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Computing $\lim_{x \to 0} \frac{\cos x - \sqrt{2 - e^{x^2}}}{\ln{(\cos x) + \frac{1}{2} x \sin x}} \cdot \frac{(x+2)^{2017}}{(x-2)^{2015}}$ I'm studying for an exam, but I have trouble with computing the following limit:
$$\lim_{x \to 0} \frac{\cos x - \sqrt{2 - e^{x^2}}}{\ln{(\cos x) + \frac{1}{2} x \sin x}} \cdot \frac{(x+2)^{2017}}{(x-2)^{2015}}$$
I tried directly plugging in a $0$, but that just results in $\frac{0}{0}$. Using L'Hospital's rule doesn't seem like it would help simplify this. Any help would be appreciated.
| This is another case of intimidation via use of large numbers. The part $(x + 2)^{2017}/(x - 2)^{2015}$ is a rational function which is defined for $x = 0$ and hence its limit as $x \to 0$ is same as its value at $x = 0$ and thus the limit of this part is $-4$. We can thus proceed as follows
\begin{align}
L &= \lim_{x \to 0}\dfrac{\cos x - \sqrt{2 - e^{x^{2}}}}{\log\cos x + \dfrac{1}{2}x \sin x}\cdot\frac{(x + 2)^{2017}}{(x - 2)^{2015}}\notag\\
&= -4\lim_{x \to 0}\dfrac{2\cos x - 2\sqrt{2 - e^{x^{2}}}}{2\log\cos x + x \sin x}\notag\\
&= -8\lim_{x \to 0}\dfrac{\cos x - \sqrt{2 - e^{x^{2}}}}{2\log\cos x + x \sin x}\notag\\
&= -8\lim_{x \to 0}\dfrac{\cos^{2} x - (2 - e^{x^{2}})}{\{2\log\cos x + x \sin x\}\{\cos x + \sqrt{2 - e^{x^{2}}}\}}\notag\\
&= -4\lim_{x \to 0}\dfrac{\cos^{2} x - (2 - e^{x^{2}})}{2\log\cos x + x \sin x}\tag{1}
\end{align}
Now we can see that $$\frac{d}{dx}\log\cos x = -\tan x = -x - \frac{x^{3}}{3} + o(x^{3})$$ and hence via integration it follows that $$\log\cos x = - \frac{x^{2}}{2} - \frac{x^{4}}{12} - o(x^{4})$$ so that $$2\log\cos x + x\sin x = - \frac{x^{4}}{3} + o(x^{4})\tag{2}$$ and clearly $$\cos^{2}x - 2 + e^{x^{2}} = \frac{\cos 2x + 2e^{x^{2}} - 3}{2} = \frac{5x^{4}}{6} + o(x^{4})\tag{3}$$ It should now be obvious from equations $(1), (2)$ and $(3)$ that the desired limit is $$-4\cdot\frac{5/6}{-1/3} = 10$$ I have tried to get the required Taylor series expansion for $\log\cos x$ without using any tedious calculation.
| {
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"url": "https://math.stackexchange.com/questions/1920673",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Prove: $\cos^3{A} + \cos^3{(120°+A)} + \cos^3{(240°+A)}=\frac {3}{4} \cos{3A}$ Prove that:
$$\cos^3{A} + \cos^3{(120°+A)} + \cos^3{(240°+A)}=\frac {3}{4} \cos{3A}$$
My Approach:
$$\mathrm{R.H.S.}=\frac {3}{4} \cos{3A}$$
$$=\frac {3}{4} (4 \cos^3{A}-3\cos{A})$$
$$=\frac {12\cos^3{A} - 9\cos{A}}{4}$$
Now, please help me to continue from here.
| Consider the following picture. We see that the triangle with vertices $(\cos A, \sin A), (\cos(A+120^\circ), \sin(A+120^\circ)), (\cos(A+240^\circ), \sin(A+240^\circ))$ are the vertices of an equilateral triangle with centriod at the origin.
Thus if we set $a = \cos A, b = \cos(A+120^\circ), c = \cos(A+240^\circ)$, then
\begin{align*}
a+b+c = 0
\end{align*}
and hence
\begin{align*}
a^3+b^3+c^3 = 3abc
\end{align*}
and
\begin{align*}
\cos^3{A} + \cos^3{(120°+A)} + \cos^3{(240°+A)} &= 3\cos A \cos(A+120^\circ)\cos(A+240^\circ) \\
&= \frac{3}{2}(2\cos A \cos(A+120^\circ))\cos(A+240^\circ) \\
&= \frac{3}{2}[\cos(2A+120^\circ)+\cos(120^\circ)]\cos(A+240^\circ)\\
&=\frac{3}{2}\cos(2A+120^\circ)\cos(A+240^\circ)-\frac{3}{4}\cos(A+240^\circ)\\
&=\frac{3}{4}\cos(3A+360^\circ)+\frac{3}{4}\cos(A-120^\circ) -\frac{3}{4}\cos(A+240^\circ)\\
&=\frac{3}{4}\cos(3A)
\end{align*}
since $\cos(A+240^\circ) = \cos(360^\circ - (A+240^\circ)) = \cos(120^\circ -A) = \cos(A-120^\circ)$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Epsilon-delta limit definition I got $\lim_{x \rightarrow 3} \frac{x^2-2x+1}{x-2} = 4$
I need to prove that by delta epsilon. I came to delta ={1/2epsilon,1/2}
Is that right? If not can you explain me how?
| |$\frac {(x-3)(x-3)}{x-2}|$< $\epsilon$
$\delta$ <= $\frac{1}{2}$
|x-3|<$\delta$ <=$\frac{1}{2}$
-$\frac{1}{2}$ < x-3 < $\frac{1}{2}$
$\frac{1}{2}$ < x-2 < $\frac{3}{2}$
2>$\frac {1}{x-2}$>$\frac{2}{3}$
$\delta$ = min{$\frac{1}{2}$$\epsilon$, $\frac{1}{2}$}
|$\frac {(x-3)(x-3)}{x-2}|$ < $\delta$*2 = $\epsilon$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1921647",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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Find function $f(x)$, such that $f(x+ \frac{1}{x}) = x^2 + \frac{1}{x^2}$. The question is,
Find function $f(x)$, if $f(x+ \frac{1}{x}) = x^2 + \frac{1}{x^2}$.
What does this mean?
Do I have to find $x$ in $x+ \frac{1}{x} = x^2 + \frac{1}{x^2}$?
In this case (not counting solutions in the complex plane), $x = 1; f(x) = 2$.
Or replace all $x$'s in $x+ \frac{1}{x}$, so that it would equal $x^2 + \frac{1}{x^2}$?
In that case, $f(x) = x^2$
Or some other option?
| Observe that
$$\left(x+\frac1x\right)^2=x^2+\frac1{x^2}+2$$
so in fact
$$f(x):=x^2-2\;\;\;\text{gives}\;\;\;f\left(x+\frac1x\right)=\left(x+\frac1x\right)^2-2=x^2+\frac1{x^2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1924927",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Distributing 4 distinct balls between 3 people
In how many ways can you distribute 4 distinct balls between 3 people such that none of them gets exactly 2 balls?
This is what I did (by the inclusion–exclusion principle) and I'm not sure, would appreciate your feedback:
$$3^4-\binom{3}{1}\binom{4}{2}\binom{2}{1}^2+\binom{3}{2}\binom{4}{2}\binom{2}{1}$$
| Method 1: If no person receives exactly two balls, there are two possible cases. One person receives all four balls or one person receives three balls and another person receives one ball.
Case 1: One person receives all four balls.
There are three ways of selecting the person who receives all the balls.
Case 2: There are three ways of selecting the person who receives three balls, $\binom{4}{3}$ ways of selecting which three of the four balls that person receives, and two ways of choosing the person who receives the remaining ball. Consequently, there are
$$\binom{3}{1}\binom{4}{3}\binom{2}{1}$$
of distributing three of the balls to one person and one ball to another person.
Note that Parcly Taxel and Jon Mark Perry obtained the factor of $$3! = \binom{3}{1}\binom{2}{1}$$ in this case by making the observation that there are $3!$ ways of distributing different numbers of balls to three different people.
Total: Since these cases are disjoint, the number of ways of distributing four distinct balls to three people so that no person receives exactly two balls is
$$\binom{3}{1} + \binom{3}{1}\binom{4}{3}\binom{2}{1}$$
Method 2: Exclude those cases in which one person receives exactly two balls from the total.
Since both the people and the balls are distinct, there would be $3^4$ ways of distributing the balls to the people if there were no restrictions.
There are two cases in which a person receives exactly two balls. Either one person receives two balls, while the others receive one each, or two people receive two balls each.
Case 1: One person receives two balls, while the others receive one each.
There are three ways to select the person who receives two balls, $\binom{4}{2}$ ways of selecting the balls that person receives, two ways of selecting a ball for the remaining person whose name appears first alphabetically, and one way to give the remaining person the remaining ball.
$$\binom{3}{1}\binom{4}{2}\binom{2}{1}\binom{1}{1}$$
Case 2: Two people each receive two balls.
There are $\binom{3}{2}$ ways to select which two people receive two balls, $\binom{4}{2}$ ways to choose which two balls are given to the selected person whose name appears first alphabetically, and one way to give the remaining two balls to the other selected person.
$$\binom{3}{2}\binom{4}{2}\binom{2}{2}$$
Total: Since the cases are disjoint, the number of ways of distributing four distinct balls to three people so that no person receives exactly two balls is
$$3^4 - \binom{3}{1}\binom{4}{2}\binom{2}{1} - \binom{3}{2}\binom{4}{2}$$
Method 3: We use the Inclusion-Exclusion Principle.
There are $3^4$ ways of distributing four distinct balls to three people.
We must exclude those cases in which a person receives exactly two balls.
There are three ways of selecting a person to receive exactly two balls, $\binom{4}{2}$ ways of choosing which balls that person receives, and $2^2$ ways of distributing the remaining two balls to the other people. Hence, there are
$$\binom{3}{1}\binom{4}{2}2^2$$
ways of distributing the balls so that a person receives exactly two of them.
However, we have counted those distributions in which two people receive two balls each twice, once for each of the people we designated as the person who receives two balls. In Method 2, we calculated that the number of distributions in which two people each receive two balls is
$$\binom{3}{2}\binom{4}{2}$$
Hence, by the Inclusion-Exclusion Principle, the number of ways of distributing four distinct balls to three people so that no person receives exactly two balls is
$$3^4 - \binom{3}{1}\binom{4}{2}2^2 + \binom{3}{2}\binom{4}{2}$$
You calculated the last term incorrectly.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1926503",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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$P:\mathbb{R}^2 \to \mathbb{R}, P(x,y) = x.y$ is continuous. I need to prove $P:\mathbb{R}^2 \to \mathbb{R}, P(x,y) = x.y$ is continuous.
I just proved that the sum is continuous, but I'm lost at how to manipulate the inequalities in this case.
My attempt:
Let $\epsilon>0$.
$d((x,y),(a,b)) = \sqrt{(x-a)^2 + (y-b)^2} $. So I need to find some $\delta$ such that $\sqrt {(x-a)^2 + (y-b)^2} < \delta \Rightarrow |xy-ab| < \epsilon $.
I verified that $\sqrt{(x-a)^2 + (y-b)^2} \geq |x-a| $ and $\sqrt{(x-a)^2 + (y-b)^2} \geq |y-b|$, but can't see if that helps. I don't know how can I find $\delta$ in this case, can someone give me a hint?
Thanks.
| Fix $a,b\in\mathbb{R}$, then we have the following:
Scratch work: If $d((x,y),(a,b))=|x-a|+|y-b|<\delta$ then $|x-a|<\delta$ and $|y-b|<\delta$. So we have:
\begin{align*}
x-a&<\delta\\
x+a&<\delta+2a\\
\end{align*}
and so we have:
\begin{align*}
|(x+a)(y-b)|&=|xy-ab + (ay-bx)|\geq |xy-ab|\\
&\Downarrow\\
|xy-ab|&\leq |(x+a)(y-b)|\\
&|xy-ab|<(\delta+2a)\cdot \delta :=\epsilon
\end{align*}
So, the appropriate $\delta$ might be the solution of the equation
\begin{align*}
\delta^2+2a\delta&=\epsilon\\
(\delta+a)^2&=\epsilon+a^2\\
\delta&=\sqrt{\epsilon+a^2}-a
\end{align*}
Proof of continuity: Take $\epsilon>0$ arbitrary, and put $\delta=\sqrt{\epsilon+a^2}-a$. Then if $d((x,y),(a,b))=|x-a|+|y-b|<\delta$ we have $|y-b|<\delta$ and $|x+a|<\delta+2a$, which implies:
\begin{align*}
|xy-ab|&\leq |xy-ab+ay-bx| &\text{(by triangle inequality)}\\
&=|(x+a)(y-b)|<(\delta+2a)\cdot \delta &\\
&=\left(\left(\sqrt{\epsilon+a^2}-a\right)+2a\right)\cdot \left(\sqrt{\epsilon+a^2}-a\right) &\text{(replacing $\delta=\sqrt{\epsilon+a^2}-a$)}\\
&=\left(\sqrt{\epsilon+a^2}+a\right)\cdot \left(\sqrt{\epsilon+a^2}-a\right)\\
&=\left(\sqrt{\epsilon+a^2}\right)^2 - a^2\\
&=\epsilon+a^2-a^2=\epsilon.
\end{align*}
So, since for every $\epsilon>0$ there is $\delta>0$ such that $d((x,y),(a,b))<\delta$ implies $|xy-ab|<\epsilon$, we conclude that the function $P(x,y)=x\cdot y$ is continuous in $(a,b)$.
Answer after edited question:
Notice that $d((x,y),(a,b))=\sqrt{(x-a)^2+(y-b)^2}<\delta$ implies
\begin{align*}
(x-a)^2<\delta^2 &\Rightarrow |x-a|<\delta\\
(y-b)^2<\delta^2 &\Rightarrow |y-b|<\delta
\end{align*}
So the previous proof still applies for the edited question.
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove inequality $2e^x>x^3+x^2$ If $x \in \Bbb R$, show that
$$2e^x>x^3+x^2$$
This inequality is right, see
Own ideas: If $x\in \Bbb R$,
$$f(x)=2e^x-x^3-x^2$$
$$f'(x)=2e^x-3x^2-2x$$
$$f''(x)=2e^x-6x-2$$
$$f'''(x)=2e^x-6$$
$$f''''(x)=2e^x>0$$
Like the symbol cannot judge $f$ sign.
So how can we show this $$f(x)>0 \text{ for } x \in \Bbb R?$$
| *
*For $x \leq 1$, notice that
$$ x^3 + x^2 \leq x + 1 \leq e^x \leq 2e^x. $$
*For $x \geq 1$, it suffices to prove that $f(x) := \log(2e^x) - \log(x^3 + x^2) > 0$. Differentiating twice,
$$ f'(x) = 1 - \frac{2}{x} - \frac{1}{x+1}, \qquad f''(x) = \frac{2}{x^2} + \frac{1}{(x+1)^2} $$
This shows that $f$ is strictly convex and attains global minimum on $[1, \infty)$ at $x = 1+\sqrt{3}$. Now the conclusion follows from
$$ f(1+\sqrt{3}) = 1+\sqrt{3} - 2\log(2+\sqrt{3}) \approx 0.098135 > 0. $$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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A problem about quadratic function and inequality Let $f(x) = ax^2 + bx + c(c \neq 0)$. If
\begin{equation*}
|f(0)| \leqslant 1, |f(1)| \leqslant 1, |f(-1)| \leqslant 1,
\end{equation*} try to show that
\begin{equation*}
\text{for each } |x| \leqslant 1, \text{ we have }|f(x)| \leqslant \frac{5}{4}.
\end{equation*}
Apparently we can verify this by express $a$, $b$, $c$ in terms of $f(0)$, $f(1)$ and $f(-1)$,
\begin{equation*}
\left\{ \begin{array}{l} f(0)=c, \\ f(1)=a+b+c, \\ f(-1)=a-b+c. \end{array} \right. \Longrightarrow \left\{ \begin{array}{l}a=\frac{f(1)+f(-1)}{2}-f(0), \\ b=\frac{f(1)-f(1)}{2}, \\ c=f(0). \end{array} \right.
\end{equation*}
I know we can solve it in the above way, but I do wonder if there is any other ideas ??? Is this the only way ???
| We have that
$$f(x)=f(0)(1-x)(1+x)+\frac{f(1)}{2}x(1+x)-\frac{f(-1)}{2}x(1-x)$$
then for $x\in [-1,1]$,
$$|f(x)|\leq |f(0)||1-x^2|+\frac{|f(1)|}{2}|x(1+x)|+\frac{|f(-1)|}{2}|x(1-x)|\\
\leq |1-x^2|+\frac{1}{2}|x(1+x)|+\frac{1}{2}|x(1-x)|\\
=(1-x^2)+\frac{1}{2}|x|(1+x)+\frac{1}{2}|x|(1-x)$$
Now the RHS is an even function such that for $x>0$ is equal to $1+x-x^2$. Its maximum value in $[0,1]$ is $5/4$.
| {
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Convergence/divergence of $\sum_{k=1}^\infty\frac{2\times 4\times 6\times\cdots\times(2k)}{1\times 3\times 5\times\cdots\times(2k-1)}$ A problem asks me to determine if the series
$$\sum_{k=1}^\infty \frac{2 \times 4 \times 6 \times \cdots \times (2k)}{1 \times 3 \times 5 \times \cdots \times (2k-1)}$$
converges or diverges.
(from the textbook Calculus by Laura Taalman and Peter Kohn (2014 edition); section 7.6, p. 639, problem 33)
I am allowed to use the ratio test first and then any other convergence/divergence test if the former test does not work. In my original work, I attempted the ratio test and it was rendered inconclusive.
$$ a_k = \frac{2 \times 4 \times 6 \times \cdots \times (2k)}{1 \times 3 \times 5 \times \cdots \times (2k-1)} $$
$$ a_{k + 1} = \frac{2 \times 4 \times 6 \times \cdots \times (2k) \times (2(k+1))}{1 \times 3 \times 5 \times \cdots \times (2k-1) \times (2(k+1)-1)} = \frac{2 \times 4 \times 6 \times \cdots \times (2k) \times (2k+2)}{1 \times 3 \times 5 \times \cdots \times (2k-1) \times (2k+1)} $$
$$ \frac{a_{k + 1}}{a_k} = \frac{\frac{2 \times 4 \times 6 \times \cdots \times (2k) \times (2k+2)}{1 \times 3 \times 5 \times \cdots \times (2k-1) \times (2k+1)}}{\frac{2 \times 4 \times 6 \times \cdots \times (2k)}{1 \times 3 \times 5 \times \cdots \times (2k-1)}} = \frac{2 \times 4 \times 6 \times \cdots \times (2k) \times (2k+2)}{1 \times 3 \times 5 \times \cdots \times (2k-1) \times (2k+1)} \times \frac{1 \times 3 \times 5 \times \cdots \times (2k-1)}{2 \times 4 \times 6 \times \cdots \times (2k)} = \frac{2k+2}{2k+1}$$
Evaluating $\rho = \lim_{x \to \infty} \frac{a_{k + 1}}{a_k}$ will determine if $\sum_{k=1}^\infty \frac{2 \times 4 \times 6 \times \cdots \times (2k)}{1 \times 3 \times 5 \times \cdots \times (2k-1)}$ converges or diverges. The conclusions for the ratio test are as follows:
$\circ$ If $\rho < 1$, then $\sum_{k=1}^\infty a_k$ converges.
$\circ$ If $\rho > 1$, then $\sum_{k=1}^\infty a_k$ diverges.
$\circ$ If $\rho = 1$, then the test is inconclusive.
$$ \rho = \lim_{x \to \infty} \frac{a_{k + 1}}{a_k} = \lim_{x \to \infty} \frac{2k+2}{2k+1} = 1$$
Since $\rho = 1$, the ratio test is rendered inconclusive, as I stated earlier. I will have to use other convergence/divergence tests to solve the problem.
My issue is that I'm not sure which other convergence/divergence test to use. Any suggestions?
Many thanks for the help.
| The series is divergent since $a_k>1$ for every $k$.
| {
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Prove the inequality $\sqrt{a^2+b^2}+\sqrt{c^2+b^2}+\sqrt{a^2+c^2}<1+\frac{\sqrt2}{2}$ Let $a,b,c -$ triangle side and $a+b+c=1$. Prove the inequality
$$\sqrt{a^2+b^2}+\sqrt{c^2+b^2}+\sqrt{a^2+c^2}<1+\frac{\sqrt2}{2}$$
My work so far:
1) $a^2+b^2=c^2-2ab\cos \gamma \ge c^2-2ab$
2) $$\sqrt{a^2+b^2}+\sqrt{c^2+b^2}+\sqrt{a^2+c^2}\le3\sqrt{\frac{2(a^2+b^2+c^2)}3}$$
| A similar approach to the one suggested by Jack to obtain the loose inequality: the function $\sqrt{a^2+b^2}+\sqrt{c^2+b^2}+\sqrt{a^2+c^2}$ is convex in $a,b,c$ being the sum of three convex functions (they're the restriction of the Euclidean norm to the planes $\{a=0\}$ etc.). The domain of the function is
$$\{a,b,c\geq 0\}\cap\{a+b+c=1\}\cap \{a\leq b+c\}\cap\{b\leq c+a\}\cap\{c\leq a+b\}$$
which is a polygon in $\mathbb{R}^3$, actually an equilateral triangle inside the plane $\{a+b+c=1\}$ if you want to visualize it, with vertices $(0,\frac12,\frac12)$ and cyclicals. Therefore to know the maximum it is sufficient to look at the extremal points of the domain, which are the three vertices.
In this way you obtain the loose inequality. The strict one comes from the fact that you want nondegenerate triangles, and you might have to work a bit to prove that restricted to the sides of the triangle the function is strictly convex.
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "3",
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not able to prove a trig identity For the identity,
$$\dfrac{\cos A}{1 - \tan A} + \dfrac{\sin A}{1 - \cot A}=\sin A + \cos A$$
What I have been able to perform,
$$
\dfrac{\cos A}{\dfrac{\cos A}{\cos A}-\dfrac{\sin A}{\cos A}} + \dfrac{\sin A}{\dfrac{\sin A}{\sin A} - \dfrac{\cos A}{\sin A}}$$
$$
\dfrac{\cos^2 A}{\cos A - \sin A} + \dfrac{\sin^2 A}{\sin A - \cos A}
$$
I am unable to prove this identity
| Okay, so we want to verify that $$\frac{\cos{A}}{1 - \tan{A}} + \frac{\sin{A}}{1 - \cot{A}} = \sin{A} + \cos{A}.$$
Staring with the left hand side, we have (based on what you've already done):
\begin{split} \frac{\cos{A}}{1 - \tan{A}} + \frac{\sin{A}}{1 - \cot{A}} &= \frac{\cos{A}}{\left(\frac{\cos{A} - \sin{A}}{\cos{A}}\right)} + \frac{\sin{A}}{\left(\frac{\sin{A} - \cos{A}}{\sin{A}}\right)} \\ &= \frac{\cos^{2}{A}}{\cos{A}-\sin{A}} + \frac{\sin^{2}{A}}{\sin{A}-\cos{A}} \end{split}
where we got the last step by inverting and multiplying the fractions in the denominators. Now, it should be clear to you that $\sin{A}- \cos{A}= -(\cos{A} - \sin{A})$ so we can make the denominators of the two fractions the same by pulling out a minus sign from the second denominator, to get:
\begin{split} \frac{\cos^{2}{A}}{\cos{A}-\sin{A}} + \frac{\sin^{2}{A}}{\sin{A}-\cos{A}} &= \frac{\cos^{2}{A}}{\cos{A}-\sin{A}} - \frac{\sin^{2}{A}}{\cos{A}-\sin{A}} \\ &= \frac{\cos^{2}{A} - \sin^{2}(A)}{\cos{A}-\sin{A}} \end{split}
and recognizing the numerator of that last fraction is the difference of squares! So it factors into $(\cos{A} - \sin{A})(\cos{A} + \sin{A})$, giving us:
\begin{split}\frac{\cos^{2}{A} - \sin^{2}(A)}{\cos{A}-\sin{A}} &= \frac{(\cos{A} - \sin{A})(\cos{A} + \sin{A})}{\cos{A}-\sin{A}} \\ &= \cos{A} + \sin{A} \end{split}
where we cancelled the common factor of $(\cos{A} - \sin{A})$ from the numerator and denominator in the last step.
| {
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"url": "https://math.stackexchange.com/questions/1929323",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
limit of $a_n=\sqrt{n^2+2} - \sqrt{n^2+1}$ as $n$→∞ $a_n=\sqrt{n^2+2} - \sqrt{n^2+1}$ as $n$→∞
Both limits tend to infinity, but +∞ −(+∞) doesn't make sense. How would I get around to solving this?
| Note that:
$$
a_n=\sqrt{n^2+2} - \sqrt{n^2+1}=\frac{\big(\sqrt{n^2+2}-\sqrt{n^2+1}\big)\big(\sqrt{n^2+2}+\sqrt{n^2+1}\big)}{\sqrt{n^2+2} + \sqrt{n^2+1}}= \\
=\frac{n^2+2-n^2-1}{\sqrt{n^2+2} + \sqrt{n^2+1}}=\frac{1}{\sqrt{n^2+2} + \sqrt{n^2+1}}
$$
Thus:
$$
\lim_{n\to\infty}a_n=\lim_{n\to\infty}\big(\sqrt{n^2+2} - \sqrt{n^2+1}\big)=\\
=\lim_{n\to\infty}\frac{n^2+2-n^2-1}{\sqrt{n^2+2} + \sqrt{n^2+1}}=\lim_{n\to\infty}\frac{1}{\sqrt{n^2+2} + \sqrt{n^2+1}}=0
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1929548",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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Find the equation of two ...
Find the single equation of two straight lines that pass through the point $(2,3)$ and parallel to the line $x^2 - 6xy + 8y^2 = 0$.
My Attempt:
Let, $a_1x+b_1y=0$ and $a_2x+b_2y=0$ be the two lines represented by $x^2-6xy+8y^2=0$.
then,
$$(a_1x+b_1y)(a_2x+b_2y)=0$$
$$(a_1a_2)x^2+(a_1b_2+b_1a_2)xy+(b_1b_2)y^2=0$$
Comparing with $x^2-6xy+8y^2=0$,
$a_1a_2=1, -(a_1b_2+b_1a_2)=6, b_1b_2=8$.
I got stuck at here. Please help me to continue this.
| Note that $x^2-6xy+8y^2=(x-2y)(x-4y)$, so the lines represented by the equation $x^2-6xy+8y^2=0$ are
$$
x-2y=0 \qquad \text{and} \qquad x - 4y = 0.
$$
The line parallel to $x-2y=0$ and passing through $(2,3)$ is
$$
(x-2) - 2(y-3) =0,
$$
which in expanded form is $x - 2y + 4 = 0$. Similarly, the other line is $x-4y+10=0$. The single equation is then
$$
(x - 2y + 4)(x - 4y + 10) = 0.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1930128",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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Polar to cartesian form of r=sin(4θ)? The Polar to cartesian form of $ r = \sin(2\theta)$ is fairly simple.
What is the Cartesian form of the polar equation r=sin(4θ)?
[edit]
$$r=4sin(θ)cos(θ)(cos(θ)^2-sin(θ)^2)$$, so $$r^5=4rsin(θ)rcos(θ)(r^2cos(θ)^2-r^2sin(θ)^2)$$,so $$r^5=4xy(x^2-y^2)$$, so $$(x^2+y^2)^{5/2}=4xy(x^2-y^2)$$, so $$(x^2+y^2)^{5}=(4xy(x^2-y^2))^{2}$$, so $$(x^2+y^2)^{5}=16x^2y^2(x^2-y^2)^{2}$$
| Hint.
$$\sin(4\theta) = 2\sin(2\theta)\cos(2\theta)$$
Then again you may use
$$\sin(2\theta) = 2\sin\theta\ \cos\theta$$
and
$$\cos(2\theta) = \cos^2\theta - \sin^2\theta$$
Can you go on?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1930325",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Integral of polynomial times rational function of trig function over multiple periods. $$\int_0^{12\pi} \frac{x}{6+\cos 8x} dx$$
I tried all kinds of stuff but end up with $\arctan\frac{5^{1/2} \tan(x/2)}{7^{1/2}}$
Please help :)
| It looks like the OP is trying to compute the antiderivative and use the fundamental theorem of calculus. With multiple periods, that approach is paved with all sorts of difficulty that is really an artifice related to the functional form of the antiderivative. In truth, there should be no such difficulty. A better approach involves the residue theorem, which I will outline below. Note that the presence of the linear term in the numerator provides a slight complication, but one that has been treated before in this site several times.
$$\int_{0}^{12 \pi} dx \frac{x}{6+\cos{8 x}} = \frac1{64} \int_0^{96 \pi} du \frac{u}{6+\cos{u}}$$
$$\int_{n 2 \pi}^{(n+1) 2 \pi} du \frac{u}{6+\cos{u}} = \int_0^{2 \pi} dv \frac{v+2 \pi n}{6+\cos{v}} = \int_0^{2 \pi} dv \frac{v}{6+\cos{v}} + 2 \pi n \int_0^{2 \pi} \frac{dv}{6+\cos{v}}$$
Thus, summing over $n$:
$$\int_{0}^{12 \pi} dx \frac{x}{6+\cos{8 x}} = \frac{48}{64} \int_0^{2 \pi} dv \frac{v}{6+\cos{v}} + \frac{48 (47) \pi}{64} \int_0^{2 \pi} \frac{dv}{6+\cos{v}}$$
We may now compute each of these integrals. The second integral is straightforward by the residue theorem, i.e., let $z=e^{i v}$, then
$$\int_0^{2 \pi} \frac{dv}{6+\cos{v}} = -i 2 \oint_{|z|=1} \frac{dz}{z^2+12 z+1}$$
The only pole of the integrand inside the unit circle is at $z=-6+\sqrt{35}$. The integral is then $i 2 \pi$ times the residue of the integrand at this pole, or $2 \pi/\sqrt{35}$.
To compute the first integral, we consider the complex integral
$$ \oint_C dz \frac{\log{z}}{z^2+12 z+1} $$
where $C$ is the unit circle with a detour up and back about the positive real axis. (A circular piece about the origin vanishes.) This integral is equal to
$$-\frac12 \int_0^{2 \pi} dv \frac{v}{6+\cos{v}} + \int_1^0 dx \frac{\log{x}+i 2 \pi}{x^2+12 x+1} + \int_0^1 dx \frac{\log{x}}{x^2+12 x+1}$$
or, simplifying,
$$-\frac12 \int_0^{2 \pi} dv \frac{v}{6+\cos{v}} - i 2 \pi \int_0^1 \frac{dx}{x^2+12 x+1} $$
The contour integral is also equal to $i 2 \pi$ times the residue at the pole $z=-6+\sqrt{35}$. Note that the negative sign is taken to be $e^{i \pi}$. Thus we have
$$-\frac12 \int_0^{2 \pi} dv \frac{v}{6+\cos{v}} = i 2 \pi \int_0^1 \frac{dx}{x^2+12 x+1} + i 2 \pi \frac{-\log{\left ( 6+\sqrt{35} \right )}+i \pi}{2 \sqrt{35}} $$
Now,
$$\frac1{x^2+12 x+1} = \frac1{2 \sqrt{35}} \left (\frac1{x+6-\sqrt{35}} - \frac1{x+6+\sqrt{35}} \right ) $$
so that
$$\int_0^1 \frac{dx}{x^2+12 x+1} = \frac1{2 \sqrt{35}} \log{\left (\frac{7+\sqrt{35}}{7-\sqrt{35}} \right )} = \frac1{2 \sqrt{35}} \log{\left (6+\sqrt{35}\right )}$$
Note the cancellation with the real part of the residue. Thus,
$$\int_0^{2 \pi} dv \frac{v}{6+\cos{v}} = \frac{2 \pi^2}{\sqrt{35}} $$
Putting these results altogether, we have
$$\int_0^{12 \pi} dx \frac{x}{6+\cos{8 x}} = \frac{72 \pi^2}{\sqrt{35}} $$
ADDENDUM
Using the above analysis, we can easily generalize the above result:
$$\int_0^{2 \pi P} dx \frac{x}{a + \cos{N x}} = \frac{2 P^2 \pi^2}{\sqrt{a^2-1}} $$
where $a \gt 1$ and $N$ and $P \in \mathbb{N}$, i.e., other than $N$ being an integer, the result is independent of $N$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1931893",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Does $\sum \frac{n^2 + 1}{n^{3.5} -2}$ converge? Does $\sum \frac{n^2 + 1}{n^{3.5} -2}$ converge? I think it does. But I cannot find a series to compare. I tried to compare it with $\sum \frac{1}{n^{1.5}}$, but I do not know how.
$$\sum \frac{n^2 + 1}{n^{3.5} -2}?\sum\frac{n^2}{n^{3.5}-2}?\sum\frac{n^2}{n^{3.5}}$$
| Beside the good and simple solutions you already received, setting $x=\frac 1 n$ and using Taylor series around $x=0$, you could show that $$\frac{n^2+1}{n^{7/2}-2}=x^{3/2}+x^{7/2}+2 x^5+2 x^7+4 x^{17/2}+O\left(x^{21/2}\right)$$ Back to $n$ $$\frac{n^2+1}{n^{7/2}-2}=\frac{1}{n^{3/2}}+\frac{1}{n^{7/2}}+\frac{2}{n^5}+\frac{2
}{n^7}+
\frac{4}{n^{17/2}}+O\left(\frac{1}{n^{21/2}}\right)$$ and then $$\sum_{n=2}^\infty\frac{n^2+1}{n^{7/2}-2}=-10+\zeta \left(\frac{3}{2}\right)+\zeta \left(\frac{7}{2}\right)+2 \zeta (5)+2 \zeta
(7)+4 \zeta \left(\frac{17}{2}\right)+\cdots$$ which is $\approx 1.8411$ while the exact summation would lead to $\approx 1.8469$.
Using the ratio test and Taylor again, $$u_n=\frac{n^2+1}{n^{7/2}-2}\implies \frac{u_{n+1}}{u_n}=1-\frac{3}{2 n}+O\left(\frac{1}{n^2}\right)$$ which is inconclusive. But, using Raabe's test and Taylor again, $$n\left(\frac{u_n}{u_{n+1}}-1 \right)=\frac{3}{2}+\frac{3}{8 n}+O\left(\frac{1}{n^{3/2}}\right)$$ the limit is $>1$ then convergence.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1932198",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
5 odd-numbered taxis out of 9 to 3 airports
A fleet of 9 taxis must be dispatched to 3 airports: three to airport A, five to B and one to C. If the cabs are numbered 1 to 9, what is the probability that all odd-numbered cabs are sent to airport B?
What I have have come up with so far:
*
*Probability the cabs are odd is $\frac59$
*Probability they are sent to only airport B is $\frac13$
*$\frac59 + \frac13 = \frac89$
Is this the right answer?
| If I understand your question correctly, you can rephrase it as, "What's the probability of selecting all odd numbers when sampling uniformly from the collection $\{1,2, \ldots, 9\}$?" Well, the probability of selecting an odd number out of these $9$ possible numbers is $P(1) = \frac{5}{9}$. Having selected this first odd number, the probability of selecting another odd number out of the remaining $8$ numbers (for two odd numbers total) is $P(2 \mid 1) = \frac{4}{8}$. Given we've selected two odd numbers, the probability of selected a third out of the remaining $7$ numbers is $P(3 \mid 1,2) = \frac{3}{7}$. See where this is going? The last two will be $P(4 \mid 1,2,3) = \frac{2}{6}$ and $P(5 \mid 1,2,3,4) = \frac{1}{5}$.
To figure out how to combine these, use the law of total probability repeatedly:
\begin{align*}
P(5) & = P(5 \mid 1) \, P(1) \\
& = P(5 \mid 1, 2) \, P(2 \mid 1) \, P(1) \\
& = P(5 \mid 1,2,3) \, P(3 \mid 1, 2) \, P(2 \mid 1) \, P(1) \\
& = P(5 \mid 1,2,3,4) \, P(4 \mid 1,2,3) \, P(3 \mid 1, 2) \, P(2 \mid 1) \, P(1) \\
& = \frac{1}{5}\frac{2}{6}\frac{3}{7}\frac{4}{8}\frac{5}{9} \\
& = \frac{1}{126}
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1932330",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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Recurrence relation of type $a_{n+1} = a^2_{n}-2a_{n}+2$
A sequence $\{a_{n}\}$ is defined by $a_{n+1} = a^2_{n}-2a_{n}+2\forall n\geq 0$ and $a_{0} =4$
And another sequence $\{b_{n}\}$ defined by the formula $\displaystyle b_{n} = \frac{2b_{0}b_{1}b_{2}..........b_{n-1}}{a_{n}}\forall n\geq 1$ and $\displaystyle b_{0}=\frac{1}{2}$,Then
$(a)$ value of $a_{10}$
$(b)\;\; $ The value of $n$ for which $\displaystyle b_{n} = \frac{3280}{3281}$
$(c)$ The Sequence $\{b_{n}\}$ satisfy the recurrence formula
$\bf{Options::}$
$(1)\; \displaystyle b_{n+1} = \frac{2b_{n}}{1-b^2_{n}}\;\;\;\;\;\; (b)\; \displaystyle b_{n+1} = \frac{2b_{n}}{1b^2_{n}}\;\;\;\;\;\; (c)\; \displaystyle b_{n+1} = \frac{b_{n}}{1+2b^2_{n}}\; (d)\; \displaystyle b_{n+1} = \frac{b_{n}}{1-2b^2_{n}}$
$\bf{My\; Try::}$ Given $a_{n+1} = a^2_{n}-2a_{n}+2 = \left(a_{n}-1\right)^2+1$
So $a_{1} = (a_{0}-1)^2+1=(4-1)^2+1=10$
Similarly $a_{2} = (a_{1}-1)^2+1 = 9^2+1 = 82$
Similarly $a_{3} = (81)^2+1 = $
But Calculation like this is very complex, Plz help me how can i solve above problems, Thanks
| For the first problem, as LaloVelasco answered, you have $$c_{n+1}=c_{n}^{2}$$ Take logarithms $$\log(c_{n+1})=2\log(c_n)$$ Define $d_n=\log(c_n)$ which makes $$d_{n+1}=2d_n\implies d_n=k\, 2^{n-1}\implies c_n=e^{k\, 2^{n-1}}\implies a_n=1+e^{k\, 2^{n-1}}$$ and $k=2 \log (a_0-1)$ which makes $$a_n=(a_0-1)^{2^n}+1$$ and then the result for any $n$.
Then, $a_{10}=(a_0-1)^{1024}+1$.
| {
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"url": "https://math.stackexchange.com/questions/1932519",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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For how many positive integers $a$ is $a^4−3a^2+9$ a prime number? I understand that there are many posts on the problems similar to mine. I have tried my best, but still get different answers from the answer sheet. Can anyone help me? Also is there a simple way to find $a$?
For how many positive integers $a$ is $a^4-3a^2+9$ a prime number?
Here is what I did: $$a^4-3a^2+9=(a^2+3+3a)(a^2+3-3a)$$ To find $a$, I looked at the following situations:
$$a^4-3a^2+9=1$$
$$a^4-3a^2+9=3$$
$$a^4-3a^2+9=5$$
$$a^4-3a^2+9=7$$
$$a^4-3a^2+9=11$$
$$a^4-3a^2+9=13$$
$$....$$
| One must have $$a^2+3a+3=\pm1\\a^2-3a+3=\pm1$$ From these we get, discarding the constant term $4$, only two equations
$$a^2+3a+2=0=(a+1)(a+2)\\a^2-3a+2=0=(a-1)(a-2)$$ which give
both the primes $7$ and $13$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1933220",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
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Easier way to calculate the derivative of $\ln(\frac{x}{\sqrt{x^2+1}})$? For the function $f$ given by
$$
\large \mathbb{R^+} \to \mathbb{R} \quad x \mapsto \ln \left (\frac{x}{\sqrt{x^2+1}} \right)
$$
I had to find $f'$ and $f''$.
Below, I have calculated them.
But, isn't there a better and more convenient way to do this?
My method:
$$
{f'(x)}=\left [\ln \left (\frac{x}{(x^2+1)^\frac{1}{2}} \right) \right ]'=\left (\frac{(x^2+1)^\frac{1}{2}}{x} \right)\left (\frac{x}{(x^2+1)^\frac{1}{2}} \right)'=\left (\frac{(x^2+1)^\frac{1}{2}}{x} \right) \left (\frac{(x^2+1)^\frac{1}{2}-x[(x^2+1)^\frac{1}{2}]'}{[(x^2+1)^\frac{1}{2}]^2} \right)=\left (\frac{(x^2+1)^\frac{1}{2}}{x} \right) \left (\frac{(x^2+1)^\frac{1}{2}-x[\frac{1}{2}(x^2+1)^{-\frac{1}{2}}(x^2+1)']}{\left | x^2+1 \right |} \right)=\left (\frac{(x^2+1)^\frac{1}{2}}{x} \right) \left (\frac{(x^2+1)^\frac{1}{2}-x[\frac{1}{2}(x^2+1)^{-\frac{1}{2}}(2x)]}{x^2+1} \right)=\left (\frac{(x^2+1)^\frac{1}{2}}{x} \right) \left (\frac{(x^2+1)^\frac{1}{2}-x^2(x^+1)^{-\frac{1}{2}}}{x^2+1} \right)=\frac{(x^2+1)^{(\frac{1}{2}+\frac{1}{2})}-x^2(x^2+1)^{\frac{1}{2}+-\frac{1}{2}{}}}{x(x^2+1)}=-\frac{x^2}{x}=-x
$$
and
$$
f''(x)=(-x)'=-1\
$$
This took me much more than 1.5 hours just to type into LaTex :'(
| Hint:
\begin{align}
f(x) = \log|x| - \frac{1}{2}\log|x^2+1|.
\end{align}
| {
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"url": "https://math.stackexchange.com/questions/1933460",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Why $\forall n\gt2\in\Bbb N:\lfloor\sqrt{n\cdot\sqrt{n\cdot\lfloor\sqrt{...\lfloor\sqrt{n\cdot\lfloor\sqrt{n}\rfloor}}\rfloor\rfloor}}\rfloor=n-2$? I have found this just by chance, but can not understand why the limit turns out to be $n-2, \forall n\gt 2 $.
Def:
$a_0 = \lfloor \sqrt{n} \rfloor$
$a_1 = \lfloor \sqrt{n \cdot a_0} \rfloor$
...
$a_k = \lfloor \sqrt{n \cdot a_{k-1}} \rfloor$
And apparently:
$\forall n\gt 2:\ lim_{k\to \infty} a_k = \lfloor \sqrt{n \cdot \sqrt{ n \cdot \lfloor \sqrt{...\lfloor \sqrt{n\cdot \lfloor \sqrt{n} \rfloor }} \rfloor \rfloor}} \rfloor = n-2$
I understand that $(n-2) \cdot n = n^2-2n \lt (n-1)^2 = n^2+1-2n$ and for that reason if one of the elements of the sequence is $n-2$ then can not grow up more because $\lfloor \sqrt {n \cdot (n-2)} \rfloor = n-2$. But what I do not understand is why the sequence exactly arrives to $n-2$. Why not $n-1$ or $n-3$ for instance?
I would like to ask the following questions:
*
*Is the observation correct? Are there counterexamples?
*Why does exactly arrive to $n-2$ instead of any other value like, for instance, $n-3$? Probably the reason is quite trivial but I can not see the property behind the behavior. Thank you!
| All $n\gt 2$ converge to $n-2$ as you surmise.
The easiest way to see it is to look what happens if $a_n=n-3$. Then $a_{n+1}=\lfloor \sqrt {n(n-3)} \rfloor = \lfloor \sqrt {n^2-3n} \rfloor $. If $n \ge 4, n^2-3n \ge n^2-4n+4 = (n-2)^2$ so you will climb from $n-3$ to $n-2$ As you have pointed out, once you get to $n-2$ you have stability because $n^2-2n \lt n^2-2n+1=(n-1)^2$ This argument does not work for $n=3,$ but $\lfloor \sqrt 3 \rfloor =1$ and we start out at $n-2$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Solving a Coupled Second-Order Differential Equation I'm trying to solve the real-valued differential equation
$$xy\ddot{y}+2y\dot{x}\dot{y}+\alpha\ddot{x}+x\dot{y}^2=0$$
where $\alpha\in\mathbb{R}$ and $\alpha>0$ for both $x(t)$ and $y(t)$. How would I even approach this problem?
EDIT: Sorry, I forgot a conservation law:
$$\alpha^2\dot{x}^2+x^4\dot{y}^2 = P^2$$
where $P$ is a conserved quantity.
EDIT 2: The original Lagrangian this comes from is
$$\mathcal{L}=\frac{1}{2}\left(-\alpha\dot{x}^2+x^2\dot{y}^2+x^2y^2\dot{\theta}^2+x^2y^2\sin^2\theta\dot{\phi}^2\right)$$
| Is this a homework problem or is it some sort of practice question? If not, any possible symmetries coming from the formulation of the problem? If not, I don't know how far one can go, but let's see. It may help if you tell us how you derived the equation, weather it is an Euler-Lagrange equation of some sort. Maybe it's coming from a problem in differential geometry related to curves on surfaces?
Invert the function $x=x(t)$ and write it as a function $t=t(x)$. Then
$$\frac{d}{dt} = \frac{dx}{dt} \, \frac{d}{dx}$$
Form the metric $$\alpha^2 \left(\frac{dx}{dt}\right)^2 + x^4 \, \left(\frac{dy}{dt}\right)^2 = P^2$$ one gets
$$P^2 = \alpha^2 \left(\frac{dx}{dt}\right)^2 + x^4 \, \left(\frac{dx}{dt}\right)^2\left(\frac{dy}{dx}\right)^2 = \left(\frac{dx}{dt}\right)^2 \, \left(\alpha^2 + x^4 \, \left(\frac{dy}{dx}\right)^2\right)$$ and so
$$ \left(\frac{dx}{dt}\right)^2 = \frac{P^2}{\alpha^2 + x^4 \, \left(\frac{dy}{dx}\right)^2}$$
$$\frac{dx}{dt}= \frac{P}{\sqrt{\alpha^2 + x^4 \, \left(\frac{dy}{dx}\right)^2}}$$ Thus
$$\frac{d}{dt} = \left(\frac{P}{\sqrt{\alpha^2 + x^4 \, \left(\frac{dy}{dx}\right)^2}} \right)\,\, \frac{d}{dx}$$
Now the equation $$x y \, \frac{d^2 y}{dt^2} + 2 \, y \, \frac{dx}{dt}\frac{dy}{dt} + \alpha \frac{d^2x}{dt^2} + x \left(\frac{dy}{dt}\right)^2 = 0$$ can be written as
$$\frac{d}{dt}\left(x y \, \frac{dy}{dt} + \alpha \frac{dx}{dt}\right) + y \, \frac{dx}{dt}\frac{dy}{dt} = 0$$ Writing the latter equation in terms of $x$ as an independent variable
$$\left(\frac{P^2}{\sqrt{\alpha^2 + x^4 \, \left(\frac{dy}{dx}\right)^2}} \right)\,\, \frac{d}{dx} \left(\frac{x y \frac{dy}{dx} + \alpha}{\sqrt{\alpha^2 + x^4 \, \left(\frac{dy}{dx}\right)^2}}\right) + \frac{P^2 \, y \, \frac{dy}{dx}}{\alpha^2 + x^4 \, \left(\frac{dy}{dx}\right)^2} = 0$$
$$ \frac{d}{dx} \left(\frac{x y \frac{dy}{dx} + \alpha}{\sqrt{\alpha^2 + x^4 \, \left(\frac{dy}{dx}\right)^2}}\right) + \frac{y \, \frac{dy}{dx}}{\sqrt{\alpha^2 + x^4 \, \left(\frac{dy}{dx}\right)^2}} = 0$$ In addition to that, one can also write $z(x) = \frac{1}{2} y(x)^2$ and so the equation becomes
$$ \frac{d}{dx} \left(\frac{x \frac{dz}{dx} + \alpha}{\sqrt{\alpha^2 + \frac{x^4}{2 \, z} \, \left(\frac{dz}{dx}\right)^2}}\right) + \frac{ \frac{dz}{dx}}{\sqrt{\alpha^2 + \frac{x^4}{2 \, z} \, \left(\frac{dz}{dx}\right)^2}} = 0$$
Alternatively, you can multiply the initial equation by $x$ and get
$$x^2 y \ddot{y} + 2xy \, \dot{x} \dot{y} + x^2 \dot{y}^2 + \alpha \, x \ddot{x} = 0$$ which becomes
$$0 = \frac{d}{dt} \big(x^2 y \dot{y}\big) + \alpha \, x \ddot{x} = \frac{d}{dt} \big(x^2 y \dot{y}\big) + \alpha \,\big( x \ddot{x} + \dot{x}^2\big) - \alpha \, \dot{x}^2 = \frac{d}{dt} \big(x^2 y \dot{y} + \alpha \, x \dot{x} \big) - \alpha \, \dot{x}^2$$
Now had your second equation been $$\alpha^2 \, \dot{x}^2 + x^4 y^2 \dot{y}^2 = P^2$$ then a miracle could have happened and since $x^2y\dot{y} = \sqrt{P^2 - \alpha^2 \, \dot{x}^2}$ the equation would have become
$$\frac{d}{dt} \Big(\sqrt{P^2 - \alpha^2 \, \dot{x}^2} + \alpha \, x \dot{x}\Big) - \alpha \, \dot{x}^2 = 0$$ which would have been a whole new story. This equation would have had much more potential.
Currently, you can end up with $$\frac{d}{dt} \Big( y \sqrt{P^2 - \alpha^2 \, \dot{x}^2} + \alpha \, x \dot{x}\Big) - \alpha \, \dot{x}^2 = 0$$ Maybe now one can write $y=y(t)$ as a function $t=t(y)$ and $x=x(y)$. Then
$$\frac{d}{dt} = \dot{y} \frac{d}{dy}$$ so
$$\dot{y} \frac{d}{dy}\left( y \sqrt{P^2 - \alpha^2 \, \dot{x}^2} + \alpha \, x \dot{x}\right) - \alpha \, \dot{x}^2 = 0$$
$$\dot{y} \frac{d}{dy}\left( y \sqrt{P^2 - \alpha^2 \, \dot{y}^2 \left(\frac{dx}{dy}\right)^2} + \alpha \, x \dot{y} \, \frac{dx}{dy}\right) - \alpha \, \dot{y}^2 \left(\frac{dx}{dy}\right)^2 = 0$$ Express $$\dot{y} = \frac{P}{\sqrt{\alpha^2\left(\frac{dx}{dy}\right)^2 + x^4}} = f\left(x,\frac{dx}{dy}\right)$$ so we get
$$f \frac{d}{dy}\left( y \sqrt{P^2 - \alpha^2 \, f^2 \left(\frac{dx}{dy}\right)^2} + \alpha \, x f \, \frac{dx}{dy}\right) - \alpha \, f^2 \left(\frac{dx}{dy}\right)^2 = 0$$
$$\frac{d}{dy}\left( y \sqrt{P^2 - \alpha^2 \, f^2 \left(\frac{dx}{dy}\right)^2} + \alpha \, x f \, \frac{dx}{dy}\right) - \alpha \, f \, \left(\frac{dx}{dy}\right)^2 = 0$$
$$\frac{d}{dy}\left( y \sqrt{P^2 - \alpha^2 \, f\left(x,\frac{dx}{dy}\right)^2 \left(\frac{dx}{dy}\right)^2} + \alpha \, x f\left(x,\frac{dx}{dy}\right) \, \frac{dx}{dy}\right) - \alpha \, f\left(x,\frac{dx}{dy}\right) \, \left(\frac{dx}{dy}\right)^2 = 0$$ This is again a $y$ inhomogeneous second order ODE of type $F\big(y, x, x'(y), x''(y)\big) = 0$.
Are you sure that the restriction is not $$\alpha^2 \, \dot{x}^2 + x^4 y^2 \dot{y}^2 = P^2 \, ? :)$$
| {
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"url": "https://math.stackexchange.com/questions/1934186",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
Solving system of $9$ linear equations in $9$ variables I have a system of $9$ linear equations in $9$ variables:
\begin{array}{rl}
-c_{1}x_{1} + x_{2} + x_{3} + x_{4} + x_{5} + x_{6} + x_{7} + x_{8} + x_{9} &= 0 \\
x_{1} - c_{2}x_{2} + x_{3} + x_{4} + x_{5} + x_{6} + x_{7} + x_{8} + x_{9} &= 0 \\
x_{1} + x_{2} - c_{3}x_{3} + x_{4} + x_{5} + x_{6} + x_{7} + x_{8} + x_{9} &= 0 \\
x_{1} + x_{2} + x_{3} - c_{4}x_{4} + x_{5} + x_{6} + x_{7} + x_{8} + x_{9} &= 0 \\
x_{1} + x_{2} + x_{3} + x_{4} - c_{5}x_{5} + x_{6} + x_{7} + x_{8} + x_{9} &= 0 \\
x_{1} + x_{2} + x_{3} + x_{4} + x_{5} - c_{6}x_{6} + x_{7} + x_{8} + x_{9} &= 0 \\
x_{1} + x_{2} + x_{3} + x_{4} + x_{5} + x_{6} - c_{7}x_{7} + x_{8} + x_{9} &= 0 \\
x_{1} + x_{2} + x_{3} + x_{4} + x_{5} + x_{6} + x_{7} - c_{8}x_{8} + x_{9} &= 0 \\
x_{1} + x_{2} + x_{3} + x_{4} + x_{5} + x_{6} + x_{7} + x_{8} - c_{9}x_{9} &= 0
\end{array}
I want to find a general non-trivial solution for it. What would be the easiest and least time consuming way to find it by hand? I don't have a lot of background in maths, so I would very much appreciate if you actually found the solution and explained briefly.
Thanks in advance!
EDIT: Very important to mention is that always any $c_{i} > 1$ and any $x_{i} \geq 20$. Also it would be nice if someone posted how would a general non-trivial solution look in the form of $$S = \left \{( x_{1}, x_{2}, x_{3}, x_{4}, x_{5}, x_{6}, x_{7}, x_{8}, x_{9}\right )\}$$
| If $T = x_1 + x_2 + \ldots + x_9$, you can write this as
$$ T - (1+c_1) x_1 = T - (1 + c_2) x_2 = \ldots = T - (1+c_9) x_9 = 0$$
If any $c_i = -1$, then $T = 0$, and any $x_j$ for which $c_j \ne -1$ must be $0$, while those for which $c_j = -1$ must add to $0$.
On the other hand, if all $c_i \ne -1$,
then each $x_i = T/(1+c_i)$, and then
$$T = x_1 + \ldots + x_9 = T \left( \dfrac{1}{1+c_1} + \ldots + \dfrac{1}{1+c_9}\right) $$
Since you want a nontrivial solution, you don't want $T=0$ which would make all $x_i = 0$.
So then you need $$ \dfrac{1}{1+c_1} + \ldots + \dfrac{1}{1+c_9} = 1$$ and $T$ can be anything.
EDIT: The added condition that all $c_i > 1$ rules out the case where some $c_i = -1$, so you need $1/(1+c_1) + \ldots + 1/(1+c_9) = 1$. You want
$x_i \ge 20$, and since $x_i = T/(1+c_i)$ that says $T \ge 20 (1+c_i)$.
So now the solutions are
$$ (x_1, \ldots, x_9) = \left(\dfrac{T}{1+c_1}, \ldots, \dfrac{T}{1+c_9}\right)$$
where $1/(1+c_1) + \ldots + 1/(1+c_9) = 1$ and $T \ge 20 (1 + \max(c_1, \ldots, c_9))$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1934857",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Determine $ax^4 + by^4$ for system of equations I found the following recreational problem without further specification for $a,b$.
Let $x,y$ be real numbers s.t.
$a + b = 6$,
$ax + by = 10$,
$ax^2 + by^2 = 24$,
$ax^3 + by^3 = 62$.
Determine $ax^4 + by^4$.
I am new to problem solving exercises like this and therefore appreciate diverse approaches to this problem as well as comments on how to tackle those types of exercises.
| I prefer to add another answer for the most general case.
Consider the four equations which I shall write $$ax^{i-1}+b y^{i-1}=c_i \qquad (i=1,2,3,4)$$ Manipulating them, we can show that $x y=\lambda$ and $a b=\mu$ using $$\lambda=\frac{c_3^2-c_2 c_4}{c_2^2-c_1 c_3}$$
$$\mu=-\frac{\left(c_2^2-c_1 c_3\right)^3}{(4 c_4 c_2^3-3 c_3)^2 c_2^2-6 c_1 c_2
c_3 c_4+c_1 \left(4 c_3^3+c_1 c_4^2\right)}$$ This makes $$a_{\pm}=\frac{1}{2} \left(c_1\pm\sqrt{c_1^2-4 \mu }\right)$$ Using $a_+$ , this gives $$b=c_1-a\qquad x=\frac{c_2+\sqrt{c_2^2-4 \lambda \mu }}{2 a}\qquad y=\frac \lambda x$$ Using the given values for the $c_i$'s, this leads to $$\lambda=1\qquad \mu=\frac{44}{5}$$ $$a=3+\frac{1}{\sqrt{5}}\qquad b=3-\frac{1}{\sqrt{5}}\qquad x=\frac{1}{2} \left(3+\sqrt{5}\right)\qquad y=\frac{1}{2} \left(3-\sqrt{5}\right)$$
Now, play with the numbers of your choice.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 3
} |
Lower bound of product. We have the product :
$$\prod{(1-\frac{1}{2^i})}$$ I know that product has upper bound and possibly has lower bound.
My question is : how can I get the lower bound of this product?
My idea was to estimate this product with other product , but I didn't get the product with non-zero limit.
| Observe for $n\geq 4$ we have
\begin{align}
2^n \geq n^2
\end{align}
then it follows
\begin{align}
\prod^\infty_{n=4}\left(1-\frac{1}{n^2}\right) \leq \prod^\infty_{n=4}\left( 1- \frac{1}{2^n}\right)
\end{align}
which means
\begin{align}
\prod^\infty_{n=2}\left(1-\frac{1}{n^2}\right) \leq
\frac{\left(1-\frac{1}{2^2}\right)\left(1-\frac{1}{3^2}\right)}{\left(1-\frac{1}{2} \right)\left(1-\frac{1}{2^2} \right)\left(1-\frac{1}{2^3} \right)}
\prod^\infty_{n=1}\left( 1- \frac{1}{2^n}\right).
\end{align}
Thus, we have
\begin{align}
\frac{63}{256}= \frac{63}{128}\prod^\infty_{n=2}\left( 1-\frac{1}{n^2} \right)\leq \prod^\infty_{n=1}\left( 1- \frac{1}{2^n}\right).
\end{align}
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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If $a,b,c>0$ and $a^3+b^3=c^3$ then prove that $a^2+b^2-c^2>6(c-a)(c-b)$. If $a,b,c>0$ and $a^3+b^3=c^3$ then prove that $$a^2+b^2-c^2>6(c-a)(c-b)$$.
I tried factoring but things got more complicated. I dont think standard AM-Gm can be applied directly. Can somebody help me to proceed? Thanks a lot.
| We need to prove that
$$c(a^2+b^2)-(a^3+b^3)>6c(c-a)(c-b)$$ or
$$\frac{(c^3-b^3)(c-a)}{a}+\frac{(c^3-a^3)(c-b)}{b}>6c(c-a)(c-b)$$ or
$$\frac{b^2+bc+c^2}{a}+\frac{a^2+ac+c^2}{b}>6c$$
which is AM-GM.
Done!
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Integration by completing the square I need to complete the square on the following integral. Once this is done apparently I will be able to use on of the integration tables in the back of my book.
$\int x \sqrt{x^2 + 6x +3} dx $
This is what I have come up with so far:
$\int x \sqrt{(x+3)^2 -6} $
I really am at a loss.
Any help with this would be appreciated.
Thank you
| $$ I= \int x \sqrt {x^2 +6x +3} dx $$
This is best done by substitution.
$$I= \int x \sqrt {(x+3)^2-6} dx \\ x+3= \sqrt 6 \sec \theta \implies dx=\sqrt 6 \sec \theta \tan \theta d \theta$$
$$ \therefore I= \sqrt 6 \int \sec \theta \tan \theta ( \sqrt 6 \sec \theta -3) \sqrt {6( \sec^2 \theta -1)} \\ I=6 \int \sec \theta \tan^2 \theta ( \sqrt 6 \sec \theta -3)d \theta \\ I=6 \sqrt6 \int \sec^2 \theta \tan^2 \theta d\theta-18\int \sec \theta \tan^2 \theta d \theta \\ \therefore I= 2 \sqrt6 \tan^3 \theta - 18\int \sec^3 \theta d \theta + 18\int \sec \theta d \theta \\ I =2 \sqrt6 \tan^3 \theta -9 \sec \theta \tan \theta +9 \ln | \sec \theta + \tan \theta | +C$$
Unless I screwed up the algebra I think this is correct. It could also be done using the substitution $ u=x^2+6x+3 $ and evaluating $\int \sqrt {x^2+6x+3} dx$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Calculating $ \sum_{n \ge 3|m|}^{} (-1)^n q^{(2n + 1)^2 - 32m^2} $. $ \sum_{n \ge 3|m|}^{} (-1)^n q^{(2n + 1)^2 - 32m^2} = η(8t)η(16t)$.
(Kac-Peterson)
I try to confirm the formula, however I don't know the left.
(The right is $q -q^9 -2q^{17} + q^{25} + 2q^{41} + \cdots $)
Define $a(k)$ as the following.
$ \sum_{n \ge 3|m|}^{} (-1)^n q^{(2n + 1)^2 - 32m^2} = \sum a(k) q^{k}.$
For example,
$(2n + 1)^2 - 32m^2 = 1$ and $n \ge 3|m|⇔ (n, m) = (0, 0)$. So $a(1) = (-1)^0 = 1.$
How do we calculate $a(9), a(17), \cdots$?
| $(2n + 1)^2 - 32m^2 = 9$ and $n \ge 3|m|⇔ (n, m) = (1, 0)$.
So $a(9) = (-1)^1 = -1.$
$(2n + 1)^2 - 32m^2 = 17$ and $n \ge 3|m|⇔ (n, m) = (3, -1), (3, 1)$.
So $a(17) = (-1)^3 + (-1)^3 = -2.$
The other $a(k)$ is calculated by the same way.
| {
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"url": "https://math.stackexchange.com/questions/1948046",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Find the Maclaurin Series for $\frac{x}{x^2 + 1}$ How to find the Maclaurin Series for $$\frac{x}{x^2 + 1}?$$
I know that the Maclaurin Series is given by:
$$f(x) = f(0) + f'(0)x + \dfrac{f''(0)}{2!}x^2 + \dots + \dfrac{f^{(k)}(0)}{k!}x^k.$$
So from repeatedly deriving $f(x)$ until there was a pattern. I have found that the Maclaurin Series for $\frac{1}{1+x}$ is given as $1-x + x^2 - x^3 + x^4 - x^5 + \dots$
In sigma notation:
$$\sum_{n=0}^{\infty} (-1)^n \cdot x^n.$$
Now I need to get the result for the new function $\frac{x}{x^2 + 1}$. Because $x^2+1$ is irreducible is there any way to do this without getting into imaginary numbers? Formally, the question says "Using your series from $\frac{1}{1+x}$, find the Maclaurin series for $\frac{x}{x^2+1}$" Perhaps I have to multiply through by $x^2$?
| $${\frac {1}{1+x}}=\sum _{n=0}^{\infty }(-1)^nx^{n}$$
let $x\rightarrow x^2$
$${\frac {1}{1+x^2}}=\sum _{n=0}^{\infty }(-1)^nx^{2n}$$
then multiply by $x$
$${\frac {x}{1+x^2}}=\sum _{n=0}^{\infty }(-1)^nx^{2n+1}$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Find the real solutions of the equations:$x+y=2$, $xy-z^2=1$ Can anyone help with this? I will show you what I did.
Find the real solutions of the equation:$$x+y=2\tag{1}$$ $$xy-z^2=1 \tag{2}$$
From equation (2), I have $$xy=1+z^2 \tag{3}$$ From equations (1) and (3), I have a new equation $$r^2-2r+(1+z^2)=0$$ The roots of this equation are the solutions. How do I find the root?
| Given $x+y=2$ and $xy-z^2=1$
So $xy=1+z^2\geq 1>0$.
So $x,y>0$
Using $\bf{A.M\geq G.M},$ we get $$\frac{x+y}{2}\geq \sqrt{xy}\Rightarrow (x+y)^2\geq 4xy$$ and equality hold when $x=y$
So $4\geq 4xy\Rightarrow xy\leq 1\Rightarrow 1+z^2\leq 1\Rightarrow z^2\leq 0\Rightarrow z=0$
and $x=y=1.$ So we get $(x,y,z) = (1,1,0)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1953570",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 4
} |
Square root of a $2 \times 2$ matrix
How can I find the square root of the following non-diagonalisable matrix?
$$\begin{pmatrix}
5 & -1 \\
4 & 1
\end{pmatrix}$$
I have shown that
$$A^{n}=3^{n-1}\begin{pmatrix}
2n+3 & -n \\
4n & 3-2n
\end{pmatrix}$$
though I'm not sure how to use this fact.
| The Jordan decomposition $A=SJS^{-1}$ is given by the matrices $$S = \begin{pmatrix}1&\frac12\\2&0 \end{pmatrix},\quad J = \begin{pmatrix}3&1\\0&3 \end{pmatrix}. $$
Let $\lambda = 3$ and write $J = \lambda (I+K)$, where $$K=\begin{pmatrix}0&\frac13\\0&0 \end{pmatrix}.$$ Using the Mercator series, we compute the logarithm of $J$:
\begin{align}
\log J &= \log(\lambda(I+K))\\
&= \log(\lambda I) + \log (I+K)\\
&= 0 + \sum_{n=1}^\infty (-1)^{n+1}\frac{K^n}n\\
&= \begin{pmatrix}0&\log\frac43\\0&0 \end{pmatrix}.
\end{align}
Write $J=S^{-1}AS$, then we have
\begin{align}
\log A &= S(\log J) S^{-1}\\
&= \begin{pmatrix}1&\frac12\\2&0 \end{pmatrix}\begin{pmatrix}0&\log\frac43\\0&0 \end{pmatrix}\begin{pmatrix}0&\frac12\\2&-1 \end{pmatrix}\\
&= \begin{pmatrix}\frac23+\log 3&-\frac13\\\frac43&-\frac23+\log3 \end{pmatrix}.
\end{align}
It follows that
$$A^{\frac12} = e^{\log\left(A^{\frac12}\right)}
= e^{\frac12\log A}
= \frac{1}{\sqrt{3}}\begin{pmatrix} 4 & -\frac{1}{2} \cr 2 & 2 \end{pmatrix}.
$$
By the way, there is a very similar formula for power of $\log A$:
$$(\log A)^n = \frac13(\log 3)^{n-1}\begin{pmatrix}2n+3\log 3& -n\\ 4n& -2n+3\log 3 \end{pmatrix}. $$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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"answer_count": 3,
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Prove that $(a+b+c-d)(a+c+d-b)(a+b+d-c)(b+c+d-a)\le(a+b)(a+d)(c+b)(c+d)$
Let $a,b,c,d>0$. Prove that $$(a+b+c-d)(a+c+d-b)(a+b+d-c)(b+c+d-a)\le(a+b)(a+d)(c+b)(c+d)$$
I don't know how to begin to solve this problem
| Let $a=\min\{a,b,c,d\}$, $b=a+u$, $c=a+v$ and $d=a+w$.
Hence, $(a+b)(b+c)(c+d)(d+a)-\prod\limits_{cyc}(a+b+c-d)=$
$$4(u^2+v^2+w^2-uv-vw)a^2+(4(u^3+v^3+w^3)-2u^2v-2v^2u-2v^2w-2w^2v)a+$$
$$+\sum\limits_{cyc}(u^4-2u^2v^2+u^2vw)+u^2w^2$$
Since $\sum\limits_{cyc}(2u^3-u^2v-u^2w)=\sum\limits_{cyc}(u+v)(u-v)^2\geq0$,
it remains to prove that $\sum\limits_{cyc}(u^4-2u^2v^2+u^2vw)\geq0$, which is Schur:
$$\sum\limits_{cyc}(u^4-2u^2v^2+u^2vw)=\sum\limits_{cyc}(u^4-u^3v-u^3w+u^2vw)+\sum\limits_{cyc}uv(u-v)^2\geq0$$
Done!
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove that there is a triangle with side lengths $a_n-1,a_n,a_n+1$ and that its area is equal to an integer
Given a sequence $(a_n)$, with $a_1 = 4$ and $a_{n+1} = a_n^2-2$ for all $n \in \mathbb{N}$, prove that there is a triangle with side lengths $a_n-1,a_n,a_n+1$ and that its area is equal to an integer.
We must have that \begin{align*}(a_n-1)+a_n &> a_n+1\\a_n+(a_n+1) &> a_n-1\\(a_n-1)+(a_n+1) &> a_n,\end{align*} which gives us equivalently that $a_n > 2$. Now using Heron's Formula we get \begin{align*}S = \sqrt{\dfrac{3a_n}{2}\left(\dfrac{3a_n}{2}-(a_n-1)\right)\left(\dfrac{3a_n}{2}-a_n\right)\left(\dfrac{3a_n}{2}-(a_n+1)\right)} &= \sqrt{\dfrac{3a_n}{2} \cdot \dfrac{a_n+2}{2} \cdot \dfrac{a_n}{2} \cdot \dfrac{a_n-2}{2}}\\&=\dfrac{1}{4}\sqrt{3a_n^2(a_n-2)(a_n+2)}.\end{align*} How do we continue from here?
| You have proven that $$S_n = \frac{1}{4}\sqrt{3a_n^2(a_n^2-4)}$$
Suppose $S_n$ is an integer.
$$S_n^2 = \frac{3}{16}a_n^2(a_n^2-4)$$
We have
\begin{align}
S_{n+1}^2 &= \frac{3}{16}a_{n+1}^2(a_{n+1}^2-4)\\
& = \frac{3}{16}(a_{n+1}^2-4) a_{n+1}^2\\
& = \frac{3}{16}((a_{n}^2-2)^2-4) a_{n+1}^2\\
&= \frac{3}{16}(a_n^4-4a_n^2) a_{n+1}^2\\
&= S_n^2a_{n+1}^2\\
\end{align}
That is we have $$S_{n+1}=S_na_{n+1}$$
Checking $a_n$ are integers are simple. Also, for base case, $S_1=6$ which is an integer.
| {
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Show that point $P $ and $\Delta ABC$ satisfy the given condition
Let ABC be a triangle and $P$ be any point in complex plane. Then show that
$$1.BC^2+CA^2+AB^2 \le 3 (PA^2+PB^2+PC^2)$$
$$2.BC^2+CA^2+AB^2 = 3 (PA^2+PB^2+PC^2) \iff \text {P is centroid of} \space\Delta ABC$$
I tried to solve it by assuming point $A(x_1,y_1),B(x_2,y_2),C(x_3,y_3),P (x,y)$
$$2({x_1}^2+{x_2}^2+{x_3}^2)+2({y_1}^2+{
y_2}^2+{y_3}^2)-2({x_1}{x_2}+{x_2}{x_3}+{x_1}{x_3})-2({y_1}{y_2}+{y_2}{y_3}+{y_1}{y_3})=3({x_1}^2+{x_2}^2+{x_3}^2+{y_1}^2+{
y_2}^2+{y_3}^2+3x^2+3y^2-2x({x_1}+{x_2}+{x_3})-2y({y_1}+{y_2}+{y_3}))$$
I am not able to conclude anything from this.
| Your expression is wrong. At least, you must point to what is the LHS and RHS. Let me guess. You mean LHS is $AB^2 + BC^2 + CA^2$ and RHS is $3(PA^2 + PB^2 + PC^2)$, but we have $$LHS = AB^2 + BC^2 + CA^2 = 2\sum x_i^2 + 2\sum y_i^2 -\sum 2x_ix_j - 2\sum y_iy_j,$$
$$RHS = 3(PA^2 + PB^2 + PC^2) = 3[3(x^2+y^2) + \sum x_i^2 + \sum y_i^2 - 2x\sum x_i - 2y \sum y_i]$$
Then, you can see easily that $LHS < RHS$. For example,
$$RHS - LHS = 9(x^2+y^2) + \sum x_i^2 + \sum y_i^2 + 2\sum x_ix_j + 2\sum y_iy_j - 2x\sum x_i -2y\sum y_i$$
$$ = 9(x^2+y^2) + (\sum x_i)^2 + (\sum y_i)^2 - 2 x\sum x_i - 2y\sum y_i$$
$$ = 8(x^2+y^2) + (x - \sum x_i)^2 + (y-\sum y_i)^2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1960822",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Limits at infinity by rationalizing I am trying to evaluate this limit for an assignment.
$$\lim_{x \to \infty} \sqrt{x^2-6x +1}-x$$
I have tried to rationalize the function:
$$=\lim_{x \to \infty} \frac{(\sqrt{x^2-6x +1}-x)(\sqrt{x^2-6x +1}+x)}{\sqrt{x^2-6x +1}+x}$$
$$=\lim_{x \to \infty} \frac{-6x+1}{\sqrt{x^2-6x +1}+x}$$
Then I multiply the function by $$\frac{(\frac{1}{x})}{(\frac{1}{x})}$$
Leading to
$$=\lim_{x \to \infty} \frac{-6+(\frac{1}{x})}{\sqrt{(\frac{-6}{x})+(\frac{1}{x^2})}+1}$$
Taking the limit, I see that all x terms tend to zero, leaving -6 as the answer. But -6 is not the answer. Why is that?
| it should be $$\lim _{ x\to \infty } \frac { -6x+1 }{ \sqrt { x^{ 2 }-6x+1 } +x } =\lim _{ x\to \infty } \frac { x\left( -6+\frac { 1 }{ x } \right) }{ x\left( \sqrt { 1-\frac { 6 }{ x } +\frac { 1 }{ { x }^{ 2 } } } +1 \right) } =\frac { -6 }{ 2 } =-3$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1960911",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 0
} |
Evaluating $\arccos(\cos\frac{15\pi }{4})$ I have a problem with understanding of this exercise:
$\arccos(\cos\frac{15\pi }{4})= ?$
$\cos(\frac{15\pi}{4}-2\pi)=\cos(\frac{7\pi}{4})$
$\cos(\frac{7\pi}{4}+\pi)= -\cos(\frac{3\pi}{4})$
then $\arccos(-\cos\frac{3\pi}{4})$
All above I understand pretty clearly. We did it at school, too and this is a solution. I just don`t understand where $-x$ comes from. It can possibly be $x=\cos(\frac{3\pi}{4})$ but what about the $-\frac{\pi}{2}$ then ??
$\arccos(-x)-\frac{\pi}{2} = -(\arccos x- \frac{\pi}{2})$
$\arccos(-x)=\pi-\arccos x$
$\arccos(\cos\frac{15\pi}{4})=\arccos(-\cos\frac{3\pi}{4})=\pi-\arccos(\cos\frac{3\pi}{4})=\pi-\frac{3\pi}{4}=\frac{\pi}{4}$
or do you know any other method how to solve this ??
Thank you for your time.
| Since $\arccos x: [-1, 1] \to [0, \pi]$, we need to find the angle in $[0, \pi]$ that has the same cosine as $15\pi/4$. Since coterminal angles have the same cosine and
$$\frac{15\pi}{4} = -\frac{\pi}{4} + 4\pi = -\frac{\pi}{4} + 2 \cdot 2\pi$$
we have
$$\cos\left(\frac{15\pi}{4}\right) = \cos\left(-\frac{\pi}{4}\right)$$
Since $\cos(-x) = \cos x$, we have
$$\cos\left(-\frac{\pi}{4}\right) = \cos\left(\frac{\pi}{4}\right)$$
Thus,
\begin{align}
\arccos\left[\cos\left(\frac{15\pi}{4}\right)\right] & = \arccos\left[\cos\left(-\frac{\pi}{4}\right)\right]\\
& = \arccos\left[\cos\left(\frac{\pi}{4}\right)\right]\\
& = \frac{\pi}{4}
\end{align}
where the final equality follows from the observation that if $\theta \in [0, \pi]$, then $\arccos(\cos\theta) = \theta$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1961155",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Solutions of $\sin2x-\sin x>0$ with $x\in[0,2\pi]$ What are the solutions of this equation with $x\in[0,2\pi]$?
$$\sin2x-\sin x>0$$
I took this to
$$(\sin x)(2\cos x-1)>0$$
Now I need both terms to be the same sign. Can you please help me solve this?
| $$\begin{align}\sin(2x)-\sin(x)&=2\sin(x)\cos(x)-\sin(x)\\
&=2\sin(x)(\cos(x)-\frac{1}{2})\tag{1}
\end{align}$$
We examine the two factors:
$$\cos(x)=\frac{1}{2}\Rightarrow x=\frac{\pi}{3},\frac{5\pi}{3}$$
And $$\cos(x)-\frac{1}{2}>0,\, 0<x<\frac{\pi}{3} \text{ and } \frac{5\pi}{3}<x<2\pi$$
$$\cos(x)-\frac{1}{2}<0,\, \frac{\pi}{3}<x<\frac{5\pi}{3}$$
The other term is $\sin(x)$:
$$\sin(x)>0,\, 0<x<\pi$$
$$\sin(x)<0,\, \pi<x<2\pi$$
So in the following interval the product is positive:
$$\boxed{0<x<\frac{\pi}{3} \text{ and } \pi<x<\frac{5\pi}{3}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1963495",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
How to set up equation to find the unknown values if a limit exist.
For what values of the constants $a$ and $b$ does the following limit exist?
$$\lim_{x\to0}\frac{|x+3|(\sqrt{ax+b}-2)}x$$
for this question, $$f(x) = \frac{-(x+3)(\sqrt{ax+b}-2)}x,x<-3$$
$$f(x) = \frac{(x+3)(\sqrt{ax+b}-2)}x,x>=-3$$
Firstly, I found difficulties in setting up two equation solving for a and b.
Secondly, since $|x+3|=x+3$ only when $x\to0$, I can only use the follwoing to set up the equation. $$\lim_{x\to0}\frac{(x+3)(\sqrt{ax+b}-2)}x$$
However, I cannot simplified it to cancel out the denominator $x$ by some method like rationalization. It gets $$\lim_{x\to0}\frac{(x+3)(ax+b-4)}{(\sqrt{ax+b}-2)x}$$.
Therefore, how can we set up the equation to find $a$ and $b$ or it needs other methods to do this type of questions?
| $\lim_{x\to0}\frac{|x+3|(\sqrt{ax+b}-2)}x
$
Since
$\lim_{x\to0} |x+3|
=3$,
that doesn't matter.
What is left is
$\lim_{x\to0}\frac{(\sqrt{ax+b}-2)}x
$.
For this limit to exist,
we must have
$\lim_{x\to0}\sqrt{ax+b}-2
=0
$
or
$\lim_{x\to0}\sqrt{ax+b}
=2
$.
Since
$\lim_{x\to0}ax
=0
$,
we must have
$b=4
$.
The answer is,
therefore,
$b=4$
and any value of $a$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1964631",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
if $abc=1$,show that $a^3+b^3+c^3+\frac{256}{(a+1)(b+1)(c+1)}\ge 35$
Let $a,b,c>0,abc=1$ show that
$$a^3+b^3+c^3+\dfrac{256}{(a+1)(b+1)(c+1)}\ge 35\tag{1}$$
iff $a=b=c=1$
I know use AM-GM inequality
$$a^3+b^3+c^3\ge 3abc=3$$
and
$$(a+1)(b+1)(c+1)\ge 2\sqrt{a}\cdot 2\sqrt{b}\cdot 2\sqrt{c}=8$$
In this way, will lead to inequality reverse, but $(1)$ seem is right,so How to prove this inequality
| Alternative proof
WLOG, assume $c \in (0, 1]$.
Denote the inequality by $f(a, b, c) \ge 0$. We have
\begin{align}
&f(a, b, c) - f(\sqrt{ab}, \sqrt{ab}, c)\\
=\ & (a + b - 2\sqrt{ab})\Big[(a+b)^2 + (a+b)\cdot 2\sqrt{ab} + (2\sqrt{ab})^2 - 3ab\Big]\\
&\qquad - \frac{256(a + b - 2\sqrt{ab})}{(c+1)(ab + a+b+1)(ab + 2\sqrt{ab} + 1)}\\
=\ & (a + b - 2\sqrt{ab}) \\
& \times \left(
(a+b)^2 + 2(a+b)\sqrt{ab} + ab - \frac{256}{(c+1)(ab + a+b+1)(ab + 2\sqrt{ab} + 1)}\right)\\
\ge\ & (a + b - 2\sqrt{ab}) \\
& \times \Big(
(2\sqrt{ab})^2 + 2(2\sqrt{ab})\sqrt{ab} + ab - \frac{256}{(\frac{1}{ab}+1)
(ab + 2\sqrt{ab} +1)(ab + 2\sqrt{ab} + 1)}\Big)\\
=\ & (a + b - 2\sqrt{ab})\Big(9ab - \frac{256ab}{(1+ab)(\sqrt{ab} + 1)^4}\Big)\\
\ge\ & (a + b - 2\sqrt{ab})\Big(9ab - \frac{256ab}{(1+1)(1 + 1)^4}\Big)\\
\ge\ & 0
\end{align}
where we have used $a^3 + b^3 = (a+b)^3 - 3ab(a+b)$, $a + b \ge 2\sqrt{ab}$,
and $ab = \frac{1}{c} \ge 1$.
Thus, it suffices to prove that $f(\sqrt{ab}, \sqrt{ab}, c) \ge 0$ or
$$2(ab)^{3/2} + c^3 + \frac{256}{(\sqrt{ab} + 1)^2(c+1)} - 35 \ge 0$$
or
$$2(1/c)^{3/2} + c^3 + \frac{256}{(\sqrt{1/c} + 1)^2(c+1)} - 35 \ge 0.$$
Let $c = u^2$. It suffices to prove that
$$\frac{2}{u^3} + u^6 + \frac{256}{(1/u + 1)^2(u^2 + 1)} - 35 \ge 0$$
or
\begin{align}
&\frac{(u-1)^2}{u^3(u+1)^2(u^2+1)}\\
&\cdot (u^{11}+4 u^{10}+9 u^9+16 u^8+24 u^7+32 u^6+5 u^5-92 u^4-3 u^3+18 u^2+8 u+2)\ge 0
\end{align}
which is true.
We are done.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1966306",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
Probability that a number is divisible by 11 The digits $1, 2, \cdots, 9$ are written in random order to form a nine digit number. Then, the probability that the number is divisible by $11$ is $\ldots$
I know the condition for divisibility by $11$ but I couldn't guess how to apply it here.
Please help me in this regard. Thanks.
| since $10 = -1 \pmod {11}$, a number $abcdefghi$ is a multiple of $11$ if and only if $(a+c+e+g+i)-(b+d+f+h)$ is a multiple of $11$.
Since $(a+c+e+g+i)+(b+d+f+h) = 45 = 1 \pmod {11}$, this is equivalent to $1-2(b+d+f+h) = 0 \pmod {11}$, and to $(b+d+f+h) = 6 \pmod {11}$.
So we want to know, when is the sum of $4$ numbers randomly taken in $\{1 ; \ldots ; 9 \}$ is congruent to $6$ modulo $11$.
Clearly, if we were picking our four numbers in $\{1 ; \ldots ; 11\}$ (or $\{0 ; \ldots ; 10 \}$), the sum is uniformly distributed mod $11$ (if we add $3$ to each number, it's like adding $1$ to the sum). Which means there are $\frac 1 {11}\binom {11}4 = 30$ good quadruplets there.
Out of all of those we are only interested in those that don't use $10$ nor $11$.
Let's count how many use $10$ :
A quadruplet that use $10$ and that sums to $6$ is $10$ plus a triplet that sums to $7$ and that doesn't use $10$.
Once again we count the total number of triplets that sum to $7$, but we again have extra triplets, those that contain $10$.
We can continue like this, to remove the extra triplets we have to remove pairs that sum to $8$, and finally remove from those pairs the pair $\{10 ; 9 \}$.
So we get $\frac 1 {11}(\binom {11}3 - \binom {11}2 + \binom {11}1) = 11$ quadruplets that sum to $6$ and use $10$, which means there are $\frac 1 {11}(\binom {11}4 - \binom {11}3 + \binom {11}2 - \binom {11}1) = 19$ quadruplets that sum to $6$ and don't use $10$.
Now, we count the quadruplets that use $11$. The same thing happens the same way, even at the last step (because $4 \times 11 \neq 6 \neq 4 \times 10)$.
Had we wanted to count the number of quadruplets that don't use $7$, then we would have a difference at the end (because $7 \times 4 = 28 = 6$): none of the pairs that sum to $3$ contained a $7$ in the first place, so we don't count that last $\frac 1{11}\binom {11}1$.
Or said another way, the sum of quadruplets that don't use $10$ is almost uniform : it hits every sum $19$ times except $4 \times 10 = 7$, who is hit $20$ times (for a total of $210$, and there are $210$ quadruplets that don't use $10$).
Finally we want to count how many quadruplets sum to $6$ and use both $10$ and $11$. Those are the number of pairs that sum to $7$ and don't use $10$ nor $11$.
There are $5$ pairs that sum to $7$, one of which uses $10$ and one of which uses $11$ (none use both because $10+11 \neq 7$)
So that's a total of $3$ quadruplets that sum to $6$ and use both $10$ and $11$.
The final number is $30 - 11 - 11 + 3 = 11$ quadruplets that sum to $6$ and don't use $10$ or $11$.
Since there are $126$ quadruplets that don't use $10$ or $11$, the final probability is $\frac {11}{126}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1967378",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "22",
"answer_count": 5,
"answer_id": 0
} |
A better way to evaluate a certain determinant
Question Statement:-
Evaluate the determinant:
$$\begin{vmatrix}
1^2 & 2^2 & 3^2 \\
2^2 & 3^2 & 4^2 \\
3^2 & 4^2 & 5^2 \\
\end{vmatrix}$$
My Solution:-
$$
\begin{align}
\begin{vmatrix}
1^2 & 2^2 & 3^2 \\
2^2 & 3^2 & 4^2 \\
3^2 & 4^2 & 5^2 \\
\end{vmatrix} &=
(1^2\times2^2\times3^2)\begin{vmatrix}
1 & 1 & 1 \\
2^2 & \left(\dfrac{3}{2}\right)^2 & \left(\dfrac{4}{3}\right)^2 \\
3^2 & \left(\dfrac{4}{2}\right)^2 & \left(\dfrac{5}{3}\right)^2 \\
\end{vmatrix}&\left[\begin{array}{11}C_1\rightarrow\dfrac{C_1}{1} \\
C_2\rightarrow\dfrac{C_2}{2^2}\\
C_3\rightarrow\dfrac{C_3}{3^2}\end{array}\right]\\
&=
(1^2\times2^2\times3^2)\begin{vmatrix}
1 & 0 & 0 \\
2^2 & \left(\dfrac{3}{2}\right)^2-2^2 & \left(\dfrac{4}{3}\right)^2-2^2 \\
3^2 & 2^2-3^2 & \left(\dfrac{5}{3}\right)^2-3^2 \\
\end{vmatrix} &\left[\begin{array}{11}C_2\rightarrow C_2-C_1 \\
C_3\rightarrow C_3-C_1\end{array}\right]\\
&=
(1^2\times2^2\times3^2)
\begin{vmatrix}
1 & 0 & 0 \\
2^2 & 2^2-\left(\dfrac{3}{2}\right)^2 & 2^2-\left(\dfrac{4}{3}\right)^2 \\
3^2 & 3^2-2^2 & 3^2-\left(\dfrac{5}{3}\right)^2 \\
\end{vmatrix}\\
&=(1^2\times2^2\times3^2)
\begin{vmatrix}
1 & 0 & 0 \\
2^2 & \dfrac{7}{4} & \dfrac{20}{9} \\
3^2 & 5 & \dfrac{56}{9} \\
\end{vmatrix}\\
&=(1^2\times2^2)
\begin{vmatrix}
1 & 0 & 0 \\
2^2 & \dfrac{7}{4} & 20 \\
3^2 & 5 & 56 \\
\end{vmatrix}\\
&=4\times(-2)\\
&=-8
\end{align}
$$
As you can see, my solution is a not a very promising one. If I encounter such questions again, so would you please suggest a better method which doesn't include this ridiculous amount of calculations.
| Subtract the first column from other columns to reduce $n^2$-s, then subtract twice the second column from the third one to reduce $n$:
$$\begin{vmatrix}
n^2 & (n+1)^2 & (n+2)^2 \\
(n+1)^2 & (n+2)^2 & (n+3)^2 \\
(n+2)^2 & (n+3)^2 & (n+4)^2
\end{vmatrix}
=
\begin{vmatrix}
n^2 & 2n+1 & 4n+4 \\
(n+1)^2 & 2n+3 & 4n+8 \\
(n+2)^2 & 2n+5 & 4n+12
\end{vmatrix}
=
\begin{vmatrix}
n^2 & 2n+1 & 2 \\
(n+1)^2 & 2n+3 & 2 \\
(n+2)^2 & 2n+5 & 2
\end{vmatrix}
$$
Next subtract the first row from other rows, then twice the second row from the third one:
$$\cdots
=
\begin{vmatrix}
n^2 & 2n+1 & 2 \\
2n+1 & 2 & 0 \\
4n+4 & 4 & 0
\end{vmatrix}
=
\begin{vmatrix}
n^2 & 2n+1 & 2 \\
2n+1 & 2 & 0 \\
2 & 0 & 0
\end{vmatrix}
$$
Now we have a triangular determinant with three twos on its antidiagonal, so the determinant is
$$2\cdot (-2)\cdot 2 = -8$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1969290",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 6,
"answer_id": 2
} |
find $\lim_{(x,y)\to (0,0)} \frac{x^2y^2}{x^4+y^4}$ if $$f(x,y)=\frac{x^2y^2}{x^4+y^4}$$ is a 2 variable function, find $$\lim_{(x,y)\to (0,0)} \frac{x^2y^2}{x^4+y^4}$$
I really don't understand 2 variable limits. I understand that if limits from 2 or more paths aren't the same, the limit doesn't exist, but I don't know how to find the limit since there is an infinite number of paths possible.
| Approach #1: $\epsilon$-$\delta$ proof...probably too much work for the given problem.
Approach #2: Convert to polar/spherical
Recalling that $x=r \cos \theta$ and $y= r \sin \theta$, we note that $(x,y) \to (0,0) \iff r \to 0$. As such,
$$\lim_{(x,y) \to (0,0) } \frac{ x^2 y^2}{x^4 + y^4} = \lim_{r \to 0} \frac{r^2 \cos^2 \theta \cdot r^2 \sin ^2 \theta}{r^4 ( \cos^4 \theta + \sin^4 \theta)}=\lim_{r \to 0} \frac{\cos^2 \theta \cdot \sin ^2 \theta}{ \cos^4 \theta + \sin^4 \theta} =\frac{\cos^2 \theta \cdot \sin ^2 \theta}{ \cos^4 \theta + \sin^4 \theta} $$
Since the last is expression is a non-constant function of $\theta$ (i.e., doesn't simplify to a constant using trig identities), this means that the limiting value depends on the angle taken to approach $(0,0)$, so there is no way to assign a single value to the limit along all possible paths.
Approach #3: lines
Let's look at what happens if we approach $(0,0)$ along lines $y=kx$:
$$\lim_{(x,y) \to (0,0)} \frac{x^2 y^2}{x^4 + y^4} = \lim_{(x, kx) \to (0,0)} \frac{x^2 (kx)^2}{x^4 + (kx)^4} = \lim_{x \to 0} \frac{k^2 x^4}{x^4 (1+k^4)} = \frac{k^2}{1+k^4}$$
Again, we got that the limiting value depended upon path taken to reach $(0,0)$, i.e., the slope of the particular line followed. Therefore, there is no consistent way to define what value $f(x,y)$ should approach near the origin.
Approach #4: Graph it
Notice that as we approach $(0,0)$ in different directions, we limit on different values between $0$ and $1/2$. There is no way to pick a specific value that all possible curves limit upon.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1970422",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Find the extreme values of the given function, subject to constraints $f(x,y) = y^2 - x^2 ; x^2 + y^2 = 36$
Find the extreme values of the function, subject to constraints
$$f(x,y) = y^2 - x^2 ; x^2 + y^2 = 36$$
Here is what I have done:
Using the substitution method we can say:
$$y = \sqrt{-x^2 + 36}$$
Then plugging that into our original equation:
$$f(x,y) = (\sqrt{-x^2 + 36})^2 - x^2$$ which simplifies to $$-2x^2+36$$
now $$f_x(x,y) = 2(-2x) + 0$$
set that $=0$ and get $x=0$
Now find if min/max:
$$f_{xx} = -4 $$
$$-4 < 0; maximum$$
Answer: There is a maximum point at $x=0$?
Is this right?
| Let $x = 6\cos \theta, y = 6\sin \theta \implies F(x,y) = F(\theta) = 36\cos (2\theta)\le 36$, and the max is $36$. The min is $-36$ and this occurs when $\cos (2\theta) = -1$ which corresponds to $x^2 - y^2 = -36$. Since $x^2+y^2 = 36$, we have: $x = 0, y = \pm 6$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1971382",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Proof by induction that $4^3+4^4+4^5+⋅⋅⋅+4^n = \frac{4(4^n-16)}{3}$ I am stuck on this problem for my discrete math class.
Prove the equation by induction for all integers greater than or equal to $3$:
$$4^3+4^4+4^5+⋅⋅⋅+4^n = \frac{4(4^n-16)}{3}.$$
I know that base case $n=3$:
$4^3=64$ as well as $4(4^3-16)/3 = 64$
My confusion is on the induction step where:
$4^3+4^4+4^5+⋅⋅⋅+4^n+4^{(n+1)} = 4(4^{(n+1)}-16)/3$.
I don't know what to do next.
| By the induction hypothesis
$$
4^3+4^4+\ldots+4^n=\frac{4(4^n-16)}{3}=\frac{4^{n+1}-64}{3}
$$
so
$$
4^3+4^4+\ldots+4^n+4^{n+1}=\frac{4^{n+1}-64}{3}+4^{n+1}=\frac{4^{n+1}-64+3\cdot4^{n+1}}{3}\\
=\frac{4\cdot 4^{n+1}-64}{3}=\frac{4^{n+2}-64}{3}=\frac{4(4^{n+1}-16)}{3}
$$
QED
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1972847",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Prove by induction that $\sum\limits_{i=1}^n \frac{1}{n+i} \leq \frac{3}{4}$
Prove by induction that $\sum\limits_{i=1}^n \frac{1}{n+i} \leq \frac{3}{4}$.
I have to prove this inequality using induction, I proved it for $n=1$ and now I have to prove it for $n+1$ assuming $n$ as hypothesis, but this seems impossible to me because the difference between the sum of $n$ and the sum of $n+1$ is a positive value. Adding a positive value on both sides of the inequality, I don't know how to prove that is always less than or equal to $\frac{3}{4}$.
| Define $\displaystyle{S_n=\sum_{i=1}^n \dfrac{1}{n+i}}$. Then we have for every $n$ that:
\begin{align*}
S_{n+1}-S_n&=\sum_{i=1}^{n+1}\dfrac{1}{n+1+i}-\sum_{i=1}^n \dfrac{1}{n+i}\\
&=\sum_{i=1}^{n+1}\dfrac{1}{n+1+i} - \left(\sum_{i=2}^n \dfrac{1}{n+i}\right)-\dfrac{1}{n+1}\\
&=\sum_{i=1}^{n+1}\dfrac{1}{n+1+i} - \left(\sum_{j=1}^{n-1} \dfrac{1}{n+1+j}\right)-\dfrac{1}{n+1}\\
&=\dfrac{1}{n+1+(n+1)}+\dfrac{1}{n+1+n}-\dfrac{1}{n+1}\\
&=\dfrac{1}{2n+1}-\dfrac{1}{2(n+1)}=\dfrac{1}{2(n+1)(2n+1)}
\end{align*}
In particular, the sequence $S_n$ can also be defined by recurrence with the formula $$\begin{cases}S_1=\dfrac{1}{2}\\ S_{n+1}=S_n+\dfrac{1}{2(n+1)(2n+1)}\end{cases}$$
and so, $S_n$ is simply defined by the formula $$S_n=\sum_{i=1}^n \dfrac{1}{2n(2n-1)}.$$
Finally, showing that $S_n\leq \frac{3}{4}$ for all $n$ is the same as proving that the series $\displaystyle{\sum_{n=1}^\infty \dfrac{1}{2n(2n-1)}}$ converges to a number less than or equal to $\dfrac{3}{4}$.
If we focus on the infinite series, notice that $\dfrac{1}{2n(2n-1)}=\dfrac{1}{2n-1}-\dfrac{1}{2n}$, and so we have:
$$\sum_{n=1}^\infty \dfrac{1}{2n(2n-1)}=\sum_{n=1}^\infty \left(\dfrac{1}{2n-1}-\dfrac{1}{2n} \right)= \sum_{k=1}^\infty \dfrac{(-1)^{k+1}}{k}$$
I know the last series converges (by the test of alternating series), but I am just not sure about the value.
EDIT: AS lhf points out in his comment, the value of the last sum is $\log 2=0.69314\ldots$, which is surely less than $\dfrac{3}{4}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1974177",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Can we show that all $2 \times 2$ matrices are sums of matrices with determinant 1? I came across a paper on the Sums of 2-by-2 Matrices with Determinant One. In the paper, which I have conveniently indicated here for reference, the author claims, but without proof, that a $2 \times 2$ is a sum of elements of the special linear group, $SL_2(\mathbb{F})$, whose elements, $U$, are also $2 \times 2$ matrices, such that $|U|=1$.
I was thinking of proving this by either technique. Let $A= \begin{bmatrix}a & b\\c & d\end{bmatrix}$, and $a, b, c, d \in \mathbb{F}$.
Technique 1. Consider the following 4 matrices with determinant 1: $M_1= \begin{bmatrix}e & 0\\0 & 1/e\end{bmatrix}$, $M_2 = \begin{bmatrix}0 & -f\\1/f & 0\end{bmatrix}$, $M_3 = \begin{bmatrix}1/h & 0\\0 & h\end{bmatrix}$, and $M_4 = \begin{bmatrix}0 & 1/g\\-g & 0\end{bmatrix}$.
We show that that $\sum\limits_{i=1}^4 M_i = A$.
Thus, we have $e + 1/h = a$, $1/e + h = d$, $-f + 1/g = b$, $1/f - g = c$.
Technique 2. Consider $\{U_i\}_{i=1} ^\infty \in SL_2(\mathbb{F})$. Show that the sum of a countable number of $U_i$s is $A$. The problem I have here is that I don't know how to proceed from here.
I don't know if either of these will be considered correct, though.
I hope someone could help me out here. Thanks.
| It is sufficient to show that the matrices $\left(\begin{matrix} a & 0 \\ 0 & 0 \end{matrix}\right)$, $\left(\begin{matrix} 0 & b \\ 0 & 0 \end{matrix}\right)$, $\left(\begin{matrix} 0 & 0 \\ c & 0 \end{matrix}\right)$, and $\left(\begin{matrix} 0 & 0 \\ 0 & d \end{matrix}\right)$ can be written as the sum of matrices with determinant $1$.
Now, notice that
\begin{align}\left(\begin{matrix} a & 0 \\ 0 & 0 \end{matrix}\right) &= \left(\begin{matrix} \frac{a}{2} & 1 \\ -1 & 0 \end{matrix}\right) + \left(\begin{matrix} \frac{a}{2} & -1 \\ 1 & 0 \end{matrix}\right) \\
\left(\begin{matrix} 0 & b \\ 0 & 0 \end{matrix}\right) &= \left(\begin{matrix} 1 & \frac{b}{2} \\ 0 & 1 \end{matrix}\right) + \left(\begin{matrix} -1 & \frac{b}{2} \\ 0 & -1 \end{matrix}\right) \\
\left(\begin{matrix} 0 & 0 \\ c & 0 \end{matrix}\right) &= \left(\begin{matrix} 1 & 0 \\ \frac{c}{2} & 1 \end{matrix}\right) + \left(\begin{matrix} -1 & 0 \\ \frac{c}{2} & -1 \end{matrix}\right) \\
\left(\begin{matrix} 0 & 0 \\ 0 & d \end{matrix}\right) &=\left(\begin{matrix} 0 & 1 \\ -1 & \frac{d}{2} \end{matrix}\right) + \left(\begin{matrix} 0 & -1 \\ 1 & \frac{d}{2} \end{matrix}\right).
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1976815",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
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} |
$\sum _{n=1}^{\infty} \frac 1 {n^2} =\frac {\pi ^2}{6}$ and $ S_i =\sum _{n=1}^{\infty} \frac{i} {(36n^2-1)^i}$ . Find $S_1 + S_2 $ I know to find sum of series using method of difference. I tried sum of write the term as (6n-1)(6n+1). i don't know how to proceed further.
| \begin{align*}
S_1+S_2&=\sum_{n\geq 1}\frac{1}{36n^2-1}+\frac{2}{(36n^2-1)^2}\\
&=\sum_{n\geq 1}\frac{1}{2}\frac{1}{(6n-1)^2}+\frac{1}{2}\frac{1}{(6n+1)^2}\\
&=\frac{1}{2}\sum_{\substack{n\geq 5\\n\equiv \pm1\,\!\!\!\mod 6}}\frac{1}{n^2}\\
&=\frac{1}{2}\left[\sum_{n\geq 1}\frac{1}{n^2}-\sum_{\substack{n\geq 1\\n\equiv 0\!\!\!\mod 2}}\frac{1}{n^2}-\sum_{\substack{n\geq 1\\n\equiv 0\!\!\!\mod 3}}\frac{1}{n^2}+\sum_{\substack{n\geq 1\\n\equiv 0\!\!\!\mod 6}}\frac{1}{n^2}-1\right]\\
&=\frac{1}{2}\left[\frac{\pi^2}{6}-\frac{1}{4}\frac{\pi^2}{6}-\frac{1}{9}\frac{\pi^2}{6}+\frac{1}{36}\frac{\pi^2}{6}-1\right]\\
&=\frac{\pi^2}{18}-\frac{1}{2}.
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1978142",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Solve the system of inequalities Solve the system:
$$|3x+2|\geq4|x-1|$$
$$\frac{x^{2}+x-2}{2+3x-2x^{2}}\leq 0$$
So for $|3x+2|\geq4|x-1|$ I got the solution $x=6$ and $x=2/7$ and Wolframalpha agrees with me. I'm having troubles writing the final solution for $\frac{x^{2}+x-2}{2+3x-2x^{2}}\leq 0$.
$1)$ $x^{2}+x-2\leq0$ and $-2x^{2}+3x+2>0$
$x^{2}+x-2\leq0$ $\Rightarrow x\in [-2,1]$
$-2x^{2}+3x+2>0$ $\Rightarrow x\in (-1/2,2)$
Now I need the intersection of those two sets, which is $(-1/2,1]$. From the first inequality I got $x=6$ and $x=2/7$ so the final solution is $x=2/7$.
$2)$ $x^{2}+x-2\geq 0$ and $-2x^{2}+3x+2<0$
$x^{2}+x-2\geq 0$ $\Rightarrow x\in(-\infty, -2]\cup[1,\infty]$
$-2x^{2}+3x+2<0$ $\Rightarrow x\in(-\infty, -1/2)\cup(2,\infty)$
Now I'm having troubles finding the intersection of those sets.
| If $x\ge 1$ we have $$|3x+2|\geq4|x-1|\iff 3x+2\geq 4x-4 \iff x\le 6.$$ So $[1,6]$ is solution of the first inequality. Now, if $-2/3\le x\le 1$ we have
$$|3x+2|\geq4|x-1|\iff 3x+2\geq 4-4x \iff x\ge 2/7.$$ So $[2/7,1]$ is solution of the system. Finally, if $x\le -2/3$ then
$$|3x+2|\geq4|x-1|\iff -3x-2\geq 4-4x \iff x\ge 6,$$ which is impossible. That is, the solution set of the first inequality is $[2/7,6].$
(The idea is to solve $3x+2=0$ and $x-1=0$ and study the inequality on each region you obtain.)
Now, we will work with the second inequality. Write it as
$$\dfrac{(x-1)(x+2)}{2(x-2)(x+1/2)}\ge 0.$$ Study the sign on $(-\infty,-2),$ $(-2,-1/2),$ $(-1/2,1),$ $(1/2,1)$ and $(1,\infty).$ You should obtain that the set solution is $(-\infty,-2]\cup (-1/2,1]\cup (2,\infty).$
(The idea is to solve $x^2+x-2=0$ and $2+3x-2x^2=0$ and study the inequality on each region you obtain. Note that we can't divide by $0.$ So $x\ne-1/2$ and $x\ne 2.$)
Finally, one gets the intersection to obtain $(2/7,1]\cup (2,6].$
| {
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"url": "https://math.stackexchange.com/questions/1981186",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Show $\frac1{a^2+a+1}+\frac1{b^2+b+1}+\frac1{c^2+c+1}\ge1$
Positive real numbers $a,b,c$ satisfy $abc=1$. Prove
$$\frac1{a^2+a+1}+\frac1{b^2+b+1}+\frac1{c^2+c+1}\ge1$$
I tried AM–GM, Cauchy–Schwarz and Jensen's but they all failed.
| Following stewbasic's hint in the comments:
$$\frac1{a^2+a+1}+\frac1{b^2+b+1}+\frac1{c^2+c+1}\ge1$$
$$\iff(b^2+b+1)(c^2+c+1)+(c^2+c+1)(a^2+a+1)+(a^2+a+1)(b^2+b+1)\ge(a^2+a+1)(b^2+b+1)(c^2+c+1)$$
The LHS expands as
$$a^2b^2+a^2b+ab^2+\color{blue}{a^2}+b^2+ab+a+b+1\\
+b^2c^2+b^2c+bc^2+\color{blue}{b^2}+c^2+bc+b+c+1\\
+c^2a^2+c^2a+ca^2+\color{blue}{c^2}+a^2+ca+c+a+1$$
while the RHS expands as
$$a^2b^2c^2+abc(ab+bc+ca)+\\
a^2b^2+b^2c^2+c^2a^2+abc(a+b+c)+\\
abc+a^2b+b^2a+b^2c+c^2b+c^2a+a^2c+\\
\color{green}{ab+bc+ca}+a^2+b^2+c^2+a+b+c+1$$
Cancelling and using the relation $abc=1$ we find that the inequality above is equivalent to
$$\color{blue}{a^2+b^2+c^2}\ge\color{green}{ab+bc+ca}$$
which can be seen to be true by applying the Cauchy–Schwarz inequality on $(a,b,c)$ and $(b,c,a)$. By following the biconditionals back we see that the original inequality is true.
| {
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"url": "https://math.stackexchange.com/questions/1983734",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
} |
value of $a$ for which $25^x+(a+2)5^x-(a+3)<0$for at least one real $x$
Find the values of $a$ for which the inequality is satisfied for $25^x+(a+2)5^x-(a+3)<0$
for at least one real value of $x$
$\bf{My\; Try::}$ We can write it as $a(5^x-1)<-\left[25^x+2\cdot 5^x-3\right]$
So $$a < -\left(\frac{25^x-5^x+3\cdot 5^x-3}{5^x-1}\right) = -\frac{(5^x+3)(5^x-1)}{5^x-1} = -(5^x+3)$$
Now how can i solve after that help required, Thanks
| $z=5^x \to $ $z^2+(a+2)z-(a+3) \lt 0$
now $ \\ \Delta = \to (a+2)^2-4(1)(-(a+3)) = \\a^2+4+4a+4a+12=a^2+8a+16=(a+4)^2 \geq0 \\$
now $$z=\frac{-(a+2)\pm \sqrt{(a+4)^2}}{2}=\\z=1 ,z=-(a+3)\\z=1 \to 5^x=1 \to x=0\\ (5^x-1)(5^x+a+3)<0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1983956",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Polynomial of degree 5 such that $P(x)-1$ is divisible by $(x-1)^3$ and $P(x)$ is divisible by $x^3$
$P(x)$ is a polynomial of degree 5 such that $P(x)-1$ is divisible by $(x-1)^3$ and $P(x)$ is divisible by $x^3$. Find $P(x)$.
No idea where to start, would $P(x)$ be of the form $x^3(Ax^2+Bx+C)$?
| We can say:
\begin{align}P(x)-1&=(x-1)^3(ax^2+bx+c)\\
P(x)&=x^3(dx^2+ex+f)
\end{align}
Therefore, we can say that \begin{align}(x-1)^3(ax^2+bx+c)+1&=x^3(dx^2+ex+f)\\
(x^3-3x^2+3x-1)(ax^2+bx+c)+1&=dx^5+ex^4+fx^3\\
ax^5+(b-3a)x^4+(c-3b+3a)x^3+(3b-3c-a)x^2+(3c-b)x+(1-c)&=dx^5+ex^4+fx^3
\end{align}
We can then equate coefficients
\begin{align}a&=d\tag{$x^5$ term}\\
b-3a&=e\tag{$x^4$ term}\\
c-3b+3a&=f\tag{$x^3$ term}\\
3b-3c-a&=0\tag{$x^2$ term}\\
3c-b&=0\tag{$x$ term}\\
1-c&=0\tag{constant term}\end{align}
We can use the last three equations to say that $c=1$, $b=3$, $a=6$
We can then use the top three equation to say that $d=6$, $e=-15$, $f=10$
We can use the equation from above, to give us:
\begin{align}P(x)&=x^3(dx^2+ex+f)\\
&=x^3(6x^2-15x+10)\\
&=6x^5-15x^4+10x^3\end{align}
| {
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"source": "stackexchange",
"question_score": "4",
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If $ f(x) = \frac{b+2x-x^2}{ax-1}\forall x \in \mathbb{R}$ and $\min(f(x)) = -5,$ Then $a+b$
If $\displaystyle f(x) = \frac{b+2x-x^2}{ax-1}\forall x \in \mathbb{R}$ and $\min(f(x)) = -5,$ Then $a+b=$
Let $$y = \frac{b+2x-x^2}{ax-1}\Rightarrow axy-y=b+2x-x^2$$
So $x^2+(ay-2)x-(y+b) = 0$
Now for real values of $y,$ Equation must have real roots. So $\bf{Discriminant \geq 0}$
So $(ay-2)^2+4(y+b)\geq 0\Rightarrow a^2y^2+4-4ay+4y+4b\geq 0$
So $a^2y^2-4(a-1)y+(4b+4)\geq 0$ Now if $a\neq 0,$ Then $\bf{Dicriminant \leq 0}$
So $$16(a-1)^2-4a^2(4b+4)\leq 0$$
Now how can i solve after that, Help required, Thanks
| As $x \rightarrow \infty$, $f(x)$ will be dominated by the $x^2$ term in the numerator. Hence if $a > 0$, $f(x) \rightarrow -\infty$. Again, if $a < 0$, $f(x) \rightarrow -\infty$ as $x \rightarrow -\infty$. Thus $a=0$.
Now
$$f(x) = x^2-2x-b = (x-1)^2 -b-1 $$ and if the minimum of $f(x) = -5$, it follows that $b=4$. Thus $a+b= 4$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Prove: $ \frac{1}{\sin 2x} + \frac{1}{\sin 4x } + \cdots + \frac{1 }{\sin 2^n x} = \cot x - \cot 2^n x $
Prove
$$ \frac{1}{\sin 2x} + \frac{1}{\sin 4x } + \cdots + \frac{1 }{\sin 2^n x} = \cot x - \cot 2^n x $$
where $n \in \mathbb{N}$ and $x$ not a multiple of $\frac{ \pi }{2^k} $ for any $k \in \mathbb{N}$.
My try. If $n=2$, we have
$$\begin{align}
\frac{1}{\sin 2x} + \frac{ 1}{\sin 4 x} &= \frac{1}{\sin 2x} + \frac{1}{2 \sin 2x \cos 2x } \\[6pt]
&= \frac{ 2 \cos 2x + 1 }{2 \sin 2x \cos 2x} \\[6pt]
&= \frac{2 \cos^2 x - 2 \sin^2 x + \cos^2 x + \sin^2 x}{2 \sin 2x \cos 2x} \\[6pt]
&= \frac{3 \cos^2 x - \sin^2 x}{2 \sin 2x \cos 2x}
\end{align}$$
but here I got stuck. I am on the right track? My goal is to ultimately use induction.
| Without Induction
We can werite series as $$\sum^{n}_{r=1}\frac{1}{\sin 2^{r}x} = \sum^{n}_{r=1}\frac{\sin(2^{r}x-2^{r-1}x)}{\sin 2^{r-1}x\cdot \sin 2^{r}x} = \sum^{n}_{r=1}(\cot 2^{r-1}x-\cot 2^{r}x )$$
Now use Telescopic Sum
| {
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Summation using Euler's formula I must verify the following formula:
$$
\sum _{i=1}^n \sin ^2 (2k-1) \theta = -\frac{n}{2} - \frac{\sin 4n \theta}{4 \sin 2 \theta}
$$
I believe that I must do this by using the Euler's formula,
$$
e^{i \theta} = \cos \theta + i \sin \theta
$$
and taking only the imaginary parts of it.
However, I am really stuck. Here is my work so far:
Let $x=e^{i \theta}$. Then,
$$
e^{i \theta} = \cos \theta + i \sin \theta
$$
$$
e^{2i \theta} = \cos ^2 \theta + i \sin \theta \cos \theta - \sin ^2 \theta
$$
The sum we need to verify then becomes equivalent to the imaginary part of:
$$
-e^{2i \theta} + \cos ^2 \theta-e^{6i \theta} + \cos ^2 3\theta -e^{10i \theta} + \cos ^2 5\theta -... +...-e^{(4k-2)i \theta} + \cos ^2 (2k-1)\theta.
$$
This sum doesn't really seem to help me, especially because the exponent of the $e$ terms do not increase linearly. How should I approach this problem?
| \begin{align*}
\sum _{i=1}^n \sin ^2 (2k-1) \theta &= \frac{1}{2}\sum _{k=1}^n (1-\cos 2(2k-1)\theta)\\
&= \frac{n}{2} - \frac{1}{2}\sum _{k=1}^n \cos (4k - 2) \theta\\
&= \frac{n}{2} - {\text {Real part of }}\frac{1}{2}\sum _{k=1}^n e^{(4k - 2)i\theta}\\
&=\frac{n}{2} - {\text {Real part of }}\frac{1}{2}e^{2i\theta}(1+e^{4i\theta} + \cdots + e^{4(n-1)\theta})\\
&=\frac{n}{2} - {\text {Real part of }}\frac{1}{2}e^{2i\theta}\frac{1-e^{4ni\theta}}{1-e^{4i\theta}}\\
\end{align*}
The computation can now be easily done.
| {
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Solve for $x$ where $0\leq x\leq 360$ Solve
$$4\sin x \cdot \sin 2x \cdot \sin 4x =\sin 3x$$
My Attempt :
Here,
$$4\sin x \cdot \sin 2x \cdot \sin 4x=\sin 3x$$
$$4\sin x \cdot (2\sin x \cdot \cos x ). (4 \sin x \cdot \cos x \cdot \cos 2x)=\sin3x$$
$$32\sin^3 x \cdot \cos^2 x \cdot \cos2x=\sin3x$$
How should I proceed further?
| \begin{align}
\sin 3x
&= \sin(2x + x) \\
&= \sin 2x \; \cos x + \cos 2x \; \sin x \\
&= 2 \sin x \; \cos^2 x + \cos 2x \; \sin x \\
&= \sin x \; (2 \cos^2 x + \cos 2x) \\
&= \sin x \; (2 \cos 2x + 1)
\end{align}
\begin{align}
4 \sin 2x \; \sin 4x
&= 2(\cos 4x \; \cos 2x + \sin 4x \; \sin 2x)
-2(\cos 4x \; \cos 2x - \sin 4x \; \sin 2x) \\
&= 2\cos(4x - 2x) - 2\cos(4x + 2x) \\
&= 2(\cos 2x - \cos 6x)
\end{align}
\begin{align}
4\sin x \; \sin 2x \; \sin 4x &= \sin 3x \\
2\sin x \; (\cos 2x - \cos 6x) &= \sin x \; (2 \cos 2x + 1) \\
\hline
\sin x &= 0\\
x &\in \{180^\circ n : n \in \mathbb Z\} \\
\hline
2 \cos 2x - 2 \cos 6x &= 2 \cos 2x + 1 \\
\cos 6x &= -\dfrac 12\\
6x &\in \{360^\circ n \pm 240^\circ : n \in \mathbb Z \} \\
x &\in \{60^\circ n \pm 40^\circ : n \in \mathbb Z \} \\
\end{align}
Solution set:
$$x \in (\{60^\circ n \pm 40^\circ : n \in \mathbb Z \}
\cup \{180^\circ n : n \in \mathbb Z\})
\cap [0^\circ, 360^\circ]$$
$$x \in
\left\{ \begin{array}{rrrrr}
0^\circ, & 20^\circ, & 40^\circ, & 80^\circ, & 100^\circ, \\
140^\circ, & 160^\circ, & 180^\circ, & 200^\circ, & 220^\circ, \\
260^\circ, & 280^\circ, & 320^\circ, & 340^\circ, & 360^\circ \\
\end{array} \right\} $$
| {
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"timestamp": "2023-03-29T00:00:00",
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Using induction to prove formula I am revising for my test from Discrete math. I have come to this problem.
I am to prove by using mathematical induction that
$6\times7^{n} - 2 \times 3^{n}$ is divisible by 4. for $n \ge 1$ ;
I created basic step :
$6\times7^{1} - 2\times3^{1} = 36 $
and induction step
$\forall n\ge 1, \exists K: 6\times7^{n} - 2\times3^{n} = 4K \Rightarrow \forall n \ge 1, \exists l: 6\times7^{n+1} - 2\times3^{n+1} = 4l$
we can transform the formula into
$6\times7\times7^{n} - 2\times3\times3^{n}$
which is basicly
$42\times7^{n} -6\times3^{n}$
But what is the next step? I can i prove this fact?
| First, show that this is true for $n=1$:
$6\cdot7^{1}-2\cdot3^{1}=4\cdot9$
Second, assume that this is true for $n$:
$6\cdot7^{n}-2\cdot3^{n}=4\cdot{k}$
Third, prove that this is true for $n+1$:
$6\cdot7^{n+1}-2\cdot3^{n+1}=$
$6\cdot7\cdot7^{n}-2\cdot3\cdot3^{n}=$
$7\cdot6\cdot7^{n}-3\cdot2\cdot3^{n}=$
$7\cdot6\cdot7^{n}-(7-4)\cdot2\cdot3^{n}=$
$7\cdot6\cdot7^{n}-7\cdot2\cdot3^{n}+4\cdot2\cdot3^{n}=$
$7\cdot(\color\red{6\cdot7^{n}-2\cdot3^{n}})+4\cdot2\cdot3^{n}=$
$7\cdot\color\red{4\cdot{k}}+4\cdot2\cdot3^{n}=$
$4\cdot(7\cdot{k}+2\cdot3^{n})$
Please note that the assumption is used only in the part marked red.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1990358",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 4
} |
Determine using the comparison test whether the series $\sum_{n=1}^\infty \frac{1}{\sqrt[3]{n^2 + 1}}$ diverges How can I determine whether or not the following series converges using the comparison test?
$$\sum_{n=1}^\infty \frac{1}{(n^2 + 1)^\frac{1}{3}}$$
As $n$ goes to infinity, the sum is roughly equal to $\sum_{n=1}^\infty \frac{1}{(n^2)^\frac{1}{3}} = \sum_{n=1}^\infty \frac{1}{n^\frac{2}{3}}$.
I believe that $\sum_{n=1}^\infty \frac{1}{(n^2 + 1)^\frac{1}{3}} < \sum_{n=1}^\infty \frac{1}{n^\frac{2}{3}}$ for all $n$.
Using the $p$ test, it is clear that the latter sum diverges. However I cannot say that the former also diverges as the inequality sign does not satisfy the conditions of the comparison test for divergence.
| Note that $n^2+1 \le 2 n^2$ so you have
${1 \over \sqrt[3]{n^2+1}} \ge {1 \over \sqrt[3]{2}} {1 \over \sqrt[3]{n^2}} $.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1993244",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
How to get $\frac{2}{3}(3x-5)^{2}+\frac{19}{3}$ from this expression $6x^{2}-20x+23$. I'm trying to figure out how to get $\frac{2}{3}(3x-5)^{2}+\frac{19}{3}$ from this expression $6x^{2}-20x+23$.
Hints?
| $$\begin{align}6x^2-20x+23&=\frac{2}{3}\left(\frac326x^2-\frac3220x+\frac3223\right)\\
&=\frac23\left(9x^2-15x+\frac{69}2\right)\\
&=\frac23\left(9x^2-15x+\frac{50}2+\frac{19}{2}\right)\\
&=\frac23\left(9x^2-15x+25+\frac{19}{2}\right)\\
&=\frac23\left((3x-5)^2+\frac{19}{2}\right)\\
&=\frac23\left(3x-5\right)^2+\frac23\cdot\frac{19}{2}\\
&=\frac23\left(3x-5\right)^2+\frac{19}{3}\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1994851",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
How to calculate $\int_0^\pi \frac{\cos^4(x)}{1+\sin^2(x)}dx$ using contour integrals How do I go about calculating $\int_0^\pi \frac{\cos^4(x)}{1+\sin^2(x)}dx$ using contour integrals?
| Suppose we seek to evaluate
$$J = \int_0^{\pi} \frac{\cos^4 x}{1+\sin^2 x} dx
= \frac{1}{2} \int_0^{2\pi} \frac{\cos^4 x}{1+\sin^2 x} dx.$$
Put $z = \exp(ix)$ so that $dz = i\exp(ix) dx$ and hence
$\frac{dz}{iz} = dx$ to obtain
$$\frac{1}{2} \int_{|z|=1}
\frac{(z+1/z)^4/2^4}{1+(z-1/z)^2/(2i)^2} \frac{dz}{iz}
\\ = -\frac{1}{8} \int_{|z|=1}
\frac{(z^2+1)^4}{-4z^2+(z^2-1)^2} \frac{dz}{iz^3}
\\ = \frac{i}{8} \int_{|z|=1}
\frac{(z^2+1)^4}{(z^2-1)^2-4z^2} \frac{z\; dz}{z^4}.$$
Now put $w=z^2$ so that $z\; dz = \frac{1}{2} \; dw$ and note that
this loops around the origin twice so that we must multiply a single
loop by two to get
$$2\times \frac{i}{8} \int_{|w|=1}
\frac{(w+1)^4}{(w-1)^2-4w} \frac{1}{2} \frac{dw}{w^2}
\\ = \frac{i}{8} \int_{|w|=1}
\frac{(w+1)^4}{w^2-6w+1} \frac{dw}{w^2}.$$
In addition to the double pole at $w=0=\rho$ there are two simple
poles here at
$$\rho_{0,1} = 3 \pm 2\sqrt{2}$$
and clearly $\rho_1$ is the only one inside the contour.
Converting to partial fractions we get
$$\frac{64}{w^2-6w+1} + 1 + \frac{10}{w} + \frac{1}{w^2}.$$
Hence the non-zero contribution comes from
$$\frac{64}{w^2-6w+1} + \frac{10}{w}.$$
The first term yields
$$\mathrm{Res}_{w=\rho_1}
\frac{64}{w^2-6w+1} =
\left. \frac{64}{2w-6} \right|_{w=\rho_1}
\\ = \frac{64}{-4\times\sqrt{2}}
= -\frac{16}{\sqrt{2}} = -8\sqrt{2}.$$
The second term produces $10$ by inspection. Collecting
everything we have
$$2\pi i \times \frac{i}{8}
\times(10 - 8\sqrt{2})
= \frac{\pi}{4} (8\sqrt{2}-10)$$
or
$$\bbox[5px,border:2px solid #00A000]{
\frac{\pi}{2}(4\sqrt{2}-5).}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1996021",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Finding $\lim \sup, \lim \inf$ for sequence $s_n$ defined by $s_1=0$, $s_{2m}=\frac{s_{2m-1}}{2}$ $s_{2m+1}=\frac{1}{2}+s_{2m}$ I want to find $\lim \sup, \lim \inf$ for the sequence $s_n$ defined by $s_1=0$, $s_{2m}=\frac{s_{2m-1}}{2}$ $s_{2m+1}=\frac{1}{2}+s_{2m}$. So $s_n$ depends on $s_{n-1}$, but the dependence is dependant on even and odd n.
The strategy given is to write down the 10 or so terms to figure out a conjecture about $s_n$, then use induction to prove said conjecture and then solve the problem. I really don't know where to start on this problem. Here's how I wrote out the sequence:
$s_1 = 0, s_2 = 0, s_{2(1)+1} = \frac{1}{2}, s_4=1, s_5= \frac{3}{4}, s_6= \frac{3}{8}, s_7 = \frac{7}{8}, s_8= \frac{7}{16}, s_9 = \frac{15}{16}, s_{10}= \frac{15}{32} $
I don't know what I'm supposed to make of this. I have a feeling the sequence is 'bouncing' between $\frac{1}{2}$ and 1 so I'm guessing those would be my lip sup and lim inf, but how do I prove this?
| You just have to note the following pattern in your sequence.
$$
s_n = \begin{cases}
1- \dfrac 1{2^{\frac{n-1}{2} }} & n \neq 1 \text{ odd} \\
\dfrac 12 - \dfrac 1{2^{\frac{n}{2} }} & n \text{ even}
\end{cases}
$$
To see this, the base cases are easy to see: Put $n = 3$ and $s_3 = \frac{1}{2} = 1 - \frac 1{2^1}$, and $n=4$ gives $s_4 =\frac 14$ which is also $\frac{1}{2} - \frac{1}{2^2}$.
Induction: Assume that the above is true for even numbers up till $2k$, then we have to prove it is true for $2(k+1) = 2k+2$. Note that
\begin{split}
s_{2k+2} & = \frac{s_{2k+1}}2 \\
& = \frac{\frac{1}{2} + s_{2k}}{2}\\
& = \frac{\frac{1}{2} + \frac{1}{2}- \frac{1}{2^k}}{2} \\ & = \frac{1 - \frac{1}{2^k}}{2} \\ & = \frac{1}{2} - \frac{1}{2^{k+1}}
\end{split}
Which was to be proved.
Now, assume it is true for odd numbers up till $2k-1$, then want to show it is true for $2k+1$.Note that:
\begin{split}
s_{2k+1} & = \frac{1}{2} + s_{2k} \\
& = \frac{1}{2} + \frac{s_{2k-1}}{2}\\
& = \frac{1 + \left(1 - \frac 1{2^{k-1}}\right)}{2} \\ & = \frac{2 - \frac 1{2^{k-1}}}{2}\\ & = 1 - \frac{1}{2^k} \\ & = 1-\frac{1}{2^\frac{(2k+1)-1}{2}}
\end{split}
Which was to be proved. Hence the induction is complete, and the result stands.
Now, note that $s_n-s_m$ is of the following forms:
If $n$, $m$ are both odd, then it is $-\frac 1{2^{\frac{n-1}{2} }} + \frac 1{2^{\frac{m-1}{2} }} $, which converges to zero as $n,m$ increase.
If $n$, $m$ are both even, then it is $-\frac 1{2^{\frac{n}{2} }} + \frac 1{2^{\frac{m}{2} }} $, which converges to zero as $n,m$ increase.
However, if $n$ is odd, $m$ is even, then it is $\frac 12-\frac 1{2^{\frac{n-1}{2} }} + \frac 1{2^{\frac{m}{2} }} $, which converges to half as $n,m$ increase.
Similarly, if $n$ is even, $m$ is odd, then it is $-\frac 12-\frac 1{2^{\frac{n}{2} }} + \frac 1{2^{\frac{m-1}{2} }} $, which converges to negative half as $n,m$ increase.
Now, any convergent sequence must be Cauchy. Hence, this difference must go to zero as $n,m$ increase. However, in the third and fourth case, we can clearly see that the difference doesn't go to zero.
Hence, in every convergent subsequence of $s_n$ ,$n$ must eventually have the same sign.
Now, one sign leads to $\frac 12$, the other sign leads to $1$. Hence, these are the only two limits of the sequences $s_n$. Now it is clear that $\liminf s_n = \frac 12$, $\limsup s_n = 1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1997012",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
find the volume of the region bounded by the paraboloid $z=x^2+y^2$ and the cylinder $x^2+y^2=9$ My attempt:
$\int_{x=-3}^3 \int_{y=-3}^3 x^2+y^2 dy dx$
$\int_{-3}^3 6x^2 + 18 dx$
$= 216?$
| The limits of your solution actually define the squared cylinder $[-3,3]\times[-3,3]$. You could either try by changing the limits to $-\sqrt{9 - x^2} \le y \le \sqrt{9 - x^2} $ and $-3 \le x \le 3$ or by changing your coordinate system to $x = r\cos\theta$, $y = r sin\theta$
$$
\int\int_{x^2 + y^2 < 9} dxdy\; (x^2+y^2) = \int\int_{r<3} dr r\; (r^2) = 2\pi \int_0^3dr\; r^3 = \frac{81}{2}\pi
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1998765",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Squared magnitude of two complex numbers I believe one easy questions for maths experts. If I have two complex numbers x=a+ib and y=c+di is the squared magnitude of their sum equal to:
\begin{equation}
|x+y|^2=|x|^2+2|xy|+|y|^2
\end{equation}
| Your LHS equals
$$
(a+c)^2+(b+d)^2=(a^2+b^2)+2(ac+bd)+(c^2+d^2)=|x|^2+2(ac+bd)+|y|^2
$$
so it's almost the RHS, except for the fact that
\begin{aligned}
2|xy|&=2|(ab-cd)+i(ac+bd)|=2\sqrt{(ab-cd)^2+(ac+bd)^2}\\
&=2\sqrt{a^2b^2+c^2d^2+a^2c^2+b^2d^2}\\
&\neq 2(ac+bd).
\end{aligned}
In fact,
$$
(ac+bd)^2=a^2c^2+b^2d^2+2(ab)\cdot (cd)\leq a^2b^2+c^2d^2+a^2c^2+b^2d^2.
$$
So in fact you have LHS $\leq$ RHS, not LHS $=$ RHS.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1999380",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Show that $5\cdot10^n+10^{n-1}+3$ is divisible by 9 Prove by induction the following:
$5*10^n+10^{n-1}+3$ is divisible by 9
Base case: $n=1$ $5*10+10^{1-1}+3=5*10+10^0+3=50+1+3=54$
$9|54=6$
Inductive Hypothesis: If $k$ is a natural number such that $9|5*10^k+10^{k-1} +3$
Inductive step:
Show that $S_k$ is true $\Rightarrow$ $S_{k+1}$ is true
$S_{k+1}$: $9|5*10^{k+1}+10^{k} +3$
$9|10(5*10^{k+1}+10^{k} +3)$
$9|5*10^{k+2}+10^{k+1} +10*3$
$9|5*(10^{k+1}*10^1)+(10^{k}*10^1) +(9+1)*3$
$9|5*(10^{k+1}*(9+1))+(10^{k}*10*(9+1)) +(9+1)*3$
$9|5(9*10^{k+1}+10^{k+1})+9*10^k+10^k+((9*3)+(1*3))$
This is were I am stuck for the last day try to figure out what move next would speed up the inductive proof as I have a feeling it can be finished up. Anyone help me see what I am unable to find.
| An integer is divisible by $3^k$ if and only if the sum of the digits is divisible by $3^k$. In this case, note the sum of the digits is always $9$!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2000123",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 4
} |
Fibonacci Identity with Binomial Coefficients A friend showed me this cool trick: Take any row of Pascal's triangle (say, $n = 7$):
$$1, 7, 21, 35, 35, 21, 7, 1$$
Leave out every other number, starting with the first one:
$$7, 35, 21, 1$$
Then these are backwards base-5 "digits", so calculate:
$$7 + 35 \cdot 5 + 21 \cdot 5^2 + 1 \cdot 5^3 = 7 + 175 + 525 + 125 = 832$$
and divide by $2^{7-1} = 2^6 = 64$:
$$\frac{832}{64} = 13$$
and $F_7 = 13$ (the seventh Fibonacci number)!
He said it works for any $n$. I have worked out that this would be to prove that:
$$\frac{1}{2^{n-1}}\sum_{k = 0}^{\lfloor{\frac{n}{2}}\rfloor}{\left(5^k {n \choose 2k + 1} \right)} = F_n $$
I'm not sure how to proceed from here. Is there a neat combinatoric or easy algebraic proof I am missing? Thanks!
| Suppose we seek to show that
$$\frac{1}{2^{n-1}} \sum_{k=0}^{\lfloor n/2\rfloor}
5^k {n\choose 2k+1} = F_n$$
a Fibonacci number. Introduce
$${n\choose 2k+1} = {n\choose n-2k-1} =
\frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^{n-2k}} (1+z)^n \; dz.$$
Observe that the largest index $k$ producing a non-zero value is
$\lfloor n/2\rfloor-1$ when $n$ is even and $\lfloor n/2\rfloor$ when
$n$ is odd. The integral correctly represents this
behavior. Extending the range of $k$ to infinity we thus obtain for
the sum
$$\frac{1}{2^{n-1}} \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^{n}}
(1+z)^n \sum_{k\ge 0} 5^k z^{2k} \; dz
\\ = \frac{1}{2^{n-1}} \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^{n}} (1+z)^n
\frac{1}{1-5z^2} \; dz.$$
Now put $z/(1+z) = w$ so that $z = w/(1-w)$ and $dz = 1/(1-w)^2 dw$
to get
$$\frac{1}{2^{n-1}} \frac{1}{2\pi i}
\int_{|w|=\gamma} \frac{1}{w^n}
\frac{1}{1-5w^2/(1-w)^2} \frac{1}{(1-w)^2} \; dw
\\ = \frac{1}{2^{n-1}} \frac{1}{2\pi i}
\int_{|w|=\gamma} \frac{1}{w^n}
\frac{1}{1-2w-4w^2} \; dw.$$
Extracting coefficients we find
$$\frac{1}{2^{n-1}} [w^{n-1}] \frac{1}{1-2w-4w^2}
= [w^{n-1}] \frac{1}{1-2(w/2)-4(w/2)^2}
\\ = [w^{n-1}] \frac{1}{1-w-w^2}
\\ = [w^{n}] \frac{w}{1-w-w^2}.$$
This is precisely the generating function of the Fibonacci numbers and
we are done.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2002702",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "16",
"answer_count": 4,
"answer_id": 1
} |
What is $(-1)^{\frac{1}{7}} + (-1)^{\frac{3}{7}} + (-1)^{\frac{5}{7}} + (-1)^{\frac{9}{7}} + (-1)^{\frac{11}{7}}+ (-1)^{\frac{13}{7}}$? The question is as given in the title. According to WolframAlpha, the answer is 1, but I am curious as to how one gets that. I tried simplifying the above into $$6(-1)^{\frac{1}{7}}$$ however that does not give 1.
| Notice:
$$-1=\left|-1\right|e^{\left(\arg\left(-1\right)+2\pi\text{k}\right)i}=e^{\pi\left(1+2\text{k}\right)i}$$
Where $\text{k}\in\mathbb{Z}$
So, when $\text{n}\in\mathbb{R}$:
$$\left(-1\right)^\text{n}=e^{\text{n}\pi\left(1+2\text{k}\right)i}=\cos\left(\text{n}\pi\left(1+2\text{k}\right)\right)+\sin\left(\text{n}\pi\left(1+2\text{k}\right)\right)i=\cos\left(\text{n}\pi\right)+\sin\left(\text{n}\pi\right)i$$
So (they are called the principal roots):
*
*When $\text{n}=\frac{1}{7}$:
$$\left(-1\right)^{\frac{1}{7}}=\cos\left(\frac{\pi}{7}\right)+\sin\left(\frac{\pi}{7}\right)i$$
*When $\text{n}=\frac{3}{7}$:
$$\left(-1\right)^{\frac{3}{7}}=\cos\left(\frac{3\pi}{7}\right)+\sin\left(\frac{3\pi}{7}\right)i$$
*When $\text{n}=\frac{5}{7}$:
$$\left(-1\right)^{\frac{5}{7}}=\cos\left(\frac{5\pi}{7}\right)+\sin\left(\frac{5\pi}{7}\right)i$$
*When $\text{n}=\frac{9}{7}$:
$$\left(-1\right)^{\frac{9}{7}}=\cos\left(\frac{9\pi}{7}\right)+\sin\left(\frac{9\pi}{7}\right)i$$
*When $\text{n}=\frac{11}{7}$:
$$\left(-1\right)^{\frac{11}{7}}=\cos\left(\frac{11\pi}{7}\right)+\sin\left(\frac{11\pi}{7}\right)i$$
*When $\text{n}=\frac{13}{7}$:
$$\left(-1\right)^{\frac{13}{7}}=\cos\left(\frac{13\pi}{7}\right)+\sin\left(\frac{13\pi}{7}\right)i$$
Adding those together, gives us:
$$2\left\{\cos\left(\frac{\pi}{7}\right)+\sin\left(\frac{\pi}{14}\right)-\sin\left(\frac{3\pi}{14}\right)\right\}=1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2003317",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
Integration of Erf function Let $f(x) = \int^\sqrt{x}_1 e^{-t^2}dt$. Find $\int^1_0 \frac{f(x)}{
\sqrt{x}}dx$.
Could anyone give me any hint how to start? The Erf function $f(x)$ seems not to be easily integrated.
| Brevan Ellefsen's solution is the most straightforward and efficient. Here is an alternative solution.
\begin{equation}
f(x) = \int\limits_{1}^{\sqrt{x}} \mathrm{e}^{-t^{2}} dt
\label{eq:161108-1}
\tag{1}
\end{equation}
\begin{equation}
\int\limits_{0}^{1} \frac{f(x)}{\sqrt{x}} dx
\label{eq:161108-2}
\tag{2}
\end{equation}
We need the following result
\begin{equation}
\int \mathrm{erf}(x) dx = x\,\mathrm{erf}(x) + \frac{\mathrm{e}^{-x^{2}}}{\sqrt{\pi}}
\label{eq:161108-3}
\tag{3}
\end{equation}
Proof: Integrate by parts
\begin{align}
\int \mathrm{erf}(x) dx &= x\,\mathrm{erf}(x) -\frac{2}{\sqrt{\pi}} \int x\,\mathrm{e}^{-x^{2}} dx \\
&= x\,\mathrm{erf}(x) + \frac{\mathrm{e}^{-x^{2}}}{\sqrt{\pi}}
\end{align}
we used the substitution $u=x^{2}$.
Now we evaluate equation \eqref{eq:161108-1}
\begin{equation}
f(x) = \frac{\sqrt{\pi}}{2} \mathrm{erf}(x) \Big|_{1}^{\sqrt{x}} = \frac{\sqrt{\pi}}{2} [\mathrm{erf}(\sqrt{x}) - \mathrm{erf}(1)]
\label{eq:161108-4}
\tag{4}
\end{equation}
Substitute equation \eqref{eq:161108-4} into equation \eqref{eq:161108-2} and evaluate the following two integrals.
\begin{align}
I_{1} &= \int\limits_{0}^{1} \frac{\mathrm{erf}(\sqrt{x})}{\sqrt{x}} dx \\
&= 2\int\limits_{0}^{1} \mathrm{erf}(z) dz \\
&= 2\,\mathrm{erf}(1) + \frac{2}{\mathrm{e}\sqrt{\pi}} - \frac{2}{\sqrt{\pi}}
\end{align}
we used the substitution $z=\sqrt{x}$.
\begin{equation}
I_{2} = \int\limits_{0}^{1} \frac{1}{\sqrt{x}} dx = 2
\end{equation}
Putting all of the pieces together yields our final result
\begin{align}
\int\limits_{0}^{1} \frac{f(x)}{\sqrt{x}} dx
&= \frac{\sqrt{\pi}}{2} \left(2\,\mathrm{erf}(1) + \frac{2}{\mathrm{e}\sqrt{\pi}} - \frac{2}{\sqrt{\pi}} - 2\,\mathrm{erf}(1) \right) \\
&= \frac{1}{\mathrm{e}} - 1
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2004424",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Finding Polynomial Equations for $1^4 + 2^4 + 3^4 + \ldots n^4$ Find a polynomial expression for
$$ 1^4 + 2^4 + 3^4 + ... + n^4 $$
I know you have to use the big theorem but I can't figure out how you would start to compute the differences. Suggestions?
| Note that
\begin{aligned}
\sum_{i=1}^{n}i^5 &= \sum_{i=0}^{n-1}(i+1)^5 \\
& = \sum_{i=0}^{n-1}\left(i^5 + 5i^4 + 10i^3 + 10i^2+5i+1 \right) \\
&= \sum_{i=0}^{n-1}i^5 + \sum_{i=0}^{n-1}\left(5i^4 + 10i^3 + 10i^2+5i+1 \right) \\
&= -n^5 + \sum_{\bbox[yellow]{i=1}}^{\bbox[yellow]{n}}i^5 + \sum_{i=0}^{n-1}\left(5i^4 + 10i^3 + 10i^2+5i+1 \right) \\
\end{aligned}
After canceling out the expression $\sum i^5$ from the sides, we'll get
$$n^5=\sum_{i=0}^{n-1}\left(5i^4 + 10i^3 + 10i^2+5i+1 \right)$$
Which can be rewritten as
\begin{aligned}
5\sum_{i=0}^{n-1}i^4 &= n^5 - \sum_{i=0}^{n-1}\left(10i^3 + 10i^2+5i+1 \right)\\
\sum_{i=0}^{n-1}i^4 &=\frac15\left(n^5- 10\sum_{i=0}^{n-1}i^3 -10\sum_{i=0}^{n-1}i^2 -5\sum_{i=0}^{n-1}i -n \right)
\end{aligned}
By defining $S_k=\sum_{i=0}^{n-1}{i^k}$, this is equivalent to
$$S_4 =\frac15\left(n^5- 10S_3 -10S_2 -5S_1 -n \right) $$
or in the form below to resemble the similarity with the binomial expansion ($S_0=n$)
$$ 5S_4 +10S_3+10S_2+5S_1 + S_0 =n^5 $$
So, to find $S_4$, you just need to replace the expressions for $S_3$, $S_2$ and $S_1$, or repeat the above process to manually derive them.
| {
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Sum of the digits of $N=5^{2012}$ The sum of the digits of $N=5^{2012}$ is computed.
The sum of the digits of the resulting sum is then computed.
The process of computing the sum is repeated until a single digit number is obtained.
What is this single digit number?
| You want to know the value of $5^{2012} \pmod 9$.
Since $\varphi(9) = 3^2 - 3 = 6$ and $\gcd(5,9) = 1$, then, by Euler's theorem, $5^6 \equiv 1 \pmod 9$.
Since $2012 = 335 \times 6 + 2$,
$$5^{2012} \equiv (5^6)^{335} \times 5^2
\equiv 1^{335} \times 25 \equiv 7 \pmod 9.$$
| {
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How to show this equality? This is a equation from a book of control theory:
$$c(sI-A)^{-1}b=\frac{\det(sI-A)-\det(sI-A-bc)}{\det(sI-A)}$$
$I$ is identity matrix, $A$ is $n\times n$, $b$ is $n \times 1$ vector, $c$ is $1 \times n$ vector.
I was trying to use induction, but seems not work. I would very appreciate some advice.
| Note that $bc$ is a rank-1-matrix. We will find a formula for $(B - bc)^{-1}$, where $B$ is any $n\times n$-matrix ($sI - A$ in your case). We will prove
$$ \det (B - bc) = (1 - cB^{-1}b)\det(B) $$
To prove that, note
\begin{align*}
\det(B- bc) &= \det B\det(I - B^{-1}bc)\\
&= \det B \det\begin{pmatrix} I- B^{-1}bc & -B^{-1}b\\ 0 & 1 \end{pmatrix}\\
&= \det B \det \begin{pmatrix} I & 0\\ c & 1\end{pmatrix}
\det\begin{pmatrix} I- B^{-1}bc & -B^{-1}b\\ 0 & 1 \end{pmatrix}\begin{pmatrix} I & 0\\ -c & 1\end{pmatrix}\\
&= \det B \cdot \det\begin{pmatrix} I & -B^{-1}b\\ 0 & 1 - cB^{-1}b\end{pmatrix}\\
&= \det B \cdot (1 - cB^{-1}b)
\end{align*}
Hence
$$ -\det(sI - A - bc) = -\det(sI - A) \cdot \bigl(1 - c(sI-A)^{-1}b\bigr) $$
Now solve for $c(sI- A)^{-1}b$.
| {
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Does the following series converges or diverges? $\sum_{n=1}^{+\infty}\left[n \log\left ( \frac{2n+1}{2n-1} \right )-1\right]$ ,converges or not
My attempt
$a_n=n \log\left ( \frac{2n+1}{2n-1} \right )-1 \\ \\
\log \left ( \frac{2n+1}{2n-1} \right )=\log \left ( 1+\frac{2}{2n-1} \right ) \sim \frac{2}{2n-1} \ (n\rightarrow +\infty) \\ \\
a_n=n\log \left ( 1+\frac{2}{2n-1} \right )-1 \sim n \frac{2}{2n-1}-1=\frac{2n}{2n-1}-1=\frac{1}{2n-1} \\ \\ $
$\sum_{n=1}^{+\infty}\frac{1}{2n-1} $ diverges ,then $\sum_{n=1}^{+\infty}a_n $ diverges
but according to Wolfram $\sum_{n=1}^{+\infty}a_n=\frac{1}{2}-\frac{\log 2}{2}$
where is the mistake in my solution please?
thanks
| Your mistake was assuming this: If $x_n \sim y_n,$ then $x_n - 1 \sim y_n -1.$ A counterexample is $x_n =1+ 1/n, y_n = 1+1/n^2.$
| {
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Find a limit without l'Hospital: $\lim_{x\to -\infty}\left(\frac{\ln(2^x + 1)}{\ln(3^x+ 1)}\right)$ Find $\displaystyle \lim_{x\to -\infty}\left(\frac{\ln(2^x + 1)}{\ln(3^x+ 1)}\right)$.
$\displaystyle \lim_{x\to -\infty}\left(\frac{\ln(2^x + 1)}{\ln(3^x+ 1)}\right)$ =
$\displaystyle \lim_{x\to -\infty}\left(\frac{\ln2^x + \ln(1+\frac{1}{2^x})}{\ln3^x + \ln(1+\frac{1}{3^x})}\right)$ =
$\displaystyle \lim_{x\to -\infty}\left(\frac{\ln2^x + \frac{1}{2^x}\ln(1+\frac{1}{2^x})^{2^{x}}}{\ln3^x + \frac{1}{3^x}\ln(1+\frac{1}{3^x})^{3^{x}}}\right)$ = $\cdots$
This method is not working.
| We have
$$
\ln(1+t)=t+o(t^2).
$$
Thus, since $2^x\to0$ and $3^x\to0$ when $x\to-\infty$,
$$
\frac{\ln(1+2^x)}{\ln(1+3^x)}=\frac{2^x+o(2^{2x})}{3^x+o(3^{2x})}
=\left(\frac23\right)^x\,\frac{1+o(2^x)}{1+o(3^x)}.
$$
The second fraction goes to $1$ when $x\to-\infty$. So the limit will be equal to
$$
\lim_{x\to-\infty}\left(\frac23\right)^x=\lim_{x\to-\infty}\exp\left(x\ln\left(\frac23\right)\,\right)=+\infty.
$$
| {
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Showing $\frac{x+y+z}{\sqrt 2}\le\sqrt{x^2+y^2}+\sqrt{y^2+z^2}+\sqrt{x^2+z^2}$
Showing $\frac{x+y+z}{\sqrt 2}\le\sqrt{x^2+y^2}+\sqrt{y^2+z^2}+\sqrt{x^2+z^2}$ $(x,y,z>0)$
Can I apply Jensen here, sum of the squareroots is greater then squareroot of the sum, so I make the RHS smaller then show also then the inequality is valid, i.e.
$\frac{x+y+z}{\sqrt 2}\le\sqrt{2(x^2+y^2+z^2)}$
squaring both sides we have on the left,
$\left(\frac{x+y+z}{\sqrt 2}\right)^2=\frac{x^2+y^2+z^2+2(xy+yz+xz)}{2}\le\frac{3(x^2+y^2+z^2)}{2}$
and on the right we have $2(x^2+y^2+z^2)$, so it is true
If it is OK, can you suggest another proof related more to Cauchy-Schwarz
| Recall $\|(x,y,z)\| = \sqrt{x^2+y^2+z^2}$.
By C-S
$$x+y+ z= (x,y,z)\cdot (1,1,1) \le \|(x,y,z)\|\|(1,1,1)\|=\|(x,y,z)\|\sqrt{3}.$$
But by triangle inequality (follows from C-S):
$$ \|(x,y,z)\| = \frac 12 \|(x,y,0) +(0,y,z)+ (x,0,z)\|\le \frac 12 \left(\|(x,y,0)\|+\|(0,y,z)\|+\|(x,0,z)\|\right).$$
Summarizing:
$$(*)\quad x+y+z \le \frac{\sqrt{3}}{2}\left ( \|(x,y,0)\|+\|(0,y,z)\|+\|(x,0,z)\|\right).$$
But $\frac{\sqrt{3}}{2}<1<\sqrt{2}$, and we're all set.
| {
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A proof of fraction inequality by induction Prove that $$\left(\frac{a+b}{2}\right)^n\le\frac{a^{n}+b^{n}}{2}$$ I have checked the base case and assumed the inequality is valid and continued like this $$\left(\frac{a+b}{2}\right)^{n+1}\le\frac{a^{n+1}+b^{n+1}}{2}\Rightarrow \left(\frac{a+b}{2}\right)^n\left(\frac{a+b}{2}\right)\le\frac{a^{n+1}+b^{n+1}}{2}$$ Multiplying by $2$ I got $$\frac{(a+b)^{n+1}}{2^n}\le a^{n+1}+b^{n+1}$$ and I don't know how to continue from here. I've also tried using the binomial theorem to get something different but couldn't get anything useful.
| Work:
$\left( \dfrac{a+b}{2}\right)^{n+1}=\left(\dfrac{a+b}{2}\right)^n\left(\dfrac{a+b}{2}\right)$
$ \hspace{2.6 cm}\leq \dfrac{a^n+b^n}{2}\cdot \left(\dfrac{a+b}{2}\right)$
$\hspace{2.6 cm} = \dfrac{a^{n+1}+a^nb+ab^n+b^{n+1}}{4}$
Now write what you want to obtain and subtract extras:
$\hspace{2.6 cm} = \dfrac{a^{n+1}+b^{n+1}}{2}-\dfrac{a^{n+1}+b^{n+1}}{4}+\dfrac{a^nb+ab^n}{4}$
$\hspace{2.6cm} = \dfrac{a^{n+1}+b^{n+1}}{2}-\left(\dfrac{a^{n+1}-a^nb-ab^n-b^{n+1}}{4}\right)$
$\hspace{2.6cm} = \dfrac{a^{n+1}+b^{n+1}}{2}-\left(\dfrac{a^{n}(a-b)-b^n(a-b)}{4}\right)$
$\hspace{2.6cm} = \dfrac{a^{n+1}+b^{n+1}}{2}-\dfrac{(a-b)(a^n-b^n)}{4}=*$
No matter if $a\geq b$ or $b> a$ expression $ \dfrac{(a-b)(a^n-b^n)}{4}$ is positive. So, in conclusion:
$\hspace{2.6cm}*\leq \dfrac{a^{n+1}+b^{n+1}}{2}$
| {
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Let $k$ be an odd positive integer. Prove that $(1+2+3...+n)|(1^k+2^k+3^k+...+n^k)$ for all positive integers $n$. Problem:Let $k$ be an odd positive integer. Prove that $(1+2+3...+n)|(1^k+2^k+3^k+...+n^k)$ for all positive integers $n$.
My Attempt: Proof (By Induction on $k$): Let $P(k)$ be the proposition that $(1+2+3...+n)|(1^k+2^k+3^k+...+n^k)$ for all positive integers $n$ when $k$ is odd. Then $P(1)$ and $P(3)$ are trivially true. So for $k\geq 5$, we assume that $P(k)$ holds. We would like to show that $P(k+2)$ also holds. In order to do that consider the following sums
$$1^{2}(1^k+2^k+3^k+...+n^k)=1^{k+2}+2^k+3^k+...+n^k$$
$$2^{2}(1^k+2^k+3^k+...+n^k)=2^2*1^k+2^{k+2}+2^2*3^k+...+2^2*n^k$$
$$..$$
$$n^2(1^k+2^k+3^k+...+n^k)=n^2*1^k+n^2*2^k+n^2*3^k+...+n^{k+2}$$
Note that $$1^2(1^k+2^k+3^k+...+n^k)+2^2(1^k+2^k+3^k+...+n^k)+...+n^2(1^k+2^k+3^k+...+n^k)\equiv 0\pmod {\frac{n(n+1)}{2}}$$
Therefore, if we denote $\sum_{i=1}^{n}i^2=s$, then
$(t)(1^k+2^k+3^k+...+n^k)-t+\sum_{i=1}^{n}i^{k+2}\equiv 0\pmod {\frac{n(n+1)}{2}}$ which implies that $P(k+2)$ holds. Hence by PMI we are done.
Now I am a bit hesitant in convincing myself whether this proof is correct or not. This is because I have never seen induction performed in this manner. Any reasons as to why this argument falls would be much appreciated.
| Your identity is wrong so your solution is wrong.
Here is one way to attempt the problem. We know $1+2+ \cdots +n=\dfrac{n(n+1)}{2}$ and $\gcd (n,n+1)=1$. We consider following two cases:
Case 1. If $n$ is even then there are $n/2$ pairs of $i^k,(n-i+1)^k$ and see that $n+1 \mid i^k+(n-i+1)^k$ since $k$ is odd. Thus, $n+1 \mid 1^k+ \cdots + n^k$.
Next, we group $i^k,(n-i)^k$ then there are $(n-2)/2$ pairs and there is two numbers $n^k,(n/2)^k$ left. Note that $n \mid i^k+(n-i)^k$ since $k$ is odd so $$n/2 \mid \sum_{i=1}^{(n-2)/2} \left( i^k+(n-i)^k \right)+n^k+(n/2)^k.$$
Since $\gcd \left( n/2,n+1 \right)=1$ so $\dfrac{n(n+1)}{2} \mid 1^k+ \cdots + n^k$ or $1+2+ \cdots +n \mid 1^k+2^k+ \cdots + n^k$.
We do the similar thing to the case $n$ is odd.
| {
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Solve the equation $x^2+\frac{9x^2}{(x+3)^2}=27$
Problem Statement:-
Solve the equation $$x^2+\dfrac{9x^2}{(x+3)^2}=27$$
I have tried to turn it into a quadratic equation so as to be saved from solving a quartic equation, but have not been able to come up with anything of much value.
These are the things that I have tried to turn the given equation into a quadratic equation.
$$x^2+\dfrac{9x^2}{(x+3)^2}=27\implies 1+\dfrac{1}{\left(\dfrac{x}{3}+1\right)^2}=3\left(\dfrac{3}{x}\right)^2$$ $$\text{OR}$$
$$x^2+\dfrac{9x^2}{(x+3)^2}=27\implies 1+\dfrac{\left(\dfrac{3}{x}\right)^2}{\left(1+\dfrac{3}{x}\right)^2}=3\left(\dfrac{3}{x}\right)^2$$
| HINT:
Using $a^2+b^2=(a-b)^2+2ab,$
$$x^2+\left(\dfrac{3x}{x+3}\right)^2=\left(x-\dfrac{3x}{x+3}\right)^2+2\cdot x\cdot\dfrac{3x}{x+3}=\left(\dfrac{x^2}{x+3}\right)^2+6\cdot\dfrac{x^2}{x+3}$$
Generalization :
For $a^2+b^2=k$
If $\dfrac{ab}{a+b}=c$ where $c$ is a non-zero finite constant,
$$\implies k=(a+b)^2-2ab=(a+b)^2-2(a+b)c$$ $$\iff(a+b)^2-2(a+b)c-k=0$$ which isa Quadratic equation in $a+b$
Can you recognize $a,b$ here?
If $\dfrac{ab}{a-b}=c$ use $a^2+b^2=(a-b)^2+2ab$
| {
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How can I derive what is $1\cdot 2\cdot 3\cdot 4 + 2\cdot 3\cdot 4\cdot 5+ 3\cdot 4\cdot 5\cdot 6+\cdots + (n-3)(n-2)(n-1)(n)$ ?? I'd like to evaluate the series $$1\cdot 2\cdot 3\cdot 4 + 2\cdot 3\cdot 4\cdot 5+ 3\cdot 4\cdot 5\cdot 6+\cdots + (n-3)(n-2)(n-1)(n)$$
Since I am a high school student, I only know how to prove such formula's (By principal of mathematical induction). I don't know how to find result of such series.
Please help. I shall be thankful if you guys can provide me general solution (Since I have been told that there exist a general solution by my friend who gave me this question).
| Let us conjugate freak_warrior's tip with a telescopic sum to find the closed formula.
Let us set $a_n = n(n+1)(n+2)(n+3) $
You want to find
$$\sum_{i = 1}^{k} a_i $$
Rewrite
$$a_n = \frac{n(n+1)(n+2)(n+3)(n+4)}{5} - \frac{(n-1)n(n+1)(n+2)(n+3)}{5} $$
One can prove this is true by factoring out $n(n+1)(n+2)(n+3)$.
Set $l_n = \frac{n(n+1)(n+2)(n+3)(n+4)}{5}$ and $r_n = \frac{(n-1)n(n+1)(n+2)(n+3)}{5}$. Now we have
$$a_n = l_n - r_n $$
But as you can see, $r_{n+1} = l_n $ thus
$$\sum_{i = 1}^{k} a_i = \sum_{i = 1}^{k} l_i - r_i = \sum_{i = 1}^{k} r_{i+1} - r_i$$
If you apply the telescopic sum, you get $r_{k+1} - r_1$
The telescopic sum is noticing that two adjacent terms always cancel:
$$\sum_{i = 1}^{k} r_{i+1} - r_i = (r_2 - r_1) + (r_3 - r_2) + \cdots + (r_k - r_{k-1}) + (r_{k+1} - r_k) = r_{k+1} - r_1 $$
Now all you have to do is substitute the desired value of $k $.
| {
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Find all primes $p$ such that $p^3-4p+9=x^2$, where $x$ is a positive integer. Problem:Find all primes $p$ such that $p^3-4p+9=x^2$, where $x$ is a positive integer.
My Attempt: I have made the following observations while trying tot solve this problem:
*
*$p=2$ yields $x=3$.
*For $p>2$ the following is true: $x^2\equiv 0\pmod 4$ and $p\equiv 3\pmod 4.$
*It seems that $p=2,7$ and $11$ are the only primme numbers that satisfy the conditions of this problem. Below are the values of $p^3-4p+9$ evaluated for $1\leq p\leq 100.$
9
24
114
324
1296
2154
4854
6792
12084
24282
29676
50514
68766
79344
103644
148674
205152
226746
300504
357636
388734
492732
571464
704622
912294
I would like to know how one can prove that $p=2,7$ and $11$ are the only solutions to the above-mentioned equation.
| If $p^3-4p+9=n^2$ then $n^2\equiv 9 \pmod{p}$ and $\pm n \equiv 3 \pmod{p}$ since $n^2-9\equiv 0$ can only have two roots modulo a prime.
We then have
$$
\begin{array}{ll}
n^2=(-n)^2 &= (kp+3)^2 \\
p^3-4p+9 &= k^2p^2+6kp+9 \\
0 &= p^2-k^2p-(6k+4) \\
p & = \frac{k^2\pm \sqrt{k^4+24k+16}}{2}
\end{array}
$$where we used $p\ne 0$ since it's a prime. This can only have an integer solution for $p$ when $k^4+24k+16$ is a square.
But
$$
\begin{array}{ll}
(k^2-1)^2 = k^4-2k^2+1 < k^4+24k+16<(k^2)^2 & \text{if }k<-11 \\
(k^2)^2 < k^4+24k+16< k^4+2k^2+1 =(k^2+1)^2 & \text{if }k>12
\end{array}
$$so if $k<-11$ or $k>12$ then $k^4+24k+16$ lies between two squares and hence is not a square. This leaves 24 values to check, which you could do by hand. It leads to solutions when $k=-3,0,3$ which give integral values for $p$ of $-2,2,7,11$, of which only $2,7,11$ are prime.
| {
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$\lim _\limits{n\to \infty }\left(4n^2\left(\frac{n^3+3n^2+3}{2n^3+n-1}\right)^n\right)$ How can I calculate this limit?
$\lim_\limits{n\to \infty }\left(4n^2\left(\frac{n^3+3n^2+3}{2n^3+n-1}\right)^n\right)\:$
Can I apply the rule $a_n+1/a_n$ to proove its convergence?
| Hint. Note that
$$a_n:=4n^2\left(\frac{n^3+3n^2+3}{2n^3+n-1}\right)^n=4n^2\left(\frac{1+3/n+3/n^3}{2+1/n^2-1/n^3}\right)^n.$$
Can we say that $a_n\sim\frac{4n^2}{2^n}$?
| {
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Show that $1^3 + 2^3 + ... + n^3 = (n(n+1)/2)^2$ by induction I have some with proving by induction. I cannot find a solution for the inductive step: $1^3 + 2^3 + ... + n^3 = (n(n+1)/2)^2$
I already did the induction steps:
Basis: P(1) = $1^3 = (1(1+1)/2)^2$ (This is true)
Inductive step: Assume $P(k) = ((k)(k+1)/2)^2$
To be proven: $((k)(k+1)/2)^2 + (k+1)^3 = ((k+1)(k+2)/2)^2$
My problem is that I do not know how I can put the $ + (k+1)^3$ inside $((k)(k+1)/2)^2$.
Simplifying the left and right part of the statement does not help:
Simplifying the left side: $((k)(k+1)/2)^2 + (k+1)^3 = ((k^2+k)/2)^2 + (k+1)^3 $
Simplifying the right side: $((k+1)(k+2)/2)^2 = ((k^2+3k+2)/2)^2$
So i am left with: $((k^2+k)/2)^2 + (k+1)^3 = ((k^2+3k+2)/2)^2$
That is the same as: $1/4 (k^2+k)^2 + (k+1)^3 = 1/4((k^2+3k+2))^2$
Going further with the left side:$1/4 (k^2+k)^2 + (k+1)^3 = (1/4)(k^4 + 2k^3 + k^2) + k^3+3 k^2+3 k+1$
Going further with the right side: $1/4((k^2+3k+2))^2 = 1/4 (k^4+6 k^3+13 k^2+12 k+4)$
Now I am stuck with: $(1/4)(k^4 + 2k^3 + k^2) + k^3+3 k^2+3 k+1 = 1/4 (k^4+6 k^3+13 k^2+12 k+4)$
Now I am kind of left with garbage. Am I missing something? What do I do wrong? Where can I find a good resources to learn how to solve this issue?
| Hint
It's easier to begin by the end
$$\left(\frac{(n+1)(n+2)}{2}\right)^2-\frac{n(n+1)}{2}^2=$$
$$\left(\frac{n+1}{2}\right)^2\left((n+2)^2-n^2\right)=$$
$$\left(\frac{n+1}{2}\right)^2(4(n+1))=$$
$$(n+1)^3.$$
qed.
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "1",
"answer_count": 6,
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Solutions of the complex equation: $(z+1)^3 = 36(z^*+1)$ today in an exam a friend had this equation with complex number:
$$(z+1)^3 = 36(z^*+1)$$
where with z* I mean conjugate z; can someone solve this?
We tried and there is probably a way to solve it doing a big system with everyterm with i = 0 and everyterm without = 0; but we think that another solution must exists; thanks a lot!
| Set $z + 1 = w$, then $\overline{z} + 1 = \overline{w} + 1$ as well, and you want to solve
$$
w^3 = 36\overline{w}.
$$
Multiplying both sides by $w$, we find
$$
w^4 = 36\left|w\right|.
$$
If $w = re^{i\theta}$, then $r^4e^{i4\theta} = 36 r^2$. It follows that either $r = 0$ (in which case $w = 0$), or $r^2 = 36$, so $r = 6$, and that $4\theta = 2k\pi$ for some $k\in\Bbb Z$, or $\theta = k\pi/2$. Since $e^{it}$ is periodic with period $2\pi$, you only need to consider $\theta = 0$, $\theta = \pi/2$, $\theta = \pi$, and $\theta = 3\pi/2$. You get $w = 0$, $w = 6$, $w = 6 i$, $w = -6$, $w = -6i$ as possible solutions. Checking:
\begin{align*}
0^3 &= 36\cdot 0\quad\checkmark\\
6^3&=36\cdot 6\quad\checkmark\\
(6i)^3&=36\cdot (-6i)\quad\checkmark\\
(-6)^3 &=36\cdot(-6)\quad\checkmark\\
(-6i)^3 &=36\cdot(6i)\quad\checkmark\\
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2024617",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Need help conjugating I need some help conjugating because I feel like I am overlooking a step:
$$\lim_{x \rightarrow 4} \frac{3(x-4)(\sqrt{x+5})}{3 - \sqrt{x+5}}$$
I've done the conjugation by multiplying by $(3 + \sqrt{x+5})$ to clear out the bottom, but how would the top look like once multiplied?
| $$\lim _{ x\rightarrow 4 } \frac { 3(x-4)(\sqrt { x+5 } ) }{ 3-\sqrt { x+5 } } =\lim _{ x\rightarrow 4 } \frac { 3(x-4)(\sqrt { x+5 } )\left( 3+\sqrt { x+5 } \right) }{ \left( 3-\sqrt { x+5 } \right) \left( 3+\sqrt { x+5 } \right) } =\\ =\lim _{ x\rightarrow 4 } \frac { 3(x-4)(\sqrt { x+5 } )\left( 3+\sqrt { x+5 } \right) }{ 4-x } =-54$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2025329",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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$ABC$ is an acute angle triangle, then point $P((\cos B-\sin A),(\sin B-\cos A))$ lies in which quadrant triangle $ABC$ is an acute angle triangle, then point $P((\cos B-\sin A),(\sin B-\cos A))$ lies in which quadrant $.........$
$\cos B-\sin A= \sin(\frac{\pi}{2}-B)-\sin A = 2\cos(\frac{\pi}{2}+A-B)\sin (\frac{\pi}{2}-B-A)$
and $\sin B-\cos A = \cos(\frac{\pi}{2}-B)-\cos A=-2\sin(\frac{\pi}{2}+A-B)\sin (\frac{\pi}{2}-B-A)$
wan,t be able to go further, could some help me with this, thanks
|
$\sin(\frac{\pi}{2}-B)-\sin A = 2\cos(\frac{\pi}{2}+A-B)\sin (\frac{\pi}{2}-B-A)$
$\cos(\frac{\pi}{2}-B)-\cos A=-2\sin(\frac{\pi}{2}+A-B)\sin (\frac{\pi}{2}-B-A)$
These are incorrect. They should be
$$\sin\left(\frac{\pi}{2}-B\right)-\sin A=2\cos\left(\frac{\frac{\pi}{2}+A-B}{\color{red}{2}}\right)\sin\left(\frac{\frac{\pi}{2}-A-B}{\color{red}{2}}\right)$$
$$\cos\left(\frac{\pi}{2}-B\right)-\cos A=-2\sin\left(\frac{\frac{\pi}{2}+A-B}{\color{red}2}\right)\sin\left(\frac{\frac{\pi}{2}-A-B}{\color{red}2 }\right)$$
Since we have that
$$0\lt A\lt \frac{\pi}{2},-\frac{\pi}{2}\lt -B\lt 0\implies 0\lt \frac{\frac{\pi}{2}+A-B}{2}\lt \frac{\pi}{2}$$
and that
$$0\lt (C=)\pi-A-B\lt\frac{\pi}{2}\implies -\frac{\pi}{4}\lt\frac{\frac{\pi}{2}-A-B}{2}\lt 0$$
we get
$$\cos B-\sin A\lt 0\quad\text{and}\quad \sin B-\cos A\gt 0$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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summation of a infinite series i want to know if this infinite sum $\sum_{n=1}^{\infty}$ $\frac{a(a+1)(a+2)......(a+n-1)}{b(b+1)(b+2)......(b+n-1)}$ converges or diverges ? where a>0 and b>a+1.
if the sum converges what is the sum i.e where it will converge? i need a concrete explanation
i found some inequalities $\frac{a(a+1)}{b(b+1)}$ <$\frac{a}{b+1}$ and $\frac{a(a+1)(a+2)}{b(b+1)(b+2)}$< $\frac{a}{b+2}$ .......and continuing $\frac{a(a+1)....(a+n-1)}{b(b+1)....(b+n-1)}$ < $\frac{a}{b+n-1}$
this inequalities can be useful for this problem
| Since Hamou has answered the question of convergence, I'll answer the question about the sum of the series. For $b - a > 1$, the sum of the series is $$-1 + \frac{\Gamma(b)\Gamma(b-a-1)}{\Gamma(b-a)\Gamma(b-1)}$$
Indeed, Abel's continuity theorem gives
$$\sum_{n = 1}^\infty \frac{a(a+1)\cdots (a + n-1)}{b(b + 1)\cdots (b + n - 1)} = \lim_{x\to 1^{-}} \sum_{n = 1}^\infty \frac{a(a+1)\cdots (a+n-1)}{b(b+1)\cdots (b+n-1)}x^n$$
For $\lvert x \rvert < 1$,
$$\sum_{n = 0}^\infty \frac{a(a+1)\cdots(a+n-1)}{b(b+1)\cdots(b+n-1)}x^n = \frac{\Gamma(b)}{\Gamma(a)}\sum_{n = 0}^\infty \frac{\Gamma(a+n)}{\Gamma(b+n)}x^n = \frac{\Gamma(b)}{\Gamma(b-a)\Gamma(a)}\sum_{n = 1}^\infty \frac{\Gamma(b-a)\Gamma(a+n)}{\Gamma(b+n)}x^n$$
The quotients $\Gamma(b-a)\Gamma(a+n)/\Gamma(b+n)$ are represented by Beta integrals
$$\int_0^1 (1 - t)^{b-a-1}t^{a+n-1}\, dt$$
Thus
$$\sum_{n = 0}^\infty \frac{\Gamma(b-a)\Gamma(a+n)}{\Gamma(b+n)}x^n = \int_0^1 \sum_{n = 0}^\infty (tx)^n(1-t)^{b-a-1}t^{a-1}\, dt = \int_0^1 (1 - tx)^{-1}(1-t)^{b-a-1}t^{a-1}\, dt$$
This gives the integral representation
$$\sum_{n = 0}^\infty \frac{a(a+1)\cdots(a+n-1)}{b(b+1)\cdots(b+n-1)}x^n = \frac{\Gamma(b)}{\Gamma(b-a)\Gamma(a)}\int_0^1 (1 - tx)^{-1}(1-t)^{b-a-1}t^{a-1}\, dt$$
Taking the limit as $x\to 1^{-}$ results in
$$\frac{\Gamma(b)}{\Gamma(b-a)\Gamma(a)} \int_0^1 (1 - t)^{b-a-2}t^{a-1}\, dt = \frac{\Gamma(b)}{\Gamma(b-a)\Gamma(a)}\frac{\Gamma(b-a-1)\Gamma(a)}{\Gamma(b-1)} = \frac{\Gamma(b)\Gamma(b-a-1)}{\Gamma(b-a)\Gamma(b-1)}$$
So
$$\sum_{n = 1}^\infty \frac{a(a+1)\cdots(a+n-1)}{b(b+1)\cdots(b+n-1)} = -1 + \sum_{n = 0}^\infty \frac{a(a+1)\cdots(a+n-1)}{b(b+1)\cdots(b+n-1)} = -1 + \frac{\Gamma(b)\Gamma(b-a-1)}{\Gamma(b-a)\Gamma(b-1)}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2034148",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Regular non-orthodox semigroups Is there a resource somewhere with smallest finite and other examples of regular semigroups that are not orthodox? I want concrete examples for my private research. I once installed GAP and two semigroup packages too but I have no idea how difficult it would be to calculate such examples there, I haven't used GAP before.
| There is only one finite semigroup of order 4 whose idempotents don't form a subsemigroup (up to anti-isomorphism), and has the following Cayley table:
\begin{array}{l|llll}
& 1 & 2 & 3 & 4 \\ \hline
1 & 1 & 1 & 1 & 1 \\
2 & 1 & 1 & 1 & 2 \\
3 & 1 & 2 & 3 & 2 \\
4 & 1 & 1 & 1 & 4
\end{array}
Sadly it is not regular. It has 3 idempotents and one non-regular element (the 2).
However it might be possible to extend this semigroup by adding another element that should become an inverse of 2:
\begin{array}{l|lllll}
& 1 & 2 & 3 & 4 & 5 \\ \hline
1 & 1 & 1 & 1 & 1 & \\
2 & 1 & 1 & 1 & 2 & 3 \\
3 & 1 & 2 & 3 & 2 & \\
4 & 1 & 1 & 1 & 4 & 5 \\
5 & & 4 & 5 & &
\end{array}
As I was writing this, I did a quick program and indeed this is possible: there is only one way to extend it and the resulting regular non-orthodox semigroup is this one:
\begin{array}{l|lllll}
& 1 & 2 & 3 & 4 & 5 \\ \hline
1 & 1 & 1 & 1 & 1 & 1 \\
2 & 1 & 1 & 1 & 2 & 3 \\
3 & 1 & 2 & 3 & 2 & 3 \\
4 & 1 & 1 & 1 & 4 & 5 \\
5 & 1 & 4 & 5 & 4 & 5
\end{array}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2035590",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove $4a^2+b^2+1\ge2ab+2a+b$ Prove $4a^2+b^2+1\ge2ab+2a+b$
$4a^2+b^2+1-2ab-2a-b\ge0$
$(2)^2(a)^2+(b)^2+1-2ab-2a-b\ge0$
Any help from here? I am not seeing how this can be factored
| Multiply by $2$, then we get:
$$8a^2+2b^2+2 \ge 4ab+4a+2b$$
and rearranging we get:
$$(4a^2-4ab+b^2)+(b^2-2b+1)+(4a^2-4a+1) \ge 0$$
and then
$$(2a-b)^2+(b-1)^2+(2a-1)^2 \ge 0$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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} |
Probability only 2 blue balls are selected out of 6 different colored balls? We don't have an answer key for this test questions and I just wanted to confirm my answers.
We have a bag with 6 balls, 3 blue, 2 green, and 1 red.
If we select 3 at random, what is the chance we select only 2 of 3 are blue balls?
So I have there are a total of $\binom{6}{3} = 20$ different ways to select 3 balls. There is $\binom{3}{2} * \binom{2}{1} = 6$ ways of selecting 2 blue and 1 green, and $\binom{3}{2} * \binom{1}{1} = 3$ of selecting 2 blue and 1 red.
So we have a probability of $\frac{6 + 3}{20} = \frac{9}{20} $ chance of selecting only 2 blue balls.
Also, what is the chance of selecting only 1 of each color?
$\binom{3}{1} * \binom{2}{1} * \binom{1}{1} = 3*2*1 = 6$ different ways of selecting one of every color. So we have a probability of $\frac{6}{20} $ of getting 1 of each color.
Is this correct?
| Yes, that is okay. $9/20$ and $6/20$ are the answers for the reasons you gave.
The probability of selecting exactly $2$ from $3$ blue ball (and $1$ from $3$ non-blue) when selecting $3$ from all $6$ balls is more simply $\dbinom 3 2\dbinom 31\big/\dbinom 6 3$.
But yes, $\dbinom 3 2\Big(\dbinom 2 1 +\dbinom 1 1\Big)\big/\dbinom 6 2$ is also valid. It counts the same thing.
The probability of selecting $1$ ball from each group of sizes $3,2,1$ when selecting $3$ from all $6$ balls, is indeed $\dbinom 3 1\dbinom 2 1\dbinom 1 1\big/\dbinom 63$
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "1",
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Show that if $x^2 + y^2 = z^2$, then at leeast one of x and y is a multiple of 3. Show that if $x^2 + y^2 = z^2$, then at least one of $x$ and $y$ is a multiple of 3.
Attempt:
Given $x,y,z$ is pyhagorean triple,
$$
x^2+y^2=z^2.
$$
Assume neither $x$ nor $y$ is divisible by $3$. Since $x^2+y^2=z^2$, $x^2+y^2 \equiv z^2 \pmod 3$. But neither $x$ nor $y$ is congruent to $0 \pmod 3$. Therefore,
\begin{align}
x \equiv \pm 1 &\pmod 3 \\
y \equiv \pm 1 &\pmod 3.
\end{align}
Hence,
\begin{align}
x^2 \equiv 1 &\pmod 3 \\
y^2 \equiv 1 &\pmod 3,
\end{align}
and so
$$
x^2+y^2 \equiv 2 \pmod 3.
$$
But $2$ is not a square modulo $3$. Therefore,
$$
2 \not\equiv z^2 \pmod 3,
$$
which is a contradiction.
| Suppose none of $x,y,z$ is divisible by $3$.
then
$$x\equiv \pm 1\mod 3,$$
$$y\equiv \pm 1 \mod 3,$$
and
$$z\equiv \pm 1 \mod 3$$
$\implies$
$$x^2+y^2-z^2\equiv 1+1-1\mod 3$$
$$\implies x^2+y^2\neq z^2.$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Calculate: $\lim\limits_{n\to\infty} \sum_{k=0}^{n} \frac{2n+k}{n^2+(2n+k)^2}$ Calculate: $$\lim\limits_{n\to\infty} \sum_{k=0}^{n} \frac{2n+k}{n^2+(2n+k)^2}$$
I thought a Riemann sum could lead to something, but couldn't find a suitable partition.
Hint, please?
| One may recognize a Riemann sum, by writing
$$
\sum_{k=0}^{n} \frac{2n+k}{n^2+(2n+k)^2}=\frac1n \cdot\sum_{k=0}^{n} \frac{2+\frac{k}n }{1+(2+\frac{k}n)^2},
$$ then letting $n \to \infty$, to obtain
$$
\frac1n \cdot\sum_{k=0}^{n} \frac{2+\frac{k}n}{1+(2+\frac{k}n)^2} \to \int_0^1 \frac{2+x }{1+(2+x)^2}\:dx.\tag1
$$
Add-on. Since $f:[0, 1] \rightarrow [0, 1]$ with $f(x)=\frac{2+x}{1+(2+x)^2}$ satisfies $f \in \mathcal{C}^1([0,1])$, then one is allowed to apply the standard result
$$
\frac1n\sum_{k=0}^{n} f\left(\frac{k}{n}\right) =\int_0^1 f(x)\,dx + \frac{f(0) + f(1)}{2n}+o\left(\frac1n \right) \tag2
$$ giving, as $n \to \infty$,
$$
\sum_{k=0}^{n} \frac{2n+k}{n^2+(2n+k)^2}=\frac{\ln 2}2+\frac{7}{20\: n}+o\left(\frac1n \right). \tag3
$$
One may in fact express the given sum in terms of the digamma function, using
$$
\sum_{k=0}^{n} \frac{1}{k+b}=\psi\left(n+b+1\right)-\psi\left(b\right), \qquad \text{Re}\:b>0,
$$ and writing
$$
\sum_{k=0}^{n} \frac{2n+k}{n^2+(2n+k)^2} = \text{Re}\:\sum_{k=0}^{n} \frac{1}{k+(2+i)n}
$$ then recalling the asymptotics of the digamma function, as $n \to \infty$, one obtains
$$
\sum_{k=0}^{n} \frac{2n+k}{n^2+(2n+k)^2}=\frac{\ln 2}2+\frac7{20\:n}+\frac1{300\:n^2}+\frac7{60\:000 \:n^4}+o\left(\frac1{n^5} \right).\tag4
$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Showing $\Gamma\left(x+1\right) \ge x^{x/2}$, when $x\ge2$ It is easy to show that $(2n)! \ge (2n)^n$.
Indeed, if we have $ 1\le k\le n$, for integer $k,n$ , we can write:
$$
\begin{aligned}
0\lt n+\frac{1}{2}-k &\le n - \frac{1}{2} \\
\left(n+\frac{1}{2}-k\right)^2 &\le \left(n - \frac{1}{2}\right)^2 \\
k\cdot(2n-k+1)&\ge2n
\end{aligned}
$$
and so
$$
(2n)! = 1\cdot2\cdots (2n) = \prod_{k=1}^{n} {k\cdot(2n-k+1)} \ge (2n)^n
$$
A variation of the above argument can be used to prove the statement for odd integers.
Is this statment true for all real $x\ge2$?
| Let
$$f(x) = \log \Gamma(x + 1) - \frac{x \log(x)}{2} = \log(x) + \log \Gamma(x) - \frac{x \log(x)}{2}$$
We will prove that $f$ is a convex function with $f(2) = 0$ and $f'(2) > 0$. Using the subderivative property of convex functions we then get
$$f(x) \geq f(2) + (x - 2) f'(2) \ge 0 \qquad \text{for } x \ge 2,$$
which is what we wanted to prove.
The first two derivatives of $f$ are
$$\begin{align*}
f'(x) &= \psi(x) + \frac{1}{x} - \frac{1}{2} - \frac{\log(x)}{2} \\
f''(x) &= \psi'(x) - \frac{1}{x^2} - \frac{1}{2x}
\end{align*},$$
where $\psi$ resp. $\psi'$ are the di- resp. trigamma function. We will now use the formulae from Abramowitz and Stegun.
First of all, formula 6.3.2 tells us
$$f'(2) = \psi(2) - \frac{\log(2)}{2} = 1 - \gamma - \frac{\log(2)}{2} = 0.076... > 0$$
To prove convexity, we will show $f''(x) \ge 0$ for all $x \ge 2$. From 6.4.10 we get the representation
$$\psi'(x) = \sum \limits_{k = 0}^\infty \frac{1}{(x + k)^2} = \frac{1}{x^2} + \sum \limits_{k = 1}^\infty \frac{1}{(x + k)^2}.$$
Now it only remains to prove the inequality
$$\sum \limits_{k = 1}^\infty \frac{1}{(x + k)^2} \ge \frac{1}{2x}.$$
As pointed out by Jack Lam in the comments (many thanks!) this can be proven using telescoping:
$$\sum \limits_{k = 1}^\infty \frac{1}{(x + k)^2} \ge \sum \limits_{k = 1}^\infty \frac{1}{(x + k)(x + k + 1)} = \sum \limits_{k = 1}^\infty \left(\frac{1}{x + k} - \frac{1}{x + k + 1}\right) = \frac{1}{x + 1} \ge \frac{1}{2x}.$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Number of occurrences of k consecutive 1's in a binary string of length n (containing only 1's and 0's) Say a sequence $\{X_1, X_2,\ldots ,X_n\}$ is given, where $X_p$ is either one or zero ($0 < p < n$). How can I determine the number of strings, which do contain at least one occurrence of consequent $1$'s of length $k$ ($0 < k < n$).
For example, a string $\{1, 0, 1, 1, 1, 0\}$ is such a string for $n = 6$ and $k = 3$.
Here I have found an answer for arbitrary $n$ and $k = 2$, ($k = 1$ is trivial), but I need a more general answer for any natural number $k$ smaller than $n$.
| One systematic way(not very cool to do it by hand tho) to do it is using automatons and the Chomsky-Schûtzenberger theorem in the following way.
Case $k = 2$:
The automata that accepts your language(namely $F = \{x\in \{0,1\}^*:\underbrace{11\cdots 11}_{\text{$k$ times}}\in Sub(x)\}$) is described by the image below($S_0$ is the initial state, $S_2$ is the final state and you can just reach that state if you have read $11$ as substring.). From there, by the C-S theorem you have the following set of equations( the equations relate the transition of the automaton, for example, if you see in $S_1$ there is one arrow going out to $S_0$ and the other one to $S_2$ and $x$ measures the number of letters of the transition) $$S_0 = xS_0+xS_1$$$$S_1 = xS_2+xS_0$$$$S_2 = 1+2xS_2,$$
and you want to recover $S_0$ as a power series. Solving for $S_2,$ we get $S_2=\frac{1}{1-2x},$ so $S_1 = xS_0+\frac{x}{1-2x}$ and finally $S_0 = xS_0+x(xS_0+\frac{x}{1-2x})=xS_0+x^2S_0+\frac{x^2}{1-2x}$ which implies $S_0(1-x-x^2)=\frac{x^2}{1-2x},$ so $S_0 = \frac{x^2}{(1-2x)(1-x-x^2)}.$ To recover the numbers you have can do partial fractions and you will end up with $S_0=\frac{1}{1-2x}-\frac{x+1}{1-x-x^2}$ which agrees with the answer in the post you have linked.
General Case:
In the general case you have $k+1$ states, the initial one, say $S_0 = xS_0+xS_1,$ the intermediate states i.e., $1\leq j<k$ $S_j = xS_{j+1}+xS_0$ and the final state $S_k = 1+2xS_k,$ from this $k+1$ equations, you can deduce, first that $S_k = \frac{1}{1-2x}$ and that $$S_{k-1} =xS_k+xS_0=\frac{x}{1-2x}+xS_0 $$
$$S_{k-2} =xS_{k-1}+xS_0=\frac{x^2}{1-2x}+x^2S_0+xS_0,$$
$$\vdots$$
$$S_{k-j}=\frac{x^j}{1-2x}+S_0\sum _{i=1}^{j}x^i=\frac{x^j}{1-2x}+S_0(\frac{1-x^{j+1}}{1-x}-1),$$
and so $$S_1 = S_{k-(k-1)}=\frac{x^{k-1}}{1-2x}+S_0(\frac{1-x^{k}}{1-x}-1),$$ therefore $$S_0 = xS_0+x(\frac{x^{k-1}}{1-2x}+S_0(\frac{1-x^{k}}{1-x}-1))=xS_0+\frac{x^{k}}{1-2x}+S_0(\frac{x^2-x^{k+1}}{1-x}),$$
concluding $$S_0(1-x-\frac{x^2-x^{k+1}}{1-x})=\frac{x^k}{1-2x},$$
So $$S_0=\frac{x^k}{1-2x}(\frac{(1-2x+x^2-x^2+x^{k+1})}{1-x})^{-1}=\frac{x^k(1-x)}{(1-2x)(1-2x+x^{k+1})}.$$
You can extract the numbers from there by saying that $$\frac{x^k(1-x)}{(1-2x)(1-2x+x^{k+1})}=\sum _{i=0}^{\infty}A_ix^i,$$
where $A_i = |\{x\in \{0,1\}^i:\underbrace{11\cdots 11}_{\text{$k$ times}}\in Sub(x)\}|$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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How many solutions does $x^2 + 3x +1 \equiv 0\, \pmod{101}$ have? $x^2 + 3x +1 \equiv 0 \pmod{101}$. To solve this I found the determinant $D = 5 \pmod{101}$). Using the Legendre symbol,
$$\left(\frac{5}{101}\right) = \left(\frac{101}{5}\right) \equiv \left(\frac{1}{5}\right) \equiv 1,$$
$\therefore$ The equations have a solution.
My question is how I can find out how many solutions it has?
| Since
$$
(x-49)^2-77\equiv x^2+3x+1\pmod{101}
$$
we are looking for solutions to $(x-49)^2\equiv77\pmod{101}$. You have verified that there is a root, so $77$ is a quadratic residue mod $101$, thus, there are two solutions for
$$
(x-49)^2\equiv77\pmod{101}
$$
Alternatively, working mod $101$, by squaring and multiplying
$$
77^2\equiv71\\
77^3\equiv13\\
77^6\equiv68\\
77^{12}\equiv79\\
77^{24}\equiv80\\
77^{25}\equiv100\\
77^{50}\equiv1
$$
Therefore, $77$ is a quadratic residue mod $101$. Thus, there are two solutions to $x^2+3x+1\pmod{101}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2047575",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
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Let $a,b,c$ be the length of sides of a triangle then prove that $a^2b(a-b)+b^2c(b-c)+c^2a(c-a)\ge0$
Let $a,b,c$ be the length of sides of a triangle then prove that:
$a^2b(a-b)+b^2c(b-c)+c^2a(c-a)\ge0$
Please help me!!!
| let
$a^2b(a-b)+b^2c(b-c)+c^2a(c-a)=0$ (1)
let
$x=-a+b+c; y=a-b+c; z=a+b-c$
$z,y,x$ are twice the length of the segments between the vertices and the touching point of the incircles. so
$a=\frac{y+z}{2}, b=\frac{z+x}{2}, c=\frac{x+y}{2}$
substitute them to (1) and multiply the inequality by 16
$(y+z)^2(z+x)(y-x)+(z+x)^2(x+y)(z-y)+(x+y)^2(y+z)(x-z)\geqslant 0$
$x^3z+y^3x+z^2y\geqslant x^2yz+y^2zx+z^2xy$
and so
$x^3z+y^3x+z^3y-x^2yz-y^2zx-z^2xy$
$=zx(x-y)^2+xy(y-z)^2+yz(z-x)^2\geqslant 0$
as $x>0, y>o, z>0$ equality holds if and only if $x=y=z$.
example if $a=b=c$ then triangle is equilateral.
| {
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"url": "https://math.stackexchange.com/questions/2055559",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 0
} |
If $\frac{\sqrt{31+\sqrt{31+\sqrt{31+ \cdots}}}}{\sqrt{1+\sqrt{1+ \sqrt{1+ \cdots}}}}=a-\sqrt b$, find the value of $a+b$
$\frac{\sqrt{31+\sqrt{31+\sqrt{31+ \cdots}}}}{\sqrt{1+\sqrt{1+ \sqrt{1+ \cdots}}}}=a-\sqrt b$ where $a,b$ are natural numbers. Find the value of $a+b$.
I am not able to proceed with solving this question as I have no idea as to how I can calculate $\frac{\sqrt{31+\sqrt{31+\sqrt{31+ \cdots}}}}{\sqrt{1+\sqrt{1+ \sqrt{1+ \cdots}}}}$. A small hint would do.
| To find $\sqrt{a+\sqrt{a+\cdots}} $, solve the equation
$x = \sqrt{a+x}$
The solution of $\sqrt{31+\sqrt{31+\cdots}}$ gives us $x = \frac{1\pm 5\sqrt{5}}{2}$.
The solution of $\sqrt{1+\sqrt{1+\cdots}}$ gives us $y = \frac{1\pm \sqrt{5}}{2}$.
Thus, we have $$\frac{x}{y} = 6-\sqrt{5}$$ Hence, $a=6, b=5 \Rightarrow a+b = 11$. Hope it helps.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
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All fractions which can be written simultaneously in the forms $\frac{7k-5}{5k-3}$ and $\frac{6l-1}{4l-3}$ Find all fractions which can be written simultaneously in the forms $\frac{7k-5}{5k-3}$ and $\frac{6l-1}{4l-3}$
for some integers $k,l$.
Please check my answer and tell me is correct or not....
$$\frac{43}{31},\frac{31}{27},1,\frac{55}{39},\frac{5}{3},\frac{61}{43},\frac{19}{13},\frac{13}{9}$$
| $$\frac{7k-5}{5k-3}=\frac{6l-1}{4l-3}$$
$$28kl-20l-21k+15=30kl-18l-5k+3$$
$$2kl+2l+16k-12=0$$
$$kl+l+8k-6=0$$
either:
$$l(1+k)=2(3-4k)$$
so
$$l=2\frac{3-4k}{1+k}$$
or:
$$k(l+8)=6-l$$
so
$$k=\frac{6-l}{l+8}$$
Let's go with this second one to complement the other answer.
Then $$k=-\frac{l-6}{l+8}=-\left(\frac{l+8-14}{l+8}\right)=-\left(1-\frac{14}{l+8}\right)$$
We want $l+8=\pm(1,2,7,14)$
So $l=-7,-9,-6,-10,-1,-15,6,-22$ or more nicely ordered $$l=-22,-15,-10,-9,-7,-6,-1,6$$
The pairs are then $(l,k)=$ $(-22,-2)$, $(-15,-3)$, $(-10,-8)$, $(-9,-15)$, $(-7,13)$, $(-6,6)$, $(-1,1)$, $(-22,-2)$, $(6,0)$,
or that the fractions are
$$\frac{19}{13},\frac{13}{9},\frac{61}{43},\frac{55}{39},\frac{43}{31},\frac{37}{27},1,\frac{5}{3}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2060154",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
} |
$\sqrt{3-x}-\sqrt{-x^2+8x+10}=1$ Consider the equation
$$
\sqrt{3-x}-\sqrt{-x^2+8x+10}=1.
$$
I have solved it in a dumb way by solving the equation of degree four. So, the
only real solution is $x = -1$. Can you please suggest, maybe there are better or easier ways to solve it?
| This is a simple way used when I was in high school.
Basically, all you have to do is draw out the minimal polynomial of the known solutions (which make this even useful with most algebraic solutions, not only integers).
Firstly, match each square root with its real value.
\begin{align} (\sqrt{3-x}-2)-(\sqrt{-x^2+8x+10}-1)=0 \end{align}
Then multiply each with its conjugate.
\begin{align} \frac{-1-x}{\sqrt{3-x}+2}-\frac{-x^2+8x+9}{\sqrt{-x^2+8x+10}+1}=0
\end{align}
\begin{align} (1+x)\left (\frac{1}{\sqrt{3-x}+2}+\frac{9-x}{\sqrt{-x^2+8x+10}+1} \right )=0
\end{align}
From the fact that $3-x>0$ and $-x^2+8x+10>0$, it can easily be checked that $9-x>0$
Therefore, the only solution would be $x=-1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2061205",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
How can we compute $\lim_{x\rightarrow 0}\frac{\sin (x)-x+x^3}{x^3}$ and $\lim_{x\rightarrow 0}\frac{e^x-\sin (x)-1}{x^2}$? Could we compute the limits
$$\lim_{x\rightarrow 0}\frac{\sin (x)-x+x^3}{x^3} \\ \lim_{x\rightarrow 0}\frac{e^x-\sin (x)-1}{x^2}$$
without using the l'Hospital rule and the Taylor expansion?
| The first one is equal to $1+\lim\limits_{x\to0}\dfrac{\sin x - x}{x^3}.$ We can apply to this limit the technique in this answer of mine.
Let $L=\lim\limits_{x\to0}\dfrac{\sin x - x}{x^3}$. This is not infinite if it exists because near $0$ one has $$-\frac12\leftarrow\frac{\cos x-1}{x^2}<\frac{\frac{\sin x}{x}-1}{x^2}<\frac{1-1}{x^2}=0.$$ Then, considering a similar limit, $$\lim_{x\to0}\frac{\sin x\cos x-x}{x^3}=\lim_{x\to0}\frac{\sin(2x)-2x}{2x^3}=4L$$ we can deduce$$\lim_{x\to0}\frac{\sin x\cos x-\sin x}{x^3}=\lim_{x\to0}\frac{\sin x}{x}\cdot\lim_{x\to0}\frac{\cos x-1}{x^2}=-\frac12=3L,$$that is $L=-\frac16.$ So your limit equals $\frac56$.
Getting back to this; due to the finiteness of the first limit, the second one is the same as $$\lim_{x\to0}\frac{e^x-1-x}{x^2}+\lim_{x\to0}\frac{x-\sin x}{x^2}=\lim_{x\to0}\frac{\frac{e^x-1}{x}-1}{x},$$i.e. $f'(0)$ where $f(x)=\frac{e^x-1}{x}, f(0)=1.$ One has $$f'(0)=\lim_{x\to0}\frac{e^x(x-1)+1}{x^2}=\lim_{x\to0}\frac{e^x-1}{x}-\lim_{x\to0}\frac{e^x-1-x}{x^2}=1-f'(0),$$ so it equals $\frac12$.
Alternatively, let $L_2$ be this limit. It is largely known that $e^x>1+x$; this also means $e^{-x}>1-x$, or equivalently for $x<1$, $e^x<\frac1{1-x},$ and it's easy to prove $\frac1{1-x}<1+x+2x^2 $ for $0\ne x<\frac12.$ Thus, $0\le L_2\le2.$ Assuming it exists, we have $$L_2=\lim_{x\to0}\frac{e^x-1-x}{x^2}=\lim_{x\to0}\frac{e^{2x}-(1+x)^2}{x^2(e^x+1+x)}=\frac12\lim_{x\to0} \frac{e^{2x}-1-2x-x^2}{x^2}=2L_2-\frac12,$$whence $L_2=\frac12$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2061655",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
show $\frac{y}{x^2+y^2} $ is harmonic except at $y=0,x=0$ Let $f(z)=u(x,y)+iv(x,y) $
where $$ f(z)=u(x,y)=\frac{y}{x^2+y^2}$$
show $u(x,y)$ is harmonic except at $z=0$
Attempt
$$ u=\frac{y}{x^2+y^2}=y(x^2+y^2)^{-1} $$
Partial derivatives with x
$$\begin{aligned}
u_x&= y *(x^2+y^2)^{-2}*-1*2x
\\ &= -y*2x(x^2+y^2)^{-2}=-2xy(x^2+y^2)^{-2}
\\ u_{xx} &= -2xy*(x^2+y^2)^{-3}*-2*2x+-2y*(x^2+y^2)^{-2}
\\&= \frac{-2xy}{(x^2+y^2)^3}*-4x +\frac{-2y}{(x^2+y^2)^2}
\\&=\frac{8x^2y}{(x^2+y^2)^3}+\frac{-2y}{(x^2+y^2)^2}
\\
\end{aligned} $$
Partial Derivatives with y
$$\begin{aligned}
u_y&=1(x^2+y^2)^{-1}+y*(x^2+y^2)^{-2}*-1*2y
\\ &=(x^2+y^2)^{-1}-2y^2(x^2+y^2)^{-2}
\\ u_{yy}&=-1(x^2y^2)^{-2}*2y -2*2y(x^2+y^2)^{-2}-2y^2*-2(x^2+y^2)^{-3}*2y
\\ &=-2y(x^2+y^2)^{-2}-4y(x^2+y^2)^{-2}+8y^3(x^2+y^2)^{-3}
\\ &=\frac{-6y}{(x^2+y^2)^2} + 8y^3(x^2+y^2)^{-3}
\end{aligned} $$
From here need to show that $u_{xx}+u_{yy}=0$ and technically say why the other partials are continous right?? This was a test question whith 3 lines of paper by the way
| Let $F(x,y) = \frac{1}2 \log(x^2+y^2)$. Then $f$ is harmonic everywhere away from the origin, and so are its derivatives $\partial_1 F$ and $\partial_2 F$. To prove the first claim, note that
$$\partial_1 F = \frac{x}{x^2+y^2}$$
$$\partial_{11} F = \frac{y^2-x^2}{(x^2+y^2)^2}$$
and by symmetry
$$\partial_{22} F = \frac{x^2-y^2}{(x^2+y^2)^2} = -\partial_{11}F.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2062523",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
Struggling with a strong induction problem involving recurrence relations The problem states:
Let $b_0$ = $12$ and $b_1$ = $29$, and for all integers $k ≥ 2$, let $b_k$ = $5b_{k-1}-6b_{k-2}$.
Prove that for all $n ≥ 0$, $b_n$ = $5\:\cdot \:3^n\:+\:7\:\cdot\:2^n$
I'm familiar with questions where I'm supposed to find the close form of a recurrence relation, and then prove that formula via induction. This question is more strangely structured, however, and I'm not sure how to proceed. Am I supposed to show that:
$b_k$ = $5b_{k-1}-6b_{k-2}$ = $b_n$ = $5\:\cdot \:3^n\:+\:7\:\cdot
\:2^n$
i.e. use strong induction to show both statements are equivalent?
If that's the case, do I show $b_n$ = $5\:\cdot \:3^n\:+\:7\:\cdot \:2^n$ holds for n = 0 and 1 (establishing base cases) and then manipulate what emerges algebraically to equal $b_k$ = $5b_{k-1}-6b_{k-2}$ ? Or should I do that to $b_k$ = $5b_{k-1}-6b_{k-2}$? Advice and insight is most welcome!
| I think you just have to show that $b_n = 5\cdot 3^n + 7\cdot 2^n$ fulfills the recurrence relation and initial conditions for every instance $n$.
As the relation involves up to two prior sequence elements I would use strong induction.
You start with direct proofs for $b_0$, then $b_1$ and the induction from $n=2$.
Update: It seems I need to provide a full solution:
The initial conditions hold by evaluation and comparison.
$$
S(0): b_0 = 12 \wedge
b_0 = 5\cdot 3^0 + 7\cdot 2^0 = 5 + 7 = 12 \\
S(1): b_1 = 29 \wedge
b_1 = 5\cdot 3^1 + 7\cdot 2^1 = 15 + 14 = 29 \\
$$
Where the $\wedge$ means logical "and".
The statement for $n$, $n\ge 2$ is
$$
S(n): b_n = 5 b_{n-1} - 6 b_{n-2} \wedge
b_n = 5\cdot 3^n + 7\cdot 2^n
$$
Base case $n=2$:
$$
S(2):b_2 = 5 b_1 - 6 b_0 \wedge b_2 = 5 \cdot 3^2 + 7 \cdot 2^2
$$
This evaluates to
$$
S(2): b_2 = 5 \cdot 29 - 6 \cdot 12 = 145-72=73 \wedge b_2 = 45 + 28 = 73
$$
which is a true statement.
Induction Step:
Assuming $S(k)$ is true for $k\in\{2,\dotsc, n\}, n\ge 2$ we have
$$
S(k+1): b_{k+1} = 5 b_k - 6 b_{k-1} \wedge
b_{k+1} = 5\cdot 3^{k+1} + 7\cdot 2^{k+1}
$$
where
\begin{align}
b_{k+1}
&= 5 b_k - 6 b_{k-1} \\
&= 5 \left( 5\cdot 3^k + 7\cdot 2^k \right) -
6 \left( 5\cdot 3^{k-1} + 7\cdot 2^{k-1} \right)
\end{align}
The first term in parentheses is justified by the truth of $S(k)$.
The second term needs a case distinction:
For $k=2$ it relies on the truth of the fulfillment of the base condition $b_1$, otherwise we can build on the true statements $S(k-1)$.
Then:
$$
\begin{align}
b_{k+1}
&= (25 - 10) 3^k + (35-21) 2^k \\
&= 15\cdot 3^k + 14\cdot 2^k \\
&= 5\cdot 3^{k+1} + 7\cdot 2^{k+1}
\end{align}
$$
so we end up with
$$
S(k+1): b_{k+1} = 5\cdot 3^{k+1} + 7\cdot 2^{k+1} \wedge
b_{k+1} = 5\cdot 3^{k+1} + 7\cdot 2^{k+1}
$$
which is a true statement.
By the principle of strong induction we conclude the truth of $S(n)$ for all integer $n$ with $n>=2$.
Adding the truth of the initial conditions we can claim the truth of
$S(n)$ for all integer $n$ with $n \ge 0$.
Now
\begin{align}
S(0)&: b_0 = 12 \wedge
b_0 = 5\cdot 3^0 + 7\cdot 2^0 \\
S(1)&: b_1 = 29 \wedge
b_1 = 5\cdot 3^1 + 7\cdot 2^1 \\
S(n)&: b_n = 5 b_{n-1} - 6 b_{n-2} \wedge
b_n = 5\cdot 3^n + 7\cdot 2^n \quad (n \ge 2)
\end{align}
means that for all integer $n$ with $n\ge 0$ the sequence $b_n = 5\cdot 3^n + 7\cdot 2^n$ is the solution of the recurrence relation.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2063747",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Hall and Knight- Higher Algebra Problem 23, Example V. b If $r < 1$ and positive and m is a positive integer, show that $(2m+1)r^m(1-r)<1-r^{2m+1}$. Hence show that $nr^n$ is indefinitely small when $n$ is indefinitely great.
I have tried by taking $\frac{1-r^{2m+1}}{1-r}$ as the sum of the series $1 + r + r^2 ... + r^{2m}$ and then we may try to prove that $(2m+1)r^m > 1 + r + r^2 ... + r^{2m}$.
I have proved the inequality but am unable to prove the limit. The solution book says something which I do not understand. The solution given is
$(2m+1)r^m(1-r)<1-r^{2m+1}$
(multiplying by $r^{m+1}$ on both sides)
$=>(2m+1)r^{2m+1}(1-r)<r^{m+1}(1-r^{2m+1})$
putting $n=2m+1$
$=>nr^n(1-r)<r^{\frac{n+1}{2}}(1-r^{n})$
this much I understand but then it says that if $n$ is made indefinitely great, then $r^{\frac{n+1}{2}}$ becomes indefinitely small and $nr^n$ becomes indefinitely small.
| You have
$$\begin{aligned}
\frac{1 - r^{2m+1}}{1-r}&= 1+r+ \dots +r^{2m}\\
&= r^m(\frac{1}{r^m} + \frac{1}{r^{m-1}} + \dots + \frac{1}{r} + 1+ r +r^2 + \dots + r^{m-1}+r^m)\\
&= r^m[1 + (r^m+\frac{1}{r^m}) + (r^{m-1} + \frac{1}{r^{m-1}}) + \dots +(r + \frac{1}{r})]\\
&> r^m(1 + \underbrace{2 + \dots + 2}_{m \text{ times}}) = (2m + 1)r^m
\end{aligned}$$
because for $0<x<1$ you have $x + \frac{1}{x} >2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2064803",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Prob. 3, Chap. 3 in Baby Rudin: If $s_1 = \sqrt{2}$, and $s_{n+1} = \sqrt{2 + \sqrt{s_n}}$, what is the limit of this sequence? Here's Prob. 3, Chap. 3 in the book Principles of Mathematical Analysis by Walter Rudin, 3rd edition:
If $s_1 = \sqrt{2}$, and $$s_{n+1} = \sqrt{2 + \sqrt{s_n}} \ \ (n = 1, 2, 3, \ldots),$$ prove that $\left\{ s_n \right\}$ converges, and that $s_n < 2$ for $n = 1, 2, 3, \ldots$.
My effort:
We can show that $\sqrt{2} \leq s_n \leq 2$ for all $n = 1, 2, 3, \ldots$. [Am I right?]
Then we can also show that $s_n < s_{n+1}$ for all $n = 1, 2, 3, \ldots$. [Am I right?]
But how to calculate the exact value of the limit? Where does this sequence occur in applications?
| If the limit $s$ of $s_n$ exists, then
$$
s=\lim s_n=\lim s_{n+1}=\lim \sqrt{2+\sqrt{s_n}}=\lim \sqrt{2+\sqrt{s}}.
$$
Hence, the limit satisfies the equation
$$
s=\sqrt{2+\sqrt{s}}.
$$
Thus,
$$
s^2=2+\sqrt{s}\qquad\text{or equivalently}\qquad (s^2-2)^2=s.
$$
Thus the limit is the unique solution of $s^4-4s^2-s+4=0$ which lies in the interval $[\sqrt{2},2]$.
To solve this equation, first observe that $s=1$ is a solution, and hence
$$
s^4-4s^2-s+4=(s-1)(s^3+s^2-3s-4).
$$
We can use this method to solve $\,s^3+s^2-3s-4=0,\,$ and obtain that
$$
s=-\frac13+\frac{1}{\sqrt[3]{54}} \Big(\left(79+3 \sqrt{249}\right)^{1/3}+ \left(79-3 \sqrt{249}\right)^{1/3}\Big).
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2066320",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 2
} |
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