Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Prove that there are only 3 integer solutions to $2^a+3 ^b=5^c$ I have figured out that $a \pmod{2} = b\pmod{2} =c \pmod{2} $ by making a few modulo tables and using that both sides of the equation must be divisible by $5$, and I have found the three solutions, I'm just at a loss as to how to prove that those are the only three. I struggle specifically with how to move from something being divisible by $5$ to something being a power of $5$. I have also tried to use the binomial expansion by noting that $5=2+3$, but I quickly ran out of ideas what to do with it.
Any help would be appreciated!
| There are three solutions which can all be found by elementary means.
If $b$ is odd
$$2^a+3\equiv 1 \bmod 4$$
Therefore $a=1$ and $b$ is odd.
If $b>1$, then $2\equiv 5^c \bmod 9$ and $c\equiv 5 \bmod 6$
Therefore $2+3^b\equiv 5^c\equiv3 \bmod 7$ and $b\equiv 0 \bmod 6$, a contradiction.
The only solution is $(a,b,c)=(1,1,1)$.
If $b$ is even, $c$ is odd
$$2^a+1\equiv 5 \bmod 8$$
Therefore $a=2$.
If $b\ge 2$, then $4\equiv 5^c \bmod 9$ and $c\equiv 4 \bmod 6$, a contradiction.
The only solution is $(a,b,c)=(2,0,1)$.
If $b$ and $c$ are even
Let $b=2B$ and $c=2C$. Then
$$2^a=5^{2C}-3^{2B}=(5^C-3^B)(5^C+3^B)$$
Therefore $5^C-3^B$ is also a (smaller) power of 2.
A check of $(B,C)=(0,1)$ and $(1,1)$ yields the third solution $(a,b,c)=(4,2,2)$.
$(B,C)=(2,2)$ does not yield a further solution and we are finished.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3358683",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Partial fraction expansion - handling multiple roots I am trying to solve the partial fraction expansion of the following:
$X(z) = \dfrac{3-15z}{9z^2-6z+1}$.
So I calculated the roots of $9z^2 - 6z + 1$.
This gives $n_1 = n_2 = \frac{1}{3}$
and $9z^2-6z+1 = 9(z-\frac{1}{3})(z-\frac{1}{3})$.
Now I don't know how to deal with it.
If there were no factor $9$, I would take the approach $X(z) = \frac{A}{z-\frac{1}{3}} + \frac{B}{(z-\frac{1}{3})^2}$.
And if there were two different roots, $z_1$ and $z_2$, I'd take the approach $X(z) = \frac{A}{z-z_1} + \frac{B}{z-z_2}$.
But since neither of these two cases is what we have here, I don't know what to do.
| You could absolutely do it your way: \begin{align}
X(z)=\frac{3-15z}{9z^2-6z+1}&=\frac{3-15z}{9(z-\frac13)^2}\\
&=\frac19\bigg(\frac{3-15z}{(z-\frac13)^2}\bigg)\\
&=\frac19\bigg(\frac{A}{z-\frac13}+\frac{B}{(z-\frac13)^2}\bigg)
\end{align}
Not impossible, but not very nice.
Alternatively:
\begin{align}
X(z)=\frac{3-15z}{9z^2-6z+1}&=\frac{3-15z}{(3z-1)^2}\\
&=\frac{A}{3z-1}+\frac{B}{(3z-1)^2}\\
&=\frac{A(3z-1)+B}{(3z-1)^2}\\
&=\frac{3Az-A+B}{(3z-1)^2}\Rightarrow \left\{\begin{array}
& 3A &=-15\\
-A+B&=3\end{array} \right. \Rightarrow (A,B)=(-5,-2)\\
&=-\frac{5}{3z-1}-\frac{2}{(3z-1)^2}
\end{align}
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Finding $f\in\mathbb{Q}[x]$ such that $f(\sqrt{2}+\sqrt{3})=\sqrt{2}$ and $\deg(f)\leq 3$. What's wrong with my approach?
I'd like to find a polynomial $f(x) \in \mathbb{Q}[x]$ satisfying
$$f(\sqrt{2}+\sqrt{3})=\sqrt{2}$$
and $\deg(f) \leq 3$.
What I've been trying is the following:
Since $f(\sqrt{2}+\sqrt{3})=\sqrt{2}$,
then $f(\sqrt{2}+\sqrt{3})-\sqrt{2}=0$.
so think of $g(x)=f(x+\sqrt{3})-x$ as a polynomial over $\mathbb{Q}(\sqrt{3})$.
And I've already known that $x^2 -2$ is irreducible over $\mathbb{Q}(\sqrt{3})$
$g(x)$ has a $\sqrt{2}$ as a root of itself, $x^2 -2$ divides $g(x)$ in $\mathbb{Q}(\sqrt{3})[x]$.
$g(x)$ must be of the form $(x^2 -2)(ex+f)$ where $e, f \in \mathbb{Q}(\sqrt{3})$
and also of the form $a(x+\sqrt{3})^3 +b(x+\sqrt{3})^2 +c(x+\sqrt{3}) +d -x$, where $f(x)=ax^3 +bx^2 +cx+d \in \mathbb{Q}[x]$.
after comparing the coefficients of two polynomials, I found that $f(x)=\frac{1}{4}x^3-\frac{9}{4}x$.
But the actual polynomial is $\frac{1}{2}x^3-\frac{9}{2}x$.
There must be a flaw in the above reasoning.
Where did I do a mistake? Could you point it out?
Thank you.
| The error appears to be somewhere in the unwritten details of "after comparing the coefficients of two polynomials, I found that...".
Here's a lowbrow approach:
Hint Denote $\beta := \sqrt{2} + \sqrt{3}$. We're looking to solve $\sum_{i = 0}^3 a_i \beta^i = \sqrt{2}$ in rational numbers $a_i$. Computing gives that
$$\beta^2 = 5 + 2 \sqrt{6}, \qquad \beta^3 = 11 \sqrt{2} + 9 \sqrt{3} .$$ Now, $1, \sqrt{2}, \sqrt{3}, \sqrt{6}$ are linearly independent over $\Bbb Q$, but the only power in $\beta^0, \ldots, \beta^3$ in which a nonzero rational multiple of $\sqrt{6}$ occurs is $\beta^2$, so $a_2 = 0$.
Among $\beta^0, \beta^1, \beta^3$ only $\beta^0$ has a nonzero rational summand, so $a_0 = 0$. Thus, we are solving $$a_1 (\sqrt{2} + \sqrt{3}) + a_3 (11 \sqrt{2} + 9 \sqrt{3}) = \sqrt{2} .$$ Linear independence now implies that we can compare coefficients of $\sqrt{2}$ and $\sqrt{3}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3363107",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
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Use $T(x,y)=\begin{pmatrix}\frac{\sin x}{\cos y}\\ \frac{\sin y}{\cos x}\end{pmatrix}$ to evaluate $\sum_{n=1}^{\infty}\frac{1}{n^2}$ Let $O=\{(x,y) \in \mathbb R^2: x,y>0, x+y<\frac{\pi}{2}\}$ and $P=(0,1)\times (0,1)$.
Then $T:O \to \mathbb R^2, T(x,y)=\begin{pmatrix}\frac{\sin x}{\cos y}\\ \frac{\sin y}{\cos x}\end{pmatrix}$ is a diffeomorphism with $T(O)=P$
How can I use $T$ to evaluate $\sum_{n=1}^{\infty}\frac{1}{n^2}$?
The hints of the exercise are:
$3\sum_{n=1}^{\infty}\frac{1}{n^2}=4\sum_{n=1}^{\infty}\frac{1}{(2n-1)^2}$ and for $n \in \mathbb N \cup \{0\}$: $\frac{1}{(n+1)^2}=\int_{(0,1)\times(0,1)}(xy)^n d\lambda_2(x,y)$
| Well,
$$ \frac{1}{(n+1)^2}=\iint_{(0,1)^2}(xy)^n\,dx\,dy $$
implies
$$ \sum_{n\geq 0}\frac{1}{(2n+1)^2}=\iint_{(0,1)^2}\frac{dx\,dy}{1-x^2 y^2}.$$
Since the diffeomorphism $T$ maps the triangle $O=\{(a,b):a,b>0,a+b<\frac{\pi}{2}\}$ into the square $P=(0,1)^2$, by letting $x=\frac{\sin a}{\cos b},y=\frac{\sin b}{\cos a}$ the RHS of the previous line is converted (considering the Jacobian) into
$$ \iint_{O} \frac{1-\tan^2(a)\tan^2(b)}{1-\tan^2(a)\tan^2(b)}\,da\,db $$
so $\sum_{n\geq 0}\frac{1}{(2n+1)^2}$ turns out to be just the area of $O$, i.e. $\frac{\pi^2}{8}$. After this we get
$$ \zeta(2)=\sum_{n\geq 1}\frac{1}{n^2}=\sum_{m\geq 1}\frac{1}{(2m)^2}+\sum_{m\geq 0}\frac{1}{(2m+1)^2} = \frac{1}{4}\zeta(2)+\frac{\pi^2}{8}$$
so we have a solution of Basel problem, $\zeta(2)=\color{red}{\frac{\pi^2}{6}}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3364583",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Limit of $f(x, y)$ at $(0, 0)$ I need to show
$$\lim_{(x, y) \rightarrow (0,0)} \frac{xy(x^2-y^2)}{(x^2+y^2)^{3/2}} = 0
$$
Not really sure how to go about this without using the epsilon delta definition, which I would prefer not to. Any sort of help is appreciated.
Edit: I do have the inequality:
$$\frac{xy(x^2-y^2)}{(x^2+y^2)^{3/2}} \le \frac{xy(x^2+y^2)}{(x^2+y^2)^{3/2}} = \frac{xy}{(x^2+y^2)} $$
| Let $(x,y) \not =(0,0)$.
$0\le |\dfrac{xy(x^2-y^2)}{(x^2+y^2)^{3/2}}| \le \dfrac{|xy||x^2+y^2|}{(x^2+y^2)^{3/2}} $
$\le \dfrac{(x^2+y^2)^2}{(x^2+y^2)^{3/2}}= (x^2+y^2)^{1/2}.$
Used:
1)$|x^2-y^2| \le |x^2+y^2|;$
2)$x^2+y^2 \ge |xy|$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3365419",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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a) Find a base for $W_1 \cap W_2$ and b) Find a base for $W_1 + W_2$ Let
$$W_1 = \left\{ \begin{pmatrix}
a & b & c \\
-b & a & b \\
c & b & a \\
\end{pmatrix} :\, a, b, c \in \Bbb R \right\}$$
$$W_2 = \left\{
\begin{pmatrix}
a & 0 & 0 \\
b & 0 & c \\
c & b & a \\
\end{pmatrix} :\, a, b, c \in \Bbb R \right\}$$
a) Find a basis for $W_1 \cap W_2$
b) Find a basis for $W_1 + W_2$
For a) I`m not sure if the set for $W_1 \cap W_2 = $ $ \{
\begin{pmatrix}
0 & 0 & 0 \\
0 & 0 & 0 \\
0 & 0 & 0 \\
\end{pmatrix}\}
$
so the base is only the zero matrix.
For b) I belive the set is: $W_1 + W_2 = $ $ \{
\begin{pmatrix}
2a & b & c \\
0 & a & b+c \\
2c & 2b & 2a \\
\end{pmatrix}
$:$ a, b, c \in R\}$
But don`t know how to give an
appropriate base for this set.
| a) Since $W_1\cap W_2=\{0\}$, the only basis of $W_1\cap W_2$ is the empty set.
b) Since $W_1\cap W_2=\{0\}$, a way of finding a basis $B$ of $W_1+W_2$ is to take a basis $B_1$ of $W_1$ and a basis $B_2$ of $W_2$ and then to take $B=B_1\cup B_2$. It is natural to take$$B_1=\left\{\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix},\begin{bmatrix}0&1&0\\-1&0&1\\0&1&0\end{bmatrix},\begin{bmatrix}0&0&1\\0&0&0\\1&0&0\end{bmatrix}\right\}$$and$$B_2=\left\{\begin{bmatrix}1&0&0\\0&0&0\\0&0&1\end{bmatrix},\begin{bmatrix}0&0&0\\1&0&0\\0&1&0\end{bmatrix},\begin{bmatrix}0&0&0\\0&0&1\\1&0&0\end{bmatrix}\right\}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3365840",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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A First Course in Mathematical Analysis, Ch 1, 1.6 Exercises, Section 1.3, ex 2 Prove that
\begin{align*}
(a^{2}+b^{2})(c^{2}+d^{2}) & \ge(ac+ad)^{2}\\
a^{2}c^{2}+a^{2}d^{2}+b^{2}c^{2}+b^{2}d^{2} & \ge a^{2}c^{2}+2a^{2}cd+a^{2}d^{2}\\
b^{2}c^{2}+b^{2}d^{2} & \ge 2a^{2}cd\\
b^{2}c^{2}-2a^{2}cd+b^{2}d^{2} & \ge 0\\
b^{2}(c^{2}+d^{2})-2a^{2}cd & \ge 0\\
\end{align*}
Any hints on things to try? It seems trickier than say
$$
(a+b)^{2}(c+d)^{2}\ge(ac+bd)^{2}
$$
Proof (direct):
\begin{align*}
a^{2}c^{2}+a^{2}d^{2}+b^{2}c^{2}+b^{2}d^{2} & \ge a^{2}c^{2}+2abcd+b^{2}d^{2}\\
a^{2}d^{2}+b^{2}c^{2} & \ge2abcd\\
a^{2}d^{2}-2abcd+b^{2}c^{2} & \ge0\\
a^{2}d^{2}-2(ad)(bc)+b^{2}c^{2} & \ge0\\
(ad-bc)^{2} & \ge0
\end{align*}
| As written, the inequality is false.
$$(ac+ad)^2=a(c+d)^2=a^2c^2+2a^2cd+a^2d^2.$$
If $b=0 \text{ but } acd \neq 0 \text{ with } cd \gt 0$, then
$$(a^2+b^2)(c^2+d^2) = a^2(c^2+d^2) = a^2c^2+a^2d^2 \lt a^2c^2+2a^2cd+a^2d^2.$$
As noted in the comments, I suspect the right-hand side is supposed to be $(ac+\color{red} b d)^2.$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Rational $a$ such that $\sin(a\pi)$ or $\tan(a\pi)$ have the form $\pm\sqrt{a_1}\pm\sqrt{a_2}\pm\cdots\pm\sqrt{a_n}$ for rational $a_i$ I'm interested in simple values of trigonometric functions, e.g., $\sin 30^{\circ} = 1/2$ or $\sin 18^{\circ} = \frac{\sqrt{5}-1}{4}$.
Let $S$ be the set of real numbers that can be written of the form $\pm \sqrt{a_1} \pm \sqrt{a_2} \cdots \pm \sqrt{a_k}$ (here $a_i \in \mathbb{Q}$).
*
*Can we determine all $a \in \mathbb{Q}$ that satisfies $\sin(a \pi) \in S$?
*Can we determine all $a \in \mathbb{Q}$ that satisfies $\tan(a \pi) \in S$?
I've heard of this, but here nested square roots are allowed.
| Most probably this is the complete list (between 0 and 90 degrees):
$$\begin{array}{|c|c|c|} \hline
&\sin &\tan\\ \hline
0^{\circ} &0 &0\\ \hline
7.5^{\circ} & &(\sqrt{2}-1)(\sqrt{3}-\sqrt{2})\\ \hline
15^{\circ} &(\sqrt{6}-\sqrt{2})/4 &2-\sqrt{3}\\ \hline
18^{\circ} &(\sqrt{5}-1)/4 &\\ \hline
22.5^{\circ} & &\sqrt{2}-1\\ \hline
30^{\circ} &1/2 &\sqrt{3}/3\\ \hline
37.5^{\circ} & &(\sqrt{2}+1)(\sqrt{3}-\sqrt{2})\\ \hline
45^{\circ} &\sqrt{2}/2 &1\\ \hline
52.5^{\circ} & &(\sqrt{2}-1)(\sqrt{3}+\sqrt{2})\\ \hline
54^{\circ} &(\sqrt{5}+1)/4 &\\ \hline
60^{\circ} &\sqrt{3}/2 &\sqrt{3}\\ \hline
67.5^{\circ} & &\sqrt{2}+1\\ \hline
75^{\circ} &(\sqrt{6}+\sqrt{2})/4 &2+\sqrt{3}\\ \hline
82.5^{\circ} & &(\sqrt{2}+1)(\sqrt{3}+\sqrt{2})\\ \hline
90^{\circ} &1 &\\ \hline
\end{array}$$
Let's prove that $\sin$ and $\tan$ of other angles are not in $S$.
Since $\cos(k \theta)$ is a polynomial of $\cos(\theta)$ for $k \in \mathbb{N}$, if $\cos(\theta) \in S$, $\cos(k \theta) \in S$.
This observation significantly reduces the number of angles we need to check;
for example, once we verify that $\cos(\frac{1}{20} \times 2\pi) \notin S$, we don't need to check angles $\frac{a}{b} \times 2 \pi$ such that $a$ and $b$ are coprime and $b$ is a multiple of $20$. $\tan$ can be handled similarly by using the fact that if $\tan(\theta) \in S$, $\cos(2 \theta) \in S$.
Furthermore, from Galois theory we know which angles can be expressed with (possibly nested) square roots. By combining these results, it suffices to verify that the following values are not in $S$: $\cos(\frac{2\pi}{15}), \cos(\frac{2\pi}{16}), \cos(\frac{2\pi}{20}), \tan(\frac{2\pi}{5})$, and $\cos(\frac{2\pi}{p})$ for Fermat Primes larger than $5$. It's relatively easy to verify the first four.
The conclusion is that, if we can prove the following, the list above is complete.
For all Fermat Prime $p \geq 17$, $\cos(\frac{2 \pi}{p}) \notin S$.
| {
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"url": "https://math.stackexchange.com/questions/3370069",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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how to solve $\lim_{n\to\infty}{\left(\sum_{k=1}^{n}{\frac{1}{\sqrt{n^2+k}}}\right)^{n}}$? $\displaystyle\left(\sum_{k=1}^{n} \frac{1}{\sqrt{n^{2}+1}}\right)^{n}\ge\left(\sum_{k=1}^{n} \frac{1}{\sqrt{n^{2}+k}}\right)^{n}\ge\left(\sum_{k=1}^{n} \frac{1}{\sqrt{n^{2}+n}}\right)^{n}$
left=$\displaystyle\lim_{n\to\infty}{\left(\sum_{k=1}^{n}{\frac{1}{\sqrt{n^2+1}}}\right)^{n}}=e^{\displaystyle n \ln{\frac{n}{\sqrt{n^2+1}}} }=e^{0}=1$
right=$\displaystyle\lim_{n\to\infty}{\left(\sum_{k=1}^{n}{\frac{1}{\sqrt{n^2+n}}}\right)^{n}}=e^{\displaystyle n \ln{\frac{n}{\sqrt{n^2+n}}} }=e^{-\frac{1}{2}}$
left $\ne$ right ,what to do next?
$\displaystyle\lim_{n\to\infty}{\left(\sum_{k=1}^{n}{\frac{1}{\sqrt{n^2+k}}}\right)^{n}}=\lim _{n \rightarrow \infty} e^{\displaystyle n \ln \sum_{k=1}^{n} \frac{1}{\sqrt{n^{2}+k}}(1)}$
$(1)=\displaystyle \lim _{n \rightarrow \infty} n\left(\ln \frac{1}{n} \sum_{k=1}^{n} \frac{1}{\sqrt{1+k/n^{2}}}\right)$$=\lim _{n \rightarrow \infty} n\left(\ln \frac{1}{n} \sum_{k=1}^{n} \frac{1}{\sqrt{1+k/n \cdot 1/n}}\right)$$=\lim _{n \rightarrow \infty} n \ln \int_{0}^{1} \frac{1}{\sqrt{1+x/n}} d x$$=\lim _{n \rightarrow \infty} n \ln \int_{0}^{1} \frac{nd(x/n+1)}{\sqrt{1+x/n}}$$= \lim_{n\to\infty}{n\ln{n \cdot2 \left.\sqrt{1+\frac{x}{n} }\right|_{0}^{1}}}$$= \lim_{n\to\infty}{n\ln{n \cdot2 (\sqrt{1+\frac{1}{n}}-1)}}$$=\lim_{n\to\infty}{n\ln{n \cdot2 (\frac{1}{2n} -\frac{1}{2n^2} +o(\frac{1}{n^2}))}}$$=\lim_{n\to\infty}{n\ln{n \cdot2 (\frac{1}{2n} +\left(\frac{1}{2!}\cdot \frac{1}{2} \cdot \left(\frac{1}{2}-1\right) \right)\frac{1}{n^2} +o(\frac{1}{n^2}))}}=\lim_{n\to\infty}{n \ln{\left(1-\frac{1}{4n}\right)}}=-\frac{1}{4} $
so that $\displaystyle\lim_{n\to\infty}{\left(\sum_{k=1}^{n}{\frac{1}{\sqrt{n^2+k}}}\right)^{n}}=\lim _{n \rightarrow \infty} e^{(1)}=e^{-\frac{1}{4}}$
this solution is right.
| Another solution: Let $$S_n= \sum_{k=1}^{n} \frac{1}{n}\frac{1}{\sqrt{1+\frac{k}{n^2}}}$$
We will prove that $$ \frac{2x+2}{2+x}<\sqrt{1+x} < 1+\frac{x}{2}, x>0~~~~(1)$$
Right one: is nothing but $$2\sqrt{1+x}=-1-(1+x)-[1-\sqrt{1+x}]^2 \le 0.$$
The left one isnothing but $$\frac{2(1+x)}{1+1+x} < \sqrt{1+x} \Rightarrow 2\sqrt{1+x}<1+(1+x) \Rightarrow -[1-\sqrt{1+x}]<0.$$
From (1) it follows that $$ 1+ \frac{k}{2n^2+k} ~<~\sqrt{1+\frac{k}{n^2}}~ < ~1+\frac{k}{2n^2}$$
$$\Rightarrow 1+ \frac{k}{2n^2+\underline{2n}} ~<~\sqrt{1+\frac{k}{n^2}}~ < ~1+\frac{k}{2n^2}$$
$$\Rightarrow \left(1+ \frac{k}{2n^2+2n}\right)^{-1} ~>~\frac{1}{\sqrt{1+\frac{k}{n^2}}}~ > ~\left( 1+\frac{k}{2n^2} \right)^{-1}$$
$$\Rightarrow \left(1- \frac{k}{2n^2+2n}\right) ~>~\frac{1}{\sqrt{1+\frac{k}{n^2}}}~ > ~\left( 1-\frac{k}{2n^2} \right)$$
$$\Rightarrow 1- \sum_{k=1}^{n} \frac{k}{n(2n^2+2n)} ~>~\sum_{k=1}^{n} \frac{1}{n}\frac{1}{\sqrt{1+\frac{k}{n^2}}}~ >~ 1-\sum_{k=1}^{n}\frac{k}{2n^3}$$
$$\Rightarrow 1-\frac{1}{4n}~ > ~S_n~ >~1-\frac{1}{4n}-\frac{1}{4n^2}.$$
Now $$\ln L= n \lim_{n \rightarrow \infty} \ln S_n = \lim_{n \rightarrow \infty} n \ln \left(1-\frac{1}{4n}\right) =\lim_{n \rightarrow \infty} n \frac{-1}{4n}=\frac{-1}{4} \Rightarrow L = e^{-\frac{1}{4}}$$
| {
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"url": "https://math.stackexchange.com/questions/3370438",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 4,
"answer_id": 1
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Different proof for the inequality $b \gcd(a,c)^2+ a \gcd(b,c)^2-c \gcd(a,b)^2 \le abc$? In Encyclopedia of Distances, there is given a distance on natural numbers without zero:
$$d(a,b) = \log\left( \frac{ab}{\gcd(a,b)^2}\right)\,.$$
Taking (again for a reference see the Encyclopedia of Distances), the Schoenberg transform of this metric we get the metric:
$$D(a,b) = 1-\exp\big(-d(a,b)\big) = 1-\frac{\gcd(a,b)^2}{ab}\,.$$
Since this metric satisfies the triangle inequality, we get:
$$D(a,b) \le D(a,c) + D(c,b)\,,$$
and the inequality, after some algebra,
$$b \gcd(a,c)^2+ a \gcd(b,c)^2-c \gcd(a,b)^2 \le abc$$
follows for three natural numbers $a,b,c$.
My question is, if there is a more direct proof for this?
| Write
$$abc-b\gcd(a,c)^2-a\gcd(b,c)^2=c\left(a-\frac{\gcd(a,c)^2}{c}\right)\left(b-\frac{\gcd(b,c)^2}{c}\right)-\frac{\gcd(a,c)^2\gcd(b,c)^2}{c}.$$
Since $ac\ge\gcd(a,c)^2$ and $bc\ge\gcd(b,c)^2$, we see that
$$abc-b\gcd(a,c)^2-a\gcd(b,c)^2\ge-\frac{\gcd(a,c)^2\gcd(b,c)^2}{c}.$$
We only need to show that
$$\frac{\gcd(a,c)^2\gcd(b,c)^2}{c}\leq c\gcd(a,b)^2.$$
But this is equivalent to
$$\gcd(a,c)\gcd(b,c)\leq c\gcd(a,b).$$
However, this follows from
$$\gcd(xy,mn)\geq \gcd(x,m)\gcd(y,n)$$
by taking $x=a$, $y=c$, $m=c$, and $n=b$. From this proof, we can also see that the inequality $$b\gcd(a,c)^2+a\gcd(b,c)^2-c\gcd(a,b)^2\leq abc$$ is an equality iff $a=c$ or $b=c$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3371958",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Solve the equation $\sqrt{a(2^{x}-2)+1}=1-2^{x}$ for every value of the parameter $a$. Question : Solve the equation $\sqrt{a(2^{x}-2)+1}=1-2^{x}$ for every value of the parameter a.
I have solved the problem as follows
$\sqrt{a\left(2^{x}-2\right)+1}=1-2^{x}$
$a\left(2^{x}-2\right)+1=\left(1-2^{x}\right)^{2}=2^{2 x}+1-2 \cdot 2^{x}=2^{2 x}-2^{x+1}+1$
$a 2^{x}-2 a=2^{2 x}-2^{x+1}$
$2^{2 x}-(a-1) 2^{x}+2 a=0$
$y^{2}-(a-1) y+2 a=0$
$y=\frac{(a-1) \pm \sqrt{(a-1)^{2}-8 a}}{2}=\frac{(a-1) \pm \sqrt{a^{2}-10 a+1}}{2}$
$2^{x}=\frac{(a-1) \pm \sqrt{(a-5)^{2}-24}}{2}$
After this there are so many conditions on a. Do i need to check for each and every value ?
| With $z:=2^x-2$, $$\sqrt{az+1}=-z-1$$
or after rewriting,
$$z(z+2-a)=0,z\le-1.$$
Hence
$$z=a-2\le-1,$$ which is $$2^x=a\le1$$
or
$$x=\log_2a,\\0<a\le1.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3376613",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
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} |
Interesting function : $f(x)=\frac{\ln (\frac{x+2}{x+1})}{\ln (1+\frac{1}{x})}$ , $x>1$ Question :
Prove this function :
$$f(x)=\dfrac{\ln \left(\dfrac{x+2}{x+1}\right)}{\ln \left(1+\dfrac{1}{x}\right)},\quad x>1$$
is increasing.
I don't know how I solve by my effort is :
Derivative of $f$ is :
$$f'(x)=\dfrac{(x+2)\ln \left(\frac{x+2}{x+1}\right)-x\ln \left(\frac{x+1}{x}\right)}{x(x+2)(x+1)\ln^{2} \left(1+\dfrac{1}{x}\right)}$$
Then we will prove that $f'(x)≥0$ for any $x>1$
mean that But I don't know how to prove :
$$(x+2)\ln \left(\frac{x+2}{x+1}\right)-x\ln \left(\frac{x+1}{x}\right)\ge0$$
I need some ideas here if any one have.
Thanks!
| EDIT
Here's a proof without expansions.
We wish to prove that
$$f(x) = \frac{\log (\frac{x+2}{x+1})}{\log (\frac{x+1}{x})}$$
is an increasing function, i.e. that
$$f'(x) = \frac{(x+2) \log \left(\frac{1}{x+1}+1\right)-x \log \left(\frac{1}{x}+1\right)}{x (x+1) (x+2) \log ^2\left(\frac{1}{x}+1\right)}>0$$
Since the denominator is positive we need to show that the numerator is positive, i.e.
$$(x+2) \log \left(\frac{1}{x+1}+1\right)-x \log \left(\frac{1}{x}+1\right)>0\tag{1}$$
Now the l.h.s. of $(1)$ can easily be shown to be identical to the integral
$$g= \int_x^{x+1} \log \left(\frac{1}{t}+1\right) \, dt\tag{2}$$
which is a positive quantity since the integrand is positive.
This completes the proof. Q.E.D.
Corollary
Using the same technique we can easily prove that
$$h(x) = \left(1+\frac{1}{x}\right)^x$$
a strictly increasing function for $x>0$.
Indeed, the derivative is
$$h'(x) = \frac{\left(\frac{1}{x}+1\right)^x \left((x+1) \log \left(\frac{1}{x}+1\right)-1\right)}{x+1}$$
Hence we need to show that
$$\log \left(\frac{1}{x}+1\right)-\frac{1}{x+1}>0 $$
But this is obvious since the l.h.s. is equal to the positive integral
$$\int_x^{\infty } \frac{1}{t (t+1)^2} \, dt$$
Original post
For $x\gt0$ we have
$$f(x) = \frac{\log (\frac{x+2}{x+1})}{\log (\frac{x+1}{x})}\\=\frac{\log (\frac{x+1+1}{x+1})}{\log (\frac{x+1}{x})}\\=\frac{\log \left(1+\frac{1}{x+1}\right)}{\log \left(1+\frac{1}{x}\right)}\\\simeq \frac{{\frac{1}{x+1}}}{\frac{1}{x}}=\frac{x}{x+1}=1-\frac{1}{x+1}$$
In the last line we have assumed that $x>>1$ and have used the expansion for small $|\epsilon|<<1$
$$\log(1+\epsilon)= \epsilon-\frac{1}{2}\epsilon^2+ ...$$
Hence
$$\frac{d}{dx}f(x)\simeq \frac{d}{dx}(1-\frac{1}{x+1})= +\frac{1}{(x+1)^2}) \gt 0$$
showing that the function is increasing. Q.E.D.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3379519",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
How to find number of distinct terms in a multinomial expansion?
Find the number of distinct terms in the expansion of
$$\left(x+\frac{1}{x}+\frac{1}{x^2}+x^2\right)^{15}$$
(with respect to powers of $x$)
I saw that the formula for the number of distinct terms (or dissimilar) in a multinomial expansion $(x_1+x_2+x_3+...+x_k)^n$ is $$\binom{n+k-1}{k-1}$$
But applying that here means $$\binom{15+4-1}{4-1}= \binom{18}{3} = 816$$
But the answer says 61.
Is there a difference between the situations for which that formula is meant to be used and that in which I am using??
| Let us equivalently find the number of different monomials in
$$
f=(x^2)^{15}\left(\frac{1}{x^2}+\frac{1}{x}+x+x^2\right)^{15}
=
\left(1 + x + x^3 + x^4\right)^{15}
$$
(after collecting them and writing the polynomial w.r.t. the basis $1,x,x^2,\dots$), which is a reciprocal polynomial of degree $4\cdot 15=60$. It is clear that there are at most $61$ terms (in the degrees $0,1,2,\dots,60$), so let us show that each degree is indeed taken. It is enough to check each degree up to $30$, the polynomial $f$ being reciprocal. We compute
$$
(x^4 + x^3 + x + 1)^2
=x^{8} + 2 x^{7} + x^{6} + 2 x^{5} + 4 x^{4} + 2 x^{3} + x^{2} + 2 x + 1
$$
and see that each degree in the range $0,1,\dots,8$ is taken. So each degree in $\color{red}1\cdot(x^4 + x^3 + x + 1)^{2\cdot 7}$, and in particular also in
$$
f= (x^4 + x^3 + x + \color{red}1)\cdot(x^4 + x^3 + x + 1)^{2\cdot 7}$$ between $0$ and $8\cdot 7$ is taken.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3380253",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
Ramanujan's Nested Radical By noting Ramanujan's Nested Radical, we have
$3 = \sqrt{1+2\sqrt{1+3\sqrt{1+4\sqrt{1+\cdots}}}}$
On the other hand, we can manipulate the number $4$ by applying the similar principle. Here we have
$\begin{aligned} 4 & = \sqrt{16} \\ & = \sqrt{1+15} \\ & = \sqrt{1+2 \cdot \dfrac{15}{2}} \\ & = \sqrt{1+2\sqrt{\dfrac{225}{4}}} \\ & = \sqrt{1+2\sqrt{1+\dfrac{221}{4}}} \\ & = \sqrt{1+2\sqrt{1+3 \cdot \dfrac{221}{12}}} \\ & = \sqrt{1+2\sqrt{1+3\sqrt{\dfrac{44841}{144}}}} \\ & \vdots \\ & = \sqrt{1+2\sqrt{1+3\sqrt{1+4\sqrt{1+\cdots}}}} \end{aligned}$
How can it be? Something contradicts?
| By Ramanujan's nested radical, we can also get this curious identity;
$$n+m=\sqrt{m^2+n\sqrt{m^2+(n+m)\sqrt{m^2+(n+2m)\sqrt{..}}}}$$
Ramanujan proved this identity, and if you want a look at how it's done in a simple manner (not-rigorous), check out this blog. And by plugging in $4$, we get;
$$3+1=\sqrt{1+3\sqrt{1+4\sqrt{1+5\sqrt{1+6...}}}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3382458",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
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Solution of a limit of a sequence $\frac{\sqrt{4n^2+1}-2n}{\sqrt{n^2-1}-n}$ I'm trying to solve the limit of this sequence without the use an upper bound o asymptotic methods:
$$\lim_{n\longrightarrow\infty}\frac{\sqrt{4n^2+1}-2n}{\sqrt{n^2-1}-n}=\left(\frac{\infty-\infty}{\infty-\infty}\right)$$
Here there are my differents methods:
*
*assuming $f(n)=\sqrt{4n^2+1}, \,$$\ g(n)=2n$, $h(n)=\sqrt{n^2-1}$, $\ \psi(n)= n$ $$f(n)-g(n)=\frac{\dfrac{1}{g(n)}-\dfrac{1}{f(n)}}{\dfrac{1}{f(n)\cdot g(n)}}, \quad h(n)-\psi(n)=\frac{\dfrac{1}{\psi(n)}-\dfrac{1}{h(n)}}{\dfrac{1}{h(n)\cdot \psi(n)}}$$
I always have an undetermined form.
*I've done some rationalizations:
$$\lim_{n\longrightarrow\infty}\frac{\sqrt{4n^2+1}-2n}{\sqrt{n^2-1}-n}=\lim_{n\longrightarrow\infty}\frac{\sqrt{4n^2+1}-2n}{\sqrt{n^2-1}-n}\cdot \frac{\sqrt{4n^2+1}+2n}{\sqrt{4n^2+1}+2n}$$ where to the numerator I find $1$ and to the denominator an undetermined form. Similar situation considering
$$\lim_{n\longrightarrow\infty}\frac{\sqrt{4n^2+1}-2n}{\sqrt{n^2-1}-n}=\lim_{n\longrightarrow\infty}\frac{\sqrt{4n^2+1}-2n}{\sqrt{n^2-1}-n}\cdot \frac{\sqrt{n^2-1}+n}{\sqrt{n^2-1}+n}$$
*$$\frac{\sqrt{4n^2+1}-2n}{\sqrt{n^2-1}-n}=\frac{n\left(\sqrt{4+\dfrac{1}{n^2}}-2\right)}{n\left(\sqrt{1-\dfrac{1}{n^2}}-1\right)}\rightsquigarrow \left(\frac{0}{0}\right)$$
At the moment I am not able to think about other possible simple solutions.
| Note
\begin{eqnarray}
\lim_{n\to\infty}\frac{\sqrt{4n^2+1}-2n}{\sqrt{n^2-1}-n}&=&\lim_{n\to\infty}\frac{(\sqrt{4n^2+1}-2n)(\sqrt{4n^2+1}+2n)}{(\sqrt{n^2-1}-n)(\sqrt{n^2-1}+n)}\cdot\frac{\sqrt{n^2-1}+n}{\sqrt{4n^2+1}+2n}\\
&=&-\lim_{n\to\infty}\frac{\sqrt{n^2-1}+n}{\sqrt{4n^2+1}+2n}\\
&=&-\lim_{n\to\infty}\frac{\sqrt{1-\frac1{n^2}}+1}{\sqrt{4+\frac1{n^2}}+2}\\
&=&-\frac12.
\end{eqnarray}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3384280",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Substitution integral with $t=x^3$
$$
\int \frac{3}{t(3+\sqrt[3]{t})} \mathrm{d} t . \text { Substitution } t=x^{3}
$$
$$
\int \frac{3}{x^3(3+\sqrt[3]{x^3})}3x^2
$$
$$
\int \frac{9}{x(3+{x})}
$$
$$
9\int \frac{1}{3x+{x^2}}
$$
Now i'm stuck
my calculation so far
| You could do partial fraction decomposition, but instead let's multiply the top and bottom by $\frac{1}{x^2}$:
$$3\int\frac{3}{3x+x^2}dx = 3\int \frac{\frac{3}{x^2}}{\frac{3}{x}+1}dx$$
Notice that the numerator is the derivative of the denominator times $-1$, so we have an integral of the form $\int\frac{du}{u}$, which integrates to
$$ -3\log\left(\frac{3}{x}+1\right)+C$$
Then substitute back in for $t$:
$$ -3\log\left(\frac{3}{\sqrt[3]{t}}+1\right)+C$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3386227",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Find the roots of $z^4-3z^2+1=0$ in polar form. Question :
Prove that the solutions of $z^4-3z^2+1=0$ are given by :
$$z=2\cos{36^\circ},2\cos{72^\circ},2\cos{216^\circ},2\cos{252^\circ}$$
My work :
First of all, i want ro find the roots with quadratic formula
$\begin{align}
&(z^2)^2-3z^2+1=0\\
&z^2=\dfrac{3\pm \sqrt{5}}{2}\\
&z_{1,2}=\pm\dfrac{\sqrt{3+\sqrt{5}}}{\sqrt{2}}\\
&z_{3,4}=\pm\dfrac{\sqrt{3-\sqrt{5}}}{\sqrt{2}}\\
\end{align}$
And i'm stuck. I don't know how to transform this complicated roots into polar form. Bcz, i'm not sure that the modulus for each number is exactly $2$?
Btw, i've found and read some possible duplicates
Here's one of the links :
Roots of $z^4 - 3z^2 + 1 = 0$.
But it seems doesn't answer my question...
Please give me a clear hint or another way to solve this without quadratic formula or something else.
| Note that
$$\cos36 = -\cos216 = \frac{1+\sqrt5}{4}$$
$$\cos72 = -\cos252 = \frac{1-\sqrt5}{4}$$
Then,
$$(x-2\cos36)(x-2\cos216) = x^2 - \frac{3+\sqrt5}{2}=0$$
$$(x-2\cos72)(x-2\cos252) = x^2 - \frac{3-\sqrt5}{2}=0$$
Together,
$$(x-2\cos36)(x-2\cos216)(x-2\cos72)(x-2\cos252)=x^4-3z^2+1=0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3386733",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 4
} |
Domain of inequation inside squared root How should I go when restricting the roots of the inequation:
$\sqrt {x^2+5x+6} - \sqrt {x^2-x+1} \lt 1$?
By restricting both the squared roots, I know that:
$x \le 3$ and $x \ge -2$
However when simplifying the whole inequation, I get the two roots:
$\frac{-13-\sqrt{73}}{16}$ and $\frac{-13+\sqrt{73}}{16}$.
Both roots are valid when swapping them in the first inequation, so how should I restrict my $x$? Should the final answer be: $x \le 3$ and $x \ge -2$?
| As $x^2-x+1 > 0$ it has no effect on the domain. However, $x^2+5x+6$ is negative in the interval $(-3,-2)$. So we must keep this in mind when we are crafting a solution
Consider $$\sqrt{(x+2)(x+3)} < 1+\sqrt{x^2-x+1}$$
$$x^2+5x+6 < x^2-x+2 + 2\sqrt{x^2-x+1}$$
$$3x+2 < \sqrt{x^2-x+1}$$
$3x+2$ being linear and $\sqrt{x^2-x+1}$ being a function with only one critical point there are at most two points where these expressions are equal (note that both their squares are quadratic)
Consider the points where
$$9x^2 + 4 + 12x =x^2 -x+1$$
$$\implies 8x^2 + 13x +3 = 0$$
$$\implies x = {-13\pm\sqrt{73}\over16}$$
As you have found.
(Note that $\sqrt{73}$ is between 8 and 9)
So, everywhere that is not between these points and is in the domain is in our solution. Again using our estimate for the roots, this gives:
$$\boxed{x\in (-\infty,-3]\cup\left[-2,{-13+\sqrt{73}\over 16}\right)}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3388042",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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If $\alpha$, $\beta$, $\gamma$ and $\delta$ be the roots of the polynomial $x^4+px^3+qx^2+rx+s$, prove the following.
If $\alpha$, $\beta$, $\gamma$ and $\delta$ be the roots of the polynomial $x^4+px^3+qx^2+rx+s$, show that $$(1+\alpha^2)(1+\beta^2)(1+\gamma^2)(1+\delta^2)=(1-q+s)^2+(p-r)^2.$$
I have tried putting
\begin{align}P(1)&=(\alpha-1)(\beta-1)(\gamma-1)(\delta-1)=1+p+q+r+s\\P(-1)&=(\alpha+1)(\beta+1)(\gamma+1)(\delta+1)=1-p+q-r+s.\end{align}
Then \begin{align}P(1)P(-1)&=(\alpha^2-1)(\beta^2-1)(\gamma^2-1)(\delta^2-1)\\&=(1+p+q+r+s)(1-p+q-r+s)=(1+q+s)^2-(p+r)^2\end{align}
Somehow it does not match the statement given.
| Another way using Transformation of equation:
We have $$x^4+qx^2+s=-x(px^2+r)$$
Squaring both sides
$$(x^2)^4+q^2(x^2)+s^2+2q(x^2)^3+2qs(x^2)+2s(x^2)^2=x^2(p^2(x^2)^2+r^2+2pr(x^2))$$
Replacing $x^2+1=y$
$$(y-1)^4+q^2(y-1)^2+s^2+2q(y-1)^3+2qs(y-1)+2s(y-1)^2=(y-1)(p^2(y-1)^2+r^2+2pr(y-1))$$
$$\iff y^4+\cdots+1+q^2+s^2-2q-2qs+2s+p^2+r^2-2pr=0$$
Using Vieta's formula $$\prod_{j=1}^4y_j=\dfrac{1+q^2+s^2-2q-2qs+2s+p^2+r^2-2pr}1=(1-q+s)^2+(p-r)^2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3390796",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 2
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How many different ways to select a subset
We want to select three subsets A, B, and C of {1, 2, . . . , n} so that A ⊆ C,
B ⊆ C, and A ∩ B $\ne$ ∅. In how many different ways can you do this?
This is my attempt:
Let A= $A_1$, B= $A_2$ and C = $A_1 \cup A_2$. The set = $A_1 \cup A_2 \cup A_3$. Then $3^n$ ways?
| We want to ensure that $A \subset C$ and $B \subset C$ such that $A \cap B \neq \emptyset$.
We'll suppose that $A \cap B = \emptyset$, which are the cases that we don't want to account... Having in hands that formula, we can just subtract from all possible cases of sets $A$ and $B$ without restriction.
So suppose that the condition is: $A \subset C$ and $B \subset C$ such that $A \cap B = \emptyset$.
First suppose that $|A| = 0 \implies A = \emptyset$. Therefore $B$ can be any subset of $C$. It'll follow that for $A = \emptyset$ you can choose $B$ in $2^n$ different ways.
Now suppose that $|A| = 1$. The way how you pick $B$ defines $A$, so you only need to think about defining $B$. For that, $B$ can have any element different than the element that you picked for $A$. It can be of cardinality $0,1,\cdots,n-1$. If it's of cardinality $0$ than $B = \emptyset$. If it's of cardinality $1$ then you have $n-1 \choose 1$ ways to define it. If it's of cardinality $2$ then you have $n-1 \choose 2$ ways to define it... Continuing it in the same manner it'll follow that for $|A| = 1$ you could pick $B$ in:
$$
{n-1 \choose 0} + {n-1 \choose 1} + {n-1 \choose 2} + \cdots + {n-1 \choose n-1}
$$ways.
But you can pick $A$ for $|A| = 1$ in $n \choose 1$ different ways, so it'll follow that if $|A| = 1$ you would have:
$$
{n \choose 1} \cdot \bigg( {n-1 \choose 0} + {n-1 \choose 1} + {n-1 \choose 2} + \cdots + {n-1 \choose n-1} \bigg)
$$
ways to pick $A$ and $B$ such that the condition holds.
Using the same argument for the cardinality of $A$ over $\{0,1,2,3,\cdots,n\}$, it follows that the final answer is:
$$
\sum_{0 \leq j \leq n}{n \choose j} \cdot \bigg(\sum_{0 \leq i \leq j} {n-j \choose i}\bigg)
$$
If you use the following identity:
$$
\sum_{0 \leq i \leq n} {n \choose i} = 2^n
$$
then you can simplify the result as:
$$
\sum_{0 \leq j \leq n}{n \choose j} \cdot 2^{n-j}
$$
And that is the number for the cases where $A \cap B = \emptyset$. All cases are given by $R = \{(a,b) \mid a \subset C \text{ and } b \subset C\}$. It's clear to see that $|R| = 2^n \cdot 2^n = 2^{2n}$. So our final answer is going to be:
$$
2^{2n} - \sum_{0 \leq j \leq n}{n \choose j} \cdot 2^{n-j}
$$
Quick sanity check.
Suppose that $n = 3$, therefore $C = \{1,2,3\}$. Let's list all the possible combinations of the sets $A$ and $B$ such that $A \cap B = \emptyset$ and see if the first formula gives us the same result. A solution is going to be written as $(A,B)$.
\begin{align*}
S_1 &= \{(\emptyset,\emptyset),(\emptyset,\{1\}),(\emptyset,\{2\}),(\emptyset,\{3\}),(\emptyset,\{1,2\}),(\emptyset,\{1,3\}),(\emptyset,\{2,3\}),(\emptyset,\{1,2,3\})\}\\
S_2 &= \{(\{1\},\emptyset),(\{1\},\{2\}),(\{1\},\{3\}),(\{1\},\{2,3\})\}\\
S_3 &= \{(\{2\},\emptyset),(\{2\},\{1\}),(\{2\},\{3\}),(\{2\},\{1,3\})\}\\
S_3 &= \{(\{3\},\emptyset),(\{3\},\{1\}),(\{3\},\{2\}),(\{3\},\{1,2\})\}\\
S_5 &= \{(\{1,2\},\emptyset),(\{1,2\},\{3\})\}\\
S_6 &= \{(\{1,3\},\emptyset),(\{1,3\},\{2\})\}\\
S_7 &= \{(\{2,3\},\emptyset),(\{2,3\},\{1\})\}\\
S_8 &= \{(\{1,2,3\},\emptyset)\}\\
\end{align*}
The number of solutions is the sum of the cardinalities of all the sets above. So it'll follow that it is:
$$
8 + (4 + 4 + 4) + (2 + 2 + 2) + 1 = 27
$$
which is indeed every term in the sum of our first formula:
\begin{align*}
{3 \choose 0} 2^3 + {3 \choose 1} 2^2 + {3 \choose 2} 2^1 + {3 \choose 3} 2^0 &= 1\cdot 8 + 3 \cdot 4 + 3 \cdot 2 + 1 \cdot 1\\
&= 8 + (4 + 4 + 4) + (2 + 2 + 2) + 1\\
&= 27
\end{align*}
So, using our last formula, it follows that $2^{6} - 27 = 37$ is the required number for $A \subset C$, $B \subset C$ and $A \cap B \neq \emptyset$.
Hope it helped :)
Doyun Nam's answer
It's a combinatorial identity that
$$
{n \choose 0} + 2{n \choose 1} + \cdots + 2^n{n \choose n} = 3^n
$$
So simplyfing even more my answer (last formula), we'll get:
$$
2^{2n} - \sum_{0 \leq j \leq n}{n \choose j} \cdot 2^{n-j} = 4^n - 3^n
$$
which is Doyun Nam corrected answer as pointed out by Ross Millikan in the comments.
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "3",
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If $f(x)=x^2+ax+b$ divides $f(x+2)f(x-2)$, find the minimum value of $f$. I am stuck with this Precalculus problem about polynomial functions. The problem:
Consider $f(x)=x^2+ax+b$ with $a^2-4b>0$. Let $\alpha$ and $\beta$ be the roots of $f$. Assume that $f(x)$ divides $f(x+2)f(x-2)$. Then
*
*Show that $\alpha=\beta+2$ or $\alpha=\beta-2$
*Using $1.$, find the minimum value of $f(x)$.
Part $1$ is easy: write $f(x+2)f(x-2)=f(x)g(x)$ and substitute $x=\alpha$ to obtain $f(\alpha+2)f(\alpha-2)=0$, so $\alpha+2$ or $\alpha-2$ is a root of $f$. This root cannot be $\alpha$, so $\alpha+2=\beta$ or $\alpha-2=\beta$.
I am stuck with part $2$. Any idea?
The minimum is obtained at $-\frac{a}{2}=\frac{\alpha+\beta}{2}=\beta+1$ or $\beta-1$ but I don't know how to continue.
| $ f(x) = ( x - \alpha) ( x - \beta) = (x - c + 1)( x - c - 1) = (x-c)^2 - 1$
Hence, the minimum value is -1.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
Taylor series of $\sinh{(x)}$ at $\ln{(2)}$. Determine the Taylor series of $\sinh{(x)}$ about $x = \ln{(2)}$.
Equating each derivative at $x = \ln{(2)}$ gives:
\begin{equation*}
\begin{split}
f'(\ln{(2)}) &= \frac{2+\frac{1}{2}}{2} = \frac{5}{4} \\
f''(\ln{(2)}) &= \frac{2-\frac{1}{2}}{2} = \frac{3}{4} \\
f^{(3)}(\ln{(2)}) &= \frac{2+\frac{1}{2}}{2} = \frac{5}{4} \\
f^{(4)}(\ln{(2)}) &= \frac{2-\frac{1}{2}}{2} = \frac{3}{4} \\
f^{(5)}(\ln{(2)}) &= \frac{2+\frac{1}{2}}{2} = \frac{5}{4}
\end{split}
\end{equation*}
and so on. This means the Taylor series is
\begin{equation*}
\begin{split}
\sum_{n=0}^{\infty} \frac{f^{(n)}(\ln{(2)})}{n!}(x-\ln{(2)})^n &= f(\ln{(2)}) + \frac{f'(\ln{(2)})}{1!}(x-\ln{(2)}) + \frac{f''(\ln{(2)})}{2!}(x-\ln{(2)})^2 \\
&+ \frac{f^{(3)}(\ln{(2)})}{3!}(x-\ln{(2)})^3 + \frac{f^{(4)}(\ln{(2)})}{4!}(x-\ln{(2)})^4 + \frac{f^{(5)}(\ln{(2)})}{5!}(x-\ln{(2)})^5 + \ldots \\
&= \frac{3}{4}+\frac{5}{4\cdot1!}(x-\ln{(2)})^1+\frac{3}{4\cdot 2!}(x-\ln{(2)})^2 \\
&+\frac{5}{4\cdot 3!}(x-\ln{(2)})^3+\frac{3}{4\cdot 4!}(x-\ln{(2)})^4+\frac{5}{4\cdot 5!}(x-\ln{(2)})^5+\ldots
\end{split}
\end{equation*}
How do I find the general sum of this? Has it got something to do with the Maclaurin series which is $\sum_{n=0}^{\infty} \frac{x^{2n+1}}{(2n+1)!}$.
| $$\sinh (\ln 2)= \sum_{k=0}^{\infty} \left(\frac{2-(-1)^k\frac{1}{2}}{k!}\right) (x-\ln 2)^k$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 2
} |
Calculating Legendre symbol : $\left(\frac{3}{2^{2^n}+1}\right)$ , n being positive. I want to find the Legendre symbol : $\left(\frac{3}{2^{2^n}+1}\right)$
for any positive n.
Also I want to find the Legendre symbol for: $\left(\frac{5}{2^{2^n}+1}\right)$, when $n>1$
Solution:
Using Gaussian reciprocity law: $\left(\frac{p}{q}\right)\left(\frac{q}{p}\right)=(-1)^{\frac{p-1}{2}\frac{q-1}{2}}$, we get
$\left(\frac{3}{2^{2^n}+1}\right)$=$\left(\frac{2^{2^n}+1}{3}\right)$.Similarly, $\left(\frac{5}{2^{2^n}+1}\right)$=$\left(\frac{2^{2^n}+1}{5}\right)$.
But I am unable to proceed any further. Thanks in advance.
| Wikipedia's Legendre symbol page says the definition is
$$\left({\frac {a}{p}}\right)={\begin{cases}1&{\text{if }}a{\text{ is a quadratic residue modulo }}p{\text{ and }}a\not \equiv 0{\pmod {p}},\\-1&{\text{if }}a{\text{ is a non-quadratic residue modulo }}p,\\0&{\text{if }}a\equiv 0{\pmod {p}}.\end{cases}} \tag{1}\label{eq1A}$$
Since $2^2 = 4 \equiv 1 \pmod 3$ and $2^{2^n} = 4^{2^{n-1}}$, then $2^{2^n} + 1 \equiv 1 + 1 \equiv 2 \pmod{3}$, with $2$ being a non-quadratic residue (as only $1$ is a quadratic residue), you have that
$$\left(\frac{2^{2^n}+1}{3}\right) = -1 \tag{2}\label{eq2A}$$
Similarly, since $2^{4} = 16 \equiv 1 \pmod 5$ and $2^{2^n} = 16^{2^{n-2}}$ for $n \gt 1$, you have that $2^{2^n} + 1 \equiv 1 + 1 \equiv 2 \pmod{5}$, with $2$ being a non-quadratic residue (as only $1$ and $4$ are quadratic residues), you get
$$\left(\frac{2^{2^n}+1}{5}\right) = -1 \tag{3}\label{eq3A}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3397034",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Find $x$ that minimizes $\max_{1 \leq i \leq 3} f_i(x)$ My question is:
For $0 < a_1 \leq a_2 \leq a_3$ fixed, find $x>0$ that minimizes
\begin{align*}
g(x) = \max_{1\leq i \leq 3} f_i(x).
\end{align*}
where
\begin{align*}
f_i(x) = \left(\frac{a_i - x}{a_i + x}\right)^2
\end{align*}
I do not know how to start. Any idea, reference is welcome. Thank you!
| I think I have a solution to my question.
Answer: the minimizer is $x = \sqrt{a_1 a_3}$.
Let $0 < \underline{a} < \overline{a}$ be fixed. We study the relative position of $f_{\underline{a}}$ and $f_{\overline{a}}$ defined by
\begin{align*}
f_{\underline{a}}(x) = \left(\frac{\underline{a} - x}{\underline{a} + x}\right)^2, \qquad
f_{\overline{a}}(x) = \left(\frac{\overline{a} - x}{\overline{a} + x}\right)^2
\end{align*}
for $x>0$. Let $h$ be defined by $h = f_{\underline{a}} - f_{\overline{a}}$. Then,
\begin{align*}
h(x) &= \frac{(\underline{a}-x)^2(\overline{a} + x)^2 - (\overline{a} - x)^2(\underline{a}+x)^2}{(\underline{a}+x)^2(\overline{a}+x)^2}\\
&= \frac{\big[(\underline{a}-x)(\overline{a} + x) - (\overline{a} - x)(\underline{a}+x)\big]\big[(\underline{a}-x)(\overline{a} + x) + (\overline{a} - x)(\underline{a}+x)\big]}{(\underline{a}+x)^2(\overline{a}+x)^2}\\
&= \frac{4 x (\overline{a} - \underline{a})(x^2 - \underline{a}\overline{a})}{(\underline{a}+x)^2(\overline{a}+x)^2}.
\end{align*}
Hence, $f_{\underline{a}} < f_{\overline{a}}$ for $x \in (0, \sqrt{\underline{a}\overline{a}})$, $f_{\underline{a}} = f_{\overline{a}}$ for $x = \sqrt{\underline{a}\overline{a}}$, and $f_{\underline{a}} > f_{\overline{a}}$ for $x \in (\sqrt{\underline{a}\overline{a}}, +\infty)$. Since $0 < a_1 < a_2 < a_3$, we deduce that
\begin{align*}
&f_1 \leq f_2 \quad \text{if} \quad x \leq \sqrt{a_1 a_2}, \quad\text{otherwise} \quad f_1>f_2,\\
&f_1 \leq f_3 \quad \text{if} \quad x \leq \sqrt{a_1 a_3}, \quad\text{otherwise} \quad f_1>f_3, \\
&f_2 \leq f_3 \quad \text{if} \quad x \leq \sqrt{a_2 a_3}, \quad\text{otherwise} \quad f_2 > f_3.
\end{align*}
Consequently, the maximum is given by
\begin{align*}
g(x) = \begin{cases}
f_3(x) & \text{if }x \leq \sqrt{a_1 a_3}\\
f_1(x) & \text{if }x > \sqrt{a_1 a_3}.
\end{cases}
\end{align*}
On the other hand, for $i \in \{1, 2, 3\}$,
\begin{align*}
f_i'(x) = \frac{4 a_i (x- a_i)}{(a_i+x)^3}.
\end{align*}
Hence, $f_i$ is decreasing on $(0, a_i]$, and increasing on $(a_i, + \infty)$. This implies that $f_3$ is decreasing on $(0, \sqrt{a_1 a_3}]$, and $f_1$ is increasing on $(\sqrt{a_1a_3}, + \infty)$. Hence, $g(x)$ is the smallest for $x = \sqrt{a_1 a_3}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3399153",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
Which is bigger 5^(√3) and 4^(√5)? How do you know which one is bigger between $5^\sqrt{3}$ and $4^\sqrt{5}$?
For my method I used in $2^\sqrt{3}$ and $3^\sqrt{2}$ I put both numbers in the function $f(x)=x^\sqrt{3}$ so $f(2^\sqrt{3})$ become $2^3$ which is equal to 8 and $f(3^\sqrt{2})$ become $3^\sqrt{6}$ and $3^\sqrt{6}>3^\sqrt{2}$ which is equal to 9 and 9>8 so $3^\sqrt{2}>2^\sqrt{3}$
But for this problem I don't know what I should multiply or is there any method besides that? Please kindly help. Thank you
| Take both to the power of $\sqrt{5}$
$\left(5^\sqrt{3}\right)^\sqrt{5}=5^{\sqrt{15}}<5^{\sqrt{16}}=5^4=625$
$\left(4^\sqrt{5}\right)^\sqrt{5}=4^5=1024$
so we have that $4^\sqrt{5}>5^\sqrt{3}$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
a line tangent to curve $y = \frac{x}{2−2x}$ goes through (1, -1) $y = \frac{x}{2−2x}$
$y = \frac{f(x)}{g(x)} \rightarrow y' = \frac{f'(x) \cdot g(x) - g'(x)\cdot f(x)}{{g(x)}^2}$
$y = \frac{x}{2−2x}$
$y' = \frac{(2-2x)-(x \cdot (-2))}{{(2−2x)}^2}$ $\rightarrow$ $y' = \frac{(2-2x+2x)}{{(2−2x)}^2} = \frac{2}{{(2-2x)}^2}$
why cant i find the gradient with derivative?
| Your derivative is absolutely correct.
Equation of a line is $(y-y_1)=m(x-x_1)$ where $(x_1,y_1) $ is a point that lies on the line
Here $m = \frac{2}{(2-2x)^2}$
Now we know that $(1,-1)$ lies on the line and the gradient of the line is $\frac{2}{(2-2k)^2}$
Substitute everything
We get $y=\frac{2}{(2-2k)^2}(x-1)-1$
Now this line also passes from point and $(k,\frac{k}{2-2k})$
We get $\frac{k}{2-2k}=\frac{2}{(2-2k)^2}(k-1)-1$
Now solve this equation for k and you'll get $k=3$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3399947",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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If $A$ is a rotation matrix, then $||Ax||=||x||$. Attempt:
Let $$
A =
\begin{bmatrix}
\cos\theta & -\sin\theta \\
\sin\theta & \cos\theta
\end{bmatrix},
x = \begin{bmatrix} x_1 \\ x_2 \end{bmatrix}
$$
Then,
$$Ax =
\begin{bmatrix}
\cos\theta & -\sin\theta \\
\sin\theta & \cos\theta
\end{bmatrix} \cdot \begin{bmatrix} x_1 \\ x_2 \end{bmatrix} = \begin{bmatrix} \cos\theta \cdot x_1 -\sin\theta\cdot x_2 \\ \sin\theta \cdot x_1 +\cos\theta\cdot x_2 \end{bmatrix} $$
$$
\begin{align}
||Ax|| &= \sqrt{(\cos\theta \cdot x_1 -\sin\theta\cdot x_2)^2 + (\sin\theta \cdot x_1 +\cos\theta\cdot x_2)^2} \\
&=\sqrt{(\cos^2\theta \cdot x^2_1 -2\cos\theta \cdot x_1\sin\theta\cdot x_2+ \sin^2\theta\cdot x^2_2) + (\sin^2\theta \cdot x^2_1 +2\sin\theta \cdot x_1\cos\theta\cdot x_2+ \cos^2\theta\cdot x^2_2)} \\
&=\sqrt{x^2_1+x^2_2}
\end{align}
$$
However, my text seems to suggest that this is only true for $0\leq\theta\leq\pi$ here:
So where in my attempt did I go wrong?
| What you have done is correct. The reason for putting the condition that $0\leq \theta \leq \pi$ is that the 'rotation of the plane' plays the role here. To make reader understand better, they have put the condition. For example , Let $\theta = 359^o$ then one might get confused whether rotation is by $1^o$ or $359^0$. To avoid this, they must have put the condition on $\theta$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3401051",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
help solving for delta in epsilon delta proof $$\lim_{x\to 9} \frac{1}{\sqrt{x}} = \frac{1}{3} $$
so the two statements are
$$0<|x-9|<\delta$$
$$\left|\frac{1}{\sqrt{x}}-\frac{1}{3}\right|<\epsilon$$
I've tried to multiply by the conjegate to get
$$\frac{\frac{1}{x}-\frac{1}{9}}{\sqrt{\frac{1}{x}}+\frac{1}{3}}<\epsilon$$
I'm unsure where to go from here to get that x-9 factor i need
| Note that$$\left\lvert\frac1{\sqrt x}-\frac13\right\rvert=\frac{\left\lvert3-\sqrt x\right\rvert}{3\sqrt x}=\frac{\lvert9-x\rvert}{3\sqrt x\left(3+\sqrt x\right)}.$$Now, if $\lvert x-9\rvert<5$, then $x>4$ and therefore $\sqrt x>2$. Also, $3+\sqrt x>5$. So $3\sqrt x\left(3+\sqrt x\right)>30$. Therefore, if you take $\delta=\min\left\{5,30\varepsilon\right\}$, then$$\lvert x-9\rvert<\delta\implies\left\lvert\frac1{\sqrt x}-\frac13\right\rvert=\frac{\lvert9-x\rvert}{3\sqrt x\left(3+\sqrt x\right)}<\frac{\lvert9-x\rvert}{30}<\varepsilon.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3403270",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Diophantine equations with cubic roots.
Find all positive, integral $x, y$ such that
$$x^3-y^3=xy+61$$
Little work:
$$(x-y)(x^2+xy+y^2)-xy=61$$
Also $x^3-xy-y^3$ seems to to be unable to be "completed by the square," or factored, so I'm stuck. Thanks!
| Draw some pictures. There is an integral point on it $(6,5)$
As soon as real $x > 6,$ we get
$$ x-1 < y < x $$
so that $y$ cannot be an integer when $x > 6$
Tuesday:
We know that $x > y$ are positive. So,
$(x-y)^2 > 0$ or $x^2 - 2xy + y^2 > 0,$ or $x^2 + y^2 > 2xy.$ Add $xy$ to both sides with positive $x,y$ we get
$$x^2 + xy + y^2 > 3xy > 0$$ and
$$ \frac{xy}{x^2 + xy + y^2} < \frac{1}{3}. $$
Since $$ (x-y)(x^2 + xy + y^2) = xy + 61, $$ we have
$$ x-y = \frac{xy}{x^2 + xy + y^2} + \frac{61}{x^2 + xy + y^2} < \frac{1}{3} + \frac{61}{x^2 } $$
since $x^2 < x^2 + xy + y^2$ when $x,y>0.$
$$ \color{magenta}{ x-y < \frac{1}{3} + \frac{61}{x^2 }} $$
Therefore, when $x \geq 10$ is an integer, we find $\frac{61}{x^2} < \frac{2}{3} \; , \;$ so that $0 <x-y < 1$
and $y$ is not an integer.
We need only check $x$ values up to $9,$ and $x=6,y=5$ is the only one that works.
There are just the two integer points, but infinitely many rational points. The tangent line at $(-5,-6)$ intersects the curve in a third point, that being
$$ \left( \frac{1385}{341}, \frac{384}{341} \right) $$
Given a rational point on the curve, the tangent line there meets the curve in another rational point. Given two distinct rational points on the curve, the line through them meets the curve in a third rational point.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3403526",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
$\int \frac{\sin x}{1+2\sin x}dx$ calculate:
$$\int \frac{\sin x}{1+2\sin x}dx$$
I tried using $\sin x=\dfrac{2u}{u^2+1}$, $u=\tan \dfrac{x}2$ and after Simplification:
$$\int \frac{2u}{u^2+4u+1}×\frac{2}{u^2+1}du$$
and I am not able to calulate that.
| Simplify first before substituting,
$$\int \frac{\sin x}{1+2\sin x}dx
=\frac12\int dx -\frac12 \int \frac{1}{1+2\sin x}dx$$
$$=\frac12 x -\frac12 \int \frac{1+\tan^2\frac x2}{\tan^2\frac x2 +4\tan\frac x2 +1}dx$$
$$=\frac12 x -\int \frac{d(\tan \frac x2)}{\left(\tan^2\frac x2 +2\right)^2 -3}$$
$$=\frac12 x +\frac{1}{\sqrt3}\tanh^{-1}\left( \frac{
\tan\frac x2 +2}{\sqrt3}\right)+C$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 3
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How many ways from point A to point B? I am trying to solve this problem
In how many ways can we go from point $A$ to point $B$ if we are only allowed to move along the arrows?
I use combinations and got an answer of $20$.
Since this is 3D grid
The height is $3$ units, length is $1$ unit and width is $1$ unit. Thus we can model the path to letters
$D$- downs,$L$-Left and $R$-Right
$$DDDLR$$
That is $5!$/$(3!*1!*1!)$
I would like to verify my answer.
| Graphical method:
We can easily calculate the number of different paths from $A$ to $B$ by adding the numbers of related sub-paths in the graphic.
Algebraic method:
We introduce a coordinate system and look for the number of different paths to go from $A=(0,0,0)$ to $B=(1,-1,-2)$ using steps $(1,0,0), (0,-1,0)$ and $(0,0,-1)$.
*
*We encode the step $(1,0,0)$ with $x$, $(0,-1,0)$ with $\frac{1}{y}$ and $(0,0,-1)$ with $\frac{1}{z}$.
*Each step has the form $x+\frac{1}{y}+\frac{1}{z}$.
*Since paths from $A$ to $B$ have length $4$ we are looking for the coefficient of $x\cdot\frac{1}{y}\cdot\frac{1}{z^2}$ in $\left(x+\frac{1}{y}+\frac{1}{z}\right)^4$.
We use the coefficient of operator $[x^n]$ to denote the coefficient of $x^n$ and obtain
\begin{align*}
\color{blue}{[xy^{-1}z^{-2}]}&\color{blue}{\left(x+\frac{1}{y}+\frac{1}{z}\right)^4}\\
&=[xy^{-1}z^{-2}]\sum_{j=0}^4\binom{4}{j}x^j\left(\frac{1}{y}+\frac{1}{z}\right)^{4-j}\tag{1}\\
&=\binom{4}{1}[y^{-1}z^{-2}]\left(\frac{1}{y}+\frac{1}{z}\right)^{3}\tag{2}\\
&=\binom{4}{1}[y^{-1}z^{-2}]\sum_{j=0}^3\binom{3}{j}\left(\frac{1}{y}\right)^j\left(\frac{1}{z}\right)^{3-j}\tag{3}\\
&=\binom{4}{1}\binom{3}{1}\tag{4}\\
&\,\,\color{blue}{=12}
\end{align*}
in accordance with the graphical solution above.
Comment:
*
*In (1) we expand the binomial.
*In (2) we select the coefficient of $x^1$.
*In (3) we expand the binomial.
*In (4) we select the coefficient of $y^{-1}$ (which is also the coefficient of $z^{-2}$).
| {
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"timestamp": "2023-03-29T00:00:00",
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A bunch of lilac A bunch of lilac consists of flowers with $4$ or $5$ petals. The number of flowers and the total number of petals are perfect squares. Can the number of flowers with $4$ petals be divisible by the number of flowers with $5$ petals?
What I produced:
Total Flowers = $x + y$
Petal Total $5x + 4y$
$x | y$ can?
Like, be $x, y, k, q$ integers
$x + y = k^2$
$5x + 4y = q^2$
Prove that $x does not divide y
$x + 4k^2 = q^2$
$y = k^2 - x$
$\frac{y}{x} = \frac{k^2}{x -1}$
$x$ only divides $y$ divides $k^2$
x only splits k² splits q²
So divide $q^2$ and $k^2$
But it implies that there are two positive integers, say r and s such that
$1 + r^2x = s^2x$
Thus,
$1 = x (r-s) (r + s)$
Which is absurd, because they are all natural numbers, and there is no other number that divides $1$ apart from itself.
So $x, r-s$ and $ r + s$ are equal to $ 1$
| You are correct in the first steps.
Continuing your argument: $x = n^2 - 4m^2$ and $y = 5m^2 - n^2$.
Hence we have $5m^2 - n^2 = k(n^2 - 4m^2)$. This can be rewritten as $\frac{4k + 5}{k + 1} = \frac{n^2}{m^2}$.
Now notice that, since $k$ is integer, we have $\gcd(4k + 5, k + 1) = \gcd(1, k + 1) = 1$, hence both $4k + 5$ and $k + 1$ are actually square of integers.
This leads to $4k + 5 = u^2$ and $k + 1 = v^2$, or $(2v)^2 + 1 = u^2$.
This is impossible, since two positive squares cannot differ by $1$.
| {
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"timestamp": "2023-03-29T00:00:00",
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formula for $\left(1\cdot2\cdot...\cdot k\right)+...+\left(n\left(n+1\right)...\left(n+k- 1\right)\right)$ proof if $k$ is a constant then for every $n$ natural numbers we have:
$$\left(1\cdot2\cdot...\cdot
k\right)+\left(2\cdot3\cdot...\cdot\left(k+1\right)\right)+...+\left(n\left(n+1\right)...\left(n+k-
1\right)\right)=\frac{n\left(n+1\right)\left(n+2\right)...\left(n+k\right)}{k+1}.$$
clearly $$\left(1\cdot2\cdot...\cdot
k\right)+\left(2\cdot3\cdot...\cdot\left(k+1\right)\right)+...+\left(n\left(n+1\right)...\left(n+k-
1\right)\right)=\sum_{m=1}^{n}\left(\prod_{j=0}^{k-1}\left(m+j\right)\right)$$
$$=\sum_{m=1}^{n}\left(\Gamma\left(m+j+1\right)\right)$$where$\left(0\le j\le k-1\right)$ and
$\Gamma(x)$ is Gamma function,
or equivalently $$=\sum_{m=1}^{n}\int_{0}^{∞}x^{\left(m+j\right)}e^{-x}dx$$
but I could not find the given formula.
| Note that by the Hockey-stick identity we have that
$$1\cdot2\cdots k+2\cdot3\cdots(k+1)+\dots+n\cdot(n+1)\cdots(n+k-1)$$
$$\begin{align}
&=k!\binom{k}{k}+k!\binom{k+1}{k}+\dots+k!\binom{n+k-1}{k}\\
&=k!\sum_{i=k}^{n+k-1}\binom{i}{k}\\
&=k!\binom{n+k}{k+1}\\
&=k!\cdot\frac{n\cdot(n+1)\cdots(n+k)}{(k+1)!}\\
&=\frac{n\cdot(n+1)\cdots(n+k)}{k+1}\\
\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3407338",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Find $\lim_{x \to \ 0}{\frac{1-\sqrt{\cos x}}{1-\cos\sqrt{x}}}$ without using the l'Hospital rule The task is to evaluate
$$\lim_{x \to \ 0}{\frac{1-\sqrt{\cos x}}{1-\cos \sqrt{x}}}.$$
I tried to use some trigonometric identities such as
$$\lim_{x \to \ 0}{\frac{1-\sqrt{\cos\left(x\right)}}{1-\cos\left(\sqrt{x}\right)}}= \lim_{x \to \ 0} \frac{1-
\sqrt{\cos\left(x\right)}}{1-
\cos\left(\sqrt{x}\right)}\cdot\frac{1+\sqrt{\cos\left(x\right)}}{1+\sqrt{\cos\left(x\right)}}$$$$= \lim_{x \to \ 0} \frac{1-
\cos\left(x\right)}{2\sin^{2}\left(\frac{\sqrt{x}}{2}\right)}\cdot\frac{1}{1+\sqrt{\cos\left(x\right)}}$$$$=
\lim_{x \to \ 0} \left(\frac{\sin\left(\frac{\sqrt{x^{2}}}{2}\right)}{\sin\left(\frac{\sqrt{x}}{2}\right)}
\right)^{2}\cdot\frac{1}{1+\sqrt{\cos\left(x\right)}}$$$$=\frac{1}{2}\lim_{x \to \
0}\left(\frac{\sin\left(\frac{\sqrt{x^{2}}}{2}\right)}{\sin\left(\frac{\sqrt{x}}{2}\right)}\right)^{2}$$
and this is where I have a problem.
| Using the standard Taylor expansion of $\cos(x)$, you should have
$${\frac{1-\sqrt{\cos\left(x\right)}}{1-\cos\left(\sqrt{x}\right)}}=\frac{\frac{x^2}{4}+O\left(x^4\right) } {\frac{x}{2}+O\left(x^2\right) }=\frac{x}{2}+O\left(x^2\right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3408177",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 7,
"answer_id": 0
} |
Show that $\frac{3\> + \>\cos x}{\sin x}$ cannot have any value between $-2\sqrt2$ and $2\sqrt2$ Show that
$$\dfrac{3+\cos x}{\sin x}\quad \forall \quad x\in R $$
cannot have any value between $-2\sqrt{2}$ and $2\sqrt{2}$.
My attempt is as follows:
There can be four cases, either $x$ lies in the first quadrant, second, third or fourth:-
First quadrant: $\cos x$ will decrease sharply and sinx will increase sharply, so $y_{min}=3$ at $x=\dfrac{\pi}{2}$.
$y_{max}$ would tend to $\infty$ near to $x=0$
Second quadrant: $\cos x$ will increase in magnitude and sinx will decrease sharply, so $y_{min}=3$ at $x=\dfrac{\pi}{2}$.
$y_{max}$ would tend to $\infty$ near to $x=\pi$
Third quadrant: $\cos x$ will decrease in magnitude and sinx will increase in magnitude but negative, so $y_{min}$ would tend to $-\infty$ near to $x=\pi$
$y_{max}$ would be $-3$ at $x=\dfrac{3\pi}{2}$
Fourth quadrant: $\cos x$ will increase sharply and sinx will decrease in magnitude, so $y_{min}$ would tend to $-\infty$ near to $x=2\pi$
$y_{max}$ would be $-3$ at $x=\dfrac{3\pi}{2}$
So in this way I have proved that $\dfrac{3+\cos x}{\sin x}$ cannot lie between $-2\sqrt{2}$ and $2\sqrt{2}$, but is their any smart solution so that we can calculate quickly.
| Use the half-angle expressions
$\cos x = \frac{1-\tan^2\frac x2}{1+\tan^2\frac x2}$ and
$\sin x = \frac{2\tan\frac x2}{1+\tan^2\frac x2}$ to express
$$I=\frac{3+\cos x}{\sin x}=
\frac{2}{\tan\frac x2} +\tan \frac x2$$
Note
$$I^2=\left(\frac{2}{\tan \frac x2} +\tan \frac x2\right)^2
=\left(\frac{2}{\tan \frac x2} -\tan \frac x2\right)^2+8 \ge 8$$
Thus, $I^2$ can not have values within $[0,8)$, which means that $I=\frac{3+\cos x}{\sin x}$ can not have values within $(-2\sqrt2, \>2\sqrt2)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3411788",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 0
} |
A simpler way to evaluate $\int {dx\over \sin^2x\cos^4x}$
Evaluate:
$$
\int {dx\over \sin^2x\cos^4x}
$$
I've started by using identities:
$$
{1\over \sin^2x} = 1+\cot^2x\\
{1\over \cos^2x} = 1+\tan^2x
$$
So the integral becomes:
$$
\begin{align}
I &= \int (1+\cot^2x)(1+\tan^2x)^2dx \\
&=\int (1+\cot^2x)(1+2\tan^2x + \tan^4x)dx \\
&=\int (1 + 2\tan^2x + \tan^4x+\cot^2x + 2 + 2\tan^2x)dx \\
&=\int (3 + 4\tan^2x+\tan^4x+\cot^2x)dx
\end{align}
$$
I know the integral of $\tan^2x$ and $\cot^2x$, while for $\tan^nx$ one could use a reduction formula. But I find this approach too complicated and seek for a simpler one. Another approach I've tried out is expanding $1$ in the nominator to:
$$
\int {\sin^2x + \cos^2x\over \sin^2x\cos^4x}dx
$$
But that was also clumsy.
Is there another approach simpler than the one suggested above? Could this approach be generalized for integrals of the form:
$$
\int {dx\over \sin^{2k}x\cos^{2p}x}
$$
where $k, p \in\Bbb N$?
Thank you!
| $I = \displaystyle\int \dfrac{1}{\sin^2{x}\cos^4{x}} dx = \displaystyle\int \sec^2{x}\dfrac{\left(\tan^2{x}+1\right)^2}{\tan^2{x}} dx$. Now substitute $y = \tan{x}$.
$I= \displaystyle\int \dfrac{(y^2+1)^2}{y^2} dy = \frac{y^3}{3}+2y-\frac{1}{y} +C$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3414034",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Find the smallest integer $n$ such that $5x^5 + (\log x)^{5^5} \in O(x^n).$ Find the smallest integer $n$ such that $5x^5 + (\log_2 x)^{5^5}$ is $O(x^n).$
I know that $5x^5$ is $O(x^5),$ and that $\log_2 x$ is $O(x).$ Since $5^5=3,125$ and $\log_2 x$ is $O(x),$ I know that I can use a Theorem to get
$$
\begin{aligned}
(\log_2 x)^{5^5} &= \underbrace{\log_2 x\cdot\log_2 x\cdots\log_2 x}_{3,125\text{ times}}\\
&{}\\
&= O(\underbrace{x\cdot x\cdots x}_{3,125\text{ times}})\\
&{}\\
&= O(x^{3,125}).
\end{aligned}
$$
Then I can use another theorem to get
$$
5x^5 + (\log_2x)^{5^5} = O(\max(|x^5|,|x^{3,125}|) = O(x^{3,125}).
$$
The issue that I'm having is showing that $3,125$ is the smallest $n.$ I'm pretty sure that it's not the smallest $n$, but how do I find the smallest one if it isn't? If it is the smallest one, how do I prove it?
| The following shows that $(\log x)^{5^5} \in O(x)$: For positive $y$ we have
$$
\begin{aligned}
y &< 2^y &&\text{can prove by induction}\\
\sqrt[5^5]{x} &< 2^{\sqrt[5^5]{x}} &&\text{set }y = \sqrt[5^5]{x}\text{ for positive } x\\
\log_2\left(\sqrt[5^5]{x}\right) &<\log_2\left(2^{\sqrt[5^5]{x}}\right) &&\text{take $\log_2$ of both sides}\\
\frac{1}{5^5}\log_2x &< \sqrt[5^5]{x} &&\text{simplify}\\
\left(\frac{1}{5^5}\log_2x\right)^{5^5} &< \left(\sqrt[5^5]{x}\right)^{5^5} &&\text{raise to the }5^5\text{ power}\\
\left(\frac{1}{5^5}\right)^{5^5}(\log_2x)^{5^5}&< x\\
(\log_2x)^{5^5} &< (5^5)^{5^5}x &&\text{multiply by}(5^5)^{5^5}.
\end{aligned}
$$
Hence, $(\log_2x)^{5^5}\in O(x).$
Combining this with $5x^5\in O(x^5)$, we have
$$
5x^5+(\log_2x)^{5^5}\in O(\max(|x^5|,|x|)) = O(x^5).
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3414382",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Divisibility by 7 of a number consisting of 0 and 1's A decimal number is of arbitrary length consisting of 0 and 1 only i.e. (10,111,10000,11111,1010, number of digits in the number can be upto 100 )
Can this number ever be divisible by 7 if yes, is there any efficient way to list all those numbers
| $10 \equiv 3 \pmod 7$ and so $10^2 \equiv 2\pmod 7$ and $10^3 \equiv -1\pmod 7$ and $10^4 \equiv -3 \pmod 7$ and $10^5 \equiv -2 \pmod 7$ and $10^6 \equiv 1 \pmod 7$.
Let $N = 10^{a_n} + 10^{a_{n-1}} + ....... + 10^{a_0}$ where no $a_i = a_j$ if $i\ne j$ be a number consisting of only $0$s and $1$.
Then $N \equiv b_n + b_{n-1} + ...... + b_0 \pmod 7$ where
where $b_k = \begin{cases} 1 &\text{if }a_k\equiv 0 \pmod 6\\ 3 &\text{if }a_k\equiv 1 \pmod 6\\ 2 &\text{if }a_k\equiv 2 \pmod 6\\ -1 &\text{if }a_k\equiv 3 \pmod 6\\ -3 &\text{if }a_k\equiv 4 \pmod 6\\ -2 &\text{if }a_k\equiv 5 \pmod 6\\ \end{cases}$
So Choose a bunch of $a_k$ where the sum of the $b_k$ add to $0$ or a multiple of $7$.
Example If $a_0 =0$ then $b_0=1$ and so if $a_1=3$ then $b_1=-1$ and so $10^{a_1} + 10^{a_0} =10^3+ 10^0 = 1001$ will have be divisible by $7$. And indeed $1001 = 7*143$.
Another example if $a_0=1$ and $a_1=2$ then $b_0=3$ and $b_1=2$ so $b_0 + b_1=5$ so we need the rest to and to $-5$ or $2$ or $2$ more than multiple of $7$.
So we could do $a_2=4$ and $a_3=5$ to get $b_2=-3; b_3=-5$ and so $10^1 + 10^2 + 10^4 + 10^5 = 110110$ is divisible by $7$.
Or we could to $a_3=2+6=8$ and $b_3=2$ so $10^1 + 10^2 + 10^8 = 100000110$ is divisible by $7$.
Of course if you get any $N$ divisible by $7$ then $10^k*N$ will also be divisible by $7$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3417330",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
} |
Given a point and a line in homogeneous coordinates, what is the coordinate of the projection of the point on the line? Knowing $M$ and $L$ in homogeneous coordinates, how can we determine the coordinates of the point $P$?
| it still depends on what kind of projection is required; Given the constraints in the original post, it looks like an orthographic parallel projection. Both central projection and parallel projection (oblique or orthographic) are computation-friendly redefined as a common algebraic formulation equation 3.1 in Householder elemenetary matrix form this arXiv.org article: Unified Framework of Elementary Geometric Transformation
Suppose the homogeneous coordinates of the point $(x,y,1)^T$ and the line $(a,b,c)^T$; then an orthographic parallel projection onto the line $(a,b,c)^T$ uniquely determined by $(a,b,c)^T$ and its normal direction per definition 3.6, definition 3.7 and equation 3.1 (line No. 2 of table 1 in page 9) for 2D projective transformations in Unified Framework of Elementary Geometric Transformation has the following form:
$$\left[
\begin{array}{ccc}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1 \\
\end{array}
\right]-\dfrac{\left[
\begin{array}{c}
a \\
b \\
0 \\
\end{array}
\right]\cdot \left[a,b,c\right]}{\left[a,b,0\right]\cdot \left[
\begin{array}{c}
a \\
b \\
c \\
\end{array}
\right]}
=\dfrac{1}{a^2+b^2}\left[
\begin{array}{ccc}
b^2 & -a b & -a c \\
-a b & a^2 & -b c \\
0 & 0 & a^2+b^2 \\
\end{array}
\right]$$
Then the orthographic parallel projection of $(x,y,1)^T$ onto the line $(a,b,c)^T$ is:
$$\dfrac{1}{a^2+b^2}\left[
\begin{array}{ccc}
b^2 & -a b & -a c \\
-a b & a^2 & -b c \\
0 & 0 & a^2+b^2 \\
\end{array}
\right]\cdot \left[
\begin{array}{c}
x \\
y \\
1 \\
\end{array}
\right]= \dfrac{1}{a^2+b^2}\left[
\begin{array}{c}
b^2 x-a b y-a c \\
a^2 y-a b x-b c \\
a^2+b^2 \\
\end{array}
\right]$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3419542",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Solve : $\sin\left(\frac{\sqrt{x}}{2}\right)+\cos\left(\frac{\sqrt{x}}{2}\right)=\sqrt{2}\sin\sqrt{x}$ Solve : $\sin\left(\dfrac{\sqrt{x}}{2}\right)+\cos\left(\dfrac{\sqrt{x}}{2}\right)=\sqrt{2}\sin\sqrt{x}$
$$\dfrac{1}{\sqrt{2}}\cdot\sin\left(\dfrac{\sqrt{x}}{2}\right)+\dfrac{1}{\sqrt{2}}\cos\left(\dfrac{\sqrt{x}}{2}\right)=\sin\sqrt{x}$$
$$\sin\left(\dfrac{\sqrt{x}}{2}+\dfrac{\pi}{4}\right)=\sin\sqrt{x}$$
$$\sin\left(\dfrac{\sqrt{x}}{2}+\dfrac{\pi}{4}\right)-\sin\sqrt{x}=0$$
$$2\sin\left(\dfrac{\dfrac{\sqrt{x}}{2}-\dfrac{\pi}{4}}{2}\right)\cos\left(\dfrac{\dfrac{3\sqrt{x}}{2}+\dfrac{\pi}{4}}{2}\right)=0$$
$$\left(\dfrac{\sqrt{x}}{2}-\dfrac{\pi}{4}\right)=2n\pi \text { or }
\left(\dfrac{3\sqrt{x}}{2}+\dfrac{\pi}{4}\right)=(2n+1)\pi$$
$$\sqrt{x}=4n\pi+\dfrac{\pi}{2} \text { or } \sqrt{x}=\dfrac{4n\pi}{3}+\dfrac{\pi}{2}$$
$$x=\left(4n\pi+\dfrac{\pi}{2}\right)^2 \text { or } x=\left(\dfrac{4n\pi}{3}+\dfrac{\pi}{2}\right)^2 \text { where n $\in$ I}$$
But actual answer is $$x=\left(4n\pi+\dfrac{\pi}{2}\right)^2 \text { or } x=\left(\dfrac{4m\pi}{3}+\dfrac{\pi}{2}\right)^2 \text { where n,m $\in$ W}$$
I tried to find the mistake but didn't get any breakthroughs.
| First,
$\sin \alpha = \sin \beta \iff
\beta \equiv \alpha \pmod{2 \pi}
\quad \text{or} \quad
\beta \equiv \pi - \alpha \pmod{2 \pi}$
So, from
$$ \sin\left(\dfrac{\sqrt{x}}{2}+\dfrac{\pi}{4}\right) = \sin\sqrt{x} $$
You could have argued
$$ \sqrt x = 2m\pi + \dfrac{\sqrt{x}}{2}+\dfrac{\pi}{4}
\qquad \text{or} \qquad
\pi - \sqrt x = -2n\pi + \dfrac{\sqrt{x}}{2}+\dfrac{\pi}{4}$$
$$ \dfrac{\sqrt x}{2} = 2m\pi +\dfrac{\pi}{4}
\qquad \text{or} \qquad
\sqrt x = \dfrac{8n+3}{6}\pi$$
$$ x = \left( 4m\pi + \dfrac{\pi}{2} \right)^2
\qquad \text{or} \qquad
x = \left( \dfrac{8n+3}{6}\pi \right)^2$$
Note that, when $x = \left( 4m\pi + \dfrac 12\pi \right)^2$, then
$\sqrt x = \left| m\pi + \dfrac 12\pi \right|$ but we need to have $\sqrt x = m\pi +\dfrac 12\pi$. So we need to require $m \in \mathbb W$. Similarly, we will need to require $n \in \mathbb W$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3420908",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
A general formula to generate functions of power series I have been playing with Maclaurin series lately, I have been able to come across this:
$\dfrac{1}{1+x}=1-x+x^2-x^3+x^4-x^5...$
$\dfrac{1}{(1+x)^2}=1-2x+3x^2-4x^3+5x^4-6x^5+7x^6...$
I found out by accident that:
$\dfrac{1-x}{(1+x)^3}=1-2^2x+3^2x^2-4^2x^3+5^2x^4+6^2x^5...$
I found in an old paper of Euler that this can continue on with these functions:
$\dfrac{1-4x+x^2}{(1+x)^4}=1-2^3x+3^3x^2-4^3x^3+5^3x^4+6^3x^5...$
$\dfrac{1-11x+11x^2-x^3}{(1+x)^5}=1-2^4x+3^4x^4-4^4x^3+5^4x^4+6^4x^5...$
$\dfrac{1-26x+66x^2-26x^3+x^4}{(1+x)^5}=1-2^5x+3^5x^4-4^5x^3+5^5x^4+6^5x^5...$
$\dfrac{1-57x+320x^2-302x^3+57x^4-x^5}{(1+x)^5}=1-2^6x+3^6x^4-4^6x^3+5^6x^4+6^6x^5...$
and so on...
Is there a general formula to generate the functions on the right hand side? How did Euler calculate these series? I have to say I deeply respect him since only he and Ramanujan know how to play with series.
| The most efficient way to obtain such formulae is to compute the Newton series for $k$-th powers, and then use the fact that $1/(1-x)^{k+1} = \sum_{n=0}^∞ \binom{n+k}{k} x^n$ for $|x| < 1$, which is easy to prove by induction (or by observing that the coefficients for the series are a column of Pascal's triangle).
For example, we can easily get $n^3 = 1 \binom{n}{1} + 6 \binom{n}{2} + 6 \binom{n}{3}$, as shown in the linked post, and hence
$\sum_{n=0}^∞ (n+1)^3 x^n = \sum_{n=0}^∞ \left( 1 \binom{n+1}{1} + 6 \binom{n+1}{2} + 6 \binom{n+1}{3} \right) x^n$
$ = \sum_{n=0}^∞ \left( 1 \binom{n+1}{1} + 6 \binom{n+2}{2} x + 6 \binom{n+3}{3} x^2 \right) x^n$ [since $\binom{n}{k} = 0$ for $0 ≤ n < k$]
$ = 1/(1-x)^2 + 6x/(1-x)^3 + 6x^2/(1-x)^4$ [for $|x| < 1$]
$ = \left( (1-x)^2 + 6x(1-x) + 6x^2 \right) / (1-x)^4$
$ = \left( 1 + 4x + x^2 \right) / (1-x)^4$
(Substituting $x$ with $-x$ gives the first series cited in the question from "an old paper of Euler".)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3422095",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Find the sum of series: $\frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{3}+\sqrt{4}}+...+\frac{1}{\sqrt{97}+\sqrt{98}}+\frac{1}{\sqrt{99}+\sqrt{100}}$ Find the sum of series:
$$\frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{3}+\sqrt{4}}+\frac{1}{\sqrt{5}+\sqrt{6}}+...+\frac{1}{\sqrt{97}+\sqrt{98}}+\frac{1}{\sqrt{99}+\sqrt{100}}$$
My Attempt:
I tried to go by telescopic method but nothing appears to be cancelling.
Something similar was given in a book by Titu Andreescu. I will try to reproduce
Let $S=\frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{3}+\sqrt{4}}+\frac{1}{\sqrt{5}+\sqrt{6}}+...+\frac{1}{\sqrt{97}+\sqrt{98}}+\frac{1}{\sqrt{99}+\sqrt{100}}$
Further let $T=\frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{3}+\sqrt{4}}+\frac{1}{\sqrt{5}+\sqrt{6}}+...+\frac{1}{\sqrt{97}+\sqrt{98}}+\frac{1}{\sqrt{99}+\sqrt{100}}$
Clearly $S+T=\sqrt{100}-1=9$
Also $S-T=2S+1-\sqrt{100}$
and $S>T$
$\Rightarrow 2S>S+T$
$\Rightarrow S>4.5$
This is the best I could come up with
| Let $S_m$ denote $\sum_{j=1}^m\sqrt{j}$. (In terms of Hurwitz eta functions $S_n=\zeta(-\frac12)-\zeta(-\frac12,\,m+1)$.) Take $n=50$ in $$\sum_{k=1}^n\frac{1}{\sqrt{2k-1}+\sqrt{2k}}=\sum_{k=1}^n(\sqrt{2k}-\sqrt{2k-1})=\sqrt{2}\sum_{k=1}^n\sqrt{k}-\sum_{2|k,\,1\le k\le 2n}\sqrt{k}\\=2\sqrt{2}\sum_{k=1}^n\sqrt{k}-\sum_{k=1}^{2n}\sqrt{k}=2\sqrt{2}S_n-S_{2n}.$$I think that's about the best we can do.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3422567",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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How do I eliminate the repeating cases? Let $K$ be the field with exactly $7$ elements. Let $\mathscr M$ be the set of all $2×2$ matrices with entries in $K$. How many elements of $\mathscr M$ are similar to the following matrix?
$ \begin{pmatrix}
0 & 0 \\
0 & 1
\end{pmatrix}$
My attempt: Answer is given is $56$. We need to find the cardinality of $\{\begin{pmatrix}
a & b \\
c & d
\end{pmatrix} a,b,c,d\in K:a+d=1 \wedge ad=bc\}=\{\begin{pmatrix}
a & b \\
c & 1-a
\end{pmatrix} a,b,c,d\in K: ad=bc\}.$ I got $ad=bc \implies a(1-a)=bc \implies a-a^2=bc\implies $ $a$ as a function of $bc$. So, there are $7^3$ possibilities.How do I eliminate the repeating cases?
| You can continue as below:
$$a-a^2=bc \Longrightarrow a^2-a+bc=0 \Longrightarrow a=\frac{1\pm\sqrt{1-4bc}}{2}\overset{1/2 = 4}{===}4\pm4\sqrt{1-4bc}$$
So $(1-4bc)$ must be square:
$$(1-4bc) \in \{ 0^2,(\pm)1^2,(\pm2)^2,(\pm3)^2 \} = \{ 0,1,4,2 \} \Longrightarrow 4bc \in \{ 1,0,-3,-1 \} \overset{1/4=2}{=\Longrightarrow}$$
$$\overset{1/4=2}{=\Longrightarrow} bc \in \{ 2,0,-6,-2 \}$$
Now we can count:
$$\begin{array}{l}
bc \overset{\text{Has 6 solutions}}{=======} {\color{white}-}2 & \Longrightarrow & a\overset{\text{Has 1 solution}}{=======}{\color{white}-}4 & \Longrightarrow & \text{Has ${\color{white}6}$6 Solutions} \\
bc \overset{\text{Has 13 solutions}}{=======} {\color{white}-}0 & \Longrightarrow & a\overset{\text{Has 2 solutions}}{=======}{\color{white}-}0,1 & \Longrightarrow & \text{Has 26 Solutions} \\
bc \overset{\text{Has 6 solutions}}{=======} -6 & \Longrightarrow & a\overset{\text{Has 2 solutions}}{=======}{\color{white}-}3,5 & \Longrightarrow & \text{Has 12 Solutions} \\
bc \overset{\text{Has 6 solutions}}{=======} -2 & \Longrightarrow & a\overset{\text{Has 2 solutions}}{=======}-1,2 & \Longrightarrow & \text{Has 12 Solutions} \\
\end{array}$$
So totally we have:
$$6+26+12+12=56 \ \text{ Solutions }$$
Also you may don't calculate $a$ and just notice the relation between "Number of solution & Sign of $\Delta$" is hold in any field.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3423194",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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If $a, b, c, d \in \mathbb{N}$ and $ab = cd \ne 0$ show that $a^2+b^2+c^2+d^2$ is composite A problem I saw on quora,
slightly modified.
If $a, b, c, d \in \mathbb{N}$
and
$ab = cd \ne 0$
show that
$a^2+b^2+c^2+d^2$
is composite.
My solution uses the theorem that
a number that can be written as the sum of two squares in two different ways is composite,
and I wondered if there is
a simpler proof
(probably involving
the difference of squares).
Here is my proof.
$a^2+b^2+c^2+d^2
=a^2+2ab+b^2+c^2–2cd+d^2
=(a+b)^2+(c-d)^2
$
and
$a^2+b^2+c^2+d^2
=a^2-2ab+b^2+c^2+2cd+d^2
=(a-b)^2+(c+d)^2
$
and it is known that
a number that can be written as the sum of two squares in two different ways is composite.
If the two ways are the same
then either
$a+b=a-b, c-d=c+d$
or
$a+b=c+d, c-d=a-b$.
In the first case,
$b=d=0$
which is not allowed.
In the second case,
adding them
$2a = 2c$ so $a=c$
and then $b=d$
so
$a^2+b^2+c^2+d^2
=2a^2+2b^2
$
which is composite.
| standard is to let $g = \gcd(a,c)$ so that $a = g \alpha$ and $c = g \gamma,$ with
$$ \gcd(\alpha, \gamma) = 1 $$
As $b \alpha = d \gamma$ and $ \gcd(\alpha, \gamma) = 1, $ we find $\gamma|b,$ let
$ b = h \gamma,$ which leads to $d = h \alpha$
$$ a^2 + c^2 + d^2 + b^2 = g^2 \alpha^2 + g^2 \gamma^2 + h^2 \alpha^2 + h^2 \gamma^2 = (g^2 + h^2) (\alpha^2 + \gamma^2) $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3423653",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Is $\frac{x^2}{a^2}+\frac{y^2}{b^2} = \frac{y}{b}$ the equation of an ellipse? Shouldn't it be $\frac{x^2}{a^2}+\frac{y^2}{b^2} = 1$?
While solving a question today I came across a locus of the form $$\frac{x^2}{a^2}+\frac{y^2}{b^2} = \frac{y}{b}$$ It was told in my book that it is the equation of an ellipse. How is that so?
Forgive me if I ask this, this may seem really silly (I'm still in high school) but isn't the general equation of an ellipse of this form?
$$\frac{x^2}{a^2}+\frac{y^2}{b^2} = 1$$
Or is it of the form
$$ax^2+by^2+2hxy+2gx+2fy+c=0$$ Can someone please confirm this for me.
| You can manipulate your equation as so:
$$\frac{x^2}{a^2} + \frac{y^2}{b^2} - \frac y b = 0$$
Then we see by completing the square on top of the $y$ terms as below,
$$\begin{align}
\frac{y^2}{b^2} - \frac y b &= \frac{y^2 - by}{b^2} \\
&= \frac{y^2 - by + b^2/4 - b^2/4}{b^2} \\
&= \frac{(y-b/2)^2 - b^2/4}{b^2} \\
&= \frac{(y-b/2)^2}{b^2} - \frac 1 4
\end{align}$$
Therefore, the first euqation and your equation is equivalent to
$$\frac{x^2}{a^2} + \frac{(y-b/2)^2}{b^2} = \frac 1 4$$
We can multiply throughout by $4$, and in the fractions this is equivalent to multiplying the denominator by $1/4$. Then we have
$$\frac{x^2}{(a/2)^2} + \frac{(y-b/2)^2}{(b/2)^2} = 1 \tag{1 }$$
A slight amending of your definition of an ellipse: an ellipse, centered at $(h,k)$, with semi-major and semi-minor axes $a,b$ which are parallel to the $x$ and $y$ axes, has the form
$$\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1$$
In this case, then, your equation - and thus $(1)$ - describes an ellipse, with center $(0,b/2)$, and semi-major and semi-minor axes $a/2,b/2$.
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "5",
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Let A,B be two $4\times4$ real commuting matrices and $\det(A^2-AB+B^2) = 0$ show that $\det(A+B) + 3\det(A-B) = 6\det(A) + 6\det(B)$ Let $A$, $B$ be two $4\times 4$ matrices with real enteries, such that $AB = BA$ and $\det(A^2-AB+B^2) = 0$. Show that $\det(A+B) + 3\det(A-B) = 6\det(A) + 6\det(B)$.
How can I get started on this question?
| We can exploit the fact that each expression is homogeneous on $A$ and $B$.
Assume that $\det(B) \neq 0$. Multiply both sides of the first equation by $\det(B)^{-2}$, and both sides of the second equation by $\det(B)^{-1}$. Using the fact that $A$ and $B$ commute, we have
$$
\det(A^2 - AB + B^2)\det(B)^{-2} = 0 \implies\\
\det(B^{-2}(A^2 - AB + B^2)) = 0 \implies\\
\det((AB^{-1})^2 - (AB^{-1}) + I) = 0.
$$
Similarly, multiply both sides of the second equation by $\det(B)^{-1}$ to get
$$
\det((AB^{-1}) + I) + 3 \det((AB^{-1}) - I) = 6 \det(AB^{-1}) + 6.
$$
So, let $X = AB^{-1}$. It suffices to show that
$$
\det(X^2 - X + I) = 0 \implies \det(X + I) + 3 \det(X - I) = 6 \det(X) + 6.
$$
From there, the trick is hidden below (put your mouse over it to see). This is enough to get started though, and I recommend that you try to figure this part out for yourself.
Hint:
$X$ is a matrix with real entries, so its complex eigenvalues must come in conjugate pairs.
Solution:
Thus, we can guarantee that the roots of $x^2 - x + 1 = 0$,
namely $\lambda_\pm = \frac 12 \pm \frac{\sqrt{3}}2 i$, must be eigenvalues of
$X$. Let $a,b$ denote the remaining two eigenvalues. We can now rewrite each
expression as follows:
$$ \det(X+I) = (\lambda_+ +1)(\lambda_- + 1)(a+1)(b+1) = 3(a+1)(b+1)\\ \det(X-I) = (\lambda_+ - 1)(\lambda_- - 1)(a-1)(b-1) = (a-1)(b-1)\\ \det(X) = \lambda_+ \lambda_- ab = ab$$
We can now show by routine simplification that the desired equality holds in the case that $\det(B) \neq 0$.
For the case that $\det(B) = 0$, it suffices to makes an argument by continuity. Since
$$
\det(A+(B + \epsilon I)) + 3\det(A-(B + \epsilon I)) = 6\det(A) + 6\det(B + \epsilon I)
$$
will hold for all sufficiently small $\epsilon > 0$ (which will force $B + \epsilon I$ to be invertible), it must also hold that the two sides are equal for $\epsilon = 0$.
| {
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"timestamp": "2023-03-29T00:00:00",
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closed form for this recurrence I am having trouble figuring out if the following recurrence has a closed form:
$$f(n)=2f(n-1)+(n-1)2^{n-2}$$
I have never really done such problems, so I dont know what is a good strategy. I will appreciate any hints!
Edit: $f$ is defined for $n\geq 1$ and $f(1)=0$.
| I was thinking on using the Z-transform here:
$$ f(n) = 2f(n-1) + (n-1)2^{n-2} $$
using $f(n) = x_n$ and $ f(n-1) = x_{n-1}$ we get:
$$ x_n = 2x_{n-1}+n\frac{2^n}{4}-\frac{2^n}{4} $$
Taking the transform on both sides:
$$ X(z) = 2(z^{-1}X(z)+x_{-1}) + \frac{2z}{4(z-2)^2}-\frac{z}{4(z-2)} $$
$$ X(z) = 2(z^{-1}X(z)+\frac{1}{8}) + \frac{2z}{4(z-2)^2}-\frac{z}{4(z-2)} $$
$$ X(z)(1-\frac{2}{z}) = \frac{2z-z(z-2)}{4(z-2)^2} + \frac{1}{4}$$
$$ X(z) = \frac{z^2(4-z)}{4(z-2)^3} + \frac{z}{4(z-2)}$$
Taking the inverse Z transform we get a result:
$$ x_n = n(n-1)2^{n-3} $$
| {
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"timestamp": "2023-03-29T00:00:00",
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Find $4x \equiv 7 \pmod{15}$ and $3x \equiv 5 \pmod{16}$ (different exercises) This is how I solved each:
$$4x \equiv 7\pmod {15} \\
4x - 7 \equiv 0\pmod{15} \\
4x-7 = 15k \Leftrightarrow 4x-15k= 7 \\
$$
$$15 = 4*3+3 \\
4=3*1+1\\
3=3*1+0$$
$$1 = 4-3*1 \\
3 = 15-4*3 \\
1 = 4 - (15-4*3)*1 \\
1 = 4-15+4*3 \\
1 = 4*4-15*1 \\
$$
$$7 = (7*4)*4-(7*1)*15$$
$7*4*4 = 112$ which is congruent with 7 in mod 15. Yet my book says the answer is 13. What went wrong?
The second one:
$$3x - 5 \equiv 0 \pmod{16} \\
3x-5=16k\\
3x-16k=5 \\$$
$$16=3*5+1\\
3=1*3+0$$
$$1 = 16-3*5 \\
5=16*5-3*5*5$$
Yet $3*5*5-5$ is not congruent with zero in mod 16. What went wrong?
| Don't do that with such computations! For each of these congruence equations, you just have to find the modular inverse of the coefficient (which happens to be a modular unit since it is coprime to the modulus).
Example for the first congruence:
The modular inverse of $4\bmod 15$ is none other than itself since $4^2=16\equiv 1\mod 15$. So multiply both sides of the congruence equation by $4$:
$$4x\equiv 7\mod 15\implies 4^2x\equiv \color{red}x\equiv 4\cdot 7=28\color{red}{\equiv 13\mod 15}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3431776",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How do I solve this: I was doing my homework and I stumbled over this particular exercise. I would've known how to solve it, if it had been the same thing under square root in both cases.
$$
\lim_{n\to \infty} \left(\sqrt{4n^2+3n+2}-\sqrt{4n^2+n-1}\right)
$$
| By binomial approximation
$$\sqrt{4n^2+3n+2}=2n\left(1+\frac3{4n}+\frac1{2n^2}\right)^\frac12 \approx 2n+\frac34+\frac1{2n}$$
$$\sqrt{4n^2+n-1}=2n\left(1+\frac1{4n}-\frac1{4n^2}\right)^\frac12 \approx 2n+\frac14-\frac1{4n}$$
therefore
$$\sqrt{4n^2+3n+2}-\sqrt{4n^2+n-1}\approx 2n+\frac34+\frac1{2n}-2n-\frac14+\frac1{4n}=\frac12+\frac3{4n}\to \frac12$$
or as alternative, we can use the standard trick
$$A-B=(A-B)\dfrac{A+B}{A+B}=\frac{A^2-B^2}{A+B}$$
to obtain
$$(\sqrt{4n^2+3n+2}-\sqrt{4n^2+n-1})=$$
$$=(\sqrt{4n^2+3n+2}-\sqrt{4n^2+n-1})\dfrac{\sqrt{4n^2+3n+2}+\sqrt{4n^2+n-1}}{\sqrt{4n^2+3n+2}+\sqrt{4n^2+n-1}}=$$
$$=\dfrac{4n^2+3n+2-4n^2-n+1}{\sqrt{4n^2+3n+2}+\sqrt{4n^2+n-1}}=$$
$$=\dfrac{2n+3}{\sqrt{4n^2+3n+2}+\sqrt{4n^2+n-1}}=$$
$$=\dfrac{2+\frac3n}{\sqrt{4+\frac3n+\frac2{n^2}}+\sqrt{4+\frac1n-\frac1{n^2}}} \to \frac2 4 = \frac12$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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$x,y,z > 0$, prove that $ 3xyz + x^{3}+ y^{3} + z^{3} \ge 2 \left[ (xy)^{3/2} + (yz)^{3/2} + (xz)^{3/2} \right] $ $x,y,z > 0$, prove that
$$ 3xyz + x^{3}+ y^{3} + z^{3} \ge 2 \left[ (xy)^{3/2} + (yz)^{3/2} + (xz)^{3/2} \right] $$
Without using Schur's inequality,
Attempt:
By $C.S$:
$$ (xy)^{3/2} + (yz)^{3/2} + (xz)^{3/2} = x^{3/2}y^{3/2} + y^{3/2}z^{3/2} + z^{3/2}x^{3/2} $$
$$ \le \sqrt{ x^{3} + y^{3} +z^{3}} \sqrt{ y^{3} + z^{3} + x^{3}} = x^{3} + y^{3} + z^{3} $$
then I have to prove
$$ (xy)^{3/2} + (yz)^{3/2} + (xz)^{3/2} \le 3xyz$$
this one is difficult. Another thing that we know by AM-GM: $ x^{3} + y^{3}+ z^{3} \ge 3xyz$.
| First, Schur's inequality provides that
$$
x^3+y^3+z^3+3xyz\ge xy(x+y)+yz(y+z)+zx(z+x).
$$
Then,
$$
xy(x+y)\ge 2xy\sqrt{xy}
$$
and hence
$$
x^3+y^3+z^3+3xyz\ge 2\big(xy\sqrt{xy} +yz\sqrt{yz} +zx\sqrt{zx} \big)
$$
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "4",
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How big is the smallest triangle inscribed in a square A square is divided in 7 areas as show on the figure.
The dots show the corners of the square and the middle points on the edges.
How large a fraction is area $D$ and how do I work it out?
I have tried using trigonometry to calculate the area A. If we say each side of the square is $1$, and look at the triangle $ABC$ using Pythagoras, it's hypotenuse must be $ \sqrt {1 \cdot 1 + 0.5 \cdot 0.5} = 1.118$.
Then using the sine relation we see that the $ \hat A$ is $\sin A = 1/1.118 = 63.43°$. It then follows the other angles must be $90°$ and $71.57°$. If I use Heron's formula I can calculate the area of $A = \sqrt {p(p−a)(p−b)(p−c)} = \frac 1 {12}$.
I know $C$ is $ \frac1 {16}$ just by looking at the figure.
The area of $ABC$ is $ \frac 1 4 $, so $B$ must be $\frac 1 4 - C - A = \frac 5 {48}$.
Now the area of $BD$ must be $ \frac 1 8$. It therefore follows that $D = \frac 1 8 - \frac 5 {48} = \frac 1 {48}$.
The trigonometry part just seems too elaborate, and I was wondering if there is a much more simple solution I am missing?
| Alternatively, we can find that the triangle containing $D$ and $E$ has a base of $1$ and a height of $\frac{2}{3}$ (obtained by solving $1-x = \frac{1}{2} + \frac{1}{2}x$), so it has an area of $\frac{1}{2} \times 1 \times \frac{2}{3} = \frac{1}{3}$.
The height of $D$ is $\frac{2}{3} - \frac{1}{2} = \frac{1}{6}$, so the sides are $\frac{1}{4}$ of the big triangle. Since triangles $D$ and $D + E$ are similar by AA, the area of $D$ is $\frac{1}{16}$ times smaller. This gives the area of triangle $D$ as $\frac{1}{3} \times \frac{1}{16} = \frac{1}{48}$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Find the smallest value of the following expression $\sqrt{(x-9)^2 +4} + \sqrt{x^2+y^2}+ \sqrt{(y-3)^2 +9}$
Find the smallest value of the following expression:
$$\sqrt{(x-9)^2 +4} + \sqrt{x^2+y^2} + \sqrt{(y-3)^2 +9}$$
I tried to derive the expression with respect to $x$ and $y$ and then equal the derivative to zero and find the critical points, but I could not do that. It is complicated. Is there a way?
| Proceeding as your first idea by derivative, we have that for
$$f(x,y)=\sqrt{(x-9)^2 +4} + \sqrt{x^2+y^2} + \sqrt{(y-3)^2 +9}$$
in order to minimize we can guess that $0<x<9$ and $0<y<3$ and then
$$f_x(x,y)=\frac{x}{\sqrt{x^2 +y^2}}+ \frac{x-9}{\sqrt{(x-9)^2+4}}=0 \implies y^2=\frac{4x^2}{(9-x)^2}\implies y=\frac{2x}{9-x}$$
$$f_y(x,y)=\frac{y}{\sqrt{x^2 +y^2}}+ \frac{y-3}{\sqrt{(y-3)^2+9}}=0\implies x^2=\frac{9y^2}{(3-y)^2}\implies x=\frac{3y}{3-y}$$
from which we obtain
$$(9-x)(3-y)=6$$
and then
$$y=\frac{2x}{(9-x)}\implies 3-\frac6{9-x}=\frac{2x}{(9-x)}$$
$$\implies {5x-21}{x-9}=0 \implies x=\frac{21}5$$
and
$$y=\frac{2x}{(9-x)}=\frac{\frac{42}5}{\frac{24}5}=\frac 7 4$$
therefore the solution is $(x,y)=\left(\frac{21}5,\frac 7 4\right)$.
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "10",
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General formula for the power series of $\dfrac{1}{(1+x)^3}$ I wish to find the general formula for the following power series:
$\dfrac{1}{(1+x)^3}=1-3x+6x^2-10x^3+15x^4-21x^5+28x^6...$
The difference between the first and second term is $2$
The difference between the second and third term is $3$
The difference between the third and fourth term is $4$
The difference between the fourth and fifth term is $5$
and so on.
How do you find the general formula for this series? It is not quite an arithmetic series, and definitely not a geometric series, so what should I do?
| If you know calculus, then
$\frac {1}{(1+x)^3} = \frac {d^2}{dx^2}\frac {1}{2(1+x)} = \frac {d^2}{dx^2} \sum_\limits{n=0}^\infty \frac {(-1)^nx^n}{2} = \sum_\limits{n=0}^\infty \frac {(-1)^n(n+1)(n+2) x^n}{2}$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "1",
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$\cot \pi iy$ tends to $-i$ as $y \to \infty$ For a fixed real number $x$, why does $\cot \pi z$ tends to $-i$ as $y \to \infty$? (Here, $z=x+iy$.) Can this be proved simply?
| Note the following:
$\cot(z) = \frac{\cos(z)}{\sin(z)}$
$\cos(x+iy) = \cos(x) \cosh(y) - i \sin(x) \sinh(y)$
$\sin(x+iy) = \sin(x) \cosh(y) + i \cos(x) \sinh(y)$
so
\begin{equation}
\cot(z) = \frac{\cos(x) \cosh(y) - i \sin(x) \sinh(y)}{\sin(x) \cosh(y) + i \cos(x) \sinh(y)}
\end{equation}
and multiply the numerator and denominator by the conjugate of the denominator to arrive at
\begin{equation}
\frac{(\cos(x) \cosh(y) - i \sin(x) \sinh(y))(\sin(x) \cosh(y) - i \cos(x) \sinh(y))}{\sin^2(x) \cosh^2(y) + \cos^2(x) \sinh^2(y)}
\end{equation}
and distribute and factor a little to arrive at
\begin{equation}
\frac{\cos(x)\sin(x)(\cosh^2(x) - \sinh^2(x)) - i \cosh(y)\sinh(y)(\cos^2(x)+\sin^2(x))}{\sin^2(x) \cosh^2(y) + \cos^2(x) \sinh^2(y)}
\end{equation}
this simplifies by the Pythagorean identities: $\cos^2(x) + \sin^2(x) = 1$ and $\cosh^2(x) - \sinh^2(x) = 1$:
\begin{equation}
\frac{\cos(x)\sin(x) - i \cosh(y)\sinh(y)}{\sin^2(x) \cosh^2(y) + \cos^2(x) \sinh^2(y)}
\end{equation}
We now focus on the denominator, expressing the $\sinh^2(y) = \cosh^2(y) - 1$ and $\sin^2(x) = 1 - \cos^2(x)$ we have
\begin{equation}
\frac{\cos(x)\sin(x) - i \cosh(y)\sinh(y)}{(1-\cos^2(x)) \cosh^2(y) + \cos^2(x) (\cosh^2(y)-1)}
\end{equation}
and distributing we notice two cross terms cancel giving:
\begin{equation}
\frac{\cos(x)\sin(x) - i \cosh(y)\sinh(y)}{\cosh^2(y) - \cos^2(x)}
\end{equation}
Finally we ask about the limit, in the denominator $\cosh^2(y)$ grows without bound as $\pi y \rightarrow \infty$ and dominates the bounded $\cos^2(x)$. It also dominates the real part of the numerator. So we must only consider the ratio:
\begin{equation}
\lim_{y \rightarrow \infty} -i\frac{\cosh(y)\sinh(y)}{\cosh^2(y)} = \lim_{y \rightarrow \infty} -i\tanh(y)
\end{equation}
Finally $\tanh(y) = \frac{e^y - e^{-y}}{e^y + e^{-y}}$ when $y$ is large and positive only the positive exponents matter. It should be clear that $\tanh(y)$ approaches one in this limit. So in fact the solution should be:
\begin{equation}
\lim_{y \rightarrow \infty} \cot(x+iy) = \lim_{y \rightarrow \infty} -i\tanh(y) = -i
\end{equation}
not $i$ as you suggested.
| {
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Why is the graph of $8x^2+8xy+2y^2+2x+y+5=0$ empty? For the equation
$$8x^2+8xy+2y^2+2x+y+5=0$$
Comparing it with standard equation of conic section we get that
$h^2-ab=0$ and
$$\Delta= \left|\begin{array}{ccc}
a & h & g \\
h & b & f \\
g & f & c \end{array}\right| = 0$$
So as per I know it should represent a pair of straight line which are real and coincident but after plotting it on desmos( an app to plot graphs) it showed nothing. Please help me.
| (as Oscar Lanzi started...)
\begin{align}
2(2x+y)^2 + (2x+y) + 5 &= 0 \\
16(2x+y)^2 + 8(2x+y) &= -40 \\
[4(2x+y)+1]^2 &= -39 \\
\end{align}
| {
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Prove $S=\{a+b\sqrt[3]{2} +c\sqrt[3]{4}:a,b,c \in \mathbb{Q}\}$ is a ring. Prove $S=\{a+b\sqrt[3]{2} +c\sqrt[3]{4}:a,b,c \in \mathbb{Q}\}$ is a ring.
My professor has gone over how to prove a subring, but I am not sure how to prove a ring.
This is what I have so far, but I am not sure if it is similar to proving a subring, or what I have left to prove.
To prove this set is a ring, we must prove
(i) $S$ is closed under addition, and
(ii) $S$ is closed under multiplication, and
(iii)$S$ is associative on the operations, and
(iv) $S$ is commutative on the operations, and
(v) $S$ is distributive on the operations, and
(vi) $S$ has an additive identity ($0_R$), and
(vii) All operations in $S$ have an additive inverse.
We will begin by proving part (i). That is, we will prove $S$ is closed under addition. We will take $a+b\sqrt[3]{2}+c\sqrt[3]{4} \in S$ and add it to an arbitrary element $d+e\sqrt[3]{2}+f\sqrt[3]{4} \in S$ to get $(a+b\sqrt[3]{2}+c\sqrt[3]{4})+(d+e\sqrt[3]{2} +f\sqrt[3]{4})$. We will then factor out $\sqrt[3]{2}$ and $\sqrt[3]{4}$ using the distributive property to get $(a+d+(b+e)\sqrt[3]{2}+(c+f)\sqrt[3]{4})$. Since $b,e,c,f \in \mathbb{Q}$, $(b+e),(c+f) \in \mathbb{Q}$, we know $S$ is closed under addition which proves part (i) of the proof.
We will now prove part (ii). That is, we will prove $S$ is closed under multiplication. We will take $a+b\sqrt[3]{2}+c\sqrt[3]{4} \in S$ and multiply it to an arbitrary element $d+e\sqrt[3]{2}+f\sqrt[3]{4} \in S$ to get $(a+b\sqrt[3]{2}+c\sqrt[3]{4})(d+e\sqrt[3]{2} +f\sqrt[3]{4})$. We will then multiply to get $ad+ae\sqrt[3]{2}+af\sqrt[3]{4}+bd\sqrt[3]{2}+be\sqrt[3]{2}\sqrt[3]{2}+bf\sqrt[3]{2}\sqrt[3]{4}+cd\sqrt[3]{4}+ce\sqrt[3]{2}\sqrt[3]{4}+cf\sqrt[3]{4}\sqrt[3]{4}$. We know $\sqrt[3]{4}=\sqrt[3]{2}\sqrt[3]{2}$, so we can make a substitution to get $ad+ae\sqrt[3]{2}+af\sqrt[3]{2}\sqrt[3]{2}+bd\sqrt[3]{2}+be\sqrt[3]{2}\sqrt[3]{2}+bf\sqrt[3]{2}\sqrt[3]{2}\sqrt[3]{2}+cd\sqrt[3]{2}\sqrt[3]{2}+ce\sqrt[3]{2}\sqrt[3]{2}\sqrt[3]{2}+cf\sqrt[3]{2}\sqrt[3]{2}\sqrt[3]{2}$. We can then simplify this to $ad+ae\sqrt[3]{2}+af(2\sqrt[3]{2})+bd\sqrt[3]{2}+be(2\sqrt[3]{2})+bf(3\sqrt[3]{2})+cd(2\sqrt[3]{2})+ce(3\sqrt[3]{2})+cf(4\sqrt[3]{2})$ factor out $\sqrt[3]{2}$ using the distributive property to get $ad+(ae+bd+af(2)+bd+be(2)+bf(3)+cd(2)+ce(3)+cf(4))\sqrt[3]{2}$. Since $a,b,c,d,e,f,2,3,4 \in \mathbb{Q}$, $(ae+bd+af(2)+bd+be(2)+bf(3)+cd(2)+ce(3)+cf(4)) \in \mathbb{Q}$. Thus, we know $S$ is closed under multiplication which proves part (ii) of the proof.
We will now prove part (vi). That is, we will prove $S$ has an additive identity, or in other words has $0_R$. We will take $a+b\sqrt[3]{2}+c\sqrt[3]{4} \in S$. Since $a,b,c \in \mathbb{Q}$ and $0 \in Q$, we will let $a$, $b$, and $c$ be $0$ to get $0+0\sqrt[3]{2}+0\sqrt[3]{4}$. We can simplify this to get $0+0\sqrt[3]{2}+0\sqrt[3]{4}=0+0+0=0_R$. Since the ring contains 0, we know it has an additive identity and proves part (vi) of the proof.
Can anyone help me get on track, please?
| Alternative proof: $\mathbb{Q}[x]$ is a ring and let $I$ be the ideal generated by $x^3 -2$, then we have that $R = \mathbb{Q}[x]/ I$ is a ring.
We have a homomorphism of rings $\phi: \mathbb{Q}[x] \to \mathbb{R}$ given by $\phi(1) = 1$ and $\phi(x) = \sqrt[3]{2}$, since $\phi(I) \subseteq \{0\}$ this defines a homomorphism $\phi: R \to \mathbb{R}$, now $R$ is generated by $1,x$ and the image of $\phi$ contains, $1, \sqrt[3]{2},\sqrt[3]{2} = \phi(x^2)$ (the image of a map is generated by generators of its domain) so in fact $S = \text{im}(\phi)$, as it is the image of a ring under a homomorphism of rings, it is a ring, a posteriori it is an isomorphism.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3446309",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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If $n\geq 3$, $9^n \equiv a (\mod 100)$ and $9^{n+1} \equiv b (\mod 100)$, then $a+b=90$. I noticed a pattern in the powers of 9 modulo 100.
$9^1 \equiv 9 \pmod{100}$
$9^2 \equiv 81 \pmod{100}$
$9^3 \equiv 29 \pmod{100}$
$9^4 \equiv 61 \pmod{100}$
.
.
.
and conjectured the following:
If $n\geq 3$ is an odd integer where $9^n \equiv a \pmod{100}$ and $9^{n+1} \equiv b \pmod{100}$, then $a+b=90$.
How do I prove this?
| $\ \overbrace{ \color{#c00}{10}\cdot 9^n}^{\textstyle 9^n\!+\!9\cdot9^n\!\!\!\!\!\!\!\!}\bmod \color{#c00}{10}0\, =\, \color{#c00}{10}\,(\!\!\!\overbrace{9^n}^{\large (-1)^{\Large n}}\!\!\!\bmod 10)\, = \,\left\{\begin{align}&10,\,\ n\rm\ even\\ &90,\,\ n\ \rm odd\end{align}\right.$
$\text{using}\,\ \color{#c00}ab\bmod \color{#c00}ac\, =\, \color{#c00}a\,(b\bmod c)\, = $ $\bmod\!$ Distributive Law to factor out $\,\color{#c00}{a = 10}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3446410",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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What are the values of the parameters $a,b \in \mathbb{R}$ such that $\lim\limits_{x \to \infty}(\sqrt{x^2+x+1}+\sqrt{x^2+2x+2}-ax-b)=0.$ I have to find the values of $a$ and $b$ (with $a,b \in \mathbb{R}$) such that the following is true:
$$\lim\limits_{x \to \infty} (\sqrt{x^2+x+1} + \sqrt{x^2+2x+2}-ax-b)=0$$
This is what I did:
$$\lim\limits_{x \to \infty} (\sqrt{x^2+x+1} + \sqrt{x^2+2x+2}-ax-b)=0$$
$$\lim\limits_{x \to \infty} \bigg (x\sqrt{1+\dfrac{1}{x}+\dfrac{1}{x^2}} + x\sqrt{1+\dfrac{2}{x}+\dfrac{2}{x^2}}-ax-b \bigg )=0$$
$$\lim\limits_{x \to \infty} x\bigg (\sqrt{1+\dfrac{1}{x}+\dfrac{1}{x^2}} + \sqrt{1+\dfrac{2}{x}+\dfrac{2}{x^2}}-a-\dfrac{b}{x} \bigg )=0$$
So, we'd have something like:
$$\infty \cdot(2-a) = 0$$
$(*)$If $a \in (-\infty, 2)$ $\Rightarrow$ $(2-a) > 0$, which means:
$$\infty \cdot(2-a) = \infty$$
$(*)$If $a\in (2, +\infty)$ $\Rightarrow$ $(2-a) < 0$, which means:
$$\infty \cdot(2-a) = -\infty$$
So the only option left for the limit to have any chance of being true is:
$$a=2$$
which would result in the indeterminate form:
$$\infty \cdot (2-a) = \infty \cdot 0$$
And now that we have $a=2$, we need to find the value of $b$ for the limit to be true. Substituting $a$ with $2$ in the initial limit we get:
$$\lim\limits_{x \to \infty} (\sqrt{x^2+x+1} + \sqrt{x^2+2x+2}-2x-b)=0$$
Since $b$ is a constant we can pull it out of the limit and get:
$$b=\lim\limits_{x \to \infty} (\sqrt{x^2+x+1} + \sqrt{x^2+2x+2}-2x)$$
And this is where I got stuck. I tried a bunch of methods and tricks for finding this limit and I got nowhere. That leads me to think that either I made some mistake/mistakes along the way, or I simply don't know how to solve this limit. So, how can I find $b$?
| \begin{align*}
&\sqrt{x^{2}+x+1}-x+\sqrt{x^{2}+2x+2}-x\\
&=\dfrac{x+1}{\sqrt{x^{2}+x+1}+x}+\dfrac{2x+2}{\sqrt{x^{2}+2x+2}+x}\\
&\rightarrow \dfrac{1}{2}+\dfrac{2}{2}\\
&=\dfrac{3}{2}.
\end{align*}
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Computing the cycle decomposition of the composition of permutations. Let $\sigma,\tau\in S_5$ be given by
$$
\sigma = \begin{pmatrix}1&3&5\end{pmatrix}\begin{pmatrix}2&4\end{pmatrix},\quad \tau = \begin{pmatrix}1&5 \end{pmatrix}\begin{pmatrix}2&3\end{pmatrix}.
$$
I want to find the cycle decomposition of $\tau\sigma$. So I write
$$
\tau\sigma = \begin{pmatrix}1&5 \end{pmatrix}\begin{pmatrix}2&3\end{pmatrix}\begin{pmatrix}1&3&5\end{pmatrix}\begin{pmatrix}2&4\end{pmatrix}.
$$
Reading from right to left and starting with the element $1$, I see that $1$ is taken to $3$ and $3$ is taken to $2$. Now, $2$ is taken to $4$, and $4$ is fixed in the other cycles, so our first cycle is $\begin{pmatrix}1&2&4\end{pmatrix}$. Since $3$ is taken to $5$ and $5$ is taken to $1$, the next cycle would be $\begin{pmatrix}3&1\end{pmatrix}$. But these cycles aren't disjoint, so I must have done something wrong. Where is my error?
| in Simple Form :
${ \begin{bmatrix}{1} && {2} && {3} && {4} && {5} \\ {5} && {3} && {2} && {4} && {1}\end{bmatrix} }$ ${ \begin{bmatrix}{1} && {2} && {3} && {4} && {5} \\ {3} && {4} && {5} && {2} && {1}\end{bmatrix} }$
= ${ \begin{bmatrix}{1} && {2} && {3} && {4} && {5} \\ {2} && {4} && {1} && {3} && {5}\end{bmatrix} } = (1\ 2\ 4\ 3)\, (5) = (1\ 2\ 4\ 3)$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Triangle sides $a,b,c$ are in arithmetic progression. Show $\sin^2(A/2)\csc2A$, $\sin^2(B/2)\csc2B$, $\sin^2(C/2)\csc2C$ are in harmonic progression
If sides $a,b,c$ of the triangle ABC are in arithmetic progression, prove that $\sin^2\frac{A}{2}\mathrm{cosec}(2A),\sin^2\frac{B}{2}\mathrm{cosec}(2B),\sin^2\frac{C}{2}\mathrm{cosec}(2C)$ are in harmonic progression.
My attempt is as follows:-
$$T_1=\dfrac{1-\cos A}{2\sin2A}$$
$$T_1=\dfrac{1}{4}\dfrac{\sec A-1}{\sin A}$$
$$T_1=\dfrac{2R}{4}\left(\dfrac{\dfrac{2bc}{b^2+c^2-a^2}-1}{a}\right)$$
$$T_1=\dfrac{R}{2}\left(\dfrac{\dfrac{2bc-b^2-c^2+a^2}{b^2+c^2-a^2}}{a}\right)$$
$$T_1=\dfrac{R}{2}\left(\dfrac{\dfrac{a^2-(b-c)^2}{b^2+c^2-a^2}}{a}\right)$$
$$T_1=\dfrac{R}{2}\left(\dfrac{(a+c-b)(a+b-c)}{(b^2+c^2-a^2)a}\right)$$
By symmetry, we can say $T_2=\dfrac{R}{2}\left(\dfrac{(b+c-a)(b+a-c)}{(a^2+c^2-b^2)b}\right)$
$T_3=\dfrac{R}{2}\left(\dfrac{(c+a-b)(c+b-a)}{(a^2+b^2-c^2)c}\right)$
For $T_1,T_2,T_3$ to be in HP, $\dfrac{1}{T_1},\dfrac{1}{T_2},\dfrac{1}{T_3}$ should be in A.P
$$\dfrac{1}{T_1}+\dfrac{1}{T_3}-\dfrac{2}{T_2}=\dfrac{2}{R}\left(\dfrac{(b^2+c^2-a^2)a}{(a+c-b)(a+b-c)}+\dfrac{(a^2+b^2-c^2)c}{(c+a-b)(c+b-a)}-\dfrac{2(a^2+c^2-b^2)b}{(b+c-a)(a+b-c)}\right)$$
$$\dfrac{1}{T_1}+\dfrac{1}{T_3}-\dfrac{2}{T_2}=\dfrac{2}{R}\left(\dfrac{a(b^2+c^2-a^2)(b+c-a)+c(a^2+b^2-c^2)(a+b-c)-2b(a^2+c^2-b^2)(c+a-b)}{(b+c-a)(a+b-c)(c+a-b)}\right)$$
$$\dfrac{1}{T_1}+\dfrac{1}{T_3}-\dfrac{2}{T_2}=\dfrac{2}{R}\left(\dfrac{a(b^3+b^2c-ab^2+bc^2+c^3-ac^2-a^2b-a^2c+a^3)+c(a^3+a^2b-a^2c+b^2a+b^3-b^2c-c^2a-c^2b+c^3)-2b(a^2c+a^3-a^2b+c^3+ac^2-bc^2-b^2c-ab^2+b^3)}{(b+c-a)(a+b-c)(c+a-b)}-\right)$$
$$\dfrac{1}{T_1}+\dfrac{1}{T_3}-\dfrac{2}{T_2}=\dfrac{2}{R}\left(\dfrac{a^4+c^4-2b^4+ab^3-a^3b+ac^3-a^3c+a^3c+b^3c-ac^3-bc^3+ab^2c+abc^2+a^2bc+ab^2c-2a^2bc-2abc^2\cdot\cdot}{(b+c-a)(a+b-c)(c+a-b)}\right)$$
It was getting very difficult to solve from here, is there any other method in which we can solve this question?
| You have not used the fact that $a$, $b$, $c$ are in arithmetic progression. WLOG write $a=b-d$, $c=b+d$; then
\begin{align*}
\dfrac{1}{T_1}+\dfrac{1}{T_3}-\dfrac{2}{T_2}&=\dfrac{2}{R}\left(\dfrac{(b^2+c^2-a^2)a}{(a+c-b)(a+b-c)}+\dfrac{(a^2+b^2-c^2)c}{(c+a-b)(c+b-a)}-\dfrac{2(a^2+c^2-b^2)b}{(b+c-a)(a+b-c)}\right)\\
&= \frac{2}{R}\left(\frac{b^2+3 b d-4 d^2}{b-2 d}+\frac{b^2-3bd-4d^2}{b+2d}-\frac{2b^3+4bd^2}{(b-2d)(b+2d)}\right).
\end{align*}
To see that the parenthesized expression is zero, just compute the combined numerator:
$$(b^2+3bd-4d^2)(b+2d)+(b^2-3bd-4d^2)(b-2d)-2b^3-4b d^2=0.$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Find an eigenvector I am trying to find an eigenvector of the following matrix, where I get an eigenvector zero. But, to my knowledge, an eigenvector cannot be zero.
$$ \pmatrix{0&1\\0&i-1}\pmatrix{x\\y} = \pmatrix{0\\0}.$$
| You are looking for an eigenvector $(x,y)$ associated to the eigenvalue $\lambda=0$.
$$
\begin{pmatrix}0&1\\0&i-1\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix} = 0\begin{pmatrix}x\\y\end{pmatrix} = \begin{pmatrix}0\\0\end{pmatrix}
$$
On the other hand,
$$\begin{pmatrix}0&1\\0&i-1\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix} = \begin{pmatrix}y\\(i-1)y\end{pmatrix}$$
This implies $y=0$. So, an eigenvector is given by $\alpha\begin{pmatrix}1\\0\end{pmatrix}$ with $\alpha\in\mathbb{R}^*$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3451605",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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Generating function of $a_{n+1} = 4a_{n} - a_{n-1}$ I need to find the generating function of this recurrence when $a_0 = 4$, $a_1$ = 8
$f(x) = \sum_{i=0}^\infty a_ix^i = 4 + 8x + (4a_1 - a_0)x^2 + (4a_2 - a_1)x^3 + \dots$
$f(x) = 4 + 8x + 4a_1x^2 - a_0x^2 + 4a_2x^3 - a_1x^3 + \dots$
$f(x) = 4 + 8x + 4x(a_1x + a_2x^2 + \dots) - x^2(a_0 + a_1x + \dots)$
$f(x) = 4 + 8x + 4x(f(x) - 4) - x^2f(x)$
$f(x) = \dfrac{4 - 8x}{x^2 - 4x + 1}$
I dont know if this is the correct generating function, but either way I dont know how to interpret the answer, like for instance $f(1)$ is the sum of all terms of the recurrence? then it seems strange I get $f(1) = 2$. Did I do a mistake?
| To rewrite your work more clearly:
\begin{align}f(x)&=\sum_{n=0}^\infty a_nx^n\\&=4+8x+\sum_{n=0}^\infty a_{n+2}x^{n+2}\\&=4+8x+\sum_{n=0}^\infty(4a_{n+1}-a_n)x^{n+2}\\&=4+8x+4x(f(x)-4)-x^2f(x)\\&=4-8x-(x^2-4x)f(x)\end{align}
$$f(x)=\frac{4-8x}{x^2-4x+1}$$
As noted in the comments, this representation is valid provided that it converges, which is only for $|x|<2-\sqrt3$.
From this form, we can then derive the coefficients by expanding it out as follows:
\begin{align}f(x)&=\frac{4-8x}{x^2-4x+1}\\&=\frac{4-8x}{(x-2)^2-3}\\&=\frac{4-8x}{(x-2-\sqrt3)(x-2+\sqrt3)}\\&=\frac{-4-2\sqrt3}{x-2-\sqrt3}+\frac{-4+2\sqrt3}{x-2+\sqrt3}\\&=\frac2{1-\frac x{2+\sqrt3}}+\frac2{1-\frac x{2-\sqrt3}}\\&=2\sum_{n=0}^\infty\underbrace{\left[\frac1{(2+\sqrt3)^n}+\frac1{(2-\sqrt3)^n}\right]}_{a_n/2}x^n\end{align}
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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$k=-\sqrt{3}(\sqrt{5+2\sqrt{6}}+\sqrt{5-2\sqrt{6}})$
Let $k$ be equal to:
$$k=-\sqrt{3} \left(\sqrt{5+2\sqrt{6}}+\sqrt{5-2\sqrt{6}} \right)$$
I am trying to simplify the expression and express $5+2\sqrt{6}$ and $5-2\sqrt{6}$ as squares. I don't think it's smart to multiply the brackets by $-\sqrt{3}$ at first. Is there any algorithm that I should know to express the expressions as squares?
| $\sqrt{5 \pm 2 \sqrt{6}} = \sqrt{3} \pm \sqrt{2}$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Minimise $2x+y$ subject to $x^2+y^2=1$ using KKT I'm doing this exercise in preparing for the final exam in optimization:
$$\begin{align*}
\text{min} &\quad 2x+y \\
\text{s.t} & \quad x^2+y^2=1
\end{align*}$$
Could you please verify if I correctly understanding the KKT's theorem? Thank you so much for your help!
My attempt:
Let $f(x,y)=2x+y$ and $h(x,y)=x^2+y^2-1$. The problem becomes $$\begin{align*}
\text{min} &\quad f(x,y) \\
\text{s.t} & \quad h(x,y)=0
\end{align*}$$
Let $\mathcal K = \{(x,y) \in \mathbb R^2 \mid h(x,y)=0\}$. Then $\mathcal K$ is compact and $(0,0) \notin \mathcal K$. Moreover, $f$ is continuous. Hence the problem has at least one solution. All solutions $(\bar x,\bar y)$ are different from $(0,0)$.
We have $\mu \nabla h(\bar x,\bar y) = \begin{pmatrix}0 \\0 \\ \end{pmatrix} \iff \mu \begin{pmatrix}2 \bar x \\2\bar y \\ \end{pmatrix} = 0 \iff \mu =0$. So the constraint qualification is satisfied. Consider $$\begin{cases} \nabla f( x, y) + \mu \nabla h( x, y) = \begin{pmatrix}0 \\0 \\ \end{pmatrix} \\ { x}^2+{ y}^2-1=0 \end{cases} \iff \begin{cases} \begin{pmatrix}2 \\1 \\ \end{pmatrix} + \mu \begin{pmatrix}2 x \\2 y \\ \end{pmatrix} = \begin{pmatrix}0 \\0 \\ \end{pmatrix} \\{ x}^2+{ y}^2=1 \end{cases}$$
Solving this system, we get $$(\mu, x, y) = (\sqrt{5}/2,-2/\sqrt{5} , -1/\sqrt{5})$$ and $$(\mu, x, y) = (-\sqrt{5}/2,2/\sqrt{5} ,1/ \sqrt{5})$$
As such, $f(-2/\sqrt{5} , -1/\sqrt{5}) = -\sqrt{5}$ and $f(2/\sqrt{5} , 1/\sqrt{5}) = \sqrt{5}$.
Hence $(-2/\sqrt{5} , -1/\sqrt{5})$ is the solution to the problem.
| Using that $$(ax+by)^2\le (a^2+b^2)(x^2+y^2)$$ we get
$$(2x+y)^2\le (4+1)(x^2+y^2)=5$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Find $\lim_{n \to \infty}\frac1{\ln^2n}\left( \frac{\ln 2}{2} + \frac{\ln 3}{3} +\cdots + \frac{\ln n}{n}\right)$ I have to find the following limit:
$$\lim_{n \to \infty} \frac{\frac{\ln 2}{2} + \frac{\ln 3}{3} + \cdots + \frac{\ln n}{n}}{\ln^2n}$$
I tried splitting this limit like so:
$$\lim\limits_{n \to \infty} \dfrac{\frac{\ln 2}{2} + \frac{\ln 3}{3} + \cdots + \frac{\ln n}{n}}{\ln n} \cdot \dfrac{1}{\ln n}$$
Because $\dfrac{1}{\ln n} \rightarrow 0$ as $n \rightarrow \infty$, I concluded that the limit is $0$. I know that in order to use this I firstly would have to show that
$$\lim\limits_{n \to \infty} \dfrac{\frac{\ln 2}{2} + \frac{\ln 3}{3} + \cdots +\frac{\ln n}{n}}{\ln n}$$
is bounded, but I didn't know how to do that and kinda hoped for the best. It turns out that my hopes were in vain, since the limit is actually $\dfrac{1}{2}$ and not $0$, like I got. I also tried using Stolz-Cesaro, resulting in:
$$\lim\limits_{n \to \infty} \dfrac{\frac{\ln (n+1)}{n+1}}{\ln^2 (n + 1) - \ln^2 n} = \lim\limits_{n \to \infty} \dfrac{\frac{\ln (n+1)}{n + 1}}{(\ln (n+1)-\ln n)(\ln (n + 1) + \ln n)}$$
$$= \lim\limits_{n \to \infty} \dfrac{\frac{\ln (n + 1)}{n + 1}}{\ln (\frac{n + 1}{n}) \cdot \ln(n(n + 1))}$$
Aaand I got stuck.
So how should I approach this and get $\dfrac{1}{2}$ as the final answer?
| Define
$$
S_n=\frac{\ln 2}{2} + \frac{\ln 3}{3} + ... + \frac{\ln n}{n}
$$
The function $\ln(x)/x$ is decreasing for $x\geq 3$, so
$$
\frac{\ln 2}{2} + \int_3^n\frac{\ln(x)}{x}\,dx \leq S_n\leq \frac{\ln 2}{2} + \frac{\ln 3}{3} + \int_3^{n}\frac{\ln(x)}{x}\,dx,
$$
$$
\frac{\ln(2)}{2} + \frac{\ln(n)^2}{2}-\frac{\ln(3)^2}{2}\leq S_n\leq\frac{\ln(2)}{2}+\frac{\ln(3)}{3} + \frac{\ln(n)^2}{2}-\frac{\ln(3)^2}{2}.
$$
Now divide through by $\ln(n)^2$ and apply the squeeze theorem to conclude
$$
\lim_{n\to\infty}\frac{S_n}{\ln(n)^2}=\frac{1}{2}.
$$
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "9",
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Let $f(x)$ be a polynomial such that $f(x)*f(1/x) +3f(x)+3f(1/x)=0$ and $f(3) =24$. What is $f(2) +f(-2)$? Let $f(x)$ be a polynomial such that $f(x)*f(1/x) +3f(x)+3f(1/x)=0$ and $f(3) =24$. What is $f(2) +f(-2)$?
| Let $g(x)=f(x)+3$, then $g(x)g(\frac{1}{x})=9$ and $g(3)=27$.
Let $g(x)=ax^i+... +kx^j$ where $a$ and $k$ are the non-zero coefficients of the lowest and highest powers of $x$ in $g(x)$. Then
$$(ax^i+... +kx^j)(ax^{-i}+... +kx^{-j})=9$$
$$(ax^i+... +kx^j)(k+ ...+ax^{j-i})=9x^j$$
The LHS has a non-zero coefficient of $x^i$ and so we have $i=j$ i.e. $g(x)=ax^i$.
Then $ax^i\times ax^{-i}=9$ and therefore $a=3$ or $-3$. Also, $a3^i=27$ and so $a=3,i=2$.
$f(x)=3x^2-3$ and $f(2)+f(-2)=18$.
| {
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How can I prove analytically that the golden ratio is less then $\frac{\pi^2}{6}$. Or stated in other terms, prove that
$$\frac{1+\sqrt{5}}{2} < \sum_{n=1}^{\infty}\frac{1}{n^2}$$
| First, note that $\frac{1+\sqrt{5}}{2}$ is a root of the quadratic equation
$$f(x)=x^2-x-1$$
This equation is increasing on $(1/2,\infty)$ which is easily shown as
$$f'(x)=2x-1$$
which has one zero at $x=\frac{1}{2}$ and $f'(1)=1>0$. Now, note that
$$f\left(\frac{81}{50}\right)=\left(\frac{81}{50}\right)^2-\left(\frac{81}{50}\right)-1=\frac{11}{2500}>0.$$
Since $81/50>1/2$, and $f(81/50)>0$, we can conclude that
$$\frac{81}{50}>\frac{1+\sqrt{5}}{2}.$$
However, if we take the first $40$ terms of the infinite sum, we get
$$\frac{\pi^2}{6}=\sum_{n=1}^\infty \frac{1}{n^2}>\sum_{n=1}^{40} \frac{1}{n^2}=\frac{46252969210499754415427421586309}{28546916554875489385168794240000}=1.62024>1.62=\frac{81}{50}>\frac{1+\sqrt{5}}{2}.$$
| {
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$f(x)=x^3-21x+35$ is irreducable over $Q$. Show if $f(\alpha)=0$ then $f(\alpha^2+2\alpha-14)=0$ I am not sure how to do this in a short, nice way.
I have proved it by factoring $f(x)=(x-\alpha)$ and plugging in, but that is very very messy. Is there a slick way?
Thank you!
| We can prove that is $\alpha$ is a root of $x^3-21x+35=0$ then $\alpha^2+2\alpha-14$ is a root of the same equation.
First use completion of squares to render
$\alpha^2+2\alpha-14=(\alpha+1)^2-15$
Now suppose that $\alpha^3-21\alpha+35=0$. Define $A=\alpha+1$:
$(A-1)^3-21(A-1)+35=0$
$A^2-3A^2+3A-1-21A+21+35=0$
$A^3-3A^2-18A+55=0$
Next separate even and odd degree terms and square both sides:
$A^3-18A=3A^2-55$
$A^6-36A^4+324A^2=9A^4-330A^2+3025$
$B^3-45B^2+654B-3025=0$
where $B=A^2$.
Now define $B=C+15$ where $C=\alpha^2+2\alpha-14$ is your target function:
$(C+15)^3-45(C+15)^2+654(C+15)-3025=0$
$C^3+45C^2+675C+3375-45C^2-1350C-10125+654C+9810-3025=0$
And somewhat magically, when you combine like terms the big numbers mostly cancel:
$\color{blue}{C^3-21C+35=0}$
| {
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Limit of $\lim_{x \to 0} \frac{(1+x^5)^{10} -1}{(\sqrt{1+x^3}-1)(\sqrt[5]{1+x^2}-1) }$ I tried using symbolab to get the limit of
$$\lim_{x \to 0} \frac{(1+x^5)^{10} -1}{(\sqrt{1+x^3}-1)(\sqrt[5]{1+x^2}-1) }$$
but it couldn't solve it. With WolframAlpha I got $100$ for the limit.
Can someone show me how it's done "per hand"? I tried, but couldn't figure it out.
| Note that
*
*$\frac{({1+x^k})^{\alpha} - 1}{x^k} \stackrel{t=x^k}{=}\frac{(1+t)^{\alpha} - 1}{t}\stackrel{t\to 0}{\longrightarrow} \left.\left((1+t)^{\alpha}\right)' \right|_{t=0} = \alpha$
Hence,
$$\frac{(1+x^5)^{10} -1}{(\sqrt{1+x^3}-1)(\sqrt[5]{1+x^2}-1) } =\frac{(1+x^5)^{10} -1}{x^{5}}\cdot \frac{1}{\frac{\sqrt{1+x^3}-1}{x^3}}\cdot \frac{1}{\frac{\sqrt[5]{1+x^2}-1}{x^2}}$$
$$\stackrel{x\to 0}{\longrightarrow}10\cdot \frac{1}{\frac 12}\cdot \frac{1}{\frac 15} = 100$$
| {
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How to find the limit of $a_1=1\:,\:a_{n+1}=\frac{a_n+2}{a_n+1}$ How to find the limit of $a_1=1\:,\:a_{n+1}=\frac{a_n+2}{a_n+1}$?
I think to find the limit, first I need to prove that the sequence is convergent then it is easy to find the limit :
$L = \frac{L+2}{L+1} \to L = \sqrt{2}$ because the sequence $>0$
any hint to prove that the sequence is convergent?
thanks
| First, suppose the sequence converges to some $L$. We have
$$a_{n+1}=\frac{a_n+2}{a_n+1}=1+\frac{1}{1+a_n},$$
and we can take the limit of both sides. On the left,
$$\lim_{n\to\infty}(a_{n+1}-L+L)=\lim_{n\to\infty}(a_{n+1}-L)+L=L.$$
Now since $a_n>0$ (or even $a_n>1$) for all $n$, we know that $\lim\limits_{n\to\infty}a_n\neq -1$. Thus the limit of the right hand side is
$$\lim_{n\to\infty}\frac{a_n-L+L+2}{a_n-L+L+1}=\frac{\lim\limits_{n\to\infty}(a_n-L+L+2)}{\lim\limits_{n\to\infty}(a_n-L+L+1)}=\frac{L+2}{L+1}.$$
As you noted, this gives $L^2=2$. Of course $L\ge 0$, so $L=\sqrt 2$.
As you also noted, this is not a proof, since we began by assuming $L$ exists. We can show this sequence is Cauchy as follows. Write
\begin{align}
a_{n}-a_{n-1}&=1+\frac{1}{1+a_{n-1}}-1-\frac{1}{1+a_{n-2}}\\
&=\frac{a_{n-2}-a_{n-1}}{(1+a_{n-1})(1+a_{n-2})}.
\end{align}
In particular, $a_n>1$ for all $n$, and so
$$|a_{n}-a_{n-1}|=\left|\frac{a_{n-2}-a_{n-1}}{(1+a_{n-1})(1+a_{n-2})}\right|<\frac12|a_{n-1}-a_{n-2}|.$$
By induction,
\begin{align}
|a_n-a_{n-1}|&<\frac12|a_{n-1}-a_{n-2}|\\
&<\frac{1}{2^2}|a_{n-2}-a_{n-3}|\\
&\hspace{3.5pt}\vdots\\
&<\frac{1}{2^{n-2}}|a_2-a_1|\\
&=\frac{1}{2^{n-1}}
\end{align}
which shows the sequence is Cauchy.
| {
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Partial fraction decomposition of $\frac{1}{x^a(x+c)^b}$ We have the partial fraction decomposition
$$\frac{1}{x^a(x+c)^b}=\frac{d_a}{x^a}+\frac{d_{a-1}}{x^{a-1}}+...+\frac{d_{1}}{x}+\frac{e_b}{(x+c)^b}+\frac{e_{b-1}}{(x+c)^{b-1}}+...+\frac{e_{1}}{x+c},$$
where $a,b,c\in\mathbb{N}$.
Is there anything particular we can say about the nature of coefficients $d_i$ and $e_i$? Any references would be greatly appreciated.
P.S. Sorry for being so open-ended.
| Multiply both sides by $x^a$ and $(x+ c)^b$ to get $$1= d_a(x+ c)^b+ d_{a-1}x(x+ c)^b+ \cdots+ d_1x^{a-1}(x+ c)^b+ e_bx^a+ e_{b+1}x^a(x+ c)+ \cdots+e_1x^a(x+ c)^{b-1}~.$$
Taking $~x= 0~$ gives immediately $~1= c^b~d_a~$ so $~d_a= 1/c^b~$. Taking $~x= -c~$ immediately gives $~1= e_b~c^a~$ so $~e_b= 1/c^a~$.
That leaves $~a-1+ b-1= a+ b- 2~$ values to be determined. Although there are no more "easy" equations but taking $~a+ b- 2~$ different values for $~x~$ will give $~a+ b- 2~$ equations for the $~a+ b- 2~$ unknown values.
| {
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Why should it be $\sqrt[3]{6+x}=x$? Find all the real solutions to:
$$x^3-\sqrt[3]{6+\sqrt[3]{x+6}}=6$$
Can you confirm the following solution? I don't understand line 3. Why should it be $\sqrt[3]{6+x}=x$?
Thank you.
$$
\begin{align}
x^3-\sqrt[3]{6+\sqrt[3]{x+6}} &= 6 \\
x^3 &= 6+ \sqrt[3]{6+\sqrt[3]{x+6}} \\
x &= \sqrt[3]{6+ \sqrt[3]{6+\sqrt[3]{x+6}}} \\
\sqrt[3]{6+x} &= x \\
x^3 &= 6+x \\
x^3-2x^2+2x^2-4x+3x-6 &= 0 \\
(x-2)(x^2+2x+3) &= 0 \\
x &= 2
\end{align}
$$
| I think the step that is is missing is
$$x = \sqrt[3]{6+ \sqrt[3]{6+\sqrt[3]{\color{red}x+6}}}$$
$$=\sqrt[3]{6+ \sqrt[3]{6+\sqrt[3]{6+\color{red}x}}}$$
$$= \sqrt[3]{6+ \sqrt[3]{6+\sqrt[3]{6+\color{red}{\sqrt[3]{6+ \sqrt[3]{6+\sqrt[3]{6+\color{blue}x}}} }}}}, $$
and we can continue to obtain the infinite expression
$$x = \sqrt[3]{6 + \sqrt[3]{6+\sqrt[3]{6+\cdots }}}$$
(This is actually invalid and handwavey and not legitimate... but let's go with it.)
Then, we can do
$$x = \sqrt[3]{6 + \color{blue}{\sqrt[3]{6 + \sqrt[3]{6+\sqrt[3]{6+\cdots }}}}}=\sqrt[3]{6 + \color{blue}x}.$$
And then, we continue.
The problem is, of course, what the #@%! is $\sqrt[3]{6 + \sqrt[3]{6+\sqrt[3]{6+\cdots }}}$ supposed to mean? Is that even well defined and can we do math on it?
And the answer is, yes.
If $a_0=6$ and $a_k = \sqrt[3]{6 + a_{k-1}}$ for $k>1$, then we can prove by induction that $0 < a_{k+1} < a_k\le 6$, with equality holding if and only if $k =0$. Therefore, $\{a_k\}$ is monotonically decreasing and bounded below, so $\lim_{n\to \infty} a_n = x$ for some real $x$, which if we wanted to, we could express as $$\sqrt[3]{6 + \sqrt[3]{6+\sqrt[3]{6+\cdots }}}$$ if we took $$\sqrt[3]{6 + \sqrt[3]{6+\sqrt[3]{6+\cdots }}}:= \lim_{n\to \infty} a_n$$ as a definition.
Now, for converging limits, $$x = \lim_{n\to \infty} a_n = \lim_{n\to \infty}a_{n+1} = \lim_{n\to\infty}\left(\sqrt[3]{6+a_n}\right) = \sqrt[3]{6 + \lim_{n\to \infty}a_n} = \sqrt[3]{6 + x}.$$
So we can do it.
I guess we can also do it without resorting to the infinite.
Let $?$ be notation for $<$ or for $>$ or for $+$, where we do not know which. However, since each of $<,>,=$ are transitive and preserved via "adding to both sides" and "cubing both sides", we may do the following manipulations.
If $x \enspace?\enspace \sqrt[3]{6+x}$, then
$$6+ x \enspace?\enspace 6 +\sqrt[3]{6+x}\Rightarrow$$
$$\sqrt[3]{6 + x} \enspace?\enspace \sqrt[3]{6 +\sqrt[3]{6+x}}\Rightarrow$$
$$x\enspace?\enspace\sqrt[3]{6 + x}\enspace?\enspace\sqrt[3]{6 +\sqrt[3]{6+x}}\Rightarrow$$
$$\sqrt[3]{6+x} \enspace?\enspace \sqrt[3]{6 +\sqrt[3]{6+x}} \enspace?\enspace \sqrt[3]{6+\sqrt[3]{6 +\sqrt[3]{6+x}}}\Rightarrow$$
$$x\enspace?\enspace\sqrt[3]{6+x} \enspace?\enspace \sqrt[3]{6 +\sqrt[3]{6+x}} \enspace?\enspace \sqrt[3]{6+\sqrt[3]{6 +\sqrt[3]{6+x}}}.$$
But we showed $$x = \sqrt[3]{6+\sqrt[3]{6 +\sqrt[3]{6+x}}}.$$
So by transitivity, $$x \enspace?\enspace x.$$ But by trichotomy, $x =x$ and $x \not < x$ and $x \not > x$. So it must be that $?$ is notation for $=$, and $x = \sqrt[6]{x+6}$.
| {
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$\sqrt{m^2 - x^2} \leq x , m > 0$ the range for $x$ is? $\sqrt{m^2 - x^2} \leq x $
$\sqrt{{m}^{2} - {x}^{2}} \geq 0$
$x \geq 0$
$\sqrt{m^2 - x^2} \leq x $
$m^2 - x^2 \leq x^2 $
$m^2 \leq 2x^2$
$\frac{m}{\sqrt 2} \leq x$
$\sqrt{{m}^{2} - {x}^{2}} \geq 0$
${m}^{2} \geq {x}^{2}$
$m > 0, x > 0$
$m \geq x$
$x \leq m$
$\frac{m}{\sqrt 2} \leq x \leq m$
is it right?
| Let $0<x<m^2$, tthe squaring we get
$$m^2-x^2 \ge x^2 \implies 0\le x \le m/\sqrt{2}$$
| {
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Find value of $a_1$ such that $a_{101}=5075$ Let $\{a_n\}$ be a sequence of real numbers where $$a_{n+1}=n^2-a_n,\, n=\{1,2,3,...\}$$
Find value of $a_1$ such that $a_{101}=5075$.
I have
$$a_2=1^2-a_1$$
$$a_3=2^2-a_2=2^2-1^2+a_1$$
$$a_4=3^2-2^2+1^2-a_1$$
$$a_5=4^2-3^2+2^2-1^2+a_1$$
$$\vdots$$
$$a_{101}=100^2-99^2+98^2-97^2+\ldots+2^2-1^2+a_1.$$
Therefore, $$a_{101}=\sum_{i=1}^{50}(2i)^2-\sum_{i=1}^{50}(2i-1)^2.$$
Thus, $$5075=\sum_{i=1}^{50}(4i^2-4i^2+4i-1)+a_1,$$
and, $$a_1=5075-4\sum_{i=1}^{50}(i)+\sum_{i=1}^{50}(1).$$
Hence, $$a_1=5075-4(\frac{50}{2})(51)+50=25,$$
and $a_1=25$.
Is it correct? Do you have another way? Please check my solution, thank you.
| What if at the point where:
$a_{101} = 100^2 - 99^2 + 98^2 - 97^2 \cdots 2^2 - 1^2 + a_1$
You opted for a clever factorization of squares:
$a_{101} = (100 - 99)(100 + 99) + (98 - 97)(98 + 97) \cdots (2 - 1)(2 + 1) + a_1$
$a_{101} = 199 + 195 + 191 \cdots 3 + a_1$
Therefore those numbers become a simple arithmetic series with the first term as 3 and the common ratio as 4.
But how many terms exactly?
$3 + 4(n - 1) = 199$
$4(n - 1) = 196$
$n - 1 = 49$
$n = 50$
Therefore with the knowledge of evaluating the sum this comes down to:
$a_{101} = 25(199 + 3) + a_1$
$a_{101} = 5050 + a_1$
And by substituting the choice for a_101:
$5075 = 5050 + a_1$
$a_1 = 25$
Thanks to @Zera for recommending the edit. I'm learning how to use these markup languages
| {
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Find the maximum of a sum Given $x + y + z = 0, x + 1 > 0, y + 1 > 0, z + 4 > 0$, find the maximum of $$Q=\frac{x}{x+1} + \frac{y}{y+1} + \frac{z}{z+4}$$
The answer is $\frac{1}{3}$
I've tried inverting each of the fractions then using AM-GM on them, but it doesn't give me the correct answer
| First set
*
*$a=x+1, b=y+1, c= z+4 \Rightarrow a,b,c >0$ and $a+b+c = 6$
Hence, to maximize is
$$Q = 3-\left(\frac 1a + \frac 1b + \frac 4c \right)$$
So, minimize
$$\left(\frac 1a + \frac 1b + \frac 4c \right) = \frac{a+b+c}6\left(\frac 1a + \frac 1b + \frac 4c\right)$$
$$= \frac 16+\frac 16 + \frac 23 + \frac 16\left(\frac{b+c}{a} + \frac{a+c}{b} + 4\frac{a+b}{c}\right)$$
$$=1 + \frac 16\left( \frac ba+\frac ab + \frac ca + 4 \frac ac + \frac cb + 4 \frac bc\right)$$
$$\stackrel{3\times AM-GM}{\geq} 1 + \frac 16\left( 2+ 4 + 4 \right)= \frac 83 $$
with equality if $\frac ab= 1$ and $\frac ca = \frac cb = 2$. With $a+b+c = 6$ it follows the minimum is achieved for $c=3, a = b=\frac 32$.
So, maximum of $Q$ is $3-\frac 83 = \boxed{\frac 13}$
| {
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What's wrong in my calculation of $\int \frac{\sin x}{1 + \sin x} dx$? I have the following function:
$$f: \bigg ( - \dfrac{\pi}{2}, \dfrac{\pi}{2} \bigg ) \rightarrow \mathbb{R} \hspace{2cm} f(x) = \dfrac{\sin x}{1 + \sin x}$$
And I have to find $\displaystyle\int f(x) dx $. This is what I did:
$$\int \dfrac{\sin x}{1 + \sin x}dx=
\int \dfrac{1+ \sin x - 1}{1 + \sin x}dx =
\int dx - \int \dfrac{1}{1 + \sin x}dx =
$$
$$ = x - \int \dfrac{1 - \sin x}{(1 + \sin x)(1 - \sin x)} dx$$
$$= x - \int \dfrac{1 - \sin x}{1 - \sin ^2 x} dx$$
$$= x - \int \dfrac{1 - \sin x}{\cos^2 x} dx$$
$$= x - \int \dfrac{1}{\cos^2x}dx + \int \dfrac{\sin x}{\cos^2 x}dx$$
$$= x - \tan x + \int \dfrac{\sin x}{\cos^2 x}dx$$
Let $u = \cos x$
$du = - \sin x dx$
$$=x - \tan x - \int \dfrac{1}{u^2}du$$
$$= x - \tan x + \dfrac{1}{u} + C$$
$$= x - \tan x + \dfrac{1}{\cos x} + C$$
The problem is that the options given in my textbook are the following:
A. $x + \tan {\dfrac{x}{2}} + C$
B. $\dfrac{1}{1 + \tan{\frac{x}{2}}} + C$
C. $x + 2\tan{\dfrac{x}{2}} + C$
D. $\dfrac{2}{1 + \tan{\frac{x}{2}}} + C$
E. $x + \dfrac{2}{1 + \tan{\frac{x}{2}}} + C$
None of them are the answer I got solving this integral. What is the mistake that I made and how can I find the right answer? By what I've been reading online, you can get different answers by solving an integral in different ways and all of them are considered correct. They differ by the constant $C$. I understand that, but I don't see how to solve this integral in such a way to get an answer among the given $5$. And, even more importantly, how can I recognize the right answer in exam conditions if the answer provided by my solution is not present among the given options? Is solving in a different manner my only hope?
| Beside verifying your answer is equivalent to one of the listed results, you may also integrate in $\tan\frac x2$ since all the choices are in terms of it.
So, use $\sin x =\frac{2\tan\frac x2}{1+\tan^2\frac x2}$ to integrate,
$$\int \dfrac{1}{1 + \sin x}dx = \int \dfrac{1}{1 + \frac{2\tan\frac x2}{1+\tan^2\frac x2}}dx
=\int \dfrac{1+\tan^2\frac x2}{1+\tan^2\frac x2 + 2\tan\frac x2}dx$$
$$=\int \dfrac{\sec^2\frac x2}{(1+\tan\frac x2)^2}
=\int \dfrac{2d(\tan\frac x2)}{(1+\tan\frac x2)^2}
=-\dfrac{2}{1+\tan\frac x2}$$
| {
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number of ways of arranging four pairs of socks in a line so that no adjacent socks form a pair
Suppose there are four pairs of socks, let's label them $AA,BB,CC,DD$. The left and right socks of the pair are indistinguishable. How many ways are there to arrange them in a line such that no adjacent socks are a pair, e.g. $ABABCDCD$. In extension, how about if there are $n$ pairs?
My attempt is as follows, but I feel it is too complicated. I am seeking an easier method.
I tried to partition into four distinct cases:
$$ABCDXXXX, 432$$
$$ABACXXXX,120$$
$$ABCAXXXX,288$$
$$ABABXXXX, 24$$
So in total there are $864$ ways. This is the correct answer but it is complicated.
| We can use the Inclusion-Exclusion Principle.
We have eight positions to fill. We can fill two of them with A's in $\binom{8}{2}$ ways, two of the remaining six positions with B's in $\binom{6}{2}$ ways, two of the remaining four positions with C's in $\binom{4}{2}$ ways, and fill the final two positions with D's in $\binom{2}{2}$ ways. Hence, there are
$$\binom{8}{2}\binom{6}{2}\binom{4}{2}\binom{2}{2}$$
distinguishable arrangements of four pairs of socks.
From these, we must subtract those arrangements in which two adjacent socks form a pair.
Two adjacent socks form a pair: There are $\binom{4}{1}$ ways to choose which of the four pairs of socks form an adjacent pair. Say the two socks labeled A are adjacent. Then we have seven objects to arrange: AA, B, B, C, C, D, D. Choose one of the seven positions for the pair AA, two of the remaining six positions for the B's, two of the remaining four positions for the C's, then fill the final two positions with the D's. The socks can be arranged in
$$\binom{7}{1}\binom{6}{2}\binom{4}{2}\binom{2}{2}$$
ways. Hence, there are
$$\binom{4}{1}\binom{7}{1}\binom{6}{2}\binom{4}{2}\binom{2}{2}$$
arrangements in which two adjacent socks form a pair.
However, if we subtract this from the total, we will have subtracted too much since we will have subtracted each case in which there are two pairs in which two adjacent socks form a pair twice, once for each way we could have designated one of those pairs as the pair of adjacent socks that form a pair. We only want to subtract them once, so we must add them back.
Two pairs in which two adjacent socks form a pair: There are $\binom{4}{2}$ ways to select two pairs of socks. Say the socks labeled A are adjacent and the socks labeled B are adjacent. Then we have six objects to arrange: AA, BB, C, C, D, D. They can be arranged in
$$\binom{6}{1}\binom{5}{1}\binom{4}{2}\binom{2}{2}$$
ways. Hence, there are
$$\binom{4}{2}\binom{6}{1}\binom{5}{1}\binom{4}{2}\binom{2}{2}$$
in which there are two pairs in which two adjacent socks form a pair.
If we add this to our running total, our answer will be too large. This is because we first added each case in which there are three pairs in which adjacent socks form a pair three times, once for each way of designating one of those three pairs as the pair of adjacent socks that form a pair of socks, and subtracted them three times, once for each of the $\binom{3}{2}$ ways we could designate two of those three pairs as the pairs of adjacent socks that form a pair of socks. Thus, we have not subtracted cases in which there are three pairs of socks in which adjacent pairs of socks form a pair.
Three pairs in which two adjacent socks form a pair: There are $\binom{4}{3}$ ways of selecting the three pairs of socks. Say the two socks labeled A are adjacent, the two socks labeled B are adjacent, and the two socks labeled C are adjacent. Then we have five objects to arrange: AA, BB, CC, D, D. They can be arranged in
$$\binom{5}{1}\binom{4}{1}\binom{3}{1}\binom{2}{2}$$
ways, so there are
$$\binom{4}{3}\binom{5}{1}\binom{4}{1}\binom{3}{1}\binom{2}{2}$$
arrangements in which there are three pairs in which two adjacent socks form a pair of socks.
If we subtract this amount from our total, our count will be too small. This is because we have counted each case in which there are four pairs in which adjacent socks form a pair of socks twice. We subtracted them four times, once for each way we could designate one of those four pairs as the pair of adjacent socks that form a pair; added them six times, once for each of the $\binom{4}{2}$ ways we could designate two of those four pairs as the pairs of adjacent socks that form a pair of socks; then subtracted them four times, once for each of the $\binom{4}{3}$ ways we could designate three of those four pairs as the pairs of adjacent socks that form a pair of socks. We only want to subtract them once, so we must add them back.
Four pairs of adjacent socks in which adjacent socks form a pair: We have four objects to arrange: AA, BB, CC, DD. They can be arranged in $4!$ ways.
By the Inclusion-Exclusion Principle, the number of admissible arrangements is
$$\binom{8}{2}\binom{6}{2}\binom{4}{2} - \binom{4}{1}\binom{7}{1}\binom{6}{2}\binom{4}{2} + \binom{4}{2}\binom{6}{1}\binom{5}{1}\binom{4}{2} - \binom{4}{3}\binom{5}{1}\binom{4}{1}\binom{3}{1} + 4!$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3486226",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Suggestions on solving a system of equations How do you suggest solving the following sytem?
$$\begin{cases} \dfrac{12}{x}+\dfrac{12}{y}=1 \\ \dfrac{6}{x-2}+\dfrac{8}{y-6}=\dfrac{2}{3} \end{cases}$$
After simplifying the equations, I got $xy-12y=12x$ and $xy-11y=18x-90$.
Should I continue from here or there's a better solution?
| Alternatively:
$$\begin{cases} \dfrac{12}{x}+\dfrac{12}{y}=1 \\ \dfrac{6}{x-2}+\dfrac{8}{y-6}=\dfrac{2}{3} \end{cases} \Rightarrow
\begin{cases} \dfrac{12}{x}=1-\dfrac{12}{y} \\ \dfrac{6}{x-2}=\dfrac23-\dfrac{8}{y-6} \end{cases} \Rightarrow
\begin{cases} x=\dfrac{12y}{y-12} \\ x=\dfrac{18(y-6)}{2y-36}+2=\dfrac{22y-180}{2y-36} \end{cases}$$
Now equate them:
$$12y(2y-36)=(y-12)(22y-180) \Rightarrow y^2+6y-1080=0 \Rightarrow \\
y_{1,2}=-36;30 \Rightarrow x_{1,2}=9;20.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3486452",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Find min and max of $A = \sqrt{x-2} + 2\sqrt{x+1} + 2019 -x$
Find minimum and maximum of $$A = \sqrt{x-2} + 2\sqrt{x+1} + 2019 -x$$
How to solve this problem without using derivatives?
Michael Rozenberg's edit:
Because with a derivative it's not so easy:
$$A'(x)=\frac{1}{2\sqrt{x-2}}+\frac{1}{\sqrt{x+1}}-1=\frac{1}{2\sqrt{x-2}}-\frac{1}{2}+\frac{1}{\sqrt{x+1}}-\frac{1}{2}=$$
$$=\frac{1-\sqrt{x-2}}{2\sqrt{x-2}}+\frac{2-\sqrt{x+1}}{2\sqrt{x+2}}=$$
$$=(3-x)\left(\frac{1}{2(1+\sqrt{x-2})\sqrt{x-2}}+\frac{1}{2(2+\sqrt{x+1})\sqrt{x+2}}\right),$$
which gives $x_{max}=3.$
| Let $f(x) = \sqrt{x-2} + 2\sqrt{x+1} + 2019 -x$ then
$$
f'(x) = \frac{1}{2\sqrt{x - 2}} + \frac{1}{\sqrt{x+1}} - 1
$$
Equating this to zero for real $x$, we get $x = 3$. The second derivative is negative hence the global maximum is $f(3) = 2021.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3487710",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
} |
Prove $2\cos5B+2\cos4B+2\cos3B+2\cos2B+2\cos B+1=\frac{\sin(11B/2)}{{\sin(B/2)}}$ Working through a book, I have stumbled upon a question I don't know how to solve:
Prove
$$2\cos(5B) + 2\cos(4B) + 2\cos(3B) + 2\cos(2B) + 2\cos(B) + 1 = \dfrac{\sin\left(\frac{11B}2\right)}{{\sin\left(\frac B2\right)}}$$
(this is a smaller problem I have reduced the larger problem down from)
I don't know how to simply the cosine functions into a sine functions, as using the identity $\cos A + \cos B = 2\cos\left(\dfrac{A+B}2\right)\cos\left(\dfrac{A-B}2\right)$ simply gives me a result with even more cosine functions. I can't think of other identities that are helpful.
| \begin{align}
& 1 + 2\cos B + 2\cos(2B) + 2\cos(3B) + 2\cos(4B) + 2\cos(5B) \\[8pt]
= {} & \cos(-5B) + \cos(-4B) + \cdots + \cos(0B) + \cdots + \cos(4B) + \cos(5B) \\
& \text{(since cosine is an even function)} \\[10pt]
= {} & \operatorname{Re} \big( e^{-5iB} + e^{-4iB} + \cdots + e^{0iB} + \cdots + e^{4iB} + e^{5iB} \big) \\[8pt]
= {} & e^{-5iB} + e^{-4iB} + \cdots + e^{0iB} + \cdots + e^{4iB} + e^{5iB} \\
& \text{(since the imaginary parts cancel out)} \\[10pt]
= {} & \text{a finite geometric series} \\[8pt]
= {} & \text{(first term)} \times \frac{1 - (\text{common ratio})^\text{number of terms}}{1 - (\text{common ratio})} \\[8pt]
= {} & e^{-5iB} \cdot \frac{1 - e^{11iB}}{1 - e^{iB}} = \frac{e^{-5iB} - e^{6iB}}{1 - e^{iB}} \\[10pt]
= {} & \frac{e^{-11iB/2} - e^{11iB/2}}{e^{-iB/2} - e^{iB/2}} \quad \left( \text{We multiplied by } \frac{e^{-iB/2}}{e^{-iB/2}}. \right) \\[10pt]
= {} & \frac{\sin(11B/2)}{\sin(B/2)}.
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3490414",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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} |
Po-Shen Loh's new way of solving quadratic equation Quadratic equation, $ax^2+bx+c=0$ and its solution is quadratic equation, $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$
Now setting $a=1$ then we have $x^2+bx+c=0$
$$x=\frac{-b\pm \sqrt{b^2-4c}}{2}$$ rewrite as
$$x=-\frac{b}{2}\pm \sqrt{\left(\frac{b}{2}\right)^2-c}$$
In this new video Dr. Loh claims to discover a new way of solving the quadratic equation! How? It is the same as the above formula, by using the quadratic formula, the only thing I see different, is he rewrite it in the above form!
Can someone please explain to me how this is a new way?
| The teacher's conclusion at the end "guesswork has been replaced by a clever trick" implies that the main result of the proposed new method is the clever trick (change of unknowns $x_1$ and $x_2$) of solving the system of equations (which is the Vieta's theorem):
$$\begin{cases}x_1+x_2=-b\\ x_1x_2=c\end{cases} \stackrel{x_1=\frac{-b}{2}-t\\x_2=\frac{-b}{2}+t}\Rightarrow x_1x_2=\frac{b^2}{4}-t^2=c \Rightarrow t=\pm \sqrt{\frac{b^2}{4}-c} \Rightarrow \\
x_1=\frac{-b}{2}-\sqrt{\frac{b^2}{4}-c}=\frac{-b-\sqrt{b^2-4c}}{2}\\
x_2=\frac{-b}{2}+\sqrt{\frac{b^2}{4}-c}=\frac{-b+\sqrt{b^2-4c}}{2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3490752",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 8,
"answer_id": 5
} |
Find $\lim\limits_{n \to \infty} \int_0^1 \sqrt[n]{x^n+(1-x)^n} \,dx$. I have the following limit to find:
$$\lim_{n \to \infty} \int_0^1 \sqrt[n]{x^n+(1-x)^n} \, dx$$
And I have to choose between the following options:
A. $0$
B. $1$
C. $\dfrac{3}{4}$
D. $\dfrac{1}{2}$
E. $\dfrac{1}{4}$
This is what I did:
$$\int_0^1 \sqrt[n]{x^n+(1-x)^n} \, dx =
\int_0^{\frac{1}{2}} \sqrt[n]{x^n+(1-x)^n} \, dx +
\int_{\frac{1}{2}}^1 \sqrt[n]{x^n+(1-x)^n} \, dx$$
*
*If we consider the first term of this sum:
$$\int_0^{\frac{1}{2}} \sqrt[n]{x^n+(1-x)^n} \, dx \ge
\int_0^\frac{1}{2} \sqrt[n]{x^n+x^n} \, dx =
\int_0^\frac{1}{2}2^{\frac{1}{n}}x \, dx$$
And that means:
$$\lim_{n \to \infty} \int_0^{\frac{1}{2}} \sqrt[n]{x^n+(1-x)^n} \, dx \ge
\lim_{n \to \infty} \int_0^\frac{1}{2}2^{\frac{1}{n}}x \, dx =
\int_0^\frac{1}{2}x \, dx =
\frac{1}{8} \tag 1$$
*
*If we consider the second term of that sum we have:
$$\int_{\frac{1}{2}}^1 \sqrt[n]{x^n+(1-x)^n} \, dx \ge
\int_\frac{1}{2}^1 \sqrt[n]{(1-x)^n+(1-x)^n} \, dx =
\int_\frac{1}{2}^1 2^{\frac{1}{n}}(1-x) \, dx$$
And that means:
$$\lim_{n \to \infty} \int_{\frac{1}{2}}^1 \sqrt[n]{x^n+(1-x)^n} \, dx \ge \lim_{n \to \infty} \int_\frac{1}{2}^1 2^{\frac{1}{n}}(1-x) \, dx = \lim_{n \to \infty} \int_\frac{1}{2}^1 (1-x) \, dx = \frac{1}{8} \tag 2$$
Now we can sum $(1)$ and $(2)$ and we have:
$$\lim_{n \to \infty} \int_0^1 \sqrt[n]{x^n+(1-x)^n} \, dx \ge \frac{1}{4}$$
So now we have a lower bound. We can do something similar for the upper bound.
*
*Again, let's consider the first term of that sum:
$$\int_0^{\frac{1}{2}} \sqrt[n]{x^n+(1-x)^n} \, dx \le
\int_0^{\frac{1}{2}} \sqrt[n]{(1-x)^n+(1-x)^n} \, dx =
\int_0^{\frac{1}{2}} 2^\frac{1}{n}(1-x) \, dx$$
That means:
$$\lim_{n \to \infty} \int_0^{\frac{1}{2}} \sqrt[n]{x^n+(1-x)^n} \, dx \le
\lim_{n \to \infty} \int_0^{\frac{1}{2}} 2^\frac{1}{n}(1-x) \, dx =
\int_0^{\frac{1}{2}} (1-x) \, dx =
\frac{3}{8} \tag 3$$
*
*And if we consider the second part of that sum:
$$\int_{\frac{1}{2}}^1 \sqrt[n]{x^n+(1-x)^n} \, dx \le
\int_{\frac{1}{2}}^1 \sqrt[n]{x^n+x^n} \, dx =
\int_{\frac{1}{2}}^1 2^\frac{1}{n} x \, dx$$
And that means:
$$\lim_{n \to \infty} \int_{\frac{1}{2}}^1 \sqrt[n]{x^n+(1-x)^n} \, dx \le
\lim\_{n \to \infty} \int_{\frac{1}{2}}^1 2^\frac{1}{n} x \, dx =
\lim_{n \to \infty} \int_{\frac{1}{2}}^1 x \, dx =
\frac{3}{8} \tag 4$$
And now if we sum $(3)$ and $(4)$ we get:
$$\lim_{n \to \infty} \int_0^1 \sqrt[n]{x^n+(1-x)^n} \, dx \le
\frac{3}{4}$$
So after all of that, we have:
$$\frac{1}{4} \le
\lim_{n \to \infty} \int_0^1 \sqrt[n]{x^n+(1-x)^n} \, dx \le \frac{3}{4}$$
But this didn't help me all that much. Choices C, D and E are still consistent with this inequality that I got. So what should I do to find the exact answer? Or did I do something wrong in my calculations?
| Let the integrand be $f(x)$. Obviously $f(0)=f(1)=1$ and the function achieves a single minimum $\left(\frac12,\frac{\sqrt[n]2}2\right)$. Hence we have certainly $\frac12<I_n<1$.
If you want to compute the limit anyway,
$$\sqrt[n]{x^n+(1-x)^n}dx=x\sqrt[n]{1+\left(\frac{1-x}x\right)^n}\to x$$ when $1-x<x$, and symmetrically. Hence the limit function is simply
$$f(x):=\begin{cases}x\le\frac12\to1-x,\\x\ge\frac12\to x\end{cases}$$ and the area under it is $\frac34$.
Plot for $n=15$:
Update:
The initial argument is wrong, as it only proves $$\frac12\le I\le 1.$$
For a better proof, observe that for all $x$, $$\sqrt{x^2+(1-x)^2}\ge f(x)\ge\max(1-x,x),$$ which excludes $\frac12$ and $1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3492509",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Evaluate: $\lim\limits_{x \to \infty} (\sqrt{x+2}-\sqrt{x})$
Evaluate: $\displaystyle\lim_{x \to \infty} (\sqrt{x+2}-\sqrt{x})$
I was given this problem. I'm not sure how to tackle it. $\infty-\infty$ is in indeterminate form, so I need to get it into a fraction form in order to solve. I did this by pulling an $x^2$ out of it: $$\lim_{x\to\infty}x\left(\sqrt{\frac1x+\frac2{x^2}}-\sqrt{\frac1x}\right)\\=\lim_{x\to\infty}\frac{\left(\sqrt{\frac1x+\frac2{x^2}}-\sqrt{\frac1x}\right)}{\frac1x}$$ Now taking the derivative: $$\lim_{x\to\infty}\Large\frac{\frac{\frac{-1}{x^2}+\frac{-4}{x^3}}{2\sqrt{\frac1x+\frac2{x^2}}}-\frac{-\frac1{x^2}}{2\sqrt{\frac1x}}}{-\frac1{x^2}}\\=\lim_{x\to\infty}\frac{-x^2\left(\frac{-1}{x^2}+\frac{1}{x^2}\right)}{2\sqrt{\frac1x+\frac2{x^2}}-2\sqrt{\frac1x}}\\=\lim_{x\to\infty}\frac{-x^2\left(0\right)}{2\sqrt{\frac1x+\frac2{x^2}}-2\sqrt{\frac1x}}\\=\lim_{x\to\infty}0=0$$ Since my numerator becomes zero, my entire fraction becomes zero, making zero the answer.
Is this correct? This was very messy - did I miss the easy way to do it? Is there a better way to do problems of this sort?
| We have that $\sqrt{x+2} - \sqrt{x} = \dfrac{(\sqrt{x+2}-\sqrt{x})(\sqrt{x+2}+\sqrt{x})}{\sqrt{x+2}+\sqrt{x}} = \dfrac{2}{\sqrt{x-2}+\sqrt{x}}.$ Hence $\lim\limits_{x\to\infty} \sqrt{x+2}-\sqrt{x} = \lim\limits_{x\to\infty}\dfrac{2}{\sqrt{x}(\sqrt{1-\frac{2}{x}}+1)}=0.$
If you were looking for a nonzero answer, you could have tried something like $\lim\limits_{x\to\infty} \sqrt{x^2+x}-x$, which evaluates to $\dfrac{1}{2}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3492588",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
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Why does this pattern occur: $123456789 \times 8 + 9 = 987654321$ I came across the following:
$\begin{align} 1 \times 8 + 1 &= 9 \\ 12 \times 8 + 2 & = 98 \\ 123 \times 8 + 3 & = 987 \\ 1234 \times 8 + 4 & = 9876 \\ 12345 \times 8 + 5 & = 98765 \\ 123456 \times 8 + 6 & = 987654 \\ 1234567 \times 8 + 7 & = 9876543 \\ 12345678 \times 8 + 8 & = 98765432 \\ 123456789 \times 8 + 9 & = 987654321. \\ \end{align}$
I'm looking for an explanation for this pattern. I suspect that there is some connection to the series $\frac{1}{(1 - x)^2} = 1 + 2x + 3x^2 + \cdots$.
This post asks the same question but has no answers posted.
| $$\left\lfloor {10^n\over (1-x)^2} \right\rfloor \cdot 8+n= 9\cdot \left\lfloor {10^n\over (1-x)^2} \right\rfloor -\left\lfloor {10^n\over(1-x)^2} \right\rfloor +n$$
With $x=1$ is what you've observed ( yes I realize the division by 0, just don't know a better way yet to present what the OP sees). The real question though is what makes it work.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3492701",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
"answer_count": 5,
"answer_id": 2
} |
Value of expression in the roots of the cubic obtained by shifting $x^3-3x^2+x+2$ to eliminate the $x^2$ term
For a given function $f(x)=x^3-3x^2+x+2$, roots are shifted by $\lambda$ to eliminate $x^2$ terms and $\alpha$, $\beta$, $\gamma$ be the roots of new function. Evaluate
$$\begin{align}
\frac{\alpha^3}{(\alpha-\beta)(\alpha-\gamma)(\alpha-\lambda)}&+\frac{\beta^3}{(\beta-\gamma)(\beta-\alpha)(\beta-\lambda)}+\frac{\gamma^3}{(\gamma-\alpha)(\gamma-\beta)(\gamma-\lambda)} \\
&+\frac{\lambda^3}{(\lambda-\alpha)(\lambda-\beta)(\lambda-\gamma)}
\end{align}$$
What I try
Replace $x\rightarrow x+\lambda$ in $f(x)=x^3-3x^2+x+2$
$f(x+\lambda) = (x+\lambda)^3-3(x+\lambda)^2+(x+\lambda)+2$
$x^3+\lambda^3+3x^2\lambda+3x\lambda^2-3x^2-3\lambda^2-6x\lambda+x+\lambda+2$
$x^3+(3\lambda-3)x^2+(3\lambda^2-6\lambda+1)x+\lambda^3-3\lambda^2+\lambda+2=0$
put $3\lambda-3=0\Rightarrow \lambda = 1$
So equation is $x^3-2x+1=0$ has roots $\alpha,\beta,\gamma$
So we have $\alpha+\beta+\gamma=0,\sum \alpha \beta=-2,\alpha\beta\gamma=-1$
How do I solve it Help me please
| Hint: $(x^3-2x+1)$=$(x-1)(x^2+x-1)$
| {
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"url": "https://math.stackexchange.com/questions/3494812",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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Finding kernel and image of endomorphism. I have been trying to solve this, but I can't. I am given the endorphism: $$f(X) = A\cdot X - X\cdot A$$
Where $X$ is any real-valued $2 \times 2$ matrix and $$A = \begin{pmatrix}1 & 2 \\ 2 & -1\end{pmatrix}$$
And I have to find a basis of the kernel and a basis of the image of $f$. Finding a basis for the kernel was easy. By definition, the basis is the set of matrices $X$ that verify $f(X) = (0)$, therefore:
$$A\cdot X - X \cdot A = 0$$
$$A\cdot X = X \cdot A$$
$$X = A^{-1}\cdot X \cdot A$$
So, since X must be a diagonal matrix with its two elements equal to its eigenvalues, we know it must be a matrix of the form: $$X =\begin{pmatrix} \lambda_1 & 0 \\ 0 & \lambda_2 \end{pmatrix}$$
And since both $\lambda_1$ and $\lambda_2$ must verify $A-\lambda I_2 = 0$. Writing out the just mentioned equation in matrix form, one gets a matrix with only one non-zero entry for each eigenvalue, being the result in both cases that $\lambda_1= \lambda_2$. Therefore, a basis for the kernel could be: $$B_{ker(f)} = \Big{\{} \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}\Big{\}}$$
But I am stuck here. Since this is an endomorphism on the space of square matrices of order $2$, the dimension equation tells us that $$\dim(Im(f)) + \dim(ker(f)) = \dim(Im(f)) + 1 = 4 \Rightarrow \dim(Im(f)) = 4$$
I am not sure I am right and I don't know how to find a basis for the image of $f$, though.
| Hint: Observe
\begin{align}
AX-XA =&\
\begin{pmatrix}
1 & 2\\
2 & -1
\end{pmatrix}
\begin{pmatrix}
x & y\\
z & w
\end{pmatrix}
-
\begin{pmatrix}
x & y\\
z & w
\end{pmatrix}
\begin{pmatrix}
1 & 2\\
2 & -1
\end{pmatrix}\\
=&
\begin{pmatrix}
x+2z & y+2w\\
2x-z & 2y-w
\end{pmatrix}
-
\begin{pmatrix}
x+2y & 2x-y\\
z+2w & 2z-w
\end{pmatrix}\\
=&\
\begin{pmatrix}
2(z-y) & 2(w-x)+2y\\
2(x-w)-2z & 2(y-z)
\end{pmatrix}.
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3495524",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Find the $f^{(3)}(0)$ of $f(x) = \sin^{3}(\ln(1+x))$ using Taylor expansion For the Taylor expansions I used:
$\ln(1+x) = \sum_{k=1}^\infty (-1)^{k+1} \frac{x^{k}}{k}$
$\sin(x) = \sum_{k=0}^\infty (-1)^{k+1} \frac{x^{2k+1}}{(2k+1)!}$
So $\sin(x) = x - \frac{x^{3}}{3!} + O(x^{5})$
and $\ln(1+x) = x - \frac{x^2}{2} + \frac{x^3}{3} + O(x^4)$
Then the composite expansion (for rest of question dropping H.O.T.):
$\sin(\ln(1+x)) \approx (x - \frac{x^2}{2} + \frac{x^3}{3}) - \frac{(x - \frac{x^2}{2} + \frac{x^3}{3})^3}{3!} $
Reducing:
$\approx (x - \frac{x^2}{2} + \frac{x^3}{3}) - \frac{x^3}{6} \approx x - \frac{x^2}{2} + \frac{x^3}{6}$
Cube the result and simplify again:
$(x - \frac{x^2}{2} + \frac{x^3}{6})^3 \approx x^3$ keeping the third-order term.
At this point, my understanding is that the coefficient of the third-order term in the formula for the general Taylor series is $f^{(3)}$
$\frac{f^{(3)}(a)}{3!}(x-a)^3$
So I believe I can infer that the coefficient of the composite Taylor expansion, 1, also relates to $f^{(3)}$, and then solving:
$\frac{f^{(3)}}{3!} = 1$
$f^{(3)} = 3! = 6$
Have I understood this question and the right approach for solving it?
| This is correct, though take note that you should be more careful with your remainders on each step to ensure you didn't miss anything that might contribute to the $x^3$ coefficient:
\begin{align}\sin(\ln(1+x))&=\sin\bigg(x-\frac{x^2}2+\frac{x^3}3+\mathcal O(x^4)\bigg)\tag{$\star$}\\&=\left[\color{#2255FF}{x-\frac{x^2}2+\frac{x^3}3+\mathcal O(x^4)}\right]-\frac16(\color{#228855}{x+\mathcal O(x^2)})^3+\mathcal O(\color{#FF9944}{x}^4)\\&=x-\frac{x^2}2+\frac16x^3+\mathcal O(x^4)\end{align}
$(\star)$: $\color{#2255FF}{x-\frac{x^2}2+\frac{x^3}3+\mathcal O(x^4)}=\color{#228855}{x+\mathcal O(x^2)}=\color{#FF9944}{\mathcal O(x)}$
| {
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"question_score": "2",
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How to factorize $X^5-1$ into irreductible factors? I have to factorize the polynomial $P(X)=X^5-1$ into irreducible factors in $\mathbb{C}$ and in $\mathbb{R}$, this factorisation happens with the $5$th roots of the unity.
In $\mathbb{C}[X]$ we have $P(X)=\prod_{k=0}^4 (X-e^\tfrac{2ki\pi}{5})$.
In $\mathbb{R}[X]$ the solution states that by gathering all complex conjugate roots we find that $P(X)=(X-1)(X^2-2\cos(\frac{2\pi}{5})+1)(X^2-2\cos(\frac{4\pi}{5})+1)$, but I can't figure out how. Another problem I ran into was trying to figure out where the $2\cos(\frac{2\pi}{5})$ and $2\cos(\frac{4\pi}{5})$ come from so I tried these two methods:
The sum of the roots of unity is zero so we have:
$1+e^\tfrac{2i\pi}{5}+e^\tfrac{4i\pi}{5}+e^\tfrac{6i\pi}{5}+e^\tfrac{8i\pi}{5}=0$
In the $5$th roots of unity circle P3 and P4 are images according to the x-axis the same goes to P2 and P5, therefore $e^\tfrac{6i\pi}{5}=e^\tfrac{-4i\pi}{5}$ and $e^\tfrac{8i\pi}{5}=e^\tfrac{-2i\pi}{5}$ afterwards by using Euler's formula we find $1+2\cos(\frac{2\pi}{5})+2\cos(\frac{4\pi}{5})=0$.
Another method is that $\cos(6\pi/5) = \cos(-6\pi/5) = \cos(-6\pi/5 + 2\pi) = \cos(4\pi/5)$ $\cos(8\pi/5) = \cos(-8\pi/5) = \cos(-8\pi/5 + 2\pi) = \cos(2\pi/5)$
therefore $1 + \cos(2\pi/5) + \cos(4\pi/5) + \cos(4\pi/5) + \cos(2\pi/5) = 0$ and we find $1+2\cos(\frac{2\pi}{5})+2\cos(\frac{4\pi}{5})=0$
I don't know if both of these methods are correct on their own and I don't know if they will help in the factorisation since I don't know how to go from there and find $P(X)=(X-1)(X^2-2\cos(\frac{2\pi}{5})+1)(X^2-2\cos(\frac{4\pi}{5})+1)$
| In $\mathbb C[x]$ we can write it as $$p(x)=(x-1)(x-\color{blue}{e^{2\pi i/5}})(x-\color{blue}{e^{-2\pi i/5}})(x-\color{red}{e^{4\pi i/5}})(x-\color{red}{e^{-4\pi i/5}})$$ where the blue and red are conjugate root pairs. Multiplying these conjugate root pairs together gives (i.e. multiply the second and third terms together, and then the fourth and fifth together)
$$\begin{align}p(x)&=(x-1)(x^2-(e^{2\pi i/5}+e^{-2\pi i/5})x+e^{2\pi i/5}e^{-2\pi i/5})(x^2-(e^{4\pi i/5}+e^{-4\pi i/5})x+e^{4\pi i/5}e^{-4\pi i/5}) \\&=(x-1)\left(x^2-2\cos\left(\frac{2\pi}{5}\right)x+1\right)\left(x^2-2\cos\left(\frac{4\pi}{5}\right)x+1\right) \end{align}$$
using Euler's formula for $e^{2k\pi i/5}+e^{-2k\pi i/5}=2\cos\left(\frac{2k\pi}{5}\right)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3496594",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to evaluate $\int_{0}^{\pi}\frac{2x^3-3\pi x^2}{(1+\sin{x})^2}\,\textrm{d}x\,$? Problem:
Let $f$ be a bounded continuous function on the interval [0,1].
(a) show that $\int_{0}^{\pi} xf\,(\sin{x})\,\textrm{d}x = \frac{\pi}{2}\int_{0}^{\pi}f\,(\sin{x})\,\textrm{d}x\,$.
(b) Hence evaluate $\int_{0}^{\pi} \dfrac{x}{1+\sin{x}}\,\textrm{d}x\,$.
(c) Hence deduce that $\int_{0}^{\pi} \dfrac{2x^3-3\pi x^2}{(1+\sin{x})^2}\,\textrm{d}x = -\dfrac{2\pi ^3}{3}\,$.
I have completed Part (a) and Part (b). The answer of Part (b) is $\pi\,$. I have tried to do Part (c) in the following way:
$\textrm{Let }I=\int_{0}^{\pi} \dfrac{2x^3-3\pi x^2}{(1+\sin{x})^2}\,\textrm{d}x\\
\quad\;\;\;\,=\int_{0}^{\pi} \dfrac{2(\pi-x)^3-3\pi (\pi-x)^2}{(1+\sin{(\pi-x)})^2}\,\textrm{d}x\quad(\textrm{by letting }u=\pi-x\textrm{)}\\
\quad\;\;\;\,=\int_{0}^{\pi} \dfrac{2(\pi^3-3\pi^{2}x+3\pi x^2-x^3)-3\pi (\pi-x)^2}{(1+\sin{(\pi-x)})^2}\,\textrm{d}x\\
\quad\;\;\;\,=\int_{0}^{\pi} \dfrac{-\pi^3+3\pi^{2}x-2x^3}{(1+\sin{x})^2}\,\textrm{d}x\\
\quad\;\;\;\,=-\pi^3\int_{0}^{\pi}\dfrac{1}{(1+\sin{x})^2}\,\textrm{d}x-I\\
\quad\;\;\;\,=-\dfrac{\pi^3}{2}\int_{0}^{\pi}\dfrac{1}{(1+\sin{x})^2}\,\textrm{d}x\\
\quad\;\;\;\,=-\dfrac{\pi^3}{2}\cdot\dfrac{4}{3} \quad\textrm{(by letting }t=\tan{(\frac{x}{2})}\textrm{)}\\
\quad\;\;\;\,=-\dfrac{2\pi^3}{3}\\ \textrm{However, I did not make use of any results from Part (a) and Part (b).}\\
\textrm{I hope there is another solution that follows the hints of the problem.}$
| Here is to derive the last integral below using the result in Part (b).
$$J=\int_0^\pi \frac{dx}{(1 + \sin x)^2}
=\int_0^\pi d\left(-\frac{\cos x}{1 + \sin x}\right)\frac{1}{1 + \sin x}$$
Integrate by parts,
$$J=2-\int_0^\pi \frac{\cos^2 x}{(1 + \sin x)^3}dx
= 2- \int_0^\pi \frac{1 - \sin x}{(1 + \sin x)^2}
=2+\int_0^\pi \frac{dx}{1 + \sin x}-2J
$$
Now use the result$\int_0^\pi \frac{dx}{1 + \sin x} = 2$ from Part (b) to obtain
$$J= \frac13\left(2+\int_0^\pi \frac{dx}{1 + \sin x} \right)=\frac43$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3498840",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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} |
sum of coefficient of all even power of $x$
The sum of all Coefficient of even power of $x$ in
$(1-x+x^2-x^3+\cdots +x^{2n})(1+x+x^2+x^3+\cdots +x^{2n}),n\in \mathbb{N}$
what i try
for $n=1,$ we have
$(1-x+x^2)(1+x+x^2)=(1+1)-(1)+(1+1)=3$
for $n=2,$ we have
$(1-x+x^2-x^3+x^4)(1+x+x^2+x^3+x^4)$
$=(1+1+1)-(1+1)+(1+1+1)-(1+1)+(1+1+1)=5$
so in this way , get sum of coefficient in original expression is $2n+1$
but How do i solve it without substituting value of $n$, Help me
| Let's call sum of odd coefficients $C_O$ and sum of even coefficients $C_E$. You can write your polynomial as $$P(x)=O(x)+E(x)$$, where $O(x)$ is the sum of odd powered terms and $E(x)$ is the sum of even powered terms. Note that if you plug in $x=1$, you get $$C_O=O(x=1)$$ and $$C_E=E(x=1)$$ Similarly, if you plug in $x=-1$, the odd terms will acquire a minus sign, while the even ones will be unchanged. So $$O(x=-1)=-C_O$$ and $$E(x=-1)=C_E$$
Then $$C_E=\frac12 (E(1)+E(-1))=\frac12(E(1)+E(-1))+\frac12(O(1)+O(-1))=\frac12(P(1)+P(-1))$$
If you plug in $x=1$ in your original form, the second parenthesis will evaluate to $2n+1$, and the first parenthesis will be $1$. If you put in $x=-1$, you have the first parenthesis $2n+1$ and the second one $1$. Then $$C_E=2n+1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3500198",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
$\begin{cases} x^3-y^3=19(x-y) \\ x^3+y^3=7(x+y) \end{cases}$
I should solve the following system: $$\begin{cases} x^3-y^3=19(x-y) \\
x^3+y^3=7(x+y) \end{cases}$$
by reducing the system to a system of second degree.
We can factor:
$$\begin{cases} (x-y)(x^2+xy+y^2)=19(x-y) \\
(x+y)(x^2-xy+y^2)=7(x+y) \end{cases}$$
I really don't want to divide the equations by $x-y$ and $x+y$, respectively. I am taught to divide by expressions containing variables only in special cases. Is there any other way here?
| To avoid all possible problems, add the two equations to get
$$x^3-13 x+6 y=0 \implies y=\frac{1}{6} \left(13 x-x^3\right)$$ Plug in the first equation to end with
$$x \left(x^8-39 x^6+507 x^4-2665 x^2+4788\right)=0$$ So $x=0$ if a root.
For the remaining, let $z=x^2$ to make
$$z^4-39z^3+507z^2-2665z+4788=0$$ By inspection, $z=4$ and $z=7$ are solutions. Now, long division
$$\frac{z^4-39z^3+507z^2-2665z+4788 } {(z-4)(z-7) }=z^2-28 z+171=(z-19) (z-9)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3500828",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Solving differential equation by separating variables Solve the following differential equation:
$$\frac yx \frac {dy}{dx}=\sqrt {1+x^2+y^2+x^2y^2}$$
I have tried like this:
$$\frac yx \frac{dy}{dx}=\sqrt {1+x^2+y^2(1+x^2)}\\
\implies \frac yx \frac {dy}{dx}=\sqrt {(1+x^2)(1+y^2)}\\
\implies \int \frac {ydy}{\sqrt {1+y^2}}=\int \frac {xdx}{\sqrt{1+x^2}}\\
\implies \log \sqrt{1+y^2}=\log \sqrt {1+x^2}+c_1\\
\implies \frac {1+y^2}{1+x^2}=c\\$$
But in my book the answer is:
$$(1+x^2)^{3/2}-3\sqrt {1+y^2}=c$$
I can't understand how they did it,please check this..
| It should be like this:
$$\int x\sqrt{1+x^2} dx=\int \dfrac{y}{\sqrt{1+y^2}}dy$$
Also,
$$\int \dfrac{y}{\sqrt{1+y^2}}dy=\sqrt{1+y^2}+C$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3508437",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
} |
Find eigenvalues of a linear map knowing its eigenvectors Let $\;f: R^{3} \rightarrow R^{3}\;$ be an endomorphism whose eigenvectors are $(0,1,-2),(1,0,4),(1,0,-2)$. Knowing that $f(0,1,0) = (2,1,2)$ find the eigenvalues of $f$.
What is the methodology to solve this problem?
Thank you.
| A more pedestrian approach is to use the matrix for $f$. The condition that $f(0,1,0)=(2,1,2)$ means that the middle column of the matrix must be $(2,1,2)$.
Also $f(1,0,4)=(.,0,.)$ and $f(1,0,-2)=(.,0,.)$ so the middle row of the matrix must be 0,1,0. So we can write the matrix as
$$\begin{pmatrix}a&2&b\\
0&1&0\\
c&2&d\end{pmatrix}$$ Hence $f(0,1,-2)=(2-2b,1,2-2d)$. But $(0,1,-2)$ is an eigenvector, so $b=1$, the eigenvalue is 1 and $d=2$.
Also $f(1,0,4)=(a+4,0,c+8)$, so $c+8=4(a+4)$. And $f(1,0,-2)=(a-2,0,c-4)$, so $c-4=-2a+4$. Hence $a=0$ and $c=8$, so the matrix is
$$\begin{pmatrix}0&2&1\\
0&1&0\\
8&2&2\end{pmatrix}$$ and it is easy to check that $f(0,1,0)=(2,1,2),f(0,1,-2)=(0,1,-2)$, $f(1,0,4)=4(1,0,4)$ and $f(1,0,-2)=-2(1,0,-2)$.
So the eigenvalues are 1,4 and -2.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3508932",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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A circle is drawn with its centre on the line $x + y = 2$ to touch the line $4x – 3y + 4 = 0$ and pass through the point $(0, 1)$. Find its equation. A circle is drawn with its centre on the line $x + y = 2$ to touch the line $4x – 3y + 4 = 0$ and pass through the point $(0, 1)$. Find its equation.
My attempt is as follows:-
Let the equation of circle be $x^2+y^2+2gx+2fy+c=0$
As center lies on the line $x+y=2$
$$-g-f=2$$
$$g+f=-2\tag{1}$$
As circle passes through the point $(0,1)$
$$1+2f+c=0$$
$$2f+c=-1\tag{2}$$
As $4x – 3y + 4 = 0$ is tangent to the circle
$$\dfrac{\left|-4g+3f+4\right|}{5}=\sqrt{g^2+f^2-c}$$
Squaring both sides
$$16g^2+9f^2-24gf+16+8(-4g+3f)=25g^2+25f^2-25c$$
$$9g^2+16f^2+24gf+32g-24f-25c-16=0$$
Eliminating $g$ with the help of equation $(1)$
$$9(-2-f)^2+16f^2+24(-2-f)f+32(-2-f)-24f-25c-16=0$$
$$9(4+f^2+4f)+16f^2-48f-24f^2-64-32f-24f-25c-16=0$$
$$f^2-68f-44-25c=0$$
Eliminating $c$ with the help of equation $(2)$
$$f^2-68f-44-25(-1-2f)=0$$
$$f^2-18f-19=0$$
$$f^2-19f+f-19=0$$
$$f=19,-1$$
$$(g,f,c)\equiv (-1,-1,1),(-21,19,-39)$$
So equations are $x^2+y^2-2x-2y+1=0$, $x^2+y^2-42x+38y-39=0$
But this got too long, any shorter method?
| Let $(a, 2 - a)$ be a center of circle A.
Line $L_0 : 4x - 3y + 4 = 0$ is tangent of A, so point of tangency P has minimum distance with A.
For all X = $(x, y) \in L_0$,
$$
\begin{align*}
d (X, A)^2 = \left( 4^2 + 3^2 \right) \left( \left( x - a \right)^2 + \left(y - 2 + a\right)^2 \right) & \geq \left( 4 \left(x - a\right) - 3 \left(y - 2 + a\right) \right)^2 \\
& = (4x - 3y + 6 - 7a )^2 \\
& = (7a - 2)^2
\end{align*}
$$
has minimum at $P$.
Circle A has two points $P$, $(0, 1)$. From ${(7a-2)^2 \over 25} = (a - 0)^2 + (2 - a - 1)^2$, we obtain $a = 1 \lor a = 21$ and ${7a - 2 \over 5} = 1 \lor {7a - 2 \over 5} = 29$.
So, $A : x^2 + y^2 - 2x - 2y + 1 = 0$ and $A : x^2 + y^2 - 42x + 38y - 39=0$ are only exist.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3510191",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
How to solve $\sin 2x \sin x+(\cos x)^2 = \sin 5x \sin 4x+(\cos 4x)^2$? How to solve $\sin 2x \sin x+(\cos x)^2 = \sin 5x \sin 4x+(\cos 4x)^2$?
\begin{align*}
2\cos x(\sin x)^2+(\cos x)^2 & = \frac{1}{2}(\cos x- \cos 9x) +(\cos 4x)^2\\
4\cos x(1-(\cos x)^2)+2(\cos x)^2 & = \cos x + \cos 9x +(2(\cos 4x)^2-1)+1\\
\cos x(4(1-(\cos x)^2 + 2\cos x -1) & = \cos 9x + \cos 8x + 1
\end{align*}
I don't see any way to get rid of $8x$ and $9x$ as arguments to have same angles from there
| You should be able to transform the equation into polynomial of $\cos(x)$.
For example, $\sin(2x)\sin(x)=2\sin^2(x)\cos(x)=2(1-\cos^2(x))\cos(x)$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
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solve for $y$ in $e^{xy}+x^2+y-1.2$
I cant figure out how to make $y$ the subject in this equation:
$$e^{xy}+x^2+y-1.2=0$$
I did these steps,
$$\begin{align}&y=1.2-x^2-e^{xy}\\
&\ln(y)=\ln(1.2)-\ln(x^2)-\ln(e^{xy})\\
&\ln(y)=\ln(1.2)-\ln(x^2)-xy\\
&\ln(y)+xy=\ln\frac{1.2}{x^2}\end{align}$$
ahead of this I am lost and would appreciate any and all help.
| Let $a=1.2$, rearrange a little and we have
\begin{eqnarray*}
e^{-xy} (a-x^2-y)=1.
\end{eqnarray*}
Multiply by $xe^{ax-x^3}$
\begin{eqnarray*}
(\color{red}{ax-x^3-xy})e^{\color{red}{ax-x^3-xy}}=xe^{ax-x^3}.
\end{eqnarray*}
Now recall the Lambert $W$ function is defined by $\color{red}we^{\color{red}w}=z$ gives $w=W(z)$. So we have
\begin{eqnarray*}
ax-x^3-xy =W(xe^{ax-x^3}) \\
y=\frac{ax-x^3-W(xe^{ax-x^3})}{x}.
\end{eqnarray*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3511979",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find the volume cut off from the sphere $x^2+y^2+z^2=a^2$ by the cylinder $x^2+y^2=ax$
Find the volume cut off from the sphere $x^2+y^2+z^2=a^2$ by the cylinder $x^2+y^2=ax$
Attempt: The projection of the Cylinder ( denoted $D$) on the $xy$ plane is a circle which has the equation: $x^2+y^2=ax ~~~\equiv~~~(x-a/2)^2+y^2=(a/2)^2 ~~~\equiv~~~~r=a \cos \theta$ ( In Polar Coordinates)
The circle has centre $(a/2,0)$ and radius $a/2$. Hence, the $y$ axis is a tangent to it.
Hence, the required volume $V = \int \int \int dv $
$= \int \int_D \int_{-\sqrt {a^2-x^2-y^2}} ^{\sqrt {a^2-x^2-y^2}}dz~~dx~dy$
$=2\int \int_D \sqrt {a^2-x^2-y^2}~~ dx dy$
Switching to Polar Coordinates :
$V=2\int_{-\pi/2}^{\pi/2} \int_{0}^{a \cos \theta} \sqrt {a^2-r^2}~~ r~dr~ d\theta=\dfrac {2}{3}a^3 \pi$
Although, the limits of $\theta$ look conceptually fine to me, my textbook uses the limits $0$ to $\pi$ and gives the result $= \dfrac {2}{3}a^3 (\pi-\dfrac{4}{3})$.
Could someone please clarify why $-\pi/2$ to $\pi/2$ usage in my expression might be incorrect?
Thanks a lot for your help!
| $V=2\int_{-\pi/2}^{\pi/2} \int_{0}^{a \cos \theta} \sqrt {a^2-r^2}~~ r~dr~ d\theta\\
V=2\int_{-\pi/2}^{\pi/2}-\frac 13 (a^2-r^2)^\frac 32|_0^{a\cos\theta}~~ d\theta$
When we evaluate the limit at $a\cos \theta$....
$(a^2-a^2\cos^2\theta)^\frac 32\\
(a^2\sin^2\theta)\frac 32\\
|a^3\sin^3\theta|$
That is we only consider the principal root.
Or, $a^2-a^2\cos^2\theta \ge 0$ for all $\theta$ and so the cube of the square root.
$\frac 23 \int_{-\pi/2}^{\pi/2}a^3 - |a^3\sin^3 \theta|~~ d\theta = \frac 23 \int_{0}^{\pi}a^3 - |a^3\sin^3 \theta|~~ d\theta = (\frac 23\pi - \frac 43) a^3$
And the limits of integration don't really matter.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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On the Diophantine equation $m^2 - p^k = 4z$, where $z \in \mathbb{N}$ and $p$ is a prime satisfying $p \equiv k \equiv 1 \pmod 4$ The title says it all:
Do there always exist $m, p, k \in \mathbb{N}$ such that $m^2 - p^k = 4z$, $z \in \mathbb{N}$ and $p$ is prime satisfying $p \equiv k \equiv 1 \pmod 4$?
MY ATTEMPT FOR $z=1$
When $z = 1$, I get
$$m^2 - p^k = 4$$
$$p^k = m^2 - 4 = (m+2)(m-2)$$
Hence, we have the simultaneous equations $p^{k-x} = m+2$ and $p^x = m-2$ where $k \geq 2x + 1$. Consequently, we have $p^{k-x} - p^x = 4$, which implies that
$$p^x (p^{k-2x} - 1) = 4$$
where $k-2x$ is odd. Since $(p-1) \mid (p^y - 1) \hspace{0.05in} \forall y \geq 1$, this last equation implies that $(p-1) \mid 4$. Likewise, the congruence $p \equiv 1 \pmod 4$ implies that $4 \mid (p-1)$. These two divisibility relations imply that $p-1=4$, or $p=5$. Hence,
$$5^x (5^{k-2x} - 1) = 4.$$
Since $5$ does not divide $4$, $x=0$ and thus $5^k - 1 = 4$, which means that $k=1$.
Therefore, $p=5$, $k=1$, and $x=0$. Consequently, $p^{k-x} = 5^{1-0} = 5 = m + 2$ and $p^x = 5^0 = 1 = m - 2$. Either way, $m=3$.
CONCLUSION
Thus, for the equation $m^2 - p^k = 4$, I get the solution $m=3$, $p=5$, and $k=1$.
QUESTION
How does one treat the case for general $z \in \mathbb{N} \setminus \{1\}$?
For one, I cannot seem to wrap my head around $m^2 - p^k = 8$.
I could, of course, rewrite it as
$$(m+3)(m-3) = m^2 - 9 = p^k - 1 = (p-1)\sigma(p^{k-1})$$
(where $\sigma$ is the classical sum-of-divisors function), but then I am no longer able to apply Euclid's Lemma.
UPDATE TO QUESTION (January 20, 2020)
The equation $m^2 - p^k = 4z$ (under the given constraints) does not have a solution when $z=4$.
Here is my additional question:
When is the equation $m^2 - p^k = 4z$, where $z \in \mathbb{N}$ and $p$ is a prime satisfying $p \equiv k \equiv 1 \pmod 4$, guaranteed to have a solution?
| Consider the case $z = 4$.
We obtain
$$m^2 - p^k = 16$$
$$m^2 - 16 = p^k$$
$$(m + 4)(m - 4) = p^k$$
$$p^{k-x} = m + 4$$
$$p^x = m - 4$$
$$p^{k-x} - p^x = 8$$
$$p^x (p^{k-2x} - 1) = 8$$
Since $p$ is prime and $p \equiv 1 \pmod 4$, then we have $\gcd(p^x, 8) = 1$, which implies that
$$p^{k-2x} - 1 = 8$$
$$p^{k-2x} = 9$$
$$p = 3 \land k - 2x = 2$$
contradicting $p \equiv 1 \pmod 4$.
Therefore, under the given constraints, the Diophantine equation
$$m^2 - p^k = 16$$
does not have any solutions.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3515496",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Finding the third roots of unity by equating $(a + bi)^3$ to $1$ I know what the third roots of unity are but I want to solve this exercise:
Complex numbers can be written as $a + bi$. Simplify $(a + bi)^3$ and
equate the real part to $1$ and the imaginary part to $0$ to find the
three roots of unity.
So I expanded $(a + bi)^3$ to get $(a^3 - 3ab^2) + (3a^2b - b^3)i$.
I know that $a = 1$ and $b = 0$ is a solution just by looking at it. Now is there some sort of clever way to get another root of unity? I tried to do it with just algebra but it ended up being very messy. I was wondering whether there is some neat solution to this.
| Factoring a difference of cubes, we have
\begin{align}
(a+bi)^3&=1\\
(a+bi)^3-1&=0\\
(a+bi-1)((a+bi)^2+(a+bi)+1)&=0\\
(a-1+bi)(a^2+2abi-b^2+a+bi+1)&=0.
\end{align}
So, we have either $$\operatorname{Re}(a-1+bi)=1\ \wedge\ \operatorname{Im}(a-1+bi)=0 \implies (a,b)=(1,0)$$
or
$$\operatorname{Re}(a^2+2abi-b^2+a+bi+1)=1\ \wedge\ \operatorname{Im}(a^2+2abi-b^2+a+bi+1)=0,$$
giving us the system of equations
\begin{align}
a^2-b^2+a+1&=1\tag1\\
2ab+b&=0.\tag2\end{align}
From $(2)$, we have $b(2a+1)=0\implies a=-\frac12$. Substituting this into $(1)$, we have
$$\left(-\frac12\right)^2-b^2+\left(-\frac12\right)=0\implies b=\pm\frac{\sqrt3}2.$$
Therefore, the three roots of unity are $(1,0), (-\frac12,\frac{\sqrt3}2), (-\frac12,-\frac{\sqrt3}2)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3517726",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 4
} |
Cyclic Inequality $(x^8+1)(x^4+1)(y^8+1)(y^4+1)(z^8+1)(z^4+1)≥((x^2+1)(y^2+1)(z^2+1))^2$ I have been challenged to prove the following:
If $xy+yz+zx=3$ for positive $x,y,z$, then $(x^8+1)(x^4+1)(y^8+1)(y^4+1)(z^8+1)(z^4+1)≥((x^2+1)(y^2+1)(z^2+1))^2$.
Initial attempts at AM-GM are not very helpful for me, I don't think.
$x^8+1≥2x^4,$ and $x^4+1≥2x^2$, so $(x^8+1)(x^4+1)≥4x^2(2x^2-1)$. It is easy to show $(x^8+1)(x^4+1)≥4x^2(2x^2-1)≥(x^2+1)^2$ if $x≥\sqrt{\frac{3}{7}},$ but that's insufficient.
I doubt that this problem turns into a Vieta bash due to the nature of having $x^8$ on the LHS. Initial attempts at Newton's sums were giving what looked like annoying numbers and something more bashy than what I would guess is a more elegant solution.
| We begin with the following observation $$(x^8+1)(x^4+1)=((x^4)^2+1^2)((x^2)^2+1^2)\geq(x^4\cdot x^2+1\cdot1)^2=(x^6+1)^2.$$
By the Cauchy-Schwarz Inequality $$\prod_{cyc}(x^8+1)(x^4+1)\geq\prod_{cyc}(x^6+1)^2.$$
Thus, it's enough to prove that
$$\prod_{cyc}(x^4-x^2+1)\geq1$$ and since $$x^4-x^2+1\geq\left(\frac{x^2+1}{2}\right)^2,$$ it's enough to prove that
$$\prod_{cyc}(x^2+1)\geq8$$ and since by the Cauchy-Schwarz Inequality again $$\prod_{cyc}(x^2+1)(1+y^2)\geq\prod_{cyc}(x+y)^2,$$ it's enough to prove that
$$(x+y)(x+z)(y+z)\geq8,$$ which is true because
$$(x+y)(x+z)(y+z)\geq\frac{8}{9}(x+y+z)(xy+xz+yz)=\frac{8}{3}(x+y+z)=$$
$$=\frac{8}{3}\sqrt{(x+y+z)^2}\geq\frac{8}{3}\sqrt{3(xy+xz+yz)}=8$$
Finally, equality is achieved when $$x=y=z=1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3521631",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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