Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Two methods lead to different answers Question:
Evaluate limit
$$\lim_{x\to 0}\frac{10^x-2^x-5^x+1}{x\tan x}$$
My attempt:
$$\lim_{x\to 0}\frac{10^x-2^x-5^x+1}{x\tan x}$$
$$= \lim_{x\to 0}\frac{(10^x-1)-(2^x-1)-(5^x-1)}{x\tan x}$$
$$= \lim_{x\to 0}\frac{(10^x-1)/x-(2^x-1)/x-(5^x-1)/x}{\tan x}$$
Now using identity $\lim_{x\to 0} \frac{a^x-1}{x}=\ln a$, we get:
$$= \lim_{x\to 0}\frac{\ln 10-\ln5-\ln2}{\tan x}= \lim_{x\to 0}\frac{\ln 1}{\tan x}= \lim_{x\to 0}\frac{0}{\tan x}= 0$$
The answer however is obviously wrong. I have the correct solution in my school book. But I wish to know why my solution is wrong. I have consulted my classmates but none of them can see the problem here either.
| This may be solved with Taylor approximations. Recall that
$$e^x=1+x+\frac12x^2+\mathcal O(x^3)$$
It thus follows that
$$\frac{a^x-1}x=\ln(a)+\frac{\ln^2(a)}2x+\mathcal O(x^2)$$
And
$$\frac{10^x-1}x-\frac{2^x-1}x-\frac{5^x-1}x=\left[\frac{\ln^2(10)-\ln^2(2)-\ln^2(5)}2\right]x+\mathcal O(x^2)$$
Now recall that $\lim_{x\to0}\frac{\tan(x)}x=1$, so
$$\begin{align}\frac{10^x-2^x-5^x+1}{x\tan(x)}&=\frac{\frac{10^x-1}x-\frac{2^x-1}x-\frac{5^x-1}x}{\tan(x)}\\&=\frac{\left[\frac{\ln^2(10)-\ln^2(2)-\ln^2(5)}2\right]x+\mathcal O(x^2)}{\tan(x)}\\&=\frac{\frac{\ln^2(10)-\ln^2(2)-\ln^2(5)}2+\mathcal O(x)}{\frac{\tan(x)}x}\\&\to\frac{\frac{\ln^2(10)-\ln^2(2)-\ln^2(5)}2+0}1\end{align}$$
$$L=\frac{\ln^2(10)-\ln^2(2)-\ln^2(5)}2$$
$$=\ln5\ln2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2172820",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Find the area of given triangle In a triangle $ABC$, where $a=8$, $c=1$ and $\cos (A-C) ={16\over 65}$.
Where $(a,b,c)$ are sides and $(A,B,C)$ are angles.
How can we find the area of the triangle $ABC$?
A hint will be appreciated.
| Hint: The Law of Sines says
$$
\sin(A)=8\sin(C)\tag{1}
$$
Then
$$
\begin{align}
\frac{16}{65}
&=\cos(A)\cos(C)+\sin(A)\sin(C)\tag{2}\\
\left(\frac{16}{65}-8\sin^2(C)\right)^2
&=\left(1-64\sin^2(C)\right)\left(1-\sin^2(C)\right)\tag{3}\\
\frac{256}{4225}-\frac{256}{65}\sin^2(C)+64\sin^4(C)
&=1-65\sin^2(C)+64\sin^4(C)\tag{4}\\
\sin^2(C)
&=\frac1{65}\tag{5}\\
\end{align}
$$
This means that
$$
\sin^2(A)=\frac{64}{65}\quad\cos^2(A)=\frac1{65}\quad\sin^2(C)=\frac1{65}\quad\cos^2(C)=\frac{64}{65}\tag{6}
$$
and therefore,
$$
\begin{align}
\cos(A+C)
&=\cos(A)\cos(C)-\sin(A)\sin(C)\\
&=\frac8{65}-\frac8{65}\\[2pt]
&=0\tag{7}
\end{align}
$$
What does this say about $B$, the angle between $a$ and $c$?
Note: Since $A,C\le\pi$, we have $\sin(A),\sin(C)\ge0$. However, only given $(6)$, $\cos(A),\cos(C)$ may not both be positive. Therefore, we know that $\sin(A)\sin(C)=\frac8{65}$, but we only know that $|\cos(A)\cos(C)|=\frac8{65}$. However, $(2)$ tells us that $\cos(A)\cos(C)=\frac8{65}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2174060",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Trigonometric equation with high exponents Solve the following equation:
\begin{align}
\sin^{14}x+\cos^{14}x=\frac{169}{64}\cos^{6}2x
\end{align}
I tried setting $\cos^2x=a$ such that the equation would become $a^7+(1-a)^7=\frac{169}{64}(2a-1)^6$ and then expaned both sides, but the obtained polynomial isn't pretty at all...
| \begin{align*}
2\sin^2 x &= 1-\cos 2x \\
2\cos^2 x &= 1+\cos 2x \\
128(\sin^{14} x+\cos^{14} x) &= (1-\cos 2x)^7+(1+\cos 2x)^7 \\
&= 2(1+21\cos^2 2x+35\cos^4 2x+7\cos^6 2x) \\
169\cos^6 2x &= 1+21\cos^2 2x+35\cos^4 2x+7\cos^6 2x \\
0 &= 162\cos^6 2x-35\cos^4 2x-21\cos^2 2x-1 \\
0 &= (2\cos^2 2x-1)(81\cos^4 2x+23\cos^2 2x+1) \\
\cos^2 2x &= \frac{1}{2} \quad \text{or} \quad
\frac{-23\pm \sqrt{23^2-4(81)}}{2(81)} < 0
\end{align*}
Can you proceed?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2174781",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Polynomial Factoring $(2^x)+(2^{-x})=a$. What is $(8^x)+(8^{-x})$?
I did:
$$\begin{align} (2^x)^3+(2^{-x})^3
&= [(2^x)+(2^{-x})] [(2^x)^2-(2^x)(2^{-x})+(2^{-x})^2]
\\&= a [(2^x)^2+2(2^x)(2^{-x})+(2^{-x})^2-3(2^x)(2^{-x})]
\\&= a {[(2^x)+(2^{-x})]^2-3(2^x)(2^{-x})}
\\&= a [a^2-3(2^x)(2^{-x})]
\\&= a^3-3a(2^x)(2^{-x}) \end{align}$$
The answer is $(a^3-3a)$. But I don't know how to solve now. Please, could you help me?
| The posted proof looks good. Alternatively, using the identity $(u+v)^3=u^3+v^3+3uv(u+v)$:
$$a^3=(2^x+2^{-x})^3=\left(2^x\right)^3+\left(2^{-x}\right)^3+3\cdot 2^x\cdot 2^{-x}(2^x + 2^{-x})=8^x+8^{-x}+ 3 \cdot 1 \cdot a$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2180073",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Help with a Bernoulli DE Problem:
Solve the following differential equation.
\begin{eqnarray*}
x \frac{dy}{dx} + y &=& xy^3 \\
\end{eqnarray*}
Answer:
This is a Bernoulli equation so I use the substitution $z = y^{-2}$ to
reduce it to a linear equation.
\begin{eqnarray*}
\frac{dz}{dx} &=& -2y^{-3} \frac{dy}{dx} \\
-2xy^{-3} \frac{dy}{dx} - 2y^{-2} &=& -2x \\
\frac{dz}{dz} - 2z &=& -2x \\
\end{eqnarray*}
Now, I have reduced the equation to a linear equation, so I find $I$ an
integrating factor.
\begin{eqnarray*}
I &=& e^{\int {-2 dx} } = e^{-2x} \\
e^{-2x} \frac{dz}{dx} - 2e^{-2x}z &=& -2xe^{-2x} \\
D(ze^{-2x}) &=& -2xe^{-2x} dx \\
ze^{-2x} &=& -2 \int xe^{-2x} dx \\
\end{eqnarray*}
Now I need to evaluate the following integral:
\begin{eqnarray*}
\int xe^{-2x} dx \\
\end{eqnarray*}
This can be done using integration by parts with $u = x$ and
$dv = e^{-2x} dx$.
\begin{eqnarray*}
\int xe^{-2x} dx &=& \frac{xe^{-2x}}{-2} - \int { \frac{e^{-2x}}{-2} dx} \\
\int xe^{-2x} dx &=& \frac{-xe^{-2x}}{2} - \frac{e^{-2x}}{4} + C_1
\end{eqnarray*}
Now I substitute into the differential equation.
\begin{eqnarray*}
ze^{-2x} &=& -2 (\frac{-xe^{-2x}}{2} - \frac{e^{-2x}}{4} + C_1) \\
ze^{-2x} &=& \frac{2xe^{-2x}}{2} + \frac{2e^{-2x}}{4} + C \\
ze^{-2x} &=& xe^{-2x} + \frac{e^{-2x}}{2} + C \\
z &=& x + \frac{1}{2} + Ce^{2x} \\
y^{-2} &=& x + \frac{1}{2} + Ce^{2x} \\
y^{2} &=& \frac{1}{x + \frac{1}{2} + Ce^{2x}} \\
\end{eqnarray*}
However, the back of the book gets:
\begin{eqnarray*}
y^2 &=& \frac{1}{2x + Cx^2} \\
\end{eqnarray*}
I have good reason to believe that the back of the book is right. Could
somebody please tell me where I went wrong?
Thanks,
Bob
| The Bernoulli equation of the form $y'=P(x)y+Q(x)y^{\alpha}$ can be transformed by $y=z^{\frac{1}{1-\alpha}}$ into
$$z'(x)=(1-\alpha)P(x)z(x)+(1-\alpha)Q(x).$$
For your problem that is $y'=-\frac{1}{x}y+y^3$. Hence, $P(x)=-1/x$, $Q(x)=1$ and $\alpha =3$. So we obtain the transformed ODE ($y=z^{-1/2}$):
$$z'=(1-3)(-1/x)z+(1-3)\cdot 1 \implies z'=\frac{2}{x}z-2.$$
This ODE has the solution $z(x)=cx^2+2x \implies y^2=\frac{1}{cx^2+2x}$. It seems your transformed ODE is wrong.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2184805",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
can $z=x^2 + y^2$ have more than 2 different solutions when $z$ is given and $x,y,z$ are all integers? For instance $z=50$ has two solutions: $(x,y)=(1,7)$ and $(x,y)=(5,5)$.
$z=125$ has two solutions $(x,y)=(10,5)$ and $(x,y)=(11,2)$.
Just fumbling around, I have not found any $z$ that has 3 or more solutions.
Is there a systematic way to generate solutions?
I have since made a C program to check every number between 1000 and 10000 and
found about 10% decompose into 2 squares,quite a few that decompose into 3 squares, a lesser amount into 4 and even a couple that are the sum of 5 squares.
| Yes, for example:
\begin{align}
71825&=1^2+268^2\\
&=40^2+265^2\\
&=65^2+260^2\\
&=76^2+257^2\\
&=104^2+247^2\\
&=127^2+236^2\\
&=160^2+215^2\\
&=169^2+208^2\\
&=188^2+191^2
\end{align}
However the counterexample mentioned by @lulu is the smallest counterexample;
$$325=10^2+15^2=1^2+18^2=6^2+17^2$$
We can produce infinitely many of them by setting $p,q$ primes both $1\mod 4$, then $p^2q$ has $3$ distinct ways to write as sum of two squares. For example, $p=13$ and $q=29$, then $p^2q=4901$ has three ways to write as sum of three squares, and indeed, $$4901=70^2+1^2=26^2+65^2=49^2+50^2$$
mathworld.wolfram has an excellent page about this subject (linked by @lulu).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2186729",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Alternating sum of binomial of coefficients up to $\binom{n}{r}$ Question : Evaluate - $$ \binom{n}{0}-\binom{n}{1}+\binom{n}{2}-\binom{n}{3} \dots + (-1)^r\binom{n}{r}$$
I tried to solve it but I am able to solve only when $r=n$.In that case I can put $x=-1$ in the expansion of $(1+x)^n$ and the result is $0$.
I know that for asking any question on this forum, I need to show my approach and thoughts but literally I have no idea how to solve up to $r+1$ terms.
Although I tried to add and subtract the terms to reach upto $\binom{n}{n}$, and evaluate the remaining sum, but nothing helpful.
Any help will be appreciable!
| I also apply:
$$\binom{n}{k} = \binom{n-1}{k} + \binom{n-1}{k-1}$$
But in another way, call:
$$f(n) = \binom{n}{0}-\binom{n}{1}+\binom{n}{2}-\binom{n}{3} \dots + (-1)^r\binom{n}{r} = \sum\limits_{k=0}^{r} \binom{n}{k}-1^k$$
$$f(n+1)-f(n)=\sum\limits_{k=0}^{r} -1^k (\binom{n+1}{k}-\binom{n}{k})=\sum\limits_{k=0}^{r} -1^k \binom{n}{k-1}=\sum\limits_{k=0}^{r-1} -1^{k+1} \binom{n}{k}$$
$$f(n+1)=f(n+1)-f(n)+f(n)=\sum\limits_{k=0}^{r-1} -1^{k+1} \binom{n}{k} + \sum\limits_{k=0}^{r} \binom{n}{k}-1^k = -1^r\binom{n}{r} + \sum\limits_{k=0}^{r-1} (-1^k+(-1^{k+1})) \binom{n}{k} = -1^r\binom{n}{r}$$
Because
$$f(n+1) = -1^r\binom{n}{r}$$
So
$$f(n) = -1^r\binom{n-1}{r}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2188074",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
For what integer $n$ is $1+n$ divisible by $1+n^2$? For what integer $n$ is $1+n$ divisible by $1+n^2$?
I tried it in modulo $3$ and $8$ but to no avail.
Trial and error gave me answers $0$,$1$,$-2$ and $-3$.
| HINT:
$$\dfrac{n^2+1}{n+1}=\dfrac{(n+1-1)^2+1}{n+1}=n+1-2+\dfrac2{n+1}$$
OR
$$\dfrac{n^2+1}{n+1}=\dfrac{(n+1)(n-1)+2}{n+1}=n-1+\dfrac2{n+1}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2191170",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Determine all real numbers x > 0 for which log equation is true.
Determine all real numbers x > 0 for which
$$\log_4 x - \log_x 16 = \frac{7}{6} - \log_x 8$$
Attemp at solution:
$$\frac{1}{\log_x 4} -\log_x 16 + \log_x 8 = \frac{7}{6}$$
$$\frac{1}{\log_x 4} + \log_x \frac{8}{16} = \frac{7}{6}$$
$$\frac{1+\log_x \frac{1}{2} \cdot \log_x 4}{\log_x 4} = \frac{7}{6}$$
$$\frac{1+(-2\log_x 2\cdot \log_x 2 )}{\log_x 4} = \frac{7}{6}$$
$$\frac{1+(-2\log_x 2\cdot \log_x 2 )}{2\log_x 2} = \frac{7}{6}$$
Not sure how to solve further
| $1-2{(\log_x2) }^2=\frac73\cdot \log_x2$
Solve the resulting quadratic using $y= \log_x2$
$$1-2y^2=\frac73y$$
$$2y^2 +\frac73y-1=0$$
$$6y^2+7y-3=0$$
$$6y^2 +9y-2y-3=0$$
$$(3y-1)(2y+3)=0$$
Which gives
$$x=2^{-2/3} or \ \ x=8$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2194487",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How form a linear mapping $f:\mathbb{R}^{4} \rightarrow \mathbb{R}^{3}$ to a matrix correctly?(solved) Given is $f: \mathbb{R}^4 \rightarrow \mathbb{R}^3$
$f(x)= \begin{pmatrix}
x_1-2x_2+x_4\\
-2x_1+5x_2+x_3-4x_4\\
x_1+2x_3-3x_4
\end{pmatrix}$
How can I form this to a matrix correctly?
We have $f: \mathbb{R}^4 \rightarrow \mathbb{R}^3$, and we have $x_1,x_2,x_3,x_4$
I think because we go to $\mathbb{R}^3$, we will only have $x_1,x_2,x_3$
So when I form a matrix, I will ignore $x_4$:
\begin{pmatrix}
1 & -2 & 0\\
-2 & 5 & 1\\
1 & 0 & 2
\end{pmatrix}
Is it fine like that?
| Once your function is from $\Bbb R^4$ to $\Bbb R^3$ your matrix has to able to accepty a vector in $\Bbb R^4$ and give as an answer a vector in $\Bbb R^3$. It means that your matrix must have a dimension $3\times4$.
$A\begin{pmatrix}
x_1\\
x_2\\
x_3\\
x_4
\end{pmatrix}=\begin{pmatrix}
x_1-2x_2+x_4\\
-2x_1+5x_2+x_3-4x_4\\
x_1+2x_3-3x_4
\end{pmatrix}=\begin{pmatrix}
1 & -2 & 0 & 1\\
-2 & 5 & 1 &-4\\
1 & 0 & 2 & -3
\end{pmatrix}.\begin{pmatrix}
x_1\\
x_2\\
x_3\\
x_4
\end{pmatrix}$
so
$$A=\begin{pmatrix}
1 & -2 & 0 & 1\\
-2 & 5 & 1 &-4\\
1 & 0 & 2 & -3\end{pmatrix}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2198254",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Prove that $\sqrt{\frac{2}{a}}+\sqrt{\frac{2}{b}}+\sqrt{\frac{2}{c}}\le\sqrt{\frac{a+b}{ab}}+\sqrt{\frac{b+c}{bc}}+\sqrt{\frac{c+a}{ca}}$ For $a,b,c$ are positive real number. Prove that
$$\sqrt{\frac{2}{a}}+\sqrt{\frac{2}{b}}+\sqrt{\frac{2}{c}}\le\sqrt{\frac{a+b}{ab}}+\sqrt{\frac{b+c}{bc}}+\sqrt{\frac{c+a}{ca}}$$
Let $\left(\dfrac{1}{a};\dfrac{1}{b};\dfrac{1}{c}\right)\rightarrow\left(x;y;z\right)$
We need prove $\sqrt{2x}+\sqrt{2y}+\sqrt{2z}\le\sqrt{x+y}+\sqrt{y+z}+\sqrt{z+x}$
We have: $\left(\sqrt{x+y}+\sqrt{y+z}+\sqrt{z+x}\right)^2=2\left(x+y+z\right)+2\left[\sqrt{\left(x+y\right)\left(y+z\right)}+\sqrt{\left(y+z\right)\left(z+x\right)}+\sqrt{\left(z+x\right)\left(x+y\right)}\right]$
By C-S we have: $\sqrt{\left(x+y\right)\left(y+z\right)}\ge\sqrt{xy}+\sqrt{yz}$
$\Rightarrow 2\sum\sqrt{\left(x+y\right)\left(y+z\right)}\ge4\left(\sqrt{xy}+\sqrt{yz}+\sqrt{xz}\right)$
$\Rightarrow LHS^2\ge 2(x+y+z+2\sqrt {xy}+2\sqrt {yz}+2\sqrt {xz})=2(\sqrt{x}+\sqrt{y}+\sqrt{z})=RHS^2$
Can do other way ?
| we know that:
\begin{align*}
\left(\frac{1}{\sqrt{a}}-\frac{1}{\sqrt{b}}\right)^2&\geq0\\
or,\ \frac{2}{\sqrt{ab}}&\leq\left(\frac{1}{a}+\frac{1}{b}\right)\\
or,\ \frac{1}{a}+\frac{1}{b}+\frac{2}{\sqrt{ab}}&\leq2\left(\frac{1}{a}+\frac{1}{b}\right)\\
or,\ \left(\frac{1}{\sqrt{a}}+\frac{1}{\sqrt{b}}\right)^2&\leq2\left(\frac{1}{a}+\frac{1}{b}\right)\\
or,\ \frac{1}{\sqrt{a}}+\frac{1}{\sqrt{b}}&\leq\sqrt{2\left(\frac{1}{a}+\frac{1}{b}\right)}
\end{align*}
Similarly, $\dfrac{1}{\sqrt{b}}+\dfrac{1}{\sqrt{c}}\leq\sqrt{2\left(\dfrac{1}{b}+\dfrac{1}{c}\right)}\ $ and, $\ \dfrac{1}{\sqrt{c}}+\dfrac{1}{\sqrt{a}}\leq\sqrt{2\left(\dfrac{1}{c}+\dfrac{1}{a}\right)}$
By adding them,
\begin{align*}
2\left(\dfrac{1}{\sqrt{a}}+\dfrac{1}{\sqrt{b}}+\dfrac{1}{\sqrt{c}}\right)&\leq \sqrt{2}\left[\sqrt{\left(\dfrac{1}{a}+\dfrac{1}{b}\right)}+\sqrt{\left(\dfrac{1}{b}+\dfrac{1}{c}\right)}+\sqrt{\left(\dfrac{1}{c}+\dfrac{1}{a}\right)}\right]\\
or,\ \sqrt{\dfrac{2}{a}}+\sqrt{\dfrac{2}{b}}+\sqrt{\dfrac{2}{c}}&\leq\sqrt{\dfrac{a+b}{ab}}+\sqrt{\dfrac{b+c}{bc}}+\sqrt{\dfrac{c+a}{ca}}
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2203382",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
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How do I prove that $\frac{(3^k-1)+6k}{2} $ is the same as $\frac{3^{k+1}-1}{2}$? Use mathematical induction to prove that the following statement is true for every positive integer $n$
$1+3+3^2+...+3^{n-1}=\frac{3^n-1}{2}$
Here are my steps:
*
*Show that $S_1$ is true
$S_1: 1=\frac{3^{(1)}-1}{2}$
$S_1: 1=\frac{3 -1}{2}$
$S_1: 1=\frac{2}{2}$
$S_1: 1=1$
*Show that if $S_k$ is assumed to be true, then $S_{k+1}$ is also true, for every positive integer k.
$S_k:1+3+3^2+...+3^{k-1}=\frac{3^k-1}{2}$
$S_{k+1}:1+3+3^2+...+3^{(k+1)-1}=\frac{3^{k+1}-1}{2}$
So I begin by adding $3^{(k+1)-1}$ to both sides of $S_k$, and then simplifying from that point on until the final result is the statement $S_{k+1}$.
$S_k:1+3+3^2+...+3^{k-1}+3^{(k+1)-1}=\frac{3^k-1}{2} + 3^{(k+1)-1}$
$S_k:1+3+3^2+...+3^{k-1}+3^{(k+1)-1}=\frac{3^k-1}{2} + 3^{k}$
$S_k:1+3+3^2+...+3^{k-1}+3^{(k+1)-1}=\frac{3^k-1}{2} + \frac{6^{k}}{2}$
$S_k:1+3+3^2+...+3^{k-1}+3^{(k+1)-1}=\frac{(3^k-1)+6^k}{2}$
But this is where I get stuck...
How do I prove that $\frac{(3^k-1)+6k}{2} $ is the same as $\frac{3^{k+1}-1}{2}$?
| Proposition: $$1+3+3^2+...+3^{n-1}=\frac{3^n-1}{2}$$
Assume true for $n$
Now we must prove true for $n+1$:
$$1+3+3^2+...+3^{n-1}+3^n = \frac{3^n-1}{2} + 3^n = \frac{3^n-1+2\cdot 3^n}{2}$$
$$= \frac{3\cdot 3^n-1}{2} = \frac{3^{n+1}-1}{2}$$ as required, hence true $\forall n\in\mathbb{Z^+}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2204240",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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$f(x,y)= (x^2-y^2, 2xy)$. Show $f$ is one-to-one Let $A:=\{(x,y)\in \mathbb{R},x>0\}$. Let $f:\mathbb{R}^2 \rightarrow \mathbb{R}^2$ be defined by $f(x,y)= (x^2-y^2, 2xy$). Show that $f$ is one-to-one on $A$.
My attempt:
$$f(x_1,y_1)=f(x_2,y_2)$$ so
$$\tag{$*$} x_1^2-y_1^2=x_2^2-y_2^2$$
and
$$2x_1y_1=2x_2y_2$$
$$\tag{$**$}x_1y_1=x_2y_2$$
From $(**)$, we have two cases. Either $y=0$ or $y\neq 0$.
CASE $1$ : $y=0$. From $(*)$: $$x_1^2=x_2^2$$ Taking the square root of both sides $$x_1=x_2$$ where $x_1,x_2>0$.
CASE $2$: $y \neq 0$
I am stuck on Case $2$...
| $f(x,y)= (x^2-y^2, 2xy)
$.
If
$f(x, y) = f(u, v)$
then
$x^2-y^2 = u^2-v^2$
and
$2xy = 2uv$.
Squaring these,
$x^4-2x^2y^2 + y^4 = u^4-2u^2v^2+v^4$
and
$4x^2y^2 = 2u^2 v^2$,
Adding these,
$x^4+2x^2y^2 + y^4 = u^4+2u^2v^2+v^4$
or
$(x^2+y^2)^2 = (u^2+v^2)^2$.
Since both $x^2+y^2$
and
$u^2+v^2$
are positive,
we can take the square root
to get
$x^2+y^2 = u^2+v^2$.
Since
$x^2-y^2 = u^2-v^2$,
adding these we get
$2x^2 = 2u^2$;
subtracting them we get
$2y^2 = 2v^2$.
Therefore
$x^2 = u^2$
and
$y^2 = v^2$.
Therefore $x = \pm u$
and
$y = \pm v$.
By assumption
(on $A$),
$x \ge 0 $ and $u \ge 0$.
Therefore
$x = u$.
Since
$2xy = 2uv$,
$y = v$.
If you allow $x < 0$,
this is equivalent to
the two-valued nature of
the complex square root.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2204449",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Find the minimum value of $P=\left(3+\frac{1}{a}+\frac{1}{b}\right)\left(3+\frac{1}{b}+\frac{1}{c}\right)\left(3+\frac{1}{c}+\frac{1}{a}\right)$ For $a,b,c$ are positive numbers satisfying $a+b+c\leq \frac{3}{2}$, find the minimum value of $$P=\left(3+\frac{1}{a}+\frac{1}{b}\right)\left(3+\frac{1}{b}+\frac{1}{c}\right)\left(3+\frac{1}{c}+\frac{1}{a}\right)$$
We have: $\frac{3}{2}\ge a+b+c \ge 3\sqrt[3]{abc}\Rightarrow abc\leq\frac{1}{8}$
We have $3+\frac{1}{a}+\frac{1}{b}\ge 7\sqrt[7]{\frac{1}{16a^2b^2}}$
$\Rightarrow P=\prod 7\sqrt[7]{\frac{1}{16a^2b^2}}=7^3$
When $a=b=c=\frac{1}{2}$
It looks wrong for $abc \le \frac{1}{8}$, and I need another way.
| If you simply write out the inequality you obtained for the product explicitly, what you've got, rewriting $abc\le{1\over8}$ as ${1\over abc}\ge8$, is
$$P\ge\left(7\sqrt[7]{1\over16a^2b^2}\right)\left(7\sqrt[7]{1\over16b^2c^2}\right)\left(7\sqrt[7]{1\over16c^2a^2}\right)
=7^3\sqrt[7]{1\over16^3(abc)^4}\ge7^3\sqrt[7]{8^4\over16^3}=7^3\sqrt[7]{2^{12}\over2^{12}}=7^3$$
with equality achieved when $a=b=c={1\over2}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2205321",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
proving $\cos (A+B)>0$, if given angles $A$ and $B$ If $\displaystyle A=3\sin^{-1}\left(\frac{6}{11}\right)$ and $\displaystyle B = 3\cos^{-1}\left(\frac{4}{9}\right),$ then proving $\cos (A+B)>0$
Attempt: $$ 3\sin^{-1}\left(\frac{1}{2}\right)<3\sin^{-1}\left(\frac{6}{11}\right)<3\sin^{-1}\left(\frac{\sqrt{3}}{2}\right)\Rightarrow \frac{\pi}{2}<A<\pi$$
same way $$\displaystyle 3\cos^{-1}\left(0\right)<3\cos^{-1}\left(\frac{4}{9}\right)<3\cos^{-1}\left(\frac{1}{2}\right)\Rightarrow \frac{\pi}{2}<B<\frac{3\pi}{2}$$
so $$\pi<A+B<\frac{5\pi}{2}$$
could some help me how to prove $\cos (A+B)>0,$ thanks
| Hint:
$\cos(3(A/3+B/3)) = 4 \cos^3 (A/3+B/3) - 3 \cos (A/3+B/3) $
Now expand this and you will gets terms only in $\cos (A/3), \cos (B/3),\sin (A/3), \sin (B/3)$ which we can get from given hypothesis.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2208004",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Coefficient of $x^5$ in $(4x^2- \frac{1}{x^3})^8$ I took one $x$ out of bracket, and I got $\frac{1}{x^{24}} (4x^5-1)^8.$ Now, we get $x$ to the power of if we get $x$ to the power $29,$ then we divide it by $24,$ we get $x$ to the power $5,$ but i can't get $29,$ by multiplying $5.$
| There are no odd powers in this expansion.
The generic term will look like
$$
a_{p,q} \left( 4x^{2} \right)^{p} \left( \frac{1}{x^{3}} \right)^{q}
$$
where $p+q=8$.
The powers in the generic term reduce to
$$
x^{2p-3q} = x^{2p-3(8-p)} = x^{5p-24}
$$
The question is this: For what integer is this equation valid:
$$
5p = 24?
$$
There is no such integer. Therefore, there is no term like $x^{5}$.
A brute force method computes all the powers:
$$
\begin{array}{ccr}
p & q & 2p-3q \\\hline
0 & 8 & -24 \\
1 & 7 & -19 \\
2 & 6 & -14 \\
3 & 5 & -9 \\
4 & 4 & -4 \\
5 & 3 & 1 \\
6 & 2 & 6 \\
7 & 1 & 11 \\
8 & 0 & 16 \\
\end{array}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2209204",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
How am I computing $\int \frac{x^{3}}{\sqrt{x^{2} + 9}}dx$ incorrectly? $ \displaystyle\int \frac{x^{3}}{\sqrt{x^{2} + 9}}dx$
doing a trig sub:
$x = 3\tan(\theta)$
$dx = 3\sec^{2}(\theta)d\theta$
$\displaystyle\int \frac{(3\tan(\theta))^{3}}{\sqrt{3\tan(\theta)^{2} + 9}}d\theta$
$\displaystyle\int \frac{ ( 27\tan^{3}(\theta) } { \sqrt{ 3\tan(\theta)^{2} + 9 } }d\theta$
$27\displaystyle\int \frac{\tan^{3}(\theta)\cdot\sec^{2}(\theta)}{\sqrt{\sec^{2}(\theta)}}d\theta$
$27\displaystyle\int \tan(\theta)\cdot\tan(\theta)\cdot\sec(\theta)\cdot d(\theta)$
Got stuck here and think I am doing this wrong.
Please help
Thank you
| Your approach is correct except that the last step is wrong. It can be fixed and completed as below:
$$
\begin{aligned}
&\text { Let } x=3 \tan \theta \quad d x=3 \sec ^2 \theta d \theta\\
&I=\int \frac{27 \tan ^2 \theta\cdot 3 \sec ^2 \theta d \theta}{\sqrt{9 \tan ^2 \theta+9}}\\
&=27 \int \tan ^3 \theta \sec \theta d \theta\\
&=27 \int \tan ^2 \theta d(\sec \theta)\\
&=27 \int\left(\sec ^2 \theta-1\right) d(\sec \theta)\\
&=27\left(\frac{\sec ^3 \theta}{3}-\sec \theta\right)\\
&=9 \sec \theta\left(\sec ^2 \theta-3\right)+C\\&=9 \sec \theta\left(\tan ^2 \theta-2\right)+C\\&= \frac{\sqrt{x^2+9}}{3}\left(x^2-18\right)+C
\end{aligned}
$$
One more quicker method is integration by parts.
$$
\begin{aligned}
\int \frac{x^3}{\sqrt{x^2+9}} d x &=\int x^2 d\left(\sqrt{x^2+9}\right) \\
&=x^2 \sqrt{x^2+9}-\int \sqrt{x^2+9} d\left(x^2\right) \\
&=x^2 \sqrt{x^2+9}-\frac{2}{3}\left(x^2+9\right)^{\frac{3}{2}}+C \\
&=\frac{1}{3} \sqrt{x^2+9}\left(3 x^2-2 x^2-18\right)+C \\
&=\frac{1}{3}\left(x^2-18\right) \sqrt{x^2+9}+C
\end{aligned}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2211018",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
If $a^3+a^2+a=9b^3+b^2+b$ and $a,b$ are integers then show $a-b$ is a perfect cube. If $a^3+a^2+a=9b^3+b^2+b$ and $a,b$ are integers then show $a-b$ is a perfect cube.
My attempt:I factorized it like below:
$(a-b)(a^2+b^2+ab+a+b+1)=8b^3=(2b)^3$
I take $gcd(a-b,a^2+b^2+ab+a+b+1)=d$
If $d=1$ then it is clear that $a-b$ is a perfect cube then consider $d>1$ then there is a $p$ that is prime and $p \mid d$.We have :
$p\mid a-b \Rightarrow p \mid (2b)^3 \Rightarrow p \mid 2b \Rightarrow p\mid 2$ or $p\mid b$
If $p\mid b$ then also $p\mid a$ (as $p\mid a-b$ holds). Then we will get to $p \mid 1$ because:
$ p \mid a^2+b^2+ab+a+b,p \mid a^2+b^2+ab+a+b+1 \Rightarrow p \mid 1$
which is clearly wrong then we have $p\mid 2$ so $p=2$ means $d=2^k$ where $k$ is a natural number including $0$.In the case $d=1$ we have the right result.So assemble $k \ge 1$.Because $2 \mid a-b$ we can conclude that $a,b$ have the same parity @Ghartal showed in his answer that if $a,b$ are both even we have a right result but if $a,b$ are both odd we don't.So maybe we have to prove $a,b$ can,t be both odd.
| Note: This is not a complete solution. It may give others some clues.
We have that $$(a-b)(a^2+b^2+ab+a+b+1)=8b^3 \tag{1}$$ Let
$$d=\gcd(a-b, a^2+b^2+ab+a+b+1).$$
And let $p$ be an odd prime divisor of $d$. Then $p \mid a-b$ and $p \mid b$. It follows that $p|a$. Since $p \mid a^2+b^2+ab+a+b+1$ we get $p \mid 1$. Contradiction. Thus only prime divisor of $d$ is $2$.
From equation $(1)$ observe that $a$ and $b$ have the same parity. Thus $a-b$ is even. So let $a-b=c=2^k (2l+1)$, where $l$ is an integer. Equation $(1)$ becomes.
$$c(3b^2+c^2+3bc+2b+c+1)=8b^3 \tag{2}$$
Notice that if $b$ is an even number then $3b^2+c^2+3bc+2b+c+1$ will be odd. Hence
$$d=\gcd\big(c, 3b^2+c^2+3bc+2b+c+1\big)=1.$$ And $c$ is a perfect cube. Now, suppose that $b$ is odd. Then, since $3b^2+1 \equiv 0 \pmod{4}$ we get
$$3b^2+c^2+3bc+2b+c+1=3b^2+1+c(c+1+3b)+2b \\ =4m+2b=2(2m+b)$$Thus the highest power of $2$ dividing LHS of equation $(2)$ is $k+1$. And since $v_2($RHS$)=3$ we get that $k=2$. Thus $a-b=4(2l+1)$ cannot be a cube.
Maybe in this case there is no solution for the equation at all.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2212824",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "26",
"answer_count": 4,
"answer_id": 0
} |
How can one evaluate the basic Tornheim Sum? I recently came across the claim that the double sum from $1$ to $\infty$ of $\frac{1}{mn(m+n)} = 2 \zeta(3)$.
I can show it equals the sum from $1$ to $\infty$ of $\frac{H_n}{n^2}$, where $H_n$ is the $n^{th}$ Harmonic number.
Does anyone know the full proof?
| Consider the following three sums
\begin{eqnarray*}
T(1,1,1) & =& \sum_{m=1}^{\infty} \sum_{n=1}^{\infty} \frac{1}{mn(m+n)} \\
\zeta(2,1) & =& \sum_{m=1}^{\infty} \sum_{n=1}^{\infty} \frac{1}{m(m+n)^2} \\
\zeta(3) & =& \sum_{m=1}^{\infty} \frac{1}{m^3}. \\
\end{eqnarray*}
Using partial fractions, you already know
\begin{eqnarray*}
\sum_{m=1}^{\infty} \sum_{n=1}^{\infty} \frac{1}{mn(m+n)} = \sum_{m=1}^{\infty} \sum_{n=1}^{\infty} \frac{1}{m^2} \left(\frac{1}{n} -\frac{1}{m+n} \right)= \sum_{m=1}^{\infty} \frac{H_m}{m^2} = \zeta(3) + \zeta(2,1) .
\end{eqnarray*}
Now the trick you need is this
\begin{eqnarray*}
\sum_{m=1}^{\infty} \sum_{n=1}^{\infty} \frac{1}{mn(m+n)} = \sum_{m=1}^{\infty} \sum_{n=1}^{\infty} \color{red}{\frac{1}{(m+n)^2} \left(\frac{1}{m} +\frac{1}{n} \right) } \\ = \sum_{m=1}^{\infty} \sum_{n=1}^{\infty} \frac{1}{m(m+n)^2} +\sum_{m=1}^{\infty} \sum_{n=1}^{\infty} \frac{1}{n(m+n)^2} = 2 \zeta(2,1)
\end{eqnarray*}
So $\zeta(2,1)=\zeta(3)$ & your sum is $T(1,1,1) =2 \zeta(3)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2214973",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Prove that $\lim\limits_{x\rightarrow 4} \sqrt{2x+7} = \sqrt{15}$. Prove that $\displaystyle \lim_{x\rightarrow 4} \sqrt{2x+7} = \sqrt{15}$ using the epsilon-delta definition.
This is what I have, but I know my delta value is incorrect. My professor said that it was the right path but my delta is incorrect.
Proof: Let $\varepsilon>0$. Choose $\delta$ such that $0<\delta<\min(\varepsilon,1)$. This means that both $\delta<1$ and $\delta<\varepsilon$. Let $x\in\mathbb{R}$ such that $0<|2x-8|<\delta$. Since $\delta<1$, we have
$$\begin{array}{cccccc}
&-1 &< & 2x-8 & < & 1\\
\Rightarrow & 7 &<& 2x &<& 9 \\
\Rightarrow & 7/2 & < & x & < & 9/2
\end{array}$$
Since $7/2<x<9/2$,
$$\begin{array}{cccccc}
&7/2 & < & x & < & 9/2\\
\Rightarrow & 7 &<& 2x &<& 9 \\
\Rightarrow & 7+7 & < & 2x+7 & < & 9+7 \\
\Rightarrow & \sqrt{14} & < & \sqrt{2x+7} & < & \sqrt{16} \\
\Rightarrow & \sqrt{14} + \sqrt{15} & < & \sqrt{2x+7}+\sqrt{15} & < & \sqrt{16}+\sqrt{15}\\
\Rightarrow & \displaystyle \frac{1}{\sqrt{14} + \sqrt{15}} & > & \displaystyle \frac{1}{\sqrt{2x+7}+\sqrt{15}} & > & \displaystyle \frac{1}{\sqrt{16} + \sqrt{15}}\\
\end{array}$$
This implies $$\left|\frac{1}{\sqrt{2x+7}+\sqrt{15}}\right|< \frac{1}{\sqrt{14} + \sqrt{15}}<1.$$
Therefore,
$$\begin{align*}
\left|\sqrt{2x+7}-\sqrt{15}\right|
&= \left|\left(\sqrt{2x+7}-\sqrt{15}\right) \cdot \left(\frac{\sqrt{2x+7}+\sqrt{15}}{\sqrt{2x+7}+\sqrt{15}}\right)\right| \\
&= \left|2x+7-15\right| \cdot \left| \frac{1}{\sqrt{2x+7}+\sqrt{15}}\right|\\
&=\left|2x-8\right|\cdot \left|\frac{1}{\sqrt{2x+7}+\sqrt{15}}\right|\\
&< \delta \cdot 1 \\
&< \varepsilon \cdot 1\\
\end{align*}$$
Thus, $|\sqrt{2x+7}-\sqrt{15}|<\varepsilon$. So, $\displaystyle \lim_{x\rightarrow 4} \sqrt{2x+7} = \sqrt{15}$.
| Hint:
$$
\sqrt{2x+7} - \sqrt{15} = \frac {(\sqrt{2x+7} - \sqrt{15} )(\sqrt{2x+7} + \sqrt{15} )}{\sqrt{2x+7} + \sqrt{15} } = \frac{2x-8}{\sqrt{2x+7} + \sqrt{15}}
$$
To remove the dependency in $x$ in the denominator, use that square roots are positive:
$$
|\sqrt{2x+7} - \sqrt{15}| = \frac {|2x-8|}{\sqrt{2x+7} + \sqrt{15}}
\le \frac {|2x-8|}{\sqrt{15}}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2215161",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 2,
"answer_id": 0
} |
Solve $\sum_{k=0}^{\infty}\frac{1}{1-16k^2}$ I could use some help on calculating this infinite sum: $\sum_{k=0}^{\infty}\frac{1}{1-16k^2}$. Included was that I had to start with a Fourier series for the function $f:\Re \to \Re: x \mapsto \sin(x)$ for $x\in[0, \frac{\pi}{2}[$, so let's start with that.
Let
\begin{eqnarray*}
g(x) = \frac{a_0}{2} + \sum_{k=1}^{\infty}a_k\cos(4kx) + \sum_{k=1}^{\infty}b_k\sin(4kx).
\end{eqnarray*}
This is the Fourier series for $f$.
With $a_k = \frac{4}{\pi}\int_{0}^{\frac{\pi}{2}}\sin(x)\cos(4kx)dx$ and $b_k = \frac{4}{\pi}\int_{0}^{\frac{\pi}{2}}\sin(x)\sin(4kx)dx$. Solving this leads to (or at least I found that): $a_k = \frac{4}{(1-16k^2)\pi}$, $b_k = \frac{16k}{(1-16k^2)\pi}$ for $k\geq1$ and $a_0 = \frac{4}{\pi}$. Bringing this to $g(x)$ gives:
\begin{eqnarray*}
g(x) = \frac{2}{\pi} + \sum_{k=1}^{\infty}\frac{4}{(1-16k^2)\pi}\cos(4kx) + \sum_{k=1}^{\infty}\frac{16k}{(1-16k^2)\pi}\sin(4kx).
\end{eqnarray*}
Since $f(x) \approx g(x)$, we can say that $f(0) = g(0)$. We get
\begin{eqnarray*}
\sin(0) = \frac{2}{\pi} + \sum_{k=1}^{\infty}\frac{4}{(1-16k^2)\pi}\cos(0) + \sum_{k=1}^{\infty}\frac{16k}{(1-16k^2)\pi}\sin(0),
\end{eqnarray*}
this becomes
\begin{eqnarray*}
0 = \frac{2}{\pi} + \sum_{k=1}^{\infty}\frac{4}{(1-16k^2)\pi}.
\end{eqnarray*}
We get
\begin{eqnarray*}
\frac{-2}{\pi} = \frac{4}{\pi}\sum_{k=1}^{\infty}\frac{1}{1-16k^2}
\end{eqnarray*} so
\begin{eqnarray*}
\sum_{k=1}^{\infty}\frac{1}{1-16k^2} = \frac{-1}{2}.
\end{eqnarray*} We need the sum from k = 0. The term $\frac{1}{1-16k^2}$ for k = 0 gives 1, so we add 1 to both sides. This leads to my solution
\begin{eqnarray*}
\sum_{k=0}^{\infty}\frac{1}{1-16k^2} = \frac{1}{2}.
\end{eqnarray*}
However, when approaching this sum numerically and using Wolfram, I find that the sum should be $\frac{4+\pi}{8}$. Could some help and point out where I went wrong with my approach? Thanks in advance
| The reason that the approach in the OP is flawed is that the Fourier series for $\sin(x)$ for $x\in [0,\pi/2]$ is discontinuous at the end points. This is Gibb's Phenomenon.
In fact, we have
$$\sin(x)= \frac2\pi +\frac4\pi\sum_{n=1}^\infty \frac{\cos(4nx)}{1-16n^2}+\frac{16}{\pi}\sum_{n=1}^\infty\frac{n\sin(4nx)}{1-16n^2}$$
for $x\in (0,\pi/2)$ pointwise, but the convergence is $L^2[0,\pi/2]$. We do not have pointwise convergence at the end points. This is not surprising given $\sin(0)=0\ne \sin(\pi/2)=1$.
And since $\sin(0)=0$ and $\sin(\pi/2)=1$ we have
$$\frac{1+0}{2}=\frac2\pi +\frac4\pi\sum_{n=1}^\infty \frac{1}{1-16n^2} \tag 1$$
Solving $(1)$ the series of interest yields
$$\sum_{n=0}^\infty \frac{1}{1-16n^2} =1+\frac\pi4\left(\frac{1}{2}-\frac{2}{\pi}\right)=\frac{4+\pi}{8}$$
as expected!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2216441",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 0
} |
Prove $\sum\limits_{n = 1}^\infty \frac{( - 1)^n}{\ln n + \sin n} $ is convergent. Help prove the alternating series $\sum\limits_{n = 1}^\infty \frac{(-1)^n}{\ln n + \sin n}$ is convergent. $\frac 1 {\ln n + \sin n}$ is a decreasing sequence but it is not motonically decreasing. I am not sure how to deal with this situation.
My failed attempt..
For even terms,
$$\sum_{n = 1}^\infty \frac{( - 1)^{2n}}{\ln 2n + 1} \le \sum_{n = 1}^\infty \frac{( - 1)^{2n}}{\ln 2n + \sin 2n} \leqslant \sum_{n = 1}^\infty \frac{( - 1)^{2n}}{\ln 2n - 1} $$
where the two "bound" series do not converge
For odd terms
$$\sum_{n = 1}^\infty \frac{( - 1)^{2n + 1}}{\ln (2n + 1) - 1} \leqslant \sum_{n = 1}^\infty \frac{(-1)^{2n + 1}}{\ln (2n + 1) + \sin (2n + 1)} \leqslant \sum_{n = 1}^\infty \frac{( - 1)^{2n + 1}}{\ln 2n + 1 + 1} $$
where the two "bound" series do not converge.
| Does
$\sum\limits_{n = 1}^\infty \frac{(-1)^n}{\ln n + \sin n}
$
converge or diverge?
I think that it probably diverges,
but I can show that
the similar sum
$\sum\limits_{n = 1}^\infty \frac{(-1)^n}{\ln n + (-1)^n}
$
diverges
and
$\sum\limits_{n = 1}^\infty \frac{(-1)^n}{\ln n + c}
$
converges
for any real $c$
such that
$\ln(n)+c
\ne 0$
for all $n$
(in particular
for
$c > 0$).
If $f(n)$ is bounded,
let
$s_m
=\sum\limits_{n = 1}^m \frac{(-1)^n}{\ln n + f(n)}
$
and let
$s
=\lim_{m \to \infty} s_m
$
if the limit exists.
$\begin{array}\\
s_{2m}
&=\sum\limits_{n = 1}^{2m} \frac{(-1)^n}{\ln n + f(n)}\\
&=\sum\limits_{n = 1}^{m} (\frac{-1}{\ln (2n-1) + f(2n-1)}+\frac{1}{\ln (2n) + f(2n)})\\
&=\sum\limits_{n = 1}^{m} \frac{-(\ln (2n) + f(2n))+(\ln (2n-1) + f(2n-1))}{(\ln (2n-1) + f(2n-1))(\ln (2n) + f(2n))}\\
&=\sum\limits_{n = 1}^{m} \frac{\ln (1-1/(2n)) + f(2n-1)-f(2n)}{(\ln (2n-1) + f(2n-1))(\ln (2n) + f(2n))}\\
&=\sum\limits_{n = 1}^{m} \frac{\ln (1-1/(2n))}{(\ln (2n-1) + f(2n-1))(\ln (2n) + f(2n))}+\sum\limits_{n = 1}^{2m} \frac{f(2n-1)-f(2n)}{(\ln (2n-1) + f(2n-1))(\ln (2n) + f(2n))}\\
&=u_m+v_m\\
\end{array}
$
where
$u_m
=\sum\limits_{n = 1}^{2m} \frac{\ln (1-1/(2n))}{(\ln (2n-1) + f(2n-1))(\ln (2n) + f(2n))}
$
and
$v_m
=\sum\limits_{n = 1}^{2m} \frac{f(2n-1)-f(2n)}{(\ln (2n-1) + f(2n-1))(\ln (2n) + f(2n))}
$.
Each term in $u_m$
is,
for large enoough $n$,
less in absolute value than
$\frac1{n\ln^2(n)}
$
(because
$\ln(1-1/(2n))
\approx -\frac1{2n}$
and
$(\ln (2n-1) + f(2n-1))(\ln (2n) + f(2n))
\approx \frac1{\ln^2(n)}
$)
and the sum of these converges.
Therefore convergence
depends on $v_m$.
Each term in $v_m$
is
$ \frac{f(2n-1)-f(2n)}{(\ln (2n-1) + f(2n-1))(\ln (2n) + f(2n))}
\approx \frac{f(2n-1)-f(2n)}{\ln^2(2n)}
$.
If
$f(n) = c$,
where $c$ is a constant,
then
$f(2n-1)-f(2n) = 0$,
so $v_m = 0$
and the sum converges.
If $f(n) = (-1)^n$,
then
$f(2n-1)-f(2n)
= (-1)-(1)
= -2
$
and the sum of
$\frac{-2}{\ln^2(n)}
$
diverges.
If
$f(n) = \sin(n)$,
$\begin{array}\\
f(2n-1)-f(2n)
&=\sin(2n-1)-\sin(2n)\\
&=2\sin((2n-1-2n)/2)\cos((2n-1+2n)/2)\\
&=2\sin(-1/2)\cos(2n-1/2)\\
\end{array}
$
so the sum depends
on the sum
$\sum_{n=1}^m \frac{\cos(2n-1/2)}{\ln^2(2n)}
$.
According to Wolfy,
$\sum_{n=1}^m \cos(2 n - 1/2)
= \csc(1) \sin(m) \cos(m + 1/2)
$
so this sum is bounded
but not convergent.
I think that
this implies that
$\sum_{n=1}^m \frac{\cos(2n-1/2)}{\ln^2(2n)}
$
does not converge.
This might be proved
using a variation of
summation by parts,
but I am not sure
how to do this,
so I will leave it
at this.
| {
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"url": "https://math.stackexchange.com/questions/2218454",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "35",
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} |
the sum of 3 numbers on the second power to be sub 0 This is my exercise
$x^3 - 3 x^2 + 6 x - a$, the polynom is from $\mathbb{R}[X]$.
And my job is to calculate $x_1^2+x_2^2+x_3^2$. I solved it and I get $-3$ as a result.
My question is: can 3 numbers $a^2+b^2+c^2$ be lower than $0$? Because the power is supposted to make all 3 of them positive.
| your answer is correct
your polynomial is : $x^3 - 3 x^2 + 6 x - a $
and its coefficients are : $a_0=a$, $a_1=6$, $a_2=-3$, $a_3=1$
$\sigma_k = (-1)^k \frac{a_{3-k}}{a3} =(-1)^ka_{3-k} $ for $k=1,2,3$
$$x_1+x_2+x_3=\sigma_1 = -a_2=3$$
$$x_1x_2+x_1x_3+x_2x_3 =\sigma_2 = a_1=6$$
$$(x_1+x_2+x_3)^2= x_1^2+x_2^2+x_3^2+2(x_1x_2+x_1x_3+x_2x_3) \implies x_1^2+x_2^2+x_3^2 = \sigma_1^2 - 2\sigma_2= 9-12 =-3 $$
and to answer your question yes it can be less than $0$ if $a,b,c \in \mathbb{C}$
for example : $a = i, b = 2i, c = -1$ $a^2+b^2+c^2= i^2+4i^2+(-1)^2=-1-4+1=-4$
| {
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"source": "stackexchange",
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Directrix and focus of $ax^2 + bx + c$ How can you find the directrix and focus of a parabola (quadratic function) $$ax^2 + bx + c,$$ where $a \neq 0?$ I mean, given the focus $x, y$ and directrix (I'll use a horizontal line for simplicity) $y = k$ you can find the equation of the quadratic; how do you do this backwards?
| The goal is essentially to get $y = ax^2 + bx + c$ into the form $4p(y - k) = (x - h)^2$. This can be done by completing the square:
\begin{align}
y &= ax^2 + bx + c \\
\frac{y - c}{a} &= x^2 + \frac{b}{a} x \\
\frac{y - c}{a} + \left(\frac{b}{2a} \right)^2 &= x^2 + \frac{b}{a}x + \left(\frac{b}{2a} \right)^2 \\
\frac{1}{a} \left(y - c + \frac{b^2}{4a} \right) &= \left(x + \frac{b}{2a} \right)^2.
\end{align}
Thus the parabola can be written as
$$
4 \cdot \frac{1}{4a} \left(y - \left(c - \frac{b^2}{4a} \right) \right) = \left(x - \left(-\frac{b}{2a} \right) \right)^2.
$$
The focus is
$$
\left(-\frac{b}{2a}, c - \frac{b^2}{4a} + \frac{1}{4a} \right) = \left(-\frac{b}{2a}, \frac{1 - b^2}{4a} + c \right)
$$
and the directrix is
$$
y = c - \frac{b^2}{4a} - \frac{1}{4a} = c - \frac{1 + b^2}{4a}.
$$
Update: Why $4p(y - k) = (x - h)^2$ is a useful form.
Starting from scratch, suppose we want to construct an upward or downward opening parabola (i.e., one of the form $y = ax^2 + bx + c$). Let its vertex be $(h, k)$. The definition of a parabola is that it is the set of all points equidistant to a point (the focus) and a line (the directrix). So let $(h, k + p)$ be the focus and $y = k - p$ be the directrix for some $p$. (Note that the vertex $(h, k)$ is midway between the focus and directrix, as required.)
Let $(x, y)$ be a point on the parabola. This means that its distance to the point $(h, k + p)$ is equal to its (shortest) distance to the line $y = k - p$. The point $(x, k - p)$ on the directrix is the closest point to $(x, y)$ (draw a picture to see why). So we equate distances:
\begin{align}
\sqrt{(x - h)^2 + (y - (k + p))^2} &= \sqrt{(x - x)^2 + (y - (k - p))^2} \\
(x - h)^2 + (y^2 - 2(k + p)y + (k + p)^2) &= y^2 - 2(k - p)y + (k - p)^2 \\
(x - h)^2 + (y^2 - 2ky - 2py + k^2 + 2kp + p^2) &= y^2 - 2ky + 2py + k^2 - 2kp + p^2 \\
(x - h)^2 &= 4py - 4kp \\
&= 4p(y - k).
\end{align}
Therefore, if we have a parabola with equation $4p(y - k) = (x - h)^2$, its focus will be $(h, k + p)$ and its directrix will be $y = k - p$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2225539",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How do I to find the limits of these? $$\lim _{x\to \frac{\pi }{4}}\left(\frac{1-\sin 2x}{\cos x -\sin x}\right)$$
$$\lim _{x\to \frac{\pi}{4}}\left(\frac{\cos^5x - \sin^5x}{\cos\:2x}\right)$$
I tried to solve but I get 0/0.
Help me to find the limit.
| For the first limit write the numerator as $\cos^2{x}+\sin^2{x}-2\sin{x}\cos{x}=\left(\cos{x}-\sin{x}\right)^2$ where we have used $\sin{2x}=2\sin{x}\cos{x}$ and $\cos^2{x}+\sin^2{x}$. The fraction simplify to
$$\cos{x}-\sin{x}\to 0$$
when $x\to\pi/4$
For the second limit we use
$$\cos{2x}=\cos^2{x}-\sin^2{x}=(\cos{x}-\sin{x})(\cos{x}+\sin{x})$$
and
$$a^5-b^5=(a-b)(a^4+a^3b+a^2b^2+ab^3+b^4)$$
to simplify the fraction to
$${\cos^4{x}+\cos^3{x}\sin{x}+\cos^2{x}\sin^2{x}+\cos{x}\sin^3{x}+\sin^4{x}\over \cos{x}+\sin{x}}$$
and the limit of this fraction is
$${5\sqrt{2}\over 8}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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A line through $P=(\sqrt3,0)$ and at an angle of $\pi/3$ with the positive direction of x axis cuts the parabola $y^2=x+2$ at A and B
A line passing through $P=(\sqrt3,0)$ and making an angle of $\pi/3$ with the positive direction of x axis cuts the parabola $y^2=x+2$ at A and B, then:
(a)$PA+PB=2/3$
(b)$|PA-PB|=2/3$
(c)$(PA)(PB)=\frac{4(2+\sqrt3)}{3}$
(d)$\frac{1}{PA}+\frac{1}{PB}=\frac{2-\sqrt3}{2}$
Equation of line:
$y=\sqrt3(x-\sqrt3)$
$y=\sqrt3x-3$
substituting $y=\sqrt3x-3$ in $y^2=x+2$
$(\sqrt3x-3)^2=x+2$
$3x^2-6\sqrt3x+9=x+2$
$3x^2-(6\sqrt3+1)x+7=0$
$x_A=\frac{(6\sqrt3+1)+\sqrt{(6\sqrt3+1)^2-84}}{6}$
$x_B=\frac{(6\sqrt3+1)-\sqrt{(6\sqrt3+1)^2-84}}{6}$
But this gives a very complicated value of $x$ and that makes me feel that there ought to be a shorter way to solve this question, just can't figure out what it is.
| Let the feet of A and B on the x-axis be A’ and B’ respectively.
Compute $A’P$. It is … $\dfrac {1}{6} + \dfrac {\sqrt(\delta)}{6}$.
Similarly, $PB’ = -\dfrac {1}{6} + \dfrac {\sqrt(\delta)}{6}$.
Then, $|PA – PB| = 2 \times |A’P – PB’| = \dfrac {2}{3}$.
| {
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"timestamp": "2023-03-29T00:00:00",
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How do we integrate $x\sqrt{x^3 + 1}$? Tried to use integration by parts but just keep going round in circles. Any help appreciated, thanks.
| Alternatively, and perhaps a little more transparent than just stating the result in terms of elliptic functions, we can find an antiderivative by expanding the square root and integrating the powers of x.
This leads to
$$a(x) = \sum _{k=0}^{\infty } \frac{\binom{\frac{1}{2}}{k} x^{3 k+2}}{3 k+2}$$
This infinite sum can be expressed by a hypergeometric function
$$a(x) =\frac{1}{2} x^2 \, _2F_1\left(-\frac{1}{2},\frac{2}{3};\frac{5}{3};-x^3\right)$$
Hence the function $a(x)$ is analytic in $x$ except for a branch cut from $x = -1$ to $x = -\infty$.
EDIT
We can obtain a possibly different antiderivative by writing
$$\sqrt{x^3+1}=x^{3/2} \sqrt{\frac{1}{x^3}+1}$$
expanding into negative powers and integrating, viz.
$$b(x)=\sum _{k=0}^{\infty } \frac{\binom{\frac{1}{2}}{k} x^{\frac{7}{2}-3 k}}{\frac{7}{2}-3 k}$$
which can be expressed as
$$b(x) = \frac{2}{7} x^{7/2} \, _2F_1\left(-\frac{7}{6},-\frac{1}{2};-\frac{1}{6};-\frac{1}{x^3}\right)$$
This expression holds even for $x\gt 0$.
Corollary
There should be a standard transformation of the hypergeometric function showing some equivalence of $a(x)$ and $b(x)$. Or, inverting the reasoning, we have just derived this transformation.
Not really, we must be careful here. In fact, $a(x)$ and $b(x)$ are not identical in the overlaping region of definition $x > 0$ but differ by a constant (which is permissible for antiderivatives):
$$b(x) = a(x) + c$$
where
$$ c = \lim_{x\to 0} \, b(x) = -\frac{2 \left(\Gamma \left(-\frac{1}{3}\right)-6 \Gamma \left(\frac{2}{3}\right)\right) \Gamma \left(\frac{5}{6}\right)}{21 \sqrt{\pi }} \simeq 0.739174$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2231076",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Is the set $M = \{d(n) \mid n \in S\}$ unbounded?
Consider the set $S = \{c \in \mathbb{Z}^+ \mid a^2+b^2 = c^2, \text{for some} \text{ } \text{coprime positive integers} \text{ } a,b\}$. Let $d(n)$ denote the number of positive integer divisors of a positive integer $n$. Is the set $M = \{d(n) \mid n \in S\}$ unbounded?
In other words, I am wondering if the hypotenuse of a primitive right triangle can have an arbitrarily large amount of divisors. How can we relate the divisor function to a primitive right triangle?
| If you take $n = 5 \cdot 13 \cdot 17 \cdot \cdots \cdot p_r$ is the product of $r$ distinct primes, all $p \equiv 1 \pmod 4,$ then $n$ is the hypoteneuse of a primitive Pythagorean triangle. Also $d(n) = 2^r$
Fermat: every prime $p \equiv 1 \pmod 4$ can be written as $p = a^2 + b^2$
I can't seem to remember the full result on coprime stuff, so I'm going the cheap way here. From the Brahmagupta identity we have $n = A^2 + B^2.$ As $n$ is squarefree we know that $\gcd(A,B) = 1.$ As $n \equiv 1 \pmod 4$ we know that one of $A,B$ is even and the other odd. We then get
$$ (2AB)^2 + (A^2 - B^2)^2 = n^2. $$
We have $A^2 - B^2$ odd. If we have some prime $q | \gcd(2AB, A^2 - B^2),$ we know $q$ is odd. We also know $q | n.$ more to come... Oh, I see, as $q$ is odd and $q | (A-B)(A+B),$ and $q$ divides one of $A,B,$ but $q$ also divides at least one of $A-B, A+B,$ it folows that $q$ actually divides both $A,B.$ This contradicts $\gcd(A,B) = 1.$ Therefore, $ \gcd(2AB, A^2 - B^2) = 1$ and
$$ 2AB, A^2 - B^2,n $$
is primitive.
Brahmagupta's identity
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Determinant of a $5\times 5$ matrix using properties Apparently the determinant of the following matrix can be transformed into the determinant of a $2\times 2$ matrix multiplied by a scalar. Would anyone show me how?
$$ \left( \begin{array}{ccccc} 0 & 0 & 1 & 0 & 0 \\
2 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 3 & 0 \\
0 & 0 & 0 & 0 & 4 \\
0 & 5 & 0 & 0 & 0\end{array} \right)$$
| I agree with the rest that permutations of rows and columns is sufficient.
Another way is by doing Laplace expansion.
\begin{align} \begin{vmatrix} 0 & 0 & 1 & 0 & 0 \\
2 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 3 & 0 \\
0 & 0 & 0 & 0 & 4 \\
0 & 5 & 0 & 0 & 0\end{vmatrix} &= \begin{vmatrix}
2 & 0 & 0 & 0 \\
0 & 0 & 3 & 0 \\
0 & 0 & 0 & 4 \\
0 & 5 & 0 & 0\end{vmatrix} \\ &= 2\begin{vmatrix}
0 & 3 & 0 \\
0 & 0 & 4 \\
5 & 0 & 0\end{vmatrix} \\ &=
2(5)\begin{vmatrix}
3 & 0 \\
0 & 4 \\
\end{vmatrix} \end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2232720",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Prove this $(ab)^2+(bc)^2+(ca)^2\le ab+bc+ac$
Let$a$, $b$ and $c$ be non-negative numbers such that
$$a^2+b^2+c^2=a+b+c.$$
Show that
$$(ab)^2+(bc)^2+(ca)^2\le ab+bc+ac.$$
I tried the uvw's technique and BW and more but without some success.I think can use C-S solve it?
| we have to show that
$$(a+b+c)^2((ab)^2+(bc)^2+(ca)^2)\le (ab+bc+ca)(a^2+b^2+c^2)^2$$
multiplying this out we get:
$${a}^{5}b+{a}^{5}c-{a}^{4}{b}^{2}+{a}^{4}bc-{a}^{4}{c}^{2}-{a}^{2}{b}^{
4}-3\,{a}^{2}{b}^{2}{c}^{2}-{a}^{2}{c}^{4}+a{b}^{5}+a{b}^{4}c+ab{c}^{4
}+a{c}^{5}+{b}^{5}c-{b}^{4}{c}^{2}-{b}^{2}{c}^{4}+b{c}^{5}
\geq 0$$
with
$$b=a+u,c=a+u+v$$ we obtain
$$\left( 9\,{u}^{2}+9\,uv+9\,{v}^{2} \right) {a}^{4}+ \left( 20\,{u}^{3
}+30\,{u}^{2}v+42\,u{v}^{2}+16\,{v}^{3} \right) {a}^{3}+ \left( 15\,{u
}^{4}+30\,{u}^{3}v+57\,{u}^{2}{v}^{2}+42\,u{v}^{3}+9\,{v}^{4} \right)
{a}^{2}+ \left( 4\,{u}^{5}+10\,{u}^{4}v+28\,{u}^{3}{v}^{2}+32\,{u}^{2}
{v}^{3}+14\,u{v}^{4}+2\,{v}^{5} \right) a+3\,{u}^{4}{v}^{2}+6\,{u}^{3}
{v}^{3}+4\,{u}^{2}{v}^{4}+u{v}^{5}
\geq 0$$ which is true.
| {
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"url": "https://math.stackexchange.com/questions/2233876",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 1
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Analyzing quadratic forms $x^2-3xy+y^2$, $2xy+yz-3xz$ and $x^2+y^2+2xy-xt+2yt$. Positive, negative or indefinite?
By means of successive coordinate changes, write each one of the
quadratic forms below as a sum of terms of the type $\pm u^2$ and
decide wich ones are positive, negative or indefinite:
$$A(x,y) = x^2-3xy+y^2$$
$$B(x,y,z) = 2xy+yz-3xz$$
$$C(x,y,z,t) = x^2+y^2+2xy-xt+2yt$$
For $A$ I did $A(x,y) = x^2-3xy+y^2 = x^2-2xy+y^2-xy = (x-y)^2-xy$ which is indefinite.
For $B(x,y) = 2xy+yz-3xz$ I remembered that $(x+y+z)^2 = x^2 + 2 x y + y^2 + 2 x z + 2 y z + z^2$ so $2xy+yz-3xz = (x+y+z)^2 -x^2-y^2-z^2-yz+xz$ which won't help anything.
For $C(x,y,z,t)$ there's too much terms, how should I do it?
| Just use Sylvester's criterion. For example: $$A(x,y) = \begin{pmatrix} x & y \end{pmatrix} \begin{pmatrix} 1 & -3/2 \\ -3/2 & 1 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix}.$$The minors of that middle matrix are $1$ and $1 - 9/4 <0 $, so $A$ is indefinite. Writing the expression as the sum of square terms amount to diagonalizing that matrix, but you don't need this to classify the quadratic form. You'll also have $$B = \begin{pmatrix} 0 & 1 & -3/2 \\ 1 & 0 & 1/2 \\ -3/2 & 1 & 0
\end{pmatrix}, \quad C = \begin{pmatrix}1 & 1 & 0 & -1/2 \\ 1 & 1 & 0 & 1 \\ 0 & 0 & 0 & 0 \\ -1/2 & 1 & 0 & 0 \end{pmatrix}.$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Find major and minor axis of given conic. Find the equation of major and minor axis of the given conic.Also find length of major and minor axis.
$$5x^2 + 5y^2 +6xy + 22x - 26y + 29$$ =0
I tried finding the nature of conic and centre of conic by partially differentiating it but could not think ahead.
| Rewrite the equation using matrices:
$\begin{pmatrix}x&y\end{pmatrix}\begin{pmatrix}5&3\\3&5\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix}+\begin{pmatrix}22&-26\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix}+29=0$
$M=\begin{pmatrix}5&3\\3&5\end{pmatrix}, K=\begin{pmatrix}22&-26\end{pmatrix}$
Diagonalising $M$ using the eigenvalues and eigenvectors we get:
$P=\begin{pmatrix}\frac{\sqrt{2}}{2}&-\frac{\sqrt{2}}{2}\\\frac{\sqrt{2}}{2}&\frac{\sqrt{2}}{2}\end{pmatrix}$
$\begin{pmatrix}x\\y\end{pmatrix}=P\begin{pmatrix}x'\\y'\end{pmatrix}, P^tMP=\begin{pmatrix}8&0\\0&2\end{pmatrix}$
$\begin{pmatrix}x'&y'\end{pmatrix}P^tMP\begin{pmatrix}x'\\y'\end{pmatrix}+KP\begin{pmatrix}x'\\y'\end{pmatrix}+29=0$
that is:
$8x'^2+2y'^2-2\sqrt{2}x'-24\sqrt{2}y'+29=0$
or completing the squares and rearranging:
$(\frac{x'-\frac{\sqrt{2}}{8}}{\frac18 \sqrt{922}})^2+(\frac{y'-6\sqrt{2}}{\frac14\sqrt{922}})^2=1$
Making the semi-major axis $\frac14\sqrt{922}$ and the semi-minor axis $\frac18 \sqrt{922}$.
Using the inverse transformation
$\begin{pmatrix}x'\\y'\end{pmatrix}=P^t\begin{pmatrix}x\\y\end{pmatrix}:$
$x'-\frac{\sqrt{2}}{8}=0, \frac{\sqrt{2}}{2}x+\frac{\sqrt{2}}{2}y-\frac{\sqrt{2}}{8}=0$
we get the equation of one of the axes: $y=\frac14-x,$ and
$y'-6\sqrt{2}=0, -\frac{\sqrt{2}}{2}x+\frac{\sqrt{2}}{2}y-6\sqrt{2}=0$
the equation of the other axis $y=12+x.$
Edit:
In differentiating to get the center we get the following system:
$10x+6y+22=0$
$6x+10y-26=0$
Since by the comments we get a rotation by $\frac{\pi}{4}$ ($\tan(2\theta)=\frac{B}{A-C}$), let's try and find the two equations in this system that is rotated to that angle (adding and subtracting):
$16x+16y=4$
$-4x+4y=48$
These are the same two lines on the figure.
| {
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Differentiating $\sin(x+y)+ \cos(x+y) = \log(x+y)$ with respect to $x$. It was given that $$\sin\left(x+y\right)+ \cos\left(x+y\right) = \log\left(x+y\right)$$ and I was asked to find $\frac{dy}{dx}$ for this.
I attempted it as:
$1-$ $$\sin\left(x+y\right)+ \cos\left(x+y\right) = \log\left(x+y\right)$$
$2-$ $$\frac{d}{dx} \left(\sin\left(x+y\right)+ \cos\left(x+y\right)\right) =\frac {d}{dx} \log\left(x+y\right)$$
$3-$ $$\cos\left(x+y\right)\left(1+\frac{dy}{dx}\right) - \sin\left(x+y\right)\left(1+\frac{dy}{dx}\right) = \frac{ \left(1+\frac{dy}{dx}\right) }{x+y}$$
And now in step $4$, term $\left(1+\frac{dy}{dx}\right)$ gets cancelled from both sides and we are left with
$$\cos\left(x+y\right) - \sin\left(x+y\right)= \frac{1}{x+y}.$$
So we don't get any term containing $\frac{dy}{dx}$. So how can I find out $\frac{dy}{dx}$ if I am not getting it, same problem arises if I differentiate it again.
| you can cancel $\left(1+\dfrac{dy}{dx}\right)$ only when $\left(1+\dfrac{dy}{dx}\right)\neq0$. But in this case:
\begin{align*}
\sin(x+y)+\cos(x+y)=\log(x+y)\\
\Rightarrow \dfrac{d}{dx}[\sin(x+y)]+\dfrac{d}{dx}[\cos(x+y)]&=\dfrac{d}{dx}[\log(x+y)]\\
\Rightarrow \cos(x+y)\cdot\left(1+\dfrac{dy}{dx}\right)-\sin(x+y)\cdot\left(1+\dfrac{dy}{dx}\right)&=\dfrac{1}{x+y}\cdot\left(1+\dfrac{dy}{dx}\right)\\
\Rightarrow\left(1+\dfrac{dy}{dx}\right)\left[\cos(x+y)-\sin(x+y)-\dfrac{1}{x+y}\right]&=0
\end{align*}
So either $\left(1+\dfrac{dy}{dx}\right)=0\quad$ or, $\quad\cos(x+y)-\sin(x+y)=\dfrac{1}{x+y}$.
Therefore, $\dfrac{dy}{dx}=-1\quad$ or, $\quad\cos(x+y)-\sin(x+y)=\dfrac{1}{x+y}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2236285",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
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} |
If $x^2 - 3x + 2$ is a factor of $x^4 - px^2 +q$, then find the value of $p$ and $q$ If $x^2 - 3x +2$ is a factor of $x^4-px^2+q$ then find the value of $p$ and $q$.
My attempt:
$$x^2-3x+2$$
$$x^2-2x-x+2$$
$$x(x-2)-1(x-2)$$
$$(x-1)(x-2)$$
How do I proceed further?
P.S: Edit after Deepak's comment!
| Since $1$ and $2$ are zeros of $x^2 - 3x + 2$, they must also be zeros of $x^4-px^2+q.$
Substituting $x=1\,$ into $x^4-px^2+q\,$ yields $1-p+q = 0$.
Substituting $x=2\,$ into $x^4-px^2+q\,$ yields $16-4p+q = 0$.
So now you have two equations in two unknowns.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2236641",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
Summing 1d discrete random walks in opposite directions Fix some time $T$. Suppose we have $A$ with $A_0 = 0$, and $A_t = A_{t-1} + X_{t - 1}$ for $T \geq t > 0$ where $X_t$ takes value either $+1/2$ or $-1/2$ with equal probability; and that we have $B$ with $B_T = 0$, and $B_{t-1} = B_t + X'_t$, with $X'_t$ defined as $X_t$.
That is to say, a random walk $A$, starting at time $0$, running to time $T$; and $B$ starting at $T$ running backwards to time $0$.
I'm interested in $C_t = A_t + B_t$. How much time does it spend with $|C| > v$ for some value $v$, and how can I calculate $P(C_T > a | C_t = b)$
I suspect there aren't nice answers with $X$ so defined, so if there are nicer results when $X \sim N(0,1)$, then let that be my question instead.
| Writing down the combined step and the corresponding probability we get a random walk $C = A - B$ with
$$C_t = C_{t-1} + Y$$
where
$$Y=\left(
\begin{array}{cc}
+1 & p=\frac{1}{4} \\
0 & p=\frac{1}{2} \\
-1 & p=\frac{1}{4} \\
\end{array}
\right)$$
This gives a generating function
$$g(z,t) = \left(\frac{1}{4}\left(z+2+\frac{1}{z}\right)\right)^t$$
and the final probability for the distance between the two Walkers to be $x$ at the time $t$ is
$$p(t,x) =\frac{1}{4^t}\binom{2 t}{t+x}$$
The first few values in the format $(t,p(t,x)$ with $( x=-t..+t))$ are
$$p(t,x) = \left(
\begin{array}{cc}
1 & \left\{\frac{1}{4},\frac{1}{2},\frac{1}{4}\right\} \\
2 & \left\{\frac{1}{16},\frac{1}{4},\frac{3}{8},\frac{1}{4},\frac{1}{16}\right\} \\
3 & \left\{\frac{1}{64},\frac{3}{32},\frac{15}{64},\frac{5}{16},\frac{15}{64},\frac{3}{32},\frac{1}{64}\right\} \\
4 & \left\{\frac{1}{256},\frac{1}{32},\frac{7}{64},\frac{7}{32},\frac{35}{128},\frac{7}{32},\frac{7}{64},\frac{1}{32},\frac{1}{256}\right\} \\
5 & \left\{\frac{1}{1024},\frac{5}{512},\frac{45}{1024},\frac{15}{128},\frac{105}{512},\frac{63}{256},\frac{105}{512},\frac{15}{128},\frac{45}{1024},\frac{5}{512},\frac{1}{1024}\right\} \\
\end{array}
\right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2238119",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Number Theory with Pythagorean Triples For $n\ge3$ a given integer, find a Pythagorean Triple having n as one of its members.
Hint: For n an odd integer, consider the triple $$\left(n, \frac 12\left(n^2-1\right), \frac 12(n^2+1)\right);$$ For n even, consider the triple $$\left(n, \left(\frac{n^2}{4}\right)-1, \left(\frac{n^2}{4}\right)+1 \right)$$
I have been trying to solve this problem by letting $n=x-y$ or $n=2xy$ but have been unable to use the hints properly. Do I just add the squares of the first 2 and see if it equals the square of the last? I am just missing the concept here. Any suggestions would be appreciated.
$$n^2+\left(\frac{n^2-1}{2}\right)^2=n^2+\frac{n^4-2n^2+1}{4}=\frac{n^4+2n^2+1}{4}$$
$$\left( \frac{n^2+1}{2}\right)^2=\frac{n^4+2n+1}{4}$$
Is this all I have to do for odd?
| $\\ \textbf{Matching sides using Euclid's F(m,n)}: $ Solving for $n$, any values of $m$ that yield integers provide the $F(m,n)$ to identify a triple. Examples follow the solutions
$$\mathbf{A=m^2-n^2\Rightarrow n=\sqrt{m^2-A}\qquad\qquad \lceil\sqrt{A+1}\rceil \le m \le \biggl\lceil\frac{A}{2}\biggr\rceil}$$
The lower limit ensures $m^2>A$ and the upper limit ensures $m-n\ge 1$.
$$\text{Example:}\qquad A=21\Rightarrow \lceil\sqrt{21+1}=5 \le m \le \biggl\lceil\frac{21}{2}\biggr\rceil =11\quad and \quad m\in\{5,11\}\Rightarrow n\in\{2,10\} $$
$F(5,2)=(21,20,29)\qquad\qquad F(11,10)=(21,220,221)\\ $
$$\mathbf{B=2mn\Rightarrow n=\frac{B}{2m}\qquad\qquad \bigl\lceil\sqrt{B}\bigr\rceil\le m \le \frac{B}{2}}$$
The lower limit ensures $m>n$ and the upper limit ensures $m-n\ge 1$.
$$\text{Example:}\qquad B=44\Rightarrow \lceil\sqrt{88}\rceil =10 \le m \le \frac{44}{2}=22\quad and \quad m\in\{11,22\}\Rightarrow n\in\{2,1\}$$
$F(11,2)=(117,44,125)\qquad\qquad F(22,1)=(483,44,485)\\$
$$\mathbf{C=m^2+n^2\Rightarrow n=\sqrt{C-m^2}\qquad\qquad \biggl\lceil\sqrt{\frac{C}{2}}\biggr\rceil \le m < \sqrt{C}}$$
$$\text{Example:}\qquad C=1105\Rightarrow \biggl\lfloor\sqrt{\frac{1105}{2}}\biggr\rfloor=23 \le m < \lfloor\sqrt{1105}=33\quad and \quad m\in\{24,31,32,33\}\Rightarrow n\in\{23,12,9,4\}\\$$
$F(24,23)=(47,1104,1105)\quad F(31,12)=(817,744,1105)\\ $
$F(32,9)=(943,576,1105)\quad F(33,4)=(1073,264,1105)\\$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2238614",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Better way of finding $\int {x^2 + 1 \over x^4 + x^2 + 1} dx$.
$$\int {x^2 + 1 \over x^4 + x^2 + 1} dx$$
By separating partial fractions,
$${Ax + B \over x^2 - x + 1} + {Cx + D \over x^2 + x + 1} = {x^2 + 1 \over x^4 + x^2 + 1} \\ \implies (Ax + B)(x^2 + x + 1) + (Cx + D)(x^2 - x + 1) = x^2 + 1$$
I get $$\begin{cases} A = -1/2 \\
B = 1/2\\
C = 1/2 \\
D = 1/2\end{cases}$$
For which the integrand becomes
$${1 \over 2}\int {1 - x\over x^2 - x + 1} dx + {1\over2} \int {x + 1\over x^2 + x + 1} dx$$
Now these two are easy enough to solve but still very tedious, not to mention that partial fractions was also very tedious.
Is there a less cumbersome way to solve this ? I tried to change the integrand of form $1 + 1/x^2$ so that I substitute $u = 1 - 1/x$ but was unsuccessful .
| Here is a funny way that I noticed by chance. I don't see really why it would work, it popped out when I played with the function. In any case, I think someone might get something out of it, so I post it.
We first notice that
$$
\frac{1+x^2}{1+x^2+x^4}-\frac{a}{b+x^2}=\frac{b-a+(1+b-a)x^2+(1-a)x^4}{b+(1+b)x^2+(1+b)x^4+x^6}
$$
If we choose $b=3$ then the denominator becomes
$$
3+(2x+x^3)^2,
$$
and the right-hand side becomes
$$
\frac{3-a+(4-a)x^2+(1-a)x^4}{3+(2x+x^3)^2}
$$
If we next choose $a=1$ the numerator magically becomes the derivative of $2x+x^3$!
Thus, we conclude that
$$
\begin{aligned}
\int\frac{1+x^2}{1+x^2+x^4}\,dx
&=\int\Bigl(\frac{1}{3+x^2}+\frac{2+3x^2}{3+(2x+x^3)^2}\Bigr)\,dx\\
&=\frac{1}{\sqrt{3}}\Bigl(\arctan\frac{x}{\sqrt{3}}+\arctan\frac{2x+x^3}{\sqrt{3}}\Bigr)+C.
\end{aligned}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2240614",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Find that probability that exactly $ 4 $ students get candy Question: A teacher gives $ 5 $ candies to $ 8 $ students. She gives each candy to a randomly chosen student, without regard to whether the student has received candy.
a) What is the probability that exactly $ 4 $ of the students get candy?
Here is my approach, I treat the candies to be identical and the students to be distinguishable: There are total of $ \displaystyle \binom{8 + 5 - 1}{5 - 1} = \binom{12}{4} $ ways to distribute $ 5 $ identical candies to $ 8 $ students. Next, there are $ \displaystyle \binom{8}{4} $ ways to choose $ 4 $ students to receive $ 5 $ candies so that each student among the $ 4 $ chosen must receive at least $ 1 $ candy, yielding a total of $ \displaystyle 4 .\binom{8}{4} $ possibilities. So the answer for part a) is $ \displaystyle \frac{4 .\binom{8}{4}}{\binom{12}{4}} $.
b) What is the probability that $ 5 $ different students get the candy?
I interpret this question as the same as: what is the probability that at least $ 5 $ students get the candy? Again, there are $ \displaystyle \binom{8 + 5 - 1}{5 - 1} = \binom{12}{4} $ ways to distribute $ 5 $ identical candies to $ 8 $ students. Next, I will find the number of possibilities that exactly $ 5, 6, 7, $ and $ 8 $ students get candies, respectively. There are $ \displaystyle \binom{8}{5} $ ways that exactly $ 5 $ students get candies. Note that there is no way more than $ 5 $ students get the candy if each student gets at least $ 1 $ candy. So the answer for part b) is $ \displaystyle \frac{\binom{8}{5}}{\binom{12}{4}} $.
Am I approaching the problem correctly? Any comment is appreciated.
| I would look at it this way, starting with b:
b) In order for five students to receive the five sweets, one student must receive one sweet each. The probability of the teacher selecting a different student each time equals:
$$\frac{8}{8} \cdot \frac{7}{8} \cdot \frac{6}{8} \cdot \frac{5}{8} \cdot \frac{4}{8} = \frac{6720}{32768} \approx 0.205$$
a) In order for four students to receive five sweets, one student must receive two sweets and three others must receive one. Let the teacher randomly select a first student. The probability of the teacher selecting this student again in the next turn equals $\frac{1}{8}$. If this happens, the teacher must select three different students afterwards, since one student already got two sweets. If this does not happen (i.e. two different students were selected), there are two options in the third turn: one of the two students who already got a sweet are selected, or a new student is selected. The former case occurs with a probability of $\frac{2}{8}$, in which case two different students must be selected afterwards. If the teacher however selects a third unique student, the probability of the teacher selecting one of them on the fourth term equals $\frac{3}{8}$ (in which case a different student must be selected in the last term). If a new student is however selected, one of the four selected students must be selected again in order to receive two sweets, which happens with a probability of $\frac{4}{8}$. All in all, the probability is thus:
$$\frac{8 \cdot 1 \cdot 7 \cdot 6 \cdot 5}{8^5} + \frac{8 \cdot 7 \cdot 2 \cdot 6 \cdot 5}{8^5} + \frac{8 \cdot 7 \cdot 6 \cdot 3 \cdot 5}{8^5} + \frac{8 \cdot 7 \cdot 6 \cdot 5 \cdot 4}{8^5} \approx 0.512$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2240900",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Coefficient of $x^5$ in binomial expansion. We have to find coefficient of $x^5$ in the expansion of $$(1+5x)^5 +(1+5x)^6 +\cdots+(1+5x)^{19}$$
My try
from first expansion I get $(^5 _5)5^5$ from second $(^6 _5)5^5$ .
but from this I can prove the answer as $(^{20}_{14})5^5$
| Better to first factor the expression and recognize it as the sum of a finite geometric series with common ratio $r = 1+5x$; thus $$\sum_{k=5}^{19} (1+5x)^k = (1+5x)^5 \sum_{k=0}^{14} (1+5x)^k = (1+5x)^5 \frac{(1+5x)^{15} - 1}{(1+5x) - 1} = \frac{(1+5x)^5 ((1+5x)^{15} - 1)}{5x}.$$ Letting $y = 5x$ and further expanding we get $$\frac{(1+y)^{20}}{y} - \frac{(1+y)^5}{y},$$ and if we want the coefficient of $x^5$, this corresponds to finding the coefficient of $y^5$ and multiplying by $5^5$. The second term contributes no terms of this form, as the highest power is $y^4$. Thus we are looking for the coefficient of $y^6$ in $(1+y)^{20}$, which is simply $\binom{20}{6} = \binom{20}{14}$, hence the coefficient of $x^5$ in the original expression is $$5^5 \binom{20}{6}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2242128",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove if $B^2=I+BA$ and $A^2=AB$ then $A=0$ I need to prove that if $B^2=I+BA$ and $A^2=AB$ then $A=0$, $A$ and $B$ are square matrices. I'm not sure if my answer is correct but I thought of this:
$$
A^2-B^2=A^2+AB-BA-B^2=A^2+A^2-(B^2-I)-B^2=2(A^2-B^2)+I
$$
$$
\Rightarrow B^2-A^2=I=(B-A)(B+A)
$$
This means that $(B-A)$ is invertible. It is also given that $I=B^2-AB$ then:
$$
B^2-AB=B^2-A^2
$$
$$
\Leftrightarrow B(B-A)=(B+A)(B-A)
$$
Because we proved that $(B-A)$ is invertible then we can simplify and get $B=B+A \Rightarrow A=0$
| $B^2 = I +BA\implies B(B - A) = I$, so $B-A$ is invertible; hence $A-B$ is invertible. Now:
$A^2 = AB \implies A(A-B) = 0 \implies A(A-B)(A-B)^{-1} = 0 \implies A = 0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2243423",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How to solve the homogeneous differential equation? EDITED WITH FINAL ANSWER:
Solve the following differential equation:
$$y' = \frac{2xy}{x^2-y^2}$$
Someone please help me to finish this problem.
My solution so far:
$$\frac{dy}{dx} = \frac{\frac{1}{x^2}(2xy)}{\frac{1}{x^2}(x^2-y^2)}$$
$$\frac{dy}{dx}= 2\frac{y}{x} * \frac{1}{1-{\frac{y^2}{x^2}}}$$
Let $v = \frac{y}{x}$, $y=vx$ then $\frac{dy}{dx} = v+x\frac{dv}{dx}$
$$\frac{dy}{dx} = \frac{2v}{1-v^2}$$
Setting the two equations equal to one another:
$$v+x\frac{dv}{dx} = \frac{2v}{1-v^2}$$
$$x\frac{dv}{dx} = \frac{2v}{1-v^2} - \frac{v-v^3}{1-v^2}$$
$$x\frac{dv}{dx} = \frac{v+v^3}{1-v^2}$$
$$xdv = \frac{v+v^3}{1-v^2}dx$$
$$\frac{1-v^2}{v+v^3}dv = \frac{1}{x}dx$$
$$\int\frac{1-v^2}{v+v^3}dv = \int\frac{1}{x}dx$$
$$\ln \left|v\right|-\ln \left|v^2+1\right| = ln|x| + c$$
Substituting $\frac{y}{x}$ back for $v$:
$$\ln \left|\frac{y}{x}\right|-\ln \left|\frac{y^2}{x^2}+1\right| = ln|x| + c$$
$$\ln \left|{y}\right|-\ln|x|-\ln \left|\frac{y^2}{x^2}+1\right| = ln|x| + c$$
Taking $e$ to everything we obtain:
$$y - x - (\frac{y^2}{x^2}+1) = x + e^c$$
$$y - (\frac{y^2}{x^2}+1) = 2x + e^c$$
$$y - \frac{y^2}{x^2} - 1= 2x + e^c$$
$$y - \frac{y^2}{x^2} = 2x + e^c + 1$$
$$\frac{x^2y-y^2}{x^2} = 2x + e^c + 1$$
$$x^2y-y^2 = 2x^3 + x^2e^c + x^2$$
$$0 = y^2 - x^2y + 2x^3 + x^2e^c + x^2$$
Using the quadratic formula we obtain
$$y = \frac{x^2±\sqrt{x^4-8x^3-4x^2e^c-4x^2}}{2}$$
$$y = \frac{x^2±\sqrt{x^2(x^2-8x-4e^c-4)}}{2}$$
$$y = \frac{x^2±x\sqrt{x^2-8x-4e^c-4}}{2}$$
| In such problems it pays to describe the solution curves in polar coordinates. If
$$\phi\mapsto\bigl(r(\phi)\cos\phi,\> r(\phi)\sin\phi\bigr)$$
is a solution curve then $y'={\displaystyle{\dot y\over\dot x}}$, where the dot denotes differentiation with respect to $\phi$. It follows that $$y'={2xy\over x^2-y^2}$$
translates into
$${\dot r\sin\phi+r\cos\phi\over\dot r\cos\phi-r\sin\phi}={\sin(2\phi)\over\cos(2\phi)}\ .$$
This is equivalent to
$${\dot r\over r}={\cos\phi\over\sin\phi}\ ,$$
so that the solution curves turn out to have the polar description
$$r(\phi)=C\>\sin\phi\ .$$
It is easy to verify that these curves are circles through the origin.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2244315",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 1
} |
calculus - solving limit with complex polynomial I'm try to solve $\lim \limits_{x \to -2} \dfrac{x^3 - x^2 - x + 10}{x^2 + 3x + 2}$.
I have simplified it down to $\lim \limits_{x \to -2} \dfrac{(x + 2)(x^2 - 3x + 5)}{(x + 2)(x + 1)}$.
The solution is $-15$.
However, I can't wrap my mind around it. $(x^2 - 3x + 5)$ has only complex roots because $\Delta < 0$, so how am I supposed to reach the real number $-15$ as the solution?
| "Solve" is the wrong word here. You're trying to evaluate a limit, not to "solve" a limit. One solves equations; one solves problems; one evaluates expressions.
To evaluate $\lim \limits_{x \to -2} \dfrac{(x + 2)(x^2 - 3x + 5)}{(x + 2)(x + 1)},$ you don't need to know the zeros of $x^2-3x+5.$ The reason for being concerned with zeros of $x^3 - x^2 - x + 10$ and $x^2 + 3x + 2$ is only that $-2$ is such a zero, so you get $0$ in both the numerator and the denominator. If substitution into polynomials yields a nonzero number in the denominator, then you're done with the "hard" part.
You have
$$
\lim_{x \to -2} \frac{(x + 2)(x^2 - 3x + 5)}{(x + 2)(x + 1)} = \lim_{x\to-2} \frac{x^2 - 3x+5}{x+1} = \frac{(-2)^2 - 3\cdot(-2) + 5}{-2+1} = \frac {15} {-1} = -15.
$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How can we show that $\int_{-\infty}^{+\infty}{b\over 1+2ax+(cx)^2}\cdot{\mathrm dx\over 1-2ax+(cx)^2}={\pi\over 2}?$ Consider this integral
$$\int_{-\infty}^{+\infty}{b\over 1+2ax+(cx)^2}\cdot{\mathrm dx\over 1-2ax+(cx)^2}={\pi\over 2}\tag1$$
Where $a^2+b^2=c^2$; Pythagoras theorem.
Making an attempt:
Complete the square of $(cx)^2+2ax+1=c^2\left(x+{a\over c^2}\right)^2+{c^2-a^2\over c^2}$, I can't see how this would be of any useful.
$1+2x^2(c^2-2a^2)+c^4x^4=(1+2ax+c^2x^2)(1-2ax+c^2x^2)$
I am even sure how to go about making an attempt to find the equivalent of $(1)$.
How may we go about and prove $(1)?$
| I will assume $b > 0$ (in which case $a^2$ can never equal $c^2$).
Employing a partial fraction decomposition we have
$$\frac{1}{1 + 2ax + c^2 x^2} \cdot \frac{1}{1 - 2ax + c^2 x^2} = \frac{1}{4a} \cdot \frac{2a + c^2 x}{1 + 2ax + c^2 x^2} + \frac{1}{4a} \cdot \frac{2a - c^2 x}{1 - 2ax + c^2 x^2}.$$
Now noting that
$$\int \frac{2a + c^2 x}{1 + 2ax + c^2 x^2} \, dx = \frac{a}{\sqrt{c^2 - a^2}} \tan^{-1} \left (\frac{c^2 x + a}{\sqrt{c^2 - a^2}} \right ) + \frac{1}{2} \ln |1 + 2ax + c^2 x^2| + \cal{C}_1,$$
and
$$\int \frac{2a - c^2 x}{1 - 2ax + c^2 x^2} \, dx = \frac{a}{\sqrt{c^2 - a^2}} \tan^{-1} \left (\frac{c^2 x - a}{\sqrt{c^2 - a^2}} \right ) - \frac{1}{2} \ln |1 - 2ax + c^2 x^2| + \cal{C}_2.$$
(note that each of these integrals can be solved by elementary means). Thus
\begin{align*}
\int_{-\infty}^{\infty}\frac{b}{1+2ax+c^2x^2}\cdot \frac{dx}{1-2ax+c^2 x^2} &= \left [\frac{b}{4 \sqrt{c^2 - a^2}} \left \{\tan^{-1} \left (\frac{c^2 x + a}{\sqrt{c^2 - a^2}} \right ) \right. \right.\\
& \left. \left. + \tan^{-1} \left (\frac{c^2 x - a}{\sqrt{c^2 - a^2}} \right ) \right \}+ \frac{b}{8a} \ln \left |\frac{1 + 2ax + c^2 x^2}{1 - 2ax + c^2 x^2} \right | \right ]^\infty_{-\infty}\\
&= \frac{b}{4 \sqrt{c^2 - a^2}} \left [\frac{\pi}{2} + \frac{\pi}{2} - \left (-\frac{\pi}{2} - \frac{\pi}{2} \right ) \right ]\\
&= \frac{\pi b}{2 \sqrt{c^2 - a^2}}.
\end{align*}
But given $a^2 + b^2 = c^2$, then $\sqrt{c^2 -a^2} = \sqrt{b^2} = |b| = b$, assuming $b > 0$ giving
$$\int_{-\infty}^\infty \frac{b}{1+2ax+c^2x^2} \cdot \frac{dx}{1-2ax+c^2 x^2} = \frac{\pi}{2},$$
as required to show
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2250435",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Easiest way to find characteristic polynomial for this 4x4 matrix I have been given the matrix
$$
\begin{bmatrix}
1 & 3 & 0 & 3 \\
1 & 1 & 1 & 1 \\
0 & 4 & 2 & 8 \\
2 & 0 & 3 & 1 \\
\end{bmatrix}
$$
and told I must find the characteristic polynomial. I began by applying cofactor expansion along the top row of the matrix
$$
\begin{bmatrix}
1-\lambda & 3 & 0 & 3 \\
1 & 1-\lambda & 1 & 1 \\
0 & 4 & 2-\lambda & 8 \\
2 & 0 & 3 & 1-\lambda \\
\end{bmatrix}
$$
and attempting to multiply out my results to get the correct answer of $\lambda^4 -5\lambda^3 - 28\lambda^2 + 58\lambda - 8$. However, this takes several pages of work and I keep making calculation errors and ending up with the wrong answer.
My question is, is there an easier way to find the determinant of this specific matrix, or, once the determinant is found, to multiply out the result to find the polynomial?
The only methods I have been taught have been to either try to find or create a row with several 0's to make the cofactor expansion easier, or to get an upper or lower triangular matrix, however, those seem equally as messy here.
| Let $\chi_A(t)$ be our characteristic polynomial.
Add $\DeclareMathOperator{Row}{Row}-2\cdot\Row_2$ to $\Row_4$ to obtain
$$
\chi_A(t)=
\det
\left[\begin{array}{rrrr}
-t + 1 & 3 & 0 & 3 \\
1 & -t + 1 & 1 & 1 \\
0 & 4 & -t + 2 & 8 \\
0 & 2 \, t - 2 & 1 & -t - 1
\end{array}\right]
$$
Then add $-(-t+1)\cdot\Row_1$ to $\Row_2$ to obtain
\begin{align*}
\chi_A(t)
&=
\det
\left[\begin{array}{rrrr}
0 & -{\left(t - 1\right)}^{2} + 3 & t - 1 & t + 2 \\
1 & -t + 1 & 1 & 1 \\
0 & 4 & -t + 2 & 8 \\
0 & 2 \, t - 2 & 1 & -t - 1
\end{array}\right] \\
&=
-
\det
\left[\begin{array}{rrrr}
-{\left(t - 1\right)}^{2} + 3 & t - 1 & t + 2 \\
4 & -t + 2 & 8 \\
2 \, t - 2 & 1 & -t - 1
\end{array}\right]
\end{align*}
This gives a relatively simple way to reduce your $4\times 4$ determinant to a $3\times 3$ determinant.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2252444",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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If one root of the equation $ax^2+bx+c=0$ be the square of the other then which is true? If one root of the equation $ax^2+bx+c=0$ be the square of the other then which is true?
$1$. $a^3+b^3+c^3-3abc=0$
$2$. $a^3+b^3+bc^2=3abc$
$3$. $b^3+a^2c+ac^2=3abc$
$4$. none.
My Attempt:
Let one root be $\alpha $ then the other root will be $\alpha^2$. Then,
$$(x-\alpha)(x-\alpha^2)=0$$
$$x^2-x(\alpha^2+\alpha)+\alpha^3=0$$ Comparing with $ax^2+bx+c=0$ we get,
$$a=1$$
$$b=-(\alpha^2+\alpha)$$
$$c=\alpha^3$$
| Please, in your future exercises, try not to name different things with the same variable ($a$ in this example), because it causes a huge confusion. Also, remember than the quadratic factorization has the coefficient of $x^2$.
We have the equation : $ax^2 + bx + c = 0$
Now, let's assume it has two roots, $ρ_1 , ρ_2 $ with $ρ_2 = ρ_1^2$
Then, the polynomial will be factorized as :
$a(x-ρ_1)(x-ρ_2) = 0 \Leftrightarrow a(x-ρ_1)(x-ρ_1^2)=0 \Leftrightarrow ax^2 - aρ_1 (ρ_1 + 1) x + aρ_1 ^3 = 0 $
You can continue on from there.
Alternativelly, you can take on Vietta's Formulas :
$$\alpha+\alpha^2=-\dfrac ba$$
$$\alpha\cdot\alpha^2=\dfrac ca$$
and continue on from there.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2252741",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
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} |
Evaluate $\int \frac{\log(1+x^2)}{\sqrt{1-x^2}} \,\mathrm dx$
$\def\d{\mathrm{d}}$Evaluate $$\int \frac{\log(1+x^2)}{\sqrt{1-x^2}}\,\d x.$$
I have used Integration by parts as follows:
$$I=\log(1+x^2) \: \sin^{-1}x-\int \frac{2x \sin^{-1}x}{1+x^2}\,\d x=\log(1+x^2) \: \sin^{-1}x-J,$$ where
$$J=\int \frac{2x \sin^{-1}x}{1+x^2}\,\d x$$ In this put $x=\sin y$ we get
$$J=\int \frac{y \sin2y}{1+\sin^2y}\,\d y.$$
Any clue here?
| We can also do this.
Consider the double integral:
\begin{align}I=\int_{0}^{1}\int_{0}^{1} \frac{2x^2y}{\sqrt{1-x^2}(1+x^2y^2)} \ dy \ dx. \end{align}
Integrating with respect to $y$ first, we have
\begin{align}I=\int_{0}^{1} \frac{\ln(1+x^2)}{\sqrt{1-x^2}} \ dx. \end{align}
On the other hand, we reverse the order of integration
\begin{align}I=\int_{0}^{1}\int_{0}^{1} \frac{2x^2y}{\sqrt{1-x^2}(1+x^2y^2)} \ dx \ dy. \end{align}
For this, we use the antiderivative of $ \frac{2x^2y}{\sqrt{1-x^2}(1+x^2y^2)}$ is \begin{align} \frac{2}{y} \left(\sin^{-1}(x) - \frac{\tan^{-1} \left( \frac{x\sqrt{1+y^2}}{\sqrt{1-x^2}}\right)}{\sqrt{y^2+1}} \right), \end{align}
so we upon plugging in endpoints of integration that
\begin{align} I= \int_{0}^{1} \frac{\pi}{y} - \frac{\pi}{y\sqrt{1+y^2}} \ dy = \int_{0}^{1} \frac{\pi \sqrt{y^2+1} - \pi}{y\sqrt{1+y^2}} \ dy= \pi \int_{0}^{1}\frac{ y}{\sqrt{1+y^2}(1+\sqrt{1+y^2})} \ dy, \end{align} which we get by simplifying with common denominators and multiplying by the conjugate of the numerator. Another u-substitution $u=1+\sqrt{1+y^2}, \ du = \frac{y}{\sqrt{1+y^2}(1+\sqrt{1+y^2})}$ gives us
\begin{align} I= \pi \ln(1+\sqrt{2})-\pi\ln(2) \end{align}
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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} |
$1 + \frac 64 + \frac {14}8 + \frac {30}{16} + \ldots$ I have to calculate the sum upto n terms of the following:
$$1 + \frac 64 + \frac {14}8 + \frac {30}{16} + \ldots$$
I found the general term as:
$$T_n = \frac {an^2+bn+c}{2^{n}},$$ where a, b and c are determined as follows:
$$a+b+c=2; 4a+2b+c=6; 9a+3b+c=14.$$
Solving and applying:
$$T_n = \frac {2n^2-2n+2}{2^{n}} = \frac {n^2-n+1}{2^{n-1}}.$$
Now, how to compute:
$$\sum_{k=1}^{n} \frac {k^2-k+1}{2^{k-1}}$$
| How about choosing $$T_n=\frac{\frac23n^3-2n^2+\frac{16}3n-2}{2^n}$$ whose elements match the few elements you mentioned and $$\sum_{n=1}^{\infty}T_n=14.$$
Here is an interesting discussion.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Find $2^2 + 5^2 + 8^2 + ... + (3n-1)^2 $
First prove that $1^2 + 2^2 + 3^2 +...+ n^2 = \frac{n(n+1)(n+2)}{6}$, then find $$2^2 + 5^2 + 8^2 + ... + (3n-1)^2.$$
I can prove the first part but I have no idea about the second part.
| Since
$$\sum_{k=1}^n (3n-1)^2 = 9\sum_{k=1}^n k^2-6\sum_{k=1}^n k + \sum_{k=1}^n 1$$
and
$$\sum_{k=1}^n k = \frac{n(n+1)}{2}$$, use your previous result to get
$$\begin{align}
\sum_{k=1}^n (3n-1)^2 &= 9\sum_{k=1}^n k^2-6\sum_{k=1}^n k + \sum_{k=1}^n 1 \\
&=9\left(\frac{n(n+1)(n+2)}{6}\right)-6\left( \frac{n(n+1)}{2}\right)+n \\
\end{align}$$ and work from there
| {
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"source": "stackexchange",
"question_score": "2",
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How to find all integer solutions of an equation which are divisible by 8? I have to find out all integer solutions of $n^2+15$ which are divisible by 8. So my idea was to solve the following equation.
$$n^2+15=8\cdot k \to n^2=8\cdot k-15 \to n=\sqrt{8\cdot k-15}$$
But this doesn't work.I tried some numbers $n=1, n=3, n=5, n=7$ for these numbers I get integer solutions of $n^2+15$ which are divisible by 8. It seems to be that $n$ must be odd. How do I solve this correctly?
Best regards
| $n^2 + 15 = 8k \implies$
$n^2 -1 = 8(k-2) \implies$
$(n + 1)(n-1) = 8(k-2)$
If $n$ is even then $n+1$ and $n-1$ are odd and this is impossible.
If $n$ is odd then $n+1$ and $n-1$ are even. If one of them is not divisible by $4$ than the other one is. So if $n$ is odd hen $(n+1)(n-1)=n^2 -1$ is divisible by $8$ and $n^2 +15$ is divisible by $8$.
..or ..
Postactively: If $n = 2k+1$ than $n^2 + 15 = (2k+1)^2 + 4k^2 + 4k + 16 = 4(k^2 + k + 2)= 8(\frac {k(k+1) }2 + 1)$
... or ...
Let $n = 4k + i; i = 0, \pm 1, \pm 2$
Then $n^2 + 15 = 16k + 8ki + i^2 + 15$ so $n^2 + 15$ is divisible by $8$ iff $15+i^2$ is. $15+ 0^2$ and 15+ (\pm 2)^2$ are not. $15 + (\pm 1)^2 = 16$ is.
So $n^2 + 15$ is divisible by $8$ iff $n \equiv \pm 1 \mod 4$ iff $n$ is odd.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2258135",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
Define a set X of integers recursively as follows: Define a set X of integers recursively as follows:
Base: 5 is in X
Rule 1: If x is in X and x>0, then x+3 is in X
Rule2: If x is in X and x>0, then x+5 is in X
Show that every integer n>7 is in X
I am pretty new to this stuff and I am very lost on this. Please help!
| Let $n \ge 13$. Divide $n$ by $5$ and take the remainder to get $n = k*5 + r$ and $k$ is at least $3$.
Case 0: if $r = 0$ then $n = 5+5+5+5 ..... = 5*k + 0 = n$ and $n $ is in $X$.
Case 3: if $r = 3$ then $n = 5+ 5+ 5+ .... + 3 = 5+ 3 + 5(k-1)$ is in $X$.
Case 1: if $r = 1$ then $n = 5+5+5+5+ ..... + 5 + 1 = 5+5+5+..... + 6 = 5(k-1)+3+3$ is in $X$
Can you do Case 2 and 4?
=====
This is a cute way.
Let $n = ......ba$ be the number written in digits. $a$ is the last digt and $b$ is the second to last digit.
If $a = 0$ or $5$ then $n = 5+5+5+5+..... $ for some number of $5$ so $n $ is in $x$.
If $a = 1$ or $6$ then $n = 5+5+5+5+ .... + 5 + 3+3$ for some number of $5$ so $n$ is in $X$ if $n \ge 11$.
If $a = 2$ or $7$ then $n = 5+5+5+5+ ..... + 5 + 3+3+3+3$ for some number of $5$ so $n$ is in $n$ if $n > 12$.
If $a = 3$ or $8$ then $n = 5+5+5+... + 3$ and $n$ is in $X$ for $n \ge 8$.
If $a = 4$ or $9$ then $n = 5+5+5+5 + .... + 3+3+3$ and $n$ is in $X$ for $n \ge 14$.
So all integers greater than $12$ are in $X$.
====
Ah, This is supposed to be a proof by induction! (I was worried that might be advanced.)
1: All numbers in the form $5k + 3j; k \ge 1; j \ge 0$ are in $X$.
Base case: It is true for $k = 1$ and $j=0$ as $5 = 5*1 + 0 \in X$.
Induction case: Assume it is true for $n = 5k + 3*0$. Then by Rule one it is true for $n = 5k + 3*0 + 5 = 5(k+1) + 3*0$. So it is true for all $n = 5k + 3*0$.
Assume it is true $n = 5k + 3j$. Then by Rule two it is true for $n = 5k + 3j + 3 = 5k + 3(j+1)$ . So it is true for all $n = 5k + 3j$.
2) Prove that for all $n\ge 13$ that $n = 5k + 3j$ for some $k \ge 1$ and $j \ge 0$ and either $k >1$ or $j > 2$.
Base case: $13 = 5*2 + 3$
Induction: If $n = 5k + 3j$ and $k > 1$ then$n+1 = 5(k-1) + 3(j+2)$ and $k \ge 1;j > 2$. If $n = 5k + 3j$ and $j > 2$ then $n+1 = 5(k+2) + 3(j-3)$ and $k >1; j \ge 0$.
So All $n \ge 13$ are $n = 5k + 3j$ for some $k \ge 1$ and $j \ge 0$ and either $k > 1$ or $j > 2$.
| {
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"url": "https://math.stackexchange.com/questions/2258720",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How does one evaluate $ \int _2^4\:\frac{x^4-x^2-4x-6}{\:x^4-x^2-2x}dx $? I tried this : $$\begin{align}&\int _2^4\:\frac{x^4-x^2-4x-6}{\:x^4-x^2-2x}dx\\&=\int _2^4\:dx\:+\:\int _2^4\:\frac{-2x-6}{x^4-x^2-2x}dx\\&=2-\int _2^4\:\frac{4x^3-2x-2-4x^3-4}{x^4-x^2-2x}dx\\&=\:2-\:\ln\left(29\right)\:-4\int _2^4\:\frac{x^3+1}{x^4-x^2-2x}dx\end{align}$$ but I'm stuck at the last integral.
| Hint : $${\displaystyle\int}\dfrac{x^3+1}{x^4-x^2-2x}\,\mathrm{d}x=\class{steps-node}{\cssId{steps-node-1}{\dfrac{1}{2}}}{\displaystyle\int}\dfrac{3x^2-1}{x^3-x-2}\,\mathrm{d}x-\class{steps-node}{\cssId{steps-node-2}{\dfrac{1}{2}}}{\displaystyle\int}\dfrac{1}{x}\,\mathrm{d}x$$
And use the change of variable $u=x^3-x-2$ for the first integral.
| {
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Prove that $n$ is prime given two conditions. $$\text{Suppose that $d$ is a divisor of $n$ and $d+1$ is a divisor of $n+1$. Prove that $n$ is prime.}$$
| First Assume $p$ is prime then the only divisors $d=1,p$ so $d+1 =2,p+1$ which both divides $p+1$ if $p+1$ is even,which means that $p$ is odd prime.
So it works for all odd primes $p\not=2$ that if $d|p$ then $d+1|p+1$.
On the other hand assume $n=a b$ not a prime then $1<a,b<n$, so there are at least the divisors $d=1,a,b,ab$ which means that $d+1=2,a+1,b+1,ab +1$ that must divide $n+1=ab+1$ sure the $2,ab+1$ divides $n+1$ if $n+1$ is even and $n$ is odd, we need to see if $a+1,b+1$ divides $ab+1$.
Which means that $ab+1=0 (\mod a+1)$ and $ab+1=0(\mod b+1)$, we can see that $ab+1 = (a+1)b+1-b =1-b (\mod a+1)$ and the same to the second modular equation $ab+1=1-a (\mod b+1)$, which means that $1-a=0 (\mod b+1)$ and $1-b=0 (\mod a+1)$ which means that $a=1 (\mod b+1)$ and $b=1 (\mod a+1)$, and since $a,b >1$ so $a=(b+1)c_1 +1$ and $b=(a+1)c_2 +1$ with $c_1,c_2 \geq 1$.
Now in the first equation substitute instead of $b$ the value $(a+1)c_2+1$, we arrive at $a=(b+1)c_1+1=((a+1)c_2+1)c_1+1$ when trying to solve it with the conditions that $a>1$ and $c_1,c_2 \geq 1$ you get contradictions,which conclude the proof.
if d|n and d+1|n+1 then n is a odd prime.
| {
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"timestamp": "2023-03-29T00:00:00",
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$5x^2+4e^{7-x}=x^2e^{7-x}+20$, find $x$ Given that $5x^2+4e^{7-x}=x^2e^{7-x}+20$.
The given solution is the equation is factorised to $e^{-x}(5e^x-e^7)(x^2-4)=0$, but I've no idea how it's factorised. Can anyone show it?
| \begin{align*}
5x^2 + 4e^{7-x} &= x^2e^{7-x} + 20\\
\\
5x^2 + 4e^7e^{-x} - x^2 e^7e^{-x} - 20 &= 0\\
\\
5x^2e^xe^{-x} + 4e^7e^{-x} - x^2e^7e^{-x} - 20e^xe^{-x} &= 0\\
\\
e^{-x}(5x^2e^x + 4e^7 - x^2e^7 - 20e^x) &= 0\\
\\
e^{-x}(5x^2e^x - 4\cdot 5e^x - x^2e^7 + 4e^7) &= 0\\
\\
e^{-x}[5e^x(x^2 - 4) - e^7(x^2+4)] &= 0\\
\\
e^{-x}[(5e^x - e^7)(x^2-4)] &= 0\\
\\
e^{-x}(5e^x - e^7)(x-2)(x+2) &= 0
\\
\end{align*}
I guess they didn't factor the $x^2-4$ into $(x-2)(x+2)$ but there you have it. It's basically just factoring out $e^{-x}$ from everything (and noting that if a term doesn't have an $e^{-x}$ in it then it will have $e^x$ in it after factoring) and then factoring by grouping.
| {
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"timestamp": "2023-03-29T00:00:00",
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Parameterize Intersection of Surfaces I need to parameterize the intersection of $$4x^2 + y^2 + z^2 = 9\tag{1}$$ and $$z=x^2+y^2\tag{2}$$.
First, I'll solve (2) for $y^2$ and substitute the result into (1):
$$3x^2+z+z^2 = 9 \tag{3}$$
Next, I'll make the substitution $u=\sqrt{3}x$, such that we can complete the square in (3) by adding $1/4$ to each side and arrive at
$$\frac{u^2}{r^2} + \frac{(z+\frac{1}{2})^2}{r^2} = 1$$
where $r^2 = 9 + \frac{1}{4} = \frac{37}{4}$
Now I'll write a parameterization:
$$u = r\cos \phi \implies x(\phi) = \frac{1}{\sqrt{3}}r\cos\phi$$
$$z(\phi) = r\sin \phi -\frac{1}{2}$$
$$y(\phi) = \pm \sqrt{z-x^2} = \pm\left(\sqrt{r\sin \phi - \frac{1}{2} - \left(\frac{1}{\sqrt{3}}r\cos\phi\right)^2}\right)$$
such that we have two branches:
$$\mathbf{r}(\phi)_1 = \big<x(\phi), y(\phi), z(\phi)\big>$$
$$\mathbf{r}(\phi)_2 = \big<x(\phi), -y(\phi), z(\phi)\big>$$
Is this correct?
| Yes your equation is correct. It was smart of you to do a substitution.
If you [I] need a visual verification:
| {
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Find $b$ in $2\sin(\frac{x}{2}+b)$ given the graph Following is the graph of the function. $2\sin \left(\dfrac{x}{2}+b\right)$
I tried solving the equation using the point $\left (\dfrac{\pi}{2},2\right)$:
$$2 = 2\sin\left(\dfrac{x}{2}+b\right) \Leftrightarrow \\
1 = \sin\left(\dfrac{x}{2}+b\right)\Leftrightarrow \\
\frac{x}{2}+b = k\pi + (-1)^k\arcsin(1) \Leftrightarrow \\
b = 2k\pi$$
But my book $b = \dfrac{\pi}{4}$. How do I solve this?
|
How did you get $b=2k\pi$?
Put $x=\dfrac{\pi}{2}$ in the equation.
$$\frac{x}{2}+b = k\pi + (-1)^k\arcsin(1)$$
Since $\arcsin(1)=\dfrac{\pi}{2}$
You'll get $b+\dfrac{\pi}{4}=k\pi+
(-1)^k\dfrac{\pi}{2}$
Since we have added $b$ in the argument of sine, that is $2b$ in the $x$, the graph has been shifted backwards by $2b$ units.
Now put $k=1$ , you'll get $b=\dfrac{\pi}{4}$You can try $k=0$ and $k=2$ also, but that all will result in a value of $b$ which isn't in the range $(0,\pi/2)$
(Since any other value of $k$ will lead $b$ to get out of the range $(0,\pi/2)$)
Alternatively since the first negative $x$-intercept of graph is $2b$, and it is clearly visible that it is $\pi/2$, we can say that $$2b=\dfrac{\pi}{2} \implies b=\dfrac{\pi}{4}$$
| {
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Can't compute this integral I would appreciate your help solving this integral:
$$\int_2^3 \frac{1+x^3}{(x^2+a^2)^\frac{3}{2}}\mathrm{d}x$$
I tried using linear substitution with $ t = x/a $ and then trying to bring it to some combination of the known integral of $\arctan (x) = \frac{1}{x^2+1}$ but I'm not sure it will be helpful because there isn't just $1 $ in the numerator.
Basically, I got stuck very early in the process:
$$\int_2^3 \frac{1+x^3}{(a^2(\frac{x^2}{a^2}+1))^\frac{3}{2}}\mathrm{d}x$$
Thank you.
| I suggest breaking it into two integrals, and using the trig substitution $x=a\tan t$ to turn each one into a trigonometric integral. Thus:
$$\begin{align}
\int_2^3 \frac{1+x^3}{(x^2+a^2)^\frac{3}{2}}\mathrm{d}x &= \int_2^3 \frac{dx}{(x^2+a^2)^{3/2}} + \int_2^3\frac{x^3}{(x^2+a^2)^{3/2}}dx\\
&=\int_{\tan^{-1}(2/a)}^{\tan^{-1}(3/a)}\frac{a\sec^2 t}{a^3\sec^3 t}dt + \int_{\tan^{-1}(2/a)}^{\tan^{-1}(3/a)}\frac{a^3 \tan^3 t \cdot a\sec^2 t}{a^3\sec^3 t}dt\\
&=\int_{\tan^{-1}(2/a)}^{\tan^{-1}(3/a)}\frac{1}{a^2}\cos t dt + \int_{\tan^{-1}(2/a)}^{\tan^{-1}(3/a)}\frac{a\sin^3 t}{\cos^2 t}dt
\end{align}$$
Can you get it from there?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2269067",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 0
} |
integrate $\int_D e^{x^2+3y^2}$ Evaluate $\int_D e^{x^2+3y^2}$, where $D$ is the region bounded in the first quadrant by the lines $y=0, y=x, x^2+3y^2=1$.
My method is as follows, and I am not sure if it is correct.
Let $u=x, v=\sqrt{3}y$. Then $D$ becomes bounded by $v=0, \frac{\sqrt{3}}{3}v=u, u^2+v^2=1$.
$u=r\cos\theta, v=r\sin\theta$, so $\int_D e^{x^2+3y^2} = \int_D e^{r^2}r = \int^{\pi/6}_0e^{r^2}r$
Is this correct? If not, can you tell me where I got this wrong? Thank you.
| Let $u=x$ and $v=\sqrt{3}y$.
Then the three constraints become $v=0$, $v=\sqrt{3}u$, and $u^2+v^2=1$. Using the equations $x=u$ and $y=\frac{1}{\sqrt{3}}v$,
we obtain the absolute value of the Jacobian to be:
\begin{align*}
\Bigg| \frac{\partial(x,y)}{\partial(u,v)} \Bigg| &=
\Bigg|
\det
\begin{pmatrix}
x_u & x_v \\
y_u & y_v \\
\end{pmatrix}
\Bigg|
= \Bigg|
\det
\begin{pmatrix}
1 & 0 \\
0 & \frac{1}{\sqrt{3}} \\
\end{pmatrix}
\Bigg|
= \frac{1}{\sqrt{3}}.
\end{align*}
So we have
\begin{align*}
\iint_{D} e^{x^2+3y^2}dA
&= \int_{0}^{\frac{1}{2}}\int_{y}^{\sqrt{1-3y^2}} e^{x^2+3y^2}dxdy \\
&= \int_{0}^{\frac{\sqrt{3}}{2}}\int_{\frac{1}{\sqrt{3}}v}^{\sqrt{1-v^2}} e^{u^2+v^2}\Bigg| \frac{\partial(x,y)}{\partial(u,v)} \Bigg|dudv \\
&= \int_{0}^{\frac{\sqrt{3}}{2}}\int_{\frac{1}{\sqrt{3}}v}^{\sqrt{1-v^2}} e^{u^2+v^2}\frac{1}{\sqrt{3}} dudv \\
&= \frac{1}{\sqrt{3}} \int_{0}^{\frac{\pi}{3}} \int_{0}^{1} e^{r^2} r drd\theta \hspace{4mm}\mbox{ since } u = r \cos \theta\mbox{ and }v = r\sin \theta \\
&= \frac{1}{\sqrt{3}}\frac{\pi}{3} \frac{1}{2}\int_{0}^{1} e^{w} dw \hspace{4mm}\mbox{ since } w=r^2, \mbox{ so } dw = 2rdr \\
&=\frac{\pi}{6\sqrt{3}}(e-1).
\end{align*}
| {
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"url": "https://math.stackexchange.com/questions/2270729",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Sum $\sum_{n=1}^\infty (\log 2)^n \sum_{r=1}^n \frac {r^2}{r!(n-r)!},$ The general term of a sequence is given as
$$\displaystyle T_n = (\log 2)^n \sum_{r=1}^n \frac {r^2}{r!(n-r)!},$$ $n\in \mathbb N$.
How can I find an expression for the below given summation?
$$ \sum_{k=1}^\infty T_k$$
Multiplying and dividing by $n!$ the first expression is reduced to $$ T_n= \frac{(\log2)^n}{n!} \sum_{r=1}^{n} {^nC_r} r^2$$
$$\implies T_n=2^{n-2} (\log2)^n \frac{ (n+1)}{(n-1)!}$$
But I cannot think of any method to evaluate $\sum_{k=1}^{\infty} T_k$.
Is there some hidden telescoping I am missing?
Is it possible that this sum to infinity be reduced to an integral? If so, how can it be done?
| We have
$$e^x = \sum_{n=0}^{\infty} \frac{1}{n!} x^n$$
This implies that
$$e^x = \frac{d}{dx}e^x = \sum_{n=1}^{\infty} \frac{1}{(n-1)!} x^{n-1}$$
Thus,
$$x^2 e^x = x^2 \frac{d}{dx}e^x = \sum_{n=1}^{\infty} \frac{1}{(n-1)!} x^{n+1}$$
Taking a second derivative gives
$$(2x+x^2) e^x = \frac{d}{dx}(x^2 e^x) = \sum_{n=1}^{\infty} \frac{n+1}{(n-1)!} x^{n}$$
Now,
$$\begin{align}
\sum_{n=1}^{\infty} T_n & = \sum_{n=1}^{\infty} 2^{n-2} (\ln 2)^n \frac{n+1}{(n-1)!} \\
& = \frac14 \sum_{n=1}^{\infty} \frac{n+1}{(n-1)!} (2 \ln 2)^n\\
\end{align}$$
Thus,
$$\sum_{n=1}^{\infty} T_n = \frac14 \left.(2x+x^2) e^x \right|_{x=2\ln 2} = \frac14(2\cdot 2\ln 2+(2\ln 2)^2)e^{2\ln 2} = (\ln 2 + (\ln 2)^2) \cdot 4$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2273207",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Find the sum of the infinite series $\frac{1}{1\cdot 2}+\frac{1\cdot3}{1\cdot2\cdot3\cdot4}+\frac{1\cdot3\cdot5}{1\cdot2\cdot3\cdot4\cdot5\cdot6}+...$ Find the sum of the series $\frac{1}{1\cdot 2}+\frac{1\cdot3}{1\cdot2\cdot3\cdot4}+\frac{1\cdot3\cdot5}{1\cdot2\cdot3\cdot4\cdot5\cdot6}+...$.
This type of questions generally require a trick or something and i am not able to figure that out. My guess is that it has something to do with exponential series or binomial series. Any help?
| Sorry guys, got it.
$\frac{1}{1\cdot 2}+\frac{1\cdot3}{1\cdot2\cdot3\cdot4}+\frac{1\cdot3\cdot5}{1\cdot2\cdot3\cdot4\cdot5\cdot6}+...=\frac{1}{2}\cdot\frac{1}{1!}+\frac{1}{2^2}\cdot\frac{1}{2!}+\frac{1}{2^3}\cdot\frac{1}{3!}+... = e^\frac{1}{2}-1.$
The first equality holds after cancelling the common terms in the numerator and denominator
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2274577",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
} |
Find $A \left(\begin{smallmatrix} 3 \\ -11 \\ -1 \end{smallmatrix}\right).$ Let $A$ be a $3 \times 3$ matrix such that
$$A \begin{pmatrix} 3 \\ 4 \\ 5 \end{pmatrix} = \begin{pmatrix} 2 \\ 7 \\ -13 \end{pmatrix}, \quad A \begin{pmatrix} 4 \\ 5 \\ 6 \end{pmatrix} = \begin{pmatrix} -6 \\ 0 \\ 4 \end{pmatrix}, \quad A \begin{pmatrix} 5 \\ -9 \\ 1 \end{pmatrix} = \begin{pmatrix} 3 \\ 3 \\ -11 \end{pmatrix}.$$
Find
$$A \begin{pmatrix} 3 \\ -11 \\ -1 \end{pmatrix}.$$
I'm sure there's a way to arrange the equations so that we get the result, but I've tried adding and subtracting but none leads to $\begin{pmatrix} 3 \\ -11 \\ -1 \end{pmatrix}$.
| You're thinking the right thing. There is a way to do it systematically. Essentially you want to find $x,y,z$ such that
$$x\begin{pmatrix} 3 \\ 4 \\ 5 \end{pmatrix}+ y\begin{pmatrix} 4 \\ 5 \\ 6 \end{pmatrix} + z\begin{pmatrix} 5 \\ -9 \\ 1 \end{pmatrix} = \begin{pmatrix} 3 \\ -11 \\ -1 \end{pmatrix}$$
This amounts to solving this system of equations for $x,y,z$:
\begin{align*}
3x + 4y + 5z &= 3\\
4x + 5y - 9z &= -11\\
5x + 6y + z &= -1
\end{align*}
Once you have $x,y,z$ you'll be able to express $\begin{pmatrix} 3 \\ -11 \\ -1\end{pmatrix}$ as a linear combination of those other $3$ vectors above. Then use linearity of matrix-vector multiplication to get the desired result. Specifically, you'll want to use the fact that if $A$ is a matrix, $a,b,c$ are scalars, and $\vec u, \vec v, \vec w$ are vectors, then $$A \left[a\vec u + b\vec v + c\vec w\right] = a\cdot A\vec u + b\cdot A\vec v + c\cdot A\vec w.$$
Let me know if you require further assistance.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2275624",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Solve $\frac{1}{x-1}\geq\frac{a}{x+1}$, provided $a>1$ Moving everything to the left,
$$\frac{1}{x-1}-\frac{a}{x+1}\geq0\Longleftrightarrow\frac{ax-x-a-1}{x^2-1}\leq 0.$$
Finding roots to nominator and denominator:
$$\begin{array}{lcl}
ax-x-a-1 & = & 0 \Leftrightarrow \frac{a+1}{a-1}\\
x^2-1 & = & 0 \Leftrightarrow x_1 =1 \ \text{och} \ x_2=-1.\\
\end{array}$$
And that's it. How to proceed?
| $$\frac{(x+1-ax+1)}{x^2-1}\ge0$$
$$\frac{x(1-a)+2}{x^2-1}\ge0$$
Since $a>1$
$$x\ge\frac{-2}{a-1}$$
$$x\le\frac{2}{a-1}\tag{x not equal to +,-1}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2276309",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
$\int \frac{x^4-1}{x^2\sqrt{x^4+x^2+1}}dx$ $$\int \frac{x^4-1}{x^2\sqrt{x^4+x^2+1}}\,dx$$
My attempt,
I changed it to $\int \frac{x^2}{\sqrt{x^4+x^2+1}} \, dx-\int \frac 1 {x^2\sqrt{x^4+x^2+1}} \, dx$, but I stuck here.
Any method to solve this integral? Thanks in advance.
| Everytime I see an integrand that look sort of "symmetric" under $x \to x^{-1}$,
I will try to see whether it can be reexpressed in terms of $x \pm x^{-1}$.
One useful identity to remember is
$$\frac{dx}{x} = \frac{d(x+x^{-1})}{x - x^{-1}} = \frac{d(x-x^{-1})}{x + x^{-1}}$$
This trick does work for the integral at hand.
$$\begin{align}
\int \frac{x^4-1}{x^2\sqrt{x^4+x^2+1}} dx &= \int \frac{x^2-x^{-2}}{\sqrt{x^2+x^{-2}+1}} \frac{dx}{x} \\
&= \int \frac{x^2-x^{-2}}{\sqrt{x^2+x^{-2}+1}} \frac{d(x-x^{-1})}{x+x^{-1}} \\
&= \int \frac{(x-x^{-1}) \, d(x - x^{-1})}{\sqrt{(x-x^{-1})^2+3}} \\
&= \int d\sqrt{(x-x^{-1})^2+3} \\
&= \sqrt{x^2+x^{-2}+1} + \text{const.}
\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2276721",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
} |
Regular Tetrahedron rotation problem Say that I have a regular tetrahedron with vertices $a$, $b$, $c$ and $d$ where $a$, $b$ and $c$ sit on a plane. The height of $d$, above the plane, is trivial to obtain given the length of any edge.
If I were now to re-orient the tetrahedron such that vertex $a$ remains in contact with the plane but $b$ and $c$ are raised from it, how can I now obtain the height of $d$ above the plane, given the elevations of $b$ and $c$?
Presumably this is relatively trivial, but it is outside my area of expertise.
One thing to note is that I need to be able calculate the final algorithm fairly quickly in real time using the 'C' language on a fairly low power processor.
| Suppose an edge of the tetrahedron has length $1$, and let the heights of $b$ and $c$ after elevation be $j$ and $k$, respectively.
Place vertex $a$ at the origin of a system of coordinates. Let vertex $b$ be located in the $xz$-plane, so its coordinates, after elevation, are $(\sqrt{1-j^2},0,j)$. We can find the coordinates of $c$ by specifying that it is $1$ unit away from $a$ and from $b$, and $k$ units above the $xy$-plane.
$x^2+y^2+z^2=1$
$(x-\sqrt{1-j^2})^2 + y^2 + (z-j)^2 = 1$
$z=k$
Solving these equations, we get the coordinates of $c$ as $\left(\sqrt{\frac{1-4jk(1-jk)}{4(1-j^2)}},\sqrt{\frac{3-4(j^2+k^2-jk)}{4(1-j^2)}},k\right)$. As a reality-check, note that when $j=k=0$, these expressions reduce to the appropriate numbers.
Finally, the vertex $d$ is at the intersection of the three spheres centered at $a$, $b$ and $c$, each with radius $1$.
$x^2 + y^2 + z^2 = 1$
$\left(x-\sqrt{1-j^2}\right)^2 + y^2 + (z-j)^2 = 1$
$\left(x-\sqrt{\frac{1-4jk(1-jk)}{4(1-j^2)}}\right)^2 + \left(y-\sqrt{\frac{3-4(j^2+k^2-jk)}{4(1-j^2)}}\right)^2 + (z-k)^2 = 1$
The height you're looking for is the $z$-coordinate of the intersection of those three spheres. The top equation can be solved for $y$: $y=\sqrt{1-x^2-z^2}$, and that can be substituted into the other two. Then the second equation can be solved for $x$, resulting in $x=\frac{1-2jz}{2\sqrt{1-j^2}}$. Finally, this can be plugged into the third equation, resulting in a horrific mess, which is nevertheless, after some simplifying, a quadratic equation in $z$.
Indeed:
$$\frac{2j\sqrt{A}\sqrt{1-j^2}+j-2k(1-j^2)}{1-j^2}z+A+\frac{\sqrt{A}}{\sqrt{1-j^2}}+k^2=2\sqrt{B}\sqrt{C(z)}$$,
where $A=\frac{1-4jk(1-jk)}{4(1-j^2)}$, $B=\frac{3-4(j^2+k^2-jk)}{4(1-j^2)}$, and $C(z)=1-z^2-\frac{1-4jz+4j^2z^2}{4(1-j^2)}$
If you square both sides of this, it's quadratic in $z$. Unfortunately, I have to stop here. If you have specific values for $j$ and $k$, you can plug them in, and then it won't be so bad.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2277084",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 3,
"answer_id": 2
} |
What's the trick to get the limit of a sequence with roots like $a_n = -3n + \sqrt{9n^2+11n+16}$? I know of some "tricks" for limits of sequences like
$$b_n = \frac{n^2+7}{3n+1} = \frac{n^2+7}{3n+1} \cdot \frac{\frac{1}{n^2}}{\frac{1}{n^2}} = \frac{\frac{n^2}{n^2}+\frac{7}{n^2}}{\frac{3n}{n^2}+\frac{1}{n^2}}$$
Surely, there has to be a trick like this for the above example $a_n = -3n + \sqrt{9n^2+11n+16}$? Thanks. :-)
| \begin{align*}
-3n + \sqrt{9n^2+11n+16}&=\frac{(-3n + \sqrt{9n^2+11n+16})(-3n - \sqrt{9n^2+11n+16})}{-3n - \sqrt{9n^2+11n+16}}\\
&=\frac{9n^2-(9n^2+11n+16)}{-3n - \sqrt{9n^2+11n+16}}
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2278235",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Calculate the number of real numbers $k$ such that $f(k) = 2$ if $f(x) = x^4 − 3x^3 − 9x^2 + 4$.
Calculate the number of real numbers $k$ such that $f(k) = 2$ if $f(x) = x^4 − 3x^3 − 9x^2 + 4$.
How do I calculate the number of values for $k$? I can't seem to understand. Please help me solve this problem.
| This is to find the number of real roots of $f(x)-2$.
$$\frac{d}{dx}(f(x)-2)=4x^3-9x^2-18x$$
which is equal to $0$ when $x=0$ or $\displaystyle \frac{9\pm4\sqrt{41}}{8}$.
$$f\left(\frac{9-4\sqrt{41}}{8}\right)-2=\frac{-23195-3321\sqrt{41}}{512}<0$$
$$f(0)-2=2>0$$
$$f\left(\frac{9+4\sqrt{41}}{8}\right)-2=\frac{-23195-3321\sqrt{41}}{512}<0$$
So $f(x)-2$ has a zero in each of the intervals $(-\infty,\frac{9-4\sqrt{41}}{8})$, $(\frac{9-4\sqrt{41}}{8},0)$, $(0,\frac{9+4\sqrt{41}}{8})$ and $(\frac{9+4\sqrt{41}}{8},\infty)$.
There are $4$ solutions.
You can also try the Descartes' rule of signs
https://en.wikipedia.org/wiki/Descartes%27_rule_of_signs
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2280275",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find the interval of $a^2+b^2+c^2$ Given that the system of equations $x=cy+bz,y=az+cx,z=bx+ay$ has non-zero solutions and at least one of a,b,c is a proper fraction, then find the interval in which value of $a^2+b^2+c^2$ lies($a,b,c\in R$).
MY APPROACH:
I applied the cramer's rule and got the equation as $a^2+b^2+c^2=1-2abc$. Now how do i find the range of $a^2+b^2+c^2?$
| Let $a=b=c=\frac{1}{2}$.
Hence, $a^2+b^2+c^2=\frac{3}{4}$.
We'll prove that $\frac{3}{4}$ is a minimal value.
Indeed, Let $a^2+b^2+c^2=\frac{3}{4}x^2$, where $x\geq0$.
Thus, by AM-GM
$$1=a^2+b^2+c^2+2abc\leq\frac{3}{4}x^2+2|abc|\leq$$
$$\leq\frac{3}{4}x^2+2\left(\sqrt{\frac{a^2+b^2+c^2}{3}}\right)^3=\frac{3}{4}x^2+\frac{1}{4}x^3,$$
which gives $x\geq1$ and we are done!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2280413",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Stronger than Nesbitt's inequality $\frac{a}{\sqrt[4]{8(b^4+c^4)}}+\frac{b}{a+c}+\frac{c}{a+b}\geq\frac{3}{2}$
Let $a$, $b$ and $c$ be non-negative numbers such that $ab+ac+bc\neq0$. Prove that:
$$\frac{a}{\sqrt[4]{8(b^4+c^4)}}+\frac{b}{a+c}+\frac{c}{a+b}\geq\frac{3}{2}$$
Nesbitt's inequality is the following:
Let $a$, $b$ and $c$ be non-negative numbers such that $ab+ac+bc\neq0$. Prove that:
$$\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}\geq\frac{3}{2},$$
which follows from C-S:
$$\sum_{cyc}\frac{a}{b+c}=\sum_{cyc}\frac{a^2}{ab+ac}\geq\frac{(a+b+c)^2}{\sum\limits_{cyc}(ab+ac)}=\frac{(a+b+c)^2}{2(ab+ac+bc)}\geq\frac{3}{2},$$
but this way does not help for the starting inequality.
There is a nice solution for the following inequality, which was in our test six months ago.
Let $a$, $b$ and $c$ be positive numbers. Prove that:
$$\frac{a}{\sqrt[3]{4(b^3+c^3)}}+\frac{b}{c+a}+\frac{c}{a+b}\ge \frac{3}{2}$$
By C-S $$\frac{b}{c+a}+\frac{c}{a+b}\geq\frac{(b+c)^2}{ab+ac+2bc}.$$
Thus, it's enough to prove that
$$\frac{a}{\sqrt[3]{4(b^3+c^3)}}+\frac{(b+c)^2}{ab+ac+2bc}\geq\frac{3}{2}$$ or
$$2(b+c)a^2-(3(b+c)\sqrt[3]{4(b^3+c^3)}-4bc)a+2(b^2-bc+c^2)\sqrt[3]{4(b^3+c^3)}\geq0.$$
Thus, it's enough to prove that
$$16(b+c)(b^2-bc+c^2)\sqrt[3]{4(b^3+c^3)}\geq\left(3(b+c)\sqrt[3]{4(b^3+c^3)}-4bc\right)^2$$ or
$$16\sqrt[3]{4(b^3+c^3)^4}\geq\left(3(b+c)\sqrt[3]{4(b^3+c^3)}-4bc\right)^2$$ and since
$$3(b+c)\sqrt[3]{4(b^3+c^3)}>4bc,$$
it remains to prove that
$$4\sqrt[3]{2(b^3+c^3)^2}\geq3(b+c)\sqrt[3]{4(b^3+c^3)}-4bc$$ and after using
$$x^3+y^3+z^3-3xyz=(x+y+z)(x^2+y^2+z^2-xy-xz-yz),$$ where
$$x^2+y^2+z^2-xy-xz-yz\neq0$$ we need to prove that
$$128(b^3+c^3)^2-108(b+c)^3(b^3+c^3)+64b^3c^3+288(b+c)(b^3+c^3)bc\geq0.$$
Now, let $b^2+c^2=2kbc$.
Hence, we need to prove that:
$$128(2k+2)(2k-1)^2-108(2k+2)^2(2k-1)+64+288(2k+2)(2k-1)\geq0$$ or
$$(k-1)^2(10k+11)\geq0.$$
Done!
But this way gives a wrong inequality again.
| Your inequality is equivalent to :
$$f(x)=\frac{1}{(8(x^4+h^4))^{0.25}}+\frac{x}{h+1}+\frac{h}{x+1}$$
Here we assume that $x\geq h \geq 1$
It's easy to see that the function $f(x)$ is convex with $h$ fixed (as the sum of convexs functions) and increasing for $x\geq h$
Furthermore for $h$ fixed we have that the minimum is inferior to $h$ (in abscissa)
So we have :
$$f(x)\geq f(h)$$
We get :
$$f(x)=\frac{1}{(8(x^4+h^4))^{0.25}}+\frac{x}{h+1}+\frac{h}{x+1}\geq\frac{1}{(8(h^4+h^4))^{0.25}}+\frac{h}{h+1}+\frac{h}{h+1} $$
So the inequality is verified in this case because it's easy to threat this one variable inequality.
The case where $h\geq x \geq 1$ is the same (because of the symetry) and the cases $x\geq 1 \geq h$ and
$h\geq 1 \geq x$ are similar for the same reasons
Now we continue with the case $x\leq h \leq 1$
Remark that if we inverse each variable it becomes :
$$\frac{xh}{(8(x^4+h^4))^{0.25}}+\frac{x}{h+1}+\frac{h}{x+1}$$
With $h\geq 1$ and $x\geq 1$ but we have :
$$\frac{xh}{(8(x^4+h^4))^{0.25}}+\frac{x}{h+1}+\frac{h}{x+1}\geq \frac{1}{(8(x^4+h^4))^{0.25}}+\frac{x}{h+1}+\frac{h}{x+1}\geq 1.5$$
So this case is verified .
Done !
Edit :My proof is incomplete like this because the case where $x\geq 1 \geq h$ is not solved . So I purpose to you a solution :
The main idea is to use the three chord lemma related to the convexity of the function $f(x)$ so we have for $x\geq \sqrt{x}\geq x^{0.25}$ :
$$\frac{f(x)-f(\sqrt{x})}{x-\sqrt{x}}\geq \frac{f(x)-f(x^{0.25})}{x-x^{0.25}} \geq \frac{f(\sqrt{x})-f(x^{0.25})}{\sqrt{x}-x^{0.25}}$$
We can apply this inequality $n$ times to get :
$$\frac{f(x)-f(\sqrt{x})}{x-\sqrt{x}}\geq \frac{f(x^{1/2^n})-f(x^{1/2^{n+1}})}{x^{1/2^n}-x^{1/2^{n+1}}}$$
But we have to prove that the RHS is positive for a variable wich tends to $1$ so we get the following result to evaluate :
$$\lim_{x\to 1} \frac{\dfrac{1}{(8(x^{1/2^n}+h^4))^{0.25}}+\dfrac{x^{1/2^n}}{h+1} + \dfrac{h}{x^{1/2^n}+1}- \left(\dfrac{1}{(8(x^{1/2^{n+1}}+h^4))^{0.25}} + \dfrac{x^{1/2^{n+1}}}{h+1}+\dfrac{h}{x^{1/2^{n+1}}+1} \right)}{x^{1/2^n}-x^{1/2^{n+1}}}$$
For that see the the interesting answer of G.Cab here
With this answer we know that the result does not depends on $n$ and is equal to :
$$
\eqalign{
& \mathop { }
& {1 \over {h + 1}} - {1 \over {2^{\,11/4} \left( {h^{\,4} + 1} \right)^{\,5/4} }} - {h \over 4} \cr}
$$
Quantity wich is positive for $h\leq 1$
So we get :
$$\frac{f(x)-f(\sqrt{x})}{x-\sqrt{x}}\geq 0$$
Or :
$$f(x)\geq f(\sqrt{x})$$
If we apply this inequality $n$ times we get :
$$f(x)\geq f(x^{\frac{1}{2^n}})$$
But the variable tends $1$ and the RHS to $1.5$ so we get the desired result !
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Volume of a solid 2 I've tried to solve this exercise but the answer is different from the one on my notes: using cilindrical coordinates compute the volume of the subset of $\mathbb{R}^3$ defined by:
$z \geq 0$, $x^2+y^2+z^2 \leq 4$, $x^2-2x+y^2 \leq 0$.
My solution:
$$
\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\left ( \int_{0}^{2\cos \theta} \rho \left ( \int_{0}^{\sqrt{4-\rho^2}}dz \right )d\rho \right )d\theta=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \left ( -\frac{1}{3}\sqrt{(4-\rho^2)^3}|^{2 \cos \theta}_0 \right )d \theta= -\frac{8}{3}\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sqrt{(1- \cos^2 \theta)^3} +\frac{8}{3} \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} d \theta =
-\frac{8}{3} \left [ -\cos \theta \sin^2 \theta -\frac{2}{3} \cos^3 \theta\right ]^{\frac{\pi}{2}}_{-\frac{\pi}{2}}+\frac{8}{3}\pi=\frac{8}{3}\pi
$$
Can anyone please check my solution?
| We have $$\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sqrt{(1- \cos^2 \theta)^3}d\theta=2\int_{0}^{\frac{\pi}{2}} \sin\theta(1- \cos^2 \theta)d\theta=2\left[-\cos\theta+\frac{\cos^3\theta}{3}\right]_{0}^{\frac{\pi}{2}}=\frac{4}{3}\not=0.$$
Hence the volume of the solid should be $\frac{8}{3}\pi-\frac{32}{9}$.
P.S. Note that $\sqrt{(1- \cos^2 \theta)^3}=|\sin\theta|(1- \cos^2 \theta)$.
| {
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"timestamp": "2023-03-29T00:00:00",
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$\pi$ as an Infinite Series using Taylor Expansion on Equation of a Circle If we write $\ y = \sqrt{r^2-x^2}$ for the equation of a circle in Cartesian coordinates with radius $r$ and perform a Taylor expansion on this equation, then integrate term by term from $0$ to $r$, giving the area of a quarter of a circle. Then, isolating $\pi$, all $r$ terms conveniently drop out and the expression becomes:
$$\pi = 4 - \frac{4}{3!}- \frac{12}{5!}- \frac{180}{7!}- \frac{6300}{9!}- \frac{396900}{11!} - ...$$
This can be written as:
$$\ \pi = \sum_{n=1}^\infty \frac{-4[(2n-3)!!]^2}{(2n-3)(2n-1)!}$$
I attach here a WolframAlpha Link of this evaluated to 2700 terms.
I haven't seen this expression anywhere else. I was wondering if you have, or if you know if this is valid?
| You can do the same thing using a unit circle (this is basic dimensionality analysis).
\begin{align}\frac{\pi}{4} &= \int_0^1 \sqrt{1-x^2} \,dx
= \int_0^1 \left((1+(-x^2)\right)^{1 /2}\, dx\\
&= \int_0^1 \sum_{k=0}^{\infty}\binom{1 /2}{k} (-x^2)^k\, dx
\overset{(*)}{=} \sum_{k=0}^{\infty} (-1)^k\binom{1 /2}{k}\int_0^1 x^{2k}\, dx\\
&= \sum_{k=0}^{\infty} (-1)^k \frac{\binom{1 /2}{k}}{2k+1}
\end{align}
Now you can show that the mess of double factorials simplifies to
$$\binom{1/ 2}{k} = -\frac{1}{2k-1}\binom{2k}{k}\left(-\frac{1}{4}\right)^k,\quad\text{for $k> 0$, and}\quad \binom{1/ 2}{0} = 1$$
hence
$$\sum_{k=0}^{\infty} (-1)^k \frac{\binom{1 /2}{k}}{2k+1}
= 1 - \sum_{k=1}^{\infty} \frac{\binom{2k}{k}}{(2k-1)(2k+1)4^k}
= 1 - \sum_{k=1}^{\infty} \frac{\binom{2k}{k}}{(4k^2-1)4^k}$$
So we get a form that is similar to yours:
$$
\pi = 4 - \sum_{k=1}^{\infty} \frac{\binom{2k}{k}}{(4k^2-1)4^{k-1}}
$$
The central binomial coefficients $\binom{2k}{k}$ satisfy:
$$\binom{2k}{k} \leq \frac{4^k}{\sqrt{3k+1}},\quad\text{for $k\geq 1$}$$
which may be proven by induction. Then
\begin{align}
\sum_{k=1}^{\infty} \frac{\binom{2k}{k}}{(4k^2-1)4^{k-1}}
&\leq\sum_{k=1}^{\infty} \frac{4^k}{\sqrt{3k+1}(4k^2-1)4^{k-1}}
= \frac{1}{4} \sum_{k=1}^{\infty} \frac{1}{\sqrt{3k+1}(2k-1)(2k+1)}\\
&\leq \frac{1}{4} \sum_{k=1}^{\infty} \frac{1}{\sqrt{3k+1}(2k-1)(2k+1)}
\leq \frac{1}{4} \sum_{k=1}^{\infty} \frac{1}{\sqrt{2k-1}(2k-1)(2k-1)}\\
&\leq \frac{1}{4} \sum_{k=1}^{\infty} \frac{1}{(2k-1)^{5/2}}
\leq \frac{1}{4} \sum_{k=1}^{\infty} \frac{1}{k^{5/2}}
< \infty
\end{align}
with the latter convergence given by the $p$-series test.
As the partial series are positively increasing and bounded above, they must converge. This proves the convergence for the expression we gave.
Finally I somewhat skipped over the $(*)$ way back at the beginning of this answer, where we assume that we can swap the integral and summation. We eventually showed the terms are non-negative and converge, hence by Tonelli's Theorem we were justified in swapping the order of integrals (and sums).
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Polynomial sum relation Given the following polynomial: $$P(x)=(x^2+x+1)^{100}$$ How do I find : $$\sum_{k=1}^{200} \frac{1}{1+x_k} $$ Is there a general solution for this type of problem cause I saw they tend to ask the same thing for $\sum_{k=1}^{200} \frac{1}{x_k}$? Also how do I find the coefficient of $a_1$ and the remainder for $$P(x)/(x^2+x)$$ (/=divided) (I found the coef of a1 is 100 adn the remainder is 1 but im not sure)
| Note that the roots of $x^2+x+1$ are
$$\zeta_3 =e^{2\pi i/3} = -\frac{1}{2}+i\frac{\sqrt{3}}{2}$$
and
$$\zeta_3^2 =e^{4\pi i/3} = -\frac{1}{2}-i\frac{\sqrt{3}}{2}$$
Both of these are roots of $P(x)$ with multiplicity $100$. Therefore,
$$\sum_{k=1}^{200}\frac{1}{1+x_k} = 100\left( \frac{1}{1+\zeta_3}+\frac{1}{1+\zeta_3^2}\right)$$
$$ = 100\left(\frac{1+\zeta_3+1+\zeta_3^2}{1+\zeta_3+\zeta_3^2+\zeta_3^3} \right) = 100$$
As for the polynomial division, suppose
$$P(x) = g(x)(x^2+x)+r(x)$$
where $r$ is linear. We have $P(0) = 1$, which implies that $r(0) = 1$. Further, $P(-1) = 1$, and so $r(-1)=1$. Consequently, $r(x)=1$ for all $x$, so $a_1=0$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Why is any prime factor of $(3)$ also a prime factor of $(2 + \sqrt[3]{19})$ in the ring of integers of $\mathbb{Q}(\sqrt[3]{19})$ I am reading this blurb from K. Conrad's expository notes on the conductor ideal: http://www.math.uconn.edu/~kconrad/blurbs/gradnumthy/conductor.pdf
In page 3 (example 2.3) he shows that $(2 + \root 3 \of{19})$ has norm $3^3 = 27$ and that $(-1 + \root 3 \of{19})$ has norm $2 \times 3^2 = 18$, and then states that any prime factor of $(3)$ is a factor of both $(2 + \root 3 \of{19})$ and $(-1 + \root 3 \of{19})$.
Of course, if a prime factor of $(3)$ divides any one of these ideals, it will divide the other too since their generators have a difference of 3. But I am not sure why a prime divisor of $(3)$ necessarily divides any of the two ideals. Clearly $3$ itself does not divide them because $\frac{2 + \root 3 \of{19}}{3}$ is not an algebraic integer.
I suspect it must have something to do with the norm of $(2 + \root 3 \of{19})$ being a power of $3$, but perhaps I am missing something obvious and can't figure out what's happening here.
| After a bit of experimentation, I noticed that
$$ (2 + \sqrt[3]{19})(-1 + \sqrt[3]{19})^2 = 3 \times (7 - \sqrt[3]{19}).$$
So $(3)$ divides $(2 + \sqrt[3]{19})(-1 + \sqrt[3]{19})(-1+\sqrt[3]{19})$.
Therefore, any prime ideal $\mathfrak p$ that divides $(3)$ must divide at least one of $(2 + \sqrt[3]{19})$ or $(-1 + \sqrt[3]{19})$.
But as you pointed out, the difference between $2 + \sqrt[3]{19}$ and $-1 + \sqrt[3]{19}$ is $3$, so every such $\mathfrak p$ will actually divide both $(2 + \sqrt[3]{19})$ and $(-1 + \sqrt[3]{19})$.
| {
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Rewrite complex number using de Moivre We have the following number $$z = (\dfrac{1}{2}-i)(1+i)^n + (\dfrac{1}{2}+i)(1-i)^n$$ and we want to rewrite it using de Moivre, in the neatest way I guess.
What I did is $$(1+i)^n = (\cos(\pi) + i \sin(\dfrac{1\pi}{2}))^n =(\cos(n\pi) + i \sin(\dfrac{n\pi}{2})) $$
Doing it for $(1-i)^n$ we obtain $$cos(n\pi) + i \sin (\dfrac{3n\pi}{2})$$
If we then multiply by the remaining factors, I get the form $$\cos(\pi n) + i \sin (\dfrac{\pi n}{2}) + \sin (\dfrac{\pi n}{2}) - \sin (\dfrac{3\pi n}{2})$$
However I'm wondering if this is the best way to use de Moivre here to clean the expression up. Wolfram alpha seems to give a different result.
| We first recall de Moivre's formula:
$(\cos \phi + i\sin \phi)^n = \cos (n \phi) + i \sin(n \phi). \tag{1}$
Also
$z = (\dfrac{1}{2}-i)(1+i)^n + (\dfrac{1}{2}+i)(1-i)^n. \tag{2}$
If we set
$w = (\dfrac{1}{2} - i)(1 + i)^n, \tag{3}$
then we see that
$z = w + \bar w. \tag{4}$
We can evaluate $w$ with the aid of the deMoivre formula; we have
$1 + i = \sqrt 2 (\dfrac{1}{\sqrt 2} + i \dfrac{1}{\sqrt 2}); \tag{5}$
noting that
$\cos \dfrac{\pi}{4} = \sin \dfrac{\pi}{4} = \dfrac{1}{\sqrt 2}, \tag{6}$
we write (5) as
$1 + i = \sqrt 2 (\cos \dfrac{\pi}{4} + i \sin \dfrac{\pi}{4}), \tag{7}$
whence
$(1 + i)^n = (\sqrt 2)^n(\cos \dfrac{\pi}{4} + i \sin \dfrac{\pi}{4})^n.\tag{8}$
We apply de Moivre to (8), yielding
$(1 + i)^n = (\sqrt 2)^n(\cos \dfrac{\pi}{4} + i \sin \dfrac{\pi}{4})^n = (\sqrt 2)^n(\cos \dfrac{n\pi}{4} + i \sin \dfrac{n\pi}{4}). \tag{9}$
As for the factor of $\frac{1}{2} - i$, we may write
$\dfrac{1}{2} - i = \dfrac{1}{2}(1 - 2i) = \dfrac{\sqrt 5}{2}(\dfrac{1}{\sqrt5} - \dfrac{2}{\sqrt 5}i); \tag{10}$
At this point we can follow Donald Splutterwit's answer and introduce the angle $\phi$ in the first quadrant such that $\tan \phi = 2$, so we have
$\dfrac{1}{2} - i = \dfrac{\sqrt 5}{2}(\cos \phi - i\sin \phi), \tag{11}$
but to my taste this doesn't gain us much; thus I prefer to write
$w = (\dfrac{1}{2} - i)(1 + i)^n = \dfrac{\sqrt 5}{2}(\dfrac{1}{\sqrt5} - \dfrac{2}{\sqrt 5}i)(\sqrt 2)^n(\cos \dfrac{n\pi}{4} + i \sin \dfrac{n\pi}{4})$
$ = \dfrac{(\sqrt 2)^n \sqrt 5}{2}(\dfrac{1}{\sqrt5} - \dfrac{2}{\sqrt 5}i)(\cos \dfrac{n\pi}{4} + i \sin \dfrac{n\pi}{4})$
$ = \dfrac{(\sqrt 2)^n \sqrt 5}{2}((\dfrac{1}{\sqrt 5} \cos \dfrac{n\pi}{4} + \dfrac{2}{\sqrt 5} \sin \dfrac{n\pi}{4}) + i(\dfrac{1}{\sqrt 5}\sin \dfrac{n\pi}{4} - \dfrac{2}{\sqrt 5}\cos \dfrac{n\pi}{4}))$
$= \dfrac{(\sqrt 2)^n }{2}(( \cos \dfrac{n\pi}{4} + 2 \sin \dfrac{n\pi}{4}) + i(\sin \dfrac{n\pi}{4} - 2\cos \dfrac{n\pi}{4})); \tag{12}$
thus
$\bar w = \dfrac{(\sqrt 2)^n }{2}(( \cos \dfrac{n\pi}{4} + 2 \sin \dfrac{n\pi}{4}) - i(\sin \dfrac{n\pi}{4} - 2\cos \dfrac{n\pi}{4})), \tag{13}$
whence
$z = w + \bar w =(\sqrt 2)^n ( \cos \dfrac{n\pi}{4} + 2 \sin \dfrac{n\pi}{4}), \tag{14}$
and I'll leave it at that.
NB: In preparing this derivation, I have tried to stick as much as possible to elementary algebraic operations and trigonometric identities, avoiding the use of fractional exponents and the exponential identity ($e^{i\theta} = \cos \theta + i \sin \theta$). In this context it is worth notting that de Moivre's formula (1) may be proved by a simple induction, for
if, for $k \in \Bbb Z$, $k \ge 1$, we assume
$(\cos \theta + i \sin \theta)^k = \cos (k\theta) + i\sin (k \theta), \tag{15}$
then
$(\cos \theta + i \sin \theta)^{k + 1} = (\cos \theta + i \sin \theta)(\cos (k\theta) + i\sin (k \theta))$
$ = \cos((k + 1)\theta) + i\sin((k + 1)\theta) \tag{16}$
follows from the elementary angle addition formulas
$\sin (a + b) = \cos a \sin b + \cos b \sin a, \tag{17}$
$\cos (a + b) = \cos a \cos b - \sin a \sin b. \tag{18}$
Finally, I must admit that the introduction of (10), with its factors of $\sqrt 5$, which subsequently cancels out, is a bit overdone and inelegant, but I'm not about to edit his answer any more now.
| {
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"timestamp": "2023-03-29T00:00:00",
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Limit with 2 variables: $\frac{x^2y}{x^4 +y^2}$, $\frac{e^{xy^3}-1}{x^2 +y^4}$ and $(x^2 +y^2)^{xy}$ at $(0,0)$ I am new to this topic so I appreciate any help on this.
a)$$\lim_{(x,y) \rightarrow (0,0)} f(x,y) = \frac{x^2y}{x^4 +y^2}$$
For $x=0 \lor y=0 : f(x,y) \longrightarrow 0$ but for $y = x^2$ the limit is $\frac{1}{2}$ so the limit does not exist.
b)
$$\lim_{(x,y) \rightarrow (0,0)} f(x,y) = \frac{e^{xy^3}-1}{x^2 +y^4}$$
I don't have any ideas here
c)
$$\lim_{(x,y) \rightarrow (0,0)} f(x,y) = (x^2 +y^2)^{xy}$$
Let $x \geq y$ then $|x^2 +y^2|^{xy} \leq |2y^2|^{y^2} \longrightarrow 1 \quad $for $y \longrightarrow 0$ but for $x = \frac{1}{y} \Longrightarrow f(x,y) = (\frac{1}{y^2} + y^2)^1 \longrightarrow \infty \quad$ for $y \longrightarrow 0 $ so the limit does not exist.
| For b):
we have that $f(0,y) = f(x, 0) = 0$ for every $x,y\neq 0$.
If $x$ and $y$ are both non zero, we have that
$$
\frac{e^{xy^3} - 1}{x^2+y^4} = \frac{e^{xy^3} - 1}{xy^3}\cdot \frac{xy^3}{x^2+y^4}.
$$
The first factor at the r.h.s. is bounded (for $x,y$ bounded), since $(e^t-1)/t \to 1$ as $t\to 0$.
The second factor can be estimated by
$$
\left|\frac{xy^3}{x^2+y^4}\right| =
|y| \cdot \frac{|x| y^2}{x^2+y^4}\leq \frac{|y|}{2}.
$$
Hence the limit exists and is equal to $0$.
| {
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When is $a^2 + b^2 + c^2 > ac\cos{x} + bc\sin{x}$? I was hoping someone could help me with finding the values of $a$, $b$, $c$ and $x$ for which the following inequation is true: $$a^2 + b^2 + c^2 > ac\cos{x} + bc\sin{x}$$
where $a$, $b$ and $c$ are real numbers and $0 \le x < 2\pi.$
I don't really know how to start and I think this problem is more complicated than it looks.
Thank you very much for reading.
| By C-S and AM-GM we obtain:
$$ac\cos{x}+bc\sin{x}\leq\sqrt{(a^2c^2+b^2c^2)(\cos^2x+\sin^2x)}=\sqrt{(a^2+b^2)c^2}\leq$$
$$\leq\frac{1}{2}(a^2+b^2+c^2)\leq a^2+b^2+c^2.$$
The equality occurs for $a=b=c=0$, which gives the answer:
$$\{(a,b,c,x)|a^2+b^2+c^2\neq0\}$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Confusion about proving $\frac{1}{1\cdot2} + \frac{1}{2\cdot3} + \frac{1}{3\cdot4} + \dotsb + \frac{1}{n(n+1)} = \frac{n}{n+1}$ by induction I know what the answer to this question is, but I am not sure how the answer was reached and I would really like to understand it! I am omitting the base case because it is not relevant for my question.
Inductive hypothesis:
$$\frac{1}{1\cdot2} + \frac{1}{2\cdot3} + \frac{1}{3\cdot4} + \dotsb + \frac{1}{n(n+1)} = \frac{n}{n+1}$$ is true when $n = k$ and $k > 1$
Therefore: $$\frac{1}{1\cdot2} + \frac{1}{2\cdot3} + \frac{1}{3\cdot4} + \dotsb + \frac{1}{k(k+1)} = \frac{k}{k+1}$$
Inductive step:
Prove that $$\frac{1}{1\cdot2} + \frac{1}{2\cdot3} + \frac{1}{3\cdot4} + \dotsb + \frac{1}{k(k+1)} = \frac{k+1}{k+1+1} = \frac{k+1}{k+2}$$
$$\frac{1}{1\cdot2} + \frac{1}{2\cdot3} + \frac{1}{3\cdot4} + \dotsb + \frac{1}{k(k+1)} = \left[\frac{1}{1\cdot2} + \frac{1}{2\cdot3} + \frac{1}{3\cdot4} + \dotsb + \frac{1}{k(k+1)}\right] + \frac{1}{(k+1)(k+2)}$$
$$\frac{1}{1\cdot2} + \frac{1}{2\cdot3} + \frac{1}{3\cdot4} + \dotsb + \frac{1}{k(k+1)} = \frac{k}{k+1} + \frac{1}{(k+1)(k+2)}$$
$$\frac{1}{1\cdot2} + \frac{1}{2\cdot3} + \frac{1}{3\cdot4} + \dotsb + \frac{1}{k(k+1)} = \frac{k+1}{k+2}$$
What I am confused about is where the $\frac{1}{(k+1)(k+2)}$ comes from in the first line of the inductive step. Can someone please explain this in a little more detail? The source of the answer explains it as "break last term from sum", but I am unclear on what that means.
| The base case $n=1$ is true. Assume that for some $n> 1$
$$\sum_{k=1}^n\frac{1}{k(k+1)}=\frac{n}{n+1}$$
Then, we have
$$\begin{align}
\sum_{k=1}^{n+1}\frac{1}{k(k+1)}&=\sum_{k=1}^n\frac{1}{k(k+1)}+\frac{1}{(n+1)(n+2)}\\\\
&=\frac{n}{n+1}+\frac{1}{(n+1)(n+2)}\\\\
&=\frac{n(n+2)+1}{(n+1)(n+2)}\\\\
&=\frac{(n+1)^2}{(n+1)(n+2)}\\\\
&=\frac{n+1}{n+2}
\end{align}$$
which completes the proof by induction.
| {
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How to take $\int_0^{+\infty} \frac{x^2+1}{x^4+1}dx$? The integral:
$$\int_0^{+\infty} \frac{x^2+1}{x^4+1}dx$$
If num were greater than denum I would just devide it normally with long division, but it is not, how should I handle it then?
| Whenever I see an integrand that is sort of symmetric under $x \to x^{-1}$,
I will try to see whether I can express the intergrand in terms of $x \pm x^{-1}$.
Using following identities,
$$\frac{dx}{x} = \frac{d(x-x^{-1})}{x+x^{-1}} = \frac{d(x+x^{-1})}{x-x^{-1}}$$
we find
$$\begin{align}
\int_0^\infty \frac{x^2+1}{x^4+1}dx
&= \int_0^\infty \frac{x+x^{-1}}{x^2+x^{-2}} \frac{dx}{x} = \int_0^\infty \frac{x+x^{-1}}{x^2+x^{-2}} \frac{d(x-x^{-1})}{x+x^{-1}} = \int_0^\infty \frac{d(x-x^{-1})}{(x-x^{-1})^2+2}\\
&= \int_{-\infty}^\infty \frac{dy}{y^2+2} = \left[\frac{1}{\sqrt{2}}\tan^{-1}\frac{y}{\sqrt{2}}\right]_{-\infty}^\infty
= \frac{\pi}{\sqrt{2}}
\end{align}
$$
| {
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Show that the vector field does not have periodic orbit Show that the vector field $F(x,y)=(2x-x^5-xy^4,y-y^3-x^2y)$ defined in $R^2$ does not have periodic orbits; the Bendixson criterion is not useful.
| @Robert Israel provides the best answer. This post is an elaboration of his remarks.
Dynamical System
Find the periodic solution for the dynamical system:
$$
%
\begin{align}
%
\dot{x} &= -x \left(x^4+y-2\right) \\
%
\dot{y} &= -y \left(x^2+y^2-1\right) \\
%
\end{align}
\tag{1}
%
$$
Poincaré–Bendixson Theorem
Use the theorem of Poincare and Bendixson to identify a trapping region, an area where the sign of the radial time derivative can change.
The trapping region must
*
*Be closed and bounded,
*Not contain any critical points.
Fixed points
Proceed by locating fixed points.
Nullclines
Find the nullclines.
$$
\begin{array}{ccccl}
\dot{x} = 0 &\implies &y &= &\left\{ 2-x^4 \right\} \\
\dot{y} = 0 &\implies &x &= &\left\{ 0, \pm\sqrt{1-x^2} \right\}
\end{array}
$$
Identify fixed points
The fixed points are:
$$
\left[ \begin{array}{c}
\dot{x} \\ \dot{y}
\end{array} \right]_{(0,0)}
=
\left[ \begin{array}{c}
\pm\sqrt[4]{2} \\ 0
\end{array} \right]
$$
At this juncture, invoke the arguments of @Robert Israel and you are done.
To reinforce the lesson, the answer continues with the canonical approach. The plot below shows the flow and the $\dot{x}$ nullcline in black and the $\dot{y}$ nullcline in red.
Compute $\dot{r}$
The polar coordinate transform
$$
%
\begin{align}
%
x &= r \cos \theta \\
%
y &= r \sin \theta \\
%
\tag{2}
\end{align}
%
$$
implies
$$
r^{2} = x^{2} + y^{2}
\tag{3}
$$
Differentiate (3) with respect to time:
$$
2r\dot{r} = 2x \dot{x} + 2y \dot{y}
$$
Therefore
$$
\dot{r} = \frac{x \dot{x} + y \dot{y}} {r}
\tag{4}
$$
Transform $\dot{x}$ and $\dot{y}$ to $r$ and $\theta$ using (2):
$$
%
\begin{align}
%
\dot{x}
&= -x \left(x^4+y-2\right)
= -r \cos \theta \left(r^4 \cos ^4\theta+r \sin \theta-2\right) \\
%
\dot{y}
&= -y \left(x^2+y^2-1\right)
= -r \left(r^2-1\right) \sin \theta \\
%
\end{align}
%
$$
Inserting these identities in $(4)$ produces the final differential equation:
$$
\dot{r} = -r \left(-\sin ^2\theta+r^4 \cos ^6\theta+r^2 \sin ^4\theta+\cos ^2\theta (r \sin \theta-1) (r \sin \theta+2)\right)
\tag{5}
$$
The fact that this differential equation is so difficult is a sign that you have either made a mistake, or missed the reasoning on stationary points.
To close the discussion, examine the plot below which shows $\dot{r}\left(r, \theta\right)$. The purple line is the nullcline $\dot{r}=0$. Below the line $\dot{r}>0$, above the line $\dot{r}<0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2292711",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
If $a>1$ and $b \mid a^2+1$ and $b>a$.Prove that $b-a > \sqrt{a}$. If $a>1$ and $b \mid a^2+1$ and $b>a$.Prove that $b-a > \sqrt{a}$.
I take $b=a+m$ where $m$ is a natural number then we have:
$b \mid a^2+1 \Rightarrow b \mid m^2+1 \Rightarrow b-1 \le m^2 \Rightarrow a \le m^2 \Rightarrow \sqrt{a} \le b-a$
But I don't know why I get the equality case with it.I can't even prove or disprove that there is an equality case.
| With $b=a+m$ we have $b| b(a-m)=(a+m)(a-m)=a^2-m^2=(a^2+1)-(m^2+1),$ so $b|m^2+1,$ so $b\leq m^2+1.$
Now if $m\leq \sqrt a$ then $a<b\leq m^2+1\leq (\sqrt a)^2+1=a+1 ,$ implying $b=a+1.$ But $a+1$ cannot divide $a^2+1.$ Otherwise $a+1| a(a+1)-(a^2+1)=a-1>0,$ which implies $a+1\leq a-1,$ a paradox. Therefore $m>\sqrt a.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2293476",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
If $ { a\over a+1} + { b\over b+1 } + { c\over c+1 } = 1 $ prove $ abc \le 1/8 $ If $ a,b,c $ are positive real numbers and $ { a\over a+1} + {b\over b+1} + {c\over c+1} = 1 $ then prove that $ abc \le {1\over 8} $
Could I get some help with this?
What I have tried-
Making the denominators equal and adding to get another equation of the form $ {N\over M}= 1 $ then multiplying by $ M $ both the sides and simplifying to get $ 2abc + ab + bc + ac = 1 $ , from which we have,
$ abc \lt 1/2 $
But which obviously isn't enough to prove that $ abc \le 1/8 $
I also tried using Jensen's inequality which states that
$ f(E(X)) \ge E(f(X)) $ when $ f $ is a concave function and $ f(E(X)) \le E(f(X)) $ when $ f $ is a convex function, and where $ E(X) $ denotes the expectation of X ,
but that didn't really help.
| If you multiply both sides by $(a+1)(b+1)(c+1)$ you'll obtain $$3abc+2ab+2ac+2bc+a+b+c=(a+1)(b+1)(c+1),$$
which simplifies to $$3abc+2ab+2ac+2bc+a+b+c=1 + a + b + a b + c + a c + b c + a b c,$$
rearranging to $$abc=\frac{1-(ab+ac+bc)}{2}.$$
Using AM-GM you get
$$1-2abc=ab+ac+bc\geq3\sqrt[3]{a^2b^2c^2}.$$ Now set $u=abc$ and you have $$u=\frac{1-\sqrt[3]{u^2}}{2}$$ which leads to $$1\geq2u+3u^{2/3}.$$
Now you see that $u=abc>1/8$ violates the inequality.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2294748",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 4
} |
Solving a 7th degree polynomial using De Moivre's theorem Use De Moivre Theorem to show that
$$\cos 7θ=64\cos^7θ-112\cos^5θ+56\cos^3θ-7\cosθ$$ *Done
Hence obtain the roots of the equation
$$128x^7-224x^5+112x^3-14x+1=0$$
in the form $\cos q\pi$
Attempt
$$\cos7θ=-1/2$$
$$θ=2π/21, 4π/21,8π/21,10π/21,14π/21,16π/21,20π/21$$
$$x=\cos θ$$
However the answer provided is
$\cos (\frac{2π}{21}+\frac{2kπ}{7})$ where $k=0,1,2,3,4,5,6$
Can somebody tell me what's wrong in my approach?
| You are correct.
When $k=0$, $\cos(\frac{2\pi}{21}+\frac{2k\pi}{7})=\cos\frac{2\pi}{21}$.
When $k=1$, $\cos(\frac{2\pi}{21}+\frac{2k\pi}{7})=\cos\frac{8\pi}{21}$.
When $k=2$, $\cos(\frac{2\pi}{21}+\frac{2k\pi}{7})=\cos\frac{14\pi}{21}$.
When $k=3$, $\cos(\frac{2\pi}{21}+\frac{2k\pi}{7})=\cos\frac{20\pi}{21}$.
When $k=4$, $\cos(\frac{2\pi}{21}+\frac{2k\pi}{7})=\cos\frac{26\pi}{21}=\cos\frac{16\pi}{21}$.
When $k=5$, $\cos(\frac{2\pi}{21}+\frac{2k\pi}{7})=\cos\frac{32\pi}{21}=\cos\frac{10\pi}{21}$.
When $k=6$, $\cos(\frac{2\pi}{21}+\frac{2k\pi}{7})=\cos\frac{38\pi}{21}=\cos\frac{4\pi}{21}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2295604",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Without the aid of a calculator, compute $\sin\frac{5\pi}{12}+\sin\frac{\pi}{12}$
Without the aid of a calculator, compute $\sin\frac{5\pi}{12}+\sin\frac{\pi}{12}$
The only method in doing this that I know was to just plug it in into the calculator, but I really had no clue on how to do this without the aid of a calculator,
I was thinking more along the lines of using the Taylor polynomial series but that really didn't work out for me though.
| Use the product formula:
$$\sin(a)\cos(b) = \frac{1}{2}(\sin(a+b) + \sin(a-b))$$
let $a = \frac{3\pi}{12}$ and $b = \frac{2\pi}{12}$
Then
$$\sin(\frac{5\pi}{12}) + \sin(\frac{\pi}{12})=
\sin(\frac{3\pi}{12}+\frac{2\pi}{12}) + sin(\frac{3\pi}{12}-\frac{2\pi}{12}) $$
$$=2\sin(\frac{3\pi}{12})\cos(\frac{2\pi}{12})=2\sin(\frac{\pi}{4})\cos(\frac{\pi}{6})$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2295696",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
Calculate $\int_0^\pi{x\sin{x} \over 1+\cos^2x}dx$ I am trying to get started on
$$\int_0^\pi{x\sin{x} \over 1+\cos^2x}dx$$
The usual trick I am familiar with would be to substitute $y=\tan{x \over 2}$. This doesn't seem to work in this case.
| Use the trick that $\int_a^bf(x)=\int_a^bf(a+b-x)$ we get
$$\int_0^\pi\frac{x\sin x}{1+\cos^2x}=\int_0^\pi\frac{(\pi-x)\sin(\pi-x)}{1+\cos^2(\pi-x)}=\int_0^{\pi}\frac{(\pi-x)\sin x}{1+\cos^2 x}$$
From this $$2\int_0^\pi\frac{x\sin x}{1+\cos^2 x}=\int_0^{\pi}\frac{(\pi-x)\sin x}{1+\cos^2 x}+\int_0^\pi\frac{x\sin x}{1+\cos^2 x}=\pi\int_0^\pi\frac{\sin x}{1+\cos^2 x}$$
With the substitution $u=\cos x$ we get
$$\pi\int_{-1}^1\frac{1}{1+u^2}=\pi(\arctan(1)-\arctan(-1))=\frac{\pi^2}{2}$$
So $$2\int_0^\pi\frac{x\sin x}{1+\cos^2 x}=\frac{\pi^2}{2}\\\int_0^\pi\frac{x\sin x}{1+\cos^2 x}=\frac{\pi^2}{4}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2296874",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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Show that $(1+u) \log (1+u) - u \ge \frac{u^2}{2(1+u/3)} $ This is used to go from Bennett's inequality to Bernstein's inequality.
Yet I don't understand how to prove the inequality.
Assume that $u > 0$, define
$$
h(u) = (1+u) \log (1+u) - u
$$
show that
$$
h(u) \ge \frac{u^2}{2(1+u/3)}
$$
My research :
Decomposing the function $h$ as a power series show that it is equivalent to
$$
\sum_{n=1}^\infty (-1)^n \frac{u^n}{n+1} \frac{n-1}{n (n+2)} \ge 0
$$
Sadly, I see no reason for this series to be positive.
SOLUTION
The twice differenciation technique given below works. However, I don't agree with the calculus, only on small things that don't change the result. Define
$$
g(u) = (1+u)\log (1+u) - u - \frac{3u^2}{2(3+u)}
$$
$$
g'(u) = \log (1+u) + 1 - 1 - \frac{3}{2} \frac{2u (3+u) - u^2}{(3+u)^2}\\
$$
Simplifying,
$$
g'(u) = \log (1+u) - \frac{3}{2} \frac{u(u+6 )}{(3+u)^2}
$$
$$
g''(u) = \frac{1}{1+u} - \frac{3}{2} \frac{(2u +6)(3+u)^2 -
2 (u+3)(u^2 + 6u) }{(3+u)^4}
$$
Simplifying,
$$
g''(u) = \frac{1}{1+u} - \frac{3}{2} \frac{2u^2 +6u + 6u + 18 -
2u^2 - 12u }{(3+u)^3} =
\frac{1}{1+u} - \frac{ 27 }{(3+u)^3}
$$
Simplifying again,
$$
g''(u)= \frac{ u^3 + 3\times 3u^2 + 3 \times 9u + 27 - 27(1+u)}{(1+u)(3+u)^3}
= \frac{ u^2(u+9)}{(1+u)(3+u)^3} > 0
$$
And the result is given by the reasoning of Clement.
| Your inequality is related to the Padé approximation $$\frac{(x-1)(x+5)}{4x+2}$$ of $\log(x)$. In fact one can show (e.g. see below) that $$\log(x) \leq \frac{(x-1)(x+5)}{4x+2}$$ for all $x>0$. (This is a useful inequality to remember.) Substituting $x\leftarrow 1/x$ reverses the inequality: $$\log(x)= -\log\left(\frac1x\right) \geq -\frac{(\frac1x-1)(\frac1x+5)}{\frac4x+2} = \frac{(x-1)(5x + 1)}{x(2x+4)}.$$ As a direct consequence $$ x \log(x) - (x-1) \geq \frac{3(x-1)^2}{6+2(x-1)}.$$
Here is one way to prove the inequality for the Padé approximation. Fix $x \geq 1$ and let $f(t) = (t-1)^2(x-t)$. Then $f(1)=f(x)=0$ and $f(t) \geq 0$ for $t \in[1, x]$. Integration by parts shows $$\int_1^x \frac{f'(t)}t \mathrm{d}t = \int_1^x \frac{f(t)}{t^2}\mathrm{d}t \geq 0.$$ Now the integral on the left can be easily computed: $$\int_1^x \frac{f'(t)}t \mathrm{d}t = (x-1)(x+5)/2-(2x+1)\log(x).$$ A similar argument works for $0<x\leq 1$ (now $f(t)\leq 0$ on the interval $[x,1]$).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2297875",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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How to compute 2-adic square roots? I know that, for a $2$-adic unit to be a perfect square, it must be of the form $\cdots001.$, for example the number $17$ ($10001.$) is a $2$-adic square. How would I go about finding the $2$ adic expansion of its square roots? There ought to be two, either of which is $-1$ times the other, but I don't know how to find either one.
I've tried setting up long multiplication and guessing digits that work, but there seem to be too many degrees of freedom. Any insights are appreciated.
| Because the derivative of $x^2-17$, i.e. $2x$ is $0 \bmod{2}$ Hensel's Lemma doesn't work very cleanly. In this situation when going from $p$ to $p^2$ either there is no lift, or every lift will work$\bmod p^2$. Let's look at what happens here -
$x^2\equiv 17 \bmod 2 \text{ has the solution }x\equiv 1 \bmod 2$
$(2y+1)^2 \equiv 17 \bmod 4 \text { is always true, telling us } x\equiv 1,3 \bmod 4 \text{ both work}$
When we lift to$\bmod 8$ we find $1$ and $5$ (lifts of $1 \bmod 4\,$) both work$\bmod 8$ as well as $3$ and $7$ (the lifts of $3 \bmod 4$). Note that we seem to have 4 solutions! Let's look at$\bmod 16$ and beyond.
$$
\begin{array}\\
1,5\pmod 8 & 1^2 \equiv (1+16) \equiv 17 \pmod{16} & 5^2\equiv 9 \not \equiv 17 \pmod{16} \\
3,7\pmod{ 8} & 3^2 \equiv 9 \not\equiv 17 \pmod{16} & 7^2\equiv 49 \equiv 17 \pmod{16} \\
\end{array}
$$
So of our 4 solutions only $1$ and $7\bmod 8$ will lift to$\bmod 16$. We lift those and try$\bmod 32$.
$$
\begin{array}\\
1,9\pmod{16} & 1^2 \not\equiv 17 \pmod{32} & 9^2\equiv 81 \equiv 17 \pmod{32} \\
7,15\pmod{16} & 7^2 \equiv 49 \equiv 17 \pmod{32} & 15^2\equiv 225 \not\equiv 17 \pmod{32} \\
\end{array}
$$
So of our 4 solutions only $9$ and $7\bmod 16$ will lift to$\bmod 32$. We lift those and try$\bmod 64$.
\begin{array}\\
9,25\pmod{32} & 9^2 \equiv 81 \equiv 17 \pmod{64} & 25^2\equiv 625 \not\equiv 17 \pmod{64} \\
7,23\pmod{32} & 7^2 \equiv 49 \not\equiv 17 \pmod{64} & 23^2\equiv 529 \equiv 17 \pmod{64}\end{array}
Fairly tedious stuff for humans, but nothing a computer algebra system won't whip out in no time. We have found 2 roots, $1 + 2^3 + O(2^5)$ and $1 + 2+ 2^2 + 2^4 + O(2^5)$.
When Doing the calculations by hand it would probably make more sense to find only one root and multiply by $-1=\frac{1}{1-2}=1+2+2^2+...$ for the other root.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2298779",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
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"answer_id": 0
} |
Function sequence $ \left( \cos\left( \frac{x}{\sqrt{n}} \right) \right)^n $ I'm studying about uniform convergence of function sequences. I haven't been able to prove that
$$\lim_{n \to \infty} \left( \cos\left( \frac{x}{\sqrt{n}} \right) \right)^n=e^{-\frac{x^2}{2}}.$$
Can you help me, please?
| Alternatively $$\lim _{ n\to \infty } \left( \cos \left( \frac { x }{ \sqrt { n } } \right) \right) ^{ n }=exp\left( \lim _{ n\to \infty } n\ln { \left( \cos \left( \frac { x }{ \sqrt { n } } \right) \right) } \right) =\\ =exp\left( \lim _{ n\to \infty } \frac { \ln { \left( \cos \left( \frac { x }{ \sqrt { n } } \right) \right) } }{ \frac { 1 }{ n } } \right) \overset { L'Hopital }{ = } exp\left( \lim _{ n\to \infty } \frac { -\frac { \sin { \left( \frac { x }{ \sqrt { n } } \right) } }{ \cos \left( \frac { x }{ \sqrt { n } } \right) } }{ -\frac { 1 }{ { n }^{ 2 } } } \left( -\frac { x }{ 2n\sqrt { n } } \right) \right) =\\ =exp\left( -\lim _{ n\to \infty } \tan { \left( \frac { x }{ \sqrt { n } } \right) } \frac { \sqrt { n } }{ 2x } { x }^{ 2 } \right) ={ e }^{ -\frac{x^2}{2} }$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2300330",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 4
} |
Convex Matrix function $f$ Let $g$ be the function the takes the sum of the squares of the elements in a $2\times 2$ matrix. And let $f$ be the function $$f(A) = g(A) + [\det(A)]^2.$$
Is it possible to find 2 matrices $A$ and $B$ such that $f(tA + (1-t)B) > tf(A) + (1-t)f(B)$? In other words, can we show that $f$ is not convex?
| Function is not convex! Take $A=\begin{bmatrix}
a & 0\\
0& 0
\end{bmatrix}$ ,and $B=\begin{bmatrix}
0 & 0\\
0& a
\end{bmatrix}$ we'll show that for large enough $a$, convexity of $f$ fails for these two matrix. Note that $f(A) =f(B) = a^2$
$$f(tA+(1-t)B) = a^2t^2 + a^2 (1-t)^2+a^4t^2(1-t)^2$$ and, therefore for $t=\frac{1}{2}$ we have
$$f(\frac{1}{2}A+\frac{1}{2}B)-\frac{1}{2}f(A)- \frac{1}{2}f(B) = \frac{a^4}{16} - \frac{a^2}{2}$$ so for example for $a=10$ above is positive number which shows $f$ is NOT convex!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2300958",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
How to solve an inequality involving the floor function? I am trying to solve the following disequation:
$$\left \lfloor \frac{(n+1)(n+2)}{6}\right\rfloor < \frac{n^2+3n}{8}.$$
Which is the right method to solve that?
My thougths so far: I wrote the term
$$\left \lfloor \frac{(n+1)(n+2)}{6}\right\rfloor = \frac{(n+1)(n+2)-r}{6}$$
where $r=0, \dots, 5$, but I was wondering if there is another method to solve that.
| \begin{align}
\left \lfloor \frac{(n+1)(n+2)}{6}\right\rfloor &< n^2+4n \\
\left \lfloor \frac{(n+1)(n+2)}{6} - ( n^2+4n)\right\rfloor &< 0 \\
\left \lfloor \frac{-5n^2-21n+2}{6} \right\rfloor &< 0 \\
\frac{-5n^2-21n+2}{6} &< 0 \\
\end{align}
The rest is standard stuff.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2301344",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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For what values of $\alpha > 0$ the equation $p(x) = x^3-9x^2+26x-\alpha =0$ has three positive real roots?
For what values of $\alpha > 0$ the equation $p(x) = x^3-9x^2+26x-\alpha =0$ has three positive real roots?
The options given are as follows :
$(A)$ $\alpha \ge 27$
$(B)$ $\alpha > 81$
$(C)$ $27 < \alpha <81$
$(D)$ $54 < \alpha \le 81$
What I have tried is that I first transform the equation to the form $y^3 + 3Hy + G = 0$ and then I have applied the fact that this equation has three positive real roots if $G^2 + 4H^3 < 0$.Here I found $G = 24 - \alpha$ and $H = -\frac {1} {3}$ and then calculating we get $\alpha \in (24-\frac {2 {\sqrt 3}} {9} , 24 + \frac {2 {\sqrt 3}} {9})$ which is not similar to any of the given options.Please help me in finding the right option.
Thank you in advance.
| Consider the roots are $a$,$b$, $c$.
Consider
$$(x-a)(x-b)(x-c)=x^3-(a+b+c)x^2+(ab+bc+ac)x-abc$$
So
$$ab+bc+ac=26,\alpha=abc$$
By AM-GM inequality
$$\sqrt[3]{ab\cdot bc\cdot ac}\le\frac{ab+bc+ac}{3}\implies\alpha^{2/3}\le\frac{26}3<9\implies\alpha<27$$
So no options are correct.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2302450",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove inequality $a^3b^3+2b^3c^3+3a^3c^3\le 0$ For $a^3+b^3+c^3=0$. Prove that $a^3b^3+2b^3c^3+3a^3c^3\le 0$
i think from $a^3+b^3+c^3=0$ we have one of the three numbers is zero, the other two are opposites.Suppose $a;b$ opposites and $c=0$
| $$a^3b^3+2b^3c^3+3a^3c^3=a^3b^3-(a^3+b^3)(3a^3+2b^3)=$$
$$=-(3a^6+4a^3b^3+2b^6)=-a^6-2(a^3+b^3)^2\leq0.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2304936",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
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angle of intersection between two lines Show that the angle between the tangent at any point $p$ and the line joining $p$ to the origin $O$ is same at all the points of the curve $\log(x^2+y^2)=c\tan^{-1}\left(\dfrac{y}{x}\right)$ where $c$ is any arbitrary constant .
| Denote $r = \sqrt{x^2 + y^2}$ and $\theta = \arctan(y/x)$ as the polar coordinates, then the given equation
$$\log(x^2+y^2)=c\tan^{-1}\left(\dfrac{y}{x}\right) \implies 2\log r = c \theta \implies r = \mathrm{Exp}\left(\frac{c}2 \theta \right)$$
is readily recognized as the logarithmic spiral.
The requested "constant angle" property is well known and easily proven, see for example Eq.(5) of this page.
If one wants to start from scratch and deal with the equation in the given form, then read on.
The title of the original post "angle of intersection between two lines", suggests that the asker was having difficulties determining the angle between the tangent direction $(1, y')$ and radial direction $(x,y)$.
The standard way to do this is to consider the inner product. Denote $\alpha$ as the angle between the two lines (or local angle between two local directional vectors), then the task is to show
$$\cos\alpha = \frac{ \text{inner product} }{ \text{product of norms} } =\frac{ (1, y') \bullet (x,y) }{ \lVert (1,y')\rVert\,\cdot\, \lVert(x,y) \rVert} \overset{?}{=} \text{constant}$$
Here $\displaystyle y' \equiv \frac{ \mathrm{d} y }{ \mathrm{d}x}$. Let's take the implicit derivative on both sides of the entire give equation.
\begin{gather*}
&\frac1{x^2 + y^2} \left( 2x + 2yy'\right) = c \frac1{ 1 + (y/x)^2} \frac{ y' x - y }{ x^2 } \\
&\implies 2\left( x + yy'\right) = c(y' x - y) \implies y' = \frac{ 2x + cy }{ cx - 2y} \tag{1} \label{1}
\end{gather*}
This gives the inner product as
$$(1, y') \bullet (x,y) = x+yy' = x + y\frac{ 2x + cy }{ cx - 2y} = \frac{ x(cx - 2y) + y (2x + cy) }{ cx - 2y} = \frac{ c(x^2 + y^2)}{ cx - 2y} \tag{2} \label{2}$$
Now, also make use of Eq.\eqref{1} to obtain the product of the norms:
\begin{gather*}
\lVert(1,y')\rVert\,\cdot\, \lVert(x,y)\rVert = \sqrt{ \left( 1+(y')^2 \right) \left( x^2 + y^2\right)} \\
\text{with} \quad 1+(y')^2 = \frac{ (cx - 2y)^2 + (2x + cy)^2 }{ (cx - 2y)^2 } \implies 1+(y')^2 = \frac{ (c^2 + 4)( x^2 + y^2) }{ (cx - 2y)^2 } \\
\implies \lVert(1,y')\rVert\,\cdot\, \lVert (x,y) \rVert = \sqrt{(c^2 + 4)}\frac{ x^2 + y^2 }{ |cx - 2y| } \tag{3} \label{3}
\end{gather*}
Finally, putting Eq.\eqref{2} and Eq.\eqref{3} together we have
$$\cos\alpha = \frac{ \pm c }{ \sqrt{(c^2 + 4)} } \tag{4} \label{4}$$
where the sign is determined by the sign of $cx - 2y$.
We know that a cosine flipping sign means the angle flips to its complement to $\pi$. That is, Eq.\eqref{4} correctly gives the cosine of a oriented angle of a constant magnitude between two oriented directions.$~~$Q.E.D.
| {
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"url": "https://math.stackexchange.com/questions/2307717",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "1",
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How to prove that $\sqrt{-3-2i}+\sqrt{-3+2i} = \sqrt{2(\sqrt{13}-3)}$? Is there a trick to show that
$$\sqrt{-3-2i}+\sqrt{-3+2i} = \sqrt{2(\sqrt{13}-3)}$$
is true ?
| Just power the left equation like the following:
$$(\sqrt{-3-2i} + \sqrt{-3 + 2i})^2 = -3 - 2i - 3 + 2i + 2\sqrt{(-3 + 2i)\times(-3 - 2i)} = -6 + 2 \sqrt{9 - (2i)^2} = -6 + 2\sqrt{13} = 2(\sqrt{13}-3)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2310053",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Prove that there exists an integer $k$ such that $x+y\alpha={\alpha}^k$
Let $x,y,z,w$ be integers such that $(x+y\alpha)(z+w\alpha)=1$ and $x+y\alpha>0$. Prove that there exists an integer $k$ such that $x+y\alpha={\alpha}^k$ where $\alpha=\frac{1+\sqrt{5}}{2}$.
Note that $\alpha^2 = \alpha+1$. We have $$(x+y\alpha)(z+w\alpha) = xz+wx\alpha+yz\alpha+yw\alpha^2-1 = 0.$$ This is the same as $xz+yw-1+\alpha(wx+yz+yw) = 0$. Therefore, $wx+yz+yw = 0$ and $xz+yw = 1$. Thus from the first equation we get $w = -\dfrac{yz}{x+y}$. Then substituting this into the equation $xz+yw = 1$ we get $$xz-\dfrac{y^2z}{x+y} = \dfrac{z(x(x+y)-y^2)}{x+y} = 1.$$ Therefore, $z(x(x+y)-y^2) = x+y$.
I didn't see how to continue.
| As you've already written, we have
$$wx+yz+yw = 0\quad\text{and}\quad xz+yw = 1$$
Representing $z,w$ by $x$ and $y$, we get
$$z=\frac{x+y}{x^2+xy-y^2},\quad w=\frac{-y}{x^2+xy-y^2}$$
implying that we have to have
$$\frac{x}{x^2+xy-y^2},\frac{y}{x^2+xy-y^2}\in\mathbb Z$$
which implies that
$$\small\left(\frac{x}{x^2+xy-y^2}\right)^2+\frac{x}{x^2+xy-y^2}\cdot \frac{y}{x^2+xy-y^2}-\left(\frac{y}{x^2+xy-y^2}\right)^2=\frac{1}{x^2+xy-y^2}\in\mathbb Z$$
from which we have to have
$$|x^2+xy-y^2|=1\tag1$$
Now, it is known that $(x,y)=(F_n,F_{n+1})$ with $n\ge 1$, i.e. two successive terms of the Fibonacci sequence, are the only positive solutions for $(1)$. (see, for example, here for a proof)
*
*Case 1 : If both $x,y$ are positive, then we get $(x,y)=(F_n,F_{n+1})$ where $F_n=\frac{\alpha^n-\left(-\frac{1}{\alpha}\right)^n}{\sqrt 5}$ with $n\ge 1$, so$$x+y\alpha=\frac{\alpha^n-\left(-\frac{1}{\alpha}\right)^n}{\sqrt 5}+\frac{\alpha^{n+1}-\left(-\frac{1}{\alpha}\right)^{n+1}}{\sqrt 5}\cdot \alpha=\alpha^n\cdot\frac{\alpha^2+1}{\sqrt 5}=\alpha^{n+1}$$
*Case 2 : If $xy=0$, then with $x+y\alpha\gt 0$, we have $(x,y)=(0,1),(1,0)$, and$$x+y\alpha=0+1\cdot\alpha=\alpha^1,\qquad x+y\alpha=1+0\cdot\alpha=\alpha^0$$respectively.
*Case 3 : If $x\gt 0$ and $y\lt 0$, then we get$$(1)\iff |(-y)^2+x(-y)-x^2|=1$$So, $(x,y)=(F_{n+1},-F_n)$ are the only solutions for $(1)$ in this case, and we get$$x+y\alpha=\frac{\alpha^{n+1}-\left(-\frac{1}{\alpha}\right)^{n+1}}{\sqrt 5}-\frac{\alpha^{n}-\left(-\frac{1}{\alpha}\right)^{n}}{\sqrt 5}\cdot \alpha=(-1)^n\alpha^{-n}$$From $x+y\alpha\gt 0$, $n$ has to be even, and if $n$ is even, then $x+y\alpha$ is of the form $\alpha^k$ where $k$ is an integer.
*Case 4 : If $x\lt 0$ and $y\gt 0$, then we get$$(1)\iff |y^2+(-x)y-(-x)^2|=1$$So, $(x,y)=(-F_{n+1},F_n)$ are the only solutions for $(1)$ in this case, and we get $$x+y\alpha=-\frac{\alpha^{n+1}-\left(-\frac{1}{\alpha}\right)^{n+1}}{\sqrt 5}+\frac{\alpha^{n}-\left(-\frac{1}{\alpha}\right)^{n}}{\sqrt 5}\cdot \alpha=(-1)^{n+1}\alpha^{-n}$$From $x+y\alpha\gt 0$, $n$ has to be odd, and if $n$ is odd, then $x+y\alpha$ is of the form $\alpha^k$ where $k$ is an integer.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Minimal area ellipse containing a given circumference
Out of all the ellipses
$$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$$
that contain the following circumference
$$x^2 + y^2 = 2y$$
determine the one that has the minimal area.
So far I've done this (in the picture), but now I'm kinda lost on using Lagrange multipliers!
| We want to minimize $f(a,b)=\pi ab$ subject to the constraint that the circle $x^2+y^2=2y$ lies inside the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$.
Letting $x^2=2y-y^2$ the two figures will intersect for any $a,\,b$ satisfying
$$ \frac{2y-y^2}{a^2} + \frac{y^2}{b^2}=0 $$
The discriminant of this quadratic is
$$ 4b^4+4a^2b^2(a^2-b^2) $$
Since we wish for no intersection or at most two points of tangency we require that
$$ 4b^4+4a^2b^2(a^2-b^2) \le0$$
which simplifies to the requirement
$$ g(a,b)=b^2+a^4-a^2b^2\le0 \tag{1}$$
One may then apply
$$\nabla f(a,b)=\lambda\nabla g(a,b)$$
together with $(1)$ to obtain the result that
$$ a=\frac{\sqrt{6}}{2}\quad b=\frac{3\sqrt{2}}{2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2311128",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Minimal distance between the origin and intersection of $x^2 = 2yz$ and $x^2+3y^2+2z^2 = 30$
Find the minimal distance between the origin and intersection of $x^2 = 2yz$ and $x^2+3y^2+2z^2 = 30$
Attempt:
$2yz+3y^2+2z^2 = 30$
I called $f(y,z) = 2yz+3y^2+2z^2$
Then, $\nabla f(y,z) = (2z+6y,2y+4z) = 0 \Rightarrow y=z=0$
So the critical point is $(0,0,0)$, which is clearly a minimum. But, I think this is wrong, because the minimum point is not necessarily the minimal distance to the origin. How do I get the closest point to the origin?
| I'll try to answer my own question based on the hints.
Define $f(x,y,z)=x^2+y^2+z^2$, $\varphi_1(x,y,z)=x^2-2yz$ and $\varphi_2(x,y,z)=x^2+3y^2+2z^2$
By doing $\nabla f = \lambda_1\varphi_1+\lambda_2\varphi_2$, we have
$(2x,2y,2z)=\lambda_1(2x,-2z,-2y)+\lambda_2(2x,6y,4z)$
$2x = \lambda_12x+\lambda_22x$
$2y =-\lambda_12z+\lambda_26y$
$2z=-\lambda_12y+\lambda_24z$
Together with $\varphi_1 = 0$ and $\varphi_2=30$, calculating this system of equation, we have:
$y=(3/2)z$ and $x =\pm\sqrt3z$, so $z=\pm\sqrt{60/19}$
And the points are $(\pm\sqrt3\sqrt{60/19},(3/2)\sqrt{60/19},\sqrt{60/19})$ and $(\pm\sqrt3\sqrt{60/19},-(3/2)\sqrt{60/19},-\sqrt{60/19})$
which are all minimum points as $\det Hf(x,y,z) = 8>0$, as
$Hf(x,y,z)=\begin{bmatrix}2&0&0\\0&2&0\\0&0&2\end{bmatrix}$
| {
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"source": "stackexchange",
"question_score": "1",
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Prove that $\frac{(\sin 20^\circ + \cos 20^\circ)^2}{\cos 40^\circ} = \cot 25^\circ$ So I'm trying to come up with an answer to this question for hours now. I don't know what I'm doing wrong and none of the calculators on the internet couldn't help so I figured I should ask people.
What have I done so far:
$\frac{(\sin 20^\circ + \cos 20^\circ)^2}{\cos 40^\circ} = \frac{(\frac{2\sin(45^\circ+20^\circ)}{\sqrt{2}})^2}{\cos 40^\circ} = \frac{\sin^2 65^\circ}{\cos 40^\circ} = \frac{1 - \cos 130^\circ}{\cos 40^\circ} = \frac{1 + \cos 50^\circ}{\cos 40^\circ} = \frac{2\cos^2 25^\circ}{\cos 40^\circ}$ ... etc.
I can't seem to figure out where to go from here so I'm just stuck.
I also tried the classic approach:
$\frac{(\sin 20^\circ + \cos 20^\circ)^2}{\cos 40^\circ} = \frac{1 + \sin 40^\circ}{\cos 40^\circ} = \frac{1}{\cos 40^\circ} + {\tan 40^\circ} = \sec 40^\circ + \tan 40^\circ$
But how can I prove that
$\sec 40^\circ + \tan 40^\circ = \cot 25^\circ$ ?
What am I doing wrong? Any hints or solutions would be great. Thanks in advance.
| Notice that $\cos 40 =\sin 50$ and $ \sin 40 = \cos 50$. So we have
\begin{eqnarray*}
\frac{1+\sin(40)}{ \cos 40} = \frac{1+\cos(50)}{\sin 50}
\end{eqnarray*}
Now use the double angle formulea $ \sin 2 \alpha = 2 \sin \alpha \cos \alpha$ and $ cos 2 \alpha =2 \cos^2 \alpha -1$.
So
\begin{eqnarray*}
\frac{1+\cos(50)}{\sin 50} = \frac{1+2 \cos^2 25 -1}{2 \sin 25 \cos 25 } = \frac{\cos 25}{\sin 25} = \color{red}{ \cot 25}.
\end{eqnarray*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2312805",
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"question_score": "5",
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Vectors Year 11
This is a vectors question I got, could any confirm my answer? I got $6.066\underline i - 279.28\underline j$.
$AB = \underline B - \underline A $
(Working on Vector $\underline B$)
\begin{align}x &= |B|\cos \theta \\
x &= 150\cos 315 \\
x &= 75\sqrt 2 \end{align}
\begin{align}y &= |B|\sin \theta \\
y &= 150\sin 315 \\
y &= -75\sqrt 2 \end{align}
(Working on Vector $\underline A$)
\begin{align}x = |A|\cos \theta \\
x = 200\cos 60 \\
x = 100 \end{align}
\begin{align}y &= |A|\sin \theta \\
y &= 200\sin 60 \\
y &= -173.21 \text{(2 decimals)} \end{align}
$\underline B - \underline A = 75\sqrt 2\underline i - 75\sqrt 2\underline j + -(100\underline i + 173.21\underline j)$
$\underline B - \underline A = 6.066\underline i - 279.28\underline j$
| $$ \vec{200} =(200\cos(60))i + (200\sin(60))j=100i+100\sqrt3j$$
$$ \vec{150} = (150\cos(45))i-(150\sin(45))j=75\sqrt{2}i-75\sqrt{2}j$$
$$ R= \vec{200} + \vec{150} =?$$
| {
"language": "en",
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"source": "stackexchange",
"question_score": "2",
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Point out my fallacy, in sequence and series.
The sum of the first $n$--terms of the series $1^2+2\cdot2^2+3^2+2\cdot4^2+\cdots$ is $\dfrac{n(n+1)^2}{2}$, when $n$ is even. When $n$ is odd, the sum is?
I got the correct answer when is replaced $n\rightarrow (n+1)$ to make above valid for odd, but when I tried the different approach then something following had happened.
For $n$ even, last term $=n$ which is even and term before it $=n-1$ which is odd. Clubbing all odds and evens separately as follows:
$\big(1^2+3^2+\cdots +(n-1)^2\big)+2\big(2^2+4^2+\cdots+n^2\big)=\dfrac{n(n+1)^2}{2}\tag{1}$
For $n$ odd, last term $=n$ which is odd and term before it $=n-1$ which is even. Clubbing all odds and evens separately as follows:
$\big(1^2+3^2+\cdots +n^2\big)+2\big(2^2+4^2+\cdots+(n-1)^2\big)\tag*{}$
$=\big(1^2+3^2+\cdots +(n-1)^2\big)+2\big(2^2+4^2+\cdots+n^2\big)-n^2+(n-1)\tag*{}$
From equation $(1)$
$=\dfrac{n(n+1)^2}{2}-n^2+(n-1)\tag*{}$
And answer given is: $\dfrac{n^2(n+1)}{2}$
please help.
| Let $n$ be even. Then:
$$S(n+1)-S(n)=(n+1)^2 \Rightarrow$$
$$S(n+1)=\frac{n(n+1)^2}{2}+(n+1)^2=\frac{(n+1)^2(n+2)}{2}.$$
| {
"language": "en",
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"question_score": "2",
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Compute $\sum\limits_n(-1)^{n-1}\frac{2^{n+1}}{2^{2n}-1}$ in terms of $\sum\limits_n\frac1{2^n-1}$ and $\sum\limits_n\frac1{2^n+1}$ Question:
let $$\sum_{n=1}^{+\infty}\dfrac{1}{2^n-1}=E,\sum_{n=1}^{+\infty}\dfrac{1}{2^n+1}=F$$ where $E,F$ are constant,(in fact,$E$ is Erdős-Borwein Constant ),Find the sum
$$f=\sum_{n=1}^{+\infty}(-1)^{n-1}\dfrac{2^{n+1}}{2^{2n}-1}$$
I tried this which I found in my texbook which seems relative, but Im not sure how to apply it to the problem:
$$f=\sum_{n=1}^{+\infty}(-1)^{n-1}\left(\dfrac{1}{2^n+1}+\dfrac{1}{2^n-1}\right)=\sum_{n=1}^{+\infty}\dfrac{(-1)^{n-1}}{2^n-1}+\sum_{n=1}^{+\infty}\dfrac{(-1)^{n-1}}{2^n+1}$$ following can't try
| Let $d(m),d_1(m),d_0(m)$ be the number of divisors, odd divisors, even divisors of $m$. We have:
$$ F=\sum_{n\geq 1}\left(\frac{1}{2^n}-\frac{1}{2^{2n}}+\frac{1}{2^{3n}}-\ldots\right)=\sum_{m\geq 1}\frac{d_1(m)-d_0(m)}{2^m}=2\sum_{m\geq 1}\frac{d_1(m)}{2^m}-E $$
$$ E = \sum_{m\geq 1}\frac{d(m)}{2^m} $$
On the other hand:
$$\begin{eqnarray*}f=\sum_{n\geq 1}(-1)^{n+1}\frac{2^{n+1}}{(2^n-1)(2^n+1)}&=&\sum_{n\geq 1}(-1)^{n+1}\left(\frac{1}{2^n-1}+\frac{1}{2^n+1}\right)\\&=&2\sum_{n\geq 1}(-1)^{n+1}\left(\frac{1}{2^n}+\frac{1}{2^{3n}}+\frac{1}{2^{5n}}+\ldots\right)\\&=&2\sum_{m\geq 1}\frac{1}{2^m}\sum_{\substack{d\mid m \\ d\text{ odd}}}(-1)^{\frac{m}{d}+1}\\&=&2\sum_{\substack{m\geq 1\\m\text{ odd}}}\frac{d(m)}{2^m}-2\sum_{\substack{m\geq 1\\m\text{ even}}}\frac{d_1(m)}{2^m}\\&=&4\sum_{\substack{m\geq 1\\m\text{ odd}}}\frac{d(m)}{2^m}-2\sum_{m\geq 1}\frac{d_1(m)}{2^{m}}\end{eqnarray*} $$
and
$$ E+F=2\sum_{m\geq 1}\frac{d_1(m)}{2^m}=2\sum_{\substack{m\geq 1\\m\text{ odd}}}d(m)\left(\frac{1}{2^m}+\frac{1}{2^{2m}}+\frac{1}{2^{4m}}+\frac{1}{2^{8m}}+\ldots\right) $$
so
$$ f=\sum_{\substack{m\geq 1 \\ m\text{ odd}}}d(m)\left(\frac{2}{2^m}-\frac{2}{2^{2m}}-\frac{2}{2^{4m}}-\frac{2}{2^{8m}}-\frac{2}{2^{16m}}-\ldots\right) $$
$$ E=\sum_{\substack{m\geq 1 \\ m\text{ odd}}}d(m)\left(\frac{1}{2^m}+\frac{2}{2^{2m}}+\frac{3}{2^{4m}}+\frac{4}{2^{8m}}+\frac{5}{2^{16m}}+\ldots\right) $$
$$ F=\sum_{\substack{m\geq 1 \\ m\text{ odd}}}d(m)\left(\frac{2}{2^m}+\frac{0}{2^{2m}}-\frac{1}{2^{4m}}-\frac{2}{2^{8m}}-\frac{3}{2^{16m}}+\ldots\right) $$
and I do not see any way to write $f$ in terms of $E$ and $F$ only, without involving $\sum_{\substack{m\geq 1\\m\text{ odd}}}\frac{d(m)}{2^m}$ too.
In any case the numerical evaluation of such (Lambert) series is pretty simple: an acceleration technique is outlined here.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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prove that $ x-\frac{1}{6}x^3<\sin(x)
prove that :
there exists a deleted neighborhood of $x=0$
such that :
$$
x-\frac{1}{6}x^3<\sin(x)<x-\frac{1}{6}x^3+\frac{1}{120}x^5
$$
MyTry:
let:$$f(x):=\sin x-x+\dfrac{1}{6}x^3$$
And :
$$g(x):=\sin x-x+\dfrac{1}{6}x^3-\dfrac{1}{125}x^5$$
Now what ?
| Taking my answer in
Is this really equal to sin x?
one more step:
If you start with
$\sin'(x) = \cos(x),
\cos'(x) = -\sin(x),
\sin(0) = 0,
\cos(0) = 1,
\sin^2(x)+\cos^2(x) = 1$,
you can proceed like this
(not original with me):
$$\sin(x)
=\int_0^x \cos(t)dt
\le\int_0^x dt
=x
$$
$$\cos(x)-\cos(0)
=\int_0^x -\sin(t) dt
=-\int_0^x \sin(t) dt
\ge-\int_0^x t dt
=-\frac{x^2}{2}\\
\text{ so }
\cos(x)
\ge 1-\frac{x^2}{2}
$$
$$\sin(x)
=\int_0^x \cos(t)dt
\ge\int_0^x (1-\frac{t^2}{2})dt
=x-\frac{x^3}{6}
$$
$$\cos(x)-\cos(0)
=\int_0^x -\sin(t) dt
=-\int_0^x \sin(t) dt
\ge-\int_0^x (t-\frac{t^3}{6}) dt
=-\frac{x^2}{2}+\frac{x^4}{24}\\
\text{ so }
\cos(x)
\le 1-\frac{x^2}{2}+\frac{x^4}{24}
$$
$$\sin(x)
=\int_0^x \cos(t)dt
\le\int_0^x (1-\frac{x^2}{2}+\frac{x^4}{24})dt
=x-\frac{x^3}{6}+\frac{x^5}{120}
$$
By induction you can
derive the power series
for sin and cos
and show that they are
enveloping
(the sum is between any two
consecutive sums).
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Creating a series that skips every $n^\text{th}$ term? I want to write a function using sigma notation that could represent an arbitrary number of terms of, for example, $1+2+4+5+7+8+10+11+13\ldots$, skipping every third term. I think one would need functions like floor and mod, but I'm not certain.
| The $r$-th roots of unity $\exp\left(\frac{2\pi ij}{r}\right), 0\leq j<r$ have the nice property to filter elements. For $r>0$ we obtain
\begin{align*}
\frac{1}{r}\sum_{j=0}^{r-1}\exp\left(\frac{2\pi ij n}{r}\right)=
\begin{cases}
1&\qquad r\mid n\\
0& \qquad otherwise
\end{cases}
\end{align*}
In case $r=3$ we obtain the sequence
\begin{align*}
\left(\frac{1}{3}\sum_{j=0}^2\exp\left(\frac{2\pi ijn}{3}\right)\right)_{n=0}^{\infty}
=(1,0,0,1,0,0,1,0,0,1,0,0,1,\ldots)
\end{align*}
which generates a $1$ at each $3$-rd position and $0$ otherwise.
If we need the positions of $1$ shifted by $k$ positions to the right, we subtract $k$ from the index $j$. With $k=2$ we get
\begin{align*}
\left(\frac{1}{3}\sum_{j=0}^2\exp\left(\frac{2\pi i(j-2)n}{3}\right)\right)_{n=0}^{\infty}
=(0,0,1,0,0,1,0,0,1,0,0,1,0,\ldots)
\end{align*}
and reverting the sequence by subtracting the elements from $1$ we finally get
\begin{align*}
\left(1-\frac{1}{3}\sum_{j=0}^2\exp\left(\frac{2\pi i(j-2)n}{3}\right)\right)_{n=0}^{\infty}
=(1,1,\color{blue}{0},1,1,\color{blue}{0},1,1,\color{blue}{0},1,1,\color{blue}{0},1,\ldots)
\end{align*}
Multiplying this sequence elementwise with $(a_n)_{n=0}^\infty$ and summing up the values gives the wanted series with each third element skipped.
Hint: Some instructive examples can be found in H.S. Wilf's book generatingfunctionology, (2.4.5) to (2.4.9).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2316742",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 5
} |
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