\over n}}
$$
So $a_k$ is clearly bounded. Which means:
$$
\sqrt[^n]{(n+1)^2} < 1 + \sup\{a_n\} = 3
$$
Also $\sqrt[^n]{(n+1)^2} > 1$. So:
$$
\ln1 < \ln \sqrt[^n]{(n(n+1))} < \ln3
$$
Now going back to initial expression:
$$
\ln1 \cdot \ln2 < \ln\left({1 + {1\over n}}\right)^n \cdot\ln \sqrt[^n]{(n(n+1))} < \ln3 \cdot \ln3
$$
Meaning $x_n$ is bounded. Have I missed something?
| Lagrange's theorem provides a one-liner:
$$ \log^2(n+1)-\log^2(n) = \frac{d}{dx}\left.\log^2(x)\right|_{x=\xi\in(n,n+1)},\quad \frac{2\log\xi}{\xi}\to 0\text{ as }n\to +\infty. $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2972286",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 4
} |
find the sum to n term of $\frac{1}{1\cdot2\cdot3} + \frac{3}{2\cdot3\cdot4} + \frac{5}{3\cdot4\cdot5} + \frac{7}{4\cdot5\cdot6 } + ... $ $$\frac{1}{1\cdot2\cdot3} + \frac{3}{2\cdot3\cdot4} + \frac{5}{3\cdot4\cdot5} + \frac{7}{4\cdot5\cdot6 } + ... $$
$$=\sum \limits_{k=1}^{n} \frac{2k-1}{k\cdot(k+1)\cdot(k+2)}$$ $$= \sum \limits_{k=1}^{n} - \frac{1}{2}\cdot k + \sum \limits_{k=1}^{n} \frac{3}{k+1} - \sum \limits_{k=1}^{n}\frac{5}{2\cdot(k+2)} $$
I do not know how to get a telescoping series from here to cancel terms.
| HINT:
Note that we have
$$\begin{align}
\frac{2k-1}{k(k+1)(k+2)}&=\color{blue}{\frac{3}{k+1}}-\frac{5/2}{k+2}-\frac{1/2}{k}\\\\
&=\color{blue}{\frac12}\left(\color{blue}{\frac{1}{k+1}}-\frac1k\right)+\color{blue}{\frac52}\left(\color{blue}{\frac{1}{k+1}}-\frac{1}{k+2}\right)
\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2972505",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 4,
"answer_id": 0
} |
Reflection of point in barycentric coordinates I have a triangle $ABC$ and a point $P$ with barycentric coordinates ($\alpha, \beta, \gamma)$ that I want to reflect about the sides $a,b$ and $c$.
Calculating the general expression for a displacement vector perpendicular to $c$ and then using $|PB|=|P'B|$, I got
$$P'=\left(\alpha+x, \beta-\frac{S_A}{c^2}x, \gamma-\frac{S_B}{c^2}x\right)$$
for the reflection about $c$, where
$$x=\frac{-a^2\left((\beta-1)(-\frac{S_B}{c^2})+\gamma(-\frac{S_A}{c^2})\right)-b^2\left(\gamma+\alpha(-\frac{S_B}{c^2})\right)-c^2\left(\alpha(-\frac{S_A}{c^2})+(\beta-1)\right)}{\frac{S_A}{c^2}+\frac{S_B}{c^2}-\frac{S_A S_B}{c^4}}$$
and
$$S_A=\frac{-a^2+b^2+c^2}{2}, S_B=\frac{a^2-b^2+c^2}{2}$$
is Conway's Notation.
Can anyone confirm this or provide an easier formula? Any help is appreciated.
| Let's try some brute force with a special case. Assign Cartesian coordinates
$$A := (0,0) \qquad B := (c,0) \qquad C := (b\cos A, b\sin A) \tag{1}$$
and define $P$ with barycentric coordinates $(\alpha, \beta, \gamma)$, so that
$$P = \frac{\alpha A + \beta B + \gamma C}{\alpha + \beta + \gamma} = \frac{(\beta c+\gamma b \cos A, \gamma b \sin A )}{\alpha+\beta+\gamma} =: (P_x,P_y) \tag{2}$$
Let $P^\prime = (\alpha^\prime, \beta^\prime, \gamma^\prime)$ be the reflection of $P$ in $\overline{AB}$ (the $x$-axis). Then $P^\prime_x=P_x$ and $P^\prime_y=-P_y$, giving the equations
$$\begin{align}
(\beta\,c+\gamma\,b \cos A)(\alpha^\prime+\beta^\prime+\gamma^\prime)&=(\beta^\prime\,c+\gamma^\prime\,b\cos A)(\alpha+\beta+\gamma) \\[4pt]
\gamma\,(\alpha^\prime+\beta^\prime+\gamma^\prime)&= - \gamma^\prime\,(\alpha+\beta+\gamma)
\end{align}\tag{3}$$
Solving the system for, say, $\alpha^\prime$ and $\beta^\prime$ gives
$$\begin{align}
\alpha^\prime &= -\frac{\gamma^\prime}{\gamma\,c} \left(\;\alpha\,c + 2 \gamma\,(c - b \cos A)\;\right) = -\frac{\gamma^\prime}{\gamma\,c} \left(\;\alpha\,c + 2 \gamma\,a \cos B\;\right) \\[4pt]
\beta^\prime &= -\frac{\gamma^\prime}{\gamma\,c}\left(\;\beta\,c + 2 \gamma\,b \cos A\;\right)
\end{align}\tag{4}$$
from which we can deduce barycentric coordinates, in a smattering of variants,
$$\begin{align}
\alpha^\prime:\beta^\prime:\gamma^\prime\quad&=\quad
\alpha\,c + 2 \gamma\,a \cos B\;:\;
\beta\,c + 2 \gamma\,b \cos A\;:\;
-\gamma\,c \\[8pt]
\quad&=\quad
\alpha+\frac{2\gamma\,a\cos B}{c}:\beta+\frac{2\gamma\,b \cos A}{c} : \gamma - 2\gamma \\[8pt]
\quad&=\quad
\alpha+2\gamma\,\sin A\cos B \csc C:\beta+2\gamma\,\cos A\sin B\csc C : \gamma - 2\gamma \\[8pt]
\quad&=\quad
\alpha+\frac{2\gamma\,S_B}{c^2}:\beta+\frac{2\gamma\,S_A}{c^2} : \gamma - 2\gamma
\end{align} \tag{5}$$
As a sanity check:
*
*Any point on the $x$-axis has $\gamma=0$; reflection fixes the such a point, and we see from $(5)$ that, indeed $\alpha^\prime : \beta^\prime : 0 = \alpha:\beta:0$.
*The reflection of $C$, which has barycentric coordinates $(0,0,1)$, should have Cartesian coordinates $(b\cos A,-b\sin A)$; from the first form in $(5)$,
$$\begin{align}
\frac{\alpha^\prime A + \beta^\prime B + \gamma^\prime C}{\alpha^\prime+\beta^\prime+\gamma^\prime}
&= \frac{A\cdot 2a\cos B + B\cdot 2b\cos A -C\cdot c}{2a\cos B+2b\cos A-c} \\[8pt]
&=\frac{(2bc\cos A-bc \cos A, -bc\sin A)}{2c-c} \\[8pt]
&= (b\cos A,-b\sin A)
\end{align} \tag{6}$$
as expected.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2972761",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Divisibility 1,2,3,4,5,6,7,8,9,&10 Tried:
Seems the ten-digit number ends with $240$ or $640$ or $840$ (Is not true, there are more ways the number could end)
$8325971640,$
$8365971240,$
$8317956240,$
$8291357640,$
$8325971640,$
$8235971640,$
$1357689240,$
$1283579640,$
$1783659240,$
$1563729840,$
$1763529840,$
$1653729840,$
$7165239840,$
$7195236840,$
$2165937840,$
$9283579640$
| Suppose the number is of form $N=jihgfedcba$. We may write:
$a+b+c+d+e+f+g+h+i+j+(10-1)b+(10^2-1)c+(10^3-1)d+(10^4-1)e+(10^5-1)f+(10^6-1)g+(10^7-1)h+(10^8-1)i+(10^9-1)j$ | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2972869",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 0
} |
Trigonometric Identities: Given $\cot(b)=-2$ find $\sin(4b)$ and $\cos(4b)$
Given $\cot(b)=-2$ find $\sin(4b)$ and $\cos(4b)$
I found $\sin(4b)$ as $-24/25$ but I always get a very bad answer for $\cos(4b)$
Any hints?
edit:
I made $sin(4b)$ into a expanded form.
$4sin(b)cos^3(b)-4sin^3(b)cos(b)$
And then I made a triangle using $cot(b)=-2$ for information.
I got from that triangle that, $sin(b)=\frac{\sqrt{5}}{5}$ and that $cos(b)=\frac{-2\sqrt{5}}{5}$
And from their I simplified.
But whenever I expand $cos(4b)$ and plug in I get a bad answer, $\frac{353}{25}$ and I do not know if it is correct or wrong.
| The result $353/25$ is surely wrong, as a cosine cannot be $>1$.
You're complicating your own life! ;-)
There are standard formulas:
\begin{align}
\sin 2x&=\frac{2\tan x}{1+\tan^2x}=\frac{2\cot x}{\cot^2x+1}
\\[4px]
\cos 2x&=\frac{1-\tan^2x}{1+\tan^2x}=\frac{\cot^2x-1}{\cot^2x+1}
\end{align}
Thus
$$
\sin 2b=\frac{-4}{4+1}=-\frac{4}{5}
$$
and
$$
\cos 2b=\frac{4-1}{4+1}=\frac{3}{5}
$$
Now you know that
\begin{align}
\sin 4b&=2\sin2b\cos2b=-\frac{24}{25} \\[4px]
\cos 4b&=\cos^22b-\sin^22b=…
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2973963",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Inequality sign changing when multiplying with a unknown variable? This is the inequality:
$$\frac{(2x-1)}{(3-5x)} < 2$$
Next step is:
$$\frac{(2x-1)}{(3-5x)} \cdot (3-5x) < 2\cdot (3-5x)$$
Now is the sign $<$ or $>$?
The end solution is $x < \frac{7}{12}.$
| In general, when multiplying an inequality by an unknown quantity, we can't tell whether the sign changes or not. To solve the problem, then, we must consider both cases, i.e. whether the unknown quantity is positive or negative. In this example, we would do the following:
CASE 1: $3-5x > 0$
\begin{align*}
\frac{2x-1}{3-5x} (3-5x) < 2(3-5x)
\quad \Rightarrow \quad x < \frac{7}{12}
\end{align*}
CASE 2: $3-5x < 0$
\begin{align*}
\frac{2x-1}{3-5x} (3-5x) > 2(3-5x)
\quad \Rightarrow \quad x > \frac{7}{12}
\end{align*}
Note: I have left out the case $3-5x = 0$ because it is in the denominator of a fraction. So, there are in fact two solutions: we require $3-5x>0$ and $x<\frac{7}{12}$, OR we require $3-5x<0$ and $x>\frac{7}{12}$.
However, we can simplify this, since $3-5x>0$ implies $x<\frac{3}{5}$, and similarly $3-5x<0$ implies $x>\frac{3}{5}$. Then since $\frac{3}{5} > \frac{7}{12}$, the two possibilities become:
\begin{align*}
x < \frac{3}{5} \text{ and } x < \frac{7}{12}
\quad \Rightarrow \quad x < \frac{7}{12} \\
x > \frac{3}{5} \text{ and } x > \frac{7}{12}
\quad \Rightarrow \quad x > \frac{3}{5} \\
\end{align*}
So, the solution set is $x < \frac{7}{12}$ OR $x > \frac{3}{5}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2974876",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Proving $\sum\limits_{\text{cyc}} \frac{a}{b^2+c^2+d^2} \geq \frac{3\sqrt{3}}{2}\frac{1}{\sqrt{a^2+b^2+c^2+d^2}}$ for $a, b, c, d >0$
Let $a, b, c, d > 0$. Prove that
$$\frac{a}{b^2+c^2+d^2}+\frac{b}{a^2+c^2+d^2}+\frac{c}{a^2+b^2+d^2}+\frac{d}{a^2+b^2+c^2}$$
$$\geq\frac{3\sqrt{3}}{2}\frac{1}{\sqrt{a^2+b^2+c^2+d^2}}.$$
What I tried, was to say that $a^2+b^2+c^2+d^2=1$ and so
$$\frac{a}{1-a^2}+\frac{b}{1-b^2}+\frac{c}{1-c^2}+\frac{d}{1-d^2}≥\frac{3\sqrt{3}}{2}$$ and
$$-\frac{a}{(a+1)(a-1)}-\frac{b}{(b+1)(b-1)}-\frac{c}{(c+1)(c-1)}-\frac{d}{(d+1)(d-1)}≥\frac{3\sqrt{3}}{2}$$
and so
$$\frac{a}{a+1}-\frac{a}{a-1}+\frac{b}{b+1}-\frac{b}{b-1}+\frac{c}{c+1}-\frac{c}{c-1}+\frac{d}{d+1}-\frac{d}{d-1}≥\frac{3\sqrt{3}}{2}$$
Perhaps this leads somewhere?
| Since the desired inequality is homogeneous, assume that $a^2 + b^2 + c^2 + d^2 = 1$. The desired inequality becomes
$$\sum_{\mathrm{cyc}}\frac{a}{1 - a^2} \ge \frac{3\sqrt 3}{2}.$$
Note that
\begin{align*}
\frac{a}{1 - a^2} - \frac{3\sqrt 3}{2}a^2
&= \frac{a}{2(1 - a^2)}
[2 - 3\sqrt 3\, a(1 - a^2)]\\
&= \frac{a}{2(1 - a^2)}\frac{4 - 27a^2(1 - a^2)^2}{2 + 3\sqrt 3\, a(1 - a^2)}\\
&\ge 0
\end{align*}
where we have used AM-GM to get
$2a^2(1 - a^2)^2 \le \left(\frac{2a^2 + 1-a^2 + 1 - a^2}{3}\right)^3 = \frac{8}{27}$.
Thus, we have
$$\sum_{\mathrm{cyc}} \frac{a}{1 - a^2}
\ge \frac{3\sqrt 3}{2}(a^2 + b^2 + c^2 + d^2) = \frac{3\sqrt 3}{2}.$$
We are done.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2975426",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
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Formal proof of the sequence involving double factorials $x_n = \frac{(2n)!!}{(2n-1)!!}$ is not bounded. I'm trying to prove the following:
Let $n\in \mathbb N$ and
$$
x_n = \frac{(2n)!!}{(2n-1)!!}
$$
Where:
$$
(2n)!! = 2\cdot 4 \cdot 6 \cdot \dots \cdot 2n \\
(2n - 1)!! = 1\cdot 3 \cdot 5 \cdot \dots \cdot (2n-1)
$$
Prove $\{x_n\}$ is not bounded.
Intuitively it feels like it is unbounded. Let's consider the following fraction:
$$
x_n = \frac{2\cdot 4 \cdot 6 \cdot \dots \cdot (2n)}{1\cdot 3 \cdot 5 \cdot \dots \cdot (2n-1) }
$$
Now if we consequently take a number from nominator and denominator we get:
$$
\frac{2}{1} > 1 \\
\frac{4}{3} > 1 \\
\frac{6}{5} > 1 \\
\dots \\
\frac{2n}{2n-1} > 1
$$
So the fraction is a product of rational numbers each of which is greater than $1$ and the product of rational numbers greater than $1$ is increasing.
I've tried to formalize that by expanding $(2n)!!$ and $(2n-1)!!$:
$$
(2n)!! = (2n)(2n-2)(2n-4)\cdots(4)(2) = 2^kn!\\
(2n-1)!! = (2n-1)(2n-3)\cdots(5)(3)(1) = \\
= \frac{(2n-1)(2n-2)(2n-3)\cdots(5)(4)(3)(2)(1)}{(2n-2)(2n-4)\dots(4)(2)} = \\
\frac{(2n-1)!}{2^{n-1}(n-1)!}
$$
So using the above:
$$
x_n = \frac{2^nn!2^{n-1}(n-1)!}{(2n-1)!} = \frac{2^{2n-1}n!(n-1!)}{(2n-1)!}
$$
Here is where I got stuck. How do i proceed with the proof using some constant $M$ and some number $N > n$ such that $x_N > M$? Should i introduce some inequality?
I've seen a similar question, but the sequence there is proven to be divergent which i guess is more a calculus concept (yet very similar to (un)boundedness), and i'm in search of a precalculus solution.
| Without integrals or derivatives:
$$ x_n=\frac{(2n)!!}{(2n-1)!!}= \prod_{k=0}^{n-1}\left(1+\frac{1}{2k+1}\right)=\prod_{k=1}^{n}\left(1-\frac{1}{2k}\right)^{-1}$$
implies
$$ x_n^2 = \prod_{k=1}^{n}\left(1-\frac{1}{k}+\frac{1}{4k^2}\right)^{-1}=\frac{1}{4}\prod_{k=2}^{n}\left(1-\frac{1}{k}\right)^{-1}\prod_{k=2}^{n}\left(1-\frac{1}{(2k-1)^2}\right) $$
or
$$ x_n^2 = \frac{n}{4}\prod_{k=1}^{n-1}\left(1-\frac{1}{(2k+1)^2}\right). $$
On the other hand the infinite product $\prod_{k\geq 1}\left(1-\frac{1}{(2k+1)^2}\right)$ is convergent to a positive constant since $\sum_{k\geq 1}\frac{1}{(2k+1)^2}$ is convergent to a positive constant, hence $x_n\geq \sqrt{K n}$ for some mysterious constant $K>0$, which actually equals $\pi$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2976028",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
Power Series (Cambridge Tripos 1900) If
$$a/(a+bz+cz^2)=1+p_1z+p_2z^2
+\dots$$
then
$$1+p_1^2z+p_2^2z^2+\dots=\frac{a+cz}{a-cz}\frac{a^2}{
a^2-(b^2-2ac)z+c^2z^2}$$
A tricky problem from G.H.Hardy's "A Course in Pure Mathematics", or maybe I'm missing the obvious.Any help will be greatly appreciated.
| If $\alpha,\beta$ are roots of $az^2+bz+c$ then we have $(a+bz+cz^2)/a=(1-\alpha z)(1-\beta z)$, so
$$
p_n=\alpha^n+\alpha^{n-1}\beta+\dots+\alpha\beta^{n-1}+\beta^n=\frac{\alpha^{n+1}-\beta^{n+1}}{\alpha-\beta}
$$
Thus
$$
p_n^2=\frac{1}{(\alpha-\beta)^2}\left[\alpha^{2n+2}-2\alpha^{n+1}\beta^{n+1}+\beta^{2n+2}\right]
$$
so
$$
\begin{align*}
\color{red}{(\alpha-\beta)^2}\sum_{n=0}^\infty p_n^2z^n
&=\sum_{n=0}^\infty \alpha^{2n+2}z^n-2\sum_{n=0}^\infty (\alpha\beta)^{n+1}z^n+\sum_{n=0}^\infty\beta^{2n+2}z^n\\
&=\frac{\alpha^2}{1-\alpha^2z}-\frac{2\alpha\beta}{1-\alpha\beta z}+\frac{\beta^2}{1-\beta^2z}\\
&=\frac{\alpha^2+\beta^2-2\alpha^2\beta^2z}{(1-\alpha^2z)(1-\beta^2z)}-\frac{2\alpha\beta}{1-\alpha\beta z}\\
&=\frac{\color{red}{(\alpha-\beta)^2}(1+\alpha\beta z)}{(1-\alpha^2z)(1-\beta^2z)(1-\alpha\beta z)}
\end{align*}
$$
So
$$
\sum_{n=0}^\infty p_n^2z^n=\frac{1+\alpha\beta z}{(1-(\alpha^2+\beta^2)z+\alpha^2\beta^2z^2)(1-\alpha\beta z)}
$$
and this works even for the limiting case $\alpha=\beta$.
Substituting $\alpha\beta=\dfrac ca$ and $\alpha^2+\beta^2=\dfrac{b^2-2ac}{a^2}$ gives
$$
\begin{align*}
\sum_{n=0}^\infty p_n^2z^n&=\frac{1+\dfrac{c}{a}z}{\left(1-\left(\dfrac{b^2-2ac}{a^2}\right)z+\dfrac{c^2}{a^2}z^2\right)\left(1-\dfrac{c}{a} z\right)}\\
&=\frac{a+cz}{a-cz}\frac{a^2}{a^2-(b^2-2ac)z+c^2z^2}\\
\end{align*}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2977675",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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} |
Numerical method to find the roots of the derivative of $y = (x^3 - 4x) \cdot \frac{\sqrt{13+4x^2}}{2}$ (Newton's Method) I am working on an Optimization problem and I would like to suggest finding the solution (i.e. the extrema) of the function below using two methods: the first one being calculating the derivative and making it equal to zero and the second one using a numerical method.
The equation of the model is shown below:
\begin{align*}
y = (x^3 - 4x) \cdot \frac{\sqrt{13+4x^2}}{2}
\end{align*}
(the derivative being the following)
\begin{align*}
y' = \frac{16x^4 +7x^2 - 52}{2 \sqrt{13+4x^2}}
\end{align*}
I have two questions about using it:
*
*Based on some research, I found it is possible to use Newton's Method on functions that are not polynomials. Is that correct?
*Is the Newton's Method the most adequate in this case?
| We want to find the extrema of the function
$$\begin{align*}
y(x) = (x^3 - 4x) \cdot \frac{\sqrt{13+4x^2}}{2}
\end{align*}$$
One approach is to use Newton's Method on $y'(x)$, where
$$y'(x) = \frac{16 x^4+7 x^2-52}{2 \sqrt{4 x^2+13}}$$
We start by plotting $y(x)$ and see two extrema to try and find using Newton's
The iteration formula for Newton's Method is given by
$$x_{n+1} = x_n-\dfrac {f(x_n)}{f'(x_n)} = x_n - \dfrac{16x_n^4 + 7x_n^2 - 52}{2 \sqrt{4 x_n^2 + 13}\left( \dfrac{64 x_n^3 + 14 x_n}{2 \sqrt{4 x_n^2 +13}} - \dfrac{2 x_n (16 x_n^4 + 7 x_n^2 - 52)}{(4 x_n^2 + 13)^{3/2}}\right) }$$
This can be simplified to remove the square roots and nasty divisions as
$$x_{n+1} = x_n - \dfrac{\left(4 x^2+13\right) \left(16 x^4+7 x^2-52\right)}{2 x \left(96 x^4+430 x^2+195\right)}$$
If we initialize Newton's method using $x_0 = -1.4$, we arrive at $x_4 = -1.26382$ in $4$ steps.
If we initialize Newton's method using $x_0 = 1.4$, we arrive at $x_4 = 1.26382$ in $4$ steps.
Update
There is an easier approach to find where $y'(x) = 0$, we need only find the roots of the numerator of $y'(x)$, so the Newton iteration can be simplified to
$$x_{n + 1} = x_n - \dfrac{16 x_n^4 + 7 x_n^2 - 52}{64 x_n^3 + 14 x_n}$$
This gives the same results as before. Also, comparing this with the above iteration, the quadratic term $(4 x^2 + 13)$, leads to imaginary roots that we don't care about.
Lastly, it is worth noting that you can use a simple transform, $t = x^2$, on the iterations' numerator to get $16 t^2 + 7 t - 52 = 0$, and then solve a quadratic formula and eliminate Newton's Method altogether!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2978945",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
Prove that $10|n+3n^3+7n^7+9n^9$ Prove that $10|n+3n^3+7n^7+9n^9$ for every $n\in \mathbb N$
Only what i see that 10=5*2 and both number is free numbers, and if I show that $5|n+3n^3+7n^7+9n^9$ and $2|n+3n^3+7n37+9n^9$ that I prove, since 5 and 2 is free numbers i can use Fermat's little theorem such that $5|n^5-n$ and $2|n^2-n$ but i can not see that somehow help me, do you have some idea?
| Hint:
$$n+3n^3+7n^7+9n^9\equiv n+3n^3-3n^7-n^9\equiv -(n^4-1)(n)(n^4+3n^2+1)$$
mod 10, so prove
$$2|n(n^4-1)$$
and
$$5|n^4-1$$
when $n\neq5k$ and $5|n$ when $n=5k$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2979515",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 3
} |
$2 < (1 + \frac{1}{n})^n < 3$ for all $n > 2$ Show by induction that $2 < (1 + \frac{1}{n})^n < 3$ for all natural n > 2.
*
*Induction base. For $n = 3$ the inequality is obviously true.
*Assume that for $n = k$ inequality is true, than i can prove $(1 + \frac{1}{n+1})^{n +1} < 3$.
$$\frac{(n + 1)^{n}}{n^{n}} < 3 \\
1 < \frac{3n^{n}}{(n+1)^{n}} \\
\frac{(n+2)^{n+1}}{(n+1)^{n+1}} < \frac{3 n^{n} (n + 2)^{n+1} }{(n+1)^{2n +1}}, but \\
\frac{3 n^{n} (n + 2)^{n+1} }{(n+1)^{2n +1}} < 3, then \\
\frac{(n+2)^{n+1}}{(n+1)^{n+1}} < 3
$$
How can i prove, that $(1 + \frac{1}{n+1})^{n +1} > 2$
| HINT
What about Bernoulli's inequality?
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove the inequality: $\left(\frac{a}{a+2b}\right)^2+\left(\frac{b}{b+2c}\right)^2+\left(\frac{c}{c+2a}\right)^2 \geq \frac{1}{3}$
Let $a,b,c$ are all positive real numbers. Prove
$$\left(\frac{a}{a+2b}\right)^2+\left(\frac{b}{b+2c}\right)^2+\left(\frac{c}{c+2a}\right)^2 \geq \frac{1}{3}$$
Can anyone give a hint?
| By Cauchy-Schwarz,
$$\sum \limits_{\text{cyc}} \left ( \dfrac{a}{a+2b} \right )^2 \geq \dfrac{1}{3} \left(\sum \limits_{\text{cyc}} \left ( \dfrac{a}{a+2b} \right )\right)^2 = \dfrac{1}{3}\left( \sum \limits_{\text{cyc}} \left ( \dfrac{a^2}{a^2+2ab} \right) \right)^2 \geq \dfrac{1}{3} \left ( \dfrac{(a+b+c)^2}{(a+b+c)^2} \right )^2 = \dfrac{1}{3}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2983036",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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What mistake am I making when trying to do REF? I hope this kinda of question is allowed. I know that you're all uh.. very.. particular, for a lack of a better term, about the kinds of questions you get. I am just practicing a the most basic of reducing a matrix into row echelon form. I am getting a similar, but different answer in the book, but I cannot identify what I am doing incorrectly. Please help.
Matrix A = $ \begin{pmatrix}
2 & 2 & -5 & 2 \\
1 & 1 & -2 & 1 \\
1 & 0 & -3 & 4
\end{pmatrix} $
I start by permuting rows 1 & 3 then proceeding to multiply R1 by -1 and adding it to row 2 giving:
Matrix A = $ \begin{pmatrix}
1 & 0 & -3 & 4 \\
0 & 1 & 1 & -3 \\
3 & 2 & -5 & 2
\end{pmatrix} $
From here I multiply R1 by -3 and adding R3:
Matrix A = $ \begin{pmatrix}
1 & 0 & -3 & 4 \\
0 & 1 & 1 & -3 \\
0 & 2 & 4 & -10
\end{pmatrix} $
Followed by multiplying row 2 by -2 and adding row 3:
Matrix A = $ \begin{pmatrix}
1 & 0 & -3 & 4 \\
0 & 1 & 1 & -3 \\
0 & 0 & 2 & -4
\end{pmatrix} $
Lastly, I multiply row 3 by 1/2 giving:
Matrix A = $ \begin{pmatrix}
1 & 0 & -3 & 4 \\
0 & 1 & 1 & -3 \\
0 & 0 & 1 & -2
\end{pmatrix} $
The book says the answer is:
Matrix A = $ \begin{pmatrix}
1 & 1 & -1 & 1 \\
0 & 1 & 2 & 1 \\
0 & 0 & 0 & 1
\end{pmatrix} $
One problem may be that I don't have the pivot column correct so I am not doing it correctly because 1 should be in the last column of the last row, but I tried another example of a 4x4 matrix, using the same operations, and I definitely know the correct pivot column in that situation, but the answer was still different. Are any of these operations now allowed? Anyone notice what I am doing wrong here? Thanks!
| In the first step we should have
$$\begin{pmatrix}
1 & 0 & -3 & 4 \\
0 & 1 & 1 & -3 \\
\color{red}2 & 2 & -5 & 2
\end{pmatrix}$$
moreover we don't need to exchange rows at first, we can proceed as follow
$$\begin{pmatrix}
2 & 2 & -5 & 2 \\
1 & 1 & -2 & 1 \\
1 & 0 & -3 & 4
\end{pmatrix} \stackrel{R3-R2}\to \begin{pmatrix}
2 & 2 & -5 & 2 \\
1 & 1 & -2 & 1 \\
0 & -1 & -1 & 3
\end{pmatrix}$$$$\stackrel{2R2-R1}\to \begin{pmatrix}
2 & 2 & -5 & 2 \\
0 & 0 & 1 & 0 \\
0 & -1 & -1 & 3
\end{pmatrix}\stackrel{exhange\,R3,R2}\to \begin{pmatrix}
2 & 2 & -5 & 2 \\
0 & -1 & -1 & 3 \\
0 & 0 & 1 & 0
\end{pmatrix}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2983349",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find the value of $(\frac{\alpha}{\alpha +1})^3+(\frac{\beta}{\beta +1})^3+(\frac{\gamma}{\gamma +1})^3$ Find the value of
$(\frac{\alpha}{\alpha +1})^3+(\frac{\beta}{\beta +1})^3+(\frac{\gamma}{\gamma +1})^3$ where $\alpha,\beta,\gamma$ are roots of the equation $x^3+2x^2+3x+3=0$.
I tried to use the formula which is wrong
$(\frac{\alpha}{\alpha +1})^3+(\frac{\beta}{\beta +1})^3+(\frac{\gamma}{\gamma +1})^3=((\frac{\alpha}{\alpha +1})+(\frac{\beta}{\beta +1})+(\frac{\gamma}{\gamma +1}))^3-3((\frac{\alpha}{\alpha +1})(\frac{\beta}{\beta +1})(\frac{\gamma}{\gamma +1}))$
And then I broke the terms to get
$(\frac{13}{5})^3-\frac{36}{5}$
but this is $\ne 44$, which should be the answer.
I know where I was wrong, the formula is $(a+b+c)^3=a^3+b^3+c^3-3(a+b)(b+c)(c+a)$
| Substituting $x=\frac y{1-y}$, which is the inverse of $y=\frac x{x+1}$, yields
$$y^3-5y^2+6y-3=0$$
and $a,b,c=\frac\alpha{1+\alpha},\frac\beta{1+\beta},\frac\gamma{1+\gamma}$ are its roots. Then the desired expression is
$$a^3+b^3+c^3=(a+b+c)^3-3(a+b+c)(ab+bc+ca)+3abc$$
and we can extract the values of the symmetric polynomials from the coefficients: $a+b+c=5,ab+bc+ca=6,abc=3$ and
$$a^3+b^3+c^3=5^3-3\cdot5\cdot6+3\cdot3=44$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2983790",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Is $x(1 - 2x) \le \frac{1}{8}$ and further, is $x(1 - ax) \le \frac{1}{4a}$ It is clear that $x(1-x) \le \frac{1}{4}$
Does it likewise follow that $x(1-2x) \le \frac{1}{8}$?
Here's my reasoning:
(1) For $x < \frac{1}{4}$, $x(1-2x) < \frac{1}{8}$
(2) For $\frac{1}{4} < x < \frac{1}{2}$, $x(1-2x) < \frac{1}{8}$
(3) For $\frac{1}{2} < x$, $x(1-2x) < 0$
Further, can this be generalized to $x(1-ax) \le \frac{1}{4a}$
Since:
(1) For $x < \frac{1}{2a}$, $x(1-ax) < \left(\frac{1}{2a}\right)\left(\frac{1}{2}\right) = \frac{1}{4a}$
(2) For $\frac{1}{2a} < x < \frac{1}{a}$, $x(1-ax) < \left(\frac{1}{2a}\right)\left(\frac{1}{2}\right) = \frac{1}{4a}$
(3) For $\frac{1}{a} < x$, $x(1-ax) < 0$
Are both of these observations correct? Is only one correct? Is there an exception that I am missing?
| The first it's
$$16x^2-8x+1\geq0$$ or
$$(4x-1)^2\geq0.$$
The second for $a>0$ it's
$$4a^2x^2-4ax+1\geq0$$ or $$(2ax-1)^2\geq0.$$
For $a<0$ the second inequality is reversed.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Number of solutions in degree four Find number of postive integra solutions of the equation
$ x^4+ 4y^4 + 16z^4 +64= 32xyz$.
I could just proceed till that x cant be odd.
| There are no integer solutions.
As you remark, $x$ must be even. Write $x=2x_1$. Then we see $$16x_1^4+4y^4+16z^4+64=64x_1yz\implies 16\,|\,4y^4\implies y=2y_1$$
Continuing we get $$4x_1^4+16y_1^4+4z^4+16=16x_1y_1z\implies x_1^4+4y_1^4+z^4+4=4x_1y_1z$$
Thus, $x_1\equiv z\pmod 2$. If they were both odd, we'd have $x_1^4,z^4\equiv 1 \pmod 4$ in which case out last equation becomes $2\equiv 0 \pmod 4$, a contradiction. Thus they are both even. Write $x_1=2x_2, z=2z_1$. We have $$16x_2^4+4y_1^4+16z_1^4+4=16x_2y_1z_1\implies 4x_2^4+y_1^4+4z_1^4+1=4x_2y_1z_1$$
Thus $y_1$ must be odd. But in that case $y_1^4\equiv 1\pmod 4$ in which case our last equation becomes $2\equiv 0 \pmod 4$, a contradiction, and we are done.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to solve this radical expression I've been trying to solve this expression for at least two hours now... And I always get stuck towards the end, I don't know what I'm missing.
$\frac 1{xy} \times (\sqrt{xy} - \frac{xy}{x-\sqrt{xy}})\times (\sqrt{xy} + \frac{xy}{x+\sqrt{xy}})$
My first step was to rationalize the fractions inside the parenthesis like so
$\frac 1{xy} \times (\sqrt{xy} - \frac{xy(x+\sqrt{xy})}{(x-\sqrt{xy})(x+\sqrt{xy})})\times (\sqrt{xy} + \frac{xy(x-\sqrt{xy})}{(x+\sqrt{xy})(x-\sqrt{xy})})$
to get
$\frac 1{xy} \times (\sqrt{xy} - \frac{xy(x+\sqrt{xy})}{(x^2-xy)})\times (\sqrt{xy} + \frac{xy(x-\sqrt{xy})}{(x^2-xy)})$
then
$\frac 1{xy} \times (\sqrt{xy} - \frac{x^2y+ xy\sqrt{xy}}{(x^2-xy)})\times (\sqrt{xy} + \frac{x^2y-xy\sqrt{xy}}{(x^2-xy)})$
and then I'm kind of lost, nothing I've tried works.
I tried grouping each fraction by x and simplyfing removing it, and then computing the lcm inside the parenthesis in order to subtract the $\sqrt{xy}$. Or the other way around, first I did the lcm and subtracted and then simplyfied. I even tried multiplying the first factor by the second and simplifying as I went on. I tried using Wolfram Alpha to help me with each step, too. I think my calculations are correct, I'm just missing some simplification or something similar. The result should be
$\frac{x-4y}{x-y}$
I was able to get the $x-y$ but not the $x-4y$. I hope I didn't mess up the equations in the question, I'm really tired.
| We have
$$\frac 1{xy} \times (\sqrt{xy} - \frac{xy}{x-\sqrt{xy}})\times (\sqrt{xy} + \frac{xy}{x+\sqrt{xy}})
=\frac 1{xy} \times \left(xy - \frac{x^2y^2}{x^2-xy}+\frac{xy\sqrt{xy}}{x+\sqrt{xy}}-\frac{xy\sqrt{xy}}{x-\sqrt{xy}}\right)=$$
$$=1 - \frac{xy}{x^2-xy}+\frac{\sqrt{xy}}{x+\sqrt{xy}}-\frac{\sqrt{xy}}{x-\sqrt{xy}}=$$
$$=1 - \frac{xy}{x^2-xy}+\frac{x\sqrt{xy}-xy}{x^2-xy}-\frac{x\sqrt{xy}+xy}{x^2-xy}=$$
$$=\frac{x^2-4xy}{x^2-xy}=\frac{x-4y}{x-y}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2986340",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find the final tableau knowing the optimal solution
Consider the following linear program $$\displaystyle \max z=5x_1+2x_2+3x_3\\ s.t. x_1+5x_2+2x_3\le b_1\\ x_1-5x_2-6x_3\le b_2 \\
x_1,x_1,x_3\ge0$$
If the optimal solution is reached at $x_1=30,x_5=10,$ write directly the complete optimal tableau, whitout executing the simplex method. And then find the values of $b_1$ and $b_2$.
Attempt
The optimal tableau should have the following structure
\begin{array}{r|rrrrr|r}
& x_1 & x_2 & x_3 & x_4 & x_5 & \\ \hline
z & 0 & & & & 0 & 150 \\ \hline
x_1 & 1 & & ? & & 0 & 30 \\
x_5 & 0 & & & & 1 & 10 \\
\end{array}
I don't know how to calculate what is inside the tableau. How will I calculate $B^{-1}$? Please help me.
I know the last simplex tableau in matrix form is this
\begin{array}{|r r|r}
c_BB^{-1}N-c & c_BB^{-1} & c_BB^{-1}b \\ \hline
B^{-1}N & B^{-1} & B^{-1}b
\end{array}
| From the given optimal solution $z(x_1,x_2,x_3,x_4,x_5)=z(30,0,0,0,10)=150$, you can find:
$$x_1+5x_2+2x_3+x_4=b_1=\color{red}{30},\\ x_1-5x_2-6x_3+x_5=b_2=\color{red}{40},
$$
where $x_4$ and $x_5$ are the slack variables.
Since in the optimal table only the slack variable $x_4$ is replaced with $x_1$, then the intersection of $x_4$ row and $x_1$ column was the pivot element, which corresponds to the original value $1$. Hence the row remained from the initial table:
$$\begin{array}{r|rrrrr|r}
& x_1 & x_2 & x_3 & x_4 & x_5 & \\ \hline
z & 0 & & & & 0 & 150 \\ \hline
x_1 & \boxed 1 & \color{red}5 & \color{red}2 & \color{red}1 & 0 & 30 \\
x_5 & 0 & & & & 1 & 10 \\
\end{array}$$
In order to get $0$ as the first element in the row $z=5x_1+2x_2+3x_3$, where the first element was $-5$ initially, pivot row must have been multiplied by $5$ and added to it:
$$\begin{array}{r|rrrrr|r}
& x_1 & x_2 & x_3 & x_4 & x_5 & \\ \hline
z & 0 & \color{red}{23} & \color{red}7 & \color{red}5 & 0 & 150 \\ \hline
x_1 & \boxed 1 & \color{red}5 & \color{red}2 & \color{red}1 & 0 & 30 \\
x_5 & 0 & & & & 1 & 10 \\
\end{array}
$$
In order to get $0$ as the first element in the row $x_5$, where the first element was $1$ initially, pivot row must have been multiplied by $-1$ and added to it:
$$\begin{array}{r|rrrrr|r}
& x_1 & x_2 & x_3 & x_4 & x_5 & \\ \hline
z & 0 & \color{red}{23} & \color{red}7 & \color{red}5 & 0 & 150 \\ \hline
x_1 & \boxed 1 & \color{red}5 & \color{red}2 & \color{red}1 & 0 & 30 \\
x_5 & 0 & \color{red}{-10} & \color{red}{-8} & \color{red}{-1} & 1 & 10 \\
\end{array}
$$
Note: I am not sure if this is ejecuting (executing). But it is reverse engineering (backwards working).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2986579",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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} |
ّFind $x$ such that $ \frac{1}{x^2} + \frac{1}{(3-x)^2} = \frac{104}{25}$.
ّFind $x$ such that
$$ \frac{1}{x^2} + \frac{1}{(3-x)^2} = \frac{104}{25}\,.$$
My attempt:
After clearing the denominators, I obtain this quartic equation
$$104 x^{4} -624 x^{3} +886 x^{2} +150x-225=0.$$
I don't know how to proceed from here.
| Now, you can make the following.
Easy to see that $\frac{1}{2}$ and $\frac{5}{2}$ they are roots of the equation,
which gives a factor $$(2x-1)(2x-5)=4x^2-12x+5$$ and
$$104x^4-624x^3+886x^2+150x-225=$$
$$=104x^2-312x^3+130x^2-312x^3+936x^2-390x-180x^2+540x-225=$$
$$=26x^2(4x^2-12x+5)-78(4x^2-12x+5)-45(4x^2-12x+5)=$$
$$=(4x^2-12x+5)(26x^2-78x-45),$$ which gives the answer:
$$\left\{\frac{1}{2},\frac{5}{2},\frac{3}{2}\left(1+\sqrt{\frac{23}{13}}\right),\frac{3}{2}\left(1-\sqrt{\frac{23}{13}}\right)\right\}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2988342",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 3
} |
What are the coefficients in the expansion of $(x+y)(x+2y) \cdots (x+ny)$? Are the numbers appearing as coefficients in the following sequence of polynomials known? Is there a known recurrence relation to compute them?
\begin{align*}
(x+y) &= x+y \\
(x+y)(x+2y) &= x^2+3yx+2y^2 \\
(x+y)(x+2y)(x+3y) &= x^3+6yx^2+11y^2x+6y^3 \\
(x+y)(x+2y)(x+3y)(x+4y) &= x^4+10yx^3+35y^2x^2+50y^3x+24y^4 \\
&\text{etc.}
\end{align*}
| The On-Line Encyclopedia of Integer Sequences® is a
searchable database of many integer sequences, and (as @Kevin also said in a comment) can be a valuable
tool to identify a particular sequence. Searching for
$$
1, 1, 1, 3, 2, 1, 6, 11, 6, 1, 10, 35, 50, 24
$$
returns as the top result
A094638 Triangle read by rows: T(n,k) = |s(n,n+1-k)|, where s(n,k) are the signed Stirling numbers of the first kind (1 <= k <= n; in other words, the unsigned Stirling numbers of the first kind in reverse order).
So your numbers are Unsigned Stirling numbers of the first kind ${ n \brack k }$, which can be defined as the coefficients in the expansion
$$ \tag{*}
x(x+1)(x+2) \cdots (x+n-1) = \sum_{k=0}^n { n \brack k } x^k
$$
They satisfy various recurrence relations, such as
$$
{n+1 \brack k} = n { n \brack k } + { n \brack k-1 }
$$
for $k > 0$, with the initial conditions
$$
{ 0 \brack 0 } = 1 \, , \quad { n \brack 0 } = { 0 \brack n } = 0
$$
for $n > 0$. For more resources about the Stirling numbers and their
relations, see @Michael's comment below.
The precise connection with your polynomial expansion is
$$
(x+y)(x+2y) \cdots (x+ny) = \sum_{k=0}^n { n+1 \brack k+1 } x^k y^{n-k}
$$
This follows from $(*)$ with $n+1$ instead of $n$ and $\frac xy$ instead of $x$.
| {
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"url": "https://math.stackexchange.com/questions/2988624",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "21",
"answer_count": 1,
"answer_id": 0
} |
Problem in reducing a expression In my previous question, I was able to solve it using Weierstrass substitution as the answers pointed out.
But I was keen to know if my expression can be reduced to the given answer.
Desmos proves that the two integrals are indeed equal, here's the proof.
Now how do I reduce this expression, $$\int_{0}^{\sqrt{2}-1}\frac{2}{(1+t)\bigg(\sqrt{(1+t)+\sqrt{(1+t)^2-1}}+\sqrt{(1+t)-\sqrt{(1+t)^2-1}}\bigg)}dt$$
to this,
$$\int_{0}^{\sqrt{2}-1}\frac{4t}{(1+t^2)\sqrt{1-t^2}}dt$$
Any help would be really appreciated. Thank you.
| Let's denote $$I_1=\int_{0}^{\sqrt{2}-1}\frac{2}{(1+t)\bigg(\sqrt{(1+t)+\sqrt{(1+t)^2-1}}+\sqrt{(1+t)-\sqrt{(1+t)^2-1}}\bigg)}dt$$ and $$I_2=\int_{0}^{\sqrt{2}-1}\frac{4t}{(1+t^2)\sqrt{1-t^2}}dt$$
First let's simplify the denominator in the first integral. Since both terms are positive we are able to do the following trick
$$\sqrt{(1+t)+\sqrt{(1+t)^2-1}}+\sqrt{(1+t)-\sqrt{(1+t)^2-1}}=\left(\left(\sqrt{(1+t)+\sqrt{(1+t)^2-1}}+\sqrt{(1+t)-\sqrt{(1+t)^2-1}}\right)^2\right)^{1/2}=\left((1+t)+\sqrt{(1+t)^2-1}+(1+t)-\sqrt{(1+t)^2-1}+2\sqrt{(1+t)^2-(1+t)^2+1}\right)^{1/2}=\left(2(1+t)+2\right)^{1/2}=\sqrt{2(t+2)}$$
Thus the first integral $I_1$ is simplified to $$I_1=\int_0^{\sqrt{2}-1}\frac{2}{(1+t)\sqrt{2(t+2)}}~dt=\left[
\begin{matrix}
x=\sqrt{t+2} & t=0~\rightarrow x=\sqrt{2}\\
t=x^2-2 & t=\sqrt{2}-1 \rightarrow x=\sqrt{\sqrt{2}+1}\\
dt=2x~dx &
\end{matrix}
\right]=
$$
$$=\int_\sqrt{2}^\sqrt{\sqrt{2}+1}\frac{2\sqrt{2}x}{(x^2-1)x}~dx=2\sqrt{2}\int_\sqrt{2}^\sqrt{\sqrt{2}+1}\frac{dx}{x^2-1}$$
Now let's transform $I_2$: $$I_2=\int_{0}^{\sqrt{2}-1}\frac{4t}{(1+t^2)\sqrt{1-t^2}}dt=
\left[\begin{matrix}
t^2=1-\frac{2}{x^2} & t=0 \rightarrow x=\sqrt{2}\\
2t~dt=\frac{4}{x^3}~dx & t=\sqrt{2}-1 \rightarrow x=\sqrt{\sqrt{2}+1}
\end{matrix}\right]=$$
$$=\int_\sqrt{2}^\sqrt{\sqrt{2}+1}\frac{\frac{8}{x^3}}{\left(2-\frac2{x^2}\right)\sqrt{1-\left(1-\frac2{x^2}\right)}}~dx=2\sqrt{2}\int_\sqrt{2}^\sqrt{\sqrt{2}+1}\frac{dx}{x^2-1}
$$
That's it. If we are just to prove that $I_1=I_2$ everything is done. But it's very simple to compute them.
| {
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"url": "https://math.stackexchange.com/questions/2989841",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Proving continuity in the origin $\frac{xy}{\sqrt{x^2 + y^2}}$ Let $g: \mathbb{R^2} \to \mathbb{R}$.
How can I prove that $g$ is continuous in its origin, but not totally differentiable?
If I take
$$g(\frac{1}{n},\frac{1}{n}) = \frac{1}{n\sqrt{2}} \to 0 \text{ for } n \to \infty $$
Or rather:
$$|x,y| \leq \frac{1}{2} (x^2 + y^2) $$
from which we can follow
$$|g(x,y)| \leq \frac{1}{2} \sqrt{x^2 + y^2}$$
which proves continuity of $g$ in the origin $(0,0)$.
But how can I show that this function is not total differentiable?
Can I do the following estimation?
$$|g(x,y) - 0| = |y| \cdot \frac{x \cdot y}{\sqrt{x^2 + y^2}} \leq |y| \cdot 1 = |y| $$
From which it follows, that
$$|y| \leq |\sqrt{x^2 + y^2}| = ||(x,y)|| $$
| Note that $x^2+y^2\ge 2|xy|$. Hence, we have
$$\left|\frac{xy}{\sqrt{x^2+y^2}}\right| \le \frac{\sqrt{|xy|}}{\sqrt 2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2990614",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Polynomial division with remainder
If the polynomial
$$x^4-x^3+ax^2+bx+c$$
divided by the polynomial
$$x^3+2x^2-3x+1$$
gives the remainder $$3x^2-2x+1$$
then how much is (a+b)c?
So what I know, and how I solved these problems before, I can write this down like this:
$x^4-x^3+ax^2+bx+c=Q(x)(x^3+2x^2-3x+1)+3x^2-2x+1$
And from here I should get the roots of $x^3+2x^2-3x+1$ so I could use the remainder only since I don't know $Q(x)$
$x^3+2x^2-3x+1=0$
but I can't find the roots of this by hand, so I used WolframAlpha to get the roots, but they are not whole numbers so I couldn't write this in the form of $(x-x_1)(x-x_2)(x-x_3)$ and then just use $P(x_1)$, $P(x_2)$, $P(x_3)$ to get $a,b,$ and $c$
I tried regular dividing on these polynomials to get $Q(x)$ but I couldn't divide them because of the $a,b$ and $c$ present
| Hint: Compute the remainder of $x^4-x^3$ when divided by $x^3+2x^2-3x+1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2993694",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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$M$ is a point in an equaliateral $ABC$ of area $S$. $S'$ is the area of the triangle with sides $MA,MB,MC$. Prove that $S'\leq \frac{1}{3}S$.
$M$ is a point in an equilateral triangle $ABC$ with the area $S$. Prove that $MA, MB, MC$ are the lengths of three sides of a triangle which has area $$S'\leq \frac{1}{3}S$$
|
Let $N$ be the rotation of $M$ about $C$ by an angle of $60^\circ$, then $\triangle AMN$ has three sides equal to $MA,MB, MC$ respectively. Now
$$\begin{aligned}
P &= \frac12 MN\cdot MA\cdot \sin(\angle AMN)\\
&= \frac12 MC \cdot MA\cdot\sin(\angle AMC - 60)\\
&= \frac12 MC \cdot MA\cdot (\sin(\angle AMC)\cos 60 - \cos(\angle AMC)\sin 60)\\
&= \frac12 S_{MCA} - \frac{\sqrt3}{4}MC\cdot MA\cdot \cos(\angle AMC),
\end{aligned}$$
where $P$ denotes the area of $\triangle AMN$. The cosine law in $\triangle AMC$ gives us
$$MC\cdot MA \cdot \cos(\angle AMC) = \frac12 (MA^2+MC^2 - a^2),$$
where $a$ is the side of $\triangle ABC$. Substitute that to the above, we have
$$ P = \frac12 S_{MCA} - \frac{\sqrt3}8(MA^2 +MC^2 - a^2).$$
Take cyclic sum of the above with respect to $A,B,C$, we get
$$3P = \frac12S - \frac{\sqrt3}8\sum_{cyclic}(MA^2 + MC^2 -a^2),$$
where $S$ is the area of $\triangle ABC$, which is also equal to $\frac{\sqrt3}4a^2$. Thus we have
$$\begin{aligned}
3P &= \frac12S + \frac{3\sqrt3}8a^2 - \frac{\sqrt3}4\sum MA^2\\
&= 2S - \frac{\sqrt3}4\sum MA^2.
\end{aligned}$$
The fact $P\le \frac13S$ follows from the fact that $\sum MA^2 \ge 4\sqrt3 P$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2993965",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
} |
Prove $\int^{\infty}_{-\infty}\frac{1}{(x^2+x+1)^3}dx=\frac{4\pi}{3\sqrt3}$. I have no idea how to do this question.
I'm given $\int^{\infty}_{-\infty}\frac{1}{x^2+2ax+b^2}dx=\frac{\pi}{\sqrt{b^2-a^2}}$ if $b>|a|$ and I'm asked to prove $\int^{\infty}_{-\infty}\frac{1}{(x^2+x+1)^3}dx=\frac{4\pi}{3\sqrt3}$.
What I've tried:
$\int^{\infty}_{-\infty}\frac{1}{(x^2+x+1)}dx=\frac{2\pi}{\sqrt3}$. I've tried setting a variable as the power, ie: $I(r)=\int^{\infty}_{-\infty}\frac{1}{(x^2+x+1)^r}dx$ but differenciating the inside w.r.t to $r$ doesn't seem to form any differential equation that can be solved.
Answers and hints appreciated!
| For any $A>0$ we have $\int_{-\infty}^{+\infty}\frac{dz}{z^2+A} = \frac{\pi}{\sqrt{A}}$, hence by applying $\frac{d^2}{dA^2}$ to both sides we get
$$ \int_{-\infty}^{+\infty}\frac{dz}{(z^2+A)^3}=\frac{3\pi}{8A^2\sqrt{A}}.\tag{1} $$
On the other hand
$$ \int_{-\infty}^{+\infty}\frac{dx}{(x^2+x+1)^3}\stackrel{x\mapsto z-\frac{1}{2}}{=}\int_{-\infty}^{+\infty}\frac{dz}{\left(z^2+\tfrac{3}{4}\right)^3}=\frac{4\pi}{3\sqrt{3}}.\tag{2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2994751",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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} |
Show that $\det\left[\begin{smallmatrix}1&\cos a&\cos b\\ \cos a&1&\cos(a+b) \\ \cos b&\cos(a+b)&1 \end{smallmatrix}\right]=0$ I am unable to show - without expanding, by using determinant properties - that
$$\det\begin{bmatrix} 1 &\cos a &\cos b\\
\cos a &1 &\cos(a+b) \\
\cos b &\cos(a+b) &1
\end{bmatrix}=0$$
I am using trigonometric identities to solve this but I don't understand what would be the next step.
| Here we go: directly
$$\begin{align}\begin{vmatrix} 1 &\cos a &\cos b\\
\cos a &1 &\cos(a+b) \\
\cos b &\cos(a+b) &1
\end{vmatrix}&=\\&=
[1+2\cos a\cos b\cos(a+b)]-[\cos^2b+\cos^2a+\cos^2(a+b)]\\&=
1+\cos(a+b)\cdot [2\cos a\cos b-\cos (a+b)]-\cos^2a-\cos^2b\\&=
1+\cos(a+b)\cos(a-b)-\cos^2a-\cos^2b\\&=
1+\frac12[\cos(2a)+\cos(2b)]-\cos^2a-\cos^2b\\&=
1+\frac12[2\cos^2a-1+2\cos^2b-1]-\cos^2a-\cos^2b\\&=0\end{align}$$
Another method: triangulation
$$\begin{vmatrix} 1 &\cos a &\cos b\\
\cos a &1 &\cos(a+b) \\
\cos b &\cos(a+b) &1
\end{vmatrix}\stackrel{(-\cos a)R_1+R_2\to R_2; \\(-\cos b)R_1+R_3\to R_3}{=}\\
\begin{vmatrix} 1 &\cos a &\cos b\\
0 &1-\cos^2a &\cos(a+b)-\cos a\cos b \\
0 &\cos(a+b)-\cos a\cos b &1-\cos^2b
\end{vmatrix}=\\
\begin{vmatrix} 1 &\cos a &\cos b\\
0 &\sin^2a &-\sin a\sin b \\
0 &-\sin a\sin b &\sin^2b
\end{vmatrix}\stackrel{\frac{\sin b}{\sin a}\cdot R_2+R_3\to R_3}{=}\\
\begin{vmatrix} 1 &\cos a &\cos b\\
0 &\sin^2a &-\sin a\sin b \\
0 &0 &0
\end{vmatrix}=0.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2994980",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "1",
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A question about the accumulation points of a weird sequence. Recently I came across a strange problem asking me to find the accumulation points of the sequence: $a_n$=$\sqrt[3]{n}\cdot\sin\sqrt{n}$
I really don't know how to write in detail, although I'm sort of sure that the accumulation points should be all $\mathbb{R}$.
I sincerely hope that you can help me solve the problem!!
| Fix any $x \in \mathbb{R}$ and $\varepsilon > 0$. WLOG, assume $\varepsilon < |x|/2$. We want to show that there is some $n$ such that $|a_n - x| < \varepsilon$.
*
*We intend to describe the ''distribution'' of $\sqrt{n}$. The distance between two adjacent points is $$\sqrt{n+1} - \sqrt{n} = \frac{1}{\sqrt{n} + \sqrt{n+1}} < \frac{1}{2\sqrt{n}}.$$
Thus, $$|\sin\sqrt{n+1} -\sin\sqrt{n}| \leq 1\cdot|\sqrt{n+1}-\sqrt{n}| < \frac{1}{2\sqrt{n}}.$$
*Similarly, $$\sqrt[3]{n+1}-\sqrt[3]{n} < \frac{1}{3\sqrt[3]{n}}.$$
*Hence, $$|a_{n+1} - a_n| < \sqrt[3]{n+1} \frac{1}{2\sqrt{n}}. + \frac{1}{2\sqrt[3]{n}} =: \delta_n.$$
Clearly $\delta_n \to 0$ as $n \to \infty$.
*Let $I_k = \left\{ n \in \mathbb{N}: \left(k-\frac{1}{2}\right)\pi < \sqrt{n} \leq \left(k+\frac{1}{2}\right)\pi \right\}$. Thus in each $I_k$, where $k$ is even, $\sin\sqrt{n}$ is increasing, so $a_n$ is an increasing sequence in $I_k$ where $k$ is even. Note also that as $k\to \infty$, $$\eta^1_k := \sqrt[3]{\min\{I_k\}}-\sqrt[3]{\left(k-\frac{1}{2}\right)^2\pi^2 } \to 0, $$
$$
\eta^2_k := \sqrt{\min\{I_k\}}-\sqrt{\left(k-\frac{1}{2}\right)^2\pi^2 } \to 0,
$$ and similarly $$\eta^3_k := \sqrt[3]{\left(k+\frac{1}{2}\right)^2\pi^2 } - \sqrt[3]{\max\{I_k\}}\to 0,
$$
$$\eta^4_k := \sqrt{\left(k+\frac{1}{2}\right)^2\pi^2 } - \sqrt{\max\{I_k\}}\to 0.
$$
*Choose an even $k$ large such that $$ a_{\min\{I_k\}} < -|x|, \quad a_{\max\{I_k\}} >|x|$$ and $\delta_n < \varepsilon/2$ for all $n \in I_k$. The claimed result thus follows.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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What am I doing wrong finding the limit of $\left(\frac{3x^2-x+1}{2x^2+x+1}\right)^\left(\frac{x^3}{1-x}\right)$? $$\lim_{x\to\infty}\left(\frac{3x^2-x+1}{2x^2+x+1}\right)^\left(\frac{x^3}{1-x}\right)$$
Divide by $x^2$, get
$$\lim_{x\to\infty}(1,5)^\infty=\infty$$
The answer in the book is $0$.
I've also tried substituting $x$ for $\frac{1}{h}$ where $h$ tends to zero and using some form of $(1+\frac{1}{n})^n$, and using the exponent rule, but everything lands me at infinity.
Do I misunderstand something fundamentally?
| Note that
$$\frac{3x^2-x+1}{2x^2+x+1} \to \frac32$$
but
$$\frac{x^3}{1-x}=-\frac{x^3}{x-1}\to -\infty$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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If $a$, $b$, $c$ are positive real numbers such that $a^2 + b^2 + c^2 = 27$. Find the least value of $a^3 + b^3 + c^3$? If $a$, $b$, $c$ are positive real numbers such that $a^2 + b^2 + c^2 = 27$. Find the least value of $a^3 + b^3 + c^3$ ?
I tried with Tchebyshev inequality on sets $\{a, b, c\}$ and $\{a^2, b^2 , c^2\}$ but was blocked. I also tried with mean of $m-$the power with $m=3$ but again could not proceed further.
I only know elementary inequalities like Cauchy-Schwarz and AM-GM apart from above mentioned inequalities. Please help.
| From the power mean inequality we know that
$$\sqrt[q]{\sum_{i=1}^nw_ix_i^q} \geq \sqrt[p]{\sum_{i=1}^nw_ix_i^p} \quad \forall \ \ p <q$$
where $ \sum_{i=1}^n w_i=1 $
In your case $p=2, q=3$ and $n=3$. Hence,
$$\left(\dfrac{a^3+b^3+c^3}{3}\right)^{1/3}\ge \left(\dfrac{a^2+b^2+c^2}{3}\right)^{1/2}=3$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3001046",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Evaluating $\int_0^1\frac{\ln(1+x-x^2)}xdx$ without using polylogarithms.
Evaluate $$I=\int_0^1\frac{\ln(1+x-x^2)}xdx$$ without using polylogarithm functions.
This integral can be easily solved by factorizing $1+x-x^2$ and using the values of dilogarithm at some special points.
The motivation of writing this post is someone said that this integral cannot be solved without using special functions.
Another alternative solutions will be appreciated.
| $$\begin{aligned}
I&=\int_0^1\frac{\ln(1+x-x^2)}x\mathrm{d}x\\
&\overset{(1)}{=}\int_0^1\sum_{n=1}^\infty\frac{(-1)^{n-1}(x-x^2)^n}{nx}\mathrm{d}x\\
&\overset{(2)}{=}\sum_{n=1}^\infty\frac{(-1)^{n-1}}n\int_0^1x^{n-1}(1-x)^n\mathrm{d}x\\
&\overset{(3)}{=}\sum_{n=1}^\infty\frac{(-1)^{n-1}}n\frac{(n-1)!n!}{(2n)!}\\
&=\sum_{n=0}^\infty\frac{(-1)^{n}(n!)^2}{(2n+2)!}\\
&=\sum_{n=0}^\infty\frac{(-1)^{n}(1\times2\times\cdots\times n)(1\times2\times\cdots\times n)}{1\times2\times\cdots\times (2n+2)}\\
&=\sum_{n=0}^\infty\frac{(-1)^{n}(1\times2\times\cdots\times n)}{1\times3\times5\times\cdots\times(2n+1)\times (2n+2)2^n}\\
&=\sum_{n=0}^\infty\frac{(-1)^nn!}{(2n+1)!!(2n+2)2^n}
\end{aligned}$$
Explanation
(1) Using the Maclaurin series of $\ln(1+w)$, where $w=x-x^2$.
(2) It is legal to change the position of $\sum$ and $\int$.
(3) Integrate by parts $n-1$ times.
Notice that $$\sum_{n=0}^\infty\frac{(2n)!!}{(2n+1)!!}x^{2n+1}=\frac{\arcsin x}{\sqrt{1-x^2}},$$
integrate both sides from $0$, we have $$\sum_{n=0}^\infty\frac{(2n)!!}{(2n+1)!!(2n+2)}x^{2n+2}=\frac12\arcsin^2x.$$
Letting $x=\frac i2$ leads to
$$\sum_{n=0}^\infty\frac{(-1)^{n+1}(2n)!!}{(2n+1)!!(2n+2)}2^{-2n-2}=\frac12\arcsin^2\frac i2.$$
Combining with $(2n)!!=2^{n}n!$, we have $$-\frac14I=-\frac12\operatorname{arccsch}^22,$$
or $I=2\ln^2\varphi,$ where $\varphi$ denotes the golden ratio.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How do I simplify this expression? How do I simplify:
$$d=\frac{-(x+\alpha x)^2+(y+\alpha y)^2+2+x^2-y^2-2}{\sqrt{\alpha^2 x^2+\alpha^2 y^2}}$$
If simplification is possible, it should be possible with elementary algebra, but I'm completely lost as to how to go about it.
What I've done so far:
$$d=\frac{-(x+\alpha x)^2+(y+\alpha y)^2+2+x^2-y^2-2}{\sqrt{\alpha^2 x^2+\alpha^2 y^2}}$$
$$=\frac{-(x+\alpha x)^2+(y+\alpha y)^2+x^2-y^2}{\sqrt{\alpha^2 x^2+\alpha^2 y^2}}=\frac{-α^2 x^2 - 2 α x^2 + α^2 y^2 + 2 α y^2}{\sqrt{\alpha^2 x^2+\alpha^2 y^2}}$$
$$\implies d^2=\frac{(-α^2 x^2 - 2 α x^2 + α^2 y^2 + 2 α y^2 - 2)^2}{\alpha^2x^2+\alpha^2y^2}$$
| I believe your third equality is incorrect. Numerator distribution won't lead to any terms with x and y to the first power.
Then when you squared d below, you also have additional squares added to a in denominator where there were none as well as a double square for the entire numerator and denominator - essentially you have d to the 4th power with a being squared on bottom where it wasn't in the initial equation.
$$d=\frac{-(x+\alpha x)^2+(y+\alpha y)^2+2+x^2-y^2-2}{\sqrt{\alpha^2 x^2+\alpha^2 y^2}}$$
$$=\frac{-x^2(1+\alpha)^2+y^2(1+\alpha)^2+x^2-y^2}{\sqrt{\alpha x^2+\alpha y^2}}=\frac{(y^2-x^2)(1+ 2\alpha + \alpha^2)+ x^2 - y^2}{\sqrt{\alpha^2} \sqrt{x^2+ y^2}}$$
$$ =\frac{(2\alpha+\alpha^2)(y^2-x^2)}{\alpha \sqrt{x^2+ y^2}}$$
Can continue from there perhaps, factor an alpha, square both sides if that helps.
| {
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"url": "https://math.stackexchange.com/questions/3003860",
"timestamp": "2023-03-29T00:00:00",
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Binomial Theorem with Three Terms $(x^2 + 2 + \frac{1}{x} )^7$
Find the coefficient of $x^8$
Ive tried to combine the $x$ terms and then use the general term of the binomial theorem twice but this does seem to be working.
Does anyone have a method of solving this questions and others similar efficiently?
Thanks.
| Let $$R(x)= \left(x^2+2+{1\over x}\right)^7$$ then we need to find a coeficient at $x^{15}$ for$$x^7R(x)= (x^3+2x+1)^7 $$ $$= \sum _{k=0}^7 {7\choose k}x^{21-3k}(2x+1)^k$$
Clearly if $21-3k\geq 16$ there is no term with $x^{15}$ so $21-3k\leq 15$ so $k\geq 2$.
Also if $21-3k\leq 7$ we have no term with $x^{15}$ so $21-3k\geq 8$ so $3k\leq 13$ so $k\leq 4$.
If $k=2$ we have $${7\choose 2}x^{15}(2x+1)^2$$ so the term is $21$
If $k=3$ we have $${7\choose 3}x^{12}(2x+1)^3$$ so the term is $35\cdot 8= 280$
If $k=4$ we have $${7\choose 4}x^{9}(2x+1)^4$$ there is no trem with $x^{15}$
so the answer is $301$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Double Infinite sum of $1/n^2$ I am trying to use an identity we showed on our homework:
$$ \sum_{-\infty}^{\infty} \frac{1}{(n+a)^2} = \frac{\pi^2}{\sin^2(\pi a)} $$
to show that $$ \sum_{1}^{\infty} \frac{1}{n^2} = \frac{\pi^2}{6}.$$
I have broken the first double infinite sum into the sum from $-\infty$ to $1$ plus the the $0^{th}$ term, plus the sum from $1$ to $+\infty$ and then I want to take the limit of $a$ going to $0$.
This results in taking the limit of the following:
$$\lim_{a\to 0} \frac{\pi^2}{\sin^2(\pi a)} - \frac{1}{a^2} .$$
Which I know should result in $\frac{\pi^3}{3}$ from Wolfram Alpha, as desired, but I am struggling with showing it analytically.
My idea was to try and find the Maclaurien Expansion of $\sin^2(\pi a)$, but then taking that series to the exponent of negative 1 since it is in the denominator is causing issues.
Is there a trick I am not seeing or a possible better way to use the above property to show the other infinite sum?
This question is also for a complex analysis class, so perhaps there is a way to use complex Laurent or power series?
| Your idea of using Taylor expansions is good.
Let us compose the series around $a=0$
$$\sin(\pi a)=\pi a-\frac{\pi ^3 a^3}{6}+\frac{\pi ^5 a^5}{120}+O\left(a^7\right)$$
$$\sin^2(\pi a)=\pi ^2 a^2-\frac{\pi ^4 a^4}{3}+\frac{2 \pi ^6 a^6}{45}+O\left(a^8\right)$$
$$\frac{\pi^2}{\sin^2(\pi a)}=\frac{\pi^2}{\pi ^2 a^2-\frac{\pi ^4 a^4}{3}+\frac{2 \pi ^6 a^6}{45}+O\left(a^8\right) }=\frac{1}{a^2}+\frac{\pi ^2}{3}+\frac{\pi ^4 a^2}{15}+O\left(a^4\right)$$
$$\frac{\pi^2}{\sin^2(\pi a)} - \frac{1}{a^2} =\frac{\pi ^2}{3}+\frac{\pi ^4 a^2}{15}+O\left(a^4\right)$$ which shows both the limit and also how it is approached.
Try with $a=\frac 16$ which is "large". The exact value would be $4 \pi ^2-36\approx 3.47842$ while the expansion would give $\frac{\pi ^2}{3}+\frac{\pi ^4}{540}\approx 3.47026$.
| {
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Does $\left\lfloor\frac{x^2+x}{i}\right\rfloor - \left\lfloor\frac{x^2}{i}\right\rfloor = \left\lfloor\frac{x}{i}\right\rfloor$? Does $\left\lfloor\dfrac{x^2+x}{i}\right\rfloor - \left\lfloor\dfrac{x^2}{i}\right\rfloor = \left\lfloor\dfrac{x}{i}\right\rfloor$?
where $i \le x^2$ and $x$ are any positive integer.
Intuitively, this doesn't seem correct to me but here's my argument which appears valid:
(1) There exists an integer $a$ such that: $x \equiv a \pmod i$ where $0 \le a < i$
(2) $\left\lfloor\dfrac{x^2+x}{i}\right\rfloor - \left\lfloor\dfrac{x^2}{i}\right\rfloor = \dfrac{x^2 + x - a^2 - a}{i} - \dfrac{x^2 - a^2}{i} = \dfrac{x-a}{i} = \left\lfloor\dfrac{x}{i}\right\rfloor$
Is my argument wrong? Is my intuition wrong?
| Consider x = 7 and i = 5.
you get
$\left\lfloor\dfrac{49+7}{5}\right\rfloor = 11$
$\left\lfloor\dfrac{49}{5}\right\rfloor = 9$
$\left\lfloor\dfrac{7}{5}\right\rfloor = 1$
as you see the equation doesn't hold.
Your intuition is right and the argument is wrong. The same argument should follow the following logic
$\dfrac{x}{i} = k + \dfrac{a}{i}$ where $k$ is highest possible positive integer without making term a negative
$\dfrac{x^2}{i} = l + \dfrac{a^2}{i}$
$\left\lfloor\dfrac{x^2+x}{ i}\right\rfloor = k + l + \left\lfloor\dfrac{a+a^2}{i}\right\rfloor$
$\left\lfloor\dfrac{x^2}{i}\right\rfloor = l + \left\lfloor\dfrac{a^2}{i}\right\rfloor$
$\left\lfloor\dfrac{x}{i}\right\rfloor = k + \left\lfloor\dfrac{a}{i}\right\rfloor$
as you see the floor terms are not necessarily equal. There could be cases when $a+a^2$
is higher than $i$ when both $a$ and $a^2$ are less than $i$.
Hope it helps
| {
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"timestamp": "2023-03-29T00:00:00",
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Test three randomly selected blocks of a program There is one error in one of five blocks of a program. To find the error, we test three randomly selected
blocks. Let $$ be the number of errors in these three blocks. Compute $E(X)$ and $Var(X)$.
That's what i've tried:
$p = 0.6$
$n = 1$
$E(X) = n\cdot p = 1\cdot (0.6) = 0.6$
$Var(X) = n \cdot p\cdot q = 1\cdot (0.6)\cdot (0.4) = 0.24$
If we were using $n = 2$ with success probability of $0.6$ we would have:
$P(X = 0) = 0.16$
$P(X = 1) = 0.48$
$P(X = 2) = 0.36$
$p = 0.6$
$n = 2$
$E(X) = 2\cdot (0.6) = 1.2$
$Var(X) = 2\cdot (0.6)\cdot (0.4) = 0.48$
| Since there is one error in the 5 blocks of code, $X \in \{0,1\}$, where $X=0$ if all 3 blocks of code being tested are correct.
Assuming the error is equally likely to be in any of the 5 blocks of code, the probability of drawing three correct blocks of code is $\frac{4}{5} \times \frac{3}{4} \times \frac{2}{3} = \frac{2}{5}$, i.e. $P(X=0) = \frac{2}{5}$ and $P(X=1)=\frac{3}{5}$.
$\mathbb{E}[X] = 0 \times \frac{2}{5} + 1 \times \frac{3}{5} = \frac{3}{5}$.
$\mathbb{E}[X^2] = 0^2 \times \frac{2}{5} + 1^2 \times \frac{3}{5} = \frac{3}{5}$.
$Var(X) = \mathbb{E}[X^2] - \mathbb{E}[X]^2 = \frac{3}{5}-\left( \frac{3}{5} \right)^2 = \frac{6}{25}$.
| {
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"timestamp": "2023-03-29T00:00:00",
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How to show $\frac{\Gamma((n-1)/2)}{\Gamma(n/2)} \approx \frac{\sqrt{2}}{\sqrt{n-2}}$
Show $\frac{\Gamma((n-1)/2)}{\Gamma(n/2)} \approx \frac{\sqrt{2}}{\sqrt{n-2}}$
Try
Using the facts:
*
*$(1 + \alpha/m)^m = e^\alpha ( 1+ r_m)$, where $\lim_{m \to \infty} \sqrt{m}r_m = 0$
*$\Gamma(n+1) = n^{n + 1/2} e^{-n} \sqrt{2 \pi} (1 + r_n)$, where $\lim_{n \to \infty} \sqrt{n}r_n = 0$
Note :
$$
\begin{aligned}
\frac{\Gamma((n-1)/2)}{\Gamma(n/2)} &= \frac{\left(\frac{n-3}{2}\right)^{(n-2)/2} e^{-(n-3)/2}(\sqrt{2\pi}(1 + r_{1n})}{\left(\frac{n-2}{2}\right)^{(n-1)/2} e^{-(n-2)/2}(\sqrt{2\pi}(1 + r_{2n})} \\
&=\left( \left(\frac{n-3}{n-2}\right)^{(n-2)/2} \sqrt{e} \frac{1+r_{1n}}{1 + r_{2n}} \right) \times \frac{\sqrt{2}}{\sqrt{n-2}}
\end{aligned}
$$
But I'm stuck at how I should proceed to eliminate the $\left( \left(\frac{n-3}{n-2}\right)^{(n-2)/2} \sqrt{e} \frac{1+r_{1n}}{1 + r_{2n}} \right)$ term.
| A possible approach:
$$ \frac{\Gamma\left(\tfrac{n-1}{2}\right)}{\Gamma\left(\tfrac{n}{2}\right)}=\tfrac{1}{\sqrt{\pi}}\,B\left(\tfrac{n-1}{2},\tfrac{1}{2}\right)=\frac{1}{\sqrt{\pi}}\int_{0}^{1}x^{\frac{n-3}{2}}(1-x)^{-1/2}\,dx=\frac{2}{\sqrt{\pi}}\int_{0}^{1}\frac{x^{n-2}}{\sqrt{1-x^2}}\,dx $$
gives
$$ \frac{\Gamma\left(\tfrac{n-1}{2}\right)}{\Gamma\left(\tfrac{n}{2}\right)}= \frac{2}{\sqrt{\pi}}\int_{0}^{\pi/2}\left(\cos\theta\right)^{n-2}\,d\theta\sim\frac{2}{\sqrt{\pi}}\int_{0}^{+\infty}\exp\left[-(n-2)\frac{\theta^2}{2}\right]d\theta=\sqrt{\frac{2}{n-2}}. $$
Notice that the integral representation (as a moment) instantly gives that the LHS is a log-convex function. The asymptotic equivalence $\sim$ can be seen as an instance of Laplace/Hayman's method.
| {
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"timestamp": "2023-03-29T00:00:00",
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Find $\alpha$,$\beta$ if $\lim_{x→∞}[ \sqrt {ax^2+2bx+c} - \alpha x -\beta] = 0$ Here is my approach.
Consider;
$$ax^2 + 2bx + c = 0$$
or;
$$ x_{±} = \frac {-b±\sqrt {b^2-ac}}{a} = \frac {-b±\sqrt D}{a}$$
Hence;
$$\sqrt {ax^2+2bx+c} = \sqrt {(x+\frac {b-√D}{a})(x+\frac {b+√D}{a})}$$
$$=x\sqrt {(1+\frac {b-√D}{ax})(1+\frac {b+√D}{ax})}$$
For large values of $x$ we may apply the binomial approximation, so that;
$$\sqrt {ax^2+2bx+c} ≈ x(1+\frac {b-√D}{2ax})(1+\frac {b+√D}{2ax})$$
$$=x + \frac {b}{a} + \frac {c}{4ax}$$
As $x→∞$ the final term in the above expression vanishes. Hence;
$$\lim_{x→∞}[ \sqrt {ax^2+2bx+c} - \alpha x - \beta ] = 0,$$
gives;
$$x+\frac {b}{a} - \alpha x - \beta = 0,$$
or;
$$(1-\alpha)x + (\frac {b}{a} - \beta) = 0$$
As $x→∞$, $1-\alpha$ must be $0$ for the former term to vanish, hence,
$$\alpha = 1, \beta = \frac {b}{a}$$
But I doubt it is hardly correct. Is there any better method for the problem?
| Set $1/x=h$ to find $$\lim_{h\to0^+}\dfrac{\sqrt{a+bh+ch^2}-(\alpha+\beta h)}h$$
$$=\lim_{h\to0^+}\dfrac{(a+bh+ch^2)-(\alpha+\beta h)^2}h\cdot\lim_{h\to0^+}\dfrac1{\sqrt{a+bh+ch^2}-(\alpha+\beta h)}$$
$(a+bh+ch^2)-(\alpha+\beta h)^2=a-\alpha^2+h(b-2\alpha\beta)+h^2(c-\beta^2)$
As the denominator $\to0,$ $$a-\alpha^2=0$$
as the denominator is $O(h)$ $$b-2\alpha\beta=0$$
| {
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"timestamp": "2023-03-29T00:00:00",
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All real number for which $n$ in $5^n+7^n+11^n=6^n+8^n+9^n$
Finding all real number $n$ in
$$5^n+7^n+11^n=6^n+8^n+9^n$$
Try: From given equation
$n=0,1$ are the solution
But i did not understand any other solution exists or not
Although i have tried like this way
$$\bigg(\frac{5}{9}\bigg)^n+\bigg(\frac{7}{9}\bigg)^n+\bigg(\frac{11}{9}\bigg)^n = \bigg(\frac{6}{9}\bigg)^n+\bigg(\frac{8}{9}\bigg)^n+1$$
Right side is strictly increasing function. but i have a confusion whether left side is strictly increasing or not
could some help me how to solve it, thanks
| Consider the function $f(x)=x^n$ for positive $x$. Its second derivative is $n(n-1)x^{n-2}$ and therefore for $n > 1$ or $n<0$ $f$ is strictly convex while for $0 < n < 1$ f is strictly concave.
Our equation is equivalent to $f(5) + f(7) + f(11) = f(6) + f(8) + f(9)$
In the convex case we have:
$f(6) = f(\frac{5+7}{2}) < \frac{f(5)+f(7)}{2}$,
$f(9) = f(\frac{7+11}{2}) < \frac{f(7)+f(11)}{2}$,
and that
$f(8) = f(\frac{11+5}{2}) < \frac{f(11)+f(5)}{2}$
So $ f(5) + f(7) + f(11) > f(6) + f(8) + f(9)$
In the concave case the same strict inequalities hold, but reversed.
Thus only n=0 or n=1 could be solutions and they both work.
| {
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"timestamp": "2023-03-29T00:00:00",
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Calculate the following limit without L'Hopital $\lim\limits_{x\to 0}\frac{e^x-\sqrt{1+2x+2x^2}}{x+\tan (x)-\sin (2x)}$ I know how to count this limit with the help of l'Hopital rule. But it is very awful, because I need 3 times derivate it. So, there is very difficult calculations. I have the answer $\frac{2}{5}$.
I want to know if there is other ways to calculate it, without 3 times using l'Hopital rule? (I could write my steps, but they are very big. I just took third derivative of numerator and denominator)
| $$\begin{eqnarray*} \frac{e^x-\sqrt{1+2x+2x^2}}{x+\tan (x)-\sin (2x)}
& = & \underbrace{\frac{1}{e^x+\sqrt{1+2x+2x^2}}}_{\mbox{harmless}}\cdot \frac{e^{2x}-(1+2x+2x^2)}{x+\tan (x)-\sin (2x)} \\
& \stackrel{\mbox{Taylor}}{=} & \frac{1}{e^x+\sqrt{1+2x+2x^2}}\cdot \frac{\frac{4}{3}x^3+o(x^4)}{\frac{5}{3}x^3+o(x^4)} \\
&\stackrel{x \to 0}{\longrightarrow} & \frac{1}{2}\cdot \frac{4}{5} = \frac{2}{5}
\end{eqnarray*}$$
| {
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"url": "https://math.stackexchange.com/questions/3007785",
"timestamp": "2023-03-29T00:00:00",
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An equilateral triangle is inscribed in a circle of radius $r$. If $P$ is any point on the circumference, find the value of $PA^2 + PB^2 + PC^2$.
An equilateral triangle is inscribed in a circle of radius $r$. If $P$ is any point on the circumference find the value of $PA^2 + PB^2 + PC^2$.
I have managed to solve this problem using co-ordinate geometry by taking a triangle with its centroid at (0,0) and assuming a point at $(r\cos\theta , r\sin \theta)$.
A simpler proof using geometry however eludes me. Is it possible to prove this using Euclidean geometry or without a calculation intensive approach?
The answer is
$6r^2$
| Without the loss of generality, we can assume that $P$ is between $A$ and $B$. Thus $m\angle{AOP}+m\angle{BOP}=120°$. Next, either $m\angle{AOP} \ge m\angle{BOP}$ or $m\angle{AOP} < m\angle{BOP}$. Let's assume $m\angle{AOP} \ge m\angle{BOP}$ (this we can do due to the symmetry of the following proof). Then $m\angle{COP}=120°+m\angle{BOP}$.
Let $m\angle{BOP}/2=x$, then
$PA=2r\sin (60°-x)$, $PB=2r\sin x$, $PC=2r\sin (60°+x)$ and we have $$PA^2+PB^2+PC^2=4r^2 \bigl( \sin^2 (60°-x)+\sin^2 x+ \sin^2 (60°+x) \bigr)=4r^2 \bigl( \sin^2 60° \cos^2 x -2\sin 60° \cos 60° \sin x \cos x + \cos^2 60° \sin^2 x+\sin^2 x+ \sin^2 60° \cos^2 x +2\sin 60° \cos 60° \sin x \cos x + \cos^2 60° \sin^2 x \bigr)=4r^2 \bigl( \frac{3}{2} \cos^2 x +\sin^2 x+ \frac{1}{2} \sin^2 x \bigr)=6r^2$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Contour intergal of a rational trigonometric function, can't find my mistake. This is the integral :
$$I = \int_{0}^{2\pi} \frac{dx}{(5+4\cos x)^2}\ $$
Which, according to wolfram alpha, should evaluate to $\frac{10\pi}{27} $,
but the value i find is $\frac{20\pi}{27} $.
These are my calculations :
Using complex form of cosine i get $ \cos x = \frac{t^2+1}{2t}$
where $ t = e^{ix}$ and $ dx = \frac{dt}{it}$.
If $\partial{D} $ is the unit circle
$$ I = \int_{\partial{D}} \frac{-i}{t(\frac{5t+2t^2+2}{t})^2} dt =
-i \int_{\partial{D}}\frac{t}{(t+2)^2(t+1/2)^2}dt = -i I_a$$
$I_a = \pi i $ times the residue in -$\frac{1}{2}$
The residue is equal to
\begin{align}
\lim_{z\to-\frac{1}{2}} \frac{d}{dz} \left( \frac{t(t+1/2)^2}{(t+2)^2(t+1/2)^2}\right)
&= \lim_{z\to-\frac{1}{2}} \frac{d}{dz} \frac{t}{(t+2)^2}
\\&= \lim_{z\to-\frac{1}{2}} \frac{(t+2)^2-2t(t+2)}{(t+2)^4}
\\&= \lim_{z\to-\frac{1}{2}} \frac{2-t}{(t+2)^3}
= \frac{2+1/2}{(-1/2+2)^3}
\\&= \frac{5/2}{27/9} = \frac{20}{27}
\end{align}
so $ I_a = i\pi\frac{20}{27}$
So our original integral $$I = -i I_a = -i^2 \pi\frac{20}{27} = \frac{20\pi}{27}$$
| Your mistake is when using the residue theorem:
$$\oint_{C^+} f(z) dz = 2\pi i\sum_{singularities} Res(f,singularity)$$
You forgot the factor $2$ in this formula.
Another mistake: $$5t + 2t^2 + 2 = 2(t+2)(t+1/2)$$
You again forgot a factor 2 (which will yield a factor $4$ eventually)
You also made a mistake in calculating the residue.
Note that $(-1/2 + 2)^3 = (3/2)^3 = 27/8$
Correcting these two mistakes, we easily find the correct answer.
| {
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Condition in terms of b and a if $ax^2+bx+c=0$ has two consecutive odd positive integers as roots
The roots of the equation$$ax^2+bx+c=0$$, where $a \geq 0$, are two consecutive odd positive integers, then
(A) $|b|\leq 4a$
(B) $|b|\geq 4a$
(C) $|b|=2a$
(D) None of these
My attempt
Let p and q be the roots then if they are consecutive positive integers (q>p) then $$ pq=\frac{c}{a} \geq 0$$
So, $$c \geq 0$$ and $$q-p=2$$
So, $$\frac{\sqrt{b^2-4ac}}{a}=2$$
So,$$|b|>2a$$
$(Since, a>0,c>0)$
But I know that 4ac should be taken in consideration since its not equal to zero. But I don't know how to use it.
Any hints and suggestions are welcome!
| Using the quadratic formula, we have that (using $p<q$ as the roots):
$$p+2=q$$
$$\frac{-b-\sqrt{b^2-4ac}}{2a}+2=\frac{-b+\sqrt{b^2-4ac}}{2a}$$
$$\frac{-b}{2a}-\frac{\sqrt{b^2-4ac}}{2a}+2=\frac{-b}{2a}+\frac{\sqrt{b^2-4ac}}{2a}$$
$$2=2\frac{\sqrt{b^2-4ac}}{2a}$$
$$2a=\sqrt{b^2-4ac}$$
$$4a^2=b^2-4ac$$
$$b^2=4a^2+4ac$$
$$\frac{b^2}{a^2}=4(1+\frac{c}{a})=4(1+pq)$$
$$\frac{|b|}{a}=2\sqrt{1+pq}$$
Then, given that $p,q$ are different odd integers, you can show that $pq\geq3$, so that $2\sqrt{1+pq}\geq2\sqrt{1+3}=4$
| {
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How to solve $\int_0^{\infty}\frac{\log^n(x)}{1+x^2}dx$? As an exercise for myself I constructed the Integral
$$
\int_0^{\infty}\frac{\log^n(x)}{1+x^2}dx
$$
with $n\in \mathbb{N}$. With the help of Mathematica I found the analytical result
$$
\int_0^{\infty}\frac{\log^n(x)}{1+x^2}dx=\frac{1+(-1)^n}{4^{n+1}}\Gamma(n+1)\left[\zeta\left(n+1,\frac{1}{4}\right)-\zeta\left(n+1,\frac{3}{4}\right)\right].
$$
For $n=1$ (and probably $n\in \mathbb{N}$) one can employ the methods of complex analysis and find a result. For $n\in \mathbb{N}$ I encountered a nasty recursion relation. I can provide details if needed. Is there another way how to solve the integral at hand?
| Let, for $n$, a natural integer,
\begin{align}J_n=\int_0^\infty \frac{\ln^n x}{1+x^2}\,dx\end{align}
First, observe that if $n$ is odd then $J_n=0$ (perform the change of variable $y=\dfrac{1}{x}$ )
Consider, for $n$, a natural integer,
\begin{align}K_n=\int_0^\infty \int_0^\infty\frac{\ln^n(xy)}{(1+x^2)(1+y^2)}\,dx\,dy\end{align}
Observe that,
\begin{align}K_{2n}&=\sum_{k=0}^{2n}\binom{2n}{k} J_kJ_{2n-k}\\
&=\sum_{k=0}^{n}\binom{2n}{2k} J_{2k}J_{2(n-k)}
\end{align}
On the other hand,
Perform the change of variable $u=xy$ ($x$ variable, $y$, a "parameter"),
\begin{align}K_n&=\int_0^\infty \int_0^\infty\frac{y\ln^n u}{(1+y^2)(u^2+y^2)}\,du\,dy\\
&=\frac{1}{2}\int_0^\infty \left[\ln\left(\frac{1+y^2}{u^2+y^2}\right)\right]_{y=0}^{y=\infty}\frac{\ln^{n} u}{u^2-1}\,du\\
&=\int_0^\infty \frac{\ln^{n+1} u}{u^2-1}\,du\\
&=\int_0^1 \frac{\ln^{n+1} u}{u^2-1}\,du+\int_1^\infty \frac{\ln^{n+1} u}{u^2-1}\,du\\
\end{align}
In the second integral perform the change of variable $v=\dfrac{1}{u}$,
\begin{align}K_n&=\int_0^1 \frac{\ln^{n+1} u}{u^2-1}\,du+\int_0^1 \frac{\ln^{n+1} u}{u^2-1}\,du\\
&=\int_0^1 \frac{\left(1+(-1)^n\right)\ln^{n+1} u}{u^2-1}\,du\\
&=\left(1+(-1)^n\right)\int_0^1 \frac{\ln^{n+1} u}{u-1}\,du-\left(1+(-1)^n\right)\int_0^1\frac{u\ln^{n+1} u}{u^2-1}\,du\\
\end{align}
In the latter integral perform the change of variable $y=u^2$,
\begin{align}K_n&=\left(1+(-1)^n\right)\int_0^1 \frac{\ln^{n+1} u}{u-1}\,du-\frac{1+(-1)^n}{2^{n+2}}\int_0^1 \frac{\ln^{n+1} u}{u-1}\,du\\
&=\left(1+(-1)^n\right)\left(1-\frac{1}{2^{n+2}}\right)\int_0^1 \frac{\ln^{n+1} u}{u-1}\,du\\
&=\left(1+(-1)^n\right)\left(1-\frac{1}{2^{n+2}}\right)(-1)^{n+2}(n+1)!\zeta(n+2)
\end{align}
Therefore, one obtains a recursion relation,
\begin{align}
\boxed{\sum_{k=0}^{n}\binom{2n}{2k} J_{2k}J_{2(n-k)}=2\left(1-\frac{1}{2^{2n+2}}\right)(2n+1)!\zeta(2n+2)}
\end{align}
Observe that,
\begin{align}J_0=\frac{\pi}{2}\end{align}
| {
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} |
How do I simplify $\frac{\sin(2x)}{1-\cos (2x)}$? $$\dfrac{\sin(2x)}{1-\cos (2x)}$$
How do I simplify given expression?
My attempt:
We know the double-angle identity for $\sin(2x)$ and $\cos(2x)$ as shown below
$$\sin(2x) = 2\sin (x)\cos(x)$$
$$\cos(2x) = 2\cos(x)-1$$
So we have that
$$\dfrac{2\sin (x)\cos(x)}{1-2\cos(x)-1} = \dfrac{2\sin(x)\cos(x)}{-2\cos^2(x)} = \dfrac{2\sin(x)}{-2\cos(x)} = -\tan (x)$$
I believe I've gone wrong somewhere.
| We have
$$\cos(2x) = \cos^2x-\sin^2x=2\cos^\color{red}2(x)-1=1-2\sin^2 x$$
and then
$$\dfrac{\sin(2x)}{1-\cos (2x)}=\dfrac{2\sin x \cos x}{1-1+2\sin^2 x}=\cot x$$
| {
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"source": "stackexchange",
"question_score": "2",
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} |
Recurrence k-th pattern I am trying to solve this recurrence $T(n) = 6 T(\frac{n}{3}) + n$.
1st recurrence: $6^2T(\frac{n}{3^2}) + \frac{6n}{3} + n$
2nd: $6^3T(\frac{n}{3^3}) + \frac{6^3n}{3^2} + n$
3rd: $6^4T(\frac{n}{3^4}) + \frac{6^6n}{3^3} + n$
I am having trouble describing the general pattern after the k-th iteration.
| I assume you are looking to find $T(n)$ after K iterations. First
$$ T(n) = 6 T(n/3) + n$$ then rewriting $T(n/3)$ in terms of $T(n/3^2)$ we conclude:
$$
T(n) = 6^2 T(n/3^2) + 6n/3 +n
$$
similarly $T(n)$ after K iterations becomes:
$$
T(n) = 6^K T(n/ 3^K) + \sum_{i=1} ^K 6^{i-1}n/3^{i-1}
$$
or
$$
T(n) = 6^K T(n/ 3^K) + (2^{K}-1)n
$$
| {
"language": "en",
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} |
Proving $(n+1)^2+(n+2)^2+...+(2n)^2= \frac{n(2n+1)(7n+1)}{6}$ by induction My question: $(n+1)^2+(n+2)^2+(n+3)^2+...+(2n)^2= \frac{n(2n+1)(7n+1)}{6}$
My workings
*
*LHS=$2^2$ =$4$ RHS= $\frac{24}{6} =4 $
*$(k+1)^2+(k+2)^2+(k+3)^2+...+(2k)^2= \frac{k(2k+1)(7k+1)}{6}$
*LHS (subsituting $n= k+1$)----> $(k+2)^2+(k+3)^2+(k+4)^2+...+(2k)^2+(2k+1)^2$
Now, this is where my problem has started. When I substitute $n=k+1$ in the third step, I do not have the $(k+1)^2$ anymore. So I cannot use the statement in second step. So my question is what should i do next?
| $(k+2)^2+\dots+(2k)^2+(2k+1)^2+(2k+2)^2=\frac{k(2k+1)(7k+1)}6-(k+1)^2+(2k+1)^2+(2k+2)^2=\frac{k(14k^2+9k+1)+6(7k^2+10k+4)}6=\frac{(k+1)(2k+3)(7k+8)}6$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "1",
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Solve the Differential equation $\left(y e^{\sin x}\cos x-y^3+2xy\right)dx+\left(2e^{\sin x}-4y^2(x+1)+2x^2\right)dy=0$ Solve the equation
$$\left(y e^{\sin x}\cos x-y^3+2xy\right)dx+\left(2e^{\sin x}-4y^2(x+1)+2x^2\right)dy=0$$
My try:
Letting $e^{\sin x}=t$ we have $e^{\sin x}\cos xdx=dt$
so have modified the equation as:
$$ydt+2tdy-y^3dx-3y^2xdy+2xydx+x^2dy-y^2xdy+x^2dy-4y^2dy=0$$ $\implies$
$$ydt+tdy+tdy-d\left(y^3x\right)+d\left(x^2y\right)+(x^2-xy^2)dy=4y^2dy$$ $\implies$
$$d(ty)+tdy-d\left(y^3x\right)+d\left(x^2y\right)+(x^2-xy^2)dy=4y^2dy$$
any way to proceed here?
| If you multiply the expression on the left-hand side by $y$, you get the exact differential,
$$ y \times \left(\left(y e^{\sin x}\cos x-y^3+2xy\right)dx+\left(2e^{\sin x}-4y^2(x+1)+2x^2\right)dy \right) =d\left( y^2 e^{\sin x} - y^4 (x + 1) + x^2 y^2\right).$$
So the solution is
$$ y^2 e^{\sin x} - y^4 (x + 1) + x^2 y^2 = {\rm constant}.$$
| {
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Find the Limit $\lim_{n \to \infty}\sum_{k=1}^{\infty}\frac{k^{n}}{1+k^{n+2}}$ Find $\lim_{n \to \infty}\sum_{k=1}^{\infty}\frac{k^{n}}{1+k^{n+2}}$
My ideas: let $ n \in \mathbb N$ be constant, looking at $\frac{k^{n}}{1+k^{n+2}}$, we know
$$\frac{k^{n}}{1+k^{n+2}}\leq\frac{k^{n}}{k^{n+2}}=\frac{1}{k^{2}}$$
but this does not help me because $\sum_{k=1}^{\infty}\frac{1}{k^{2}}=\pi^{2}/6$
any ideas?
| The series equals
$$\frac{1}{2}+ \sum_{k=2}^{\infty}\frac{1}{k^{-n} + k^2}.$$
For each $k\ge 2,$ the terms in the last series increase to $1/k^2$ as $n\to \infty.$ By the monotone convergence theorem, the desired limit is
$$\frac{1}{2} +\sum_{k=2}^{\infty}\frac{1}{k^2} = \frac{1}{2}+\left (\frac{\pi^2}{6} - 1\right) = \frac{\pi^2}{6} - \frac{1}{2}.$$
| {
"language": "en",
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"source": "stackexchange",
"question_score": "1",
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Seeking Methods to solve $\int_{0}^{\frac{\pi}{2}} \ln\left|\sec^2(x) + \tan^4(x) \right|\:dx $ After weeks of going back and forth I've been able to solve the following definite integral:
$$I = \int_{0}^{\frac{\pi}{2}} \ln\left|\sec^2(x) + \tan^4(x) \right|\:dx $$
To solve this I employ Feynman's Trick with Glasser's Master Theorom but I'm excited to learn of other methods that can be employed. Are there any other 'tricks' that can be used? or alternatively series based solutions? or transformations? (or anything for that matter).
For those who may be interested my process was:
(1) First make the substitution: $u = \tan(x)$
$$I = \int_{0}^{\infty} \frac{\ln\left|u^2 + 1 + u^4 \right|}{u^2 + 1}\:du = \int_{0}^{\infty} \frac{\ln\left|1 + u^2\left(u^2 + 1\right) \right|}{u^2 + 1}\:du$$
(2) Now employ Feynman's Trick by introducing a new parameter:
$$I(t) = \int_{0}^{\infty} \frac{\ln\left|1 + t^2u^2\left(u^2 + 1\right) \right|}{u^2 + 1}\:du$$
Note here that $I = I(1)$ and $I(0) = 0$
(3) Take the derivative w.r.t 't'
$$I'(t) = \int_{0}^{\infty} \frac{2tu^2\left(u^2 + 1\right)}{1 + t^2u^2\left(u^2 + 1\right)}\frac{1}{u^2 + 1}\:du = \frac{1}{t} \int_{-\infty}^{\infty} \frac{1}{\left(u - \frac{1}{tu}\right)^2 + \frac{2}{
t} + 1}\:du$$
(4) Employ Glasser's Master Theorem:
$$I'(t) = \frac{1}{t} \int_{-\infty}^{\infty} \frac{1}{\left(u - \frac{1}{tu}\right)^2 + \frac{2}{t} + 1} \:du= \frac{1}{t}\int_{-\infty}^{\infty}\frac{1}{u^2 + \frac{2}{t} + 1} \:du$$
As: $\frac{2}{t} + 1 > 0 $ we arrive at
$$I'(t) = \frac{1}{t}\left[\frac{1}{\sqrt{\frac{2}{t} + 1}}\arctan\left(\frac{u}{\frac{2}{t} + 1}\right)\right]_{-\infty}^{\infty}= \frac{\pi}{\sqrt{t\left(t + 2\right)}}$$
(5) We now integrate w.r.t 't'
$$I(t) = \int \frac{\pi}{\sqrt{t\left(t + 2\right)}}\:dt = 2\pi\sinh^{-1}\left(\frac{t}{\sqrt{2}} \right) + C$$
Where $C$ is the constant of integration. As above $I(0) = 0 \rightarrow C = 0$ and so, our final solution is given by:
$$I = I(1) = 2\pi\sinh^{-1}\left(\frac{1}{\sqrt{2}} \right)$$
| Here's an alternative solution. First, notice that if $I(a,b) = \int_0^\pi \ln\left(a + b\cos(x)\right) \mathrm{d}x$ then $I(a,0) = \pi \ln(a)$. Thus
\begin{align}
\frac{\partial I}{\partial a}
&=\int_0^\pi \frac{\mathrm{d}x}{a + b\cos(x)} \overset{t=\tan\left(\frac{x}{2}\right)}{=}\frac{2}{a-b}\int_0^\infty \frac{\mathrm{d}t}{t^2 + \frac{a+b}{a-b}} \overset{t =\sqrt{\frac{a+b}{a-b}}\tan(u)}{=} \frac{\pi }{\sqrt{a^2-b^2}}
\end{align}
So integrating the above equation w.r.t. $a$ we get
$$
I(a,b) = \pi\int\frac{\mathrm{d}a}{\sqrt{a^2-b^2}} = \pi\int\frac{1 + \frac{a}{\sqrt{a^2-b^2}}}{a+\sqrt{a^2-b^2}} \,\mathrm{d}a \overset{u =a+\sqrt{a^2-b^2} }{=} \pi\ln\left(a+\sqrt{a^2-b^2}\right) + C
$$
Recalling our initial condition, by plugging in $b=0$ into the previous equation we get $\pi \ln(a) = I(a,0) = \pi \ln(2a) + C$. Hence $C = - \pi \ln(2)$ and
$$\int_0^\pi \ln\left(a + b\cos(x)\right) \,\mathrm{d}x = \pi\ln\left(\frac{a+\sqrt{a^2-b^2}}{2}\right) \tag{1}$$
Lastly, using basic integral manipulations we can obtain $\int_0^{\frac{\pi}{2}}\ln(\cos(x)) \mathrm{d}x = - \frac{\pi}{2}\ln(2)$. Combining everything we conclude the problem as follows
\begin{align}
\int_0^{\frac{\pi}{2}} \ln\left(\sec^2(x) + \tan^4(x)\right) \mathrm{d}x & =\int_0^{\frac{\pi}{2}} \ln\left(\frac{7+\cos(4x)}{8\cos^4(x)}\right) \mathrm{d}x \\
& = \frac{1}{2}\int_0^{\pi}\ln(7+\cos(u))\, \mathrm{d}u - 4\int_0^{\frac{\pi}{2}} \ln(\cos(x)) \mathrm{d}x - \frac{\pi}{2}\ln(8)\\
& \overset{(1)}{=}\frac{\pi}{2}\ln\left( \frac{7+4 \sqrt{3}}{2}\right) + 2\pi \ln(2) - \frac{3\pi}{2}\ln(2)\\
& = \boxed{\frac{\pi}{2} \ln\left(7 + 4 \sqrt{3}\right)}
\end{align}
which agrees with the other answers since $\sqrt{7 + 4 \sqrt{3}} = 2 + \sqrt{3}$.
| {
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"question_score": "10",
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What is the area shaded in this figure? I was sent this problem by a friend, they say it is a year 6 question. Using trigonometry I got the shaded area as $4 - \frac{\pi}{3}$.
Now Year six, I doubt if they do trigonometry. Is there another approach out there?
|
Label the vertices as shown in the diagram.
$\tan\alpha=\frac{1}{2}\implies \tan 2\alpha=\frac{4}{3}\implies 2\alpha=\tan^{-1}(\frac{4}{3})$
$\therefore$ Area of circular sector$ BAG=8\tan^{-1}(\frac{4}{3})$
Area of $\triangle AGC=16\cos\alpha\sin\alpha=8\sin2\alpha=\frac{32}{5}$
Thus, area of curved figure $BEG=\frac{48}{5}-8\tan^{-1}(\frac{4}{3})$
Therefore, Area of shaded part$=\frac{1}{2}$ (Area of $BEFC$)$-\frac{1}{2}$ (Area of semicircle)$-$Area of curved figure $BEG$.
This gives $\frac{32}{5}-4\pi+8\tan^{-1}(\frac{4}{3})=\frac{32}{5}-4\tan^{-1}(\frac{3}{4})\approx1.252$
| {
"language": "en",
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Average power of 2 in all even natural numbers Consider all even natural numbers.
Every 4th number has a power of 4 (or $2^2$)
Every 8th number has a power of 8 (or $2^3$)
Every 16th number has a power of 16 (or $2^4$)
What is the average number of power of 2 in any random even number?
Since every 2nd number has a power of 2,and every 4th number is a power of 4 and so on
$\left(2 \times \dfrac{1}{2}\right) + \left(4 \times \dfrac{1}{4}\right) + \left(8 \times \dfrac{1}{8}\right) + \ldots$
This goes all the way to Infinity and hence the answer is $\infty$. Is this correct?
| A multiple of $4$ only has two powers of $2$ in its factorization and a multiple of $8$ only has three. Your sum should therefore be
$$\left(1 \times \dfrac{1}{2}\right) + \left(2 \times \dfrac{1}{4}\right) + \left(3 \times \dfrac{1}{8}\right) + \ldots=2$$
because half the evens have exactly one factor of two, one quarter have exactly two factors of two and so on. Although it you can't pick a random natural number, this makes sense as the limit of the average number of factors of two in the even numbers up to $n$ as $n \to \infty$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3032706",
"timestamp": "2023-03-29T00:00:00",
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Calculate $\sin\frac{\pi}{16}$ from given trigonometric identities There is the choice between four trigonometric identities
$$\sin(4\phi)=8\cos^{3}(\phi)\sin(\phi)-4\cos(\phi)\sin(\phi)$$
$$\cos(4\phi)=8\cos^{4}(\phi) -8\cos^{2}(\phi)+1$$
$$\sin^{4}(\phi)=1/8(\cos(4\phi) -4\cos(2\phi)+3)$$
$$\cos^{4}(\phi)=1/8(\cos(4\phi)+4\cos(2\phi)+3)$$
to calculate $\sin\frac{\pi}{16}$. When using the first I got stuck trying to eliminate cos expressions and bring $\sin(4\phi)$ and $\sin(\phi)$ on different sides of the equation, as in
$$\sin(4\phi)=\sin (2\phi)(2-4\sin^{2}(\phi))$$
What way to go ?
| Use the second identity:
$$ 8\cos^4\left(\frac{\pi}{16}\right) - 8\cos^2\left(\frac{\pi}{16}\right) + 1 = \cos\left(\frac{\pi}{4}\right) = \frac{\sqrt2}{2} $$
Let $x = \cos^2\left(\frac{\pi}{16}\right)$ then
$$ 8x^2-8x = \frac{\sqrt{2}-2}{2} $$
Completing the square:
$$ 4x^2 - 4x + 1 = \frac{\sqrt{2}-2}{4}+1 = \frac{2+\sqrt{2}}{4} $$
$$ (2x-1)^2 = \frac{2+\sqrt{2}}{4} $$
Note that $2x-1 = 2\cos^2\left(\frac{\pi}{16}\right)-1 = \cos \left(\frac{\pi}{8}\right) > 0$ so we take the positive root
$$ 2x-1 = \frac{\sqrt{2+\sqrt{2}}}{2} $$
$$ x = \frac{2+\sqrt{2+\sqrt{2}}}{4} $$
Then we have
$$ \sin\left(\frac{\pi}{16}\right) = \sqrt{1-x} = \frac{\sqrt{2-\sqrt{2+\sqrt{2}}}}{2} $$
| {
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"source": "stackexchange",
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Finding a recurrence relation I need to find a recurrence relation or an exact formula to the sequence
$$1,2,4,8,12,16,24,32,40,48,64,80,96,112,128,160,192,224,256,288,\ldots$$
Well, considering $a_0=1$, $a_1=2$ and $a_2=4$, the terms $a_n$ for $n\geq3$ is obtained by adding a power of $2$. In fact the sequence is:
$$1\xrightarrow{+1}2\xrightarrow{+2}4\xrightarrow{+4}8\xrightarrow{+4}12\xrightarrow{+4}16\xrightarrow{+8}24\xrightarrow{+8}32\xrightarrow{+8}40\xrightarrow{+8}48\xrightarrow{+16}64\xrightarrow{+16}80\xrightarrow{+16}96\xrightarrow{+16}112\xrightarrow{+16}128\xrightarrow{+32}160\ldots$$
and we will have $5+1$ addition of $32=2^5$ and then $6+1$ addition of $64=2^6$ and so on.
Any, solutions, suggestions,... please.
| To compute the $n^{th}$ term, find $k$ such that $\frac{k(k-1)}{2} \leq n \leq \frac{k(k+1)}{2}$. Then:
$x(n) = x(n-1) + 2^{k-1}$.
For example, if $n=9$, $k=4$ and $x(9) = x(8) + 8$.
Let's first look at the roots of $k^2-k -2n=0$. They are given by $r_1 = \frac{1-\sqrt{1+8n}}{2}$ and $r_2 = \frac{1+\sqrt{1+8n}}{2}$.
The inequality $\frac{k(k-1)}{2} \leq n$ implies $k^2-k-2n \leq 0$, i.e. $(k-r_1)(k-r_2) \leq 0$, which is true when $r_1 \leq k \leq r_2$. Since $k >0$ and $r_1<0$, the condition reduces to $0 < k \leq r_2$. For example, when $n=9$, $0 < k \leq 4.77$.
Now let's look a the roots of $k^2+k-2n=0$. They are given by $r_3 = \frac{-1-\sqrt{1+8n}}{2}$ and $r_4 = \frac{-1+\sqrt{1+8n}}{2}$.
The inequality $n \leq \frac{k(k+1)}{2}$ implies $k^2+k-2n \geq 0$, i.e. $(k-r_3)(k-r_4) \geq 0$, which is true when $k \leq r_3$ or $k \geq r_4$. Since $k>0$, this reduces to $k \geq r_4$. For example, when $n=9$, $k \geq 3.77$.
Combining the two, we have:
$\frac{-1+\sqrt{1+8n}}{2} \leq k \leq \frac{1+\sqrt{1+8n}}{2}$.
One may be able to write this as $k = \text{round} \left( \frac{\sqrt{1+8n}}{2}\right)$ and $x(n) = x(n-1) + 2^{k-1}$. The ambiguity when $8n+1$ is a perfect square needs to be resolved (e.g. when $n=10$).
| {
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"timestamp": "2023-03-29T00:00:00",
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How many pairs $(x,y)$ such that $x+y=n$ have an $x$ or $y$ divisible by 3 but $x$ and $y$ are not equal to 3? Let the set $S_{n}$ = {$(x,y):x,y \in \mathbb{O}$} such that $x+y=n$ where $\mathbb{O}$ is set of odd integers > 1.
Let us define the function $f(n) = |S_{n}|$ that counts the number of pairs in $S_{n}$.
Examples: $S_{10} =\{(3,7),(5,5),(7,3)\}$ and $f(10) = 3$.
$S_{14} = \{(3,11), (5,9), (7,7), (9,5),(11,3)\}$ and $f(14) = 5$
$S_{32} = \{(3,29), (5,27), (7,25), (9,23), (11,21), (13,19), (15,17), (17,15), (19,13), (21,11), (23,9), (25,7), (27,5), (29,3)\}$
and $f(32) = 14$
It turns out that $f(n) = ((n/2) - 2)$.
Let us define another function $g(n)$ that counts the number of pairs $(x,y)$ such that either $x$ or $y$ is divisible by 3, but $x \neq 3$ and $y\neq 3$.
Example: $g(32) = 8$ because there are 8 pairs where $x$ or $y$ is divisible by 3 but not equal to 3.
They are (5,27), (9,23), (11,21), (15,17), (17,15), (21,11), (23,9) and (27,5).
There are two cases: Case 1, $n$ is divisible by 3 and case 2, $n$ is not divisible by 3. Let us only consider case where $n$ is not divisible by 3, since if $n$ is divisible by 3, any pair where $x$ is divisible by 3, $y$ will also be divisible by 3.
What is the formula for $g(n)$ in terms of $f(n)$ for values of $n$ not divisible by 3?
What is the formula for $g(n)$ in terms of $f(n)$ for limit $n \to\infty$?
My guess is that as $n \to \infty$ for $n$ not divisible by 3, the value of $g(n)$ approaches $(2/3)f(n)$.
| The sum of $2$ multiples of $k$ is also a multiple of $k$. Therefore, if $n$ is not divisible by $3$, we know that the $\left\lfloor \frac{n}{6}\right\rfloor -1$ values of $p$ where $p$ is divisible by $3$ will not overlap any of the $\left\lfloor \frac{n}{6}\right\rfloor -1$ values of $q$ where $q$ is divisible by $3$.
So the formula for $g(n)$ in terms of $n$ for values of $n$ not divisible by $3$ is:
$$g(n)=2\left\lfloor \frac{n}{6}\right\rfloor -2$$
The formula for $g(n)$ in terms of $f(n)$ for values of $n$ not divisible by $3$ is:
$$g(n)=2\left\lfloor \frac{f\left(n\right)+2}{3}\right\rfloor -2$$
Your guess about the limit for $\frac{g(n)}{f\left(n\right)}$ is also right:
$$\frac{g(n)}{f\left(n\right)}=\frac{2\left\lfloor \frac{n}{6}\right\rfloor -2}{\frac{n}{2}-2}=\frac{\frac{n}{3}-2}{\frac{n}{2}-2}=\frac{2n-12}{3n-12}$$
$$\lim_{n\rightarrow\infty}\frac{2n-12}{3n-12}=\frac{2\infty-12}{3\infty-12}=\frac{2\infty}{3\infty}=\frac{2}{3}\left(\frac{\infty}{\infty}\right)=\frac{2}{3}$$
| {
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Show that the $L^{p}$ norm $\|f\|_{L^{p}} := \big( \int^{b}_{a} |f(x)|^p\big)^{1/p}$ is not induced by a scalar product for $p \neq 2$.
On $X = C^0\big([a,b]\big)$, for any $p \in \mathbb{R}$, $p>1$, we define the $L^p$ norm by,
$$\|f\|_{L^{p}}:=\big(\int^{b}_{a}|f(x)|^{p}dx \big)^{1/p}.$$
Show that for $p\neq 2$, this norm is not induced by a scalar product.
My method of trying to prove this was to prove a contradiction to the parallelogram rule,
$$ \|f+g\|^{2}_{p} + \|f-g\|^{2}_{p} = 2\|f\|^{2}_{p} + 2\|g\|^{2}_{p}, \tag{$1$}$$
where $f,g \in C^{0}([a,b])$.
So I defined the following functions;
$$f(x):=\frac{a+b}{2}-x$$
$$g(x) := \begin{cases}\frac{a+b}{2}-x, \ \ for \ \ a \leq x \le \frac{a+b}{2}. \\
x-\frac{a+b}{2}, \ \ for \ \ \frac{a+b}{2} < x \le b \end{cases}$$
which gives
$$f(x)+g(x) = \begin{cases} a+b-2x, \ \ & for \ \ a\le x \le \frac{a+b}{2}. \\ 0, & for \ \ \frac{a+b}{2} < x \le b\end{cases}$$
$$f(x)-g(x) = \begin{cases} 0, & for \ \ a \le x \le \frac{a+b}{2}. \\
2x - (a+b), \ \ & for \ \ \frac{a+b}{2} < x \le b \end{cases}$$
Then I proceeded to calculate each term of the parallelogram rule,
$$\|f+g\|^{2}_{p} = \bigg( \int^{\frac{a+b}{2}}_{a}|a+b-2x|^{p}\bigg)^{2/p} = \frac{(b-a)^{\frac{2(p+1)}{p}}}{(2(p+1))^{2/p}} $$
$$ \|f-g\|^{2}_{p} = \bigg( \int_{\frac{a+b}{2}}^{b}|2x- (a+b)|^{p}\bigg)^{2/p} = \frac{(b-a)^{\frac{2(p+1)}{p}}}{(2(p+1))^{2/p}}$$
$$2\|f\|^{2}_{p} = 2 \bigg( \int^{b}_{a}| \frac{a+b}{2}-x|^{p} dx \bigg)^{2/p} = 2 \cdot \frac{2^{2/p}(\frac{b-a}{2})^{\frac{2(p+1)}{p}}}{(p+1)^{2/p}} $$
$$\begin{align}2 \|g\|^{2}_{p} & = 2 \bigg(\int^{\frac{a+b}{2}}_{a} |\frac{a+b}{2} - x|^{p} dx \ + \ \int^{b}_{\frac{a+b}{2}}|x- \frac{a+b}{2}|^{p} dx\bigg)^{2/p} \\ & =2 \cdot \frac{2^{2/p}(\frac{b-a}{2})^{\frac{2(p+1)}{p}}}{(p+1)^{2/p}} \end{align}$$
Plugging into $(1)$ we then get
$$2 \cdot \frac{(b-a)^{\frac{2(p+1)}{p}}}{(2(p+1))^{2/p}} = 4 \cdot \frac{2^{2/p}(\frac{b-a}{2})^{\frac{2(p+1)}{p}}}{(p+1)^{2/p}}$$
which simplifies quite nicely to
$$2^{p} = 4.$$
So the equality only holds for $p = 2$.
Is what i've done correct? is there another way of proving the question which is better?
| Here is a similar idea with simpler computation. Define
$$f(x) = \begin{cases} \left(\frac{a+b}2-x\right)^{1/p},&\text{ if } x \in \left[a, \frac{a+b}2\right] \\
0,&\text{ if } x \in \left[\frac{a+b}2,b\right]
\end{cases}$$
$$g(x) = \begin{cases} 0,&\text{ if } x \in \left[a, \frac{a+b}2\right] \\
\left(x-\frac{a+b}2\right)^{1/p},&\text{ if } x \in \left[\frac{a+b}2,b\right]
\end{cases}$$
Notice that $f$ and $g$ have disjoint supports so $$f(x)+g(x) = \left|\frac{a+b}2-x\right|^{1/p}$$ and $$f(x)-g(x) = \operatorname{sign}\left(\frac{a+b}2-x\right)\left|\frac{a+b}2-x\right|^{1/p}$$
We get
$$\|f\|_p^2 = \left(\int_{\left[a, \frac{a+b}2\right]} \left(\frac{a+b}2-x\right)\,dx\right)^{2/p} = \left[\frac{(b-a)^2}8\right]^{2/p}$$
Similarly we also see $\|g\|_p^2 = \left[\frac{(b-a)^2}8\right]^{2/p}$, $\|f+g\|_p^2 = \left[\frac{(b-a)^2}4\right]^{2/p}$ and $\|f-g\|_p^2 = 0$.
Therefore
$$\left[\frac{(b-a)^2}4\right]^{2/p} = \|f+g\|_p^2 = 4\|f_p\|^2 = 4\left[\frac{(b-a)^2}8\right]^{2/p}$$
or $2 = 4^{p/2}$. This implies $p=2$.
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Compute the limit $\lim_{n\to\infty} I_n(a)$ where $ I_n(a) :=\int_0^a \frac{x^n}{x^n+1}\,\mathrm{d}x, n\in N$. For $a>0$ we define
$$\space I_n(a)=\int_0^a\frac{x^n}{x^n+1}\,\mathrm{d}x , n\in N.$$
*
*Prove that $0\le I_n(1) \le \frac{1}{n+1}$
*Compute $\lim_{n\to\infty} I_n(a)$
My attempt:
*
*I regard $I_n(1)=\int_0^1\frac{x^n}{x^n+1}$. If $x\in (0,1)$ then $x^n\in(0,1)$ and $x^n+1\in(1,2)$.
$$x^n>0 \Rightarrow x^n+1>1 \Rightarrow 1>\frac{1}{1+x^n }\Rightarrow x^n>\frac{x^n}{x^n+1}\Rightarrow \int_0^1\frac{x^n }{x^n+1}dx<\int_o^1 x^n \mathrm{d}x\\ \Rightarrow \int_0^1\frac{x^n }{x^n+1}dx<\frac{1}{n+1} \\ 0\le\frac{x^n}{x^n+1} \\ \text{In concusion } 0\le I_n(1) \le \frac{1}{n+1}.$$
*first case $a\in(0,1) \Rightarrow \lim_{n\to\infty} I_n(a) =0$. $I_n(a)\le\frac{1}{n+1})\text{case 2 . }a\in(1,\infty) \Rightarrow$ ???????
I don't believe the limit is $\infty$ because $\frac{x^n }{x^n+1}\le 1$.
I would appreciate some hints.
| Certainly not the most compact approach, but:
\begin{equation}
I_n(a) = \int_{0}^{a} \frac{w^n}{w^n + 1}\:dw = \int_{0}^{a}\left[ 1 - \frac{1}{w^n + 1}\right]\:dw = a - \int_{0}^{a}\frac{1}{w^n + 1}\:dw
\end{equation}
Now:
\begin{equation}
J_n(a) = \int_{0}^{a} \frac{1}{w^n + 1}\:dw
\end{equation}
With $n \geq 1$ and $x \geq 0$
Here, let $t = a^n$ to arrive at:
\begin{equation}
J_n(a) = \frac{1}{n}\int_{0}^{x^n} \frac{1}{t + 1}t^{\frac{1}{n} - 1}\:dt
\end{equation}
Now let $u = \frac{1}{1 + t}$ to arrive at:
\begin{align}
J_n(a) &= \frac{1}{n}\int_{0}^{a^n} \frac{1}{t + 1}t^{\frac{1}{n} - 1}\:dt = \frac{1}{n}\int_{1}^{\dfrac{1}{a^n + 1}} u \left(\frac{1 - u}{u} \right)^{1 - \frac{1}{n} }\frac{-1}{u^2}\:du \\
&= \frac{1}{n}\int_{\dfrac{1}{a^n + 1}}^{1} u^{-\frac{1}{n}}\left(1 - u\right)^{\frac{1}{n} - 1}\:du
\end{align}
Here, as $x \geq 0$ and $n > 1$, we see that $\dfrac{1}{a^n + 1} < 1$ and thus,
\begin{align}
J_n(a) &= \frac{1}{n}\int_{\dfrac{1}{a^n + 1}}^{1} u^{-\frac{1}{n}}\left(1 - u\right)^{\frac{1}{n} - 1}\:du \\
&= \frac{1}{n}\left[\int_{0}^{1} u^{-\frac{1}{n}}\left(1 - u\right)^{\frac{1}{n} - 1}\:du - \int_{0}^{\dfrac{1}{a^n + 1}} u^{-\frac{1}{n}}\left(1 - u\right)^{\frac{1}{n} - 1}\:du\right] \\
&= \frac{1}{n}\left[B\left(1 - \frac{1}{n}, \frac{1}{n} \right) - B\left(1 - \frac{1}{n}, \frac{1}{n}, \frac{1}{a^n + 1}\right)\right]
\end{align}
Where $B(a,b)$ is the Beta function and $B(a,b,x)$ is the Incomplete Beta function
Using the relationship between the Beta and Gamma functions we arrive at:
\begin{align}
J_n(a) &= \int_{0}^{a} \frac{1}{w^n + 1}\:dw = \frac{1}{n}\left[\Gamma\left(1 - \frac{1}{n} \right)\Gamma\left(\frac{1}{n} \right)- B\left(1 - \frac{1}{n}, \frac{1}{n}, \frac{1}{a^n + 1}\right)\right]
\end{align}
For $a \geq 0$ and $n \geq 1$. Returning to $I(a)$ we have
\begin{align}
I_n(a) = a - J_n(a) = a - \frac{1}{n}\left[\Gamma\left(1 - \frac{1}{n} \right)\Gamma\left(\frac{1}{n} \right)- B\left(1 - \frac{1}{n}, \frac{1}{n}, \frac{1}{a^n + 1}\right)\right]
\end{align}
From here you can attempt your direct questions.
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Proof verification: For $a$, $b$, $c$ positive with $abc=1$, show $\sum_{\text{cyc}}\frac{1}{a^3(b+c)}\geq \frac32$ I would like to have my solution to IMO 95 A2 checked. All solutions I've found either used Cauchy-Schwarz, Chebyshevs inequality, the rearrangement inequality or Muirheads inequality. Me myself, I've used Jensens inequality, and I am a bit unsure if my solution holds. Here it is:
Let $a,b,c$ be positive real numbers with $abc=1$. Show that $$\frac{1}{a^3(b+c)} + \frac{1}{b^3(a+c)} + \frac{1}{c^3(a+b)}\geq \frac32$$
First, substitute $\frac1a = x, \frac1b = y, \frac1c = z$, giving the new constraint $xyz=1$, and transforming the inequality into $$\frac{x^2}{z+y} + \frac{y^2}{x+z} + \frac{z^2}{x+y} \geq \frac32$$ Now, let $f(x)=\dfrac{x^2}{S-x}$ where $S=x+y+z$. We have that $f''(x)=\dfrac{2S^2}{(S-x)^2}$, and $S>0$ beacuse $x,y,z>0$, thus $f''(x)>0$ for all $x,y,z\in \mathbb{R}^+$. Thus we see that $f(x)$ is convex, so by Jensens inequality we have that $$\begin{alignat*}{2}\frac{x^2}{y+z} + \frac{y^2}{x+z} + \frac{z^2}{x+y} & =\frac{x^2}{S-x} + \frac{y^2}{S-y} + \frac{z^2}{S-z} \\
&\geq 3f\left(\frac{x+y+z}{3} \right) \\
&= 3\frac{\left(\frac{x+y+z}{3}\right)^2}{S-\frac{x+y+z}{3}} = \frac{1}{3}\frac{(x+y+z)^2}{\frac{2x+2y+2z}{3}} = \frac13 \frac{(x+y+z)^2}{\frac23 (x+y+z)} \\
&= \frac12 (x+y+z) \geq \frac32\end{alignat*}$$
The final inequality follows by AM-GM ($x+y+z\geq 3\sqrt[3]{xyz}=3$)
| Your proof is nearly correct: $$f''(x)=\frac{2S^2}{(S-x)^3}$$ (the exponent is $3$, not $2$). This is still okay because $x,y,z>0$, but you need a bit more justification.
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