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Proving an inequality(with a term in cosine) using the mean value theorem Given that $x\in \mathbb{R}$ and
$-2\leq x\leq 2$
prove:
$|cos(x) -1 +\frac{x^2}{2}| \leq\frac{2}{3} $
What I have so far:
for all x in between -2 and 2, $cos(x)\in[0, 1] $ and $-1 +\frac{x^2}{2}\in[-1, 1] $
I've tried to solve for this inquality using the mean value theorem and just using the fact that the terms $cos(x)$ and $-1 +\frac{x^2}{2}$ are bounded. But i can't seem to get to the number 2/3
|
It's easier.
$$
\cos(x)-1+\frac{x^2}{2} = \sum_{k = 0}^\infty \frac{(-1)^k x^{2k}}{(2k)!} -1+\frac{x^2}{2} = \sum_{k = 2}^\infty \frac{(-1)^k x^{2k}}{(2k)!}
$$
So:
$$
\left \lvert \cos(x)-1+\frac{x^2}{2} \right \rvert \leq \left \lvert \sum_{k=2}^\infty \frac{ x^{2k}}{(2k)!} \right \rvert = \left \lvert \frac{x^4}{4!}-\frac{x^6}{6!} + \frac{x^8}{8!} - \frac{x^{10}}{10!}+...\right \rvert \leq \left \lvert \frac{x^4}{4!}-\frac{x^6}{6!} \right \rvert + \underbrace{\sum_{k = 4}^\infty \frac{2^{2k}}{(2k)!}}_{=\cosh(2)-\sum_{k = 0}^3 \frac{2^{2k}}{(2k)!}} \leq \frac{26}{45} + \cosh(2)-1-\frac{2^2}{2!}-\frac{2^4}{4!}-\frac{2^6}{6!} \approx 0.5844
$$
It is a little delicate but works (only with a calculator).
|
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"url": "https://math.stackexchange.com/questions/4415528",
"timestamp": "2023-03-29T00:00:00",
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|
Find the side length of an inscribed tilted cube in a spherical octant
A cube of unknown side length is to be inscribed in the unit spherical octant where all coordinates are nonnegative. The cube is tilted, and has four vertices belonging to the same cube face on the curved surface of the unit sphere, and two vertices lying on the $xy$ plane, and in addition one vertex on the $xz$ plane, and one vertex on the $yz$ plane. This is shown in the figure above. Calculate the tilt angle and the side length of this cube.
My attempt: Knowing that the four vertices of a cube face lie on the surface of the sphere, it follows that the axis of the cube passing through its center and the center of this cube face, passes through the origin. Using this fact, we can attach a local coordinate reference frame to the cube with it origin at the center of the cube, and parameterize it using the tilt angle, and the distance of the center of the cube from the world origin. Then we can express the coordinates of vertices of the cube, using a third variable to represent the side length, and apply the conditions on their coordinates, as set by the question. From here, the tilt angle and the side length can be found. I got a tilt angle of $30^\circ$ and a side length of $\approx 0.5011$
|
Attaching a coordinate reference frame to the cube with its $z'$ axis pointing from the origin towards the center of the cube, and its $x'$ axis horizontal (parallel to the $xy$ plane), we can express the coordinates of the vertices.
We have
$\hat{z'} = (\cos(\theta) \cos(\frac{\pi}{4}) , \cos(\theta) \sin(\frac{\pi}{4}), \sin(\theta)) $
where $\theta$ is the tilt angle.
$\hat{x'} = (-\cos(\frac{\pi}{4}), \sin(\frac{\pi}{4}), 0) $
It follows that
$\hat{y'} = (-\sin(\theta) \cos(\frac{\pi}{4}), - \sin(\theta) \sin(\frac{\pi}{4}) , \cos(\theta) ) $
The origin $O'$ of the this $O'x'y'z'$ frame is at the cube center $C$ where
$C = r $\hat{z'} $
If the side length of the cube is $s = 2a$ then, we have the following vertices coordinates expressed in the coordinate system $O'x'y'z'$
$P_1 = a (1, -1, -1 ), P_2 = a (1, 1, -1), P_3 = (1, 1, 1) $
The corresponding world coordinates of these points are
$Q_1 = C + R P_1, Q_2 = C + R P_2 , Q_3 = C + RP_3 $
where $R = [\hat{x'} , \hat{y'} , \hat{z'} ] $
Hence,
$Q_1 = r \hat{z'} + a (\hat{x'} - \hat{y'} - \hat{z'} ) $
Plugging the expressions for the unit vectors above and letting $c_1 = \cos(\frac{\pi}{4}) = \frac{1}{\sqrt{2}}$, then
$Q_1 = \begin{bmatrix} c_1( r \cos \theta + a (-1 + \sin \theta - \cos \theta)) \\
c_1( r \cos \theta + a ( 1 + \sin \theta - \cos \theta ) ) \\
r \sin \theta + a (- \cos\theta - \sin \theta ) \end{bmatrix}$
Similarly, for $Q_2$
$Q_2 = r \hat{z'} + a (\hat{x'} + \hat{y'} - \hat{z'} ) $
$Q_2 = \begin{bmatrix} c_1( r \cos \theta + a (-1 - \sin \theta - \cos \theta)) \\
c_1( r \cos \theta + a ( 1 - \sin \theta - \cos \theta ) ) \\
r \sin \theta + a ( \cos\theta - \sin \theta ) \end{bmatrix}$
And,
$Q_3 = r \hat{z'} + a (\hat{x'} + \hat{y'} + \hat{z'} ) $
$Q_3 = \begin{bmatrix} c_1( r \cos \theta + a (-1 - \sin \theta + \cos \theta)) \\
c_1( r \cos \theta + a ( 1 - \sin \theta + \cos \theta ) ) \\
r \sin \theta + a ( \cos\theta + \sin \theta ) \end{bmatrix}$
Now we will impose the conditions on $Q_1$ , $Q_2$ and $Q_3$. We have that the $z$ coordinate of $Q_1$ is zero, therefore,
$ r \sin \theta + a (- \cos\theta - \sin \theta ) = 0 \hspace{20pt}(1)$
Next, we know that the $x$ coordinate of $Q_2$ is zero
$ r \cos \theta + a (-1 - \sin \theta - \cos \theta) = 0 \hspace{20pt}(2)$
Finally, point $Q_3$ is on the unit sphere. Equations $(1)$ and $(2)$, (since $r, a$ are non-zero), imply that
$ \sin \theta (-1 - \sin \theta - \cos \theta) - \cos \theta( -\cos\theta - \sin \theta ) = 0 $
And this reduces to
$ - \sin \theta - \sin^2 \theta + \cos^2 \theta = 0 $
So,
$ 1 - 2 \sin^2 \theta = \sin \theta $
whose solution is $\theta = \dfrac{\pi}{6} $.
Using the found value of $\theta$ in equation $(1)$
$ r - a (\sqrt{3} + 1 ) = 0 \hspace{20pt}(3)$
so that, $r = (1 + \sqrt{3}) a \hspace{20pt}(4) $
Using the value of $\theta$ and equation $(4)$ into the expression for $Q_3$, it becomes
$Q_3 = a \begin{bmatrix} \frac{1}{2} c_1( 2 \sqrt{3}) \\
\frac{1}{2} c_1( 4 + 2 \sqrt{3} ) \\
1 + \sqrt{3} \end{bmatrix}$
Adding the squares of the components of $Q_3$ and equating that to $1$,
$ a^2 (\frac{1}{8} ( 12 + 16 + 12 + 16 \sqrt{3} ) + 4 + 2 \sqrt{3} ) = 1 $
Simplifying,
$a^2 ( 9 + 4 \sqrt{3} ) = 1 $
Hence
$ a = \dfrac{1}{\sqrt{ 9 + 4 \sqrt{3} }} $
Hence the side length is
$ s = 2 a = \dfrac{2}{\sqrt{9 + 4 \sqrt{3}}} \approx 0.50112561417 $
|
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"timestamp": "2023-03-29T00:00:00",
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|
Evaluate $\int_{0}^{\pi/2} \frac{y \sqrt{\cos(y)}}{\sqrt{\cos(y)} + \sqrt{\sin(y)}} dy$ This integral:
$$\int_{0}^{\pi/2} \frac{y \sqrt{\cos(y)}}{\sqrt{\cos(y)} + \sqrt{\sin(y)}} dy$$
came up when I was playing around with trying to solve this similar problem:
Is there a closed form for $\int_0^{\pi/2} \dfrac{e^{-x}\sqrt{\cos x}}{\sqrt{\cos x}+\sqrt{\sin x}}dx $?
for which I had only a partial amount of success by Taylor-expanding $e^{-x}$ about $\frac{\pi}{4}$ - integrals of more generally
$$\int_{0}^{\pi/2} \frac{y^{2k+1} \sqrt{\cos(y)}}{\sqrt{\cos(y)} + \sqrt{\sin(y)}} dy$$
, $k = 0, 1, 2, \cdots$ are involved, so it'd be also interesting to hear about the general type too, but I'm wondering if cracking the simplest one wouldn't yield some hints at least.
|
Disclaimer : this is not a full answer.
Let
$$I_{n} = \int_{0}^{\frac{\pi}{2}} \frac{y^{n} \sqrt{\cos y}}{\sqrt{\cos y} + \sqrt{\sin y}} dy$$
and
$$J_{n} = \int_{0}^{\frac{\pi}{2}} \frac{y^{n} \sqrt{\sin y}}{\sqrt{\cos y} + \sqrt{\sin y}} dy.$$
Substitute $y' = \frac{\pi}{2} - y$ then
$$I_{n} = \int_{0}^{\frac{\pi}{2}} \frac{(\pi/2 - y)^{n} \sqrt{\sin x}}{\sqrt{\cos y} + \sqrt{\sin x}} dy = (-1)^{n} \int_{0}^{\frac{\pi}{2}} \left( \sum_{i=0}^{n} \binom{n}{i} \left( - \frac{\pi}{2} \right)^{n-i} y^{i} \right) \frac{\sqrt{\sin y}}{\sqrt{\cos y} + \sqrt{\sin y}} dy$$
If $2 \:|\: n$ then we can take the $J_{n}$ of the RHS to the LHS, and we can evaluate $I_{n} - J_{n}$ using $J_{i} \:\:(i < n)$.
Meanwhile
$$I_{n} + J_{n} = \int_{0}^{\frac{\pi}{2}} y^{n} dy = \frac{\pi^{n+1}}{(n+1) \cdot 2^{n+1}}$$
so now we can evaluate $I_{n}$ using some recurrence relations of $J_{n}$.
Unfortunately, this method is useless for $2 \not|\:n$, because when we take the $J_{n}$ to the LHS then it just becomes $I_{n} + J_{n}$. (It's the same reason why it doesn't work for $I_{1}$, it becomes an obvious identity and there's nothing we can do with it.)
Or maybe we can integrate using parts, where
$$\int \frac{y^{2k+1} \sqrt{\cos y}}{\sqrt{\cos y} + \sqrt{\sin y}} \cdot 1 dy = \frac{1}{2k+2} \cdot \frac{y^{2k+2} \sqrt{\cos y}}{\sqrt{\cos y} + \sqrt{\sin y}} - \int y \cdot \left( \frac{y^{2k+1} \sqrt{\cos y}}{\sqrt{\cos y} + \sqrt{\sin y}} \right)' dy$$
since we know $I_{n}$ for $2\:|\:n$. Still very complicated.
|
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"timestamp": "2023-03-29T00:00:00",
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|
How to find the closed form of $\int_{0}^{\frac{\pi}{2}} \frac{d x}{\left(p \cos ^{2} x+q \sin ^{2} x+r\right)^{n}}$, where $n\in N$? In my post, I found the integral
$$\int_{0}^{\frac{\pi}{2}} \frac{d x}{\left(a^{2} \cos ^{2} x+b^{2} \sin ^{2} x\right)^{2}}=\frac{\pi\left(a^{2}+b^{2}\right)}{4 a^{3} b^{3}}$$
Then I want to find the generalized integral using similar technique by using the integral
$$
I(p, q, r)=\int_{0}^{\frac{\pi}{2}} \frac{d x}{p \cos ^{2} x+q \sin ^{2} x+r} \quad \text{ where }p+r> 0 \textrm{ and } q+r>0.
$$
Letting $t=\tan x$ yields
$$
\begin{aligned}
I(p, q, r)&=\int_{0}^{\infty} \frac{d t}{p+qt^{2}+r\left(1+t^{2}\right)}\\
&=\int_{0}^{\infty} \frac{d t}{(q+r) t^{2}+p+r}\\&=\frac{1}{\sqrt{(p+r)(q+r)}} \left.\tan ^{-1}\left(\frac{t\sqrt{q+r} }{\sqrt{p+r}}\right)\right]_{0}^{\infty}\\
&=\frac{\pi}{2 \sqrt{(p+r)(q+r)}}
\end{aligned}
$$
Differentiating it w.r.t. $r$ by $n$ times yields
$$
\begin{aligned}\int_{0}^{\frac{\pi}{2}} \frac{(-1)^{n}n! d x}{\left(p \cos ^{2} x+q \sin ^{2} x+r\right)^{n+1}}=\frac{\pi}{2 }\cdot \frac{\partial^{n}}{\partial r^{n}}\left(\frac{1}{\sqrt{(p+r)(q+r)}}\right) \\
\boxed{\int_{0}^{\frac{\pi}{2}} \frac{d x}{\left(p \cos ^{2} x+q\sin ^{2} x+r\right)^{n+1}}=\frac{(-1)^{n} \pi}{2n !} \frac{\partial^{n}}{\partial r^{n}}\left(\frac{1}{\sqrt{(p+r)(q+r)}}\right)}
\end{aligned}
$$
Using the formula, we can evaluate
$$
\int_{0}^{\frac{\pi}{2}} \frac{d x}{\left(p \cos ^{2} x+q \sin ^{2} x+r\right)^{2}}=\frac{-\pi}{2} \frac{\partial}{\partial r}\left(\frac{1}{\sqrt{(p+r)(q+r)}}\right)= \frac{\pi(p+q+2r)}{4\left[(p+r)( q+r)\right]^{\frac{3}{2}}}
$$
For higher derivative, applying Leibniz’s Rule gives
$$
\begin{aligned}\frac{d^{n}}{d r^{n}}\left[(p+r)^{-\frac{1}{2}}(q+r)^{-\frac{1}{2}}\right]&= (-1)^n \sum_{k=0}^{n}\left(\begin{array}{l}
n \\
k
\end{array}\right)\left(\frac{1}{2} \right)_k(p+r)^{-\frac{1}{2}-k}\left(\frac{1}{2} \right)_{n-k}(q+r)^{-\frac{1}{2}-n+k} \\&=\frac{(-1)^n}{(q+r)^{n} \sqrt{(p+r)(q+r)}} \sum_{k=0}^{n} \binom{n}{k} \left(\frac{1}{2}\right)_{k}\left(\frac{1}{2}\right)_{n-k}\left(\frac{q+r}{p+r}\right)^{k}\end{aligned}
$$
where $\displaystyle (\alpha)_n=\frac{\Gamma(\alpha+n)}{\Gamma(n)} $.
Now we can conclude that for any natural number $n\geq 2$,
$$
\int_{0}^{\frac{\pi}{2}} \frac{d x}{\left(p\cos ^{2} x+q\sin ^{2} x+r\right)^{n+1}}\\=\boxed{\frac{\pi}{2n!(q+r)^{n} \sqrt{(p+r)(q+r)}} \sum_{k=0}^{n} \binom{n}{k}\left(\frac{1}{2}\right)_{k}\left(\frac{1}{2}\right)_{n-k}\left(\frac{q+r}{p+r}\right)^{k}}
$$
Question:
Is there any other solution?
|
Let $a=p+r$, $b=q+r$ and, for sure, as you did $x=\tan^{-1}(t)$
$$I_n=\int_{0}^{\frac{\pi}{2}} \frac{d x}{\left(p \cos ^{2} (x)+q \sin ^{2} (x)+r\right)^{n}}=\int_0^\infty \frac{(1+ t^2)^{n-1} }{(a+b\,t^2)^n }\,dt$$There is an antiderivative
$$\int\frac{(1+ t^2)^{n-1} }{(a+b\,t^2)^n }\,dt=\frac t {a^{n}}\, F_1\left(\frac{1}{2};1-n,n;\frac{3}{2};-t^2,-\frac{b }{a}t^2\right)$$ where appears the Appell hypergeometric function of two variables.
Using the bounds and simplifying, this gives
$$I_n=(-1)^n \frac{\pi ^{3/2}}{2 a^n}\Bigg[\frac{\, _2F_1\left(\frac{1}{2},n;n+\frac{1}{2};\frac{b}{a}\right)}{\Gamma (1-n)
\Gamma \left(n+\frac{1}{2}\right)}-\left(\frac{a}{b}\right)^{n-\frac{1}{2}}\frac{\, _2F_1\left(\frac{1}{2},1-n;\frac{3}{2}-n;\frac{b}{a}\right)}{\Gamma
\left(\frac{3}{2}-n\right) \Gamma (n)} \Bigg]$$ where appear two Gaussian hypergeometric functions.
Edit
Let $b=k\,a$ to get
$$I_n=\frac{\pi}{a^n \,k^{n-\frac{1}{2}} }\,\frac {P_n(k)}{c_k}$$ The $c_k$'s form sequence $A101926$ in $OEIS$ and the first polynomials are
$$\left(
\begin{array}{cc}
n & P_n(k) \\
1 & 1 \\
2 & k+1 \\
3 & 3 k^2+2 k+3 \\
4 & 5 k^3+3 k^2+3 k+5 \\
5 & 35 k^4+20 k^3+18 k^2+20 k+35 \\
6 & 63 k^5+35 k^4+30 k^3+30 k^2+35 k+63 \\
7 & 231 k^6+126 k^5+105 k^4+100 k^3+105 k^2+126 k+231 \\
8 & 429 k^7+231 k^6+189 k^5+175 k^4+175 k^3+189 k^2+231 k+429
\end{array}
\right)$$ which are interesting (it would be worth to analyse the patterns using $OEIS$).
|
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"timestamp": "2023-03-29T00:00:00",
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|
Is there positive integer solution of $a^b - c^d = 6$? Question: is there any integer solution of
$$a^b - c^d = e$$
where $e=6$, and $a,b,c,d \geq 2$? Any idea to attack this type of problem? Thanks.
Adam Rubinson comments that $a$ and $c$ are both odd by considering on $\mathbb Z/2 \mathbb Z$. We have $b \neq d$ since otherwise $a-c = 2$ which is impossible because the value of $(a+2)^n - a^n$ increases with $n$ and $a$, and $4^2 - 2^2 = 12 > 6$.
What I've done: for $1 \leq e \leq 9$ and $e \neq 6$, there does exist a solution. But for $e = 6$, I've searched through all integers under $10^{13}$ of the form $a^b$ with no luck. I'll list below the solutions found:
$$
\begin{align}
1 & = 3^2 - 2^3 \\
2 & = 3^3 - 5^2 \\
3 & = 2^7 - 5^3 \\
4 & = 2^3 - 2^2 \\
5 & = 2^5 - 3^3 \\
7 & = 2^4 - 3^2 \\
8 & = (2^3 \times 3 \times 13)^2 - (2 \times 23)^3 \\
9 & = 5^2 - 2^4 \\
\end{align}
$$
|
THIS IS NOT AN ANSWER, BUT IS TOO LONG FOR THE COMMENTS SECTION
Does $a^b-c^d=6$ have solutions in the integers, $a,b,c,d \ge 2$?
$\{a,c,6\}$ are pairwise coprime, because any prime factor of two of them is a factor of all three of them, and $6$ has only $2,3$ as prime factors. For $b,d \ge 2$, any prime factor of $a,c$ is present in $a^b,c^d$ at least twice, but $6$ is square-free.
Since $2\not \mid ac$ and $3\not \mid ac$, $a$ and $c$ have the form $6m\pm 1 \Rightarrow a,c \ge 5$. Furthermore, since both $a,c$ are odd, they also can be expressed in the form $4n\pm 1$
In general, $x^{yz}$ can be expressed as $(x^y)^z=X^z$. Accordingly, $b,d$ are coprime; if they share any factor $q$, then the equation can be rewritten $A^q-C^q=6$. In that case, $A>C$, so we can let $A=C+k$, where $C\ge 5, k\ge 2$. There are no $q$th powers of such numbers that differ by less than $24$
For prime number $p$, if $p\mid a$ and $(p-1)\mid d$, or if $p\mid c$ and $(p-1)\mid b$, then $a^b-c^d \equiv \pm 1 \bmod p$. But $6 \equiv \pm 1 \bmod p \iff p \in \{5,7\}$. Thus for $p\ge 11$, $p\mid a \Rightarrow (p-1)\not \mid d$ and $p\mid c \Rightarrow (p-1)\not \mid b$
Unless one of $b,d$ is of the form $2^k$, both $b,d$ will each have at least one odd factor, and we can consider the equation to be expressible in a form with $b,d$ both odd.
For odd exponent $y$, odd number $x$ has the property $x^y\equiv x \bmod 4$ and $x^y\equiv x \bmod 6$. Assuming odd exponents, the equation $a^b-c^d=6$ implies $a \not \equiv c \bmod 4$ and $a \equiv c \bmod 6$. This leads to four permissible cases:
$$a \equiv 1 \bmod 12 \text{ and } c \equiv 7 \bmod 12 \\
a \equiv 5 \bmod 12 \text{ and } c \equiv 11 \bmod 12 \\
a \equiv 7 \bmod 12 \text{ and } c \equiv 1 \bmod 12 \\
a \equiv 11 \bmod 12 \text{ and } c \equiv 5 \bmod 12$$
If $2\mid bd$, then the equation can be written either as $A^2-c^d=6$ or $a^b-C^2=6$. Since all bases are of the form $6m \pm 1$, the square term in each equation is $\equiv 1 \bmod 24$ (as is true of the even part $2k$ of the odd exponent $2k+1$) leading to the respective cases
$$A^2 \equiv 1, c\equiv c^d \equiv 19 \bmod 24 \\
a\equiv a^b \equiv 7, C^2 \equiv 1 \bmod 24$$
As a suggestion for a focused search (with no guarantee of success), one might look at the equation $A^4-(5k)^d=6$, since that reduces to $1-0 \equiv 1 \bmod 5$. Bearing in mind that $k$ must be of the form $6m \pm 1$, one could then generate numbers of the form $(30m \pm 5)^d+6$ for various values of $m,d$ and determine whether they have integer fourth roots. Not a high chance of success, but less work than looking at every number up to $10^{13}$ as OP reports having done.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Is this interval notation for the solution of an inequality problem correct? Is this notationally correct?
$$
\begin{align*}
5 - x^2 &< -2, \\[1ex]
5 - x^2 + (-5) &< -2 + (-5), \\[1ex]
-x^2 &< -7, \\[1ex]
(-x^2)(-1) &> (-7)(-1), \\[1ex]
x^2 &> 7, \\[1ex]
x^2 - 7 &> 0, \\[1ex]
(x + \sqrt{7})(x - \sqrt{7}) &> 0, \\[1ex]
x < -\sqrt{7} \,\lor\, x &> \sqrt{7}.
\; \llap{\mathrel{\boxed{\phantom{\;x < -\sqrt{7} \,\lor\, x > \sqrt{7}.}}}}\end{align*}
$$
or would the following be better notation?
$$
\begin{align*}
(x < -\sqrt{7}) \,\lor\, (x &> \sqrt{7}).
\; \llap{\mathrel{\boxed{\phantom{\;(x < -\sqrt{7}) \,\lor\, (x > \sqrt{7}).}}}}
\end{align*}
$$
Thank you.
|
I suggest either spelling out that logical symbol (for readability and because some readers may not know what it means) $$x < -\sqrt{7} \quad\text{or}\quad x > \sqrt{7},$$ or actually adopting interval notation $$(-\infty,-\sqrt{7})\cup (\sqrt{7},\infty)$$ as suggested by paw88789.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Finding the solutions of the PDE $x^2\frac{\partial u}{\partial x}+y^2\frac{\partial u}{\partial y}=(x+y)u$ How to find all solutions of the PDE
$x^2\frac{\partial u}{\partial x}+y^2\frac{\partial u}{\partial y}=(x+y)u$?
My way was:
The Lagrange-Charpit equations are: $\frac{dx}{x^2}=\frac{dy}{y^2}=\frac{du}{(x+y)u}$.
There are two linear independent solutions. For the first one I got:
$\frac{du}{(x+y)u}=\frac{dx-dy}{x^2-y^2}=\frac{dx-dy}{(x+y)(x-y)} \Leftrightarrow \frac{1}{u}du-\frac{1}{x-y}d(x-y)=0 \Leftrightarrow d\mathrm{ln}(|\frac{u}{x-y}|)=0$.
So the first solution is $f(x,y,u)=\frac{u}{x-y}$.
Now I don't know how to find the second solution. I tried to manipulate the terms $\frac{dx}{x^2}, \frac{dy}{y^2}$ and $\frac{du}{u(x+y)}$ in a similar way as above but I don't get any simplification .
Can it be done in the same way or how can the second solution be found?
|
$\frac{\frac{1}{x}dx}{x}=\frac{\frac{1}{y}dy}{y}=\frac{\frac{1}{u}du}{(x+y)}$
$\implies\frac{\frac{1}{x}dx+\frac{1}{y}dy}{x+y}=\frac{\frac{1}{u}du}{x+y} $
$\implies \frac{1}{x}dx+\frac{1}{y}dy=\frac{1}{u}du$
Solve and get second integral curve to form the solution surface.
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
Calculate double integral $\int_Se^{\frac{x}{y}}dxdy$ Calculate double integral $\int_Se^{\frac{x}{y}}dxdy$ for the region $1 \le y \le2$ and $y \le x \le y^3$
What I have tried:
$y = 1, y=2 \\ x = y, x=y^3$
$1 \le x \le 2, \text{when }x=y \\
1 \le x \le 2^{\frac{1}{3}}, \text{when }x=y^3 \\
2 \le x \le 8, \text{ when} y=2, \text{when y=1, we have x=1}$
However, we want to integrate with respect to $x$ first and then $y$. How can I correctly derive the calculations with respect to $x$?
From looking at a graph of the bounds, I have got the following:
$$\int_1^2\int_y^2e^{\frac{x}{y}}dxdy+\int_{2^{\frac{1}{3}}}^2 \int_{2}^{y^3}e^{\frac{x}{y}}dxdy$$
I cannot seem to figure out the calculation to get these bounds without the need for visualisation.
|
Integrating wrt $x$ first (the inner integral in equation below),
\begin{equation}
\int_1^2 dy \int_y^{y^3} dx \, e^{x/y} = \int_1^2 dy \, y \, \left[ e^{x/y} \right]_{y}^{y^3} = \int_{1}^2 dy \, y \, [e^{y^2} - e] = \frac{e^4 - e }{2} -\frac{3}{2}e = \frac{1}{2} e^4 - 2 e \,.
\end{equation}
We can see that $x\le y^3$ is the same as $y \ge x^{1/3}$, so we are interested in the region enclosed by $y=x^{1/3}, y = x, y=1,y=2$. Three of these curves meet at $(x=1,y=1)$, so the other boundaries are $y=x^{1/3}, y = x, x=2$. We have integrated wrt to $x$ first so that $y$ varies freely over the interval $[1,2]$.
|
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|
True or False: There is a $6\times 6$ matrix $A$ with $\text{Rank}(A)=4$ and $A^3 =0$ I understand how to do it if the question changed $A^3$ to $A^2$, because then you can just use the rank–nullity theorem. $\text{Rank}+\text{Nullity}=6$, $\text{Rank}=4$ so $\text{Nullity}=2$ so of the $6$ column vectors $v_1,v_2,\dots,v_6,$ only two of them satisfy $Av_i=0$. I don't know what to do with $A^3$ however because I'm not sure what nullity says about $A^2v_i$.
|
If $A = \begin{bmatrix} 0 && 0 && 1 && 0 && 0 && 0 \\
0 && 0 && 0 && 1 && 0 && 0 \\
0 && 0 && 0 && 0 && 1 && 0 \\
0 && 0 && 0 && 0 && 0 && 1 \\
0 && 0 && 0 && 0 && 0 && 0 \\
0 && 0 && 0 && 0 && 0 && 0 \end{bmatrix}$
Then
$A^2 = \begin{bmatrix} 0 && 0 && 0 && 0 && 1 && 0 \\
0 && 0 && 0 && 0 && 0 && 1 \\
0 && 0 && 0 && 0 && 0 && 0 \\
0 && 0 && 0 && 0 && 0 && 0 \\
0 && 0 && 0 && 0 && 0 && 0 \\
0 && 0 && 0 && 0 && 0 && 0 \end{bmatrix}$
And
$A^3 = \begin{bmatrix} 0 && 0 && 0 && 0 && 0 && 0 \\
0 && 0 && 0 && 0 && 0 && 0 \\
0 && 0 && 0 && 0 && 0 && 0 \\
0 && 0 && 0 && 0 && 0 && 0 \\
0 && 0 && 0 && 0 && 0 && 0 \\
0 && 0 && 0 && 0 && 0 && 0 \end{bmatrix} $
|
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|
$I(x) = -\int_0^1 \frac{1}{z}\ln\left(\frac{1-x z + \sqrt{1-2 x z+ z^2}}{2}\right)\,dz$ Is there a closed form integral for $$I(x) = -\int_0^1 \frac{1}{z}\ln\left(\frac{1-x z + \sqrt{1-2 x z+ z^2}}{2}\right)\,dz$$ for $-1 < x < 1$?
This integral is related to Legendre polynomials as
$$\frac{-1}{z}\ln\left(\frac{1-x z + \sqrt{1-2 x z+ z^2}}{2}\right)=\sum_{n=1}^{\infty} \frac{P_{n}(x)}{n} z^{n-1}$$
$$I(x) = \sum_{n=1}^{\infty} \frac{P_{n}(x)}{n^2}$$ where $P_{n}(x)$ satisfies $\sum_{n=0}^{\infty} P_n(x) z^2 = (1-2 x z + z^2)^{-
1/2}$ for all $-1< x< 1$ and $-1< z< 1$. Integrating twice with respect to $z$ gives a relation
\begin{align}\sum_{n=0}^{\infty} \frac{P_{n}(x)}{(n+1)(n+2)} &= \int_0^1 dz \int_0^z dz'\frac{1}{\sqrt{1 - 2 z' x +x^2}}\,.
\\
&=
1 - \sqrt{2(1-x)} +(1-x) \ln\left(1+\sqrt{\frac{2}{1-x}}\right) \\&= 1-2s+2s^2 \ln\frac{1+s}{s}
\end{align}
where $s=\sin(\theta/2)$ and $x=\cos\theta$. The elements of this sum are asymptotically equal to the sum defining $I$, but I am not sure if this is useful to get a closed form for $I$. Another related relation is
$$I(x) = \int_0^1 \frac{dz}{z} \int_0^z \frac{dz'}{z'} \left[\frac{1}{\sqrt{1 - 2 z' x +x^2}}-1\right]\,.
$$
|
Here's an approach using differentiation under the integral sign:
$$\begin{align}I'(x) &= \int_{0}^{1}\frac{\frac{1}{\sqrt{-2xz+z^2+1}}+1}{\sqrt{-2xz+z^2+1}-xz+1}\, dz\\
&=\left.\frac{(x+1)\ln\left(\sqrt{z^2-2xz+1}+x-z\right)-2x\ln\left(\sqrt{z^2-2xz+1}-z+1\right)}{x^2-1}\right]_{z=0}^{z=1}\\
&=\frac{x\ln(2)-x\ln(1-x)-(x+1)\ln(x+1)+(x+1)\ln\left(x+\sqrt{2-2x}-1\right)}{x^2-1}\end{align}$$
Mathematica now gives:
$$\begin{align}I(x) &= \int \frac{x\ln(2)-x\ln(1-x)-(x+1)\ln(x+1)+(x+1)\ln\left(x+\sqrt{2-2x}-1\right)}{x^2-1}\,dx\\
&=\frac{1}{2}\left(4\operatorname{Li}_2\left(\frac{1}{2}\left(2-\sqrt{2-2x}\right)\right)+\operatorname{Li}_2\left(\frac{1}{2}-\frac{x}{2}\right)-\ln^2(1-x)+2\ln\left(2\left(x+\sqrt{2-2x}-1\right)\right)\ln(2-2x)-\ln\left(2-\sqrt{2-2x}\right)\ln(8-8x)+\ln\left(\sqrt{2-2x}+2\right)(\ln(2)-\ln(1-x))-3\ln^2(2)\right)+C\end{align}$$
Now using, as @Steven Clark in the comments said
$$I(0) = \frac{1}{4} \left(2 \operatorname{Li}_2 \left(\frac{1}{2}-\frac{1}{\sqrt{2}}\right)-\left(\operatorname{arsinh} (1)-\ln (2)\right)^2\right)$$
we can solve for $C$.
Upon doing so, one determines the rather horrible expression (which might be possible to simplify)
$$\boxed{I(x) = 2\operatorname{Li}_2\left(1-\frac{1}{2}\sqrt{2-2x}\right)+\frac{1}{2}\text{Li}_2\left(\frac{1}{2}-\frac{x}{2}\right)-2\operatorname{Li}_2\left(1-\frac{1}{\sqrt{2}}\right)+\frac{1}{2}\operatorname{Li}_2\left(\frac{1}{2}-\frac{1}{\sqrt{2}}\right)-\frac{1}{2}\ln^2(2-2x)+\ln\left(2x+2\sqrt{2-2x}-2\right)\ln(2-2x)+\frac{1}{2}\ln(2)\ln\left(\sqrt{2-2x}+2\right)-\frac{1}{2}\ln\left(2-\sqrt{2-2x}\right)\ln(8-8x)-\frac{1}{2}\ln\left(\sqrt{2-2x}+2\right)\ln(1-x)-\frac{\pi^2}{24}+\frac{\ln^2(2)}{4}-\frac{1}{2}\ln(2)\ln\left(2+\sqrt{2}\right)-\frac{1}{4}\operatorname{arsinh}^2(1)}$$
EDIT
Through some dilogarithm identities, one can simplify this expression to $$I(x)=\frac{\pi^2}{6}-\operatorname{Li}_{2}{\left(p\right)}+\operatorname{Li}_{2}{\left(-p\right)}-\ln{\left(p\right)}\ln{\left(1-p^{2}\right)}$$
with $p=\sqrt{\frac{1-x}{2}}$ as mentioned in the comments below by @David H
|
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|
Prove $\sum \frac{n^{p+1}}{a_1+2^pa_2+\cdots+n^pa_n}$ is convergent.
Assume $\sum\limits_{n=1}^{\infty} \dfrac{1}{a_n}$ is a convergent positive term series and
$p>0$. Prove $$ \sum_{n=1}^{\infty} \frac{n^{p+1}}{a_1+2^pa_2+\cdots+n^pa_n}$$ is
convergent.
Since
$$a_1+2^pa_2+\cdots+k^pa_k\ge \sqrt[k]{a_1\cdot2^pa_2\cdots k^pa_k}=\sqrt[k]{a_1a_2\cdots a_k}\cdot \sqrt[k]{(k!)^p},$$
then
\begin{align*} \frac{k^{p+1}}{a_1+2^pa_2+\cdots+k^pa_k}&\le \frac{k^{p+1}}{\sqrt[k]{a_1a_2\cdots a_k}\cdot \sqrt[k]{(k!)^p}}\sim \frac{k^{p+1}}{\sqrt[k]{a_1a_2\cdots a_k}\cdot \frac{k^p}{e^p}}= e^p\cdot \frac{k}{\sqrt[k]{a_1a_2\cdots a_k}}. \end{align*}
This perhaps can not work.
|
A solution (in Chinese) is as follows. Is it correct?
\begin{align*} \sum_{k=1}^n\frac{k^{p+1}}{a_1+2^pa_2+\cdots+k^pa_k}&\le \sum_{k=1}^n\frac{k^{p+1}}{k\sqrt[k]{a_1\cdot2^pa_2\cdots k^pa_k}}\\ &=\sum_{k=1}^n\left(\frac{k}{\sqrt[k]{k!}}\right)^p\frac{1}{\sqrt[k]{a_1a_2\cdots a_k}}\\& \le e^p\sum_{k=1}^n\frac{1}{\sqrt[k]{a_1a_2\cdots a_k}}\\ &\le e^{p+1}\cdot \sum_{k=1}^n\frac{1}{a_k}.~~~~~~~~\color {blue}{\text{(Carleman's Inequality)}} \end{align*}
|
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|
Remarquable identities $f(n) = \frac{a^n}{(a-b)(a-c)} + \frac{b^n}{(b-a)(b-c)} + \frac{c^n}{(c-a)(c-b)}$ Let $n$ be an integer, and
\begin{equation}
f(n) = \frac{a^n}{(a-b)(a-c)} + \frac{b^n}{(b-a)(b-c)} + \frac{c^n}{(c-a)(c-b)}
\end{equation}
\begin{equation}
g(n) = \frac{(bc)^n}{(a-b)(a-c)} + \frac{(ac)^n}{(b-a)(b-c)} + \frac{(ab)^n}{(c-a)(c-b)}
\end{equation}
We have the following impressive identities, for all $a,b,c$,
\begin{align}
f(0) &= 0 \\
f(1) &= 0 \\
f(2) &= 1 \\
f(3) &= a+b+c \\
f(4) &= a^2 + b^2 + c^2 + ab + ac + bc \\
f(5) &= a^3 + b^3 + c^3 + a^2b + a^2c + b^2c + ab^2 + ac^2 + bc^2 \\
f(6) &= a^4 + b^4 + c^4 + a^3b + a^3c + b^3c + ab^3 + ac^3 + bc^3 + a^2bc + ab^2c + abc^2 +a^2b^2 + a^2c^2 + b^2c^2 \\
\\
g(0) &= 0 \\
g(1) &= 1 \\
g(2) &= ab + ac + bc \\
g(3) &= a^2b^2 + a^2c^2 + b^2c^2 + a^2bc + ab^2c + abc^2 \\
g(4) &= a^3b^3 + a^3c^3 + b^3c^3 + a^3b^2c + a^3bc^2 + a^2b^3c + ab^3c^2 + a^2bc^3 + ab^2c^3 + a^2b^2c^2
\end{align}
which I have verified by plugging the expressions into Wolfram Alpha. It seems that the general form should be, for $n > 2$.
\begin{align}
f(n) &= \sum_{i+j+k = n-2}a^ib^jc^k \\
g(n) &= \sum_{\substack{i+j+k = 2(n-1)\\1\leq i,j,k \leq n-1}}a^ib^jc^k
\end{align}
The questions are :
*
*How to demonstrate the statements for $n$ general using induction. Intuitively, we should use induction, however I do not see the induction step.
*Could we demonstrate the general case without using induction ? There is a link, between these formulas and Vandermondt matrices (see below), would there be a nice demonstration using matrices ?
=======================================================================
I arrived at such identities when working with partial fractions decomposition,
and after some related work I realized that the Vandermondt matrices where almost the inverse of the matrices which appear when we do partial fractions decomposition,
Then I realised that the Vandermondt matrices have very nice inverse :
\begin{equation}
\begin{pmatrix}
1&1 \\
a&b
\end{pmatrix}
\begin{pmatrix}
-\frac{b}{(a - b)} & \frac{1}{(a - b)} \\
-\frac{a}{(b - a)} & \frac{1}{(b - a)} \\
\end{pmatrix}
= I_{2}
\end{equation}
\begin{equation}
\begin{pmatrix}
1&1&1 \\
a&b&c \\
a^2&b^2&c^2
\end{pmatrix}
\begin{pmatrix}
\frac{b c}{(a - b) (a - c)} & -\frac{b + c}{(a - b) (a - c)} & \frac{1}{(a - b) (a - c)} \\
\frac{a c}{(b - a) (b - c)} & -\frac{a + c}{(b - a) (b - c)} & \frac{1}{(b - a) (b - c)} \\
\frac{a b}{(c - a) (c - b)} & -\frac{a + b}{(c - a) (c - b)} & \frac{1}{(c - a) (c - b)}
\end{pmatrix}
= I_{3}
\end{equation}
\begin{equation}
\begin{pmatrix}
1&1&1&1 \\
a&b&c&d \\
a^2&b^2&c^2&d^2 \\
a^3&b^3&c^3&d^3
\end{pmatrix}
\begin{pmatrix}
-\frac{bcd}{(a - b) (a - c)(a-d)} & \frac{bc + cd + bd}{(a - b) (a - c)(a-d)} &-\frac{b+c+d}{(a - b) (a - c)(a-d)} & \frac{1}{(a - b) (a - c)(a-d)}\\
-\frac{a cd}{(b - a) (b - c)(b-d)} & \frac{ac + ad + cd}{(b - a) (b - c)(b-d)} & -\frac{a + c + d}{(b - a) (b - c)(b-d)}& \frac{1}{(b - a) (b - c)(b-d)}\\
-\frac{a bd}{(c - a) (c - b)(c-d)} & \frac{ab + ad + bd}{(c - a) (c - b)(c-d)} & -\frac{a + b + d}{(c - a) (c - b)(c-d)}&\frac{1}{(c - a) (c - b)(c-d)}\\
-\frac{a bc}{(d - a) (d - b)(d-c)} & \frac{ab + ac + bc}{(d - a) (d - b)(d-c)} & -\frac{a + b + c}{(d - a) (d - b)(d-c)}&\frac{1}{(d - a) (d - b)(d-c)}
\end{pmatrix}
= I_{4}
\end{equation}
The identities $f(0),f(1), f(2)$ are the last column of the inverse equation for 3-dim matrices. However, it seems that such matrix argument is not sufficient to prove the case for $n$ general, and that many similar identities (the other places in the matrices) should exist.
|
It should be mentioned that this is one of the Schur polynomials.
|
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|
Find the equation of the straight line that contains a point, is perpendicular to a line , and is parallel to a plane. Find the equation of the straight line that contains the point $(3,7,5)$, is perpendicular to line $x-1=\frac{y+7}{-2}=\frac{z-2}{3}$, and is parallel to plane $2x+y-z-2=0$.
To make this possible, should the plane must be either perpendicular also or parallel to the given line?
I am trying to solve it by illustrating it first using graphing software. It seems that the plane and the given line are neither perpendicular nor parallel.
If the said condition is not needed, how can I solve the problem?
Note: This is not an exam or homework. I am reviewing for an upcoming test.
Thank you in advance for those who will answer my question.
|
If your searched for line contains the point $(3,7,5)$, we can write it as
$$f=\begin{pmatrix}3\\7\\5\end{pmatrix}+r\begin{pmatrix}a\\b\\c\end{pmatrix}$$
With a free variable $t$ the given line can be rewritten to
$$g=\begin{pmatrix}t\\-2t-5\\3t+1\end{pmatrix}=\begin{pmatrix}0\\-5\\1\end{pmatrix}+t\begin{pmatrix}1\\-2\\3\end{pmatrix}$$
To get there, set $x=t$ and rewrite the equations to get a value for $y$ and $z$ in regard to $t$.
The given plane has the normal vector
$$ n=\begin{pmatrix}2\\1\\-1\end{pmatrix} $$
as can be seen directly from its coordinate form.
To fulfill the requirement that the line is perpendicular to the given line and parallel to the given plane we compute the cross product of the two vectors
$$\begin{pmatrix}1\\-2\\3\end{pmatrix} * \begin{pmatrix}2\\1\\-1\end{pmatrix} =\begin{pmatrix} -2\cdot (-1)-1\cdot 3\\ 3\cdot 2 - 1\cdot(-1)\\1\cdot 1- 2\cdot (-2)\end{pmatrix}=\begin{pmatrix}-1\\7\\5\end{pmatrix}$$
Thus, the final solution is
$$f=\begin{pmatrix}3\\7\\5\end{pmatrix}+r\begin{pmatrix}-1\\7\\5\end{pmatrix}$$
The only requirement here is that the cross product exist, that is that the plane is not exactly parallel to the given line.
|
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|
Are $\frac{p^2+1}{2}$ and $\frac{p^{5n}(p^5-1)}{2}$ are coprime to each other, $n \in \mathbb{N}$? Let $p$ be a prime integer greater than $2$.
Then I want to prove the followings:
$(1)$ $\frac{p^2+1}{2}$ and $\frac{p^5-1}{2}$ are coprime to each other.
$(2)$ $\frac{p^2+1}{2}$ and $\frac{p^{5n}(p^5-1)}{2}$ are coprime to each other, $n \in \mathbb{N}$
$(3)$ Are $\frac{p^2+1}{2}$ and $\frac{p^{5n-3}(p^5-1)}{2}$ coprime to each other, $n \in \mathbb{N}$ ?
$(1)$
For $p=3$, $(p^2+1)/2=5$ and $(p^5-1)/2=141$ while $\gcd(5,141)=1$.
For $p=5$, $(p^2+1)/2=13$ and $(p^5-1)/2=1562$ while $\gcd(13,1562)=1$.
For $p=7$, $(p^2+1)/2=25$ and $(p^5-1)/2=8403$ while $\gcd(25,8403)=1$.
But how to prove the general case ?
From here, it is known that $a$ and $b$ are coprime if and only if $a^n$ and $b^n$ are coprime.
Thanks
|
This is for the question in the title. Let $q$ be a prime dividing both numbers. Then $q\mid p^2+1$, say $p^2+1 = Kq.$ Also note that $q\neq p$, so we must have $$q\mid p^5 -1 = p^5 + p^3 - p^3 - p +p -1 $$ $$= p^3(p^2+1) - p(p^2+1) + p-1$$
$$=p^3Kq -pKq + p-1.$$
Therefore $q\mid p-1$. And so $q\mid p^2-1$. Since it also divides $p^2+1$ it must divide $2$. So $q=2$.
But if $p$ is an odd prime then $p^2 = 4m+1$ for some $m$. So $(p^2+1)/2$ is odd, and so $2$ can't divide it.
Since the two numbers have no common factor, they are coprime.
|
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|
Proving that $a_i,b_i \to 0$ where $a_{i+1} := |b_i - a_i|$ and $b_{i+1} := |a_i - a_{i+1}|$? Let $a_0 := 1$ and $b_0 := \sqrt{2}$, and define
\begin{align*}
a_{i+1} &:= |b_i - a_i| \\
b_{i+1} &:= |a_i - a_{i+1}|
\end{align*}
Prove that $\lim_{i \to \infty}{a_i} = 0$ and $\lim_{i \to \infty}{b_i} = 0$.
Context: I am given the set $L := \{a+b\sqrt{2} : a,b\in \mathbb{Z}\} \subset \mathbb{R}$, and I want to prove it is NOT a lattice over $\mathbb{R}$, since it is not discrete.
My idea to solve it was to build a sequence converging to $0$ and I came up with this sequence.
Intuitively/empirically seems to be converging to $0$, but I am having trouble proving it formally. Does anyone have any idea or hint?
|
Prove, by induction that
$$a_n = \left(\sqrt{2} - 1\right)^n, ~~b_n = \left(\sqrt{2} - 1\right)^{n-1} - \left(\sqrt{2} - 1\right)^n. \tag1 $$
Empirically true for $n=1$.
$$a_{n+1} = |b_n - a_n| = |\left(\sqrt{2} - 1\right)^{n-1} \times [ 1 - 2(\sqrt{2} - 1) ]|$$
$$ = |\left(\sqrt{2} - 1\right)^{n-1} \times (3 - 2\sqrt{2})|$$
$$ = |\left(\sqrt{2} - 1\right)^{n-1} \times \left(\sqrt{2}
- 1\right)^2| = \left(\sqrt{2} - 1\right)^{n+1}.$$
Therefore,
$$b_{n+1} = |a_{n+1} - a_{n}| = \left(\sqrt{2} - 1\right)^n - \left(\sqrt{2} - 1\right)^{n+1}.$$
So, (1) above is proven.
Further, it is immediate from (1) above that as $n \to \infty, a_n \to 0.$
Finally, $b_n$ may be re-written
$$\left(\sqrt{2 - 1}\right)^{n-1} [1 - (\sqrt{2} - 1)] =
(2-\sqrt{2})a_{n-1}.$$
So, as $n \to \infty, b_n \to 0.$
|
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|
Prove that $\frac{a}{\sqrt{a^2+b^2}}+\frac{b}{\sqrt{b^2+c^2}}+\frac{c}{\sqrt{c^2+d^2}}+\frac{d}{\sqrt{d^2+a^2}}\leq3$ Prove that if $a,b,c,d$ are positive reals we have:
$$\frac{a}{\sqrt{a^2+b^2}}+\frac{b}{\sqrt{b^2+c^2}}+\frac{c}{\sqrt{c^2+d^2}}+\frac{d}{\sqrt{d^2+a^2}}\leq3.$$
I think that I have found a equality case, for example, when $a=x^4,b=x^3,c=x^2,d=x$, and as $x$ tends to $\infty$, the LHS tends to $3$, but this means that the inequality is very unlikely to be solved with traditional methods, such as Cauchy-Schwartz (my starting idea), so I got stuck.
|
Partial hint :
We have a two first not hard to show :
Let $x>0$ then define :
$$f\left(x\right)=\frac{x}{\sqrt{x^{2}+1}}$$
Then :
$$f''(x)\leq 0$$
Second fact :
Let $x>0$ then define :
$$(f(e^x))''<0$$
So we use Jensen's inequality with the constraint $c\geq d\ge a\geq 1$ we need to show for $x,y\leq 1$ :
$$2f\left(\frac{\left(x+y\right)}{2}\right)+2f\left(\sqrt{z}\right)\le3$$
Where $x=\frac{a}{b},y=\frac{b}{c},z=\frac{c}{a}$ and $xyz=1$
Wich is easier and true.
We have the easy inequalities firstly for $0<x\leq y$ :
$$\frac{\sqrt{2}xy}{x^{2}+y^{2}}-\frac{x}{\sqrt{x^{2}+y^{2}}}\geq 0$$
And for $0<y\leq x$ :
$$\frac{\sqrt{2}x^{2}}{x^{2}+y^{2}}-\frac{x}{\sqrt{x^{2}+y^{2}}}\geq 0$$
So assuming $y\geq z\geq x> 0$ and $d\geq 0$ we have the inequality :
$$g\left(x,y\right)+g\left(y,z\right)+h\left(z,d\right)+h\left(d,x\right)\leq 3$$
$$g\left(x,y\right)=\frac{\sqrt{2}xy}{x^{2}+y^{2}}\quad\quad,h\left(x,y\right)=\frac{x}{\sqrt{x^{2}+y^{2}}}$$
|
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|
Proof verification that $ \lim_{n\to \infty} \frac{n^2+n-1}{n^2 + 2n +2}=1$ EDIT: I've had some problems uploading this question today as I initially used the mobile verision, hence the quite absurd first proof if you saw it. Here is the full one:
We do this using the epsilon-N definition of the limit of :
$$\forall \varepsilon>0 ,\hspace{1mm} \exists N>0 \hspace{1mm}\text{s.t}\hspace{2mm} n\geq N \implies |a_{n}-L|<\varepsilon$$
Now,
\begin{align}
|a_{n}-L| & = \left| \frac{n^2+n-1}{n^2 + 2n +2}-1 \right| \\ &=\left| \frac{-n-3}{n^2 + 2n +2} \right|
\end{align}
We now consider the definition of $|x|$ for q quick moment, where we have that:
$$|x|=\left\{
\begin{aligned}
&x \hspace{2mm}\text{if} \hspace{2mm}x\geq0\\
-&x \hspace{2mm}\text{if} \hspace{2mm}x<0
\end{aligned} \right\}
$$
Since we can split the absolute value into the numerator and denominator separately, we have that:
$$|-n-3|=n+3 \hspace{2mm} \text{if} \hspace{2mm} -n-3<0 \iff -3<n$$
However, in our proof, we require $N>0$, thereby $n>0$, meaning that $|-n-3|=n+3$. A similar argument shows that $|n^2 +2n +2|=n^2+2n+2$ for $n>0$. Thereby, for $n>0$:
\begin{align}
|a_{n}-L| &=\frac{n+3}{n^2 + 2n +2} \\
&\leq \frac{n+3}{n^2+2n}
\end{align}
For $n\geq 3$,
\begin{align}|a_n-L| &\leq \frac{n+3}{n^2+2n} \\
&\leq \frac{2n}{n^2+2n} \\
&= \frac{2}{n+2} \\
&\leq \frac{2}{n} < \varepsilon
\end{align}
Thus, to complete our proof, we choose $N=\max\{{\frac{2}{\varepsilon},3}\}$.
Does this look good?
|
Do
$\frac{n^2+n-1}{n^2+2n+2}=\frac{1+\frac{1}{n}-\frac{1}{n^2}}{1+\frac{2}{n}+\frac{2}{n^2}}$
Make use of the fact that this is each $1$ plus something that is getting smaller and smaller the bigger $n$ gets.
$\frac{1}{1+x}\approx 1-x$
in first order for small $x$.
So the division goes over into a simple binomial multiplication for larger $n$.
$(1+\frac{1}{n}-\frac{1}{n^2})(1-\frac{2}{n}-\frac{2}{n^2})$
The leading term is already the desired $1$ and the work is done without conducting the multiplication.
One can do
$(1+\frac{1}{n})(1-\frac{2}{n})=1-\frac{1}{n}-\frac{2}{n^2}$
To show that the approximation towards one is for large natural $n$ dominated by $-\frac{1}{n}$.
The trick with $\frac{1}{1+x}\approx 1-x$ is usually or at least often introduced in courses facing this kind of tasks. An appropriate reference will be a plus in the answer.
Representate with a value table:
or the corresponding plot:
|
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"timestamp": "2023-03-29T00:00:00",
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|
Find residue of $f(z) = \frac{\sin z}{(z^2+1)^2}$ at $z = \infty$ Find residue of $f(z) = \dfrac{\sin z}{(z^2+1)^2}$ at $z = \infty$. Then this is the same as finding the residue at $z=0$ for $\dfrac{-1}{z^2}f(1/z)= \dfrac{-z^2 \sin 1/z}{(z^2+1)^2}$
$z = 0$ is a pole. But $z^n \sin 1/z \to \infty$ when $z \to 0, n = 1,2,3...$ Also the denominator is not $0$ when $z=0$. So I can't use famous formulas. Now, $-z^2\sin 1/z = \sum_{n = 0}^\infty {\dfrac{(-1)^{n-1}}{(2n+1)!z^{2n-1}}}$. But what I should do with $(z^2+1)^2$ I don't know. Also, the answer is $1/2e$ which suggests me that there is something tricky. Should I evaluate an integral over some circle? Also, I know that the residue at $z=\infty$ of $g(z) = \dfrac{\cos z}{(z^2+1)^2}$ is $0$, so $-\dfrac{1}{2\pi i}\int_{|z|=R}{g(z)dz}=0$, can this fact be of help? Any hints will be appreciated, thanks in advance.
|
The function $\frac{-z^2}{(z^2+1)^2}$ is holomorphic near $z=0$. So you can expand it as a power series at $z=0$:
$$
\frac{-z^2}{(z^2+1)^2}=a_0+a_1z+a_2z^2+\cdots\tag{1}
$$
where you can work out the coefficients by looking at the Taylor expansion of $\frac{z}{(z^2+1)^2}$ at $z=0$ and then multiply by $-z$.
Using the expansion of $\sin(z)$, you can write the Laurent series for $\sin(\frac1z)$ as
$$
\frac1z-\frac{1}{3!z^3}+\frac{1}{5!z^5}-\cdots\tag{2}
$$
Now multiply the two series (1) and (2) to find the coefficient of $\frac1z$:
$$
(a_0-\frac{a_2}{3!}+\frac{a_4}{5!}-\frac{a_6}{7!}+\cdots)\frac1z
$$
To find out the exact coefficients, observe that:
\begin{align}
\frac{-z^2}{(1+z^2)^2} &= \frac{z}2(\frac{1}{1+z^2})'
=\frac{z}2(1-z^2+z^4+\cdots)'\\
&=\frac{z}2(-2z+4z^3-6z^5+\cdots)\\
&=\frac12(-2z^2+4z^4-6z^6+\cdots)
\end{align}
So
$$
\begin{align}
(a_0-\frac{a_2}{3!}+\frac{a_4}{5!}-\frac{a_6}{7!}+\cdots)
&=\frac12\left(\frac{2}{3!}+\frac{4}{5!}+\frac{5}{6!}+\cdots\right)\\
&=\frac12\left(\frac{3-1}{3!}+\frac{5-1}{5!}+\frac{6-1}{6!}+\cdots\right)\\
&=\frac12\left(\frac{1}{2!}-\frac{1}{3!}
+\frac{1}{4!}-\frac{1}{5!}
+\frac{1}{6!}-\frac{1}{7!}
+\cdots\right)\\
&=\frac12\sum_{n=0}^\infty\frac{(-1)^n}{n!}=\frac{1}{2e}\;.
\end{align}
$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Find the locus of a point, sum of whose distances from $(c, 0)$ and $(-c, 0)$ is constant and greater than $2c$ Find the locus of a point, sum of whose distances from $(c, 0)$ and $(-c, 0)$ is constant and greater than $2c$.
Try
Let the point be $(h, k)$. Then by given condition
$\sqrt {(h-c) ^2+k^2} + \sqrt {(h+c) ^2 +k^2} = 2c + \alpha$, where $2c +\alpha$ is a constant. We consider $2c+\alpha$ because given that the sum is a constant greater than $2c$.So $\alpha \in \Bbb {R^+} $.
After this stage I have squared this equation to make it free of square root but unable to determine the locus. Please help and thanks in advance.
|
Yeah, basically it s ellipse.. Let $2a=2c+\alpha$ then we can derive the standatd ellipse equation from it. The trick to deal with the sqrt is to square twice.
$$
\sqrt {(x-c) ^2+y^2} + \sqrt {(x+c) ^2 +y^2} = 2c + \alpha := 2a\\
\sqrt {x^2+y^2+c^2-2cx} + \sqrt {x^2+y^2+c^2+2cx} = 2a
$$
Let $A=x^2+y^2+c^2$ Square once
$$
A-2cx+A+2cx+2\sqrt{A^2-4c^2x^2}=4a^2\\
2a^2-A=\sqrt{A^2-4c^2x^2}\\
A^2-4a^2A+4a^4=A^2-4c^2x^2
$$
Then you get the ellipse equation
$$
a^4=a^2A-c^2x^2\\
a^2(a^2-c^2)=(a^2-c^2)x^2+a^2y^2\\
1=\frac{x^2}{a^2}+\frac{y^2}{a^2-c^2}\\
$$
|
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "1",
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|
Let $a,b,c \in [0,1[$ such that $a+b+c=2$. Prove that $a^3+b^3+c^3+2abc\leq 2$
Let $a,b,c \in [0,1[$ such that $a+b+c=2$. Prove that
$$a^3+b^3+c^3+2abc\leq 2$$
My attempt:
put $a=x^{\frac{2}{3}}$,$b=y^{\frac{2}{3}}$,$c=z^{\frac{2}{3}}$
\begin{align*}
&x^{2}+y^{2}+z^{2}+2x^{\frac{2}{3}}y^{\frac{2}{3}}z^{\frac{2}{3}} \\
&\quad \leq x^2+y^2+z^2+\frac{2}{3}(x^{2}+y^{2}+z^{2})\\
&\quad =\frac{5}{3}(x^{2}+y^{2}+z^{2}),
\end{align*}
because $\frac{a+b+c}{3}\geq (abc)^{\frac{1}{3}}$. And
\begin{align*}
\frac{5}{3}(x^{2}+y^{2}+z^{2})
&= \frac{5}{3}(x^{\frac{1}{3}} \cdot x^{\frac{5}{3}} + y^{\frac{1}{3}} \cdot y^{\frac{5}{3}} + z^{\frac{1}{3}} \cdot z^{\frac{5}{3}}) \\
&\leq \frac{5}{3} \sqrt{x^{\frac{10}{3}} + y^{\frac{10}{3}} + z^{\frac{10}{3}}}\sqrt{x^{\frac{2}{3}} + y^{\frac{2}{3}} + z^{\frac{2}{3}}}\\
&= \frac{5\sqrt{2}}{3}\sqrt{x^{\frac{10}{3}} + y^{\frac{10}{3}} + z^{\frac{10}{3}}}
\end{align*}
Now I need to find a good frame to $\sqrt{x^{\frac{10}{3}}+y^{\frac{10}{3}}+z^{\frac{10}{3}}}$, or $\sqrt{a^5+b^5+c^5}$ for make
$$ \frac{5\sqrt{2}}{3}\sqrt{x^{\frac{10}{3}}+y^{\frac{10}{3}}+z^{\frac{10}{3}}}\leq 2. $$
I have two question:
*
*Is my attempt correct? (I mean if my algebraic manipulation is true.)
*Can you find a good frame to $\sqrt{x^{\frac{10}{3}}+y^{\frac{10}{3}}+z^{\frac{10}{3}}}$, or $\sqrt{a^5+b^5+c^5}$ for make
$$\frac{5\sqrt{2}}{3}\sqrt{x^{\frac{10}{3}}+y^{\frac{10}{3}}+z^{\frac{10}{3}}}\leq 2 \quad ?$$
If you have an other method, you can post it, but it will be nice if you can complete my attempt.
|
Hint :
Using :
$$a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)$$
And $$(a+b+c)^2=a^2+b^2+c^2+2ab+2bc+2ca$$
We got :
$$8-6(ab+bc+ca)+5abc\leq 2$$
You can conclude using uvw's method .
|
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"timestamp": "2023-03-29T00:00:00",
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|
Minimum of $\begin{aligned}\frac{a^2+b^2}{c^2}\end{aligned}$ in $\Delta ABC$
In $\triangle ABC$, $\sin B=-\cos C$. Find the minimum of $\begin{aligned}\frac{a^2+b^2}{c^2}\end{aligned}$.
According to the law of sines, $\begin{aligned}\frac{a^2+b^2}{c^2}=\frac{\sin^2A+\sin^2B}{\sin^2C}\end{aligned}$.
Solution $1$
Let $\sin B=-\cos C=k$, then $\sin B=k>0$, so $\cos C=-k<0$, meaning that $\begin{aligned}C>\frac\pi2\end{aligned}$. Thus $\begin{aligned}B<\frac\pi2\end{aligned}$, so $\cos B=\sqrt{1-k^2}$.
Now $\sin A=\sin(B+C)=k(-k)+\sqrt{1-k^2}\times\sqrt{1-k^2}=1-2k^2$. So $$\frac{\sin^2A+\sin^2B}{\sin^2C}=4-4k^2+\frac{10}{1-k^2}-13\ge2\sqrt{40}-13=4\sqrt{10}-13.$$
Solution $2$
From $\sin B=-\cos C>0$ we have $$B=C-\frac{\pi}{2}, \sin B=\sin \left(C-\frac{\pi}{2}\right)=-\cos C ,
\sin A=\sin (B+C)=\sin \left(2 C-\frac{\pi}{2}\right)=-\cos 2 C$$. So \begin{aligned}\frac{\sin ^{2} A+\sin ^{2} B}{\sin ^{2} C}&=\frac{\cos ^{2} 2 C+\cos ^{2} C}{\sin ^{2} C} \\
&=\frac{\left(1-2 \sin ^{2} C\right)^{2}+\left(1-\sin ^{2} C\right)}{\sin ^{2} C} \\
&=\frac{2+4 \sin ^{4} C-5 \sin ^{2} C}{\sin ^{2} C}=\frac{2}{\sin ^{2} C}+4 \sin ^{2} C-5 \\
& \geqslant 2 \sqrt{\frac{2}{\sin ^{2} C} \cdot 4 \sin ^{2} C}-5=4 \sqrt{2}-5,
\end{aligned}
|
Everything should be correct in solution $2$ because AM-GM can be used as $\sin^2 C$ is always non-negative.
For solution $1$ however, after $\sin A = 1 - 2k^2$ your steps don't follow: something must have been gone catastrophically wrong there. Using the values that we obtained for $\sin A, \sin B$ and that $\sin^2 C = 1 - \cos^2 C$:
$$\frac{\sin^2A+\sin^2B}{\sin^2C}= \frac{(1 - 2k^2)^2 + k^2}{1-k^2} = \frac{1 - 3k^2 + 4k^4}{1 - k^2}.$$
After doing some synthetic division on $\frac{4x^2 - 3x + 1}{x - 1}$ (or polynomial long division if you prefer):
$$
\begin{array}{c|cc}
1 & 4 & -3 & 1\\
\ & \ & 4 & 1 \\
\hline
\ & 4 & 1 & 2 \\
\end{array}
$$
and this implies $(4x^2 - 3x + 1) = (4x + 1)(x - 1) + 2$ or that $(4k^4 - 3k^2 + 1)$ $ = -(4k^2 + 1)(1 - k^2) + 2$. Thus we have:
$$\frac{1 - 3k^2 + 4k^4}{1 - k^2} = -(4k^2 + 1) + \frac{2}{1 - k^2} = 4 - 4k^2 + \frac{2}{1 - k^2} - 5.$$
instead of what you had there.
Now as $-1 ≤ \sin B ≤ 1$ and also $k \ne -1, 1$ due to the denominator, we can apply AM-GM since $1 - k^2$ will be non-negative.
This gives the minimum as $2 \sqrt{8} - 5 = 4 \sqrt{2} - 5$ like in your other answer. Thus solution $2$ is correct, but we can use AM-GM as well on your solution attempt $1$ with a bit more effort.
Addendum: the two functions which are in terms of $k$ or $\sin^2 C$ are remarkably similar. However, applying the transformation $x^2 \to 1 - x^2$ in both directions shows that the two functions do indeed have the same minimum, which is not obvious before polynomial division.
|
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|
If $A=\begin{bmatrix}2&1&1\\0&1&0\\1&1&2\end{bmatrix}$, find $A^8-5A^7+7A^6-3A^5+A^4-5A^3+8A^2-2A+I$
If $A=\begin{bmatrix}2&1&1\\0&1&0\\1&1&2\end{bmatrix}$, find $A^8-5A^7+7A^6-3A^5+A^4-5A^3+8A^2-2A+I$
Attempt $1:$ $A=I+B$, where $B=\begin{bmatrix}1&1&1\\0&0&0\\1&1&1\end{bmatrix}$
$B^n$ comes out to be $2^{n-1}B$
After that I tried doing $A^8=(I+B)^8$ and then opening RHS with binomial. But it didn't help.
Attempt $2:$ Finding characteristic equation and putting $A$ in it. It didn't help either.
The equation I got here is $A^3-5A^2+7A-3I=0$
Note: Concept of minimal polynomial is not in syllabus. Also, if we can do this without characteristic polynomial, that would be great.
|
We have, $$A=\begin{bmatrix}2&1&1\\0&1&0\\1&1&2\end{bmatrix}$$
According to Caley-Hamilton Theorem,
$$|A - \lambda I| = 0$$
So we have,
$$A - \lambda I = \begin{bmatrix}2&1&1\\0&1&0\\1&1&2\end{bmatrix} - \lambda \begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix} = \begin{bmatrix}2 - \lambda&1&1\\0&1 - \lambda &0\\1&1&2 - \lambda\end{bmatrix}$$
Now,
$$|A - \lambda I| = 0 $$$$\implies \begin{vmatrix}2 - \lambda&1&1\\0&1 - \lambda &0\\1&1&2 - \lambda\end{vmatrix} = 0$$$$\implies \lambda^3 - 5\lambda^2 + 7 \lambda - 3 = 0\tag{1}$$
Now, since every matrix is the root of its Eigen matrix, we have:
$$A^3 - 5A^2 + 7A - 3 = 0 \tag{2}$$
Now consider,
$$A^8-5A^7+7A^6-3A^5+A^4-5A^3+8A^2-2A+I$$
$$ = (A^3- 5A^2 + 7A - 3) (A^5 + A) + (A^2 + A + I) $$
Using equation $(2)$ this equation will be changed into,
$$ (0) (A^5 + A) + (A^2 + A + I) $$
$$ = (A^2 + A + I)$$
Now put values of $A^2, A$ and $I$, it will be done!
Factorization step:
$$A^8-5A^7+7A^6-3A^5+A^4-5A^3+8A^2-2A+I$$
$$A^5(A^3-5A^2+7A-3) +A^4-5A^3+8A^2-2A+I$$
$$A^5(A^3-5A^2+7A-3) +A^4-5A^3+7A^2 + A^2 - 3A + A+I$$
$$A^5(A^3-5A^2+7A-3) +A^4-5A^3+7A^2- 3A + A^2 + A+I$$
$$A^5\underbrace{(A^3-5A^2+7A-3)} +A\underbrace{(A^3-5A^2+7A - 3)} + A^2 + A+I$$
$$(A^5 + A)(A^3-5A^2+7A-3) + (A^2 + A+I)$$
|
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|
$\sqrt[3]{10-x}+\sqrt[3]{30-x}=\sqrt[3]{15-x}+\sqrt[3]{25-x}$ I just happened to find a problem and an elegant solution.
The question asks us to solve the following equation
$$\sqrt[3]{10-x}+\sqrt[3]{30-x}=\sqrt[3]{15-x}+\sqrt[3]{25-x}$$
I am answering this question below but I would love if you can also share a different solution.
P.S: I composed this problem by myself. I do not know if this problem is available anywhere. I would love to get some feedback about the same. It motivates me to create problems and discuss with others.
|
Suppose $$A=\sqrt[3]{10-x}+\sqrt[3]{30-x}=\sqrt[3]{15-x}+\sqrt[3]{25-x}$$
I will use the following
\begin{align}
A^3=&(p+q)^3 \\
=&p^3+q^3+3pq(p+q) \\
=&p^3+q^3+3pq(A)
\end{align}
Then
\begin{align}
A^3=&(10-x)+(30-x)+3(\sqrt[3]{10-x})(\sqrt[3]{30-x})(A) \\
=&(40-2x)+3(\sqrt[3]{10-x})(\sqrt[3]{30-x})(A)
\end{align}
\begin{align}
A^3=&(15-x)+(25-x)+3(\sqrt[3]{15-x})(\sqrt[3]{25-x})(A) \\
=&(40-2x)+3(\sqrt[3]{15-x})(\sqrt[3]{25-x})(A)
\end{align}
and hence we have
$$(\sqrt[3]{10-x})(\sqrt[3]{30-x})(A)=(\sqrt[3]{15-x})(\sqrt[3]{25-x})(A)$$
which is impossible unless $A=0$
Hence $$A=\sqrt[3]{10-x}+\sqrt[3]{30-x}=0$$ and
I conclude $x=20$
|
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"timestamp": "2023-03-29T00:00:00",
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|
find the minimum of $A=\frac{x^3}{3y+1}+\frac{y^3}{3z+1}+\frac{z^3}{3x+1}$ with $x^3+y^3+z^3=3$ With $x,y,z \ge 0, x^3+y^3+z^3=3$: find the minimum of $A=\dfrac{x^3}{3y+1}+\dfrac{y^3}{3z+1}+\dfrac{z^3}{3x+1}$
My attempts: $A=\dfrac{x^6}{3yx^3+x^3}+\dfrac{y^6}{3zy^3+y^3}+\dfrac{z^6}{3xz^3+z^3} \ge \dfrac{(x^3+y^3+z^3)^2}{3(yx^3+zy^3+xz^3)+x^3+y^3+z^3}=\dfrac{3}{yx^3+zy^3+xz^3+1}$ and I don't know what to do next (I think my approach is not right)
|
Note: The following answer is incorrect, see comments. I don't have a correct version yet. However since the other answer mentions this, I've marked it as community wiki and made it visible.
Let
$$
L
=
\lambda (x^3+y^3+z^3-3)+{x^3\over 3y+1}+{y^3\over 3z+1} +{z^3\over 3x+1}\\
$$
Differentiating with respect to $\lambda, x, y$, and $z$ gives
$$
{\partial L\over\partial\lambda}=x^3+y^3+z^3-3
$$
and three symmetric equations of the form
$$
{\partial L\over\partial x}=3x^2(\lambda+{1\over 3y+1})
$$
Setting $\partial L/\partial\lambda=\partial L/\partial x=...=0$ gives $\lambda=-1/(3y+1)=-1/(3z+1)=-1/(3x+1)$, so $x=y=z$ and since $x^3+y^3+z^3=3$, $x=y=z=1$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Prove decay function concaved-up Below decay function were used for solving TVM, for interest rate.
$\displaystyle f_n(x) = \frac{n\;x}{(1+x)^n-1}\qquad$, where $n>1,\;x>-1$
I wanted to show decay function is concaved-up.
In other words, I wanted to show $f_n^{''}(x) > 0$
This is what I get for second derivative, confirmed by CAS.
$\displaystyle f_n^{''}(x) =
\frac{f_n(x)^2 × (1+x)^n × 2 × (f_n(x) + ({n-1 \over 2})x-1)} {(x+x^2)^2} $
Details not shown, but if we take limit at $x=0$, we have:
$f_n^{''}(0) = \frac{n^2-1}{6} > 0$
For $x≠0$, all we need is to show
$(f_n(x) + ({n-1 \over 2})x-1) > 0$
Or, $f_n(x)$ is above its tangent line, at $x=0$
Is there a simple proof for decay function concaved up ?
Perhaps induction proof, without doing second derivative ?
|
For $x≠0$, all we need is to show
$(f_n(x) + ({n-1 \over 2})x-1) > 0$
Let $R=1+x>0$
$\displaystyle f_n(x) = \frac{n\;(R-1)}{R^n-1} \stackrel{?}{>}\; \left(1 - \left(\frac{n-1}{2}\right)\;(R-1)\right)$
Or, $\displaystyle \; n \stackrel{?}{>}
\left( \frac{R^n-1}{R-1} \right)
\left(1 - \left(\frac{n-1}{2}\right)\;(R-1)\right)$
We want to locate RHS extremum.
$\displaystyle \frac{d(RHS)}{dR} =
\frac{1 - ({n^2+n \over 2})\;R^{n-1} + (n^2\!-\!1)\;R^n - ({n^2-n \over 2})\;R^{n+1}} { (R-1)^2}$
From numerator, with $n>1$, we have 3 sign changes, thus 1 or 3 roots.
However, RHS is a polynomial; so does all its derivatives.
Numerator must have 3 roots, with 2 roots cancelled by denominator.
⇒ RHS has only 1 extremum, at the $x=0$ tangent.
All is require is to check curve above, or below tangent line.
For $x → ∞$:
Decay function $f_n(∞) = 0$
Tagent line: $1-\left({n-1 \over 2}\right)(∞) = -∞$
Thus, decay function above tangent line.
For $x≠0:\;(f_n(x) + ({n-1 \over 2})x - 1) > 0\qquad$QED
We can also move everything to one side, and expand polynomial.
$\begin{align}\displaystyle g_n(R) &=
\left( \frac{R^n-1}{R-1} \right)
\left(1 - \left(\frac{n-1}{2}\right)\;(R-1)\right) - n \\
&= -\left(\frac{n-1}{2}\right)\;R^n + R^{n-1} + R^{n-2} \;+\;...\;+\;R - \left(\frac{n-1}{2}\right)
\end{align}$
$g(1) = -\left(\frac{n-1}{2}\right) + (n-1) -\left(\frac{n-1}{2}\right)= 0$
$g'(1) = -n\left(\frac{n-1}{2}\right) + (n-1)+(n-2)\;+\;...\;+\;1= 0$
$g_n(R)$ have 2 sign changes, 2 repeated roots at R=1
⇒ Only 1 extremum, at the $\;x=0\;$ tangent line.
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/4475305",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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|
Explaining symmetry in sequences defined by $\frac{a_{n+1}}{n+1}=\frac{a_n}{n}-\frac12$. (Eg, for $a_1=5$, we get $5,9,12,14,15,15,14,12,9,5$) I've noticed recently the following fact: consider ratios $\frac{a_n}{n}$ with $n = 1,\dots,10$. If we require that
$$
\frac{a_{n+1}}{n+1} = \frac{a_n}{n} - \frac12
$$
that is, each successive ratio drops by $\frac12$ in value, then for $a_1 = 5$ we get $a_2 = 9$, $a_3 = 12$, $a_4 = 14$, $a_5 = 15$, $a_6 = 15$, $a_7 = 14$, $a_8 = 12$, $a_9 = 9$ and $a_{10} = 5$. That is, the numerators show a symmetric behvaior $a_{11 - n} = a_n$. This looks pretty random to me, so I wonder whether there's something obvious I've missed why such symmetry may arise, and perhaps a more general case where it happens exists.
|
$b_n = \dfrac{a_n}{n}\,$ is an arithmetic progression with starting value $\,a_1\,$ and common difference $\,-\dfrac{1}{2}\,$, so:
$$
b_n = \frac{2a_1 - n + 1}{2} \quad\implies\quad a_n = n \cdot \frac{2a_1 - n + 1}{2}
$$
For $\,2a_1 \in \mathbb N\,$, it follows that:
$$
\require{cancel}
\begin{align}
a_{2a_1+1 - n} &= (2a_1+1 - n) \cdot \frac{\cancel{2a_1} - (\cancel{2a_1}+\bcancel{1} - n) + \bcancel{1}}{2}
\\ &= n \cdot \frac{2a_1 - n + 1}{2}
\\ &= a_n
\end{align}
$$
OP's problem is the case $\,a_1=5\,$.
[ EDIT ] More generally, consider the recurrence $\,\dfrac{a_{n+1}}{n+1}=\dfrac{a_n}{n} - a\,$, then $b_n = \dfrac{a_n}{n}\,$ is an arithmetic progression with starting value $\,a_1\,$ and common difference $\,-a\,$, so:
$$
b_n = a_1 - (n - 1)a \quad\implies\quad a_n = n \cdot \left(a_1 - (n - 1)a\right)
$$
For $\,\dfrac{a_1}{a} \in \mathbb N\,$, it follows that:
$$
\require{cancel}
\begin{align}
a_{a_1/a + 1 - n} &= \left(\frac{a_1}{a} + 1 - n\right) \cdot \left(\cancel{a_1} - \left(\cancel{\frac{a_1}{a}} + \bcancel{1} - n - \bcancel{1}\right)a\right)
\\ &= n \cdot \left(a_1 - (n-1) a\right)
\\ &= a_n
\end{align}
$$
This reduces to the previous case when $\,a = \dfrac{1}{2}\,$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4476240",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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|
Volume of $y=x^2+1; y=-x^2+2x+5; x=0; x=3$ about $x$ axis (Shell Method). I was working on this exercise for an assignment. However, I get stuck in the following part.
$
y=-x^2+2x+5
$
Complete the square
$y=-(x^2-2x)+5$
$(b/2)^2=(-2/2)^2=1$
$y=-(x^2-2x+1-1)+5$
$y=-(x^2-2x+1)+5+1$
$y=-(x-1)^2+6$
$y-6=-(x-1)^2$
$\sqrt{6-y}+1=x$
For $x=0$, $\sqrt{6-y}+1=x$ has no real answer and I was wondering, is it possible to solve this with the Shell Method about the $x$-axis? I tried with Washer Method, and the expected volume is $277π/3$.
|
Yes, it can be solved through the shell method. Consider the picture below:
Let us first see what is the volume of the region for which $0\leqslant x\leqslant2$. The shell method tells us that it is equal to\begin{multline}2\pi\left(\int_1^5y\sqrt{y-1}\,\mathrm dy+\int_5^6y\left(\left(1+\sqrt{6-y}\right)-\left(1-\sqrt{6-y}\right)\right)\,\mathrm dy\right)\\=2\pi\left(\int_1^5y\sqrt{y-1}\,\mathrm dy+\int_5^62\sqrt{6-y}\,\mathrm dy\right)=\frac{152}3\pi.\end{multline}And the volume of the region for which $2\leqslant x\leqslant3$ is (again, by the shell method)$$2\pi\left(\int_2^5y\left(3-\left(1+\sqrt{6-y}\right)\right)\,\mathrm dy+\int_5^{10}y\left(3-\sqrt{y-1}\right)\,\mathrm dy\right)=\frac{125}3\pi.$$And $\dfrac{152}3\pi+\dfrac{125}3\pi=\dfrac{177}3\pi$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4476593",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Flux of $\mathbf{F}=(xz,yz,x^2+y^2)$ across paraboloid $z=4-x^2-y^2,\ z\geq 0$ I have calculated the flux of the vector field $\mathbf{F}=\begin{bmatrix}xz\\ yz\\ x^2+y^2\end{bmatrix}$ outward across the surface of the paraboloid $S$ given by $z=4-x^2-y^2,\ z\geq 0$ (with outward-pointing normal having positive $\mathbf{e}_3$ component but my answer doesn't agree with the one given by the book but I can't see what I am doing wrong. I would thus appreciate some feedback about what I am missing in my solution, thanks.
EDIT: there was an error in $D\mathbf{g}$ that's why the result was wrong. Now the calculation is correct.
What I have done:
We parametrize the surface by $\mathbf{g}:(0,2)\times (0,2\pi),\ \mathbf{g}\begin{pmatrix}r\\ \theta\end{pmatrix}=\begin{bmatrix}r\cos(\theta)\\ r\sin(\theta)\\ 4-r^2\end{bmatrix}$ so $D\mathbf{g}=\begin{bmatrix}\cos(\theta) & -r\sin(\theta)\\ \sin(\theta) & r\cos(\theta)\\ -2r & 0\end{bmatrix}$ and
\begin{align}
\text{flux}_{S}(\mathbf{F})&=\int_{S}\eta=\int_{(0,2)\times (0,2\pi)}\mathbf{g}^*\left(xz\ dy\wedge dz+yz\ dz\wedge dx+(x^2+y^2)\ dx\wedge dy\right)\\&=\int_{\theta=0}^{\theta=2\pi}\left(\int_{r=0}^{r=2}\left( r\cos(\theta)(4-r^2)(2r^2\cos(\theta))+r\sin(\theta)(4-r^2)(2r^2\sin(\theta))+r^2\cdot r \right)dr\right)d\theta\\&=\int_{\theta=0}^{\theta=2\pi} \left(\int_{r=0}^{r=2}(2r^3(4-r^2)+r^3dr\right)d\theta=\int_{\theta=0}^{\theta=2\pi} \left(\int_{r=0}^{r=2}(9r^3-2r^5)dr\right)d\theta\\ &=2\pi\cdot \left[\frac{9}{4}r^4-\frac{1}{3}r^6\right]_{r=0}^{r=2}=2\pi\left[36-\frac{64}{3}\right]=\frac{88\pi}{3}.
\end{align}
|
I got a different result,
$$\begin{align}
\iint_S \mathbf{F}\cdot d \mathbf{S}&=\iint_{x^2+y^2\leq 4}
\langle (xz,yz,x^2+y^2),(2x,2y,1)\rangle dx dy\\
&=\int_{0}^{2\pi}\int_{r=0}^2(2r^2(4-r^2)+r^2)r drd\theta\\
&=2\pi\int_{r=0}^2(9r^3-2r^5)\,dr=2\pi\left[9r^4/4-r^6/3\right]_0^2=\frac{88\pi}{3}.
\end{align}$$
The result is confirmed by applying the divergence theorem:
$$\iint_S \mathbf{F}\cdot d \mathbf{S}=
\iiint_D \text{div}(\mathbf{F})\,dxdydz-\iint_{S_2} \mathbf{F}\cdot d \mathbf{S}=\frac{64\pi}{3}-(-8\pi)=\frac{88\pi}{3}$$
where $D=\{(x,y,z):0\leq z \leq 4-x^2-y^2\}$ and $S_2=\{(x,y,0):x^2+y^2\leq 4\}$ is oriented downward.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4478000",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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|
Why are the two calculations of $ \int \frac{1}{\sqrt{x(1-x)}} \mathrm{d} x $ give two distinct answers? One of the calculations:
$$
\begin{aligned}
\int \frac{1}{\sqrt{x(1-x)}} \mathrm{d} x &=\int \frac{1}{\sqrt{1-(\sqrt{x})^{2}}} \cdot \frac{1}{\sqrt{x}} \mathrm{~d} x \\
&=2 \int \frac{1}{\sqrt{1-(\sqrt{x})^{2}}} \mathrm{~d}(\sqrt{x})=2 \arcsin \sqrt{x}+C .
\end{aligned}
$$
The other:
$$
\int \frac{1}{\sqrt{x(1-x)}} \mathrm{d} x = \int \frac{d\left(x-\frac{1}{2}\right)}{\sqrt{\left(\frac{1}{2}\right)^{2}-\left(x-\frac{1}{2}\right)^{2}}} = \arcsin \frac{x-1 / 2}{1 / 2}+C=\arcsin(2x-1)+C
$$
|
The answers are actually the same.
$\begin{align*}
\theta = 2\arcsin \sqrt x &\implies \sin \frac \theta 2 = \sqrt x \\
&\implies \sin^2 \frac \theta 2 = x \\
&\implies \frac{1-\cos \theta}{2} = x \\
&\implies \cos \theta = 1-2x \\
&\implies \sin(\theta+\pi/2) = 2x-1 \\
&\implies \theta = \arcsin(2x-1) - \pi/2
\end{align*}$
And $\pi/2$ gets absorbed into $+C$
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/4478651",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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|
Let $A$ be a diagonal matrix and $B$ be a Hermitian and Idempotent matrix, then is $(I + ABA^H)^{-1} = (I - \frac{1}{2}ABA^H)$? Let $A$ be a diagonal matrix and $B$ be a Hermitian and Idempotent matrix, then is $(I + ABA^H)^{-1} = (I - \frac{1}{2}ABA^H)$, where $I$ is an identity matrix?
I am reading a paper, which states the above equality. I do not understand how this is true. Can someone please enlighten me?
EDIT: Constraint $A^H A = I$ should also be considered
|
This can be shown by a direct computation as such:
\begin{align}
(I + ABA^H)(I - \frac{1}{2}ABA^H) &= I - \frac{1}{2}ABA^H + \frac{1}{2}(ABA^H)^2 \\
&\stackrel{\spadesuit}{=} I - \frac{1}{2}ABA^* + \frac{1}{2}AB^2A^* \\
&= I - \frac{1}{2}ABA^H + \frac{1}{2}ABA^H = I
\end{align}
Assuming that $A^HA = I$ at $\spadesuit$, otherwise equality isn't assured. See this example:
\begin{align}
&A = \begin{bmatrix}1 & 0 \\ 0 & 2\end{bmatrix} &&B = \frac{1}{2}\begin{bmatrix}1 & 1 \\ 1 & 1\end{bmatrix}
\end{align}
Then:
$$(I + ABA^H)(I - \frac{1}{2}ABA^H) = \begin{bmatrix}\frac{5}{8} & -\frac{3}{4} \\ -\frac{3}{4} & -\frac{1}{2}\end{bmatrix} \neq I$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4480616",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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|
Smith Normal forms in a polynomial ring
Let $M$ be a $\mathbb{C[x]}$ module with generators $m_1,m_2$ and relations : $$(x^2+ix)m_1+(x+i)m_2=0 \\ (-2x+2i)m_1+ (x^2+1)m_2=0 $$
Find integers $t,n_1,...,n_s \in \mathbb{N_0}$ and $\lambda_1,\lambda_2,...,\lambda_s\in \mathbb{C}$ such that the following holds: $$\mathbb{C[x]^t}\oplus\mathbb{C[x]}/((x-\lambda_1)^{n_1})\oplus...\oplus\mathbb{C[x]}/((x-\lambda_s)^{n_s}) \cong M $$
Unless my Work so far is completely wrong I started with the presentation matrix $\begin {pmatrix} x^2+ix & x+i \\-2x+2i & x^2+1\end {pmatrix}$ and by using column/row operations got to the point where $\begin {pmatrix} x+i & 4i \\0 & -x^3-3x+2i\end {pmatrix}$ but I haven't been able to get it into proper Smith normal form so I don't know how to proceed from this point.
|
You're very close to finishing it. Once you have a unit, like $4i$, you can rescale a row or column to make it $1$ and then begin canceling all the other entries in its row and column.
\begin{align*}
\begin {pmatrix} x+i & 4i \\0 & -x^3-3x+2i\end {pmatrix} &\leadsto
\left(\begin{array}{rr}
4 i & x + i \\
-x^{3} - 3 x + 2 i & 0
\end{array}\right)\\
&\leadsto
\left(\begin{array}{rr}
1 & -\frac{1}{4} i x + \frac{1}{4} \\
-x^{3} - 3 x + 2 i & 0
\end{array}\right)\\
&\leadsto
\left(\begin{array}{rr}
1 & -\frac{1}{4} i x + \frac{1}{4} \\
0 & -\frac{1}{4} i x^{4} + \frac{1}{4} x^{3} - \frac{3}{4} i x^{2} + \frac{1}{4} x - \frac{1}{2} i
\end{array}\right)\\
&\leadsto
\left(\begin{array}{rr}
1 & 0 \\
0 & -\frac{1}{4} i x^{4} + \frac{1}{4} x^{3} - \frac{3}{4} i x^{2} + \frac{1}{4} x - \frac{1}{2} i
\end{array}\right)\\
&\leadsto
\left(\begin{array}{rr}
1 & 0 \\
0 & x^{4} + i x^{3} + 3 x^{2} + i x + 2
\end{array}\right)
\end{align*}
|
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"url": "https://math.stackexchange.com/questions/4482158",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Prove that $|z_{1}+z_{2}| < |1+\overline{z_{1}}\cdot z_{2}|\;$ if $|z_{1}| < 1$ and $|z_{2}| < 1$. Given $z_{1},z_{2} \in \mathbb{C}$
Prove that $|z_{1}+z_{2}| < |1+\overline{z_{1}}\cdot z_{2}|\;$, if $|z_{1}| < 1$ and $|z_{2}| < 1$.
My Professor gave my classmates and I, the following answer:
$|z_{1} + z_{2}| \leq |z_{1} + z_{2}|\cdot |\overline{z_{1}}|\;$ $\Leftarrow$ The part I'm having trouble with
$|z_{1} + z_{2}|\cdot |\overline{z_{1}}| = |z_{1}\overline{z_{1}} + z_{2}\overline{z_{1}}| = ||z_{1}|^{2} + z_{2}\overline{z_{1}}| < |1+\overline{z_{1}}z_{2}|$
$\therefore |z_{1}+z_{2}| < |1+\overline{z_{1}}z_{2}|$
I don't understand why the line highlighted is correct.
Since $|\overline{z_{1}}| = |z_{1}| < 1$, $\;|\overline{z_{1}}|\cdot |z_{1} + z_{2}|< 1\cdot |z_{1} + z_{2}|$ which contradicts the proof above.
If my reasoning is correct, How could I prove this statement?
|
Square both sides:
\begin{align*}
|z + w| < |1 + \overline{z}w| & \Longleftrightarrow |z + w|^{2} < |1 + \overline{z}w|^{2}\\\\
& \Longleftrightarrow (z + w)(\overline{z} + \overline{w}) < (1 + \overline{z}w)(1 + z\overline{w})\\\\
& \Longleftrightarrow z\overline{z} + z\overline{w} + \overline{z}w + w\overline{w} < 1 + z\overline{w} + \overline{z}w + z\overline{z}w\overline{w}\\\\
& \Longleftrightarrow |z|^{2} + |w|^{2} < 1 + |z|^{2}|w|^{2}\\\\
& \Longleftrightarrow |z|^{2}(|w|^{2} - 1) + 1 - |w|^{2} > 0\\\\
& \Longleftrightarrow (1 - |z|^{2})(1 - |w|^{2}) > 0
\end{align*}
which clearly holds based on the assumption that $|z| < 1$ and $|w| < 1$.
Hopefully this helps!
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/4483122",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "2",
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|
$y''+y=x^2+1, y(\pi)=\pi^2, y'(\pi)=2\pi$ - By Laplace Transform I am solving the following IVP by Laplace Transform:
$$y''+y=x^2+1,\qquad y(\pi)=\pi^2, \qquad y'(\pi)=2\pi$$
Let $f(x)=u_{\pi}(x)y(x-\pi).$ Then,
$$f''(x)+f'(x)=u_{\pi}(x)(x-\pi)^2+u_{\pi}(x), \qquad f(0)=\pi ^2, \qquad f'(0)=2\pi.$$
Using the Laplace Transform and writing $F:=\mathcal{L}\{f\}(s)$, we have
$$s^2F-sf(0)-f'(0)+F=e^{-\pi s}\dfrac{s+1}{s^2},$$
$$(s^2+1)F-(s+1)\pi^2-2\pi=e^{-\pi s}\dfrac{s+1}{s^2},$$
$$F=e^{-\pi s}\dfrac{s+1}{s^2(s^2+1)}+\dfrac{(s+1)\pi^2}{(s^2 +1)}+\dfrac{2\pi}{(s^2+1)}.$$
Be
$$\dfrac{s+1}{s^2(s^2+1)}=\dfrac{As+B}{s^2}+\dfrac{Cs+D}{s^2+1}. $$
We have
$$As^3+As+Bs^2+B+Cs^3+Ds^2=s+1,$$
$$A+C=0, B+D=0, A=1, B=1,$$
$$C=-1, D=-1, A=1, B=1.$$
So,
$$\dfrac{s+1}{s^2(s^2+1)}=\dfrac{s+1}{s^2}-\dfrac{s+1}{s^2+1}= \dfrac{1}{s}+\dfrac{1}{s^2}-\dfrac{s}{s^2+1}-\dfrac{1}{s^2+1}.$$
Then,
$$F=e^{-\pi s}\left(\dfrac{1}{s}+\dfrac{1}{s^2}-\dfrac{s}{s^2+1}-\dfrac {1}{s^2+1}\right)+\pi^2\left(\dfrac{s}{s^2+1}+\dfrac{1}{s^2+1}\right)+ 2\pi\dfrac{1}{s^2+1}.$$
By the inverse transform,
$$f(x)=u_\pi(x)\left(1+(x-\pi)-\cos(x-\pi)-\sin(x-\pi)\right)+\pi^2 \left(\cos(x)+\sin(x)\right)+2\pi\sin(x).$$
Returning to the variable $y$ and remembering that $\cos(x+\pi)=-\cos(x)$ and $\sin(x+\pi)=-\sin(x)$, we have
$$y(x)=1+x-\cos(x)-\sin(x)+\pi^2(-\cos(x)-\sin(x))-2\pi \sin (x) .$$
However, the Wolfram's solution is $y(x)=-1 + x^2 - \cos(x)$.
Thank you in advance!
|
After Laplace transformation we have
$$
Y(s) = \frac{s^2+y_0s^4+s^3y'_0+2}{s^3(s^2+1)}
$$
with inverse
$$
y(x) = x^2+\sin (x) y'_0+y_0 \cos (x)+\cos (x)-1
$$
here $y_0$ and $y'_0$ are generic constants so imposing the initial conditions we have
$$
\cases{
y(\pi) = \pi^2-y_0-2=\pi^2\\
y'(\pi) = 2\pi-y'_0=2\pi
}
$$
and solving we have $y_0 = -2,\ y'_0 = 0$
hence
$$
y(x) = x^2-\cos x-1
$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4483871",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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|
When not to use $ \lim\limits_{x\rightarrow 0}\frac{\log\left( 1+x\right) }{x}=1 $ My Teacher has told to use this as a standard result so when question is in this I directly apply this result as in this
$$\lim\limits_{x\rightarrow 0}\dfrac{\log\left( 5+x\right) }{x}-\dfrac{\log\left( 5-x\right) }{x}=2/5$$
but here
$$\lim\limits_{x\rightarrow 0}\dfrac{\log\left( 1+2x\right) -2{\log\left( 1+x\right)} }{x^2}=0 $$
doesn't work because solution is give using expansion of $\log\left( 1+x\right)$
I am interested to know why we cant simply multiply and divide 2x to log(1+2x) to convert it in known form like log(1+2x) / 2x *2x yielding 2x
So what is limitation of result when to use it and when to not?
|
The proposed result is not valid.
Having said that, you may be also interested in the power series method:
\begin{align*}
\ln(1 + x) = x - \frac{x^{2}}{2} + \frac{x^{3}}{3} - \ldots
\end{align*}
whenever $|x| < 1$. If we restrict the values of $x$ to $(-1/2,1/2)$, we may claim as well that
\begin{align*}
\ln(1 + 2x) = 2x - 2x^{2} + \frac{8x^{3}}{3} - \ldots
\end{align*}
Therefore the proposed limit is given by
\begin{align*}
\lim_{x\to 0}\frac{\ln(1 + 2x) - 2\ln(1 + x)}{x^{2}} & = \lim_{x\to 0}\frac{2x - 2x^{2} - 2x + x^{2} + o(x^{2})}{x^{2}} = \lim_{x\to 0}\frac{-x^{2} + o(x^{2})}{x^{2}} = -1
\end{align*}
Hopefully this helps!
|
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"timestamp": "2023-03-29T00:00:00",
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|
Limit question involving the the greatest integer function and fractional part The value of $$\mathop {\lim }\limits_{n \to \infty } {\left( {\frac{{\sin \left\{ {\frac{2}{n}} \right\}}}{{\left[ {2n\tan \frac{1}{n}} \right]\left( {\tan \frac{1}{n}} \right)}} + \frac{1}{{{n^2} + \cos n}}} \right)^{{n^2}}},$$ where $[.]$ denotes greatest integer function and $\{.\}$ denotes fractional part function, is _________.
My approach is as follow $\mathop {\lim }\limits_{n \to \infty } \sin \left\{ {\frac{2}{n}} \right\} = {0^ + };\mathop {\lim }\limits_{n \to \infty } \left[ {2n\tan \frac{1}{n}} \right] = \mathop {\lim }\limits_{n \to \infty } \left[ {2\frac{{\tan \frac{1}{n}}}{{\frac{1}{n}}}} \right] = \mathop {\lim }\limits_{n \to \infty } \left[ {{2^ + }} \right] = 2$
$\mathop {\lim }\limits_{n \to \infty } \tan \frac{1}{n} = 0;\mathop {\lim }\limits_{n \to \infty } \frac{1}{{{n^2} + \cos n}} = \mathop {\lim }\limits_{n \to \infty } \frac{1}{{n\left( {1 + \frac{{\cos n}}{n}} \right)}} = 0.$
Cannot approach from here.
|
Recall the Maclaurin series of $\sin x$ and $\tan x$:
$$\sin x= x-\frac{x^3}6+O(x^5),\qquad x\to0,$$
$$\tan x= x+\frac{x^3}3+O(x^5),\qquad x\to0.$$
As a result,
$$\sin\left\{\frac 2n\right\}=\sin\left(\frac2n\right)=\frac2n-\frac4{3n^3}+O\left(\frac1{n^5}\right),\qquad n\to\infty,$$
$$\tan\left(\frac1n\right)=\frac1n+\frac1{3n^3}+O\left(\frac1{n^5}\right),\qquad n\to\infty.$$
Since
$$2n\tan\left(\frac1n\right)=2+\frac2{3n^2}+O\left(\frac1{n^4}\right),\qquad n\to\infty,$$
we know $\left[2n\tan\left(\frac1n\right)\right]=2$ for $n$ sufficiently large. Therefore, as $n\to\infty$, we have
\begin{align*}
\frac{{\sin \left\{ {\frac{2}{n}} \right\}}}{{\left[ {2n\tan \frac{1}{n}} \right]\left( {\tan \frac{1}{n}} \right)}}&=\frac{\frac2n-\frac4{3n^3}+O\left(\frac1{n^5}\right)}{\frac2n+\frac2{3n^3}+O\left(\frac1{n^5}\right)}\\
&=\frac{1-\frac2{3n^2}+O\left(\frac1{n^4}\right)}{1+\frac1{3n^2}+O\left(\frac1{n^4}\right)}\\
&=\left(1-\frac2{3n^2}+O\left(\frac1{n^4}\right)\right)\left(1-\frac1{3n^2}+O\left(\frac1{n^4}\right)\right)\\
&=1-\frac1{n^2}+O\left(\frac1{n^4}\right).
\end{align*}
Now, using $\log(1+x)\sim x$ as $x\to 0$ gives that, as $n\to\infty$
\begin{align*}
n^2\log\left( {\frac{{\sin \left\{ {\frac{2}{n}} \right\}}}{{\left[ {2n\tan \frac{1}{n}} \right]\left( {\tan \frac{1}{n}} \right)}} + \frac{1}{{{n^2} + \cos n}}} \right)&=n^2\log\left(1-\frac1{n^2}+\frac1{n^2+\cos n}+O\left(\frac1{n^4}\right)\right)\\
&=n^2\left(-\frac1{n^2}+\frac1{n^2+\cos n}+O\left(\frac1{n^4}\right)\right)\\
&=-1+\frac{n^2}{n^2+\cos n}+O\left(\frac1{n^2}\right);
\end{align*}
Since $\frac{n^2}{n^2+1}\leq\frac{n^2}{n^2+\cos n}\leq \frac{n^2}{n^2-1}$, by squeezing we have $\lim_{n\to\infty}\frac{n^2}{n^2+\cos n}=1$, and thus
$$\lim_{n\to\infty}n^2\log\left( {\frac{{\sin \left\{ {\frac{2}{n}} \right\}}}{{\left[ {2n\tan \frac{1}{n}} \right]\left( {\tan \frac{1}{n}} \right)}} + \frac{1}{{{n^2} + \cos n}}} \right)=0.$$
Therefore,
$$\lim_{n\to\infty}{\left( {\frac{{\sin \left\{ {\frac{2}{n}} \right\}}}{{\left[ {2n\tan \frac{1}{n}} \right]\left( {\tan \frac{1}{n}} \right)}} + \frac{1}{{{n^2} + \cos n}}} \right)^{{n^2}}}=1.$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4484674",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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|
What am I doing wrong in integrating $\frac1{\sqrt{x} + \sqrt[4]{x}}$ $\displaystyle\int \dfrac{1}{\sqrt{x} + \sqrt[4]{x}}\mathrm dx$
Putting $x = t^4$ and $dx=4t^3\mathrm dt$,
$$ = \int \dfrac{4 t^3 \mathrm dt}{t^2 + 1}\\ = 4\int \dfrac{t^3 + t - t \mathrm dt}{t^2 + 1} \\= 4\int \dfrac{t^3 + t}{t^2 + 1}\mathrm dt - 4\int \dfrac{t }{t^2 + 1}\mathrm dt\\= 4\int \dfrac{t(t^2 + 1)}{t^2 + 1}\mathrm dt - 2\int \dfrac{2t }{t^2 + 1}\mathrm dt$$
$$= 4\int t\ \mathrm dt - 2\ln|t^2 + 1| + C$$
$$= 2t^2 - 2\ln|t^2 + 1| + C \\= \color{red}{2\sqrt{x} - 2\ln|\sqrt{x} + 1| + C}$$
|
The problem lies in the first step. After the substitution, you should have obtained$$\int\frac{4t^3}{t^2+t}\,\mathrm dt\left(=4\int t-1+\frac1{t+1}\,\mathrm dt\right).$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4485266",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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|
when ring with roots is a field I encountered the following problem in abstract algebra textbook. To determine if the set
$$\{a+b\times 3^{1/3}+c \times 3^{2/3},\quad a,b,c \in \mathbb{Q}\}$$
is a ring, and, if so, a field as well.
I think it's clearly a ring because the product of two such element is also in the set, and each element has additive inverse in the set, for example.
Based on experimenting, I think it is also a field. For example,
$$\frac{1}{1-3^{1/3}} = -\frac{1}{2}-\frac{1}{2}3^{1/3}-\frac{1}{2}3^{2/3},$$
which shows that $(1-3^{1/3})$ has multiplicative inverse.
However, I do not know how to prove it. The interesting this is that if I set up the above problem as a linear system, the determinant of the coefficient matrix is zero, but the solution of course is $a = b = c = -1/2$. So perhaps linear algebra not the correct way to approach linear combinations like these.
Please help me understand this if you can.
|
Hint: $\,$ let $\,\omega\,$ be a primitive cube root of unity so that $\,\omega^3=1\,$ and $\,\omega^2+\omega+1=0\,$, then:
$$
(a + b t + ct^2)(a + b\omega t + c\omega^2 t^2)(a + b \omega^2 t + c\omega t^2) = c^3\, t^6 + (b^3 - 3 a b c)\, t^3 + a^3
$$
With $\,t = \sqrt[3]{3}\,$:
$$
\frac{1}{a + b \sqrt[3]{3} + c \sqrt[3]{9}} = \frac{1}{9 c^3 + 3(b^3 - 3 a b c) + a^3}(a + b\omega \sqrt[3]{3} + c\omega^2 \sqrt[3]{9})(a + b \omega^2 \sqrt[3]{3} + c\omega \sqrt[3]{9})
$$
[ EDIT ] $\,$ Worked out example for the inverse of $\,1 - \sqrt[3]{3}\,$ $\iff a=1, b=-1, c=0\,$.
*
*$9 c^3 + 3(b^3 - 3 a b c) + a^3 = 3 \cdot (-1) + 1 = -2$
*$(a + b\omega \sqrt[3]{3} + c\omega^2 \sqrt[3]{9})(a + b \omega^2 \sqrt[3]{3} + c\omega \sqrt[3]{9}) = (1-\omega\sqrt[3]{3})(1-\omega^2\sqrt[3]{3})\\ = 1 - (\omega+\omega^2) \sqrt[3]{3} + \sqrt[3]{9} = 1 + \sqrt[3]{3} + \sqrt[3]{9}$
Then: $\displaystyle\quad
\frac{1}{1-\sqrt[3]{3}} = \frac{1}{-2}\left(1 + \sqrt[3]{3} + \sqrt[3]{9}\right)\,$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4486341",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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|
Evaluate $\underset{z=0}{\text{Res}} \; \frac{(z^6-1)^2}{z^5(2z^4 -5z^2 + 2)}$ . Problem:
Evaluate
$$
\underset{z=0}{\text{Res}} \; \frac{(z^6-1)^2}{z^5(2z^4 -5z^2 + 2)}.
$$
My question:
This is a question from a previous complex analysis qualifying exam that I am trying to work through. I know that the straightforward formula for calculating this residue would be
$$
\frac{1}{4!} \lim_{z\to 0} \frac{d^4}{dz^4} \left(z^5 \cdot \frac{(z^6-1)^2}{z^5(2z^4 -5z^2 + 2)} \right)
$$
However, I'm guessing there is a better method in this case than taking four derivatives or trying to turn this into a Laurent series. Can someone point me in the right direction here?
|
A variation: It is convenient to use the coefficient of operator $[z^n]$ to denote the coefficient of $z^n$ of a series.
We obtain
\begin{align*}
\color{blue}{\underset{z=0}{\text{Res}}}&\color{blue}{ \; \frac{(z^6-1)^2}{z^5(2z^4 -5z^2 + 2)}}
=[z^{-1}]\frac{\left(z^6-1\right)^2}{z^5\left(2z^4-5z^2+2\right)}\\
&=[z^{4}]\frac{\left(z^6-1\right)^2}{2z^4-5z^2+2}\tag{1}\\
&=\frac{1}{2}[z^{4}]\frac{1}{1+\left(z^4-\frac{5}{2}z^2\right)}\tag{2}\\
&=\frac{1}{2}[z^{4}]\left(1-\left(z^4-\frac{5}{2}z^2\right)+\left(z^4-\frac{5}{2}z^2\right)^2\right)\tag{3}\\
&=\frac{1}{2}\left(-1+\frac{25}{4}\right)\tag{4}\\
&\,\,\color{blue}{=\frac{21}{8}}
\end{align*}
in accordance with other given answers.
Comment:
*
*In (1) we apply the rule $[z^{p-q}]A(z)=[z^p]z^qA(z)$. Here we have $$[z^{-1}]\frac{P(z)}{z^5Q(z)}=[z^{-1}]z^{-5}\frac{P(z)}{Q(z)}=[z^4]\frac{P(z)}{Q(z)}$$
*In (2) we take only the constant term $1$ from the numerator, since other terms do not contribute to $[z^4]$. We also factor out $2$ from the denominator as preparation for the geometric series expansion in the next step.
*In (3) we expand the series up to the third term, since other terms do not contribute to $[z^4]$.
*In (4) we select the coefficient of $z^4$.
Note: The usage of the coefficient of operator $[z^n]$ can be found for instance in section 5.4 in Concrete Mathematics by R.L. Graham, D. Knuth and O. Patashnik or in Generatingfunctionology by H.S. Wilf.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Finding a closed form of $f(n) = \frac{n-4}{n}f(n-1)+1\ ,\ n \geq 5$ where $f(4) = 1$. I'm trying to find the closed form of
$f(n) = \frac{n-4}{n}f(n-1)+1\ ,\ n \geq 5$ where $f(4) = 1$.
Empirically, it turns out the answer is simply $f(n)=1+\frac{n-4}{5}\ ,\ n \geq 4$, but I'm having a hard time getting there. I've tried two ways but neither is very successful. Could someone please suggest a way through?
Attempt 1: I try to set up a generating function
$$G(x)=f(4)x^4+f(5)x^5+\dots=\sum_{k=4}^\infty f(k)x^k$$
Substituting in the recurrence (edit: noticed I forgot to add the +1)
$$\begin{align}G(x)&=x^4+\sum_{k=5}^\infty(1-\frac{4}{k})f(k-1)x^k \\
&=x^4+x\sum_{k=5}^\infty (1-\frac{4}{k}) f(k-1)x^{k-1} \\
&= x^4+x\sum_{k=4}^\infty (1-\frac{4}{k+1})f(k)x^k \\
&= x^4+xG(x)-x\sum_{k=4}^\infty \frac{4}{k+1}f(k)x^k
\end{align} $$
But I don't know how to express the last part in terms of G(x).
Attempt 2: Just brute force it:
With arbitrary $k$,
$$\begin{align}f(n) &= \frac{(n-4)(n-5) \dots (n-(k+3))}{n(n-1)\dots(n-(k-1))}f(n-k)\\
&\ \ + \left(1+\frac{n-4}{n}+\frac{(n-4)(n-5)}{n(n-1)}+\dots+\frac{(n-4)\dots(n-(k+2))}{n\dots(n-(k-2))}\right)\end{align}$$
Setting $n-k=4 \implies k = n -4$ and using $f(4)=1$,
$$\begin{align} f(n) &= \frac{(n-4)!}{n!/4!}+\left(1+\frac{n-4}{n}+\frac{(n-4)(n-5)}{n(n-1)}+\dots+\frac{(n-4)\dots2}{n \dots 6}\right) \\
&= {n \choose 4}^{-1}+\left(1+\frac{(n-4)!}{n!}\left(\frac{(n-1)!}{(n-5)!} + \frac{(n-2)!}{(n-6)!}+\dots +\frac{5!}{1!}\right)\right)
\\
&= {n \choose 4}^{-1}+\frac{(n-4)!}{n!}\sum_{k=1}^{n-5}\frac{(k+4)!}{k!} \\
&= {n \choose 4}^{-1}+\frac{1}{n(n-1)(n-2)(n-3)}\sum_{k=1}^{n-5}(k+4)(k+3)(k+2)(k+1) \\
&= {n \choose 4}^{-1}+\frac{1}{n(n-1)(n-2)(n-3)}\sum_{k=5}^{n-1}k(k-1)(k-2)(k-3)\end{align}$$
Empirically this seems to be correct, but even Mathematica refuses to simplify it all the way down. Surely there is a better way?
|
Hint
If you solve
$$f(n) = \frac{n-4}{n}f(n-1)+1$$ the solution is given by
$$f(n)=\frac {n+1} 5+\frac C{(n-3) (n-2) (n-1) n}$$ Then, ..., try to find the mistake.
|
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"timestamp": "2023-03-29T00:00:00",
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|
$\left\lfloor\frac{8n+13}{25}\right\rfloor-\left\lfloor\frac{n-12-\left\lfloor\frac{n-17}{25}\right\rfloor}{3}\right\rfloor$ is independent of $n$
If $n$ is a positive integer, prove that
$$\left\lfloor\frac{8n+13}{25}\right\rfloor-\left\lfloor\frac{n-12-\left\lfloor\frac{n-17}{25}\right\rfloor}{3}\right\rfloor$$
is independent of $n$.
Taking
$$f(n)=\left\lfloor\frac{8n+13}{25}\right\rfloor-\left\lfloor\frac{n-12-\left\lfloor\frac{n-17}{25}\right\rfloor}{3}\right\rfloor$$
I could prove that
$$f(n+25)=f(n)$$
But, I have no idea how to show
$$f(n+k)=f(n)\\\forall k\in \{1,2,\dots 24\}$$
Of course, we can check these finite number of cases by force. But is there any other elegant method to handle this?
|
Here's an approach that involves checking considerably fewer than $24$ cases, with it also having the advantage that it helps to provide a somewhat intuitive explanation of why the relation holds.
First, note that $\left\lfloor\frac{n-12-\left\lfloor\frac{n-17}{25}\right\rfloor}{3}\right\rfloor = \left\lfloor\frac{n-\left\lfloor\frac{n-17}{25}\right\rfloor}{3}\right\rfloor - 4$. Since $-4$ doesn't affect whether or not your $f(n)$ is always a constant, I'll ignore it. Next, set
$$g(n) = \left\lfloor\frac{8n+13}{25}\right\rfloor, \; \; h(n) = \left\lfloor\frac{n-\left\lfloor\frac{n-17}{25}\right\rfloor}{3}\right\rfloor, \; \; f_1(n) = g(n) - h(n) \tag{1}\label{eq1A}$$
As you've indicated, $f(n)$ (and, thus, my $f_1(n)$) has a period of $25$. Thus, we just need to prove that $f_1(n)$ doesn't change for any consecutive set of $25$ integers, so induction can then be used for any larger values, & also smaller values, to show $f_1(n)$ is constant for all integers, not only just the positive ones.
Note the numerators inside the floor functions in $g(n)$ and $h(n)$ are both increasing functions, both $g(n)$ and $h(n)$ are integral valued step functions, and the increase of each step is just one. I'll show that $f_1(n)$ is a constant for $0 \le n \le 24$ by checking the end-points and each value of $n$ where $h(n)$ increases. At those points, I'll show that
$$8n + 13 = 25(h(n)) + r, \; \; 0 \le r \le 12 \tag{2}\label{eq2A}$$
to confirm $g(n) = h(n)$ and $g(n-1) = g(n) - 1$.
For $0 \le n \le 16$, we have $h(n) = \left\lfloor\frac{n+1}{3}\right\rfloor$, and for $17 \le n \le 24$, we have $h(n) = \left\lfloor\frac{n}{3}\right\rfloor$. Next, $g(0) = h(0) = 0$ and
$$h(2) = 1, \; 8(2) + 13 = 25(1) + 4 \tag{3}\label{eq3A}$$
For $n$ up to $16$, the value of $h(n)$ increases by $1$ for each increase in $n$ by $3$, with $8n + 13$ increasing by $8(3) = 25 - 1$ each time. This means $g(n)$ also increases by $1$, but with the value of $r$ decreasing by $1$ each time, starting from $4$ as indicated in \eqref{eq2A}. Thus, we get a match up to $h(14) = 5, \; 8(14) + 13 = 25(5) + 0$.
The next $h(n)$ increase is at $n = 18$, with $h(18) = 6, \; 8(18) + 13 = 25(6) + 7$. Thus, as discussed above, we'll continue to have matching increases with $g(21) = h(21)$ and $g(24) = h(24)$. This confirms $g(n) = h(n)$ for $0 \le n \le 24$.
Note: Since the remainder $r$ is $0$ at $n = 14$, the next increase in $h(n)$ must occur when $n$ increases by $4$ instead of $3$, with this accomplished by the increase in value of $\left\lfloor\frac{n-m}{25}\right\rfloor$ at $n = m$ for any $15 \le m \le 17$, not just $m = 17$ as used in the question.
|
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|
Linear upper bound for function involving Lambert W I have a function
$$f(x)=\frac{2-2x}{\ln x}W_{-1}\left(\frac{\ln x}{(2-2x)\sqrt x}\right)$$
Looking at the plot, it appears somewhat linear in shape and empirically, it seems that we have $\lim_{x\to\infty}f(x)/x=3$. Can this limit be proven, and if so, is there a simpler function that can act as an upper bound for $f(x)$? There is a related problem in which I need to check all integer values less than or equal to $f(x)$ for a given $x$ and I would like to find a simpler function for an upper bound that is still closely fitting enough to not drastically increase computation time.
Update: I have found a function that supplies an upper bound.
$$f(x)\leq2x-2+\frac{(4-4x)[\ln(\ln x)-\ln(x-1)+1-\ln4]}{\ln x}$$
I found this with the observation that $xe^x\geq\frac4{xe^2}$ and thus that $W_{-1}(x)\geq2-\ln4+2\ln(-x)$. Does this help prove the limit and is there a better upper bound still?
|
Fact 1: It holds that, for all $x > 2$,
$$f(x) \le 3x \cdot \frac{(x - 1)(3\ln x + 2\ln 3 - 2)}{3x\ln x - 2x + 2}.$$
Fact 2: It holds that $f(x) \ge 3x$ for all $x > 2$.
By Facts 1-2, using the squeeze theorem, we have
$$\lim_{x\to \infty} \frac{f(x)}{3x} = 1.$$
Proof of Fact 1:
Let
$$y := \frac{1}{3x} \cdot \frac{2-2x}{\ln x}W_{-1}\left(\frac{\ln x}{(2-2x)\sqrt x}\right).$$
Then we have
$$\frac{3xy\ln x}{2 - 2x} = W_{-1}\left(\frac{\ln x}{(2-2x)\sqrt x}\right)$$
or
$$\frac{3xy\ln x}{2 - 2x}
\mathrm{exp}\left({\frac{3xy\ln x}{2 - 2x}}\right)
= W_{-1}\left(\frac{\ln x}{(2-2x)\sqrt x}\right)\mathrm{exp}\left(W_{-1}\left(\frac{\ln x}{(2-2x)\sqrt x}\right)\right)$$
or
$$\frac{3xy\ln x}{2 - 2x}\,
\mathrm{exp}\left({\frac{3xy\ln x}{2 - 2x}}\right) = \frac{\ln x}{(2-2x)\sqrt x}$$
or
$$y\,\mathrm{exp}\left({\frac{3xy\ln x}{2 - 2x}}\right)
= \frac{1}{3x^{3/2}}$$
or
$$\ln y + \frac{3xy\ln x}{2 - 2x} = -\ln (3x^{3/2}). \tag{1}$$
Using $\ln y \le y - 1$ for all $y > 0$, we have
$$y - 1 + \frac{3xy\ln x}{2 - 2x} \ge -\ln (3x^{3/2})$$
which results in
$$y \le \frac{(x - 1)(3\ln x + 2\ln 3 - 2)}{3x\ln x - 2x + 2}.$$
Thus, we have
$$f(x) \le 3x \cdot \frac{(x - 1)(3\ln x + 2\ln 3 - 2)}{3x\ln x - 2x + 2}.$$
We are done.
$\phantom{2}$
Proof of Fact 2:
Let
$$z := W_{-1}\left(\frac{\ln x}{(2-2x)\sqrt x}\right).$$
According to definition, we have $z < -1$ and
$$z \mathrm{e}^{z} = \frac{\ln x}{(2-2x)\sqrt x}. $$
It suffices to prove that
$$\frac{1}{3x} \cdot \frac{2-2x}{\ln x} z \ge 1$$
or
$$z \le \frac{3x\ln x}{2 - 2x}.$$
Since $\frac{3x\ln x}{2 - 2x} < -1$ and $z\mapsto z\mathrm{e}^z$ is strictly decreasing on $(-\infty, -1)$, it suffices to prove that
$$z \mathrm{e}^z \ge \frac{3x\ln x}{2 - 2x}\mathrm{exp}\left(\frac{3x\ln x}{2 - 2x}\right)$$
or
$$\frac{\ln x}{(2-2x)\sqrt x} \ge \frac{3x\ln x}{2 - 2x}\mathrm{exp}\left(\frac{3x\ln x}{2 - 2x}\right)$$
or
$$\frac{1}{3x^{3/2}} \le \mathrm{exp}\left(\frac{3x\ln x}{2 - 2x}\right)$$
or
$$-\ln 3 - \frac32\ln x \le \frac{3x\ln x}{2 - 2x}$$
or
$$\ln x\le \frac23(x - 1)\ln 3$$
which is true (easy).
We are done.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Polynomial division by polynomial without remainder I have two polynomials:
$$
Q(x)=x^{624} + x^{524} + x^{424} + x^{324} + x^{224} + x^{124} + x^{n}
$$
$$
P(x) = x^{6} + x^{5} + x^{4} + x^{3} + x^{2} + x + 1
$$
And the question: what is the largest natural value of the number $n \leq 100 $ at which the polynomial $Q(x)$ is divisible without reminder by $P(x)$?
I have no idea about solution, I even don't know what area of math it is.
|
First, using the geometric series formula, it is easy to see that
$$
P(x) = \sum_{i=0}^6 x^i = \frac{x^7-1}{x-1}
$$
If we let $H(x) = Q(x)/P(x)$ for some polynomial $H(x)$, then rearranging the equation gives the following:
$$
(x-1)Q(x) = (x^7-1)H(x)\\
x^{625}-x^{624} + x^{525}-x^{524} + \dots+ x^{125}-x^{124} + x^{n+1}-x^n = (x^7-1)H(x)
$$
Notice that we can reduce the powers in the left side. Pick any of the terms, for example, $x^{625}$. Then, by letting $H(x) = x^{617} + J(x)$, we get
$$
x^{625}-x^{624} + x^{525}-x^{524} + \dots+ x^{125}-x^{124} + x^{n+1}-x^n = x^{625}-x^{618} + (x^7-1)J(x)\\
x^{618}-x^{624} + x^{525}-x^{524} + \dots+ x^{125}-x^{124} + x^{n+1}-x^n = (x^7-1)J(x)
$$
The right side of the equation hasn't changed since $J(x)$ is a polynomial just like $H(x)$. So, we managed to reduce the term $x^{625}$ to $x^{618}$. Since nothing is special about $x^{625}$, we could have reduced the power of each term in the left side similarly. And every time we do so, the power of the term decreases by 7.
Once we realize this, we can keep decreasing the powers of the terms until they cancel out. For example, notice that $625\equiv 324\pmod 7$, so if we decrease decrease the power of $x^{625}$ $ 43$ times, we would get $x^{324}$, which will cancel out with $-x^{324}$. Similarly, $x^{525}$ will cancel out with $-x^{224}$, $x^{425}$ will cancel out with $-x^{124}$, $-x^{624}$ will cancel out with $x^{225}$, $-x^{524}$ will cancel out with $x^{125}$. We are left with $-x^{424} + x^{325} + x^{n+1}-x^{n}$. We need to find the largest $n<100$ to make everything cancel out. Since $424\equiv 4\pmod 7$ and $325\equiv 3\pmod 7$, we need $n+1\equiv 4\pmod 7$, $n\equiv 3\pmod 7$. And the largest $n<100$ with a remainder of $3$ when divided by $7$ is $94$.
|
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|
show that $0\le \frac a{1+b}+\frac b{1+a} \le 1$ Let $0\le a < b \le 1$
prove that $0\le \frac a{1+b}+\frac b{1+a} \le 1$
This is the answer from book :
Obviously $0\le \frac a{1+b}+\frac b{1+a}$ so :
\begin{align}
&1\ge \frac a{1+b}+\frac b{1+a} \\
\iff & (1+a)(1+b) \ge a(1+a)+b(1+b)\\
\iff & 1-a^2\ge b^2-ab \\
\iff &(1-a)(1+a)\ge b(b-a) \tag{*}
\\ \iff & 1+a\ge b-a
\end{align}
which is obvious. Q.E.D.
But I can't understand how we get last inequality from $(*)$.
|
I guess this is a typo, they meant to say
$$ 1-a \ge b-a,$$
which is true from
$$ (1-a)(1+a) \ge b(b-a)$$
since $1+a \ge 1 \ge b$.
(Of course, $1-a\ge b-a$ is the same as $1\ge b$)
|
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|
Asymptotic expansion of solutions of $x = \tan\left(x\right)$ I'm working on having an asymptotic expansion of the solution of $\tan\left(x\right)=x$ for $x \in I_n$ where $I_n = \left]-\frac{\pi}{2}+n\pi ; \frac{\pi}{2} + n\pi\right[$. Let's note $x_n$ this solution.
I've used that
$$
\tan\left(x_n\right)=x_n \Leftrightarrow x_n = \tan_n^{-1}\left(x_n\right)
$$
where $\tan_n^{-1}$ denotes the inverse function of $\tan$ on $I_n$. I've shown that
$$
\tan_n^{-1}\left(x\right) = \text{tan}_0^{-1}\left(x\right)+n\pi
$$
And then i've used that
$$
x_n = n\pi+\left(\frac{\pi}{2}-\text{tan}_0^{-1}\frac{1}{x_n}\right)=n\pi + \frac{\pi}{2}-\tan_0^{-1}\left(\frac{1}{n\pi+o\left(n\right)}\right)
$$
and by expanding the classical $\tan_0^{-1}$ I've got
$$
x_n \underset{(+\infty)}{=}n\pi + \frac{\pi}{2}-\frac{1}{n\pi}+\frac{1}{3}\frac{1}{n^3\pi^3}+o\left(\frac{1}{n^3}\right)
$$
Is my expansion true ? I explain : In a "famous" french website, I've encountered this question :
This suggests that my expansion is wrong as a second order term is missing. I'm wondering if my reasoning is wrong... Any help ?
|
Say that you are looking for the $n^{\text{th}}$ zero of function
$$f(x)=x-\tan(x)$$ which is discontinuous at any $x=(2 n+1)\frac{\pi }{2}$. Since this cannot be a root, consider instead
$$g(x)=x\cos(x)-\sin(x)$$ and expand as series around $x=(2 n+1)\frac{\pi }{2}$.
This gives
$$g(x)=-(-1)^n-\frac{1}{2} \left(\pi (-1)^n (2 n+1)\right) \left(x-\pi
\left(n+\frac{1}{2}\right)\right)-\frac{1}{2} (-1)^n \left(x-\pi
\left(n+\frac{1}{2}\right)\right)^2+\frac{1}{12} \pi (-1)^n (2 n+1) \left(x-\pi
\left(n+\frac{1}{2}\right)\right)^3+O\left(\left(x-\pi
\left(n+\frac{1}{2}\right)\right)^4\right)$$
Use series reversion to obtain
$$x_n=\pi \left(n+\frac{1}{2}\right)-\frac{2 (-1)^{-n} \left(g(x)+(-1)^n\right)}{\pi (2
n+1)}-\frac{4 (-1)^{-2 n} \left(g(x)+(-1)^n\right)^2}{\pi ^3 (2 n+1)^3}-\frac{4
\left((-1)^{-3 n} \left(4 \pi ^2 n^2+4 \pi ^2 n+\pi ^2+12\right)\right)
\left(g(x)+(-1)^n\right)^3}{3 \left(\pi ^5 (2
n+1)^5\right)}+O\left(\left(g(x)+(-1)^n\right)^4\right)$$ Since you want $g(x)=0$, this gives
$$x_n=q-\frac{1}{q}-\frac{2}{3 q^3}-\frac{1}{2
q^5}+O\left(\frac{1}{q^6}\right)\qquad \text{with} \qquad q=(2 n+1)\frac{\pi }{2}$$
Now, if you want to expand the above, you effectively have
$$x_n=\pi n+\frac{\pi }{2}-\frac{1}{\pi n}+\frac{1}{2 \pi n^2}-\frac{8+3 \pi ^2}{12
\pi ^3 n^3}+\frac{8+\pi ^2}{8 \pi ^3 n^4}-\frac{8+16 \pi ^2+\pi ^4}{16 \pi ^5 n^5}+O\left(\frac{1}{n^6}\right)$$
|
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|
Understanding the alternate method for evaluating the number of ways for below question Number of positive ($0$ excluded) integral solutions which contains only $1,2,3$ as variable values to the equation $a_1 +a_2 + a_3 + \ldots + a_6 = 12$ are?
My method was this: It's equivalent to finding coefficient of $x^{12}$ in expansion of $(x+x^2 +x^3)^6$, from that I managed to attain the value as $141$, but I saw a method which I don't get how they managed to do it. They wrote this : $${11 \choose 2} - 5{8 \choose 2} + 15{5 \choose 2} - 15{2 \choose 2} = 141$$ cases. It might be related to PIE but I am not getting how. Can anyone elaborate how they are doing it?
|
As JMoravitz indicated in the comments, the correct expression should be
$$\binom{11}{5} - \binom{6}{1}\binom{8}{5} + \binom{6}{2}\binom{5}{5}$$
A particular solution of the equation
$$a_1 + a_2 + a_3 + a_4 + a_5 + a_6 = 12 \tag{1}$$
corresponds to placing $6 - 1 = 5$ addition signs in the $12 - 1 = 11$ spaces between successive ones in a row of $12$ ones. For instance,
$$1 + 1 1 + 1 1 1 + 1 1 1 1 + 1 + 1$$
corresponds to the solution $x_1 = 1, x_2 = 2, x_3 = 3, x_4 = 4, x_5 = x_6 = 1$. The number of such solutions is
$$\binom{12 - 1}{6 - 1} = \binom{11}{5}$$
since we must select which five of the eleven spaces between successive ones in which to place an addition sign.
The number of solutions of the equation
$$x_1 + x_2 + x_3 + \cdots + x_k = n$$
in the positive integers is
$$\binom{n - 1}{k - 1}$$
since a solution corresponds to placing $k - 1$ addition signs in the $n - 1$ spaces between successive ones in a row of $n$ ones.
We wish to exclude solutions in which at least one of the variables exceeds $3$. There can be at most two such variables since $3 \cdot 4 + 3 \cdot 1 = 15 > 12$.
One variable exceeds $3$: There are six choices for the variable which exceeds $3$. Suppose it is $a_1$. Then $a_1 \geq 4$, so $a_1' = a_1 - 3$ is a positive integer. Substituting $a_1' + 3$ for $a_1$ in equation 1 yields
\begin{align*}
a_1' + 3 + a_2 + a_3 + a_4 + a_5 + a_6 & = 12\\
a_1' + a_2 + a_3 + a_4 + a_5 + a_6 & = 9 \tag{2}
\end{align*}
Equation 2 is an equation in the positive integers with $\binom{9 - 1}{6 - 1} = \binom{8}{5}$ solutions. Hence, there are
$$\binom{6}{1}\binom{8}{5}$$
cases in which a variable exceeds $3$.
However, if we subtract this amount from the total, we will have subtracted too much. That is because we have subtracted cases in which two variables each exceed $3$ twice, once for each variable we could designate as the one that exceeds $3$. We only want to subtract those cases once, so we must add them to the total.
Two variables each exceed $3$: There are $\binom{6}{2}$ ways to select the two variables which exceed $3$. Suppose they are $a_1$ and $a_2$. Then $a_1' = a_1 - 3$ and $a_2' = a_2 - 3$ are positive integers. Substituting $a_1' + 3$ for $a_1$ and $a_2' + 3$ for $a_2$ in equation 1 yields
\begin{align*}
a_1' + 3 + a_2' + 3 + a_3 + a_4 + a_5 + a_6 & = 12\\
a_1' + a_2' + a_3 + a_4 + a_5 + a_6 & = 6 \tag{3}
\end{align*}
Equation 3 is an equation in the positive integers with one solution. Hence, there are
$$\binom{6}{2}\binom{5}{5}$$
cases in which two variables exceed $3$.
By the Inclusion-Exclusion Principle, the number of solutions of equation 1 in which no variable exceeds $3$ is
$$\binom{11}{5} - \binom{6}{1}\binom{8}{5} + \binom{6}{2}\binom{5}{5}$$
|
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|
pseudo C-S inequality? Problem :
For $x,y,z\in\mathbb{R}$,
Find minimum of $$8x^4+27y^4+64z^4$$
where $$x+y+z=\frac{13}{4}$$
I tried to apply C-S inequality but it has little difference,
The form what I know is :
$$(a^4+b^4+c^4)(x^4+y^4+z^4)\ge (ax+by+cz)^4$$
But in this problem, coefficient is form of $()^3$, not a $()^4$.
I tried to rewrite $8x^4=(2x)^4\times\frac{1}{2}$, but It wasn't helpful :
$$\left(\frac{1}{2^4}+\frac{1}{3^4}+\frac{1}{4^4}\right)((2x)^4+(3y)^4+(4z)^4)\ge (x+y+z)^4$$
So, is there any nice transform to apply C-S inequality?
or should I apply Lagrange Multiplier?
|
Hint: You can also use Hölder's Inequality, like so:
$$\left(8x^4+27y^4+64z^4 \right) \cdot \left(\frac12+\frac13+\frac14 \right)^3 \geqslant \left(|x|+|y|+|z|\right)^4 \geqslant (x+y+z)^4$$
|
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|
Prove $f(-\frac12) \le \frac{3}{16}$ if all roots of $f(x) = x^4 - x^3 + a x + b$ are real
Let $a, b$ be real numbers such that all roots of $f(x) := x^4 - x^3 + ax + b$ are real. Prove that $f(-1/2) \le 3/16$.
The question was posted recently which was closed,
due to missing of contexts etc.
My attempt: $f(-1/2) \le 3/16$ is equivalent to $a \ge 2b$.
Let $x_1, x_2, x_3, x_4$ be the real roots of $f(x)$.
By Vieta, we have
\begin{align*}
x_1 + x_2 + x_3 + x_4 &= 1, \\
x_1 x_2 + x_1x_3 + x_1x_4 + x_2 x_3 + x_2x_4 + x_3x_4 &= 0, \\
x_1x_2x_3 + x_1x_2x_4 + x_1 x_3 x_4 + x_2x_3 x_4 &= -a, \\
x_1x_2x_3x_4 &= b.
\end{align*}
The problem becomes:
If $x_1, x_2, x_3, x_4$ are real numbers such that
$x_1 + x_2 + x_3 + x_4 = 1$ and $x_1 x_2 + x_1x_3 + x_1x_4 + x_2 x_3 + x_2x_4 + x_3x_4 = 0$,
prove that
$$x_1x_2x_3 + x_1x_2x_4 + x_1 x_3 x_4 + x_2x_3 x_4 + 2x_1x_2x_3x_4 \le 0.$$
This is true (e.g. verified by Mathematica). Is there a nice proof for it (or the original problem)?
|
Suppose $b= a/2+c$ where $c>0$. Then we have
$$\underbrace{x^4-x^3+ax+{a\over 2}}_{p(x)} =-c$$
This means that graph of $p$ and line $y=-c<0$ have at least two different common points (if it has only one then $p(x)=(x-d)^4-c$ for some real $d$, this case is not possible) so $p$
has exactly two minimums with negative value, say at $x_1$ and $x_3$ where
$x_1<x_3$, so $p(x_1)<0$ and $p(x_3)<0$ i.e.
$$x_i^4-x_i^3+ax_i+{a\over 2} <0\;\;\;(*)$$
for $i=1,3$. Since $x_1,x_3$ satisfies also $p'(x_i)=0$ we have $$a=-4x_i^3+3x_i^2\;\;\;(**)$$ so we
get plugging $(**)$ in inequality $(*)$ for $i=1,3$:
$$-6x_i^4+3x_i^2<0\implies x_i^2>{1\over 2}$$
If we subtract equations $(**)$, we get
$$(x_1-x_3)\Big(4(x_1^2+x_1x_3+x_3^2)-3(x_1+x_3)\Big)=0$$
so $$4(x_1^2+x_1x_3+x_3^2)=3(x_1+x_3)$$ and from here we get
$$2(x_1^2+x_3^2)+2(x_1+x_3)^2=3(x_1+x_3)$$ which means that $$
2+2t^2<3t$$ where $t=x_1+x_3$. But this inequality does not have a
solution so we have a contradiction.
|
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|
How many methods are there to find $\int_{0}^{\pi} \frac{d \theta}{a+b \cos \theta}, \textrm{ where }a>b>0$ When I deal with this simple integral, I found there are several methods. Now I share one of them.
Letting $x\mapsto \pi-x $ yields $$I=\int_{0}^{\pi} \frac{d \theta}{a+b \cos \theta}=\int_{0}^{\pi} \frac{d \theta}{a-b \cos \theta}$$
Adding two versions together yields
$$
\begin{aligned}
2 I &=2 a \int_{0}^{\pi} \frac{d \theta}{a^{2}-b^{2} \cos ^{2} \theta} \\
&=4 a\int_0^{\frac{\pi}{2}} \frac{\sec ^{2} \theta}{a^{2} \sec ^{2} \theta-b^{2}} d \theta \quad \textrm{( By symmetry )}\\
&=4 a{\int_{0}^{\infty}} \frac{d t}{\left(a^{2}-b^{2}\right)+a^{2} t^{2}} \\
&=\frac{4}{\sqrt{a^{2}-b^{2}}}\left[\tan^{-1} \left(\frac{at}{\sqrt{a^{2}-b^{2}}}\right)\right]_{0}^{\infty} \\
&=\frac{2 \pi}{\sqrt{a^{2}-b^{2}}}
\end{aligned}
$$
We can now conclude that $$
\boxed{I=\frac{\pi}{\sqrt{a^{2}-b^{2}}}}
$$
Are there any other methods to deal with the integral?
|
Given $I(a)=\int_{0}^{\pi} \ln (a+b \cos x)dx
=\pi \ln\frac{a+\sqrt{a^2-b^2}}2$
$$\int_{0}^{\pi} \frac{1}{a+b \cos \theta}d\theta
=\frac{dI(a)}{da}=\frac{\pi}{\sqrt{a^{2}-b^{2}}}
$$
|
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|
Parameterizing the parabola $9x^2 +y^2-6xy+4x-4y+1=0$
Find parametrization of curve given by equation:
$$9x^2 +y^2-6xy+4x-4y+1=0$$
My attempt:
Let's notice that \begin{split} 9x^2 +y^2-6xy+4x-4y+1=0 & \iff (3x)^2 -6xy + y^2 +4x -4y +1=0\\
& \iff (3x-y)^2 +6x -2y + 1 -2x -2y=0\\
& \iff (3x-y)^2 +2(3x -y) + 1^2 -2x -2y=0\\
& \iff (3x-y+1)^2-2(x+y)=0.\end{split}
So this look like some crazy parabola, right? But I don't know how to derive parametric equation of this?
|
I give here another parametrization, a little harder and less practical but somewhat enlightening for the beginners. The focus is $F=\left(\dfrac{-7}{40},\dfrac{11}{40}\right)$ and the directrice is $4x+12y-1=0$ then if $(x,y)$ is any point of the parabola then its distance to the focus and to the directrice are equal by definition of parabola. This gives the equation
$$(40x+7)^2+(40y-11)^2=10(4x+12y-1)^2$$
Because $10=3^2+1$ we have by the known parametrization of solutions of the equation $x^2+y^2=z^2+w^2$
$$\begin{cases}40x+7=rX+sY\\40y-11=rY-sX\\12x+36y-3=rX-sY\\4x+12y-1=rY+sX\end{cases}$$ from which
$$\begin{cases}52x+36y+4=2rX\\4x+52y-12=2rY\\28x-36y+10=2sY\\4x-28y+10=2sX\end{cases}$$ so
$$\frac{13x+9y+1}{x+13y-3}=\frac{2x-14y+5}{14x-18y+5}=t$$ Solving this system we get for $t$ real distinct of $1$
$$x=\frac{-11t^2+86t-59}{200(t-1)^2}$$
$$y=\frac{47t^2-62t+63}{200(t-1)^2}$$
This parametrization is equivalent to the simpler one suggested by the O.P.
|
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How do I make a change of variable for $\;\lim_{x\to 0}\frac{\sqrt{x^2+p^2}-p}{\sqrt{x^2+q^2}-q}$? I can't use l'hopital, so change of variable is the only way.
$$\lim_{x\to 0}\frac{\sqrt{x^2+p^2}-p}{\sqrt{x^2+q^2}-q}$$
|
We can do
$$\frac{\sqrt{x^2+p^2}-p}{\sqrt{x^2+q^2}-q} \times \frac{\sqrt{x^2+q^2}+q}{\sqrt{x^2+q^2}+q} = \frac{(\sqrt{x^2+p^2}-p)(\sqrt{x^2 + q^2}+q}{x^2}$$
First $$\frac{\frac{\sqrt{x^2+p^2}-p}{1}\times \sqrt{x^2 + q^2} +q}{x^2} \tag{1}$$
and
$$\frac{\frac{\sqrt{x^2 + p^2}+p}{\sqrt{x^2 + p^2}+p}\times \frac{\sqrt{x^2+p^2}-p}{1}\times \sqrt{x^2 + q^2} +q}{x^2} \tag{2}$$
and finally
$$\frac{\frac{x^2}{\sqrt{x^2 + p^2}+p}\times \sqrt{x^2 + q^2} +q}{x^2} = \frac{\frac{x^2( \sqrt{x^2 + q^2}+q)}{\sqrt{x^2 + p^2}+p}}{x^2} \tag{3}$$
to get
$$\frac{x^2(\sqrt{x^2 + q^2}+q)}{x^2 (\sqrt{x^2 + p^2}+p)} = \frac{\sqrt{x^2 + q^2}+q}{\sqrt{x^2 + p^2}+p}$$
|
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|
Minimize $f(a,b,c) = \dfrac{1}{a²+b²+c²} +\dfrac{1}{9abc}$ subject to $a+b+c=1$ and $a,b,c>0$
Minimize $f(a,b,c) = \dfrac{1}{a²+b²+c²} +\dfrac{1}{9abc}$ subject to
$a+b+c=1$ and $a,b,c>0$
My attempt:
Since $f(a,b,c) \ge 0$, then we can minimizing the addends individually. Furthermore, minimizing fractions is the same as maximizing their denominators.
Apply the Lagrange Multiplier on ${a²+b²+c²}$ with the contraints $\implies a=b=c=\frac{1}{3}.$
Similarly, apply the Lagrange Multiplier on ${9abc}$ with the contraints $\implies a=b=c=\frac{1}{3}.$
Therefore, the minimum of $f\!\left(\frac{1}{3},\frac{1}{3},\frac{1}{3}\right)=6.$
I'm looking for a better solution not using the Lagrange Multiplier.
|
This solution was posted by Arkady Alt on LinkedIn. Let $t:=ab+bc+ca.$ Since $3abc=3abc(a+b+c)\leq(ab+bc+ca)^2=t^2$,
$a^2+b^2+c^2=1-2t$ and $3t\leq(a+b+c)^2=1$ then
$$(1/(a^2+b^2+c^2))+(1/(9abc))-6≥(1/(1-2t))+(1/(3t^2))-6=$$
$$((4t+1)(1-3t)^2)/(3t^2(1-2t))≥0.$$
|
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|
Limit of $u_{n}=\frac{1}{n !} \sum_{k=0}^{n} k !$ I encountered a sequence $(u_n)_{n \in \mathbb{N}} $ defined as
$$ u_{n}=\frac{1}{n !} \sum_{k=0}^{n} k ! $$
And I wonder what is the limit. It seems to be 1 but even Wolfram Alpha cannot figure it out.
My first idea was to write $$ \frac{k!}{n!} = \frac{1}{(k+1)...(n-1)n}$$ and
$$ (k+1)^{n-k} \leq (k+1)...(n-1)n \leq n^{n-k} $$
EDIT : I found a solution but Yanko found something more elementary.
My solution :
\begin{align}
u_n &= 1 + \frac{1}{n} + \frac{1}{n(n-1)} + \frac{1}{n(n-1)(n-2)} + ... + \frac{1}{n!} \\
&= 1 + \frac{1}{n} + (n-1) \times o(\frac{1}{n^2}) \\
&= 1 + \frac{1}{n} + o(\frac{1}{n}) \\
\end{align}
Thus all terms converge towards 0 except the first one, and the limit is 1.
|
Even more elementary:
Note that $x_{n+1} = {1 \over n+1} x_n +1$. Note that
$x_{n+1} \le x_n$ iff $x_n \ge {n+1 \over n}$. Also note that $x_1 = 2$.
If $x_n \ge {n+1 \over n}$ then $x_{n+1} \ge {n+1 \over n} \ge {n+2 \over n+1}$. Hence $x_n$ is non increasing and bounded below by $0$. Hence it has a limit $L$ and from continuity we have $L = 1$. Hence $x_n \downarrow 1$.
|
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|
Real solutions to the depressed cubic equation How can I find the allowed domain to this depressed cubic inequality
$$x^3 - 3 x + 2 \cos(\frac{3 \sqrt{3} n}{2}) \geq 0$$
where $n$ is a real non-negative number. Using Cardano's method, I can obtain some solutions for $x$ ($x_1$, $x_2$, and $x_3$) that should be real and not complex.
How can the solutions $x_j$ to define the feasible region satisfying the inequality for fixed $n$?
|
Let $\alpha = \frac{n\sqrt 3}{2}$.
Using $\cos 3\alpha = 4\cos^3 \alpha - 3 \cos \alpha$, the equation is written as
$$x^3 - 3x + 2(4\cos^3 \alpha - 3 \cos \alpha) = 0$$
or
$$(x^3 + (2\cos \alpha)^3) - 3(x + 2\cos \alpha) = 0$$
or
$$(x + 2\cos\alpha)(x^2 - 2x\cos\alpha + 4\cos^2\alpha - 3) = 0$$
or
$$(x + 2\cos\alpha)((x - \cos \alpha)^2 - 3\sin^2\alpha) = 0$$
or
$$(x + 2\cos\alpha)(x - \cos \alpha - \sqrt 3 \sin\alpha)(x - \cos \alpha + \sqrt 3\sin\alpha) = 0.$$
Thus, the roots are $x_1 = -2\cos \alpha, ~ x_2 = \cos \alpha + \sqrt 3 \sin\alpha, ~ x_3 = \cos \alpha - \sqrt 3 \sin \alpha$.
Note that $x_1 = x_2 = x_3$ implies
$\cos \alpha = \sin \alpha = 0$ which is impossible.
To find the domain, we need to sort $x_1, x_2, x_3$.
For example, if $x_1 < x_2 < x_3$,
then the allowed domain of $(x-x_1)(x-x_2)(x-x_3)\ge 0$ is $[x_1, x_2]\cup [x_3, \infty)$;
if $x_1 = x_2 < x_3$, then the allowed domain is $[x_3, \infty)$.
|
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Convergence of $1-\frac{1 \times 3}{3!} + \frac{1 \times 3 \times 5}{5!} +....+ \frac{1 \times 3 \times 5 \times ... \times(2n-1)}{(2n-1)!} - ...$ I was asked to prove whether
$$1-\frac{1 \times 3}{3!} + \frac{1 \times 3 \times 5}{5!}- \frac{1 \times 3 \times 5 \ \times 7}{7!} +....+ \frac{1 \times 3 \times 5 \times ... \times(2n-1)}{(2n-1)!} - ...$$
converges absolutely, conditionally or diverges, and was wondering whether my proof is correct or not.
I began by noticing that
$$1-\frac{1 \times 3}{3!} + \frac{1 \times 3 \times 5}{5!}- \frac{1 \times 3 \times 5 \ \times 7}{7!} +....+ \frac{1 \times 3 \times 5 \times ... \times(2n-1)}{(2n-1)!} -...$$
$$= 1- \frac{1}{2} + \frac{1}{2\times4} - \frac{1}{2 \times 4 \times 6}+...+\frac{1}{(2n)\times(2n-2)... \times2} -...$$
In other words, each $n$th term is the inverse of the product of all even natural numbers, and thus the initial sum is simply $ 1 + \sum a_n$ with
$$a_n := (-1)^n \frac{1}{\prod_{j=1}^n2j}$$
We can show that
$$\lim_{n\to\infty} \frac{|a_{n+1}|}{|a_n|} = \lim_{n\to\infty} \frac{\prod_{j=1}^{n}2j}{\prod_{j=1}^{n+1}2j}=\lim_{n\to\infty} \frac{1}{2(n+1)} = 0 < 1$$
and therefore the series converges absolutely according to the ratio test.
Is this proof correct?
|
Possibly adding to the confusion, but
$$
x+{1\over 3!}x^3+{1\over 5!}x^5+...
$$
is the Taylor series for $\sinh x$ at $x=0$ and
$$
1+{1\over 2!}x^2+{1\over 4!}x^4+...
$$
is the Taylor series for $\cosh x$ at $x=0$ so your sum is
$$
\cosh\frac12-\sinh\frac12
$$
which is of course $\exp(-1/2)$ as noted in the other answer.
|
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|
Limit of $(\sqrt{4x^2+2x+1}-ax-b) = -\frac{1}{2}$ Find $a,b$ of:
$$\lim_{x \to \infty}(\sqrt{4x^2+2x+1}-ax-b) = -\frac{1}{2}$$
I can't use L'hopital, I tried multiplying by the conjugate, and solving it,
$$\lim_{x \to \infty}(\sqrt{4x^2+2x+1}-(ax+b))\cdot \frac{\sqrt{4x^2+2x+1}+(ax+b)}{\sqrt{4x^2+2x+1}+(ax+b)}$$
$$=\lim_{x \to \infty}\frac{4x^2+2x+1-(ax+b)^2}{\sqrt{4x^2+2x+1}+(ax+b)} = \lim_{x \to \infty}\frac{x^2\left(4+\frac{2}{x}+\frac{1}{x^2}-a^2-\frac{2ab}{x}-\frac{b^2}{x^2}\right)}{x^2\left(\sqrt{\frac{4x^2+2x+1}{x^4}}+\frac{a}{x}+\frac{b}{x^2}\right)}$$
Applying limit on the numerator and denominator
$$\frac{\lim_{x \to \infty}\left(4-a^2+\frac{2-2ab}{x}+\frac{1-b^2}{x^2}\right)}{\lim_{x \to \infty}\left(\sqrt{\frac{4x^2+2x+1}{x^4}}+\frac{a}{x}+\frac{b}{x^2}\right)}$$
It can be seen that the denominator tends to $0$
|
From
$$\lim_{x \to \infty}(\sqrt{4x^2+2x+1}-ax-b) = -\frac{1}{2}$$
we have $$\lim_{x \to \infty}(\frac{\sqrt{4x^2+2x+1}-ax-b}{x}) = 0$$
hence
$$\lim_{x \to \infty}(\frac{\sqrt{4x^2+2x+1}}{x})=a=2$$
$$b=\lim_{x \to \infty}(\sqrt{4x^2+2x+1}-2x) +\frac{1}{2}$$
and
$$\sqrt{4x^2+2x+1}-2x=\frac{2x+1}{\sqrt{4x^2+2x+1}+2x}$$
using L'hopital Rule we get $b=1$.
|
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|
Prove the following equality in a triangle
If $A,B,C$ are angles of a triangle then prove that $$\frac{\operatorname{tan}A}{\operatorname{tan}B}=\frac{c^2+a^2-b^2}{c^2+b^2-a^2}$$ where $\angle A$ is opposite to side $a$ , $\angle B$ to side $b$ and $\angle C$ to side $c$.
I wasn't having a clue of proceeding. So, I applied another method. If the above expression is true then this expression must also be true $$\frac{\operatorname{tan}A+\operatorname{tan}B}{\operatorname{tan}A-\operatorname{tan}B}=\frac{c^2+a^2-b^2+c^2+b^2-a^2}{c^2+a^2-b^2-c^2-b^2+a^2}$$ $$\frac{\operatorname{tan}A+\operatorname{tan}B}{\operatorname{tan}A-\operatorname{tan}B}=\frac{c^2}{a^2-b^2}$$ Now, I am stuck here.
I applied sine rule and got the result $$\frac{c^2+a^2-b^2}{c^2+b^2-a^2}=\frac{\operatorname{sin}^2C+\operatorname{sin}^2A-\operatorname{sin}^2B}{\operatorname{sin}^2C+\operatorname{sin}^2B-\operatorname{sin}^2A}$$
Any help is greatly appreciated.
|
Use the sine and cosine rules: $$\tan A =\frac{\sin A}{\cos A}, \tan B=\frac{\sin B}{\cos B}$$
$$\text{Sine Rule:}\quad \sin A=\frac{a}{2R}, \sin B=\frac{b}{2R}$$
$$\text{Cosine Rule:}\quad \cos A=\frac{c^2+b^2-a^2}{2bc}, \cos B=\frac{c^2+a^2-b^2}{2ac}$$
Thus, $$\tan A=\frac{abc}{R}\frac{1}{c^2+b^2-a^2}, \tan B= \frac{abc}{R}\frac{1}{c^2+a^2-b^2} $$ Divide. Hence proved.
|
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|
Solve in exact form: $x^6 - x^5 + 4 x^4 - 4 x^3 + 4 x^2 - x + 1=0$ ( WolframAlpha failed)
Solve the polynomial in closed form:
$$x^6 - x^5 + 4 x^4 - 4 x^3 + 4 x^2 - x + 1=0$$
WolframAlpha obviously failed.
I tried several ways:
*
*I tried the Rational Root Thereom, but there is no rational root.
*I tried possible factorisations
$$x^6-x^5+4x^4-4x^3+4x^2-x+1= (x^2+a_1x+a_2)(x^4+a_3x^3+a_4x^2+a_5x+a_6)$$
and
$$x^6-x^5+4x^4-4x^3+4x^2-x+1= (x^3+a_1x^2+a_2x+a_3)(x^3+a_4x+a_5)$$
But, the expansions are terrible!
|
(Too long for a comment.)
While the other answers addressed how to express the sextic in radicals, I was also curious about what other contexts it may arise. When faced with an equation, one trick is to look at its discriminant $d$ since it is an important invariant and may give clues. The OP's original equation is
$$x^6 - x^5 + 4 x^4 - 4 x^3 + 4 x^2 - x + 1 = 0\tag1$$
But this is palindromic so its degree can be reduced in half (as was done by S.Iyer's answer) to the cubic (which looks awfully familiar)
$$z^3-z^2+z-2=0\tag2$$
The discriminant of the sextic is $d_6 = -2^6\cdot83^2$ and the cubic is $d_3=83$. This is the clue. We have the class number $h(-83) = 3.$ Thus, the j-function $j(\tau)$ with $\tau = \tfrac{1+\sqrt{-83}}{2}$ is an algebraic integer of degree $3$ and it may be possible to express $j(\tau)$ using the real root $z$ of the cubic. In fact, one can do so and we have the cube,
$$j\left(\tfrac{1+\sqrt{-83}}{2}\right)=-32^3\left(\frac{25z+20}{23z-31}\right)^3\tag3$$
Therefore,
$$\quad\quad e^{\pi\sqrt{83}} \approx 32^3\left(\frac{25z+20}{23z-31}\right)^3 + 743.999999926\dots$$
just like its more famous cousin,
$$e^{\pi\sqrt{163}}\;\approx\; 640320^3 + 743.9999999\dots\quad$$
So it is curious in what context the OP found the sextic in the first place.
P.S. I may have seen $(2)$ in a table of cubics with small coefficients relating to class invariants.
|
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|
Olympiad Math Problem About Simultaneous Equations Given the question:
*
*$x + y = 1$
*$x^2 + y^2 = 2$
*$x^5 + y^5 = ?$
After a bunch of manipulation of the above equations we find the solution to be 19/4. Yet could the above be solved by simultaneous equations?
From $1)$ we can conclude that $x = 1 - y$.
Substituting this into $2)$ we get a quadratic $2y^2 - 2y - 1 = 0$.
However, when we solve for $y$ we get two possible solutions. Hence how do we proceed from here? Why doesn't simultaneous equations work to solve the above problem?
|
First note that we can get
$$2xy = (x+y)^2 - x^2-y^2 = 1 - 2 = -1$$
We can then apply binomial theorem again to obtain $x^3+y^3$ and $x^5+y^5$
$$
\begin{align*}
x^3 + y^3 &= (x+y)^3 - 3x^2y - 3xy^2 \\
&= 1 + \frac{3}{2}(x+y) = \frac{5}{2} \\
x^5 + y^5 &= (x+y)^5 - 5x^4y - 10x^3y^2 - 10x^2y^3 - 5xy^4 \\
&= 1 + \frac{5}{2}(x^3+y^3) - \frac{5}{2}(x+y) \\
&= 1 + \frac{25}{4} - \frac{5}{2} = \frac{19}{4}
\end{align*}
$$
|
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|
prove $\sum_{i=1}^n \frac{x_i^2}{\sum_{j=1}^i x_j}\ge \frac{2}{n}-\frac{1}{n^2}$
Let $x_1,x_2,\cdots,x_n$ be $n$ non-negative numbers such that
$x_1+x_2+\cdots+x_n=1$. Show $$\sum_{i=1}^n \frac{x_i^2}{\sum_{j=1}^i
x_j}\ge \frac{2}{n}-\frac{1}{n^2}.$$
We may consider using Cauchy. That is
$$\sum_{i=1}^n\sum_{j=1}^i x_j\cdot \sum_{i=1}^n \frac{x_i^2}{\sum_{j=1}^i
x_j}\ge \left(\sum_{i=1}^nx_i\right)^2=1.$$
How to go on?
|
Fact 1: Let $n \ge 2$. Let $x_1, x_2, \cdots, x_n \ge 0$ and $x_1 > 0$. Then
$$\sum_{i=1}^n \frac{x_i^2}{\sum_{j=1}^i
x_j} \ge \frac{2n - 1}{n^2}(x_1 + x_2 + \cdots + x_n).$$
(The proof is given at the end.)
By Fact 1, we are done.
Proof of Fact 1:
We use Mathematical Induction.
It is easy to verify the case $n = 2$.
Assume that it is true for $n$.
For $n+1$, by the inductive hypothesis, we have
\begin{align*}
\sum_{i=1}^{n+1} \frac{x_i^2}{\sum_{j=1}^i
x_j}
&= \sum_{i=1}^{n} \frac{x_i^2}{\sum_{j=1}^i
x_j} + \frac{x_{n+1}^2}{\sum_{j=1}^{n+1} x_j}\\
&\ge
\frac{2n - 1}{n^2}(x_1 + x_2 + \cdots + x_n)
+ \frac{x_{n+1}^2}{\sum_{j=1}^{n+1} x_j}.
\end{align*}
It suffices to prove that
$$\frac{2n - 1}{n^2}(x_1 + x_2 + \cdots + x_n)
+ \frac{x_{n+1}^2}{\sum_{j=1}^{n+1} x_j} \ge \frac{2n+1}{(n+1)^2}(x_1 + x_2 + \cdots + x_{n+1}).$$
Since the inequality is homogeneous, assume that $x_1 + x_2 + \cdots + x_n = 1$.
Letting $y = x_{n+1}$, it suffices to prove that
$$\frac{2n - 1}{n^2}
+ \frac{y^2}{1+y} \ge \frac{2n+1}{(n+1)^2}(1 + y)$$
or
$$n^4y^2 + (-2n^3 + n^2 - 1)y + 2n^2 - 1 \ge 0$$
which is true (easy).
We are done.
|
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|
Limit of 2 variable function I'm trying to determine the limit of this function:
$$\lim_{(x,y)\to (1,2)} \frac{xy^2-4xy-y^2+4x+4y-4}{x^2+y^2-2x-4y+5}$$
I tried to approach in many different ways, such as
$$\lim_{t\to 1} f(t,2t) \quad, \quad\lim_{t\to 1} f(t,2) \quad, \quad \lim_{t\to 2} f(1,t) $$
But i got that the limit is 0 for all of them, tried with polar coordinates but it seems hopeless to get the limit!
How should I think there? Thanks in advance!
|
Let $f(x,y)=xy^2-4xy-y^2+4x+4y-4$ and $g(x,y)=x^2+y^2-2x-4y+5$. Then, for each $(x,y)\in\Bbb R^2$,\begin{align}f(x,y)&=f\bigl((x-1)+1,(y-2)+2\bigr)\\&=(x-1)(y-2)^2\end{align}and\begin{align}g(x,y)&=g\bigl((x-1)+1,(y-2)+2\bigr)\\&=(x-1)^2+(y-2)^2.\end{align}Therefore\begin{align}\lim_{(x,y)\to(1,2)}\frac{xy^2-4xy-y^2+4x+4y-4}{x^2+y^2-2x-4y+5}&=\lim_{(x,y)\to(1,2)}\frac{f(x,y)}{g(x,y)}\\&=\lim_{(x,y)\to(1,2)}\frac{(x-1)(y-2)^2}{(x-1)^2+(y-2)^2}\\&=\lim_{(x,y)\to(1,2)}(x-1)\overbrace{\frac{(y-2)^2}{(x-1)^2+(y-2)^2}}^{\phantom{[0,1]}\in[0,1]}\\&=0.\end{align}
|
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|
How to differentiate $y=\sqrt{\frac{1-x}{1+x}}$? It is an example question from "Calculus Made Easy" by Silvanus Thompson (page 68-69). He gets to the answer $$\frac{dy}{dx}=-\frac{1}{(1+x)\sqrt{1-x^2}}$$ The differentiation bit of the question is straightforward, but I'm having trouble simplifying it to get the exact answer. My working is the following:
$$y=\sqrt{\frac{1-x}{1+x}}$$
This can be written as $$y=\frac{(1-x)^\frac{1}{2}}{(1+x)^\frac{1}{2}}$$
Using the quotient rule we get $$\frac{dy}{dx}=\frac{(1+x)^\frac12\frac{d(1-x)^\frac12}{dx}-(1-x)^\frac12\frac{d(1+x)^\frac12}{dx}}{1+x}$$
Hence $$\frac{dy}{dx}=-\frac{(1+x)^\frac12}{2(1+x)\sqrt{1-x}}-\frac{(1-x)^\frac12}{2(1+x)\sqrt{1+x}}$$
$$=-\frac{1}{2\sqrt{1+x}\sqrt{1-x}}-\frac{\sqrt{1-x}}{2\sqrt{(1+x)^3}}$$
What is the next step?
|
It's easier to use the chain rule, that is, if $h=g\circ f$, then $$\frac{dh}{dx} = \frac{dg}{df}\cdot \frac{df}{dx}$$ or, written in another way, $$(g\circ f)' = g'\circ f\cdot f'$$
In this case, you can take $g$ as the square root funcion that is $g(x)=\sqrt{x}$, and $f$ as $f(x)=\frac{1-x}{1+x}$.
This quickly leads to the solution, since:
*
*The derivative of $g$ is $g'(x)=\frac{1}{2\sqrt{x}}$
*The derivative of $f$ is $$\frac{-1\cdot(1+x)-1\cdot(1-x)}{(1+x)^2} = \frac{-1-x-1+x}{(1+x)^2} = -\frac{2}{(1+x)^2}$$that the derivative is
which means the derivative of the compositum is
$$\frac{1}{2\sqrt{f(x)}}\cdot f'(x) = \frac{\sqrt{1+x}}{2\sqrt{1-x}}\cdot\frac{-2}{(1+x)^2} = -\frac{1}{\sqrt{1-x}\cdot\sqrt{1+x}\cdot(1+x)} =-\frac{1}{\sqrt{1-x^2}\cdot(1+x)}$$
That said, you can also continue from what you wrote. Finding the least common denominator will get you the following:
$$\begin{align}
\frac{dy}{dx}&=-\frac{(1+x)^\frac12}{2(1+x)\sqrt{1-x}}-\frac{(1-x)^\frac12}{2(1+x)\sqrt{1+x}}\\
&=-\frac{\sqrt{1+x}\cdot[(1+x)^\frac12]}{2(1+x)\sqrt{1+x}\sqrt{1-x}}-\frac{\sqrt{1-x}\cdot [(1-x)^\frac12]}{2(1+x)\sqrt{1+x}\sqrt{1-x}}\\
&=-\frac{1-x}{2(1+x)\sqrt{1+x}\sqrt{1-x}} - \frac{1+x}{2(1+x)\sqrt{1+x}\sqrt{1-x}}\\
&=-\frac{1-x+1+x}{2(1+x)\sqrt{(1+x)(1-x)}}\\
&=-\frac{2}{2(1+x)\sqrt{1-x^2}}
\end{align}$$
which is the same result (as it should be :).
|
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|
For all $x \in \Bbb R$ there exists a $y \in (-\infty, 1)$ such that $3yx^2+y-x=0$ I have this question:
Prove that for all real numbers $x$ there exist a number $y$ in the interval $(-\infty, 1)$ such that $$3x^2y+y-x=0$$
I proved that the range of the function
$$y = \frac{x}{3x^2 + 1}$$
is $(-1/2\sqrt{3},1/2\sqrt{3})$.
But the teacher said I should use another way.
I didn't find any.
|
Observe that, the given polynomial is exactly quadratic with respect to the variable $x$. You have:
$$
\begin{align}
&3x^2y+y-x=0\\
\iff &3yx^2-x+y=0\\
&\Delta_x=1-12y^2\ge 0\\
\iff &-\frac {1}{2\sqrt 3}\leq y\leq \frac {1}{2\sqrt 3}
\end{align}
$$
This result tells us, $\forall x\in\mathbb R$ if $3x^2+y-x=0$, then $-\frac {1}{2\sqrt 3}\leq y\leq \frac {1}{2\sqrt 3}$ which is logically equivalent to the original statement.
Finally, we see that
$$
\begin{align}
-&\frac {1}{2\sqrt 3}\leq y\leq \frac {1}{2\sqrt 3}\\
\implies &y\in\left(-\infty, 1\right).
\end{align}
$$
This completes the proof.
You can also get a stronger result here:
$\forall x \in \mathbb R$, there exists a $y \in \left[-\frac {1}{2\sqrt 3},\frac {1}{2\sqrt 3}\right]$, such that
$$3yx^2+y-x=0$$ holds.
Remember that, you can also reach the result in a shorter way:
We observe that, if $x≤0$ the statement is obviously trivial. Therefore, we analyze the case $x>0$. Then applying the AM-GM inequality, you have:
$$
\begin{align}
y=\frac {x}{3x^2+1}&<\frac{x}{2\sqrt 3x}\\
&=\frac {1}{2\sqrt 3}\\
&<1.\end{align}
$$
This also proves the original statement.
|
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|
How to solve this quadratic system of equations? The equations are:
$$ \begin{aligned} b + d &= c^2 - 6 \\ b - d &= -\frac{1}{c} \\ b d &= 6 \end{aligned} $$
and I want integer solutions for this.
I tried using various methods such as using $(a+b)^2-(a-b)^2=4ab$ or trying to solve for $b+d$ using the last two equations and substituting into the first. None of these methods worked. How do I do this?
|
Since OP said "I want integer solutions for this," meaning $b, c,$ and $d$ should be integers. In that case, $c$ can only be $1$ or $-1$ since the second eqiation implies:
$$c(b−d) = −1 \tag{i}$$
Also, in either case of $c = \pm 1$, the first eqiation implies:
$$b + d = (\pm 1)^2 - 6 = -5 \tag{ii}$$
From the equation $(i)$, you find that if $c=+1$, $b = d−1$ and when $c=-1$, $b = d+1$. Thus, for in general, when $c=\pm 1$:
$$b = d \mp 1\tag{iii}$$
If you apply these findings on the third eqiation, you can obtain numerical values for $b,d$. For example: When $c=+1$, $b = d−1$ and if you substitute those values on the third eqiation:
$$bd = d(d-1) = d^2 - d = 6 \ \Rightarrow \ d^2 - d - 6= 0$$
Factors for that equation is $(d - 3)(d + 2) = 0$.
As a result, $d=+3 $ or $d=-2 $. If $d=+3 $, then $b=+3 -1 = +2$. If $d=-2 $, then $b=-2 -1 = -3$.
Now check these two sets $(b,d = 2,3$ and $b,d = -3,-2)$ with second equation:
$$b-d = \frac{-1}{c} \ \Rightarrow \ 2-3 = \frac{-1}{+1} = -1$$
$$b-d = \frac{-1}{c} \ \Rightarrow \ -3-(-2) = \frac{-1}{+1} = -1$$
Thus both sets satisfied $c=1$ conditions. Therefore, half of solutions for $(b,d,c)$ set is $(2,3,1)$ and $(-3,-2,1)$.
You can solve for other two sets using $c= -1$ easily.
Following is my solution for other two sets using $c= -1$:
When $c=-1$, $b = d+1$ and when we substitute those values on the third eqiation:
$$bd = d(d+1) = d^2 + d = 6 \ \Rightarrow \ d^2 + d - 6= 0$$
Factors for above equation is $(d + 3)(d - 2) = 0$.
As a result, $d=-3 $ or $d=+2 $. If $d=-3 $, then $b=-3 +1 = -2$. If $d=+2 $, then $b=+2 +1 = +3$.
Now check these two sets $(b,d = -2,-3$ and $b,d = +3,+2)$ with the second equation:
$$b-d = \frac{-1}{c} \ \Rightarrow \ -2-(-3) = \frac{-1}{-1} = +1$$
$$b-d = \frac{-1}{c} \ \Rightarrow \ +3-(+2) = \frac{-1}{-1} = +1$$
Thus both sets satisfied $c= -1$ conditions. Therefore, other half of solutions for $(b,d,c)$ set is $(-2,-3,-1)$ and $(+3,+2,-1)$.
|
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|
Check if this series is convergent or not I've been going crazy for a long time in determining if this series converges or diverges. Most likely it converges.
$\displaystyle \sum_{n=1}^\infty \left(\frac{n}{2}\, \sin\frac{1}{n}\right)^\frac{n^2+1}{n+2}$
I am stuck in the necessary condition.
I tried to solve in this way :
$\displaystyle \frac{1}{2}\lim_{n \to \infty} \left(\frac{\sin\frac{1}{n}}{\frac{1}{n}}\right)^\frac{n^2+1}{n+2} = \frac{1}{2}\lim_{n \to \infty} (1)^\infty$
To solve this indeterminate form, I tried to use the exponential in this way
$\displaystyle \frac{1}{2}\lim_{n \to \infty} \left(e^{\frac{n^2+1}{n+2}log \left(\frac{sin \frac{1}{n}}{\frac{1}{n}}\right)}\right)$
In this way I always get an indeterminate form
$\displaystyle \frac{1}{2}\lim_{n \to \infty} (e^{\infty *0})$
could you kindly give me support for the resolution of the character of this series?
|
Letting
$$
a_n
=
\left(
\frac{n}{2}
\cdot
\sin\Bigl(\frac{1}{n}\Bigr)
\right)
^{\large{\frac{n^2+1}{n+2}}}
$$
we have $a_n > 0$ for all positive integers $n$, so we get
$$
\lim_{n\to\infty}
|a_n|
^
\frac{1}{n}
=
\lim_{n\to\infty}
(a_n)
^
\frac{1}{n}
=
\lim_{n\to\infty}
\left(
\frac{1}{2}
\cdot
\frac
{\sin\Bigl({\large{\frac{1}{n}}}\Bigr)}
{\Bigl({\large{\frac{1}{n}}}\Bigr)}
\right)
^{{\large{\frac{n^2+1}{n^2+2n}}}}
=
\Bigl(
\frac{1}{2}
\cdot
1
\Bigr)
^1
=
\frac{1}{2}
$$
hence by the root test, ${\displaystyle{\sum_{n=1}^\infty a_n}}$ converges.
|
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|
How do I show that $\sin(x^2+3x+2)$ is not periodic? I know that if $g(x)$ is periodic then $f(g(x))$ is periodic.
This is a sufficient condition but not necessary as $\sin(x)=f(g(x))$ is periodic where $g(x)=x$(non periodic) and $f(x)=\sin(x)$.
Can the above condition be made into a necessary and sufficient condition (assuming that $g(x)$ is not the identity function) or is there a better way to find the solution?
This is from a previous year college entrance exam. Any short trick rather to check for the periodicity of $f(g(x))$ when $g(x)$ is a complicated function would be helpful.
|
Suppose it is periodic with period $2\pi T$ for some $T>0$. This means that for all $x$, $\sin(x^2+3x+2)=\sin((x+2\pi T)^2+3(x+2\pi T)+2)$. Expanding the RHS gives $\sin(x^2+3x+2)=\sin(x^2+4\pi Tx+4\pi^2T^2+3x+6\pi T+2)$.
If $\sin(u)=\sin(v)$, then either $u=v-2\pi n$ or $u=\pi-v+2\pi n$ for some constant integer $n$. We will consider each of these cases:
*
*Suppose $x^2+3x+2=x^2+4\pi Tx+4\pi^2T^2+3x+6\pi T+2 - 2\pi n$. Then:
$$2\pi n=4\pi Tx+4\pi^2T^2+6\pi T$$
$$n=Tx+\pi T+3T$$
$n$ is a constant, so $\frac\partial{\partial x}(Tx+\pi T+3T)=0$ (as is the case for all higher derivatives). Calculating the derivative shows that $T=0$, but we previously defined that $T>0$, a contradiction.
*Suppose $x^2+3x+2=\pi-x^2-4\pi Tx-4\pi^2T^2-3x-6\pi T-2 + 2\pi n$. Then:
$$2x^2+6x+4-\pi+4\pi Tx+4\pi^2T^2+6\pi T = 2\pi n$$
$$n=\frac 1\pi x^2+\frac3\pi x+\frac 2\pi-\frac 12+2 Tx+2\pi T^2+3T$$
As in the previous case, $n$ is constant, so $\frac{\partial^2}{\partial x^2}(\frac 1\pi x^2+\frac3\pi x+\frac 2\pi-\frac 12+2 Tx+2\pi T^2+3T)=0$. Calculating this derivative gives the false statement $\frac 2\pi=0$, a contradiction.
|
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|
Find numbers divisible by 6 Find the number of all $n$, $1 \leq n \leq 25$ such that $n^2+15n+122$ is divisible by 6.
My attempt. We know that:
\begin{align*}
n^2+15n+122 & \equiv n^2+3n+2 \pmod{6}
\end{align*}
But $n^2+3n+2=(n+1)(n+2)$, then $n^2+15n+122 \equiv (n+1)(n+2) \pmod{6}$, now we have
\begin{align*}
n(n^2+15n+122) & \equiv n(n+1)(n+2)\pmod{6} \\
n^3+15n^2+122n & \equiv 0 \pmod{6}
\end{align*}
I have done this and I think I have complicated the problem even more.
|
If $n$ is even then $n^2+15n$ is even. If $n$ is odd, then $n^2+15n$ is still even. So $n^2+15n+122$ is even for every $n$.
So we just need to determine when the expression is divisible by $3$.
$$n^2+15n+122 \equiv n^2 +2 \pmod{3}.$$
It's easy to check that $n=1$ and $n=2$ are the only solutions. So the answer is "all non multiples of $3$". There are $8$ multiples of $3$ between $1$ and $25$, so the final number is $25-8 = 17.$
|
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|
Prove that $ \sum_{n=0}^{\infty} 2^{-n(n+1)/2} = \prod_{k=1}^{\infty} (1 - \frac{1}{2^k})^{(-1)^k}$ Prove the following equality:
$$ \sum_{n=0}^{\infty} 2^{-n(n+1)/2} = \prod_{k=1}^{\infty} (1 - \frac{1}{2^k})^{(-1)^k}$$
|
We can write
\begin{align*}
\sum_{n=0}^{\infty} 2^{-n(n+1)/2} &= \sum_{n=0}^{\infty} 2^{-n(n+1)/2} \prod_{k=1}^{\infty} (1 - \frac{1}{2^k})^{(-1)^k}\\
&= \prod_{k=1}^{\infty} (1 - \frac{1}{2^k})^{(-1)^k} \sum_{n=0}^{\infty} 2^{-n(n+1)/2} \prod_{k=1}^{\infty} (1 - \frac{1}{2^k})^{(-1)^k}\\
&= \prod_{k=1}^{\infty} (1 - \frac{1}{2^k})^{(-1)^k} \prod_{k=1}^{\infty} (1 - \frac{1}{2^k})^{(-1)^k}\\
&= \prod_{k=1}^{\infty} (1 - \frac{1}{2^k})^{(-1)^k + (-1)^k}\\
&= \prod_{k=1}^{\infty} (1 - \frac{1}{2^k})^{(-2)^k}\\
&= \prod_{k=1}^{\infty} (1 - \frac{1}{2^k})^{(-1)^k}
\end{align*}
where we have used the fact that $\sum_{n=0}^{\infty} 2^{-n(n+1)/2} = \prod_{k=1}^{\infty} (1 - \frac{1}{2^k})^{(-1)^k}$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Determination of complex logarithm: $\mathrm{Log}(z) = i$ I read that the determination of complex logarythm $\mathrm{Log}(z)$ is defined as:
$$\mathrm{Log}(z) := \ln|z| + i\mathrm{Arg}(z)$$
with $\mathrm{Arg}(z)$ the main argument of the complex number. I need to solve:
$$\mathrm{Log}(z) = i$$
Assuming $z =: x + iy$, the right answer is $\cos 1 + i \sin 1$. I found:
$$\mathrm{Log}(z) = i$$
$$\Longleftrightarrow \ln|z| + i\mathrm{Arg}(z) = i$$
\begin{equation}
\Longleftrightarrow
\begin{cases}
\ln|z| = 0\\
\arctan(y/x) = 1
\end{cases}
\end{equation}
\begin{equation}
\Longleftrightarrow
\begin{cases}
|z| = 1\\
y = x
\end{cases}
\end{equation}
\begin{equation}
\Longleftrightarrow
\begin{cases}
x^2 + y^2 = 1\\
y = x
\end{cases}
\end{equation}
\begin{equation}
\Longleftrightarrow
\begin{cases}
x^2 = 1/2\\
y = x
\end{cases}
\end{equation}
\begin{equation}
\Longleftrightarrow
z = \frac{\sqrt{2}}{2} + i\frac{\sqrt{2}}{2}
\end{equation}
Where is the problem in my calculations ?
|
Assuming $x + y \cdot \mathrm{i} := z$, the right answer is $\cos(1) + \sin(1) \cdot \mathrm{i} $. I found: [...]
Where is the problem in my calculations?
The problems are the step from $\arctan\left( \frac{y}{x} \right) = 1$ to the next one and $x = y$.
The argument is not always the arctangent of $\frac{y}{x}$, e.g. when $x = 0$! In addition, $x \ne y$! Accordingly, their premises are wrong and so is the calculation.
E.G. If you say $\arctan\left( \frac{y}{x} \right) = 1$ and $x = y$ it follows:
$$
\begin{align*}
\arctan\left( \frac{y}{x} \right) &= 1\\
\arctan\left( \frac{y}{y} \right) &= 1\\
\arctan\left( 1 \right) &= 1\\
\frac{\pi}{4} + 2 \cdot k \cdot \pi &= 1\\
\end{align*}
$$
And $1$ is not the same as $\frac{\pi}{4} + 2 \cdot k \cdot \pi$!
Correct step by step solution with your method
If you go through it correctly you will get:
$$
\begin{align*}
\ln(z) &= \mathrm{i} \quad\mid\quad \text{use polar form } z = |z| \cdot (\cos(\arg(z)) + \sin(\arg(z)) \cdot \mathrm{i})\\
\ln(|z| \cdot (\cos(\arg(z)) + \sin(\arg(z)) \cdot \mathrm{i})) &= \mathrm{i} \quad\mid\quad \text{find } |z| \text{ and find } \arg{z}\\
\ln(|z|) + \arg(z) \cdot \mathrm{i} &= \mathrm{i}\\
\begin{cases} (1.) \qquad \ln(|z|) &= 0 \\ (2.) \qquad \arg(z) \cdot \mathrm{i} &= \mathrm{i}\end{cases}\\
(1.) \qquad \ln(|z|) &= 0 \quad\mid\quad \exp(~~)\\
(1.) \qquad |z| &= 1\\
(2.) \qquad \arg(z) \cdot \mathrm{i} &= \mathrm{i} \quad\mid\quad \div \mathrm{i}\\
(2.) \qquad \arg(z) &= 1\\
\\
z &= |z| \cdot (\cos(\arg(z)) + \sin(\arg(z)) \cdot \mathrm{i}) \quad\mid\quad |z| = 1 \wedge \arg(z) = 1\\
z &= 1 \cdot (\cos(1) + \sin(1) \cdot \mathrm{i})\\
z &= \cos(1) + \sin(1) \cdot \mathrm{i}\\
\end{align*}
$$
But there is also a much easier way to solve for $z$:
$$
\begin{align*}
\ln(z) &= \mathrm{i} \quad\mid\quad \exp(~~)\\
\exp(\ln(z)) &= \exp(\mathrm{i})\\
z &= \exp(\mathrm{i})\\
z &= \operatorname{cis}(1) \quad\mid\quad \text{use euler's formula } \operatorname{cis}(x) = \cos(x) + \sin(x) \cdot \mathrm{i}\\
z &= \cos(1) + \sin(1) \cdot \mathrm{i}\\
\end{align*}
$$
|
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|
find a closed form formula for $\sum_{k=1}^n \frac{1}{x_{2k}^2 - x_{2k-1}^2}$
Let $\{x\} = x-\lfloor x\rfloor$ be the fractional part of $x$. Order the (real) solutions to $\sqrt{\lfloor x\rfloor \lfloor x^3\rfloor} + \sqrt{\{x\}\{x^3\}} = x^2$ with $x\ge 1$ from smallest to largest by $x_1,x_2,\cdots$. Provide a closed form formula for $\sum_{k=1}^n \frac{1}{x_{2k}^2 - x_{2k-1}^2}.$
First, I'm not even sure why there are infinitely many solutions to $\sqrt{\lfloor x\rfloor \lfloor x^3\rfloor} + \sqrt{\{x\}\{x^3\}} = x^2$. One inequality that might be useful is the Cauchy-Schwarz inequality. There's also the AM-GM inequality. We have $ac + bd \leq \sqrt{a^2+b^2}\sqrt{c^2+d^2}$ for all real numbers $a,b,c,d$ where $ac,bd\ge 0$ with equality iff $(a,b),(c,d)$ are proportional vectors in $\mathbb{R}^2$. Observe that $\lfloor x\rfloor$ always has the same sign as $x$, so $\lfloor x\rfloor \lfloor x^3\rfloor$ has the same sign as $x^4 \ge 0$. It might be useful to substitute $\{x\} = x-\lfloor x\rfloor$ into the original equation and simplify the result somehow. Also, it could be possible to write the given sum as a telescoping sum.
|
Let $u$ be a noncubic integer with $\lfloor\sqrt[3]{u}\rfloor = n \in \mathbb{Z}^+$. Assuming $\sqrt[3]{u} \le x < \sqrt[3]{u+1}$, we can reformulate the equation as follows.
$$ \sqrt{nu} + \sqrt{(x-n)(x^3-u)} = x^2 $$
Squaring both sides, we have
$$(\sqrt{n} x - \sqrt{u})^2 = 0$$
Hence the equation holds for $x = \sqrt{u/n}$ if $\sqrt[3]{u} \le \sqrt{u/n} < \sqrt[3]{u+1}$. Since the first inequality always holds, we only need to check which $u$ satisfies the second inequality. After calculating some examples, it is easy to guess that
$$u = n^3, n^3+1$$
are only $u$ we find. You can show this guess by substituting $u$ by $n^3+k$ for $k=0,1,2$, and checking that
$$f(u) = u^3 - n^3(u+1)^2$$
is increasing for $u \ge n^3+1$.
|
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|
Integration of $\int \frac{x^2}{(x-3)(x+2)^2}$ by partial fractions
Integration of $\int \frac{x^2}{(x-3)(x+2)^2}$
I know that we can use substitution to make the expression simplier before solving it, but I am trying to solve this by using partial fractions only.
$\frac{x^2}{(x-3)(x+2)^2} = \frac{A}{x-3} + \frac{B}{x+2} + \frac{C}{(x+2)^2} = \frac{A(x+2)^2+B(x-3)(x+2) + C(x-3)}{(x-3)(x+2)} = \frac{Ax^2 + 4Ax + 4A + Bx^2 - Bx - 6B +Cx - 3C}{(x-3)(x+2)}$
Looking at the numerator: $x^2(A+B) + x(4A-B+C) + (4A-6B-3C)$
So, comparing coefficients:
$A+B=1$,
$4A-B+C=0$
$4A-6B-3C=0$
I am struggling to solve these 3 equations to find A,B,C
|
You have a good start!
I'd suggest trying to reduce the number of variables. You can solve the first equation for $B=1-A$. Then try multiplying the second equation by 3 to get $12A-3B+3C=0$. Add this to the 3rd equation. Combine with the substitution for $B$ and solve for $A$. Then you can work backwards through the equations to solve for the other variables.
For a more systematic way to solve such things you may be interested in studying linear algebra.
|
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|
how do I evaluate this definite integral? $I=\int_3^5{\left( \frac{x^2+x}{\sqrt[3]{(x-3)(5-x)}} \right)dx}$ $$I=\int_3^5{\left( \frac{x^2+x}{\sqrt[3]{(x-3)(5-x)}} \right)dx}$$
I tried using substitution and some common definite integration results like
$$\int_a^bf(x)dx=\int_a^bf(a+b-x)dx \\
\int_a^bf(x)dx = (b-a)\int_0^1f((b-a)x+a)dx$$
and done substitutions like
$$x=3\cos^2(t)+5\sin^2(t)$$
but it just becomes messy.
I have tried adding the different results hoping that stuff would cancel out or simplify,
but I'm not able to figure it out.
Please help. Thanks!
|
Let $x=2t+3$, then $dx=2dt$. The bounds of the integral change like this: $x=3=2t+3\implies t=0$ and $x=2t+3=5\implies t=1$. Hence,
$\begin{align}
\int_3^5\frac{x^2+x}{\sqrt[3]{(x-3)(5-x)}}&=\int_0^1\frac{4t^2+14t+12}{\sqrt[3]{2t(2-2t)}}2dt\\
&=\sqrt[3]{2}\int_0^1\frac{4t^2+14t+12}{\sqrt[3]{t(1-t)}}dt\\
&=\sqrt[3]{2}\int_0^1 (4t^2+14t+12)t^{-\frac{1}{3}}(1-t)^{-\frac{1}{3}}dt\\
&=4\sqrt[3]{2}\int_0^1 t^{\frac{5}{3}}(1-t)^{-\frac{1}{3}}dt+
14\sqrt[3]{2}\int_0^1 t^{\frac{2}{3}}(1-t)^{-\frac{1}{3}}dt+
12\sqrt[3]{2}\int_0^1 t^{-\frac{1}{3}}(1-t)^{-\frac{1}{3}}dt\\
&=4\sqrt[3]{2}B(\frac{8}{3},\frac{2}{3})
+14\sqrt[3]{2}B(\frac{5}{3},\frac{2}{3})
+12\sqrt[3]{2}B(\frac{2}{3},\frac{2}{3})
\end{align}$
|
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|
A not straightforward limit The original limit is:
$$\lim_{x \to +\infty} \frac{e^{\sin \frac{1}{x}}-1-\frac{1}{x}}{\ln \left(1+\frac{x^2}{(1+x)^3} \right )-\frac{x^2}{(1+x)^3}}$$
I performed the substitution $y=\frac{1}{x}$ which leads to
$$\lim_{y \to 0^{+}} \frac{e^{\sin y}-1-y}{\ln \left ( 1+\frac{y}{(1+y)^3} \right )-\frac{y}{(1+y)^3 }}$$
now using the Maclaurin polynomials it should lead to the more simple limit
$$\displaystyle \lim_{ y\to 0^{+}} \left ( -\frac{1}{6} \frac{y + o(y^3)}{\frac{1}{(1+y)^6}+o\left (\frac{y}{(1+y)^3}\right )} \right ) $$
which is zero.
Could anyone tell me if this solution is right please?
|
That is not what I have.
$\lim_\limits{y\to 0} \frac {e^{\sin y} -1 - y}{\ln (1+ f(y)) - f(y)}$
The Talor expansion of the numerator is
$1 + \sin y + \frac 12 \sin ^2 y+\cdots - 1 - y$
Expanding each of those sine functions
$1+ y - \frac 16y^3 +\cdots + \frac 12 y^2 - \frac 16 y^4+\cdots -1-y$
Which will ultimately be
$\frac 12 y^2 + o(y^3)$
For the denominator, the Taylor expanson of $\ln (1+f(y)) = f(y) - \frac 12 (f(y))^2 + \cdots$
giving us $\frac {\frac 12 y^2 + o(y^3)}{-\frac 12 \frac {y^2}{(1+y)^6} + o(y^3)}$
As the limit approaches $0$ we will get $-1$
|
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|
Compute the limit of a determinant of a given $n \times n$ matrix Compute:
$\displaystyle \lim_{n \rightarrow \infty} \begin{vmatrix}
1+x & -x & 0 & 0 & \cdots & 0 & 0 \\
-\frac{1}{2} & 1+\frac{x}{2} & -x & 0 & \cdots & 0 & 0 \\
0 & -\frac{1}{3} & 1+\frac{x}{3} & -x & \cdots & 0 & 0 \\
0 & 0 & -\frac{1}{4} & 1+\frac{x}{4} & \cdots & 0 & 0 \\
\vdots & \vdots & \vdots & \cdots & \vdots & \vdots \\
0 & 0 & 0 & 0 & \cdots & 1+\frac{x}{n-1} & -x \\
0 & 0 & 0 & 0 & \cdots & -\frac{1}{n} & 1+\frac{x}{n} \\
\end{vmatrix}$
Let $D_n$ be the given determinant. I solved it via mathematical induction and got the following solution:
$\displaystyle \lim_{n \rightarrow \infty} D_n = \lim_{n \rightarrow \infty} \sum_{i=0}^n \frac{x^i}{i!} = \sum_{i=0}^{\infty} \frac{x^i}{i!} = \mathrm{e}^x$
I want to solve it by using recurrence relations:
We know that $D_1 = 1+x$ and $D_2 = 1+x+\frac{x^2}{2}$. If I develop the determinant $D_n$ by the last column I get:
$D_{n+2} = \left( 1+\frac{x}{n+2} \right) D_{n+1} - \left( \frac{x}{n+2} \right) D_{n} \text{ if } n \geq 1$.
How can I solve this linear recurrence relation with variable coefficients?
|
Hint : $D_{n+2} = \left( 1+\dfrac{x}{n+2} \right) D_{n+1} - \left( \dfrac{x}{n+2} \right) D_{n}$ rewrites as $$D_{n+2} -D_{n+1} = \frac{x}{n+2} (D_{n+1} - D_{n})$$
Let $A_n = D_{n+1}-D_n$ : you get $$A_{n+1} = \dfrac{x}{n+2}A_n$$
Can you proceed from here ?
|
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|
Why can't we multiply both sides of $x-1=0$ by $x$ to create another solution $x=0$? Consider the equation $x-1=0$. The obvious solution is $x=1$. However, what is stopping us from creating more solutions by multiplying both sides by arbitrary values? For example:
$$ \begin{align}
x-1 & = 0 \\
x\cdot(x-1) &= 0\cdot x \\
x\cdot(x-1) &= 0 \end{align} $$
If we substitute $x=0$ into this equation now, it gives us a solution. But 0 is not a solution to $x-1=0$. Does multiplying both sides by $x$ imply that there is a restriction $x\neq0$ or something? Why?
We could(?) go further with this, for example, multiply by $(x-2)$ giving $(x-2)(x-1)=0$, meaning $x=2$ is now an (incorrect) solution.
|
Taking your question, every one of your steps is valid.
$$\begin{equation}x - 1 = 0\end{equation}$$
has one solution, $x = 1$, and
$$x(x-1) = 0$$
has two solutions, $x = 0$ and $x = 1$.
Where this goes wrong is in assuming that this holds for the original equation: you are implicitly saying
$$\begin{aligned}0(0 - 1) &= 0\\
\therefore \, 0-1&=\frac{0}{0} =0\end{aligned}$$
However, this fails to hold, as division by zero is undefined. It is this undefined division by zero which lets you accomplish this for any polynomial $p(x)$, to which you wish to add the solution $x = c$:
$$\begin{aligned} p(x) &= 0\\
(x -c)p(x) &= 0\\
(c - c)p(c)&=0\\
p(c) &= \frac{0}{c-c} = \frac{0}{0} \neq 0\end{aligned}$$
|
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|
Find a complex number that satisfies both $|z-3+i|=3$ and $|z+2-i|=|z-3+2i|$. What I did:
Let $z=x+yi$, Then:
$$|(x-3)+(y+1)i|=3$$
$$|(x+2)+(y-1)i|=|(x-3)+(y+2)i|$$
Since $|x+iy|=\sqrt{x^2+y^2}$:
$$(x-3)^2+(y+1)^2=9$$
$$y=\frac{5x-4}{3}$$ Hence:
$$(x-3)^2+\bigg(\frac{5x-1}{3}\bigg)^2=9$$
Or
$$34x^2-64x+1=0$$
So
$$x_{1,2}=\frac{32\pm3\sqrt{110}}{34}$$
And
$$y_{1,2}=\frac{8\pm5\sqrt{110}}{34}$$
Hence
$$z_1=\frac{32+3\sqrt{110}}{34}+\frac{8+5\sqrt{110}}{34}i$$
$$z_2=\frac{32-3\sqrt{110}}{34}+\frac{8-5\sqrt{110}}{34}i$$
But, when I looked up the answer in the book:
I think there is a misprint in the book as this is the question:
I think that in part c they meant $|z+2-i|=|z-1+2i|$ and not $|z+2-i|=|z-3+2i|$, becuase then the answer would make sense, but I thought I would double check. Thanks for any help
|
Geometrically u are looking for the intersection of the circle centered at $(3,-1)$ with radius 3 and the perpendicular bisector of the line segment ralating the points $(-2,1)$ and $(3,-2)$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4568220",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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|
Prove $f(r, \theta) = (\cos\theta, \sin\theta)$ is continuous I am trying to prove that $f(r, \theta)= (\cos\theta, \sin\theta)$ is continuous using an $\varepsilon - \delta$ definition. Here is my work so far.
Let $f(r, \theta) = (\cos\theta, \sin\theta)$ and $(r_0, \theta_0) \in \mathbb{R}^2$. We will show that $f_r$ is continuous at any point $(r_0, \theta_0) \in \mathbb{R}^2$. Given $\varepsilon > 0$, let $\delta = ??$. Then we have that
\begin{align*}
||f_r(r, \theta) - f_r(r_0, \theta_0)|| &= ||(\cos\theta,
\sin\theta) - (\cos\theta_0,
\sin\theta_0)||\\
&= \sqrt{(\cos\theta_0 - \cos\theta)^2 + (\sin \theta_0 - \sin
\theta)^2}\\
&= \sqrt {(\cos\theta - \cos\theta_0)^2 + (\sin\theta -
\sin\theta_0)^2}\\
&= \sqrt{\cos^2\theta + \cos^2\theta_0 - 2\cos\theta\cos\theta_0 +
\sin^2\theta + \sin^2\theta_0 - 2\sin\theta\sin\theta_0}\\
&=\sqrt {2 - 2\cos(\theta - \theta_0)}\\
&= 2\sin\frac {\theta - \theta_0}{2} \text{ (Thanks to Doug's answer)}
\end{align*}
whenever $||(r, \theta) - (r_0, \theta_0)|| < \delta$. However, the problem is that the $\delta$-inequality is dependent on $r$ and $\theta$, while the $\varepsilon$-inequality is only dependent on $\theta$. I can't see how to manipulate this so we can use the $\delta$-inequality to complete the proof.
|
$\sqrt {(\cos\theta - \cos\theta_0)^2 + (\sin\theta - \sin\theta_0)^2}\\
\sqrt{\cos^2\theta + \cos^2\theta_0 - 2\cos\theta\cos\theta_0 + \sin^2\theta + \sin^2\theta_0 - 2\sin\theta\sin\theta_0}\\
\sqrt {2 - 2\cos(\theta - \theta_0)}\\
2\sin\frac {\theta - \theta_0}{2}$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4571663",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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|
Arrangements of $a,a,b,b,b,c,c,c,c$ in which no two consecutive letters are the same I have the following question:
We are going to generate permutations from a,a,b,b,b,c,c,c,c. Please compute the number of permutations such that:
(a) for any consecutive 4 elements, they are not all the same;
(b) for any consecutive 3 elements, they are not all the same;
(c) for any consecutive 2 elements, they are not all the same.
See Arrangements of $a,a,b,b,b,c,c,c,c$ in which no four/three/two consecutive letters are the same for the first two parts.
I got an answer for part c but I am not sure if it is correct or not.
My step is the following.
I found that there are six pairs containing adjacent identical letters. I will define them as$S_1, S_2,..., S_6$ the small number is the number of pairs of adjacent identical letters.
$$S_0 =\binom{9}{2,3,4}$$
$$S_1 =\binom{8}{3,4} + \binom{8}{2,4} + \binom{8}{2,3,2}$$
$$S_2 = \frac{7!}{4!}+\binom{7}{3,2}+\binom{7}{4,2} + \binom{7}{3,2} + \binom{6}{3,2}+\binom{7}{2,2}$$
$$S_3 = \frac{6!}{4!}+\frac{6!}{3!}+\frac{5!}{3!} + \binom{6}{2,2} + \frac{6!}{2!} + \frac{5!}{2!}$$
$$S_4 = \frac{5!}{2!}+\frac{5!}{3!}+\frac{5!}{2!} + \frac{5!}{2!} + {5!}+{4!}$$
$$S_5 = {4!}+{3!}+\frac{4!}{2!} $$
$$S_6 = 3!$$
I sum these up by the inclusion-exclusion principle, and I got the answer of the following:
$$N = S_0 - S_1+ S_2- S_3 +S_4 -S_5+ S_6 = 473$$
I want to know if there is any miscalculation or if the wrong steps exist. Could anyone please help?
|
Such problems become intractable very rapidly, I prefer to use a form of the generalized Laguerre polynomial as described by Jair Taylor
Define polynomials for $k\geq 1$ by $q_k(x) =
\sum_{i=1}^k \frac{(-1)^{i-k}}{i!} {k-1 \choose i-1}x^i$.
The number of permutations will be given by
$$\int_0^\infty \prod_j q_{k_j}(x)\, e^{-x}\,dx.$$
We get $q_2(x) = (x^2-2x)/2!$
$q_3(x) = (x^3-6x^2+6x)/3!$
$q_4(x) = (x^4-12x^2+36x^2-24x)/4!$
Using the above method, I get an answer of 79
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/4572019",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 0
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|
Generalise a function, $f\left(x\right)=x^4+ax^3+bx^2$ such that $f$ and $f'$ only has integer roots.
Consider a function, $f:\mathbb{R}\rightarrow \mathbb{R},\:f\left(x\right)=x^4+ax^3+bx^2$ where $a,b\in \mathbb{Z}\setminus \left\{0\right\}$ and $b\ne \left(\frac{a}{2}\right)^2$ with three distinct real roots whose roots and stationary points have integer $x$ coordinates. Generalise such a function, $f(x)$.
My initial thoughts were to factorise to enforce integer requirement such that $f\left(x\right)=x^2\left(x^2+ax+b\right)$ and finding $x^2+ax+b=\left(x-c\right)\left(x-d\right)$ for some $c,d\in \mathbb{Z}$ but that leads to complications when trying to enforce an integer requirement for $f'(x)$
Any help would be greatly appreciated.
|
Since $$f(x)=x^2(x^2+ax+b),\qquad f'(x)=x(4x^2+3ax+2b)$$
it is necessary that there are integers $c,d$ such that
$$a^2-4b=c^2,\qquad 9a^2-32b=d^2$$
Eliminating $b$ gives
$$a^2+2(2c)^2=d^2$$
which is of the form $x^2+2y^2=z^2$, so
$$a= \pm k|B^2-2A^2|,\quad 2c=\pm 2ABk,\quad d=\pm k(B^2+2A^2)$$
(where $A,B$ are positive integers and $k$ is a non-negative integer) giving
$$b=\frac{a^2-c^2}{4}=\frac{k^2(B^2-A^2)(B^2-4A^2)}{4}$$
Now, for $a=k(B^2-2A^2)$, we have
$$x^2+ax+b=0\implies x=A^2k-\frac{kB^2}{2}\pm\frac{kAB}{2}$$
$$4x^2+3ax+2b=0\implies x=A^2k-\frac{kB^2}{4},-\frac{kB^2}{2}+\frac{kA^2}{2}$$
For $a=-k(B^2-2A^2)$, we have
$$x^2+ax+b=0\implies x=-\bigg(A^2k-\frac{kB^2}{2}\pm\frac{kAB}{2}\bigg)$$
$$4x^2+3ax+2b=0\implies x=-\bigg(A^2k-\frac{kB^2}{4}\bigg),-\bigg(-\frac{kB^2}{2}+\frac{kA^2}{2}\bigg)$$
Therefore, in conclusion,
$$\color{red}{a=\pm k|B^2-2A^2|,\qquad b=\frac{k^2(B^2-A^2)(B^2-4A^2)}{4}}$$
where $k,A,B$ are positive integers satisfying
$$(B-A)(B-2A)\not=0,$$
$$\frac{kB^2}{4},\frac{kA}{2}\in\mathbb Z$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4572405",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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|
Showing that $\prod_{k=1}^{n} \left( 3 + 2\cos\left(\frac{2\pi}{n+1}k\right) \right)$ is the square of a Fibonacci number I was experimenting with products of the form
$$\prod_{k=1}^{n} \left( a + b\cos(ck) \right)$$
when I found that the expression
$$\prod_{k=1}^{n} \left( 3 + 2\cos\left(\frac{2\pi}{n+1}k\right) \right)$$
seems to always return a perfect square. (verified numerically).
$$\left(3 + 2\cos\left(\frac{2\pi}{2}\right)\right) = 1 = F_2^2$$
$$\left(3 + 2\cos\left(\frac{2\pi}{3}\right)\right) \left(3 + 2\cos\left(\frac{4\pi}{3}\right)\right) = 4 = F_3^2$$
$$\left(3 + 2\cos\left(\frac{2\pi}{4}\right)\right)\left(3 + 2\cos\left(\frac{4\pi}{4}\right)\right) \left(3 + 2\cos\left(\frac{6\pi}{4}\right)\right) = 9 = F_4^2$$
$$\dots$$
Furthermore, it appears these squares are the squares of Fibonacci numbers. ($F_1 = F_2 = 1$, $F_n = F_{n-1} + F_{n-2})$
I've tried to prove that this expression always gives a perfect square by considering the identity (Cassini's identity)
$$F_{n}^2 = F_{n+1}F_{n-1} + (-1)^{n-1}$$
and using induction to show that the above product satisfies this relation, and since the base cases ($F_1, F_2...$) are that of the Fibonacci sequence then this would prove that the expression is the square of Fibonacci numbers. However, this approach led nowhere.
Is there a clever way to show that the above expression is related to Fibonacci numbers?
|
Changing notation slightly (my $n$ is your $n + 1,$ and my $F_n$ is your $F_{n + 1},$ if that's not too confusing a way to put it!), let $\omega$ be a primitive $n$th root of unity. Then
$$
\sqrt{5}F_n = \varphi^n - \frac{(-1)^n}{\varphi^n} = \frac{(\varphi^2)^n - (-1)^n}{\varphi^n} = \prod_{k=0}^{n-1}\frac{\varphi^2 + \omega^k}{\varphi} = \prod_{k=0}^{n-1}\left(\varphi + \frac{\omega^k}{\varphi}\right),
$$
therefore
$$
5F_n^2 = \prod_{k=0}^{n-1}\left(\varphi + \frac{\omega^k}{\varphi}\right)\left(\varphi + \frac{\omega^{-k}}{\varphi}\right) = \prod_{k=0}^{n-1}\left(\varphi^2 + \frac1{\varphi^2} + 2\cos\left(\frac{2\pi k}{n}\right)\right).
$$
But
$$
\varphi^2 = \left(\frac{\sqrt5 + 1}{2}\right)^2 = \frac{3 + \sqrt5}{2}, \quad \therefore \ \frac1{\varphi^2} = \frac{3 - \sqrt5}{2}, \quad \therefore \ \varphi^2 + \frac1{\varphi^2} = 3.
$$
Cancelling out the factor for $k = 0,$ which is equal to $5,$ we get:
$$
F_n^2 = \prod_{k=1}^{n-1}\left(3 + 2\cos\left(\frac{2\pi k}{n}\right)\right).
$$
|
{
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"url": "https://math.stackexchange.com/questions/4573706",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
"answer_count": 3,
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|
Struggles solving : $\frac23xyy'=\sqrt{x^6-y^4}+y^2.$
Solve $\frac23xyy'=\sqrt{x^6-y^4}+y^2.$
(Hint: the equation can be boiled down to a homogeneous one.)
My attempt:
I didn't know how to obtain a homogeneous equation, so I'm going to show what I did instead.
We can write $\frac23xyy'=|x|^3\sqrt{1-\frac{y^4}{x^6}}+y^2.$ Let $z=\frac{y^2}{x^3}=y^2x^{-3}.$ Then
$$\begin{aligned}z'&=2yy'x^{-3}-3y^2x^{-4}\\ \implies x^4z'&=2xyy'-3y^2\\ \implies\frac23xyy'&=\frac{x^4z'+3y^2}3=\frac{x^4z'}3+y^2\\ \implies\frac{x^4z'}3+y^2&=|x|^3\sqrt{1-z^2}+y^2\\ \implies\frac{|x|z'}3&=\sqrt{1-z^2}\\ \implies\frac{dz}{\sqrt{1-z^2}}&=3\frac{dx}{|x|}\\ \implies\arcsin z&=3\ln|x|+C, C\in\Bbb R\\ \implies z&=\sin\ln(C|x|^3), C>0\\ \implies y^2&=x^3\sin\ln(C|x|^3)C>0.\end{aligned}$$
I'm not sure my answer is correct. I didn't get far plugging the result into the equation.
How do we solve the given ODE? Or should I better ask, how do we obtain a homogeneous ODE from it?
|
We can obtain the homogenious equation as follows.
$$\begin{align}
&\Longleftrightarrow \frac{x}{3}\cdot \frac{d(y^2)}{dx} = \sqrt{(x^3)^2-y(^2)^2}+y^2\\
&\Longleftrightarrow x^3\cdot \frac{d(y^2)}{3x^2dx} = \sqrt{(x^3)^2-y(^2)^2}+y^2\\
&\Longleftrightarrow x^3\cdot \frac{d(y^2)}{d(x^3)} = \sqrt{(x^3)^2-y(^2)^2}+y^2\\
&\Longleftrightarrow \frac{d(y^2)}{d(x^3)} = \frac{\sqrt{(x^3)^2-y(^2)^2}+y^2}{x^3} \tag{1}\\
\end{align}$$
Let's denote $(p,t) = (y^2,x^3)$ then
$$(1) \Longleftrightarrow \frac{dp}{dt} = \frac{\sqrt{t^2-p^2}+p}{t} \tag{2}$$
$(2)$ is homogenious and you can solve it easily (for example, transform $(2)$ to an equation of $\frac{p}{t}$).
|
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"timestamp": "2023-03-29T00:00:00",
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|
Are there any other decent methods to evaluate $\int_0^1 \frac{\ln \left(1-x^4\right)}{1+x^2} d x?$ We first split the integrand into 3 parts as
\begin{aligned}
\int_0^1 \frac{\ln \left(1-x^4\right)}{1+x^2} d x &= \underbrace{\int_0^1 \frac{\ln \left(1+x^2\right)}{1+x^2} d x}_J+\underbrace{\int_0^1 \frac{\ln (1+x)}{1+x^2} d x}_K+ \underbrace{\int_0^1 \frac{\ln (1-x)}{1+x^2} d x}_L
\end{aligned}
Denotes the Catalan’s constant by $G$.
By my post, $$J=\frac{\pi}{2}\ln2-G$$
Dealing with the last $2$ integrals, we use a powerful substitution $x=\frac{1-t}{1+t} ,$ then $dx=-\frac{2dt}{(1+t)^2}.$
$$
\begin{aligned}
K&=\int_1^0 \frac{\ln 2-\ln (1+t)}{\frac{2+2 t^2}{(1+t)^2}} \frac{-2 d t}{(1+t)^2} \\
&=\ln 2 \int_0^1 \frac{d t}{1+t^2}-\int_0^1 \frac{\ln (1+t)}{1+t^2}
\end{aligned}
$$
Hence $$K=\frac{\pi}{8} \ln 2 $$
$$
\begin{aligned}
L=& \int_0^1 \frac{\ln 2+\ln t-\ln (1+t)}{1+t^2} d t \\
&=\frac{\pi}{4} \ln 2+\int_0^1 \frac{\ln t}{1+t^2}-\int_0^1 \frac{\ln (1+t)}{1+t^2} d t . \\
&=\frac{\pi}{4} \ln 2-G-\frac{\pi}{8} \ln 2 \\
&=\frac{\pi}{8} \ln 2-G
\end{aligned}
$$
Combining them to get
$$
\begin{aligned}
I &=\left(\frac{\pi}{2} \ln 2-G\right) +\frac{\pi}{8} \ln 2 +\left(\frac{\pi}{8} \ln 2-G\right)\\
&=\frac{3 \pi}{4} \ln 2-2 G
\end{aligned}
$$
I do want to know if it can be solved by any other elegant methods. Your comments and methods are highly appreciated.
|
This approach is somewhat similar to yours. It is more of a comment than an answer, but it is too long to fit in the comments section.
Let the integral in question be $I$. Letting $x \to \dfrac{1-x}{1+x}$ at the beginning yields
$$I = 2\int_{0}^{1}\frac{\ln\left(1-\left(\frac{1-x}{1+x}\right)^{4}\right)}{1+\left(\frac{1-x}{1+x}\right)^{2}}\left(\frac{1}{\left(1+x\right)^{2}}\right)dx$$
which simplifies down to
$$3\ln\left(2\right)\int_{0}^{1}\frac{1}{1+x^{2}}dx+\int_{0}^{1}\frac{\ln\left(x\right)}{1+x^{2}}dx+\int_{0}^{1}\frac{\ln\left(1+x^{2}\right)}{1+x^{2}}dx-4\int_{0}^{1}\frac{\ln\left(1+x\right)}{1+x^{2}}dx.$$
Evaluating the first integral trivially and using the assumptions from your query, we get
$$I = \frac{3}{4}\pi\ln\left(2\right) - G + \frac{\pi}{2}\ln\left(2\right)-G - 4\left(\frac{\pi}{8}\ln\left(2\right)\right)$$
Therefore, the integral $I$ is
$$\int_{0}^{1}\frac{\ln\left(1-x^{4}\right)}{1+x^{2}}dx = \frac{3\pi}{4}\ln\left(2\right)-2G.$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4579291",
"timestamp": "2023-03-29T00:00:00",
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|
Solve the equation $\sqrt{x^2-9x+24}-\sqrt{6x^2-59x+149}=|5-x|$ Solve the equation $$\sqrt{x^2-9x+24}-\sqrt{6x^2-59x+149}=|5-x|$$
$x=5$ is the only real solution
We can note that $x^2-9x+24>0$ and $6x^2-59x+149>0$ for all $x$. The first thing I decided to try: as $$|5-x|=\begin{cases}5-x,x\le5\\x-5,x>5\end{cases},$$ we can look at two different cases based on the sign of $5-x$: let's $x>5$. Then the equation becomes $$\sqrt{x^2-9x+24}-\sqrt{6x^2-59x+149}=x-5\\\sqrt{x^2-9x+24}=x-5+\sqrt{6x^2-59x+149}\\x^2-9x+24=(x-5)^2+6x^2-59x+149+2(x-5)\sqrt{6x^2-59x+149}\\30x-3x^2-7=(x-5)\sqrt{6x^2-59x+149}$$ We have to raise both sides to the power of 2 again, so I decided to stop here.
Something else we can try is to raise both sides of the initial equation to the power of 2, as $|5-x|^2=(5-x)^2,$ so we have $$7x^2-68x+173-2\sqrt{(x^2-9x+24)(6x^2-59x+149)}=x^2-10x+25\\3x^2-29x+72=\sqrt{(x^2-9x+24)(6x^2-59x+149)}$$ I think it's obvious I am missing something.
|
Since, $|5-x|\geqslant 0$, then you have:
$$\begin{align}&\sqrt{x^2-9x+24}-\sqrt{6x^2-59x+149}\geqslant 0\\
\iff &x^2-9x+24\geqslant 6x^2-59x+149\geqslant 0\\
\iff &-5(x-5)^2\geqslant 0\\
\iff &x=5. \end{align}$$
This implies that, if there's a solution, then it is $5$. Indeed, $x=5$ is a solution.
Remember that, sometimes it is easier to find the domain of the equation than to solve the equation itself. That's exactly what we're doing in this equation.
|
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|
How many solutions are there to the equation? I need to find the number of solutions in positive integers to the following equation:
$$x_1 + x_2 + x_3 + x_4 = 19$$
where $x_2 \neq 2x_3$ and $x_1 \neq x_2$
I know to solve the cases like $x_1 \ge 2x_3$, but I don't understand how to do the jump from this inequality to the case where they're not equal.
Any hint or guide would be much appreciated!
|
A generating function approach. We start with the number of solutions of
\begin{align*}
&\color{blue}{x_1+x_2+x_3+x_4=19}\tag{1}\\
&\color{blue}{x_1,x_2,x_3,x_4\geq 1}
\end{align*}
We represent the solutions of a variable $x_j\geq 1, 1\leq j\leq 4$ as generating function for $x_j$ which is
\begin{align*}
z+z^2+z^3+\cdots=\frac{z}{1-z}
\end{align*}
Using the coefficient of operator $[z^n]$ to denote the coefficient of $z^n$ of a series, the number of solutions of (1) is
\begin{align*}
[z^{19}]\left(\frac{z}{1-z}\right)^4&=[z^{15}]\frac{1}{(1-z)^4}=\binom{-4}{15}(-1)^{15}\color{blue}{=816}
\end{align*}
Condition $x_2\ne 2x_3$:
*
*We respect this condition by subtracting from (1) all solutions with $\color{blue}{x_2=2x_3}$. This gives
\begin{align*}
\color{blue}{x_1+3x_3+x_4=19}\tag{2}
\end{align*}
A generating function for $3x_3$ is
$z^3+z^6+\cdots=\frac{z^3}{1-z^3}$
The number of positive solutions of (2) is
\begin{align*}
[z^{19}]\left(\frac{z}{1-z}\right)^2\left(\frac{z^3}{1-z^3}\right)=[z^{14}]\frac{1}{(1-z)^2}\,\frac{1}{1-z^3}\color{blue}{=45}
\end{align*}
Condition $x_1\neq x_2$:
*
*We respect this condition by subtracting from (1) all solutions with $x_1=x_2$. This gives
\begin{align*}
\color{blue}{2x_2+x_3+x_4=19}\tag{3}
\end{align*}
The number of positive solutions of (3) is
\begin{align*}
[z^{19}]\frac{z^2}{1-z^2}\left(\frac{z}{1-z}\right)^2=[z^{15}]\frac{1}{(1-z)^2}\,\frac{1}{1-z^2}\color{blue}{=72}
\end{align*}
Conditions $x_2\ne 2x_3$ and $x_1\neq x_2$:
*
*In (2) and (3) we did some overcounting, since we counted $x_2=2x_3$ and $x_1= x_2$ twice. We have to compensate this by subtracting once the number of solutions of (1) with $x_1=x_2=2x_3$, which gives
\begin{align*}
\color{blue}{5x_3+x_4=19}\tag{4}
\end{align*}
The number of positive solutions of (4) is
\begin{align*}
[z^{19}]\frac{z^5}{1-z^5}\,\frac{z}{1-z}=[z^{14}]\frac{1}{1-z^5}\,\frac{1}{1-z}\color{blue}{=3}
\end{align*}
Combining (1) to (4) we find the number of wanted solutions is
\begin{align*}
\color{blue}{816-45-72+3=702}
\end{align*}
|
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"timestamp": "2023-03-29T00:00:00",
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|
Why $\lim\limits_{n \to \infty} \frac{\frac1{\sqrt{n+1}}}{\sqrt{n+1}-\sqrt{n}} = \lim\limits_{n \to \infty}\frac{\sqrt{n+1} + \sqrt{n}}{\sqrt{n+1}}$? I just need clarification about answer to this question:
$$\lim_{n \to \infty} \frac{\sum\limits_{i = 1}^{n} \frac{1}{\sqrt{i}}}{\sqrt{n}} = \lim_{n \to \infty} \frac{\sum\limits_{i = 1}^{n+1} \frac{1}{\sqrt{i}} - \sum\limits_{i = 1}^{n} \frac{1}{\sqrt{i}}}{\sqrt{n+1} - \sqrt{n}} = \lim_{n \to \infty} \frac{\frac{1}{\sqrt{n+1}}}{\sqrt{n+1}-\sqrt{n}} = \lim_{n \to \infty}\frac{\sqrt{n+1} + \sqrt{n}}{\sqrt{n+1}} = 2$$
$$\lim_{n \to \infty} \left(\frac{1}{\sqrt{n}} \sum\limits_{i = 1}^{n} \frac{1}{\sqrt{i}}\right)^k = 2^k$$
In this case I don't understand why
$$\lim_{n \to \infty} \frac{\frac{1}{\sqrt{n+1}}}{\sqrt{n+1}-\sqrt{n}} = \lim_{n \to \infty}\frac{\sqrt{n+1} + \sqrt{n}}{\sqrt{n+1}}$$
My reasoning. $$\frac{ \frac{1}{a} }{b - c} = \frac{1}{a(b-c)}$$Hence, we get:
$$\lim_{n \to \infty} \frac{\frac{1}{\sqrt{n+1}}}{\sqrt{n+1}-\sqrt{n}}=\lim_{n \to \infty} \frac{1}{n+1 - \sqrt{n}\sqrt{n+1}} = \lim_{n \to \infty} \frac{ \frac{1}{n} }{ \frac{n}{n} + \frac{1}{n} + \sqrt{\frac{n^2}{n^2}+\frac{n}{n^2} } } = \lim_{n \to \infty}\frac{0}{1+0+\sqrt{1+0}}=0$$
While $\lim_{n \to \infty}\frac{\sqrt{n+1} + \sqrt{n}}{\sqrt{n+1}} = \lim_{n \to \infty}\frac{ \sqrt{\frac{n}{n} + \frac{1}{n}} + \sqrt{\frac{n}{n}} }{ \sqrt{\frac{n}{n} + \frac{1}{n} } } = \lim_{n \to \infty}\frac{\sqrt{1+0} + \sqrt{1} }{\sqrt{1+0}} = 2$
Any explanation, what's going on? Thank you
|
We have,
$$\lim_{n \to \infty} \frac{\frac{1}{\sqrt{n+1}}}{\sqrt{n+1}-\sqrt{n}} = \lim_{n \to \infty}\left(\frac{1}{\sqrt{n+1}}\cdot \frac{1}{\sqrt{n+1}-\sqrt{n}}\cdot \frac{\sqrt{n+1} + \sqrt{n}}{\sqrt{n+1} + \sqrt{n}}\right)$$
To rationalize the denominator $\sqrt {n+1}-\sqrt n$, we need to use the conjugate based on the formula: $a^2-b^2=(a+b)(a-b)$, which gives
$$\left(\sqrt{n+1} - \sqrt{n}\right)\cdot \left(\sqrt{n+1} + \sqrt{n}\right)=n+1-n=1.$$
Hence, we have
$$
\begin{align}\lim_{n \to \infty} \frac{\sqrt{n+1} + \sqrt{n}}{\sqrt{n+1}}=\lim_{n \to \infty} \frac{\sqrt{n+1}}{\sqrt{n+1}}+\lim_{n \to \infty} \frac{\sqrt{n}}{\sqrt{n+1}}=1+1=2.\end{align}
$$
|
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|
Diophantine equation ${x ^ 2}+{y ^ 2}+{z ^ 2}={m ^ 2}+{n ^ 2} $ has infinite solutions. What is its general solution formula? Diophantine equation ${x ^ 2}+{y ^ 2}+{z ^ 2}={m ^ 2}+{n ^ 2} $ has infinite solutions. For examples,
$\begin{array}{l}
{1^2}+ {3^2} + {4^2} = {1^2} + {5^2}\\
{1^2} + {2^2} + {6^2} = {4^2} + {5^2}\\
{322^2} + {562^2} + {597^2} = {601^2} + {644^2}\\
{938^2} + {1063^2} + {4722^2} = {536^2} + {4901^2}\\
\vdots
\end{array}$
What is its general solution formula? How to get the general solution formula?
|
When $z\neq 0,$ that means we want all rational solutions to:
$$m^2+n^2-x^2-y^2=1\tag 1$$
We pick one solution, $\mathbf s_0=(m_0,n_0,x_0,y_0)=(-1,0,0,0)$ and any directional vector $\mathbf p=(p_1,p_2,p_3,p_4)$ with the $p_i$ integers with $\gcd(p_1,p_2,p_3,p_4)=1.$
Then $\mathbf s_0+t\mathbf p$ yields a solution for $t=0$ trivially, and we get another value $t$ where you get another solution.
Specifically, you want $$(-1+p_1t)^2+(p_2t)^2-(p_3t)^2-(p_4t)^2=1$$ which, when simplified and factoring out $t,$ gives you:
$$-2p_1+(p_1^2+p_2^2-p_3^2-p_4^2)t=0\\
t=\frac{2p_1}{p_1^2+p_2^2-p_3^2-p_4^2}$$ where now we assume $p_1^2+p_2^2\neq p_3^2+p_4^2.$
Then our integer solution formula is $$\begin{align}m&=p_1^2-p_2^2+p_3^2+p_4^2\\
n&=2p_1p_2\\
x&=2p_1p_3\\
y&=2p_1p_4\\
z&=p_1^2+p_2^2-p_3^2-p_4^2.
\end{align}$$
This won't give primitive solutions, in that we won't always have $\gcd(m,n,x,y,z)=1,$ but we'll get one from every equivalence class of solutions with $z\neq 0.$
This will also give solutions with $z=0,$ but we don't know yet if it will give all (equivalence classes) of solutions for $z=0.$ But we can show if $z=0,$ the above solution is a scalar multiple of $(m,n,x,y,z)=(p_1,p_2,p_3,p_4,0),$ which gives all solutions.
The fact that three of the numbers are even makes it clear that we can't get all the primitive solutions with $z\neq 0.$ For example, we can get values $m$ odd, so we should be able to get $n$ odd.
For example, $(p_i)=(2,1,1,1)$ gives a solution $(m,n,x,y,z)=(5,4,4,4,3).$
This means we need a solution with $(m,n)$ switched, and possibly scaled .
The corresponding rational solution to $(1)$ is $(m,n,x,y)=(\frac43,\frac53,\frac43,\frac43)$ which is in the direction $p=(7,5,4,4)$ from $s_0.$
That gives $m=56, n=70, x=56,y=56, z=42.$ And that is $14$ times our goal $(m,n,x,y,z)=(4,5,4,4,3).$
This technique works for all homogenous quadratic Diophantine equations: once you find one non-zero solution, you can find an infinite class. But the formula won't general all solutions - it will only generate all solutions of the corresponding rational equation, which means it will generate all solutions modulo scalar multiplication, and with one fixed variable chosen to be non-zero.
In general, we would first assume $z\neq 0,$ apply the above technique, then assume $z=0$ and solve the resulting smaller Diophantine equation using the same technique. But often, the resulting formula for $z\neq 0$ would give (up to equivalence) all solutions for $z=0.$
Sometimes, we get an obvious variable to assume non-zero. $x^2+xy+y^2=z^2$ has only the trivial solution when $z=0,$ so we can assume $z\neq 0$ for all others.
|
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|
Prove $\cos(2\pi/7)$, $\cos(4\pi/7)$, and $\cos(6\pi/7)$ are roots of the equation $8x^3 + 4x^2 - 4x - 1 = 0$. The first part of the question asked you to show that $\theta = 0, 2\pi/7, 4\pi/7, 6\pi/7$ when $\cos(4\theta) = \cos(3\theta)$. By using $\cos(\theta) = \frac{1}{2} \left(z + \frac{1}{z} \right)$, I have managed to show that $$(z - 1)\left(z^3 - \frac{1}{z^4} \right) = 0$$ which gives you the seventh roots of unity. It is showing that those cosines are roots of the equation that I'm stuck on. The textbook has done similar problems, but those are conveniently much easier.
First, I tried expanding $$\cos(4\theta) = 8\cos^4(\theta) - 8\cos^2(\theta) + 1$$ and $$\cos(3\theta) = 4\cos^3(\theta) - 3\cos(\theta)$$ as powers of cosine, but am unable to get the cubic equation above (by substituting $x = \cos(\theta)$).
(Those cosines of multiples of $2\pi/7$are indeed roots of the cubic, I checked them.)
Next, I attempted to brute-force the cubic by expanding $(x - \cos(2\pi/7))(x - \cos(4\pi/7))(x - \cos(6\pi/7))$, giving you $$x^3 - \left(\cos\left(\frac{2\pi}{7}\right) + \cos\left(\frac{4\pi}{7}\right) + \cos\left(\frac{6\pi}{7}\right)\right)x^2 + \left(\cos\left(\frac{2\pi}{7}\right)\cos\left(\frac{4\pi}{7}\right) + \cos\left(\frac{2\pi}{7}\right)\cos\left(\frac{6\pi}{7}\right) + \cos\left(\frac{4\pi}{7}\right)\cos\left(\frac{6\pi}{7}\right)\right)x - \left(\cos\left(\frac{2\pi}{7}\right)\cos\left(\frac{4\pi}{7}\right)\cos\left(\frac{6\pi}{7}\right)\right).$$ The coefficient of $x^2$ equals $-0.5$, but again, the values of the coefficient of $x$ and the constant term evade me (not allowed to use decimals).
How could I find their values? Or is there another, simpler way of solving the question?
|
Using your work, note that $\cos(4 \theta) = \cos(3 \theta)$ gives:
$$(8u^4 - 8u^2 + 1) - (4u^3 - 3u) = 0 \implies 8u^4 - 4u^3 - 8u^2 + 3u + 1 = 0$$
where $u = \cos \theta$. $u = 1$ is a root, and hence dividing through by $u - 1$ we get:
$$\begin{array}{c|rrrr} 1 & 8 & -4 & -8 & 3 & 1 \\ & \downarrow & 8 & 4 & -4 & -1 \\ \hline & 8 & 4 & -4 & -1 & 0 \end{array}$$
or $8u^3 + 4u^2 - 4u - 1 = 0$.
|
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|
Show that the equation $X^2=J_n$has no solution if $n\geq2$, if $J_n$ is the jordan block of size $n$ with zeros on diagonal. More clearly:
$$
J_n=
\begin{pmatrix}
0 & 1 & 0 &\cdots & 0 & 0 \\
0 & 0 & 1 &\cdots & 0 & 0 \\
\vdots&\vdots&\vdots&\ddots&\vdots&\vdots\\
0 & 0 & 0 &\cdots & 1 & 0 \\
0 & 0 & 0 &\cdots & 0 & 1 \\
0 & 0 & 0 &\cdots & 0 & 0
\end{pmatrix}
$$
The solutions manual says: $X$ is nilpotent, thus $\operatorname{dim}(\operatorname{ker} X^2) \geq 2$ and $X^2$ cannot have rank $n - 1$.
I understand that $X$ is nilpotent, but I do not understand how that implies $\operatorname{dim}(\operatorname{ker} X^2) \geq 2$.
|
Let $y$ be any $n$-dimensional vector not in Ker$(X)$, such that $X^2y$ is nonzero. [That $X^2y$ is nonzero for some $y$ is what would have to be for $X^2$ to be $J$, and furthermore, for Dim Ker$(X^2)$ to be only $1$.] Then the fact that $X$ is nilpotent gives $X^ny=0$ for some positive integer $n$. Thus, let $n$ now be the smallest integer such that $X^ny=0$. Then $n$ is at least $3$. Furthermore:
*
*Then $X^ny=X(X^{n-1}y)$ is $0$, which gives $X^{n-1}y \in$ Ker$(X)$. That $n$ is the smallest integer and at least $3$ such that $X^{n}y=0$ gives $X^{n-1}y$ a nonzero vector in Ker$(X)$.
*Furthermore, $X^{n-1}y$ a nonzero vector in Ker$(X)$ gives $z=X^{n-2}y$ a nonzero vector not in Ker$(X)$ satisfying $X^2z = X^2(X^{n-2}y) = 0$. [Indeed, $z$ cannot be in Ker$(X)$, lest $Xz=$ $X(X^{n-2}y)= X^{n-1}y$ is $0$, which would contradict that $n$ is the smallest integer such that $X^ny=0$.]
But then this gives $X^2z = X^2(X^{n-2}y) = 0$ for some $z$ not in Ker$(X)$. But also $X^2v = X(Xv)$ $= X(0) = 0$ for all nonzero $v \in$ Ker$(X)$. So then Ker$(X^2)$ is a subset of $\langle v,z \rangle$, and $z \not \in$ Ker$(X)$ and $v \in$ Ker$(X)$ gives $2$ $\le$ Dim$(\langle v,z \rangle)$ $\le$ Dim Ker$(X^2)$. Which is what you want.
|
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|
Without calculus, what is the maximum value of the rational function : $f(t)= \frac {30t} {t^2 + 2}$ Source : Stewart, Precalculus.
The original question is to graph the function : $f(t)= \frac {30t} {t^2 + 2}$.
Desmos construction : https://www.desmos.com/calculator/19kybcl3hc
It can be shown that $f(x)$ tends to $0$ as $t$ goes to infinity ( in both directions), so $y= 0$ is a horizonal asymptote.
Also, near zero, the function looks like $y = t$.
The part of the question I am interested in here is : what is the
maximum value of $f(t)$, not using calculus?
Following a method shown by Sybermath ( ' Finding the maximum value of a rational function", YT ) , I attempted this approach:
(1) Set $f(t)= M$ , that is $f(t)= \frac {30t} {t^2 + 2}=M$.
The question becomes: what is the maximum value of $M$?
(2) $\frac {30t} {t^2 + 2}=M$
$ \iff 30t = M(t^2 +2)$
$\iff 30t = Mt^2 +2M$
$\iff -Mt^2 +30 t -2M = 0$
(3) Since we only want real values of $t$ , we require
$\Delta = b^2 - 4 ac \geq 0 $
$\iff 30^2 - 4(-M)(-2M) \geq 0 $
$\iff 30^2 -8M^2 \geq 0$
$\iff M^2 \leq 30^2/8$
$\iff \sqrt{M^2} \leq \sqrt {30^2 / 8}$
$\iff |M|\leq \sqrt {30^2 / 8}$
$\iff -\sqrt {30^2 / 8} \leq M \leq \sqrt {30^2 / 8} \approx 10.6$
(3) So, the maximum value of $f(x)= M $ is $\sqrt {30^2 / 8} \approx 10.6.$
Is this answer correct? Are there other , desirably quicker methods to answer the question ( without calculus)?
|
We can also work out a generalization in this way. We can establish by the "zero-discriminant" quadratic polynomial method you used or by application of the AM-GM inequality (Feng's approach) that $ \ x + \frac{1}{x} \ $ has its relative minimum of $ \ +2 \ $ at $ \ x \ = \ 1 \ $ for $ \ x \ > \ 0 \ $ and its relative maximum of $ \ -2 \ $ at $ \ x \ = \ -1 \ $ for $ \ x \ < \ 0 \ \ . $ We can "rescale" the variable to find that $ \frac{x}{a} + \frac{a}{x} \ $ also has its relative minimum of $ \ 2 \ $ at $ \ x \ = \ a \ $ for $ \ \frac{x}{a} \ > \ 0 \ \ . \ $ Finally, multiplying this function by $ \ a \ $ produces $ a· \left(\frac{x}{a} + \frac{a}{x} \right) \ = \ \frac{x^2 \ + \ a^2}{x} \ \ , \ $ which for $ \ a \ \neq \ 0 \ $ has its relative minimum of $ \ 2a \ $ at $ \ x \ = \ a \ \ . $ (This of course can also be arrived at by the methods mentioned above. An alternative expression is to write $ \ \frac{x^2 \ + \ a^2}{x} \ = \ c \ $ as $ \ x^2 - cx \ = \ -a^2 \ \ ; \ $ "completing the square" yields $ \ \left( x - \frac{c}{2} \right)^2 \ = \ \frac{c^2}{4} - a^2 \ \ , \ $ for which there is a single root at $ \ x \ = \ \pm a \ $ when $ \ c \ = \ \pm \ 2a \ \ . \ ) $
We then conclude that $ \ \large{\frac{x}{x^2 \ + \ a^2} } \ $ has its relative maximum of $ \ \large{ \frac{1}{+2|a|} } \ $ at $ \ x \ = \ +|a| \ $ (and its relative minimum of $ \ \frac{1}{-2|a|} \ $ at $ \ x \ = \ -|a| \ ) \ $ for $ \ a \ \neq \ 0 \ \ . \ $ The function $ \ f(x) \ = \ \large{ \frac{30x}{x^2 \ + \ 2} \ = \ 30·\frac{x}{x^2 \ + \ 2} \ } \ \ $ therefore has its maximal value of $ \ 30·\large{ \frac{1}{2·\sqrt2} \ = \ \frac{15·\sqrt2}{2} } \ $ at $ \ x \ = \ +\sqrt2 \ $ (and its minimal value of $ \ -\frac{15·\sqrt2}{2} \ $ at $ \ x \ = \ -\sqrt2 \ \ ) \ . $ The symmetry of the extrema about the origin is to be expected for an odd function.
|
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Solve $x\left(1+\sqrt{1-x^2}\right)=\sqrt{1-x^2}$ I would like to solve the equation $x\left(1+\sqrt{1-x^2}\right)=\sqrt{1-x^2}$ analytically, without using Wolfram Alpha. I have tried several substitutions, including $x=\cos(t)$, $\sqrt{1-x^2}=t$, but they all fail and eventually result in a quartic polynomial which is not easily solved. Any suggestions would be helpful.
|
I wanted to provide the trigonometric approach as a $3$rd answer, thinking it might be useful to some readers.
Substitute $x=\sin\theta$, where $0\leqslant \theta\leqslant \frac {\pi}{2}$ based on the fact $x\geqslant 0$, then we have:
$$\begin{align}&\sin\theta+\sin \theta\cos \theta=\cos \theta\\
\implies &\cos \theta-\sin \theta=\sin \theta\cos \theta\\
\implies &1-\sin (2\theta)=\frac{\sin ^2 (2\theta)}{4}\end{align}$$
Since $\sin (2\theta)\geqslant 0$ we obtain:
$$\begin{align}&\sin^2(2\theta)+4\sin (2\theta)-4=0\\
\implies &\sin (2\theta)=2\sqrt 2-2\\
\implies &\theta=\frac 12\arcsin \left(2\sqrt 2-2\right)\end{align}$$
Finally, the answer is:
$$\bbox[5px,border:2px solid #C0A000]{x=\sin\left(\frac 12\arcsin \left(2\sqrt 2-2\right)\right)}$$
Important note:
The final answer assume that, for $0\leqslant \theta\leqslant \frac {\pi}{2}$ the inequality $\cos\theta\geqslant \sin\theta$ holds to avoid extraneous roots. No other root satisfies the inequality $\cos\theta\geqslant \sin\theta$, under the restriction $0\leqslant \theta\leqslant \frac {\pi}{2}.$
|
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|
In $\triangle ABC$, if $AC=4$ and $BC=5$, and $\cos(A-B)=\frac{7}{8}$, find $\cos(C)$ The problem is as the title suggests, in the given figure below, the goal is to find the Cosine of $\angle C$. I tried multiple ways of approaching this, such as with the Law of Sines, area formula, etc but none of them seemed to lead anywhere. My actual approach, which I will post as an answer below, uses the law of Cosines. Please share your own approaches especially if they use a different method!
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A novel way to solve the problem is to change the question, with equvalent answer.
$\cos(A-B) \;=\; \sin(A)\sin(B) \;+\; \cos(A)\cos(B)$
$\cos(\quad C \quad) \;=\; \sin(A)\sin(B) \;-\; \cos(A)\cos(B)$
Only difference between the two are the sign of cosine product term.
We change the question! $\; a=5,\; b=4,\;\cos(C)=7/8,\;\text{find}\;\cos(A-B)$
$c^2 = a^2 + b^2 - 2\,a\,b\,\cos(C) = 5^2+4^2-2×5×4×\frac{7}{8} = 6$
$\displaystyle 2R = \frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}
= \sqrt{\frac{6}{1-(7/8)^2}} = \sqrt{\frac{128}{5}}$
Compare with original question, we have the same $R$
$\displaystyle \cos(A-B) + \cos(C) = 2\,\sin(A)\sin(B)$
$\displaystyle \cos(A-B) = -\cos(C) + \frac{2\,a\,b}{4R^2}
= -\frac{7}{8} \;+\; \frac{2×5×4}{128/5}
= \frac{11}{16}$
--> Original question, $\displaystyle\;\cos(C) = \frac{11}{16}$
|
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Prove $a^x+a^{1/x}\le 2a$ ($\frac{1}{2}\le a < 1$) for any $x\in (0,\infty)$
Prove $a^x+a^{1/x}\le 2a$ (with $\frac{1}{2}\le a < 1$) for any $x\in (0,\infty)$.
Is it true? I can't figure out how to prove it.
It may use the conclusion $\ln x+\frac{1-x}{x}\ge 0, \forall x\in (0,+\infty) $ as this question is followed.
UPDATED:
I have find a complicated method to prove it.
Let $f(x) = a^x+a^{1/x}-2a$, with $\frac{1}{2}\le a<1, x\in(0,\infty)$.
$\therefore f'(x)=a^{1/x}\ln \frac{1}{a}(\frac{1}{x^2}-a^{x-1/x})$
we have $a^{1/x}\ln\frac{1}{a}>0$, so we can only consider
$g(x)=\frac{1}{x^2}-a^{x-1/x}, x>0, \frac{1}{2}\le a<1$
we can make the observation that $g(1)=0$, but we are not sure if it has any other roots. So we have to do some careful analysis.
If we keep taking derivative, we will get into endless trouble.
Note that $g(x)=\frac{1}{x^2}-a^{x-1/x}=a^{-2\log_a x}-a^{x-1/x}$
because $a<1$, so $a^x$ is decreasing, we can consider $h(x)=-2\log_a x-(x-\frac{1}{x})=-2\log_a x-x+\frac{1}{x},x>0,\frac{1}{2}\le a<1$
This is much easier. $h'(x)=-\dfrac{x^2+\dfrac{2}{\ln a}\cdot x+1}{x^2}$
because $\frac{1}{2}\le a<1$, so $\frac{2}{\ln a}\le -\frac{2}{\ln 2}\approx -2.9$. For the quadratic functions in the numerator, $\Delta >0$, there are two positive zeros $x_1,x_2$ and $0<x_1<1<x_2$, we have:
In $(0,x_1)$ and $(x_2,\infty)$，$h'(x)<0$, so $h(x)$ is decreasing.
In $(x_1, x_2)$, $h'(x)>0$, so $h(x)$ is increasing.
Note that $h(1)=0$, and when $x\to 0^+$, $h(x)\to +\infty$, when $x\to +\infty, h(x)\to -\infty$, therefore, the graph of $h(x)$ is illustrated as follows:
$h(x)$" />
Therefore, there is three zeros for $h(x)$, i.e. $x_3, x_4=1, x_5$.
In $(0,x_3)$ and $(1,x_5)$, $h(x)>0$, so $g(x)<0$, so $f'(x)<0$, so $f(x)$ is decreasing.
In $(x_3,1)$ and $(x_5,+\infty)$, $h(x)<0$, so $g(x)>0$, so $f'(x)>0$, so $f(x)$ is increasing.
Note that $f(1)=0$, and when $x\to 0^+$, $f(x)\to 1-2a\le 0$; when $x\to +\infty$, $f(x)\to 1-2a\le 0$. The graph of $f(x)$ can be illustrated as follows:
$f(x)$" />
Now, we can surely say, the maximum of $f(x)$ is $f(1)=0$, which means $a^x+a^{1/x}\le 2a$, with $\frac{1}{2}\le a<1$, $\forall x\in(0,+\infty)$.
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Yes, it is true. Fix $a$ in the range $1/2 \le a < 1$. It suffices to prove
$$
a^x + a^{1/x} \le 2a
$$
for $0 < x \le 1$, because the left-hand side is invariant under the substitution $x \leftrightarrow 1/x$. This inequality can be rearranged to
$$
g(x) := a^{-1/x} (2a - a^x) \ge 1 \, .
$$
We have $g(1) = 1$, and we will now show that $g$ is decreasing on $(0, 1]$. The derivative is
$$
g'(x) = \frac{a^{-1/x} \ln(a)}{x^2} \left( 2a - (x^2+1) a^x\right) \, .
$$
The first factor is negative, so we need to show that
$$
2a - (x^2+1) a^x \ge 0
$$
for $0 \le x \le 1$. It turns out to be convenient to set $y = 1-x$. With that substitution we need to show that
$$
h(y) := 2a^y - 1- (1-y)^2 \ge 0
$$
for $0 \le y \le 1$. We have $h(0) = 0$ and $h(1) = 2a-1 \ge 0$, and the second derivative
$$
h''(y) = 2 \left( \ln^2(a) \cdot a^y - 1\right)
$$
is negative. So $h$ is concave and therefore attains its minimum on the interval $[0, 1]$ at one of the boundary points.
We have thus shown that $h(y) \ge 0$, so that $g'(x) \le 0$, and therefore $g(x) \ge g(1) = 1$ for $0 < x \le 1$, and that completes the proof.
|
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Show that $\int_{0}^{\infty}x^{-x}dx<\pi-\ln\pi$ I ask for an inequality which is a follow up of this question: Prove that $\int_0^\infty\frac1{x^x}\, dx<2$ :
$$\int_{0}^{\infty}x^{-x}dx<\pi-\ln\pi$$
You can find a nice proof @RiverLi among others and also an attempt of mine.
My try
We have:
$$\int_{1}^{\phi}\phi^{\left(a-ax^{2}\right)}dx+\int_{0}^{1}x^{-x}dx+\int_{b}^{\infty}x^{-x}dx>\int_{0}^{\infty}x^{-x}dx$$
Where $\phi=\frac{1+\sqrt{5}}{2},a=1$ because we have the inequality for $x\in[1,b]$:
$$b^{a-ax^{2}}\geq x^{-x}$$
Unfortunetaly it's already bigger...
We also have:
$$\int_{1}^{\frac{1}{2}+\sqrt{\frac{5}{4}}}\left(\frac{1+\sqrt{5}}{2}\right)^{1-x^{2}}dx+\int_{0}^{1}x^{-x}dx+\int_{\frac{1}{2}+\sqrt{\frac{5}{4}}}^{1.8}f\left(x+\frac{1}{9}\right)dx+\int_{1.8}^{\infty}x^{-x}dx<2$$
Where:
$$f\left(x\right)=2^{\frac{1+\sqrt{5}}{2}}\left(x^{\frac{1}{3}}\right)e^{-x^{\frac{4}{3}}}$$
$$f(x+1/9)>x^{-x},x\in[\phi,\infty)$$
Until now all the derivatives are easy and there is one trick .
How to show it analytically?
|
I put here my progress so far we have :
$$\int_{2}^{\pi}k\left(x\right)dx+\int_{1}^{2}f\left(x\right)dx+\int_{0}^{1}h\left(x\right)dx+\int_{\pi}^{4}\frac{3+\frac{3.5}{1000}}{\ln3}x^{\frac{12}{1000}+\frac{2}{5}}\left(\frac{3+\frac{3.5}{1000}}{\ln3}\right)^{-x^{1+\frac{2}{5}+\frac{12}{1000}}}dx+\int_{4}^{5}ex^{\frac{12}{1000}+\frac{2}{5}}e^{-x^{\left(1+\frac{2}{5}+\frac{12}{1000}\right)}}dx+\int_{5}^{\infty}5.25^{-x}e^{\left(-x+5+0.25\right)}dx\simeq 1.996912$$
Where :
$$h\left(x\right)=x^{-x}$$
$$f\left(x\right)=a\left(\left((1-c)^{2}+ac(2-c)-ac(1-c)\ln a\right)\right)$$
$$k\left(x\right)=a^{2}\left(\left((1-d)^{2}+ad(2-d)-ad(1-d)\ln a\right)\right)$$
$$a=1/x,c=x-1,d=x-2$$
For some references see my first answer on TheSimplifire's question .
I got it !!!
$$\int_{0}^{1}x^{-x}dx+\int_{1}^{2}f\left(x\right)dx+\int_{2}^{3-0.75}k\left(x\right)dx+\int_{3-0.75}^{4}\frac{3+\frac{3.5}{1000}}{\ln3}x^{\frac{12}{1000}+\frac{2}{5}}\left(\frac{3+\frac{3.5}{1000}}{\ln3}\right)^{-x^{1+\frac{2}{5}+\frac{12}{1000}}}dx+\int_{4}^{4.8}ex^{\frac{13}{1000}+\frac{2}{5}}e^{-x^{\left(1+\frac{2}{5}+\frac{13}{1000}\right)}}dx+\int_{4.8}^{\infty}5^{-x}e^{-x+5}dx<\pi-\ln \pi$$
|
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"url": "https://math.stackexchange.com/questions/4607028",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
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|
Help Solve $\int \sqrt{x^{2}+x-2}\,dx$ I have this integral $\int \sqrt{x^{2}+x-2}\,dx=\int |x-1|\sqrt{\frac{x+2}{x-1}}\,dx$.
My textbook advices to use the substitution $t^{2}=\frac{x+2}{x-1}$.
Observation: This integral should be solvable without using integral of modules as it is placed before moduled function integrals chapter.
I did the substitution and i obtaind $-18\int \frac{{t^{2}}}{|(t^{2}-1)^{3}|}\,dt$ . At this point i thought that since the substitution has to be an invertible function, i may consider a restrinction to cancel the absolute value.
I checked on wolfram alpha both integrals but it gives different results. Wolfram solution:
$$\int \sqrt{(x^2 + x - 2)}\,dx = \frac14 (2 x + 1) \sqrt{(x^2 + x - 2)} - \frac98 \log\left(2 \sqrt{(x^2 + x - 2)} + 2 x + 1\right) + c$$
Any help?
The purpose of the exercise is to trasform the irrational function into a rational one and use the Hermite's equation to solve it
|
Another standard approach, set $t = x + \sqrt{x^2 +x -2}$, then
\begin{align*}
& (t -x)^2 = x^2 +x -2 \\
& t^2 -2tx = x -2 \\
& x = \frac{t^2 +2}{1 +2t} \\
& dx = 2\frac{t^2 +t -2}{(1 +2t)^2}dt \\
\end{align*}
moreover, express the square root in terms of $t$
\begin{align*}
\sqrt{x^2 +x -2} &= t - x = \\
& = t - \frac{t^2 +2}{1 +2t} = \\
& = \frac{t^2 +t -2}{1 +2t} \\
\end{align*}
Then
\begin{align*}
\int \sqrt{x^2 +x -2} \, dx &= \int \frac{t^2 +t -2}{1 +2t} \cdot 2\frac{t^2 +t -2}{(1 +2t)^2}dt = \\
&= 2\int \frac{(t^2 +t -2)^2}{(1 +2t)^3} dt = \\
&= \int \left(\frac{1}{4}t +\frac{1}{8} +\frac{81}{16}\cdot\frac{2}{(1 +2t)^3} -\frac{9}{8}\cdot\frac{2}{1 +2t} \right)dt = \\
&= \frac{1}{8}t^2 +\frac{1}{8}t -\frac{81}{32}\cdot\frac{1}{(1 +2t)^2} -\frac{9}{8}\log{(1 +2t)} +C \\
\end{align*}
and this can be simplified, with some efforts, to
$$
\frac{1}{4}(2x+1)\sqrt{x^2+x-2} -\frac{9}{8}\log\left(2x+1+2\sqrt{x^2+x-2}\right) +C
$$
|
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"url": "https://math.stackexchange.com/questions/4608272",
"timestamp": "2023-03-29T00:00:00",
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|
Any better way to find $\cot \left(10^{\circ}\right)+\cot \left(70^{\circ}\right)-\cot \left(50^{\circ}\right)$
Find the value of $\cot \left(10^{\circ}\right)+\cot \left(70^{\circ}\right)-\cot \left(50^{\circ}\right)$
My Method:
I used the following Identities:
\begin{aligned}
& \sin A \cos B-\cos A \sin B=\sin (A-B) \\
& 2 \sin A \sin B=\cos (A-B)-\cos (A+B) \\
& \cos (2 A)=2 \cos ^2 A-1 \\
& \cos (3 A)=4 \cos ^3 A-3 \cos A \\
& \sin (2 A)=2 \sin A \cos A \\
& 2 \sin A \cos B=\sin (A+B)+\sin (A-B) \\
&
\end{aligned}
$$\begin{aligned}
S_1 & =\cot \left(10^{\circ}\right)+\cot \left(70^{\circ}\right)-\cot \left(50^{\circ}\right)=\cot(10^{\circ})-\cot \left(50^{\circ}\right)+\tan(20^{\circ}) \\
\\
\Rightarrow S_1 & =\frac{\cos \left(10^{\circ}\right)}{\sin \left(10^{\circ}\right)}-\frac{\cos \left(50^{\circ}\right)}{\sin \left(50^{\circ}\right)}+\frac{\sin \left(20^{\circ}\right)}{\cos \left(20^{\circ}\right)} \\
\\
\\
\Rightarrow S_1 & =\frac{2 \sin \left(40^{\circ}\right)}{2 \sin \left(10^{\circ}\right) \sin \left(50^{\circ}\right)}+\frac{\sin \left(20^{\circ}\right)}{\cos \left(20^{\circ}\right)} \\
\\
\Rightarrow S_1 & =\frac{2 \sin \left(40^{\circ}\right)}{\cos \left(40^{\circ}\right)-\frac{1}{2}}+\frac{\sin \left(20^{\circ}\right)}{\cos \left(20^{\circ}\right)} \\ \\
\Rightarrow S_1 & =\frac{4 \sin 40^{\circ}}{4 \cos ^2\left(20^{\circ}\right)-3}+\frac{\sin \left(20^{\circ}\right)}{\cos \left(20^{\circ}\right)}=\frac{6 \sin \left(40^{\circ}\right) \cos \left(20^{\circ}\right)-3 \sin \left(20^{\circ}\right)}{0.5}
\end{aligned}$$
$$\begin{aligned}
& \Rightarrow S_1=\frac{3}{0.5}\left(2 \sin \left(40^{\circ}\right) \cos \left(20^{\circ}\right)-\sin \left(20^{\circ}\right)\right) \\ \\
& \Rightarrow S_1=6\left(\sin \left(60^{\circ}\right)\right)=3 \sqrt{3}
\end{aligned}$$
|
I think that I got a different way
Another identity
$$\tan(x+y+z) = \frac{\tan x + \tan y + \tan z - \tan x \tan y \tan z}{1 - \tan y \tan z - \tan x \tan y - \tan x \tan z}$$
Now let $x = 20^{\circ}, y = -40^{\circ}, z=80^{\circ}$, so that $x+y+z = 60^{\circ}$
Let's do some math, using your identities in your post
\begin{align*}
&\tan x \tan (-y) \tan z \\
&= \frac{\sin (20^{\circ})\sin(40^{\circ})\sin(80^{\circ})}{\cos(20^{\circ})\cos(40^{\circ})\cos(80^{\circ})} \\
&= \frac{\sin(40^{\circ})\sin(80^{\circ})\sin(160^{\circ})}{\cos(20^{\circ})\cos(40^{\circ})\cos(80^{\circ})} \\
&= 8\sin(20^{\circ})\sin(40^{\circ})\sin(80^{\circ}) \\
&= 4(\cos(20^{\circ}) - \cos(60^{\circ}))\sin(80^{\circ}) \\
&= 4\sin(70^{\circ})\sin(80^{\circ}) - 2\sin(80^{\circ}) \\
&= 2(\cos(10^{\circ}) - \cos(150^{\circ})) - 2\sin(80^{\circ}) \\
&= 2\cos(30^{\circ}) \\
&= \sqrt{3}
\end{align*}
Using the identity
$$
\tan (a-b) = \frac{\tan a - \tan b}{1+\tan a \tan b}
$$
we obtain
\begin{align*}
&\tan x \tan z + \tan x \tan y + \tan x \tan z \\
=& -3 + (1 + \tan x \tan z) + (1 + \tan x \tan y) + (1 + \tan z \tan y) \\
=& -3
+ \frac{\tan (20^{\circ}) - \tan (80^{\circ})}{\tan(-60^{\circ})}+ \frac{\tan(20^{\circ}) + \tan(40^{\circ})}{\tan(60^{\circ})} + \frac{\tan(80^{\circ}) + \tan(40^{\circ})}{\tan(120^{\circ})} \\
=& -3
\end{align*}
By the identity at the beginning, we obtain
\begin{align*}
\tan(60^{\circ}) &= \frac{\tan x + \tan y + \tan z + \sqrt{3}}{1 + 3} \\
4\sqrt{3} &= \tan x + \tan y + \tan z + \sqrt{3} \\
3\sqrt{3} &= \tan(20^{\circ}) - \tan(40^{\circ}) + \tan(80^{\circ})
\end{align*}
Thus, $\cot(10^{\circ}) + \cot(70^{\circ}) - \cot(50^{\circ}) = 3\sqrt{3}$
|
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"url": "https://math.stackexchange.com/questions/4608452",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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|
Solve the equation $\log_x(9x^2)\cdot\log_3^2(x)=4$ Solve the equation $$\log_x(9x^2)\cdot\log_3^2(x)=4$$
The answers are $x_1=\dfrac19$ and $x_2=3$.
For the range we have: $$D_x:\begin{cases}x>0\\x\ne1\\9x^2>0\iff x\ne0\end{cases}\iff x\in(0;1)\cup(1;+\infty)$$
I wrote the equation as follows using $\log_a(b)=\dfrac{1}{\log_b(a)}:$ $$\log_x(9x^2)\cdot\dfrac{1}{\log_x^2(3)}=4\\\iff \log_x(9x^2)=4\log_x^2(3)=\log_x^2(3^4)$$ Is this reasonable and how can one continue?
|
$$\log_x(9x^2)\cdot\log_3^2(x)=4$$
$$\Rightarrow \dfrac{\log_3(9x^2)}{\log_3(x)}\cdot\log_3^2(x)=4$$
$$\Rightarrow \dfrac{\log_3(9)+\log_3(x^2)}{\log_3(x)}\cdot\log_3^2(x)=4$$
$$\Rightarrow \dfrac{2+2\log_3(x)}{\log_3(x)}\cdot\log_3^2(x)=4$$
$$\Rightarrow 2\log_3 (x)\left[1+\log_3(x)\right]=4$$
$$\Rightarrow \left[\log_3 (x)\right]^2+\log_3(x) - 2=0$$
Using the quadratic formula, we get $\log_3(x) = \dfrac{-1\pm\sqrt{1+8}}{2}=\dfrac{-1\pm3}{2}=1,-2$
This gives $x=3$ and $x=\dfrac{1}{9}$.
|
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"url": "https://math.stackexchange.com/questions/4608594",
"timestamp": "2023-03-29T00:00:00",
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|
Evaluate $\int \frac{\sqrt{1+x^4}}{1-x^4}dx$ How to evaluate
$$\int \frac{\sqrt{1+x^4}}{1-x^4}dx.$$
If we substitute $x=\sqrt{\tan{\theta}}$, it becomes
$$\int \frac{1}{2\cos{\theta}\cos{2\theta}\sqrt{\tan{\theta}}}d\theta.$$
What can I do next ?
Edit
Are these steps correct ?
$$=\int \frac{1}{2\cos{\theta}\cos{2\theta}\sqrt{\frac{\sin\theta}{\cos\theta}}}d\theta$$
$$=\int \frac{1}{2\cos{2\theta}\sqrt{\cos^2 \theta\frac{\sin\theta}{\cos\theta}}}d\theta$$
$$=\int \frac{1}{2\cos{2\theta}\sqrt{\cos^2 \theta\frac{\sin\theta}{\cos\theta}}}d\theta$$
$$=\int \frac{1}{\sqrt{2}\cos{2\theta}\sqrt{\sin{2 \theta}}}d\theta$$
Now If we substitute $t=\sin{2\theta}$, it becomes
$$=\int \frac{1}{\sqrt{2}(1-t^2)\sqrt{t}}dt$$
$$=\int \frac{\frac{1}{t^2}}{\sqrt{2}\frac{(1-t^2)}{t}\sqrt{\frac{t}{t^2}}}dt$$
$$=\int \frac{\frac{1}{t^2}}{\sqrt{2}(\frac{1}{t} -t)\sqrt{\frac{1}{t}}}dt$$
Now If we substitute $u^2=\frac{1}{t}$, it becomes
$$=\int \frac{-2u}{\sqrt{2}(u^2 -\frac{1}{u^2})u}du$$
$$=\int \frac{-u^2}{\sqrt{2}(u^4 -1^2)}du$$
$$=\int \frac{-u^2}{\sqrt{2}(u^2 +1)(u^2-1)}du$$
$$=-\frac{1}{2\sqrt{2}}\int \frac{u^2+1+u^2-1}{(u^2 +1)(u^2-1)}du$$
This can be solved easily now and we will get an elementary solution.
But Wolfram alpha gives a solution in non elementary functions.
Therefore I am confused. I guess the above solution is correct only for some restricted values of x.
|
Maple also gives a non-elementary answer to the original integral in terms of elliptic integral functions. However, for the integral over $\theta$ it does give an elementary answer, which translated back to the original integral becomes
$$ \frac{\sqrt{2}\, \mathrm{arctanh}\! \left(\sqrt{\sin\left(2 \arctan \! \left(x^{2}\right)\right)}\right)}{4}+\frac{\sqrt{2}\, \arctan \! \left(\sqrt{\sin\left(2 \arctan \! \left(x^{2}\right)\right)}\right)}{4}$$
which is correct for $x > 0$ (for $x < 0$ you want to multiply that by $-1$).
I believe both Maple and Wolfram Alpha use some form of the Risch algorithm to decide whether an integral is elementary, however in the algebraic case, which the original integral is, the implementation seems not to be complete.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Find the symmetric matrix given its eigenvalues and eigenvector.
$A$ is a $3 \times 3$ symmetric matrix. It has the eigenvalue $\lambda_1 = 3$ with the eigenvector
$$
\begin{bmatrix}
1 \\ 0 \\ -1
\end{bmatrix}
$$
and a double eigenvalue $\lambda_2 = -1$.
Find the matrix $A$.
Since $A$ is symmetric, it can be diagonalized orthogonally, with an orthogonal matrix $P$ as $A = PDP^T$. The matrix is made of the eigenvectors that are orthogonal to each other. What I then did was with the help of inner product, I got a vector that is orthogonal to $v_1$ and with the help of cross product I got another orthogonal vector. Those vectors are
$$
\begin{bmatrix}
1 \\ 1 \\ 1
\end{bmatrix}
\quad\text{and}\quad
\begin{bmatrix}
1 \\ -2 \\ 1
\end{bmatrix}
$$
so
$$
A = PDP^{-1}
= \begin{bmatrix}
1 & 1 & -5 \\
1 & -5 & 1 \\
-5 & 1 & -5
\end{bmatrix}.
$$
But my answer is wrong obviously since $Av \neq \lambda v$. The answer is
$$
\begin{bmatrix}
1 & 0 & -2 \\
0 & -1 & 0 \\
-2 & 0 & 1
\end{bmatrix}
$$
where the vectors are
$v_2 = \begin{bmatrix} 1 & 0 & 1 \end{bmatrix}^T$ and
$\begin{bmatrix} 0 & 1 & 0 \end{bmatrix}^T$. The answer sheet says that these vectors are orthogonal to $v_1$. Also they didn't do $A = PDP^T$ instead they did $A = PDP^{-1}$. Can anyone help me understand?
|
(I will use row vectors because they are easier to type and use up less space; interpret them as transposes of the vectors you want)
Symmetric matrices are always orthogonally diagonalizable. So we know that the eigenspace of $-1$ will be the orthogonal complement of the eigenspace of $3$; you know the eigenspace of $3$ is $\mathrm{span}\bigl((1,0,-1)\bigr)$. So a basis of eigenvectors is obtained by finding the orthogonal complement of this span, and taking an basis for it (and you might as well make it orthogonal). There are multiple ways of doing this. The most mindless/automatic is to start with $(1,0,-1)$, complete to a basis, and then apply the orthogonalization part of the Gram-Schmidt process. Doing slightly less work and more thought, we might notice that $(0,1,0)$ and $(1,0,1)$ are both orthogonal to the vector we already had, and also orthogonal to each other, so these two will do, as will any two vectors that are orthogonal to each other and to $(1,0,-1)$. Your choice of $(1,1,1)$ and $(1,-2,1)$ is certainly a valid choice.
Now, if we let $\beta=[(1,0,-1),(1,1,1),(1,-2,1)]$ be an ordered orthogonal basis for $\mathbb{R}^3$, then we know that if $Q$ is the change-of-basis matrix from the standard basis to $\beta$, then
$$QAQ^{-1} = \left(\begin{array}{rrr}
3 & 0 & 0\\
0 & -1 & 0\\
0 & 0 & -1
\end{array}\right).$$
That means that if we let $P$ be the inverse of $Q$, so that $P=Q^{-1}$ is the change-of-basis matrix fvrom $\beta$ to the standard basis, then
$$A = P\left(\begin{array}{rrr}
3 & 0 & 0\\
0 & -1 & 0\\
0 & 0 & -1
\end{array}\right)P^{-1}.$$
I don't know how you did the calculations. The matrix $P$ has the vectors of $\beta$ as columns,
$$P = \left(\begin{array}{rrr}
1 & 1 & 1\\
0 & 1 & -2\\
-1 & 1 & 1
\end{array}\right).$$
Its inverse is given by
$$P^{-1} = \left(\begin{array}{rrr}
\frac{1}{2}_{\vphantom{2_2}} & 0 & -\frac{1}{2}\\
\frac{1}{3}_{\vphantom{2_2}} & \frac{1}{3} & \frac{1}{3}\\
\frac{1}{6} & -\frac{1}{3} & \frac{1}{6}
\end{array}\right),$$
which can be computed by hand (or with a calculator), and verified by multiplying them together. Note that because the columns of $P$ are not an orthonormal basis (although they are mutually orthogonal, the vectors do not have size $1$), the inverse of $P$ is not merely $P^T$.
So we get
$$\begin{align*}
A &= P\left(\begin{array}{rrr}
3 & 0 & 0\\
0 & -1 & 0\\
0 & 0 & -1
\end{array}\right)P^{-1}\\
&= \left(\begin{array}{rrr}
1 & 1 & 1\\
0 & 1 & -2\\
-1 & 1 & 1
\end{array}\right) \left(\begin{array}{rrr}
3 & 0 & 0\\
0 & -1 & 0\\
0 & 0 & -1
\end{array}\right)
\left(\begin{array}{rrr}
\frac{1}{2}_{\vphantom{2_2}} & 0 & -\frac{1}{2}\\
\frac{1}{3}_{\vphantom{2_2}} & \frac{1}{3} & \frac{1}{3}\\
\frac{1}{6} & -\frac{1}{3} & \frac{1}{6}
\end{array}\right)\\
&= \left(\begin{array}{rrr}
3 & -1 & -1\\
0 & -1 & 2\\
-3 & -1 & -1
\end{array}\right)
\left(\begin{array}{rrr}
\frac{1}{2}_{\vphantom{2_2}} & 0 & -\frac{1}{2}\\
\frac{1}{3}_{\vphantom{2_2}} & \frac{1}{3} & \frac{1}{3}\\
\frac{1}{6} & -\frac{1}{3} & \frac{1}{6}
\end{array}\right)\\
&=\left(\begin{array}{rrr}
1 & 0 & -2\\
0 & -1 & 0\\
-2 & 0 & 1
\end{array}\right),
\end{align*}$$
which is the same as the answer they gave.
From what I can tell, you mistakenly compute $PDP^T$ instead of $PDP^{-1}$. Note that because you do not have an orthonormal basis, the inverse of $P$ is not $P^T$, as noted above. That is the source of your error. Had you used $P^{-1}$ instead of $P$, you would have obtained the same matrix $A$.
|
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|
Prove a binomial identity: $\sum_{i=1}^n i \binom{2n}{n-i}=\frac12(n+1) \binom{2n}{n-1}$ I want to prove the product of even/odd power with a combinatorial number:
\begin{aligned}
\sum_{i=1}^n i \binom{2n}{n-i}&=\frac12(n+1) \binom{2n}{n-1}, \\
\sum_{i=1}^n i^2 \binom{2n}{n-i}&=2^{2n-2} n.
\end{aligned}
I am sure these results are correct since WolframAlpha verifies them. I wonder if the first formula can be proved, and also the general version can be proved: Formula related to combinatorial number.
Below is proof for the second formula:
First, note that $r\binom{n}{r}=n\binom{n-1}{r-1}$, and $\sum_{i=0}^n\binom{n}{i}=2^n$. We thus have
\begin{aligned}
\sum_{i=0}^n i\binom{n}{i}&=2^{n-1}n, \\
\sum_{i=0}^n i^2\binom{n}{i}
&= \sum_{i=0}^n i(i-1)\binom{n}{i}+\sum_{i=0}^n i\binom{n}{i} \\
&= 2^{n-2}(n^2-n) + 2^{n-1} n \\
&= n(n+1)2^{n-2}.
\end{aligned}
Therefore,
\begin{aligned}
\sum_{i=0}^{2n}(n-i)^2\binom{2n}{i}
&= n^2\sum_{i=0}^{2n}\binom{2n}{i}-2n\sum_{i=0}^{2n}i\binom{2n}{i}+\sum_{i=0}^{2n}i^2 \binom{2n}{i} \\
&= 2^{2n}n^2-2^{2n+1}n^2+n(2n+1)2^{2n-1} \\
&= 2^{2n-1} n,
\end{aligned}
and thus
\begin{aligned}
\sum_{i=0}^n i^2 \binom{2n}{n-i}
= \sum_{i=0}^n (n-i)^2 \binom{2n}{i}
= \frac12 \sum_{i=0}^{2n} (n-i)^2 \binom{2n}{i}
= 2^{2n-2} n.
\end{aligned}
|
\begin{align}
\sum_{i=1}^n i\binom{2n}{n-i}
&= \sum_{i=0}^n i\binom{2n}{n-i} \\
&= \sum_{i=0}^n (n-i)\binom{2n}{i} \\
&= n\sum_{i=0}^n \binom{2n}{i} - \sum_{i=1}^n i\binom{2n}{i} \\
&= \frac{n}{2}\left(\binom{2n}{n}+\sum_{i=0}^{2n} \binom{2n}{i}\right) - 2n\sum_{i=1}^n \binom{2n-1}{i-1} \\
&= \frac{n}{2}\left(\binom{2n}{n}+2^{2n}\right) - 2n\sum_{i=0}^{n-1}\binom{2n-1}{i} \\
&= \frac{n}{2}\left(\binom{2n}{n}+2^{2n}\right) - \frac{2n}{2}\sum_{i=0}^{2n-1}\binom{2n-1}{i} \\
&= \frac{n}{2}\left(\binom{2n}{n}+2^{2n}\right) - n2^{2n-1}\\
&= \frac{n}{2}\binom{2n}{n}\\
&= \frac{n+1}{2}\binom{2n}{n-1}
\end{align}
|
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"timestamp": "2023-03-29T00:00:00",
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|
Find $\frac {dy}{dx}$ if $y=\frac{x\sin^{-1}x}{\sqrt{1-x^2}}$ taking JDs advice i used $(fg)'=f'g+fg'$ rule
$$f=\frac x{\sqrt{(1-x^2)}}$$ $$f'=\frac 1{\sqrt{(1-x)}^3}$$
$$g=sin^{-1}x$$ $$g'=\frac{1}{\sqrt{1-x^2}}$$
so anyway adding together
we get
$$\frac 1{(\sqrt{(1-x)}^3}*sin^-x+\frac x{\sqrt{(1-x^2)}}*\frac{1}{\sqrt{1-x^2}}$$
which can be simplified
$$\frac {sin^{-1}x}{(1-x)^3}+\frac x{\sqrt{(1-x^2)^2}}$$
i then multiplied $$\frac x{\sqrt{(1-x^2)^2}}$$ with $\sqrt{(1-x^2)}$ on both numerator and denominator
getting
$$\frac {x(\sqrt{(1-x^2)})}{\sqrt{(1-x^2)^3}}$$
and combining them both we get
$$\frac{sin^{-1}x+x(\sqrt{(1-x^2)}}{\sqrt{(1-x^2)^3}}$$
ps
i never used arcsin before and am compeletely unfamilliar with it
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*
*No.
*Since it is false, this is irrelevant.
*I would write the function like this $y=x\arcsin x(1-x^2)^{-1/2}$. By triple product rule,
$$y'=\arcsin x(1-x^2)^{-1/2}+x\frac{1}{\sqrt{1-x^2}}(1-x^2)^{-1/2}+\\x\arcsin x(-1/2)(1-x^2)^{-3/2}(-2x)$$
$$y'=\frac{\arcsin x}{(1-x^2)^{1/2}}+\frac{x}{{1-x^2}}+\frac{x^2\arcsin x}{(1-x^2)^{3/2}}$$
First and third terms can be combined to give:
$$y'=\frac{x}{{1-x^2}}+\frac{\arcsin x}{(1-x^2)^{3/2}}.$$ This is WA.
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"language": "en",
"url": "https://math.stackexchange.com/questions/4614715",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Integrating $\frac{x^3}{\sqrt{x^2 + 4x + 6}}$ Question:
Evaluate $\displaystyle\int \frac{x^3}{\sqrt{x^2 + 4x + 6}}\ dx $.
My attempt:
$\begin{align} \int \frac{x^3}{\sqrt{x^2 + 4x + 6}}\ dx & = \int \frac{x^3}{\sqrt{(x+2)^2 + 2}}\ dx \\& \overset{(1)}= \int \frac{(\sqrt{2} \tan(t) - 2 )^3 \sqrt{2} \sec^2(t)\ }{\sqrt{2\tan^2(t) + 2}}\ dt\\& = \frac{1}{\sqrt{2}}\int\frac{(\sqrt{2} \tan(t) - 2 )^3 \sqrt{2} \sec^2(t)}{\sec(t)}\ dt\\& = \int(\sqrt{2} \tan(t) - 2 )^3 \sec(t)\ dt\\& = \int -8 \sec(t) + 2 \sqrt2 \tan^3(t) \sec(t) - 12\tan^2(t) \sec(t)\\&\qquad + 12 \sqrt2 \tan(t) \sec(t)\ dt\\\\& = -8\ln|\sec(t) + \tan(t)| + 12\sqrt{2} \sec{(t)} \\
&\qquad + 2\sqrt{2} \int \tan^3(t) \sec(t) \ dt - 12\int\tan^2(t) \sec(t) \ dt\end{align}$
Now both of these integrals can be evaluated using some sort of substitution and integration by parts rule.
This would give us,
$$\boxed{-8\ln|\sec(t) + \tan(t)| + 12\sqrt{2} \sec{(t)} + 2\sqrt{2}\left[\frac{\sec^3(t)}{3} - \sec(t)\right] - 12\left[\frac{-1}{2} \ln|\sec t + \tan t| + \frac12 \sec t \tan t \right] + C}$$
$(1)$ Here I've made a substitution $t = \tan^{-1}\left(\frac{x+ 2}{\sqrt 2}\right)$.
Wolframalpha gives answer as $$\boxed{\frac{1}{3}(x^2 - 5x + 18) \sqrt{x^2 +4x + 6} - 2 \sinh^{-1}\left(\frac{x+2}{\sqrt{2}}\right) + C}$$
How would I simplify my answer equal to this? Undoing my substitution $t = \tan^{-1}\left(\frac{x+ 2}{\sqrt 2}\right)$ is also not an easy task I think.
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Undoing the substitution is exactly how you would match the two results.
By the Pythagorean theorem, a right triangle with a reference angle $t$ such that $\tan(t)=\frac{x+2}{\sqrt2}$ has its sides occurring in a ratio of $\sqrt2$ (leg adjacent to $t$) to $x+2$ (leg opposite $t$) to $\sqrt{(x+2)^2+2}=\sqrt{x^2+4x+6}$ (hypotenuse). It follows that
$$t = \tan^{-1}\left(\frac{x+2}{\sqrt2}\right) \\
\implies
\begin{cases} \tan(t) = \frac{x+2}{\sqrt2} \\
\sec(t)=\frac{\sqrt{x^2+4x+6}}{\sqrt2} \\
\sec(t)+\tan(t) = \frac{\sqrt{x^2+4x+6}+x+2}{\sqrt2} \\
\sec(t)\tan(t) = \frac{(x+2)\sqrt{x^2+4x+6}}2 \\
\ln\left|\sec(t)+\tan(t)\right| = \ln\left(\frac{\sqrt{x^2+4x+6}+x+2}{\sqrt2}\right)
\end{cases}$$
After making the replacements, your antiderivative reduces to
$$-2 \ln\left(\frac{\sqrt{x^2+4x+6}+x+2}{\sqrt2}\right) + (4-3x) \sqrt{x^2+4x+6} + \frac13 (x^2+4x+6)^{3/2} + C$$
Factorize the last two terms as
$$\bigg(4-3x + \frac13 (x^2+4x+6)\bigg) \sqrt{x^2+4x+6} = -\frac13 (x^2+5x-18) \sqrt{x^2+4x+6}$$
Finally, the logarithm can be rewritten using the definition of the inverse hyp. sine:
$$\begin{align*}
\ln\left(\frac{\sqrt{x^2+4x+6}+x+2}{\sqrt2}\right) &= \ln\left(\sqrt{\frac{(x+2)^2+2}2} + \frac{x+2}{\sqrt2}\right) \\
&= \ln\left(\sqrt{\left(\frac{x+2}{\sqrt2}\right)^2+1} + \frac{x+2}{\sqrt2}\right) \\[1ex]
&= \sinh^{-1}\left(\frac{x+2}{\sqrt2}\right)
\end{align*}$$
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"language": "en",
"url": "https://math.stackexchange.com/questions/4615616",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 2
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