Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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If $xy+yz+zx=1$ , show that $\dfrac x{1-x^2}+\dfrac y{1-y^2}+\dfrac z{1-z^2}=\dfrac{4xyz}{(1-x^2)(1-y^2)(1-z^2)}$. It is to be solved using trigonometry. I tried taking $x$, $y$ and $z$ as $\sin(a)$, $\sin(b)$ and $\sin(c)$ respectively, but could get no further.
| $$x(1 - y^2)(1 - z^2) + y(1 - x^2)(1 - z^2) + z(1 - x^2)(1 - y^2)$$
$$=$$
$$x - xy^2 - xz^2 + xy^2z^2 + y - yx^2 - yz^2 + yx^2z^2 + z - zx^2 - zy^2 + zx^2y^2$$
$$=$$
$$(x + y + z)\cdot 1 - xy^2 - xz^2 - yx^2 - yz^2 - zx^2 - zy^2 + \frac{xy^2z^2 + yx^2z^2 + zx^2y^2}{1}$$
$$=$$
$$4xyz$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1894688",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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If $\frac{\sin A}{\sin B}=\frac{\sqrt{3}}{2}$ and $\frac{\cos A}{\cos B}=\frac{\sqrt{5}}{2}$, then $\tan A+\tan B = $???
$$\text{If}\quad \frac{\sin A}{\sin B}=\frac{\sqrt{3}}{2} \quad\text{and}\quad \frac{\cos A}{\cos B}=\frac{\sqrt{5}}{2}\,, \quad\text{then}\quad \tan A+\tan B = \text{???}$$
Here, $0<A,B<\frac\pi2$.
I tried many times but I am getting no result. Please help with some hint.
| hint: $1 = \dfrac{3\sin^2B+5\cos^2B}{4}=\dfrac{5-2\sin^2B}{4}\implies \sin B = \dfrac{1}{\sqrt{2}}$. Can you continue to the finish line?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1896720",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
Find the co-efficient of $x^3y^3zw^2$ in the expression of $(x-y+2z-2w)^9$.
Find the co-efficient of $x^3y^3zw^2$ in the expression of $(x-y+2z-2w)^9$.
I have clearly understood how to use the bin theorem for $(a+b)^n$, but am unable to extend the theorem to $4$ terms .
| We have $9$ copies of $x-y+2z-2w$. We need to choose $3$ factors for the $x^3$, $3$ factors for the $(-y)^3$, and $1$ factor for the $(2z)$; the remaining two factors will be for the $(-2w)^2$. This produces a term $-8 x^3 y^3 z w^2$.
It remains to count how many ways to make the choices above. This is $\binom{9}{3} \binom{6}{3} \binom{3}{1} = \frac{9!}{3!3!1!2!} = 7! = 5040$. So the coefficient is $5040 \cdot (-8) = -40320$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1897814",
"timestamp": "2023-03-29T00:00:00",
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Can we prove this geometry question without too much calculation? Can we prove the following geometry question without too much calculation?
$ABCD$ is a square, $AC=AE$, $DF=DE$, $AC$ and $BF$ meet at $G$, $BG=CE$.
Prove: $BG \parallel CE$.
| It looks to be related with a dissection of a dodecagon in three squares:
Let we assume that the square has a unit side, $\widehat{CAE}=\theta$ and $H$ is the projection of $D$ on $AE$.
If we prove $\theta=30^\circ$, the claim easily follows. We have $AE=\sqrt{2}$ and
$$ AH = \cos\left(\frac{\pi}{4}-\theta\right) = \frac{\cos\theta+\sin\theta}{\sqrt{2}}, $$
so $AF=\sqrt{2}(\cos\theta+\sin\theta-1)$. By the cosine theorem
$$ BF^2 = 1+AF^2-2 AF\cos\left(\theta+\frac{\pi}{4}\right) $$
and
$$ \frac{BG}{GF}=\frac{[BAG]}{[GAF]}=\frac{\sin\frac{\pi}{4}}{AF\sin\theta}, $$
hence
$$ BG^2 = \left(\frac{\sin\frac{\pi}{4}}{\sin\frac{\pi}{4}+AF\sin\theta}\right)^2 BF^2 = \frac{5-2 \cos\theta-2\cos(2\theta)-6\sin\theta+2\sin(2\theta)}{(2-\cos(2\theta)-2\sin(\theta)+\sin(2\theta))^2}$$
and by equating $BG^2$ and $CE^2=4(1-\cos\theta)$
$$ \sin\theta=\frac{1}{2} $$
follows (not easily - better to have the help of a CAS).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1898122",
"timestamp": "2023-03-29T00:00:00",
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Given the Dirichlet generating function for the divisor sum case, what is the Dirichlet generating function for the Möbius function case? From a search in the OEIS I found from the comment by Antonio G. Astudillo that:
$$\sum _{n=1}^{\infty} \frac{\left(\sum\limits_{d|n}d^{c}\right) \left(\sum\limits_{d|n}d^{z}\right)}{n^{\text{s}}}=\frac{\zeta (s) \zeta (s-c) \zeta (s-z) \zeta (-c+s-z)}{\zeta (-c+2 s-z)}$$
What is then the Dirichlet generating function for:
$$\sum _{n=1}^{\infty} \frac{\left(\sum\limits_{d|n}\mu(d) d^{c}\right) \left(\sum\limits_{d|n}\mu(d)d^{z}\right)}{n^{\text{s}}}=?$$
where $\mu(d)$ is the Möbius function of a divisor $d$. Above $\zeta(s)$ is of course the Riemann zeta function.
| First let us try to evaluate
$$L_1(s) = \sum_{n\ge 1} \frac{1}{n^s} q_{a,b}(n) =
\sum_{n\ge 1} \frac{1}{n^s}
\left(\sum_{d|n} d^a\right) \left(\sum_{d|n} d^b\right).$$
With the prime factorization of $n$ being
$$n = \prod_p p^v$$
we get for
$$q_{a,b}(n) =
\prod_p \frac{p^{av+a}-1}{p^a-1}
\frac{p^{bv+b}-1}{p^b-1}.$$
This yields for the Euler product
$$L_1(s) = \prod_p
\left(1 + \sum_{v\ge 1} \frac{1}{p^{vs}}
\frac{p^{av+a}-1}{p^a-1}
\frac{p^{bv+b}-1}{p^b-1}\right)
\\ = \prod_p
\left(1 + \frac{1}{p^a-1} \frac{1}{p^b-1}
\sum_{v\ge 1} \frac{1}{p^{vs}}
(p^{(a+b)v+a+b} - p^{av+a} - p^{bv+b} + 1)\right)
\\ = \prod_p
\left(\frac{1}{p^a-1} \frac{1}{p^b-1}
\sum_{v\ge 0} \frac{1}{p^{vs}}
(p^{(a+b)v+a+b} - p^{av+a} - p^{bv+b} + 1)\right)
\\ = \prod_p
\left(\frac{1}{p^a-1} \frac{1}{p^b-1}
\\ \times
\left(\frac{p^{a+b}}{1-1/p^{s-(a+b)}}
- \frac{p^{a}}{1-1/p^{s-a}}
- \frac{p^{b}}{1-1/p^{s-b}}
+ \frac{1}{1-1/p^s}
\right)\right)
\\ = \prod_p
\left(\frac{1}{p^a-1} \frac{1}{p^b-1}
\frac{1}{1-1/p^{s-(a+b)}}
\frac{1}{1-1/p^{s-a}}
\frac{1}{1-1/p^{s-b}}
\frac{1}{1-1/p^s}
\\ \times \left(p^{a+b}-p^a-p^b+1
- (p^{2a+2b} - p^{2a+b} - p^{a+2b} + p^{a+b})/p^{2s}
\right)\right)
\\ = \prod_p
\left(
\frac{1}{1-1/p^{s-(a+b)}}
\frac{1}{1-1/p^{s-a}}
\frac{1}{1-1/p^{s-b}}
\frac{1}{1-1/p^s}
\left(1-p^{a+b}/p^{2s}\right)\right).$$
We now obtain by inspection that
$$L_1(s) = \frac{\zeta(s-(a+b))\zeta(s-a)\zeta(s-b)\zeta(s)}
{\zeta(2s-(a+b))}$$
as claimed. Continuing we introduce
$$L_2(s) = \sum_{n\ge 1} \frac{1}{n^s} r_{a,b}(n) =
\sum_{n\ge 1} \frac{1}{n^s}
\left(\sum_{d|n} \mu(d) d^a\right)
\left(\sum_{d|n} \mu(d) d^b\right).$$
This time we have
$$r_{a,b}(n) =
\prod_p (1-p^a) (1-p^b).$$
We get the Euler product
$$L_2(s) = \prod_p
\left(1 + \sum_{v\ge 1} \frac{1}{p^{vs}}
(1-p^a)(1-p^b)\right)
\\= \prod_p
\left(1 + (1-p^a)(1-p^b) \frac{1/p^s}{1-1/p^s}\right)
\\ = \zeta(s)
\prod_p (1 - 1/p^s +
1/p^s - 1/p^{s-a} - 1/p^{s-b} + 1/p^{s-(a+b)})
\\ = \zeta(s)
\prod_p \left(1 + \frac{p^{a+b}-p^a-p^b}{p^s}\right)
\\ = \zeta(s)
\sum_{n\ge 1} \frac{1}{n^s} \mu(n)^2
\prod_{p|n} (p^{a+b}-p^a-p^b).$$
The inner product runs over genuine prime divisors excluding one.
This seems to be the best we can do at this point. What we have shown
here is that
$$\sum_{d|n} \mu(d)^2 \prod_{p|d} (p^{a+b}-p^a-p^b)
= \left(\sum_{d|n} \mu(d) d^a\right)
\left(\sum_{d|n} \mu(d) d^b\right).$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1899914",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Proof that $\sum_{n=1}^{\infty}2^{-3n} = \frac{1}{7}$ Exactly as it says in the title. A proof that:
$$\sum_{n=1}^{\infty}2^{-3n} = \frac{1}{7}$$
I first realised this when I attempted to find the binary representation of $\frac{1}{7}$ and found that it was $1.\overline{001}$ (where the line denotes the recurring part). Since every bit at a position which is a multiple of $-3$ is a $1$, I derived that series. I would like a more mathematical proof; preferably as easy to understand as possible.
EDIT: The series given actually equals $\frac{8}{7}$, I made an error.
| Here is easier solution...
$A=\frac {1}{8} + \frac {1}{64} + ..... = \frac {1}{8} + \frac {1}{8} \cdot (\frac {1}{8} + \frac {1}{64} + .....) = \frac {1}{8} + \frac {A}{8} $
$ \frac {7A}{8} = \frac {1}{8} $
$ A = \frac {1}{7} $
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1900139",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 4
} |
Finding the determinant of a skew-symmetric matrix $K$
Find the determinant of the skew-symmetric matrix $K$
$$K = \begin{bmatrix}
0 & 1 & 3\\
-1 & 0 & 4 \\
-3 & -4 & 0 \\
\end{bmatrix}$$
My Attempted Solution:
I performed the following row operations to reduce $K$ into upper-triangular form $U$
$R_2 \leftrightarrow R_1$
$R_3 - (l_{31} = 3)R_1$
$R_3 \leftrightarrow R_2$
$R_3 - (l_{32} = -4) R_2$
$$U = \begin{bmatrix}
-1 & 0 & 4 \\
0 & -4 & -12 \\
0 & 0 & -45
\end{bmatrix}$$
From this I got
$$\begin{align}
\det(K) &= \pm \ \det(U) \\
&= + \det(U) & \text{(Even no. of row exchanges)} \\
& = (-1)(-4)(-45) \\
&= 180
\end{align}$$
However the correct answer is $\det(K) = 0$. What could I have done wrong, I wouldn't think it would've been the row operations as the row operations apart from the row exchanges don't affect the $\det(K)$? Any hints or suggestiong are greatly appreciated
| You can develop the first row to calculate the determinant :
$$\begin{vmatrix} 0 & 1 & 3 \\ -1 & 0 & 4 \\ -3 & -4 & 0 \end{vmatrix}=0-\begin{vmatrix} -1 & 4 \\ -3 & 0 \end{vmatrix}+3\begin{vmatrix} -1 & 0 \\ -3 & -4 \end{vmatrix}=12-12=0.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1904093",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 2
} |
Find the coefficient of $x^3y^2z^3w$ in the expansion of $(2x+3y-4z+w)^9$
Find the coefficient of $x^3y^2z^3w$ in the expansion of $(2x+3y-4z+w)^9$
Using the formula of multinational coefficients
$$
\begin{pmatrix}
n \\
r_1,r_2,...,r_k \\
\end{pmatrix}= \ \frac{n!}{r_1! \cdot r_2! \cdot \ldots \cdot r_k!} \
$$
$$\Rightarrow \begin{pmatrix}
9 \\
3, 2, 3, 1 \\
\end{pmatrix}= \frac {9!}{3!2!3!1! } = \bbox[yellow]{5040} $$
Would the above be the answer or does one have to take this one step further by further expanding?
$$ 5040 *(2)(3)(-4)(1) = -120960$$
| $5040 \times 2^3 \times 3^2 \times (-4)^3 \times 1 =-23224320$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1904889",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove that $\cos^6 \beta - \sin^6 \beta=\frac {1}{16} (15 \cos 2\beta + \cos 6\beta)$.
Prove the given trigonometric identity:
$$\cos^6 \beta - \sin^6 \beta=\frac {1}{16} (15 \cos 2\beta + \cos 6\beta)$$
My Approach:
\begin{align*}
\text{L.H.S.} &=\cos^6 \beta - \sin^6 \beta\\
&=(\cos^2 \beta)^3 - (\sin^2 \beta)^3\\
&=\cos^32\beta+3\cos^2\beta\cdot\sin^2\beta\cdot\cos2\beta
\end{align*}
Please help me to continue further.
| $$\cos^{6} (x)-\sin^{6}(x)=(\cos^{2}(x)-\sin^{2}(x))(\cos^{4} (x)+\cos^{2}(x)\sin^{2}(x)+\sin^{4}(x))$$
$$a^{2}+b^{2}=(a+b)-2ab.$$
$$=\cos (2 x)(1-\cos^{2} (x)\sin^{2}(x))$$
$$=\cos 2x-\dfrac{1}{2}\cos (2x)\sin^{2}(2x)$$
$$=\dfrac{\cos 2x(1+\cos^{2}(2x))}{2}....؟$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1905142",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
$\int \cos^2(x)$ by parts $$\int \cos^2(x)\ dx = \int \cos x \cos x \ dx$$
By parts, we have:
$$ \int \cos x \cos x \ dx = \cos x \sin x - \int-\sin x\ \sin x dx = \cos x \sin x + \color{Red}{\int \sin x \ \sin x \ dx}$$
but
$$\color{Red}{\int \sin x \ \sin x \ dx} = \sin x(-\cos x)-\int\cos x \ (-\cos x) \ dx = -\sin x \ \cos x +\int \cos^2 x \ dx$$
So:
$$\int \cos^2 x \ dx = \cos x \ \sin x-\sin x \ \cos x + \int \cos^2 x dx \implies$$
$$\int \cos^2 x\ dx = 0$$
Where's the error?
| Using integration by parts, you have
$\int\cos^2x\,dx=\cos x\sin x+\int\sin^2 x\,dx=\cos x\sin x+\int(1-\cos^2 x)\,dx=\cos x\sin x+x-\int\cos^2 x\,dx$.
Now you can solve this equation for $\int\cos^2 x\,dx$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Find $\frac{x^{3333}+x^{333}+x^{33}+x^{3}+1996}{100(x^2+x)}$ if $x^2+x+1=0$
Find $$\frac{x^{3333}+x^{333}+x^{33}+x^{3}+1996}{100(x^2+x)}$$ if $x^2+x+1=0$
My work so far:
1)$x^2+x=-1$, then $100(x^2+x)=-100$
2)$x^2+x+1=0$
$x=\frac{-1\pm\sqrt{3}i}{2}$
| $$x^2=-x-1 \Rightarrow x^3=-x^2-x=(x+1)-x=1 \ \ \& \ \ \Rightarrow x^{33}=(x^3)^{11}=1$$
$$\& \ \ \ x^{333}=(x^3)^{111}=x^{3333}=(x^3)^{1111}=x^3=1.$$
Also:
$$x^{2}+x=x(x+1)=x(-x^{2})=-x^3=-1.$$
So:
$$\dfrac{x^{3333}+x^{333}+x^{33}+x^{3}+1996}{100(x^2+x)}=\dfrac{1+1+1+1+1996}{100\times (-1)}=\dfrac{2000}{-100}=-20.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1906012",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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Complex image of $\left(\frac{\sqrt{2}}{2}+i\frac{\sqrt{2}}{2}\right)z^2$ over a line I need to take the image of
$$f(z) = \left(\frac{\sqrt{2}}{2}+i\frac{\sqrt{2}}{2}\right)z^2$$
over the set $S=\{z=(x,y)\in \mathbb{C}; y=3x+1\}$.
I suspect that $z^2$ might be a parabola but I don't see how. When I have such parabola, I just multiply by the constant $\left(\frac{\sqrt{2}}{2}+i\frac{\sqrt{2}}{2}\right)$, that is, I rotate the parabola by the angle of this number, and stretch by its absolute value.
I must see that $(x+i(3x+1))^2$ is a parabola:
$$x^2+2i(3x+1)+(3x+1)^2 = x^2+6ix+2i+9x^2+6x+1 = 10x^2+6x+1+i(6x+2)$$
$$u = 10x^2+6x+1$$
$$v = 6x+2$$
Then we can see that $u = 10x^2+v-1\implies \frac{u-v+1}{10} = x^2\implies$
$$x = \sqrt{\frac{u-v+1}{10}}$$
but $x$ gets ugly, I don't know how to make $u$ dependent on $v$ only. Can somebody help me?
Is there a generalization for $az^2$?
| $(x+(3x+1)i)^2=x^2 +2xi(3x+1)-(3x+1)^2=-8x^2 -6x-1+i(6x^2+2x)$ Now $$-u=8x^2+6x+1$$ and $$v=6x^2+2x$$ completing the square by adding and subtracting $\frac{1}{6}$ you get $v+\frac{1}{6}=(\sqrt{6}x+\frac{1}{\sqrt{6}})^2$
for $v \geq -1/6$ $$6x=\sqrt{6v+1}-1=$$ now going back to the first you get $$-u=8/6(\sqrt{6v+1}-1)^2 +\sqrt{6v+1}$$ which is a parabola can you see why?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1910206",
"timestamp": "2023-03-29T00:00:00",
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Let $f(x) =x^5+x^2+1$ with $a, b, c, d, e$ as zeros. . . Let $f(x) =x^5+x^2+1$ with $a, b, c, d, e$ as zeros and $g(x)=x^2-2.$ Show that $$g(a)g(b)g(c)g(d)g(e)=-23. $$
I have no idea how to solve it.
Please help.
Thanks in advance.
| Let $a_1,\ldots,a_5$ be the roots of $f(x)=x^5+x^2+1$. Then for every real $x$ we have:
$$f(x)=x^5+x^2+1=\prod_{i=1}^5(x-a_i).$$
So, $g(a_1)\cdot g(a_2)\cdot g(a_3)\cdot g(a_4)\cdot g(a_5)$ is equal to:
\begin{align*}
\prod_{i=1}^5 g(a_i)&=\prod_{i=1}^5 (a_i^2-2)=\prod_{i=1}^5 (a_i-\sqrt{2})(a_i+\sqrt{2})\\
&=\prod_{i=1}^5 (a_i+\sqrt{2}) \cdot \prod_{i=1}^5 (a_i-\sqrt{2})\\
&=\prod_{i=1}^5 (a_i+\sqrt{2}) \cdot \prod_{i=1}^5 [(-1)(\sqrt{2}-a_i)]\\
&=\prod_{i=1}^5 (a_i+\sqrt{2}) \cdot (-1)^5 \prod_{i=1}^5 (\sqrt{2}-a_i)\\
&=\prod_{i=1}^5 [(-1)(a_i+\sqrt{2})] \cdot \prod_{i=1}^5 (\sqrt{2}-a_i)\\
&=\prod_{i=1}^5 (-\sqrt{2}-a_i) \cdot \prod_{i=1}^5 (\sqrt{2}-a_i)\\
&=f(-\sqrt{2})\cdot f(\sqrt{2})\\
&=(-4\sqrt{2}+3)(4\sqrt{2}+3)\\
&=-23
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1911020",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Does $\sum_{n=1}^{\infty} (\frac{1}{n}-\frac{1}{\sqrt{n(n+1)}})$ converge $$\sum_{n=1}^{\infty} \left ( \frac{1}{n}-\frac{1}{\sqrt{n(n+1)}}\right )=\sum_{n=1}^{\infty} \left ( \frac{\sqrt{n(n+1)}-n}{n\sqrt{n(n+1)}}\right )$$
By ratio test I get $$ \lim_{n\to \infty} \frac{\frac{\sqrt{n(n+1)}-n}{n\sqrt{n(n+1)}}}{\frac{1}{n}}=\frac{0}{0}$$
So i can't conclude anything. I tried comparison tests, but i get $\le$ something divergent.
| \begin{align*}
\sum_{n=1}^{\infty}\frac{\sqrt{n(n+1)}-n}{n\sqrt{n(n+1)}}
&=\sum_{n=1}^{\infty}\frac{(\sqrt{n(n+1)}-n)(\sqrt{n(n+1)}+n)}{n\sqrt{n(n+1)}(\sqrt{n(n+1)}+n)}\\
&=\sum_{n=1}^{\infty}\frac1{\sqrt{n(n+1)}(\sqrt{n(n+1)}+n)}\\
&=\sum_{n=1}^{\infty}\frac1{n(n+1)+n\sqrt{n(n+1)}}\\
&\le\sum_{n=1}^{\infty}\frac1{n^2}.
\end{align*}
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Positive integers $a$ and $b$ are such that $a + b =\frac{ a}{b} +\frac{ b}{a}$. What is the value of $a^2 + b^2$? I tried to take $a = 1$ and $b = 2$ and arrive at the conclusion that $a^2 + b^2 = 2$ which indeed is the answer.
However, I wanted to prove this equation algebraically without using the replacement technique.
| Either $a \ge b$ or $b \ge a$. Without loss of generality (wolog) we might as well assume $b \ge a$, as were we to relabel variable $a$ with $b$ and $b$ with $a$ the expression would remain the same. The proof will follow the same whichever $a$ or $b$ we assume to be the larger. So let's assume $b \ge a$.
$a+b = \frac ab + \frac ba \le 1 + \frac ba \le 1 + b \le a + b $.
So $a+b = 1+b $ and $a = 1$.
So $1+ b = \frac 1b + b $ so $b = 1$.
So $a^2 +b^2 =2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1912782",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
What is the sum of the squares of the roots of the equation $x^2 − 7[x] + 5 = 0?$ (Here $[x]$ denotes the greatest integer less than or equal to $x$) I tried many different approaches to this question,
$α^2 = 7[α] - 5$ and
$β^2 = 7[β] - 5$
So combining both the equations we get,
$α^2 + β^2 = 7([α] + [β]) - 10$,
Which is the answer. However I am unable to simplify the LHS.
| $x^2–7[x]+5=0$
$\Rightarrow x^2=7[x]–5 \hspace{1cm}(1)$
$\Rightarrow x^2≤7x–5$ [Since $[x]≤x$]
$\Rightarrow x^2–7x+5≤0$
Therefore, $0.8≤x≤6.2$
i.e., $0≤[x]≤6$
Since for $[x]=0$, equation $(1)$ becomes, $x²=–5$, which is not possible.
Therefore, $1≤[x]≤6$.
In equation $(1)$
For $[x]=1, x^2=2 \Rightarrow x=\pm\sqrt{2}$
For $[x]=2, x²=9 \Rightarrow x=\pm\sqrt{3}$ [not possible because $x=\pm3$ and $[x]=2$ doesn’t make any sense]
For $[x]=3, x²=16 \Rightarrow x=\pm\sqrt{4}$ [not possible because $x=±4$ and $[x]=3$ doesn’t make any sense]
For $[x]=4, x²=23 \Rightarrow x=\pm\sqrt{23}$
For $[x]=5, x²=30 \Rightarrow x=\pm\sqrt{30}$
For $[x]=6, x²=37 \Rightarrow x=\pm\sqrt{37}$
Therefore, sum of all the roots of the given equation $=2+23+30+37=92$.
Hope it helps!!!!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1913250",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
Real-Analysis Methods to Evaluate $\int_0^\infty \frac{x^a}{1+x^2}\,dx$, $|a|<1$.
In THIS ANSWER, I used straightforward contour integration to evaluate the integral $$\bbox[5px,border:2px solid #C0A000]{\int_0^\infty \frac{x^a}{1+x^2}\,dx=\frac{\pi}{2}\sec\left(\frac{\pi a}{2}\right)}$$for $|a|<1$.
An alternative approach is to enforce the substitution $x\to e^x$ to obtain
$$\begin{align}
\int_0^\infty \frac{x^a}{1+x^2}\,dx&=\int_{-\infty}^\infty \frac{e^{(a+1)x}}{1+e^{2x}}\,dx\\\\
&=\int_{-\infty}^0\frac{e^{(a+1)x}}{1+e^{2x}}\,dx+\int_{0}^\infty\frac{e^{(a-1)x}}{1+e^{-2x}}\,dx\\\\
&=\sum_{n=0}^\infty (-1)^n\left(\int_{-\infty}^0 e^{(2n+1+a)x}\,dx+\int_{0}^\infty e^{-(2n+1-a)x}\,dx\right)\\\\
&=\sum_{n=0}^\infty (-1)^n \left(\frac{1}{2n+1+a}+\frac{1}{2n+1-a}\right)\\\\
&=2\sum_{n=0}^\infty (-1)^n\left(\frac{2n+1}{(2n+1)^2-a^2}\right) \tag 1\\\\
&=\frac{\pi}{2}\sec\left(\frac{\pi a}{2}\right)\tag 2
\end{align}$$
Other possible ways forward include writing the integral of interest as
$$\begin{align}
\int_0^\infty \frac{x^a}{1+x^2}\,dx&=\int_{0}^1 \frac{x^{a}+x^{-a}}{1+x^2}\,dx
\end{align}$$
and proceeding similarly, using $\frac{1}{1+x^2}=\sum_{n=0}^\infty (-1)^nx^{2n}$.
Without appealing to complex analysis, what are other approaches one can use to evaluate this very standard integral?
EDIT:
Note that we can show that $(1)$ is the partial fraction representation of $(2)$ using Fourier series analysis. I've included this development for completeness in the appendix of the solution I posted on THIS PAGE.
| This is an elaboration of the last step in Dr. MV's original solution.
The Mittag-Leffler expansion of the secant function is
$$\sec(z) = \pi \sum\limits_{n = 0}^{\infty} \frac{(-1)^{n}(2n+1)}{(\pi /2)^{2}(2n+1)^{2} - z^{2}}$$
Thus we have
\begin{align}
\frac{\pi}{2} \sec\left(\frac{\pi a}{2} \right)
&= \frac{\pi ^{2}}{2} \sum\limits_{n = 0}^{\infty} \frac{(-1)^{n}(2n+1)}{(\pi /2)^{2}(2n+1)^{2} - (\pi a/2)^{2}} \\
&= 2 \sum\limits_{n = 0}^{\infty} \frac{(-1)^{n}(2n+1)}{(2n+1)^{2} - a^{2}}
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1913325",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "28",
"answer_count": 5,
"answer_id": 4
} |
RMO inequality problem If $a,b,c,d,e>1$, then prove that:
$$\frac{a^2}{b-1} + \frac{b^2}{c-1} + \frac{d^2}{e-1} + \frac{c^2}{a-1} + \frac{e^2}{d-1} \ge 20. $$
I don't know how to begin. What should be the approach?
| By Cauchy Schwarz, $($given expression$)(a+b+c+d+e-5)\geq(a+b+c+d+e)^2$
Now this shows $($given expression$)\geq \dfrac{(a+b+c+d+e)^2}{a+b+c+d+e-5}$
Now $(a+b+c+d+e)^2=(a+b+c+d+e-5)^2+2\times(a+b+c+d+e-5)\times 5+25$ hence on division by $a+b+c+d+e-5$ we get $a+b+c+d+e-5+10+\dfrac{25}{a+b+c+d+e-5}=(a+b+c+d+e-5)+\dfrac{25}{a+b+c+d+e-5}+10$
Letting $x=a+b+c+d+e-5$ we note $x\geq0$ and the above is $x+\dfrac{25}{x}+10\geq 10+10=20$ by AM-GM.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1913873",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 5,
"answer_id": 3
} |
Determine d-b if $a^5=b^4,c^3=d^2,c-a=19$ if $a,b,c,d$ are integers. My Attempt:
Let $a^5=b^4=x^{20}$ and let $c^3=d^2=y^6$. Therefore, $a=x^4$ and $c=y^2$. $ \implies 19 = c-a=y^2-x^4=(y-x^2)(y+x^2)$. Since $a,\ b,\ c,\ d$ are integers, $x,\ y$ are also integers $\implies y-x^2$ and $y+x^2$ are also integers. And since $-x^2<x^2$ for all integers $x$, $y-x^2<y+x^2$.
Therefore there are only two cases:
1) $\ y-x^2=1,\ y+x^2=19$. In this case, x=3 and y=10.
2) $\ y-x^2=-19,\ y+x^2=-1$. This case is not possible since $y+x^2$ is always positive because $y,\ x$ are both positive.
So $b=x^5=3^5=243$ and $d=y^3=1000$. Therefore, $d-b=1000-243=757$.
Is my attempt to the problem correct?
| The comments basically sum it up, the answer is correct. And yes, it is the only solution. The numbers $a$ and $c$ are squares differing by an odd prime, so must be consecutive squares, and $a$ must further be a fourth power. This forces $a=81=9^2=3^4, c=100=10^2$, which gives the quoted solution only.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1914161",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 1,
"answer_id": 0
} |
If $A+B+C+D=2\pi$, prove that: If $A+B+C+D=2\pi$, prove that: $$\cos A+\cos B+\cos C+\cos D=4\cos\frac {A+B}{2}\cdot\cos\frac {A+C}{2}\cdot\cos\frac {B+C}{2}$$.
My Approach:
Here,
$$A+B+C+D=2\pi$$
$$A+B=2\pi - (C+D)$$
$$ \sin(A+B)=\sin(2\pi-(C+D))$$
$$\sin(A+B)=-\sin(C+D)$$
Again,
$$\cos(A+B)=\cos(2\pi-(C+D))$$
$$\cos(A+B)=\cos(C+D)$$
Now,
$$L.H.S=\cos A+\cos B+\cos C+\cos D$$
$$=2 \cos\frac {A+B}{2}\cdot\cos\frac {A-B}{2} + 2 \cos\frac {C+D}{2}\cdot \cos\frac {C-D}{2}$$.
I got stuck at here. Please help me to complete the proof.
| We will use $2\cos X \cos Y = \cos(X+Y) + \cos(X-Y)$.
\begin{align*}
4\cos \frac{A+B}{2}\cos \frac{A+C}{2} \cos\frac{B+C}{2} &= 2\left(\cos\left(A + \frac{B+C}{2}\right) + \cos\frac{B-C}{2}\right)\cos\frac{B+C}{2}\\
&= 2\cos\left(A + \frac{B+C}{2}\right)\cos\frac{B+C}{2} + 2 \cos\frac{B-C}{2}\cos\frac{B+C}{2}\\
&= \cos(A+B+C) + \cos A + \cos B + \cos C \\
&= \cos(2\pi - D) + \cos A + \cos B + \cos C \\
&= \cos A + \cos B + \cos C + \cos D
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1916560",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Convergence of the series $\sum\limits_{n=0}^{\infty }\left ( \tan\frac{\pi n+1}{4n+2}-2\sin\frac{\pi n+1}{6n+1} \right )$
Does the series: $\sum\limits_{n=0}^{\infty }\left ( \tan\frac{\pi n+1}{4n+2}-2\sin\frac{\pi n+1}{6n+1} \right )$ converge or diverge?
I've tried to apply Taylor expansion, but I wasn't able to find a proper way to apply it.
| We have
$$ \tan\left(\frac{\pi}{4}+x\right)=1+2x+O(x^2),\qquad 2\sin\left(\frac{\pi}{6}+x\right)= 1+\sqrt{3}\,x+O(x^2)\tag{1}$$
$$ \frac{\pi n+1}{4n+2}=\frac{\pi}{4}+\left(\frac{1}{4}-\frac{\pi}{8}\right)\frac{1}{n}+O\left(\frac{1}{n^2}\right)\tag{2}$$
$$ \frac{\pi n+1}{6n+1}=\frac{\pi}{6}+\left(\frac{1}{6}-\frac{\pi}{36}\right)\frac{1}{n}+O\left(\frac{1}{n^2}\right)\tag{3}$$
hence by combining $(1),(2)$ and $(3)$ it follows that:
$$ \tan\left(\frac{\pi n+1}{4n+2}\right)-2\sin\left(\frac{\pi n+1}{6n+1}\right) = \frac{C}{n}+O\left(\frac{1}{n^2}\right)\tag{4} $$
with $C\approx -0.42292335$, so the given series is divergent by comparison with the harmonic series.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1917991",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Prove that $\lfloor 2x \rfloor + \lfloor 2y \rfloor \geq \lfloor x \rfloor + \lfloor y \rfloor + \lfloor x+y \rfloor$ for all real $x$ and $y$. Prove that $\lfloor 2x \rfloor + \lfloor 2y \rfloor \geq \lfloor x \rfloor + \lfloor y \rfloor + \lfloor x+y \rfloor$ for all real $x$ and $y$.
How should I solve this? I can't think of a way with casework and I can't really simplify it more. Thanks in advance for posting a proof!
| In another way, premised that:
$$\begin{gathered}
\left\lfloor {x + y} \right\rfloor = \left\lfloor {\left\lfloor x \right\rfloor + \left\{ x \right\} + \left\lfloor y \right\rfloor + \left\{ y \right\}} \right\rfloor = \hfill \\
= \left\lfloor x \right\rfloor + \left\lfloor y \right\rfloor + \left\lfloor {\left\{ x \right\} + \left\{ y \right\}} \right\rfloor = \hfill \\
= \left\lfloor x \right\rfloor + \left\lfloor y \right\rfloor + \left[ {1 \leqslant \left\{ x \right\} + \left\{ y \right\}} \right] \hfill \\
\end{gathered}$$
where the square brackets indicate the Iverson bracket
($\left[ {FALSE} \right] = 0,\;\left[ {TRUE} \right] = 1$)
then
$$\begin{gathered}
\left\lfloor {2x} \right\rfloor + \left\lfloor {2y} \right\rfloor = 2\left\lfloor x \right\rfloor + 2\left\lfloor y \right\rfloor + \left[ {1 \leqslant 2\left\{ x \right\} + 2\left\{ y \right\}} \right] \hfill \\
\left\lfloor x \right\rfloor + \left\lfloor y \right\rfloor + \left\lfloor {x + y} \right\rfloor = 2\left\lfloor x \right\rfloor + 2\left\lfloor y \right\rfloor + \left[ {1 \leqslant \left\{ x \right\} + \left\{ y \right\}} \right] \hfill \\
\end{gathered}$$
and, clearly
$$\left[ {1 \leqslant \left\{ x \right\} + \left\{ y \right\}} \right] \leqslant \left[ {1/2 \leqslant \left\{ x \right\} + \left\{ y \right\}} \right]$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1920924",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Finding the real and imary part of complex number using sigma notation I am trying to figure out the follwoing
\begin{align*}
2^{\frac{n}{2}}\Big[\cos(\frac{n\pi}{4})+i\sin(\frac{n\pi}{4})\Big]&=2^{\frac{n}{2}}\Big[\cos(\frac{\pi}{4})+i\sin(\frac{\pi}{4})\Big]^n\\
&=(1+i)^n\\
&=1+\begin{pmatrix}n\\1\end{pmatrix}i-\begin{pmatrix}n\\2\end{pmatrix}-\begin{pmatrix}n\\3\end{pmatrix}i+\begin{pmatrix}n\\4\end{pmatrix}+........
\end{align*}
I am trying to find the real and imaginary part in sigma notation. Any help would be appreciated.
| We know
$$e^{ix}=\exp(ix)=\cos(x)+i\sin(x)$$
so
\begin{align*}
2^{\frac{n}{2}}\Big[\cos(\frac{n\pi}{4})+i\sin(\frac{n\pi}{4})\Big]
&=2^{\frac{n}{2}}\Big[\exp\Big(i\frac{n\pi}{4}\Big)\Big]\\
&=2^{\frac{n}{2}}\Big[\exp\Big(i\frac{\pi}{4}\Big)\Big]^{n}\\
&=2^{\frac{n}{2}}\Big[\cos(\frac{\pi}{4})+i\sin(\frac{\pi}{4})\Big]^n\\
&=\Big(2^{\frac{1}{2}}\Big)^{n}\Big[\cos(\frac{\pi}{4})+i\sin(\frac{\pi}{4})\Big]^n\\
&=\Big(\sqrt2\ \Big)^{n}\Big[\frac{1}{\sqrt2}+i\frac{1}{\sqrt2}\Big]^n\\
&=\Big(\sqrt2\cdot\Big[\frac{1}{\sqrt2}+i\frac{1}{\sqrt2}\Big]\Big)^{n}\\
&=(1+i)^n
\end{align*}
Now I use Binomial Theorem
$$(a+b)^n=\sum_{k=0}^{n}{\binom{n}{k}a^k b^{n-k}}$$
if $a=i$ and $b=1$ then
$$
\begin{align}
(a+b)^n&=(i+1)^n\\
&=\sum_{k=0}^{n}{\binom{n}{k}i^k 1^{n-k}}\\
&=\sum_{k=0}^{n}{\binom{n}{k}i^k}\\
&=\binom{n}{0}i^0+\binom{n}{1}i^1+\binom{n}{2}i^2+\binom{n}{3}i^3+ \dots
\end{align}
$$
$i^k$ has 4 values for k, be aware '$k \equiv 1 \pmod 4$' means remainder of division of $k$ by $4$ is $1$.
$$
\begin{align}
k \equiv 0 \pmod 4 & \Rightarrow i^k=i^0=i^4=\dots=1 \\
k \equiv 1 \pmod 4 & \Rightarrow i^k=i^1=i^5=\dots=i\\
k \equiv 2 \pmod 4 & \Rightarrow i^k=i^2=i^6=\dots=-1\\
k \equiv 3 \pmod 4 & \Rightarrow i^k=i^3=i^7=\dots=-i
\end{align}
$$
so
$$
\begin{align}
\sum_{k=0}^{n}{\binom{n}{k}i^k}&=\binom{n}{0}+\binom{n}{1}i-\binom{n}{2} -\binom{n}{3}i+\binom{n}{4}+\dots
\end{align}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1922344",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Which one is the correct way to compute the change of basis matrix? I was looking for a way to compute the change of basis matrix (given the old basis and the new basis), but I found two methods that lead to different results, and I can't understand which one is correct.
Method 1:
$$
B = \left\{
\begin{pmatrix}
1 \\ 2
\end{pmatrix} ,
\begin{pmatrix}
3 \\ 4
\end{pmatrix}
\right\} ~ , ~
D= \left\{
\begin{pmatrix}
1 \\ 4
\end{pmatrix} ,
\begin{pmatrix}
2 \\ 3
\end{pmatrix} \right\}
$$
The vectors in $D$ are expressed as a linear combination of the ones in $B$, and then the coefficients are used to construct the change of basis matrix $S$, i.e.:
$$
\begin{pmatrix}1 \\ 4 \end{pmatrix} = s_{11}\cdot \begin{pmatrix}1 \\ 2 \end{pmatrix} + s_{12} \cdot \begin{pmatrix}3 \\ 4 \end{pmatrix} \ \ \Rightarrow s_{11}=4 \ \ \text{and} \ \ s_{12}=-1 \\
\begin{pmatrix}2 \\ 3 \end{pmatrix} = s_{21}\cdot \begin{pmatrix}1 \\ 2 \end{pmatrix} + s_{22} \cdot \begin{pmatrix}3 \\ 4 \end{pmatrix} \Rightarrow s_{21}=\frac{1}{2} \ \ \text{and} \ \ s_{22}=\frac{1}{2} \\
\Rightarrow S =\begin{pmatrix}4 & -1 \\ \frac{1}{2} & \frac{1}{2} \end{pmatrix}
$$
Method 2:
$$\begin{pmatrix}s_{11} & s_{12} \\ s_{21} & s_{22} \end{pmatrix} \cdot \begin{pmatrix}1 & 3 \\ 2 & 4 \end{pmatrix} = \begin{pmatrix}1 & 2 \\ 4 & 3 \end{pmatrix} \\
\Rightarrow s_{11}=0 \ \ ;\ \ s_{12}= \frac{1}{2} \ \ ; \ \ s_{21}= -5 \ \ ; \ \ s_{22}=\frac{9}{2}\\
\Rightarrow S =\begin{pmatrix}0 & \frac{1}{2} \\ -5 & \frac{9}{2} \end{pmatrix}
$$
| The method 2) is correct. It is the same as to find a matrix
$$
S=\begin{pmatrix}s_{11} & s_{12} \\ s_{21} & s_{22} \end{pmatrix}
$$
such that
$$
S\begin{pmatrix}1 \\2 \end{pmatrix}=\begin{pmatrix}1 \\4 \end{pmatrix}
$$
and
$$
S\begin{pmatrix}3 \\4 \end{pmatrix}=\begin{pmatrix}2 \\3 \end{pmatrix}
$$
that means exactly to transform the vectors of the first basis to the vectors of the second basis.
I don't see how the method 1) comes from...but it is wrong.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1922568",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
partial derivative of cosine similarity I asked a question about
derivative of cosine similarity.
But no one has answered my question.
Therefore I tried to do it my self as bellow.
$$
cossim(a,b)=\frac{a\cdot{b}}{\sqrt{a^2\cdot{b^2}}}
\\\frac{cossim(a,b)}{\partial{a_1}}=\frac{\partial}{\partial{a_1}} \frac{a_1\cdot{b_1}+...+a_n\cdot{b_n}}{|a|\cdot|b|}
\\=\frac{\partial}{\partial{a_1}}a_1\cdot{b_1}\cdot{(a_1^2+a_2^2+...a_n^2)^{-1/2}}\cdot{|b|^{-1}}
\\= {b_1}\cdot{(a_1^2+a_2^2+...a_n^2)^{-1/2}}\cdot{|b|^{-1}}-a_1^2b_1(a_1^2+a_2^2+...a_n^2)^{-3/2} {|b|^{-1}}
\\=\frac{b_1}{|a|\cdot{|b|}}-\frac{a_1|a|^{-2}\cdot{a_1b_1}}{|a|\cdot{|b|}}
\\=\frac{b_1}{|a|\cdot{|b|}}-\frac{a_1\cdot{b_1}}{|a|\cdot{|b|}}\cdot{\frac{a_1}
{|a|^2}}
\\\therefore \frac{\partial}{\partial{a}}cossim(a,b)= \frac{b_1}{|a|\cdot{|b|}}-cossim(a,b)\cdot{\frac{a_1}
{|a|^2}}
$$
Is this correct?
| Putting
$$
\cos (\mathbf{v},\mathbf{w}) = \frac{{\mathbf{v} \cdot \mathbf{w}}}
{{\left| {\mathbf{v} } \right|\;\left| \mathbf{w} \right|}}
$$
I would develop the required derivative as follows:
$$
\cos (\mathbf{v} + d\mathbf{v},\mathbf{w}) = \frac{{\mathbf{v} \cdot \mathbf{w} + d\mathbf{v} \cdot \mathbf{w}}}
{{\left| {\mathbf{v} + d\mathbf{v}} \right|\;\left| \mathbf{w} \right|}}
$$
Now $\left| {\mathbf{v} + d\mathbf{v}} \right|$ can be rewritten as:
$$
\begin{gathered}
\left| {\mathbf{v} + d\mathbf{v}} \right| = \sqrt {\left( {\mathbf{v} + d\mathbf{v}} \right) \cdot \left( {\mathbf{v} + d\mathbf{v}} \right)} = \sqrt {\left| \mathbf{v} \right|^2 + \left| {d\mathbf{v}} \right|^2 + 2\mathbf{v} \cdot d\mathbf{v}} = \hfill \\
= \left| \mathbf{v} \right|\sqrt {1 + 2\frac{\mathbf{v}}
{{\left| \mathbf{v} \right|^2 }} \cdot d\mathbf{v} + \frac{{\left| {d\mathbf{v}} \right|^2 }}
{{\left| \mathbf{v} \right|^2 }}} \approx \left| \mathbf{v} \right|\left( {1 + \frac{\mathbf{v}}
{{\left| \mathbf{v} \right|^2 }} \cdot d\mathbf{v}} \right) \hfill \\
\end{gathered}
$$
hence:
$$
\begin{gathered}
\cos (\mathbf{v} + d\mathbf{v},\mathbf{w}) = \frac{{\mathbf{v} \cdot \mathbf{w} + d\mathbf{v} \cdot \mathbf{w}}}
{{\left| {\mathbf{v} + d\mathbf{v}} \right|\;\left| \mathbf{w} \right|}} \approx \frac{{\mathbf{v} \cdot \mathbf{w} + d\mathbf{v} \cdot \mathbf{w}}}
{{\left| \mathbf{v} \right|\left( {1 + \frac{\mathbf{v}}
{{\left| \mathbf{v} \right|^2 }} \cdot d\mathbf{v}} \right)\;\left| \mathbf{w} \right|}} \approx \hfill \\
\approx \frac{{\mathbf{v} \cdot \mathbf{w} + \mathbf{w} \cdot d\mathbf{v}}}
{{\left| \mathbf{v} \right|\;\left| \mathbf{w} \right|}}\left( {1 - \frac{\mathbf{v}}
{{\left| \mathbf{v} \right|^2 }} \cdot d\mathbf{v}} \right) \approx \frac{{\mathbf{v} \cdot \mathbf{w}}}
{{\left| \mathbf{v} \right|\;\left| \mathbf{w} \right|}} + \left( {\frac{\mathbf{w}}
{{\left| \mathbf{v} \right|\;\left| \mathbf{w} \right|}} - \frac{{\mathbf{v} \cdot \mathbf{w}}}
{{\left| \mathbf{v} \right|\;\left| \mathbf{w} \right|}}\frac{\mathbf{v}}
{{\left| \mathbf{v} \right|^2 }}} \right) \cdot d\mathbf{v} = \hfill \\
= \cos (\mathbf{v},\mathbf{w}) + \left( {\frac{\mathbf{w}}
{{\left| \mathbf{v} \right|\;\left| \mathbf{w} \right|}} - \cos (\mathbf{v},\mathbf{w})\frac{\mathbf{v}}
{{\left| \mathbf{v} \right|^2 }}} \right) \cdot d\mathbf{v} \hfill \\
\end{gathered}
$$
Therefore, apart from a typo, your derivation
$$
\frac{\partial}{\partial{a_1}}cossim(a,b)= \frac{b_1}{|a|\cdot{|b|}}-cossim(a,b)\cdot{\frac{a_1}
{|a|^2}}
$$
looks to be correct.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1923613",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
limit of $\lim_{x\to 0} \frac{(\sin x)^2 - (x \cos x) ^ 2}{x^4}$ I have question. I want to solve this limit. it's $\frac{0}{0}$ so we have to change it. there is two way with two different value.
$\lim_{x\to 0} \frac{(\sin x)^2 - (x \cos x) ^ 2}{x^4}$
First way:
before that we know that $\lim_{x\to 0} \frac{\sin x}{x}$ or $\lim_{x\to 0} \frac{(\sin x) ^ 2}{x ^ 2}$ is equal to 1 so:
$\lim_{x\to 0} \frac{(\sin x)^2 - (x \cos x) ^ 2}{x^4} = \lim_{x\to 0} \frac{x^2 - x ^ 2 (\cos x) ^ 2}{x^4} = \lim_{x\to 0} \frac{x^2 (1 - (\cos x) ^ 2)}{x^4} = \lim_{x\to 0} \frac{(\sin x) ^ 2}{x^2} = 1$
second way:
before that we know that $\sin x \sim x - \frac{x^3}{6}$ and $\cos x \sim 1 - \frac{x^2}{2}$ so:
$\lim_{x\to 0} \frac{(\sin x)^2 - (x \cos x) ^ 2}{x^4} = \lim_{x\to 0} \frac{(\sin x) - (x \cos x)}{x^3} * \frac{(\sin x) + (x \cos x)}{x} = \lim_{x\to 0} \frac{x - \frac{x^3}{6} - x + \frac{x^3}{2}}{x^3} * \frac{x - \frac{x^3}{6} + x - \frac{x^3}{2}}{x} = (\frac{1}{2} - \frac{1}{6}) * 2 = \frac{2}{3}$
Update:
Is it possible to explain more? We have limit and we solve like this and that's work but in this limit we can't use $\lim_{x\to0}\left(\frac{\sin x}x\right)^2=1$.
this in another limit:
$\lim_{x\to0} \frac{1 - \cos 2x}{x^2}= \lim_{x\to0} \frac{2 (\sin x)^2}{x^2} = 2 * \lim_{x\to0} (\frac{\sin x}{x})^2 = 2 * 1 = 2$
Which way is true? Is it possible to help me?
I'm sorry for bad English.
Thanks.
| The correct way is the second one. The limit is indeed $\frac{2}{3}$.
Since$x$ is approaching to zero, you're lead to use Taylor expansion which is a cool tool to evaluate some limits.
| {
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"url": "https://math.stackexchange.com/questions/1924213",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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} |
expanding and factoring in algebra Expanding very simple process, just open round brackets and multiply each with each for example
$(x-6)(x+3) = x^2 + 3x - 6x -18 = x^2 -3x -18$
Factoring is confused me a little for example I have $x^2 -3x -18$ I try to brute force all combinations in my mind and get the result like this
$(x-6)(x+3)$
Cold you please recommend some algorithm, how to find factoring in the easy way.
Thanks.
| Here's an algorithm to factor any quadratic (so $a\neq 0$):
\begin{align}ax^2+bx+c&=a\left(x^2+\frac{b}{a}x+\frac{c}{a}\right)\\&=a\left(\left(x+\frac{b}{2a}\right)^2-\frac{b^2}{4a^2}+\frac{c}{a}\right)\\&=a\left(\left(x+\frac{b}{2a}\right)^2-\frac{\Delta}{4a^2}\right) &(\Delta=b^2-4ac)\\&=a\left(\left(x+\frac{b}{2a}\right)^2-\left(\frac{\sqrt{\Delta}}{2a}\right)^2\right)\\&=a\left(x+\frac{b-\sqrt{\Delta}}{2a}\right)\left(x+\frac{b+\sqrt{\Delta}}{2a}\right)\end{align}
Note that we used $\sqrt{\Delta}$. If you know complex numbers, that shouldn't be a problem. If not, note that when $\Delta<0$, the equation $ax^2+bx+c=0$ has no solutions in $\mathbb{R}$ because it's equivalent to $$\left(x+\frac{b}{2a}\right)^2-\frac{\Delta}{4a^2}=0$$ which has clearly no solutions in $\mathbb{R}$ as $\left(x+\frac{b}{2a}\right)^2\ge 0$ and $-\frac{\Delta}{4a^2}>0$. Thus, if $\Delta<0$, you can't factor over $\mathbb{R}$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Solve for $x$ in $\frac{\sqrt x-1}{x-1}>\frac{4^{3/2}}{2^4}$ I am trying to solve for $x$ but am very unsure of the steps. If someone could walk me through this I would really appreciate it.
$$\frac{\sqrt x-1}{x-1}>\frac{4^{3/2}}{2^4}$$
| $$\frac{\sqrt x-1}{x-1}>\frac{4^{3/2}}{2^4}$$
Clearly we can see that $4^{3/2}=\left(4^{\frac{1}{2}}\right)^3=2^3=8$. So
$$\frac{\sqrt x-1}{x-1}>\frac{2^3}{2^4}$$
$$\frac{\sqrt x-1}{x-1}>\frac{1}{2}$$
$$\frac{2\sqrt x-2}{x-1}>1$$
$$\frac{2\sqrt x-2}{x-1}-1>0$$
$$\frac{2\sqrt x-2-x+1}{x-1}>0$$
$$\frac{2\sqrt x-x-1}{x-1}>0$$
Multiply both sides by $(-1)$ . Then ,
$$\frac{x-2\sqrt x+1}{x-1}<0$$
$$\frac{(\sqrt{x}-1)(\sqrt{x}-1)}{(\sqrt{x}-1)(\sqrt{x}+1)}<0$$
Thus for $\sqrt{x} \neq1$ ,
$$\frac{\sqrt{x}-1}{\sqrt{x}+1}<0$$
Since $\sqrt{x}+1$ is always positive as $\sqrt{x}>0$ for all $x$ ,
$$\sqrt{x}-1<0$$
That is $$\sqrt{x}<1$$
$$0\leq x<1$$
| {
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"url": "https://math.stackexchange.com/questions/1926061",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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What's my mistake in this integration??? $\int \sqrt{3-2x-x^2} dx$ This is how I'm trying to integrate this function:
\begin{align*}
\int \sqrt{3-2x-x^2}\, dx &= \int \sqrt{4-(x+1)^2}\, dx \\
&= \int \sqrt{2^2-(x+1)^2}\, dx
\end{align*}
Here I make the substitution:
$$ u=x+1 $$
$$du=dx$$
So the integral is now:
$$ \int \sqrt{2^2-u^2}\, du $$
I make a trigonometric substitution thinking about a right triangle where the hypothenuse is $2$, the adjacent side is $u$, the opposite side is $\sqrt{2^2-u^2}$, and the angle is called $\theta$.
$$ \sin(\theta)= \frac{\sqrt{2^2-u^2}}{ 2}$$
$$\bbox[2px,border:2px solid red]
{ 2\sin(\theta)= \sqrt{2^2-u^2}\qquad
}$$
$$\frac{u}{2} =\cos(\theta)$$
$$ u=2 \cos(\theta)$$
$$ \bbox[2px,border:2px solid red]
{du=-2 \sin(\theta)\,d\theta\qquad
}$$
So I write the integral as:
\begin{align*}
\int 2\sin(\theta)(-2)\sin(\theta)\,d\theta &= \int (-4)\sin(\theta)\sin(\theta)\,d\theta \\
&= \int (-4){\sin}^2(\theta)\,d\theta \\
&= (-4)\int {\sin}^2(\theta)\,d\theta \\
&= (-4)\int \frac{1}{2}(1-\cos(2\theta))\,d\theta \\
&= (-4)\frac{1}{2}\int (1-\cos(2\theta))\,d\theta \\
&= (-2)\int (1-cos(2\theta))\,d\theta \\
&= (-2)\left[\int d\theta-\int \cos(2\theta)\,d\theta \right]\\
&= (-2)\left[\theta-\int \cos(2\theta)\,d\theta \right]\\
&= (-2)\left[\theta-\frac{1}{2} \sin(2\theta) \right]\\
&= \sin(2\theta) -2\theta
\end{align*}
And since $\cos(\theta) = u/2$, I know that $\theta =\arccos(u/2)$.
Therefore I have:
$$ \sin(2\arccos(u/2)) -2\arccos(u/2)$$
According to my first substitution $u=x+1$ so the final result is:
$$\bbox[2px,border:2px solid red] {\sin\left(2\arccos\left(\frac{x+1}{2}\right)\right) -2\arccos\left(\frac{x+1}{2}\right) + constant }\qquad$$
Can anyone help me? I don't understand what I'm doing wrong. Thanks!
| Answer is correct, if you want it to match with one that Wolfram produce, there should be done following:
$$
\sin(2\theta)=2\sin(\theta)\cos(\theta)=2\frac{1}{2}\sqrt{2^2-u^2}\frac{u}{2}=\frac{u}{2}\sqrt{4-u^2}\\
u\to x+1\\
\sin(2\theta)=\frac{1}{2}(x+1)\sqrt{3-2x-x^2}
$$
And for second addendum
$$
\arccos(u/2)=\frac{\pi}{2}-\arcsin(u/2)\\
2\arccos(u/2)=\pi-2\arcsin\left(\frac{x+1}{2}\right)
$$
Altogether (hiding $\pi/2$ in constant):
$$
\int \sqrt{3-2x-x^2} dx = \frac{1}{2}(x+1)\sqrt{3-2x-x^2}+2\arcsin\left(\frac{x+1}{2}\right)+const
$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Prove that $(b+c)^2\geq a^2+4\cdot AD^2$
Let $ABC$ be a triangle and $AD$ be the altitude through $A$. Prove that $$(b+c)^2\geq a^2+4\cdot AD^2$$
(where $a=BC$, $b=CA$, $c=AB$).
I used Apollonius theorem and Pythagoras theorem every where. I guess that we can do it using these two theorems but I can't process.
| Let $h_a=AD$
Heron's formula
$$\sqrt{s(s-a)(s-b)(s-c)}=\frac{1}{2}h a$$
where
$$s=\frac{a+b+c}{2}$$
written in terms of $a,b,c$ gives
$$\frac{(b+c)^2-a^2}{4h^2}=\frac{a^2}{a^2-(b-c)^2}$$
The right hand side is $\geq 1$ therefore
$$\frac{(b+c)^2-a^2}{4h^2}\geq 1$$
so
$$(b+c)^2\geq a^2+ 4h^2$$
Equality holds if and only if $b=c$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Evaluation of $\int^{\frac{\pi}{2}}_{0}\frac{\cos^6 x}{\sin x+\cos x}dx$
Evaluation of $\displaystyle \int^{\frac{\pi}{2}}_{0}\frac{\cos^6 x}{\sin x+\cos x}dx$
$\bf{My\; Try::}$ Let $\displaystyle I = \int^{\frac{\pi}{2}}_{0}\frac{\cos^6 x}{\sin x+\cos x}dx = \int^{\frac{\pi}{2}}_{0}\frac{\sin^6 x}{\sin x+\cos x}dx$
Above we have used $\displaystyle \int^{a}_{0}f(x)dx = \int^{a}_{0}f(a-x)dx$
So $$2I = \int^{\frac{\pi}{2}}_{0}\frac{\sin^6 x+\cos^6 x}{\sin x+\cos x}dx = \int^{\frac{\pi}{2}}_{0}\frac{(\sin^2 x+\cos^2 x)(\sin^4 x+\cos^4 x-\sin^2 x\cos^2 x)}{\sin x+\cos x}dx$$
So $$2I = \int^{\frac{\pi}{2}}_{0}\frac{\sin^4 x+\cos^4 x-\sin^2 x\cos^2 x}{\sin x+\cos x}dx = \int^{\frac{\pi}{2}}_{0}\frac{1-3\sin^2 x\cos^2 x}{\sin x+\cos x}dx$$
Now How can i solve it after that, Help Required, Thanks
| A little bit expansion of my above comment. Use $\cos x+\sin x=\sqrt{2}\cos(x-\pi/4) $ and changing variable, we get
$$I=\int_{-\pi/4}^{\pi/4}\frac{\cos^6(x+\pi/4)}{\sqrt{2}\cos (x)}dx.$$
Since $\cos(x+\pi/4)=\frac{1}{\sqrt{2}}(\cos x-\sin x)$, one has
$$I=\int_{-\pi/4}^{\pi/4}\frac{(\cos x-\sin x)^6}{(\sqrt 2)^7 \cos x}dx.$$
The following is easy.
| {
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"timestamp": "2023-03-29T00:00:00",
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Polar to cartesian form of r=sin(3θ) $$\sin(2θ) = \sinθ\cosθ + \cosθ\sinθ = 2\sinθ\cosθ $$
$$\cos(2θ) = \cosθ\cosθ - \sinθ\sinθ = \cos²θ - \sin²θ $$
$$\sin(3θ) = \sin(2θ+θ) $$
$$= \sin(2θ)\cosθ + \cos(2θ)\sinθ $$
$$= 2\sinθ\cos²θ + \cos²θ\sinθ - \sin³θ
$$ $$= 2\sinθ(1-\sin²θ) + (1-\sin²θ)\sinθ - \sin³θ
$$ $$= 2\sinθ - 2\sin³θ + \sinθ - \sin³θ - \sin³θ
$$ $$r=\sin(3θ)= 3\sinθ - 4\sin³θ$$
How to deal with the differing exponents? I could add an $r$ to both sides or $r^3$ but not both.
$$r^4= 3r^3\sinθ - 4r^3\sin^{3}θ $$
$$r^4= 3r^2y - 4y^3$$
$$(x^2+y^2)^{4/2}= 3(x^2+y^2)^{2/2}y - 4y^3$$
$$(x^2+y^2)^{2}= 3(x^2+y^2)y - 4y^3$$
$$(x^2+y^2)^{2}= 3yx^2-y^3$$
| Solution:
$$r=3\sin(\theta)-4\sin^3(\theta)=\sin(\theta)\{ 4\cos^2(\theta)-1\}$$
Now put: $\cos(\theta)=x/r$ , $\sin(\theta)=y/r$ and $r=(x^2+y^2)^{1/2}$.
This will give you: $$r=\frac{y(4x^2-r^2)}{r^3}$$ which implies $$(x^2+y^2)^2=4x^2y-y(x^2+y^2)=3x^2y-y^3$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1930388",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
Is $b=\left(\begin{smallmatrix}2\\-1\end{smallmatrix}\right)$ in the null space of $A=\left(\begin{smallmatrix}1&2\\3&4\\5&6\end{smallmatrix}\right)$?
Is vector $\mathrm b = \begin{pmatrix} 2 \\ -1 \end{pmatrix}$ in the null space of $\mathrm A = \begin{pmatrix} 1 & 2 \\ 3 & 4 \\ 5 & 6\end{pmatrix}$?
My solution:
I took matrix A and put it in reduced echelon form. I calculated this to be:
$$\begin{pmatrix} 1 & 0 & | & 0 \\ 0 & 1 & | & 0 \\ 0 & 0 & | & 0\end{pmatrix}$$
So $x_1= 0$ and $x_2 = 0$. But from here I am not sure how to answer the question
| We have that $b \in \operatorname{Null} (A)$ if and only if $Ab =0$, so in this case we have
$$Ab = \begin{pmatrix} 1 & 2 \\ 3 & 4 \\ 5 & 6 \end{pmatrix}\begin{pmatrix} 2 \\ -1 \end{pmatrix} = \begin{pmatrix} 0 \\ 2 \\ 4 \end{pmatrix} \neq \vec{0}$$
Hence $b \not \in \operatorname{Null}(A)$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Does the same determinant make two matrices equal to each other?
Does the same determinant make two matrices equal to each other?
If I have:
Find all values of $x$ that make
$$\begin{pmatrix}2 & -1 &4\\3 & 0 & 5\\4 & 1 & 6\end{pmatrix}=\begin{pmatrix} x & 4\\5 & x\end{pmatrix} $$
Would I calculate and equate the determinants of both matrices to solve this problem?
Edit: Below is the exact question. Do the style of brackets refer to the determinants?
| If the problem is about an equality of the determinant, all you have to do is compute the determinants separately. The determinant of the $3\times 3$ matrix is
$$
(2)(0)(6) + (-1)(5)(4) + (4)(3)(1) - (4)(0)(4) - (1)(5)(2) - (6)(3)(-1) = 0 - 20 + 12 - 0 - 10 + 18 = 0.
$$
The $2\times 2$ determinant is just $x^{2} - 20$. Then, we arrive at the equation
$$
0 = x^{2} - 20
$$
which has two possible solutions: $x=\sqrt{20}$ or $x=-\sqrt{20}$. Thus, the answer is (D) if the question refers to determinants.
If not, then there is no solution.
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove by induction (Fibonacci) $F_n=\frac{\left(\frac{1+\sqrt 5}{2}\right)^n-\left(\frac{1-\sqrt 5}{2}\right)^n}{\sqrt5}$ Can someone advise on proving the following by induction. I'm assuming strong induction should be used here?
Here $F_n$ is the $n$th Fibonacci number. Prove that
$$F_n=\frac{\left(\frac{1+\sqrt 5}{2}\right)^n-\left(\frac{1-\sqrt 5}{2}\right)^n}{\sqrt5}$$
| Yes, go with induction.
First, check the base case $$F_1=1$$ That should be easy.
For the inductive step, consider, on the one hand:
(1) $$F_{n+1}= F_n+F_{n-1}$$
Then, write what you need to prove, to have it as a guidance of what you need to get to. That is:
$$F_{n+1}=\frac{\left(\frac{1+\sqrt 5}{2}\right)^{n+1}-\left(\frac{1-\sqrt 5}{2}\right)^{n+1}}{\sqrt5}$$
Use (1) and your hypothesis and write
$$F_{n+1}= \frac{\left(\frac{1+\sqrt 5}{2}\right)^n-\left(\frac{1-\sqrt 5}{2}\right)^n}{\sqrt5} + $$ $$\frac{\left(\frac{1+\sqrt 5}{2}\right)^{n-1}-\left(\frac{1-\sqrt 5}{2}\right)^{n-1}}{\sqrt5}$$
this translates to
$$= {\frac {1} {\sqrt5}}[{\left(\frac{1+\sqrt 5}{2}\right)^{n}\left(1+\frac{2}{1+\sqrt 5}\right)}-{\left(\frac{1-\sqrt 5}{2}\right)^{n}\left(1+\frac{2}{1-\sqrt 5}\right)}]$$
To finish, note that (a) $${\left(1+\frac{2}{1+\sqrt 5}\right)}{\left(\frac{1-\sqrt 5}{1-\sqrt 5}\right)}={\frac {1+\sqrt 5}{2}}$$
and (b)
$${\left(1+\frac{2}{1-\sqrt 5}\right)}{\left(\frac{1+\sqrt 5}{1+\sqrt 5}\right)}={\frac {1-\sqrt 5}{2}}$$
which is exactly what you need to end the proof.
| {
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Evaluating $\lim_{x\to 0} \frac{1-\cos x\sqrt{\cos 2x}\sqrt[3]{\cos 3x}}{x^2}$ I want to evaluate:
$$\lim_{x\to 0} \frac{1-\cos x\sqrt{\cos 2x}\sqrt[3]{\cos 3x}}{x^2}.$$
Here's what I did. We know that as soon as ${x\to 0}$
$$1 - \cos x = \frac{x^2}{2} + O(x^4)$$
Therefore
$$\cos x = 1 - \frac{x^2}{2} + O(x^4)$$
Now we apply this to cosines in limit
$$\cos 2x = 1 - 2x^2 + O(x^4)$$
$$\cos 3x = 1 - \frac{9x^2}{2} + O(x^4)$$
Then use equality
$$ (1 + x)^n = 1 + xn + o(x)$$
It yields us
$$\lim_{x\to 0} \frac{1 - (1 - \frac{x^2}{2})(1 - x^2)(1 - \frac{3x^2}{2})}{x^2}$$
After simplification 1s cancel and we get $-(-1/2 - 1 - 3/2) = 3$.
| The $x^2$ at the denominator suggests we can use the fact that
$$
\lim_{x\to0}\frac{1-\cos x}{x^2}=\frac{1}{2}
$$
Rewrite the numerator as
$$
(1-\sqrt{\cos2x}\sqrt[3]{\cos3x})+\sqrt{\cos2x}\sqrt[3]{\cos3x}(1-\cos x)
$$
Since
$$
\lim_{x\to0}\frac{\sqrt{\cos2x}\sqrt[3]{\cos3x}(1-\cos x)}{x^2}=
\frac{1}{2}
$$
we are reduced to compute
$$
\lim_{x\to0}\frac{1-\sqrt{\cos2x}\sqrt[3]{\cos3x}}{x^2}
$$
Let's retry the trick, rewriting the numerator as
$$
(1-\sqrt[3]{\cos3x})+\sqrt[3]{\cos3x}(1-\sqrt{\cos2x}\,)
$$
Now
$$
\lim_{x\to0}\frac{1-\sqrt{\cos2x}}{x^2}=
\lim_{x\to0}4\frac{1-\cos2x}{(2x)^2}\frac{1}{1+\sqrt{\cos2x}}=
4\cdot\frac{1}{2}\cdot\frac{1}{2}=1
$$
This leaves us with
$$
\lim_{x\to0}\frac{1-\sqrt[3]{\cos3x}}{x^2}=
\lim_{x\to0}9\frac{1-\cos3x}{(3x)^2}\frac{1}{1+\sqrt[3]{\cos3x}+\sqrt[3]{\cos^23x}}=9\cdot\frac{1}{2}\cdot\frac{1}{3}=\frac{3}{2}
$$
Thus our limit is
$$
\frac{3}{2}+1+\frac{1}{2}=3
$$
Alternative method. Define
$$
g_n(t)=\log\sqrt[n]{\cos(n\sqrt{t})}=\frac{1}{n}\log\cos(n\sqrt{t})
$$
defined in a suitable neighborhood of $0$.
For $t\ne0$, the derivative is
$$
g_n'(t)=
\frac{1}{n}
\frac{1}{\cos(n\sqrt{t})}
(-\sin(n\sqrt{t}))\frac{n}{2\sqrt{t}}
$$
and the limit at $0$ is easily seen to be $-n/2$, so by l'Hôpital the function $g_n$ is also differentiable at $0$.
The function you have to compute the limit of is even, so we can as well compute the limit from the right and do the substitution $x=\sqrt{t}$, so we have
$$
\lim_{t\to0^+}\frac{1-\cos\sqrt{t}\sqrt{\cos(2\sqrt{t})}\sqrt[3]{\cos(3\sqrt{t})}}{t}
$$
which is the negative of the derivative at $0$ of the function
$$
f(t)=\cos\sqrt{t}\sqrt{\cos(2\sqrt{t})}\sqrt[3]{\cos(3\sqrt{t})}
$$
and we can do the logarithmic derivative, so
$$
\log f(t)=g_1(t)+g_2(t)+g_3(t)
$$
and so
$$
\frac{f'(t)}{f(t)}=g_1'(t)+g_2'(t)+g_3'(t)
$$
Since $f(0)=1$, our limit is
$$
-f'(0)=-(g_1'(t)+g_2'(t)+g_3'(t))=\frac{1}{2}+\frac{2}{2}+\frac{3}{2}=3
$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove or disprove: $\sum_{k=1}^{\infty}\frac{\sqrt{k+1}-\sqrt{k}}{k\sqrt{k}}$ is convergent
Prove or disprove: The following series is convergent
$$\sum_{k=1}^{\infty}\frac{\sqrt{k+1}-\sqrt{k}}{k\sqrt{k}}$$
$$\frac{\sqrt{k+1}-\sqrt{k}}{k\sqrt{k}}= \frac{\left(\sqrt{k+1}-\sqrt{k}\right)\cdot \left(\sqrt{k+1} + \sqrt{k}\right)}{k\sqrt{k} \cdot \left(\sqrt{k+1} + \sqrt{k}\right)}= \frac{k+1-k}{k\sqrt{k} \cdot \left(\sqrt{k+1}+\sqrt{k}\right)}$$
$$=\frac{1}{k\sqrt{k} \cdot \left(\sqrt{k+1}+\sqrt{k}\right)}=\frac{1}{\left(k\sqrt{k}\right)\cdot \left(\sqrt{k+1}\right)+k\sqrt{k}\cdot\sqrt{k}}= \frac{1}{k\sqrt{k}\cdot \left(\sqrt{k+1}\right)+k^{2}}< \frac{1}{k^{2}}$$
$$\Rightarrow\sum_{k=1}^{\infty}\frac{1}{k^{2}}$$
This is a convergent series and thus the original series is convergent as well.
*
*Did I do everything correcty (I'm especially not sure about the last step where I used "<")?
*Is there another way of proofing convergence here without that much work? I have tried ratio test too but it got so complicated and I couldn't solve it
| $$\sum_{k=1}^{\infty}\frac{\sqrt{k+1}-\sqrt{k}}{k\sqrt{k}}< \sum_{k=1}^{\infty}\frac{1}{k\sqrt{k}}$$
$$\sum_{k=1}^{\infty}\frac{1}{k\sqrt{k}}=\sum_{k=1}^{\infty}\frac{1}{k^{\frac{3}{2}}}=\zeta (\frac{3}{2})$$
| {
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} |
Prove that $\frac1n\sum\limits^n_{i=1}(X_i-\overline{X})^2=\overline{X^2}-\overline{X}^{\ 2}$
Prove that $\frac{1}{n}\sum\limits^n_{i=1}(X_i-\overline{X})^2=\overline{X^2}-\overline{X}^{\ 2}$ where $\overline{X}=\frac{1}{n}\sum\limits^n_{i=1}X_i$ and $\overline{X^2}=\frac{1}{n}\sum\limits^n_{i=1}X_i^2$
From the left, I can see that $$\frac{1}{n}\sum^n_{i=1}(X_i-\overline{X})^2=\frac{(X_1-\overline{X})^2+\cdots+(X_n-\overline{X})^2}{n} = \frac{(X_1-\frac{X_1+\cdots+X_n}{n})^2+\cdots+(X_n-\frac{X_1+\cdots+X_n}{n})^2}{n} = \cdots$$
I don't see how we can get to the right. Any suggestions?
| Observe
\begin{align}
\frac{1}{n}\sum^n_{i=1}(X_i-\bar X)^2 =&\ \frac{1}{n}\sum^n_{i=1} (X_i^2-2\bar XX_i +\bar X^2)\\
=&\ \frac{1}{n}\sum^n_{i=1}X_i^2 -2\bar X \frac{1}{n}\left(\sum^n_{i=1}X_i\right)+\bar X^2 \\
=&\ \overline{X^2}-2\bar X^2+\bar X^2 = \overline{X^2}-\bar X^2.
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1935086",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Least value of a determinant If $a^2-b^2=2$ then what is the least possible value of: \begin{vmatrix} 1+a^2-b^2 & 2ab &-2b\\ 2ab & 1-a^2+b^2&2a\\2b&-2a&1-a^2-b^2 \end{vmatrix}
I tried to express the determinant as a product of two determinants but could not do so. Seeing no way out, I tried expanding it but that took too long and was difficult to evaluate. Please help me with this one, thanks.
| \begin{align*}
A = \begin{vmatrix}
1+a^2-b^2 & 2ab & -2b \\
2ab & 1-a^2+b^2 & 2a \\
2b & -2a & 1-a^2-b^2
\end{vmatrix} &= (1+a^2+b^2)^3
\end{align*}
To get $1+a^2+b^2$ in the first row, we can apply $R_1 \rightarrow R_1+bR_3$. This will give
\begin{align*}
A = \begin{vmatrix}
1+a^2+b^2 & 0 & -b(1+a^2+b^2) \\
2ab & 1-a^2+b^2 & 2a \\
2b & -2a & 1-a^2-b^2
\end{vmatrix}
\end{align*}
Taking $1+a^2+b^2$ out from the first row, we get
\begin{align*}
A = (1+a^2+b^2)\begin{vmatrix}
1 & 0 & -b \\
2ab & 1-a^2+b^2 & 2a \\
2b & -2a & 1-a^2-b^2
\end{vmatrix}
\end{align*}
We need one more zero in the first row. This can be obtained by taking $C_3 \rightarrow C_3+bC_1$. This gives
\begin{align*}
A = (1+a^2+b^2)\begin{vmatrix}
1 & 0 & 0 \\
2ab & 1-a^2+b^2 & 2a(1+b^2) \\
2b & -2a & 1-a^2+b^2
\end{vmatrix}
\end{align*}
Now, expanding by the first row, we obtain
\begin{align*}
A &= (1+a^2+b^2)[(1-a^2+b^2)^2 + 4a^2(1+b^2)] \\
&= (1+a^2+b^2)[(1+b^2-a^2)^2 + 4a^2(1+b^2)] \\
&= (1+a^2+b^2)[(1+b^2)^2 + a^4 -2a^2(1+b^2)+ 4a^2(1+b^2)] \\
&\qquad \qquad \qquad \qquad \text{expanding } ((1+b^2) - a^2)^2 \\
&= (1+a^2+b^2[(1+b^2)^2 + a^4 +2a^2(1+b^2)] \\
&= (1+a^2+b^2)(1+a^2+b^2)^2 \\
&= (1+a^2+b^2)^3
\end{align*}
Since $a^2-b^2 = 2$, the determinant can be written as
$$ (1+ b^2 + 2 + b^2)^3 = (3+2b^2)^3 $$
This is least when $b^2$ is least. The least value of $b^2$ is 0 and hence the least value is 27.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1935320",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Prove that $\sin^4\theta+\cos^4\theta=\frac{3+\cos4\theta}{4}$ I have a question that goes exactly like this:
By considering $(\sin^2\theta+\cos^2\theta)^2$ and $(\sin^2\theta+\cos^2\theta)^3$ prove that
$$
\text{a) }\sin^4\theta+\cos^4\theta=\frac{3+\cos4\theta}{4},\qquad\text{b) }\sin^6\theta+\cos^6\theta=\frac{5+3\cos4\theta}{8}
$$
I have not idea how to do this. Please help.
| Write $c=\cos\theta,\,s=\sin\theta$ so $c^2+s^2=1$ and $$c^2s^2=\frac{1}{4}\sin^22\theta=\frac{1}{4}\left( 1-\cos^22\theta\right)=\frac{1}{8}\left( 1-\cos 4\theta\right).$$Hence $c^4+s^4=1-2c^2s^2=\frac{1}{4}\left( 3+\cos 4\theta\right)$.
Multiplying by $c^2+s^2$ gives $c^6+s^6+c^2s^2\left( c^2+s^2\right)$. Hence $$c^6+s^6=\frac{1}{4}\left( 3+\cos 4\theta\right)-c^2s^2=\frac{1}{8}\left( 5+3\cos 4\theta\right).$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1936121",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 8,
"answer_id": 3
} |
Combinatorial proof of q-binomial recurrence relation I have no idea how to proof combinatorially the following recurrence relation:
$${n+1\choose k}_q = q^k {n \choose k}_q + {n \choose k-1}_q$$
Could anyone guide me through a combinatorial proof or give me hints on how to prove it?
| Here is a combinatorial approach to show the $q$-binomial identity
\begin{align*}
\binom{n}{k}_q=q^k\binom{n-1}{k}_q+\binom{n-1}{k-1}_q\tag{1}
\end{align*}
by counting $k$-dimensional subspaces of the vector space $\mathbb{F}_q^n$ over the finite field $\mathbb{F}_q$.
This is done in two steps. At first we show the number of subspaces is $\binom{n}{k}_q$. Then we use the figure showing a lattice path from the other answer as inspiration and prove (1).
Let $q$ be a prime power and consider the $\mathbb{F}_q$-vector space
$$\mathbb{F}_q^n=\{(\alpha_1,\ldots,\alpha_n):\alpha_i\in\mathbb{F}_q\}$$
Let $G(n,k)$ equal the number of $k$-dimensional subspaces of $\mathbb{F}_q^n$.
$$ $$
Step 1: $G(n,k)=\binom{n}{k}_q$
For the proof of this, we WLOG assume $0 \leq k \leq n$, as the claim is otherwise trivial.
Let $N=N(n,k)$ be the number of ordered $k$-tuples $\{v_1,\ldots,v_k\}$ of linearly independent vectors of $\mathbb{F}_q^n$.
We may choose $v_1$ in $q^n-1$ ways, then $v_2$ in $q^n-q$ ways, and so on, yielding
\begin{align*}
N=(q^n-1)(q^n-q)\cdots(q^n-q^{k-1})\tag{2}
\end{align*}
We may also count $N=N(n,k)$ in a different way. We choose $(v_1,\ldots,v_k)$ by first selecting a $k$-dimensional subspace $W$ of $\mathbb{F}_q^n$ in $G(n,k)$ ways, and then choosing $v_1\in W$ in $q^k-1$ ways, $v_2\in W$ in $q^k-q$ ways, and so on. Hence
\begin{align*}
N=G(n,k)(q^k-1)(q^k-q)\cdots(q^k-q^{k-1})\tag{3}
\end{align*}
We obtain from (2) and (3)
\begin{align*}
G(n,k)&=\frac{(q^n-1)(q^n-q)\cdots (q^n-q^{k-1})}{(q^k-1)(q^k-q)\cdots(q^k-q^{k-1})}\\
&=\frac{(n!)_q}{(k!)_q((n-k)!)_q}\\
&=\binom{n}{k}_q
\end{align*}
and the claim follows.
Note: This is the proof of Proposition 1.3.18 in the first edition of Enumerative Combinatorics by R.P. Stanley.
$$ $$
Step 2: The $q$-binomial identity
Let's take a look at the $q$-binomial identity
\begin{align*}
\binom{n}{k}_q=q^k\binom{n-1}{k}_q+\binom{n-1}{k-1}_q
\end{align*}
We know $\binom{n}{k}_q$ is the number of $k$-dimensional subspaces of the $n$-dimensional vector space $\mathbb{F}_q^n$. If we consider the figure showing the lattice path, we partitioned the paths according to the first step $(1,0)$ resp. $(0,1)$. We will do here something similar.
We fix a dimension and partition the number of $k$-dimensional subspaces according to it. Let's say when considering an $n$-tuple $(\alpha_1,\ldots,\alpha_n)\in\mathbb{F}_q^n$, we fix the first dimension indexed with $1$.
When we select a $k$-dimensional subspace then there are two possibilities with respect to this fixed dimension.
*
*The dimension is part of the $k$-dimensional subspace
*The dimension is not part of the $k$-dimensional subspace and therefore part of the $n-k$-dimensional complement space.
$$ $$
Case 1: The dimension is part of the $k$-dimensional subspace
*
*This corresponds to step $(1,0)$. We observe, that one dimension of a $k$-dimensional subspace is already consumed, leaving $k-1$ dimensions which are to select out of $n-1$ dimensions. So, the number of $k-1$ dimensional subspaces which is to select is
\begin{align*}
\binom{n-1}{k-1}_q
\end{align*}
Case 2: The dimension is not part of the $k$-dimensional subspace.
*
*This corresponds to step $(0,1)$. The fixed dimension is now part of the $n-k$-dimensional complement space.
We still have to select $k$ dimensions for the $k$-dimensional subspace, but now out of $n-1$ dimensions.
This implies that the entry with index $1$ has no influence to the $k$-dimensional subspace and we are free to put any of $q$ values for it. Since we select a $k$-dimensional subspace where the first index in each of the $k$ base vectors is free to select, we have $q^k$ possibilities to do so, giving a total of
\begin{align*}
q^k\binom{n-1}{k}_q
\end{align*}
and the claim (1) follows.
$$ $$
Example: $\mathbb{F}_2^3$
To better see what's going on in case 1 and case 2 we look at a small example. Here we set
$$q=2,n=3,k=2$$
and consider
\begin{align*}
\binom{3}{2}_2=\color{blue}{2^2}\cdot\binom{2}{2}_2+\binom{2}{1}_2=\color{blue}{4}\cdot 1+3=7
\end{align*}
We have
\begin{align*}
\mathbb{F}_2^3=\left\{
\begin{pmatrix}0\\0\\0\end{pmatrix},
\begin{pmatrix}1\\0\\0\end{pmatrix},
\begin{pmatrix}0\\1\\0\end{pmatrix},
\begin{pmatrix}1\\1\\0\end{pmatrix},
\begin{pmatrix}0\\0\\1\end{pmatrix},
\begin{pmatrix}1\\0\\1\end{pmatrix},
\begin{pmatrix}0\\1\\1\end{pmatrix},
\begin{pmatrix}1\\1\\1\end{pmatrix}
\right\}
\end{align*}
We now list the $3$ subspaces which belong to case $1$ and the $4$ subspaces which belong to case $2$. To simplify the notation and save space, we do not write the zero vector and omit commas. We obtain
Case 1:
The entries with index $1$ are part of the two-dimensional subspace.
\begin{align*}
W_1=\left\{
\begin{pmatrix}1\\0\\0\end{pmatrix}
\begin{pmatrix}0\\1\\0\end{pmatrix}
\begin{pmatrix}1\\1\\0\end{pmatrix}
\right\}
&\qquad W_2=\left\{
\begin{pmatrix}1\\0\\0\end{pmatrix}
\begin{pmatrix}0\\0\\1\end{pmatrix}
\begin{pmatrix}1\\0\\1\end{pmatrix}
\right\}\\
W_3=\left\{
\begin{pmatrix}1\\0\\0\end{pmatrix}
\begin{pmatrix}0\\1\\1\end{pmatrix}
\begin{pmatrix}1\\1\\1\end{pmatrix}
\right\}
\end{align*}
$$ $$
Case 2:
The entries with index $1$ are not part of the two-dimensional subspace. To better identify the four subspaces see the blue colored components $\color{blue}{00},\color{blue}{01},\color{blue}{10}$ and $\color{blue}{11}$.
\begin{align*}
W_4=\left\{
\begin{pmatrix}\color{blue}{0}\\1\\0\end{pmatrix}
\begin{pmatrix}\color{blue}{0}\\0\\1\end{pmatrix}
\begin{pmatrix}0\\1\\1\end{pmatrix}
\right\}
&\qquad W_5=\left\{
\begin{pmatrix}\color{blue}{0}\\1\\0\end{pmatrix}
\begin{pmatrix}\color{blue}{1}\\0\\1\end{pmatrix}
\begin{pmatrix}1\\1\\1\end{pmatrix}
\right\}\\
W_6=\left\{
\begin{pmatrix}\color{blue}{1}\\1\\0\end{pmatrix}
\begin{pmatrix}\color{blue}{0}\\0\\1\end{pmatrix}
\begin{pmatrix}1\\1\\1\end{pmatrix}
\right\}
&\qquad W_7=\left\{
\begin{pmatrix}\color{blue}{1}\\1\\0\end{pmatrix}
\begin{pmatrix}\color{blue}{1}\\0\\1\end{pmatrix}
\begin{pmatrix}0\\1\\1\end{pmatrix}
\right\}
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1940441",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
How can I evaluate: $\lim_{x \to \infty} \sqrt{x^2 + 4x} - \sqrt{x^2 - 5x}$ I need to find this limit:
$$\lim_{x \to \infty} \sqrt{x^2 + 4x} - \sqrt{x^2 - 5x}$$
The answer I got from using the limit laws is $\sqrt{\infty} - \sqrt{\infty}$.
How do I proceed now?
Added I took the conjugate of the function and I got a new and probably better function to work with: $$\lim_{x\to \infty}\frac{9x}{\sqrt{x^2 + 4x} + \sqrt{x^2-5x}}$$
| $$\lim_{x \to \infty} \sqrt{(x^2)+4x} - \sqrt{(x^2)-5x}$$
$$\lim_{x \to \infty} {(x^2)+4x - (x^2)+5x\over\sqrt{(x^2)+4x} +\sqrt{(x^2)-5x}}$$
$$\lim_{x \to \infty} {9x\over\sqrt{(x^2)+4x} +\sqrt{(x^2)-5x}}$$
$$\lim_{x \to \infty} {9\over\sqrt{(x^2)/x^2+4x/x^2} +\sqrt{(x^2)/x^2-5x/x^2}}$$
$$\lim_{x \to \infty} {9\over\sqrt{1+4/x} +\sqrt{1-5/x}}$$
$${9\over\sqrt{1} +\sqrt{1}}={9\over 2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1941003",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 1
} |
Estimate $ \int_\sqrt{n}^{\sqrt{n+1}} \sin (2\pi x^2) \; dx$ using the Trapezoid rule As a follow-up to my previous question. What is the integral of $\sin (2\pi x^2)$ as $x \in [\sqrt{n}, \sqrt{n+1}]$ we get this:
$$ \int_\sqrt{n}^{\sqrt{n+1}} \sin (2\pi x^2) \; dx \tag{$\ast$}$$
Since $\sin 2\pi n = 0$ for all $n \in \mathbb{Z}$ this integral should be close to zero. I could approximate by adding over the four corners:
$$
\int_\sqrt{n}^{\sqrt{n+1}} \sin (2\pi x^2) \; dx \approx \frac{1}{4} \left[ \sin (2\pi n) + \sin 2\pi (\sqrt{n}+ \frac{1}{4})^2
+ \sin 2\pi (\sqrt{n}+ \frac{1}{2})^2 + \sin 2\pi (\sqrt{n}+ \frac{3}{4})^2\right]
$$
This might be hard to estimate since we get unpredictible terms like. And I may have misaplied the trapezoid rule
$$ \sin 2\pi (n + \frac{1}{2} \sqrt{n} + \frac{1}{4})$$
if we place the mesh and $\sqrt{n} < \sqrt{n + \frac{1}{4}} < \sqrt{n + \frac{1}{2}}< \sqrt{n + \frac{3}{4}}< \sqrt{n + 1}$ the trapeoid rule is:
$$\tiny \sin(2\pi n ) \frac{ \sqrt{n+\tfrac{1}{4}} - \sqrt{n}}{2}
+ \sin (2\pi \sqrt{n + \frac{1}{4}})\left[ \sqrt{n+ \frac{1}{2} }- \sqrt{n}\right]
+ \sin (2\pi \sqrt{n + \frac{1}{2}})\left[ \sqrt{n+ \frac{3}{4} }- \sqrt{n+ \frac{1}{4}} \right]
+ \sin (2\pi \sqrt{n + \frac{3}{4}})\left[ \sqrt{n+ \frac{1}{2} }- \sqrt{n+ 1}\right]
+ \sin (2\pi \sqrt{n + 1})\left[ \frac{\sqrt{n+ \frac{3}{4} }- \sqrt{n+ 1}}{2} \right] $$
I think there is a mistake what should the trapezoid rule be? Maybe if I said:
$$ \sqrt{n+ \frac{1}{4}} - \sqrt{n} = \sqrt{n} \left( \sqrt{1 + \frac{1}{4} } - 1 \right)= \frac{\sqrt{n}}{8} $$
Then after fixing all my mistakes I get something of order $\frac{1}{\sqrt{n}}$. How does this look?
\begin{eqnarray}\small \int \dots dx &\approx& \frac{1}{8} \left[ \sqrt{n }\cdot 0
+ 2\sqrt{n + \tfrac{1}{4}}\cdot 1
+ 2\sqrt{n + \tfrac{1}{2}}\cdot 0
+ 2\sqrt{n + \tfrac{3}{4}}\cdot (-1)
+ \sqrt{n+1}\cdot 0\right] \\ \\
&=& \frac{1}{4} \left[ \sqrt{n+ \frac{1}{2}} - \sqrt{n + \frac{3}{4}}\right] = O(\frac{1}{\sqrt{n}})\end{eqnarray}
| Setting $t=x^2$, we obtain the integral to be
$$I = \int_n^{n+1} \sin(2\pi t) \dfrac{dt}{2\sqrt{t}}$$
We then have
$$\vert 2I \vert = \left \vert \int_n^{n+1} \sin(2\pi t) \dfrac{dt}{\sqrt{t}}\right \vert \leq \int_n^{n+1} \dfrac{dt}{\sqrt{t}} = 2\left.\sqrt{t} \right \vert_{n}^{n+1} = 2\left(\sqrt{n+1}-\sqrt{n}\right)$$
Hence, we obtain that
$$\vert I \vert \leq \dfrac1{\sqrt{n}+\sqrt{n+1}}$$
A better estimate can be determined as follows. We have
$$I = \int_0^1 \dfrac{\sin\left(2\pi(n+t)\right)}{2\sqrt{n+t}}dt = \int_0^1 \dfrac{\sin(2\pi t)}{2\sqrt{n+t}}dt$$
Hence, we obtain
$$\sqrt{n}I = \int_0^1 \dfrac{\sin(2\pi t)}{2\sqrt{1+t/n}}dt = \dfrac12\sum_{k=0}^{\infty} \dbinom{2k}k \left(\dfrac{-1}{4n}\right)^k \int_0^1 t^k \sin(2\pi t) dt = \dfrac1{8n\pi} - \dfrac3{32n^2 \pi} + \cdots$$
Hence, infact $I \sim \dfrac1{8\pi n^{3/2}} \left(1-\dfrac3{4n}\right)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1943837",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Is 2 always a primitive root of 3ˣ? That is, is it always that $$2^{3^x}\equiv -1\pmod{3^{x+1}}\large?$$
| Another way to solve this question is by induction.
►The statement holds for $n=1$ because $2^3=8=9-1\equiv -1\pmod {3^2}$.
►Suppose it is true for $n$, that is $2^{3^n}\equiv-1\pmod{3^{n+1}}$.
►Proof it is true for $n+1$.
$$2^{3^n}\equiv-1\pmod{3^{n+1}}\iff2^{3^n}=3^{n+1}M_n-1$$ It follows
$$2^{3^{n+1}}=(2^{3^n})^3=(3^{n+1}M_n-1)^3=3^{3n+3}M_n^3-3\cdot3^{2n+2}M_n^2+3\cdot3^{n+1}M_n-1$$ Hence
$$2^{3^{n+1}}=3^{n+2}(3^{2n+1}M_n^3-3^{n+1}M_n^2+M_n)-1\Rightarrow\color{red}{2^{3^{n+1}}\equiv-1\pmod{3^{n+2}}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1944806",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
How to evaluate the integral $\int_0^{\infty} e^{-\frac{1}{2}\left(x^2+ \frac{c}{x^2}\right)}dx$? I am wondering how one would calculate the integral:
$$
\int_0^{\infty} e^{-\frac{1}{2}\left(x^2+ \frac{c}{x^2}\right)}dx
$$
where $c$ is a constant. I have tried to reparametrize by letting $u = x^2$ to get:
$$
\int_0^{\infty} \frac{1}{2\sqrt{u}} e^{-\frac{1}{2}\left(u+\frac{c}{u}\right)}du
$$
and then trying to use integration by parts. However, I am getting nowhere with that approach.
| We can also use the Cauchy-Schlomilch transformation. For $a,b \gt 0$
\begin{equation}
\int\limits_{0}^{\infty} f\Big[\left(ax - \frac{b}{x} \right)^{2} \Big] \mathrm{d} x = \frac{1}{a} \int\limits_{0}^{\infty} f(y^{2}) \mathrm{d}y
\tag{1}
\label{eq:1}
\end{equation}
Expanding the term inside of the integrand, we have
\begin{equation}
\left(ax - \frac{b}{x} \right)^{2} = a^{2} x^{2} - 2ab + \frac{b^{2}}{x^{2}}
\end{equation}
\begin{equation}
-a^{2} x^{2} - \frac{b^{2}}{x^{2}} = -2ab - \left(ax - \frac{b}{x} \right)^{2}
\end{equation}
Matching variables with our problem, we have $a^{2} = 1/2$ and $b^{2} = c/2$ and the term in the
exponential of our problem becomes
\begin{equation}
-\frac{1}{2}x^{2} - \frac{c}{2}\frac{1}{x^{2}} = -\sqrt{c} - \left(\frac{x}{\sqrt{2}} - \frac{1}{x}\sqrt{\frac{c}{2}} \right)^{2}
\end{equation}
Substituting this into our integral, yields
\begin{align}
\int\limits_{0}^{\infty} \mathrm{exp}\left( -\frac{1}{2}x^{2} - \frac{c}{2}\frac{1}{x^{2}} \right) \mathrm{d} x
&= \mathrm{e}^{-\sqrt{c}} \int\limits_{0}^{\infty}
\mathrm{exp}\Big[-\left( \frac{x}{\sqrt{2}} - \frac{1}{x}\sqrt{\frac{c}{2}} \right)^{2} \Big] \mathrm{d} x \\
\tag{a}
&= \sqrt{2}\,\mathrm{e}^{-\sqrt{c}} \int\limits_{0}^{\infty} \mathrm{e}^{-y^{2}} \mathrm{d} y \\
\tag{b}
&= \sqrt{2}\,\mathrm{e}^{-\sqrt{c}} \,\frac{\sqrt{\pi}}{2} \mathrm{erf}(y) \Big|_{0}^{\infty} \\
&= \sqrt{\frac{\pi}{2}}\mathrm{e}^{-\sqrt{c}}
\end{align}
Notes:
a. Use the Cauchy-Schlomilch transformation, equation $\eqref{eq:1}$.
b. $\mathrm{erf}(z)$ is the error function.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1947697",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 4
} |
Limit involving exponentials and arctangent without L'Hôpital $$\lim_{x\to0}\frac{\arctan x}{e^{2x}-1}$$
How to do this without L'Hôpital and such? $\arctan x=y$, then we rewrite it as $\lim_{y\to0}\frac y{e^{2\tan y}-1}$, but from here I'm stuck.
| Another way using Taylor series $$\tan^{-1}(x)=x-\frac{x^3}{3}+O\left(x^4\right)$$ $$e^{2x}-1=2 x+2 x^2+\frac{4 x^3}{3}+O\left(x^4\right)$$ $$\frac{\tan^{-1}(x) } {e^{2x}-1 }=\frac{x-\frac{x^3}{3}+O\left(x^4\right) } {2 x+2 x^2+\frac{4 x^3}{3}+O\left(x^4\right) }=\frac{1-\frac{x^2}{3}+O\left(x^3\right) } {2 +2 x+\frac{4 x^2}{3}+O\left(x^3\right) }$$ Performing the long division $$\frac{\tan^{-1}(x) } {e^{2x}-1 }=\frac{1}{2}-\frac{x}{2}+O\left(x^3\right)$$ which shows the limit and how it is approached.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1949286",
"timestamp": "2023-03-29T00:00:00",
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} |
$\lim_{x\to 0}\left[1^{\frac{1}{\sin^2 x}}+2^{\frac{1}{\sin^2 x}}+3^{\frac{1}{\sin^2 x}}+\cdots + n^{\frac{1}{\sin^2 x}}\right]^{\sin^2x}$
$$\lim_{x\to 0}\left[1^{\frac{1}{\sin^2x}}+2^{\frac{1}{\sin^2x}}+3^{\frac{1}{\sin^2x}}+\cdots + n^{\frac{1}{\sin^2x}}\right]^{\sin^2x}$$
Limit is of form $(\infty)^{0} $
$$\lim_{x\to 0}e^{\sin^2x\log{ {\left[1^{\frac{1}{\sin^2x}}+2^{\frac{1}{\sin^2x}}+3^{\frac{1}{\sin^2x}}+\cdots + n^{\frac{1}{\sin^2x}}\right]}}}$$
I don't know how to proceed further.
| $$
\begin{align}
&\lim_{x\to0}\left[1^{\frac1{\sin^2(x)}}+2^{\frac1{\sin^2(x)}}+3^{\frac1{\sin^2 (x)}}+\cdots+n^{\frac1{\sin^2(x)}}\right]^{\sin^2(x)}\\
&=\lim_{x\to\infty}\left[1^x+2^x+3^x+\cdots+n^x\right]^{1/x}\\
&=n\lim_{x\to\infty}\left[\left(\frac1n\right)^x+\left(\frac2n\right)^x+\left(\frac3n\right)^x+\cdots+\left(\frac{n-1}n\right)^x+1^x\right]^{1/x}\\[4pt]
&=n\,[0+0+0+\cdots+0+1]^0\\[8pt]
&=n
\end{align}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1949375",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 2
} |
What will be the value of the following determinant without expanding it? $$\begin{vmatrix}a^2 & (a+1)^2 & (a+2)^2 & (a+3)^2 \\ b^2 & (b+1)^2 & (b+2)^2 & (b+3)^2 \\ c^2 & (c+1)^2 & (c+2)^2 & (c+3)^2 \\ d^2 & (d+1)^2 & (d+2)^2 & (d+3)^2\end{vmatrix} $$
I tried many column operations, mainly subtractions without any success.
| The determinant is a polynom of the second degree with respect to $a$, on the other hand, it has three different (in general) roots $b$, $c$, and $d$ (as, e.g., if $a=b$, the determinant has two identical rows), therefore, the determinant vanishes.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "27",
"answer_count": 7,
"answer_id": 0
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What is the probability that a 5-card poker hand contains 3 of a kind? I tried searching on the net and it shows an answer of
$\frac{ \binom{13}{1} \times \binom{4}{3} \times \binom{12}{2} \times \binom{4}{1} \times \binom{4}{1} }{ \binom{52}{5} }$
Why is it not the same as:
$\frac{ \binom{13}{1} \times \binom{4}{3} \times \binom{48}{1} \times \binom{44}{1}}{ \binom{52}{5} }$
This answer seems to be doubled of the correct answer based on the net.
I also tried different approach and it resulted as this one which is also doubled of the correct answer based on the net.
$\frac{\binom{13}{1} \times \binom{4}{3} \times \binom{12}{1} \times \binom{11}{1} \times \binom{4}{1} \times \binom{4}{1} }{\binom{52}{5}}$
| Let's say our 3-of-a-kind is made of 5s. When you say " $\binom{48}{1} \times \binom{44}{1}$ " , what you mean is "pick a card that isn't a $5$" (say, a $6$), and then "pick a card that isn't a $5$ or $6$".
The problem is that this counts every pair twice: if your final cards are $(4,7)$ then you count $(7,4)$ a separate time.
To get around this, select two cards $\binom{12}{2}$ and two suits $\binom{4}{1}^2$.
| {
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"source": "stackexchange",
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Solve this system equation $\lfloor xy\rfloor=12,\lfloor x^2\rfloor+\lfloor y^2\rfloor=25$ Let $x,y\in R$.solve this system equation
$$\begin{cases}
\lfloor xy\rfloor=12\\
\lfloor x^2\rfloor+\lfloor y^2\rfloor=25
\end{cases}$$
I have found integer solution $(x,y)=(3,4)$ How to Find the other
| There is a non-countable solution set $S$ and we expose here a subset $T$ of $S$.
We have
$$\begin{cases}x=\lfloor x\rfloor+\{x\}\\y=\lfloor y\rfloor+\{y\}\end{cases}\Rightarrow\begin{cases} xy=\lfloor x\rfloor\lfloor y\rfloor+\lfloor x\rfloor\{y\}+\lfloor y\rfloor\{x\}+\{x\}\{y\}\\x^2=\lfloor x\rfloor^2+2\lfloor x\rfloor\{x\}+\{x\}^2\\y^2=\lfloor y\rfloor^2+2\lfloor y\rfloor\{y\}+\{y\}^2\end{cases}$$
A clear solution is the Pythagorean $(x,y)=(3,4)$. Taking $x=3+\{x\}$ and $y=4+\{y\}$ one has
$$\begin{cases}xy=12+3\{y\}+4\{x\}+\{x\}\{y\}\\x^2=9+6\{x\}+\{x\}^2\\y^2=16+8\{y\}+\{y\}^2\end{cases}$$ We still have the two given equations when $\{x\}$ and $\{y\}$ satisfy the system
$$\begin{cases}0\lt3\{y\}+4\{x\}+\{x\}\{y\}\lt 1\\0\lt6\{x\}+\{x\}^2+ 8\{y\}+\{y\}^2\lt 1\end{cases}$$ In particular it is satisfied making $\{x\}=\{y\}$ which gives
$7\{x\}+\{x\}^2\lt1$ and $14\{x\}+2\{x\}^2\lt1$ which reduces to $7\{x\}+\{x\}^2\lt \frac 12$ so we can choose $\{x\}\le0.07$. Hence our subset $T$ is formed by all the positive
ordered pair $$(x,y)=(3+\{x\},4+\{y\})\text { with } \{x\}, \{y\}\le 0.07$$
Example.- $x=3.07$ and $y=4.07$ hence $7(0.07)+(0.07)^2=0.4949\lt \frac 12$
consequently we can choose arbitrarily $ \{x\}, \{y\}\le 0.07$.
Note.- There are other subsets without the integer part $3$ and $4$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Equation for Distance of the Straight line from the Origin. By reduction of the equation $ax + by + c = 0 $ of a straight line to the normal form , we get
$$\left(\frac{-a}{\sqrt{a^2 + b^2}}\right)x + \left(\frac{-b}{\sqrt{a^2 + b^2}}\right)y = \frac{c}{\sqrt{a^2 + b^2}}$$
And,
$$p= \frac{∣c∣}{\sqrt{a^2 + b^2}}$$
And my textbook says that $p$ is the distance of the straight line from the origin. I don't know why we are getting it as a distance from origin?
I know $p= x\cos\theta + y\sin\theta$ ( where $p$ is distance of line from origin).
Also I want to know how can we relate both equations?
| For this derivation, we won’t be needing any trigonometric equation nor you have to prior study trigonometry.
For this derivation we just need the following things:
*
*Pythagoras Theorem
*Algebra
*Area of triangle $$\frac{1}{2} \times \text{height} \times \text{base}$$
Let us take the line equation $Ax+By+C = 0$.
So by putting first $y = 0$ solving for $x$ and then putting $x = 0$ and solving for $y$.
*
*By putting $y = 0$ and solve for $x$:
\begin{align}
Ax+By+C &= 0, \quad y = 0 \\
Ax+B(0)+C &= 0 \\
Ax+C &= 0 \\
Ax &= -C \\
x &= -\frac C A
\end{align}
*By putting $x = 0$ and solve for $y$:
\begin{align}
Ax+By+C &= 0, \quad x = 0 \\
A(0)+By+C &= 0 \\
By+C &= 0 \\
By &= -C \\
y &= -\frac C B
\end{align}
By doing the above we get the base distance and perpendicular distance of the triangle formed by the line equation $Ax+By+C = 0$.
Since the triangle formed is a right-angled triangle, we can get its area and hypotenuse with no problem.
*
*So first, deriving the hypotenuse:
$$\text{Rise} = \left|\frac{-C}{A}\right|,\ \text{Run} = \left|\frac{-C}{B}\right|$$
Pythagoras Thm.: Hyp.${}^2$ = Rise${}^2$ + Run${}^2$:
\begin{align}
\text{Hyp.}^2 &= \left|\frac{-C}{A}\right|^2 + \left|\frac{-C}{B}\right|^2 \\
&= \frac{C^2}{A^2} + \frac{C^2}{B^2} \\
&= \frac{C^2(A^2+B^2)}{(AB)^2}; \\
\text{Hyp.} &= \sqrt\frac{C^2(A^2+B^2)}{(AB)^2} \\
&= \frac{|C|\sqrt{A^2+B^2}}{|AB|}.
\end{align}
*Deriving the Area:
\begin{align}
\text{Base} &= \left|\frac{-C}{A}\right|, \\
\text{Height} &= \left|\frac{-C}{B}\right|; \\
\text{Area} &= \frac 1 2 \times \left|\frac{-C}{A}\right| \times \left|\frac{-C}{B}\right| \\
&= \frac{C^2}{2|AB|}.
\end{align}
We can also write the area as half the hypotenuse times the height that is perpendicular to the hypotenuse (Note: the height perpendicular to the hypotenuse is the shortest distance to the origin, so let us call this height $w$).
\begin{align}
\text{Area} &= \frac 1 2 \times \text{hypotenuse} \times \text{perp. height} \\
\frac{C^2}{2|AB|} &= \frac 1 2 \times \frac{|C| \sqrt{A^2+B^2}}{|AB|} \times w \\
\frac{C^2}{|AB|} &= \frac{|C|\sqrt{A^2+B^2}}{|AB|} \times w \\
|C| &= \sqrt{A^2+B^2} \times w \\
\frac{|C|}{\sqrt{A^2+B^2}} &= w.
\end{align}
This is the answer.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1958856",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
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A problem on inequality and number theory Suppose $n \in \mathbb{Z}^+$;
$$\sum_{i=1}^n\sqrt{1-x_ix_{i+1}}\ge\sqrt{n(n-1)}\tag{1}$$
where $, x_1= x_{n+1}, \displaystyle \quad\forall x_i\in\mathbb{R} \text{ and }\sum_{i=1}^nx_i^2=1 $
Is it possible to find all the positive integer numbers $n$ which satisfy $(1)$ ?
| A proof for $n=4$.
Id est, for $a^2+b^2+c^2+d^2=1$ we need to prove that:
$$\sqrt{1-ab}+\sqrt{1-bc}+\sqrt{1-cd}+\sqrt{1-da}\geq2\sqrt3$$ or
$$\sum\limits_{cyc}(1-ab)+2\sum\limits_{cyc}\sqrt{(1-ab)(1-bc)}+2\sqrt{(1-ab)(1-cd)}+2\sqrt{(1-bc)(1-da)}\geq12$$ or
$$\sum\limits_{cyc}\sqrt{(1-ab)(1-bc)}+\sqrt{(1-ab)(1-cd)}+\sqrt{(1-bc)(1-da)}\geq4+\frac{1}{2}\sum\limits_{cyc}ab$$
Now by C-S we obtain
$$\sum\limits_{cyc}\sqrt{(1-ab)(1-bc)}=\sum\limits_{cyc}\sqrt{(a^2-ab+b^2+c^2+d^2)(b^2-bc+c^2+a^2+d^2)}=$$
$$=\sum\limits_{cyc}\sqrt{\left(\left(b-\frac{a}{2}\right)^2+\frac{3}{4}a^2+\frac{3}{4}c^2+\frac{1}{4}c^2+d^2\right)\left(\left(b-\frac{c}{2}\right)^2+\frac{3}{4}a^2+\frac{3}{4}c^2+\frac{1}{4}a^2+d^2\right)}\geq$$
$$\sum\limits_{cyc}\left(\left(b-\frac{a}{2}\right)\left(b-\frac{c}{2}\right)+\frac{3}{4}a^2+\frac{3}{4}c^2+\frac{1}{4}ac+d^2\right)=$$
$$=\frac{7}{2}-ab-bc-cd-da+ac+bd$$
In another hand, by C-S again we obtain
$$\sqrt{(1-ab)(1-cd)}=\sqrt{(a^2-ab+b^2+c^2+d^2)(c^2-cd+d^2+a^2+b^2)}=$$
$$=\sqrt{\left(\frac{(a-b)^2}{2}+\frac{1}{2}+\frac{1}{2}c^2+\frac{1}{2}d^2\right)\left(\frac{(c-d)^2}{2}+\frac{1}{2}+\frac{1}{2}a^2+\frac{1}{2}b^2\right)}\geq$$
$$\geq\frac{(a-b)(d-c)}{2}+\frac{1}{2}+\frac{1}{2}(ad+bc)=\frac{1}{2}+ad+bc-\frac{1}{2}(ac+bd)$$
similarly we'll obtain
$$\sqrt{(1-ad)(1-bc)}\geq\frac{1}{2}+ab+cd-\frac{1}{2}(ac+bd)$$
Thus,
$$\sum\limits_{cyc}\sqrt{(1-ab)(1-bc)}+\sqrt{(1-ab)(1-cd)}+\sqrt{(1-bc)(1-da)}\geq\frac{9}{2}$$
and it remains to prove that
$$\frac{9}{2}\geq4+\frac{1}{2}\sum\limits_{cyc}ab$$ or
$$a^2+b^2+c^2+d^2\geq ab+bc+cd+da$$ or
$$(a-b)^2+(b-c)^2+(c-d)^2+(d-a)^2\geq0$$
Done!
| {
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"url": "https://math.stackexchange.com/questions/1958986",
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"source": "stackexchange",
"question_score": "1",
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Find the parametric representation of a line
Find the parametric representation of a line $l$ so that $l$ is in $\pi:\overline X=(3,0,0)+t(2,3,1)+s(6,-2,3)$ and $l$ intersects $l_1:\overline X=(6,3,0)+t(1,1,2)$ with a 60° angle.
I did find the intersection point at $(7,4,2)$ by equating the coordinates of said point expressed once as a point on $\pi$ and again as a point on $l_1$.
I tried using the dot product, since it has $\cos\theta$ as the angle between the two vectors, but I get stuck since I don't know the length of any of the vectors. I'm not even sure I need their length. I'd love some help please.
| First of all, rewrite $\pi$ in normal form:
$$(2,3,1)×(6,-2,3)=(11,0,-22)=11(1,0,-2)$$
$$(3,0,0)\cdot(1,0,-2)=3$$
$$\pi:\mathbf r\cdot(1,0,-2)=3$$
The equation of $l$ is $(7,4,2)+\lambda\mathbf b$ where $\lambda\in\Bbb R$ and $\mathbf b$ is a vector with the following properties:
*
*It is perpendicular to $\pi$'s normal: $\mathbf b\cdot(1,0,-2)=0$. Writing $\mathbf b=(a,b,c)$ we have
$$a=2c\tag1$$
*It makes a 60° angle with $l_1$:
$$\mathbf b\cdot(1,1,2)=\sqrt{1^2+1^2+2^2}\sqrt{a^2+b^2+c^2}\cos60^\circ$$
$$a+b+2c=\frac{\sqrt6}2\sqrt{a^2+b^2+c^2}\tag2$$
Now assume $\sqrt{a^2+b^2+c^2}=a^2+b^2+c^2=1$, i.e. $\mathbf b$ is a unit vector. (2) then simplifies, and (1) removes $a$ from it:
$$b+4c=\frac{\sqrt6}2$$
$$b=\frac{\sqrt6}2-4c\tag3$$
$b^2$ can be written in terms of $c$:
$$a^2+b^2+c^2=1$$
$$(2c)^2+b^2+c^2=5c^2+b^2=1$$
$$b^2=1-5c^2$$
Squaring (3) we get
$$1-5c^2=16c^2-4\sqrt6c+\frac32$$
$$21c^2-4\sqrt6c+\frac12=0$$
Solving for $c$:
$$c=\frac{4\sqrt6\pm\sqrt{54}}{42}=\frac{\sqrt6}6\text{ or }\frac{\sqrt6}{42}$$
From here two solutions for $\mathbf b$ may be constructed:
$$\mathbf b=\left(\frac{\sqrt6}3,-\frac{\sqrt6}6,\frac{\sqrt6}6\right)\text{ or }\left(\frac{\sqrt6}{21},\frac{17\sqrt6}{42},\frac{\sqrt6}{42}\right)$$
Removing the factor of $\sqrt6$ and common denominators leads us to the two solutions for $l$:
$$l=(7,4,2)+\lambda(2,-1,1)\text{ or }(7,4,2)+\lambda(2,17,1),\ \lambda\in\Bbb R$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1959210",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
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Prove a combinatoric sum: $\sum_{j=0}^{k}{{2k-j}\choose{j}}2^j=\frac{1}{3}(1+2^{2k+1})$ $$\sum_{j=0}^{k}{{2k-j}\choose{j}}2^j=\frac{1}{3}\large(1+2^{2k+1})$$
I'm 99% certain it's correct, and I also ran a first few examples with python (up to $k = 0$), but so far I haven't been able to prove it.
update:
I have tried to use induction, but going from $k$ to $k+1$ didnt work. I also tried multiplying by 3, and then splitting the sum ($rhs$) into two sums $\sum_{j=0}^{k}{{2k-j}\choose{j}}2^j + \sum_{j=0}^{k}{{2k-j}\choose{j}}2^{j+1}$. Then I tranformed the second one to $\sum_{j=1}^{k+1}{{2k-j+1}\choose{j-1}}2^j$. This would be helpfull if I could somehow calculate ${{2k-j}\choose{j}}+{{2k-j+1}\choose{j-1}}$, but I couldnt do that either.
thanks
| Evaluating
$$\sum_{q=0}^n {2n-q\choose q} 2^q$$
we write this as
$$\sum_{q=0}^n {2n-q\choose 2n-2q} 2^q$$
and introduce
$${2n-q\choose 2n-2q} =
\frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{2n-2q+1}} (1+z)^{2n-q} \; dz.$$
Observe that this vanishes for $q\gt n$ so we may extend the sum to
infinity, getting
$$\frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{2n+1}} (1+z)^{2n}
\sum_{q\ge 0} 2^q \frac{z^{2q}}{(1+z)^q}
\; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{2n+1}} (1+z)^{2n}
\frac{1}{1-2z^2/(1+z)}
\; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{2n+1}} (1+z)^{2n+1}
\frac{1}{1+z-2z^2}
\; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{2n+1}} (1+z)^{2n+1}
\frac{1}{(1+2z)(1-z)}
\; dz.$$
The residues at the poles sum to zero and we have three potential
poles other than zero which are at $z=-1/2$, $z=1$ and $z=\infty.$ The
integral equals the negative of the residues at these poles. We get
for $z=1$
$$-\frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{2n+1}} (1+z)^{2n+1}
\frac{1}{(1+2z)(z-1)}
\; dz$$
for a residue of
$$- 2^{2n+1} \frac{1}{3}.$$
We get for $z=-1/2$
$$\frac{1}{2} \frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{2n+1}} (1+z)^{2n+1}
\frac{1}{(1/2+z)(1-z)}
\; dz$$
for a residue of
$$\frac{1}{2} (-2)^{2n+1} \frac{1}{2^{2n+1}}
\frac{1}{3/2} = \frac{1}{3} (-1)^{2n+1} = - \frac{1}{3}.$$
Finally we have
$$\mathrm{Res}_{z=\infty}
\frac{1}{z^{2n+1}} (1+z)^{2n+1}
\frac{1}{(1+2z)(1-z)}
\\ = - \mathrm{Res}_{z=0} \frac{1}{z^2}
z^{2n+1} \frac{(1+z)^{2n+1}}{z^{2n+1}}
\frac{1}{(1+2/z)(1-1/z)}
\\ = - \mathrm{Res}_{z=0} \frac{1}{z^2}
(1+z)^{2n+1}
\frac{1}{(1+2/z)(1-1/z)}
\\ = - \mathrm{Res}_{z=0}
(1+z)^{2n+1}
\frac{1}{(z+2)(z-1)} = 0.$$
Summing the negative of these three contributions yields
$$\frac{1}{3} 2^{2n+1} + \frac{1}{3}
= \frac{1}{3} (1+2^{2n+1}).$$
| {
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"url": "https://math.stackexchange.com/questions/1959426",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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A function with domain that is asking for sum of a+b+c+d Let $f$ be a function such that $$ \sqrt {x - \sqrt { x + f(x) } } = f(x) , $$for $x > 1$. In that domain, $f(x)$ has the form $\frac{a+\sqrt{cx+d}}{b},$ where $a,b,c,d$ are integers and $a,b$ are relatively prime. Find $a+b+c+d.$
| Let $y$ equal $f(x)$. Therefore, we have $$y=\sqrt{x-\sqrt{x+y}}\tag{1}$$
Repeatedly squaring both sides gives us $$y^4-2xy^2-y+x^2-x=0\implies x^2+(-2y^2-1)+y^4-y=0\tag2$$
And using the quadratic formula on $(2)$, we get $$x=\frac {2y^2+1\pm(2y+1)}{2}\tag3$$
Simplifying $(3)$ gives us $$x_1=y^2+y+1\\x_2=y^2-y\tag4$$
So now, we have $2$ cases to consider.
Case 1: $x_1=y^2+y+1$
When $x=y^2+y+1$, we have $$x+y=(y+1)^2\\\therefore \sqrt{x+y}=|y+1|$$
Since the square root of a number cannot be less than $0$, we know that $y\geq 0$. Thus, $y+1\geq 1$ and we see that $\sqrt{x+y}=y+1$. Plugging that back into $(1)$ gives us $$\sqrt{x-\sqrt{x+y}}=\sqrt {y^2}=y$$
And from $(4)$, we see that $$y=\frac {-1\pm\sqrt{4x-3}}{2}$$
Case 2: $x_2=y^2-y$
Since $\sqrt{x+y}=x-y^2$, substituting gives us $$\sqrt{y^2}=-y$$ which has no solution because we take the $x$ values larger than $1$. So therefore, Case 1 is correct and $$a+b+c+d=2$$
Using the factoring method I mentioned in the comments:
From $(1)$, we factor it into $$(y^2-y-x)(y^2+y-x+1)=0$$ which has roots $$x=\frac {1\pm\sqrt{4x+1}}{2}\\x=\frac {-1\pm\sqrt{4x-3}}{2}$$
with the latter of the two being the right answer.
| {
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} |
Integrating $\frac{x}{1+x^2}$ I need to integrate $\frac{x}{1+x^2}$ and I tried it like this:
$$f(x)=\frac{x}{1+x^2}=\frac{x}{(x+1)(x-1)}=\frac{x+1-1}{(x+1)(x-1)}= \frac{x+1}{(x+1)(x-1)}-\frac{1}{(x+1)(x-1)}=\frac{1}{x-1}-\frac{1}{1+x^2}$$
After I simplified it I integrate it!
$$\int{f(x)}=\int{\frac{1}{x-1}}-\int{\frac{1}{1+x^2}}$$
$\int{\frac{dx}{1+x^2}}$ becomes arctan(x) and I substitute $\int{\frac{dx}{x-1}}$ to $\int{\frac{du}{u}}$ now it becomes $ln(x-1)$ after this been done I get:
$$\int{f(x)}=ln(x-1)-arctan(x)$$
but when I type integrate x/(1+x^2) into wolframAlpha I it tells me the answer is: $$\int{f(x)}=\frac{1}{2}log(x^2+1)$$
where did I go wrong or what should I have done instead ?
Thank you for your help in advance,
Raavgo
| I would integrate 'by inspection', using the standard result stated above:
$$ \int \frac{f'(x) } {f(x) } \quad dx= ln(f(x)) +c$$
Therefore, answering your question;
$$\int \frac{x} {1+x^2} \quad dx, $$
If $f(x) =x^2 \, then, \, \, f'(x) =2x$
Another equivalent form of the integral becomes;
$${1\over2} \cdot \int \frac{2x} {1+x^2} \quad dx $$
The integral is in the form of $\frac{f' (x) } {f(x) }, $
So the integral may be evaluated as;
$${1 \over 2} \cdot ln(1+x^2)+c$$
Hope this helps.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
Consecutive Matrices Clarification $$
\begin{vmatrix}
1 & 2 & 3 & \cdots & n \\
n+1 & n+2 & n+3 & \cdots & 2n \\
2n+1 & 2n+2 & 2n+3 & \cdots & 3n \\
\vdots & \vdots & \vdots & \vdots & \vdots \\
n^2-n+1 & n^2-n+2 & n^2-n+3 & \cdots & n^2
\end{vmatrix}
$$
My instructor gave me this matrix in this form to find the determinant of it. My question is not to how find the determinant, but what or how to figure out the the fourth row down? Or what is the pattern of the fourth row?
| Is this clearer?
$$
\begin{vmatrix}
1 & 2 & 3 & \cdots & n-2 & n-1 & n \\
n+1 & n+2 & n+3 & \cdots & 2n-2 & 2n- 1 &2n \\
2n+1 & 2n+2 & 2n+3 & \cdots& 3n-2 & 3n- 1 & 3n \\
\vdots & \vdots & \vdots & \vdots & \vdots& \vdots& \vdots \\
n^2-n+1 & n^2-n+2& n^2-n+3 & \cdots & n^2-2 & n^2-1 & n^2
\end{vmatrix}
$$
The pattern for each entry is $a_{ij}= (i-1)n+j$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1962593",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Inequality with the constraint $xy=x+y$ I have to prove that if $x$ and $y$ are two positive numbers such that $xy=x+y$, the following inequality
$$\frac{x}{y^2+4}+\frac{y}{x^2+4}\geq \frac{1}{2}$$
holds. Any help is appreciated, thank you in advance.
| By renaming $x$ as $\frac{1}{u}$ and $y$ as $\frac{1}{v}$ we have to prove that $u,v>0$ and $u+v=1$ grant
$$ \frac{u^2}{v+4u^2 v}+\frac{v^2}{u+4uv^2}\geq \frac{1}{2} $$
that is the same as showing that for any $x\in(0,1)$ the inequality
$$ \frac{x^2}{(1-x)(1+4x^2)}+\frac{(1-x)^2}{x(1+4(1-x)^2)} \geq \frac{1}{2} $$
holds. That is almost trivial since $f(x)=\frac{x^2}{(1-x)(1+4x^2)}$ is a convex function on $(0,1)$, hence
$$ f(x)+f(1-x) \geq 2\cdot f\left(\frac{1}{2}\right) = \frac{1}{2} $$
as wanted.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1965351",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Need help in indefinite integral elimination I'm trying to get a easy and short answer for that indefinite integral:
$$\int\frac{\sqrt{x^2+x+1}}{(x+1)^2}\,dx$$
But everything I find is very big and difficult to understand (too much so it seems like a wrong result). Is there any way to solve it in a more beautiful way?
| If we let $$p=2\frac{\sqrt{x^2+x+1}}{x+1},q=\frac{x-1}{x+1}
$$So $p^2=3+q^2$ with $p\text{d}p=q\text{d}q$.And
\begin{aligned}\int{\frac{x\mathrm{d}x}{\left( 1+x \right) ^2\sqrt{x^2+x+1}}}&=\frac{1}{2}\int{\frac{q+1}{p}\mathrm{d}q}
\\
&=\frac{1}{2}\int{\mathrm{d}p}+\frac{1}{2}\int{\frac{\mathrm{d}q}{p}}
\\
&=\frac{1}{2}p+\frac{1}{2}\int{\frac{\mathrm{d}q+\mathrm{d}p}{p+q}}
\\
&=\frac{1}{2}p+\frac{1}{2}\ln \left( p+q \right)
\end{aligned}
At this time,indefinite integral can be easily done:
\begin{align}&\int{\frac{\sqrt{x^2+x+1}}{\left( 1+x \right) ^2}\mathrm{d}x}=\int{\frac{x^2+x+1}{\left( 1+x \right) ^2\sqrt{x^2+x+1}}\mathrm{d}x}
\\
&=\int{\frac{\mathrm{d}x}{\sqrt{x^2+x+1}}}-\int{\frac{x\mathrm{d}x}{\left( 1+x \right) ^2\sqrt{x^2+x+1}}}
\\
&=\ln \left( x+\frac{1}{2}+\sqrt{x^2+x+1} \right) -\frac{\sqrt{x^2+x+1}}{x+1}-\frac{1}{2}\ln \left( \frac{x-1+2\sqrt{x^2+x+1}}{x+1} \right)
\end{align}
| {
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"timestamp": "2023-03-29T00:00:00",
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Find the range of $a$ such $|f(x)|\le\frac{1}{4},$ $|f(x+2)|\le\frac{1}{4}$ for some $x$.
Let $$f(x)=x^2-ax+1.$$
Find the range of all possible $a$ so that there exist $x$ with
$$|f(x)|\le\dfrac{1}{4},\quad |f(x+2)|\le\dfrac{1}{4}.$$
A sketch of my thoughts: I write
$$f(x)=\left(x-\dfrac{a}{2}\right)^2+1-\dfrac{a^2}{4}\ge 1-\dfrac{a^2}{4}$$
so if $1-\dfrac{a^2}{4}>\dfrac{1}{4}$ or $-\sqrt{3}<a<\sqrt{3}$
this case impossible
But I don't know how to prove the other case, or if this there are better ideas.
| We first see that the case for $a$ is identical to the case for $-a$, so we consider only $a\geq 0$. Let's find the solutions to
$$|f(x)|=\frac{1}{4}.$$
If $f(x)=\frac{1}{4}$ we have
$$x^2-ax+\frac{3}{4}=0$$
$$x = \frac{a\pm\sqrt{a^2-3}}{2},$$
which has real solutions iff $a^2\geq 3$. If $f(x)=-\frac{1}{4}$ we have
$$x^2-ax+\frac{5}{4}=0$$
$$x = \frac{a\pm\sqrt{a^2-5}}{2},$$
which has real solutions iff $a^2\geq 5$. We may now, as $a>0$, write the solution set to $|f(x)|\leq \frac{1}{4}$ as
$$\left[\frac{a-\sqrt{a^2-3}}{2},\frac{a+\sqrt{a^2-3}}{2}\right]$$
if $3\leq a^2<5$, and
$$\left[\frac{a-\sqrt{a^2-3}}{2},\frac{a-\sqrt{a^2-5}}{2}\right]\bigcup\left[\frac{a+\sqrt{a^2-5}}{2},\frac{a+\sqrt{a^2-3}}{2}\right].$$
We have if $a^2<5$ a solution iff
$$\frac{a+\sqrt{a^2-3}}{2}-\frac{a-\sqrt{a^2-3}}{2}\geq 2$$
$$\sqrt{a^2-3}\geq 2$$
$$a^2\geq 7,$$
a contradiction. So, $a^2\geq 5$, and we need either the length of one interval to be $\geq 2$, or the distance between the "innermost" points to be $\leq 2$ and the distance between the "outermost" points to be $\geq 2$. In the first case, we have
$$\sqrt{a^2-3}-\sqrt{a^2-5}\geq 2$$
$$\frac{2}{\sqrt{a^2-3}+\sqrt{a^2-5}}\geq 2$$
$$\sqrt{a^2-3}+\sqrt{a^2-5}\leq 1.$$
However, $\sqrt{a^2-3}\geq \sqrt{5-3}=\sqrt{2}>1$, so this cannot occur. The other case is
$$\frac{a+\sqrt{a^2-5}}{2}-\frac{a-\sqrt{a^2-5}}{2}\leq 2$$
$$\sqrt{a^2-5}\leq 2$$
$$a^2\leq 9$$
$$a\leq 3,$$
as well as the first calculation we did (the distance between the two outermost points is algebraically identical to the distance between the two points in our $a^2<5$ case, which reduced to $a^2\geq 7$. So, our solution set is $\left[\sqrt{7},3\right]$ for $a>0$, and correspondingly $\left[-3,-\sqrt{7}\right]$ for $a<0$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Solve for x: $\sqrt{x+3-4\sqrt{x-1}}$ + $\sqrt{x+8-6\sqrt{x-1}}$ = $1$. How do i solve this equation?
$\sqrt{x+3-4\sqrt{x-1}}$ + $\sqrt{x+8-6\sqrt{x-1}}$ = $1$.
i do not know how to start?
Thanks
| Hint:
$$\sqrt{x+3-4\sqrt{x-1}}=\sqrt{x-1-4\sqrt{x-1}+4}=$$
$$=\sqrt{(\sqrt{x-1}-2)^2}=|\sqrt{x-1}-2|$$
Then:
$$\sqrt{x+3-4\sqrt{x-1}} + \sqrt{x+8-6\sqrt{x-1}} = 1$$
$$\sqrt{x-1-4\sqrt{x-1}+4} + \sqrt{x-1-6\sqrt{x-1}+9} = 1$$
$$\sqrt{(\sqrt{x-1}-2)^2}+\sqrt{(\sqrt{x-1}-3)^2}=1$$
$$|\sqrt{x-1}-2|+|\sqrt{x-1}-3|=1$$
$$2\le\sqrt{x-1}\le3$$
$$5\le x \le 10$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1972372",
"timestamp": "2023-03-29T00:00:00",
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Recurrence formula for a series $x_n$ -- very stuck!!! The Problem:
$x_n$ is defined as $$\sum_{k=0}^\infty \frac{(n+k)!}{k!(2n+2k)!}(\frac{1}{2})^{2k} $$
and satisfies the recurrence: $x_n-2(2n+1)x_{n+1}=x_{n+2}$
But I cannot show that this is the case...
My Attempt:
$$LHS=\sum_{k=0}^\infty\frac{(n+k)!}{k!(2n+2k)!}(\frac{1}{2})^{2k}-2(2n+1)\frac{(n+1+k)!}{k!(2n+2+2k)!}(\frac{1}{2})^{2k}$$
$$=\sum_{k=0}^\infty\frac{(n+k)!(2n+1+2k)(2n+2+2k)}{k!(2n+2+2k)!}(\frac{1}{2})^{2k}-2(2n+1)\frac{(n+1+k)!}{k!(2n+2+2k)!}(\frac{1}{2})^{2k}$$
$$=\sum_{k=0}^\infty\frac{2(n+k)!(n+1+k)(2n+1+2k)-2(2n+1)(n+1+k)!}{k!(2n+2+2k)!}(\frac{1}{2})^{2k}$$
$$=\sum_{k=0}^\infty\frac{2(n+1+k)!( (2n+1+2k)-(2n+1))}{k!(2n+2+2k)!}(\frac{1}{2})^{2k}$$
$$=\sum_{k=0}^\infty\frac{2(n+1+k)!( 2k)}{k!(2n+2+2k)!}(\frac{1}{2})^{2k}$$
$$=\sum_{k=0}^\infty\frac{2(n+2+k)!( 2k)}{k!(2n+2+2k)!(n+2+k)}(\frac{1}{2})^{2k}$$
$$=\sum_{k=0}^\infty\frac{(n+2+k)!}{(k-1)!(2n+2+2k)!(2n+4+2k)}(\frac{1}{2})^{2(k-1)-1}$$
But from this point, I have no idea how to obtain $x_{n+2}$ from this...
| My Solution:
To verify that the recurrence is true: we simply test the LHS and the RHS $\Rightarrow$
$$LHS=x_n - 2(2n+1)x_{n+1}$$
$$=\sum_{k=0}^\infty \frac{2(n+k+1)!( (2n+2k+1)-(2n+1) )}{k!(2n+2+2k)!}\times(1/2)^{2k}$$
Note that this is equal to $0$ when $k=0$ so we may just consider the sum starting at $k=1$
$$=\sum_{k=1}^\infty\frac{(n+k+1)!}{(k-1)!( 2(n+2)+2(k-1))!}\times(1/2)^{2(k-1)}$$
$$=\sum_{k=1}^\infty\frac{((n+2)+(k-1))!}{(k-1)!( 2(n+2)+2(k-1))!}\times(1/2)^{2(k-1)}$$
$$=\sum_{k=0}^\infty\frac{((n+2)+(k))!}{(k)!( 2(n+2)+2(k))!}\times(1/2)^{2(k)}$$
$$=x_{n+2} \text{ as required}$$
| {
"language": "en",
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"source": "stackexchange",
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Simplify rational fraction
$$\large \frac {bx(a^2x^2+2a^2y^2+b^2y^2)+ay(a^2x^2+2b^2x^2+b^2y^2)}{bx+ay}$$
My attempt:
$$\dfrac{ba^2x^3+2a^2bxy^2+b^3xy^2+a^3x^2y+2b^2x^2ay+b^2ay^3}{bx+ay}$$
I multiplied it out like this, then got stuck...
| Like @RossMillikan's comment stated, you can switch the second and the fifth terms and then factor out $bx$ and $ay$.
$$\frac {bx(a^2x^2+2a^2y^2+b^2y^2)+ay(a^2x^2+2b^2x^2+b^2y^2)}{bx+ay}=$$
Distribute
$$\frac{ba^2x^3+2a^2bxy^2+b^3xy^2+a^3x^2y+2b^2x^2ay+b^2ay^3}{bx+ay}=$$
Switch the second and fifth terms
$$\frac{ba^2x^3+2b^2x^2ay+b^3xy^2+a^3x^2y+2a^2bxy^2+b^2ay^3}{bx+ay}=$$
Factor out $bx$ and $ay$
$$\frac{bx(a^2x^2+2bxay+b^2y^2)+ay(a^2x^2+2abxy+b^2y^2)}{bx+ay}=$$
Simplify
$$\frac{(bx+ay)(a^2x^2+2abxy+b^2y^2)}{bx+ay}=$$
$$a^2x^2+2abxy+b^2y^2=$$
$$(ax+by)^2$$
| {
"language": "en",
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$a+b+c=3\Rightarrow\sum\limits_{cyc}\frac{a}{b\sqrt{c^2+3}}\geq\frac{a^2+b^2+c^2}{2}$ Let $a$, $b$ and $c$ be positive numbers such that $a+b+c=3$. Prove that:
$$\frac{a}{b\sqrt{c^2+3}}+\frac{b}{c\sqrt{a^2+3}}+\frac{c}{a\sqrt{b^2+3}}\geq\frac{a^2+b^2+c^2}{2}$$
I tried C-S, AM-GM, Holder and more, but without success.
Thank you!
| Your inequality is equivalent to :
$$\sum_{cyc}\frac{a}{b\sqrt{c^2+\frac{(a+b+c)^2}{3}}}\geq\frac{a^2+b^2+c^2}{(a+b+c)^3}\frac{27}{2}$$
We begin with a first substitution we put :
$\frac{b}{2}-a=-u$
$\frac{b}{2}+a=v$
$c+a=w$
The initial inequality become :
$$\sum_{cyc}\frac{\frac{v+u}{2}}{(v-u)\sqrt{(\frac{2w-u-v}{2})^2+\frac{(v-u+w)^2}{3}}}\geq \frac{(\frac{v+u}{2})^2+(v-u)^2+(\frac{2w-u-v}{2})^2}{(v-u+w)^3}\frac{27}{2}$$
So we start here with a second substitution we put :
$-u=\frac{p}{\sqrt{-r^2+q^2-p^2}}$
$v=\frac{r}{\sqrt{-r^2+q^2-p^2}}$
$w=\frac{q}{\sqrt{-r^2+q^2-p^2}}$
We get this :
$p+r+q=3\sqrt{-r^2+q^2-p^2}$
The initial inequality become
$$\sum_{cyc}\frac{\frac{-p+r}{2}}{(r+p)\sqrt{(\frac{2q+p-r}{2})^2+\frac{(r+p+q)^2}{3}}}\geq \frac{(\frac{-p+r}{2})^2+(r+p)^2+(\frac{2q-r+p}{2})^2}{(r+p+q)^3}\frac{27}{2}$$
We make a last substitution :
$r=R$
$p=RP$
$q=RQ$
So the initial condition become :
$P+1+Q=3\sqrt{-1+Q^2-P^2}$
Wich is equivalent to :
$Q=\frac{3}{8}\sqrt{(9P^2+2P+9)}+\frac{(1+P)}{8}$
The initial inequality become :
$$\sum_{cyc}\frac{\frac{-P+1}{2}}{(1+P)\sqrt{(\frac{2Q-1+P}{2})^2+\frac{(1+P+Q)^2}{3}}}\geq \frac{(\frac{-P+1}{2})^2+(1+P)^2+(\frac{2Q-1+P}{2})^2}{(1+P+Q)^3}\frac{27}{2}$$
So if you replace the value of $Q$ the inequality is just with the variable $P$
So you have many ways to treat it .
Edit :
We treat the case where $b>2a$ and $a\leq c$
It's easy to see that we have :
$$\sum_{cyc}\frac{a}{b\sqrt{c^2+3}}\geq 3(0.75-a)+0.5\frac{1}{\sqrt{(3-3a)^2+3}}+\frac{2a}{(3-3a)\sqrt{a^2+3}}+\frac{3-3a}{a\sqrt{4a^2+3}}\geq 3(0.75-a)+\frac{a^2+4a^2+(3-3a)^2}{2}\geq\frac{a^2+b^2+c^2}{2}$$
So it's conclude the proof .
| {
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Discrete mathematics, divisibility How can I prove that for all $n\in\mathbf{N}$ that $6 | n^5 + 5n$?
I tested for $n = 2$ and got $6 | 32 + 10 = 42$.
| Work modulo 6:
$$\begin{align*}
\text{if $n \equiv 0 \mod 6$,}&\qquad n^5+5n\equiv 0,\\
\text{if $n \equiv 1 \mod 6$,}&\qquad n^5+5n\equiv 1+5=6\equiv0,\\
\text{if $n \equiv 2 \mod 6$,}&\qquad n^5+5n\equiv 32+10\equiv2+4\equiv0,\\
\text{if $n \equiv 3 \mod 6$,}&\qquad n^5+5n\equiv 3+3\equiv0,\\
\text{if $n \equiv 4 \mod 6$,}&\qquad n^5+5n\equiv 4+2\equiv0,\\
\text{if $n \equiv 5 \mod 6$,}&\qquad n^5+5n\equiv 5+1\equiv0.\\
\end{align*}
$$
So $n^5+5n$ is always a multiple of 6.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Compute $ \sqrt{n^2 +3} - \sqrt{n^2 +2} $. I have a question concerning determining a basic limit of $ \sqrt{n^2 +3} - \sqrt{n^2 +2} $ as $n \rightarrow \infty $. Upon using the difference of squares, this yields: $ \frac{1}{\sqrt{n^2 +3} + \sqrt{n^2 +2}} $Now the limit seems intuitive, but I cant find a formal way of determining it. Any hints about what the next step should be would be appreciated.
| Hint :
Use Squeeze Principle.
$0 \le \dfrac{1}{\sqrt{n^2 +3} + \sqrt{n^2 +2}} \le \dfrac{1}{2n}$.
| {
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How to prove this $\sum\limits_{cyc}a\sqrt{b^2+(n^2-1)c^2}\le (a+b+c)^2+(n-3)(ab+bc+ca)$ Let $a,b,c>0$ and $n$ be postive integer,show that
$$a\sqrt{b^2+(n^2-1)c^2}+b\sqrt{c^2+(n^2-1)a^2}+c\sqrt{a^2+(n^2-1)b^2}\le (a+b+c)^2+(n-3)(ab+bc+ac)$$
For $n=1$ it suffices to show that
$$ab+bc+ac\le (a+b+c)^2-2(ab+bc+ac)=a^2+b^2+c^2$$ this is equivalent to the obvious inequality
$$(a-b)^2+(b-c)^2+(c-a)^2\ge 0$$
But How to prove by $n\ge 2?$
| For $n\geq2$ we have $\sqrt{a^2+(n^2-1)b^2}\leq\frac{a^2+(n-1)ab+(n^2-n)b^2}{a+(n-1)b}\Leftrightarrow b^2(a-b)^2\geq0$.
Thus, it remains to prove that
$$\frac{c(a^2+(n-1)ab+(n^2-n)b^2)}{a+(n-1)b}\leq\sum\limits_{cyc}(a^2+(n-1)ab)$$ or
$$n(n-2)^2\sum\limits_{cyc}(a^3bc-a^2b^2c)+(n-1)\sum\limits_{cyc}(a^4b+(n-1)a^4c+(n-3)a^3b^2-(2n-3)a^3c^2)\geq0$$
for which it's enough to prove that
$$\sum\limits_{cyc}(a^4b-a^4c-3a^3b^2+3a^3c^2)+n\sum\limits_{cyc}(a^4c+a^3b^2-2a^3c^2)\geq0$$
and since $\sum\limits_{cyc}(a^4c+a^3b^2-2a^3c^2)=\sum\limits_{cyc}(b^4a-2b^3a^2+a^3b^2)=\sum\limits_{cyc}ab^2(a-b)^2\geq0$,
it remains to prove that
$$\sum\limits_{cyc}(a^4b-a^4c-3a^3b^2+3a^3c^2)+2\sum\limits_{cyc}(a^4c+a^3b^2-2a^3c^2)\geq0$$ or
$$\sum\limits_{cyc}(a^4b+a^4c-a^3b^2-a^3c^2)\geq0$$
which is Muirhead.
Done!
| {
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How to prove the equation is correct I now the Lagrange identity: $$\sqrt{a+\sqrt b}=\sqrt\frac{{a+\sqrt{a^2-b}}}{2}+\sqrt\frac{{a-\sqrt{a^2-b}}}{2}$$
but i didnt know how to prove that the equation
$$\sqrt[3]{2+\sqrt 5}+\sqrt[3]{2-\sqrt 5}=1$$
I do not think that should be used above identity. I tried to resolve the draw as irrational but appears more complicated, so please help me to solve.
Previously, thank you for the helping.
| Observe that
$$\left(\frac{1+\sqrt 5}{2}\right)^3=\frac{1+3\sqrt 5+3\cdot5+5\sqrt 5}{8} = \frac{16+8\sqrt 5}{8}=2+\sqrt 5$$
Similarly,
$$\left(\frac{1-\sqrt 5}{2}\right)^3=\frac{1-3\sqrt 5+3\cdot5-5\sqrt 5}{8} = \frac{16-8\sqrt 5}{8}=2-\sqrt 5$$
Your equation transforms into
$$\sqrt[3]{2+\sqrt 5}+\sqrt[3]{2-\sqrt 5}=\sqrt[3]{\left(\frac{1+\sqrt 5}{2}\right)^3}+\sqrt[3]{\left(\frac{1-\sqrt 5}{2}\right)^3} = \frac{1+\sqrt 5}{2}+\frac{1-\sqrt 5}{2}=1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1978850",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 4
} |
Prove that $ 2/1!+4/3!+6/5!+8/7!+\dots = e$
Prove that $ 2/1!+4/3!+6/5!+8/7!+\dots = e$
My attempt:
I googled the problem, and I found that $\sum_{(n=1)}^∞ \cfrac{2 n}{(2 n-1)!} = e$
I also found that $\sum _{n=0}^{\infty \:}\cfrac{2n+2}{\left(2n+1\right)!}$ is equal to $e$.
How can I prove this?
| Note that
$$\begin{align}
\frac 2{1!}&=\frac 1{0!}+\frac 1{1!}\\
\frac 4{3!}&=\frac 1{2!}+\frac 1{3!}\\
\frac 6{5!}&=\frac 1{4!}+\frac 1{5!}\\
\frac 8{7!}&=\frac 1{8!}+\frac 1{7!}\\
&\vdots\\
\frac {2n}{(2n-1)!}&=\frac 1{(2n-2)!}+\frac 1{(2n-1)!}
\end{align}$$
Summing for $n=1$ to $\infty$ for RHS gives $e$ which is a well-known result.
Hence the sum for LHS also equals $e$, which is the required proof.
Alternatively
$$\begin{align}
e=\sum_{k=0}^\infty \frac 1{k!}&=\sum_{n=0}^\infty \frac 1{(2n)!}+\frac 1{(2n+1)!}\\
&=\sum_{n=0}^\infty \frac 1{(2n)!}+\frac 1{(2n+1)(2n)!}\\
&=\sum_{n=0}^\infty \frac 1{(2n)!}\left(1+\frac 1{2n+1}\right)\\
&=\sum_{n=0}^\infty \frac {2n+2}{(2n+1)(2n)!}\\
&=\sum_{n=\color{red}0}^\infty \frac {2n\color{red}{+2}}{(2n\color{red}{+1})!}\\
&=\sum_{n=\color{red}1}^\infty \frac {2n}{(2n\color{red}{-1})!}\quad\blacksquare\\
\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1979848",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 3
} |
Which number is greater, $2^\sqrt2$ or $e$?
Claim: $\color{red}{2^\sqrt2<e}$
Note: $2^\sqrt2=e^{\sqrt2\ln2}$
Different approach: We show $e^{x-1}>x^\sqrt x$ for $x>2$.
Let $f(x) = x -1 - \sqrt{x} \ln x $. We have $f'(x) = 1 - \frac{ \ln x }{2 \sqrt{x} } - \frac{1}{\sqrt{x}}$. By inspection, note that $f'(1)=0$ and since for $x>1$, pick $x=4$, for instance, we have $f'(4)=1- \frac{ \ln 4 }{4} - \frac{1}{2} = \frac{1}{2} - \frac{ \ln 2 }2 > 2$, then we know $f(x)$ is increasing for $x> 1$, Thus
$$ f(x) > f(1) \implies x - 1 - \sqrt{x} \ln x > 0 \implies x - 1 \geq \sqrt{x} \ln x \implies e^{x-1} > x^{\sqrt{x}} $$
Putting $x=2$, we obtain
$$e>2^\sqrt2$$
What do you think about this approach? Is there an easier way?
| I have an approach take logs we get $1,\sqrt {2 }ln (2) $ then using series expansion of $\ln (\frac{1+x}{1-x}) $ for $ln (2) $ we have $x=1/3$ the series is $2 (x+\frac {x^3}{3} ...) $ we get $\sqrt {2}ln (2) =1.4(2)(\frac {1}{3}+\frac {1}{81} ) <1$ thus $e>2^{\sqrt {2} }$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1980947",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 6,
"answer_id": 3
} |
solve for $A$ in $\cos A+\cos2A+\cos3A=0$ Solve for $A$ where, $0°\leq A\leq 360°$
$\cos A+\cos2A+\cos3A=0$
My attempt;
Here,
$$\cos A+\cos2A+\cos3A=0$$
$$\cos A+2\cos^2A-1+4\cos^3A-3\cos A=0$$
$$4\cos^3A+2\cos^2A-2\cos A-1=0$$.
Now, what should I do to move on?
| $$4cos^{ 3 }A+2cos^{ 2 }A-2cosA-1=0\\ 2\cos ^{ 2 }{ A } \left( 2\cos { A } +1 \right) -\left( 2\cos { A } +1 \right) =0\\ \left( 2\cos { A } +1 \right) \left( 2\cos ^{ 2 }{ A } -1 \right) =0\\ \cos { A } =-\frac { 1 }{ 2 } ,\cos ^{ 2 }{ A } =\frac { 1 }{ 2 } \Rightarrow \cos { A } =\pm \frac { \sqrt { 2 } }{ 2 } \\ $$ can you proceed?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1981591",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Quadratic equation with absolute values Solve $|x^{2}+x-2|+|x^{2}-x-2|=2$
My attempt:
$|x^{2}+x-2|=
x^{2}+x-2, x\in (-\infty,-2]\cup[1,\infty)$
$|x^2+x-2|=-x^{2}-x+2, x\in(-2,1)\\$
$|x^{2}-x-2|=x^{2}-x-2, x\in (-\infty,-1]\cup[2,\infty)$
$|x^{2}-x-2|=-x^{2}+x+2, x\in (-1,2)\\$
$1)$ $x\in(-\infty,-2]\cup[1,\infty)$ and $x\in(-\infty,-1]\cup[2,\infty)$
$x^{2}+x-2+x^{2}-x-2=2$
$\Rightarrow 2x^{2}-6=0$
$\Rightarrow x^{2}=3$
$\Rightarrow x_1=\sqrt{3}, x_2=-\sqrt{3}$
$2)$ $x\in(-\infty,-2]\cup[1,\infty)$ and $x\in(-1,2) \Rightarrow x\in[1,2)$
$x^{2}+x-2-x^{2}+x+2=2$
$\Rightarrow 2x=2$
$\Rightarrow x=1$
$3)$ $x\in(-2,1)$ and $x\in(-\infty,-1]\cup[2,\infty)$ $\Rightarrow x\in(-2,-1]$
$-x^{2}-x+2+x^{2}-x-2=2$
$\Rightarrow x=-1$
$4)$ $x\in(-2,1)$ and $x\in(-1,2)$ $\Rightarrow x\in(-1,1)$
$-x^{2}-x+2-x^{2}+x+2=2$
$\Rightarrow -2x^{2}+2=0$
$\Rightarrow -2x^{2}=-2$
$\Rightarrow x^{2}=1$
$\Rightarrow x=\pm{1}$ but that's not in the interval $(-1,1)$ so I can throw away that solution.
My question is how do I find the intersection in $1)$ $x\in(-\infty,-2]\cup[1,\infty)$ and $x\in(-\infty,-1]\cup[2,\infty)$ since $-\sqrt{3}$ and $\sqrt{3}$ should not be the solutions.
| It is a bit hard to follow. Perhaps we could write
$$
\begin{align}
f(x)&=x^2+x-2\\
g(x)&=x^2-x-2
\end{align}
$$
and solve $|f(x)|+|g(x)|=2$ as you did, case by case:
$$
\begin{align}
+f(x)+g(x)&=2\implies x\in\{-\sqrt3,\sqrt3\}\\
+f(x)-g(x)&=2\implies x=1\\
-f(x)+g(x)&=2\implies x=-1\\
-f(x)-g(x)&=2\implies x\in\{-1,1\}
\end{align}
$$
And then since
$$
\begin{align}
f(x)&=0\implies x\in\{-2,1\}\\
g(x)&=0\implies x\in\{-1,2\}
\end{align}
$$
where $f,g$ are quadratic functions pointing in the positive $y$-direction, we get the following sign table
$$
\begin{array}{|c|c|}
\hline
x&&-2&&-1&&1&&2&\\
\hline
f(x)&+&0&-&-&-&0&+&+&+\\
\hline
g(x)&+&+&+&0&-&-&-&0&+\\
\hline
\end{array}
$$
and then we can simply check the solutions up against this table directly. Since $\pm\sqrt3$ fall in between $-2,-1$ or $1,2$ where $f$ and $g$ have opposite signs, they cannot be valid solutions, since they are solutions to the first equation, where $f,g$ both have positive sign.
Note: Zero can be taken as having any sign you want since $+0=-0$, so all the other solutions are valid, which makes the solution set $x\in\{-1,1\}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1981701",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
How to Solve Summation by Hand I am having trouble solving the following summation by hand.
$$ \sum_{i=0}^{n-1} 2^i (n-i) $$
Can someone guide me in the right direction, especially for the $i \cdot 2^i$ part?
| Alternatively, one can use generating function method to find the sum: We know that
$$ \sum_{n=0}^{\infty} 2^n x^n = \frac{1}{1-2x}
\qquad \text{and} \qquad
\sum_{n=0}^{\infty} n x^n = \frac{x}{(1-x)^2}. $$
Thus it follows that
\begin{align*}
\sum_{n=0}^{\infty} \bigg( \sum_{i=0}^{n} 2^i (n-i) \bigg) x^n
&= \frac{1}{1-2x} \cdot \frac{x}{(1-x)^2} \\
&= \frac{2}{1-2x} - \frac{1}{1-x} - \frac{1}{(1-x)^2} \\
&= \sum_{n=0}^{\infty} ( 2^{n+1} - 1 - (n+1) ) x^n.
\end{align*}
Therefore
$$ \sum_{i=0}^{n} 2^i (n-i) = 2^{n+1} - n - 2. $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1982265",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 2
} |
$\lim \limits_{x \to 0} \frac{1}{x^2}-\csc^2x $ using L'Hôpital's rule This is from an exercise set on L'Hôpital's rule. I've never seen anything so stubborn and recalcitrant. Anybody have suggestions?
$\lim \limits_{x \to 0} \frac{1}{x^2}-\csc^2x $
| As $x\to0$,
$$\frac{1}{x^2}-\frac{1}{\sin^2 x}=\frac{1}{x^2}-\frac{1}{(x-x^3/6+o(x^3))^2}=\frac{1}{x^2}-\frac{1}{x^2-x^4/3+o(x^4)}\\=\frac{1}{x^2}-\frac{1}{x^2}\left(\frac{1}{1-x^2/3+o(x^2)}\right)=\frac{1}{x^2}-\frac{1}{x^2}\left(1+x^2/3+o(x^2)\right)=-\frac{1}{3}+o(1)\to-\frac{1}{3}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1983697",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
What is the remainder when $P(x^{12})$ is divided by $P(x)$?
Let $P(x) = 1 + x + x^2 + x^3+ x^4 + x^5$. What is the remainder when $P(x^{12})$ is divided by $P(x)$?
(a) $0$
(b) $6$
(c) $1+x$
(d) $1 + x + x^2 + x^3+ x^4 $
The way I adopted to solve this question was by putting x=1 which gives remainder= $0$ and this matches with option (a). But the answer to this question is 6. I am not able to understand the solution given:
It would be great if someone could explain the solution in simpler or more intuitive terms. Another way of solving the question is also welcome.
| $P(x) = \frac{x^6-1}{x-1}$
$$x^{6n}-1 = (x^6-1)(1+x^6+x^{12}+\cdots +x^{6(n-1)})$$
Thus $$P(x^{12}) - 6 = x^{60} -1 + x^{48} - 1 + x^{36}-1 + x^{24}-1 + x^{12}- 1 $$ and hence $P(x^{12})-6$ is a multiple of $x^6-1$ and hence $P(x)$ divides $P(x^{12})-6$. Thus the reminder is 6.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1987288",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
If $a+b+c=3$ so $\sum\limits_{cyc}\frac{a}{\sqrt{b^2+c^2}}\geq\sum\limits_{cyc}\frac{1}{\sqrt{a^2+bc}}$ Let $a$, $b$ and $c$ be positive numbers such that $a+b+c=3$. Prove that:
$$\frac{a}{\sqrt{b^2+c^2}}+\frac{b}{\sqrt{a^2+c^2}}+\frac{c}{\sqrt{a^2+b^2}}\geq\frac{1}{\sqrt{a^2+bc}}+\frac{1}{\sqrt{b^2+ac}}+\frac{1}{\sqrt{c^2+ab}}$$
I tried to use C-S, Holder, SOS, Rearrangement and more, but without some success.
| We begin with a first substitution we put :
$\frac{b}{2}-a=x-2\epsilon<0=-u$
$\frac{b}{2}+a=y+\epsilon=v$
$c+a=z+\epsilon=w$
The initial inequality become :
$$\frac{v-u}{\sqrt{(\frac{v+u}{2})^2+(\frac{2w-v-u}{2})^2}}+\frac{\frac{v+u}{2}}{\sqrt{(v-u)^2+(\frac{2w-v-u}{2})^2}}+\frac{\frac{2w-v-u}{2}}{\sqrt{(v-u)^2+(\frac{u+v}{2})^2}}\geq \frac{1}{\sqrt{(v-u)^2+(\frac{2w-v-u}{2})(\frac{v+u}{2})}}+\frac{1}{\sqrt{(\frac{v+u}{2})^2+(\frac{2w-v-u}{2})(v-u)}}\frac{1}{\sqrt{(\frac{2w-v-u}{2})^2+(v-u)(\frac{v+u}{2})}}$$
We get $w+v-u=3$
Now we make a second substitution :
$-u=\frac{-p}{\sqrt{r^2+q^2-p^2}}$
$v=\frac{r}{\sqrt{r^2+q^2-p^2}}$
$w=\frac{q}{\sqrt{r^2+q^2-p^2}}$
We get this :
$-p+r+q=3\sqrt{r^2+q^2-p^2}$
The initial inequality become
$$\frac{2\frac{r-p}{2}}{\sqrt{(\frac{r+p}{2})^2+(\frac{2q-r-p}{2})^2}}+\frac{\frac{r+p}{2}}{\sqrt{(2\frac{r-p}{2})^2+(\frac{2q-r-p}{2})^2}}+\frac{\frac{2q-r-p}{2}}{\sqrt{(2\frac{r-p}{2})^2+(\frac{r+p}{2})^2}}\geq \frac{\sqrt{r^2+q^2-p^2}}{\sqrt{(2\frac{r-p}{2})^2+(\frac{2q-r-p}{2})(\frac{r+p}{2})}}+\frac{\sqrt{r^2+q^2-p^2}}{\sqrt{(\frac{r+p}{2})^2+(\frac{2q-r-p}{2})(2\frac{r-p}{2})}}\frac{\sqrt{r^2+q^2-p^2}}{\sqrt{(\frac{2q-r-p}{2})^2+(2\frac{r-p}{2})(\frac{r+p}{2})}}$$
We make a last substitution :
$r=R$
$-p=-RP$
$q=RQ$
With $P\geq 1$ if we put $p\geq q \geq r$
So the initial condition become :
$-P+1+Q=3\sqrt{1+Q^2-P^2}$
Wich is equivalent to :
$Q=\frac{3}{8}\sqrt{(9P^2-2P-7)}+\frac{(1-P)}{8}$
The initial inequality become :
$$\frac{2\frac{1-P}{2}}{\sqrt{(\frac{1+P}{2})^2+(\frac{2Q-1-P}{2})^2}}+\frac{\frac{1+P}{2}}{\sqrt{(2\frac{1-P}{2})^2+(\frac{2Q-1-P}{2})^2}}+\frac{\frac{2Q-1-P}{2}}{\sqrt{(2\frac{1-P}{2})^2+(\frac{1+P}{2})^2}}\geq \frac{\sqrt{1+Q^2-P^2}}{\sqrt{(2\frac{1-P}{2})^2+(\frac{2Q-1-P}{2})(\frac{1+P}{2})}}+\frac{\sqrt{1+Q^2-P^2}}{\sqrt{(\frac{1+P}{2})^2+(\frac{2Q-1-P}{2})(2\frac{1-P}{2})}}\frac{\sqrt{1+Q^2-P^2}}{\sqrt{(\frac{2Q-1-P}{2})^2+(2\frac{1-P}{2})(\frac{1+P}{2})}}$$
So to conclude it sufficient to combine the condition with the inequality to obtain an inequality with one variable .
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1991054",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
Prime number problem Find natural numbers $x$,$y$ and prime $p$ so that $y^4+4=p^x$.
In my opinion, i think we should consider cases y=0 and y=1. Hence (x;y;p)=(2;0;2), (1;1;5). Then we proof: if y>1 =>there's no solution.
| We have $y^4+4=(y^2+2y+2)(y^2-2y+2).$ We shall consider cases on $y.$
Case $(i).$ $y$ even. Since $y^2+2y+2\equiv y^2-2y+2\equiv0\pmod2$ and $y^4+4=p^x$ then $p=2;$ moreover, clearly, this implies that $x\geqslant2.$ Thus $y$ is even. Let $y=2k.$ Then $4(4k^4+1)=2^x$ so $4k^4+1=2^{x-2}.$ If $x>2$ then $2\mid1$ which is absurd. Hence $x=2$ so $4k^4=0$ which implies that $k=0.$ Hence $y=0,\;x=2$ and $p=2.$
Case $(ii).$ $y$ odd. Since $y^4+4=(y^2+2y+2)(y^2-2y+2)=p^x$ then $y^2+2y+2=p^a$ and $y^2-2y+2=p^b,$ where $a$ and $b$ are non-negative integers such that $a>b$ and $a+b=x.$ Clearly, since $y$ is a natural number, $b>0.$ Thus solving both quadratic equations for $y$ we have $(y+1)^2=p^a-1$ and $(y-1)^2=p^b-1.$ Thus $p^b(p^{a-b}-1)=4y.$ Since $p$ is an odd prime then $p^b=y$ and $p^{a-b}-1=4.$ Hence $p^{a-b}=5$ which implies that $p=5$ and $a=b+1.$ As $5^a-1=5^{b+1}-1=(5^b-1)\cdot5+4$ then $(y+1)^2=5(y-1)^2+4$ so $y\in\{1,2\}.$ A quick calculation shows that only $y=1$ works. Hence the only solution in this case is $y=1,$ $p=5$ and $x=1.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1992477",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Evaluation of $\int_\limits{-1}^{3}\frac{x\ln |x|}{1+x^4}dx$
Evaluation of $$\int_\limits{-1}^{3}\frac{\arctan(1+x^2)}{x}dx$$
$\bf{My\; Try::}$ Let $$I = \int_\limits{-1}^3\arctan(1+x^2)\cdot \frac{1}{x}dx$$
Using By parts, We get
$$I = \left[\arctan (1+x^2)\cdot \ln|x|\right]^{3}_{-1}-2\int_\limits{-1}^3\frac{x\ln |x|}{1+x^4}dx$$
Now How can i solve it , Help required, Thanks
| By continuity, there exists a neighborhood about zero such that
\begin{align}
|\arctan 1 -\arctan (1+x^2)| < \frac{1}{4} \ \ \Rightarrow \ \ \arctan1<\frac{1}{4}+\arctan (1+x^2)
\end{align}
which means
\begin{align}
0<\frac{\frac{\pi}{4}-\frac{1}{4}}{x} < \frac{\arctan(1+x^2)}{x}
\end{align}
whenever $r<x<0$ for some $r$.
Now, observe
\begin{align}
\infty =\int^0_{-1} \frac{\frac{\pi-1}{4}}{x}\ dx< \int^0_{-1} \frac{\arctan(1+x^2)}{x}\ dx.
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1992589",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Solving non linear congruence with a large mod Eg, solve $x^{11} \equiv 7 \pmod{61}$
The solution is $x \equiv 31 \pmod{61}$. But how do I get there? Brute checking every #, from $0 \to 60$ is very difficult?
| All equivalences $\bmod 61$
\begin{align}
x^{11} &\equiv 7 \\
x^{22} &\equiv 49 \equiv -12 \\
x^{44} &\equiv 144 \equiv 22 \\
x^{66} &\equiv -264 \equiv -20 \quad \equiv x^{6} \text{ by Fermat's Little Theorem}\\
x^{12} &\equiv 400 \equiv 34 \\
x &\equiv 34 \cdot 7^{-1} \equiv 34\cdot 35 \equiv 1190 \equiv 31
\end{align}
The only involved part there being finding $7^{-1} \equiv 35 \bmod 61$, which I quickly got by observing $7\cdot9 \equiv 2 $ and so $7\cdot 36 \equiv 8 $ .
Check: $31^2 \equiv 961 \equiv 46 \equiv -15 \\
31^4 \equiv 225 \equiv 42 \equiv -19\\
31^5 \equiv -589 \equiv 21 \\
31^{10} \equiv 441 \equiv 14 \\
31^{11} \equiv 434 \equiv 7 \quad \checkmark$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1995311",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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Compute the sum of binomial probability I have a sum that is $$\sum_{k=\lfloor{\frac{n}{2}+1}\rfloor}^n\binom{n}{k}p^n$$
I try $n=3$ and $n=4$, I got
$$\sum_{k=\lfloor{\frac{3}{2}+1}\rfloor}^3\binom{n}{k}p^3=4p^3$$
$$\sum_{k=\lfloor{\frac{4}{2}+1}\rfloor}^4\binom{n}{k}p^4=5p^4$$
How do I do it for all $n$?
| For the sake of brevity let's define:
\begin{equation}
A = \sum_{k=\lfloor\frac{n}{2}\rfloor + 1}^n \binom{n}{k}p^n.
\end{equation}
First of all notice that by the Binomial Theorem we have:
\begin{equation}
2^n = (1+1)^n = \sum_{k=0}^n \binom{n}{k}
\end{equation}
and so multiplying by $p^n$ we get
\begin{equation}
(2p)^n = \sum_{k=0}^n \binom{n}{k}p^n.
\end{equation}
Notice that when $n$ is odd, the binomial coefficients $\binom{n}{0},\binom{n}{1},\dots,\binom{n}{n}$ are symmetric in the following sense:
\begin{equation}
\sum_{k=0}^{\lfloor\frac{n}{2}\rfloor}\binom{n}{k} =
\sum_{\lfloor\frac{n}{2}\rfloor + 1}^n\binom{n}{k}
\end{equation}
(think of Pascal's triangle: an example of an odd row is $1, 3, 3, 1$, and $1 + 3 = 3 + 1 = 4$). Thus, if $n$ is odd we have
\begin{equation}
\sum_{\lfloor\frac{n}{2}\rfloor+1}^n\binom{n}{k} = \frac{2^n}{2} = 2^{n-1}
\end{equation}
and hence $A = 2^{n-1}p^n$ whenever $n$ is odd. Can you see how to do the case where $n$ is even?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1998863",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Does series $(\cosh\frac{1}{n} - 1)^p$ converge? Given: series $\sum^\infty_{n=1}(\cosh\frac{1}{n} - 1)^p$.
Question: Consider different $p$ and find out if it converges.
I tried to represent $\cosh$ as $\frac{e^x - e^{-x}}{2}$. What I got as $n$-th member:
$$\left(\frac{\sqrt[n]{e} + \frac{1}{\sqrt[n]{e}} - 2}{2}\right)^p$$
Notice that degree next to $e$ approaches to $0$ as n approaches to $\infty$:
$$\lim_{n\to\infty} 1/n = 0 \text{ and } \lim_{n\to\infty} -1/n = 0$$
So that I can use Maclaurin series for $e^x$ and get:
$$1 + \frac{1}{n+1} + \frac{1}{2(n+1)^2} + \ldots + \frac{1}{k!(n+1)^k} + \ldots + 1 - \frac{1}{n+1} + \frac{1}{2(n+1)^2} + \ldots + (-1)^k\frac{1}{k!(n+1)^k} + \ldots - 2 = \{\forall k: k \text{ is even}\} \frac{2}{2(n+1)^2} + \ldots + \frac{2}{k!(n+1)^k} + \ldots$$
So put it all together:
$$\left(\frac{1}{2(n+1)^2} + \ldots + \frac{2}{k!(n+1)^k} + \ldots\right)^p \forall k: k \text{ is even}$$
As I know, the sum of convergent series is sum of their sums. But how to prove it strictly for infinite number of series? Induction, started from two series? Is my idea even right?
| You can use comparison test. As
$$
\lim_{x\to 0}\frac{\cosh x-1}{x^2}=\frac12,
$$
the convergence of your series is equivalent to the convergence of
$$
\sum_{n=1}^\infty\frac1{n^{2p}}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1999104",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Find the volume of the regions enclosed by $z = x^2+y^2-2$ and $z = 30-x^2-y^2$
Find the volume of the regions enclosed by $z = x^2+y^2-2$ and $z = 30-x^2-y^2$
I set up a triple integral with the bounds of the inmost as $x^2 + y^2 - 2$ to $30 - x^2 - y^2$. The two outer integrals both had the bounds from $-4$ to $4$. When I solved it I got $1024$ as the volume, but this isn't correct. Can someone please show me the steps to finding the bounds, and the correct answer?
| This is where it goes wrong:
The two outer integrals both had the bounds from $-4$ to $4$.
If you let $x$ and $y$ run from $-4$ to $4$, you are integrating over a square in the $xy$-plane, but you want to integrate over the projection of the given region onto the $xy$-plane. Observe that the two surfaces intersect at:
$$x^2+x^2-2 = 30-x^2-y^2 \iff x^2+y^2 = 16$$
and this is a circle of radius $4$, centered at the origin.
You can keep $x:-4\to4$ but then you need $y$ as a function of $x$ to integrate over this circle (or vice versa). It works, but the calculations can become a bit messy.
Hint: polar coordinates.
Can you take it from here?
Addendum after comments.
In polar coordinates, you have $r^2=x^2+y^2$ so for the integrand:
$$\left( 30-x^2-y^2 \right) - \left( x^2+x^2-2 \right) = 32 - 2 \left( x^2 +y^2\right)
\to 32-2r^2$$
Integrating over the circle $x^2+y^2 = 4^2$ is done by letting $r:0\to 4$ and $t:0\to 2\pi$. Note though that $\mbox{d}x\,\mbox{d}y \to \color{red}{r}\,\mbox{d}r\,\mbox{d}t$.
You can verify that:
$$\int_{-4}^{4} \int_{-\sqrt{16-x^2}}^{\sqrt{16-x^2}} 32 - 2 \left( x^2 +y^2\right) \,\mbox{d}y\,\mbox{d}x = \int_{0}^{4} \int_{0}^{2\pi} r \left( 32 - 2r^2\right) \,\mbox{d}t\,\mbox{d}r = 256\pi$$
| {
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"answer_count": 3,
"answer_id": 0
} |
How to show that $\sqrt[3]{26+15\sqrt{3}}+\sqrt[3]{26-15\sqrt{3}} \in \mathbb{Z}$? How to show that the following is true? $$\sqrt[3]{26+15\sqrt{3}}+\sqrt[3]{26-15\sqrt{3}} \in \mathbb{Z}$$
I have tried to set $$\sqrt[3]{26+15\sqrt{3}}+\sqrt[3]{26-15\sqrt{3}} = r,$$ $$a=\sqrt[3]{26+15\sqrt{3}},$$ $$b=\sqrt[3]{26-15\sqrt{3}},$$ and used the identity $$(a^{1/3} + b^{1/3})^3 = a + b + 3(ab)^{1/3}(a^{1/3} + b^{1/3})$$ but I got nowhere. I am stuck at $$\left(a^{1/3}+b^{1/3}\right)^3=52+ 3 \cdot \left(a^{1/3}+b^{1/3}\right)$$
I'd be glad at any help.
| We have
$$
26 \pm 15\sqrt{3} = (2 \pm \sqrt{3})^3
$$
(by educated guessing, see note below) which gives us
$$
\sqrt[3]{26 + 15\sqrt{3}} + \sqrt[3]{26-15\sqrt3} = 2+\sqrt3+2-\sqrt3 = 4
$$
Note on guessing: I guessed that $26 + 15\sqrt3$ would be a nice cube of the form $(n + m\sqrt3)^3$ for integers (or at least rational numbers) $n, m$. If that were indeed the case, then we would have
$$
26 + 15\sqrt3 = (n + m\sqrt3)^3 = n^3 + 3n^2m\sqrt3 + 9nm^2 + 3m^3\sqrt3
$$
which then becomes
$$
n^3 + 9nm^2 = 26 \qquad\bigwedge\qquad 3(n^2m + m^3) = 15
$$
The second equation becomes $(n^2 + m^2)m = 5$ which has $n = 2, m = 1$ as the nicest solution. It turns out that this solution also solves the first equation, so we're done.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1999767",
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"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Find 6 real numbers that satisfy these systems of equations and then compute a value.
$a \neq 17$ should be $r \neq 17$.
I came across this problem in a past math competition test and wanted to tackle it just for fun.
How would you solve this?
| The equations given are:
\begin{gather}
17x+ sy + tz=0 \\
rx+34y+tz=0 \\
rx+sy+87z=0
\end{gather}
When we attempt to rewrite this is matrix form, we should remind ourselves of how matrices are multiplied with vectors:
$$
\begin{pmatrix}a & b & c \\ d & e & f \\ g & h & i\end{pmatrix} \begin{pmatrix} x \\y\ \\z\end{pmatrix} = \begin{pmatrix} ax+by+cz \\dx+ey+fz\ \\gx+hy+iz\end{pmatrix}
$$
You can imagine it like this: the rows on the left matrix, sink one-by-one (like divers) into the vector on the right (like an ocean), so that they come side by side, like $\begin{pmatrix} ax \\by\ \\cz\end{pmatrix}$. Then, we add all the products to get the entry, in this case $ax+by+cz$. Do the same with the other rows.
Now,with this background, I request you to verify that the above equations can be rewritten in the following form:
$$
\begin{pmatrix}17 & s & t \\ r & 34 & t \\ r & s & 87\end{pmatrix} \begin{pmatrix} x \\y\ \\z\end{pmatrix} = \begin{pmatrix} 0 \\0\ \\0\end{pmatrix}
$$
The above exercise will be beneficial to you, you can familiarize yourself with how matrices are multiplied.
There is something very special about the relation between equation solving and matrix representations of equations. This is the determinant. Every square matrix has associated to it, something called a determinant, which is a real number.
While the determinant of a matrix can be defined in terms of it's entries directly (I request you to read this up from Wikipedia, but if you don't understand please get back), I'll get to the salient features of the determinant:
1) Suppose you have a matrix, and you subtract a row from another row, leaving the subtracted row unchanged. Then, the determinant of the new matrix is the same. In fact, if you add/subtract multiples of rows from one another, the determinant remains unchanged.
2) If you interchange the rows of a matrix, the determinant changes sign.
3) Suppose there is a constant $c$ that divides a whole row or column of a matrix. Then,suppose we have a new matrix where we divide that row by the constant and leave everything else unchanged. Then, the ratio of the two determinants is $c$.
And the most important property: invertibility.
We have the number $1$, which if you multiply with any number,it remains unchanged. Similarly, we have the identity matrix, which if you multiply with any matrix, leaves it unchanged. This matrix has $1$ s on the diagonal, and zero elsewhere, like:
$$
I=\begin{pmatrix} 1&0&0 \\0&1&0\\ 0&0&1\end{pmatrix}
$$
Now, for every real number $a \neq 0$, we have this reciprocal $\frac{1}{a}$, which when we multiply by $a$, we get $1$, right?So we ask a similar question: Given a matrix $A$, can we find a matrix $B$, such that $B \cdot A = A \cdot B = I$? (i.e. does $\frac{I}{A}$ exist, in an odd sense?). $A$ is said to be invertible if this is possible. The answer to that question is non-trivial in the least:
4) A matrix is invertible if and only if it's determinant is non-zero.
That means, that the determinant actually gives a description of invertible matrices!
Finally, we return to our question after the preliminaries. Our question is:
$$
\begin{pmatrix}17 & s & t \\ r & 34 & t \\ r & s & 87\end{pmatrix} \begin{pmatrix} x \\y\ \\z\end{pmatrix} = \begin{pmatrix} 0 \\0\ \\0\end{pmatrix}
$$
Suppose that the large square matrix (call it $A$)is invertible. Then, there is some matrix, let's call it $\frac{I}{A}$, such that $frac{I}{A} \cdot A = A \cdot \frac{I}{A} = I$. Let us multiply by this matrix on both sides of the equation:
$$
\frac{I}{A}\begin{pmatrix}17 & s & t \\ r & 34 & t \\ r & s & 87\end{pmatrix} \begin{pmatrix} x \\y\ \\z\end{pmatrix} = \frac{I}{A}\begin{pmatrix} 0 \\0\ \\0\end{pmatrix} = \begin{pmatrix} 0 \\0\ \\0\end{pmatrix}
$$
(Zero matrix times any matrix is zero matrix, you can check this very easily from the "diving into the ocean" argument I gave above).
Now, by definition of $\frac{I}{A}$, the right side simplifies to:
$$
\begin{pmatrix} x \\y \\z\end{pmatrix}
=\begin{pmatrix} 0 \\0\ \\0\end{pmatrix}$$
But this means that $x=0$! We have assumed that is not the case. Hence, $A$ is not invertible, hence it's determinant is zero. We write $\det$ for determinant.
Hence:
$$
0=\det \begin{pmatrix}17 & s & t \\ r & 34 & t \\ r & s & 87\end{pmatrix}
$$
Subtract the third row from the second:
$$
0=\det \begin{pmatrix}17 & s & t \\ 0 & 34-s & t-87 \\ r & s & 87\end{pmatrix}
$$
Now, subtract the first row from the third:
$$
0=\det \begin{pmatrix}17 & s & t \\ 0 & 34-s & t-87 \\ r-17 & 0 & 87-t\end{pmatrix}
$$
Now, we can remove $17-r$,$34-s$ and $87-t$ outside, while ensuring that their columns have been respectively divided by the right constants. Since the left side is zero, and we have assumed $r \neq 17, s \neq 34, t \neq 87$, we can remove these constants from the equation, because they are non-zero. Hence, we would get:
$$
0=\det \begin{pmatrix}\frac{17}{17-r} & \frac{s}{34-s} & \frac t{87-t} \\ 0 & 1 & -1 \\ -1 & 0 & 1\end{pmatrix}
$$
Now, we can evaluate the determinant directly, using any rule we know (read up the ordinary rule from Wikipedia, if you like).
Ishall expand by the first row:
$$
\det \begin{pmatrix}\frac{17}{17-r} & \frac{s}{34-s} & \frac t{87-t} \\ 0 & 1 & -1 \\ -1 & 0 & 1\end{pmatrix} \\ = \frac{17}{17-r} \det \begin{pmatrix}1&-1 \\0&1\end{pmatrix} - \frac{s}{34-s} \det \begin{pmatrix}0&-1 \\-1&1\end{pmatrix} + \frac{t}{87-t} \det \begin{pmatrix}0&1 \\-1&0\end{pmatrix}
$$
Simplifying the $2 \times 2$ determinants (exercise, if you like), we get:
$$
2 = \frac{17}{17-r} + \frac{34}{34-s} + \frac{87}{87-t}
$$
Hence the answer follows. If you do not understand any of the above steps, please get back, I am very happy to help.
| {
"language": "en",
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"source": "stackexchange",
"question_score": "1",
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How to relate the $|z^3+z^{-3}|$ with $|z+z^{-1}|$? I got stuck on this seemingly simple question:
If $z$ is a complex number satisfying $|z^3+z^{-3}| \le 2$, then the maximum possible value of $|z+z^{-1}|$ is:
(A) $2$
(B) $2^{1/3}$
(C) $2\sqrt 2$
(D) $1$
Using the AM-GM inequality, I showed that $|z|^3 + |z|^{-3} \ge 2$ and so I got:
$$ |z^3+z^{-3}| \le 2 \le |z|^3 + |z|^{-3}$$
I don't inderstand how I can remove the cubic power, and bring any of this in terms of $|z|$ and $|z|^{-1}$. What am I missing?
Thank you.
| We can write $$z^3+\frac{1}{z^3} = \left(z+\frac{1}{z}\right)^3-3\left(z+\frac{1}{z}\right).$$
Given $$2 \geq \left|z^3+\frac{1}{z^3}\right| = \left|\left(z+\frac{1}{z}\right)^3-3\left(z+\frac{1}{z}\right)\right|\geq \left|z+\frac{1}{z}\right|^3-3\left|z+\frac{1}{z}\right|.$$
Now put $$\left|z+\frac{1}{z}\right|=y.$$
So we get $y^3-3y\leq 2\Rightarrow y^3-3y-2\leq 0\Rightarrow y^3-4y+y-2\leq0$.
So $$\displaystyle y(y-2)(y+2)+1(y-2)\leq 0\Rightarrow (y-2)(y+1)^2\leq0.$$
So we get $$y\leq 2\Rightarrow \left|z+\frac{1}{z}\right|\leq 2$$
because $$(y+1)^2\geq 0.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2000322",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Determining Runge-Kutta Order Given the Runge-Kutta Method given by:
$y_{i+1} = y_{i} + \frac{h}{9}(2 K_1 + 3 K_2 + 4K_3)$
with:
$K_1 = f(x_i, y_i) \qquad K_2 = f(x_i + \frac{h}{2}, y_i + \frac{h}{2} K1) \qquad K_3 = f(x_i + \frac{3}{4}h, y_i + \frac{3}{4}h K_2)$
Prove that it's order 3.
If it's order 3 we'll have:
$T_3 - \Phi(xi,yi,h) = O(h^3)$
Where $T_3 = f(x,y(x)) + \frac{h}{2} [f_t + f_x f] + \frac{h^2}{2} [f_{tt} + 2f_{tx} +f_x f_t + f_{xx}f^2 + f_x f_x f]$ is the order 3 taylor expansion.
So my idea was to taylor expand each of the $K_i$:
$K_2 = f + f_t \frac{h}{2} + f_x K_1 \frac{h}{2} \qquad K_3 = f + f_t \frac{3}{4} h + f_x K_2 $
And in the expansion of $K_3$ plug in the expansion of $K_2$, doing that I can match between the two methods the first terms but I'm missing the terms that have $f_{tt}$. So what is missing?. Should I expand $K_3 = f + f_t \frac{3}{4} h + f_x K_2 + (\frac{3}{4}h)^2 [f_{tt} + f_{tx}f + K_2(f_{xt} + f_{xx}f)]$??
| You should expand more as follows:
$$K_2=f+\frac{h}{2}f_x+\frac{h}{2}ff_y+\frac{1}{2}\left[\frac{h^2}{4}f_{xx}+2\cdot\frac{h}{2}\frac{h}{2}ff_{xy}+\frac{h^2}{4}f^2f_{yy}\right]+O(h^3)\\
K_3=f+\frac{3}{4}hf_x+\frac{3}{4}hf_yK_2+\frac{1}{2}\left[\frac{9}{16}h^2f_{xx}+2\cdot\frac{3}{4}h\frac{3}{4}hK_2f_{xy}+\frac{9}{16}h^2K_2f_{yy}\right]+O(h^3).$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2005711",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Four distinct integers a,b,c,d are in A.P. If $a^2+b^2+c^2=d$, then find$ a+b+c+d$. Four distinct integers a,b,c,d are in arithmetic progression. If $a^2+b^2+c^2=d$, then find $a+b+c+d$.
My attempt:-
| Let $a=A-X$, $b=A$, $c=A+X$, $d=A+2X$, with $A,X$ integers. Then $a^2+b^2+c^2=d$ becomes:
$3A^2+2X^2=A+2X$
$3A^2-A+2X^2-2X=0$
Multiply by 2:
$6A^2-2A+4X^2-4X=0$
$5A^2+(A-1)^2+(2X-1)^2=2$
Since $5A^2$ is divisible by $5$, we can only have $A=0$. Otherwise the sum of the squares exceeds $2$
Now $2X-1=\pm 1$ which gives $X=0$ or $X=2$
So we have either:
$a=b=c=d=0$, so $a+b+c+d=0$
or
$a=-1$, $b=0$, $c=1$, $d=2$, so $a+b+c+d=2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2008921",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Positive integers $(a,b, c)$ such that $\frac{1}{a}+ \frac{1}{b} + \frac{1}{c}$ is an integer Are $$(a, b, c) = (\mid 1 \mid, \mid 2 \mid, \mid 2 \mid), (\mid 2 \mid, \mid 4 \mid, \mid 4 \mid), (\mid 2 \mid, \mid 3\mid, \mid 6 \mid), (\mid 1 \mid, \mid 1 \mid, \mid 1 \mid)$$ the only integers such that
$$\frac{1}{a} + \frac{1}{b} + \frac{1}{c} $$ is an integer ?
| Assuming we only care about positive integers for the time being, notice that if $\frac{1}{a} + \frac{1}{b} + \frac{1}{c}$ is an integer then it must be at least $1$. If three numbers add up to at least $1$, then at least one of them must be at least $\frac{1}{3}$ - so at least one of $a$, $b$, and $c$ must be no more than $3$.
If none of them is $2$ or less, then $\frac{1}{a}$, $\frac{1}{b}$, and $\frac{1}{c}$ are all no more than a third. But if they add up to at least $1$, then they must all then be exactly $\frac{1}{3}$. So $a = b = c = 3$.
Say the smallest number we have is $2$. Then we can't have two of them (because if $\frac{1}{2} + \frac{1}{2} + \frac{1}{c}$ is a whole number, then $c = 1$ and that's smaller than $2$). But $\frac{1}{2} + \frac{1}{b} + \frac{1}{c}$ can only be at least $1$ if $\frac{1}{b} + \frac{1}{c}$ is at least $\frac{1}{2}$, so we need $\frac{1}{b},\frac{1}{c}$ to be at least $\frac{1}{4}$. Therefore at least one of $b$ and $c$ is at most $4$. So we have either $\frac{1}{2} + \frac{1}{3} + \frac{1}{c}$ (in which case $c = 6$) or $\frac{1}{2} + \frac{1}{4} + \frac{1}{c}$ (in which case $c = 4$).
So far we have $(3, 3, 3)$, $(2, 3, 6)$, and $(2, 4, 4)$, and we've found all of the ones that don't involve a $1$. I'll leave it to you to try to apply this approach to the case when we do have a $1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2010423",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Solving System of Linear Equations with LU Decomposition of $4 \times 3$ matrix The following is all confirmed to be true:
Matrix A =
$
\begin{bmatrix}
0 & 1 & -2 \\
-1 & 2 & -1 \\
2 & -4 & 3 \\
1 & -3 & 2 \\
\end{bmatrix}
$
U =
$
\begin{bmatrix}
-1 & 2 & -1 \\
0 & 1 & -2 \\
0 & 0 & 1 \\
0 & 0 & 0 \\
\end{bmatrix}
$
L =
$
\begin{bmatrix}
1 & 0 & 0 & 0\\
0 & 1 & 0 & 0\\
-2 & 0 & 1 & 0\\
-1 & -1 & -1 & 0\\
\end{bmatrix}
$
Okay so using that I need to solve the following system:
$
x_2 - 2x_3 = 0 \\
-x_1 + 2x_2 - x_3 = -2 \\
2x_1 -4x_2 + 3x_3 = 5 \\
x_1 - 3x_2 + 2x_3 = 1
$
So step one is solving $Ly = b$, where $y = Ux$
So that is:
$
y_1 = 0\\
y_2 = -2\\
-2y_1 + y_3 = 5 \\
-y_1 - y_2 -y_3 = 1 \\
$
How can we find $y_3$ in the last two equations? Because,
$
-2(0) + y_3 = 5 \\
-(0) - (-2) - y_3 = 1 \\
$
So in the second to last equation $y_3 = 5$, but in the last equation $y_3 = 1$. Very confused.
| Problem
$$
\mathbf{A} =
\left[
\begin{array}{rrr}
0 & 1 & -2 \\
-1 & 2 & -1 \\
2 & -4 & 3 \\
1 & -3 & 2 \\
\end{array}
\right]
$$
Associated Permutation Matrix
Don't start with a $0$ pivot element. Move the first row down. The permutation matrix interchanges the first two rows.
$$ \mathbf{P} =
\left[
\begin{array}{cccc}
0 & 1 & 0 & 0 \\
1 & 0 & 0 & 0 \\
0 & 0 & 1 & 0 \\
0 & 0 & 0 & 1 \\
\end{array}
\right]
$$
Input
$$
\begin{align}
\mathbf{P} \mathbf{A} =
% P
\left[
\begin{array}{cccc}
0 & \boxed{1} & 0 & 0 \\
\boxed{1} & 0 & 0 & 0 \\
0 & 0 & 1 & 0 \\
0 & 0 & 0 & 1 \\
\end{array}
\right]
% A
\left[
\begin{array}{rrr}
0 & 1 & -2 \\
-1 & 2 & -1 \\
2 & -4 & 3 \\
1 & -3 & 2 \\
\end{array}
\right]
%
&=
% L
\left[
\begin{array}{rrr}
-1 & 2 & -1 \\
0 & 1 & -2 \\
2 & -4 & 3 \\
1 & -3 & 2 \\
\end{array}
\right]
\end{align}
$$
Decomposition
$$
\begin{align}
\mathbf{P} \mathbf{A} &= \mathbf{L} \mathbf{U} \\
% PA
\underbrace{\left[
\begin{array}{rrr}
-1 & 2 & -1 \\
0 & 1 & -2 \\
2 & -4 & 3 \\
1 & -3 & 2 \\
\end{array}
\right]}_{\color{blue}{m}\times \color{red}{n}}
%
&=
% L
\underbrace{\left[
\begin{array}{rrrc}
1 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 \\
-2 & 0 & 1 & 0 \\
-1 & -1 & -1 & 1 \\
\end{array}
\right]}_{\color{blue}{m}\times \color{blue}{m}}
% U
\underbrace{\left[
\begin{array}{rrr}
-1 & 2 & -1 \\
0 & 1 & -2 \\
0 & 0 & 1 \\
0 & 0 & 0 \\
\end{array}
\right]}_{\color{blue}{m}\times \color{red}{n}}
\end{align}
$$
| {
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"url": "https://math.stackexchange.com/questions/2012532",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
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Value of $a$ for which two integrals are equal Determine $a$ such that $$\int _0 ^a ([\arctan \sqrt {x}]dx=\int _0 ^a [\frac {\pi}{2}-\arctan\sqrt {x}]dx$$ where $[.] $ is greatest integer function so i first took the first integral to get integral as $a [\arctan (a)] $ as I think derivative of integer function is 0 as its always an integer. But I dont know how to proceed.
| $\tan^{-1} \sqrt{x}$ is defined for $x\ge 0$, so we need to consider $a\ge 0$ only.
*
*$\tan^{-1} \sqrt{x}$ is an increasing function and
$$0\le \tan^{-1} \sqrt{x} \le \frac{\pi}{2} < 2$$
\begin{align*}
\left \lfloor \tan^{-1} \sqrt{x} \right \rfloor &=
\left \{
\begin{array}{lr}
0 \, , & 0\le x < \tan^2 1 \\
1 \, , & x \ge \tan^2 1
\end{array}
\right. \\[5pt]
\int_0^a
\left \lfloor
\tan^{-1} \sqrt{x}
\right \rfloor \, dx &=
\left \{
\begin{array}{lr}
0 \, , & 0\le a < \tan^2 1 \\
a-\tan^2 1 \, , & a \ge \tan^2 1
\end{array}
\right.
\end{align*}
*
*$\dfrac{\pi}{2}-\tan^{-1} \sqrt{x}$ is a decreasing function and
$$0\le \frac{\pi}{2}-\tan^{-1} \sqrt{x} \le \frac{\pi}{2} < 2$$
\begin{align*}
\left \lfloor
\frac{\pi}{2}-\tan^{-1} \sqrt{x}
\right \rfloor &=
\left \{
\begin{array}{lr}
1 \, , & 0\le x < \cot^2 1 \\
0 \, , & x \ge \cot^2 1
\end{array}
\right. \\[5pt]
\int_0^a
\left \lfloor
\frac{\pi}{2}-\tan^{-1} \sqrt{x}
\right \rfloor \, dx &=
\left \{
\begin{array}{lr}
a \, , & 0\le a < \cot^2 1 \\
\cot^2 1 \, , & a \ge \cot^2 1
\end{array}
\right.
\end{align*}
Note that $\cot^2 1 <\tan^2 1$,
$$\int_0^a
\left \lfloor
\tan^{-1} \sqrt{x}
\right \rfloor \, dx =
\int_0^a
\left \lfloor
\frac{\pi}{2}-\tan^{-1} \sqrt{x}
\right \rfloor \, dx$$
$$\implies a=0 \quad \text{or} \quad a=\tan^2 1+\cot^2 1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2013642",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
How many ways are there to distribute 12 distinguishable balls into 6 distinguishable bins? How many ways are there to distribute 12 distinguishable balls into 6 distinguishable bins so that 2 balls are placed in each bin ?
I solved it in the following way =>
$C(12,2) * C(10,2) * C(8,2) * C(6,2) * C(4,2) * C(2,2)$ and got the correct answer.
Can I apply Sterling formula of 2nd kind here ?
I want to know how Sterling formula of 2nd kind is applied here ?
| Applying $S_2$ here will be quite convoluted.
As explained in another question of yours,
The number of ways of getting exactly k different values when you throw n fair r-sided dice is
$\binom{r}{k}\cdot S_2(n,k)\cdot k! $
so to get exactly $6$ faces in $12$ throws of a normal die, $\binom66\cdot S_2(12,6)\cdot6!$ ways,
but this would include $11$ patterns, viz.
$\boxed 7\boxed 1\boxed 1\boxed 1\boxed 1\boxed 1\;\;\boxed 6\boxed 2\boxed 1\boxed 1\boxed 1\boxed 1\;\;\boxed 5\boxed 3\boxed 1\boxed 1\boxed 1\boxed 1\;\; \boxed 5\boxed 2\boxed 2\boxed 1\boxed 1\boxed 1\;\;\boxed 4\boxed 4\boxed 1\boxed 1\boxed 1\boxed 1$
$\boxed 4\boxed 3\boxed 2\boxed 1\boxed 1\boxed 1\;\;\boxed 4\boxed 2\boxed 2\boxed 2\boxed 1\boxed 1\;\;\boxed 3\boxed 3\boxed 3\boxed 1\boxed 1\boxed 1 \;\;\boxed 3\boxed 3\boxed 2\boxed 2\boxed 1\boxed 1\;\;\boxed 3\boxed 2\boxed 2\boxed 2\boxed 2\boxed 1$
$\boxed 2\boxed 2 \boxed 2 \boxed 2 \boxed 2 \boxed 2$
and we are interested only in the last pattern.
Now the number of ways for placing $12$ distinguishable balls in $6$ bins, the bins being indistinguishable except by occupancy, the multinomial coefficient has to be divided by ${p! q!..}$ where $p$ bins each have $p_n$ balls, $q$ bins each have $q_n$ balls, etc, e.g. for $\boxed 4\boxed 2\boxed 2\boxed 2\boxed 1\boxed 1$, there are $\binom{12}{4,2,2,2,1,1}/(3!2!)$ ways.
$S_2(12,6)= 1,323,652$ From this, if we subtract number of ways for $10$ unwanted patterns, we get
$1323652 - \left[\binom{12}{7,1,1,1,1,1}/ 5!+\binom{12}{6,2,1,1,1,1,}/4! ...+\binom{12}{5,2,2,1,1,1}/(2!3!) + ... \binom{12}{3,2,2,2,2,1}/4!\right] = 10,395$
$10,395$ is the number of ways to place$2$ each of $12$ distinguishable balls in $6$ indistinguishable bins,
So $10,395\times 6! = 7,484,400$, the desired answer.
But this looks like trying to catch your nose by bringing your hand from behind your neck !
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2014650",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Subtract two exponents with same base How do I solve
First I thought the difference is $D - A = 9^{12} - 9^9$, but I didn't know the rules for subtracting exponents with different bases.
But the question states that the numbers are evenly spaced, so I guess I need to calculate $9^{13} - 9^{11}$. Since I know that the distance between A and D is 1.5 times the distance between C and E, then the correct answer must be $\frac{3}{2} \cdot (9^{13} - 9^{11})$.
Would it be easier if I say something like
$$
\frac{3}{2} \cdot (9^{13} - 9^{11}) = \frac{3}{2} \cdot (9^3 \cdot 9^{10} - 9 \cdot 9^{10}) = \frac{3}{2} \cdot (9^3 - 9) \cdot 9^{10} = \frac{3}{2} \cdot 720 \cdot 9^{10}
$$
Edit
I guess what I should have done was to say
$$
\frac{3}{2} \cdot (9^{13} - 9^{11}) = \frac{3}{2} \cdot (9^{2} \cdot 9^{11} - 9^{11}) = \frac{3}{2} \cdot (9^{2} - 1) \cdot 9^{11} = \frac{3}{2} \cdot 80 \cdot 9^{11} = 120 \cdot 9^{11}
$$
which is the fifth (last) answer in the image.
| Yes, it seems you have the correct answer, but it can be done even more simply:
$$\frac32(9^{13}-9^{11}) = \frac32(9^2\cdot9^{11} - 1\cdot9^{11}) = \frac32\cdot 9^{11}\cdot (81-1)=80\cdot\frac32\cdot9^{11} = 120\cdot 9^{11}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2014835",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Improper double integrals $$
\iint_{\mathbb{R}^{2}}{\ln\left(1 + 2x^{2} + y^{2}\right) \over 1+x^{2}+y^{2}}\,\mathrm{d}x\,\mathrm{d}y
$$
using
$\ln\left(1 + 2x^{2} + y^{2}\right) \leq
\,\sqrt{\, 1 + x^{2} + y^{2}\,}\,$ and showed convergence by the comparison test, but according to my friends it diverges.
Who is right ?. Thanks in advance.
| The integral diverges as $x^2+y^2\to \infty$. To see this, we note that
$$\begin{align}
\iint_{1\le x^2+y^2\le R}\frac{\log(1+2x^2+y^2)}{x^2+y^2+1}\,dx\,dy &\ge \frac{\log(2)}{2}\iint_{1\le x^2+y^2\le R}\frac{1}{x^2+y^2}\,dx\,dy\\\\
&=\frac{\log(2)}{2}\int_0^{2\pi}\int_1^R\frac{1}{r}\,dr\,d\theta\\\\
&=\pi \log(2R)\\\\
&\to \infty \,\,\text{as}\,\,R\to \infty
\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2019201",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Different results implicit differentiation When trying to differentiate $\frac{x+y}{xy}=x$, I get different results. If I use the quotient rule:
$$
\begin{array}{rcl}
\dfrac{xy(1+y´)-(x+y)(xy´+y)}{(xy)^2} &=& 1 \\
\dfrac{xy+xyy´-x^2y´-xy-xyy´-y^2}{(xy)^2} &=& 1 \\
-x^2y´ -y^2 &=& x^2y^2 \\
y´ &=& \dfrac{-(y^2+x^2y^2)}{x^2}
\end{array}
$$
On the other hand if I do the following
$$
\begin{array}{rcl}
\dfrac{x+y}{xy} &=& x \\
x+y &=& x^2y \\
1+y´ &=& x^2y´ +2xy \\
y´ -x^2y´ &=& 2xy-1 \\
y´ &=& \dfrac{2xy-1}{1-x^2}
\end{array}
$$
I don't understand what's happening.
| $\frac{x+y}{xy}=x$ so that $y=\frac{x}{x^2-1}$ If we use this we can show that two terms for $y'$ are equal
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2019620",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
How should this fraction involving powers be solved
$$\sqrt { 5 } \cdot { \left( \frac { 5 }{ 4 } \right) }^{ \frac { 1 }{ 2 } }$$
All I know is that this can be written as
$\sqrt { 5 } \cdot { \left( \frac { 5 }{ { 2 }^{ 2 } } \right) }^{ \frac { 1 }{ 2 } }$
Any Ideas?
| $$\sqrt { 5 } { \left( \frac { 5 }{ 4 } \right) }^{ \frac { 1 }{ 2 } }={ 5 }^{ \frac { 1 }{ 2 } }\cdot { \left( \frac { 5 }{ { 2 }^{ 2 } } \right) }^{ \frac { 1 }{ 2 } }={ 5 }^{ \frac { 1 }{ 2 } }\cdot \frac { { 5 }^{ \frac { 1 }{ 2 } } }{ { \left( { 2 }^{ 2 } \right) }^{ \frac { 1 }{ 2 } } } =\frac { { 5 }^{ \frac { 1 }{ 2 } +\frac { 1 }{ 2 } } }{ { 2 } } =\frac { 5 }{ 2 } $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2020757",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Is $(-1)^n$ needed in the power series of this function? To represent this function $$f(x)=\frac{2 x + 3}{x^{2} + 3 x + 2}$$
as the sum of a power series.
My answer is $$\sum^{\infty}_{n=0}(-1)^{n}\left(\frac{1}{2^{n+1}}+1\right)x^{n}$$
$R=1$ and $I =(-1,1)$.
But the solution manual from Chegg is $$\sum^{\infty}_{n=0}\left(\frac{1}{2^{n+1}}+1\right)x^{n}$$ and $R = 2$, $I=(-2,2)$. Which I think it is wrong.
But just want to double check with the people here.
| One has, by a partial fraction decomposition,
$$
\frac{2 x + 3}{x^{2} + 3 x + 2}=\frac{1}{x+1}+\frac{1}{x+2}
$$ then
$$
\frac{1}{1+x}=\sum_{n=0}^\infty(-1)^n x^n,\qquad |x|<1,
$$
$$
\frac{1}{2+x}=\frac{1}{2}\cdot\frac{1}{1+\frac{x}2}=\sum_{n=0}^\infty\frac{(-1)^n}{2^{n+1}} x^n,\qquad |x|<2,
$$ giving
$$
\frac{2 x + 3}{x^{2} + 3 x + 2}=\sum_{n=0}^\infty (-1)^n\left(1+\frac{1}{2^{n+1}}\right) x^n,\qquad |x|<1.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2023518",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Finding the range of $\frac{a}{b+c} + \frac{b}{c+a} + \frac{c}{a+b}$, for $a$, $b$, $c$ the sides of a triangle
If $a$, $b$, and $c$ are the three sides of a triangle, then
$$\frac{a}{b+c} + \frac{b}{c+a} + \frac{c}{a+b}$$
lies in what interval?
| Since $a<b+c, b<c+a,c<a+b$ we have that
$$
\frac{1}{b+c}\leq \frac{1}{a}, \frac{1}{c+a}\leq \frac{1}{b}, \leq \frac{1}{a+b}\leq \frac{1}{c},
$$
and
$$
\frac{a}{b+c} + \frac{b}{c+a} + \frac{c}{a+b}\leq \frac{a}{a}+\frac{b}{b}+\frac{c}{c}=3
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2023828",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 3
} |
Can the antiderivate of $\frac{1}{(x^2+1)^2}$ be calculated without using the iteration formula? To calculate the antiderivate of $$\frac{1}{(x^2+1)^2}$$ , we can either use the iteration formula reducing the exercise to the integral of $\ \frac{1}{x^2+1}\ $ or we can use $$(\frac{x}{x^2+1})'=\frac{1-x^2}{(x^2+1)^2}$$ , but I do not see how we can get the antiderivate of $\ \frac{1-x^2}{(x^2+1)^2}\ $ either without any guess (If we know this antiderivate , we can express $\frac{1}{(x^2+1)^2}$ as a linear combination of $\ f(x):=\frac{1}{x^2+1}\ $ and $\ g(x):=\frac{1-x^2}{(x^2+1)^2}\ $ , namely $\ \frac{1}{(x^2+1)^2}=\frac{f(x)+g(x)}{2}\ $)
How can I calculate $\int \frac{dx}{(x^2+1)^2}$ only by using intgration by parts and the substitution rule as well as other basic integration rules ? I am looking for a solution not containing a guess or the iteration formula.
| By the change of variable
$$
x=\tan t, \quad dt= \frac{dx}{1+x^2},
$$ one gets
$$
\int\frac{dx}{(1+x^2)^2}=\int\cos^2 t\:dt=\int \left(\frac12+\frac{\cos 2t}2\right)dt
$$ which is easier.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2025833",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Solve for $x$ in $\log_4{(x+4)} \le \log_2{(2x+5)}$
I would like to solve $$\log_4{(x+4)} \le \log_2{(2x+5)}$$ for $x$.
I did:
$$\log_4{(x+4)} \le \log_2{(2x+5)} \Leftrightarrow \frac{\log(x+4)}{\log(4)} \le \frac{\log(2x+5)}{\log(2)}\\ \Leftrightarrow \frac{\log(x+4)}{2\log(2)} \le \frac{\log(2x+5)}{\log(2)} \Leftrightarrow \frac{1}{2} \cdot \frac{\log(x+4)}{\log(2)} \le \frac{\log(2x+5)}{\log(2)} \Leftrightarrow\ ???$$
What do I do next?
| Since the following identity holds for all $u>0$:
$$
\log_{2}\left(u\right) = 2 \log_{4}\left(u\right) = \log_{4}\left(u^2\right)
$$
The original inequality $\log_{4}\left(x+4\right) \leqslant \log_{2}\left(2 x + 5\right)$ can be rewritten as $$\log_{4}\left(x+4\right) \leqslant \log_{4}\left(\left(2 x + 5\right)^2 \right)$$
Since $\log_{4}\left(u\right)$ is an increasing function for $u>0$, this implies
$$
x + 4 \leqslant \left(2 x+5\right)^2
$$
which has to be supplemented with auxiliary conditions $x+4 > 0$ and $2 x+5 > 0$. This results in the answer
$$
x \geqslant - \frac{7}{4}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2028048",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
If $x>0$, then find the greatest value of the expression $ \frac{x^{100}}{1+x+x^2+x^3+\cdots+x^{200}}$
If $x>0$, then find the greatest value of the expression $ \dfrac{x^{100}}{1+x+x^2+x^3+\cdots+x^{200}}$
This expression simplifies to $ \frac{(x^{100})(x-1)}{x^{200}-1}$ using sum of n terms of GP. Now one can find the maxima by equating the derivative to zero. But is there any other way not involving calculus to get the maximum value?
| Hint: write it as
$$
\begin{align}
& \quad \frac{1}{\cfrac{1}{x^{100}} + \cfrac{1}{x^{99}} + \cdots + \cfrac{1}{x} + 1 + x + \cdots + x^{99} + x^{100}} \\
&= \frac{1}{1 + \left(x+\cfrac{1}{x}\right) + \left(x^2+\cfrac{1}{x^2}\right) + \cdots + \left(x^{100}+\cfrac{1}{x^{100}}\right)}
\end{align}
$$
and use the fact that $a+\frac{1}{a} \ge 2$ for $a \gt 0$, with equality iff $a=1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2028324",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
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.