Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Why does differentiating $-\frac{2}{x+1}$ or $\frac{2x}{x+1}$ result in the same function $\frac{2}{(x+1)^2}$? I have some trouble understanding the integration of $\frac{2}{(x+1)^2}$.
When integrated using $u$-substitution I get:
$$-\frac{2}{x+1}$$
However, I also know that:
$$\frac{2x}{x+1}$$
is a solution as well, due to the fact when you take the derivative of either one of the solutions mentioned above you end up with the same answer:
$$\frac{2}{(x+1)^2}$$
I want to understand how the relation between the both integrated solutions works and why. I can't find any proof or examples that might guide me in the right way. Any help in understand this phenomena would be very much appreciated.
| Don't forget that there should be a constant of integration, $C$, at the end of your answer that always differentiates to zero: $\int\frac{2}{(x+1)^2}\,dx=-\frac{2}{x+1}+C$. The integral of a function is not just one single function, but a whole family of infinitely many functions that differ only by a constant. And the second function is nothing more than the first function plus a $2$ (a constant):
$$
-\frac{2}{x+1}+2=
-\frac{2}{x+1}+2\frac{x+1}{x+1}=\\
\frac{-2}{x+1}+\frac{2x+2}{x+1}=
\frac{-2+2x+2}{x+1}=\\
\frac{2x}{x+1}.
$$
Since they're the same function, their derivatives should be equal:
$$
\left(-\frac{2}{x+1}+2\right)'=\frac{2}{(x+1)^2}+0=\frac{2}{(x+1)^2}.\\
\left(\frac{2x}{x+1}\right)'=\frac{2}{(x+1)^2}.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3278976",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 1
} |
Evaluating $\lim\limits_{x\to 2} \frac{\sqrt[3]{x} - \sqrt[3]{2}}{x^3 - 8}$ without l'Hospital's rule
How to evaluate the following limit?
$$
\lim\limits_{x\to 2} \dfrac{\sqrt[3]{x} - \sqrt[3]{2}}{x^3 - 8}
$$
I factored the denominator into $(x-2)(x^2+2x+4)$, but I couldn't go on from there.
| $$\lim_{x\to 2}\frac{\sqrt[3]{x}-\sqrt[3]{2}}{x^3-8}=$$
$$\lim_{x\to 2}\frac{\sqrt[3]{x}-\sqrt[3]{2}}{(x-2)(x^2+2x+4)}= $$
$$\lim_{x\to 2}\frac{(\sqrt[3]{x}-\sqrt[3]{2})}{(\sqrt[3]{x}-\sqrt[3]{2})(\sqrt[3]{x^2}+\sqrt[3]{4} +\sqrt[3]{x}\sqrt[3]{2})(x^2+2x+4)}= $$
$$\lim_{x\to 2}\frac{1}{(\sqrt[3]{x^2}+\sqrt[3]{4} +\sqrt[3]{x}\sqrt[3]{2})(x^2+2x+4)}=\frac {1}{36\sqrt[3] 4} $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3279312",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 4
} |
Prove that $a^{2n} -b^{2n}$ is divisible by $a+b$
Prove that for all $n\in\Bbb{Z}$, $a^{2n}-b^{2n}$ is divisible by $a+b$ using induction.
I know that if $a$ is divisible by $b$, then $a=kb$, where $k\in\Bbb{Z}$. Here we have that $a^{2n}-b^{2n}=(a+b)k$, with $k\in\Bbb{Z}$.
For the base case I set $n=1$, so $a^2-b^2=(a+b)(a-b)=(a+b)k$, where $k=a-b\in\Bbb{Z}$.
Now the inductive step (where I have doubts): $$a^{2n}-b^{2n}=(a+b)k\implies a^{2(n+1)}-b^{2(n+1)}=(a+b)m,\;k,m\in\Bbb{Z}.$$ We start from $a^{2(n+1)}-b^{2(n+1)}$. Then $$a^{2n+2}-b^{2n+2}=(a+b)\color{red}{(a^{2n+1}-a^{2n}b+a^{2n-1}b^2-\dots-a^2b^{2n-1}-ab^{2n}+b^{2n+1})},$$ so $a^{2(n+1)}-b^{2(n+1)}=(a+b)m$, where $m=a^{2n+1}-a^{2n}b+a^{2n-1}b^2-\dots-a^2b^{2n-1}-ab^{2n}+b^{2n+1}\in\Bbb{Z}.\qquad\square$
I have two questions:
*
*Is the math in $\color{red}{\text{red}}$ a correct descomposition of $a^{2(n+1)}-b^{2(n+1)}$?
*We have not used the inductive hypothesis. Could we use it?
| Another way:
We know by inductive hypothesis $a^{2n}=(a+b)k+b^{2n}$, with $k\in\Bbb{Z}$. Then, \begin{align*}a^{2n+2}-b^{2n+2}&=a^{2n}a^2-b^{2n}b^2\\
&=a^{2}((a+b)k+b^{2n})-b^{2n}b^2\\
&=a^{2}(a+b)k+a^{2}b^{2n}-b^{2n}b^2\\
&=a^{2}(a+b)k+b^{2n}(a^2-b^2)\\
&=a^{2}(a+b)k+b^{2n}(a-b)(a+b)\\
&=(a+b)(a^2k+b^{2n}(a-b))\\
&=(a+b)m,\quad m=a^2k+b^{2n}(a-b)\in\Bbb{Z}.\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3279380",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
Show that $x^2+2 \equiv 3 \mod 4$ and deduce that there exists a prime $p$ with $p|x^2+2$ and $p \equiv 3 \mod 4$. Let $x$ be an odd natural number. Show that $x^2+2 \equiv 3 \mod 4$ and deduce that there exists a prime $p$ with $p|x^2+2$ and $p \equiv 3 \mod 4$.
For the first part I would assume
$$x \equiv 1 \mod 2$$
$$x^2 \equiv 1^2 \mod 2^2$$
$$x^2 +2 \equiv 1^2+2 \mod 2^2$$
$$x^2 +2 \equiv 3 \mod 4$$
Can someone briefly tell me if this is correct and if it is a rule that squaring the number on the left means you square both numbers on the right?
Also I don't know how to start with the second part.
Note: This question is to do with rings in general.
| For a more quick way: Notice, $x^2+2$ is always odd, as $x$ is odd. Therefore, all prime divisors of $x^2+2$ are either of form $4k+1$ or $4k+3$. Now, if all of them are of form $4k+1$, then simply check that $x^2+2\equiv 1\pmod{4}$. However, since $x$ is odd, $x^2+2\equiv 3\pmod{4}$, which is yielding a contradiction.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3279998",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Tricky trigonometric problem I can not seem to solve this type of easy problem, it seems somewhat tricky, can I have a hint or an explanation.
The problem: $sin\left(2x\right)\:=\:\sqrt{3}cos\left(\:\frac{3\pi \:}{2}-x\right)$ on the interval $[ ; \frac{7}{2}
]$
| $$
\sqrt{3}\cdot \cos(\frac{3\pi}{2}-x)\\
=\sqrt{3}\cdot \sin(\frac{3\pi}{2}+\frac{\pi}{2}-x)\\
= \sqrt{3}\cdot \sin(2\pi-x)\\
= \sqrt{3}\cdot \sin(-x)\\
= -\sqrt{3}\cdot \sin(x)\\
$$
$$
\sin 2x = 2\sin x\cos x
$$
$$
\therefore,\\\
2\sin x\cos x = -\sqrt{3}\cdot \sin(x)\\
$$
Now don't go cancelling stuff away first, acknowledge $\bf{\sin x=0}$ as a solution first.
$$
2\cos x = -\sqrt{3}\\
\bf{\cos x = -\frac{\sqrt{3}}{2}}
$$
The two bolded parts give solutions and they are probably easy for you to work with.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3281678",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Integrating the trigonometric function So the problem goes as follows:-
$$\int{{\cos^2x+\sin2x}\over{(2\cos x-\sin x)^2}}~dx$$
My attempt is as follows:-
\begin{align*}\int{{\cos^2x+\sin2x}\over{(2\cos x-\sin x)^2}}~dx&=\int{{\cos^2x+2\sin x\cos x}\over{(2\cos x-\sin x)^2}}~dx\\
&=
\int{{(\cos x)(\cos x+2\sin x)}\over{(2\cos x-\sin x)^2}}~dx
\end{align*}
Now i could see that $\cos x+2\sin x $ is $-1$ times derivative of the denominator:-
$$\cos x+2\sin x=(-1){{d(2\cos x-\sin x)}\over{dx}}$$
And to handle the $\cos x$ term in the numerator i tried the following:-
$$2\cos x=t+\sin x \implies \cos x={{t+\sin x}\over{2}}$$
But the $\sin x$ term in the numerator is annoying.
| Integrate by parts
With $v=\dfrac{\cos x+2\sin x}{(2\cos x-\sin x)^2}=\dfrac{d(2\cos x-\sin x)^{-1}}{dx}$
the required integration $$=\cos x\int v\ dx-\int\left(\dfrac{d(\cos x)}{dx}\cdot \int v\ dx\right)$$
Finally for $\int\dfrac{a\cos x+b\sin x}{c\sin x+d\sin x)}dx,$
write $a\cos x+b\sin x=p(c\sin x+d\cos x)+q\cdot\dfrac{d(c\sin x+d\cos x)}{dx}$
Can you identify $a,b,c,d$ here and find $p,q$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "2",
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If $q$ is prime, can $\sigma(q^{k-1})$ and $\sigma(q^k)/2$ be both squares when $q \equiv 1 \pmod 4$ and $k \equiv 1 \pmod 4$?
This is related to this earlier MSE question. In particular, it appears that there is already a proof for the equivalence
$$\sigma(q^{k-1}) \text{ is a square } \iff k = 1.$$
Let $\sigma(x)$ denote the sum of divisors of the positive integer $x$.
Here is my question:
If $q$ is prime, can $\sigma(q^{k-1})$ and $\sigma(q^k)/2$ be both squares when $q \equiv 1 \pmod 4$ and $k \equiv 1 \pmod 4$?
MY ATTEMPT
Suppose that
$$\sigma(q^{k-1}) = a^2$$
and
$$\frac{\sigma(q^k)}{2} = b^2$$
for $q \equiv 1 \pmod 4$ and $k \equiv 1 \pmod 4$.
Since $\sigma(q^k) = q^k + \sigma(q^{k-1})$, it follows that
$$2b^2 = \sigma(q^k) = q^k + \sigma(q^{k-1}) = q^k + a^2.$$
Additionally, congruence-wise we obtain
$$a^2 = \sigma(q^{k-1}) \equiv 1 + (k-1) \equiv k \equiv 1 \pmod 4,$$
from which it follows that $a$ is odd, and
$$2b^2 = \sigma(q^k) = q^k + \sigma(q^{k-1}) \equiv 1^1 + 1 \equiv 2 \pmod 4,$$
which implies that $b$ is likewise odd.
Now, using the definition of $\sigma(q^k)$ and $\sigma(q^{k-1})$ for $q$ prime, we derive
$$\frac{1}{2}\cdot\frac{q^{k+1} - 1}{q - 1} = b^2$$
and
$$\frac{q^k - 1}{q - 1} = a^2.$$
Assume to the contrary that
$$\frac{1}{2}\cdot\frac{q^{k+1} - 1}{q - 1} = b^2 \leq a^2 = \frac{q^k - 1}{q - 1}.$$
This assumption leads to
$$q^{k+1} - 1 \leq 2(q^k - 1)$$
which implies that
$$16 = {5^1}(5-2) + 1 \leq q^k(q - 2) + 1 = q^{k+1} - 2q^k + 1 \leq 0,$$
since $q$ is a prime satisfying $q \equiv k \equiv 1 \pmod 4$. This results in the contradiction $16 \leq 0$. Consequently, we conclude that $a < b$.
Furthermore, I know that
$$(q+1) = \sigma(q) \mid \sigma(q^k) = 2b^2$$
so that
$$\frac{q+1}{2} \leq b^2.$$
Finally, I also have
$$\frac{q^{k+1} - 1}{2b^2} = q - 1 = \frac{q^k - 1}{a^2}.$$
Alas, here is where I get stuck.
CONJECTURE (Open)
If $q$ is a prime satisfying $q \equiv k \equiv 1 \pmod 4$, then $\sigma(q^{k-1})$ and $\sigma(q^k)/2$ are both squares when $k = 1$.
SUMMARY OF RESULTS SO FAR
zongxiangyi appears to have proven the implication
$$\sigma(q^k)/2 \text{ is a square} \implies k = 1.$$
The proof of the following implication is trivial
$$k = 1 \implies \sigma(q^{k-1}) \text{ is a square}.$$
The truth value of the following implication is currently unknown:
$$\sigma(q^{k-1}) \text{ is a square} \implies k = 1.$$
Together, the two results give
$$\sigma(q^k)/2 \text{ is a square} \implies k = 1 \iff \sigma(q^{k-1}) \text{ is a square},$$
so that $\sigma(q^{k-1})$ is a square if $\sigma(q^k)/2$ is a square.
Therefore, $\sigma(q^{k-1})$ and $\sigma(q^k)/2$ are both squares (given $q \equiv 1 \pmod 4$ and $k \equiv 1 \pmod 4$) when $\sigma(q^k)/2$ is a square.
| Here are a couple of other approaches to consider which may be useful. First, your equation of
$$2b^2 = \sigma(q^k) = q^k + \sigma(q^{k-1}) = q^k + a^2 \tag{1}\label{eq1}$$
can be rewritten as
$$2b^2 - a^2 = q^k \tag{2}\label{eq2}$$
This is in the generalized Pell equation form of $x^2 - Dy^2 = N$. The blog Solving the generalized Pell equation explains how to solve this.
Next, note that
$$\sigma(q^{k-1}) = \sum_{i=0}^{k-1} q^i \tag{3}\label{eq3}$$
$$\sigma(q^{k}) = \sum_{i=0}^{k} q^i \tag{4}\label{eq4}$$
Thus, you can express $\sigma(q^{k})$ in terms of $\sigma(q^{k-1})$ as
$$\sigma(q^{k}) = q\sigma(q^{k-1}) + 1 \tag{5}\label{eq5}$$
As you stated, suppose
$$\sigma(q^{k-1}) = a^2 \tag{6}\label{eq6}$$
$$\frac{\sigma(q^k)}{2} = b^2 \iff \sigma(q^k) = 2b^2 \tag{7}\label{eq7}$$
Substituting \eqref{eq6} and \eqref{eq7} into \eqref{eq5} gives
$$2b^2 = qa^2 + 1 \iff 2b^2 - qa^2 = 1 \iff (2b)^2 - (2q)a^2 = 2 \tag{8}\label{eq8}$$
Wikipedia's Pell's equation page's Transformations section gives a related equation of
$$u^{2}-dv^{2}=\pm 2 \tag{9}\label{eq9}$$
and how it can be transformed into the Pell equation form of
$$(u^{2}\mp 1)^{2}-d(uv)^{2}=1 \tag{10}\label{eq10}$$
Here, $u = 2b$, $v = a$, $d = 2q$ and the right side of \eqref{eq8} is $2$, so \eqref{eq10} becomes
$$((2b)^2 - 1)^2 - (2q)(2ba)^2 = 1 \tag{11}\label{eq11}$$
This is in Pell's equation form of $x^2 - ny^2 = 1$. Since $n = 2q$ is not a perfect square, there are infinitely many integer solutions. However, among these solutions, you first need to check that $x$ is in the form $4b^2 - 1$, the determined $b$ divides $y = 2ba$ and then that $a$ and $b$ satisfy \eqref{eq6} and \eqref{eq7} for some $k \equiv 1 \pmod 4$.
As for your open conjecture, if $k = 1$, then isn't $\sigma(q^{k-1}) = \sigma(q^{0}) = 1$ and $\frac{\sigma(q^{k})}{2} = \frac{\sigma(q)}{2} = \frac{1 + q}{2}$, so having both of them be squares requires $q = 2b^2 - 1$ for some $b$ and, thus, is not always true for all primes $q \equiv 1 \mod 4$, e.g., for $q = 5$, you get $5 = 2b^2 - 1 \implies 6 = 2b^2 \implies b = \sqrt{3}$?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3284709",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Is $\frac{\arccos\left((\sqrt{r}+1)/(r+1/\sqrt{r})\right)}{\pi \left|1-r^{-3/2}\right|}$ analytic at $r=1$? Is the function $f:(0,\infty)\rightarrow(0, 1)$, defined below, analytic at $r=1$?
$$f(r) := \frac{\arccos\left(\frac{\sqrt{r}+1}{r+\frac{1}{\sqrt{r}}}\right)}{\pi
\left|1-\frac{1}{r^{3/2}}\right|}\quad
\mathrm{if\ } r>0\mathrm{\ and\ } r\neq1,
$$
and $f(1) :=\frac{\sqrt{2}}{3 \pi }$ where $\arccos(x)\in [0,\pi].$
If you are interested, this function arises from retrograde motion.
The following statements seem to be true:
*
*$\lim_{r\rightarrow\infty} f(r) = 1/2$,
*$\lim_{r\rightarrow 0} \frac{f(r)}{r^{3/2}}=1/2$,
*$f(r) = f(1/r)r^{(3/2)}$,
*$g(r) = \frac{\sqrt{r}+1}{r+\frac{1}{\sqrt{r}}}$ is analytic at $r=1$,
*$g(r) = 1-\frac{1}{4} (r-1)^2+\frac{1}{4} (r-1)^3-\frac{11}{64} (r-1)^4+ \frac{3}{32} (r-1)^5 -\frac{21}{512} (r-1)^6+\frac{7}{512}
(r-1)^7+ O((r-1)^8),$
*$1-g(r) = \frac{(r-1)(1-1/\sqrt{r})}{r+1/\sqrt{r}}\geq 0$,
*$\mathrm{sgn}(x)\arccos(1-x^2) = \sqrt{2} x + \frac{x^3}{6\sqrt{2}} +\frac{3 x^5}{80 \sqrt{2}} + \frac{5 x^7}{448 \sqrt{2}}+O(x^9)$, and
*$1-1/r^{3/2} = \frac{3 (r-1)}{2}-\frac{15}{8} (r-1)^2+\frac{35}{16}
(r-1)^3-\frac{315}{128} (r-1)^4+\frac{693}{256} (r-1)^5-\frac{3003
(r-1)^6}{1024}+\frac{6435 (r-1)^7}{2048}+O\left((r-1)^8\right).$
| The answer is yes.
1) Consider $f$ as a function of a complex variable $z$. If $g(z)$ is analytic at $z=1$ then $g(\sqrt{z})$ is also analytic as a composition of holomorphic functions. Thus it's enough to establish analyticity of the function
$$
g(z)=f(z^2)=\frac{\cos ^{-1}\left(\frac{z^2+z}{z^3+1}\right)}{\pi\sqrt{\left(\frac{1}{z^3}-1\right)^2}}.
$$
2) Denote for convenience $h(z)=g(1+z)$ so the point of interest is $z=0$ now. Using the equality
$$
\cos ^{-1}(z)=2 \tan ^{-1}\left(\frac{\sqrt{1-z^2}}{z+1}\right)
$$
gives
$$
h(z)=
\frac{2 \tan ^{-1}\left(\frac{\left(z^2+z+1\right) \sqrt{1-\frac{(z+1)^2}{\left(z^2+z+1\right)^2}}}{z^2+2z+2}\right)}{\pi \sqrt{\left(\frac{1}{(z+1)^3}-1\right)^2}}=
\frac{2 \tan ^{-1}\left(\sqrt{z^2}\frac{\left(z^2+z+1\right) \sqrt{\frac{\left(z^2+2 z+2\right)}{\left(z^2+z+1\right)^2}}}{z^2+2z+2}\right)}{\pi \sqrt{z^2} \sqrt{\frac{ \left(z^2+3 z+3\right)^2}{(z+1)^6}}}.
$$
All functions in the rhs are analytic at $z=0$ except $\sqrt{z^2}$. Since $\tan^{-1}(z)$ is odd, function $\tan^{-1}(\sqrt{z^2}u(z))/\sqrt{z^2}$ is analytic for $u$ analytic at $z=0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3286306",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Given $\triangle ABC$ with $C=60^\circ$, show that $\frac{1}{a+c}+\frac{1}{b+c}=\frac{3}{a+b+c}$
Given that $C=60^\circ$ on a triangle $ABC$, prove the following relation: $$\frac{1}{a+c}+\frac{1}{b+c}=\frac{3}{a+b+c}$$
P.S. Maybe this info could be of help: I used the cosine rule of triangles, given that $C=60°$.
| Let's try to simplify the expression $\frac{1}{a+c}+\frac{1}{b+c}=\frac{3}{a+b+c}$.
Multiply through by $a + b + c$ and you obtain the equivalent expression
$$\frac{a+b+c}{a+c} + \frac{a + b + c}{b + c} = 3$$
This is equivalent to
$$\frac{b}{a+c} + \frac{a}{b + c} = 1$$
Multiplying through by $(a+c)(b+c)$ this is the same as
$$b(b+c) + a(a + c) = (a + c)(b+c)$$
Expanding out this becomes
$$b^2 + bc + a^2 + ac = ab + cb + ac + c^2$$
Cancelling terms this becomes
$$b^2 + a^2 - ab = c^2$$
This is exactly what the law of cosines gives you.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3286942",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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z the root of $z+\frac{1}{z}=2\cos\frac{\pi}{2018} $ then the value of $z^{2018}+\frac{1}{z^{2018}}$ is.. $z \in\mathbb{C}^{*}$ is a root of the equation $z+\frac{1}{z}=2\cos\frac{\pi}{2018} $ then $z^{2018}+\frac{1}{z^{2018}}$ has the value...the right answer is -2.
| If $z+\dfrac1z=2\cos t$
$z^2+\dfrac1{z^2}=\left(z+\dfrac1z\right)^2-2=\cdots=2\cos2t$
Similarly, $z^3+\dfrac1{z^3}=\left(z+\dfrac1z\right)^3-3\left(z+\dfrac1z\right)=\cdots=2\cos3t$
Like Proving that $\frac{\phi^{400}+1}{\phi^{200}}$ is an integer.
using strong induction
$$z^{n+1}+\dfrac1{z^{n+1}}=\left(z^n+\frac1{z^n}\right)\left(z+\frac1z\right)-\left(z^{n-1}+\frac1{z^{n-1}}\right) =2\cos nt\cdot2\cos t-2\cos(n-1)t$$
By Werner Formulas
$$z^{n+1}+\dfrac1{z^{n+1}}=2[\cos(n-1)t+\cos(n+1)t]-2\cos(n-1)t$$
$$\implies z^{n+1}+\dfrac1{z^{n+1}}=2\cos(n+1)t$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Let $S$ be the subset of $M(\mathbb{R})$ consisting of matrices of the form: The proof I read on slader seemed like it only used either $1$ or $2$ matrices to prove this when the question asks for all matrices of a certain form (when $x+y = 1$).
Let $S$ be the subset of $M(\mathbb{R})$ consisting of matrices of the form: $\begin{pmatrix}a & a\\\ b & b\end{pmatrix}$.
If $x + y = 1$, show that $\begin{pmatrix}x & x\\\ y & y\end{pmatrix}$ is not a left identity in $S$.
If $\begin{pmatrix}x & x\\\ y & y\end{pmatrix}$ is a left identity, then we have
$\begin{pmatrix}x & x\\\ y & y\end{pmatrix}$$\begin{pmatrix}0 & 0\\\ 1 & 1\end{pmatrix}$ = $\begin{pmatrix}x & x\\\ y & y\end{pmatrix}$ = $\begin{pmatrix}0 & 0\\\ 1 & 1\end{pmatrix}$.
But $\begin{pmatrix}x & x\\\ y & y\end{pmatrix}$$\begin{pmatrix}1 & 1\\\ 0 & 0\end{pmatrix}$ = $\begin{pmatrix}x & x\\\ y & y\end{pmatrix}$ = $\begin{pmatrix}1 & 1\\\ 0 & 0\end{pmatrix}$
which implies the contradiction that $\begin{pmatrix}x & x\\\ y & y\end{pmatrix}$ = $\begin{pmatrix}0 & 0\\\ 1 & 1\end{pmatrix}$ = $\begin{pmatrix}1 & 1\\\ 0 & 0\end{pmatrix}$.
Hence $\begin{pmatrix}x & x\\\ y & y\end{pmatrix}$ is not a left identity in $S$.
Is this a correct proof without using the fact that $x+y = 1$?
| Your proof is correct and you haven't used the condition that $x+y=1$. In fact, the set of matrices in question has not any left identity. So, whether $x+y=1$ is irrelevant.
I believe that the "left" in the question is a typo. The question makes more sense if it is talking about right identities.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3291630",
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Question about general solution of inhomogeneous ODE $u''+4u= 8x^2 +13e^{3x}+16\cos(2x)$ Given the initial conditions: $u(0)=0$ and $u'(0)=0,$
I need to solve:
$u''+4u= 8x^2 +13e^{3x}+16\cos(2x)$
The solution from the book is : $ u = A\cos 2x + B\sin 2x + 2x^
2 − 1 + e^{3x} + 4x\sin(2x), A = 0, B = -\frac{3}{2}.$
And the solution I got was:
$ u = A\cos 2x + B\sin 2x + 2x^
2 − 1 + e^{3x} + Ex\cos(2x) + Fx\sin(2x)$
$Ex\cos(2x) + Fx\sin(2x)$ was found by applying annihilator method on $16\cos(2x)$: $$(D^2+4)(D^2+4)=0$$
Now since $u(0)=0$,
you get $A=0$.
$u'(x):$ $$=-2A\sin(2x) +2B\cos(2x)+4x+3e^{3x}+E\cos(2x) +F\sin(2x)+x[-2E\sin(2x) +2F\cos(2x)].$$
$u'(0)= 2B+E+3=0$ .
So just by observing, this $E$ is somehow $0$, and $F$ is $4$. I realised if I did the initial conditions separately, then $E$ is indeed $0$, but why is $F=4?$ (also by doing it separately for the homogeneous solution, then $A=0$ and $B=-\frac{3}{2}$
| $$u''+4u= 8x^2 +13e^{3x}+16\cos~2x $$
$$\implies (D^2+4)u=8x^2 +13e^{3x}+16\cos~2x \qquad \text{where $~D\equiv\frac{d}{dx}~$}$$
Clearly, complementary function (C.F.) is $$C_1 ~\cos 2x~+~C_2~\sin~2x\qquad \text{where $~C_1~, ~ C_2~$are arbitrary constants.}$$
Particular integral (P.I.) is
P.I.$~~=~~\frac{1}{D^2+4}~\left\{8x^2 +13e^{3x}+16\cos~2x \right\}$
$~~~~~~~=~\frac{8}{4}~\left(1+\frac{D^2}{4}\right)^{-1}~x^2~+~13~\frac{1}{D^2+4}~e^{3x}~+~16~\frac{1}{D^2+4}~\cos~2x$
$~~~~~~~=~2~\left(1-\frac{D^2}{4}+\cdots\right)~x^2~+~13~\frac{e^{3x}}{3^2+4}~+~16~\frac{x}{4}~\sin~2x$
$~~~~~~~=~2~\left(x^2-\frac{1}{2}\right)~+~e^{3x}~+~4~x~\sin~2x$
$~~~~~~~=(2~x^2-1)~+~e^{3x}~+~4~x~\sin~2x$
So the general solution is
$$u(x)=\text{C.F.$~+~$P.I.}$$
$$\implies u(x)~=C_1 ~\cos 2x~+~C_2~\sin~2x~+~(2~x^2-1)~+~e^{3x}~+~4~x~\sin~2x$$
Given that $~u(0)=0~\implies C_1~-1+1=0\implies C_1=0~,$
$~u'(0)=0~\implies 2~C_2~+~3=0\implies C_2=-\frac{3}{2} $
So the required solution of the given solution is $$u(x)~=~-\frac{3}{2}~\sin~2x~+~(2~x^2-1)~+~e^{3x}~+~4~x~\sin~2x$$
which is of the form $$u = A\cos 2x + B\sin 2x + 2x^
2 − 1 + e^{3x} + 4x\sin(2x), \qquad \text{where}~~~A = 0,~~ B = -\frac{3}{2}.$$
*
*If $f(D)$ can be expressed as $\phi(D^2)$ and $\phi(-a^2)\neq 0$, then
$1.$ $\frac{1}{f(D)} \sin ax=\frac{1}{\phi(D^2)} \sin ax = \frac{1}{\phi(-a^2)} \sin ax$
$2.$ $\frac{1}{f(D)} \cos ax=\frac{1}{\phi(D^2)} \cos ax = \frac{1}{\phi(-a^2)} \cos ax$
Note: If $f(D)$ can be expressed as $\phi(D^2)=D^2+a^2$, then $\phi(-a^2)= 0$.
$1.$ $\frac{1}{f(D)} \sin ax =\frac{1}{\phi(D^2)} \sin ax=x\frac{1}{\phi'(D^2)} \sin ax= x \frac{1}{2D} \sin ax= -\frac{x}{2a} \cos ax$.
$2.$ $\frac{1}{f(D)} \cos ax =\frac{1}{\phi(D^2)} \cos ax=x\frac{1}{\phi'(D^2)} \cos ax= x \frac{1}{2D} \cos ax= \frac{x}{2a} \sin ax$.
${}$
Consider a differential equation of the form $f(D)y=X$
If $X=e^{ax}$, then
$1.$ P.I.$\quad = \frac{1}{f(D)}e^{ax}=\frac{e^{ax}}{f(a)}$, if $f(a)\neq 0$
$2.$ P.I.$\quad =\frac{1}{(D-a)^n}e^{ax}=\frac{x^n}{n!}e^{ax}$
| {
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Solving $x^5+px^2y^3+p^2y^5 = 0$ where $p$ is prime and $x,y \in \mathbb{Z}$ How do you solve $x^5+px^2y^3+p^2y^5 = 0$ where $p$ is prime and $x,y \in \mathbb{Z}$?
Working
in modulo $p$ we have $x^5 = 0 \pmod{p}$ and $x = 0$ which is the only solution in modulo $p$ since $a^{p} \equiv a \mod{p}$ for any prime $p$ and integer $a$. Taking this back we get $p^2y^5 = 0$ which implies $y=0$. Hence the only solution is $(x,y) = (0,0)$.
Edit/attempt 2:
Since $x = 0 \mod{p}$ we can let $x = pm$ where $m<x$ then we get $p^2 (m^5 p^3 + m^2 p y^3 + y^5) = 0 \implies m^5 p^3 + m^2 p y^3 + y^5$ since $p$ can't be zero. But now $y^5 = 0 \mod{p}$ ; suppose $y = pn$ where $n < y$. Then $p^3 (m^5 + m^2 n^3 p + n^5 p^2) =0$ and now $m^5 + m^2 n^3 p + n^5 p^2=0.$
| I would like to emphasize the fact that the equation is a homogeneous polynomial in x,y, all "total" exponents are 5. What that means is that, if we began with $\gcd(x,y) = g > 1,$ we are allowed to doivide both $x,y$ by the same $g,$ arriving at a coprime pair still giving a solution.
In contrast: if we began with $p s^3 = p^2 t^2,$ variables $s,t,$ we get equality with $s=p, t=p$ or with $s=p^3, t=p^4,$ but no solution with $s,t$ coprime. Generally $s = p^{1+2k}, t = p^{1+3k}$ as well as $s = a^2 p^{1+2k}, t = a^3 p^{1+3k}$
| {
"language": "en",
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Factorization of cubic $x^3 + x^2 - 2x -1\pmod p$ Show that $x^3 + x^2 - 2x -1$ factorizes mod $p$ as a product of distinct linear factors if $p \equiv \pm 1 \pmod 7$, is a cube for $p=7$ and irreducible otherwise.
I was told there are both a solution with splitting fields and an elementary solution but I can't think of any of them.
Any help appreciated!
| To test whether $f$ has multiple roots, we perform the Euclidean algorithm for $\gcd(f,f')$. This gives
$$
7 = (2 x^2 + x - 3)(x^3 + x^2 - 2x -1)' + (-6 x - 1)(x^3 + x^2 - 2x -1)
$$
Therefore, $x^3 + x^2 - 2x -1$ has a multiple root iff $p=7$.
Indeed, $x^3 + x^2 - 2x -1=(x+5)^3$. This also follows from the Euclidean algorithm. The first step tells us a double root, which in this case turns out to be a triple root:
$$
x^3 + x^2 - 2x -1=5(x+5)(x^3 + x^2 - 2x -1)' \bmod 7
$$
Here is a proof that if $x^3 + x^2 - 2x -1$ has a root, then it splits.
If $a$ is a root of $x^3 + x^2 - 2x -1$, then the other roots are $b= a^2 - 2$ and $c = -a^2 - a + 1$. This can also be expressed as follows: if $u$ is a root, then so is $g(u)=u^2-2$.
Indeed, the discriminant of $x^3 + x^2 - 2x -1$ is $\Delta=49$.
On the other hand, $\Delta=d^2$, where
$$d=(a-b)(a-c)(b-c)=(a-(b+c)a+bc)(b-c)=(3a^2+2a-2)(b-c)$$
since $b+c=-1-a$ and $bc=1/a=a^2+a-2$.
Therefore, $b-c=\dfrac{14 a^2 + 7 a - 21}{49}d$. Using $b+c=-1-a$ and $d=\pm 7$ gives the result.
| {
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Block symmetric matrix expressed with Kronecker product and its determinant My ultimate goal is to find a way to compute the determinant of the following block symmetric matrix:
$$
\underset{np\times np}H=\begin{bmatrix}
\frac{1}{n}A+\frac{2(n-1)}{n^3s^2}I_p & -\frac{2}{n^3s^2}I_p & \dots & -\frac{2}{n^3s^2}I_p \\
-\frac{2}{n^3s^2}I_p & \frac{1}{n}A+\frac{2(n-1)}{n^3s^2}I_p & \dots & -\frac{2}{n^3s^2}I_p \\
\dots & \dots & \dots & \dots\\
-\frac{2}{n^3s^2}I_p & -\frac{2}{n^3s^2}I_p & \cdots & \frac{1}{n}A+\frac{2(n-1)}{n^3s^2}I_p\\
\end{bmatrix},
$$
where $A$ is a $p\times p$ symmetric matrix and $n,s\in \mathbb{R}$.
As a first step, I would like to express this matrix with a unique expression envolving Kronecker product. Any idea of how to achieve such expression? Or any suggestion to compute the determinant of $H$ in another way?
| We can write
$$
H = \frac 1n I_n \otimes A + M \otimes I_p
$$
where the entries of $M$ are the coefficients to your $I_p$. That is,
$$
M = \pmatrix{
\frac{2(n-1)}{n^3s^2} & -\frac{2}{n^3s^2} & \cdots & -\frac{2}{n^3s^2} \\
-\frac{2}{n^3s^2} & \frac{2(n-1)}{n^3s^2} & \cdots & -\frac{2}{n^3s^2} \\
\vdots & \vdots & \ddots & \vdots\\
-\frac{2}{n^3s^2} & -\frac{2}{n^3s^2} & \cdots & \frac{2(n-1)}{n^3s^2}\\
}
$$
You could say that your $H$ is the Kronecker sum $M \oplus \frac{A}{n}$.
One approach to finding the determinant is as follows: you can find the eigenvalues of $M$ with one of the methods outlined here. If $\lambda_1,\dots,\lambda_n$ are the eigenvalues of $M$ and $\mu_1,\dots,\mu_p$ are the eigenvalues of $\frac An$, then the eigenvalues of $H$ will be $\lambda_i + \mu_j$ for every pair $i,j$ with $1 \leq i \leq n$ and $1 \leq j \leq p$. So, the determinant of $H$ is the product of the eigenvalues of $H$, that is the product of all such sums.
Here is a reference for the Kronecker sum. See section 2.1.3 in particular.
| {
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Prove That $3^n + 8^n$ is Not Divisible by $5$ (Using Induction) Prove that $3^n+8^n$ is not divisible by 5.
I know that this can be proved by using congruence and I am providing the proof by congruence below. But is there any way to Prove It By Induction.
The proof by congruence goes like this:
$3\equiv 3\pmod 5 \\ 3^2 \equiv 4\pmod 5 \\ 3^3\equiv 7\pmod 5 \\ 3^4\equiv 1\pmod 5 \\ 3^5\equiv 3\pmod 5$
Also,
$8\equiv 3\pmod 5 \\ 8^2 \equiv 4\pmod 5 \\ 8^3\equiv 7\pmod 5 \\ 8^4\equiv 1\pmod 5 \\ 8^5\equiv 3\pmod 5$
Adding the congruence up (since the same cycle repeats after the 4th power) none of them are divisible by 5 or equal to 0.
But I need a proof by Induction.
Any help will be appreciated.
| COMMENT.- Another way very easy: Take modulo $10$
If $n=4k,\space 4k+1,\space 4k+2,\space 4k+3$ we have respectively
$3^n+8^n\equiv7,\space 1,\space 3,\space 9\pmod{10}$ It is never congruent to $5$ because all multiple of $5$ ends in $5$ or $0$.
| {
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Number of nine digits numbers whose sum of the digits is even I am reading Mathematical Circle. Problem $48$ in chapter two says that
How many nine-digit numbers have an even sum of their digits?
I am trying in this way, that we can divide the problem in four cases.
*
*$1$ even digit and $8$ odd digits
*$3$ even digits and $6$ odd digits
*$5$ even digits and $4$ odd digits
*$7$ even digits and $2$ odd digits
For the first case we get $4 \cdot 5^8 +5\cdot 5^7 \cdot 5$ number of solution. Because if the even digit is placed in first place (left to right) then we get $4\cdot 5^8$ ways to write the number and if an odd digit is placed in first place then we get $5\cdot 5^7 \cdot 5$ ways to write the number. Similarly for the second case we get $4\cdot 5^8+5^9$ , for the third case we get $4\cdot 5^8+5^9$ and for the fourth case we get $4\cdot 5^8+5^9$ ways to write the number. So total number is $ 4 \cdot (4\cdot 5^8+5^9)$.
The answer is different. So Where I have made a mistake?
Thanks.
| $9$ digit numbers range from $10^8$ to $10^{9}-1$. There are exactly $9 \cdot 10^8$ numbers, that have nine digits.
$10^8 + 0$ doesn't have an even sum of its digits $(=1)$.
$10^8+1$ has an even sum of its digits $(=2)$.
$10^8+2$ doesn't have an even sum of its digits $(=3)$.
$10^8+3$ has an even sum of its digits $(=4)$.
$10^8+4$ doesn't have an even sum of its digits $(=5)$.
$10^8+5$ has an even sum of its digits $(=6)$.
$10^8+6$ doesn't have an even sum of its digits $(=7)$.
$10^8+7$ has an even sum of its digits $(=8)$.
$10^8+8$ doesn't have an even sum of its digits $(=9)$.
$\vdots \hspace{3cm} \vdots \hspace{3cm} \vdots$
$10^9-3$ has an even sum of its digits$(=-2)$.
$10^9-2$ doesn't have an even sum of its digits $(=-1)$.
$10^9-1$ has an even sum of its digits $(=0)$.
See the pattern? Here is a C++ Code that finds that out for you:
int main()
{
unsigned long int x;
x = 0;
for(unsigned long int i = 100000000; i <= 1000000000-1; i++)
if (i % 2 == 0)
x++;
std::cout << x;
}
This yields $x = (9 \cdot 10^8)/2 = 45\cdot 10^7$. Half of the numbers with nine digits have an even digit sum.
| {
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What about $[n,k,7]$ binary perfect code? (Exercise from the book Fundamentals of Error Correcting Codes) I'm trying to solve an exercise from the book Fundamentals of Error Correcting Codes, can somebody help me?
The problem is the following:
Let $C$ be an $[n, k, 7]$ perfect binary code.
(a) Using equality in the Sphere Packing Bound, prove that
$$(n+1)[(n+1)^2 −3(n+1)+8]=3·2^{n−k+1}.$$
(b) Prove that $n+1$ is either $2^b$ or $3·2^b$ where, in either case,
$b≤n−k+1$.
(c) Prove that $b < 4$.
(d) Prove that $n=23$ or $n=7$.
(e) Name two codes that are perfect $[n, k, 7]$ codes, one with $n = 7$
and the other with $n = 23$.
For part a) I could easily prove the claim :)
For part b) I answered in the following way:
From part a) we know $n+1=(3·2^{n−k+1})/(n^2-n+6)$, so since $n$ is a
natural number, then so is $n+1$ and thus we can consider three cases:
1) $(n^2-n+6)$ divides $3$ (it does not divide $2^{n-k+1}$) -> $n+1=2^{n-k+1}$
OK
2) $(n^2-n+6)$ divides both $3$ and $2^{n-k+1}$ -> $n+1=2^b$ with $b<=n-k+1$ OK
3) $(n^2-n+6)$ divides $2^{n-k+1}$ (it does not divides $3$) -> $n+1=3·2^b$ with $b<=n-k+1$ OK
I'm really not sure if it's enough to prove the claim or if there is some "more mathematically" proof. any suggestion?
For part c) I'm completely lost. I cannot find a way to prove that $b<4$.
(I tried to use the information that $C$ is a perfect code. but I'm stuck!)
& also for part d) & e) I need any suggestions.
| The starting point is the identity (your part a) gotten by having an equality in the sphere packing bound
$$(n+1)[(n+1)^2 −3(n+1)+8]=3·2^{n−k+1}.\qquad(*)$$
*
*Equation $(*)$ tells us that $n+1$ is a factor of the integer $3\cdot 2^{n-k+1}$. By the uniqueness of integer factorization it follows that $n+1=3^\epsilon\cdot 2^b$ with $\epsilon\in\{0,1\},0\le b\le n-k+1$. This settles part (b).
*Assume that $b\ge4$. It follows that both $(n+1)^2$ and $-3(n+1)$ are divisible by $2^4$. A key observation is that the third term, $+8$, in square brackets is not. This means that the quantity in square brackets is divisible by $8$ but is not divisible by $16$. Let's take advantage by writing $(n+1)=16m$, $m$ a positive integer. Then
$$[(n+1)^2-3(n+1)+8]=256m^2-48m+8=8(32m^2-6m+1).$$ Here the factor $32m^2-6m+1$ is an odd integer $\ge27$. But the integer $3\cdot2^{n-k+1}$ has no odd factors larger than $3$, so this contradicts $(*)$. Therefore $b\in\{0,1,2,3\}$.
*If $b=0$ or $b=1$ then $n+1=3^\epsilon\cdot2^b<7$ making it obvious that a code of length $n$ cannot have minimum distance $7$. If $b=2$ then $n+1=4\cdot3^{\epsilon}$ is either $4$ or $12$. The former is similarly impossible. If $n+1=12$ then $(n+1)^2-3(n+1)+8=116=2^2\cdot29$ violating $(*)$. Therefore the only possibility is $b=3$, when either $n+1=2^3=8$ or
$n+1=3\cdot2^3=24$.
*It is possible that $n+1=8$ for the repetition code of length $7$ is perfect. Similarly, $n+1=24$ is possible, as the binary Golay code of length $23$ is a perfect code with minimum distance $7$.
| {
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$\sqrt2 x^2 - \sqrt3 x + k = 0$ with solutions $\sin \theta , \cos \theta$
$\sqrt2 x^2 - \sqrt3 x + k = 0$ with solutions $\sin \theta , \cos \theta $, $\enspace0\leq\theta\leq2\pi$.
$(x-\sin \theta)(x-\cos \theta)=0$
$(\sin \theta + \cos \theta) = \sqrt3/ \sqrt2$
$(\sin \theta \cdot \cos \theta) = k/\sqrt2$
But how to find $k$?
| $(\sin\theta+\cos\theta)^2=1+2\cos\theta\sin\theta=1+2\frac{k}{\sqrt{2}}$.
On the other hand, $(\sin\theta+\cos\theta)^2=\frac32$, hence
$$1+\frac{2k}{\sqrt{2}}=\frac32$$
namely $$\frac{2k}{\sqrt{2}}=\frac12 \quad \Longrightarrow k=\frac{\sqrt{2}}{4}$$
Notice that this solution is acceptable. Since both $\sin\theta$ and $\cos\theta$ are bounded by one, hence their product is subject to the same bound.
| {
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Evaluate $\sec \frac{2\pi}{7}+\sec \frac{4\pi}{7}+\sec \frac{6\pi}{7}$ The problem is as follows:
Find the value of $\textrm{H}$ which belongs to a certain vibration coming from a magnet.
$$H=\sec \frac{2\pi}{7}+\sec \frac{4\pi}{7}+\sec \frac{6\pi}{7}$$
It was easy to spot that each term was related to multiples of two and three of the first angle. So I rewrote that equation like this:
$$H=\sec \frac{2\pi}{7}+\sec \frac{2\times 2\pi}{7}+\sec \frac{3\times 2\pi}{7}$$
One method which I tried was to transform the multiples of each angle into their equivalents as a single one as shown below:
$$\cos^{2}\omega=\frac{1+\cos 2\omega}{2}$$
$$\cos 2\omega= 2 \cos^{2}\omega - 1$$
$$\cos^{3}\omega=\frac{1}{4}\left(3cos\omega+\cos 3\omega \right)$$
$$\cos 3\omega = 4 \cos^{3}\omega - 3 cos\omega$$
Therefore by plugin these expressions into the above equation would become into (provided that secant function is expressed in terms of secant):
$$H=\frac{1}{\cos \frac{2\pi}{7}}+\frac{1}{2\cos^{2}\frac{2\pi}{7}-1}+\frac{1}{4\cos^{3}\frac{2\pi}{7}-3\cos\omega}$$
But from here on it looks convoluted or too algebraic to continue. My second guess was it could be related to sum to product identity but I couldn't find one for the secant.
Does it exist a shortcut or could it be that am I missing something? Can somebody help me to find the answer?
Can this problem be solved without requiring to use Euler's formulas?
| If $z=e^{2\pi i/7}$, then
$$
\cos\frac{2n\pi}{7}=\frac{z^n+z^{-n}}{2}=\frac{z^{2n}+1}{2z^n}
$$
so your expression becomes
$$
\frac{2z}{z^2+1}+\frac{2z^2}{z^4+1}+\frac{2z^3}{z^6+1}
$$
We get the numerator
$$
2z(z^{10}+z^4+z^6+1+z^9+z^3+z^7+z+z^8+z^4+z^6+z^2)
$$
Now we can note that $z^7=1$ and $z^6+z^5+z^4+z^3+z^2+z+1=0$, so the expression becomes
$$
4z(z^6+z^4+z^3+z^2+z+1)=-4z^6
$$
The denominator is
\begin{align}
(z^2+1)(z^{10}+z^6+z^4+1)
&=z^{12}+z^8+z^6+z^2+z^{10}+z^6+z^4+1\\
&=z^5+z+z^6+z^2+z^3+z^6+z^4+1\\
&=z^6
\end{align}
| {
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Complex limit question
*
*Calculate $\displaystyle\lim_{x\to\infty} \frac{x^2e^{-4x}+xe^{-3x}}{e^{-2x}+xe^{-3x}}$.
*Calculate $\displaystyle\lim_{x\to 0} \frac{x^2e^{x^4}-\sin{(x^2)}}{1-\cos{(x^3)}}$.
For the first question I got \begin{equation*}
\begin{split}
\lim_{x\to\infty} \frac{x^2e^{-4x}+xe^{-3x}}{e^{-2x}+xe^{-3x}} &= \lim_{x\to\infty} \frac{xe^{-3x}\left(xe^{-x}+1\right)}{e^{-2x}\left(xe^{-x}+1\right)} \\
&= \lim_{x\to\infty} \frac{xe^{-3x}}{e^{-2x}} \\
&= \lim_{x\to\infty} \frac{x}{e^x}.
\end{split}
\end{equation*}
This has an indeterminate form $\frac{\infty}{\infty}$. So we can use L'Hopital's rule to get $\lim_{x\to\infty} \frac{x}{e^x} = \lim_{x\to\infty} \frac{1}{e^x} = 0$.
For the second question I tried L'Hopital's rule but that didn't work and I also tried multiplying by the conjugate $\left(1+\cos{(x^3)}\right)$. Any help would be great!!!
| Near $0$, you have\begin{align}x^2e^{x^4}-\sin(x^2)&=x^2\left(1+x^4+\frac{x^8}{2!}+\cdots\right)-\left(x^2-\frac{x^6}6+\cdots\right)\\&=\frac76x^6+O(x^7)\end{align}and$$1-\cos(x^3)=\frac{x^6}2+O(x^7)$$and therefore$$\lim_{x\to0}\frac{x^2e^{x^4}-\sin(x^2)}{1-\cos(x^3)}=\frac{7/6}{1/2}=\frac73.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3307810",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Friedberg Linear algebra exercise 3.4.6 Let the reduced row echelon form of $A$ be
\begin{pmatrix}
1 & -3 & 0 & 4 & 0 & 5\\
0 & 0 & 1 & 3 & 0 & 2 \\
0 & 0 & 0 & 0 & 1 & -1 \\
0 & 0 & 0 & 0 & 0 & 0
\end{pmatrix}
Determine $A$ if the first, third, and sixth colum of $A$ are
\begin{pmatrix}
1 \\
-2 \\
-1 \\
3
\end{pmatrix}
\begin{pmatrix}
-1 \\
1 \\
2 \\
-4
\end{pmatrix}
\begin{pmatrix}
3 \\
-9 \\
2 \\
5
\end{pmatrix}
Using (d), I can easily find the first and fourth column of $A$ (for example, the second column of $A$ is the first column of $A$ multiplied by $-3$, similarly I can compute the 4th column), but the 5th column cannot be found in this way. Could you give some help?
| Let $B$ the reduced row echelon form of $A$ and let $a_1,a_2,a_3,a_4,a_5,a_6$ and $b_1,b_2,b_3,b_4,b_5,b_6$ be the ordered columns of $A$ and $B$ respectively.
It is easy to see that $b_2=-3b_1$, so, by $(\textrm d)$ we conclude that also
$$a_2=-3a_1=\begin{pmatrix} -3\\6\\3\\-9 \end{pmatrix}$$
Similarly, the fourth column of $B$, $b_4$, is equal to $4b_1+3b_3$. Hence, the fourth column of $A$ is
$$a_4=4a_1+3a_3=4 \begin{pmatrix} 1\\-2\\-1\\3 \end{pmatrix} + 3 \begin{pmatrix} -1\\1\\2\\-4 \end{pmatrix} = \begin{pmatrix} 1\\-5\\2\\0 \end{pmatrix}$$
Finally, to get the fifth column of $ A $ note that $b_5=5b_1+2b_3-b_6$, that is
$$a_5=\begin{pmatrix} 0\\1\\-3\\2 \end{pmatrix}$$
Therefore, we already have all the columns of $ A $ and then it would look
$$A=\begin{pmatrix}
1&-3&-1&1&0&3 \\
-2&6&1&-5&1&-9 \\
-1&3&2&2&-3&2 \\
3&-9&-4&0&2&5
\end{pmatrix}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3309244",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
How to integrate $\sec^{-1}\sqrt\frac{x}{a-x}$ $\int\sec^{-1}\sqrt\frac{x}{a-x}dx$
How to solve the above integration? Please give me some hint about what substitution I should make.
Thanks in advance.
| \begin{align}
I&=\int\sec^{-1}\sqrt\frac{x}{a-x}\ dx,\quad \color{red}{\sqrt\frac{x}{a-x}=y}\\
&=\int\sec^{-1}(y)\frac{2ay}{(1+y^2)^2}\ dy\quad \color{red}{\text{apply IBP}}\\
&=-\frac{a\sec^{-1}(y)}{1+y^2}+a\int\frac{dy}{y(1+y^2)\sqrt{y^2-1}},\quad \color{red}{y=\sec\theta}\\
&=-\frac{a\sec^{-1}(y)}{1+y^2}+a\int\frac{d\theta}{1+\sec^2\theta},\quad \color{red}{y=\sec\theta}\\
\end{align}
Lets take care of the last integral:
\begin{align}
K&=\int\frac{d\theta}{1+\sec^2\theta}\\
&=\int\frac{1}{2+\tan^2\theta}\ d\theta, \quad\color{red}{1=\sec^2\theta-\tan^2\theta}\\
&=\int\frac{\sec^2\theta-\tan^2\theta}{2+\tan^2\theta}\ d\theta\\
&=\int\frac{\sec^2\theta}{2+\tan^2\theta}\ d\theta-\int\frac{\tan^2\theta\color{red}{+2-2}}{2+\tan^2\theta}\ d\theta\\
&=\frac1{\sqrt{2}}\tan^{-1}\left(\frac{\tan\theta}{\sqrt{2}}\right)-\int\left(1-\frac{2}{2+\tan^2\theta}\ d\theta\right)\\
&=\frac1{\sqrt{2}}\tan^{-1}\left(\frac{\tan\theta}{\sqrt{2}}\right)-\theta+2K\\
-K&=\frac1{\sqrt{2}}\tan^{-1}\left(\frac{\tan\theta}{\sqrt{2}}\right)-\theta\\
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3309697",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Solve the following equation: $\sin x \cos x = \frac{1}{2}$ I am required to solve the following equation:
$$\sin x \cos x = \frac{1}{2}$$
My attempt:
Rewriting $\cos x$
$$\sin x \sqrt{1 - \sin^2 x} = \frac{1}{2}$$
Squaring both sides
$$\bigl(\sin x \sqrt{1 - \sin^2 x}\bigr)^2 = \bigl(\frac{1}{2}\bigr)^2$$
$$\sin^2 x (1 - \sin^2 x) = \frac{1}{4}$$
Expanding left side and multiplying both sides by 4
$$\sin^2 x - \sin^4 x = \frac{1}{4}$$
$$4\sin^2 x - 4\sin^4 x = 1$$
$$4\sin^2 x - 4\sin^4 x -1 = 0$$
Reordering left side
$$- 4\sin^4 x + 4\sin^2 x -1 = 0$$
$$4\sin^4 x - 4\sin^2 x + 1 = 0$$
Expression above can be factored as
$$(2\sin^2 x - 1)(2\sin^2 x - 1) = 0$$
$$(2\sin^2 x - 1)^2 = 0$$
It follows that
$$2\sin^2 x - 1 = 0 $$
$$\sin^2 x = \frac{1}{2} $$
$$\sin x = ± \frac{1}{\sqrt{2}} $$
So the resulting angles are: $45^{\circ},135^{\circ},225^{\circ},315^{\circ}$
Is my solution correct?
The reason why I am asking is, the author of the book used different method, and the end result he got was:
$$\sin2x = 1$$
So $2x = \sin^{-1}(1) = 90^{\circ},450$, and thus $x = 45^{\circ},225^{\circ}$
| $$2\sin(x)\cos(x)=1$$
$$\sin(2x)=1$$
$$2x=\arcsin(1)+2k\pi = \dfrac{\pi}{2}+2k\pi$$
therefore $x=\dfrac{\pi}{4}+k\pi,~k\in\mathbb{Z}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3314378",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 8,
"answer_id": 3
} |
Solving $(xy)y'= x^2+3y^2$ I am having a very frustrating time with the back book that says my answer is way off but to me everything looks fine:
\begin{align*}
(xy)y'&= x^2+3y^2\\
y' &= \frac{x^2}{xy} + \frac{3y^2}{xy}\\
y' &= \frac{x}{y} + \frac{3y}{x}\\
y' &= \frac{1}{v} + 3v\\
y' &= \frac{1 + 3v^2}{v}\\
v+\frac{dv}{dx}x &= \frac{1+3v^2}{v}\\
\frac{dv}{dx}x&= \frac{1+3v^2-v^2}{v}\\
\frac{dv}{dx}x &= \frac{1+2v^2}{v}\\
\int \frac{v}{2v^2+1}\,dv &= \int\frac{1}{x}\,dx\\
u &= 2v^2+1\\
du &= 4v\,dv\\
dv &= \frac{1}{4v}\,du\\
\int \frac{v}{u} \frac{1}{4v}\,du &= \int \frac{1}{x} \,dx\\
\int \frac{1}{4u}\,du &= \ln|x| + c\\
\frac{1}{4} \int \frac{1}{u}\,du &= \ln|x| +c\\
\frac{1}{4} \ln|2v^2 + 1| &= \ln |x| + c\\
\ln|2v^2 + 1|&= 4\ln|x|+c\\
2v^2 + 1 &= e^{4\ln|x|}e^c\\
2v^2 + 1 &= Cx^4\\
2v^2 &= Cx^4\\
v^2 &= Cx^4\\
\frac{y}{x} &= \sqrt{Cx^4}\\
y &= x\sqrt{Cx^4}
\end{align*}
However the book says the answer is $x^2 + 2y^2 = Cx^6.$ I am fairly sure there are no mistakes.
| Where did the $+1$ go?
$$2v^2+1=Cx^4$$
$$\frac{2y^2}{x^2}+1=Cx^4$$
$$2y^2+x^2=Cx^6$$
$$x^2+2y^2=Cx^6$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3315558",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
} |
Let $a_1,a_2,a_3....$ be an arithmetic progression. Question is
Let $a_1,a_2,a_3....$ be an arithmetic progression. If $\frac{a_1+a_2+a_3+...+a_p}{a_1+a_2+a_3+...+a_q}=\frac{p^2}{q^2},p\not=q$, then find $\frac{a_6}{a_{21}}$
Answer: $\frac{11}{41}$
All options: $\frac{7}{2}$, $\frac{2}{7}$, $\frac{11}{41}$, $\frac{41}{11}$
My approach to the question is
Let $S_r$ be sum of first $r$ terms of this AP
Method 1
we know
$\frac{S_p}{S_q}=\frac{p^2}{q^2}$
So,
$\frac{\frac{p}{2}(a_1+a_p)}{\frac{q}{2}(a_1+a_q)}=\frac{p^2}{q^2}$
$\frac{a_1+a_p}{a_1+a_q}=\frac{p}{q}$
After that I am stuck.
Method 2
But after taking another look at the question, I observed
$\frac{S_6}{S_{21}}=\frac{6^2}{21^2}=\frac{36}{441}$
Also,
$\frac{S_5}{S_{20}}=\frac{5^2}{20^2}=\frac{25}{400}$
And we know $S_r-S_{r-1}=a_r$
Then subtracting numerators and denominators of both
$\frac{S_6-S_5}{S_{21}-S_{20}}=\frac{a_6}{a_{21}}=\frac{36-25}{441-400}=\frac{11}{41}$
which we can't always do in maths, I cannot understand why Method 2 gave correct answer.
| I also got $\frac{11}{41}.$
We have:
$$\frac{(2a_1+(p-1)d)p}{(2a_1+(q-1)d)q}=\frac{p^2}{q^2},$$ which gives $$d=2a_1$$ and
$$\frac{a_6}{a_{21}}=\frac{a_1+5\cdot2a_1}{a_1+20\cdot2a_1}=\frac{11}{41}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3316407",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Can't find the Jordan form of this 3x3 I have the matrix
$$\begin{pmatrix}
2 & 2 & -1 \\
-1 & -1 & 1 \\
-1 & -2 & 2
\end{pmatrix}$$
and need to find its Jordan canonical form. I can find that the only eigenvalue is 1 with algebraic multiplicity 3, and I can find two independent eigenvectors,
$$\begin{pmatrix}1\\0\\1\end{pmatrix},\qquad \begin{pmatrix}0\\1\\2\end{pmatrix}$$
I am told that to find a generalized eigenvector I need to solve
$$A\vec{x} = \vec\xi$$
where $\vec\xi$ is an eigenvector. However, when I try to solve in this case,
$$\begin{pmatrix}
1 & 2 & -1 \\
-1 & -2 & 1 \\
-1 & -2 & 1
\end{pmatrix}\vec{x} = \begin{pmatrix}1\\0\\1\end{pmatrix}$$
I find that there is no solution, and likewise for the other eigenvector.
I've also seen instruction that to find the generalized eigenvector you can solve $(A-\lambda I)^2\vec x=\vec0$ which in this case is
$$\begin{pmatrix}
1 & 2 & -1 \\
-1 & -2 & 1 \\
-1 & -2 & 1
\end{pmatrix}^2\vec{x} = \vec{0}$$
But the square of this matrix is the zero matrix. I thought perhaps that means I can select any independent vector I want, but when I select $\begin{pmatrix}0\\0\\1\end{pmatrix}$ I find that the resulting transformation matrix does not produce a matrix in Jordan form. That is to say, with
$$T = \begin{pmatrix}
1 & 0 & 0 \\
0 & 1 & 0 \\
1 & 2 & 1 \end{pmatrix}$$
If I compute $J=T^{-1}AT$ I get
$$J = \begin{pmatrix} 1 & 0 & -1 \\
0 & 1 & 1 \\
0 & 0 & 1 \end{pmatrix}$$
Since that's not in Jordan form, I'm not sure what I'm doing wrong.
| Since the only eigenvalue is $1$ and it has multiplicity $3$, you know that the Jordan normal form is one of a short list:
$$\begin{pmatrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix},\hspace{1cm}
\begin{pmatrix}1 & 1 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix},\hspace{0.5cm}\text{or}\hspace{0.5cm}
\begin{pmatrix}1 & 1 & 0 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{pmatrix}.$$
Call these $A_1,A_2$, and $A_3$. As you say, we can determine the size of the Jordan blocks by looking at $\dim\ker(A-I)$, where $I$ is the $3\times 3$ identity matrix. Notice that
$$A-I=\begin{pmatrix}1 & 2 & -1 \\ -1 & 2 & 1 \\ -1 & -2 & 1 \end{pmatrix}$$
from which, by row operations for example, we see $\dim\ker(A-I)=2$. This means that $A$ has two Jordan blocks, so we have ruled out $A_1$. Next, compute $\dim\ker(A-I)^2$. Actually, $(A-I)^2$ is the $0$ matrix, so $\dim\ker(A-I)^2=3$. This tells us that the biggest Jordan block has rank $2$, so the correct answer is $A_2$.
As confirmation, Mathematica returns the Jordan normal form
$$J=\left(
\begin{array}{ccc}
1 & 0 & 0 \\
0 & 1 & 1 \\
0 & 0 & 1 \\
\end{array}
\right)$$
with the conjugating matrix as
$$T=\left(
\begin{array}{ccc}
0 & -1 & 0 \\
1 & 1 & 0 \\
2 & 1 & 1 \\
\end{array}
\right)$$
That is, $T^{-1}AT=J$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3316602",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Derivative for $x^3+ax^2y+bxy^2+y^3=0$ Let $x^3+ax^2y+bxy^2+y^3=0$. Find $y'$ in point $M(0;0)$
My work:
$$3x^2+2axy+ax^2y'+by^2+2bxyy'+3y^2y'=0$$
$$y'=-\frac{3x^2+2axy+by^2}{ax^2+2bxy+3y^2}$$
How find $y'(0;0)$?
| Hint. Let $y=mx$ be the equation of the tangent line at $(0,0)$ where $m=y'(0)$, then $m$ satisfies cubic equation
$$1+am+ bm^2+m^3=0.$$
P.S. Note that the number of tangent line through the origin can be $1$, $2$ or $3$ (it depends on the number of real roots of the cubic). For example if $a=0$ and $b=-2$ then
$$x^3+ax^2y+bxy^2+y^3=(y-x)(y-\frac{1+\sqrt{5}}{2}\, x)(y-\frac{1-\sqrt{5}}{2}\, x)=0$$
and we have three tangents at $(0,0)$ (actually the curve is the union of those three lines)
$$y=x,\quad y=\frac{1+\sqrt{5}}{2}\, x,\quad y=\frac{1-\sqrt{5}}{2}\, x.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3317130",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Find the surface area of part of cone $x^{2} = y^{2} + z^{2}$ which lies inside the sphere $x^{2} + y^{2} + z^{2} = 2z$. Find the surface area of part of cone $x^{2} = y^{2} + z^{2}$ which lies inside the sphere $x^{2} + y^{2} + z^{2} = 2z$
Intersection of both surfaces on $yz$ plane will be $ y^{2} + (z-1/2)^{2} = 1/4$
Surface area $= \int \sqrt{1 + (y/x)^{2} + (z/x)^{2}} dydz$
$= \int (\sqrt{ (x^{2} + y^{2} + z^{2}})/x) dydz$
$= \int ( \sqrt{ 2(y^{2} + z^{2})})/(\sqrt{y^{2} + z^{2}}) dydz$
$= \int \sqrt{2} dydz$
Since area of the circle $ y^{2} + (z-1/2)^{2} = 1/4$ is $ \pi /4$ so the surface area should be $( \sqrt{2} \pi)/4$
But answer is given $ \sqrt{2} \pi$
So where did I go wrong$?$
| The condition $ x^2 = y^2 + z^2 $ implies the existence of two cones
$x = \sqrt {y^2 + z^2}$ and $x = -\sqrt {y^2 + z^2}$ so the result should be doubled, as you only used the first cone.
However, the expected answer still remains different.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3318484",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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If $a$, $b$, $c$ are distinct sides of a triangle and the roots of $x^2+2(a+b+c)x+3m(ab+bc+ca)=0$ are real, show that $m<4/3$ if $m$ is real
Let $a$, $b$, $c$ be the sides of a triangle such that all are distinct. If the roots of the equation $$x^2+ 2(a+b+c)x +3m (ab+bc+ac) =0$$ are real, then show that $m<\frac43$ if $m $ is a real number.
I've been trying to solve this question since a long time but I'm unable to do so. I don't understand how to use the "sides of the triangle condition" except that they show $a$, $b$, $c$ are positive real numbers.
Would someone please help me to prove the required result?
| The roots of the equation$$x^2+ 2(a+b+c)x +3m (ab+bc+ac) =0$$are real, so we have
$$4(a+b+c)^2-4\cdot 3m(ab+bc+ca)\ge 0,$$
i.e.
$$m\le \frac{(a+b+c)^2}{3(ab+bc+ca)}\tag1$$
Let $$x=a+b-c\gt 0,\ \ y=b+c-a\gt 0,\ \ z=c+a-b\gt 0$$
Then, $(1)$ can be written as
$$m\le \frac 43\cdot \frac{(x+y+z)^2}{x^2+y^2+z^2+3xy+3yz+3zx}\tag2$$
Now, we have
$$x^2+y^2+z^2+3xy+3yz+3zx-(x+y+z)^2=xy+yz+zx\gt 0$$
from which we have
$$\frac{(x+y+z)^2}{x^2+y^2+z^2+3xy+3yz+3zx}\lt 1\tag3$$
It follows from $(2)(3)$ that
$$m\lt\frac 43$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3319100",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
$\lim\limits_{n \to \infty} \cos^2\left(\pi \sqrt[3]{n^3+n^2+2n}\right) $, where $n \in \mathbb{N}$ $$
\lim\limits_{n \to \infty} \cos^2\left(\pi \sqrt[3]{n^3+n^2+2n}\right)
$$
where $n \in \mathbb{N}$.
In this question, what I thought was, since $n \to \infty$ and $\cos ^2x$ is periodic , all I need is actually the fractional part of this. And it easy to say that $n+1> \sqrt[3]{n^3+n^2+2n}> n$.
But evaluation of $\sqrt[3]{n^3+n^2+2n} - n$ , is getting tricky. I'm sure there must be a short way to solve it. Can someone help me with it?
| Try with $$\cos(a+b)=\cos(a) \cos(b) - \sin(a) \sin(b) \tag{1}$$
In this case
$$\cos^2\left(\pi \sqrt[3]{n^3+n^2+2n}\right)=\cos^2\left(\pi \sqrt[3]{n^3+n^2+2n}-\pi n+\pi n\right)=\\
\left(\cos\left(\pi \sqrt[3]{n^3+n^2+2n}-\pi n\right) \cdot \color{red}{\cos(\pi n)}-\sin\left(\pi \sqrt[3]{n^3+n^2+2n}-\pi n\right)\cdot \color{red}{\sin(\pi n)}\right)^2=
\left(\cos\left(\pi \sqrt[3]{n^3+n^2+2n}-\pi n\right)\right)^2=\\
\left(\cos\left(\pi \frac{n^2+2n}{\left(\sqrt[3]{n^3+n^2+2n}\right)^2+n\sqrt[3]{n^3+n^2+2n}+n^2}\right)\right)^2$$
from $$a^3-b^3=(a-b)(a^2+ab+b^2) \tag{2}$$
leading to
$$\color{red}{\cos^2\left(\frac{\pi}{3}\right)}$$
because
$$\frac{n^2+2n}{\left(\sqrt[3]{n^3+n^2+2n}\right)^2+n\sqrt[3]{n^3+n^2+2n}+n^2}=\\
\frac{1+\frac{2}{n}}{\left(\sqrt[3]{1+\frac{1}{n}+\frac{2}{n^2}}\right)^2+\sqrt[3]{1+\frac{1}{n}+\frac{2}{n^2}}+1} \to \frac{1}{3}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3321399",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
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How to derive $ \frac ab -x= \frac {c-xd}{b}+\left(\frac{b-d}{b}\right) \left(\frac{a-c}{b-d} - x \right) $ $$ \frac ab -x= \frac {c-xd}{b}+\left(\frac{b-d}{b}\right) \left(\frac{a-c}{b-d} - x \right) $$
It is easy to check it by computing right hand side. It feels unnatural and a little magical. I could't derive it starting from LHS
This identity is used in the proof of Stolz Cesaro theorem (https://ru.wikipedia.org/wiki/Теорема_Штольца). It is in russian I understood with the help of google translate .
| Let
$$
Z=\:\frac{\left(c-xd\right)}{b}+\left(\frac{b-d}{b}\right)\:\left(\frac{a-c}{b-d}\:-\:x\:\right)\:$$
Let
$$ K=\frac{\left(c-xd\right)}{b}$$
Let
$$ N=\left(\frac{b-d}{b}\right)\:\left(\frac{a-c}{b-d}\:-\:x\:\right)\:$$
$$Z=K+N$$
We will rewrite $\left(\frac{a-c}{b-d}\:-\:x\:\right)\:$ as:
$$\left(\frac{a-c}{b-d}\:-\:x\:\right)\:=\left(\frac{a-c}{b-d}\:-\:x\:\frac{\left(b-d\right)}{\left(b-d\right)}\:\right)\:$$
$$N\:=\left(\frac{b-d}{b}\right)\left(\frac{a-c}{b-d}\:-\:x\:\frac{\left(b-d\right)}{\left(b-d\right)}\:\right)\:=\left(\frac{b-d}{b}\right)\left(\frac{a-c\:-x\left(b-d\right)}{b-d}\:\right)$$
$$N\:=\left(\frac{a-c\:-x\left(b-d\right)}{b}\:\right)$$
$$N\:=\left(\frac{a-c\:-xb+xd}{b}\:\right)$$
Using $Z=K+N$ and substituting the last expression for $N$ we get:
$$z\:=\frac{\left(c-xd\right)}{b}+\left(\frac{a-c\:-xb+xd}{b}\:\right) $$
$$z\:=\left(\frac{a\:-xb}{b}\:\right)$$
$$ z\:=\frac{a\:}{b}-x$$
| {
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"url": "https://math.stackexchange.com/questions/3322232",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 2
} |
Rate of change with $a(t)=\frac{1}{t+4}$ on $[9,9+h]$ I am working on an exercise to find the rate of change between points $[9, 9+h]$ with the function $a(t)=\frac{1}{t+4}$.
The solution provided is $\frac{-1}{13(13+h)}$ whereas I arrive at $\frac{\frac{1}{h}}{h}$.
My working:
$a(t_1)$ = $\frac{1}{9+4}$ = $\frac{1}{13}.$
$x(t_2)$ = $\frac{1}{9+h+4}$ = $\frac{1}{13+h}.$
The rate of change is: $\frac{a(t_2)-a(t_1)}{t_2-t_1}.$
So: $\dfrac{\frac{1}{13+h}-\frac{1}{13}}{9+h-9}$ = $\dfrac{\frac{1}{13}+\frac{1}{h}-\frac{1}{13}}{h}$ = $\dfrac{\frac{1}{h}}{h}.$
Where did I go wrong and how can I arrive at $\frac{-1}{13(13+h)}$?
| Because$$\frac1{13+h}\neq\frac1{13}+\frac1h.$$In fact\begin{align}\frac1{13+h}-\frac1{13}&=\frac{13}{(13+h)13}-\frac{13+h}{(13+h)13}\\&=\frac{13-13-h}{(13+h)13}\\&=-\frac h{(13+h)13}.\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3323150",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
$d(x,y) = | x^{2} - y^{2}|$ and $ d(x,y) = | x^{3} - y^{3}|$ are metrics on $\mathbb R$ or not$?$ $d(x,y) = | x^{2} - y^{2}|$ and $ d(x,y) = | x^{3} - y^{3}|$ are metrics on $\mathbb R$ or not$?$
Clearly,
*
*$d(x,x) = 0$
*$d(x,y) = d(y,x)$
*$d(x,y) ≥ 0$ , for all $x,y$ for both.
Now, for fourth axiom.
*$d(x,z) = |x^{2} - z^{2}| = |x^{2} - y^{2} + y^{2} - z^{2}| ≤ |x^{2} - y^{2}| + |y^{2} - z^{2}| = d(x,y) + d(y,z)$
Similarly,
$d(x,z) = |x^{3} - z^{3}| = |x^{3} - y^{3} + y^{3} - z^{3}| ≤ |x^{3} - y^{3}| + |y^{3} - z^{3}| = d(x,y) + d(y,z)$
So both should be metrics. But answer is given that only the second one in metric. So what's wrong with my proof. I think something is wrong with my proof of fourth axiom.
I've just started metric spaces today.
Please suggest..
| Recall the definition of a metric. A function $d: X \to \mathbb{R}$ on a space $X$ is a metric if:
1) $d(x,y) = 0$ if and only if $x=y$.
2) $d(x,y) =d(y,x)\ \forall\ x,y \in X.$
3) $d(x,y) \le d(x,z) + d(y,z) \ \forall \ x,y,z \in X.$
Now we check above conditions on two of our metrics:
Case 1 : $d(x,y) = |x^2-y^2|$.
$(i)\ d(x,y) = 0 \Rightarrow |x^2-y^2| = 0 \Rightarrow x^2-y^2 = 0 \ (\text{as modulus is always} \ge 0) \Rightarrow x^2 =y^2.$
So any $x,y \in \mathbb{R}$ such that $x^2 =y^2$ gives $d(x,y) = 0.$ This contradicts our first condition for a metric. Hence $d(x,y) = |x^2-y^2|$ is not a metric on $\mathbb{R}$.
Case 2 : $d(x,y) = |x^3-y^3|$.
$(i)\ d(x,y) = 0 \Rightarrow |x^3-y^3| = 0 \Rightarrow x^3 - y^3 = 0 \Rightarrow (x-y)(x^2+xy+y^2) = 0 \Rightarrow x = y$ (as second term is never zero under the condition given). The other side is trivial as if $ x= y $ then $d(x,y) = |x^3-y^3| = 0$.
$(ii)\ d(x,y) = |x^3-y^3| = |y^3-x^3| = d(y,x).$
$(iii)\ d(x,y) = |x^3- y^3| = |x^3-z^3+ z^3-y^3| \le |x^3- z^3|+ |z^3-y^3|$ (by the triangular inequality in modulus.)
$\Rightarrow d(x,y) \le d(x,z)+d(z,y). $
Hence $d(x,y) = |x^3-y^3|$ is a metric on $\mathbb{R}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3323377",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Taylor series of $(1+3x) \cdot \ln(1+x)$ I have to find Taylor series of $(1+3x) \cdot \ln(1+x)$. I know Taylor series of $(1+3x) \cdot \ln(1+x)$ but I do not know hot to simplify.
Any help?
| Alternatively, you can calculate derivatives:
$$y=(1+3x) \cdot \ln(1+x) \quad (=0)\\
y'=3\ln (1+x)+\frac{1+3x}{1+x}=3\ln (1+x)+3-\frac2{1+x}\quad (=1)\\
y''=\frac3{1+x}+\frac2{(1+x)^2} \quad (=5)\\
y'''=-\frac3{(1+x)^2}-\frac{2\cdot 2}{(1+x)^3} \quad (=-7)\\
\vdots \\
y^{(n)}=\frac{(-1)^{n}\cdot 3\cdot (n-2)!}{(1+x)^{n-1}}+\frac{(-1)^n\cdot 2\cdot (n-1)!}{(1+x)^n}=\\
\frac{(-1)^{n}(n-2)!(3(x+1)+2(n-1))}{(1+x)^n},n\ge 2 \qquad (=(-1)^{n}(n-2)!(1+2n))$$
Hence:
$$y=\sum_{n=0}^{\infty} \frac{y^{(n)}(0)}{n!}x^n=y(0)+y'(0)x+\sum_{n=2}^{\infty} \frac{y^{(n)}(0)}{n!}x^n=\\
x+\sum_{n=2}^{\infty} \frac{(-1)^n(n-2)!(1+2n)}{n!}x^n=x+\sum_{n=2}^{\infty} \frac{(-1)^n(1+2n)}{n(n-1)}x^n.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3323693",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 4
} |
Integration of $ \int x^{2} \sqrt{2x-6} dx $
$$ \int x^{2} \sqrt{2x-6} dx = ?$$
My Attempt:
by partial integration
$$ \int x^{2} \sqrt{2x-6} dx = \frac{x^{2} (2x-6)^{3/2}}{3}- \frac{2}{3} \int x(2x-6)^{3/2}dx$$
continuing partial integration
$$ = \frac{x^{2} (2x-6)^{3/2}}{3}- \frac{2}{3} \left[ \frac{x(2x-6)^{5/2}}{5} - \int \frac{(2x-6)^{5/2}}{5} dx\right] $$
$$ = (x^{2}/3)(2x-6)^{3/2} - (2x/15)(2x-6)^{5/2} + (2/105)(2x-6)^{7/2} + C$$
Is this the correct and best/simplest answer?
Strangely, the multiple choices only include answers in the form:
$$ A(2x+6)^{7/2} + B(2x+6)^{5/2} + C(2x+6)^{3/2} + D$$
where $A,B,C,D$ are constants.
| Your result is correct as soon as you change a sign (typo?):
the result can be written as
$$A(2x-6)^{7/2} + B(2x-6)^{5/2} + C(2x-6)^{3/2} + D$$
Note that
$$\frac{2x}{15}(2x-6)^{5/2}=\frac{2x-6+6}{15}(2x-6)^{5/2}=\frac{1}{15}(2x-6)^{7/2}+\frac{2}{5}(2x-6)^{5/2}.$$
Similarly, you can write $\frac{x^2}{3}(2x−6)^{3/2}$ as a linear combination of $(2x-6)^{7/2}$, $(2x-6)^{5/2}$, and $(2x-6)^{3/2}$:
$$\frac{x^2}{3}(2x−6)^{3/2}=\frac{(2x-6)^2/4+3(2x-6)+9}{3}(2x−6)^{3/2}
\\=\frac{1}{12}(2x−6)^{7/2}+(2x−6)^{5/2}+3(2x−6)^{3/2}.$$
Hence, your result can be written as
$$\underbrace{\left(\frac{1}{12}-\frac{1}{15}+\frac{2}{105}\right)}_{1/28}(2x-6)^{7/2} + \underbrace{\left(1-\frac{2}{5}\right)}_{3/5}(2x-6)^{5/2} + 3(2x-6)^{3/2} + C.$$
However, the final result in that form can be obtained more easily by applying the substitution $t=\sqrt{2x-6}$. Then $x=\frac{t^2}{2}+3$, $dx=tdt$ and
$$\int x^{2} \sqrt{2x-6} dx =\frac{1}{4}\int (t^2+6)^2 t (t dt)=\int \left(\frac{t^6}{4}+3t^4+9t^2\right) dt\\
=\frac{t^7}{28}+\frac{3t^5}{5}+3t^3+C
=\frac{1}{28}(2x-6)^{7/2} + \frac{3}{5}(2x-6)^{5/2} + 3(2x-6)^{3/2} + C.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3323783",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
How to compute $\sum_{k=1}^{\infty}{(\zeta(2k)-1)}$
How to compute $\sum_{k=1}^{\infty}{(\zeta(2k)-1)}$, where $\zeta(s) :=\sum_{k=1}^{\infty} \frac{1}{n^s}$ with $s>1$.
Here's my process, what am I doing wrong?:
| Using $~\displaystyle\prod\limits_{k=1}^\infty\left(1-\left(\frac{x}{k}\right)^2\right)=\frac{\sin(\pi x)}{\pi x}~$ we have:
$$\sum\limits_{k=1}^\infty\frac{x^{2k}}{2k}\zeta(2k) = -\frac{1}{2}\ln\frac{\sin(\pi x)}{\pi x}~,~~|x|<1$$
Differentiation and then substraction of $~\displaystyle\sum\limits_{k=1}^\infty x^{2k-1} = \frac{x}{1-x^2}~$ and after that the limit for $~x\uparrow 1~$ gives us the result $~\displaystyle\frac{3}{4}~$.
Hint: $~~\displaystyle\lim_{x\to 0}\left(\frac{1}{x}+\pi\cot(\pi(1-x))\right)=0$
Note: $~~$ But of course, using the result of e.g. Jam, it's easier to write
$\displaystyle\sum\limits_{n=2}^\infty\frac{1}{n^2-1} = \frac{1}{2}\sum\limits_{n=2}^\infty \left(\frac{1}{n-1} - \frac{1}{n+1}\right) = \frac{1}{2}\left(1+\frac{1}{2}\right)=\frac{3}{4}\,$ .
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3328325",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
How can calculate $ a \mod n^2 $ be computed smartly if we know $a \mod n$? By example:
How can $ 2^{65536} \mod 7^2 $ be computed smartly?
I think that there is a faster method than writing next element and find cycles. I found that
$\mod 7 :$
$$ \color{red}{ 2^{0} \equiv 1\\2^{1} \equiv 2\\2^{2} \equiv 4}\\2^{3} \equiv 1 $$
and by these equations each number can be fastly computed for example:
$$2^{65536} = 2^{21845\cdot 3 +1 } \equiv 2^1 \equiv 2 $$
but how can it be applied for $\mod 49$?
| Since, as you noticed, $2^3\equiv 1\pmod 7$, let's look at what multiplying by $2^3$ does to a number mod $7^2$. We'll write our congruence class in the form $7a+b$ where $0\leq a, b\lt 7$; then $2^3(7a+b)$ $=(7+1)(7a+b)$ $=7^2a+7a+7b+b$ $\equiv 7(a+b)+b$. By iterating this, we see that $2^{3i}(7a+b)$ $\equiv 7(a+ib)+b$, so $2^{21}(7a+b)$ $\equiv 7(a+7b)+b$ $\equiv 7a+7^2b+b$ $\equiv 7a+b$. In other words, $2^{21}$ here is the identity mod $7^2$ just as $2^3$ was mod $7$ (this is another way of looking at Arthur's answer), so we can remove all the factors of $2^{21}$ from $2^{65536}$ and wind up with $2^{65536}\equiv 2^{16}$ $\equiv 2\cdot2^{15}$ $\equiv 2\cdot (2^3)^5$ $\equiv 2\cdot (5\cdot 7+1)$ $\equiv 3\cdot7+2 \equiv 23$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3331214",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 4
} |
Derive $f(i,j) = \frac{(i+j-2) (i+j-1)}{2} + i$ given values table Given
\begin{array}{ll|lllll}
& & & & j & & \\
& f(i,j) & 1 & 2 & 3 & 4 & 5 \\ \hline
& 1 & 1 & 2 & 4 & 7 & 11 \\
& 2 & 3 & 5 & 8 & 12 & \\
i & 3 & 6 & 9 & 13 & & \\
& 4 & 10 & 14 & & & \dots \\
& 5 & 15 & & & &
\end{array}
How to derive the formula $f(i,j) = \frac{(i+j-2) (i+j-1)}{2} + i$ ?
Not sure if there is any trick other than trial and error.
| Let's change the coordinates from $(i,j)$ to $(i,u)$, where $u$ is the diagonal number.
Number of elements in $k$th diagonal is $k$.
That means total number of elements from $1$st to $u$th diagonal is $\sum\limits_{k=1}^{u} k$.
Then $g(i,u) = i+\sum\limits_{k=1}^{u-1}k$ gives the integer value at row $i$, diagonal $u$.
$
\begin{array}{ll|lllll}
& & & & j & & \\
& f(i,j) & 1 & 2 & 3 & 4 & 5 \\ \hline
& 1 & 1 & 2 & 4 & 7 & 11 \\
& 2 & 3 & 5 & 8 & 12 & \\
i & 3 & 6 & 9 & 13 & & \\
& 4 & 10 & 14 & & & \dots \\
& 5 & 15 & & & &
\end{array}
$
Seeing $u=i+j-1$ mustn't be hard. In first quadrant on xy plane, visualize the straight line $x+y=u+1$. On this line, adding up the $x, y$ coordinates of any point gives the constant $u+1$. Flipping the table horizontally makes it easy to see:
$\begin{array}{ll|lllll}
& 5 & 15 & & & & \\
& 4 & 10 & 14 & & & \dots \\
i & 3 & 6 & 9 & 13 & & \\
& 2 & 3 & 5 & 8 & 12 & \\
& 1 & 1 & 2 & 4 & 7 & 11 \\\hline
& f(i,j) & 1 & 2 & 3 & 4 & 5 \\
& & & & j & & \\
\end{array}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3332002",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
find all complex numbers $z$ such that $\mid 2z/(z^2+1) \mid < 1$ I need to find all complex numbers $z$ such that $\mid 2z/(z^2+1) \mid < 1$. If $z$ were real, it would be easy, but now that $z$ being complex numbers makes things so twisted...I cannot find a way to approach. Could anyone please help me?
| If you substitute $z=x+iy$ into this inequality then square both sides you should get
$$x^4+y^4+2x^2y^2-2x^2-6y^2+1\gt0$$
$$(x^2+(y-1)^2-2)(x^2+(y+1)^2-2)\gt0$$
So we need
$$z\in\{x+iy:x^2+(y-1)^2-2\gt0\land x^2+(y+1)^2-2\gt0\}\cup\{x+iy:x^2+(y-1)^2-2\lt0\land x^2+(y+1)^2-2\lt0\}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3333942",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Solve the equation $x^{x^5} = 5$ over $\Bbb R.$ Find real numbers $x$ (if any) such that $x^{x^5} = 5.$
I have shown that $x \notin \Bbb Z.$ Does there exist any $x \in \Bbb R \setminus \Bbb Z$ which satisfies the above equation? Any help will be highly appreciated.
| \begin{align}
x^{x^5}&=5
\tag{1}\label{1}
\end{align}
Equation \eqref{1} can be solved formally
with the help of the Lambert W function:
\begin{align}
x^5\ln x&=\ln 5
,\\
5x^5\ln x&=5\ln 5
,\\
x^5\ln (x^5)&=5\ln 5
,\\
\ln (x^5)\,\exp(\ln (x^5))&=\ln 5\exp(\ln5)
.
\end{align}
At this point we can apply the Lambert W function:
\begin{align}
\operatorname{W}(\ln (x^5)\,\exp(\ln (x^5)))
&=
\operatorname{W}(\ln 5\exp(\ln5))
,
\end{align}
and since the argument
$\ln 5\exp(\ln5)>0$,
there is just one solution
\begin{align}
\ln (x^5)&=\ln5
,\\
x^5&=5
,\\
x&=5^{\tfrac15}\approx 1.379729661
.
\end{align}
Edit
Or, less formally, without Lambert W function:
\begin{align}
(x^{x^5})^5&=5^5
,\\
x^{5x^5}&=5^5
,\\
(x^5)^{(x^5)}&=(5)^{(5)}
,\\
x^5&=5
.
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3335552",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
Can there be two inequalities? $$x+a+\sqrt{x^2+a^2}>b$$
I can consider this inequality for $a,b,x>0$.
1) $\sqrt{x^2+a^2}>b-(x+a)$
$$x^2+a^2>b^2-2b(x+a)+(x+a)^2$$
$$0>b^2-2b(x+a)+2ax$$
$$x(2b-2a)>b^2-2ba$$
$$x>\frac{b^2-2ba}{(2b-2a)}$$
2) $(x+a)-b>-\sqrt{x^2+a^2}$
$$x^2+a^2<b^2-2b(x+a)+(x+a)^2$$
$$0<b^2-2b(x+a)+2ax$$
$$x(2b-2a)<b^2-2ba$$
$$x<\frac{b^2-2ba}{(2b-2a)}$$
I get two inequalities. What is Im doing wrong?
| You should use this fundamental fact about irrational inequalities:
On its domain, $\;\sqrt{A}>B\iff A>B^2\: \textbf{ or }\: B<0.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3336347",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
Show convergence of $\int_1^\infty{\frac{dx}{\sqrt{x^3-x}}}$ How can I show that $I=\int_1^\infty{\frac{dx}{\sqrt{x^3-x}}}$ converges. I've had no success with finding any bounding function.
I also tried tried trig-substitution so $I=\int_0^{\pi/2}{\frac{dx}{\sqrt{\cos{x}}}}=\int_0^{\pi/2}{\frac{dx}{\sqrt{\sin{x}}}}$. But this did not help me.
| I believe the simplest way is just to enforce the substitution $x\mapsto x+1$ and then break the integral into simpler pieces:
$$\int_{1}^{+\infty}\frac{dx}{\sqrt{x^3-x}} = \int_{0}^{+\infty}\frac{dx}{\sqrt{x^3+3x^2+2x}} \leq \int_{0}^{1}\frac{dx}{\sqrt{2x}}+\int_{1}^{+\infty}\frac{dx}{\sqrt{x^3}} = \sqrt{2}+2.$$
You may also notice that the second term is an elliptic integral:
$$ \int_{0}^{+\infty}\frac{dx}{\sqrt{x(x^2+3x+2)}}\stackrel{x\mapsto x^2}{=}2\int_{0}^{+\infty}\frac{dx}{\sqrt{x^4+3x^2+2}}=2\int_{0}^{+\infty}\frac{dx}{\sqrt{(x^2+a^2)(x^2+b^2)}} $$
and the last term equals $\frac{\pi}{\text{AGM}(a,b)} $ where $a=1$ and $b=\sqrt{2}$, so
$$\boxed{2.60258\ldots= \frac{2\pi}{1+\sqrt{2}}\leq\int_{1}^{+\infty}\frac{dx}{\sqrt{x^3-x}}\leq \frac{\pi}{2^{1/4}}=2.64175\ldots.}$$
By the relations between $K$ (the complete elliptic integral of the first kind) and the $\text{AGM}$ the exact value of the integral is $\frac{\Gamma\left(\frac{1}{4}\right)^2}{2\sqrt{2\pi}}=2.62205755429211981\ldots.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3337215",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 4
} |
Limit development for $f(x)=\frac{x}{\arctan(x^{2})}-\frac{1}{x}$. I'm trying to solve this problem: Find a limit development of order
7 for the function $f(x)=\frac{x}{\arctan(x^{2})}-\frac{1}{x}$. Where
we use the next definition:
Definition. Let $f\colon I\to\mathbb{R}$ be a function and $x_{0}\in I$. We
say that $f$ has a limit development of order $n$ in $x_{0}$ provided
that there exist $a_{0},a_{1},\ldots,a_{n}\in\mathbb{R}$ such that
for $x\in I$
$$
f(x)=a_{0}+a_{1}(x-x_{0})+\ldots+a_{n}(x-x_{0})^{n}+o((x-x_{0})^{n}) \text{ (small $o$)}.
$$
Using Taylor series for $\arctan x$ at 0, we have that $\arctan x=x-\frac{x^{3}}{3}+\frac{x^{5}}{5}-\frac{x^{7}}{7}+\ldots$,
and therefore
\begin{align*}
\arctan(x^{2}) & =x^{2}-\frac{(x^{2})^{3}}{3}+\frac{(x^{2})^{5}}{5}-\frac{(x^{2})^{7}}{7}+\ldots\\
& =x^{2}-\frac{x^{6}}{3}+\frac{x^{10}}{5}-\frac{x^{14}}{7}+\ldots
\end{align*}
So, replacing for $f(x)=\frac{x}{\arctan(x^{2})}-\frac{1}{x}$ I have
that:
\begin{align*}
f(x) & =\frac{x}{x^{2}-\frac{x^{6}}{3}+\frac{x^{10}}{5}-\frac{x^{14}}{7}+\ldots}-\frac{1}{x}\\
& =\frac{x}{x\left(x-\frac{x^{5}}{3}+\frac{x^{9}}{5}-\frac{x^{13}}{7}+\ldots\right)}-\frac{1}{x}\\
& =\frac{1}{x}\cdot\left(1-\frac{x^{4}}{3}+\frac{x^{8}}{5}-\frac{x^{12}}{7}+\ldots\right)^{-1}-\frac{1}{x}
\end{align*}
But I couldn't obtain the form that is necessary for the limit development
because I have that part with the inverse of $1-\frac{x^{4}}{3}+\frac{x^{8}}{5}-\frac{x^{12}}{7}+\ldots$.
Could you help me or give me some suggestion?
Thanks.
| $$f(x)=\dfrac{x^2-\arctan(x^2)}{x^6}\cdot\dfrac{x^6}{x\arctan(x^2)}$$
Now use series expansion for $\arctan(x^2)$ for the first ratio or use the last problem of Are all limits solvable without L'Hôpital Rule or Series Expansion
Can you take up the second?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3337730",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
How to solve the following differential equation $(y^3-2yx^2)dx+(2xy^2-x^3)dy=0.$ I'd love your help with solving this following differential equation: $$(y^3-2yx^2)dx+(2xy^2-x^3)dy=0.$$
I am trying to use the method of finding integrating factor, that is $\frac{1}{Mx+Ny}$. But it is not coming.
Any suggestions?
| $$Mx+Ny=3xy(y^2-x^2)$$
$$\frac{y(y^2-2x^2)dx+x(2y^2-x^2)dy}{3xy(y^2-x^2)}=0$$
$$\frac{(y^2-x^2)(ydx+xdy)}{3xy(y^2-x^2)}+\frac{xy(ydy-xdx)}{3xy(y^2-x^2)}=0$$
$$\frac{log{xy}}{3}+\frac{\log(y^2-x^2)}{6}=\frac{logc}{3}$$
$$xy(y^2-x^2)^{1/2}=c$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3338186",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Evaluate the following integral : $I=\int_{0}^{1}\frac{x^{a-1}}{1+x}\ dx, a>0$? Here is my attempt:
$$\begin{split}\int_{0}^{1}\frac{x^{a-1}}{x+1}\ dx&=\sum_{k=0}^{\infty }(-1)^k\int_{0}^{1}\ x ^{a+k-1}dx\\
&=\sum_{k=0}^{\infty }\frac{(-1)^k}{a+k}\\
&=\sum_{k=0}^{\infty }\left [ \frac{1}{2k+a} -\frac{1}{2k+1+a}\right ]\\
&=\frac{1}{2}\sum_{k=1}^{\infty }\left [ \frac{1}{k+\frac{a-2}{2}}-\frac{1}{k+\frac{a-1}{2}} \right ]\\
&=\frac{1}{2}\left [ \Psi \left (1+\frac{a-1}{2} \right ) -\Psi \left (1+\frac{a-2}{2} \right )\right ]\\
&=\frac{1}{2}\left [ \Psi \left (\frac{a+1}{2} \right )-\Psi \left (\frac{a}{2} \right ) \right ]
\end{split}$$
Can we compute the integral above in different methods?
All approaches are highly appreciated.
| A substitution of $u=x+1$ makes the process much easier using the binomial expansion on the top.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3340359",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
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How to prove that $ \prod_{k=1}^{m} (cos(\frac{(2k-1)\pi}{2m+1} )+1) = \frac{2m+1}{2^{m}} $ $$
\mbox{How to prove that
}\quad\prod_{k = 1}^{m}\left\{%
\cos\left(\left[2k - 1\right]\pi \over 2m+1 \right)+1\right\} =
{2m + 1 \over 2^{m}}\quad {\Large ?}
$$
where $m$ is an integer and $1 < m < 90$.
I try to use the identity that
$\cos\left(2\theta\right) + 1 = 2\cos^{2}\left(\theta\right)$, but I'm stuck with
$\prod_{k = 1}^{m}\cos^{2}\left(\left[2k - 1\right]\pi \over 4m + 2\right)$ part.
Please give me an advice, I appreciate it and thank you for every reply.
| If $\cos(2m+1)x=0,$
$(2m+1)x=(2n+1)\dfrac\pi2$ where $n$ is any integer
$x=\dfrac{(2n+1)\pi}{2(2m+1)}$ where $0\le n\le2m$
Now using this
$\cos(2m+1)x=2^{2m}\cos^{2m+1}x+\cdots+(-1)^m(2m+1)\cos x$
So, the roots of $$2^{2m}c^{2m+1}+\cdots+(-1)^m(2m+1)c=0$$ are $\cos\dfrac{2(2n+1)\pi}{(2m+1)},0\le n\le2m$
If $\cos x=0,n=m$
So, the roots of $$2^{2m}c^{2m}+\cdots+(-1)^m(2m+1)=0$$ are $\cos\dfrac{(2n+1)\pi}{2(2m+1)},0\le n\le2m, m\ne n$
$\implies2^{2m}\prod_{n=0,n\ne m}^{2m}\cos\dfrac{(2n+1)\pi}{2(2m+1)}=(-1)^m(2m+1)$
As $\pi-\dfrac{(2n+1)\pi}{2(2m+1)}=\dfrac{\pi(1+2(2m-n))}{2(2m+1)}$
$\implies2^{2m}\prod_{n=0}^{m-1}(-1)^m\cos^2\dfrac{(2n+1)\pi}{2(2m+1)}=(-1)^m(2m+1)$
Can you take it from here?
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove that $1^2 + 2^2 + {...}+ {(n - 1)}^2 < \frac{n^3}{3} < 1^2 + 2^2 + {...}+ {n}^2 $ Prove that $1^2 + 2^2 + {...}+ {(n - 1)}^2 < \frac{n^3}{3} < 1^2 + 2^2 + {...}+ {n}^2 $
I know I need to use induction for this proof, but it feels like a pretty complicated one.
Basis: For $n = 2$,
$$1^2 < \frac{8}3 < 1^2 + 2^2$$
Induction Hypothesis: Assume $P(n)$ holds for $n=k$, that is,
$$1^2 + 2^2 + {...}+ {(k - 1)}^2 < \frac{k^3}{3} < 1^2 + 2^2 + {...}+ {k}^2$$
We need to show that $P(n)$ also holds for $n=k+1$
Proof:
$$1^2+2^2+{...}+{(k)}^2=1^2+2^2+{...}+{(k-1)}^2+{k}^2$$
After this, I'm not sure how to use the assumed inequality to prove it because it's a less than inequality. If I could get a hint that'd be awesome.
| HINT: Since we know that $1^2 + 2^2 + {...}+ {(k - 1)}^2 < \frac{k^3}{3} < 1^2 + 2^2 + {...}+ {k}^2$ by induction hypothesis, what we need to prove is that
$$k^2 \le \frac{3k^2+3k+1}{3} \le (k+1)^2$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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} |
Find $k^{th}$ power of a square matrix I am trying to find the $A^{k}$, for all $k \geq 2$ of a matrix, \begin{pmatrix} a & b \\ 0 & 1 \end{pmatrix}
My approach:
$A^{2}=\begin{pmatrix} a^2 & ab+b \\ 0 & 1 \end{pmatrix}$
$A^{3}=\begin{pmatrix} a^3 & a^{2}b+ab+b \\ 0 & 1 \end{pmatrix}$
$A^{4}=\begin{pmatrix} a^4 & a^{3}b+a^{2}b+ab+b \\ 0 & 1 \end{pmatrix}$
$A^{5}=\begin{pmatrix} a^5 & a^{4}b+a^{3}b+a^{2}b+ab+b \\ 0 & 1 \end{pmatrix}$
Continuing this way, we obtain
$A^{k}=\begin{pmatrix} a^k & (a^{k-2}+a^{k-3}+a^{k-4}+.....+1)b \\ 0 & 1 \end{pmatrix}$
I am stuck here! I was wondering if you could give me some hints to move further. I appreciate your time.
| Hint: Write $A=D+B$ here $D$ is diagonal. Use that $B^2=0$, $DB=aB$, $BD=B$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3345771",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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"answer_id": 0
} |
Evaluating integral delta function I am trying to evaluate the integral below (delta function) and not sure if I evaluated correctly? The integral is the following:
$$\int^{100}_{-100}x^3\sin(2x)\delta(x^2-9)dx$$
I have the following $x=\pm 3$, $$\therefore \int^{100}_{-100}x^3\sin(2x)\delta(x^2-9)dx = \int^{0}_{-100}x^3\sin(2x)\delta(x^2-9)dx + \int^{100}_{0}x^3\sin(2x)\delta(x^2-9)dx= \ (-3)^3\sin(2\cdot(-3)) + (3)^3\sin(2\cdot(3)) = \ -27\sin(-6) + 27\sin(6)$$
| Start from from property of delta function:
$\delta(x^2-3^2)=\frac{1}{2*3}[\delta(x+3)+\delta(x-3)]$
Your function is even, $f(x)=f(-x)$ due to symmetry you can write
$\int_{-100}^{100} x^3 \sin(2x) \delta(x^2-9)dx = 2\int_{0}^{100} x^3 \sin(2x) \delta(x^2-9)dx = 2\frac{1}{2*3}\left(\int_{0}^{100} x^3 \sin(2x)\delta(x+3)dx+\int_{0}^{100} x^3 \sin(2x)\delta(x-3)dx\right)=2\frac{1}{2*3}\left(0+27\sin(6)\right)=9\sin(6)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3348897",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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What criterion of convergence should I use here? I'm trying to check if the series $$\sum_{n=1}^{\infty} \sqrt[3]{n^3+n}-n$$ does converge and I don't know which criterion should I use. D'Alembert doesn't help, what's more I can't limit this series with a sense to use comparative criterion. The necessary condition is fullfiled, but it is unsufficient. Thanks for help :)
| From $a^3-b^3=(a-b)(a^2+ab+b^2)$ you get
$$
\sqrt[3]{a} - \sqrt[3]{b} = \frac{{a - b}}
{{\sqrt[3]{{a^2 }} + \sqrt[3]{{ab}} + \sqrt[3]{{b^2 }}}}
$$
Therefore
$$
\begin{gathered}
\sqrt[3]{{n^3 + n}} - \sqrt[3]{{n^3 }} = \frac{{n^3 + n - n^3 }}
{{\sqrt[3]{{\left( {n^3 + n} \right)^2 }} + \sqrt[3]{{n^3 \left( {n^3 + n} \right)}} + \sqrt[3]{{n^6 }}}} = \hfill \\
\hfill \\
= \frac{n}
{{\sqrt[3]{{\left( {n^3 + n} \right)^2 }} + \sqrt[3]{{n^3 \left( {n^3 + n} \right)}} + \sqrt[3]{{n^6 }}}} \geqslant \hfill \\
\hfill \\
\geqslant \frac{n}
{{\sqrt[3]{{\left( {n^3 + n^3 } \right)^2 }} + \sqrt[3]{{n^3 \left( {n^3 + n^3 } \right)}} + \sqrt[3]{{n^6 }}}} = \hfill \\
\hfill \\
= \frac{n}
{{\sqrt[3]{{4n^6 }} + \sqrt[3]{{2n^3 }} + \sqrt[3]{{n^6 }}}} > \frac{n}
{{3n^2 + 3n^2 + 3n^2 }} = \frac{1}
{{9n}} \hfill \\
\end{gathered}
$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 2
} |
Sum of increasing geometric series Sum of the series
$1 + \frac{1+3}{2!}+ \frac{1+3+3^2}{3!}+....... $
The series becomes $ \sum_{k=1}^{\infty} \frac {3^{k-1}}{k!}$.
How to calculate it's sum? Is it divergent due to the numerator?
| We have series of type $\sum_{n = 1}^\infty a_n$, where $a_n = \frac{\sum_{j=1}^n 3^{j-1}}{n!} = \frac{1}{3} \frac{\sum_{j=1}^n 3^j}{n!}$
Since $\sum_{j=1}^n 3^j = 3 \cdot \frac{1 - 3^n}{1-3} = \frac{3}{2}(3^n - 1)$
So, we get $a_n = \frac{1}{2} (\frac{3^n}{n!} - \frac{1}{n!})$
And our series look like:
$\frac{1}{2} \sum_{n=1}^\infty ( \frac{3^n}{n!} - \frac{1}{n!})$
Since both $\sum_{n=1}^\infty \frac{3^n}{n!}$ and $\sum_{n=1}^\infty \frac{1}{n!} $ are convergent ( u can use D'alambert test), we can split our infinite sum into $2$ sums, getting:
$\sum_{n=1}^\infty a_n = \frac{1}{2} \sum_{n=1}^\infty \frac{3^n}{n!} - \frac{1}{2} \sum_{n=1}^\infty \frac{1}{n!}$
Now, we should recall that $e^x = \sum_{n=0}^\infty \frac{x^n}{n!}$, so we have:
$\sum_{n=1}^\infty \frac{3^n}{n!} = e^3 - 1$ and similarly $\sum_{n=1}^\infty \frac{1}{n!} = e -1$. We finally get:
$\sum_{n=1}^\infty a_n = \frac{1}{2}(e^3 - 1 - e+ 1) = \frac{e}{2}(e^2-1)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3357453",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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limit superior for given recurrence for sequence Sequence ${a_n}$ is defined as follows:
$a_1 = 1$ and for $n \geq 1,\, a_{n+1} = (-1)^n \left( \frac {1}{2} \right) \left(|a_n|+ \frac{2}{|a_n|}\right).$ Then find $\lim \sup a_n$ and $\lim \inf a_n .$
How to consider subsequences here? And how to obtain a simplified recurrence? The sequence has alternate positive and negative terms. Please help.
| An induction shows that $|a_{n+1}| = \left( \frac {1}{2} \right) \left(|a_n|+ \frac{2}{|a_n|}\right)\le \left( \frac {1}{2} \right) \left(2+ \frac{2}{1}\right)=2$ and $|a_{n+1}| = \left( \frac {1}{2} \right) \left(|a_n|+ \frac{2}{|a_n|}\right)\ge \left( \frac {1}{2} \right) \left(1+ \frac{2}{2}\right)=1$ so $1\le |a_n|\le 2.$
Thus, $\limsup a_n=L$ and $\liminf a_n=l$ are real numbers.
Now, $(a_n)$ is alternating, the even terms are negative and the odd terms are positive. Therefore,
$L=\limsup a_n=\limsup a_{2n+1}=\limsup \left(\frac{|a_{2n}|}{2}+\frac{1}{|a_{2n}|}\right )=$
$\limsup \left(\frac{-a_{2n}}{2}+\frac{1}{-a_{2n}}\right )=-\liminf \left(\frac{a_{2n}}{2}+\frac{1}{a_{2n}}\right ).$
But $\liminf a_n=\liminf a_{2n}$ so the above becomes $-L=\frac{l}{2}+\frac{1}{l}.$ Similarly, $-l=\frac{L}{2}+\frac{1}{L}.$ Solving these equations gives $L=-l=\sqrt 2.$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Determine $\left(1 - \frac{x_{1}}{y_{1}} \right)\left(1 - \frac{x_{2}}{y_{2}} \right)\left(1 - \frac{x_{3}}{y_{3}} \right)$ $(x_{i},y_{i}), i=1,2,3$ are solutions for
$$ x^{3} - 3xy^{2} = 2010 $$
$$ y^{3} - 3yx^{2} = 2009 $$
What is $\left(1 - \frac{x_{1}}{y_{1}} \right)\left(1 - \frac{x_{2}}{y_{2}} \right)\left(1 - \frac{x_{3}}{y_{3}} \right)$?
Attempt:
My approach is that we should look for $\frac{y-x}{y}=$something.
Notice that
$$ x^{3} - y^{3} + 3(yx^{2} - xy^{2}) = 1 $$
$$ (x-y)(x^{2} + y^{2} + xy) + 3xy(x-y) = 1 $$
$$ (y-x)(x^{2} + y^{2} + 4xy) = -1$$
$$ \frac{y-x}{y} = -\frac{1}{y(x^{2} + y^{2} + 4xy)} (\text{does not seemt to go anywhere})$$
Anther approach if I sum the equations:
$$ x^{3} + y^{3} - 3(xy^{2} + yx^{2}) = 4019 $$
$$ (x + y)(x^{2} + y^{2} - xy) - 3(xy)(x + y) = 4019 $$
$$ (x+y)(x^{2} + y^{2} - 4xy) = 4019 $$
Multiply both we get
$$ (x^{2}-y^{2})((x^{2}+y^{2})^{2} -16 (xy)^{2}) = 4019 $$
???
| We have $xy\ne0$
Let $x=my$
$$\dfrac{2010}{2009}=\dfrac{m^3y^3-3(my)y^2}{y^3-3(my)^2y}=\dfrac{m^3-3m}{1-3m^2}$$
Now set $1-m=n\iff m=1-n$
Replace $m$ with $1-n$ to form a cubic equation in $n$
Now use Vieta's formula
| {
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"timestamp": "2023-03-29T00:00:00",
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Find a plane which is orthogonal to P1 and P2 and state its general equation I'm having a bit of trouble approaching this problem
Let P1 and P2 be planes with the general equations
P1 = -2x + y - 4z =2
P2 = x + 2y = 7
I understand that the unit vectors are (-2, 1, -4) & (1, 2, 0)
And I know that the unit vector of the general equation of P3 must dot product with the 2 above unit vectors to give 0.
Where do I go from the information I have?
| You don't have to use the cross product if you're not comfortable with it. You're looking for a normal vector, and as you say, it must dot product with the two given vectors to produce $0$. That is, if $(a, b, c)$ is our normal vector, then
\begin{align*}
0 &= (-2, 1, 4) \cdot (a, b, c) = -2a + b + 4c \\
0 &= (1, 2, 0) \cdot (a, b, c) = a + 2b.
\end{align*}
We can solve this system of linear equations. Note that these equations are homogeneous, meaning that $(a, b, c) = (0, 0, 0)$ is definitely a solution (though useless for our purposes). Also, there are only two equations and three variables, so we can expect infinitely many solutions to the system. This should make sense, as we can scale any non-zero normal vector as we like, and still get a normal vector.
As an augmented matrix, we get
\begin{align*}\left(\begin{array}{ccc|c}-2 & 1 & 4 & 0 \\ 1 & 2 & 0 & 0\end{array}\right) &\sim \left(\begin{array}{ccc|c}1 & 2 & 0 & 0 \\ -2 & 1 & 4 & 0\end{array}\right) \\
&\sim \left(\begin{array}{ccc|c}1 & 2 & 0 & 0 \\ 0 & 5 & 4 & 0\end{array}\right) \\
&\sim \left(\begin{array}{ccc|c}1 & 2 & 0 & 0 \\ 0 & 1 & 4/5 & 0\end{array}\right).
\end{align*}
Note that the third column has no pivot, so $c$ shall be our free variable. The second row implies
$$b + \frac{4}{5}c = 0 \implies b = -\frac{4}{5}c.$$
The first row implies
$$a + 2b = 0 \implies a = -2b = \frac{8}{5}c.$$
Thus, the general solution is
$$(a, b, c) = \left(\frac{8}{5}c, -\frac{4}{5}c, c\right) = \frac{c}{5}(8, -4, 1).$$
We can choose any $c \neq 0$ to obtain a valid normal direction. I suggest choosing $c = 5$, so that you get all integers (for the sake of tidiness). This gives us the normal direction $(8, -4, 5)$. So, any plane of the form
$$(8, -4, 5) \cdot (x, y, z) = ?$$
will do the trick, e.g.
$$8x - 4y + 5z = 0.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3361842",
"timestamp": "2023-03-29T00:00:00",
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Prove that the sequence $C_{n+1} = 1 + \frac{C_n}{C_n + 1}$ is monotonically increasing by induction I have a sequence
$$C_{n+1} = 1 + \frac{C_n}{C_n + 1}$$
With a base case $C_1 = 3/2$ and want to prove that it's monotonically increasing by induction.
Whenever I try to prove $C_n < C_{n+1} \implies C_{n+1} < C_{n+2}$
where $$C_{n+2} = 1 + \frac{2C_n + 1}{3C_n + 2} $$
I get to the inequality $$ \frac{C_n}{C_n + 1} < \frac{2C_n + 1}{(C_n + 1)^2} $$
but don't know how to get to the point where that implies that $C_{n+1} < C_{n+2}$
Thank
| As $C_1=\frac 32$, it is clear that every consecutive term $C_2, C_3, \dots, C_n$ is positive as $C_{n+1}$ is the sum of two positive terms for all $n>0$. We have
\begin{align}C_{n+2}-C_{n+1}&=\Big(1 + \frac{C_{n+1}}{C_{n+1} + 1}\Big) - \Big(1- \frac{C_n}{C_n + 1}\Big)\\&=
\Big(1 + \frac{1 + \frac{C_n}{C_n + 1}}{2 + \frac{C_n}{C_n + 1}}\Big) - \Big(1- \frac{C_n}{C_n + 1}\Big)\\&=
\Big(1 + \frac{2C_n+1}{3C_n+2}\Big) - \Big(1- \frac{C_n}{C_n + 1}\Big)\\&=
\frac{2C_n+1}{3C_n+2}+\frac{C_n}{C_n + 1}\\&>0
\end{align}
which shows that $C_{n+2}>C_{n+1}$.
| {
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Find equations of all parabolas passing through points (0,3), (2,0) and tangential to $x+y=0$ I need to find equations of all parabolas, that pass through points $(0,3)$ and $(2,0)$ on affine plane $\mathbb{R}^{2}$ and tangent to line $x+y=0$ at origin.
I am trying to solve this problem using conformal geometry. Any hints?
| Parabola is a curve that has equal distance from given point $A$ and line $L$. Since parabola pass the origin, so that line can't, so we can suppose $L: ax+by=1$ and $A: (c,d)$. Now write the equation explicitly:
$$\frac{(ax_0+by_0-1)^2}{a^2+b^2}=(x_0-c)^2+(y_0-d)^2$$
We have $3$ points:
$$\left\{\begin{array}{c} \frac{(3b-1)^2}{a^2+b^2}=c^2+(d-3)^2 \\ \frac{(2a-1)^2}{a^2+b^2}=(c-2)^2+d^2 \\ \frac{(-1)^2}{a^2+b^2}=c^2+d^2 \\ \end{array}\right\}$$
It tangent to $x+y=0$ at origin:
$$-\frac{2(a^2+b^2)(x_0-c)-2a(ax_0+by_0-1)}{2(a^2+b^2)(y_0-d)-2b(ax_0+by_0-1)}(x_0=0,y_0=0)=-1 \Longrightarrow$$
$$\Longrightarrow (a^2+b^2)(d-c)=(b-a)$$
So we get $4$ equation of degree $2$ on $4$ variables that you can solve it on:
https://www.wolframalpha.com/widgets/view.jsp?id=ae438682ce61743f90d4693c497621b7
And fine $2$ possible parabola.
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "3",
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Prove that $\tan\frac{\pi}{9}+ 4\sin\frac{\pi}{9}= \sqrt{3}$ .
Prove that $$\tan\frac{\pi}{9}+ 4\sin\frac{\pi}{9}= \sqrt{3}$$
I think the best solution here is using right triangle . . . I have one too, but not pretty.
| $$(\tan x+ 4\sin x)^{2}- 3= - \frac{1}{4}(2\cos x+ 1)(2\cos 3x- 1)\csc^{2}\left ( \frac{\pi}{4}- \frac{x}{2} \right )\csc^{2}\left ( \frac{x}{2}+ \frac{\pi}{4} \right )$$
Therefore
$$\tan\frac{\pi}{9}+ 4\sin\frac{\pi}{9}= \sqrt{3}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "3",
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Integrating $\int_0^\pi x^4\cos(nx)\,dx$ using the Feynman trick I should solve the following integral
$$\displaystyle\int_0^\pi x^4\cos(nx)\,dx$$
Usually you would integrate 4 times by parts. I was wondering if there is a more direct way, something like the Leibniz rule (aka Feynman trick).
| Take $\int$ as operator D^(-1)
(1/D)e^(ax)x^4=e^(ax)(1/(D+a))x^4=e^(ax)(1/a)(1+D/a)^-1 {x^4}
=(1/a)e^(ax){1-D/a+D^2 /a^2 +D^3/a^3-d^4/a^4}x^4
We can take a as $in$ and proceed to obtain the real part of the expression.
[\begin{gathered}
\int {{e^{ax}}} {x^n}dx = \frac{1}{D}\left\{ {{e^{ax}}.{x^4}} \right\} = {e^{ax}}\frac{1}{{D + a}}\left\{ {{x^4}} \right\} \hfill \\
= \frac{{{e^{ax}}}}{a}\frac{1}{{\left( {1 + \frac{D}{a}} \right)}}\left\{ {x4} \right\} = \frac{{{e^{ax}}}}{a}{\left( {1 + \frac{D}{a}} \right)^{ - 1}}\left\{ {{x^4}} \right\} \hfill \\
= \frac{{{e^{ax}}}}{a}\left( {1 - \left( {\frac{D}{a}} \right) + {{\left( {\frac{D}{a}} \right)}^2} - {{\left( {\frac{D}{a}} \right)}^3} + {{\left( {\frac{D}{a}} \right)}^4} - \cdots } \right)\left\{ {{x^4}} \right\} \hfill \\
= \frac{{{e^{ax}}}}{a}\left\{ {{x^4} - \frac{{4{x^3}}}{a} + \frac{{4.3{x^2}}}{{{a^2}}} - \frac{{4.3.2x}}{{{a^3}}} + \frac{{4.3.2.1}}{{{a^4}}}} \right\} \hfill \\
Let\,a = j.n;and\,obtain\,the\,real\,part\,of\,the\,result. \hfill \\
\end{gathered} ]
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3372041",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
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Evaluate $\int \cos^2x\sin^4x\mathrm{d}x$
Evaluate integral $$\int \cos^2x\sin^4x\mathrm{d}x.$$
Attempt. Setting $\tan x=t$, gives:
$$\int \cos^2x\sin^4x\mathrm{d}x =\int \frac{1}{1+t^2} \,\left(\frac{t^2}{1+t^2}\right)^2 \frac{\mathrm{d}t}{1+t^2}=\int \frac{t^4}{(1+t^2)^4} \mathrm{d}t,$$
which does not seem to be elementary.
Thank in advance for the help.
| Alternative solution:
$$\int \cos^2x\sin^4x\mathrm{d}x=\\
\int(\frac{1+\cos (2x)}{2})(\frac{1+\cos (2x)}{2})^2\mathrm{d}x=\\\frac{1}{8}
\int (1+\cos (2x))(1-2\cos (2x)+\cos^2(2x))\mathrm{d}x=
\\\frac{1}{8}
\int (1+\cos (2x))(1-2\cos (2x)+\frac{1+\cos(4x)}{2})\mathrm{d}x=$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3372371",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Why does this trick for solving this equation work? The question is to
solve the equation $$\frac{4x}{\left ( 2x-2 \right )^{2}+3} + \frac{12x}{\left ( 4x-5 \right )^{2}+3} = 1.$$
Now if we solve $ \left | 2x-2 \right | = \left | 4x-5 \right |$ first (that is, setting the denominators equal) we find that $x=\frac{3}{2},\frac{7}{6}.$ If we now go back to substitute these values of $x$ in LHS of the original equation, we get that $$\frac{ 4x\left ( \left ( 4x-5 \right )^{2}+3 \right ) + 12x\left ( \left ( 2x-2 \right )^{2}+3 \right ) }{\left ( \left ( 4x-5 \right )^{2}+3 \right )\left ( \left ( 2x-2 \right )^{2}+3 \right )} = 6.$$
Thus, we see that $x=\frac{3}{2},\frac{7}{6}$ both satisfy the equation $$\frac{4x}{\left ( 2x-2 \right )^{2}+3} + \frac{12x}{\left ( 4x-5 \right )^{2}+3} = 6.$$
However, I don't know how this is relevant to the fact that $x=\frac{1}{2},\frac{7}{2}$ are solutions to the original equation $$\frac{4x}{\left ( 2x-2 \right )^{2}+3} + \frac{12x}{\left ( 4x-5 \right )^{2}+3} = 1.$$
That is, what is the relationship between the respective solutions of $\text{original LHS}=6$ and $\text{original LHS}=1$?
Please tell me the mystery behind this process, and can we use this technique on other equations like this? Thank you.
| As the LHS is not a homogeneous function, there is no particular relation between the solutions of
$$f(x)=1$$ and $$f(x)=6.$$
By the way, there is no clear relation between $(\frac32,\frac76)$ and $(\frac12,\frac72)$ besides the coincidental equality of the products.
I doubt that this pseudo-property generalizes.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Non-calculus ways to get higher-order terms in $(1 + k/n)^n$? There's a well-known limit $$\lim_{n \to \infty} \left( 1 + \frac{k}{n} \right)^n = e^k $$ that can be proved either by computing limits of individual terms in the binomial expansion, or by defining $$f(x) = (1 + kx)^{1/x}$$ and using calculus to compute $$\lim_{x \to 0} \ln f(x) = \lim_{x \to 0} \frac{\ln (1+kx)}{x} = \lim_{x \to 0} \frac{k}{1+kx} = k.$$
I was working on a problem for which the rate of convergence of this limit is relevant: computing the expected size of the image of a random function from $\{1, \ldots, n\}$ to itself. The chance that any particular number is included in the image is $1 - \left(\frac{n-1}{n}\right)^n$, so the expected size, by linearity of expectation, is $n - n\left( \frac{n-1}{n} \right)^n = \left( 1 - \frac{1}{e} \right) n + O(1).$ WolframAlpha showed that the error term was a constant $\frac{2}{e} + O(n^{-1})$, and I proved this myself by developing $f$ as a power series:
\begin{align*}
\ln f(x) &= \frac{\ln (1+kx)}{x} \\
\frac{d}{dx} \ln f(x) &= -\frac{\ln (1+kx)}{x^2}+ \frac{k}{x (1+kx)} \\
f'(x) &= f(x) \frac{d}{dx} \ln f(x) \\
&= (1 + kx)^{1/x} \left[-\frac{\ln (1+kx)}{x^2} + \frac{k}{x (1+kx)}\right] \\
\lim_{x \to 0} f(x) &= e^k \lim_{x \to 0} \left[-\frac{\ln (1+kx)}{x^2} + \frac{k}{x (1+kx)}\right] \\
&= e^k \lim_{x \to 0} \frac{-(1+kx) \ln (1+kx) + kx}{x^2 (1+kx)} \\
&= e^k \lim_{x \to 0} \frac{-k \ln (1+kx) }{3kx^2 + 2x} \\
&= e^k \lim_{x \to 0} \frac{-k^2}{(6kx+2) (1+kx)} \\
&= \frac{-e^k k^2}{2}
\end{align*}
and therefore $$\left(1 + \frac{k}{n}\right)^n = e^k - \frac{e^k k^2}{2} \frac{1}{n} + O\left( \frac{1}{n^2} \right).$$
(This procedure can be continued, cumbersomely, to get higher-order terms, each of which is a $e^k$ times a polynomial in $k$ with rational coefficients that don't seem to match any OEIS sequence.) I'd like to know, though: Is there any way of getting the $-e^k k^2/2n$ term purely through series manipulation, without resorting to calculus? I tried finding a way for a bit, but the computations quickly get unpleasant.
| Consider
$$a_n=\left(1 + \frac{k}{n}\right)^n \implies \log(a_n)=n \log\left(1 + \frac{k}{n}\right)$$ Using Taylor for large values of $n$
$$\log(a_n)=k-\frac{k^2}{2 n}+\frac{k^3}{3 n^2}+O\left(\frac{1}{n^3}\right)$$ Continue with Taylor
$$a_n=e^{\log(a_n)}=e^k\left(1-\frac{k^2}{2 n}+\frac{k^3 (3 k+8)}{24
n^2}+O\left(\frac{1}{n^3}\right)\right)$$
| {
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Best game strategy in draws from a bin A bin has 2 white balls and 3 black balls. You play a game as follows: you draw balls one at a time without replacement. Every time you draw a white ball , you win a dollar, but every time you draw a black ball , you loose a dollar . You can stop the game at any time.Devise a strategy for playing this game which results in an expected profit.
According to my reasoning the best strategy is not to play the game at all : the expected value at every extraction remains the same and it's always negative $$E[X]=\left(\frac{2}{5}\right)*1 +\left(\frac{3}{5}\right)*(-1) =-\frac{1}{5}$$ So since I can treat it as a sum of expectations of random variables ,the best strategy is not to play…so the best expected profit is zero dollars right?
| There may be a more elegant and/or general way to do this, but here's a brute-force approach.
Consider the following strategy: Draw balls until you've drawn more white balls than black, or if that's no longer possible, until you've drawn both white balls.
Under this strategy, only the following outcomes are possible:
$\begin{align}\\[1ex]
\text{W:} \quad & \text{win \$1} && p=\dfrac{2}{5}\\[1ex]
\text{BWW:} \quad & \text{win \$1} && p= \dfrac{3}{5}\cdot\dfrac{1}{2}\cdot\dfrac{1}{3}=\dfrac{1}{10}\\[6ex]
\text{BWBW:} \quad & \text{break even} && p= \dfrac{3}{5}\cdot\dfrac{1}{2}\cdot\dfrac{2}{3}\cdot\dfrac{1}{2}=\dfrac{1}{10}\\[1ex]
\text{BBWW:} \quad & \text{break even} && p= \dfrac{3}{5}\cdot\dfrac{1}{2}\cdot\dfrac{2}{3}\cdot\dfrac{1}{2}=\dfrac{1}{10}\\[6ex]
\text{BWBBW:} \quad & \text{lose \$1} && p= \dfrac{3}{5}\cdot\dfrac{1}{2}\cdot\dfrac{2}{3}\cdot\dfrac{1}{2}\cdot1=\dfrac{1}{10}\\[1ex]
\text{BBWBW:} \quad & \text{lose \$1} && p= \dfrac{3}{5}\cdot\dfrac{1}{2}\cdot\dfrac{2}{3}\cdot\dfrac{1}{2}\cdot1=\dfrac{1}{10}\\[1ex]
\text{BBBWW:} \quad & \text{lose \$1} && p= \dfrac{3}{5}\cdot\dfrac{1}{2}\cdot\dfrac{1}{3}\cdot1\cdot1=\dfrac{1}{10}\\[1ex]
\\
\end{align}$
Then your expected winnings are
$\left( \dfrac{2}{5} + \dfrac{1}{10} \right) * (1) + \left( \dfrac{1}{10} + \dfrac{1}{10} + \dfrac{1}{10}\right) * (-1) = \boxed{\dfrac{1}{5}}$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Convert $31213111332_{4}$ to hexadecimal I did
$$31213111332_{4} = \\
2*4^0+3*4^1+3*4^2+1*4^3+1*4^4+1*4^5+3*4^6+1*4^7+2*4^8+1*4^9+3*4^{10} = \\
2*16^0+3*4^1+3*16^1+1*4*16^1+1*16^2+1*4*16^2+3*16^3+1*4*16^3+2*16^4+1*4*16^4+3*16^5 = \\
2*16^0 + 3*4^1+7*16^1+5*16^2+7*16^3+6*16^4+3*16^5 = \\
36757E_{16}$$
Which is correct but I have a question. What if $2*16^0+3*4^1$ was greater than 15 (F) ?
| Don’t worry. You would have at most $3*4^0+3*4^1=15\;(F)$, since base $4$ digits are at most $3.$
| {
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Number of ways of arranging $5$ distinct rings on $5$ fingers As a task after Combinatorics, I was asked to calculate the number of ways of arranging $5$ distinct rings on $5$ fingers (counting the thumb as a finger). The emphasis on arranging is indicating that the order of which ring was put on in what order matters for a particular finger. Note that all $5$ rings are to be used in the arrangement.
Here's what I attempted. I counted few cases like those of putting on all $5$ in $1$ finger, all $5$ rings in selected $2$ fingers and so on.
$$\text{ways}=\underbrace{{5\choose 1 }5!}_{\text{all 5 in 1}}+\underbrace{\left[{5\choose 2}2!\left({5\choose 1}4!+{5\choose 2}2!3!\right)\right]}_{\text{all 5 in 2}}+\underbrace{\left[{5\choose 3}\left\{{5\choose 1}{4\choose 1}3!+{5\choose 1}{4\choose 2}2!2!+{5\choose 2}{3\choose 1}2!2!+{5\choose 2}{3\choose 2}2!2!+{5\choose 3}{2\choose 1}3!+{5\choose 1}{4\choose 3}3!\right\}\right]}_{\text{all 5 in 3}}+\underbrace{\left[{5\choose 4}\left\{{5\choose 1}{4\choose 1}{3\choose 1}2!+{5\choose 1}{4\choose 1}{3\choose 2}2!+{5\choose 1}{4\choose 2}{2\choose 1}2!+{5\choose 2}{3\choose 1}{2\choose 1}2!\right\}\right]}_{\text{all 5 in 4}}+\underbrace{5!}_{\text{all 5 in 5 }}$$
All this evaluates to $15120$. Could you check my work? Also suggestions to shorten the approach are welcome. Thanks
Edit $1$: I've corrected the mistake I was making. The answer now matches with the expected answer. Thanks @N. F. Taussig for taking the time to explain what the discrepancy was.
Edit $2$: I also came to know about another method. So for the first ring say $\text{R}_1$ there are $5$ places, for the second ring there's $6$ places, the $4$ remaining fingers and for the case of that finger being the same on which we put the first ring on there's $2$ options, either above it or below it, and since the job of placing the second ring would be said to be completed in either way the addition principle applies, thus amounting to $6$ possible places. So forth and so on till we've accounted for all the rings. This gives us the number of ways as $5\times 6\times 7\times 8\times 9=9!/4!={9\choose 4}\times 5!$.
| Method 1: We choose how many rings to place on each of the five fingers, then multiply by the $5!$ of arranging the rings from bottom left to top right. Let $x_i$ be the number of rings placed on the $i$th finger. Then
$$x_1 + x_2 + x_3 + x_4 + x_5 = 5$$
is an equation in the nonnegative integers. A particular solution of the equation corresponds to the placement of four addition signs in a row of five ones. For instance,
$$+ 1 + 1 + 1 1 + 1$$
corresponds to the solution $x_1 = 0$, $x_2 = 1$, $x_3 = 1$, $x_4 = 2$, $x_5 = 1$ (no rings on the thumb, one ring each on the index, middle, and pinky fingers, and two rings on the ring finger). The number of such solutions is
$$\binom{5 + 5 - 1}{5 - 1} = \binom{9}{4}$$
since we must choose which four of the nine positions required for five ones and four addition signs will be filled with addition signs. Multiplying by the $5!$ ways of arranging the rings from bottom left to top right yields
$$\binom{9}{4}5!$$
Method 2: We arrange five distinct rings and four dividers, which correspond to the decision to jump from one finger to the next, in a row. Choose four of the nine positions for the dividers, then arrange the five distinct rings in the remaining five positions. Again, we obtain
$$\binom{9}{4}5!$$
ways of arranging five distinct rings on five fingers.
Where did you make a mistake?
Your calculations for distributing five distinct rings to one finger, two fingers, and five fingers are correct.
I do not follow your calculations for three fingers or four fingers.
Distributing five rings to three fingers: Either one of the three selected fingers receives three rings with the others each receiving one or two of the three selected fingers each receive two rings with the other receiving one.
One of the three selected fingers receives three rings with the others each receiving one: There are $\binom{5}{3}$ ways to select which three fingers receive a ring, $\binom{3}{1}$ ways to select which of those fingers receives three rings, $\binom{5}{3}$ ways to select three rings for that finger, $3!$ ways to arrange the rings on that finger, and $2!$ ways to arrange the remaining two rings on the remaining two fingers, giving
$$\binom{5}{3}\binom{3}{1}\binom{5}{3}3!2!$$
such distributions.
Two of the three selected fingers each receive two rings with the other receiving one: There are $\binom{5}{3}$ ways to select which three fingers receive a ring, $\binom{3}{2}$ ways to select which two of those fingers receive two rings, $\binom{5}{2}$ ways to choose which two rings are placed on the leftmost of those fingers, $2!$ ways to arrange the ring on that finger, $\binom{3}{2}$ ways to choose which two the remaining rings are placed on the other finger that receives two rings, $2!$ ways to arrange those rings on that finger, and one way to place the remaining ring on the remaining finger. There are
$$\binom{5}{3}\binom{3}{2}\binom{5}{2}\binom{3}{2}2!2!$$
such distributions.
Therefore, the number of ways to distributing five distinct rings so that exactly three of the five fingers receive a ring is
$$\binom{5}{3}\binom{3}{1}\binom{5}{3}3!2! + \binom{5}{3}\binom{3}{2}\binom{5}{2}\binom{3}{2}2!2!$$
Distributing five rings to four fingers: It must be the case that one of the four selected fingers receives two rings, while each of the others receive one ring. There are $\binom{5}{4}$ ways to select the fingers that receive a ring, $\binom{4}{1}$ ways to choose which of those fingers receives two rings, $\binom{5}{2}$ ways to choose which two rings are placed on that finger, $2!$ ways to arrange those rings on that finger, and $3!$ ways to arrange the remaining three rings on the remaining three fingers. There are
$$\binom{5}{4}\binom{4}{1}\binom{5}{2}2!3!$$
such distributions.
With these corrections, you will find that there are $15,120$ ways to distribute five distinct rings to five fingers when the order of the rings on each finger matters, in agreement with the answer we obtained above.
| {
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Expressing the vector between a vertex of a triangle and a point on the opposite edge? Given a triangle $ABC$, and a point $D$ on the edge $BC$, is it possible to express vector $\overrightarrow{AD}$ with positive integers $x,y$ and vectors $\overrightarrow{AB}$, $\overrightarrow{AC}$ such that:
$$\overrightarrow{AD} = \frac{x\overrightarrow{AB}+y\overrightarrow{AC}}{x+y}$$
This was an explanation used in Solution $3$ in the 2019 AIME Problem $4$, but I can't seem to find any justification or proof for the claim.
Any help would be appreciated.
| The solution doesn't claim that $ x $ and $ y $ are positive integers in the beginning, but only positive (real numbers). The fact that we can find such integers is only shown later, and is because the other ratios given in the question are rational numbers, and rational numbers are closed under addition, subtraction, multiplication and division (by nonzero rationals). But if your question is also about why there are such positive reals $ x $ and $ y $, the following may help.
Consider any point $ D $ on the segment $ B C $. Let $ x $ be the length of the segment $ C D $ and $ y $ be length of the segment $ B D $, so that the length of $ B C $ is $ x + y $ (and thus $ x + y > 0 $). We have $ 0 \le x , y \le x + y $ (with either $ x > 0 $ or $ y > 0 $), and in case $ D $ is strictly between $ B $ and $ C $, we have $ 0 < x , y < x + y $. It's then fairly easy to see that
$$ \overrightarrow { B D } = \frac y { x + y } \overrightarrow { B C } \quad \text {and} \quad \overrightarrow { C D } = \frac x { x + y } \overrightarrow { C B } \text . $$
The rest is simple algebraic manipulation of vectors:
$$ x \overrightarrow { B D } + y \overrightarrow { C D } = \frac { x y } { x + y } \left( \overrightarrow { B C } + \overrightarrow { C B } \right) = \vec 0 \text ; $$
$$ \therefore \ x \left( \overrightarrow { A D } - \overrightarrow { A B } \right) + y \left( \overrightarrow { A D } - \overrightarrow { A C } \right) = \vec 0 \text ; $$
$$ \therefore \ ( x + y ) \overrightarrow { A D } = x \overrightarrow { A B } + y \overrightarrow { A C } \text ; $$
$$ \therefore \ \overrightarrow { A D } = \frac { x \overrightarrow { A B } + y \overrightarrow { A C } } { x + y } \text . $$
Note that we could get the same results if we took $ x $ and $ y $ as not necessarily the length of $ C D $ and $ B D $, but $ \alpha $ times each of the lengths, for some positive real $ \alpha $. This shows that if the ratio between the lengths of the segments is a rational number, we can take $ x $ and $ y $ to be integers (since either $ x > 0 $ or $ y > 0 $, at least one of the ratios $ \frac y x $ and $ \frac x y $ can be defined).
| {
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Showing that two combinatorial expressions are equal Is there an algebraic way of showing that $$\sum_{m=\lceil p / b\rceil}^{c} \left(-1\right)^m \frac{\binom{b\cdot m}{p}}{\binom{b\cdot c}{p}} \sum_{k=m}^{c}k\cdot(-1)^{k} \binom{c}{k} \binom{k}{m} = c\cdot\left[1 - \frac{\binom{b\cdot[c - 1]}{p}}{\binom{b\cdot c}{p}}\right], $$
for all positive integers $b$, $c$, $p$?
Background:
A friend sent me a puzzle, which I've solved in two different ways, and I'd like to convince myself that they're algebraically the same.
Specifically, I was asked to consider the following problem: Suppose we have 80 balls in a bowl of 8 different colours, with 10 of each colour. If we draw 15 of them without replacement, how many colours would we expect to see?
Method 1:
We can find the expected number of colours as the weighted average of the possible number of colours, with the weights being the probabilities: $\bar{k} = \sum_{k} k \cdot P_k$. To do this we need to calculate the probabilities of seeing exactly $k$ colours for all possible values of $k$.
The probability of using just one colour is 0, as $15 > 10$.
The probability of using exactly two colours is $\binom{8}{2} \cdot \frac{\binom{20}{15}}{\binom{80}{15}}$.
The probability of using exactly at most three colours is $\binom{8}{3} \cdot \frac{\binom{30}{15}}{\binom{80}{15}}$.This probability however includes some cases where we've picked only two colours. To get the probability of picking exactly three colours, we need to subtract these cases, giving: $\binom{8}{3} \cdot \left[\frac{\binom{30}{15}}{\binom{80}{15}} - \binom{3}{2}\frac{\binom{20}{15}}{\binom{80}{15}}\right]$. Using the inclusion-exclusion principle, the general probability of getting exactly $k$ colours is
$$
P_k = (-1)^{k} \frac{\binom{8}{k}}{\binom{80}{15}} \sum_{m=2}^{k}\left(-1\right)^m \binom{k}{m}\binom{10m}{15}
$$
The expected number of colours can then be written as
$$ \begin{align}
\bar{k} &= \sum_{k=2}^8k\cdot(-1)^{k} \frac{\binom{8}{k}}{\binom{80}{15}} \sum_{m=2}^{k}\left(-1\right)^m \binom{k}{m}\binom{10m}{15}\\
&= \sum_{m=2}^{8} \left(-1\right)^m \frac{\binom{10m}{15}}{\binom{80}{15}} \sum_{k=m}^8k\cdot(-1)^{k} \binom{8}{k} \binom{k}{m}.
\end{align} $$
Method 2
Alternatively, I reasoned that for any given colour, the probability of having that colour in the draw is $P_{\mathrm{c}} = 1 - \frac{\binom{70}{15}}{\binom{80}{15}}$. If the colour is included, it contributes 1 to the total number of colours, and if not, it contributes 0. The expected value of the number of colours is then
$$
\bar{k} = \sum_{\mathrm{colours}} P_{\mathrm{c}} = 8 - 8 \frac{\binom{70}{15}}{\binom{80}{15}}
$$
I can calculate both of these numbers, and for this particular case, they are the same.
What I'd like is to understand why they should be equal in general -- there's nothing in the derivations which limits us to 8 colours, 10 balls of each colour and 15 picks.
It's been a while since I had any combinatorics, so I'm a bit stuck
| We seek to show that
$$\sum_{m=\lceil p/b \rceil}^c (-1)^m {bm\choose p}
\sum_{k=m}^c (-1)^k k {c\choose k} {k\choose m}
= c{bc\choose p} - c{b(c-1)\choose p}.$$
We have
$${c\choose k} {k\choose m}
= \frac{c!}{(c-k)! \times m! \times (k-m)!}
= {c\choose m} {c-m\choose c-k}$$
and we get for the LHS
$$\sum_{m=\lceil p/b \rceil}^c (-1)^m {bm\choose p}
{c\choose m}
\sum_{k=m}^c (-1)^k k {c-m\choose c-k}
\\ = \sum_{m=\lceil p/b \rceil}^c (-1)^m {bm\choose p}
{c\choose m}
\sum_{k=0}^{c-m} (-1)^{k+m} (k+m) {c-m\choose c-m-k}
\\ = \sum_{m=\lceil p/b \rceil}^c {bm\choose p}
{c\choose m}
\sum_{k=0}^{c-m} (-1)^{k} (k+m) {c-m\choose k}.$$
Now we have
$$m \sum_{k=0}^{c-m} (-1)^{k} {c-m\choose k}
= m [[ c = m ]]$$
so that this becomes
$$ c {bc\choose p} +
\sum_{m=\lceil p/b \rceil}^c {bm\choose p}
{c\choose m}
\sum_{k=0}^{c-m} (-1)^{k} k {c-m\choose k}
\\ = c {bc\choose p} +
\sum_{m=\lceil p/b \rceil}^c {bm\choose p}
(c-m) {c\choose m}
\sum_{k=1}^{c-m} (-1)^{k} {c-m-1\choose k-1}
\\ = c {bc\choose p} -
\sum_{m=\lceil p/b \rceil}^c {bm\choose p}
(c-m) {c\choose m}
\sum_{k=0}^{c-m-1} (-1)^{k} {c-m-1\choose k}.$$
Again we have
$$\sum_{k=0}^{c-m-1} (-1)^{k} {c-m-1\choose k}
= [[c-1 = m]]$$
so we find
$$c {bc\choose p} -
{b(c-1)\choose p} (c-(c-1)) {c\choose c-1}
= c {bc\choose p} - c {b(c-1)\choose p}.$$
This is the claim.
| {
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Convergence of series $x+\frac{x^3}{3}+\frac{x^5}{5}+\cdots$ Convergence of series $$S=x+\frac{x^3}{3}+\frac{x^5}{5}+\cdots$$
When $x=\frac{1}{2}$
I used ratio test as:
$$a_{n}=\frac{\left(\frac{1}{2}\right)^{2n-1}}{2n-1}$$
Then
$$a_{n+1}=\frac{\left(\frac{1}{2}\right)^{2n+1}}{2n+1}$$
Then we get:
$$\frac{a_{n+1}}{a_n}=\frac{2n-1}{4(2n+1)}$$
Hence we get:
$$\lim_{n \to \infty}\frac{a_{n+1}}{a_n}=0.25 \lt 1$$
Hence $\sum a_n$ Converges.
But how to find its value, i guess $\arctan x$ is not useful here
| You could aswell use the sandwich-lemma:
$$0 < x + \dfrac{x^3}{3} + \dfrac{x^5}{5} + ... < x+x^3+x^5+...<x+x^2+x^3+x^4+...=\sum_n^\infty 1\cdot x^n = \dfrac{1}{1-x}=2$$
Furthermore note that the sum is strictly increasing, hence the sum is convergent.
| {
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If $f(2x-3) = 4x-2$, then what is $f(x)$? I have this statement:
If $f(2x-3) = 4x-2,$ the function $f(x)$ is ...?
My attempt was:
Move the function $3$ units to the left
$f(2x) = 4(x+3) -2 = 4x+10$
Divide $x$ by $2$
$f(x) = 2x+10$
Verifiy $f(x) = 2x+10 \to f(2x) = 4x+10 \to f(2x-3) = 4(x-3)+10 = \underbrace{4x-2}_{f(2x-3)}$
But according to the guide the correct answer is $2x+4$ and i don't know why. Thans in advance.
| If $y=2x-3$, what is $x$? We have $x = {1 \over 2} (y+3)$.
If $x = {1 \over 2} (y+3)$, what is $4x-2$? We have $4x-2=2y+4$.
This will give $f(y)$ in terms of $y$.
Explicitly, $f(y) = 2y+4$.
| {
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Polynomials that induce the zero function mod $n$
*
*Which polynomials induce the zero function mod $n$?
In particular:
*
*What is the polynomial of least degree that induces the zero function mod $n$?
*What is the monic polynomial of least degree that induces the zero function mod $n$?
These are not vacuous questions because of the following general result:
If $r$ is the maximum exponent in the prime factorization of $n$, then $x \mapsto x^{r+\lambda (n)}-x^r$ is the zero function mod $n$. [Wikipedia]
Here, $\lambda$ is the Carmichael function.
*
*When is $x^{r+\lambda (n)}-x^r$ the monic polynomial of least degree that induces the zero function mod $n$?
Fermat's theorem implies that $x^n-x$ is the answer for $n$ prime: all polynomials that induce the zero function mod $n$ are a multiple of $x^n-x$.
How can this be generalized to composite $n$?
Here are some other examples:
$$
\begin{array}{rll}
n & L_n: \text{least degree} & M_n: \text{least degree monic}
\\ 2 & x^2+x
\\ 3 & x^3-x
\\ 4 & 2(x^2+x) & x^4-x^2
\\ 5 & x^5-x
\\ 6 & 3(x^2+x) & x^3-x
\\ 7 & x^7-x
\\ 8 & 4(x^2+x) & x^4+2x^3+3x^2+2x = x(x+1)(x^2+x+2)
\\ 9 & 3(x^3-x) & x^8-x \quad (???)
\\ 10 & 5(x^2+x) & x^5-x
\\ 11 & x^{11}-x
\\ 12 & 6(x^2+x) & x^4+5x^2+6x = x(x+1)(x^2-x+6)
\\ 13 & x^{13}-x
\\ 14 & 7(x^2+x) & ???
\\ 15 & 5(x^3-x) & x^5-x
\\ 21 & 7(x^3-x) & ???
\\ 24 & 12(x^2+x) & x^4+2x^3+11x^2+10x = x(x+1)(x^2+x+10)
\end{array}
$$
It seems clear that $L_{2m}=m(x^2+x)$, because $x^2+x$ is always even.
More generally,
*
*Is $L_{pq} = qL_p$ and $M_{pq}=L_q$ for $p<q$ primes?
*If $n=pm$ and $p$ is smallest prime divisor of $n$, then is $L_{pm}=mL_p$?
Corrections and additions welcome.
Please collect partial results as answers.
| Edited and improved
Using the fact that the product of any $n$ consecutive integers is divisible by $n!$ we immediately get
Lemma 1 Let $R_n(X)=X(X-1)(X-2)...(X-n+1)$. Then $R(X)$ is trivial $\pmod{n!}$.
Lemma2: Let $P(X)=a_kX^k+..+a_1X+a_0$ and let $p$ be prime. If $P(X)$ is trivial modulo $p$ and $p$ does not divide all coefficients of $P(X)$ then $k \geq p$.
Proof: Since $P(X)$ has $p$ roots in the field $\ZZ_p$, it has degree at least $p$ modulo $p$. Then $k \geq p$.
As consequences we get immediately:
Lemma 3: If $p$ is prime, and $n$ is so that $p|n$ and $n |p!$ then
$$
M_n=X(X-1)(X_2)...(X-p+1)=: R_p
$$
[Or another polynomial of same degree, namely $M_n=R_p+Q(X)L_n(X)$ for some $Q(X)$.]
Proof: Since $n|p!$, by Lemma 1 $R_p(X) $ is trivial modulo $n$. By Lemma2, $deg(M_n) \geq p$.
\qed
Lemma 4 If $p$ is the smallest prime dividing $n$, and $n$ is square free, then
$$L_n=\frac{n}{p}(X^p-X)$$
Proof It is clear that this polynomial works. We show next $\deg(L_n) \geq p$.
Let $L_n=a_kX^k+....a_1X+a_0$. Take the first coefficient $a_l$ which is not divisible by $n$. Then, there exists a prime $q|n$ such that $q \nmid a_l$.
By Lemma 2$deg(L_n) \geq q \geq p$.
Final Note: Your $M_8$ is wrong. Note that if $n$ is even then $8|n(n-2)$ and if $n$ is odd $8|(n-1)(n+1)$.
I think that it is easy to show that $M_8=(X-2)(X-1)X(X+1)$ or something equivalent $\pmod{8}$. THis also works for $M_{24}$.
| {
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Sum of a Sequence based on prime numbers I was playing around with prime numbers and arrived at this sequence:
$$
\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^5}+\frac{1}{2^{11}}+\frac{1}{2^{13}}+\frac{1}{2^{17}}+\frac{1}{2^{19}}+\frac{1}{2^{23}}+\frac{1}{2^{27}} ... = 0.414682388306 \approx \sqrt{2}-1
$$
Although the sum does not approach $\sqrt2 - 1$ it is very similar. Is there any reason or is this just a coincidence.
| Here is an explanation why $\dfrac{209}{504}$ is a much better approximation than $\sqrt{2} - 1$ and also shows why the latter must be a co-incidence.
The number $\sum_{n = 1}^{\infty} \frac{1}{2^{p_n}}$ is known as the prime constant and its value is approximately $0.414682509...$.
$$
s = \sum_{n = 1}^{\infty} \frac{1}{2^{p_n}} < \frac{1}{2^2} + \frac{1}{2^3}
+ \sum_{n = 1}^{\infty} \frac{1}{2^{6n-1}} + \sum_{n = 1}^{\infty} \frac{1}{2^{6n+1}} = \frac{3}{8} + \frac{5}{126} = \frac{209}{504} \approx 0.41468253...
$$
Each term in the RHS decreases roughly by a factor of $1/64$ and among the first few natural numbers, we have a low density of numbers of the form $6n \pm 1$ which are not primes; $25$ is the only number of this form below $30$ which is not a prime. Hence the actual sum must be very close to the above upper bound because $1/64$ raised to non primes of the $6n \pm 1$ will not effect the first few decimal places.
In fact $\dfrac{209}{504}$ agrees to seven decimal places with the actual sum and is therefore a much better estimate that $\sqrt{2}-1$ which agrees only to $3$ decimal places.
| {
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Prove that $a = b$ (Leningrad 1990) Let a and b be natural numbers that $b ^ 2 + ba + 1$ divides $a ^ 2 + ab +1$. Prove that $a = b$
I thought of the limitation property $b ^ 2 + ba + 1 ≤ a ^ 2 + ab + 1$ and hence $b ≤ a$, but I don't know how to process. How can I prove it differently?
| If $b^2 + ba + 1$ divides $a^2 + ab+1$, then it also divides $(a^2+ab+1)-(b^2+ba+1) = a^2-b^2$, which factors as $(a+b)(a-b)$.
However, it can't share any divisors with the first factor: $\gcd(b^2+ba+1, a+b) = \gcd(b^2+ba+1 - b(a+b), a+b) = \gcd(1, a+b) = 1$.
Therefore $b^2+ba+1$ also divides $\frac{a^2-b^2}{a+b} = a-b$.
However, $b^2+ba+1 > a > a-b$ (assuming $a,b\ge 1$) so the only way this can hold is if $a-b=0$, or $a=b$.
(If we allow $b=0$, then $a$ can be anything and $b^2+ba+1 =1$ still divides $a^2+ab+1=a^2+1$, so it appears our definition of "natural number" excludes $0$ here.)
| {
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I'm stuck with this differential equations Mainly I need to get the real fundamental system of solutions of $y^{vi}+729y=0$.
As well, I'm stuck in getting the solution for $4y''+36y=\operatorname{cosc}(3x)$. Basically because I don't know how to operate it with the $\operatorname{cosc}(3x)$.
Thanks in advance.
| For $y^{vi} + 729y = 0:$
Consider the characteristic equation $$c^6 + 729 = 0$$ or $$c^6 = -729$$
We can find the sixth roots of $-729$, and then convert those roots to the form $$c_n e^{\lambda t} \cos (\mu t) + c_{n+1} e^{\lambda t} \sin (\mu t)$$
To begin, let $z = -729 + 0i.$ then $|z| = |-729 + 0i| = 729$ and $\arg |z| = \arctan (\frac {0}{-729}) = \pi$.
Using DeMoivre's Theorem for roots $$z^{1/n} = |z|^{1/n} \left (\cos \frac {\theta k}{n} + i \sin \frac {\theta k}{n}\right)$$
we have $$729^{1/6} \left (\cos \frac {\pi k}{6} + i \sin \frac {\pi k}{6}\right)$$ which reduces to $$3 \left (\cos \frac {\pi k}{6} + i \sin \frac {\pi k}{6}\right)$$
Letting $k$ = $0$ through $5$ we get the following six roots:
$$\pm 3i, \pm \frac {3}{2} (\sqrt{3} + i), \pm \frac {3}{2} (1 + i \sqrt{3})$$
For each of the roots above, we can now use $c_{n} e^{\lambda t} \cos (\mu t) + c_{n+1} e^{\lambda t} \sin (\mu t)$ to find the solutions of the differential equation.
$3i: c_0 \cos 3t + c_1 \sin 3t$
$-3i: c_2 \cos 3t - c_3 \sin 3t$
(using $\lambda = 0, \mu = 3$)
$\frac {3}{2} (\sqrt{3} + i): c_4 e^{\frac {3t\sqrt {3}}{2}} \cos \frac {3t}{2} + c_5 e^{\frac {3t\sqrt {3}}{2}} \sin \frac {3t}{2}$
$\frac {3}{2} (\sqrt{3} - i): c_6 e^{\frac {3t\sqrt {3}}{2}} \cos \frac {3t}{2} - c_7 e^{\frac {3t\sqrt {3}}{2}} \sin \frac {3t}{2}$
(using $\lambda = \frac {3\sqrt {3}}{2}, \mu = \frac {3}{2}$)
$\frac {3}{2} (1 + i\sqrt{3}): c_8 e^{\frac {3t}{2}} \cos \frac {3\sqrt{3}}{2} t + c_{9} e^{\frac {{3t}}{2}} \sin \frac {3\sqrt{3}}{2} t$
$\frac {3}{2} (1 - i\sqrt{3}): c_{10} e^{\frac {3t}{2}} \cos \frac {3\sqrt{3}}{2} t - c_{11} e^{\frac {{3t}}{2}} \sin \frac {3\sqrt{3}}{2} t$
(using $\lambda = \frac {3}{2}, \mu = \frac {3\sqrt {3}}{2}$)
The solution is then
$$(K_0) \cos 3t + (K_1) \sin 3t + (K_2)e^{\frac {3t\sqrt {3}}{2}} \cos \frac {3t}{2} + (K_3) e^{\frac {3t\sqrt {3}}{2}} \sin \frac {3t}{2} + (K_4) e^{\frac {3t}{2}} \cos \frac {3\sqrt{3}}{2} t + (K_5) e^{\frac {3t}{2}} \sin \frac {3\sqrt{3}}{2} t$$
where
$$\begin{matrix}
K_0 = (c_0 + c_2) \\
K_1 = (c_1 - c_3) \\
K_2 = (c_4 + c_6) \\
K_3 = (c_5 - c_7) \\
K_4 = (c_8 + c_{10})\\
K_5 = (c_{9} - c_{11})\\
\end{matrix}$$
| {
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Prove geometric sum with induction I'm not certain how to complete the proof:
Question:
Prove by induction that $1 − \frac{1}{3} + \frac{1}{9} − · · · + (\frac{−1}{3})^n = (\frac{3}{4})[1 − (\frac{−1}{3})^{n+1}]$, for every non negative integer $n$.
Solution
Base Step:
Verify that:
$LHS = 1 = (\frac{3}{4})[1 − (\frac{−1}{3})^{0+1}] = RHS$.
$RHS = (\frac{3}{4})[1 − (\frac{−1}{3})^1]\\
= (\frac{3}{4})[1 − (\frac{−1}{3})]\\
= (\frac{3}{4})(1 + \frac{1}{3})\\
= (\frac{3}{4})(\frac{4}{3}) = 1 = LHS.$
Inductive Step:
Assume that:
$1 − \frac{1}{3} + \frac{1}{9} − · · · + (\frac{−1}{3})^k = (\frac{3}{4})[1 − (\frac{−1}{3})^{k+1}]$, for some integer k.
We try to deduce that:
$1 − \frac{1}{3} + \frac{1}{9} − · · · + (\frac{−1}{3})^{k+1} = (\frac{3}{4})[1 − (\frac{−1}{3})^{k+2} ]$.
$LHS
= 1 − \frac{1}{3} + \frac{1}{9} − · · · + (\frac{−1}{3})^{k+1} \\
= 1 − \frac{1}{3} + \frac{1}{9} − · · · + (\frac{−1}{3})^{k} + (\frac{−1}{3})^{k+1}\\
= \frac{3}{4}[1-(\frac{-1}{3})^{k+1}] + (\frac{-1}{3})^{k+1}\\
... $
Lost from this point onwards.
| So far so good. Let $q = -\frac{1}{3}$. Thus, following your lines, you have shown
$$ \sum_{j=1}^{k+1} q^j = \sum_{j=1}^{k} q^j + q^{k+1} = \frac{1-q^{k+1}}{1-q} + q^{k+1}.$$
To finish the proof, note that
$$q^{k+1} = \frac{1-q}{1-q} q^{k+1}.$$
| {
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Two values of minima of a function by two methods.
I had a problem of finding minima of a function
$$f(x)=2^{x^2}-1+\frac{2}{2^{x^2}+1}$$
I solved it using AM-GM inequality,
$$2^{x^2}-1+\frac{2}{2^{x^2}+1}$$
$$2^{x^2}+1+\frac{2}{2^{x^2}+1}-2$$
$$2^{x^2}+1+\frac{2}{2^{x^2}+1}\ge\ 2\sqrt2$$
$$2^{x^2}-1+\frac{2}{2^{x^2}+1}\ge\ 2\sqrt2-2$$
But in the solution answer was given as 1 and it was solved using differentiation,
$$f'(x)=\frac{2x.ln2.2^{x^2}(2^{x^2}+1-\sqrt2)(2^{x^2}+1+\sqrt2)}{(2^{x^2}+1)^2}$$
$$2^{x^2}\ge1$$
$$2^{x^2}+1-\sqrt2\ge2-\sqrt2>0$$
At $x=0$ ,$f(x)$ is least.
Least value = $f(0)$ $=1$
I cannot understand how can there be two values by two different methods,please help me in the problem.
| A much simpler method is to define $$y(x)=2^{x^2}$$ and $$g(y) = y-1+\frac{2}{y+1}$$
Such that $f(x) = g(y(x))$.
Then, rewrite g : $$g(y) = \frac{y^2+1}{y+1}$$
So, it's easy to see that g is increasing on $[1,+\infty)$, and $y\geq1$, therefore, the minimum of $f$ is $g(1)=1$ with $x=0$
| {
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Intersection of curve $ y = x^4 – 6x^3 + 12x^2 + cx + 1$ The number of integers in the range of 'c' such that there exists a line which intersects the curve $ y = x^4 – 6x^3 + 12x^2 + cx + 1$ at four distinct points.
My approach we need to intersect with line $y=mx+C$
Substituting we get $x^4 – 6x^3 + 12x^2 + cx + 1-mx-C=0$
Now this is an equation of polynomial of degree 4
$x^4 – 6x^3 + 12x^2 + (c-m)x + 1-C=0$
All four roots needs to be real.
I don't have any idea how to check whether all the roots are real or not.
| The second derivative $y''(x)=12x^2-36x+24=12(x-1)(x-2)$ has roots $\{1,2\}$ independently of $c$. Thus, $y(x)$ has two points of inflection at $(1, 8+c)$ and $(2, 17+2c)$. A line that goes through these two points will always intersect $y(x)$ at 4 points (two points of indflection and $\{(3\pm\sqrt5)/2\}$).
| {
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Show that there exists no solution with $(x, y, z)$ ∈ $\mathbb N^+ × \mathbb N^+ × \mathbb N^+$
Show that there exists no solution with $(x, y, z)$ ∈ $\mathbb N^+ × \mathbb N^+ × \mathbb N^+$ for
(a) $5x^2 + 2y^2 = z^2$
(b) $17x^2 + 6y^2 = z^2$
My attempt:
a) if $5x^2 + 2y^2 = z^2$ has a solution in $\mathbb Z/n \mathbb Z$, then it has a solution in $\mathbb Z$.
mod $5$ : $2y^2 = z^2$
mod $2$ : $5x^2 = z^2$
Then in mod $5$: we have $y = 0$ mod $5$ and $z = 0$ mod $5$ iff $5|y$ in $\mathbb Z$ and $5|z$ in $\mathbb Z$
and $5^2|z^2-2y^2$ then $5^2|5x^2$ then $5|x^2$ then $5|x$
Assume that $(x,y,z)$ is a solution with $x>=0$, $y>=0$, $z>=0$ and $(x,y,z) \not= (0,0,0)$, then $\frac x5$, $\frac y5$, $\frac z5$ is a smaller positive solution. This contradicts the well-ordering principle. Thus $5x^2 + 2y^2 = z^2$ can't have a solution $\not= (0,0,0)$
Is my attempt correct?
And do I solve part b the same as part a?
| Well, for part (b) the method is the same. Suppose that $17^2+6^2=^2$ is the solution with smallest $x \in \mathbb{N}$. Then, the equation tells us that $2x^2 \equiv z^2 \pmod{3}$, and since any square is congruent to $0$ or $1$ modulo 3, we have that $x \equiv 0 \pmod{3} \Rightarrow z \equiv 0 \pmod{3}$, then $9\vert x^2$ and $9\vert z^2$. So, replacing $x^2=9a^2$ and $z^2=9b^2$, we obtain
$
17\cdot9a^2+6y^2=9b^2
$,
which gives us that $3\vert y^2 \Rightarrow 9\vert y^2$. Replacing $y^2=9\cdot c^2$, we have $17\cdot9a^2+9\cdot 6 \cdot c^2=9b^2$, and by dividing by 9, we get a new solution $17a^2+6b^2=c^2$ to the given diophantine equation, where $a<x$. So, the conclusion follows.
| {
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Argument of a complex fraction, why different answers? I want to take the argument of the following complex fraction. Using the second method I get a different answer, why is that?
$$
G(\omega)=
\frac{1}{(1+2\omega i)^2}
\tag 1
$$
Method 1:
\begin{align}
\arg\frac{1}{(1+2\omega i)^2}
&=\arg1-\arg\Big((1+2\omega i)^2\Big) \tag 2\\
&=\arg1-\arg\Big((1+2\omega i)(1+2\omega i)\Big) \tag 3\\
&=\arg1-\arg(1+2\omega i)-\arg(1+2\omega i) \tag 4\\
&=\arctan\frac{0}{1}-\arctan\Big(\frac{2\omega}{1}\Big)-\arctan\Big(\frac{2\omega}{1}\Big) \tag 5\\
&=-2\arctan(2\omega) \tag 6
\\
\end{align}
Method 2:
Expand the denominator:
$$
(1+2\omega i)^2=1-4\omega^2+4\omega i
$$
So I have
\begin{align}
\arg \frac{1}{(1+2\omega i)^2}
&=\arg\frac{1}{1-4\omega^2+4\omega i} \tag 7\\
&=\arg1-\arg(1-4\omega^2+4\omega i)\tag 8\\
&=-\arctan\bigg (\frac{4\omega}{1-4\omega^2}\bigg) \tag 9
\end{align}
So $(6)$ is not equal to $(9)$, What is wrong with method 2?
| The problem is that given $x = x + i y$ you are using $\arctan(y/x)$ with one argument. You should use instead $\arctan(x,y)$. You can find those options in matlab or MATHEMATICA. Note that $\arctan(y/x)=\arctan((-y)/(-x))$ and $\arctan((-y)/x)=\arctan(y/(-x))$
| {
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$a(x)= x^4 - b^2x^3 - bx^2 -2x$ is divisible by $x-2$. Find $b$. $a(x)= x^4 - b^2x^3 - bx^2 -2x$ is divisible by $x-2$ for certain value $b$. Calculate all possible values $b$ can assume.
According to the factor theorem $x=2$ is a root when $a(x)=0$
$a(x)= x^4 - b^2x^3 - bx^2 -2x=0$
$a(2)= 2^4 - b^22^3 - b2^2 -2*2=0$
$8b^2 + 4b- 12=0$
$b^2 + 0.5b- 1.5=0$
After solving the quadratic equation, I get $b_1=-1,5$ and $b_2=1$
Is this right? Would you have solved it differently?
| "$x^4 - b^2x^3 - bx^2 -2x$ divisible by $(x-2)$" is equivalent to
"$x^3 - b^2x^2 - bx -2$ divisible by $(x-2)$".
\begin{align}
&\quad\; x^3 - b^2x^2 - bx -2 \\
&=(x^3-8) - b^2x^2 - bx +6 \\
&=(x^3-8) - (bx-2)(bx+3)
\end{align}
$(x^3-8)$ is divisible by $(x-2)$ apparently.
For $(bx-2)(bx+3)$ divisible by $(x-2)$, there are two possibilities:
$\qquad (bx-2)$ divisible by $(x-2)$, $\qquad b=1$
$\qquad (bx+3)$ divisible by $(x-2)$, $\qquad b=-\frac32$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3406520",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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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.
| Hint:
In a limit to $0$, you can replace $\sin(x)$ by $x$ in a ratio because
$$\sin(x)=x\frac{\sin(x)}x.$$
And for this reason, you can replace $1-\cos(x)$ by $\dfrac{x^2}2$.
| {
"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": 5
} |
what is $\lim\limits_{x\to\infty}x\left(\sqrt{x^{2}+2x}- 2\sqrt{x^{2}+x}+x\right)$? $$\lim\limits_{x\to\infty}x\left(\sqrt{x^{2}+2x}-2\sqrt{x^{2}+x}+x\right)$$$$=\lim\limits_{x\to\infty}x\left(\sqrt{x^{2}+2x}-1-2\sqrt{x^{2}+x}+1+x\right)$$$$=\lim\limits_{x\to\infty}x\left(\sqrt{x^{2}+2x}-1\right)+\lim\limits_{x\to\infty}\left(-2x\right)\left(\sqrt{x^{2}+x}+1\right)+\lim\limits_{x\to\infty}x^{2}$$$$=\lim\limits_{x\to\infty}x\lim\limits_{x\to\infty}\left(x\sqrt{1+\frac{2}{x}}-1\right)+\lim\limits_{x\to\infty}\left(-2x\right)\lim\limits_{x\to\infty}\left(x\sqrt{1+\frac{1}{x}}+1\right)+\lim\limits_{x\to\infty}x^{2}$$$$=\lim\limits_{x\to\infty}x\left(1\right)+\lim\limits_{x\to\infty}\left(-2x\right)\left(\frac{1}{2}\right)+\lim\limits_{x\to\infty}x^{2}= ∞$$
another way I tried:$$\lim\limits_{x\to\infty}x\left(\sqrt{x^{2}+2x}-2\sqrt{x^{2}+x}\right)+\lim\limits_{x\to\infty}x^{2}$$$$=\lim\limits_{x\to\infty}x\left(\sqrt{x^{2}+2x}-2\sqrt{x^{2}+x}\right)\cdot\frac{\sqrt{x^{2}+2x}+2\sqrt{x^{2}+x}}{\sqrt{x^{2}+2x}+2\sqrt{x^{2}+x}}+\lim\limits_{x\to\infty}x^{2}$$$$=\lim\limits_{x\to\infty}\frac{x\left(-2x^{2}-2x\right)}{\sqrt{x^{2}+2x}+2\sqrt{x^{2}+x}}+\lim\limits_{x\to\infty}x^{2}$$$$=\lim\limits_{x\to\infty}\frac{x\left(-2x^{2}-2x\right)}{x\left(\sqrt{1+\frac{2}{x}}+2\sqrt{1+\frac{1}{x}}\right)}+\lim\limits_{x\to\infty}x^{2}$$$$=\lim\limits_{x\to\infty}\frac{\left(-2x^{2}-2x\right)}{\left(\sqrt{1+\frac{2}{x}}+2\sqrt{1+\frac{1}{x}}\right)}+\lim\limits_{x\to\infty}x^{2}$$$$=\lim\limits_{x\to\infty}\frac{-2x^{3}-2x}{3}+\lim\limits_{x\to\infty}x^{2}=\lim\limits_{x\to\infty}\frac{x^{3}\left(-2+\frac{3}{x}-\frac{2}{x^{2}}\right)}{3}=-∞$$
and none of them are the answer, why the solutions are not right and can someone use an elementary way to solve the limit?
| Here is an elementary way by substitution $x = \frac{1}{t}$ and considering $t\to 0^+$:
\begin{eqnarray*} x\left(\sqrt{x^{2}+2x}- 2\sqrt{x^{2}+x}+x\right)
& \stackrel{x=\frac{1}{t}}{=} & \frac{1}{t}\frac{\sqrt{1+2t} - 2\sqrt{1+t} + 1}{t} \\
&= & \left(\frac{\sqrt{1+2t}-1}{t^2} + 2\frac{1-\sqrt{1+t}}{t^2}\right)\\
& = & \frac{2}{t(\sqrt{1+2t}+1)}- \frac{2}{t(1+\sqrt{1+t})} \\
& = & \frac{2}{(1+\sqrt{1+t})((\sqrt{1+2t}+1))}\underbrace{\frac{\sqrt{1+t}-\sqrt{1+2t}}{t}}_{= -\frac{1}{\sqrt{1+t}+\sqrt{1+2t}}}\\
&\stackrel{t \to 0^+}{\longrightarrow} &\frac{2}{4}\cdot \frac{-1}{2} = -\frac{1}{4}
\end{eqnarray*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3408457",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Integrate $\int_{0}^{\infty}{\frac{(x^2+4)\ln(x)}{x^4+16}}dx$
$$\int_{0}^{\infty}{\dfrac{\left(x^2+4\right)\ln\left(x\right)}{x^4+16}}dx$$
I tried indefinite integration and got struck at $$\int_{0}^{\infty}{\dfrac{\left(u^2+1\right)\ln\left(u\right)}{u^4+1}}du.$$
I tried by-parts now taking $\ln(u)$ as my first function, but it doesn't seem to help. How to solve it and also is there an easy way out to solve it, as my method isn't that good.
| Start by letting $x\to \frac{4}{x}$ then you get:
$$I=\int_{0}^{\infty}{\dfrac{\left(x^2+4\right)\ln\left(x\right)}{x^4+16}}dx=\int_{0}^{\infty}{\dfrac{\left(x^2+4\right)\ln\left(\frac{4}{x}\right)}{x^4+16}}dx$$
Now add them up to get:
$$2I=\ln 4 \int_0^\infty \frac{x^2+4}{x^4+16}dx\Rightarrow I=\ln 2\int_0^\infty \frac{1+\frac{4}{x^2}}{x^2+\frac{4}{x^2}}dx$$
$$=\ln 2\int_0^\infty \frac{d\left(x-\frac{4}{x}\right)}{\left(x-\frac{4}{x}\right)^2+8}=\ln 2 \int_{-\infty}^\infty \frac{dx}{x^2+8}=\frac{\pi}{2\sqrt 2}\ln 2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3414066",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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A number with the form $n^2 +1$ , every of its odd divisor has the form $4k+1$ I dont know if it’s correct, so I ask you.
If we show, that every odd p which divides $n^2 +1$, has the form $4k+1$, then we are done.
So:
$$n^2\equiv -1\pmod{p} \implies (n^2)^{(p-1)/2}\equiv (-1)^{(p-1)/2}\pmod{p}\implies n^{p-1}\equiv 1\pmod{p}$$
If we substitute $p=4k+3$,
$$n^{4k+2}\equiv (n^2)^{2k+1}\equiv 1\pmod{4k+3}$$ which isn’t true, because $n^2\equiv -1\pmod{4k+3}$, and $2k+1$ is odd.
And if we subtitute $p=4k+1$ we get
$$n^{4k}\equiv (n^2)^{2k}\equiv (-1)^{2k}\equiv 1\pmod{4k+1}$$ which is true.
| If $p=4k+3$ then your first line is not correct because $\frac{p-1}{2}$ is odd.
Corrected proof
We can use Fermat's Little Theorem, $n^{p-1}\equiv 1\pmod{p}$.
If we substitute $p=4k+3$, then
$n^{4k+2}\equiv (n^2)^{2k+1}\equiv 1\pmod{4k+3}$
which isn’t true, because $n^2\equiv -1\pmod{4k+3}$, and $2k+1$ is odd.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3418917",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to prove this beautiful series by using Taylor and Maclaurin series I have been playing with Taylor and Maclaurin series lately and stumble on this beautiful identity. I don't know to expand the left hand side to yield the right hand side:
How to prove: $\dfrac{1}{\sqrt{1-x^2}} =1+\dfrac{1}{2}x^2+\dfrac{1 \cdot3}{2\cdot4}x^4+\dfrac{1\cdot 3\cdot 5}{2\cdot 4\cdot 6}x^6+\dfrac{1\cdot 3\cdot 5\cdot 7}{2\cdot 4\cdot 6\cdot 8}x^8...$
I can only expand this as followed:
$$\frac{1}{\sqrt{1-x^2}} = 1+\frac{1}{2}x^2+\frac{3}{8}x^4+\frac{5}{16}x^6...,$$
How can you prove this by using Maclaurin series?
I need two proofs, one in Maclaurin series and one in binomial theorem. Please don't use the sigma notation too much as I cannot see the pattern.
| I have finally been able to derive the series after some manipulations:
The general formula for binomial series is:
$(1+x)^k=1+kx+\dfrac{k(k-1)}{2!}x^2+\dfrac{k(k-1)(k-2)}{3!}x^3+\dfrac{k(k-1)(k-2)(k-3)}{4!}x^4...$
$
\dfrac{1}{\sqrt{1-x^2}}=(1-x^2)^{-\frac{1}{2}}=1+(-\dfrac{1}{2})(-x^2)+\dfrac{(-\dfrac{1}{2})(-\dfrac{1}{2}-1)}{2!}(-x^2)^2+\dfrac{(-\dfrac{1}{2})(-\dfrac{1}{2}-1)(-\dfrac{1}{2}-2)}{3!}(-x^2)^3+\dfrac{(-\dfrac{1}{2})(-\dfrac{1}{2}-1)(-\dfrac{1}{2}-2)(-\dfrac{1}{2}-3)}{4!}(-x^2)^4...$
$=1+\dfrac{1}{2}x^2+\dfrac{(-\dfrac{1}{2})(-\dfrac{3}{2})}{2!}(x^4)+\dfrac{(-\dfrac{1}{2})(-\dfrac{3}{2})(-\dfrac{5}{2})}{3!}(-x^6)+\dfrac{(-\dfrac{1}{2})(-\dfrac{3}{2})(-\dfrac{5}{2})(-\dfrac{7}{2})}{4!}(x^8)...$
$=1+\dfrac{1}{2}x^2+\dfrac{1 \cdot3}{2^2\cdot2!}x^4+\dfrac{1\cdot 3\cdot 5}{2^3\cdot 3!}x^6+\dfrac{1\cdot 3\cdot 5\cdot 7}{2^4\cdot 4!}x^8...$
$=1+\dfrac{1}{2}x^2+\dfrac{1 \cdot3}{2\cdot4}x^4+\dfrac{1\cdot 3\cdot 5}{2\cdot 4\cdot 6}x^6+\dfrac{1\cdot 3\cdot 5\cdot 7}{2\cdot 4\cdot 6\cdot 8}x^8...$
$$=\sum_{n=0}^{\infty} \dfrac{(2n-1!!)}{2^nn!}x^{2n}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3419594",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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$2 \times 2$ matrix transformations on the unit circle I have been given the following problems to solve:
*
*Find all $2 \times 2$ matrices that transform each point on the circle $x^2 + y^2 = 1$ into a point on the same circle. What happens to points on other circles?
*Find all $2 \times 2$ matrices that transform each point on the hyperbola $x^2 - y^2 = 1$ into a point on the same hyperbola. What happens to points on other hyperbolas?
*How would you extend your methods to other sets of points in the plane?
I am really struggling to prove that multiplying the rotation by any angle theta matrix with any point on the unit circle results in a point on the same circle, and likewise with the hyperbola question.
| To prove that rotation of a point on the unit circle by any angle $\theta$ results in a point on the same circle,
note the following.
If $x^2+y^2=1$ and $\begin{pmatrix}x' \\ y'\end{pmatrix}=\begin{pmatrix}\cos\theta &-\sin\theta\\ \sin\theta & \cos\theta \end{pmatrix}\begin{pmatrix}x \\y\end{pmatrix}=\begin{pmatrix}x\cos\theta-y\sin\theta\\ x\sin\theta+y\cos\theta\end{pmatrix}$,
then $x'^2+y'^2=(x\cos\theta-y\sin\theta)^2+(x\sin\theta+y\cos\theta)^2$
$=x^2\cos^2\theta-2xy\cos\theta\sin\theta+y^2\sin^2\theta+x^2\sin^2\theta+2xy\sin\theta\cos\theta+y^2\cos^2\theta$.
Can you now combine terms and simplify using $\cos^2\theta+\sin^2\theta=1$?
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Coefficient of generating function (hard) How to find the coefficient of $x^{46}$ in $\dfrac{1}{1 - x^3 -x^4 -x^{20}}$ without software like Maple?
I tried everything... :(
| $$\frac{1}{1-x^3-x^4-x^{20}}=\sum_{k\geq 0}(x^3+x^4+x^{20})^k $$
and the coefficient of $x^{46}$ in $(x^3+x^4+x^{20})^k$ is the cardinality of the $k$-uples with coordinates in $\{3,4,20\}$ such that the sum of the coordinates equals $46$. We either use two $20s$ and two $3$s (which can be arranged in six ways), a single $20$ and a representation of $46-20=26$ as a sum of $3s$ and $4s$, or just a representation of $46$ with $3$s and $4$s only. In the last cases we need an even number of $3$s since both $26$ and $46$ are even, but at least two $3s$ since neither $26$ or $46$ is a multiple of $4$. The count can be performed by hand now:
$$(3,3,20,20)$$
$$(3,3,4,4,4,4,4,20)$$
$$(3,3,3,3,3,3,4,4,20)$$
$$(3,3,4,4,4,4,4,4,4,4,4,4)$$
$$(3,3,3,3,3,3,4,4,4,4,4,4,4)$$
$$(3,3,3,3,3,3,3,3,3,3,4,4,4,4)$$
$$(3,3,3,3,3,3,3,3,3,3,3,3,3,3,4)$$
together with their anagrams give all the chances and
$$ [x^{46}]\frac{1}{1-x^3-x^4-x^{20}}=\color{red}{3224}.$$
| {
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"url": "https://math.stackexchange.com/questions/3423393",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Prove the divisibility test by $7,11,13$ for numbers more than six digits Prove the divisibility test by $7,11,13$ for numbers more than six digits
Attempt:
We know that $7\cdot 11 \cdot 13 = 1001$. The for a six-digit number, for example, $120544$, we write it as
$$ 120544 = 120120 + 424 = 120\cdot1001 + 424 $$
thus we just check the divisibility of $424$ by $7,11,13$.
Know for a number with more than six digits, for example: $270060340$,
$$270060340 = 270270270 - 209930$$
$$ = 270 \cdot (1001001) - 209930 $$
$$ = 270 \cdot (1001000) + (270 - 209930) =270 \cdot (1001000) - 209660$$
so we check the divisibility of $209660 = 209209 + 451$, or just $451$.
But the test states that: for $270060340$, we group three digits from the right:
$$ 270, 60, 340$$
then check divisibility of $340+270 - (60)$.
How to prove this?
| For 9-digit number: $abcdefghi$,
$$ abcdefghi = abc000000 + defghi = abc000000 + def \cdot 1001 - def + ghi $$
so we check divisibility of
$$ abc000000 - def + ghi = abc \cdot 1001000 - abc000 - def + ghi$$
so we cehck divisibility of
$$ -abc000 - def + ghi = -(abc \cdot 1001 - abc) - def + ghi$$
in the end we only check $abc + ghi - def$.
If 12-digit number:
$$ a_{1} a_{2} ... a_{11}a_{12} = a_{1} a_{2} a_{3} 000000000 + \underbrace{a_{4} ... a_{12}}_{9-digit}$$
so we check the divisibility of
$$ a_{1} a_{2} a_{3} 000000000 + ( a_{10}a_{11}a_{12} + a_{4} a_{5}a_{6} - a_{7}a_{8}a_{9}) $$
or just
$$ = (a_{10}a_{11}a_{12} + a_{4} a_{5}a_{6} - a_{7}a_{8}a_{9} - a{1} a_{2} a_{3}) $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3425319",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find maxima and minima of $xyz$ given $x+y+z=0$ and $x^2+y^2+z^2=1$ Find the maximum and minimum value of the product of three real numbers $x,y,z$. If the sum of these is equal to zero and the sum of its squares is equal to one.
I supose the following equations: $$x+y+z=0$$ $$x^2+y^2+z^2=1$$ $$xyz=k$$
with $k\in \mathbb R$ the number to find.
So I think that I have to solve the system of equations. The problem is from a chapter of implicit derivates in vector calculus, but I don't even know if I started well.
| By $(x+y+z)^2=x^2+y^2+z^2+2(xy+yz+zx)$, we find $xy+yz+zx=-\dfrac{1}{2}$.
Now let's take a polinomial function $P(t)=t^3+bt^2+ct+d$ such that real numbers $x,y,z$ are roots of $P$.
By Vieta formulas, we yields $b=0$, $c=-\dfrac{1}{2}$, $d=-k$ and $P(t)=t^3-t\dfrac{1}{2}-k$.
Now let's define $R(t)=t^3-t\dfrac{1}{2}t$. Hence, $R'(t)=3t^2-\dfrac{1}{2}=0$ and $t=\pm\dfrac{1}{\sqrt{6}}$. Therefore local extreme values of $R$:
$R(\pm\dfrac{1}{\sqrt{6}})=\mp\dfrac{\sqrt{6}}{18}$.
Thus; $x,y,z$ will be real roots of $P$ for $ -\dfrac{\sqrt{6}}{18}\leq k \leq \dfrac{\sqrt{6}}{18}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3426644",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Can someone help me with this algebraic inequality? I found this inequality on twitter and I can't seem to prove the statement.
Prove that for $a,b,c > 0$ that
$$
\frac{a+b+c}{2} \geq \frac{ab}{a+b} + \frac{ac}{a+c} + \frac{bc}{b+c}
$$
After an hour (and a crick in my neck) I've only been able to turn it into
$$
a^3(b+c)+b^3(a+c)+c^3(a+b)-2abc(a+b+c) \geq 0
$$
and I'm not even sure if that's much better.
| We have that by HM-AM inequality
$$\frac{ab}{a+b} + \frac{ac}{a+c} + \frac{bc}{b+c}=\frac{1}{\frac1a+\frac1b} + \frac{1}{\frac1a+\frac1c}+ \frac{1}{\frac1b+\frac1c} \le$$ $$\le\frac{a+b}4+\frac{a+c}4+\frac{b+c}4=\frac{a+b+c}2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3428995",
"timestamp": "2023-03-29T00:00:00",
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Show that if $G$ is free abelian with basis $\{x, y\}$, show that $\{2x+3y, x+y\}$ is also a basis for $G$ This is from Munkres' Section 67, but there is a typo in the original question which is as follows:
Show that if $G$ is free abelian with basis $\{x, y\}$, show that $\{2x+3y, x-y\}$ is also a basis for $G$.
This is indeed a typo and has been disucssed here: Let $x$ , $y$ be the basis of a free abelian group of rank 2, prove that $2x+3y$ and $ x-y$ generate a free subgroup
However, this link https://dbfin.com/topology/munkres/chapter-11/section-67-direct-sums-of-abelian-groups/problem-3-solution/ gives an alternative version, which he claims doable:
Show that if $G$ is free abelian with basis $\{x, y\}$, show that $\{2x+3y, x+y\}$ is also a basis for $G$.
However, I did not follow the proof given in this link.
$2x+3y$ and $x+y$ are clearly independent, since if $$A(2x+3y)+B(x+y)=0,$$ then we have $$(2A+B)x+(3A+B)y=0,$$ since $\{x,y\}$ is a basis, we must have $2A+B=0$ and $3A+B=0$, which implies $A=B=0$.
Thus, $\{2x+3y, x+y\}$ clearly generates a group with them as basis.
However, how could I show this group generated by them is indeed $G$?
I tried to identify $g=AX+BY=C(2X+3Y)+D(X-Y)$ for $g\in G$, and in the end I have $A=2C+D$ and $B=3C-D$, but it seems that the link suggest another set of solutions, but I don't really understand where it comes from..
Any idea? Thank you!
| Let $u=2x+3y$ and $v=x+y$. You want to show that $u$ and $v$ generate $G$. It suffices to show that there exist integers $A$, $B$, $C$, and $D$ such that
\begin{align*}
Au + Bv &= x &\implies A(2x+3y) + B(x+y) &= x &\implies (2A+B)x + (3A+B)y &= x\\
Cu + Dv &= y &\implies C(2x+3y) + D(x+y) &= y &\implies (2C+D)x + (3C+D)y &= y\\
\end{align*}
The first line gives equations
\begin{align*}
2A+B &= 1 \\
3A+B &= 0
\end{align*}
Apparently $A=-1$ and $B=3$ satisfy the system.
The second line gives equations
\begin{align*}
2C+D &= 0 \\
3C+D &= 1
\end{align*}
This time $C=1$ and $D=-2$ work.
What's really going on here is that the matrix which expresses $u$ and $v$ in terms of $x$ and $y$ is invertible over $\mathbb{Z}$:
$$
\begin{bmatrix}u \\ v \end{bmatrix}
= \begin{bmatrix} 2 & 3 \\ 1 & 1 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix}
\implies
\begin{bmatrix} x \\ y \end{bmatrix}
= \begin{bmatrix} 2 & 3 \\ 1 & 1 \end{bmatrix}^{-1}
\begin{bmatrix}u \\ v \end{bmatrix}
= \begin{bmatrix} -1 & 3 \\ 1 & -2 \end{bmatrix}
\begin{bmatrix}u \\ v \end{bmatrix}
$$
| {
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"url": "https://math.stackexchange.com/questions/3431784",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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if $s=\sqrt{x^2+6x+9} +\sqrt{x^2+24x+144}$ and $-10 < x < -8$ there follows that...
if $s=\sqrt{x^2+6x+9} + \sqrt{x^2+24x+144}$ and $-10 < x < -8$ there follows that $s=9$
After noticing that the first radical is $x+3$ and the second one is $x+12$ all I get is $s=2x+15$, which btw. is suggested as an alternative answer in the exam.
So, how to go about this?
| Note that $$x^2+6x+9=(x+3)^2$$ and $$x^2+24x+144=(x+12)^2$$, so you will get
$$s=|x+3|+|x+12|$$, now consider the inequality $$-10<x<-8$$
| {
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Proof $a_{n+1} \geq a_n$ with $a_n := (1+1/n)^n$ for $n \in \mathbb{N}$ Let $a_n := (1+1/n)^n$ for $n \in \mathbb{N}$
How can one prove that $a_{n+1} \geq a_n$ for all $n \in \mathbb{N}$ with Bernoulli's inequality?
I know that the inequality states that $(1+x)^r \geq 1+rx$ for every integer $r \geq 0$ and every real number $x \geq -2$. And if the exponent $r$ is even, then the inequality is valid for all real numbers x.
So I have to use induction and for $n=1$ we would get $(1+1/1)^1 = (1+1/(1+1))^2$, and that would give $2 < 2,25$. But wouldn't that imply that all numbers $> 1$ would make $a_{n+1} > a_n$? At which case is it equal? Can someone show me where I can find an induction proof for this?
| $$
a_{n+1}\ge a_n \quad\Longleftrightarrow\quad
\frac{(n+2)^{n+1}}{(n+1)^{n+1}}\ge\frac{(n+1)^n}{n^n}
\quad\Longleftrightarrow\quad \frac{n+2}{n+1}\ge\left(\frac{n^2+2n+1}{n^2+2n}\right)^n
\\ \quad\Longleftrightarrow\quad \left(1-\frac{1}{n^2+2n+1}\right)^n\ge 1-\frac{1}{n+2}
$$
The last inequality holds since
$$
\left(1-\frac{1}{n^2+2n+1}\right)^n\ge 1-\frac{n}{n^2+2n+1}
$$
and
$$
\frac{1}{n+2}\ge\frac{n}{n^2+2n+1}
$$
| {
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"timestamp": "2023-03-29T00:00:00",
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$2\arctan(\phi^{-n})=\arctan\frac{p}{q}$ or $\arctan\frac{p\sqrt{5}}{q}$, where $\phi$ is the Golden Ratio. Is there a pattern in the $\frac{p}{q}$s? It is very interesting to know that
$$\arctan\frac{1}{\phi} + \arctan\frac{1}{\phi^3}= \arctan 1 = \frac{\pi}{4}$$
where Golden ratio $\phi = \frac12(\sqrt5 +1)$ is in association with circle constant $\pi$.
More interesting phenomenon is evaluation of inverse tan functions of inverse of $\phi$ in its consecutive powers as follows
$$\begin{align}
2\arctan\frac{1}{\phi} &= \arctan 2 &
2\arctan\frac{1}{\phi^2} &= \arctan\frac{2\sqrt{5}}{5}\\
2\arctan\frac{1}{\phi^3} &= \arctan\frac{1}{2} &
2\arctan\frac{1}{\phi^4} &= \arctan\frac{2\sqrt{5}}{15}\\
2\arctan\frac{1}{\phi^5} &= \arctan\frac{2}{11} &
2\arctan\frac{1}{\phi^6} &= \arctan\frac{\sqrt5}{20} \\
2\arctan\frac{1}{\phi^7} &= \arctan\frac{2}{29} &
2\arctan\frac{1}{\phi^8} &= \arctan\frac{2\sqrt5}{105} \\
2\arctan\frac{1}{\phi^9} &= \arctan\frac{1}{38} &
2\arctan\frac{1}{\phi^{10}} &= \arctan\frac{2\sqrt5}{275} \\
2\arctan\frac{1}{\phi^{11}} &= \arctan\frac{2}{199}
\end{align}$$
Here are the observations
*
*Odd powers of inverse $\phi$ in double arctan functions lead to arctan of well defined fractions
*Even powers of inverse $\phi$ in double arctan functions lead to arctan of fractions involving $\sqrt5$.
My curiosity is to know, is there any pattern in these interesting series?
I will be grateful to understand more, if anyone has come across such evaluations.
| The explanation is that $$\tan(2\theta) = \frac{2 \tan(\theta)}{1 - \tan^2(\theta)}$$
so, for $-1 < x < 1$,
$$2 \arctan(x) = \arctan\left(\frac{2x}{1-x^2}\right)$$
Thus
$$ 2 \arctan(1/\phi^n) = = \arctan \left( \frac{2 \phi^{-n}}{1-\phi^{-2n}}\right)
= \arctan\left(\frac{2}{\phi^{n} - \phi^{-n}}\right)$$
Now the Fibonacci numbers $$F_n = \frac{\phi^n - (-1/\phi)^n}{\sqrt{5}}$$
so if $n$ is even, $$2 \arctan(1/\phi^n) = \arctan\left(\frac{2}{\sqrt{5} F_n}\right)$$
On the other hand, the Lucas numbers $$L_n = \phi^n + (-1/\phi)^n$$
so if $n$ is odd, $$2 \arctan(1/\phi^n) = \arctan\left(\frac{2}{L_n}\right)$$
| {
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$(p-1)(p+1)/24 \in \mathbb N$ for all primes $p \geq 5$ I want to show
\begin{align}
\frac{(p-1)(p+1)}{24} \in \mathbb N \quad \text{for all primes} \quad p \geq 5 \tag{1}.
\end{align}
I can show $(1)$, if the following statement is true.
Let $a,b,c,d \in \mathbb N$ and $a \geq b \cdot c \cdot d$.
\begin{align}
\text{If} \quad \frac{a}{b},\frac{a}{c},\frac{a}{d} \in \mathbb N, \quad \text{then} \quad \frac{a}{b \cdot c \cdot d} \in \mathbb N \tag{2}.
\end{align}
Given $(2)$ we show that $(1)$ is true for $a = (p-1)(p+1)$, $b = 2$, $c = 3$ and $d = 4$. Since $p$ is a prime $(p-1)$ and $(p + 1)$ are even, implying $(p-1)/2 \in \mathbb N$, $(p+1)/2 \in \mathbb N$ and thus $(p-1)(p+1)/2 \in \mathbb N$ and $(p-1)(p+1)/4 \in \mathbb N$. One of the three numbers $(p-1)$, $p$ and $(p+1)$ must be divisible by 3. Since $p$ is a prime either $(p-1)$ or $(p+1)$ is divisible by 3, implying $(p-1)(p+1)/3 \in \mathbb N$.
Question Is $(2)$ true?
| As written $(2)$ is not true. Consider $(a,b,c,d) = (36,2,3,4)$.
Clearly, $36 \ge 24 = 2 \cdot 3 \cdot 4$ and $\dfrac{36}{2} = 18$, $\dfrac{36}{3} = 12$, $\dfrac{36}{4} = 9$ are all positive integers.
However, $\dfrac{36}{24} = \dfrac{3}{2}$, which is not an integer.
If you know that $\gcd(b,c) = \gcd(c,d) = \gcd(b,d) = 1$, then $\dfrac{a}{b},\dfrac{a}{c},\dfrac{a}{d} \in \mathbb{N} \implies \dfrac{a}{bcd} \in \mathbb{N}$ is true.
For the problem you are given, notice that $24 = 2^3 \cdot 3$. So if you can show that $(p-1)(p+1)$ is divisible by $2^3 = 8$ and $(p-1)(p+1)$ is divisible by $3$, then it must hold that $(p-1)(p+1)$ is divisible by $\text{lcm}(8,3) = 24$.
| {
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Issue with the following limit $\bigg(2 \sqrt{1 + \frac{1}{n}}\bigg)^n$ Calculate the following limit:
$\bigg(2 \sqrt{1 + \frac{1}{n}}\bigg)^n$
When I calculate it I get to different answers.
First way (Edit: this is where I did the mistake): $$\bigg(2 * \sqrt{1 + \frac{1}{n}}\bigg)^n = \bigg({4 + \frac{4}{n} \bigg)^\frac{n}{2}} = \bigg({4 + \frac{4}{n} \bigg)^{\frac{n}{4} \cdot \frac{4}{n}\cdot\frac{n}{2}}}$$
When we do $\lim_{n \to \infty}\bigg(\bigg({4 + \frac{4}{n} \bigg)^{\frac{n}{4}\cdot \frac{4}{n}\cdot\frac{n}{2}}}\bigg)$ we get $e^2$
Now the second way:
$$\bigg(2 \cdot \sqrt{1 + \frac{1}{n}}\bigg)^n = 2^n\cdot (1 + \frac{1}{n})^{n \cdot \frac{1}{2}}$$
When we do limit out of this we get $2^\infty \cdot \sqrt{e}$ which is of course $\infty$.
Could someone point out the mistake I made?
Edit:
I just realised where my mistake lies! I mistakenly thought that $(4 + \frac{4}{n})^\frac{n}{4} = e$ which is false, actually $(1 + \frac{4}{n})^\frac{n}{4} = e$. The second way of calculating this limit is the correct one!
| What about using the limit chain rule.
$$\lim_{n\to \infty} \left(4+\frac{4}{n}\right)^{\frac{n}{2}} =\lim_{n\to \infty} e^{\frac{n}{2}\ln\left(4+\frac{4}{n}\right)}$$
Here you can see why the limit diverges (how does $\frac{n}{2}\ln\left( 4+\frac{4}{n}\right)$ behave as $n$ tends to infinity), as opposed to something like $\left(1+\frac{a}{n}\right)^{\frac{n}{2}}$.
| {
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Find the 4 roots of the equation: $z^{4}+4=0$ Find the 4 roots of the equation: $z^{4}+4=0$
Note: I can't use the form: $e^{i\theta}$
My attempt:
Note we have: $z=(-4)^{\frac{1}{4}}$, consider $w=-4$, then $w^{1/4}=z$ this implies $w=z^4$.
Consider the polar form of $z$ and $w$:
$|w|=r=4$,
$Arg(w)=\pi=\theta$
Then,
$$w=4\cos\pi+i\sin\pi$$
$$z=p\cos\phi+i\sin\phi$$
This implies: $w=z^4$ iff $4=p^4$ and $4\phi=\pi+2k\pi$ iff $p=4^{1/4}$ and $\phi=\frac{\pi+2k\pi}{4}$ with $k=0,1,2,3$
Then the roots are:
$$z_k=4^{1/4}(\cos{\frac{\pi+2k\pi}{4}}+i\sin{\frac{\pi+2k\pi}{4}})$$
with $k=0,1,2,3$
is correct this?
| You solution looks OK; the writing can be improved slightly though. I will follow your argument and rewrite it as follows.
The equation is equivalent to $z^4=-4$.
Now let $z=r(\cos \theta+i\sin\theta)$ and write
$$
-4=4(\cos \pi +i\sin \pi)\tag{1}
$$
By the De Moivre's formula,
$$
z^4=r^4(\cos (4\theta)+i\sin (4\theta))\tag{2}
$$
Comparing (1) and (2), we have
$$
r=\sqrt[4]{4},\quad 4\theta=(2k+1)\pi,\quad k\in\mathbb{Z}.
$$
Thus the four roots are
$$
z_k=4^{1/4}(\cos{\frac{\pi+2k\pi}{4}}+i\sin{\frac{\pi+2k\pi}{4}})$$
with $k=0,1,2,3$.
[Added:] The solution can be simplified further:
*
*Observe that $4^{1/4}=2^{2/4}=\sqrt{2}$.
*The expression $\cos{\frac{\pi+2k\pi}{4}}+i\sin{\frac{\pi+2k\pi}{4}}$ can be found exactly by hand and thus can be used to simply the solution. For instance,
$$
z_0=\sqrt{2}(\frac{\sqrt{2}}{2}+i\frac{\sqrt{2}}{2})=1+i.
$$
I will leave the cases $k=1,2,3$ to you.
| {
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Suppose $a,b$ are real numbers and $0 < a < b$. Prove $ba^n + ab^n < a^{n+1} + b^{n+1}$
$a,b$ are real numbers and $0 < a < b$
Prove that for all $n ≥ 1$, $ba^n + ab^n < a^{n+1} + b^{n+1}$
My attempt:
By contradiction. Suppose
$$\tag{* }ba^n + ab^n ≥ a^{n+1} + b^{n+1}$$
Rearranging $(*)$ gives
$$\begin{align}
0 & ≥ a^{n+1} + b^{n+1} - ba^n - ab^n \\
& ≥ a^{n+1} - ba^n + b^{n+1} - ab^n \\
& ≥ a^n(a - b) - b^{n}(a - b) \\
& ≥ (a - b)(a^n - b^n)
\end{align}$$
But since $0 < a < b$, we have $(a-b) < 0$ and $(a^n - b^n) < 0$, and thus $(a-b)(a^n - b^n) > 0$.
Hence a contradiction. $\Box$
*
*Is it correct?
*Is there a better way? (Preferably direct proof)
| Yes, it is correct. But it is more natural to apply the same idea to get a direct proof:$$a^{n+1}+b^{n+1}-ba^n-ab^n=(b^n-a^n)(b-a)>0.$$
| {
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Prove that if $f(x) = \frac{\sqrt{x} - \sqrt{a}}{x-a}$, then $f$ has a limit at a. Prove that if $f(x) = \frac{\sqrt{x} - \sqrt{a}}{x-a}$, then $f$ has a limit at a.
I have written a very basic level proof here:
Given $\epsilon > 0$, we want $\delta > 0$ such that $|(\frac{1}{\sqrt{x} + \sqrt{a}}) - (\frac{1}{2\sqrt{a}})| < \epsilon$ whenever $0<|x - a|<\delta$. So for $\delta = g(\epsilon)$, $0<|x - a|<\delta \implies |(\frac{1}{\sqrt{x} + \sqrt{a}}) - (\frac{1}{2\sqrt{a}})| < \epsilon$.
I tried and failed to find a way to represent delta in terms of epsilon. Saying, $\delta = g(\epsilon)$, is wrong. I saw some examples where they set $\delta=1$ and the used the $min$ function. I tried this but was not successful.
Edit: I need to find an answer using the epsilon-delta form. just solving for the limit is not sufficient.
Anyone have any tips?
| We have that for $a> 0$
$$f(x) = \frac{\sqrt{x} - \sqrt{a}}{x-a}= \frac{\sqrt{x} - \sqrt{a}}{x-a} \frac{\sqrt{x} + \sqrt{a}}{\sqrt{x} + \sqrt{a}}=$$
$$=\frac{x-a}{(x-a)(\sqrt{x} + \sqrt{a})}=\frac{1}{\sqrt{x} + \sqrt{a}} \to \frac1{2\sqrt a}$$
As an alternative assuming wlog $|x-a|<\frac a 2$
$$ \left|\frac{\sqrt{x} - \sqrt{a}}{x-a}-\frac1{2\sqrt a}\right|=\left|\frac{1}{\sqrt{x} + \sqrt{a}}-\frac1{2\sqrt a}\right|=\left|\frac{2\sqrt a-\sqrt{x} - \sqrt{a}}{2\sqrt a(\sqrt{x} + \sqrt{a})}\right|=$$
$$=|x-a|\left|\frac{\sqrt a-\sqrt{x} }{2\sqrt a(\sqrt{x} + \sqrt{a})(x-a)}\right|<|x-a|\left|\frac{\sqrt a }{2\sqrt a(3 \sqrt{a})\frac a 2}\right|=\frac{|x-a|}{3a^\frac32}$$
therefore assuming $\delta=3\epsilon a^\frac32$ we have
$$ \left|\frac{\sqrt{x} - \sqrt{a}}{x-a}-\frac1{2\sqrt a}\right|<\frac{|x-a|}{3a^\frac32}<\epsilon$$
| {
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Friend squares numbers
Two perfect squares are friends if one is obtained from the other
adding the digit $1$ at the left.
For instance, $1225 = 35^2$ and $225 = 15^2$ are friends.
Prove that there are infinite pairs of odd perfect squares that are
friends.
Solution: Suppose that $n^2$ and $m^2$ are friends and both odd, with the former being greater than the latter. Then $n$ and $m$ are both odd, and $n^2 - m^2$ is a power of ten.
Then, because of the difference-of-squares identity $n^2 - m^2 = (n-m)(n+m)$, this means that $n+m = 2^a 5^b$ for some $a, b$ and $n-m = 2^c 5^d$ for some $c, d$, where $a + c = b + d$.
Adding the two equations, we obtain $2n = 2^a 5^b + 2^c 5^d$. The right hand side has to be even, but not a multiple of 4. This means that $a$ and $c$ have to be both positive or zero. If they are both positive, they cannot be both at least two (since the right hand side would be a multiple of 4), but they cannot be both one (since $2^a 5^b + 2^c 5^d = 2(5^b + 5^d)$ would be a multiple of 4), so one of them has to be one and the other has to be at least two. If $a$ and $c$ are both zero, then $2^a 5^b + 2^c 5^d = 5^b + 5^d \equiv 1^b + 1^d \equiv 2\pmod{4}$, so all such cases work.
Similarly, subtracting the two equations, we obtain $2m = 2^a 5^b - 2^c 5^d$. The right hand side has to be even, but not a multiple of 4. This means that $a$ and $c$ have to be both positive or both zero. If they are both positive, they cannot be both at least two, and they cannot be both one, so one of them has to be one and the other has to be at least two. However, if $a$ and $c$ are both zero, then $2^a 5^b - 2^c 5^d = 5^b - 5^d \equiv 1^b - 1^d \equiv 0\pmod{4}$, so no such cases work.
In conclusion, we have demonstrated that $n^2$ and $m^2$ are friends and both odd iff $n+m = 2^a 5^b$ for some $a, b$ and $n-m = 2^c 5^d$ for some $c, d$, where $a + c = b + d$ and where $a = 1, c \geq 2$ or vice versa. This clearly covers infinitely many cases, so we are done.
Why is $ n ^ 2 - m ^ 2 $ a power of ten?
|
Why is $ n ^ 2 - m ^ 2 $ a power of ten?
Because they are friends: the power of ten is the digit $1$ on the left, followed by as many zeros as there are digits in the smallest of the numbers.
In more detail, if $n^2=\overline{n_1\cdots n_r}$ and $m^2=\overline{m_1\cdots m_s}$ are friends and $n>m$, where the bar represents the decimal expansion of the number at hand, then $r=s+1$, $n_1=1$, and $m_i=n_{i+1}$ for all $i\in\{1, \dots , s\}$. Hence
$$\begin{align}
n^2-m^2&=\overline{1m_1\cdots m_{s}}-\overline{m_1\cdots m_s}\\
&=\overline{1\underbrace{0\cdots 0}_{s\text{ times}}}\\
&=10^s.
\end{align}$$
| {
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Find the given limit of $S$?
Find the given limit of $S$ ?
$$S=\lim _{n \rightarrow \infty} \left (\frac{\sin \frac{\pi}{ n+1}} { 1} +\frac{\sin \frac{2\pi}{ n+1}} { 2} + + ....+ \frac{\sin \frac{n\pi}{ n+1}} { n} \right )$$
My attempt :
i construct modified the given series
$\frac{\pi}{ n+1}\left (\frac{\sin \frac{\pi}{ n+1}} { \frac{1\pi}{ n+1}} +\frac{\sin \frac{2\pi}{ n+1}} { \frac{2\pi}{ n+1}} + ... + ....+ \frac{\sin \frac{n\pi}{ n+1}} { \frac{n\pi}{ n+1}} \right )$
After that im not able to proceed further
Any hints/solution will be appreciated
thanks u
| The limit exists and is bounded above as
$$\lim_{n \to \infty} \sum_{k=1}^n \frac{1}{k}\sin \left( \frac{k\pi}{n+1}\right) \leqslant \lim_{n \to \infty} \sum_{k=1}^n \frac{\pi}{n+1} = \lim_{n \to \infty} \frac{n}{n+1} \pi = \pi$$
| {
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Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.