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Is the series $\sum_{n=1}^\infty\frac{1}{n}\left(\sum_{k=1}^n\frac{1}{k}\left(\frac{1}{2}\right)^{n-k}\right)$ convergent? $$
\sum_{n = 1}^\infty\dfrac{1}{n}\left(\sum_{k = 1}^n\dfrac{1}{k}\left(\dfrac{1}{2}\right)^{n - k}\right)
$$
Does the series converge?
I calculate it using Matlab, and it seems that the sum converges to 2.4673. I also tried to use the ratio test to prove the convergent, but stopped at to calculate the sum $\sum_{k = 1}^n\dfrac{1}{k}(1/2)^{n - k}$.
|
Let
$$S_n = \sum_{k=1}^n \frac{1}{k} \left( \frac{1}{2} \right)^{n-k}$$
Let's show that $(nS_n)$ is eventually decreasing. By direct computation, for $n > 2$
\begin{align*} (n-1)S_{n-1} – nS_n &= \left( \sum_{k=1}^{n-1} \frac{n-1}{k} \left( \frac{1}{2} \right)^{n-1-k}\right)- \left(\sum_{k=1}^n \frac{n}{k} \left( \frac{1}{2} \right)^{n-k} \right)\\
&= \left( \sum_{k=1}^{n-1} \frac{n-1}{k} \left( \frac{1}{2} \right)^{n-1-k}\right)- \left(\sum_{k=0}^{n-1} \frac{n}{k+1} \left( \frac{1}{2} \right)^{n-k-1} \right)\\
&=\left[ \sum_{k=1}^{n-1} \left(\frac{n-1}{k} - \frac{n}{k+1}\right)\left( \frac{1}{2} \right)^{n-1-k}\right]- \frac{n}{2^{n-1}}\\
&=\left( \sum_{k=1}^{n-1} \frac{n-k-1}{k(k+1)2^{n-1-k}} \right)- \frac{n}{2^{n-1}} \\
&=\left( \sum_{k=0}^{n-2} \frac{k}{2^k(n-k)(n-k-1)} \right) - \frac{n}{2^{n-1}}\quad \quad (\text{substitution } k \rightarrow n-1-k)\\
&= \left( \sum_{k=2}^{n-2} \frac{k}{2^k(n-k)(n-k-1)} \right) + \left(\frac{1}{2(n-1)(n-2)}- \frac{n}{2^{n-1}} \right)
\end{align*}
The second parenthesis is clearly positive for $n$ sufficiently large, hence $(n-1)S_{n-1} – nS_n > 0$ for $n$ sufficiently large. In particular, the sequence $(nS_n)$ is bounded : there exists $M \geq 0$ such that $nS_n \leq M$. In particular, one has
$$0 \leq \frac{1}{n} S_n \leq \frac{M}{n^2}$$
By comparison, you get that the series $\displaystyle{\sum\frac{1}{n} S_n}$ converges.
Remark : Actually, one can prove using the recursion
$$S_n = \frac{1}{2}S_{n-1} + \frac{1}{n}$$
that the sequence $(nS_n)$ converges to $2$, so $\displaystyle{S_n \sim \frac{2}{n}}$.
|
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"url": "https://math.stackexchange.com/questions/4344760",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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|
Find function $ f $ such that $f(\frac{x-3}{x+1})+f(\frac{x+3}{1-x})=x$ I am looking for functions $ f:\Bbb R \to \Bbb R $ satisfying
$$f\Big(\frac{x-3}{x+1}\Big)+f\Big(\frac{x+3}{1-x}\Big)=x$$
I used the substitution $ x=\cos(2t) $ for $ x\in (0,2\pi) $, to use the identities
$$1+x=2\cos^2(t) \text{ and } 1-x=2\sin^2(t)$$
The new equation will be
$$f\Big(1-\frac{2}{\cos^2(t)}\Big)+f\Big(\frac{2}{\sin^2(t)}-1\Big)=$$
$$\cos^2(t)-\sin^2(t)$$
I think that this approach won't allow me the get the answer.
Any idea will be appreciated.
|
Set $u= \frac{x-3}{x+1}$ the equation becomes $f(u)+f(\frac{u+3}{1-u})=\frac{u+3}{1-u}$
Set $v=\frac{x+3}{1-x}$ the equation becomes $f(\frac{v-3}{v+1})+f(v)=\frac{v-3}{v+1}$
Now replace all dummy variables with $y$ and solve for $f(x)$,
*
*$$A+B=y$$
2.$$B+C=\frac{y+3}{1-y}$$
*$$A+C=\frac{y-3}{y+1}$$
Can you finish it?
|
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"timestamp": "2023-03-29T00:00:00",
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|
How does $ 1 - \frac{1}{\sqrt{n+1}} + \frac{1}{(n+2)\sqrt{(n+1)} + (n+1) \sqrt{(n+2)}}$ reduce to $ 1 - \frac{1}{\sqrt{n+2}}\;$? $$ 1 - \frac{1}{\sqrt{n+1}} + \frac{1}{(n+2)\sqrt{(n+1)} + (n+1) \sqrt{(n+2)}}$$
Reduces to:
$$ 1 - \frac{1}{\sqrt{n+2}} $$
I have no clue how. What is the exact trick here and how can I practice this? I can't properly google the problem.
|
HINT
Here is a different approach for the sake of curiosity (as suggested by Paul in the comments).
Precisely, we shall multiply the numerator and the denominator by the conjugate of the denominator:
\begin{align*}
\frac{1}{(n+2)\sqrt{n+1} + (n+1)\sqrt{n+2}} & = \frac{(n+2)\sqrt{n+1} - (n+1)\sqrt{n+2}}{(n + 2)^{2}(n+1) - (n+1)^{2}(n+2)}\\\\
& = \frac{(n+2)\sqrt{n+1} - (n+1)\sqrt{n+2}}{(n+1)(n+2)}\\
& = \frac{1}{\sqrt{n+1}} - \frac{1}{\sqrt{n+2}}
\end{align*}
Can you take it from here?
|
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"timestamp": "2023-03-29T00:00:00",
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|
Let $a$ and $b$ be positive integers such that $an + 1$ s a cube if and only if $ bn + 1$ is a cube. Prove that $a = b.$ Let $a$ and $b$ be positive integers such that $an + 1$ is a cube if and only if $bn + 1$ is a cube. Prove that $a = b.$
By choosing $p_n^3 \equiv 1 \mod b$ , we find that there are infinitely many numbers $n$ such that $:bn+1=p_n^3\Rightarrow an+1=q_n^3$
So we have : $(a-b)n = q_n^3-p_n^3 \Rightarrow (a-b)$ is a divisor of an infinite number of numbers $T$ of the form$:T=p^3-q^3$
This is a quite hard problem for me. Any assistance would be appreciated.
|
As noted in the answer above by @richrow, the crux is showing that, if $a \not =b$, there is an integer $M$ such that $ab^2M^3+3abM^2+3aM+1$ is not a perfect cube. Note also that we may also assume that $a<b$. We give an alternative proof of this, that does not use the Kronecker Theorem.
Case 1: $ab^2$ is not a perfect cube. To this end first let $M_1,M_2,M_3\ldots$ be an infinite sequence of positive integers satisfying $\phi(M_k)$ not divisible by $3$ for each $k$, where $\phi(\cdot)$ is the Euler totient function; each $M_k$ a product of the first $k$ primes $p_1,p_2$ where each $p_i =2\pmod 3$ will do--or even simpler $M_k =2^k$ for each integer $k$ will work. Then that $3$ does not divide $\phi(M_k)$ implies the following:
For each $k$: If an integer $c$ satisfies $c^3 \equiv_1 M_k$, then $c \equiv_1 M_k$.
Thus, for each $k$, let us write $f(M_k)=ab^2M_k^3+3abM_k^2+3aM_k$. Then the following holds:
Claim 1: For each $k=1,2,\ldots$, if $f(M_k)$ is a perfect cube, then $f(M_k)$ must then be of the form $f(M_k) = (c_kM+1)^3$ for some integer $c_k$.
Furthermore, as $ab^2$ is integral and not a perfect cube, iy follows that the inequality $|c^3_k -ab^2| \ge 1$ holds. Also clearly $c_k \le ab^2+3ab+3a+1$ $=O(1)$.
Then suppose that each $f(M_k)$ is a perfect cube. Then by Claim 1, $f(M_k)=(c_kM_k+1)^3$ for each integer $k$. So on the one hand, the following is true:
$$f(M_k)-c^3_kM^3_k$$ $$= 3c^2_kM^2_k + 3c_kM_k +1$$
$$= O(M_k^2) \quad \forall k=1,2, \ldots.$$ However, the inequality $|ab^2-c^3_k| \ge 1$ holds for all $k$. So on the other hand, the following is also true:
$$|f(M_k)-c^3_kM^3_k| \ge M^3_k\pm O(M^2_k)$$ $$=\theta(M^3_k) \quad \forall k=1,2,3,\ldots.$$ Now as both of the above cannot be simultaneously true for an infinite number of $k$ however, so we arrive at a contradiction. So it follows that $f(M_k)$ can be a perfect cube for only finitely many of the $M_k$s, and thus there exists a $k$ such that $f(M_k)$ is not a perfect cube for the case where $ab^2$ is not a perfect cube. So the result follows if $a$ and $b$ satisfy Case 1.
Case 2: $ab^2$ is a perfect cube. Then assume $a<b$ and so writing $ab^2=c^3$, it follows that $c<b$ and so $3ab=3c^3/b <3c^2$, and similarly $3a < 3c$. So then for each integer $M$:
$$c^3M^3 < ab^2M^3 + 3abM^2+3aM+1$$ $$ < c^3M^3+3c^2M^2+3cM+1 = (cM+1)^3.$$
So here $ab^2M^3+3abM^2+3aM+1$ cannot be a perfect cube in the case where $ab^2$ is a perfect cube, so the result follows if $a$ and $b$ satisfy Case 2 as well.
|
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"timestamp": "2023-03-29T00:00:00",
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|
How to prove this inequality with the simplest means? $$ x^2+5y^2+6z^2 \geq 4\sqrt{5xyz^2},\ \text{ if }\ xy>0 $$
I was trying to prove it. The right hand side should be nonnegative, so I can square it on both sides.
But once I have done it I get to a point, where I do not see how to show, that the statement is greater or equal to $0.$
$$x^4+25y^4+36z^4+10x^2y^2+12x^2z^2+60y^2z^2-80xyz^2 \geq 0. $$
Thank you
|
Looks like the AM.GM inequality will by useful. We might as well assume $x,y>0$. (As $xy>0$ then either $x,y > 0$ or $x,y < 0$. If $x,y < 0$ then we can replace $x,y$ with $|x|, |y|$ and the result will be the same.... Or... I'll just use $|x|,|y|$).
By AM.GM we have $x^2 + 5y^2 \ge 2\sqrt{x^2\cdot 5y}=2|x||y|\sqrt 5= 2xy\sqrt 5$.
So $x^2 + 5y^2 + 6z^2 \ge 2xy\sqrt 5 + 6z^2$.
Apply AM.GM on $2xy\sqrt 5$ and $6z^2$ and we have $2xy\sqrt 5 +6z^2 \ge 2\sqrt{2xy\sqrt 5\cdot 6z^2}=2\sqrt{12xyz^2\sqrt 5}=4\sqrt{3\sqrt 5xyz^2}$.
And as $3 > \sqrt 5$ we have $4\sqrt{3\sqrt 5xyz^2}>4\sqrt{\sqrt 5^2xyz^2}=4\sqrt{5xyz^2}$
And that does it.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4361823",
"timestamp": "2023-03-29T00:00:00",
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|
Tiling of a deficient $7\times7$ chessboard with L trominoes
Prove that a $7\times7$ chessboard with one square removed can always be tiled by $L$ trominoes.
I'm looking for a reasonably simple proof. I was able to prove some specific cases, For instance, when the central square is deleted, the chessboard can be partitioned into four $4\times3$ rectangles, which can be easily tiled.
However, I was unable to prove the general case. Any help would be greatly appreciated.
|
Consider this (i am going draw some formal pictures after I have taken a nap.)
$$\begin{array}{|c|c|c|c|c|c|}
\hline
X&X&\circ&\circ&\triangle&\triangle&\circ\\
\hline
X&X&\circ&\square&\triangle&\circ&\circ\\
\hline
\circ&\circ&\square&\square&\blacksquare &\blacksquare &\square\\
\hline
\circ&\triangle&\circ&\circ&\blacksquare&\square&\square\\
\hline
\triangle&\triangle&\circ&\square&\square&\triangle&\triangle\\
\hline
\circ&\circ&\triangle&\square&\circ&\triangle&\square\\
\hline
\circ&\triangle&\triangle&\circ&\circ&\square&\square\\
\hline
\end{array}$$
$$\begin{array}{|c|c|c|c|c|c|}
\hline
\square&\square&\circ&\triangle&\triangle&\circ&\circ\\
\hline
\square&\circ&\circ&\triangle&\blacksquare&\blacksquare&\circ\\
\hline
X&X&\square&\square&\blacksquare &\square &\square\\
\hline
X&X&\circ&\square&\triangle&\triangle&\square\\
\hline
\triangle&\triangle&\circ&\circ&\triangle&\circ&\circ\\
\hline
\triangle&\circ&\triangle&\triangle&\square&\circ&\triangle\\
\hline
\circ&\circ&\triangle&\square&\square&\triangle&\triangle\\
\hline
\end{array}$$
$$\begin{array}{|c|c|c|c|c|c|}
\hline
\square&\square&\circ&\triangle&\triangle&\circ&\circ\\
\hline
\square&\circ&\circ&\triangle&\blacksquare&\blacksquare&\circ\\
\hline
\triangle&\triangle&X&X&\blacksquare &\square &\square\\
\hline
\triangle&\square&X&X&\triangle&\triangle&\square\\
\hline
\square&\square&\circ&\circ&\triangle&\circ&\circ\\
\hline
\circ&\circ&\triangle&\circ&\square&\circ&\triangle\\
\hline
\circ&\triangle&\triangle&\square&\square&\triangle&\triangle\\
\hline
\end{array}$$
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{
"language": "en",
"url": "https://math.stackexchange.com/questions/4362756",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "3",
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|
Finding integer solutions to $(x-y)^2+(y-z)^2+(z-x)^2=2022$ This was from round A of the Awesomemath Summer Program application. The deadline was yesterday and we can discuss the problems now.
Find all integer triples $(x,y,z)$ which satisfy
$$(x-y)^2+(y-z)^2+(z-x)^2=2022$$
I tried a couple of things. First I noticed $x-y+y-z+z-x=0,$ so if we let $a=x-y, b=y-z,c=z-x$ than we have
$$\begin{align}
a^2+b^2+c^2&=2022 \tag1\\
a+b+c &=0 \tag2\\
(a+b+c)^2-2ab-2bc-2ca &=2022 \tag3
\end{align}$$ So,
$$-2ab-2bc-2ca=2022 \tag4$$
so
$$ab+bc+ca=-1011 \tag5$$
However, it is tricky to proceed from here.
I also recalled the factorization
$$x^2+y^2+z^2-xy-yz-zx=\frac{1}{2}((x-y)^2+(y-z)^2+(z-x)^2)=1011 \tag6$$ This also equals $$x(x-y)+y(y-z)+z(z-x) \tag7$$
but this doesn't seem promising.
I also tried to factorize $2022=2 \cdot 3 \cdot 337$, but I am not sure what to do after prime factorizing.
So far, I feel like my first approach is the most promising, but I am not sure how to finish with it.
|
We have:
$$(x-y)^2+(y-z)^2+(z-x)^2=2(x^2+y^2+z^2)-2(xy+yz+zx)$$
$$(x+y+z)^2=x^2+y^2+z^2+2(xy+yz+zx)$$
Summing these relations we get:
$$(x-y)^2+(y-z)^2+(z-x)^2+(x+y+z)^2=3(x^2+y^2+z^2)$$
$(x-y)^2+(y-z)^2+(z-x)^2=3\times 2\times 337$
$\Rightarrow 3\times 2\times 337+(x+y+z)^2=3(x^2+y^2+z^2)\space\space\space\space\space\space(1)$
Which indicate that:
$3\big|(x+y+z)$
Relation (1) also says not all x , z and z can be a multiple of 3,because $2022$ has only 3 as a factor, but one of them, say x can be, so the sum of other two i.e y and z must be a multiple of 3. Also $\sqrt {2022}=44.96$. So $x<50$:
Now letting $x=3k$, equation (1) can be reduced as:
$3\times 2\times 337+(3k+y+z)^2=3[(3k)^2+y^2+z^2]\space\space\space\space\space\space (2)$
We value of x as:
$x=3k=3, 6, 9, 12, 15, 18, 30, 33, 36,39, 42, 45, 48$
Putting these in (2) we get an equation in terms of y and z with following restrictions:
$\begin{cases}3\big|y+z\\y<50, z<50\end{cases}$
which is easy to solve. Some examples:
$(x, y, z)=(3, 37, 32), (6,11, 40), (9,14, 43), (12, 7, 41), (15, 20, 49), (18, 13, 47), (30, 1, 35), (33, 38, 4), (36, 7, 41), (39, 10, 44), (42, 8, 37), (45, 11, 40), (48, 14, 43)$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Prove that $\sum \frac{a^3}{a^2+b^2}\le \frac12 \sum \frac{b^2}{a}$
Let $a,b,c>0$. Prove that
$$ \frac{a^3}{a^2+b^2}+\frac{b^3}{b^2+c^2}+\frac{c^3}{c^2+a^2}\le \frac12 \left(\frac{b^2}{a}+\frac{c^2}{b}+\frac{a^2}{c}\right).\tag{1}$$
A idea is to cancel the denominators, but in this case Muirhead don't work because the inequality is only cyclic, not symmetric.
Another idea would be to apply Cauchy reverse technique:
$$(1)\iff\sum \left(a-\frac{a^3}{a^2+b^2}\right)\ge \frac12 \sum (2a-b^2/a)\iff \sum\frac{ab^2}{a^2+b^2}\ge \frac12\sum\frac{2a^2-b^2}{a}$$
$$\iff \sum \frac{(ab)^2}{a^3+ab^2}\ge \frac12\sum \frac{2a^2-b^2}a.$$
Now we can apply Cauchy-Schwarz, and the problem reduces to
$$\frac{(\sum ab)^2}{\sum a^3+\sum ab^2}\ge \frac12\sum \frac{2a^2-b^2}{a},$$
and at this point I am stuck. Here the only idea is to cancel the denominators, but as I say above it can't work.
|
A funny solution (which seems correct): note that for $t>0$
$$
\frac{1}{1+t^2} \leq \frac{1}{2}t^2 + \frac{3}{2}(1-t).
$$
To prove this note first that $p(t) = (t^2 - 3t + 3)(t^2 + 1)$ has no real roots. Next, $p''(t) = 12t^2 -18t + 8 > 0$, implying that $p$ is strictly convex. Finally $p'(t) = 4t^3 - 9t^2 + 8t -3$, and $p'(1) = 0$ implying that $p$ has a minimum $2=p(1)\leq p(t)$.
Next, substitute $t\mapsto \frac{b}{a}$ and multiply both sides by $a$, then we get
$$
\frac{a^3}{a^2+b^2}\leq \frac{1}{2}\frac{b^2}{a} + \frac{3}{2}(a - b).
$$
Finally cyclically sum, and done.
|
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"timestamp": "2023-03-29T00:00:00",
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|
solving $\sin(5x)+\sin(x)=0$ using the sum formulas so symbolab says its just
$$2 \sin(3x) \cos(2x)$$
But after applying the angle sum formula I get
$$\sin(5x)+\sin(x)=\sin(3x+2x)+\sin(x)=\sin(3x)\cos(2x)+\sin(2x)\cos(3x)+\sin(x)$$
How does this reduce more?
|
Using the compound angle formula, $$
\sin A+\sin B=2 \sin \frac{A+B}{2} \cos \frac{A-B}{2},
$$
we have
$$
\begin{array}{l}
\sin 5 x+\sin x=0 \\
2 \sin 3 x \cos 2 x=0 \\
\sin 3 x=0 \text { or } \cos 2 x=0 \\
\displaystyle 3 x=n \pi \text { or } 2 x=n \pi+\frac{\pi}{2} \\
\displaystyle x=\frac{n \pi}{3} \text { or } \frac{(2 n+1) \pi}{4},
\end{array}
$$
where $n\in Z.$
|
{
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"url": "https://math.stackexchange.com/questions/4374025",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "1",
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|
Six people (half are female, half are male) for seven chairs. Problem:
Suppose there are $7$ chairs in a row. There are $6$ people that are going to randomly
sit in the chairs. There are $3$ females and $3$ males. What is the probability that
the first and last chairs have females sitting in them?
Answer:
Let $p$ be the probability we seek. Out of $3$ females, only $2$ can be sitting at the end of the row. I consider the first and last chairs to be at the end of the row.
\begin{align*}
p &= \dfrac{ {3 \choose 2 } 3(2) (4)(3)(2) } { 7(6)(5)(4)(3)(2) } \\
p &= \dfrac{ 3(3)(2) (4)(3)(2) } { 7(6)(5)(4)(3)(2) } \\
p &= \dfrac{ 3(4)(3)(2) } { 7(5)(4)(3)(2) } = \dfrac{ 3(3)(2) } { 7(5)(3)(2) } \\
p &= \dfrac{ 18 } { 35(3)(2) } \\
p &= \dfrac{ 3 } { 35 }
\end{align*}
Am I right?
Here is an updated solution.
Let $p$ be the probability we seek. Out of $3$ females, only $2$ can be sitting at the end of the row. I consider the first and last chairs to be at the end of the row.
\begin{align*}
p &= \dfrac{ 3(2) (5)(4)(3)(2) } { 7(6)(5)(4)(3)(2) } \\
p &= \dfrac{ (5)(4)(3)(2) } { 7(5)(4)(3)(2) } \\
p &= \dfrac{1}{7}
\end{align*}
Now is my answer right?
|
An alternative approach is that you have the outer seats both occupied with probability $\frac57$ and the conditional probability that they both have a woman is $$\frac{\binom32}{\binom62}=\frac15.$$
So the probability is $\frac57\cdot \frac15=\frac17.$
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
What is wrong with this projection onto two basis vectors? Given
\begin{align}
b_1 &= \begin{bmatrix} 2 \\ -1 \end{bmatrix} \\
b_2 &= \begin{bmatrix} 1 \\ 3 \end{bmatrix} \\
v &= \begin{bmatrix} 3 \\ 2 \end{bmatrix}
\end{align}
it is clear that
\begin{align}
v &= b_1 + b_2 \\
&= \begin{bmatrix} b_1 & b_2 \end{bmatrix} \begin{bmatrix} 1 \\ 1 \end{bmatrix}
\end{align}
However, I want to compute the vector $\begin{bmatrix} 1 \\ 1 \end{bmatrix}$ manually using projections. I know that
$$
v = \left(v \cdot \frac{b_1}{||b_1||}\right)\frac{b_1}{||b_1||} + \left(v \cdot \frac{b_2}{||b_2||}\right)\frac{b_2}{||b_2||}
$$
and so I expect that
\begin{align}
v \cdot \frac{b_1}{||b_1||^2} &= 1 \\
v \cdot \frac{b_2}{||b_2||^2} &= 1
\end{align}
However, what I get instead is
\begin{align}
v \cdot \frac{b_1}{||b_1||^2} &= \begin{bmatrix} 3 & 2 \end{bmatrix} \begin{bmatrix} \frac{2}{5} \\ \frac{-1}{5} \end{bmatrix} \\
&= \frac{4}{5} \\
v \cdot \frac{b_2}{||b_2||^2} &= \begin{bmatrix} 3 & 2 \end{bmatrix} \begin{bmatrix} \frac{1}{10} \\ \frac{3}{10} \end{bmatrix} \\
&= \frac{9}{10}
\end{align}
Where am I going wrong?
|
Your formula will only work when $b_1$ and $b_2$ are orthogonal. Here they are not, so you get an incorrect result. In general the projection matrix is $A(A^\top A)^{-1} A^\top$ where the columns of $A$ are $b_1$ and $b_2$.
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|
To find the number of ways to put $14$ identical balls into $4$ bins with the condition that no bin can hold more than $7$ balls. To find the number of ways to put $14$ identical balls into $4$ bins with the condition that no bin can hold more than $7$ balls.
I have tried the following:
The total no of ways to distribute $14$ identical balls into $4$ bins without any restriction is $$\binom{14+4-1}{4-1}= \binom{17}{3}.$$
Note that there can't be two bins with more than $7$ balls since we have only $14$ identical balls only.
Now, we count the no. of ways so that one bin has more than $7$ balls. So, it has at least $8$ balls and the remaining $6$ can be distributed in $\binom{6+4-1}{4-1}= \binom{9}{3}$ ways. We can choose one bin out of $4$ in $4$ ways.
Hence the reqd number of ways = $$\binom{17}{3} - 4 \times \binom{9}{3}.$$
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What you have is correct. Here’s a slightly alternative approach. You want to count the nonnegative integer solutions to $x_1+x_2+x_3+x_4=14$ such that $x_i\le 7$ for all $i$. By inclusion-exclusion and stars-and-bars, we have
$$\binom{14+4-1}{4-1}-\binom{4}{1}\sum_{k=8}^{14}\binom{14-k+3-1}{3-1}$$ solutions. By the hockey-stick identity, the sum reduces to $\binom{9}{3}$ as in your answer.
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Solve for theta , $0<\theta<90^\circ$
Solve $$\cos60=\frac{\cos(2θ)+\frac32\sin(2θ))}{\sqrt{4-\sin^2 (2θ)}},$$ where $0<θ<90^\circ $
Question: Is the following solution correct?
Solution attempt:
$$\sqrt{4-\sin^2 (2θ)}=2\cos(2θ)+3\sin(2θ)$$
$$4-\sin^2 (2θ)=(2\cos(2θ)+3\sin(2θ))^2$$
$$4-\sin^2 (2θ)=4\cos^2 (2θ)+12 \sin(2θ) \cos(2θ)+9\sin^2 (2θ)$$
$$4=4\cos^2 (2θ)+12\sin(2θ) \cos(2θ)+10\sin^2 (2θ)$$
$$4=4(1-\sin^2 (2θ) +12 \sin(2θ) \cos(2θ)+10\sin^2 (2θ)$$
$$6\sin^2 (2θ)+12 \sin(2θ) \cos(2θ)=0$$
$$\sin(2θ) (\sin(2θ)+2 \cos(2θ) )=0$$
$$\sin(2θ)=0 \text{ or } \sin(2θ)+2 \cos(2θ)=0$$
$$θ=0 \ \text{or} \ \tan(2θ)+2=0$$
$θ=0$ or $\tan^{-1} (-2)=-63.4$
$θ=0$ or $2θ=-63.4, \theta=-31.7$
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Noting that $\mathbf{0^\circ<2\theta<180^\circ}$ and continuing from
$$\sin(2θ)=0 \text{ or } \sin(2θ)+2 \cos(2θ)=0$$
$2\theta=\pi\;$ or $\;\big[\tan(2θ){=}-2\;$ or $\;\cos2\theta=0\big]$
$\displaystyle\theta=\frac\pi2\;$ or $\;\big[2\theta=180^\circ{-}\arctan2\;$ or $\displaystyle\;\frac\pi2\big]$
$\theta\displaystyle=\frac\pi4\;$ or $\displaystyle\;90^\circ{-}\frac12\arctan2\;$ or $\displaystyle\;\frac\pi2$
Of these three candidate solutions, only the middle one, $$58.3^\circ,$$ actually satisfies the given equation.
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$(n^1+n)/2$ sequence This is going to be hard to explain so I'll just give an example
Let's say we have a standard arithmetic sequence that goes up by 1 each time
1, 2, 3, 4, 5, 6, 7, 8, 9, 10
The last number (in this case, 10) is n
Scenario z:
Pick 2 numbers in the sequence and multiply them. The product should be equal to the rest of the numbers added together. For example, 6 and 7 (let's call 6 and 7 x and y for now) are between 1 and 10 and if multiplied together, we get 42. Add 1+2+3+4+5+8+9+10 (don't count the numbers that were multiplied together) and we also get 42.
n can only be an integer, no decimals involved here.
x and y has to be an integer between 0 and and n, and x cannot be equal to y
The question is, is there a mathematical way for figuring out the different numbers that can represent n that satisfies scenario z? And if n does satisfy the given scenario, is there a way to figure out x and y without having to try every possible combination?
$n^2+1$ numbers seems to always work (10, 17, 26, 37 etc), and I have no clue why.
I know I explained this pretty badly, feel free to ask any questions regarding this.
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$\left(n^2 + n + 2\right) \times \left(n^2 - n + 1\right) =
n^4 + 2n^2 - n + 2.$
Therefore
$$\left[\frac{n^4 + 3n^2 + 2}{2} - \frac{n^2 + n}{2}\right] \div \frac{n^2 + n + 2}{2} = (n^2 - n + 1).$$
$$S = \sum_{i=1}^{n^2 + 1} i = \frac{(n^2 + 1)(n^2 + 2)}{2} = \frac{n^4 + 3n^2 + 2}{2}.$$
So, for a particular value of $n$, you take $~\displaystyle y = \sum_{i=1}^n i = \frac{n^2 + n}{2}.$
Then,
$$\frac{S - y}{y+1} = \frac{n^4 + 2n^2 - n + 2}{n^2 + n + 2} = n^2 - n + 1.$$
Taking $n^2 - n + 1 = z$ you have that
$\displaystyle \frac{S - y}{y+1} = z \implies
S - y = yz + z \implies S = y + z + yz.$
Note that both $~\displaystyle y = \frac{n^2 + n}{2}$ and $z = (n^2 - n + 1)$
will be elements in $\{1,2,\cdots, (n^2 + 1)\}.$
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Find all intgers $a,b,c$ such that $a\cdot 4^n+b\cdot 6^n+c\cdot 9^n$ is a perfect square Find all intgers $a,b,c$ such that
$$a\cdot 4^n+b\cdot 6^n+c\cdot 9^n$$ is a perfect square for all sufficiently large $n$
This problem is creat by Vesselin Dimitrov,From the (Problems from the books chapter 17).
My approach:it is clear $a=1,b=2,c=1$ such it.lf let $a_{n}=a\cdot 4^n+b\cdot 6^n+c\cdot 9^n$,then we have
$$a_{n+3}=19a_{n+2}-114a_{n+1}+216a_{n}$$
and $a_{0}=a+b+c,a_{1}=4a+6b+9c,a_{2}=16a+36b+81c$, then this problem it suffices find $a,b,c$ such $a_{n}(n\to +\infty)$ is square
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We’re given that $x_n = \sqrt{a4^n + b6^n + c9^n}$ is an integer for sufficiently large $n$. We have
\begin{multline*}
(6x_n + 5x_{n+1} + x_{n+2})(6x_n + 5x_{n+1} - x_{n+2}) \\ (-6x_n + 5x_{n+1} + x_{n+2})(-6x_n + 5x_{n+1} - x_{n+2}) = 900(b^2 - 4ac)36^n,
\end{multline*}
and since $x_n = \sqrt c⋅3^n\left[1 ± O{\left(\left(\tfrac23\right)^n\right)}\right]$, we can divide this by
\begin{align*}
6x_n + 5x_{n+1} + x_{n+2} &= 30\sqrt c⋅3^n\left[1 ± O{\left(\left(\tfrac23\right)^n\right)}\right], \\
6x_n + 5x_{n+1} - x_{n+2} &= 12\sqrt c⋅3^n\left[1 ± O{\left(\left(\tfrac23\right)^n\right)}\right], \\
-6x_n + 5x_{n+1} + x_{n+2} &= 18\sqrt c⋅3^n\left[1 ± O{\left(\left(\tfrac23\right)^n\right)}\right],
\end{align*}
leaving
$$y_n = -6x_n + 5x_{n+1} - x_{n+2} = k\left(\tfrac43\right)^n\left[1 ± O{\left(\left(\tfrac23\right)^n\right)}\right], \quad \text{where } k = \frac{5(b^2 - 4ac)}{36\sqrt{c^3}}.$$
So $y_n - k\left(\frac43\right)^n = ±O{\left(\left(\tfrac89\right)^n\right)}$ converges to zero. Hence the integer $4y_n - 3y_{n+1}$ converges to zero and must eventually remain zero. Since a nonzero integer is only divisible by finitely many powers of $3$, this can only happen if $y_n$ eventually remains zero, $k = 0$, and $b^2 = 4ac$.
We can therefore write $x_n = \sqrt{\frac{b^2}{4c}4^n + b6^n + c9^n} = \frac{\lvert b2^{n-1} + c3^n\rvert}{\sqrt c}$. If this is to be an integer, $c$ must be a perfect square $s^2$ and $a = \frac{b^2}{4c}$ must be a perfect square $r^2$. So the solution must be of the form $x_n = \lvert r2^n + s3^n\rvert$, $a = r^2$, $b = 2rs$, $c = s^2$ for integers $r, s$, which always works.
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|
Proof that $\exp(x)-1 > x^2 ,\forall x>0$ I'm trying to prove that
$\exp(x)-1 > x^2 ,\forall x>0$
What I've tried so far:
for $0<x<1$ I know that the equation is valid because:
$\exp(x)-1= x + \frac{x^2}{2!} + \frac{x^3}{3!} + ...$
so considering only the first term ($x$) I compared $x^2$ with it by subtracting both
$x^2-x = x(x-1)$
so for $0<x<1$ this is always positive.
Then I tried to compare the first four terms of $\exp(x)$ with $x^2$ by dividing it by $x^2$ trying to find where it is greater than one
$\exp(x)-1 > x + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!}$
$\frac{(x + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!})}{x^2} = \frac{1}{x} + \frac{1}{2} + \frac{x}{6} + \frac{x^2}{24}$
I ignored the first term $\frac{1}{x}$ because if the rest of the equation was already bigger than $1$ for $x>1$ then the problem would be solved.
Solving the equation:
$\frac{1}{2} + \frac{x}{6} + \frac{x^2}{24} > 1$ with $x>0$
$\frac{x^2}{24} + \frac{x}{6} - \frac{1}{2}> 0$ with $x>0$
Its roots are at $x=-6$ and $x=2$. So for $x>2$ the equation is also valid
I still do not know how to prove that $\exp(x)-1>x^2$ for $1<x<2$.
Edit: the solution from rogerl is much better than mine, but just for completeness using the comment from JetfiRex
considering the first two terms:
$x + \frac{x^2}{2!} > x^2$
$x - \frac{x^2}{2} > 0$
$x(1 - \frac{x}{2}) > 0$
x=0 and x=2 are solutions of this downward parabola, so the equation is also valid from 0 to 2
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$x+\frac{x^2}{2} + \frac{x^3}{6} > x^2$ reduces to
$$\frac{1}{6}x(x^2-3x+6) > 0,$$
which is true for all $x>0$ since the discriminant of the quadratic is negative.
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How to solve simultaneous equations of the form $\frac{a}{x}+\frac{b}{y}=1$ really fast? Context:
*
*Taking the major and minor axes of an ellipse as the $x$ and $y$ axes respectively find the equation of the ellipse passing through the points $(1,\sqrt{6})$ and $(3,0)$.
*Taking the major and minor axes of an ellipse as the $x$ and $y$ axes respectively find the equation of the ellipse passing through the points $(2,4)$ and $(5,\sqrt{2})$.
In problems like these, I frequently find myself solving simultaneous equations of the form $\frac{a}{x}+\frac{b}{y}=1$. For example, when doing math 2, I had to solve the following system of equations (I assumed $p=a^2$, $q=b^2$).
$$\frac{4}{p}+\frac{16}{q}=1\tag{1}$$
$$\frac{25}{p}+\frac{2}{q}=1\tag{2}$$
I'll now show how I solved them.
$$\frac{4}{p}+\frac{16}{q}=1\tag{1}$$
$$\frac{25}{p}+\frac{2}{q}=1\tag{2}$$
Now in $(1)$,
$$\frac{4q+16p}{pq}=1$$
$$4q+16p=pq$$
Again in $(2)$,
$$\frac{25}{p}+\frac{2}{q}$$
$$\frac{25q+2p}{pq}=1$$
$$25q+2p=pq$$
Now,
$$4q+16p=25q+2p$$
$$21q-14p=0$$
$$21q=14p$$
$$q=\frac{14p}{21}$$
Now,
$$\frac{4}{p}+\frac{16}{q}=1$$
$$\frac{4}{p}+\frac{16}{\frac{14p}{21}}=1$$
$$\frac{4}{p}+\frac{16}{\frac{14p}{21}}=1$$
$$\frac{4}{p}+\frac{336}{14p}=1$$
$$\frac{56+336}{14p}=1$$
$$14p=392$$
$$p=28$$
Now,
$$\frac{4}{28}+\frac{16}{q}=1$$
$$\frac{1}{7}+\frac{16}{q}=1$$
$$\frac{16}{q}=\frac{6}{7}$$
$$6q=112$$
$$q=\frac{56}{3}$$
So, the equation of the ellipse is $$\frac{x^2}{28}+\frac{y^2}{\frac{56}{3}}=1\ \text{(Ans.)}$$
As you can see, the process is pretty long and time-taking. Is there any way to find the values of $p,q$ in 2-3 lines?
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Use the substitution $x = \dfrac{1}{p} , y = \dfrac{1}{q} $, then your equations become:
$ 4 x + 16 y = 1 $
$ 25 x + 2 y = 1 $
And these can solved very easily, using for example, elimination
$(1) - 8 \times (2) $:
$- 196 x = - 7 \Rightarrow x = \dfrac{7}{196} \Rightarrow p = \dfrac{196}{7} = 28 $
And then, from equation $(2)$,
$y = \dfrac{1}{2} ( 1 - 25 x ) = \dfrac{3}{56} $
Therefore, $q = \dfrac{56}{3}$
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|
$y'=2-\frac{3}{x}y+\frac{2}{x^2}y^2$ (Riccati) $y'=2-\frac{3}{x}y+\frac{2}{x^2}y^2$ (Riccati)
(a) Find the solutions.
(b) $y(x_0)=y_0$, prove two cases:
$$0<y_0<x_0 \implies \text{solution's domain is} [x_0,\infty) $$
$$0<x_0<y_0 \implies \text{solution's domain is} [x_0,x_0+\alpha) , \alpha \in \mathbb {R}.$$
I will be grateful for help in $(b)$.
My solution for (a):
$y_1(x)=x$
Denote $y=x+z$
$$(x+z)'=2-\frac{3}{x}(x+z)+\frac{2}{x^2}(x+z)^2$$
$$1+z'=-1-\frac{3z}{x}+\frac{2}{x^2}(x^2+2xz+z^2)$$
$$z'=\frac{z}{x}+\frac{2z^2}{x^2}$$
This is a Bernoulli differential Equation.
Denote $u=z^{-1} \implies -\frac{z'}{z^2} \implies z'=-u'z^2$
Then,
$$u'=\frac{u}{x}-\frac{2}{x^2}$$
Integration factor is $\mu=x$
$$\int(xu)'=\int-\frac{2}{x} \implies xu=-2\ln|x|+c \implies u=\frac{-2\ln|x|+c}{x} \implies z=\frac{x}{-2\ln|x|+c}$$
The solution is $y=x+\frac{x}{-2\ln|x|+c}$
Is my solution correct ?
Thanks !
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$$z'=\frac{z}{x}+\frac{2z^2}{x^2}$$
$$\dfrac {z'}{z^2}=\frac{1}{zx}+\frac{2}{x^2}$$
$$-\left(\dfrac {1}{z}\right)'=\frac{1}{zx}+\frac{2}{x^2}$$
Looks like there is a little sign mistake. But all what you did looks really good.
$$u'=\color{red}{-\frac{u}{x}}-\frac{2}{x^2}$$
$$(ux)'=-\frac{2}{x}$$
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Minimize $xy$ over $x^2+y^2+z^2=7$ and $xy+xz+yz=4$.
Let $x,$ $y,$ and $z$ be real numbers such that $x^2 + y^2 + z^2 = 7$ and $xy + xz + yz = 4.$ Find the smallest possible value of $xy.$
I used Cauchy to get $$(x^2+y^2+z^2)(1^2+1^2+1^2)\geq (x+y+z)^2$$ and $$(x^2+y^2+z^2)(y^2+z^2+x^2)\geq(xy+xz+yz)^2,$$ but this doesn't do much as both of the inequalities already work. :( Any guidance would be appreciated!!
Thanks in advance!!!
P.S. I've been given that this can be solved by Lagrange multipliers, but I haven't learned it yet. Hopefully, someone can give me hints on a method that doesn't involve Lagrange multipliers. :)
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I would start with noticing that
\begin{align*}
(x + y + z)^{2} = x^{2} + y^{2} + z^{2} + 2(xy + xz + yz) = 15
\end{align*}
Therefore $x + y + z = \pm\sqrt{15} =: k$. Better saying, $z = k - x - y$.
Then we arrive at the relation
\begin{align*}
xy + xz + yz & = xy + z(x + y)\\\\
& = xy + (k - (x + y))(x + y)\\\\
& = xy + k(x + y) - (x + y)^{2} = 4\\\\
& \Rightarrow xy = (x + y)^{2} - k(x + y) + 4
\end{align*}
Let us make the change of variable $t = x + y$.
Then we are interested in the minimum value of the RHS above, which is given by $1/4$ as follows:
\begin{align*}
t^{2} - kt + 4 & = \left(t^{2} - kt + \frac{k^{2}}{4}\right) + 4 - \frac{k^{2}}{4}\\\\
& = \left(t - \frac{k}{2}\right)^{2} + 4 - \frac{15}{4} \geq \frac{1}{4}
\end{align*}
Hence the minimum value of $xy$ satisfying the proposed constraints equals $1/4$.
Hopefully this helps !
EDIT
As an answer to the (well posed) comment of @RobPratt, we must show such value is attained.
Indeed, this is the case as the following system of equations show:
\begin{align*}
\begin{cases}
xy = \dfrac{1}{4}\\\\
x + y = \dfrac{\sqrt{15}}{2}
\end{cases} & \Longleftrightarrow
\begin{cases}
x = \dfrac{\sqrt{15} \mp \sqrt{11}}{4}\\\\
y = \dfrac{\sqrt{15} \pm \sqrt{11}}{4}
\end{cases}
\end{align*}
as well as the following system of equations:
\begin{align*}
\begin{cases}
xy = \dfrac{1}{4}\\\\
x + y = -\dfrac{\sqrt{15}}{2}
\end{cases} & \Longleftrightarrow
\begin{cases}
x = -\dfrac{\sqrt{15} \mp \sqrt{11}}{4}\\\\
y = -\dfrac{\sqrt{15} \pm \sqrt{11}}{4}
\end{cases}
\end{align*}
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What am I doing wrong with this Euler substitution? I want to evaluate the integral $\displaystyle\int\frac{1}{x\sqrt{x^2+2}}dx$ using the substitution $\sqrt{x^2+2}=-x+t$.
So I get from the substitution $x=\frac{t^2-2}{2t}$, $dx=\frac{t^2+2}{2t^2}dt$ and $\sqrt{x^2+2}=\frac{t^2+2}{2t}$. Substituting all this in I get:
$\displaystyle\int\frac{1}{x\sqrt{x^2+2}}dx=\int\frac{1}{\frac{t^2-2}{2t}\frac{t^2+2}{2t}}\frac{t^2+2}{2t^2}dt$
$\displaystyle= 2\int\frac{1}{t^2-2}dt$
$\displaystyle=\frac{1}{\sqrt{2}}\int\frac{1}{t-\sqrt{2}}-\frac{1}{t+\sqrt{2}}dt$
$\displaystyle=\frac{1}{\sqrt{2}}\ln{\left(\frac{t-\sqrt{2}}{t+\sqrt{2}}\right)}+c$
$\displaystyle=\frac{1}{\sqrt{2}}\ln{\left(\frac{x+\sqrt{x^2+2}-\sqrt{2}}{x+\sqrt{x^2+2}+\sqrt{2}}\right)}+c$
However the answer on Wolfram Alpha seems to be $\displaystyle=\frac{1}{2\sqrt{2}}\ln{\left(\frac{\sqrt{x^2+2}-\sqrt{2}}{\sqrt{x^2+2}+\sqrt{2}}\right)}+c$. I'm am completely new to these Euler substitutions so could someone check if I'd done it right?
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Both the answers are correct as $
{\left(\frac{x+\sqrt{x^2+2}-\sqrt{2}}{x+\sqrt{x^2+2}+\sqrt{2}}\right)}^2={\left(\frac{x^2+2+x\sqrt{x^2+2}-\sqrt{2x^2+4}-\sqrt{2}x}{x^2+2+x\sqrt{x^2+2}+\sqrt{2x^2+4}+\sqrt{2}x}\right)}={\left(\frac{\sqrt{x^2+2}-\sqrt{2}}{\sqrt{x^2+2}+\sqrt{2}}\right)}$
It can be proved by using compendo-dividendo: ($\frac{a}{b}=\frac{c}{d}\implies \frac{a+b}{a-b}=\frac{c+d}{c-d})$
${\left(\frac{x^2+2+x\sqrt{x^2+2}-\sqrt{2x^2+4}-\sqrt{2}x}{x^2+2+x\sqrt{x^2+2}+\sqrt{2x^2+4}+\sqrt{2}x}\right)}={\left(\frac{\sqrt{x^2+2}-\sqrt{2}}{\sqrt{x^2+2}+\sqrt{2}}\right)}$
$\implies $
${\left(\frac{x^2+2+x\sqrt{x^2+2}}{\sqrt{2x^2+4}+\sqrt{2}x}\right)}={\left(\frac{\sqrt{x^2+2}}{\sqrt{2}}\right)}$
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How do I determine the value of $d$ from the tangent graph of the form $y = a\times \tan (bx + c) + d$?
I understand that $d$ is the vertical shift, and I have already worked out the values of $a$, $b$ and $c$ to be the following (please tell me if I am wrong):
$\pmb{a}\quad$ Given point is $(-1, -7)$ so I substituted it in as the values of $x$ and $y$ ; $-7= a\times \tan\left(\dfrac{\pi}{4}\left(-1\right)+5\right)$ which I solved for $a$ and that gave me$\space$ -3.8040.
$\pmb{b}\quad$ Period so $\dfrac{\pi}{4}$
$\pmb{c}\quad$ $y$-intercept so $5$
But how do I find $d$ for this type of graph with only vertical asymptotes?
|
You are correct in that $b = \frac{\pi}{4}$, but your reasoning isn't so clear. The function's period is $4$, and since $f(x) = \tan (\alpha x)$ has period $\pi / \alpha$, $b$ is $\frac{\pi}{4}$ by simple arithmetic.
So far we have $f(x) = a\tan(\frac{\pi}{4}x + c) + d$.
The function $g(x) = \tan(\frac{\pi}{4}x)$ has asymptotes at $x = 4n + 2$ e.g $x = 2$ where $n$ is an integer, while $f(x)$ has asymptotes at $x = 4n +1$. This means in the simplest case that $f(x)$ is $\tan(\frac{\pi}{4}x)$ translated $1$ unit to the left.
Hence $$ f(x) = a\tan\left(\frac{\pi}{4}(x+1)\right) + d \tag{1}\label{eq1}$$
Substitute the value $f(-1) = -7$ into \eqref{eq1} to obtain that $d = -7$, and substitute $f(0) = -5$ to obtain $a = 2$.
Finally we arrive at
$$f(x) = 2 \tan \left(\frac{\pi}{4}x + \frac{\pi}{4}\right) -7$$
I hope this answer was clear.
This is my first time giving an answer on this website, if you are a more experienced user and have any constructive criticisms feel free to share them with me.
|
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|
Finding max{f(x)} without derivative Consider the function $f(x)=\frac{\sqrt{x^3+x}}{x^2+x+1}$
the question is about: to find max{f} without using derivative.
I can find max with derivation and it is not hard to find. it is $f'(x)=0 \to x=1 $ so $max\{f\}=\frac{\sqrt 2}{3}$
but I am looking for an idea to do as the question said.
I am thankful for any help.(because I got stuck on this problem)
|
Using AM-GM, we have
\begin{align*}
\frac{\sqrt{x^3 + x}}{x^2 + x + 1} &= \frac{\sqrt{2}\sqrt x \sqrt{\frac{x^2 + 1}{2}}}{x^2 + x + 1}\\
&\le \frac{\sqrt{2}\cdot\frac12\left(x + \frac{x^2 + 1}{2}\right)}{x^2 + x + 1}\\
&= \frac{\sqrt 2 (x^2 + 2x + 1)}{4(x^2 + x + 1)}\\
&= \frac{\sqrt{2}}{3} - \frac{\sqrt 2 (x - 1)^2}{12(x^2 + x + 1)}\\
&\le \frac{\sqrt{2}}{3}.
\end{align*}
Also, when $x = 1$,
we have $\frac{\sqrt{x^3 + x}}{x^2 + x + 1} = \frac{\sqrt 2}{3}$.
Thus, the maximum of $\frac{\sqrt{x^3 + x}}{x^2 + x + 1}$ is $\frac{\sqrt 2}{3}$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
How to evaluate $\sum_{n=0}^{\infty} \frac{(-1)^{n}}{12n+1} $? First of all, I am going to convert this series into a definite integral.
For $|x|<1,$ we have
$\displaystyle \quad \frac{1}{1+x^{12}}=\displaystyle \sum_{n=0}^{\infty}\left(-x^{12}\right)^{n}=\displaystyle \sum_{n=0}^{\infty}(-1)^{n} x^{12 n} \tag*{} $
Integrating both sides from $x=0$ to $1$, we can relate the target series and the definite integral.
$$\displaystyle \int_{0}^{1} \frac{d x}{1+x^{12}} =\displaystyle \sum_{n=0}^{\infty} \int_{0}^{1}(-1)^{n} x^{12 n} d x
\displaystyle =\sum_{n=0}^{\infty}\left[(-1)^{n} \frac{x^{12 n+1}}{12n+1}\right]_{0}^{1} \displaystyle =\sum_{n=0}^{\infty} \frac{(-1)^{n}}{12 n+1}$$
By my post,
$\displaystyle \sum_{n=0}^{\infty} \frac{(-1)^{n}}{12 n+1}$
$\displaystyle \\ \displaystyle =\frac{1}{24 \sqrt{2}}\left[2 \ln \left(\frac{2+\sqrt{2}}{2-\sqrt{2}}\right)+(1+\sqrt{3}) \ln \left(\frac{2 \sqrt{2}+\sqrt{3}+1}{2 \sqrt{2}-\sqrt{3}-1}\right)\right.\\$
$\displaystyle \qquad\qquad \qquad\qquad \left.+(\sqrt{3}-1) \ln \left(\frac{2 \sqrt{2}+\sqrt{3}-1}{2 \sqrt{2}-\sqrt{3}+1}\right)+2(1+\sqrt{3}) \pi\right] \\$
(checked by Wolframalpha)
My question
How to generalize the result for
$$\sum_{k=0}^{\infty} \frac{(-1)^{k}}{k n+1}$$
where $n\in \mathbb N$?
|
Similar to the original integral, using the same technique yields
$\displaystyle \quad \frac{1}{1+x^{n}}=\displaystyle \sum_{k=0}^{\infty}\left(-x^{n}\right)^{k}=\displaystyle \sum_{k=0}^{\infty}(-1)^{k} x^{nk} ,\tag*{}
$
we have
Integrating both sides from $x=0$ to $1$, we can relate the target series and the definite integral.
$$\displaystyle \int_{0}^{1} \frac{d x}{1+x^{n}} =\displaystyle \sum_{k=0}^{\infty} \int_{0}^{1}(-1)^{k} x^{nk} d x
\displaystyle =\sum_{k=0}^{\infty}\left[(-1)^{k} \frac{x^{nk+1}}{nk+1}\right]_{0}^{1} \displaystyle =\sum_{k=0}^{\infty} \frac{(-1)^{k}}{nk+1}$$
By the Dr Brian Sittinger’s post,
A. when $n=2m$ is even, we obtain the series$$
\sum_{k=0}^{\infty} \frac{(-1)^{k}}{k n+1}\\=\boxed{\frac{1}{2 m} \sum_{k=0}^{m-1}\left[\frac{(2 k+1) \pi}{2 m} \sin \left(\frac{(2 k+1) \pi}{2 m}\right)-\cos \left(\frac{(2 k+1) \pi}{2 m}\right) \ln \left(2-2 \cos \left(\frac{(2 k+1) \pi}{2 m}\right)\right)\right]}
$$
B. when $n=2m+1$ is odd, we obtain the series$$
\sum_{k=0}^{\infty} \frac{(-1)^{k}}{k n+1}\\ =\boxed{\frac{\ln 2}{2 m+1}+\frac{1}{2 m+1} \sum_{k=0}^{m-1}\left[\frac{(2 k+1) \pi}{2 m+1} \sin \left(\frac{(2 k+1) \pi}{2 m+1}\right)-\cos \left(\frac{(2 k+1) \pi}{2 m+1}\right) \ln \left(2-2 \cos \left(\frac{(2 k+1) \pi}{2 m+1}\right)\right)\\+2 \sin \left(\frac{(2 k+1) \pi}{2 m+1}\right) \cdot \arctan \left(\cot \left(\frac{(2 k+1) \pi}{2 m+1}\right)\right)\right]}
$$
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"language": "en",
"url": "https://math.stackexchange.com/questions/4404528",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Volume enclosed by the surface $\left(\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}\right)^2=\frac{x^2}{a^2}+\frac{y^2}{b^2}-\frac{z^2}{c^2}.$
Compute the volume enclosed by the surface $$\left(\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}\right)^2=\frac{x^2}{a^2}+\frac{y^2}{b^2}-\frac{z^2}{c^2}.$$
My attempt:
I used the following change of coordinates:
$$\begin{aligned}x&=ar\sin\theta\cos\varphi\\y&=br\sin\theta\sin\varphi\\z&=cr\cos\theta,\\\varphi&\in[0,2\pi]\\r&\in[0,1]\\\text{ Jacobian: }J_\psi&=abcr^2\sin\theta.\end{aligned}$$
The equation of the curve becomes: $$\begin{aligned}r^4=r^2-2\frac{c^2r^2\cos^2\theta}{c^2}\\\iff r^2=1-2\cos^2\theta\\r^2=-\cos(2\theta)\end{aligned}$$ so $$0\le r^2\le -\cos(2\theta)\\\implies 0\ge -r^2\ge\cos(2\theta)\ge -1\\\implies \theta\in\left[-\pi,-\frac{\arccos(-r^2)}2\right]\cup\left[\frac{\arccos(-r^2)}2,\pi\right],$$ but $\theta$ must be in the interval $[0,\pi],$ so I eliminated the first possibility. I get the integral $$\begin{aligned}\int_0^{2\pi}\int_0^1\int_{\arccos(-r^2)/2}^\pi abcr^2\sin\theta d\theta drd\varphi&=2abc\pi\int_0^1 r^2\left(-\cos\theta\Big|_{\arccos(-r^2)/2}^\pi\right)dr\\&=-2abc\pi\int_0^1r^2\left(-1-\cos\left(\frac{\arccos(-r^2)}2\right)\right) dr\\&=2abc\pi\int_0^1\left(r^2+r^2\sqrt{\frac{1-\cos(\arccos(-r^2))}2}\right)dr\\&=2abc\pi\left(\int_0^1r^2dr+\frac1{\sqrt 2}\int_0^1r^2\sqrt{1\color{red}+r^2}dr\right)\\&=2abc\pi\left(\frac13+\frac1{\sqrt 2}\frac13 \int_0^1r\sqrt{1+r^2}3rdr\right)\\&=2abc\pi\left(\frac13+\frac1{3\sqrt 2} r\left(1+r^2\right)^{3/2}\Big|_0^1-\int_0^1(1+r^2)^{3/2}dr\right)\\&=2abc\pi\left(1-\frac1{3\sqrt 2}\int_0^1(1+r^2)^{3/2}dr\right)\end{aligned}$$
EDIT:
I found one mistake (marked now with red plus, I had put minus instead). Now, if I proceed with substitution $r=\tan\alpha,$ the procedure continues. Is there a way to avoid all of this?
I compared my procedure with this and mine looks suspicious. I thought this integral would be easier to compute if I changed the order of integration in $\theta$ and $r$. Can anybody tell me what I did wrong?
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$$\cos\left(\pi-\frac12\cos^{-1}(-\rho^2)\right) = $$
If $0 \le \rho^2 \le -\cos(2\theta)$, then $\frac{(4n-3)\pi}4 \le \theta \le \frac{(4n-1)\pi}4 \text{ for } n\in\mathbb Z$.
The polar angle must generally be between $0$ and $\pi$, so we take $\theta\in\left[\frac\pi4,\frac{3\pi}4\right]$. Then the volume is
$$\int_0^{2\pi} \int_{\frac\pi4}^{\frac{3\pi}4} \int_0^{\sqrt{-\cos(2\theta)}} abc\rho^2 \sin(\theta) \, d\rho \, d\theta \, d\varphi = \frac{abc\pi^2}{4\sqrt2}$$
Or, preserving the order you chose, we have
$$\rho^2 = -\cos(2\theta) \implies \theta = \frac12 \cos^{-1}(-\rho^2)$$
but this is only true if $\theta\in\left[0,\frac\pi2\right]$. For $\theta\in\left[\frac\pi2,\pi\right]$, we instead would have
$$\rho^2 = -\cos(2\theta) \implies \theta = \pi - \frac12 \cos^{-1}(-\rho^2)$$
Now, if $\theta\le\frac\pi2$, then
$$\rho=0 \implies \theta = \frac\pi4 \text{ and } \rho=1\implies\theta=\frac\pi2$$
Otherwise, if $\frac\pi2<\theta$, then
$$\rho=0\implies\theta=\frac{3\pi}4 \text{ and } \rho=1\implies\theta=\frac\pi2$$
All this tells us that if $\rho\in[0,1]$, then $\theta\in\left[\frac12\cos^{-1}(-\rho^2),\pi-\frac12\cos^{-1}(-\rho^2)\right]$.
The subsequent integral agrees with the previous result:
$$\begin{align}
&\int_0^{2\pi} \int_0^1 \int_{\frac12\cos^{-1}(-\rho^2)}^{\pi-\frac12\cos^{-1}(-\rho^2)} abc\rho^2\sin(\theta) \, d\theta \, d\rho \, d\varphi \\[1ex]
&= 2abc\pi \int_0^1 \rho^2\left(\cos\left(\frac12\cos^{-1}(-\rho^2)\right) - \cos\left(\pi-\frac12\cos^{-1}(-\rho^2)\right)\right) \, d\rho \\[1ex]
&= 4abc\pi \int_0^1 \rho^2\cos\left(\frac12\cos^{-1}(-\rho^2)\right) \, d\rho \\[1ex]
&= 2\sqrt2\,abc\pi \int_0^1 \rho^2 \sqrt{1-\rho^2} \, d\rho \\[1ex]
&= 2\sqrt2\,abc\pi \times \frac\pi{16} = \frac{abc\pi^2}{4\sqrt2}
\end{align}$$
|
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|
Taylor series of $\sqrt{1-x}$ This is problem 6.7.6 in Abbott's Understanding Analysis 2nd ed. The section is called The Weierstrass approximation theorem.
a) Let $c_{n} = \frac{1\cdot 3\cdot 5\cdots (2n-1)}{2\cdot 4\cdot 6\cdots 2n}$ for $n\geq 1$. Show $c_{n} < \frac{2}{\sqrt{2n+1}}$.
b) Use a) to show that $\sum_{n=0}^{\infty}a_{n}$ converges (absolutely, in fact) where $a_{n}$ is the sequence of Taylor coefficients generated in Exercise 6.7.4. $\left[a_{n}=\frac{\prod_{k=1}^{n}(2k-3)}{\prod_{k=1}^{n}(2k)}=\frac{-1\cdot 3\cdot 5\cdots (2n-3)}{2\cdot 4\cdot 6\cdots 2n}\right]$
c) Carefully, explain how this verifies that equation (1) $\left[\sqrt{1-x}=\sum_{n=0}^{\infty}a_{n}x^{n}\right]$ holds for all $x\in [-1,1]$.
With help of this MSE answer I was able to answer a). For c) I think the point is that if $\sum_{n=0}^{\infty}a_{n}$ converges absolutely, then the power series converges too, because $\sqrt{1-x}=\sum_{n=0}^{\infty}a_{n}x^{n}=\sum_{n=0}^{\infty}a_{n}$ at $x=1$. And for $x=-1$ we could use the alternating series to prove that $\sum_{n=0}^{\infty}a_{n}(-1)^{n}$ converges. I have already proved in a previous exercise that it converges for $x=(-1,1)$.
My problem is with b). I suppose that Abbott wants us to use the Weierstrass M-test and maybe a comparison test. However, $\sum_{n=0}^{\infty}c_{n}$ diverges, so I don't know how to go from there. In other words, I feel I didn't get much information from a). What am I missing?
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Using the hint given by Gary in the comments of the question and the MSE linked in the question, I will answer the whole exercise.
a) First, let's start with a lemma.
For all $k \in \mathbb{N}$, $\frac{k}{k+1} < \frac{k+1}{k+2}$ holds.
Proof:
We can write $k^{2} +2k < k^{2} + 2k + 1$. This implies $k(k+2) < (k+1)^{2}$. Dividing both sides of the inequalities by $k+2$ and $k+1$ gives us the desired result.
Proof of the main result:
Consider
$
\begin{align}
c_{n}^{2} &= \frac{\prod^{n}_{k=1}(2k-1)^2}{\prod^{n}_{k=1}(2k)^2} \\
&< \frac{\prod^{n}_{k=1}(2k-1)(2k)}{\prod^{n}_{k=1}(2k)(2k+1)} \quad \text{by the lemma}\\
&= \frac{\prod^{n}_{k=1}(2k-1)}{\prod^{n}_{k=1}(2k+1)} = \frac{1}{2n+1}
\end{align}
$
So $c_{n} < \frac{1}{\sqrt{2n+1}} < \frac{2}{\sqrt{2n+1}}$, as desired.
b) We can write $a_{n} = -\frac{c_{n}}{2n-1}$, because $\frac{c_{n}}{2n-1} = \frac{\prod^{n}_{k=1}(2k-1)}{\prod^{n}_{k=1}(2k)}\frac{1}{2n-1} = \frac{\prod^{n}_{k=1}(2k-3)}{\prod^{n}_{k=1}(2k)} = -a_{n}$.
So for all $n$, we have $a_{n} < \frac{1}{(2n-1)\sqrt{2n+1}} < b_{n} := \frac{1}{\sqrt{2n(2n-1)^{2}}}$. Let's introduce a third sequence $c_{n} = \frac{1}{\sqrt{2}n^{3/2}}$. If we prove that $\sum_{n=0}^{\infty} b_{n}$ converges, then by the comparison test $\sum_{n=0}^{\infty} a_{n}$ converges too.
We prove that $\sum_{n=0}^{\infty} b_{n}$ converges using the limit comparison test:
$
\begin{align}
\lim_{n\rightarrow \infty} \left\lvert \frac{c_{n}}{b_{n}} \right\rvert
&= \lim_{n\rightarrow \infty} \left\lvert \frac{\sqrt{2n(2n-1)^{2}}}{\sqrt{2}n^{3/2}}\right\rvert \\
&= \lim_{n\rightarrow \infty} \left\lvert \frac{\sqrt{2n(2n-1)^{2}/n^{3}}}{\sqrt{2}n^{3/2}/n^{3/2}}\right\rvert \\
&= \lim_{n\rightarrow \infty} \left\lvert \frac{\sqrt{2n(4n^{2} - 4n+1)/n^{3}}}{\sqrt{2}}\right\rvert \\
&= \lim_{n\rightarrow \infty} \left\lvert \frac{\sqrt{8n^{3}/n^{3} - 8n^{2}/n^{3}+2n/n^{3}}}{\sqrt{2}}\right\rvert \\
&= \lim_{n\rightarrow \infty} \left\lvert \frac{\sqrt{8 - 8/n+2/n^{2}}}{\sqrt{2}}\right\rvert \\
&= \left\lvert \frac{2\sqrt{2}}{\sqrt{2}}\right\rvert \\
&= 2
\end{align}
$
Since $\sum_{n=1}^{\infty} c_{n}$ converges as it is a p-series with $p>1$, then $\sum_{n=1}^{\infty} b_{n}$ converges, proving that $\sum_{n=0}^{\infty} a_{n}$ converges too.
c) Let's return to the Taylor series of $\sqrt{1-x}$.
Let $-1<x<1$. Then we have that $\left\lvert \sum_{n=0}^{\infty} a_{n} x^{n}\right\rvert \leq \sum_{n=0}^{\infty} \left\lvert a_{n} x^{n} \right\rvert = \sum_{n=0}^{\infty} \left\lvert a_{n} \right\rvert \left\lvert x^{n} \right\rvert = \sum_{n=0}^{\infty} \left\lvert a_{n} \right\rvert$.
Since $\sum_{n=0}^{\infty} \left\lvert a_{n} \right\rvert$ converges, as proved before, then the power series converges. Proving the desired result.
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|
Idempotent relations in a ring Let $(A,+,.)$ be a ring such that, if $x \in A$ with $6x = 0$, then $x=0$. Let $a,b,c \in A$ such that $a-b$ , $b-c$ , $c-a$ are idempotent. Prove that $a=b=c$.
Unfortunately, I haven't made any big progress on this one. I noticed that $(a+b+c)^2 = 3(a^2+b^2+c^2)$ and I tried finding an expression $E$ with $6E =0$, but with no success. I haven't encountered many problems with rings, so I am quite a beginner in this area.
Can you help me on this?
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Note that
$\begin{align*}
(a-c)^2=(a-c)&=((a-b)+(b-c))^2\\
&=(a-b)^2+(b-c)^2+(a-b)(b-c)+(b-c)(a-b)\\
&=a-b+b-c+ab-ac-b^2+bc+ba-b^2-ca+cb
\end{align*}$
It follows that
$\begin{align*}
0=&ab-ac-b^2+bc+ba-b^2-ca+cb\\
=&\color{blue}{(ab+ba)-b^2} +\color{red}{(bc+cb)-b^2}-(ac+ca)\\
=&\color{blue}{a^2-a+b}+\color{red}{c^2-c+b}-(ac+ca)\\
=&(a^2+c^2-ac-ca)-(a+c)+2b\\
=&(a-c)^2-a-c+2b\\
=&2b-2c
\end{align*}$
It follows that $6(b-c)=0$, whence $b=c$.
Similarly, show that $a=b$
Explanation for blue/red colored parts:
$(a-b)^2=a-b\implies a^2+b^2-ab-ba=a-b\implies ab+ba-b^2=a^2-a+b$
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|
Finding the equation of the Parabola The parabola $y = x^2 + bx + c$ has the following properties:
*
*The point on the parabola closest to $(12,3)$ is the intersection with the $y$ axis of the parabola.
*The parabola passes through $(-5,0).$
How can I find $(b, c)$?
Here is my attempt:
The point $(0, c)$ is the intersection with the $y$ axis of the parabola. The distance from $(12, 3)$ is $\sqrt{144 + (c-3)^2}$, and we have the equation $-5b +25+c =0$.
But if we don't know vertex of the parabola how do we find $b$ & $c$?
|
The parabola $y = x^2 + bx + c$
*
*The point on the parabola closest to $(12,3)$ is the intersection of the $y$ axis and the parabola
and we have the equation $-5b +25+c =0$.
*
*Let $P$ be the parabola's $y$-intercept $(0,c)$ and $Q$ be $(12,3).$
Then the parabola's tangent at $P$ is perpendicular to $PQ,$ and,
since $P$ and $Q$ have different $x$-coordinates, $PQ$ is not
vertical. So,
$$(2x+b)\left(\frac{c-3}{0-12}\right)=-1\quad\text{and}\quad
x=0\\b(c-3)=12.$$ And as $c=5b-25,$
$$\\5b^2-28b-12=0\\b=6\quad\text{or}\quad
-\frac25\\c=5\quad\text{or}\quad -27.$$
*If $(b,c)=\left(-\frac25,-27\right),$ then the parabola is
$\displaystyle y=x^2-\frac25x-27,$ and its points whose tangents are
perpendicular to $PQ$ are given by
$$\left(2x-\frac25\right)\left(\frac{x^2-\frac25x-27-3}{x-12}\right)=-1\\x=-5.13\quad\text{or}\quad
0 \quad\text{or}\quad 5.73,$$ and the corresponding distances from
$(12,3)$ are approximately $17,32$ and $7.$ We eliminate this
case since here the point on the parabola closest to $(12,3)$ is
not the $y$-intercept.
SIMPLER ALTERNATIVE (suggested by Oscar Lanzi below): If $b=-\frac25,$ then the parabola's axis $x={-}\frac b2$ is $x=\frac15,$ and $P$ and $Q$ lie on opposite sides of it, which means that some point on the parabola is closer than $P$ to $Q;$ so, we eliminate this case.
*Hence, the required
parabola must be
$y=x^2+6x+5.$
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|
Given $z^2-1\mid x^2z^2-1$, prove $\frac{x^2z^2-1}{z^2-1}$ is never prime, for $x$, $z$ integers such that $x>z>1$.
Given $z^2-1\mid x^2z^2-1$, prove that $\frac{x^2z^2-1}{z^2-1}$ can never be prime, assuming $x$, $z$ are integers such that $x>z>1$.
So far I have tried taking mod a lot of different numbers, but I cannot find solution. I also tried writing as a quadratic and using quadratic formula, but that doesn't work either. Please help.
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Answered for my own benefit, as I already had seen the above two answers first******
Note that $x^2z^2-1 = (xz-1)(xz+1)$. Then, as $\frac{x^2z^2-1}{z^2-1} = \frac{(xz-1)(xz+1)}{z^2-1}$ is an integer, it follows that $z^2-1$ can be written as follows: $z^2-1 = ab$ for some positive integers $a$ and $b$, that satisfy the following:
*
*$a|(xz-1)$ say $xz-1=ca$ and
*$b|(xz+1)$ say $xz+1=bd$.
[One of $a,b$ may be $1$]. But then, as $xz-1 > z^2-1$ [because $x>z$], it follows that $c$ as in 1. above must be an integer greater than $1$, and similarly, $d$ as in 2. above must be an integer greater than $1$. But then that gives:
$$\frac{x^2z^2-1}{z^2-1} = \frac{(xz-1)(xz+1)}{z^2-1} = \frac{(ca)(db)}{ab} = cd,$$ where $c$ and $d$ are both integers greater than $1$.
Thus, as $cd$ is clearly compositive for such $c$ and $d$, it follows that $\frac{x^2z^2-1}{z^2-1}=cd$ cannot be prime.
In general, the following is true:
Let $A,B,C$ be positive integers such that both $\frac{AB}{C}$ is integral, and $C<\min\{A,B\}$. Then $\frac{AB}{C}$ is always not prime.
|
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|
On the log-cosine integral $\int\limits_0^{\pi/2}\ln(\cos(x))\,dx$ I know that
$$\mathcal I = \int\limits_0^{\pi/2}\ln(\cos(x))\,dx = -\frac\pi2 \ln(2)$$
and I am aware of a few different clever ways to demonstrate this result (e.g. MSE 4065767 and MSE 1992462). I would like to know if there's any way to wrap up the method I outline below.
Substituting $t=\tan\left(\frac x2\right)$ yields
$$\mathcal I= 2\int_0^1 \frac{\ln\left(\frac{1-t^2}{1+t^2}\right)}{1+t^2} \, dt = 2 \left(\underbrace{\int_0^1\frac{\ln(1-t^2)}{1+t^2}\,dt}_{\mathcal J^-} - \underbrace{\int_0^1\frac{\ln(1+t^2)}{1+t^2} \, dt}_{\mathcal J^+} \right)$$
Exploiting the series expansion for $\frac1{1+t^2}$, we have
$$\mathcal J^- = \int_0^1 \ln(1-t^2) \sum_{n=0}^\infty (-t^2)^n = \sum_{n=0}^\infty (-1)^n \underbrace{\int_0^1 t^{2n} \ln(1-t^2) \, dt}_{{J_n}^-}$$
and similarly,
$$\mathcal J^+ = \sum_{n=0}^\infty \underbrace{\int_0^1 t^{2n} \ln(1+t^2) \, dt}_{{J_n}^+}$$
I derive the following recurrences for ${J_n}^{\pm}$:
$$\begin{cases}{J_0}^- = 2\ln(2) - 2 \\ {J_n}^- = \frac{2n-1}{(2n+1)^2} - \frac1{2n+1} + \frac{2n-1}{2n+1} {J_{n-1}}^- & \text{for }n\ge1\end{cases}$$
$$\begin{cases}{J_0}^+ = \ln(2) - 2 + \frac\pi2 \\ {J_n}^+ = \frac{2n-1}{(2n+1)^2} - \frac{2\ln(2)-1}{2n+1} - \frac{2n-1}{2n+1}{J_{n-1}}^+ & \text{for }n\ge1\end{cases}$$
where ${J_0}^\pm$ are found by exploiting the series for $\ln(1\pm t)$ and several others derived from the expansion of $\frac1{1-t}$. I managed to solve these for ${J_n}^{\pm}$, so that
$$\mathcal J^- = \frac{(\ln(2)-1)\pi}2 + \sum_{n=0}^\infty \frac{(-1)^n}{2n+1} \left(\sum_{i=0}^{n-1} \frac{2i+1}{2i+3} - n\right)$$
$$\mathcal J^+ = \frac{\pi^2-3\pi}8 + \sum_{n=0}^\infty \frac{(-1)^n}{2n+1} \sum_{i=0}^{n-1} (-1)^i \frac{2i+1}{2i+3}$$
Barring any mistakes, the remaining sum seems to be
$$2 \sum_{n=0}^\infty \frac{(-1)^n}{2n+1} \left(\sum_{i=0}^{n-1} (1-(-1)^i) \frac{2i+1}{2i+3} - n\right) = \frac{3\pi}2\ln(2) - \frac{\pi^2}4 - \frac\pi4$$
but I have no idea where to go with this.
Is there anything I can do to further refine $\mathcal J^- - \mathcal J^+$? I do see a bit of cancellation of terms when $i$ is even, which lets me simplify the inner sum to
$$2\sum_{i=0}^{\left\lfloor\frac{n-1}2\right\rfloor} \frac{4i+1}{4i+3} - n$$
but that doesn't seem helpful.
We do have, with $H_n$ the $n$-th harmonic number,
$$\sum_{i=0}^{n-1} \frac{2i+1}{2i+3} = \sum_{i=0}^{n-1}\left(1 - \frac2{2i+3}\right) = n - 2\left(H_{2n+1} - \frac12 H_n - 1\right)$$
though I'm not so sure I can write its alternating counterpart in $\mathcal J^+$ in a similar way. Then
$$\mathcal J^- = \frac12\ln(2) - \pi - 2 \sum_{n=0}^\infty \frac{(-1)^n H_{2n+1}}{2n+1} + \sum_{n=0}^\infty \frac{(-1)^n H_n}{2n+1} = \frac{2\ln(2)-\pi}4 - G$$
|
With $x\to (1-x)/(1+x)$,
$$\int_0^1 \frac{\ln\left(\frac{1-x^2}{1+x^2}\right)}{1+x^2} dx=\int_0^1 \frac{\ln\left(\frac{2x}{1+x^2}\right)}{1+x^2}dx$$
$$=\int_0^1\frac{\ln(2)}{1+x^2}dx+\int_0^1\frac{\ln(x)}{1+x^2}dx-\int_0^1\frac{\ln(1+x^2)}{1+x^2}dx$$
$$=\frac{\pi}{4}\ln(2)-G+\sum_{n=0}^\infty\frac{(-1)^nH_n}{2n+1}.\tag{1}$$
To get this sum, multiply both sides of
$$\int_0^1\frac{x^{2n}}{1+x}dx=\ln(2)+\frac1{2n+1}+H_n-H_{2n+1},$$
by $\frac{(-1)^n}{2n+1}$ then consider the summation over $n\ge 0$,
$$\underbrace{\sum_{n=0}^\infty\frac{\ln(2)(-1)^n}{2n+1}}_{\frac{\pi}{4}\ln(2)}+\underbrace{\sum_{n=0}^\infty\frac{(-1)^n}{(2n+1)^2}}_{G}+\sum_{n=0}^\infty\frac{(-1)^nH_n}{2n+1}-\sum_{n=0}^\infty\frac{(-1)^nH_{2n+1}}{2n+1}$$
$$=\int_0^1\frac{1}{1+x}\sum_{n=0}^\infty \frac{(-x^2)^n}{2n+1}dx=\int_0^1\frac{\arctan x}{x(1+x)}dx$$
Since
$$\sum_{n=0}^\infty\frac{(-1)^nH_{2n+1}}{2n+1}=-\int_0^\infty (-1)^n \int_0^1 x^{2n}\ln(1-x)dx$$
$$=- \int_0^1 \ln(1-x)\sum_{n=0}^\infty (-x^2)^n dx=-\int_0^1\frac{\ln(1-x)}{1+x^2}dx=\int_0^1\frac{\arctan x}{x(1+x)}dx,$$
where the last equality was proved by @Sangchul Lee here, we have
$$\sum_{n=0}^\infty\frac{(-1)^n H_n}{2n+1}=-\frac{\pi}{4}\ln(2)-G+2\int_0^1\frac{\arctan x}{x(1+x)}dx$$
$$=-\frac{\pi}{4}\ln(2)-G+2\underbrace{\int_0^1\frac{\arctan x}{x}dx}_{G}-2\int_0^1\frac{\arctan x}{1+x}dx$$
$$=-\frac{\pi}{4}\ln(2)+G-2\int_0^1\frac{\arctan x}{1+x}dx$$
where
$$\int_0^1\frac{\arctan x}{1+x}dx\overset{x\to (1-x)/(1+x)}{=}\int_0^1\frac{\pi/4-\arctan x}{1+x}dx$$
$$\Longrightarrow\int_0^1\frac{\arctan x}{1+x}dx=\frac{\pi}{8}\ln(2).$$
Therefore
$$\sum_{n=0}^\infty \frac{(-1)^n H_n}{2n+1}=G-\frac{\pi}{2}\ln(2).$$
Plug this result in $(1)$, we have
$$\int_0^1 \frac{\ln\left(\frac{1-x^2}{1+x^2}\right)}{1+x^2} dx=-\frac{\pi}{4}\ln(2)\Longrightarrow \int\limits_0^{\pi/2}\ln(\cos(x))dx = -\frac\pi2 \ln(2).$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4411338",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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|
Integral $\int( 3x^2 +5x + 1 )\sqrt{2x^2 + 2x + 1}dx$ I tried to integrate
$$\int( 3x^2 +5x + 1 )\sqrt{2x^2 + 2x + 1}dx$$
by multipling by $\sqrt{2x^2 + 2x + 1}$ in the numerator and the denominator to break it into $5$ fractions
The answer is possible but it's too long
Another better solution
|
Notice that
$$
\left(3x^2 + 5x +1\right)\sqrt{2x^2 + 2x+1} = \frac{1}{4\sqrt{2}}\left(3(2x+1)^2 + 4(2x+1) -3\right)\sqrt{(2x+1)^2+1}
$$
under substitution $2x +1 = \tan(\alpha)$ simplifies your integral to
\begin{align*}
&\int\left(3x^2 + 5x +1\right)\sqrt{2x^2 + 2x+1} \, \mathrm{d}x \\
&=\frac{1}{8\sqrt{2}} \int \left(3 \tan^2(\alpha) + 4\tan(\alpha) -3\right)\sec^3(\alpha) \, \mathrm{d}\alpha \\
& = \frac{3}{8\sqrt{2}} \int \left(\sec^2(\alpha) -2\right)\sec^3(\alpha) \, \mathrm{d}\alpha + \frac{1}{2\sqrt{2}} \int \left[\tan(\alpha)\sec(\alpha)\right]\sec^2(\alpha) \, \mathrm{d}\alpha\\
\end{align*}
Since the last integral is immediate under susbtitution $u = \sec(\alpha)$, you just need to solve $\int \sec^n(\alpha) \mathrm{d} \alpha$ for $n =3,5$. For this use the reduction formula for secant until you get to $
\int\sec(x) \mathrm{d}x = \ln|\sec(x) + \tan(x)| $, and afer returning the resulting expression to be written in terms of $x$ this concludes the problem.
|
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"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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|
Calculate $\gcd(a^2b^2, a^2 + ab + b^2)$ Given $\gcd(a, b) = 1$, calculate $d =\gcd(a^2b^2, a^2 + ab + b^2)$ in terms of $a$ and $b$.
I have tried some manipulations of the terms arriving to some expressions such as that $d$ divides $a^4 + b^4$ or that $d$ divide s $(a+b)^4$ but those haven't given me much help. I have also tried dividing the problem in cases: $d$ divides $a$ but not $b$, $d$ divides $a$ and $b$, $d$ doesn't divide neither $a$ or $b$, etc
The first statements come from:
$d\vert a^2(a^2+ab+b^2) - (a^2b^2) = a^4 +a^3b$
$d\vert b^2(a^2+ab+b^2) - (a^2b^2) = ab^3 +b^4$
$d\vert ab(a^2+ab+b^2) - (a^2b^2) = a^3b+ab^3$
$d\vert a^4+a^3b - (a^3b+ab^3) + ab^3 +b^4= a^4 +b^4$
$d\vert a^4 +b^4= (a+b)^4-4(a^3b+ab^3)-6a^2b^2 \to d\vert (a+b)^4$
|
It is easier in such problems to consider first the case when $d=p$ is a prime number (instead of an arbitary postive integer).
Indeed, suppose that $p\mid\gcd(a^2b^2,a^2+ab+b^2)$. Then, $p\mid a^2b^2$ and $p\mid a^2+ab+b^2$. Can you now find out for which prime $p$ it is possible given that $\gcd(a,b)=1$?
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4414353",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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|
Range of $\sqrt{a^2 + (1 - b)^2} + \sqrt{b^2 + (1 - c)^2} + \sqrt{c^2 + (1 - d)^2} + \sqrt{d^2 + (1 - a)^2}$ on $0 \le a,$ $b,$ $c,$ $d \le 1.$ Let $0 \le a,$ $b,$ $c,$ $d \le 1.$ Find the possible values of the expression
$$\sqrt{a^2 + (1 - b)^2} + \sqrt{b^2 + (1 - c)^2} + \sqrt{c^2 + (1 - d)^2} + \sqrt{d^2 + (1 - a)^2}.$$
I tried to use some inequalities to find the bounds of the expression, but it didn't really work. Also, I don't know calculus yet, so please keep the responses and hints non-calc.
Thanks in advance!!
|
The max must be $4$ (while $a=b=c=d=1$) and min be $a=c$ and $b=d$).
Define
$$L(a_1,a_2,b_1,b_2,c_1,c_2,d_1,d_2,\lambda_1,\lambda_2,\lambda_3,\lambda_4)=\sqrt{a_1^2+b_2^2}+\sqrt{b_1^2+c_2^2}+\sqrt{c_1^2+d_2^2}+\sqrt{d_1^2+a_2^2}+\lambda_1(a_1+a_2-1)+\lambda_2(b_1+b_2-1)+\lambda_3(c_1+c_2-1)+\lambda_4(d_1+d_2-1).$$
Let $\nabla L=0$, we find $a_1=b_2=c_1=d_2,a_2=b_1=c_2=d_1$ are stationary points of L. Hence it could be the extremum of the function in question.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4420384",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
calculate one determinant to show 【National College Entrance Exam, old China】
$ \omega =\frac{1 \pm \sqrt{3} i } {2}$
Please calculate and show:
$ \begin{vmatrix} 1 &\omega& \omega^2 & 1\\ \omega& \omega^2 & 1 &1\\ \omega^2& 1& 1 & \omega \\ 1& 1& \omega& \omega^2\\ \end{vmatrix} =3\sqrt{-3} $
Here's what I have done until now
$$
D=\begin{vmatrix}
1 & \omega & {{\omega }^{2}} & 1 \\
\omega & {{\omega }^{2}} & 1 & 1 \\
{{\omega }^{2}} & 1 & 1 & \omega \\
1 & 1 & \omega & {{\omega }^{2}} \\
\end{vmatrix} \\= \begin{vmatrix}
1 & \omega & {{\omega }^{2}} & 1+1+\omega +{{\omega }^{2}} \\
\omega & {{\omega }^{2}} & 1 & 1+1+\omega +{{\omega }^{2}} \\
{{\omega }^{2}} & 1 & 1 & 1+1+\omega +{{\omega }^{2}} \\
1 & 1 & \omega & 1+1+\omega +{{\omega }^{2}} \\
\end{vmatrix} \\= \begin{vmatrix}
1 & \omega & {{\omega }^{2}} & 1 \\
\omega & {{\omega }^{2}} & 1 & 1 \\
{{\omega }^{2}} & 1 & 1 & 1 \\
1 & 1 & \omega & 1 \\
\end{vmatrix} \\= \begin{vmatrix}
1 & \omega & {{\omega }^{2}} & 1 \\
\omega -1 & {{\omega }^{2}}-\omega & 1-{{\omega }^{2}} & 0 \\
{{\omega }^{2}}-1 & 1-\omega & 1-{{\omega }^{2}} & 0 \\
0 & 1-\omega & \omega -{{\omega }^{2}} & 0 \\
\end{vmatrix}
$$
|
You typed $\omega=\frac{1\pm\sqrt3\,i}{2}$ but I think you mean
$$\omega=\frac{-1+\sqrt3\,i}{2}\ .$$
In that case, subtracting a suitable multiple of row 1 from every other row (and not bothering to calculate irrelevant entries) gives
$$\eqalign{\det\pmatrix{1&\omega&\omega^2&1\cr \omega&\omega^2&1&1\cr \omega^2&1&1&\omega\cr 1&1&\omega&\omega^2\cr}
&=\det\pmatrix{1&\omega&\omega^2&1\cr 0&0&0&1-\omega\cr 0&0&1-\omega&\cdot\cr 0&1-\omega&\cdot&\cdot\cr}\cr
&=-(1-\omega)^3\cr &=3(\omega-\omega^2)\cr &=3\sqrt{-3}\ .\cr}$$
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/4421875",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Can we prove that $\int_0^{2\pi} e^{r \cos \theta} \cos (r \sin \theta+n \theta) d \theta=0$ without complex analysis? In textbooks on complex analysis, a typical example using Cauchy Integral Theorem is shown that
$$
\int_{0}^{2 \pi} e^{r \cos \theta} \cos (r \sin \theta+\theta) d \theta=0
$$
Later, I generalized it in my answer that $$\int_{0}^{2 \pi} e^{r \cos \theta} \cos (r \sin \theta+n\theta) d \theta=0,$$
where $r\in R^+,n\in N.$
However, I want to investigate the integral without complex analysis. So I started to try some simple cases using compound angle formula by integration by parts. $$
\begin{aligned}
I_1=\int e^{r \cos \theta} \cos (r\sin \theta+\theta) d \theta =& \int e^{r \cos \theta}[\cos (r\sin \theta) \cos \theta-\sin (r \sin \theta) \sin \theta] d \theta \\
=& \int e^{r \cos \theta} \cos (r\sin \theta) \cos \theta d \theta- \int e^{r\cos \theta} \sin (r \sin \theta) \sin \theta d \theta
\end{aligned}
$$
Fortunately,
\begin{aligned} \int e^{r\cos \theta} \sin (r \sin \theta) \sin \theta d \theta &\stackrel{IBP}{=}
-\frac{1}{r} \int \sin (r \sin \theta) d\left(e^{r \cos \theta}\right)\\&=-\frac{e^{r \cos \theta} \sin (r\sin \theta)}{r}+\int e^{r \cos \theta} \cos (r\sin \theta)\cos \theta d\theta\\
\therefore \int e^{r\cos \theta} \sin (r \sin \theta) \sin \theta d \theta&=\frac{e^{r\cos \theta} \sin (r\sin \theta)}{r}+C_1
\end{aligned}
Applying the same technique decreases $n$ to $n-1$ yields
\begin{aligned}
I_{n} &=\int e^{r \cos \theta} \cos (r \sin \theta+n \theta) d \theta\\
&=\int e^{r \cos \theta}[\cos (r \sin \theta+(n-1) \theta) \cos \theta-\sin (r \sin \theta+(n-1) \theta)\sin \theta] d \theta\\
&=\int e^{r \cos \theta}[\cos (r \sin \theta+(n-1) \theta) \cos \theta d \theta-\int e^{r \cos \theta} \sin (r \sin \theta+(n-1) \theta) \sin \theta d \theta
\end{aligned}
Applying integration by parts to the last integral yields
\begin{aligned}&\int e^{r\cos \theta} \sin (r \sin \theta+(n-1) \theta) \sin \theta d \theta\\=& -\frac{1}{r} \int \sin (r \sin \theta+(n-1) \theta) d\left(e^{r \cos \theta}\right)\\=&-\frac{1}{r}\left[e^{r \cos \theta} \sin (r \sin \theta+(n-1) \theta)\right] +\frac{1}{r} \int e^{r \cos \theta} \cos (r\sin \theta+(n-1) \theta)(r\cos \theta+(n-1)) d \theta \\=&-\frac{1}{r}\left[e^{r \cos \theta} \sin (r \sin \theta+(n-1) \theta)\right] + \int e^{r \cos \theta} \cos (r \sin \theta+(n-1) \theta) \cos \theta d \theta+\frac{n-1}{r} I_{n-1}\end{aligned}
Putting it back yields the reduction formula
$$\boxed{I_n = \frac{1}{r}\left[e^{r \cos \theta} \sin (r \sin \theta+(n-1) \theta)-(n-1)I_{n-1}\right]}$$
Now let’s use the reduction formula and Mathematical Induction to prove that $$J_n:=\int_0^{2\pi} e^{r \cos \theta} \cos (r \sin \theta+n \theta) d \theta=0$$
First of all, $$J_1=\left[I_1\right]_0^{2\pi}= \left[ \frac{e^{r \cos \theta} \sin (r \sin \theta)}{r}\right] _0^{2\pi}=0$$
Now assume it is true for some $k$ i.e. $J_k=0$, then
$$
J_{k+1}=\frac{1}{r}\left(\left[e^{r \cos \theta} \sin (r\sin \theta+k \theta)\right]_{0}^{2 \pi}-k J _k\right)=0
$$
Therefore it is also true for $n=k+1$ and hence by the principle of Mathematical Induction, we have $$
\boxed{\int_{0}^{2 \pi} e^{r \cos \theta} \cos (r \sin \theta+n \theta) d \theta
=0}$$
For reference, $$
\begin{aligned}
I_{2} &=\frac{1}{r}\left[e^{r \cos \theta} \sin (r \sin \theta+\theta)-I_{1}\right] \\
&=\frac{1}{r}\left[e^{r \cos \theta} \sin (r \sin \theta+\theta)-e^{r \cos \theta} \sin (r \sin \theta)\right]+C_2
\end{aligned}
$$
$$
\begin{aligned}
I_{3} &=\frac{1}{r}\left[e^{r\cos \theta}(\sin (r \sin \theta+2 \theta))-2 I_{2}\right] \\
& =\frac{1}{r^{2}}\left[r e^{r \cos \theta}(\sin (r\sin \theta+2 \theta))-2 e^{r \cos \theta } \sin (r \sin \theta+\theta)-2 e^{r \cos \theta} \sin (r\sin \theta)\right]+C_3
\end{aligned}
$$
Question: Is there an alternative proof other than MI?
|
With
$$I_n(r)=\int_{0}^{2\pi}e^{r\cos\theta}\cos(n\theta+r\sin\theta)d\theta
$$
it is straightforward to establish $I_{n+1}(r) =I_n’(r)$, along with
$$I_1(r)= \int_{0}^{2\pi}e^{r\cos\theta}\cos(\theta+r\sin\theta)d\theta=\frac1r e^{r\cos\theta}\cos(r\sin\theta)\bigg|_{0}^{2\pi} =0$$
Then
\begin{align}
I_{n+1}(r) = \frac{d^{n}}{dr^n}I_1(r)=0
\end{align}
|
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"timestamp": "2023-03-29T00:00:00",
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|
If $|ax^2+bx+c|\leq 2\ \ \forall x\in[-1,1]$ then find the maximum value of $|cx^2+2bx+4a|\ \ \forall x\in [-2,2]$.
If
$$\left|ax^2+bx+c\right|\leq 2\quad \forall x\in[-1,1]$$
then find the maximum value of
$$\left|cx^2+2bx+4a\right|\quad \forall x\in [-2,2].$$
My Attempt
Let $f(x)=ax^2+bx+c$, then
$$|cx^2+2bx+4a|=x^2|4\frac{a}{x^2}+2\frac{b}{x}+c|=x^2\left|f\left(\frac{2}{x}\right)\right|$$
But then I couldn't make any headway.
I also tried to express $a,b,c$ in terms of $f(-1),f(0),f(1)$ i.e $$c=f(0);b=\frac{f(1)-f(-1)}{2};a=\frac{f(-1)-2f(0)+f(1)}{2}$$ but again couldn't go further
|
We have
\begin{align*}
|a - b + c| &\le 2, \tag{1}\\
|a + b + c| &\le 2, \tag{2}\\
|c| &\le 2. \tag{3}
\end{align*}
Using (1) and (2), we have
$$|a + c| + |b| \le 2. \tag{4}$$
(Note: If $(a + c)b \ge 0$, then $|a + c| + |b| = |a + c + b| \le 2$. If $(a + c)b < 0$, then $|a + c| + |b| = |a + c - b| \le 2$.)
Using (1) and (3), we have
$$|a - b| \le |a - b + c| + |-c| \le 4. \tag{5}$$
Using (2) and (3), we have
$$|a + b| \le |a + b + c| + |-c| \le 4. \tag{6}$$
Using (5) and (6), we have
$$|a| + |b| \le 4. \tag{7}$$
(Note: If $ab \ge 0$, then $|a| + |b| = |a + b| \le 4$.
If $ab < 0$, then $|a| + |b| = |a - b| \le 4$.)
Using (4) and (7), we have, for all $x\in [-2, 2]$,
\begin{align*}
|cx^2 + 2bx + 4a|
&\le |cx^2 + 4a| + |2bx| \\
&\le \max(|4c + 4a|, |4a|) + 4|b| \tag{8}\\
&\le 16
\end{align*}
where in (8) we have used
$$\max_{x\in [-2, 2]} |cx^2 + 4a|
= \max(|c\cdot (-2)^2 + 4a|, |c \cdot 2^2 + 4a|, |c \cdot 0^2 + 4a|).$$
On the other hand, when $a = -4, b = 0, c = 2$, we have $|ax^2 + bx + c| = |-4x^2 + 2|\le 2$ for all $x\in [-1, 1]$, and
the maximum of $|cx^2 + 2bx + 4a|= |2x^2 - 16|$ on $[-2, 2]$ is $16$.
Thus, the maximum of $|cx^2 + 2bx + 4a|$ on $[-2, 2]$ is $16$.
|
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|
Find $k\in\mathbb{R}$ given $w = k+i$ and $z=-4+5ki$ and $\arg(w+z)$ I am working on problem 2B-11 from the book Core Pure Mathematics, Edexcel 1/AS. The question is:
The complex numbers $w$ and $z$ are given by $w = k + i$ and $z = -4 + 5ki$ where $k$ is a real constant. Given at $\arg(w + z) = \frac{2\pi}{3}$, find the exact value of $k$.
I have attempted this question, but by answer is different from the given one and I can't figure out why. My answer:
Let $w + z = a + bi$.
$a = k - 4 \\
b = 1 + 5k$
The angle of the triangle formed by the vector $w + z$ (on Argand diagram) and the x-axis is $\pi -\frac{2\pi}{3} = \frac{\pi}{3}$
$\tan\frac{\pi}{3} = \frac{b}{a} \\
\tan\frac{\pi}{3} = \frac{1 + 5k}{k - 4} \\
k\tan\frac{\pi}{3} - 4\tan\frac{\pi}{3} = 1 + 5k \\
k\tan\frac{\pi}{3} - 5k = 1 + 4\tan\frac{\pi}{3} \\
k = \frac{1 + 4\tan\frac{\pi}{3}}{\tan\frac{\pi}{3} - 5} = -2.42$
However, the given answer is:
$\frac{4\sqrt{3} - 1}{5 + \sqrt{3}} = 0.88$ where $\tan\frac{\pi}{3} = \sqrt{3}$
I have just noticed that assuming the angle is $-\tan\frac{\pi}{3}$, my answer is correct. How can you tell the quadrant of $w + z$ when the constant $k$ appears in both $a$ and $b$?
|
$k\in\Bbb R,\,w:=k+i,\,z:=-4+i5k$ and $\arg(w+z)=\frac{2\pi}{3}$
Of course $w+z=(k-4)+i(5k+1)$. Since $0\lt\frac{2\pi}{3}\le\pi$ the (principal) argument tells us it is in the second quadrant (the top left), where $\arg(x+iy)=\pi-\arctan\frac{y}{-x}$ ("$-x$" because $x$ is negative and $-\arctan$ since we begin at $\pi$, on the negative real axis, and start turning clockwise - for this quadrant).
This means $k\lt 4$ but $5k+1\gt0$, so $-\frac{1}{5}\lt k\lt 4$. Let's compute the argument:
$$\begin{align}\arg(w+z)&\overset{\color{red}{1}}{=}\pi-\arctan\frac{5k+1}{4-k}=\frac{2\pi}{3}\\&\implies\frac{\pi}{3}=\arctan\frac{5k+1}{4-k}\\&\overset{\color{red}{2}}{\implies}\sqrt{3}=\frac{5k+1}{4-k}\\&\implies\sqrt{3}(4-k)=5k+1\\&\implies(5+\sqrt{3})k=4\sqrt{3}-1\\&\implies k=\frac{4\sqrt{3}-1}{5+\sqrt{3}}\end{align}$$
$1)$: It is also worth noting that we need $k\in\Bbb R$. If $k$ had an imaginary component, then $(k-4)$ would not be the real part and $(5k+1)$ would not be the imaginary part - they'd be a mix of both.
$2)$: It is worth noting that $\arctan(x)=y$ does not mean $x=\tan(y)$ in general. However, the way that the principal argument works, and by the fact we were careful with the quadrants, we know that it will hold for our purposes.
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/4423608",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
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|
If $f(x)=\frac {x^ 2 -2x +4}{ x^ 2 +2x+4}$ for $x \in \mathbb{R}$, prove that the range of $f(x)$ is $[1/3, 3]$ One method to solve it would by putting $y = f(x)$ then multiplying the denominator with $y$ hence making a quadratic equation in x then we can just use the inequalities for $x$ being real to prove it.
For an alternative what I did is
$y = \frac{x^2 + 2x + 4}{x^2 - 2x + 4}$
Then$ \frac{1+y}{1-y} =\frac{x^2+4}{2x} $
Therefore$ \frac{1+y}{1-y} = {x/2+2/x}$
If we go further putting rhs range to lhs the answer gets altered . Is this method wrong if so why if not where did the process go wrong.
|
HINT
I would recommend you to notice that
\begin{align*}
\frac{x^{2} - 2x + 4}{x^{2} + 2x + 4} & = \frac{(x^{2} + 2x + 4) - 4x}{x^{2} + 2x + 4}\\\\
& = 1 - \frac{4x}{x^{2} + 2x + 4}
\end{align*}
Consequently, the problem reduces to study the critical points of the last expression.
In order to do so, let us take its derivative:
\begin{align*}
f(x) = 1 - \frac{4x}{x^{2} + 2x + 4} & \Rightarrow f'(x) = -\frac{4(x^{2} + 2x + 4) - 4x(2x + 2)}{(x^{2} + 2x + 4)^{2}}\\\\
& \Rightarrow f'(x) = \frac{4x^{2} - 16}{(x^{2} + 2x + 4)^{2}}
\end{align*}
Based on its expression, we conclude the critical points are given by $x = -2$ and $x = 2$.
Can you take it from here?
EDIT
Let us approach it by parts. Firstly, let us set that $y = x + \frac{4}{x}$, where $x > 0$.
Then one concludes that $y\geq 4$ based on AM-GM inequality, for example.
Hence we obtain the lower bound according to:
\begin{align*}
\frac{x^{2} - 2x + 4}{x^{2} + 2x + 4} & = \frac{x - 2 + \frac{4}{x}}{x + 2 + \frac{4}{x}} = \frac{y - 2}{y + 2} = 1 - \frac{4}{y + 2} \geq 1 - \frac{2}{3} = \frac{1}{3}
\end{align*}
Now we can suppose that $x < 0$. If we set the same change of variable $y = x + \frac{4}{x}$, one gets $y\leq -4$.
Consequently, we obtain the upper bound as follows:
\begin{align*}
\frac{x^{2} - 2x + 4}{x^{2} + 2x + 4} & = \frac{x - 2 + \frac{4}{x}}{x + 2 + \frac{4}{x}} = \frac{y - 2}{y + 2} = 1 - \frac{4}{y + 2} \leq 1 + 2 = 3
\end{align*}
and we are done.
Hopefully this helps!
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4424161",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
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|
How do I find the parabola parameter equation? Show that $x(t)=\cos^4t, y(t)=\sin^4t$ is a parametrization of the parabola $(x-y-1)^2=4y$.
I think that to solve this problem, we need to know how to find the parametric equation of the parabola. So I searched through books and internet search, but I haven't found out yet.
If my method is wrong, I will need some guide to solve this problem.
|
Given the conic of Cartesian equation:
$$
(x-y-1)^2 = 4\,y
\quad \quad \Rightarrow \quad \quad
x^2+y^2-2\,x\,y-2\,x-2\,y+1=0
$$
we can rewrite this polynomial equation in the following two ways:
$$
\small
\begin{bmatrix}
x & y & 1 \\
\end{bmatrix}
\underbrace{\begin{bmatrix}
1 & -1 & -1 \\
-1 & 1 & -1 \\
-1 & -1 & 1 \\
\end{bmatrix}}_{A_1}
\begin{bmatrix}
x \\
y \\
1 \\
\end{bmatrix}
= 0\,,
\quad \quad
\begin{bmatrix}
x & y \\
\end{bmatrix}
\underbrace{\begin{bmatrix}
1 & -1 \\
-1 & 1 \\
\end{bmatrix}}_{A_2}
\begin{bmatrix}
x \\
y \\
\end{bmatrix} + 2
\begin{bmatrix}
-1 & -1 \\
\end{bmatrix}
\begin{bmatrix}
x \\
y \\
\end{bmatrix} + 1
= 0
$$
where being:
$$
\det(A_1) = \color{blue}{-4}\,,
\quad \quad \quad
\det(A_2) = 0
$$
we deduce that the conic is non-degenerate, in particular it's a parabola.
That done, it's time to compute the eigenvalues and eigenvectors of $A_2$:
$$
\lambda_1 = 0,
\;
\mathbf{v}_1 =
\begin{bmatrix}
\color{green}{1} \\
\color{green}{1} \\
\end{bmatrix};
\quad \quad \quad
\lambda_2 = \color{red}{2},
\;
\mathbf{v}_2 =
\begin{bmatrix}
\color{green}{-1} \\
\color{green}{1} \\
\end{bmatrix};
$$
from which we can deduce that:
*
*the axis of the parabola is parallel to $\mathbf{v}_1$;
*the directrix of the parabola is parallel to $\mathbf{v}_2$;
*the canonical equation of the parabola is $x' = \pm\sqrt{\frac{-(\color{red}{2})^3}{4\,(\color{blue}{-4})}}\;{y'}^2 = \pm\color{orange}{\frac{1}{\sqrt{2}}}\,{y'}^2$.
In particular, the Cartesian equation of a line parallel to the directrix is of the type $y = -x + q$, which turns out to be tangent to the parabola if and only if $q = 1/2$, line that intersected in turn with the parabola allows to calculate the coordinates of the vertex: $(1/4,\,1/4)$.
In light of all this, a parameterization of the parabola can be obtained by roto-translation:
$$
\begin{bmatrix}
x \\
y \\
\end{bmatrix}
=
\begin{bmatrix}
\color{green}{\frac{1}{\sqrt{2}}} & \color{green}{-\frac{1}{\sqrt{2}}} \\
\color{green}{\frac{1}{\sqrt{2}}} & \color{green}{\frac{1}{\sqrt{2}}} \\
\end{bmatrix}
\begin{bmatrix}
\color{magenta}{+}\color{orange}{\frac{1}{\sqrt{2}}}\,u^2 \\
u \\
\end{bmatrix}
+
\begin{bmatrix}
\frac{1}{4} \\
\frac{1}{4} \\
\end{bmatrix}
\quad
\Rightarrow
\quad
\begin{cases}
x = \frac{u^2}{2}-\frac{u}{\sqrt{2}}+\frac{1}{4} \\
y = \frac{u^2}{2}+\frac{u}{\sqrt{2}}+\frac{1}{4} \\
\end{cases}
\quad \text{with} \; u \in \mathbb{R}
$$
where the magenta sign has been chosen positive so that the signs of the coefficients of $u^2$ are in agreement with the position of the parabola placed above the tangent line in the vertex. Finally, wanting to make the parameterization cleaner, it's sufficient to set $u=\sqrt{2}\;v$ from which it follows:
$$
\begin{cases}
x = v^2-v+\frac{1}{4} \\
y = v^2+v+\frac{1}{4} \\
\end{cases}
\quad \text{with} \; v \in \mathbb{R}\,.
$$
This is a parameterization that covers every point $(x,\,y) \in \mathbb{R}^2$ of the parabola, instead the parameterization you propose only covers the points $(x,\,y) \in [0,\,1] \times [0,\,1]$, which is restrictive!
|
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"url": "https://math.stackexchange.com/questions/4425338",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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|
Limit/asymptotic of a modification of the exponent's Taylor series Consider the following function on $\left[0,\infty\right)\times\left[0,1\right]$
$$f\left(x,y\right)=\sum_{n=0}^{\infty}\dfrac{x^{n}}{n!}y^{n^{2}}$$
I am interested in the limit/asymptotic behavior of the following expression for $x\rightarrow\infty$
$$g\left(x,y\right)=e^{x}\dfrac{y^{2}f\left(xy^{3},y\right)}{f^{2}\left(xy,y\right)}$$
I encountered these summations in the context of the $g^{\left(2\right)}$ correlation function of some quantum state of light. On physical grounds, I suspect it might converge asymptotically to $1$ from below, but not completely sure.
Any suggestion on how to evaluate this will be appreciated. Thanks in advance!
EDIT I:
I found a mistake in my calculation, which after fixing gives a slightly different function $g\left(x,y\right)$
$$g\left(x,y\right)=\dfrac{y^{2}f\left(xy^{3},y\right)f\left(xy^{-1},y\right)}{f^{2}\left(xy,y\right)}$$
Plugging inside @SangchulLee asymptotic expansion indeed gives $g\left(x\rightarrow\infty,y\right)=1$. Since there is no dependence on $x$, I would also be interested in the next order of the asymptotic expansion. Thanks!
|
Step 1. Let us parametrize $x = e^{s}$ and $y = e^{-\varepsilon}$ for $s \in \mathbb{R}$ and $\varepsilon > 0$. Fix the values of $k$ and $\varepsilon$, and then write
$$ f(xy^k, y) = \sum_{n=0}^{\infty} e^{-h(n)}, \qquad \text{where} \quad h(n) = \varepsilon n^2 + k\varepsilon n - sn + \log n! $$
Regarding $h(n)$ as an analytic function of $n$ and differentiating twice,
\begin{align*}
h'(n) &= 2\varepsilon n + k\varepsilon - s + \psi(n+1), \\
h''(n) &= 2\varepsilon + \psi'(n+1),
\end{align*}
where $\psi$ is the digamma function. Since $h''(n) > 0$ for $n > 0$, it follows that $h$ is strictly concave. Moreover, for each fixed $k$ and $\varepsilon$ and for large $s$, we have $h'(0) < 0$ and $h'(n) \to \infty$ as $n\to\infty$. So it follows that the equation $h'(n) = 0$ has a unique zero $n = n_*$. Then, regarding $n_*$ as a function of $s$, the Laplace's method suggests that
$$ f(xy^k, y) \sim \sqrt{\frac{2\pi}{h''(n_*)}} \, e^{-h(n_*)} \quad\text{as}\quad s\to\infty. $$
Step 2. The above observation leads us to study the asymptotic behavior of $n_*$ as $ s \to \infty $. To this end, note that $s$ and $n_*$ are related via the equation
$$ s = 2\varepsilon n_* + \psi(n_* + 1) + k\varepsilon. \tag{1} $$
Using this, we will progressively improve the asymptotic formula for $n_*$ as $s \to \infty$:
*
*Since $s$ as a function of $n_*$ is strictly increasing and unbounded, the same is true for $n_*$ as a function of $s$.
*Dividing both sides of $\text{(1)}$ by $n_*$ and letting $n_* \to \infty$, we get $s/n_* \to 2 \varepsilon$ as $n_* \to \infty$. This then implies that
$$ n_* = \frac{s}{2\varepsilon}(1 + o(1)) \qquad\text{and}\qquad \psi(n_* + 1) = \log\left(\frac{s}{2\varepsilon}\right) + o(1) \tag{2} $$
as $s \to \infty$, where we utilized the asymptotic relation $\psi(z+1) \sim \log z$ as $z \to \infty $ for the second one.
*Rearranging $\text{(1)}$, we get
$$ n_* = \frac{s}{2\varepsilon} - \frac{k}{2} - \frac{1}{2\varepsilon}\psi(n_* + 1). $$
In light of $\text{(2)}$, we define the quantity $s_*$ as
$$ s_* := s - \log\left(\frac{s}{2\varepsilon}\right) - k\varepsilon $$
and use this to recast $ n_* $ as
$$ n_* = \frac{1}{2\varepsilon}\left[s_* - \psi(n_* + 1) + \log\left(\frac{s}{2\varepsilon}\right) \right]. \tag{3} $$
*Using $\text{(3)}$ and the asymptotic expansion $\psi(z+1) = \log z + (2z)^{-1} + \mathcal{O}(z^{-2})$ as $z \to \infty$,
\begin{align*}
&\psi(n_* + 1) - \log\left(\frac{s}{2\varepsilon}\right) \\
&= \biggl(\log n_* + \frac{1}{2n_*} + \mathcal{O}(s^{-2}) \biggr) - \log\left(\frac{s}{2\varepsilon}\right) \\
&= \log\left(\frac{2\varepsilon n_*}{s}\right) + \frac{1}{2n_*} + \mathcal{O}(s^{-2}) \\
&= \log \left(\frac{s_*}{s}\right) + \log \biggl( 1 - \frac{\psi(n_* + 1) - \log\left(\frac{s}{2\varepsilon}\right)}{s_*} \biggr) + \frac{1}{2n_*} + \mathcal{O}(s^{-2})
\end{align*}
In light of the asymptotic formulas in $\text{(2)}$, this further reduces to
\begin{align*}
&= -\log\left(1 + \frac{s - s_*}{s_*}\right) + \log \biggl( 1 - \frac{o(1)}{s_*} \biggr) + \frac{\varepsilon}{s_*(1 + o(1))} + \mathcal{O}(s^{-2}) \\
&= -\frac{1}{s_*}\log\left(\frac{s_*}{2\varepsilon}\right) - \frac{(k-1)\varepsilon}{s_*} + o(s^{-1})
\end{align*}
Therefore we obtain
\begin{align*}
n_*
&= \frac{1}{2\varepsilon}\left[ s_* + \frac{1}{s_*}\log\left(\frac{s_*}{2\varepsilon}\right) + \frac{(k-1)\varepsilon}{s_*} + o(s^{-1}) \right], \tag{5}
\end{align*}
Step 3. Now we are ready to estimate $h(n_*)$. Using $\text{(1)}$ and the Stirling's approximation together,
\begin{align*}
h(n_*)
&= \varepsilon n_*^2 - (s - k\varepsilon)n_* + \log (n_*!) \\
&= \varepsilon n_*^2 - (2\varepsilon n_* + \psi(n_* + 1))n_* + \log (n_*!) \\
&= -\varepsilon n_*^2 - n_* [\psi(n_* + 1) - \log n_*] - n_* + \frac{1}{2}\log n_* + \log\sqrt{2\pi} + o(1) \\
&= -\varepsilon n_*^2 - n_* + \frac{1}{2}\log n_* - \frac{1}{2} + \log\sqrt{2\pi} + o(1)
\end{align*}
Then plugging $\text{(4)}$ into the last line,
\begin{align*}
&= -\frac{1}{4\varepsilon}\left[ s_*^2 + 2\log\left(\frac{s_*}{2\varepsilon}\right) + 2(k-1)\varepsilon \right] - \frac{s_*}{2\varepsilon} + \frac{1}{2}\log \left(\frac{s_*}{2\varepsilon}\right) - \frac{1}{2} + \log\sqrt{2\pi} + o(1) \\
&= -\frac{1}{4\varepsilon}s_*^2 - \frac{1}{2\varepsilon}\left[ s_* + \log\left(\frac{s_*}{2\varepsilon}\right) + k\varepsilon \right] + \frac{1}{2}\log \left(\frac{s_*}{2\varepsilon}\right) + \log\sqrt{2\pi} + o(1) \\
&= -\frac{1}{4\varepsilon}s_*^2 - \frac{s}{2\varepsilon} + \frac{1}{2}\log \left(\frac{\pi s}{\varepsilon}\right) + o(1)
\end{align*}
On the other hand, using $\psi'(z+1) \sim \frac{1}{z}$, we get $h''(n_*) \sim 2\varepsilon$. Therefore,
\begin{align*}
f(e^{s-k\varepsilon}, e^{-\varepsilon})
\sim \sqrt{\frac{\pi}{\varepsilon}} e^{-h(n_*)}
\sim \exp\left\{ \frac{1}{4\varepsilon}s_*^2 + \frac{s}{2\varepsilon} - \frac{1}{2}\log s + o(1) \right\}
\end{align*}
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4425581",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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|
minimizing the distance between values
Find all values of $a$ that minimize the expression $|a- x_1| + |a-x_2|+\cdots + |a-x_n|$, where $x_1\leq x_2\leq \cdots \leq x_n$.
I know the minimum is achieved when $a=x_{\lfloor n/2\rfloor + 1}$ if $n$ is odd and when $a\in [x_{\lfloor n/2\rfloor}, x_{\lfloor n/2\rfloor + 1}]$ when $n$ is even.
I think one approach is mostly "brute force" and is relatively complicated. If I let $f(a) = |a-x_1|+\cdots + |a-x_n|$. Or maybe I could use derivatives? For $a$ so that $x_i < a<x_{i+1}$, $f'(a)$ equals $i - (n-i)$, which is less than $0$ for $i < n/2$ and $>0$ for $i > n/2$.
I think the following approach might work, but even if it does, it's way too tedious.
I could show directly that $f(x_i) < f(a) $ or $f(x_{i+1}) < f(a)$, depending on whether $i < \frac{n}2$ or $i > \frac{n}2$. I could then evaluate $f(a)$ directly by setting $a = x_k$ for some $k < \lfloor n/2\rfloor, x_k < x_{\lfloor n/2\rfloor}$ and for $k > \lfloor n/2\rfloor + 1, x_k > x_{\lfloor n/2\rfloor + 1}$. Then I could show the distance would be decreased by choosing $a=x_{\lfloor n/2\rfloor }$ or $x_{\lfloor n/2\rfloor + 1}.$ Finally, I could show that when $n$ is even, the distance will be the same if we choose any $a\in [x_{\lfloor b/2\rfloor}, x_{\lfloor n/2\rfloor + 1}$.
Here are some more details regarding my approach:
First observe that there cannot exist an index i so that $x_i < a < x_{i+1}$ unless $i = \frac{n}2$ (in which case of course n has to be even).
To see why, observe that for such an a, we have
\begin{align}
f(x_i) - f(a) &= \sum_{j=1}^i (x_i - a) + \sum_{j=i+1}^n (a- x_i)\\
&= (a- x_i)(n- 2i)\tag{1}
\end{align}
and
\begin{align}
f(x_{i+1}) - f(a) &= \sum_{j=1}^i (x_{i+1} - a) + \sum_{j=i+1}^n (a - x_{i+1})\\
&= (a-x_{i+1})(n-2i) \tag{2}
\end{align}
Clearly, $(1)$ shows $f(x_i) - f(a) < 0$ for $i > \frac{n}2$ and $(2)$ shows that $f(x_{i+1} ) - f(a) < 0$ for $i < \frac{n}2$. Thus we know that if $a$ lies strictly between two indices $x_i$ and $x_{i+1}$ and $f(a)$ is minimum, then $i = \frac{n}2$. Now we find the values of $i$ so that $f(x_i)$ is minimum.
Now we claim that if $n$ is odd, and $x_k < x_{\lfloor \frac{n}2\rfloor + 1}$ then $f(x_{\lfloor \frac{n}2\rfloor + 1}) < f(x_k)$.
Observe that $f(x_k) = \sum_{i=1}^k (x_k - x_i) + \sum_{i=k+1}^n (x_i - x_k)$. Then if $x_k < x_{\lfloor \frac{n}2\rfloor}$, we have
\begin{align}f(x_{\lfloor \frac{n}2\rfloor}) - f(x_k) &= \sum_{j=1}^k (x_{\lfloor \frac{n}2\rfloor} - x_k) + \sum_{j=k+1}^{\lfloor \frac{n}2\rfloor - 1} (x_k + x_{\lfloor \frac{n}2\rfloor } - 2x_j) + \sum_{j= \lfloor \frac{n}2\rfloor}^n (x_k - x_{\lfloor \frac{n}2\rfloor })\\
&\leq (k- (n-\lfloor \frac{n}2\rfloor + 1)) (x_{\lfloor \frac{n}2\rfloor} - x_k) + (x_{\lfloor \frac{n}2\rfloor} - x_k) (\lfloor \frac{n}2\rfloor - k - 1) \\
&= (2 \lfloor \frac{n}2\rfloor - n - 2)(x_{\lfloor \frac{n}2\rfloor }- x_k)
< 0,\end{align}
Now if $x_k > x_{\lfloor \frac{n}2\rfloor + 1}$, then we have
\begin{align}
f(x_{\lfloor \frac{n}2\rfloor + 1}) - f(x_k) &= \sum_{j=1}^{\lfloor \frac{n}2\rfloor +1} (x_{\lfloor \frac{n}2\rfloor + 1} - x_k) + \sum_{j=\lfloor \frac{n}2\rfloor + 2}^{k} (2x_j - (x_k + x_{\lfloor \frac{n}2\rfloor + 1} )) + \sum_{j= k + 1}^n (x_k - x_{\lfloor \frac{n}2\rfloor + 1})\\
&\leq (\lfloor\frac{n}2 \rfloor + 1- (n-k )) (x_{\lfloor \frac{n}2\rfloor + 1} - x_k) - (x_{\lfloor \frac{n}2\rfloor + 1} - x_k) (k - \lfloor\frac{n}2 \rfloor - 1) \\
&= (2 \lfloor \frac{n}2\rfloor - n + 2)(x_{\lfloor \frac{n}2\rfloor + 1}- x_k)
< 0\end{align}
So to minimize $f(a)$, we must have $x_{\lfloor \frac{n}2 \rfloor }\leq a \leq x_{\lfloor \frac{n}2 \rfloor + 1}$. Then we have for $k=\lfloor \frac{n}2\rfloor$ that (3)
\begin{align}f(x_{\lfloor \frac{n}2\rfloor + 1}) - f(a) &= \sum_{j=1}^k (x_{\lfloor \frac{n}2\rfloor + 1} - a) + \sum_{j=k+1}^{\lfloor \frac{n}2\rfloor} (a + x_{\lfloor \frac{n}2\rfloor + 1} - 2x_j) + \sum_{j= \lfloor \frac{n}2\rfloor + 1}^n (a - x_{\lfloor \frac{n}2\rfloor + 1})\\
&\leq (k- (n-\lfloor \frac{n}2\rfloor)) (x_{\lfloor \frac{n}2\rfloor + 1} - a) + (x_{\lfloor \frac{n}2\rfloor + 1} - a) (\lfloor \frac{n}2\rfloor - k) \\
&= (2 \lfloor \frac{n}2\rfloor - n)(x_{\lfloor \frac{n}2\rfloor + 1}- a)
\leq 0,\end{align}
with equality when $n$ is even. Since the minimum cannot be achieved for $x < \lfloor n/2\rfloor$ by a proof above, this shows that if $n$ is odd we cannot have $x_{\lfloor n/2\rfloor + 1} > a$ if $f(a)$ is minimum (as either $a < x_{\lfloor n/2\rfloor}$, in which case $f(a)$ is not minimum, or $x_{\lfloor n/2\rfloor} \leq a < x_{\lfloor n/2\rfloor + 1}$, in which case by the proof above ((3)) $f(a)$ is not minimum. If n is even, we thus have the the minimum is achieved for $a \in [x_{n/2}, x_{n/2+1}]$.
|
Note that
$$ |x - c| = \begin{cases}
c - x & \text{for $x \leq c$} \\
x - c & \text{for $x \geq c$}
\end{cases} $$
So, adopting the conventions that $x_0 = -\infty$ and $x_{n+1} = \infty$, then for any $x$ satisfying $ x_k \leq x \leq x_{k+1} $,
$$ f(x) = \sum_{i=1}^{n} |x - x_i| = \sum_{i=1}^{k} (x - x_i) + \sum_{i=k+1}^{n} (x_i - x) = (k - 2n)x + \text{[constant]}.$$
Using this and the continuity of $f$ together, we find that
*
*$f(x)$ is decreasing for $ x \leq x_{\lceil n/2 \rceil} $;
*$f(x)$ is increasing for $ x_{\lfloor n/2\rfloor+1} \leq x $;
*$f(x)$ is constant for $ x_k \leq x \leq x_{k+1} $, if $k = \frac{n}{2}$;
Therefore we conclude that $f$ is minimized precisely at the points $x$ satisfying
$$ x_{\lceil n/2 \rceil} \leq x \leq x_{\lfloor n/2\rfloor+1}. $$
|
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"url": "https://math.stackexchange.com/questions/4429742",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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|
Finding closed form expression for $n^{th}$ term of sequence with generating function $F(x)$? I have been asked the above question regarding the generating function
$$F(x) = \frac{x^2(1-x)}{(1-x)^3}$$
I have no idea what procedure this type of question follows. The solution gives that $F(x)$ can be written as
$$\frac{x^2}{(1-x)^3}-\frac{x^3}{(1-x)^3}$$
This I understand. But then it then says for $F(x) = \dfrac{1}{(1-x)^3}$ we get
$$\frac{F^n(0)}{n!} = \frac{(n+1)(n+2)}{2} \tag1$$
From which we can get the coefficients.
Can someone help me understand what is computed here? Why do we use $F(x) = \dfrac{1}{(1-x)^3}$, and how is $(1)$ computed? All help appreciated!
|
Denoting with $[x^n]$ the coefficient of $x^n$ of a series we obtain
\begin{align*}
\color{blue}{[x^n]F(x)}&=[x^n]\frac{x^2(1-x)}{(1-x)^3}\\
&=[x^n]\left(\frac{x^2}{(1-x)^3}-\frac{x^3}{(1-x)^3}\right)\tag{1}\\
&\,\,\color{blue}{=[x^{n-2}]\frac{1}{(1-x)^3}-[x^{n-3}]\frac{1}{(1-x)^3}}\tag{2}
\end{align*}
In the last line (2) we applied the rule $[x^p]x^qF(x)=[x^{p-q}]F(x)$.
We see in (2) what we essentially need is
\begin{align*}
[x^n]G(x)=[x^n]\frac{1}{(1-x)^3}
\end{align*}
Recalling the Taylor series expansion of $G(x)$ at $x=0$ we have
\begin{align*}
\color{blue}{[x^n]\frac{1}{(1-x)^3}}=[x^n]\sum_{k=0}^\infty \frac{G^{(k)}(0)}{k!}x^k
\color{blue}{=\frac{G^{(n)}(0)}{n!}}\tag{3}
\end{align*}
In the right-hand part of the equality chain we select the coefficient of $[x^n]$ which is the middle step of the hint to this problem. Here we use $G$ instead of $F$ to avoid using the same name for different functions.
We can easily calculate the $n$-th derivative of $G(x)$ as
\begin{align*}
G^{(n)}(x)&=\left(\frac{1}{(1-x)^3}\right)^{(n)}=\left((1-x)^{-3}\right)^{(n)}\\
&=(3)(4)\cdots(3+(n-1))(1-x)^{-3-n}\\
&=3\cdot4\cdots(n+2)\frac{1}{(1-x)^{n+3}}\\
&=\frac{(n+2)!}{2}\,\frac{1}{(1-x)^{n+3}} \tag{4}
\end{align*}
Combining (3) and (4) we see
\begin{align*}
\color{blue}{\frac{G^{(n)}(0)}{n!}}=\frac{(n+2)!}{2n!}\,\,\color{blue}{=\frac{(n+1)(n+2)}{2}}\tag{5}
\end{align*}
which was the final hint to this problem.
We see the Taylor series of $G(x)$ has the triangle numbers as coefficients
\begin{align*}
G(x)=1+3x+6x^2+10x^3+15x^4+\cdots\tag{6}
\end{align*}
We are now well prepared to derive the wanted coefficient $[x^n]F(x)$. We obtain from (2) and (5) for $n\geq 3$
\begin{align*}
\color{blue}{[x^n]F(x)}&=[x^{n-2}]\frac{1}{(1-x)^3}-[x^{n-3}]\frac{1}{(1-x)^3}\\
&=\frac{(n-1)n}{2}-\frac{(n-2)(n-1)}{2}\\
&=\frac{n-1}{2}(n-n+2)\\
&\,\,\color{blue}{=n-1}
\end{align*}
We finally need to determine the coeffcients of $[x^n]F(x)$ for $0\leq n\leq 2$. This is easy, since we already know $G(x)$ and we conclude from (2) and (6)
\begin{align*}
F(x)&=\left(x^2+3x^3+6x^4+\cdots\right)-\left(x^3+3x^4+6x^5+\cdots\right)\\
&=x^2+2x^3+3x^4+\cdots
\end{align*}
from which
\begin{align*}
\color{blue}{[x^n]F(x)=\begin{cases}
n-1&\qquad n\geq 1\\
0&\qquad n=0
\end{cases}}
\end{align*}
follows in accordance with the answer from @RobPratt.
|
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|
How can I find this limit that is supposed to converge to e? I made a formula to get an approximation $x-1$ based on $x$; $f(x) = \frac{(-2 + x)^{3/2 - x} x^{-1/2 + x}}{e^2}$. I got this formula by looking at $\frac{n!}{n(n-2)!}$ which equals $n-1$, then substituting the factorials with Stirling approximants. After simplifying through Wolfram Alpha, I got my function. Then, I realized I could solve for $e$, and I got $$\lim_{x \rightarrow \infty} \frac{(x - 2)^{3/4} \sqrt{x^x}}{\sqrt{{(x - 2)^x}{(x - 1)}{\sqrt{x}}}} = e$$. The problem is, I cannot figure out how to solve this limit. It should converge to e (albeit very slowly) but I do not know how to make progress or even simplify this limit significantly. I want to know, if I were presented this limit without knowing how it was obtained, how I could solve it.
|
$$ f(x)=\frac{(x-2)^{\frac{3}{2}-x} x^{x-\frac{1}{2}}}{e^2}\implies \log[f(x)]=\left(\frac{3}{2}-x\right)\log(x-2)-\left(x-\frac{1}{2}\right)\log(x)-2$$ Expanding as series for large values of $x$
$$\log[f(x)]=\log (x)-\frac{1}{x}-\frac{1}{3 x^2}+O\left(\frac{1}{x^4}\right)$$
$$f(x)=e^{\log[f(x)]}=x-1+\frac{1}{6 x}+\frac{1}{6 x^2}+O\left(\frac{1}{x^3}\right)$$
Edit
For the second one
$$g(x)= \frac{(x - 2)^{3/4} \sqrt{x^x}}{\sqrt{{(x - 2)^x}{(x - 1)}{\sqrt{x}}}} \implies g^2(x)=\frac{(x-2)^{\frac{3}{2}-x} x^{x-\frac{1}{2}}}{x-1}$$
$$\log[g^2(x)]=\left(\frac{3}{2}-x\right) \log (x-2)+\left(x-\frac{1}{2}\right)\log(x)-\log(x-1)$$
$$\log[g^2(x)]=2+\frac{1}{6 x^2}+\frac{1}{3 x^3}+O\left(\frac{1}{x^4}\right)$$
$$\log[g(x)]=1+\frac{1}{12 x^2}+\frac{1}{6 x^3}+O\left(\frac{1}{x^4}\right)$$
$$g(x)=e^{\log[g(x)]}=e \Big[1+\frac{1}{12 x^2}+\frac{1}{6 x^3} \Big]+O\left(\frac{1}{x^4}\right)$$
|
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|
Explanation for a solution: Howard Anton, Elementary Linear Algebra I am currently working with the 1st edition of Howard Anton's "Elementary Linear Algebra". I tried the following problem:
Excercise Set 1.2 (p. 17), Problem 12:
For which values of $a$ will the following system have no solutions? Exactly one solution? Infinitely many solutions?
$$\begin{array}{rccccl}
x &+& 2y &-& 3z &=& 4 \\
3x &-& y &+& 5z &=& 2 \\
4x &+& y &+& (a^2 - 14)z &=& a + 2\end{array}$$
By using Gauss-Jordan-Elimination (and Gaussian for a double check), I found the following solution set:
$$\begin{align*} x &= \frac{8}{7} + \frac{-a+4}{a^2-16} \\
y &= \frac{10}{7} + \frac{2a-8}{a^2-16} \\
z &= \frac{a-4}{a^2-16}\end{align*}$$
The solution in the textbook says, that the system has no solution if $a=-4$ an one solution if $a\neq\pm4$. This part I understand, since $z=\frac{a-4}{a^2-16}$ as well as others terms in the formulas for $y$ and $z$ are not defined for $a=-4$ but the formulas for $x,y,z$ will yield unambiguous values for $a\neq\pm4$. However, the solution also says, that the system has infinitely many solutions for $a=4$ and this is the point, which I don't understand. Isn't for instance $z=\frac{a-4}{a^2-16}$ still undefined for $a=-4$ or can I simply put in $z = \frac{0}{0} = 0$. And if I can, isn't $z=0$ still a unique value. I can find no room for different values of $x,y,z$ to satisfy the system of equations.
Can somebody explain this to me? Thanks in advance!
|
Alternative approach:
Compute the determinant of the matrix.
If the determinant is non-zero, then there will automatically be exactly $1$ solution.
If the determinant is $0$, then there will either be $0$ solutions or an infinite number of solutions, depending on whether there is an inconsistency (explained at the end of this answer) among the values to the right of the equal signs.
The determinant is
$(14 - a^2 - 5) + [(-2)(3a^2 - 62)] + [(-3)(7)] = 112 - 7a^2 = 7(16 - a^2).$
So, if $(16 - a^2) \neq 0$, then you know immediately that there is exactly one solution.
The problem then reduces to a consideration of $a^2 = 16.$
Here, I will explain the idea of inconsistency among linear equations as follows:
Consider the following two pairs of linear equations:
*
*$x + 2y = 6, ~2x + 4y = 12.$
*$x + 2y = 6, ~2x + 4y = 13.$
Both pairs of equations will have a determinant of $(1 \times 4) - (2 \times 2) = 0.$
However, the first pair of equations above is consistent, and therefore permits an infinite number of solutions. The second pair of equations above is inconsistent, so there will be $0$ solutions.
The first thing to notice is that the superficial judgement that the evaluation of $a=4$ and $a=-4$ will yield identical results is wrong. It is true that the LHS makes no distinction between $a=4,$ and $a=-4$, since the LHS only features an $a^2$ item.
However the RHS features an $(a+2)$ value which requires that $a=4$ and $a=-4$ be evaluated separately, to determine which values of $a$ (if any) results in consistent values that yield an infinite number of solutions, and which values of $a$ (if any) result in inconsistent values that yield $0$ solutions.
With $a$ equal to either $+4$ or $-4$, the equations become
$1x + 2y - 3z = 4$
$3x - 1y + 5z = 2$
$4x + 1y + 2z = 6 ~\text{or}~ -2.$
Adding the 2nd and 3rd equations above together yields:
$7x + 7z = 8 ~\text{or}~ 0.$
Multiplying the 2nd equation above by $(2)$ and adding it to the 1st equation yields:
$7x + 7z = 8.$
At this point, you know immediately that $a=-4$ causes an inconsistency, and that therefore, if $a=-4$, there are no solutions.
However, the value of $a=4$ must be explored further, to determine whether it yields consistent results.
Treating $z$ as a fixed (unknown) value, results in :
$1x + 2y = 4 + 3z$
$3x - 1y = 2 - 5z$
$4x + 1y= 6 - 2z$.
Adding the 2nd and 3rd of these revised equations together yields
$\displaystyle 7x = 8 - 7z \implies x = \frac{8}{7} - z.$
Using the 3rd equation above, this implies that
$\displaystyle y = (6 - 2z) - 4\left(\frac{8}{7} - z\right) = \frac{10}{7} + 2z.$
It only remains to verify that the following values also satisfy the 1st and 2nd equations above, which they do:
*
*$\displaystyle x = \frac{8}{7} - z, ~y = \frac{10}{7} + 2z.$
Therefore, $a=4$ yields consistent results.
Therefore, $a=4$ yields an infinite number of solutions.
|
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|
Is there a mistake in solving this limit? I want to solve this:
\begin{equation}
L=\lim_{x\rightarrow 0} \frac{\sum\limits_{m=1}^{M}a_{m}\exp\left\{\frac{bm^2}{ (m^2+2x^2)x^2}\right\}{(m^2+2x^2)^{-3/2}} } {\sum\limits_{m=1}^{M} c_{m}\exp\left\{\frac{bm^2}{ (m^2+2x^2)x^2}\right\}{(m^2+2x^2)^{-3/2}}}
\end{equation}
where
\begin{equation}
\begin{aligned}
m=1,2,\cdots,M\\
a_m,b,c_m \neq 0 \quad \text{and are all constants}
\end{aligned}
\end{equation}
Here is my approach:
Since $x\rightarrow 0$, I ignore the term $2x^4$ in the denominator of $\exp\left\{\frac{b_m}{ (m^2+2x^2)x^2}\right\}$ and I get:
\begin{equation}
\begin{aligned}
L&=\lim_{x\rightarrow 0} \frac{\exp\left\{\frac{b}{x^2}\right\}\sum\limits_{m=1}^{M}a_{m}{(m^2+2x^2)^{-3/2}} } {\exp\left\{\frac{b}{x^2}\right\}\sum\limits_{m=1}^{M} c_{m}{(m^2+2x^2)^{-3/2}}}\\
&=\lim_{x\rightarrow 0} \frac{\sum\limits_{m=1}^{M}a_{m}{(m^2+2x^2)^{-3/2}} } {\sum\limits_{m=1}^{M} c_{m}{(m^2+2x^2)^{-3/2}}}\\
&=\frac{\sum\limits_{m=1}^{M}\frac{a_{m}}{m^3} } {\sum\limits_{m=1}^{M} \frac{c_{m}}{m^3}}
\end{aligned}
\end{equation}
My question is:
*
*Is there a mistake in my derivation?
*If I made mistakes in the derivation, then, what is the correct derivation?
Thanks for helpful comments and answers!
|
If $x^2 \to 0$ then you certainly can't say that $\exp(\frac{b}{x^2}) \to 1$. Quite the contrary, this term goes to infinity very fast, the larger $b$ the faster.
You started with neglecting a smaller order term, that's all and well to get a rough idea, but that's certainly not rigorous, all the more inside an exponential. So let's factor out that $\exp(b/x^2)$ growth rate and see what's left.
$\exp(\frac{bm^2}{ (m^2+2x^2)x^2}) = \exp(\frac{b}{x^2})\exp(\frac{bm^2}{ (m^2+2x^2)x^2} - \frac{b}{x^2})
= \exp(\frac{b}{x^2})\exp(\frac{bm^2-b(m^2+2x^2)}{ (m^2+2x^2)x^2} )
= \exp(\frac{b}{x^2})\exp(\frac{-2b}{m^2+2x^2}).
$
Now you can plug that in your sum and cancel out the $\exp(\frac{b}{x^2})$'s. You will end up with something nicely convergent.
|
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|
Diagonal element of traceless hermitian matrix? In physics, we are familiar with a set of traceless hermitian matrices named Pauli matrices:
$$
{\displaystyle {\begin{aligned}\sigma _{1}=\sigma _{\mathrm {x} }&={\begin{pmatrix}0&1\\1&0\end{pmatrix}}\\\sigma _{2}=\sigma _{\mathrm {y} }&={\begin{pmatrix}0&-i\\i&0\end{pmatrix}}\\\sigma _{3}=\sigma _{\mathrm {z} }&={\begin{pmatrix}1&0\\0&-1\end{pmatrix}}\\\end{aligned}}}
$$
I notice that the above matrices are written in basis $\left( \begin{array}{c}
1\\
0\\
\end{array} \right) $ and $\left( \begin{array}{c}
0\\
1\\
\end{array} \right) $, but if we change basis into $\frac{1}{\sqrt{2}}\left( \begin{array}{c}
1\\
1\\
\end{array} \right) $ and $\frac{1}{\sqrt{2}}\left( \begin{array}{c}
1\\
-1\\
\end{array} \right) $, then $\sigma_z$ become $$\frac{1}{2}\left( \begin{array}{c}
1\\
1\\
\end{array} \right) \left( \begin{matrix}
1& 1\\
\end{matrix} \right) -\frac{1}{2}\left( \begin{array}{c}
1\\
-1\\
\end{array} \right) \left( \begin{matrix}
1& -1\\
\end{matrix} \right) =\left( \begin{matrix}
0& 1\\
1& 0\\
\end{matrix} \right) ,$$showing that the diagonal elements vanish.
My question is, for a single traceless hermitian matrix $H$, can we always find a unitary $u$ such that diagonal element of $u^{\dagger}Hu$ are all zeros? Furthermore, can the diagonal elements run over all possibilities as long as they satisfy sum up to zero?
|
Yes. Let $H$ be any $n\times n$ traceless Hermitian matrix and $UDU^\ast$ be its unitary diagonalisation. Let $Q$ be a real orthogonal matrix whose last column is $\frac{1}{\sqrt{n}}(1,1,\ldots,1)^T$. The last diagonal element of $H':=Q^TU^\ast HUQ=Q^TDQ$ is then the mean of all diagonal elements of $H$, which is zero. Since $H'$ is real symmetric, we can do the similar to the leading principal $(n-1)$-rowed submatrix of $H'$ and continue recursively to obtain ultimately a real symmetric matrix with a zero diagonal.
|
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|
Integral Basis of $O_k$ Let $K=Q(\sqrt 6,\sqrt{11})$. Write $α ∈ O_K$ and its conjugates in terms of a $Q$-basis.
And show that an integral basis of $O_K$ is given by
${1,\sqrt 6,\sqrt {11},\frac{\sqrt 6+\sqrt{66}}2 }$, from first principles.
I'm really not sure how to do this, if anyone could help I'd really appreciate it!
|
You can consider the traces of $\alpha = a+b\sqrt{6}+c\sqrt{11}+d\sqrt{66}$ over the 3 quadratic subfields, noting that this gives an algebraic integer (in that subfield).The three subfields are $L_1 = \mathbb{Q}(\sqrt{6})$, $L_2=\mathbb{Q}(\sqrt{11})$ and $L_3=\mathbb{Q}(\sqrt{66})$ having integers $\mathbb{Z}[6]$, $\mathbb{Z}[11]$, $\mathbb{Z}[66]$.
$Tr_{K/L_1}(\alpha)=2a+2b\sqrt{6}$, so $a=A/2, b=B/2,$ where $A,B$ are integers. Similarly, we get $c=C/2,d=D/2$. Then from here, we can check which of the 16 possibilities are algebraic integers, for $A,B,C,D \in \{0,1\}$. To do this, we can compute the norm over the quadratic subfields. For example,
$N_{K/L_1}(\alpha)= \frac{1}{4} N_{K/L_1}((A+B\sqrt{6})+\sqrt{11}(C+D\sqrt{6})=\frac{1}{4}(A^2+2AB\sqrt{6}+6B^2-11(C^2+2CD\sqrt{6}+6D^2)) = \frac{A^2+6B^2-11C^2-66D^2}{4} + \sqrt{6}\frac{AB-11CD}{2}$
Thus $$A^2+6B^2-11C^2-66D^2=A^2+2B^2+C^2+2D^2=0 \mod 4$$, $$AB-11CD=AB+CD=0\mod 2$$
For, $N_{K/L_2}$ swap $B,C$ and $6,11$. So
$A^2+11C^2-6B^2-66D^2=0 \mod 4$, $$AC-6BD=0 =AC\mod 2$$.
Now do case work. $AC=0 \mod 2$, so $A=0$ or $C=0$.
If $A=0,C=0$, then we must have $B=D=1,0$.
If $A=0, C=1$, then contradiction since $A^2+2B^2+C^2+2D^2$ is odd.
Similarly for $A=1,C=0$.
Now we verify that the thing we found $\frac{\sqrt{6}+\sqrt{66}}{2}$ is an alg integer, as you've done. And then this gives an integral basis after adding the 3 other "non-halved" basis vectors.
Would love to know if there's a nicer way to do this.
|
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To prove the inequality of positive rational numbers Show that: $$ \left(\frac{a+b}{a+b+c}\right)^{c} \left(\frac{b+c}{a+b+c}\right)^{a} \left(\frac{a+c}{a+b+c}\right)^{b}< \left(\frac{2}{3}\right)^{a+b+c} ,a\ne b\ne c$$
PS: I am supposed to use weighted AM > GM > HM for this problem.
I tried by expressing $$\frac{a+b}{a+b+c} = 1- \frac{c}{a+b+c} $$ but I cannot get required expression.
|
Edit: Fixed algebra error.
Consider the weighted AM-GM with weights: $\frac{a}{a+b+c}, \frac{b}{a+b+c}, \frac{c}{a+b+c}$. We attain:
$$(\frac{a(b+c) + b(a+c) + c(a+b)}{a+b+c})^{a+b+c} > (a+b)^c (b+c)^a (a+c)^b.$$
And thus:
$$(\frac{a(b+c) + b(a+c) + c(a+b)}{(a+b+c)^2})^{a+b+c} > (\frac{a+b}{a+b+c})^c (\frac{b+c}{a+b+c})^a (\frac{a+c}{a+b+c})^b.$$
So it's sufficient to show $\frac{2}{3} > (\frac{a(b+c) + b(a+c) + c(a+b)}{(a+b+c)^2})$. Note that
$$\frac{a(b+c) + b(a+c) + c(a+b)}{(a+b+c)^2} = \frac{(a + b+c)^2 - (a^2 + b^2 + c^2)}{(a+b+c)^2}.$$
But by AM-QM: $a^2 + b^2 + c^2 \geq \frac{(a+b+c)^2}{3}$, so the RHS is less than or equal to
$$\frac{(a+b+c)^2 - \frac{(a+b+c)^2}{3}}{(a+b+c)^2} = \frac{2}{3}.$$
|
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How to factorize $2ab^2+2b^2c-ac^2-bc^2-a^2b-a^2c$? If $a,b,c$ are in AP $\implies \frac{1}{bc}, \frac{1}{ca}, \frac{1}{ab}$ are in AP $\implies
\frac{ab+bc+ca}{bc}, \frac{ab+bc+ca}{ca}, \frac{ab+bc+ca}{ab}$ are in AP $\implies \frac{bc+ca}{bc},\frac{ab+bc}{ca},\frac{bc+ca}{ab}$ are in AP $\implies \frac{bc}{bc+ca}, \frac{ca}{ab+bc}, \frac{ab}{bc+ca}$
are in HP.
If $P,Q,R$ are in HP $\implies Q=\frac{2PR}{P+R}.$ By this method proving that $\frac{bc}{bc+ca}, \frac{ca}{ab+bc}, \frac{ab}{bc+ca}$, if $a,b,c$ are in AP; is not straight forward.
Curiously, then we need to factorize $(2ab^2+2b^2c-ac^2-bc^2-a^2b-a^2c)$.
The question is: How do you factorize the last expression?
|
It's tricky!
\begin{array}{l}
2ab^2 + 2b^2 c - ac^2 - bc^2 - a^2 b - a^2 c + 2abc- 2abc = \\
\\
= (2ab^2 + 2b^2 c + 2abc) - \left( {a^2 b + a^2 c + abc} \right) - \left( {ac^2 + bc^2 + abc} \right) = \\
\\
= 2b(ab + bc + ac) - a(ab + bc + ac) - c(ab + bc + ac) = \\
\\
= - (ab + bc + ac)\left( {a - 2b + c} \right) \\
\end{array}
|
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|
Algebra of o-symbols In theorem $7.8$, Tom Apostol in his Calculus Vol. $1$ gave and proved the following basic rules for algebra of o-symbols
*
*$o(g(x)) \pm o(g(x)) = o(g(x)) $
*$o(cg(x)) = o(g(x)) = o(g(x)) $, if $c \ne 0$
*$f(x) \cdot o(g(x)) = o(f(x)g(x)) $
*$o(o(g(x))) = o(g(x)) $
*$\frac{1}{1 + g(x)} = 1 - g(x) + o(g(x))$
Now when proving that $\tan x = x + \frac{1}{3}x^3 + o(x^3)$ as $x \to 0$, he used the following equality, only mentioning that $g(x) = -\frac{1}{2}x^2 + o(x^3)$ and that he is applying the 5th case of the theorem above:
$$
Equation 1: \frac{1}{1 - \frac{1}{2}x^2 + o(x^3)} = 1 + \frac{1}{2}x^2 + o(x^2)
$$
By just applying that 5th case of the theorem, we get
$$
Equation 2: \frac{1}{1 - \frac{1}{2}x^2 + o(x^3)} = 1 + \frac{1}{2}x^2 - o(x^3) + o(-\frac{1}{2}x^2 + o(x^3))
$$
Now, I can prove that the right handside of Eq1 is equal to the right hand side of Eq2 (e.g. using this method), by proving that:
*
*$o(-\frac{1}{2}x^2 + o(x^3)) = o(x^2)$
*$o(x^2) + o(x^3) = o(x^2)$
But I wonder if there is an obvious reason why the above two equalities directly follow from the Theorem $7.8$?
Edit, to add a suggested proof for $o(x^2) + o(x^3) = o(x^2)$, which I hope was not even necessary, because it could be inferred trivially from the theorem $7.8$, potentially by the application of step $4$ $n-m$ times.
Let $m < n \land f(x) = o(x^m) \land g(x) = o(x^n)$.
Then we have:
$f(x) + g(x) = o(x^m) + o(x^n) = o(x^m) \cdot (1 + x^{n - m})$, because of T7.8:3.
\begin{equation}
\implies \frac{f(x) + g(x)}{1 + x^{n - m}} = o(x^m) \\
\implies 0 = \lim_{x \to 0} \frac{f(x) + g(x)}{x^m + x^n} \\
\implies 0 = \lim_{x \to 0} \frac{f(x) + g(x)}{x^m + x^n} \cdot (1 + 0) \\
\implies 0 = \lim_{x \to 0} \frac{f(x) + g(x)}{x^m + x^n} \cdot (\lim_{x \to 0} 1 + \lim_{x \to 0} x^{n-m})
\end{equation}
The last equality holds because $\lim_{x \to 0} x^{n-m} = 0$. Then further we have:
\begin{equation}
\implies 0 = \lim_{x \to 0} \frac{f(x) + g(x)}{x^m + x^n} \cdot (\lim_{x \to 0} \frac{x^m + x^n}{x^m}) \\
\implies 0 = \lim_{x \to 0} \frac{f(x) + g(x)}{x^m + x^n} \cdot \frac{x^m + x^n}{x^m} \\
\implies 0 = \lim_{x \to 0} \frac{f(x) + g(x)}{x^m} \\
\implies f(x) + g(x) = o(x^m)
\end{equation}
The last equality follows by definition of little-o. By just substituting $f$ and $g$, we get $o(x^m) + o(x^n) = o(x^m)$.
|
The best way to deal with the $"o"$ notation is to write it as
$$F(x)=o(g(x)) \; (x\to 0)$$
means that
$$F(x)=g(x)\epsilon(x)$$
with
$$\lim_{x\to 0}\epsilon(x)=0$$
So, your last equality gives
$$o(x^2)+o(x^3)=$$
$$x^2\epsilon_1(x)+x^3\epsilon_2(x)=$$
$$x^2(\epsilon_1(x)+x\epsilon_2(x))=$$
$$x^2\epsilon(x)=o(x^2)$$
|
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|
If $abc=1$ ,and $n$ is a natural number,prove $ \frac{a^n}{(a-b)(a-c)}+\frac{b^n}{(b-a)(b-c)}+\frac{c^n}{(c-a)(c-b)} \geqslant \frac{n(n-1)}{2} $ If $a, b, c$ are distinct positive real numbers such that $abc=1$,and $n$ is a natural number,prove
$$
\frac{a^n}{(a-b)(a-c)}+\frac{b^n}{(b-a)(b-c)}+\frac{c^n}{(c-a)(c-b)} \geqslant \frac{n(n-1)}{2}
$$
I know for $n=3$,the answer is here,and how to go further?
|
Hint :
Using Fuchs's inequality wich is an extension of Karamata's inequality one can show that for ($f(x)=x^n$),$a\geq b \geq c>0$ such that $abc=1$:
$$\left(\frac{a^{n}}{\left(a+c\right)\left(a+c\right)}-1\right)a\cdot\frac{1}{\left(a-b\right)\left(a-c\right)}\geq \left(\frac{b^{n}}{\left(b+c\right)\left(b+a\right)}-1\right)\cdot\frac{b}{\left(b-a\right)\left(b-c\right)}\geq \left(\frac{c^{n}}{\left(a+c\right)\left(b+c\right)}-1\right)\cdot\frac{c}{\left(c-a\right)\left(c-b\right)}$$
We have the inequalities :
$$\left(\frac{a^{n}}{\left(a+c\right)\left(a+c\right)}-1\right)a\cdot\frac{1}{\left(a-b\right)\left(a-c\right)}\ge 0$$
$$\left(\frac{a^{n}}{\left(a+c\right)\left(a+c\right)}-1\right)a\cdot\frac{1}{\left(a-b\right)\left(a-c\right)}+\left(\frac{b^{n}}{\left(b+c\right)\left(b+a\right)}-1\right)\cdot\frac{b}{\left(b-a\right)\left(b-c\right)}\ge 0$$
And :
$$\left(\frac{a^{n}}{\left(a+c\right)\left(a+c\right)}-1\right)a\cdot\frac{1}{\left(a-b\right)\left(a-c\right)}+ \left(\frac{b^{n}}{\left(b+c\right)\left(b+a\right)}-1\right)\cdot\frac{b}{\left(b-a\right)\left(b-c\right)}+ \left(\frac{c^{n}}{\left(a+c\right)\left(b+c\right)}-1\right)\cdot\frac{c}{\left(c-a\right)\left(c-b\right)} \ge 0$$
So using Fuchs's inequality we have shown that :
$$\frac{a^{n}}{(a-b)(a-c)}+\frac{b^{n}}{(b-a)(b-c)}+\frac{c^{n}}{(c-a)(c-b)}\le \frac{a^{2n}}{(a^{2}-b^{2})(a^{2}-c^{2})}+\frac{b^{2n}}{(b^{2}-a^{2})(b^{2}-c^{2})}+\frac{c^{2n}}{(c^{2}-a^{2})(c^{2}-b^{2})}$$
Now conclude is easy in fact .
Hope it helps .
|
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|
How does this expression simplification make sense? My simplification does not equal what is given. I attached an image of the expression I'm working with
I have no idea how $A(x)$ is simplified to $\displaystyle 5x - \left(\frac{1+\pi}{2}\right) x^2$.
Every time I work out the equation I get $$A(x) = 5x - x^2 - (\pi x^2/2) + \pi(x/2)^2$$
|
So, step-by-step
\begin{align*}
A(x)&=xy+\pi\left(\frac{x}{2}\right)^{2},\\&=x\left(5-x-\frac{\pi}{2}x\right)+\pi\left(\frac{x}{2}\right)^{2},\\&=5x-x^{2}-\frac{\pi}{2}x^{2}+\frac{\pi}{2^{2}}x^{2},\\&=5x+\left(\frac{\pi}{2^{2}}-\frac{\pi}{2}-1\right)x^{2},\\&=5x+\left(\frac{\pi}{2^2}-\frac{2\pi}{2^{2}}-\frac{2^{2}\cdot 1}{2^{2}\cdot 1}\right)x^{2},\\&=5x+\left(\frac{\pi-2\pi-2^{2}}{2^{2}}\right)x^{2},\\&=5x+\left(\frac{-2^{2}-\pi}{2^{2}}\right)x^{2},\\&=\boxed{5x-\left(\frac{2^{2}+\pi}{2^{2}}\right)x^{2}},\\&\not=5x-\left(\frac{1+\pi}{2}\right)x^{2}
\end{align*}
Hence the answer in your picture is wrong which is clear because in the second sign "$=$" they did not add the term $\pi\left(\frac{x}{2}\right)^{2}$ . You're correct in the simplification.
|
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|
Showing the value the expression takes using another approach Let $1, \omega, \omega^{2}$ be the cube roots of unity. Then the product
$$
\left(1-\omega+\omega^{2}\right)\left(1-\omega^{2}+\omega^{2^{2}}\right)\left(1-\omega^{2^{2}}+\omega^{2^{3}}\right) \cdots\left(1-\omega^{2^{9}}+\omega^{2^{10}}\right)
$$ is equal to ?
what i considered was all 10 epxression can be written as $\frac{2}{1+w} * \frac{2}{1+w^2}... $ , From that expanding two term wise the below product i got all equal to 1 and hence product of the required sum being $2^{10}$ , is there a another way of doing it ? Like considering a polynomial which when given a value w will give that required multiplication , or might the product of $(w+1)(w^2+1)(w^4 +1) ...$ ?
|
Note that, $\omega^2 + \omega +1=0$. Also note that,
\begin{equation*}
2^n \equiv \begin{cases}
1 (\text{ mod } 3), \hspace{1cm} \text{if $n$ is even}\\
2 (\text{ mod } 3), \hspace{1cm} \text{if $n$ is odd}
\end{cases}
\end{equation*}
Establishing above statement is easy. Using this and the fact that $\omega^3 =1$ we get,
\begin{equation*}
\omega^{2^n} = \begin{cases}
\omega, \hspace{1cm} \text{ if $n$ is even}\\
\omega^2, \hspace{1cm} \text{if $n$ is odd}\\
\end{cases}
\end{equation*}
Let, $S_n = 1- \omega^{2^n}+ \omega^{2^{n+1}}$ for all $n \geq 0$. Then We required the find the value of $S$. Where,
\begin{equation*}
S = S_0.S_1.S_3.....S_9
\end{equation*}
But using the above discussion notice that,
\begin{equation*}
S_n = \begin{cases}
1-\omega+\omega^2, \hspace{1cm} \text{if $n$ is even}\\
1+\omega-\omega^2, \hspace{1cm} \text{if $n$ is odd}
\end{cases}
\end{equation*}
So,
\begin{align*}
S&= S_0.S_1.S_2.....S_8.S_9\\
&= (1-\omega+\omega^2)^5(1+\omega-\omega^2)^5\\
&= (-2\omega)^5.(-2\omega^2)^5 &&[\omega^2 + \omega + 1 =0]\\
&= (-2)^{10}. \omega^{15}\\
&= 2^{10}. (\omega^3)^5\\
&= 2^{10}
\end{align*}
So the answer is $2^{10}$.
|
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|
Find $ k $ if both roots of this equation $ x^2-2kx+k^2+k-5=0 $ are less than 5 If both roots of this equation $ x^2-2kx+k^2+k-5=0 $ are less than $ 5$, then find the limit of $ k$.
Method-1
Let's say,
\begin{align} f(x)=x^2-2kx+k^2+k-5 \end{align}
\begin{align} f(0)=0 →\alpha,\beta \end{align}
$ a=1 \gt 0 ,$ So graph is upward. Minimal point $ x_{m} \lt 5$.
Since, the roots$ \alpha,\beta \lt 5, f(5) \gt 0$. And also to have the roots, $ D \geq 0 $ must be followed.
(i)
\begin{align} x_{m} \lt 5 \end{align}
\begin{align} \Rightarrow -\frac{(-2k)}{2×1} \lt 5 \end{align}
\begin{align} \Rightarrow k \lt 5 \end{align}
(ii)
\begin{align} D \geq 0 \end{align}
\begin{align} \Rightarrow (-2k)^2-4(k^2+k-5) \geq 0 \end{align}
\begin{align} \Rightarrow k \leq 5 \end{align}
(iii)
\begin{align} f(5) \gt 0 \end{align}
\begin{align} \Rightarrow 25-9k+k^2-5 \gt 0 \end{align}
\begin{align} \Rightarrow (k-4)(k-5) \gt 0 \end{align}
\begin{align} \Rightarrow k\lt 4,
k \gt 5 \end{align}
Now, from (i),(ii),(iii) we can say, $ k \lt 4 $ .
Method-2
Given that , $ \alpha,\beta \lt 5$
So, \begin{align} \alpha+\beta \lt 10 \qquad{(1)} \end{align} \begin{align} \alpha\beta\lt 25 \qquad (2) \end{align}
From, (1)
\begin{align} 2k \lt 10 \end{align}
\begin{align} \Rightarrow k \lt 5\end{align}
From (2)
\begin{align} \Rightarrow k^2+k-5 \lt 25\end{align}
\begin{align} \Rightarrow (k+6)(k-5) \lt 0\end{align}
\begin{align} \Rightarrow -6 \lt
k \lt 5 \end{align}
Now, from (1),(2) also including (ii) we can say, $-6 \lt
k \lt 5$
Two different answers!
|
Find $ k $ if both roots of this equation $ x^2-2kx+k^2+k-5=0 $ are less than 5
Alternative approach:
Use the quadratic equation directly, calculate the value of the two roots, in terms of $k$, and then apply the constraint that both roots are less than $5$.
The roots are
$~\displaystyle \frac{1}{2}\left[2k \pm \sqrt{4k^2 - 4(k^2 + k - 5)}\right]$
$= ~\displaystyle k \pm \sqrt{k^2 - (k^2 + k - 5)} = k \pm \sqrt{5-k}.$
At this point, the first observation is that the equation will have one or more real roots if and only if
$$(5 - k) \geq 0. \tag1 $$
The larger root is taken by adding, rather than subtracting the square root, which is always non-negative.
Therefore, the problem reduces to finding all values of $k$ such that
$$k + \sqrt{5 - k} < 5. \tag2 $$
Warning
The following conclusion to this problem is a one-off. Apparently, the problem composer wanted the Math student to have a fairly easy time of it, from this point forward. Normally, you would reason that $\displaystyle ~\sqrt{5-k} < 5-k,~$ square both sides, and see where that leads.
For this particular problem, there happens to be a shortcut. The constraint in (2) above will be satisfied if and only if
$$\sqrt{5-k} < (5-k). \tag3 $$
Here, it helps to know that for any non-negative real number $r$,
*
*if $r = 0$ then $\sqrt{r} = r.$
*if $0 < r < 1$, then $\sqrt{r} > r.$
This is a direct consequence of the fact that
if $0 < r < 1$, then $r = \sqrt{r} \times \sqrt{r} < \sqrt{r} \times 1 = \sqrt{r}.$
*if $r = 1$, then $\sqrt{r} = r.$
*if $r > 1$, then $\sqrt{r} < r.$
This is a direct consequence of the fact that
if $1 < r$, then $r = \sqrt{r} \times \sqrt{r} > \sqrt{r} \times 1 = \sqrt{r}.$
So, under the constraint imposed by (1) above, that $(5 - k) \geq 0$, the constraint in (3) is equivalent to the constraint that $1 < (5 - k).$
Therefore, since the constraint that $(5 - k) \geq 0$ is subordinate to the constraint that $(5 - k) > 1$, all satisfying values of $k$ are given by
$$1 < (5-k) \iff k < 4. $$
|
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|
Solving $\tan ^{-1}(\frac{1-x}{1+x})=\frac{1}{2}\tan ^{-1}(x)$ I was solving the following equation,$$\tan ^{-1}\left(\frac{1-x}{1+x}\right)=\frac{1}{2}\tan ^{-1}\left(x\right)$$
But I missed a solution (don't know where's the mistake in my work).
Here's my work:
$$\tan ^{-1}\left(\frac{1-x}{1+x}\right)=\frac{1}{2}\tan ^{-1}\left(x\right)$$
Putting $x = \tan(\theta)$
$$\begin{align} \tan ^{-1}\left(\frac{1-\tan\theta}{1+\tan\theta}\right)&=\frac{1}{2}\tan ^{-1}\left(\tan\theta\right)\\
\tan ^{-1}\left(\tan\left(\frac\pi 4 - \theta\right)\right)&=\frac{1}{2}\theta\\
\frac\pi 4 - \theta&=\frac{1}{2}\theta\\
\frac\pi 4&=\frac{1}{2}\theta + \theta\\
\frac\pi 4&=\frac{3}{2}\theta\\
\frac\pi 6&=\theta\\
\frac\pi 6&=\tan^{-1}(x)\\
\tan\frac\pi 6&=(x)\\
\frac1{\sqrt{3}}&=x\end{align}$$
I got only one solution, but the answer in my textbook is $\pm\frac{1}{\sqrt{3}}$. Where is the mistake?
|
$$\tan ^{-1}\left(\frac{1-x}{1+x}\right)=\frac{1}{2}\tan ^{-1}\left(x\right)$$
Putting $x = \tan(\theta)$
Here, it is useful to impose the restriction $$\theta\in\left(-\frac\pi2,-\frac\pi4\right)\cup\left(-\frac\pi4,\frac\pi2\right);\tag1$$ since $\tan\theta$ for which the given equation is defined is surjective on this interval, this restriction is valid and does not contract the solution set.
\begin{align} \tan ^{-1}\left(\frac{1-\tan\theta}{1+\tan\theta}\right)&=\frac{1}{2}\tan ^{-1}\left(\tan\theta\right)\\
\tan ^{-1}\left(\tan\left(\frac\pi 4 - \theta\right)\right)&=\frac{1}{2}\theta\\
\frac\pi 4 - \theta&=\frac{1}{2}\theta\end{align}
Note that $$\arctan(\tan \alpha)\not\equiv \alpha;\tag{*}$$ instead, for each $\alpha\in\mathbb R,$ there is some $k\in\mathbb Z$ such that $$\arctan(\tan \alpha)=\alpha+k\pi.$$ By condition $(1),$ $$\arctan \left(\tan\theta\right)=\theta$$ while \begin{align}\arctan\left(\tan\left(\frac\pi4 -\theta\right)\right)&=\frac\pi4 -\theta \quad\text{or}\quad \left(\frac\pi4 -\theta\right)-\pi\\
&=\frac\pi4 -\theta \quad\text{or}\quad -\frac34\pi -\theta.\end{align}
\begin{align} \tan ^{-1}\left(\frac{1-\tan\theta}{1+\tan\theta}\right)&=\frac{1}{2}\tan ^{-1}\left(\tan\theta\right)\end{align}
Continuing with your substitution method:
\begin{align}
\frac12\arctan \left(\tan\theta\right) &=\arctan\left(\tan\left(\frac\pi4 -\theta\right)\right)\\
\frac12\big(\theta\big) &= \left(\frac\pi4 -\theta\right) \quad\text{or}\quad \left(-\frac34\pi -\theta\right)\\
\theta &= \frac\pi6 \quad\text{or}\quad -\frac\pi2\\
&= \frac\pi6\\
x&=\frac1{\sqrt3}.\end{align} Plugging this value into the given equation reveals that it is the only solution. So, while you had obtained the correct solution set by serendipity (without proper justification), your textbook is actually wrong too in neglecting to eliminate its extraneous solution $x=-\dfrac1{\sqrt3}.$
|
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Regarding the solution of finding the remainder of $g(x^{12})$ divided by $g(x)$
Let $g(x) = x^5 + x^4 + x^3 + x^2 + x + 1.$ What is the remainder when the polynomial $g(x^{12})$ is divided by the polynomial $g(x)$?
I'm reading the solution for this and I don't understand how can someone come up with it.
It states
We have that $g(x^{12}) = x^{60} + x^{48} + x^{36} + x^{24} + x^{12} + 1$
which is still very clear. Then it goes to say that
Note that $(x - 1)g(x) = (x - 1)(x^5 + x^4 + x^3 + x^2 + x + 1) = x^6 - 1$
but I am not "noting" this. Where is this expression coming form? There is nothing natural to it.
Furhtermore they state that
Also,
\begin{align*}
g(x^{12}) - 6 &= (x^{60} + x^{48} + x^{36} + x^{24} + x^{12} + 1) - 6 \\
&= (x^{60} - 1) + (x^{48} - 1) + (x^{36} - 1) + (x^{24} - 1) + (x^{12} - 1).
\end{align*}
which is also not very natural to come up with. Where is this $6$ coming from?
The rest follows from $g(x^{12})-6$ being a multiple of $g$ and the remainder is thusfore $6$. I hate reading these kinds of solutions there is nothing to be learnt from this. It's just a bunch of sentences with not substance to it.
|
$(x - 1)g(x) = (x - 1)(x^5 + x^4 + x^3 + x^2 + x + 1) = x^6 - 1$
That product is "telescoping":
$$\begin{array}{ccccccc}
(x - 1)(x^5 + x^4 + x^3 + x^2 + x + 1)
&= &x^6 &+ x^5 &+ x^4 &+ x^3 &+ x^2 &+ x \\
& & &- x^5 &- x^4 &- x^3 &- x^2 &- x &-1 \\
&=& x^6&&&&&&-1
\end{array}$$
More generally: $$\sum_{k=0}^n x^k = \frac{x^{n+1}-1}{x-1}$$
|
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|
Simpler proof that $y^3[d^2y/dx^2]$ is a constant if $y^2=ax^2+bx+c$? here's my question
If $y^2=ax^2+bx+c$ then prove that $y^3[d^2y/dx^2]$ is a constant .
I have solved this using the conventional method, taking square root, differentiating w.r.t to x and using chain and quotient rule
But can't think of some alternative smaller and more efficient method ??
Can you ?
Also can someone suggest me any alternative to quotient rule
I need something more non conventional, time saving and vastly
Applicable method .
|
Using differentials, it is easy to see that
$$
2y dy = (2ax+ b) dx \tag{1}
$$
from which we can write
$$
u = \frac{dy}{dx} = \frac{2ax+b}{2y}
$$
Using again differentials and product rule,
\begin{eqnarray}
du
&=& \frac{2a}{2y} dx - \frac{2ax+b}{2y^2} dy
= \left[ \frac{a}{y} - \frac{(2ax+b)^2}{4y^3} \right] dx
\end{eqnarray}
where we have used (1).
Finally
\begin{eqnarray}
y^3\frac{du}{dx}
= y^3\frac{d^2y}{dx^2}
&=&
\left[ ay^2 - \frac{(2ax+b)^2}{4} \right]
=
\left[ a(ax^2+bx+c) - \frac{(2ax+b)^2}{4} \right]
=
ac-\frac{b^2}{4}
\end{eqnarray}
|
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|
What did I do wrong in solving $\int\sec^{-1} x\,{dx}$? I used integration by parts:
let u=$\sec^{-1}\,x$, dv=dx,
then du=$\frac{1}{|x|\sqrt{x^2-1}}$, v=x.
I = $x\sec^{-1}\,x\;-\;\int\frac{x}{|x|\sqrt{x^2-1}}dx\\$
Integration of $\int\frac{x}{|x|\sqrt{x^2-1}}dx$:
Let x=$\sec\theta$, then dx = $\sec\theta\tan\theta\,d\theta$. $\theta\in(0, \frac{\pi}2)\cup(\frac{\pi}2, \pi)\\$.
$\int\frac{x}{|x|\sqrt{x^2-1}}dx\,=\,\int\frac{\sec\theta}{|\sec\theta|\sqrt{\sec^2\theta-1}}\sec\theta\tan\theta\,d\theta\\\qquad\qquad\qquad=\,\int\frac{\sec\theta}{|\sec\theta|}\frac{\tan\theta}{|\tan\theta|}\sec\theta\,d\theta$
When $\theta\in(0, \frac{\pi}2),\;\sec\theta=x\text{, which is}\gt0$. $\tan\theta\gt0\text{, and }\tan\theta=\sqrt{x^2-1}$,
$\int\frac{x}{|x|\sqrt{x^2-1}}dx\,=\,\int\frac{\sec\theta}{\sec\theta}\frac{\tan\theta}{\tan\theta}\sec\theta\,d\theta\\\qquad\qquad\qquad=\,\int\sec\theta\,d\theta\\\qquad\qquad\qquad=\,\ln|\sec\theta+\tan\theta|+c\\\qquad\qquad\qquad=\,\ln|x+\sqrt{x^2-1}|+c$.
Then, I = $x\sec^{-1}\,x\;-\;\ln|x+\sqrt{x^2-1}|+c$.
I know that this is the right answer, but as I continue, I get a different answer for $\theta\in(\frac{\pi}2, \pi)$.
When $\theta\in(\frac{\pi}2, \pi)$, $\sec\theta=x\text{, which is }\lt0$, $\tan\theta\lt0\text{, and }\tan\theta=-\sqrt{x^2-1}$,
$\int\frac{x}{|x|\sqrt{x^2-1}}dx\,=\,\int\frac{\sec\theta}{-\sec\theta}\frac{\tan\theta}{-\tan\theta}\sec\theta\,d\theta\\\qquad\qquad\qquad=\,\int\sec\theta\,d\theta\\\qquad\qquad\qquad=\,\ln|\sec\theta+\tan\theta|+c\\\qquad\qquad\qquad=\,\ln|x-\sqrt{x^2-1}|+c\\\qquad\qquad\qquad=\,-\ln|x+\sqrt{x^2-1}|+c$
Then, I = $x\sec^{-1}\,x\;\boldsymbol{+}\;\ln|x+\sqrt{x^2-1}|+c$
The answer is $x\sec^{-1}\,x\;\boldsymbol{-}\;\ln|x+\sqrt{x^2-1}|+c\\$, what did I do wrong in the last part?
|
When $\theta\in(0, \frac{\pi}2),$
Then I = $x\sec^{-1}\,x\;-\;\ln|x+\sqrt{x^2-1}|+c$.
Small correction: $\theta\in[0, \frac{\pi}2).$ This corresponds to $x\ge1.$
When $\theta\in(\frac{\pi}2, \pi),$
Then I = $x\sec^{-1}\,x\;\boldsymbol{+}\;\ln|x+\sqrt{x^2-1}|+c$
Small correction: $\theta\in(\frac{\pi}2, \pi].$ This corresponds to $x\le-1.$
what did I
do wrong in the last part?
Nothing: you appear to have correctly shown that $$\int\operatorname{arcsec} x\,\mathrm dx= \begin{cases}x\operatorname{arcsec} x+\ln\left|x+\sqrt{x^2-1}\right|+C_1&\text{if }x\le-1;\\x\operatorname{arcsec} x-\ln\left|x+\sqrt{x^2-1}\right|+C_2&\text{if }x\ge1\end{cases}.$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Let $a$,$b$ be the roots of $x^2-6cx-7d=0$, and let $c$, $d$ be the roots of $x^2-6ax-7b=0$. Find $b+d$.
Let $a$,$b$ be the roots of the equation $x^{2}-6 c x-7 d=0$ and let $c$,$d$ be the roots of the equation $x^{2}-6ax-7 b=0$. $a,b,c,d$ are distinct. Find value of $b+d$.
Using the relation between roots and coefficients of an equation, I got $a+b+c+d=6(a+c)$, and hence getting $b+d= 5(a+c)$, but I don't know what to do next.
Also, $abcd= 49bd$, but I don't know how to use it here.
|
I like to work with a bit of generality whenever possible, so take the equations to be
$$x^2+pcx+qd=0 \qquad x^2+pax+qb=0$$
By Vieta's formula for the linear coefficients, we have a solvable linear system in $a$ and $c$:
$$
\left.\begin{align}
a+b=-pc \\ c+d=-pa
\end{align}\right\}\quad\to\quad a=\frac{b-dp}{p^2-1}\qquad c = \frac{d-bp}{p^2-1}\tag1$$
Then, by Vieta for the constant terms,
$$qd = ab=\frac{b-dp}{p^2-1}\,b \qquad\qquad qb=cd=\frac{d-bp}{p^2-1}\,d \tag2$$
Subtracting, and recalling the requirement that $b\neq d$,
$$q(d-b) = \frac{b^2-d^2}{p^2-1}=-\frac{(d-b) (b + d)}{p^2-1} \quad\to\quad b+d=-q(p^2-1) \tag3$$
For $p=-6$ and $q=-7$, the sum becomes $245$, in agreement with @user2661923's answer.
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|
Evaluate: $\prod_{n=1}^{n=9}(\sin(20n-10^\circ))$ The given answer is $2^{-8}$
My attempt
$$=\sin(10^\circ) \sin(30^\circ) \sin(50^\circ)\cdots\sin(170^\circ)$$
after this I multiply and divide to fill in the even multiples of $10^\circ$
$\dfrac{\sin(10)\sin(20)\sin(30) \sin(40)\cdots \sin(160) \sin(170)}{\sin(20)\sin(40)\sin(60)\cdots\sin(160)}$
after this I know there is some identity to simplify the numerator, and a slight modification of that also helps to solve the denominator, but I'm unsure what it is, and how to apply it.
Any help or hint will be appreciated.
Also, note that all angles I've used are measured in degrees.
|
$$A=\sin(10^\circ) \sin(30^\circ) \sin(50^\circ) \sin(70^\circ) \sin(90^\circ) \sin(110^\circ) \sin(130^\circ) \sin(150^\circ) \sin(170^\circ)$$
$$\sin(180^\circ-n)=\sin n \Rightarrow A=\sin^2(10^\circ)\sin^2(30^\circ)\sin^2(50^\circ)\sin^2(70^\circ)\sin(90^\circ)$$
$$\sin(30^\circ)=\frac12,\sin(90^\circ)=1\Rightarrow A=\frac14 \sin^2(10^\circ)\sin^2(50^\circ)\sin^2(70^\circ)$$
$$B=\sin(10^\circ)\sin(50^\circ)\sin(70^\circ)\Rightarrow A=\frac14 B^2$$
$$\sin(90^\circ-n)=\cos n\Rightarrow B=\sin(10^\circ)\cos(20^\circ)\cos(40^\circ)$$
$$2 \sin n \cos n = \sin(2n)\Rightarrow 8B\cos(10^\circ)=8 \sin(10^\circ)\cos(10^\circ)\cos(20^\circ)\cos(40^\circ)=\\
4\sin(20^\circ)\cos(20^\circ)\cos(40^\circ)=2\sin(40^\circ)\cos(40^\circ)=\sin(80^\circ)$$
$$\sin(90^\circ-n)=\cos n \Rightarrow 8B\cos(10^\circ)=\cos(10^\circ) \Rightarrow B=\frac18$$
$$A=\frac14 B^2 = \frac1{4\cdot 8^2}=2^{-8}$$
|
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|
The integral $\int_{0}^{1} \frac{ \log (1-x)}{1+x^2}dx$ Recently a very interesting result
$\int_0^1\frac{\ln(1-x)}{1+x^2}\mathrm{d}x+\int_0^1\frac{\arctan x}{x(1+x)}\mathrm{d}x=0$
has been proved in a more than elegant way. See Show that $\int_0^1\frac{\ln(1-x)}{1+x^2}\mathrm{d}x+\int_0^1\frac{\arctan x}{x(1+x)}\mathrm{d}x=0$
Here we find the value of the integral
$I=\int_{0}^1 \frac{\log(1-x)}{1+x^2} dx$
Let $x=\tan t$, then
$I=\int_{0}^{\pi/4} \log(1-\tan t) dt.$
By IVth property, we get
$I=\int_0^{\pi/4} \log (1-\tan(\pi/4-t)) dt=\int_{0}^{\pi/4} \log \left( 1-\frac{1-\tan x}{1+\tan x}\right) dt=\int_{0}^{\pi/4} \log (2 \tan x)~dx-\int_{0}^{\pi/4}\log(1+\tan x)dx=\frac{\pi}{4} \log 2+J-K.$
$J=\int_{0}^{\pi/4} \log \tan x ~dx=-C,$ see Definite integral $\int_0^{\pi/4}\log\left(\tan{x}\right)\ dx$
Let us work for $K=\int_{0}^{\pi/4} \log(1+\tan x) dx$, by IV property, again we get
$K=\int_0^{\pi/4} \log (1+\tan(\pi/4-t)) dt=\int_{0}^{\pi/4} \log \left( 1+\frac{1-\tan x}{1+\tan x}\right) dt=\int_{0}^{\pi/4} \log 2~ dx-K \implies K=\frac{\pi}{8} \log 2.$
Finally, we get $I=\frac{\pi}{8}\log 2-C,$ where $C$ is the Catalan constant.
What could be other interesting ways of finding $I$?
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Split the integrand\begin{align}
\int_{0}^1 \frac{\ln(1-x)}{1+x^2} dx
=& \int_{0}^1 \underset{= \frac\pi8 \ln2} {\frac{\ln \sqrt2}{1+x^2} }dx
+ \int_{0}^1 \underset{=-G}{\frac{\ln \frac{1-x}{1+x}}{1+x^2}}\overset{x\to\frac{1-x}{1+x}}{dx}
+ \int_{0}^1 \underset{K=- K=0}{\frac{\ln \frac{1+x} {\sqrt2 }}{1+x^2} }\overset{x\to\frac{1-x}{1+x}}{dx}
= \frac\pi8 \ln2-G
\end{align}
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|
Number of ways to form $24$ using $7$,$3$ and $2$
$\text{What is the number of ways to form }24\text{ using }7,2,\text{ and }3,\text{ zero or more times?}$
We can write $24$ as $7a+3b+2c$ . Since $\lfloor\frac{24}{7}\rfloor = 3,$ $\lfloor\frac{24}{3}\rfloor = 8$ and $\lfloor\frac{24}{2}\rfloor = 12$ ,We can say $0 \le a \le 3$ , $0 \le b \le 8$ and $0 \le c \le 12$.Then I put $a = 0,1,2,3$ and I got $11$ ways.
They are
$1. 7*3+ 2*0 +3*1$
$2. 7*2+ 2*5+ 3*0$
$3. 7*2+ 2*2 +3*2$
$4. 7*1 +2*7+ 3*1$
$ 5. 7*1 +2*4 +3*3$
$ 6. 7*1+ 2*1+ 3*5$
$ 7. 7*0+ 2*12 +3*0$
$8. 7*0 +2*9 +3*2$
$9. 7*0+ 2*6+ 3*4$
$10. 7*0 +2*3+ 3*6$
$11. 7*0 +2*0 +3*8$
Is there any efficient way of calculating this kind of problem??
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Note that once you've found $a$ and $b$, you can calculate $c$ directly:
$$c = \frac{24-7a-3b}{2}$$
This limits your search space to the 36 possible combinations of $(a, b)$ with $0 \le a \le 3$ and $0 \le b \le 8$.
Also, we need $c \ge 0$, which means that once we've set a value for $a$, we can constrain $7a + 3b \le 24$, or $b \le \frac{24 - 7a}{3}$. This reduces the number of valid $(a, b)$ combinations to 21.
For $c$ to be an integer, we must have $24 - 7a - 3b$ be even. This means that $a$ and $b$ must be either both even or both odd.
This gives us a reasonably efficient algorithm for enumerating the solutions.
for a = 0 to 3:
for b = (a mod 2) to floor((24 - 7a) / 3) step 2:
c = (24 - 7a - 3b) / 2
print(a, b, c)
Where a mod 2 equals 0 if a is even, or 1 if a is odd, thus ensuring that b has the same parity as a.
This can straightforwardly be converted to a computer program, or even done manually.
|
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|
Classification of singularity and Laurent series We have the following complex function:
$$f(z) = \frac{1}{z^2+4}$$
We need to get the classification of singularities and also the main part (The one with negative coefficients) of Laurent series around the point $2i$.
What we have done is the following:
$$z^2 +4 = (z+2i)(z-2i) \implies h \,\text{has a pole in 2i of degree 1 }$$
$$m = 1 \implies c_{-1} = \lim_{z\rightarrow 2i}\frac{1}{z+2i} = -\frac{i}{4} \implies $$
The main part of Laurent is $-\frac{i}{4}(z-2i)^{-1}$
However I don't really understand the procedure above. And usually when we solved problems like that we did something like $w = z-3$ if we would do it around the point $3$. Why don't we use $w = z-2i$ here ?
|
This technique is a convenient shortcut to determine the residuum of a Laurentseries expansion at a simple pole $z_0$. Doing it the long way we obtain
\begin{align*}
\color{blue}{f(z)=\frac{1}{z^2+4}}&=\frac{1}{(z-2i)(z+2i)}\tag{1}\\
&=\frac{1}{z-2i}\,\frac{1}{4i+(z-2i)}\\
&=\frac{1}{z-2i}\,\frac{1}{4i\left(1+\frac{z-2i}{4}\right)}\\
&\color{blue}{=\frac{1}{z-2i}\,\frac{1}{4i}}\left(1-\frac{z-2i}{4i}+\left(\frac{z-2i}{4i}\right)^2-\cdots\right)\tag{2}
\end{align*}
We determine from (2) the residuum of $f$ at $z=2i$ and get
\begin{align*}
\color{blue}{\lim_{z\to 2i}f(z)(z-2i)}=\frac{1}{4i}\left(1-0+0-0+\cdots\right)\color{blue}{=-\frac{i}{4}}
\end{align*}
Since $z_0$ is a simple pole, the main part consists of just one term and we have
\begin{align*}
\color{blue}{-\frac{i}{4}\frac{1}{z-2i}}
\end{align*}
Comparing (1) with (2) we observe we can also calculate the residuum by using as shortcut the representation (1) and obtain
\begin{align*}
\color{blue}{\lim_{z\to 2i}f(z)(z-2i)}=\lim_{z\to 2i}\frac{1}{z+2i}=\frac{1}{4i}\color{blue}{=-\frac{i}{4}}
\end{align*}
|
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|
Let $a$ and $b$ be roots of $x^2-7x+2$. Find the value of $a^6 + b^6$.
Let $a$ and $b$ be roots of $x^2-7x+2$. Find the value of $a^6 + b^6$.
Answer:
$a+b = 7, ab = 2$
$$\begin{align}
(a+b)^6 &= a^6 + 6a^5b+15a^4b^2+20a^3b^3+15a^2b^4+6ab^5+b^6 \\[4pt]
a^6 + b^6 &= (a+b)^6 - (6a^5b+15a^4b^2+20a^3b^3+15a^2b^4+6ab^5) \\
&= (a+b)^6 - (6ab(a^4 + b^4) + 15a^2b^2 (a^2 + b^2) + 20(ab)^3)
\end{align}$$
now,
$$\begin{align}
a^4 + b^4 &= (a+b)^4 - (4a^3b + 6a^2b^2 + 4ab^3) \\
&= (a+b)^4 - (4ab(a^2 + b^2) + 6(ab)^2) \\
&= (a+b)^4 - (4ab((a + b)^2 - 2ab) + 6(ab)^2) \\
&= 7^4 - (4(2)(7^2 - 2(2)) + 6(2)^2) \\
&= 2017
\end{align}$$
so
$$\begin{align}
&\phantom{=}\; (a+b)^6 - (6ab(a^4 + b^4) + 15a^2b^2 (a^2 + b^2) + 20(ab)^3)\\
&= 7^6 - (6\cdot2\cdot(2017) + 15(2)^2 (7^2 - 2(2)) + 20(2)^3) \\
&= 90585
\end{align}$$
correct?
|
Given $a$ and $b$ be the roots of the equation $x^2-7x+2=0$. Sum of roots, $$a+b=7$$
Product of roots, $$ab=2.$$
Now, $(a-b)^2=(a+b)^2-4ab=49-8=41$.
$\Rightarrow a-b=\pm\sqrt{41}$.
Taking $a-b=\sqrt{41}$.
On solving equation $a+b=7$ and $a-b=\sqrt{41}$.
We obtain, $a=\frac{7+\sqrt{41}}{2}$ and $b=\frac{7-\sqrt{41}}{2}.$
|
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|
Find the minimum value of a trigonometric function If the minimum value of $f\left(x\right)=\left(1+\frac{1}{\sin ^6\left(x\right)}\right)\left(1+\frac{1}{\cos ^6\left(x\right)}\right),\:x\:∈\:\left(0,\:\frac{\pi }{2}\right)$ is $m$, find $\sqrt m$.
How do I differentiate this function without making the problem unnecessarily complicated? If there are any other methods to finding the minimum value I am open to those too.
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Without calculus, let $\,a = \sin^2(x)\,$, $\,b = \cos^2(x)\,$, then $\,f(x) = \left(1+\dfrac{1}{a^3}\right)\left(1+\dfrac{1}{b^3}\right)\,$, where $\,a,b \ge 0\,$ and $\,a+b=1\,$. By the generalized means inequalities:
*
*$\sqrt{ab} \le \dfrac{a+b}{2} = \dfrac{1}{2} \quad\implies\quad \dfrac{1}{ab} \ge 4 \quad\implies\quad \dfrac{1}{a^3b^3} \ge 4^3 = 64\;$;
*$\sqrt[3]{\dfrac{2}{\frac{1}{a^3}+ \frac{1}{b^3}}} \le \dfrac{a+b}{2} = \dfrac{1}{2} \quad\implies\quad \dfrac{1}{a^3}+ \dfrac{1}{b^3} \ge 16\;$.
Then the following inequality holds, with equality iff $a=b = \dfrac{1}{2}$, or $|\sin(x)| = |\cos(x)| = \dfrac{1}{\sqrt{2}}$:
$$
f(x) = \left(1+\frac{1}{a^3}\right)\left(1+\frac{1}{b^3}\right) \;=\; 1 + \frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{a^3b^3} \;\ge\; 1 + 16 + 64 = 81
$$
|
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|
Getting two different answers on differentiating $\cos^{-1}(\frac{3x+4\sqrt{1-x^2}}{5})$
Question given in my book asks to find $\frac{dy}{dx} $ from the following equation.$$y=\cos^{-1}\left(\frac{3x+4\sqrt{1-x^2}}{5}\right)$$
My Attempt:
Starting with substitutions,
*
*Putting $\frac35=\cos\alpha\implies \frac45 = \sin\alpha$.
*Putting $x = \cos\beta\implies \sqrt{1-x^2} = \sin\beta$.
$$\begin{align}&y=\cos^{-1}\Big(\frac{3x+4\sqrt{1-x^2}}{5}\Big)\\ \implies& y = \cos^{-1}\Big(\frac{3}{5}x + \frac{4}{5}\sqrt{1-x^2}\Big)\\\implies& y = \cos^{-1}(\cos\alpha\cos\beta + \sin\alpha \sin \beta)\\\implies& y = \cos^{-1}[\cos(\alpha-\beta )] \tag{1}\\\implies& y = \alpha-\beta\\\implies& y = \cos^{-1}\left(\frac35\right) - \cos^{-1}(x)\\\implies&\color{blue}{\boxed{ \dfrac{dy}{dx} =\frac{1}{\sqrt{1-x^2}}}}.\end{align}$$
But, if I consider $(1.)$ again,
$$y = \cos^{-1}\left[\cos\color{red}{(\alpha-\beta)}\right]$$
This is also equals to, $$\cos^{-1}\left[\cos\color{red}{(\beta - \alpha)}\right].$$differentiating this, will give the negative of the answer which I got earlier.
My book shows that $\frac{-1}{\sqrt{1-x^2}}$ is correct. But why?
I think the mistakes lie in the very first step of substitution i.e.,$\frac35=\cos\alpha$ doesn't imply $\frac45 = \sin\alpha$. It should be $\sin\alpha = \pm \frac45$.
Similary $x = \cos\beta$ doesn't imply $\sqrt{1-x^2} = \sin\beta$ instead $\sin\beta = \pm \sqrt{1-x^2}$.
But how can I make sure that in the equation $(1.)$, $(\alpha - \beta)$ lies in the principal branch of the inverse cosine function?
|
$\cos^{-1}(\cos x)$ needn’t always be equal to $x$. See the graph below for $\cos^{-1}(\cos x)$ :
When $x < 0$ then $\cos^{-1}(\cos x) = -x$.
Now if $x < 0.6$ then $\beta >\alpha$ because $\cos^{-1} x$ is a decreasing function in $[-1, 1]$. So ideally there should be two cases, $x < 0.6$ and $x > 0.6$. Also note that in your substitution, you are perfectly entitled to restrict the values of $\alpha$ to $(0,\frac{\pi}{2})$ so there is nothing wrong in taking $\sin x =0.8$.
As you will notice, there is a discontinuity in the graph at $x= 0.6$:
|
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|
Find $\int \sin^4x dx$ In my textbook, I came upon the following problem:
Problem: Find $\int \sin^4x dx$
Since there was only given a solution sketch and I came to a different conclusion in one step, I ask if my following solution is correct.
Solution: In my solution, I use the following identities:
$$\sin^2x = \frac{1-\cos2x}{2} \tag{1} \label{1}$$
$$\cos^2x = \frac{1+\cos2x}{2} \tag{2} \label{2}$$
Now we can write
$$\int \sin^4x dx = \int \left( \frac{1-\cos2x}{2} \right)^2 dx = \int\frac{1}{4}dx - \frac{1}{2}\int \cos2xdx + \frac{1}{4} \int cos^22xdx$$
My textbook gave the last term in this equation as $\int\frac{1}{4}dx - \int \cos2xdx + \frac{1}{4} \int cos^22xdx$. But, since $(1-\cos2x)^2 = 1 - 2\cos2x + \cos^22x$ the middle integral should simplify to $- \frac{1}{2}\int \cos2xdx$. Is this correct? If not, how do we come to the conclusion of $-\int \cos2xdx$?
To complete the integration, we have
$$\int \frac{1}{4} dx = \frac{x}{4}$$
$$ - \frac{1}{2} \int \cos 2xdx = - \frac{1}{4}\int \cos u du = - \frac{1}{4} \sin2x = - \frac{\sin2x}{4}$$
and
$$\frac{1}{4} \int cos^22xdx = \frac{1}{4} \int \frac{1 + cos4x}{2}dx = \frac{4x + \sin 4x}{32}$$
so
$$\int \sin^4x dx = \frac{x}{4} - \frac{\sin2x}{4} + \frac{4x + \sin 4x}{32}$$
Are these conclusions and computations correct? Thanks for any help!
EDIT: According to the correction of @Eevee Trainer I now use the correct identity $\cos^2x = \frac{1+\cos2x}{2}$. The rest of the computation should be correct. Thanks all for your help and comments!.
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I like to rewrite things so as to reduce errors; as $\sin^4 x$ is even of small degree (as a polynomial in $\sin, \cos$), it can be written in terms of $\cos 2 x$ and $\cos 4x.$ After that the integral is immediate, finally we may wish to expand that back into a polynolial in sine and cosine.
First
$$ \cos 2x = 2 \cos^2 x - 1 = 1 - 2 \sin^2 x$$
Good start. Next $ \cos 4x = 2 \cos^2 2 x - 1 .$ However we know $ \cos 2x = 1 - 2 \sin^2 x$ and find $\cos^2 2x = 1 - 4 \sin^2 x + 4 \sin^4 x.$ Plug that into $ \cos 4x = 2 \cos^2 2 x - 1 ,$ I get
$$ \cos 4x = 8 \sin^4 x - 8 \sin^2 x + 1$$
Anyway, my calculations may be checked for themselves, before starting any calculus, and any necessary corrections made.
Again, I get
$$ \sin^4 x = \; \frac{1}{8} \; \left( \cos 4x - 4 \cos 2 x + 3 \right) = \frac{\cos 4x}{8} - \frac{\cos 2x}{2} + \frac{3}{8} $$
The integral...
$$ \int \sin^4 x dx = \frac{\sin 4x}{32} - \frac{\sin 2x}{4} + \frac{3x}{8} $$
I note that you did not combine the two $x$ terms in your last line
|
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|
When are matrices $AB$ and $BA$ similar? If $A$ or $B$ is invertible, it is easy to see that $AB$ and $BA$ are similar. I'm curious about when (maybe a strong sufficient or necessary condition) they become similar.
Here's what I have done:
There exists invertible matrices $P, Q$ such that $A=P\begin{pmatrix} I_r &\\ &0\end{pmatrix}Q$ where $r=\text{rank} A$. Let $B=Q^{-1}\begin{pmatrix} B_{11} & B_{12}\\ B_{21}&B_{22}\end{pmatrix}P^{-1}$.Then $AB=P\begin{pmatrix} B_{11} & B_{12}\\ 0&0\end{pmatrix}P^{-1}$ and $BA=Q^{-1}\begin{pmatrix} B_{11} & 0\\ B_{21}&0\end{pmatrix}Q$. Hence we just need to find when $\begin{pmatrix} B_{11} & B_{12}\\ 0&0\end{pmatrix}$ and $\begin{pmatrix} B_{11} & 0\\ B_{21}&0\end{pmatrix}$ are similar.
Now if $B_{11}$ is invertible, let $R=\begin{pmatrix} C &0\\D&F\end{pmatrix}$ and then $R^{-1}\begin{pmatrix} B_{11} & 0\\ B_{21}&0\end{pmatrix}R=\begin{pmatrix} C^{-1}B_{11}C & 0\\ -F^{-1}DC^{-1}B_{11}C+F^{-1}B_{21}C&0\end{pmatrix}$. Known that $\begin{pmatrix} B_{11} & B_{12}\\ 0&0\end{pmatrix}$ is similar to $\begin{pmatrix} B_{11}^t & 0\\ B_{12}^t&0\end{pmatrix}$, we just check when the matrices above are equal.
First, there exists $C$ s.t. $C^{-1}B_{11}C=B_{11}^t$ since $B_{11}$ and $B_{11}^t$ are similar.
Second, $-F^{-1}DC^{-1}B_{11}C+F^{-1}B_{21}C=B_{12}^t\Leftrightarrow DB_{11}^t+FB_{12}^t=B_{21}C$. Hence if $\text{rank} \begin{pmatrix} B_{11} &B_{12} \end{pmatrix} = r$ then $D, F$ exist. Disappointingly, however, we need $F$ to be invertible. I cannot overcome this problem but I still have the romantic guess that $\text{rank}AB=\text{rank}BA=\text{rank}A$ even $\text{rank} AB=\text{rank} BA$ might be a sufficient condition (please tell me a counter-example so I could give up).
Of course, $F$ might not be invertible really. So if there is a way then that way must consider both $AB$ and $BA$. A weaker result we can get is if $B_{11}$ is invertible then $F=I, D=(B_{21}C-B_{12}^t){B_{11}^t}^{-1}$ satisfies the equation. But I cannot tell what $B_{11}$'s invertibility actually means. I wonder if there is a direct relation from $A$, $B$ (or $AB$, $BA$) to $B_{11}$.
I'll appreciate any discussions about this topic.
P.S. I know $AB$ and $BA$ are not similar when $A=\begin{pmatrix} 0 &1\\0&0\end{pmatrix}$ and $B=\begin{pmatrix} 1 &0\\0&0\end{pmatrix}$. That's just when $F$ is not invertible. Thanks for commenting anyway.
|
In general, two matrices $X$ and $Y$ are similar over an algebraically closed field $F$ if an only if $\operatorname{rank}\left((\lambda I-X)^k\right)=\operatorname{rank}\left((\lambda I-Y)^k\right)$ for all $\lambda\in F$ and $k\ge1$. However, in case $X=AB$ and $Y=BA$ for some square matrices $A$ and $B$, one only needs to check the previous condition for $\lambda=0$. In other words, $AB$ and $BA$ are similar if and only if $\operatorname{rank}\left((AB)^k\right)=\operatorname{rank}\left((BA)^k\right)$ for every integer $k\ge1$.
Merely $\operatorname{rank}(AB)=\operatorname{rank}(BA)$ is not enough to guarantee that $AB$ is similar to $BA$. Here is a random counterexample where $\operatorname{rank}(AB)=\operatorname{rank}(BA)=2$ but $(AB)^2=0\ne(BA)^2$:
\begin{aligned}
AB&=\pmatrix{1&1&0&0\\ 0&0&1&0\\ 0&0&1&0\\ 0&1&0&0}
\pmatrix{0&1&0&0\\ 0&0&1&0\\ 0&0&0&0\\ 0&1&1&1}
=\pmatrix{0&1&1&0\\ 0&0&0&0\\ 0&0&0&0\\ 0&0&1&0},\\
BA&=\pmatrix{0&1&0&0\\ 0&0&1&0\\ 0&0&0&0\\ 0&1&1&1}
\pmatrix{1&1&0&0\\ 0&0&1&0\\ 0&0&1&0\\ 0&1&0&0}
=\pmatrix{0&0&1&0\\ 0&0&1&0\\ 0&0&0&0\\ 0&1&2&0}.\\
\end{aligned}
However, that $\operatorname{rank}(AB)=\operatorname{rank}(BA)=\operatorname{rank}(A)$ is sufficient for the similarity between $AB$ and $BA$. In this case, we may continue from your approach. By Roth's removal rule, both $\pmatrix{B_{11}&B_{12}\\ 0&0}$ and $\pmatrix{B_{11}&0\\ B_{21}&0}$ and similar to $\pmatrix{B_{11}&0\\ 0&0}$. In fact, we can exhibit the similarity transforms explicitly:
\begin{aligned}
&\pmatrix{I&B_{11}^{-1}B_{12}\\ 0&I}
\pmatrix{B_{11}&B_{12}\\ 0&0}
\pmatrix{I&-B_{11}^{-1}B_{12}\\ 0&I}\\
&=\pmatrix{B_{11}&0\\ 0&0}\\
&=\pmatrix{I&0\\ -B_{21}B_{11}^{-1}&I}
\pmatrix{B_{11}&0\\ B_{21}&0}
\pmatrix{I&0\\ B_{21}B_{11}^{-1}&I}.
\end{aligned}
|
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|
Show $2\mid (n^7 -n), \forall n \in\mathbb{N}$. Show $2\mid (n^7 -n), \forall n \in\mathbb{N}$.
$n^7 \equiv n \bmod 2\implies n^6 \equiv 1 \bmod 2$
Using Fermat's little theorem, it is easy to see that:
$n^1 \equiv 1\bmod 2\implies n^a \equiv 1^a\bmod 2, \forall a \in \mathbb{N}$
Algebraic approach:
$(n^3-1)(n^3+1)\equiv 0\bmod 2$.
So, if either of the two factors is divisible by $2$, then fine.
So, now have:
*
*$n^3 \equiv 1 \pmod 2$
*$n^3 \equiv -1 \pmod 2$
So, how to pursue? Cannot get anything by stating $n^3 = 1 + 2k, k\in \mathbb {Z}$, or $n^3 = -1 + 2l, l\in \mathbb{Z}$.
Nor, will gain anything by assuming a form of $n$.
|
$n^7-n = n(n^6-1) = n(n^3-1)(n^3+1)=n(n-1)(n^2+n+1)(n^3+1)$. Observe the product consists of $2$ consecutive integers $n,n-1$ hence divisible by $2$.
|
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|
A specific binomial summation identity Let $m$ and $n$ be positive integers with $m < n$. Prove
\begin{equation}
\left(\sum_{k=0}^m \binom{m}{k}\frac{(-1)^k}{n-k} \right)\left(\sum_{k=0}^m \binom{n}{k}\frac{(-1)^k}{k+1} \right) =\sum_{k=0}^m \binom{m}{k}\frac{(-1)^k}{(n-k)(k+1)}
\end{equation}
I'm not sure how to approach this problem. A Hint would be greatly appreciated!
|
We seek to prove for $n,m\ge 1$ with $m\lt n$ that
$$\sum_{k=0}^m {m\choose k} \frac{(-1)^k}{n-k}
\sum_{k=0}^m {n\choose k} \frac{(-1)^k}{k+1}
= \sum_{k=0}^m {m\choose k} \frac{(-1)^k}{(n-k)(k+1)}.$$
The RHS is
$$\frac{1}{n+1} \sum_{k=0}^m {m\choose k} \frac{(-1)^k}{n-k}
+ \frac{1}{n+1} \sum_{k=0}^m {m\choose k} \frac{(-1)^k}{k+1}.$$
The second term is
$$\frac{1}{n+1} \frac{1}{m+1} \sum_{k=0}^m
{m+1\choose k+1} (-1)^k
= \frac{1}{n+1} \frac{1}{m+1}.$$
This means we evidently require
$$S_{m, n} = \sum_{k=0}^m {m\choose k} \frac{(-1)^k}{n-k}.$$
With this in mind we introduce
$$f(z) = m! (-1)^m \frac{1}{n-z} \prod_{q=0}^m \frac{1}{z-q}.$$
This has the property that for $0\le k\le m$
$$\mathrm{Res}_{z=k} f(z)
= m! (-1)^m \frac{1}{n-k} \prod_{q=0}^{k-1} \frac{1}{k-q}
\prod_{q=k+1}^m \frac{1}{k-q}
\\ = m! (-1)^m \frac{1}{n-k}
\frac{1}{k!} \frac{(-1)^{m-k}}{(m-k)!}
= {m\choose k} \frac{(-1)^k}{n-k}.$$
Now with residues summing to zero and the residue at infinity being
zero we get
$$S_{m,n} = -\mathrm{Res}_{z=n} f(z)
= m! (-1)^m \prod_{q=0}^m \frac{1}{n-q}
\\ = (-1)^m \frac{m! \times (n-m-1)!}{n!}
= \frac{(-1)^m}{m+1} {n\choose m+1}^{-1}.$$
This means we need to show
$$\frac{(-1)^m}{m+1} {n\choose m+1}^{-1}
\sum_{k=0}^m {n\choose k} \frac{(-1)^k}{k+1}
= \frac{1}{n+1} \frac{(-1)^m}{m+1} {n\choose m+1}^{-1}
+ \frac{1}{n+1} \frac{1}{m+1}$$
which is
$$(-1)^m
\sum_{k=0}^m {n\choose k} \frac{(-1)^k}{k+1}
= \frac{1}{n+1} (-1)^m
+ \frac{1}{n+1} {n\choose m+1}.$$
Note however that the LHS is
$$\frac{(-1)^m}{n+1} \sum_{k=0}^m {n+1\choose k+1} (-1)^k
= \frac{1}{n+1} (-1)^m +
\frac{1}{n+1} (-1)^m \sum_{k=-1}^m {n+1\choose k+1} (-1)^k
\\ = \frac{1}{n+1} (-1)^m
- \frac{1}{n+1} (-1)^m \sum_{k=0}^{m+1} {n+1\choose k} (-1)^k.$$
Working with the sum we find
$$[z^{m+1}] \frac{1}{1-z}
\sum_{k\ge 0} {n+1\choose k} z^k (-1)^k
= [z^{m+1}] \frac{1}{1-z}
(1-z)^{n+1}
\\ = (-1)^{m+1} {n\choose m+1}.$$
We finally obtain
$$\frac{1}{n+1} (-1)^m
- \frac{1}{n+1} (-1)^m
(-1)^{m+1} {n\choose m+1}
= \frac{1}{n+1} (-1)^m
+ \frac{1}{n+1} {n\choose m+1}$$
which is the RHS as desired.
|
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|
Quick evaluation of definite integral $\int_0^\pi \frac{\sin 5x}{\sin x}dx$
Find$$\int_0^{\pi}\frac{\sin 5x}{\sin x}dx$$
I can solve it by involving polynomials in sine and cosine as shown in the links below, but it’s huge (doing double angle formulas twice; I noticed that using polynomials in cosine is better because the integral spits out sines which are 0 between the limits) so I want a faster method, if it exists. Please don’t use contour integration:)
The only thing I noticed is that the integrand is symmetric about the midpoint in the given interval, i.e.$$\frac{\sin 5x}{\sin x}= \frac{\sin 5(\pi-x)}{\sin(\pi- x)}.$$
Determine the indefinite integral $\int \frac{\sin x}{\sin 5x}dx$
Integral of $\int \frac{\sin(3x)}{\sin(5x)} \, dx$
Expressing $\frac {\sin(5x)}{\sin(x)}$ in powers of $\cos(x)$ using complex numbers
|
We can generalized for positive integer k:
$\displaystyle \frac{\sin((2k-1)\,x)}{\sin(x)}
= 1 + 2\cos(2x) + 2\cos(4x) \;+\;...\;+\; 2\cos((2k-2)\,x)
$
Prove by induction:
Base case, $k=1$, is obviously true: $\displaystyle \frac{\sin x}{\sin x} = 1$
Assume formula is true for $n=k$, prove it is true for $n=k+1$
$\begin{align} \displaystyle
\frac{\sin((2k-1)\,x)}{\sin(x)} + 2\cos((2k)\,x)
&=\frac{\sin((2k-1)\,x) + 2\cos((2k)\,x)×\sin(x)}{\sin(x)} \\
&=\frac{\sin((2k-1)\,x) + \sin((2k+1)\,x) - \sin((2k-1)\,x)}{\sin(x)} \\
&= \frac{\sin((2(k+1)-1)\,x)}{\sin(x)}
\end{align}$
QED
$\displaystyle ⇒ \int_0^\pi \frac{\sin((2k-1)\,x)}{\sin(x)}\;dx = \pi$
More direct way for proof, express as telescoping series.
$\sin((2k-1)\,x)$
$= \sin(x) + [\sin(3x)-\sin(x)] + [\sin(5x)-\sin(3x)]
\;+\; ... \;+\; [\sin((2k-1)\,x) - \sin((2k-3)\,x)]$
$= \sin(x) + 2\cos(2x)\sin(x) + 2\cos(4x)\sin(x)
\;+\; ... \;+\; 2\cos((2k-2)\,x)\sin(x)
$
Divide both side by $\sin(x)$, we have the required proof.
Proof using complex numbers, $z = e^{\,i\,x}$
$\begin{align} \displaystyle
\frac{\sin((2k-1)\,x)}{\sin x} &=
\frac{z^{2k-1}-1/z^{2k-1}}{z-1/z} \\
&= \left(\frac{z^{4k-2}-1}{z^2-1}\right)\bigg/z^{2k-2} \\
&= \frac{1+z^2+z^4 \;+\;...\;+\;z^{4k-4}}{z^{2k-2}}\\
&= 1+(z^2+1/z^2)+(z^4+1/z^4) \;+\;...\;+\;(z^{2k-2}+1/z^{2k-2}) \\
&= 1 + \;\;2\cos(2x)\;\, + \;\;2\cos(4x)\;\, \;+\;...\;+\;\;2\cos((2k-2)\,x)
\end{align}$
|
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|
From 7 consonants and 5 vowels, how many words can be formed consisting of 4 consonants and 3 vowels if letters may be repeated? cannot be repeated?
From $7$ consonants and $5$ vowels, how many words can be formed consisting of $4$ consonants and $3$ vowels if:
*
*Any letter can be repeated.
*No letter can be repeated.
My answer for the 1st part was $7 \cdot 7 \cdot 7 \cdot 7 \cdot 5 \cdot 5 \cdot 5$ (or $7^4 \cdot 5^3$). The way I thought about it is there are $4$ available slots for $7$ consonants that can be repeated so despite taking a consonant which we call it $c_1$, the second/third/fourth slot might also be $c_1$ so each of the $4$ slots has $7$ consonants to choose from so its $7 \cdot 7 \cdot 7 \cdot 7$. Same case for vowels which is $5 \cdot 5 \cdot 5$. Multiply those together to get total number of ways where the letter is repeated in a word which is $7^4 \cdot 5^3$.
However it turned out that the correct answer is actually $7^7 (7C4 \cdot 5C3)$. I don't understand the logic behind this. I don't get why not only the combinations nCr is used which is supposed to be used for the cases that don't require order and repetition, but also it is multiplied by $7^7$.
For the 2nd part, since the letter cannot be repeated then the available consonants and vowels decreases every time you choose from them so its $7 \cdot 6 \cdot 5 \cdot 4 \cdot 5 \cdot 4 \cdot 3$ (or $7P4 \cdot 5P3$).
However that's not the case because the correct answer is $7! \cdot (7C4 \cdot 5C3)$.
|
From $7$ consonants and $5$ vowels, how many words can be formed consisting of $4$ consonants and $3$ vowels if any letter can be repeated?
Your answer $7^4 \cdot 5^3$ is the number of ways the first four positions can be filled with consonants and the last three positions can be filled with vowels when letters may be repeated. However, the problem does not specify the positions of the consonants and vowels, so we must select them.
Notice that selecting which four of the seven positions will be filled with consonants also determines the positions of the three vowels since each of the remaining three positions must be filled with vowels.
We must select which four of the seven positions must be filled with consonants, which can be done in $\binom{7}{4}$ ways. Each of those four selected positions can be filled with a consonant in seven ways. Each of the remaining three positions can be filled with a vowel in five ways. Hence, there are
$$\binom{7}{4}7^4 \cdot 5^3$$
words that can be formed consisting of four consonants and three vowels selected from an alphabet of seven consonants and three vowels with repetition.
The stated answer $7^7\dbinom{7}{4}\dbinom{5}{3}$ is nonsense.
From $7$ consonants and $5$ vowels, how many words can be formed consisting of $4$ consonants and $3$ vowels if no letter can be repeated?
Your answer $P(7, 4)P(5, 3)$ is the number of ways of filling the first four positions with distinct consonants and the last three positions with distinct vowels. However, the problem does not specify the positions of the consonants and vowels, so we must select them.
As above, we select four of the seven positions for the consonants. There are $P(7, 4)$ ways to arrange four of the seven consonants in those positions and $P(5, 3)$ ways to arrange three of the five vowels in the remaining three positions. Hence, there are
$$\binom{7}{4}P(7, 4)P(5, 3)$$
words that can be formed consisting of four distinct consonants and three distinct vowels selected from an alphabet of seven consonants and five vowels.
Notice that
$$\binom{7}{4}P(7, 4)P(5, 3) = \frac{7!}{4!3!} \cdot \frac{7!}{3!} \cdot \frac{5!}{2!} = 7! \cdot \frac{7!}{4!3!} \cdot \frac{5!}{2!3!} = 7!\binom{7}{4}\binom{5}{3}$$
The author of your book arrived at the answer by selecting four of the seven consonants in $\binom{7}{4}$ ways, selecting three of the five vowels in $\binom{5}{3}$ ways, and then arranging the seven distinct selected letters in $7!$ ways.
|
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|
Expand the function $f(z) = \frac{1+2z^2}{z^2+z^4} $ into power series of $z$ in all areas of convergence. I'm studying for my upcoming complex analysis qualifying exam by working through problems in past exams. For this problem, I'd like to know (1) if my answer is correct and complete (i.e. whether I've made any errors/omissions), and (2) if there are any better ways of doing this. I'm specifically curious about whether or not there are other areas of convergence that I need to address. Thanks!
Question:
Expand the function
$$
f(z) = \frac{1+2z^2}{z^2+z^4}
$$
into power series of $z$ in all areas of convergence.
Attempted Solution:
We begin by noting that this function has poles at $z=0$ and $z=\pm i$. Then we proceed by recalling the geometric series
$$
\sum_{k=0}^{\infty} z^k = \frac{1}{1-z},
$$
which converges for $|z|<1$. Using this form, we can rewrite $f(z)$ as follows
\begin{align*}
f(z) &= \frac{1+2z^2}{z^2+z^4} \\
&= \frac{1}{z^2+z^4} + \frac{2z^2}{z^2+z^4} \\
&= \frac{1}{z^2(1-(-z^2))} + \frac{2z^2}{z^2(1-(-z^2))} \\
&= \frac{1}{z^2}\left(\frac{1}{1-(-z^2)}\right) + 2\left(\frac{1}{1-(-z^2)}\right)\\
&= \left(\frac{1}{z^2}+2\right)\left(\frac{1}{1-(-z^2)}\right)\\
&= \left(\frac{1}{z^2}+2\right) \sum_{k=0}^{\infty} (-z^2)^k \\
&= \sum_{k=0}^{\infty} \left(z^{-2}+2\right)(-1)^k z^{2k}
\end{align*}
for $0<|z|<1$, since there is a pole at $z=0$.
|
Since we are looking for a power series we are looking for a representation
\begin{align*}
f(z)=\sum_{k=0}^\infty a_k\left(z-z_0\right)^k
\end{align*}
evaluated at center $z_0$.
We start with a partial fraction decomposition
\begin{align*}
\color{blue}{f(z)}&=\frac{1+2z^2}{z^2\left(1+z^2\right)}\\
&\,\,\color{blue}{=\frac{1}{z^2}+\frac{i}{2}\,\frac{1}{z+i}-\frac{i}{2}\,\frac{1}{z-i}}\tag{1}
\end{align*}
We have poles at $0$ and $\pm i$. So we can expand each of the terms at $z_0\in\mathbb{C}\setminus\{0,\pm i\}$ in a power series. We obtain
\begin{align*}
\frac{1}{z^2}&=\frac{1}{\left(z-z_0+z_0\right)^2}=\frac{1}{z_0^2}\,\frac{1}{\left(1+\frac{z-z_0}{z_0}\right)^2}\\
&=\frac{1}{z_0^2}\sum_{k=0}^\infty (-1)^k(k+1)\left(\frac{z-z_0}{z_0}\right)^k\qquad\qquad |z-z_0|<|z_0|\tag{2.1}
\\
\frac{i}{2}\,\frac{1}{z+i}&=\frac{i}{2}\,\frac{1}{(z-z_0)+(z_0+i)}\\
&=\frac{i}{2}\,\frac{1}{z_0+i}\,\frac{1}{1+\frac{z-z_0}{z_0+i}}\\
&=\frac{i}{2}\,\frac{1}{z_0+i}\sum_{k=0}^\infty(-1)^k\left(\frac{z-z_0}{z_0+i}\right)^k\qquad\qquad |z-z_0|<|z_0+i|\tag{2.2}
\\
\frac{i}{2}\,\frac{1}{z-i}&=\frac{i}{2}\,\frac{1}{(z-z_0)+(z_0-i)}\\
&=\frac{i}{2}\,\frac{1}{z_0-i}\,\frac{1}{1+\frac{z-z_0}{z_0-i}}\\
&=\frac{i}{2}\,\frac{1}{z_0-i}\sum_{k=0}^\infty(-1)^k\left(\frac{z-z_0}{z_0-i}\right)^k\qquad\qquad |z-z_0|<|z_0-i|\tag{2.3}
\end{align*}
Putting (1) and (2.1) to (2.3) together we conclude $f(z)$ admits the power series representation
\begin{align*}
\color{blue}{f(z)}&=\frac{1+2z^2}{z^2\left(1+z^2\right)}\\
&\,\,\color{blue}{=\sum_{k=0}^\infty(-1)^k\left(\frac{k+1}{z_0^{k+2}}+\frac{i}{2}\frac{1}{(z_0+i)^{k+1}}
-\frac{i}{2}\frac{1}{(z_0-i)^{k+1}}\right)\left(z-z_0\right)^k}\\
\end{align*}
evaluated at $z_0\in\mathbb{C}\setminus\{0,\pm i\}$ and convergent in
\begin{align*}
\{z\in\mathbb{C}: |z-z_0|<|z_0| \text{ and } |z-z_0|<|z_0+i| \text{ and } |z-z_0|<|z_0-i|\}
\end{align*}
|
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|
Finding the range of $\frac {2x^2+x-3}{x^2+4x-5}$ I solved for the range as follows:
Setting $f(x)=\frac {2x^2+x-3}{x^2+4x-5} =y$, I rearranged it to get a quadratic in x.
$$(y-2)x^{2}+ (4y-1)x +(3-5y)=0$$
Next, using $\Delta \ge 0$, I got
$$(4y-1)^2-4(y-2)(3-5y) \ge 0$$
Which boiled down to
$$(6y-5)^2 \ge 0$$
And this gave me $$ y \in R$$
Next, the value of $x$ for which $y=2$ is $x=1$, for which the function isn't defined, so that gives me $$y \in R - \{2\}$$
However, the solution is $$y \in R- \{\frac{5}{6}, 2\}$$
I understand that the original function is identical to $$g(x) =\frac{2x+3}{x+5}
\forall x \in R - \{1\}$$
and that $\frac{5}{6} =g(1)$, but in the original function $f(x)$ I found $y=2$ corresponded to $x=1$ and therefore excluded it, but if $x= 1$ also corresponds to $y=\frac{5}{6}$ then wouldn't this $not$ be a function, as $x=1$ would then be associated with two different $y$ values?
I've read the answers in Why D≥0 while finding the range of rational functions and Finding the range of $y =\frac{x^2+2x+4}{2x^2+4x+9}$ (and $y=\frac{\text{quadratic}}{\text{quadratic}}$ in general) but I'm still unsure of how to apply the information from those to figure out what values of $y$ need to be excluded when dealing with such questions.
While I am familiar with derivatives and limits to a certain degree, we were assumed to $\underline {not}$ know calculus when we were taught this and solved such problems.
I apologise if there are any issues with formatting, this is my first time using Latex.
|
Alternative approach, along the lines of R. J. Mathar in the comments.
For all $x\in\mathbb{R}\setminus\{1,-5\},$
$$ \frac{2x^2 + x -3}{x^2 + 4x - 5} \equiv \frac{2(x^2 + 4x - 5) - 7x + 7}{x^2 + 4x - 5} \equiv 2 - \frac{7x-7}{x^2 + 4x - 5} $$
$$ \equiv 2 - \frac{7(x-1)}{(x+5)(x-1)} \equiv 2 - \frac{7}{(x+5)}. $$
Define $g:x\to 2 - \frac{7}{(x+5)}\ $ and $\ f:x\to\frac{2x^2 + x -3}{x^2 + 4x - 5}.$
We showed above that $f$ and $g$ are equivalent for all $x\in\mathbb{R}\setminus\{1,-5\}.$
The function $g:x\to 2 - \frac{7}{(x+5)}$ has domain $\mathbb{R}\setminus\{-5\}$ and range $\mathbb{R}\setminus\{2\}.$
But the function $ f:x\to\frac{2x^2 + x -3}{x^2 + 4x - 5}$ has domain $\mathbb{R}\setminus\{1,-5\}.$ Therefore the range of $f$ is the range of $g$ with the point $ g(1)$ (if it exists) removed.
|
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|
Find the number of solutions to the equation $\sin\left(\frac{\pi\sqrt x}4\right)+\cos\left(\frac{\pi\sqrt {2-x}}4\right)=\sqrt 2$ The following question appeared in a JEE Mock exam, held two days ago.
Question:
Find the number of solutions to the equation $\sin\left(\frac{\pi\sqrt x}4\right)+\cos\left(\frac{\pi\sqrt {2-x}}4\right)=\sqrt 2$
My Attempt:
I think if each term is $\frac1{\sqrt 2}$ then it would add upto $\sqrt 2$. So, $x=1$ is a solution.
How can we tell about other solutions, if any?
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From the comments, the domain is $[0,2]$.
Let $f(x)=\sin\left(\frac{\pi\sqrt x}4\right)+\cos\left(\frac{\pi\sqrt {2-x}}4\right)-\sqrt 2$
$f'(x)=\frac1{2\sqrt x}\frac\pi4\cos\left(\frac{\pi\sqrt x}4\right)+\frac1{2\sqrt{2-x}}\frac\pi4\sin\left(\frac\pi4\sqrt{2-x}\right)$
Angles of sine and cosine are varying from $0$ to $\frac\pi{2\sqrt2}$. That means $f'(x)\gt0$. Thus it crosses x-axis either once or never.
Also, since $f(x)$ is increasing, we only need to check values at domain end points.
$f(0)=\cos\left(\frac\pi{2\sqrt2}\right)-\sqrt2\lt0$
$f(2)=\sin\left(\frac\pi{2\sqrt2}\right)+1-\sqrt2$
$4\gt2\sqrt2\implies\frac\pi4\lt\frac\pi{2\sqrt2}\implies \sin\left(\frac\pi{2\sqrt2}\right)\gt\frac1{\sqrt2}\gt0.7$
Thus, $f(2)\gt0$
Thus, the given equation has one and only one solution.
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"url": "https://math.stackexchange.com/questions/4492881",
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|
Why is $ \frac{x}{\sqrt{x^{2}+1}}=x^{4}-x$ considered impossible to solve? This is from Stewart's Transcendentals.
If we were to try to find the exact intersection points, we
would have to solve the equation $$ \frac{x}{\sqrt{x^{2}+1}}=x^{4}-x$$
This looks like a very difficult equation to solve exactly (in
fact, it's impossible), so instead we use a graphing device to draw
the graphs of the two curves in Figure 7.
Why is this equation considered imposible to solve?
What is the factor that indicates you to not even try and just go directly with a graphing program.
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Of course it is possible to solve this equation, but answer cannot be written in closed form.
$$\frac{x}{\sqrt{x^{2}+1}}=x^{4}-x,x\in\mathbb{R}$$
$$\frac{x}{\sqrt{x^2+1}}\left((x^3-1)\sqrt{x^2+1}-1\right)=0$$
$$x=0\lor (x^3-1)\sqrt{x^2+1}=1$$
Let find real root other than $x=0$.
$$(x^3-1)\sqrt{x^2+1}=1,x\in\mathbb{R}$$
$$(x^6-2x^3+1)(x^2+1)=1 \land x^3-1>0$$
$$x^8-2x^5+x^2+x^6-2x^3+1-1=0 \land x>1$$
$$x^2(x^6+x^4-2x^3-2x+1)=0 \land x>1$$
$$x^6+x^4-2x^3-2x+1=0 \land x>1$$
Real root of original equation other than $x=0$ is real root of $x^6+x^4-2x^3-2x+1=0$ which is greater than $1$. This root can be written in Wolfram Root notation (which is not part of commonly used closed form arrangement) as $Root(x^6+x^4-2x^3-2x+1,i)$, where $i$ is some integer number from $1$ to $6$. Numerical value of this root can be found using numerical methods or plotting.
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|
What is the probability of drawing balls until you get 3 matching colors? A bag has $50$ balls - $37$ red, $10$ white, and $3$ blue. I draw balls without replacement until I pull out $3$ of the same color. What is the probability of each color that it is the first to be drawn 3 times?
My attempt: I know I have at least $3$ draws and at most $7$. I used hypergeometric distribution to figure the probability of the first three balls being the same color (red $39.6\%$, white $0.612\%$, blue $0.0051\%$). I figured the same with $4$, $5$, $6$, and $7$ balls but I do not know if I'm going in the right direction, and if so, how to figure the total probability of each color?
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You want the probability of, say, the third red being drawn before the third white, or third blue ; when there are 37 red, 10 white, 3 blue.
The probability that the third ball is red and the first two also red is $\left.{\binom{37}1\binom{36}{2}}\middle/{\binom{50}{1}\binom{49}{2}}\right.$
The probability that the fourth ball is red and two among the first three also red is $\left.{\binom{37}{1}\binom32\binom{36}{2}\binom{13}{1}}\middle/{\binom{50}{1}\binom{49}{3}}\right.$
The probability that the fifth ball is red and two among the first four also red is $\left.{\binom{37}{1}\binom42\binom{36}{2}\binom{13}{2}}\middle/{\binom{50}{1}\binom{49}{4}}\right.$
The probability that the sixth ball is red and two among the first five also red (yet not more than two of each other colour among the remaining 3) is: $\left.{\binom{37}{1}\binom52\binom{36}{2}\left[\binom 32\binom{10}2\binom 31+\binom31\binom{10}1\binom 32\right]}\middle/{\binom{50}{1}\binom{49}{5}}\right.$
The probability that the seventh ball is red and two among the first six also red (and only two of each of the other colour among the remaining 4) is: $\left.{\binom{37}{1}\binom62\binom{36}{2}\binom42\binom{10}2\binom 32}\middle/{\binom{50}{1}\binom{49}{6}}\right.$
$$\dfrac{37 \binom{36}2}{50}\left[\dfrac{1}{\binom{49}2}+\dfrac{\binom 31\binom{13}1}{\binom{49}3}+\dfrac{\binom42\binom{13}2}{\binom{49}{4}}+\dfrac{\binom 53\binom 32\left[\binom{10}2\binom 31+\binom{10}1\binom 32\right]}{\binom{49}5}+\dfrac{\binom64\binom 42\binom{10}2\binom32}{\binom{49}6}\right]$$
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$\int_0^1 \frac{\arcsin x\arccos x}{x}dx$ Someone on Youtube posted a video solving this integral.
I can't find on Math.stack.exchange this integral using search engine https://approach0.xyz
It is related to $\displaystyle \int_0^\infty \frac{\arctan x\ln x}{1+x^2}dx$
Following is a solution that is not requiring the use of series:
\begin{align}J&=\int_0^1 \frac{\arcsin x\arccos x}{x}dx\\
&\overset{\text{IBP}}=\underbrace{\Big[\arcsin x\arccos x\ln x\Big]_0^1}_{=0}-\underbrace{\int_0^1 \frac{\arccos x\ln x}{\sqrt{1-x^2}}dx}_{x=\cos t }+\underbrace{\int_0^1 \frac{\arcsin x\ln x}{\sqrt{1-x^2}}dx}_{x=\sin t}\\
&=\int_0^{\frac{\pi}{2}} t\ln(\tan t)dt\\
&\overset{u=\tan t}=\int_0^\infty \frac{\arctan u\ln u}{1+u^2}du\\
&\overset{\text{IBP}}=\underbrace{\left[\arctan u\left(\int_0^u \frac{\ln t}{1+t^2}dt\right)\right]_0^\infty}_{=0}-\int_0^\infty \frac{1}{1+u^2}\left(\underbrace{\int_0^u \frac{\ln t}{1+t^2}dt}_{y(t)=\frac{t}{u}}\right)du\\
&=-\int_0^\infty \left(\int_0^1 \frac{u\ln(uy)}{(1+u^2)(1+u^2y^2)}dy\right)du\\
&=-\int_0^\infty \left(\int_0^1 \frac{u\ln u}{(1+u^2)(1+u^2y^2)}dy\right)du-\int_0^1 \left(\int_0^\infty \frac{u\ln y}{(1+u^2)(1+u^2y^2)}du\right)dy\\
&=-\int_0^\infty \left[\frac{\arctan(uy)}{1+u^2}\right]_{y=0}^{y=1}\ln udu-\frac{1}{2}\int_0^1 \left[\frac{\ln\left(\frac{1+u^2}{1+u^2y^2}\right)}{1-y^2}\right]_{u=0}^{u=\infty}\ln ydy\\
&=-J+\int_0^1 \frac{\ln^2 y}{1-y^2}dy\\
&=\frac{1}{2}\int_0^1 \frac{\ln^2 y}{1-y}dy-\frac{1}{2}\underbrace{\int_0^1 \frac{y\ln^2 y}{1-y^2}dy}_{z=y^2}\\
&=\frac{7}{16}\int_0^1 \frac{\ln^2 y}{1-y}dy\\
&=\frac{7}{16}\times 2\zeta(3)=\boxed{\frac{7}{8}\zeta(3)}
\end{align}
NB:I assume $\displaystyle \int_0^1 \frac{\ln^2 y}{1-y}dy=2\zeta(3)$
Feel free to post your solution.
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Let $J(a)=\int_0^\infty \frac{\tan^{-1}(ax)\ln x}{1+x^2}$
\begin{align}
J’(a)=& \int_0^\infty \frac{x\ln x}{(1+x^2)(1+a^2x^2)} \overset{x\to \frac1{ax}}{dx}\\
= &
-\frac1{2}\int_0^\infty \frac{x\ln a}{(1+x^2)(1+{a^2}x^2)} {dx}
= \frac{\ln^2 a}{2(1-a^2)}
\end{align}
which leads to
\begin{align}\int_0^\infty \frac{\arctan x\ln x}{1+x^2}dx
=\int_0^1J’(a)da= \frac12 \int_0^1 \frac{\ln^2 a}{1-a^2}da
=\frac78\zeta(3)
\end{align}
Besides, the integral is also related to
$$\int_0^1 \frac{\sin^{-1} x\cos^{-1} x}{x}dx
=\frac14\int_0^\infty \frac{x^2}{\sinh x}dx
=\frac18\int_0^\pi\frac{x(\pi-x)}{\sin x}dx=\frac78\zeta(3)$$
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|
$a, b, c, d, e$ are distinct real numbers. Prove that this equation has distinctive 4 real roots.
$a, b, c, d, e$ are distinct real numbers. Prove that this equation has distinctive four real roots.
$$(x-a)(x-b)(x-c)(x-d) \\ +(x-a)(x-b)(x-c)(x-e) \\ +(x-a)(x-b)(x-d)(x-e) \\ +(x-a)(x-c)(x-d)(x-e) \\ + (x-b)(x-c)(x-d)(x-e) =0.$$
My attempt:
\begin{align}
& f(x)=\sum_{cyc} (x-a)(x-b)(x-c)(x-d). \\
& f(a)=(a-b)(a-c)(a-d)(a-e). \\
& f(b)=(b-a)(b-c)(b-d)(b-e). \\
& f(c)=(c-a)(c-b)(c-d)(c-e). \\
& f(d)=(d-a)(d-b)(d-c)(d-e). \\
& f(e)=(e-a)(e-b)(e-c)(e-d). \\
\ \\
& f(0)=\sum_{cyc} abcd.\\
\end{align}
I put some arbitrary integers in $a, b, c, d, e$ and drew a graph of $\displaystyle f(x)=\sum_{cyc} (x-a)(x-b)(x-c)(x-d).$
It sure makes 4 distinct roots... How can we prove this?
Values in the graph above:
$a=-1.34, b=-4.67, c=-2.91, d=0.33, e=-6.09$
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I found an answer from the comment.
\begin{align}
& \text{WLOG } a<b<c<d<e. \\
\ \\
& f(a)>0, f(b)<0, f(c)>0, f(d)<0, f(e)>0. \\
\ \\
& \therefore \text{ There exists 4 roots, one each in } [a, b], [b, c], [c, d], [d, e] \qquad \text{by the Intermediate Value Theorem}.
\end{align}
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Generating irreducible polynomials for binary numbers From a paper, there is a discussion about generating irreducible polynomials for a certain degree as can be seen below.
The reference is wolfram, but it is not clear how those polynomials are generated. For example, 37 and 41 are shown below.
5 4 3 2 1 0
37 = 1 0 0 1 0 1 (x5+x2+1)
41 = 1 0 1 0 0 1 (x5+x3+1)
With this tool, I can verify that x5+x2+1 is irreducible. But how those numbers are generated is a question for me. Unfortunately, the author didn't answer that. I would like to generate other degrees, e.g. 6 or 7.
Any help on that?
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For small degrees you can basically do this by hand with a variant of the sieve of Eratosthenes, starting with all polynomials and then successively eliminating those that are divisible by polynomials of lower degree. For polynomials of degree $6$ it suffices to eliminate polynomials which are divisible by irreducible polynomials of degree $\le 3$. These are easy to list: $x, x + 1, x^2 + x + 1, x^3 + x + 1, x^3 + x^2 + 1$. To check our work we can use the fact that there are
$$\frac{1}{n} \sum_{d \mid n} \mu(d) 2^{\frac{n}{d}}$$
monic irreducible polynomials of degree $n$ over $\mathbb{F}_2$, which handily reduces to $\frac{2^p - 2}{p}$ when $n = p$ is prime. So there is $\frac{2^2 - 2}{2} = 1$ irreducible of degree $2$ and $\frac{2^3 - 2}{3} = 2$ irreducibles of degree $3$, which we've listed. You can check that the cubic polynomials are irreducible by verifying that they have no linear factors, which just means they have to end in $1$ (to not be divisible by $x$) and have an odd number of terms (to not be divisible by $x + 1$).
So, there are $2^6 = 64$ monic polynomials of degree $6$. Of these, half do not end with $1$ so are divisible by $x$ and we can eliminate them, giving $2^5 = 32$ monic polynomials of degree $6$ ending with $1$. Of these, half have an even number of terms so are divisible by $x + 1$ and we can eliminate them too, giving $2^4 = 16$ monic polynomials of degree $6$ ending with $1$ with an odd number of terms as our remaining candidates. Via the necklace formula above we know that there are
$$\frac{2^6 - 2^3 - 2^2 + 2}{6} = 9$$
monic irreducibles of degree $6$, so we have to eliminate $7$ of the remaining candidates. Testing for divisibility by $x^2 + x + 1$ is easy because $x^3 \equiv 1 \bmod (x^2 + x + 1)$ so we can reduce exponents $\bmod 3$. This eliminates the following $4$ polynomials:
$$x^6 + x^3 + x^2 + x + 1 \equiv x^2 + x + 1 \bmod x^2 + x + 1$$
$$x^6 + x^4 + x^3 + x^2 + 1 \equiv x^2 + x + 1 \bmod x^2 + x + 1$$
$$x^6 + x^5 + x^4 + x^3 + 1 \equiv x^2 + x + 1 \bmod x^2 + x + 1$$
$$x^6 + x^5 + x^3 + x + 1 \equiv x^2 + x + 1 \bmod x^2 + x + 1.$$
The condition these polynomials satisfy is that the exponents which occur in the middle, between $x^6$ and $1$, are in arithmetic progression; that's easy enough to remember. The only polynomials left to eliminate after this are ones which have only irreducible cubic factors. There are $3$ of these and we can just compute them:
$$(x^3 + x + 1)^2 = x^6 + x^2 + 1$$
$$(x^3 + x^2 + 1)^2 = x^6 + x^4 + 1$$
$$(x^3 + x + 1)(x^3 + x^2 + 1) = x^6 + x^5 + x^4 + x^3 + x^2 + x + 1.$$
The remaining $9$ polynomials are all irreducible. Here they are:
$$\boxed{ \begin{eqnarray*} x^6 + x + 1 &=& \text{Poly}(67) \\
x^6 + x^3 + 1 &=& \text{Poly}(73) \\
x^6 + x^4 + x^2 + x + 1 &=& \text{Poly}(87) \\
x^6 + x^4 + x^3 + x + 1 &=& \text{Poly}(91) \\
x^6 + x^5 + 1 &=& \text{Poly}(97) \\
x^6 + x^5 + x^2 + x + 1 &=& \text{Poly}(103) \\
x^6 + x^5 + x^3 + x^2 + 1 &=& \text{Poly}(109) \\
x^6 + x^5 + x^4 + x + 1 &=& \text{Poly}(115) \\
x^6 + x^5 + x^4 + x^2 + 1 &=& \text{Poly}(117) \end{eqnarray*} }$$
which you can confirm via the OEIS. It may also be a bit easier and more enlightening to do this computation by writing everything in terms of binary strings.
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|
Abel's binomial theorem Abel's generalization of the binomial theorem goes as follows
$$(a+x)^n=x^n+ \sum_{i=1}^n \binom{n}{i} a(a+iz)^{i-1}(x-iz)^{n-i}$$
The objective is to prove the generalization by induction. My first step following the textbook was to set $n=k$ and integrate both sides with respect to $x$, in order to obtain; $$\frac {(a+x)^{k+1}}{k+1}=\frac {x^{k+1}}{k+1}+ \sum_{i=1}^k \binom{k}{i} \frac {a(a+iz)^{i-1}(x-iz)^{k-i+1}}{k-i+1} + C$$
The next step was to find the value of the integration constant $C$ by setting $x=(k+1)z$ in the formula above and use the inductive hypothesis. This is the part where I got stuck and cannot seem to figure it out...
$$\frac {(a+z(k+1))^{k+1}}{k+1}=\frac {(z(k+1))^{k+1}}{k+1}+ \sum_{i=1}^k \binom{k}{i} \frac {a(a+iz)^{i-1}(z(k+1)-iz)^{k-i+1}}{k-i+1} + C$$
I see how multiplying both sides by $k+1$ would give us $$(a+z(k+1))^{k+1}= (z(k+1))^{k+1}+\sum_{i=1}^k \binom{k+1}{i} a(a+iz)^{i-1}(z(k+1)-iz)^{k-i+1}+ C$$
where, again, $(k+1)z=x$. But I still am confused as to determining $C$. It was previously tried by setting $z=0$, that is $C(a,0)$ for $C(a,z)$, where $a,z \in \mathbb {R}$. But this computes only for $z$, and therefore the answer could not hold...
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Here we complete OPs proof by induction. Note, we will employ the induction hypothesis twice. Assuming $a\ne 0$ we can write OPs version of Abel's binomial theorem somewhat more convenient as:
\begin{align*}
(a+x)^k&=\sum_{q=0}^k\binom{k}{q}a(a+qz)^{q-1}(x-qz)^{k-q}\tag{1.1}
\end{align*}
The induction step we want to show is
Induction step:
\begin{align*}
(a+x)^{k+1}&=\sum_{q=0}^{k+1}\binom{k+1}{q}a(a+qz)^{q-1}(x-qz)^{k+1-q}\tag{1.2}
\end{align*}
OP already integrated (1.1) and obtained after some simplification
\begin{align*}
(a+x)^{k+1}&=\sum_{q=0}^k\binom{k+1}{q}a(a+qz)^{q-1}(x-qz)^{k+1-q}\tag{2}\\
&\qquad+(k+1)C(a,z)
\end{align*}
with $C(a,z)$ an integration constant dependent on the constants $a$ and $z$ . Comparison of (2) with (1.2) shows that $(k+1)C(a,z)$ is the summand with index $q=k+1$, so that we have to show
\begin{align*}
\color{blue}{C(a,z)=\frac{a\left(a+(k+1)z\right)^k}{k+1}}\tag{3}
\end{align*}
The representation of $C(a,z)$ makes it plausible, that we start using the induction hypothesis (1.1) by letting $x=(k+1)z$. We multiply (1.1) also with $(a+(k+1)z)$ and obtain
\begin{align*}
&(a+(k+1)z)^{k+1}\\
&\qquad=(a+(k+1)z)\sum_{q=0}^k\binom{k}{q}a(a+qz)^{q-1}((k+1-q)z)^{k-q}\\
&\qquad=\sum_{q=0}^k\binom{k}{q}a(a+qz)^{q}((k+1-q)z)^{k-q}\\
&\qquad\qquad+\sum_{q=0}^k\binom{k}{q}a(a+qz)^{q-1}((k+1-q)z)^{k+1-q}\tag{4.1}\\
\end{align*}
where we use $a+(k+1)z=(a+qz)+((k+1-q)z)$. On the other hand we obtain from (2) by using $\binom{k+1}{q}=\binom{k}{q}+\binom{k}{q-1}$
\begin{align*}
&(a+(k+1)z)^{k+1}\\
&\qquad=\sum_{q=0}^k\binom{k+1}{q}a(a+qz)^{q-1}(x-qz)^{k+1-q}\\
&\qquad=\sum_{q=0}^k\binom{k}{q}a(a+qz)^{q-1}((k+1-q)z)^{k+1-q}\\
&\qquad\qquad+\sum_{q=0}^k\binom{k}{q-1}a(a+qz)^{q-1}((k+1-q)z)^{k+1-q}\\
&\qquad\qquad+(k+1)C(a,z)\tag{4.2}
\end{align*}
Equating (4.1) and (4.2) one of the sums cancels and we get after rearranging
\begin{align*}
\color{blue}{(k+1)}&\color{blue}{C(a,z)}=\sum_{q=0}^k\binom{k}{q}a(a+qz)^q((k+1-q)z)^{k-q}\\
&\qquad-\sum_{q=0}^{k}\binom{k}{q-1}a(a+qz)^{q-1}((k+1-q)z)^{k+1-q}\\
&=\sum_{q=0}^k\left(\binom{k}{q}-\binom{k}{q-1}\frac{(k+1-q)z}{a+qz}\right)
a(a+qz)^q((k+1-q)z)^{k-q}\\
&=\sum_{q=0}^k\left(\binom{k}{q}-\binom{k}{q}\frac{qz}{a+qz}\right)
a(a+qz)^q((k+1-q)z)^{k-q}\\
&=\sum_{q=0}^k\binom{k}{q}\left(1-\frac{qz}{a+qz}\right)
a(a+qz)^q((k+1-q)z)^{k-q}\\
&=a\sum_{q=0}^k\binom{k}{q}a(a+qz)^{q-1}((k+1-q)z)^{k-q}\\
&\,\,\color{blue}{=a(a+(k+1)z)^k}
\end{align*}
and the claim (3) and so the induction step (1.2) follows. Note, in the last line we again used the induction hypothesis (1.1) with $x=(k+1)z$.
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|
Prove that $a^{4/a} + b^{4/b} + c^{4/c} \ge 3$
Let $a, b, c > 0$ with $a + b + c = 3$. Prove that
$$a^{4/a} + b^{4/b} + c^{4/c} \ge 3.$$
This question was posted recently, closed and then deleted, due to missing of contexts etc.
By https://approach0.xyz/, the problem was proposed by Grotex@AoPS.
My strategy is to split into many cases.
WLOG, assume that $a \ge b \ge c$.
If $a \ge 8/5$, true.
If $a \le 10/7$, let
$f(x) = x^{4/x} - 1 - 4(x - 1)$.
We have $f(x) \ge 0$ for all $x \in (0, 10/7)$.
If $10/7 < a < 8/5$ and $b \ge 4/5$, true.
(I stopped here since this approach is ugly. Actually, the proof of $x^{4/x} - 1 - 4(x - 1) \ge 0$ for all $x\in (0, 10/7)$ is complicated.)
I hope to see nice proofs.
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Before the actual proof, let's start with a discussion of $f(x) = x^{4/x}$ which is needed later.
We have
$ df(x) /dx = 4 x^{4/x - 2} (1 - \log(x))$ so $f(x)$ is rising for all $0 < x < e$ which is all we need to consider below. Further,
$$ d^2f(x) /dx^2 = 4 x^{4/x - 4} (-3 x + 4 \log^2(x) + 2 (x - 4) \log(x) + 4)$$
To interpret this result for positive and negative ranges, consider $ h(x) = -3 x + 4 \log^2(x) + 2 (x - 4) \log(x) + 4$. Clearly, $h(x) {{(x\to 0) } \atop{\longrightarrow}} \infty $. Further,
$$ h'(x) = (-x + 2 (x + 4) \log(x) - 8)/x
$$
which remains negative for at least $0 < x < 2$. So if we have a value where $d^2f(x) /dx^2$ changes sign in the range $0 < x < 2$, this will be the only sign change there. The sign change indeed occurs near $x= 1.125$.
We can now describe the behavior of $f(x) = x^{4/x}$ for $0 < x < 2$. It is rising, it is convex for $0 < x < x_0 $ and concave for $ x_0 < x < 2$ where $x_0≈1.125$ is the unique real solution of $h(x)=0$ on $(0,2)$.
Now consider $g(x) = -3 + 4x$. Note $g(x)$ touches $f(x) = x^{4/x}$ at $x=1$, since $g(1) =f(1) = 1$ and $g'(1) = f'(1) = 4$. Since $x^{4/x}$ is convex at $x=1$ and turns concave at $x \simeq 1.125$, we have an "S-shape" and hence, there is at most one intersection $g(x) = f(x)$ in the range $1.125 < x < 2$. By inspection we have that this occurs near $\bar{x} = 1.429$ . So $x^{4/x} \ge g(x)$ for $x < \bar{x} = 1.429$ and hence, for $a,b,c < \bar{x}$, $a^{4/a} + b^{4/b} + c^{4/c} \ge -9 + 4(a+b+c) =3$.
As OP already noted, only $1.52 > a \ge b \ge c$ must then be considered (since $1.52^{4/1.52} > 3$) . Now if $1.52 > a > \bar{x} = 1.429$, then $a^{4/a} \ge \bar{x}^{4/\bar{x}} \simeq 2.71$ and $b +c \ge 3 - 1.52 = 1.48$. Hence it remains to be shown that for $b +c = 1.48$, that $b^{4/b} + c^{4/c} \ge 3 - 2.71 = 0.29$. This can be established by direct calculation, since indeed $b^{4/b} + (1.48-b)^{4/(1.48-b)} > 0.29$ for all $0 \le b \le 1.48$. $\qquad \Box$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Finding the maximum value of $f(x)=\frac{|x|-2-x^2}{|x|+1},x\in\mathrm R$ Question:
Find the maximum value of $f(x)=\frac{|x|-2-x^2}{|x|+1},x\in\mathrm R$
My Attempt:
I took two cases. One where $x\ge0$, so, $|x|=x$ and another where $x\le0$, so, $|x|=-x$. Then I cross multiplied, made a quadratic in $x$ and wrote discriminant greater than equal to zero. Accordingly, I got range of $f(x)$.
I wonder if there is another way to solve this question, where we don't have to take two cases.
|
Perhaps surprisingly, $ \ f(x) \ = \ \large{\frac{|x| \ - \ 2 \ - \ x^2}{|x| \ + \ 1} } \ $ is an even function(!). If we consider $ \ f_{+}(x) \ = \ \large{\frac{ x \ - \ 2 \ - \ x^2}{ x \ + \ 1} } $ $ = \ (-x + 2) - \frac{4}{x + 1} \ \ , \ $ the corresponding curve has a vertical asymptote of $ \ x \ = \ -1 \ $ and an "oblique" asymptote $ \ y \ = \ -x + 2 \ \ ; \ $ the branch for $ \ x \ > \ -1 \ $ lies "below" the oblique asymptote, making $ \ f_{+}(x) \ < \ 0 \ \ , \ $ while the branch for $ \ x \ < \ -1 \ $ lies "above" that asymptote and so $ \ f_{+}(x) \ > \ 0 \ \ $ there. The function $ \ f_{-}(x) \ = \ \large{\frac{ -x \ - \ 2 \ - \ x^2}{ -x \ + \ 1} } \ = \ \normalsize{( x + 2) - \frac{4}{1 - x} } \ \ $ is the "horizontal reflection" of $ \ f_{+}(x) \ \ , \ $ thus $ \ f_{-}(x) \ > \ 0 \ \ $ for $ \ x \ > \ +1 \ $ and $ \ f_{-}(x) \ < \ 0 \ \ $ for $ \ x \ < \ +1 \ \ . $ [These individual functions have no symmetry, but the presence of the absolute-value brackets leads to $ \ f(x) \ = \ f(-x) \ \ . \ ] $
Our function curve is then the union of the portions of $ \ f_{+}(x) \ \ $ "to the right" of the $ \ y-$axis and of $ \ f_{-}(x) \ \ $ "to its left", which produces a symmetrical curve "below" the two asymptote lines with its $ \ y-$intercept at $ \ (0 \ , \ \frac{0 \ - \ 2 \ - \ 0^2}{0 \ + \ 1} = -2) \ \ . \ $ (The "positive" branches and vertical asymptotes of the individual curves are excluded.)
To find the relative (and also global) extrema of this curve, we may solve either
$$ f'_{+}(x) \ \ = \ \ \frac{ -x^2 \ - \ 2x \ + \ 3}{ (x \ + \ 1)^2} \ \ = \ \ -\frac{ (x - 1)·(x + 3)}{ (x \ + \ 1)^2} \ \ = \ \ 0 \ \ \ \text{or} $$
$$ f'_{-}(x) \ \ = \ \ \frac{ -x^2 \ + \ 2x \ + \ 3}{ (1 \ - \ x)^2} \ \ = \ \ -\frac{ (x + 1)·(x - 3)}{ (1 \ - \ x)^2} \ \ = \ \ 0 \ \ . $$
[We can alternatively differentiate the "quotient-remainder" expressions for the functions, producing $ \ f'_{+}(x) \ \ = \ \ -1 \ + \frac{4}{(x + 1)^2} \ = \ 0 \ $ or $ \ f'_{-}(x) \ \ = \ \ 1 \ - \frac{4}{(1 - x)^2} \ = \ 0 \ \ , \ $ again needing to apply the appropriate interval restriction to remove the "spurious" solution.]
[ADDENDUM (8/5) -- We can also find this maximum without calculus by asking what "horizontal line" $ \ y \ = \ c \ $ intersects, say, $ \ f_{+}(x) \ $ at a single point:
$$ \frac{ x \ - \ 2 \ - \ x^2}{ x \ + \ 1} \ \ = \ \ c \ \ \Rightarrow \ \ x^2 \ + \ (c - 1)·x \ + \ ( c + 2 ) \ \ = \ \ 0 \ \ , $$
which has the determinant $ \ (c - 1)^2 - 4·(c + 2) \ = \ c^2 - 6c - 7 \ = \ (c + 1)·(c - 7) \ \ . \ $ This equals zero for $ \ c \ = \ -1 \ \ , \ $ which is the maximum at $ \ x \ = \ 1 \ \ , \ $ and at $ \ c \ = \ 7 \ \ , \ $ which is the minimum at $ \ x \ = \ -3 \ \ $ on the "excluded" branch of the function curve.]
Within their applicable intervals, the derivative for $ \ f_{+}(x) \ $ is zero for $ \ x \ = \ +1 \ $ and that for $ \ f_{-}(x) \ $ is zero for $ \ x \ = \ -1 \ \ . \ $ The (global) maximum for $ \ f(x) \ $ is therefore
$$ f_{+}(1) \ \ = \ \ (-1 + 2) - \frac{4}{1 + 1} \ \ = \ \ f_{-}(-1) \ \ = \ \ ([-1] + 2) - \frac{4}{1 - [-1]} \ \ = \ \ -1 \ \ . $$
Knowing where the curve is relative to its asymptote lines, we can be assured that these are global maxima. (The second derivatives are a bit complicated, but do prove to be negative.)
|
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|
Summation of reciprocal products When studying summation of reciprocal products I found some interesting patterns.
$$\sum_{k=1}^{N} \frac{1}{k\cdot(k+1)}=\frac{1}{1\cdot1!}-\frac{1}{1\cdot(N+1)}$$
$$\sum_{k=1}^{N} \frac{1}{k\cdot(k+1)(k+2)}=\frac{1}{2\cdot2!}-\frac{1}{2\cdot(N+1)(N+2)}$$
$$\sum_{k=1}^{N} \frac{1}{k\cdot(k+1)(k+2)(k+3)}=\frac{1}{3\cdot3!}-\frac{1}{3\cdot(N+1)(N+2)(N+3)}$$
The pattern is $$\sum_{k=1}^{N} \frac{1}{k\cdot(k+1)(k+2)(k+3)\cdot\cdot\cdot\cdot(k+i)}\space=\space\frac{1}{i\cdot i!}-\frac{1}{i\cdot(N+1)(N+2)(N+3)\cdot\cdot\cdot\cdot(N+i)}$$
which is easy to prove by induction.
As an easy consequence it follows that $$\lim_{N \to \infty} \sum_{k=1}^{N} \frac{1}{k\cdot(k+1)(k+2)(k+3)\cdot\cdot\cdot\cdot(k+i)}\space=\space\frac{1}{i\cdot i!}$$
I did not find this summation of reciprocal products in my mathbooks. Is this a known theorem?
Can anyone point me in the right direction for further study?
|
This identity is a matter of telescoping. Let $q$ be a positive integer.
We obtain
\begin{align*}
\color{blue}{\sum_{k=1}^N}&\color{blue}{\frac{1}{k(k+1)\cdots(k+q)}}\\
&=\frac{1}{q}\sum_{k=1}^N\frac{1}{k(k+1)\cdots(k+q)}\left((k+q)-k\right)\tag{1.1}\\
&=\frac{1}{q}\sum_{k=1}^N\frac{1}{k(k+1)\cdots(k+q-1)}
-\frac{1}{q}\sum_{k=1}^N\frac{1}{(k+1)\cdots(k+q)}\tag{1.2}\\
&=\frac{1}{q}\sum_{k=1}^N\frac{1}{k(k+1)\cdots(k+q-1)}\\
&\qquad\quad-\frac{1}{q}\sum_{k=2}^{N+1}\frac{1}{k(k+1)\cdots(k+q-1)}\tag{1.3}\\
&=\frac{1}{q}\left(\frac{1}{q!}-\frac{1}{(N+1)(N+2)\cdots(N+q)}\right)\tag{1.4}\\
&\,\,\color{blue}{=\frac{1}{q}\left(\frac{1}{q!}-\frac{N!}{(N+q)!}\right)}\tag{1.5}\\
\end{align*}
and the claim follows.
Comment:
*
*In (1.1) we expand numerator and denominator with $q$.
*In (1.2) we split the sums and cancel terms accordingly.
*In (1.3) we shift the index of the right-hand sum and start with $k=2$.
*In (1.4) we do the cancellation thanks to telescoping.
Note: A theory behind it is the calculus of finite differences. In fact this identity is an application of a discrete analogon of the fundamental theorem of calculus. An application of it is given for instance here.
Calculus of finite differences: Let's do the calculation from above a bit more sophisticated.
*
*In terms of calculus of finite differences we consider the falling factorial operator
\begin{align*}
x^{\underline{q}}=x(x-1)\cdots(x-q+1)
\end{align*}
*which can be expanded to negative integers, written as $-q$ with $q> 0$ which is then defined as
\begin{align*}
\color{blue}{x^{\underline{-q}}:=\frac{1}{(x+1)(x+2)\cdots(x+q)}}\tag{2.1}
\end{align*}
*The forward difference operator $\Delta$ defined as
\begin{align*}
\left(\Delta F\right)(x):=F(x+1)-F(x)
\end{align*}
applied to $x^{\underline{-q}}$ gives
\begin{align*}
\color{blue}{\Delta x^{\underline{-q}}=-qx^{\underline{-q-1}}}\tag{2.2}
\end{align*}
*A discrete analogon to the fundamental theorem of calculus is given as
\begin{align*}
\color{blue}{\sum_{k=0}^{N-1}\left(\Delta F\right)(k)}&\color{blue}{=F(N)-F(0)}\tag{2.3}\\
\int_{0}^xF^{\prime}(t)\,dt&=F(x)-F(0)
\end{align*}
We can now perform the calculation from above as
\begin{align*}
\color{blue}{\sum_{k=1}^N}&\color{blue}{\frac{1}{k(k+1)\cdots(k+q)}}
=\sum_{k=1}^{N}(k-1)^{\underline{-\left(q+1\right)}}\tag{$\to\ $2.1}\\
&=\sum_{k=0}^{N-1}k^{\underline{-\left(q+1\right)}}\tag{shift $k$}\\
&=-\frac{1}{q}\sum_{k=0}^{N-1}\left(\Delta k^{\underline{-q}}\right)\tag{$\to\ $2.2}\\
&=-\frac{1}{q}\left(N^{\underline{-q}}-0^{\underline{-q}}\right)\tag{$\to\ $2.3}\\
&=-\frac{1}{q}\left((N+1)(N+2)\cdots(N+q)-1\cdot 2\cdots q\right)\tag{$\to\ $2.1}\\
&\,\,\color{blue}{=\frac{1}{q}\left(\frac{1}{q!}-\frac{N!}{(N+q)!}\right)}
\end{align*}
and the claim follows again in accordance with the result (1.5).
Note: A generalisation of this identity even more compactly derived is given as formula (1.16) in Differenzen und Summen in Konkrete Analysis by J. Cigler.
|
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|
Series expansion of $(1-cx)^{1/x}$ I am trying to understand the series expansion of $$(1-cx)^{1/x}$$ The wolframalpha seems to solve the problem by using taylor series for $ x\rightarrow 0$ and Puiseux series for $ x\rightarrow \infty$. Any ideas how can I calculate them ?
https://www.wolframalpha.com/input?i=%281-cx%29%5E%281%2Fx%29+series
Here is the link of the problem
|
As already said in comments
$$y=(1-c x)^{\frac{1}{x}}\quad \implies \quad \log(y)=\frac{1}{x}\log(1-cx)$$ For small values of $cx$
$$\log(y)=-\sum_{n=1}^\infty \frac {c^n} n\,x^{n-1}$$ Now, using $y=e^{\log(y)}$
$$\color{blue}{y=1 +c\,e^{-c}\sum_{n=1}^\infty(-1)^n \frac{ c^{n}}{\alpha_n}\,P_n(c)\,x^n}$$ The $\alpha_n$ form the sequence $A053657$ in $OEIS$ (have a look here) and the first $P_n(c)$ are listed below
$$\left(
\begin{array}{cc}
n & P_n(c) \\
1 & 1 \\
2 & 3 c-8 \\
3 & (c-6) (c-2) \\
4 & 15 c^3-240 c^2+1040 c-1152 \\
5 & (c-4) \left(3 c^3-68 c^2+408 c-480\right) \\
6 & 63 c^5-2520 c^4+35280 c^3-211456 c^2+526176 c-414720 \\
7 & (c-6) \left(9 c^5-450 c^4+7800 c^3-55792 c^2+151104 c-120960\right)
\end{array}
\right)$$
Looking separately at $P_{2n}(c)$ and $P_{2n+1}(c)$, it seems that there are some interesting patterns (worth to be explored ?).
For large values of $x$, using $L=\log(-cx)$
$$\color{blue}{y=1+ c \sum_{n=1}^\infty \frac 1{n!\,c^n} Q_n(c,L) \frac 1 {x^n}}$$ where the first $Q_n(c,L)$ are
$$\left(
\begin{array}{cc}
n & Q_n(c,L) \\
1 & L \\
2 & c L^2-2 \\
3 & c^2 L^3-6 c L-3 \\
4 & c^3 L^4-12 c^2 L^2-12 c L+12 c-8 \\
5 & c^4 L^5-20 c^3 L^3-30 c^2 L^2+60 c^2 L-40 c L+60 c-30
\end{array}
\right)$$
Again, possible interesing patterns.
|
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "2",
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|
How many methods to tackle the integral $\int_{0}^{\frac{\pi}{4}} \frac{\sin x+\cos x}{9+16 \sin 2 x} d x ?$ $ \text{We are going to evaluate the integral}$
$\displaystyle \int_{0}^{\frac{\pi}{4}} \frac{\sin x+\cos x}{9+16 \sin 2 x} d x \tag*{} \\$
by letting $ y=\frac{\pi}{4}-x. $ Then
$$\begin{aligned} \displaystyle \int_{0}^{\frac{\pi}{4}} \frac{\sin x+\cos x}{9+16 \sin 2 x} d x \displaystyle &=\int_{\frac{\pi}{4}}^{0} \frac{\sin \left(\frac{\pi}{4}-y\right)+\cos \left(\frac{\pi}{4}-y\right)}{9+16 \sin 2\left[(\frac{\pi}{4}-y)\right]}(-d y) \\
\displaystyle &=\int_{0}^{\frac{\pi}{4}} \frac{\frac{1}{\sqrt{2}}(\cos y-\sin y)+\frac{1}{\sqrt{2}}(\cos y+\sin y)}{9+16 \cos 2 y} d y \\
\displaystyle &=\sqrt{2} \int_{0}^{\frac{\pi}{4}} \frac{\cos y}{9+16\left(1-2 \sin ^{2} y\right)} d y \\
\displaystyle &=\sqrt{2} \int_{0}^{\frac{1}{\sqrt 2} } \frac{d z}{25-32 z^{2}} \text { , where } z=\sin y\\
\displaystyle &=\frac{\sqrt{2}}{10} \int_{0}^{\frac{1}{\sqrt 2} }\left(\frac{1}{5-4 \sqrt{2}z}+\frac{1}{5+4 \sqrt{2} z}\right) d z \\
\displaystyle &=\frac{\sqrt{2}}{10(4 \sqrt{2})}\left[\ln \left|\frac{5+4 \sqrt{2} z}{5-4 \sqrt{2} z}\right|\right]_{0}^{\frac{1}{\sqrt 2} } \\
\displaystyle &=\frac{1}{40}\ln 9 \end{aligned}$$
|
This is similar to your original solution but a bit quicker using sum-to-product formula:$$\begin{aligned} \displaystyle \int_{0}^{\frac{\pi}{4}} \frac{\sin x+\cos x}{9+16 \sin 2 x} d x \displaystyle &=\sqrt2\int_{0}^{\frac{\pi}{4}} \frac{\sin \left(\frac{\pi}{4}+x\right)}{9+16 \sin 2x}d x \\
\displaystyle &=\sqrt2\int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \frac{\sin y}{9-16 \cos 2 y} d y \\
\displaystyle &=-\sqrt{2} \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \frac{d(\cos y)}{9-16\left(2 \cos ^{2} y-1\right)} \\
\end{aligned}$$Following with the substitution $z=\cos y$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Upper triangular determinants Given that
$$A=
\begin{bmatrix}
1&211&311&327&337\\
0&2&427&438&491\\
0&0&3&547&551\\
0&0&0&2&672\\
0&0&0&0&3
\end{bmatrix}
$$
Find $|2A^{-1}+I_5|$. The solution is $\frac{100}{3}$ but I can't figure out how to do it. Any hints? My best guess is to use properties of the determinants of upper triangular matrices, but then I got lost in the algebra...
|
Hints:
*
*Eigenvalues of an upper triangular matrix are precisely the entries of the main diagonal.
*If $A$ is invertible then $\lambda$ is an eigenvalue of $A$ iff $\frac{1}{\lambda}$ is an eigenvalue of $A^{-1}$
*Eigenvalues are polynomial invariant i.e $\lambda$ is an eigenvalue of $A$ implies $p(\lambda) $ is an eigenvalue of $p(A) $ where $p(x) \in K[x]$
*$\det A$ is the product of all eigenvalues taking care of multiplicity.
Eigenvalues of $A$ are $1, 2,3,2,3$.
Hence eigenvalues of $A^{-1}$ are $1, \frac{1}{2}, \frac{1}{3}, \frac{1}{2}, \frac{1}{3}$
Now eigenvalues of $2A^{-1}+I_5$ are $2\cdot 1+1, 2\cdot\frac{1}{2}+1, 2\cdot \frac{1}{3}+1, 2\cdot\frac{1}{2}+1, 2\cdot\frac{1}{3}+1$
$\begin{align}&|2A^{-1}+I_5|\\&=3\cdot 2\cdot \frac{5}{3}\cdot 2\cdot \frac{5}{3}\\&=\frac{100}{3}\end{align}$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Finding all possible complex solutions to the system $xy = \sqrt 2(x + y)$, $yz = \sqrt 3(y + z)$, $zx = \sqrt 5(z + x)$
Find all possible complex solutions to:
$$\begin{cases}xy = \sqrt 2(x + y) \\
yz = \sqrt 3(y + z)\\
zx = \sqrt 5(z + x)\end{cases}$$
I have some ideas:
$$xy-\sqrt 2y=\sqrt 2x\implies y=\frac{\sqrt 2x}{x-\sqrt 2} $$
and
$$yz-\sqrt 3y=\sqrt 3z\implies y=\frac{\sqrt 3z}{z-\sqrt 3} $$
and
$$xz-\sqrt 5x=\sqrt 5z\implies x=\frac{\sqrt 5z}{z-\sqrt 5} $$
So I got
$$y=\frac{\sqrt 2x}{x-\sqrt 2}=\frac{\sqrt 3z}{z-\sqrt 3}=\frac{\sqrt 2×\frac{\sqrt 5z}{z-\sqrt 5}}{\frac{\sqrt 5z}{z-\sqrt 5}-\sqrt 2}$$
From here we can go to a solution, but that brings up really scary equations. Is there any clever way to solve this? That is a minor Olympiad problem.
|
The problem is not so bad.
From the first equation $y=\frac{\sqrt{2} x}{x-\sqrt{2}}$; from the third $z=\frac{\sqrt{5} x}{x-\sqrt{5}}$
Plug in the second
$$\frac{x \left(\left(\sqrt{6}-\sqrt{10}+\sqrt{15}\right) x-2
\sqrt{30}\right)}{\left(x-\sqrt{2}\right) \left(x-\sqrt{5}\right)}=0$$ So, if $x\neq 0$, then
$$x=\frac{2 \sqrt{30}}{\sqrt{6}-\sqrt{10}+\sqrt{15}}\qquad y=\frac{2 \sqrt{30}}{-\sqrt{6}+\sqrt{10}+\sqrt{15}}\qquad z=\frac{2 \sqrt{30}}{\sqrt{6}+\sqrt{10}-\sqrt{15}}$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
If $\vec a,\vec b,\vec c$ be three vectors such that $|\vec a|=1,|\vec b|=2,|\vec c|=4$ and then find the value of $|2\vec a+3\vec b+4\vec c|$ If $\vec a,\vec b,\vec c$ be three vectors such that
$\vert \vec a\vert =1,\vert \vec b\vert =2,\vert \vec c\vert=4$
and
$\vec a \cdot \vec b+\vec b \cdot \vec c+\vec c \cdot\vec a=-10$
then find the value of $\vert 2\vec a+3\vec b+4\vec c \vert$
My Attempt
$\vert \vec a+\vec b+\vec c \vert^2=a^2+b^2+c^2+2(\vec a \cdot \vec b+\vec b \cdot \vec c+\vec c \cdot \vec a)=1+4+16-20=1$
So, $\vert \vec a+\vec b+\vec c \vert=1$
Further by hit and trial I could see that if $\vec a=\vec i,\vec b=2\vec i,\vec c=-4\vec i$ (where $\vec i$ is unit vector along x-axis) satisfies all conditions.
So, $\vert 2\vec a+3\vec b+4\vec c \vert =\vert 2\vec i+6\vec i-16\vec i\vert =8$
But can there be a better way to do this.
|
Recall that $\|\vec x\|^2 = \vec x \cdot \vec x$. Then
$$\begin{align*}
\|2\vec a + 3\vec b + 4\vec c\|^2 &= (2\vec a + 3\vec b + 4\vec c) \cdot (2\vec a + 3\vec b + 4\vec c) \\
&= 4\|\vec a\|^2 + 9\|\vec b\|^2 + 16\|\vec c\|^2 + 2 \left(6\vec a\cdot\vec b + 8\vec a\cdot\vec c + 12\vec b\cdot \vec c\right)\\
&= (4\cdot1)+(9\cdot4)+16^2 + (12\cdot(-10)) + 4\vec a\cdot\vec c + 12\vec b\cdot\vec c \\
&= 176 + 4 (\vec a\cdot \vec c + 3\vec b \cdot \vec c) \\
&= 176 + 16 \cos(\theta) + 96 \cos(\phi)
\end{align*}$$
where $\theta$ is the angle between $\vec a$ and $\vec c$, and $\phi$ is the angle between $\vec b$ and $\vec c$. If we take $\theta=\phi=\pi$ as you've done, the minimum value of $\|2\vec a+3\vec b + 4\vec c\|$ is $\sqrt{176 - 16 - 96} = 8$.
|
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|
Implicit Differentiation - $x^my^n = (x+y)^{m+n}$
Use implicit differentiation to find $\frac{\mbox{dy}}{\mbox{dx}}$ if $$x^my^n = (x+y)^{m+n}$$
Differentiation both sides with respect to $x$:
$$mx^{m-1}y^n + x^mny^{n-1}y' =(m+n)(x+y)^{m+n-1}(1 + y')$$
$$y' = \frac{(m+n)(x+y)^{m+n-1} - mx^{m-1}y^n}{x^mny^{n-1}-(m+n)(x+y)^{m+n-1}}$$
$$y'= \frac{nxy-my^2}{nx^2-mxy}$$ after using given $x^my^n = (x+y)^{m+n}$. The answer given to me is $\frac{y}{x}$. So, it seems $y'$ can be simplified even more.
How do we do this?
Thanks
|
Simpler way: Take logarithms (base e unless specified). $$m\log x+n\log y=(m+n)\log (x+y)$$ Now differentiate: $$\frac mx +\frac ny\frac{dy}{dx}=\frac{m+n}{x+y}\left(1+\frac{dy}{dx}\right)$$ Therefore $$\frac mx-\frac{m+n}{x+y}=\left(\frac{m+n}{x+y}-\frac ny\right)\frac{dy}{dx}$$$$\implies \frac{my-nx}{x(x+y)}= \frac{my-nx}{y(x+y)} \cdot \frac{dy}{dx}$$ so $$\frac{dy}{dx}=\frac yx.$$
|
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|
Find the sum of radicals without squaring, Is that impossible?
Find the summation:
$$\sqrt {3-\sqrt 5}+\sqrt {3+\sqrt 5}$$
My attempts:
\begin{align*}
&A = \sqrt{3-\sqrt{5}}+ \sqrt{3+\sqrt{5}}\\
\implies &A^2 = 3-\sqrt{5} + 3 + \sqrt{5} + 2\sqrt{9-5}\\
\implies &A^2 = 6+4 = 10\\
\implies &A = \sqrt{10}
\end{align*}
So I was wondering about a way to find this sum without squaring? It seems impossible, but I still want to ask.
|
Trick:
$$
\sqrt {3-\sqrt{5}}=\frac{\sqrt {6-2\sqrt{5}}}{\sqrt 2}=\frac{\sqrt {(\sqrt 5)^2-2\sqrt{5}+1}}{\sqrt 2}=\frac{\sqrt {(\sqrt 5-1)^2}}{\sqrt 2}
=\frac{\sqrt 5-1}{\sqrt 2}.
$$
Similarly we see $\sqrt {3+\sqrt{5}}
=\frac{\sqrt 5+1}{\sqrt 2}$. Thus the sum is $2\cdot \frac{\sqrt 5}{\sqrt 2}=\sqrt {10}$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Proving $30x + 3y^2 + \frac{2z^3}{9} + 36 (\frac{1}{xy} + \frac{1}{yz} + \frac{1}{zx}) \ge 84$
If $x, y, z$ are positive real numbers, prove that $$30x + 3y^2 + \frac{2z^3}{9} + 36 \left(\frac{1}{xy} + \frac{1}{yz} + \frac{1}{zx}\right) \ge 84.$$
I genuinely have no clue on how to proceed. Is it proved using repeated CS?
|
Partial answer:
If we want to find the minimum, I think we can create a system from the implicit derivatives. We can multiply through by $xyz$, then subtract the right side to get:
$$f(x,y,z) = 30x^2yz+3xy^3z+\textstyle{\frac29}xyz^4+36x+36y+36z-84xyz \geqslant 0$$
And we want to show that this function has a minimum of $0$. Then we have:
$$
\begin{align}
\frac{\partial f}{\partial x} &= 60xyz+3y^3z+ \textstyle{\frac{2}{9}}yz^4-84yz+36 \\
\frac{\partial f}{\partial y} &= 30x^2z+9xy^2z+ \textstyle{\frac{2}{9}}xz^4 -84xz+36\\
\frac{\partial f}{\partial z} &= 30x^2y+3xy^3+ \textstyle{\frac{8}{9}}xyz^3-84xy+36
\end{align}
$$
To find minima, we want all of these partial derivatives to be equal to $0$. This is a system of three equations in three variables:
$$
\left\{
\begin{aligned}
60xyz+3y^3z+ \textstyle{\frac{2}{9}}yz^4-84yz &= -36 \\
30x^2z+9xy^2z+ \textstyle{\frac{2}{9}}xz^4 -84xz &= -36 \\
30x^2y+3xy^3+ \textstyle{\frac{8}{9}}xyz^3-84xy &= -36
\end{aligned}
\right.
$$
That's a bit... ugly. Let's try something:
$$
\left\{
\begin{aligned}
60x^2yz+3xy^3z+ \textstyle{\frac{2}{9}}xyz^4-84xyz +36x &= 0 \\
30x^2yz+9xy^3z+ \textstyle{\frac{2}{9}}xyz^4-84xyz +36y &= 0 \\
30x^2yz+3xy^3z+ \textstyle{\frac{8}{9}}xyz^4-84xyz +36z &= 0
\end{aligned}
\right.
$$
Now set $xyz=a$ and we have:
$$
\left\{
\begin{aligned}
60ax+3ay^2z+ \textstyle{\frac{2}{9}}az^3 +36x &= 84a \\
30ax+9ay^2z+ \textstyle{\frac{2}{9}}az^3 +36y &= 84a \\
30ax+3ay^2z+ \textstyle{\frac{8}{9}}az^3 +36z &= 84a
\end{aligned}
\right.
$$
I do believe we'll need to solve the system by substitution. I'll leave this here for the moment and return.
|
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|
If $\frac{x^2+y^2+x+y-1}{xy-1}$ is an integer for positive integers $x$ and $y$, then its value is $7$. I saw this on quora
and haven't been able to solve it.
If $\dfrac{x^2+y^2+x+y-1}{xy-1}$
is an integer for positive integers
$x$ and $y$,
then its value is $7$.
If
$y=1$ this is
$\dfrac{x^2+x+1}{x-1}
= x+2+\dfrac{3}{x-1}$
which is an integer only when
$x=2$ or $x=4$ and has value $7$.
Looking at the values from 2 to 20
for $x$ and $y$,
this is an integer only for
$x=2, y=12$ (and $x=12, y=2$).
This value is 7.
So it looks like this might be correct,
and I don't know how to show it.
|
If $f(x,y)=\dfrac{x^2+y^2+x+y-1}{xy-1}$ when integer must be equal to $7$ then
$$g(x,y)=\frac{x^2+y^2+x+y-6xy+5}{xy-1}$$ must be equal to $1$ because $f(x,y)=6+g(x,y)$ so we can solve the equation
$$x^2+y^2+x+y-7xy+6=0$$ which admits several infinite sets of solutions $(x,y)$ like to them of Pell-Fermat equations (look at Wolfram about this). One of these sets is given by
$$15x=(6+\sqrt5)(161-72\sqrt5)^n+(6-\sqrt5)(161+72\sqrt5)^n+3\\ \\30y=(57+25\sqrt5)(161-72\sqrt5)^n+(57-25\sqrt5)(161+72\sqrt5)^n+6$$ For example for $n=2$ we get the solution $(x,y)=(26017,3796)$ which certainly satisfies $f(x,y)=7$ as well as all the other ones.
|
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|
If the matrix A is given, then find the value of $A^{2022}$ If it is given that matrix $ A = \begin{bmatrix}
\dfrac{5}{2} & \dfrac{3}{2}\\
-\dfrac{3}{2} & -\dfrac{1}{2}
\end{bmatrix}$ then find the value of $A^{2022}$.
Here is my try on it : $$\\$$
If we look at the pattern while multliplying A with itself it comes out to be something like this,
$$A^2 = \begin{bmatrix}
4 & 3\\
-3 & -2
\end{bmatrix}$$
$$ A^3 = \begin{bmatrix}
\dfrac{11}{2} & \dfrac{9}{2}\\
-\dfrac{9}{2} & -\dfrac{7}{2}
\end{bmatrix}$$
$$A^4 = \begin{bmatrix}
7 & 6\\
-6 & -5
\end{bmatrix}$$
so $$ A^n = \begin{bmatrix}
\dfrac{3n}{2}+1 & \dfrac{3n}{2}\\
-\dfrac{3n}{2} & -\dfrac{3n}{2}+1
\end{bmatrix} \tag{1}\label{eq1}$$
thus after substituting $n=2022$ we get $ A^{2022} = \begin{bmatrix}
3034 & 3033\\
-3033 & -3032
\end{bmatrix}$
$$\\$$
Is there is any other way to find the value of $A^{2022}$ without finding out the pattern and then general equation \eqref{eq1} like I did above?
|
The matrix has determinant 1 and trace 2, so it is parabolic, so a conjugate of an upper triangular matrix (with diagonal elements equal to 1). Computing powers of parabolic matrices is fun and easy (though your solution is essentially that).
|
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|
Calculate $\sin^5\alpha-\cos^5\alpha$ if $\sin\alpha-\cos\alpha=\frac12$ Calculate $$\sin^5\alpha-\cos^5\alpha$$ if $\sin\alpha-\cos\alpha=\dfrac12$.
The main idea in problems like this is to write the expression that we need to calculate in terms of the given one (in this case we know $\sin\alpha-\cos\alpha=\frac12$).
I don't see how to even start to work on the given expression as we cannot use $a^2-b^2=(a-b)(a+b)$ or $a^3-b^3=(a-b)(a^2+ab+b^2)$. So in other words, I can't figure out how to factor the expression (even a little).
The given answer is $\dfrac{79}{128}$.
|
We will use a few identities.
First is the Pythagorean Identity:
$$\sin^2\alpha +\cos^2\alpha = 1$$
Second is a simple factorization:
$$a^5 - b^5 = (a-b)(a^4 + a^3b + a^2b^2 + ab^3 + b^4) = (a-b)(a^4 + b^4 + ab(a^2+b^2+ab))$$
Third will help us write things of fourth degree in terms of second and first degree terms:
$$a^4 + b^4 = (a^2+b^2)^2-2(ab)^2$$
So we can combine them and get the following:
$$a^5 - b^5 = (a-b)((a^2+b^2)^2-2(ab)^2 + ab(a^2+ab+b^2))$$
Now, all we have to do is calculate $\sin\alpha \cos\alpha$ and plug the values in.
$$\sin\alpha - \cos\alpha = \frac{1}{2}$$
Squaring both sides, we get
$$\sin^2\alpha + \cos^2\alpha - 2\sin\alpha\cos\alpha = \frac{1}{4}$$
$$\Rightarrow 1 - 2\sin\alpha\cos\alpha = \frac{1}{4}$$
$$\Rightarrow\sin\alpha\cos\alpha = \frac{3}{8}$$
Then, if we let $a = \sin\alpha$ and $b = \cos\alpha$, we get
$$\sin^5\alpha - \cos^5\alpha = (\sin\alpha - \cos\alpha)((\sin^2\alpha + \cos^2\alpha)^2 - 2(\sin\alpha\cos\alpha)^2 \\+ \sin\alpha\cos\alpha (\sin^2\alpha + \cos^2\alpha + \sin\alpha\cos\alpha))$$
$$= \frac{1}{2}\left(1^2 - 2\left(\frac{3}{8}\right)^2 + \frac{3}{8}\left(1+\frac{3}{8}\right)\right) = \frac{79}{128}$$
|
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|
Prove that: $\sum_{\text {sym}} a^3+3\sum_{\text{sym}} a^2b\ge 18$ Let $a,b,c$ be the sides of triangle, such that $abc=1$, then prove that:
$$\sum_{\text{sym}} a^3+3\sum_{\text{sym}} a^2b\ge 18.$$
Find at least one case where equality is possible.
My attempts:
I tried the well-known formula with triangle inequalities
$$a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ac)>0$$
But, this formula didn't help.
|
$$3(a^2b+c^2b+b^2c+a^2c+c^2a+b^2a)=$$ $$=3abc((a/c+c/a)+(b/a+a/b)+(c/b+b/c))\ge$$ $$\ge 3(6)=18$$ because $abc=1$ and because if $x\in\{a/c,b/a,c/b\}$ then $x+1/x=(\sqrt x\,-1/\sqrt x)^2+2\ge 2.$
|
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"timestamp": "2023-03-29T00:00:00",
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|
$\frac{3}{b+c+d}+ \frac{3}{c+d+a}+\frac{3}{d+a+b}+\frac{3}{a+b+c} \ge \frac{16}{a+b+c+d}$ for $a,b,c,d>0$ Prove that
$$\frac{3}{b+c+d}+ \frac{3}{c+d+a}+\frac{3}{d+a+b}+\frac{3}{a+b+c}
\ge \frac{16}{a+b+c+d}$$
if $a,b,c,d>0$.
My attempt:
I've put this in the answers.
See also comments under this question.
|
My attempt:
Let p=a+b+c+d then
$$
\frac{1}{b+c+d}+ \frac{1}{c+d+a}+\frac{1}{d+a+b}+\frac{1}{a+b+c}\equiv\\{1\over p-a}+{1\over p-b}+{1\over p-c}+{1\over p-d}
$$
By the AM-GM inequality,
$$
{1\over p-a}+{1\over p-b}+{1\over p-c}+{1\over p-d} \geq4\sqrt[4]{1\over(p-a)(p-b)(p-c)(p-c)}.
$$
Also by AM-GM inequality,
$$
\sqrt[4]{(p-a)(p-b)(p-c)(p-c)}\leq\frac14 \left((p-a)+(p-b)+(p-c)+(p-d)\right)\\
=\frac14(4p-p)=\frac34p\\
$$
so
$$
{1\over p-a}+{1\over p-b}+{1\over p-c}+{1\over p-d} \geq {16\over 3p}
$$
|
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|
Analytic guess for integral $\int_0^\infty \frac{\ln^2(\sqrt{x^2+1}+x) }{\sqrt{x^2 + 1}} \ln \frac{\cosh w+\sqrt{x^2+1} }{1+\sqrt{x^2+1} } dx$ (I was directed here from mathoverflow.)
I am not able to do the following integral analytically, but after numerical evaluations was able to guess the result for all real $w$ values. Not only is the full $w$-dependence clear numerically, the actual coefficients are also correct to very high precision.
$$\int_0^\infty \frac{1}{\sqrt{x^2 + 1}} \log^2(\sqrt{x^2+1}+x) \log\left(\frac{\cosh(w)+\sqrt{x^2+1} }{1+\sqrt{x^2+1} } \right) dx = \frac{1}{12} \left( 2\pi^2 w^2 + w^4 \right)$$
The fact that the result is a simple low order polynomial in $w$ is surprising enough to me.
I tried all kinds of tricks (new variables, partial integrations, etc.) to prove the above but was not able to. How would I go about proving it analytically?
|
Substitute $x=\sinh t$ to reduce the integral
\begin{align}
I(w)=&\int_0^\infty \frac{\ln^2(\sqrt{x^2+1}+x) }{\sqrt{x^2 + 1}} \ln \frac{\cosh w+\sqrt{x^2+1} }{1+\sqrt{x^2+1} } dx\\
= &\int_0^\infty t^2 \ln \frac{\cosh w+\cosh t}{1+\cosh t}dt
\end{align}
and evaluate
\begin{align}
I’(w) =& \int_0^\infty \frac{t^2 \sinh w}{\cosh w+\cosh t}\overset{y=e^{-t}}{dt}\\
=& \ \sinh w\int_0^\infty \frac{\ln^2 y}{y^2 +2y \cosh w+1}dy
= \frac w3(\pi^2+w^2)
\end{align}
Then
$$I(w) =\int_0^w I’(a)da =\frac13 \int_0^w a(\pi^2+a^2)da
=\frac{1}{12} \left( 2\pi^2 w^2 + w^4 \right)$$
|
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|
If $A^2=A,\,B^2=B,\,C^2=C$ and $A+B+C=I,$ prove $AB=BC=AC=\mathbb{O}$.
Let $A,\,B,\,C$ be square matrices such that $A^2=A,\,B^2=B,\,C^2=C$ and $A+B+C=I.$
Prove that $AB=BC=CA=\mathbb{O}.$
Attempt. We have $A(B+C)=A(I-A)=A-A^2=\mathbb{O}$ and similarly we get
$B(A+C)=C(A+B)=\mathbb{O}$, but this is where i get so far.
Thanks in advance for the help.
|
Roland's answer gives a great and simple linear algebra argument.
Here is a purely algebraic development:
\begin{align*}
(A+B+C)^2 & = A^2 + B^2 + C^2 + AB + AC + BC + BA + CA + CB \\
& = I + A(B+C) + BC + (B+C)A + CB \\
& = I + A(I - A) + (I-A)A + BC + CB \\
& = I + A - A^2 + A - A^2 + BC + CB \\
& = I + BC + CB \\
I & = I + BC + CB \\
0 & = BC + CB \\
\end{align*}
By permuting the roles of $A,B,C$, we get $BC + CB = AC + CA = AB + BA = 0$.
This would allow us to conclude if we knew that $AB = BA$. It is not one of the hypotheses, but at this point we can deduce it:
\begin{align*}
AB + BA & = 0 \\
AB & = - BA \\
A^2B & = - ABA \\
AB & = - ABA \\
AB & = BA^2 \\
AB & = BA \\
\end{align*}
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|
How to solve this logarithm problem If $x$ and $y$ are positive real numbers and $x^{\log _yx}=2$ And $y^{\log _xy}=16$ Find $\log _yx$ and find $x$.
The way I tried to solve it was to take $\log _x$ on the base $x$ one and $\log _y$ for the $y$ one. Then something really messy came up. I got that $\log _yx=\sqrt[3]{\dfrac{1}{4}}$ but I don't know if it's correct or not.
|
Note the following:
$$x^{\log_y x} = 2 \implies x^{\frac{\log_2 x}{\log_2 y}} = 2$$
$$\implies 2^{\frac{\log^2_2 x}{\log_2 y}}=2 \implies \frac{\log^2 x}{\log_2 y} = \log_2 2 = 1,$$ or in particular,
$$\frac{\log^2_2 x}{\log_2 y} = 1.$$
Similarly,
$$y^{\log_x y} = 16 \implies \frac{\log^2_2 y}{\log_2 x} = \log_2 16 = 4,$$
or in particular, $$\frac{\log^2_2 y}{\log_2 x} = 4.$$
So write $a = \log_2 x$ and $b=\log_2 y$. Then on the one hand, the above derived equations become $$\frac{a^2}{b} = 1,$$ and $$\frac{b^2}{a} = 4.$$ On the other hand, $\log_y x = \frac{a}{b}$. So let us now calculate $\frac{a}{b}$. Note that
$$\frac{a^2}{b} = 1 \implies a^2 = b.$$ Plugging $a^2=b$ into $\frac{b^2}{a} = 4$ gives $\frac{a^4}{a}=a^{3} = 4$, and so plugging this into $\frac{a^2}{b}=1$ [and dividing both sides by $a$ to get $\frac{a}{b}=\frac{1}{a}$] gives
$$\frac{a}{b}=\frac{1}{a} = a^{-1} = (4)^{-1/3}.$$ So you arrived at the correct answer.
|
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|
Reducing $ax^6-x^5+x^4+x^3-2x^2+1=0$ to a cubic equation using algebraic substitutions
Use algebraic substitutions and reduce the sextic equation to the cubic equation, where $a$ is a real number:
$$ax^6-x^5+x^4+x^3-2x^2+1=0$$
My attempts.
First, I tried to use the Rational root theorem, when $a$ is an integer $x=\pm 1$, but this implies $a=0$ and this is not always correct. Then I realized that,
$$x^4-2x^2+1=(x^2-1)^2$$
is a perfect square. So, I tried to write the original equation as
$$ax^4-x^3+x+\bigg(x-\frac 1x\bigg)^2=0$$
$$x^2\bigg(ax^2-x+\frac 1x\bigg)+\bigg(x-\frac 1x\bigg)^2=0$$
But I failed again. I couldn't spot the palindromic property.
|
Define $t,u=\frac{1\pm\sqrt{1-4a}}{2a}$ as the roots of $ax^2-x+1$. Then the sextic splits into two cubics over $\mathbb Q(t)$:
$$ax^6-x^5+x^4+x^3-2x^2+1=a(x^3-tx^2+t)(x^3-ux^2+u)$$
This can be verified by re-expanding. Here the substitution is $a=\frac{t-1}{t^2}$ and not in $x$.
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Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.