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Prove that $ y = x^{\frac{1}{n}} \Rightarrow y^n = x $ Tao, Analysis I, exercise 5.6.1
I have to prove the following where $x,y \in \Bbb R^+, \ n \in \Bbb Z^+$
$$ y = x^{\frac{1}{n}} \Rightarrow y^n = x $$
Hints: review the proof of Proposition
5.5.12. Also, you will find proof by contradiction a useful tool, especially when
combined with the trichotomy of order in Proposition 5.4.7 and Proposition
5.4.12.
My attempt:
Assume that $ y = x^{\frac{1}{n}} \Rightarrow y^n > x $. By definition:
$$x^{\frac 1n}= \sup \{ y \in \mathbb R, s.t. y \geq 0, y^n \leq x \}$$
Thus $ y^n > x $ contradicts the above definition, since $x$ is upper bound of $y^n$.
Does this part seem ok? I next need to get to a contradiction starting from here:
Now assume that $ y = x^{\frac{1}{n}} \Rightarrow y^n < x $
| Let $a := x^{1/n}$.
Then $a^n > x$ does not conflict with the definition of the set, since it merely states that $a^n$ is an upper bound.
Here is my version of the proof following the same mechanism than for
Proposition 5.5.12. There exists a positive real number $x$ such that
$x^2 = 2$.
and using the result of
Exercise 5.4.4. Show that for any positive real number $x > 0$ there
exists a positive integer $N$ such that $x > 1/N > 0.$
Assume that $a^n > x$. Then there exists $\varepsilon > 0$ such that $(a - \varepsilon)^n > x$ and $a - \varepsilon$ is not in the set $\{y \in \mathbb{R}: y \ge 0 \; \wedge \; y^n \le x\}$. This contradicts the fact that $a - \varepsilon$ is not an upper bound, since $a$ is least upper bound, by definition.
Assume that $a^n < x$. Then there exists $\varepsilon > 0$ such that $(a + \varepsilon)^n < x$, and therefore there is an element in $\{y \in \mathbb{R}: y \ge 0 \; \wedge \; y^n \le x\}$ larger than $a$. But this contradicts that $a + \varepsilon > a$ is an upper bound.
In effect, $a^n \le x$ and $a^n \ge x$, equivalent to $a^n = x$.
To see that there exists an $\varepsilon > 0$ such that $(a + \varepsilon)^n < x$, note that $a$ is bounded by $x$, if $x > 1$. And it is bounded by $1$ otherwise. This yields
$$
(a + \varepsilon)^n = \sum_{k=0}^n
\left(
\begin{array}{c}
n \\
k
\end{array}
\right)
a^{n-k}
\varepsilon^k
= a^n + \varepsilon
\left(
\sum_{k=1}^n
\left(
\begin{array}{c}
n \\
k
\end{array}
\right)
a^{n-k}
\varepsilon^{k-1}
\right)
<
\begin{cases}
a^n + \varepsilon
\left(
\sum_{k=1}^n
\left(
\begin{array}{c}
n \\
k
\end{array}
\right)
x^{n-k}
\varepsilon^{k-1}
\right) \; \text{if} \; x > 1, \\
a^n + \varepsilon
\left(
\sum_{k=1}^n
\left(
\begin{array}{c}
n \\
k
\end{array}
\right)
\varepsilon^{k-1}
\right) \; \text{otherwise}.
\end{cases}
$$
In both cases there exists a positive real number $y$ such that $(a + \varepsilon)^n < a^n + \varepsilon y$.
Now if $a^n < x$, we have that $x - a^n =: \delta > 0$. But for every real number $\delta$ there exists a natural number $N$ such that $\delta > N^{-1}$. Choosing $\varepsilon = y^{-1}N^{-1}$ yields $x - a^n > \varepsilon y \Rightarrow x > (a + \varepsilon)^n$.
Equivalently, to see that there exists an $\varepsilon > 0$ such that $(a - \varepsilon)^n > x$, expand $(a - \varepsilon)^n$:
$$
(a - \varepsilon)^n = a^n + (-\varepsilon)\sum_{k = 1}^n
\left(
\begin{array}{c}
n \\
k
\end{array}
\right)
a^{n-k}(-\varepsilon)^{k-1}
=
a^n +
\varepsilon
\sum_{k = 1}^n
(-1)^k
\left(
\begin{array}{c}
n \\
k
\end{array}
\right)
a^{n-k}\varepsilon^{k-1}.
$$
Splitting the sum in a positive and a negative part:
$$
(a - \varepsilon)^n = a^n + \varepsilon
\left(
\sum_{i=2k}^n
\left(
\begin{array}{c}
n \\
i
\end{array}
\right)
a^{n-i}\varepsilon^{i-1}
- \sum_{j=2k - 1}^n
\left(
\begin{array}{c}
n \\
j
\end{array}
\right)
a^{n-j}\varepsilon^{j-1}
\right),
$$
$$
(a - \varepsilon)^n
>
a^n - \varepsilon
\left(
\sum_{j=2k - 1}^n
\left(
\begin{array}{c}
n \\
j
\end{array}
\right)
a^{n-j}\varepsilon^{j-1}
\right)
$$
But, again, since $a$ is bounded by $\max(1, x)$, there is a positive real number $y$ such that $(a - \varepsilon)^n > a^n - \varepsilon y$.
Now if $a^n > x$, we have that $a^n - x =: \delta > 0$. Following the same argument than above, $a^n - x > \varepsilon y \Rightarrow -x > \varepsilon y - a^n \Leftrightarrow x < a^n - \varepsilon y < (a - \varepsilon)^n$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3257098",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
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"answer_id": 0
} |
How to tackle this squaring of inequality problem
If the roots of quadratic equation $$x^2 β 2ax + a^2 + a β 3 = 0$$
are real and less than $3$, find the range of $a$.
The roots are $a \pm \sqrt {3 β a}$
For the roots to be real, we must have a < 3.
Also, for the roots to be less than 3, we must have $\pm \sqrt {3 β a } \lt 3 β a $
If squaring both sides is allowable, I will get $(a β 2)(a β 3) > 0$. Then the problem is solved.
The question is:- how to convince others that the squaring of both sides of $\pm \sqrt {3 - a } \lt 3 - a $ is allowable?
| $y=(x-a)^2+(a-3)$;
$a-3 >0:$ This parabola does not cut the $x-$axis ($(x-a)^\ge 0)$.
Hence $a-3\le 0$.
Roots:
$y=(x-a)^2 +(a-3) = 0;$
$x_{1,2}=a\pm \sqrt{3-a} <3$.
1)Let $0\le a \lt 3$ ($a=3$ is ruled out)
$\sqrt{3-a} <3-a=\sqrt{3-a}\sqrt{3-a}$.
This is true for $\sqrt{3-a} >1$, or $ 3-a >1$.
Hence $2>a \ge 0$.
2) Let $a <0$;
Then $-|a| +\sqrt{3+|a|} <3$.
$\sqrt{3+|a|} <3+|a|=\sqrt{3+|a|}\sqrt{3+|a|}.$
This is true for $\sqrt{3+|a|} >1$, or $3+|a|>1$.
Hence $a<0$;
Combining:
$a \in (-\infty, 2)$
| {
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"source": "stackexchange",
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} |
Evaluate $\int_{0}^{1}\frac{x(1-x)^2}{1+x+x^2}\frac{\mathrm dx}{\ln x}$ I would like to evaluate this integral.
$$\int_{0}^{1}\frac{x(1-x)^2}{1+x+x^2}\frac{\mathrm dx}{\ln x}$$
$1+x+x^2=(x+\frac{1}{2})^2+1-\frac{1}{4}=(x+\frac{1}{2})^2-\frac{3}{4}$
$$\int_{0}^{1}\frac{x(1-x)^2}{(x+\frac{1}{2})^2-\frac{1}{4}}\frac{\mathrm dx}{\ln x}$$
$$-\int_{0}^{\infty}\frac{e^{2y}(1-e^y)^2}{(e^y+\frac{1}{2})^2-\frac{1}{4}}\frac{\mathrm dy}{y}$$
I did a useless substitution, I was hoping to get a simpler integral.
| Consider the function
$$ I(s) = \int_{0}^{1} \frac{x^{s+1} (1 - x)^2}{1+x+x^2}\,\frac{\mathrm{d}x}{\log x}. $$
Then
\begin{align*}
I'(s)
&= \int_{0}^{1} \frac{x^{s+1} (1 - x)^3}{1-x^3}\,\mathrm{d}x \\
&= \frac{1}{3}\int_{0}^{1} \frac{u^{(s-1)/3} (1 - u^{1/3})^3}{1-u}\,\mathrm{d}u \tag{$x=u^{1/3}$} \\
&= \frac{1}{3} \sum_{k=0}^{3} \binom{3}{k} (-1)^{k-1} \int_{0}^{1} \frac{1-u^{(s+k-1)/3}}{1-u}\,\mathrm{d}u \\
&= \frac{1}{3} \sum_{k=0}^{3} \binom{3}{k} (-1)^{k-1} \psi((s+k+2)/3),
\end{align*}
where we utilized the identity $\sum_{k=0}^{3}\binom{3}{k}(-1)^k = 0$. Also, $\psi$ is the digamma function and we utilized the identity that $\int_{0}^{1} \frac{1-u^z}{1-u} \, \mathrm{d}z = \gamma + \psi(z+1)$ in the final step. Together with $I(\infty) = 0$, we get
\begin{align*}
I(0)
&= -\lim_{R\to\infty}\int_{0}^{R} I'(s) \, \mathrm{d}s \\
&= \lim_{R\to\infty}\sum_{k=0}^{3} \binom{3}{k} (-1)^{k} \left[ \log\Gamma((R+k+2)/3) - \log\Gamma((k+2)/3) \right] \\
&= - \sum_{k=0}^{3} \binom{3}{k} (-1)^{k} \log\Gamma((k+2)/3) \\
&= \log(18) - 3\log\Gamma(1/3).
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3260167",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
maximum and minimum value of $\frac{x^2+y^2}{x^2+xy+4y^2}$
If $x,y\in\mathbb{R}$ and $x^2+y^2>0.$ Then maximum and minimum value of $\displaystyle \frac{x^2+y^2}{x^2+xy+4y^2}$
Plan
Let $$K=\frac{x^2+y^2}{x^2+xy+4y^2}$$
$$Kx^2+Kxy+4Ky^2=x^2+y^2\Rightarrow (4K-1)y^2+Kxy+(K-1)x^2=0$$
put $y/x=t$ and equation is $(4K-1)t^2+Kt+(K-1)=0$
How do i solve it Help me plesse
| $K=\frac{1+t^2}{1+t+4t^2}$ To find min and max, set $K'(t)=0$ or $t^2-6t-1=0$ and solve for $t$. Answer $t=3\pm \sqrt{10}$ leading to max and min $K=\frac{20\pm 6\sqrt{10}}{80\pm 25\sqrt{10}}$. or $1.088303688022443$ and $0.245029645310883$
Corrected (stupid error in solving quadratic).
| {
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"answer_id": 0
} |
Find residues of a function $f(z) = z^{10}e^{\frac{1}{2z}} + \frac{z+4\pi}{z-4\pi}\cot{\frac{z}{2}}$ I am trying to find residues at singularity points of this function, but I cannot wrap my head around its Laurent series. Singularity points are $2k\pi, k \in \mathbb{Z}.$ $f$ is a sum of two functions and $0$ is an essential singularity for both of them. Should I conclude it is an essential singularity for $f$? Not sure, since some terms could cancel out. It is easy to show that $-4\pi$ is a removable singularity. For $k \neq -2$, I have no idea.
| Actually, $0$ is an essential singularity of $z^{10}e^{1/2z}$, but not of $\frac{z+4\pi}{z-4\pi}\cot\left(\frac z2\right)$. So, $0$ is an essential singularity of $f$.
Anyway,$$\cot\left(\frac z2\right)=\frac2{z-4\pi}-\frac16(z-4\pi)+\cdots,$$and so$$\frac{\cot\left(\frac z2\right)}{z-4\pi}=\frac2{(z-4\pi)^2}-\frac16+\cdots$$It follows from this that\begin{align}\frac{z+4\pi}{z-4\pi}\cot\left(\frac z2\right)&=\bigl((z-4\pi)+8\pi\bigr)\left(\frac2{(z-4\pi)^2}-\frac16+\cdots\right)\\&=\frac{16\pi}{(z-4\pi)^2}+\frac2{z-4\pi}-\frac{4\pi}3+\cdots\end{align}and therefore$$\operatorname{rez}_{z=4\pi}\frac{z+4\pi}{z-4\pi}\cot\left(\frac z2\right)=2.$$So, the residue of $f$ at $4\pi$ is $2$.
On the other hand.$$\operatorname{res}_{z=0}z^{10}\exp\left(\frac1{2z}\right)=\frac1{2^{11}11!}$$and\begin{align}\operatorname{res}_{z=0}\frac{z+4\pi}{z-4\pi}\cot\left(\frac z2\right)&=\lim_{z\to0}z\frac{z+4\pi}{z-4\pi}\cot\left(\frac z2\right)\\&=\lim_{z\to0}\frac{z+4\pi}{z-4\pi}\frac{\cos\left(\frac z2\right)}{\sin\left(\frac z2\right)/z}\\&=\frac{0+4\pi}{0-4\pi}\frac{\cos(0)}{1/2}\\&=-2.\end{align}Therefore,$$\operatorname{res}_{z=0}f(z)=\frac1{2^{11}11!}-2.$$
| {
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"source": "stackexchange",
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Show that $\frac{(m!)^{1/m}}{m/e}$ is a decreasing function of $m$ Show that
$\dfrac{(m!)^{1/m}}{m/e}$
is a decreasing function of $m$.
Here is my proof.
I would like to see others,
preferably simpler.
I have shown in
Proof explanation $\lim\limits_{n\to\infty}\frac{n}{\sqrt[n]{n!}}=e$
that,
if
$r_m
=\dfrac{(m!)^{1/m}}{m/e}
$,
then
$r_m > 1+1/m
$.
This was based on
$(1+1/m)^m < e
\lt (1+1/m)^{m+1}
$,
which was also shown there.
Since
$r_m \to 1$,
this leads to the conjecture that
$r_m$ is a decreasing function of $m$.
Let
$s_m = \dfrac{r_{m+1}}{r_m}$.
Then,
since $r_m > 1+1/m$
implies that
$m! > (1+1/m)^mm^m/e^m$,
$\begin{array}\\
s_m
&= \dfrac{r_{m+1}}{r_m}\\
&= \dfrac{\dfrac{((m+1)!)^{1/(m+1)}}{(m+1)/e}}{\dfrac{(m!)^{1/m}}{m/e}}\\
&= \dfrac{\dfrac{((m+1)!)^{1/(m+1)}}{(m+1)}}{\dfrac{(m!)^{1/m}}{m}}\\
&= \dfrac{m((m+1)!)^{1/(m+1)}}{(m+1)(m!)^{1/m}}\\
\text{so}\\
s_m^{m(m+1)}
&= \dfrac{m^{m(m+1)}((m+1)!)^{m}}{(m+1)^{m(m+1)}(m!)^{m+1}}\\
&= \dfrac{m^{m(m+1)}m!^m(m+1)^m}{(m+1)^{m(m+1)}m!^{m}m!}\\
&= \dfrac{m^{m(m+1)}}{(m+1)^{m^2}m!}\\
&= \dfrac{m^{m}}{(1+1/m)^{m^2}m!}\\
&< \dfrac{m^{m}}{(1+1/m)^{m^2}(1+1/m)^mm^m/e^m}\\
&= \dfrac{e^m}{(1+1/m)^{m^2+m}}\\
\text{so}\\
s_m^{m+1}
&< \dfrac{e}{(1+1/m)^{m+1}}\\
&< 1\\
\end{array}
$
I find it pleasing that
the proof ends up
depending on the
upper bound for $e$.
| I guess the following is "quicker", but relies on some heavy machinery. We have
\begin{align*}
\frac{d}{dm} \log r_m &= -\frac{1}{m^2} \log m! + \frac{1}{m} \frac{\Gamma'(m+1)}{m!} - \frac{1}{m} \\
&\le -\frac{\log\sqrt{2\pi}-m+(m+\frac{1}{2})\log m}{m^2} + \frac{H_m - \gamma - 1}{m}
\end{align*}
where we used the fact that $\sqrt{2\pi}m^{m+\frac{1}{2}}e^{-m} \le m!$ and $\Gamma'(m+1) = m!(-\gamma + H_m)$, where $H_m$ is the $m$th harmonic number. So
\begin{align*}
\frac{d}{dm} \log r_m &\le \frac{1}{m}\left(H_m - \left(1 + \frac{1}{2m}\right)\log m - \gamma - \frac{1}{m}\log\sqrt{2\pi}\right)
\end{align*}
But since
\begin{align*}
H_m &= \log m + \gamma + \frac{1}{2m} - \frac{1}{12m^2} + \cdots \\
&< \log m + \gamma + \frac{1}{2m} \\
&< \left(1 + \frac{1}{2m}\right)\log m + \gamma + \frac{1}{m}\log\sqrt{2\pi}
\end{align*}
we are done.
| {
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Show that a following equation has no solution in integers: $x^3-x+9=5y^2$
Show that a following equation has no solution in integers: $$x^3-x+9=5y^2$$
Clearly we see that $y$ is odd, so $y^2\equiv_8 1$ and thus $8\mid x^3-x-4$.
So if $x$ is odd, then $x-1$ and $x+1$ one is divisible by $2$ and other by $4$ so $8\mid x^3-x$ and thus $8\mid 4$ which is not true. So $x$ must be also even and even divisible by $4$ but not by $8$.
Also if we take mod 5 we get $x^3\equiv_5 x+1$ and thus $$x^2 \equiv_5 x^6 \equiv_5 (x+1)^2 = x^2+2x+1\implies 2x\equiv_5 -1 \implies x\equiv_5 -3$$
If we look at modulo 3 we have $$x^3-x+9\equiv_3 0\implies 3\mid y \implies 9\mid (x-1)x(x+1)$$
so $x \equiv_9 0,\pm1$.
but I can not go any further.
| Aqua, here it is.
As you've obtained, $x\equiv 2\pmod{5}$. Now, $x(x-1)(x+1)=5y^2-9$. Note that, $x+1\equiv 3\pmod{5}$, and is odd. I now claim that, there is a prime $p\mid x(x-1)(x+1)$ such that $p\equiv 2,3\pmod{5}$, and $p\neq 3$. If $x\equiv 1,2\pmod{3}$, then the object $x$ clearly has such a prime divisor (indeed, if all prime divisors of $x$ are of form $5k\pm 1$ then it cannot be congruent to $2$ in modulo $5$). Similarly, if $x\equiv 0\pmod{3}$, then $x+1$ is not divisible by $3$, and with the same logic, has such a prime divisor.
Now, isolate such a prime divisor, $p\neq 3$ and $p\equiv \pm 2\pmod{5}$. Observe that, $p\mid 5y^2-9$, that is,
$$
5y^2\equiv 3^2\pmod{p}\Rightarrow (\frac{5}{p})=1,
$$
namely, $5$ is a quadratic residue, modulo $p$. Next, by the quadratic reciprocity law,
$$
(\frac{5}{p})(\frac{p}{5})=(-1)^{\frac{p-1}{2}\cdot 2}=1\Rightarrow (\frac{p}{5})=1
$$
But since $p\equiv \pm 2\pmod{5}$, this is clearly impossible!
| {
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U~$(0,i)$ where $i$ is a fair dice roll.
If I throw a fair dice and the result is $i$ then I choose a point $X$~$(0,i)$
1st) What is the expected value and standard deviation of $X$?
2nd) if $X > 3$ then what is the probability that the result of the dice was $6$?
I think I got the 1st part correct but the second part I am confused on what method to use.
My attempt:
1st)
expected value of a dice roll = $\frac{1}{6} *(1+2+3+4+5+6) = 3.5 $
so, $i = 3.5$ and $X$~Uni$(0,3.5)$
By definition of the Uniform distribution :
$$E(X)=\frac{3.5}{2}=1.75 $$
$$D(X)= \frac{3.5}{\sqrt{12}} $$
2nd)
This is the part am not sure of:
if $X$~Uni$(0,3.5)$ and $P(X>3)= 1-P(X \leq3) = 1- \frac{3}{3.5}= \frac{1}{7}$
we want $$P(i=6|X>3) = \frac{P(i=6,X>3)}{P(X>3)} $$but idk how to get the numerator here, also I may be completely wrong about both parts so please point that out if it happens to be the case!
ALso, this might be right:
$$ P(i=6,X>3) = P(X>3|i=6)* P(i = 6) = \frac{3}{6}* \frac{1}{6}$$
This would imply $$ P(i=6|X>3) = \frac{1/6 * 1/2}{1/7} = \frac{7}{12}$$
Any help is appreciated!
| a)
Note that, for any $\sigma$-field $\mathcal G$, we have $E[X] = E[E[X|\mathcal G]]$
Let us consider $\mathcal G$ = $\sigma(Y)$, where Y is rv. that characterises the number we've rolled. Due to Y being discrete, we only need to find $E[X|Y=k]$ to know $E[X|Y]$.
So take $k\in\{1,2,3,4,5,6\}$
$E[X|Y=k]$ = $\frac{E[X\cdot \chi _{\{Y=k\}}]}{P(Y=k)}$ = $\frac{E[X_k*\chi _{\{Y=k\}}]}{\frac{1}{6}}$ = $E[X_k]$ = $\frac{k}{2}$, where $\chi$ is the characteristic function, $X_k$ is the uniform rv on $[0,k]$ (third equality is due to independence), so
$E[E[X|Y]]$ = $E[\frac{Y}{2}]$ = $\frac{7}{4}$
To find $D(X)$ note that, $D(X) = \sqrt{ E[X^2] - (E[X])^2 } $, and that we already have $E[X]$, to find $E[X^2]$, we proceed similarly, considering $E[X^2|Y=k]$ we find it is equal to $E[X_k^2]$ = $\frac{1}{k} \cdot \frac{k^3}{3}$ = $\frac{k^2}{3}$, so again
$E[X^2] = E[E[X^2|Y]] = E[\frac{Y^2}{3}] = \frac{1^2+2^2+3^2+4^2+5^2+6^2}{18} = \frac{91}{18}$, so plugging that: $ D(X) = \sqrt{ \frac{91}{18} - \frac{49}{16} }
$=$ \sqrt{\frac{91\cdot 8 - 49 \cdot 9}{16\cdot9}} = \frac{\sqrt{287}}{12}$
b)
As you noted: $P(Y = 6 | X>3) = \frac{P(X>3)|Y=6)P(Y=6)}{P(X>3)}$
$P(X>3|Y=6) = P(X_6 >3) = \frac{1}{2}$
$P(Y=6) = \frac{1}{6}$
The problematic one is $P(X>3) = \sum_{k=1}^6 P(X>3|Y=k)P(Y=k) = \frac{1}{6}\sum_{k=4}^6 P(X_k>3) = \frac{1}{6}\sum_{k=4}^6\frac{k-3}{k}$ =$ \frac{1}{6}(\frac{1}{4} + \frac{2}{5} + \frac{1}{2}) = \frac{1}{6}(\frac{5+4+10}{20}) = \frac{19}{120}$
So, $P(X>3) = \frac{\frac{1}{6}\cdot \frac{1}{2}}{\frac{19}{120}} = \frac{10}{19}$
| {
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Find minimum value of $P=\sum \frac{a}{a^2+b^3}$ If $a,b,c$ be non-negative numbers and $a+b+c=3$, then find minima value of function $$P=\frac{a}{a^2+b^3}+\frac{b}{b^2+c^3}+\frac{c}{c^2+a^3}$$
Here is my idea.
By WA i see that if the equality occurs $a=b=c=1$, we will get $\text {Min}_P=1.5$. Indeed i will prove $P\ge 1.5$ by BW and C-S
We have:
$$\text{L.H.S}=3\sum \frac{a^2}{a^3\sum a+3ab^3}\ge \frac{3\left(\sum a\right)^2}{\left(a+b+c\right)\left(\sum a^3\right)+3\sum ab^3}$$
Or $$2\left(\sum a\right)^4\ge 9\left(\sum a\right)\left(\sum a^3\right)+27\left(\sum ab^3\right)$$
I tried BW here but failed.The last inequality is wrong. Help me.
| By the Andreas's hint it remains to prove that
$$\sum_{cyc}\frac{a}{a^2+b^3}\geq\frac{1}{3}.$$
Now, by C-S
$$\sum_{cyc}\frac{a}{a^2+b^3}\geq\frac{(a+b+c)^2}{\sum\limits_{cyc}(a^3+b^3a)}.$$
Id est, it's enough to prove that
$$(a+b+c)^4\geq(a+b+c)(a^3+b^3+c^3)+3(a^3c+b^3a+c^3b)$$ or
$$\sum_{cyc}(3a^3b+6a^2b^2+12a^2bc)\geq0,$$ which is obvious.
Thus, $\frac{1}{3}$ is a minimal value.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3271709",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
If $\tan x=3$, then what is the value of ${3\cos{2x}-2\sin{2x}\over4\sin{2x}+5\cos{2x}}$?
If $\tan x=3$, then what is the value of
$${3\cos{2x}-2\sin{2x}\over4\sin{2x}+5\cos{2x}}$$
So what I did is change all the $\sin{2x}$ and $cos{2x}$ with double angle formulas, getting
$${3\cos^2{x}-3\sin^2{x}-4\sin{x}\cos{x}\over5\cos^2{x}-5\sin^2{x}+8\sin{x}\cos{x}}$$
Now I thought of changing the top part to $\sin{x}$ and bottom part to $\cos{x}$ hoping to somehow get $\tan{x}$ in this way, but I ultimately got just
$${3-6\sin^2{x}-4\sin{x}\cos{x}\over-5+10\cos^2{x}+8\sin{x}\cos{x}}$$
Had really no ideas what to either do after this, seems pretty unusable to me. Was there possibly a mistake I made in the transformation or maybe another way of solving this?
| Since $\tan x = {\sin x \over \cos x}$ we have $\sin x = 3\cos x$ so
$${3\cos{2x}-2\sin{2x}\over4\sin{2x}+5\cos{2x}}= {3\cos^2x-3\sin^2x-4\sin x \cos x\over 8\sin x \cos x +5\cos^2-5\sin^2 x}$$
$$ {3-27-12\over 24+5-45} = {9\over 4}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3273674",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Radical equation solve $\sqrt{3x+7}-\sqrt{x+2}=1$. Cannot arrive at solution $x=-2$ I am to solve $\sqrt{3x+7}-\sqrt{x+2}=1$ and the solution is provided as -2.
Since this is a radical equation with 2 radicals, I followed suggested textbook steps of isolating each radical and squaring:
$\sqrt{3x+7}-\sqrt{x+2}=1$
$(3x+7=(1-\sqrt{x+2})^2$ # square both sides
(Use perfect square formula on right hand side $a^2-2ab+b^2$)
$3x+7=1^2-2(1)(-\sqrt{x+2})+x+2$ # lhs radical is removed, rhs use perfect square formula
$3x+7=1+2(\sqrt{x+2})+x+2$ # simplify
$3x+7=x+3+2\sqrt{x+2}$ # keep simplifying
$2x+4=2\sqrt{x+2}$ # simplify across both sides
$(2x+4)^2=(2\sqrt{x+2})^2$
$4x^2+16x+16=4(x+2)$ # now that radical on rhs is isolated, square both sides again
$4x^2+12x+14=0$ # a quadratic formula I can use to solve for x
For use int he quadratic function, my parameters are: a=4, b=12 and c=14:
$x=\frac{-12\pm\sqrt{12^2-(4)(4)(14)}}{2(4)}$
$x=\frac{-12\pm{\sqrt{(144-224)}}}{8}$
$x=\frac{-12\pm{\sqrt{-80}}}{8}$
$x=\frac{-12\pm{i\sqrt{16}*i\sqrt{5}}}{8}$
$x=\frac{-12\pm{4i*i\sqrt{5}}}{8}$
$x=\frac{-12\pm{-4\sqrt{5}}}{8}$ #since $4i*i\sqrt{5}$ and i^2 is -1
This is as far as I get:
$\frac{-12}{8}\pm\frac{4\sqrt{5}}{8}$
I must have gone of course somewhere further up since the solution is provided as x=-2.
How can I arrive at -2?
| In addition to the possible typo $\sqrt{3x+7}=1+\sqrt{x+2}$ not $1-\sqrt{x+2}$ in the RHS, you made an arithmetic error later.
From $4x^2+16x+16=4(x+2)$, you should get $4x^2+12x+8=0$, not $4x^2+12x+14=0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3273876",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 9,
"answer_id": 4
} |
Find $ \frac{1}{2^2 β1} + \frac{1}{4^2 β1} + \frac{1}{6^2 β1} + \ldots + \frac{1}{20^2 β1} $ Find the following sum
$$
\frac{1}{2^2 β1} + \frac{1}{4^2 β1} + \frac{1}{6^2 β1} + \ldots + \frac{1}{20^2 β1}
$$
I am not able to find any short trick for it.
Is there any short trick or do we have to simplify and add it?
| $$\sum_{n=1}^{10}\frac{1}{4n^2-1}=\frac12\sum_{n=1}^{10}\frac{1}{2n-1}-\frac{1}{2n+1}=\frac12\left(1-\frac{1}{21}\right)=\frac{10}{21}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3274605",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
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Evaluate $\frac{9}{1!}+\frac{19}{2!}+\frac{35}{3!}+\frac{57}{4!}+\frac{85}{5!}+......$
Prove that $$\frac{9}{1!}+\frac{19}{2!}+\frac{35}{3!}+\frac{57}{4!}+\frac{85}{5!}+......=12e-5$$
$$
e=1+\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+\frac{1}{4!}+\frac{1}{5!}+......
$$
I have no clue of where to start and I am not able to find the general expression of the nth term.
| Look at the sequence of numerators and their differences
\begin{array}
n &1 & 2 &3 &4& 5 \\
u_n & 9 & 19 &35&57&85 \\
1st diff. & 10&16&22&28& \\
2nd diff. &6&6&6&&
\end{array}
The first differences show the sequence is not of the form $an+b$, i.e., not a linear sequence. Working out the $2$nd differences gives all the same difference of $6$ and this means our sequence is of the form $an^2+bn+c$, i.e., a quadratic sequence.
Now $a=1/2 * 2$nd difference, which for us is $a = 3$.
Our sequence is then of the form: $3n^2+bn+c$.
Plug in $n=1$ to get $3*1^2+b*1+c = 3+b+c=9$, so $b+c=6$.
Plug in $n=2$ to get $3*2^2+b*2+c = 12+2b+c = 19$, so $2b+c=7$. Hence $b=1$, $c=5$. So our sequence is $3n^2+n+5$.
Set this series equal to the first and rearrange
\begin{align}
\sum_{n=1}^{\infty} \frac{3n^2+n+5}{n!}&=3\sum_{n=1}^{\infty} \frac{n}{(n-1)!}+\sum_{n=1}^{\infty} \frac{1}{(n-1)!}+5\sum_{n=1}^{\infty} \frac{1}{n!}\\
&=12+12\sum_{n=1}^{\infty} \frac{1}{n!}-5=7+12\sum_{n=1}^{\infty} \frac{1}{n!}
\end{align}
Rearrange to
$$3\sum_{n=1}^{\infty} \frac{n}{(n-1)!}+\sum_{n=1}^{\infty} \frac{1}{(n-1)!}=7+7\sum_{n=1}^{\infty} \frac{1}{n!}=7e$$
now subtract
$$3\sum_{n=1}^{\infty} \frac{n}{(n-1)!}=7e-\sum_{n=0}^{\infty} \frac{1}{n!}=6e$$
and divide
\begin{align}
\sum_{n=1}^{\infty} \frac{n}{(n-1)!}=\sum_{n=0}^{\infty} \frac{1+n}{n!}=\sum_{n=0}^{\infty} \frac{1}{n!}+\sum_{n=0}^{\infty} \frac{n}{n!}=\sum_{n=0}^{\infty} \frac{1}{n!}+\sum_{n=1}^{\infty} \frac{1}{(n-1)!}=2e
\end{align}
and everything matches up as required.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3275356",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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For which of the following functions $f$ is $f(x)=f(2-x)$ for all $x$? The options are:
$A)$ $f(x) = x(x+2)$
$B)$ $f(x) = x-2$
$C)$ $f(x) = 2-x$
$D)$ $f(x) = 3x(x-2)$
$E)$ $f(x) = x^2(2-x)^2$
The answer given in my worksheet is $E)$ and I got it. My doubt is why it can't be $D)$ also? Solving thus:
$f(x) = 3x(x-2)$
$f(2-x) = 3(2-x)(2-x-2) = 3(2-x)(-x) = 3[-(2-x)](x) = 3(x-2)x = 3x(x-2) = f(x)$
If I substitute some random values for $x$, then it works as well.
Example $1$: $x = 3 \rightarrow f(x) = f(3) = 3(3)(3-2) = 9$
$f(2-x) = f(2-3) = f(-1) = 3(-1)(-1-2) = 9$
Example $2$: $x = -1/2 \rightarrow f(x) = f(-1/2) = 3(-1/2)(-1/2-2) = 15/4$
$f(2-x) = f(2-(-1/2)) = f(5/2) = 3(5/2)(5/2-2) = 3(5/2)(1/2) = 15/4$
Thus, I think $D)$ should be the answer as well. Please correct me if I am missing something.
| The functions such that $f(x)=f(2-x)$ for all $x$ are the functions with a symmetry around the vertical line $x=1$ (we can see this solving $x=2-x$, giving us $x=1$).
Therefore we can exclude $B)$ and $C)$ since the graph is a line and they can't have this kind of symmetry (unless they are horizontal lines, which is not the case).
$A)$ has as symmetry axis the line $x=-1$
The symmetry axis for $D$ is indeed $x=1$ and if you calculate $f(2-x)=3(2-x)(2-x-2)=3(2-x)(-x)=3x(x-2)=f(x)$
For $E)$ $f(2-x)=(2-x)^2(2-(2-x))^2=(2-x)^2x^2=f(x)$
So the correct answers are $D)$ and $E)$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Solving a second-order homogeneous differential equation I got into this ODE which looks simple but I have a hard time figuring out how to solve:
$$a\frac{d^2 y}{dx^2}+\frac{da}{dx}\frac{dy}{dx}-\frac{l(l+1)}{x^2}y(x)=0$$
where $a=(1-k/x)^2$, $l\geq 0$, and $k>0$. Can anyone help me? I just want the general solution. I tried using series solution but I guess there must be a simpler way. Thanks!
| Let $z'(x)=y'(x)\,a(x).$ Then $z''(x)=y''(x)\,a(x)+y'(x)\,a'(x).$ This allows us to write the DE as $z''-l(l+1)y/x^2=0.$ Now
$a(x)=(1-k/x)^2=(x-k)^2/x^2;$ our goal is to write $y/x^2$ in terms of $z$ and $x,$ if possible. We have
\begin{align*}
z'&=y'a \\
\frac{z'}{a}&=y' \\
\int\frac{z'}{a}\,dx&=y \\
z''-l(l+1)\,\frac{1}{x^2}\,\int\frac{z'}{a}\,dx&=0.
\end{align*}
Next, we would like to differentiate this expression to get rid of the integral, but we don't want to use the quotient rule, because that would leave the integral intact. We multiply through by $x^2$ to obtain
\begin{align*}
x^2\,z''-l(l+1)\int\frac{z'}{a}\,dx&=0 \\
x^2\,z'''+2xz''-l(l+1)\,\frac{z'}{a}&=0.
\end{align*}
There is an obvious substitution to reduce the order: $w=z',$ to obtain
$$x^2\,w''+2xw'-l(l+1)\,\frac{w\,x^2}{(x-k)^2}=0. $$
This is exactly solvable:
$$w(x)=\frac{c_1 (k-x)^{\frac{1}{2}-\frac{1}{2} |2
l+1|}}{x}-\frac{c_2 (k-x)^{\frac{1}{2} |2
l+1|+\frac{1}{2}}}{|2 l+1|\,x}. $$
Now $w=y'a,$ so that we must perform
\begin{align*}
y'\,\frac{(x-k)^2}{x^2}&=\frac{c_1 (k-x)^{\frac{1}{2}-\frac{1}{2} |2
l+1|}}{x}-\frac{c_2 (k-x)^{\frac{1}{2} |2
l+1|+\frac{1}{2}}}{|2 l+1|\,x} \\
y'&=\frac{c_1 x(k-x)^{\frac{1}{2}-\frac{1}{2} |2
l+1|}}{(x-k)^2}-\frac{c_2 x(k-x)^{\frac{1}{2} |2
l+1|+\frac{1}{2}}}{|2 l+1|\,(x-k)^2}\\
y&=\frac{2c_1 (k-x)^{-\frac{1}{2} \left| 2 l+1\right| -\frac{1}{2}} (x \left| 2
l+1\right| -2 k+x)}{\left| 2 l+1\right| ^2-1}
+\frac{2c_2 (k-x)^{\frac{1}{2} (\left| 2 l+1\right| -1)} (x \left| 2 l+1\right|
+2 k-x)}{\left| 2 l+1\right| \left(\left| 2 l+1\right| ^2-1\right)}.
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3279845",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
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On the rate of convergence of nested radicals The sequence of nested radicals $\sqrt {a+\sqrt{a+\sqrt{a+...}}}$ is defined by
β $x_1=\sqrt a$, β $x_{n+1}=\sqrt{a+x_n}$, β where $a>0$.
Here are three questions:
(a) Show that $x=\displaystyle\lim_{n\to\infty} x_n=\frac{1+\sqrt{1+4a}}{2}$.
(b) Prove $\displaystyle x-x_n\sim \frac{C_a}{(2x)^n}$ for some $C_a>0.$
(c) For $a=2$, what is the value of $C_2$ in (b)?
My attempt:
(a) is easy. It is easy to prove $\{x_n\}$ is monotone and bounded, hence $x=\displaystyle\lim_{n\to\infty} x_n$ exists. Then we have the equation $x^2-x-a=0$.
For (b), I start by writing:
\begin{align}
x-x_n &= \frac{x^2-x_n^2}{x+x_n} \\
&= \frac{(x+a)-(x_{n-1}+a)}{x+x_n} \\
&= \frac{x-x_{n-1}}{x+x_n}
\end{align}
By induction,
$$x-x_n= \frac{x}{\prod_{k=1}^n (x+x_k)}$$
and I got stuck at this step.
For (c), once we prove $\displaystyle x_n=2\cos\left(\frac{\pi}{2^{n+1}}\right)$, it is not hard to see the constant $C_2$ is equal to $\displaystyle \frac{\pi^2}{4}$. This is a standard example of trigonometric substitution.
Any idea on how to tackle (b)?
Notation: $f(n)\sim g(n)$ if and only if $\displaystyle\lim_{n\to \infty} \frac{f(n)}{g(n)}=1$.
| Let $x_n:=x-h_n$, so that $$(x-h_{n+1})^2=a+x-h_n$$ which simplifies to $$h_{n+1}^2-2xh_{n+1}+h_n=0$$ \begin{align*}h_{n+1}&=x-\sqrt{x^2-h_n}\\
&=\frac{h_n}{2x}(1+\frac{h_n}{4x^2}+\frac{h_n^2}{8x^4}+\cdots)\\
\end{align*}
So multiplying out gives $$h_n=\frac{h_1}{(2x)^n}\prod_{k=1}^n(1+\frac{h_k}{4x^2}+\frac{h_k^2}{8x^4}+\cdots)$$ so the required constant is $$C_a=h_1\prod_{k=1}^n(1+\frac{h_k}{4x^2}+\frac{h_k^2}{8x^4}+\cdots)$$
This constant need not have a closed formula but it exists by standard results on convergence of products.
| {
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"url": "https://math.stackexchange.com/questions/3280493",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Evaluate $S=\sum_{n=2}^\infty\frac{_nC_2}{(n+1)!}.$
Prove that $$S=\sum_{n=2}^\infty\frac{_nC_2}{(n+1)!}=\frac{e}{2}-1.$$
$$
S=\sum_{n=2}^\infty\frac{_nC_2}{(n+1)!}=\sum_{n=2}^\infty\frac{n!}{2(n-2)!(n+1)!}=\sum_{n=2}^\infty\frac{1}{2(n+1)(n-2)!}\\
=\frac{1}{2}\bigg[\frac{1}{3.0!}+\frac{1}{4.1!}+\frac{1}{5.2!}+\frac{1}{6.3!}+\dots\bigg]=\frac{1}{2}\bigg[\Big(\frac{1}{0!}+\frac{1}{1!}+\frac{1}{2!}+\dots\Big)-\Big(\frac{2}{3.0!}+\frac{5}{4.1!}+\dots\Big)\bigg]\\=\frac{e}{2}-\frac{1}{2}\Big(\frac{2}{3.0!}+\frac{5}{4.1!}+\dots\Big)
$$
How do I proceed further as I am stuck with the last infinite series.
| $$\dfrac{\binom n2}{(n+1)!}=\dfrac{n(n-1)}{2(n+1)!}$$
As $n(n-1)=(n+1)n-2(n+1)+2$
$$ \dfrac{\binom n2}{(n+1)!}=\dfrac12\cdot\dfrac1{(n-1)!}-\left(\underbrace{\dfrac1{n!}-\dfrac1{(n+1)!}}\right)$$
The terms under brace telescopes
Use $e^y=\displaystyle\sum_{r=0}^\infty\dfrac{y^r}{r!}$ for the first term
See also : Evaluate the series $\lim\limits_{n \to \infty} \sum\limits_{i=1}^n \frac{n+2}{2(n-1)!}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3280661",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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Finding zero divisors in a polynomial quotient ring
Is $x^2+x+I$ a zero divisor in $\mathbb{Z}_7[x]/I$, where $I=(x^3+5x^2+2x+5)$?
I know that $\gcd(x^3+5x^2+2x+5,x^2+x)=x+1$, and that it means that $x^2+x$ is indeed a zero divisor.
What I struggle with is finding $g(x)$ such that $$(f(x)+\langle x^3+5x^2+2x+5\rangle)(g(x)+\langle x^3+5x^2+2x+5\rangle)=\langle x^3+5x^2+2x+5\rangle$$ in the quotient ring.
I know how to find an inverse, is this any similar? How is it done?
| The long division of $x^3+5x^2+2x+5$ by the GCD $x+1$ yields $x^2+4x+5$, so we have (in the given ring)
$$(x^2+x)(x^2+4x+5)=x(x+1)(x^2+4x+5)=x(x^3+5x^2+2x+5)=x\cdot0=0$$
Thus if $f(x)=x^2+x$, $g(x)$ may be $x^2+4x+5$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3283085",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Find $\sum_{n=0}^\infty\frac{S_n}{(n+1)!}$ if $S_n$ is the sum of the products of the first $n$ natural numbers taken two at a time
If $S_n$ denotes the sum of the products of the first $n$ natural numbers taken two at a time, then find $$\sum_{n=0}^\infty\frac{S_n}{(n+1)!}$$
$$
S_n=\sum_{1\leq i<j\leq n} a_ia_j=\frac{1}{2}\bigg[(\sum a_i)^2-\sum a_i^2\bigg]\\
=\frac{1}{2}\bigg[\frac{n^2(n+1)^2}{4}-\frac{n(n+1)(2n+1)}{6}\bigg]$$
Then
$$\sum_{n=0}^\infty\frac{S_n}{(n+1)!}=\frac{1}{24}\bigg[\sum_{n=0}^\infty\frac{3}{(n-3)!}+\sum_{n=0}^\infty\frac{8}{(n-2)!}\bigg]\\
=T_0+T_1+T_2+\frac{1}{24}\bigg[\sum_{n=3}^\infty\frac{3}{(n-3)!}+\sum_{n=3}^\infty\frac{8}{(n-2)!}\bigg]\\
=0+0+2+\frac{1}{8}\Big[1+\frac{1}{1!}+\dots\Big]+\frac{1}{3}\Big[\frac{1}{1!}+\frac{1}{2!}+\dots\Big]\\=2+\frac{e}{8}+\frac{1}{3}\big[e-1\big]=2+\frac{11e}{24}-\frac{1}{3}=\frac{11e}{24}+\frac{5}{3}
$$
But my reference gives the solution $\frac{11e}{24}$, so what is going wrong with my attempt ?
| You are almost correct, just pay attention to the starting index of each sum at each step:
$$\begin{align}
\sum_{n=0}^\infty\frac{S_n}{(n+1)!}&=\frac{1}{24}\sum_{n=1}^\infty\frac{3n(n+1)-2(2n+1)}{(n-1)!}\\
&=\frac{1}{24}\sum_{n=1}^\infty\frac{3(n-1)(n-2)+8(n-1)}{(n-1)!}\\
&=\frac{1}{24}\sum_{n=3}^\infty\frac{3}{(n-3)!}+\frac{1}{24}\sum_{n=2}^\infty\frac{8}{(n-2)!}\\
&=\frac{e}{8}+\frac{e}{3}=\frac{11e}{24}.
\end{align}$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Show that $\sum_{cyc} \sqrt{8a+b^3}\ge 9$ Prove that $$\sqrt{8a+b^3}+\sqrt{8b+c^3}+\sqrt{8c+a^3}\ge 9 \text{
for } a,b,c\ge0 \text{
and } a+b+c=3.$$
By Holder $$\left(\sum _{cyc}\sqrt{8a+b^3}\right)^2\left(\sum _{cyc}\frac{1}{8a+b^3}\right)\ge 27$$
Or $$\sum _{cyc}\frac{1}{8a+b^3}\le \frac{1}{3} \text{ WLOG } 0\le a\le b =a+u\le c=a+v\le 3$$
By full expanding it's obvious, because:
$$u^2-uv+v^2\ge 0$$
$$\cdots $$
$$72u^9-80u^8v-776u^7v^2-591u^6v^3+683u^5v^4+1403u^4v^5+1569u^3v^6+1168u^2v^7+424uv^8+72v^9\ge $$
$$\ge uv(v-u)(80u^6+\cdots 80v^6)\ge 0$$
My solution is very ugly. Can i solve it by Holder but more beautiful than it does? Help me
| By Holder $$\left(\sum_{cyc}\sqrt{8a+b^3}\right)^2\sum_{cyc}\frac{(2a^2+2b^2+3ab+4ac+bc)^3}{8a+b^3}\geq$$
$$\geq\left(\sum_{cyc}(2a^2+2b^2+3ab+4ac+bc)\right)^3=64(a+b+c)^6.$$
Thus, it's enough to prove that
$$64(a+b+c)^3\geq27\sum_{cyc}\frac{(2a^2+2b^2+3ab+4ac+bc)^3}{8(a+b+c)^2a+9b^3},$$
which is obviously true by BW and computer.
It's interesting that Hoder with $(ka+mb+c)^3$ does not help.
I got these coefficients by the following way.
Holder for two sequences it's the following.
Let $a_i>0$, $b_i>$, $\alpha>0$ and $\beta>0$. Prove that:
$$(a_1+a_2+...+a_n)^{\alpha}(b_1+b_2+...+b_n)^{\beta}\geq\left(\left(a_1^{\alpha}b_1^{\beta}\right)^{\frac{1}{\alpha+\beta}}+\left(a_2^{\alpha}b_2^{\beta}\right)^{\frac{1}{\alpha+\beta}}+...+\left(a_n^{\alpha}b_n^{\beta}\right)^{\frac{1}{\alpha+\beta}}\right)^{\alpha+\beta}$$
The equality occurs for
$$(a_1,a_2,...,a_n)||(b_1,b_2,...,b_n).$$
Now, by Holder
$$\left(\sum_{cyc}\sqrt{8a+b^3}\right)^2\sum_{cyc}\frac{(ma^2+nb^2+kab+lac+bc)^3}{8a+b^3}\geq$$
$$\geq\left(\sum_{cyc}(ma^2+nb^2+kab+lac+bc)\right)^3=\left(\sum_{cyc}((m+n)a^2+(k+l+1)ab)\right)^3.$$
The equality occurs when
$$\left(\sqrt{8a+b^3},\sqrt{8b+c^3},\sqrt{8c+a^3}\right)||$$
$$||\left(\tfrac{(ma^2+nb^2+kab+lac+bc)^3}{8a+b^3},\tfrac{(mb^2+nc^2+kbc+lab+ca)^3}{8b+c^3},\tfrac{(mc^2+na^2+kca+lbc+ab)^3}{8c+a^3}\right),$$
which after substitution $$(a,b,c)=(1,2,0)$$ gives
$$(4,4,1)||\left(\frac{(m+4n+2k)^3}{16},\frac{(4m+2l)^3}{16},(n+2)^3\right),$$
which gives
$$m+4n+2k=4m+2l$$ and
$$m+4n+2k=4(n+2).$$
Now, we need to find values of $m,$ $n$, $k$ and $l$ such that they are solutions of this system and the inequality
$$\left(\sum_{cyc}((m+n)a^2+(k+l+1)ab)\right)^3\geq81\sum_{cyc}\frac{(ma^2+nb^2+kab+lac+bc)^3}{8a+b^3}$$
is true.
For $$(m,n,k,l)=(2,2,3,4)$$ it's indeed true and we are done!
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Solving equation with two unknown degrees I sticked at this problem:
$3^x-7^y=2$ where it should be solved for natural $x$ and $y$.
I made the conclusion that $x=12a+2$ and $y=6b+1$ after playing with different mods. I am sure that the only solution is $x=2$ and $y=1$ as wolfram alpha confirmed it. Can you help with proof for this?
| Let $x>2$ and $y>1$.
Thus, our equation it's $$3^x-9=7^x-7$$ or
$$9\left(3^{x-2}-1\right)=7\left(7^{y-1}-1\right),$$
which says that $3^{x-2}-1$ is divisible by $7$, which gives $x-2$ is divisible by $6,$
which gives $3^{x-2}-1$ is divisible by $3^6-1=8\cdot7\cdot13,$ which gives $7^{y-1}-1$ is divisible by $13$,
which says $y-1$ is divisible by $12$, which says $7^{y-1}-1$ is divisible by $7^{12}-1=2^5\cdot3^2\cdot5^2\cdot13\cdot19\cdot181,$
which says $3^{x-2}-1$ is divisible by $19$, which gives $x-2$ is divisible by $18$, which says
$3^{x-2}-1$ is divisible by $3^{18}-1,$ which is divisible by $37$, which gives $7^{y-1}-1$ is divisible by $37$,
which gives $y-1$ is divisible by $9$, which says that $7^{y-1}-1$ is divisible by $7^9-1$, which is divisible by $27$,
which says $9\left(3^{x-2}-1\right)$ is divisible by $27$, which is a contradiction.
Id est, our equation has no solutions for $x\geq3$ and $y\geq2$.
Can you end it now?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3287777",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Solving a Limit Problem if the Limit Exists My question is, consider $\lim [(x^{2} +2x +1) / (x^{4} -1)]$ for $x \to -1$ (just arrow sign to the right)
What I have done so far is
$(x+1)^{2}/[(x^{2} +1)(x+1)(x-1)]$
$(x+1)/[(x^{2}+1)(x-1)]$
I'm not too sure what to do at this point. If I substitute in $x=-1$, I would get $0$.
So in this case would the limit not exist?
| Note that $x^2+2x+1=(x+1)^2$ and $x^4-1=(x^2+1)(x+1)(x-1),$
so that $$\frac{x^2+2x+1}{(x^2+1)(x+1)(x-1)} = \frac{x+1}{x^2+1)(x-1)},$$ so $$\lim_{x \to -1} \frac{x^2+2x+1}{(x^2+1)(x+1)(x-1)} = \frac{0}{2 \cdot -2} = 0.$$
$0$ is a perfectly fine answer. You may be confusing this with getting 0 in the denominator, which either means the limit doesn't exist or more work needs to be done.
| {
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"url": "https://math.stackexchange.com/questions/3290505",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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minimum value of of $(5+x)(5+y)$
If $x^2+xy+y^2=3$ and $x,y\in \mathbb{R}.$
Then find the minimum value of $(5+x)(5+y)$
What I try
$$(5+x)(5+y)=25+5(x+y)+xy$$
$x^2+xy+y^2=3\Rightarrow (x+y)^2-3=xy$
I am finding $f(x,y)=22+5(x+y)+(x+y)^2$
How do I solve it? Help me please
| Let $x=y=-1$.
Thus, we obtain the value $16$.
We'll prove that it's a minimal value.
Indeed,
$$(5+x)(5+y)-16=xy+5(x+y)+9=3x^2+4xy+3y^2+5(x+y)\sqrt{\frac{x^2+y^2+xy}{3}}\geq$$
$$\geq3x^2+4xy+3y^2-5\sqrt{\frac{(x+y)^2(x^2+y^2+xy)}{3}}=$$
$$=\frac{3(3x^2+4xy+3y^2)^2-25(x+y)^2(x^2+y^2+xy)}{\sqrt3(\sqrt{3}(x^2+y^2+xy)+5\sqrt{(x+y)^2(x^2+y^2+xy)})}=$$
$$=\frac{(x-y)^2(2x^2+xy+2y^2)}{\sqrt3(\sqrt{3}(x^2+y^2+xy)+5\sqrt{(x+y)^2(x^2+y^2+xy)})}\geq0$$ and we are done!
| {
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"timestamp": "2023-03-29T00:00:00",
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Discuss a set of three numbers $x, y, z β N$ such that $x^2+5^4=5^y+z$ Discuss a set of three numbers $x, y, z β N$ such that $x^2+5^4=5^y+z$ .
What about the possible pairs of numbers $x, y β N$, $y$ being an even number, such that $x^2+5^4=5^y$ ? What if $y$ being odd? What if we have another prime number instead of $5$?
About the first question, I have found that for $x=0, y=4, z=0$ and $x=50, y=5, z=0$ the equation meet its conditions. I know it is possibile to find even more solutions, but I can't really find a proper way to "discuss" all of them.
About the second one, if $y$ is an even number, the solution I've found is $x=0, y=4$, while if $y$ is an odd number it's $x=50, y=5$.
About the third question, it seems that for every prime number the only thing we can say for sure is that $yβ₯4$.
| Consider $x^2+5^4 = 5^y+z \implies z = x^2-5^y+5^4$ sub to $x,y,z \in N$. There is one equation and three unknowns, which means degree of freedom is two (possibly infinite solutions). Choose any two variables and the third one will be constrained by the equation. Since $x, y$ are appear in square and exponential terms, choose these two first. It makes easy to get the value of z. Since $z \in N$ choose $x, y$ such that $z > 0 \implies 5^y < x^2 + 5^4$. The procedure is
*
*Choose any $x \in N$
*Choose y, $\mathrm{s.t}\hspace{1ex} 5^y < x^2+5^4$ i.e
*Determine $z = x^2 + 5^4-5^y$
Consider $x^2+5^4=5^y \implies x^2 = 5^y-5^4 = 5^4(5^{y-4}-1) = 125^2 (5^{y-4}-1)$. Things get tricky here, because we have to choose y s.t $z = 5^{y-4}-1$ is a perfect square. Just like you mentioned, when $y=0$ or $y=1$, $z$ is a perfect square. If $y$ is even, then $y-4$ is also even, $(5^{y-4}-1)$ can be written as $(a^2-1)$ where $a = 5^{\frac{y-4}{2}} $. $a^2-1$ will be a perfect square only when $y=0$. Therefore $y=0$ is the only even solution. I'm not sure about the no of solutions, when $y$ is odd. Regarding your third question $y$ should be greater than or equal to 4, because if $ y < 4 \implies 5^{y-4}-1 < 0$. Therefore $y \geq 4$ for any number, not just 5.
| {
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"timestamp": "2023-03-29T00:00:00",
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$ \int_0^\frac{\pi}{2}\ln^n\left(\tan(x)\right)\:dx$ I'm currently working on a definite integral and am hoping to find alternative methods to evaluate. Here I will to address the integral:
\begin{equation}
I_n = \int_0^\frac{\pi}{2}\ln^n\left(\tan(x)\right)\:dx
\end{equation}
Where $n \in \mathbb{N}$. We first observe that when $n = 2k + 1$ ($k\in \mathbb{Z}, k \geq 0$) that,
\begin{equation}
I_{2k + 1} = \int_0^\frac{\pi}{2}\ln^{2k + 1}\left(\tan(x)\right)\:dx = 0
\end{equation}
This can be easily shown by noticing that the integrand is odd over the region of integration about $x = \frac{\pi}{4}$. Thus, we need only resolve the cases when $n = 2k$, i.e.
\begin{equation}
I_{2k} = \int_0^\frac{\pi}{2}\ln^{2k}\left(\tan(x)\right)\:dx
\end{equation}
Here I have isolated two methods.
Method 1:
Let $u = \tan(x)$:
\begin{equation}
I_{2k} = \int_0^\infty\ln^{2k}\left(u\right) \cdot \frac{1}{u^2 + 1}\:du = \int_0^\infty \frac{\ln^{2k}\left(u\right)}{u^2 + 1}\:du
\end{equation}
We note that:
\begin{equation}
\ln^{2k}(u) = \frac{d^{2k}}{dy^{2k}}\big[u^y\big]_{y = 0}
\end{equation}
By Leibniz's Integral Rule:
\begin{align}
I_{2k} &= \int_0^\infty \frac{\frac{d^{2k}}{dy^{2k}}\big[u^y\big]_{y = 0}}{u^2 + 1}\:du = \frac{d^{2k}}{dy^{2k}} \left[ \int_0^\infty \frac{u^y}{u^2 + 1} \right]_{y = 0} \nonumber \\
&= \frac{d^{2k}}{dy^{2k}} \left[ \frac{1}{2}B\left(1 - \frac{y + 1}{2}, \frac{y + 1}{2} \right) \right]_{y = 0} =\frac{1}{2}\frac{d^{2k}}{dy^{2k}} \left[ \Gamma\left(1 - \frac{y + 1}{2}\right)\Gamma\left( \frac{y + 1}{2} \right) \right]_{y = 0} \nonumber \\
&=\frac{1}{2}\frac{d^{2k}}{dy^{2k}} \left[ \frac{\pi}{\sin\left(\pi\left(\frac{y + 1}{2}\right)\right)} \right]_{y = 0} = \frac{\pi}{2}\frac{d^{2k}}{dy^{2k}} \left[\operatorname{cosec}\left(\frac{\pi}{2}\left(y + 1\right)\right) \right]_{y = 0}
\end{align}
Method 2:
We first observe that:
\begin{align}
\ln^{2k}\left(\tan(x)\right) &= \big[\ln\left(\sin(x)\right) - \ln\left(\cos(x)\right) \big]^{2k} \nonumber \\
&= \sum_{j = 0}^{2k} { 2k \choose j}(-1)^j \ln^j\left(\cos(x)\right)\ln^{2k - j}\left(\sin(x)\right)
\end{align}
By the linearity property of proper integrals we observe:
\begin{align}
I_{2k} &= \int_0^\frac{\pi}{2} \left[ \sum_{j = 0}^{2k} { 2k \choose j}(-1)^j \ln^j\left(\cos(x)\right)\ln^{2k - j}\left(\sin(x)\right) \right]\:dx \nonumber \\
&= \sum_{j = 0}^{2k} { 2k \choose j}(-1)^j \int_0^\frac{\pi}{2} \ln^j\left(\cos(x)\right)\ln^{2k - j}\left(\sin(x)\right)\:dx \nonumber \\
& = \sum_{j = 0}^{2k} { 2k \choose j}(-1)^j F_{n,m}(0,0)
\end{align}
Where
\begin{equation}
F_{n,m}(a,b) = \int_0^\frac{\pi}{2} \ln^n\left(\cos(x)\right)\ln^{m}\left(\sin(x)\right)\:dx
\end{equation}
Utilising the same identity given before, this becomes:
\begin{align}
F_{n,m}(a,b) &= \int_0^\frac{\pi}{2} \frac{d^n}{da^n}\big[\sin^a(x) \big] \cdot \frac{d^m}{db^m}\big[\cos^b(x) \big]\big|\:dx \nonumber \\
&= \frac{\partial^{n + m}}{\partial a^n \partial b^m}\left[ \int_0^\frac{\pi}{2} \sin^a(x)\cos^b(x)\:dx\right] = \frac{\partial^{n + m}}{\partial a^n \partial b^m}\left[\frac{1}{2} B\left(\frac{a + 1}{2}, \frac{b + 1}{2} \right)\right] \nonumber \\
&= \frac{1}{2}\frac{\partial^{n + m}}{\partial a^n \partial b^m}\left[\frac{\Gamma\left(\frac{a + 1}{2}\right)\Gamma\left(\frac{b + 1}{2}\right)}{\Gamma\left(\frac{a + b}{2} + 1\right)}\right]
\end{align}
Thus,
\begin{equation}
I_{2k} = \sum_{j = 0}^{2k} { 2k \choose j}(-1)^j \frac{1}{2}\frac{\partial^{2k }}{\partial a^j \partial b^{2k - j}}\left[\frac{\Gamma\left(\frac{a + 1}{2}\right)\Gamma\left(\frac{b + 1}{2}\right)}{\Gamma\left(\frac{a + b}{2} + 1\right)}\right]_{(a,b) = (0,0)}
\end{equation}
So, I'm curious, are there any other Real Based Methods to evaluate this definite integral?
| Using the Reflection property of Beta function, we have
$$
\begin{aligned}
\int_0^{\frac{\pi}{2}} \tan ^a x d x=\int_0^{\frac{\pi}{2}} \sin ^a x \cos ^{-a} x d x =& \frac{1}{2} B\left(\frac{a+1}{2}, \frac{1-a}{2}\right)=\frac{\pi}{2 \sin \left(\frac{a+1}{2} \pi\right)}=\frac{1}{2} \sec \frac{a \pi}{2}
\end{aligned}
$$
$$
\begin{aligned}
\int_0^{\frac{\pi}{2}} \ln ^n(\tan x) d x =& \frac{\partial^n}{\partial a^n}\left.\left(\frac{\pi}{2} \sec \frac{a \pi}{2}\right) \right|_{x=0} \\
=&\left.\frac{\pi^{n+1}}{2^{n+1}} \frac{\partial^n}{\partial x^n}(\sec x)\right|_{x=0}\\
\end{aligned}
$$
Now we can conclude that
$$\boxed{\int_0^{\frac{\pi}{2}} \ln ^n(\tan x) d x =\left\{\begin{array}{l}
0 \qquad\qquad \text { if } n \text { is odd } \\
\dfrac{\pi^{n+1}\left|E_{n}\right|}{2^{n+1}} \quad \text { if } n \text { is even }
\end{array}\right.}$$
where $E_n$ is the Euler Number whose formula comes from WA.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
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Proving $\frac{\sqrt{k(k+1)} }{k-1} \leq1 + \frac{2}{k} + \frac{3}{4k^2}$ for $k \geq 3$. Could you please give me a hint on how to prove the inequality below for $k \geq 3$?
$$\frac{\sqrt{k(k+1)} }{k-1} \leq1 + \frac{2}{k} + \frac{3}{4k^2} $$
Thank you in advance.
| If $k \geq 3, $ then...
$$\begin{align} \frac{\sqrt{k(k+1)} }{k-1} &=\frac{\sqrt{k^2+k}}{k-1} \\
&\lt \frac{\sqrt{k^2+k+\frac 1 4}}{k-1}= \frac{k+ \frac 1 2}{k-1}=1+\frac{3}{2k-2} \\
\end{align}$$
Have to compare $\frac{3}{2k-2}$ with $\frac{8k+3}{4k^2}$ when $ \ \ k \geq 3$
$$\begin{align}\frac{3}{2k-2}&=\frac{12k^2}{4k^2\cdot (2k-2)} \\
\frac{8k+3}{4k^2}&=\frac{(8k+3)(2k-2)}{4k^2 \cdot (2k-2)}=\frac{16k^2-10k-6}{4k^2 \cdot (2k-2)} \\
&=\frac{12k^2+4(k-3)^2+14(k-3)}{4k^2 \cdot (2k-2)}\end{align}$$
therefore when $ \ \ k \geq 3$
$$\frac{3}{2k-2} \leq \frac{8k+3}{4k^2}$$
And
$$\begin{align} \frac{\sqrt{k(k+1)} }{k-1} & \lt 1+\frac{3}{2k-2} \\
&=1+ \frac{12k^2}{4k^2\cdot (2k-2)}\\
&\lt 1+\frac{16k^2-10k-6}{4k^2 \cdot (k-1)} \\
&=1+\frac{8k+3}{4k^2}=1+\frac{2}{k}+\frac{3}{4k^2}
\end{align}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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1025th term of the sequence $ 1,2,2,4,4,4,4,8,8,8,8,8,8,8,8, ... $ Consider the following sequence - $$ 1,2,2,4,4,4,4,8,8,8,8,8,8,8,8, ... $$
In this sequence, what will be the $ 1025^{th}\, term $
So, when we write down the sequence and then write the value of $ n $ (Here, $n$ stands for the number of the below term) above it We can observe the following -
$1 - 1$
$2 - 2 $
$3 - 2$
$4 - 4$
$5 - 4$
. . .
$8 - 8$
$9 - 8$
. . .
We can notice that $ 4^{th}$ term is 4 and similarly, the $ 8^{th}$ term is 8.
So the $ 1025^{th}$ term must be 1024 as $ 1024^{th} $ term starts with 1024.
So the value of $ 1025^{th}$ term is $ 2^{10} $ .
Is there any other method to solve this question?
| The given sequence is equivalent to
$$ 1 + 2(2) + 4(2^{2}) + 8(2^{3}) + 16(2^{4}) + \ldots $$
$$ = 1 + (2^{2})^{1} + (2^{2})^{2} + (2^{3})^{3} + \ldots + (2^{2})^{k-1} + \ldots $$
Now, since this is a geometric series, we may solve
$$ s_{k} = \frac{a(1 - r^{k})}{1-r} = \frac{1(1-4^{k})}{1-4} = 1025 $$
which easily can be shown to give $ 5 < k < 6$.
Hence, the integer $k$ we seek is $5$; and so, the $1025th$ term of the sequence is $4^{k} = 4^{5} = 1024.$
| {
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"url": "https://math.stackexchange.com/questions/3299825",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "17",
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Alternative methods for solving a system of one linear one non linear simultaneous equations Take the equations $$x+y=5$$ $$x^2 + y^2 =13$$
The most basic method to solve this system is to first express the linear equation in terms of one of the variables and then sub that into the non-linear equation.
But I am curious if there are other methods to solve such a system ?
| The 'ac' method can be used.
From above:
2x^2 + (-10x) + 12 = 0
a = 2, c = 12
a*c = 24
The possible factors of 24 are: (24*1),(12*2),(6*4),(3*8)
In the 'ac method' the sets of factors must multiply to 24 and sum to -10.
By examination we find that this is true of only one of the above factors, (6*4) where (-6*-4) = 24 and (-6 + -4) = -10.
2x^2 + (-10x) + 12 = 0
but, as we found, -10x = (-6x + -4x) so by substitution..
2x^2 + (-6x + -4x) + 12 = 0
Regrouping we get..
[2x^2 - 6x] + [-4x + 12] = 0
Factoring like terms out we get..
2x(x-3) + -4(x-3) = 0
Factoring (x-3) out we get..
(x-3)(2x-4) = 0
The zeros occur where x-3 = 0 and where 2x-4 = 0
x-3 = 0, x = 3
2x - 4 = 0, 2x = 4, x = 4/2, x = 2
Given x+y = 5
when x = 3, 3+y=5, y = 5-3, y = 2
when x = 2, 2+y=5, y = 5-2, y = 3
Answers:
x = 3, y = 2 and x = 2, y = 3
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
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Find range of $m$ such that $|x^2-3x+2 | = mx$ has 4 distinct solutions
$|x^2-3x+2 | = mx$
I think that $(x^2-3x+2) = mx$ or $-(x^2-3x+2) = mx$
For $-(x^2-3x+2) = mx$, $D > 0$,
$m = 3 \pm \sqrt{8}$
For $(x^2-3x+2) = mx$, $D > 0$,
$m = -3 \pm \sqrt{8}$
Now. What do you think it's the range?
Note that $ |x^2-3x+2 | = mx$ which means $mx > 0$ isn't it?
| If $f(x) = x^2-(3+m)x+2$, the discriminant is
$D = (3+m)^2-4\cdot 2 = (m+3)^2-8.$
This gives you a quadratic polynomial in $m$. If you have two solutions for $m$, you get in total four solutions of $f(x)$.
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "2",
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How to show that $\sqrt{2+\sqrt{3}} = \dfrac{\sqrt{6}+\sqrt{2}}{2}$
$\sqrt{2+\sqrt{3}} = \dfrac{\sqrt{6}+\sqrt{2}}{2}$
How to change $\sqrt{2+\sqrt{3}}$ into $\dfrac{\sqrt{6}+\sqrt{2}}{2}$
|
$$\sqrt{2+\sqrt3}=\sqrt{\frac{4+2\sqrt3}{2}}=\sqrt{\frac{(1+\sqrt3)^2}{2}}=\frac{1+\sqrt3}{\sqrt2}$$
Can you end it now?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3303949",
"timestamp": "2023-03-29T00:00:00",
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Evaluating $\sqrt{a\pm bi\sqrt c}$ I recently encountered this problem
$$\sqrt{10-4i\sqrt{6}}$$
To witch I set the solution equal to $a+bi$ squaring both sides leaves
$${10-4i\sqrt{6}}=a^2-b^2+2abi$$
Obviously $a^2-b^2=10$ and $2abi=-4i\sqrt{6}$, using geuss and check, the solution is $a=\sqrt12, b=\sqrt2$
But I was wondering if there is a faster way to solve these types of problems or a method that doesn't involve guess and check (since it can get tedious at times) for the basic form
$$\sqrt{a\pm bi\sqrt c}=x+yi$$ Where you're solving for $x$ and $y$.
I've attempted but failed since it gets pretty messy. All formula will be very much appreciated. (not all equations of that form can reduce)
| One thing right off the bat. Every non-zero complex number will have two square roots so there wont be just $x + yi$ there will also be $-x -yi$
" using geuss and check" Why use guess and check when you can use the quadratic formula?
$a^2 - b^2 = 10$ and $2abi = -4\sqrt 6i$
$a=\frac{-2\sqrt 6}b$
$\frac {4*6}{b^2} - b^2 = 10$
$24 - b^4 = 10b^2$
$b^4 + 10b^2 - 24 = 0$
$b^2 = \frac {-10 \pm \sqrt{100+96}}2= -5 \pm \sqrt{25+ 24}=-5\pm 7$.
As $b$ is presumed to be real $b^2 = 2$ and $b=\pm \sqrt 2$
And $a=\frac{-2\sqrt 6}b= \mp \frac {2\sqrt 6}{\sqrt{2}}=\pm 2\sqrt 3(= \pm \sqrt {12})$
So $\sqrt{10-4\sqrt 6i} = \pm(-2\sqrt 3 + \sqrt 2i)$
No guessing. No checking.
| {
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Simplify $\frac{1}{h}\Big( \frac{1}{(x-a-h)^2} - \frac{1}{(x-a)^2}\Big)$
Simplify $\frac{1}{h}\Big( \frac{1}{(x-a-h)^2} - \frac{1}{(x-a)^2}\Big)$
I would like a little help, so that I can finish solving this exercise.
So far, I've got
$$
\frac{\frac{1}{(x-a-h)^2} - \frac{1}{(x-a)^2}}{h}
= \frac{\frac{(x-a)^2 - (x-a-h)^2}{(x-a-h)^2(x-a)^2}}{h}
= \frac{\frac{(x-a)^2 - (x-a-h)^2}{(x-a-h)^2(x-a)^2}}{ \frac{h}{1}}
= \frac{(x-a)^2 - (x-a-h)^2}{h(x-a-h)^2(x-a)^2}
$$
In the numerator maybe I can continue with a difference of squares, but I'm a little confused.
| $$\frac{(x-a)^2 - (x-a-h)^2}{h(x-a-h)^2(x-a)^2}=\frac{\left((x-a)-(x-a-h)\right)\left((x-a)+(x-a-h)\right)}{h(x-a-h)^2(x-a)^2}=$$
$$=\frac{h(2x-2a-h)}{h(x-a-h)^2(x-a)^2}=\frac{(2x-2a-h)}{(x-a-h)^2(x-a)^2}$$
If now you want to take the limit when $\;h\to0\;$ , for example, the above equals
$$\frac2{(x-a)^3}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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How to calculate $\lim_{x \to - 1} \frac{2}{(x+1)^4}$ Calculate $$\lim_{x \to - 1} \frac{2}{(x+1)^4}$$
a) $0$
b) $\infty$
c) $-\infty$
d) $2$
I am able to see that it is equivalent the limit as $x$ approaches $-1$ of $\frac{2}{(x^2+2x+1)^2}$.
I know that when doing limits to infinity this would be $0$ because the denominator has the highest exponent, but I am confused for $x$ approaches $-1$.
Is this a limit from the right and left kind of thing? Would the process be the same if I changed $-1$ to another number?
| Let's construct tables of values.
$$
\begin{array}{c c}
x & f(x) = \dfrac{2}{(x + 1)^4}\\ \hline
0 & 2\\
-0.9 & 20,000\\
-0.99 & 200,000,000\\
-0.999 & 2,000,000,000,000
\end{array}
\qquad
\begin{array}{c c}
x & f(x) = \dfrac{2}{(x + 1)^4}\\ \hline
-2 & 2\\
-1.1 & 20,000\\
-1.01 & 200,000,000\\
-1.001 & 2,000,000,000,000
\end{array}
$$
The values of $x$ have been chosen so that the $x$-coordinates of the points in the table at left are, respectively, $1$, $0.1$, $0.01$, and $0.001$ units greater than $-1$ and so that the $x$-coordinates of the points in the table at right are, respectively, $1$, $0.1$, $0.01$, and $0.001$ units less than $-1$.
Examining the values in the table suggests that as $x \to -1$, $f(x)$ increases without bound.
Let $M > 0$. We will show that we can find $x$ sufficiently close to $-1$ such that $f(x) > M$.
\begin{align*}
f(x) & > M\\
\frac{2}{(x + 1)^4} & > M\\
2 & > M(x + 1)^4\\
\frac{2}{M} & > (x + 1)^4\\
\sqrt[4]{\frac{2}{M}} & > |x + 1|
\end{align*}
Since the steps are reversible (the solution set of each step of the inequality is the same), the final inequality is equivalent to the initial one. Since $|x + 1| = |x - (-1)|$ is the distance of $x$ from $-1$, we may conclude that whenever the distance of $x$ from $-1$ is smaller than $\sqrt[4]{\frac{2}{M}}$, then $f(x) > M$. Since $M$ is arbitrary, $f(x)$ grows larger than any finite number $M$ as $x \to -1$. Therefore,
$$\lim_{x \to -1} f(x) = \lim_{x \to - 1} \frac{2}{(x + 1)^4} = \infty$$
The function
$$f(x) = \frac{2}{(x + 1)^4}$$
is a rational function. Its implicit domain is the set of all real numbers except those where the denominator is zero, which is the set of all real numbers except $-1$. Rational functions are continuous on their domains. Since a function is equal to its limit at a point of continuity, if $x_0 \in (-\infty, -1) \cup (-1, \infty)$, then
$$\lim_{x \to x_0} f(x) = f(x_0)$$
which means we can simply substitute the value of $x_0$ into the function to find the limit at $x = x_0$. For example,
$$\lim_{x \to 0} f(x) = f(0) = 2$$
This is not true at $x = -1$ since it lies outside the domain of the function. Consequently, we must check what value, if any, the function approaches as $x \to - 1$.
| {
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"timestamp": "2023-03-29T00:00:00",
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The Curve $y = \frac{x^2}2$ divides the curve $x^2 + y^2 = 8$ in two partss. Find the area of smaller part. Question: The Curve $y = \frac{x^2}2$ divides the curve $x^2 + y^2 = 8$ in two parts. Find the area of smaller part.
This question is from Basic Mathematics. Please explain how I can solve it according to class 11th student.
My work is below.
$\implies y = \frac{x^2}2 \\
\implies 2y = x^2 \\
\implies -x^2 +2y = 0 \ \ \ \ (1)$
and
$x^2 + y^2 = 8 \ \ \ \ (2) $
Adding (1) and (2)
$y^2 + 2y = 8 \\
(y-2)(y+4) =0 \\
\therefore y = 2 or -4$
But when I plot it in the plain with help of https://www.desmos.com/ value of y can only be 2 for the point of intersections.
I'm unable to understand how I can eliminate -4 from the results.
| Find the intersection of the curves $y=\frac{x^2}{2}$ with $x^2+y^2=8$
From the parabola $y=\frac{x^2}{2}$ we have $2y=x^2$. Pluging this result $x^2=2y$ in the equation function of the circle gives $2y+y^2=8$. The equation $2y+y^2=8$ is a quadratic equation $y^2+2y-8=0$ and can be solved by many methods. One of these methods is completing the square: $y^2+2y-8=(y+1)^2-9=0$.This last equation can be brought to the form $(y+1)^2=9$. Taking the square root of both sides, we get $y_{1,2}+1 = \pm 3$ and subtracting one from both sides of the equation we get $y_{1,2} = -1\pm 3$. This gives finally two possible solutions for the original problem: $y_1 = -1+3=+2 $ and $y_2 =-1-3= -4$. A possible solution is a "maybe" solution, but not yet a solution!
Then check, if $y_1=+2$ is a solution: Plug $y_1=+2$ in the parabola equation $y=\frac{x^2}{2}$ to get $2=\frac{x^2}{2} \Leftrightarrow x^2 = 4\Rightarrow x_{1,2} = \pm 2$. Plug $y_1$ in the circle equation $x^2+y^2=8$ to get $x^2+4=8 \Leftrightarrow x^2 = 4 \Rightarrow x_{1,2} = \pm 2$ (or $x_1 = +2$ and $x_2 = -2$). This means that the points $(x_{1},y_1)=(+2,+2)$ and $(x_{2},y_1)=(+2,-2)$ lie on both curves, so these are intersection points of the parabola and the circle. Then also check, if $y_2=-4$ is a solution: Plug $y_2=-4$ in the parabola equation $y=\frac{x^2}{2}$ to get $-4=\frac{x^2}{2} \Leftrightarrow x^2 = -8\Rightarrow x = \sqrt{-8} = \sqrt{8}~i= 2\sqrt{2}~i$ with the imaginary unit number $i$. In this problem we are not dealing with the complex plane, that's why, $y_2=-4$ is no good solution.
Now if you want the smaller area $A$, then substract the $y-$coordinates ($f(x)=y_{\text{circle}}-y_{\text{parabola}}$) and integrate over this function $f(x)$ from the left to the right intersection point along the $x$-axis or from $x=-2$ to $x=+2$ like this:
$$A=\int_{-2}^{2}f(x)~dx =\int_{-2}^{2}(y_{\text{circle}}-y_{\text{parabola}})~dx=\int_{-2}^{2}(\sqrt{8-x^2}-\frac{x^2}{2})~dx = 2\pi+\frac{4}{3}.$$
| {
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"timestamp": "2023-03-29T00:00:00",
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How to get the value of $A + B ?$ I have this statement:
If $\frac{x+6}{x^2-x-6} = \frac{A}{x-3} + \frac{B}{x+2}$, what is the value of $A+B$ ?
My attempt was:
$\frac{x+6}{(x-3)(x+2)} = \frac{A(x+2) + B(x-3)}{(x-3)(x+2)}$
$x+6=(x+2)A + B(x-3)$: But from here, I don't know how to get $A + B$, any hint is appreciated.
| By equating coefficients, you have the system of equations
$$A+B = 1,$$
$$2A-3B = 6.$$
| {
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"timestamp": "2023-03-29T00:00:00",
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$E(x)=Β |x|-|x+1|+|x+2|-|x+3|+\dots+|x+2016|$. Find the minimum value of $E$. Can any one help me finding the minimum value of the following expression:
$$
E(x)=Β |x|-|x+1|+|x+2|-|x+3|+\dots+|x+2016|
$$
where $x$ is a real number.
| Note that $$\begin{align}E(x+2)&=|x+2|-|x+3|
\mp\cdots+|x+2016|-|x+2017|+|x+2018|\\&=E(x)-|x|+|x+1|-|x+2017|+|x+2018|\end{align} $$
and $$|x+1|-|x|=\begin{cases}1&x\ge 0\\-1&x\le -1\\2x+1&-1\le x\le0\end{cases} $$
$$|x+2017|-|x+2018|=\begin{cases}1&x\ge -2017\\-1&x\le -2018\\2(x+2017)+1&-2018\le x\le2017\end{cases} $$
From this, $$\tag1E(x+2)\le E(x) \qquad \text{for }x\le -1$$
$$\tag2E(x+2)\ge E(x) \qquad \text{for }x\ge -2017$$
so that the minimum of $E$ on the interval $[-2017,-1]$ (which exists because $E$ is continuous and the interval is compact) is also the global minimum of $E$. Combining $(1)$ and $(2)$,
$$\tag3E(x+2)= E(x) \qquad \text{for }-2017\le x\le -1$$
so that we can look for the minimum on any interval of length $2$ within $[-2017,-1]$, for example on $[-3,-1]$. There,
$$ E(x)=|x|-|x+1|+|x+2|-(x+3)+(x+4)\mp\cdots+(x+2016)=|x|-|x+1|+|x+2|+const$$
and the rest is easy.
| {
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"timestamp": "2023-03-29T00:00:00",
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Determining multiplicity of 1 as an eigenvalue for a certain matrix By Matlab, I know that the eigenvalues of the matrix $B^{-1}A$ are 2.457, 0.542, and 1 (multiplicity 3) where $A$ and $B$ are defined as:
\begin{equation}
A=
\begin{pmatrix}
1 & 0 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 & 0 \\
0 & 0 & 2 & 1 & 1 \\
0 & 0 & 1 & 2 & 1 \\
0 & 0 & 1 & 1 & 2 \\
\end{pmatrix},
B=
\begin{pmatrix}
1 & 0 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 & 0 \\
0 & 0 & 1 & 0 & 0 \\
0 & 0 & 0 & 2 & 1 \\
0 & 0 & 0 & 1 & 2 \\
\end{pmatrix}
\end{equation}
Similarly, the eigenvalues of the matrix $B^{-1}A$ are 4.56, 0.43, and 1 (multiplicity 4) where $A$ and $B$ are defined as:
\begin{equation}
A=
\begin{pmatrix}
1 & 0 & 0 & 0 & 0 & 0\\
0 & 2 & 1 & 1 & 1 & 1 \\
0 & 1 & 2 & 1 & 1 & 1 \\
0 & 1 & 1 & 2 & 1 & 1\\
0 & 1 & 1 & 1 & 2 & 1\\
0 & 1 & 1 & 1 & 1 & 2\\
\end{pmatrix},
B=
\begin{pmatrix}
1 & 0 & 0 & 0 & 0 & 0\\
0 & 1 & 0 & 0 & 0 & 0\\
0 & 0 & 1 & 0 & 0 & 0\\
0 & 0 & 0 & 1 & 0 & 0\\
0 & 0 & 0 & 0 & 2 & 1\\
0 & 0 & 0 & 0 & 1 & 2\\
\end{pmatrix}
\end{equation}
In general, given $n$, the matrices are defined as follows:
\begin{equation}
A =
\begin{pmatrix}
I_{n-m_A} & 0 \\
0 & I_{m_A} + J_{m_A} \\
\end{pmatrix},
B =
\begin{pmatrix}
I_{n-m_B} & 0 \\
0 & I_{m_B} + J_{m_B} \\
\end{pmatrix},
\end{equation}
where $m_A \ne m_B$ and they can be $1,...,n-1$ (so it can be that $m_A < m_B$). $J_m$ is a $m \times m$ matrix of ones.
Is there any explanation to why the multiplicity of 1 as an eigenvalue is always $n-2$ where $n$ is the dimension of the matrices?
| Here's one.
Since $A$ and $B$ must always be invertible, the following conditions are equivalent:
*
*$1$ is an eigenvalue of $B^{-1}A$ for eigenvector $x$
*$B^{-1}Ax=x$
*$Ax = Bx$
*$(A-B)x=0$
*$x \in \ker(A-B)$
So the multiplicity of the eigenvalue $1$ of $B^{-1}A$ is equal to the dimension of $\ker(A-B)$.
Edit: As pointed out by Theo Bendit, it's worth showing that $B^{-1}A$ must be diagonalizable in order to rule out any issues of a discrepancy between algebraic and geometric multiplicity of the eigenvalue. One way to see that $B^{-1}A$ must be diagonalizable is to note that $A$ and $B$ are clearly symmetric and positive definite and then apply this result.
Let's now assume $m_A>m_B$; if the opposite is true, we can easily switch the roles of $A$ and $B$. If we examine $A-B$, we find that it has $n-m_A$ zero rows, followed by $m_A-m_B$ rows consisting of $n-m_A$ zeros and $m_A$ ones, followed by $m_B$ rows consisting of $n-m_A$ zeros, $m_A-m_B$ ones, and then $m_B$ zeros. Since $m_B$ and $m_A-m_B$ are both nonzero by assumption, $A-B$ has exactly two distinct nonzero rows, so has rank $2$; thus $\ker(A-B)$ has dimension $n-2$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Can i factor this expression: $x^3+y^3+z^3$ I have the following numerical expression, which is exactly equal to $1$
Text version:
(-(7 2^(2/3))/38 + 3^(2/3)/19 + (23 5^(2/3))/95 + (5 6^(1/3))/19 + 3/38 (5^(1/3) 6^(2/3)) - 10^(1/3)/19 - 3/190 (3^(1/3) 10^(2/3)) - (4 15^(1/3))/19 - 6/95 (2^(1/3) 15^(2/3)))*(2^1/3+3^1/3+5^1/3)
Image version:
according to this fact, $x^3+y^3+z^3$ would be factorized at all and first factor of it should be $x+y+z$ if we set
$
x^3=2,y^3=3,z^3=5
$
If so then how to factor $x^3+y^3+z^3$ ?
PS. I tried to divide $x^3+y^3+z^3$ by $x+y+z$ and failed (got remainder).
| You have the identity:
$$x^3+y^3+z^3-3xyz=(x+y+z)(x^2+y^2+z^2-xy-yz-xz)$$
As far as I know, this is the best you can do, though I'd be happy if someone proves me wrong.
| {
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Modular arithmetic $(3n - 2)x + 5n \equiv 0 \pmod{9n-9}$ $(3n - 2)x + 5n \equiv 0 \pmod{9n-9}$
I know that the congruence has a solution if $gcd((3n - 2), (9n -9)) \mid -5n$ And it seems to have solution because $gcd((3n - 2), (9n -9)) = 1$
I tried with several $x$βs in the equation $(3n-2)x +5n$ but couldnβt found one that would be divisible by $(9n -9)$
Iβm lost. Before, I posted a similar problem Modular arithmetic $(2n+1)x \equiv -7 \pmod 9$
The similarity I see is that in the module the number $9$ Appears again. Still I donβt know if itβs useful.
Thanks in advance.
| Okay... this is a little symbol heavy but you are correct.
This will have a solution if and only if $\gcd(3n-2,9n-9)|-5n$.
Let's work with that first. Remember Euclids algorithm for finding $\gcd$s: $\gcd(a, b) = \gcd(a, b - ka)$ and repeat until you can go any further. Also bear in mind $\gcd(a,b) = \gcd (b,a)$ and $\gcd(a,b)=\gcd(a,-b)$.
$\gcd(3n-2, 9n-9) = \gcd(3n-2, (9n-9)-(9n-6))=$
$\gcd(3n-2, -3) = \gcd(3, 3n-2)=$
$\gcd(3, (3n-2)-3n) = \gcd(3,-2)=$
$\gcd(2,3)\equiv \gcd(2,3-2)=\gcd(2,1)=\gcd(1,2)=\gcd(1,2-1)=\gcd(1,1)=1$.
So this will always have a solution. Hallelujah!
(Okay, I went a little overboard with figuring the $\gcd$ but... always good to hammer ideas home.)
To solve we must figure what $(3n-2)^{-1} \pmod{9n-9}$ is.
$(3n -2)* K \equiv 1\pmod{ 9n-9}$ what is $K$?
Well, we sort of did that we did Euclids Algorithm. Let's call $3n-2=a$ and $9n-9 = b$
$(9n-9) -3(3n-2) = -3\implies -3 = b-3a$
$(3n-2) +(-3)n = -2 \implies -2=a+(b-3a)n= (-3n+1)a + bn$
$-2 = -3 + 1\implies 1 = -2-(-3) = [(-3n+1)a + bn] -[b-3a]=(-3n+4)a + (n-1)b$.
So $(-3n+4)a \equiv 1 \pmod b$ or
$(-3n+4)(3n-2) \equiv 1 \pmod {9n-9}$
So $(3n-2)^{-1} \equiv -3n+4\pmod {9n-9}$and we can verify:
$(-3n+4)(3n-2) \equiv -9n^2 + 18n -8\equiv $
$-9n^2 + 9n + 9n -8 \equiv$
$-n(9n-9) + (9n-9)+1\equiv 1 \pmod{9n-9}$.
Double Hallelujah!.
So finally:
$(3n - 2)x + 5n \equiv 0 \pmod{9n-9}$
$(3n-2)x \equiv -5n \pmod{9n-9}$ so
$(-3n+4)(3n-2)x \equiv -5n(-3n+4) \pmod{9n-9}$ so
$x \equiv 15n^2 - 20n \equiv 6n^2-11n\equiv -3n^2-2n\pmod {9n-9}$
Or $-3n^2 +7n-9$ if that's any easier.
We can verify:
$(3n - 2)(-3n^2-2n) + 5n \equiv$
$-9n^3 +9n \equiv $
$-9n^3 + 9n^2 -9n^n + 9n\equiv$
$-n^2(9n-9) - n(9n -9) \equiv 0 \pmod{9n-9}$.
Triple Hallelujah! Let's go home and drink coffee.
| {
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"timestamp": "2023-03-29T00:00:00",
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Computation Of Integrals Computer the Integral: $$\int\frac{2x+1}{(x-1)(x-2)}dx$$
Now using partial fraction we can write $$\frac{2x+1}{(x-1)(x-2)}=\frac{A}{x-1}+\frac{B}{x-2}$$, So we get $$\frac{2x+1}{(x-1)(x-2)}=\frac{A(x-2)+B(x-1)}{(x-1)(x-2)}$$ Now for all $x$ not equal to $1, 2$ we can cancel out the denominator to get $$2x+1=A(x-2)+B(x-1)$$ Now to find $A$ and $B$ how can we put $x=1$ and $x=2$ in this identity as this identity is valid if and only if $x$ is not equal to $1, 2$
| \begin{align*}
\frac{2x+1}{(x-1)(x-2)} & = \frac{(2x-2) + 3}{(x-1)(x-2)} = \frac{2(x-1)}{(x-1)(x-2)} + \frac{3}{(x-1)(x-2)}\\\\
& = \frac{2}{x-2} + 3\times\frac{(x-1) - (x-2)}{(x-1)(x-2)} = \frac{2}{x-2} + \frac{3}{x-2} - \frac{3}{x-1}\\\\
& = \frac{5}{x-2} - \frac{3}{x-1}
\end{align*}
| {
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"timestamp": "2023-03-29T00:00:00",
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Inscribed Equilateral Triangle Trig - Expression for P/A Problem: The vertices of an equilateral triangle, with perimeter P and area A, lie on a circle with radius r. Find an expression for $\frac{P}{A}$ in the form $\frac{r}{k}$, where k β Z+.
Hi, I'm having a bit of trouble solving this problem. What I've tried is using the formula for the area of an equilateral triangle (a = $\frac{3\sqrt{3}}{4}$$r^2$). Since one side is equal to $\frac{P}{3}$, I inserted that into the formula a = $2rsin(60)$ to get $P=3(2rsin(60))$. That meant that $\frac{P}{A}$ -> $\frac{3(2rsin(60))}{\frac{3\sqrt{3}}{4}r^2}$ = $\frac{2\sqrt{3}}{3\sqrt{3}r}$ = $\frac{2}{3r}$. But I don't think this is the correct answer because when I put them back into the formulas I get different values for the radius! If anyone can help and explain what I did wrong it would help a lot. Thanks!
| I get that $A = \frac{P^2\sqrt{3}}{36}$. Also, $r = \frac{P}{3\sqrt{3}} \implies P = 3\sqrt{3}r$. This effectively matches your results. Thus,
$$\frac{P}{A} = \frac{36}{\sqrt{3}P} = \frac{36}{9r} = \frac{4}{r} \tag{1}\label{eq1A}$$
In your calculations, since $\sin(60^{\circ}) = \frac{\sqrt{3}}{2}$, then $P = 3(2r\sin(60^{\circ})) = 3\sqrt{3}r$. You multiplied by $4$ from the denominator, which should give $12\sqrt{3}$, so perhaps you just dropped the initial $1$ digit to get $2\sqrt{3}$. Multiplying your result by $6$ to compensate gives $\frac{12}{3r} = \frac{4}{r}$, i.e., my result in \eqref{eq1A}.
Since you confirmed in your comment below that, instead of $\frac{r}{k}$, the source question actually asks for the form of $\frac{k}{r}$, then \eqref{eq1A} shows that $k = 4$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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Inequality $\frac{\ln(7a+b)}{7a+b}+\frac{\ln(7b+c)}{7b+c}+\frac{\ln(7c+a)}{7c+a}\leq \frac{3\ln(8\sqrt{3})}{8\sqrt{3}}$ I'm interested by the following problem :
Let $a,b,c>0$ such that $abc=a+b+c$ then we have :
$$\frac{\ln(7a+b)}{7a+b}+\frac{\ln(7b+c)}{7b+c}+\frac{\ln(7c+a)}{7c+a}\leq \frac{3\ln(8\sqrt{3})}{8\sqrt{3}}$$
I have tried to use convexity and Jensen but the result is weaker . I try also Karamata's inequality but it fails totaly so I'm a bit lost .
If you have a hint it would be great .
Thank you .
| Using the fact that $\ln x \le \ln (8\sqrt{3}) + \frac{1}{8\sqrt{3}}(x-8\sqrt{3})$ for $x > 0$
(the proof is simple and thus omitted),
it suffices to prove that
$$\sum_{\mathrm{cyc}}\frac{\ln (8\sqrt{3}) + \frac{1}{8\sqrt{3}}(7a+b-8\sqrt{3})}{7a+b}
\le \frac{3\ln (8\sqrt{3})}{8\sqrt{3}}$$
or
$$\frac{1}{7a+b} + \frac{1}{7b+c} + \frac{1}{7c+a} \le \frac{\sqrt{3}}{8}.$$
This has been solved in the link below:
If $a+b+c=abc$ then $\sum\limits_{cyc}\frac{1}{7a+b}\leq\frac{\sqrt3}{8}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3319146",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Maximizing $\frac{a^2+6b+1}{a^2+a}$, where $a=p+q+r=pqr$ and $ab=pq+qr+rp$ for positive reals $p$, $q$, $r$
Given $a$, $b$, $p$, $q$, $r \in\mathbb{R_{>0}}$ s.t.
$$\begin {cases}\phantom{b}a=p+q+r=pqr \\ab =pq+qr+rp\end{cases} $$
Find the maximum of $$\dfrac{a^2+6b+1}{a^2+a}$$
This question is terrifying that I even don't know how to start it. I've found different ways such as multiplying the numerator and denominator by $a$ and substitute, but it is not useful (maybe?). Anyway, please comment or answer if you solve it or having clues that may help solve this question.
| Hint.
Using the AM-GM we get
$$
a \ge 3\sqrt[3]{pqr}=3\sqrt[3]{a}\Rightarrow a\ge \sqrt{27}
$$
$$
ab\ge 3\sqrt[3]{a^2}\ge 3\sqrt[3]{27}
$$
now we can solve
$$
\max \frac{a^2+6b+1}{a^2+a}\ \ \ \text{s. t.}\ \ \ (a\ge \sqrt{27})\cap (ab \ge 9)
$$
and the solution is
$$
a = 3\sqrt 3,\ \ b = \sqrt 3
$$
with
$$
\frac{a^2+6b+1}{a^2+a} = 1+\frac{1}{3 \sqrt{3}}
$$
NOTE
Using the Lagrangian to obtain the stationary points
$$
L(a,b,\lambda,\mu,s_1,s_2) = \frac{a^2+6b+1}{a^2+a}+\lambda (a-\sqrt 27-s_1^2)+\mu(a b-9-s_2^2)
$$
the stationary points are the solutions for
$$
\nabla L = 0 \Rightarrow\left\{
\begin{array}{rcl}
\frac{2 a}{a^2+a}-\frac{(2 a+1) \left(a^2+6 b+1\right)}{\left(a^2+a\right)^2}+\lambda +b \mu & = & 0 \\
a \mu +\frac{6}{a^2+a} & = & 0\\
-s_1^2+a-3 \sqrt{3} & = & 0 \\
-s_2^2+a b-9 & = & 0 \\
\lambda s_1 & = & 0\\
\mu s_2 & = & 0 \\
\end{array}
\right.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3319843",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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} |
Determine points on line with specific distance from plane There is a line $$p: \dfrac{x-1}{2} = \dfrac{y+1}{3} = \dfrac{z+1}{-1}$$
and plane $$\pi : x+y+2z-3=0.$$
I need to find points $T_1, T_2 \in p$. Requirement when finding those points are that they have a set distance from plane $\pi$, with distance being $\sqrt{6}$, and that the points are on the line.
The solution for this problem is
$T_1=\left( \dfrac 13, -2, -\dfrac 23\right)$
$T_2 = \left( \dfrac{25}{3}, 10, -\dfrac{14}{3} \right)$
Problem is I don't have idea how to get there, so I will apreciate any advice/guidance for solving this problem.
EDIT1
Equation for distance between point and plane is:
$d(pi,T_\pi')=\dfrac{|AX_0+BY_0+CZ_0+D|}{\sqrt{(A^2+B^2+C^2)}})$
so I get:
$\sqrt6=\dfrac{|X_0+Y_0+2Z_0-3|}{\sqrt6}$
$6=|X_0+Y_0+2Z_0-3|$
since the requirement for parallel planes is: $\dfrac{A_1}{A_2}=\dfrac{B_1}{B_2}=\dfrac{C_1}{C_2}$ I deduced that $A_2=B_2=2C_2$
returning to where I left:
$6=|4X_0-3|$
$4X_0-3=6$
and
$4X_0-3=-6$
$X_01=\dfrac{9}{4}$
and
$X_02=\dfrac{-3}{4}$
Finally I have two planes but I will focus on one because I messed up something:
$\pi .. \dfrac{9}{4}X+\dfrac{9}{4}Y+\dfrac{9}{2}-3=0$
If I fit vector of line direction into this equation we get that $t=\dfrac{10}{9}$
Now if try to fit $t$ back in the equation I will get wrong results since x will be $\dfrac{29}{9}$.
What am I currently missing?
| Let $ \dfrac{x-1}{2} = \dfrac{y+1}{3} = \dfrac{z+1}{-1} = t$ for some point $(x,y,z)$ on the line $p$.
Then $p(t) = (1+2t, -1+3t, -1-t)$
Using your distance formula, we get
\begin{align}
\dfrac{|1(1+2t) + 1(-1+3t) + 2(-1-t) - 3|}{\sqrt{(1^2 + 1^2 + 2^2)}} &= \sqrt 6 \\
|3t-5| &= 6 \\
3t - 5 &= \pm 6 \\
t &\in \left\{ -\dfrac 13, \dfrac{11}{3} \right\}
\end{align}
So the points are
$p\left(-\dfrac 13 \right) = T_1$
$p\left( \dfrac{11}{3} \right) = T_2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3320539",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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equation simplification. $(5y-1)/3 + 4 =(-8y+4)/6$ Simplification of this equation gives two answers when approched by two different methods.
Method 1 Using L.C.M( least common multiple)
$(5y-1)/3 + 4 =(-8y+4)/6$
$(5y-1+12)/3 = (-8y+4)/6$
$5y-11 = (-8y+4)/2$
$(5y-11)2= (-8y+4)$
$10y-22 = -8y+4$
$18y=26$
$y = 26/18=13/9$
Method 2 multiplying every term by 3
$3(5y-1)/3 + 4*3 = 3(-8y+4)/6$
$5y-1 + 12 = (-8y+4)/2$
$2(5y-1 + 12) = -8y+4$
$10y-2+24 = -8y+4$
$18y + 22 = 4$
$18y = -18$
$y = -1$
The correct method is method 2 and the correct answer is y = -1
Why is method 1 is incorrect? Could anyone explain why the answer is wrong when using the L.C.M( method 1)?
| We have $5y-1+12=5y+11$ and not $=5y-11.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3323817",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Line$ 3x-4y+k$ touches a circle $x^2+y^2-4x-8y-5 = 0$ at $(a,b)$, then find $k+a+b =?$
The line $3x-4y+\lambda=0, (\lambda > 0)$ touches the circle $x^2+y^2-4x-8y-5 = 0$ at $(a,b)$. Find the value of $\lambda+a+b$.
I tried solving in the following manner:
Clearly the circle has center at $(2,4)$ with a radius of $5$.
So, the perpendicular distance from the center onto the given line must be the radius (Radius is perpendicular to tangent)
So, using the formula for perpendicular distance
$$\frac{|3\times2-4\times4+\lambda|}{(3^2+4^2)^{1/2}}= 5$$
$$\implies\frac{|3\times2-4\times4+\lambda|}{5} = 5$$
$$\implies|\lambda-10| = 25$$
From this, I get $\lambda = 35,-15$, but since in the question it is given $\lambda>0$, I'm taking $\lambda = 35$
Now, the line is $3x-4y+35 = 0$
Given, it is the tangent at $(a,b)$
Writing equation of tangent at (a,b), I get,
$$ax+by+(-2)(x+a)+(-4)(y+b)-5 = 0$$
$$\implies (a-2)x+(b-4)y+-2a-4b-5 = 0$$
which should be identical to the line $$3x-4y+35= 0$$
So, I write
$$\frac{(a-2)}{3} = \frac{b-4}{-4}=\frac{-(2a+4b+5)}{35}$$
Solving these, I obtained that a = -1 and b = 8
So $\lambda+a+b = 42$
But the answer says 20. Is there a problem with my above procedure?
| Here is the picture for the problem, could not insert it as a comment, so it became an answer:
$5$ through the points $(2,-1)$, $(5,0)$, $(6,1)$, $(7,4)$, $(6,7)$, $(5,8)$, $(2,9)$, $(-1,8)$, $(-2,7)$, $(-3,4)$, $(-2,1)$, and $(-1,0)$">
In the picture, we see the circle with center $C(2,4)$ and radius $5$, it passes through the many points $(2,-1)$, $(5,0)$, $(6,1)$, $(7,4)$, $(6,7)$, $(5,8)$, $(2,9)$, $(-1,8)$, $(-2,7)$, $(-3,4)$, $(-2,1)$, and $(-1,0)$, and in the points $A(5,0)$ and $B(-1,8)$ we have a tangent with the same slope as the given line $3x-4y+k=0$. (Because such a line has the slope of $3x-4y=0$, or to have the diametral line passing through the center, $3x-4y=-10$, which also passes through the points $D(-2,1)$ and $E(6,7)$. Note that $DE\perp AB$. The parallel lines to $3x-4y=0$ going through $A$, respectively $B$, are
$$
\begin{aligned}
3x-4y-15 &=0 &&\text{ through $A(5,0)$,}\\
3x-4y+35 &=0 &&\text{ through $B(-1,8)$.}
\end{aligned}
$$
This is all that can be said mathematically.
My speculation is that the "given line $3x-4y+\lambda$" (which is not an equation, well...), should have been the line with equation $3x-4y\color{red}{=}\lambda$. (There is also a comment with a red minus, it is the same story.) Then the numbers game applies, and we would have $a,b;\lambda$ to have the values either $5,0;15$, with the sum $20$, or...
| {
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"url": "https://math.stackexchange.com/questions/3324454",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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find min of $a+b$ given sum of $a\geq b\geq c\geq d$ is 9 and square sum is 21 Suppose $a\geq b\geq c\geq d>0$ and all are real numbers, and $a+b+c+d=9,a^2+b^2+c^2+d^2=21$, how to find the minmum of $a+b$?
What I attempted:
I can show $b\geq 1.5$ and $a\leq 3$.for $r\geq 0$, I consider $\sum(a-r)^2=\sum a^2-2r\sum a+4r^2=21-18r+4r^2$, then $a\leq \sqrt{21-18r+4r^2}+r$ which implies $a\leq 3$. Also I guess the min shoul be 5 and there are two solutions: 3,2,2,2 and $2.5,2.5,2.5,1.5$. But I cannot prove it.
New attempts:
*
*$6-2a=(a-3)^2+\sum(b-2)^2$
*$2d-3=\sum(a-2.5)^2+(d-1.5)^2$
then $2(d-a)+3\geq 1$ which implies that $d\leq c\leq b\leq a\leq d+1$
| Apply the following identity involving an orthogonal array of sign patterns:
$4(a^2+b^2+c^2+d^2)=(a+b+c+d)^2+(a-b-c+d)^2+(a-b+c-d)^2+(a+b-c-d)^2$
Put in the given values and simplify giving
$3=(a-b-c+d)^2+(a-b+c-d)^2+(a+b-c-d)^2$
The condition $a\ge b\ge c\ge d>0$ forces $a+b-c-d$ to have an absolute value greater than or equal the other two squared terms on the right side, forcing $a+b-c-d\ge 1$. This together with $a+b+c+d=9$ then guarantees that $a+b\ge 5$.
This does not assure that $5$ is the true minimum. We have to be able to reach this value. Render the following which satisfies the last equation and all squares the right side being equal:
$a-b-c+d=\pm 1, a-b+c-d=a+b-c-d=1$
With either choice of sign in the first relation, combine with the given equality $a+b+c+d=9$ to form a linear system with one of the two solutions
$a=3,b=c=d=2$ if $a-b-c+d=1$
$a=b=c=5/2, d=3/2$ if $a-b-c+d=-1$
Both of these realize $a+b=5$ making that the true minimum.
| {
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"url": "https://math.stackexchange.com/questions/3327246",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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How can I solve prove that $8(1-a)(1-b)(1-c)\le abc$ with the conditions below? There was a homework about inequalities (that why I ask a bunch of inequality problems). But I couldn't solve the following:
If $0<a,b,c<1$ and $a+b+c=2$, prove that $8(1-a)(1-b)(1-c)\le abc$
I tried many times, and finally I used Muirhead, but it failed!
$\begin{split}L.H.S.-R.H.S.&=8(1-a)(1-b)(1-c)-abc\\&=8-8(a+b+c) +8(ab+bc+ca)-9abc\\&=-(a^3+b^3+c^3)+(a^2b+b^2c+c^2a+ab^2+bc^2+ca^2)-3abc\\&=\frac{1}{2}\Bigg(\sum_{sym}a^2b-\sum_{sym}a^3\Bigg)+\frac{1}{2}\Bigg(\sum_{sym}a^2b-\sum_{sym}abc\Bigg)\end{split}$
But as $(3,0,0)$ majorizes $(2,1,0)$ but $(2,1,0)$ majorizes $(1,1,1)$, so it fails.
Could someone help? Any help is appreciated!
| By AM-GM you have
$$(1 - a)^{1\over 2}(1 - b)^{1 \over 2} \leq {1 - a + 1 - b \over 2} = {c \over 2}$$
$$(1 - b)^{1\over 2}(1 - c)^{1 \over 2} \leq {1 - b + 1 - c \over 2} = {a \over 2}$$
$$(1 - c)^{1\over 2}(1 - a)^{1 \over 2} \leq {1 - c + 1 - a \over 2} = {b \over 2}$$
Multiplying these three inequalities together gives the desired inequality.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Study the convergence of the series I need to study the convergence of the following series:
$$\sum_{n=1}^{\infty}(\frac{1\cdot3\cdot5\cdot\cdots\cdot(2n+1)}{1\cdot4\cdot7\cdot\cdots\cdot(3n+1)}x^n)^2$$ I tried to expand that term and reduce some terms but didn't succeed, I also tried to apply the ratio test but still nothing. Any help is much appreciated.
| The ratio test gives you
$${\left(\frac{1 \cdot 3\cdot 5\cdot ...\cdot (2n+1)\cdot (2n+3)}{1\cdot 4\cdot 7\cdot ...\cdot (3n+1)\cdot (3n+4)}x^{n+1}\right)^2\over \left(\frac{1\cdot 3\cdot 5\cdot ...\cdot (2n+1)}{1\cdot 4\cdot 7\cdot ...\cdot (3n+1)}x^n\right)^2}=\frac{(2n+3)^2}{(3n+4)^2} x^2$$
and here you need to take the limit of the ratio. Note that the final result also depends on the value of $x$.
| {
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"url": "https://math.stackexchange.com/questions/3331096",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "1",
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How to refactor $\pi / 12$ into $\pi /4 - \pi /3$? How to refactor $\pi / 12$ into $\pi /4 - \pi /3$ (standard calculus angles)? How do you know that $\pi / 12$ converts exactly into $\pi /4 - \pi /3$? Do I need to memorize that or is there a simple procedure to follow in order to refactor any angle into combination of standard calculus angles ($\pi /4, \pi /3, \pi /6$) for which sin/cos values are well known?
Similar for $7\pi / 12 = \pi /4 + \pi /3$
| A good place to start is by expressing each of the standard angles as multiples of $\frac{\pi}{12}$.
\begin{align*}
\frac{\pi}{6} & = \frac{2\pi}{12}\\
\frac{\pi}{4} & = \frac{3\pi}{12}\\
\frac{\pi}{3} & = \frac{4\pi}{12}
\end{align*}
From there, you can immediately see that
\begin{align*}
\frac{\pi}{12} & = \frac{\pi}{4} - \frac{\pi}{6}\\
& = \frac{\pi}{3} - \frac{\pi}{4}
\end{align*}
and that
\begin{align*}
\frac{5\pi}{12} & = \frac{\pi}{4} + \frac{\pi}{6}\\
\frac{7\pi}{12} & = \frac{\pi}{3} + \frac{\pi}{4}
\end{align*}
For a larger angle such as $17\pi/12$, look for linear combinations of the special angles. For instance,
\begin{align*}
\frac{17\pi}{12} & = \frac{9\pi}{12} + \frac{8\pi}{12}\\
& = \frac{3\pi}{4} + \frac{2\pi}{3}
\end{align*}
It is less useful to express $17$ as $10 + 7$ since $7$ is not a multiple of $2$, $3$, or $4$. Similarly, it is not helpful to express $17$ as $11 + 6$, $12 + 5$, $13 + 4$, $14 + 3$, or $16 + 1$ since one of the summands is not a multiple of $2$, $3$, or $4$. On the other hand, we can express $17$ as $15 + 2$, which yields
\begin{align*}
\frac{17\pi}{12} & = \frac{15\pi}{12} + \frac{2\pi}{12}\\
& = \frac{5\pi}{4} + \frac{\pi}{6}
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3332551",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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A problem on inequality If real numbers $a >b >1$ satisfy the inequality
$$ ( ab+1)^{2} +( a+b)^{2} \leqslant 2( a+b)\left( a^{2} -ab+b^{2} +1\right) $$
What is the minimum possible value of $\frac{\sqrt{a-b}}{b-1}$?
My approach:
$$ ( a+b)\left( a^{2} -ab+b^{2} +1\right) =( a+b)\left( a^{2} -ab+b^{2}\right) +( a+b) =a^{3} +b^{3} +a+b $$
So, the question becomes $ ( ab+1)^{2} +( a+b)^{2} \leqslant 2\left( a^{3} +b^{3}\right) +2( a+b) $
| The condition gives
$$2a^3-(b^2+1)a^2-2a(2b-1)+2b^3-b^2+2b-1\geq0$$ or
$$a^2(2a-b^2-1)-2a(2b-1)+(2b-1)(b^2+1)\geq0$$ or
$$(2a-b^2-1)(a^2-2b+1)\geq0$$ and since $$a^2-2b+1>a^2-2a+1=(a-1)^2>0,$$ we obtain $$2a\geq b^2+1$$ or
$$2(a-b)\geq(b-1)^2$$ or
$$\frac{\sqrt{a-b}}{b-1}\geq\frac{1}{\sqrt2}.$$
The equality occurs for $(a,b)=(2.5,2)$ for example, which says that $\frac{1}{\sqrt2}$ it's a minimal value.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3333212",
"timestamp": "2023-03-29T00:00:00",
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Find real numbers c and d from $\frac{1}{a+bi}$
Suppose $a$ and $b$ are real numbers, both not 0. Find real numbers $c$ and $d$ such that
$$ \frac{1}{a+bi} = c + di$$
I am not really sure what the question is asking me to do. Am I supposed to represent $c$ and $di$ both in terms of $a$ and $b$ since $a$ and $b$ are real numbers?
I multiplied the fraction by its conjugate, but that didn't give me any hints.
$$\frac{1}{a+bi} \bigg( \frac{a-bi}{a-bi}\bigg) = \frac{a-bi}{a^2+b^2}$$
A nudge in the correct direction would be helpful, thanks
| Notice that
$$
\frac{1}{a+bi}=\frac{a-bi}{a^2+b^2}=\frac{a}{a^2+b^2}-\frac{b}{a^2+b^2}i=c+di.
$$
Hence
$$
c=\frac{a}{a^2+b^2}\,\, \text{ and }\,\,d=-\frac{b}{a^2+b^2}.
$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Use formal definition of a limit to find the largest value of $\delta > 0$ Given $f(x)= {\sqrt x}$, L = 2, $\epsilon = {\frac 1 5}$ and C=4.
I know that I start with
$$|{\sqrt x} -2| \le {\frac 1 5}$$
If I rewrite that as
$$-{\frac 1 5} \le {\sqrt x} -2 \le {\frac 1 5}$$
I can then add 2 to both sides and square everything to get rid of the radical which would give me
$$\frac{81}{25} \lt x \lt \frac{121}{25}$$
but something doesn't seem right with that approach.
| Nothing wrong with the approach so far, aside from mixing up $<$ with $\le$ sometimes. What you have shown so far is
$$\frac{81}{25} < x < \frac{121}{25} \iff |\sqrt{x} - 2| < \frac{1}{5}.$$
Further, we can say,
$$-\frac{19}{25} < x - 4 < \frac{21}{25} \iff |\sqrt{x} - 2| < \frac{1}{5}.$$
Our largest possible value of $\delta$ is therefore $\frac{19}{25}$; any larger value of $\delta$ will allow for $x - 4$ to be too large and negative. We now have
\begin{align*}
|x - 4| < \frac{19}{25} &\iff -\frac{19}{25} < x - 4 < \frac{19}{25} \\
&\implies -\frac{19}{25} < x - 4 < \frac{21}{25} \\
&\iff |\sqrt{x} - 2| < \frac{1}{5},
\end{align*}
completing the proof for this particular $\varepsilon$.
So, why aren't $\varepsilon$-$\delta$ proofs this straightforward, generally speaking? Well, the idea here is to examine the condition $|f(x) - L| < \varepsilon$, and figure out the set of $x$ for which this condition holds true. In our case, we found it was the open interval $\left(\frac{81}{25}, \frac{121}{25}\right)$. Most of the time, this method turns out to either be either intractable, or at least, a far more difficult way of going about things.
Consider, for example, proving $f(x) = x^4 + 2x$ is continuous at $x = 1$. If we fix $\varepsilon = \frac{1}{5}$ as before, then now we need to solve the inequality:
$$-\frac{1}{5} < x^4 + 2x - 3 < \frac{1}{5}$$
Solving this involves finding the roots of the two quartics $x^4 + 2x - 3 \pm \frac{1}{5}$, a problem that is possible to do, but horribly, horribly complicated. If we consider higher degree polynomials, it can easily be impossible to do with exact values.
It's much easier instead to make estimates. You don't need to find all $x$ such that $|f(x) - L| < \varepsilon$, but just an interval an interval around $x = c$. This can yield much more straightforward proofs that are easier to read, write, and come up with!
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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How to calculate this definite integral $\int_0^1\frac{{\ln^4 x }}{1+x^2}dx=\frac{5\pi^5}{64}$ $$\displaystyle\int_0^1\dfrac{{\ln^4 x}}{1+x^2}\text{d}x=\dfrac{5\pi^5}{64}$$
let $x=e^{-t}$οΌ
$$
\displaystyle\int_0^1\dfrac{({\ln x})^4}{1+x^2}\text{d}x=\displaystyle\int_0^{+\infty}\dfrac{t^4\text{e}^{-t}}{1+\text{e}^{-2t}}\text{d}t=\displaystyle\sum_{k=0}^{\infty}(-1)^k\displaystyle\int_0^{+\infty}t^4\text{e}^{-(2k+1)t}\text{d}t.
$$
let $u=(2k+1)t$οΌ
$$
\displaystyle\sum_{k=0}^{\infty}(-1)^k\displaystyle\int_0^{+\infty}t^4\text{e}^{-(2k+1)t}\text{d}t=\displaystyle\sum_{k=0}^{\infty}\dfrac{(-1)^k}{(2k+1)^5}\Gamma(5)=24\displaystyle\sum_{k=0}^{\infty}\dfrac{(-1)^k}{(2k+1)^5}.
$$
I don't know how to solve this series.
So how can I solve this series? And is there any other ways to solve this definite integral. Thank you.
| \begin{align}
J&=\int_0^1 \frac{\ln^4 x}{1+x^2}\,dx\\
&=\frac{1}{2}\int_0^\infty \frac{\ln^4 x}{1+x^2}\,dx\\
J_n&=\int_0^\infty \frac{\ln^{2n} x}{1+x^2}\,dx\\
J&=\frac{1}{2}J_2\\
K_n&=\int_0^\infty \int_0^\infty\frac{\ln^{2n}(xy)}{(1+x^2)(1+y^2)}\,dx\,dy\\
&\overset{u=yx}=\int_0^\infty\left(\int_0^\infty\frac{y\ln^{2n} u}{(u^2+y^2)(1+y^2)}\,du\right)\,dy\\
&=\frac{1}{2}\int_0^\infty\left[\ln\left(\frac{u^2+y^2}{1+y^2}\right)\right]_{y=0}^{y=\infty}\frac{\ln^{2n} u}{1-u^2}\,du\\
&=\int_0^\infty \frac{\ln^{2n+1}u}{u^2-1}\,du\\
&=2\int_0^1 \frac{\ln^{2n+1}u}{u^2-1}\,du\\
&=2\int_0^1 \frac{\ln^{2n+1}u}{u-1}\,du-2\int_0^1 \frac{u\ln^{2n+1}u}{u^2-1}\,du\\
&=\left(2-\frac{1}{2^{2n+1}}\right)\int_0^1 \frac{\ln^{2n+1}u}{u-1}\,du\\
&=(2n+1)!\left(2-\frac{1}{2^{2n+1}}\right)\zeta(2n+2)
\end{align}On the other hand,
\begin{align}
K_1&=2\int_0^\infty \frac{\ln^2 x}{1+x^2}\,dx\int_0^\infty \frac{1}{1+y^2}\,dy\\
&=\pi J_1\\
K_2&=6J_1^2+\pi J_2\\
&=\frac{6K_1^2}{\pi^2}+\pi J_2\\
J_2&=\frac{K_2}{\pi}-\frac{6K_1^2}{\pi^3}\\
&=\frac{945\zeta(6)}{4\pi}-\frac{6075\zeta^2(4)}{8\pi^3}\\
\end{align}Moreover,\begin{align}\zeta(4)&=\frac{\pi^4}{90}\\
\zeta(6)&=\frac{\pi^6}{945}
\end{align}Therefore,\begin{align}\boxed{J=\frac{5}{64}\pi^5}\end{align}
NB: for $n\geq 0$ integer \begin{align}\int_0^\infty \frac{\ln^{2n+1} x}{1+x^2}\,dx=0\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3339783",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
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Stem and leaf diagrams I have the following data:
$2.6$ $ $ $3.3$ $ $ $2.4$ $ $ $1.1$ $ $ $0.8$ $ $ $3.5$ $ $ $3.9$ $ $ $1.6$ $ $ $2.8$ $ $ $2.6$ $ $ $3.4$ $ $ $4.1$ $ $ $2.0$ $ $ $1.7$ $ $ $2.9$ $ $ $1.9$ $ $ $2.9$ $ $ $2.5$ $ $ $4.5$ $ $ $5.0$
Built stem and leaf plot:
$Stem$ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $Leaf$ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $f$
$0$ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $8$ $ $ $ $ $ $$ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $$ $$ $ $ $$ $$ $ $ $
$ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $$1$
$1$ $ $ $ $ $ $ $ $$ $ $ $ $ $ $ $ $ $ $ $ $1$ $ $ $ $ $ $ $6$ $ $ $ $ $ $ $7$ $ $ $ $ $ $ $9$ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $4$
$2$ $ $ $ $ $ $ $ $ $ $ $ $ $0$ $ $ $4$ $ $ $5$ $ $ $6$ $ $ $6$ $ $ $8 \ $ $9$ $ $ $9$ $ $ $ $ $ $ $ $ $ $ $8$
$3$ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $3$ $ $ $ $ $4$ $ $ $ $ $5$ $ $ $ $ $9$ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $4$
$4$ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $1$ $ $ $ $ $ $ $ $ $ $ $5$ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $$ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $2$
$5$ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $0$ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $1$
My lecturer said that this is true:
$P$ $(2\le$$X\le3)$ = ${8\over 20}$
$P$ $(1\le$$X\le6)$ = $1$ - ${1\over 20}$ = ${19\over 20}$
So, my question is:
Why is he not adding the last number into the range?
To me, since it is $"less $ $ than $ $ or $ $ equal $ $ to"$ this has to be:
$P$ $(2\le$$X\le3)$ = ${12\over 20}$
$P$ $(1\le$$X\le6)$ = $1$ - ${0\over 20}$ = ${20\over 20}$
Does it have something to do with continuous or discrete data? Cause in another example he added the last number into the range.
| It's important to note that (for example) $2\le X\le 3$ translates to "$2$ is less than or equal to $X,$ and $X$ is less than or equal to $3.$" Thus, $2\le X\le 3$ will be true if and only if both $2\le X$ and $X\le 3$ are true. Put another way, $2\le X\le 3$ will be true if and only if neither $X<2$ nor $3<X$ is true.
Thus, the reason that $P(2\le X\le 3)\ne\frac{12}{20}$ is because each of the following four statements is false: $$2\le 3.3\le 3\\2\le 3.4\le 3\\2\le 3.5\le 3\\2\le 3.9\le 3.$$ After all, $3$ is strictly less than each of $3.3,3.4,3.5,3.9.$
The reason that $P(1\le X\le 6)\ne\frac{20}{20}$ is because the following statement is false: $$1\le 0.8\le 6.$$ After all, $0.8$ is strictly less than $1.$
As a side note, you omitted $2.8$ from your stem-and-leaf plot.
| {
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"url": "https://math.stackexchange.com/questions/3340924",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Quartic polynomial roots The equation
$x^4 - x^3-1=0$
has roots Ξ±, Ξ², Ξ³, Ξ΄. By using the substitution $y=x^3$ find the exact value of $Ξ±^6+Ξ²^6+Ξ³^6+Ξ΄^6$ .
The solution is
$x=y$ (1/3)
$y^4=(1+y)^3$
$y^4 -y^3 -3y^2-3y^2-1=0$
$S$N+4 $=$ $S$N + $S$N+3
$S$-1 = $\frac {0}{1} =0$
$S$2 = $ 1^2 -2*0 =1$
$S$3 = $0+1 =1$
$S$4 = $1+4 =5$
$S$5 = $5+1=6$
$S$6 = $6+1 =7$
Therefore, $Ξ±^6+Ξ²^6+Ξ³^6+Ξ΄^6 = 7$
Can someone explain what is this $S$n is and how did they work out solution from that.
| This looks like an application of the Newton identities which are a generalization of Viete's identities connecting roots and coefficients. If $S_k=\sum_{j=1}^d Ξ±_j^k$ is the degree $k$ power sum of the roots of the polynomial equation $0=x^4+c_1x^3+c_2x^2+c_3x+c_4$, then
$$
c_1+S_1=0\\
2c_2+c_1S_1+S_2=0\\
3c_3+c_2S_1+c_1S_2+S_3=0\\
4c_4+c_3S_1+c_2S_2+c_1S_3+S_4=0\\
c_4S_1+c_3S_2+c_2S_3+c_1S_4+S_5=0\\
c_4S_2+c_3S_3+c_2S_4+c_1S_5+S_6=0\\
$$
For the given polynomial with $(c_1,c_2,c_3,c_4)=(-1,0,0,-1)$ this gives
$$
S_1=1,\\
S_2=S_1=1,\\
S_3=S_2=1,\\
S_4=4+S_4=5,\\
S_5=S_1+S_5=6,\\
S_6=S_2+S_6=7.
$$
Or using the proposed approach, $S_6$ is the square sum of the roots of the polynomial for $y$ which was found as $$0=y^4-y^3-3y^2-3y-1=y^4+C_1y^3+C_2y^2+C_3y+C_4,$$ where again one finds $S_3=-C_1=1$ and $$S_6=-C_1S_3-2C_2=7.$$
| {
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"url": "https://math.stackexchange.com/questions/3341019",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Ramanujan's series for the exponential integral According to Wikipedia, Ramanujan came up with the following series expansion of the exponential integral:
$$\operatorname{Ei}(x)=\gamma+\ln|x|+e^{x/2}\sum_{n=1}^\infty\frac{(-1)^{n-1}x^n}{n!2^{n-1}}\sum_{k=0}^{\lfloor(n-1)/2\rfloor}\frac1{2k+1}$$
My first instinct was to try and verify the derivative:
$$\frac{e^x}x\stackrel?=\frac1x+\frac12e^{x/2}\left(\sum_{n=1}^\infty\frac{(-1)^{n-1}x^n}{n!2^{n-1}}\sum_{k=0}^{\lfloor(n-1)/2\rfloor}\frac1{2k+1}\right)+e^{x/2}\left(\sum_{n=0}^\infty\frac{(-1)^nx^n}{n!2^n}\sum_{k=0}^{\lfloor n/2\rfloor}\frac1{2k+1}\right)$$
and then perhaps apply Cauchy products? If so, then this doesn't seem very nice at all.
How can I show this equality?
Also as an aside, but below the formula on Wikipedia are some claimed bounds, which no matter how I look at it, are wrong, or if perhaps a typo, what should they be?
| You can subtract $\frac{1}{x}$ and multiply by $xe^{-x/2}$ to avoid the Cauchy product and instead use the known series expansions of $e^x$. You would get $$e^{x/2} -e^{-x/2} \stackrel?= \frac{1}{2} \sum_{n=1}^\infty\frac{(-1)^{n-1}x^{n+1}}{n!2^{n-1}}\sum_{k=0}^{\lfloor(n-1)/2\rfloor}\frac1{2k+1} + \sum_{n=0}^\infty\frac{(-1)^nx^{n+1}}{n!2^n}\sum_{k=0}^{\lfloor n/2\rfloor}\frac1{2k+1}$$
Write out the series expansion of the left hand side to get $$\sum_{n=0}^{\infty} \frac{x^n}{2^n n!} - \sum_{n=0}^{\infty} \frac{(-1)^n x^n}{2^n n!} \stackrel?= \frac{1}{2} \sum_{n=1}^\infty\frac{(-1)^{n-1}x^{n+1}}{n!2^{n-1}}\sum_{k=0}^{\lfloor(n-1)/2\rfloor}\frac1{2k+1} + \sum_{n=0}^\infty\frac{(-1)^nx^{n+1}}{n!2^n}\sum_{k=0}^{\lfloor n/2\rfloor}\frac1{2k+1}$$
From here, you can just show that the $x^n$ terms are the same on both sides. It's clearly equal for $n = 0$ and $n = 1$ ($[x^0] = 0$ and $[x^1] = 1$). Other than that, you would need to show the relation $$\frac{1}{2^n n!} - \frac{(-1)^n}{2^n n!} \stackrel?= \frac{1}{2}\frac{\left(-1\right)^{n-2}}{\left(n-1\right)!2^{n-2}}\sum_{k=0}^{\lfloor (n-2)/2 \rfloor}\frac{1}{2k+1}+\frac{\left(-1\right)^{n-1}}{\left(n-1\right)!2^{n-1}}\sum_{k=0}^{\lfloor (n-1)/2 \rfloor}\frac{1}{2k+1}$$
Multiplying by $2^n n!$ and factoring the RHS, this simplifies to $$1 - (-1)^n \stackrel?= 2n\left(-1\right)^{n}\left( \sum_{k=0}^{\lfloor (n-2)/2 \rfloor}\frac{1}{2k+1}-\sum_{k=0}^{\lfloor (n-1)/2 \rfloor}\frac{1}{2k+1}\right)$$
Splitting this up into even and odd cases makes it easier. If $n$ is even, then the LHS is $0$ and $\lfloor (n-2)/2 \rfloor = \lfloor (n-1)/2 \rfloor$, so the RHS is $0$ as well. If $n$ is odd, then the LHS is $2$. The RHS would be $$-2n\left( -\frac{1}{2\lfloor (n-1)/2 \rfloor + 1} \right) = -2n \cdot \frac{-1}{n} = 2$$
Therefore, the left hand side and the right hand side are both equal.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3343633",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Is my solution of $\lim_{(x, y)\rightarrow (0, 0)} \left( \frac{\cos (x^2+y)}{1-x^2y} \right) ^{\frac{1}{x^4+y^2}}$ correct? Is this solution correct?
If not, why?
Alternative (especially less wordy) proofs are also welcome.
$$\lim_{(x, y)\rightarrow (0, 0)} \left( \frac{\cos (x^2+y)}{1-x^2y} \right) ^{\frac{1}{x^4+y^2}}$$
$$y=x^2$$
$$\lim_{(x, x^2)\rightarrow (0, 0)}\left( \frac{\cos (2x^2)}{1-x^4}\right) ^{\frac{1}{2x^4}}$$
$$x^2=a$$
$$\lim_{a\rightarrow 0}\left( \frac{\cos (2a)}{1-a^2}\right) ^{\frac{1}{2a^2}}$$
$$\lim_{a\rightarrow 0} e^{\ln \left( \frac{\cos (2a)}{1-a^2}\right) ^{\frac{1}{2a^2}}}$$
Zooming in on the power:
$$\lim_{a\rightarrow 0} {\ln \left( \frac{\cos (2a)}{1-a^2}\right) {\frac{1}{2a^2}}}.$$
In the neighborhood of $0$, $\cos(2a)$ is barely distinguishable from $1$:
$$ \lim_{a\rightarrow 0} \frac{\ln\frac{1}{1-a^2}}{2a^2} = \langle \frac{0}{0} \rangle,$$
so L'HΓ΄pital:
$$ \lim_{a\rightarrow 0} \frac{\frac{-(-2a)}{(1-a^2)^2}(1-a^2)}{4a} = \frac{1}{2} .$$
The answer: $\sqrt{e}.$
| You have approached this by looking at one curve, namely $y=x^2.$ And that curve should have yielded the limit $e^{-1/2}.$ So there's that mistake. The bigger mistake is: One curve is never enough.
We can write
$$\tag 1 \cos (x^2+y) = 1-\frac{(x^2+y)^2}{2} +o(x^4+y^2)$$
and
$$\tag 2 \frac{1}{1-x^2y} = 1 + x^2y + o(x^4+y^2).$$
The product of $(1)$ and $(2)$ is then
$$\tag 3 1-\frac{x^4+y^2}{2}+ o(x^4+y^2).$$
Rasing $(3)$ to the power $\dfrac{1}{x^4+y^2}$ and taking the limit as $(x,y)\to (0,0)$ shows the desired limit is $e^{-1/2}.$
Note This is edited from the original, which contained a mistake that Peter Foreman pointed out to me.
| {
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"timestamp": "2023-03-29T00:00:00",
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If $a$, $b$, $c$ are sides of a triangle, prove $2(a+b+c)(a^2+b^2+c^2)\geq3(a^3+b^3+c^3+3abc)$
$a$, $b$, $c$ are sides of a triangle, prove:
$$2(a+b+c)(a^2+b^2+c^2)\geq3(a^3+b^3+c^3+3abc)$$
What I have tried:
$$
β2\sum (a+b)ab\geq \sum a^3+9abc
$$
so I can't use
$$\sum a^3+3abc\geq \sum ab(a+b)$$
| Let $a=y+z$, $b=x+z$ and $c=x+y$.
Thus, $x$, $y$ and $z$ are positives and we need to prove that
$$\sum_{cyc}(x^3-x^2y-x^2z+xyz)\geq0,$$ which is Schur.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3345936",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How to solve $x(3x+3)(x+5)(2x+12)+576 = 0$?
How to solve $$x(3x+3)(x+5)(2x+12)+576 = 0?$$
This was a question on a test i recently took, And i wasn't able to solve it. I later tried to solve it using online calculators and it turns out this doesn't have any real solutions.
I know there's a general formula for quartic polynomials that can work but we were only taught two methods, modifying the equation to a quadratic one using substitution or guessing some of the solutions. Both of these didn't work for me.
Are there any ways to prove that this doesn't have real solutions without the quartic formula?
| Another possible way is using the symmetry of $x(3x+3)(x+5)(2x+12) = 6x(x+1)(x+5)(x+6)$ around $x=\color{blue}{3}$:
$$6x(x+1)(x+5)(x+6) = 6(x+\color{blue}{3}-3)(x+\color{blue}{3}+3)(x+\color{blue}{3}-2)(x+\color{blue}{3}+2)$$ $$=6(\underbrace{(x+\color{blue}{3})^2}_{y:=}-9)((x+\color{blue}{3})^2-4)$$
The minimum value of $(y-9)(y-4)$ is $-\frac{25}{4}$. Hence,
$$6x(x+1)(x+5)(x+6)+576 \geq 6(-\frac{25}{4})+576 >0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3347022",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Prove that $0<\sum_{k=1}^n \frac{g(k)}{k} - \frac{2n}{3} < \frac{2}{3}$ Prove
$$0<\sum_{k=1}^n \frac{g(k)}{k} - \frac{2n}{3} < \frac{2}{3}$$
where $g(k)$ is the greatest odd divisor of k
Please Find Holes in my Proof.
Let $k=2m+1$ if we show that the right hand side of the equation is true for odd numbers, then it is true for even numbers since there is a net total of $1/3$ since $g(k)/k = 1$ for odd numbers.
$$\sum_{k=1}^n\frac{g(k)}{k} <\frac{2n}{3} + \frac{2}{3}$$
$$\sum_{k=1}^n\frac{g(k)}{k} <\frac{2(2m+1)}{3} + \frac{2}{3}$$
$$\sum_{k=1}^n\frac{g(k)}{k} <\frac{4m}{3} + \frac{4}{3}$$
From $1$ to $2m+1$ there are $m$ even numbers and $m+1$ even numbers, the value of $\frac{g(k)}{k}$ for even numbers is $\frac{1}{2^{V2(k)}}$ where $V2(k)$ is the exponent of 2 in the factorization of k
$$\sum_{k=1}^n\frac{g(k)}{k} = m+1 + (m)\frac{1}{2} -((\lfloor{\frac{m}{2}\rfloor \frac{1}{4}) +(\lfloor{\frac{m}{4}\rfloor \frac{1}{8}}})...)$$
$$\leq m+1 + (m)\frac{1}{2} - (\frac{m}{8} + \frac{m}{32}...) = \frac{4m}{3} + 1 <\frac{2(2m+1)}{3} + \frac{2}{3}$$
Solving for the left hand side of the inequality
$$0<\sum_{k=1}^n \frac{g(k)}{k} - \frac{2n}{3}$$ Let $k = 2m$ with the same reasoning as above
$$\sum_{k=1}^n\frac{g(k)}{k} = m + (m)\frac{1}{2} -((\lfloor{\frac{m}{2}\rfloor \frac{1}{4}) +(\lfloor{\frac{m}{4}\rfloor \frac{1}{8}}})...)$$
$$\geq m + (m)\frac{1}{2} - (\frac{m}{8} + \frac{m}{32}...) = \frac{4m}{3} + \frac{1}{3} >\frac{2(2m)}{3}$$
I'm not sure about the solution since it was a strict inequality to begin with. Is this a correct solution? Any other solutions are welcome
| $$g(2^m (2k+1))=2k+1$$ Let $$h(n) = \sum_{2k+1 \le n} 1=\lfloor (n+1)/2\rfloor$$ then $$\sum_{k=1}^n \frac{g(k)}{k} =\sum_{m\ge 0} \frac{h(n/2^m)}{2^m}=\sum_{m\ge 0} \frac{\lfloor (n/2^m+1)/2\rfloor}{2^m}=\sum_{m\ge 0} \frac{ (n/2^m+1)/2-O(1)}{2^m}\\ = \frac{n/2}{1-2^{-2}}+\frac{1/2}{1-2^{-1}}-\frac{O(1)}{1-2^{-1}}=\frac23 n+1-2 O(1)$$
where here $O(1) \in [0,1)$
| {
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"timestamp": "2023-03-29T00:00:00",
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Number of words with no adjacent vowels I am faced with the following problem:
Given $C = \{B,C,D,F\}$ and $V = \{A, E, I, O, U\}$ find the number of 9-letter words with elements from $C$ and $V$ such that no two vowels (elements of V) are adjacent.
Following this answer about a very similar question I get that I should express the problem as a double recurrence:
$a_{n+1} = 5b_n$, $a_0 = 1$
$b_{n+1} = 4(a_n + b_n)$, $a_0 = 1$
where $a_n$ is the number of string starting by a vocal and $b_n$ is the number of string starting by a consonant.
Expressing it as a matrix I get $\begin{pmatrix} a_{n+1} \\ b_{n+1} \end{pmatrix} = \begin{pmatrix} 0 & 5 \\ 4 & 4 \end{pmatrix} \begin{pmatrix} a_n \\ b_n \end{pmatrix}$
And I get that the eigenvalues of this matrix are $\{2+\frac{\sqrt{96}}{2}, 2-\frac{\sqrt{96}}{2}\}$. Is this the correct approach?
I don't really know where to go from here or how to solve the recurrence for any given $k$. Thank you in advance
| You boil your problem down to a linear recurrence relation, which is nice!
There's a little issue with $a_0$ and $b_0$, though : how can the empty word start by anything? One should start the recurrence at $n=1$, we have $a_1 = 5$ and $b_1 = 4$.
Call
$A = \begin{pmatrix} 0 & 5 \\ 4 & 4 \end{pmatrix}$, then
$$\begin{pmatrix} a_n \\ b_n \end{pmatrix} = A^{n-1} \begin{pmatrix} 5 \\ 4 \end{pmatrix} $$
The point of computing the eigenvalues is to put $A$ in a nicer form so that $A^n$ is easy to compute. Since you have two distinct eigenvalues, you can diagonalize $A$ which will lead you to a nice expression of $a_n$ and $b_n$.
Diagonalization of $A$ :
Edit : in fact $96 = 4 \cdot 24$ and not $4\cdot 19$ !
One finds two distinct eigenvalues, $\lambda_1 = 2 + \sqrt{24}$ and $\lambda_2 = 2 - \sqrt{24}$.
Let $i=1,2$, an eigenvector $v_i$ corresponding to $\lambda_i$ is a nonzero solution to the equation $$(A - \lambda_i \operatorname{Id}) v_i = 0$$
Write $v_i = \begin{pmatrix} x \\ y \end{pmatrix}$, we get
$$\left\{\begin{matrix} -\lambda_i x + & 5 y = & x \\
4 x + & (4-\lambda_i)y = & y \end{matrix} \right.$$
The first line implies $y = \frac{\lambda_i}{5} x$, after which the second line vanishes. Now chose an arbitrary $x \neq 0$, let's take $x = 5$ for convenience.
Now $v_1 = \begin{pmatrix} 5 \\ 2 + \sqrt{24} \end{pmatrix}$ and $v_2 = \begin{pmatrix} 5 \\ 2 - \sqrt{24} \end{pmatrix}$ form a basis of eigenvectors. Take
$$P = \begin{pmatrix} 5 & 5 \\ 2 + \sqrt{24} & 2 - \sqrt{24} \end{pmatrix}$$
One has
$$ P^{-1}AP = \begin{pmatrix} 2 + \sqrt{24} & 0 \\ 0 & 2 - \sqrt{24}\end{pmatrix} $$
Hence,
$$ A^k = P \begin{pmatrix} (2 + \sqrt{24})^k & 0 \\ 0 & (2 - \sqrt{24})^k\end{pmatrix} P^{-1}$$
| {
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"url": "https://math.stackexchange.com/questions/3352601",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Ways to prove the area of an ellipse formula One can prove the ellipse area formula $A=\pi a b$ ($a$, $b$ the major and minor semi-axis) either by integration or by the stretched-circle argument. See for instance here: https://proofwiki.org/wiki/Area_of_Ellipse
Are there any other proofs of this formula?
| Let's get creative. For any $R>1$ we have
$$ f_R(\theta)=\frac{1}{R+e^{i\theta}} = \sum_{n\geq 0}(-1)^n \frac{e^{ni\theta}}{R^{n+1}} $$
hence by Parseval's identity
$$\int_{0}^{2\pi}\frac{d\theta}{(R^2+1)+2R\cos\theta}= \int_{0}^{2\pi}f_R(\theta)\overline{f_R}(\theta)\,d\theta = 2\pi\sum_{n\geq 0}\frac{1}{R^{2n+2}}=\frac{2\pi}{R^2-1}$$
and by multiplying both sides by $(R^2+1)$ we get
$$ \int_{0}^{2\pi}\frac{d\theta}{1+\frac{2R}{R^2+1}\cos\theta}=2\pi\frac{R^2+1}{R^2-1}. $$
Through a substitution we get that for any $B>1$ the identity
$$ \int_{0}^{2\pi}\frac{d\theta}{B+\cos\theta} = \frac{2\pi}{\sqrt{B^2-1}} $$
holds. By differentiating both sides with respect to $B$ we get
$$\forall B>1,\quad \int_{0}^{2\pi}\frac{d\theta}{(B+\cos\theta)^2}=\frac{2\pi B}{(B^2-1)^{3/2}}.\tag{1} $$
Let us assume that an ellipse has semiaxis $a>b$. How can we write its polar equation with respect to a focus?
By imposing that the property $PF_1+PF_2=2a$ holds, i.e. by imposing that
$$ \rho(\theta) + \sqrt{\rho^2(\theta)\sin^2(\theta)+(\rho(\theta)\cos(\theta)-2c)^2} = 2a, $$
$$ \rho^2(\theta)-4c\rho(\theta)\cos(\theta)+4c^2 = 4a^2-4a\rho(\theta)+\rho^2(\theta), $$
$$ -c\rho(\theta)\cos(\theta) = b^2-a\rho(\theta), $$
$$ \rho(\theta) = \frac{b^2/a}{1-e\cos\theta} \tag{2}$$
where $e=\frac{c}{a}=\sqrt{1-\frac{b^2}{a^2}}$. The area enclosed by the ellipse is
$$ A(a,b)=\frac{1}{2}\int_{0}^{2\pi}\rho^2(\theta)\,d\theta = \frac{b^4}{2a^2}\int_{0}^{2\pi}\frac{d\theta}{(1-e\cos\theta)^2}\stackrel{\theta\mapsto\pi+\theta}{=}\frac{b^4}{2a^2 e^2}\int_{0}^{2\pi}\frac{d\theta}{\left(\frac{1}{e}+\cos\theta\right)^2}\tag{3}$$
and by invoking $(1)$ we have
$$ A(a,b) = \frac{2\pi b^4}{2a^2(1-e^2)^{3/2}}=\large{\color{red}{\pi a b}}. $$
| {
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"timestamp": "2023-03-29T00:00:00",
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Olympiad Inequality $\frac{ab+bc+ca}{a+b+c}(\sum_{cyc}\frac{1}{2a+b})\leq 1$ with proof (long post) It takes me a lot a time to solve this :
Let $a,b,c>0$ with $a\geq b\geq c$ then we have :
$$\frac{ab+bc+ca}{a+b+c}(\sum_{cyc}\frac{1}{2a+b})\leq 1$$
My proof :
We need some lemmas to begin :
First lemma :
Let $x,y>0$ then we have :
$$\frac{x}{x+\frac{1}{2}}+\frac{y}{y+\frac{1}{2}}=\frac{\frac{x+y}{2}}{\frac{x+y}{2}+\frac{1}{2}}+\frac{p}{p+\frac{1}{2}}$$
Where $p$ is equal to:
$$\frac{\frac{x}{(x+\frac{1}{2})^2}+\frac{y}{(y+\frac{1}{2})^2}}{\frac{1}{(x+\frac{1}{2})^2}+\frac{1}{(y+\frac{1}{2})^2}}$$
the proof is easy it's just calculus .
Second lemma :
With the notation of the first lemma we have :
$$\frac{\frac{x+y}{2}}{\frac{x+y}{2}+\frac{1}{2}}\geq \frac{p}{p+\frac{1}{2}}$$
Proof :
With the first lemma and the concavity of $\frac{x}{x+\frac{1}{2}}$ and the use of Jensen's inequality :
$$\frac{x}{x+\frac{1}{2}}+\frac{y}{y+\frac{1}{2}}-\frac{\frac{x+y}{2}}{\frac{x+y}{2}+\frac{1}{2}}=\frac{p}{p+\frac{1}{2}}\leq \frac{\frac{x+y}{2}}{\frac{x+y}{2}+\frac{1}{2}} $$
Now we attack the problem :
We prove a stronger result :
We have with $x=\frac{1}{a}$ and $y=\frac{1}{b}$ and $z=\frac{1}{c}$:
$$\sum_{cyc}\frac{1}{2a+b}+\sum_{cyc}\frac{1}{2b+a}=\sum_{cyc}\frac{xy}{2x+y}+\sum_{cyc}\frac{xy}{2y+x}$$
But :
$$\frac{xy}{2x+y}+\frac{yz}{2z+y}=y(\frac{\frac{1}{2}\frac{x}{y}}{\frac{x}{y}+\frac{1}{2}}+\frac{\frac{1}{2}\frac{z}{y}}{\frac{z}{y}+\frac{1}{2}})$$
Now we apply the first lemma to get :
$$y(\frac{\frac{1}{2}\frac{x}{y}}{\frac{x}{y}+\frac{1}{2}}+\frac{\frac{1}{2}\frac{z}{y}}{\frac{z}{y}+\frac{1}{2}})=\frac{\frac{x+z}{4}}{\frac{\frac{x}{y}+\frac{x}{y}}{2}+\frac{1}{2}}+Q$$
Finally we have :
$$\sum_{cyc}\frac{xy}{2x+y}+\sum_{cyc}\frac{xy}{2y+x}=\sum_{cyc}\frac{\frac{x+z}{4}}{\frac{\frac{x}{y}+\frac{x}{y}}{2}+\frac{1}{2}}+Q+R+T$$
We multiply by $\frac{x+y+z}{xy+yz+zx}$ :
$$\frac{x+y+z}{xy+yz+zx}(\sum_{cyc}\frac{xy}{2x+y}+\sum_{cyc}\frac{xy}{2y+x})=\frac{x+y+z}{xy+yz+zx}(\sum_{cyc}\frac{\frac{x+z}{4}}{\frac{\frac{x}{y}+\frac{x}{y}}{2}+\frac{1}{2}})+\frac{x+y+z}{xy+yz+zx}(Q+R+T)$$
But :
$$\frac{x+y+z}{xy+yz+zx}(\sum_{cyc}\frac{\frac{x+z}{4}}{\frac{\frac{x}{y}+\frac{x}{y}}{2}+\frac{1}{2}})=1$$
And with the second lemma we have :
$$\frac{x+y+z}{xy+yz+zx}(\sum_{cyc}\frac{\frac{x+z}{4}}{\frac{\frac{x}{y}+\frac{x}{y}}{2}+\frac{1}{2}})=1\geq\frac{x+y+z}{xy+yz+zx}(Q+R+T)$$
So we prove :
$$\frac{x+y+z}{xy+yz+zx}(\sum_{cyc}\frac{xy}{2x+y}+\sum_{cyc}\frac{xy}{2y+x})\leq 2$$
Or
$$\frac{ab+bc+ca}{a+b+c}(\sum_{cyc}\frac{1}{2a+b}+\sum_{cyc}\frac{1}{2b+a})\leq 2$$
But :
$$\sum_{cyc}\frac{1}{2a+b}-\sum_{cyc}\frac{1}{2b+a}= -\frac{(2 (a - b) (a - c) (b - c) (2 a^2 + 7 a b + 7 a c + 2 b^2 + 7 b c + 2 c^2))}{((2 a + b) (a + 2 b) (2 a + c) (a + 2 c) (2 b + c) (b + 2 c))}$$
And with the condition $a\geq b\geq c$ it's negative so :
$$2\frac{ab+bc+ca}{a+b+c}(\sum_{cyc}\frac{1}{2a+b})\leq\frac{ab+bc+ca}{a+b+c}(\sum_{cyc}\frac{1}{2a+b}+\sum_{cyc}\frac{1}{2b+a})\leq 2$$
Done !
My question is have you an alternative proof ?
Thanks a lot for your time !
| As an alternative, while not shorter, it can be brute-forced with just elementary algebra . . .
Since the $\text{LHS}$ is homogeneous, we can assume $c=1$, and $a\ge b \ge 1$.
Replacing $c$ by $1$, and simplifying, we get
$$
1-\text{LHS}
=
\frac
{-a^2b+2a^3b-a^2-ab^2-a^2b^2-ab+2b^3-b^2+2a}
{(a+b+1)(2a+b)(2b+1)(2+a)}
$$
so it remains to show
$$
-a^2b+2a^3b-a^2-ab^2-a^2b^2-ab+2b^3-b^2+2a\ge 0
$$
Replacing $b$ by $1+x$ and $a$ by $1+x+y$, the $\text{LHS}$ of the above inequality simplifies to
$$
3x^2+3xy+3y^2+9x^2y+9y^2x+4x^3y+5y^2x^2+2y^3x+4x^3+2y^3+x^4
$$
which is nonnegative since all coefficients are positive and $x,y\ge 0$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Find all $x\in \mathbb{Z}$ with $x \equiv 2 \mod 4$ , $x \equiv 8\mod9$ and $x \equiv 1\mod5$
Find all $x\in \mathbb{Z}$ with $x \equiv 2 \mod 4$ , $x \equiv 8\mod9$ and $x \equiv 1\mod5$
My attempt:
$\gcd\left(9,5,4\right) = 1 = 9r+5s+4\times0$
$9 = 1\times5+4$
$5 = 1\times4+1$
so $1 = 5-(9-5) = 2\times5 - 1\times9$
thus $r=-1$ and $s = 2$
so $x = 2\times5\times8 -1\times9\times1=80 - 9=71\mod 180$
But the problem is that $71\equiv8\mod9$ and $71\equiv1\mod5$, but $71 \ne 2 \mod 4$. What is wrong?
| Here's an elementary way to do it: Since $x\equiv 2\mod 4$ then $x$ is even. Since $x\equiv 1\mod 5$ then the decimal expansion of $x$ has $1$ or $6$ in the units. But $x$ is even, so it must be $6$.
So $x=10k+6$ for some $k$. But also, $x\equiv 8\mod 9$, so the sum of the algarisms of $x$ is of the form $9p+8$. But since $x=10k+6$, the sum of the algarisms of $x$ is $k+6$, i.e.
$$9p+8=k+6$$
so $k=9p+2$. Substituting, we have $x=10(9p+2)+6=90p+26$ for some $p$.
But now we use again that $x\equiv 2\mod 4$, which means that the remainder of the division of $x=90p+26$ by $4$ is $2$. From this, $90p$ should be divisible by $4$, so $p$ is even, say $p=2q$.
Therefore, $x=180q+26$, for some $q=0,1,2,\ldots$. Conversely, you can check that any number of this form satisfies all desired congruences.
| {
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"timestamp": "2023-03-29T00:00:00",
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Proving $n^2(n^2-1)(n^2+1)=60\lambda$ such that $\lambda\in\mathbb{Z}^{+}$ I'm supposed to prove that the product of three successive natural numbers the middle of which is the square of a natural number is divisible by $60$. Here's my attempt.
My Attempt:
$$\text{P}=(n^2-1)n^2(n^2+1)=n(n-1)(n+1)[n(n^2+1)]$$
It is now enough to prove that $n(n^2+1)$ is divisible by $10$. But for $n=4$, $4(17)\ne10\lambda$ but for $n=4$, $\text{P}$ is $4080=60\cdot68$ which means apart from just being a multiple of $6$, $n(n-1)(n+1)$ is actually helping $n(n^2+1)$ with a $5$ to sustain divisibility by $60$.
How to tackle this? Thanks
| Note that we have $$n^2(n^2-1)(n^2+1)=n^2(n^2-1)(n^2-4)+5n^2(n^2-1)$$ and that $n^2$ is divisible by $4$ or $n^2-1$ is divisible by $8$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3356512",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Find all solutions of $a,b,c,d$ if $a+b+c+d=12$ and $abcd = 27 +ab+ac+ad+bc+bd+cd$ Find all solutions of $a,b,c,d>0$ if
$$a+b+c+d=12$$
$$abcd = 27 +ab+ac+ad+bc+bd+cd$$
Attempt:
$a + b + c + d = 12 \implies 3 \ge \sqrt[4]{abcd} \implies 81 \ge abcd$ by AM-GM, eqyality when $a=b=c=d$. Also
$$ abcd - 27 = ab + ac + ad + bc + bd + cd \ge 6 \sqrt[6]{a^{3} b^{3} c^{3} d^{3}} = 6 \sqrt{a b c d} $$
$$ (abcd - 27)^{2} \ge 36 a b c d$$
dari AM-GM, equality when $ab=ac=ad=bc=bd=cd$. then we must have
$$ (abcd - 27)^{2} = 36 \times 81 \implies abcd = 54 + 27 = 81 $$
atau
$$ (abcd - 27)^{2} = 36 \times 81 \implies abcd = -54 + 27 = -27 (\text{not possible}) $$
So $abcd = 81$. Since $a=b=c=d$ then $a=b=c=d=3$.
Is this the only solution? if it is, why?
| I think it's better from $$abcd-27\geq6\sqrt{abcd}$$ to write
$$abcd-6\sqrt{abcd}-27\geq0$$ or
$$(\sqrt{abcd}-9)(\sqrt{abcd}+3)\geq0$$ or
$$abcd\geq81$$ and since by your work
$$abcd\leq81,$$ it's enough to check an equality case.
Another solution.
By AM-GM we obtain:
$$abcd=27\left(\frac{a+b+c+d}{12}\right)^4+(ab+ac+bc+ad+bd+cd)\left(\frac{a+b+c+d}{12}\right)^2\geq$$
$$\geq27\left(\frac{4\sqrt[4]{abcd}}{12}\right)^4+6\sqrt[6]{a^3b^3c^3d^3}\left(\frac{4\sqrt[4]{abcd}}{12}\right)^2=abcd,$$
which says that the inequality is equality, which occurs for $a=b=c=d,$ which gives only
$$(a,b,c,d)=(3,3,3,3).$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove $\frac{a}{b^{2}+1} + \frac{b}{c^{2}+1} + \frac{c}{a^{2} + 1} \ge \frac{3}{2}$ $a,b,c > 0$ and $a+b+c=3$, prove
$$ \frac{a}{b^{2} + 1} + \frac{b}{c^{2}+1} + \frac{c}{a^{2}+1} \ge 3/2 $$
Attempt:
Notice that by AM-Gm
$$\frac{a}{b^{2} + 1} + \frac{b}{c^{2}+1} + \frac{c}{a^{2}+1} \ge 3\frac{\sqrt[3]{abc}}{\sqrt[3]{(a^{2}+1)(b^{2}+1)(c^{2}+1)}} $$
Now, AM-GM again
$$ a^{2}+b^{2}+c^{2} + 3 \ge 3 \sqrt[3]{(a^{2}+1)(b^{2}+1)(c^{2}+1)} ... (1)$$
Then $a+b+c = 3 \ge 3 \sqrt[3]{abc} \implies 1 \ge \sqrt[3]{abc}$. Also
$$a^{2} + b^{2} + c^{2} \ge 3 \sqrt[3]{(abc)^{2}}$$
multiply by $1 \ge \sqrt[3]{abc}$ and will get
$$ a^{2} + b^{2} + c^{2} \ge 3 abc ... (2)$$
subtract $(1)$ with $(2)$ and get
$$ 3 \ge 3 \sqrt[3]{(a^{2}+1)(b^{2}+1)(c^{2}+1)} - 3 abc$$
$$ 3 + 3 abc \ge \sqrt[3]{(a^{2}+1)(b^{2}+1)(c^{2}+1)} $$
$$ \frac{3abc}{\sqrt[3]{(a^{2}+1)(b^{2}+1)(c^{2}+1)}} \ge 1 - \frac{3}{\sqrt[3]{(a^{2}+1)(b^{2}+1)(c^{2}+1)}} $$
How to continue..?
| $$\sum_{cyc}\frac{a}{b^2+1}=3+\sum_{cyc}\left(\frac{a}{b^2+1}-a\right)=3-\sum_{cyc}\frac{ab^2}{b^2+1}\geq$$
$$\geq3-\sum_{cyc}\frac{ab^2}{2b}=3-\frac{1}{2}(ab+ac+bc).$$
Can you end it now?
Since by your work $$3-\frac{1}{2}(ab+ac+bc)=3-\frac{1}{2}\cdot\frac{9-a^2-b^2-c^2}{2},$$ it's enough to prove that
$$3-\frac{1}{2}\cdot\frac{9-a^2-b^2-c^2}{2}\geq\frac{3}{2}$$ or
$$a^2+b^2+c^2\geq3,$$ which is true by C-S:
$$a^2+b^2+c^2=\frac{1}{3}(1^2+1^2+1^2)(a^2+b^2+c^2)\geq\frac{1}{3}(a+b+c)^2=3.$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove that $\sqrt{x^{3} + y^{3}+1} + \sqrt{z^{3} + y^{3}+1} + \sqrt{x^{3} + z^{3}+1} \ge 2 + \sqrt{2(x^{3} + y^{3}+z^{3})+1}$ Given $x, y,z \ge 0$. Prove that
$$\sqrt{x^{3} + y^{3}+1} + \sqrt{z^{3} + y^{3}+1} + \sqrt{x^{3} + z^{3}+1} \ge 2 + \sqrt{2(x^{3} + y^{3}+z^{3})+1} $$
Attempt
Notice that $$(x^{3} + y^{3}+1) + (z^{3} + y^{3}+1) + (x^{3} + z^{3}+1) = 2 + 2(x^{3} + y^{3}+z^{3})+1 $$
So the sum of squares must be between two forms as below:
$$2 + \sqrt{2(x^{3} + y^{3}+z^{3})+1} \le \sqrt{x^{3} + y^{3}+1} + \sqrt{z^{3} + y^{3}+1} + \sqrt{x^{3} + z^{3}+1} \le 2 + 2(x^{3} + y^{3}+z^{3})+1$$
If I work backwards I will get, by letting the form in each root be $a,b,c$ respectively:
$$\sqrt{a} + \sqrt{b} + \sqrt{c} \ge 2 + \sqrt{a+b+c-2} $$
$$ a + b +c + 2(\sqrt{ab} + \sqrt{bc} + \sqrt{ca}) \ge 4 + (a+b+c-2) + 4\sqrt{a+b+c-2}$$
$$ (\sqrt{ab} + \sqrt{bc} + \sqrt{ca}) \ge 1 + 2\sqrt{a+b+c-2}$$
$$ ((\sqrt{ab} + \sqrt{bc} + \sqrt{ca}) -1))^{2} \ge 4(a+b+c-2)$$
$$ (ab + bc + ca + 2a^{2}\sqrt{bc} +2b^{2}\sqrt{ac} +2c^{2}\sqrt{ab}) - 2(\sqrt{ab} + \sqrt{bc} + \sqrt{ca}) + 1 \ge 4(a+b+c-2)$$
How to continue?
Apart from the attempt above, I have tried:
*
*AM-GM for the sum of the 3 squares
*AM-GM for the sum of the 3 squares, then using Holder's inequality from the GM
*AM-GM for each square
| The inequality $$\sum_{cyc}\sqrt{a+1}\geq2+\sqrt{a+b+c+1}$$ we can prove also by the following way:
$$\sum_{cyc}\sqrt{a+1}-2-\sqrt{a+b+c+1}=$$
$$=\sqrt{a+1}+\sqrt{b+1}-2-\left(\sqrt{a+b+c+1}-\sqrt{c+1}\right)=$$
$$=\sqrt{a+1}+\sqrt{b+1}-2-\frac{a+b}{\sqrt{a+b+c+1}+\sqrt{c+1}}\geq$$
$$\geq\sqrt{a+1}+\sqrt{b+1}-2-\frac{a+b}{\sqrt{a+b+1}+1}=$$
$$=\sqrt{a+1}+\sqrt{b+1}-2-(\sqrt{a+b+1}-1)=$$
$$=\sqrt{a+1}-1-\left(\sqrt{a+b+1}-\sqrt{b+1}\right)=$$
$$=\sqrt{a+1}-1-\frac{a}{\sqrt{a+b+1}+\sqrt{b+1}}\geq\sqrt{a+1}-1-\frac{a}{\sqrt{a+1}+1}=0.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3363591",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find equation of the circle whose diameter is the common chord of two other circles? Circle 1:
$$x^2 + y^2 +6x + 2y +6 = 0$$
Circle 2:
$$x^2 + y^2 + 8x + y + 10 = 0$$
My attampt:
From circle 1 and 2, I found
$$ y = 2x + 4 $$
which is the common chord.
Pluging that in equation 1 I got
$$5x^2 + 26x + 30 = 0$$
here I got stuck for too much complication if I use quadratic formula.Is there anything wrong?
| You are on the right track. The numbers cancel out nicely when you sum them. Indeed:
$$5x^2 + 26x + 30 = 0 \Rightarrow x_1=\frac{-13-\sqrt{19}}{5},x_2=\frac{-13+\sqrt{19}}{5}\\
y_1=\frac{-6-2\sqrt{19}}{5}, y_2=\frac{-6+2\sqrt{19}}{5}$$
The center of the new circle:
$$\frac{x_1+x_2}{2}=-\frac{13}{5},\frac{y_1+y_2}{2}=-\frac{6}{5}$$
The diameter of the new circle:
$$d=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}=\sqrt{\frac{4\cdot 19}{25}+\frac{16\cdot 19}{25}}=\sqrt{\frac{76}{5}} \Rightarrow \\
r=\frac12d=\sqrt{\frac{76}{4\cdot 5}}=\sqrt{\frac{19}{5}}$$
Thus:
$$\left(x+\frac{13}{5}\right)^2+\left(y+\frac65\right)^2=\frac{19}{5}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3363697",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "5",
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If $A \in \mathbb{R}^{n\times n}$ and $\textbf{x}\in \mathbb{R}^{n}$why $A = \frac{1}{2} A + \frac{1}{2} A^\top $ is not always true? If $A \in \mathbb{R}^{n\times n}$ and $\textbf{x}\in \mathbb{R}^{n}$,
It is possible to prove that
$$
\textbf{x}^\top A \textbf{x} = \textbf{x}^\top(\frac{1}{2} A + \frac{1}{2} A^\top)\textbf{x}
$$
That could let us think that
$$
A = \frac{1}{2} A + \frac{1}{2} A^\top ?
$$
However if we take
$$A = \begin{bmatrix}1&2\\ 3&4 \end{bmatrix}$$
and compute
$$B = \frac{1}{2} A + \frac{1}{2} A^\top$$
we get $$B = \begin{bmatrix}1&2.5\\ 2.5&4 \end{bmatrix}\ne A$$
Although
$$
\textbf{x}^\top A \textbf{x} = \textbf{x}^\top B \textbf{x}
$$
Here is my question:
How can we explain that the following is not always true?
$$
A = \frac{1}{2} A + \frac{1}{2} A^\top
$$
| In the product $\mathbf{x}^\top A \mathbf{x}$ we get terms like $x_1 A_{12} x_2 + x_2 A_{21} x_1 = (A_{12}+A_{21}) x_1 x_2$ so that for example $\begin{bmatrix}1&2.5\\ 2.5&4 \end{bmatrix}$, $\begin{bmatrix}1&5\\ 0&4 \end{bmatrix}$ and $\begin{bmatrix}1&3\\ 2&4 \end{bmatrix}$ give the same result since $2.5+2.5 = 5+0 = 3+2$. Thus only the symmetric part $\frac12(A+A^\top)$ matters; the antisymmetric part $\frac12(A-A^\top)$ does not matter.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3368309",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Solve the equations and find $x,y,z$ Given:
$$x^3+y^3+z^3=x+y+z$$
And:
$$x^2+y^2+z^2=xyz$$
Find all real and positive solutions to these equations, if any.
So most probably, we'll factorise in this way:
$$x^3+y^3+z^3-3xyz=x+y+z-3xyz
\rightarrow (x+y+z)(x^2+y^2+z^2-xy-yz-zx)=(x+y+z)-3(x^2+y^2+z^2)$$
But I'm having trouble in proceeding further.
Kindly help.
| You're given
$$x^3+y^3+z^3=x+y+z \tag{1}\label{eq1A}$$
$$x^2+y^2+z^2=xyz \tag{2}\label{eq2A}$$
As you've shown, you have
$$x^3 + y^3 + z^3 - 3xyz = (x+y+z)(x^2 + y^2 + z^2 - xy - yz - zx) \tag{3}\label{eq3A}$$
Next, note
$$(x-y)^2 + (y-z)^2 + (x-z)^2 = 2(x^2 + y^2 + z^2 - xy - yz - zx) \tag{4}\label{eq4A}$$
Since $x,y,z$ are positive values, this means the RHS of \eqref{eq3A} is non-negative (note you can also determine the LHS is non-negative by the Inequality of arithmetic and geometric means which gives $\frac{x^3 + y^3 + z^3}{3} \ge \sqrt[3]{x^3 y^3 z^3} = xyz$), so using this and \eqref{eq2A} gives
$$\begin{equation}\begin{aligned}
x^3 + y^3 + z^3 - 3xyz & \ge 0 \\
x^3 + y^3 + z^3 - 3(x^2 + y^2 + z^2) & \ge 0 \\
x^2(x - 3) + y^2(y - 3) + z^2(z - 3) & \ge 0
\end{aligned}\end{equation}\tag{5}\label{eq5A}$$
This shows at least one of $x,y,z$ is $\ge 3$. Due to the symmetry of \eqref{eq1A} and \eqref{eq2A}, WLOG assume $x \le y \le z$, so $z \ge 3$. From \eqref{eq1A}, moving the terms on the right to the left & factoring gives
$$x(x^2 - 1) + y(y^2 - 1) + z(z^2 - 1) = 0 \tag{6}\label{eq6A}$$
With $z \ge 3$, then $z(z^2 - 1) \ge 3(8) = 24$. However, the minimum values of $x(x^2 - 1)$ and $y(y^2 - 1)$ are each $-\frac{2}{3\sqrt{3}} \approx -0.385$ (this comes from $f(a) = a^3 - a$, with $f'(a) = 3a^2 - 1 = 0 \implies a = \frac{1}{\sqrt{3}}$ gives the minimum value of $f(a)$ for $a \gt 0$), so the sum of those $2$ terms is much greater than $-24$. This means the LHS of \eqref{eq6A} can never be $0$ and, thus, the equation never holds. As such, there are no positive, real solutions to \eqref{eq1A} and \eqref{eq2A}.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Calculating the value of a triple integral I would like to evaluate the following integral
$$
I=\int_{\mathbb{R}^3}\frac{x^2+y^2+1-z^2}{|(x,y,z-1)|^3|(x,y,z+1)|^3}\,dx\,dy\,dz.
$$
My attempt: with $R^2=x^2+y^2$ we have
$$
I=2\pi\int_{\mathbb{R}_+\times\mathbb{R}}\frac{R^2+1-z^2}{[(R^2+(z-1)^2)(R^2+(z+1)^2)]^{3/2}}R\,dR\,dz.
$$
The $[\cdots]$ expression in the denominator can be rearranged:
$$
\cdots=R^4+(1-z^2)^2+2R^2z^2+2R^2=(R^2+1-z^2)^2+4R^2z^2.
$$
In addition, one can do another change of variables $D=R^2$, and so
$$
I=\pi\int_{\mathbb{R}_+\times\mathbb{R}}
\frac{D+1-z^2}{[(D+1-z^2)^2+4Dz^2]^{3/2}}\,dD\,dz.
$$
Any thoughts on how to proceed?
| Probably the fastest method: Do not rewrite the denominator and integrate with respect to $D$ directly:
\begin{align}
I &= \pi \int \limits_{-\infty}^{\infty} \int \limits_0^\infty \frac{D + 1 - z^2}{[(D + (1-z)^2)(D + (1 + z)^2)]^{3/2}} \, \mathrm{d} D \, \mathrm{d} z \\
&= \pi \int \limits_{-\infty}^{\infty} \left[\frac{D - (1 - z^2)}{2\sqrt{(D + (1-z)^2)(D + (1 + z)^2)}}\right]_{D=0}^{D=\infty} \, \mathrm{d} z \\
&= \pi \int \limits_{-\infty}^{\infty} \frac{1 + \operatorname{sgn}(1-z^2)}{2} \, \mathrm{d} z = \pi \int \limits_{-1}^1 \, \mathrm{d} z = 2 \pi \, .
\end{align}
However, it seems difficult to come up with this antiderivative without a CAS or a lucky guess.
Alternatively, we can use Feynman parameters to write
\begin{align} I &= \pi \int \limits_{-\infty}^{\infty} \int \limits_0^\infty (D + 1 -z^2)\left[\frac{8}{\pi} \int \limits_0^1 \frac{\sqrt{u(1-u)}}{[(D+(1-z)^2)u + (D+(1+z)^2)(1-u)]^3} \, \mathrm{d} u\right] \, \mathrm{d} D \, \mathrm{d} z \\
&= 8 \int \limits_{-\infty}^{\infty} \int \limits_0^\infty \int \limits_0^1 \frac{(D + 1 -z^2)\sqrt{u(1-u)}}{[D + (1+z)^2 - 4 z u]^3} \, \mathrm{d} u \, \mathrm{d} D \, \mathrm{d} z \, .
\end{align}
Fubini's theorem allows us to integrate with respect to $D$ first, which is now straightforward (partial fractions). We find
\begin{align}
I &= 8 \int \limits_0^1 \sqrt{u(1-u)} \int \limits_{-\infty}^\infty \frac{1 + z - 2 z u}{[(1+z)^2 - 4 z u]^2} \, \mathrm{d} z \, \mathrm{d} u \\
&= 8 \int \limits_0^1 \sqrt{u(1-u)} \int \limits_{-\infty}^\infty \frac{1 + (1-2u) z}{[(z + 1-2u)^2 + 4 u (1-u)]^2} \, \mathrm{d} z \, \mathrm{d} u \\
&= 8 \int \limits_0^1 \sqrt{u(1-u)} \int \limits_{-\infty}^\infty \frac{4 u (1-u) \color{red}{+ (1-2u) \zeta}}{[\zeta^2 + 4 u (1-u)]^2} \, \mathrm{d} \zeta \, \mathrm{d} u
\end{align}
after introducing $\zeta = z + 1 - 2u$ . The integral of the red term vanishes by symmetry, so the substitution $\zeta = 2 \sqrt{u(1-u)} \tau$ yields
$$ I = 4 \int \limits_0^1 \int \limits_{-\infty}^\infty \frac{\mathrm{d} \tau}{(1+\tau^2)^2} \, \mathrm{d} u = 4 \int \limits_{-\infty}^\infty \frac{\mathrm{d} \tau}{(1+\tau^2)^2} \, .$$
The final integral is a typical exercise in residue calculus, but integration by parts also does the trick:
$$ I = 4 \int \limits_{-\infty}^\infty \left[\frac{1}{1+\tau^2} - \tau \frac{\tau}{(1+\tau^2)^2}\right] \, \mathrm{d} \tau = 4 \left[\pi - \frac{1}{2} \int \limits_{-\infty}^\infty \frac{\mathrm{d}\tau}{1+\tau^2}\right] = 2 \pi \, .$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Polynomial of 4th degree I would like to ask if someone could help me with the following equation.
\begin{equation}
x^4+ax^3+(a+b)x^2+2bx+b=0
\end{equation}
Could you first solve in general then $a=11$ and $b=28$.
I get it to this form but I stuck.
\begin{equation}
1+a\left(\frac{1}{x}+\frac{1}{x^2}\right)+b\left(\frac{1}{x^2}+\frac{1}{x^3}+\frac{1}{x^3}+\frac{1}{x^4}\right) = 0
\end{equation}
Thank you in advance.
| The polynomial equation for $(a, b)=(11,28)$ is given by
$$
x^4+11x^3+39x^2+56x+28=(x^2 + 7x + 7)(x + 2)^2=0.
$$
"How do we come up wit this layout"? By the rational root test, we find the factor $x-2$ twice, and dividing gives the factor $x^2+7x+7$. Hence the roots are $x=-2,-2,\frac{\sqrt{21}-7}{2},\frac{-\sqrt{21}-7}{2}$.
In general, the polynomial will not have any integral roots. A good example is the case $(a, b)=(1,1)$, where the polynomial
$$
x^4+x^3+2x^2+2x+1
$$
is irreducible over $\Bbb Q$. It has no real root at all.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Diophantine equation from the Latvian Baltic Way team selection competition 2019 So here is the problem statement:
Find all integer triples $(a, b, c)$ such that
$(a-b)^3(a+b)^2 = c^2 + 2(a-b) + 1$
The only things I have so far figured out is that (-1, 0, 0) and (0, 1, 0) are solution, gcd((a-b), c) = 1 and that c must be even.
Any ideas? Maybe even a general solution to the problem?
| Consider the equation already found:-
$$(a-b)\left((a^2-b^2)^2-2\right)=c^2+1,$$
where $c^2+1$ is odd or singly even.
If $a-b$ is even then the LHS would be divisible by 4. Therefore $a-b$ is odd and $c$ is even.
Then $(a^2-b^2)^2-2$ is of the form $4k-1$ and it is therefore either $-1$ or has a prime divisor of the form $4k-1$. However, $c^2+1$ cannot have a prime factor of the form $4k-1$ and so $$(a^2-b^2)^2-2=-1.$$
Therefore $a^2-b^2$ is $-1$ or $1$ and then one of $a$ and $b$ is $0$. We also have $b-a=c^2+1$ and so the solutions for $(a,b,c)$ are $(0,1,0)$ or $(-1,0,0).$
| {
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"timestamp": "2023-03-29T00:00:00",
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Find all real matrices such that $X^{3}-4X^{2}+5X=\begin{pmatrix} 10 & 20 \\ 5 & 10 \end{pmatrix}$ The following question come from the 1998 Romanian Mathematical Competition:
Find all matrices in $M_2(\mathbb R)$ such that $$X^{3}-4X^{2}+5X=\begin{pmatrix} 10 & 20 \\ 5 & 10 \end{pmatrix}$$
Can you guy please help me? Thanks a lot!
| There is a matrix $U$ such that
$$U^{-1}\begin{pmatrix} 10 & 20 \\ 5 & 10 \end{pmatrix} U=\begin{pmatrix} 20 & 0 \\ 0&0 \end{pmatrix}.$$
Let $Y=U^{-1}XU$, then $Y^{3}-4Y^{2}+5Y=\begin{pmatrix} 20 & 0 \\0 & 0 \end{pmatrix}$. The matrix $Y$ then commutes with $\begin{pmatrix} 20 & 0 \\ 0&0 \end{pmatrix}$ and so is diagonal. Let $Y=\begin{pmatrix} a & 0 \\ 0&b \end{pmatrix}.$
Then $a^{3}-4a^{2}+5a=20,b^{3}-4b^{2}+5b=0.$ The only real solutions are $a=4,b=0.$
Then $X$ is uniquely determined as $UYU^{-1}$. If we did not know that $X=\begin{pmatrix} 2 & 4 \cr 1 & 2 \end{pmatrix}$ we could find it using the matrix $U$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Simplifying $i \left[ \ln(x+i)-\ln(x-i)-\ln(1+ix)+\ln(1-ix) \right]$ Can the following expression further be simplified and expressed in terms of usual functions such as inverse hyperbolic or inverse trigonometric functions?
$$
f(x) = i \left[ \ln(x+i)-\ln(x-i)-\ln(1+ix)+\ln(1-ix) \right] \, ,
$$
where $x\ge 0$ is a real number.
Inputs and ideas welcome.
Thanks
| Another approach to simplifying $f(x)$ is to simply use the expansion in power series of $x$.
Specifically,
\begin{align}
i \left( \ln(x+i)-\ln(x-i) \right) &=-\pi+ 2\left( x - \frac{x^3}{3} + \frac{x^5}{5} + \cdots \right) , \\
i \left(-\ln(1+ix)+\ln(1-ix) \right) &= 2\left( x - \frac{x^3}{3} + \frac{x^5}{5} + \cdots \right).
\end{align}
Thus
$$
f(x) = -\pi + 4 \left( x - \frac{x^3}{3} + \frac{x^5}{5} + \cdots \right) = -\pi+4\arctan x.
$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Functional equation problem: $ f \left( y ^ 2 - f ( x ) \right) = y f ( x ) ^ 2 + f \left( x ^ 2 y + y \right) $ This functional equation problem is from the Latvian Baltic Way team selection competition 2019:
Find all functions $ f : \mathbb R \to \mathbb R $ such that for all real $ x $ and $ y $,
$$ f \left( y ^ 2 - f ( x ) \right) = y f ( x ) ^ 2 + f \left( x ^ 2 y + y \right) \text . \tag 1 \label {eqn1} $$
OK, so I think that the only answer is $ f ( x ) = 0 $.
I just want to see if my proof that it is the only solution is correct.
So we start off by plugging $ y = - y $. We get that
$$ f \left( y ^ 2 - f ( x ) \right) = - y f ( x ) ^ 2 + f \Big( - y \left( x ^ 2 + 1 \right) \Big) \text . $$
Then we add the two equations together getting that
$$ 2 f \left( y ^ 2 - f ( x ) \right) = f \Big( - y \left( x ^ 2 + 1 \right) \Big) + f \Big( y \left( x ^ 2 + 1 \right) \Big) \text . $$
From the above equations we get that
$$ \frac { f \Big( - y \left( x ^ 2 + 1 \right) \Big) + f \Big( y \left( x ^ 2 + 1 \right) \Big) } 2 = y f ( x ) ^ 2 + f \left( x ^ 2 y + y \right) \text. \tag 2 \label {eqn2} $$
Now, if we plug $ x = - x $ then we will get that the LHS is the same and that the RHS is $ y f ( - x ) ^ 2 + f \left( x ^ 2 y + y \right) $.
So we proceed by subtracting the two and getting that
$$ 0 = y f ( x ) ^ 2 - y f ( - x ) ^ 2 \text . $$
So, lets assume that $ y \ne 0 $ getting that
$$ 0 = \big( f ( x ) - f ( - x ) \big) \big( f ( x ) + f ( - x ) \big) \text . $$
Now we do a two case analysis, 1) the function is even and 2) the function is odd.
Lets start with the function being even then from \eqref{eqn2} we get that
$$ 0 = y f ( x ) ^ 2 \text , $$
which of course implies that the function is just $ 0 $.
OK, now the odd case. Since the function is odd, $ f ( 0 ) = 0 $. Then plugging $ x = 0 $ in \eqref{eqn1} we get that
$$ f \left( y ^ 2 \right) = f ( y ) \text , $$
which implies that $ f $ is also an even function. Since $ f $ is both even and odd, it can only be $ 0 $.
Since we got that $ f $ is zero in both cases, the only solution to the equation is $ f ( x ) = 0 $.
| You can show that the only function $ f : \mathbb R \to \mathbb R $ satisfying
$$ f \left( y ^ 2 - f ( x ) \right) = y f ( x ) ^ 2 + f \left( x ^ 2 y + y \right) \tag 0 \label 0 $$
for all $ x , y \in \mathbb R $ is the constant zero function, by continuing your own argument. You've established showing
$$ f ( - x ) ^ 2 = f ( x ) ^ 2 \text . \tag 1 \label 1 $$
Now, take the "$ y f ( x ) ^ 2 $" term in \eqref{0} to the left-hand side, square both sides and rearrange the terms to get
$$ f \left( y ^ 2 - f ( x ) \right) ^ 2 + y ^ 2 f ( x ) ^ 4 - f \left( x ^ 2 y + y \right) ^ 2 = 2 y f ( x ) ^ 2 f \left( y ^ 2 - f ( x ) \right) \text . \tag 2 \label 2 $$
Substituting $ - y $ for $ y $ in \eqref{2} and using \eqref{1}, the left-hand side will be equal to the left-hand side of \eqref{2}, while the right-had side will be the opposite of the right-hand side of \eqref{2}. This in particular implies that
$$ y f ( x ) ^ 2 f \left( y ^ 2 - f ( x ) \right) = 0 \text . \tag 3 \label 3 $$
Now, if for some $ a \in \mathbb R $ we have $ f ( a ) \ne 0 $, then setting $ x = a $ and considering all nonzero values for $ y $ in \eqref{3}, we can see that for all $ z > - f ( a ) $ we have $ f ( z ) = 0 $. But then letting $ x = a $ and $ y = 1 + \frac { | f ( a ) | } { 1 + a ^ 2 } $ in \eqref{0} leads to a contradiction with $ f ( a ) \ne 0 $, since we have $ y > 0 $, $ y ^ 2 - f ( a ) > - f ( a ) $ and $ a ^ 2 y + y > - f ( a ) $. Therefore $ f ( x ) $ must be equal to $ 0 $ for all $ x \in \mathbb R $.
| {
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"timestamp": "2023-03-29T00:00:00",
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Find $\lim_{x \to 0} \frac{(\tan(\tan x) - \sin (\sin x))}{ \tan x - \sin x}$ Find $$\lim_{x\to 0} \dfrac{\tan(\tan x) - \sin (\sin x)}{ \tan x - \sin x}$$
$$= \lim_{x \to 0} \dfrac{\frac{\tan x \tan (\tan x)}{\tan x}- \frac{\sin x \sin (\sin x)}{\sin x}}{ \tan x - \sin x} = \lim_{x \to 0} \dfrac{\tan x - \sin x}{\tan x - \sin x} = 1$$
But the correct answer is $2$. Where am I wrong$?$
| Your second step is invalid. In general one can't replace a sub-expression by its limit while evaluating limit of a bigger expression in step by step fashion. You can get more details in this answer.
You can approach this problem by adding and subtracting $\tan(\sin x) $ in numerator. The numerator can thus be expressed as $$\{\tan \tan x-\tan \sin x\} +\{\tan \sin x-\sin\sin x\} $$ which can be further rewritten as $$\tan(\tan x - \sin x) (1+\tan\tan x\cdot\tan\sin x)+\tan\sin x-\sin\sin x$$ Now the first term divided by $\tan x - \sin x$ (denominator) tends to $1$ and hence the desired limit is equal to $$1+\lim_{x\to 0}\frac{\tan\sin x - \sin\sin x} {\tan x - \sin x} $$ The expression under limit above can be written as $$\frac{\sin\sin x} {\sin x} \cdot \frac{1-\cos\sin x} {1-\cos x}\cdot\frac{\cos x} {\cos\sin x} $$ The first and last factors tend to $1$ and the middle factor is equal to $$\frac{1-\cos\sin x} {\sin^2x}\cdot\frac{\sin^2x}{x^2}\cdot\frac{x^2}{1-\cos x} $$ and the above clearly tends to $(1/2)\cdot 1\cdot 2=1$. The desired answer is thus $2$.
| {
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"timestamp": "2023-03-29T00:00:00",
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$\sqrt[3]{2}$ satisfies $x^3-2=0$ Show that there is no polynomial $P(x)$ of degree less than 3 with $P(\sqrt[3]{2})=0$ $\sqrt[3]{2}$ satisfies $x^3-2=0$
Show that there is no polynomial $P(x)$ of degree less than $3$ with $P(\sqrt[3]{2})=0$ All coefficients are rational numbers.
Is it by induction? Say, if $x$ has degree of $1$, it doesn't work; and for $x$ having degree of $2$, am I applying Polynomial remainder theorem?
| You didn't precised what are the nature of the coefficients of P, so we can build a polynomial $P(x) = x - 3\sqrt{2}$ such that $P(3\sqrt{2}) = 0$.
And, we can build an infinity of polynomials of degree 2 such that $P(2 \sqrt{3}) = 0$, these polynomials are : $$P(x) = (x - 2 \sqrt{3})(x - k), k \in \mathbb{C}$$
We all must be very careful to mathematical rigor.
If we consider that $P \in \mathbb{Z}[X]$, it is obvious that there is no polynomials of degree 1 such that $P(2 \sqrt{3}) = 0$ because $\forall P \in \mathbb{Q}_1[X], P(X) = 0 \implies X \in \mathbb{Q}$, and, because $\mathbb{Z}_1[X] \subset \mathbb{Q}_1[X]$, this property is true even for polynomials with relative coefficients (and $2 \sqrt{3} \notin \mathbb{Q}$).
Then, for polynomials with degree 2, we use the quadratic formula :
$$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$, so we search solution to the equation :
$$\frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = 2 \sqrt{3}, (a,b,c) \in \mathbb{Z}^3$$
We can directly see that we must have $b = 0$ because if that was not the case, we should have $x = k + l \sqrt{m}, k \neq 0$ knowing that b must be a relative number. Or, we see that the solution we are searching for is not looking like that.
So, because we have $b = 0$, we have $$x = \frac{\pm \sqrt{-4ac}}{2a}$$ with one of these two solutions equal to $2 \sqrt{3}$. Because we have $2\sqrt{3} \notin \mathbb{C} - \mathbb{R}$, we must have $-4ac > 0$ so $4ac < 0$.
We have so $$x = \pm 2 \frac{\sqrt{-ac}}{2a} = \pm \frac{\sqrt{-ac}}{a}$$
Then, we suppose $a > 0$ so we can write $$x = \pm \frac{ \sqrt{a} }{a} \times \sqrt{-c} = \pm \frac{\sqrt{-c}}{\sqrt{a}}$$ so we must have $\sqrt{-c} = \sqrt{3}$ and $1/\sqrt{a} = \pm 2$ which is not possible with $a \in \mathbb{Z}$
And, if we have $a < 0$, we have $$x = \pm \frac{\sqrt{c}}{\sqrt{-a}}$$
So we must have $c = 3$ and $1/\sqrt{-a} = \pm 2$ which is not possible with $a \in \mathbb{Z}$, QED.
EDIT : The proof is also valable for $x = 2^{1/3}$ and $3\sqrt{2}$
| {
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What is the average area of all triangles that can be inscribed in a unit circle? There is only one circle that passes through any three given points. Hence by suitable scaling, we can inscribe every triangle inside a unit circle of radius $1$. We define distinct triangles as triangles which have different sides regardless of of the order. Hence a triangle with sides $(a,b,c)$ and a triangle with sides $(b,c,a)$ are not distinct.
Question 1: What is the average area of all distinct triangles that can be inscribed in a unit circle?
Question 2: What is the average perimeter of all distinct triangles that can be inscribed in a unit circle?
Motivation: This question was motivated by this related question where it was proved that the average perimeter of all right triangles inscribed in a semi-circle of unit diameter is $1+4/\pi$.
| Comment:
As can be seen in figure the area and perimeter of right angled triangle in half circle is maximum when the height is maximum , i.e. $h=r$ where h is height of triangle and r is the radius of circle. The area and perimeter are minimum when $h β 0$. So we can write:
$a/2 = r= 1$
$S_{max}=(\frac{(\sqrt 2)^2}{2})=1$
$S_{min}= 0$
$S_{ave.}=\frac{1+2\times 0}{3}=\frac{1}{3}$
and perimeter:
$P_{max}=2+2\sqrt2$
$P_{min}=0$
$P_{ave}=\frac{2(1+\sqrt2)+2\times 0}{3} β 2.27 β 1+\frac{4}{\pi}$
Now suppose we want to find the average area and perimeter of isosceles triangles which can be inscribed in circle with radius unit. In triangle $OHC_1$ we have:
$(\frac{a}{2})^2+(dr)^2=r^2=1$
$S_{AB_1C_1}=\frac{a}{2}\times (r+dr)=\frac{a}{2}(1+dr)$
Eliminating $dr$ and letting $\frac{a}{2}=x$ we get:
$S=x(1+\sqrt{1-x^2})$
β$S'=\frac{\sqrt{1-x^2}+1-2x^2}{\sqrt{1-x^2}}=0$
β$x=0$ and $x=\frac{\sqrt 3}{2}$
β$a= \sqrt 3$
β$S_{max}=\frac{3}{2}\times\frac{\sqrt3}{2}=\frac{3\sqrt3}{4}$
That is S is maximum when $a=b=c=\sqrt 3$, i.e. when the triangle is equilateral.The minimum value is when $h β0$ or $a β0$ , so the average of area can be:
$S_{ave.}=\frac{\frac{3\sqrt3}{4}+2\times 0}{3}=\frac{\sqrt 3}{4}$
Similarly you can find the average of perimeter:
$P=a+2\sqrt{2+2\sqrt{1-(a/2)^2}}$
Now take derivative and so on.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3386054",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Prove $\tan(\frac{x}{2}) = \frac{\sin x}{1 + \cos x} $ using the quadratic formula I am trying to prove the fact that $\tan \frac{x}{2} = \frac{1-\cos x}{\sin x}$ or alternatively $\tan \frac{x}{2} = \frac{1- \cos x}{\sin x}$. (I understand that it can be proved using the half-angle identities of $\sin$ and $\cos$ but I want to understand how to get to the solution from this specific method of derivation.)
\begin{align*}
\tan(2x) &= \frac{2\tan(x)}{1-\tan(x)^2} \\
\tan(x) &= \frac{2\tan(\frac{x}{2})}{1-\tan(\frac{x}{2})^2} \\
\end{align*}
I now let $A=\tan x$ and $B=\tan \frac{x}{2}$
\begin{align*}
A\cdot(1-B^2) &= 2B\\
AB^2+2B-A &= 0 \\
\end{align*}
Now I solve for B using the quadratic formula.
\begin{align*}
B &= \frac{-2\pm \sqrt{4+4A^2}}{2A} \\
B &= \frac{-1\pm \sqrt{1+A^2}}{A} \\
\tan(\frac{x}{2}) &= \frac{-1\pm \sqrt{1+\tan(x)^2}}{\tan(x)}\\
\tan(\frac{x}{2}) &= \frac{-1\pm \sqrt{(\sec x)^2}}{\tan(x)}\\
\tan(\frac{x}{2}) &= \frac{-1\pm |\sec x|}{\tan(x)}
\end{align*}
I am confused as to how to continue at this point (firstly, not sure how to deal with the absolute value, and secondly not sure how to deal with the plus-minus).
Any help is greatly appreciated, as I feel like I do not fully understand how to manipulate absolute values and the meaning of the plus-minus.
| Because of the $\pm$, the absolute value is superfluous.
$\begin{align}
B &= \frac{-1\pm |\sec x|}{\tan x} \\
B &= \frac{-1\pm \sec x}{\tan x} \\
B &= \frac{(-1\pm \sec x)(\cos x)}{(\tan x)(\cos x)} \\
B &= \frac{-\cos x\pm 1}{\sin x} \\
\end{align}$
We know that $\dfrac{1 - \cos x}{\sin x} = \tan \dfrac x2$
Also
$\begin{align}
\dfrac{-1 - \cos x}{\sin x}
&= -\dfrac{1 + \cos x}{\sin x}\\
&= -\dfrac{1 + (2 \cos^2 \frac x2 - 1))}
{2 \sin \frac x2 \cdot \cos \frac x2} \\
&= - \cot \frac x2
\end{align}$
So the roots of the quadratic $AB^2 + 2B -A = 0$ are $B = \tan \frac x2$
and $B =-\cot \frac x2$.
That is to say
$$\tan x \cdot \left(\tan \frac x2\right)^2 + 2\tan \frac x2 - \tan x = 0$$
and
$$\tan x \cdot \left(-\cot \frac x2\right)^2 - 2\cot \frac x2 - \tan x = 0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3388767",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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} |
Roots over $\mathbb{C}$ equation $x^{4} - 4x^{3} + 2x^{2} + 4x + 4=0 $. I need roots over $\mathbb{C}$ equation $$x^{4} - 4x^{3} + 2x^{2} + 4x + 4 = 0$$
From Fundamental theorem of algebra we have statement that the equation have 4 roots over complex.
But I prepare special reduction:
$$ \color{red}{ x^{4} - 4x^{3} + 2x^{2} + 4x + 4} = (x-1)^{4}-4(x-1)^{2} + 7 $$
for substitution $y = (x-1)^{2}$ we have:
$y^{2} - 4y + 7 = 0 $
$y_{0} = 2+i\sqrt{3}$
$y_1 = 2 - i\sqrt{3}$
and we have
$y_{0}^{1/2} + 1 = x_{0} $, $-y_{0}^{1/2} + 1 = x_{1}$,
$y_{1}^{1/2} + 1 = x_{2}$, $-y_{1}^{1/2} +1 = x_{3} $
I am not sure what is good results.
Please check my solution.
EDIT:
The LHS is not correct, I modify this equation. We should have $p(x) = x^{4} - 4x^{3} + 2x^{2} + 4x + 4 = (x-1)^{4}-4(x-1)^{2} + 7 $
EDIT2:
I need show that the $p(x)$ is reducible (or not) over $\mathbb{R}[x]$ for two polynomials of degrees 2.
But I am not sure how show that $\left(x-1-\sqrt{2-i\sqrt{3} }\right) \left(x-1+\sqrt{(2+i\sqrt3}\right)$ is (not) polynomial of degree 2.
| The equality that you have used is not correct.$$x^{4}-4x^{3}+2x^{2}+4 \ne (x-1)^{4}-4(x-1)^{2} + 7$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3390052",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Number of 6-character passwords containing at least one number - why is my answer wrong? Here's the question:
How many six-character passwords can be built with lowercase letters and numbers, given that at least one of its characters is a number?
Here's my answer:
$$10\dbinom{6}{1}36^5$$
*
*$10$ for the number possible choices for digits
*$\binom{6}{1}$ where one location in the $6$-character string is being chosen for the digit to be placed
*$36^5$ for remaining character combinations ($26$ letters + $10$ digits)
Here's the actual answer:
$$36^6 - 26^6$$
*
*$36^6$ for all strings of length $6$
*minus $26^6$ for the number of strictly alphabetical passwords
I get why the solution works, but I can't see where my solution went wrong; it's quite a bit bigger than the answer. Can someone explain why my answer is wrong?
| The number of passwords with exactly $k$ digits is
$$\binom{6}{k}10^k26^{6 - k}$$
since there are $10$ choices for each of the $k$ digits, $26$ choices for each of the $6 - k$ lowercase letters, and $\binom{6}{k}$ ways to select $k$ of the $6$ positions in the password for the letters to appear. Hence, the number of passwords with at least one digit is
\begin{align*}
\sum_{k = 1}^{6} \binom{6}{k}10^k26^{6 - k} & = \binom{6}{1} 10^1 26^5 + \binom{6}{2} 10^2 26^4 + \binom{6}{3} 10^3 26^3 + \binom{6}{4} 10^4 26^2\\
& \qquad + \binom{6}{5} 10^5 26^1 + \binom{6}{6} 10^6 26^0\\
& = 36^6 - 26^6
\end{align*}
as you can verify.
By designating a particular digit as the digit that appears in the password, you count each password with $k$ digits $k$ times, once for each way you could designate one of the digits as the digit that appears in the password. For instance, the password 1a2b3c is counted three times, once when you designate 1 as the digit that appears in the password, once when you designate the 2 as the digit that appears in the password, and once when you designate 3 as the digit that appears in the password. Notice that
\begin{align*}
\sum_{k = 1}^{6} \color{red}{k}\binom{6}{k}10^k26^{6 - k}
& = 1\binom{6}{1} 10^1 26^5 + \color{red}{2}\binom{6}{2} 10^2 26^4
+ \color{red}{3}\binom{6}{3} 10^3 26^3 + \color{red}{4}\binom{6}{4} 10^426^2\\
& \qquad + \color{red}{5}\binom{6}{5} 10^5 26^1 + \color{red}{6}\binom{6}{6} 10^6 26^0\\
& = \color{red}{10\binom{6}{1}36^5}
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3391612",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Find all $x\in \mathbb R$ such that $\sqrt[3]{x+2}+\sqrt[3]{x+3}+\sqrt[3]{x+4}=0$ The question is as the title says:
Find all $x\in \mathbb R$ such that $\sqrt[3]{x+2}+\sqrt[3]{x+3}+\sqrt[3]{x+4}=0$.
I struggle to even start this question.
By inspection, I see that $x$ must be negative. Playing around yields $x=-3$ as a solution, though I do not know how to prove that there are no other solutions.
Upon differentiation, I obtain:
\begin{align}
\frac{d}{dx} (\sqrt[3]{x+2}+\sqrt[3]{x+3}+\sqrt[3]{x+4})&= \frac{1}{3}(\sqrt[3]{(x+2)^{-2}}+\sqrt[3]{(x+3)^{-2}}+\sqrt[3]{({x+4})^{-2}})\\
\end{align}
which is always positive for all real values of $x$, implying that the function defined as $f(x)=\sqrt[3]{x+2} + \sqrt[3]{x+3} + \sqrt[3]{x+4}$ is strictly increasing.
Is there a better way to solve this equation?
| $$\sqrt[3]{x+2}+\sqrt[3]{x+4}=-\sqrt[3]{x+3}$$
Take Cube in both sides
$$2(x+3)-3\sqrt[3]{(x+2)(x+4)}\sqrt[3]{x+3}=-(x+3)$$
Set $\sqrt[3]{x+3}=y\implies x+3=y^3, (x+2)(x+4)=(y^3-1)(y^3+1)$
$$2y^3-3y\sqrt[3]{y^6-1}=-y^3$$
$$y(y^2-\sqrt[3]{y^6-1})=0$$
If $y\ne0$ $$y^2=\sqrt[3]{y^6-1}$$
Take cube in both sides $$y^6=y^6-1\iff 0=-1$$ So no finite solutions expect $y=0$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3394661",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 3
} |
Evaluate $\sum _{n=1}^{\infty } \frac{\left(\frac{4}{9}\right)^n \beta (2 n+1)}{n+1}$ I was given that
$$\sum _{n=1}^{\infty } \frac{\left(\frac{4}{9}\right)^n \beta (2 n+1)}{n+1}=\frac{-9 \left(-\frac{2 C}{3}-1+\frac{1}{3} \pi \log \left(\sqrt{3}+2\right)\right)}{2 \pi }$$
Here $\beta$ denotes Dirichlet Beta. Unfortunately, this result seems to be numerically incorrect. So what is the closed-form of this sum? Any help will be appreciated.
| Here is a sketch. Using, say, the link with Euler numbers, one gets
$$\sum_{n=0}^{\infty}\beta(2n+1)x^{2n}=\frac{\pi}{4}\sec\frac{x\pi}{2},$$
which implies
$$\sum_{n=0}^{\infty}\frac{\beta(2n+1)}{2n+2}x^{2n+2}=\frac{1}{\pi}f\left(\frac{x\pi}{2}\right),\quad f(y)=\int_0^y\frac{x\,dx}{\cos x}.$$
The given sum is then equal to $\dfrac{9}{2\pi}f\left(\dfrac{\pi}{3}\right)-\dfrac{\pi}{4}$, with the known value of
$$f\left(\frac{\pi}{3}\right)=\frac{\pi}{3}\ln(2+\sqrt{3})-\frac{2}{3}G,$$
where $G$ is Catalan's constant. The latter formula can be derived from
$$\int_0^y\ln\left|\tan\frac{x}{2}\right|\,dx=-2\sum_{n=0}^{\infty}\frac{\sin(2n+1)y}{(2n+1)^2}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3397355",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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} |
Find $[z^N]$ for $\frac{1}{1-z} \left(\ln \frac{1}{1-z}\right)^2$ Find $[z^N]$ for $\frac{1}{1-z} \left(\ln \frac{1}{1-z}\right)^2$. Here generalized harmonic numbers should be used.
$$H^{(2)}_n = 1^2 + \frac{1}{2}^2 + \dots + \frac{1}{N}^2.$$
For now, I was able to find $[z^N]$ for $\left(\ln \frac{1}{1-z}\right)^2$ which is the following convolution for $N \ge 2$:
$$\sum_{1\le k \le N-1} \frac{1}{N-k}\frac{1}{k}$$
Next step would be partial sum but I don't see how all this leads me to generalized harmonic numbers.
| Using the generalization:
$$\sum_{n=1}^\infty a_nx^n=\frac1{1-x}\sum_{n=1}^\infty (a_n-a_{n-1})x^n,\quad a_{0}=0$$
Let $a_{n}=H_n^2$ to have
\begin{align}
\sum_{n=1}^\infty H_n^2x^n&=\frac1{1-x}\sum_{n=1}^\infty \left(H_n^2-H_{n-1}^2\right)x^n\\
&=\frac1{1-x}\sum_{n=1}^\infty \left(\frac{2H_n}{n}-\frac1{n^2}\right)x^n\\
&=\frac1{1-x}\cdot 2\sum_{n=1}^\infty\frac{H_n}{n}x^n-\frac{\operatorname{Li}_2(x)}{1-x}\\
&=\frac1{1-x}\cdot 2\left(\operatorname{Li}_2(x)+\frac12\ln^2(1-x)\right)-\frac{\operatorname{Li}_2(x)}{1-x}\\
&=\frac{\ln^2(1-x)}{1-x}+\frac{\operatorname{Li}_2(x)}{1-x}\\
&=\frac{\ln^2(1-x)}{1-x}+\sum_{n=1}^\infty H_n^{(2)}x^n
\end{align}
Thus
$$\frac{\ln^2(1-x)}{1-x}=\sum_{n=1}^\infty (H_n^2-H_n^{(2)})x^{n}$$
The proof of the generalization together with other identities can be found here.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Show that $\sum_{k=0}^{n+1}\left(\binom{n}{k}-\binom{n}{k-1}\right)^2 = \frac{2}{n+1}\binom{2n}{n}$
Show that $$\displaystyle\sum_{k=0}^{n+1}\left(\dbinom{n}{k}-\dbinom{n}{k-1}\right)^2 = \dfrac{2}{n+1}\dbinom{2n}{n}$$ where $n \in \mathbb{N}$
Consider $\dbinom{n}{r}=0 $ for $r<0 $ and $r>n$. Using $\displaystyle\sum_{k=0}^{n}\dbinom{n}{k}^2 = \dbinom{2n}{n}$, we get $\displaystyle\sum_{k=0}^{n+1}\left(\dbinom{n}{k}-\dbinom{n}{k-1}\right)^2 = \displaystyle\sum_{k=0}^{n+1} \dbinom{n}{k}^2 + \displaystyle\sum_{k=0}^{n+1} \dbinom{n}{k}^2 - 2\displaystyle\sum_{k=0}^{n+1}\dbinom{n}{k}\dbinom{n}{k-1} \\ \hspace{47mm} = 2\dbinom{2n}{n}-2\displaystyle\sum_{k=1}^{n}\dbinom{n}{k}\dbinom{n}{k-1} \\ \hspace{47mm} = 2\dbinom{2n}{n}-2\displaystyle\sum_{k=1}^{n}\dfrac{1}{k(n-k+1)}\dbinom{n}{k-1}^2 \\ \hspace{47mm} = 2\dbinom{2n}{n}-\dfrac{2}{n+1}\displaystyle\sum_{k=1}^{n}\left(\dfrac{1}{k}+\dfrac{1}{n-k+1} \right)\dbinom{n}{k-1}^2 $
Now I am stuck at this point, how do I further progress in this proof? Is there a better way to solve this problem ?
| The way you simplify $\displaystyle{\sum_{k=0}^{n+1}\binom nk^2}$ also works for $\displaystyle{\sum_{k=0}^{n+1}\binom nk \binom n{k-1}}$.
Rewriting $\binom n{k-1} = \binom n{n-k+1}$ we can turn the second sum into $$\sum_{i+j=n+1} \binom ni \binom nj.$$ This is the coefficient of $x^{n+1}$ in $(1+x)^n (1+x)^n$, or $\binom{2n}{n+1}$ - but the generating function approach has already been done in the other answer to this question. We can also think of this sum as counting the number of lattice paths from $(0,0)$ to $(n+1,n-1)$ by partitioning them according to which point $(i,j-1)$ with $i+j=n+1$ they pass through: there are $\binom ni$ ways to get from $(0,0)$ to $(i,j-1)$ and $\binom n{n-j}=\binom nj$ ways to get from $(i,j-1)$ to $(n+1,n-1)$.
Anyway, once you arrive at $$\sum_{k=0}^{n+1} \binom{n}{k}^2 + \sum_{k=0}^{n+1} \binom{n}{k}^2 - 2\sum_{k=0}^{n+1}\binom{n}{k}\binom{n}{k-1}$$ we can handle all three sums in this way: you've already done the first two and my previous paragraph handles the third. We get $$\binom{2n}{n} + \binom{2n}{n} - 2\binom{2n}{n+1}$$ and since $\binom{2n}{n+1} = \frac{n}{n+1} \binom{2n}{n}$, this simplifies to $(1 + 1 - \frac{2n}{n+1}) \binom{2n}{n} = \frac{2}{n+1} \binom{2n}{n}$, as desired.
By the way, the general identity at work here is Vandermonde's identity $$\sum_{i+j=k} \binom ni \binom mj = \binom{n+m}{k}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3400677",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Ramanujan's Nested Radicals: evaluating $\sqrt{4+\sqrt{16+\sqrt{64+\sqrt{\cdots}}}}$
Find the exact value of
$$\sqrt{4+\sqrt{16+\sqrt{64+\sqrt{\cdots}}}}$$
My approach:
Suppose
$$\sqrt{4+\sqrt{4^2+\sqrt{4^3+\sqrt{\cdots}}}} = p \tag{1}$$
By multiplying each side by $2$, we have
$$2\sqrt{4+\sqrt{4^2+\sqrt{4^3+\sqrt{\cdots}}}} = 2p \tag{2}$$
and this equation is equivalent to
$$\sqrt{4^2 + \sqrt{4^3 + \sqrt{4^4 + \sqrt{\cdots}}}} = 2p \tag{3}$$
Back to our original $p$. By squaring each side, we have
$$4 + \sqrt{4^2 + \sqrt{4^3 + \sqrt{4^4 + \sqrt{\cdots}}}} = p^2 \tag{4}$$
It means, $4 + 2p = p^2$.
Solving this quadratic equation, I got $p = 1 + \sqrt5$, taking only the positive value. However, if I take account to scientific calculator approach, the answer tends to the surprisingly nice integer, that is $3$. How can it be? Did I do some mistakes?
Adding note:
My friend said that it holds a theorem.
$$\sqrt{x + \sqrt{4x + \sqrt{4^2x + \sqrt{\cdots}}}} = 1 + \sqrt{x} \tag{5}$$
for every positive integer $x$.
| The mistake is:
$2\sqrt{4+\sqrt{4^2+\sqrt{4^3+\sqrt{\cdots}}}} \neq \sqrt{4^2 + \sqrt{4^3 + \sqrt{4^4 + \sqrt{\cdots}}}} $
$2\sqrt{4+\sqrt{4^2+\sqrt{4^3+\sqrt{\cdots}}}} = \sqrt{4^2 + \sqrt{4^4 + \sqrt{4^7 + \sqrt{\cdots}}}} $
I think that an a way is:
$$2^n+1=\sqrt{(2^n+1)^2}$$
$$2^n+1=\sqrt{4^n+2\cdot2^n+1}$$
$$2^n+1=\sqrt{4^n+\sqrt{(2\cdot2^n+1)^2}}$$
$$2^n+1=\sqrt{4^n+\sqrt{4^{n+1}+4\cdot2^n+1}}$$
$$2^n+1=\sqrt{4^n+\sqrt{4^{n+1}+\sqrt{(4\cdot2^n+1)^2}}}$$
$$2^n+1=\sqrt{4^n+\sqrt{4^{n+1}+\sqrt{4^{n+2}+8\cdot2^n+1}}}$$
$$2^n+1=\sqrt{4^n+\sqrt{4^{n+1}+\sqrt{4^{n+2}+\sqrt{(8\cdot2^n+1)^2}}}}$$
$$2^n+1=\sqrt{4^n+\sqrt{4^{n+1}+\sqrt{4^{n+2}+\sqrt{4^{n+3}+\sqrt{\cdots}}}}}$$
If $n=1$, Then
$$2^1+1=\sqrt{4^1+\sqrt{4^2+\sqrt{4^3+\sqrt{4^4+\sqrt{\cdots}}}}}=3$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3400889",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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$x_1$ and $x_2$ are the solutiton of $\frac{2\cdot \sin(x) \cdot \cos(2x)}{\cos (x) \cdot \sin (2x)} - 5 \tan(x) + 5 = 0$ $x_1$ and $x_2$ are the solutiton of $\frac{2\cdot \sin(x) \cdot \cos(2x)}{\cos (x) \cdot \sin (2x)} - 5 \tan(x) + 5 = 0$
then, $\tan(x_1 + x_2) = ....$
i can do it by doing it
$\dfrac{2\cdot \sin(x) \cdot (\cos^2x - \sin^2x)}{\cos (x) \cdot 2 \sin x \cos x} - 5 \tan(x) + 5 = 0$
leads to $(\sin x - \cos x)(\sin x + 6 \cos x) = 0$
but it's complicated, do you know the less complicated way to solve it?
| Convert all to $\tan$:
$$\frac{2 \sin(x) \cos(2x)}{\cos(x) \sin(2x)} - 5 \tan(x) +5 = \frac{2 \tan(x)}{\tan(2x)} - 5 \tan(x) +5 = (1 - \tan^2(x)) - 5 \tan(x) + 5$$
$$ \Longrightarrow \tan^2(x) + 5 \tan(x) - 6 = 0$$
Now we have:
$$\tan(x_1 + x_2) = \frac{\tan(x_1) + \tan(x_2)}{1 - \tan(x_1)\tan(x_2)} = \frac{-b/a}{1-c/a} = \frac{-5/1}{1-(-6/1)} = \frac{-5}{1+6} = -\frac{5}{7}$$
Without need to compute the roots!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3401102",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
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Finding the value of C in this function If $$x^2 + (c-2)x -c^2 -3c + 5$$ is divided by $x + c$, the remainder is $-1$. find the value of c
I replaced all the x value to -c and set it to an equation which equated to $-1$
I am confused what to do after that, show me the steps how I can retrieve the value of $c$
| Given that:
$$\begin{align}x^2 + (c-2)x -c^2 -3c + 5 & = (-c)^2 + (c-2)(-c) -c^2 -3c + 5 \\
& = c^2 - c^2 +2c -c^2 -3c + 5 \\
c^2 +c -5& = 1 \\
c^2 +c-6 &= 0 \\
(c+3)(c-2) & = 0 \implies \color {blue}{ \boxed {\text { c = 2 , -3}} }\end{align} $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3401466",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove the given inequality if $a+b+c=1$ Let $a,b,c$ be positive real numbers such that $a+b+c=1$, then prove that
$\frac{a}{a^2 +b^3 +c^3}+\frac{b}{a^3 +b^2 +c^3}+\frac{c}{a^3 +b^3 +c^2} \leq \frac{1}{5abc}$
Please provide some hint to proceed. I have used Arithmetic Mean-Geometric Mean inequality to proceed in such questions, but I am not getting how to use it if it is to be used. Kindly provide some directon.
Usually the questions I have dealt so far were like if $a+b+c=1$, find maximum value of $a^3b^4c^2$ which I find using weighted A.M-G.M, but here I am not able to gather the thoughts to proceed
| We need to prove that:
$$\sum_{cyc}\frac{a}{a^3+b^3+c^3+a^2b+a^2c}\leq\frac{a+b+c}{5abc}$$ or
$$\sum_{cyc}\left(\frac{a}{5abc}-\frac{a}{a^3+b^3+c^3+a^2b+a^2c}\right)\geq0$$ or
$$\sum_{cyc}\frac{a(a^3+b^3+c^3+a^2b+a^2c-5abc)}{a^3+b^3+c^3+a^2b+a^2c}\geq0$$ or
$$\sum_{cyc}\frac{a((a-b)(a^2+5ac-2ab-2b^2)-(c-a)(a^2+5ab-2ac-2c^2))}{a^3+b^3+c^3+a^2b+a^2c}\geq0$$ or
$$\sum_{cyc}(a-b)\left(\frac{a^3+5a^2c-2a^2b-2b^2a}{a^3+b^3+c^3+a^2b+a^2c}-\frac{ b^3+5b^2c-2b^2a-2a^2b}{a^3+b^3+c^3+b^2c+b^2a}\right)\geq0$$ or
$$\sum_{cyc}\tfrac{(a-b)^2(a^5+a^4b+5a^4b^2+4a^2b^3+ab^4+b^5+(5a^4+7a^3b+10a^2b^2+7ab^3+5b^4)c+(a^2+ab+b^2)c^3+5(a+b)c^4)}{(a^3+b^3+c^3+a^2b+a^2c)(a^3+b^3+c^3+b^2c+b^2a)}\geq0.$$
Done!
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Prove that integer Let $a, b,$ and $c$ be real such that $ a ^ 2b + b ^ 2 c + c ^ 2a - ab ^ 2 - bc ^ 2 - ca ^ 2 = 6 $ and $ a ^ 2 + b ^ 2 + c ^ 2 -ab - ac - bc = 7$. Prove that if a is integer, then $b$ and $c$ are also integers.
What I tried:
$ ab (a-b) + bc (b-c) + ca (c-a) = 6 $
$a (a-b) -c (c-a) + b (b-c) = 7 $
$ca (a-b) -c^2 (c-a) + bc (b-c) = 7c$
$ca (a-b) -c^2 (c-a) -ca (c-a) -ab (a-b) = k $
($k$ integer)
$a (c-b) (a-b) -c (c^2-a^2) = k$
$a (c-b) (a-b) -ca^2 = k '$
$a ((c-b) (a-b) -ca) = k '$
$a (-cb-ba + b^2) = k '$
What I tried, but I think something is missing
| Put $x = a - b$, $y = b - c$, $z = c - a$. Then:
*
*$x + y + z = 0$;
*$x^2 + y^2 + z^2 = 2(a^2 + b^2 + c^2 - ab - bc - ca) = 14$;
*$xyz = -(a^2b + b^2c + c^2a - ab^2 - bc^2 - ca^2) = -6$.
From 1. and 2., we get:
*$xy + yz + zx = \frac{1}{2}((x + y + z)^2 - (x^2 + y^2 + z^2)) = -7$.
Therefore, the equations 1. 3. 4. tell us that $x, y, z$ are roots of the polynomial $T^3 - 7 T + 6$, which factors as $(T - 2)(T - 1)(T + 3)$.
Hence $x, y, z$ are all integers and we are done.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3402314",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
$10000=abc$ with $a,b,c$ natural numbers don't have digit $0$. Smallest possible value of $a+b+c$ is? Let $10000=abc$ with $a,b,c$ natural numbers don't have digit $0$. If $a,b,c$ does not have to be all distinct, then the smallest possible value of $a+b+c$ is?
Attempt:
First write as prime factors: $10000 = 2^{4} 5^{4}$. The possible triples are:
$$ 2, 2^{3}, 5^{4} $$
$$ 2^{2}, 2^{2}, 5^{4} $$
$$ 2^{3}, 2, 5^{4} $$
$$ 2^{4}, 2^{4}, 5^{4} $$
$$ 5, 5^{3}, 2^{4}$$
$$ 5^{2}, 5^{2}, 2^{4}$$
$$ 5^{3}, 5, 2^{4}$$
The smallest sum is $5^{2} + 5^{2} + 2^{4}$. Are there better approaches?
| Recognize that you cannot have a factor that includes both a power of $2$ and $5$ at the same time, and recognize that (obviously) $5^2+5^2<5^4$. We would try to cover all the $2$s required, then split up the $5$s as evenly as possible.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3405223",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
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Sum of floor function of general term While doing some solution to get a general term for some series I encountered at the end with this.
$S(n,i) = \left(\dfrac{n^2+n-2i}{n} - \lfloor\dfrac{n^2+n-2i}{2n}\rfloor \right) \left(\lfloor\dfrac{n^2+n-2i}{2n}\rfloor +1 \right)$
, where $i> \dfrac{n}{2}$. How can I get the general term from this summation by geting rid of the floor function.
| What's your problem with floor function?
However you can't get rid of it, but this may help:
$$S(n,i) = \left(\dfrac{n^2+n-2i}{n} - \lfloor\dfrac{n^2+n-2i}{2n}\rfloor \right) \left(\lfloor\dfrac{n^2+n-2i}{2n}\rfloor +1 \right)$$
$$ = \left(\dfrac{n^2+n-2i}{2n} + \dfrac{n^2+n-2i}{2n} - \lfloor\dfrac{n^2+n-2i}{2n}\rfloor \right) \left(\lfloor\dfrac{n^2+n-2i}{2n}\rfloor +1 \right)$$
$$ = \left(\dfrac{n^2+n-2i}{2n} + \epsilon({n^2+n-2i}{2n})\right) \left(\dfrac{n^2+n-2i}{2n} - \epsilon({n^2+n-2i}{2n}) +1 \right)$$
When the $\epsilon$ function returns the decimal part instead of integer part of given number.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3405387",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to find x from the equation $\sqrt{x + \sqrt{x}} - \sqrt{x - \sqrt{x}} = m \sqrt{\frac{x}{x+\sqrt{x}}}$ For m are real number,
Find x from the equation
$$\sqrt{x + \sqrt{x}} - \sqrt{x - \sqrt{x}} = m \sqrt{\frac{x}{x+\sqrt{x}}}$$
I tried to multiply $\sqrt{x + \sqrt{x}}$ to the both sides and I get
$$x + \sqrt{x} - \sqrt{x^{2} - x} = m \sqrt{x}$$
What should I do now to get x? Can anyone show me a hint please?
| HINT
Divide by $\sqrt{x}$ if $x \ne 0$ to get
$$
\sqrt{x} + 1 - \sqrt{x-1} = m
$$
move $1$ to the other side and square.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3405511",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 0
} |
How to prove $\lim_{n \to \infty} \left(\frac{3n-2}{3n+1}\right)^{2n} = \frac{1}{e^2}$?
It's known that $\lim_{n \to \infty} \left(1 + \frac{x}{n} \right)^n = e^x$.
Using the above statement, prove $\lim_{n \to \infty} \left(\frac{3n-2}{3n+1}\right)^{2n} = \frac{1}{e^2}$.
My attempt
Obviously, we want to reach a statement such as $$\lim_{n \to \infty} \left(1 + \frac{-2}{n}\right)^n \quad \text{ or } \quad \lim_{n \to \infty} \left(1 + \frac{-1}{n}\right)^n \cdot \lim_{n \to \infty} \left(1 + \frac{-1}{n}\right)^n$$
in order to be able to apply the above condition. However, I was unable to achieve this. The furthest I've got was the following:
\begin{align}
\left(\frac{n-2}{3n+1}\right)^{2n} &= \left( \frac{9n^2 - 12n + 4}{9n^2 + 6n + 1} \right)^n\\
&= \left(1 + \frac{-18n+3}{9n^2+6n+1}\right)^n\\
f(n)&= \left(1 + \frac{-2 + \frac{3}{n}}{n+\frac{2}{3} + \frac{1}{9n}} \right)^n
\end{align}
It seems quite obvious that $\lim_{n \to \infty} \left(f(n)\right) = \left(1 + \frac{-2}{n + \frac{2}{3}}\right)^n$, however, this is not exactly equal to the statement given above. Are you able to ignore the constant and apply the condition regardness? If so, why? How would you go about solving this problem?
| We can use that
$$\left(\frac{3n-2}{3n+1}\right)^{2n}=\left(1-\frac{3}{3n+1}\right)^{2n}=\left[\left(1-\frac{3}{3n+1}\right)^{-\frac{3n+1}3}\right]^{-\frac{6n}{3n+1}}\to e^{-2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3405914",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 3
} |
Is there any. elementary formula for the sequence$\sum_{k=1}^{n}\left(2k-1\right)\left(\frac{1}{2}\right)^{k}$ Is there any formula for the following sequence which does not use any derivative and also is less
advanced:
$$\sum_{k=1}^{n}\left(2k-1\right)\left(\frac{1}{2}\right)^{k}$$
I've calculated the general formula
and here is a photo of the solution
1:
| It's a fairly simple proof by induction.
$$\mathcal{H}_n \ : \ \left\{ \sum_{k=1}^{n}\left(2k-1\right)\left(\frac{1}{2}\right)^{k} = -2^{-n}(2n+3)+3\right\}$$
Base case is obvious, $\mathcal{H}_1$ is true.
Induction. Suppose $\mathcal{H}_n$ true. Now
$$\sum_{k=1}^{n+1}\left(2k-1\right)\left(\frac{1}{2}\right)^{k} = \sum_{k=1}^{n}\left(2k-1\right)\left(\frac{1}{2}\right)^{k} +(2n+1)\left(\frac{1}{2}\right)^{n+1}$$
By induction
$$\begin{align*}
\sum_{k=1}^{n+1}\left(2k-1\right)\left(\frac{1}{2}\right)^{k} &= -2^{-n}(2n+3) +3+(2n+1)\left(\frac{1}{2}\right)^{n+1}\\
&= \underbrace{-2^{-n}(2n+1)}_{A} -\underbrace{2^{-n}2}_{B} +\underbrace{(2n+1)2^{-n}\left(\frac{1}{2}\right)}_{C} + 3\\
\end{align*}
$$
Summing $A$ and $C$, and rewriting $B$ as $2^{-n}2=2^{-(n+1)}4$ we get
$$\sum_{k=1}^{n+1}\left(2k-1\right)\left(\frac{1}{2}\right)^{k} = -2^{-(n+1)}(2n+5)+3$$
Hence $\mathcal{H}_{n+1}$ is true, and by induction principle,
$$\forall n\geq 1, \ \sum_{k=1}^{n}\left(2k-1\right)\left(\frac{1}{2}\right)^{k} = -2^{-n}(2n+3)+3$$
Edit - more direct approach Using inspiration from this question
Let
$$F_n(x)=\sum_{k=1}^nkx^k=x+2x^2+\ldots+nx^n$$
Then
$$F_n(x)-xF_n(x) = x+x^2+x^3+\ldots+x^n-nx^{n+1}$$
$$F_n(x)-xF_n(x) = \sum_{k=1}^nx^k-nx^{n+1}$$
$$F_n(x)-xF_n(x) = x\frac{1-x^n}{1-x}-nx^{n+1}$$
So that
$$F_n(x) = x\frac{1-x^n}{(1-x)^2}-\frac{nx^{n+1}}{1-x}$$
Now, using @user remark, we have
$$\sum_{k=1}^{n}\left(2k-1\right)\left(\frac{1}{2}\right)^{k}=2\sum_{k=1}^{n}k\left(\frac{1}{2}\right)^{k}-\sum_{k=1}^{n}\left(\frac{1}{2}\right)^{k}$$
And
$$\sum_{k=1}^{n}(2k-1)\left(\frac{1}{2}\right)^{k}= 2F_n(1/2)-(1-2^{-n})$$
Using our formula for $F_n(x)$,
$$\sum_{k=1}^{n}(2k-1)\left(\frac{1}{2}\right)^{k}= 2\cdot\left(\frac{1}{2}\frac{1-2^{-n}}{(1/2)^2}-\frac{n2^{-(n+1)}}{1/2}\right) - (1-2^{-n})$$
$$\sum_{k=1}^{n}(2k-1)\left(\frac{1}{2}\right)^{k}= 4 (1-2^{-n}) - 4n2^{-(n+1)} - (1-2^{-n})$$
$$\sum_{k=1}^{n}(2k-1)\left(\frac{1}{2}\right)^{k}= 3 (1-2^{-n}) - 4n2^{-(n+1)}$$
And finally
$$\sum_{k=1}^{n}\left(2k-1\right)\left(\frac{1}{2}\right)^{k} = -2^{-n}(2n+3)+3$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3407263",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
A pair of limits involving sums Find the following pair of limits:
$$\lim_{r \to \infty} 2 \lfloor r \rfloor \sum_{n=2\lfloor r \rfloor+1}^\infty \frac{(-1)^{n+1}}{n}$$
$$\lim_{r \to \infty} (2 \lfloor r \rfloor +1) \sum_{n=2 \lfloor r \rfloor+2}^\infty \frac{(-1)^{n+1}}{n}$$
Plugging in large values of $r$, I would guess that the first limit tends to $1/2$ and that the second limit tends to $-1/2$.
The floors are intended to require that $r$ be an integer. The inside of the first limit is just the following function for even $r$ and the inside of the second is the function for odd $r$.
$$r \sum_{n=r+1}^\infty \frac{(-1)^{n+1}}{n}$$
Normally for a problem like this, I'd try L'Hopital's rule, however I don't know how that would work because the obvious way would require taking the derivative
$$\frac{d}{dr} \sum_{n=2\lfloor r \rfloor+1}^\infty \frac{(-1)^{n+1}}{n}$$
which doesn't seem to make very much sense.
Any hints or solutions on how to evaluate these limits?
| For $r\in \Bbb N$ we have $$\sum_{n=2r+1}^{\infty}(-1)^{n+1}/n=$$ $$=(\frac {1}{2r+1}-\frac {1}{2r+2})+(\frac {1}{2r+3}-\frac {1}{2r+4})+(\frac {1}{2r+5}-\frac {1}{2r+6})+...=$$ $$=\frac {1}{(2r+1)(2r+2)}+\frac {1}{(2r+3)(2r+4)}+\frac {1}{(2r+5)(2r+6)}+...$$
Now for $n>1$ we have $$\frac {1}{4}\int_{n+1}^{n+2}(1/x^2)dx<\frac {1}{4(n+1)^2}<$$ $$<\frac {1}{(2n+1)(2n+2)}<$$ $$<\frac {1}{(2n+1)^2}<\frac {1}{4n^2}<$$ $$<\frac {1}{4}\int_{n-1}^n(1/x^2)dx.$$
Summing this for $n$ from $r$ to $\infty$ ( with $1<r\in \Bbb N$ ) we have $$\frac {1}{4}\int_{r+1}^{\infty}(1/x^2)dx<\sum_{n=2r+1}^{\infty}(-1)^{n+1}/n<\frac {1}{4}\int_{r-1}^{\infty}(1/x^2)dx.$$
The far left and far right sides in the line above are $\frac {1}{4(r+1)}$ and $\frac {1}{4(r-1)}.$
The second limit can be done similarly but it can also be obtained from the first limit, for if $F(s)=\sum_{n=s}^{\infty}(-1)^{n+1}/n$ then $$(2r+1)F(2r+2)=\frac {2r+1}{2r}\cdot 2r\left(F(2r+1)-\frac {1}{2r+1}\right)= \frac {2r+1}{2r}\cdot 2rF(2r+1)-1.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3407889",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
If $x$ and $y$ are real number such that $x^2+2xy-y^2=6$, then find the minimum value of $(x^2+y^2)^2$ If $x$ and $y$ are real number such that $x^2+2xy-y^2=6$, then find the maximum value of $(x^2+y^2)^2$
My attempt is as follows:
$$(x-y)^2\ge 0$$
$$x^2+y^2\ge 2xy$$
$$2(x^2+y^2)\ge x^2+y^2+2xy$$
\begin{equation}
2(x^2+y^2)\ge (x+y)^2\tag{1}
\end{equation}
Solving the given equation:
$$x^2+2xy-y^2=6$$
$$(x+y)^2=2y^2+6$$
So putting the value of $(x+y)^2$ in equation $1$
$$2(x^2+y^2)\ge 2y^2+6$$
$$x^2+y^2\ge y^2+3$$
\begin{equation}
x^2\ge 3\tag{2}
\end{equation}
So $x\in \left(-\infty,-\sqrt{3}\right) \cup \left(\sqrt{3},\infty\right)$
But how to proceed from here?
| The function to be minimized is the fourth power of the distance from the origin, so will have its extrema at the same points at which the distance is also an extremum. The constraint is a hyperbola, so the minimum distance occurs at its vertices, therefore this problem is equivalent to computing the fourth power of a semiaxis length of this hyperbola.
One approach is to compute the eigenvalues of the matrix $\frac16\begin{bmatrix}1&1\\1&-1\end{bmatrix}$, which work out to be $\pm\frac1{3\sqrt2}$. We want the positive eigenvalue. It is the reciprocal of the square of the semi-transverse axis length, so the minimum value of $(x^2+y^2)^2$ on this hyperbola is $(3\sqrt2)^2=18$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3411531",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 3
} |
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