Q
stringlengths 70
13.7k
| A
stringlengths 28
13.2k
| meta
dict |
|---|---|---|
$\lim_{n\to \infty} \frac{3}{n}(1+\sqrt{\frac{n}{n+3}}+\sqrt{\frac{n}{n+6}}+...+\sqrt{\frac{n}{4n-3}})$ fine the limits :
$$\lim_{n\to \infty} \frac{3}{n}(1+\sqrt{\frac{n}{n+3}}+\sqrt{\frac{n}{n+6}}+...+\sqrt{\frac{n}{4n-3}})$$
My Try :
$$\lim_{n\to \infty} \frac{3}{n}(1+\frac{1}{\sqrt{1+\frac{3}{n}}}+\frac{1}{\sqrt{1+\frac{6}{n}}}+...+\frac{1}{\sqrt{1+\frac{3(n-1)}{n}}})\\$$
Now ?
| $$\frac3n\sum_{k=0}^{n-1}\sqrt{\frac n{n+3k}}=\frac3n\sum_{k=0}^{n-1}\sqrt{\frac1{1+\frac{3k}n}}\xrightarrow[n\to\infty]{}3\int_0^1\frac1{\sqrt{1+3x}}\,dx$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2281138",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
express $a$ in terms of $b$ and $c$
Given that $$c=\frac{\sqrt{a+3b}}{a-3b}$$ express $a$ in terms of $b$ and $c$
My attempt,
\begin{align}c^2(a^2-6ab+9b^2)&=a+3b\\
c^2a^2+(-6bc^2-1)a+9b^2c^2-3b&=0\\
a&=\frac{-(-6bc^2-1)\pm \sqrt{(-6bc^2-1)^2-4c^2(9b^2c^2-3b)}}{2c^2}\\
a&=\frac{6bc^2+1\pm \sqrt{24bc^2+1}}{2c^2}\end{align}
My question: Is my answer correct?
| A slightly different take on Yves Daoust's approach:
Let $\sqrt{a+3b}=u\ge0$, so that $a=u^2-3b$. Then
$$c={u\over u^2-6b}\implies cu^2-u-6bc=0$$
The quadratic formula gives
$$u={1\pm\sqrt{1+24bc^2}\over2c}$$
as the formal solutions. It is convenient to rewrite them as
$$u={1+\sqrt{1+24bc^3}\over2c}\qquad\text{and}\qquad u={-12bc\over1+\sqrt{1+24bc^2}}$$
From this it's easy to see that $u$ has no (real) solutions if $24bc^2\lt-1$, while
$$u=
\begin{cases}
0&\quad\text{if}\quad c=0\\\\
\displaystyle{1+\sqrt{1+24bc^3}\over2c}&\quad\text{if}\quad b,c\gt0\\\\
\displaystyle{1-\sqrt{1+24bc^3}\over2c}&\quad\text{if}\quad c\lt0\le b\\\\
\displaystyle{1\pm\sqrt{1+24bc^3}\over2c}&\quad\text{if}\quad\displaystyle{-1\over24c^2}\le b\lt0\lt c
\end{cases}$$
Note that the final case describes two solutions except when $b={-1\over24c^2}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2281781",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
logarithm proof I'm trying to prove the following inequalities:
\begin{align}
\frac{b+c}{b} \geq \frac{\log(\frac{a}{b})}{\log(\frac{a+c}{b+c})}, c \in (0,1)
\end{align}
I know that $\log(\frac{a}{b}) > \log(\frac{a+c}{b+c})$ for $ c > 0$, but I'm stuck as to how to proceed with the proof. This is not a homework and is a part of larger proof. This inequality may turn out to be false.
Additional constraint is that $a > b$.
Any hints will be appreciated!
| Let $a,b,c \in \mathbb{R}^+$ and $a>b$.
Assume
$$\left(\frac{a+c}{b+c}\right)^{b+c}>\left(\frac{a}{b}\right)^{b}$$
Then
$$\log\left(\left(\frac{a+c}{b+c}\right)^{b+c}\right)>\log\left(\left(\frac{a}{b}\right)^{b}\right)$$(as both sides are positive and $\log$ is an increasing function)
$$\implies (b+c)\log\left(\frac{a+c}{b+c}\right)>b\log\left(\frac{a}{b}\right)$$
$$\implies \frac{b+c}{b}>\frac{\log\left(\frac{a}{b}\right)}{\log\left(\frac{a+c}{b+c}\right)}$$
(note for the rearrangement on the last line we have to make sure $\log\left(\frac{a+c}{b+c}\right)$ is positive which is true as $\log\left(\frac{a+c}{b+c}\right) = \log\left(a+c\right)-\log\left(b+c\right)$ and $a>b$ and $\log$ is an increasing function)
All we need to show now is that
$$\left(\frac{a+c}{b+c}\right)^{b+c}>\left(\frac{a}{b}\right)^{b}$$
I haven't been able to find an elegant way, so I've resorted to calculus (the idea being to write the LHS as $y(c)$ and the RHS as $y(0)$ for some function $y$, and then showing that $y(c)>y(0)$ by showing y is a strictly increasing function).
Let $y(x)=$$\left(\frac{a+x}{b+x}\right)^{b+x}$, $x \in \mathbb{R}^+$
Then
$$\log y=(b+x)\left(\log(a+x) - \log(b+x)\right)$$
$$\implies y'y^{-1}=(b+x)((a+x)^{-1} - (b+x)^{-1})+(\log(a+x) - \log(b+x))$$
$$\implies y'=y((b+x)(a+x)^{-1} - 1 +\log(a+x) - \log(b+x))$$
$$=y\left(\frac{(b+x) - (a+x)}{a+x} +\log(a+x) - \log(b+x)\right)$$
$$y>0 \,\,\,\forall\,x\in\mathbb{R}^+\implies y' > \frac{(b+x)-(a+x)}{a+x} +\log(a+x) - \log(b+x))$$
Let $g(x)=\log x$
Then
$$y' > ((a+x)-(b+x))(-g'(a+x)) + g(a+x)-g(b+x)$$
$$a-b>0\implies y'((a+x)-(b+x))^{-1} >\frac{g(a+x)-g(b+x)}{(a+x)-(b+x)}-g'(a+x)>0$$
Observe on the graph of $y=\log x$ that the slope of the line between 2 points ($(b+x)$ and $(a+x)$ in this case) is greater than the slope at the rightmost point. Hence $\frac{g(a+x)-g(b+x)}{(a+x)-(b+x)}>g'(a+x)$. (Alternatively, an algebraic argument can be made using the intermediate value theorem on $g'(x)$ and that $g''(x)<0\,\,\,\forall\,x\in\mathbb{R}^+$)
$$\implies y' > 0$$
Hence $y(x)$ is a strictly increasing function
$$\implies y(c)>y(0)$$
$$\implies \left(\frac{a+c}{b+c}\right)^{b+c}>\left(\frac{a}{b}\right)^{b}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2282446",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
How to evaluate the closed form for $\int_{0}^{\pi/2}{\mathrm dx\over \sin^2(x)}\ln\left({a+b\sin^2(x)\over a-b\sin^2(x)}\right)=\pi\cdot F(a,b)?$ Proposed:
$$\int_{0}^{\pi/2}{\mathrm dx\over \sin^2(x)}\ln\left({a+b\sin^2(x)\over a-b\sin^2(x)}\right)=\pi\cdot F(a,b)\tag1$$
Where $a\ge b$
Examples:
Where $F(1,1)= \sqrt{2}$, $F(2,1)=\sqrt{2-\sqrt{3}}$, $F(3,1)={1\over \sqrt{3}}(2-\sqrt{2})$, $...$
How do we evaluate the closed form for $(1)?$
| Let $u=\cot{x}$. Then $du = dx/\sin^2{x}$ and $\sin^2{x}=1/(1+u^2)$, and the limits become $\infty$ and $0$, so the integral becomes
$$ \int_0^{\infty} \log{\left( \frac{a+b/(1+u^2)}{a-b/(1+u^2)} \right)} \, du = \int_0^{\infty} \log{\left( \frac{a+b+au^2}{a-b+au^2} \right)} du $$
One can now integrate this by parts to get a couple of easy rational integrals, although we should be a bit careful about the upper limit:
$$ \int_0^{M} \log{\left( \frac{a+b+au^2}{a-b+au^2} \right)} du = \left[ u\log{\left( \frac{a+b+au^2}{a-b+au^2} \right)} \right]_0^M - \int_0^M u \left( \frac{2au}{a+b+au^2}-\frac{2au}{a-b+au^2} \right) du $$
The boundary term disappears: the bottom limit is clear, and the top limit follows because
$$ M\log{(a+(a+b)M^{-2})} - M\log{(a+(a-b)M^{-2})} \sim \frac{2b}{aM} \to 0 $$
as $M \to \infty$. The remaining integral is just a rational function; some partial fractions trickery lets us rewrite it as
$$ \int_0^M \left( \frac{2(a+b)}{a+b+au^2}-\frac{2(a-b)}{a-b+au^2} \right) du, $$
which we can then use the arctangent integral on to find the final answer
$$ \left( \sqrt{1+\frac{b}{a}} - \sqrt{1-\frac{b}{a}} \right)\pi $$
or
$$ \frac{\sqrt{a+b}-\sqrt{a-b}}{\sqrt{a}}\pi $$
or
$$ \frac{2b\pi}{\sqrt{a(a+b)}+\sqrt{a(a-b)}}, $$
depending on preference.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2283669",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 2
} |
How to find the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ which goes through $(3,1)$ and has the minimal area?
How to find the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ which goes through $(3,1)$ and has the minimal area? Ellipse area is given as $\pi ab$.
My approach is to use Lagrange method where the constraint function $g=\pi ab$ while the minimization function $f=\frac{x^2}{a^2}+\frac{y^2}{b^2}-1=\frac{3^2}{a^2}+\frac{1^2}{b^2}=1$ to account for the given point.
We can then calculate $b=\pm\sqrt 2$ and $a=\pm \sqrt{18}$.
Area can't be negative so we have that the minimal values of $g$ will occur at $(\sqrt 2, \sqrt{18})$ and $(-\sqrt 2, -\sqrt{18})$.
As far as I understand Lagrange multipliers gives us just the stationary points. In order to check that they really are min values we can use the second derivatives test:
$$
g_{aa}=0, g_{bb}=0, g_{ab}=\pi
$$
Then:
$$
D=g_{aa}\cdot g_{bb}-g^2_{ab}=-\pi^2<0
$$
But in order to have a min point I need $g_{aa}>0$ while it's exactly $0$.
What am I doing wrong?
| We have the general equation of an ellipse:
$$ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 $$
whose area is:
$$ f(a,b) = \pi a b$$
We want to minimize the area $f(a,b)$ subject to the constraint that the ellipse passes through the point $(3,1)$, that is:
$$ g(a,b) = \frac{9}{a^2} + \frac{1}{b^2} = 1 $$
Applying the method of Lagrange multipliers, and noting that both $a$ and $b$ are nonzero and we only care about positive $a,b$, we get that the contrained extrema occur at:
\begin{align*}
\nabla f(a,b) &= \lambda \cdot \nabla g(a,b) \\
g(a,b) &=1 \\[15pt]
\pi b &= -\lambda \frac{18}{a^3} \\
\pi a &= -\lambda \frac{2}{b^3} \\
\frac{9}{a^2} + \frac{1}{b^2} &=1 \\[15pt]
a^3 b &= 9ab^3 \\
\frac{9}{a^2} + \frac{1}{b^2} &=1 \\[15pt]
a^2 &= 9b^2 \\
\frac{9}{a^2} + \frac{1}{b^2} &=1 \\[15pt]
a &= 3b \\
\frac{2}{b^2} &=1 \\[15pt]
\end{align*}
\begin{align*} a &= 3\sqrt{2} \\ b &= \sqrt{2} \end{align*}
This choice of $(a,b)$ gives the ellipse area
$$A = 6\pi$$
We know this must be the minimum, since the choice of $(a,b) = (6,2/\sqrt{3})$ satisfies the constraint, yet gives a greater ellipse area since $2/\sqrt{3} > 1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2288101",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
There exists a $2 \times 2$ matrix $R$ such that $r = R v$ for all 2-dimensional vectors $v$. Find $R$. For a vector $v$, let $r$ be the reflection of $v$ over the line
$$x = t \begin{pmatrix} 2 \\ -1 \end{pmatrix}.$$
There exists a $2 \times 2$ matrix $R$ such that
$$r = R v$$
for all 2-dimensional vectors $v$. Find $R$.
I know that $$\text{proj}_{w} v = \begin{pmatrix} \frac{4}{5} & -\frac{2}{5} \\ -\frac{2}{5} & \frac{1}{5} \end{pmatrix} v$$ for all 2 dimensional vectors $v$ and where $w=\begin{pmatrix} 2 \\ -1 \end{pmatrix}$. But how is that going to help me?
| If $u$ is the projection of $v$ onto $w$, the reflection of $v$ over $w$ is given by $2u-v$. See the diagram below.
Hence $$v'=2u-v=2\begin{pmatrix}\frac{4}{5}&-\frac{2}{5}\\-\frac{2}{5}&\frac{1}{5}\end{pmatrix}v-\begin{pmatrix}1&0\\0&1\end{pmatrix}v = \begin{pmatrix}\frac{3}{5}&-\frac{4}{5}\\-\frac45&-\frac35\end{pmatrix}v.\ \blacksquare$$
Another way of doing the problem, if you are given that such a matrix $R$ exists: if the matrix is $\begin{pmatrix}a&b\\c&d\end{pmatrix}$, then $$\begin{pmatrix}a&b\\c&d\end{pmatrix}\begin{pmatrix}1\\0\end{pmatrix} = \begin{pmatrix}a\\c\end{pmatrix},$$ and $$\begin{pmatrix}a&b\\c&d\end{pmatrix}\begin{pmatrix}0\\1\end{pmatrix} = \begin{pmatrix}b\\d\end{pmatrix}.$$ So all you need to do is figure out where $(1,0)$ and $(0,1)$ go after the reflection, and the transformation matrix can be formed using the two results! This method works in general for transformation matrices.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2288500",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Find $\sin(A)$ and $\cos(A)$ given $\cos^4(A) - \sin^4(A) = \frac{1}{2}$ and $A$ is located in the second quadrant.
Question: Find $\sin(A)$ and $\cos(A)$, given $\cos^4(A)-\sin^4(A)=\frac{1}{2}$ and $A$ is located in the second quadrant.
Using the fundamental trigonometric identity, I was able to find that:
• $\cos^2(A) - \sin^2(A) = \frac{1}{2}$
and
• $$ \cos(A) \cdot \sin(A) = -\frac{1}{4} $$
However, I am unsure about how to find $\sin(A)$ and $\cos(A)$ individually after this.
Edit: I solved the problem through using the Fundamental Trignometric Identity with the difference of $\cos^2(A)$ and $\sin^2(A)$.
| Hint
$$\left( \cos(A)+ \sin(A) \right)^2 = 1+2 \sin(A) \cos(A)=\frac{1}{2} \\
\left( \cos(A)- \sin(A) \right)^2 = 1-2 \sin(A) \cos(A)=\frac{3}{2} $$
Take the square roots, and pay attention to the quadrant and the fact that $\cos^4(A) >\sin^4(A)$ to decide is the terms are positive or negative.
Alternate simpler solution
$$2 \cos^2(A)= \left( \cos^2(A)+\sin^2(A)\right)+\left( \cos^2(A)-\sin^2(A)\right)=1+\frac{1}{2} \\
2 \sin^2(A)= \left( \cos^2(A)+\sin^2(A)\right)-\left( \cos^2(A)-\sin^2(A)\right)=1-\frac{1}{2} \\$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2290899",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
From $a_{n+1}= \frac{3a_n}{(2n+2)(2n+3)}$ to $a_n$, Case 2
Find and prove by induction an explicit formula for $a_n$ if $a_1=1$ and, for $n \geq 1$,
$$P_n: a_{n+1}= \frac{3a_n}{(2n+2)(2n+3)}$$
Checking the pattern:
$$a_1=1 $$
$$a_2= \frac{3}{4 \cdot 5}$$
$$a_3= \frac{3^2} { 4 \cdot 5 \cdot 6 \cdot 7}$$
$$a_4= \frac{3^3} { 4 \cdot 5 \cdot 6 \cdot 7 \cdot 8 \cdot 9 }$$
$$a_n = \frac{3^{n-1}}{ \frac {(2n+1)!}{3!} }$$
$$a_n = \frac {3! \cdot 3^{n-1}} {(2n+1)!}$$
Proof by induction:
$$a_1 = \frac {3! \cdot 3^{0}} {3!} =1$$
$$a_{n+1} = \frac {3! \cdot 3^{n}} {(2n+3)!}$$
Very grateful for the feedback given before. I am new to this. I am a bit stock at the end, what is the most efficient way to reach back to $a_n$?
| $a_{n+1} = \frac{3a_n}{(2n+2)(2n+3)} = \frac{3\cdot 3!\cdot 3^{n-1}}{(2n+2)(2n+3)(2(n-1)+3)!} = \frac{3!3^n}{(2n+3)!}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2291696",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
Show that for every integer $n$, $n^3 - n$ is divisible by 3 using modular arithmetic Problem:
Show that for every integer $n$, $n^3 - n$ is divisible by 3 using modular arithmetic
I was also given a hint:
$$n \equiv 0 \pmod3\\n \equiv 1 \pmod3\\n \equiv 2 \pmod3$$
But I'm still not sure how that relates to the question.
| Using the hint is to try the three cases:
Case 1: $n \equiv 0 \mod 3$
Remember if $a \equiv b \mod n$ then $a^m \equiv b^m \mod n$ [$*$]
So $n^3 \equiv 0^3 \equiv 0 \mod 3$
Remember if $a \equiv c \mod n$ and $b \equiv d \mod n$ then $a+b \equiv c + d \mod n$ [$**$]
So $n^3 - n\equiv 0 - 0 \equiv 0 \mod n$.
Case 2: $n \equiv 1 \mod 3$
Then $n^3 \equiv 1^3 \mod 3$ and $n^3 - n \equiv 1^3 - 1 \equiv 0 \mod n$.
Case 3: $n \equiv 2 \mod 3$
Then $n^3 \equiv 2^3 \equiv 8 \equiv 2 \mod 3$.
So $n^3 - n \equiv 2- 2 \equiv 0 \mod 3$.
So in either of these three cases (and there are no other possible cases [$***$]) we have $n^3 - n \equiv 0 \mod 3$.
That means $3\mid n^3 - n$ (because $n^3 - n \equiv 0 \mod 3$ means there is an intger $k$ so that $n^3 - n = 3k + 0 = 3k$.)
I find that if I am new to modulo notation and I haven't developed the "faith" I like to write it out in terms I do have "faith":
Let $n = 3k + r$ where $r = 0, 1$ or $2$
Then $n^3 - n = (3k+r)^3 -(3k+r) = r^3 - r + 3M$
where $M = [27k^3 + 27k^2r + 9kr^2 - 3k]/3$ (I don't actually have to figure out what $M$ is... I just have to know that $M$ is some combination of powers of $3k$ and those must be some multiple of $3$. In other words, the $r$s are the only things that aren't a multiple of three, so they are the only terms that matter. )
and $r^3 -r$ is either $0 - 0$ or $1 - 1 = 0$ or $8 - 2 = 6$. So in every event $n^3 - n$ is divisible by $3$.
That really is the exact same thing that the $n^3 - n^2 \equiv 0 \mod 3$ notation means.
[$*$] $a\equiv b \mod n$ means there is an integer $k$ so that $a = kn + b$ so $a^m = (kn + b)^m = b^m + \sum c_ik^in^ib^{m-i} = b^m + n\sum c_ik^{i}n^{i-1} b^{m-i}$. So $a^m \equiv b^m \mod n$.
[$**$] $a\equiv c \mod n$ means $a= kn + c$ and $b\equiv d \mod n$ means $b = jn + d$ for some integers $j,k$. So $a + b = c+ d + n(j+k)$. So $a+b =c + d \mod n$.
[$***$]. For any integer $n$ there are unique integers $q, r$ such that $n = 3q + r ; 0 \le r < 3$. Basically this means "If you divide $n$ by $3$ you will get a quotient $q$ and a remainder $r$; $r$ is either $0,1$ or $2$".
In other words for any integer $n$ then $n \equiv r \mod 3$ where $r$ is either $0,1,$ or $2$. These are the only three cases.
P.S. That is how I interpretted the hint. As other have pointed out, a (arguably) more elegant proof is to note $n^3 - n = (n-1)n(n+1)$ For any value of $n$ one of those three, $n$, $n-1$, or $n+1$ must be divisible by $3$.
This actually proves $n^3 - n$ is divisible by $6$ as one of $n$, $n-1$ or $n+1$ must be divisible by $2$.
There is also induction. As $(n+ 1)^3 - (n+1) = n^3 - n + 3n^2 + 3n$, this is true for $n+1)$ if it is true for $n$. As it is true for $0^3 - 0 = 0$ it is true for all $n$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2292586",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Find the simplified form of $\frac {1}{\cos x + \sin x}$ Find the simplified form of $\dfrac {1}{\cos x + \sin x}$.
a). $\dfrac {\sin (\dfrac {\pi}{4} +x)}{\sqrt {2}}$
b). $\dfrac {\csc (\dfrac {\pi}{4} + x)}{\sqrt {2}}$
c). $\dfrac {\sin (\dfrac {\pi}{4} + x)}{2}$
d). $\dfrac {\csc (\dfrac {\pi}{4} + x)}{2}$
My Attempt:
$$=\dfrac {1}{\cos x + \sin x}$$
$$=\dfrac {1}{\cos x + \sin x} \times \dfrac {\cos x - \sin x}{\cos x - \sin x}$$
$$=\dfrac {\cos x - \sin x}{\cos^2 x - \sin^2 x}$$
$$=\dfrac {\cos x - \sin x}{\cos 2x}$$
|
This is the graph of the function $\sin x+\cos x$.
Notice that itself looks like a wave.
So, we should be able to deduce that(since only the amplitude and phase seem to have changed),
$$\sin x+\cos x=A\sin( x+\phi)$$
Expanding using the identity for $\sin(A+B)$,
$$\sin x+\cos x=A\sin( x)\cos\phi+A\cos(x)\sin\phi$$
Comparing the coefficients of $\sin $ and $\cos,
$$1=A\cos\phi$$
$$1=A\sin\phi$$
Assuming non-zero $\cos$,dividing the second equation by the first gives:
$$1=\tan\phi\implies \phi=\frac{\pi}{4}$$
We only need one solution so we won"t worry about other solutions.
Now squaring the equations and adding,
$$2=A^2\implies A=\sqrt{2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2294276",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Is $f(x) = x^{10}-x^5+1$ solvable by radicals?
Is $f(x) = x^{10}-x^5+1$ solvable by radicals?
So far I've showed that $f$ is irreducible because if we let $y=x^5$ then $f(y)=y^2-y+1$ which is irreducible because it has a negative discriminant. I also know that $f$ has no real roots so I've concluded that $Gal(L_f/\mathbb{Q}) \subset S_{10}$ and that there is a 10-cycle and a transposition in $Gal(L_f/\mathbb{Q})$. However I have no clue as to what $Gal(L_f/\mathbb{Q})$ could be.
| Yes, of course.
Let $x+\frac{1}{x}=a$.
Hence,
$$x^{10}-x^5+1=x^{10}+x^7-x^7+x^6-x^5-x^6+1=$$
$$=(x^2-x+1)(x^7(x+1)-x^5-(x^3-1)(x+1))=$$
$$=(x^2-x+1)(x^8+x^7-x^5-x^4-x^3+x+1)=$$
$$=(x^2-x+1)x^4\left(x^4+\frac{1}{x^4}+x^3+\frac{1}{x^3}-x-\frac{1}{x}-1\right)=$$
$$=(x^2-x+1)x^4(a^4-4a^2+2+a^3-3a-a-1)=$$
$$=(x^2-x+1)x^4(a^4+a^3-4a^2-4a+1)=$$
$$=(x^2-x+1)x^4\left(a^2+\frac{1-\sqrt5}{2}a+\frac{3+\sqrt5}{2}\right)\left(a^2+\frac{1+\sqrt5}{2}a+\frac{-3+\sqrt5}{2}\right)...$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2295462",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Show that if $(x_n) \rightarrow x$ then $(\sqrt{x_n}) \rightarrow \sqrt{x}$. Let $x_n \ge 0$ for all $n \in \mathbf{N}$ and $x>0$, show that if $(x_n) \rightarrow x$ then $(\sqrt{x_n}) \rightarrow \sqrt{x}$.
My textbook does the following proof:
Let $\epsilon >0$, we must find an $N$ such that $n \ge N$ implies $|\sqrt{x_n} - \sqrt{x}|< \epsilon$ for all $n \ge N$.
\begin{align}
|\sqrt{x_n} - \sqrt{x}| &= |\sqrt{x_n} - \sqrt{x}|\left(\frac{\sqrt{x_n} + \sqrt{x}}{\sqrt{x_n} + \sqrt{x}}\right) \\
& = \frac{|x_n - x|}{\sqrt{x_n} + \sqrt{x}} \\
& \le \frac{|x_n - x|}{\sqrt{x}} \ \ \ \ \ \ \cdots \ \ \ \ \ \ (1)
\end{align}
Since $(x_n) \rightarrow x$ and $x>0$, we can choose $N$ such that $|x_n - x| < \epsilon\sqrt{x}$ whenever $n \ge N$ and so for all $n \ge N$, $|\sqrt{x_n} - \sqrt{x}| < \frac{\epsilon \sqrt{x}}{\sqrt{x}} = \epsilon$.
What I'm wondering is, in Eqn.$(1)$, why is $\sqrt{x}$ kept in the denominator? Couldn't one just have simply $\le |x_n - x|$ and choose $N$ such that $|x_n - x| < \epsilon$ for $n \ge N$ and the rest follows?
| Because $\sqrt{x_n}+\sqrt{x}\geq \sqrt x$ and thus $$\frac{1}{\sqrt{x_n}+\sqrt x}\leq \frac{1}{\sqrt x}.$$
When you'll see continuous function, such a proof is easier using the continuity of $x\longmapsto \sqrt x$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2297395",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Limit of function calculations I must solve limit of next function:
$$\lim_{x\to \infty}\frac{2x^3+x-2}{3x^3-x^2-x+1}$$
Does my calculations are proper? If not where is my mistake?
$$=\lim_{x\to \infty}\frac{x^3\left(2+\frac{1}{x^2}-\frac{2}{x^3}\right)}{x^3\left(3-\frac{1}{x}-\frac{1}{x^2}+\frac{1}{x^3}\right)} \\
\ =\lim_{x\to \infty}\frac{x^3\left(2+0-0\right)}{x^3\left(3-0-0+0\right)} \\
\ =\frac{2}{3}$$
| You are correct, if you have the ratio of two polynomial of the same degree $n$ then
$$\lim_{x\to +\infty}\frac{a_nx^n+a_{n-1}x^{n-1}+\dots +a_0}{b_nx^n+b_{n-1}x^{n-1}+\dots +b_0}=\lim_{x\to +\infty}\frac{x^n(a_n+\frac{a_{n-1}}{x}+\dots +\frac{a_0}{x^n})}{x^n(b_n+\frac{b_{n-1}}{x}+\dots +\frac{b_0}{x^n})}\\=\lim_{x\to +\infty}\frac{a_n+\frac{a_{n-1}}{x}+\dots +\frac{a_0}{x^n}}{b_n+\frac{b_{n-1}}{x}+\dots +\frac{b_0}{x^n}}=\frac{a_n+0+\dots +0}{b_n+0+\dots +0}=\frac{a_n}{b_n}$$
where $a_n$ and $b_n$ are different from zero.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2297710",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
Finding solutions to $2^x+17=y^2$
Find all positive integer solutions $(x,y)$ of the following equation:
$$2^x+17=y^2.$$
If $x = 2k$, then we can rewrite the equation as $(y - 2^k)(y + 2^k) = 17$, so the factors must be $1$ and $17$, and we must have $x = 6, y = 9$.
However, this approach doesn't work when $x$ is odd.
| It looks like I need to spell out the details for insipidintegrator.
If $x$ is even, the prime $17$ is the product of $y+2^{x/2}$ and $y-2^{x/2}.$ Averaging, we find $y=9$ whence $x=6.$
If $x$ is odd, write $y-2^{x/2}=\frac{17}{y+2^{x/2}}$. Letting $x=2n+1$, we have
$\Big|\frac{y}{2^n}-\sqrt{2}\Big|=\frac{17}{2^n(y+2^{n+.5})}.$
From Beuker's thesis, If $q=2^k, \ \Big|\frac{p}{q}-\sqrt{2}\Big|>2^{-43.9}q^{-1.8}$ Thus $\Big|p-q\sqrt{2}\Big|>2^{-43.9}q^{.2}$
In our case, $p=y,q=2^n$ so $p+q\sqrt{2}>2q\sqrt{2}.$ Multiplying, $$17=y^2-2^x>2^{-43.9}q^{.2}\cdot 2q\sqrt{2}$$ or $$q^{1.2}<17\cdot2^{42.4}$$
and $$n<38.73955$$ A computer check shows the only solutions are $n=1,2,4.$ These values correspond to $x=3,5,9.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2298402",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 5,
"answer_id": 0
} |
Degenerate eigenvalues problem for a 4x4 system In summary, my question is whether or not I'm allowed to have the zero vector as my generalised eigenvector. I'm given the system $$x'=\begin{bmatrix} 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ -2 & 2 & -3 & 1 \\ 2 & -2 & 1 & -3 \end{bmatrix}x$$ I also found two eigenvalues: 0 & -2. For 0, I have an eigenvector $$ \begin{bmatrix} 1 \\ 1 \\ 0 \\ 0 \end{bmatrix}$$
and for -2, I have $$ \begin{bmatrix} -1 \\ 0 \\ 2 \\ 0 \end{bmatrix} \ \text{and}\ \begin{bmatrix} 0 \\ -1 \\ 0 \\ 2 \end{bmatrix}$$
I tried finding a third generalised eigenvector using eigenvalue 2, but it just doesn't exist (I get a row of zeros equals 4). What's going on and how do I proceed?
| Let's name your matrix A.
The matrix $(A + 2E) = \begin{pmatrix} 2 & 0 & 1 & 0 \\ 0 & 2 & 0 & 1 \\ -2 & 2 & -1 & 1 \\ 2 & -2 & 1 & -1 \end{pmatrix}$ has two linearly independent ordinary eigenvectors with eigenvalue 0: $\begin{pmatrix} -1 \\ 0 \\ 2 \\ 0 \end{pmatrix}$ and $\begin{pmatrix} 0 \\ -1 \\ 0 \\ 2 \end{pmatrix}$, that you have already found. They are A's eigenvectors with eigenvalue -2. There are no other ordinary eigenvectors of A with eigenvalue -2 that are linearly independent with these two, because rk(A + 2E) = 2.
The matrix $(A + 2E)^2 = \begin{pmatrix} 2 & 2 & 1 & 1 \\ 2 & 2 & 1 & 1 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{pmatrix}$ has three linearly independent ordinary eigenvectors with eigenvalue 0: $\begin{pmatrix} -1 \\ 0 \\ 2 \\ 0 \end{pmatrix}$, $\begin{pmatrix} 0 \\ -1 \\ 0 \\ 2 \end{pmatrix}$ and $\begin{pmatrix} 1 \\ -1 \\ 0 \\ 0 \end{pmatrix}$. The first two ones are A's ordinary eigenvectors with eigenvalue -2, that you have already found. And the vector $\begin{pmatrix} 1 \\ -1 \\ 0 \\ 0 \end{pmatrix}$ is a generalised eigenvector of rank 2 of the matrix A with eigenvalue 2, making it the thing you were looking for.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2300725",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Calculating $\sum_{k=1}^\infty \frac{k^2}{2^k}=\frac{1}{2}+\frac{4}{4}+\frac{9}{8}+\frac{16}{16}+\frac{25}{32}+\cdots+\frac{k^2}{2^k}+\cdots$ I want to know the value of $$\sum_{k=1}^\infty \frac{k^2}{2^k}=\frac{1}{2}+\frac{4}{4}+\frac{9}{8}+\frac{16}{16}+\frac{25}{32}+\cdots+\frac{k^2}{2^k}+\cdots$$
I added up to $k=50$ and got the value $5.999999999997597$, so it seems that it converges to $6.$ But, I don't know how to get the exact value. Is there any other simple method to calculate it?
| If we start with the power series
$$ \sum_{k=0}^{\infty}x^k=\frac{1}{1-x} $$
(valid for $|x|<1$) and differentiate then multiply by $x$, we get
$$ \sum_{k=1}^{\infty}kx^k=\frac{x}{(1-x)^2}$$
If we once again differentiate then multiply by $x$, the result is
$$ \sum_{k=1}^{\infty}k^2x^k=\frac{x(x+1)}{(1-x)^3}$$
and setting $x=\frac{1}{2}$ shows that
$$ \sum_{k=1}^{\infty}k^22^{-k}=\frac{\frac{3}{4}}{\frac{1}{8}}=6 $$
as you guessed.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2300889",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
How to find $\lim\limits_ {n\to\infty}n^5\int_n^{n+2}\frac{{x}^2}{{ {2+x^7}}}\ dx$? How to find $$\displaystyle\lim_ {n\to\infty}n^5\int_n^{n+2}\dfrac{{x}^2}{{ {2+x^7}}}\ dx$$Can I use Mean Value Theorem? Someone suggested I should use Lagrange but I don't know how it would help.
| $$
\begin{align}
2n^5\frac{n^2}{2+(n+2)^7}&\le n^5\int_n^{n+2}\frac{x^2}{2+x^7}\,\mathrm{d}x\le2n^5\frac{(n+2)^2}{2+n^7}\\[12pt]
\frac2{\frac2{n^7}+\left(1+\frac2n\right)^7}&\le n^5\int_n^{n+2}\frac{x^2}{2+x^7}\,\mathrm{d}x\le\frac{2\left(1+\frac2n\right)^2}{\frac2{n^7}+1}
\end{align}
$$
Apply the Squeeze Theorem.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2302388",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Find the range of $f(x) = 3x^4 - 16x^3 + 18x^2 + 5$ without applying differential calculus
Find the range of $f(x) = 3x^4 - 16x^3 + 18x^2 + 5$ without applying differential calculus.
I tried to express $$f(x)=3x^4-16x^3+18x^2+5=A(ax^2+bx+c)^2+B(ax^2+bx+c)+C $$ which is a quadratic in $ax^2+bx+c$ which itself is quadratic in $x$. Comparing coefficients, we get
$$Aa^2=3 \tag{1}$$
$$2abA=-16$$
$$A(b^2+2ac)+aB=18$$
$$2bcA+bB=0$$
$$Ac^2+Bc+C=5$$
But I felt its very lengthy to solve these equations. Any hints?
| The range is $[k,+\infty)$ where $k$ is the minimum value such that the inequality
$$
3x^4-16x^3+18x^2+5\ge k
$$
is true for any $x \in \mathbb{R}$ and this is the minimum value $k$ such that the equation
$$
3x^4-16x^3+18x^2+5- k=0
$$
has a double solution, that is the discriminant of $3x^4-16x^3+18x^2+5- k$ is null.
The calculation of the discriminant for a quartic polynomial is a bit ''heavy'', but Wolfram Alpha gives:
$$\Delta= -6912(k-5)(k-10)(k+22)$$
so the minimum value is $k=-22$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2302781",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 1
} |
Improper integral $\int \limits_{2}^{4}\frac{\sqrt{(16 - x^2)^5}}{(x^2 - 9x + 20)^3}dx$ I can't figure out how to solve (say whether it converges or diverges) the following improper integral:
$$
\int \limits_{2}^{4}\frac{\sqrt{(16 - x^2)^5}}{(x^2 - 9x + 20)^3}dx
$$
I've tried to simplify this and got:
$$
\int \limits_{2}^{4}\frac{(4 + x)^{5/2}}{(4 - x)^{1/2}(5 - x)^3}
$$
But I don't know what to do next. Could you please give me some hint
| An alternate substitution is $x = 4 \, \sin(t)$ which leads to, with some difficulty,
\begin{align}
\int \frac{(16 - x^2)^{5/2}}{(x^2 - 9 x + 20)^3} \, dx = \frac{81}{2} \, \frac{x-6}{(x-5)^2} \, \sqrt{16 - x^2} - 63 \, \tan^{-1}\left(\frac{16 - 5x}{3 \, \sqrt{16 - x^2}}\right) - \sin^{-1}\left(\frac{x}{4}\right) + c_{0}.
\end{align}
Evaluating the integral at the limits (2,4) then provides the integrals value:
\begin{align}
\int_{2}^{4} \frac{(16 - x^2)^{5/2}}{(x^2 - 9 x + 20)^3} \, dx = 36 \, \sqrt{3} + \frac{125 \, \pi}{3}.
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2305911",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Existence of square root matrix $B \in \mathbb{C}^{2\times 2}$ for any $A \in \mathbb{C}^{2\times 2}$, where $A^2\neq 0$ I am trying to prove that for any $A \in \mathbb{C}^{2\times 2}$ with $A^2\neq 0$, there exists $B \in \mathbb{C}^{2\times 2}$ with $BB=A$.
I have tried the approach of a general matrix A andB with variable entries
$$B = \begin{bmatrix}
a & b \\
c & d
\end{bmatrix}$$
$$A = \begin{bmatrix}
\alpha & \beta \\
\gamma & \delta
\end{bmatrix}$$
and assuming $BB=A$, I get the equations
$$a^2+bc=\alpha$$
$$b(a+d)=\beta$$
$$c(a+d)=\gamma$$
$$d²+cb=\delta$$
However, here I am stuck since I do not know whether any of those variables is $0$, so I cannot operate with those equations.
I have seen a solution on Wikipedia, however to me it seems to fall from the sky, especially the restrictions it makes.
I have also found a thread, which shows that without the restriction $A^2\neq0$ this statement is false, however I fail to see how this is the critical restriction.
Explanations, clarifications or hints on any of the things I mentioned are most welcome.
| We have the equations
\begin{eqnarray*}
a^2+bc= A \\
b(a+d)=B \\
c(a+d)=C \\
bc+d^2=D
\end{eqnarray*}
Multiply the first equation by $(a+d)^2$ and use the second & third we have
\begin{eqnarray*}
a^2(a+d)^2 +BC=A(a+d)^2 \\
d= -a +\sqrt{\frac{BC}{A-a^2}}.
\end{eqnarray*}
Now subtract the first & the fourth
\begin{eqnarray*}
a^2-d^2=A-D \\
\end{eqnarray*}
Substitute for $d$ and we have
\begin{eqnarray*}
a^2-A+D= \left( -a +\sqrt{\frac{BC}{A-a^2}} \right)^2 \\
D-A-\frac{BC}{A-a^2}=-2a \sqrt{\frac{BC}{A-a^2}}
\end{eqnarray*}
Square this & we have a quadratic in $a^2$
\begin{eqnarray*}
a^4((D-A)^2+4BC)+a^2(-2A(D-A)^2-2BC(A-D)-4ABC)+A^2(D-A)^2-2ABC(D-A)+B^2C^2=0
\end{eqnarray*}
Note that this has discriminant $ \Delta=4B^2C^2(AD-BC)$. This gives
\begin{eqnarray*}
a^2= \frac{A(A-D)^2+BC(3A-D) \mp 2BC \sqrt{AD-BC}}{(a-d)^2+4BC} \\
=A-\frac{BC}{A+D \pm 2 \sqrt{AD-BC}}
\end{eqnarray*}
Once the dust has settled ...
\begin{eqnarray*}
\sqrt{\left[
\begin{array}{cc}
A & B \\
C & D\\
\end{array}
\right]}=\left[
\begin{array}{cc}
\sqrt{A-\frac{BC}{A+D \pm 2 \sqrt{AD-BC}} } & \frac{B}{\sqrt{A+D \pm 2 \sqrt{AD-BC}}} \\
\frac{C}{\sqrt{A+D \pm 2 \sqrt{AD-BC}}} & \sqrt{D-\frac{BC}{A+D \pm 2 \sqrt{AD-BC}} } \\
\end{array}
\right]
\end{eqnarray*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2306227",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Area between $r=4\sin(\theta)$ and $r=2$ I'm trying to find the area between $r=4\sin(\theta)$ and $r=2$.
I found the points of intersections to be $\pi/6,5\pi/6$. Which implies the area is $$A=\frac{1}{2}\int_{\pi/6}^{5\pi/6}(4\sin(\theta))^2-2^2d\theta.$$
Is this correct? Or did I find the area for the following region
| The desired red region is just the area of a circle with radius $2$ minus the area of the blue region:
$$
\pi(2)^2 - A
= 4\pi - \frac{1}{2}\int_{\pi/6}^{5\pi/6}(4\sin(\theta))^2-2^2d\theta
= 4\pi - \left( 2\sqrt 3 + \frac{4\pi}{3} \right)
= \frac{8\pi}{3} - 2\sqrt 3
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2307272",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
Find a basis for orthogonal complement in R⁴
How do I approach part 2? I found the projection of 1. to be (6,-2,2,-2) but what do I do now?
| For vector $\mathbf v = (x_1, x_2, x_3,x_4)$, the dot products of $\mathbf v$ with the two given vectors respectively are zero.
$$\begin{align*}
\begin{bmatrix}1&2&3&4\\2&5&0&1\end{bmatrix}
\begin{bmatrix}x_1\\x_2\\x_3\\x_4\end{bmatrix} &= \begin{bmatrix}0\\0\end{bmatrix}\\
\begin{bmatrix}1&2&3&4\\0&1&-6&-7\end{bmatrix}
\mathbf v &= \begin{bmatrix}0\\0\end{bmatrix}\\
\begin{bmatrix}1&0&15&18\\0&1&-6&-7\end{bmatrix}
\mathbf v &= \begin{bmatrix}0\\0\end{bmatrix}\\
\end{align*}$$
Let $x_3 = a$, $x_4 = b$, then $x_1 = -15a - 18b$, and $x_2 = 6a + 7b$.
$$\mathbf v = \begin{bmatrix}-15a - 18b\\6a+7b\\a\\b\end{bmatrix}
= a\begin{bmatrix}-15\\6\\1\\0\end{bmatrix} + b\begin{bmatrix}-18\\7\\0\\1\end{bmatrix}$$
So $(-15,6,1,0)$ and $(-18, 7,0,1)$ together is a basis.
Setting $a=1, b=-1$ gives $(3,-1,1,-1)$, which is one of the vectors in your basis above.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2307669",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
Calculus Spivak Chapter 2 problem 16(c) The question asks to prove that if $\frac mn \lt \sqrt{2}$, then there is another rational number $\frac {m'}{n'}$ with $\frac mn \lt \frac {m'}{n'} \lt \sqrt{2}$.
Intuitively, it's clear that such a number exists, but I don't understand the solution to this problem. It states: let $m_1 = m + 2n$ and $n_1 = m + n$, and choose $m' = m_1 + 2n_1 = 3m + 4n$, and $n' = m_1 + n_1 = 2m + 3n$.
Apparently $\frac {(m + 2n)^2}{(m+n)^2} \gt 2$, but can someone explain why and how plugging in those equations for $m'$ and $n'$ ensures that $\frac {m'}{n'}$ lies between $ \frac mn$ and $\sqrt {2}$?
| I actually have a slightly different answer to the above, which I think is closer to the book as it relies directly on parts (a) and (b).
If anyone spots anything wrong I'd appreciate if you comment below and point out any mistakes:
We have proven in part (a) that:
$\frac{m^2}{n^2} < 2 \implies \frac{(m + 2n)^2}{(m + n)^2}>2$
or equivalently since $m,n \in \mathbb{N}$ and therefore all terms are positive:
$\frac{m}{n} < \sqrt{2} \implies \frac{m + 2n}{m + n}>\sqrt{2}$
Similarly in part (b) we have proven that:
$\frac{m}{n} > \sqrt{2} \implies \frac{m + 2n}{m + n}<\sqrt{2}$
Therefore if we start with a ratio $\frac{m}{n} < \sqrt{2}$ we can use (a) to get a ratio $\frac{m + 2n}{m + n}>\sqrt{2}$ and then use (b) in sequence to get a ratio $\frac{3m + 4n}{2m + 3n}<\sqrt{2}$.
We just need to prove that $\frac{m}{n} < \frac{3m + 4n}{2m + 3n}$.
We have:
$
\begin{aligned}
\frac{m}{n}<\frac{3m+4n}{2m+3n} &\iff \\
m(2m+3n)<n(3m+4n) &\iff \\
2m^2 +3mn < 3mn + 4n^2 &\iff \\
2m^2 < 4n^2 & \iff \\
\frac{m^2}{n^2} < 2 & \iff \\
\frac{m}{n} < \sqrt{2} &
\end{aligned}
$
Please do let me know if this is a proper solution.
I also should note that the idea to apply (a) and (b) in sequence and thus "try" the ratio $\frac{3m+4n}{2m+3n}$ originated from the second item proven in (a) and (b). Specifically we had proven:
(a): $\quad \frac{m^2}{n^2} < 2 \implies \frac{(m + 2n)^2}{(m+n)^2} - 2 < 2 - \frac{m^2}{n^2}$
(b): $\quad \frac{m^2}{n^2} > 2 \implies \frac{(m + 2n)^2}{(m+n)^2} - 2 > 2 - \frac{m^2}{n^2}$
Notice that the expressions in the inequalities to the right of the "implies" sign can be considered as "measures of the distance". For example $\frac{(m + 2n)^2}{(m+n)^2} - 2$ is a measure of the distance of point $\frac{m+2n}{m+n}$ from the point $\sqrt{2}$. The same can be said of $2 - \frac{m^2}{n^2}$ which can be considered as a measure of the distance of point $\frac{m}{n}$ from point $\sqrt{2}$.
This means that starting from any ratio $r_1=\frac{m}{n}<\sqrt{2}$ and creating a series of ratios $r_2=\frac{m+2n}{m+n}$, $r_3=\frac{3m+4n}{2m+3n}$, ... you would be "hopping" from one side of $\sqrt{2}$ to the other (on the number line). Hopping to the right of it will take you a bit closer to it, then hopping to the left in succession will get you slightly further away. In part (c) I think we are essentially proving that the decrease when hopping to the right is more than the increase when hopping back to the left, and thus we are getting "ever closer" to the point $\sqrt{2}$ with every pair of successive "hops".
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2308909",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 0
} |
Find $\lim_{x\rightarrow \frac{\pi}{4}}\frac{\cos(2x)}{\sin(x-\frac{\pi}{4})}$ without L'Hopital I tried:
$$\lim_{x\rightarrow \frac{\pi}{4}}\frac{\cos(2x)}{\sin(x-\frac{\pi}{4})} =
\frac{\frac{\cos(2x)}{x-\frac{\pi}{4}}}{\frac{\sin(x-\frac{\pi}{4})}{x-\frac{\pi}{4}}}$$ and $$\begin{align}\frac{\cos(2x)}{x-\frac{\pi}{4}} &=
\frac{\cos(2x)}{\frac{4x-\pi}{4}} \\&=
\frac{4\cos(2x)}{4x-\pi} = \,\,???\end{align}$$
What do I do next?
| Just as an alternate approach:
I said: $\frac{cos(2\theta)}{sin\theta*cos(\frac{\pi}{4})-cos\theta*(\frac{\pi}{4})}$ =
$\frac{cos(2\theta)}{1/\sqrt(2)*sin\theta-cos\theta}$ Noting that $cos(2\theta)=cos^2\theta-sin^2\theta$ and $sin\theta-cos\theta=-(cos\theta-sin\theta)$
So, we have
$\frac{cos^2\theta-sin^2\theta}{-1/\sqrt(2)*(cos\theta-sin\theta)}$ = $-2*\frac{(cos\theta-sin\theta)*(\cos\theta+sin\theta)}{(cos\theta-sin\theta)}$=$-2*cos(\theta)+sin(\theta)$=-2 when you take the limit. Note that I just reciprocated the $-\frac{1}{\sqrt(2)}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2311265",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 4
} |
Can the following trigonometric equation be transformed into the other? Can $$16\sec^2(x)\tan^4(x)+88\sec^4(x)\tan^2(x)+16\sec^6(x)$$ be proven equal to
$$24\sec^6(x)-8\sec^4(x)+96\sec^4(x)\tan^2(x)-16\sec^2(x)\tan^2(x)$$
I have made about six attempts, but I keep getting stuck. I thought I'd ask, maybe someone else can figure it out before me or verify that we cannot transform from one equation to another.
The reason why this is important to me is because I met a question asking for the differentiation of $\tan^2(x)$ 4 times. Both expressions above are different ways to express the SAME first order derivative, so they are indeed equal. However, the second expression has been derived by replacing $\tan^2(x)$ by $\sec^2(x)-1$, and then carrying on with differentiating to get the third and fourth derivatives. However, I didn't make such a substitution, hence I ended up with the first derivative. So, I'm trying to figure out a strategy to get the derivative right in the exam. It starts with knowing whether one of these expressions can somehow be converted into the other.
| Sometimes the easiest thing to do is convert everything into sines and cosines.
\begin{array}{l}
16 \sec^2(x) \tan^4(x) + 88 \sec^4(x) \tan^2(x) + 16 \sec^6(x) \\
=\dfrac{16\sin^4(x)}{\cos^6(x)}+\dfrac{88\sin^2(x)}{\cos^6(x)}
+\dfrac{16}{\cos^6(x)} \\
=\dfrac{16\sin^4(x) + 88 \sin^2(x) + 16}{\cos^6(x)}
\end{array}
\begin{array}{l}
24\sec^6(x)-8\sec^4(x)+96\sec^4(x)\tan^2(x)-16\sec^2(x)\tan^2(x) \\
=\dfrac{24}{\cos^6(x)}-\dfrac{8}{\cos^4(x)}+\dfrac{96\sin^2(x)}{\cos^6(x)}
-\dfrac{16 \sin^2(x)}{\cos^4(x)} \\
=\dfrac{24-8\cos^2(x)+96\sin^2(x)-16\sin^2(x)\cos^2(x)}{\cos^6(x)} \\
=\dfrac{24-(8-8\sin^2(x))+96\sin^2(x)-(16\sin^2(x) - 16\sin^4(x))}{\cos^6(x)} \\
=\dfrac{16\sin^4(x) + 88\sin^2(x) + 16}{\cos^6(x)}
\end{array}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2312031",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Show that if $\frac1a+\frac1b+\frac1c = a+b+c$, then $\frac1{3+a}+\frac1{3+c}+\frac1{3+c} \leq\frac34$
Show that if $a,b,c$ are positive reals, and
$\frac1a+\frac1b+\frac1c = a+b+c$, then
$$\frac1{3+a}+\frac1{3+b}+\frac1{3+c} \leq\frac34$$
The corresponding problem replacing the $3$s with $2$ is shown here:
How to prove $\frac 1{2+a}+\frac 1{2+b}+\frac 1{2+c}\le 1$?
The proof is not staggeringly difficult: The idea is to show (the tough part) that if
$\frac1{2+a}+\frac1{2+c}+\frac1{2+c}=1$ then
$\frac1a+\frac1b+\frac1c \ge a+b+c$. The result then easily follows.
A pair of observations, both under the constraint of $\frac1a+\frac1b+\frac1c = a+b+c$:
*
*If $k\geq 2$ then $\frac1{k+a}+\frac1{k+c}+\frac1{k+c}\leq\frac3{k+1}$.
This is proven for $k=2$ but I don't see how for $k>2$
*If $0<k<2$ then there are positive $(a,b,c)$ satisfying the constraint such that $\frac1{k+a}+\frac1{k+c}+\frac1{k+c}>\frac3{k+1}$.
So one would hope that the $k=3$ case, lying farther within the "valid" region, might be easier than $k=2$ but I have not been able to prove it.
Note that it is not always true, under our constraint, that
$\frac1{2+a}+\frac1{2+c}+\frac1{2+c}\geq \frac1{3+a}+\frac1{3+c}+\frac1{3+c}+\frac14$, which if true would prove the $k=3$ case immediately.
| For the case $k≥3$, $(k\in\mathbb Z)$
Given equality implies $ab+bc+ca=abc(a+b+c)$
We know that $(x+y+z)^2≥3(xy+yz+zx)$
put $x=ab,y=bc,z=ca\Rightarrow (ab+bc+ca)^2\geq3abc(a+b+c)=3(ab+bc+ca)$
$\Rightarrow abc(a+b+c)=ab+bc+ca\geq3$ and $a+b+c\geq3$ (obvious)
$\dfrac {1}{k+a}+\dfrac {1}{k+b}+\dfrac {1}{k+c}≤\dfrac {3}{k+1} \Leftrightarrow\dfrac {3k^2+2k(a+b+c)+(ab+bc+ca)}{k^3+k^2(a+b+c)+k(ab+bc+ca)+abc}\leq\dfrac {3}{k+1}$
$\Leftrightarrow3k^3+3k^2+(2k^2+2k)(a+b+c)+(k+1)(ab+bc+ca)\leq3k^3+3k^2(a+b+c)+3k(ab+bc+ca)+3abc$
$\Leftrightarrow3k^2\leq(k^2-2k)(a+b+c)+(2k-1)(ab+bc+ca)+3abc$
Last inequality is true by Arithmetic-Geometric mean (and by $k^2-2k\geq1$)
$ \underbrace {(a+b+c)+...+(a+b+c)}_{k^2-2k\geq3 \text{ times}}+\underbrace {(ab+bc+ca)+...+(ab+bc+ca)}_{2k-1 \text{ times}}+\underbrace {3abc}_{1 \text{ times}}\geq$
$k^2\sqrt[k^2]{3abc(a+b+c)^{k^2-2k}(ab+bc+ca)^{2k-1}}=k^2\sqrt[k^2]{3(a+b+c)^{k^2-2k-1}(ab+bc+ca)^{2k}}\geq3k^2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2312672",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
How can I find maximum and minimum modulus of a complex number? I have this problem. Let be given complex number $z$ such that
$$|z+1|+ 4 |z-1|=25.$$
Find the greastest and the least of the modulus of $z$.
I tried with minimum.
Put $A(-1,0)$, $B(1,0)$ and $M(x,y)$ present of $z$.
We have $O(0,0)$ is the midpoint of the segment $AB$. Therefore
$$OM^2 = \dfrac{AM^2 + BM^2}{2}-\dfrac{AB^2}{4}.$$
Another way
$$25=AM+4BM \leqslant \sqrt{(1^2 + 4^2)(AM^2 + BM^2)},$$
Therefore
$$AM^2 + BM^2 \geqslant \dfrac{625}{17}.$$
$$OM^2 \geqslant \dfrac{625}{17} -1 = \dfrac{591}{17}.$$
Thus, minimum of $z$ is $\sqrt{\dfrac{591}{17}}$.
This answer is not true with Mathematica. Mathematica give $\dfrac{22}{5}$.
Where is wrong in my solution and how can I find the maximum?
| For maximum $|z|$, we have
\begin{align}
|5z|&=|(z+1)+4(z-1)+3|\\
&\le|z+1|+4|z-1|+|3|\\
&\le25+3\\
|z|&\le \frac{28}{5}
\end{align}
with the equality holds if and only if $\displaystyle z=\frac{28}{5}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2314488",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Solve for $x$ in $\cos(2 \sin ^{-1}(- x)) = 0$
Solve $$\cos(2 \sin ^{-1}(- x)) = 0$$
I get the answer $\frac{-1}{ \sqrt2}$
By solving like this
\begin{align}2\sin ^{-1} (-x )&= \cos ^{-1} 0\\
2\sin^{-1}(- x) &= \frac\pi2\\
\sin^{-1}(- x) &= \frac\pi4\\
-x &= \sin\left(\frac\pi4\right)\end{align}
Thus $x =\frac{-1}{\sqrt2}$
But correct answer is $\pm\frac{1}{\sqrt2}$
Where am I going wrong?
| You need to solve $\cos \left(2 \arcsin(-x) \right) = 0$. Let $y = 2 \arcsin(-x)$ then $\cos y = 0$ so $y = \pi/2 \pm n\pi$. Then,
$$
2 \arcsin(-x) = \frac{pi}{2} \pm n\pi
$$
which implies
$$
x = -\sin \left( \frac{\pi}{4} \pm \frac{n\pi}{2} \right)
$$
Can you simplify this?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2314829",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
Find this limit. Compute the value of the limit :
$$
\lim_{x\to\infty}{\frac{1-\cos x\cos2x\cos3x}{\sin^2x}}
$$
I've tried simplifying the expression to
$$
\lim_{x\to\infty}\frac{-8\cos^6x+10\cos^4x-3\cos^2x+1}{\sin^2x}
$$
But I don't know what to do after this.
| Using the identities $\displaystyle \sin^2(x)=\frac{1-\cos(2x)}{2}$, $\displaystyle \cos(4x)=2\cos^2(2x)-1$, and $\displaystyle \cos(x)\cos(3x)=\frac12(\cos(2x)+\cos(4x))$, we obtain
$$\begin{align}
\frac{1-\cos(x)\cos(2x)\cos(3x)}{\sin^2(x)}&=\frac{1-\frac{\cos(2x)\left(\cos(2x)+\overbrace{(2\cos^2(2x)-1)}^{=\cos(4x)}\right)}{2}}{\frac{1-\cos(2x)}{2}}\\\\
&=\frac{-2\cos^3(2x)-\cos^2(2x)+\cos(2x)+2}{1-\cos(2x)}\\\\
&=2\cos^2(2x)+3\cos(2x)+2
\end{align}$$
whence taking the limit as $x\to 0$ yields the coveted result
$$\lim_{x\to 0}\left(\frac{1-\cos(x)\cos(2x)\cos(3x)}{\sin^2(x)}\right)=7$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2325217",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 7,
"answer_id": 1
} |
equation of ellipse after projection If I have the intersection of $x+z=1$ and $$x^2 +y^2 +z^2=1$$ which is a circle in $O'xyz$. Then I do a projection of this circle on the $O'xy$ plane, it'll be an ellipse. How can I then find the equation of this ellipse?
| If $(x,y,z)$ is a point on the circle, then $(x,y,z)$ satisfy both $x+z=1$ and $x^2+y^2+z^2=1$. It's equations are
$$\begin{cases} z=1-x \\ x^2+y^2+(1-x)^2=1\end{cases}$$
Its projection has $z$-coordinate equals $0$ and keep the $x$ and $y$-coordinates. So the equation of the projection on the $xy$-plane is
\begin{align}
x^2+y^2+(1-x)^2&=1\\
2x^2+y^2-2x&=0
\end{align}
Another way to think of the problem.
The centre $C$ of the circle is a point on the plane $x+z=0$. The line joining $C$ and the origin is orthogonal to the plane. It is easy to see that $\displaystyle C=\left(\frac{1}{2},0,\frac{1}{2}\right)$. $C$ is $\displaystyle \frac{\sqrt{2}}{2}$ unit from the origin. So the radius of the circle is $\displaystyle\sqrt{1-\left(\frac{\sqrt{2}}{2}\right)^2}=\frac{\sqrt{2}}{2}$.
The projection of the circle is an ellipse with centre $\displaystyle \left(\frac{1}{2},0,0\right)$. The major axis of the ellipse is parallel to the $y$-axis, which has the same length as the radius of the circle ($\displaystyle =\frac{\sqrt{2}}{2}$). As the plane makes a $45^\circ$ with the $xy$-plane, the minor axis has length $\displaystyle \frac{\sqrt{2}}{2}\cos45^\circ=\frac{1}{2}$, and it is parallel to the $x$ axis. The equation of the ellipse is
\begin{align}
\frac{(x-\frac{1}{2})^2}{(\frac{1}{2})^2}+\frac{y^2}{(\frac{\sqrt{2}}{2})^2}&=1\\
(2x-1)^2+2y^2&=1
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2326177",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Evaluate the following Determinant of $12$th degree polynomial Evaluate $$\Delta=\begin{vmatrix}
\frac{1}{(a+x)^2} & \frac{1}{(b+x)^2} & \frac{1}{(c+x^2)}\\
\frac{1}{(a+y)^2} & \frac{1}{(b+y)^2} & \frac{1}{(c+y)^2}\\
\frac{1}{(a+z)^2} & \frac{1}{(b+z)^2} & \frac{1}{(c+z)^2}\\
\end{vmatrix}$$
My Try: I have taken all the denominators out and we obtain
$$\Delta=f(a,b,c,x,y,z)\times \begin{vmatrix}
(b+x)^2(c+x)^2 & (a+x)^2(c+x)^2 & (a+x)^2(b+x)^2\\
(b+y)^2(c+y)^2 & (a+y)^2(c+y)^2 & (a+y)^2(b+y)^2\\
(b+z)^2(c+z)^2 & (a+z)^2(c+z)^2 & (a+z)^2(b+z)^2\\
\end{vmatrix}$$ where
$$f(a,b,c,x,y,z)=\frac{1}{\left((a+x)(b+x)(c+x)(a+y)(b+y)(c+y)(a+z)(b+z)(c+z)\right)^2} $$
By factor theorem we observe that $a-b$,$b-c$,$c-a$,$x-y$,$y-z$ and $z-x$ are factors of the new Determinant above.
But how to find remaining factors?
| It's obvious that we have a factor $(a-b)(b-c)(c-a)(x-y)(y-z)(z-x)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2326385",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Equation of the line through the point $(\frac{1}{2},2)$ and tangent to the parabola $y=\frac{-x^2}{2}+2$ and secant to the curve $y=\sqrt{4-x^2}$ Find the equation of the line through the point $(\frac{1}{2},2)$ and tangent to the parabola $y=\frac{-x^2}{2}+2$ and secant to the curve $y=\sqrt{4-x^2}$
Let the required line is tangent to the parabola at the point $(x_1,y_1)$.It passes through $(\frac{1}{2},2)$.Its equation is $y-2=-x_1(x-\frac{1}{2})$.
This line is also secant to the the curve $y=\sqrt{4-x^2}$.
I solved $y=\sqrt{4-x^2}$ and the line $y-2=-x_1(x-\frac{1}{2})$.
I am stuck here.
| Let $\left(t,-\frac{t^2}{2}+2\right)$ be a tangent point.
Since $\left(-\frac{x^2}{2}+2\right)'=-x$, we get an equation of the tangent line:
$$y+\frac{t^2}{2}-2=-t(x-t).$$
Now, substitute $x=\frac{1}{2}$ and $y=2$, find a values of $t$ (I got $t=0$ or $t=1$) and choose a value, which you need.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2326909",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Simplification of Trigo expression
Simplify $$\frac{\tan^2 x+\cos^2 x}{\sin x+ \sec x}$$
My attempt,
$$=\frac{(\cos^2x+\frac{\sin^2x}{\cos^2x})}{\frac{\sin x \cos x+1}{\cos x}} $$
$$=\frac{\cos^4 x+\sin^2 x}{\cos^2 x}\cdot \frac{\cos x}{\sin x \cos x+1}$$
$$=\frac{\cos^4 x+\sin^2 x}{\cos x(\sin x \cos x+1)}$$
I'm stuck at here. The given answer is $\sec x -\sin x$
| Hint
Use $\tan^2x=\sec^2x -1$
$$\frac{\tan^2 x+\cos^2 x}{\sin x+ \sec x}=\frac{(\sec^2 x-1)+\cos^2 x}{\sin x+ \sec x}=\frac{\sec^2 x-\sin^2 x}{\sin x+ \sec x}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2328573",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
If $x^3 - 5x^2+ x=0$ then find the value of $\sqrt {x} + \dfrac {1}{\sqrt {x}}$ If $x^3 - 5x^2+ x=0$ then find the value of $\sqrt {x} + \dfrac {1}{\sqrt {x}}$
My Attempt:
$$x^3 - 5x^2 + x=0$$
$$x(x^2 - 5x + 1)=0$$
Either,
$x=0$
And,
$$x^2-5x+1=0$$
??
| $x^3-5x^2+x$ gives $x=0$ or $x^2+1=5x$.
For $x\leq0$ the needed value does not exist.
For $x>0$ we have $\sqrt{x}+\frac{1}{\sqrt{x}}>0$.
Thus,
$$x+\frac{1}{x}=5$$
or $$\left(\sqrt{x}+\frac{1}{\sqrt{x}}\right)^2=7,$$
which gives $\sqrt{x}+\frac{1}{\sqrt{x}}=\sqrt7.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2329371",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 0
} |
Calculate $\tan^2{\frac{\pi}{5}}+\tan^2{\frac{2\pi}{5}}$ without a calculator The question is to find the exact value of:
$$\tan^2{\left(\frac{\pi}{5}\right)}+\tan^2{\left(\frac{2\pi}{5}\right)}$$
without using a calculator.
I know that it is possible to find the exact values of $\tan{\left(\frac{\pi}{5}\right)}$ and $\tan{\left(\frac{2\pi}{5}\right)}$ to find that the answer is $10$. However, I want to know whether there is a faster way that does not involve calculating those values.
So far, I have this: let $a=\tan{\left(\frac{\pi}{5}\right)}$ and $b=\tan{\left(\frac{2\pi}{5}\right)}$; then, $b=\frac{2a}{1-a^2}$ and $a=-\frac{2b}{1-b^2}$, so multiplying the two and simplifying gives:
$$a^2+b^2={\left(ab\right)}^2+5$$
Any ideas? Thanks!
| Let $\tan\left(\frac{\pi}{5}\right)=x$
$$\tan\left(\frac{\pi}{5}+\frac{\pi}{5}+\frac{\pi}{5}+\frac{\pi}{5}+\frac{\pi}{5}\right)=0$$
$$\implies S_1-S_3+S_5=0$$
($S_k$ represents sum of tangents taken $k$ at a time)
$$\implies 5x-10x^3+x^5=0$$
Now the roots of this equation are $\tan\left(\frac{\pi}{5}\right),\tan\left(\frac{2\pi}{5}\right),\tan\left(\frac{3\pi}{5}\right),\tan\left(\frac{4\pi}{5}\right),\tan\left(\frac{\pi}{5}\right)$. Now sum of squares of the roots is $$\tan^2\left(\frac{\pi}{5}\right)+\tan^2\left(\frac{2\pi}{5}\right)+\tan^2\left(\frac{3\pi}{5}\right)+\tan^2\left(\frac{4\pi}{5}\right)+\tan^2\left(\frac{\pi}{5}\right)=2\left(\tan^2\left(\frac{\pi}{5}\right)+\tan^2\left(\frac{2\pi}{5}\right)\right)$$ Now this when calculated from basic concept of theory of equations applying to given polynomial comes to be $20$.
Hence $$2\left(\tan^2\left(\frac{\pi}{5}\right)+\tan^2\left(\frac{2\pi}{5}\right)\right)=20 \implies \tan^2\left(\frac{\pi}{5}\right)+\tan^2\left(\frac{2\pi}{5}\right)=10$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2336186",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 6,
"answer_id": 3
} |
Find $\arctan(\frac{1}{3})+\arctan(\frac{1}{9})+\arctan(\frac{7}{19})$ Firstly used this formula $$ \begin{align} \arctan(\alpha)+\arctan(\beta)
& =\arctan(\frac{1-xy}{x+y}),\quad x\gt0,y\gt0 \\ &=\arctan(\frac{1-\frac{1}{3}\frac{1}{9}}{\frac{1}{3}+\frac{1}{9}}) \\ &=\arctan(2) \end{align}$$ So it is $\arctan(2)+\arctan(\frac{7}{19}).$ Here I don't know what is the next step to solve it completely.
A SIDE NOTE: AN EDIT HAS BEEN MADE TO THIS POST, I HAVE FOUND MY MISTAKE! NOW IT IS CLEAR TO ME, THANKS!
| Just calculate:
$$\tan\left(\arctan\frac{1}{3}+\arctan\frac{1}{9}+\arctan\frac{7}{19}\right)=$$
$$=\frac{\frac{\frac{1}{3}+\frac{1}{9}}{1-\frac{1}{3}\cdot\frac{1}{9}}+\frac{7}{19}}{1-\frac{\frac{1}{3}+\frac{1}{9}}{1-\frac{1}{3}\cdot\frac{1}{9}}\cdot\frac{7}{19}}=1,$$
which gives the answer: $45^{\circ}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2337146",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Proving trigonometric identity $\frac{\sin(A)}{1+ \cos(A )}+\frac{1+ \cos(A )}{\sin(A)}=2 \csc(A)$
$$
\frac{\sin(A)}{1+\cos(A)}+\frac{1+\cos(A)}{\sin(A)}=2\csc(A)
$$
\begin{align}
\mathrm{L.H.S}&= \frac{\sin^2A+(1+\cos^2(A))}{\sin(A)(1+\cos(A))} \\[6px]
&= \frac{\sin^2A+2\sin(A)\cos(A)+\cos^2(A)+1}{\sin(A)(1+\cos(A))} \\[6px]
&= \frac{2+2\sin(A)\cos(A)}{\sin(A)(1+\cos(A))}
\end{align}
What should be done from here?
| You make several mistakes, the main one being
$$
(a+b)^2=a^2+b^2
$$
The mistake is $(1+\cos(A))^2=1+\cos^2(A)$, whereas it should be
$$
(1+\cos(A))^2=1+2\cos(A)+\cos^2(A)
$$
Note that
$$
\frac{a}{b}+\frac{b}{a}=\frac{a^2+b^2}{ab}
$$
where $a=\sin(A)$ and $b=1+\cos(A)$.
In the second step you also arbitrarily insert a term $2\sin(A)\cos(A)$ with no justification.
Start again.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2339468",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 7,
"answer_id": 3
} |
Finding the volume of the solid generated by revolving the given curve. The objective is to find the volume of the solid generated by revolving the curve $y=\dfrac{a^3}{a^2+x^2}$ about its asymptote.
Observing the given function yields that $y\ne0$, hence $y=0$ is the asymptote to the given curve. Thus, the volume of the solid formed by revolving the given curve about $x-axis$ is given as $$V=2\pi\int_0^{\infty}(f(x)^2-0)dx$$
Which gives: $2\pi\int_0^\infty\dfrac{a^6}{(a^2+x^2)^2}dx$
Now, this integral is quite tedious and I don't know why the result tends to infinity. The integral takes the form $\dfrac{1}{x^4+2x^2+1}$ for $a=1$, which is transformed into $\dfrac{\frac{1}{x^2}}{x^2+2+\frac{1}{x^2}}=2[\dfrac{1+\frac{1}{x^2}}{x^2+2+\frac{1}{x^2}}-\dfrac{(1-\frac{1}{x^2})}{x^2+2+\frac{1}{x^2}}]$ which can be integrated, but this integral is tending to infinity. Can anyone help ? IS there a simpler way of doing this problem ?
| Your integral is equal $\frac{\pi^2}{2}( \frac{1}{a^2})^{\frac{3}{2}}a^6$ according to wolfram alpha. Your algebra must be wrong somewhere. I recommend trying trigonometric substitutions.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2339578",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 1
} |
Find $a^2+b^2+2(a+b)$ minimum if $ab=2$ Let $a,b\in R$,and such
$$ab=2$$
Find the minimum of the $a^2+b^2+2(a+b)$.
I have used $a=\dfrac{2}{b}$, then
$$a^2+b^2+2(a+b)=\dfrac{4}{b^2}+b^2+\dfrac{4}{b}+2b=f'(b)$$
Let $$f'(b)=0,\,b=-\sqrt{2}$$
So $$a^2+b^2+2(a+b)\ge 4-4\sqrt{2}$$
I wanted to know if there is other way to simplify the function and find the required value without using messy methods. Can we cleanly use AM-GM inequality?
| For $a=b=-\sqrt2$ we get a value $4-4\sqrt2$.
We'll prove that it's a minimal value.
Indeed, let $a+b=2k\sqrt{ab}$.
Hence, $|k|=\left|\frac{a+b}{2\sqrt{ab}}\right|\geq1$ and we need to prove that
$$a^2+b^2+2(a+b)\geq4-4\sqrt2$$ or
$$a^2+b^2+\sqrt{2ab}(a+b)\geq(2-2\sqrt2)ab$$ or
$$(a+b)^2+\sqrt{2ab}(a+b)\geq(4-2\sqrt2)ab$$ or
$$4k^2+2\sqrt2k\geq4-2\sqrt2$$ or
$$(k+1)(2k-2+\sqrt2)\geq0,$$
which is obvious for $|k|\geq1$.
Done!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2341443",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Evaluate an indefinite integral Find the value of
$$\int{\frac{x^2e^x}{(x+2)^2}} dx$$
My Attempt: I tried to arrange the numerator as follows:
$$ e^xx^2 = e^x(x+2-2)^2 $$ but that didn't help.
Any guidance on this problem will be very helpful.
| Another method:
\begin{align}
\int \frac{x^2 \, e^{x}}{(x+2)^2} \, dx &= - \int x^2 \, e^{x} \, \frac{d}{dx} \left(\frac{1}{x+2}\right) \, dx \\
&= - \left[ \frac{x^2 \, e^{x}}{x + 2} \right] + \int x(x+2) \, e^{x} \cdot \frac{1}{x+2} \, dx \\
&= - \frac{x^2 \, e^{x}}{x + 2} + \int x \, e^{x} \, dx \\
&= - \frac{x^2 \, e^{x}}{x + 2} + (x-1) \, e^{x} + c_{0}\\
&= \left(\frac{x-2}{x+2} \right) \, e^{x} + c_{0}.
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2342042",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 2
} |
Hyperbolas: Deriving $\frac{x^2}{a^2} + \frac{y^2}{a^2 - c^2} = 1$ from $\sqrt{(x + c)^2 + y^2} - \sqrt{(x - c)^2 + y^2} = \pm2a$ My textbook's section on Hyperbolas states the following:
If the foci are $F_1(-c, 0)$ and $F_2(c, 0)$ and the constant difference is $2a$, then a point $(x, y)$ lies on the hyperbola if and only if $\sqrt{(x + c)^2 + y^2} - \sqrt{(x - c)^2 + y^2} = \pm2a$.
To simplify this equation, we move the second radical to the right-hand side, square, isolate the remaining radical, and square again, obtaining $\dfrac{x^2}{a^2} + \dfrac{y^2}{a^2 - c^2} = 1$.
I've attempted to "move the second radical to the right-hand side, square, isolate the remaining radical, and square again", but I cannot derive $\dfrac{x^2}{a^2} + \dfrac{y^2}{a^2 - c^2} = 1$. I could be misunderstanding the instructions, but my attempts to derive the textbook's solution by precisely following the instructions have not been successful.
I would greatly appreciate it if people could please take the time to demonstrate the derivation of $\dfrac{x^2}{a^2} + \dfrac{y^2}{a^2 - c^2} = 1$ from $\sqrt{(x + c)^2 + y^2} - \sqrt{(x - c)^2 + y^2} = \pm2a$, as mentioned in the textbook. Are the textbook's instructions incorrect/insufficient or am I simply misunderstanding them?
| You have\begin{multline*}\sqrt{(x+c)^2+y^2}-\sqrt{(x-c)^2+y^2}=\pm2a\Longleftrightarrow\\\Longleftrightarrow(x+c)^2+y^2+(x-c)^2+y^2-2\sqrt{(x+c)^2+y^2}\sqrt{(x-c)^2+y^2}=4a^2.\end{multline*}This is the same thing as saying that$$\sqrt{(x+c)^2+y^2}\sqrt{(x-c)^2+y^2}=-2a^2+c^2+x^2+y^2.$$Squaring both sides, one gets$$\bigl((x+c)^2+y^2\bigr)\bigl((x-c)^2+y^2\bigr)=(-2a^2+c^2+x^2+y^2)^2$$This is equivalent to$$-a^4+c^2 a^2+x^2 a^2+y^2 a^2-c^2 x^2=0$$or$$(a^2-c^2)x^2+a^2y^2=a^2(a^2-c^2)\text,$$and this means that$$\frac{x^2}{a^2}+\frac{y^2}{a^2-c^2}=1.$$Of course, a little extra work is required in order to prove that the two expressions are equivalent.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2342742",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Find the largest constant $k$ such that $\frac{kabc}{a+b+c}\leq(a+b)^2+(a+b+4c)^2$
Find the largest constant $k$ such that $$\frac{kabc}{a+b+c}\leq(a+b)^2+(a+b+4c)^2$$
My attempt,
By A.M-G.M, $$(a+b)^2+(a+b+4c)^2=(a+b)^2+(a+2c+b+2c)^2$$
$$\geq (2\sqrt{ab})^2+(2\sqrt{2ac}+2\sqrt{2bc})^2$$
$$=4ab+8ac+8bc+16c\sqrt{ab}$$
$$\frac{(a+b)^2+(a+b+4c)^2}{abc}\cdot (a+b+c)\geq\frac{4ab+8ac+8bc+16c\sqrt{ab}}{abc}\cdot (a+b+c)$$
$$=(\frac{4}{c}+\frac{8}{b}+\frac{8}{a}+\frac{16}{\sqrt{ab}})(a+b+c)$$
$$=8(\frac{1}{2c}+\frac{1}{b}+\frac{1}{a}+\frac{1}{\sqrt{ab}}+\frac{1}{\sqrt{ab}})(\frac{a}{2}+\frac{a}{2}+\frac{b}{2}+\frac{b}{2}+c)$$
$$\geq 8(5\sqrt[5]{\frac{1}{2a^2b^2c}})(5\sqrt[5]{\frac{a^2b^2c}{2^4}})=100$$
Hence the largest constant $k$ is $100$
Is my answer correct? Is there another way to solve it? Thanks in advance.
| One more way...
Noting that replacing $a, b$ with $\frac{a+b}2, \frac{a+b}2$ leaves RHS unchanged but increases the LHS, we have to only check for the case $a=b$. Further as the inequality is homogeneous in $a, b, c$; WLOG we may set $a=1$. Hence we need only look for the minimum of the univariate
$$f(c) = (4+(2+4c)^2)\cdot \frac{2+c}c$$
This is easily done using calculus, or by noting $f(c) = \dfrac{4(4+c)(2c-1)^2}c + 100 \geqslant 100$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2343592",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Hoffman and Kunze, Linear algebra sec 3.5 exercise 9
Let $V$ be the vector space of all $2\times 2$ matrices over the field of real numbers and let $$B=\begin{pmatrix}2&-2\\-1&1\end{pmatrix}.$$
Let $W$ be the subspace of $V$ consisting of all $A$ such that $AB=0.$ Let $f$ be a linear functional on $V$ which is in the annihilator of $W.$ Suppose that $f(I)=0$ and $f(C)=3,$ where $I$ is the $2\times 2$ identity matrix and $$C=\begin{pmatrix}0&0\\0&1\end{pmatrix}.$$
Find $f(B).$
Attempt: Observing that $W=$ span$\{\begin{pmatrix}1&2\\0&0\end{pmatrix},\begin{pmatrix}0&0\\1&2\end{pmatrix}\}:=$ span$\{P,Q\}$ and $$B=(-1)P+(-1)Q+(3)I$$
we get $f(B)=(-1)f(P)+(-1)f(Q)+(3)f(I)=(-1)(0)+(-1)(0)+(3)(0)=0.$
The fact that I haven't used $C$ is bothering me. Is the matrix $C$ for display or is there anything wrong in what I have done?
| There is nothing wrong with your solution.
Perhaps the author had the following approach in mind for which $C$ could be useful. Also it helps to determine the functional completely. Let us consider the following standard basis of $V$:
$$\mathcal{B}=\{E_1,E_2,E_3,E_4\}=\left\{\begin{bmatrix}1&0\\0&0\end{bmatrix}, \begin{bmatrix}0&1\\0&0\end{bmatrix}, \begin{bmatrix}0&0\\1&0\end{bmatrix}, \begin{bmatrix}0&0\\0&1\end{bmatrix}\right\}.$$
Then
$$E_1+C=I \implies f(E_1)=f(I)-f(C) \implies f(E_1)=-3.$$
Using your $P$ and $Q$ we get the following:
\begin{align*}
\because P=E_1+2E_2 && \implies &f(E_2)=\frac{3}{2}\\
\because Q=E_3+2E_4=E_3+2C && \implies &f(E_3)=-6
\end{align*}
We can express $B$ in terms of the vectors in the basis $\mathcal{B}$, therefore
\begin{align*}
B & = 2E_1-2E_2-E_3+E_4\\
f(B) & = 2f(E_1)-2f(E_2)-f(E_3)+f(E_4) && (\text{use }E_4=C)\\
f(B) & = 2(-3)-2\left(\frac{3}{2}\right)-(-6)+3\\
&=0.
\end{align*}
It further helps you identify the functional uniquely:
\begin{align*}
f\left(\begin{bmatrix}a&b\\c&d\end{bmatrix}\right) & = af(E_1)+bf(E_2)+cf(E_3)+df(E_4)\\
& = -3a+\frac{3b}{2}-6c+3d
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2344231",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
If $\frac {\sin A + \tan A}{\cos A}=9$, find the value of $\sin A$. If $\dfrac {\sin A + \tan A}{\cos A}=9$, find the value of $\sin A$.
My Attempt:
$$\dfrac {\sin A+\tan A}{\cos A}=9$$
$$\dfrac {\sin A+ \dfrac {\sin A}{\cos A}}{\cos A}=9$$
$$\dfrac {\sin A.\cos A+\sin A}{\cos^2 A}=9$$
$$\dfrac {\sin A(1+\cos A)}{\cos^2 A}=9$$
$$\tan A.\sec A(1+\cos A)=9$$
$$\tan A(1+\sec A)=9$$
How do I go further?
| Hint: using the tangent half-angle formulas, let $\,t=\tan(A/2)\,$, then the equation becomes:
$$
\frac{2t}{1+t^2} + \frac{2t}{1-t^2}=9 \,\frac{1-t^2}{1+t^2} \;\;\iff\;\; 9 t^4 - 18 t^2 - 4 t + 9 = 0
$$
The quartic has $2$ real roots which can be solved in radicals, but the calculations are not pretty.
[ EDIT ] Once $\,t\,$ is determined, $\,\sin A = 2t/(1+t^2)\,$ follows. Or, to determine $\,x = \sin A\,$ directly, one can eliminate $t$ between the equation above and $\,(1+t^2)x-2t=0\,$ using resultants:
$$
1312 x^4 + 288 x^3 - 2592 x^2 - 288 x + 1296 = 0 \;\;\iff\;\; 82 x^4 + 18 x^3 - 162 x^2 - 18 x + 81 = 0
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2344532",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Find the Roots of $(x+1)(x+3)(x+5)(x+7) + 15 = 0$ Once I came across the following problem: find the roots of $(x+1)(x+3)(x+5)(x+7) + 15 = 0$.
Here it is how I proceeded:
\begin{align*}
(x+1)(x+3)(x+5)(x+7) + 15 & = [(x+1)(x+7)][(x+3)(x+5)] + 15\\
& = (x^2 + 8x + 7)(x^2 + 8x + 15) + 15\\
& = (x^2 + 8x + 7)[(x^2 + 8x + 7) + 8] + 15\\
& = (x^2 + 8x + 7)^2 + 8(x^2 + 8x + 7) + 15
\end{align*}
If we make the substitution $y = x^2 + 8x + 7$, we get
\begin{align*}
y^2 + 8y + 15 = (y^2 + 3y) + (5y + 15) = y(y+3) + 5(y+3) = (y+5)(y+3) = 0
\end{align*}
From whence we obtain that:
\begin{align*}
y + 5 = 0\Leftrightarrow x^2 + 8x + 12 = 0 \Leftrightarrow (x+4)^2 - 4 = 0\Leftrightarrow x\in\{-6,-2\}\\
\end{align*}
Analogously, we have that
\begin{align*}
y + 3 = 0\Leftrightarrow x^2 + 8x + 10 = 0\Leftrightarrow (x+4)^2 - 6 = 0\Leftrightarrow x\in\{-4-\sqrt{6},-4+\sqrt{6}\}
\end{align*}
Finally, the solution set is given by $S = \{-6,-2,-4-\sqrt{6},-4+\sqrt{6}\}$.
Differently from this approach, could someone provide me an alternative way of solving this problem? Any contribution is appreciated. Thanks in advance.
| HINT.-Looking about integer solutions for $f(x)=(x+1)(x+3)(x+5)(x+7) + 15 = 0$ possible values should be even and negative so the only candidates are $-2,-4$ and $-6$. We verified that $-2$ and $-6$ are roots. The other two roots are solutions of $$\frac{x^4+16x^3+86x^2+176x+120}{x^2+8x+12}=x^2+8x+10$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2345609",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 3,
"answer_id": 2
} |
Closed form of function composition Given that $f(x)=\dfrac{x+6}{x+2}$, find $f^{n}(x)$ where $f^{n}(x)$ indicates the $n$th iteration of the function.
I first tried to find a pattern but there didn't seem to be an obvious one:
$$f(x) = \dfrac{x+6}{x+2}$$
$$f^2(x) = \dfrac{7x + 18}{3x + 10}$$
$$f^3(x) = \dfrac{25x + 78}{13x + 28}$$
$$\vdots$$
Then I tried to find a recurring sequence and find its closed form by induction:
I know that if $f^n(x)=\dfrac{ax+b}{cx+d}$, then substituting and expanding gives $$f^{n+1}(x) = \frac{(a+b)x + 6a + 2b}{(c+d)x + 6c + 2d}$$
How do I continue from here? Thanks.
| \begin{align}
a_{n+1} &= a_n + b_n \\
b_{n+1} &= 6a_n + 2b_n\\
c_{n+1} &= c_n + d_n \\
d_{n+1} &= 6c_n + 2d_n
\end{align}
Let's write it in matrix form:
$$\begin{bmatrix} a_{n+1} & c_{n+1}\\ b_{n+1} & d_{n+1} \end{bmatrix}= \begin{bmatrix} 1 & 1 \\6 & 2 \\ \end{bmatrix}\begin{bmatrix} a_{n} & c_{n}\\ b_{n} & d_{n} \end{bmatrix}$$
Hence in general $$\begin{bmatrix} a_{n} & c_{n}\\ b_{n} & d_{n} \end{bmatrix}= \begin{bmatrix} 1 & 1 \\6 & 2 \\ \end{bmatrix}^{n-1}\begin{bmatrix} a_{1} & c_{1}\\ b_{1} & d_{1} \end{bmatrix}=\begin{bmatrix} 1 & 1 \\6 & 2 \\ \end{bmatrix}^{n}$$
Using diagonalizaation:
\begin{align}\begin{bmatrix} a_{n} & c_{n}\\ b_{n} & d_{n} \end{bmatrix}&= \begin{bmatrix} -\frac12 & \frac13 \\ 1 & 1 \end{bmatrix}\begin{bmatrix}(-1)^{n} & 0\\ 0 & 4^{n} \end{bmatrix} \begin{bmatrix} -\frac65 & \frac25 \\ \frac65 & \frac35\end{bmatrix}
\\\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2345817",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
Solving equations with complex numbers I have to solve:
$Z^3+\bar{\omega^7} = 0$ and $Z^5\omega^{11}=1$
From the second equation, I got $Z^5=\omega$ and from the first I got $Z^3=-\omega^2$. I plugged in omega from the first result into $Z^3=-\omega^2$, giving me $Z=0$ or $Z^7=-1$, finally giving me 8 solutions:$0,-1,-\alpha...-\alpha^6$.
This is incorrect. Why?
Also, what is the correct way to solve complex equations?
| Let's assume that $\omega$ is a root of unity, but not necessarily a cube root of unity, and see what happens.
It's clear that $Z\not=0$. The equation $Z^3+\overline\omega^7=0$ can be rewritten as $Z^3\omega^7=-1$, which, by squaring both sides, implies $Z^6\omega^{14}=1$. Combining with the other equation, $Z^5\omega^{11}=1$, gives $Z\omega^3=1$, or $Z=\overline\omega^3$.
Plugging $Z=\overline\omega^3$ into $Z^3\omega^7=-1$, we find $\overline\omega^2=-1$, which means $\omega=\pm i$, and this implies $Z=\pm(-i)^3=\pm i$.
In other words, what we've shown is that in order for the simultaneous equations $Z^3+\overline\omega^7=0$ and $Z^5\omega^{11}=1$ to have a solution with $\omega$ a root of unity, we must have $Z=\omega=\pm i$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2346016",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Riccati D.E., vertical asymptotes
For the D.E.
$$y'=x^2+y^2$$
show that the solution with $y(0) = 0$ has a vertical asymptote at some point $x_0$. Try to find upper and lower bounds for $x_0$:
$$y'=x^2+y^2$$
$$x\in \left [ a,b \right ]$$
$$b> a> 0$$
$$a^2+y^2\leq x^2+y^2\leq b^2+y^2$$
$$a^2+y^2\leq y'\leq b^2+y^2$$
$$y'\geq a^2+y^2$$
$$\frac{y}{a^2+y^2}\geq 1$$
$$\int \frac{dy}{a^2+y^2}\geq \int dx=x+c$$
$$\frac{1}{a}\arctan \frac{y}{a}\geq x+c$$
$$\arctan \frac{y}{a}\geq a(x+c)$$
$$\frac{y}{a}\geq\tan a(x+c)$$
$$y\geq a\tan a(x+c)$$
$$a(x+c)\simeq \frac{\pi}{2}$$
But where to from here?
| 1. $x_0$ exists
First note that $y'''(x)$ is increasing$^{[1]}$. It is also easy to see that $y'(0)=y''(0)=0$ but $y'''(0)=2$$^{[2]}$, so by Taylor's theorem$^{[3]}$,
$$
y(x)=\frac{x^3}{6}y'''(c)\ge \frac{x^3}{3},\qquad (*)
$$
for all $x>0$ such that $y$ is defined. Choose one such $x=\epsilon>0$. Then if $x>\epsilon$, we get
$$
y'(x)\ge \epsilon^2+y(x)^2,
$$
which, since $y(\epsilon)>0$, implies $y(x)\to\infty$ as $x\to x_0<\infty$ for some $x_0>\epsilon$.
Edits:
$[1]$: Since $y'(x)=x^2+y(x)^2\ge 0$, $y$ is increasing. Since $y\ge 0$ and $x\ge 0$, we have $y''(x)=2x+2y(x)y'(x)\ge 0$, so $y'$ is also increasing. In a similar way, we deduce that $y'''(x)\ge 0$ and $y^{(4)}(x)\ge 0$.
$[2]$: Since $y(0)=0$, we have $y'(0)=0$. Therefore, $y''(0)=2x+2y(x)y'(x)|_{x=0}=0$. On the other hand, $y'''(0)=2+2y'(x)^2+2y(x)y''(x)|_{x=0}=2$.
$[3]$: First note that $y$ is smooth. Indeed, since $y$ is continuous and $y'(x)=x^2+y(x)^2$, we see that $y'(x)$ is continuous. Since $y''(x)=2x+2y(x)y'(x)$ and the right hand side is continuous, so is $y''$. In a similar way, all derivatives of $y$ are continuous. Since $y$ is smooth, Taylor's theorem can be applied:
$$
y(x)=y(0)+xy'(0)+\frac{1}{2}x^2y''(0)+\frac{1}{6}x^3 y'''(c),\qquad x>0,
$$
where $c\in(0,x)$. But the first three terms are zero by [2], so (*) holds.
2. Lower bound:
Since a finite $x_0>0$ exists, we get
$$
y'(x)\le x_0^2+y(x)^2,
$$
which, since $y(0)=0$, implies
$$
y(x)\le x_0 \tan (x_0\,x).
$$
If it were true that $x_0^2<\pi/2$, then $y(x_0)<\infty$, so $x_0\ge\sqrt{\pi/2}=:z$.
3. Upper bound
For $x>z$, where $z$ is the lower bound, we have
$$
y'(x)\ge z^2+y(x)^2,
$$
which implies
$$
y(x)\ge z\,\tan z(x+c),
$$
where
$$
c=-z+\frac{1}{z}\arctan\frac{y(z)}{z}\ge-z+\frac{1}{z}\arctan\frac{z^2}{3}
$$
by inequality (*). Let
$$
\zeta=\frac{\pi}{2z}-c\le \frac{\pi}{2z}+z-\frac{1}{z}\arctan\frac{z^2}{3}\approx 2.12.
$$
Then $y(\zeta)$ does not exist, so $x_0<\zeta$. Note that $z\approx 1.25$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2348022",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Prove that $\sum_{x=0}^{n}(-1)^x\binom{n}{n-x} (n+1-x)^n=n!$ I figure out these thing when "playing" with numbers:
$$3^2-2.2^2+1^2=2=2!$$
$$4^3-3.3^3+3.2^3-1^3=6=3!$$
$$5^4-4.4^4+6.3^4-4.2^4+1^4=24=4!$$
So I go to the conjecture that:
$$\binom{n}{n}(n+1)^n-\binom{n}{n-1}n^n+\binom{n}{n-2}(n-1)^n-...=n!$$
or
$$\sum_{x=0}^{n}(-1)^x\binom{n}{n-x} (n+1-x)^n=n!$$
Now, how can I prove this conjecture? I've tried a lot, but still couldn't have any idea.
| First of all note that $$\sum_{k=0}^{n}\dbinom{n}{n-k}\left(-1\right)^{k}\left(n-k+1\right)^{n}=\sum_{k=0}^{n}\dbinom{n}{k}\left(-1\right)^{k}\left(n-k+1\right)^{n}$$ then from the special case of the Melzak's identity: $$\sum_{k=0}^{n}\left(-1\right)^{k}\dbinom{n}{k}\frac{f\left(y-k\right)}{x+k},=\frac{f\left(x+y\right)}{x\dbinom{x+n}{n}}-n!a_{n+1}\,x,y\in\mathbb{R},\,x\neq-k$$ where $f $ is an algebraic polynomial of degree $n+1$ and $a_{n+1}$ is the coefficient of the $n+1-$th power, we have, taking $f\left(z\right)=\left(z+1\right)^{n+1},\,y=n$ $$\sum_{k=0}^{n}\dbinom{n}{k}\left(-1\right)^{k}\frac{\left(n-k+1\right)^{n+1}}{-x-k}=n!-\frac{\left(n+x+1\right)^{n+1}}{x\dbinom{x+n}{n}}$$ and so $$\sum_{k=0}^{n}\dbinom{n}{k}\left(-1\right)^{k}\left(n-k+1\right)^{n}=\lim_{x\rightarrow-n-1}\left(n!-\frac{\left(n+x+1\right)^{n}}{x\dbinom{x+n}{n}}\right)={\color{red}{n!}}$$ as wanted.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2348444",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 6,
"answer_id": 4
} |
Other ways to evaluate the integral $\int_{-\infty}^{\infty} \frac{1}{1+x^{2}} \, dx$?
$$\int_{-\infty}^{\infty}\frac{1}{x^2+1}\,dx=\pi $$
I can do it with the substitution $x= \tan u$ or complex analysis. Are there any other ways to evaluate this?
| You can use partial fractions:
$$
\begin{align}
\int_{-\infty}^\infty \frac{dx}{1+x^2}
& = \int_{-\infty}^\infty \frac{1}{2} \left( \frac{1}{1+ix} + \frac{1}{1-ix} \right) dx \\
& = \frac{1}{2i} \bigg[\log(1+ix) - \log(1-ix)\bigg]_{-\infty}^\infty \\
& = \frac{1}{2i} \left[ \lim_{x\to\infty} \log\left( \frac{1+ix}{1-ix} \right)
- \lim_{x\to-\infty} \log\left(\frac{1+ix}{1-ix}\right)\right] \\
& = \frac{1}{2i}\bigg[ i\pi - (-i\pi)\bigg] = \pi
\end{align}
$$
Note that the expression on the second line obviously must be equal to
$$
\bigg[ \arctan x\bigg]_{-\infty}^\infty
$$
because we know the antiderivative of $1/(1+x^2)$ is $\arctan x$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2349224",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 9,
"answer_id": 1
} |
let $f$ be a differentiable function. Compute $\frac{d}{dx}g(2)$, where $g(x) = \frac{f(2x)}{x}$. let $f$ be a differentiable function and $$\lim_{x\to 4}\dfrac{f(x)+7}{x-4}=\dfrac{-3}{2}.$$
Define $g(x)=\dfrac{f(2x)}{x}$. I want to know the derivative
$$\dfrac{d}{dx}g(2)=?$$
I know that :
$$\dfrac{d}{dx}g(2)=\dfrac{4(\dfrac{d}{dx}f(4))-4f(4)}{4}$$
and :
$$\lim_{x\to 4}\dfrac{f(x)-f(4)}{x-4}=a\in\mathbb{R}$$
so :
$$\lim_{x\to 4}\dfrac{f(x)-f(4)+7+f(4)}{x-4}=\dfrac{-3}{2}$$
$$\lim_{x\to 4}\dfrac{f(x)-f(4)}{x-4}+\dfrac{f(4)+7}{x-4}=\dfrac{-3}{2}$$
now what ?
| HINT
\begin{align*}
\frac{dg}{dx}(2) & = \lim_{x\rightarrow 2} = \frac{g(x) - g(2)}{x - 2} = \lim_{x\rightarrow 2}\frac{\displaystyle\frac{f(2x)}{x} - \frac{f(4)}{2}}{x - 2} = \lim_{x\rightarrow 2}\frac{2f(2x)-xf(4)}{2x(x-2)}\\
& = \lim_{u\rightarrow 4}\frac{2f(u) - u\displaystyle\frac{f(4)}{2}}{u\left(\displaystyle\frac{u}{2}-2\right)} = \lim_{u\rightarrow 4}\frac{4f(u)-uf(4)}{u(u-4)} = \lim_{u\rightarrow 4}\frac{4f(u)-uf(4)}{u^2 - 4u}\\
& \overset{\mathrm{L'H}}{=}\lim_{u\rightarrow 4}\frac{4f'(u)-f(4)}{2u-4} = \frac{4f'(4)-f(4)}{4}
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2351494",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 3
} |
Prove that number of non-isomorphic ordered tree with 'n' vertices is nth catalan number. according to wikipedia $C$n is the number of non-isomorphic ordered trees with n vertices. But I can't seem to be able to prove this result. How do we do that?
where $n$th catalan number is:
$$
C_n = \frac 1{n+1} \binom{2n}{n}
$$
| We have from basic principles for the species of ordered trees the
species equation
$$\mathcal{T} = \mathcal{Z} +
\mathcal{Z} \mathfrak{S}_{\ge 1}(\mathcal{T}).$$
This yields the functional equation for the generating function $T(z)$
$$T(z) = z + z \frac{T(z)}{1-T(z)}$$
which is
$$T(z) (1-T(z)) = z (1-T(z)) + z T(z) = z.$$
We claim that
$$[z^n] T(z) = \left.\frac{1}{m+1} {2m\choose m}\right|_{m=n-1}
= \frac{1}{n} {2n-2\choose n-1}.$$
The shift in index of the Catalan numbers represents the species
$\mathcal{T}$ which has one ordered tree on two nodes and not two
(root node with one child node).
We then have
$$[z^n] T(z)
= \frac{1}{2\pi i} \int_{|z|=\epsilon}
\frac{1}{z^{n+1}} T(z) \; dz.$$
Put $w = T(z)$ so that $w (1-w) = z$ and $dz = (1-2w) \; dw$
to get
$$\frac{1}{2\pi i} \int_{|w|=\gamma}
\frac{1}{w^{n+1} (1-w)^{n+1}} w (1-2w) \; dw.$$
This yields
$$[w^{n-1}] \frac{1}{(1-w)^{n+1}}
- 2 [w^{n-2}] \frac{1}{(1-w)^{n+1}}
\\ = {n-1+n\choose n} - 2 {n-2+n\choose n}
\\ = {2n-1\choose n} - 2 {2n-2\choose n}
= \left(\frac{2n-1}{n}-2\frac{n-1}{n}\right) {2n-2\choose n-1}
\\ = \frac{1}{n} {2n-2\choose n-1}.$$
Remark. If an alternate approach is desired
we can use formal power series on
$$T(z) = \frac{1-\sqrt{1-4z}}{2}$$
where we have solved the functional equation taking the branch that
has $T_0 = 0$ as in the given problem. Coefficient extraction then
yields
$$-\frac{1}{2} (-1)^n 4^n {1/2\choose n}
= - 2^{2n-1} \frac{(-1)^n}{n!} \prod_{q=0}^{n-1} (1/2-q)
\\ = - 2^{n-1} \frac{(-1)^n}{n!} \prod_{q=0}^{n-1} (1-2q)
= - 2^{n-1} \frac{(-1)^n}{n!} \prod_{q=1}^{n-1} (1-2q)
\\ = 2^{n-1} \frac{1}{n!} \prod_{q=1}^{n-1} (2q-1)
\\ = 2^{n-1} \frac{1}{n!} \frac{(2n-2)!}{(n-1)! 2^{n-1}}
= \frac{1}{n} {2n-2\choose n-1}.$$
Returning to the complex variables approach and taking the branch of
the logarithm with the branch cut on the negative real axis we obtain
that $T(z)$ is analytic in a neighborhood of the origin (branch point
$z=1/4$ and branch cut $[1/4,\infty)$) with series expansion
$z+z^2+\cdots$ so that the image of $|z|=\epsilon$ is a closed
near-circle (modulus of the first term dominates) that may be deformed
to a circle $|w|=\gamma.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2351676",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Calculate Derivative of a map
Consider the maps from $R^2 \to R^2$ such that $F(u, v) = (e^{u + v}, e^{u - v})$ and $G(x, y) = (xy, x^2 - y^2)$. Calculate $D(F \circ G)(1, 1)$ by directly composing.
I got
$F \circ G = (e^{x^2 +xy - y^2}, e^{y^2 + xy - x^2})$
But how do I get the derivative matrix?
| Remember that the derivative of a vector field is its Jacobian. If $F:\mathbb{R}^2\to\mathbb{R}^2$ is differentiable function with $F(x,y)=(f_1(x,y),f_2(x,y))$ where both $f_i$ are differentiable, then we have: $$D_{(x,y)}F = \left(
\begin{array}{cc}
\frac{\partial}{\partial x}f_1(x,y) & \frac{\partial}{\partial y}f_1(x,y) \\
\frac{\partial}{\partial x}f_2(x,y) & \frac{\partial}{\partial y}f_2(x,y) \\
\end{array}
\right)$$
For this problem, to calculate $D(F\circ G)$ you can either use the chain rule or calculate $F\circ G$ and then calculating the Jacobian directly like you are asking.
Solution
There are two ways to calculate the Jacobian for $F\circ G$.
First by calculating directly $D(F\circ G)$ which is:$$D(F\circ G) = D(e^{x^2 + xy - y^2},e^{y^2 + xy - x^2}) = \left(
\begin{array}{cc}
e^{x^2+y x-y^2} (2 x+y) & e^{x^2+y x-y^2} (x-2 y) \\
e^{-x^2+y x+y^2} (-2x+y) & e^{-x^2+y x+y^2} (x+2 y) \\
\end{array}
\right)$$
And we can also apply the chain rule by noticing that $D(F\circ G)=D(F(G))D(G) $ where $D(F(G))$ is $D(F)$ evaluated at $G$. We have:
$$D(F)= \left(
\begin{array}{cc}
e^{x+y} & e^{x+y} \\
e^{x-y} & -e^{x-y} \\
\end{array}
\right)\text{, } D(G) = \left(
\begin{array}{cc}
y & x \\
2 x & -2 y \\
\end{array}
\right)\text{ and }$$
$$D(F(G))= \left(
\begin{array}{cc}
e^{x^2+y x-y^2} & e^{x^2+y x-y^2} \\
e^{-x^2+y x+y^2} & -e^{-x^2+y x+y^2} \\
\end{array}
\right)$$
We multiply $D(F(G))D(G) $ to obtain:
$$\left(
\begin{array}{cc}
e^{x^2+y x-y^2} & e^{x^2+y x-y^2} \\
e^{-x^2+y x+y^2} & -e^{-x^2+y x+y^2} \\
\end{array}
\right)\left(
\begin{array}{cc}
y & x \\
2 x & -2 y \\
\end{array}
\right) = \left(
\begin{array}{cc}
e^{x^2+y x-y^2} (y+2x) & e^{x^2+y x-y^2} (x-2 y) \\
e^{-x^2+y x+y^2} (y-2 x) & e^{-x^2+y x+y^2} (x+2 y) \\
\end{array}
\right)$$
All that is left is to evalute $(x,y) = (1,1)$ which is:
$$D_{(1,1)}(F\circ G) = \left(
\begin{array}{cc}
3 e & -e \\
-e & 3 e \\
\end{array}
\right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2352064",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
If $a + \frac{1}{a} = -1$, then the value of $(1-a+a^2)(1+a-a^2)$ is?
If $a + \frac{1}{a} = -1$ then the value of $(1-a+a^2)(1+a-a^2)$ is?
Ans. 4
What I have tried:
\begin{align}
a + \frac{1}{a} &= -1 \\
\implies a^2 + 1 &= -a \tag 1 \\
\end{align}
which means
\begin{align}
(1-a+a^2)(1+a-a^2) &=(-2a)(-2a^2) \\
&=4a^{3}
\end{align}
as $1 + a^{2} = -a$ and $1 + a = -a^{2}$ from $(1)$.
| Without solving the quadratic:
$$ a^3 = - a^2 - a = a + 1 - a = 1 $$
which was found by using the equation $a^2 + a + 1 = 0$ twice. This means that $a$ is a non-real cube root of unity.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2353620",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
} |
If $ a=\frac{\sqrt{x+2} + \sqrt {x-2}}{\sqrt{x+2} -\sqrt{x-2}}$ then what is the value of $a^2-ax$ is equal to
If $$ a=\frac{\sqrt{x+2} + \sqrt {x-2}}{\sqrt{x+2} -\sqrt{x-2}}$$ then
the value of $a^2-ax$ is equal to:
a)2 b)1 c)0 d)-1
Ans. (d)
My attempt:
Rationalizing $a$ we get,
$ x+ \sqrt {x^2-4}$
$a^2=(x+\sqrt{x^2-4)^2}=2x^2-4+2x\sqrt{x^2-4}$
Now,
$a^2-ax=2x^2-4+2x\sqrt{x^2-4}-x^2-x\sqrt{x^2-4}=x^2+x\sqrt{x^2-4}-4=xa-4$
Why am I not getting the intended value?
| $$\dfrac a1=\dfrac{\sqrt{x+2}+\sqrt{x-2}}{\sqrt{x+2}-\sqrt{x-2}}$$
calling for Componendo and Dividendo
$$\dfrac{a+1}{a-1}=\dfrac{\sqrt{x+2}}{\sqrt{x-2}}$$
Squaring we get $$\dfrac{a^2+1+2a}{a^2+1-2a}=\dfrac{x+2}{x-2}$$
Again apply componendo and dividendo, $$\dfrac{a^2+1}{2a}=\dfrac x2$$
Now simplify
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2356126",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 7,
"answer_id": 0
} |
What is the number of triples (a, b, c) of positive integers such that the product $a.b.c=1000$ , and $a \leq b \leq c $?
What is the number of triples (a, b, c) of positive integers such that the product $a.b.c=1000$, and $a \leq b \leq c$?
My try:
The prime factorization of $1000$ is $2^3\cdot 5^3$
$a\cdot b \cdot c = 2^3\cdot 5^3$
$a=2^{a_1}\cdot 5^{a_2}$
$b=2^{b_1}\cdot 5^{b_2}$
$c=2^{c_1}\cdot 5^{c_2}$
$abc=2^{a_1+b_1+c_1}\cdot 5^{a_2+b_2+c_2}=2^3\cdot 5^3 $
$a_1+b_1+c_1=3$
How many ways are there such that $a_1+b_1+c_1=3$
Star's and Bar's method: -
Number of ways to chose $2$ separators($0s$) in a string of $5 $$ = {5\choose 2 }=10$
$N(a_1+b_1+c_1=3)=10$
Similarly, $N(a_2+b_2+c_2=3)=10$
$N(abc=1000)=10\cdot 10=\boxed{100}$
Is that okay ? Please write down any notes
| Your computation of $N=10$ is correct and $100$ is the number of ordered triples that have product $1000$. You have failed to account for the condition that $a \le b \le c$. All of the unordered triples that have three distinct elements have shown up six times, so you have overcounted. Those that have two or three equal elements have been counted differently. Keep going.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2356648",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
proof that $n^4+22n^3+71n^2+218n+384$ is divisible by $24$ Let $n$ be a positive integer number.
How do we show that the irreducible polynomial $n^4+22n^3+71n^2+218n+384$ is divisible by $24$ for all $n$?
| Another way to do it is to note:
$n^4+22n^3+71n^2+218n+384 \equiv $
$n^4 - 2n^3 - n^2 + 2n \mod 8$.
If $n \equiv 0, \pm 1, \pm 2, \pm3, 4 \mod 8$ then $n^4 - 2n^3 - n^2 + 2n\equiv$
$0, 1\mp 2 - 1 \pm 1, 16 \mp 16 -4 \pm 4, 81 \mp 54 - 9 \pm 6, 4^4 \mp 2*4^3 - 16 + 8 \equiv$
$0, 0, 72 \mp 48, 0 \equiv 0 \mod 8$.
So $8$ divides $n^4+22n^3+71n^2+218n+384$
and $n^4+22n^3+71n^2+218n+384 \equiv $
$n^4 + n^3 - n^2 -n \mod 3$
If $n \equiv 0, \pm 1 \mod 3$ then $n^4 + n^3 - n^2 - n \equiv 0, 1 \pm 1 -1 \mp 1 \equiv 0 \mod 3$.
So $3$ divides $n^4+22n^3+71n^2+218n+384$.
So $3*8 =24$ divides $n^4+22n^3+71n^2+218n+384$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2356982",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
How to calculate $\lim_{x\to 0^+} \frac{x^x- (\sin x)^x}{x^3}$ As I asked, I don't know how to deal with $x^x- (\sin x)^x$.
Please give me a hint. Thanks!
| $$\lim_{x\rightarrow0^+}\frac{\ln{x}-\ln{\sin{x}}}{x^2}=\lim_{x\rightarrow0^+}\left(\frac{\ln\left(1+\frac{x}{\sin{x}}-1\right)}{\frac{x}{\sin{x}}-1}\cdot\frac{\frac{x}{\sin{x}}-1}{x^2}\right)=$$
$$=\lim_{x\rightarrow0}\left(\frac{x-\sin{x}}{x^3}\cdot\frac{x}{\sin{x}}\right)=\lim_{x\rightarrow0}\frac{1-\cos{x}}{3x^2}=\frac{1}{6}\lim_{x\rightarrow0}\frac{\sin^2\frac{x}{2}}{\frac{x^2}{4}}=\frac{1}{6}$$
and since
$$\frac{x^x-(\sin{x})^x}{x^3}=\frac{1+x\ln{x}+\frac{x^2\ln^2x}{2!}+...-1-x\ln\sin{x}-\frac{x^2\ln^2\sin{x}}{2!}-...}{x^3}=$$
$$=\frac{\ln{x}-\ln{\sin{x}}}{x^2}+O(x^2),$$
we obtain that our limit is $\frac{1}{6}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2358173",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
How does the trigonometric identity $1 + \cot^2\theta = \csc^2\theta$ derive from the identity $\sin^2\theta + \cos^2\theta = 1$? I would like to understand how would the original identity of $$ \sin^2 \theta + \cos^2 \theta = 1$$ derives into
$$ 1 + \cot^2 \theta = \csc^2 \theta $$
This is my working:
a) $$ \frac{\sin^2 \theta}{ \sin^2 \theta } + \frac{\cos^2 \theta }{ \sin^2 \theta }= \frac 1 { \sin^2 \theta } $$
b) $$1 + \cot^2 \theta = \csc^2 \theta
$$
How did the $ \tan^2 \theta + 1 = \sec^2 \theta$ comes into the picture?
| The same way; you start with $\sin^2\theta + \cos^2\theta = 1$ and divide both sides by $\cos^2\theta$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2359280",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
Find the number of points of differentiability for the following function. If
$$f(x)=\begin{cases}
\cos x^3&;x\lt0\\
\sin x^3 - |x^3-1|&;x\ge0
\end{cases}$$
then find the number of points where $g(x)=f(|x|) \text { is non differentiable.}$
| The question is asking about $f(|x|)$, by symmetry, since $g(x)$ is not differentiable at $x=1$, it is not differentiable at $x=-1$ as well.
To show that it is not differentiable at $x=1$:
If it is differentiable at $x=1$, then the following limit exists.
\begin{align}\lim_{x \rightarrow 1} \frac{f(x)-f(1)}{x-1}&= \lim_{x \rightarrow 1} \frac{\sin x^3 - \sin 1-|x^3-1|}{x-1}\end{align}
since $\sin x^3$ is differentiable everywhre, the limit exists if and only if the following limit exists.
$$\lim_{x \rightarrow 1} \frac{|x^3-1|}{x-1}=\lim_{x \rightarrow 1} sign(x-1)|x^2+x+1|=\lim_{x \rightarrow 1}3sign(x-1)$$
Since $\lim_{x \rightarrow 1^+} sign(x-1) = 1 \neq -1 = \lim_{x \rightarrow 1^-} sign(x-1)$
The function is not differentiable at $x=1$.
Also, we have to verify that the function is differentiable at $x=0$.
$$f(|x|) = \begin{cases} \sin x^3-|x^3-1| & x \geq 0 \\ -\sin x^3 -|x^3+1| & x<0\end{cases}$$
When $|x|<1$,
$$f(|x|) = \begin{cases} \sin x^3-1+x^3 & 0 < x < 1 \\ -\sin x^3 -x^3-1 & -1<x<0\end{cases}$$
$$\lim_{x \rightarrow 0}\frac{f(|x|)-f(0)}{x}=\lim_{x\rightarrow 0}\frac{sign(x)(\sin x^3+x^3)}{x}=0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2360233",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Lengthy integration problem $$ \int\frac{x^3+3x+2}{(x^2+1)^2(x+1)} \, dx$$
I managed to solve the problem using partial fraction decomposition.
But that approach is pretty long as it creates five variables. Is there any other shorter method to solve this problem(other than partial fractions)?
I also tried trigonometric substitution and creating the derivative of the denominator in the numerator... but it becomes even longer. Thanks in advance!!
Just to clarify: (My partial fraction decomposition)
$$\frac{x^3+3x+2}{(x^2+1)^2(x+1)}= \frac{Ax+B}{x^2+1} + \frac{Cx+D}{(x^2+1)^2} + \frac{E}{x+1}$$
| There are much better ways to find the coefficients in partial fractions than solving five equations in five variables.
Writing your function as
$$ \frac{x^3 + 3 x + 2}{(x^2+1)^2 (x+1)} = \frac{Q(x)}{(x^2+1)^2} + \frac{E}{x+1} $$
multiply both sides by $x+1$ and substitute $x=-1$. We get
$$ \frac{-2}{2^2} = 0 + E$$
so $E = -1/2$, and
$$ \eqalign{\frac{Q(x)}{(x^2+1)^2} &= \frac{x^3 + 3 x + 2}{(x^2+1)^2 (x+1)} + \frac{1/2}{x+1}\cr &= \frac{x^2 + 3 x + 2 + (1/2)(x^2+1)^2)}{(x^2+1)^2(x+1)}\cr
&= \frac{x^3 + x^2 + x + 5}{2(x^2 + 1)^2} }$$
Now you want this in the form
$$\frac{Ax+B}{x^2+1} + \frac{Cx+D}{(x^2+1)^2} $$
Multiply by $(x^2 + 1)^2$ and substitute $x=i$ (yes, $\sqrt{-1}$).
We get $$ 2 = C i + D $$
But since we want $C$ and $D$ to be real, we must have $C = 0$, $D = 2$. This leaves
$$ \eqalign{\frac{Ax+B}{x^2+1} &= \frac{x^3 + x^2 + x + 5}{2(x^2 + 1)^2} - \frac{1}{(x^2+1)^2} \cr
&= \frac{x + 1}{2(x^2 + 1)}} $$
and we're done: $A = 1/2$, $B = 1/2$, $C = 0$, $D = 2$, $E = -1/2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2363482",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
Find number of real solutions of $3x^5+2x^4+x^3+2x^2-x-2=0$
Find a number of real roots of $$f(x)=3x^5+2x^4+x^3+2x^2-x-2$$
I tried using differentiation:
$$f'(x)=15x^4+8x^3+3x^2+4x-1=0$$ and I found number of real roots of $f'(x)=0$ by drawing graphs of $g(x)=-15x^4$ and $h(x)=8x^3+3x^2+4x-1$ and obviously from graph there are two real solutions as $h(x)$ is an increasing function.
But now how to proceed?
| $f'(x)=15x^4+8x^3+3x^2+4x-1>0$ for $x>\frac{1}{2}$ and $f\left(\frac{1}{2}\right)<0$.
Hence, since $\lim\limits_{x\rightarrow+\infty}f(x)=+\infty$, we see that there is unique root for $x>\frac{1}{2}$.
Now, prove that $f(x)<0$ for all $x\leq\frac{1}{2}$.
For example, for $0\leq x\leq\frac{1}{2}$ we have
$$3x^5+2x^4+x^3+2x^2-x-2=$$
$$=\left(3x^5-\frac{3}{4}x^3\right)+\left(2x^4-x^3\right)+(2x^2-x)+\left(\frac{11}{4}x^3-2\right)<0.$$
For $x<0$ we can replace $x$ at $-x$ and it's enough to prove that
$$-3x^5+2x^4-x^3+2x^2+x-2<0$$ for all $x>0$ or
$$3x^5-2x^4+x^3-2x^2-x+2>0$$ or
$$3x^5-3x^4-3x^3+3x^2+x^4+4x^3-5x^2-x+2>0$$ or
$$3x^2(x+1)(x-1)^2+x^4-2x^2+1+4x^3-3x^2-x+1>0$$ or
$$3x^2(x+1)(x-1)^2+(x^2-1)^2+4x^3-3x^2-x+1>0$$ or
$$3x^2(x+1)(x-1)^2+(x^2-1)^2+(x-0.5)^2+4x^3-4x^2+0.75>0$$ and it remains to prove that
$$4x^3-4x^2+0.75>0,$$
which is AM-GM:
$$4x^3-4x^2+0.75=2x^3+2x^3+0.75-4x^2\geq3\sqrt[3]{(2x^3)^2\cdot0.75}-4x^2=\left(3\sqrt[3]3-4\right)x^2>0,$$
which gives that our equation has one real root.
Done!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2363898",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
} |
Evaluate $\lim_{ x\to \infty} \left( \tan^{-1}\left(\frac{1+x}{4+x}\right)-\frac{\pi}{4}\right)x$
Evaluate
$$\lim_{ x\to \infty} \left( \tan^{-1}\left(\frac{1+x}{4+x}\right)-\frac{\pi}{4}\right)x$$
I assumed $x=\frac{1}{y}$ we get
$$L=\lim_{y \to 0}\frac{\left( \tan^{-1}\left(\frac{1+y}{1+4y}\right)-\frac{\pi}{4}\right)}{y}$$
using L'Hopital's rule we get
$$L=\lim_{y \to 0} \frac{1}{1+\left(\frac{1+y}{1+4y}\right)^2} \times \frac{-3}{(1+4y)^2}$$
$$L=\lim_{y \to 0}\frac{-3}{(1+y)^2+(1+4y)^2}=\frac{-3}{2}$$
is this possible to do without Lhopita's rule
| Well, we can start as you, by setting $y=\frac{1}{x}$. Now, our limits transforms to:
$$L=\lim_{y\to0}\frac{\tan^{-1}\left(\frac{1+y}{1+4y}\right)-\frac{\pi}{4}}{y}$$
Now, let $f:\mathbb{R}\to\mathbb{R}$ with
$$f(x)=\tan\left(\frac{1+x}{1+4x}\right)$$
Note that $f(0)=\tan^{-1}(1)=\frac{\pi}{4}$. So, we have:
$$L=\lim_{x\to0}\frac{f(x)-f(0)}{x-0}=f'(0)$$
since $f$ is differentiable.
Now, $f$'s formula is:
$$f(x)=\int_0^\frac{1+x}{1+4x}\frac{1}{1+t^2}dt$$
So, we have:
$$f'(x)=\frac{1}{1+\left(\frac{1+x}{1+4x}\right)^2}\left(\frac{1+x}{1+4x}\right)'=\frac{(1+4x)^2}{1+(1+x)^2}\frac{-3}{(1+4x)^2}=\frac{-3(1+4x)^2}{(1+(1+x)^2)(1+4x)^2}$$
So
$$L=f'(0)=-\frac{3}{2}$$
As you have already proved.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2365914",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 8,
"answer_id": 3
} |
Unable to reach the desired answer in trigonometry. The question is:
If $\sin x + \sin y = \sqrt3 (\cos y - \cos x)$
show that $\sin 3x + \sin 3y= 0 $
This is what I have tried:
*
*Squaring of the first equation (Result: Failure)
*Tried to use the $\sin(3x)$ identity but got stuck in the middle steps because I couldn't simplify it any further.
Can someone provide any hint/ suggestion?
| I was trying to find out how the condition was conceived.
$$\sin3x+\sin3y=0\implies\sin3x=\sin(-3y)$$
$\implies3x=180^\circ n+(-1)^n(-3y)$ where $n$ is any integer
$\iff x=60^\circ n+(-1)^{n+1}y$
If $n$ is even $=2m$(say), $x=120^\circ m-y$
$$\implies x+y\equiv\begin{cases}0 &\mbox{if }3\mid m\\120^\circ& \mbox{if } n \equiv1\pmod3\\240^\circ& \mbox{if } n\equiv2\pmod3 \end{cases}\pmod{360^\circ}$$
$\implies\sin\dfrac{x+y}2=\tan\dfrac{x+y}2=0$ or $\tan\dfrac{x+y}2=\pm\sqrt3$
Similarly for odd $n=2m+1$(say),
$\cos\dfrac{x-y}2=\cot\dfrac{x-y}2=0$ or $\cot\dfrac{x-y}2=\pm\sqrt3$
Here the condition chosen $$0=\sin\dfrac{x+y}2\left(\cot\dfrac{x-y}2-\sqrt3\right)=\dfrac{2\sin\dfrac{x+y}2\cos\dfrac{x-y}2-\sqrt3\cdot2\sin\dfrac{x+y}2\sin\dfrac{x-y}2}{2\sin\dfrac{x-y}2}=\dfrac{\sin x+\sin y-\sqrt3(\cos y-\cos x)}{2\sin\dfrac{x-y}2}$$ with $\sin\dfrac{x-y}2\ne0$ as $\cot\dfrac{x-y}2=\sqrt3$
which could easily be $$\sin\dfrac{x+y}2\left(\cot\dfrac{x-y}2+\sqrt3\right)=0\iff\sin x+\sin y=-\sqrt3(\cos y-\cos x)$$
Or $$\cos\dfrac{x-y}2\left(\tan\dfrac{x+y}2\pm\sqrt3\right)=0\iff\sin x+\sin y=\pm\sqrt3(\cos y+\cos x)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2368879",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Symbolic Notation for $\theta "=" \arcsin(-.5)$? I'm teaching PreCalculus and the following issue has always bugged me.
Problem: Solve $\sin\theta = -.5$ for $0 \le \theta \le 2\pi$.
Solution:
\begin{align*}
\sin\theta &= -.5\\
\theta &= \arcsin(-.5) = -\frac\pi6
\end{align*}
But to get this into our desired domain, our solutions are $\boxed{\theta = \frac{7\pi}6 \text{ or }\frac{11\pi}6}$.
So my objection is the line $\theta = \arcsin(-.5)$, because that's really not true. $\theta$ can be a whole lot of things! So does there exist some symbol or notation that expresses this? Something a-la "If $x^2=9$, then $x = \pm 3$." Like
$$\theta \stackrel{\text{is related to}}{\sim} \arcsin(-.5) = -\frac\pi6 ?$$
| The way I always explained it to my students was "First, you find a solution, then you find the solution."
Let $\theta_0 = \arcsin(-\frac 12)$.
The range of $\arcsin$ is $-\frac{\pi}{2} \le \theta_0 \le \frac{\pi}{2}$. Since, on the unit circle, $\sin \theta = y=-\frac 12$, we see quickly that
$\theta_0 = -\frac{\pi}{6}$. The two red dots indicate the two points on the unit circle for which $\sin \theta = -\frac 12$. One way to express this is
$\theta \in \{ -\frac{\pi}{6} + 2n\pi : n \in \mathbb Z\} \cup
\{ \frac{7\pi}{6} + 2n\pi : n \in \mathbb Z\}$
To achieve $0 \le \theta \le 2\pi$ we find that
$\theta \in \{\frac{7\pi}{6}, \frac{11\pi}{6} \}$
where $ \frac{11\pi}{6} = -\frac{\pi}{6} + 2\pi$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2370728",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
$x^5 + y^2 = z^3$ While waiting for my döner at lunch the other day, I noticed my order number was $343 = 7^3$ (surely not the total for that day), which reminded me of how $3^5 = 243$, so that $$7^3 = 3^5 + 100 = 3^5 + 10^2.$$
Naturally, I started wondering about nontrivial integer solutions to $$x^5 + y^2 = z^3 \tag{*}$$
("nontrivial" meaning $xyz \ne 0$). I did not make much progress, though apparently there are infinitely many solutions: this was Problem 1 on the 1991 Canadian Mathematical Olympiad. The official solutions (at the bottom of this page) only go back to 1994. A cheap answer is given by taking $x = 2^{2k}$ and $y = 2^{5k}$ so that the l.h.s. is $2^{10k + 1}$. This is a cube iff $10k + 1 \equiv 0 \,(3)$ i.e. $k \equiv 2\,(3)$ thus giving an arithmetic progression's worth of solutions, starting with $$(x, y, z) = (16, 1024, 128)$$ corresponding to $k = 2$
and $$(x, y, z) = (1024, 33554432, 131072)$$
coming from $k = 5$.
What else is known about the equation $(*)$? In particular, are there infinitely many solutions with $x$, $y$, $z$ relatively prime? The one that caught my attention was $(x, y, z) = (3, 10, 7)$. Another one is $(-1, 3, 2)$ because $-1 + 9 = 8$. By Catalan's conjecture (now a theorem), this is the only solution with $x = \pm 1$ or $y = \pm 1$ or $z = 1$. Are there any solutions with $z = -1$? In this case, $(*)$ reduces to $x^5 + y^2 = -1$ and Mihăilescu's theorem does not apply.
Update. This question was essentially already asked here, since the equation $a^2 + b^3 = c^5$ is equivalent to $(-c)^5 + a^2 = (-b)^3$.
| There is a beautiful connection between $a^5+b^3=c^2$ and the icosahedron. Consider the unscaled icosahedral equation,
$$\color{blue}{12^3u v(u^2 + 11 u v - v^2)^5}+(u^4 - 228 u^3 v + 494 u^2 v^2 + 228 u v^3 + v^4)^3 = (u^6 + 522 u^5 v - 10005 u^4 v^2 - 10005 u^2 v^4 - 522 u v^5 + v^6)^2\tag1$$
By scaling $u=12x^5$ and $v=12y^5$ (or various combinations thereof like $u=12^2x^5$, etc), we then get a relation of form,
$$12^5a^5+b^3=c^2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2373028",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 1,
"answer_id": 0
} |
Minimum of the given expression
For all real numbers $a$ and $b$ find the minimum of the following expression.
$$(a-b)^2 + (2-a-b)^2 + (2a-3b)^2$$
I tried expressing the entire expression in terms of a single function of $a$ and $b$. For example, if the entire expression reduces to $(a-2b)^2+(a-2b)+5$ then its minimum can be easily found. But nothing seems to get this expression in such a form, because of the third unsymmetric square.
Since there are two variables here we can also not use differentiation.
Can you please provide hints on how to solve this?
| Let $a=\frac{17}{15}$ and $b=\frac{4}{5}$.
Hence, we get a value $\frac{2}{15}$.
Thus, it remains to prove that
$$(a-b)^2 + (2-a-b)^2 + (2a-3b)^2\geq\frac{2}{15}$$ or
$$10(3a-3b-1)^2+3(5b-4)^2\geq0$$
Done!
I got my solution by the following way.
We need to find a maximal $k$ for which the following inequality is true for all reals $a$, $b$ and $c$.
$$(a-b)^2 + (2-a-b)^2 + (2a-3b)^2\geq k$$ or
$$6a^2-4(3b+1)a+11b^2-4b+4-k\geq0,$$ for which we need
$$4(3b+1)^2-6(11b^2-4b+4-k)\leq0$$ or
$$15b^2-24b+10-3k\geq0,$$
for which we need $$12^2-15(10-3k)\leq0$$ or
$$k\leq\frac{2}{15}.$$
The equality occurs for $k=\frac{2}{15}$, $b=\frac{24}{2\cdot15}$, which is $b=\frac{4}{5}$ and for these values we obtain
$$(a-b)^2 + (2-a-b)^2 + (2a-3b)^2\geq \frac{2}{15}$$ it's
$$6a^2-4(3b+1)a+11b^2-4b+4-\frac{2}{15}\geq0$$ or
$$90a^2-60(3b+1)a+165b^2-60b+58\geq0$$ or
$$10(9a^2-6(3b+1)a+(3b+1)^2)-10(3b+1)^2+165b^2-60b+58\geq0$$ or
$$10(3a-3b-1)^2+75b^2-120b+48\geq0$$ or
$$10(3a-3b-1)^2+3(5a-4)^2\geq0.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2374768",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 0
} |
how to prove $\frac{b^2}{b_1^2}=\frac{ac}{a_{1c_1}}$? If the Ratio of the roots of $ax^2+bx+c=0$ be equal to the ratio of the roots of $a_1x^2+b_1x+c_1=0$, then how one prove that $\frac{b^2}{b^2_1}=\frac{ac}{a_1 c_1}$?
| Hint :
let $\alpha$ and $\beta$ be the roots of $ax^2+bx+c=0$ & let $\gamma$ and $\delta$ be the roots of $a_1 x^2+b_1 x+c_1 =0$.
The ratio of their roots are equal if
\begin{eqnarray*}
\frac{\alpha}{\beta} = \frac{\gamma}{\delta}.
\end{eqnarray*}
Further hint : $\color{red}{\alpha+\beta=-\frac{b}{a}}$ & $\alpha \beta=\frac{c}{a}$
\begin{eqnarray*}
\frac{b^2}{ac} = \color{red}{\frac{b^2}{a^2}} \frac{a}{c} = \frac{\color{red}{(\alpha+\beta)^2}}{\alpha \beta}=\frac{\alpha}{\beta}+2+\frac{\beta}{\alpha} = \cdots
\end{eqnarray*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2375736",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Prove that the equation $x^3+y^3+z^3-(x^2z+y^2x+z^2y)=2$ has no solution in natural numbers I asked myself which primes $p$ can be written as $p=x^3+y^3+z^3-(x^2z+y^2x+z^2y)$ with $x,y,z \in \mathbb{N}$.
But for $p \neq 2$ we have the solution $x=y=\frac{p-1}{2}$ and $z=\frac{p+1}{2}$. So the only prime for which I can not find a solution is $p=2$. But I can not prove that there is not a solution.
Any ideas?
| EDIT: You're right stackExchangeUser; my proof doesn't work. With a similar tack, we can still salvage this:
\begin{align*}
&x^3 + y^3 + z^3 - (x^2 z + y^2 x + z^2 y) \\
= ~ &(x + y + z)^3 - 4(x^2 z + y^2 x + z^2 y) - 3(x^2y + y^2 z + z^2 x) - 6xyz
\end{align*}
So, we are solving,
$$(x + y + z)^3 = 2 + 4(x^2 z + y^2 x + z^2 y) + 3(x^2y + y^2 z + z^2 x) + 6xyz$$
Suppose first the left side is divisible by $2$. Again, at least one of $x, y, z$ must be even. If all of them are even, we see that the left side is $0$ mod $8$, but the right side is $2$ mod $8$. Thus, exactly one must be even. But then, the $3(x^2y + y^2 z + z^2 x)$ term is odd, which makes the right hand side odd, and we get a contradiction again.
Thus, the left side is odd. Similarly, either all of $x, y, z$ are odd, or exactly one is. If exactly one of them is odd, then the right hand side is even, hence all $x, y, z$ are odd.
Finally, considering the original formulation, and the fact that $x^2 \equiv 1$ mod $8$ for all odd $x$, we get,
$$x^3 + y^3 + z^3 - (x^2 z + y^2 x + z^2 y) \equiv x + y + z - (z + x + y) \equiv 0$$
mod $8$, which cannot equal $2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2375828",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 1,
"answer_id": 0
} |
Finding the minimum value of $\cot^2A + \cot^2B+ \cot^2C$ where $A$, $B$ and $C$ are angles of a triangle. The question is:
If $A+B+C= \pi$, where $A>0$, $B>0$, $C>0$, then find the minimum value of $$\cot^2A+\cot^2B +\cot^2C.$$
My solution:
$(\cot A + \cot B + \cot C)^2\ge0$ // square of a real number
$\implies \cot^2A +\cot^2B + \cot^2 +2 \ge0 $ //Conditional identity used: $\cot A \cot B + \cot B \cot C + \cot A \cot C =1$
$\implies \cot^2A +\cot^2B + \cot^2 C \ge -2$
Thus according to me the answer should be $-2$. However, the answer key states that the answer is $1$. Where have I gone wrong?
| it is equivalent to
$$\frac{1}{\sin(A)^2}+\frac{1}{\sin(B)^2}+\frac{1}{\sin(C)^2}\geq 4$$
with $$\sin(A)=\frac{a}{2R}$$ etc and $$S=\sqrt{s(s-a)(s-b)(s-c)}$$ and $$S=\frac{abc}{4R}$$ we get
$$b^2c^2+c^2a^2+a^2b^2-(-a+b+c)(a-b+c)(a+b-c)(a+b+c)\geq 0$$
and this is equivalent to
$$a^4+b^4+c^4\geq a^2b^2+b^2c^2+c^2a^2$$
which is true.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2376094",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Recurrence relation $a_n = 11a_{n-1} - 40a_{n-2} + 48a_{n-3} + n2^n$ I didn't do a lot of maths in my career, and we asked me to solve the following recurrence relation:
$$a_{n} = 11a_{n-1} - 40a_{n-2} + 48a_{n-3} + n2^n$$
with
$a_0 = 2$, $a_1 = 3$ and $a_2 = 1$
What is the procedure to solve such relation? So far, I just know that $a_n$ could be split into two difference recurrence relations (homogeneous and non-homogeneous) as $a_n = b_n + c_n$, where
$$b_n = 11b_{n-1} - 40b_{n-2} + 48b_{n-3}$$
and
$$c_n = n2^n$$
| Just to offer another approach, generating functions can be used as well:
$\begin{align}
G(x) &= \sum_{n=0}^{\infty} a_n x^n \\
G(x) &= 2x^0 + 3x^1 + x^2 + \sum_{n=3}^{\infty}(11a_{n-1} - 40a_{n-2} + 48a_{n-3} + n2^n)x^n \\
G(x) &= 2 + 3x + x^2 + 11\sum_{n=3}^{\infty}a_{n-1}x^n - 40\sum_{n=3}^{\infty}a_{n-2}x^n + 48\sum_{n=3}^{\infty}a_{n-3}x^n + \sum_{n=3}^{\infty}n2^nx^n \\
G(x) &= 2 + 3x + x^2 + 11x\sum_{n=2}^{\infty}a_{n}x^{n} - 40x^2\sum_{n=1}^{\infty}a_{n}x^{n} + 48x^3\sum_{n=0}^{\infty}a_{n}x^{n} + \sum_{n=3}^{\infty}n2^nx^n \\
G(x) &= 2 + 3x + x^2 + 11x(-a_{0}x^{0} -
a_{1}x^{1} + \sum_{n=0}^{\infty}a_{n}x^{n}) - 40x^2(-a_{0}x^{0} + \sum_{n=0}^{\infty}a_{n}x^{n}) + 48x^3\sum_{n=0}^{\infty}a_{n}x^{n} + (- 2x -2 \cdot 2^2x^2 + \sum_{n=0}^{\infty}n2^nx^n) \\
G(x) &= 2 + 3x + x^2 + 11x(-2 -
3x + G(x)) - 40x^2(-2 + G(x)) + 48x^3G(x) + (- 2x -2 \cdot 2^2x^2 + (2 x)/(2 x - 1)^2)
\end{align}$
Solve for $G(x)$:
$$G(x) = \frac{-160 x^4 + 244 x^3 - 132 x^2 + 27 x - 2}{(3 x - 1) (8 x^2 - 6 x + 1)^2}$$
Apply partial fraction decomposition:
$$G(x) = 49 \cdot \frac{1}{1 -3 x} - 38 \cdot \frac{1}{1 - 4 x} + 5 \cdot \frac{1}{(1 - 4 x)^2} - 12 \cdot \frac{1}{1 - 2 x} - 2 \cdot \frac{1}{(1 - 2 x)^2}$$
Take the $n$th coefficient of the resulting generating function, which will bring us to the final result:
$$a_n = 49 \cdot 3^n - 38 \cdot 4^n + 5 \cdot 4^n(n+1) - 12 \cdot 2^n - 2 \cdot 2^n (n+1)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2376294",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
Express a formula in terms of trigonometric expressions I am studying Kerr Black holes using Hobson's General relativity an introduction for physicists book.
In order to find circular radius for photons, two conditions need to be satisfied:
$$r_c=3\mu\frac{b-a}{b+a}$$ and
$$(b+a)^3=27\mu^2(b-a)$$
According to the book: The equations may be solved by setting y=a+b in the second condition and substituting the resulting value of b into the first. This is what I did and I obtained:
$$r_c=\frac{\mu\alpha^{1/3}(3\mu^2+3\alpha^{1/3}-2a)}{\mu^2+\alpha^{2/3}}$$
where for simplification I defined $\alpha=\sqrt{a^2\mu^4-\mu^6}-a\mu^2$.
However, the book further says that one can easily simplify $r_c$ as:
$$r_c=2\mu\Big(1+\cos\Big[\frac{2}{3}\cos^{-1}\Big(\pm\frac{a}{\mu}\Big)\Big]\Big)$$
and $$b=3\sqrt{\mu{r_c}}-a$$
I am stuck and don't know how is it possible to simplify my expression for $r_c$ into the elegant expression as given by the book. This seems to be a physics question but I am stuck with the algebraic manipulation.
| \begin{align}
r_c&=3\mu\frac{b-a}{b+a} \tag{1}\label{1}
\\
(b+a)^3&=27\mu^2(b-a) \tag{2}\label{2}
\end{align}
To get the expression for b from \eqref{1},
\begin{align}
b-a&=\frac{r_c}{3\mu}(b+a)
,
\end{align}
combined with \eqref{2},
\begin{align}
(b+a)^3&=27\mu^2\frac{r_c}{3\mu}(b+a)
,\\
(b+a)^2&=9\mu{r_c}
,\\
b&=3\sqrt{\mu{r_c}}-a
.
\end{align}
From \eqref{2}
\begin{align}
3\,\mu\,\frac{b-a}{b+a}
&=\frac{(b+a)^2}{9\,\mu}=r_c
\tag{3}\label{3}
.
\end{align}
Now we have two expressions for $r_c$.
One has a factor $\frac{\mu}{(b+a)}$,
the other has its reciprocal $\frac{(b+a)}{\mu}$,
and they both are begging to be canceled.
When we multiply them, we'll get a nice simplified
expression for $r_c^2$:
\begin{align}
r_c^2&=
\tfrac13\,(b+a)(b-a)
,\\
r_c^2&=\sqrt{\mu\,r_c}(3\,\sqrt{\mu\,r_c}-2\,a)
,\\
\left(\frac{r_c}{\mu}\right)^2
&=
\sqrt{\frac{r_c}{\mu}}
\left(
3\,\sqrt{\frac{r_c}{\mu}}-\frac{2\,a}{\mu}
\right)
,\\
\left(\sqrt{\frac{r_c}{\mu}}\right)^3
-
3\,\sqrt{\frac{r_c}{\mu}}
&=
-\frac{2\,a}{\mu}
,\\
4\,\left(\tfrac12\sqrt{\frac{r_c}{\mu}}\right)^3
-
3\,\left(\tfrac12\sqrt{\frac{r_c}{\mu}}\right)
&=
-\frac{a}{\mu}
\end{align}
Recall that
\begin{align}
4\,\cos^3 x-3\,\cos x=\cos3x.
\end{align}
So, we have
\begin{align}
\cos3x&=-\frac{a}\mu
;\\
3x&=\arccos\left(-\frac{a}\mu\right)+2\,\pi k,\quad k=0,1,2
;\\
x&=\tfrac13\arccos\left(-\frac{a}\mu\right)+\tfrac23\,\pi k,\quad k=0,1,2
;\\
\end{align}
Hence
\begin{align}
\cos x=
\tfrac12\sqrt{\frac{r_c}{\mu}}
&=
\cos\left(
\tfrac13\,\arccos\left( -\frac{a}{\mu} \right)
+\tfrac23\,\pi\,k
\right)
,\quad k=0,1,2
;\\
\tfrac12\frac{r_c}{\mu}
&=
2\,
\cos^2\left(
\tfrac13\,\arccos\left( -\frac{a}{\mu} \right)
+\tfrac23\,\pi\,k
\right)
,\quad k=0,1,2
;\\
r_c&=2\,\mu
\left(
1+\cos\left(
\tfrac23\,\arccos\left( -\frac{a}{\mu} \right)
+\tfrac43\,\pi\,k
\right)
\right)
,\quad k=0,1,2
.
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2376551",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Need help integrating $\int e^x(\frac{x+2}{x+4})^2 dx $ I have simplified the problem a bit using integration by parts, with $u = e^x(x+2)^2$ and $v = 1/(x+4)^2$ but I'm then stuck with how to integrate this:
$$\int\frac{e^x(x^2+4x+8)}{x+4}dx. $$
I've considered substituting $t = e^x$, but this doesn't seem to make the problem any easier.
| \begin{align*}\int e^x\left(\frac{x+2}{x+4}\right)^2\,\mathrm dx&=\int\frac1{(x+4)^2}e^x(x+2)^2\,\mathrm dx\\&=-\frac{e^x(x+2)^2}{x+4}+\int e^x(x+2)\,\mathrm dx\\&=-\frac{e^x(x+2)^2}{x+4}+e^x(x+2)-\int e^x\,\mathrm dx\\&=-\frac{e^x(x+2)^2}{x+4}+e^x(x+1)\\&=\frac{xe^x}{x+4}.\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2376729",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Find the maximum positive integer that divides $n^7+n^6-n^5-n^4$
Find the maximum positive integer that divides all the numbers of the form $$n^7+n^6-n^5-n^4 \ \ \ \mbox{with} \ n\in\mathbb{N}-\left\{0\right\}.$$
My attempt
I can factor the polynomial
$n^7+n^6-n^5-n^4=n^4(n-1)(n+1)^2\ \ \ \forall n\in\mathbb{N}.$
If $n$ is even then exists $\,k\in\mathbb{N}-\left\{0\right\}\,$ such that $n=2k.\;$ so:
$n^4(n-1)(n+1)^2=2^4 k^4(2k-1)(1+2k)^2$
If $n$ is odd then exists $k\in\mathbb{N}$ such that $n=2k+1$ so
$$n^4(n-1)(n+1)^3=2^3(2k+1)^4(k+1)^2$$
Can I conclude that the maximum positive integer that divides all these numbers is $N=2^3?$ (Please, help me to improve my english too, thanks!)
Note: I correct my "solution" after a correction... I made a mistake :\
| Without using the link I gave you (which is overkilling the problem, by the way), you can see that $2^4$ divides $$f(n):=n^7+n^6-n^5-n^4=(n-1)\,n^4\,(n+1)^2$$
by considering the case $n$ is odd and the case $n$ is even. Since $n-1$, $n$, and $n+1$ are consecutive integers, $3$ must divide $f(n)$. That is, $2^4\cdot 3=48$ must divide $f(n)$ for all $n\in\mathbb{Z}_{>0}$. Using Will Jagy's answer, you should be able to deduce that $48$ is indeed the GCD of $f(n)$ for all $n\in\mathbb{Z}_{>0}$.
However, if you want to use the method given in that link, you can show that $f(n)$ is equal to
$$72\cdot 2!\cdot\binom{n}{2}+360\cdot 3!\cdot\binom{n}{3}+404\cdot4!\cdot\binom{n}{4}+154\cdot5!\cdot\binom{x}{5}+22\cdot6!\cdot\binom{n}{6}+7!\cdot\binom{n}{7}\,.$$
Then, the greatest common divisor of the coefficients $72\cdot 2!$, $360\cdot 3!$, $404\cdot 4!$, $154\cdot 5!$, $22\cdot 6!$, and $7!$ is equal to $$\gcd\big(72\cdot 2!,360\cdot3!,404\cdot 4!\big)=48\,.$$
This is the modified content of $f(n)$ as defined in the link above.
P.S.: I know other people have more or less answered your question, but I am illustrating how to use the more general method given in my link. You can use this solution for other integer-valued polynomials.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2377112",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 4
} |
Definite integral for a 4 degree function
The integral is:
$$\int_0^a \frac{x^4}{(x^2+a^2)^4}dx$$
I used an approach that involved substitution of x by $a\tan\theta$. No luck :\ . Help?
| $\displaystyle\int_0^a \frac{x^4}{(x^2+a^2)^4}dx$
Where do we get with the substitution you have suggested?
$x = a\tan\theta\\
dx = a\sec^2\theta\\
\displaystyle\int_0^{\frac \pi 4} \frac{(a^4\tan^4\theta)(a\sec^2\theta)}{(a^2\tan^2\theta+a^2)^4}d\theta\\
$
Looks promising:
Keep simplifying
$\displaystyle\int_0^{\frac \pi 4} \frac{a^5\tan^4\theta\sec^2\theta}{a^8\sec^8\theta}d\theta\\
\displaystyle\int_0^{\frac \pi 4} \frac{\tan^4\theta}{a^3\sec^6\theta}d\theta$
Let's state this into terms of $\sin\theta, \cos\theta$
$\displaystyle\frac 1{a^3}\int_0^{\frac \pi 4} \sin^4\theta\cos^2\theta\ d\theta$
You need to apply your half angle identities, perhaps repeatedly.
$\displaystyle\sin^2\theta = \frac 12 (1-\cos 2\theta), \cos^2\theta = \frac 12 (1+\cos\theta) $
$\displaystyle\frac 1{8a^3}\int_0^{\frac \pi 4} (1-\cos 2\theta)^2(1+\cos 2\theta)\ d\theta\\
\displaystyle\frac 1{8a^3}\int_0^{\frac \pi 4} [1-\cos 2\theta -\cos^2 2\theta + \cos^3 2\theta] \ d\theta\\
\displaystyle\frac 1{8a^3}\int_0^{\frac \pi 4} [1-\cos 2\theta -\frac 12 (1+\cos 4\theta) + \cos 2\theta (1-\sin^2 2\theta)]\ d\theta\\$
And that looks pretty straight-forward.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2377946",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
Prove this inequality $2(a+b+c)\ge\sqrt{a^2+3}+\sqrt{b^2+3}+\sqrt{c^2+3}$ For $a,b,c$ are positive real numbers satisfy $a+b+c=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}$. Prove that $$2\left(a+b+c\right)\ge\sqrt{a^2+3}+\sqrt{b^2+3}+\sqrt{c^2+3}$$
We have:$a+b+c=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge \frac{9}{a+b+c}\Leftrightarrow \:\left(a+b+c\right)^2\ge \:9\Leftrightarrow \:a+b+c\ge \:3$
By Cauchy-Schwarz: $R.H.S^2\le \left(1+1+1\right)\left(a^2+b^2+c^2+27\right)$
$=3\left(a^2+b^2+c^2+9\right)\Rightarrow R.H.S\le \sqrt{3\left(a^2+b^2+c^2\right)+27}$
Need to prove $4(a+b+2)^2\ge 3(a^2+b^2+c^2)+27$
$\Leftrightarrow \left(a+b+c\right)^2+3\left(a+b+c\right)^2\ge 3\left(a^2+b^2+c^2\right)+27$
$\Leftrightarrow \left(a+b+c\right)^2\ge 3\left(a^2+b^2+c^2\right)$ It's wrong. Help me
| Since the function $f(t) := \sqrt{1+t}$ is concave in $[-1, +\infty)$, we have that
$$
f(t) \leq f(3) + f'(3) (t-3)
\qquad \forall t\geq -1,
$$
i.e.
$$
f(t) \leq 2 + \frac{1}{4}(t-3) = \frac{5}{4} + \frac{1}{4} t
\qquad \forall t\geq -1.
$$
Using this inequality we have that
$$
\sqrt{a^2+3} = a \sqrt{1+ 3/a^2} \leq
a \left[\frac{5}{4} + \frac{1}{4}\cdot\frac{3}{a^2}\right]
= \frac{5}{4} a + \frac{3}{4}\cdot \frac{1}{a},
$$
and a similar inequality holds also for $\sqrt{b^2+3}$ and $\sqrt{c^2+3}$.
Finally, using the condition $a+b+c = \frac{1}{a}
+ \frac{1}{b} + \frac{1}{c}$,
$$
\sqrt{a^2+3} + \sqrt{b^2+3} + \sqrt{c^2+3}
\leq
\frac{5}{4} (a + b + c) + \frac{3}{4}\left(\frac{1}{a}
+ \frac{1}{b} + \frac{1}{c}\right)
= 2(a+b+c).
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2378941",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Inequality $\frac{x_1^2}{x_1^2+x_2x_3}+\frac{x_2^2}{x_2^2+x_3x_4}+\cdots+\frac{x_{n-1}^2}{x_{n-1}^2+x_nx_1}+\frac{x_n^2}{x_n^2+x_1x_2}\le n-1$
Show that for all $n\ge 2$
$$\frac{x_1^2}{x_1^2+x_2x_3}+\frac{x_2^2}{x_2^2+x_3x_4}+\cdots+\frac{x_{n-1}^2}{x_{n-1}^2+x_nx_1}+\frac{x_n^2}{x_n^2+x_1x_2}\le n-1$$
where $x_i$ are real positive numbers
I was going to use
$\frac{x_{1}^{2}}{x_{1}^{2}+x_{2}x_{3}}=\frac{1}{1+\frac{x_{2}x_{3}}{x_{1}^{2}}}\le \frac{x_{1}}{2}\cdot \frac{1}{\sqrt{x_{2}x_{3}}}\le \frac{x_{1}}{4}\left( \frac{1}{x_{2}}+\frac{1}{x_{3}} \right)$
$\frac{x_{1}^{2}}{x_{1}^{2}+x_{2}x_{3}}+\frac{x_{2}^{2}}{x_{2}^{2}+x_{3}x_{4}}+...+\frac{x_{n-2}^{2}}{x_{n-2}^{2}+x_{n-1}x_{n}}+\frac{x_{n-1}^{2}}{x_{n-1}^{2}+x_{n}x_{1}}+\frac{x_{n}^{2}}{x_{n}^{2}+x_{1}x_{2}}$
$=\frac{1}{1+\frac{x_{2}x_{3}}{x_{1}^{2}}}+\frac{1}{1+\frac{x_{3}x_{4}}{x_{2}^{2}}}+...+\frac{1}{1+\frac{x_{n-1}x_{n}}{x_{n-2}^{2}}}+\frac{1}{1+\frac{x_{n}x_{1}}{x_{n-1}^{2}}}+\frac{1}{1+\frac{x_{1}x_{2}}{x_{n}^{2}}}$
$\le \frac{x_{1}}{4}\left( \frac{1}{x_{2}}+\frac{1}{x_{3}} \right)+\frac{x_{2}}{4}\left( \frac{1}{x_{3}}+\frac{1}{x_{4}} \right)+...+\frac{x_{n-2}}{4}\left( \frac{1}{x_{n-1}}+\frac{1}{x_{n}} \right)+\frac{x_{n-1}}{4}\left( \frac{1}{x_{n}}+\frac{1}{x_{1}} \right)+\frac{x_{n}}{4}\left( \frac{1}{x_{1}}+\frac{1}{x_{2}} \right)$
$=\frac{1}{4}\left( \left( \frac{x_{1}}{x_{2}}+\frac{x_{1}}{x_{3}} \right)+\left( \frac{x_{2}}{x_{3}}+\frac{x_{2}}{x_{4}} \right)+\left( \frac{x_{3}}{x_{4}}+\frac{x_{3}}{x_{5}} \right)+...+\left( \frac{x_{n-2}}{x_{n-1}}+\frac{x_{n-2}}{x_{n}} \right)+\left( \frac{x_{n-1}}{x_{n}}+\frac{x_{n-1}}{x_{1}} \right)+\left( \frac{x_{n}}{x_{1}}+\frac{x_{n}}{x_{2}} \right) \right)$
$=\frac{1}{4}\left( \left( \frac{x_{1}+x_{2}}{x_{3}} \right)+\left( \frac{x_{2}+x_{3}}{x_{4}} \right)+\left( \frac{x_{3}+x_{4}}{x_{5}} \right)+...+\left( \frac{x_{n-3}+x_{n-2}}{x_{n-1}} \right)+\left( \frac{x_{n-1}+x_{n-2}}{x_{n}} \right)+\left( \frac{x_{1}+x_{n}}{x_{2}} \right)+\left( \frac{x_{n-1}+x_{n}}{x_{1}} \right) \right)$
....I thought about using Cauchy's inequality, but that would only increase the problem
| Let $\frac{x_2x_3}{x_1^2}=\frac{a_1}{a_2}$,... and similar, where $a_i>0$ and $a_{n+1}=a_1$.
Thus, we need to prove that:
$$\sum_{i=1}^n\frac{1}{1+\frac{a_i}{a_{i+1}}}\leq n-1$$ or
$$\sum_{i=1}^n\left(\frac{1}{1+\frac{a_i}{a_{i+1}}}-1\right)\leq-1$$ or
$$\sum_{i=1}^n\frac{a_i}{a_i+a_{i+1}}\geq1,$$
which is true because
$$\sum_{i=1}^n\frac{a_i}{a_1+a_{i+1}}\geq\sum_{i=1}^n\frac{a_i}{a_1+a_2+...+a_n}=1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2379113",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
If $ x \in \left(0,\frac{\pi}{2}\right)$. Then value of $x$ in $ \frac{3}{\sqrt{2}}\sec x-\sqrt{2}\csc x = 1$
If $\displaystyle x \in \left(0,\frac{\pi}{2}\right)$ then find a value of $x$ in $\displaystyle \frac{3}{\sqrt{2}}\sec x-\sqrt{2}\csc x = 1$
$\bf{Attempt:}$ From $$\frac{3}{\sqrt{2}\cos x}-\frac{\sqrt{2}}{\sin x} = 1.$$
$$3\sin x-2\cos x = \sqrt{2}\sin x\cos x$$
$$(3\sin x-2\cos x)^2 = 2\sin^2 x\cos^2 x$$
$$9\sin^2 x+4\cos^2 x-12 \sin x\cos x = 2\sin^2 x\cos^2 x$$
Could some help me to solve it, thanks
| I think it's better to make the following.
Let $x=\frac{\pi}{4}+t$, where $t\in\left(-\frac{\pi}{4},\frac{\pi}{4}\right)$.
Hence, we need to solve that
$$3\sin{x}-2\cos{x}=\sqrt2\sin{x}\cos{x}$$ or
$$3(\sin{t}+\cos{t})-2(\cos{t}-\sin{t})=\cos^2t-\sin^2t$$ or
$$\sin{t}(5+\sin{t})+(1-\cos{t})\cos{t}=0$$ or
$$\sin\frac{t}{2}\left(\cos\frac{t}{2}(5+\sin{t})+\sin\frac{t}{2}\cos{t}\right)=0$$ and since
$$\cos\frac{t}{2}(5+\sin{t})+\sin\frac{t}{2}\cos{t}=5\cos\frac{t}{2}+\sin\frac{3x}{2}>5\cos\frac{\pi}{8}-1>0,$$
we obtain $\sin\frac{t}{2}=0$, which gives $t=0$ and $x=\frac{\pi}{4}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2379567",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 3
} |
If $0^\circ\leqslant x<360^\circ$, what is the maximum number of solutions to the equation $\sin x = a$ where a is a real number? I tried solving the question, but I kept getting $5$ solutions. My book only has $4$ choices: $0$, $1$, $2$, or $3$ solutions. My solutions were $0^\circ$, $90^\circ$, $150^\circ$, $180^\circ$, and $270^\circ$. What did I do wrong? Why is $2$ solutions the correct answer?
| Clearly, we need $-1\le a\le1$ for at least one real solution
If $\sin x_1=\sin x_2$
Using Prosthaphaeresis Formulas, $$\sin x_1-\sin x_2=2\sin\dfrac{x_1-x_2}2\cos\dfrac{x_1+x_2}2.$$
If $\sin\dfrac{x_1-x_2}2=0\implies\dfrac{x_1-x_2}2=m180^\circ\iff x_1\equiv x_2\pmod{360^\circ}.$
If $\cos\dfrac{x_1+x_2}2=0\implies\dfrac{x_1+x_2}2=(2n+1)90^\circ\iff x_1\equiv 180^\circ- x_2\pmod{360^\circ}$
Now $x_1,x_2$ will coincide if $x_1\equiv 180^\circ- x_1\pmod{360^\circ}\iff x_1\equiv90^\circ\pmod{180^\circ}.$
In that case, we shall have only one in-congruent solution $\pmod{360^\circ}.$
Otherwise, there will be distinct two namely, $x_1,180^\circ- x_1.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2380284",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Showing the Fourier sine series converges
The Fourier sine series for $f(x) = x$, $-2 < x < 2$ is
$$f(x) = \frac{4}{\pi}\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n}\sin \frac{n\pi x}{2}$$
For each $x$ in the interval to what does the Fourier since series for $f(x)$ converge, can we prove pointwise convergence, convergence in $L^2$, and/or uniform convergence?
Attempted solution - We have
$$\int_{-2}^{2}x^2 dx = \frac{16}{3} < \infty$$ so the function $f(x) = x$ is in $L^2$ and the Fourier series converges in $L^2$. Now,
$$\Bigg|\frac{(-1)^{n+1}}{n}\sin \frac{n\pi x}{2}\Bigg| \leq \frac{(-1)^{n+1}}{n}$$
Take $M_n = \frac{(-1)^{n+1}}{n}$. Since $\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n}$ is convergent then by the Weirstrass M-test the original series is uniformly convergent and thus pointwise convergent.
Assuming the latter above is correct, can I state that $S_n(f)\to f(x) = x$?
| *
*Your proof of convergence in $L^2$ is correct.
*Let us prove the pointwise convergenge in the open interval $(-2,2)$.
Consider, for $-2 < x < 2$,
$$ F(x) = \frac{4}{\pi}\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n}e^{ \frac{n\pi x}{2}i}$$
If $x\in (-2,2)$, then we have that $\sum_{n=1}^{\infty}(-1)^{n+1}e^{ \frac{n\pi x}{2}i}$ converges. In fact,
$$\sum_{n=1}^{\infty}(-1)^{n+1}e^{ \frac{n\pi x}{2}i}=
\left (\frac{e^{ \frac{\pi x}{2}i}}{1+e^{ \frac{\pi x}{2}i}} \right )$$
So, there is $M>0$ such that, for all $N\geqslant 1$,
$$\left | \sum_{n=1}^{N}(-1)^{n+1}e^{ \frac{n\pi x}{2}i}\right| \leqslant M $$
And since, $\{\frac{1}{n}\}_{n\geqslant 1}$ is a non-increasing sequence of real number such that, as $n \to \infty$, $ \frac{1}{n} \to 0$, we can apply Dirichlet's test to conclude that, if $x\in (-2,2)$, then $F(x)$ converges.
So we have prove that, for all $x\in (-2,2)$,
$$ F(x) = \frac{4}{\pi}\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n}e^{ \frac{n\pi x}{2}i}$$
converges pointwisely.
Now, note that
$$f(x) = \frac{4}{\pi}\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n}\sin \frac{n\pi x}{2}$$
is the imaginary part of $F(x)$. So we have that, for all $x\in (-2,2)$,
$$f(x) = \frac{4}{\pi}\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n}\sin \frac{n\pi x}{2}$$
converges pointwisely.
*Uniform convergence. Note that
$$f(x) = \frac{4}{\pi}\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n}\sin
\frac{n\pi x}{2}$$
does NOT converge uniformly on $(-2,2)$.
To see it, consider
$$f_N(x) = \frac{4}{\pi}\sum_{n=1}^{N}\frac{(-1)^{n+1}}{n}\sin
\frac{n\pi x}{2}$$
Note that for all $N \geqslant 1$, $f_N$ is continuous on $[-2,2]$. Note also that $f$ (defined by $f(x)=x$ is also continuous on $[-2,2]$.
From item 2, we know that, for all $x\in (-2,2)$, $f_N(x)$ converges to $f(x)$. However, for $x=-2$, for all $N\geqslant 1$, $f_N(-2)=0$ and $f(-2)=-2$. In a similar way, for $x=2$, for all $N\geqslant 1$, $f_N(2)=0$ and $f(2)=2$.
So $f_N$ does not converge uniformly to $f$ on $(-2,2)$.
In fact, suppose $f_N$ converges uniformly to $f$ on $(-2,2)$. Then there is $N_0 \in \mathbb{N}$ such that for any $N>N_0$, and any $x \in (-2,2)$,
$$|f_N(x)-f(x)| <1/2$$
In particular, for any $x \in (-2,2)$,
$$|f_{N_0+1}(x)-f(x)| <1/2 \tag{1}$$
But, since $f_{N_0+1}$ is continuous on $[-2,2]$, there is $\delta_1>0$ such that, for all $x \in (-2,-2+\delta_1)$
$$|f_{N_0+1}(-2)-f_{N_0+1}(x)| <1/2 \tag{2} $$
Since $f$ is continuous on $[-2,2]$, there is $\delta_2>0$ such that, for all $x \in (-2,-2+\delta_2)$
$$|f(x)-f(-2)| <1/2 \tag {3}$$
Take $\delta = \min\{\delta_1, \delta_2\}$. Combining $(1)$, $(2)$, $(3)$, we have, for all $x \in (-2,-2+\delta)$
\begin{align*} 2 = &|f_{N_0+1}(-2) - f(-2)| \leqslant \\ & \leqslant |f_{N_0+1}(-2)- f_{N_0+1}(x)|+ |f_{N_0+1}(x)-f(x)| + |f(x)-f(-2)|< \\ &< (1/2)+ (1/2)+(1/2) =3/2
\end{align*}
Contradiction. So we have proved that $f_N$ does not converge uniformly to $f$ on $(-2,2)$.
It means the series $$\frac{4}{\pi}\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n}\sin \frac{n\pi x}{2}$$ does not converge uniformly to $f$ on $(-2,2)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2380643",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Element of certain order in special linear space. What would be the conditions (if any) on the trace of an element in $SL(2,p)$ in order for it to have order 5 ? (assuming $p= \pm1 mod 10$)
For example, any traceless element in $SL(2,p)$ has order 4 (straightforward proof).
Any suggestion or comment is tremendously valuable.
| Suppose $A$ is an element of order $5$ in $\DeclareMathOperator{\SL}{SL} \SL_2(\mathbb{F}_p)$ and let $m(x)$ be its minimal polynomial. If $m$ has degree $1$, then $A$ is a scalar matrix hence must be of the form
$$
\begin{pmatrix}
c & 0\\
0 & c
\end{pmatrix}
$$
for some $c$. But then $c^2 = \det(A) = 1$, so $A$ has order at most $2$, contradiction. Thus $m$ must have degree $2$, so we can write $m(x) = x^2 + ax + 1$ for some $a$. Since $A$ satisfies $x^5 - 1$, then we must have that $x^2 + ax + 1$ divides $x^5 - 1$.
First, assume that $x-1 \nmid x^2 + ax + 1$. Then $x^2 + ax + 1$ divides $x^4 + x^3 + x^2 + x + 1$. Using division with remainder, we find that
\begin{align*}
x^4 + x^3 + x^2 + x + 1 &= (x^2 + (1-a)x + a(a-1))m(x) + (-a^3 + a^2 + a)x + (-a^2 + a + 1)
\end{align*}
Thus we must have that $a$ satisfies $a^2 - a - 1 = 0$. The quadratic equation $t^2 - t - 1$ has discriminant $5$, hence has a solution iff $5$ is a square mod $p$. Since $(5|p) = (-1)^{p-1} (p|5) = (p|5)$ by quadratic reciprocity and the only squares mod $5$ are $0,1,4$, then we can find such an $a$ iff $p=5$ or $p \equiv 1, 4 \pmod{5}$.
(Here $(\cdot | \cdot)$ denotes the Legendre symbol.)
In this case, then $a = \frac{1 \pm \sqrt{5}}{2}$ so
$$
\begin{pmatrix}
0 & -1\\
1 & -\frac{1 \pm \sqrt{5}}{2}
\end{pmatrix}
$$
is an element of order $5$. (This is the companion matrix for $x^2 + ax + 1$ for the two possible values of $a$. In fact, by uniqueness of rational canonical form, every such matrix must be similar to one of these matrices.)
If $x-1 \mid x^2 + ax + 1$, then $x^2 + ax + 1 = x^2 - 2x + 1$. Again using polynomial division, we find
$$
x^5 - 1 = (x^3 + 2x^2 + 3x + 4)(x^2-2x+1) + 5x-5
$$
so we must have $5 = 0$ in $\mathbb{F}_p$, i.e., $p=5$. (Thus we discover no new primes in this latter case.) In this case, $a = -2 = 3$ is also equal to $\frac{1 \pm \sqrt{5}}{2} = \frac{1}{2}$.
Thus in all cases, we find that $\operatorname{Tr}(A) = -a = -\frac{1 \pm \sqrt{5}}{2}$.
For instance, for $p=11$ we have $4^2 = 16 = 5$ and
$$
\frac{1 \pm \sqrt{5}}{2} = \frac{1 \pm 4}{2} = \frac{5}{2}, \, \frac{8}{2} = 8, 4
$$
are the roots of $t^2 - t - 1$. Then
$$
\begin{pmatrix}
0 & -1\\
1 & -4
\end{pmatrix}
\qquad
\begin{pmatrix}
0 & -1\\
1 & -8
\end{pmatrix}
$$
have order $5$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2380745",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Pythagorean Triple: $\text{Area} = 2 \cdot \text{perimeter}$ Find the unique primitive Pythagorean triple whose area is equal to twice the perimeter.
So far I set the sides of the triangle to be $a, b,~\text{and}~c$ where $a$ and $b$ are the legs of the triangle and c is the hypotenuse.
I came up with 2 equations which are:
$\dfrac{ab}2 = 2(a+b+c)\;\;$ and $\;\;a^2+b^2=c^2$
but I'm not sure how to proceed and solve for $a, b, c$.
| Rewrite the first equation as $c = \frac{ab}{4} - a - b$. Square it to get $$c^2 = a^2 + b^2 + \frac{a^2b^2}{16} - \frac{a^2b}{2} - \frac{ab^2}{2} + 2ab$$
Now using the other equation, we see that
$$\frac{a^2b^2}{16} - \frac{a^2b}{2} - \frac{ab^2}{2} + 2ab = 0$$
Since $a,b > 0$ divide by $ab$ and multiply by $16$ to get
$$ ab - 8a - 8b + 32 = 0$$
Use Simon's Favorite Factoring Trick to get $(a-8)(b-8) = 32$.
Now note that $a$ and $b$ are integers, so $(a-8)$ and $(b-8)$ must be factors of $32$. But factoring $32$ into anything except for $\{1,32\}$ gives you two even numbers - these can't be the legs of a primitive Pythagorean triple. Thus, we must have $a=9,b=40$, giving us the $(9,40,41)$ triangle.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2381494",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
Formula for consecutive residue of primitive modulo n. \begin{align*}
3^0 \equiv 1\mod 7\\
3^1 \equiv 3\mod 7\\
3^2 \equiv 2\mod 7\\
3^3 \equiv 6\mod 7\\
3^4 \equiv 4\mod 7\\
3^5 \equiv 5\mod 7\\
3^6 \equiv 1\mod 7\\
3^7 \equiv 3\mod 7\\
\end{align*}
Now just focusing on 1, 3, 2, 6, 4, 5, 1....
How to devise a formula to find the next number.
Like if 2 is given how to find 6 or if 4 is given how to find 5?
I am looking for an explicit function.
| Fermat says $3^{6k+r}$mod$7=3^r$mod$7,0\le r \le 5$. Up to $r=5$ the calculation is very simple, no?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2381921",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Find minimum value that the trigonometric expression may take For $x\in\left(0, \frac{\pi}{2}\right)$ find a minimal value, which the expression
$$\sec x+\csc x+\sec^{2}x+\csc^{2}x$$
can take.
My attempt:
I followed the trigonometrical approach and obtained
$$\sec x+\csc x+\sec^{2}x+\csc^{2}x=\sqrt{\left(2\csc 2x+1\right)^2-1}+4\csc^{2}2x\geq \sqrt{(2+1)^2-1}+4(1)=4+2\sqrt{2}$$
Above was obtained after lot of manipulations with the trigonometrical identities so I am looking for an easy approach to this problem.
| Let $\sin{x}=a$ and $\cos{x}=b$.
Hence, $a^2+b^2=1$ and by AM-GM we obtain:
$$\sec x+\csc x+\sec^{2}x+\csc^{2}x=$$
$$=\frac{a+b}{ab}+\frac{1}{a^2b^2}\geq\frac{2\sqrt2}{\sqrt{a^2+b^2}}+\frac{4}{(a^2+b^2)^2}=4+2\sqrt2.$$
The equality occurs for $a=b=\frac{1}{\sqrt2}$, which says that we got a minimal value.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2382235",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Finding the value of a given trigonometric series.
Find the value of $\tan^2\dfrac{\pi}{16}+\tan^2\dfrac{2\pi}{16}+\tan^2\dfrac{3\pi}{16}+\tan^2\dfrac{4\pi}{16}+\tan^2\dfrac{5\pi}{16}+\tan^2\dfrac{6\pi}{16}+\tan^2\dfrac{7\pi}{16}.$
My attempts:
I converted the given series to a simpler form:
$\tan^2\dfrac{\pi}{16}+\cot^2\dfrac{\pi}{16}+\tan^2\dfrac{2\pi}{16}+\cot^2\dfrac{2\pi}{16}+\tan^2\dfrac{3\pi}{16}+\cot^2\dfrac{3\pi}{16}+1.$
Then I found the following values because I already knew the values of $\sin22.5^{\circ}$ and $\cos22.5^{\circ}$:
$\cos^2(\frac{\pi}{16})= \dfrac{2+\sqrt{2+\sqrt2}}{4}$
$\sin^2(\frac{\pi}{16})= \dfrac{2-\sqrt{2+\sqrt2}}{4}$
$\sin^2(\frac{\pi}{8})= \dfrac{2-\sqrt2}{4}$
$\cos^2(\frac{\pi}{8})= \dfrac{2+\sqrt2}{4}$
However, at this stage I feel that my method of solving this problem is unnecessarily long and complicated. Could you guide me with a simpler approach to this question?
| $$\tan^2\dfrac{\pi}{16}+\tan^2\dfrac{2\pi}{16}+\tan^2\dfrac{3\pi}{16}+\tan^2\dfrac{4\pi}{16}+\tan^2\dfrac{5\pi}{16}+\tan^2\dfrac{6\pi}{16}+\tan^2\dfrac{7\pi}{16}=$$
$$=\tan^2\dfrac{\pi}{16}+\cot^2\dfrac{\pi}{16}+\tan^2\dfrac{3\pi}{16}+\cot^2\dfrac{3\pi}{16}+\tan^2\dfrac{\pi}{8}+\cot^2\dfrac{\pi}{8}+1=$$
$$=\left(\tan\frac{\pi}{16}+\cot\frac{\pi}{16}\right)^2+\left(\tan\frac{3\pi}{16}+\cot\frac{3\pi}{16}\right)^2+\left(\tan\frac{\pi}{8}+\cot\frac{\pi}{8}\right)^2-5=$$
$$=\frac{1}{\sin^2\frac{\pi}{16}\cos^2\frac{\pi}{16}}+\frac{1}{\sin^2\frac{3\pi}{16}\cos^2\frac{3\pi}{16}}+\frac{1}{\sin^2\frac{\pi}{8}\cos^2\frac{\pi}{8}}-5=$$
$$=\frac{4}{\sin^2\frac{\pi}{8}}+\frac{4}{\sin^2\frac{3\pi}{8}}+\frac{4}{\sin^2\frac{\pi}{4}}-5=$$
$$=\frac{4}{\sin^2\frac{\pi}{8}}+\frac{4}{\cos^2\frac{\pi}{8}}+3=\frac{4}{\sin^2\frac{\pi}{8}\cos^2\frac{\pi}{8}}+3=\frac{16}{\sin^2\frac{\pi}{4}}+3=35$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2383169",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Why should we have $\sin^2(x) = \frac{1-\cos(2x)}{2}$ knowing that $\sin^2(x) = 1 - \cos^2(x)$?
Why should we have $\sin^2(x) = \frac{1-\cos(2x)}{2}$ knowing that $\sin^2(x) = 1 - \cos^2(x)$?
Logically, can you not subtract $\cos^2(x)$ to the other side from this Pythagorean identity $\sin^2(x)+\cos^2(x)=1?$
When I look up trig identities, however, it says $\sin^2(x) = \frac{1-\cos(2x)}{2}$.
Why is this?
| Both formulas are true, however, both are useful in different contexts (applications).
*
*You use $\sin^2(x) = \frac{1-\cos(2x)}{2}$ for integrating $\sin^2(x)$.
*You use $\sin^2(x) = 1 - \cos^2(x)$, for example, when solving $\sin^2(x) = 2\cos(x)$.
Note that it is just in some way more "natural" to write $\sin^2(x) + \cos^2(x)=1$, because this gives both $\sin^2(x) = 1 - \cos^2(x)$ and $\cos^2(x) = 1 - \sin^2(x)$ in one "natural looking" formula.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2383695",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
Find $\cos2\theta+\cos2\phi$, given $\sin\theta + \sin\phi = a$ and $\cos\theta+\cos\phi = b$
If
$$\sin\theta + \sin\phi = a \quad\text{and}\quad \cos\theta+\cos\phi = b$$
then find the value of $$\cos2\theta+\cos2\phi$$
My attempt:
Squaring both sides of the second given equation:
$$\cos^2\theta+ \cos^2\phi + 2\cos\theta\cos\phi= b^2$$
Multiplying by 2 and subtracting 2 from both sides we obtain,
$$\cos2\theta+ \cos2\phi = 2b^2-2 - 4\cos\theta\cos\phi$$
How do I continue from here?
PS: I also found the value of $\sin(\theta+\phi)= \dfrac{2ab}{a^2+b^2}$
Edit: I had also tried to use $\cos2\theta + \cos2\phi= \cos(\theta+\phi)\cos(\theta-\phi)$ but that didn't seem to be of much use
| HINT: use that $$\cos(2\theta)+\cos(2\phi)=\cos(\theta-\phi)\cos(\theta+\phi)$$
and $$\sin(\theta)+\sin(\phi)=2\cos\left(\frac{\theta-\phi}{2}\right)\sin\left(\frac{\theta+\phi}{2}\right)$$
and
$$\cos(\theta)+\cos(\phi)=2\cos\left(\frac{\theta-\phi}{2}\right)\cos\left(\frac{\theta+\phi}{2}\right)$$
so another idea, and this works:
use that
$$\sin(\theta)+\sin(\phi)=2\,{\frac {\tan \left( \theta/2 \right) }{1+ \left( \tan \left( \theta
/2 \right) \right) ^{2}}}+2\,{\frac {\tan \left( \phi/2 \right) }{1+
\left( \tan \left( \phi/2 \right) \right) ^{2}}}
$$
and $$\cos(\theta)+\cos(\phi)={\frac {1- \left( \tan \left( \theta/2 \right) \right) ^{2}}{1+
\left( \tan \left( \theta/2 \right) \right) ^{2}}}+{\frac {1-
\left( \tan \left( \phi/2 \right) \right) ^{2}}{1+ \left( \tan
\left( \phi/2 \right) \right) ^{2}}}
$$
and $$\cos(2\theta)+\cos(\phi)={\frac {1- \left( \tan \left( \theta \right) \right) ^{2}}{1+ \left(
\tan \left( \theta \right) \right) ^{2}}}+{\frac {1- \left( \tan
\left( \phi \right) \right) ^{2}}{1+ \left( \tan \left( \phi
\right) \right) ^{2}}}
$$
now convert $$\tan(x)$$ into $\tan(x/2)$ and solve the equations above for $$\tan(\phi/2)$$ respective $$\tan(\theta/2)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2383791",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
Polynomial $ax^2 + (b+c)x + (d+e)$ Let $a, b, c, d$ be real number such that polynomial $ax^2 + (b+c)x + (d+e)$ has real roots greater than $1$. Prove that polynomial $ax^4+bx^3+cx^2+dx+e$ has at least one real root.
Is my work correct ?
Let $r$ be real root of $ax^2+(c+b)x+(e+d)$, so $ar^2+cr+e=(br+d)(-1)$.
Let $P(x) = ax^4+bx^3+cx^2+dx+e$
so $P(\sqrt{r}) = ar^2+cr+e + br\sqrt{r}+d\sqrt{r}= (br+d)(\sqrt{r}-1)$
$P(-\sqrt{r}) = ar^2+cr+e - br\sqrt{r}-d\sqrt{r}= (br+d)(-\sqrt{r}-1)$
Since $\sqrt{r}>1$, so $P(\sqrt{r})>0>P(-\sqrt{r})$
By Intermediate value theorem, $P(x) = ax^4+bx^3+cx^2+dx+e$ has at least one real root.
| Assume the roots are $r_1,r_2$. Then:
$$a(x-r_1)^2(x-r_2)^2=ax^2+(-ar_1-ar_2)x+ar_1r_2=0.$$
Hence the second equation:
$$f(x)=ax^4-ar_1x^3-ar_2x^2+(ar_1r_2-e)x+e=0.$$
Note:
$$f(r_1)=-er_1+e$$
$$f(0)=e$$
Now IVT is applicable.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2384344",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Partial fractions and linear vs quadratic factors I was watching some videos on partial fraction decompistion and I got confused on one of the examples:
Say for example you have $$\frac{x+4}{x^2(x^2 +3)^2}.$$
The partial fraction equation of this is apparently:
$$\frac{A}{x} + \frac{B}{x^2} + \frac{Cx+E}{x^2 +3} + \frac{Dx+F}{(x^2 +3)^2}$$
My question is why $A/x+B/x^2$ do not have numerators with an $ax+b$ form, cause $x^2$ is a quadratic not a linear right? Is it because the $x^2$ is in brackets, so you can perceive it as $(x+0)^2$?
| Hint: $$\frac{ax+b}{x^2} = \frac{a}{x}+\frac{b}{x^2}.$$ Therefore, $$\frac{A}{x}+\frac{ax+b}{x^2}=\frac{A'}{x}+\frac{b}{x^2}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2384930",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Sum of series of fractions I am trying to find the $f$ formula that returns the sum of the series created by fractions that have constant nominator and shifting by one denominator.
Here are some examples:
$$f(3) = \frac{3}{1} + \frac{3}{2} + \frac{3}{3} = 5.5$$
or
$$f(4) = \frac{4}{1} + \frac{4}{2} + \frac{4}{3} + \frac{4}{4} = 8.33$$
or
$$f(5) = \frac{5}{1} + \frac{5}{2} + \frac{5}{3} + \frac{5}{4} + \frac{5}{5} = 11.4166$$
or
$$f(200) = \frac{200}{1} + \frac{200}{2} + \frac{200}{3} + ... + \frac{200}{200} = 1175.6062$$
Does anyone have an idea on how to calculate the sum of this series?
| The numbers
$$1+\frac12+\frac13+\cdots+\frac1n$$
are called the harmonic numbers and often denoted $H_n$. There is
no simple closed formula for $H_n$, but $H_n$ is approximately $\ln n+\gamma$ for large $n$, where $\gamma$ is Euler's constant.
You are considering $nH_n$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2385514",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Find the real and imaginary part of z let $z=$ $$ \left( \frac{1 + \sin\theta + i\cos\theta}{1 + \sin\theta - i\cos\theta} \right)^n$$
Rationalizing the denominator:
$$\frac{1 + \sin\theta + i\cos\theta}{1 + \sin\theta - i\cos\theta}\cdot\left( \frac{1 + \sin\theta + i\cos\theta}{1 + \sin\theta + i\cos\theta}\right) = \frac{(1 + \sin\theta + i\cos\theta)^2}{(1 + \sin\theta)^2 + \cos^2\theta}$$
$$=\frac{(1 + \sin\theta)^2 + \cos^2\theta + 2i(1 + \sin\theta)\cos\theta }{(1 + \sin\theta)^2 + \cos^2\theta}$$
thus
$$x = \frac{(1 + \sin\theta)^2 + \cos^2\theta }{(1 + \sin\theta)^2 + \cos^2\theta} $$
$$y= \frac{2i(1 + \sin\theta)\cos\theta }{(1 + \sin\theta)^2 + \cos^2\theta}$$
According to the binomial theorem,
$$(x+y)^n = \sum_{k=0}^n \binom{n}{k} x^{n-k}y^k$$
we get
$$z = \frac{1}{(1 + \sin\theta)^2 + \cos^2\theta}\sum_{k=0}^n \binom{n}{k} ((1 + \sin\theta)^2 + \cos^2\theta)^{n-k}\cdot(2i(1 + \sin\theta)\cos\theta)^k$$
...and that is where I'm stuck. What do you think? Thanks for the attention.
| HINT: Express the fraction as $r e^{i\theta}$ and compute $r^n e^{i n\theta}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2387172",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 8,
"answer_id": 0
} |
Is there a geometric method to show $\sin x \sim x- \frac{x^3}{6}$ I found a geometric method to show $$\text{when}\; x\to 0 \space , \space \cos x\sim 1-\frac{x^2}{2}$$ like below :
Suppose $R_{circle}=1 \to \overline{AB} =2$ in $\Delta AMB$ we have $$\overline{AM}^2=\overline{AB}\cdot\overline{AH} \tag{*}$$and
$$\overline{AH}=\overline{OA}-\overline{OH}=1-\cos x$$ when $x$ is very small
w.r.t. * we can conclude $$x^2 \approx 2(1-\cos x) \\\to \cos x \sim 1- \frac{x^2}{2}$$
Now I have two question :
$\bf{1}:$ Is there other idea(s) to prove (except Taylor series) $x\to 0 , \space \cos x\sim 1-\frac{x^2}{2}\\$
$\bf{2}:$ How can show $\sin x \sim x- \frac{x^3}{6}$ with a geometric concept ?
Thanks in advance.
| Not a complete answer, I am afraid. Approximating the arc $AB$ as a line segment leads to the correct approximation of $\cos(x) \sim 1 - \frac{x^2}{2}$ but leads to the inaccurate result for the corresponding sine as $ \sin(x) \sim x - \frac{x^3}{8}$. Nevertheless, I post my approach, since it has not been covered in the preceding answers.
In the above diagram, we have $OA=OB=1$ and $\angle BOA = x$
Let the co-ordinates of point $B$ be $(h,k)$
Since $B$ lies on the unit circle, $$h^2 + k^2 =1\tag{1}$$
The equation of the line $AB$ is given as $y^*=m(x^*-1)$ and since B lies on the line $AB$, we have $$ k=m(h-1\tag{2})$$ where $m$ is the slope of $AB$.
From $(1)$ and $(2)$, we have
$$\Rightarrow h^2 + m^2(h-1)^2=1 $$
$$\Rightarrow h^2(1+m^2) - h(2m^2) + (m^2 -1)=0 $$
which gives two values of $h$, of which one value $h=1$ is for the point $A$
$$\Rightarrow h=\frac{m^2 -1}{m^2 +1}, \; k =\frac{-2m}{m^2 +1} \tag{3}$$
The length of the line segment $AB$ is assumed to be $x$, hence
$$\Rightarrow (h-1)^2 + k^2 =x^2 \tag{4}$$
Using $(2)$ and $(4)$, we get
$$ \Rightarrow (1+m^2)(h-1)^2 =x^2$$
$$\Rightarrow x=|h-1|\sqrt{1+m^2}=(1-h)\sqrt{1+m^2}$$
From $(3)$, we have
$$ \Rightarrow x=\frac{2}{\sqrt{1+m^2}}$$
$$ \Rightarrow 1+m^2 = \frac{4}{x^2}$$
$$ \Rightarrow m = \sqrt{\frac{4-x^2}{x^2}} \tag{5}$$
Now, using $(5)$ and $(3)$,
$$ h=OD=\frac{4-2x^2}{4}, \; k=DB= \frac{x\sqrt{4-x^2}}{2}$$
Thus $OD = \cos(x) \sim 1-\frac{x^2}{2}$ and $DB= \sin(x) \sim x(1-\frac{x^2}{4})^{1/2}$
This gives $\sin(x) \sim x - \frac{x^3}{8}$, which isn't quite correct.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2389537",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "20",
"answer_count": 6,
"answer_id": 4
} |
Finding a solution to a system Let $i$ be of the form $i=2^a3^b5^c$, where $a,b,c\ge 0$ are integers.
Consider numbers $x_{i,4},x_{i,6}$ where $x_{i,6}$ is defined when $2$ or $3$ divides $i$, $x_{i,4}$ is defined only when $2$ divides $i$.
The constraints are
*
*If $2$ divides $i$, then $$x_{i, 4} + x_{i,6} \le \frac{\frac{1}{2}\frac{2}{3}\frac{4}{5}}{i}$$
*If $3$ divides $i$ but $2$ doesn't divide $i$, then $$x_{i,6} \le \frac{\frac{1}{2}\frac{2}{3}\frac{4}{5}}{i}$$
*$$\sum x_{i,4} = \sum x_{i,6} = \frac{1}{6}.$$
I know that this system has a solution, but the technique doesn't work when I include more $x_{i,j}$'s. I've never solved systems of this type; what's a good way to approach this problem?
| Actually this system have no solution, first $x_2,y_2,x_3,y_3,x_5,y_5 \in \mathbb{R} \geq 0$.
And $x_2+y_2=1 = \sum \limits_{k=1}^{\infty} \frac{1}{2^k}$
And $x_3+y_3 = \frac{1}{2} = \sum \limits_{k=1}^{\infty} \frac{1}{3^k}$
And $x_5 +y_5 = \frac{1}{4} = \sum \limits_{k=1}^{\infty} \frac{1}{5^k}$.
So $\sum x_{i,4} = \frac{1}{2}\frac{2}{3}\frac{4}{5}* \frac{1}{2} * ((x_2+1)(x_3+1)(x_5+1)-(x_3+1)(x_5+1))$ because if $2\not|i$ then its not defined for $x_{i,4}$.
And $\sum x_{i,6} = \frac{1}{2}\frac{2}{3}\frac{4}{5}* \frac{1}{3} * ((y_2+1)(y_3+1)(y_5+1)-(y_5+1))$ because if $2\not|i$ and $3\not|i$ then its not defined for $x_{i,6}$.
Solving this system of equations using Wolfram|Alpha (also simplex method works) yield that there is no solution for your system.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2390917",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
How to change the appearence of the correct answer of $\cos55^\circ\cdot\cos65^\circ\cdot\cos175^\circ$ I represented the problem in the following view and solved it: $$\begin{align}-\sin35^\circ\cdot\sin25^\circ\cdot\sin85^\circ\cdot\sin45^\circ&=A\cdot\sin45^\circ\\ -\frac{1}{2}(\cos20^\circ-\cos70^\circ)\cdot\frac{1}{2}(\cos50^\circ-\cos120^\circ)&=A\cdot\sin45^\circ\\ \cos20^\circ\cdot\cos50^\circ-\cos50^\circ\cdot\cos70^\circ+\frac{\cos20^\circ}{2}-\frac{\cos70^\circ}{2}&=A\cdot(-4)\cdot \sin45^\circ\\ \frac{1}{2}(\cos30^\circ+\cos70^\circ)-\frac{1}{2}(\cos20^\circ+\cos120^\circ)&=A\cdot(-4)\cdot \sin45^\circ\\ \frac{\sqrt{3}}{4}+\frac{\cos70^\circ}{2}-\frac{\cos20^\circ}{2}+\frac{1}{4}+\frac{\cos20^\circ}{2}-\frac{\cos70^\circ}{2}&=-2\sqrt{2}A\\ A&=-\frac{\sqrt{6}+\sqrt{2}}{16} \end{align}$$ I believe that the above answer is true. But that didn't match a variant below:
A) $-\frac{1}{8}$
B)$-\frac{\sqrt{3}}{8}$
C) $\frac{\sqrt{3}}{8}$
D) $-\frac{1}{8}\sqrt{2-\sqrt{3}}$
E) $-\frac{1}{8}\sqrt{2+\sqrt{3}}$
I did the problem over again. After getting the same result, I thought that the apperance of my answer could be changed to match one above, so I tried to implement one of formulae involving radical numbers: all to no avail. How to change that?
| As $\cos175^\circ=\cos(180^\circ-5^\circ)=-\cos5^\circ,$
Like prove that : cosx.cos(x-60).cos(x+60)= (1/4)cos3x
$$4\cos(60^\circ-5^\circ)\cos5^\circ\cos(60^\circ+5^\circ)=\cos(3\cdot5^\circ)$$
Now use $15=60-45$ or $=45-30$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2391564",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
If $\frac {a_{n+1}}{a_n} \ge 1 -\frac {1}{n} -\frac {1}{n^2}$ then $\sum\limits_na_n$ diverges
Let $(a_n)_{n \ge 1}$ be a sequence of positive real numbers such that, for every $n\ge1$,
$$\frac {a_{n+1}}{a_n} \ge 1 -\frac {1}{n} -\frac {1}{n^2} \tag 2$$ Prove that $x_n=a_1 + a_2 + .. + a_n$ diverges.
It is clear that $x_n$ is increasing, so it has to have a limit. I tried to prove the limit is $+\infty$ but without success. No divergence criteria from series seems to work here.
UPDATE
Attempt:
Suppose a stronger inequality holds, namely that, for every $n\ge1$, $$\frac{a_{n+1}}{a_n}\geqslant1-\frac1n \tag 1$$ Then:
$$\frac {a_3}{a_2} \ge \frac 1 2\qquad
\frac {a_4}{a_3} \ge \frac 2 3\qquad
\ldots\qquad
\frac {a_{n-1}}{a_{n-2}} \ge \frac {n-3}{n-2}\qquad
\frac {a_n}{a_{n-1}} \ge \frac {n-2}{n-1}$$
Multiplying all the above yields
$$\frac {a_n}{a_2} \ge \frac 1 {n-1}$$
The last inequality proves the divergence.
| It's easy to show that, for every $n\ge3$,
$$ 1 -\frac {1}{n} -\frac {1}{n^2}
\ge \frac {n-2}{n-1}$$ It follows that, for every $n\ge3$,
$$\frac{a_{n+1}}{a_n}\ge \frac {n-2}{n-1}$$
Thus,
$$\frac {a_4}{a_3} \ge \frac 1 2\qquad
\frac {a_5}{a_4} \ge \frac 2 3\qquad
\ldots\qquad
\frac {a_{n-1}}{a_{n-2}} \ge \frac {n-4}{n-3}\qquad
\frac {a_n}{a_{n-1}} \ge \frac {n-3}{n-2}$$
Multiplying all the above, one gets:
$$\frac {a_n}{a_3} \ge \frac 1 {n-2}$$
hence $$a_n\ge \frac{a_3}{n-2}$$
The last inequality together with the comparison criteria for series proves the divergence.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2392220",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
"answer_id": 1
} |
Find $x\in \Bbb Z$ such that $x=\sqrt[3]{2+\sqrt{5}}+\sqrt[3]{2-\sqrt{5}}$
Find $x\in \Bbb Z$ such that $x=\sqrt[3]{2+\sqrt{5}}+\sqrt[3]{2-\sqrt{5}}$
Tried (without sucess) two different approaches: (a) finding $x^3$ by raising the right expression to power 3, but was not able to find something useful in the result that simplifies to an integer; (b) tried to find $a$ and $b$ such that $(a+\sqrt{b})^3=2+\sqrt{5}$ without success.
The answer stated for the problem in the original source (a local Math Olympiad Constest) is $x=1$.
| $$x=\sqrt[3]{2+\sqrt{5}}+\sqrt[3]{2-\sqrt{5}}\\x^3=2+\sqrt5+2-\sqrt5+3\sqrt[3]{2+\sqrt{5}}\cdot\sqrt[3]{2-\sqrt{5}}(\sqrt[3]{2+\sqrt{5}}+\sqrt[3]{2-\sqrt{5}})\\x^3=4+3\cdot(-1)\cdot(x)$$so $$x^3+3x-4=0 \\(x-1)(x^2+x+4)\to\\ x=1,x^2+x+4=0 ,\Delta <0\\x=1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2395060",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 7,
"answer_id": 0
} |
Taylor expansion of $\cos^2(\frac{iz}{2})$
Expand $\cos^2(\frac{iz}{2})$ around $a=0$
We know that $$\cos t=\sum_{n=0}^{\infty}(-1)^n\frac{t^{2n}}{{2n!}}$$
So $$\cos^2t=[\sum_{n=0}^{\infty}(-1)^n\frac{t^{2n}}{{2n!}}]^2=\sum_{n=0}^{\infty}(-1)^{2n}\frac{t^{4n}}{{4n^2!}}$$
We have $t=\frac{iz}{2}$
$$\sum_{n=0}^{\infty}(-1)^{2n}\frac{(\frac{zi}{2})^{4n}}{{4n^2!}}=\sum_{n=0}^{\infty}(-1)^{2n}\frac{(\frac{zi}{2})^{4n}}{{4n^2!}}=\sum_{n=0}^{\infty}(-1)^{2n}\frac{({zi})^{4n}}{2^{4n}{4n^2!}}=\sum_{n=0}^{\infty}(-1)^{2n}\frac{({z})^{4n}}{2^{4n}{4n^2!}}$$
But the answer in the book is $$1+\frac{1}{2}\sum_{n=1}^{\infty}\frac{({z})^{2n}}{{2n!}}$$.
| $$ \cos^2(\frac{iz}{2}) = \frac{1}{2}(1+\cos(2*\frac{iz}{2})) = \frac{1}{2}(1+\cos(iz)) = \frac{1}{2}(1+\sum_{n=0}^{+\infty} (-1)^n\frac{(iz)^{2n}}{(2n)!}) = \frac{1}{2}(1+\sum_{n=0}^{+\infty} (-1)^n (i)^{2n}\frac{(z)^{2n}}{(2n)!}) = \frac{1}{2}(1+\sum_{n=0}^{+\infty} (-1)^{2n}\frac{(z)^{2n}}{(2n)!}) = \frac{1}{2}(1+\sum_{n=0}^{+\infty} \frac{(z)^{2n}}{(2n)!}) = \frac{1}{2}(1+1+\sum_{n=1}^{+\infty}\frac{(z)^{2n}}{(2n)!}) = 1+\frac{1}{2}\sum_{n=1}^{+\infty} \frac{(z)^{2n}}{(2n)!} $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2396742",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 1
} |
If $A$ is diagonalizable, find $\alpha$ and $\beta$
Let $A$ be a $5 \times 5$ matrix whose characteristic polynomial is given by
$$p_A(\lambda)=(λ + 2)^2 (λ − 2)^3$$
If $A$ is diagonalizable, find $\alpha$ and $\beta$ such that
$$A^{-1} = \alpha A + \beta I$$
I am unable to find the inverse of $5\times 5$ matrix, I only know how to invert $3\times 3$ matrices. I don't know how to find the values of $α$ and $β$.
If anybody can help me I would be very thankful to them.
| The fact that A is diagonalizable means that there exist an invertible matrix, P, such that $PAP^{1}= \begin{bmatrix}2 & 0 & 0 & 0 & 0 \\ 0 & 2 & 0 & 0 & 0 \\ 0 & 0 & -2 & 0 & 0 \\ 0 & 0 & 0 & -2 & 0 \\ 0 & 0 & 0 & 0 & -2 \end{bmatrix}$. So $(PAP^{1})^{-1}= PA^{-1}P^{-1}= \begin{bmatrix}\frac{1}{2} & 0 & 0 & 0 & 0 \\ 0 & \frac{1}{2} & 0 & 0 & 0 \\ 0 & 0 & -\frac{1}{2} & 0 & 0 \\ 0 & 0 & 0 & -\frac{1}{2} & 0 \\ 0 & 0 & 0 & 0 & -\frac{1}{2} \end{bmatrix}$.
So $A^{-1}= P^{-1}\begin{bmatrix}\frac{1}{2} & 0 & 0 & 0 & 0 \\ 0 & \frac{1}{2} & 0 & 0 & 0 \\ 0 & 0 & -\frac{1}{2} & 0 & 0 \\ 0 & 0 & 0 & -\frac{1}{2} & 0 \\ 0 & 0 & 0 & 0 & -\frac{1}{2} \end{bmatrix}P$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2396841",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Evaluate $\int_{-\infty}^{\infty}\frac{1}{(x^2+4)^5}dx$ $$\int_{-\infty}^{\infty}\frac{1}{(x^2+4)^5}dx$$
I am trying to use residue.
We first need to find the singularities, $x^2+4=0\iff x=\pm 2i$
Just $2i$ is in the positive part of $i$ so we take the limit
$lim_{z\to 2i}\frac{1}{(z^2+4)^5}$ but the limit is $0$
| For any $a>0$ we have
$$ \int_{-\infty}^{+\infty}\frac{dx}{x^2+a}=\frac{\pi}{\sqrt{a}} $$
and by applying $\frac{d^4}{da^4}$ to both sides we get:
$$ 24\int_{-\infty}^{+\infty}\frac{dx}{(x^2+a)^5}=\frac{105 \pi}{16 a^4\sqrt{a}} $$
so by rearranging and evaluating at $a=4$ we get:
$$ \int_{-\infty}^{+\infty}\frac{dx}{(x^2+4)^5} = \color{red}{\frac{35 \pi}{2^{16}}}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2396943",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
Simplify $\sqrt{6-\sqrt{20}}$ My first try was to set the whole expression equal to $a$ and square both sides. $$\sqrt{6-\sqrt{20}}=a \Longleftrightarrow a^2=6-\sqrt{20}=6-\sqrt{4\cdot5}=6-2\sqrt{5}.$$
Multiplying by conjugate I get $$a^2=\frac{(6-2\sqrt{5})(6+2\sqrt{5})}{6+2\sqrt{5}}=\frac{16}{2+\sqrt{5}}.$$
But I still end up with an ugly radical expression.
| In this particular problem, you can pretty much guess the answer.
$$\sqrt{6-\sqrt{20}}=\sqrt{6-2\sqrt{5}}$$
Now, suppose that the $-2\sqrt{5}$ was the middle term of a perfect square trinomial, where $x = \sqrt{5}$. In other words, that middle term is $-2x$.
What would the first and last term look like? Obviously it would be $x^2$ and $1$ respectively.
$x^2 - 2x +1 = (x-1)^2$
Substituting $\sqrt{5}$ for $x$ we have...
$$x^2 - 2x +1 = (x-1)^2$$
$$\sqrt{5}^2 - 2\sqrt{5} +1 = (\sqrt{5}-1)^2$$
$$5 - 2\sqrt{5} +1 = (\sqrt{5}-1)^2$$
$$6 - 2\sqrt{5} = (\sqrt{5}-1)^2$$
and taking the square root of both sides we have $$\sqrt{6-2\sqrt{5}} = \sqrt{5}-1$$
It's almost like somebody just make up that problem to work out cleanly like that. ;)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2400336",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 5
} |
Subsets and Splits