Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
line through $P(l,3)$ intersect ellipse at $A$ and $D$ and axis at $B$ and $D$. then min. of $|l|$ A line through $P(l,3)$ meets the ellipse $\displaystyle \frac{x^2}{16}+\frac{y^2}{9}=1$ at $A$ and $D$ and meets the $x$ axis and $y$ axis at $B$ and $C$ so that $PA.PD =PB.PC.$ find minimum value of $|l|$
equation of line through $P(l,3)$ is $y-3=m(x-l)$
intersection with $x$ axis $B(l-\frac{3}{m},0)$ and $y$ axis $C(0,3-ml)$
i wan,t be able to go further, could some help me with this, thanks
| Eliminating $y$ from $\frac{x^2}{16}+\frac{y^2}{9}=1$ and $y-3=m(x-l)$ gives
$$(16m^2+9)x^2+16(-2m^2l+6m)x+16(m^2l^2-6ml)=0$$
Letting $\alpha,\beta$ be the $x$-coordinate of $A,D$ respectively, we get
$$\alpha+\beta=-\frac{16(-2m^2l+6m)}{16m^2+9},\quad \alpha\beta=\frac{16(m^2l^2-6ml)}{16m^2+9}\tag1$$
Since we have
$$PA:PC=PB:PD$$
we get
$$|l-\alpha|:|l-0|=\left|l-\left(l-\frac 3m\right)\right|:|l-\beta|,$$
and so
$$|l^2-(\alpha+\beta)l+\alpha\beta|=\left|\frac{3l}{m}\right|$$
From $(1)$,
$$\left|l^2+\frac{16(-2m^2l+6m)}{16m^2+9}l+\frac{16(m^2l^2-6ml)}{16m^2+9}\right|=\left|\frac{3l}{m}\right|$$
Multiplying the both sides by $16m^2+9$ gives
$$|9l^2|=(16m^2+9)\left|\frac{3l}{m}\right|,$$
i.e.
$$|l|=\frac{16|m|^2+9}{3|m|}$$
Now by AM-GM inequality, we get
$$|l|=\frac{1}{3}\left(16|m|+\frac{9}{|m|}\right)\ge 3\cdot 2\sqrt{16|m|\times\frac{9}{|m|}}=\color{red}{72}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2059644",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
group 12 boys and 8 girls with restrictions
There're 12 boys and 8 girls in a class. The teacher wants to randomly split them into 3 groups: 5 kids in group A, 11 kids in group B and 4 kids in group C.
1) What is the probability that John and Peter will not be in the same group?
2) What is the probability that in each group there will be at least one boy?
It looks like both questions can be solved by using an event's complement.
First let $|T|$ be the total possible ways to split kids into groups. $|T| = \binom{20}{5}\binom{15}{11}\binom{4}{4}$.
1) Let the event be named $E_1$, then its complement is $E_1^c$ which is an event such that John and Peter will be in the same group. If they're both in group A then there're 18 kids left to split and 3 kids to choose in A. In such scenario there're options to split the kids which is $\binom{18}{3}\binom{15}{11}\binom{4}{4}$. If John and Peter are in group B then there're $\binom{18}{9}\binom{9}{5}\binom{4}{4}$ ways to choose and if John and Peter are together in group C then there're $\binom{18}{2}\binom{16}{11}\binom{4}{4}$ ways to choose. Hence $|E_1^c| = \binom{18}{3}\binom{15}{11}\binom{4}{4} + \binom{18}{9}\binom{9}{5}\binom{4}{4} + \binom{18}{2}\binom{16}{11}\binom{4}{4}$ and $P(E_1) = 1 - \frac{P(E_1^c)}{P(T)}$.
2) Let the event that in each group there's at least one boy be $E_2$. Then the event such that there's not at least one boy in each group is its compliment $E_2^c$. Suppose that in A all kids are girls. There're $\binom{8}{5}\binom{15}{11}\binom{4}{4}$ ways to split the kids into groups with such restriction. Suppose that in C all kids are girls, then there're $\binom{8}{4}\binom{16}{11}\binom{4}{4}$ ways to split them. Lastly, there group B is composed of 11 kids so there will always be boys and girls in that group. Hence $|E_2^c| = \binom{8}{4}\binom{16}{11}\binom{4}{4} + \binom{8}{5}\binom{15}{11}\binom{4}{4}$ and $P(E_2) = 1 - \frac{P(E_2^c)}{P(T)}$.
The last one is really tricky so not sure I got it.
| The total number of ways to split them into groups is:
$$\frac{(12+8)!}{5!\times11!\times4!}=21162960$$
Question #$1$:
The number of combinations with John and Peter in the 1st group is:
$$\frac{(12+8-2)!}{(5-2)!\times11!\times4!}=1113840$$
The number of combinations with John and Peter in the 2nd group is:
$$\frac{(12+8-2)!}{5!\times(11-2)!\times4!}=6126120$$
The number of combinations with John and Peter in the 3rd group is:
$$\frac{(12+8-2)!}{5!\times11!\times(4-2)!}=668304$$
So the probability that John and Peter will not be in the same group is:
$$\frac{21162960-(1113840+6126120+668304)}{21162960}\approx62.63\%$$
Question #$2$:
The number of combinations with the 1st group consisting of girls only is:
$$\binom{8}{5}\times\frac{(12+(8-5))!}{11!\times4!}=76440$$
The number of combinations with the 3rd group consisting of girls only is:
$$\binom{8}{4}\times\frac{(12+(8-4))!}{5!\times11!}=305760$$
So the probability that there will not be a group consisting of girls only is:
$$\frac{21162960-(76440+305760)}{21162960}\approx98.19\%$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2059862",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How can I find the quadratic equation from highest point on parabola? If I have a parabola, where the vertex is in $P=(2,0)$, and a point on the parabola is $Q=(-1,-6)$, how can I find the quadratic equation of the form:
$$f(x) = ax^2 + bx + c$$
| Vertex form of a quadratic is: $f(x)=a(x-h)^2+k$, where $(h,k)$ is the vertex. So, in our case the vertex is $(2,0)$ so $h=2$ and $k=0$. So we have, $f(x)=a(x-2)^2$. Now, find $a$ by using the point $(-1,-6$). So we have, $-6=a(-1-2)^2$ which gives us $-6=9a$ so $a=-\frac{2}{3}$. So our equation is $f(x)=-\frac{2}{3}(x-2)^2$. Now, multiply this out to get $$f(x)=-\frac{2}{3}x^2+\frac{8}{3}x-\frac{8}{3}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2060393",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 5
} |
Prove that $(\frac{bc+ac+ab}{a+b+c})^{a+b+c} \geq \sqrt{(bc)^a(ac)^b(ab)^c}$
Prove that $$\left( \frac{bc+ac+ab}{a+b+c} \right)^{a+b+c} \ge \sqrt{(bc)^a(ac)^b(ab)^c}$$
where $a,b,c>0$
My attempt:
I couldn't proceed after that. Please help me in this regard, thanks!
| We need to prove that
$$(a+b+c)\ln\frac{ab+ac+bc}{a+b+c}\geq\frac{1}{2}\sum\limits_{cyc}a\ln{bc}$$ or
$$\ln\left(\frac{ab+ac+bc}{a+b+c}\right)^2\geq\sum\limits_{cyc}\frac{a}{a+b+c}\ln{bc}.$$
But by Jensen $$\sum\limits_{cyc}\frac{a}{a+b+c}\ln{bc}\leq\ln\sum\limits_{cyc}\frac{abc}{a+b+c}$$
Thus, it remains to prove that $$\frac{3abc}{a+b+c}\leq\left(\frac{ab+ac+bc}{a+b+c}\right)^2,$$
which is
$$\sum\limits_{cyc}c^2(a-b)^2\geq0$$
Done!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2061981",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
Value of $\int_1^\infty \frac{dx}{x\sqrt{x^2+x+1}}$? Not so thrilling... An exercise of one of my daughters.
How to evaluate
$$\int_1^\infty \frac{dx}{x\sqrt{x^2+x+1}}?$$ I made several substitution namely:
*
*Factorisation of $x^2+x+1$
*Then use of $\sinh t$
*Then substitution by $e^u$
*To get a rational fraction with at the denominator a degree two polynomial with two real roots that can be integrated with partial fraction decomposition.
Is there something more straight forward?
| $$
\begin{aligned}
\int_1^\infty \frac{dx}{x\sqrt{x^2+x+1}} & \stackrel{x\mapsto\frac{1}{x}}{=}\int_0^1 \frac{d x}{\sqrt{1+x+x^2}} \\
&=\int_0^1 \frac{d x}{\sqrt{\left(x+\frac{1}{2}\right)^2+\left(\frac{\sqrt{3}}{2}\right)^2}}
\end{aligned}
$$
Letting $x+\frac{1}{2} =\frac {\sqrt 3}{ 2 }\sinh \theta$ transforms the integral into
$$
\begin{aligned}
\int_1^\infty \frac{dx}{x\sqrt{x^2+x+1}} &=\int_{\sinh ^{-1}\left(\frac{1}{\sqrt{3}}\right)}^{\sinh ^{-1}(\sqrt{3})} \frac{\frac{\sqrt{3}}{2} \cosh \theta}{\frac{\sqrt{3}}{2} \cosh \theta} d \theta \\
&=\sinh ^{-1}(\sqrt{3})-\sinh ^{-1}\left(\frac{1}{\sqrt{3}}\right) \\
& \approx 0.76765
\end{aligned}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2062625",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Find the value of csc$\frac{16\pi}{3}$. Q. Find the value of csc$\frac{16\pi}{3}$.
A. This angle is equal to two full revolutions plus $\frac{4\pi}{3}$. I got this by subtracting, $\frac{16\pi}{3} - 4\pi = \frac{16\pi}{3} - \frac{12\pi}{3} = \frac{4\pi}{3}$. The terminal side is in Quadrant III. The reference angle is $\frac{4\pi}{3} - \pi = \frac{\pi}{3}$. The cosecant of $\frac{\pi}{3}$ is $\frac{2\sqrt{3}}{3}$. Using the negative, the answer is $-\frac{2\sqrt{3}}{3}$.
My questions:
is how do one tell how many revolutions it takes, based off the fraction they give?
And how does one find the cosecant of $\frac{\pi}{3}$? On the calculator I use sine and then take the reciprocal of sine, but sine come out to $.018276028$. And I'm not sure on how to get the reciprocal from that or even find the cosecant of that number.
| A formula for determining the quadrant containing an angle $\theta$ when $\theta$ is not an integer multiple of $\frac{\pi}{2}$ (the quadrantal angles) is
\begin{equation}
Q\left(\theta\right)=\left(\left\lfloor\frac{2\theta}{\pi} \right\rfloor\mod{4}\right)+1
\end{equation}
For example,
\begin{eqnarray}
Q\left(\frac{16\pi}{3}\right)&=&\left(\left\lfloor\frac{32\pi}{3\pi} \right\rfloor\mod{4}\right)+1\\
&=&(10\mod{4})+1\\
&=&2+1=3
\end{eqnarray}
and
\begin{eqnarray}
Q\left(-\frac{16\pi}{3}\right)&=&\left(\left\lfloor-\frac{32\pi}{3\pi} \right\rfloor\mod{4}\right)+1\\
&=&(-11\mod{4})+1\\
&=&1+1=2
\end{eqnarray}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2063947",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
P(X > Y - 1/2), given X and Y are uniform(0,1) I'm trying to calculate this, using the law of total probability. My understanding is that the answer should be 1/2.
$\displaystyle P\left(X > Y - \frac{1}{2}\right) = \int_{0}^{1} P\left(X > Y - \frac{1}{2} \middle| Y = y\right) \cdot f_Y(y) \cdot dy$
Solving the first term on the integral:
$\displaystyle P\left(X > Y - \frac{1}{2} \middle| Y = y\right) = 1 - F_x\left(y-\frac{1}{2}\right) = 1 - \int_{0}^{y-\frac{1}{2}} dx = \frac{3}{2} - y$
And combining it with the second:
$\displaystyle P\left(X > Y - \frac{1}{2}\right) = \int_{0}^{1} \left(\frac{3}{2} - y\right) \cdot 1 \cdot dy = 1$
Which seems incorrect.. could someone please point out what I'm missing?
| For $0<y<1$ we have
$$P\left(X>y-\frac{1}{2}\right)=
\begin{cases}
1 & \text{ if } y\le \frac{1}{2}\\
\frac{3}{2}-y & \text{ if } y>\frac{1}{2}
\end{cases}
$$
So
$$P\left(X>Y-\frac{1}{2}\right)=\int_0^{1/2}1\,dy+\int_{1/2}^1 \left(\frac{3}{2}-y\right) dy=\frac{1}{2}+\frac{3}{8}=\frac{7}{8}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2065250",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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Proof that $\sum\limits_{k=1}^{n}(n+1-k)^2F_k^2=F_{n+1}(F_n+F_{n+2})-n-1$ $1,1,2,3,5,8,...$ for $n=1,2,3,4,...$ It is n-th Fibonacci.
Prove this sum
$$\sum_{k=1}^{n}(n+1-k)^2F_k^2=F_{n+1}(F_n+F_{n+2})-n-1$$
I try:
We know that $\sum_\limits{k=1}^{n}F_n^2=F_nF_{n+1}$
$$\sum_{k=1}^{n}(n^2+1+k^2+2n-2nk-2k)F_k^2=\sum_{k=1}^{n}F_n^2+F_{n+1}F_{n+2}-n-1$$
$$(n^2+2n)F_nF_{n+1}+\sum_{k=1}^{n}[k^2-k(2n+1)]F_k^2=F_{n+1}F_{n+2}-n-1$$
I am not sure what to do next, please any help be nice!
| As noted in a comment, this sequence is the convolution of the two sequences $a_n=F_n^2$ and $b_n=(n+1)^2$, so we can multiply the two (standard) generating functions.
First, the generating function for $a_n$ — that is, $\displaystyle\sum_{n=0}^\infty F_n^2 x^n$ — is $\dfrac{x(1-x)}{(1+x)(1-3x+x^2)}$ (see, for instance, this math.SE answer ). Likewise, the generating function for $b_n$ is $\displaystyle\sum_{n=0}^\infty (n+1)^2x^n$ $=\dfrac{(x+1)}{(1-x)^3}$ (start with $\displaystyle\sum_{i=0}^\infty x^i=\frac1{1-x}$ and then differentiate, multiply by $x$, and differentiate again).
Now, just multiply the series together: $\displaystyle\left(\sum_{n=0}^\infty a_n\right)\left(\sum_{n=0}^\infty b_n\right)$ $\displaystyle=\sum_{n=0}^\infty\left(\sum_{i=0}^na_ib_{n-i}\right)x^n$, so
$$\begin{align}
\sum_{n=0}^\infty\left(\sum_{i=0}^n F_i^2(n+1-i)^2\right)x^n &=\left(\sum_{n=0}^\infty F_n^2\right)\left(\sum_{n=0}^\infty (n+1)^2\right) \\
&= \left(\frac{x(1-x)}{(1+x)(1-3x+x^2)}\right)\left(\frac{(x+1)}{(1-x)^3}\right) \\
&= \frac{x}{(1-3x+x^2)(1-x)^2}
\end{align}$$
(Note that the sum here goes from $i=0$ rather than $i=1$, but since $F_0=0$ it doesn't make any difference.)
Now, $F_nF_{n+1} = \sum_{i=0}^nF_i^2$ (easy to prove: $F_nF_{n+1}$ $= F_n(F_n+F_{n-1})$ $=F_n^2+F_{n-1}F_n$; now just induct), so $\displaystyle\sum_{n=0}^\infty F_nF_{n+1}x^n = \frac{1}{1-x}\sum_{n=0}^\infty F_n^2x^n$ (this is another easy convolution) $=\dfrac{x}{(1+x)(1-3x+x^2)}$. Likewise, we get the generating function $\displaystyle\sum_{n=0}^\infty F_{n+1}F_{n+2} x^n$ by shifting indices here (i.e. dividing by $x$), so $$\begin{align}
\sum_{n=0}^\infty(F_nF_{n+1}+F_{n+1}F_{n+2})x^n &=\dfrac{x}{(1+x)(1-3x+x^2)} + \dfrac{1}{(1+x)(1-3x+x^2)}\\
&=\frac{x+1}{(1+x)(1-3x+x^2)} \\
&=\frac{1}{(1-3x+x^2)}
\end{align}$$
From here, the rest is pretty easy to show; just perform a partial-fractions decomposition on $\dfrac{x}{(1-3x+x^2)(1-x)^2}$ to write it as $\dfrac{A+Bx}{(1-3x+x^2)}+\dfrac{C}{1-x}+\dfrac{D}{(1-x)^2}$. The functions $\dfrac{1}{1-x}$ and $\dfrac{1}{(1-x)^2}$ are generating functions for the constant sequence $c_n=1$ and the linear sequence $d_n=(n+1)$, respectively, so you'll be able to translate the coefficients $C$ and $D$ back to those terms in your sequence.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2067641",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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Proving $\sqrt{x+y-x^2}+\sqrt{y+z-y^2}+\sqrt{z+w-z^2}+\sqrt{w+x-w^2}\le4\sqrt2-3$ I found this inequality using unusual calculations in maths Olympics
and I wonder if some clever teenager could prove it using their elementary knowledge of mathematics.
Let $x,y,z,w$ be non-negative numbers such that $$x+y+z+w=1$$ Prove that
$$\sqrt{x+y-x^2}+\sqrt{y+z-y^2}+\sqrt{z+w-z^2}+\sqrt{w+x-w^2}\le4\sqrt2-3$$
| By Jensen $\sum\limits_{cyc}\sqrt{x+y-x^2}\geq2\sqrt{\sum\limits_{cyc}(2x-x^2)}=2\sqrt{2-\sum\limits_{cyc}x^2}\leq2\sqrt{2-\frac{1}{4}}=\sqrt7<4\sqrt2-3$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2069671",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Limits - a different approach $\lim_{x \to\infty }(\frac{x^3+4x^2+3x+5}{x^2+2x+3}+lx+m)=10$.
How do I calculate the value of l and m?
My try: I know questions having limit tending to infinity can be solved by dividing the numerator and denominator by greatest power of $x$.But it got me nowhere in this question. Any help appreciated.
| $$\lim_{x \to\infty }\left(\dfrac{x^3+4x^2+3x+5}{x^2+2x+3}+\ell x + m\right)=10$$
$$\lim_{x \to\infty }\left(\dfrac{x^3+4x^2+3x+5}{x^2+2x+3}+\ell x\right)=10-m$$
$$\lim_{x \to\infty }\left(x + \dfrac{2x^2+5}{x^2+2x+3}+\ell x\right)=10-m$$
$$\lim_{x \to\infty }(x +\ell x)=0 \implies \ell = -1$$
$$\lim_{x \to\infty }\left(\dfrac{2x^2+5}{x^2+2x+3}\right)=10-m
\implies 2=10-m \implies m=8$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2071302",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 4
} |
How does this separation of the variable from the fraction result in $3\cdot \frac{1}{x}$ & not $\frac{3}{x}$ I wish to write the following expression as the product of a whole number or a fraction and variable expression. The answer given in my textbook is as follows:
$\frac{3}{x} = \frac{3\cdot 1}{1\cdot x} = \frac{3}{1} \cdot \frac{1}{x} = 3\cdot \frac{1}{x} $
However, I am not sure how to get to $3\cdot \frac{1}{x} $. When I complete the multiplication I get to $\frac{3}{1} \cdot \frac{1}{x} = \frac{3}{x}$, which of course loops me back round!
| As you stated we have
$\frac{3}{x} = \frac{3\cdot 1}{1\cdot x} = \frac{3}{1} \cdot \frac{1}{x}$
Now we know that $\frac{3}{1}=3$ (dividing by 1 does not change a number).
Hence, we have
$ \frac{3}{1} \cdot \frac{1}{x}=3\cdot \frac{1}{x}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2072585",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
How to show $y^2,z^2,u^2$ are roots of $(1)$ Euler's solution for the general quartic are as follows:
From the depressed quartic, $x^4+px^2+qx+r=0$, assume $x=y+z+u$ and it may be shown that $u^2,y^2,z^2$ are roots of the cubic$$t^3+\dfrac q2t^2+\dfrac {q^2-4s}{16}t-\dfrac {r^2}{64}=0\tag1$$
$$\vdots$$
However, how would you go about showing that $u^2,z^2,y^2$ are roots of $(1)$? I originally started by writing $(1)$ as$$(t-u^2)(t-y^2)(t-z^2)\tag2$$
And expanding to get $t^3-(u^2+y^2+z^2)t^2+(u^2y^2+u^2z^2+y^2z^2)t-u^2y^2z^2$. However, now I don't know what to do. Any pointers on what to do now?
| That's Decartes–Euler solution
\begin{align*}
x &= u+v+w \\[5pt]
x^2 &= u^2+v^2+w^2 + 2(uv+vw+wu) \\[5pt]
x^4 &= (u^2 + v^2 + w^2)^2 + 4(u^2+v^2+w^2)(uv+vw+wu) \\
&\quad +4(u^2 v^2+v^2 w^2+w^2 u^2)+8uvw(u+v+w) \\[5pt]
0 &= x^4+px^2+qx+r \\[5pt]
0 &= (u^2+v^2+w^2)^2+(uv+vw+wu)[\color{blue}{4(u^2+v^2+w^2)+2p}] \\
& \quad +(u+v+w)(\color{red}{8uvw+q})+4(u^2 v^2+v^2 w^2+w^2 u^2) \\
& \quad +p(u^2+v^2+w^2)+r \\
\end{align*}
You have two degree of freedom on choosing the combination of $u$, $v$ and $w$.
Eliminate the terms with factor of $\, u+v+w \,$ or $\, uv+vw+wu \,$ by taking
$$
\left \{
\begin{array}{rcL}
\color{blue}{4(u^2+v^2+w^2)+2p} &=& 0 \\[8pt]
\color{red}{8uvw+q} &=& 0
\end{array}
\right. \quad \implies \quad
\left \{
\begin{array}{rcL}
u^2+v^2+w^2 &=& -\dfrac{p}{2} \\[5pt]
u^2v^2w^2 &=& \dfrac{q^2}{64}
\end{array}
\right.$$
Also from the original quartic,
$$(u^2+v^2+w^2)^2+4(u^2 v^2+v^2 w^2+w^2 u^2)+p(u^2+v^2+w^2)+r=0$$
$$u^2 v^2+v^2 w^2+w^2 u^2=\frac{p^2}{16}-\frac{r}{4}$$
Now we can form an equation with roots $u^2$, $v^2$ and $w^2$, namely
$$t^3+\frac{p}{2}t^2+\frac{p^2-4r}{16}t-\frac{q^2}{64}=0$$
| {
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"url": "https://math.stackexchange.com/questions/2072773",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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} |
The decay of the series given by $x_n=x_{n-1}+\cos x_{n-1}$. Let $x_n=x_{n-1}+\cos x_{n-1}$, $x_1=1$. It is easy to see that $x_n\to \dfrac{\pi}{2}$. However, how can we show that $n^n(x_n-\dfrac{\pi}{2})\to 0\ (n\to\infty)$?
I find Stolz formula hard to use.
| We can manipulate the recursion as follows:
\begin{align}
x_n &= x_{n-1}+\cos x_{n-1}
\\
\dfrac{\pi}{2} - x_n &= \dfrac{\pi}{2} - x_{n-1} - \cos x_{n-1}
\\
\dfrac{\pi}{2} - x_n &= \dfrac{\pi}{2} - x_{n-1} - \sin\left(\dfrac{\pi}{2}-x_{n-1}\right)
\end{align}
Since $|\theta-\sin \theta| \le \dfrac{1}{6}|\theta|^3$ for all $\theta \in \mathbb{R}$, we have that $$\left|\dfrac{\pi}{2}-x_n\right| \le \dfrac{1}{6}\left|\dfrac{\pi}{2}-x_{n-1}\right|^3$$ for all integers $n \ge 2$. We can then use induction to show that $$\left|\dfrac{\pi}{2}-x_n\right| \le \sqrt{6}\left(\dfrac{\pi/2-1}{\sqrt{6}}\right)^{3^{n-1}}.$$
Note that $\dfrac{\pi/2-1}{\sqrt{6}} \approx 0.233 < 1$. From here, it is easy to see that $$\ln\left[n^n\left|\dfrac{\pi}{2}-x_n\right|\right] \le 3^{n-1} \cdot \ln\left(\dfrac{\pi/2-1}{\sqrt{6}}\right)+n\ln n - \dfrac{1}{2}\ln 6 \to -\infty,$$ and thus, $n^n\left|\dfrac{\pi}{2}-x_n\right| \to 0$ as $n \to \infty$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2073259",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Show that :$\int_{0}^{1}(1-x^2)^ndx={(2n)!!\over (2n+1)!!}$ How do I show that
Let $n\ge0$. Then,
$$I:=\int_{0}^{1}(1-x^2)^ndx={(2n)!!\over (2n+1)!!}$$
Here, $m!!$ denotes the product of all positive integers $i \in \left\{1,2,\ldots,m\right\}$ that have the same parity as $m$.
My try:
Using Binomial theorem
$$(1-x^2)=1-nx^2+{n(n+1)\over2!}x^4-{n(n+1)(n+2)\over 3!}x^6+\cdots$$
$$\int_{0}^{1}\left(1-nx^2+{n(n+1)\over2!}x^4-{n(n+1)(n+2)\over 3!}x^6+\cdots\right)dx$$
$$I=x-n{x^3\over 3}+{n(n+1)\over 2!}{x^5\over 5}-{n(n+1)(n+2)\over 3!}{x^7\over 7}+\cdots|_{0}^{1}$$
$$I=1-{n\over 1\cdot 3}+{n(n+1)\over 2!\cdot 5}-{n(n+1)(n+2)\over 3!\cdot 7}+\cdots$$
I need help can't see how it will simplify to ${(2n)!!\over (2n+1)!!}$
| Let $z=x^{2}$
\begin{align}
\int\limits_{0}^{1} (1-x^{2})^{n} dx &= \frac{1}{2} \int\limits_{0}^{1} (1-z)^{n} z^{-1/2} dz \\
&= \frac{1}{2} \mathrm{B}(1/2,n+1) \\
&= \frac{\Gamma(1/2)\Gamma(n+1)}{2\Gamma(n+3/2)} \\
\tag{1}
&= \frac{\sqrt{\pi}n!}{(2n+1)\Gamma(n+1/2)} \\
\tag{2}
&= \frac{\sqrt{\pi}n!}{(2n+1)} \frac{2^{n}}{\sqrt{\pi}(2n-1)!!} \\
&= \frac{n!2^{n}}{(2n+1)} \frac{n!2^{n}}{(2n)!} \\
&= \frac{(2n)!!n!2^{n}}{(2n+1)!} \\
&= \frac{(2n)!!n!2^{n}}{(2n+1)!!n!2^{n}} \\
&= \frac{(2n)!!}{(2n+1)!!}
\end{align}
*
*$\Gamma(n+3/2) = \Gamma((n+1/2)+1) = (n+1/2)\Gamma(n+1/2)$
*$$\Gamma(n+1/2) = \frac{(2n-1)!!\sqrt{\pi}}{2^{n}}$$
*All double factorial identities used can be found here.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2074931",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 0
} |
in a triangle $ABC,$ if $2c^2=a^2+b^2,$ then largest possible degree measure of angle $C$ is In a triangle $ABC,$ if $2c^2=a^2+b^2,$ then largest possible degree measure of angle $C$ is
cosine formula $\displaystyle \cos C = \frac{a^2+b^2-c^2}{2ab} = \frac{c^2}{2ab}<\frac{(a+b)^2}{2ab}=\frac{a^2+b^2}{2ab}+1$
wan,t be able to process after that, could some help me
| $$\cos{C}=\dfrac{a^2+b^2-c^2}{2ab}=\dfrac{a^2+b^2}{4ab}\ge\dfrac{1}{2}$$
so
$$C\le \dfrac{\pi}{3}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2076874",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Maximum value of $f(x)=2(a-x)(x+\sqrt{x^2+b^2})$ If $a,b,x$ are real and $$f(x)=2(a-x)(x+\sqrt{x^2+b^2}),$$ then find the maximum value of $f(x)$.
Is there any method to solve this question without differentiation because using differentiation I am getting an ugly expression.
| Starting with $f(x)=2(a-x)(x+\sqrt{x^2+b^2})$, take $x=b\sinh\theta$:
$$f(x)=2(a-b\sinh\theta)(b\sinh\theta+b\cosh\theta)=2b(a-b\sinh\theta)\exp\theta$$
Now, let $\gamma=\exp\theta$:
$$f(x)=2b\gamma\left(a-b\frac{1}{2}\left(\gamma-\frac{1}{\gamma}\right)\right)=-b^2\left(\gamma^2-2\frac{a}{b}\gamma-1\right)$$
Now, complete the square:
$$f(x)=-b^2\left(\left(\gamma-\frac{a}{b}\right)^2-1-\frac{a^2}{b^2}\right)=(a^2+b^2)-(a-b\gamma)^2$$
The result is clear from here.
Post-mortem: One might now note that we could have looked at the expansion of $(a-x-\sqrt{x^2+b^2})^2$ and compare it to our $f$ to get straight to this inequality (though it is perhaps unreasonable to expect us to intuit it a priori).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2077530",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Diophantine system of equations I'm trying to know if there is an efficient way to find the smallest (i.e lexicographically) triplet $(a,b,c)$ of integers verifying $$a^2+b^2+c^2 = x$$ $$a^3+b^3+c^3=y$$ $$a^4+b^4 + c^4 = z$$ if $(x,y,z)$ is known.
We assume that a solution exists for that triplet $(x,y,z)$.
Originally, this question is part of an algorithmic problem. So what I want is a fast way to find the triplet without having to brute-force.
What I've tried (the brute-force approach): going through all the possible values of $a$ since we can deduce an upper bound for its value, we remain with three equations where it's easy to find $b$ and $c$ and see if they are integers.
EDIT: We may assume as well that $a<b<c$.
| If $a,b,c$ are integers such that
$$a^2 + b^2 + c^2 = x$$
$$a^3 + b^3 + c^3 = y$$
$$a^4 + b^4 + c^4 = z$$
then the following divisibility conditions must hold:
$$
\left(
a + b + c
\right)
\mid
\left(
x^2 - 2z
\right)
$$
$$
\left(
4\left(ab + bc + ca\right)
\right)
\mid
\left(
x^4 + 6x^2z - 16xy^2 + 9z^2
\right)
$$
$$
\left(
2abc
\right)^2
\mid
\left(
x^6 - 4x^3y^2 - 3x^2z^2 + 12xy^2z - 4y^4 - 2z^3
\right)
$$
Of these, the last will cut down your search considerably.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2077969",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
} |
A simpler proof of $(1-x)^n<\frac{1}{1+nx}$ I proved the following inequality:
Let $x\in\mathbb{R}, 0<x<1, n\in\mathbb{N}$, then $(1-x)^n<\frac{1}{1+nx}$
however, judging from the context in the exercise book, I feel like there is a much simpler way to prove it, but I can't see it. So I'm asking for that simpler alternative proof, or at least a hint to it. My proof as follows:
Bernoulli inequality states that for $-1<x, x\neq 0, n\in \mathbb{N},n>1$ the following is true:$(1+x)^n>1+nx$.
Thus, $\frac{1}{(1+x)^n}<\frac{1}{1+nx}$ is also true.
Then I need to show that $(1-x)^n<\frac{1}{(1+x)^n}$, which is equivalent to $\frac{1}{(1-x)^n}>(1+x)^n$, which I prove by induction:
Basecase: $n=1$ $1=\frac{1-x}{1-x}\Leftrightarrow 1=\frac{1}{1-x}-\frac{x}{1-x}\Leftrightarrow 1+\frac{x}{1-x}=\frac{1}{1-x}$
Let $a,b\in \mathbb{R_{>0}}$ and $0>b>1$ Then $\left[ a>ab \right]\Leftrightarrow \left[a<\frac{a}{b}\right]$. Thus $\left[ 0<x<1\right] \Rightarrow \left[ x<\frac{x}{1-x}\right]$
Thus $1+\frac{x}{1-x}=\frac{1}{1-x} \Rightarrow 1+x<\frac{1}{1-x} \square$
Inductive step: Assume $(1+x)^n<\frac{1}{(1-x)^n}$. Need to show $(1+x)^{n+1}<\frac{1}{(1-x)^{n+1}}$.
$(1+x)^{n+1}<\frac{1}{(1-x)^{n+1}}\Leftrightarrow (1+x)^n\cdot(1+x)<\frac{1}{(1-x)^n} \cdot \frac{1}{1-x}$
Let $a,b,c,d \in \mathbb{R_{>0}}$. Then $[a>c]\wedge[b>d] \Rightarrow [ab>cd]$. $(1+x)^n<\frac{1}{(1-x)^n}$ was the assumption and $1+x>\frac{1}{1-x}$ was the basecase, therefore $(1+x)^n\cdot(1+x)<\frac{1}{(1-x)^n} \cdot \frac{1}{1-x} \Leftrightarrow (1+x)^{n+1}<\frac{1}{(1-x)^{n+1}} \square$
Thus $\frac{1}{(1+x)^n}<\frac{1}{1+nx}$ and $(1-x)^n<\frac{1}{(1+x)^n}$ are both true, which implies the original statement $(1-x)^n<\frac{1}{1+nx} \square$ If I were to count the proof of the Bernoulli inequality by induction, it would mean that I used induction twice in order to prove something that basic, which to me doesn't seem to be a sensible thing to do.
| Hint: for $0<x<1$, the inequality
$$
(1-x)^n<\frac{1}{(1+x)^n}
$$
is equivalent to
$$
(1-x^2)^n<1
$$
Or, mimicking the proof of Bernoulli’s inequality, we have to prove that
$$
(1-x)^{-n}>1+nx
$$
The statement is true for $n=1$, because it is $1-x^2<1$. Suppose it holds for $n$. Then
$$
(1-x)^{-n-1}=(1-x)^{-n}(1-x)^{-1}>\frac{1+nx}{1-x}
$$
by the induction hypothesis. We are reduced to prove that
$$
\frac{1+nx}{1-x}\ge 1+(n+1)x
$$
which becomes
$$
1+nx\ge1+(n+1)x-x-(n+1)x^2
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2078342",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 0
} |
Infinitely nested radical formulas for $\pi$ Has anyone come by the infinitely nested radical formula:
$$\Bigg\{\pi=\frac{12}{5}\cdot \lim\limits_{n \to \infty} 2^{n}\cdot\frac{1}{2^{\frac{2^{n}+1}{2^n}}} \sqrt{2^{\frac{2^{n-1}+1}{2^{n-1}}} -\cdot\cdot\cdot\cdot\sqrt{ 2^{\frac{3}{2}} +\sqrt{ 2^2 + (\sqrt{ 6}- \sqrt{ 2})}}}\Bigg\}$$ And can anyone prove it?
| From here:
If there exists an infinite iterative sequence $A=\{a_1, a_2,\cdots a_n\}$, and a sequence $B=\{c_1, c_2,\cdots c_n\}$ obtainable from $A$ such that for all $b, a_n =\cos b$, $a_{n+1} =\cos b/2, c_{n+1}=\sin b/2$. Then $$a_{n+1} =\sqrt{\frac{1+a_n}{2}} \text{ and } c_n =\sqrt{\frac{1-a_n}{2}}$$
If $a_1 =\frac{\sqrt{6}-\sqrt{2}}{4}$, then $$c_2 = \frac{1}{2^{3/2}}\sqrt{2^2-(\sqrt{6}-\sqrt{2})}, c_3 = \frac{1}{2^{5/4}}\sqrt{2^{3/2}-\sqrt{2^2-(\sqrt{6}-\sqrt{2})}}$$ And thus $$c_n = \frac{1}{2^{\frac{2^n+1}{2^n}}} \sqrt{2^{\frac{2^{n-1}+1}{2^{n-1}}}-\cdots \sqrt{2^{5/4}+\sqrt{2^{3/2}+\sqrt{2^2 +(\sqrt{6}-\sqrt{2}})}}}$$ If $k = \frac{12}{5}\lim_{n \to \infty} 2^{n-1} c_n$, we get the required nested radical. Hope it helps.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2079281",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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Why is $\lim\limits_{x\to -\infty} \frac{1}{\sqrt{x^2+2x}-x}$ equal to $0$? So I made one exercise, which was $\lim_{x\to +\infty} \frac{1}{\sqrt{x^2-2x}-x}$ I solved this one by:
$\lim \limits_{x\to +\infty} \frac{1}{\sqrt{x^2-2x}-x} \frac{\sqrt{x^2-2x}+x}{\sqrt{x^2-2x}+x} $
$\lim \limits_{x\to +\infty} \frac{\sqrt{x^2}\sqrt{1-\frac{2}{x}}+x}{-2x}$
$\lim \limits_{x\to +\infty} \frac{\sqrt{1-\frac{2}{x}}+1}{-2} = -1 $
The next exercise is $\lim \limits_{x\to -\infty} \frac{1}{\sqrt{x^2+2x}-x}$
I thought that I could solve this one in the same way by:
$\lim \limits_{x\to -\infty} \frac{1}{\sqrt{x^2+2x}-x} \frac{\sqrt{x^2+2x}+x}{\sqrt{x^2+2x}+x} $
$\lim \limits_{x\to -\infty} \frac{\sqrt{x^2}\sqrt{1+\frac{2}{x}}+x}{2x}$
$\lim \limits_{x\to -\infty} \frac{\sqrt{1+\frac{2}{x}}+1}{2} = 1 $
But apparently, the answer is $0$...
Why can't the second one be solved in the same way as the first?
And why is the answer $0$?
| When $x\rightarrow-\infty$, you don't have $0$ as denominator, but $+\infty$, because when $x \rightarrow-\infty, \sqrt{x^2 -2x} \sim |x|$ and, as $x < 0$, $|x| = -x$, so you have $\lim\limits_{x\to -\infty}\frac {1}{-x-x} = \lim\limits_{x\to -\infty}\frac {1}{-2x} = 0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2079417",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
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Show that that $a>\frac{3}{4}$ Let $x^4+2ax^3+x^2+2ax+1=0$ .
If this equation has at least two real negative roots then show that that $a>\frac{3}{4}$.
I tried to solve the equation first by using the sub $t=x+ \frac{1}{x}$.
Then I got $t^2+2at-1=0$.
Thereafter how can I proceed to get the required inequality ?
| Let $x < 0$ be a real root of $x^4+2ax^3+x^2+2ax+1=0$. Since there are two different
negative real roots we can assume that $x \ne -1$.
You already found out that $t=x+ \frac{1}{x}$ satisfies $t^2+2at-1=0$. Then
$$
-t = \lvert x \rvert + \frac{1}{\lvert x \rvert} > 2
$$
from the AM-GM inequality, with strict inequality since $\lvert x \rvert \ne 1$, therefore $t < -2$.
It follows that
$$
a = \frac{1-t^2}{2t} = \frac 12 \left( \frac 1t - t \right)
> \frac 34
$$
because the expression is strictly decreasing for $t < 0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2079944",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
Solving a system of equations involving an absolute value Solve the following system:
$$
\begin{cases}
\text{a}\cdot\text{c}+\text{b}\cdot\text{d}\cdot\epsilon^2=\epsilon\cdot\left(\text{b}\cdot\text{c}-\text{a}\cdot\text{d}\right)\\
\\
\left|\epsilon\right|=\left|-\frac{\text{c}}{\text{d}}\right|
\end{cases}
$$
I don't know how to proceed?!
With the help of the comments and the answer I got:
$$
\begin{cases}
\text{a}\cdot\text{c}+\text{b}\cdot\text{d}\cdot\epsilon^2=\epsilon\cdot\left(\text{b}\cdot\text{c}-\text{a}\cdot\text{d}\right)\\
\\
\epsilon=\pm\frac{\text{c}}{\text{d}}
\end{cases}\to\text{a}\cdot\text{c}+\text{b}\cdot\text{d}\cdot\frac{\text{c}^2}{\text{d}^2}=\pm\frac{\text{c}}{\text{d}}\cdot\left(\text{b}\cdot\text{c}-\text{a}\cdot\text{d}\right)
$$
| If $c/d$ is positive, you will have $|-c/d|=c/d$. Thus
$$
ac+bd\frac{c^2}{d^2}=\dfrac{c}{d}(bc-ad),
$$
or
$$
ac+\frac{bc^2}{d}=\dfrac{bc^2}{d}-ac,
$$
or
$$ac=0.$$
Now if $c/d$ is negative, you will have $|-c/d|=-c/d$. Thus
$$
ac+bd\frac{c^2}{d^2}=-\dfrac{c}{d}(bc-ad),
$$
or
$$
ac+\frac{bc^2}{d}=-\dfrac{bc^2}{d}+ac,
$$
or
$$bc^2/d=0.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2080545",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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On $\int_0^1\arctan\,_6F_5\left(\frac17,\frac27,\frac37,\frac47,\frac57,\frac67;\,\frac26,\frac36,\frac46,\frac56,\frac76;\frac{n}{6^6}\,x\right)\,dx$ Reshetnikov gave the remarkable evaluation,
\begin{align}
I&= \int_0^1\arctan{_4F_3}\left(\frac15,\frac25,\frac35,\frac45;\frac24,\frac34,\frac54;\frac{1}{64}\,x\right)\,dx \\
&=\frac{3125}{48}\left(5+3\pi+6\ln2-3\alpha^4+4\alpha^3+6\alpha^2-12\alpha\\-12\left(\alpha^5-\alpha^4+1\right)\arctan\frac1\alpha-6\ln\left(1+\alpha^2\right)\right)\\
&=0.7857194\dots
\end{align}
where $\alpha$ is a quartic root. However, it seems this can be simplified a bit and generalized.
I. $p=5$: Given,
$$I(n)=\int_0^1\arctan\,_4F_3\left(\frac15,\frac25,\frac35,\frac45;\,\frac24,\frac34,\frac54;\,\frac{n}{4^4}\,x\right)\,\mathrm dx$$
Is it true that, in general,
$$\frac{12n}{5^5}\, I(n) = -12\Big(1\color{blue}-\frac{n}{5^5}\Big)\arctan\frac1\alpha+6\ln\Big(\frac{2}{1+\alpha^2}\Big)+3\pi+P(\alpha)$$
where $P(\alpha)=(1-\alpha)^3(5+3\alpha)$ and $\alpha$ is the largest positive root of the quintic,
$$\alpha^5-\alpha^4+\frac{n}{5^5}=0$$
provided real number $n<4^4\,$? (Note: By sheer coincidence, the choice of $n=4$ in the other post caused the quintic to factor.)
II. $p=7$: Given,
$$J(n)=\int_0^1\arctan\,_6F_5\left(\frac17,\frac27,\frac37,\frac47,\frac57,\frac67;\,\frac26,\frac36,\frac46,\frac56,\frac76;\,\frac{n}{6^6}\,x\right)\,\mathrm dx$$
is it true that,
$$\frac{60n}{7^7}\, J(n) = 60\Big(1\color{blue}+\frac{n}{7^7}\Big)\arctan\frac1\alpha-30\ln\Big(\frac{2}{1+\alpha^2}\Big)-15\pi-P(\beta)$$
where $P(\beta) = (1-\beta)^2(27-6\beta-9\beta^2+8\beta^3+10\beta^4)$ and $\beta$ is the largest positive root of,
$$\beta^7-\beta^6+\frac{n}{7^7}=0$$
provided real $n<6^6\,$?
Questions:
*
*How do we prove the two conjectured equalities?
*What's the formula for $p=3$? (Mathematica takes too long to evaluate the integral that I couldn't use an integer relations subroutine on it.)
| I think you should start by the following general form
$${}_{k}F_{k-1}\left(\frac{1}{k+1} ,\cdots ,\frac{k}{k+1};\frac{2}{k}
\cdots ,\frac{k-1}{k},\frac{k+1}{k};\left(
\frac{m(1-m^k)}{f_k}\right)^k \right) = \frac{1}{1-m^k}$$
Where
$$f_k \equiv \frac{k}{(1+k)^{1+1/k}}$$
Put $k=4$
$${}_{4}F_{3}\left(\frac{1}{5} ,\cdots ,\frac{4}{5};\frac{2}{4}
\cdots ,\frac{3}{4},\frac{5}{4};\left(
\frac{m(1-m^4)}{f_4}\right)^4 \right) = \frac{1}{1-m^4}$$
The argument simplifies to
$$ \left(\frac{m(1-m^4)}{\frac{4}{5^{1+1/4}}}\right)^4 = \frac{5^5 \,m^4(1-m^4)^4}{4^4}$$
Hence we have
$${}_{4}F_{3}\left(\frac{1}{5} ,\cdots ,\frac{4}{5};\frac{2}{4}
\cdots ,\frac{3}{4},\frac{5}{4};\frac{5^5 (1-m)m^4}{4^4} \right) =
\frac{1}{m}$$
Now suppose we want to find
$$\int \arctan\,_4F_3\left(\frac15,\frac25,\frac35,\frac45;\,\frac24,\frac34,\frac54;\,\frac{n}{4^4}\,x\right)\,\mathrm dx$$
Use $nx = 5^5m^4(1-m)$
$$\frac{5^5}{n}\int (4m^3(1-m)-m^ 4)\arctan\,_4F_3\left(\frac15,\frac25,\frac35,\frac45;\,\frac24,\frac34,\frac54;\,\frac{5^5 (1-m)m^4}{4^4}\right)\,\mathrm dm$$
This simplifies to
$$\frac{5^5}{n}\int (4m^3(1-m)-m^ 4)\arctan\,\frac{1}{m}\,\mathrm dm$$
The anti-derivative of this is elementary.
Note that when we use substittution we will need the roots of
$$m^5-m^4+\frac{n}{5^5}= 0$$
As conjectured by the OP.
Using the value $n=4$ we can verify Reshetnikov result
\begin{align}\int_0^1\arctan{_4F_3}\left(\frac15,\frac25,\frac35,\frac45;\frac24,\frac34,\frac54;\frac{1}{64}\,x\right)\,dx
&= \frac{5^5}{4}\int^{\alpha}_1 (4x^3(1-x)-x^ 4)\arctan
\left(\frac{1}{x} \right)\,\mathrm dx\\&=0.7857194\dots \end{align}
Where $\alpha$ is the largest positive root of
$$m^5-m^4+\frac{4}{5^5}= 0$$
| {
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"url": "https://math.stackexchange.com/questions/2081467",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "16",
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Factorization of quartic polynomial. I want to know other ways of factorization to get quadratic factors in this polynomial:
$$x^4+2x^3+3x^2+2x-3$$
Thanks in advance for your suggestions.
The original polynomial is $$x^6-2x^3-4x^2+8x-3$$ where the two found factors are $(x+1)$ and $(x-1)$ by synthetic division.
| Note that $$(x^2 + x + 1)^2 = x^4 + 2x^3 + 3x^2 + 2x + 1$$which means that your polynomial is equal to
$$
(x^2 + x + 1)^2 - 4 \\
= (x^2 + x + 1)^2 - 2^2\\
= (x^2 + x + 1 - 2)(x^2 + x + 1 + 2)
$$
| {
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"timestamp": "2023-03-29T00:00:00",
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I want to Find $3^{2x} + 7^{\frac{1}{x}}$ when $7^{x+1} = 21^{x}$ I figured that
\begin{align*}7^{x+1} &= 21^{x}\\
7^{x+1} &= 3^{x} \times 7^{x} \\
7 &= 3^{x}
\end{align*}
but I can't go further at the moment.
| You have : $7^{\frac{1}{x}} = 3$ and $3^{2x} = 7^2$. Thus $7^{\frac{1}{x}} + 3^{2x} = 52$
| {
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Find the sum of the reciprocals
Let $A$ be the sum of the reciprocals of the positive integers that can be formed by only using the digits $0,1,2,3$. That is, $$A = \dfrac{1}{1}+\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{10}+\dfrac{1}{11}+\dfrac{1}{12}+\dfrac{1}{13}+\dfrac{1}{20}+\dfrac{1}{21}+\dfrac{1}{22}+\dfrac{1}{23}+\dfrac{1}{30}+\cdots.$$ What is $\lfloor A\rfloor$?
The sum is greater than $2$ since $\dfrac{1}{1}+\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{10}+\dfrac{1}{11} > 2$. Can we show it is less than $3$?
| Although this is clearly a faulty approach, take the first 15 formable numbers.
These are 1,2,3,10,11,12,13,20,21,22,23,30,31,32,33.
The decimal reciprocals are roughly equal to $1+0.5+0.35+0.1+0.09+0.08+0.08+0.05+0.05+0.04+0.04+0.03+0.03+0.03+0.03$, more or less by a very small amount.
This is slightly less than 2.5
Using three digits, you can form somewhere around (4)(4)(3)=48 numbers.
Because these are 3 digit numbers, the reciprocal of each must be less than 0.01. About 0.004 on average.
So adding this, the sum is way less than 2.98 (closer to 2.7) because only 1/3 of the denominators are closer to 100.
Using 4 digits, we can create (4)(4)(4)(3)= 192 numbers. Each reciprocal is less than 0.001. About 0.0004 on average.
The final sum is about 2.841.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find $x^2+y^2+z^2$ if $x+y+z=0$ and $\sin{x}+\sin{y}+\sin{z}=0$ and $\cos{x}+\cos{y}+\cos{z}=0$ for any $x,y,z \in [-\pi,\pi]$ Find $x^2+y^2+z^2$ if $x+y+z=0$ and $\sin{x}+\sin{y}+\sin{z}=0$ and $\cos{x}+\cos{y}+\cos{z}=0$ for any $x,y,z \in [-\pi,\pi]$.
My attempt:I found one answer $x=0,y=\frac{2\pi}{3},z=-\frac{2\pi}{3}$ which gives the answer $x^2+y^2+z^2=\frac{8{\pi}^2}{9}$.But I want a way to find the answer using equations.ANy hints?
| You can try this method:
$$x+y+z=0 \implies x+y=-z\tag1$$
And $$\sin{x}+\sin{y}+\sin{z}=0$$
$$\implies \sin{x}+\sin{y}=-\sin{z}\tag2$$
And $$\cos{x}+\cos{y}+\cos{z}=0$$
$$\cos{x}+\cos{y}=-\cos{z}\tag3$$
So $$(2)^2+(3)^2 \implies 2+2\cos(x-y)=1$$
$$\implies \cos(x-y)=-\frac12\tag4$$
Similarly we get $$\cos(y-z)=-\frac12\tag5$$
$$\cos(z-x)=-\frac12\tag6$$
And from $(1)$, we get $\cos(x+y)=\cos z$
Can you proceed now?
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How to prove this sum $\sum_{k=1}^{n-1}\frac{1}{k(n-k)}\binom{2(k-1)}{k-1}\binom{2(n-k-1)}{n-k-1}=\binom{2(n-1)}{n-1}$ Let $n\ge 2$ postive integer, show that
$$I=\sum_{k=1}^{n-1}\dfrac{1}{k(n-k)}\binom{2(k-1)}{k-1}\binom{2(n-k-1)}{n-k-1}=\binom{2(n-1)}{n-1}$$
I have done this works
$$I=\sum_{k=1}^{n-1}\dfrac{1}{k}\binom{2(k-1)}{k-1}\binom{2(n-k-1)}{n-k-1}+\sum_{k=1}^{n-1}\dfrac{1}{n-k}\binom{2(k-1)}{k-1}\binom{2(n-k-1)}{n-k-1}=\sum_{k=1}^{n-1}\dfrac{2}{k}\binom{2(k-1)}{k-1}\binom{2(n-k-1)}{n-k-1}$$
| It is a convolution identity. You may consider that:
$$ \sum_{k\geq 1}\frac{x^k}{k}\binom{2k-2}{k-1} = \frac{1-\sqrt{1-4x}}{2}\tag{1}$$
holds as a consequence of the generating function of Catalan numbers. If you square both sides of $(1)$ and consider the coefficient of $x^n$, you get:
$$ \sum_{k=1}^{n-1}\frac{1}{k(n-k)}\binom{2k-2}{k-1}\binom{2(n-k)-2}{(n-k)-1} = [x^n]\left(\frac{1-\sqrt{1-4x}}{2}-x\right) \tag{2}$$
and for $n\geq 2$, the RHS of $(2)$ is just $\color{red}{\frac{1}{n}\binom{2n-2}{n-1}}=\color{blue}{\frac{1}{2n-1}\binom{2n-1}{n}}$, by $(1)$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Determine whether the following sequence is increasing or decreasing $\frac{(n-8)^2}{(1-n)^2}, n\geq 2$ Determine whether the following sequence is increasing or decreasing:
$$\frac{(n-8)^2}{(1-n)^2}, n\geq 2$$
So the first few terms are: $36,\frac{25}{4},\frac{16}{9},...$ so let's assume the sequence is decreasing.
$a_n\geq a_{n+1} \Leftrightarrow \frac{(n-8)^2}{(1-n)^2} \geq \frac{(n-7)^2}{n^2} \Leftrightarrow \frac{(n-8)^2n^2-(n-7)^2(1-n)^2}{(1-n)^2n^2}\geq 0$
The denominator is positive $\forall n\in\Bbb N$ so now I only have to prove that the numerator is non-negative. In other words, I have to prove that $(n-8)^2n^2\geq (n-7)^2(1-n)^2 $ What would be some smart move now to prove this?
| You can rewrite your expression using polynomial division:
$$\frac{(n-8)^2}{(n-1)^2} = \left(\frac{n-8}{n-1} \right)^2 = \left(1-\frac{7}{n-1} \right)^2$$
since $\frac{7}{n-1}$ is a positive but decreasing sequence, $1-\frac{7}{n-1}$ is a increasing sequence. But since $1 < \frac{7}{n-1}$ for $n<8$ we need to consider the sign of the first few terms: We see that $\frac{(n-8)^2}{(n-1)^2}$ is first decreasing, but then obviously increasing after $n\geq 8$.
| {
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"timestamp": "2023-03-29T00:00:00",
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How do we show that $\int_{0}^{1}{\arctan(x\sqrt{x^2+2})\over \sqrt{x^2+2}}\cdot{dx\over x^2+1}=\left({\pi\over 6}\right)^2$ I was going through the the top votes questions and I saw this quite of interesting post posted by @Sangchul Lee. I was just messing around with it and found a slightest Variation of Ahmed's integral.
$$\int_{0}^{1}{\arctan(x\sqrt{x^2+2})\over \sqrt{x^2+2}}\cdot{dx\over x^2+1}=\left({\pi\over 6}\right)^2\tag1$$
May be substitution might work
$u=x(x^2+2)^{1/2}$ the $du=x(x^2+2)^{-1/2}+(x^2+2)^{1/2}dx$
$$\int_{\sqrt{2}}^{4\sqrt{3}\over3}{x^2\arctan{u}\over x^3+u^2}\cdot{x^2\over u^2-x^2}du$$
More difficult then before. I can't remove the x variable from the integral.
Any suggestion on how to prove $(1)$?
| Using your approach $$u=x(x^2+2)^{1/2}\implies x=\sqrt{\sqrt{u^2+1}-1}\implies dx=\frac{u}{2 \sqrt{u^2+1} \sqrt{\sqrt{u^2+1}-1}}\,du$$ and, after simplifications $$\int{\arctan(x\sqrt{x^2+2})\over \sqrt{x^2+2}}\cdot{dx\over x^2+1}=\int\frac{\tan ^{-1}(u)}{2 u^2+2}\,du=\frac{1}{4} \tan ^{-1}(u)^2$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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} |
Cute limit: $\lim\limits_{x \to 0^{+}}\frac{\sin(x)^x − x^ {\sin(x)}}{\tan(x) ^x − x^{\tan (x)}}$ What's the fastest way to calculate the below limit (ideally, without resorting to Taylor expansions)?
$$\lim_{x \to 0^{+}}\frac{\sin(x)^x
− x^ {\sin(x)}}{\tan(x)
^x − x^{\tan (x)}}$$
| To expand on my comment, I will use the following two limits apart from the standard lmits: $$\lim_{x \to 0}\frac{x - \sin x}{x^{3}} = \frac{1}{6},\,\lim_{x \to 0}\frac{x - \tan x}{x^{3}} = -\frac{1}{3}$$ Both the above are easily obtained either via Taylor series or via L'Hospital's Rule.
We have then
\begin{align}
L &= \lim_{x \to 0^{+}}\frac{\sin^{x}x - x^{\sin x}}{\tan^{x}x - x^{\tan x}}\notag\\
&= \lim_{x \to 0^{+}}\frac{\exp(\sin x\log x)}{\exp(\tan x\log x)}\cdot\frac{\exp(x\log\sin x - \sin x\log x) - 1}{\exp(x\log\tan x - \tan x\log x) - 1}\notag\\
&= \lim_{x \to 0^{+}}\frac{\exp(x\log\sin x - \sin x\log x) - 1}{\exp(x\log\tan x - \tan x\log x) - 1}\notag\\
&= \lim_{x \to 0^{+}}\frac{\exp(x\log\sin x - \sin x\log x) - 1}{x\log\sin x - \sin x\log x}\cdot\frac{x\log\sin x - \sin x\log x}{x\log\tan x - \tan x\log x}\notag\\
&\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\cdot\frac{x\log\tan x - \tan x\log x}{\exp(x\log\tan x - \tan x\log x) - 1}\notag\\
&= \lim_{x \to 0^{+}}\frac{x\log\sin x - \sin x\log x}{x\log\tan x - \tan x\log x}\notag\\
&= \lim_{x \to 0^{+}}\dfrac{(x - \sin x)\log x + x\log\dfrac{\sin x}{x}}{(x - \tan x)\log x + x\log\dfrac{\tan x}{x}}\notag\\
&= \lim_{x \to 0^{+}}\frac{x - \sin x}{x - \tan x}\cdot\dfrac{1 - \dfrac{1}{\log x}\dfrac{x}{\sin x - x}\log\dfrac{\sin x}{x}}{1 - \dfrac{1}{\log x}\dfrac{x}{\tan x - x}\log\dfrac{\tan x}{x}}\notag\\
&= \lim_{x \to 0^{+}}\frac{x - \sin x}{x - \tan x}\cdot\dfrac{1 - 0\cdot 1}{1 - 0\cdot 1}\notag\\
&= \lim_{x \to 0^{+}}\dfrac{\dfrac{x - \sin x}{x^{3}}}{\dfrac{x - \tan x}{x^{3}}}\notag\\
&= -\frac{1}{2}\notag
\end{align}
We have used the following standard limits in the derivation above $$\lim_{x \to 0}\frac{\sin x}{x} = \lim_{x \to 0}\frac{\tan x}{x} = \lim_{x \to 0}\frac{\log(1 + x)}{x} = \lim_{x \to 0}\frac{\exp(x) - 1}{x} = 1,\lim_{x \to 0^{+}}x\log x = 0$$ Note that $\sin x\log x = \dfrac{\sin x}{x}\cdot x \log x \to 1\cdot 0 = 0$ as $x \to 0^{+}$ and similarly $\tan x\log x \to 0$. Further $x\log\sin x = x\log x + x\log\dfrac{\sin x}{x} \to 0 + 0\cdot \log 1 = 0$ as $x \to 0^{+}$ and similarly $x\log\tan x \to 0$. Also note that if $t = (\sin x - x)/x$ then $t \to 0$ as $x \to 0^{+}$ and hence $$\frac{x}{\sin x - x}\log\frac{\sin x}{x} = \frac{\log(1 + t)}{t} \to 1$$ and in similar manner $$\frac{x}{\tan x - x}\log\frac{\tan x}{x} \to 1$$
The above is a reasonably fast way (though not as good as the use of Taylor series) to evaluate this cute limit and the real effort is in typing the full solution. The limits mentioned in the beginning of the answer are already famous and play a key role here apart from the standard limits mentioned in the later part of the answer.
| {
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"timestamp": "2023-03-29T00:00:00",
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How can I solve the following sequence Let we have the following sequence
$$y_n= \frac{1}{\sqrt{n^2+1}}+\frac{1}{\sqrt{n^2+2}}+\cdots+\frac{1}{\ \sqrt{n^2+n}}$$ find the limit of the sequence $y_n$ decide whether it is increasing or decreasing
| Hint
Prove that $$\frac{n}{\sqrt{n^2+n}}\leq \sum\frac{1}{\sqrt{n^2+k}}\leq\frac{n}{\sqrt{n^2}}$$
$$\frac{n}{\sqrt{n^2+n}}\leq y_n\leq\frac{n}{\sqrt{n^2}}$$
Then use squeeze lemma
It is a increasing sequence
Knowing that $$\frac{n}{\sqrt{n^2+n}}\leq y_n\leq\frac{1}{n}$$
We have that $y_n\leq\frac{1}{n}$ and $\frac{n+1}{\sqrt{(n+1)^2+n+1}}\leq y_{n+1}$
And easy you can show that $$\frac{1}{n}\leq \frac{n+1}{\sqrt{(n+1)^2+n+1}}$$
So $$y_
n\leq y_{n+1}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Solve $\cos (2x+3y)=\frac{1}{2}$ and $\cos (3x+2y)=\frac{\sqrt{3}}{2}$ We have to solve for $x$ and $y$:
$$\left \{
\begin{align*}
\cos (2x+3y) &= \frac{1}{2} \\
\cos (3x+2y) &= \frac{\sqrt{3}}{2}
\end{align*}
\right.$$
I got
$$\left \{
\begin{align*}
2x+3y &= 2n\pi \pm\frac{\pi}{3} \\
3x+2y &= 2m\pi \pm\frac{\pi}{6}
\end{align*}
\right.$$
The final answer in the book is
$x=\dfrac{1}{5}
\left[ (6m-4n)\pi \pm \dfrac{\pi}{2} \mp \dfrac{2\pi}{3} \right]$
Why is it $\mp \dfrac{2\pi}{3}$?
If for some value we had taken
$$2x+3y=2n\pi -\frac{\pi}{3}$$
and
$$3x+2y=2m\pi +\frac{\pi}{6}$$
The value of $x$ would be
$$\frac{1}{5}[(6m-4n)\pi+\frac{\pi}{2}+\frac{2\pi}{3}]$$
which is not included in the general solution.
| To get rid of $y$,
the first equation
is multiplied by 2
and the second by 3
giving
$\pm 2\pi/3\pm\pi/2$.
To get rid of $x$,the first equation is multiplied by 3
and the second by 2
giving
$\pm \pi\pm\pi/3$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove that $\tan 7°30' = \sqrt {6} - \sqrt {3} + \sqrt {2} - 2$ Prove that:
$$\tan 7°30' = \sqrt {6} - \sqrt {3} + \sqrt {2} - 2$$
My Work:
I guess that I have to use the formula :
$$\tan A = \frac {2 \tan(\frac {A}{2})}{1-\tan^2 (\frac {A}{2})}$$
But, I am not being able to use it. Please help me.
| Hint 1:
For $x\in [0,\frac{\pi}{2}]$:
$$\tan \frac{x}{2} = \frac{\sin x}{1+\cos x}$$
$$\cos \frac{x}{2} = \sqrt{\frac{1+\cos x}{2}}$$
$$\sin \frac{x}{2} = \sqrt{\frac{1-\cos x}{2}}$$
Hint 2:
$$7^{\circ}30' = 15^{\circ}\cdot \frac{1}{2} =30^{\circ}\cdot \frac{1}{2}\cdot \frac{1}{2} $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2093253",
"timestamp": "2023-03-29T00:00:00",
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Prove the following inequality $x^2+y^2+1>x\sqrt{y^2+1}+y\sqrt{x^2+1}$ Can anybody help me prove this inequality ?
| \begin{align*}
x\sqrt{y^{2}+1}+y\sqrt{x^{2}+1}&=\left(x,y\right)\cdot\left(\sqrt{y^{2}+1},\sqrt{x^{2}+1}\right)\le\sqrt{(x^{2}+y^{2})(x^{2}+y^{2}+2)}\\
&\le\frac{2x^{2}+2y^{2}+2}{2}=x^{2}+y^{2}+1
\end{align*}
where I have used Cauchy-Schwarz followed by the arithmetic-geometric inequality. Note that equality cannot occur due to the equality case of the arithmetic geometric inequality.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Inequality solution with absolute value I have the following inequality:
$$ \left\rvert x + \frac{1}{x} \right\lvert \ge 2$$
can I say that the distance of the expression inside the absolute value from $0$ is greater or equal to 2 (from the positive side of the real axis) and smaller or equal to $-2$ (from the negative side) so:
$$ \left\rvert x + \frac{1}{x} \right\lvert \ge 2 \implies -2\ge x + \frac{1}{x} \ge 2$$
can I use it like that to solve it?
| This line is incorrect:
$$\rvert x + \frac{1}{x} \lvert \ge 2 \implies -2\ge x + \frac{1}{x} \ge 2$$
The absolute value means the value inside could have been positive or negative so it actually becomes:
$$\pm\left(x+\frac{1}{x}\right)\ge2$$
When you go to move the negative sign to the other side you must split this into two separate equations:
$$x+\frac{1}{x}\ge2\text{ and }x+\frac{1}{x}\le-2$$
You can then use these two inequalities to finish solving the actually problem (however there are other easier ways to solve it).
First inequality:
$$x+\frac{1}{x}-2\ge0$$
$$\frac{x^2-2x+1}{x}\ge0$$
$$\frac{(x-1)^2}{x}\ge0$$
The numerator is always positive so we require $x>0$. Note we can not have $x=0$ as we can not divide by 0.
Second inequality:
$$x+\frac{1}{x}+2\le0$$
$$\frac{x^2+2x+1}{x}\le0$$
$$\frac{(x+1)^2}{x}\le0$$
The numerator is always positive so we require $x<0$. Note we can not have $x=0$ as we can not divide by 0.
Combined solution
We can have either $x>0$ or $x<0$ so the solution is $x\ne0$
| {
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If function $f$ satisfies given conditions, then find $\frac{1}{3} f(1,98)-f(1,99)$ Let $f$ be a function defined on $\{(m,n):$ $m$ and $n$ are positive integers $\}$ satisfying:
$$1. f(m,m+1)=\frac{1}{3}$$, for all m
*$$f(m,n)=f(m,k)+f(k,n)-2f(m,k) \cdot f(k,n)$$ for all $k$ such that $m<k<n$, then find the value of
$$\frac{1}{3} f(1,98)-f(1,99)$$
Could someone give me hint as how to initiate this problem?
| First note the factor of $\frac{1}{3}$ and try to exploit it with the first defining equation:
$$\begin{align}S &= \frac{1}{3}f(1, 98) - f(1, 99)\\
&= f(98, 99)f(1, 98) - f(1, 99)\\
&= f(1, 98)f(98, 99) - f(1, 99)\end{align}$$
Let $m = 1, n = 99, k = 98$. Then,
$$\begin{align}S &=\frac{f(1,98) + f(98, 99) - f(1, 99)}{2} - f(1, 99)\\
&= \frac{1}{2}f(1, 98) - \frac{3}{2}f(1, 99) + \frac{1}{6}\\
&= \frac{3}{2}\left(\frac{1}{3}f(1, 98) - f(1, 99)\right) + \frac{1}{6}\\
&= \frac{3}{2}S + \frac{1}{6}\end{align}$$
so that
$$-\frac{1}{2}S = \frac{1}{6}$$
$$S = -\frac{1}{3}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Find the value of $\alpha + \beta + \alpha\beta$ from the given data Suppose the quadratic polynomial
$P(x) = ax^2 + bx +c $
has positive coefficients $ a, b, c $ in an arithmetic progression in that order. If $P(x) = 0$ has integer roots $\alpha $ and $\beta $, then $\alpha + \beta + \alpha\beta $ equals?
1) $3$
2) $5$
3) $7$
4) $14$
It's a single choice correct problem. All genuine answers are welcome :)
The question is from a famous Indian Scholarship test 'KVPY' for high school students. The official website doesn't provide any solutions to it that is why I am asking for help.
My go on the question
we know that in a quadratic equation $ax^2 + bx +c$ with roots $\alpha, \beta$ the sum of roots is $\alpha +\beta = \frac{(-b)}{a} $ and product is $\alpha\beta = \frac{c}{a} $
So $\alpha + \beta + \alpha\beta = \frac{(-b+c)}{a}$
and since $a,b,c$ are in arithmetic progression $\frac{(a+c)}{2} = b$
Something that I tried was also this -
$ a, b, c $ are in AP (arithmetic progression)
$a-b , 0, c-b$ are in AP (subtracting by $b$)
$1-\frac{b}{a}, 0, \frac{(c-b)}{a}$ are in AP (dividing by $a$)
Now I can't proceed further. After a bit of plugging and chugging I still cant get an integer as an answer...
| Any quadratic (or for that matter polynomial) can be written in terms of its roots:
$$P(x)=a(x-\alpha)(x-\beta)=ax^2-a(\alpha+\beta)x+a\alpha\beta$$
So $\alpha+\beta+\alpha\beta=-\frac{b}{a}+\frac{c}{a}=\frac{-b+c}{a}$
We also know $a,b,c$ form an AP so lets rewritten them as $a,a+d,a+2d$ where $d$ is the common difference.
So $\alpha+\beta+\alpha\beta=\frac{-(a+d)+a+2d}{a}=\frac{d}{a}$
As the roots are integers then $d$ is a multiple of $a$ and hence so is $b$ and $c$. Without loss of generality we can let $a=1$.
Also we need $b^2-4ac=n^2,n\in\mathbb{Z}$
Subbing in $a=1$ and $d$ into this gives:
$$(1+d)^2-4(1+2d)=n^2$$
$$1+2d+d^2-4-8d=n^2$$
$$d^2-6d-3=n^2$$
$$d^2-6d-9+12=n^2$$
$$(d-3)^2+12=n^2$$
$$(d-3)^2-n^2=12$$
$$(d-3+n)(d-3-n)=12$$
Let the factors of $12$ be $\left(f,\frac{12}{f}\right)$.
So $d-3+n=f$ and $d-3-n=\frac{12}{f}$ hence $2(d-3)=f+\frac{12}{f}$
So $d=\frac{f}{2}+\frac{6}{f}+3$
Possible values of $f$ can only be $2$,$6$ as $d$ is a positive integer.
Both leads to $d=1+3+3=7$ and $d=3+1+3=7$.
Hence $\alpha+\beta+\alpha\beta=\frac{-(a+d)+a+2d}{a}=\frac{d}{a}=7$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2097100",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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if $x$ is rational and $x^2$ is natural, prove that $x$ is integer If $x$ is rational: There exists $\frac{a}{b}$ such that $a$, $b$ are integers.
If $x^2$ is natural:
$\left(\frac{a}{b}\right)^2$ is natural => $\frac{a^2}{b^2}$ is natural
Then $a^2$ divides $b^2$ => $a$ divides $b$
If $a$ divides $b$ and $a$, $b$ are integers, $\frac{a}{b}$ is integer
so $x$ is integer.
| As noted in the comments, you mean to write "$b^2$ divides $a^2$" in your third line. I wouldn't jump immediately from that to concluding though. So $b^2|a^2$ so there exists a $k$ such that $b^2k=a^2$.
Lemma: Let $x$ and $y$ be natural numbers. If $xy$ is a perfect square, then it is not the case that exactly one of $x$ and $y$ are perfect squares.
From this lemma, we see that $k$ must be a perfect square, so $k=\ell^2$. Now we can write $(b\ell)^2=a^2$ and conclude that $b\ell=a$, so $b|a$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2101905",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Integral related to Pythagoras theorem ${2\over \pi}\int_{0}^{\infty}{(ab)^3\over (a^2+b^2x^2)(b^2+a^2x^2)}\mathrm dx=h^2$. Integral related to Pythagoras theorem
Triangle ABC is a right angle triangle, where Angle $ABC=90^o$.
$h$ is perpendicular to the hypotenuse AC and meet at angle ABC.
Where $a$ and $b$ are two small sides
How can I Show that h can be represented in term of this integral $(1)$
$${2\over \pi}\int_{0}^{\infty}{(ab)^3\over (a^2+b^2x^2)(b^2+a^2x^2)}\mathrm dx=h^2\tag1$$.
Any hints on this can be relate to Pythagoras theorem
Basic formulas : $AC^2=AB^2+BC^2$ and area, $A={bh\over 2}$
| Considering the integral $$I=\int{(ab)^3\over (a^2+b^2x^2)(b^2+a^2x^2)}\, dx$$ using partial fraction decomposition, we have $$I=\int\left(\frac{a^5 b^3}{\left(a^4-b^4\right) \left(a^2 x^2+b^2\right)}-\frac{a^3
b^5}{\left(a^4-b^4\right) \left(a^2+b^2 x^2\right)}\right)\,dx$$ from which $$I=\frac{a^3 b^3 \left(a^2 \tan ^{-1}\left(\frac{a x}{b}\right)-b^2 \tan
^{-1}\left(\frac{b x}{a}\right)\right)}{a^5 b-a b^5}$$ Using the given bounds $$J=\int_{0}^{\infty}{(ab)^3\over (a^2+b^2x^2)(b^2+a^2x^2)}\, dx=\frac{\pi a^3 b^3}{2 a^3 b+2 a b^3}=\frac{\pi a^2 b^2}{2 \left(a^2+b^2\right)}$$
$$K={2\over \pi}\int_{0}^{\infty}{(ab)^3\over (a^2+b^2x^2)(b^2+a^2x^2)}\, dx=\frac{a^2 b^2}{a^2+b^2}$$ from which you can easily finish, I am sure.
| {
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"url": "https://math.stackexchange.com/questions/2105390",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Solutions to $|z+1-i\sqrt{3}|=|z-1+i\sqrt{3}|$ Find all the complex numbers $z$ satisfying
\begin{equation}
|z+1-i\sqrt{3}|=|z-1+i\sqrt{3}|
\end{equation}
I tried using $z=a+bi$, then using the formula for absolute value:
\begin{equation}
|(a+1)+i(b-\sqrt{3})|=|(a-1)+i(b+\sqrt{3})|\\
\sqrt{(a+1)^2+(b-\sqrt{3})^2}=\sqrt{(a-1)^2+(b+\sqrt{3})^2}
\end{equation}
Is this the right way to go? Should I solve for b or a, and in that case what does that tell me about possible solutions? Any help would be appreciated.
Update:
Solved the equation for $y$, got
\begin{equation}
y=\frac{x}{\sqrt{3}}
\end{equation}
Does this mean that the equation is satisfied for all complex numbers on the line $y=\frac{x}{\sqrt{3}}$?
| $$
|z + 1 - i\sqrt{3}| = |z - 1 + i\sqrt{3}| \\
\implies |z - (-1 + i\sqrt{3})| = |z - (1 - i\sqrt{3})|
$$
If I take $z_0 = 1 - i\sqrt{3}$, then the equation
$$
|z - (-z_0)| = |z - z_0|
$$
describes the perpendicular bisector of the straight line joining the points $P(z_0)$ and $Q(-z_0)$. What is clear is that the required line passes through the origin. The slope of $PQ$ is
$$
m_{PQ} = \frac{\sqrt{3} + \sqrt{3}}{-1-1} = -\sqrt{3}.
$$
Hence a line perpendicular to $PQ$ has a slope $1/\sqrt{3}$. The required line is thus
$$
y = \frac{x}{\sqrt{3}} \\
\implies \sqrt{3}y = x.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2105893",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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determinant of 4x4-matrix occuring in Zarhin's trick What's the easiest/fastest way of calculating the determinant of $$\begin{pmatrix}a & b & c & d\\ -b & a & -d & c\\ -c & d & a & -b \\ -d & -c & b & a\end{pmatrix}$$? The result is $(a^2+b^2+c^2+d^2)^2$. This determinant occurs in Zarhin's trick (if $A$ is an Abelian variety, $(A \times A^t)^4$ is principally polarised).
| Write $X$ for the matrix in question. Then
$$
X \cdot X^T = (a^2+b^2+c^2+d^2) I.
$$
This implies that
$$
\det(X) =f(a,b,c,d)= \pm (a^2+b^2+c^2+d^2)^2 .
$$
Now as
$$
f(a,0,0,0)=a^4
$$
for all $a \in \mathbf R,$ we have that
$$
\det(X)=(a^2+b^2+c^2+d^2)^2
$$
for all $a,b,c,d \in \mathbf R.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2108463",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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For all triangle prove that $\sum\limits_{cyc}\frac{a}{(a+b)^2}\geq\frac{9}{4(a+b+c)}$
Let $a$, $b$ and $c$ be a sides-lengths of the triangle. Prove that:
$$\frac{a}{(a+b)^2}+\frac{b}{(b+c)^2}+\frac{c}{(c+a)^2}\geq\frac{9}{4(a+b+c)}$$
The Buffalo way kills it, but I am looking for a nice proof for this nice inequality.
SOS (sums of squares) gives $\sum\limits_{cyc}\frac{(a-b)^2(ab+b^2-ac)}{(a+b)^2}\geq0$ and I don't see what comes next.
| After using Ravi subtitution, we need to prove
$$\sum \frac{x+y}{(2y+z+x)^2} \geqslant \frac{9}{8(x+y+z)}.$$
By the Cauchy-Schwarz inequality we get
$$\sum \frac{x+y}{(2y+z+x)^2} \geqslant \frac{\left[\displaystyle \sum (x+y)(42x+3y+55z) \right]^2}{\displaystyle \sum (2y+z+x)^2(x+y)(42x+3y+55z)^2}.$$
Therefore, we will show that
$$8(x+y+z)\left[\displaystyle \sum (x+y)(42x+3y+55z) \right]^2 \geqslant 9\sum (2y+z+x)^2(x+y)(42x+3y+55z)^2,$$
or
$$11700 \sum xy^2(y-z)^2 + \sum \left(1954xyz+6723x^2y+\frac{3104xy^2}{3}+\frac{5053yz^2}{3}\right)(z+x-2y)^2$$
$$\geqslant 0.$$
Done.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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integral calculus problem involving infinity problem
The value of the integral $\displaystyle \int_{0}^{\infty}\int_{x}^{\infty} \frac{e^{-y}}{y} \, dydx$ is?
The value of the integral $\displaystyle \int_{-\infty}^{\infty}\int_{-\infty}^{\infty} e^{-(3x^2+2\sqrt{2}xy + 3y^2)} \, dxdy$
I couldn't figure out the trick here. Looks like both of them are similar. Please help if anyone.
Original images:
| Here are solutions.
Solution of 1. Apply the Fubini's theorem to interchange the order of integration. The domain of integration is specified by the inequality $0 \leq x \leq y$, which tells you that
$$ \int_{0}^{\infty}\int_{x}^{\infty} \frac{e^{-y}}{y} \, dydx
= \int_{0}^{\infty}\int_{0}^{y} \frac{e^{-y}}{y} \, dxdy
= \int_{0}^{\infty} e^{-y} \, dy
= 1. $$
Solution of 2. There are severay ways of solving this. Let me first show you a solution using the gaussian integral:
$$ \int_{-\infty}^{\infty} e^{-ax^2} \, dx = \sqrt{\frac{\pi}{a}}, \qquad a > 0. \tag{*} $$
Completing the square, you have
$$
\int_{-\infty}^{\infty}\int_{-\infty}^{\infty} e^{-(3x^2+2\sqrt{2}xy + 3y^2)} \, dxdy
= \int_{-\infty}^{\infty}\int_{-\infty}^{\infty} e^{-3(x+\frac{\sqrt{2}}{3}y)^2 - \frac{7}{3}y^2} \, dxdy. $$
Integrating w.r.t. $x$ first, we may apply the substitution $x+\frac{\sqrt{2}}{3}y = \tilde{x}$. Then
\begin{align*}
\int_{-\infty}^{\infty}\int_{-\infty}^{\infty} e^{-(3x^2+2\sqrt{2}xy + 3y^2)} \, dxdy
&= \int_{-\infty}^{\infty}\int_{-\infty}^{\infty} e^{-3\tilde{x}^2 - \frac{7}{3}y^2} \, dxdy \\
&= \sqrt{\frac{\pi}{3}} \cdot \sqrt{\frac{3\pi}{7}} \\
&= \frac{\pi}{\sqrt{7}}.
\end{align*}
Solution of 2, ver.2. Let me give an alternative answer which does not require the knowledge on $\text{(*)}$. Using the polar coordinates,
\begin{align*}
&\int_{-\infty}^{\infty}\int_{-\infty}^{\infty} e^{-(3x^2+2\sqrt{2}xy + 3y^2)} \, dxdy \\
&= \int_{-\pi}^{\pi}\int_{0}^{\infty} e^{-r^2(3\cos^2 \theta + 2\sqrt{2}\cos\theta\sin\theta + 3\sin^2 \theta)} \, rdrd\theta \\
&= \frac{1}{2} \int_{-\pi}^{\pi} \frac{1}{3\cos^2 \theta + 2\sqrt{2}\cos\theta\sin\theta + 3\sin^2 \theta} \, d\theta \\
&= \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{\sec^2\theta}{3+ 2\sqrt{2}\tan\theta + 3\tan^2 \theta} \, d\theta \\
&= \int_{-\infty}^{\infty} \frac{dt}{3+ 2\sqrt{2}t + 3t^2}. \qquad (t = \tan\theta)
\end{align*}
This integral can be evaluated multiple ways.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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2nd order linear differential equation with piece wise continuous in homogeneous term. $y^{ii}=f(x)$ where
$f(x)=\begin{cases}
0 ,& \text{ if } x \in [0,1/2]\\
1, & \text{ if } x \in (1/2,1]
\end{cases} $\
subject to $y'(0)=0,y(1)=1$
We can solve this differential equation by finite difference method.But how can we find exact solution of this differential equation ? Any body please help me.
| $y^{''}=f(x)$ where
$f(x)=\begin{cases}
0 ,& \text{ if } x \in [0,\frac{1}{2}]\\
1, & \text{ if } x \in (\frac{1}{2},1]
\end{cases} $\
subject to $y^{'}(0)=0,y(1)=1$
The function must be continuous and differentiable. In $[0,\frac{1}{2}]$ we have $ y^{'} $ is constant: $ y = a + rx $. Then $ y^{'}(0)=0 $ gives: $ y=a , x \in [0,\frac{1}{2}]$.Similarly $ y=b+cx^2+dx , x \in (\frac{1}{2},1] $.
Continuity in $x=\frac{1}{2} \implies a=b+\frac{1}{4}c+\frac{1}{2}d $.
Differentiability in $x=\frac{1}{2} \implies d=-c $.
$y^{''}=1 \in (\frac{1}{2},1] \implies c=\frac{1}{2} $.
$y(1)=1 \implies b=1$. So the solution becomes:
$y=\frac{7}{8} , x \in [0,\frac{1}{2}] \\ y=1+\frac{1}{2}x^2-\frac{1}{2}x , x \in (\frac{1}{2},1]$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2109903",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Sum calculation with binomial coefficients Knowing that $$\sum_{k} \binom{a+b}{a+k} \binom{b+c}{b+k} \binom{c + a}{c + k} (-1)^k = \frac{(a + b + c)!}{a!b!c!}$$ where $a,b,c ∈ \mathbb Z $ solve the following sum
$$\sum_{k} \binom{2a}{a+k} \binom{2b}{b+k} \binom{2c}{c + k} (-1)^k $$
| We have
$$
\sum_k\binom{a+b}{a+k}\binom{b+c}{b+k}\binom{c+a}{c+k}(-1)^k
=\sum_k\frac{(a+b)!(b+c)!(c+a)!(-1)^k}{(a+k)!(b-k)!(b+k)!(c-k)!(c+k)!(a-k)!} \\
=(a+b)!(b+c)!(c+a)!\sum_k\frac{(-1)^k}{(a+k)!(b-k)!(b+k)!(c-k)!(c+k)!(a-k)!}
$$
and
$$
\sum_k\binom{2a}{a+k}\binom{2b}{b+k}\binom{2c}{c+k}(-1)^k
=\sum_k\frac{(2a)!(2b)!(2c)!(-1)^k}{(a+k)!(a-k)!(b+k)!(b-k)!(c+k)!(c-k)!} \\
=(2a)!(2b)!(2c)!\sum_k\frac{(-1)^k}{(a+k)!(a-k)!(b+k)!(b-k)!(c+k)!(c-k)!}
$$
The term behind the sum is the same in both cases, thus we have
$$
\sum_k\binom{2a}{a+k}\binom{2b}{b+k}\binom{2c}{c+k}(-1)^k \\
= \frac{(2a)!(2b)!(2c)!}{(a+b)!(b+c)!(c+a)!}\sum_k\binom{a+b}{a+k}\binom{b+c}{b+k}\binom{c+a}{c+k}(-1)^k \\
= \frac{(2a)!(2b)!(2c)!(a+b+c)!}{(a+b)!(b+c)!(c+a)!a!b!c!}
$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Convexity of $x\left(1+\frac1x\right)^x,\ x\ge 0$ This may turn out to be really simple but I do not see a quick way to the proof. How would one show $\displaystyle x\Big(1+\frac1x\Big)^x,\ x\ge 0$ is convex?
I derived the second derivative. It has a negative term. I suppose I could combine certain terms to make the negativeness disappear. But I hope there is a really clever way to see it right away.
| Let $f(x)=x\left(1+\frac{1}{x}\right)^x$, where $x>0$.
Hence,
$$f''(x)=\frac{\left(1+\frac{1}{x}\right)^x\left(x(x+1)^2\ln^2\left(1+\frac{1}{x}\right)+2(x+1)\ln\left(1+\frac{1}{x}\right)-x-3\right)}{(x+1)^2}.$$
Thus, we need to prove that
$$\ln\left(1+\frac{1}{x}\right)>\frac{\sqrt{x^2+3x+1}-1}{x(x+1)}.$$
Let $g(x)=\ln\left(1+\frac{1}{x}\right)-\frac{\sqrt{x^2+3x+1}-1}{x(x+1)}.$
Hence, $$g'(x)=\frac{2x^3+9x^2+7x+2-2\sqrt{(x^2+3x+1)^3}}{2x^2(x+1)^2\sqrt{x^2+3x+1}}=$$
$$=-\frac{11x^3+46x^2+35x+8}{2x(x+1)^2\sqrt{x^2+3x+1}\left(2x^3+9x^2+7x+2+2\sqrt{(x^2+3x+1)^3}\right)}<0,$$
which says that $g(x)>\lim\limits_{x\rightarrow+\infty}g(x)=0$ and we are done!
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Prove the inequality ${1 \over a^3 (b + c)} + {1 \over b^3 (a+c)} + {1 \over c^3 (a+b)} \ge \frac 32 $ given that $abc = 1$ Prove the inequality $${1 \over a^3 (b + c)} + {1 \over b^3 (a+c)} + {1 \over c^3 (a+b)} \ge \frac 32 $$ So, I know a proof for this, but I basically memorized it without understanding. It's
$$ {(\frac 1a + \frac 1b + \frac 1c)}^2 + {(a(b+c) + b(a+c) + c(a+b))}^-1 = {ab + bc + ac \over 2} \ge {\frac 32} $$
I don't know what inequalities were applied here or how. Another proof would also be appreciated.
| we have $$\frac{1}{a^3(b+c)}=\frac{bc}{a^2(b+c)}=\frac{(bc)^2}{a(b+c)}$$ then we get
$$\frac{(bc)^2}{a(b+c)}+\frac{(ac)^2}{b(a+c)}+\frac{(ab)^2}{c(a+b)}\geq \frac{(ab+bc+ac)^2}{2(ab+bc+ac)}\geq \frac{3}{2}$$ by $AM-GM$
| {
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"timestamp": "2023-03-29T00:00:00",
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Solve the equation $2\arcsin x=\arcsin(\frac{3}{4}x)$ $$2\arcsin x=\arcsin(\frac{3}{4}x)$$ so $x\in[-1,1]$
so we have:
$2\arcsin x=y\Rightarrow\sin\frac{y}{2}=x$ and $\arcsin x=y \Rightarrow \sin y=\frac{3}{4}x\Rightarrow\frac{4}{3}\sin y=x$ ,
$y\in[-\frac{\pi}{2},\frac{\pi}{2}]$
$$\sin\frac{y}{2}-\frac{4}{3}\sin y=0$$$$\sin\frac{y}{2}-\frac{4}{3}\cdot 2\sin\frac{y}{2}\cos\frac{y}{2}=0$$ $$\sin\frac{y}{2}\cdot (1-\frac{8}{3}\cos\frac{y}{2})=0$$
Is it done properly at this point?
1.$\sin \frac{y}{2}=0\Rightarrow x=0$
And what else?
| If $2 \arcsin(x) = \arcsin(3x/4)$, we apply $\sin$ to both sides, and use the double-angle formula and the fact that $\cos(\arcsin(x)) = \sqrt{1-x^2}$ to get
$$ 2 x \sqrt{1-x^2} = 3x/4 $$
thus either $x=0$ or $\sqrt{1-x^2} = 3/8$. Squaring both sides of the latter,
$ 1 - x^2 = 9/64$, so $x = \pm \sqrt{55}/8$. Note that each of these is in the correct interval for $\arcsin(x)$ and $\arcsin(3x/4)$ to be defined. But they are not solutions, because $2 \arcsin(\pm \sqrt{55}/8) \approx 2.372799104$ is outside of the interval $[-\pi/2, \pi/2]$ of allowed values for arcsin.
So the only correct solution is $x=0$.
| {
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Limit of a sequence by Cauchy second test The sequence is
$$ \left[ \bigg(1+\frac{1}{n}\bigg)\bigg(1+\frac{2}{n}\bigg)\bigg(1+\frac{3}{n}\bigg)\cdots\bigg(1+\frac{n}{n}\bigg) \right]^{1/n} $$
as $n$ goes to infinity. By Cauchy second test it's pretty clear that it's limit will be equal to $1+\frac{n}{n}$ which is $2$. But the answer is $4/e$. I don't know what part I'm doing wrong.
| we can write $$\left[ \bigg(1+\frac{1}{n}\bigg)\bigg(1+\frac{2}{n}\bigg)\bigg(1+\frac{3}{n}\bigg)\cdots\bigg(1+\frac{n}{n}\bigg) \right] = n!\binom{2n}{n}\cdot \frac{1}{n^n}$$
Using $$\binom{2n}{n}\leq \binom{2n}{0}+\binom{2n}{1}+\cdots \cdots \binom{2n}{2n}=2^{2n} = 4^n$$
And Using Power mean Inequality $$\frac{\binom{n}{0}^2+\binom{n}{0}^2+\cdot \cdots \binom{n}{n}^2}{n+1}\geq \bigg[\frac{\binom{n}{0}+\binom{n}{1}+\cdots \cdots \binom{n}{n}}{n+1}\bigg]^2 = \frac{4^n}{(n+1)}$$
So $$\binom{2n}{n}=\binom{n}{0}^2+\binom{n}{0}^2+\cdot \cdots \binom{n}{n}^2>\frac{4^n}{n+1}$$
So $$\frac{4^n}{n+1}<\binom{2n}{n}\leq 4^n\Rightarrow \left(\frac{4}{(n+1)^{\frac{1}{n}}}\right)<\binom{2n}{n}^{\frac{1}{n}}\leq 4$$
So using Sequeeze Theorem $$\lim_{n\rightarrow \infty}\left(\frac{4}{(n+1)^{\frac{1}{n}}}\right)<\lim_{n\rightarrow \infty}\binom{2n}{n}^{\frac{1}{n}}\leq \lim_{n\rightarrow \infty}4$$
So we get $\displaystyle \lim_{n\rightarrow \infty}\binom{2n}{n}^{\frac{1}{n}} = 4$ and Since, $\displaystyle \lim\limits_{n \to \infty} \frac{(n!)^{1/n}}{n} = e^{-1}$ we get $$\displaystyle \lim\limits_{n \to \infty}\left[ \bigg(1+\frac{1}{n}\bigg)\bigg(1+\frac{2}{n}\bigg)\bigg(1+\frac{3}{n}\bigg)\cdots\bigg(1+\frac{n}{n}\bigg) \right]^{\frac{1}{n}} = \frac{4}{e}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Solve $\lim_{x\to 0} \frac{\sin(2x)-\sin(Ax)}{x+x^3} = A^2$ How do I solve for $A$? $$\lim_{x\to 0} \frac{\sin(2x)-\sin(Ax)}{x+x^3} = A^2$$
Since the denominator evaluates to $0$, I tried doing $$\lim_{x\to 0} [\sin(2x)-\sin(Ax)]=A^2 \cdot \lim_{x\to0}[x+x^3]$$ but it would go into $0=0$.
If I checked from the graph, then I believe that $A=1$.
Any advice on how to do this? I haven't learned l'Hopital's rule yet so I would rather avoid using it for now.
Thanks!
| Hint: $\sin 2x -\sin Ax = 2\sin {\frac{2-A}{2}x} \cos {\frac{2+A}{2}x}$
Solution: if this hint is used, $$\lim_{x\to 0}{ \frac{2\sin \big( {\frac{2-A}{2}x}\big) \cos \big( {\frac{2+A}{2} x}\big) } {x(1+x^2)}}$$
$$2 \frac{2-A}{2} \lim_{x\to 0} {\frac{\cos \big( {\frac{2+A}{2}x}\big) }{1+x^2}}=2-A$$
And this is equal to $A^2$. Then $A=1$ or $A=-2$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Question on proof of $1+2+\dots+n=\frac{n(n+1)}{2}$ by induction. I saw some video where it needs to prove $1+2+\dots+n=\frac{n(n+1)}{2}$ inductively. So it has to be true if $k=1$ and $k+1$ are true.
So, for $k=1$:
$$1=\frac{1(1+1)}{2}=\frac{1(2)}{2}=\frac{2}{2}=1$$
it is valid.
For $k+1$ here is the proof he does:
$$
\begin{align}
1+2+\dots+k+k+1&=\frac{(k+1)(k+2)}{2} &(1)\\
\frac{k(k+1)}{2}+k+1&=\frac{(k+1)(k+2)}{2} &(2)\\
&=\frac{k^2+2k+k+2}{2}&(3)\\
&=\frac{k^2+3k+2}{2}&(4)\\
&=\frac{(k+1)(k+2)}{2}&\text{factoring (4)}
\end{align}$$
Therefore this formula is valid for $k+1$.
But is this true? I think not. He is just undoing what he have just done. To prove it I think I need to do this:
$$
\begin{align}
1+2+\dots+k+k+1&=\frac{(k+1)(k+2)}{2}\\
\frac{k(k+1)}{2}+k+1&=\frac{(k+1)(k+2)}{2}\\
\frac{k(k+1)+2(k+1)}{2}&=\frac{(k+1)(k+2)}{2}\\
\frac{(k+1)(k+2)}{2}&=\frac{(k+1)(k+2)}{2}\\
(k+1)(k+2)&=(k+1)(k+2)\\
k+1&=k+1\\
k&=k\\
0&=0\text{ or }1=1
\end{align}$$
Therefore, $k+1$ is valid in this formula.
Is this right or am I making a mistake somewhere?
| This proof is at least unlucky written down.
Normally, we start with the claim for $n$ (Here $1+2+\cdots n=\frac{n(n+1)}{2}$) and then proof the claim for $n+1$.
The correct way is to start with $1+2+\cdots n+n+1=\frac{n(n+1)}{2}+n+1$ (we assume that the formula is correct for $n$) and show that this is equal to $\frac{(n+1)(n+2)}{2}$ by simply adding the two fractions.
It is unusual to start with the aim of the induction step and then going back to show that it is valid.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Using induction to show that if $x_n^2 \geq 2$, then $x_{n+1}^2 \geq 2$ Let $x_1=2$. For all $n \geq 1$, define
$$x_{n+1}=\frac{1}{2} \left( x_n + \frac{2}{x_n} \right) $$
Prove by induction that $x_n^2 \geq 2$ for any $n \geq 1$.
I have a way to solve the question, which is to use AM-GM inequaity to conclude in the inductive step.
But I don't want to apply the inequality to show it. Indeed, I want to show it without using any known inequality.
I try the following in inductive step, but couldn't show what I want:
$$x_{n+1}^2 = \frac{1}{4} \left( x_n^2 + 4 + \frac{4}{x_n^2} \right) \geq \frac{1}{4} \left( 2 + 4 + \frac{4}{x_n^2} \right) = \frac{3}{2} + \frac{1}{x_n^2}$$
| Hint: Use induction to show $x_n > 0$ always and then
$$x_{n+1}=\frac{1}{2}\left(x_n + \frac{2}{x_n}\right) \geq \sqrt{2} \Leftrightarrow \\x_n + \frac{2}{x_n} \geq 2\sqrt{2} \Leftrightarrow x_n^2 + 2 \geq 2x_n\sqrt{2} \Leftrightarrow \\ x_n^2 - 2x_n\sqrt{2} + 2 \geq 0 \Leftrightarrow \left(x_n - \sqrt{2} \right)^2 \geq 0 $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2116406",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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} |
System of two cubic equations: $x+y^2 = y^3$, $y+x^2=x^3$ I got stuck on this system of equations. Could you help and tell me how should I approach this problem?
\begin{align*}
x+y^2 &= y^3\\
y+x^2 &= x^3
\end{align*}
These are the solutions:
\begin{align*}
(0, 0); \\((1+\sqrt{5})/2, (1+\sqrt{5})/2); \\((1-\sqrt{5})/2, (1-\sqrt{5})/2);
\end{align*}
| Here's your equations:
$x+y^2 = y^3$
$y+x^2 = x^3$
I could substitute
$x=y^3-y^2$
and get a degree 9 equation
in $y$,
but I'll try something else.
Looking at these,
I notice that if I subtract them,
I get something
in which everything
is divisible by
$x-y$.
Subtracting the second from the first,
I get
$(x-y)+(y^2-x^2)
=y^3-x^3
$
or
$(x-y)+(y-x)(y+x)
=(y-x)(y^2+xy+x^2)
$.
If $y \ne x$,
dividing by
$y-x$
gives
$-1+(y+x)
=y^2+xy+x^2
$
or
$0
=y^2+y(x-1)+x^2-x+1
$.
Solving this,
$\begin{array}\\
y
&=\dfrac{-(x+1)\pm \sqrt{(x-1)^2-4(x^2-x+1)}}{2}\\
&=\dfrac{-(x+1)\pm \sqrt{x^2-2x+1-4x^2+4x-4}}{2}\\
&=\dfrac{-(x+1)\pm \sqrt{-3x^2+2x-3}}{2}\\
&=\dfrac{-(x+1)\pm \sqrt{-2x^2-x^2+2x-1-2}}{2}\\
&=\dfrac{-(x+1)\pm \sqrt{-2x^2-(x-1)^2-2}}{2}\\
\end{array}
$
Since the discriminant is negative,
there are no real values of $y$.
Therefore the only real solution is
$x=y$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2117378",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 5,
"answer_id": 0
} |
Problem of Complex Numbers in Geometry using Roots of Unity Let $\omega$ be a complex number such that $\omega^7 = 1$ and $\omega \neq 1$. Let $\alpha = \omega + \omega^2 + \omega^4$ and $\beta = \omega^3 + \omega^5 + \omega^6$. Then $\alpha$ and $\beta$ are roots of the quadratic
$x^2 + px + q = 0$
for some integers $p$ and $q$. Find the ordered pair $(p,q)$.
I am unable to find a good starting place for this problem. Should I start by working backwards using the roots?
| From the development of the geometric series with reason $\omega$ we get $1 + \omega + \omega^2 + \dots + \omega^6 = (\omega^7 - 1) / (\omega - 1) = 0$, from which we deduce $1 + \alpha + \beta = 0$.
Using the same development again, $\alpha \beta = 1 + \omega + \omega^2 + \omega^3 + \omega^4 + \omega^5 + \omega^6 + 1 + 1 = 2$.
$\alpha$ and $\beta$ being roots of $x^2 + px + q$ gives $x^2 + px + q = (x - \alpha)(x - \beta) = x^2 - (\alpha + \beta) x + \alpha \beta$ from which we deduce the equations
$\alpha + \beta = -p = 1$ and $\alpha \beta = q = 2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2118914",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
} |
Why does the discriminant tell us how many zeroes a quadratic equation has? The quadratic formula states that:
$$x = \frac {-b \pm \sqrt{b^2 - 4ac}}{2a}$$
The part we're interested in is $b^2 - 4ac$ this is called the discriminant.
I know from school that we can use the discriminant to figure out how many zeroes a quadratic equation has (or rather, if it has complex, real, or repeating zeroes).
If $b^2-4ac > 0$ then the equation has 2 real zeroes.
If $b^2-4ac < 0$ then the equation has 2 complex zeroes.
If $b^2-4ac = 0$ then the equation has repeating zeroes.
But I don't uderstand why this works.
| If the equation is $ax^2+bx+c=0$, with $a\ne0$, it is equivalent to
$$
4a^2x^2+4abx+4ac=0
$$
that can also be rewritten as
$$
4a^2x^2+4abx+b^2=b^2-4ac
$$
or, recognizing the square on the left-hand side,
$$
(2ax+b)^2=b^2-4ac
$$
Now, if $b^2-4ac<0$, we cannot find a real number $x$ such that $(2ax+b)^2=b^2-4ac$, because $(2ax+b)^2\ge0$.
If $b^2-4ac=0$, the equation becomes $(2ax+b)^2=0$, that is, $2ax+b=0$, which has a single solution $x=-\frac{b}{2a}$.
If $b^2-4ac>0$, then the equation splits into two:
$$
2ax+b=\sqrt{b^2-4ac},\qquad 2ax+b=-\sqrt{b^2-4ac}
$$
so we have two distinct real solutions.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2121646",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "37",
"answer_count": 9,
"answer_id": 3
} |
Evaluate: $\frac{1}{2^1}-\frac{2}{2^2}+\frac{3}{2^2}-\frac{4}{2^4}+\ldots-\frac{100}{2^{100}}$ The problem is to evaluate the following sum:
$$\frac{1}{2^1}-\frac{2}{2^2}+\frac{3}{2^3}-\frac{4}{2^4}+\ldots-\frac{100}{2^{100}}$$
My approach was to find the common denominator ($2^{100}$), then the series becomes:
$$ \frac{1\cdot 2^{99}-2\cdot 2^{98}+3\cdot 2^{97}-4\cdot 2^{96}+\ldots -100\cdot 2^0}{2^{100}}$$
And split it this way:
$$ \frac{(1\cdot 2^{99}+3\cdot 2^{97}+\ldots+99\cdot 2^{1})-(2\cdot 2^{98}+4\cdot 2^{96}+\ldots+100\cdot 2^{0})}{2^{100}}$$
But it seems that cacelling out the terms is not a promising approach. Any hints?
| Try to multiply the given sum by $\frac{3}{2}=1+\frac{1}{2}$:
$$\begin{eqnarray*}\left(1+\frac{1}{2}\right)\sum_{n=1}^{100}\frac{(-1)^{n+1}n}{2^n}&=&\sum_{n=1}^{100}\frac{(-1)^{n+1}n}{2^n}+\sum_{n=1}^{100}\frac{(-1)^{n+1}n}{2^{n+1}}\\&=&\sum_{n=1}^{100}\frac{(-1)^{n+1}n}{2^n}+\sum_{n=2}^{101}\frac{(-1)^{n}(n-1)}{2^{n}}\\&=&\frac{1}{2}-\frac{100}{2^{101}}-\sum_{n=2}^{100}\frac{(-1)^n}{2^n}\end{eqnarray*}$$
then perform the same manipulation with $\sum_{n=2}^{100}\frac{(-1)^n}{2^n} = \frac{1}{6}\left(1+\frac{1}{2^{99}}\right)$. That leaves you with:
$$ \sum_{n=1}^{100}\frac{(-1)^{n+1}n}{2^n} = \color{red}{\frac{1}{9}\left(2-\frac{302}{2^{100}}\right)}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2122943",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
How can I show that modul of this complex number equal to 1? Let $z$ be a complex number such that
$$(1+2 i) \left| z\right| -\frac{\sqrt{10}}{z}+2-i=0.$$
Prove that $\left| z\right|=1$.
I tried
Put $z = a + bi$ $(a,b \in \mathbb{R})$, we have
$$2-\frac{\sqrt{10}a}{a^2+b^2}+\sqrt{a^2+b^2}+i \left(-1+\frac{\sqrt{10}b}{a^2+b^2}+2\sqrt{a^2+b^2}\right) = 0.$$
Then
$$\begin{cases}
2-\dfrac{\sqrt{10}a}{a^2+b^2}+\sqrt{a^2+b^2}=0,\\
-1+\dfrac{\sqrt{10}b}{a^2+b^2}+2\sqrt{a^2+b^2}=0
\end{cases}
\Leftrightarrow
\begin{cases}
2(a^2+b^2) +(a^2+b^2) \sqrt{a^2+b^2} - \sqrt{10}a=0,\\
-(a^2+b^2) +2(a^2+b^2) \sqrt{a^2+b^2} + \sqrt{10}b=0.\\
\end{cases}$$
$$
5(a^2+b^2) \sqrt{a^2+b^2} = (2a+b)\sqrt{10}.
$$
From here, we have $2 a + b >0$.
I think, I prove
\begin{equation}
(2a+b)\sqrt{10} \leqslant 5 \sqrt{a^2+b^2}
\end{equation}
Or
$$5 (3 a-b) (a+3 b)\leqslant 0.$$
But, I can't above inequality.
$$5(a^2+b^2) \sqrt{a^2+b^2} \leqslant 5\sqrt{a^2 + b^2}\Leftrightarrow a^2 + b^2 \leqslant 1.$$
| Hint: note that $1+2i= i(2-i)$ then collect and rewrite:
$$
(2-i)(1+i|z|)= \frac{\sqrt{10}}{z}
$$
Note that $|2-i|^2=5$ and take the square of the modulus on both sides:
$$
5(1+|z|^2) = \frac{10}{|z|^2}
$$
The latter is a simple quadratic in $|z|^2$ which gives $|z|=1$ in the end.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2125275",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
find the roots of the following: $z^3=2-2i$ in $Re^{i\theta}$ form Question
find the roots of the following: $z^3=2-2i$ in $Re^{i\theta}$ form
my steps:
So i first find the magnitude of the root to get $|2-2i|^{\frac{1}{3}}$
and the argument of the root is $\frac{\frac{-\pi}{4}}{3}$
thus the first root must be $|2-2i|e^{i*\frac{\frac{-\pi}{4}}{3}}$
But then i was confused in finding the other roots as well
| Cube roots are equally distributed around a circle centered at the pole and having radius equal to the cube root of the modulus (absolute value).
In the case of $z^3=2-2\,i$ the modulus is, as you stated, $\vert 2-2\,i\vert=2\sqrt{2}$. So the modulus of the roots is $\sqrt[3]{2\sqrt{2}}=\sqrt{2}$.The argument for $2-2\,i$ is $-\frac{\pi}{4}$ so the argument for one of the cube roots is one-third of that, namely as you state, $-\frac{\pi}{12}$.
All the three cube roots of $2-2\,i$ are equally spaced about the circle containing them, so they are separated by angles of $\frac{2\pi}{3}=\frac{8\pi}{12}$. So the three arguments of the roots are $-\frac{\pi}{12},\,\frac{7\pi}{12}$ and $\frac{15\pi}{12}=\frac{5\pi}{4}$.
Thus, the three cube roots of $2-2\,i$ are
$$ \sqrt{2}\exp\left(-\frac{\pi}{12}\right),\quad \sqrt{2}\exp\left(\frac{7\pi}{12}\right),\quad \sqrt{2}\exp\left(\frac{5\pi}{4}\right) $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2131110",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Finding the area between a line and a curve
The two equations are $x+1$ and $4x-x^2-1$.
The answer is $\frac{1}{6}$, but I've done it 4 different times and gotten -$\frac{15}{2}$ each time.
My working:
*
*$x+1$ = $4x-x^2-1$
*$x^2-3x+2 = 0$
*$(x-1)(x-2)$ means $x=1$ or $x=2$
*$\int_1^2$ $3x-x^2$
*$[\frac{3x^2}{2}-\frac{x^3}{3}]_1^2$
*$\frac{3(1)^2}{2}-\frac{(1)^3}{3}$ = $\frac{3}{2}-\frac{1}{3}$
*$\frac{3(2)^2}{2}-\frac{(2)^3}{3}$ = $\frac{12}{2}-\frac{8}{3}$
*($\frac{3}{2}-\frac{1}{3}$)-($\frac{12}{2}-\frac{8}{3}$) = $-\frac{9}{2}-\frac{9}{3}$
*-$\frac{15}{2}$
| There is a mistake in the line below:
*$[\frac{3x^2}{2}-\frac{x^3}{3}]_1^2$
Actually you should be integrating the difference of the $2$ curves within that limit, i.e. $(4x-x^2-1)-(x+1)=3x-x^2-2$, since each of them represents the area under the curve and bounded by the x-axis.
The magnitude of the enclosed area will, accordingly, be
$$\begin{align} & \int_1^2 \{(4x-x^2-1)-(x+1)\} \mathrm{dx} \\ &
=\int_1^2 (3x-x^2-2) \mathrm{dx} \\ &
=\left[\frac{3x^2}{2}-\frac{x^3}{3}-2x\right]_1^2 \\ &
=\frac{3(2^2-1)}{2}-\frac{2^3-1}{3}-2(2-1) \\ &
=\frac{9}{2}-\frac{7}{3}-2 \\ &
=\frac{27-14-12}{6} \\ &
=\color{blue}{\frac{1}{6}}\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2131864",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 3
} |
Which integration techniques I should use for $\int{\frac{\sqrt{x^2-3}-3\sqrt{x^2+3}}{\sqrt{x^4-9}}}dx$
$$\int{\frac{\sqrt{x^2-3}-3\sqrt{x^2+3}}{\sqrt{x^4-9}}}dx$$
I can simplify it to:
$$\int{\frac{dx}{\sqrt{x^2+3}}} - 3\int{\frac{dx}{\sqrt{x^2-3}}}$$
but I can't go from here.
| 1.$\displaystyle\int{\frac{1}{\sqrt{x^2+3}}}dx=\int\frac{1}{\sqrt3\sqrt{\frac{x^2}{\sqrt3^2}+1}}dx $
Substitute $\tan u=\frac{x}{\sqrt3}\rightarrow dx=\sqrt3\sec^2udu$, Then
$$\begin{align}\int\frac{1}{\sqrt3\sqrt{\frac{x^2}{\sqrt3^2}+1}}dx&=\int\frac{\sec^2u}{\sqrt{\tan^2u+1}}du\\&=\int \sec udu\\&=\ln(\vert \sec u+\tan u\vert)+C\\&=\ln(\vert\sec(\tan^{-1}\frac{x}{\sqrt3})+\tan(\tan^{-1}\frac{x}{\sqrt3}) \vert)+C\\&=\ln(\vert x+\sqrt{x^2+3} \vert)+C\end{align}$$
2.$\displaystyle\int{\frac{1}{\sqrt{x^2-3}}}dx=\frac{1}{\sqrt3}\cdot\int{\frac{1}{\sqrt{\frac{x^2}{\sqrt3^2}-1}}}dx$
Substitute $\sec v=\frac{x}{\sqrt3}\rightarrow dx=\sqrt3\tan v\sec vdv$, Then
$$\begin{align} \frac{1}{\sqrt3}\cdot\int{\frac{1}{\sqrt{\frac{x^2}{\sqrt3^2}-1}}}dx&=\int\frac{\tan v\sec v}{\tan v}dv\\&=\int\sec v dv\\&=\ln(\vert\sec v+\tan v\vert)+C\\&=\ln(\vert\sec(\sec^{-1} \frac{x}{\sqrt3})+\tan(\sec^{-1}\frac{x}{\sqrt3})\vert)+C\\ &=\ln(\vert x+\sqrt{x^2-3}\vert)+C \end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2133187",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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Compute $\int_{0}^{\infty} \frac{dx}{(a^2+x^2)(b^2+x^2)} $ using fourier transform of $e^{-a|x|}$ Compute $\int_{0}^{\infty} \frac{dx}{(a^2+x^2)(b^2+x^2)} $ using fourier transform of $e^{-a|x|}$.
I computed the fourier transform of $e^{-a|x|}$, which is $\frac{a}{\pi (a^2+w^2)}$
I'm not sure how to continue from here, I tried using this formula $f(x) = \int_{-\infty}^{\infty} \frac{dx}{(a^2+x^2)}e^{iwx}dw $. But it didn't work. Any suggestions?
I looked at this post: but I still didn't understand how parseval's theorem is used there. How do I show $\int_{-\infty}^\infty \frac 1{(a^2+s^2)(b^2+s^2)} ds=\frac {\pi}{ab(a+b)}$ using the solution to the following Fourier transform?
This wont work: Parseval's theorem
| Actually:
$$ \mathcal{F} \left [ e^{- a |x|} \right ] (\omega) = \frac{ 2a}{a^2 + \omega^2} $$
As for the integral, you could use one of Plancherel's formulae:
$$ \int_{-\infty}^{+\infty} f(t) \overline{g (t)} \ \mathrm dt = \frac{1}{2\pi}\int_{-\infty}^{+\infty} \widehat{f} (\omega) \overline{ \widehat{g} (\omega)} \ \mathrm d \omega $$
In this case:
$$ \begin{aligned} \int_{0}^{+\infty} \frac{\mathrm dx}{ \left ( a^2 + x^2 \right ) \left ( b^2 + x^2 \right )} &= \frac{1}{2} \int_{-\infty}^{+\infty} \frac{\mathrm dx}{ \left ( a^2 + x^2 \right ) \left ( b^2 + x^2 \right )} \\ &= \frac{\pi}{4ab} \int_{-\infty}^{+\infty} e^{- (a + b) |\omega |} \ \mathrm d \omega \\ &= \frac{\pi}{4ab} \left [ \int_{-\infty}^{0} e^{ (a+b) \omega} \ \mathrm d\omega + \int_{0}^{+\infty} e^{-(a+b) \omega} \ \mathrm d\omega \right ] \\ &= \frac{\pi}{4ab} \left [ \frac{1}{a + b} + \frac{1}{a+b} \right ] \\ &= \frac{ \pi}{2ab (a+b)} \end{aligned} $$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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What are the solutions to $a^2+ab+b^2$ $=$ $3^n$? What are all known integer solutions ($a, b, n$) to $a^2+ab+b^2$ $=$ $3^n$ besides ($1, 1, 1$) and ($-2, 1, 1$)? Do any others even exist?
This question comes from the identity that ($a^3±b^3$)/($a±b$) $=$ $0, 1$ $\pmod 3$. If ($a^3±b^3$)/($a±b$) $=$ $0$ $\pmod 3$, then let $3^n$ be highest power of $3$ dividing ($a^3±b^3$)/($a±b$). ($a^3±b^3$)/(($a±b$)$*3^n$) $=$ $1$ $\pmod 3$. When does ($a^3±b^3$)/(($a±b$)$*3^n$) $=$ $1$?
| The set of solutions to $$ x^2 + xy + y^2 = 1 $$ in integers is finite (6).
x = 1, y = 0 target 1
x = -1, y = 0 target 1
x = 1, y = -1 target 1
x = -1, y = 1 target 1
x = 0, y = 1 target 1
x = 0, y = -1 target 1
The set of solutions to $$ x^2 + xy + y^2 = 3 $$ in integers is finite(6).
x = 2, y = -1 target 3
x = -2, y = 1 target 3
x = 1, y = 1 target 3
x = -1, y = -1 target 3
x = 1, y = -2 target 3
x = -1, y = 2 target 3
If $$ x^2 + xy + y^2 \equiv 0 \pmod 9 $$ then both $x,y$ are divisible by $3.$
This means that all solutions to your $3^n$ thing are $3^w$ times the items in the first two (finite) sets.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2136104",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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Find the Derivative of $f(x)=\frac{7}{\sqrt {x}}$ using the definition. I get that $$\frac{d}{dx}\left(7\times\dfrac{1}{\sqrt{x}}\right)=\frac{d}{dx}(7x^{.5})=\dfrac{7}{2}x^{-.5}$$ is the derivative, but I can't ever use $\lim_{h \to 0} \dfrac{f(x+h)-f(x)}{h}$.
If someone or anyone could go step by step and do the problem, I would be eternally grateful.
| $$\begin{align*}f'(x)&=\lim_{h\to0}\frac{f(x+h)-f(x)}{h}\\
&= \lim_{h\to 0} \frac7h\cdot \left( \frac1{\sqrt{x+h}} -\frac1{\sqrt{x}} \right)\\
&= \lim_{h\to 0} \frac7h\cdot\left(\frac{\sqrt x-\sqrt{x+h}}{\sqrt{x(x+h)}} \right)\\
&= \lim_{h\to 0} \frac7h\cdot\left(\frac{\sqrt x-\sqrt{x+h}}{\sqrt{x(x+h)}} \cdot \frac{\sqrt{x}+\sqrt{x+h}}{\sqrt{x}+\sqrt{x+h}} \right)\\
&= \lim_{h\to 0} \frac7h\cdot\left(\frac{x-(x+h)}{\sqrt{x(x+h)}\cdot\left(\sqrt{x+h}+\sqrt x\right)}\right)\\
&=\lim_{h\to 0}\frac{-7}{\sqrt{x(x+h)}\cdot\left(\sqrt{x+h}+\sqrt x\right)}\\
&=\frac{-7}{x\cdot 2\sqrt x}
\end{align*}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2139090",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
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Finding the solution of equation I have to find out the integral solution of a equation $ax+by=c$ such that $x \geq 0$ and $y \geq 0$ and value of $(x+y)$ is minimum.
I know if $c \equiv 0 \pmod{\gcd(a,b)}$ then it's always possible. How to find the values of $x$ and $y$?
| From the method of solving a linear Diophantene Equation we have that the solutions are $x = x_0 \pm n\cdot\frac{b}{(a,b)}$ and $y = y_0 \mp n \cdot \frac{a}{(a,b)}$, where the pair $x_0, y_0$ is a solution of the equation.
Now assume that $x+y \le x_0 + y_0$. If $x = x_0 + n\cdot\frac{b}{(a,b)}$, then $y = y_0 - n\cdot\frac{a}{(a,b)}$, this is equivalent to $b \le a$. If $x = x_0 - n\cdot\frac{b}{(a,b)}$, then $y = y_0 + n\cdot\frac{a}{(a,b)}$, this is equivalent to $a \le b$. So hence find one solution and if $a \le b$ then choose $n = \left\lfloor \frac{x_0(a,b)}{b} \right\rfloor$. Then $x = x_0 - n\cdot\frac{b}{(a,b)}$ and $y = y_0 + n\cdot\frac{a}{(a,b)}$ give you the minumum sum of $x+y$. Similarly if $b \le a$ then choose $n = \left\lfloor \frac{y_0(a,b)}{a} \right\rfloor$ so $x = x_0 + n\cdot\frac{b}{(a,b)}$ and $y = y_0 - n\cdot\frac{a}{(a,b)}$ will give you the minumum sum of $x+y$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2139172",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How to prove in this trig problem i have to prove this
$$\frac{\cos 3x}{\sin 2x \sin 4x}+\frac{\cos 5x}{\sin 4x \sin 6x}+\frac{\cos 7x}{\sin 6x \sin 8x}+\frac{\cos 9x}{\sin 8x \sin 10x} = \frac{1}{2}\csc x(\csc 2x - \csc 10x)$$
i tried taking lcm but does not leads to anything. i believe i have to write numerator as something in terms of denominator which i do not know how. Help.
Thanks
| Hint: Observe
\begin{align}
\frac{\cos 3x}{\sin 2 x \sin 4x } = \frac{1}{2\sin x}\left(\frac{1}{\sin 2x}-\frac{1}{\sin 4x} \right)
\end{align}
since
\begin{align}
\frac{1}{\sin 2x}-\frac{1}{\sin 4x}=&\ \frac{\sin 4x-\sin 2x}{\sin 2x \sin 4x}\\
=&\ \frac{\sin 3x\cos x+\sin x\cos 3x-\sin 3x \cos x+\sin x\cos 3x}{\sin 2x \sin 4x}\\
=&\ \frac{2\cos 3x \sin x}{\sin 2x \sin 4x}.
\end{align}
Telescoping sum.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Inequality trouble: $(a^2+ab+b^2)(b^2+bc+c^2)(c^2+ca+a^2) \geq (ab+bc+ca)^3$ The following inequality is exercise 1.8 from this book.
For any real $a,b,c$, prove the following $$(a^2+ab+b^2)(b^2+bc+c^2)(c^2+ca+a^2) \geq (ab+bc+ca)^3.$$
I've managed to prove this via brut-force and Muirhead's inequality (Very unsatisfying). However I'm having difficulty understanding the solution in the back of the book. I'm also interested if anyone else knows of any nice proofs of this inequality?
In particular, the solution at the back says the following (word for word):
If we can show that $\frac{27}{64}(a+b)^2(b+c)^2(c+a)^2\geq (ab+bc+ca)^2$, then the conclusion follows. Denote $S_1=a+b+c$, $S_2=ab+bc+ca$ and $S_3=abc$. We need to show that $27(S_1S_2-S_3)^2\geq 64S_2^3.$
It then goes onto prove the last inequality via cases. However I'm struggling to see how it is sufficient to prove either of those two inequalities.
We can rewrite $a^2+ab+b^2 = (\frac{a+b}{2})^2+\frac{3}{4}(a^2+b^2)$ and so by AM-GM it is sufficient to prove $$\sqrt{\frac{27}{64}(a+b)^2(b+c)^2(c+a)^2}\sqrt{(a^2+b^2)(b^2+c^2)(c^2+a^2)} \geq (ab+bc+ca)^3 .$$ However $$\sqrt{(a^2+b^2)(b^2+c^2)(c^2+a^2)} \geq (ab+bc+ca)^2 $$ is clearly wrong and so this does not justify the first claim in the solution. This does give me an idea of where the $\frac{27}{64}$ term comes from however. Other than this I'm a bit stuck and any help would be appreciated.
Edit: I realised straight after posting that $a^2+ab+b^2 \neq (\frac{a+b}{2})^2+\frac{3}{4}(a^2+b^2)$. So in fact my work is nonsense and can be ignored.
| I found this identity: $$\text{LHS-RHS} = \frac{1}{6} \sum\, \left( a-b \right) ^{2}\Big[ \left( a+b+c \right) ^{2}{c}^{2}+2\,
\left( ab+ca+bc \right) ^{2}\Big] \geqq 0$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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What is the $\prod \frac{1}{n-1}$ I am trying to compute the probability of none the events occurs where the probability for each event is $Pr[A_i]=\frac{1}{n-1}$ for all i and these events are independent.
What is the $\prod_{i=3}^{n} \frac{1}{n-1}$ when n >= 3
I know that the Pr(none event occur) = 1 - Pr(at least one occur)
= 1 - $\prod_{i=3}^{n} \frac{1}{n-1}$
I want to proof that the probability that none of them occur is ≥1/8
| *
*$\prod_{1}^{n} \frac{1}{n-1} = (\frac{1}{n-1})^{n}$
*$\Pr($none event occur$) = 1-(\frac{1}{n-1})^{n}$
For any $n\geq 3$, we have
$log(\frac{8}{7}) < n.log(n-1) \Rightarrow -log(\frac{8}{7}) > -n.log(n-1) \Rightarrow log(\frac{8}{7})^{-1} > n.log(n-1)^{-1} \Rightarrow log(\frac{7}{8}) > n.log(\frac{1}{n-1}) \Rightarrow \frac{7}{8} > (\frac{1}{n-1})^{n} \Rightarrow 1-\frac{1}{8}>(\frac{1}{n-1})^{n} \Rightarrow 1-(\frac{1}{n-1})^{n} >\frac{1}{8} \Rightarrow \Pr(none\text{ }event\text{ }occur) > \frac{1}{8}$
(All logarithms are in base 2)
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Creating a non-recursive formula I'm a bit stuck on how to convert a non-recursive formula.
Assuming:
f(0) = 0
f(1) = 0
f(n) = f(n - 2) + $2^{n-1}\ if\ n \ge 1$
I can see that:
f(2) = 0 + $2^{2-1}$
f(3) = 0 + $2^{3-1}$
f(4) = (0 + $2^{2-1})\ +\ 2^{4-1}$
f(5) = (0 + $2^{3-1})\ +\ 2^{5-1}$
f(6) = ((0 + $2^{2-1})\ +\ 2^{4-1}) + 2^{6-1}$
f(7) = ((0 + $2^{3-1})\ +\ 2^{5-1}) + 2^{7-1}$
therefore the pattern appears to fit:
f(n) = $2^1 + 2^3 + ... + 2^{n-3} + 2^{n - 1} \ \ \ \ \ if\ n = even$
f(n) = $2^2 + 2^4 + ... + 2^{n-3} + 2^{n - 1} \ \ \ \ \ if\ n = odd$
I can understand how to write this as a sum of $\sum_{k=1}^n2^k$ or $\sum_{k=2}^n2^k$, however how do I convert this to a non-recursive formula if the variable being changed is used as the exponent.
| See, that you create geometric series, so
$$f(n)=\begin{cases}2\frac{1-4^{\frac{n}{2}}}{1-4} &, n\in 2\mathbb{Z}\\
4\frac{1-4^{\frac{n-1}{2}}}{1-4} &, n\not\in 2\mathbb{Z}\end{cases}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2145750",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Compute the following without the calculator
$$4\left(5+3\sqrt2\over 2\right)^4-16\left(5+3\sqrt2\over 2\right)^3-17\left(5+3\sqrt2\over 2\right)^2+27\left(5+3\sqrt2\over 2\right)-3$$
Please solve the following equation without using calculator.
Substituting $\left(5+3\sqrt2\over 2\right)$ to x must be the first step, but then I don't know how to factor it.
| We know that $$4x^4-16x^3-17x^2+27x-7 =(4x^2-20x+7)(x^2+x-1) $$ $$=(x- \frac{5+3\sqrt {2}}{2})(x-\frac {5-3\sqrt {2}}{2})(x-\frac {-1 +\sqrt {5}}{2})(x+\frac {-1-\sqrt {5}}{2}) $$
Observe that $\frac {5+3\sqrt {2}}{2} $ is a root of this polynomial. Also observe that the above polynomial is $4$ less than the required polynomial in question. Thus, the answer is $4$. Hope it helps.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Evaluate the given limit Given a function $f : R → R$ for which $|f(x) − 3| ≤ x^2$. Find
$$\lim_{ x\to0}\frac{f(x) - \sqrt{x^2 + 9}}{x}$$
Can the function $f(x)$ be considered as $x^2 + 3$ and go about evaluating the limit using the Limit laws?
|
"Can the function $f(x)$ be considered as $x^2 + 3$ and go about solving the limit using the Limit laws?"
No, since we have only that $|f(x)-3|\le x^2\implies 3-x^2\le f(x)\le 3+x^2$.
But we can proceed by using $\color{blue}{f(x)-3=O(x^2)}$, where we are using the ("Big O notation").
Then, we can evaluate the limit of interest by writing
$$\begin{align}
\frac{f(x)-\sqrt{x^2+9}}{x}&=\frac{f(x)-3\left(1+\frac{x^2}{9}\right)^{1/2}}{x}\\\\
&=\frac{f(x)-3\left(1+\color{red}{\frac12 \frac{x^2}{9}+O(x^4)}\right)}{x}\\\\
&=\frac{\color{blue}{\left(f(x)-3\right)}+\color{red}{O(x^2)}}{x}\\\\
&=\frac{\color{blue}{O(x^2)}+\color{red}{O(x^2)}}{x}\\\\
&=O(x)\to 0\,\,\text{as}\,\,x\to 0
\end{align}$$
And we are done!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2147380",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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"answer_id": 0
} |
How to use powers on matrices In the questions compute $\begin{pmatrix} \sqrt{3} & -1 \\ 1 & \sqrt{3} \end{pmatrix}^6$ and $\begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}^{99}$, how would you solve these?
| The complex numbers are isomorphic to the set of matrices of the form
$$
a+bi \sim
\begin{pmatrix} a & -b \\ b & a \end{pmatrix}
=a I + b J,
\text{ where }
J=\begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix},
J^2=-I
$$
Your first matrix corresponds to $z=\sqrt3 +i$.
Now $\dfrac{iz}{2}=\dfrac{-1+\sqrt3}{2}$ is a third root of unity and so $\left(\dfrac{iz}{2}\right)^6=1$.
Therefore $z^6=-64$ and the corresponding matrix is $-64I$.
The second matrix can be written $A=I+N$ where $N^2=0$.
Therefore, $A^k=I+kN$ by the binomial theorem.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2148061",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
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Solve $\lim_{x \rightarrow 1}\frac{3\ln(x) - x^2+1}{x-1}$ without using L'Hôpital's I tried:
$$\lim_{x \rightarrow 1}\frac{3\ln(x) - x^2+1}{x-1} = \\
\frac{3\ln(x) + (1-x^2)}{-1(1-x)} = \\
\frac{3\ln(x)+ (1-x)(1+x)}{-(1-x)} = \\
\frac{3\ln(x)}{x-1} + \frac{1+x}{-1} = \\
\frac{\ln{x^3}}{x-1} - 1-x = \\
???$$
What do I do next? Remember, I can't use L'Hôpital.
| Substitute $$y=x-1$$ to get $$\frac{\ln(x^3)}{x-1}=\frac{3\ln(y+1)}{y}$$
and now use $$\ln(y+1)=y+O(y^2)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2150751",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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Prove that $0 \le \frac{1+\cos\theta}{2+\sin\theta} \le \frac{4}{3}$ for all real $\theta$. I've tried substituting $1+\cos\theta=2\cos^2(\frac{\theta}{2})$ but it didn't give me the answer.
| Let $\dfrac{1+\cos\theta}{2+\sin\theta}=y$
$$\iff2y-1=\cos\theta-y\sin\theta=\sqrt{y^2+1}\cos\left(\theta+\arccos\dfrac1{\sqrt{y^2+1}}\right)$$
$$-\sqrt{y^2+1}\le2y-1\le\sqrt{y^2+1}$$
$$\implies(2y-1)^2\le y^2+1\iff0\ge3y^2-4y=3y\left(y-\dfrac43\right)$$
$$\implies0\le y\le\dfrac43$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2150994",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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For $a,b,c>0$. Minimize $P=a+b+c$ For $a,b,c>0$ and $\frac{2}{a}+\frac{5}{b}+\frac{3}{c}=1$, minimize $$P=a+b+c$$
| By Cauchy Schwarz:
$$
1\times P=\left(\frac{2}{a}+\frac{5}{b}+\frac{3}{c}\right)(a+b+c)\geq(\sqrt{2}+\sqrt{5}+\sqrt{3})^2
$$
so $\min P=(\sqrt{2}+\sqrt{5}+\sqrt{3})^2$ when
$$
\frac{2}{a^2}=\frac{5}{b^2}=\frac{3}{c^2}\quad\text{and}\quad\frac{2}{a}+\frac{5}{b}+\frac{3}{c}=1;
$$
i.e.
$$
a=2+\sqrt{6}+\sqrt{10},\quad b=5+\sqrt{10}+\sqrt{15},\quad c=3+\sqrt{6}+\sqrt{15}.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2152933",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove that $a^2 + b^2 + c^2 \lt 2(ab + bc + ca)$ Given that $a, b$ and $c$ are the sides of a triangle.
How to prove that $a^2 + b^2 + c^2 \lt 2(ab + bc + ca)$?
Maybe any hint? Am I going to wrong direction?
$$2(ab + bc + ca)-a^2 + b^2 + c^2>0$$
$$2ab + 2bc + 2ca-a^2 + b^2 + c^2>0$$
$$2b(a+c) + 2ca-a^2 + b^2 + c^2>0$$
...?
| Note that we have $$a^{2}>(b-c)^{2}\;,b^{2}>(a-c)^{2}\;,c^{2}>(b-a)^{2}$$
As $a,b,c$ are sides of a trinagle. Adding these, we get $$a^{2}+b^{2}+c^{2}>2a^{2}+2b^{2}+2c^{2}-2ab-2bc-2ca$$
Which is equivalent to $$a^2+b^2+c^2<2ab+2bc+2ca$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2154850",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 1
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Producing circles with $r^2=x^2+y^2$ $r^2=x^2+y^2\ +\pi $ produces graph with two circles :
$r = 2$
This graph was produced using Desmos, is this correct or a quirk of the graph software ?
Ive noticed $\forall n \epsilon N [r^2=x^2+y^2+n\pi]$ appears to produce same circle, so $r^2=x^2+y^2\ +3\pi $ produces same circle as $r^2=x^2+y^2\ +4\pi $ :
Why is this the case, should each circle not be differing sizes ?
| The equation $2^2=x^2+y^2+\pi$ is equivalent to $x^2+y^2=4-\pi=0.8584\ldots$, which produces a circle of radius $\sqrt{4-\pi}=\sqrt{0.8584\ldots} = 0.9265\ldots$, which is the red circle in the top figure.
The equation $2^2=x^2+y^2+3\pi$ is equivalent to $x^2+y^2=4-3\pi=-5.4247\ldots$. Since that's a negative number, there are no solutions to the equation for real numbers $x$ and $y$. So there is no red circle in the bottom figure.
I don't know what the orange circle means in each figure. Try consulting the software's manual?
| {
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"timestamp": "2023-03-29T00:00:00",
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How to prove the two identities in number theory are equivalent? Let
\begin{align}
Li_2(z) = \sum_{n=1}^{\infty} \frac{z^n}{n^2}.
\end{align}
There are two different forms of Abel identities for polylogarithms:
1.
\begin{align}
& Li_2(-x) + \log x \log y \\
& + Li_2(-y) + \log ( \frac{1+y}{x} ) \log y \\
& + Li_2(-\frac{1+y}{x}) + \log ( \frac{1+y}{x} ) \log (\frac{1+x+y}{xy}) \\
& + Li_2(-\frac{1+x+y}{xy}) + \log ( \frac{1+x}{y} ) \log (\frac{1+x+y}{xy}) \\
& + Li_2(-\frac{1+x}{y}) + \log ( \frac{1+x}{y} ) \log x \\
& = - \frac{\pi^2}{2}.
\end{align}
2.
\begin{align}
& Li_2(x) + Li_2(y) - Li_2(xy) \\
& = Li_2(\frac{x(1-y)}{1-xy}) + Li_2(\frac{y(1-x)}{1-xy}) + \log(\frac{1-x}{1-xy}) \log(\frac{1-y}{1-xy}).
\end{align}
Are there some references of proving that these two identities are equivalent? Thank you very much.
| My comment is a bit too long therefore I write it as an answer.
Thanks for the link, it makes the question much clearer. With $(6.63)$ follows directly $2.$ by the substitution $x\to\frac{x}{1-y}$ and $y\to\frac{y}{1-x}$ . The script uses the symbol-technique $S$ to transform the terms of $(6.63)$ into other terms getting an equation which is somehow equivalent to $1.$ . This equation now has to be retransformed to $1.$ and the constant must be still determined. So it depends on the definition of "equivalent" if $1.$ und $2.$ are equivalent. It's the same question, if e.g. $\frac{1}{(1-x)^2}=\sum\limits_{n=1}^\infty nx^{n-1}$ and $\frac{1}{1-x}=\sum\limits_{n=0}^\infty x^n$ are equivalent. I would say: Because of the difference with the constant it's an incomplete Yes (that $1.$ and $2.$ are equivalent). I hope you will find a better expression than me. :-)
| {
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Prove that $(n^3 - 1)n^3(n^3 + 1)$ is divisible by $504$ How to prove that $(n^3 - 1)n^3(n^3 + 1)$ is divisible by $504$? Factoring $504$ yields $2^3 \cdot 3^2 \cdot 7$. Divisibility by $8$ can be easily seen, because if $n$ is even then $8 | n^3$, else $(n^3 - 1)$ and $(n^3 + 1)$ are both even and one of these is divisible by $4$ so $8|(n^3 - 1)(n^3 + 1)$. I'm stuck at proving divisibility by $9$ and $7$
| $(n^3 - 1)n^3(n^3 + 1) =n^3(n^6 -1)$
Since $\phi(7)=\phi(9)=6$, we get that $7$ and $9$ always divide $n^3(n^6 -1)$, by Euler–Fermat.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Finding null space of matrix. I need to make sure I'm understanding this correctly.
I skipped a few steps to reduce typing, but let me know if I need to clarify something.
Question asks:
Find $N(A)$ for $A$ = \begin{bmatrix}
-3 & 6 & -1 & 1 & -7 \\
1 & -2 & 2 & 3 & -1\\
2 & -4 & 5 & 8 & -4 \\
\end{bmatrix}
First thing I did was put the augmented matrix into reduced echelon row:
$\begin{bmatrix}
1 & -2 & 0 & -1 & 3 & 0 \\
0 & 0 & 1 & 2 & -2 & 0\\
0 & 0 & 0 & 0 & 0 & 0 \\
\end{bmatrix}$
$(1)$
So then...
$x=\begin{bmatrix}
x_1\\
x_2 \\
x_3\\
x_4\\
x_5\\
\end{bmatrix}
=
\begin{bmatrix}
2x_2 + x_4 - 3x_5\\
x_2 \\
-2x_4 + 2x_5\\
x_4\\
x_5\\
\end{bmatrix}$
$(2) $
Since $x_2, x_4$ and $x_5$ are free variables..
$
x_2
\begin{bmatrix}
2\\
1 \\
0\\
0\\
0\\
\end{bmatrix}
+
x_4
\begin{bmatrix}
1\\
0 \\
-2\\
1\\
0\\
\end{bmatrix}
+
x_5
\begin{bmatrix}
-3\\
0 \\
2\\
0\\
1\\
\end{bmatrix}$
$(3)$
Resulting in..
$N(A)= \left(
\begin{bmatrix}
2\\
1 \\
0\\
0\\
0\\
\end{bmatrix}
,
\begin{bmatrix}
1\\
0 \\
-2\\
1\\
0\\
\end{bmatrix}
,
\begin{bmatrix}
-3\\
0 \\
2\\
0\\
1\\
\end{bmatrix}
\right)$
$(4)$
| As mentioned in the comments, provided your arithmetic is accurate, this is the correct response. The idea behind the null space of a matrix is that it is precisely those vectors in the domain being sent to the $\mathbf{0}$ vector in the codomain. So, what you have (correctly) done, is determined the solution set of $A\mathbf{x}=\mathbf{0}$.
You did this by finding the null space of a reduced row echelon form of $A$, which has the same null space as $A$. That is, if $B$ is the reduced row echelon form for $A$ that you found, $A\mathbf{x}=\mathbf{0}$ if and only if $B\mathbf{x}=\mathbf{0}$. So, $N(B)=N(A)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2158772",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 1,
"answer_id": 0
} |
How to find all solutions for $x^3=6x+6$ Could anyone help me to find how to find all solution for $x^3=6x+6$?
| Set $x=ay$; the equation becomes $a^3y^3-6ay=6$. We want that
$$
\frac{a^3}{6a}=\frac{4}{3}
$$
so we can take $a=2\sqrt{2}$. Then we have
$$
16\sqrt{2}y^3-12\sqrt{2}y=6
$$
that becomes
$$
4y^3-3y=\frac{3}{4}\sqrt{2}>1
$$
Now set $y=\cosh z$, so the equation becomes
$$
4\cosh^3z-3\cosh z=\frac{3}{4}\sqrt{2}
$$
that is,
$$
\cosh3z=\frac{3}{4}\sqrt{2}
$$
Now solve
$$
e^{3z}+e^{-3z}=\frac{3}{2}\sqrt{2}
$$
and you'll have $z$ and so $y$ and finally $x$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2159298",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
How to solve this trigonometric equation Plese help to solve this equation:
$$ \sin x=2\sin20^{\circ}\sin\left(170^{\circ}-x\right)$$
I tried to convert this equation:
$$\sin x=2\sin20^{\circ}\left(\sin170^{\circ}\cos x-\cos170^{\circ}\sin x\right)$$
$$\sin x\left(1+2\sin20^{\circ}cos170^{\circ}\right)=2\sin20^{\circ}\sin170^{\circ}\cos x$$
$$\tan x\left(1+2\sin20^{\circ}cos170^{\circ}\right)=2\sin20^{\circ}\sin170^{\circ}$$
It will be $$\tan x=\frac{2\sin20^{\circ}\sin170^{\circ}}{1+2\sin20^{\circ}\cos170^{\circ}}$$
But how to solve it without calculator?
| $\sin170^\circ=\sin(90+80)^\circ=\cos80^\circ(?)$
$\cos170^\circ=\cos(90+80)^\circ==-\sin80^\circ(?)$
$$\implies\frac{2\sin(20^{\circ})\sin(170^{\circ})}{1+2\sin(20^{\circ})\cos(170^{\circ})}=\dfrac{2\sin20^\circ\cos80^\circ}{1-2\sin20^\circ\sin80^\circ}$$
Now, $1-2\sin20^\circ\sin80^\circ=1-(\cos60^\circ+\cos80^\circ)=\dfrac{1-2\cos80^\circ}2$
As $1-2\cos2y=1-2(2\cos^2y-1)=-\dfrac{\cos3y}{\cos y}$
$\implies1-2\cos80^\circ=-\dfrac{\cos120^\circ}{\cos40^\circ}=\dfrac1{2\cos40^\circ}$
$\implies1-2\sin20^\circ\cos10^\circ=\dfrac1{4\cos40^\circ}$
$$F=8\sin20^\circ\cos80^\circ\cos40^\circ=4\cos40^\circ\cos80^\circ\cdot\dfrac{2\cos20^\circ\sin20^\circ}{\cos20^\circ}$$
$$=2\cos80^\circ\cdot\dfrac{2\sin40^\circ\cos40^\circ}{\cos20^\circ}=\dfrac{2\sin80^\circ\cos80^\circ}{\cos20^\circ}=\dfrac{\sin160^\circ}{\cos20^\circ}=\dfrac{\sin(180-20)^\circ}{\cos20^\circ}=?$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2159710",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Evaluate $\lim_{t\to0}\frac{-(t-2)(\sin t +1)-(t+2)\cos t}{(\sin t + \cos t - 1)t^2}$ without L'hopital
Question:
Let
\begin{align}
&S(t):=\int_{\pi/4}^t (\sin t-\sin\left(\frac\pi4\right))dt\\
&T(t):=\frac{\left(\sin t-\sin\left(\frac\pi4\right)\right)\left(t-\frac\pi4\right)}2\\
\end{align}
Using $$\lim_{t\to0}\frac{\tan t - t}{t^3}=\frac13\tag1$$
Evaluate the following (without L'hopital)
\begin{align}
&\quad\lim_{t\to\frac\pi4}\frac{S(t)-T(t)}{T(t)\left(t-\frac\pi4\right)}
\end{align}
What I've done so far is:
$$\lim_{t\to\frac\pi4}\frac{S(t)-T(t)}{T(t)\left(t-\frac\pi4\right)}=\lim_{t\to0}\frac{-\cos\left(t+\frac\pi4\right)+\frac{\sqrt2}2-\left(\sin\left(t+\frac\pi4\right)+\frac{\sqrt2}2\right)\frac t2}{\left(\sin\left(t+\frac\pi4\right)-\frac{\sqrt2}2\right)\frac {t^2}2}$$
$$=\lim_{t\to0}\frac{-(t-2)(\sin t +1)-(t+2)\cos t}{(\sin t + \cos t - 1)t^2}$$
But I don't know how to go from here to the Eq.$(1)$.
Thanks.
| Based on your calculations, the given limit and the sum to product formulae,
\begin{align}
& \lim_{t\to\frac\pi4}\frac{S(t)-T(t)}{T(t)\left(t-\frac\pi4\right)} \\
=&\lim_{t\to0}\frac{-\cos\left(t+\frac\pi4\right)+\frac{\sqrt2}2-\left(\sin\left(t+\frac\pi4\right)+\frac{\sqrt2}2\right)\frac
t2}{\left(\sin\left(t+\frac\pi4\right)-\frac{\sqrt2}2\right)\frac
{t^2}2} \\
=&
\lim_{t\to0}\frac{-\cos\left(t+\frac\pi4\right)+\cos(\frac{\pi}{4})-\left(\sin\left(t+\frac\pi4\right)+\sin(\frac{\pi}{4})\right)\frac
t2}{\left(\sin\left(t+\frac\pi4\right)-\sin(\frac{\pi}{4})\right)\frac
{t^2}2} \\
=&
\lim_{t\to0}\frac{-2\sin\left(\frac{t}{2}+\frac\pi4\right)\sin(\frac{-t}{2})-\left(2\sin\left(\frac{t}{2}+\frac\pi4\right)\cos(\frac{t}{2})\right)\frac
t2}{\left(2\cos\left(\frac{t}{2}+\frac\pi4\right)\sin(\frac{t}{2})\right)\frac
{t^2}2} \\
=&
\lim_{t\to0}\frac{2\sin\left(\frac{t}{2}+\frac\pi4\right)\sin(\frac{t}{2})-t\sin\left(\frac{t}{2}+\frac\pi4\right)\cos(\frac{t}{2})}
{t^2\cos\left(\frac{t}{2}+\frac\pi4\right)\sin(\frac{t}{2})} \\
=& \lim_{t\to0} \frac{\sin(\frac{t}{2} +
\frac{\pi}{4})}{\cos(\frac{t}{2} + \frac{\pi}{4})} \cdot
\frac{2\sin\left( \frac{t}{2} \right) - t \cos\left(
\frac{t}{2}
\right)}{t^2 \sin\left( \frac{t}{2} \right)} \\
=& \lim_{t\to0} \tan\left(\frac{t}{2} +
\frac{\pi}{4}\right) \cdot
\frac{2\sin\left( \frac{t}{2} \right) - t \cos\left(
\frac{t}{2}
\right)}{t^2 \sin\left( \frac{t}{2} \right)} \\
=& \lim_{t\to0} \tan\left(t + \frac{\pi}{4}\right) \cdot
\frac{2\sin\left( t \right) - 2t \cos\left( t \right)}{(2t)^2 \sin\left( t \right)} \\
=& \lim_{t\to0} \tan\left(t + \frac{\pi}{4}\right) \cdot
\frac{\tan\left( t \right) - t}{2t^2 \tan\left( t \right)} \\
=& \lim_{t\to0} \frac{1}{2} \tan\left(t + \frac{\pi}{4}\right) \cdot
\frac{\tan\left( t \right) - t}{t^3} \frac{t}{\tan\left( t \right)} \\
=& \lim_{t\to0} \frac{1}{2} \tan\left(t + \frac{\pi}{4}\right) \cdot
\frac{\tan\left( t \right) - t}{t^3} \frac{t}{\sin\left( t \right)}
\cos\left( t \right) \\
=& \frac{1}{2} \tan\left( 0 + \frac{\pi}{4} \right) \cdot
\frac{1}{3} \cdot 1 \cdot \cos(0) \\
=& \frac{1}{6}
\end{align}
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Limit of a sequence of fractions I have that:
$$S_n=\frac{b\cdot S_{n-1}}{b+S_{n-1}}+a$$
where $n\in\mathbb{N}^+$ and $S_1=a+b$.
So, for $S_2$ we will get:
$$S_2=\frac{b\cdot S_{2-1}}{b+S_{2-1}}+a=\frac{b\cdot S_1}{b+S_1}+a=\frac{b\cdot(a+b)}{b+(a+b)}+a$$
And for $S_3$:
$$S_3=\frac{b\cdot S_{3-1}}{b+S_{3-1}}+a=\frac{b\cdot S_2}{b+S_2}+a=\frac{b\cdot\left(\frac{b\cdot(a+b)}{b+(a+b)}+a\right)}{b+\left(\frac{b\cdot(a+b)}{b+(a+b)}+a\right)}+a$$
And so on. Now my question is: what happens when $n\to\infty$?
| Assume that $S_n = \frac{p_n}{q_n}$ is associated with $(p_n,q_n)\in\mathbb{R}^2$. The recurrence
$$ S_{n} = \frac{(a+b)S_{n-1}+ab}{S_{n-1}+b}\tag{1} $$
can be written in the following form
$$ \begin{pmatrix}p_n \\ q_n \end{pmatrix} = \begin{pmatrix} a+b & ab \\ 1 & b\end{pmatrix} \begin{pmatrix}p_{n-1} \\ q_{n-1} \end{pmatrix} \tag{2}$$
hence the closed form of both the $\{p_n\}_{n\geq 0}$ and the $\{q_n\}_{n\geq 0}$ sequences just depends on the diagonalization of $M=\begin{pmatrix} a+b & ab \\ 1 & b\end{pmatrix}$, whose eigenvalues are given by $\zeta_{\pm}=\frac{(a+2b)\pm\sqrt{a^2+4ab}}{2}$.
In particular
$$ S_n = \frac{ A \zeta_+^n + B \zeta_-^n}{C \zeta_+^n + D \zeta_-^n}\tag{3}$$
for some constants $(A,B,C,D)$ depending on the initial values. By taking the limit as $n\to +\infty$, we simply get $\color{red}{\frac{A}{C}}$. On the other hand, by assuming $S_n\to L$ as $n\to \infty$, such limit has to fulfill
$$ L = a + \frac{bL}{b+L} \tag{4}$$
hence the only chances are $L=\frac{1}{2}\left(a\pm\sqrt{a^2+4ab}\right)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2160961",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Proof that $CE=2AD$ I already have a proof, but if you can please give another:
Let $ABC$ be an isosceles triangle with $AB=AC$ and point $D$ be on segment $AB$. The line perpendicular to $AB$ which passes $D$ intersects $BC$ (extended) and $AC$ at $E$ and $F$ respectively. $C$ is on segment $BE$, between $B$ and $E$. If the area of $CEF$ is twice that of $ADF$, prove that $CE=2AD$.
My proof involves the Pythagorean theorem and the similar triangles property.
| $\dfrac{CF}{FA}\cdot\dfrac{EF}{FD} = 2$ and $\dfrac{DF}{FE}\cdot \dfrac{EC}{CB}\cdot \dfrac{BA}{AD} = 1.$ This implies that $\dfrac{CE}{AD} = 2\dfrac{BC}{BA}\cdot\dfrac{AF}{FC}$. But $\dfrac{AF}{FC} = \dfrac{AD}{DB}\cdot \dfrac{BE}{CE} = \dfrac{BE}{BD}\cdot\dfrac{AD}{CE} = \dfrac{2AB}{BC}\cdot \dfrac{AD}{CE}$. Multiply these two to get $\dfrac{CE}{AD} = 2\cdot 2\dfrac{AD}{CE}$ or $\dfrac{CE}{AD} = 2.$
Here, the only tricky part was $\dfrac{BE}{BD} = 2\dfrac{AB}{BC}$, which follows from the fact that $\angle BAC = 2\angle BED.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2162279",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove that the sum of pythagorean triples is always even Problem: Given $a^2 + b^2 = c^2$ show $a + b + c$ is always even
My Attempt, Case by case analysis:
Case 1: a is odd, b is odd. From the first equation,
$odd^2 + odd^2 = c^2$
$odd + odd = c^2 \implies c^2 = even$
Squaring a number does not change its congruence mod 2.
Therefore c is even
$ a + b + c = odd + odd + even = even$
Case 2: a is even, b is even. Similar to above
$even^2 + even^2 = c^2 \implies c$ is even
$a + b + c = even + even + even = even$
Case 3:
One of a and b is odd, the other is even
Without loss of generality, we label a as odd, and b as even
$odd^2 + even^2 = c^2 \implies odd + even = c^2 = odd$
Therefore c is odd
$a + b + c = odd + even + odd = even$
We have exhausted every possible case, and each shows $a + b + c$ is even. QED
Follow Up:
Is there a proof that doesn't rely on case by case analysis?
Can the above be written in a simpler way?
| Consider $(a+b+c)^2$
Which is $a^2 + b^2 + c^2 + 2(ab+bc+ca)$
Since $c^2 = a^2 + b^2$ (c being the hypotenuse), $(a+b+c)^2 = 2(c^2 + ab + bc + ca)$ - which is an even number.
and since squares of odd is odd and evens is even $a+b+c$ has to be even.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2162888",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "54",
"answer_count": 6,
"answer_id": 1
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Factorising Non-Monic Quadratic Equations I am trying to find a way to factorise a non-monic quadratic using a system of linear equations. I know there exist various algorithms which can help with this, however I'm not fond of such methods.
I first started by creating a general equation with 4 unkown variables excluding x :
$(ax+b)(cx+d)= acx^2+bcx+adx+bd=acx^2+(bc+ad)x+bd$
Now lets say I was given the quadratic $6x^2-19x+15$. Using the general formula I made, I end up with the simultaneous equations:
$ac=6$
$bc+ad=-19$
$bd=15$
Obviously this system of equations cannot be solved as there must be one more equation. Is there a fourth relationship between the variables which I can add or are there variables which I can eliminate? Any comments or suggestions are much appreciated :)
EDIT:
I have tried to remove one variable by using the standard equation:$$a(x-b)(x-c)=ax^2+x(ab+ac)+abc$$
Now using the previous quadratic $6x^2-19x+15$ I end up with 3 equations:
$$a=6$$$$ab+ac=-19$$$$abc=15$$
However after solving these by hand I end up with another quadratic equation similar to my original expression:
$$6b^2+19b+15$$
| You can solve it in 2 ways:
FIRST WAY (longer version):
$6x^2-19x+15=0$
$x=\frac{-b\pm\sqrt{b^2-4\cdot a\cdot c}}{2\cdot a}=\frac{-(-19)\pm\sqrt{(-19)^2-4\cdot 6\cdot 15}}{2\cdot 6}$
$x=\frac{19\pm\sqrt{361-360}}{12}=\frac{19\pm\sqrt{1}}{12}=\frac{19\pm 1}{12}$
$x_1=\frac{19+1}{12}=\frac{20}{12}=\frac{5}{3} \quad \text{or} \quad x_2=\frac{19-1}{12}=\frac{18}{12}=\frac{3}{2}$
Formula of factorization by using zero points: $ax^2+bx+c=a(x-x_1)(x-x_2)$
In the given equation:
$a=6\quad x_1=\frac{5}{3}\quad x_2=\frac{3}{2}$
So we get:
$6(x-\frac{5}{3})(x-\frac{3}{2})$
$2\cdot 3\cdot(x-\frac{5}{3})(x-\frac{3}{2})$
$3(x-\frac{5}{3})\cdot 2(x-\frac{3}{2})$
$(3x-5)(2x-3)$
SECOND WAY (shorter version):
$6x^2-19x+15$
$6x^2-9x-10x+15$
$(6x^2-9x)+(-10x+15)$
$3x(2x-3)-5(2x-3)$
$(2x-3)(3x-5)$
In both ways as answer you will get: $(2x-3)(3x-5)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2171246",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Equation of circles tangent to $x$-axis with a given radius and point Find the equations of the circles which are tangent to the x-axis, with radius of 5 units and passing through the point $(0,8)$.
I know how to formulate an equation with the given radius of 5 and then substitute the point $(0,8)$, but then I get stuck at finding their different centers.
| The equation of circle is $(x-a)^2+(y-b)^2=r^2$ centered ar point $(a,b)$.
The equation of line is $y=mx+c$. The x axis is the line $y=0$.
All circles having tangent to x axis implies it have point $y=0$ and only one solution for x.
$(x-a)^2+b^2=5^2$
$x^2-2ax+a^2+b^2-25=0$
We want solutions for x.
We thus require the discriminant to be zero. $b^2-4ac$ is
$26-a^2-b^2=0$.
These are the equation which is to be solved for to find $a,b$ for single solutions.
$b^2=26-a^2$
Inserted into equation of circle with $y=0$
$x^2-2ax+a^2+26-a^2-25=0$
$x^2-2ax+1=0$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2171436",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 2
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Critical points of $f(x,y)=\sin(x)+\sin(y)-\sin(x+y)$ Domain: $0 \le x<\pi$ and $0 \le y<\pi$
After setting the gradient of $f(x,y)$ equal to zero, I obtained the following system:
$\cos(x)=\cos(x+y),$ $\cos(y)=\cos(x+y)$
I am not sure how to solve this system but what I tried was:
On the given domain, the first equation implies $x=x $ and $y=0$.
The second equation implies $y=y $ and $x=0$ so the system is consistent when $(x,y)=(0,0)$?
And thus $(0,0)$ is the only critical point in the given domain?
| Setting the gradient to zero yields
$$
\begin{cases}
\cos x - \cos(x+y)=0 \\[6px]
\cos y - \cos(x+y)=0
\end{cases}
$$
Therefore $\cos x=\cos y$. If we are interested in the whole plane, this means
$$
y=x+2k\pi
\qquad\text{or}\qquad
y=-x+2k\pi
$$
With the second set of solutions we get $x+y=2k\pi$, so $\cos(x+y)=1$ and we obtain $\cos x=1$, so
$$
x=2a\pi \qquad y=2b\pi
$$
for arbitrary integers $a$ and $b$.
With the first set of solutions we get $x+y=2x+2k\pi$, so $\cos(x+y)=\cos x$ becomes
$$
\cos 2x=\cos x
$$
This reduces to $2x=x+2h\pi$ or $2x=-x+2h\pi$. In the first case we get solutions we have already found, so it remains to consider $3x=2h\pi$.
This gives
$$
x=\frac{2h\pi}{3}\qquad y=\frac{2h\pi}{3}+2k\pi \tag{*}
$$
again for arbitrary integers $h$ and $k$.
The only solutions that satisfy $0\le x<\pi$ and $0\le y<\pi$ are
$$
\begin{cases}
x=0\\[6px]
y=0
\end{cases}
\qquad
\begin{cases}
x=\dfrac{2\pi}{3}\\[6px]
y=\dfrac{2\pi}{3}
\end{cases}
$$
If you want the solutions in (*) to be “symmetric” in $x$ and $y$, you can write them in the forms
$$
\begin{cases}
x=\dfrac{2\pi}{3}+2a\pi \\[6px]
y=\dfrac{2\pi}{3}+2b\pi
\end{cases}
\quad
\begin{cases}
x=\dfrac{4\pi}{3}+2a\pi \\[6px]
y=\dfrac{4\pi}{3}+2b\pi
\end{cases}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2173444",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "2",
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Co-ordinates of the vertices an icosahedron relative to its centroid.
This is a picture of an icosahedron. I need to know the coordinates of the vertices of the icosahedron relative to it's centroid in order to programme a projection of one on a three dimensional plane. (By the way, it has twelve vertices, so it is going to be a very long process)
Unfortunately, I have not been able to find one vertices.
| PolyhedronData["Icosahedron", "VertexCoordinates"]
$$
\left(
\begin{array}{ccc}
0 & 0 & -\frac{5}{\sqrt{50-10 \sqrt{5}}} \\
0 & 0 & \frac{5}{\sqrt{50-10 \sqrt{5}}} \\
-\sqrt{\frac{2}{5-\sqrt{5}}} & 0 & -\frac{1}{\sqrt{10-2 \sqrt{5}}} \\
\sqrt{\frac{2}{5-\sqrt{5}}} & 0 & \frac{1}{\sqrt{10-2 \sqrt{5}}} \\
\frac{1+\sqrt{5}}{2 \sqrt{10-2 \sqrt{5}}} & -\frac{1}{2} &
-\frac{1}{\sqrt{10-2 \sqrt{5}}} \\
\frac{1+\sqrt{5}}{2 \sqrt{10-2 \sqrt{5}}} & \frac{1}{2} &
-\frac{1}{\sqrt{10-2 \sqrt{5}}} \\
-\frac{1+\sqrt{5}}{2 \sqrt{10-2 \sqrt{5}}} & -\frac{1}{2} &
\frac{1}{\sqrt{10-2 \sqrt{5}}} \\
-\frac{1+\sqrt{5}}{2 \sqrt{10-2 \sqrt{5}}} & \frac{1}{2} &
\frac{1}{\sqrt{10-2 \sqrt{5}}} \\
-\frac{-1+\sqrt{5}}{2 \sqrt{10-2 \sqrt{5}}} & -\frac{1}{2}
\sqrt{\frac{5+\sqrt{5}}{5-\sqrt{5}}} & -\frac{1}{\sqrt{10-2 \sqrt{5}}}
\\
-\frac{-1+\sqrt{5}}{2 \sqrt{10-2 \sqrt{5}}} & \frac{1}{2}
\sqrt{\frac{5+\sqrt{5}}{5-\sqrt{5}}} & -\frac{1}{\sqrt{10-2 \sqrt{5}}}
\\
\frac{-1+\sqrt{5}}{2 \sqrt{10-2 \sqrt{5}}} & -\frac{1}{2}
\sqrt{\frac{5+\sqrt{5}}{5-\sqrt{5}}} & \frac{1}{\sqrt{10-2 \sqrt{5}}}
\\
\frac{-1+\sqrt{5}}{2 \sqrt{10-2 \sqrt{5}}} & \frac{1}{2}
\sqrt{\frac{5+\sqrt{5}}{5-\sqrt{5}}} & \frac{1}{\sqrt{10-2 \sqrt{5}}}
\end{array}
\right)
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2174594",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 2
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Can't calculate the integral $\int_0^{\pi/2}\ln(a^2\cos^2x+b^2\sin^2x)\,dx$ Let $I(a,b):= \int_0^{\pi/2}\ln(a^2\cos^2x+b^2\sin^2x)\,dx$
Calculate $I(a,b)$.
My attempt:
Define function $F(a,b,x)$ as following $\frac{\partial f}{\partial a}F(a,b,x)=\ln(a^2\cos^2x+b^2\sin^2x)$
Then $I(a,b):=
\int_0^{\pi/2}\ln(a^2\cos^2x+b^2\sin^2x)dx=
\int_0^{\pi/2}\frac{\partial f}{\partial a}F(a,b,x)=
\frac{\partial f}{\partial a}\int_0^{\pi/2}F(a,b,x)$
The last equality is done by using Leibniz integral rule, I must prove that the function is well defined in a 3-dimension cube.
Calculating F:
$F(a,b,x)=\int \ln(a^2\cos^2x+b^2\sin^2x)\,da$
Using the integral by parts, $u'=1, u=a, v=\ln(a^2\cos^2x+b^2\sin^2x), v'=\frac{2a\cos^2x}{a^2\cos^2x+b^2\sin^2x}$
$F(a,b,x)=uv-v'u=a\ln(a^2\cos^2x+b^2\sin^2x)\space-\space \int\frac{2a^2\cos^2x}{a^2\cos^2x+b^2\sin^2x}\,da=
a{\ln(a^2\cos^2x+b^2\sin^2x)}
\space-\space \int\frac{2a^2\cos^2x}{a^2\cos^2x+b^2\sin^2x}da=
a{\ln(a^2\cos^2x+b^2\sin^2x)}\space-\space \int\frac{2a^2\cos^2x+b^2\sin^2x-b^2\sin^2x}{a^2\cos^2x+b^2\sin^2x}da=
a{\ln(a^2\cos^2x+b^2\sin^2x)}-\int 1-\frac{b^2\sin^2x}{(a^2\cos^2x+b^2\sin^2x)}da$
And I am stuck.. I don't know how to handle that integral.
I am not even sure I am solving it correctly..
Any tips?
| Hint. One may set
$$
f(s):=\int_0^{\pi/2}\ln(s+\sin^2 x)dx, \qquad s\geq0.
$$
Then differentiating under the integral sign with respect to $s$ you get
$$
\begin{align}
f'(s)&=\int_0^{\pi/2}\frac1{s+\sin^2 x}dx\\\\
&=\int_0^{\infty}\frac1{s+\dfrac{t^2}{t^2+1}}\dfrac{dt}{t^2+1}\quad (t=\tan x)\\\\
&=\int_0^{\infty}\frac1{(s+1)t^2+s}dt\\\\
&=\frac{\pi}2\frac{1}{\sqrt{s(s+1)}}\\\\
&=\pi \left.\left(\ln \left(\sqrt{s}+\sqrt{s+1}\right)\right)\right|_s^{'}
\end{align}
$$ Thus
$$
\int_0^{\pi/2}\ln(s+\sin^2 x)dx=\pi \ln \left(\sqrt{s}+\sqrt{s+1}\right)+C
$$ with $C=f(0)=-\pi \ln 2$ (this one is standard) giving
$$
\int_0^{\pi/2}\ln(s+\sin^2 x)dx=\pi \ln \left(\frac{\sqrt{s}+\sqrt{s+1}}2\right), \quad s\geq0.$$
Assume without loss of generality that $b^2>a^2$, then your initial integral is obtained by writing
$$
\ln\left(a^2\cos^2x+b^2\sin^2x\right)=\ln(b^2-a^2)+\ln\left(s+\sin^2x\right).
$$ with $s=\dfrac{a^2}{b^2-a^2}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2175287",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
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How to integrate fractions of polynomials with complex roots $$\int\frac{4x^3-2x^2+60x+10}{x^4+30x^2+125} dx $$
The denominator has no rational roots. It can be factored as $(x^2+5)(x^2+25)$ which gives roots of $\pm i \sqrt{5}$ and $\pm 5i$.
How can these complex roots be used to integrate the function?
| First split up the integral into two parts and then use the natural logarithm and arctan.
$$\begin{align*} P = &\frac{4x^3-2x^2+60x+10}{x^4+30x^2+125} \\
= &\frac{4x^3-2x^2+60x+10}{(x^2+25)(x^2+5)} \\
= &\frac{2x-3}{x^2+25} + \frac{2x+1}{x^2+5}\end{align*}$$
So $\int P \,\mathrm{d}x = \int \frac{2x-3}{x^2+25}\mathrm{d}x + \int\frac{2x+1}{x^2+5}\mathrm{d}x $ :
$$ \int \frac{2x-3}{x^2+25}\mathrm{d}x = \int \left(\frac{2x}{x^2+25}-\frac{3}{x^2+25}\right)\mathrm{d}x = \ln(x^2+25)-\frac{3}{5}\arctan{\left(\frac{x}{5}\right)} + c$$
$$ \int\frac{2x+1}{x^2+5}\mathrm{d}x = \int \left(\frac{2x}{x^2+5} + \frac{1}{x^2+5}\right)\mathrm{d}x = \ln(x^2+5) + \frac{1}{\sqrt{5}}\arctan{\left(\frac{x}{\sqrt{5}}\right)} + c$$
Now add these to get (which could simplify further, but I'll leave as is):
$$ \int P \,\mathrm{d}x = \ln(x^2+25) + \ln(x^2+5) + \frac{1}{\sqrt{5}}\arctan{\left(\frac{x}{\sqrt{5}}\right)} - \frac{3}{5}\arctan{\left(\frac{x}{5}\right)} + c$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2176958",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Envelope of isoperimetric ellipses Find envelope of ellipses $ \frac{x^2}{a^2} + \frac{y}{b^2}=1 $ which have $a,b$ and its associated eccentricity $e$ variable while holding its perimeter
$$ p= 4 a E(e) $$
as constant.
Expected to be asteroid-like, passing through
$$
\begin{pmatrix} 0 \\ \frac{\pi a}{2} \end{pmatrix},
\begin{pmatrix} \frac{a}{\sqrt{2}} \\ \frac{a}{\sqrt{2}} \end{pmatrix},
\begin{pmatrix} \frac{\pi a}{2} \\0 \end{pmatrix}$$
| \begin{align*}
p &= 4a E(k) \\
a(k) &= \frac{p}{4E(k)} \\
F(x,y,k) &=
x^2+\frac{y^2}{1-k^2}-\frac{p^2}{16E^2(k)} \\
\frac{\partial F}{\partial k} &=
\frac{2ky^2}{(1-k^2)^2}-
\frac{p^2}{8E^3(k)} \left[ \frac{K(k)-E(k)}{k} \right]
\end{align*}
The envelope is given by $$F=\dfrac{\partial F}{\partial k}=0$$
On solving,
$$
\begin{pmatrix}
x \\ y
\end{pmatrix}=
\frac{p}{4kE(k)}
\begin{pmatrix}
\pm \sqrt{1-\dfrac{(1-k^2)K(k)}{E(k)}} \; \\[5pt]
\pm (1-k^2)\sqrt{\dfrac{K(k)}{E(k)}-1} \;
\end{pmatrix} \, , \quad 0<k<1 \tag{1}
$$
If we adimit $a<b$, then
$$
\begin{pmatrix}
x \\ y
\end{pmatrix}=
\frac{p}{4kE(k)}
\begin{pmatrix}
\pm (1-k^2)\sqrt{\dfrac{K(k)}{E(k)}-1} \; \\[5pt]
\pm \sqrt{1-\dfrac{(1-k^2)K(k)}{E(k)}} \;
\end{pmatrix} \, , \quad 0<k<1 \tag{2}
$$
A plot of the envelope for $p=4$ with the ellipses is shown below:
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Maximum/Minimum Find the Max/Min values:
$f(x)=x^3-x^2-8x+1$ on $[-2,2]$
$$\begin{align}f'(x)&=3x^2-2x-8\\&=3x^2+4x-6x-8\\&=x(3x+4)-2(3x+4)\\&=(x-2)(3x+4)\end{align}$$
Thus, $x=-2,0.75,2$
Now sub these into the original function
$f(2)=2^3-2^2-8(2)+1=-11$
$f(0.75)=0.75^3-0.75^2-8(0.75)+1=-5.14$
$f(-2)=(-2)^3-(-2)^2-8(-2)+1=5$
Is the minimum $-11$ and the maximum $5$?
Any idea as to how I would go about $f(x)=x^5+x+1$ on $[-1,1]$?
$f'(x)=5x^4+1=0$, then I become unsure... any advice is greatly appreciated!
| You've got it right so far- we wish to solve $5x^4+1=0$, and this will give us the location of the maximum and minimum of the function.
But consider, if we simplify $5x^4+1=0$, we get $x^4=-1/5$... This obviously has no real solutions, and so there is no point at which a maximum or minimum can exist.
However, we are given an interval over which to check- there can be no point between $-1$ and $1$ that is a maximum or minimum (b/c there are no solutions to $5x^4+1=0$), and therefore one of $1$ and $-1$ will be our maximum, and the other our minimum. So, we plug $-1$ and $1$ back into the equation: $(-1)^5 + (-1) + 1 = -1$, and $1^5 + 1 + 1 = 3$.
Hence $-1$ is our minimum, with $f(-1) = -1$, and $1$ is our maximum, with $f(1) = 3$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2180022",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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} |
Largest square that can fit into region. Let $R$ be the region bounded by the lines $y=cx+b$ and $y=cx+d$. Find the sides of largest square, with sides parallel to the $x$ and $y$ axis in $R$.
Here is what I know. The distance between the two lines is $\frac{|b-d|}{ \sqrt{c^2+1}}$. Without the requirement, this is the side length. What trips me up is the requirement that the sides be parallel to the $x$ and $y$ axis how do I deal with this?
| Suppose without loss of generality that $b> d$, so that line $l_1: y=cx+b$ lies above line $l_2:y=cx+d$. Moreover, in this answer we will assume that $c>0$, but the solution can be easily adapted to the case $c<0$. The case $c=0$ is trivial.
Choose any point on $l_1$, say $A=(0,b)$. Our square will be $ABCD$. We will find the point opposite $A$, that is, $C$. To this end, notice that the diagonal $AC$ will make an angle of $135^\circ$ with the $x$-axis, and that $C$ will lie on $l_2$. That is, $AC$ belongs to the line $r(t)=(0,b)+t\cdot(1,-1)$, in parametric equation. In choosing $(1,-1)$ we have used that $c>0$, and we will hence have $t>0$ in the intersection. If we had $c<0$, we'd need to choose $(-1,-1)$ for the same effect.
To find the intersection we solve
$$b-t=tc+d\implies t(1+c)=b-d\implies t=\frac{b-d}{1+c}$$
Hence, $C=\left(\frac{b-d}{1+c},\frac{bc+d}{1+c}\right)$. We have that
$$AC={\lVert C-A\rVert}={\left\lVert\left(\frac{b-d}{1+c},\frac{bc+d}{1+c}\right)-(0,b) \right\rVert}={\left\lVert\left(\frac{b-d}{1+c},-\frac{b-d}{1+c}\right) \right\rVert}=\frac{b-d}{1+c}\sqrt{2}.$$
Since $AC=e\sqrt{2}$, where $e$ is the length of the side of the square, we get that
$$e=\frac{b-d}{1+c}$$
If $c$ were negative, we'd have $r:y=(0,b)+t\cdot(-1,-1)$ so that $b-t=-tc+d$ and hence $t=\frac{b-d}{1-c}$. Then $C=\left(\frac{b-d}{1-c},\frac{d-bc}{1-c}\right)$ and $AC=\frac{b-d}{1-c}\sqrt{2}$, so that $e=\frac{b-d}{1-c}$.
Hence, in general, we have
$$e=\frac{|b-d|}{1+|c|}$$
Notice that this also works for when $c=0$, ie, $l_1$ and $l_2$ are vertical.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2182277",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
} |
Limits at infinity, in terms of another function Suppose:
$$\lim_{x\to\infty}\; f(x) + 2x^2 =1$$
Determine if possible:
$$\lim_{x\to\infty}\; \frac{f(x)}{x^2+1}$$
I determined the answer to be $-2$, through substituting for $f(x)$:
$$\frac { 1 - 2x^2 }{x^2 + 1}$$
and I solved for the infinite limit.
however I feel like that substitution is illegal, or is it?
| I think it's okay, but you're skipping a lot of steps that may not work in other problems. To be thorough (and to see exactly what's going on), I'd write it like so:
\begin{align}
\frac{f(x)}{x^2+1} &= \frac{f(x)+2x^2-2x^2}{x^2+1} \\
&= \frac{f(x)+2x^2}{x^2+1} - \frac{2x^2}{x^2+1}
\end{align}
By the laws of limits, we can split the limits across this difference as long as the separated limits converge (which they will):
\begin{align}
&\lim_{x \to \infty} \frac{f(x)+2x^2}{x^2+1} - \lim_{x \to \infty} \frac{2x^2}{x^2+1} \\
&= \lim_{x \to \infty} \frac{1}{x^2+1} - 2 \\
&= 0 - 2 \\
&= \boxed{-2}
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2182727",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Sixth degree polynomial problem If the graph of $$y = x^6 - 10x^5 + 29x^4 - 4x^3 + ax^2$$ always lies above the line $y = bx + c$, except for $3$ points where the curve intersects the line.
What is the largest value of $x$ for which the line intersects the curve?
*
*A) 4
*B) 5
*C) 6
*D) 7
Through general idea of graphs and with the help of a graphing calculator I have concluded the approximate look of such a curve and a line.
However, I would like to know a proper mathematical solution to this problem.
| As there are three double roots, $$x^6 - 10x^5 + 29x^4 - 4x^3 + ax^2-bx-c$$ is a perfect square, and we can evaluate its square root.
Looking at the first two terms,
$$x^6-10x^5\leftrightarrow(x^3-5x)^2=x^6-10x^5\cdots$$
Next
$$x^6-10x^5+29x^4\leftrightarrow(x^3-5x^2+px)^2=x^6-10x^5+(2p+25)x^4\cdots$$ so that $p=2$.
Then
$$x^6-10x^5+29x^4-4x^3\leftrightarrow(x^3-5x^2+2x+q)^2=x^6-10x^5+29x^4+(2q-20)x^3\cdots$$ and $q=8$.
The roots of
$$x^3-5x^2+2x+8$$ are $-1, 2$ and $\color{green}4$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2183127",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
How to see if this functions are linearly independent?
Is this set linearly independent? $A=\left\{ \arctan(x), \arctan(2x), \arctan\big(\frac{3x}{1-2x^2}\big)\right\}$
I've tried using the Wronskian
$$W(A)= \left| \begin{array}{ccc}
\arctan(x) & \arctan(2x) & \arctan\big(\frac{3x}{1-2x^2}\big) \\
\frac{1}{1+x^2} & \frac{2}{1+4x^2} & \frac{6x^2+3}{4x^2+5x^2+1} \\
-\frac{2x}{(x^2+1)^2} & -\frac{16x}{(4x^2+1)^2} & -\frac{6x(8x^4+8x^2+3)}{(4x^2+5x^2+1)^2} \end{array} \right|,$$
but expanding the determinant gives an extensive term difficult to simplify. So, does there exist an easier method to see whether the set is linearly independent?
| Hint:
$$\arctan\frac{\tan a+\tan b}{1-\tan a\tan b}=a+b.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2183961",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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Make $\$100$ by taking $\$1$, $\$5$, and $\$10$... but we can take only $21$ notes I am weak in mathematics, but I need to know if this is possible. I will take $\$100$ from my friend, but he will give me only $21$ notes of $\$1$, $\$5$, $\$10$. I need to tell him the numbers of each notes that will make $100$ dollars.
| Expanding the answer by @user3558 , we can show, that these two solutions are the only solutions.
We have:
$$\begin{cases}x,y,z \geq 0 \\
x,y,z \in \mathbb{Z} \\
z=t\\
y=19+\frac{3-9t}{4}\\
x=2-\frac{3-5t}{4}
\end{cases}$$
We want $y$ to be integer, so if $z$ is an integer, then we have
$3-9t \equiv 0 \mod 4$
$t \equiv -1 \mod 4$
$t=4k-1, \, t \in \mathbb{Z}$
Because $z\geq 0$:
$z=t=4k-1\geq0$
$k \geq \frac{1}{4}$
Because $y \geq 0$:
$y=19+\frac{3-9t}{4}=19+\frac{3-9(4k-1)}{4} \geq 0$
$-9k +22 \geq 0$
$ k\leq \frac{22}{9} (=3+\frac{4}{9})$
Because $x\geq 0$:
$x=2-\frac{3-5t}{4} = 2-\frac{3-5(4k-1)}{4}$
$ 2- 5k-2\geq 0$
$k \geq 0$
So if we now get these three conditions, we have
$$k\in \{1,2\}$$
*
*k=1: (x,y,z)=(5,13,3)
*k=2: (x,y,z)=(10,4,7)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2188980",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Real numbers $a$, $b$, $c$ and $d$ satisfy the inequality $abcd > a^2 + b^2 + c^2 + d^2$. Prove that $abcd > a + b + c + d + 8$. Real numbers $a$, $b$, $c$ and $d$ satisfy the inequality $abcd > a^2 + b^2 + c^2 + d^2$. How to prove that $abcd > a + b + c + d + 8$? I've tried using AM-GM inequality but with no outcome. The problem was taken from 57-th Belarusian Mathematical Olympiad available here
| It suffices to check the case $a, b, c, d>0$. Indeed, $abcd>0$, and so the problem becomes strictly stronger upon the transformation $(a, b, c, d)\to (|a|, |b|, |c|, |d|)$ if any of the variables are negative.
Now let $\frac{a+b+c+d}{4}=t$. Then $t^4\ge abcd>a^2+b^2+c^2+d^2\ge 4t^2$ due to AM-GM and Cauchy Schwarz, so $t> 2$ since $t>0$. Then we have:
$$abcd>a^2+b^2+c^2+d^2\ge 4t^2=2t^2+2t^2>4t+8$$
Hence $abcd>4t+8$, which is exactly what we need to show.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2189808",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.