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Finding the remainder when $5^{55}+3^{55}$ is divided by $16$
Find the remainder when $5^{55}+3^{55}$ is divided by $16$.
What I try
$a^{n}+b^{n}$ is divided by $a+b$ when $n\in $ set of odd natural number.
So $5^{55}+3^{55}$ is divided by $5+3=8$
But did not know how to solve original problem
Help me please
| $3^4 = 81 = 16\cdot 5 + 1 \equiv 1 \mod(16)$
Then
$3^{52} = (3^{4})^{13} \equiv 1^{13} \mod(16) $
Then
$3^{55} = 3^3 \cdot 3^{52} \equiv 27 \cdot 1 \mod(16)$
And $27 \equiv 11 \mod(16)$
Thus
$3^{55} \equiv 11 \mod(16)$
Now,
$5^2 \equiv 9 \mod(16)$
So $(5^2)^{26} \equiv (3^{2})^{26} \mod(16)$
Then
$5^{52} \equiv 3^{52} \mod(16)$
And as before, $3^{52} \equiv 1 \mod(16)$
Thus
$5^{52} \equiv 1 \mod(16)$
So
$5^{55} \equiv 5^{3} \mod(16)$
And $5^{3} = 125 = 16 \cdot 7 + 13$
Then $5^{55} \equiv 13 \mod(16)$
And thus $5^{55} + 3^{55}\equiv 13 + 11 \mod(16)$
$24 \equiv 8 \mod(16)$
Finally $5^{55} + 3^{55} \equiv 8 \mod(16)$
So $16 \not| (5^{55} + 3^{55})$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3490156",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Relation between two rational sequences approximating square root 2 We define recursive sequences $a_{n+1}=1+\frac 1{1+a_n}$, $a_1=1$ and $b_{n+1}=\frac{b_n^2+2}{2b_n}$, $b_1=1$. I wish to show that $b_{n+1}=a_{2^n}$.
This can be proven using closed forms expressions related to continued fractions. I know that $a_n$ can be expressed as $$a_n=\sqrt2\cdot \frac{(1+\sqrt 2)^n +(1-\sqrt 2)^n}{(1+\sqrt 2)^n - (1-\sqrt 2)^n}$$
On the other hand, we can prove inductively that $$\frac{b_{n+1}-\sqrt 2}{b_{n+1}+\sqrt 2}=\left(\frac{1-\sqrt 2}{1+\sqrt 2}\right)^{2^n}$$
So the relation $a_{2^n}=b_{n+1}$ can be deduced by expanding the fractions. However the computation is rather tedious, I am looking for a proof that does not involve expanding everything into closed form expressions. Thanks.
| Let $\left(\frac{1-\sqrt 2}{1+\sqrt 2}\right)^{2^n} = t$ for simplicity. Then,
$$\frac{b_{n+1}-\sqrt 2}{b_{n+1}+\sqrt 2} = t$$ implies
\begin{align}
b_{n+1} &= \sqrt 2 \frac{1+t}{1-t} \\
&=\sqrt2 \frac{(1+\sqrt 2)^{2^n} +(1-\sqrt 2)^{2^n}}{(1+\sqrt 2)^{2^n} - (1-\sqrt 2)^{2^n}}\\
&= a_{2^n}
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3491183",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
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If $f(x)=f(1-x)$, then $f'(x)=-f'(1-x)$ If $f(x)=f(1-x)$, then
$$\frac{df(x)}{dx}=\frac{df(1-x)}{dx}=\frac{df(1-x)}{d(1-x)}\frac{d(1-x)}{dx}$$
and since $\frac{d(1-x)}{dx}=-1\implies d(1-x)=-dx$,
$$\frac{df(x)}{dx}=\frac{df(1-x)}{-dx}\cdot (-1)=\frac{df(1-x)}{dx}\ne -\frac{df(1-x)}{dx},$$
where's the mistake?
| $f'(1-x)$ is $\left.\dfrac{df(t)}{dt}\right|_{t = 1-x} = \dfrac{df(1-x)}{d(1-x)}$, not $\dfrac{df(1-x)}{dx}$.
e.g. for $f(x) = x(x-1) = x^2 -x$,
$$\begin{align*}
f(1-x) &= (1-x)^2 - (1-x)\\
&= x^2 - 2x + 1 - 1 + x\\
&= f(x)\\
f'(x) &= 2x - 1\\
f'(1-x) &= 2(1-x) - 1\\
&= 1 - 2x\\
&= -f'(x)\\
\frac{df(1-x)}{dx} &= \frac{d(x^2 - x)}{dx}\\
&= 2x - 1\\
&= f(x)
\end{align*}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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If $\alpha, \beta, \gamma, \delta$ are distinct roots of equation $x^4 + x^2 + 1 = 0$ then $\alpha^6 + \beta^6 + \gamma^6 + \delta^6$ is I tried to find discriminant first and got two roots $$\frac{-1 + \sqrt{3}i}{2}$$ and $$\frac{-1 - \sqrt{3}i}{2}$$
I tried taking $x^2 = t$ and solving equation for root $w$ and $w^2$ but got stucked, and made it more complex, any help?
Sorry if I made any silly mistake, it's been while since I practiced complex equation and finding roots. Was helping my brother with his doubts :)
| An alternative approach. Your polynomial is
$$ x^4+x^2+1 = (x^4+2x^2+1)-x^2 = (x^2+1)^2-x^2 = (x^2-x+1)(x^2+x+1) = \Phi_3(x)\Phi_6(x) $$
so for any root $\xi$ of the LHS, $\xi^6=1$. This implies $p_6=4$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3494035",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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$I_{n}=\int_{-n}^{n}\{x+1\}\{x^2+2\}+\{x^2+2\}\{x^3+4\}dx$. Find $I_1$ where $\{\}$ denotes fractional part. $$I_{n}=\int_{-n}^{n}\{x+1\}\{x^2+2\}+\{x^2+2\}\{x^3+4\}dx$$. Find $I_1$ where $\{\}$ denotes fractional part.
$$I_1=\int_{-1}^{1}\{x+1\}\{x^2+2\}+\{x^2+2\}\{x^3+4\}dx$$
As $\{x+1\}=\{x\}$
$$I_1=\int_{-1}^{1}\{x\}\{x^2+2\}dx+\int_{-1}^{1}\{x^2+2\}\{x^3+4\}dx$$
$$I_1=I_2+I_3$$
$$I_2=\int_{-1}^{1}\{x\}\{x^2+2\}dx$$
$$I_2=\int_{-1}^{1}\{-x\}\{x^2+2\}dx$$
$$2I_2=\int_{-1}^{1}\{x^2+2\}\left(\{x\}+\{-x\}\right)dx$$
As $\{x\}+\{-x\}=1$ for $x\notin Z$. There are only finite number of integers in the interval $[-1,1]$, so we can ignore the area under them
$$2I_2=\int_{-1}^{1}\{x^2+2\}dx$$
$$I_2=\int_{0}^{1}\{x^2+2\}dx$$
$$I_2=\int_{0}^{1}x^2+2-\left[x^2+2\right]dx$$
$$I_2=\dfrac{1}{3}+2-2$$
$$I_2=\dfrac{1}{3}$$
$$I_3=\int_{-1}^{1}\{x^2+2\}\{x^3+4\}dx$$
$$I_3=\int_{-1}^{0}\{x^2+2\}\{x^3+4\}dx+\int_{0}^{1}\{x^2+2\}\{x^3+4\}dx$$
$$I_3=\int_{-1}^{0}\left(x^2+2-\left[x^2+2\right]\right)\left(x^3+4-\left[x^3+4\right]\right)dx+\int_{0}^{1}\left(x^2+2-\left[x^2+2\right]\right)\left(x^3+4-\left[x^3+4\right]\right)dx$$
$$I_3=\int_{-1}^{0}x^2\cdot (x^3+1)dx+\int_{0}^{1}x^5dx$$
$$I_3=\int_{-1}^{0}x^5dx+\int_{-1}^{0}x^2dx+\int_{0}^{1}x^5dx$$
$$I_3=\int_{-1}^{1}x^5dx+\int_{-1}^{0}x^2dx$$
$$I_3=0+\dfrac{1}{3}$$
$$I_1=I_2+I_3$$
$$I_1=\dfrac{2}{3}$$
But actual answer is $\dfrac{4}{3}$. What am I missing here?
| Your calculations are correct. By employing a slightly different technique,
$$\int_0^1 \{x+1\}\{x^2+2\}dx= \int_0^1 \{x\}\{x^2\}dx= \int_0^1x^3dx = \frac 14;$$
$$\int_{-1}^0\{x+1\}\{x^2+2\}dx=\int_{-1}^0\{x+1\}\{(x+1)^2+1-2x\}dx=$$
$$\int_0^1\{x\}\{x^2+3-2x\}dx=\int_0^1\{x\}\{x^2+1-2x\}dx=\int_0^1\{x\}\{(1-x)^2\}dx=$$
$$\int_0^1x(1-x)^2dx = \frac{1}{12}.$$
Similarly,
$$\int_{-1}^0\{x^2+2\}\{x^3+4\}dx =\frac 16$$
and
$$\int^{1}_0\{x^2+2\}\{x^3+4\}dx =\frac 16.$$
In total this gives, as you already found, $\frac 23$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3495953",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Trouble finding the solution of $\int_0^{\pi/2}\sin^mx\cdot\cos^nx\,dx$ from the recurrence relation
Prove
\begin{align}
& \int_0^{\pi/2}\sin^mx \cos^nx\,dx \\[12pt]
= {} & \begin{cases}\dfrac{[(m-1)(m-3)\cdots][(n-1)(n-3)\cdots]}{(m+n)(m+n-2)\cdots}\cdot\dfrac{\pi}{2} & m,n\text{ even integers}\\[10pt]
\dfrac{[(m-1)(m-3)\cdots][(n-1)(n-3)\cdots]}{(m+n)(m+n-2)\cdots} & \text{elsewhere}
\end{cases}
\end{align}
My Attempt
$$
I_{m,n}=\int_0^{\pi/2}\sin^mx\cdot\cos^n x\,dx=\int_0^{\pi/2}\sin^{m-1}x \cos^nx \sin x \, dx\\
\int\cos^nx\cdot \sin x\,dx=\dfrac{-\cos^{n+1}x}{n+1}+K\\
I_{m,n}=\bigg[\sin^{m-1}x\cdot \dfrac{-\cos^{n+1}x}{n+1}\bigg]_0^{\pi/2}-(m-1)\int_0^{\pi/2}\sin^{m-2}x\cdot \cos x\cdot \dfrac{-\cos^{n+1}x}{n+1}\,dx\\
=0+\frac{m-1}{n+1}\int_0^{\pi/2}\cos^nx\cdot\sin^{m-2}x(1-\sin^2x)\cdot dx\\
=\frac{m-1}{n+1}\int_0^{\pi/2}\cos^nx\cdot\sin^{m-2}x\,dx-\frac{m-1}{n+1}\int_0^{\pi/2}\cos^nx\cdot\sin^{m}x\,dx\\
I_{m,n}=\frac{m-1}{n+1}\cdot I_{m-2,n}-\frac{m-1}{n+1}\cdot I_{m,n}\implies
\color{blue}{\boxed{I_{m,n}=\frac{m-1}{m+n}\cdot I_{m-2,n}}}
$$
case 1: $m,n$ even:
$$
I_{m,n}=\frac{(m-1)(m-3)\cdots 1}{(m+n)(m+n-2)\cdots 2}\cdot I_{0,n}\\
I_{0,n}=\int_0^{\pi/2}\cos^nx.dx=\frac{(n-1)(n-3)...1}{n(n-2)...2}.\frac{\pi}{2}\quad;\text{ since $n$ is even}\\
I_{m,n}=\frac{(m-1)(m-3)\cdots 1}{(m+n)(m+n-2)\cdots 2} \cdot \frac{(n-1)(n-3)...2}{n(n-2)...2}.\frac{\pi}{2}\\
=\frac{[(m-1)(m-3)\cdots 1][(n-1)(n-3)...2]}{[(m+n)(m+n-2)\cdots 2]\color{red}{[n(n-2)...2]}}.\frac{\pi}{2}
$$
Why am I not getting the solution given in my reference, more precisely, why do I get the additional term, marked in red, in the final solution ?
How do I reach the final solution from the recurrence relation ?
Similar question is asked before in Reduction formula for integral $\sin^m x \cos^n x$ with limits $0$ to $\pi/2$, but it does not seem to get the same recurrence relation as in my attempt.
Note: I am having trouble proceeding from the recurrence relation, post mentioned below are attempts to derive the recurrence relation, which does not address the problem, that i have already derived it in my attempt.
| Your error is here:
$$I_{m,n}=\frac{(m-1)(m-3)\cdots 1}{(m+n)(m+n-2)\cdots\color {red}{2}}\cdot I_{0,n}.$$
The correct expression is:
$$I_{m,n}=\frac{(m-1)(m-3)\cdots 1}{(m+n)(m+n-2)\cdots\color {red}{(n+2)}}\cdot I_{0,n}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3497590",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Given $a, b, c > 0$ such that $a^2 + b^2 + c^2 + abc = 4$, prove that $\sum_{cyc}\frac{b}{\sqrt{(c^2 + 2)(a^2 + 2)}} \ge 1$.
Given positives $a, b, c$ such that $a^2 + b^2 + c^2 + abc = 4$, prove that $$\large \frac{a}{\sqrt{(b^2 + 2)(c^2 + 2)}} + \frac{b}{\sqrt{(c^2 + 2)(a^2 + 2)}} + \frac{c}{\sqrt{(a^2 + 2)(b^2 + 2)}} \ge 1$$
We have that $$\sin^2\gamma + \cos^2\gamma = \sin^2\alpha + \cos^2\alpha = 1$$
$$\iff \cos^2\gamma(1 - \sin^2\alpha) = \cos^2\alpha(1 - \cos^2\gamma) = (\cos\gamma\cos\alpha)^2$$
$$\iff \cos^2\gamma + \cos^2\alpha - 2(\cos\gamma\cos\alpha)^2 = (\cos\gamma\sin\alpha)^2 + (\cos\alpha\sin\gamma)^2$$
$$\iff \cos^2\gamma + \cos^2\alpha - 2\cos\gamma\cos\alpha(\cos\gamma\cos\alpha - \sin\gamma\sin\alpha) = (\cos\gamma\sin\alpha +\cos\alpha\sin\gamma)^2$$
$$\iff \cos^2\gamma + \cos^2\alpha - 2\cos\gamma\cos\alpha\cos(\gamma + \alpha) = \sin^2(\gamma + \alpha)$$
$$\iff \cos^2\alpha + \cos^2\beta + \cos^2\gamma + 2\cos\alpha\cos\beta\cos\gamma = 1 \ (\beta = 180^\circ - \gamma - \alpha)$$
If we let $\dfrac{\cos\alpha}{a} = \dfrac{\cos\beta}{b} = \dfrac{\cos\gamma}{c} = 2$ then $a^2 + b^2 + c^2 + abc = 4$
It needs to be sufficient to prove that $$\frac{\cos\alpha}{\sqrt{(\cos^2\beta + 2)(\cos^2\gamma + 2)}} + \frac{\cos\beta}{\sqrt{(\cos^2\gamma + 2)(\cos^2\alpha + 2)}} + \frac{\cos\gamma}{\sqrt{(\cos^2\alpha + 2)(\cos^2\beta + 2)}} \ge 1$$
Another replacement is that $$\frac{\sqrt{\dfrac{yz}{(z + x)(x + y)}}}{a} = \frac{\sqrt{\dfrac{zx}{(x + y)(y + z)}}}{b} = \frac{\sqrt{\dfrac{xy}{(y + z)(z + x)}}}{c} = 2$$
We have that $$\sum_{cyc}\frac{zx}{(x + y)(y + z)} + \prod_{cyc}\sqrt{\frac{zx}{(x + y)(y + z)}} = \frac{\displaystyle\sum_{cyc}zx(z + x)}{(y + z)(z + x)(x + y)} + \prod_{cyc}\frac{y}{z + x} = 1$$
It needs to be sufficient to prove that $$\sum_{cyc}\sqrt\frac{zx(z + x)^2}{[2(y + z)(z + x) + xy][2(z + x)(x + y) + yz]} \ge 1$$
Neither of the above does I know how to solve.
| I like the following way.
Let $a=\frac{2x}{\sqrt{(x+y)(x+z)}}$ and $b=\frac{2y}{\sqrt{(x+y)(y+z)}},$ where $x$, $y$ and $z$ are positives.
Thus, $c=\frac{2z}{\sqrt{(x+z)(y+z)}}$ and we need to prove that:
$$\sum_{cyc}\frac{\frac{2x}{\sqrt{(x+y)(x+z)}}}{\sqrt{\left(\frac{4y^2}{(x+y)(y+z)}+2\right)\left(\frac{4z^2}{(x+z)(y+z)}+2\right)}}\geq1$$ or
$$\sum_{cyc}\frac{x(y+z)}{\sqrt{(3y^2+xy+xz+yz)(3z^2+xy+xz+yz)}}\geq1$$ or
$$\sum_{cyc}x(y+z)\sqrt{3x^2+xy+xz+yz}\geq\sqrt{\prod_{cyc}(3x^2+xy+xz+yz)}$$ or
$$\sum_{cyc}x^2(y+z)^2(3x^2+xy+xz+yz)+$$
$$+2\sum_{cyc}xy(x+z)(y+z)\sqrt{(3x^2+xy+xz+yz)(3y^2+xy+xz+yz)}\geq$$
$$\geq\prod_{cyc}(3x^2+xy+xz+yz).$$
But by C-S $$\sqrt{(3x^2+xy+xz+yz)(3y^2+xy+xz+yz)}\geq3xy+xy+xz+yz=4xy+xz+yz.$$
Id est, it's enough to prove that:
$$\sum_{cyc}x^2(y+z)^2(3x^2+xy+xz+yz)+$$
$$+2\sum_{cyc}xy(x+z)(y+z)(4xy+xz+yz)\geq$$
$$\geq\prod_{cyc}(3x^2+xy+xz+yz),$$ which is identity!!!
Done!
| {
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"timestamp": "2023-03-29T00:00:00",
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No positive integer solution for $x^2 + y^3 = z^6$? I am thinking about the following problem:
Is there any solution to $x^2 + y^3 = z^6$, where $x, y, z$ are positive integers?
I searched all possible solutions for $1 \leq z \leq 1000$ and there were no solution. So I think there is no solution, but I cannot prove it.
Here, one possible approach is to transform the equation to $y^3 = (z^3 - x)(z^3 + x)$. I tried in this way but cannot reach the proof.
The main reason is because we cannot assume that $x, y, z$ are coprime without loss of generality.
Of course I think there are other approach of proving. Can you tell me how to prove it?
| This is the proof referenced in the comments above.
The only solutions of this equation in nonzero integers are $(x,y,z)=(\pm 3k^3,-2k^2,\pm k)$, where $k$ is nonzero.
Let $(x,y,z)$ be a solution with minimal $|z|$. from equation we derive that $(z^3-x)(z^3+x)=y^3$. Obviously $\gcd (x,y)=\gcd(y,z)=\gcd(z,x)=1$. Thus there are coprime integers $a,b$ such that $z^3-x=a^3$ and $z^3+x=b^3$ and therefore $a^3+b^3=2z^3$, or there are coprime integers $a,b$ such that $z^3-x=2a^3$ and $z^3+x=4b^3$ and therefore $z^3+(-a)^3=2b^3$.
In both cases we are left with an equation of type $a^3+b^3=2z^3$ with $\gcd(a,b)=1$. let $|z|$ be minimal (assume that $|z|\ne0$). $a$
and $b$ must be odd. now substitute $a=u+v$ and
$b=u-v$, where $u$ and $v$ are coprime integers and from different parity. then $u (u^2+3v^2)= z^3$ and so there are integers $tr,s$ such that $u = r^3$ and $u^2+3v^2= s^3$.
From this, we have $(u+\sqrt{-3}v)(u-\sqrt{-3}v)=s^3$. treating this equation in $\Bbb Z[\sqrt{-3}]$, you can reach a equation of the form $a_0^3+b_0^3=2z_0^3$, where $|z_0|<|z|$, unless $a=b$ (or $|z|=0$) which yields the solutions $(x,y,z)=(\pm 3k^3,-2k^2,\pm k)$.
| {
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"timestamp": "2023-03-29T00:00:00",
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A ring with $8$ elements
Let $(A,+,\cdot)$ be a unitary ring with 8 elements. Prove that :
$a)8=0$ and $k\neq 0$, for any odd $k$.
b)If $\exists a\in A$ such that $a^3+a+1=0$, then $a\neq 0$, $a\neq 1$, $2=0$, $a^7=1$ and $A$ is a field.
a) is pretty straightforward, it is obvious that $8=0$ from Lagrange's theorem in the additive group $(A,+)$, and then if we had some odd $k$ such that $k=0$ then we would have that $1=0$, which is a contradiction to $|A|=8$.
For b) the most difficult part is showing that $\operatorname{char}A=2$.
The fact that $a\neq 0$ and $a\neq 1$ is really trivial.
Proving that $a^7=1$ is easy as well if we know that $2=0$. From here we easily get that $A$ is a field and we are done.
The only progress I made towards showing that $2=0$ was that $a$ is invertible(I guess it might help) because we have that $a(-a^2-1)=(-a^2-1)a=1$.
Obviously, we also have that $\operatorname{char}A\in \{2,4,8\}$, but I didn't get any further.
| Suppose that $a = n \cdot 1$ for some $n \in \mathbb{Z}$. Then $\left( n^{3} +n +1 \right) \cdot 1 = 0$, and hence $n^{3} +n +1$ is even since it is a multiple of $\text{car}(A) \in \lbrace 2, 4, 8 \rbrace$. As this is not possible, $a \notin \mathbb{Z} \cdot 1$. In particular, $\mathbb{Z} \cdot 1 \neq A$, and so $\text{car}(A) \in \lbrace 2, 4 \rbrace$.
Now, assume that $\text{car}(A) = 4$. Then $A = \lbrace 0, 1, 2 \cdot 1, 3 \cdot 1, a, a +1, a +2 \cdot 1, a +3 \cdot 1 \rbrace$ as these elements are pairwise distinct. It follows that $2 \cdot a \in \mathbb{Z} \cdot 1$ since $a \notin \mathbb{Z} \cdot 1$, and hence $2 \cdot a \in \lbrace 0, 2 \cdot 1 \rbrace$ because $4. a = 0$. Therefore, we have either $2 \cdot a = 0$ or $2 \cdot (a +1) = 0$. This is impossible since both $a$ and $a +1 = -a^{3}$ are invertible and $\text{car}(A) = 4$.
Thus, we have proved that $\text{car}(A) = 2$.
| {
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"timestamp": "2023-03-29T00:00:00",
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We roll a fair die until a $5$ appears. What is the expected value of the minimum value rolled?
Question: Given a fair dice, we roll until we get a $5.$ What is the expected value of the minimum value rolled?
Answer is $\frac{137}{60}.$
There is a similar question asked in MSE but I do not understand the method used by Henry.
In particular, if we let $X$ be the minimum value rolled up to and including $5$, then
$$E(X) = \sum_{x=1}^5 xP(X=x) = 1 \times \frac12 + 2 \times \frac16 + 3 \times \frac1{12}+4 \times \frac{1}{20}+5 \times \frac15 = \frac{137}{60}.$$
It seems that we are using the fact that
$$P(X=x) = \frac{1}{x(x+1)}.$$
I do not understand how to obtain the equation above.
| The probability that the minimum is $X\in\{1,2,3,4,5\}$ can be found as follows:
The probability of rolling a sequence of length $k$ using numbers from the set $\{X,X+1,\dots,6\}-\{5\}$ is $\left(\frac{6-X}{6}\right)^k$. Similarly, the probability of getting a sequence with values in $\{X+1,\dots,6\}-\{5\}$ is $\left(\frac{5-X}{6}\right)^k$. So the probability of a sequence of length $k$ with all elements in $\{X,\dots,6\}-\{5\}$ and at least one instance of $X$ is $\left(\frac{6-X}{6}\right)^k-\left(\frac{5-X}{6}\right)^k$. The probability of rolling a $5$ right after this sequence is $1/6$.
We sum over $k$ to find the probability of getting a sequence of any length with minimum $X$. This works out to $$P(X)=\frac{1}{6}\sum_{k=1}^\infty \left(\frac{6-X}{6}\right)^k-\left(\frac{5-X}{6}\right)^k =
\left\{
\begin{array}{ll}
\frac{1}{X} - \frac{1}{1+X} = \frac{1}{X(X+1)} & \text{for } X=1,2,3,4\\
\frac{1}{X} = \frac{1}{5} & \text{for } X=5\\
\end{array}\right.
$$ Which is what we wanted.
| {
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Proving the inequality $\prod_{n=1}^\infty \left( 1+\frac1{n^2+\ln n} \right) < \frac72$ I am struggling to prove
$$\prod_{n=1}^k \left( 1+\frac1{n^2+\ln n} \right) < \frac72$$
For all $k \geq 1$. Clearly, this is true for $k=1$, so it suffices to show
$$\prod_{n=1}^\infty \left( 1+\frac1{n^2+\ln n} \right) < \frac72$$
The obvious way to show this would be to evaluate the product directly, though I don't think that is feasible. I am not sure how else to approach this product. This would be a (much) tighter upper bound than
$$\exp\left ({\sum_{n=1}^k} \frac1{n^2+\ln n} \right )$$
Which comes from the monotone convergence theorem.
| Proof: We will use the following bound:
$$\frac{1}{x^2 + \ln x} < \frac{1}{x^2} - \frac{3}{2x^4}, \quad \forall x\ge 5.$$
proof: It suffices to prove that $f(x) = x^2 + \ln x - \frac{2x^4}{2x^2-3} > 0$ for $x\ge 5$.
We have $f'(x) = \frac{4x^4+6x^2+9}{x(2x^2-3)^2} > 0$ for $x\ge 5$. Note also that
$f(5) > 0$. The desired result follows.
With the bound above, noting also that $\ln (1+x) \le x$ for $x\ge 0$, we have
\begin{align}
&\prod_{n=1}^\infty \Big(1 + \frac{1}{n^2 + \ln n}\Big)\\
=\ & \prod_{n=1}^4 \Big(1 + \frac{1}{n^2 + \ln n}\Big)
\cdot \mathrm{exp}\left(\sum_{n=5}^\infty \ln\Big(1 + \frac{1}{n^2 + \ln n}\Big) \right)\\
\le\ & \prod_{n=1}^4 \Big(1 + \frac{1}{n^2 + \ln n}\Big)
\cdot \mathrm{exp}\left(\sum_{n=5}^\infty \Big(\frac{1}{n^2} - \frac{3}{2n^4}\Big) \right)\\
=\ & \prod_{n=1}^4 \Big(1 + \frac{1}{n^2 + \ln n}\Big)
\cdot \mathrm{exp}\left(\frac{2689}{13824}+\frac{\pi^2}{6}-\frac{\pi^4}{60} \right)\\
< \frac{7}{2}
\end{align}
where we have used $\sum_{n=1}^\infty \frac{1}{n^2} = \frac{\pi^2}{6}$
and $\sum_{n=1}^\infty \frac{1}{n^4} = \frac{\pi^4}{90}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3506406",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 2
} |
Integration by substitution, why is the $u$ this value? $$\int\frac{1}{(x+3)\sqrt{x}}~dx$$
I was wondering how you integrate this. I know you use substitution however I think of using $\sqrt x~$ for $u$, however on the integration calculator it says to use $u = \dfrac{\sqrt x}{\sqrt 3}~.$ I don't understand why and what to do once that is the substitution value.
Please help
| Let $x=u^2$ so that $dx=2u\cdot du$. Then your integral is transformed into $$=\int\frac{2u\cdot du}{(u^2+3)\sqrt{u^2}}=2\int\frac{du}{u^2+3}$$
where the latter can be handled as an arctangent integral.
To flesh it out completely, $$=2\int\frac{du}{u^2+3}=2\int\frac{du}{3\left(\frac{u^2}{3}+1\right)}=\frac{2}{3}\int\frac{du}{\left(\frac{u}{\sqrt{3}}\right)^2+1}$$ with $t=\frac{u}{\sqrt{3}}\implies dt=\frac{du}{\sqrt{3}}\implies\sqrt{3}dt=du\implies $ $$=\frac{2}{3}\int\frac{\sqrt{3}}{t^2+1}dt=\frac{2\sqrt{3}}{3}\int\frac{dt}{t^2+1}=\frac{2\sqrt{3}}{3}\arctan(t)+C$$
Since $x=u^2$ and $t=\frac{u}{\sqrt{3}}$ we have $t=\frac{\sqrt{x}}{\sqrt{3}}=\sqrt{\frac{x}{3}}\implies$ $$=\frac{2\sqrt{3}}{3}\arctan\left(\sqrt{\frac{x}{3}}\right)+C$$
| {
"language": "en",
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"source": "stackexchange",
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Find the limit of $a_n = n \sqrt{n}(\sqrt{n + 1} - a\sqrt{n} + \sqrt{n - 1})$ given that the sequence $(a_n)$ is convergent. I am given the sequence:
$$a_n = n\sqrt{n}(\sqrt{n + 1} - a\sqrt{n} + \sqrt{n - 1})$$
with $n \in \mathbb{N}^*$ and $a \in \mathbb{R}$. I have to find the limit of $(a_n)$ given that the sequence $(a_n)$ is convergent.
This is what I did:
We have:
$$a_n = n\sqrt{n}(\sqrt{n + 1} - a\sqrt{n} + \sqrt{n - 1})$$
$$a_n = n\sqrt{n^2+n}-an^2+n\sqrt{n^2-n}$$
$$a_n = n^2\sqrt{1+\dfrac{1}{n}} + n^2 \sqrt{1 - \dfrac{1}{n}} - an^2$$
$$a_n = n^2 \bigg ( \sqrt{1 + \dfrac{1}{n}} + \sqrt{1-\dfrac{1}{n}} - a \bigg )$$
The only way we could have $a_n$ convergent is if we would have the limit result in an indeterminate form. In this case, we need $\infty \cdot 0$. So we have:
$$\lim_{n \to \infty} \bigg ( \sqrt{1 + \dfrac{1}{n}} + \sqrt{1-\dfrac{1}{n}} - a \bigg ) = 0$$
And from that we can conclude that:
$$a = 2$$
So now that I found $a$, I must find the limit of the sequence. So this limit:
$$\lim_{n \to \infty} a_n = \lim_{n \to \infty} n\sqrt{n}(\sqrt{n + 1} - 2\sqrt{n} + \sqrt{n - 1})$$
I tried this:
$$\lim_{n \to \infty} a_n = \lim_{n \to \infty} n\sqrt{n^2+n}-2n^2+n\sqrt{n^2-n}$$
$$= \lim_{n \to \infty}( n\sqrt{n^2+n} - n^2) + \lim_{n \to \infty} (n\sqrt{n^2-n} -n^2)$$
And then I multiplied both of those limits with its respective conjugate, but after my calculations, it still results in an indeterminate form, only it's $\infty - \infty$ this time.
So, if my previous calculations aren't wrong, my question is how can I find this limit:
$$\lim_{n \to \infty} n\sqrt{n}(\sqrt{n + 1} - 2\sqrt{n} + \sqrt{n - 1})$$
| Your initial working leads well into using the generalized binomial expansion which gives the asymptotic behaviour as $n\to\infty$
\begin{align}
a_n
&=n^2\left(\sqrt{1+\frac1n}+\sqrt{1-\frac1n}-2\right)\\
&\sim n^2\left(1+\frac1{2n}-\frac1{8n^2}+\dots+1-\frac1{2n}-\frac1{8n^2}-\dots-2\right)\\
&=-\frac14\\
\end{align}
Further terms of this expansion gives the more precise asymptotic behaviour of
\begin{align}
a_n
&\sim-\frac14-\frac5{64n^2}-\frac{21}{512n^4}-\dots\\
&=-\sum_{k=0}^\infty\frac{\binom{4k+1}{2k}}{4^{2k+1}(k+1)n^{2k}}\qquad(n\ge1)\\
\end{align}
| {
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Why is $(1+\frac{1}{n})^n < (1+\frac{1}{m})^{m+1}$? Why is it that
$$
\left(1+\frac{1}{n}\right)^n < \left(1+\frac{1}{m}\right)^{m+1},
$$
for any natural numbers $m, n$?
I have tried expanding using the binomial series and splitting into cases. I understand why it is trivially true when $m=n$ but I am not sure if there is a rigorous proof for other cases?
| For a simple answer to why is $\left(1+\frac{1}{n}\right)^n < \left(1+\frac{1}{m}\right)^{m+1}$? -- it is because of the inequality of geometric and arithmetic means. In particular
$a_n = \big(1 + \frac{1}{n}\big)^n \lt \big(1 + \frac{1}{n+1}\big)^{n+1} = a_{n+1}$
or equivalently, taking n+1 th roots
$a_{n}^\frac{1}{n+1}= \big(1 + \frac{1}{n}\big)^\frac{n}{n+1} = \big(1 + \frac{1}{n}\big)^\frac{n}{n+1}\cdot 1^\frac{1}{n+1} \lt \frac{n}{n+1}\big(1 + \frac{1}{n}\big) + \frac{1}{n+1}\big(1\big) = 1 + \frac{1}{n+1} = a_{n+1}^\frac{1}{n+1}$
by $\text{GM}\leq \text{AM}$ (which holds with equality iff all items in the geometric mean are identical)
and by the same manipulation we get $c_r = \big(1 - \frac{1}{r}\big)^r \lt \big(1 + \frac{1}{r+1}\big)^{r+1} = c_{r+1}$
so $a_n$ and $c_r$ are strictly monotone increasing
For the question posed: by monotone behavior of $a_n$, if $n \leq m$ then we automatically have
$(1+\frac{1}{n})^n \leq (1+\frac{1}{m})^{m}\lt (1+\frac{1}{m})^{m}\cdot (1+\frac{1}{m}) = (1+\frac{1}{m})^{m+1}$
it remains to consider the case of $m \lt n$. In this case we can divide out $(1+\frac{1}{m})^{m+1}$ and prove the equivalent
$\left(1+\frac{1}{n}\right)^n \left(1-\frac{1}{m+1}\right)^{m+1} = \left(1+\frac{1}{n}\right)^n \left(\frac{m}{m+1}\right)^{m+1}=\left(1+\frac{1}{n}\right)^n \left(1+\frac{1}{m}\right)^{-(m+1)}\lt 1 $
but
$\left(1+\frac{1}{n}\right)^n \left(1-\frac{1}{m+1}\right)^{m+1} \leq \left(1+\frac{1}{n}\right)^n \left(1-\frac{1}{n}\right)^{n}\lt 1$
by the monotone behavior of $c_r$ and $\text{GM}\leq \text{AM}$ which proves the claim
note: if you don't like splitting this into 2 cases, then the claim is proven directly by considering monotone behavior in $a_n$ and $c_r$, and for any $m,n$ select natural number $k \gt \max(m+1,n)$ and the direct proof is
$\left(1+\frac{1}{n}\right)^n \left(1-\frac{1}{m+1}\right)^{m+1} \lt \left(1+\frac{1}{k}\right)^k \left(1-\frac{1}{k}\right)^{k}\lt 1$
and of course
$\left(1+\frac{1}{k}\right)^k \left(1-\frac{1}{k}\right)^{k}\lt 1$ is equivalent to proving the claim with 2kth roots of each side but
$\left(1+\frac{1}{k}\right)^\frac{1}{2} \left(1-\frac{1}{k}\right)^\frac{1}{2}\lt \frac{1}{2}\left(1+\frac{1}{k}\right)+ \frac{1}{2}\left(1-\frac{1}{k}\right) = 1$ by $\text{GM}\leq \text{AM}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3512177",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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Find locus of $\Delta ABC$ centroid with orthocentre at origin and side slopes 2, 3 and 5 Let $ABC$ be a triangle with slopes of the sides $AB$, $BC$, $CA$ are $2,3,5$ respectively. Given origin is the orthocentre of the triangle $ABC$. Then find the locus of the centroid of the triangle $ABC$.
Since sides have slopes $2,3,5$ then altitudes must have slopes $\frac{-1}{2}, \frac{-1}{3}$ and $\frac{-1}{5}$ respectively. Then equations of the altitudes are respectively \begin{align*} & 2y+x=0 \\ & 3y+x=0 \\ & 5y+x=0. \end{align*} If $(\alpha, \beta)$ be the coordinates of the centroid, how can I find the locus of the point $(\alpha, \beta)$ from here?
| Hint: Fix a triangle $ABC$ with the required property. Show that any other triangle $A'B'C'$ with the required property is obtained from $\triangle ABC$ by a dilation about the origin. This proves that the locus of the centroid is a line passing through the origin (and the centroid $G$ of the fixed triangle $ABC$).
For example, let $D$ be the base of the altitude from $A$ of $\triangle ABC$. If $D$ has coordinates $(x_D,y_D)$, then you know that $3y_D+x_D=0$. So you may assume wlog that $x_D=3\cdot 168=504$ and $y_D=-168$ (the number $168$ is chosen so that $A$, $B$, $C$, $D$, and $G$ have integer coordinates). Therefore the equation for $BC$ is $y-y_D=3(x-x_D)$ or $y=3x-10$. Thus, the points $B$ and $C$ are the intersections of $y=3x-10$ with $5y+x=0$ and with $2y+x=0$, respectively. Now you know where $B$ and $C$ are, it should be easy to find $A$, and then $G$.
If $D$ is chosen as above, then $A=(495,-165)$, $B=(525,-105)$, and $C=(480,-240)$. The centroid is then $$G=\frac{A+B+C}{3}=\left(500,-170\right).$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Determine the conditions for $n$ and $\theta\neq0+2k\pi$ such that $(\frac{1+\cos\theta+i\sin\theta}{1-\cos\theta+i\sin\theta})^n \in\mathbb{R}$ Let $z=1+\cos\theta+i\sin\theta=|z|(\cos\alpha+i\sin\alpha)$ and $z'=1-\cos\theta+i\sin\theta=|z'|(\cos\alpha'+i\sin\alpha')$.
My line of reasoning is to convert the numerator and denominator to polar form, then knowing that in order for $(\frac{z}{z'})^n$ to be real, we must have $\arg(\frac{z}{z'})^n \equiv 0 \pmod{\pi}$ or more succinctly $$n(\arg(z)-\arg(z')) \equiv 0 \pmod{\pi}$$.
Determining $\arg(z) = \alpha$ and $\arg(z') = \alpha'$: $$\tan{\alpha}=\frac{\sin{\theta}}{1+\cos{\theta}}=\frac{2\sin{\frac{\theta}{2}\cos{\frac{\theta}{2}}}}{2\cos^2{\frac{\theta}{2}}}=\frac{\sin{\frac{\theta}{2}}}{\cos{\frac{\theta}{2}}}=\tan{\frac{\theta}{2}}$$ thus $\alpha = \frac{\theta}{2}$. $$\tan{\alpha'}=\frac{\sin{\theta}}{1-\cos{\theta}}=\frac{2\sin{\frac{\theta}{2}\cos{\frac{\theta}{2}}}}{2\sin^2{\frac{\theta}{2}}}=\frac{\cos{\frac{\theta}{2}}}{\sin{\frac{\theta}{2}}}=\frac{\sin{\alpha'}}{\cos{\alpha'}}$$ thus $$\sin{\alpha'}\sin{\frac{\theta}{2}}=\cos{\alpha'}\cos{\frac{\theta}{2}} \Leftrightarrow \frac{1}{2}(\cos(\alpha'-\frac{\theta}{2})-\cos(\alpha'+\frac{\theta}{2}))=\frac{1}{2}(\cos(\alpha'-\frac{\theta}{2})+\cos(\alpha'+\frac{\theta}{2}))$$ therefore $$-\cos(\alpha'+\frac{\theta}{2})=\cos(\alpha'+\frac{\theta}{2})$$. Knowing that $$\cos(\pi+\alpha'+\frac{\theta}{2})=\cos(\alpha'+\frac{\theta}{2})$$ is impossible, we take $$\cos(\pi-\alpha'-\frac{\theta}{2})=\cos(\alpha'+\frac{\theta}{2})$$ hence $$\pi-\alpha'-\frac{\theta}{2} = \alpha'+\frac{\theta}{2} \Leftrightarrow \alpha' = \frac{\pi-\theta}{2}$$.
So now we have $$\frac{1+\cos\theta+i\sin\theta}{1-\cos\theta+i\sin\theta} = \frac{|z|(\cos{\frac{\theta}{2}}+i\sin{\frac{\theta}{2}})}{|z'|(\cos{\frac{\pi-\theta}{2}}+i\sin{\frac{\pi-\theta}{2}})}$$
Here's where I have some difficulties. Supposing
$$\frac{\theta}{2}-\frac{\pi-\theta}{2} \equiv 0 \pmod{\pi}$$ therefore $$\frac{2\theta - \pi}{2} \equiv 0 \pmod{\pi} \Leftrightarrow 2\theta - \pi \equiv 0 \pmod{2\pi}$$ thus $$\theta \equiv \frac{\pi}{2} \pmod{\pi}$$.
However, for all $n$ and using the above condition, $$(\frac{1+\cos\theta+i\sin\theta}{1-\cos\theta+i\sin\theta})^n = 1$$
It can't be for all $n$, there's something here I'm missing.
| $$\frac{1+\cos\theta+i\sin\theta}{1-\cos\theta+i\sin\theta}=\frac{2\cos^2\frac{\theta}{2}+2i\sin\frac{\theta}{2}\cos\frac{\theta}{2}}{2\sin^2\frac{\theta}{2}+2i\sin\frac{\theta}{2}\cos\frac{\theta}{2}}=-i\cot\frac{\theta}{2}(\cos\frac{\theta}{2}+i\sin\frac{\theta}{2})^2=$$
$$=\cot\frac{\theta}{2}(-\sin\theta+i\cos\theta)=\cot\frac{\theta}{2}\left(\cos\left(270^{\circ}+\theta\right)+i\sin\left(270^{\circ}+\theta\right)\right).$$
Can you end it now?
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How to evaluate the following surface integral?
Evaluate $$\iint_S \left (x^4+y^4+z^4 \right )\ dS,$$ where $S=\left \{(x,y,z)\ :\ x^2+y^2+z^2 = a^2 \right \},\ a > 0.$
My attempt $:$ I have tried to find the integral using Gauss' divergence theorem which states that
Suppose $V$ is a subset of $\Bbb R^n$ (in this case $n=3,$ which represents the volume of the sphere of radius $a$ centered at the origin) which is compact and has a piecewise smooth boundary $S.$ If $\textbf {F}$ is a continuously differentiable vector field defined on a neighbourhood of $V,$ then
$$\begin{align*} \iiint_V\left(\mathbf{\nabla}\cdot\mathbf{F}\right)\,dV & = \iint_{S}(\mathbf{F}\cdot\mathbf{n})\,dS . \end{align*} $$
In this case $\textbf {F} = x^3 \hat {i} + y^3 \hat {j} + z^3 \hat {k},$ which is a continuously differentiable vector field on a neighbourhood of $V$ and $\textbf {n} (x,y,z) = \frac {x \hat i + y \hat j + z \hat k} {a}.$ Hence we have \begin{align*} \iint_S \left (x^4 + y^4 + z^4 \right )\ dS & = a \iiint_V \left (\mathbf {\nabla} \cdot \mathbf {F} \right)\ dV\\ & = a \iiint_V 3 \left (x^2+y^2+z^2 \right )\ dV\\ & = 3a^3 \iiint_V dV \\ & = 3a^3 \times \frac 4 3 \pi a^3 \\ & = 4 \pi a^6. \end{align*}
But the answer given as $\frac {12 \pi} {5} a^6 \neq 4 \pi a^6.$ Where did I do mistake? Any help in this regard will be highly appreciated.
Thank you very much.
EDIT $:$ Here is another way to find the integral. Let $\varphi : [0, \pi] \times [0, 2 \pi] \longrightarrow \Bbb R^3$ be the continuously differentiable parameterization of the given sphere into spherical coordinates i.e. $$\varphi (u,v) = \left (a \sin u \cos v , a \sin u \sin v , a \cos u \right ),\ (u,v) \in [0, \pi] \times [0, 2 \pi].$$ Then the above surface integral can be evaluated as follows $:$
Let $F(x,y,z) = x^4 + y^4 + z^4 ,\ (x,y,z) \in \Bbb R^3.$ Then we have
$$\begin{align*} \iint_S F(x,y,z)\ dS & = \int_{0}^{2 \pi} \int_{0}^{\pi} F \left (\varphi (u,v) \right ) \left \| \varphi_u \times \varphi_v \right \|\ du\ dv \\ & = a^6 \int_{0}^{2 \pi} \int_{0}^{\pi} \left (\sin^5 u \cos^4 v + \sin^5 u \sin^4 v + \sin u \cos^4 u \right )\ du\ dv. \end{align*}$$
The computation of the last integral is not so easy though.
| Recognize the symmetry to simplify the integral first,
$$I=\int_S \left (x^4+y^4+z^4 \right )\ dS,=3\int_S z^4 \ dS$$
Then, integrate with spherical coordinates $z=a\cos\theta$, $dS = a^2\sin\theta d\theta d\phi$,
$$I=3 \int_0^{2\pi}\int_0^{\pi}(a\cos\theta)^4 a^2\sin\theta d\theta d\phi
=-6\pi a^6 \int_0^{\pi}\cos^4\theta d(\cos\theta)
=\frac{12\pi}5a^6$$
Edit:
$$I= 3a \int_V \left (x^2+y^2+z^2 \right )dV =9a\int_V z^2 dV$$
$$=9a\int_0^{2\pi}\int_0^{\pi}\int_0^a (r\cos\theta)^2 r^2\sin\theta drd\theta d\phi=\frac{12\pi}5a^6$$
| {
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Can the quadratic formula be explained intuitively? Most people know what the quadratic formula is so I won’t post it here (Besides, I don’t know how to properly post formulas in general).
I was wondering if there is an intuitive explanation as to why the quadratic formula is structured the way it is.
| Using the factor theorem, suppose that you have an equation of the form
$$x^2 + Bx + C = 0.$$
If it has roots $r$ and $s$, then we have
$$x^2+Bx+C = (x-r)(x-s).$$
(This comes because you can write $x^2+Bx+C = (x-r)q(x)+t$, where $t$ is a remainder that must be constant by using Long Division, and then when you plug in $r$ for $x$, you get $0$ on the left and $t$ on the right. So in fact you get $x^2+Bx+C=(x-r)q(x)$. Doing the same with $x-s$ gives you the result).
Multiplying out you get
$$x^2 +Bx + C = x^2 - (r+s)x + rs.$$
So we have that $r+s = -B$ and $rs=C$. But what we actually want are $r$ and $s$, even though what we know are $B$ and $C$.
Now notice that
$$\begin{align*}
(r-s)^2 & = r^2 - 2rs + s^2\\
&= (r^2+2rs+s^2)-4rs\\
&= (r+s)^2 - 4rs\\
&= B^2-4C.
\end{align*}$$
So $|r-s| = \sqrt{B^2-4C}$. By exchanging $r$ and $s$ if necessary, we may assume that $r\geq s$, so that $r-s\geq 0$. So we have $r-s = \sqrt{B^2-4C}$.
But that means that $2r = (r+s)+(r-s) = -B + \sqrt{B^2-4C}$, and that $2s = (r+s)-(r-s) = -B-\sqrt{B^2-4C}$. Thus, the two roots are
$$ r = \frac{-B+\sqrt{B^2-4C}}{2} \qquad\text{and}\qquad s=\frac{-B-\sqrt{B^2-4C}}{2}.$$
Now, if you have an arbitrary quadratic,
$$ax^2 + bx+c = 0,\qquad a\neq 0$$
then dividing through by $a$ you get one of the form $x^2+Bx+C$ with $B=\frac{b}{a}$ and $C=\frac{c}{a}$. Plugging into the formulas, you get
$$\begin{align*}
r &= \frac{-B+\sqrt{B^2-4C}}{2} \\
&= \frac{ -\frac{b}{a} + \sqrt{\frac{b^2}{a^2} - 4\frac{c}{a}}}{2}\\
&=
\frac{-\frac{b}{a} + \sqrt{\frac{b^2-4ac}{a^2}}}{2}\\
&= \frac{-\frac{b}{a}+\frac{1}{|a|}\sqrt{b^2-4ac}}{2}.
\end{align*}$$
and similarly
$$s = \frac{-\frac{b}{a} - \frac{1}{|a|}\sqrt{b^2-4ac}}{2}.$$
Now, if $a\gt 0$, you can just get rid of the absolute value; and if $a\lt 0$, then you just end up exchanging $r$ and $s$, and so we may just ignore the absolute value as long as we keep both roots. We get:
$$\begin{align*}
r &= \frac{-\frac{b}{a}+\frac{1}{a}\sqrt{b^2-4ac}}{2} &\quad\text{and}\quad s&= \frac{-\frac{b}{a}-\frac{1}{a}\sqrt{b^2-4ac}}{2}\\
&= \frac{-b+\sqrt{b^2-4ac}}{2a} & &= \frac{-b-\sqrt{b^2-4ac}}{2a}.
\end{align*}$$
Combining the two you get that the roots are given by
$$\frac{-b\pm\sqrt{b^2-4ac}}{2a}.$$
These ideas can be used to obtain the Cardano formulas for solving cubics; they lead to solving a quadratic equation in order to “solve” for the roots. And can also be used to deduce the Ferrari formulas for a quartic, which lead to a cubic equation to solve for the roots. However, if you try to do something similar to solve the quintic, you end up obtaining a degree six equation that must be solved... which eventually leads to the proof that it is impossible to obtain a formula similar these for the general quintic or higher degree equations. The ideas that the roots play symmetric roles and that it may be possible to obtain expressions for them in terms of the coefficients eventually leads you to Galois Theory and the study of the symmetries of the roots.
| {
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Asymmetric inequality in three variables $\frac{3(a+b)^2(b+c)^2}{4ab^2c} \geq 7+\frac{5(a^2+2b^2+c^2)}{(a+b)(b+c)}$ Consider three real positive variables $a,\ b$ and $c$. Prove that the following inequality holds:
$$\frac{3(a+b)^2(b+c)^2}{4ab^2c} \geq 7+\frac{5(a^2+2b^2+c^2)}{(a+b)(b+c)}$$
My progress: We can prove that both sides are greater than $12$ using AM-GM:
$$LHS \geq \frac{3 \cdot 4ab \cdot 4bc}{4a^2bc} = 12$$
and
$$RHS \geq 7+\frac{5[(a+b)^2+(b+c)^2]}{2(a+b)(b+c)} \geq 7+5 = 12$$
So, substract $12$ from both sides and write the inequality into:
$$3\cdot \frac{(a+b)^2(b+c)^2-16ab^2c}{4ab^2c} \geq 5 \cdot \frac{a^2+b^2+c^2-ab-bc-ca}{(a+b)(b+c)}$$
or
$$3\cdot \frac{(b+c)^2(a-b)^2+4ab(b-c)^2}{4ab^2c} \geq \frac{5}{2}\cdot \frac{(a-b)^2+(b-c)^2+(c-a)^2}{(a+b)(b+c)}$$
My next idea was to use $(c-a)^2\leq 2[(a-b)^2+(b-c)^2]$ and write it into a sum of square form with only $(a-b)^2$ and $(b-c)^2$. However, I couldn't reach significant progress.
| Partial solution. I hope, it can help.
Let $a+c=2p$ and $ac=q^2,$ where $q>0.$
Thus, by AM-GM $p\geq q$ and we need to prove that
$$\frac{3(b^2+2bp+q^2)^2}{4q^2b^2}\geq7+\frac{5(b^2+4p^2-2q^2)}{b^2+2pb+q^2}.$$
Now, consider two cases:
*
*$b\geq q$, $p=q+u$ and $b=q+v$.
Thus, we need to prove that:
$$72uq^5+4(16u^2+94uv+19v^2)q^4+8(3u^3+34u^2v+85uv^2+19v^3)q^3+$$
$$+4v(18u^3+97u^2v+130uv^2+28v^3)q^2+18v^2(2u+v)^2(u+2v)q+3v^3(2u+v)^3\geq0,$$
which is obviously true.
*$b\leq q$, $q=b+u$, $p=b+u+v$.
Thus, we need to prove that:
$$72vb^5+4(19u^2-4uv+16v^2)b^4+8(19u^3-13u^2v-2uv^2+3v^3)b^3+$$
$$+4u^2(28u^2-4uv-11v^2)b^2+18u^4(2u+v)b+3u^6\geq0.$$
Now, we see that it's a cubic inequality of $v$ and after using a derivative we can get a minimal point and to end a proof.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3519062",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Solving a limit by two methods with different results I'm considering this limit
$$\lim_{x\to 0}\frac{(1-\cos x^2)\arctan x}{e^{x^2}\sin 2x - 2x + \frac{2}{3}x^3}.$$
My first attempt was using the following equivalent infinitesimals
$$1-\cos x^2 \sim \frac{x^4}{2},\quad \arctan x \sim x, \quad \sin 2x \sim 2x, \quad e^{x^2} - 1 \sim x^2 \,\,\,\,\text{when}\,\,\,\,x\rightarrow 0,$$
and then
\begin{align*}
\lim_{x\to 0}\frac{(1-\cos x^2)\arctan x}{e^{x^2}\sin 2x - 2x + \frac{2}{3}x^3}&=\lim_{x\to 0}\frac{x^5/2}{2xe^{x^2}-2x+\frac{2}{3}x^3}=\lim_{x\to 0}\frac{x^4}{4(e^{x^2}-1)+\frac{4}{3}x^2}\\
&=\lim_{x\to 0}\frac{x^4}{4x^2+\frac{4}{3}x^2}=\lim_{x\to 0}\frac{3}{16}x^2=0.
\end{align*}
Later, inspecting this other approach combining infinitesimals and the Taylor expansions
\begin{align*}
e^{x^2} &= 1+x^2+\frac{x^4}{2}+o(x^4),\\
\sin 2x &= 2x - \frac{4x^3}{3} + \frac{4x^5}{15} + o(x^5),\\
e^{x^2}\sin 2x &= 2x - \frac{2x^3}{3} - \frac{x^5}{15} + o(x^5),
\end{align*}
I get this other result
\begin{align*}
\lim_{x\to 0}\frac{(1-\cos x^2)\arctan x}{e^{x^2}\sin 2x - 2x + \frac{2}{3}x^3}&=\lim_{x\to 0}\frac{x^5/2}{2x - \frac{2x^3}{3} - \frac{x^5}{15} + o(x^5)-2x+\frac{2}{3}x^3}\\
&=\lim_{x\to 0}\frac{x^5/2}{-x^5/15 + o(x^5)}=-\frac{15}{2}.
\end{align*}
Can someone help me to identify what am I getting wrong?
Thanks.
| Your first method is correct and the answer is indeed $0$. The problem with your second approach is that the Taylor series of $e^{x^2}\sin(2x)$ centered at $0$ begins with $2x\color{red}+\frac23x^3$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Maximize product $(a^3-a^2+2)(b^3-b^2+2)(2c^3+5c^2+9)$ Consider three real variables such that $a^2+5b^2+5c^2 = 21$. Maximize the product:
$$P = (a^3-a^2+2)(b^3-b^2+2)(2c^3+5c^2+9)$$
My attempt (ideas): I suppose the first step is to give an argument that the maximum is reached when the variables are positive (or at least that's what I suspect). $a$ lies in $[-\sqrt{21}, \sqrt{21}]$, while $b$ and $c$ lie in $\left[-\sqrt{\dfrac{21}{5}}, \sqrt{\dfrac{21}{5}}\right]$. So I guess studying how the functions $f(x) = x^3-x^2+2$, $g(x)=2x^3+5x^2+9$ behave on those ranges should help, but I don't how to give a solid argument for this.
After this, I guess I should factor out:
$$P = (a+1)(b+1)(c+3)(a^2-2a+2)(b^2-2b+2)(2c^2-c+3)$$
and find some tricky way to use AM-GM, but it's difficult to find the correct way, without knowing the point for which the maximum is attained.
| Incomplete solution: (Updated 2020-01-25)
Step 1 (incomplete): Establish that the maximum of $P$ occurs when all three factors of $P$ are non-negative.
First, let's observe that from the condition $a$ lies in $[-\sqrt{21},\sqrt{21}]$, while $b$ and $c$ lie in $\left[-\sqrt{\dfrac{21}{5}}, \sqrt{\dfrac{21}{5}}\right]$, so let's define the functions:
$$f_1 : \left[-\sqrt{21}, \sqrt{21}\right] \to \mathbb{R},\ f_1(x) = x^3-x^2+2$$
$$f_2 : \left[-\sqrt{\frac{21}{5}}, \sqrt{\frac{21}{5}}\right]\to \mathbb{R},\ f_2(x) = x^3-x^2+2$$
$$g : \left[-\sqrt{\frac{21}{5}}, \sqrt{\frac{21}{5}}\right]\to \mathbb{R},\ g(x) = 2x^3+5x^2+9 $$
We have
$$P = f_1(a) \cdot f_2(b)\cdot g(c)$$
The maximum of $P$ is a positive value, so we need either all three factors of $P$ to be positive or two of them negative and the third one positive. Studying $g(x)$, we can see that this function is positive over it's entire definition domain. So $f_1(a)$ and $f_2(b)$ must both be either positive or negative.
Assuming that $f_1(a)$ and $f_2(b)$ are both negative, we should prove that the maximum reachable value is lower than if $f_1(a)$ and $f_2(b)$ are both positive.
Step 2.1 (thanks to @Maximilian Janisch comment): Checking $(a,b,c) = (4,0,1)$, we can see that $P$ equals $1600$. We will prove that $P \leq 1600$.
Step 2.2: Notice that using AM-GM:
$$x^3-x^2+2=(x+1)(x^2-2x+2)=\frac{1}{2}(2x+2)(x^2-2x+2)\leq \frac{1}{8}(x^2+4)^2$$
This usage of AM-GM is satisfactory because it conserves both equality cases for $a$ and $b$ ($4$ and $0$). Similarly:
$$2x^3+5x^2+9=(x+3)(2x^2-x+3) \leq \frac{1}{4}(2x^2+6)^2=(x^2+3)^2$$
Therefore, using these observations and AM-GM again:
$$
\begin{aligned}P &\leq \frac{1}{64} \left[(a^2+4)(b^2+4)(c^2+3)\right]^2 \\
&= \frac{1}{64} \left[\frac{1}{25}(a^2+4)\cdot 5(b^2+4)\cdot 5(c^2+3) \right]^2 \\
&\leq \frac{1}{64} \left[\frac{1}{25}\cdot \frac{1}{27} (a^2+5b^2+5c^2+39)^3 \right]^2\\
&= 1600\end{aligned} $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3521250",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Solution to two non polynomial equation If
$$x+\frac1x=-1$$
Find the value of
$$x^{99}+\frac{1}{x^{99}}$$
Is there any formal (traditional) way to solve these problems using only elementary algebra or high school math
| $$(x+\frac{1}{x})(x^2+\frac{1}{x^2})\Rightarrow x^3+\frac{1}{x^3}=2$$
$$(x+\frac{1}{x})(x^3+\frac{1}{x^3})\Rightarrow x^4+\frac{1}{x^4}=-1$$
$$(x+\frac{1}{x})(x^4+\frac{1}{x^4})\Rightarrow x^5+\frac{1}{x^5}=-1$$
$$(x+\frac{1}{x})(x^5+\frac{1}{x^5})\Rightarrow x^6+\frac{1}{x^6}=2
\\ \vdots$$
$$x^{99}+\frac{1}{x^{99}}=2$$
and so you can obtain the result.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3524359",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Find the function $f(x)$ if $f(x+y)=f(x-y)+2f(y)+xy$ I'm suffering from the procedure I found, which is contradictory.
The original question is:
Let $f$ be differentiable. For all $x, y \in \mathbb R$,
$$f(x+y)=f(x-y)+2f(y)+xy$$
Suppose $f'(0)=1$ and find $f(x)$.
My procedure was:
Put $x=y=0$:
$$f(0)=f(0)+2f(0)+0$$
$$f(0)=0$$
Put $x=y=\frac{h}{2}$:
$$f(h)=2f\left(\frac{h}{2}\right)+\frac{h^2}{4}$$
Substituting $x=x+\frac{h}{2}, y=\frac{h}{2}$ brings:
$$f'(x)=\lim_{h\to0} \frac{f(x+h)-f(x)}{h}=\lim_{h \to 0}\frac{2f\left(\frac{h}{2}\right)+\frac{h}{2}\left(x+\frac{h}{2}\right)}{h}=\lim_{h\to0}\frac{f(h)+\frac{h}{2}x}{h}\\=\lim_{h\to0}\frac{f(h)-f(0)}{h-0}+\frac{x}{2}=\frac{x}{2}+1$$
$$f(x)=\frac{x^2}{4}+x+c$$ where $c$ is constant.
Since $f(0)=0,\space\space c=0$
However, when we substitute $x=0$ to the given condition,
$$f(y)=f(-y)+2f(y)+0(y)$$
$$-f(y)=f(-y)$$
which informs that $f$ is odd: contradiction!
Could you please help me to find the error? Thanks.
| In addition to the accepted answer, there is just no function $f$ whatsoever that satisfies the given functional equation, differentiable or otherwise. Consider the following special cases:
\begin{align}
f(1) &= f(1) + 2f(0) & x=1 \;& y=0 \\
f(2) &= f(0) + 2f(1) + 1 & x=1\;&y=1 \\
f(4) &= f(2) + 2f(1) + 3 & x=3\;&y=1 \\
f(4) &= f(0) + 2f(2) + 4 & x=2\;&y=2
\end{align}
From the first equation, we get $f(0) =0$; let $f(1)=c$. From the second, $f(2)=2c+1$. From the third, $f(4)=(2c+1)+2c+3 = 4c+4$, and from the fourth, $f(4)=2(2c+1)+4 = 4c+6$.
This gives us $4c+4 = 4c+6$, a contradiction no matter what the value of $c$ is.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3524457",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
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Product of multiple sines Is there an identity for the following equation:
$\ f(x) = \sin(x.\pi/2)\cdot \sin(x.\pi/3)\cdot \sin(x.\pi/4) $
I am looking for an equation similar to this equation
| Express everything in terms of the trigonometric functions of $\frac {\pi x}{12}$ and working a little
$$4\sin \left(\frac{\pi x}{2}\right) \sin \left(\frac{\pi x}{3}\right) \sin
\left(\frac{\pi x}{4}\right)=\sin \left(\frac{\pi x}{12}\right)+\sin \left(\frac{5 \pi x}{12}\right)+\sin \left(\frac{7 \pi x}{12}\right)-\sin \left(\frac{13 \pi
x}{12}\right)$$ It would be much less funny do add the next term; the common factor being $\frac {\pi x}{60}$, we should have
$$8\sin \left(\frac{\pi x}{2}\right) \sin \left(\frac{\pi x}{3}\right) \sin
\left(\frac{\pi x}{4}\right)\left(\frac{\pi x}{5}\right)=\cos \left(\frac{7 \pi x}{60}\right)+\cos \left(\frac{13 \pi x}{60}\right)-$$ $$\cos
\left(\frac{17 \pi x}{60}\right)+\cos \left(\frac{23 \pi x}{60}\right)-\cos
\left(\frac{37 \pi x}{60}\right)-\cos \left(\frac{47 \pi x}{60}\right)-\cos
\left(\frac{53 \pi x}{60}\right)+\cos \left(\frac{77 \pi x}{60}\right)$$
All the trick is to expand $\sin(nt)$ in terms of powers of $\sin(t)$ and $\cos(t)$ and then the power reduction formulae (look here).
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
Given that $z = i + i^{2016} + i^{2017}$, find $|z^{10}|$. I am told that
$$z = i + i ^ {2016} + i ^ {2017}$$
and I have to find $|z^{10}|$. This is what I tried:
$$i ^ {2016} = (i^4)^{504} = 1 ^ {504} = 1$$
$$i ^ {2017} = i \cdot (i^4)^{504} = i \cdot 1 = i$$
So we have that:
$$z = 1 + 2i$$
And then I worked towards finding $z ^ {10}$.
$$z = 1 + 2i$$
$$z^2 = 1 + 4i + 4i^2 = -3 + 4i$$
$$z^4 = ... = -7 - 24 i$$
$$z^5 = z^4 \cdot z = (-7-24i)(1 + 2i) = ... = 41 - 38i$$
$$z^{10} = (z^5)^2 = (41 - 38i)^2 = ... = 237 -3116i $$
So we then have:
$$|z^ {10} | = \sqrt{237^2 + 3116^2} = \sqrt{56169 + 964656} = \sqrt{1020825} = 15\sqrt{4537}$$
Is this correct? Is there a better way to solve this? More efficient/faster way?
| $z=1+2i$; $\overline{z}=1-2i$;
$|z|^2=z\cdot \overline{z}=1-4i^2=5$;
$|z|^{10}=(|z|^2)^5=5^5$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3527365",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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How many different solutions for this $\sqrt{3x^2 + 2x + 5} + \sqrt{x^2-4x+5} = \sqrt{2x^2 - 2x + 2} + \sqrt{2x^2+8}$
My attempt :
All the equation inside sqrt has $D<0$.
Let it be
$\sqrt A + \sqrt B = \sqrt C + \sqrt D$
2 possibilities
$$\sqrt A = \sqrt C \cap \sqrt B = \sqrt D$$
Find $x$ that match this condition.
$\sqrt A = \sqrt C $ i get $x=-3 \cup x=-5$
$\sqrt B = \sqrt D $ i get $x=-3 \cup x=-1$
$$\sqrt A = \sqrt D \cap \sqrt B = \sqrt C$$
Find $x$ that match this condition.
$\sqrt A = \sqrt D $ i get $x=-3 \cup x=1$
$\sqrt B = \sqrt C $ i get $x=-3 \cup x=1$
2 solutions.
Another posibility
$\sqrt A + \sqrt B = \sqrt C + \sqrt D$
${[\sqrt A + \sqrt B]}^2 = {[\sqrt C + \sqrt D]}^2$
This part is complicated.
Is there easier way to solve it?
| Let $a=\sqrt{3x^2+2x+5}, b=\sqrt{2x^2+8}, c=\sqrt{2x^2-2x+2}, d=\sqrt{x^2-4x+5}$
$\implies a-b=c-d$
Observe that $$3x^2+2x+5-(2x^2+8)=x^2+2x-3=2x^2-2x+2-(x^2-4x+5)$$
$\implies a^2-b^2=c^2-d^2$
$\implies a+b=\dfrac{c-d}{a-b}\cdot c+d=c+d$
Now add and subtract to find $$a=c$$
$$\implies\sqrt{3x^2+2x+5}=\sqrt{2x^2-2x+2}$$
Square both sides
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "2",
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} |
integration of. $\int \frac{x}{x^3-3x+2}$ I am trying to integrate :
$\Large \int \frac{x}{x^3-3x+2}dx$
I decomposed the fraction and got :
$\Large \frac{x}{x^3-3x+2} = \frac {x}{(x-1)^2(x+2)}$
Then I tried to get two different fractions:
$ \Large \frac {x}{(x-1)^2(x+2)} = \frac {Ax}{(x-1)^2} + \frac {B}{(x+2)}$
Well I got A=1/3 and B=-1/3. as possible values. but that was not correct. I guess I missed something.
| $$ I=\int \frac{x}{x^3-3x+2}dx$$
$$I= \int \frac {xdx}{(x-1)^2(x+2)}$$
Since $\frac {-1}{(x-1)^2}$ is an obvious derivative substitute $$u=\dfrac {1}{x-1} \implies -du=\dfrac {dx}{(x-1)^2}$$
The integral becomes:
$$I=-\int \dfrac {u+1}{3u+1}du$$
Which is easier to integrate.
$$u+1=u+\frac 1 3 +\frac 2 3$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3529446",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find $P$ such that $P^{-1}AP$ is a given matrix
Let
$$A = \begin{pmatrix} 1 & -3 & 0 \\ 3 & 4 & -3 \\ 3 & 3 & -2\end{pmatrix}$$
Find an invertible $P \in M_3(\mathbb{R})$ such that
$$B = P^{-1}AP = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 3 \\ 0 & -3 & 1\end{pmatrix}$$
The problem is that the characteristic polynomial of both $A$ and $B$ is $-(\lambda-1)(\lambda^2 - 2\lambda+10)$, which has imaginary root, so I can't just straight up find the diagonalization decomposition of $A$ and $B$ and multiply them, as $P$ might contain complex number, which isn't what I need.
Rational canonical form might work, but that is beyond my ability.
Any solution/helps would be appreciated.
| Let $P=\begin{pmatrix} 1 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 1 & 0 \end{pmatrix}$. $\;$ You can check that $P^{-1}AP=\begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 3 \\ 0 & -3 & 1 \end{pmatrix}=B$.
The way I found this matrix was as follows: in order to have $P^{-1}AP=B$, $P$ must send the standard basis vector $e_1$ to an eigenvector of the eigenvalue $1$. The eigenvectors for the eigenvalue $1$ are elements of $\ker(A-I)$. Note that $A-I$ is what you get when you plug $A$ into $x-1$.
Note that $A-I=\begin{pmatrix} 0 & -3 & 0 \\ 3 & 3 & -3 \\ 3 & 3 & -3 \end{pmatrix}$, $\;$ and $\text{rref}(A-I)=\begin{pmatrix} 1 & 0 & -1 \\ 0 & 1 & 0 \\ 0 & 0 & 0 \end{pmatrix}$.
So I let $P(e_1)=\begin{pmatrix} 1 \\ 0 \\ 1 \end{pmatrix}$.
It remains to find out where $P$ should send $e_2$ and $e_3$. I found these vectors as follows: since the characteristic polynomial of $A$ is
$$c_A(x)=(x-1)(x^2-2x+10)$$
I found $P(e_2)$, $P(e_3)$ by finding a basis for $\ker(A^2-2A+10I)$. Note that $A^2-2A+10I$ is what you get when you plug $A$ into $x^2-2x+10$.
Note that $A^2-2A+10I=\begin{pmatrix} 0 & -9 & 9 \\ 0 & 0 & 0 \\ 0 & -9 & 9 \end{pmatrix}$, $\;$ and $\text{rref}(A^2-2A+10I)=\begin{pmatrix} 0 & 1 & -1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix}$.
So I let $P(e_2)=\begin{pmatrix} 0 \\ 1 \\ 1 \end{pmatrix}$ and $P(e_3)=\begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix}$. $\;$ Hence $P=\begin{pmatrix} 1 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 1 & 0 \end{pmatrix}$, $\;$ and $P^{-1}AP=B$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Find the remainder when $x^{100}$ is divided by $x^8 - x^6 + x^4 - x^2 + 1.$ I need help in the problem:
Find the remainder when $x^{100}$ is divided by $x^8 - x^6 + x^4 - x^2 + 1.$
I have tried factoring $x^{100}-1$ and adding 1 to that, but that hasn't helped. Could someone please help with this?
| $x^{100}-1=(x^{10}+1)(x^{90}-x^{80}+x^{70}-x^{60}+x^{50}-x^{40}+x^{30}-x^{20}+x^{10}-1)$
$=(x^8-x^6+x^4-x^2+1)(x^2+1)(x^{90}-x^{80}+x^{70}-x^{60}+x^{50}-x^{40}+x^{30}-x^{20}+x^{10}-1)$,
so $x^{100}=$
$(x^8-x^6+x^4-x^2+1)(x^2+1)(x^{90}-x^{80}+x^{70}-x^{60}+x^{50}-x^{40}+x^{30}-x^{20}+x^{10}-1) + 1.$
| {
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"timestamp": "2023-03-29T00:00:00",
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Proof by induction:$\frac{3}{5}\cdot\frac{7}{9}\cdot\frac{11}{13}\cdots\frac{4n-1}{4n+1}<\sqrt{\frac{3}{4n+3}}$ In the very beginning I'm going to refer to similar posts with provided answers:
Induction Inequality Proof with Product Operator $\prod_{k=1}^{n} \frac{(2k-1)}{2k} \leq \frac{1}{\sqrt{3k+1}}$ (answered by Özgür Cem Birler)
Prove that $\prod\limits_{i=1}^n \frac{2i-1}{2i} \leq \frac{1}{\sqrt{3n+1}}$ for all $n \in \Bbb Z_+$
I examined the solutions and tried to apply the methods used there to make sure whether I understand it or not. I'm concerned about the step of induction.
A task from an earlier exam at my university:
Prove by induction:
$$\frac{3}{5}\cdot\frac{7}{9}\cdot\frac{11}{13}\cdots\frac{4n-1}{4n+1}<\sqrt{\frac{3}{4n+3}}$$
Attempt:
rewritten:
$$\prod_{i=1}^n\frac{4i-1}{4i+1}<\sqrt{\frac{3}{4n+3}}$$
$(1)$ base case: $\tau(1)$
$$\frac{3}{5}=\sqrt{\frac{4}{7}}<\sqrt{\frac{3}{7}}$$
$(2)$ assumption:
Let:$$\frac{3}{5}\cdot\frac{7}{9}\cdot\frac{11}{13}\cdots\frac{4n-1}{4n+1}<\sqrt{\frac{3}{4n+3}}$$
hold for some $n\in\mathbb N$
$(3)$ step: $\tau(n+1)$
$$\frac{4n+3}{4n+5}\cdot\prod_{i=1}^n\frac{4i-1}{4i+1}<\frac{4n+3}{4n+5}\cdot\sqrt{\frac{3}{4n+3}}=\frac{\sqrt{3(4n+3)}}{4n+5}<\sqrt{\frac{3}{4n+7}}$$
$$\frac{12n+9}{16n^2+40n+25}<\frac{3}{4n+7}\iff\frac{48n^2+120n+63-48n^2-120n-75}{\underbrace{(4n+5)^2(4n+7)}_{>0}}<0\iff-\frac{12}{(4n+5)^2(4n+7)}<0$$
Is this combined legitimate?
| Base case:
$n = 1$
$\frac {3}{5} < \sqrt {\frac {3}{7}}$
Since both are positive, we can square both sides.
$\frac {9}{25} < \frac {3}{7}$
Cross multiply
$63 < 75$
Inductive hypothesis
Suppose, $\prod_\limits{i=1}^{n} \frac {4i-1}{4i+1} < \sqrt {\frac {3}{4n+3}}$
We must show that when the hypothesis holds, $\prod_\limits{i=1}^{n+1} \frac {4i-1}{4i+1} < \sqrt {\frac {3}{4n+7}}$
$\prod_\limits{i=1}^{n+1} \frac {4i-1}{4i+1} = \frac {4n+3}{4n+5}\prod_\limits{i=1}^{n} \frac {4i-1}{4i+1} < \frac {4n+3}{4n+5}\sqrt {\frac {3}{4n+3}}$ based on the inductive hypothesis.
$\frac {4n+3}{4n+5}\sqrt {\frac {3}{4n+3}} = \sqrt {\frac {3(4n+3)}{(4n+5)^2}}$
$(4n+5)^2 = (4n+3)(4n+7) + 4$
$\sqrt {\frac {3(4n+3)}{(4n+5)^2}} = \sqrt {\frac {3(4n+3)}{(4n+3)(4n+7) + 4}} < \sqrt {\frac {3(4n+3)}{(4n+3)(4n+7)}} = \sqrt {\frac {3}{4n+7}}$
Which is what we were required to show.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3537475",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 2
} |
Obtain an expression for the nth term of the Geometric Progression. The 2nd, 6th and 8th terms of an Arithmetic progression are three successive terms of a Geometric progression.
Find the common ratio of the Geometric progression and obtain an expression for the $n$th term of the geometric progression.
Working so far:
Let $x, xr$ and $xr^2$ be the three terms of the G.P..
Let $a$ be the first term for the A.P..
$\begin{align} a+d&=x\tag {1}\\a+5d&=xr\tag{2}\\a+7d&=xr^2 \tag{3}\end{align}$
$\begin{align}(2)-(1): 4d&=x(r-1)\tag{4}\\(3)-(2):2d&=xr(r-1)\tag{5}\\\frac{(5)}{(4)}: r&=\frac{1}{2}\end{align}$
So, I found the common ratio but I don’t know how to obtain an expression for the $n$th term of the G.P. because there are four unknowns (one already solved) but I only have 3 equations.
The answer from the book is $\frac {16}{9}a(\frac{1}{2})^n$ but I don’t know how to get the $\frac{16}{9}a$ . I’m assuming the $a$ given by the book is the first term of the A.P., that’s why I used $a$ for the first A.P. term and $x$ as the first term of G.P..
| Working backwards from the answer,
$\begin{align}\frac {16}{9}a(\frac{1}{2})^n&=xr^{n-1}\\ \frac {16}{9}a(\frac{1}{2})^n&=\frac{x(\frac{1}{2})^n}{\frac{1}{2}}\\\frac{16}{9}a&=2x\\\frac{8}{9}a&=x\end{align}$
So, that would mean $x=\frac{8}{9}a$, where $x$ is the first term of the G.P..
However, taking $\frac{(2)}{(1)}$:
$\frac{a+5d}{a+d}=\frac{1}{2}$
$2a+10d=a+d$
$d=\frac{-1}{9}a$
Substituting $d=\frac{-1}{9}a$ into $(1)$:
$x=\frac{8}{9}a$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3541404",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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} |
Smarter way to solve $ \int_0^1\arctan(x^2)\,dx$ I'm trying to solve the following definite integral:
\begin{equation}
\int_0^1 \arctan(x^2)dx
\end{equation}
I tryed first by parial integration, finding:
\begin{equation}
x\arctan(x^2)\Bigl|_0^1-\int_0^1 \dfrac{2x^2}{1+x^4}dx
\end{equation}
Then:
\begin{equation}
\int_0^1 \dfrac{2x^2}{1+x^4}dx=\int_0^1 \dfrac{(x^2+1)+(x^2-1)}{1+x^4}dx=\int_0^1 \dfrac{x^2+1}{1+x^4}dx+\int_0^1 \dfrac{x^2-1}{1+x^4}dx
\end{equation}
and i introduced the following substitution:
\begin{equation}
t=x-\dfrac{1}{x}\qquad s=x+\dfrac{1}{x}
\end{equation}
Than I used this weird substitution:
\begin{equation}
\dfrac{\sqrt{2}}{2}\arctan\left( \dfrac{t}{\sqrt{2}} \right)\Bigl|_0^1+\int_0^1\dfrac{ds}{(s-2)(s+2)}
\end{equation}
In the end simple fraction:
\begin{equation}
\left[\dfrac{\sqrt{2}}{2}\arctan\left( \dfrac{t}{\sqrt 2} \right)+\dfrac{1}{4}\log(2-s)-\dfrac{1}{4}\log(s+2)\right]_0^1
\end{equation}
and that is the solution.
Now: does it exist a simpler path to solve this integral?
| I wouldn't necessarily say it is simpler, but it is more systematic and the same method can be applied to integrating general rational functions of $x$.
You already got to the point of calculating $$x\arctan(x^2)\bigg\vert_0^1 - \int\limits_0^1\frac{2x^2}{1+x^4}dx = \frac{\pi}{4} - \int\limits_0^1\frac{2x^2}{1+x^4}dx$$
Now we use the factorization $x^4 + 1 = (x^2 + \sqrt{2}x + 1)(x^2 - \sqrt{2}x + 1)$, which allows writing: $$\frac{2x^2}{1+x^4} = \frac{2x^2}{(x^2 + \sqrt{2}x + 1)(x^2 - \sqrt{2}x + 1)} = \frac{Ax+B}{x^2 + \sqrt{2}x + 1} + \frac{Cx+D}{x^2 - \sqrt{2}x + 1}$$
We can solve for $A, B, C, D$ and find $$\frac{2x^2}{1+x^4} = \frac{\frac{\sqrt{2}}{2}x}{x^2 - \sqrt{2}x + 1} - \frac{\frac{\sqrt{2}}{2}x}{x^2 + \sqrt{2}x + 1} = \frac{\frac{\sqrt{2}}{4}(2x - \sqrt{2}) + \frac{1}{2}}{x^2 - \sqrt{2}x + 1} - \frac{\frac{\sqrt{2}}{4}(2x + \sqrt{2}) - \frac{1}{2}}{x^2 + \sqrt{2}x + 1}$$
After separating terms and completing the squares in the denominators of the last two terms we get:
$$\frac{\sqrt{2}}{4}\frac{2x - \sqrt{2}}{x^2 - \sqrt{2}x + 1} - \frac{\sqrt{2}}{4}\frac{2x + \sqrt{2}}{x^2 + \sqrt{2}x + 1} + \frac{1}{2}\frac{1}{\left(x - \frac{\sqrt{2}}{2}\right)^2+\frac{1}{2}} + \frac{1}{2}\frac{1}{\left(x + \frac{\sqrt{2}}{2}\right)^2+\frac{1}{2}}$$
The first two terms can be integrated easily with the substitution $u = x^2 - \sqrt{2}x + 1$ or $u = x^2 + \sqrt{2}x + 1$ respectively, and the standard integral $\int\frac{1}{u}du = \ln u + C$.
The last two terms are integrated with the substitution $u = x - \frac{\sqrt{2}}{2}$ or $u = x + \frac{\sqrt{2}}{2}$ and the standard integral $\int\frac{1}{u^2 + a^2}du = \frac{1}{a}\arctan\left(\frac{u}{a}\right) + C$ with $a = \frac{\sqrt{2}}{2}$.
The result is: $$\int\limits_0^1\frac{2x^2}{1+x^4}dx = $$
$$\frac{\sqrt{2}}{4}\ln\left(x^2 - \sqrt{2}x + 1\right)\bigg\vert_0^1 - \frac{\sqrt{2}}{4}\ln\left(x^2 + \sqrt{2}x + 1\right)\bigg\vert_0^1 + $$ $$\frac{\sqrt{2}}{2}\arctan\left(\sqrt{2}x-1\right)\bigg\vert_0^1 + \frac{\sqrt{2}}{2}\arctan\left(\sqrt{2}x+1\right)\bigg\vert_0^1$$
$$ = \frac{\sqrt{2}}{4}\ln(2 - \sqrt{2}) - \frac{\sqrt{2}}{4}\ln(2 + \sqrt{2}) + \frac{\sqrt{2}}{2}\arctan(\sqrt{2}-1) + \frac{\sqrt{2}}{2}\arctan(\sqrt{2}+1) $$
$$ = \frac{\sqrt{2}}{2}\ln\left(\sqrt{2} - 1\right) + \frac{\sqrt{2}\pi}{4}$$
So finally the integral is $$\int_0^1\arctan\left(x^2\right)dx = \frac{(1 - \sqrt{2})\pi}{4} - \frac{\sqrt{2}}{2}\ln\left(\sqrt{2}-1\right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3544203",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 0
} |
Sum of the series: $\sum_{n=2}^\infty\frac{(-1)^n}{n^2+n-2}$ is? This was a question I confronted in JAM 2016. I tried the following steps:
$$\displaystyle\sum_{n=2}^\infty\frac{(-1)^n}{n^2+n-2}
\implies\sum_{n=2}^\infty\frac{(-1)^n}{(n-1)(n+2)}\implies\sum_{n=2}^\infty(-1)^n\frac{1}{3}\cdot[\frac{1}{(n-1)}-\frac{1}{(n-2)}]$$$$\implies\lim_
{n\to\infty}(-1)^n\cdot\frac{1}{3}[1-\frac{1}{4}+\frac{1}{2}-\frac{1}{5}+...] $$
From the options it was evident that the sum had a relation with $\log_{e}2$, however I couldn't manage to solve the problem. What was I missing?
| $$\frac{1}{3}\sum_{n=2}^{+\infty}(-1)^n\left(\frac{1}{n-1}-\frac{1}{n+2}\right)=$$
$$=\frac{1}{3}\sum_{n=1}^{+\infty}\frac{(-1)^{n+1}}{n}+\frac{1}{3}\sum_{n=1}^{+\infty}\frac{(-1)^{n+1}}{n}-\frac{1}{3}\left(1-\frac{1}{2}+\frac{1}{3}\right)=$$
$$=\frac{1}{3}\ln2+\frac{1}{3}\ln2-\frac{1}{3}\left(1-\frac{1}{2}+\frac{1}{3}\right)=\frac{2}{3}\ln2-\frac{5}{18}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3544400",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Converge / Diverge $\sum_{n=1}^{\infty}\big( \frac{1}{\sqrt[n]{n!}}\big)^2$ $$\sum_{n=1}^{\infty}\big( \frac{1}{\sqrt[n]{n!}}\big)^2$$
$$2^n\leq n!$$
As $2\cdot 2 \cdots2\leq 1 \cdot 2 \cdots n!$
Therefore:
$$2\leq \sqrt[n]{n!}$$
$$2^2\leq \sqrt[n]{n!}^2$$
So:
$$\frac{1}{\sqrt[n]{n!}^2}\leq \frac{1}{4}$$
$$\sum_{n=1}^{\infty}\frac{1}{\sqrt[n]{n!}^2}\leq \sum_{n=1}^{\infty}\frac{1}{4}=\frac{1}{4}\cdot \frac{1}{1-\frac{1}{4}}=\frac{1}{3}$$
Is it correct?
| Since
$$
e^n = 1 + n + \cdots + \frac{{n^n }}{{n!}} + \cdots > \frac{{n^n }}{{n!}},
$$
it holds that
$$
\sum\limits_{n = 1}^\infty {\left( {\frac{1}{{\sqrt[n]{{n!}}}}} \right)^2 } < \sum\limits_{n = 1}^\infty {\frac{{e^2 }}{{n^2 }}} = \frac{{(\pi e)^2 }}{6}.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3545317",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
If $h(x) = \dfrac{f(x)}{g(x)}$, then find the local minimum value of $h(x)$ Let $f(x) = x^2 + \dfrac{1}{x^2}$ and $g(x) = x – \dfrac{1}{x}$
, $x\in R – \{–1, 0, 1\}$. If $h(x) = \dfrac{f(x)}{g(x)}$, then the local minimum value
of $h(x)$ is :
My attempt is as follows:-
$$h(x)=\dfrac{x^2+\dfrac{1}{x^2}}{x-\dfrac{1}{x}}$$
$$h(x)=\dfrac{x^2+x^2-2+2}{x-\dfrac{1}{x}}$$
$$h(x)=x-\dfrac{1}{x}+\dfrac{2}{x-\dfrac{1}{x}}$$
Case $1$: $x-\dfrac{1}{x}>0$
$$\dfrac{x^2-1}{x}>0$$
$$x\in(-1,0) \cup (1,\infty)$$
$$AM\ge GM$$
$$\dfrac{x-\dfrac{1}{x}+\dfrac{2}{x-\dfrac{1}{x}}}{2}>\sqrt{2}$$
$$h(x)\ge 2\sqrt{2}$$
$$x-\dfrac{1}{x}=\dfrac{2}{x-\dfrac{1}{x}}$$
$$x^2+\dfrac{1}{x^2}-2=2$$
$$x^2+\dfrac{1}{x^2}=4$$
$$x^4-4x^2+1=0$$
$$x^2=2\pm\sqrt{3}$$
Only $x=\sqrt{2+\sqrt{3}},-\sqrt{2-\sqrt{3}}$ are the valid solutions.
Case $2$: $x-\dfrac{1}{x}<0$
$$x\in(-\infty,-1) \cup (0,1)$$
$$h(x)=-\left(\dfrac{1}{x}-x+\dfrac{2}{\dfrac{1}{x}-x}\right)$$
By $AM\ge GM$, $h(x)\ge-2\sqrt{2}$
We will get this minimum value at $-\sqrt{2+\sqrt{3}},\sqrt{2-\sqrt{3}}$
So answer should have been $-2\sqrt{2}$ but actual answer is $2\sqrt{2}$. What am I missing here.
| Assuming that
$$h(x)=\frac{f(x)}{g(x)}\implies h'(x)=\frac{\left(x^2+1\right) \left(x^4-4 x^2+1\right)}{x^2 \left(x^2-1\right)^2}$$ Assuming $x\neq 0$ and $x\neq 1$ you are left with
$$x^4-4x^2+1=0 \implies (x^2)^2-4(x^2)+1=0$$ which is a quadratic in $x^2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3546598",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 5
} |
Finding $ \lim_{x\to1/2}\frac{f(x)}{x-1/2}$ given that $ \lim_{x\to1}\frac{f(x)}{x-1}=2$ and $\lim_{x\to-1}\frac{f(x)}{x+1}=6$ Consider the following question:
Find $\displaystyle \lim_{x\to1/2}\frac{f(x)}{x-1/2}$
given that $\displaystyle \lim_{x\to1}\frac{f(x)}{x-1}=2$ and
$\displaystyle \lim_{x\to-1}\frac{f(x)}{x+1}=6$
Attempt
Let $f(x)=(x-1)(x+1)(x-1/2)q(x)$. Here I have $q(1)=2$ and $q(-1)=2$.
\begin{align}
\lim_{x\to1/2}\frac{f(x)}{x-1/2}
&=\lim_{x\to1/2}\frac{(x-1)(x+1)(x-1/2)q(x)}{x-1/2}\\
&=\lim_{x\to1/2}\left((x-1)(x+1)q(x)\right)\\
&= -\frac{3}{4}q(1/2).
\end{align}
Question
Is it possible to find $q(1/2)$?
| Not enough information given.
For example, the polynomial $f(x)=ax^4+(1-2a)x^2+a-1$ with $a \in \mathbb{R}$ checks the hypothesis. The limit though, is finite only when $f\left(\frac{1}{2}\right) = 0$, which gives $a = \frac{4}{3}$ and the limit is $-1$,otherwise the limit can be $\pm \infty$. If we go for a polynomial of higher degree, similar discussion arises. Not to mention there are more complex functions that can satisfy the requirements.
If we're given that $f$ is a third degree polynomial, then we can solve the question:
$$f(x)=ax^3+bx^2+cx+d$$
and we need:
$$f(1)=a+b+c+d=0$$
$$f(-1)=-a+b-c+d=0$$
$$f'(1)=3a+2b+c=2$$
$$f'(-1)=3a-2b+c=6$$
Solving the system gives $f(x)=2x^3-x^2-2x+1=(x-1)(x+1)(2x-1)$, and the limit is finite:
$$\lim_{x\to \frac{1}{2}}\frac{f(x)}{x-\frac{1}{2}}=\lim_{x\to \frac{1}{2}}2(x-1)(x+1)=-\frac{3}{2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3547336",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Summation of finite series: Let $f(r)$ be what? Find the sum of the first $n$ terms of
$\displaystyle \frac{1}{1\times4\times7}+\frac{1}{4\times7\times10}+\frac{1}{7\times10\times13}+...$
My working:
Let $\begin{align}\displaystyle\frac{1}{(3x-2)(3x+1)(3x+4)} &\equiv \frac{A}{3x-2}+\frac{B}{3x+1}+\frac{C}{3x+4}\\1&\equiv A(3x+1)(3x+4)+B(3x-2)(3x+4)+C(3x-2)(3x+1)\end{align}$
Let $\displaystyle x=\frac{-1}{3}, B=\frac{-1}{9}$
Let $\displaystyle x=\frac{-4}{3}, A=\frac{1}{18}$
Let $\displaystyle x=\frac{2}{3}, C=\frac{1}{18}$
Thus, $\displaystyle\frac{1}{(3x-2)(3x+1)(3x+4)}\equiv \frac{1}{18(3x-2)}-\frac{1}{9(3x+1)} +\frac{1}{18(3x+4)}$
So, I thought of using the method $$\sum_{r=1}^n u_r= \sum_{r=1}^n [f(r+1)-f(r)]=f(n+1)-f(1)$$ but I’m not sure if finding such a function is possible in this case.
Carrying on using the above method,
$$\sum_{r=1}^n \frac{1}{(3r-2)(3r+1)(3r+4)}$$
$$=\sum_{r=1}^n \Bigl(\frac{1}{18(3r-2)} -\frac{1}{9(3r+1)} + \frac{1}{18(3r+4)}\Bigr)$$
$$=\sum_{r=1}^n \Bigl(\frac{1}{18(3r-2)} -\frac{1}{18(3r+1)}\Bigr) -\sum_{r=1}^n\Bigl(\frac{1}{18(3r+1)} -\frac{1}{18(3r+4)}\Bigr)$$
But now I don’t know what to let $f(r)=$?
How to proceed?
| In your final line, the $r$ should be replaced by $x$ to get
$$\sum_{x=1}^n \Bigl(\frac{1}{18(3x-2)} - \frac{1}{18(3x+1)}\Bigr) -\sum_{x=1}^n\Bigl(\frac{1}{18(3x+1)} - \frac{1}{18(3x+4)}\Bigr) \tag{1}\label{eq1A}$$
In each summation, the term being subtracted (i.e., $\frac{1}{18(3x+1)}$ and $\frac{1}{18(3x+4)}$) is the same as the first term being added for the next summation value, so they cancel, showing each is a Telescoping series. Thus, all of the terms cancel except for the first term minus the last term, giving
$$\begin{equation}\begin{aligned}
& \left(\frac{1}{18(3(1) - 2)} - \frac{1}{18(3n + 1)}\right) - \left(\frac{1}{18(3(1) + 1)} - \frac{1}{18(3n + 4)}\right) \\
& = \frac{1}{18} - \frac{1}{72} + \frac{1}{54n + 72} - \frac{1}{54n + 18} \\
& = \frac{1}{24} - \frac{54}{(54n + 72)(54n + 18)} \\
& = \frac{1}{24} - \frac{1}{6(3n + 4)(3n + 1)} \\
& = \frac{(3n + 4)(3n + 1)}{24(3n + 4)(3n + 1)} - \frac{4}{24(3n + 4)(3n + 1)} \\
& = \frac{9n^2 + 3n + 12n + 4 - 4}{24(3n + 4)(3n + 1)} \\
& = \frac{9n^2 + 15n}{24(3n + 4)(3n + 1)} \\
& = \frac{3n^2 + 5n}{8(3n + 4)(3n + 1)}
\end{aligned}\end{equation}\tag{2}\label{eq2A}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3550816",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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$abcd=6(1-a)(1-b)(1-c)(1-d)$ Positive integers $a \geqslant b \geqslant c \geqslant d$ satisfy:
$$abcd=6(1-a)(1-b)(1-c)(1-d)$$
$1.$ Prove that $d=2$
$2.$ Find all possible values of $a,b,c$.
What I know:
Since $x$ and $x-1$ are relatively prime for all positive integers $x$ :
$$\gcd(a,1-a)=\gcd(b,1-b)=\gcd(c,1-c)=\gcd(d,1-d)=1$$
| We can rewrite this as:
$$\prod_{cyc} \frac{a}{a-1} =6$$
We clearly have $a \geqslant b \geqslant c \geqslant d > 1$. Also, we can see that $\frac{x}{x-1}$ is a decreasing function. If $d>2$ then,
$$\prod_{cyc} \frac{a}{a-1} < (1.5)^4<6$$
This forces $d=2$. We now have:
$$\frac{a}{a-1} \cdot \frac{b}{b-1} \cdot \frac{c}{c-1}=3$$
Similarly assume that $c>3$. We have $(1.33)^3<3$ which is a contradiction and we will then have $c \in {2,3}$.
If $c=3$, we will have:
$$\frac{a}{a-1} \cdot \frac{b}{b-1} = 2$$
Continuing in the similar fashion, we can force $b=3$ which yields $a=4$ and thus we have $(a, b, c, d) =(4,3,3,2)$.
If $c=2$, we will have:
$$\frac{a}{a-1} \cdot \frac{b}{b-1} =\frac{3}{2}$$
Since $\frac{a}{a-1}>1$ we have $\frac{b}{b-1}<\frac{3}{2}$ which gives $b>3$. If $b>5$, we get $(1.2)^2<1.5$. We thus need to have $3<b \leqslant 5$. Although $b=3$ doesn't give integer $a$, the other cases gives us $(a, b, c, d) =(9,4,2,2)$ and $(a, b, c, d) =(6,5,2,2)$.
| {
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"url": "https://math.stackexchange.com/questions/3551012",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Let $\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}.... =A$. What is $\frac {1}{2^2}+\frac{1}{4^2}+\frac{1}{6^2}....$ in terms of A? Find the value of $\frac {1}{2^2}+\frac{1}{4^2}+\frac{1}{6^2}....\infty$ in terms of A
For the first expression, the $T_n$ term will be
$$T_n=\frac{1}{\frac{((n)(n+1))^2}{4}}$$
Now $$T_n=4\left[\sum \frac{1}{(n)^2(n+1)^2}\right]$$
How should I proceed?
| Given expression$$\frac{1}{2^2}+\frac{1}{4^2}+\frac{1}{6^2}…$$
Which is equal to
$$\frac1{2^2}\left[\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}…\right]$$
Hence the given Series sums up to $$\frac{A}4$$
| {
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"url": "https://math.stackexchange.com/questions/3551477",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "1",
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Prove that $n^4 \mod 8$ is identically equal to either 0 or 1, $\forall \ n\in \mathbb{N}$. I feel pretty confident about the first half of my solution for this, however I don't like how I used the induction hypothesis on the case for even integers, it feels like it isn't doing anything useful since it is really easy to show that $P(2x+2) \mod 8 \equiv 0$ without using the inductive hypothesis.
This is also the first semester I've been doing proofs, so am I going about this right?
Proof. Since $1^4 \mod 8 = 1$, $P(1)$ holds. Also, $2^4 \mod 8 = 16 \mod 8 = 0$ so, $P(2)$ holds.
We claim that $P(2x-1)$ holds, then $(2x-1)^4 \mod 8\equiv 1$ for any odd integer $n=2x-1$. So,
$$(2x-1)^4 = 16x^4-32x^3+24x^2-8x+1 \mod 8\equiv 1,$$
now consider
$$(2x+1)^4 = 16x^4+32x^3+24x^2+8x+1 = (16x^4-32x^3+24x^2-8x+1)+64x^3+16x,$$
then since
$$64x^3+16x \mod 8 = 8(8x^3+2x) \mod 8 \equiv 0,$$
we conclude that
$$(16x^4-32x^3+24x^2-8x+1)+64x^3+16x \mod 8 \equiv 1+0 = 1.$$
Thus, $P(2x+1)$ holds, and all odd integers $n$ hold.
Next, we claim that $P(2x)$ holds, then $(2x)^4 \mod 8\equiv 0$ for any even integer $n=2x$. So,
$$(2x)^4 \mod 8 = 16x^4 \mod 8 \equiv 0,$$
now consider
$$(2x+2)^4 = 16x^4+64x^3+96x^2+64x+16,$$
then
$$64x^3+96x^2+64x+16 \mod 8 = 8(8x^3+12x^2+8x+2) \mod 8 \equiv 0,$$
we conclude that
$$16x^4+64x^3+96x^2+64x+16 \mod 8 \equiv 0+0 = 0.$$
Thus, $P(2x)$ holds, and all even integers $n$ hold.
By mathematical induction, if $n=2x-1$ and $n=2x$ hold for the given statement, $n=2x+1$ and $n=2x+2$ also hold. Therefore, the statement holds for all $n \in \mathbb{N}$.
$\mathbb{QED}$
| The proof looks good, although as Doug M suggests, there are much easier ways to reach the same result. In particular, you're right to observe that you don't actually need the inductive hypothesis.
If you do want to apply induction, my advice is to start with a summary of your argument. Something like: "Let $P(n)$ be the claim. We will prove $P(1)$, $P(2)$, $P(2x-1)\implies P(2x+1)$, and $P(2x)\implies P(2x+2)$, completing a proof by induction."
| {
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"url": "https://math.stackexchange.com/questions/3552154",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Given a sequence $a_n =\frac{n^2 - 3n + (-1)^n}{3n^2 - 7n + 5}$, prove that $\lim_{n \to \infty} a_{n} = g$. The given Sequence is $a_n = \frac{n^2 - 3n + (-1)^n}{3n^2 - 7n + 5}$.
I showed that the Sequence $a_n$ converges towards a Value $g = \frac{1}{3}$.
How do I determine for each $\epsilon > 0$ an $n_0$ so that: $|a_n - g| < \epsilon \ \forall n > n_0$?
This is what I tried so far:
$\begin{array}{rcl}
\left|a_n - g\right| & < & \epsilon \\
\left|\frac{n^2 - 3n + (-1)^n}{3n^2 - 7n + 5} - \frac{1}{3}\right| & < & \epsilon \\
\left|\frac{3n^2 - 9n + 3 \cdot (-1)^n}{9n^2 - 21n + 15} - \frac{3n^2 - 7n + 5}{9n^2 - 21n + 15}\right| & < & \epsilon \\
\left|\frac{-2n - 5 + 3 \cdot (-1)^n}{9n^2 - 21n + 15}\right| & < & \epsilon \ \text{ Denominator > 0, Numerator < 0 } \forall n \in \mathbf{N} \\
\frac{2n + 5 - 3 \cdot (-1)^n}{9n^2 - 21n + 15} \leq \frac{2n + 5 + 3}{9n^2 - 21n + 15} = \frac{2n + 8}{9n^2 - 21n + 15} & < & \epsilon \\
\end{array}$
Are my Steps correct so far?
How do I proceed from here?
| Your work is correct. You can continue with more inequalities. For $n>8$, we have $2n+8<3n$ and $9n^2-21n+15>n^2$ since $8x^2-21x+15=8\left(x-\frac{21}{16}\right)^2+\frac{39}{32}>0$ for all $x\in\mathbb{R}$. Hence,
$$\frac{2n+8}{9n^2-21n+15}<\frac{3n}{n^2}=\frac{3}{n} $$
Now you just need to find $n_1$ such that when $n>n_1$, we have $\frac{3}{n}<\varepsilon$. This shouldn't be too hard. Since we also want to incorporate $n>8$, we choose $n_0=\max(n_1,8)$.
| {
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"source": "stackexchange",
"question_score": "2",
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Integrating $\frac{1}{\sqrt{2-\frac{1}{x^2}}}dx$ using two substitutions $z = x^2$ and $u = 2z-1$ An exercise asks to calculate $$\int_1^y\frac{1}{\sqrt{2-\frac{1}{x^2}}}dx$$
The solution manual states that with the two substitutions $z = x^2$ and then $u = 2z -1$ we get
$$\int_1^y\frac{1}{\sqrt{2-\frac{1}{x^2}}}dx = \int_1^{y^2}\frac{1}{2\sqrt{2z-1}}dz = \int_1^{2y^2-1}\frac{1}{4\sqrt{u}}du$$
I don't understand the first substitution. $\frac{dz}{dx} = 2x$, but I just don't get how the root in the denominator was transformed! Shouldn't $2x$ be a factor in the integrand for this substitution to work?
| $$z = x² \rightarrow dz = 2 x dx \rightarrow \frac{dz}{dx} = 2x$$
$$dx = \frac{dz}{2x} \rightarrow \int_1^y\frac{1}{\sqrt{2-\frac{1}{x^2}}}dx = \int_1^{y²}\frac{1}{\sqrt{2-\frac{1}{x^2}}} \frac{dz}{2x} = \int_1^{y²}\frac{1}{\sqrt{8x²-4}} dz = \int_1^{y²}\frac{1}{\sqrt{8z-4}} dz = \int_1^{y^2}\frac{1}{2\sqrt{2z-1}}dz$$
The top boundary changes because you are in the $z$ world in which $z = x²$ so $y \rightarrow y²$.
Then with $u= 2z-1$ I assume you've understood.
| {
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Find the relative maximum and relative minimum of the function $f(x) = 49x + \frac{4}{x}$ Find the relative maximum and relative minimum of the function $f(x) = 49x + \frac{4}{x}=49x+4x^{-1}$
Solution:
Step 1: Find the values of $x$ where $f'(x)=0$ and $f'(x)$ DNE.
$f'(x)=49-4x^{-2}=49-\frac{4}{x^2}$
We can see that $f'(x)$ DNE when $x=0$. Now lets find the values of $x$ that make $f'(x)=0$.
$f'(x)=49-4x^{-2}=49-\frac{4}{x^2}=0$
$\rightarrow x^2=\frac{4}{49}$
$\rightarrow x= \pm \sqrt{\frac{4}{49}}$
$\rightarrow x= \pm \frac{2}{7}$
Thus our critical points are $x=0, x= \frac{2}{7}, x= -\frac{2}{7}$
Step 2: Make a number line and plot the sign of $f'(x)$ to find where $f(x)$ is increasing and decreasing.
$f'(-3)= 49-\frac{4}{9}>0$
$f'(\frac{-1}{7})=49-\frac{4}{(\frac{-1}{7})^2}=49-4(49)=-3(49)<0$
$f'(\frac{1}{7})=49-\frac{4}{(\frac{1}{7})^2}=49-4(49)=-3(49)<0$
$f'(3)=49-\frac{4}{3^2} >0$
Thus $f(x)$ is increasing as $x$ increases towards $\frac{-2}{7}$ and then $f(x)$ is decreasing as $x$ increases towards $0$. Therefore $x=\frac{-2}{7}$ corresponds to a relative maximum.
Thus $f(x)$ is decreasing $x$ increases towards $\frac{2}{7}$ and then $f(x)$ is increases as $x$ increases towards $\infty$. Therefore $x=\frac{2}{7}$ corresponds to a relative minimum.
Step 3: find the corresponding $y$ values.
$f(\frac{2}{7})=28$
$f(\frac{-2}{7})=-28$
Therefore $(\frac{2}{7},28)$ is a relative minimum and $(\frac{-2}{7},-28)$ is a relative maximum.
This may be weird that our relative maximum is smaller than our relative minimum, but since these are "relative" maximum and minimum, they are only being compared to points on the graph very near to it, so this is okay.
| First note that $f(-x)=-f(x)$, so the problem is essentially reduced to $x\ge0$.
That can be solved without calculus: by the AM-GM inequality, $49x+\dfrac4x\ge2\sqrt{49\times4}=28,$
and equality holds when $x=\dfrac27$.
| {
"language": "en",
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What can be inferred from this determinant regarding a triangle?
If
$$\Delta =
\begin{array}{|ccc|}
\sin A & \sin B & \sin C \\
\cos A & \cos B & \cos C \\
\cos^3 A & \cos^3 B & \cos^3 C \\
\end{array} = 0$$
Where A, B, C are the angles of a triangle. What can we say about the triangle? (Is the triangle equilateral, isosceles or scalene)?
My Attempt:
If the triangle is isosceles, then either A=B or B=C or C=A. In this case, two of the rows of the determinant would be equal to one another and hence ∆ would be 0. Hence the triangle would be an isosceles triangle.
The Answer:
The answer key for this question states that we cannot infer anything about the given triangle.
Why is this so? Is the answer key wrong? Clearly, if the triangle is not isosceles and other cases exist, it would be a scalene triangle. Could someone explain the case in which a scalene triangle gives the above result?
| \begin{align}
\Delta&=
\sin\alpha \, \cos\beta \, \cos^3\gamma -\sin\alpha \, \cos\gamma \, \cos^3\beta
\\
&
-\cos\alpha \, \sin\beta \, \cos^3\gamma
+\cos\alpha \, \sin\gamma \, \cos^3\beta
\\
&+\cos^3\alpha \, \sin\beta \, \cos\gamma
-\cos^3\alpha \, \sin\gamma \, \cos\beta
=0
\tag{1}\label{1}
.
\end{align}
Substitutions
\begin{align}
\sin\alpha&=\frac a{2R},\quad\dots
,\\
\cos\alpha&=\frac{-a^2+b^2+c^2}{2bc},\quad\dots
\end{align}
result in
\begin{align}
\Delta=&
\frac{(a+b)(b+c)(c+a)}{8\,R\,a^3b^3c^3}
\\
&\times
(a+b+c)(-a+b+c)(a-b+c)(a+b-c)
\\
&\times
(a^2-b^2)(a^2-c^2)(b^2-c^2)
\\
&=
\frac{2S^2}{R\,a^3 b^3 c^3}\cdot(a^2-b^2)(a^2-c^2)(b^2-c^2)
=0
\tag{2}\label{2}
,
\end{align}
where $R$ and $S$ are the circumradius and the area of the triangle, respectively.
Obviously, excluding the case of the degenerate triangle for which $S=0$,
$\Delta$ can be zero iff at least one pair of the sides has the same length,
hence the non-degenerate triangle definitely must be isosceles, which also includes equilateral
as a special case.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3563878",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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How to solve matrix polynomial, that is $A^{15}$ $A$ is a matrix that suffice $A^2 = A - I$ ($I$ is identity matrix.)
Of course, $A^{15} = {(A^2)}^7A$
I think i have to find the elements of matrix A. I tried using completing the square, $$A^2 - A + I = 0$$
$${(A-I/2)}^2 + (3/4) I = 0$$
Which leads to imaginary number.
| $A$ is a zero of $x^2 -(x - 1)=x^2-x+1$, which you may recognize as the $6$-th cyclotomic polynomial. Therefore, $A^6 = I$ and so $A^{15} = A^{12} A^3 = A^3$. Now, $A^3 = A^2-A = -I$.
A systematic way that does not require bright ideas is to use polynomial division. The quotient is irrelevant; only the remainder matters:
$$
x^{15} = (x^{13} + x^{12} - x^{10} - x^9 + x^7 + x^6 - x^4 - x^3 + x + 1) (x^2 -(x - 1))-1
$$
Therefore, $A^{15} = -I$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Does $\int_{0}^{\infty} \frac{2x +3}{\sqrt {x^3 + 2x + 5}} \,dx $ converge? Does the following integral converge? I will post my solution, but I am unsure if it is true.
$$\int_{0}^{\infty} \frac{2x +3}{\sqrt {x^3 + 2x + 5}} \,dx $$
My solution
Let
$$ g(x) = \frac{2x}{\sqrt {x^3}}$$
$$ f(x) = \frac{2x +3}{\sqrt {x^3 + 2x + 5}} $$
Then $$\lim_{k \to \infty} \frac{f(x)}{g(x)} = 1$$
Therefore whatever one does, so does the other.
$$ \int_{0}^{\infty} g(x) = \int_{0}^{\infty} \frac{2}{\sqrt {x}} = +\infty $$
Therefore g(x) diverges, thus
$$\int_{0}^{\infty} f(x) = \int_{0}^{\infty} \frac{2x +3}{\sqrt {x^3 + 2x + 5}} \,dx = + \infty$$
diverges too
| $$\int_{3}^{\infty} \frac{2x +3}{\sqrt {x^3 + 2x + 5}} \,dx > \int_{3}^{\infty} \frac{x}{\sqrt {2x^3 }} \,dx= \bigg| \sqrt {2x} \bigg|_{3}^{\infty}=\infty$$
You're correct. The integral diverges.
| {
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"timestamp": "2023-03-29T00:00:00",
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Area of triangle in a circle. [Edexcel Specimen Papers Set 2, Paper 1H Q22]
The line $l$ is a tangent to the circle
$^2 + ^2 = 40$ at the point $A$. $A$ is the point $(2,6)$.
The line $l$ crosses the $x$-axis at the point $P$.
Work out the area of triangle $OAP$.
Any help is appreciated.Thank you.
I have attempted this - but I get the result $48$ (2sf) when the answer is $60$ units$^2$.
| The slope of the radius from $O(0,0)$ to $A(2,6)$ is $\frac{6}{2}=3$. The perpendicular slope of the tangent line at $(2,6)$ is the opposite reciprocal of the slope of $\overline{OA}$. Therefore the slope of $\overline{AP}=-\frac{1}{3}$.
The equation of the tangent line is found by using the slope and the coordinate $A(2,6)$.
$$y-6=-\frac{1}{3}(x-2)$$
Plug in $y=0$ to find x-intercept (Point $P$):
\begin{align*}
0-6=-\frac{1}{3}(x-2)&\implies -6=-\frac{1}{3}(x-2)\\
&\implies (-3)(-6)=(x-2)\\
&\implies 18=x-2\\
&\implies x=20\\
\end{align*}
Therefore, point $P$ on the triangle is located at $P(20,0)$.
Take base of triangle to be $\overline{OP}=20~\mbox{units}$ and height of $\triangle{OAP}=6~\mbox{units}$.
Apply area formula: $A=\frac{1}{2}bh=\frac{1}{2}(20)(6)=60~\mbox{units}^2$
| {
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how to prove $x^2\ln{x}+x+e^x-3x^2>0$ Let $x>0$; show that:
$$f(x)=x^2\ln{x}+x+e^x-3x^2>0$$
It seem this inequality
$$f'(x)=2x\ln{x}+1+e^x-5x$$
$$f''(x)=2\ln{x}+e^x-3$$
$$f'''(x)=\dfrac{2}{x}+e^x>0$$
| We need to prove that $$\ln{x}+\frac{1}{x}+\frac{e^x}{x^2}-3>0$$ for $x>0$.
But $$\left(\ln{x}+\frac{1}{x}+\frac{e^x}{x^2}-3\right)'=\frac{x^2-x+e^x(x-2)}{x^3}$$ increases for $x>0$ because easy to show that
$$\left(\frac{x^2-x+e^x(x-2)}{x^3}\right)'=\frac{(x^2-4x+6)e^x-(x^2-2x)}{x^4}>0,$$
which says that the minimum occurs for the unique positive root of the equation:
$$x^2-x+e^x(x-2)=0.$$
The inequality $(x^2-4x+6)e^x-(x^2-2x)>0$ we can prove by the following way:
We need to prove that:
$$e^x>\frac{x^2-2x}{x^2-4x+6},$$ which is true because
$$e^x-\frac{x^2-2x}{x^2-4x+6}>1+x-\frac{x^2-2x}{x^2-4x+6}=\frac{x^3-4x^2+4x+6}{x^2-4x+6}>0.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3582813",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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The circumradius of an isosceles triangle ABC is four times as that of inradius and A=B condition The circumradius of an isosceles triangle ABC is four times as that of inradius of the triangle, if A = B. Then
(1) $8 cos^2A – 8cosA + 1 = 0$
(2) $4 cos^2A – 10cosA + 1 = 0$
(3) $cos^2A – cosA – 3 = 0$
(4) $cos^2A – cosA – 8 = 0$
I am trying to use the following approach
$R=\frac{abc}{4A}$, where R is Circum-radius and r is inradius and A is the area of inscribed triangle
$A=rs$ ahre r=inradius and s= semiperimeter
a=a, b=a{ISOSCELES TRIANGLE} & c=c
$A=\frac{c}{2} \sqrt{(a^2-\frac{c^2}{4})}$
$s=\frac{2a+c}{2}$
$\frac{R}{r}=4$
$abcs=16A^2$
$R=\frac{abc}{4A}=\frac{a^2c}{4A}$
$16A^2=c^2(4a^2-c^2)$
$abc(\frac{2a+c}{2})=c^2(4a^2-c^2)$
$a^2(\frac{2a+c}{2})=c(4a^2-c^2)$
$2a^3+a^2c=8a^2c-2c^3$
$7a^2c-2a^3-2c^3=0$
2A+C=180
C=180-2A
sinC=sin2A
$\frac{sinA}{a}=\frac{sinC}{c}$
$c=\frac{asinC}{sinA}=2acosA$
$7a^2*2acosA-2a^3-2(2acosA)^3=0$
$7cosA-1-8cos^3A=0$
Still not able to get the answer , I presume that I am making a mistake
| Hint:
Using this
$$r=R(\cos A+\cos B+\cos C-1)$$
We have
$A=B\implies \cos B=\cos A,\cos C=\cos(\pi-A-A)=-\cos2A$
and $R=4r$
$\implies r=(4r)(\cos A+\cos A-\cos2A-1)$
Hope you can take it home from here?
| {
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If $A, B, C$ are angles of $\Delta ABC$ and $\sin (A-\pi /4) \sin (B-\pi/4) \sin (C-\pi/4)=\frac{1}{2\sqrt 2}$...
If $A, B, C$ are angles of $\Delta ABC$ and $\sin (A-\pi /4) \sin (B-\pi/4) \sin (C-\pi/4)=\frac{1}{2\sqrt 2}$, then prove that $\sum \tan A \tan B=\sum \tan A$
Solving the given equation, we get$$(\sin A-\cos A)(\sin B -\cos B)(\sin C -\cos C)=1$$
$$(\tan A-1)(\tan B-1)(\tan C-1)=\sec A \sec B \sec C$$
$$\sum \tan A -\sum \tan A \tan B +\sum \tan A-1=\sec A \sec B \sec C$$
How should I proceed?
| From the given
\begin{align}0=&\frac{1}{2\sqrt 2}- \sin (A-\pi /4) \sin (B-\pi/4) \sin (C-\pi/4)\\
=&\frac{1}{2\sqrt 2}-\frac12\left[ \cos(A-B)-\cos(A+B-\frac\pi2)\right] \sin (C-\pi/4)\\
=&\frac{1}{2\sqrt 2}[ 1- (\cos(A-B)-\sin C) (\sin C - \cos C)]
\end{align}
Note that $\cos(A-B)\le 1$ and
\begin{align}
0 \ge & \frac{1}{2\sqrt 2}[1-( 1 -\sin C ) (\sin C - \cos C)]\\
=& \frac{1}{2\sqrt 2}[(1+\cos C) - \sin C (1+\cos C - \sin C) ]\\
=& \frac{1}{2\sqrt 2}[2\cos^2 \frac C2- \sin C (2\cos^2 \frac C2 - 2\sin \frac C2\cos\frac C2)] \\
=&\frac{1}{\sqrt 2}\cos^2 \frac C2\left[1- 2\sin\frac C2(\cos \frac C2 - \sin \frac C2)\right] \\
=& \frac{1}{\sqrt 2}\cos^2 \frac C2(2- \sin C - \cos C )
\end{align}
which leads to
$\cos\frac C2=0$, i.e.
$C = \pi$ and $A=B = 0$. Thus,
$$
\tan A\tan B+\tan B\tan C+\tan C\tan A = \tan A+\tan B+\tan C =0
$$
| {
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Trig substitution for $\sqrt{9-x^2}$ I have an integral that trig substitution could be used to simplify.
$$ \int\frac{x^3dx}{\sqrt{9-x^2}} $$
The first step is where I'm not certain I have it correct. I know that, say, $\sin \theta = \sqrt{1-cos^2 \theta}$, but is it correct in this case $3\sin \theta = \sqrt{9 - (3\cos \theta)^2}$?
Setting then $x = 3\cos \theta; dx = -3\sin \theta d\theta$
$$-\int \frac{(3\cos\theta)^3}{3\sin\theta}3\sin\theta d\theta$$
$$-27\int\cos^3\theta d\theta$$
$$-27\int(1-\sin ^2\theta)\cos \theta d\theta$$
Substituting again, $u=\sin \theta; du=\cos \theta d\theta$
$$-27\int(1-u^2)du $$
$$-27u + 9u^3 + C$$
$$-27\sin \theta + 9 \sin^3 \theta + C$$
$$-9\sqrt{9-x^2} + 3\sin\theta\cdot 3\sin\theta\cdot \sin \theta + C$$
$$-9\sqrt{9-x^2} + (\sqrt{9-x^2})^2 \cdot \frac{\sqrt{9-x^2}}{3} + C$$
$$-9\sqrt{9-x^2} + \frac{1}{3}(9-x^2)(9-x^2)^{\frac{1}{2}} + C$$
$$-9\sqrt{9-x^2} + \frac{1}{3}(9-x^2)^\frac{3}{2} + C $$
I guess I have more doubts that I've done the algebra correctly than the substitution, but in any case I'm not getting the correct answer. Have I calculated correctly? Is the answer simplified completely?
EDIT
Answer needed to be simplified further:
$$-9\sqrt{9-x^2} + \frac{1}{3}(\sqrt{9-x^2}^2 \sqrt{9-x^2}) + C$$
$$-9\sqrt{9-x^2} + \frac{1}{3}((9-x^2)\sqrt{9-x^2}) + C$$
$$\sqrt{9-x^2} \left (-9 + \frac{1}{3}(9-x^2) \right ) + C$$
$$\sqrt{9-x^2} \left (-6 - \frac{x^2}{3} \right ) + C$$
$$ \bbox[5px,border:2px solid red]
{
- \left ( 6+ \frac{x^2}{3} \right ) \sqrt{9-x^2}
}
$$
This is the answer the assignment was looking for.
| Maybe the following hint, based on Differential binomial, seems to be the same as @Toby's. However, you focused just on a trig substitution. You can take $$(9-x^2)=t^2,$$ and simplify the integral...
| {
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Proving $a^2 + b^2 + c^2 \geqslant ab + bc + ca$
Let $a, b, c \in \mathbb{R}$. Show that $a^2 + b^2 + c^2 \geqslant ab + bc + ca$.
My reasoning went as follows and I would like to know if it's correct.
$a^2 + b^2 + c^2 \geqslant ab + bc + ca$
$\Leftrightarrow a^2 + b^2 + c^2 -ab - bc - ca \geqslant 0$
$\Leftrightarrow 2a^2 + 2b^2 + 2c^2 -2ab - 2bc - 2ca \geqslant 0$
$\Leftrightarrow a^2 + a^2 + b^2 + b^2 + c^2 + c^2 -2ab - 2bc - 2ca \geqslant 0$
$\Leftrightarrow (a^2 -2ab + b^2) + (b^2 - 2bc + c^2)+ (c^2 - 2ca + a^2) \geqslant 0$
$\Leftrightarrow (a -b)^2 + (b-c)^2 + (c-a)^2 \geqslant 0$.
And the last inequality holds since squares are non-negative.
I've seen couple of different ways of doing this and they seem to divide by $2$ and I didn't really understand it so instead can I just use the fact that $2a^2 = a^2+a^2$?
| Your proof is correct and so is the identity $2a^2=a^2+a^2$. Alternatively, if you want to divide by $2$, then you can write
$a^2 + b^2 + c^2 \geqslant ab + bc + ca$
$\Leftrightarrow\frac {a^2 + b^2}{2} + \frac {b^2 + c^2}{2} + \frac {a^2 + c^2}{2}\geqslant ab + bc + ca$
$\Leftrightarrow\frac {a^2 + b^2}{2} + \frac {b^2 + c^2}{2} + \frac {a^2 + c^2}{2}-ab- bc - ca\geqslant 0$
$\Leftrightarrow\frac {a^2 -2ab +b^2}{2} + \frac {b^2 -2bc +c^2}{2} + \frac {a^2 -2ac +c^2}{2}\geqslant 0$
$\Leftrightarrow(a-b)^2 + (b-c)^2 + (c-a)^2 \geqslant 0$
| {
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"source": "stackexchange",
"question_score": "3",
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If $(x^2-5x+4)(y^2+y+1)<2y$ for all real $y$, then $x$ belongs to the interval $(2,b)$, then $b$ can be? If $(x^2-5x+4)(y^2+y+1)<2y$ for all real $y$, then $x$ belongs to the interval $(2,b)$, then $b$ can be?
$$y^2(x^2-5x+4)+y(x^2-5x+2)+(x^2-5x+4)<0$$
As it is true for all real y, hence $D<0$
$$(x^2-5x+2)^2-4(x^2-5x+4)^2<0$$
Let $x^2-5x=u$
$$(u+2)^2-4(u+4)^2<0$$
$$u^2+4+4u-4(u^2+16+8u)<0$$
$$-3u^2-28u-60<0$$
$$3u^2+28u+60>0$$
$$3u^2+18u+10u+60>0$$
$$(3u+10)(u+6)>0$$
Back substituting u
$$(3x^2-15x+10)(x^2-5x+6)>0$$
$$\left(x-\dfrac{15+\sqrt{225-120}}{6}\right)\left(x-\dfrac{15-\sqrt{225-120}}{6}\right)(x-2)(x-3)>0$$
$$x\in\left(-\infty,\dfrac{15+\sqrt{105}}{6}\right)\cup(2,3)\cup\left(\dfrac{15+\sqrt{105}}{6},\infty\right)$$
But in the question it is given $x\in(2,b)$, what mistake am I making here.
| From $$y^2+y+1=(y+\tfrac12)^2+\tfrac34>0,$$
we know that we may divide by $y^2+y+1$.
What can we say about $\frac{2y}{y^2+y+1} $?
For $y\ge0$, this will be positive, but what about negative $y$?
We have $$y^2+y+1=(y-1)^2-y\ge-y$$
with equality iff $y=-1$. Hence $$\frac{2y}{y^2+y+1}\ge-2 $$with equality iff $y=-1$.
So what you really want is
$$x^2-5x+4<-2. $$
| {
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Compute $\left[\begin{smallmatrix}1-a & a \\ b & 1-b\end{smallmatrix}\right]^n$ Compute $\begin{bmatrix}1-a & a \\ b & 1-b\end{bmatrix}^n$, where the power of $n\in\mathbb N$ denotes multiplying the matrix by itself $n$ times; $a,b\in[0,1]$.
Edit:
I considered using induction, computed the desired matrix:
$$\begin{bmatrix}(1-a)^2+ab & a(2-a-b) \\ b(2-a-b) & (1-b)^2+ab\end{bmatrix}$$ and $$\begin{bmatrix}(1-a)^3+ab(1-a)+ba(2-a-b) & a(1-a)^2+a^2b+a^2(2-a-b) \\ (1-a)b(2-a-b)+b(1-b)^2+ab^2 & ab(2-a-b)+(1-b)^3+ab(1-b)\end{bmatrix}$$ for $n=2$ and $3$, respectively. However, I fail to see a pattern here that can be used as an induction hypothesis.
| Diagonalize the matrix first.
Let $A$ be your matrix, and you get $A = S\cdot J \cdot S^{-1}$, where
$S = \begin{bmatrix}
1 & -a/b \\
1 & 1 \\
\end{bmatrix}$
and
$J = \begin{bmatrix}
1 & 0 \\
0 & 1-a-b \\
\end{bmatrix}$.
Then calculate $A^n = (S\cdot J \cdot S^{-1})^n = S \cdot J^n \cdot S^{-1}$ (just write things down and you will see this expression holds).
So the problem reduces to computing $J^n$, which is automatic since it's a diagonal matrix.
| {
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One of my old inequality (very sharp) I'm proud to present one of my old inequality that I can't solve :
Let $a,b,c>0$ such that $a+b+c=1$ and $a\ge b \geq c $ then we have :$$\sqrt{\frac{a}{a^a+b^b}}+\sqrt{\frac{b}{b^b+c^c}}+\sqrt{\frac{c}{c^c+a^a}}\geq \sqrt{\frac{a}{a^a+c^c}}+\sqrt{\frac{c}{c^c+b^b}}+\sqrt{\frac{b}{b^b+a^a}}$$
The equality case is obvious .
The first reflex for me is to use rearrangement inequality but it gives just a little part of the inequality .The second reflex is to use power series of $x^x$ at $x=1$ . We get a polynomial and it's useful . Furthermore I have tried to denested the radical unsuccessfully.Finally my goal was to use with all of this the Buffalo's way but it's far .
Some remarks :
The inequality is very sharp because I think that we have $LHS-RHS\leq 10^{-2}$
Each coefficient under the root are $1$ behind the variable so maybe it's easier.
We have also :
Let $a,b,c>0$ such that $a+b+c=1$ and $a\ge b \geq c $ then we have :$$\frac{a}{a^a+b^b}+\frac{b}{b^b+c^c}+\frac{c}{c^c+a^a}\geq \frac{a}{a^a+c^c}+\frac{c}{c^c+b^b}+\frac{b}{b^b+a^a}$$
So the idea will be to put a power $2$ to each side and use rearrangement inequality to get other cases(maybe I have not checked that).
My last idea is in this link
If you have nice idea you are welcome .
Thanks a lot .
| We will apply Ji Chen's Symmetric Function Theorem for $n=3$
(see https://artofproblemsolving.com/community/c6h194103p1065812):
Symmetric Function Theorem: Let $d\in (0,1)$. Let $x, y, z, u, v, w$ be non-negative real numbers satisfying
$$x+y+z \ge u+v+w, \quad
xy+yz+zx \ge uv+vw+wu, \quad
xyz \ge uvw.$$
Then $x^d + y^d+z^d \ge u^d + v^d+w^d$.
Denote the six terms inside the root sign by $X, Y, Z, U, V, W$ respectively.
Clearly, $XYZ = UVW$.
We need to prove that $X+Y+Z\ge U+V+W$ and $XY+YZ+ZX \ge UV+VW+WU$.
1) To prove $X+Y+Z\ge U+V+W$, it suffices to prove that
$$\frac{a-b}{a^a + b^b} + \frac{b-c}{b^b+c^c} \ge \frac{a-c}{c^c + a^a}$$
or
$$\frac{a-b}{a^a + b^b} - \frac{a-b}{c^c + a^a}
\ge \frac{b-c}{c^c + a^a} - \frac{b-c}{b^b+c^c}$$
or
$$\frac{(a-b)(c^c - b^b)}{(a^a + b^b)(c^c + a^a)}
\ge \frac{(b-c)(b^b-a^a)}{(c^c+a^a)(b^b+c^c)}$$
or
$$(a-b)(c^{2c} - b^{2b}) \ge (b-c)(b^{2b} - a^{2a})$$
or
$$\frac{b^{2b}-c^{2c}}{b-c} \le \frac{a^{2a}-b^{2b}}{a-b}.$$
This inequality is true since $a\ge b \ge c > 0$ and $x\mapsto x^{2x}$ is convex on $x > 0$.
2) To prove $XY+YZ+ZX \ge UV+VW+WU$, it suffices to prove that
$$ab(a^a - b^b) + bc(b^b - c^c) + ca(c^c - a^a) \ge 0$$
or
$$a^{a+1}(b-c) + c^{c+1}(a-b) \ge b^{b+1}(a-c)$$
or
$$(a^{a+1}-b^{b+1})(b-c) \ge (b^{b+1} - c^{c+1})(a-b)$$
or
$$\frac{a^{a+1}-b^{b+1}}{a-b} \ge \frac{b^{b+1} - c^{c+1}}{b-c}.\tag{1}$$
If $a\ge b\ge c > \mathrm{e}^{-3/2}$, the inequality is true since $x\mapsto x^{x+1}$ is convex on $x > \mathrm{e}^{-3/2} \approx 0.2231301602$.
If $c \le \mathrm{e}^{-3/2}$, (1) is written as
$$\frac{b^b - c^c}{b-c} c \le \frac{a^a-b^b}{a-b} a.\tag{2}$$
Since $x\mapsto x^x$ is convex on $x > 0$, we have
$$\frac{b^b - c^c}{b-c} \le \frac{a^a-b^b}{a-b}.$$
Thus, it suffices to prove that $a^a \ge b^b$.
It is true. Omitted.
| {
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"timestamp": "2023-03-29T00:00:00",
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Maxima and minima of $f(x) = \frac{\sqrt{x}(x-5)^2}{4}.$
Determine the local minima and maxima of the function $f:[0, \infty) \to \mathbb{R}$ $$f(x) = \frac{\sqrt{x}(x-5)^2}{4}.$$
Does $f$ have a maximum or minimum?
Computing the derivative gets me here:
$f'(x) = \frac14(2(x-5)\sqrt{x}+\frac{(x-5)^2}{2\sqrt{x}}),$
but this is not an easy function to find the roots and then determine the maxima and minima of $f$. Is there an alternative approach here, or how should I proceed?
| Hint:
By factoring, using a common denominator and combining terms, you get
$$\begin{equation}\begin{aligned}
f'(x) & = \frac{1}{4}\left(2(x-5)\sqrt{x}+\frac{(x-5)^2}{2\sqrt{x}}\right) \\
& = \frac{x-5}{4}\left(2\sqrt{x}+\frac{x-5}{2\sqrt{x}}\right) \\
& = \frac{x-5}{4}\left(\frac{4x}{2\sqrt{x}}+\frac{x-5}{2\sqrt{x}}\right) \\
& = \frac{x-5}{4}\left(\frac{4x + x - 5}{2\sqrt{x}}\right) \\
& = \frac{x-5}{4}\left(\frac{5x - 5}{2\sqrt{x}}\right) \\
& = \frac{5(x-5)(x-1)}{8\sqrt{x}}
\end{aligned}\end{equation}\tag{1}\label{eq1A}$$
Can you finish the rest yourself?
| {
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On the integral $\int_0^{\sqrt{2}/2} \frac{\arctan \sqrt{1-2t^2}}{1+t^2} \, \mathrm{d}t$ I'm having a difficult time evaluating the integral
$$\mathcal{J} = \int_0^{\sqrt{2}/2} \frac{\arctan \sqrt{1-2t^2}}{1+t^2} \, \mathrm{d}t$$
This is integral arose after simplifying the integral $\displaystyle \int_{0}^{\pi/4 } \arctan \sqrt{\frac{1-\tan^2 x}{2}} \, \mathrm{d}x$;
\begin{align*}
\require{cancel.js}
\int_{0}^{\pi/4} \arctan \sqrt{\frac{1-\tan^2 t}{2}}\, \mathrm{d}t &\overset{1-\tan^2 t \mapsto 2t^2}{=\! =\! =\! =\! =\! =\!=\!=\!} \int_{0}^{\sqrt{2}/2} \frac{t \arctan t}{\sqrt{1-2t^2} \left ( 1-t^2 \right )} \, \mathrm{d}t \\
&=\cancelto{0}{\left [ - \arctan \sqrt{1-2t^2} \arctan t \right ]_0^{\sqrt{2}/2}} + \int_{0}^{\sqrt{2}/2} \frac{\arctan \sqrt{1-2t^2}}{1+t^2} \, \mathrm{d}t
\end{align*}
My main guess is that differentiation under the integral sign is the way to go here. Any ideas?
| Continue with\begin{align}
& \int_{0}^{\pi/4 } \tan^{-1} \sqrt{\frac{1-\tan^2 x}{2}}dx
\>\>\>\>\>\>\> \tan x =\sin t\\
=& \int_0^{\pi/2}\frac{\cos t}{2-\cos^2t}\ \tan^{-1}\frac{\cos t}{\sqrt2} \ dt\\
=& \int_0^{\pi/2}\int_0^{\pi/4} \frac{\sqrt2 \cos^2t}{(2-\cos^2t)(2\cos^2y+\sin^2 y\cos^2t)}dy\ dt\\
=& \ \frac\pi2\int_0^{\pi/4} \bigg(1-\frac{\cos y}{\sqrt{2-\sin^2y}}\bigg)dy=\frac\pi2\bigg(\frac\pi{4}-\frac\pi6\bigg)=\frac{\pi^2}{24}
\end{align}
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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In triangle $ABC$, $AA_1$, $BB_1$, $CC_1$ divide sides in ratio of $1: 2$ and meet at $M$, $K$, $L$. Find area relation of $KLM$ and $ABC$
Points $A_1$, $B_1$, $C_1$ divide sides $BC$, $CA$, $AB$ equilateral triangle $ABC$ in a ratio of $1: 2$.
The line segments $AA_1$, $BB_1$, $CC_1$ determine the triangle $KLM$.
Is the triangle $KLM$ also an equilateral side? In what relation are the area of triangle $KLM$ the area of triangle$ ABC$?
My attempt:
I can see that $KLM$ is an equilateral triangle. But, why the fraction $\frac{1}{7}$ is the area ratio of the triangle $KLM$ to the triangle $ABC$?
| According to Routh’s theorem
the ratio of the areas in general case is
\begin{align}
\frac{S_{KLM}}{S_{ABC}}
&=
\frac{(xyz-1)^2}{(xy+y+1)(yz+z+1)(zx+x+1)}
.
\end{align}
The question is a special case, for which
$x=y=z=\tfrac12$, so
\begin{align}
\frac{S_{KLM}}{S_{ABC}}
&=
\frac{(\tfrac18-1)^2}{(\tfrac14+\tfrac12+1)^3}
=\tfrac17
.
\end{align}
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Evaluate discriminant of $x^3 +px+q$ given its roots? Consider the cubic polynomial $x^3 +px+q$ with $p, q ∈ \mathbb{Q}$. Suppose $α_1, α_2, α_3$ are the roots and thus $$x^3 + px + q = (x − α_1)(x − α_2)(x − α_3).$$
Let $D = (α_1 − α_2)^2(α_1 − α_3)^2(α_2 − α_3)^2$.
(a) Express $p$ and $q$ in terms of $α_1, α_2$ and $α_3$.
(b) Show that $D = −4p^3 − 27q^2$
I have that $p= α_1α_2 + α_1α_3 + α_2α_3 $ and $ q= -α_1α_2α_3$
Is there a more succinct way to prove part (b) without expanding out a ridiculously large polynomial by brute force?
| Per $α_1 + α_2 +α_3=0$ and $α_1 α_2 α_3=-q$,
$$(α_1 − α_2)^2 = (α_1 + α_2)^2 - 4α_1 α_2=α_3^2+\frac{4q}{α_3}
=\frac p{α_3}\left(\frac{3q}p-α_3\right)$$
Likewise, $(α_2 − α_3)^2 = \frac p{α_1}\left(\frac{3q}p-α_1\right)$ and $
(α_3 − α_1)^2 = \frac p{α_2}\left(\frac{3q}p-α_2\right)$. Then,
$$\begin{align}
D
& = (α_1 − α_2)^2 (α_2 − α_3)^2 (α_3 − α_1)^2\\
& = \frac{p^3}{α_1 α_2 α_3} \left(\frac{3q}p-α_3\right)\left(\frac{3q}p-α_1\right)\left(\frac{3q}p-α_2\right) \\
& =- \frac{p^3}q \left(\left(\frac{3q}p\right)^3 + p\left(\frac{3q}p\right)+q \right) \\
& =- 27q^2-4p^3 \\
\end{align}$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Tokyo university entrance exam - mathematics No.5 - (2)
Circle $C$ which radious is 1 and center is origin on $xy$-plane, We could imagine a cone $S$ that bottom is $C$ and top is $(0,0,2)$. Point $A$ is $(1,0,2)$ and point $P$ is on $C$, $T$ is a cone that $\overline{AP}$'s set
Calculate the volume of $R$ which shares $S$ and $T$
Figure of setup
I think both $S$ and $T$'s section with $z=2-t, (0<t<2)$ is circle and they meet when $\frac{2}{3}<t<2$.
But I can't calculate area of circular sector...
I hear that answer is $\frac{2\pi}{3}+\frac{2}{3}$
My first try:
Section with $z=2-t$ and $S$ is $D_2:x^2+y^2=\frac{t^2}{4} $,
$T$ is $D_1:(x-\frac{2-t}{2})^2+y^2=\frac{t^2}{4} $
Section with $D_1, D_2$ is $E_1,E_2$
Than $\angle E_1OE_2=2\theta$
$\cos\theta=\frac{(2-t)/4}{t/2}$
$\theta=\cos^{-1}(\frac{(2-t)/4}{t/2})$
So area of circular sector is $\frac{t^2}{4}\cos^{-1}(\frac{2-t}{2t})$ and triangle $E_1OE_2$ is $\frac{(2 - t) \sqrt{3 t^2 + 4 t - 4}}{16}$...
I can't move forward after here
| I am going to assume that $$S = \{ (x,y,z) \in \mathbb R^3 : x^2 + y^2 \le (z/2 - 1)^2 \cap 0 \le z \le 2 \}$$ and $$T = \{ (x,y,z) \in \mathbb R^3 : (x-z/2)^2 + y^2 \le (z/2-1)^2 \cap 0 \le z \le 2 \}$$ and you want to find the common volume of intersection of $S$ and $T$. It is immediately obvious that for a given $z_0 \in [0,2]$, the two cones intersect the plane $z = z_0$ in two disks: $$S(z_0) : x^2 + y^2 \le (z_0/2 - 1)^2$$ and $$T(z_0) : (x - z_0/2)^2 + y^2 \le (z_0/2 - 1)^2.$$ Notably, the radii of the two disks are equal. We can readily solve for the point of intersection of the boundary circles: $$x^2 = (x - z_0/2)^2 = x^2 - z_0 x + \frac{z_0^2}{4},$$ hence $x = z_0/4$, and $$y = \pm\sqrt{(z_0/2-1)^2 - (z_0/4)^2} = \pm \sqrt{\frac{3z_0^2}{16} - z_0 + 1}.$$ On $z \in [0,2]$, we find that $y \in \mathbb R$ if and only if $0 \le z_0 \le 4/3$. The area of the intersection of the disks $S(z_0) \cap T(z_0)$ is given by $$A(z_0) = 2(z_0/2 - 1)^2 \tan^{-1} \frac{\sqrt{\frac{3z_0^2}{16} - z_0 + 1}}{z_0/4} - 2 \cdot \frac{z_0}{4} \sqrt{\frac{3z_0^2}{16} - z_0 + 1}.$$ The desired volume is then given by $$V = \int_{z=0}^{4/3} A(z) \, dz = \frac{2\pi}{3} - \frac{4}{27} \left(6 + \sqrt{3} \log (2 - \sqrt{3}) \right).$$
| {
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Find the subgroup of $S_4$ generated by $(1,2,4)$. What is the order of $(1,2,4)$? I am literally just being introduced to group theory this is all new. I know that the permutations that arise are $(4,1),(1,2),(2,4),(1,2,4)$ This is the same as saying that $4$ goes to $1$ then $1$ goes to $2$ and then $2$ goes to $4$ and then we have the trivial case which is itself $(1,2,3,4)$ is this what I am supposed to show? Isn't the order just the number of subgroups?
| I would like to explain very simply.
$$\sigma=(1,2,4)=\left( \begin{array}{cccc} 1 & 2 & 3 &4 \\
2 &4&3&1 \end{array} \right)$$$$$$then we must calculate $\sigma, \sigma^2,...$ until we reach $$e=\left( \begin{array}{cccc} 1 & 2 & 3 &4 \\
1 &2&3&4 \end{array} \right)=(1) $$$$not $$$$(1,2,3,4)=\left( \begin{array}{cccc} 1 & 2 & 3 &4 \\
2 &3&4&1 \end{array} \right)$$$$$$ $$\sigma=(1,2,4) , \sigma^2=\left( \begin{array}{cccc} 1 & 2 & 3 &4 \\
2 &4&3&1 \end{array} \right)\left( \begin{array}{cccc} 1 & 2 & 3 &4 \\
2 &4&3&1 \end{array} \right)=\left( \begin{array}{cccc} 1 & 2 & 3 &4 \\
4 &1&3&2 \end{array} \right)=(1,4,2)$$$$ $$ $$\sigma^3=\sigma.\sigma^2=\left( \begin{array}{cccc} 1 & 2 & 3 &4 \\
2 &4&3&1 \end{array} \right)\left( \begin{array}{cccc} 1 & 2 & 3 &4 \\
4 &1&3&2 \end{array} \right)=\left( \begin{array}{cccc} 1 & 2 & 3 &4 \\
1 &2&3&4 \end{array} \right)=(1) $$ $$ $$ so the subgroup generated with $\sigma=(1,2,4)$ is: $$$$ $$\left\{ \begin{array}{c}\sigma=(1,2,4),\sigma^2=(1,4,2),\sigma^3=(1)\end{array}\right\}.$$ $$$$ the order of $(1,2,4)$ is the number of elements of subgroup generated by $(1,2,4)$ so $O(\sigma)=3$
| {
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Lemniscate - two parametrizations I have a Lemniscate defined as $$ \|{F_{1}-Z}||^{2} ||F_{2}-Z||^{2}=1 $$ for $ F_1 = (-1,0), F_2=(1,0)$ in $\mathbb{R}^2$
And I should find the parametric equations using these two parametrizations
a) $x=r \cos(\phi)$, $y=r\sin(\phi)$
b) $x=x, y=x \sin(t)$
I understand that the result should be (right?) $$(x,y)=\begin{pmatrix}\frac{a \cos(\phi)}{1+\sin^2(\phi)},&\frac{a\sin(\phi)\cos(\phi)}{1+\sin^2(\phi)} \end{pmatrix} $$
So far I have managed to start from the definition
$$\begin{Vmatrix} {-1-x}\\{-y} \end{Vmatrix}^2 \begin{Vmatrix} {1-x}\\{-y} \end{Vmatrix}^2=1$$
$$(1+2x+x^2+y^2)(1-2x+x^2+y^2)=1$$
$$(x^2+y^2)^2-2(x^2-y^2)=0$$
then pluged in the first parametrization and got $$r^2=2(\cos^2(\phi)-\sin^2(\phi))=2\cos(2\phi)$$
And dont really know what to do next, to get the form that is written above. Didnt really try much with the second parametrization so far.
| A bit of algebra gives
$$
\left(y^2+x^2+1\right)^2=4x^2+1\tag1
$$
Applying $x=r\cos(\theta)$ and $y=r\sin(\theta)$, we have
$$
\left(r^2+1\right)^2=4r^2\cos^2(\theta)+1\tag2
$$
which has the solutions $r=0$ and
$$
\bbox[5px,border:2px solid #C0A000]{r^2=2\cos(2\theta)}\tag3
$$
Setting $y=x\sin(t)$ in $(1)$ gives
$$
\left(x^2\sin^2(t)+x^2+1\right)^2=4x^2+1\tag4
$$
which simplifies to
$$
x^4\left(\sin^2(t)+1\right)^2=2x^2\cos^2(t)\tag5
$$
and has the solutions $(x,y)=(0,0)$ and
$$
\bbox[5px,border:2px solid #C0A000]{(x,y)=\left(\frac{\sqrt2\cos(t)}{\sin^2(t)+1},\frac{\sqrt2\sin(t)\cos(t)}{\sin^2(t)+1}\right)}\tag6
$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Matrix rotation in space Determine the matrix of the linear transformation $F$ in the space defined by $v$ being projected onto the plane through the origin with the normal $(1,2,5)$ and then rotating $180^{\circ}$ around the vector
$e = (1,1,1)$.
Not showing all the steps, but the projected matrix should be
\begin{align*}
\frac{1}{30}\begin{pmatrix}29&-2&-5\\-2&26&-10\\-5&-10&5\end{pmatrix}
\end{align*}
How would I go about rotating this $180^{\circ}$?
| One quick answer is as follows:
The matrix of a projection onto the plane with normal $v$ is given by $I - \frac{vv^T}{v^Tv}$. The matrix of a $180^\circ$ degree rotation in the plane with normal vector $v$ is given by $2 \frac{vv^T}{v^Tv} - I$. So, the matrix $P$ of the projection, $R$ of the rotation, and $F$ of the total transformation are given by
$$
P = \frac 1{30}\pmatrix{29 & -2 & -5\\ -2 & 26 & -10\\ -5 & -10 & 5}, \quad
R = \frac 1{3} \pmatrix{-1 & 2 & 2\\ 2 & -1 & 2\\ 2 & 2 & -1},\\
F = RP = \frac 1{90}\pmatrix{-43 & 34 & -5\\ 50 & -50 & 10\\ 59 & 58 & -35}.
$$
A quick derivation of the matrix formulas:
The projection onto the plane with normal $e_1 = (1,0,0)$ is given by
$$
P_0 = \pmatrix{0&0&0\\0&1&0\\0&0&1} = \pmatrix{0&0\\0&I_2} =
I_3 - \pmatrix{1&0\\0&0_{2 \times 2}}.
$$
where $I_2$ denotes the $2 \times 2$ idenitity matrix. Let $u_1 = v/\|v\|$, and extend to an orthonormal basis $u_1,u_2,u_3$. Let $U$ denote the matrix with columns $u_1,u_2,u_3$. Via the change of basis formula, we find that the matrix of the transformation is given by
$$
P = VP_0V^{-1} = VP_0V^T.
$$
We then compute
$$
VP_0V^T =
V\left( I_3 - \pmatrix{1&0\\0&0_{2 \times 2}}\right)V^T
= VV^T - V\pmatrix{1&0\\0&0_{2 \times 2}}V^T \\
= I - u_1u_1^T = I - \frac{vv^T}{v^Tv}.
$$
We can calculate the rotation formula similarly if we note that the rotation in the plane normal to $e_1$ has matrix
$$
\pmatrix{1&0&0\\0&-1&0\\0&0&-1} = 2\pmatrix{1&0\\0&0_{2 \times 2}} - I_3.
$$
| {
"language": "en",
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Equation of the family of circle which touch the pair of lines $x^2-y^2+2y-1=0$ The question was to find the equation of the family of circles which touches the pair of lines, $x^2-y^2+2y-1 = 0$
So I tried as follows:-
The pair of lines is, by factoring the given equation, $x^2-y^2+2y-1 = 0$ is $x+y-1=0$, and $x-y+1=0$.
These are tangent to the required circle, so the center of the circle, (let that be $(h,k)$) must lie upon the angle bisector of these two lines.
Also, the distance of the center $(h,k)$ from these two lines must equal to the radius (assuming it to be $r$).
How do I use these to get the required equation?
| As yuou noted, the centre of the circle must lie on $y=1 \vee x=0$.
In the first case, we consider the centre to be $C(x_c,1)$ and we have:
$$(x-x_c)^2+(y-1)^2=r^2$$
The radius is the minimal distance from $C$ to the line $y=x+1$ or $y=-x+1$. So we have:
$$r=\frac{|x_c|}{\sqrt{2}}$$
Substituing, we have:
$$(x-x_c)^2+(y-1)^2=\frac{x_c^2}{2}$$
In the second case, we consider the centre to be $C(0,y_c)$ and we have:
$$x^2+(y-y_c)^2=r^2$$
Using the method expained before, we arrive at:
$$x^2+(y-y_c)^2=\frac{(y_c-1)^2}{2}$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Volume calculated by different orders of integration How to write volume of region
$$S=\{(x,y,z): (x-2)^2+y^2<4, z<3-x^2-y^2,z=-1\}$$ in orders x-y-z and y-z-x of integration.
I tried something but I don't get same result.
First, for x-y-z order I found proection to x-y plane
so I think integral is: $$\int\limits_{x=0}^1 \int\limits_{y=-\sqrt{4 - (x - 2)^2}}^{\sqrt{4 - (x - 2)^2}} \int\limits_{z=-1}^{3-x^2-y^2} 1\ dz\ dy\ dx + \int\limits_{x=1}^2 \int\limits_{y=-\sqrt{4 - x^2}}^{\sqrt{4 - x^2}} \int\limits_{z=-1}^{3-x^2-y^2} 1\ dz\ dy\ dx $$
For y-z-x orientation,similary I found proection and integral
$$\int\limits_{y=-2 }^2 dy\int\limits_{-1 }^{3-y^2} dz\int\limits_{x=2- \sqrt{4-y^2}}^{\sqrt{3-y^2-z}} 1\ dx $$
|
$$\int\limits_{x=0}^2 \int\limits_{y=-\sqrt{4 - (x - 2)^2}}^{\sqrt{4 - (x - 2)^2}} \int\limits_{z=-1}^{3-x^2-y^2} 1\ dx\ dy\ dz = \frac{64}{3} - 4 \pi$$
| {
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"source": "stackexchange",
"question_score": "2",
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Why is there a reduction to REF to find the characteristic polynomial of a 4x4 matrix? I'm asked to find the characteristic polynomial of the matrix: $A = \small\begin{pmatrix}
2 & 2 & 0 & 0\\
2 & 2 & 0 & 0\\
0 & 0 & 2 & 2\\
0 & 0 & 0 & 4\\
\end{pmatrix}$. I've gotten $$\begin{equation}
\begin{split}
p_A(t) &= \det (A-tI)\\
&= det \small\begin{pmatrix}
2-t & 2 & 0 & 0\\
2 & 2-t & 0 & 0\\
0 & 0 & 2-t & 2\\
0 & 0 & 0 & 4-t\\
\end{pmatrix}
\end{split}
\end{equation}$$
However, the mistake I seem to make comes from my evaluation of $det \small\begin{pmatrix}
2-t & 2 & 0 & 0\\
2 & 2-t & 0 & 0\\
0 & 0 & 2-t & 2\\
0 & 0 & 0 & 4-t\\
\end{pmatrix}$.
My way of evaluating is:
$$\begin{equation} \begin{split}
det \small\begin{pmatrix}
2-t & 2 & 0 & 0\\
2 & 2-t & 0 & 0\\
0 & 0 & 2-t & 2\\
0 & 0 & 0 & 4-t\\
\end{pmatrix} &= (2-t)(2-t)(2-t)(4-t) - 0\\
&=(2-t)^3(4-t).
\end{split} \end{equation}$$
When I use an eigenvalue calculator to check my polynomial, I see $p_A(t) = (4-t)^2(2-t)t$. The step-by-step solution says to find the determinant of $A-tI$ you have to reduce $A-tI$ to REF. My question is: Why do we have to reduce $A-tI$ to its REF?
| For an $n \times n$ matrix $A$ the best formula to use is $$\det(\lambda I-A))=\lambda^n+\sum_{i=1}^n \beta_i\lambda^{n-i} $$ where $\beta_i=(-1)^i$ sum of principal minors of order $i$. Note that since $n=4$, $$\det(A-\lambda I)=(-1)^4 \det(\lambda I-A)=\det(\lambda I-A)$$ $$\beta_1=-\text {trace} A=-(2+2+2+4)=-10,$$ $$\beta_2=0+4+8+4+8+8=32$$
$$\beta_3=-(16+16+0+0)=-32$$ $$\beta_4=\det A=\text { by generalized Laplace expansion on rows 1&2 }=0 \times 8=0.$$
$$\det(A-\lambda I=\det(\lambda I-A)=\lambda^4-10\lambda^3+32\lambda^2-32\lambda.$$
| {
"language": "en",
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How to use prove this $p^4\equiv p\pmod {13}$
Let a prime number $p$, and $n$ a positive integer such $$p\mid n^4+n^3+2n^2-4n+3.$$
Show that $$p^4\equiv p\pmod {13}.$$
A friend of mine suggested that I might be able to use the results problem.
| Proof for the case when $n\bmod 13 \neq 3$
Let $n\bmod 13 = k$. Then, by substituting $k=0,1,\ldots,12$ and $k\neq 3$, we have
$$\left[n^4+n^3+2n^2-4n+3\right]\bmod 13= \left[k^4+k^3+2k^2-4k+3\right]\bmod 13\in\{1,3,9\},$$
which are all powers of $3$. Now, notice that $$n^4+n^3+2n^2-4n+3\bmod p=0\implies p\bmod 13 \in\{ 1,3, 9\},\tag{1}$$
since $p\bmod 13\neq 0$ as $p$ is a prime number.
Finally, we deduce that
\begin{align}
p^4\bmod 13&=\left[p^4-p + p\right]\bmod 13\\
&=\left[p(p-1)(p^2+p+1)+ p\right]\bmod 13\\
&=\left[p(p-1)(p^2+p-12) + p\right]\bmod 13\\
&= \left[p(p-1)(p-3)(p+4)+ p\right]\bmod 13\\
&=\left[p(p-1)(p-3)(p-9)+ p\right]\bmod 13 \\
&= p\bmod 13.
\end{align}
where we use $(1)$.
| {
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"source": "stackexchange",
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Show that $1+2^n+2^{2n}$ is divisible by $7$, when $n$ is not a multiple of $3$ Problem taken from a paper on mathematical induction by Gerardo Con Diaz. Although it doesn't look like anything special, I have spent a considerable amount of time trying to crack this, with no luck.
Show that $1+2^n+2^{2n}$ is divisible by $7$, when $n$ is not a multiple of 3.
| Whenever a number is not divisible by 3, it leaves remainder as 1 or 2. So, any number which is not divisible by 3 is of the form $n=3l+1$ or $n=3l+2$.
Now, when $n=3l+1$, we have $1+2^{3l+1}+2^{2(3l+1)}=1+2+4$ mod $7$.
and when
$n=3l+2$, we have $1+2^{3l+2}+2^{2(3l+2)}=1+4+16$ mod $7$ and $1+4+16=0$ mod $7$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Application of Hensel's lemma $x^2 \equiv a (\mod 2^L)$ Let $a$ be an odd integer. And $L \geq 1$. I would like to know the number of solutions modulo $2^L$
to the congruence
$$
x^2 \equiv a \pmod {2^L}.
$$
Is it possible to conclude that there is number $C> 0$ such that the number of solutions modulo $2^L$ to the congruence is at most $C$ for any $L$? Thank you.
| Let us trace along an argument inspired by the proof that $a \equiv 1 \pmod{8}$ has square roots in $\mathbb{Z}_2$, in order to find out the number of square roots $\pmod{2^L}$.
First, since $a$ is odd and is a square, we know that if $L \ge 3$, then $a \equiv 1 \pmod{8}$; say $a = 8b + 1$. Then for $n^2 \equiv a \pmod{2^L}$, we must have $n$ odd, say $n = 2m + 1$. Then
\begin{align*}
(2m+1)^2 \equiv 8b+1 \pmod{2^L} & \Leftrightarrow 4m^2 + 4m + 1 \equiv 8b+1 \pmod{2^L} \\
& \Leftrightarrow m^2 + m \equiv 2b \pmod{2^{L-2}}.
\end{align*}
Now, by Hensel's lemma, since $p(x) = x^2 + x$ is a polynomial with $p(0), p(1) \equiv 0 \equiv 2b \pmod{2}$ and $p'(0), p'(1) \equiv 1 \pmod{2}$, for any given $b$, there are exactly two solutions of $m^2 + m \equiv 2b \pmod{2^{L-2}}$ for $m \pmod{2^{L-2}}$: one with $m$ even, and one with $m$ odd. Now, each of these two solutions determines $n \pmod{2^{L-1}}$; we thus get at most 4 solutions $\pmod{2^L}$.
(And in fact, this reflects the comment by J. W. Tanner that if $n$ is a solution, then the general solutions are $\pm n$ and $2^{L-1} \pm n$. To see this, note that if $m_0$ is one of the roots of $m^2 + m \equiv 2b \pmod{2^{L-2}}$, then $-1 - m_0$ is also a root. This also implies there are exactly 4 solutions in all cases with $L \ge 3$, since $n$ odd and $L \ge 3$ implies all these values are distinct.)
| {
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"url": "https://math.stackexchange.com/questions/3614073",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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An A level question about partial differentiation
The equation of a curve is $2x^4+xy^3+y^4=10$. Show that $$\frac{dy}{dx}=-\frac{8x^3+y^3}{3xy^2+4y^3}.$$
I understand that you are to work out:
\begin{align}
\frac{dz}{dx} = 8x^3 + y^3\\
\frac{dz}{dy} = 3xy^2 + 4y^3
\end{align}
and therefore, $\dfrac{dy}{dx}$.
My answer almost matches the requirement, but I don't understand how the negative symbol got there.
| You don't need to solve for $\frac{dz}{dx}$ or $\frac{dz}{dx}$.
We know $2x^4+xy^3+y^4=10$. I will take the derivative with respect to $x$ on both sides of the equation. So $\frac{d}{dx}(2x^4+xy^3+y^4)=\frac{d}{dx}10 \implies 8x^3+y^3+3xy^2\frac{dy}{dx}+4y^3\frac{dy}{dx}=0$. Trying to solving for $\frac{dy}{dx}$, we have $8x^3+y^3=\frac{dy}{dx}(-3x^2-4y^3)$. Then $\frac{8x^3+y^3}{(-3x^2-4y^3)}=\frac{dy}{dx}$, and so $-\frac{8x^3+y^3}{3xy^2+4y^3}=\frac{dy}{dx}$. Does that make any sense? I can go in depth with some steps if you want.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Prove that $\mathbb{Q}[\sqrt{3}, \sqrt{5}] = \mathbb{Q}[\sqrt{3}+ \sqrt{5}]$ Prove that $\mathbb{Q}[\sqrt{3}, \sqrt{5}] = \mathbb{Q}[\sqrt{3}+ \sqrt{5}]$
Take $x \in \mathbb{Q}[\sqrt{3}+ \sqrt{5}]. x = a_x + b_x( \sqrt{3} + \sqrt{5}) = a_x + b_x\sqrt{3} + b_x\sqrt{5} \in \mathbb{Q}[\sqrt{3}, \sqrt{5}] \Rightarrow \mathbb{Q}[\sqrt{3}+ \sqrt{5}] \subset \mathbb{Q}[\sqrt{3}, \sqrt{5}] $
I am having trouble proving the other inclusion.
| $(\sqrt 3 + \sqrt 5)(\sqrt 3-\sqrt 5) = -2$ so $\sqrt 3-\sqrt 5=\frac{-2}{\sqrt 3 + \sqrt 5}\in \mathbb Q[\sqrt{3} +\sqrt 5]$
$\sqrt 3 =\frac{(\sqrt 3-\sqrt 5)+(\sqrt 3+\sqrt 5)}2\in \mathbb Q[\sqrt{3}+\sqrt 5]$
$\sqrt 5 = (\sqrt 3 + \sqrt 5)-\sqrt 3\in \mathbb Q[\sqrt{3} + \sqrt{5}]$
So $\mathbb Q[\sqrt 3, \sqrt 5]\subset \mathbb Q[\sqrt 3+\sqrt 5]$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3617442",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Continued fraction of $π$ using sums of cubes Recently I came across this identity:
$$\pi=3+\cfrac1{6+\cfrac{1^3+2^3}{6+\cfrac{1^3+2^3+3^3+4^3}{6+\cfrac{1^3+2^3+3^3+4^3+5^3+6^3}{6+\ddots}}}},$$
thus
$$\pi=3+\cfrac{1}{6+\cfrac{(1\cdot3)^2}{6+\cfrac{(2\cdot5)^2}{6+\cfrac{(3\cdot7)^2}{6+\ddots}}}}.$$
We also know that
$$\pi=3+\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n(n+1)(2n+1)},$$
Converting the sum to a continued fraction we get:
$$\pi=3+\cfrac{1}{6+\cfrac{(1\cdot3\cdot2)^2}{6\cdot2^2+\cfrac{(2\cdot5\cdot3)^2}{6\cdot3^2+\cfrac{(3\cdot7\cdot4)^2}{6\cdot4^2+\ddots}}}}.$$
How are the two continued fractions equal? And how is the above continued fraction using sums of cubes derived?
| The (first) identity is wrong; the continued fraction with "sums of cubes" doesn't converge. This can be shown using the known criterion: for a sequence $\{a_n\}$ of positive real numbers, the continued fraction $$a_0+\cfrac{1}{a_1+\cfrac{1}{a_2+\cfrac{1}{\ldots}}}$$ converges if and only if the series $\sum_{n=0}^{\infty}a_n$ diverges. In our case, for $n>0$, we have $a_n=6c_n$ where $c_1=1$ and $c_{n+1}=1/\big(n^2(2n+1)^2 c_n\big)$, and one obtains $\color{blue}{a_n=\mathcal{O}(1/n^2)}$ using, say, the $\Gamma$-function: $$c_n=d_n\left(\frac{\Gamma\left(\frac{n}{2}\right)\Gamma\left(\frac{n}{2}+\frac14\right)}{\Gamma\left(\frac{n+1}{2}\right)\Gamma\left(\frac{n}{2}+\frac34\right)}\right)^2\quad\implies\quad d_{n+1}=\frac{1}{64d_n},$$ giving $d_n=\mathcal{O}(1)$ and $c_n=\mathcal{O}(1/n^2)$ since $\Gamma(x+a)/\big(x^a\Gamma(x)\big)\underset{x\to\infty}{\longrightarrow}1$.
For curiosity, here are computed approximate values of the lower/upper limits, respectively:
$$\mathtt{3.14221404702232210406367353362166370131484883936217-}\\\mathtt{3.15126273205858662275081482878228893534757749143403-}$$
| {
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Asymptotes of $r \theta \cos \theta = a \cos 2\theta$ Asymptotes are the lines which touch the curve at infinity.
Putting $u=\frac{1}{r}$, then
\begin{equation*}
\begin{split}
u &= \dfrac{\theta \cos \theta}{a \cos 2\theta} = F(\theta)\\
\end{split}
\end{equation*}
When $r \rightarrow \infty$, $u \rightarrow 0$, or
\begin{equation*}
\begin{split}
\dfrac{\theta \cos \theta}{a\cos 2 \theta} &= 0\\
\theta \cos \theta &= 0\\
\theta &= 0, \pm \dfrac{\pi}{2}
\end{split}
\end{equation*}
i.e. when $r \rightarrow \infty$, $\theta \rightarrow 0, \pm \dfrac{\pi}{2}$. So,
\begin{equation*}
\begin{split}
\theta_1 &= 0, \pm \dfrac{\pi}{2}
\end{split}
\end{equation*}
Differentiating $F(\theta)$ w.r.t $\theta$,
\begin{equation*}
\begin{split}
F'(\theta) &= \dfrac{a\cos 2 \theta \dfrac{\mathrm{d}}{\mathrm{d\theta}} (\theta \cos \theta) - \theta \cos \theta \dfrac{\mathrm{d}}{\mathrm{d\theta}} (a\cos 2\theta)}{a^2 \cos^2 2\theta}\\
F'(\theta) &= \dfrac{a\cos 2\theta (\cos \theta - \theta \sin \theta) + 2a\theta \cos \theta \sin 2 \theta}{a^2\cos^2 2 \theta}\\
\end{split}
\end{equation*}
The equation of the asymptote in case of polar curves is given by,
\begin{equation*}
\begin{split}
r \sin (\theta - \theta_1) &= \dfrac{1}{F'(\theta_1)}
\end{split}
\end{equation*}
So in our case, asymptotes are,
Plotting reveals that while first two are asymptotes, the third one does not seem to be.
My approach looks fine but why is this discrepancy between graph and my solution?
| You miscalculated $F'\left(-\dfrac \pi 2\right)$. It is actually $\dfrac{-\pi}{2a}$. You will find it is the same line as for $\theta = \dfrac \pi 2$.
These asymptotes are actually rays, not lines. $\dfrac \pi 2$ is the asymptote for the top side of the graph and $-\dfrac \pi 2$ is the asymptote for the bottom side of the graph.
However, there are more asymptotes that you missed. Each of those climbing curves has its own asymptote.
| {
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Shortest distance from circle to a line
Let $C$ be a circle with center $(2, 1)$ and radius $2$. Find the shortest distance from the line $3y=4x+20$.
This should be very simple, but I seem to end up with no real solutions.
The shortest distance would be from the center of the circle perpendicular to the line right?
Solving the line for $y$ we get $y=\frac{4}{3}x+\frac{20}{3}$
Substituting this to the equation of the circle we get $(x-2)^2+(\frac{4}{3}x+\frac{20}{3}-1)^2=2^2$, but solving this for $x$ ended up with no real roots. What am I missing here?
| By substituting the equation of the line into the equation of the circle you are looking for points where the line intersects the circle. The fact that the resulting equation for $x$ has no real roots means that the line does not intersect the circle i.e. the shortest distance from the line to the circle will be greater than the radius of the circle, which is $2$ units.
You can find the shortest distance from the line to the circle as follows:
(1) Note that the product of the gradients of perpendicular lines is $-1$, so the general equation of a line perpendicular to the given line is
$y = - \frac 3 4 x + m$
(2) The line with gradient $-\frac 3 4$ that passes through the centre of the circle at $(2,1)$ is
$y = - \frac 3 4 x + \frac 5 2$
(3) This line interects the original line at a point where
$\frac 4 3 x + \frac {20} 3 = - \frac 3 4 x + \frac 5 2 \\
\Rightarrow 16x + 80 = -9x + 30 \\
\Rightarrow 25x = -50 \\
\Rightarrow x=-2 \text{ and } y=4$
(4) Find the distance between $(-2, 4)$ and the centre of the circle $(2,1)$. Then subtract the radius of the circle from this distance - this is the shortest distance from the line to the circle.
| {
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Prove $x^4-18x^2+36x-27$ can never be a nonzero square rational when $x$ is rational Prove $x^4-18x^2+36x-27$ can never be a square rational (excluding 0), when x is rational
I have tried to use modulus, but didn't get anywhere, any help would be greatly appreciated.
| Say $x^4-18*x^2+36*x-27 = n^2$
The polynomial factors up
$(x-3)*(x^3+3*x^2-9*x+9) = n^2$
It means that if $x-3$ is a perfect square, then $x^3+3*x^2-9*x+9
$ is also a perfect square or both aren't perfect square
Just like $36=9×4=12×3$
If $x-3$ is not a perfect square, then polynomial $x^3+3*x^2-9*x+9$ is not also a perfect, therefore it must factor up further, Just like $36=12×3=4×3×3$
But $x^3+3*x^2-9*x+9$ does not factor
It doesn't factor as supposed, it has to factor into $(x-3)*(x-a)^2$ in my definition, but there's no value of $a$ to fix this therefore $x^3+3*x^2-9*x+9$ does not factor to $(x-3)*(x-a)^2$
Say $x-3 = k^2$, $x = 3+k^2$
$(3+k^2)^3+3*(3+k^2)^2-9*(3+k^2)+9 = (n/k)^2$
$k^6+12*k^4+36*k^2+36 = (n/k)^2$
$k^2*(k^2+6)^2+36 = (n/k)^2$
$(k\cdot(k^2+6))^2 + 6^2 = (n/k)^2$
It's easy to see that no integer $k$ exist to that make this a perfect square
Therefore $x^4-18*x^2+36*x-27$ is not a perfect square
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Prove that $1=\sqrt{2\cdot \color{red}{-1} + \sqrt{2\cdot \color{red}2 + \sqrt{2\cdot \color{red}{9}+\sqrt{2\cdot\color{red}{20}+\cdot \cdots}}}}$ I was playing around with square roots and I noticed that the number $1$ can be seemingly expressed as an infinite nested radical with an easy pattern. I then noticed that if this is true, this would mean that each nested radical inside is equal to the next odd number, respectively. I have tried to denote this using LaTeX but it did not display as I thought it might, though I hope you can understand what I am trying to concisely say.
Conjecture: $$\begin{align}\color{green}1 &= \sqrt{2\cdot \color{red}{-1} + \color{green}{3}} \\ &= \sqrt{2\cdot \color{red}{-1} + \sqrt{2\cdot \color{red}2 + \color{green}{5}}} \\ &= \sqrt{2\cdot \color{red}{-1} + \sqrt{2\cdot \color{red}2 + \sqrt{2\cdot \color{red}{9}+\color{green}7}}}\\ &= \sqrt{2\cdot \color{red}{-1} + \sqrt{2\cdot \color{red}2 + \sqrt{2\cdot \color{red}{9}+\sqrt{2\cdot\color{red}{20}+\color{green}9}}}}\\ &= \sqrt{2\cdot \color{red}{-1} + \sqrt{2\cdot \color{red}2 + \sqrt{2\cdot \color{red}{9}+\sqrt{2\cdot\color{red}{20}+\sqrt{2\cdot \color{red}{35}+\cdots}}}}}\end{align}$$ $\color{red}{-1}+(4\cdot 0+3)=\color{red}{2}$ (radical $=\color{green}1$) $\color{red}2+(4\cdot 1+3)=\color{red}9$ (radical $=\color{green}3$) $\color{red}9+(4\cdot 2+3)=\color{red}{20}$ (radical $=\color{green}5$) $\color{red}{20}+(4\cdot 3+3)=\color{red}{35}$ (radical $=\color{green}7$) $\cdots = \cdots$
Can this be proven? If it is a result of a generalised identity for odd numbers $2n+1$ (for some $n$, this case being $n=0$), please let me know.
Thanks.
Edit:
@TheSimpliFire kindly gave the recurrence relation: $1=2a_0+b_n$ where $b_n=\sqrt{2a_n+b_{n+1}}$ and $a_n+4n+3=a_{n+1}$ with $a_0=-1$ and $n>0$.
| This is the general identity:
$$2n+1=\sqrt{2(n+1)(2n-1)+\sqrt{2(n+2)(2n+1)+\sqrt{2(n+3)(2n+3)+\cdots}}}$$
Secondary (albeit similar) Proof:
This proof is my original one. @TheSimpliFire's is way more elegant and concise, but since that first proof is in the post, I will post mine here, just to contribute in benefit of Math.SE I suppose.
We will organise the key elements of the data pairwise as thus:
$$(1, -1), (3, 2), (5, 9), (7, 20), (9, 35)$$
Odd numbers are generated from $2n+1$, thus this is equivalent to
$$(0, -1), (1, 2), (2, 9), (3, 20), (4, 35)$$
Notice:
$$\begin{align} 0 &= (-1 + 1)\div 1 \\ 1 &= (2+1)\div 3 \\ 2 &= (9+1)\div 5 \\ 3 &= (20+1)\div 7 \\ 4 &= (35+1)\div 9 \\ \therefore n &= (n(2n+1)-1+1)\div (2n+1)\end{align}$$
Matching this pattern to the form of the conjecture, we get $2n+1=\sqrt{2(n(2n+1)-1)+\cdots}$ such that $n(2n+1)-1=2n^2+n-1=(n+1)(2n-1)$. $$\therefore 2n+1=\sqrt{2(n+1)(2n-1)+\cdots}$$
Since $(2n+1)^2-2(n+1)(2n-1)=2n+3=2(n+1)+1$ then, by letting $f(n)=2n+1$, we obtain that $$f(n)= \sqrt{2(n+1)(2n-1)+f(n+1)}$$ and thus the general identity follows. $\;\bigcirc$
| {
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"timestamp": "2023-03-29T00:00:00",
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Continuous functions satisfying $f(f(x))=x$, for all $x \in \mathbb{R}$, and $\int_{-x}^{0} f(t)dt - \int_{0}^{x^2}f(t)dt=x^3$ for $x>0$
Find all continuous functions $f: \mathbb{R} \to \mathbb{R}$ satisfying:
I. For all $x\in \mathbb{R}$, $f(f(x))=x$
II. For all $x>0$, $\displaystyle\int_{-x}^{0} f(t)dt - \displaystyle\int_{0}^{x^2}f(t)dt=x^3$
So far I got that $f$ is strictly decreasing. Also I differentiated both sides of the equation in II and got the equation $f(-x)-2xf(x^2)=3x^2$ for all $x>0$, but I'm stuck here. How can I solve this problem?
Edit: I found an answer, but not a proof:
$x>0$ : $f(x)=-\sqrt{x}$, $x<0$ : $f(x)=x^2$
| This is not an answer, but is too long for a comment. This is as far as I got.
Write the anti-derivative of $F$ as $f$.
Then by (2), for any $x>0$:
$$x^3 = F(0) - F(-x) - F(x^2) + F(0) = 2F(0) - F(-x) - F(x^2).$$
Taking a derivative:
$$3x^2 = f(-x) - 2xf(x^2) \iff f(-x) = 3x( x + \frac{2}{3}f(x^2)).$$
So if we know $f(t)$ for all $t>0$, we know $f(-t)$ for all $t>0$ also. Also note that $f(0) = f(-0) = 2 \cdot 0 ( 0 + \frac{2}{3}f(0^2)) = 0$, so $f(0)= 0$.
Now we consider property (1), which in other words says $f$ is a continuous involution.
Now note that $f(f(x)) =x $ implies that it must be surjective and injective, so it is a bijection, (surjective as for any $x$, $f(f(x)) =x $, so $f$ maps onto $\mathbb{R}$ and injective as $f(x) = f(y)$ implies $x=f(f(x)) = f(f(y)) = y$). Then continuous bijections are strictly monotonic.
Suppose $f$ is increasing. For any $x$ there are three cases, $f(x) < x, f(x) = x, f(x)>x$. In case (1), we then have $x=f(f(x)) <x$ a contradiction, similarly in case (3). Hence $f(x) = x$ for all $x$ if $f$ increasing. It is is easy to see this $f$ doesn't work.
So we know that $f$ is strictly decreasing.
Then $f(0) =0$, so $f(-x) >0$ for all $x>0$. Thus $3x(x+\frac{2}{3}f(x^2))>0$, so $f(x^2) > -\frac{3}{2}x$ for $x>0$, or $f(x) > - \frac{3}{2}\sqrt{x}$ for $x>0$.
We also have $x>0$ implies $f(x)<0$, so for any $x >0$, we have $- \frac{3}{2} \sqrt{x} < f(x) < 0$. Thus also for $x>0$, $f(-x) = 3x(x+ \frac{2}{3}f(x^2)) < 3x^2$.
Note then that $f|_{[0,\infty)}$ is a bijection onto $(-\infty,0]$. So for any $x>0$, $f(-x) = f^{-1}(-x)$.
Thus $$\int_{-x}^0 f(t) dt = \int_{-x} f^{-1}(t) dt = (tf^{-1}(t) - F(f^{-1}(t)))|_{-x}^0 = - F(f^{-1}(0)) - (-x f^{-1}(-x) - F(f^{-1}(-x))) = xf^{-1}(-x) + F(f^{-1}(-x)),$$
where $F$ is the antiderivative of $F$ (WLOG suppose $F(0) =0$), and we use this.
Thus $$\int_{-x}^0 f(t) dt - \int_0^{x^2} f(t) dt = xf^{-1}(-x) + F(f^{-1}(-x)) - F(x^2) = x^3.$$
A summary:
For any $f$ satisfying (1) and (2), let $F$ be it's antiderivative s.t. $F(0) =0 $.
Then:
(1) $f(0)=0$.
(2) $f$ is a strictly decreasing bijection.
(3) $f(-x) = 3x (x+\frac{2}{3}f(x^2))$
(4) For all $x>0$:
$$- \frac{3}{2} \sqrt{x} < f(x) < 0< f(-x) <3x^2$$
(5) $$xf^{-1}(-x) + F(f^{-1}(-x)) - F(x^2) = x^3$$
| {
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If $p=\frac {dy}{dx}$ then solve the equation $p^3-p(x^2+xy+y^2)+x^2y+xy^2=0$ If $p=\frac {dy}{dx}$ then solve the equation $p^3-p(x^2+xy+y^2)+x^2y+xy^2=0$
My Attempt:
$$p^3-p(x^2+xy+y^3)+x^2y+xy^2=0$$
$$p^3-p(x+y)^2+pxy+xy(x+y)=0$$
$$p(p^2-(x+y)^2)+xy(p+(x+y))=0$$
$$p\cdot (p+(x+y))\cdot (p-(x+y))+xy(p+(x+y))=0$$
$$(p+(x+y))\cdot (p(p-(x+y))+xy)=0$$
I don't get what to do after this.
| We get three equations: $p=x,y-x-y$
We get three solutions: $$y=x^2/2+C, y=Ce^x, y=1-x+Ce^{-x}$$
The total solution of this first order ODE is
$$(x^2/2-C)(ye^{-x}-C)[(y-1+x)e^{x}-C]=0,$$
with only one contant of integration: $C$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Calculate $\sum_{x=1}^{\infty} \frac{(x-1)}{2^{x}}$ I can prove it converges but I don't know at what value it converges.
$\sum_{x=1}^{\infty} \frac{(x-1)}{2^{x}}$
| Start with (for $\lvert u \rvert < 1$ the geometric series converges absolutely, we'll use $u = 1/2$ in the end, so all the manhandling is justified):
$\begin{align*}
\sum_{x \ge 1} u^x
&= \frac{u}{1 - u} \\
\frac{d}{d u} \sum_{x \ge 1} u^x
&= \frac{1}{(1 - u)^2} \\
\sum_{x \ge 1} x u^{x - 1}
&= \\
\sum_{x \ge 1} x u^x
&= \frac{u}{(1 - u)^2} \\
\sum_{x \ge 1} (x - 1) u^x
&= \frac{u}{(1 - u)^2} - \frac{u}{1 - u} \\
&= \frac{u^2}{(1 - u)^2}
\end{align*}$
Now replace $u = 1/2$ and get:
$\begin{align*}
\sum_{x \ge 1} \frac{x - 1}{2^x}
&= \frac{(1/2)^2}{(1 - 1/2)^2} \\
&= 1
\end{align*}$
| {
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"timestamp": "2023-03-29T00:00:00",
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Showing there isn't such an integer $a \not\equiv 1$ that $a^7 \equiv 1 \pmod{21}$
Establish whether there exists an integer $a \in \mathbb{Z}, a \not\equiv 1$ such that $a^7 \equiv 1 \pmod{21}$
The first thing I tried was trying to compute the result of $a^7$ for some small $a$'s.
This is what I found:
By breaking up $a^7$ into $a^2\cdot a^2 \cdot a^2 \cdot a$, I made it easier to find that:
$$2^7 = 128 \equiv 2 \pmod{21}\\
3^7 = 3^23^23^23 \equiv 81^23\equiv3 \pmod{21}\\
4^7 \equiv \cdots\equiv 4 \pmod{21}\\\vdots\\
7^7\equiv \cdots \equiv 7 \pmod{21}$$
I only did the computations until $7$ because I could clearly see a pattern.
It appears as if: $$\forall a \in \mathbb{N}, a^7\equiv a \pmod{21} \tag{conjecture}$$
I think that, if I find a way to prove my conjecture, I'd be done with the problem. I tried to prove it by induction to no avail, so maybe that's not the right path. I thought that by factorizing $a^7-a$ and trying to find a relationship between $(a-1)^7$ and $a^7$ I could have had some luck, but it didn't happen.
Any inputs?
| Well, you could just test them all. For any integer $a$ there must be an $i = 0, 1,2,....,20$ so that $a\equiv i\pmod{21}$ and $a^7\equiv i^7\pmod {21}$ and none of $0^7,2^7,....., 20^7$ are congruent to $1\pmod{21}$ and only $1^7\equiv 1 \pmod {21}$..
That's not satisfying as it's unlikely anyone want to take the effort or that anyone will be sasfied by such a brute force answer. (but it's always a last ditch option.)
....
Alternatively $a^7\equiv 1 \pmod {21}$ is only possible if $\gcd(a,21)= 1$. [Because $\gcd(a,21)|a^7$ so if $\gcd(a,21)\ne 1$ then $\gcd(a,21)\not \mid a^7-1$.]
And by Euler's Theorem $\phi(21) = \phi(3)\phi(7)=2*6=12$ and so $a^{12}\equiv 1\pmod {21}$. So if $a^7\equiv 1$ then $(a^7)^2\equiv a^{14}\equiv a^{12}a^2\equiv 1*a^2\equiv a^2 \equiv 1$ and then $(a^2)^4\equiv a^8\equiv a*a^7\equiv a*1\equiv a \equiv 1 \pmod{21}$.
A little bit more structure would be to set up an equation to find a $k$ so that $k\equiv 1 \pmod 7$ and $k \equiv 0\pmod {12}$ or an $m$ so that $m\equiv 0\pmod 7$ and $m\equiv 1\pmod {12}$ then $a^k\equiv a^1\equiv a^0\pmod{21}$.
For instance $k = 36$ or $m = 49$. Then $a^{36}\equiv (a^7)^5*a\equiv a\pmod {21}$ but $a^{36}\equiv (a^{12})^3 \equiv 1\pmod {21}$ so $a\equiv 1 \pmod{21}$. Or $a^{49}\equiv (a^7)^7\equiv 1 \pmod {21}$ but $a^{49}\equiv (a^{12})^4*a \equiv a\pmod{21}$.
....
Alterantively Fermat's Little Theorem and the Chinese Remainder Theorem lets us consider smaller numbers:
$a^7\equiv 1 \pmod {21}$ means that
$a^7 \equiv 1 \pmod 3$ and that $a^7\equiv 1 \pmod 7$.
And thus $a \not \equiv 0 \pmod 3$ and $a\not \equiv 0\pmod 7$. SO by FLT:
$a^2\equiv 1\pmod 3$ and $a^7=a^6*a\equiv a\pmod 3$ so $a \equiv 1 \pmod 3$.
And $a^6\equiv 1 \pmod 7$ and $a^7\equiv a^6*a\equiv a \pmod 7$ so $a\equiv 1 \pmod 7$.
By CRT there is only one solution $\pmod {21}$ where $a\equiv 1\pmod 7$ and $a \equiv 1\pmod 3$ and that is $a\equiv 1\pmod {21}$ is clearly one solution, and as it is the only solution we can conclude.
The onlly integers where $a^7\equiv 1\pmod {21}$ are where $a \equiv 1 \pmod {21}$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Let $s_{n}$ be a sequence such that $|s_{n+1}-s_{n}|<2^{-n}$. Prove that $s_{n}$ is Cauchy. Is this proof sound?
Proof: Let $\epsilon >0$. Since the sequence $\frac{1}{2^{n}}$ converges to $0$, there exists $N$ such that $n>N$ implies $\left|\frac{1}{2^{n}}-0\right|=\frac{1}{2^{n}}<\frac{\epsilon}{2}$. Assume without loss of generality $n,m>N$ with $n\geq m$. Then \begin{align*}
|s_{n}-s_{m}|&\leq |s_{n}-s_{n-1}|+|s_{n-1}-s_{n-2}|+...+|s_{m+2}-s_{m-1}|+|s_{m-1}+s_{m}|\\
&<\frac{1}{2^{n-1}}+\frac{1}{2^{n-2}}+...+\frac{1}{2^{m-1}}+\frac{1}{2^{m}}\\
&=\frac{\frac{1}{2^{m}}\left(1-\left(\frac{1}{2}\right)^{n-m}\right)}{\frac{1}{2}}\text{$\quad$ by the geometric series formula}\\
&=\frac{\frac{1}{2^{m}}-\frac{1}{2^{n}}}{\frac{1}{2}}\\
&=\frac{2}{2^{m}}-\frac{2}{2^{n}}\\
&<\frac{2}{2^{m}}\\
&=2\left(\frac{1}{2^{m}}\right)\\
&<2\left(\frac{\epsilon}{2}\right)\\
&=\epsilon.
\end{align*}
| It's right, and the condition can be more looser. We just need $ |s_{n+1}-s_{n}|\leqslant n^{-2} $. The key steps of proof are
\begin{align*}
|s_{n+p}-s_{n}| & \leqslant |s_{n+p}-s_{n+p-1}|+\dots+|s_{n+1}-s_{n}|\\
& \leqslant\frac{1}{n^{2}}+\frac{1}{(n+1)^{2}}+\dots+\frac{1}{(n+p-1)^{2}}\\
& \leqslant\frac{1}{(n-1)n}+\frac{1}{n(n+1)}+\dots+\frac{1}{(n+p-2)(n+p-1)}\\
& \leqslant \frac{1}{n-1}-\frac{1}{n}+\frac{1}{n}+\frac{1}{n+1}+\dots+\frac{1}{n+p-2}+\frac{1}{n+p-1}\\
& \leqslant \frac{1}{n-1}-\frac{1}{n+p-1}<\frac{1}{n-1}\to 0,\qquad n\to\infty
\end{align*}
| {
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"timestamp": "2023-03-29T00:00:00",
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Evaluate $\lim_{n\to\infty}\frac{1}{n^{p+1}}\cdot \sum_ \limits{i=1}^{n} \frac{(p+i)!}{i!} $, $p \in N$
Evaluate
$$\lim_{n\to\infty}\frac{1}{n^{p+1}}\cdot \sum_ \limits{i=1}^{n} \frac{(p+i)!}{i!} , p \in N$$
Now, I found this problem while doing some practice and I am curious on how to solve it . I have no good ideas yet, so I will appreciate any hints !
| A lower bound is given by
$$
\mathop {\lim }\limits_{n \to + \infty } \frac{1}{{n^{p + 1} }}\sum\limits_{i = 1}^n {\frac{{(p + i)!}}{{i!}}} \ge \mathop {\lim }\limits_{n \to + \infty } \frac{1}{{n^{p + 1} }}\sum\limits_{i = 1}^n {i^p } \\ = \mathop {\lim }\limits_{n \to + \infty } \frac{1}{n}\sum\limits_{i = 1}^n {\left( {\frac{i}{p}} \right)^p } = \int_0^1 {x^p dx} = \frac{1}{{p + 1}} .
$$
An upper bound:
\begin{align*}
& \mathop {\lim }\limits_{n \to + \infty } \frac{1}{{n^{p + 1} }}\sum\limits_{i = 1}^n {\frac{{(p + i)!}}{{i!}}} = \mathop {\lim }\limits_{n \to + \infty } \frac{1}{{n^{p + 1} }}\sum\limits_{i = 1}^n {(i + 1)(i + 2) \cdots (i + p)}
\\ &
\le \mathop {\lim }\limits_{n \to + \infty } \frac{1}{{n^{p + 1} }}\sum\limits_{i = 1}^n {\left( {i + \frac{{p + 1}}{2}} \right)^p } \le \mathop {\lim }\limits_{n \to + \infty } \frac{1}{{n^{p + 1} }}\sum\limits_{i = 1}^n {\int_i^{i + 1} {\left( {x + \frac{{p + 1}}{2}} \right)^p dx} }
\\ & =
\mathop {\lim }\limits_{n \to + \infty } \frac{1}{{n^{p + 1} }}\int_1^{n + 1} {\left( {x + \frac{{p + 1}}{2}} \right)^p dx} \\ & = \frac{1}{{p + 1}}\mathop {\lim }\limits_{n \to + \infty } \left( {1 + \frac{{p + 1}}{{2n}} + \frac{1}{n}} \right)^{p + 1} - \frac{1}{{p + 1}}\mathop {\lim }\limits_{n \to + \infty } \left( {\frac{{p + 1}}{{2n}} + \frac{1}{n}} \right)^{p + 1}
\\ &
= \frac{1}{{p + 1}}.
\end{align*}
I first used the inequality between the geometric and the arithmetic mean, then estimated each term by an integral taking into account the monotonicity of the power function. Thus, the limit in question is $\frac{1}{p+1}$.
| {
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Alternative methods of solving an angle in an equilateral triangle This question has already been answered here, but the OP said trigonometry was forbidden. I was thinking of different approaches that allow trigonometry, so I decided to post a new question.
An equilateral triangle is given with edges of the length $a$. Let $X\in\overline{AB}$ s.t. $|AX|=\frac{a}{3}$ and let $Y\in\overline{BC}$ s.t. $|BY|=\frac{a}{3}$. Let $T$ be the intersection point of $AY$ and $CX$. Find $\measuredangle{CTB}$.
First of all, $|BX|=2|BY|\;\&\;\measuredangle XBC=60^{\circ}\implies\Delta XBY$ is one half of an equilateral triangle$\implies\measuredangle BYX=90^{\circ}\implies\color{red}{\Delta XYC\;\text{is a right-triangle}}$.
Now, $|XY|^2=\left(\frac{2a}{3}\right)^2-\left(\frac{a}{3}\right)^2=\frac{3a^2}{9}$
$|CX|=\sqrt{|XY|^2+|YC|^2}=\frac{\sqrt{7}a}{3}$
According to the $SAS$ (Side-Angle-Side) theorem, $\Delta ABY\cong\Delta AXC\;\&\;\Delta XBC\cong\Delta AYC$.
Then,
$\measuredangle CXB=\measuredangle AYC\;\&\;\measuredangle YCT=\measuredangle BCX\implies\measuredangle CTY=\measuredangle XBC=60^{\circ}\implies\Delta TYC\sim\Delta BCX$
$$\implies\frac{|XC|}{|YC|}=\frac{|BC|}{|CT|}\implies|CT|=\frac{|BC|\cdot|YC|}{|XC|}=\frac{a\frac{2a}{3}}{\frac{\sqrt{7}a}{3}}=\frac{2a}{\sqrt{7}}$$
In $\Delta XYC$, we have:$\cos(\measuredangle YCX)=\frac{|CY|}{|CX|}=\frac{\frac{2a}{3}}{\frac{\sqrt{7}a}{3}}=\frac{2}{\sqrt{7}}$
In $\Delta BCT$, we have:
$$|BT|=\sqrt{|BC|^2+|CT|^2-2|BC|\dot|CT|\cos(\measuredangle YCX)}=\sqrt{a^2+\frac{4a^2}{7}-2a\cdot\frac{2a}{\sqrt{7}}\cdot\frac{2}{\sqrt{7}}}=\frac{\sqrt{3}a}{\sqrt{7}}$$
In $\Delta YCT$, we have: $\frac{|CY|}{\sin(\measuredangle CTY)}=\frac{|CT|}{\sin(\measuredangle TYC)}\implies \sin(\measuredangle TYC)=\frac{|CT|\sin(\measuredangle CTY)}{|CY|}=\frac{\frac{2a}{\sqrt{7}}\frac{\sqrt{3}}{2}}{\frac{2a}{3}}=\frac{3\sqrt{3}}{2\sqrt{7}}=\sin(\measuredangle BYT)$
In $\Delta BYT$, we have $\frac{|BT|}{\sin(\measuredangle BYT)}=\frac{|BY|}{\sin(\measuredangle YTB)}\implies\measuredangle YTB=\arcsin\frac{|BY|\sin(\measuredangle BYT)}{|BT|}=\arcsin\frac{\frac{a}{3}\frac{3\sqrt{3}}{2\sqrt{7}}}{\frac{\sqrt{3}a}{\sqrt{7}}}=\arcsin\frac{1}{2}\implies\measuredangle YTB=30^{\circ}$
Finally, $\measuredangle CTB=\measuredangle TYB+\measuredangle CTY=90^{\circ}$
Picture:
My question is: is there any shorter way I could solve this via trigonometry or vectors? Thank you in advance!
| Not a trigonometric solution, but I decided to post it anyway.
It is clear that triangles $CXA$ and $AYB$ are congruent, hence $\angle CXA=\angle AYB$. Hence quadrilateral $BXTY$ is cyclic. The center of this circle is the midpoint $M$ of $XB$ as $MB=MX=\frac 13 a=MY$. Hence $\angle BTY =\frac 12 \angle BMY = 30^\circ$.
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What is this mystery function that wolfram alpha says my exponential generating function is equal too? The $E_n(x^n)$ is the mystery function
$$\sum_{d=0}^{\infty}\frac{x^{dn}}{\Gamma(dn+1)}=E_n(x^n)$$
Here are the first 3 values of the function
| For $$E_n=\sum_{d=0}^{\infty}\frac{x^{dn}}{\Gamma(dn+1)}$$ the first terms are simple
$$\left(
\begin{array}{cc}
n & E_n \\
1 & e^x \\
2 & \cosh (x) \\
3 & \frac{2}{3} e^{-x/2} \cos \left(\frac{\sqrt{3} x}{2}\right)+\frac{e^x}{3} \\
4 & \frac{\cos (x)}{2}+\frac{\cosh (x)}{2}
\end{array}
\right)$$ For $n>4$, they are hypergeometric functions
$$E_5=\, _0F_4\left(;\frac{1}{5},\frac{2}{5},\frac{3}{5},\frac{4}{5};\frac{x^5}{5^5}\right)$$
$$E_6=\,
_0F_5\left(;\frac{1}{6},\frac{2}{6},\frac{3}{6},\frac{4}{6},\frac{5}{6};\frac{x^6}{6^6
}\right)$$
$$E_7=\,
_0F_6\left(;\frac{1}{7},\frac{2}{7},\frac{3}{7},\frac{4}{7},\frac{5}{7},\frac{6}{7};
\frac{x^7}{7^7}\right)$$ and the pattern is quite clear.
| {
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"url": "https://math.stackexchange.com/questions/3642591",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Prove that $7^x=1+y^2+z^2$ has no solutions in positive integers $$7^x=1+y^2+z^2$$
So far I have not gotten any remarkable results. Analyzing $mod$ $3$ I get that $y$ and $z$ must be divisible by $3$. Further analyzing $mod$ $4$ I got that $x$, $y$, $z$ must all have the same parity. Looking at the equation $mod$ $8$ I got that all the variables have to be even. Then looking at the equation $mod$ $9$ I got that $x$ is, in fact, divisible by $3$. So the most I know about the variables is that all of them are divisible by $6$.
Rewriting the equation as $(7^k-1)(7^k+1) = y^2+z^2$ we see that $(7^k-1)$ and $(7^k+1)$ greatest common divisor is $2$ so that means that the highest power of $3$ that divides the sum of squares must also divide $(7^k-1)$ because the other factor is not divisible by $3$.
That is how far I have gotten with this problem.
Any ideas?
| $(7^{2k+1}-1)/2$ is an odd integer congruent to $3\bmod 4$ and hence has a prime of the form $4k+3$ appearing an odd number of times.
This implies there are no solutions for odd values of $x$, and we can induct over $v_2(x)$ and use the factorization $(7^{k}-1)(7^k+1)$ to get there are no solutions for any $x$ (you need to use the fact that the only prime that can divide both the numbers is $2$).
To prove the claim we just write the expression as $\frac{(7-1)(7^{2k} + \dots + 7^0)}{2}=3(7^{2k} + \dots + 7^{0}) \equiv 3\times 1 \bmod 4$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Does this inequality hold with some constant factor $c>0$? Does there exist a real number $c>0$ such that
$$ (x-1)^2+(y-1)^2-2(\sqrt{xy}-1)^2\ge c\big( (x-\sqrt{xy})^2+(y-\sqrt{xy})^2 \big) \tag{*}$$
holds for every positive real numbers $x,y$ such that $xy \ge \frac{1}{4}$.
Note that the LHS vanishes exactly when
$$ (x-y)^2=2\big( (x+y)-2\sqrt{xy} \big),$$
which implies, since $xy \ge \frac{1}{4}$ that $x=y$ so the RHS vanishes as well.
Edit:
There seems to be some "symmetry imbalance" between the two sides of $(*)$. Indeed replacing $(x,y)$ by $ (\lambda x,\lambda y)$ multiplies the RHS by $\lambda^2$, but the LHS does not scale in this way exactly-some of its summands get multiplied by $\lambda$ and some $\lambda^2$. (after the $1$'s cancel each other).
Can this observation be lifted easily to a contradiction?
| Substitute
$$
r = 2\sqrt{xy} - 1
\qquad
q = x+y-2\sqrt{xy}.
$$
Thanks to AM-GM they are independent and can be any positive number.
It boils down to
$$
\frac{q+2r}{q+r+1}\ge c
$$
so $c$ cannot be positive, since for $q,r$ small enough, the expression converges to zero.
For example, if $x=1/2$ and $y=1/2+\varepsilon$, then you will find that $c(\varepsilon)$ goes to zero as $\varepsilon$ goes to zero.
Here are the details. Expand
$$
(x-1)^2+(y-1)^2-2(\sqrt{xy}-1)^2\ge c\big( (x-\sqrt{xy})^2+(y-\sqrt{xy})^2 \big) $$
and obtain
$$
x^2 + y^2 -2x-2y-2xy+4\sqrt{xy}\ge c\big( x^2 + y^2 + 2xy -2x\sqrt{xy} -2y\sqrt{xy} \big).
$$
You can now regroup and factorize as follows
$$
\big( x+y-2\sqrt{xy} \big)
(x+y+2\sqrt{xy} -2)
\ge c(x+y)\big( x+y-2\sqrt{xy} \big).
$$
If $x+y-2\sqrt{xy} = 0$, then the inequality is satisfied for every $c$, so we can semplify it. Thanks to AM-GM, $x+y-2\sqrt{xy} \ge 0$,so
$$
x+y+2\sqrt{xy} -2
\ge c(x+y).
$$
$x,y$ are positive, so we divide by $x+y$ and get
$$
\frac{x+y+2\sqrt{xy} -2}{x+y}\ge c
$$
and using the substitution
$$
r = 2\sqrt{xy} - 1
\qquad
q = x+y-2\sqrt{xy}
$$
we have
$$
\frac{q+2r}{q+r+1}\ge c.
$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Lagrange Method of Solving Cubic Equations Let $K$ be a field (for simplicity, one may assume $K\subseteq\mathbb{C}$), and let $d\in K$. Denote $w=\sqrt[3]{d}$, and $\zeta=(-1+\sqrt{-3})/2$. If $f(x)=x^3+ax^2+bx+c$ is the minimal polynomial of $x_{1}=\alpha+\beta w+\gamma w^2$, show that:
*
*The other two roots of $f$ are $x_{2}=\alpha+\zeta\beta w+\zeta^2\gamma w^2$ and $x_{3}=\alpha+\zeta^2\beta w+\zeta\gamma w^2$.
*$\alpha = (x_1+x_2+x_3)/3$, $\beta w=(x_1+\zeta x_2 +\zeta^2 x_3)/3$ and $\gamma w^2=(x_1+\zeta^2x_2+\zeta x_3).$
I tried expanding the term $f(x_1)$ (see below), using the fact that $1+\zeta+\zeta^2=0$ and $\zeta^3=1$, however I found that under the transformation $\beta \mapsto\zeta\beta, \gamma \mapsto\zeta^2\gamma$ the coefficients $a_j$ of $w^j$ ($0\le j\le 6$) are multiplied by $\zeta^j$ too. How can I show that this transformation does not change the result and deduce that $f(x_2)=f(x_3)=0$, without explicitly solving the original equation?
The expansion of $f(x_1)$ in terms of $w$ is
$$f(x_1 )=(γ^3 ) w^6+(3βγ^2 ) w^5+(aγ^2+3αγ^2+3β^2 γ) w^4
+(6αβγ+2aβγ+β^3 ) w^3+(2aαγ+aβ^2+3αβ^2+bγ+3α^2 γ) w^2
+(2aαβ+3α^2 β+bβ)w+(aα^2+αb+c+α^3 ),$$
however I do not see how it helps. Any elegant suggestions?
| For the second part, it can be shown merely by calculation:
$$\begin{align}x_1+x_2+x_3&=(α+βw+γw^2 )+(α+ζβw+ζ^2γw^2 )+(α+ζ^2βw+ζγw^2)\\& =3α+βw(1+ζ+ζ^2 )+γw^2 (1+ζ+ζ^2 ),\end{align}$$
and by using $1+ζ+ζ^2=0$, as it is a root of $\Phi_3(x)=1+x+x^2=\frac{x^3-1}{x-1}$:
$$α=\frac{1}{3} (x_1 + x_2 + x_3).$$
In a similar fashion:
$$\begin{align}x_1+ζx_2+ζ^2x_3&=(α+βw+γw^2)+ζ(α+ζβw+ζ^2 γw^2)+ζ^2 (α+ζ^2 βw+ζγw^2 )\\&=α(1+ζ+ζ^2 )+βw(1+ζ^2+ζ^4 )+γw^2 (1+ζ^3+ζ^3 )\\&=3γw^2,\end{align}$$
and
$$\begin{align}x_1+x_2ζ^2+x_3ζ&=(α+βw+γw^2)+ζ^2 (α+ζβw+ζ^2 γw^2)+ζ(α+ζ^2 βw+ζγw^2)\\&=α(1+ζ+ζ^2 )+3βw+γw^2 (1+ζ^2+ζ^4)\\&=3\beta w.\end{align} $$
| {
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"timestamp": "2023-03-29T00:00:00",
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Median of the set of numbers which consists of all positive integers whose digits strictly increase from left to right Consider the set
$$S=\{1,2,3,4,5,6,7,8,9,12,13,14,15,16,17,18,19,23,24,\ldots,123456789\},$$
which consists of all positive integers whose digits strictly increase from left to right. This set is finite. What is the median of the set?
This problem is harder than I thought at first. I first simply though the solution was $\frac{123456789+1}{2}=61728395.$ Turns out I'm wrong! Where am I going wrong? I checked my work and $61728395-1+1=61728395.$
Also,
$123456789-61728395+1=61728395.$ These are equal, so the distance should also be equal. Where am I going wrong?
| There are 9 of these numbers having 1 digit, $\binom{9}{2}$ having 2 digits, and in general $\binom{9}{k}$ having $k$ digits. The total number of elements in this set is thus $\binom{9}{1} + \binom{9}{2} + \cdots + \binom{9}{9} = 2^9-1$, by the binomial theorem, so the median is the $2^8$-th element.
The binomial theorem and symmetry of binomial coefficients also tells us $$\binom{9}{0} + \binom{9}{1} + \binom{9}{2} + \binom{9}{3} + \binom{9}{4} = \binom{9}{5} + \binom{9}{6} + \binom{9}{7} + \binom{9}{8} + \binom{9}{9} = 2^8$$
Hence $\binom{9}{1} + \binom{9}{2} + \binom{9}{3} + \binom{9}{4} = 2^8-1$. So the median is the first element having 5 digits, which is $12345$.
| {
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If $x+y+z=1$ Find the maximum of $\frac{x-y}{\sqrt{x+y}}+\frac{y-z}{\sqrt{y+z}}+\frac{z-x}{\sqrt{z+x}}$ Question -
Let $x, y, z$ be non-negative real numbers such that $x+y+z=1 .$ Find the maximum of
$$
\frac{x-y}{\sqrt{x+y}}+\frac{y-z}{\sqrt{y+z}}+\frac{z-x}{\sqrt{z+x}}
$$
My work -
first WLOG $x\ge y\ge z$
I rewrite given inequality like this
$\frac{x-y}{x+y}{\sqrt{x+y}}+\frac{y-z}{y+z}{\sqrt{y+z}}+\frac{z-x}{z+x}{\sqrt{z+x}}$
and now applying C-S and using fact that $(\frac{x-y}{x+y})^2 < 1 $ and $x+y+z=1$
i get maximum value as ${\sqrt6}$.
Is my proof correct ???
thankyou
| Your answer is not correct because your value $\sqrt6$ does not occur.
Also, you need to prove that we can assume $x\geq y\geq z$.
My solution:
Let $z=\min\{x,y,z\}$.
Now, we'll prove that if $y\geq x\geq z$ so
$$\frac{x-y}{\sqrt{x+y}}+\frac{y-z}{\sqrt{y+z}}+\frac{z-x}{\sqrt{z+x}}\leq\frac{y-x}{\sqrt{x+y}}+\frac{x-z}{\sqrt{x+z}}+\frac{z-y}{\sqrt{z+y}},$$ which will say us that we can assume that $x\geq y\geq z$.
Indeed, we need to prove that $$\frac{x-y}{\sqrt{x+y}}+\frac{y-z}{\sqrt{y+z}}+\frac{z-x}{\sqrt{z+x}}\leq0$$ or
$$\frac{x-y}{\sqrt{x+y}}+\frac{y-x+x-z}{\sqrt{y+z}}+\frac{z-x}{\sqrt{z+x}}\leq0$$ or
$$(y-x)\left(\frac{1}{\sqrt{y+z}}-\frac{1}{\sqrt{x+y}}\right)\leq(x-z)\left(\frac{1}{\sqrt{x+z}}-\frac{1}{\sqrt{y+z}}\right)$$ or
$$\frac{(y-x)(x-z)}{\sqrt{(x+y)(y+z)}(\sqrt{x+y}+\sqrt{y+z})}\leq\frac{(y-x)(x-z)}{\sqrt{(x+z)(y+z)}(\sqrt{x+z}+\sqrt{y+z})}$$ or
$$x+z+\sqrt{(x+z)(y+z)}\leq x+y+\sqrt{(x+y)(y+z)},$$ which is obvious.
Now, let $$f(z)=\frac{x-y}{\sqrt{x+y}}+\frac{y-z}{\sqrt{y+z}}+\frac{z-x}{\sqrt{z+x}},$$
where $x\geq y\geq z$.
Thus, $$f'(z)=\frac{3x+z}{2\sqrt{(x+z)^3}}-\frac{3y+z}{2\sqrt{(y+z)^3}}=$$
$$=\frac{(3x+z)^2(y+z)^3-(3y+z)^2(x+z)^3}{2\sqrt{(x+z)^3(y+z)^3}((3x+z)\sqrt{(y+z)^3}+(3y+z)\sqrt{(x+z)^3})}=$$
$$=\frac{(y-x)(9x^2y^2+6xyz(x+y)+(x^2+y^2-8xy)z^2-6(x+y)z^3-3z^4)}{2\sqrt{(x+z)^3(y+z)^3}((3x+z)\sqrt{(y+z)^3}+(3y+z)\sqrt{(x+z)^3})}\leq0.$$
Id est, $f$ decreases, which says that it's enough to assume $z=0$ and we obtain:
$$\max f=\max f(0)=\max_{x+y=1,x\geq y>0}\left(x-y+\sqrt{y}-\sqrt{x}\right)=$$
$$=\max_{\frac{1}{2}\leq x\leq 1}(2x-1+\sqrt{1-x}-\sqrt{x})=\sqrt{\frac{71-17\sqrt{17}}{32}},$$
where $$x_{max}=\frac{1}{2}+\sqrt{\frac{23-\sqrt{17}}{128}}.$$
Can you get this answer?
It's interesting, that the equation, which we obtain, when the derivative is equal to $0$, is nice enough:
$$4\sqrt{x(1-x)}=\sqrt{1-x}+\sqrt{x},$$ which we can solve by the substitution $$\sqrt{1-x}+\sqrt{x}=t.$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Second order ODE solution The task is to find a solution of second order ODE:
$$
2xx''-3(x')^2=4x^2; x(0) = 1; x'(0) = 0
$$
Now, I tried:
$$
x'' = 2x + \frac{3(x')^2}{2x}
$$
here we substitute $u(x) = x'$ to obtain:
$$
uu'=2x+\frac{3}{2}\frac{u^2}{x} \\
u' = 2\frac{x}{u} + \frac{3u}{2x}
$$
here I substitute $t = \frac{x}{u}$ to get
$$
t'=\frac{2t+\frac{3}{2}\frac{1}{t} - t}{u} \\
\frac{dt}{du} = \frac{t+\frac{3}{2t}}{u} \\
\frac{du}{u} = \frac{dt}{t + \frac{3}{2t}}
$$
integrate both sides
$$
\ln u = \frac{1}{2}\ln{(2t^2 + 3)} + c \\
u = \sqrt{2t^2+3}e^c \\
x' = e^c\sqrt{2(\frac{x}{x'})^2 + 3}
$$
Here I got stuck and I think I made a mistake somewhere, because I can't use $x'(0)=1$ and $x(0)=0$ to get the constant $c$. Could you help me?
| Let me work with $y(x)$, then the ODE is
$$yy''-(4y^2+3y'^2)=0~~~~(1)$$
Let $y'=p, y''=\frac{dp}{dx}=\frac{dp}{dy} \frac{dy}{dx}=p\frac{dp}{dy}$
Then (1) becomes
$$yp\frac{dp}{dy}-(4y^2+3p^2)=0 \implies ypdp-(4y^2+3p^2)dy=0~~~~(2)$$
The integrating factor for this in-exact ODE is
$$\mu=\exp[\int \frac{-6p-p}{py}dy=y^{-7}$$
The ODE (2) becomes
$$py^{-6} dp-y^{-7}(4y^2+3p^2) dy=0$$
Its solution i1
$$\int y^{-6} p dp-\int 4y^{-5}dy=C \implies \frac{p^2}{2y^6}+\frac{1}{y^4}-C=0$$
$$\implies p =\frac{dy}{dx}=\pm y\sqrt{2} \sqrt{Cy^4-1} \implies C=1$$ $$ \implies \int \frac{dy}{y\sqrt{y^4-1}}=\pm \int \sqrt{2} dx+B \implies -\frac{1}{2} \tan^{-1}\frac{1}{\sqrt{y^4-1}}= \pm \sqrt{2} x+B$$
$$\implies \tan^{-1}\frac{1}{\sqrt{y^4-1}}=\mp 2x\sqrt{2}+D \implies D=\pi/2$$
$$ \implies y(x)=\sqrt{\sec 2x\sqrt{2}}$$
| {
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How can I approach this inequality? Let $a, b$ and $c$ be three non-zero positive numbers. Show that:
$$\sqrt{\frac{2a}{a + b}} + \sqrt{\frac{2b}{b + c}} + \sqrt{\frac{2c}{a + c}} \leq 3$$
I know the triangular inequality would help here, but I don't know how to approach it.
I started by $a+b≥a$ then that gives $\frac{1}{a+b}≤\frac{1}{a}$ by muliplyting both sides by $2a$ we get $\frac{2a}{a+b}≤\frac{2a}{a}$ which leads eventually to $\frac{2a}{a+b}≤2$ and by adding the square root to both sides we get $\sqrt{\frac{2a}{a+b}}\leq\sqrt2$ and doing the same thing to the other terms we get $\sqrt{\frac{2a}{a+b}}+\sqrt{\frac{2b}{b+c}}+\sqrt{\frac{2c}{c+a}}\leq3\sqrt2$ beyond that I don't have any idea if that would lead to anything useful or not.
| This is not the most elegant approach, but since the inequality is homogeneous we may as well assume $a=x>0, b=1, c=y>0$ and study the behaviour of
$$ f(x,y) = \sqrt{\frac{2x}{x+1}}+\sqrt{\frac{2}{1+y}}+\sqrt{\frac{2y}{x+y}} $$
over $(0,+\infty)^2$. If $x\to 0$ or $y\to 0$ we have $f(x,y)\leq 2\sqrt{2}<3$. By solving
$\frac{\partial f}{\partial x}=0$ we find that the first partial derivative only vanishes over the curve $y=x^2$ and over the curve $y=\frac{1}{2}\left(-3x-x^2+(1+x)\sqrt{4x+x^2}\right)$. By solving $\frac{\partial f}{\partial y}=0$ we find three curves, one of them being $y=\sqrt{x}$ and the other two being defined only for $x\geq 4$. The only point in which an $f_x$-curve meets an $f_y$-curve is $(1;1)$, so $f(1,1)=3$ is the only stationary point and an actual maximum.
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove that $\sqrt{\frac{2 a^{2}+b c}{a^{2}+2 b c}}+\sqrt{\frac{2 b^{2}+c a}{b^{2}+2 c a}}+\sqrt{\frac{2 c^{2}+a b}{c^{2}+2 a b}} \geq 2 \sqrt{2}$ Question -
Prove that for all non-negative real numbers a,b, c, we have
$$
\sqrt{\frac{2 a^{2}+b c}{a^{2}+2 b c}}+\sqrt{\frac{2 b^{2}+c a}{b^{2}+2 c a}}+\sqrt{\frac{2 c^{2}+a b}{c^{2}+2 a b}} \geq 2 \sqrt{2}
$$
My work -
we may assume that $a b c=1$
The problem becomes
$$
\sqrt{\frac{2 x+1}{x+2}}+\sqrt{\frac{2 y+1}{y+2}}+\sqrt{\frac{2 z+1}{z+2}} \geq 2 \sqrt{2}
$$
where $x=a^{3}, y=b^{3}, z=c^{3}$
now i did not know where to go from here ...i tried all classic inequalities like chebyshev,re-arangement but none of them working .
can anyone solve this using classic inequalities
any help will be appreciated
thank you
| Take
$$
\sqrt{\frac{2 x+1}{x+2}}+\sqrt{\frac{2 y+1}{y+2}}+\sqrt{\frac{2 z+1}{z+2}}
$$
where $x=\frac{a^2}{bc}>0, y=\frac{b^2}{ac}>0, z=\frac{c^2}{ab}>0$.
Note that functions
$$f(w)=\sqrt{\frac{2 w+1}{w+2}}$$
are strictly increasing for $w\in[0,\infty)$.
Without the loss of generality assume that $x\geq y\geq z$. Hence
$$\frac{a^2}{bc}\geq \frac{b^2}{ac}\geq \frac{c^2}{ab},\, a\geq b\geq c.$$
Furthermore
$$x\geq\frac{a^2}{c^2}=e^2,\,y\geq\frac{c^2}{a^2}=\frac{1}{e^2},\,z\geq\frac{c^2}{a^2}=\frac{1}{e^2}.$$
Using all of this we get
$$
\sqrt{\frac{2 x+1}{x+2}}+\sqrt{\frac{2 y+1}{y+2}}+\sqrt{\frac{2 z+1}{z+2}} \geq \sqrt{\frac{2 e^2+1}{e^2+2}}+2\sqrt{\frac{2+e^2}{1+2e^2}}.
$$
Note that function on the right hand side has an infimum
$$\inf\limits_{e\rightarrow\infty}\sqrt{\frac{2 e^2+1}{e^2+2}}+2\sqrt{\frac{2+e^2}{1+2e^2}}=2\sqrt{2}$$
So, we get
$$
\sqrt{\frac{2 x+1}{x+2}}+\sqrt{\frac{2 y+1}{y+2}}+\sqrt{\frac{2 z+1}{z+2}}\geq 2\sqrt{2}.
$$
| {
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Nonlinear Differential Equation of High Degree Any help, please? How can I start to solve them? I tried to use $y'=p$. Also I tried $x=e^x$ and so many methods, but I couldn't reach them to the end. I always got blocked in the middle. Thanks in advance.
First equation: $$6x^2y-6y'^2+(12x^2-3x^3)y'+x^5-6x^4=0.$$
Second equation: $$x=\frac{y\ln y}{y'}-\frac{y'^2}{y^2}$$ with $y>0$.
| FIRST EQUATION :
$$6x^2y-6y'^2+(12x^2-3x^3)y'+x^5-6x^4=0.$$
By inspection one can see a particular solution :
$$y=\frac13 x^3\quad\implies\quad y'=x^2$$
$6x^2(\frac13 x^3)-6(x^2)^2+(12x^2-3x^3)(x^2)+x^5-6x^4=$
$=2x^5-6x^4+12x^4-3x^5+x^5-6x^4=0.$
This draw us to the change of function :
$$y(x)=\frac13 x^3+u(x)\quad\implies\quad y'=x^2+u'$$
$$6x^2(\frac13 x^3)-6(x^2+u')^2+(12x^2-3x^3)(x^2+u')+x^5-6x^4=0.$$
After simplification :
$$6x^2u-3x^3u'-6u'^2=0$$
The change of $x$ into $-x$ doesn't change the equation. This draw us to the change of variable $X=x^2$ which leads to :
$$u(X)=X\frac{du}{dX}+4\left(\frac{du}{dX}\right)^2$$
This is a Clairault's differential equation.
SECOND EQUATION :
$$x=\frac{y\ln y}{y'}-\frac{y'^2}{y^2}$$
Since Aryadeva already answered there is no need to reapeat the method of solving.
| {
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} |
Finding $\frac{c\sin(A-B)}{a^2-b^2}-\frac{b\sin(C-A)}{c^2-a^2}$
If $a, b$ and $c$ (all distinct) are the sides of a triangle ABC opposite to the angles $A, B$ and $C$, respectively, then $\frac{c\sin(A-B)}{a^2-b^2}-\frac{b\sin(C-A)}{c^2-a^2}$ is equal to $?$
By opening, $\sin(A-B)$ as $\sin A\cos B-\cos A\sin B$ and then $\sin A$ as $\frac{a}{2R}$ and $\cos A$ as $\frac{b^2+c^2-a^2}{2bc}$, I am able to get the answer as zero. But I am looking for a shorter solution, maybe by putting values of angles and sides.
My first instinct was to assume the triangle to be equilateral. But the question invalidates that case. Then I thought of a right angled triangle with pythogorean triplet as $3,4,5$. But here, I don't know other two angles.
I wonder if I could just keep a dummy triangle handy whose sides and angles I know, which I could quickly use to solve such questions. Any help please? Thanks.
| Use sine and cosine rules to evaluate
\begin{align}
& \frac{\sin(A-B)}{\sin(A-C)}
= \frac{\sin A\cos B - \cos A \sin B}{\sin A\cos C - \cos A \sin C}\\
=& \frac{a\frac{a^2+c^2-b^2}{2ac} - b\frac{b^2+c^2-a^2}{2bc} }
{a\frac{a^2+b^2-c^2}{2ab} - c\frac{b^2+c^2-a^2}{2bc} }
=\frac{\frac1c [(a^2+c^2-b^2 )- (b^2+c^2-a^2)] }
{\frac1b [(a^2+b^2-c^2 )- (b^2+c^2-a^2)] }
=\frac{b(a^2-b^2)}
{c(a^2-c^2)}\\
\end{align}
Then, rearrange to obtain
$$\frac{c\sin(A-B)}{a^2-b^2}-\frac{b\sin(C-A)}{c^2-a^2}=0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3663019",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
My first linear algebra proof - is it accurate and written correctly? This is the first proof exercise from "Linear Algebra as an Introduction to Abstract Mathematics." It also happens to be my first proof (outside of proving things like properties of integers and trigonometric identities). I'm hoping someone can critique it and let me know what I did right or what I did wrong.
Problem: Let a, b, c, and d be real numbers. Consider the system of equations (1):
$$ax_1 + bx_2 = 0$$
$$cx_1 + dx_2 = 0$$
Note that $x_1=x_2=0$ is a solution for any choice of a, b, c, and d.
Prove that, if $ad-bc\neq0$, then $x_1=x_2=0$ is the only solution.
Proof: We will prove by contrapositive. Assume $x_1=x_2=0$ is not the only solution. Assume
$$A = \begin{bmatrix}a & b \\ c & d\end{bmatrix}, X = \begin{bmatrix}x_1 \\ x_2\end{bmatrix}, 0 = \begin{bmatrix}0 \\ 0\end{bmatrix}$$
Then the system of equations (1) is equivalent to the expression $AX=0$ (2). For (2) to be true, we must have either $A=0$ or $X=0$. Since we are assuming $x_1=x_2=0$ is not the only solution, we assume $X\neq0$. Then, for (2) to be true, we must have $A=0$, and therefore, $a=b=c=d=0$. If $a=b=c=d=0$, then $ab-cd=0$. $\blacksquare$
| You can apply the following row operations in order to get that
\begin{align*}
\begin{cases}
ax_{1} + bx_{2} = 0\\\\
cx_{1} + dx_{2} = 0
\end{cases} & \Longrightarrow
\begin{cases}
acx_{1} + bcx_{2} = 0\\\\
cx_{1} + dx_{2} = 0
\end{cases}\\\\
& \Longrightarrow
\begin{cases}
(acx_{1} + bcx_{2}) - (acx_{1} + adx_{2}) = 0\\\\
cx_{1} + dx_{2} = 0
\end{cases}\\\\
& \Longrightarrow
\begin{cases}
(bc - ad)x_{2} = 0\\\\
cx_{1} + dx_{2} = 0
\end{cases}
\end{align*}
Similarly, one has that
\begin{align*}
\begin{cases}
ax_{1} + bx_{2} = 0\\\\
cx_{1} + dx_{2} = 0
\end{cases} & \Longrightarrow
\begin{cases}
adx_{1} + bdx_{2} = 0\\\\
cx_{1} + dx_{2} = 0
\end{cases}\\\\ & \Longrightarrow
\begin{cases}
(adx_{1} + bx_{2}) - (bcx_{1} + bdx_{2}) = 0\\\\
cx_{1} + dx_{2} = 0
\end{cases}\\\\ & \Longrightarrow
\begin{cases}
(ad-bc)x_{1} = 0\\\\
cx_{1} + dx_{2} = 0
\end{cases}
\end{align*}
Since $ad - bc \neq 0$, we get that the solution set $S\subseteq\{(0,0)\}$.
Once $x_{1} = x_{2} = 0$ is a solution, it is the unique solution.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3667415",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Integer solutions to multivariate polynomial I'm looking for a way to show that an equation like $$15(1+x+x^{2})(1+y+y^{2})=16(xy)^{2}-1$$ has no solutions with x and y odd positive integers. The 16 and 15 can be changed but I'm trying to generalize it. Any help is appreciated.
| Factorization should be useful, but another idea is to note that
if the equation is $a(x^2+x+1)(y^2+y+1)=b(xy)^2-1$ and $a<b$, then $min(x,y)$ is upper-bounded by a function of $a,b$ and it is sufficient to examine those values (and solve for the other).
Specifically, here is one (not necessarily sharp) bound: $min(x,y) \leq \dfrac{\sqrt{b}}{\sqrt{b}-\sqrt{a}}$.
To prove the bound, we write: $a(1+\dfrac{1}{x}+\dfrac{1}{x^2})(1+\dfrac{1}{y}+\dfrac{1}{y^2})=b-\dfrac{1}{x^2y^2}$;
thus $a\dfrac{xy}{(x-1)(y-1)}>b-\dfrac{1}{xy(x-1)(y-1)}$,
and hence $axy>b(x-1)(y-1)-1$, so that $\left(1-\dfrac{1}{x}\right)\left(1-\dfrac{1}{y}\right)\leq\dfrac{a}{b}$.
So if $x,y \geq n$, then $1-1/n \leq \dfrac{\sqrt{a}}{\sqrt{b}}$ and thus $n \leq \dfrac{\sqrt{b}}{\sqrt{b}-\sqrt{a}}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3668369",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Polynomial transformations and Vieta's formulas
Let $f(x)$ be a monic, cubic polynomial with $f(0)=-2$ and $f(1)=−5$. If the sum of all solutions to $f(x+1)=0$ and to $f\big(\frac1x\big)=0$ are the same, what is $f(2)$?
From $f(0)$ I got that $f(x)=x^3+ax^2+bx-2$ and from $f(1)=-5$ that $a+b = -4$ however I'm not sure how to use the info about the transformations to find $f(2).$ It seems that $(x+1)$ is a root for $f(x+1)$ and the same logic applies for $f\big(\frac1x\big)$?
Should I use Vieta's here or what's the appropriate way to go?
| If $\alpha$ is a root of $f(x)$ then $\alpha - 1$ is a root of $f(x+1)$ (and vice-versa). So the sum of roots of $f(x+1)$ is the sum of roots of $f(x)$, minus $3$ which by Vieta' formula is $-a$ minus $3 =-a-3$.
Now, the roots of $f(\frac 1x) = \frac{-2x^3+bx^2+ax+1}{x^3}$ are also the roots of $-2x^3+bx^2+ax+1$, the reciprocal polynomial of $f$. The sum of roots of this polynomial is $\frac b2$ by Vieta's formula.
Finally, $-a-3= \frac b2$. Conclude.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3669700",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
Prove that $x^2+y^2-\sqrt{2}xy < 4$ provided $1\le x,y \le \frac{5}{2}$. If possible I would like an elegant solution to the following problem:
Let $x,y\in\mathbb{R}$ such that $1\le x \le \frac{5}{2}$ and $1\le y \le \frac{5}{2}$. Prove that
$x^2+y^2-\sqrt{2}xy < 4$
I'm aware that you can use Lagrange multipliers but I want an elementary solution using elementary inequalities that are available up to the high school level, or inequalities used in olympiads.
Remarks
I've substituted the obvious specific values for $x,y$ (we can assume without loss of generality that $x\le y$) below
\begin{array}{|c|c|c|}
\hline
x & y &x^2+y^2-\sqrt{2}xy \text{ (rounded to 3 d.p.)}\\ \hline
1 & 1 & \approx 0.587\\ \hline
1 & \frac{5}{2} & \approx 3.714\\ \hline
\frac{5}{2} & \frac{5}{2} & \approx 3.661\\ \hline
\end{array}
Some thoughts:
*
*I've tried solving it in different equivalent forms. Using the following identity
$x^4+y^4 = \left(x^2+y^2+\sqrt{2}xy \right) \left(x^2+y^2-\sqrt{2}xy \right)$
I attempted to prove $x^4+y^4 < 4 \left(x^2+y^2 + \sqrt{2}xy \right)$ (though I'm not sure if it really helps) but to no avail.
*Also, crudely approximating via $x^2+y^2 -\sqrt{2}xy < \left(\frac{5}{2}\right)^2 + \left(\frac{5}{2}\right)^2 - \sqrt{2} \cdot 1 \cdot 1 $ doesn't work since the RHS is clearly larger than 4
| The hessian of the function $f(x,y)=x^2+y^2-\sqrt2xy$ is
$$\begin{vmatrix}
2&-\sqrt2\\
-\sqrt2 &2
\end{vmatrix}=2>0,
$$
hence $f(x,y)$ is convex and therefore takes its maximal value on an extreme point of the domain. A simple check (which you have already done) shows that the maximal value is
$$f\left(1,\frac52\right)=f\left(\frac52,1\right)<4.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3674487",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Generator of a solution space In $\mathbb{F}_3$, consider:
$a+b+2c+d=0$
$2a+c+d=0$
$2a+b+c=0$
I got $a=0$, $c=-d$ and $b=d$, so we have $\mathbb{L}=\left \{ t \begin{pmatrix} 0\\1\\-1 \\ 1 \end{pmatrix} : t \in \mathbb{F}_3
\right \}$. Does this mean that $(0,1,-1,1)^T$ is the generator of the solution space?
| And easy way: form and then bring to echelon form the coefficients' matrix:
$$\begin{pmatrix}1&1&1&1\\
2&0&1&1\\
1&1&1&0\end{pmatrix}\longrightarrow\begin{pmatrix}1&1&1&1\\
0&2&0&0\\
0&0&0&2\end{pmatrix}\implies\begin{cases}d=0=b\\{}\\a=-c=2c\end{cases}$$
so the general solution is
$$\left\{\;\begin{pmatrix}2c\\0\\c\\0\end{pmatrix}\;\right\}=t\begin{pmatrix}2\\0\\1\\0\end{pmatrix}\,,\,\,t\in\Bbb F_3=\text{Span}_{\Bbb F_3}\left\{\;\begin{pmatrix}2\\0\\1\\0\end{pmatrix}\;\right\}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3676994",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Suppose $f \circ g(x) = \frac{x^2-6x+2}{x+1}$ and $g(x) = 1 - x$. Then $f(-1)$ is equal to... I tried to substitute the value of $g(x)$ to every $x$ in $f \circ g(x) = \frac{x^2-6x+2}{x+1}$ and ended up with $\frac{x^2+4x-3}{-x+2}$.
Although honestly I do not know if that helps, or what I could do next.
I thought maybe this question just requires me to substitute $-1$ to every $x$ in $f\circ g(x) = \frac{x^2-6x+2}{x+1}$?
| We have \begin{align*}f(-1) &= (f \circ g)(g^{-1}(-1)) \\ &= (f \circ g)(1-(-1)) \qquad \text{(as $g^{-1}(x)=1-x$)} \\ &= \frac{2^2-6\cdot 2 + 2}{2+1} \\ &= -2\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3681089",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
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