Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Trace zero matrix that can't be written as $AB - BA$? According to this paper, every trace zero matrix over a field can be written in the form $AB - BA$. However, here's a basic counterexample:
Let $A = diag(a, -a)$ for some nonzero number a. Then $A = BC - CB$ if and only if a system of equations involving entries of $B,C,A$ is true. Then the entries of $A$ are:
$$\begin{align}
a = b_{1,1}c_{1,1} + b_{1,2}c_{2,1} - b_{1,1}c_{1,1} - b_{2,1}c_{1,2} \\
0 = b_{1,1}c_{1,2} + b_{1,2}c_{2,2} - b_{1,1}c_{2,1} - b_{2,1}c_{2,2} \\
-a = b_{2,1}c_{1,1} + b_{2,2}c_{2,1} - b_{1,2}c_{1,1} - b_{2,2}c_{1,2} \\
0 = b_{2,1}c_{1,2} + b_{2,2}c_{2,2} - b_{1,2}c_{2,1} - b_{2,2}c_{2,2}
\end{align}
$$
For $a$ nonzero the first and the last equations can't both be true. So where did I mess up?
| $$B=\left(\begin{array}{rr}0&1\\0&0\end{array}\right).$$
$$C=a\left(\begin{array}{rr}0&0\\1&0\end{array}\right).$$
$$BC-CB=\left(\begin{array}{rr}a&0\\0&-a\end{array}\right).$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/466018",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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How can I evaluate $\lim_{x \rightarrow 0} \frac{\left(\sqrt{2+x}-\sqrt{2} \right)}{ x}$? I am having trouble with this practice problem on limits:
$$\lim_{x \rightarrow 0} \frac{\left(\sqrt{2+x}-\sqrt{2} \right)} {x}$$
The answer is $\sqrt{2} \over 4$, but I'm having trouble seeing how the answer was reached. Any help would be appreciated.
| I assume that you made typing error (probably there should write $\sqrt{2+x}$, not $\sqrt{2x}$). Then, the result you have stated is a limit of the following expression:
\begin{eqnarray*}
\frac{\sqrt{2+x}-\sqrt{2}}{x}&=&\frac{\sqrt{2+x}-\sqrt{2}}{x}\frac{\sqrt{x+2}+\sqrt{2}}{\sqrt{x+2}+\sqrt{2}},\\
\frac{\sqrt{2+x}-\sqrt{2}}{x}&=&\frac{x+2-2}{x(\sqrt{x+2}+\sqrt{2})},\\
\frac{\sqrt{2+x}-\sqrt{2}}{x}&=&\frac{x}{x(\sqrt{x+2}+\sqrt{2})},\\
\frac{\sqrt{2+x}-\sqrt{2}}{x}&=&\frac{1}{\sqrt{x+2}+\sqrt{2}},
\end{eqnarray*}
so
$$\lim_{x\rightarrow 0} \frac{\sqrt{2+x}-\sqrt{2}}{x}=\lim_{x\rightarrow 0} \frac{1}{\sqrt{x+2}+\sqrt{2}}=\frac{\sqrt{2}}{4}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/466669",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Calculate the determinant of the matrix $(a_{ij})$ where $a_{ij}=a+b$ when $i=j$, and $a_{ij}=a$ otherwise The matrix is $n\times n$ , defined as the following:
$$
a_{ij}=\begin{cases}
a+b & \text{ when } i=j,\\
a & \text{ when } i \ne j
\end{cases}.
$$
When I calculated it I got the answer of: $b^n +nab^{n-1}$ , but I saw other solution that said the answer is $b^n + ab^{n-1}$, so I'm not sure. Can you tell me the right solution and show me the steps to calculate it if I'm wrong ? Thank you.
| Let the $n \times n$ matrix $A$ be given by $$A = \begin{pmatrix} a+b & a & a & \cdots & a & a \\ a & a+b & a & \cdots & a & a \\ a & a & a+b & \cdots & a & a \\ \vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\ a & a & a & \cdots & a+b & a \\ a & a & a & \cdots & a & a+b \end{pmatrix} $$
Subtracting row $i+1$ from row $i$ for $1 \leq i \leq n-1$ leaves the matrix:
$$A' = \begin{pmatrix} b & -b & 0 & \cdots & 0 & 0 \\ 0 & b & -b & \cdots & 0 & 0 \\ 0 & 0 & b & \cdots & 0 & 0 \\ \vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\ 0 & 0 & 0 & \cdots & b & -b \\ a & a & a & \cdots & a & a+b \end{pmatrix} $$
Adding column $j$ to column $j+1$ for $ 1 \leq j \leq n-1$ leaves the matrix:
$$A'' = \begin{pmatrix} b & 0 & 0 & \cdots & 0 & 0 \\ 0 & b & 0 & \cdots & 0 & 0 \\ 0 & 0 & b & \cdots & 0 & 0 \\ \vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\ 0 & 0 & 0 & \cdots & b & 0 \\ a & 2a & 3a & \cdots & (n-1)a & na+b \end{pmatrix} $$
As the addition or subtraction of rows and columns does not change the determinant we have $\det(A) = \det(A'')$. Finally, as $A''$ is an upper triangular matrix, it follows that:
$$ \det(A) = b^{n-1}(na+b) = b^n + nab^{n-1} $$
| {
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"timestamp": "2023-03-29T00:00:00",
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what is the taylor expansion of $\ln(\sin x/x)$ what is the Taylor expansion of the function
$$f(x) = \ln \frac{\sin x}x$$
around the point $x=0$? Ignore powers of $x$ which are greater than $6$.
Here is my method:
$$\ln(1+x)=x-\frac{x^2}2 + \frac{x^3}3 -\frac{x^4}4,$$
so we should get the function
$$g(x)=\frac{\sin x}x$$
in the form of $1 + x$ to get the answer … or is there another way to consider the Taylor expansion of $\ln (u)$ by considering $f$ to be the answer of $$\frac{\sin x}x\cdots$$
Which way is the best?? What is the final answer???
| As you suggest, it will be useful to know the expansion
$$\ln(1+t)=t-\frac{t^2}{2}+\frac{t^3}{3}-\frac{t^4}{4}+\cdots\tag{1}$$
(at least for $|t|\lt 1$).
We also need the power series expansion for $\sin x$, which is
$$\sin x=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+\cdots.$$
So the function which is $1$ at $x=0$ and $\frac{\sin x}{x}$ elsewhere has power series expansion
$$1-\frac{x^2}{3!}+\frac{x^4}{5!}-\frac{x^6}{7!}+\cdots.\tag{2}$$
Now substitute (2) for $1+t$ in (1). Thus $t=-\frac{x^2}{3!}+\frac{x^4}{5!}-\frac{x^6}{7!}+\cdots$.
Make sure not to compute too far. So throw away all terms in powers of $x$ greater than $6$.
Because the expression (2) has no $x$ term, you will only need to compute to the $\frac{t^3}{3}$ term in (1). So the calculation is less time-consuming than it seems.
Getting a general term for the power series expansion looks, however, very difficult. Luckily you were not asked to find many terms. You will end up with something of the shape $ax^2+bx^4+cx^6$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/467986",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Limits of integration, surface joined by paraboloid and plane I need help with part of this old exam question:
Q:
C is the intersection line between the paraboloid $z=x^2+y^2$ and the plane $2x-2y+z=1$ travelled counter-clockwise from $(0,0,10)$. Calculate $\oint\limits_C F\cdot dr$. $F(x,y,z)=(z-3y)i+(3x+z)j+e^{x-y}k$
A:
After some calculations the professor end up here:
$$\oint\limits_C F\cdot dr = \\\\
\iint\limits_{x^2+2x+y^2-2y\le1}(-e^{x-y}-1,1-e^{x-y},6)\cdot(2,-2,1)dxdy =\\\\= \iint\limits_{(x+1)^2+(y-1)^2\le3}-2e^{x-y}-2-2+2e^{x-y}+6dxdy=\\\\
=2\pi(\sqrt3)^2=6\pi$$
I understand he is exchanging the $z$ in $2x-2y+z=1$ for the one in $z=x^2+y^2$, but then what? I get to $(x+1)^2+1+(y-1)^2+1\le1$ but how to get to $..\le3$? It there perhaps a radius thing that I should know about?
My questions are:
How is he getting that final limit of integration?
How is he performing that integration to get to $2\pi(\sqrt3)^2$?
How would I solve the integral with polar coordinates? I end up at $2\int_0^{2\pi}\int_0^1rdrd\phi=2\pi$
Thank you for your time!
| Prof. solution is easier than the following ( see below ) !!!
You can find $x$ and $y$ in terms of $z$:
$$
z - \left(1 - z \over 2\right)^{2} = \left(x^{2} + y^{2}\right) - \left(x - y\right)^{2}
=
2xy
$$
$$
2xy = {-z^{2} + 6z -1 \over 4}
$$
$$
\left(x^{2} + y^{2}\right) + 2xy
=
z + {-z^{2} + 6z -1 \over 4}
=
{-z^{2} + 10z -1 \over 4}
$$
$$
\left.%
\begin{array}{rcl}
&&\\[1mm]
x + y & = & {1 \over 2}\,\sqrt{-z^{2} + 10z -1}
\\[2mm]
x - y & = & {1 \over 2}\,\left(1 - z\right)
\\[1mm]&&
\end{array}\right\rbrace
\ \Longrightarrow\
\left\lbrace%
\begin{array}{rcl}
&&\\[1mm]
x & = & {1 \over 4}\left(\sqrt{-z^{2} + 10z -1}\ + 1 - z\right)
\\[2mm]
y & = & {1 \over 4}\left(\sqrt{-z^{2} + 10z -1}\ - 1 + z\right)
\\[1mm]&&
\end{array}\right.
$$
Roots of $\displaystyle{-z^{2} + 10z - 1 = 0}$ are $\displaystyle{z_{\pm} = 5 \pm 2\sqrt{3}}$ such that
$$
5 - 2\sqrt{3}\quad <\ z\quad <\quad 5 + 2\sqrt{3}
$$
$$
\oint_{C}\vec{\rm F}\left(\vec{r}\right)\cdot{\rm d}\vec{r}
=
\int_{z_{-}}^{z_{+}}\left\lbrack%
{\rm F}_{x}\left(x\left(z\right),y\left(z\right),z\right)\ x'\left(z\right)
+
{\rm F}_{y}\left(x\left(z\right),y\left(z\right),z\right)\ y'\left(z\right)
+
{\rm F}_{z}\left(x\left(z\right),y\left(z\right),z\right)
\right\rbrack
{\rm d}z
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/471370",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Thinking through a Taylor error bound for arcsine In lecture, we went through solving a Taylor error bound for arcsine. I followed most of it except for where it talks about the odds divided by the evens divided by $2n+1$ gaining in accuracy by a factor of 1/10 for each successive term (see bolded sentence below), which comes from this part of the series:
$$ \sum_{x=1}^n \frac{1}{2n+1} \prod_{k=1}^n \frac{2n-1}{2n}$$
Where does the last factor of 1/10 come from? This is not obvious to me from the above expression and I am wondering whether someone could help me see what's going on.
Lecture notes:
$$\arcsin(x) = x + \sum \limits_{n=1}^{\infty} \frac{ 1 \cdot 3 \cdot 5 \cdots (2n-1) } { 2 \cdot 4 \cdot 6 \cdots (2n) } \cdot \frac{ x^{2n+1} }{ 2n+1 } $$
Asuming we have the Taylor expansion for arcsin(x),
$$\arcsin(\frac{1}{10}) = \frac{1}{10} + \frac{1}{2} \cdot \frac{1}{3} \cdot (\frac{1}{10})^3 + \frac{3}{8}\cdot \frac{1}{5} \cdot (\frac{1}{10})^5 + \frac{5}{16}\cdot\frac{1}{7}\cdot(\frac{1}{10})^7 + \cdots + E_N$$
The Taylor error bound is
$$E_N < \frac{C}{(N+1)!} (\frac{1}{10})^{N+1} $$
$$\left( \frac{d}{dx} \right)^{N+1} \arcsin(x) < C$$
$$0 \le x \le \frac{1}{10}$$
$C$ is an upper bound for the $N+1$st derivative of arcsin(x) for all $x$ between 0 and 1/10.
Even though we don't have a good bound for the $N+1$st derivative, if we look at the terms in this series, we can make a good guess.
Each step we gain 1/100 fold increase in accuracy because of $\left( \frac{1}{10} \right)^n$
If we look at the coefficients, the $2n+1$ and the product of odds over the product of evens, then we're picking up another factor of 10 in the denominator. We claim that $a_{n+2}$, the next term in the series, is less than the previous term, $a_n$, divided by 1000.
We're picking up three decimal places of accuracy with each subsequent term. That means that if we want to get within $10^{-10}$, it's going to suffice choose $N$ bigger than or equal to 7. The first four terms suffice to approximate $\arcsin(\frac{1}{10})$ within $10^{-10}$
$$N \ge 7$$
| The assertion in the notes is, roughly speaking, false.
Let $\arcsin x=\sum a_{2n+1} x^{2n+1}$. Then
$$\lim_{n\to\infty}\frac{a_{2n+3}}{a_{2n+1}}=1.$$
More informally, in the long run we are picking up nothing "in the denominator." In the case $x=\frac{1}{10}$, for large $n$, almost all the gain in accuracy comes from the extra $\frac{1}{10^2}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/473983",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Find the sum of the series $1^2-2^2+3^2-4^2+...-(2n)^2$ Find the sum of the series $$1^2-2^2+3^2-4^2+...-(2n)^2$$
I tried rewriting it as $$\sum_{r=1}^{2n}-1^{n+1}(r^2)$$ but it didn't help.
Also, looked at re-arranging as $$1^2+3^2+5^2+7^2+...+(2n-1)^2$$ and $$-2^2-4-6^2-8^2-...-(2n)^2$$
Still couldn't get to the given answer of $-n(2n+1)$
| Hint 1: We have $$\sum_{k=1}^nk^2=\frac{1}{6}n(n+1)(2n+1).$$
Hint 2: Note that
$$1^2-2^2+3^2-4^2+\cdots-(2n)^2=1^2+2^2+\cdots+(2n)^2-2(2^2+4^2+\cdots+(2n)^2)$$
$$
=1^2+2^2+\cdots+n^2-2^3(1^2+2^2+\cdots+n^2)
=\sum_{k=1}^{2n}k^2-8\sum_{k=1}^{n}k^2.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/474980",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Determinant of a $2 \times 2$ block matrix $\textbf{Problem}$: Let a $2n \times 2n$ matrix be given in the form $M=\left[ {\begin{array}{cc}
A & B \\
C & D \\
\end{array} } \right]$, where each block is an $n \times n$ matrix. Suppose that $A$ is invertible and that $AC=CA$. Use block multiplication to prove that $\det M= \det(AD-CB)$. Give an example to show that this formula need not hold if $AC \neq CA$
$\textbf{Proof}$: Let $A,B,C,D,X \in \textbf{M}_n(K)$ such that $A+BX$ is invertible.
For all $Y \in \textbf{M}_n(K)$, we have:
$$\left[ {\begin{array}{cc}
I_n & 0 \\
Y & I_n \\
\end{array} } \right] \left[ {\begin{array}{cc}
A & B \\
C & D \\
\end{array} } \right] \left[ {\begin{array}{cc}
I_n & 0 \\
X & I_n \\
\end{array} } \right]= \left[ {\begin{array}{cc}
A+BX & B \\
YA+C+(YB+D)X & YB+D \\
\end{array} } \right].$$
Let $Y=-(C+DX)(A+BX)^{-1}$. Hence:
$$YA+C+(YB+D)X=Y(A+BX)+(C+DX)=0.$$
Since $\det\left[ {\begin{array}{cc}
I_n & 0 \\
Y & I_n \\
\end{array} } \right]= \det\left[ {\begin{array}{cc}
I_n & 0 \\
X & I_n \\
\end{array} } \right]= (\det(I_n))^2=1$, we can conclude that:
\begin{align*}
\det\left[ {\begin{array}{cc}
A & B \\
C & D \\
\end{array} } \right]&=\det\left[ {\begin{array}{cc}
A+BX & B \\
0 & YB+D \\
\end{array} } \right]\\
&= \det(A+BX)\det(-(C+DX)(A+BX)^{-1}B+D).
\end{align*}
In particular for $X=0$, we have:
\begin{align*}
\det\left[ {\begin{array}{cc}
A & B \\
C & D \\
\end{array} } \right]&=\det(A)\det(-CA^{-1}B+D)=\det(-ACA^{-1}B+AD) \\
&=\det(-CAA^{-1}B+AD)=\det(AD-CB).
\end{align*}
I just wanted someone to verify my proof and help me with the second part of this question.
Thank you in advance
| Your proof seems fine to me.
As for a counterexample, consider
$$A = \begin{bmatrix}
2 & 0 & 1 & 0 \\
0 & 1 & 2 & 1 \\
2 & 1 & 1 & 1 \\
1 & 1 & 1 & 2
\end{bmatrix}.$$
In one hand, $\det A = -4$ (check here), and in the other hand, $\det (A_{11} A_{22} - A_{21}A_{12}) = 0$ (check here).
| {
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"timestamp": "2023-03-29T00:00:00",
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$\tan A + \tan B + \tan C = \tan A\tan B\tan C\,$ in a triangle I want to prove this (where each angle may be negative or greater than $180^\circ$):
When $A+B+C = 180^\circ$ \begin{equation*} \tan A + \tan B + \tan C = \tan A\:\tan B\:\tan C. \end{equation*}
We know that
\begin{equation*}\tan(A+B) = \frac{\tan A+\tan B}{1-\tan A\tan B}\end{equation*} and that \begin{equation*}\text{and that}~A+B = 180^\circ-C.\end{equation*}
Therefore $\tan(A+B) = -\tan C.$
From here, I got stuck.
| Here is a geometric proof, for the case that all three angles are acute:
$QRUV$ are collinear because $B+90^\circ+(90^\circ-B)=180^\circ$.
$STV$ are collinear because $A+B+C=180^\circ$, so $\angle QSV=\angle UTV=C$.
Similar triangles $\triangle PQR\sim\triangle TRS$ and $\triangle RTU \sim \triangle SRQ$ give $\displaystyle
\frac{QP}{RQ} = \frac{RT}{SR} = \frac{TU}{RQ}$, and therefore $TU=QP=1$.
Then,
$$\begin{align}& \tan A + \tan B + \tan C = QR+RU+UV = QV \\
&= QP \frac{QR}{QP}\, \frac{QS}{QR} \, \frac{QV}{QS} = 1 \cdot \tan(A) \tan(B) \tan(C) \end{align}$$
When one of the angles is obtuse, let it (without loss of generality) be $C$. Then a similar diagram can be drawn, except that $V$ is to the left of $Q$, and $UV$, $QV$ count as negative lengths.
| {
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Solve $x+y+z = x^3 + y^3 + z^3 = 8$ in $\mathbb{Z}$ Solve $x+y+z = x^3 + y^3 + z^3 = 8$ in $\mathbb{Z}$
First I tried to transform this equation, substituting $x = 8-y-z$. So I end up with:
$$x^3 + y^3 + z^3 = 8$$
$$(8-y-z)^3 + y^3 + z^3 = 8$$
Using Wolfram Alpha I expanded this equation and tried to factorize it so finally I got:
$$(z-8)(y^2 + y(z-8) - 8z) = 168$$
$z \in \mathbb{Z}$, which implies $z-8$ is an integer implying that the second term is also an integer.
168 has 16 positive divisors and 16 negative, which means there are 32 distinct cases to be looked, which is a painful work and made me abandon this method in search of a better and easier method.
Second method I tried is Newton Identities.
We want to find the values for the roots $x,y,z$ of the cubic function: $f(x) = x^3 + bx^2 + cx + d$
From the condition we have the following equation:
$$x + y + z = 8 = s_1\text{ and } x^3 + y^3 + z^3 = 8 = s_3$$
Using Newton Identities we obtain the following relations:
$$s_1 + b = 0 \implies b = -8$$
$$s_2 + bs_1 + 2c = 0$$
$$s_2 - 64 + 2c = 0 \implies c = \frac{64 - s_2}{2}$$
$$s_3 + bs_2 + cs_1 + 3d = 0$$
$$8 - 8s_2 + \frac{64 - s_2}{2} \times 8 + 3d = 0$$
$$8 - 8s_2 + 256 - 4s_2 + 3d = 0$$
$$3d - 12s_2 + 264 = 0$$
$$d - 4s_2 + 88 = 0 \implies s_2 = \frac{88+d}{4}$$
Substituting back we have:
$$c = \frac{64 - s_2}{2} = \frac{64 - \frac{88+d}{4}}{2} = \frac{\frac{256 - 88 - d}{4}}{2} = \frac{168 - d}{8}$$
And I'm stuck here, because I know only the ratio between $c$ and $d$, which is a consequence of not knowing the value $s_2 = x^2 + y^2 + z^2$
| Hint: Calculate $(x-8)(y-8)(z-8)$.
There are several ways to do the calculation. Working from the last equation in the first method, observe that the second term can be factorized into $(y-8)(y+z)= -(y-8)(x-8)$, which gives us
$(x-8)(y-8)(z-8) = -168 $
Use this, along with the factors of 168 and that $x-8+y-8+z-8 = -16$ to greatly restrict the cases you have to look at. For example, as Mark comments, since the sum of 3 numbers is even, so either one of them is even, all all three of them are.
Now, if you don't want an 'Wolfram Alpha' expansion of the terms, you could do the expansion yourself, it's not that hard. The way that I arrived at the calculation used the fact that $a^3 + b^3 + c^3 - 3abc = (a+b+c)( (a+b+c)^2 - 3ab -3bc - 3ca) $
Note that since you want to solve this as a Diophantine equation, using algebraic methods can only get you so far. At some point in time, you must change it into a 'number theoretic way', like analyzing the factors, or gcd of terms.
| {
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How to apply polynomial long division to $x^3/(x^2+3x+2)$ So I have a basic question. I needed to simplify this expression
$$\frac{x^{3}}{x^{2} +3x +2}$$
I was to do a polynomial long division in order to simplify it, but I still do not understand how they reached this next expression (I know it can be more simplified)
$$\frac{x^{3}}{x^{2} +3x +2}= x-3 + \frac{7x+6}{x^{2} +3x +2}$$
What was divided by what?
Any help would be appreciated. Thanks in advance!!
| $$\frac{x^3}{x^2+3x+2} = \frac{x(x^2+3x+2)-3x^2-2x}{x^2+3x+2} = x - \frac{3(x^2+3x+2)-7x-6}{x^2+3x+2}=x-3+\frac{7x+6}{x^2+3x+2}.$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Creating a degree $n$ Taylor polynomial for $\sqrt{1+x}$ I have been asked to produce a general formula for the degree $n$ Taylor polynomial for $\sqrt{1+x}$ using a=0 as the point of approximation.
Given that $p_n(x)=f(a)+(x-a)f^\prime(a)+\frac{(x-a)^2}{2!}f^{(2)}(a)+...+\frac{(x-a)^n}{n!}f^{(n)}(a)=\sum_{j=0}^{n}\frac{(x-a)^j}{j!}f^{(j)}(a)$
$f(x)=\sqrt{1+x}=(1+x)^{1/2} \\
f^\prime (x)=\left ( \frac{1}{2} \right )(1+x)^{-1/2} \\
f^{(2)}(x)=\left ( \frac{1}{2}\right )\left ( \frac{-1}{2} \right )(1+x)^{-3/2} \\
f^{(3)}(x)=\left ( \frac{1}{2}\right )\left ( \frac{-1}{2} \right )\left ( \frac{-3}{2} \right )(1+x)^{-5/2} \\
f^{(4)}(x)=\left ( \frac{1}{2}\right )\left ( \frac{-1}{2} \right )\left ( \frac{-3}{2} \right )\left ( \frac{-5}{2} \right )(1+x)^{-7/2} \\
f^{(j)}(x)=(-1)^{j+1}\left ( \frac{1}{2^{j}} \right )(1\cdot 1\cdot 3\cdot 5\cdot ... \cdot 2j-3)(1+x)^{(2j-1)/2} \\
f^{(j)}(0)=(-1)^{j+1}\left ( \frac{1}{2^{j}} \right )(1\cdot 1\cdot 3\cdot 5\cdot ... \cdot 2j-3)$
I'm thinking I have calculated the j-th derivative correctly, however I can't seem to figure out how to simplify it and the rest of the steps into a compact summation.
Added: It's been pointed out to me that the $(1\cdot 1\cdot 3\cdot 5\cdot ... \cdot 2j-3)$ above should be $\prod_{i=0}^{j-1}2i-1$
Thus $f^{(j)}(0)=(-1)^{j+1}\left ( \frac{1}{2^{j}} \right )\prod_{i=0}^{j-1}2i-1$.
If anyone can give me a hint on how to simplify this further it would be appreciated.
| Note that $$1\cdot 3\cdot 5 \cdots (2j - 3) = \dfrac{(2j-3)!}{2\cdot 4 \cdot 6 \cdots (2j -2)} = \dfrac{(2j-3)!}{2^{j-1}\cdot 1\cdot 2 \cdot 3 \cdots (j -1)} = 2^{1-j}\dfrac{(2j-3)!}{(j-1)!}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/482273",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
How find this integral limit
Find this limit $$\lim_{n\to\infty}\dfrac{\displaystyle\int_{0}^{\frac{\pi}{2}}x\left(\dfrac{\sin{(nx)}}{\sin{x}}\right)^4dx}{n^2}.$$
This problem is post my teacher(tian275461, he is very popular in China), and he told me this problem answer is $\ln{2}$, but I can't prove it. Thank you.
| With the substitution $y = nx$, we have
\begin{align}
I_n &= \int_0^{n\pi/2} \frac{\sin^4 y}{y^3} \left(\frac{\frac{y}{n}}{\sin \frac{y}{n}}\right)^4 \mathrm{d} y \\
&= \underbrace{\int_0^{\sqrt{n}} \frac{\sin^4 y}{y^3} \left(\frac{\frac{y}{n}}{\sin \frac{y}{n}}\right)^4 \mathrm{d} y}_{I_{n,1}} +
\underbrace{\int_{\sqrt{n}}^{n\pi/2} \frac{\sin^4 y}{y^3} \left(\frac{\frac{y}{n}}{\sin \frac{y}{n}}\right)^4 \mathrm{d} y}_{I_{n,2}}.
\end{align}
Since $u \mapsto \frac{u}{\sin u}$ is increasing on $(0, \frac{\pi}{2}]$, we have
$$1 \le \left(\frac{\frac{y}{n}}{\sin \frac{y}{n}}\right)^4 \le \left(\frac{\frac{1}{\sqrt{n}}}{\sin \frac{1}{\sqrt{n}}}\right)^4,
\ \forall 0 < y \le \sqrt{n}.$$
Thus, we have
$$\int_0^{\sqrt{n}} \frac{\sin^4 y}{y^3}\mathrm{d} y \le I_{n,1} \le \left(\frac{\frac{1}{\sqrt{n}}}{\sin \frac{1}{\sqrt{n}}}\right)^4 \int_0^{\sqrt{n}} \frac{\sin^4 y}{y^3} \mathrm{d} y.$$
By the squeeze theorem, we have $\lim_{n\to \infty} I_{n,1} = \int_0^{\infty} \frac{\sin^4 y}{y^3}\mathrm{d} y = \ln 2$
(see e.g. Laplace transform:$\int_0^\infty \frac{\sin^4 x}{x^3} \, dx $).
Also, we have
$$I_{n,2} \le \int_{\sqrt{n}}^{n\pi/2} \frac{1}{\sqrt{n}^3} \left(\frac{\pi}{2}\right)^4 \mathrm{d} y$$
and thus $\lim_{n\to \infty} I_{n,2} = 0$.
Thus, $\lim_{n\to \infty} I_n = \ln 2$.
| {
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"source": "stackexchange",
"question_score": "9",
"answer_count": 2,
"answer_id": 1
} |
Ancient Babylonian problem: solve the system $x + y = 50$, $x^2 + y^2 + (x - y)^2 = 1400$ $x + y = 50, x^2 + y^2 + (x - y)^2 = 1400$. [Hint: Subtract the square of the first equation from twice the second equation to get a quadratic in $x - y$.] I have gotten it reduced to $x^2 + y^2 -3xy = 550$ but am not sure how to get it in terms of $x - y$ to solve it.
| As you have now figured out the answer on your own, I'll post the way I was hinting at in a comment.
I would say your book is pointing you towards the hard way. It's easier just to solve for $y$ in the first equation and substitute that into in the second, giving you a quadratic in $x$, trivially solved by inspection or the quadratic equation.
$$
x + y = 50 \\
y = 50 - x \\
--- \\
x^2 + y^2 + (x - y)^2 = 1400 \\
x^2 + (50-x)^2 + (x - (50 - x))^2 = 1400 \\
x^2 + x^2 - 100x + 50^2 + 4x^2 - 200x + 50^2 = 1400 \\
6x^2 - 300x + 3600 = 0 \\
x^2 - 50x + 600 = 0 \\
(x - 20)(x - 30) = 0 \\
x = 20~or~30\\
y = 50 - x\\
y = 30~or~20
$$
So x and y are 20 and 30, in either combination.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
How prove this inequality $(\sum a_{1}^{1.5})^2\ge \sum a_{1}\sum a_{1}a_{2}$ Now my question let $a_{1},a_{2},\cdots,a_{n}$ are positive numbers,and $a_{n+i}=a_{i},i=1,2,\cdots$,show that
$$(\sum a_{1}^{1.5})^2\ge \sum a_{1}\sum a_{1}a_{2}$$
my teacher (tian275461) have prove this
$$(a^{1.5}+b^{1.5}+c^{1.5})^2\ge (a+b+c)(ab+bc+ac)$$
He methods:let $a\longrightarrow a^2,b\longrightarrow b^2,c\longrightarrow c^2$
then
$$\Longleftrightarrow (a^3+b^3+c^3)^2\ge (a^2+b^2+c^2)(a^2b^2+c^2a^2+b^2c^2)$$
$$\Longleftrightarrow(\sum a^3)^2\ge \sum a^2\sum a^2b^2$$
note
$$(\sum a^3)^2=\sum a^2\sum a^4-\sum b^2c^2(b-c)^2$$
$$\Longleftrightarrow \sum a^2\left(\sum a^4-\sum a^2b^2\right)-\sum b^2c^2(b-c)^2\ge 0 $$
$$\Longleftrightarrow \dfrac{1}{2}\sum a^2\sum(b^2-c^2)^2-\sum b^2c^2(b-c)^2\ge 0$$
$$\Longleftrightarrow \dfrac{1}{2}(b-c)^2 \left(\sum a^2\sum (b+c)^2-2\sum b^2c^2\right) \ge 0$$
it suffices to show that
$$\sum b^2\sum (b+c)^2-2\sum b^2c^2\ge 0$$
and note that
$$\sum b^2\sum (b+c)^2-2\sum b^2c^2=2\sum a^4+2\sum a^3b+2\sum a^3c+2\sum a^2b^2+2\sum a^2bc\ge 0$$
for n=4,it only show that
$$(a^3+b^3+c^3+d^3)^2\ge (a^2+b^2+c^2+d^2)(a^2b^2+b^2c^2+c^2d^2+d^2a^2)$$
| The conjecture is false.
(See my previous post for sufficient extra conditions such that the conjecture holds.)
Here is a counterexample. Let n even (n odd works as well).
Let $a_1=a_2 = \dots = a_{n/2} = A$ and $a_{n/2 +1 } = a_{n/2 +2 } = \dots = a_{n} = B$. Let, without loss of generality, $B<A$.
Then the conjecture is
$$
(\sum a_{1}^{1.5})^2 - \sum a_{1}\sum a_{1}a_{2} = \\
\frac{n^2}{4}(A^{1.5} +B^{1.5})^2 - \frac{n}2 (A+B) ((\frac{n}2-1)(A^2+B^2) + 2 AB) \ge 0
$$
Let $B = x^2\cdot A$, (where $0<x<1$), then this transforms into
$$
n(1 +x^{3})^2 - (1+x^2) ((n-2)(1+x^4) + 4 x^2) \ge 0
$$
Expanding this gives
$$
n(2x^3 - x^2 - x^4) - 2 x^2 - 2 x^4 + 2 x^6 + 2 \ge 0
$$
Now the factor of the leading $n$ is always negative: $2x^3 - x^2 - x^4= -( x - x^2)^2 < 0 $ for $0<x<1$. For the remaining sum terms, $2 > - 2 x^2 - 2 x^4 + 2 x^6 + 2 >0$ for $0<x<1$.
Hence the conjecture fails for large enough $n$, namely
$$
n > n_0 = \frac{- 2 x^2 - 2 x^4 + 2 x^6 + 2 }{- (2x^3 - x^2 - x^4) }
$$
$
\Box
$
The following plots illustrate the magnitude of $n_0$ for two ranges of $x$. From those plots, one could conjecture, if my example is indeed the worst case, that the proposed inequality holds for all $n<16$. ($n=3$ and $n=4$ are proven.)
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 2,
"answer_id": 0
} |
Sums of rising factorial powers Doodling in wolfram, I found that
$$
\sum^{k}_{n=1}1=k
$$
The formula is pretty obvious, but then you get
$$
\sum^{k}_{n=1}n=\frac{k(k+1)}{2}
$$
That is a very well known formula, but then it gets interesting when you calculate
$$
\sum^{k}_{n=1}n(n+1)=\frac{k(k+1)(k+2)}{3}\\
\sum^{k}_{n=1}n(n+1)(n+2)=\frac{k(k+1)(k+2)(k+3)}{4}
$$
And so on. There is an obvious pattern that I really doubt is a coincidence, but I have no idea how to prove it in the general case. Any ideas?
| Partial sums of sequences $a_n$ that can be expressed as polynomials in $n$ are easily found using discrete calculus.
We start with the discrete version of the Fundamental Theorem of Integral Calculus:
\begin{align*}
\sum\limits_{n=1}^k a_n &= \sum\limits_{n=1}^k (\Delta b)_n \\
&= (b_2 - b_1)+(b_3 - b_2)+ \ldots + (b_k - b_{k-1}) + (b_{k+1} - b_k) \\
&= b_{k+1} - b_1
\end{align*}
where $(\Delta b)_n = b_{n+1} - b_n$ is the forward difference. Finding the partial sum has now been reduced to finding a sequence $b_n$ such that $(\Delta b)_n = a_n$.
We will find $b$, the antiderivative of $a$, using falling powers, which are defined by
$$
n^{\underline{k}} = n(n-1)(n-2)\ldots (n-k+1)
$$
where $k$ is an integer and, by a second definition,
$n^{\underline{0}}=1$. For example
$$
n^{\underline{3}} = n(n-1)(n-2).
$$
We now need one more result: the discrete derivative of $n^{\underline{k}}$ is given by
\begin{align*}
\Delta n^{\underline{k}} &= (n+1)^{\underline{k}} - n^{\underline{k}} \\
&= (n+1)n^{\underline{k-1}} - n^{\underline{k-1}}(n-k+1) \\
&= kn^{\underline{k-1}}
\end{align*}
Let's now find the partial sum for a particular case:
\begin{align}
\sum^{k}_{n=1}n(n+1)(n+2) &= \sum^{k}_{n=1} (n+2)^{\underline{3}} \\
&= \sum^{k}_{n=1} \Delta \left[\frac{1}{4} (n+2)^{\underline{4}}\right] \\
&= \frac{(k+3)(k+2)(k+1)k}{4} - \require{cancel}\cancelto{0}{\frac{(1+2)(1+1)(1-0)(1-1)}{4}}
\end{align}
The general case:
\begin{align}
\sum^{k}_{n=1} (n+p)^{\underline{p+1}} &= \sum^{k}_{n=1} \Delta \left[\frac{1}{p+2} (n+p)^{\underline{p+2}}\right] \\
&= \frac{(k+1+p)(k+p)\ldots [(k+1+p)-(p+2)+1)]}{p+2} \\
&= \frac{(k+1+p)(k+p)\ldots k}{p+2}
\end{align}
where $p>0$ is an integer.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/483621",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 5,
"answer_id": 0
} |
sum of legs of inscribed right triangle Consider a right and isoceles triangle ABC inscribed in a circle such as its hypotenuse forms the diameter AB of the circle (the right vertex is thus at the "apex" of the circle). If we procede to an infinitesimal displacement of the "apex" along the circumference $\ ds$ with subsequent horizontal and vertical displacements $\ dx$ and $\ dy$ , how does the function defined as $\ D(x)=a(x)+b(x)$ (with a and b the lengths of the two legs of the triangle) change with respect to the change in abscissa $\ x$?
A crude depiction:
| Let's assume your circle to be the unit circle. Then you have
\begin{align*}
y &= \sqrt{1-x^2} \\
y^2 &= 1-x^2 \\
a(x) &= \sqrt{(1+x)^2+y^2} = \sqrt{1+2x+x^2+1-x^2} = \sqrt{2\left(1+x\right)} \\
b(x) &= \sqrt{(1-x)^2+y^2} = \sqrt{1-2x+x^2+1-x^2} = \sqrt{2\left(1-x\right)} \\
D(x) &= a(x)+b(x) = \sqrt{2\left(1+x\right)}+\sqrt{2\left(1-x\right)}
\end{align*}
I don't see a way to make this expression much easier, and neither does Wolfram Alpha. But you were writing about an infinitesimal change, so let's continue:
\begin{align*}
D'(x) &= \frac{\mathrm d\,D}{\mathrm dx} = \frac{\sqrt{1-x}-\sqrt{1+x}}{\sqrt{2-2x^2}} \\
D'(0) &= 0 \\
D''(x) &= \frac{\mathrm d^2\,D}{\mathrm dx^2} = -\frac1{2\sqrt2}\left(
\left(1+x\right)^{-\frac32}+
\left(1-x\right)^{-\frac32}
\right) \\
D''(0) &= -\frac{1}{\sqrt2} \\
D'''(0) &= 0
\end{align*}
This corresponds to the beginning of a series expansion:
\begin{align*}
D(x) &= D(0) + \frac{D''(0)}{2}x^2 + O\left(x^4\right) = 2\sqrt2 + \frac{1}{2\sqrt2}x^2 + O\left(x^4\right)
\end{align*}
If you want more terms, ask Wolfram Alpha. You can also obtain a series in terms of $\mathrm ds$ instead of $\mathrm dx$, simply by substituting $x=\sin s$. But since the first difference is in the $x^4$ (resp. $s^4$) term, the above approximation won't notice the difference.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/485089",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Electric field of a spherical shell I hope this question is appropriate for this site, if not, just leave a comment and I will delete.
I am interested in knowing how to derive the electric field due to a spherical shell by Coulomb's law without using double integrals or Gauss Law.
Relevant equations are -- Coulomb's law for electric field and the volume of a sphere:
$\vec{E} = \frac{1}{4\pi \epsilon_0}\frac{Q}{r^2}\hat{r}$, where $Q =$
charge, $r=$ distance.
$V = \frac{4}{3}\pi r^3.$
From my book, I know that the spherical shell can be considered as a collection of rings piled one above the other but with each pile of rings the radius gets smaller and smaller.
I am not interested in the final formula, just the derivation of it. Thank you!
| We assume that the sphere of radius $R$ centered at 0. Let us assume an observation
point $o$ above the north pole of the sphere (by symmetry this should provide a
good answer). We consider rings from the bottom up to the north polo $(0,0,R)$.
The ring at a high $z$, $-R \le z \le R$ has a radius
$\rho=\sqrt{R^2 - z^2}$. We prefer to see the problem as a function
of the polar angle from $-\pi/2$ to $\pi/2$. We have that $\rho=R \sin \theta$ with $\theta$
the polar angle. It is well known that for a ring with uniform charge density
$\sigma$ , radius $r$ and an observation point in the axis of the ring at a
distance $d$ from the center (in the direction of the axis of the ring) produces the field
\begin{eqnarray*}
E(d,\rho)= \frac{\sigma \rho \, d}{2 \epsilon_0(\rho^2 + d^2)^{3/2}}.
\end{eqnarray*}
The distance between the observation point $o$ and the ring at $z$
height is $d=o-z$, and $z=R \cos \theta$, then we find
\begin{eqnarray*}
E(\theta)=
\frac{\sigma (o-R \cos \theta)
R \sin \theta}{2 \epsilon_0(R^2 \sin^2 \theta + (o-R \cos \theta)^2)^{3/2}}.
\end{eqnarray*}
We need to integrate along $\theta$ between $0$ and $\pi$.
Along the polar axis the element of integration is $d \ell = R \, d \theta$,
so we will need to multiply by $R \, d \theta$.
\begin{eqnarray*}
E = \frac{\sigma}{2 \epsilon_0} \int_{-\pi/2}^{\pi/2}
\frac{(o-R \cos \theta)
R^2 \sin \theta}{(R^2 \sin^2 \theta + (o-R \cos \theta)^2)^{3/2}}
d \theta.
\end{eqnarray*}
Let us perform the following substitution
\begin{eqnarray*}
u= \cos \theta \quad , \quad du=- \sin \theta d \theta \\
\theta=0 \implies u = 1\\
\theta=\pi \implies u = -1 ,
\end{eqnarray*}
then
\begin{eqnarray*}
E = \frac{\sigma R^2}{2 \epsilon_0} \int_{-1}^1
\frac{o - u R}{(R^2 + o^2 - 2 o R u)^{3/2}} du
\end{eqnarray*}
We split the integrand in two fractions (forget the coefficient for now).
\begin{eqnarray*}
\int_{-1}^1 \frac{o} {(R^2 + o^2 - 2 o R u)^{3/2}} du
\quad \mathrm{and} \quad
-\int_{-1}^1 \frac{u R} {(R^2 + o^2 - 2 o R u)^{3/2}} du
\end{eqnarray*}
For the first integral, let us make $x=R^2 + o^2 - 2 o R u$,
then $dx=-2 o R du$, and in terms of $x$,
\begin{eqnarray*}
-\frac{1}{2 R} \int \frac{dx}{x^{3/2}} =
\frac{1}{ R \sqrt{x} },
\end{eqnarray*}
Then the first integral is
\begin{eqnarray*}
\int_{-1}^1 \frac{o} {(R^2 + o^2 - 2 o R u)^{3/2}} du &=&
\left . \frac{1}{R \sqrt{R^2 + o^2 - 2 o R u}} \right |_0^1 \\
\end{eqnarray*}
Let us do the second integral usig integration by parts.
We write
\begin{eqnarray*}
-\int_{-1}^1 \frac{u R} {(R^2 + o^2 - 2 o R u)^{3/2}} du &=&
-\frac{u }{o \sqrt{R^2 + o^2 - 2 o R u}} \\
&& + \int
\frac{1}{o \sqrt{R^2 + o^2 - 2 o R u}} du
\end{eqnarray*}
Now,
\begin{eqnarray*}
\int \frac{1}{o \sqrt{R^2 + o^2 - 2 o R u}} du =
-\frac{1}{o^2 R} \sqrt{R^2 + o^2 - 2 o R u},
\end{eqnarray*}
then
\begin{eqnarray*}
\int_{-1}^1 \frac{u R} {(R^2 + o^2 - 2 o R u)^{3/2}} du &=&
\frac{u }{o \sqrt{R^2 + o^2 - 2 o R u}}
+ \frac{ \sqrt{R^2 + o^2 - 2 o R u}}{o^2 R} \\
&=& \frac{ R^2 + o^2 - o R u}{o^2 R \sqrt{R^2 + o^2 - 2 o R u}}
\end{eqnarray*}
Putting the first and the second integrals back together we get
\begin{eqnarray*}
\frac{1}{R \sqrt{R^2 + o^2 - 2 o R u}}
- \frac{ R^2 + o^2 - o R u}{o^2 R \sqrt{R^2 + o^2 - 2 o R u}}
=
\frac{-R^2 + o Ru}{o^2 R \sqrt{R^2 + o^2 - 2 o R u}}
\end{eqnarray*}
Hence we found that
\begin{eqnarray*}
\int \frac{o - u R}{(R^2 + o^2 - 2 o R u)^{3/2}} du
= \frac{o u - R}{o^2 \sqrt{o^2 - 2 o R u + R^2}},
\end{eqnarray*}
and so
\begin{eqnarray*}
\left .
\frac{o u - R}{o^2 \sqrt{o^2 - 2 o R u + R^2}} \right |_{-1}^1
&=& \frac{o - R}{o^2 \sqrt{o^2 - 2 o R + R^2}}
+ \frac{o + R}{o^2 \sqrt{o^2 + 2 o R + R^2}} \\
&=&
\frac{o - R}{o^2 |o - R|}
+ \frac{o + R}{o^2 |o + R|}
\end{eqnarray*}
\begin{eqnarray*}
E= \frac{\sigma R^2}{2 \epsilon_0}
\left [
\frac{o - R}{o^2 |o - R|}
+ \frac{o + R}{o^2 |o + R|}
\right ].
\end{eqnarray*}
That is
\begin{eqnarray*}
E = \left \{
\begin{array}{cc}
\frac{\sigma R^2}{o^2 \epsilon_0} & o > R \\
\\
0 & o < R
\end{array}
\right .
\end{eqnarray*}
but
\begin{eqnarray*}
\frac{\sigma R^2}{o^2 \epsilon_0}
= \frac{4 \pi \sigma R^2}{4 \pi o^2 \epsilon_0}
= \frac{Q}{4 \pi o^2 \epsilon_0}
\end{eqnarray*}
where $4 \pi R^2 \sigma$ is the total charge in the sphere.
Then
\begin{eqnarray*}
E = \left \{
\begin{array}{cc}
\frac{Q}{4 \pi o^2 \epsilon_0} & o > R \\
\\
0 & o < R
\end{array}
\right .
\end{eqnarray*}
What if $o=R$?
| {
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"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Integrating $\int_0^{\frac{\pi}{2}}\cos(2x)\sin(3x)\,dx$ How would I integrate the following
$\int_0^{\frac{\pi}{2}}\cos(2x)\sin(3x)\,dx$
I did the following
$\frac{1}{2}\int \left(\sin(5x)-\sin(x)\right)\,dx$
$\frac{1}{10}-\cos(5x)+cos(x)$
$\frac{1}{10}(0)+0-0=0$
But I am not sure If I did it right.
| Not quite. Using the identity
$$2 \cos{a}\sin{b} = \sin{(a + b)} - \sin{(a - b)}$$
we find that
\begin{align}
\int_0^{\pi/2} \cos{(2x)} \sin{(3x)} &= \frac 1 2 \int_0^{\pi/2} \sin{5x} - \sin{(-x)} \\
&= \frac 1 2 \int_0^{\pi/2} \sin{5x} + \sin{x} \\
&= \frac 1 2 \left(\frac{-1}{5} \cos{5x} - \cos{x}\right)_0^{\pi/2} \\
&= \frac 1 2 \left(\frac{-1}{5} \cos{\frac{5\pi}{2}} - \cos{\frac{\pi}{2}} - (\frac{-1}{5} \cos{0} - \cos{0})\right) \\
&= \frac 1 2 \left(0 - (\frac{-1}{5} - 1)\right) \\
&= \frac 3 5
\end{align}
| {
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"source": "stackexchange",
"question_score": "2",
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Solutions to $\sqrt{x}+y=6,x^2+y^2=90$ $$\begin{gather}
\sqrt{x}+y=6 \tag{1} \\
x^2 + y^2 = 90 \tag{2}
\end{gather}$$
WE have to solve for $x$ and $y$(Note that 9 is an obvious value of x) My friend asked me this question earlier today, which he said he had made up himself. However, he himself has no idea how to solve for the two variables. I used the method of substitution to finally obtain $x(x+1)=6(21-2y)$. Then I figured that both sides must be non-negative, and hence the value of $y$ has to be less than $11$. Then we test for values less than $11$ which makes $6(21-2y)$ a product of two consecutive numbers. WE get $3$ as a value of $y$, and also $7$. But putting $7$ in (1) does not work.
But there is a flaw in my reasoning. When I get to the 'consecutive' part, I start assuming $x$ and $y$ are integers. Since the person who asked me is not familiar with complex numbers, WE can assume that the solutions are real. However, I am interested in extending the values of $x$ and $y$ beyond the reals, if there are any.
NOTE: From (2), we get the value of $y^2$ in terms of $x$. Then we figure out the value of $x$ in (1) and then substitute $y$ for $90-x^2$, which gives us $x(x+1)$.
| So we have $$\sqrt{x}=6-y$$ which we substitute into the second equation to give $$(6-y)^4+y^2=90.$$ This is equivalent to finding the roots of the polynomial $$f(y):=(6-y)^4+y^2-90.$$ The Fundamental Theorem of Algebra implies this has exactly $4$ (not necessarily distinct) roots.
You mentioned one root, $3$, in your comment. We can find the other roots comptuationally (WolframAlpha); there's closed formulas for them:
$$7+\frac{\sqrt[3]{45+\sqrt{3054}}}{3^{2/3}}-\frac{7}{\sqrt[3]{3(45+\sqrt{3054})}}$$
$$7-\frac{(1+i\sqrt{3})\sqrt[3]{45+\sqrt{3054}}}{2 \times 3^{2/3}}-\frac{7(1-i\sqrt{3})}{2\sqrt[3]{3(45+\sqrt{3054}})}$$
$$7-\frac{(1-i\sqrt{3})\sqrt[3]{45+\sqrt{3054}}}{2 \times 3^{2/3}}-\frac{7(1+i\sqrt{3})}{2\sqrt[3]{3(45+\sqrt{3054}})}$$
Then we substitute them into $\sqrt{x}=6-y$ to get the $x$-values.
The real root listed above implies $\sqrt{x} \approx 6-8.1887$ which has no solution. Thus the solutions are $(9,3)$ and two other complex roots.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/487360",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
} |
Help me solve this polynomial problem I have to find $a$,$b$ and $c$ so that this expression is a polynomial:
$$\frac{x^3+ax^2+bx+c}{x-1} + \frac{x^3+bx^2+cx+a}{x-2} + \frac{x^3+cx^2+ax+b}{x-3}$$
Please give me hints. Thanks!
| Hint: $x^3 + ax^2 + bx + c$ must have a root at $1$. This is because $(x - 1)$ must be a divisor of $x^3 + ax^2 + bx + c$, or else there would be a term involving the quantity $\frac{1}{x - 1}$. In a similar manner, $x^3 + bx^2 + cx + a$ has a root at $2$ and $x^3 + cx^2 + ax + b$ has a root at $3$. This gives us three equations, the first of which is
$$1 + a + b + c = 0$$
Now use this to determine $a, b, c$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/491923",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Proof of Bézout's Theorem Gostaria de saber como provar usando divisibilidade o teorema de Bezout
$(a,b)=d\Longrightarrow \exists f,g\in\mathbb{Z^*}$ tal que $af+gb=d$
I'd like to know how to, using divisibility, prove Bézout's Theorem:
Given integers $a,\ b$, if $(a,b)=d$ then there exist $f,g\in\mathbb{Z^*}$ such that $af + gb=d$.
| First of all, it is only an implication in one sense, in English it is incorrectly given as an equivalence.
I very much like the constructive proof. Suppose without loss of generality $a \ge 0$ and $b > 0$, and write the obvious
\begin{equation*}
\begin{matrix}
a &=& a \cdot 1 &+& b \cdot 0\\
b &=& a \cdot 0 &+& b \cdot 1\\
\end{matrix}
\end{equation*}
Let us now start Euclid's algorithm: $a = b q_{1} +
r_{1}$, with $0 \le r_{1} < b$. We can extend the previous table to:
\begin{equation*}
\begin{matrix}
a &=& a \cdot 1 &+& b \cdot 0\\
b &=& a \cdot 0 &+& b \cdot 1\\
r_{1} &=& a \cdot 1 &+& b \cdot (-q_{1})\\
\end{matrix}
\end{equation*}
Let's continue with Euclidean divisions: $b = r_{1} q_{2} + r_{2}$, with $0 \le r_{2} < r_{1}$. Thus $r_{2} = b
- r_{1} q_{2}$. Let us use the last two lines of the last table to rewrite $r_{2}$ in terms of $a$ and $b$,
\begin{equation*}
\begin{matrix}
a &=& a \cdot 1 &+& b \cdot 0\\
b &=& a \cdot 0 &+& b \cdot 1\\
r_{1} &=& a \cdot 1 &+& b \cdot (-q_{1})\\
r_{2} &=& a \cdot u_{2} &+& b \cdot v_{2}\\
\end{matrix}
\end{equation*}
Here $u_{2} = -q_2$ and $v_{2} = 1 + q_1 q_2$. But the exact values of $u_{2}$ e $v_{2}$ are not important here, what counts is that they can be calculated in terms of the coeffients in the previous two lines of the table. At the end of the algorithm the last non-zero remainder will be the $\gcd$ of $a$ and $b$, and the table will look like
\begin{equation*}
\begin{matrix}
a &=& a \cdot 1 &+& b \cdot 0\\
b &=& a \cdot 0 &+& b \cdot 1\\
r_{1} &=& a \cdot 1 &+& b \cdot (-q_{1})\\
r_{2} &=& a \cdot u_{2} &+& b \cdot v_{2}\\
& & \dots & & \\
d &=& a \cdot u &+& b \cdot v\\
\end{matrix}
\end{equation*}
We have thus found the required linear combination.
An example: $a =
24$ and $b = 14$.
The Euclidean divisions are
\begin{equation}
\begin{matrix}
\mathbf{24} &=& \mathbf{14} \cdot 1 &+& \mathbf{10}\\
\mathbf{14} &=& \mathbf{10} \cdot 1 &+& \mathbf{4}\\
\mathbf{10} &=& \mathbf{4} \cdot 2 &+& \mathbf{2}\\
\mathbf{4} &=& \mathbf{2} \cdot 1 &+& 0\\
\end{matrix}
\end{equation}
so that the $\gcd$ is (surprise!) $2$. (Remainders are given in bold here.)
Computing as above
\begin{equation}
\begin{matrix}
\mathbf{24} &=& \mathbf{24} \cdot 1 &+& \mathbf{14} \cdot 0\\
\mathbf{14} &=& \mathbf{24} \cdot 0 &+& \mathbf{14} \cdot 1\\
\mathbf{10} &=& \mathbf{24} \cdot 1 &+& \mathbf{14} \cdot (-1)\\
\mathbf{4} &=& \mathbf{24} \cdot (-1) &+& \mathbf{14} \cdot 2\\
\mathbf{2} &=& \mathbf{24} \cdot 3 &+& \mathbf{14} \cdot (-5)\\
\end{matrix}
\end{equation}
PS Linguistic remark: this is a translation of my notes, which are written in Italian.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Find all real matrices $A$ such that $A^2 = \mathrm{diag}(1,1,2,3,5,8,13)$ Let $A \in \mathcal{M}_{7 \times 7} (\mathbb{R})$ such that
$$A^2= \begin{pmatrix} 1&0&0&0&0&0&0\\0&1&0&0&0&0&0\\0&0&2&0&0&0&0 \\ 0&0&0&3&0&0&0 \\ 0&0&0&0&5&0&0 \\ 0&0&0&0&0&8&0 \\ 0&0&0&0&0&0&13\end{pmatrix} $$
How many matrices, which satisfy this condition could you find?
My friend told to me, that the correct answer is infinity. But I can find only one:
$$A^2= \begin{pmatrix} 1&0&0&0&0&0&0\\0&1&0&0&0&0&0\\0&0&\sqrt{2}&0&0&0&0 \\ 0&0&0&\sqrt{3}&0&0&0 \\ 0&0&0&0&\sqrt{5}&0&0 \\ 0&0&0&0&0&\sqrt{8}&0 \\ 0&0&0&0&0&0&\sqrt{13} \end{pmatrix} $$
I will grateful for hints, how find any other examples. I suppose that $A$ should depend from parametr which don't govern on product $A \cdot A$.
| Hint: $$\begin{pmatrix} \cos \varphi & \sin \varphi & 0 & \ldots & 0 \\ \sin \varphi & -\cos \varphi & 0 & \ldots & 0 \\ 0 & 0& 0& \ldots & 0\\ \vdots & \vdots& \vdots& \ddots & \vdots\\ 0 & 0& 0& \ldots & 0 \end{pmatrix}^2 = ?$$
| {
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"url": "https://math.stackexchange.com/questions/494276",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Limit of $s_n=\frac{1}{\sqrt{n}}\left(1+\frac{1}{\sqrt{2}}+\cdots +\frac{1}{\sqrt{n}}\right)$ \begin{align*}S_n=\frac{1}{\sqrt{n}}\left(1+\frac{1}{\sqrt{2}}+\cdots +\frac{1}{\sqrt{n}}\right)\end{align*}
how to calculate the limit $s_n$?
\begin{align*}\lim_{n\to \infty } \, S_n\end{align*}
| \begin{align}
{1 \over \sqrt{n}}\,\sum_{k = 1}^{n}{1 \over \sqrt{k\,}}
&=
{1 \over \sqrt{n}}\,\sum_{k = 1}^{n}{1 \over \sqrt{n\xi_{k}\,}}\,n\Delta\xi
=
\sum_{k = 1}^{n}{1 \over \sqrt{\xi_{k}\,}}\,\Delta\xi
\sim
\int_{1/n}^{1}{{\rm d}\xi \over \xi^{1/2}}
=
\left.\vphantom{\LARGE A}\;2\xi^{1/2}\right\vert_{1/n}^{1}
\\[3mm]&=
2\left(1 - {1 \over \sqrt{n\,}}\right)
=
2 - {2 \over \sqrt{n\,}} \to \color{#ff0000}{\Large 2}
\quad\mbox{when}\quad
n \to \infty
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/495019",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
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"answer_id": 1
} |
Prove that $(x^2-x^3)(x^4-x) = \sqrt{5}$, where $x= \cos(2\pi/5)+i\sin(2\pi/5)$
Prove $(x^2-x^3)(x^4-x) = \sqrt{5}$ if $x= \cos(2\pi/5)+i\sin(2\pi/5)$.
I have tried it by substituting $x = \exp(2i\pi/5)$
but it is getting complicated.
| Replacing the earlier sequence of hints with a solution now. The old aswer is scratched.
We know that $0=x^5-1=(x-1)(x^4+x^3+x^2+x+1)$. As $x-1\neq0$ this implies that
$$1+x+x^2+x^3+x^4=0.\qquad(*)$$
From $x^5=1$ we also deduce that $x^4=x^{-1}$. This allows a rewrite:
$$
S=(x^2-x^3)(x^4-x)=(x^2-x^3)(x^{-1}-x)=(x-x^2)(1-x^2)=x-x^2-x^3+x^4.
$$
Let's square this. We get
$$
\begin{aligned}
S^2&=(x^2+x^4+x^6+x^8)\\
&+2(-x^3-x^4+x^5+x^5-x^6-x^7).\qquad(**)
\end{aligned}
$$
Because $x^8=x^5\cdot x^3=x^3$ and $x^6=x$, the first term above is
$$
(x^2+x^4+x^6+x^8)=x^2+x^4+x+x^3=-1,
$$
by equation $(*)$. The latter term in parens is similarly simplified to
$$
(-x^3-x^4+x^5+x^5-x^6-x^7)=2-(x^3+x^4+x+x^2)=3.
$$
Plugging these both into $(**)$ gives
$$
S^2=5.
$$
So we know that $S=\pm\sqrt5$, and the remaining task is to determine the sign.
From a picture of the unit circle in the complex plane, we see that all the terms in the r.h.s. of
$$
S=x-x^2-x^3+x^4
$$
have positive real parts. Therefore $S=\sqrt5$.
No trigonometry needed. Just $(*)$ and properties of roots of unity. Study Gauss sums for generalizations to primes $>5$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Find the constant term in the development of $(2x^2 + \frac{1}{4x})^{99}$ I am struggling with the following problem:
Find the constant term in the development of $(2x^2 + \frac{1}{4x})^{99}$
This is a combinatorics problem where I should find what the constant term is when $x^0$.
I have made the following developments:
$$\sum\limits_{k=0}^{n=99} \binom {99} {k} 2^{n-k}x^{2(n-k)} \times (\frac{1}{4x})^k = \sum\limits_{k=0}^{n=99} \binom {99} {k} 2^{99-k}x^{(198-2k)} \times \frac{1}{2^{2k}x^k}$$
This sum terms could be written together like this:
$$\sum\limits_{k=0}^{n=99} \binom {99} {k} \frac{2^{99-k}x^{(198-2k)}}{{2^{2k}x^k}} = \sum\limits_{k=0}^{n=99} \binom {99} {k} 2^{99-3k}x^{(198-3k)}$$
This gives me that $x^0$ when $198-3k = 0$ But this is the wrong answer....
My textbook says that the constant term will be when $3k-99 = 0$ which gives that $k = 33$ and $\binom {99} {33} \times 2^{99-198}$
Where did I make a mistake?
Thank you!
| You found that the constant term was the $k=66$ term, which for your setup is correct; you get a coefficient of
$$\binom{99}{66}2^{-99}\;,$$
which is equal to $$\binom{99}{99-66}2^{-99}=\binom{99}{33}2^{-99}\;.$$
Don’t forget that $\dbinom{n}k=\dbinom{n}{n-k}$ always.
If you switch the rôles of the two terms before you apply the binomial theorem, you get the result directly:
$$\begin{align*}
\left(2x^2+\frac1{4x}\right)^{99}&=\sum_{k=0}^{99}\binom{99}k(2x^2)^k\left(\frac1{4x}\right)^{99-k}\\
&=\sum_{k=0}^{99}\binom{99}k2^{k-2(99-k)}x^{2k-(99-k)}\\
&=\sum_{k=0}^{99}\binom{99}k2^{3k-198}x^{3k-99}\;,
\end{align*}$$
so the constant term is the $k=33$ term, whose coefficient is
$$\binom{99}{33}4^{-99}\;.$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
How to find $\lim\limits_{n\to\infty}\sum\limits_{j=1}^{n^2}\frac{n}{n^2+j^2}$ find the limit value
$$\lim_{n\to\infty}\sum_{j=1}^{n^2}\dfrac{n}{n^2+j^2}$$
this following is my methods:
let $$S_{n}=\sum_{j=1}^{n^2}\dfrac{n}{n^2+j^2}=\sum_{j=1}^{n^2}\dfrac{1}{1+\left(\dfrac{j}{n}\right)^2}\dfrac{1}{n}$$
since
$$\int_{\dfrac{j}{n}}^{\dfrac{j+1}{n}}\dfrac{dx}{1+x^2}<\dfrac{1}{1+\left(\dfrac{j}{n}\right)^2}\cdot\dfrac{1}{n}<\int_{\dfrac{j-1}{n}}^{\dfrac{j}{n}}\dfrac{dx}{1+x^2}$$
so
$$\int_{\dfrac{1}{n}}^{\dfrac{n^2+1}{n}}\dfrac{dx}{1+x^2}<S_{n}<\int_{0}^{n}\dfrac{dx}{1+x^2}$$
and note
$$\lim_{n\to\infty}\int_{\dfrac{1}{n}}^{\dfrac{n^2+1}{n}}\dfrac{dx}{1+x^2}=\lim_{n\to\infty}\int_{0}^{n}\dfrac{dx}{1+x^2}=\int_{0}^{infty}\dfrac{dx}{1+x^2}=\dfrac{\pi}{2}$$
so
$$\lim_{n\to\infty}\sum_{j=1}^{n^2}\dfrac{n}{n^2+j^2}=\dfrac{\pi}{2}$$
I think this problem have other nice methods? Thank you
and follow other methods
$$\lim_{n\to\infty}\sum_{j=1}^{n^2}\dfrac{n}{n^2+j^2}=\lim_{n\to\infty}\int_{0}^{n}\dfrac{1}{1+x^2}dx=\dfrac{\pi}{2}$$
But there is a book say This methods is wrong,why, and where is wrong? Thank you
| Here is a method in the same spirit as what you have done. The main difference is that we explicitly estimate the difference between your sum and $\int_0^n\frac{{\rm d}x}{1+x^2}$.
Since $f$ is monotonely decreasing we have (see for example integral test on Wikipedia)
$$\left|\sum_{j=1}^n \frac{n}{n^2+(j+nm)^2} - \int_{m}^{m+1}\frac{{\rm d}x}{1+x^2}\right| < \frac{1}{n(1 + (m+1)^2)}$$
Using this we can estimate the difference between your sum and the integral of $\frac{1}{1+x^2}$ over $[0,n]$, which can be written $\sum_{m=0}^{n-1}\int_0^1\frac{{\rm d}x}{1+(x+m)^2}$, as
$$\left|\sum_{j=1}^{n^2}\frac{n}{n^2 + j^2} - \int_0^n\frac{{\rm d}x}{1+x^2}\right| \leq \sum_{m=0}^{n-1} \left|\frac{1}{n}\sum_{j=1}^n \frac{1}{1+\left(\frac{j}{n}+m\right)^2} - \int_{0}^{1}\frac{{\rm d}x}{1+(x+m)^2}\right| \\\leq \frac{1}{n}\sum_{m=1}^{n} \frac{1}{1+m^2} \leq \frac{1}{n}\sum_{m=1}^\infty \frac{1}{1+m^2}$$
This last sum can be evaluated as $\frac{\pi\coth(\pi)-1}{2}$, however all we need for this argument is that it's finite. This can be proved for example by comparison to $\sum_{m=1}^\infty \frac{1}{m^2} = \frac{\pi^2}{6}$. Taking $n\to\infty$ the result follows
$$\lim_{n\to\infty}\sum_{j=1}^{n^2}\frac{n}{n^2 + j^2} = \lim_{n\to\infty}\int_0^n\frac{{\rm d}x}{1+x^2} = \frac{\pi}{2}$$
| {
"language": "en",
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"source": "stackexchange",
"question_score": "5",
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Is this Cayley table correctly computed? If so, is it correct way of presentation of this dihedral group? Is the following table for $D_4$ correct?
$$\begin{array}{c|c|c|c|c|c|c|c|c|}
* & 1 & g & g^2& g^3& f& fg & fg^2 & fg^3 \\ \hline
1 & 1 & g & g^2 & g^3 & f & fg & fg^2 & fg^3 \\ \hline
g & g & g^2 & g^3 & 1 & fg^3 & f & fg & fg^2 \\ \hline
g^2 & g^2 & g^3 & 1 & g & fg^2 & fg^3 & f & fg \\ \hline
g^3 & g^3 & 1 & g & g^2 & fg & fg^2 & fg^3 & f \\ \hline
f & f& fg & fg^2 & fg^3 & 1 & g& g^2 & g^3 \\ \hline
fg& fg& fg^2 & fg^3 & f& g^3 & 1 & g & g^2 \\ \hline
fg^2 & fg^2 & fg^3 & f & fg & g^2 & g^3 & 1 & g \\ \hline
fg^3 & fg^3 & f & fg& fg^2 & g & g^2 & g^3 & 1 \\ \hline
\end{array}$$
| A presentation of $D_4$ is $\langle g,f \mid g^4=1, f^2=1, gf=fg^{-1}\rangle$. So we have the $8$ elements $\lbrace 1,g,g^2,g^3,f,gf=fg^3,g^2f=fg^2,g^3f=fg\rbrace$, and the multiplication table follows from the relations - as far as I can see, your table is correct.
| {
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Sum of infinite series How to find sum of the following series
$$\frac{1}{6}+\frac{5}{6\cdot12}+\frac{5\cdot8}{6\cdot12\cdot18}+\frac{5\cdot8\cdot11}{6\cdot12\cdot18\cdot24}+\cdots={1\over 6} + \sum_{n=1}^\infty{\Pi_{i=1}^n{3i+2}\over (n+1)!6^{n+1}}$$
Please give me some hints. Thanks in advance.
| The general term of the series reads:
$$
\frac{1}{2}\frac{\prod_{k=1}^n (3k-1)}{\prod_{k=1}^n 6 k} = \frac{1}{2^{n+1}} \prod_{k=1}^n \left(1-\frac{1}{3k}\right) = \frac{1}{2^{n+1}} \frac{\left(2/3\right)_n}{n!} = \frac{1}{2^{n+1}} \frac{\Gamma\left(n+2/3\right)}{n! \cdot \Gamma\left(2/3\right)}
$$
The series thus reads:
$$
\mathcal{S}= \sum_{n=1}^\infty \frac{1}{2^{n+1}} \frac{\left(2/3\right)_n}{n!} = \frac{1}{\Gamma(2/3)} \sum_{n=1}^\infty \frac{1}{2^{n+1}} \frac{\Gamma\left(n+2/3\right)}{n!}
$$
Using Euler's integral:
$$
\Gamma\left(n+2/3\right) = \int_0^\infty t^{n-1/3} \mathrm{e}^{-t} \mathrm{d}t
$$
and interchanging the summation and the integration warranted by Tonelli's theorem:
$$\begin{eqnarray}
\mathcal{S} &=& \frac{1}{\Gamma(2/3)} \sum_{n=1}^\infty \frac{1}{2^{n+1}} \frac{\Gamma\left(n+2/3\right)}{n!} = \frac{1}{\Gamma(2/3)} \int_0^\infty \left(\sum_{n=1}^\infty \frac{t^{n-1/3}}{2^{n+1} n!} \right) \mathrm{e}^{-t} \mathrm{d}t \\ &=& \frac{1}{\Gamma(2/3)} \int_0^\infty \left(\frac{\exp\left(t/2\right)-1}{2 t^{1/3}} \right) \mathrm{e}^{-t} \mathrm{d}t \\ &=& \frac{1}{2\Gamma(2/3)} \left( \int_0^\infty t^{-1/3} \mathrm{e}^{-t/2} \mathrm{d}t - \int_0^\infty t^{-1/3} \mathrm{e}^{-t} \mathrm{d}t \right) \\
&=& \frac{1}{2\Gamma(2/3)} \left( 2^{2/3} \Gamma\left(2/3\right) - \Gamma\left(2/3\right) \int_0^\infty t^{-1/3} \mathrm{e}^{-t} \mathrm{d}t \right) = \frac{2^{2/3}-1}{2}
\end{eqnarray}
$$
Alternatively, you might note that the series term can be written in terms of a binomial:
$$
\frac{\left(2/3\right)_n}{n!} = \binom{2/3}{n}
$$
and hence:
$$
\mathcal{S} = \frac{1}{2} \sum_{n=1}^\infty \frac{1}{2^n} \binom{2/3}{n} = \frac{1}{2} \left( \sum_{n=0}^\infty \frac{1}{2^n} \binom{2/3}{n} -1 \right)
$$
this can now be finished with the Newton's generalized binomial theorem.
| {
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Simultaneous Equations That Should Be Inconsistent Has a Unique Solution Find the values of $k$ for which the simultaneous equations do not have a unique solution for $x, y$ and $z$.
Also show that when $k = -2$ the equations are inconsistent
$$kx + 2y +z =0$$
$$3x + 0y -2z =4$$
$$3x - 6ky -4z =14$$
Using a determinant and setting to zero, then solving the quadratic gives
$$
\begin{vmatrix}
k & 2 & 1\\
3 & 0 & -2\\
3 & -6k & -4\\
\end{vmatrix}
=0
$$
$$
k=-2, k=\frac{1}{2}
$$
So far so good, but when subbing $-2$ for $k$
$$-2x + 2y +z =0 \;...\,eqn \,1$$
$$3x + 0y -2z =4 \;...\,eqn \,2$$
$$3x +12y -4z =14 \;...\,eqn \,3$$
subbing eqn $2$ from eqn $3$ gives
$$
12y-2z=10$$
$$\Rightarrow
y = \frac{5 +z}{6}
$$
Subbing for $y$ into eqn $1$
$$
-2x+\frac{5+z}{3}-z=0
$$
$$
\Rightarrow x = \frac{5-2z}{6}
$$
Subbing for $y$ and $x$ into eqn $3$
$$\frac{5-2z}{2}+10 -2z =14 $$
$$\Rightarrow z = \frac{-1}{2},\;
x = 1 \; and \; y = \frac{3}{4}$$
These values for $x,y$ and $z$ seem to prove unique solutions for these equations, yet from the determinant and also the question in the text book should they not be inconsistent?
Thanks in advance to anyone who helps with regards to this exercise from a slightly confused self learner!
| Firstly, a non-zero determinant doesn't imply there are no solutions. The system of equations
$$
\left[
\begin{array}{ccc|c}
1 & 1 & 1 & 1 \\
2 & 2 & 2 & 2 \\
3 & 3 & 3 & 3 \\
\end{array}
\right]
$$
has an infinite number of solutions, for example, and the relevant determinant will be zero. So, we can't be sure the system will be inconsistent from a zero determinant alone.
The matrix
$$
\left[
\begin{array}{ccc|c}
-2 & 2 & 1 & 0 \\
3 & 0 & -2 & 4 \\
3 & 12 & -4 & 14 \\
\end{array}
\right]
$$
has row echelon form
\begin{align*}
& \sim_{R_3 \gets R_3-R_2}
\left[
\begin{array}{ccc|c}
-2 & 2 & 1 & 0 \\
3 & 0 & -2 & 4 \\
0 & 12 & -2 & 10 \\
\end{array}
\right] \\
& \sim_{R_2 \gets R_2+\frac{3}{2}R_1}
\left[
\begin{array}{ccc|c}
-2 & 2 & 1 & 0 \\
0 & 3 & -0.5 & 4 \\
0 & 12 & -2 & 10 \\
\end{array}
\right] \\
& \sim_{R_3 \gets R_3-4R_2}
\left[
\begin{array}{ccc|c}
-2 & 2 & 1 & 0 \\
0 & 3 & -0.5 & 4 \\
0 & 0 & 0 & -6 \\
\end{array}
\right] \\
\end{align*}
which is inconsistent. So, no solutions exist.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/498529",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
calculate a trigonometric expression related to $\sin(\pi/5)$ I tried to solve a question which led to
$$4 \sin\frac{\pi}{5} \sin\frac{3\pi}{5} = \sqrt{5}$$
But, I got stuck to prove that.
Is there any easy solution?
| Using this,
$\sin5x=16\sin^5x-20\sin^3x+5\sin x$
Let $\sin5x=0\implies 5x=n\pi$ where $n$ is any integer
$\displaystyle\implies x=\frac{n\pi}5$
Clearly, $\displaystyle\sin\frac{\pi}5, \sin\frac{3\pi}5$ satisfy $16\sin^5x-20\sin^3x+5\sin x=0$
Again, if $\sin x=0,x=m\pi$ where $m$ is any integer
$\implies \displaystyle\sin\frac{\pi}5, \sin\frac{3\pi}5$ satisfy $16\sin^4x-20\sin^2x+5=0$
$\displaystyle\implies \sin^2x=\frac{20\pm\sqrt{20^2-4\cdot16\cdot5}}{2\cdot16}=\frac{5\pm\sqrt5}8$
Now, as $\displaystyle0<\frac\pi5,\frac{3\pi}5<\pi;\sin\frac{\pi}5, \sin\frac{3\pi}5>0,$ the values of $\displaystyle \sin\frac{\pi}5, \sin\frac{3\pi}5 $ will be $\displaystyle\sqrt{\frac{5\pm\sqrt5}8}$
$\displaystyle\implies \sin\frac{\pi}5\cdot\sin\frac{3\pi}5=\sqrt{\frac{5+\sqrt5}8}\cdot\sqrt{\frac{5-\sqrt5}8}=\frac{\sqrt{(5+\sqrt5)(5-\sqrt5)}}8=\frac{\sqrt5}4$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/499565",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Squares modulo a product of distinct odd primes Let $p,q$ be distinct odd primes. Is it true that a number $a$ is a square modulo $pq$ if and only if $a$ is a square modulo $p$ and modulo $q$?
| If $x^2\equiv a\pmod{pq}$, then it is immediate that $x^2\equiv a\pmod{p}$ and $x^2\equiv a\pmod{q}$. Thus a square modulo $pq$ is a square modulo each of $p$ and $q$.
Next we prove the converse, if $a$ is a square modulo $p$ and a square modulo $q$, then $a$ is a square modulo $pq$.
So we want to show that if there exist $x, y$ such that $x^2\equiv a\pmod{p}$ and $y^2\equiv a\pmod{q}$, then $a$ is a square modulo $pq$.
By the Chinese Remainder Theorem, there is a $z$ such that $z\equiv x\pmod{p}$ and $z\equiv y\pmod{q}$.
Then $z^2\equiv x^2\equiv a\pmod{p}$ and $z^2\equiv y^2\equiv a\pmod{q}$. It follows that $z^2\equiv a \pmod{pq}$.
Remark: Note that we only used the fact that $p$ and $q$ are relatively prime.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/499862",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Infinite Series $\sum\limits_{n=1}^{\infty}\frac{1}{4^n\cos^2\frac{x}{2^n}}$ How to prove that
$$\sum_{n=1}^{\infty}\frac{1}{4^n\cos^2\frac{x}{2^n}}=\frac1{\sin^2x}-\frac1{x^2}$$
| Often when $\sin(x)$ and $x$ appear in a formula such as this, it is based on the fact that
$$
\lim_{a\to0}\frac{\sin(ax)}{a}=x\tag{1}
$$
Since the summands are $\dfrac{4^{-k}}{\cos^2(2^{-k}x)}$ , it appears that an identity involving a double angle formula is being exploited. Therefore, with an eye toward a telescoping sum, I started with
$$
\begin{align}
\frac4{\sin^2(2x)}-\frac1{\sin^2(x)}
&=\frac{4-4\cos^2(x)}{4\sin^2(x)\cos^2(x)}\\
&=\frac{4\sin^2(x)}{4\sin^2(x)\cos^2(x)}\\
&=\frac1{\cos^2(x)}\tag{2}
\end{align}
$$
Substitute $x\mapsto 2^{-k}x$ and multiply by $4^{-k}$ and sum (and telescope)
$$
\begin{align}
\sum_{k=1}^n\frac{4^{-k}}{\cos^2(2^{-k}x)}
&=\sum_{k=1}^n\left(\frac{4^{-k+1}}{\sin^2(2^{-k+1}x)}-\frac{4^{-k}}{\sin^2(2^{-k}x)}\right)\\
&=\frac1{\sin^2(x)}-\frac{4^{-n}}{\sin^2(2^{-n}x)}\\
&\to\frac1{\sin^2(x)}-\frac1{x^2}\tag{3}
\end{align}
$$
since, by $(1)$,
$$
\lim_{n\to\infty}\frac{4^{-n}}{\sin^2(2^{-n}x)}=\frac1{x^2}\tag{4}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/502676",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 3,
"answer_id": 2
} |
A book has a few pages on which page numbers are written. Someone has torn one page out of it and now average of all page numbers is $\frac{105}{4}$ I couldn't relate this question to any of the topics specifically , I found this in a miscellaneous math problems book(non-calculus) . Here's how it goes, A book has a few pages on which page numbers are written. Someone has torn one page out of it and now average of all page numbers is $\frac{105}{4}$. Answer the following:
(i) If the total numbers of pages in book is n then find the value of
$$\sum_{r=1}^{10} \biggl\lfloor{\frac{n+r}{r+1}}\biggr\rfloor\,.$$
OPTIONS:
(A)100 (B)107 (C)105 (D)82
(ii) If the line $x+y=\bigl\lfloor\frac{n}{3}\bigr\rfloor$ is drawn ,then the total number of points with integral co-ordinates enclosed within the region bounded by $x=0,y=0$ and $x+y=\bigl\lfloor\frac{n}{3}\bigr\rfloor$ is -----?
(A)105 (B)153 (C)59 (D)78
STATUS: No clue how to start.
Any help is welcome.
| More on how to deduce the value of $n$. The average page number before a page is torn is
$$ \frac{1+2+\cdots+n}{n}=\frac{n+1}{2}$$
Since two consecutive page numbers are removed, let the two page numbers be $2k-1$ and $2k$, where $2\le 2k\le n$, then
$$\begin{align}
n\cdot\frac{n+1}{2}-(n-2)\frac{105}{4} =& 2k-1+2k\\
\frac{n}{2}(n+1)-\left(\frac{n}{2}-1\right)\left(52+\frac{1}{2}\right) =& 4k-1
\end{align}$$
Since $\frac{n}{2}$ and $k$ are integers, to make the above left hand side an integer, we must have $\frac{n}{2}-1$ even, or $\frac{n}{2}$ is an odd number. Let $n=4m+2$, where $m$ is an integer.
$$\begin{align}
(2m+1)(4m+2+1)-105m+1 =& 4k
\end{align}$$
Again, since $m$ and $k$ are integers, to make the above left hand side an even number, $m$ must be an even number. Let $m=2p$, where $p$ is an integer. Then $n=8p+2$.
$$\begin{align}
(4p+1)(8p+3)-210p+1 =& 4k\\
32p^2-190p+4 =& 4k
\end{align}$$
Again, since $p$ and $k$ are integers, to make the above left hand side a multiple of 4, $p$ must be an even number. Let $p=2q$, where $q$ is an integer. Then $n=16q+2$.
$$\begin{align}
128q^2-380q+4 =& 4k\\
32q^2-95q+1 =& k
\end{align}$$
Considering the conditions for the value of $k$,
$$\begin{array}{rcl}
2 \le& 2k &\le n\\
2 \le& 2(32q^2-95q+1) &\le 16q+2\\
0 \le& 2q(32q- 95) &\le 16q\\
0 \le& 32q- 95 &\le 8\\
95 \le& 32q &\le 103\\
3-\frac{1}{32} \le& q &\le 3+\frac{7}{32}\\
\end{array}$$
Therefore, we only have
$$\begin{align}
q =& 3\\
n =& 16q+2 = 50\\
2k=& 8\\
2k-1 =& 7
\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/502825",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
Independence of $\frac{1-\cos(x)+k\sin(x)}{\sin(x)+k(1+\cos(x))}$ from $k$. How to one can show that the value of the following expression
$$\frac{1-\cos(x)+k\sin(x)}{\sin(x)+k(1+\cos(x))}$$
doesn't depend to values of $k$?
| Writing $x = 2y$, we obtain
$$\begin{align}
\frac{1-\cos x + k\sin x}{\sin x + k(1+\cos x)} &= \frac{(1-\cos (2y)) + k\sin (2y)}{\sin(2y) + k(1+\cos(2y))}\\
&= \frac{2\sin^2 y + 2k\sin y\cos y}{2\sin y\cos y + 2k\cos^2 y}\\
&= \frac{\sin y}{\cos y}\cdot\frac{\sin y+k\cos y}{\sin y + k\cos y}\\
&= \tan y
\end{align}$$
using the double-angle formulae $\sin (2y) = 2\sin y\cos y$ and $\cos (2y) = \cos^2 y - \sin^2 y$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/503489",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
Rate of convergence for sequences I have numerically determined that the sequence $\{f_x\} = \frac{\sin(x^2)}{x^2}$ approaches $1$ (as $x$ approaches $0$) faster than the sequence $\{g_x\} = \frac{\sin^2(x)}{x^2}$. However, I am stuck for determing the rate of convergence of each sequence.
Should I use the small angle approximation $\sin(x)\approx x$ and the fact that $|\sin(x)|\leq 1$? Thanks.
| Use Taylor expansion:
$$
f_x=\frac{1}{x^2}\Bigl(x^2-\frac{x^6}{6}+O(x^{10})\Bigr)=1-\frac{x^4}{6}+O(x^8).
$$
$$
g_x=\frac{1}{x^2}\Bigl(x-\frac{x^3}{6}+O(x^{5})\Bigr)^2=1-\frac{x^2}{3}+O(x^4).
$$
Added detail on the expansion of $g_x$:
$$
\Bigl(x-\frac{x^3}{6}+O(x^{5})\Bigr)^2=x^2-2\cdot x\cdot\frac{x^3}{6}-2\cdot x\cdot O(x^5)+\frac{x^6}{36}+O(x^{10})=x^2-\frac{x^4}{3}+O(x^6)
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/504825",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 0
} |
When computing $\lim_{x\to -\infty}\frac{2x+7}{\sqrt{x^2+2x-1}}$, I don't get -2. This limit:
$\lim_{x\to -\infty}\frac{2x+7}{\sqrt{x^2+2x-1}}$
is supposed to be equal to -2. My textbook and Wolfram Alpha both state that. However, I can't seem to get same exact result.
Here's what I tried to do:
$$\lim_{x\to -\infty}\frac{2x+7}{\sqrt{x^2+2x-1}} = \lim_{x\to-\infty}\frac{2x+7}{\sqrt{x^2+2x-1}} \cdot \frac{1 \over x}{1 \over x} = \lim_{x\to -\infty}\frac{\frac{x}{x}\cdot(2+\frac7x)}{\sqrt{\frac{x^2}{x^2}\cdot(1+\frac{2}{x}-\frac{1}{x^2})}} = \frac{1\cdot(2+0)}{\sqrt{1\cdot(1+0-0)}} =\frac{2}{\sqrt{1}}=2$$
Where did I make a mistake?
| Hint:
Think about the signs of the numerator and denominator in the last step. If $x\to -\infty$, then what happens to $2x + 7$? If you take the positive square root on the denominator, then what is the sign of the quotient?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/505085",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Binomial Coefficient Identities I am trying to simplify the following fraction, which I think is equal to 1 but I am not sure.
$$\frac{\frac{\left(\begin{array}{c}
b-1\\
k-1
\end{array}\right)\left(\begin{array}{c}
r\\
n-k
\end{array}\right)}{\left(\begin{array}{c}
r+b-1\\
n-1
\end{array}\right)}}{\frac{\left(\begin{array}{c}
b\\
k
\end{array}\right)\left(\begin{array}{c}
r\\
n-k
\end{array}\right)}{\left(\begin{array}{c}
r+b\\
n
\end{array}\right)}}$$
I tried to use the identity
$$\left(\begin{array}{c}
n\\
r
\end{array}\right)=\left(\begin{array}{c}
n-1\\
r-1
\end{array}\right)+\left(\begin{array}{c}
n-1\\
r
\end{array}\right)
$$
I have done the following
Step 1:
$$\frac{\left(\begin{array}{c}
b-1\\
k-1
\end{array}\right)\left(\begin{array}{c}
r\\
n-k
\end{array}\right)}{\left(\begin{array}{c}
r+b-1\\
n-1
\end{array}\right)}\cdot\frac{\left(\begin{array}{c}
r+b\\
n
\end{array}\right)}{\left(\begin{array}{c}
b\\
k
\end{array}\right)\left(\begin{array}{c}
r\\
n-k
\end{array}\right)}$$
Step 2:
$$\frac{\left(\begin{array}{c}
b-1\\
k-1
\end{array}\right)}{\left(\begin{array}{c}
b\\
k
\end{array}\right)}\cdot\frac{\left(\begin{array}{c}
r+b\\
n
\end{array}\right)}{\left(\begin{array}{c}
r+b-1\\
n-1
\end{array}\right)}$$
From there I get stuck here
$$\frac{\left[\left(\begin{array}{c}
b\\
k
\end{array}\right)-\left(\begin{array}{c}
b-1\\
k
\end{array}\right)\right]}{\left(\begin{array}{c}
b\\
k
\end{array}\right)}\cdot\frac{\left(\begin{array}{c}
r+b\\
n
\end{array}\right)}{\left[\left(\begin{array}{c}
r+b\\
n
\end{array}\right)-\left(\begin{array}{c}
r+b-1\\
n
\end{array}\right)\right]}$$
Is there any other identity that would be more useful for this problem? If not does anyone have a useful hint for where to proceed from here?
| The Fraction above does not simplify down to 1 as I had thought it would. It simplifies down to $\frac{k}{b}\cdot\frac{r+b}{n}$.
If we go back to Step 2 of the original question, we have
$$\frac{\left(\begin{array}{c}
b-1\\
k-1
\end{array}\right)}{\left(\begin{array}{c}
b\\
k
\end{array}\right)}\cdot\frac{\left(\begin{array}{c}
r+b\\
n
\end{array}\right)}{\left(\begin{array}{c}
r+b-1\\
n-1
\end{array}\right)}$$
Now, we can express the binomial coefficients as factorials, as suggest by André Nicolas above, and then we get
$$\frac{\left(\frac{(b-1)!}{[(b-1)-(k-1)]!(k-1)!}\right)}{\left(\frac{b!}{(b-k)!k!}\right)}\cdot\frac{\left(\frac{(r+b)!}{(r+b-n)!n!}\right)}{\left(\frac{(r+b-1)!}{[(r+b-1)-(n-1)](n-1)!}\right)}$$
Which simplifies to
$$\frac{\left(\frac{(b-1)!}{(b-k)!(k-1)!}\right)}{\left(\frac{b!}{(b-k)!k!}\right)}\cdot\frac{\left(\frac{(r+b)!}{(r+b-n)!n!}\right)}{\left(\frac{(r+b-1)!}{(r+b-n)!(n-1)!}\right)}$$
Which in turn simplifies to
$$\frac{\left(\frac{(b-1)!}{(k-1)!}\right)}{\left(\frac{b!}{k!}\right)}\cdot\frac{\left(\frac{(r+b)!}{n!}\right)}{\left(\frac{(r+b-1)!}{(n-1)!}\right)}$$
Which is equal to
$$\left(\frac{(b-1)!k!}{(k-1)!b!}\right)\left(\frac{(r+b)!(n-1)!}{n!(r+b-1)!}\right)$$
Which at last reduces down to
$$\frac{k}{b}\cdot\frac{r+b}{n}$$
If you know of any other clever ways to simplify this, please post you answer. Thanks.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/505411",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Find the inverse of $x^6+x^4+x+1$ in the finite field $\mathbb{Z}_2[x]/(x^8+x^4+x^3+x+1)$
How can I find the inverse of $x^6+x^4+x+1$ in the field $\mathbb{Z}_2[x]/(x^8+x^4+x^3+x+1)$?
can someone help me please to solve the problem.
actually I want to know the process of calculation.
I know that $1+(x^8+x^4+x^3+x+1)$ is the identity element and try to calculate another polynomial $a(x) $ such that $(x^6+x^4+x+1)a(x)=1$ but this is very lengthy process.
| You can use Extended Euclidean algorithm.
It only take three steps:
$x^8+x^4+x^3+x+1=(x^2+1)(x^6+x^4+x+1)+x^2$
$x^6+x^4+x+1=(x^4+x^2)x^2+x+1$
$x^2=(x+1)(x+1)+1$
You also needs to express the remainders using $f(x)=x^8+x^4+x^3+x+1$ and $g(x)=x^6+x^4+x+1$.
Let us make this into a table.
Each line contains coefficients such that $a(x)f(x)+b(x)g(x)=r(x)$. The next line is always computed using the preceding two lines.
$$
\begin{array}{|c|c|c|c|c|}
\hline
a(x) & b(x) & r(x) \\\hline
1 & 0 & x^8+x^4+x^3+x+1 \\\hline
0 & 1 & x^6+x^4+x+1 \\\hline
1 & x^2+1 & x^2 \\\hline
x^4+x^2 & x^6+x^2+1 & x+1 \\\hline
x^5+x^4+x^3+x^2+1 & x^7+x^6+x^3+x & 1 \\\hline
\end{array}
$$
We have $(x^5+x^4+x^3+x^2+1)(x^8+x^4+x^3+x+1)+(x^7+x^6+x^3+x)(x^6+x^4+x+1)=1$ in $\mathbb Z_2[x]$.
A useful thing to do, when computing something like this, is to plug in some values into the polynomials (in this case we only have $x=0$ and $x=1$). There is at least some probability, that if you made a mistake, it might manifest already on the values you tried. (To be entirely sure, you would have to try to multiply the polynomials, which is rather cumbersome.)
You can find many other examples on this site for finding inverses in finite fields.
For example: Finding inverse of polynomial in a field.
Another post explaining table method is here: How to use the Extended Euclidean Algorithm manually? (There are certainly more such posts.)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/505580",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
How to evaluate $\int_{0}^{\pi} \log(2+\cos x)dx$? I tried integrating it by part but it didn't work. I konw it can be solved by the Gauss Mean Value theorem .Is there some elementary method for evaluate it or just some others? It seems there are at least four differential ways(page 4 (17)) but I cannot think out any(the link only says there are four but not give any way).
| Another way: differentiate under integral sign!
Let $\displaystyle\quad J(a) = \int_0^\pi \log(a + \cos x) dx,\quad$ and $t = \tan\frac{x}{2}$, we have:
$$\begin{align}
J'(a)
= & \int_0^\pi \frac{dx}{a + \cos x}
= \int_0^{\infty} \frac{\frac{2 dt}{1+t^2}}{a + \frac{1-t^2}{1+t^2}}
= \frac{2}{a+1}\int_0^{\infty}\frac{dt}{1 + \frac{a-1}{a+1}t^2}\\
= & \frac{2}{\sqrt{a^2-1}}\int_0^\infty \frac{ds}{1+s^2} = \frac{\pi}{\sqrt{a^2-1}}
\end{align}$$
Let $a = \cosh\theta$, and integrate over $a$, we get
$$\begin{align}
F(a)
= & \int F'(a) da = \pi \int \frac{d\cosh\theta}{\sqrt{\cosh^2\theta-1}}\\
= & \pi \theta + C = \pi\cosh^{-1}(a) + C = \pi \log( a + \sqrt{a^2 - 1} ) + C\tag{*1}
\end{align}$$
for some integration constant $C$. For large $a$, we have
$$\log(a + \cos x) = \log a + \frac{\cos x}{a} + O(\frac{1}{a^2})
\quad\implies\quad
J(a) = \pi \log a + O(\frac{1}{a})\tag{*2}
$$
Compare $(*1)$ with $(*2)$, we find $C = -\pi \log 2$ and hence:
$$J(a) = \pi\log\left(\frac{a + \sqrt{a^2-1}}{2}\right)$$
In particular, the integral we want to calculate is $\displaystyle\quad J(2) = \pi\log\left(\frac{2+\sqrt{3}}{2}\right)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/505714",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
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Is $\frac{a+b}{c+d}<\frac{a}{c}+\frac{b}{d}?$, for $a,b,c,d>0$? Is $$\frac{a+b}{c+d}<\frac{a}{c}+\frac{b}{d}$$
for $a,b,c,d>0$
If it is true, then can we generalize?
EDIT:typing mistake corrected.
EDIT, WILL JAGY. Apparently the real question is
Is $$\color{magenta}{\frac{a+b}{c+d}<\frac{a}{c}+\frac{b}{d}}$$
for $a,b,c,d>0,$ where letters on the left hand side and in the numerator stay in the numerator on the right-hand side, and letters on the left hand side and in the denominator stay in the denominator on the right-hand side.
| A slightly different approach:
Multiply both sides by (c+d), which we can do without altering the inequality because c and d are positive:
$$ a+b < \frac{a(c+d)}{c} +\frac{b(c+d)}{d}$$
$$ a+b < \frac{ac}{c} +\frac{ad}{c} +\frac{bc}{d} +\frac{bd}{d}$$
$$ a+b < a + \frac{ad}{c} +\frac{bc}{d} +b$$
$$ a+b < a+b +\frac{ad}{c}+\frac{bc}{d}$$
$$ \frac{ad}{c} +\frac{bc}{d} > 0$$
This is clearly always true because both terms must be > 0.
This same basic outline works for a 3 term version of this:
$$\frac{a+b+c}{d+e+f} < \frac{a}{d}+\frac{b}{e} +\frac{c}{f}$$
and will clearly work for any number of terms because after multiplying by the denominator on the left hand side, you will always spit out on the right hand side, exactly the left hand side numerator plus some additional terms which must be positive.
| {
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"url": "https://math.stackexchange.com/questions/506152",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Solution of Equation $ 1 + x + \frac{1\cdot3}{2!}x^2 + \frac{1\cdot3\cdot5}{3!}x^3 + \ldots = \sqrt{2} $ I have an equation whose left side is a infinite series . I can solve the equation if I am able to find a close form of the series . The equation is as follows :
$$ 1 + x + \frac{1\cdot3}{2!}x^2 + \frac{1\cdot3\cdot5}{3!}x^3 + \ldots = \sqrt{2} $$
I have tried to find the sum of the series by trying to apply binomial theorm of negative exponent . But I cant solve this problem . Can you help me to find the sum of series of the left hand side of the above equation ?
| Compare the series with $1+na+\frac{n(n-1)a^2}{2!}\cdots$ we get
$na=x$ or $$n^2a^2=x^2\dots(1)$$
also $$\frac{n(n-1)a^2}{2!}=\frac{1.3x^2}{2!}\cdots(2)$$
Divide $(1)$ and $(2)$ and solve for $n$, substitute in first and solve for $a$
you will get the sum to be $(1-2x)^{\frac{-1}2}$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Simplify $(A+C)(AD+AD) + AC + C$ using Boolean algebra I have solved the equation like this:
(A + C)(AD + AD) + AC + C
=(A + C)(AD) + (A + C)(AD) + AC + C
=AAD + ACD + AAD + ACD + AC + C
=AAD + AAD + ACD + ACD + AC + C
=AAD + ACD + AC + C
=AAD + ACD + C(A+1)
=AAD + ACD + C
=AD + ACD + C
=AD(1+C) + C
=AD + C
Am I right?
| I would have taken the following steps:
$$
\begin{align}
(A + C)(AD + AD) + AC + C
&= (A + C)(AD) + (AC + C)\\
&= AD + ADC + C\\
&= AD + C\\
\end{align}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/513518",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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} |
cubic equations which have exactly one real root Question is to check :
For any real number $c$, the polynomial $x^3+x+c$ has exactly one real root .
the way in which i have proceeded is :
let $a$ be one real root for $x^3+x+c$ i.e., we have $a^3+a+c=0$
i have seen that $(x-a)(x^2+ax+(a^2+1))=x^3+x-a^3-a$
But, $a^3+a+c=0$. So, $-a^3-a=c$.
so, $(x-a)(x^2+ax+(a^2+1))=x^3+x-a^3-a=x^3+x+c$
i.e, $(x-a)(x^2+ax+(a^2+1))=x^3+x+c$
Now, determinant for quadratic $x^2+ax+(a^2+1)$ is $a^2-4(a^2+1)=-3a^2-4 <0$ for any real number $a$
Thus, quadratic has no real root and so is the cubic $x^3+x+c$
I would like to know if this justification is sufficient and if this can be generalized.
I mean, can i say $x^3+ax^2+bx+c$ for $a,b,c\in \mathbb{Z}$ have exactly one real root. (Conditions apply)
can this be generalized to any odd degree polynomial (at least for some special cases)
| You essentially split off the linear factor belnging to a real root and showed per determinant that the remaining quadratic has no real root.
This is fine but does not readily generalize to higher degrees.
Instead, it is probably easier to show that $f$ is injective: Assume $a,b$ are two real roots, i.e. $f(a)=f(b)=0$.
Then by Rolle, there exists $\xi$ between $a$ and $b$ (including the case $a=\xi=b$ if you also want to show that no multiple root - $a=b$ - exists) with $f'(\xi)=0$. But for $f(x)=x^3+x$ we have $f'(x)=3x^2+1\ge1$ for all $x$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Given $x,y,z >0$, $1/x+1/y+1/z = 4$, prove that $ 1/(2x+y+z)+1/(x+2y+z) +1/(x+y+2z) \le 1$ Given $x,y,z >0$, $1/x+1/y+1/z = 4$, prove that
$$ 1/(2x+y+z)+1/(x+2y+z) +1/(x+y+2z) \le 1 .$$
Any hints or direction will be appreciated.
| By C-S
$$\sum_{cyc}\frac{1}{2x+y+z}\leq\frac{1}{(2+1+1)^2}\sum_{cyc}\left(\frac{2^2}{2x}+\frac{1^2}{y}+\frac{1^2}{z}\right)=1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/517323",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Writing u as a linear combination of the vectors in S. Write vector
u = $$\left[\begin{array}{ccc|c}2 \\10 \\1\end{array}\right]$$
as a linear combination of the vectors in S. Use elementary row operations on an augmented matrix to find the necessary coefficients.
S = {
$v1$$\left[\begin{matrix}1\\2\\2\end{matrix}\right] , v2\left[\begin{matrix}4\\2\\1\end{matrix}\right],
v2\left[\begin{matrix}5\\4\\1\end{matrix}\right]
$ }. If it is not possible, explain why?
This is what i have so far:
S = {
$v1$$\left[\begin{matrix}1\\2\\2\end{matrix}\right] , v2\left[\begin{matrix}4\\2\\1\end{matrix}\right],
v3\left[\begin{matrix}5\\4\\1\end{matrix}\right].
v4\left[\begin{matrix}2\\10\\1\end{matrix}\right]
$ }
$c1$$\left[\begin{matrix}1\\2\\2\end{matrix}\right] , c2\left[\begin{matrix}4\\2\\1\end{matrix}\right],
c3\left[\begin{matrix}5\\4\\1\end{matrix}\right].
c4\left[\begin{matrix}2\\10\\1\end{matrix}\right]
=\left[\begin{matrix}0\\0\\0\end{matrix}\right]$
$c1$$\left[\begin{matrix}1\\2\\2\end{matrix}\right] , c2\left[\begin{matrix}4\\2\\1\end{matrix}\right],
c3\left[\begin{matrix}5\\4\\1\end{matrix}\right].
c4\left[\begin{matrix}2\\10\\1\end{matrix}\right]$
$
\begin{bmatrix}
1 & 4 & 5 & 2\\
2 & 2 & 4 & 10\\
2 & 1 & 1 & 1\\
\end{bmatrix}
$
Now i don't know how to do this. Help will greatly be appreciated.
Thanks
| $a\left[\begin{matrix}1\\2\\2\end{matrix}\right] + b\left[\begin{matrix}4\\2\\1\end{matrix}\right]+
c\left[\begin{matrix}5\\4\\1\end{matrix}\right]=\left[\begin{matrix}2\\10\\1\end{matrix}\right]$.
Putting this into matrix form gives:
$\begin{bmatrix}1&4&5&2\\2&2&4&10\\2&1&1&1\end{bmatrix}\to\begin{bmatrix}1&4&5&2\\1&1&2&5\\2&1&1&1\end{bmatrix}\to\begin{bmatrix}1&4&5&2\\0&3&3&-3\\0&7&9&3\end{bmatrix}\\\to\begin{bmatrix}1&4&5&2\\0&1&1&-1\\
0&7&9&3\end{bmatrix}\to\begin{bmatrix}1&0&1&6\\0&1&1&-1\\0&0&1&5\end{bmatrix}\to\begin{bmatrix}1&0&0&1\\0&1&0&-6\\0&0&1&5\end{bmatrix}$.
So, $a=1,b=-6,c=5.$
Plug the values back in to check if this is indeed correct:
$1\left[\begin{matrix}1\\2\\2\end{matrix}\right] -6\left[\begin{matrix}4\\2\\1\end{matrix}\right]+
5\left[\begin{matrix}5\\4\\1\end{matrix}\right]=\left[\begin{matrix}2\\10\\1\end{matrix}\right]$$\implies$$\left[\begin{matrix}1\\2\\2\end{matrix}\right] + \left[\begin{matrix}-24\\-12\\-6\end{matrix}\right]+
\left[\begin{matrix}25\\20\\5\end{matrix}\right]=\left[\begin{matrix}2\\10\\1\end{matrix}\right]$.
So, it must be correct.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Proof of divergence of $1/2 + 1/3 + 1/5 + 1/7 + 1/11 +....$ What is an elementary proof to the fact that $\frac{1}{p_1} + \frac{1}{p_2} + \frac{1}{p_3} + \dots$ diverges. ($p_i$ denotes the $i$th prime.)
| $$
\begin{align}
\log \left( \sum_{n=1}^\infty \frac{1}{n}\right) & {} = \log \left( \prod_p \frac{1}{1-p^{-1}}\right)
= \sum_p \log \left( \frac{1}{1-p^{-1}}\right) = \sum_p - \log(1-p^{-1}) \\
& {} = \sum_p \left( \frac{1}{p} + \frac{1}{2p^2} + \frac{1}{3p^3} + \cdots \right) \\
& {} = \left( \sum_{p}\frac{1}{p} \right) + \sum_p \frac{1}{p^2} \left( \frac{1}{2} + \frac{1}{3p} + \frac{1}{4p^2} + \cdots \right) \\
& {} < \left( \sum_p \frac{1}{p} \right) + \sum_p \frac{1}{p^2} \left( 1 + \frac{1}{p} + \frac{1}{p^2} + \cdots \right) \\
& {} = \left( \sum_p \frac{1}{p} \right) + \left( \sum_p \frac{1}{p(p-1)} \right) \\
& {} = \left( \sum_p \frac{1}{p} \right) + C
\end{align}$$
for a fixed constant $C<1.$ Since the harmonic series diverges, the sum of the reciprocals of the prime numbers diverges.
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "3",
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How would I evaluate this limit? $\lim_{x\to 3+} \frac{x-3}{\sqrt{x^2-9}}$ How would I evaluate the following limit by hand?
$\lim_{x\to 3+} \frac{x-3}{\sqrt{x^2-9}}$
Thanks in advance.
| $$\begin{align} \lim_{x\to 3+} \frac{x-3}{\sqrt{x^2-9}} & = \lim_{x \to 3+}\; \frac{\left(\sqrt{x - 3}\;\right)^2}{\sqrt{(x - 3)(x+3)}}\\ \\
& = \lim_{x \to 3+}\;\frac{\left(\sqrt{x-3}\;\right)^2}{(\sqrt{x-3}\;)(\sqrt{x+3}\;)}\\ \\
& = \lim_{x \to 3+}\;\frac{\sqrt{x - 3}}{\sqrt{x+3}}\\ \\
& = \frac 06\; = \;0\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/524622",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Evaluate the limit $\lim_{x\to 2} \frac{x-2}{\sqrt{x^2+5}-3}$ I need to evaluate the following limit:
$\lim_{x\to 2} \frac{x-2}{\sqrt{x^2+5}-3}$
I have multiplied both sides by the conjugate $\sqrt{x^2+5}+3$ but am getting $x^2-4$ as the denominator. Is this the correct way to go about it?
| You are on the right track.
$\lim_{x\to 2} \frac{x-2}{\sqrt{x^2+5}-3}=\lim_{x\to 2} \frac{x-2}{\sqrt{x^2+5}-3} \frac{\sqrt{x^2+5}+3}{\sqrt{x^2+5}+3}=\lim_{x\to 2} \frac{(x-2)(\sqrt{x^2+5}+3)}{(x-2)(x+2)}$ simplifying
$$\lim_{x\to 2} \frac{(\sqrt{x^2+5}+3)}{(x+2)}$$ and evaluating at $2$ we get
$$\lim_{x\to 2} \frac{(\sqrt{x^2+5}+3)}{(x+2)}= \frac{(\sqrt{4+5}+3)}{(2+2)}=\frac{6}{4}=\frac{3}{2}$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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} |
Is it true that $\lfloor\sqrt{n}+\sqrt{n+1}+\sqrt{n+2} \rfloor=\lfloor\sqrt{9n+7}\rfloor$? I know it's true that
$$\lfloor\sqrt{n}+\sqrt{n+1} \rfloor=\lfloor\sqrt{4n+1}\rfloor,\forall n\in \mathbb N. \tag 1$$
Is it true that
$$\left[\sqrt{n}+\sqrt{n+1}+\sqrt{n+2} \right ]=\left[\sqrt{9n+7}\right], \tag 2$$
$$\left[\sqrt{n} +\sqrt{n+1}+\sqrt{n+2}+\sqrt{n+3}\right ]=\left[\sqrt{16n+20}\right], \tag 3$$
$$\left[\sqrt{n} +\sqrt{n+1}+\sqrt{n+2}+\sqrt{n+3}+\sqrt{n+4}\right ]=\left[\sqrt{25n+49}\right], \tag 4$$
for all $n\in \mathbb N$? I have checked $(2),(3),(4)$ through $10^6.$
(I think $(4)$ maybe has counter-examples, but I cannot find it.)
| For $0<a<b$ we have $\sqrt b-\sqrt a=\frac{b-a}{\sqrt b+\sqrt a}$ and hence the estimate
$$\frac{b-a}{2\sqrt b}<\sqrt b-\sqrt a<\frac{b-a}{2\sqrt a}.$$
Therefore
$$x:=\sqrt {n}+\sqrt {n+1}+\sqrt{n+2} =3\sqrt{n+1}+(\sqrt{n+2}-\sqrt {n+1})-(\sqrt {n+1}-\sqrt n)$$
is estimated by
$$3\sqrt {n+1}+\frac1{2\sqrt{n+2}}-\frac1{2\sqrt n}< x<3\sqrt{n+1}$$
and then (applying the same trick to $a=\frac1{n+2}$, $b=\frac1n$)
$$ x > 3\sqrt{n+1}-\frac{\frac1n-\frac1{n+2}}{4\sqrt{\frac1{n+2}}}= 3\sqrt{n+1} - \frac{1}{2n\sqrt{n+2}}.$$
Therefore
$$ 9(n+1)-\frac{3\sqrt{n+1}}{n\sqrt{n+2}}+\frac1{4n^2(n+2)}< x^2 < 9n+9$$
For $n>3$ this gives us $ 9n+8<x^2<9n+9$ and hence
$ x=\sqrt{9n+\theta}$
with $8<\theta<9$. Since $9n+8$ cannot be a perfect square(!), we conclude that $$\lfloor\sqrt{9n+7}\rfloor = \lfloor\sqrt{9n+\theta}\rfloor = \lfloor \sqrt n+\sqrt{n+1}+\sqrt {n+2}\rfloor$$
(at least for $n>3$, but one checks the rest manually).
The same approach, though with more complicated terms, should work for $(3)$ and $(4)$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Is my answer correct? (And what's the name of the rule?)$\lim_{n \to \infty} \frac{\left(n+3\right)!-n!}{n\left(n+2\right)!}$ Want to know if I solved this problem correctly:
$$\lim_{n \to \infty} \frac{\left(n+3\right)!-n!}{n\left(n+2\right)!} =\lim_{n \to \infty} \frac{1 \cdot 2 \ldots(n-1)n(n+1)(n+2)(n+3) - 1 \cdot 2 \ldots(n-1)n}{(1 \cdot 2 \ldots(n-1)n(n+1)(n+2))\cdot n}=\lim_{n \to \infty}\frac{(1 \cdot 2 \ldots(n-1) n)\cdot((n+1)(n+2)(n+3)-1)}{(1 \cdot 2 \ldots(n-1)n(n+1)(n+2))\cdot n} =\lim_{n \to \infty} \frac{(n+1)(n+2)(n+3)-1}{(n+1)(n+2)\cdot n} = \frac{1}{1}=1$$
The answer is 1 because we've got $n^3$ in numerator and $n^3 $ in denominator, and the constants are both equals 1 so the limit can be calculated as $\frac{1}{1}$. (Sorry I don't remember that rule. Could you remain me please?)
| If $m=n$, and $a_n,b_m\neq 0$, then you can say the following:$$\lim_{x \to \infty}\frac{a_nx^n+\cdots+a_0}{b_mx^m+\cdots+b_0}=\frac{a_n}{b_m}.$$
To prove this, you can factor out the $x$ terms from the numerator and denominator like so: $$\frac{a_nx^n+\cdots+a_0}{b_mx^m+\cdots+b_0}=\frac{x^n}{x^m}\cdot\frac{a_n+\cdots+\frac{a_0}{x^n}}{b_m+\cdots+\frac{b_0}{x^m}}.$$
Now since $n = m$, $x^n/x^m=1$, the limit becomes $$\lim_{x \to \infty}\frac{a_n+\cdots+\frac{a_0}{x^n}}{b_m+\cdots+\frac{b_0}{x^m}} =\frac{\lim_{x \to \infty}a_n+\cdots+\frac{a_0}{x^n}}{\lim_{x \to \infty}b_m+\cdots+\frac{b_0}{x^m}}=\frac{a_n}{b_m}.$$
since all the terms with an $x$ in the denominator go to $0$ as $x \to \infty$. You can also use a similar technique to show that if $m > n$ then the limit is $0$.
| {
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Finding the limit of $\frac{\sqrt{x}}{\sqrt{x}+\sin\sqrt{x}}$ How would one find the limit of
$\displaystyle\lim_{x\to 0}\frac{\sqrt{x}}{\sqrt{x}+\sin\sqrt{x}}$
I know I have to use the L'Hospital rule.
$\displaystyle\lim_{x\to 0}\frac{\frac{1}{2}x^{-1/2}}{\frac{1}{2}\frac{1}{\sqrt{x}}+\frac{1}{2}\frac{1}{\sqrt{x}}\cos\sqrt{x}}$
But I find myself stuck
| Continuing from where you left off:
Simply cancel the common factor of $\frac {1}{2 \sqrt x}$ from numerator and denominator:
$$\frac{\frac{1}{2\sqrt x}}{\frac{1}{2}\frac{1}{\sqrt{x}}+\frac{1}{2}\frac{1}{\sqrt{x}}\cos\sqrt{x}} = \dfrac 1{1 + \cos \sqrt x}$$ Now evaluate the limit as $x \to 0$. You should arrive at a limit of $\dfrac 12$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/528661",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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How find this maximum $S_{\Delta ABC}$ in $\Delta ABC$,and $\angle ABC=60$,such that $PA=10,PB=6,PC=7$,
find the maximum $S_{\Delta ABC}$.
My try:let $AB=c,BC=a,AC=b$, then
$$b^2=a^2+c^2-2ac\cos{\angle ABC}=a^2+c^2-2ac$$
then
$$S_{ABC}=\dfrac{1}{2}ac\sin{60}=\dfrac{\sqrt{3}}{4}ac$$
Then I can't
| let $B$ is on the $O$ of polar system, $P(6,\alpha)$
so $C$ is on circle $p^2+6^2-2*6*cos(\beta- \alpha)=7^2$ and $\beta=0$
$A$ is on circle $p^2+6^2-2*6*cos(\beta- \alpha)=10^2$ and $\beta=\dfrac{\pi}{3}$
$a=BC=6cos\alpha+\sqrt{(6cos\alpha)^2+13}, c=BA=6cos(\dfrac{\pi}{3}-\alpha)+\sqrt{(6cos(\dfrac{\pi}{3}-\alpha))^2+64}$
$ac=f(\alpha),\alpha=x, f'(x)=(\sqrt{36 cos^2(x)+13}+6 cos(x)) \left(6 cos(x+\dfrac{\pi}{6})+\dfrac{(36 sin(x+\dfrac{\pi}{6}) cos(x+\dfrac{\pi}{6})}{\sqrt{36 sin^2(x+\dfrac{\pi}{6})+64)}}\right)$$+(\sqrt{36 sin^2(x+\dfrac{\pi}{6})+64)+6 sin(x+\dfrac{\pi}{6}}) \left(-6 sin(x)-\dfrac{(36 sin(x) cos(x))}{\sqrt{36 cos^2(x)+13)}}\right)=0$
it is only numeric method can solve the solution is $x=.422064 \implies f_{max}=171.138$
BTW, this triangle also have max area without fixed $\angle B$ but it is more difficult as there is two varies.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/530887",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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How to evaluate the series: $ \frac {20} {(5-4)(5^2-4^2)} + \frac {20^2} {(5^2-4^2)(5^3-4^3)} + \frac {20^3} {(5^3-4^3)(5^4-4^4)} + \ldots $ How to evaluate this infinte summation?
$$ \dfrac {20} {(5-4)(5^2-4^2)} + \dfrac {20^2} {(5^2-4^2)(5^3-4^3)} + \dfrac {20^3} {(5^3-4^3)(5^4-4^4)} + \ldots $$
Telescopic cancellation is the way to go, I feel, but I don't know how to get this into telescopic form.
Hints?
| Notice that:
$$\frac{20^k}{(5^k-4^k)(5^{k+1}-4^{k+1})}=\frac{4^k}{5^k-4^k}-\frac{4^{k+1}}{5^{k+1}-4^{k+1}},$$
in order to prove that:
$$\sum_{k=1}^{+\infty}\frac{20^k}{(5^k-4^k)(5^{k+1}-4^{k+1})} = 4.$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Find the orthogonal trajectories of the following family of curves: $x^3y - xy^3 = \alpha$ The answer is $-6x^2y^2 + x^4 + y^4 = \beta$, but I get $-3x^2y^2 + \frac{x^4}{4} + \frac{y^4}{4} = \beta$ and I can't seem to find my mistake. My work:
Differentiaing the given equation, we get
$$3x^2y + x^3\frac{dy}{dx} - y^3 - 3y^2x\frac{dy}{dx} = 0. $$
So $$3x^2y - y^3 = (3y^2x -x^3)\frac{dy}{dx}, $$ and hence
$$ \frac{dy}{dx} = \frac{3x^2y - y^3}{3y^2x - x^3}. $$
Now, the slope of the other family of curves will be the negative inverse, so for the orthogonal trajectory, we have $$\frac{dy}{dx} = -\frac{3y^2x-x^3}{3x^2y-y^3}, $$ which we can write as
$$-(3y^2x-x^3)dx = (3x^2y-y^3)dy. $$ Integration of the last differential equation yields
$$-\frac{3}{2}y^2x^2 + \frac{x^4}{4} = \frac{3}{2}x^2y^2 - \frac{y^4}{4} + \beta, $$ $\beta$ being an arbitrary constant.
This simplifies to $$-3y^2x^2 + \frac{x^4}{4} + \frac{y^4}{4} = \beta. $$ which is not correct.
Where am I going wrong?
| I t seems that we have $$y'=\frac{3xy^2-x^3}{y^3-3x^2y}$$ instead, which is an homogenous OE. We have $$\frac{du}{\frac{3u^2-1}{u^3-3u^2}-u}=\frac{dx}{x}~\equiv~\frac{(-u^3+3u^2)du}{u^4+3u^2-3u^2+1}=\frac{dx}{x},~~~~~~u=y/x,x\neq 0$$ in which the right ODE is a separable one.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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When does the locus $|z-z_1|^2 +|z-z_2|^2=k$ describe a circle?
Prove that the locus $|z-z_1|^2 +|z-z_2|^2=k$ of a moving point $z$ on the Argand plane is a circle when $|z_1-z_2|^2\leqslant 2k$.
What i have tried out-
$|z-1|^2+|z-2|^2=k$
let $z=x+iy$
$$\\ \implies2x^2-6x+5+2y^2-k=0$$
This is the equation of a circle.
centre $=(\frac32,0)$
radius=$\sqrt{\dfrac{(2k−1)}4 }$
means $\\ k\geqslant\frac12$
| Try with $z=x+iy, z_1=a+ib, z_2=c+id$
On rearrangement we have, $$x^2+y^2-x(a+c)-y(b+d)+\frac{a^2+b^2+c^2+d^2-k}2=0$$
$$\implies \left(x-\frac{a+c}2\right)^2+\left(y-\frac{b+d}2\right)^2=\frac{2k-2a^2-2b^2-2c^2-2d^2+(a+c)^2+(b+d)^2}4=\frac{2k-\{(a-c)^2+(b-d)^2\}}4$$
For real circle we need $2k-\{(a-c)^2+(b-d)^2\}\ge0$
But here, $|z_1-z_2|=?$
| {
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Is there any simple method to calculate $\sqrt x$ without using logarithm Suppose that we don't know logarithm, then how we would able to calculate $\sqrt x$, where $x$ is a real number? More generally, is there any algorithm to calculate $\sqrt [ n ]{ x } $ without using logarithm? More simple techniques would be nice.
Here is a simple technique used to approximate square roots by Persian author Hassan be al-Hossein:
For example: $\sqrt {78}\approx 8\frac { 14 }{ 17 } $ , where $8$ is the nearest integer root of $78$, $14 = 78 - 8^2$, $17 = 2 \times 8 + 1$.
if $n=2^k$ we can use the method above.
For example, for $k=2$ Lets calculate $\sqrt [ 4 ]{ 136 } $: $$\sqrt [ 4 ]{ 136 } =\sqrt { \sqrt { 136 } } \approx \sqrt { 11\frac { 136-{ 11 }^{ 2 } }{ 11\times 2+1 } } =\sqrt { 11\frac { 15 }{ 23 } } \\ \sqrt { 11\frac { 15 }{ 23 } } \approx 3\frac { 11\frac { 15 }{ 23 } -{ 3 }^{ 2 } }{ 3\times 2+1 } =\frac { 544 }{ 161 } =3.38\\$$ The exact result is$$ \sqrt [ 4 ]{ 136 } =3.4149\cdots$$ The method approximates well, but it is working for only $n=2^k$ as I know.
| Every positive number $x$ can be always written as the sum of two other numbers (or the difference).
Say that one of the two is a perfect square $n^2$, then we can always say that
$$x = n^2 + q$$
Where $q$ is the obvious remainder. Since we can also adopt the difference convention, it's better to write
$$x = n^2 \pm q$$
When to choose the plus or the minus? Well the rule is that the smaller is $q$, the better.
After that, we can use the following approximation:
$$\sqrt{x} = \sqrt{n^2 \pm q} \approx n \pm \frac{q}{2n}$$
Example
Let's calculate $\sqrt{40}$. Either we choose $40 = 36 + 4$ or $40 = 49 - 9$. The first one is better of course, hence
$$\sqrt{40} = \sqrt{36 + 4} \approx 6 + \frac{4}{12} = 6 + \frac{1}{3} = 6.\bar 3$$
Notice that the real value is
$$\sqrt{40} = 6.324(...)$$
| {
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Solving two simultaneous equations Suppose that $x$, $y$ and $z$ are three integers (positive,negative or zero) such that we get the following relationships simultaneously
*
*$x + y = 1 - z$ and
*$x^3 + y^3= 1 - z^2$
Find all such $x$, $y$ and $z$.
I have tried to determine the values of $x$ and $y$ by fixing some values of $z$ but it is obvious that this method will not work. I also tried to determine the bounds of $x$ and $y$ and the related terms like $xy$ and ($x$$2$ +$y$$2$) but this this method failed as well. What then is the way out?
| If $z = 1$, then $y = -x$, and all triples $(x,-x,1)$ are solutions. So let's suppose $z \neq 1$. Then we can divide
$$x^2 - xy + y^2 = \frac{x^3+y^3}{x+y} = \frac{1-z^2}{1-z} = 1+z. \tag{3}$$
Adding $(3)$ to $(1)$ yields
$$x^2 - xy + y^2 + x + y = 2.$$
Multiplying by $2$ and adding $2$ to both sides yields
$$\begin{align}
6 &= 2x^2 - 2xy + 2y^2 + 2x +2y + 2\\
&= (x^2 -2xy+y^2) + (x^2+2x+1) + (y^2+2y+1)\\
&= (x-y)^2 + (x+1)^2 + (y+1)^2.
\end{align}$$
There aren't many combinations of $x$ and $y$ left to check.
| {
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system of equation (3 unknown, 3 equations) I want to solve a system of equations, but I seem to get it wrong.
Problem: see picture, and note that I should tell for which a I have
*
*no solution
*one solution
*$\infty$ solutions
Attempt:
$$\begin{matrix}i \\ ii \\ iii\end{matrix}\left\{\begin{matrix}x+y-az=3\\ ax-y+z=-2\\ -3x+y-z=-a+2\end{matrix}\right.$$
$ii+iii\to (a-3)x=-a\implies x=-\frac a{a-3}=\frac a{3-a}$
$i+ii\to (a+1)x+z(1-a)=1$ and $x=\frac a{3-a}\implies (a+1)\frac a{3-a}+z(1-a)=1\implies$
$$z=\frac {1-(a+1)\frac a{3-a}}{1-a}=\frac {a+3}{3-a}, a\ne 1,3$$
$$y=3-x+az=3-\frac a{3-a}+a\frac {a+3}{3-a}=\frac {3(3-a)-a(a+4)}{3-a},a\ne 1,3$$
| Using Gaussian Elimination (your way works too) and reducing the augmented system, we get:
$$\begin{bmatrix}1 & 0 & 0 & -\dfrac{a}{a-3}\\0 & 1 & 0 & \dfrac{-a^2+a-9}{a-3} \\ 0 & 0 & 1 & \dfrac{a+3}{-a + 3}\end{bmatrix}$$
We found that $a = 1$ and $a = 3$ appeared to be problematic.
If we dig deeper, we actually find:
*
*$a = 3$ is no solution, because we get a RREF of:
$$\begin{bmatrix}1 & 0 & -\dfrac{1}{2} & 0\\0 & 1 & -\dfrac{5}{2} & 0 \\ 0 & 0 & 0 & 1\end{bmatrix}$$
*
*$a = 1$ is infinite solutions because we get a RREF of:
$$\begin{bmatrix}1 & 0 & 0 & \dfrac{1}{2}\\0 & 1 & -1 & \dfrac{5}{2} \\ 0 & 0 & 0 & 0\end{bmatrix}$$
This gives us: $x = \dfrac{1}{2}, y = \dfrac{5}{2} + z$, where $z$ is a free variable (hence, infinite solutions).
*
*All other $a$ (from the RREF above) give us a unique (single solution).
| {
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How to prove $\cos ^6x+3\cos ^2x\space \sin ^2x+\sin ^6x=1$ Prove the following equation.
\begin{eqnarray}
\\\cos ^6x+3\cos ^2x\space \sin ^2x+\sin ^6x=1\\
\end{eqnarray}
I can't prove it by many methods I use.
Please give me some hints.
Thank you for your attention
| I'll start from your left hand side instead of R.H.S.
All terms contain integer powers of $\sin^2x$ and $\cos^2x$. Rewrite the l.h.s to
$$(\sin^2 x)^3+3(\sin^2x)(\cos^2 x)+(\cos^2x)^3$$
Cubic? It feels like $(a+b)^3=a^3+3a^2b+3ab^2+b^3$ but notice the lack of $a^2b$ and $ab^2$. To make up for the missing powers, I'm tempted to tweak the L.H.S by introducing a harmless multiplier $1=\sin^2 x+\cos^2 x$:
$$ \begin{align}
L.H.S.&=(\sin^2 x)^3+3(\sin^2x)(\cos^2 x)\cdot1+(\cos^2x)^3\\
&=(\sin^2 x)^3+3(\sin^2x)(\cos^2 x)\cdot(\sin^2x+\cos^2x)+(\cos^2x)^3\\
&=\quad ?\quad \text{(Try to expand)} \\
&=(\sin^2 x + \cos^2x)^3=1
\end{align} $$
| {
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Number Theory Contest Problem Given that $x, y$ are positive integers with $x(x + 1)\mid y(y + 1)$, but neither $x$ nor $x + 1$ divides either of $y$ or $y + 1$, and $x^2+ y^2$ as small as possible, find $x^2+ y^2$.
I have tried looking at the values, and it seems that neither $x$ or $x+1$ or the $y$'s are prime.
| Yes, we don't want $x$ or $x+1$ to be a prime power. The first candidate is $x=14$. Then $y=20$ works.
Added: Suppose we find consecutive composites integers $x$ and $x+1$ neither of which is a prime power. Let $x=ab$ and $y=cd$, where $a$ and $b$ are relaively prime, as are $c$ and $d$, and none is equal to $1$. Consider the system of congruences $y\equiv 0\pmod{ac}$, $y=\equiv -1\pmod{bd}$. By the Chinese Remainder Theorem, this has a solution. Note that $x$ does not divide $y$, for $b$ divides $y+1$, so is relatively prime to $y$. The other required "non-divisibilities" can be verified in a similar way. But $x(x+1)$ divides $y(y+1)$.
| {
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Show that $(\sqrt{y^2-x}-x)(\sqrt{x^2+y}-y)=y \iff x+y=0$
Let $x,y$ be real numbers such that
$$\left(\sqrt{y^{2} - x\,\,}\, - x\right)\left(\sqrt{x^{2} + y\,\,}\, - y\right)=y$$
Show that $x+y=0$.
My try:
Let
$$\sqrt{y^2-x}-x=a,\sqrt{x^2+y}-y=b\Longrightarrow ab=y$$
and then
$$\begin{cases}
y^2=a^2+(2a+1)x+x^2\cdots\cdots (1)\\
x^2=b^2+(2b-1)y+y^2\cdots\cdots
\end{cases}$$
$(1)+(2)$
then
$$x=-\dfrac{a^2+b^2+(2b-1)ab}{2a+1}\cdots\cdots (3)$$
so
$$x+y=ab-\dfrac{a^2+b^2+(2b-1)ab}{2a+1}=\dfrac{(a-b)(2ab-a+b)}{2a+1}$$
we take $(3)$ in $(2)$,we have
$$b^2+(2b-1)y+y^2-x^2=\dfrac{(2ab-a+b)(2a^3b+a^3+3a^2b-2ab^3+ab^2+4ab-b^3+b)}{(2a+1)^2}=0$$
so
$$(2ab-a+b)=0$$
or
$$2a^3b+a^3+3a^2b-2ab^3+ab^2+4ab-b^3+b=0$$
if
$$2ab-a+b=0\Longrightarrow x+y=\dfrac{(a-b)(2ab-a+b)}{2a+1}=0$$
and if
$$2a^3b+a^3+3a^2b-2ab^3+ab^2+4ab-b^3+b=0$$
I don't prove
$$x+y=\dfrac{(a-b)(2ab-a+b)}{2a+1}=0?$$
| We need to prove that $x=y$, where
$$\left(\sqrt{y^{2}+x}+x\right)\left(\sqrt{x^{2} + y}- y\right)=y$$ or
$$\left(\sqrt{y^{2}+x}+x\right)\left(\sqrt{x^{2} + y}- y\right)=\left(\sqrt{x^{2}+y}+x\right)\left(\sqrt{x^{2} + y}- x\right)$$ or
$$\left(\sqrt{y^{2}+x}+x\right)\left(\sqrt{x^{2} + y}- y\right)-\left(\sqrt{x^{2}+y}+x\right)\left(\sqrt{x^{2} + y}- y\right)+$$
$$+\left(\sqrt{x^{2}+y}+x\right)\left(\sqrt{x^{2} + y}- y\right)-\left(\sqrt{x^{2}+y}+x\right)\left(\sqrt{x^{2} + y}- x\right)=0$$ or
$$\left(\sqrt{y^{2}+x}-\sqrt{x^2+y}\right)\left(\sqrt{x^{2} + y}- y\right)+\left(\sqrt{x^{2}+y}+x\right)(x-y)=0,$$ which gives $x=y$ or
$$\frac{(1-x-y)\left(\sqrt{x^{2} + y}- y\right)}{\sqrt{y^{2}+x}+\sqrt{x^2+y}}+\sqrt{x^{2}+y}+x=0,$$ which is
$$\sqrt{x^2+y}-y\sqrt{x^2+y}+\sqrt{(x^2+y)(y^2+x)}+x\sqrt{y^2+x}+x^2+xy+y^2=0.$$
Now, we'll consider four cases.
*
*$x\geq0$, $y\geq 0$.
Since $$-y\sqrt{x^2+y}+\sqrt{(x^2+y)(y^2+x)}=\sqrt{x^2+y}\left(\sqrt{y^2+x}-y\right)\geq0,$$ we obtain $x=y=0.$
*$x\geq0,$ $y\leq0.$
It's obvious that this case gives $x=y=0$ again.
*$x\leq0$, $y\geq0.$
Since, $$\sqrt{(x^2+y)(y^2+x)}+x\sqrt{y^2+x}=\sqrt{y^2+x}\left(\sqrt{x^2+y}+x\right)\geq0,$$ it's enough to prove that
$$x^2+xy+y^2\geq(y-1)\sqrt{x^2+y},$$ which is obvious for $y\leq1.$
But for $y\geq1$ by AM-GM we obtain:
$$(y-1)\sqrt{x^2+y}\leq\frac{1}{2}((y-1)^2+x^2+y)$$ and it's enough to prove that
$$x^2+xy+y^2\geq\frac{1}{2}((y-1)^2+x^2+y)$$ or
$$\require{cancel} \cancel{(x+y)^2+y^2+y-1\geq0.}\\
(x+y)^2+y-1\geq0.$$
We see that for $y\geq1$ the equality does not occur and in the case $y<1$ the equality occurs for
$$x^2+xy+y^2=(y-1)\sqrt{x^2+y}=0,$$ which gives $x=y=0$ again.
*$x\leq0$ and $y\leq0.$
In this case it's enough to prove that
$$xy+x\sqrt{y^2+x}\geq0$$ or
$$x\left(y+\sqrt{y^2+x}\right)\geq0,$$ which is obvious.
The equality occurs for $x^2+y^2=0$ and we got $x=y=0$ again.
Done!
| {
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"timestamp": "2023-03-29T00:00:00",
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$\int_0^{2\pi} \sqrt{1-\cos(x)}\,dx = 4\sqrt{2}$. Why? According to the textbook, and Wolfram Alpha the above is correct.
Here is the step by step procedure from Wolfram Alpha for evaluating the indefinite integral:
Take the integral: $$\int\sqrt{1-\cos(x)}\,dx$$ For the integrand $\sqrt{1-\cos(x)}$, substitute $u=1-\cos(x)$ and $du=\sin(x)\,dx$: $$=\int-\frac{1}{\sqrt{2-u}}\,du$$ Factor out constants: $$=-\int\frac{1}{\sqrt{2-u}}\,du$$ For the integrand $1/\sqrt{2-u}$, substitute $s=2-u$ and $ds=-du$: $$=\int\frac{1}{\sqrt{s}}\,ds$$ The integral of $1/\sqrt{s}$ is $2\sqrt{s}$: $$=2\sqrt{s}+\text{constant}$$ Substitute back for $s=2-u$: $$=2\sqrt{2-u}+\text{constant}$$ Substitute back for $u=1-\cos(x)$: $$=2\sqrt{\cos(x)+1}+\text{constant}$$ Which is equivalent for restricted $x$ values to: $$\boxed{=-2\sqrt{1-\cos(x)}\cot\big(\frac{x}{2}\big)+\text{constant}}$$
I understand up to the below (which is a valid solution to the integral): $$2\sqrt{\cos(x)+1}+\text{constant}$$
However, if you evaluate this at $2\pi$ and $0$, you get the same thing, so the definite integral evaluates to zero.
After, you transform the above to: $$-2\sqrt{1-\cos(x)}\cot\big(\frac{x}{2}\big)+\text{constant}$$
The expression is indeterminate at $2\pi$ and $0$ of the form $0 \times \infty$. So I guess you would set up a limit and then use L'Hospital's rule to evaluate the expression at $2\pi$ and $0$ and get the answer to the definite integral?
In any case, all this seems strange. Why should the definite integral evaluated one way give $0$, and in another way give something else?
| Your $u$-substitutions should be injective on their interval of evaluation. Otherwise, you risk running into exactly this sort of issue.
Note that $$\begin{align}|\sin x| &= \sqrt{\sin^2 x}\\ &= \sqrt{1-\cos^2 x}\\ &= \sqrt{1-\cos x}\sqrt{1+\cos x}\\ &=\sqrt{1-\cos x}\sqrt{2-(1-\cos x)},\end{align}$$ so if you want to use $u=1-\cos x$, then $$\frac{du}{dx}=\sin x=\begin{cases}|\sin x|=\sqrt{1-\cos x}\sqrt{2-(1-\cos x)} & 0\le x\le \pi\\-|\sin x|=-\sqrt{1-\cos x}\sqrt{2-(1-\cos x)} & \pi\le x\le2\pi,\end{cases}$$ so $$\begin{align}\int_0^{2\pi}\sqrt{1-\cos x}\,dx &= \int_0^\pi\sqrt{1-\cos x}\,dx+\int_\pi^{2\pi}\sqrt{1-\cos x}\,dx\\ &= \int_0^\pi\frac{|\sin x|}{\sqrt{2-(1-\cos x)}}\,dx+\int_\pi^{2\pi}\frac{|\sin x|}{\sqrt{2-(1-\cos x)}}\,dx\\ &= \int_0^\pi\frac{\sin x\,dx}{\sqrt{2-(1-\cos x)}}-\int_\pi^{2\pi}\frac{\sin x\,dx}{\sqrt{2-(1-\cos x)}}\\ &= \int_0^2\frac{du}{\sqrt{2-u}}-\int_2^0\frac{du}{\sqrt{2-u}}\\ &= 2\int_0^2\frac{du}{\sqrt{2-u}}.\end{align}$$ At that point, we can use that antiderivative, with no need to resubstitute.
Alternately, you could note that $\cos(2\pi-x)=\cos x$, so $$\begin{align}\int_0^{2\pi}\sqrt{1-\cos x}\,dx &= \int_0^\pi\sqrt{1-\cos x}\,dx+\int_\pi^{2\pi}\sqrt{1-\cos x}\,dx\\ &= \int_0^\pi\sqrt{1-\cos x}\,dx+\int_\pi^{2\pi}\sqrt{1-\cos(2\pi-x)}\,dx\\ &= \int_0^\pi\sqrt{1-\cos x}\,dx-\int_{2\pi}^\pi\sqrt{1-\cos(2\pi-x)}\,dx\\ &= \int_0^\pi\sqrt{1-\cos x}\,dx-\int_0^\pi\sqrt{1-\cos x}\frac{d(2\pi-x)}{dx}\,dx\\ &= 2\int_0^\pi\sqrt{1-\cos x}\,dx,\end{align}$$ at which point you can use your $u$-substitution without fear, since the cosine function is injective on $[0,\pi]$.
| {
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What is the minimum value of $a+b+\frac{1}{ab}$ if $a^2 + b^2 = 1$? For the case when $a,b>0,$ I used AM-GM Inequality as follows that:
$\frac{(a+b+\frac{1}{ab})}{3} \geq (ab\frac{1}{ab})^\frac{1}{3}$
This implies that $(a+b+\frac{1}{ab})\geq 3$. Hence, the minimum value of $(a+b+\frac{1}{ab})$ is 3
But the answer is $2+\sqrt{2}$ ... how is it ?
| If you want to use AM-GM for this, you need to ensure that the equality condition can be met along with the constraint, by "balancing coefficients". Illustrated below:
$$a+b+\frac1{ab} = a+b +\frac{1}{2\sqrt2 ab}+\left(1-\frac1{2\sqrt2}\right)\frac1{ab}$$
Now by AM-GM,
$$ a+b +\frac{1}{2\sqrt2 ab}\ge \frac{3}{\sqrt2}$$
and for the remaining term we have again
$$\left(1-\frac1{2\sqrt2}\right)\frac1{ab} \ge \left(1-\frac1{2\sqrt2}\right)\frac{2}{a^2+b^2} = 2-\frac{1}{\sqrt2}$$
Combining these results, we have $a+b+\dfrac1{ab} \ge 2 + \sqrt2$, with equality iff $a=b=\frac1{\sqrt2}$.
Added: The key here is of course knowing how to split the LHS, which is by noting that if for equality we need $a=b=\dfrac1{k~ab}$ and for the constraint we need $a^2+b^2=1$, what could be the value of $k$. The rest is then easy applications of AM-GM.
| {
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How to prove this inequality?$\frac{1}{n+1} + \frac{1}{n+2} + \cdots+\frac{1}{n+n} + \frac{1}{4n} > \ln 2$ $$\frac{1}{n+1} + \frac{1}{n+2} + \cdots +\frac{1}{n+n} + \frac{1}{4n} > \ln 2$$
$n$ is positive integer.
Thank you !
| Let
$$a_n=\frac{1}{n+1} + \frac{1}{n+2} + \cdots +\frac{1}{n+n} + \frac{1}{4n}$$
Then
$$a_{n+1}-a_n=\frac{1}{2n+1}+\frac{1}{2n+2}+\frac{1}{4n+4}-\frac{1}{n+1}-\frac{1}{4n}\\
=\frac{1}{2n+1}+\frac{2}{4n+4}+\frac{1}{4n+4}-\frac{4}{4n+4}-\frac{1}{4n}\\
=\frac{2}{4n+2}-\frac{1}{4n+4}-\frac{1}{4n} <0 $$
Thus, $a_n$ is a strictly decreasing sequence. As $\lim_n a_n = \ln 2$, it follows that
$$a_n > \ln(2)$$
P.S.
$$\frac{2}{4n+2} < \frac{1}{4n+4}+\frac{1}{4n}$$
can be easily proven by bringing everything to the same denominator, but it also follows from the HM-AM inequality:
$$ \frac{2}{\frac{1}{4n+4}+\frac{1}{4n}} < \frac{4n+4n+4}{2}=4n+2$$
| {
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Limit at Infinity $\lim\limits_{m\to\infty}\frac{\sum\limits_{k=1}^m\cot^{2n+1}\left(\frac{k\pi}{2m+1}\right)}{m^{2n+1}}$ How can I prove the following equality?
$$\lim_{m\to{\infty}}\frac{\displaystyle\sum_{k=1}^m\cot^{2n+1}\left(\frac{k\pi}{2m+1}\right)}{m^{2n+1}}=\frac{2^{2n+1}\zeta(2n+1)}{\pi^{2n+1}}$$
| We can use Paramanand Singh's idea of sandwiching $\cot x$ between $\frac1x$ and something close to that regardless of whether the exponents are odd or even.
The inequality we will use is
$$\frac1x - \frac{x}{2} < \cot x < \frac1x\tag{1}$$
for $0 < x \leqslant \frac{\pi}{2}$. To see $(1)$, note
$$\sin x - x\cos x = \int_0^x y\sin y\,dy > 0$$
for $0 < x \leqslant \pi$, which yields the right hand inequality of $(1)$, and
$$x\cos x - \sin x + \frac{x^2}{2}\sin x = \int_0^x \frac{y^2}{2}\cos y\,dy > 0$$
for $0 < x \leqslant \pi/2$, which yields the left inequality of $(1)$.
So we have
$$\frac{1}{x^{2n+1}} > \cot^{2n+1} x > \frac{1}{x^{2n+1}}\left(1 - \frac{x^2}{2}\right)^{2n+1} > \frac{1}{x^{2n+1}} - \frac{2n+1}{2x^{2n-1}}\tag{2}$$
for $0 < x \leqslant \sqrt{2}$ by the monotonicity of $z\mapsto z^{2n+1}$ and Bernoulli's inequality. Since the two terms on the right hand side of $(2)$ are negative for $\sqrt{2} < x \leqslant \pi/2$, and $\cot x$ is non-negative then, we have
$$\frac{1}{x^{2n+1}} > \cot^{2n+1} x > \frac{1}{x^{2n+1}} - \frac{2n+1}{2x^{2n-1}}\tag{3}$$
for all $0 < x \leqslant \pi/2$. Thus
$$\begin{align}
\sum_{k=1}^m \left( \frac{2m+1}{c\cdot k}\right)^{2n+1} &- \left(n+\frac12\right)\sum_{k=1}^m \left(\frac{2m+1}{c\cdot k}\right)^{2n-1}\\
& < \sum_{k=1}^m \cot^{2n+1}\left(\frac{c\cdot k}{2m+1}\right) < \sum_{k=1}^m \left( \frac{2m+1}{c\cdot k}\right)^{2n+1}.
\end{align}$$
Now we have
$$
\frac{1}{m^{2n+1}}\sum_{k=1}^m \left( \frac{2m+1}{c\cdot k}\right)^{2n+1} = \frac{(2+1/m)^{2n+1}}{c^{2n+1}}\sum_{k=1}^m \frac{1}{k^{2n+1}} \xrightarrow{m\to\infty} \frac{2^{2n+1}\zeta(2n+1)}{c^{2n+1}}
$$
and
$$\frac{n+\frac12}{m^{2n+1}}\sum_{k=1}^m \left(\frac{2m+1}{c\cdot k}\right)^{2n-1} = \frac{\left(n+\frac12\right)(2+1/m)^{2n-1}}{m^2c^{2n-1}}\sum_{k=1}^m \frac{1}{k^{2n-1}} \to 0$$
for $0 < c \leqslant \pi$ and $n > 0$, which yields the desired result.
| {
"language": "en",
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How prove $\frac{2}{3}<\frac{3x^6+15x^2+2}{2x^6+15x^4+3}\le\frac{3}{2}$
Let $x\in(0,1]$, show that
$$\dfrac{2}{3}<\dfrac{3x^6+15x^2+2}{2x^6+15x^4+3}\le\dfrac{3}{2}$$
My try: since
$$\begin{align}\dfrac{3x^6+15x^2+2}{2x^6+15x^4+3} &=\dfrac{\dfrac{3}{2}(2x^6+15x^4+3)+15x^2-\dfrac{45}{2}x^4-\dfrac{5}{2}}{2x^6+15x^4+3} \\ &=\dfrac{3}{2}-\dfrac{5}{2}\cdot\dfrac{9x^4-6x^2+1}{2x^6+15x^4+3}\end{align}$$
then
I can't,Thank you for your help.
| It is enough to note that for all $x\in\mathbb{R}$
$$
\dfrac{3}{2}-\dfrac{3x^6+15x^2+2}{2x^6+15x^4+3}=\dfrac{5}{2}\cdot\dfrac{(3x^2-1)^2}{2x^6+15x^4+3}\geq 0
$$
and
$$
\dfrac{3x^6+15x^2+2}{2x^6+15x^4+3}-\dfrac{2}{3}=\dfrac{5x^2(x^2-3)^2}{3(2x^6+15x^4+3)}\geq 0
$$
| {
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Find the sum of the series $\sum \frac{1}{n(n+1)(n+2)}$ I got this question in my maths paper
Test the condition for convergence of $$\sum_{n=1}^\infty \frac{1}{n(n+1)(n+2)}$$
and find the sum if it exists.
I managed to show that the series converges but I was unable to find the sum. Any help/hint will go a long way. Thank you.
| You are on the right track. Fix $N \geq 1$. Then
$$\begin{align}
\sum_{n=1}^{N} \frac{1}{n \cdot (n+1) \cdot (n+2)}
&= \sum_{n=1}^N \left(\frac{1}{2n}-\frac{1}{n+1}+\frac{1}{2n+4}\right)
= \frac{1}{2}\sum_{n=1}^N \frac{1}{n}-\sum_{n=1}^{N}\frac{1}{n+1}+\frac{1}{2}\sum_{n=1}^{N} \frac{1}{n+2}\\
&= \frac{1}{2}\sum_{n=1}^N \frac{1}{n}-\sum_{n=2}^{N+1}\frac{1}{n}+\frac{1}{2}\sum_{n=3}^{N+2} \frac{1}{n} \\
&= \frac{1}{2}\left(1+\frac{1}{2}\right) + \frac{1}{2}\sum_{n=3}^N \frac{1}{n}-\left(\frac{1}{2}+\frac{1}{N+1}\right)-\sum_{n=3}^{N}\frac{1}{n}+\\
&\quad \left(\frac{1}{N+1}+\frac{1}{N+2}\right)+\frac{1}{2}\sum_{n=3}^{N} \frac{1}{n} \\
\end{align}$$
Can you continue from there? (the partial sums cancel out, and you only have a few remaining terms. Taking the limit $N\to\infty$ will give you the limit.)
| {
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"question_score": "18",
"answer_count": 10,
"answer_id": 4
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Does the integral $\int _0^\infty \frac 1{1+(x \cos x)^2}\,dx$ converge? Question
$$\int _0^\infty \frac 1{1+(x \cos x)^2}dx$$
Does this converge?
Thoughts
Tried bounding it with $\frac 1{\cos^2x}$ for $x>1$ but not much success.
| Note that
*
*$(x \cos x)^{2} \leq n^{2}\pi^{2} \cos^{2} x$ on each interval $[(n-1)\pi, n\pi]$, and
*$\sin^{2} x \leq x^{2} $ for any $x$.
So it follows that
\begin{align*}
\int_{0}^{\infty} \frac{dx}{1+(x \cos x)^{2}}
&\geq \sum_{n=1}^{\infty} \int_{0}^{\pi} \frac{dx}{1+n^{2}\pi^{2}\cos^{2} x}
= 2 \sum_{n=1}^{\infty} \int_{0}^{\frac{\pi}{2}} \frac{dx}{1+n^{2}\pi^{2}\sin^{2} x} \\
&\geq 2 \sum_{n=1}^{\infty} \int_{0}^{\pi/2} \frac{dx}{1+n^{2}\pi^{2}x^{2}} \\
&= \sum_{n=1}^{\infty} \frac{2}{n\pi} \int_{0}^{n\pi^{2}/2} \frac{dx}{1+x^{2}} \qquad (n\pi x \mapsto x)
\end{align*}
But since
$$ \frac{2}{\pi} \int_{0}^{n\pi^{2}/2} \frac{dx}{1+x^{2}} \xrightarrow[]{n\to\infty} 1, $$
by the Limit Comparison Test we have
$$ \sum_{n=1}^{\infty} \frac{2}{n\pi} \int_{0}^{n\pi^{2}/2} \frac{dx}{1+x^{2}} = \infty. $$
This proves the divergence of the integral.
A little bit elaborated argument shows that, for $N\pi \leq x \leq (N+1)\pi$, we have
$$ \int_{0}^{x} \frac{dt}{1+(t \cos t)^{2}} = \sum_{n=1}^{N} \int_{0}^{\pi/2} \frac{2 \, dt}{1+n^{2}\pi^{2}\sin^{2} t} + O(1) $$
By observing the following identity
$$ \int_{0}^{\pi/2} \frac{dt}{1+a^{2}\sin^{2} t} = \frac{\pi}{2\sqrt{a^{2}+1}}, $$
it follows that
$$ \int_{0}^{\pi/2} \frac{2 \, dt}{1+n^{2}\pi^{2}\sin^{2} t} = \frac{1}{n} + O\left(\frac{1}{n^{3}} \right). $$
Therefore we have the following estimate:
$$ \int_{0}^{x} \frac{dt}{1+(t \cos t)^{2}} = \log x + O(1). $$
| {
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"url": "https://math.stackexchange.com/questions/564067",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How find this $\sum_{i=1}^{5}\tan^4{\frac{i\pi}{11}}$ show that:$$\tan^4{\dfrac{\pi}{11}}+\tan^4{\dfrac{2\pi}{11}}+\tan^4{\dfrac{3\pi}{11}}+\tan^4{\dfrac{4\pi}{11}}+\tan^4{\dfrac{5\pi}{11}}=2365$$
my try:
I think first we can find this value：$$x=\tan{\dfrac{\pi}{11}}+\tan{\dfrac{2\pi}{11}}+\tan{\dfrac{3\pi}{11}}+\tan{\dfrac{4\pi}{11}}+\tan{\dfrac{5\pi}{11}}$$
But I can't.Thank you for you help.
| Using this for odd $n=2m+1$ and setting $\tan(2m+1)x=0\implies x=\frac{r\pi}{2m+1}$ where $0\le r\le 2m$
So, the roots of $$\tan^{2m+1}x-\binom{2m+1}2\tan^{2m-1}x+\binom{2m+1}4\tan^{2m-3}x-\cdots=0$$ are $\tan\frac{r\pi}{2m+1}$ where $0\le r\le 2m$
Discarding $\tan 0=0,$ the roots of $$\tan^{2m}x-\binom{2m+1}2\tan^{2m-2}x+\binom{2m+1}4\tan^{2m-4}x-\cdots=0$$ are $\tan\frac{r\pi}{2m+1}$ where $1\le r\le 2m$
Now observe that $\displaystyle \tan\left(\frac{(2m+1-u)\pi}{2m+1}\right)=\tan\left(\pi-\frac{u\pi}{2m+1}\right)=-\tan\left(\frac{u\pi}{2m+1}\right)$
$\displaystyle\implies\tan^2\left(\frac{(2m+1-u)\pi}{2m+1}\right)=\tan^2\left(\frac{u\pi}{2m+1}\right)$
So, $\displaystyle\tan^2\frac{r\pi}{2m+1}$ where $1\le r\le m$ or $m+1\le r\le2m$ (more generally we can replace $u,1\le u\le m$ with $2m+1-u$) are the $m$ roots of $$t^mx-\binom{2m+1}2t^{m-1}x+\binom{2m+1}4t^{m-2}x-\cdots=0$$
$\displaystyle\sum_{1\le r\le m}\tan^4\frac{r\pi}{2m+1}= \left(\sum_{1\le r\le m}\tan^2\frac{r\pi}{2m+1}\right)^2-2\sum_{1\le r\le m}\tan^2\frac{r_i\pi}{2m+1}\tan^2\frac{r_j\pi}{2m+1}$ where $1\le i<j\le m$
Now using Vieta's formula,
$\displaystyle\sum_{1\le r\le m}\tan^2\frac{r\pi}{2m+1}=\binom{2m+1}2$
and $\displaystyle\sum_{1\le r\le m}\tan^2\frac{r_i\pi}{2m+1}\tan^2\frac{r_j\pi}{2m+1}=\binom{2m+1}4$ where $1\le i<j\le m$
In this problem $m=5$
| {
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"source": "stackexchange",
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generating function and lucas numbers Defines the sequence $l$.
$l_n = l_{n−1} + l_{n−2}$, for all $n \ge 2$.
So the first few members of the sequence are: $2, 1, 3, 4, 7, 11, 18\;.$
Find the generating function for this sequence and deduce the following formula
$$l_n =\left(\frac{1 − \sqrt5}2\right)^n +\left(\frac{1 + \sqrt5}2\right)^n\;.$$
After some calculation I got the generating function
$g(x)={x-2\over x^2+x-1} =2+x+3x^2+4x^3\cdots$
| Let $$g(x)=h_0+h_1x+h_2x^2+\cdots+h_nx^n+\cdots$$ $$-xg(x)=-h_0x-h_1x^2-h_2x^3-\cdots-h_nx^{n+1}-\cdots$$ $$-x^2g(x)=-h_0x^2-h_1x^3-h_2x^4-\cdots-h_nx^{n+2}-\cdots.$$ Adding we obtain $$(1-x-x^2)g(x)=h_0+(h_1-h_o)x+(h_2-h_1-h_0)x^2+\cdots.$$ Since $l_0=2$ and $l_1=1$ we now have $$g(x)={2\over 1-x-x^2}-{x\over 1-x-x^2}.$$ Let $1-x-x^2=(1-q_1x)(1-q_2x)$ so that $$g(x)={2-x\over (1-q_1x)(1-q_2x)}$$ where $q_1={1+\sqrt5\over 2}$ and $q_2={1-\sqrt5\over 2}$. Using partial fractions $${2-x\over (1-q_1x)(1-q_2x)}={A\over 1-q_1x}+{B\over 1-q_2x}$$ we obtain $A=1$ and $B=1$. So $${1\over 1-q_1x}+{1\over 1-q_2x}.$$Now using the geometric series we have $$\sum_{n=0}^\infty (q_1x)^n+\sum_{n=0}^\infty (q_2x)^n=\sum_{n=0}^\infty[(q_1)^n+(q_2)^n]x^n.$$ Thus $$l_n =\left(\frac{1 − \sqrt5}2\right)^n +\left(\frac{1 + \sqrt5}2\right)^n\;.$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Solving the improper integral $\int_0^\infty \frac{x^{1/3}}{1+x^2} \mathrm dx$ I'm trying to solve:
$\displaystyle \int \limits_0^\infty \dfrac{x^{1/3}}{1+x^2} \mathrm dx$
I have tried contour integration with $C_R^+$ and the real line like this:
$\displaystyle \int \limits_T \dfrac{z^{1/3}}{1+z^2} \mathrm dz = \int \limits_{-\infty}^\infty \dfrac{z^{1/3}}{1+z^2}\mathrm dz + \int \limits_{C_R^+} \frac{z^{1/3}}{1+z^2} \mathrm dz$
Where the last integral tends to $0$ as $R \longrightarrow \infty$
$\text{Res}(f(z);i) = \dfrac{i^{1/3}}{2i}$
and
$\displaystyle \int \limits_{-\infty}^\infty \frac{z^{1/3}}{1+z^2}\mathrm dz = \int \limits_{0}^\infty \frac{z^{1/3}}{1+z^2} \mathrm dz + \int \limits_{-\infty}^0\frac{z^{1/3}}{1+z^2} \mathrm dz$
If i manipulate the last term by changing the limits and substitute $u=-t$ i get:
$\displaystyle \int \limits_{-\infty}^0\dfrac{z^{1/3}}{1+z^2} \mathrm dz = -\int \limits_{0}^{-\infty}\frac{z^{1/3}}{1+z^2} \mathrm dz$
If i now substitue $u=-z, u'=-1$
$\displaystyle \int \limits_{-\infty}^0\dfrac{z^{1/3}}{1+z^2} \mathrm dz = \int \limits_{0}^{\infty}\dfrac{(-u)^{1/3}}{1+u^2}\mathrm dz = (-1)^{1/3}\int \limits_{0}^{\infty}\dfrac{u^{1/3}}{1+u^2}\mathrm dz$
$\displaystyle \int \limits_{-\infty}^\infty \dfrac{z^{1/3}}{1+z^2}\mathrm dz = \left(1+e^\frac{i\pi}{3}\right) \int \limits_{0}^\infty \frac{z^{1/3}}{1+z^2} \mathrm dz $
So i end up with:
$\displaystyle \dfrac{2i\cdot\pi \cdot e^{i\cdot\pi /6}}{2i\cdot\left(1+e^\frac{i\pi}{3}\right)} = \dfrac{\pi\cdot e^{i\cdot \pi /6}}{\left(1+e^\frac{i\pi}{3}\right)} = \int \limits_{0}^\infty \dfrac{z^{1/3}}{1+z^2} \mathrm dz$ which is wrong answer
| Hint: using the substitution $x=u^{\frac{3}{2}}$ your integral will be equal to $$\frac{3}{2}\int\frac{u}{u^3+1}du$$ which can be evaluated by decomposing partial fractions. Additional note: if you divide the integral as an integral from $0$ to $1$ and an integral from $1$ to $\infty$, it is clearly that the first one is divergent.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove that $\sum^n_{k=1} k^2 = \binom{n+1}{2} + 2\binom{n+1}{3}$ for $n\geq 2$ Prove, for all $n\geq 2$ that
$$\sum^n_{k=1} k^2 = \binom{n+1}{2} + 2\binom{n+1}{3}$$
Let us prove the inductive base for $n = 2$
$$\rm{LHS} = 1^2 + 2^2= 1 + 4 = 5$$
$$\rm{RHS} = \binom{3}{2} + 2\binom{3}{3} = 3 + 2\cdot 1 = 5$$
$$\rm{LHS} = \rm{RHS}$$
as desired.
Now, assume for some $k$,
$$1^2 + 2^2 + \dots + k^2 = \binom{k+1}{2} + 2\binom{k+1}{3}$$
To prove the inductive step, it is sufficient to prove that,
$$\binom{k+2}{2} + 2\binom{k+2}{3} - \binom{k+1}{2} - 2\binom{k+1}{3} = (k+1)^2$$
The LHS can be simplified as:
$$\frac{(k+2)!}{2!k!} + 2\left[\frac{(k+2)!}{3!(k-1)!}\right] - \frac{(k+1)!}{2!(k-1)!} - 2\left[\frac{(k+1)!}{3!(k-2)!}\right]$$
$$=\frac{3(k+2)! + 2k(k+2)! -3k(k-1)! - 2k(k-1)(k+1)!}{6k!}$$
$$=\frac{(k+1)!(3k + 6 - 2k^2 - 2k) + (k-1)!(2k^3 + 2k^2 - 3k)}{6k!}$$
$$=\frac{(k+1)!(-2k^2+k+6) + k!(2k^2+2k - 3)}{6k!}$$
$$=\frac{(k+1)(-2k^2 + k + 6) + 2k^2 +2k - 3}{6}$$
$$=\frac{-2k^3 + k^2 + 9k + 3}{6}$$
I'm pretty sure I'm making a mistake somewhere, but I can't figure it out. If someone could complete this inductive proof for me, I will be grateful.
Also, I don't feel satisfied with this ugly proof. Please add a combinatorial proof for this in your answer if possible.
| There is a typo in one of your steps, it should be as follows (look at the red ${\color{red} + }$) :
$$\frac{3(k+2)! + 2k(k+2)! -3k(k{\color{red}+}1)! - 2k(k-1)(k+1)!}{6k!} \,\,(*)$$
And then, I do not see how you got from there to
$$\frac{(k+1)!(3k + 6 - 2k^2 - 2k) + (k-1)!(2k^3 + 2k^2 - 3k)}{6k!}$$
which in fact is not equal to the first expression above.
I would recommend you try taking out the factor $(k+1)!$ in the former expression $(*)$ above.
Another proof
Now I am going to present a more combinatorial proof (not strictly combinatorial though) :
Observation 1: We have $$ k^2 = \binom{k}{2} + \binom{k+1}{2} $$
Observation 2: To compute $ S_n = \sum_{k=0}^n k^2 = \sum_{k=0}^n\left\{\binom{k}{2} + \binom{k+1}{2} \right\} $ it is enough to compute the sum $A_n = \sum_{k=0}^n \binom{k}{2}$, because we then have $S_n = A_n + A_{n+1}$.
Observation 3: We have $ A_n = \binom{n+1}{3} $. Here comes the combinatorial part:
Note that $A_n$ is actually the number of unordered "trios" you can make out of $\{1,\ldots,n+1\}$. Indeed, $\binom{n}{2}$ is the number of trios of the form $\{i,j,n+1\}$ with $1\leq i<j\leq n$, then $\binom{n-1}{2}$ is the number of trios of the form $\{i,j,n\}$ with $1\leq i < j \leq n-1$ and so on. That is, $k$ corresponds to the choice of the maximum element in the unordered trio. Since there are $\binom{n+1}{3}$ unordered trios, we get
$$ A_n = \binom{n+1}{3} $$
Conclusion: We have $$ S_n = \binom{n+1}{3} + \binom{n+2}{3}$$ but remember that $\binom{n+2}{3} = \binom{n+1}{3} + \binom{n+1}{2}$ (this has a well-known combinatorial intepretation), and so
$$ S_n = \binom{n+1}{2} + 2\binom{n+1}{3}$$
as desired. $\square$
Remark By the same argument above, we actually prove that $$\sum_{k=0}^n \binom{k}{r} = \binom{n+1}{r+1}$$
| {
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Integer solutions of the equation $x^2+y^2+z^2 = 2xyz$
Calculate all integer solutions $(x,y,z)\in\mathbb{Z}^3$ of the equation $x^2+y^2+z^2 = 2xyz$.
My Attempt:
We will calculate for $x,y,z>0$. Then, using the AM-GM Inequality, we have
$$
\begin{cases}
x^2+y^2\geq 2xy\\
y^2+z^2\geq 2yz\\
z^2+x^2\geq 2zx\\
\end{cases}
$$
So $x^2+y^2+z^2\geq xy+yz+zx$. How can I solve for $(x,y,z)$ after this?
| Suppose that $(x,y,z)$ is a solution. An even number of these must be odd. If two are odd, say $x$ and $y$, then $x^2+y^2$ has shape $4k+2$, and therefore so does $x^2+y^2+z^2$, since $z^2$ is divisible by $4$. But $2xyz$ has shape $4k$.
So $x,y,z$ are all even, say $2u,2v,2w$. Substituting we get $u^2+v^2+w^2=4uvw$.
Again, $u,v,w$ must be all even.
Continue, forever. We conclude that $x$, $y$, and $z$ are divisible by every power of $2$.
It follows that $x=y=z=0$.
Remark: The same argument can be used for $x^2+y^2+z^2=2axyz$.
This is an instance of Fermat's Method of Infinite Descent, aka induction.
| {
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Proof: Divisible by 15 I have to proof that $16^m - 1$ is divisible by $15$. Is my following proof correct?
$$\begin{align}
16^m - 1=&\frac{16^{m+1}}{16}-1\\
=&\frac{16^{m+1}-16}{16} \\
=&(16^{m+1}-16)\cdot\frac{1}{16} \\
=&\underbrace{(16^{m+1}-16)}\cdot\frac{1}{15(1+1/15)} \\
=&\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,a\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\cdot\frac{1}{1+1/15}\cdot\frac{1}{15}\\
=&\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,a\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\cdot\frac{1}{16/15}\cdot\frac{1}{15}\\
=&\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,a\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\cdot\frac{1}{1}\cdot\frac{15}{16}\cdot\frac{1}{15}\\
=&\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\underbrace{a\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\cdot\frac{15}{16}}\cdot\frac{1}{15}\\
=&\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,b\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\cdot\frac{1}{15}\\
\end{align}$$
$$\therefore \boxed{16^m - 1=\frac{b}{15}}$$
Or is this the wrong way and I have to do it with mathematical induction?
| $$
16^{n} - 1 = 15\,{16^{n} - 1 \over 16 - 1} = 15\sum_{k = 0}^{n - 1}16^{k}
$$
$$
\color{#0000ff}{\large{16^{n} - 1 \over 15} = \sum_{k = 0}^{n - 1}16^{k}}
$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Understanding the solution of a telescoping sum $\sum_{n=1}^{\infty}\frac{3}{n(n+3)}$ I'm having trouble understanding infinite sequence and series as it relates to calculus, but I think I'm getting there.
For the below problem:
$$\sum_{n=1}^{\infty}\frac{3}{n(n+3)}$$
The solution shows them breaking this up into a sum of partial fractions. I understand how they got the first two terms, but then they show the partial fractions of the $n$ terms and I find myself lost.
Here is the what I'm talking about:
$$S_n=\sum_{i=1}^{n}\frac{3}{i(i+3)}=\sum_{i=1}^{n}\left(\frac{1}{i}-\frac{1}{i+3} \right)$$
The next few terms are shown to be this:
$$=\left(1-\frac{1}{4}\right)+\left(\frac{1}{2}-\frac{1}{5}\right)+\left(\frac{1}{3}-\frac{1}{6}\right)+\left(\frac{1}{4}-\frac{1}{7}\right)+..+$$
And it continues but this is the part where I get confused...
$$\left(\frac{1}{n-3}-\frac{1}{n}\right)+\left(\frac{1}{n-2}-\frac{1}{n+1}\right)+\left(\frac{1}{n-1}-\frac{1}{n+2}\right)+\left(\frac{1}{n}-\frac{1}{n+3}\right)$$
Where did the $n$ terms in the denominator come from?
| $$\sum_{n=1}^{\infty}\frac{3}{n(n+3)}=\lim_{k\to\infty}\sum_{n=1}^{k}\frac{3}{n(n+3)}=$$
$$=\lim_{k\to\infty}\left(\frac{11}{6}-\left(\frac{1}{k+1}+\frac{1}{k+2}+\frac{1}{k+3}\right)\right)=\frac{11}{6}$$
because
$$\sum_{n=1}^{k}\frac{3}{n(n+3)}=\sum_{n=1}^{k}\left(\frac{1}{n}-\frac{1}{n+3}\right)=\sum_{n=1}^{k}\frac{1}{n}-\sum_{n=1}^{k}\frac{1}{n+3}=$$
$$=1+\frac{1}{2}+\frac{1}{3}+\sum_{n=4}^{k}\frac{1}{n}-\sum_{n=1}^{k}\frac{1}{n+3}=\frac{11}{6}+\sum_{n=4}^{k}\frac{1}{n}-\sum_{j=4}^{k+3}\frac{1}{j}=$$
$$=\frac{11}{6}+\sum_{n=4}^{k}\frac{1}{n}-\left(\sum_{j=4}^{k}\frac{1}{j}+\frac{1}{k+1}+\frac{1}{k+2}+\frac{1}{k+3}\right)=$$
$$=\frac{11}{6}-\left(\frac{1}{k+1}+\frac{1}{k+2}+\frac{1}{k+3}\right)$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Solve for: $2\log_3\left(x^2-4\right)+3\sqrt{\log_3\left(x+2\right)^2}-\log_3\left(x-2\right)^2\leq4$ Solve for: $$2\log_3\left(x^2-4\right)+3\sqrt{\log_3\left(x+2\right)^2}-\log_3\left(x-2\right)^2\leq4$$
My try:
$2\log_3\left(x^2-4\right)+3\sqrt{\log_3\left(x+2\right)^2}-\log_3\left(x-2\right)^2\leq4\\\Leftrightarrow \log_3\left(x^2-4\right)^2+3\sqrt{\log_3\left(x+2\right)^2}-\log_3\left(x-2\right)^2\leq4\\\Leftrightarrow \log_3\left[\left(x-2\right)^2\times\left(x+2\right)^2\right]+3\sqrt{\log_3\left(x+2\right)^2}-\log_3\left(x-2\right)^2\leq4\\\Leftrightarrow \log_3\left(x-2\right)^2+\log_3\left(x+2\right)^2+3\sqrt{\log_3\left(x+2\right)^2}-\log_3\left(x-2\right)^2\leq4\\\Leftrightarrow \log_3\left(x+2\right)^2+3\sqrt{\log_3\left(x+2\right)^2}-4\leq0\,\,\,(*)$
Put: $t=\sqrt{\log_3\left(x+2\right)^2}\Rightarrow (*)\Leftrightarrow t^2+3t-4\leq0$
But I don't know Conditions defined for this math? Could help me?
| The domain of the first inequation is $x>3$ or $x\leq 3$, so the range of your $t$ is $t\geq 0$. Then you can use $t\geq 0$ as the domain to solve $t^2+3t−4\leq0$, which gets $0\leq t \leq 1$. At last, you just go back to solve $0 \leq t \leq 1$.
| {
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Calculation of $\int_{0}^{\pi}\frac{1}{(5+4\cos x)^2}dx$ Calculation of $\displaystyle \int_{0}^{\pi}\frac{1}{(5+4\cos x)^2}dx$
$\bf{My\; Try}::$ Using $\displaystyle \cos x = \frac{1-\tan^2 \frac{x}{2}}{1+\tan^2 \frac{x}{2}}$
Let $\displaystyle I = \int_{0}^{\pi}\frac{1}{\left(5+\frac{4-4\tan^2 \frac{x}{2}}{1+\tan^2 \frac{x}{2}}\right)^2}dx = \int_{0}^{\pi}\frac{1+\tan^2 \frac{x}{2}}{\left(9+\tan^2 \frac{x}{2}\right)^2}dx$
$\displaystyle I = \int_{0}^{\pi}\frac{\sec^2 \frac{x}{2}}{\left(9+\tan^2 \frac{x}{2}\right)^2}dx$
Now Let $\tan \frac{x}{2} = t$ and $\sec^2 \frac{x}{2}dx = 2dt$
$\displaystyle I = 2\int_{0}^{\infty}\frac{1}{(9+t^2)^2dt}$
Now I did not understand how can i solve after that
Help Required
Thanks
| Using Weierstrass substitution, we get
$$\int_0^\pi \frac{dx}{(5+4\cos x)^2}=\int_0^\infty\frac{1}{(5+4\frac{1-t^2}{1+t^2})^2}\frac{2dt}{1+t^2}=2\int_0^\infty \frac{t^2+1}{(t^2+9)^2}dt=\int_{-\infty}^\infty \frac{t^2+1}{(t^2+9)^2}dt$$
Now let
$$f(z)=\frac{z^2+1}{(z^2+9)^2}$$
Then, $f$ has a poles of order 2 at $\pm 3i$. We get
$$\mathrm{res}_{z=3i}f=\lim_{z\to 3i}\frac{d}{dz}\frac{z^2+1}{(z+3i)^2}=\lim_{z\to 3i}\left(\frac{-2(1+z^2)}{(z+3i)^3}+\frac{2z}{(z+3i)^2}\right)=-\frac{5i}{54}$$
Hence, letting $\gamma$ be the semi circle of radius $R>3$ centered at the origin and in the upper half-plane, we get that $\gamma$ encloses the pole $z=3i$, so
$$\int_\gamma f(z)dz=2\pi i\ \mathrm{res}_{z=3i}f=\frac{5\pi}{27}$$
Letting $R\to\infty$, we see that the integral on the arc-section vanish so that we are left with
$$\int_0^\pi \frac{dx}{(5+4\cos x)^2}=\frac{5\pi}{27}$$
| {
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Infinite series $\sum _{n=2}^{\infty } \frac{1}{n \log (n)}$ Recently, I encountered a problem about infinite series.
So my question is how to know whether the infinite series $\sum _{n=2}^{\infty } \frac{1}{n \log (n)}$ is convergent?
| $\sum_{n≥2} \frac{1}{n(\log n)}=\sum_{n≥2} \int_{1}^{\infty} \frac{1}{n^x} dx$
$=\int_{1}^{\infty} \zeta_2(x) dx$ where [$\zeta_2(x)=\zeta(x)-1$]
So, this is the area under the curve $\zeta(x), x≥1, y≥0$.
Now, for $x>1$,
$\zeta_2(x) > {\frac{1}{2^x} + 2\frac{1}{4^x}+ 4\frac{1}{8^x} +......}$
$=\frac{1}{2}(\frac{1}{2^{x-1}}+\frac{1}{4^{x-1}}+....)$
$=\frac{1}{2^x}(\frac{1}{1-\frac{1}{2^{x-1}}})$
$=\frac{1}{2}(\frac{1}{2^{x-1}-1})$.
So, $\int_{1}^{\infty} \zeta_{2}(x) dx >\int_{1}^{\infty} \frac{1}{2}(\frac{1}{2^{x-1}-1}) dx=\int_{0}^{\infty} \frac{1}{2}(\frac{1}{2^z-1})dz $ which diverges.
This means $\int_{1}^{\infty} \zeta_{2}(x) dx$ also diverges.
And this implies $\sum_{n≥2} \frac{1}{n(\log n)}$ diverges.
| {
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How prove this $\tan{\frac{2\pi}{13}}+4\sin{\frac{6\pi}{13}}=\sqrt{13+2\sqrt{13}}$ Nice Question:
show that: The follow nice trigonometry
$$\tan{\dfrac{2\pi}{13}}+4\sin{\dfrac{6\pi}{13}}=\sqrt{13+2\sqrt{13}}$$
This problem I have ugly solution, maybe someone have nice methods? Thank you
My ugly solution:
let $$A=\tan{\dfrac{2\pi}{13}}+4\sin{\dfrac{6\pi}{13}},B=\tan{\dfrac{4\pi}{13}}+4\sin{\dfrac{\pi}{13}}$$
since
$$\tan{w}=2[\sin{(2w)}-\sin{(4w)}+\sin{(6w)}-\sin{(8w)}+\cdots\pm \sin{(n-1)w}]$$
where $n$ is odd,and $w=\dfrac{2k\pi}{n}$
so
$$\tan{\dfrac{2\pi}{13}}=2\left(\sin{\dfrac{4\pi}{13}}-\sin{\dfrac{5\pi}{13}}+\sin{\dfrac{\pi}{13}}+\sin{\dfrac{3\pi}{13}}-\sin{\dfrac{6\pi}{13}}+\sin{\dfrac{2\pi}{13}}\right)$$
$$\tan{\dfrac{4\pi}{13}}=2\left(\sin{\dfrac{5\pi}{13}}-\sin{\dfrac{3\pi}{13}}-\sin{\dfrac{2\pi}{13}}-\sin{\dfrac{6\pi}{13}}-\sin{\dfrac{\pi}{13}}+\sin{\dfrac{4\pi}{13}}\right)$$
then
$$A^2-B^2=(A+B)(A-B)=16\left(\sin{\dfrac{\pi}{13}}+\sin{\dfrac{3\pi}{13}}+\sin{\dfrac{4\pi}{13}}\right)\left(\sin{\dfrac{2\pi}{13}}-\sin{\dfrac{5\pi}{13}}+\sin{\dfrac{6\pi}{13}}\right)=\cdots=4\sqrt{13}$$
$$AB=\cdots=6\left(\cos{\dfrac{\pi}{13}}+\cos{\dfrac{2\pi}{13}}+\cos{\dfrac{3\pi}{13}}-\cos{\dfrac{4\pi}{13}}-\cos{\dfrac{5\pi}{13}}+\cos{\dfrac{6\pi}{13}}\right)=\cdots=3\sqrt{3}$$
so
$$A=\sqrt{13+2\sqrt{13}},B=\sqrt{13-2\sqrt{13}}$$
Have other nice metods?
and I know this is simlar 1982 AMM problem: How to prove that: $\tan(3\pi/11) + 4\sin(2\pi/11) = \sqrt{11}$
But My problem is hard then AMM problem。Thank you very much!
| The following argument is more or less a duplicate in this paper:
Let $x=e^{2\pi i/13}$. Then $$i\tan{2\pi/13}=\frac{x^2-1}{x^2+1}=\frac{x^2-x^{26}}{x^2+1}$$
(recall that $x^{13}=1$)
$$=x^2(1-x^2)(1+x^4+x^8+x^{12}+x^3+x^7)$$
$$=(x+x^2+x^5+x^6+x^9+x^{10}-x^3-x^4-x^7-x^8-x^{11}-x^{12})$$
$$4i\sin{6\pi/13}=2(x^3-x^{10})$$
So $i\tan{2\pi/13}+4i\sin{6\pi/13}=(x+x^2+x^3+x^5+x^6+x^9-x^4-x^7-x^8-x^{10}-x^{11}-x^{12})$
Recall that $1+x+x^2+\cdots+x^{12}=0$.
After some tedious computation, we arrive at
$$(x+x^2+x^3+x^5+x^6+x^9)(x^4+x^7+x^8+x^{10}+x^{11}+x^{12})$$
$$=4+x+x^3+x^4+x^9+x^{10}+x^{12}$$
The key step in the deduction is the famous exponential sum of Gauss, which gives,
$$1+2(x+x^4+x^9+x^3+x^{12}+x^{10})=\sqrt{13}.$$
Hence $$(x+x^2+x^3+x^5+x^6+x^9)(x^4+x^7+x^8+x^{10}+x^{11}+x^{12})=(7+\sqrt{13})/2$$
Recall our formula $1+x+x^2+\cdots+x^{12}=0$ again, and
$$(x+x^2+x^3+x^5+x^6+x^9-x^4-x^7-x^8-x^{10}-x^{11}-x^{12})^2=(-1)^2-4\times(7+\sqrt{13})/2$$
$$=-13-2\sqrt{13}$$
Hence $i\tan{2\pi/13}+4i\sin{6\pi/13}=\pm i\sqrt{13+2\sqrt{13}}$
and it is obvious that $\tan{2\pi/13}+4\sin{6\pi/13}=\sqrt{13+2\sqrt{13}}$, Q.E.D.
P.S. I have a strong feeling that a generalization of such an identity to all primes is possible, but I cannot work them out right now.
| {
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All roots of the quartic equation $a x^4 + b x^3 + x^2 + x + 1 = 0$ cannot be real Problem
Prove that all roots of $a x^4 + b x^3 + x^2 + x + 1 = 0$ cannot be real. Here $a,b \in \mathbb R$, and $a \neq 0$.
Source
This is one of the previous year problem of Regional Math Olympiad (India). I had a hard time solving it, so thought I'd better ask here.
Observations
*
*Some real roots are possible: when $a<0$, the equation has two of them.
*If one more coefficient was allowed to be arbitrary: $a x^4 + b x^3 + cx^2 + x + 1 = 0$, then the roots could be all real, since every quartic can be brought into such form by scaling
| Let $f(x) = x^4 + x^3 + x^2 + bx + a$. Then the quartic
$ax^4 + bx^3 + x^2 + x + 1$ is $x^4 f(1/x)$, and has four real roots
iff $f$ does. But $f''(x) = 12x^2 + 6x + 2 > 0$ for all $x$.
Therefore $f$ is convex upwards and hence can have at most two real roots, QED.
| {
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Solve equation $ 1+2^x=3^y$ Find integers $x$ and $y$ such that$$ 1+2^x=3^y.$$
It is obvious that $x = y = 1$ and $x = 3, y = 2$ are solutions.
I think others are not.
How to show that?
| Case 1: $y$ is odd. Then $3^y \equiv 3 \mod 8$. When is $2^x \equiv 2 \mod 8$?
Case 2: $y = 2 z$ is even. Then $2^x = 3^{2z} - 1 = (3^z - 1)(3^z + 1)$, so
$3^z - 1$ and $3^z + 1$ are powers of $2$. What powers of $2$ differ by $2$?
By the way, this proof of a special case of Catalan's conjecture dates back to Gersonides (Levi ben Gershon) in 1343.
| {
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Evaluate $\int\frac{\cot{x}}{1+\sin{x}+\cos{x}} \mathrm dx$ Find this integral:
$$\int\dfrac{\cot{x}}{1+\sin{x}+\cos{x}}\mathrm dx$$
My try: since
$$1+\sin{x}+\cos{x}=2\cos^2{\dfrac{x}{2}}+2\sin{\dfrac{x}{2}}\cos{\dfrac{x}{2}}$$
$$\cot{x}=\dfrac{1-\tan^2{\dfrac{x}{2}}}{2\tan{\dfrac{x}{2}}}$$
so
$$\dfrac{\cot{x}}{1+\sin{x}+\cos{x}}=\dfrac{1-\tan^2{\dfrac{x}{2}}}{2\tan{\dfrac{x}{2}}\left(2\cos^2{\dfrac{x}{2}}+2\sin{\dfrac{x}{2}}\cos{\dfrac{x}{2}}\right)}$$
then I fell very ugly.Thank you
| Now divide the numerator & the denominator by $\displaystyle \sec^2\frac x2=1+\tan^2\frac x2$
So, we have $I=\displaystyle\int\frac{1-\tan^2\frac x2}{2\tan\frac x2(2+2\tan\frac x2)}\sec^2\frac x2dx$
Putting $\displaystyle \tan\frac x2=u$
$$I=\int\frac{1-u^2}{2u(1+u)}du=\frac12\int\frac1u du-\frac12\int du$$ assuming $1+u\ne0$
| {
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What is the $n$ times iterate of $f(x)=\frac{x}{\sqrt{1+x^2}}$? We were asked to determine the composition $f \circ f \circ f \circ \cdots \circ f $, $n$ times, where $f(x)=\dfrac{x}{\sqrt{1+x^2}}$.
Has anyone an idea?
| $$f(x)=\frac{x}{\sqrt{1+x^2}}$$
we see that
$$f^2(x)=f(f(x))=\frac{\frac{x}{\sqrt{1+x^2}}}{\sqrt{1+(\frac{x}{\sqrt{1+x^2}})^2}}=\frac{x}{\sqrt{1+2x^2}}$$
...... continuing this way
$$f^n(x)=\frac{x}{\sqrt{1+nx^2}}$$
| {
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Evaluate $\frac{1}{3}+\frac{1}{4}\frac{1}{2!}+\frac{1}{5}\frac{1}{3!}+\dots$ Question is to Evaluate :
$$\frac{1}{3}+\frac{1}{4}\frac{1}{2!}+\frac{1}{5}\frac{1}{3!}+\dots$$
what all i could do is :
$$\frac{1}{3}+\frac{1}{4}\frac{1}{2!}+\frac{1}{5}\frac{1}{3!}+\dots=\sum_{n=1}^{\infty} \frac{1}{(n+2)n!}=\sum_{n=1}^{\infty} \frac{n+1}{(n+2)!}=\sum_{n=1}^{\infty} \frac{n}{(n+2)!}+\sum_{n=1}^{\infty} \frac{1}{(n+2)!}$$
I have $$\sum_{n=1}^{\infty} \frac{1}{(n+2)!}=\sum_{n=0}^{\infty} \frac{1}{n!}-1-\frac{1}{2}=e-\frac{3}{2}$$
Now, I am not able to see what $$\sum_{n=1}^{\infty} \frac{n}{(n+2)!}$$ would be.
I would be thankful if some one can help me to clear this.
Thank you.
| Hint : $\displaystyle f(x)=\sum\frac{x^n}{(n+1)!}\iff f'(x)=\sum\frac{n\cdot x^{n-1}}{(n+1)!}\quad,\quad S=f'(1)$.
| {
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Please help me with this inequality $a,b,c > 0$ (no other conditions)
$\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}\geq\sqrt{3\left(a^2+b^2+c^2\right)}$
I tried this:
$\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}\geq\frac{\left (a+b+c\right)^2}{a+b+c}=a+b+c$
then
$a+b+c\geq\sqrt{3\left(a^2+b^2+c^2\right)}$ which is not correct.
| As the inequality is homogeneous, WLOG we can set $a^2+b^2+c^2 = 3$ to get the equivalent inequality:
$$\frac{a^2}b + \frac{b^2}c + \frac{c^2}a \ge 3$$
$$\iff \left(\frac{a^2}b + \frac{b^2}c + \frac{c^2}a \right)^2 \ge 9 \iff \sum_{cyc} \frac{a^4}{b^2} \ge 3 \iff \sum_{cyc} a^2 \left(\frac{a^2}{b^2}+\frac{b^2}{a^2}-2 \right)\ge 0$$
Another way, using Holder's Inequality:
$$\left(\frac{a^2}b + \frac{b^2}c + \frac{c^2}a \right)\left(\frac{a^2}b + \frac{b^2}c + \frac{c^2}a \right)\left(a^2b^2 + b^2c^2 + c^2a^2 \right)\ge \left(a^2+b^2+c^2\right)^3$$
Let $x=a^2, y=b^2, z=c^2$. Then it remains sufficient to show that
$$ \left(x+y+z \right)^3 \ge 3\left(xy+yz+zx \right)(x+y+z)$$
$$ \iff \left(x+y+z \right)^2 \ge 3\left(xy+yz+zx \right)$$
which is easy to show.
| {
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Find an integral solution for a polynomial Given odd prime p that $\binom{-2}{p}=1$, show that there are integers x,y such that $x^2+2y^2=p$.
What i did: I move $2y^2$ to the right, then $x^2=p-2y^2$
If $a^2=-2$ (mod p) then this equation turn into $x=ay$ (mod p)
I don't know how to continue from here, should i just say that a is an integer, so obviously there is an answer? I don't see any restriction on x or y, take y=b, then x is just ab mod p, there are p-1 solutions, that's it?
I don't think it's that easy, I'm thinking of proving it using Minkowski's theorem, but i'm not good with that. Can anyone help me?
| There are two distinct ways I know of to prove such statements.
Proof 1 (elementary, using Pigeonhole Principle):
Since $(\frac{-2}{p})=1$, there exists $t$ s.t. $t^2 \equiv -2 \pmod{p}$.
Consider the integers $a+tb$, for $0 \leq a, b \leq \lfloor \sqrt{p} \rfloor$. There are $(1+\lfloor \sqrt{p} \rfloor)^2>p$ such integers, so by the Pigeonhole Principle, we have $a+tb \equiv c+td \pmod{p}$ where $(a, b) \not =(c, d)$.
Thus $(a-c) \equiv t(d-b) \pmod{p}$ so $(a-c)^2 \equiv t^2(d-b)^2 \equiv -2(d-b)^2$, so $p \mid (a-c)^2+2(d-b)^2$.
Since $(a, b) \not =(c, d)$, we have $(a-c)^2+2(d-b)^2>0$. Also, $|a-c|, |b-d| \leq \lfloor \sqrt{p} \rfloor<\sqrt{p}$ so $(a-c)^2+2(d-b)^2<(\sqrt{p})^2=3p$.
Therefore $(a-c)^2+2(d-b)^2=p, 2p$.
If $(a-c)^2+2(d-b)^2=p$, we are done. If $(a-c)^2+2(d-b)^2=2p$, then $2 \mid (a-c)^2$ so $2 \mid a-c$, so $(d-b)^2+2(\frac{a-c}{2})^2=p$, so we are done. Therefore we can always find integers $x, y$ s.t. $x^2+2y^2=p$.
Proof 2 (Using binary quadratic forms):
Again, since $(\frac{-2}{p})=1$, there exists $t$ s.t. $t^2 \equiv -2 \pmod{p}$. We may write $t^2+2=kp$ for some integer $k$.
Observe that the binary quadratic form $px^2+2txy+ky^2$ represents $p$. It has discriminant $(2t)^2-4kp=-8$. We may apply Gauss reduction to get an equivalent reduced form $ax^2+bxy+cy^2$, where $-a<b \leq a<c$ or $0 \leq b \leq a=c$, and discriminant $b^2-4ac=-8$.
Note that $8=4ac-b^2 \geq 4a^2-a^2=3a^2$ so $a=1$. Also $|b| \leq a=1$ and $b$ is even, so $b=0$, and we get $c=2$. Thus the reduced form $x^2+2y^2$ is equivalent to the quadratic form $px^2+2txy+ky^2$, hence represents the same numbers, hence in particular represents $p$. Therefore there are integers $x, y$ s.t. $x^2+2y^2=p$.
| {
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Integration: Method of partial fractions - Any standard method of finding constants in hard to solve expressions? I've been computing many indefinite integrals using the method of partial decomposition.
The integrals are usual on the form $$\int \frac {x^2-29x+5} {(x-4)^2(x^2+3)} dx$$
which is equal to
$$\int \frac {A} {x-4} + \frac {B} {(x-4)^2} + \frac {Cx+D} {x^2+3}dx$$
and the constants are given by
$$x^2-29x+5=A(x-4)(x^2+3)+B(x^2+3)+(Cx+D)(x-4)^2 = (A+C)x^3+(-4A+B-8C+D)x^2+(3A+16C-8D)x + 12A+ 3B+16D$$
which can be solved using a matrix because no two constants are multiplied, squared or somehow messed up.
Also as I've been reading here:
Link the constants should be unique since the proof is based on polynomial division in a polynomial ring over field $F$, in this case the field is $\mathbb R$.
My question is: so far I have solved for all constants using a matrix. But what if the partial fractions include terms like $$(ax^2+bx+c)^k = \frac {A_1x + B_1} {a^2+bx+c} + .. + \frac {A_kx + B_k} {(ax^2+bx+c)^k}$$ for some very high power of $k$.
Then when multiplying all the denominators together to solve for constants, I will get some expression involving constants that cannot be solved for using a matrix. I know we don't meet these hard to solve tasks in a standard textbook, but they must arise in real.
As stated above, we know there exists values for the constants such that equality holds. However how would you solve these expressions involving constants I'm refering too ? - Is there some standard way ? Should I use some algebra system on my PC, or is there somehow to do it by hand when the power of $k$ is high ?
Thanks for your time.
| One can find partial fractions this way:
\begin{equation*}
f(x)=\frac{x^{2}-29x+5}{(x-4)^{2}(x^{2}+3)}=\frac{A}{(x-4)}+\frac{B}{%
(x-4)^{2}}+\frac{Cx+D}{x^{2}+3}.
\end{equation*}
\begin{equation*}
B=\left. (x-4)^{2}f(x)\right] _{x=4}=\left. \frac{x^{2}-29x+5}{(x^{2}+3)}%
\right] _{x=4}=-5.
\end{equation*}
\begin{eqnarray*}
Cx+D &=&\left. (x^{2}+3)f(x)\right] _{x^{2}=-3} \\
&=&\left. \frac{x^{2}-29x+5}{x^{2}-8x+16}\right] _{x^{2}=-3} \\
&=&\left. \frac{\left( -3\right) -29x+5}{(-3)-8x+16}\right] _{x^{2}=-3} \\
&=&\left. \frac{-29x+2}{-8x+13}\right] _{x^{2}=-3} \\
&=&\left. \frac{\left( -29x+2\right) \left( 8x+13\right) }{\left(
-8x+13\right) \left( 8x+13\right) }\right] _{x^{2}=-3}\ tricky \\
&=&\left. \frac{26-232x^{2}-361x}{169-64x^{2}}\right] _{x^{2}=-3} \\
&=&\frac{26-232(-3)-361x}{169-64(-3)} \\
&=&-x+2.
\end{eqnarray*}
To find $A$ multiply by $x$
\begin{equation*}
xf(x)=A\left( \frac{x}{x-4}\right) +\frac{Bx}{(x-4)^{2}}+\frac{Cx^{2}+Dx}{%
x^{2}+3}
\end{equation*}
then take the limit as $x$ tends to $+\infty $
\begin{eqnarray*}
\lim_{x\rightarrow \infty }xf(x) &=&\lim_{x\rightarrow \infty }A\left( \frac{%
x}{x-4}\right) +\frac{Bx}{(x-4)^{2}}+\frac{Cx^{2}+Dx}{x^{2}+3} \\
0 &=&A+0+C
\end{eqnarray*}
then
\begin{equation*}
A=-C=1.
\end{equation*}
Therefore,
\begin{equation*}
\frac{x^{2}-29x+5}{(x-4)^{2}(x^{2}+3)}=\frac{1}{x-4}-\frac{5}{\left(
x-4\right) ^{2}}+\frac{-x+2}{x^{2}+3}.
\end{equation*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/587129",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
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} |
$\left(\frac1\alpha-\frac1\beta\right)^2$ for $p(x)=x^2+x-2$ If $\alpha$ and $\beta$ are the zeroes of the quadratic polynomial $p(x)=x^2+x-2$,
then $\left(\frac1\alpha-\frac1\beta\right)^2 is:$
A) $\frac94$
B) $\frac{-9}4$
C) $\frac25$
D) $\frac{-2}5$
This is a homework question from our school's home assignment for class 10. I tried this:
We can write $p(x)=x^2+x-2$ as $$p(x)=\underbrace{(1)x^2}_{ax^2}
+\underbrace{(1)x}_{bx}+\underbrace{(2)}_c$$ Therefore, $$a=1\quad
b=1\quad c=2$$ We know that $$\alpha\beta=\frac ca=2$$ and
$$\alpha+\beta=\frac{-b}{a}=-1$$ Now $$\left(\frac1\alpha-\frac1
\beta\right)^2=\left(\frac{\alpha-\beta}{\alpha\beta}\right)^2=
\frac{(\alpha-\beta)^2}{(\alpha\beta)^2}=\frac{\alpha^2+\beta^2-2
\alpha\beta}4=\frac{\alpha^2+\beta^2-4}4$$
I got stuck here. What to do now?
| $\alpha^2 + \beta^2 = (\alpha+\beta)^2 - 2\alpha\beta$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
} |
Can $x^3+3x^2+1=0$ be solved using high school methods? I encountered the following problem in a high-school math text, which I wasn't able to solve using factorization/factor theorem:
Solve $x^3+3x^2+1=0$
Am I missing something here, or is indeed a more advanced method necessary to solve this particular cubic? The answer provided was $x\doteq-3.1$, which I was only able to confirm using CAS.
| You might be able to get away with using some "miracle" substitutions that solve the cubic (which can be generalised to solve a general cubic).
Begin with
$$x^3+3x^2+1=0 \tag{$\dagger$}$$
*
*Substitute $x = y - 1$ into $(\dagger)$ to obtain
$$y^3 - 3y + 3 = 0. \tag{$\star$}$$
*Let $y = z + \frac{1}{z}$ so that $(\star)$ becomes $z^3+\frac{1}{z^3}+3 = 0$ or by multiplying through by $z^3$,
$$\left(z^3\right)^2 + 3\left(z^3\right) + 1 = 0. \tag{$\ast$}$$
*Solve $(\ast)$ for $z^3$ with the quadratic equation formula,
$$z^3 = \frac{-3\pm\sqrt{5}}{2}.$$
The two real solutions of $(\ast)$ are thus $z_1 = -\sqrt[3]{(3+\sqrt{5})/2}$ and $z_2 = -\sqrt[3]{(3-\sqrt{5})/2}$.
*Unravel the substitutions: $x = y - 1 = z + \frac{1}{z} - 1$ so solutions to $(\dagger)$ are
$$\begin{align*}x_1&= z_1 + 1/z_1 - 1 = -\sqrt[3]{(3+\sqrt{5})/2} - \sqrt[3]{2/(3+\sqrt{5})} - 1\\
x_2&= z_2 + 1/z_2 - 1 =-\sqrt[3]{(3-\sqrt{5})/2} - \sqrt[3]{2/(3-\sqrt{5})} - 1\end{align*}$$
It actually turns out that $x_1 = x_2$ (another miracle!). So we may write the final solution in exact form as
$$x = -\sqrt[3]{(3+\sqrt{5})/2} - \sqrt[3]{2/(3+\sqrt{5})} - 1 \approx -3.1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/588817",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "21",
"answer_count": 6,
"answer_id": 3
} |
How to integrate $\int\frac{\sqrt{1+x^2}}{x}\,\mathrm dx$ I want to know how to integrate
$$\int\frac{\sqrt{1+x^2}}{x}\,\mathrm dx$$
Could anyone solve it?
Thanks
| $$\int\frac{\sqrt{1+x^2}}{x}\,dx=\int\frac{x\sqrt{1+x^2}}{x^2}\,dx$$
Let $u=x^2+1, du =2x dx$. Then
$$\int\frac{\sqrt{1+x^2}}{x}\,dx= \frac{1}{2}\int\frac{\sqrt{u}}{u^2-1}\,du$$
If $v=\sqrt{u}$ the n $u=v^2, du=2vdv$. Thus
$$\int\frac{\sqrt{1+x^2}}{x}\,dx=\int\frac{v^2}{v^4-1}\,dv$$
By Partial Fraction Decomposition
$$\frac{v^2}{v^4-1}=\frac{A}{v-1}+\frac{B}{v+1}+\frac{Cv+D}{v^2+1}$$
find $A,B,C,D$, and use $v=\sqrt{u}=\sqrt{x^2+1}$ and you are done.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/591073",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 2
} |
Evaluating the integral $\int\frac{2x^3+x^2+1}{x^2+x-2}\space dx$ As suggested in the title, I am given the following integral:
$$\int\frac{2x^3+x^2+1}{x^2+x-2}\space dx$$
I have tried to solve it several times, but I wind up with the wrong answer, although its close. I assume there is a blunder somewhere.
Applying long division, we have that
$$\frac{2x^3+x^2+1}{x^2+x-2} = 2x - 1 + \frac{-3x-1}{x^2+x-2}$$
I note now that I may write this as
$$\frac{2x^3+x^2+1}{x^2+x-2} = 2x - 1 + \frac{(-1)3x+1}{x^2+x-2} = 2x - 1 - \frac{3x+1}{x^2+x-2}$$
I did not notice this while doing the task originally, but I am having a hard time seeing how this will make answer different. (After calculating it on paper: it didn't.)
Thus, we have the integral
$$\int 2x - 1 \space dx - \int \frac{3x+1}{x^2+x-2}\space dx$$
The first integral is trivial. For the second one, we have that
$$\frac{3x+1}{(x-1)(x+2)} = \frac{A}{x-1} + \frac{B}{x+2} = \frac{A(x+2)+B(x-1)}{(x-1)(x+2)}$$
We have the set $A + B = 3, \space 2A - B = 1$.This gives us the solutions $A = \frac{4}{3}$ and $B = \frac{5}{3}$
Thus, we are left with
$x^2 - x - \int \frac{3x+1}{x^2+x-2}\space dx = x^2 - x - \int \frac{4/3}{x-1} + \frac{5/3}{x+2} \space dx = x^2 - x - \frac{4}{3}\ln{|x-1|}-\frac{5}{3}\ln{|x+2|} + C$
I'm having a hard time finding my mistake here.
EDIT: Seems like I made my mistake in the long division part, confirming now.
| The remainder after long division is $5x - 1$; when you do long division, make sure you account for the 'missing' $x$ in the numerator: $2x^3 + x^2 + 0x + 1$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to find the minimum of $a+b+\sqrt{a^2+b^2}$ let $a,b>0$, and such
$$\dfrac{2}{a}+\dfrac{1}{b}=1$$
Find this minimum
$$a+b+\sqrt{a^2+b^2}$$
My try: since
$$2b+a=ab$$
so
$$a+b+\sqrt{a^2+b^2}=\sqrt{a^2+b^2+2ab}+\sqrt{a^2+b^2}=\sqrt{a^2+b^2+4b+2a}+\sqrt{a^2+b^2}$$
then I can't
maybe this problem can use AM-GM or Cauchy-Schwarz inequality solve it.Thank you very much
| Now that someone has worked out the place where the minimum is achieved ($a:b = 4:3$), one can purposely use Cauchy-Schwarz/AM-GM.
$$(a^2+b^2)(16+9) \ge (4a+3b)^2 \Rightarrow \sqrt{a^2+b^2} \ge \frac{4a+3b}{5}$$
Then
$$a+b+\sqrt{a^2+b^2} \ge \frac{9a+8b}{5}$$
Again Cauchy-Schwarz,
$$(9a+8b)(\frac{4}{a} + \frac{2}{b}) \ge (6+4)^2 = 100 \Rightarrow 9a+8b \ge 50$$
Therefore
$$a+b+\sqrt{a^2+b^2} \ge \frac{9a+8b}{5} \ge 10$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/593703",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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A high school contest question Compute the largest root of $$x^4 −x^3 −5x^2 +2x+6$$
I cannot solved it by factor,what else I may do.
Such a question, how to think.
| Note that $x^4-x^3-5x^2+2x+6=(x^4-5x^2+6)-x(x^2-2)=(x^2-2)(x^2-3)-x(x^2-2)=(x^2-2)(x^2-x-3)$.
So we have two possible positive roots, $\sqrt{2}$ and $\frac{1+\sqrt{13}}{2}$. Notice that $\sqrt{13}>3$, so $\frac{1+\sqrt{13}}{2}>2>\sqrt{2}$. Thus, the largest root is $\frac{1+\sqrt{13}}{2}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/595057",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How find this integral $\int\frac{1}{1+\sqrt{x}+\sqrt{x+1}}dx$ Question:
Find the integral
$$I=\int\dfrac{1}{1+\sqrt{x}+\sqrt{x+1}}dx$$
my solution:
let $\sqrt{x}+\sqrt{x+1}=t\tag{1}$
then
$$t(\sqrt{x+1}-\sqrt{x})=1$$
$$\Longrightarrow \sqrt{x+1}-\sqrt{x}=\dfrac{1}{t}\tag{2}$$
$(1)-(2)$ we have
$$2\sqrt{x}=t-\dfrac{1}{t}\Longrightarrow x=\dfrac{1}{4}(t-\dfrac{1}{t})^2$$
so
$$dx=\dfrac{1}{2}(t-\dfrac{1}{t})(1+\dfrac{1}{t^2})dt=\dfrac{t^4-1}{2t^3}dt$$
$$I=\int\dfrac{1}{1+t}\cdot\dfrac{t^4-1}{2t^3}dt=\dfrac{1}{2}\int\left(1+\dfrac{1}{t}+\dfrac{1}{t^2}+\dfrac{1}{t^3}\right)dt=\dfrac{1}{2}\left(t+\ln{t}-\dfrac{1}{t}-\dfrac{1}{2t^2}+C\right)$$
so
$$I=\dfrac{1}{2}\left(\sqrt{x}+\sqrt{x+1}+\ln{(\sqrt{x}+\sqrt{x+1})}-\dfrac{1}{\sqrt{x}+\sqrt{x+1}}-\dfrac{1}{2(\sqrt{x}+\sqrt{x+1})^2}+C\right)$$
My question: have other methods? Thank you very much
| since
\begin{align*}I&=\int\dfrac{\sqrt{x}+1-\sqrt{x+1}}{(\sqrt{x}+1)^2-(\sqrt{x+1})^2}dx=\int\dfrac{\sqrt{x}+1-\sqrt{x+1}}{2\sqrt{x}}dx=\dfrac{1}{2}\int\left(1+\dfrac{1}{\sqrt{x}}-\sqrt{1+\dfrac{1}{x}}\right)dx\\
&=\dfrac{1}{2}\left(x+2\sqrt{x}-\int\sqrt{1+\dfrac{1}{x}}dx\right)
\end{align*}
since
$$\int\sqrt{1+\dfrac{1}{x}}dx=x\sqrt{1+\dfrac{1}{x}}+\dfrac{1}{2}\ln{\left(2\left(\sqrt{1+\dfrac{1}{x}}+1\right)x+1\right)}+C$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Identity for the sum of three squares equal to sum of three squares I have been searching on Internet about a general solution to the Diophantine equation $A^2+B^2+C^2=D^2+E^2+F^2$. However, I haven't found anything. Can anyone help me? I'm looking for a identity or something like that. Very important is that must be a general solution, it must contain all the odds with all the possible numbers.
| Diophantine equation:
$A^2+B^2+C^2=D^2+E^2+F^2$
Has a solution:
$A=-p^2+2(t+k-q-2a)ps+2(4t^2+4k^2-2a^2-5q^2+q(k+t)+7a(q-t-k)-tk)s^2$
$B=-p^2+2(t+a-q-2k)ps+2(4t^2+4a^2-2k^2-5q^2+q(t+a)+7k(q-a-t)-ta)s^2$
$C=-p^2+2(k+a-q-2t)ps+2(4k^2+4a^2-2t^2-5q^2+q(k+a)+$$7t(q-a-k)-ka)s^2$
$D=-p^2+2(t+k+a-q)ps+2(4t^2+4k^2+4a^2-5q^2+q(t+k+a)-tk-ta-ka)s^2$
$E=p^2+4(a+k+t-q)ps+2(2t^2+2k^2+2a^2-7q^2+5q(t+k+a)-5(tk+ta+ka))s^2$
$F=-p^2+2(a+k+t-4q)ps+2(4t^2+4k^2+4a^2+q^2-5q(t+k+a)-tk-ta-ka)s^2$
more:
$A=p^2+2(t+k+2a-q)ps+2(2a^2+q^2-q(t+k)+3a(t+k-q)+tk)s^2$
$B=p^2+2(t+2k+a-q)ps+2(2k^2+q^2-q(t+a)+3k(t+a-q)+ta)s^2$
$C=p^2+2(2t+k+a-q)ps+2(2t^2+q^2-q(k+a)+3t(k+a-q)+ka)s^2$
$D=p^2+2(t+k+a-q)ps+2(q^2+tk+ta+ka-q(t+k+a))s^2$
$E=p^2+4(t+k+a-q)ps+2(2t^2+2k^2+2a^2+q^2+3(tk+ta+ka)-3q(t+k+a))s^2$
$F=p^2+2(t+k+a)ps+2(tk+ta+ka-q^2+q(t+k+a))s^2$
$p,s,t,k,a,q$ - integers asked us.
| {
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"timestamp": "2023-03-29T00:00:00",
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Convert 2 + 5 +8+ . . . + (6n-1) = n(6n+1) to Sigma Notation 2 + 5 + 8 + . . . + (6n-1) = n(6n+1)
This is what I have so far.
The sum of (3j-1) from j=1 to something I`m not sure of.
| $$\sum\limits_{i=1}^{2n} (3i-1) = n(6n+1)$$
This will give you as terms of the series: \begin{align*}3(1)-1&=2\\3(2)-1&=5\\\ldots\\3(2n)-1&=6n-1\end{align*}
EDIT: the problem statement purported that $\sum_{i=1}^{2n} (3i-1) = n(6n-1)$. I will just prove below that this is wrong, and $\sum_{i=1}^{2n} (3i-1) = n(6n+1)$.
\begin{align*}
\sum\limits_{i=1}^{2n} (3i-1) &= \sum\limits_{i=1}^{2n} 3i + \sum\limits_{i=1}^{2n}-1
\\ &= \sum\limits_{i=1}^{2n} 3i + (-2n)
\\ &= 3\sum\limits_{i=1}^{2n} i + (-2n)
\end{align*}
Then we just use the summation formula: $\sum_{i=1}^{N}i = \frac{N(N+1)}{2}$.
\begin{align*}
\\ 3\sum\limits_{i=1}^{2n} i + (-2n) &= 3\left( \frac{2n(2n+1)}{2}\right) + (-2n)
\\ &= 3n(2n+1)-2n
\\ &= 6n^2+3n-2n
\\ &= 6n^2+n
\\ &= n(6n+1)
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/601951",
"timestamp": "2023-03-29T00:00:00",
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showed that $x^2 + y^2$ = $1$ if ........... Please help me .how can I solve this problem:
If $x \sin^3 \alpha$ + $y \cos^3 \alpha$ = $\sin\alpha\cos\alpha$ and $x \sin\alpha-y\cos\alpha$ = $0$ then showed that $x^2 + y^2$ = $1$
| from the second equation, you get $y = x\tan \alpha$
Substitute this in first equation and you will get $x = \cos \alpha$
which means $y= \sin \alpha$
$$x^2+y^2 = \sin^2 \alpha + \cos^2 \alpha = 1$$
proved
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove that $\frac{x_1}{1+x_2+x_3+\ldots+x_n}+\frac{x_2}{1+x_1+x_3+\ldots+x_n}+\ldots+\frac{x_n}{1+x_1+x_2+\ldots+x_{n-1}}\ge\frac{n}{2n-1}$. If $x_1,x_2,\ldots,x_n>0$ and $x_1+x_2+\ldots+x_n=1$, prove that $$\frac{x_1}{1+x_2+x_3+\ldots+x_n} + \frac{x_2}{1+x_1+x_3+\ldots+x_n} +\ldots + \frac{x_n}{1+x_1+x_2+\ldots +x_{n-1}} \ge \frac {n}{2n-1}$$
This can easily be simplified: $$\frac{x_1}{2-x_1} + \frac{x_2}{2-x_2} +\ldots + \frac{x_n}{2-x_n} \ge \frac {n}{2n-1}$$
I could try using the Cauchy-Schwarz inequality: $$\frac{x_1}{2-x_1} + \frac{x_2}{2-x_2} +\ldots + \frac{x_n}{2-x_n} = \frac{x_1^2}{2x_1-x_1^2} + \frac{x_2^2}{2x_2-x_2^2} +\ldots + \frac{x_n^2}{2x_n-x_n^2} \ge \frac{(x_1+x_2+\ldots+x_n)^2}{2(x_1+x_2+\ldots+x_n)-(x_1^2+x_2^2+\ldots+x_n^2)}=\frac{1}{2-(x_1^2+x_2^2+\ldots+x_n^2)}=\frac{n}{2n-(x_1^2+x_2^2+\ldots+x_n^2)n}$$
It's left to prove that $$(x_1^2+x_2^2+\ldots+x_n^2)n \ge 1$$
or $$(x_1^2+x_2^2+\ldots+x_n^2)n=1$$I can continue using the Cauchy-Schwarz inequality again: $$(x_1^2+x_2^2+\ldots+x_n^2)n=(x_1^2+x_2^2+\ldots+x_n^2)(1+1+\ldots+1)\ge(x_1+x_2+\ldots+x_n)^2=1$$
So I've proved it myself. (I've shown this proof after editing, I didn't post the question with the solution in the details. I found the proof a while after putting the question here).
| Hint: Use Jensen's inequality for f(x) = x/(2 -x)
| {
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integer $a$ for which $x^3-3x+a = 0$ has $3$ distinct real roots in $x\in (0,1)$ If the equation $x^3-3x+a=0$ has $3$ distinct roots between $0$ and $1$ then the integer value of $a$ is
Can we solve it without Calculus.
$\bf{My\; Try}::$ Let $x=2\cos \phi$, where $-1\leq \cos \phi\leq 1\Rightarrow 0\leq \phi \leq \pi$(Principle value, Inverse exists)
$\displaystyle 8\cos^3 \phi-6\cos \phi+a = 0\Rightarrow 4\cos^3 \phi-3\cos \phi = -\frac{a}{2}\Rightarrow \cos 3\phi = -\frac{a}{2}$
Now $-1\leq \cos 3\phi\leq 1$, So $\displaystyle -1 \leq -\frac{a}{2}\leq 1\Rightarrow -2\leq a\leq 2$
So we get $a = \left\{-2,-1,0,1,2\right\}$
But answer is no integer value of $a$
Help me
Thanks
| If we factor the polynomial as $(x - b)(x - c)(x - d)$, we have
$$a = -bcd$$
But since $0 < b, c, d < 1$, we have that $|a| < 1$; the only integer satisfying this is zero, which certainly doesn't work.
| {
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Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.