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Why is $(n+1)^{n-1}(n+2)^n>3^n(n!)^2$
Why is $(n+1)^{n-1}(n+2)^n>3^n(n!)^2$ for $n>1$
I can use $$(n+1)^n>(2n)!!=n!2^n$$ but in the my case, the exponent is always decreased by $1$, for the moment I don't care about it, I apply the same for $n+2$
$(n+2)^{n+1}>(2n+2)!!=(n+1)!2^{n+1}$
gathering everything together,
$(n+1)^{n-1}(n+2)^n=\frac{(n+1)^n(n+2)^{n+1}}{(n+1)(n+2)}>\frac{(n+1)(n!)^22^{2n+1}}{(n+1)(n+2)}$
$\iff(n+1)^{n-1}(n+2)^n>(n!)^2\times\frac{2^{2n+1}}{(n+2)}$
but $\frac{2^{2n+1}}{(n+2)}>3^n$ is not true for $n=2$
can you suggest another approach ?
| prove: $(n+1)^{nβ1}(n+2)^n>3^n(n!)^2=3^n(n!n!)$ for $n>1$
$n=2:3^14^2=48>3^2(2)(2)=36$
assume: $(n+1)^{nβ1}(n+2)^n>3^n(n!n!)$
need to arrive at: $(n+2)^{n}(n+3)^{n+1}>3^{n+1}(n+1)!(n+1)!$
for lhs need to multiply by:
${{(n+2)^{n}(n+3)^{n+1}}\over {(n+1)^{nβ1}(n+2)^n}}
={{(n+3)^n(n+3)(n+1)}\over {(n+1)^n}}=({{n+3}\over {n+1}})^n(n+3)(n+1)$
where $({{n+3}\over {n+1}})^n$ is greater than $3$ for $n=3$ and increasing for all $n\in N$.
for rhs need to multiply by: $3(n+1)(n+1)$. therefore lhs > rhs and
$(n+1)^{nβ1}(n+2)^n>3^n(n!)^2$ for $n>1$ by mathematical induction.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2067511",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
How do I solve quadratic equations when the coefficients are complex and real? I needed to solve this: $$x^2 + (2i-3)x + 2-4i = 0 $$
I tried the quadratic formula but it didn't work. So how do I solve this without "guessing" roots? If I guess $x=2$ it works; then I can divide the polynomial and find the other root; but I can't "guess" a root.
$b^2-4ac=4i-3$, now I have to work with $\sqrt{4-3i}$ which I don't know how. Apparently $4i-3$ is equal to $(1+2i)^2$, but I don't know how to get to this answer, so I am stuck.
| The quadratic formula is perfectly valid here for the same reason for which it is valid when working only with real numbers:
$$
\frac{-b\pm\sqrt{b^2 - 4ac}} 2 = \frac{-(2i-3) \pm\sqrt{-3+4i}} 2.
$$
The question now is how to find $\pm\sqrt{-3+4i}.$
In polar form, you have $-3+4i = \sqrt{3^3+4^2} \cdot(\cos\alpha+i\sin\alpha) = 5(\cos\alpha+i\sin\alpha)$ where $\cos\alpha = -3/5$ and $\sin\alpha = 4/5,$ so $\tan\alpha = -4/3.$ We then have
$$
\sqrt{5(\cos\alpha+i\sin\alpha)} = \sqrt 5 \cdot\left( \cos\frac\alpha 2 + i\sin\frac\alpha 2 \right).
$$
Now recall from trigonometry that $\tan\dfrac\alpha 2 = \dfrac{\sin\alpha}{1+\cos\alpha} = \dfrac{4/5}{1+(-3/5)} = 2.$
Since $\tan=\dfrac{\text{opposite}}{\text{adjacent}}$, we want $\tan\dfrac\alpha 2 = \dfrac 2 1.$ Since $\tan={\sin}/{\cos},$ the fact that $\tan=2$ means that $\sin = 2\cos.$ Thus we have
$$
\sqrt 5 \cdot \left( \cos\frac\alpha 2 + i\sin\frac\alpha 2 \right) = \sqrt 5 \left( f + i (2f) \right)
$$
where $f^2 + (2f)^2 = \cos^2 + \sin^2 = 1,$ so $f=\dfrac 1 {\sqrt 5},$ and we then have
$$
\sqrt 5\left(\cos\frac\alpha 2 + i\sin\frac \alpha 2\right) = \sqrt 5\left( \frac 1 {\sqrt 5} + i \frac 2 {\sqrt 5} \right) = 1 + 2i.
$$
Thus $\pm\sqrt{-3+4i} = \pm(1+2i).$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Given a positive number n, how many tuples $(a_1,...,a_k)$ are there such that $a_1+..+a_k=n$ with two extra constraints The problem was: Given a positive integer $n$, how many tuples $(a_1,...,a_k)$ of positive integers are there such that $a_1+a_2+...+a_k=n$. And $0< a_1 \le a_2 \le a_3 \le...\le a_k$. Also, $a_k-a_1$ is either $0$ or $1$.
Here is what I did:
For $n=1$, there is one way $1=1$.
For $n=2$, there are $2$ ways, $2=1+1,2=2$
For $n=3$, there are $3$ ways, $3=1+1+1,3=1+2,3=3$
For $n=4$, there are $4$ ways, $4=1+1+1+1,4=1+1+2,4=2+2,4=4$
So it seems that there are $n$ tuples that satisfies the three conditions for each $n$. But I'm not sure how to prove it.
| This can also be done with a simple generating function. Call the
desired quantity $T_n.$ We first choose the value $a_1$ and then the
gaps between successive values among the $a_q$ which are either zero
or one. We will have $1\le k\le n$ but for $k=1$ there is just one
possibility so we may assume $2\le k\le n.$ We get for $a_1,$ which is
positive,
$$\frac{z}{1-z}$$
but it contributes to all $k$ terms so we have in fact
$$\frac{z^k}{1-z^k}.$$
We combine this with $k-1$ gaps of either zero or one. The first gap
contributes to $k-1$ terms, the next one to $k-2$ and so on. Using the
variable $u$ to mark gaps we thus obtain
$$\bbox[5px,border:2px solid #00A000]{
1 + \sum_{k=2}^n
\frac{z^k}{1-z^k} \prod_{q=1}^{k-1} \left(1+uz^{k-q}\right).}$$
Now from the constraint that $a_k-a_1$ be either zero or one it
follows we need the coefficient on $[u^0]$ and $[u^1].$ We set $u=0$
for the first one and get
$$\sum_{k=2}^n \frac{z^k}{1-z^k}.$$
We differentiate and set $u=0$ for the second one and obtain
$$\left.\sum_{k=2}^n
\frac{z^k}{1-z^k} \prod_{q=1}^{k-1} \left(1+uz^{k-q}\right)
\sum_{q=1}^{k-1} \frac{z^{k-q}}{1+uz^{k-q}}\right|_{u=0}
\\ = \sum_{k=2}^n
\frac{z^k}{1-z^k}\sum_{q=1}^{k-1} z^{k-q}.$$
It follows that the desired result is given by
$$\bbox[5px,border:2px solid #00A000]{
1 + [z^n] \sum_{k=2}^n
\frac{z^k}{1-z^k}\sum_{q=1}^{k} z^{k-q}.}$$
Evaluating this we initially obtain
$$1 + [z^n] \sum_{k=2}^n
\frac{z^{2k}}{1-z^k}\sum_{q=1}^{k} z^{-q}
= 1 + [z^n] \sum_{k=2}^n
\frac{z^{2k}}{1-z^k} \frac{1}{z} \sum_{q=0}^{k-1} z^{-q}
\\ = 1 + [z^n] \sum_{k=2}^n
\frac{z^{2k}}{1-z^k} \frac{1}{z} \frac{1-1/z^k}{1-1/z}
= 1 + [z^n] \sum_{k=2}^n
\frac{z^{2k}}{1-z^k} \frac{1-1/z^k}{z-1}
\\ = 1 + [z^n] \sum_{k=2}^n
\frac{z^k}{1-z^k} \frac{z^k-1}{z-1}
= 1 - [z^n] \sum_{k=2}^n
\frac{z^k}{z-1}.$$
Now we may extend $k$ to infinity because the terms for $k\gt n$ do
not contribute to the coefficient on $[z^n],$ getting
$$1 + [z^n] \frac{1}{1-z} \sum_{k\ge 2} z^k
= 1 + [z^n] \frac{z^2}{(1-z)^2}
\\ = 1 + [z^{n-2}] \frac{1}{(1-z)^2}
= 1 + {n-2+1\choose 1}.$$
This yields the end result
$$\bbox[5px,border:2px solid #00A000]{T_n = n.}$$
I do think this computation is interesting and perhaps different from
what one might have expected.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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} |
Method of solving an Associated Legendre Equation with m=0, and m not equal to 0 with explanation I am looking to find the Legendre polynomial for the following standard Associated Legendre equation:
$$\left(\frac{d}{dx}[1-x^2]\frac{d}{dx}+\lambda-\frac{m^2}{1-x^2}\right)\Theta(x)=0$$
I have found the indicial equation for m=0, but not for m$\neq$0. How do I go about solving this case?
| What you can show is that, if $y$ is a solution of
$$
\frac{d}{dx}\left((1-x^2)\frac{dy}{dx}\right)+\mu y =0,
$$
then
$$
w=(1-x^2)^{m/2}\frac{d^my}{dx^m}
$$
is a solution of
$$
\frac{d}{dx}\left((1-x^2)\frac{dy}{dx}\right)-\frac{m^2}{1-x^2}y+\mu y = 0. \tag{$\dagger$}
$$
Therefore, if $P_n(x)$ is the Ordinary Legendre Polynomial, which is the polynomial solution of $((1-x^2)y')'+n(n+1)y=0$, then
$$
P_{n,m}(x) = (1-x^2)^{m/2}\frac{d^{m}}{dx^{m}}P_{n}(x)
$$
is a solution of the Associated Legendre Equation $(\dagger)$. This holds for a fixed $m$ and for all $n \ge m$ (for $m > n$, the above is identically $0$.) So the Associated Legendre equation has eigenvalues
$$
\mu_n=n(n+1),\;\;\; n=m,m+1,m+2,\cdots.
$$
For example: $x$ is a solution of the ordinary equation with $\mu=(1)(1+1)=2$. Therefore
$$
\sqrt{1-x^2}\frac{d}{dx}x = \sqrt{1-x^2}
$$
is a solution of the associated equation with $\mu=(2)(1)$. To check this,
$$
\frac{d}{dx}\left((1-x^2)\frac{d}{dx}\sqrt{1-x^2}\right)-\frac{1}{1-x^2}\sqrt{1-x^2} \\
= -\frac{d}{dx}\left((1-x^2)\frac{x}{\sqrt{1-x^2}}\right)-\frac{1^2}{\sqrt{1-x^2}} \\
= -\frac{d}{dx}(x\sqrt{1-x^2})-\frac{1}{\sqrt{1-x^2}} \\
= -\sqrt{1-x^2}+x\frac{x}{\sqrt{1-x^2}}-\frac{1}{\sqrt{1-x^2}} \\
= -\sqrt{1-x^2}+\frac{x^2-1+1}{\sqrt{1-x^2}}-\frac{1}{\sqrt{1-x^2}} \\
= -2\sqrt{1-x^2}
$$
Therefore, as expected, $f=\sqrt{1-x^2}$ is a solution of
$$
\frac{d}{dx}\left((1-x^2)\frac{df}{dx}\right)-\frac{1^2}{1-x^2}f+(2)(1)f=0.
$$
Using the Method of Frobenius: Start by rewriting the equation as
$$
\frac{d^2f}{dx^2}-\frac{2x}{1-x^2}\frac{df}{dx}-\frac{m^2}{(1-x^2)^2}f+\frac{\lambda}{1-x^2}f = 0.
$$
In order to consider the equation at the singular point $x=-1$, which is a regular singular point, you have to expand in powers of $(x+1)$. For $x=-1$, you only need the coefficients $A$, $B$, $C$ shown below
$$
\frac{d^2f}{dx^2}+\left[\frac{A}{x+1}+\cdots\right]\frac{df}{dx}+\left[\frac{B}{(x+1)^2}+\cdots\right]f=0.
$$
You can easily determine $A$ and $B$. For example,
$$
-\frac{2x}{1-x^2}=\frac{1}{x+1}\left[\frac{2x}{x-1}\right]
\implies A = \left.\frac{2x}{x-1}\right|_{x=-1} = 1. \\
-\frac{m^2}{(1-x^2)^2}=\frac{1}{(x+1)^2}\left[-\frac{m^2}{(x-1)^2}\right]
\implies B=-\left.\frac{m^2}{(x-1)^2}\right|_{x=-1}=-\frac{m^2}{4}.
$$
So the indicial equation comes from solving Euler's equation
$$
(x+1)^2f''+A(x+1)f'+Bf = 0,\;\;\; f=(x+1)^r \\
r(r-1)+Ar+B=0 \\
r^2-r+r-\frac{m^2}{4}=0 \\
r^2=\frac{m^2}{2^2} \\
r = \pm \frac{m}{2}.
$$
Only the root with the largest real part--in this case $\frac{m}{2}$--is guaranteed to give a solution $(x+1)^{m/2}\sum_{n=0}^{\infty}a_n(x+1)^n$. If you want to take out the singularity at both endpoints $x=\pm 1$, then you can substitute $y=(1-x^2)^{m/2}w$ into the original equation and then solve by a series. In other words, the method of Frobenius leads to the same place in terms of the substitution.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
Find integral $\int\frac{\arcsin(x)}{x^{2}}dt$ Find integral $$\int\frac{\arcsin(x)}{x^{2}}dx$$
what I've done: $$\int\frac{\arcsin(x)}{x^{2}}dx=-\int\arcsin(x)d(\frac{1}{x})=-\frac{\arcsin(x)}{x}+\int\frac{dx}{x\sqrt{1-x^{2}}}$$ I got stuck with that
| $$\int\frac{\arcsin(x)}{x^{2}}dx=-\frac{\arcsin(x)}{x}+\int\frac{dx}{x\sqrt{1-x^{2}}}$$
Let $$u=\sqrt{1-x^2}$$
Hence
$$\frac{du}{dx}=\frac{-x}{\sqrt{1-x^2}}$$
$$\int \frac{dx}{x\sqrt{1-x^2}}=-\int \frac{du}{x^2}=-\int\frac{du}{1-u^2}=-\tanh^{-1}(u)+C=-\tanh^{-1}\left(\sqrt{1-x^2}\right)+C$$
Hence
\begin{align}
\int\frac{\arcsin(x)}{x^{2}}dx &=-\frac{\arcsin(x)}{x}-\tanh^{-1}\left( \sqrt{1-x^2}\right)+C \\
&=-\frac{\arcsin(x)}{x}-\frac12\left(\ln\left(1+\sqrt{1-x^2}\right)+\ln\left(1-\sqrt{1-x^2} \right) \right)+C
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2072026",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 3
} |
Generators of the intersection of twelve ideals. I have to show that the intersection of the twelve ideals $$(X\pm 1, Y\pm 1), (X\pm 1, Z\pm 1), (Y\pm 1, Z\pm 1) \subset \mathbb{R}[X,Y,Z]$$ is the ideal $$\Big( (X^2 β1)(Y^2 β1),(X^2 β1)(Z^2 β1),(Y^2 β1)(Z^2 β1)\Big).$$ Is there an easy way to do it, maybe some combinatorial way or something?
Thank you.
| Let $m$ be an element of the intersection of the twelve ideals. At first we have $m \in (x\pm1,y\pm1)$, so minimally we must have:
$y-1|m \lor x-1|m ,y+1|m \lor x-1|m ,y-1|m \lor x+1|m ,y+1|m \lor x+1|m$
There is a symmetry among these conditions, so assume WLOG that $y-1|m$, which satisfies the first and third minimal conditions. In order to satisfy the second and fourth, we must either have $y+1|m$ or $x^2-1|m$ by inspection. Overall if $m \in (x\pm1,y\pm1)$, then $x^2-1|m$ or $y^2-1|m$. By applying the same logic to $(x\pm1,z\pm1)$ and $(y\pm1,z\pm1)$ we must have the following:
$x^2-1|m \lor y^2-1|m, z^2-1|m \lor y^2-1|m, x^2-1|m \lor z^2-1|m$
Again we can assume WLOG that $x^2-1|m$, in which case we must also have $z^2-1|m \lor y^2-1|m$ in order to satisfy the second condition. Therefore we have shown that $(y^2-1)(x^2-1)|m $ or $(z^2-1)(x^2-1)|m $ or $(y^2-1)(z^2-1)|m $ by symmetry and so $m \in ((X^2β1)(Y^2β1),(X^2β1)(Z^2β1),(Y^2β1)(Z^2β1)).$
It remains to show that if $n \in ((X^2β1)(Y^2β1),(X^2β1)(Z^2β1),(Y^2β1)(Z^2β1))$, then $n$ lies in the intersection of the twelve ideals, which is straightforward. After this we have basically established a one to one relationship between the intersection and $((X^2β1)(Y^2β1),(X^2β1)(Z^2β1),(Y^2β1)(Z^2β1))$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2073418",
"timestamp": "2023-03-29T00:00:00",
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Explicit solution of the recursion $x_n = x_{n-1}^2 - 2$ with $x_0>2$ Let $m>2$ be an integer, $x_0 = m$ and $x_n = {(x_{n-1})}^2 - 2$ for $n > 0.$ Prove that $x_n=\lceil\tau(n) \rceil$, where $\tau(n) = Ξ±^{2^n}$ and $\alpha >1$ satisfies $\alpha + \frac{1}{\alpha} = m$.
This is the problem. The only thing I can think of is that it could have something to do with $\varphi$, since for $m=3$ we have $x(n)=\lceil\left(\varphi^2\right)^n\cdot \varphi) \rceil$, $\alpha = \varphi^2$. Other than that I have no idea how to move on. Thanks in advance
| From $x_0=m=\alpha + \frac{1}{\alpha}$ we have
$$x_1=m^2-2=\alpha^2+\frac{1}{\alpha^2}$$
And from $\alpha^2-m\alpha+1=0$ we have $$\alpha=\frac{m+\sqrt{m^2-4}}{2}\Rightarrow \alpha^2=\frac{m^2-2+m\sqrt{m^2-4}}{2}$$
which satisfies, from (easy to check) $m^2-4<m\sqrt{m^2-4}<m^2-2$: $$m^2-3<\alpha^2<m^2-2$$
So, we conclude
$$x_1=m^2-2=\alpha^2+\frac{1}{\alpha^2}=\left \lceil \alpha^2 \right \rceil$$
Then, inductively and recursively, we do the same steps with $m_1=x_1=\beta+\frac{1}{\beta}>2$, where $\beta=\alpha^2$ to conclude:
$$x_2=m_1^2-2=\beta^2+\frac{1}{\beta^2}=\left \lceil \beta^2 \right \rceil$$
Or
$$x_2=x_1^2-2=\alpha^4+\frac{1}{\alpha^4}=\left \lceil \alpha^4 \right \rceil$$
$$x_3=x_2^2-2=\alpha^8+\frac{1}{\alpha^8}=\left \lceil \alpha^8 \right \rceil$$
$$...$$
$$x_n=x_{n-1}^2-2=\alpha^{2^n}+\frac{1}{\alpha^{2^n}}=\left \lceil \alpha^{2^n} \right \rceil$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2073920",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Why is the solution to $x-\sqrt 4=0$ not $x=\pm 2$? If the equation is $x-\sqrt 4=0$, then $x=2$.
If the equation is $x^2-4=0$, then $x=\pm 2$.
Why is it not $x=\pm 2$ in the first equation?
| The symbol $\sqrt{a}$ stands for the principal square root of $a$ which is always positive. Thus, the value of $x$ in the first equation is $2$.
The solution of the second equation, $x^2-4=0$, is given by
\begin{align*}
&\frac{-b\pm\sqrt{b^2-4ac}}{2a}\\[2ex]
=\ &\frac{0\pm\sqrt{(0)^2-4(1)(-4)}}{2(1)}
\end{align*}
Simplyfing the above expression, we get two answers: $+\sqrt{4}$
and $-\sqrt{4}$.
That's why the solution of the second equation is $\pm2$.
| {
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"url": "https://math.stackexchange.com/questions/2075745",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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How to simplify an expression that does not have a common factor I am trying to simplify this expression :
$$9a^4 + 12a^2b^2 + 4b^4$$
So I ended up having this :
$$(3a^2)^2 + 2(3a^2)(2b^2) + (2b^2)^2$$
However, after that I don't know how to keep on simplifying the equation, it is explained that the answer is $(3a^2 + 2b^2)^2$ because the expression is equivalent to $(x + y)^2$ but I don't understand how they get to that ?
| Using the following formula.
$x^2 + 2xy + y^2 = (x+y)^2$
Here $x = 3a^2$, $y = 2b^2$
On putting values in left hand side of above formula.
$(3a^2)^2+2(3a^2)(2b^2)+(2b^2)^2 = 9a^4+12a^2b^2+4b^4 = (3a^2+2b^2)^2$
| {
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"url": "https://math.stackexchange.com/questions/2077624",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "2",
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$\sum\limits_{k=1}^{n-1}\frac{1}{\sqrt{k(n-k)}}\geq 1$ proof
$\sum_\limits{k=1}^{n-1}\frac{1}{\sqrt{k(n-k)}}\geq 1$ for all $n\geq 2$
Basecase n=2
$\sum_\limits{k=1}^{2-1}\frac{1}{\sqrt{k(2-k)}}=1\geq 1$
Assumption
$\sum_\limits{k=1}^{n-1}\frac{1}{\sqrt{k(n-k)}}\geq 1$ holds for some $n$
Claim
$\sum_\limits{k=1}^{n}\frac{1}{\sqrt{k(n+1-k)}}\geq 1$ holds too
Step
Assume $n$ is odd
$\sum_\limits{k=1}^{n}\frac{1}{\sqrt{k(n+1-k)}}=\frac{1}{\sqrt{n}}+\frac{1}{\sqrt{2(n-1)}}+...+\frac{1}{\sqrt{n}}$
Then, to determine which of those terms is the smallest, we want to find the maximum of $k(n+1-k)$.
The deriviative would be $n+1-2k$. So $k=\frac{n+1}{2}$, that's why we assume $n$ is odd in this step.
So our smallest term looks like: $\frac{1}{\sqrt{(\frac{n+1}{2})^2}}=\frac{2}{n+1}$
And since we add this term $n$ times, the sum is bounded below by $\frac{2}{n+1}n=\frac{2n}{n+1}$.
Via induction it is very easy to see, that $\frac{2n}{n+1}>1$ for all n>1, which is all we care about.
Now, how do I proceed for even $n$?
| By AM-GM $\sqrt{k(n-k)} \le \frac{1}{2}\big(k+(n-k)\big) = \frac{n}{2}\,$ Then:
$$\sum_{k=1}^{n-1}\frac{1}{\sqrt{k(n-k)}}\geq \sum_{k=1}^{n-1}\frac{2}{n} = \frac{2(n-1)}{n} \ge 1 \;\;\text{for}\;\; n \ge 2$$
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "4",
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Geometric (Trigonometric) inequality $\frac{(a+b+c)^3}{3abc}\leq1+\frac{4R}{r}$ How can one prove/disprove that $\frac{(a+b+c)^3}{3abc}\leq1+\frac{4R}{r}$ where $R$ and $r$ denote the usual circum and inradii respectively.
I know that $R=\frac{abc}{4\Delta}$ and $r=\frac{\Delta}{s}$, where $\Delta$ denotes area of triangle, and $s$ the semi perimeter. Any ideas. Thanks beforehand.
This is problem J392 from the Problem column of Mathematical Reflections - Issue 6 2016.
| We can prove that $$\frac{(a+b+c)^3}{3abc}\leq1+\frac{4R}{r}$$
Indeed, we need to prove that
$$\frac{(a+b+c)^3}{3abc}\leq1+\frac{\frac{abc}{S}}{\frac{2S}{a+b+c}}$$ or
$$\frac{(a+b+c)^3}{3abc}\leq1+\frac{8abc(a+b+c)}{16S^2}$$ or
$$\frac{(a+b+c)^3}{3abc}\leq1+\frac{8abc}{\prod\limits_{cyc}(a+b-c)}$$ or
$$\frac{(a+b+c)^3}{3abc}\leq\frac{\sum\limits_{cyc}(-a^3+a^2b+a^2c+2abc)}{\prod\limits_{cyc}(a+b-c)}$$ or
$$\frac{(a+b+c)^3}{3abc}\leq\frac{\sum\limits_{cyc}(-a^3+abc+a^2b+a^2c+abc)}{\prod\limits_{cyc}(a+b-c)}$$
$$\frac{(a+b+c)^2}{3abc}\leq\frac{\sum\limits_{cyc}(-a^2+ab+ab)}{\prod\limits_{cyc}(a+b-c)}$$ or
$$\sum\limits_{cyc}a\sum\limits_{cyc}(2a^2b^2-a^4)\leq3\sum\limits_{cyc}(2a^2b^2c-a^3bc)$$ or
$$2\sum\limits_{cyc}(a^4b+a^4c-a^3b^2-a^3c^2-2a^3bc+2a^2b^2c)+\sum\limits_{cyc}(a^5-a^4b-a^4c+a^3bc)\geq0,$$ for which it's enough to prove that
$$\sum\limits_{cyc}(a-b)^2(ab(a+b)-abc)\geq0$$ or
$$\sum\limits_{cyc}(a-b)^2ab(a+b-c)\geq0.$$
Done!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2079194",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
} |
What is the general solution to $2 \cos^2 x-\cos x=0$? What is the general solution to this trig equation?
$$2 \cos^2 x-\cos x=0$$
Thanks.
| $2\cos^2 x-\cos x=0$
$\implies \cos x(2\cos x-1)=0$
Either
$\cos x=0$
$\implies x=\dfrac{\pi}{2}\pm 2n\pi$
Or
$2\cos x-1=0$
$\implies \cos x=\dfrac{1}{2}$
$\implies x=\dfrac{\pi}{3}\pm 2n\pi$
where
$n \in Z$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2081473",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 2
} |
If $ f(x) = \frac{\sin^2 x+\sin x-1}{\sin^2 x-\sin x+2},$ then value of $f(x)$ lies in the interval If $\displaystyle f(x) = \frac{\sin^2 x+\sin x-1}{\sin^2 x-\sin x+2},$ then value of $f(x)$ lies in the interval
assume $\sin x= t$ where $|\sin x|\leq 1$
let $\displaystyle y = \frac{t^2+t-1}{t^2-t+1}$
$\displaystyle yt^2-yt+y=t^2+t-1$
$(y-1)t^2-(y+1)t+(y-1)=0$
for real roots $(y+1)^2-4(y-1)^2\geq 0$ or $(y+1)^2-(2y-2)^2\geq0$
$(3y-3)(3y-1)\leq 0$ or $\displaystyle \frac{1}{3}\leq y\leq 1$
but walframalpha shows different answer
https://www.wolframalpha.com/input/?i=range+of+f(x)+%3D+%5Cfrac%7B%5Csin%5E2+x%2B%5Csin+x-1%7D%7B%5Csin%5E2+x-%5Csin+x%2B2%7D
could some help me with this, thanks
| hint: $y = 1+\dfrac{2t-3}{t^2-t+2}, -1 \le t \le 1, $ and proceed to find critical values of $t$ for $y'(t) = 0$. Then consider the values of $y$ at $\pm 1$, and those of critical values of $t$ above.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2081524",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Use Mean value theorem to prove the following inequality A) Use the Mean value theorem to prove that
\begin{equation}
\sqrt{1+x} < 1 + \frac{1}{2}x \text{ if } x>0
\end{equation}
B) Use result in A) to prove that
\begin{equation}
\sqrt{1+x}>1+\frac{1}{2}x-\frac{1}{8}x^2 \text{ if } x>0
\end{equation}
Can someone give an answer for part B) ?
| From (A), for $x>0$ we have:
$0<\sqrt{x+1}-1<\frac{1}{2}x\Rightarrow {(\sqrt{x+1}-1)}^2<\frac{1}{4}x^2\Rightarrow {(\sqrt{x+1})}^2+1^2-2\sqrt{x+1}<\frac{1}{4}x^2 \Rightarrow \sqrt{x+1}>1+\frac{x}{2}-\frac{x^2}{8}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2081769",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
Simple Polynomial Algebra question to complete the square $4y^2+32y = 0$
$4(y^2+8y+64-64) = 0$
$4(y+4)^2 = 64$
Is that correct?
| $4y^2+32y = 0$
$4(y^2+8y+16-16) = 0$
$4(y+4)^2 = 64$
$(y+4)^2 = 16$
$y+4= \pm4$
$y_1=-8$, $y_2=0$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2081864",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove that $(2+ \sqrt5)^{\frac13} + (2- \sqrt5)^{\frac13}$ is an integer When checked in calculator it is 1. But how to prove it?
Also it is not a normal addition like $x+ \frac1x$ which needs direct rationalization. So I just need something to proceed.
| Take the cube of the expression and expand it. You get
\begin{multline}
\left(\sqrt[3]{2 + \sqrt{5}} + \sqrt[3]{2 - \sqrt{5}}\right)^3
\\= 2 + \sqrt{5} + 2 - \sqrt{5} + 3\sqrt[3]{(2 + \sqrt{5})(2 - \sqrt{5})}\left(\sqrt[3]{2 + \sqrt{5}} + \sqrt[3]{2 - \sqrt{5}}\right)
\\ = 4 - 3 \left(\sqrt[3]{2 + \sqrt{5}} + \sqrt[3]{2 - \sqrt{5}}\right)
\end{multline}
Thus, $\sqrt[3]{2 + \sqrt{5}} + \sqrt[3]{2 - \sqrt{5}}$ solves $x^3 = 4 - 3x$. It's not too difficult to find that $1$ is a solution to this equation.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2082836",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 8,
"answer_id": 5
} |
Calculate the antiderivative of a given function Consider the function $f : \left[ 0, \frac{\pi}{4} \right)$, $f(x) = \frac{(\cos x + \sin x)^n}{(\cos x - \sin x)^{n + 2}}$, where $n \in \mathbb{N}^*$. Find the antiderivative $F$ of $f$ such that $F(0) = \frac{1}{2(n + 1)}$.
I've noticed that $(\cos x - \sin x)' = -(\cos x + \sin x)$, but I can't figure out how to use this in finding $F$.
Thank you!
| Noting
$$ \cos x+\sin x=\sqrt2\sin(x+\frac{\pi}{4}),\cos x-\sin x=\sqrt2\cos(x+\frac{\pi}{4}) $$
one has
\begin{eqnarray}
F(x)&=&\int \frac{(\cos x + \sin x)^n}{(\cos x - \sin x)^{n + 2}} dx\\
&=&\int \frac{(\sqrt2\sin(x+\frac{\pi}{4}))^n}{(\sqrt2\cos(x+\frac{\pi}{4}))^{n + 2}} dx\\
&=&\int \tan^n(x+\frac{\pi}{4})\frac{1}{2\cos^2(x+\frac{\pi}{4})} dx\\
&=&\frac12\int \tan^n(x+\frac{\pi}{4})\sec^2(x+\frac{\pi}{4}) dx\\
&=&\frac12\int \tan^n(x+\frac{\pi}{4})d\tan(x+\frac{\pi}{4})\\
&=&\frac1{2(n+1)}\tan^{n+1}(x+\frac{\pi}{4})+C.
\end{eqnarray}
Using $F(0)=\frac{1}{2(n+1)}$, it is easy to see $C=0$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
Asymptotes and focus of a conic? This is the conic $$x^2+6xy+y^2+2x+y+\frac{1}{2}=0$$
the matrices associated with the conic are:
$$
A'=\left(\begin{array}{cccc}
\frac{1}{2} & 1 & \frac{1}{2} \\
1 & 1 & 3 \\
\frac{1}{2} & 3 & 1
\end{array}\right),
$$
$$
A=\left(\begin{array}{cccc}
1 & 3 \\
3 & 1
\end{array}\right),
$$
His characteristic polynomial is: $p_A(\lambda) = \lambda^2-2\lambda-8$
A has eigenvalues discordant $(\lambda = 4, \lambda = -2)$, so it's an Hyperbole.
Then i found that the center of the conic is: $(-\frac{1}{16}, -\frac{5}{16})$
Then with the eigenvalues i found the two lines passing through the center:
$$4x-4y-1=0$$
$$8x+8y+3=0$$
Now i want to find the focus and the asymptotes but i have no idea how to do it.There is a way to find These two things through the data I have now? or do i need the canonical form of the conical? Thanks
| You've two principal axes:
\begin{align*}
x^2+6xy+y^2+2x+y+\frac{1}{2} & \equiv
A\left( \frac{4x-4y-1}{\sqrt{4^2+4^2}} \right)^2+
B\left( \frac{8x+8y+3}{\sqrt{8^2+8^2}} \right)^2+C \\
& \equiv
-2\left( \frac{4x-4y-1}{4\sqrt{2}} \right)^2+
4\left( \frac{8x+8y+3}{8\sqrt{2}} \right)^2+\frac{9}{32}
\end{align*}
Now
\begin{align*}
\frac{4x-4y-1}{4\sqrt{2}} &= x'\\
\frac{8x+8y+3}{8\sqrt{2}} &= y'\\
a &= \frac{3}{8} \\
b &= \frac{3}{8\sqrt{2}} \\
\frac{x'^2}{a^2}-\frac{y'^2}{b^2} &=1
\end{align*}
Asymptotes
$$b x' \pm a y'=0$$
Foci
$$(x',y')=(\pm \sqrt{a^2+b^2},0)$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
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Doing definite integration $\int_0^{\pi/4}\frac{x\,dx}{\cos x(\cos x + \sin x)}$ We have to solve the following integration
$$
\int_0^{\pi/4}\frac{x\,dx}{\cos x(\cos x + \sin x)}
$$
I divided both in Nr and Dr by $\cos^2 x$.
But after that I stuck.
| We have $$I=\int_{0}^{\pi/4}\frac{2x}{2\cos^2x+2\cos x\sin x}dx=\int_0^{\pi/4}\frac{2x}{\cos 2x+\sin 2x+1}dx=\frac{1}{2}\int_0^{\pi/2}\frac{xdx}{\cos x+\sin x+1}=\frac{1}{2}\int_0^{\pi/2}\frac{\pi/2-x}{\cos x+\sin x+1}dx\\I=\frac{1}{4}\int_0^{\pi/2}\frac{\pi/2}{\cos x+\sin x+1}dx$$Now substitute $x$ for $2x$ again$$=\frac{\pi}{4}\int_0^{\pi/4}\frac{dx}{2\cos^2x+2\sin x\cos x}=\frac{\pi}{8}\int_0^{\pi/4}\frac{\sec^2xdx}{1+\tan x}=\frac{\pi}{8}\int_0^{1}\frac{1}{1+u}du=\frac{\pi}{8}\log 2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2084474",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 6,
"answer_id": 0
} |
Quadratic equation system $A^2 + B^2 = 5$ and $AB = 2$ Given a system of equations
$A^2 + B^2 = 5$
$AB = 2$
what is the correct way to solve it?
I see immediately that the answers are
*
*$A=1, B=2$
*$A=2, B=1$
*$A=-1, B=-2$
*$A=-2, B=-1$
but I don't understand the correct way of getting there. I have tried to isolate one of the variables and put the resulting expression into one of the equations, but this didn't get me anywhere. What is the correct way of solving this problem?
| From $AB=2$ we know $A$ and $B$ have the same sign.
Moreover, if we find a solution for some values $A=A_1$ and $B=B_1$,
then $A=-A_1$ and $B=-B_1$ also is a solution.
So we can assume initially that $A$ and $B$ are both positive,
find all such solutions, and then simply change signs to produce all the remaining solutions (the ones in which $A$ and $B$ are both negative).
Similarly, we can assume initially that $A\geq B$, because all the solutions
in which $B \geq A$ can then be produced just by swapping the values of
the two variables.
So we now have (with these assumptions) $A \geq B > 0,$
$A^2 + B^2 = 5,$ and $AB = 2.$ Then
\begin{align}
(A + B)^2 &= A^2 + 2AB + B^2 = 5 + 2(2) = 9, \\
(A - B)^2 &= A^2 - 2AB + B^2 = 5 - 2(2) = 1. \\
\end{align}
In each of these equations, the positive square roots of both sides
must be equal, and since $A+B>0$ and $A-B\geq0$ this tells us that
\begin{align}
A + B &= 3, \\
A - B &= 1. \\
\end{align}
Therefore (adding the two equations) $2A = 4,$ so $A = 2,$ so $B = 1.$
That is, the unique solution (under the assumption $A\geq B>0$)
is $(A,B) = (2,1).$
We can now remove the assumptions that $A\geq B$ and that $B>0,$
then use the facts mentioned earlier to derive the remaining solutions:
$(A,B) = (1,2)$ (swapping values), $(A,B) = (-2,-1)$
or $(A,B) = (-1,-2)$ (changing signs).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2084539",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 10,
"answer_id": 3
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Proving $a\cos x+b\sin x=\sqrt{a^2+b^2}\cos(x-\alpha)$
Show that $a\cos x+b\sin x=\sqrt{a^2+b^2}\cos(x-\alpha)$ and find the correct phase angle $\alpha$.
This is my proof.
Let $x$ and $\alpha$ be the the angles in a right triangle with sides $a$, $b$ and $c$, as shown in the figure. Then, $c=\sqrt{a^2+b^2}$. The left-hand side is $a\cos x+b\sin x=\frac{ab}{c}+\frac{ab}{c}=2\frac{ab}{c}$. The right-hand side is $\sqrt{a^2+b^2}\cos(x-\alpha)=c\left(\cos{x}\cos{\alpha}+\sin{x}\sin{\alpha}\right)=c\left(\frac{ab}{c^2}+\frac{ab}{c^2}\right)=2\frac{ab}{c}$.
Is my proof valid? Is there a more general way to prove it?
For the second part of the question, I think it should be $\alpha=\arccos\frac{a}{\sqrt{a^2+b^2}}=\arcsin\frac{b}{\sqrt{a^2+b^2}}$. Is this correct?
| The proof is only valid for $0<x<\frac\pi2$. For a more general proof, let:
$$a\cos x+b\sin x\ \equiv\ R\cos(x-\alpha)=R\cos\alpha\cos x + R\sin\alpha\sin x$$
So $a=R\cos\alpha$ and $b=R\sin\alpha$, from which you can readily find $R$ and $\alpha$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2085687",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Provide a different method of proving:$\int_{-\infty}^{\infty}{1\over [\pi(x+e^{\pi})^2+\pi^{1/3}]^2}dx={1\over 2}$ Accidentally founded this particular integral producing a rational number
I can't be for sure it is correct, so can one provide a proof of it.
$$\int_{-\infty}^{\infty}{1\over [\pi(x+e^{\pi})^2+\pi^{1/3}]^2}dx={1\over 2}\tag
1$$
I found related to $(1)$ is this
Let enforce a substitution of $u=x+e^{\pi}$ then $du=dx$
$$\int_{-\infty}^{\infty}{1\over [\pi{u}^2+\pi^{1/3}]^2}du={1\over 2}\tag2$$
To avoid confusing with too much $\pi$ symbol, we write a general
$$\int_{-\infty}^{\infty}{1\over [A{u}^2+B]^2}du={1\over 2}\tag3$$
We could apply partial decomposition
$${au+b\over Au^2+B}+{cu+d\over (Au^2+B)^2}=1\tag4$$ then find a,b,c and d.
I found a general integral of
$$\int{dx\over(x^2+a^2)^2}={x\over2a^2(x^2+a^2)}+{1\over 2a^3}\tan^{-1}{\left(x\over a\right)}\tag5$$
I am sure this is enough to prove $(1)$
Question: what are other methods can we apply to prove $(1)?$
| To avoid confusion, $a = A = \pi$ and $b = B = \pi ^{1/3}$
Let $z=\frac{a u^{2}}{b}$
\begin{align}
\int\limits_{-\infty}^{\infty} \frac{1}{(a u^{2} + b)^{2}} du &= 2 \int\limits_{0}^{\infty} \frac{1}{(a u^{2} + b)^{2}} du \\
&= \frac{2}{b^{2}} \int\limits_{0}^{\infty} \frac{1}{(\frac{a}{b} u^{2} + 1)^{2}} du \\
&= \frac{1}{b^{2}} \sqrt{\frac{b}{a}} \int\limits_{0}^{\infty} \frac{z^{-1/2}}{(z+1)^{2}} dz \\
&= a^{-1/2} b^{-3/2} \mathrm{B}\left(\frac{1}{2},\frac{3}{2} \right) \\
&= \frac{1}{\pi} \mathrm{B}\left(\frac{1}{2},\frac{3}{2} \right) \\
&= \frac{1}{\pi} \frac{\Gamma(1/2)\Gamma(3/2)}{\Gamma(2)} \\
&= \frac{1}{2}
\end{align}
Note that we used the following integral definition of the beta function
\begin{equation}
\mathrm{B}(x,y) = \int\limits_{0}^{\infty} \frac{z^{x-1}}{(z+1)^{x+y}} dz
\end{equation}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2085788",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 4
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Determine whether the following sequence is increasing or decreasing $\frac{n^2+2n+1}{3n^2+n}$ Determine whether the following sequence is increasing or decreasing:
$$\frac{n^2+2n+1}{3n^2+n}$$
I'm not sure whether my solution is correct:
$$\frac{n^2+2n+1}{3n^2+n}=\frac{n(n+2)+1}{n(3n+1)}=\frac{n+2}{3n+1}+\frac{1}{n(3n+1)}.$$
Let's prove $\frac{n+2}{3n+1}$ is a decreasing sequence.
$$a_n>a_{n+1} \Leftrightarrow \frac{n+2}{3n+1}>\frac{n+3}{3n+4}\Leftrightarrow(n+2)(3n+4)>(n+3)(3n+1)\Leftrightarrow3n^2+10n+8>3n^2+10n+3\Leftrightarrow 8>3$$
So $\frac{n+2}{3n+1}$ is a decreasing sequence and we know that $\frac{1}{n(3n+1)}$ is also decreasing so our given sequence is a decreasing sequence as a sum of $2$ decreasing sequences.
| After breaking up the fraction using Partial Fractions, we see that $\frac3n$ is bigger than $\frac4{3n+1}$, so we give $\frac4{3n}$ of $\frac3n$ to $-\frac4{3n+1}$ to make it positive, but decreasing.
$$
\begin{align}
\frac{n^2+2n+1}{3n^2+n}
&=\frac13\left(1+\frac{5n+3}{3n^2+n}\right)\\
&=\frac13\left(1-\frac4{3n+1}+\frac3n\right)\\
&=\frac13\left(1-\frac4{3n+1}+\frac4{3n}+\frac5{3n}\right)\\
&=\frac13\left(1+\frac4{3n(3n+1)}+\frac5{3n}\right)\tag{1}
\end{align}
$$
For $n\gt0$, each non-constant term in $(1)$ is decreasing.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2086547",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 4
} |
Integral $\int \frac{\mathrm{d}x}{\sin x+\sec x}$ In the following integral
$$\int \frac{\mathrm{d}x}{\sin x+\sec x}$$
My attempt is
I first multiplied and divided by $\cos^2x$
And then substitued $\tan x = t$
But after that got stuck .
| $\displaystyle I = \int\frac{1}{\sin x+\sec x}dx = \int\frac{\cos x}{\sin x\cos x+1}dx = \int \frac{(\cos x+\sin x)+(\cos x-\sin x)}{\sin 2x +2}dx$
$\displaystyle =\int\frac{\cos x+\sin x}{\sin 2x+2}dx+\int\frac{\cos x-\sin x}{\sin 2x+2}dx = I_{1}+I_{2}$
for $\displaystyle I_{1} = \int\frac{\cos x+\sin x}{3-(\sin x-\cos x)^2}dx,$ put $(\sin x-\cos x) = t$ then $(\cos x+\sin x)dx = dt$
for $\displaystyle I_{2} = \int\frac{\cos x-\sin x}{(\sin x+\cos x)^2+1}dx,$ put $(\sin x+\cos x) = u$ then $(\cos x-\sin x)dx = du$
So $\displaystyle I = \int\frac{1}{3-t^2}dt +\int\frac{1}{1+u^2}du$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2087112",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find the $\gcd[x+y+z; x^2+xy+z^2; y^2+yz+z^2; z^2+zx+x^2]$ What I have done:
There exists a non-zero integer $t$ such:
$$x+y+z=kt$$
$$x^2+xy+y^2=ut$$
$$y^2+yz+z^2=vt$$
$$z^2+zx+x^2=wt$$
$\implies$
$$(x-y)(x+y+z)=(u-v)t$$
$$(y-z)(x+y+z)=(v-w)t$$
$$(z-x)(x+y+z)=(w-u)t$$
$\implies$
$$\dfrac{x+y+z}{t}= \dfrac{u-v}{x-y}=\dfrac{v-w}{y-z}=\dfrac{w-u}{z-x}=k$$
$\implies$
$$x+y+z=kt $$
$$k(x-y)=u-v$$
$$k(y-z)=v-w $$
$$k(z-x)=w-u $$
$\implies$
$$u=w+k(x-z) $$
$$v=w+k(y-z)$$
$$w=w $$
$\implies$
$$x+y+z=kt$$
$$x^2+xy+y^2=[w+k(x-z)]t$$
$$y^2+yz+z^2=[w+k(y-z)t]t$$
$$z^2+zx+x^2=wt$$
$\implies$
| If $x,y,z$ are not coprime, the result is trivial. So We can assume that $x,y,z$ are not all even. Then $t$ must be an odd integer since the bottom 3 numbers are always odd.
$$x^2+y^2+z^2+\dfrac{xy+xz+zx}{2}=\dfrac{t(u+v+w)}{2}$$
$$(x+y+z)^2-\dfrac{3(xy+xz+zx)}{2}=\dfrac{t(u+v+w)}{2}$$
$$(kt)^2-\dfrac{3(xy+xz+zx)}{2}=\dfrac{t(u+v+w)}{2}$$
So $t$ must divide $3(xy+yz+zx)$. Then, $t=\pm 3$ and/or divides $xy+yz+zx$
$\implies$
I still have a quadratic equation to deal with to decide that $t$ is an integer. So I will have to keep imposing restrictions to find $t$. I guess .
| {
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"timestamp": "2023-03-29T00:00:00",
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"answer_id": 1
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Sum of $n$ terms of the given series. Find the sum of $n$ terms of the series:
$$\frac{1}{x+1}+\frac{2x}{(x+1)(x+2)}+\frac{3x^2}{(x+1)(x+2)(x+3)}+\frac{4x^3}{(x+1)(x+2)(x+3)(x+4)}+.......$$
Could someone give me slight hint to proceed in this question?
| Recalling that the lower incomplete gamma function has the power series $$\gamma\left(s,x\right)=x^{s-1}\Gamma\left(s\right)e^{-x}\sum_{k\geq1}\frac{x^{k}}{\Gamma\left(s+k\right)}\tag{1}$$ we have $$\frac{d}{dx}\left(\frac{\gamma\left(s,x\right)}{x^{s-1}\Gamma\left(s\right)e^{-x}}\right)=\sum_{k\geq1}\frac{kx^{k-1}}{\Gamma\left(s+k\right)}
$$ hence $$\sum_{k\geq1}\frac{kx^{k-1}}{\Gamma\left(s+k\right)}=\frac{1+x^{-s}e^{x}\left(x-1-s\right)\gamma\left(s,x\right)}{\Gamma\left(s\right)}$$ so taking $s=x+1$ we get $$\sum_{k\geq1}\frac{kx^{k-1}}{\Gamma\left(x+1+k\right)}=\frac{1}{\Gamma\left(x+1\right)}\tag{2}
$$ and now recalling that the Pochhammer symbol can be written as $$\left(x+1\right)\cdots\left(x+n\right)=\left(x+1\right)_{n}=\frac{\Gamma\left(x+1+n\right)}{\Gamma\left(x+1\right)}\tag{3}$$ we get $$\sum_{k\geq1}\frac{kx^{k-1}}{\left(x+1\right)\cdots\left(x+k\right)}=1.
$$ Then $$\sum_{k=1}^{N}\frac{kx^{k-1}}{\left(x+1\right)\cdots\left(x+k\right)}=\sum_{k\geq1}\frac{kx^{k-1}}{\left(x+1\right)\cdots\left(x+k\right)}-\sum_{k=N+1}^{\infty}\frac{kx^{k-1}}{\left(x+1\right)\cdots\left(x+k\right)}
$$ $$=1-x^{N}\sum_{k\geq1}\frac{\left(k+N\right)x^{k-1}}{\left(x+1\right)\cdots\left(x+k+N\right)}=1-\frac{x^{N}\Gamma\left(x+1\right)}{\Gamma\left(x+N+1\right)}$$ using $(1)$, $(2)$ and $(3)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2088797",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Problem with integrating this function I have the function $$f(u) = \frac{1}{\sqrt{1-bu^2 -cu}}$$ with $b,c$ positive. I want to integrate this function from $0$ to its first root. Within this interval, the function is not complex since the polynomial under the root is positive. But the function itself is obviously complex in general and I have its complex antiderivative.
Do I just plug in the values and take the real part? How do I proceed? (The result should be real).
| As suggested in the comments section, you need to complete the square, i.e.
\begin{equation} \int f(u) du = \int \frac{1}{\sqrt{1 - bu^2 - cu}} du
= \int \frac{1}{\sqrt{-b \big( u^2 + \frac{cu}{b} + \frac{c^2}{4b^2} - \frac{c^2}{4b^2} - \frac{1}{b} \big) }} du \\
= \frac{1}{\sqrt{- b \Big( \big(u^2 + \frac{cu}{b} + \frac{c^2}{4b^2} \big) - \frac{c^2 + 4b}{4b^2} \Big) }} du
= \frac{1}{\sqrt{b}} \int \frac{1}{\sqrt{\frac{c^2 + 4b}{4b^2} - \big( u + \frac{c}{2b} \big)^2}} du \end{equation}
Since you have an expression of the form $ k^2 - g(u)^2 $, then you want to use the substitution $ g(u) = k \times sin\theta $, i.e. $ u + \frac{c}{2b} = \sqrt{\frac{c^2+4b}{4b^2}}sin\theta \rightarrow du = \sqrt{\frac{c^2+4b}{4b^2}}cos\theta $.
Substituting these expressions back into the above equation, we get:
\begin{equation} \int \frac{1}{\sqrt{\frac{c^2 + 4b}{4b^2} - \big( \sqrt{\frac{c^2+4b}{4b^2} \times sin\theta} \big)^2}} \sqrt{\frac{c^2+4b}{4b^2}} cos\theta d\theta
= \int \frac{1}{\sqrt{\frac{c^2 + 4b}{4b^2} (1 - sin^2\theta)}} \sqrt{\frac{c^2+4b}{4b^2}} cos\theta d\theta \\
= \frac{1}{\sqrt b} \int \frac{cos\theta}{\sqrt{1-sin^2\theta}} d\theta = \frac{1}{\sqrt b} \int d\theta = \frac{1}{\sqrt b} \theta + Constant \end{equation}
Since $ u+\frac{c}{2b} = \sqrt{\frac{c^2+4b}{4b^2}} sin\theta \rightarrow \frac{2ub+c}{2b} = \frac{\sqrt{c^2+4b}}{2b}sin\theta \rightarrow 2ub+c = \sqrt{c^2+4b} sin\theta $, then $ sin\theta = \frac{2ub+c}{\sqrt{c^2+4b}} $, which means that $ \theta = sin^{-1} \Big(\frac{2ub+c}{\sqrt{c^2+4b}}\Big) $.
Putting this all together, we get:
$$ \int f(u)du = \frac{1}{\sqrt{b}} sin^{-1} \Bigg(\frac{2ub+c}{\sqrt{c^2+4b}}\Bigg) + Constant $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2089310",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Remarkable limit and LβHΓ΄pital $\lim_{x\to0}\left(\frac{a^x-x\ln a}{b^x-x\ln b}\right)^{1/x^2}$
Compute
$$
\lim_{x\to0}\left(\frac{a^x-x\ln a}{b^x-x\ln b}\right)^{1/x^2}
$$
where $a$ and $b$ are positive numbers.
I came to the two different forms of this limit, as $\lim_{x\to0}$
$$e^{\frac{\ln b-\ln b}{\ln a-\ln a+\ln b-\ln b}}$$
$$\frac{x^2\cdot a^x+1-\frac{a^2}x}{x^2\cdot b^x+1-\frac{b^2}x}$$
So, what I want to say is that I can't solve this problem and I'm here for any kind of help.
| Assuming that the desired limit is $L$ we can proceed as follows:
\begin{align}
\log L &= \log\left\{\lim_{x \to 0}\left(\frac{a^{x} - x\log a}{b^{x} - x\log b}\right)^{1/x^{2}}\right\}\notag\\
&= \lim_{x \to 0}\log\left(\frac{a^{x} - x\log a}{b^{x} - x\log b}\right)^{1/x^{2}}\text{ (via continuity of log)}\notag\\
&= \lim_{x \to 0}\frac{1}{x^{2}}\log\left(1 + \frac{a^{x} - x\log a - b^{x} + x\log b}{b^{x} - x\log b}\right)\notag\\
&= \lim_{x \to 0}\frac{1}{x^{2}}\cdot\frac{a^{x} - x\log a - b^{x} + x\log b}{b^{x} - x\log b}\cdot\dfrac{\log\left(1 + \dfrac{a^{x} - x\log a - b^{x} + x\log b}{b^{x} - x\log b}\right)}{\dfrac{a^{x} - x\log a - b^{x} + x\log b}{b^{x} - x\log b}}\notag\\
&= \lim_{x \to 0}\frac{1}{x^{2}}\cdot\frac{a^{x} - x\log a - b^{x} + x\log b}{b^{x} - x\log b}\notag\\
&= \lim_{x \to 0}\frac{a^{x} - b^{x} - x\log a + x\log b}{x^{2}}\tag{1}\\
&= \lim_{x \to 0}\frac{a^{x}\log a - b^{x}\log b - \log a + \log b}{2x}\text{ (via L'Hospital's Rule)}\notag\\
&= \frac{1}{2}\lim_{x \to 0}\log a\cdot\frac{a^{x} - 1}{x} - \log b\cdot\frac{b^{x} - 1}{x}\notag\\
&= \frac{(\log a)^{2} - (\log b)^{2}}{2}\notag
\end{align}
If you prefer the easier technique of Taylor series then you need to apply it after step marked $(1)$ and use the expansion $$a^{x} = 1 + x\log a + \frac{x^{2}}{2}(\log a)^{2} + o(x^{2})$$ This way you get the answer immediately.
Thus finally $$L = \exp\left(\frac{(\log a)^{2} - (\log b)^{2}}{2}\right)$$
| {
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"url": "https://math.stackexchange.com/questions/2090142",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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A polynomial question. I came across this question in my weekly test paper and didn't know how to tackle it. The question goes as follows:
$a_0,a_1,a_2....a_{34}$ are the coefficients of $x^0,x^1,x^2......x^{34}$ of the polynomial obtained on opening the parenthesis of $(1+x+x^2)^{17}$, then which is true?
$1)~a_1+a_2+.......+ a_{34}$ is even;
$2)~a_1=17$;
$3)~a_{33}=17$;
$4)~a_2=153$.
(One or more is correct)
| Here is some info regarding (2) - (4)
*
*Ad (2): $a_1$ is the coefficient of $x^1$ in $(1+x+x^2)^{17}$. Since there are $17$ factors of the form $1+x+x^2$ we have to choose precisely from one factor $x$ whereas from all other factors we have to choose $1$. This can be done in
\begin{align*}
a_1=\binom{17}{1}=17
\end{align*}
different ways.
*Ad (3): The polynomial $1+x+x^2=x\left(\frac{1}{x}+1+x\right)$ has a nice symmetrical structure. This implies the coefficient of $x^k$ in $(1+x+x^2)^{17}$ is also the coefficient of $x^{34-k}$ with $0\leq k \leq 34$. We conclude
\begin{align*}
a_{33}=a_{34-1}=a_1=17
\end{align*}
*Ad (4): $a_2$ is the coefficient of $x^2$ in $(1+x+x^2)^{17}$. In order to get $x^2$ from $17$ factors of the form $1+x+x^2$ we have to choose one factor $x^2$ which can be done in $\binom{17}{1}$ different ways and $1$ from the remaining $16$ factors, or we have to choose two factors $x$ which can be done in $\binom{17}{2}$ different ways and $1$ from the remaining $15$ factors. We conclude
\begin{align*}
a_2=\binom{17}{1}+\binom{17}{2}=17+\frac{1}{2}\cdot 17\cdot 16=153
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2090343",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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$\forall n\in\mathbb{Z}$, find the $\gcd(n^2+1, (n+1)^2+1)$?
$\forall n\in\mathbb{Z}$, find the $\gcd(n^2+1, (n+1)^2+1)$.
I think this is a simple exercise, but I get this:
$(n+1)^2+1=n^2+2n+2$.
$n^2+2n+2 = (n^2+1)+(2n+1)$
then $\gcd(n^2+1, (n+1)^2+1)=\gcd(n^2+1, 2n+1)$
and $\displaystyle n^2+1 = \frac{n(2n+1)}{2}+\left(-\frac{n}{2}+1\right)$
then $\gcd(n^2+1, 2n+1)=\gcd(2n+1, \frac{n}{2}-1)$.
But gcd is on integer numbers and $\dfrac{n}{2}-1$ is not always an integer number, so, I need some help?
| Your step that says $\displaystyle n^2+1 = \frac{n(2n+1)}{2}+\left(-\frac{n}{2}+1\right)$ is correct but not useful. We are working in the naturals and this threatens (if $n$ is not even) to takes us out of the naturals.
If we just make a spreadsheet and try the small numbers we find that the GCD is $1$ unless $n\equiv 2 \pmod 5$, in which case it is $5$. We can see that if $n \equiv 2 \pmod 5$, $n^2+1 \equiv (n+1)^2+1 \equiv 0 \pmod 5$. Now we need to ask if there are any other numbers $k$ where $n^2\equiv (n+1)^2 \equiv -1 \pmod k$. This requires that $2n+1 \equiv 0 \pmod k$. We can factor $k$ demand that $2n+1 \equiv 0 \pmod p$ for some prime $p$ dividing $k$. If we square that we get $4n^2+4n+1\equiv 4n-3\equiv 0, n \equiv \frac 34 \pmod p$ where we take advantage of the fact that $\Bbb Z_p$ is a field to divide. Now $n^2 \equiv -1 \pmod p$ says $9 \equiv -16 \pmod p$ and the only prime is $5$. So the $\gcd$ is $1$ unless $n \equiv 2 \pmod 5$ in which case it is $5$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2091252",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Maximum of $f(x) =(1+\cos x)\cdot \sin(\frac{x}{2})$ on $x \in (0, \pi)$ I've attempted to solve this question by using $$f(x) = 2\sin\frac{x}{2}\cos^2\frac{x}{2} \leq \frac{(2\sin\frac{x}{2}+\cos^2\frac{x}{2})^2}{2}$$ but it results in the wrong answer every time. Is there another way to solve this question and is there anything I can do to change my method to make it work.
| Since $\cos x = 1-2\sin^2 \frac{x}{2}$,
$$
f(x)=\left(2-2\sin^2\frac{x}{2}\right)\sin \frac{x}{2} = 2\sin \frac{x}{2} - 2\sin^3\frac{x}{2}.
$$
Thus you can find the maximum of $f(x)$ by finding the maximum of $g(x)=2x-2x^3$ defined on $(0,1)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2091339",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Evaluate $\int_0^1 \frac{\ln \left(1+x+x^2+\cdots +x^{n} \right)}{x}dx$ How to evaluate
$$\int_0^1 \frac{\ln \left(1+x+x^2+\cdots +x^{n} \right)}{x}dx$$
My attempt:
\begin{align*}
\int_0^1 \frac{\ln \left(1+x+x^2+\cdots +x^{n} \right)}{x}dx &= \int_0^1 \frac{\ln \left(\dfrac{1-x^{n+1}}{1-x} \right)}{x}dx \\
&= \int_0^1 \frac{\ln \left({1-x^{n+1}} \right)}{x}dx -\int_0^1 \frac{\ln \left({1-x} \right)}{x}dx \\
&=\frac{1}{n+1} \int_0^1 \frac{\ln \left({1-x} \right)}{x}dx -\int_0^1 \frac{\ln \left({1-x} \right)}{x}dx
\end{align*}
but what's next?I have been stuck here for a while.
| $$\int_0^1 \dfrac{\ln(1-x)}xdx = \int_0^1 \left(-\sum_{k=1}^{\infty} \dfrac{x^{k-1}}k\right)dx = -\sum_{k=1}^{\infty} \dfrac1k\int_0^1 x^{k-1}dx = - \sum_{k=1}^{\infty} \dfrac1{k^2} = -\zeta(2)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2094327",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 2,
"answer_id": 1
} |
Number of positive integer solutions of $\frac{1}{x}+\frac{1}{y}=\frac{1}{pq}$ for distinct primes $p$ and $q$
Let $p$ and $q$ be distinct primes. Then find the number of positive integer solutions of the equation $$\frac{1}{x}+\frac{1}{y}=\frac{1}{pq}$$
We get $pq=\frac{xy}{x+y}$
Now $x+y$ must divide $xy$ as L.H.S. is a positive integer with two prime factors but how do we make sure the same on R.H.S. ?
Given options are $3$ or $4$ or $8$ or $9$.
| Since
$$
\begin{align}
(x-pq)(y-pq)
&=xy-pq(x+y)+p^2q^2\\
&=p^2q^2
\end{align}
$$
By breaking up the $9$ factors of $p^2q^2$, and solving for $x$ and $y$, we get
$$
\begin{align}
\frac1{pq}
&=\frac1{pq+p^2q^2}+\frac1{pq+1}\\
&=\frac1{pq+pq^2}+\frac1{pq+p}\\
&=\frac1{pq+q^2}+\frac1{pq+p^2}\\
&=\frac1{pq+q}+\frac1{pq+p^2q}\\
&=\frac1{pq+pq}+\frac1{pq+pq}
\end{align}
$$
So there are $9$ solutions because the last is symmetric.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 2
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Solving the differential equation $7x(x-y)dy = 2(x^2+6xy-5y^2)dx$ How do I solve the differential equation $$7x(x-y)dy = 2(x^2+6xy-5y^2)dx$$
Is it homogeneous? I have tried taking the variables from the LHS and applying them to the RHS, making $\frac{dy}{dx}$ subject and ending up with:
$$\frac{dy}{dx} = \frac{2(x^2+6xy-5y^2)}{7x(x-y)}$$
After simplifying the numerator:
$$\frac{dy}{dx} = \frac{2[(x-y)(x+5y)+2xy]}{7x(x-y)}$$
$$\frac{dy}{dx} = \frac{2(x+5y)}{7x} + \frac{4y}{7(x-y)}$$
I have no idea how to proceed from here. Is the methodology correct so far? Any input would be appreciated.
| Starting with your equation
$$\frac{dy}{dx} = \frac{2(x+5y)}{7x} + \frac{4y}{7(x-y)}$$
multiply the top and bottom of each fraction on the right, giving
$$\frac{dy}{dx} = \frac{2(x+5y)}{7x}\frac{1/x}{1/x} + \frac{4y}{7(x-y)}\frac{1/x}{1/x}$$
$$\frac{dy}{dx} = \frac{2(1+5y/x)}{7} + \frac{4y/x}{7(1-y/x)}$$
Since $y'=u'x+u$ you get
$$u'x+u = \frac{2(1+5u)}{7} + \frac{4u}{7(1-u)}$$
which is a separable differential equation.
| {
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"url": "https://math.stackexchange.com/questions/2095460",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
A Ramanujan infinite series $$ 1-5\left(\frac{1}{2}\right)^3+9\left(\frac{1 \cdot 3}{2 \cdot 4}\right)^3-13\left(\frac{1 \cdot 3 \cdot 5}{2 \cdot 4 \cdot 6}\right)^3+\cdots $$
I went on evaluating the above series and encountered that solving $\displaystyle \sum_{n\ge 0}\left(\binom{2n}{n}\right)^3x^n$ would suffice.
But how do we make a generating function for the third power of a central binomial coefficient using the fact $\displaystyle \sum_{n\ge 0}\binom{2n}{n}x^n=\frac{1}{\sqrt{1-4x}}$
| This is not an answer but just a result obtained using a CAS.
Let $$f_k=\sum_{n=0}^\infty \binom{2 n}{n}^kx^n$$ The following expressions have been obtained $$f_1=\frac{1}{\sqrt{1-4 x}}$$ $$f_2=\frac{2 }{\pi }K(16 x)$$ $$f_3=\frac{4 }{\pi ^2}K\left(\frac{1}{2} \left(1-\sqrt{1-64 x}\right)\right)^2$$
$$f_4=\, _4F_3\left(\frac{1}{2},\frac{1}{2},\frac{1}{2},\frac{1}{2};1,1,1;256 x\right)$$ $$f_5=\,
_5F_4\left(\frac{1}{2},\frac{1}{2},\frac{1}{2},\frac{1}{2},\frac{1}{2};1,1,1,1;1
024 x\right)$$ where appear, for $k=2,3$, the complete elliptic integrals of the first kind and, for $k>3$, the generalized hypergeometric functions
| {
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"url": "https://math.stackexchange.com/questions/2095780",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 0
} |
How to calculate $\lim\limits_{n\to \infty} \frac{(n+3)^{\sqrt{n+1}}}{(n+1)^{\sqrt{n}}}$ I have no idea how to find a limit of sequence
$$a_n= \frac{(n+3)^{\sqrt{n+1}}}{(n+1)^{\sqrt{n}}}$$. I think that it can be easily bounded by 1 from down, but this shows only $1β€a_n$. I think also it can be shown quite easily that we need only to show that limit of a sequence $n^{\sqrt{n+1}-\sqrt{n}}$ is equal to $1$, I am unable to show that too.
| shows that using conjugation that
\begin{split}
a_n=\frac{(n+3)^{\sqrt{n+1}}}{(n+1)^{\sqrt{n}}} &= &\exp \left( {\sqrt{n+1}}\ln(n+3)- {\sqrt{n}}\ln(n+1)\right)\\
&=& \exp \left( \ln(n+3)\color{red}{\left(\sqrt{n+1}-\sqrt{n}\right)}+\sqrt{n}[\color{blue}{\ln(n+3)-\ln(n+1)} ]\right)\\
&=& \exp \left( \frac{\ln(n+3)}{\color{red}{\sqrt{n+1}+\sqrt{n}}}+\sqrt{n}\color{blue}{\ln\frac{n+3}{n+1}}\right)\\
\end{split}
On the other hand, we have
$$ \lim_{n\to \infty}\sqrt{n}\ln\frac{n+3}{n+1} = \lim_{n\to \infty}\frac{2\sqrt{n}}{n+1}\times\frac{\ln\left(1+\frac{2}{n+1}\right)}{\frac{2}{n+1}} = 0\times1= 0$$
Given that, $\frac{\ln(x+1)}{x}\to 0$ as $x\to0$
whereas
$$\lim_{n\to \infty}\frac{\ln(n+3)}{\sqrt{(n+1)}+\sqrt{n}} =\lim_{n\to \infty} 2\frac{\ln\sqrt{n+3}}{\sqrt{n+3}}\times \frac{\sqrt{1+\frac{3}{n}}}{\sqrt{1+\frac{1}{n}}+1}= 2\times0\times \frac{1}{\sqrt{2}}=0 $$
Given that $$\lim_{x\to \infty} \frac{\ln x}{x} = 0$$
so $$\color{brown}{\lim_{n\to \infty}a_n =\lim_{n\to \infty} \frac{(n+3)^{\sqrt{n+1}}}{(n+1)^{\sqrt{n}}}=1}$$
| {
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"url": "https://math.stackexchange.com/questions/2096167",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Evaluating $\quad\lim_{ x \to -\infty }\frac{\sqrt[n]{x+1}+\sqrt[n]{x+2}-2\sqrt[n]{x-3}}{\sqrt[n]{x-1}+\sqrt[n]{x-2}-2\sqrt[n]{x+3}}$ find limit :
$$\lim_{ x \to -\infty }\frac{\sqrt[n]{x+1}+\sqrt[n]{x+2}-2\sqrt[n]{x-3}}{\sqrt[n]{x-1}+\sqrt[n]{x-2}-2\sqrt[n]{x+3}}=\;\; ? \;\quad \text {given }\,n \in \mathbb{N}, n>2 ,\text{odd}$$
I tried :
$$\lim_{ x \to -\infty }\frac{\sqrt[n]{x+1}+\sqrt[n]{x+2}-2\sqrt[n]{x-3}}{\sqrt[n]{x-1}+\sqrt[n]{x-2}-2\sqrt[n]{x+3}}=\frac{x+x-2x }{x+x-2x}=\frac{0}{0}$$
| Substitute $x=-1/t$, so you get
$$
\lim_{t\to0^+}
\frac{\sqrt[n]{t-1}+\sqrt[n]{2t-1}-2\sqrt[n]{-3t-1}}
{\sqrt[n]{-t-1}+\sqrt[n]{-2t-1}-2\sqrt[n]{3t-1}}
=
\lim_{t\to0^+}
\frac{\sqrt[n]{1-t}+\sqrt[n]{1-2t}-2\sqrt[n]{1+3t}}
{\sqrt[n]{1+t}+\sqrt[n]{1+2t}-2\sqrt[n]{1-3t}}
$$
The Taylor expansion of the numerator is
$$
1-\frac{1}{n}t+1-\frac{2}{n}t-2-\frac{6}{n}t+o(t)=-\frac{9}{n}t+o(t)
$$
The Taylor expansion of the denominator isβ¦
| {
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"timestamp": "2023-03-29T00:00:00",
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sum of series $(2^2-1)(6^2-1)+(4^2-1)(8^2-1)+\cdots \cdots +\cdots (100^2-1)(104^2-1)$ The sum of series $(2^2-1)(6^2-1)+(4^2-1)(8^2-1)+\cdots \cdots +\cdots (100^2-1)(104^2-1)$
Attempt Assume $\displaystyle S = \sum^{50}_{r=1}((2r)^2-1)((2r+4)^2-1) = \sum^{50}_{r=1}(4r^2-1)(4r^2+16r+15)$
$\displaystyle S = \sum^{50}_{r=1}(16r^4+64r^3+56r^2-16r-15)$
could some help me how to solve it, thanks
| Being lazy, we could use the classical sums of powers of integer numbers as given here. This would lead to
$$ S_n = \sum^{n}_{r=1}(16r^4+64r^3+56r^2-16r-15)=\frac{1}{5} \left(16 n^5+120 n^4+280 n^3+180 n^2-71 n\right)$$
For sure, much more elegant is mathlove's solution.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Sum of given infinite series: $\frac14+\frac2{4 \cdot 7}+\frac3{4 \cdot 7 \cdot 10}+\frac4{4 \cdot 7 \cdot 10 \cdot 13 }+....$ Find the sum of infinite series
$$\frac{1}{4}+\frac{2}{4 \cdot 7}+\frac{3}{4 \cdot 7 \cdot 10}+\frac{4}{4 \cdot 7 \cdot 10 \cdot 13 }+....$$
Generally I do these questions by finding sum of $n$ terms and then putting $ \lim{n \to \infty}$ but here I am not able to find sum of $n$ terms. Could some suggest as how to proceed?
| The partial sums, according to Maple, are
$$-{\frac {2\,{3}^{1/2-N}\pi}{27\,\Gamma \left( 4/3+N \right) \Gamma
\left( 2/3 \right) }}+\frac{1}{3}
$$
It should be possible to prove that by induction.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
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Sum of real roots of two different cubic polynomials Let $\alpha$ be the only real root of $p(x)=x^3-3x^2+5x-17$ and $\beta$ the only real root of $q(x)=x^3-3x^2+5x+11$. Compute $\alpha + \beta$.
I've noticed that the two graphs are just the same shifted vertically (the first coefficients are equal except for the last), but I don't know if it is useful.
(This is supposed to be doable without any calculator)
| Re-arrange $p(x)$ and $q(x)$:
\begin{align*}
p(x) &= x^3-3x^2+5x-17 \\
&= (x-1)^3+2(x-1)-14 \\
q(x) &= x^3-3x^2+5x+11 \\
&= (x-1)^3+2(x-1)+14
\end{align*}
Note that $$p(1+x) = -q(1-x)$$
If $p(1+x)=-q(1-x)=0$, then $$
\left \{
\begin{align*}
1+x &= \alpha \\
1-x &= \beta
\end{align*}
\right.$$
$$\fbox{$\alpha+\beta=2$}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2100829",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Explanation of Digital Root/ Sum formula The formula to find the digital root/ sum is:
digital root of n = 1 + ( (n - 1) % 9 )
Can someone explain me the intuition behind this formula? Why does this result give the sum of digits?
| Numbers in base $10$ are "arranged" into boxes of $10$ entries. For example, $21$ looks like
$$
\blacksquare \blacksquare \blacksquare \blacksquare \blacksquare \blacksquare \blacksquare \blacksquare \blacksquare \blacksquare\\
\blacksquare \blacksquare \blacksquare \blacksquare \blacksquare \blacksquare \blacksquare \blacksquare \blacksquare \blacksquare\\
\cdot\\
\blacksquare \square \square \square \square \square \square \square \square \square
$$
and $123$ looks like:
$$
\blacksquare \blacksquare \blacksquare \blacksquare \blacksquare \blacksquare \blacksquare \blacksquare \blacksquare \blacksquare\\
\blacksquare \blacksquare \blacksquare \blacksquare \blacksquare \blacksquare \blacksquare \blacksquare \blacksquare \blacksquare\\
\blacksquare \blacksquare \blacksquare \blacksquare \blacksquare \blacksquare \blacksquare \blacksquare \blacksquare \blacksquare\\
\blacksquare \blacksquare \blacksquare \blacksquare \blacksquare \blacksquare \blacksquare \blacksquare \blacksquare \blacksquare\\
\blacksquare \blacksquare \blacksquare \blacksquare \blacksquare \blacksquare \blacksquare \blacksquare \blacksquare \blacksquare\\
\blacksquare \blacksquare \blacksquare \blacksquare \blacksquare \blacksquare \blacksquare \blacksquare \blacksquare \blacksquare\\
\blacksquare \blacksquare \blacksquare \blacksquare \blacksquare \blacksquare \blacksquare \blacksquare \blacksquare \blacksquare\\
\blacksquare \blacksquare \blacksquare \blacksquare \blacksquare \blacksquare \blacksquare \blacksquare \blacksquare \blacksquare\\
\blacksquare \blacksquare \blacksquare \blacksquare \blacksquare \blacksquare \blacksquare \blacksquare \blacksquare \blacksquare\\
\blacksquare \blacksquare \blacksquare \blacksquare \blacksquare \blacksquare \blacksquare \blacksquare \blacksquare \blacksquare\\
\cdot\\
\blacksquare \blacksquare \blacksquare \blacksquare \blacksquare \blacksquare \blacksquare \blacksquare \blacksquare \blacksquare\\
\blacksquare \blacksquare \blacksquare \blacksquare \blacksquare \blacksquare \blacksquare \blacksquare \blacksquare \blacksquare\\
\cdot\\
\blacksquare \blacksquare \blacksquare \square \square \square \square \square \square \square
$$
When you take the modulo $9$, you remove all the "full" packages such as the $100$ and only $1$ remains. (Remove the first $9$ columns, and $10$ boxes remain. $10 \equiv 1\mod 9$). This builds up, so $200$ loses for each $100$ box $99$ boxes, and $2$ boxes remain.
Hope this helps to visualize the operation.
| {
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"timestamp": "2023-03-29T00:00:00",
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Compute $\lim \limits_{n \to \infty} \frac{\sqrt[4]{n^4+4n}\,-\,\sqrt[3]{n^3+3n}}{\sqrt[5]{n^5+1}\,-\, \sqrt[5]{n^5+n}}\cdot \frac{1}{n^2}$ I need to compute:
$\lim \limits_{n \to \infty} \frac{\sqrt[4]{n^4+4n}\,-\,\sqrt[3]{n^3+3n}}{\sqrt[5]{n^5+1}\,-\, \sqrt[5]{n^5+n}}\cdot\frac{1}{n^2}$
I tried this
$\lim \limits_{n \to \infty} \frac{n}{n}\frac{\sqrt[4]{n^4+4n}\,-\,\sqrt[3]{n^3+3n}}{\sqrt[5]{n^5+1}\,-\, \sqrt[5]{n^5+n}}\cdot\frac{1}{n^2}=\lim \limits_{n \to \infty} \frac{\sqrt[4]{1+\frac{4}{n^3}}\,-\,\sqrt[3]{1+\frac{3}{n^2}}}{\sqrt[5]{1+\frac{1}{n^5}}\,-\, \sqrt[5]{1+\frac{1}{n^4}}}\cdot\frac{1}{n^2}$
I get stuck and I will have to probably use different method, would someone give me an advice, how to approach to this problem?
I also thought about multiplying it by $\frac{(a+b)}{(a+b)}$, but I am not really sure how.
| I know you are all "begeistert" with taylor series, but here is another method.
Note that $$\sqrt[3]{n^3+3n}-n=\frac{3n}{(\sqrt[3]{n^3+3n})^2+(\sqrt[3]{n^3+3n})n+n^2}$$ so it follows easily that $$n(\sqrt[3]{n^3+3n}-n)\to \frac{3}{3}=1$$
in a similar fashion one sees that
$$n^2(\sqrt[4]{n^4+4n}-n)\to 1$$
$$n^4(\sqrt[5]{n^5+1}-n)\to \frac{1}{5}$$
$$n^3(\sqrt[5]{n^5+n}-n)\to \frac{1}{5}$$
Now write the original expression as
$$\frac{\frac{n^2(\sqrt[4]{n^4+4n}-n)}{n}-n(\sqrt[3]{n^3+3n}-n)}{\frac{n^4(\sqrt[5]{n^5+1}-n)}{n}-n^3(\sqrt[5]{n^5+n}-n)}\to\frac{0-1}{0-\frac{1}{5}}=5$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Does the vector belong to the column space? I have to check whether the vector
$$b = \begin{vmatrix}
-2\\
4\\
5\\
6\\
6\\
\end{vmatrix}$$
belongs to the column space of matrix
$$A = \begin{bmatrix}
1 & 1 & 1 & β2& 1\\
1 & 0 & 2 & 4 & 2\\
1 & 0 & 2 & 5 & 3\\
1 & 0 & 2 & 6 & 4\\
1 & 0 & 2 & 6 & 4\\
\end{bmatrix}$$
I know that it is a question whether the matrix $A$ extended by vector $b$ has a solution. Therefore we have
$$[A|b]= \left[
\begin{array}{ccccc|c}
1 &1& 1& β2& 1& -2\\
1 &0 &2 &4 &2&4\\
1 &0 &2 &5 &3&5\\
1 &0 &2 &6 &4&6\\
1 &0 &2 &6 &4&6\\
\end{array}
\right] $$
After Gauss-Jordan elimination:
$$\left[
\begin{array}{ccccc|c}
1& 0& 2& 0& -2& 0\\
0& 1& -1& 0& 5& 0\\
0& 0& 0& 1& 1& 1\\
0& 0& 0& 0& 0& 0\\
0& 0& 0& 0& 0& 0\\
\end{array}
\right] $$
So it proves that the vector b belongs to the column space of matrix $A$? Because my prof. in his book gave the solution it doesn't. He make mistakes often but I'm not sure about it...
| It obviously does, because the given matrix and the augmented matrix have the same rank. You even can read the solutions in the reduced row echelon form (I denote the unknowns $x, y,z,t,u$, in the order of the columns):
$$\begin{cases}x=-2z+2u,\\y=z-5u,\\t=1-u.\end{cases}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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} |
Evaluate $y'=-2-y+y^2$
$$y'=-2-y+y^2$$ and $y_{0}=2$ is a solution
So it is a riccati ODE:
$y=2+z\Rightarrow z=y-2$
$z'=y'$
substation into the original ODE gives:
$z'=-2-(2+z)+(z+2)^2=-4-z+z^2+4z+4=z^2+3z$
So we have:
$z'=z^2+3z$ which is bernoulli $z'=3z+z^{1+1}$ when $k=1$
$w=\frac{1}{z}$
$w'=-z'\cdot z^{-2}=-(3z+z^2)z^{-2}=\frac{-3}{z}-1=-3w-1$
$w'+3w=-1$
$w_{h}=e^{-3\int dx}=e^{-3x+c}=e^{-3x}k$
$w_{p}=e^{-3x}k(x)$
$w'_{p}=-3e^{-3x}k(x)+k'(x)e^{-3x}$
Substation into $w'+3w=-1$:
$-3e^{-3x}k(x)+k'(x)e^{-3x}+3e^{-3x}k(x)=-1\Rightarrow k'(x)e^{-3x}=-1\Rightarrow k'(x)=-e^{3x}$
integrating gives:
$k(x)=\frac{-e^{3x}}{3}+c$
So
$w_{p}=e^{-3x}(\frac{-e^{3x}}{3}+c)=-\frac{1}{3}+ce^{3x}$
$w=w_{h}+w_{p}=e^{-3x}k-\frac{1}{3}+ce^{3x}$
is it correct?
| Clearly $y=-1$ and $y=2$ are trivial solutions. Note that
$$ y^2-y-2=(y-2)(y+1)$$
and hence
$$ y'=(y-2)(y+1)$$
or
$$ \frac{dy}{(y-2)(y+1)}=dx.$$
Integrating both sides gives
$$ \int\frac{dy}{(y-2)(y+1)}=\int dx+C.$$
So
$$ \frac13\ln\left|\frac{y-2}{y+1}\right|=x+C$$
or
$$ \frac{y-2}{y+1}=Ce^{3x}$$
So the solution is
$$ y=\frac{2+Ce^{3x}}{1-Ce^{3x}}.$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Solve $x=\pm\frac{y'}{\sqrt{(y')^2+1}}$
$$x=\pm\frac{y'}{\sqrt{(y')^2+1}}$$
$$x=\pm\frac{y'}{\sqrt{(y')^2+1}}$$
$$x^2=\frac{(y')^2}{{(y')^2+1}}$$
$$x^2(y')^2+x^2=(y')^2$$
$$(y')^2[x^2-1]=-x^2$$
$$(y')^2=\frac{-x^2}{x^2-1}$$
$$y'=\pm \sqrt{\frac{-x^2}{x^2-1}}$$
Have I got it wrong? as there is no ODE
| It looks alright.
You're mistaken in saying there's no ODE. Every line in your posting is just an ODE.
From where you left off I'd go on to say
$$
\int\pm\sqrt{\frac{-x^2}{x^2-1}}\,dx = \int\pm\sqrt{\frac{x^2}{1-x^2}} = \int\pm \frac x {\sqrt{1-x^2}} \, dx = \int\pm \frac {du}{2\sqrt{u}} = \pm \sqrt u + C
$$
$$
= \pm \sqrt{1-x^2} + C.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2110777",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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Solve the equation: $\sin 3x=2\cos^3x$ Solve the equation :
$$\sin 3x=2\cos^3x$$
my try :
$\sin 3x=3\sin x-4\sin^3x$
$\cos^2x=1-\sin^2x$
so:
$$3\sin x-4\sin^3x=2((1-\sin^2x)(\cos x))$$
then ?
| Hint:
$$\sin 3x=2\cos^3x$$
$$3\sin x-4\sin^3x=2\cos^3x$$
$$3\sin x(\cos^2x+\sin^2x)-4\sin^3x=2\cos^3x$$
$$\sin^3x-3\sin x\cos^2x+2\cos^3x=0$$
Divide both sides by $\cos^3x$ and $\tan x=t$
$$t^3-3t+2=0$$
$$(t-1)^2(t+2)=0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2111959",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How to solve a system of 3 trigonometric equations How to solve system of three trigonometric equations:
$(\sin x)^2 (\cos y)^2 = 4 \cos x \sin y\tag1$
$(\sin y)^2 (\cos z)^2 = 4 \cos y \sin z \tag2$
$1- \sqrt{\sin z}(1+\sqrt{\cos x})=\sqrt{\frac{1-\sin y}{1+\sin y}}\tag3$
and to verify that
$\sin x=\sqrt{2}(\sqrt{2}+1)\sqrt{\sqrt{10}-3}(\sqrt{5}-2)\\
\sin y=(\sqrt{2}-1)^2(\sqrt{10}-3)\\
\sin z=(2\sqrt{2}+\sqrt{5}-\sqrt{12+4\sqrt{10}})^2(\sqrt{5}+\sqrt{2}-\sqrt{6+2\sqrt{10}})^2$
not is the only solution, but that many others exist?
The simplest solution is:
$\sin x=2^{5/4}\big(\sqrt{2}-1\big)$
$\sin y=\frac{1}{\sqrt{2}}$
$\sin z=3-2\sqrt{2}$.
Three other solutions are:
1)
$\sin x=\sqrt{2\sqrt{2}-2}$
$\sin y=\sqrt{2}-1$
$\sin z=5+4\sqrt{2}-\sqrt{56+40\sqrt{2}}$,
2)
$\sin x=\frac{\sqrt{2}}{2}$
$\sin y=3-2\sqrt{2}$
$\sin z=\big(\sqrt{2}+1\big)^{2}\big(2^{1/4}-1\big)^{4}$
3)
$\sin x=\frac{2^{5/4}a^{2}c. d^{6}\varphi^{3}}{b^{2}}$
$\sin y=\frac{d^4}{\sqrt{2}\varphi^{4}}$
$\sin z=a^{4}b^{4}c^{2}\varphi^{6}$
where
$$a=\sqrt{2}-1$$
$$b=5^{1/4}-\sqrt{2}$$
$$c=\sqrt{10}-3$$
$$d=\frac{5^{1/4}-1}{\sqrt{2}}$$
$\varphi$ is the golden ratio.
| Note that, if any of $\sin x$, $\sin y$, $\sin z$ are zero, then they all are, so we have the solution
$$\sin x = \sin y = \sin z = 0 \qquad \cos x = 1 \tag{0}$$
(where the latter condition allows $\sqrt{\cos x}$ to be real). On the other hand, if any of $\cos x$, $\cos y$, $\cos z$ are zero, then they all are, and we have
$$\cos x = \cos y = \cos z = 0 \qquad \sin y = \sin z = 1 \tag{00}$$
So, we may assume none of these sines and cosines are zero. Moreover, since $\cos x$ and $\sin z$ must then be (strictly) positive, we see that $\sin y$ and $\cos y$ must be, as well.
Substituting $\cos^2y\to 1-\sin^2 y$ into OP's equation $(1)$ gives a quadratic in $\sin y$ we can solve to get
$$\sin y = \frac{ 1 - \cos x }{ 1 + \cos x } \qquad\text{or}\qquad \sin y = -\frac{ 1 + \cos x }{ 1 - \cos x } \tag{1}$$
As $\sin y$ should be strictly positive, we discard the latter option, and we find
$$\cos y = \frac{2 \sqrt{\cos x}}{ 1 + \cos x} \qquad\qquad \frac{1-\sin y}{1+\sin y} = \cos x \tag{2}$$
Consequently, OP's equation $(3)$ becomes
$$1 - \sqrt{\sin z}(1+\sqrt{\cos x}) = \sqrt{\cos x} \quad\to\quad 1 - c - s - s c = 0 \quad\to\quad
s = \frac{1-c}{1+c}\tag{3}$$
where $c := \sqrt{\cos x}$ and $s := \sqrt{\sin z}$. But, then OP's $(2)$ is
$$\frac{(1 - c^2 )^2}{(1 + c^2)^2}(1-s^4) = \frac{8cs^2}{1+c^2} \quad\to\quad 0 = 0 \tag{4}$$
which is to say: OP's equation $(3)$ is not independent of OP's $(1)$ and $(2)$. We have only two equations in three unknowns, hence multiple solutions are not unexpected. Thus,
For $0< c < 1$ we have
$$\sin^2 x = 1 - c^4 \qquad \sin y = \frac{1 - c^2}{1+c^2} \qquad \sin z = \frac{(1-c)^2}{(1+c)^2}$$
$$\cos x= c^2 \qquad \cos y = \frac{2c}{1+c^2} \qquad \cos^2 z =\frac{8 c (1 + c^2)}{(1 + c)^4}$$
Including $c=0$ and $c=1$ gives solutions $(00)$ and $(0)$, respectively.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to factorize this cubic equation? In one of the mathematics book, the author factorized following term
$$x^3 - 6x + 4 = 0$$
to
$$( x - 2) ( x^2 + 2x -2 ) = 0.$$
How did he do it?
| There is a neat trick called the rational roots theorem. All we have to do is factor the first and last numbers, put them over a fraction, and take $\pm$. This gives us the following possible rational roots:
$$x\stackrel?=\pm1,\pm2,\pm4$$
due to the factorization of $4$. Checking these, it is clear $x=2$ is the only rational root, since
$$\begin{align}0&\ne(+1)^3-6(+1)+4\\0&\ne(-1)^3-6(-1)+4\\\color{#4488dd}0&=\color{#4488dd}{(+2)^3-6(+2)+4}\\0&\ne(-2)^3-6(-2)+4\\0&\ne(+4)^3-6(+4)+4\\0&\ne(-4)^3-6(-4)+4\end{align}$$
leaving us with
$$x^3-6x+4=(x-2)(\dots)$$
We can find the remainder through synthetic division:
$$\begin{array}{c|c c}2&1&0&-6&4\\&\downarrow&2&4&-4\\&\hline1&2&-2&0\end{array}$$
which gives us our factorization:
$$x^3-6x+4=(x-2)(x^2+2x-2)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2118291",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 5,
"answer_id": 0
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How to simplify derivatives
The math problem asks to find the derivative of the function
$$y=(x+1)^4(x+5)^2$$
I get to the part $$(x+1)^4 \cdot 2(x+5) + (x+5)^2 \cdot 4(x+1)^3$$
How do they arrive at the answer
$$2(x+1)^3(x+5)(3x+11) ?$$
| $$\begin{align}(x+1)^4 \cdot 2(x+5) + (x+5)^2 \cdot 4(x+1)^3&=(x+1)^3(x+5)\cdot\big[(x+1)2+(x+5)4\big]\\
&=(x+1)^3(x+5)\cdot\big[2x+2+4x+20 \big]\\
&=(x+1)^3(x+5)\cdot(6x+22)\\
&=(x+1)^3(x+5)\cdot 2(3x+11)\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2118826",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "3",
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Prove that; $8\cos^3 (\frac {\pi}{9})- 6\cos(\frac {\pi}{9})=1$ Prove that; $8\cos^3 (\frac {\pi}{9})- 6\cos(\frac {\pi}{9})=1$
My Attempt,
$$L.H.S=8\cos^3(\frac {\pi}{9}) - 6\cos(\frac {\pi}{9})$$
$$=2\cos(\frac {\pi}{9}) [4\cos^2(\frac {\pi}{9}) - 3]$$
$$=2\cos(\frac {\pi}{9}) [2+2\cos(\frac {2\pi}{9}) - 3]$$
$$=2\cos(\frac {\pi}{9}) [2\cos(\frac {2\pi}{9})-1]$$.
What should I do further?
| \begin{align}
8\cos^3 (\frac {\pi}{9})- 6\cos(\frac {\pi}{9})
&=8\cos(\frac{\pi}9)\left[1-\sin^2(\frac{\pi}9)\right]-6\cos^(\frac{\pi}9)\\
&=2\cos(\frac{\pi}9)-8\cos(\frac{\pi}9)\sin^2(\frac{\pi}9)\\
&=2\cos(\frac{\pi}9)-4\sin(\frac{2\pi}9)\sin(\frac{\pi}9)\tag1\\
&=2\cos(\frac{\pi}9)-2\left[\cos(\frac{\pi}9)-\cos(\frac{3\pi}9)\right]\tag2\\
&=2\cos(\frac{\pi}3)=1
\end{align}
Note:
$(1)$:(Double Angle Formula) $\sin{2t}=2\sin t\cos t$
$(2)$:(Product-to-sum-formula) $2\sin a\sin b=\cos(a-b)-\cos(a+b)$
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "1",
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Find the value of $x$ that satisfy the equation: $3^{11}+3^{11}+3^{11} = 3^x$ I have this question:
$$3^{11}+3^{11}+3^{11} = 3^x$$
Find the value of $x$
| Takking $3^{11}$ common from left hand side,
$3^{11}(1 + 1 + 1) = 3^x$
$3^{11}(3) = 3^x$
$3^{12} = 3^x$
$x = 12$
Or -
$3(3^{11})= 3^x$
$3^{12} = 3^x$
$x = 12$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2122390",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find the determinant of order $100$ Find the determinant of order $100$:
$$D=\begin{vmatrix}
5 &5 &5 &\ldots &5 &5 &-1\\
5 &5 &5 &\ldots &5 &-1 &5\\
5 &5 &5 &\ldots &-1 &5 &5\\
\vdots &\vdots &\vdots &\ddots &\vdots &\vdots &\vdots\\
5 &5 &-1 &\ldots &5 &5 &5\\
5 &-1 &5 &\ldots &5 &5 &5\\
-1 &5 &5 &\ldots &5 &5 &5
\end{vmatrix}$$
I think I should be using recurrence relations here but I'm not entirely sure how that method works. I tried this:
Multiplying the first row by $(-1)$ and adding it to all rows:
$$D=\begin{vmatrix}
5 &5 &5 &\ldots &5 &5 &-1\\
5 &5 &5 &\ldots &5 &-1 &5\\
5 &5 &5 &\ldots &-1 &5 &5\\
\vdots &\vdots &\vdots &\ddots &\vdots &\vdots &\vdots\\
5 &5 &-1 &\ldots &5 &5 &5\\
5 &-1 &5 &\ldots &5 &5 &5\\
-1 &5 &5 &\ldots &5 &5 &5
\end{vmatrix}=\begin{vmatrix}
5 &5 &5 &\ldots &5 &5 &-1\\
0 &0 &0 &\ldots &0 &-6 &6\\
0 &0 &0 &\ldots &-6 &0 &6\\
\vdots &\vdots &\vdots &\ddots &\vdots &\vdots &\vdots\\
0 &0 &-6 &\ldots &0 &0 &6\\\
0 &-6 &0 &\ldots &0 &0 &6\\
-6 &0 &0 &\ldots &0 &0 &6
\end{vmatrix}$$
Applying Laplace's method to the first column
$$D=5\begin{vmatrix}
0 &0 &\ldots &0 &-6 &6\\
0 &0 &\ldots &-6 &0 &6\\
\vdots &\vdots &\ddots &\vdots &\vdots &\vdots\\
0 &-6 &\ldots &0 &0 &6\\
-6 &0 &\ldots &0 &0 &6
\end{vmatrix}+6\begin{vmatrix}
5 &5 &\ldots &5 &5 &-1\\
0 &0 &\ldots &0 &-6 &6\\
0 &0 &\ldots &-6 &0 &6\\
\vdots &\vdots &\ddots &\vdots &\vdots &\vdots\\
0 &-6 &\ldots &0 &0 &6\\
-6 &0 &\ldots &0 &0 &6
\end{vmatrix}$$
I can see that this one is $D$ but of order $99$...Is this leading anywhere? How would you solve this?
| If we reverse the order of the rows, we end up with a matrix of the form $M = A-6I$, where $A$ has $5$ for every entry. This requires $50$ transpositions, so the determinant of this new matrix is the same. To find this determinant, we can multiply the eigenvalues of the matrix $A - 6I$, which can be found to be
$$
\{\overbrace{-6,-6,\dots,-6}^{99 \text{ times}},494\}
$$
in the method outlined here.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2122803",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 2
} |
finding limit with $\cos$ function occur $n$ times Finding $\displaystyle \lim_{x\rightarrow 0}\frac{1-\cos(1-\cos(1-\cos(1-\cdots \cdots (1-\cos x))))}{x^{2^n}}$
where number of $\cos$ is $n$ times
when $x\rightarrow 0$ then $\displaystyle 1-\cos x = 2\sin^2 \frac{x}{2} \rightarrow 2\frac{x}{2} = x$
so $1-\cos (1-\cos x) = 1-\cos x$
some help me., thanks
| Let $f^{n} $ denote the composition of $f$ with itself $n$ times and let $f^{0}(x)=x$. Then the numerator of the given expression (whose limit is to be evaluated) is equal to $f^{n} (x) $ where $f(x) = 1-\cos x$. Note that as $x\to 0$ each of the functions $f^{n} (x) \to 0$ and also note that we have $$\lim_{x\to 0}\frac{1 - \cos x} {x^{2}}=\frac{1}{2}$$ Replacing $x$ by $f^{n-1}(x)$ in the above equation we get $$\lim_{x\to 0}\frac{f^{n} (x)} {(f^{n-1}(x))^{2}} = \frac{1} {2} $$ Now we need to replace $n$ by $n-1$ in above equation and square it and continue like this and multiply resulting equations to get $$\lim_{x\to 0}\frac{f^{n}(x)}{(f^{0}(x))^{2^{n}}}= \frac{1}{2}\cdot\frac{1}{2^{2}}\cdots\frac{1}{2^{2^{n-1}}}$$ and thus the desired limit is $1/2^{1+2+2^{2}+\cdots +2^{n-1}}$ or $1/2^{2^{n} - 1}$.
See a similar answer to a related question.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2124023",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
Finding function when dx is integrated $\int (5-x^2+\frac{18}{x^4})dx$ I have a question on how to evaluate this integral:
$\int (5-x^2+\frac{18}{x^4})dx$
Is this correct?
$\int (5-x^2+\frac{18}{x^4})dx$
$ \int 5 dx = 5x + C$
$ \int -x^2 = \frac {-x^3}{3}= -\frac{1}{3}x^3 +C $
$ \int \frac {18}{x^4} = \int 18x^{-4} dx = \frac {18x^{-3}}{3}=6x^{-3} + C$
So, $\int (5-x^2+\frac{18}{x^4})dx= 5x-\frac{1}{3}x^3+6x^{-3}+C$
| This is how you would do this
$$\int (5-x^2+\frac{18}{x^4})dx$$
Integration as you have probably shown spits across addition so we have;
$$
\int5dx+\int-x^2dx+\int\frac{18}{x^4}dx
$$
now we can also take out constaints so we have:
$$
5\int1dx-\int x^2dx+18\int x^{-4}dx
$$
now by defintion we have
$$
5x-\frac{x^3}{3}+18\frac{x^{-3}}{-3}+C
$$
Not we only need one C because three constants added together are constants
$$
5x-\frac{x^3}{3}-\frac{6}{x^3}+C
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2124833",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
What real numbers do algebraic numbers cover? Hardy and Wright mention ( though don't give a proof ) that any finite combination of real quadratic surds is an algebraic number. For example
$\sqrt{11+2\sqrt{7}}$. Are all finite combinations of cube root, fourth root ... $n^{th}$ root also algebraic ? such as $\sqrt[3]{2+3\sqrt[7]{5+3\sqrt{6}}}+\sqrt[9]{2}$.
| Yes, you can unpack them into a polynomial that they satisfy. For your example, we can write $$x=\sqrt[3]{2+3\sqrt[7]{5+3\sqrt{6}}}\\x^3=2+3\sqrt[7]{5+3\sqrt{6}}\\\frac 13(x^3-2)=\sqrt[7]{5+3\sqrt{6}}\\\left(\frac 13(x^3-2)\right)^7=5+3\sqrt 6\\\frac 19\left(\left(\frac 13(x^3-2)\right)^7-5\right)^2-6=0$$
Note: this uses a previous example in the question. The approach is the same. Note that the algebraics are closed under the field operations, so the sum of algebraics is again algebraic.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2126032",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 8,
"answer_id": 3
} |
If:$(\sqrt{x + 9})^{\frac{1}{3}} - (\sqrt{x-9})^{\frac{1}{3}} = 3$, Find $x^2$
If:$$\sqrt[3]{(x + 9)} - \sqrt[3]{(x-9)} = 3$$
Find $x^2$
I can't seem to solve this question. Any hints or solutions is welcomed.
| We have this:
$$(x + 9)^{\frac{1}{3}} - (x-9)^{\frac{1}{3}} = 3$$
Then:
$$(x + 9)^{\frac{1}{3}} -3 = (x-9)^{\frac{1}{3}}$$
$$((x + 9)^{\frac{1}{3}} -3)^3 = ((x-9)^{\frac{1}{3}})^3$$
$$x + 9 -9(x+9)^{\frac{2}{3}} + 27(x+9)^{\frac{1}{3}} -27 = x -9$$
$$-9(x + 9)^{\frac{2}{3}} +27(x +9)^{\frac{1}{3}} =-9-9 +27 $$
$$-9(x + 9)^{\frac{2}{3}} +27(x +9)^{\frac{1}{3}} =9 $$
$$3(x+9)^{\frac{1}{3}} - (x+9)^{\frac{2}{3}} = 1$$
$$(x+9)^{\frac{2}{3}} - 3(x+9)^{\frac{1}{3}} + 1 = 0$$
Let $y = (x+9)^{\frac{1}{3}}$
Then we have:
$$y^2 -3y +1 - 0$$
$$y = \frac{3 \pm \sqrt{5}}{2} = (x +9)^{\frac{1}{3}}$$
$$(\frac{72+32\sqrt{5}}{8})^{\frac{1}{3}} = (x + 9)^{\frac{1}{3}}$$
$$\frac{72+32\sqrt{5}}{8} = x + 9$$
$$72 + 32\sqrt{5} = 8x + 72$$
$$8x = 32\sqrt{5}$$
Hence:
$$x = 4\sqrt{5} = \sqrt{80}$$
So $x^2$ would be $80$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2127400",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
Solving a univariable identity that satisfies the following relationship:
Question: Consider the following equations:$$\begin{align*}1^2+2^2+2^3 & =3^2\\2^2+3^2+6^2 & =7^2\\3^2+4^2+12^2 & =13^2\\4^2+5^2+20^2 & =21^2\end{align*}$$
State a one variable identity that is suggested by these examples.
Since the question asked for a uni-variable identity, I assumed the form$$(ax+b)^2+(cx+d)^2+(ex+f)^2=(gx+h)^2\tag1$$
And equated coefficients to get an undermined system. Namely,$$\begin{align*} & a^3+c^3+e^3=g^3\\ & ab+cd+ef=gh\\ & b^2+d^2+f^2=h^2\end{align*}$$
And solving for integer solutions (I set $(a,c,e,g)=(1,2,2,3)$ and solved the remaining system) to get$$(x+2)^2+(2x+4)^2+(2x+4)^2=(3x+6)^2$$
Which works for $x\in\mathbb{Z}$. However, the question asked for an identity that gave the examples listed above. Something my formula clearly isn't capable of.
So my actual question is simple: How would you go about solving this problem?
| I believe the identity you are looking for contains a quadratic for $n$, $$n^2+(n+1)^2+(n^2+n)^2=(n^2+n+1)^2$$
Putting $n=1,2,3,4$ gives $$\begin{align*}1^2+2^2+2^3 & =3^2\\2^2+3^2+6^2 & =7^2\\3^2+4^2+12^2 & =13^2\\4^2+5^2+20^2 & =21^2\end{align*}$$
Which are the examples above. These identities are well known, and can also be noticed since of our given numbers on the left hand side the rightmost number is the product of the other two.
PROOF OF THE IDENTITY
It is well known $(x+1)^2-x^2=2x+1$
Now $$n^2+(n+1)^2=2n^2+2n+1=(n^2+n+1)^2-(n^2+n)^2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2128295",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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How to integrate $\int_{0}^{1} \frac{1-x}{1+x} \frac{dx}{\sqrt{x^4 + ax^2 + 1}}$? The question is how to show the identity
$$ \int_{0}^{1} \frac{1-x}{1+x} \cdot \frac{dx}{\sqrt{x^4 + ax^2 + 1}} = \frac{1}{\sqrt{a+2}} \log\left( 1 + \frac{\sqrt{a+2}}{2} \right), \tag{$a>-2$} $$
I checked this numerically for several cases, but even Mathematica 11 could not manage this symbolically for general $a$, except for some special cases like $a = 0, 1, 2$.
Addendum. Here are some backgrounds and my ideas:
*
*This integral came from my personal attempt to find the pattern for the integral
$$ J(a, b) := \int_{0}^{1} \frac{1-x}{1+x} \cdot \frac{dx}{\sqrt{1 + ax^2 + bx^4}}. $$
This drew my attention as we have the following identity
$$ \int_{0}^{\infty} \frac{x}{x+1} \cdot \frac{dx}{\sqrt{4x^4 + 8x^3 + 12x^2 + 8x + 1}} = J(6,-3), $$
where the LHS is the integral from this question. So establishing the claim in this question amounts to showing that $J(6,-3) = \frac{1}{2}\log 3 - \frac{1}{3}\log 2$, though I am skeptical that $J(a, b)$ has a nice closed form for every pair of parameters $(a, b)$.
*A possible idea is to write
\begin{align*}
&\int_{0}^{1} \frac{1-x}{1+x} \cdot \frac{dx}{\sqrt{x^4 + ax^2 + 1}} \\
&\hspace{5em}= \int_{0}^{1} \frac{(x^{-2} + 1) - 2x^{-1}}{x^{-1} - x} \cdot \frac{dx}{\sqrt{(x^{-1} - x)^2 + a + 2}}
\end{align*}
This follows from a simple algebraic manipulation. This suggests that we might be able to apply Glasser's master theorem, though in a less trivial way.
I do not believe that this is particularly hard, but I literally have not enough time to think about this now. So I guess it is a good time to seek help.
| Enforcing the substitution $x^{-1}-x=u$ in your last integral we get:
$$ \int_{0}^{+\infty}\left(-1+\frac{2}{\sqrt{4+u^2}}\right)\frac{du}{u\sqrt{u^2+a+2}} $$
and by setting $u=\sqrt{a+2}\sinh\theta$ we get:
$$ \frac{1}{\sqrt{a+2}}\int_{0}^{+\infty}\left(-1+\frac{2}{\sqrt{4+(a+2)\sinh^2\theta}}\right)\frac{d\theta}{\sinh\theta}$$
We may get rid of the last term through the "hyperbolic Weierstrass substitution"
$$ \theta = 2\,\text{arctanh}(e^{-v}) = \log\left(\frac{e^v+1}{e^v-1}\right)$$
that wizardly gives
$$ \frac{1}{\sqrt{a+2}}\int_{0}^{+\infty}\left(-1+\frac{2}{\sqrt{4+\frac{a+2}{\sinh^2 v}}}\right)\,dv$$
i.e., finally, a manageable integral through differentiation under the integral sign.
This proves OP's initial identity. Beers on me.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2129537",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "20",
"answer_count": 2,
"answer_id": 0
} |
Vector Parameterisation of Ellipse How would you show that, if
$$\mathbf r = \mathbf A \sin \theta + \mathbf B \cos \theta$$
where $\mathbf A$ and $\mathbf B$ are arbitrary constant vectors in $\Bbb R^2$,
then $\mathbf r$ may be written as,
$$\mathbf r = \mathbf a \sin (\theta + \alpha) + \mathbf b \cos (\theta + \alpha)$$
for some constant orthogonal vectors $\mathbf a$ and $\mathbf b$ in $\Bbb R^2$, and real constant $\alpha$.
[If it helps: Note that an explicit expression for $\mathbf a$ and $\mathbf b$ is not necessarily needed, just proof of their existence.]
| Let $\boldsymbol{x}=
\begin{pmatrix}
\cos \theta \\ \sin \theta
\end{pmatrix} \in S^1$
\begin{align*}
\boldsymbol{x}^T \boldsymbol{x} &= 1 \\
\boldsymbol{y} &=
\begin{pmatrix}
\boldsymbol{a} & \boldsymbol{b}
\end{pmatrix}
\begin{pmatrix}
\cos \theta \\ \sin \theta
\end{pmatrix} \\
\boldsymbol{y}^T \boldsymbol{y} &=
\boldsymbol{x}^T
\begin{pmatrix}
\boldsymbol{a} \cdot \boldsymbol{a} &
\boldsymbol{a} \cdot \boldsymbol{b} \\
\boldsymbol{b} \cdot \boldsymbol{a} &
\boldsymbol{b} \cdot \boldsymbol{b}
\end{pmatrix}
\boldsymbol{x}
\end{align*}
Eigenvalues:
$$\lambda_{min} \le \boldsymbol{y}^T \boldsymbol{y} \le \lambda_{max}$$
$$
\frac{a^2+b^2-\sqrt{(a^2-b^2)^2+4(\boldsymbol{a} \cdot \boldsymbol{b})^2}}{2}
\le \boldsymbol{y}^T \boldsymbol{y} \le
\frac{a^2+b^2+\sqrt{(a^2-b^2)^2+4(\boldsymbol{a} \cdot \boldsymbol{b})^2}}{2}$$
Eigenvectors:
$$\boldsymbol{\alpha}=
\boldsymbol{a} \cos
\left(
\frac{1}{2} \tan^{-1} \frac{2\boldsymbol{a} \cdot \boldsymbol{b}}{a^2-b^2}
\right)+
\boldsymbol{b} \sin
\left(
\frac{1}{2} \tan^{-1} \frac{2\boldsymbol{a} \cdot \boldsymbol{b}}{a^2-b^2}
\right)$$
$$\boldsymbol{\beta}=
\boldsymbol{b} \cos
\left(
\frac{1}{2} \tan^{-1} \frac{2\boldsymbol{a} \cdot \boldsymbol{b}}{a^2-b^2}
\right)-
\boldsymbol{a} \sin
\left(
\frac{1}{2} \tan^{-1} \frac{2\boldsymbol{a} \cdot \boldsymbol{b}}{a^2-b^2}
\right)$$
In general, the angle will be distorted and the sense of rotation may not be preserved:
$$\boldsymbol{y}=
\boldsymbol{\alpha} \cos (\phi+\phi_0) \pm
\boldsymbol{\beta} \sin (\phi+\phi_0)$$
See another answer here.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2134219",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Inequality problem with 4th root. $ \frac{6x}{x-2} - \sqrt{\frac{12x}{x-2}} - 2\sqrt[4]{\frac{12x}{x-2}}>0 $
I've tried putting $t= \frac{6x}{x-2} $ and play algebraically, using square of sum, but still no luck, any help?
| Instead, let's put $$t = \sqrt[4]{\frac{12x}{x-2}}.$$
Then:
$$\frac{6x}{x-2} - \sqrt{\frac{12x}{x-2}} - 2\sqrt[4]{\frac{12x}{x-2}}>0 \Rightarrow \\
\frac{1}{2}t^4 - t^2 - 2t>0 \Rightarrow \\
t^4 - 2t^2 - 4t > 0 \Rightarrow \\
t(t-2)(t^2+2t+2) >0.
$$
Then:
$$t <0 ~ \vee t > 2, $$
or equivalently:
$$\sqrt[4]{\frac{12x}{x-2}} <0 ~ \vee \sqrt[4]{\frac{12x}{x-2}} > 2.$$
For sure, $\sqrt[4]{\frac{12x}{x-2}} >0$, then we deal only with $\sqrt[4]{\frac{12x}{x-2}} > 2$.
Furthermore, the argument of the $4$th root must be positive. That is:
$$\frac{12x}{x-2} > 0 \Rightarrow x < 0 \vee x > 2.$$
Regarding $\sqrt[4]{\frac{12x}{x-2}} > 2$, we have that:
$$\frac{12x}{x-2} > 2^4.$$
We have two cases:
*
*$x < 0$. In this case we get that $x > 8$. But this is a contradiction.
*$x > 2$. In this case we get that $x < 8$.
Finally, we found that:
$$ 2 < x < 8.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2135811",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
How can you prove $\frac{n(n+1)(2n+1)}{6}+(n+1)^2= \frac{(n+1)(n+2)(2n+3)}{6}$ without much effort?
I will keep it short and take only an extract (most important part) of
the old task.
$$\frac{n(n+1)(2n+1)}{6}+(n+1)^2= \frac{(n+1)(n+2)(2n+3)}{6}$$
What I have done is a lot work and time consuming, I have "simply" solved it. But I think with a lot less work, there would be an easier and faster way. It's just I cannot see it : /
If anyone wants see, here is my long solution which I'm not happy with:
$$\frac{n(n+1)(2n+1)+6(n+1)^2}{6}=\frac{(n^2+2n+n+2)(2n+3)}{6} \Leftrightarrow$$
$$\Leftrightarrow \frac{(2n^3+n^2+2n^2+n)+6n^2+12n+6}{6} = \frac{(n^2+3n+2)(2n+3)}{6} \Leftrightarrow$$
$$\Leftrightarrow \frac{2n^3+3n^2+n+6n^2+12n+6}{6}=\frac{2n^3+3n^2+6n^2+9n+4n+6}{6} \Leftrightarrow$$
$$\Leftrightarrow \frac{2n^3+9n^2+13n+6}{6}=\frac{2n^3+9n^2+13n+6}{6}$$
| The most general method: do all multiplications on both sides to transform both numerators to canonical form, then verify polynomials are the same degree and have respective coefficients equal.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2137292",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 5,
"answer_id": 4
} |
Prove that: $\sec^2 20^\circ +\sec^2 40^\circ +\sec^2 80^\circ = \textrm 36$ Prove that: $\sec^2 20^\circ +\sec^2 40^\circ +\sec^2 80^\circ = \textrm 36$
My Attempt:
$$L.H.S=\sec^2 20^\circ + \sec^2 40^\circ +\sec^2 80^\circ$$
$$=\dfrac {1}{\cos^2 20Β°} +\dfrac {1}{\cos^2 40Β°} +\dfrac {1}{\cos^2 80Β°}$$
$$=\dfrac {\cos^2 40Β°.\cos^2 80Β°+\cos^2 20Β°.\cos^2 80Β°+\cos^2 20Β°.\cos^2 40Β°}{\cos^2 20Β°.\cos^2 40Β°.\cos^2 80Β°}$$.
I got paused here. Please help to prove this..
| A clean start, not remotely obvious ( but easy enough to prove ), the roots of
$$ x^3 - 3 x + 1 $$ are
$$ 2 \cos \frac{2 \pi}{9}, \; \; 2 \cos \frac{4 \pi}{9}, \; \; 2 \cos \frac{8 \pi}{9}. $$
From page 174 in Reuschle (1875). The method used is due to Gauss.
I learned today that, in 1933, D. H. Lehmer published a two-page proof that $2 \cos (2k\pi/n)$ is always an algebraic integer. Mentioned in an article in the January (M. A. A.) Monthly by Y. Z. Gurtas.
We immediately get that the roots of
$$ 8 x^3 - 6 x + 1, $$ therefore
$$ x^3 - \frac{3}{4} x + \frac{1}{8} $$ are
$$ \cos \frac{2 \pi}{9}, \; \; \cos \frac{4 \pi}{9}, \; \; \cos \frac{8 \pi}{9}. $$
Compare with the three requested proofs in the answer by Michael R.
Proof comes by taking $\omega$ as a primitive ninth root of unity, then taking
$$ x = \omega + \frac{1}{\omega}. $$
Primitive means
$$ \omega \neq 1, \; \; \omega^3 \neq 1, \omega^9 = 1 $$
Then $$ x^3 = \omega^3 + 3 \omega + \frac{3}{\omega} + \frac{1}{\omega^3}, $$
$$ x^3 - 3 x + 1 = \omega^3 + 1 + \frac{1}{\omega^3}, $$
$$ \omega^3 (x^3 - 3 x + 1) = \omega^6 + \omega^3 + 1, $$
$$ (\omega^3 - 1)\omega^3 (x^3 - 3 x + 1) = (\omega^3 - 1)(\omega^6 + \omega^3 + 1) = \omega^9 - 1 = 0. $$
These can be combined with double angle formulas/half angle formulas.
| {
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"url": "https://math.stackexchange.com/questions/2137609",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Prove, by contradiction, that, if $cx^2 + bx + a$ has no rational root, then $ax^2 + bx + c$ has no rational root. Proposition: Suppose that $a$, $b$, and $c$ are real numbers with $c \not = 0$. Prove, by contradiction, that, if $cx^2 + bx + a$ has no rational root, then $ax^2 + bx + c$ has no rational root.
Hypothesis: $cx^2 + bx + a$ has no rational root where $a$, $b$, and $c$ are real numbers with $c \not = 0$.
Conclusion: $ax^2 + bx + c$ has no rational root
To form a proof by contradiction, we take the negation of the conclusion:
$\neg B$: $ax^2 + bx + c$ has a rational root.
We now have a suitable hypothesis and conclusion for proof by contradiction:
A (Hypothesis): $cx^2 + bx + a$ has no rational root where $a$, $b$, and $c$ are real numbers with $c \not = 0$.
A1: $ax^2 + bx + c$ has a rational root.
Given that this is a proof by contradiction, we can work forward from both the hypothesis and conclusion, as shown above.
My Workings
A2: Let $x = \dfrac{p}{q}$ where $p$ and $q \not = 0$ are integers. This is the definition of a rational number (in this case, $x$): A rational number is any number that can be expressed as the quotient/fraction of two integers.
A3: $a\left(\dfrac{p}{q}\right)^2 + b\left(\dfrac{p}{q}\right) + c = 0$
$\implies \dfrac{ap^2}{q^2} + \dfrac{bp}{q} + c = 0$ where $q \not = 0$.
$\implies ap^2 + bpq + cq^2 = 0$
A4: $ap^2 + bpq + cq^2 = 0$ where $c \not = 0$
$\implies ap^2 + bpq = -cq^2$ where $-cq \not = 0$ since $c \not = 0$ and $q \not = 0$.
A5: $ap^2 + bpq + cq^2 = 0$ where $ap^2 + bpq \not = 0$ and $cq^2 \not = 0$.
But $ap^2 + bpq + cq^2 = 0$? Contradiction. $Q.E.D.$
I would greatly appreciate it if people could please take the time to review my proof and provide feedback on its correctness.
| The roots of
$ax^2+bx+c = 0
$
are
$\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}
$
and
the roots of
$cx^2+bx+a = 0
$
are
$\dfrac{-b\pm\sqrt{b^2-4ac}}{2c}
$.
If the first are rational
and the second are not,
then their sum
and ratio
are irrational.
Since the roots of the first equation are rational,
their sum and product are rational.
These are
$\dfrac{b}{a}$
and
$\dfrac{c}{a}$.
Since the ratio
of the two equations' roots is irrational,
$\dfrac{a}{c}$
is irrational.
Since their sum is irrational,
$\dfrac{b}{c}$
is irrational.
But these both contradict
the previously proven
rationality
of these two ratios.
Therefore the roots
of the second equation
are also rational.
| {
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"url": "https://math.stackexchange.com/questions/2139056",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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If a,b and c are sides of a triangle, then prove that the following polynomial has no real roots This is the polynomial: $$a^2x^2+(b^2+a^2-c^2)x+b^2=0$$
Now this is my progress:
Assuming l,m, and n are sides of a triangle, then $$|m-n|\lt l\lt m+n$$
Also, if a second degree polynomial in the form $kx^2+px+q$ has real roots, then $$p^2-4kq\ge 0$$
In this case, if there are no real roots, $$(b^2+a^2-c^2)^2-4a^2b^2\lt 0$$
$$b^2+a^2-c^2\lt 2ab$$
But by the first equation, $$a+b\gt c$$
$$a^2+b^2-c^2\gt 2ab$$
Wich actually proves by contradiction there ARE real roots. What did I do wrong?
| Nice exercise. It is well known (and not difficult to prove) that a second-degree polynomial has no real roots iff its discriminant is negative. The discriminant of the given polynomial is
$$ (b^2+a^2-c^2)^2-4a^2 b^2 = (a^2+2ab+b^2-c^2)(a^2-2ab+b^2-c^2) $$
or
$$ (a+b+c)(a+b-c)\color{red}{(a-b-c)}(a-b+c) $$
that is clearly negative by the triangular inequality or, equivalently, by Heron's formula.
| {
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"source": "stackexchange",
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Computing: $\lim_{x\to\infty}\frac{\sqrt{1-\cos^2\frac{1}{x}}\left(3^\frac{1}{x}-5^\frac{-1}{x}\right)}{\log_2(1+x^{-2}+x^{-3})}$ Find the following limit:
$$\lim_{x\to\infty}\frac{\sqrt{1-\cos^2\frac{1}{x}}\left(3^\frac{1}{x}-5^\frac{-1}{x}\right)}{\log_2(1+x^{-2}+x^{-3})}$$
I'm not sure whether my solution is correct.
$t:=\frac{1}{x}$
$$\lim_{x\to\infty}\frac{\sqrt{1-\cos^2\frac{1}{x}}\left(3^\frac{1}{x}-5^\frac{-1}{x}\right)}{\log_2(1+x^{-2}+x^{-3})}=\lim_{t\to 0}\frac{\sqrt{1-\cos^2 t}\left(3^t-5^{-t}\right)}{\log_2(1+t^2+t^3)}$$
$$=\lim_{t\to 0}\frac{\frac{\sqrt{1-\cos^2t}}{\sqrt t^2}\cdot t\cdot\left(\frac{3^t-1}{t}\cdot t+(-t)\frac{(-5)^{-t}+1}{-t}\right)}{\log_2(1+t^2+t^3)}$$
$$=\frac{1}{2}(\ln 3+\ln 5)\left[\lim_{t\to 0}\log_2(1+t^2+t^3)^\frac{1}{t^2}\right]^{-1}=\frac{1}{2}(\ln3+\ln 5)\left(e^{{\lim_{t\to 0}\frac{t^2+t^3}{t^2}}^{-1}}\right)^{-1}=\frac{\ln3+\ln5}{2e}$$
| Here is a solution without using Taylor series or applying L'HΓ΄pital's rule:
$$\lim_{x\to\infty}\frac{\sqrt{1-\cos^{2}\left(\frac{1}{x}\right)}\left(3^{\frac{1}{x}}-5^{\frac{-1}{x}}\right)}{\log_{2}\left(1+x^{-2}+x^{-3}\right)}$$$$=\lim_{x\to\infty}\frac{\sin\left(\frac{1}{x}\right)}{\color{blue}{\frac{1}{x}}}\cdot\frac{\color{blue} {x^{-2}+x^{-3}}}{\log_{2}\left(1+x^{-2}+x^{-3}\right)}\cdot\left(\frac{3^{\frac{1}{x}}-1}{\color{blue} {\frac{1}{x}}}+\frac{5^{\frac{-1}{x}}-1}{\color{blue}{\frac{-1}{x}}}\right)\cdot\frac{1}{\color{blue}{x^{-1}+1}}$$$$=\ln\left(2\right)\ln\left(15\right)$$
| {
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Minimize $\big(3+2a^2\big)\big(3+2b^2\big)\big(3+2c^2\big)$ if $a+b+c=3$ and $(a,b,c) > 0$
Minimize $\big(3+2a^2\big)\big(3+2b^2\big)\big(3+2c^2\big)$ if $a+b+c=3$ and $(a,b,c) > 0$.
I expanded the brackets and applied AM-GM on all of the eight terms to get :
$$\big(3+2a^2\big)\big(3+2b^2\big)\big(3+2c^2\big) \geq 3\sqrt{3}abc$$
, which is horribly weak !
I can not use equality constraint whichever method I use. Thanks to Wolfram|Alpha, I know the answer is $125$ when $(a,b,c) \equiv (1,1,1).$
Any help would be appreciated. :)
| For $a=b=c=1$ we get the value $125$.
We'll prove that it's the minimal value.
Indeed, let $f(x)=-\ln(3+2x^2)$.
Hence, $f''(x)=\frac{4(2x^2-3)}{(2x^2+3)^2}<0$ for all $0<x<1$.
Thus, by Vasc's LCF Theorem it's enough to prove our inequality for $b=a$ and $c=3-2a.$
Hence, it's enough to prove that
$$(3+2a^2)^2(3+2(3-2a)^2)\geq125$$ or
$$(a-1)^2(8a^4-8a^3+21a^2-22a+16)\geq0$$
which is obvious.
Done!
| {
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"timestamp": "2023-03-29T00:00:00",
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Angles in pyramid
We have to find the value of $\cos\theta$.
I tried it alot. But could not able to do it.
Solution given in the book
| This works out nicely with vectors.
Place $A, B, C, D$ on the $x$ and $y$ axes. Let $A = (1, 0, 0), B = (0, -1, 0), C = (-1, 0, 0), D = (0, 1, 0)$
$O$ must be on the $z$ axis by symmetry and $O = (0, 0, h)$ The only question is the value of $h$. Since angle $AOB = 45$ degrees, $OA \cdot OB = ||OA|| ||OB|| cos 45$ degrees.
$$(1, 0, -h) \cdot (0, -1, -h) = \sqrt{1^2 + 0 + h^2} \sqrt{0 + 1^2 + h^2} \frac{1}{\sqrt{2}}$$
$$h^2 = (1 + h^2) \frac{1}{\sqrt{2}}$$
$$\sqrt{2} h^2 = 1 + h^2$$
$$(\sqrt{2} - 1) h^2 = 1$$
$$ h = \sqrt{ \frac{1}{\sqrt{2}-1} }$$
Or equivalently, rationalizing.
$$h = \sqrt{\sqrt{2} + 1}$$
We would like the normals to the triangles $OAB $ and $OBC$
$OA = (1, 0, -h), OB = (0, -1, -h), $ and $OC = (-1, 0, -h)$
$OA \cdot OB = det$
$$\pmatrix{i &j &k\\1 &0&-h\\0&-1&-h} $$
$$= (-h, h, -1)$$ = normal to face $OAB = N_1$
$$OB \cdot OC = det$$
$$\pmatrix{i &j &k\\0 & -1 &-h\\-1& 0 &-h} $$
$$ = (h, h, -1) $$ = normal to face $OBC = N_2$
Then $N_1 \cdot N_2 = (-h, h, -1) \cdot (h, h, -1) = ||N_1||||N_2|| cos \theta$
$$ -h^2 + h^2 + 1 = \sqrt{h^2 + h^2 + 1} \sqrt{h^2 + h^2 + 1}cos \theta$$
$$ cos \theta = \frac{1}{2h^2 + 1}$$
$$cos \theta = \frac{1}{2(\sqrt{2} + 1) + 1} = \frac{1}{2\sqrt{2} + 3}$$
$$ \theta \approx 80.12 degrees$$
As always I do not promise perfect accuracy -- there may be a typo or sign error in there so proofread before you trust it.
But the method is sound and the amount of calculation is reasonable.
Addenda:
Possibly -- please see comment below -- $\theta$ = 180 degrees - 80.12 degrees = 99.88 degrees,
I plan to build a model and check.
This method is very similar to the text method given ( did not look first :-) ) but my calculations came out a little differently. As always, double-check first drafts.
| {
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"timestamp": "2023-03-29T00:00:00",
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Compute $\lim\limits_{x \to 0} \frac{ \sqrt[3]{1+\sin x} - \sqrt[3]{1- \sin x }}{x}$
Compute $\lim\limits_{x \to 0} \dfrac{ \sqrt[3]{1+\sin x} - \sqrt[3]{1- \sin x }}{x}$
Original question was to solve $\lim\limits_{x \to 0} \dfrac{ \sqrt[3]{1+ x } - \sqrt[3]{1-x }}{x}$ and it was solved by adding and subtracting 1 in denominator. Making it in the form $\lim\limits_{x \to a}\dfrac {x^n - a^n}{x-a} = ax^{n-1}$
How to solve for above limit without using lhopitals rule?
| Just another way using Taylor series.
$$\sin(x)=x-\frac{x^3}{6}+O\left(x^4\right)$$ Now, using the generalized binomial theorem
$$\sqrt[3]{1+\sin(x)}=1+\frac{x}{3}-\frac{x^2}{9}+\frac{x^3}{162}+O\left(x^4\right)$$
$$\sqrt[3]{1-\sin(x)}=1-\frac{x}{3}-\frac{x^2}{9}-\frac{x^3}{162}+O\left(x^4\right)$$
Using the above,
$$\sqrt[3]{1+\sin(x)}-\sqrt[3]{1-\sin(x)}=\frac{2 x}{3}+\frac{x^3}{81}+O\left(x^4\right)$$
$$\dfrac{ \sqrt[3]{1+\sin (x)} - \sqrt[3]{1- \sin (x) }}{x}=\frac{2}{3}+\frac{x^2}{81}+O\left(x^3\right)$$ which shows the limit and also how it is approached.
Making the problem more general considering
$$A=\frac{(1+\sin (x))^k-(1-\sin (x))^k}{x}$$ and using the same process, we should obtain
$$A=2k+\frac{k \left(k^2-3 k+1\right)}{3} x^2+O\left(x^3\right)$$
| {
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"url": "https://math.stackexchange.com/questions/2143650",
"timestamp": "2023-03-29T00:00:00",
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Solutions to a Cubic Root I'm looking for a sanity check to my work.
We are supposed to find the solutions of the cubic polynomial, $x^3+3x+5=0$.
Let $x=w-\frac{1}{w}$ and substitute.
Then $(w-\frac{1}{w})^3+3(w-\frac{1}{w})+5=0$.
Then $w^3+5w^3-1=0$.
Finally $w^3=\frac{-5+\sqrt{29}}{2}$ and $\frac{-5-\sqrt{29}}{2}$.
Solution 1: $\sqrt[3]\frac{-5+\sqrt{29}}{2}-\sqrt[3]\frac{-5-\sqrt{29}}{2}$.
Solution 2: $\left(\frac{-1+\sqrt{3}i}{2}\right)$$\sqrt[3]\frac{-5+\sqrt{29}}{2}-\left(\frac{-1-\sqrt{3}i}{2}\right)$$\sqrt[3]\frac{-5-\sqrt{29}}{2}$.
Solution 3: $\left(\frac{-1-\sqrt{3}i}{2}\right)$$\sqrt[3]\frac{-5+\sqrt{29}}{2}-\left(\frac{-1+\sqrt{3}i}{2}\right)$$\sqrt[3]\frac{-5-\sqrt{29}}{2}$.
| In general Let be $f(x)=x^3+ax+b$
put $x=u-v$, then $f(u-v)=(u^3-v^3)-(3uv-a)(u-v)+b$ and if $f(u,v)=0$, then $(u^3-v^3)+b=0$ and $3uv-a=0$. From second equation we have $v=\frac{a}{3u}$ and then $u^3-(\frac{a}{3u})^3+b=0 \Rightarrow 3^3u^6-a^3+3^3u^3b=0\Rightarrow 3^3y^2+3^3by-a^3=0 ; y=u^3 $, we solve the equation for $y$, then we have $y=\frac{-b}{2}+\sqrt{(\frac{b}{2})^2+(\frac{a}{3})^3 } \Rightarrow u=\sqrt[3] {y}=\sqrt[3]{\frac{-b}{2}+\sqrt{(\frac{b}{2})^2+(\frac{a}{3})^3 }} $ and $v^3=u^3+b\Rightarrow v=\sqrt[3]{u^3+b}=\sqrt[3]{\frac{b}{2}+\sqrt{(\frac{b}{2})^2+(\frac{a}{3})^3 }} $.Therefore first solution of $f(x)$ is $u-v$
and other solutions are complex, i.e. second solution is $\zeta u-\zeta^2v$ and third solution is $\zeta^2 u-\zeta v$.
In question we have $a=3, b=5$, then $ y=\frac{-5+\sqrt{29}}{2} \Rightarrow u= \sqrt[3] {\frac{-5+\sqrt{29}}{2} }, v= \sqrt[3] {\frac{5+\sqrt{29}}{2} } $ ,so
first solution is $u-v=\sqrt[3] {\frac{-5+\sqrt{29}}{2} }- \sqrt[3] {\frac{5+\sqrt{29}}{2} } $
second solution is $\zeta u-\zeta^2v=(\frac{-1+\sqrt3i}{2})\sqrt[3] {\frac{-5+\sqrt{29}}{2} }- (\frac{-1-\sqrt3i}{2}) \sqrt[3] {\frac{5+\sqrt{29}}{2} }$
third solution is $\zeta^2 u-\zeta v=(\frac{-1-\sqrt3i}{2})\sqrt[3] {\frac{-5+\sqrt{29}}{2} }- (\frac{-1+\sqrt3i}{2}) \sqrt[3] {\frac{5+\sqrt{29}}{2} }$
| {
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"url": "https://math.stackexchange.com/questions/2146211",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How does one show that $\lim_{n\to \infty}{n\over \sqrt{2k}}\cdot\sqrt{1-\cos^k\left({2\pi\over n}\right)}=\pi$? Consider
$$\lim_{n\to \infty}{n\over \sqrt{2k}}\cdot\sqrt{1-\cos^k\left({2\pi\over n}\right)}=L\tag1$$
How does one show that $L=\pi$ for $k>0$?
An attempt:
For $k=2$
$$\lim_{n\to \infty}{n\over 2}\cdot\sqrt{1-\cos^2\left({2\pi\over n}\right)}=L\tag2$$
$$\lim_{n\to \infty}{n\over 2}\cdot\sin\left({2\pi\over n}\right)=L\tag3$$
$$\lim_{n\to \infty}\pi\cdot{\sin\left({2\pi\over n}\right)\over {2\pi\over n}}=L\tag4$$
$$L=\pi\tag5$$
| $$x\to 0 \\\cos x=1-\frac{1}2x^2+\frac{1}{4!}x^4+...\sim 1-\dfrac12x^2\\
\cos ^kx\sim (1-\frac{1}2x^2)^k \sim 1-\dfrac k2x^2\\ $$
now put into ,$\frac{2\pi}{n} \to 0$
$$\lim_{n\to \infty}{n\over \sqrt{2k}}\cdot\sqrt{1-\cos^k\left({2\pi\over n}\right)}=\\
\lim_{n\to \infty}{n\over \sqrt{2k}}\cdot\sqrt{1-(1-\dfrac k2(\frac{2\pi}{n})^2)}\\=
\lim_{n\to \infty}{n\over \sqrt{2k}}\cdot\sqrt{\dfrac k2(\frac{2\pi}{n})^2}\\=
\lim_{n\to \infty}{n\over \sqrt{2k}}\cdot\sqrt{\dfrac k2}\times (\frac{2\pi}{n})=\\
\lim_{n\to \infty}{1\over \sqrt{2}}\cdot\sqrt{\dfrac 12}\times (\frac{2\pi}{1})=\dfrac{2\pi}{2}=\pi$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2146865",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Condition of two hyperbolas do not intersect Given two hyperbolas h1(with foci and center) and h2(with foci and center), In what condition these hyperbolas will not intersect to each other?. I can get the condition when h1 and h2 are standard hyperbola (parallel to axis and the center is the origin). I want to find the condition when both of them are not standard hyperbola. Thanks
| Let $x^2-y^2=1$ be the first hyperbola.
*
*Re-scaling
$$x^2-y^2=c^2 \tag{$c^2 \ne 1$}$$
*Conjugate
$$y^2-x^2=c^2$$
*Translation
\begin{align*}
(x-h)^2-(y-k)^2 &= 1 \\
2(ky-hx)+h^2-k^2 &= 0 \\
x &= \frac{h^2-k^2+2ky}{2h} \\
(h^2-k^2+2ky)^2-4h^2 y^2 &= 4h^2 \\
4(h^2-k^2)y^2-4k(h^2-k^2)y+4h^2-(h^2-k^2)^2 &= 0 \\
\Delta & < 0 \\
k^2(h^2-k^2)^2-(h^2-k^2)[4h^2-(h^2-k^2)^2] & < 0 \\
h^2(h^2-k^2)(h^2-k^2-4) &< 0 \\
\end{align*}
$$\fbox{$0<h^2-k^2<4$}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Solve $\lim_{x \rightarrow +\infty}\frac{\sqrt{x}+x}{x+\sqrt[3]{x}}$ I tried first without L'HΓ΄pital's rule:
$$\lim_{x \rightarrow +\infty}\frac{\sqrt{x}+x}{x+\sqrt[3]{x}} =
\frac{\sqrt{x}}{\sqrt[3]{x}} \cdot \frac{1+\frac{x}{\sqrt x}}{1+ \sqrt[3] x} = \frac{\sqrt x}{\sqrt[3] x} \cdot \frac{\frac{\sqrt x+x}{\sqrt x}}{\frac{\sqrt[3]{x}+x}{\sqrt[3]{x}}} = \frac{\sqrt x}{\sqrt[3] x} \cdot \frac{\sqrt[3]{x}(\sqrt x +x)}{\sqrt x (\sqrt[3]{x}+x)} = \frac{\sqrt x \sqrt[3]{x}(\sqrt x +x)}{\sqrt[3] x \sqrt x (\sqrt[3]x +x )} = \frac{\sqrt x +x }{\sqrt[3] x +x }$$
That didn't work, so then I tried L'HΓ΄pital's:
$$\frac{\frac{1}{2}(x)^{-\frac{1}{2}}+1}{1+\frac{1}{3}(x)^{-\frac{1}{3}}} = \frac{\frac{1}{2\sqrt x }+1}{1+\frac{1}{{3\sqrt[3]{x}}}} = \frac{\frac{1+2\sqrt x}{2\sqrt x}}{\frac{1+3\sqrt[3]{x}}{3\sqrt[3]{x}}} = \frac{3\sqrt[3]{x}(1+2\sqrt x)}{2\sqrt x (1+3\sqrt[3]{x})} = \frac{3\sqrt[3]{x}+3\sqrt[3]{x}2\sqrt{x}}{2\sqrt{x} + 2\sqrt{x}3\sqrt[3]{x}} = ??? $$
How do I solve this? If you can solve this with and without L'HΓ΄pital's, please do so.
| set ${ t }^{ 6 }=x$ then we have $$\lim _{ x\rightarrow +\infty } \frac { \sqrt { x } +x }{ x+\sqrt [ 3 ]{ x } } =\lim _{ t\rightarrow +\infty } \frac { { t }^{ 3 }+{ t }^{ 6 } }{ { t }^{ 6 }+{ t }^{ 2 } } =\lim _{ t\rightarrow +\infty } \frac { { \left( \frac { 1 }{ { t }^{ 3 } } +1 \right) } }{ \left( \frac { 1 }{ { t }^{ 4 } } +1 \right) } =1$$
| {
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Geometric interpretation of complex number.
Let $z \in \mathbb{C}, z^2, z^3$ be the verticies of a right triangle. Find the geometric images of $z$.
I did not understand the question but I guess it want me to find the figure formed by $z$ under these constraints.
Let $z = x + iy$, then $z^2 = x^2 - y^2 +2xyi$ and $z^3 = x^3 - iy^3 + 3x^2iy - 3xy^2$
It forms a triangle, So, $|z - z^2|^2 + |z^3 - z^2|^2 = |z - z^3|^2$ by Pythagoras theorem.
But when I put the values in I get polynomial in two degree with 6th degree with many extra terms, in short a complete mess.
How can I do this question without actually explanding the Pythogoras theorem ?
| Hint: let $r = |z| \in \mathbb{R}^+\,$, so that $\bar z = r^2 / z\,$.
It forms a triangle, So, $|z - z^2|^2 + |z^3 - z^2|^2 = |z - z^3|^2$ by Pythagoras theorem.
This assumes that the hypotenuse is $z\,z_3\,$, and the following works on this assumption. The other cases would follow similarly.
Rewriting it in terms of conjugates:
$$
\require{cancel}
\begin{align}
0 & = |z - z^2|^2 + |z^3 - z^2|^2 - |z - z^3|^2 \\ & = (z - z^2)(\bar z - \bar z^2) + (z^3 - z^2)(\bar z^3-\bar z^2) - (z - z^3)(\bar z -\bar z^3) \\
& = \bcancel{r^2} + r^4 - z \bar z^2 - \bar z z^2 + \cancel{r^6} + r^4 - z^3 \bar z^2 - \bar z^2 z^3 - \bcancel{r^2} - \cancel{r^6} + z \bar z^3 + \bar z z^3 \\
& = r^2 (z^2 + \bar z^2) - (r^2+r^4) (z+\bar z) +2 r^4 \\
& = r^2 \left(z^2+ \frac{r^4}{z^2}\right) -r^2(r^2+1)\left(z + \frac{r^2}{z}\right) + 2 r^4 \\
& = r^2\left( \left( z + \frac{r^2}{z}\right)^2 - \bcancel{2 r^2} \right) -r^2(r^2+1)\left(z + \frac{r^2}{z}\right) + \bcancel{2 r^4} \\
& = r^2 \left( z + \frac{r^2}{z}\right) \left( z + \frac{r^2}{z} - r^2 -1 \right) \\
& = r^2 \left( z + \frac{r^2}{z}\right)\,\frac{(z-1)(z-r^2)}{z}
\end{align}
$$
The latter can be easily solved and, discarding the degenerate triangles given by $z=0,1\,$, the solutions are $z=\pm i \,r\,$ with $r \in \mathbb{R}^+\,$, which is the same as $z = \lambda i$ with $\lambda \in \mathbb{R} \setminus \{0\}\,$.
| {
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Find the coefficient of $x^{29}$ in the given polynomial. The polynomial is :
$$
\left(x-\frac{1}{1\cdot3}\right) \left(x-\frac{2}{1\cdot3\cdot5}\right) \left(x-\frac{3}{1\cdot3\cdot5\cdot7}\right) \cdots \left(x-\frac{30}{1\cdot3\cdot5\cdots61}\right)
$$
What I've done so far : The given polynomial is an expression of degree $30$. Hence, the coefficient of $x^{29}$ will be the negative of the sum of the roots. But the resulting sum is too complicated to handle and I think I'm doing it wrong.
| HINT: The sum $S_k $ given by
$$\sum_{n=1}^{k} \dfrac {n}{(2n+1)!!} $$
for $k=1,2,3....$ gives
$$S_1=\frac {1}{3} \\ =\frac {1}{2}-\frac {1/2}{ 1 \cdot 3}$$
$$S_2=\frac {1}{3} + \frac {2}{15} =\frac {7}{15} \\= \frac {1}{2}-\frac {1/2}{1 \cdot 3 \cdot 5 }$$
$$S_3=\frac {1}{3} + \frac {2}{15} +\frac {3}{105} =\frac {52}{105} \\= \frac {1}{2}-\frac {1/2}{1 \cdot 3 \cdot 5 \cdot 7 }$$
You can use induction to confirm this pattern.
| {
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Use infinite series to prove Use infinite series to prove that
$$\arcsin{x}\lt \frac{x}{1-x^2},$$ for $0\lt x\lt1$.
| From AM-GM inequality we have, for a given $t$ in the interval $(0,1)$,
$$\frac12\left[\frac{1}{(1+t)^2}+\frac{1}{(1-t)^2}\right]\ge \frac1{(1+t)(1-t)}\tag{1}$$
Now, since $1>\sqrt{(1+t)(1-t)}$, it follows
$$\frac1{(1+t)(1-t)}>\frac{\sqrt{(1+t)(1-t)}}{(1+t)(1-t)}=\frac1{\sqrt{(1+t)(1-t)}}=\frac1{\sqrt{1-t^2}}\tag{2}$$
From $(1)$ and $(2)$
$$\frac12\left[\frac{1}{(1+t)^2}+\frac{1}{(1-t)^2}\right]>\frac1{\sqrt{1-t^2}}$$
Let $0<x<1$, integration on the interval $0<t<x$ give us
\begin{align*}
\frac12\int_0^x\left[\frac{1}{(1+t)^2}+\frac{1}{(1-t)^2}\right]dt&>\int_0^x\frac1{\sqrt{1-t^2}}dt\\
\frac12\left[-\frac{1}{1+x}+\frac1{1-x}+1-1\right]&>\arcsin x-0\\
\frac{x}{1-x^2}&>\arcsin x
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2152374",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Showing that $\sum_{n=1}^{\infty}\sum_{m=1}^{x-1} \frac{m}{n^2 x^2 - m^2}=\frac{1}{2}H_{x-1}$ Consider $(1)$, $H_n$ is the nth-harmonic number
$$\sum_{n=1}^{\infty}\left({1\over (nx)^2-1}+{2\over (nx)^2-2^2}+{3\over (nx)^2-3^2}+\cdots+{x-1\over (nx)^2-(x-1)^{2}}\right)=S\tag1$$
$x\ge2$
How does one show that $$\color{blue}{S={H_{x-1}\over 2}}?$$
An attempt:[Edited]
Because $(1)$ is the difference of two squares form, so it becomes
$$\sum_{n=1}^{\infty}\left[\left({1\over nx-1}+{1\over nx-2}+{1\over nx-3}+\cdots+{x-1\over nx-(x-1)}\right)-\left({1\over nx+1}+{1\over nx+2}+{1\over nx+3}+\cdots+{x-1\over nx+(x-1)}\right)\right]=2S\tag2$$
Not sure how to proceed next
| From Cotangent identity (formula 18) and Digamma reflection formula:
$$
\begin{align}
\pi &\,\cot(\pi z)=\frac{1}{z}+2z\,\sum_{n=1}^{\infty}\,\frac{1}{z^2-n^2}=\psi(1-z)-\psi(z) \\[3mm]
S &= \sum_{n=1}^{\infty}\,\sum_{m=1}^{x-1}\,\frac{m}{(nx)^2-m^2} = \sum_{m=1}^{x-1}\,\frac{m}{x^2}\sum_{n=1}^{\infty}\,\frac{1}{n^2-(m/x)^2} \\[3mm]
&= \frac{1}{2x}\,\sum_{m=1}^{x-1}\,\left[-2\left(\frac{m}{x}\right)\sum_{n=1}^{\infty}\,\frac{1}{\left(\frac{m}{x}\right)^2-n^2}\right] \\[3mm]
&= \frac{1}{2x}\,\sum_{m=1}^{x-1}\,\left[\frac{x}{m}-\pi\,\cot\left(\pi\,\frac{m}{x}\right)\right] \\[3mm]
&= \frac{1}{2x}\,\sum_{m=1}^{x-1}\,\left[\frac{x}{m}+\psi\left(\frac{m}{x}\right)-\psi\left(1-\frac{m}{x}\right)\right] \\[3mm]
&= \frac{1}{2}\,\sum_{m=1}^{x-1}\frac{1}{m}+\frac{1}{2x}\left[\sum_{m=1}^{x-1}\psi\left(\frac{m}{x}\right)-\sum_{m=1}^{x-1}\psi\left(\frac{x-m}{x}\right)\right] \\[3mm]
&= \frac{1}{2}\,\sum_{m=1}^{x-1}\frac{1}{m} = \color{red}{\frac{H_{x-1}}{2}}
\end{align}
$$
Where:
$$ \color{blue}{\sum_{m=1}^{x-1}\psi\left(\frac{x-m}{x}\right)}=\psi\left(\frac{x-1}{x}\right)+\psi\left(\frac{x-2}{x}\right)+\cdots+\psi\left(\frac{2}{x}\right)+\psi\left(\frac{1}{x}\right)=\color{blue}{\sum_{m=1}^{x-1}\psi\left(\frac{m}{x}\right)} $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2152635",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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Finding a quartic with integer coefficients such that $\sqrt{3} β\sqrt{5}$ is a zero Find integers $a,b,c,d$ and $e$ such that $\sqrt{3} β\sqrt{5}$ is a solution to the equation:
$$ax^4 + bx^3 + cx^2 + dx + e = 0.$$
Being new to quartic equations I wasn't sure how to solve this problem, must we use the quartic formula?
Any help is greatly appreciated!
Thanks
| Since $$\sum_{cyc}(2a^2b^2-a^4)=(a+b+c)(a+b-c)(a+c-b)(b+c-a),$$
we have that $\sqrt3-\sqrt5$ is a root of the equation
$$2(3x^2+5x^2+15)-x^4-9-25=0$$ or
$$x^4-16x^2+4=0$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 3
} |
Solve the equation : $\tan \theta + \tan 2\theta + \tan 3\theta = \tan \theta \tan 2\theta \tan 3\theta $ I've been having some trouble solving this equation. (The solution in my book is given as $ \frac {n \pi}{3}, n \in Z $)
Here is what I've done
$$\frac {\sin \theta}{\cos \theta} + \frac {\sin 2\theta} {\cos 2\theta} + \frac{\sin 3\theta}{\cos 3\theta}= \frac {\sin \theta}{\cos \theta} \frac {\sin 2\theta} {\cos 2\theta} \frac{\sin 3\theta}{\cos 3\theta}$$
$$ \frac {\sin \theta \cos 2\theta \cos 3\theta + \cos \theta \sin 2\theta \cos 3\theta + \cos \theta \cos 2\theta \sin 3\theta - \sin \theta \sin 2\theta \sin 3\theta }{\cos\theta \cos 2\theta \cos 3\theta} = 0 $$
$$\cos 2\theta \{\sin\theta \cos 3\theta + cos \theta \sin 3\theta \} + \sin 2\theta \{\cos \theta \cos 3\theta - \sin \theta \sin 3\theta \} = 0 $$
$$\cos 2\theta \sin(3\theta + \theta) +\sin2\theta \cos(3\theta + \theta) = 0 $$
$$ \cos 2\theta \sin 4\theta + sin 2\theta cos 4\theta = 0$$
$$ \sin (2\theta + 4\theta) = 0$$
$$\sin 6\theta = 0 $$
$$ \theta = \frac {n\pi}{6}, n \in Z$$
I understand from this question that whatever mistake I am making is in the third step, where I remove $\cos \theta \cos 2\theta \cos 3\theta $ from the denominator. However, despite reading through the aforementioned post, I couldn't really get the intuition behind why this is wrong.
I'd like :
*
*To understand the intuition behind why removing $\cos \theta \cos 2\theta cos 3\theta $ is a mistake.
*To know how to solve this question correctly
*How do I avoid making these types of mistakes when solving trigonometric equations
| In your simplified expression there is $\cos 3 \theta $ term in the denominator that goes to zero for solution you obtained. Should be checked before accepting or discarding it as a valid solution.
I cannot resist an elementary trig approach ..
If $ (A+B+C) = 2 \pi, $ then $ {(\tan A + \tan B + \tan C) = \tan A \tan B \tan C} $
If $ {(\tan A + \tan B + \tan C) = \tan A \tan B \tan C} $ then $(A+B+C) = 2 \pi $
is among possible solutions.
In the above if $ A= t, B=2t, C=3t$ then $t= \pm 2 \pi/6= \pm \pi/3,\pm 2\pi/3, ... $
By inspection $ t= 0 ,\,2 k\pi,$ plus co-terminals
| {
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"url": "https://math.stackexchange.com/questions/2154221",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove that $\frac{a}{\sqrt{a^2+b^2}}+\frac{b}{\sqrt{9a^2+b^2}}+\frac{2ab}{\sqrt{a^2+b^2}\times \sqrt{9a^2+b^2} }\leq \frac32.$
Prove that $$\dfrac{a}{\sqrt{a^2+b^2}}+\dfrac{b}{\sqrt{9a^2+b^2}}+\dfrac{2ab}{\sqrt{a^2+b^2}\times \sqrt{9a^2+b^2} }\leq \dfrac{3}{2}.$$
When is equality attained ?
My Attempt :
I could not think of anything suitable for the entire LHS. The last term $\dfrac{2ab}{\sqrt{a^2+b^2}\times \sqrt{9a^2+b^2}} $ suggested C-S
. So I applied C-S on the values $\left(\dfrac{2a}{\sqrt{a^2+b^2}}\right)$ and $\left(\dfrac{b}{\sqrt{9a^2+b^2}}\right)$ to get :
$$\left(\dfrac{2ab}{\sqrt{a^2+b^2}\times \sqrt{9a^2+b^2}}\right)^2\leq \left(\dfrac{4a^2}{a^2+b^2}\right)\left(\dfrac{b^2}{9a^2+b^2}\right)\leq\left(\dfrac{4a^2}{b^2}\right)\times \left(\dfrac{b^2}{9a^2}\right)=\dfrac{4}{9}.\,\,\,\,(β¦)$$
For the first term $\dfrac{a}{\sqrt{a^2+b^2}}$, I applied C-S on the terms $\left(a\right)$ and $\left(\dfrac{1}{\sqrt{a^2+b^2}}\right)$ to get :
$$\left(\dfrac{a}{\sqrt{a^2+b^2}}\right)\leq \left(\dfrac{a^2}{a^2+b^2}\right)^{1/2}\leq 1.\,\,\,\,(β£)$$
Using the same logic for the second term, I get :
$$\dfrac{b}{\sqrt{9a^2+b^2}}\leq 1 \,\,\,\,(β )$$
Adding all the inequalities, I get a very "weak" inequality when compared to the problem.
What is the best way to prove this inequality ?
| setting $$x=\sqrt{\frac{a^2}{a^2+b^2}}$$ and $$y=\sqrt{\frac{b^2}{9a^2+b^2}}$$
we get further $$1+\left(\frac{b}{a}\right)^2=\frac{1}{x^2}$$ and $$9\left(\frac{a}{b}\right)^2+1=\frac{1}{y^2}$$ we can eliminate $$\frac{a}{b}$$ and we get $$y^2=\frac{1-x^2}{8x^2+1}$$ thus our inequality is equivalent to
$$x+\sqrt{\frac{1-x^2}{8x^2+1}}+2x\sqrt{\frac{1-x^2}{8x^2+1}}\le \frac{3}{2}$$
this is equivalent to
$$\sqrt{\frac{1-x^2}{8x^2+1}}(1+2x)\le \frac{3}{2}-x$$
squarint and factorizing we obtain $$1/4\,{\frac { \left( 12\,{x}^{2}-8\,x+5 \right) \left( 2\,x-1
\right) ^{2}}{8\,{x}^{2}+1}}
\geq 0$$ which is true.
We also assume that $$a,b$$ are positive.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2154453",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Considering $0Problem: Considering $0<s<1$ for the series $\sum_{i=1}^{\infty} \frac {1}{i^s}$, I want to show that $a_{2^{n+1} -1}$ $<\sum_{j=0}^{n}$ $({\frac {1} {2^{s-1}})}^j$ where $a_n$ is the partial sums of the aforementioned series..
So for $n=1$ this would yield the inequality:
$\frac {1}{1^s} +\frac {1}{2^s} + \frac {1}{3^s}+ \frac {1}{4^s} < (\frac {1} {2^{s-1}})^{0} + (\frac {1} {2^{s-1}})^{1}=1+\frac {1} {2^{s-1}}$ which would imply that
$\frac {1} {2^{s-1}} > \frac {1}{2^s} + \frac {1}{3^s} + \frac {1}{4^s} $
which I think I can show since $ \frac {1}{2^s} + \frac {1}{3^s} + \frac {1}{4^s} < \frac {3} {2^{s-1}} < \frac {2} {2^{s-1}}= \frac {1} {2^s} < \frac {1} {2^{s-1}}$.
I just need a way to generalize this notion of grouping the terms in groups of $log_2$ to complete the desired statement. Any hints/help appreciated.
Edit: It seems as if I have mis-written the question several times. I will post it from the notes I have and if it is deemed an invalid question, well then I guess we have our answer. We are answering part (b)
| $$
\begin{align}
a_{2^{n+1}} &= \sum_{i=1}^{2^{n+1}} \frac 1{i^s} \\
&= \sum_{j=0}^{n} \sum_{i=2^j}^{2^{j+1}-1} \frac 1{i^s} \\
& \leq \sum_{j=0}^{n} \sum_{i=2^j}^{2^{j+1}-1} \frac 1{2^{js}} \\
& = \sum_{j=0}^{n} \frac {2^j}{2^{js}} \\
& = \sum_{j=0}^{n} \left(\frac 1{2^{s-1}}\right)^j \\
\end{align}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2155372",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Derivation of $1/(s^2-a^2)^{3/2}$ in Laurent Series So Given modified bessel equation (t/a)*I_1(at) using the general formula
I have arrived at the answer.
$$\frac{1}{s^3}+\frac{3a^2}{2s^5}+\frac{15a^4}{8s^7}+ \cdots$$
And according to wolfram alpha its laurent series is equal to $\frac{1}{(s^2-a^2)^{3/2}}$ equal to my answer above
So the question is how my answer
$$\frac{1}{s^3}+\frac{3a^2}{2s^5}+\frac{15a^4}{8s^7} +
\cdots$$
equal to:
$$\frac{1}{(s^2-a^2)^{3/2}}$$
Im not familiar with laurent series please help me
| Using
$$(1+x)^n=1+nx+\frac{n(n-1)}{2!}x^2+\frac{n(n-1)(n-2)}{3!}x^3+\cdots$$
we have
\begin{eqnarray}
\frac{1}{\sqrt{(s^2-a^2)^3}}&=&(s^2-a^2)^{-\frac32}\\
&=&s^{-3}\left(1-\left(\dfrac{a}{s}\right)^2\right)^{-\frac32}\\
&=&\frac{1}{s^3}\left(1
-\left(-\dfrac{3}{2}\right)\left(\dfrac{a}{s}\right)^2
+\frac{(-\frac{3}{2})(-\frac{3}{2}-1)}{2!}\left(\dfrac{a}{s}\right)^4
-\frac{(-\frac{3}{2})(-\frac{3}{2}-1)(-\frac{3}{2}-2)}{3!}\left(\dfrac{a}{s}\right)^6
+\cdots\right)\\
&=&1
+\dfrac{3a^2}{2s^5}
+\dfrac{15a^4}{8s^7}
+\dfrac{35a^6}{16s^9}
+\cdots
\end{eqnarray}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2155636",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Partition of numbers : $1, 2, ..., 20$ Integers $1, 2, ..., 20$ are partitioned into $2$ groups. Sum of all integers in one group is equal to $n$ and the product of all integers in another group is also equal to $n$. Find the maximal $n$.
Since $1 + 2 + ... + 20 = 210$, so the product of all integers in another group is less than $210$.
Please suggest how to proceed.
| Here's a answer by trial and error.
Notice that the product of the smallest $5$ integers is $5!=5\cdot 4\cdot 3\cdot 2\cdot 1=120$ and the product of $6$ integers is bigger then $210$.
Now the sum will be bigger if we choose smaller numbers,lets try to make the sum the max possible value $<210$ by changing the digit $5$,testing $8$ is the biggest digit so the sum doesn't exceed $210$.We have that $1\cdot 2\cdot 3\cdot 4\cdot 8=192$ and the sum is $210-1-2-3-4-8=192$ so we have that $192$ can be $n$.
Now lets try to prove that $n\leq 192$ first $193$ is a prime $194=97\cdot 2$ where $97$ is prime so those are not solutions (because of the product).Now lets try numbers $\geq 195$ then sum of those numbers is $\leq 15$ checking products of $2$ numbers such that their sum is $\leq 15$ the greatest such number is $8\cdot 7=56$, now checking product of $3$ numbers the biggest is $5\cdot 4\cdot 6=120$ since the closer the $3$ numbers the bigger the product is,checking $4$ numbers we get that the biggest product is $2\cdot 3\cdot 4\cdot 5=120$ and at last checking five numbers it's $1\cdot 2\cdot 3\cdot 4\cdot 5=120$ this isn't the answer either so we concluded that $n=192$.
| {
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"url": "https://math.stackexchange.com/questions/2157807",
"timestamp": "2023-03-29T00:00:00",
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If $f_k(x)=\frac{1}{k}\left (\sin^kx +\cos^kx\right)$, then $f_4(x)-f_6(x)=\;?$ I arrived to this question while solving a question paper. The question is as follows:
If $f_k(x)=\frac{1}{k}\left(\sin^kx + \cos^kx\right)$, where $x$ belongs to $\mathbb{R}$ and $k>1$, then $f_4(x)-f_6(x)=?$
I started as
$$\begin{align}
f_4(x)-f_6(x)&=\frac{1}{4}(\sin^4x + \cos^4x) - \frac{1}{6}(\sin^6x + \cos^6x) \tag{1}\\[4pt]
&=\frac{3}{12}\sin^4x + \frac{3}{12}\cos^4x - \frac{2}{12}\sin^6x - \frac{2}{12}\cos^6x \tag{2}\\[4pt]
&=\frac{1}{12}\left(3\sin^4x + 3\cos^4x - 2\sin^6x - 2\cos^6x\right) \tag{3}\\[4pt]
&=\frac{1}{12}\left[\sin^4x\left(3-2\sin^2x\right) + \cos^4x\left(3-2\cos^2x\right)\right] \tag{4}\\[4pt]
&=\frac{1}{12}\left[\sin^4x\left(1-2\cos^2x\right) + \cos^4x\left(1-2\sin^2x\right)\right] \tag{5} \\[4pt]
&\qquad\quad \text{(substituting $\sin^2x=1-\cos^2x$ and $\cos^2x=1-\sin^2x$)} \\[4pt]
&=\frac{1}{12}\left(\sin^4x-2\cos^2x\sin^4x+\cos^4x-2\sin^2x\cos^4x\right) \tag{6} \\[4pt]
&=\frac{1}{12}\left[\sin^4x+\cos^4x-2\cos^2x\sin^2x\left(\sin^2x+\cos^2x\right)\right] \tag{7} \\[4pt]
&=\frac{1}{12}\left(\sin^4x+\cos^4x-2\cos^2x\sin^2x\right) \tag{8} \\[4pt]
&\qquad\quad\text{(because $\sin^2x+\cos^2x=1$)} \\[4pt]
&=\frac{1}{12}\left(\cos^2x-\sin^2x\right)^2 \tag{9} \\[4pt]
&=\frac{1}{12}\cos^2(2x) \tag{10}\\[4pt]
&\qquad\quad\text{(because $\cos^2x-\sin^2x=\cos2x$)}
\end{align}$$
Hence the answer should be ...
$$f_4(x)-f_6(x)=\frac{1}{12}\cos^2(2x)$$
... but the answer given was $\frac{1}{12}$.
I know this might be a very simple question but trying many a times also didn't gave me the right answer. Please tell me where I am doing wrong.
| Hint.
Let $x=\frac{\pi}{4}$
In your answer $f4(x)-f(6x) = 0$
By brute-forse, $f4(x)-f6(x) = \frac{0.5}{4} - \frac{0.25}{6} = \frac{1}{8}-\frac{1}{24} =\frac{1}{12}$, so, you make a typo :(
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
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Find all primes $p$ for which there are integers $n, x, y$ such that $p^n = x^3 + y^3$ I want to find all primes $p$ for which there exist integers $n, x, y$ such that $p^n = x^3 + y^3$. One solution is $p = 2$ because $2^1 = 1^3 + 1^3$. Now I can assume that $p$ is odd, so exactly one of $x$ and $y$ must be odd and another one must be even. Furhtermore $p^n = (x + y)(x^2 - xy + y^2)$, so $x + y = p^k$ and $x^2 - xy + y^2 = p^l$, where $k + l = n$ and $(x + y) | (x^2 - xy + y^2)$. Could you please help me move forward? I don't know what to do next.
| Hint
Let solution exists => exists a solution with $gcd(x,y) = 1$ => exists a solution with $gcd(x,p) = gcd(y,p) = 1$
in this case: $x+y=p^k,x^2-xy+y^2=p^m => 3xy=p^{2k}-p^m = p^m(p^{2k-m}-1)$ =>$m=0$ or $m=1, p=3$, in other cases $p|xy$
$x^2-xy+y^2=1$ => $x=1, y=1$ (you was found this solution)
$x^2-xy+y^2=3$ => $(x-0.5y)^2+0.75y^2 = 3$ => $y^2 \le 4$
Cases:
*
*$y=1,x^2-x=2 =>x=2, x^3+y^3=9$ => it is a solution
*$y=2, x=1 => $ same solution
So, (p,n,x,y) is (2,1,1,1), (3,2,2,1) and(3,2,1,2).
And multiplying x,y by p series
Explain 1:
$gcd(x,y)>1, gcd(x,y)|p^n => p|x, p|y => (p, n-3, \frac{x}{p}, \frac{y}{p}) $ is a solution too.
Let $gcd(x,y) = 1, gcd(x,p) = p. $ => $y^3=p^n - p^3c$=>$p|y$ => $gcd(x,y)|p$ -it is absurd
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Logarithmic differentiation - need help please Determine the following by using logarithmic differentiation.
$$y=\frac{\sqrt{x}(x^2-1)^5}{\cos(x))}$$
My answer is up to:
Take the ln on both sides so:
$\ln(y) = (1/2)\ln(x) + 5\ln(x^2-1) - \ln(\cos(x))$
Am I doing it right though?
| You have everything right so far.
$\ln(y) = (1/2)\ln(x) + 5\ln(x^2-1) - \ln(\cos(x))$
Now remember, that the derivative of $\ln(y) = \frac{1}{y}\times y'$
This is because $y$ is a function, whereas $x$ is just a variable.
The derivative of $\ln(x)$ in this case is just $\frac{1}{x}$
Also remember that if you have something like $\ln(x^2 + 1)$, that's chain rule.
It's the derivative of $\frac{1}{x^2-1}\times\frac{d}{dx}(x^2-1)$
Differentiating both sides, we get:
$$\frac{1}{y}\times y' = (\frac{1}{2}\cdot\frac{1}{x}) + (5\cdot\frac{1}{x^2-1}\cdot2x) - (\frac{1}{\cos(x)}\cdot-\sin(x))$$
$$y' = y(\frac{1}{2x} + \frac{10x}{x^2-1}) + \tan(x)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2163588",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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If $abc=1$ so $\sum\limits_{cyc}\frac{a}{a^2+b^2+4}\leq\frac{1}{2}$
Let $a$, $b$ and $c$ be positive numbers such that $abc=1$. Prove that:
$$\frac{a}{a^2+b^2+4}+\frac{b}{b^2+c^2+4}+\frac{c}{c^2+a^2+4}\leq\frac{1}{2}$$
This inequality is a similar to the following.
Let $a$, $b$ and $c$ be non-negative numbers. Prove that:
$$\frac{a}{a^2+b^2+2}+\frac{b}{b^2+c^2+2}+\frac{c}{c^2+a^2+2}\leq\frac{3}{4},$$
which we can prove by AM-GM:
$$\sum_{cyc}\frac{a}{a^2+b^2+2}\leq\frac{1}{2}\sum_{cyc}\frac{a}{\sqrt{(a^2+1)(b^2+1)}}\leq\frac{1}{4}\sum_{cyc}\left(\frac{a^2}{a^2+1}+\frac{1}{b^2+1}\right)=\frac{3}{4}$$
| The Buffalo Way works.
With the substitutions $a = \frac{x^2}{yz}, \ b = \frac{y^2}{zx}, \ c = \frac{z^2}{xy}; \ x, y, z > 0$, after clearing denominators,
it suffices to prove that $f(x, y, z) \ge 0$ where $f(x,y,z)$ is a homogeneous polynomial.
WLOG, assume that $z = \min(x, y, z) = 1$.
There are two possible cases:
1) $1 = z \le x \le y$: Let $x = 1+s, \ y = 1+s + t; \ s, t\ge 0$.
Then $f(1+s, 1+s+t, 1)$ is a polynomial in $s, t$ with non-negative coefficients.
The inequality is true.
2) $1 = z \le y\le x$: Let $y = 1+s, \ x = 1+s+t; \ s,t\ge 0$.
Then
\begin{align}
f(1+s+t, 1+s, 1) &= (4s^2t^{14} - 2s^4t^{13} + \frac{1}{4}s^6t^{12})
+ (\frac{3}{4}s^6 - 20s^5+ 249s^4)t^{12} \\
&\quad + (12s^7 - 72s^6 + 772s^5)t^{11}
+ (66s^8-46s^7+1834s^6)t^{10} + g(s,t)
\end{align}
where $g(s,t)$ is a polynomial with non-negative coefficients.
Clearly, $f(1+s+t, 1+s, 1)\ge 0$. We are done.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2166910",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Equation in radicals $x^7+7x^3-7x^2+7x+1=0$ Solve the following equation in radicals.
$$x^7+7x^3-7x^2+7x+1=0$$
It's obvious that this equation has no rational roots.
And what is the rest?
| The Galois group of the polynomial is $D_{14}$, thus it is solvable.
RadiRoot is an implementation of the algorithm described in this paper:
http://www.icm.tu-bs.de/ag_algebra/software/distler/Diplom.pdf. This is the output of RadiRoot:
Let $\zeta_7$ be a primitive $7$-th root of unity and
$$\omega_1 = \sqrt[7]{ - \frac{5}{7^3}\zeta_{7}^{5} - \frac{4}{7^3}\zeta_{7}^{4} - \frac{4}{7^3}\zeta_{7}^{3} - \frac{5}{7^3}\zeta_{7}^{2} - \frac{3}{7^3}}.$$
Then a root is
$$\left(-14 \zeta _7^5+28 \zeta _7^4+28 \zeta _7^3-14 \zeta _7^2+21\right) \omega
_1^6+\left(7 \zeta _7^5-7 \zeta _7^4-7 \zeta _7^3+7 \zeta _7^2\right) \omega
_1^5+\left(7 \zeta _7^4+7 \zeta _7^3+7\right) \omega _1^4+\left(\zeta _7^5-2 \zeta
_7^4-2 \zeta _7^3+\zeta _7^2-5\right) \omega _1^3+\left(\zeta _7^5-\zeta _7^4-\zeta
_7^3+\zeta _7^2\right) \omega _1^2+\omega _1.$$
Edit.
For example, let $\zeta_7 = e^{-\frac{4 i \pi }{7}}$.
Then we have
$$\omega_1 = \sqrt[7]{\frac{-3-8 \sin \left(\frac{3 \pi }{14}\right)+10 \cos \left(\frac{\pi
}{7}\right)}{7^3}} \approx 0.43566,$$
and a root is:
\begin{align}
\left(56 \sin \left(\frac{3 \pi }{14}\right)+28 \cos \left(\frac{\pi }{7}\right)+21\right) &\omega
_1^6
\\-\left(14 \sin \left(\frac{3 \pi }{14}\right)+14 \cos \left(\frac{\pi }{7}\right)\right) &\omega
_1^5
\\+\left(14 \sin \left(\frac{3 \pi }{14}\right)+7\right) &\omega _1^4
\\-\left(4 \sin \left(\frac{3 \pi }{14}\right)+2 \cos \left(\frac{\pi }{7}\right)+5\right) &\omega _1^3
\\-\left(2 \sin \left(\frac{3 \pi }{14}\right)+2 \cos \left(\frac{\pi }{7}\right)\right) &\omega _1^2
\\+\ &\omega _1 \approx -0.125215.
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2168911",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
} |
Infinitely many $n$ such that $n, n+1, n+2$ are each the sum of two perfect squares. Prove that there exist infinitely many integers $n$ such that $n$, $n+1$, $n+2$ are all the sum of two perfect squares. Induction does not seem to be yielding any results.
| One such infinite family is
$$(2a^2 + 1)^2 - 1, (2a^2 + 1)^2 + 0^2, (2a^2 + 1)^2 + 1^2$$ for all integer $a$.
This works because
$$(2a^2 + 1)^2 - 1 = 4a^4 + 4a^2 = (2a^2)^2 + (2a)^2.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2170494",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 4,
"answer_id": 3
} |
Prove by induction that $n^4+2n^3+n^2$ is divisible by 4
I'm trying to prove by induction that $n^4+2n^3+n^2$ is divisible by 4.
I know that P(1) it's true. Then $ n=k, P(k):k^4+2k^3+k^2=4w$ it's true by the hypothesis of induction.
When I tried to prove $n=k+1, P(k+1):(k+1)^4+(k+1)^3+(k+1)^2 = 4t$,
$$k^4+4k^3+6k^2+4k+1+2k^3+6k^2+6k+2+k^2+2k+1 = 4t.$$
$$(k^4+2k^3+k^2)+4k^3+6k^2+4k+1+6k^2+6k+2+2k+1 = 4t.$$
Can I replace $(k^4+2k^3+k^2)$ of the previous expression by $4w$ and then continue the proof? Is this correct?
Thanks.
| You don't need induction to prove this.
$f(n)=n^4+2n^3+n^2 = n^2\cdot(n^2+2n+1) = n^2\cdot(n+1)^2$
Case 1: If $n$ is even, $n=2k, n^2=2k\cdot2k=4k^2$, now $4k^2\cdot(n+1)^2$, which is obvious that is divisible by 4
Case 2: If $n$ is odd then $n+1$ is even, let $m=n+1$, $m=2k, \ m^2=2k\cdot2k = 4k^2$, if we change in our first formula $n^2\cdot4k^2$, which is obvious that is divisible by 4.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2173455",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
Solve quartic equation $a^4-6a^2b-8ac-3b^2=0$ Please help me to find roots of this quartic equation for a:
$$a^4-6a^2b-8ac-3b^2=0$$
Wolfram Alpha gave this result.
But may be there is simple way to get all a?
| Rewrite the equation by completing the squares
\begin{align}
a^4-6a^2b-8ac-3b^2
=(a^2+s)^2 -( 2s+6b)\left( a +\frac{4c}{2s+6b} \right)^2\tag1\\
\end{align}
where $s$ happens to satisfy $(s+b)^3=8(c^2-b^3)$, or
$s=2\sqrt[3]{c^2-b^3}-b$.
For notational convenience, denote
$$p=\frac12\sqrt{2s+6b}= \sqrt{\sqrt[3]{c^2-b^3}+b }\tag2$$
and then $s=2p^2-3b$.
Substitute them into (1) and then factorize to get
$$\left(a^2+2pa+2p^2-3b-\frac{2c}p\right) \left(a^2-2pa+2p^2-3b+\frac{2c}p\right)=0
$$
which yields the solutions
$$a=p\pm \sqrt{ 3b-p^2+\frac{2c}p },\>\>\> -p\pm \sqrt{ 3b-p^2-\frac{2c}p } $$
with $p$ given by (2).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2174599",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Does series $\sum _{n=1}^{\infty \:}\ln\left(\frac{n\left(n+2\right)}{\left(n+1\right)^2}\right)$ converge? Does series
$\sum _{n=2}^{\infty \:}\ln\left(\frac{n\left(n+2\right)}{\left(n+1\right)^2}\right)$ converge?
My idea is to apply the Cauchy test, but I dont know how to simplify it next.
Thanks
| Using the identity $$\prod_{n\geq0}\frac{\left(n+a\right)\left(n+b\right)}{\left(n+c\right)\left(n+d\right)}=\frac{\Gamma\left(c\right)\Gamma\left(d\right)}{\Gamma\left(a\right)\Gamma\left(b\right)},\,a+b=c+d$$ we have $$\prod_{n\geq2}\frac{n\left(n+2\right)}{\left(n+1\right)^{2}}=\prod_{n\geq0}\frac{\left(n+2\right)\left(n+4\right)}{\left(n+3\right)^{2}}=\frac{\Gamma^{2}\left(3\right)}{\Gamma\left(2\right)\Gamma\left(4\right)}=\frac{2}{3}.$$ so $$\sum_{n\geq2}\log\left(\frac{n\left(n+2\right)}{\left(n+1\right)^{2}}\right)=\log\left(\prod_{n\geq2}\frac{n\left(n+2\right)}{\left(n+1\right)^{2}}\right)=\color{red}{\log\left(\frac{2}{3}\right)}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2175121",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 7,
"answer_id": 6
} |
Number of real zeros of a polynomial Given a $n$-th degree polynomial with real coefficients
$$c_nx^n + c_{n-1}x^{n-1} + ... + c_1x+c_0$$
Is there any theorem that tells me how many zeros are real (zero imaginary part). Anything related would help. Thanks!
| Sturm's theorem might help!
Sturm's Theorem: If $f(x)$, freed from equal roots, be divided by $f'(x)$, and the last divisor by the last remainder, changing the sign of each remainder before dividing by it, until a remainder independent of $x$ is obtained, or else a remainder which cannot change its sign; then $f(x)$, $f'(x)$, and the successive remainders constitute Sturm's functions, and are denoted by $f(x),f_1(x),f_2(x),\,\&\text{c}\ldots\ldots f_m(x)$.
The operation may be exhibited as follows:$$\begin{align*} & f(x)=q_1f_1(x)-f_2(x),\\ & f_1(x)=q_2f_2(x)-f_3(x),\\ & f_2(x)=q_3f_3(x)-f_4(x),\\ & \ldots\hspace{7mm}\ldots\hspace{7mm}\ldots\hspace{7mm}\ldots\\ & f_{m-2}(x)=q_{m-1}f_{m-1}(x)-f_m(x).\end{align*}$$
Note: Any constant factor of a remainder may be rejected, and the quotient may be set down for the corresponding function.
Example:
To find the position of the roots of$$f(x):=x^4-4x^3+x^2+6x+2=0$$
Sturm's functions, formed as the rule above, are calculated as$$\begin{align*} & f(x)=x^4-4x^3+x^2+6x+2\\ & f_1(x)=2x^3-6x^2+x+3\\ & f_2(x)=5x^2-10x-7\\ & f_3(x)=x-1\\ & f_4(x)=12\end{align*}$$
The first terms of the functions are all positive, therefore there is no imaginary root.
There are changes of signs in the functions as $x$ passes through a specific integral.
$$\begin{array}{|r|c|c|c|c|c|c|c|}\hline\\x= & -2 & -1 & 0 & 1 & 2 & 3 & 4\\\hline\\f(x)= & + & + & + & + & + & + & +\\f_1(x)= & - & - & + & + & - & + & +\\f_2(x)= & + & + & - & - & - & + & +\\f_3(x)= & - & - & - & + & + & + & +\\f_4(x)= & + & + & + & + & + & + & +\\\hline\\\text{No. of changes} & 4 & 4 & 2 & 2 & 2 & 0 & 0\\\text{of sign}\small\ldots\ldots\\\hline\end{array}$$
There are two changes of sign lost while $x$ passes from $-1$ to $0$, and two more while $x$ passes from $2$ to $3$. Therefore, there are two roots lying between $0$ and $-1$; and two roots also between $2$ and $3$. These roots are also incommensurable.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2180189",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
A rigorous proof of a trigonometric inequality in $(0,\pi/2)$ I want to show
$$f(x)=\sin \left(\frac{2 x}{17}\right)-8 \sin
\left(\frac{4 x}{17}\right)+27 \sin
\left(\frac{6 x}{17}\right)-64 \sin
\left(\frac{8 x}{17}\right)+125 \sin
\left(\frac{10 x}{17}\right)-216 \sin
\left(\frac{12 x}{17}\right)+343 \sin
\left(\frac{14 x}{17}\right)-512 \sin
\left(\frac{16 x}{17}\right)<0,\quad x\in(0,\pi/2).$$
This trigonometric inequality has been verified by Mathematica using the Plot commend. I found $f$ can be rewritten
$$f(x)=\sin \left(\frac{2 x}{17}\right)-2^3 \sin
\left(\frac{2*2 x}{17}\right)+3^3 \sin
\left(\frac{2*3 x}{17}\right)-4^3 \sin
\left(\frac{2*4 x}{17}\right)+5^3 \sin
\left(\frac{2*5 x}{17}\right)-6^3 \sin
\left(\frac{2*6 x}{17}\right)+7^3 \sin
\left(\frac{2*7 x}{17}\right)-8^3 \sin
\left(\frac{2*8 x}{17}\right)<0,\quad x\in(0,\pi/2).$$
However, I cannot give a rigorous proof of it. Any suggestion, idea, or comment is welcome, thanks!
| All right, that is easier than expected. If we define
$$ C_n(x) = \sum_{k=1}^{n}(-1)^k k^3\sin\left(\frac{2kx}{2n+1}\right)\tag{1} $$
through $\sin(\theta)=\frac{e^{i\theta}-e^{-i\theta}}{2i}$ we have:
$$\small{ C_n(x) = -\frac{i (-1)^n}{2z^n (1+z)^4}\left(3 n^2 z (1+z)^2 \left(-1+z^{2 n}\right)+z (1-4z+z^2)\left(-1+z^{2 n}\right)-3 n z \left(-1+z^2\right) \left(1+z^{2 n}\right)+n^3 (1+z)^3 \left(-1+z^{1+2 n}\right)\right)}\tag{2} $$
with $z=\exp\left(\frac{2ix}{2n+1}\right)$. Such explicit formula gives that $C_n(x)$ and $(-1)^n$ have the same sign on the interval $(0,\pi)$. So $(1)$ really is an uglier but simpler version of the Fejer-Jackson inequality.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2180839",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
How do we find the closed form for $\int_{0}^{\pi/2}{\mathrm dx\over \sqrt{\sin^6x+\cos^6x}}?$ We would like to find out the closed form for integral $(1)$
$$\int_{0}^{\pi/2}{\mathrm dx\over \sqrt{\sin^6x+\cos^6x}}\tag1$$
An attempt:
We may write
$x^6+y^6=(x^2+y^2)(x^4-x^2y^2+y^4)$
Let $x=\sin x$ and $y=\cos x$
$x^6+y^6=(x^4-x^2y^2+y^4)$
Simplified down to
$x^6+y^6=\sin^4 x+\cos2x\cos^2x$
$x^6+y^6=\sin^2 x-{1\over 4}\sin^2 2x+\cos2x\cos^2x$
$(1)$ becomes
$$\int_{0}^{\pi/2}{\mathrm dx\over \sqrt{\sin^2 x-{1\over 4}\sin^2 2x+\cos2x\cos^2x}}\tag2$$
The power has reduced but looked more messier
How else can we evaluate $(1)$?
| Using this,$$\sin^6(x)+\cos^6(x)=1-\frac 34 \sin^2(2x)$$ makes $$\int{ dx\over \sqrt{\sin^6(x)+\cos^6(x)}}=\int{ dx\over \sqrt{1-\frac 34\sin^2(2x)}}=\frac{1}{2} F\left(2 x\left|\frac{3}{4}\right.\right)$$ where appears the elliptic integral of the first kind.
Using the bounds, the result simplifies to $$K\left(\frac{3}{4}\right)\approx 2.15652$$ which is the complete elliptic integral of the first kind.
If I may suggest, have a look here.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2184398",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Finding a basis for a set of $2\times2$ matrices
Find a basis for $M_{2\times2}$ containing the matrices $$\begin{pmatrix} 1 & 1 \\ 2 & 3 \end{pmatrix}$$ and $$\begin{pmatrix} 1 & 1 \\ 3 & 2 \end{pmatrix}$$
I know that every $2\times2$ matrix $\begin{pmatrix} a & b \\ c & d \end{pmatrix}$ can be written as:
$$a\begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix} + b \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix} + c \begin{pmatrix} 0 & 0 \\ 1 & 0 \end{pmatrix} + d \begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix}$$
so those matrices are a basis for the vector space of $2\times2$ matrices, but how do I apply this to specific matrices? I know how to find a basis for a set of vectors, but matrices confuse me.
| The two matrices are linearly independent, so they are a basis for the two dimensional vector space spanned by them.
If you want a basis for $M_{2\times2}$ add two linearly independent matrices that are linearly independent from them. As an example you can chose:
$$
\begin{pmatrix} 0 & 0 \\ 2 & 3 \end{pmatrix}
\qquad
\begin{pmatrix} 3 & 2 \\ 0 & 0 \end{pmatrix}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2184690",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Prove an identity which uses Drazin inverses and Moore-Penrose inverses. Let $A$ be $m\times n$ complex matrix. Then how can we prove the following.
$$A^+=(A^*A)^DA^*=A^*(AA^*)^D$$ where $D$ denotes the Drazin inverse and $A^+$ is the Moore-Penrose Pseudoinverse of $A$. I found this question at http://planetmath.org/drazininverse.
Thanks.
| Core-Nilpotent Decomposition
Let $\mathbf{A}$ be a square matrix of rank $\rho<n$, with index $k$ such that $\text{rank} \left( \mathbf{A}^{k} \right) = \rho$:
$$
\mathbf{A} \in \mathbb{C}^{n\times n}
$$
then there exists a nonsingular matrix $\mathbf{Q}$ such that
$$
\mathbf{A} =
\mathbf{Q}
\left[ \begin{array}{cc}
\mathbf{C} & \mathbf{0} \\ \mathbf{0} & \mathbf{N}
\end{array} \right]
\mathbf{Q}^{-1}
$$
where the nonsingular core matrix
$$
\mathbf{C} \in \mathbb{C}^{\rho \times \rho}
$$
and the nilpotent matrix $\mathbf{N}$ has index $k$. We may think of this as a dilute form of diagonalization.
Drazin inverse
The Drazin inverse is defined in terms of the core-nilpotent decomposition
$$
\mathbf{A}^{D} =
\mathbf{Q}
\left[ \begin{array}{cc}
\mathbf{C}^{-1} & \mathbf{0} \\ \mathbf{0} & \mathbf{0}
\end{array} \right]
\mathbf{Q}^{-1}
$$
Singular value decomposition
The SVD can be viewed as a tool which gets matrices as close to diagonalization as possible. For the target matrix we can lift the requirement that the matrix has as many rows as columns $m=n$:
$$
\mathbf{A} \in \mathbb{C}^{m\times n}_{\rho}
$$
The SVD provides an orthonormal basis for both domain $\mathbb{C}^{n}$ and codomain $\mathbb{C}^{m}$.
$$
\begin{align}
%
\mathbf{C}^{n} =
\color{blue}{\mathcal{R} \left( \mathbf{A} \right)} \oplus
\color{red}{\mathcal{N} \left( \mathbf{A}^{*} \right)} \\
%
\mathbf{C}^{m} =
\color{blue}{\mathcal{R} \left( \mathbf{A}^{*} \right)} \oplus
\color{red} {\mathcal{N} \left( \mathbf{A} \right)}
%
\end{align}
$$
The $\color{red}{nullspace}$ terms will be silent in the pseudoinverse, just as the nilpotent matrix vanishes in the core-nilpotent decomposition.
$$
\begin{align}
\mathbf{A} &=
\mathbf{U} \, \Sigma \, \mathbf{V}^{*} \\
%
&=
% U
\left[ \begin{array}{cc}
\color{blue}{\mathbf{U}_{\mathcal{R}}} & \color{red}{\mathbf{U}_{\mathcal{N}}}
\end{array} \right]
% Sigma
\left[ \begin{array}{cccccc}
\sigma_{1} & 0 & \dots & & & \dots & 0 \\
0 & \sigma_{2} \\
\vdots && \ddots \\
& & & \sigma_{\rho} \\
& & & & 0 & \\
\vdots &&&&&\ddots \\
0 & & & & & & 0 \\
\end{array} \right]
% V
\left[ \begin{array}{c}
\color{blue}{\mathbf{V}_{\mathcal{R}}}^{*} \\
\color{red}{\mathbf{V}_{\mathcal{N}}}^{*}
\end{array} \right] \\
%
& =
% U
\left[ \begin{array}{cccccccc}
\color{blue}{u_{1}} & \dots & \color{blue}{u_{\rho}} & \color{red}{u_{\rho+1}} & \dots & \color{red}{u_{n}}
\end{array} \right]
% Sigma
\left[ \begin{array}{cc}
\mathbf{S}_{\rho\times \rho} & \mathbf{0} \\
\mathbf{0} & \mathbf{0}
\end{array} \right]
% V
\left[ \begin{array}{c}
\color{blue}{v_{1}^{*}} \\
\vdots \\
\color{blue}{v_{\rho}^{*}} \\
\color{red}{v_{\rho+1}^{*}} \\
\vdots \\
\color{red}{v_{n}^{*}}
\end{array} \right]
%
\end{align}
$$
Let's rewrite the SVD in a form similar to the $\mathbf{CN}$ decomposition:
$$
\begin{align}
\mathbf{A} &=
\mathbf{U} \, \Sigma \, \mathbf{V}^{*} \\
%
&=
% U
\left[ \begin{array}{cc}
\color{blue}{\mathbf{U}_{\mathcal{R}}} & \color{red}{\mathbf{U}_{\mathcal{N}}}
\end{array} \right]
% Sigma
\left[ \begin{array}{cc}
\mathbf{S} & \mathbf{0} \\
\mathbf{0} & \mathbf{0}
\end{array} \right]
% V
\left[ \begin{array}{c}
\color{blue}{\mathbf{V}_{\mathcal{R}}}^{*} \\
\color{red}{\mathbf{V}_{\mathcal{N}}}^{*}
\end{array} \right]
\end{align}
$$
While the matrix of singular values $\mathbf{S}$ is diagonal, the core matrix $\mathbf{C}$ is simply nonsingular.
Manipulating the singular value decomposition
The forms needed are the Hermitian conjugate and the Moore-Penrose pseudoinverse. Recall that $\mathbf{S}^{\mathrm{T}} = \mathbf{S}$:
$$
\begin{align}
\mathbf{A} &=
\mathbf{U} \, \Sigma \, \mathbf{V}^{*} \\
%
&=
% U
\left[ \begin{array}{cc}
\color{blue}{\mathbf{U}_{\mathcal{R}}} & \color{red}{\mathbf{U}_{\mathcal{N}}}
\end{array} \right]
% Sigma
\left[ \begin{array}{cc}
\mathbf{S} & \mathbf{0} \\
\mathbf{0} & \mathbf{0}
\end{array} \right]
% V
\left[ \begin{array}{c}
\color{blue}{\mathbf{V}_{\mathcal{R}}}^{*} \\
\color{red}{\mathbf{V}_{\mathcal{N}}}^{*}
\end{array} \right] \\[5pt]
%% hc
\mathbf{A}^{*} &=
\mathbf{V} \, \Sigma^{\mathrm{T}} \mathbf{U}^{*} \\
%
&=
% U
\left[ \begin{array}{cc}
\color{blue}{\mathbf{V}_{\mathcal{R}}} & \color{red}{\mathbf{V}_{\mathcal{N}}}
\end{array} \right]
% Sigma
\left[ \begin{array}{cc}
\mathbf{S} & \mathbf{0} \\
\mathbf{0} & \mathbf{0}
\end{array} \right]
% V
\left[ \begin{array}{c}
\color{blue}{\mathbf{U}_{\mathcal{R}}}^{*} \\
\color{red}{\mathbf{U}_{\mathcal{N}}}^{*}
\end{array} \right] \\[5pt]
%% mp
\mathbf{A}^{\dagger} &=
\mathbf{V} \, \Sigma^{\dagger} \mathbf{U}^{*} \\
%
&=
% U
\left[ \begin{array}{cc}
\color{blue}{\mathbf{V}_{\mathcal{R}}} & \color{red}{\mathbf{V}_{\mathcal{N}}}
\end{array} \right]
% Sigma
\left[ \begin{array}{cc}
\mathbf{S}^{-1} & \mathbf{0} \\
\mathbf{0} & \mathbf{0}
\end{array} \right]
% V
\left[ \begin{array}{c}
\color{blue}{\mathbf{U}_{\mathcal{R}}}^{*} \\
\color{red}{\mathbf{U}_{\mathcal{N}}}^{*}
\end{array} \right]
%
\end{align}
$$
Assemble the product matrices
$$
\begin{align}
\mathbf{A}^{*} \mathbf{A} &=
\left(
\mathbf{V} \, \Sigma^{\mathrm{T}} \mathbf{U}^{*}
\right)
\left(
\mathbf{U} \, \Sigma \, \mathbf{V}^{*}
\right) \\
&= \mathbf{V} \, \Sigma^{\mathrm{T}} \Sigma \, \mathbf{V}^{*}
= \left[ \begin{array}{cc}
\color{blue}{\mathbf{V}_{\mathcal{R}}} & \color{red}{\mathbf{V}_{\mathcal{N}}}
\end{array} \right]
\,
\left[ \begin{array}{cc}
\mathbf{S}^{2} & \mathbf{0} \\ \mathbf{0} & \mathbf{0}
\end{array} \right]
\left[ \begin{array}{cc}
\color{blue}{\mathbf{V}_{\mathcal{R}}} & \color{red}{\mathbf{V}_{\mathcal{N}}}
\end{array} \right]
^{*} \\
%% Wy
\mathbf{A} \mathbf{A}^{*}
&= \mathbf{U} \, \Sigma \Sigma^{\mathrm{T}} \, \mathbf{U}^{*}
= \left[ \begin{array}{cc}
\color{blue}{\mathbf{U}_{\mathcal{R}}} & \color{red}{\mathbf{U}_{\mathcal{N}}}
\end{array} \right]
\,
\left[ \begin{array}{cc}
\mathbf{S}^{2} & \mathbf{0} \\ \mathbf{0} & \mathbf{0}
\end{array} \right]
\left[ \begin{array}{cc}
\color{blue}{\mathbf{U}_{\mathcal{R}}} & \color{red}{\mathbf{U}_{\mathcal{N}}}
\end{array} \right]
^{*}
%
\end{align}
$$
The Moore-Penrose inverses of the product matrices are
$$
\begin{align}
%% Wx
\left( \mathbf{A}^{*} \mathbf{A} \right)^{\dagger}
&= \left( \mathbf{V} \, \Sigma^{\mathrm{T}} \Sigma \, \mathbf{V}^{*} \right)^{\dagger}
%
= \left[ \begin{array}{cc}
\color{blue}{\mathbf{V}_{\mathcal{R}}} & \color{red}{\mathbf{V}_{\mathcal{N}}}
\end{array} \right]
\,
\left[ \begin{array}{cc}
\mathbf{S}^{-2} & \mathbf{0} \\ \mathbf{0} & \mathbf{0}
\end{array} \right]
\left[ \begin{array}{cc}
\color{blue}{\mathbf{V}_{\mathcal{R}}} & \color{red}{\mathbf{V}_{\mathcal{N}}}
\end{array} \right]
^{*} \\
%% Wy
\left( \mathbf{A} \mathbf{A}^{*} \right)^{\dagger}
&= \left( \mathbf{U} \, \Sigma \Sigma^{\mathrm{T}} \, \mathbf{U}^{*} \right)^{\dagger}
%
= \left[ \begin{array}{cc}
\color{blue}{\mathbf{U}_{\mathcal{R}}} & \color{red}{\mathbf{U}_{\mathcal{N}}}
\end{array} \right]
\,
\left[ \begin{array}{cc}
\mathbf{S}^{-2} & \mathbf{0} \\ \mathbf{0} & \mathbf{0}
\end{array} \right]
\left[ \begin{array}{cc}
\color{blue}{\mathbf{U}_{\mathcal{R}}} & \color{red}{\mathbf{U}_{\mathcal{N}}}
\end{array} \right]
^{*}
%
\end{align}
$$
Connecting to the Drazin inverses
After the associations:
*
*$\mathbf{Q} \to \mathbf{V}$
*$\mathbf{Q}^{-1} \to \mathbf{V}^{*}$
*$\mathbf{C} \to \mathbf{S}^{2}$
the Drazin inverse is
$$
\begin{align}
\color{green}{\left( \mathbf{A}^{*} \mathbf{A} \right)^{D}}
= \left( \mathbf{V} \, \Sigma^{\mathrm{T}} \Sigma \, \mathbf{V}^{*} \right)^{D}
%
= \left(
\mathbf{V}
\left[ \begin{array}{cc}
\mathbf{S}^{2} & \mathbf{0} \\ \mathbf{0} & \mathbf{0}
\end{array} \right]
\mathbf{V}^{*}
\right)^{D}
=
\mathbf{V}
\left[ \begin{array}{cc}
\mathbf{S}^{-2} & \mathbf{0} \\ \mathbf{0} & \mathbf{0}
\end{array} \right]
\mathbf{V}^{*}
%
&= \color{green}{\left( \mathbf{A}^{*} \mathbf{A} \right)^{\dagger}} \\
\end{align}
$$
The other Drazin inverse should be straightforward:
$$
\begin{align}
\color{green}{\left( \mathbf{A} \mathbf{A}^{*} \right)^{D}}
= \left( \mathbf{V} \, \Sigma^{\mathrm{T}} \Sigma \, \mathbf{V}^{*} \right)^{D}
= \mathbf{U}
\left[ \begin{array}{cc}
\mathbf{S}^{-2} & \mathbf{0} \\ \mathbf{0} & \mathbf{0}
\end{array} \right]
\mathbf{U}^{*}
%
&= \color{green}{\left( \mathbf{A} \mathbf{A}^{*} \right)^{\dagger}} \\
\end{align}
$$
Conclusion
Another post derives the different forms of the Moore-Penrose pseudoscience.
What forms does the Moore-Penrose inverse take under systems with full rank, full column rank, and full row rank? Two specific cases are of interest in this post.
Case (1):
The only nontrivial nullspace is $\color{red}{\mathcal{N}_{\mathbf{A}^{*}}}$: the target matrix is overdetermined, more rows than columns, full column rank. The normal equations solution is equivalent to the pseudoinverse:
$$
\mathbf{A}^{\dagger} =
\left( \mathbf{A}^{*} \mathbf{A}
\right)^{-1} \mathbf{A}^{*}
$$
Using an identity from the previous section leaves us with
$$
\mathbf{A}^{\dagger} =
\left( \mathbf{A}^{*} \mathbf{A}
\right)^{-1} \mathbf{A}^{*} =
\left( \mathbf{A}^{*} \mathbf{A}
\right)^{D} \mathbf{A}^{*}
$$
Case (2):
The only nontrivial nullspace is $\color{red}{\mathcal{N}_{\mathbf{A}}}$: the target matrix is underdetermined, more columns than rows, full row rank. The normal equations solution is equivalent to the pseudoinverse:
$$
\mathbf{A}^{\dagger} =
\mathbf{A}^{*}
\left( \mathbf{A} \mathbf{A}^{*}
\right)^{-1} =
\mathbf{A}^{*}
\left( \mathbf{A} \mathbf{A}^{*}
\right)^{D}
$$
Case (3)
Neither nullspace is trivial. To prove the conjecture with the Drazin inverse, start with the Moore-Penrose identity
$$
\mathbf{A}^{\dagger} =
%
\color{green}{\left(
\mathbf{A}^{*} \mathbf{A}
\right)^{\dagger}} \mathbf{A}^{*} =
%
\mathbf{A}^{*}
\color{green}{\left(
\mathbf{A} \mathbf{A}^{*}
\right)^{\dagger}}
$$
(The proof of this statement in on p. 27 Regression and the Moore-Penrose pseudoinverse)
The block form manipulations of the SVD verify this statement quickly. Notice there is no assumption about the invertability of the product matrices. Using the $\color{green}{equivalence}$ relationships provides
$$
\mathbf{A}^{\dagger} =
%
\color{green}{\left(
\mathbf{A}^{*} \mathbf{A}
\right)^{D}} \mathbf{A}^{*} =
%
\mathbf{A}^{*}
\color{green}{\left(
\mathbf{A} \mathbf{A}^{*}
\right)^{D}}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2186679",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Does $\int _0^{\infty }\:\frac{1}{1+x^2\left(\sin x\right)^2}\ \operatorname dx$ converge? I have been trying to prove the following integral:
$$\int _0^{\infty }\:\frac{1}{1+x^2\left(\sin x\right)^2}\ dx$$
diverges (please correct me if I am mistaken).
I have tried to use different comparison tests (as this is an integral of a positive function) with no success.
Any ideas?
| The integral would diverge if $x^2\sin^2(x)$ was "quite small" when $x$ is near a multiple of $\pi$. Using the periodicity of $\sin(x)$, we can examine the behavior near $k\pi$ by shifting and looking at
$$ (x+k\pi)^2 \sin^2(x) \approx (x+k\pi)^2 x^2$$
for $x$ near $0$. This is valid for very small $x$ through the Taylor polynomial, in a way that could be made rigorous.
For $\lvert x \rvert < \frac{1}{2k\pi}$ (for $k \geq 1$), we have that
$$(x+k\pi)^2 \sin^2(x) \leq 2.$$
So for $\lvert x \rvert < \frac{1}{2k\pi}$, we have
$$ \frac{1}{1+(x+k\pi)^2 \sin^2(x)} \geq \frac{1}{3}.$$
With this in mind, we can write
$$\begin{align}
\int_0^\infty \frac{1}{1 + x^2\sin^2(x)}dx &= \sum_{k \geq 0} \int_{k\pi}^{(k+1)\pi} \frac{1}{1 + x^2\sin^2(x)}dx \\
&= \sum_{k \geq 0} \int_{0}^{\pi} \frac{1}{1 + (x+k\pi)^2\sin^2(x)}dx.
\end{align}$$
Now as everything in sight is positive,
$$ \begin{align}
\sum_{k \geq 0} \int_0^\pi \frac{1}{1 + (x+\pi)^2\sin^2(x)}dx &\geq \sum_{k \geq 1} \int_0^{\frac{1}{2\pi k}} \frac{1}{1 + (x+k\pi)^2\sin^2(x)}dx \\
&\geq \sum_{k \geq 1} \int_0^{\frac{1}{2\pi k}} \frac{1}{3} dx\\
&\geq \sum_{k \geq 1} \frac{1}{6\pi k},
\end{align}$$
which diverges.
In other words, I believe that $x^2\sin^2(x)$ is "quite small" when $x$ is near $k\pi$, and it is "quite small" in a region that decays linearly with $k$. And this is enough to show that the integral diverges.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2186784",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 0
} |
Inequality with 4 variables, powers. $\{ a,b,c,d \in\Bbb R_+\ \}$
$a^6 + b^3 + c^2 + d \ge 2 \sqrt[3]{2} \sqrt[4]{3} \sqrt {abcd} $
I have no idea how to do this sort of inequalities, never seen it before.
| By the AM-GM inequality, one has
$$ a^6+b^3+c^2+d=a^6+\frac{1}{2}b^3+\frac{1}{2}b^3+\frac{1}{3}c^2+\frac{1}{3}c^2+\frac{1}{3}c^2+\frac{1}{6}d+\frac{1}{6}d+\frac{1}{6}d+\frac{1}{6}d+\frac{1}{6}d+\frac{1}{6}d\ge 12\sqrt[12]{a^6(\frac12b^3)^2(\frac13c^2)^3(\frac16d)^6}$$
After simple calculation you will get the answer.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2186893",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Prove that $\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+a}\ge\frac{3}{2}$ For $a\geq b\geq c >0$. Prove that $$\dfrac{a}{a+b}+\dfrac{b}{b+c}+\dfrac{c}{c+a}\ge\dfrac{3}{2}$$
$a=100;b=1;c=1/100$ it's wrong ???
$\dfrac{a}{a+b}+\dfrac{b}{b+c}+\dfrac{c}{c+a}\ge\dfrac{3}{2}$
$<=>\sum\frac{a}{a+b}-\frac{1}{2}+\frac{b}{b+c}-\frac{1}{2}+\frac{c}{c+a}-\frac{1}{2}\geq 0$
$<=>\sum \frac{a-b}{2(a+b)}\geq 0$$<=>\sum\frac{z-y}{2x}\geq 0$
$<=>xy^2+yz^2+xz^2-x^2y-y^2z-z^2x\geq 0$
$<=>(y-x)(z-y)(z-x)\geq 0$
And we have $y-x\le 0; z-x\le 0; z-y\geq 0$
we are done !?
| For $a\geq b\geq c>0$ we obtain:
$$\sum_{cyc}\frac{a}{a+b}-\frac{3}{2}=\sum_{cyc}\left(\frac{a}{a+b}-\frac{1}{2}\right)=\sum_{cyc}\frac{a-b}{2(a+b)}=$$
$$=\sum_{cyc}\frac{(a-b)(c^2+ab+ac+bc)}{2\prod\limits_{cyc}(a+b)}=\sum_{cyc}\frac{c^2a-c^2b}{2\prod\limits_{cyc}(a+b)}=$$
$$=\sum_{cyc}\frac{a^2b-a^2c}{2\prod\limits_{cyc}(a+b)}=\frac{(a-b)(a-c)(b-c)}{2\prod\limits_{cyc}(a+b)}\geq0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2192210",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Find value or $W=\sqrt[4x+4]{3^{x}\left ( \sqrt{5}-1 \right )}$ according to condition $9^{x}=6^x +4^x$ If $$9^{x}=6^x +4^x$$
Find the value of:
$$W=\sqrt[4x+4]{3^{x}\left ( \sqrt{5}-1 \right )}$$
Solving the equation arrives at:
$x = \frac{\log\left ( \sqrt{5}-1 \right ) - \log(2)}{\log(2) - \log(3)}$
But W yields a giant result,
Is there any algebraic manipulation that I do not see
| Note that
$$1=\left(\frac23\right)^x+\left(\frac49\right)^x$$
Let $a=\left(\frac23\right)^x$
$$a^2+a-1=0$$
Solving gives
$$a=\left(\frac{\sqrt5 -1}{2}\right)$$
$$\left(\frac23\right)^x=\frac{\sqrt5 -1}{2}$$
$$\left(\frac32\right)^x=\frac{2}{\sqrt5-1}$$
$$3^x=\frac{2^{x+1}}{\sqrt5-1}$$
Substitute for this in question
$$\left(\left(\frac{2^{x+1}}{\sqrt5-1}\right)(\sqrt5-1)\right)^{\frac{1}{4(x+1)}}=2^{\frac{x+1}{4(x+1)}}=2^{1/4}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2192392",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
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