Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Why does simplifying a function give it another limit I'm asked:
$$\lim_{x\to 1} \frac{x^3 - 1}{x^2 + 2x -3}$$
This does obviously not evaluate since the denominator equals $0$. The solution is to:
$$\lim_{x\to 1} \frac{(x-1)(x^2+x+1)}{(x-1)(x+3)}$$
$$\lim_{x\to 1} \frac{x^2 + x + 1}{x + 3}$$
$$\frac{1+1+1}{1+3} = \frac{3}{4}$$
My question: what is actually happening? How can simplifying a function give it another limit? Is it a complete other function and if so why would it be relevant to our original question?
| It's not that the simplified function has a different limit, it's that the limit of the original expression cannot be found by evaluation. In other words, write
$$
f(x) = \frac{x^{3} - 1}{x^{2} + 2x - 3},\qquad
g(x) = \frac{x^{2} + x + 1}{x + 3}
= \begin{cases}
f(x) & x \neq 1, \\
\frac{3}{4} & x = 1.
\end{cases}
$$
Since $g$ is a quotient of polynomials with denominator non-zero at $x = 1$, its limit at $1$ can be found by evaluation. The limit of $f$ at $1$ has the same value, since $f(x) = g(x)$ for all $x \neq 1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1822665",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "30",
"answer_count": 10,
"answer_id": 6
} |
How do you simplify this square root of sum: $\sqrt{7+4\sqrt3}$? I came around this expression when solving a problem.
$$\sqrt{7+4\sqrt{3}}$$
WolframAlpha says it equals $2+\sqrt{3}$. We can confirm it like this
$$\left(2+\sqrt{3}\right)^2 \;=\; 4+4\sqrt{3} + 3 \;=\; 7 + 4\sqrt{3}.$$
However, the only way I can think of how to simplify that expression in hand is guessing. Is there a better way of calculating square root of a sum like that one?
| In general, there is a formula where given the expression $\sqrt{X\pm Y}$, one can rewrite that into $\sqrt{\frac {X+\sqrt{X^2-Y^2}}{2}}\pm\sqrt{\frac {X-\sqrt{X^2-Y^2}}{2}}$ where $X$ and $Y$ can be any number and $X>Y$.
In your case, we have $X=7$ and $Y=4\sqrt{3}$ because $7<4\sqrt{3}$. Plugging the values into the formula, we get: $$\sqrt{\frac {7+\sqrt{7^2-(\sqrt{48})^2}}{2}}+\sqrt{\frac {7-\sqrt{7^2-(\sqrt{48})^2}}{2}}\tag{1}$$
The radical $\sqrt{7^2-(\sqrt{48})^2}$ simplifies into $1$ so $(1)$ becomes $$\sqrt{\frac {7+1}{2}}+\sqrt{\frac {7-1}{2}}$$ which further simplifies into $$\sqrt{4}+\sqrt{3}\iff\boxed{\sqrt{3}+2}$$
Extra: This works on any nested radical under a square root. For example: Denest $\sqrt{5\sqrt{3}+6\sqrt{2}}$. Setting $X=5\sqrt{3}$ and $Y=6\sqrt{2}$, gives us $$\sqrt{\frac {5\sqrt{3}+\sqrt{3}}{2}}+\sqrt{\frac {5\sqrt{3}-\sqrt{3}}{2}}\tag{1}$$
Simplifying gives us our denesting. $$\sqrt[4]{27}+\sqrt[4]{12}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1828493",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 9,
"answer_id": 4
} |
Find the integer $x$ such $x^6+x^5+x^4+x^3+x^2+x+1=y^3$ Find the equation integer solution $$\color{red}{y^3=x^6+x^5+x^4+x^3+x^2+x+1}$$
It is obvious $x=0,y=1$ or $x=-1,y=1$ are solutions.
How to find all solutions?
| Here is an elementary approach, albeit tedious if one computes everything by hand. Our strategy is to find perfect cubes that are close enough to RHS (as Jack D'Aurizio mentioned in the comment).
Denoting $~f(x)=27(x^6+x^5+x^4+x^3+x^2+x+1)$, we have
*
*Claim 1: If $x\ge3$, then
$$(3x^2+x)^3<f(x)<(3x^2+x+1)^3.$$
*
*Claim 2: If $x\ge2$, then
$$(3x^2-x)^3<f(-x)<(3x^2-x+1)^3.$$
Back to the original equation, we can easily check the case $x=-1,~0,~1,~2$; when $x\ge3$, by claim 1, we see $(3y)^3=f(x)$ lies between two consecutive cubes, therefore no integer solutions; similarly claim 2 covers the case $x\le-2$. We are done.
*
*Proof of claim 1:
*
*$$f(x)-(3x^2+x)^3=27+27x+27x^2+26x^3+18x^4>0,$$
*$$(3x^2+x+1)^3-f(x)=94+6(x-2)(7x+10)+x^2(x-3)(9x+19)>0,$$
*Proof of claim 2:
*
*$$f(-x)-(3x^2-x)^3=19+21(x-1)+(x-1)^2(18x^2+10x+29)>0;$$
*$$(3x^2-x+1)^3-f(-x)=(x-1)(9x^3+17x^2+2x+26)>0.$$
(1) In hindsight, there is no reason to use elementary methods only. Note the two claims above are essentially same:
$$(3x^2+x)^3<f(x)<(3x^2+x+1)^3$$
for $x\le-2$ or $x\ge3$.
We could simply take the derivative and prove the differences are monotonic on either range...
(2) A bit heavier but more general approach is by Runge's method, e.g., see this paper by Sankaranarayanan and Saradha.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1829370",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
} |
Prove inequality: $\frac{P+2004a}{P-2a}\cdot\frac{P+2004b}{P-2b}\cdot\frac{P+2004c}{P-2c}\ge2007^3.$
Let $a,b,c$ the sides of a triangle, $P$ its perimeter. Prove inequality:
$$\frac{P+2004a}{P-2a}\cdot\frac{P+2004b}{P-2b}\cdot\frac{P+2004c}{P-2c}\ge2007^3.$$
My attempt:
1) $P=a+b+c$. Then $\frac{P+2004a}{P-2a}=\frac{2005a+b+c}{b+c-a}$. Here $b+c-a>0 -$ triangle inequality.
2) $\sqrt[3]{xyz}\ge\frac{3}{\frac1x+\frac1y+\frac1z} \Rightarrow xyz\ge \left(\frac{3}{\frac1x+\frac1y+\frac1z}\right)^3$
| I think it has to be $3\cdot 2007$ instead of $2007^3$. Since $a,b,c$ are sides of a triangle, there exist $x,y,z>0$ such that
$a=y+z$, $b=z+x$ and $c=x+y$. Hence, we have to prove:
$$
\sum_{cyc}\frac{2005(x+y)+(y+z)+(z+x)}{2(x+y+z)-2(x+y)}≥3\cdot 2007\iff\\
\sum_{cyc}\frac{2006x+2006y+2z}{2z}≥3\cdot 2007\iff\\
3+1003\sum_{sym}\frac{x}{y}≥3\cdot 2007
$$
Now with AM-GM, we have $\sum_{sym}\frac{x}{y}≥6$ and thus:
$$
3+1003\sum_{sym}\frac{x}{y}≥3+6\cdot1003=3\cdot 2007
$$
Withe equality iff $x=y=z\iff a=b=c$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1832166",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
How can I solve $y^4 = 5 \pmod{11\times19}$ with legendre? Solve $y^4 = 5 \pmod{11\times19}$
I'm trying to let $y^2=A$ then $A^2=5 \pmod{11\times19}$.
And solve this problem then $A= 104,-104,28,-28 \pmod{11\times19}$
Then should I solve this problem for all this four $A$ ?
I want to solve this easily with Legendre.
(I think I can use this thing that $(\frac{100}{14})=1$ and $(\frac{104}{19})=1$ , $(\frac{28}{11})=-1$, $(\frac{14}{11})= 1$.)
All fractions are translated as Legendre symbol.
| Separating out the $\bmod 11$ and $\bmod 19$ strands of the solution (just looking up to $\lfloor n/2 \rfloor$, we can use symmetry after that):
\begin{array}{c|c|c} n & n^2 \bmod 11 & n^4 \bmod 11 & n^2 \bmod 19 & n^4 \bmod 19\\ \hline
1 & 1 & 1 & 1 & 1 \\
2 & 4 & \color{red}
5 & 4 & 16 \\
3 & 9 & 4 & 9 & \color{red}
5 \\
4 & 5 & 3 & 16 & 9 \\
5 & 3 & 9 & 6 & 17 \\
6 & & & 17 & 4 \\
7 & & & 11 & 7 \\
8 & & & 7 & 11 \\
9 & & & 5 & 6 \\
\end{array}
This gives us $y \equiv \pm 2\equiv (2,9) \bmod 11$ and $y \equiv \pm 3\equiv (3, 16) \bmod 19$
\begin{align}
y=19k+3 &\equiv \pm2 \bmod 11\\
8k &\equiv -1 \bmod 11 \tag{from $+2$}\\
k = 4 \implies y &=76+3 = 79 \text{ is a solution}\\
8k &\equiv -5 \bmod 11 \tag{from $-2$}\\
k = -4\times -5 \equiv 9 \implies y &= 171+3 = 174 \text{ is a solution}\\
\end{align}
So $ y \equiv (\pm 79, \pm 174) \equiv (35, 79, 130, 174) \bmod 209$ are the solutions
The numbers are small enough that we could also just step through the possibilities:
$\{2,9,13,20,24,31,\color{red}{35},42,46,53,57,64,68,75,\color{red}{79},86,90,97,101,\ldots\}$
$\{3,16,22,\color{red}{35},41,54,60,73,\color{red}{79},92,98,\ldots\}$
Then $ y \equiv (\pm 35, \pm 79) \equiv (35, 79, 130, 174) \bmod 209$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1834025",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
If xy + yz + zx = 1, ........... If $xy + yz + zx = 1$, then show that
$$\dfrac{x}{1-x^2} + \dfrac{y}{1-y^2} + \dfrac{z}{1-z^2} = \dfrac{4xyz}{(1-x^2)(1-y^2)(1-z^2)}$$
I tried doing the sum algebraically, that is, by solving the LHS but, that method isn't really getting us anywhere.
Also, I found this sum in an a book on trigonometry. So how can I solve it with trigonometry?
| For a trigonometric proof, follow Parth Kohli's suggestion which is nice. Meanwhile here's an algebraic proof.
Let $a=\dfrac{x}{1-x^2}$, $b=\dfrac{y}{1-y^2}$, $c=\dfrac{z}{1-z^2}$.
We need to prove: $a+b+c=4abc$, i.e. $\sum \dfrac{1}{ab}=4$.
We have $\sum \dfrac{1}{ab}=\sum \dfrac{(1-x^2)(1-y^2)}{xy}=1 + \sum \left( \dfrac{1}{xy}-(\dfrac{x}{y}+\dfrac{y}{x}) \right)$.
Now, $\sum \dfrac{1}{xy}=\sum \dfrac{xy+yz+zx}{xy}=3+\sum (\dfrac{x}{y}+\dfrac{y}{x})$,
which gives the desired equality.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1834114",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Prove: $|a\sin x+b \cos x|\leq \sqrt{a^2+b^2}$
$$|a\sin x+b \cos x|\leq \sqrt{a^2+b^2}$$
I have tried: $$|a\sin x+b \cos x|\leq |a+b|\leq \sqrt{a^2+b^2}$$
enough to prove: $$|a+b|\leq \sqrt{a^2+b^2}$$
But I can find how to continue from here
| $$|a \sin x + b \cos x|=\big|\sqrt{a^2+b^2}\left(\frac a{\sqrt{a^2+b^2}}\sin x+\frac b{\sqrt{a^2+b^2}}\cos x\right)\big|=$$
$$=\sqrt{a^2+b^2}|\left(\sin(x+\phi)\right)|\le\sqrt{a^2+b^2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1834356",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 0
} |
How to solve without solving by inspection? $\frac{x+5}{x+k}=\frac{-kx+5}{x-1}$ Background: This is from a test review on functions. The original problem was
Find the value of $k$ so that the function $f(x) = \frac{x+5}{x+k}$ will be its own inverse.
I found the answer by inspection, and then tried to solve it through more rigorous means. Continuing from the question title, I cross multiplied and used the quadratic formula on $k.$
$x^2+4x-5=5x+5k-kx^2-k^2x$
$k^2x+k(x^2-5)+(x^2-x-5)=0$
$k = \frac{-x+5\pm\sqrt{x^4-10x^2+25-4x(x^2-x-5)}}{2x}$
$k = \frac{-x+5\pm\sqrt{x^4-4x^3-6x^2+20x+25}}{2x}$
This is where I got stuck. Any ideas?
| You can use the fact that the $x$-intercept and the $y$-intercept get switched, OR you can use the fact that the horizontal asymptote and the vertical asymptote get switched.
(ADDED A FEW MINUTES LATER)
Since I am pretty sure the intended solution of this problem is how I suggested (and not by the heavy algebraic manipulations that others have given), I thought it best to give some more details.
The $x$- and $y$-intercepts of $\;y = \frac{x+5}{x+k}\;$ are $(-5,0)$ and $(0,\frac{5}{k}).$ Therefore, the $x$- and $y$-intercepts of the inverse will be $(\frac{5}{k},0)$ and $(0,-5).$ (Recall that if $(a,b)$ is on the graph of the original function, then $(b,a)$ will be on the graph of the inverse.) Since the function and its inverse are the same function (hence their graphs will be identical), it follows that $\;(-5,0) = (\frac{5}{k},0)\;$ and $\;(0,\frac{5}{k}) = (0,-5),\;$ and each of these equations gives us $\;-5 = \frac{5}{k},\;$ or $k = -1.$
Alternatively, the horizontal asymptote of $\;y = \frac{x+5}{x+k}\;$ is $y = 1$ and the vertical asymptote of $\;y = \frac{x+5}{x+k}\;$ is $x=-k.$ Therefore, the inverse will have a horizontal asymptote of $y = -k$ and a vertical asymptote of $x = 1.$ (When you perform the transformation that replaces $x$ with $y$ and replaces $y$ with $x,$ the line $y=1$ gets transformed into the line $x=1$ and the line $x=-k$ gets transformed into the line $y=-k.$ Also, if the graph gets close to the former lines far from the origin, the graph of the inverse gets close to the latter lines far from the origin, and thus the latter lines are also asymptotes.) Now, since the function and its inverse are the same function (hence their graphs will be identical), we immediately get that $k = -1.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1834958",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 1
} |
Find the values of $x$ such that $2\tan^{-1}x+\sin^{-1}\left(\frac{2x}{1+x^2}\right)$ is independent of $x$.
Find the values of $x$ such that
$$2\tan^{-1}x+\sin^{-1}\left(\frac{2x}{1+x^2}\right)$$ is independent of $x$.
Checking for $x\in [-1,1]$
In the taken domain $\sin^{-1}\left(\frac{2x}{1+x^2}\right)$ comes out to be $2\tan^{-1}x$ hence the taken function comes out to be equal to $4\tan^{-1}x$ hence the function is clearly dependent on $x$.
Now checking for $x\in (1,\infty)$
In the taken domain $2\tan^{-1}x$ comes out to be $\pi-\sin^{-1}\left(\frac{2x}{1+x^2}\right)$ and hence the net sum becomes independent of $x$.
Now checking for $x\in (-\infty,-1)$
In the taken domain $2\tan^{-1}x$ comes out to be $-\pi-\sin^{-1}\left(\frac{2x}{1+x^2}\right)$ and hence the net sum becomes $-\pi$ therefore becomes, independent of $x$.
But the answer has been mentioned as just $x\in [1,\infty)$ Can anybody tell me why the second set has not been included.
| Differentiate it:
$$\left(2\arctan x+\arcsin\frac{2x}{1+x^2}\right)'=\frac2{1+x^2}+\frac{2-2x^2}{(1+x^2)^2}\frac1{\sqrt{1-\frac{4x^2}{(1+x^2)^2}}}=$$
$$=\frac2{1+x^2}+\frac{2-2x^2}{(1+x^2)^2}\frac{1+x^2}{\sqrt{(1-x^2)^2}}=\frac2{1+x^2}+\frac{2(1-x^2)}{(1+x^2)|1-x^2|}$$
We can thus see that if $\;|1-x^2|=-(1-x^2)\iff 1-x^2<0\iff |x|>1\;$ , then the above last expression is zero and thus the function is a constant one and doesn't depend on $\;x\;$ .
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1838549",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
A fraction problem $$a = x + \frac1x \\b =y + \frac1y \\ c = xy +\frac1{xy} $$
Express $c$ in terms of $a$ and $b$
| Note that $c$ is not uniquely specified by $a$ and $b$, since solving $a=x+\frac{1}{x}$ for $x$ yields two solutions which are reciprocals of each other, and applying $x\mapsto\frac{1}{x}$ or $y\mapsto\frac{1}{y}$ to $xy+\frac{1}{xy}$ yields $\frac{x}{y}+\frac{y}{x}$. However, applying either substitution again gives back $xy+\frac{1}{xy}$, so $c$ can take on either of those values.
Let $c_1=xy+\frac{1}{xy}$ and $c_2=\frac{x}{y}+\frac{y}{x}$. One can check that
$$c_1+c_2=ab$$
and
$$c_1c_2=a^2+b^2-4.$$
Use the quadratic formula to solve for the possible values of $c$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1840316",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Uniform convergence of $\sum_{n=1}^{\infty}\frac{x}{n(n+x^2)}$ on [0,1] Prove that $\sum_{n=1}^{\infty}\frac{x}{n(n+x^2)}$ converges uniformly on $[0,\infty)$.
On $[1,\infty)$, we have $\frac{1}{n}\leq x$ which implies $\frac{x}{n(n+x^2)}\leq\frac{1}{n^2}$. Since $\sum_{n=1}^{\infty}\frac{1}{n^2}<\infty$, the M-test gives that $\sum_{n=1}^{\infty}\frac{x}{n(n+x^2)}$ converges uniformly on $[1,\infty)$.
I'm not sure though how to show this for $[0,1)$.
| For $x\ge0$, $x+\frac1x=\left(\sqrt{x}-\frac1{\sqrt{x}}\right)^2+2\ge2$. Therefore,
$$
\begin{align}
\frac x{n(n+x^2)}
&=\frac1{n^{3/2}}\frac1{\frac{\sqrt{n}}x+\frac x{\sqrt{n}}}\\
&\le\frac1{2n^{3/2}}
\end{align}
$$
Therefore, independent of $x$,
$$
\begin{align}
\sum_{n=N+1}^\infty\frac x{n(n+x^2)}
&\le\sum_{n=N+1}^\infty\frac1{2n^{3/2}}\\
&\le\frac12\int_N^\infty\frac{\mathrm{d}x}{x^{3/2}}\\[3pt]
&=\frac1{\sqrt{N}}
\end{align}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1840859",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
If $x,y,z\gt 0$ and $xyz=1$ Then minimum value of $\frac{x^2}{y+z}+\frac{y^2}{z+x}+\frac{z^2}{x+y}$
If $x,y,z\gt 0$ and $xyz=1$ Then find the minimum value of $\displaystyle \frac{x^2}{y+z}+\frac{y^2}{z+x}+\frac{z^2}{x+y}$
$\bf{My\; Try::}$Using Titu's Lemma $$\frac{x^2}{y+z}+\frac{y^2}{z+x}+\frac{z^2}{x+y}\ge \frac{(x+y+z)^2}{2(x+y+z)} = \frac{x+y+z}{2}\ge 3\frac{\sqrt[3]{xyz}}{2} = \frac{3}{2}$$
and equality holds when $$x=y=z=1$$
My question is how can we solve it without the above lemma, like using Jensen's Inequality or other inequality.
Please explain me.
Thanks
| This answer shows how Nesbitt's inequality can used with other proof techniques to prove the given inequality.
Nesbitt's inequality
$$\frac{x}{y+z}+\frac{y}{z+x}+\frac{z}{x+y} \geq \frac{3}{2}\tag{$*$}$$
$(1a)$ Observe that
$$\sum_{cyc}\left(\frac{x^2}{y+z}\right)+(x+y+z)=(x+y+z)\left(\frac{x}{y+z}+\frac{y}{z+x}+\frac{z}{x+y}\right)\tag{A}$$
Using $(*)$ and (A), we get
$$\sum_{cyc}\left(\frac{x^2}{y+z}\right) \geq \frac{(x+y+z)}{2}\tag{B}$$
Now, using AM-GM inequality, we get
$$\frac{(x+y+z)}{2} \geq \frac{3\sqrt[3]{xyz}}{2}=\frac{3}{2}\tag{C}$$
Using (B) and (C), we get the desired result.
$(1b)$ Observe that
$$\sum_{cyc}\left(\frac{x^2}{y+z}\right)=(x+y+z)\left[\left(\sum_{cyc}\frac{x}{y+z}\right)-1\right]\tag{D}$$
Using $(*)$ and (C) in (D), we get
$$\sum_{cyc}\left(\frac{x^2}{y+z}\right) \geq 3 \cdot \left(\frac{3}{2}-1\right)=\frac{3}{2}$$
$(2)$
Without loss of generality, let $x \geq y \geq z$
Therefore, $$\frac{x}{y+z} \geq \frac{y}{x+z} \geq \frac{z}{x+y}$$
Now, using the Chebyshev inequality for these two increasing sequences, we get
$$3 \sum_{cyc}\left(\frac{x^2}{y+z}\right) \geq (x+y+z)\sum_{cyc}\left(\frac{x}{y+z}\right)\tag{E}$$
Using $(*)$ and (C) in (E), we get the required result.
$(3)$ Let $f(x)=x^2$
Now, we can write
$$\sum_{cyc}\left(\frac{x^2}{y+z}\right)=\sum_{cyc}\left(\frac{f(x)}{y+z}\right)$$
Note, that $f''(x)=2 > 0$, so the function is convex.
Now, using the Weighted-Jensen inequality, we get
$$\sum_{cyc}\left(\frac{f(x)}{y+z}\right) \geq 3f(M)\tag{F}$$
where using $(*)$, we get
$$3M=\sum_{cyc}\left(\frac{x}{y+z}\right) \geq \frac{3}{2} \Longleftrightarrow M \geq \frac{1}{2}\tag{G}$$
Using (G) in (F), we get the desired result.
Note: Notice that this method is independent of the constraint $xyz=1$
Finally, note that equality is indeed attained when $\boxed{x=y=z=1}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1841084",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
} |
Prove: $\frac{r_a}{bc} + \frac{r_b}{ca} + \frac{r_c}{ab} = \frac{1}{r} - \frac{1}{2R}$, for circumradius R, inradius $r$, and exradii $r_x$
In $\triangle ABC$, prove:
$$\frac{r_a}{bc} + \frac{r_b}{ca} + \frac{r_c}{ab} = \frac{1}{r} - \frac{1}{2R}$$
for circumradius $R$, inradius $r$, and exradii $r_a$, $r_b$, $r_c$ in the standard arrangement.
It is known that $r_a = \sqrt{\dfrac{s\left(s-b\right)\left(s-c\right)}{s-a}}$, where $s = \dfrac12\left(a+b+c\right)$ is the semiperimeter of $\triangle ABC$. Similar formulas exist for $r_b$, $r_c$ and $r$. But how does $R$ connect with all of this?
| Using Sine Rule
and
$r_a=4R\sin\dfrac A2\cos\dfrac B2\cos\dfrac C2,$
$r=4R\sin\dfrac A2\sin\dfrac B2\sin\dfrac C2,$
$11(v)$,$12(iii)$ respectively of Properties of triangle ,
$\dfrac{r_a}{bc}=\cdots=\dfrac{\sin^2\dfrac A2}r$
Now $\cos A=1-2\sin^2\dfrac A2\iff2\sin^2\dfrac A2=1-\cos A$
Finally use $\cos A+\cos B+\cos C=1+\dfrac rR,11(vi)$ of Properties of triangle
Can you take it from here?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1842537",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
A natural number written as an arithmetic progression
Let $a$ and $n>1$ be positive integers and $$x = a+(a+1)+\cdots+(a+n-1) = \dfrac{n(2a+n-1)}{2}$$ where $x$ is also a positive integer. Prove that there exist $a,n$ if and only if $x$ has an odd divisor.
Since one of $n$ and $2a+n-1$ is odd and the other is even and both are greater than $1$, $x$ has an odd factor greater than or equal to $3$. On the other hand, for every $x$ with an odd divisor $p > 3$, there must exist $a,n$ since if $x$ doesn't contain a factor of $2$ then let $n = 2$ and we can make $2a+1$ be any odd number.
What do we do in the case that $x$ contains a factor of $2$?
| $$4\cdot 3=3+4+5$$
$$4\cdot 5=2+3+4+5+6$$
$$4\cdot 7=1+2+3+4+5+6+7$$
We are now at a roadblock. But we can continue by changing the method.
$$4\cdot 9=1+2+3+4+5+6+7+8$$
$$4\cdot 11=2+3+4+5+6+7+8+9$$
$$4\cdot 13=3+4+5+6+7+8+9+10$$
$$4\cdot 15=4+5+6+7+8+9+10+11$$
etc., etc.
Do you see how this method could be used to analyze other powers of 2?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1844023",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Each prime $p$ not $2$ or $5$ divides $F_{p-1}$ or $F_{p+1}$, where $(F_n)$ is the Fibonacci sequence with $F_1=F_2=1$
Let $\{F_n\}$ - Fibonacci sequence: $F_1=F_2=1, F_{n+1}=F_n+F_{n-1}, n\ge2$ and $p -$ prime number, $p\not =2, p \not=5$. Prove that $p|F_{p-1}$ or $p|F_{p+1}$
My work so far.
I used formula $$F_p^2-1=F_{p-1}\cdot F_{p+1}.$$
But can not solve the problem.
| An elementary proof inspired by the paper linked by Kelenner:
Lemme 1: Let $a\in \mathbb{F}_p, n\geq3 $, if the polynomial $x^2-x-1|x^{n+2}+a\pmod{p}$ then $F_{n+2}\equiv0\pmod{p}$
Proof: Calculating mod $p$: we have that $(x^2-x-1)\cdot(F_0x^n+F_1x^{n-1}+...+F_{n-1}x+F_n)=$
$=F_0x^{2n+2}+(F_2-F_1)x^{2n+1}+(F_3-F_2-F_1)x^{2n}+...+$
$+(F_{i+3}-F_{i+2}-F_{i+1})x^{2n-i}+...+(-F_{n}-F_{n+1})x-F_{n+1}=$
$=x^{2n+2}-F_{n+2}x-F_{n+1}$
So if $x^2-x-1|x^{n+2}+a$ then in particular $x^2-x-1$ divide the difference between the polynomials:
$x^2-x-1|(x^{n+2}+a-x^{2n+2}+F_{n+2}x+F_{n+1})$ so
$x^2-x-1|(F_{n+2}x+F_{n+1}+a)$
Since the polynomial on the right is of first degree it must be zero. In particular $F_{n+2}=0$ $\square$
Lemme 2: Let $p$ a prime, $p\neq5$, then either $x^2-x-1|x^{p-1}-1\pmod{p}$ either $x^2-x-1|x^{p+1}+1\pmod{p}$
*Proof:*Calculating mod $p$: Let be $\phi,\phi'$ the two roots of $x^2-x-1$ respect $\mathbb{F}_p$. If $p\neq 5$ then $\phi\neq\phi'$ and $\phi\cdot\phi'=-1$ (true in the integer case too). We have that $\phi^p,\phi'^p$ are also roots of $x^2-x-1$ since $\phi^{2p}-\phi^p-1=(\phi^2-\phi-1)^p=0$.
Then evaluating at the $p$-th power otherwise exchanges the two roots otherwise keeps them fixed. In the first case we have $\phi^p=\phi'=\phi$ so $\phi^{p+1}=\phi^{p}\cdot\phi=-1$. Then the roots of $x^2-x-1$ are roots of $x^{p+1}+1$ too so the first polyinomial divide the second one. Otherwise, in tthe second case we have $\phi^{p-1}=1$ so the roots of $x^2-x-1$ are roots of $x^{p-1}-1$ too and we conclude $\square$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1846532",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
How to solve this inequality with absolute value: $ \frac{\left|x-3\right|}{\left|x+2\right|}\le 3 $ Good morning to everyone. I have an inequality that I don't know how to solve: $$ \frac{\left|x-3\right|}{\left|x+2\right|}\le 3 $$ I tried to solve it in this way: $$ \frac{\left|x-3\right|}{\left|x+2\right|}\le 3 \rightarrow \frac{\left|x-3\right|}{\left|x+2\right|} - 3 \le 0 \rightarrow \frac{\left|x-3\right|-3\left|x+2\right|}{\left|x+2\right|}\le 0 \rightarrow $$
Case 1:
$$ \left|x+2\right| > 0 \wedge \left|x-3\right|-3\left|x+2\right| \le 0 $$
Case 1 a) $$ x+2 >0 \rightarrow x>-2$$
Case 1 b) $$ -x-2 <0 \rightarrow x>-2$$
Case 1 c) $$ \left|x-3\right|-3\left|x+2\right| \le 0 \rightarrow \left|x-3\right| \le 3\left|x+2\right| \rightarrow x-3 \le x+2 \rightarrow $$
The solution are all real numbers
Case 1 d)$$ -x+3 < x+2 \rightarrow x>\frac{1}{2} $$
Case 1 e) $$ x-3 \le -x-2 \rightarrow x \le \frac {1}{2} $$
Case 1 f) $$ x-3 < x+2 $$
The solution: all real numbers
Case 2: $$ \left|x+2\right| \le 0 \wedge \left|x-3\right|-3\left|x+2\right| > 0 $$
It'll have the same solutions because are the same inequalities. Therefore $x$ belongs to $(\frac{1}{2}, \infty)$ But my answer sheet shows that it belongs to $(-\infty, -\frac{9}{4}) \wedge (-\frac{3}{4},\infty)$
| hint: $x \neq -2$, and $|x-3| \leq 3|x+2|$, and square both sides.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1846654",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
$ \frac{5}{\cos^2x} = 7\tan x + 3 $ Prelude
I was reading a trigonometry textbook, and came across the following problem.
Problem
$ \frac{5}{\cos^2x} = 7\tan x + 3 $
Attempt
$ \frac{5}{\cos^2x} = 7\tan x + 3 $
$ \frac{5}{\cos^2x} = 7\frac{\sin x}{\cos x} + 3 $
$ 5
= \cos^2x(7\frac{\sin x}{\cos x} + 3)
= \cos x(7\sin x + 3\cos x)
= \sqrt{7^2+3^2}\cos x(\frac{7}{\sqrt{7^2+3^2}}\sin x + \frac{3}{\sqrt{7^2+3^2}}\cos x)
= \sqrt{58}\cos x(\frac{7}{\sqrt{58}}\sin x + \frac{3}{\sqrt{58}}\cos x)
= \sqrt{58}\cos x(\cos (\arctan{\frac{3}{7}})\sin x + \sin (\arctan{\frac{3}{7}}) \cos x)
= \sqrt{58}\cos x\sin(x +\arctan{\frac{3}{7}})$
$ \cos x\sin(x +\arctan{\frac{3}{7}}) = \frac{5}{\sqrt{58}} $
My attempt is the polar opposite of elegant, and I never even come to the solution.
Postscript
The book is "Trigonometry" by I. M. Gelfand.
You can find the problem under Chapter 9: Inverse Functions and Trigonometric Equations; Section 6: More complicated trigonometric equations; Problem 11.
| hint: $\dfrac{1}{\cos^2 x} = \sec^2 x = \tan^2 x + 1$. This substitution takes you back to a quadratic equation in $\tan x$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1849273",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Exact value of $\cos^2(\frac{\pi}{8})+\sin^2(\frac{15\pi}{8})?$ I tried separating it into $\cos^2(\frac{\pi}{8})+\sin^2(\frac{\pi}{8}+\frac{14\pi}{8})?$ and using the angle sum identity but it didn't help.
| Using $\cos2A=2\cos^2A-1=1-2\sin^2A$
$$2\cos^2\dfrac\pi8=1+\cos\dfrac\pi4$$
$$2\sin^2\dfrac{15\pi}4=1-\cos\dfrac{15\pi}4$$
Now $\cos\dfrac{15\pi}4=\cos\left(4\pi-\dfrac\pi4\right)=\cos\dfrac\pi4$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1850320",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Bessel Function of the first kind Could you please help me understand how to prove
$$J_{(1/2)} (x) = \sqrt{\frac2{\pi x}}\cdot \sin x$$
using,
$$J_p (x) = \sum_{(n=0)}^\infty \frac{(-1)^n}{(n! \Gamma(n+p+1) )} \left( \frac x 2 \right)^{2n+p}$$
Thank you
| The Bessel Function of the first kind and order $p$ has series representation given by
$$J_p (x) = \sum_{n=0}^\infty \frac{(-1)^n}{n! \,\Gamma(n+p+1) } \left( \frac x 2 \right)^{2n+p} $$
For $p=1/2$, we find
$$\begin{align}
J_{1/2} (x) &= \sum_{n=0}^\infty \frac{(-1)^n}{n! \,\Gamma(n+3/2) } \left( \frac x 2 \right)^{2n+1/2} \\\\
&=\sqrt{\frac {1}{2x}}\,\sum_{n=0}^\infty \frac{(-1)^n\,x^{2n+1}}{n! \,4^n\,\Gamma(n+3/2) } \tag 1
\end{align}$$
Applying recursively the functional relationship, $\Gamma(1+x)=x\Gamma(x)$, for the Gamma Function $\Gamma(n+3/2)$ reveals
$$\begin{align}
\Gamma(n+3/2)&=(n+1/2)(n-1/2)(n-3/2)\cdots (3/2)(1/2)\Gamma(1/2)\\\\
&=\frac{(2n+1)!!}{2^{n+1}}\sqrt \pi\\\\
&=\frac{(2n+1)!}{2^{2n+1}\,n!}\sqrt \pi \tag2
\end{align}$$
Substituting $(2)$ into $(1)$ yields
$$\begin{align}
J_{1/2}(x)&=\sqrt{\frac{1}{2x}}\sum_{n=0}^\infty \frac{(-1)^n\,x^{2n}}{n!\,4^n\,\frac{(2n+1)!}{2^{2n+1}\,n!}\sqrt \pi }\\\\
&=\sqrt{\frac{2}{\pi x}}\sum_{n=0}^\infty \frac{(-1)^n\,x^{2n+1}}{(2n+1)!}\\\\
&=\sqrt{\frac{2}{\pi x}}\,\sin(x)
\end{align}$$
as was to be shown!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1850941",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Prove: $\frac{x}{\sqrt{y}}+\frac{y}{\sqrt{x}}\geq \sqrt{x}+\sqrt{y}$
Prove: $$\frac{x}{\sqrt{y}}+\frac{y}{\sqrt{x}}\geq \sqrt{x}+\sqrt{y}$$ for all x, y positive
$$\frac{x}{\sqrt{y}}+\frac{y}{\sqrt{x}}-\sqrt{x}-\sqrt{y}\geq 0$$
$$\frac{x\sqrt{x}+y\sqrt{y}-x\sqrt{y}-y\sqrt{x}}{\sqrt{y}\sqrt{x}}\geq 0$$
$$\frac{x(\sqrt{x}-\sqrt{y})+y(\sqrt{y}-\sqrt{x})}{\sqrt{y}\sqrt{x}}\geq 0$$
$$\frac{x(\sqrt{x}-\sqrt{y})-y(-\sqrt{y}+\sqrt{x})}{\sqrt{y}\sqrt{x}}\geq 0$$
$$\frac{(x-y)(\sqrt{x}-\sqrt{y})}{\sqrt{y}\sqrt{x}}\geq 0$$
$$\frac{(\sqrt{x}-\sqrt{y})(\sqrt{x}+\sqrt{y})(\sqrt{x}-\sqrt{y})}{\sqrt{y}\sqrt{x}}\geq 0$$
$$\frac{(\sqrt{x}-\sqrt{y})^2(\sqrt{x}+\sqrt{y})}{\sqrt{y}\sqrt{x}}\geq 0$$
All the elements are positive and if $\sqrt{x}=\sqrt{y}$ we get $0$
Is the proof valid?
| Your proof is valid, but I suggest an alternative.
Let $a=\sqrt{x}>0$ and $b=\sqrt{y}>0$. You want to show
$$
\frac{a^2}{b}+\frac{b^2}{a}\ge a+b
$$
which becomes
$$
\frac{a^3+b^3}{ab}\ge a+b
$$
or
$$
\frac{(a+b)(a^2-ab+b^2)}{ab}\ge a+b
$$
Since $a+b>0$ and $ab>0$, we can factor out $a+b$ and remove the denominator, so the inequality becomes
$$
a^2-ab+b^2\ge ab
$$
Can you finish?
Your strategy works as well: the inequality is equivalent to
$$
\frac{a^2}{b}+\frac{b^2}{a}-a-b\ge0
$$
that becomes
$$
\frac{a^3+b^3-ab(a+b)}{ab}\ge0
$$
The numerator can be rewritten as
$$
a^3+b^3-ab(a+b)=(a+b)(a^2-ab+b^2)-ab(a+b)=
(a+b)(a-b)^2
$$
So the inequality is
$$
\frac{(a+b)(a-b)^2}{ab}\ge0
$$
which is true.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1851378",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 4
} |
Is there a nicer way to show that the series is convergent? I'd like to show that for a fixed $z\in\mathbb C\setminus\mathbb Z$ the series $$\sum_{n=1}^\infty \left| \frac{1}{z-n} + \frac{1}{n} \right|$$ is convergent.
I think, one can do it as follows. Fix some $n_0> |z|$. Then
\begin{align*}
\sum_{n=1}^\infty \left| \frac{1}{z-n} + \frac{1}{n} \right| & = \underbrace{\sum_{n=1}^{n_0-1} \left| \frac{1}{z-n} + \frac{1}{n} \right|}_{=:C} + \sum_{n=n_0}^\infty \left| \frac{1}{z-n} + \frac{1}{n} \right| \\
& = C + \sum_{n=n_0}^\infty \left| \frac{1}{z-n} + \frac{1}{n} \right| \\
& = C + \sum_{n=n_0}^\infty \left| \frac{z}{(z-n)n} \right| \\
& \leq C + |z|\sum_{n=n_0}^\infty \frac{1}{|z-n|n} \\
& \leq C + |z|\sum_{n=n_0}^\infty \frac{1}{||z|-|n||n} \\
& \leq C + |z|\sum_{n=n_0}^\infty \frac{1}{(n-|z|)n} \\
& \leq C + |z|\sum_{n=n_0}^\infty \frac{1}{n^2-|z|n} \\
& \leq C + |z|\sum_{n=n_0}^\infty \frac{1}{n^2-|z|n^2} \\
& \leq C + \frac{|z|}{1-|z|} \underbrace{\sum_{n=n_0}^\infty \frac{1}{n^2}}_{<\infty} \\
\end{align*}
That strikes me as somewhat cumbersome. Is there a nicer way? E.g. without separating the series into before and after $n_0$?
| We have
$$\left |\frac{1}{z-n} + \frac {1}{n}\right | = \frac{|z|}{n|z-n|}.$$
Divide this by $1/n^2$ to get
$$\frac{|z|}{|z/n - 1|}.$$
The last expression $\to |z|$ as $n\to \infty.$ By the limit comparison test, your series converges (absolutely).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1851934",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Prove that $4m + 1$ is a perfect square if $\{ \sqrt {n + \sqrt n}\} = \{\sqrt m\}$
Let $n,m \in \mathbb{N}-\{0\}$ so that $\{ \sqrt {n + \sqrt n}\} =
\{\sqrt m\} \tag1$ Prove that $4m + 1$ is a perfect square.
($\{x\}$ is the fractional part of $x$)
No idea how to start. Replacing $\{x\}$ by $x-[x]$ doesn't seem to help.
Trying some values for $n, m$, it seems that (1) cannot hold unless $n$ is a perfect square.
| So $$\tag1\sqrt{n+\sqrt n}-\sqrt m$$ is an integer $k\in\Bbb Z$.
Then
$\alpha_=\sqrt{n+\sqrt n}$ is a root of the polynomial
$$f(X)=X^4-2nX^2+n^2-n$$
but because $\alpha=\sqrt m+k$, it is also a root of
$$g(X)=X^2-2kX+k^2-m $$
hence also a root of
$$\tag2\begin{align}&f(X)-(X^2+2kX+3k^2+m+2n)g(X)\\&=4k(k^2+m-n)X +(-3k^4+2(n+m)k^2+(n-m)^2-n)\end{align}$$
If $4k(k^2+m-n)\ne 0$, this implies that $\sqrt m$ and $\sqrt n$ are rational, so $m,n$ are perfect squares, say $n=a^2$, $m=b^2$ with $a,b>0$, and also $n+\sqrt n=a^2+a$ is a perfect square, say $a^2+a=c^2$ with $c>0$.
But then $a=c^2-a^2=(c+a)(c-a)$, contradicting $|c+a|>|a|>0$.
If $k^2+m-n=0$, the constant term in $(2)$ must be zero, i.e.,
$$\begin{align}0&=-3k^4+2(n+m)k^2+(n-m)^2-n
\\&= -3(n-m)^2+2(n+m)(n-m)+(n-m)^2-n\\
&=(4m-1)n-4m^2 \end{align}$$
i.e., $4m^2=(4m-1)n$.
As $4$ and $m$ are coprime to $4m-1$ and $4m-1\ge 3$, we arrive at a contradiction.
Therefore, we are left only with the case $k=0$.
But then $n+\sqrt n=m$, so that $\sqrt n$ is rational, hence a perfect square, say $n=a^2$, and finally
$$4m+1=4(a^2+a)+1=(2a+1)^2. $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1856256",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Solve $3^{4x}-3^{3x}-7\times 3^{2x}+ 3^x + 6 < 0$ I'd like to solve the following inequality $3^{4x}-3^{3x}-7\times 3^{2x}+ 3^x + 6 < 0$.
I made it so that
$$y = 3^{x}$$
I then replaced $3^{x}$ with $y$:
$$y^4-y^3-7y^2+y+6<0$$
I then used Ruffini's rule to break it down a little bit more:
$$y^4-y^3-7y^2+y+6<0 \rightarrow (y+1)(y^3-2y^2-5y+6)<0$$
Though, I'm not sure using Ruffini's rule was the best approach to this problem and the result I got from isn't probably correct because $(y^3-2y^2-5y+6)$ seems to be having complex numbers as solutions. Any hints on what I could have done better?
| Notice that by rational root theorem , the roots of
$p(z)=z^4-z^3-7z^2+z+6$ are $z=-2,-1,1,3$
Hence the polynomial can be factored into
$p(z)=(z-1)(z-3)(z+2)(z+1)$
Where $z=3^x$
when $p(z)<0$
We have
$$(3^x-1)(3^x-3)(3^x+2)(3^x+1)<0$$
However $3^x$ is strictly increasing and is never negative
So we have $3^x=1 \Leftrightarrow x=0 $ and $3^x=3^1\Leftrightarrow x=1$
Hence the inequality is just $0<x<1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1856913",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Coefficient of $x^{41}$ in $(x^5 + x^6 + x^7 + x^8 + x^9)^5$
What is the coefficient of coefficient of $x^{41}$ in $(x^5 + x^6 + x^7 + x^8 + x^9)^5$?
Using summation of G.P., this is equivalent to finding the coefficient of $x^{41}$ in
$$\left(x^5 \left(\frac{1-x^5}{1-x}\right)\right)^5$$
and thus finding coefficient of $x^{16}$ in $(\frac{1-x^5}{1-x})^5$. How to proceed after this?
| Why not to apply the multinomial sum (just to study it a little bit) so we need the coefficient of $x^{16}$ in $(x^4+x^3+x^2+x+1)^5$?
Put $A=x^3+x^2+x+1$ so one has $$(x^4+A)^5=x^{20}+5x^{16}A+10x^{12}A^2+10x^8A^3+5x^4A^4+A^5$$
Hence, after to see in $A,A^2,A^3,A^4$ ($A^5$ does not participate here because its greatest exponent is $15$) the coefficients of $1,x^4,x^8,x^{12}$ respectively, the required coefficient is equal to
$$\color{red}{5\cdot1+10\cdot3+10\cdot3+5\cdot1=70}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1857004",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Closed form of recurrence relation $F(n) = 2 + F(n-1) + F(n-2)$ I was figuring out an answer to the question,
How many Boolean arrays of length $n$ could be formed if there are to
be no two falses in a row?
I could see that it boils down to a Fibonacci like equation,
$$F(n) = 2 + F(n-1) + F(n-2)$$
But I am unable to find a closed form for this recursion. May be I am following the wrong direction here.
Please help.
| I like to do these recurrences in terms of generating functions:
Let $A(x) = \sum_{n\geq1} F(n)x^n$ for some $x<1$. Now take your recurrence,
$$
F(n+2) = 2 + F(n+1) + F(n),
$$
multiply it by $x^n$ and sum over all $n$. We get
$$
\frac{A(x) - x^2 F(2) - xF(1)}{x^2} = \frac{2x}{1-x} + \frac{A(x)-xF(1)}{x} + A(x),
$$
which, after solving for $A(x)$, gives
$$
A(x) = \frac{x(x+1)}{(1-x)(1-x-x^2)} = \frac{x+2}{1-x-x^2} - \frac{2}{1-x}.
$$
It is well known that the generating functions for the Fibonacci and Lucas numbers are
$$
f(x) = \frac{x}{1-x-x^2}, \quad L(x) = \frac{2-x}{1-x-x^2}
$$
Hence,
$$
A(x) = 2f(x) + L(x) - \frac{2}{1-x}.
$$
Recall now that $F(n)$ are the coefficients of the series expansion of $A(x)$, and thus
$$
F(n) = 2f_n + L_n - 2,
$$
where $f_n$ and $L_n$ are the $n$th Fibonacci and Lucas numbers respectively.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1858087",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Find all the integer pairs $(r,s)$ that satisfy $s= (r^2 +3r +8) / (r^2 +r -2)$? I have been trying to solve this question but struggling to see where to start. Examples I've seen that works are the pairs: $(-3,2) , (4,2), (0,-4)$
| If integer $d>0$ divides both $r^2+3r+8, r^2+r-2;$
$d$ must divide $r^2+3r+8-(r^2+r-2)=2r+10$
$d$ must divide $r(2r+10)-2(r^2+3r+8)=4r-16$
$d$ must divide $2(2r+10)-(4r-16)=36$
So, $r^2+r-2=(r+2)(r-1)$ must divide $36$
Now as $r+2-(r-1)=3; 3|(r+2)\iff3|(r-1)$
Case$\#1:$ If $3|(r+2),$ let $r=3s+1\implies s(s+1)|4$
so $s(s+1)$ can not have an odd factor $>1\implies s=1,-2\implies r=?,?$
Case$\#2:$ If $3\nmid(r+2),(r+2)(r-1)$ must divide $4\implies r=2,0$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1858144",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Summation of a convergent series I have the following problem:
So I start as follows:
$B_{2}M_{2}=\frac{1}{\sqrt{3}}$ and I realize that $B_{2}M_{2} = A_{2}B_{2}$, so $B_{3}M_{3} = \frac{1}{\sqrt{3}}^{2}$. Next, I compute $A_{1}B_{2}=A_{1}B_{1}-B_{2}M_{2} = 1-\frac{1}{\sqrt{3}}$. For $A_{2}B_{3}$, we can do something similar, namely: $A_{2}B_{3} = A_{2}B_{2}-B_{3}M_{3}=\frac{1}{\sqrt{3}}-(\frac{1}{\sqrt{3}})^{2}$.
We can now derive the formula for computing the area of both triangles:
$$\frac{1}{2}\sqrt{(A_{1}B_{1})^{2}+(B_{1}M_{1})^{2}}\cdot A_{1}B_{2} \sin(30)$$
Now since $A_{1}B_{1}=B_{1}M_{1}$, we can simplify it to:
$$\frac{1}{\sqrt{2}}\sqrt{(A_{1}B_{1})^{2}}\cdot A_{1}B_{2}$$
This can be written as:
$$S_{n} = \frac{1}{\sqrt{2}}\sqrt{(A_{n}B_{n})^{2}}\cdot A_{n}B_{n+1}$$
I fill in the missing pieces and I get:
$$S_{n} = \frac{1}{\sqrt{2}} \cdot \sqrt{\left(\frac{1}{3}\right)^{n-1}} \cdot \left( \left(\frac{1}{3} \right) ^{n-1} - \left( \frac{1}{3} \right)^{n} \right) $$
Now I simplify it to:
$$S_n = \sqrt{6} \cdot 3^{-1.5n}$$
Now I have a small problem, if I try to solve $\sum_{n=1}^{\infty} \sqrt{6} \cdot 3^{-1.5n}$, I know we can rewrite it to $\sqrt{6} \sum_{n=1}^{\infty} 3^{-1.5n}$, but I have no idea how to continue to get the closed form.
In addition, I was wondering, is there a faster way to get the solution?
| Hint. One may recall the evaluation of a geometric series
$$
\sum_{n=1}^\infty r^n=\frac{r}{1-r}, \quad |r|<1.
$$ Then put $r=3^{-1.5}=0.1924\cdots$ to get the answer.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1858700",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
The notion of $s(f,P)$ in Darboux sums I am trying to understand what does inf $s(f,P)$ or sup $s(f,P)$ means.
For example if I have $P=\{0,\frac{1}{4},\frac{1}{3},\frac{2}{3},\frac{8}{9},1\}$ a partition of $[0,1]$ and $f(x)=x^2$ does:
inf $s(f,P)=0*\frac{1}{4}+\frac{1}{16}*\frac{1}{12}+\frac{1}{9}*\frac{1}{3}+\frac{4}{9}*\frac{1}{3}+\frac{64}{81}*\frac{1}{9}=\frac{12979}{46656}=0.278$
Sup $s(f,P)=\frac{1}{16}*\frac{1}{4}+\frac{1}{9}*\frac{1}{12}+\frac{4}{9}*\frac{1}{3}+\frac{64}{81}*\frac{1}{3}+1*\frac{1}{9}=\frac{8515}{1552}=0.547$
If we refine the partition of the upper/lower sums, why do we then need to define upper\lower Darboux integral and not just to use the "most refined partition"?
| Note: The left hand side of OPs equations are lower and upper Darboux integrals whereas the right hand side are lower and upper Darbous sums. These are quite different things as we will see below.
Let us consider the bounded function $f(x)=x^2$ on $[0,1]$ and the partition $P=\{0,\frac{1}{4},\frac{1}{3},\frac{2}{3},\frac{8}{9},1\}$.
At first we want to calculate the lower Darboux sum $\underline{s}(f;P)$ which approximates the function $f$ from below with rectangles and the upper Darboux sum $\overline{s}(f;P)$ which approximates the function $f$ from above with rectangles with respect to a specific partition $P$.
$$ $$
Lower and upper Darboux sum
We consider a partition $P$ with $x_0=0<x_1<\ldots<,x_{n-1}<x_n=1$. With $I_k$ we denote the interval $I_k:=[x_{k-1},x_k]$ for $1\leq k \leq n$. The length of each interval $|I_k|=|x_k-x_{k-1}|$ is the width of the rectangle we will use. The height $m_k$ of the rectangle with width $I_k$ is defined as
\begin{align*}
m_k := \inf f(I_k)
\end{align*}
The lower Darboux sum $\underline{s}(f;P)$ is then defined as
\begin{align*}
\underline{s}(f;P)=\sum_{k=1}^nm_k|I_k|\tag{1}
\end{align*}
Note that in (1) we use the infimum to define the lower Darboux sum, but they are in inner part of the sum. Analogously we define the upper Darboux sum $\overline{s}(f;P)$.
We define the height $M_k$ of the rectangle with width $I_k$ as
\begin{align*}
M_k:=\sup f(I_k)
\end{align*}
and define the upper Darboux sum $\overline{s}(f;P)$ as
\begin{align*}
\overline{s}(f;P)=\sum_{k=1}^nM_k|I_k|\tag{1}
\end{align*}
Observe that foreach partition $P$ according to the construction above we obtain
\begin{align*}
\underline{s}(f;P)\leq \overline{s}(f;P)
\end{align*}
Example: We consider the special case $P=\left\{0,\frac{1}{4},\frac{1}{3},\frac{2}{3},\frac{8}{9},1\right\}$ and $f(x)=x^2$ and calculate lower and upper Darboux sums. We obtain
\begin{align*}
\underline{s}(f;P)&=\sum_{k=1}^5 m_k|I_k|\\
&=m_1|I_1|+m_2|I_2|+m_3|I_3|+m_4|I_4|+m_5|I_5|\\
&=0^2\cdot\frac{1}{4}+\left(\frac{1}{4}\right)^2\cdot\frac{1}{12}+\left(\frac{1}{3}\right)^2\cdot\frac{1}{3}
+\left(\frac{2}{3}\right)^2\cdot\frac{2}{9}+\left(\frac{8}{9}\right)^2\cdot\frac{1}{9}\\
&=\frac{10675}{46656}\doteq 0.22880
\end{align*}
\begin{align*}
\overline{s}(f;P)&=\sum_{k=1}^5 M_k|I_k|\\
&=M_1|I_1|+M_2|I_2|+M_3|I_3|+M_4|I_4|+M_5|I_5|\\
&=\left(\frac{1}{4}\right)^2\cdot\frac{1}{4}+\left(\frac{1}{3}\right)^2\cdot\frac{1}{12}+\left(\frac{2}{3}\right)^2\cdot\frac{1}{3}
+\left(\frac{8}{9}\right)^2\cdot\frac{2}{9}+1^2\cdot\frac{1}{9}\\
&=\frac{21449}{46656}\doteq 0.45973
\end{align*}
Note that in order to calculate the lower Darboux sum $\underline{s}(f;P)$ for a specific partition $P$ we do not need any $\inf$ or $\sup$ in front of it. The same holds for the upper Darboux sum with respect to a specific partition.
Lower and upper Darboux integral
Here we consider all partitions and define the lower Darboux integral $\underline{\int_0^1}f(x)\,dx$ and the upper Darboux integral $\overline{\int_0^1}f(x)\,dx$ as
\begin{align*}
\underline{\int_0^1}f(x)\,dx:=\sup_{P} \underline{s}(f;P)\qquad \qquad\overline{\int_0^1}f(x)\,dx:= \inf_{P}\overline{s}(f;P)
\end{align*}
The lower Darboux integral is the supremum of lower Darboux sums over all partitions. Analogously is the upper Darboux integral the infimum of upper Darboux sums over all partitions.
We note that according to the construction lower and upper Darboux integral fulfil
\begin{align*}
\underline{\int_0^1}f(x)\,dx\leq\overline{\int_0^1}f(x)\,dx
\end{align*}
but equality is not always given as we can see in the next example.
Example: We consider the nowhere continuous Dirichlet function $D(x)$ on $[0,1]$ defined as
\begin{align*}
D(x)=
\begin{cases}
1&\qquad x\in [0,1]\cap\mathbb{Q}\\
0&\qquad x\in [0,1]\setminus \mathbb{Q}
\end{cases}
\end{align*}
then
\begin{align*}
0=\underline{\int_0^1}D(x)\,dx\ne\overline{\int_0^1}D(x)\,dx=1
\end{align*}
So, even if lower and upper Darboux integral exist, we cannot necessarily define them as the Darboux integral, since they need not be equal. But if they are equal, then ...
Darboux integral
Let $f$ be a bounded function defined on $[a,b]$. If the lower Darboux integral $\underline{\int_0^1}f(x)\,dx$ and the upper Darboux integral $\overline{\int_0^1}f(x)\,dx$ exist and they are equal we define them as the Darboux Integral $\operatorname{D-}\int_0^1 f(x)\,dx$.
\begin{align*}
\operatorname{D-}\int_0^1 f(x)\,dx:=\underline{\int_0^1}f(x)\,dx
\end{align*}
Note: It can be shown, that the Darboux integral and the Riemann integral coincide. Each Darboux integrable function is Riemann integrable and vice versa. Both integrals give the same value.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1859003",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Prove that $\cos (5A) = 16 \cos^5 (A) - 20 \cos^3 (A) + 5 \cos (A)$
Prove the given trigonometric identity
$$\cos (5A) = 16 \cos^5 (A) - 20 \cos^3 (A) + 5 \cos (A)$$
My attempt
L.H.S.$=\cos5A$
$$\cos(A+4A)$$
$$\cos A\cos4A-\sin A\sin4A$$
Now how should I move further?
| Here’s a “fun” way.
Recall the standard formulas for $\cos(A+B)$ and $\sin(A)\sin(B)$. We can find:
\begin{align*}
\cos(nx+x) &= \cos(nx) \cos(x) - \sin(nx) \sin(x) \\
\cos(nx+x) &= \cos(nx) \cos(x) - \tfrac 12 ( \cos(nx-x) - \cos(nx+x) ) \\
2 \cos(nx+x) &= 2 \cos(nx) \cos(x) - \cos(nx-x) + \cos(nx+x) \\
\color{blue}{\cos((n+1)x)} &= 2 \color{blue}{\cos(nx)} \cos(x) - \color{blue}{\cos((n-1)x)}
\end{align*}
This is cool: if we know $\cos((n-1)x)$ and $\cos(nx)$, we can easily compute $\cos((n+1)x)$ using this formula. We can write down $\cos(0x)=1$ and $\cos(1x)=\cos(x)$, and then repeatedly apply the formula above to get the next term in the sequence $a_n = \cos(nx)$.
If we now substitute $t=\cos(x)$, then our sequence is just:
\begin{align*}
a_0&=1 \\
a_1&=t \\
a_n&=2t \cdot a_{n-1} - a_{n-2}
\end{align*}
Calculate up to $a_5$ and you have your answer.
…What’s that? You’re allergic to polynomial multiplication? Say no more. I got you covered.
This is a sequence of polynomials (Chebyshev polynomials of the first kind) over $t = \cos(x)$.
A simple application of induction shows us that the $n$-th polynomial has degree $n$. Another shows us that each $a_{2k}$ is even, and each $a_{2k+1}$ is odd. We conclude that
$$\cos(5x) = a \cos^5 x + b \cos^3 x + c \cos x.$$
Let's sample both sides thrice:
\begin{align*}
1 = \cos(5 \cdot 0) &= a \cos^5 0 + b \cos^3 0 + c \cos 0 = a + b + c \\
-2^{-1/2} = \cos(5 \cdot \tfrac \pi 4) &= a \cos^5 \tfrac \pi 4 + b \cos^3 \tfrac \pi 4 + c \cos \tfrac \pi 4 = 2^{-1/2} \left( a/4 + b/2 + c \right) \\
\tfrac 12 = \cos(5 \cdot \tfrac \pi 3) &= a \cos^5 \tfrac \pi 3 + b \cos^3 \tfrac \pi 3 + c \cos \tfrac \pi 3 = a/32 + b/8 + c/2
\end{align*}
In other words:
\begin{cases}
a + b + c &=1 \\
a + 2b + 4c &=-4 \\
a + 4b + 16c &=16
\end{cases}
Solving the system, we find $(a,b,c)=(16,-20,5)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1859349",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 3
} |
Find $\int\frac1{\sqrt{1+x^4}}\,dx$ $$ \text{Find } \int \frac 1 {\sqrt{1+x^4}} \, dx$$
Let $x^2=\tan u$
$\implies 2x \,dx=\sec^2 u \,du$
$\implies dx=\dfrac{\sec^2 u}{2\sqrt{\tan u}}\,du$
$$= \int \frac{\sec^2 u}{2\sec u\sqrt{\tan u}} \, du $$
$$= \int \frac{\sec u}{2\sqrt{\tan u}} \, du $$
I am unsure how to continue..
| For any real number of $x$ ,
When $|x|\leq1$ ,
$\int\dfrac{1}{\sqrt{1+x^4}}dx$
$=\int\sum\limits_{n=0}^\infty\dfrac{(-1)^n(2n)!x^{4n}}{4^n(n!)^2}dx$
$=\sum\limits_{n=0}^\infty\dfrac{(-1)^n(2n)!x^{4n+1}}{4^n(n!)^2(4n+1)}+C$
When $|x|\geq1$ ,
$\int\dfrac{1}{\sqrt{1+x^4}}dx$
$=\int\dfrac{1}{x^2\sqrt{1+\dfrac{1}{x^4}}}dx$
$=\int\sum\limits_{n=0}^\infty\dfrac{(-1)^n(2n)!x^{-4n-2}}{4^n(n!)^2}dx$
$=\sum\limits_{n=0}^\infty\dfrac{(-1)^n(2n)!x^{-4n-1}}{4^n(n!)^2(-4n-1)}+C$
$=\sum\limits_{n=0}^\infty\dfrac{(-1)^{n+1}(2n)!}{4^n(n!)^2(4n+1)x^{4n+1}}+C$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1860230",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 1
} |
Prove that $\cos\frac {2\pi}{7}+ \cos\frac {4\pi}{7}+ \cos\frac {8\pi}{7}=-\frac{1}{2}$
Prove that
$$\cos\frac {2\pi}{7}+ \cos\frac {4\pi}{7}+ \cos\frac {8\pi}{7}=-\frac{1}{2}$$
My attempt
\begin{align}
\text{LHS}&=\cos\frac{2\pi}7+\cos\frac{4\pi}7+\cos\frac{8\pi}7\\
&=-2\cos\frac{4\pi}7\cos\frac\pi7+2\cos^2\frac{4\pi}7-1\\
&=-2\cos\frac{4\pi}7\left(\cos\frac\pi7-\cos\frac{4\pi}7\right)-1
\end{align}
Now, please help me to complete the proof.
| Here a solution solving a cubic equation in order to extend a bit the statement of the post.
$$\cos x+\cos 2x+\cos 4x=\cos x+(2\cos 3x\cos x)=\cos x(1+2\cos 3x)$$
When $x=\frac{2\pi}{7}$ one has $$\cos 3x=\cos(\pi-\frac{\pi}{7})=-\cos(\frac{\pi}{7})$$ so the equality becomes $$\cos(\frac{2\pi}{7})(1-2\cos(\frac{\pi}{7}))=-\frac 12\\(2\cos^2(\frac{\pi}{7})-1)(1-2\cos(\frac{\pi}{7}))=-\frac 12$$ Puting now $X=\cos x$ we get the equation $$8X^3-4X^2-4X+1=0$$ The roots are
$$X_1=-0.623489801819=\cos(\frac{5\pi}{7})\\X_2=0.222520933956=\cos(\frac{3\pi}{7})\\\color{red}{X_3=0.900968867902=\cos(\frac{\pi}{7})}$$ It follows that we also could have put, instead of the given equality $\cos\frac {2\pi}{7}+ \cos\frac {4\pi}{7}+ \cos\frac {8\pi}{7}=-\frac{1}{2}$ the two following ones
$$\cos\frac {5\pi}{7}+ \cos\frac {10\pi}{7}+ \cos\frac {20\pi}{7}=-\frac{1}{2}\\\cos\frac {3\pi}{7}+ \cos\frac {6\pi}{7}+ \cos\frac {12\pi}{7}=-\frac{1}{2}$$
(Or even put $-X_i+n\pi; i=1,2,3$ instead of $X_i$)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1860400",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
Fully simplify $\sqrt {14 + 8\sqrt {3}}$. Please help me with this problem: Fully simplify $\sqrt{14 + 8\sqrt {3}}$.
I've tried to assume that the radical is in the form of $a+b\sqrt{3}$, so $a+b\sqrt{3}=\sqrt {14 + 8\sqrt {3}}$, then I squared both sides and got $$a^2+3b^2+2ab\sqrt{3}=14+8\sqrt{3}.$$
So I assumed that $a^2+3b^2$ must be equalto $14$, and $2ab$ must be equal to $-8$. So $ab$ must equal to $-4$. So now I know that $a$ and $b$ must be factors of $-4$. But now I am stuck because the cases $a=-1$, $b=4$, $a=1$, $b=-4$, $a=4$, $b=-1$, and $a=-4$, $b=1$ do not work because of the fact that $a^2+3b^2$ must equal to $14$. (We have that $(-1)^2+3\cdot4^2=1+3\cdot16$ which is clearly not $14$, same goes with $a=1$, $b=-4$; $4^2+3(-1)^2=16+3=19$ which is also not $14$.)
So now I am out of ideas to simplify this expression, or maybe this is already fully simplified? Please help!
| Hint:
Try to write $\;\sqrt{14+8\sqrt 3}=a+b\sqrt 3$. This means
$$(a+b\sqrt3)^2=14+8\sqrt 3.$$
This relation will be satisfied if $\;\begin{cases}a^2+3b^2=14,\\ab=4.\end{cases}$
If $a^2$ and $b^2$ are integers, there are not so many possibilities to have $a^2+3b^2=14$. Try to check if any of these possibilities also satisfies $ab=4$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1863684",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 0
} |
Find point on a line that is nearest to the origin Can you help me with this exercise?
Find the nearest point to the origin $(0,0,0)$ in the line given by the intersection of planes $x+y+z=2$ and $12x+3y+3z=12$.
The intersection of the planes is the line : $x=2/3$, $3y+3z=4$.
So I restrict the function $f(x,y,z)=x^2+y^2+z^2$ to the set $A=\{x=2/3, 3y+3z=4\}$. Let $g(y,z)=f(2/3,y,z)=4/9+y^2+z^2$.
So the problem is equivalent to find the maximum of $g(y,z)=4/9+y^2+z^2$ restricted to $h(y,z)=3y+3z-4=0$. Using Lagrange multipliers, I get
$(2y,2z)=\lambda(3,3)$,
$3y+3z=4$
By the first equation I get $y=z$, then in the second I get $6y=4$, so $y=2/3$, therefore $z=2/3$. This is why I get $x=y=z=2/3$.
Is it better now?
Thanks
| You could even solve the problem using basic calculus since minimizing the distance is the same as minimizing the square of the distance.
The constraints being $$x+y+z=2\qquad , \qquad 12x+3y+3z=12$$ take advantage of their linearity and solve these two equations for $x$ and $z$ as functions of $y$. Tou will get $x=\frac 23$ and $z=\frac 43-y$.
So $$d^2=x^2+y^2+z^2=\frac{4}{9}+y^2+\left(\frac{4}{3}-y\right)^2=2 y^2-\frac{8 y}{3}+\frac{20}{9}$$ The derivative $(d^2)'=4y-\frac 83$ cancel for $y=\frac 23$ and, using $z=\frac 43-y$, $z=\frac 23$.
Then the solution $x=y=z=\frac 23$ and $d^2=\frac 43$.
Notice that the second derivative test $(d^2)''=4y$ confirms that this is a minimum.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1865861",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
What is the problem in my computation of $\sin 18^{\circ}$? I needed to compute $\sin 18^{\circ}$.
Now, these two relations hold for every $x$:
$\cos 5x=16\cos^5x-20\cos^3x+5\cos x$
$\sin5x=16\sin^5x-20\sin^3x+5\sin x$, which can be easily proved using the multiple angle formulae.
Now, one thing to observe is : $\sin5(18^{\circ})=1$ and $\cos5(18^{\circ})=0$
So,
$$\begin{align}16\cos^518^{\circ}-20\cos^318^{\circ}+5\cos18^{\circ}&=0 \\\implies16\cos^418^{\circ}-20\cos^218^{\circ}+5&=0 \tag{1}
\end{align}$$
And,
$$\begin{align}16\sin^518^{\circ}-20\sin^3+5\sin18^{\circ}&=1\\
\implies16\sin^418^{\circ}-20\sin^218^{\circ}+5&=\dfrac{1}{\sin18^{\circ}} \tag{2}
\end{align}$$
So, before moving forward with my computation, I would like to ask: Is there any problem in these two equations?
Now, equation (1) consists of squares of cosines, which clearly means that (1) can be represented consisting of sines like this:
$$\begin{align}16\cos^418^{\circ}-20\cos^218^{\circ}+5 &=0\\
\implies16(1-\sin^218^{\circ})^2 -20(1-\sin^218^{\circ})+5&=0 \\
\implies 16\sin^418^{\circ}-12\sin^218+1&=0
\end{align}$$
Now, using the above equation, we can easily deduce that $\sin^218=\dfrac{12\pm4\sqrt5}{32}=\dfrac{3\pm\sqrt5}{8}$.
Now, if I put this value in equation (2), I get something like this:
$\dfrac{1}{\sin18^{\circ}}=16\dfrac{(3\pm\sqrt5)^2}{64}-20\dfrac{(3\pm\sqrt5)}{8}+5$, which reduces to $1\mp\sqrt5$, which upon solving does not give me the correct value for $\sin18^{\circ}$. So, what mistake am I doing here?
Edit: By solving, I mean that the actual value of $\sin18^{\circ}=\dfrac{\sqrt5-1}{4}$, but in my computation, the negative $1$ and the denominator $4$ is missing.
| Your expression got reduced to $1\mp\sqrt5$. But remember, this expression is the value of $\dfrac{1}{\sin18^{\circ}}$. Since $\sin18^{\circ}$ is not negative, it's reciprocal cannot be negative. Thus, we discard the value $1-\sqrt{5}$. Thus, the correct value is -
$\dfrac{1}{\sin18^{\circ}} = 1+\sqrt{5}$
$\sin18^{\circ} = \dfrac{1}{1+\sqrt{5}} = \dfrac{1}{\sqrt{5}+1}*\dfrac{\sqrt{5}-1}{\sqrt{5}-1}$
$\sin18^{\circ} = \dfrac{\sqrt{5}-1}{4}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1866663",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Solve the equation $(x+y)^2 + 3x + y + 1=z^2$ over positive integers.
Solve the equation $(x+y)^2 + 3x + y + 1=z^2$ where $x, y, z \in
\mathbb{N}$
I've found some solutions, like $(0, 0, 1), (1, 1, 3)$ and, more general, $x=k,y=k,z=2k+1$. No idea how to prove or disprove there is no other solutions.
| Write this as
$$3x + y + 1 = z^2 - (x+y)^2 = (z-x-y)(z+x+y)$$
For any integers $a,b$, we might look for a solution with
$$ \eqalign{a &= z - x - y\cr
b &= z + x + y\cr
ab &= 3x + y + 1\cr} $$
Solving this system for $x,y,z$:
$$ \eqalign{x &= \frac{ab}{2} + \frac{a}{4} - \frac{b}{4} - \frac{1}{2}\cr
y &= \frac{-ab}{2} - \frac{3a}{4} + \frac{3b}{4} + \frac{1}{2}\cr
z &= \frac{b}{2} + \frac{a}{2}\cr}$$
In order for $x, y, z \ge 0$, we need $b > 0$ with
$$ \frac{b+2}{2b+1} \le a \le \frac{3b+2}{2b+3} $$
The only integer that fits is $a = 1$. Then for $x$ to be an integer we need $b \equiv 1 \mod 4$. The result is that all the positive integer solutions are the ones you found.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1867781",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Show $(-1)^{n}\ln\left[ \frac{n(n+2)}{n^2-n+1} \right]=3\frac{(-1)^{n}}{n}+\mathcal{O}\left( \frac{1}{n^2}\right) $
I would like to show that :
$$(-1)^{n}\ln\left[ \dfrac{n(n+2)}{n^2-n+1} \right]=3\dfrac{(-1)^{n}}{n}+\mathcal{O}\left( \dfrac{1}{n^2}\right) $$
by starting from the left side and get the right side
My Proof:
Note that :
*
*$$\ln\left(1+x \right)\underset{ \overset { n \rightarrow +\infty } {} } {=}x+\mathcal{O}\left(x^{2} \right)$$
*$$\dfrac{n(n+2)}{n^2-n+1}=1+\dfrac{3n-1}{n^2-n+1} $$
*$$\dfrac{3n-1}{n^2-n+1}\underset{ \overset { n \rightarrow +\infty } {} } {\sim}\dfrac{3}{n}\underset{ \overset { n \rightarrow +\infty } {} } {\longrightarrow}0$$
\begin{align*}
(-1)^{n}\ln\left[ \dfrac{n(n+2)}{n^2-n+1} \right]&=(-1)^{n}\ln\left[ 1+\dfrac{3n-1}{n^2-n+1} \right] \\
&\underset{ \overset { n \rightarrow +\infty } {} } {\sim} (-1)^{n}\ln\left[ 1+\dfrac{3}{n} \right] \\
&= (-1)^{n}\left(\dfrac{3}{n}+\mathcal{O}\left( \dfrac{1}{n^{2}}\right) \right) \\
&=3\dfrac{(-1)^{n}}{n}+\mathcal{O}\left( \dfrac{1}{n^2}\right)
\end{align*}
$$\fbox{$(-1)^{n}\ln\left[ \dfrac{n(n+2)}{n^2-n+1} \right]=3\dfrac{(-1)^{n}}{n}+\mathcal{O}\left( \dfrac{1}{n^2}\right) $$
$}$$
*
*Is my proof correct ?
| Your proof is correct. But going along your steps, I would rather write
$$
\begin{align}
\frac{n(n+2)}{n^2-n+1}=\frac{1+\frac2{n}}{1-\frac1n+\frac1{n^2}}=\left(1+\frac2{n}\right)\left(1+\frac1n+O\left(\frac1{n^2}\right)\right)=1+\frac3n+O\left(\frac1{n^2}\right)
\end{align}
$$ giving
$$
\ln\left[ \dfrac{n(n+2)}{n^2-n+1} \right]=\frac3n+O\left(\frac1{n^2}\right)
$$ and
$$
(-1)^n\ln\left[ \dfrac{n(n+2)}{n^2-n+1} \right]=3\frac{(-1)^{n}}n+O\left(\frac1{n^2}\right).
$$ I think it is clearer to avoid using the symbol $\sim$ within your equalities.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1868302",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
The cubic equation $x^3-5x^2+6x-3 = 0$ has solutions $\alpha$, $\beta$ and $\gamma$. The cubic equation $x^3-5x^2+6x-3 = 0$ has solutions $\alpha$, $\beta$ and $\gamma$.
Find the value of $$\frac{1}{\alpha^2}+\frac{1}{\beta^2}+\frac{1}{\gamma^2}$$
| Given
$$ x^3-5x^2+6x-3 = (x-\alpha)(x-\beta)(x-\gamma)\tag{1} $$
we have $\alpha\beta\gamma=3$, hence:
$$ \left(1-\frac{x}{\alpha}\right)\left(1-\frac{x}{\beta}\right)\left(1-\frac{x}{\gamma}\right) = 1-2x+\frac{5}{3}x^2-\frac{1}{3}x^3 \tag{2}$$
and by switching to logarithms:
$$ -\sum_{n\geq 1}\sum_{\xi\in\{\alpha,\beta,\gamma\}}\frac{x^n}{n\xi^n} = \log\left(1-2x+\frac{5}{3}x^2-\frac{1}{3}x^3\right)\tag{3} $$
hence:
$$ \frac{1}{\alpha^2}+\frac{1}{\beta^2}+\frac{1}{\gamma^2}=-\left.\frac{d^2}{dx^2}\log\left(1-2x+\frac{5}{3}x^2-\frac{1}{3}x^3\right)\right|_{x=0}=\color{red}{\frac{2}{3}}.\tag{4}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1870694",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
Prove that ${2^n-1\choose k}$ and ${2^n-k\choose k}$ ar always odd. How can I prove that ${2^n-1\choose k}$ and ${2^n-k\choose k}$ always returns odd numbers? It is possible to prove this by congruence?
by the way : $0 \leq k \leq (2^n-1)$
| For a completely different approach, you can prove it by induction on $n$ if you first prove, also by induction on $n$, that $\binom{2^n}k$ is odd if and only if $k=0$ or $k=2^n$. Assume this result for $n$. By Vandermonde’s identity we have
$$\binom{2^{n+1}}k=\sum_{\ell=0}^{2^n}\binom{2^n}\ell\binom{2^n}{k-\ell}\;,\tag{1}$$
where by symmetry we may assume that $k\le 2^n$. The $\ell$ term in the summation in $(1)$ is odd if and only if $\binom{2^n}\ell$ and $\binom{2^n}{k-\ell}$ are both odd, which by the induction hypothesis is the case if and only if $\ell=k=0$, $\ell=0$ and $k=2^n$, or $\ell=k=2^n$. Thus, $\binom{2^{n+1}}k$ is even if $0<k<2^n$,
$$\binom{2^{n+1}}{2^n}\equiv\binom{2^n}0\binom{2^n}{2^n}+\binom{2^n}{2^n}\binom{2^n}0\equiv 0\pmod2\;,$$
and
$$\binom{2^{n+1}}0\equiv\binom{2^n}0^2+\binom{2^n}0\binom{2^n}{2^n}+\binom{2^n}{2^n}\binom{2^n}0\equiv 1\pmod2\;,$$
as desired.
Now assume the actual desired result for $n$. By Vandermonde’s identity we have
$$\binom{2^{n+1}-1}k=\sum_{\ell=0}^{2^n}\binom{2^n}\ell\binom{2^n-1}{k-\ell}\;,$$
where by symmetry we may assume that $k<2^n$. By the induction hypothesis $\binom{2^n-1}{k-\ell}$ is odd for each $\ell\le k$, so
$$\binom{2^n}\ell\binom{2^n-1}{k-\ell}\equiv\binom{2^n}\ell\pmod2$$
for each $\ell\le k$, and
$$\binom{2^{n+1}-1}k\equiv\sum_{\ell=0}^k\binom{2^n}\ell\pmod2\;.\tag{2}$$
From the preliminary result we know that $\binom{2^n}\ell$ is even for $1\le\ell\le k$ (since $k<2^n$), so the only odd term in the summation in $(2)$ is $\binom{2^n}0=1$, and $\binom{2^{n+1}-1}k$ is odd.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1870782",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 7,
"answer_id": 3
} |
Prove that $\alpha^n+\beta^n+\gamma^n \equiv 4^n+5^n+(-6)^n \pmod{17}$
Let $\alpha,\beta,\gamma$ be the roots of $x^3-3x^2+1 = 0$. Prove that $$\alpha^n+\beta^n+\gamma^n \equiv 4^n+5^n+(-6)^n \pmod{17}$$ where $n$ is an integer.
It sort of makes sense why they are congruent since we can say $x^3-3x^2+1 \equiv (x-4)(x-5)(x+6) \pmod{17}$, but since $\alpha,\beta,\gamma$ aren't integers, how do we prove this?
| Hint $\ $ Both sides satisfy the associated recurrence $\,f_{n+3} = \cdots$ with same initial conditions, thus they are equal for all $\,n\,$ by a trivial induction (i.e. uniqueness theorem for recurrences).
To obtain the associated recurrence sum the following for $\,a\,$ over all $\,3\,$ roots.
$$ a^{n+3}-3a^{n+2} +a^n =\, a^n(a^3-3a^2+1) = 0,\ \ {\rm for\ any\ root}\ \ a$$
yielding $\ \ \alpha^{n+3} +\beta^{n+3} + \gamma^{n+3} - 3 (\alpha^{n+2} +\beta^{n+2} + \gamma^{n+2}) + (\alpha^{n} +\beta^{n} + \gamma^{n}) \ =\ 0$
therefore $\,\ f_{n+3} - 3f_{n+2} + f_n = 0,\ $ for $\ f_n = \alpha^{n} +\beta^{n} + \gamma^{n}$
TIP $ $ note that $\,4,5, -6\,$ are the roots of the polynomial mod $17.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1875266",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
} |
Prove that if $ a,b,c > 0 $, then $ [(1 + a) (1 + b) (1 + c)]^{7} > 7^{7} (a^{4} b^{4} c^{4}) $.
Problem. Prove that if $ a,b,c > 0 $, then $ [(1 + a) (1 + b) (1 + c)]^{7} > 7^{7} (a^{4} b^{4} c^{4}) $.
I don’t know how to solve this problem... What I can think of is to just simplify this inequality:
$$
\left[ \frac{(1 + a) (1 + b) (1 + c)}{7} \right]^{7} > a^{4} b^{4} c^{4}.
$$
How can I proceed with solving this problem?
Note: This is a question of sequence and series, specifically AM-GM-HM inequality...
| Consider the function $f(a)=\frac{(1+a)^7}{a^4}$.
We must prove that $f(a)\geq 7^{7/3}$ for all $a>0$.
to minimize this function we derive.
$f'(a)=\frac{(a+1)^6(3a-4)}{a^5}$.
So the global minimum for $f$ in $(0,\infty)$ is $f(\frac{4}{3})=\frac{(7/3)^7}{(4/3)^4}=\frac{7^7}{3^34^4}$
So we just have to prove $3^34^4\leq 7^{14/3}$:
Notice $3^34^4=27\times 256=6912$
Notice $7^{14/3}= 7^{4}\times 7^{2/3}\geq2401\times 3=7203$.
(To see $7^{2/3}\geq 3$ notice $3^3\leq 49$)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1876539",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 1
} |
A series $10^{12} + 10^7 - 45\sum_{k=1}^{999}\csc^4\frac{k\pi}{1000}.$ There's a math clock with formulas for each of $1,\ldots,12$, most of which are easy. Number 11, however, intrigues me: $$10^{12} + 10^7 - 45\sum_{k=1}^{999}\csc^4\frac{k\pi}{1000}.$$
Wolfram Alpha agrees the answer is (around) 11. How does one prove this? How does one come up with this?
| Like Trig sum: $\tan ^21^\circ+\tan ^22^\circ+...+\tan^2 89^\circ = ?$,
the roots of
$$\binom{2n}{2n-1}u^{n-1}-\binom{2n}{2n-3}u^{n-2}+\cdots+\binom{2n}3u-\binom{2n}1=0$$ are $\tan^2\dfrac{k\pi}{2n}$ where $0<k<n$
the roots of
$$\binom{2n}1v^n-\binom{2n}3v^{n-1}+\cdots+\binom{2n}{2n-3}v^2-\binom{2n}{2n-1}v=0$$ are $\cot^2\dfrac{k\pi}{2n}$ where $0<k<n$
Now as $\csc\dfrac{k\pi}{1000}=\csc\left(\pi-\dfrac{k\pi}{1000}\right)=\csc\dfrac{(1000-k)}{1000}\pi$
$$\sum_{k=1}^{999}\csc^4\frac{k\pi}{1000}=\csc^4\frac{500\pi}{1000}+2\sum_{k=1}^{499}\csc^4\frac{k\pi}{1000}$$
Let $w=v+1\iff v=w-1$
$$0=\binom{2n}1(w-1)^n-\binom{2n}3(w-1)^{n-1}+\cdots+\binom{2n}{2n-3}(w-1)^2-\binom{2n}{2n-1}(w-1)$$
$$\iff\binom{2n}1w^n-w^{n-1}\left(\binom{2n}1\cdot\binom n1+\binom{2n}3\right)+w^{n-2}\left(\binom{2n}1\cdot\binom n2+\binom{2n}3\cdot\binom{n-1}1+\binom{2n}5\right)+\cdots=0$$
whose root are $w=1+\cot^2\dfrac{k\pi}{2n}$ where $0<k<n$
Now use $\sum_{r=1}^m a_r^2=(\sum_{r=1}^m a_r)^2-2\sum_{r,s=1;r>s}^m a_ra_s$
Here $2n=1000$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1877014",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Evaluate $\int\left(\sqrt{4-x^2}+x\right)dx$ I'm looking for simple way to solve
$$\int\left(\sqrt{4-x^2}+x\right) \, dx$$
I tried substitute $x=2\sin u$ and then
$$\cdots =\frac{x^2}{2}+\frac 1 2x\sqrt{4-x^2}+2\arcsin (x/2)+c$$
I'm looking for other solution please
| Another way forward is to integrate by parts with $u=\sqrt{4-x^2}$ and $v=x$. Then, we have
$$\begin{align}
\int \sqrt{4-x^2}\,dx&=x\sqrt{4-x^2}+\int \frac{x^2}{\sqrt{4-x^2}}\,dx\\\\
&=x\sqrt{4-x^2}+\int \frac{x^2-4+4}{\sqrt{4-x^2}}\,dx\\\\
&=x\sqrt{4-x^2}-\int \sqrt{4-x^2}\,dx+4\int \frac{1}{\sqrt{4-x^2}}\,dx\\\\
2 \int \sqrt{4-x^2}\,dx&=x\sqrt{4-x^2}+4\int \frac{1}{\sqrt{4-x^2}}\,dx\\\\
\int \sqrt{4-x^2}\,dx&=\frac12 x\sqrt{4-x^2}+2\int \frac{1}{\sqrt{4-x^2}}\,dx\\\\
&=\frac12 x\sqrt{4-x^2}+2\int \frac{1}{\sqrt{1-(x/2)^2}}\,d(x/2)\\\\
&=\frac12 x\sqrt{4-x^2}+2\arcsin(x/2)+C
\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1878236",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Limits problem $\lim_{x\rightarrow \infty }{\sqrt{x^2+x}-x}$ The question is to find:
$$\lim_{x\rightarrow \infty }{\sqrt{x^2+x}-x}$$
Working number 1:
$$\lim_{x\rightarrow \infty }{\sqrt{x^2+x}-x}=\lim_{x\rightarrow \infty }(\sqrt{x^2+x}-x)\times\frac{\sqrt{x^2+x}+x}{\sqrt{x^2+x}+x} $$
$$\lim_{x\rightarrow \infty }\frac{(\sqrt{x^2+x})^2-x^2}{\sqrt{x^2+x}+x}=\lim_{x\rightarrow \infty }\frac{x^2+x-x^2}{\sqrt{x^2+x}+x}=\lim_{x\rightarrow \infty }\frac{x}{\sqrt{x^2+x}+x}$$
$$\lim_{x\rightarrow \infty }\frac{\frac{x}{\sqrt{x}^2}}{\sqrt{\frac{x^2}{{x^2}}+\frac{x}{{x}^2}}+\frac{x}{\sqrt{x}^2}}=\frac{1}{\sqrt{1+0}+1}=\frac{1}{2}$$
Working number 2:
$$\lim_{x\rightarrow \infty }{\sqrt{x^2+x}-x}=\lim_{x\rightarrow \infty }\sqrt{\frac{x^2}{{x^2}}+\frac{x}{{x}^2}}-\frac{x}{\sqrt{x}^2}=\sqrt{1+0}-1=0$$
The answer was $\frac{1}{2}$, but the question is:
What is wrong with working number 2?
| Your first solution is fine but the second solution is a bit problematic.
$$\sqrt{\frac{x^2}{{x^2}}+\frac{x}{{x}^2}}-\frac{x}{\sqrt{x}^2} = \frac{\sqrt{x^2+x}}{x} - 1 \neq {\sqrt{x^2+x}-x}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1878859",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Combining all three rotation matrices Is it possible to create one single matrix which contains all of these three rotation matrices?
$$R(x) = \begin{pmatrix}1 & 0 & 0 & 0 \\ 0 & \cos\alpha & -\sin\alpha & 0 \\ 0 & \sin\alpha & \cos\alpha & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix} \qquad R(y) = \begin{pmatrix}\cos\beta & 0 & -\sin\beta & 0\\ 0 & 1 & 0 & 0\\ \sin\beta & 0 & \cos\beta & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix} \\R(z) = \begin{pmatrix}\cos\gamma & -\sin\gamma & 0& 0 \\ \sin\gamma & \cos\gamma & 0& 0 \\ 0 & 0 & 1& 0 \\ 0 & 0 & 0 & 1\end{pmatrix}$$
If yes, how would this one (4x4-)matrix look like (written out)?
| As mentioned earlier, you just need to multiply these three matrices:
$$ R = R(\alpha) \cdot R(\beta) \cdot R(\gamma) = \begin{pmatrix}\cos\beta\cdot\cos\gamma & \cos\beta\cdot\sin\gamma & -\sin\beta & 0 \\ \sin\alpha\sin\beta\cos\gamma - \cos\alpha\sin\gamma & \sin\alpha\sin\beta\sin\gamma + \cos\alpha\cos\gamma &
\sin\alpha\cos\beta & 0 \\ \cos\alpha\sin\beta\cos\gamma + \sin\alpha\sin\gamma &
\cos\alpha\sin\beta\sin\gamma -\sin\alpha\cos\gamma & \cos\alpha\cos\beta &
0 \\ 0 &0 &0 &1\end{pmatrix}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1882276",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Evaluate $\sqrt[2]{2} \cdot \sqrt[4]{4}\cdot \sqrt[8]{8}\cdot \dots$ Evaluate: $$\lim_{n\to \infty }\sqrt[2]{2}\cdot \sqrt[4]{4}\cdot \sqrt[8]{8}\cdot \dots \cdot\sqrt[2^n]{2^n}$$
My attempt:First solve when $n$ is not infinity then put infinity in.
$$2^{\frac{1}{2}}\cdot 4^{\frac{1}{4}}\cdot \dots\cdot (2^n)^{\frac{1}{2^n}}$$
$$=2^{\frac{1}{2}}\cdot 2^{\frac{2}{4}}\cdot \dots\cdot 2^{\frac{n}{2^n}}$$
Now calculate the sum of the powers:
$$\frac{1}{2}+\frac{2}{4}+\frac{3}{8}+\dots+\frac{n}{2^n}$$
$$=\frac{2^{n-1}+2\cdot2^{n-2}+3\cdot2^{n-3}+\dots+n\cdot2^0}{2^n}$$
Now calculate the numerator:
$$2^0+2^1+2^2+\dots+2^{n-1}=2^n-1$$
$$+$$
$$2^0+2^1+\dots+2^{n-2}=2^{n-1}-1$$
$$+$$
$$2^0+2^1+\dots+2^{n-3}=2^{n-2}-1$$
$$+$$
$$\vdots$$
$$+$$
$$2^0=2^1-1$$
$$=2^1+2^2+2^3+\dots+2^n-n=2^{n+1}-n-1$$
Now put the numerator on the fraction:
$$\frac{2^{n+1}-n-1}{2^n}=2-\frac{n}{2^n}-\frac{1}{2^n}$$
Now we can easily find $\lim_{n \to \infty}\frac{1}{2^n}=0$
Then we just have to find $\lim_{n \to \infty }\frac{n}{2^n}$, that by graphing will easily give us the answer zero.
That gives the total answer is $4$.
But now they are two problems:
1.I cannot find $\lim_{n \to \infty }\frac{n}{2^n}$ without graghing.
2.My answer is too long.
Now I want you to help me with these problems.Thanks.
| $I=1+\frac 12+\frac 14+\frac 18+\frac 1{16}+\cdots=2$
$I^2 =1+\frac{2}{2}+\frac{3}{4}+\frac{4}{8}+\frac{5}{16}+\cdots$
$I^2 -I=\frac{1}{2}+\frac{2}{4}+\frac{3}{8}+\frac{4}{16}+\cdots=2^2-2=2$
Then
$\lim_{n\to \infty }\sqrt[2]{2}\cdot \sqrt[4]{4}\cdot \sqrt[8]{8}\cdot \dots\cdot \sqrt[2^n]{2^n}=2^{I^2-I}=2^2=4$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1883743",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "17",
"answer_count": 7,
"answer_id": 4
} |
$a_1 + a_2 + \dots a_n = 1$ find min of $a_1^2 +\frac{a_2^2}{2} + \dots + \frac{a_n^2}n.$ Given $n$ numbers $a_1$, $\cdots$ and such that $a_1$, $a_2$, $\cdots$, $a_n > 0$ and their sum is $1$, I want to find the minimum value of
$$a_1^2 + \frac{a_2^2}{2} + \cdots + \frac{a_n^2}{n}.$$
I have tried using weighted AM-GM inequality, like this:
$$\frac{a_1^2 + \frac{a_2^2}{2} + \cdots + \frac{a_n^2}{n}}{a_1 + a_2 + \cdots + a_n } \geqslant \frac{a_1^{a_1} \cdots a_n^{a_n}}{2^{a_2} \cdots n^{a_n}}$$
but was unable to make progress on the right hand side.
Is there a better way to apply AM-GM inequality? Or is there some different way altogether to solve this?
| I would go with Cauchy Schwarz inequality
$$\left(\sum_{cyc} \frac{a_i^2}i\right)(1+2+\dots + n) \geqslant \sum_{cyc} a_i = 1$$
Gives the minimum to be $\dfrac2{n(n+1)}$ directly, as equality is possible when $a_i$ is proportional to $i$.
In case you want to use AM-GM, you may sum $n$ equations of kind
$$\frac{a_i^2}i + \frac{4i}{n^2(n+1)^2} \geqslant \frac{4a_i}{n(n+1)} \implies \sum \frac{a_i^2}i \geqslant \frac4{n(n+1)} - \frac2{n(n+1)} = \frac2{n(n+1)}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1884317",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
"answer_id": 3
} |
solve $\tan{x} = \tan{3x}$ I'm asked to solve $\tan{x} = \tan{3x}$
Here's my attempt:
$$\tan{x} = \tan{3x}$$
$$\tan{x} = \tan{(x + 2x)}$$
$$\tan{x} = \frac{\tan{x} + \tan{2x}}{1-\tan{x}\tan{2x}}$$
Recall the identity:
$$\tan{2x} = \frac{2\tan{x}}{1-\tan^2{x}}$$
So then we have:
$$\tan{x} = \frac{\tan{x} + \frac{2\tan{x}}{1-\tan^2{x}}}{1-\tan{x}\frac{2\tan{x}}{1-\tan^2{x}}}$$
$$\tan{x} - \tan^2{(x)} \cdot \frac{2\tan{x}}{1-\tan^2{x}} = \tan{x} + \frac{2\tan{x}}{1-\tan^2{x}}$$
$$-\tan^2{(x)} \cdot \frac{2\tan{x}}{1-\tan^2{x}} = \frac{2\tan{x}}{1-\tan^2{x}}$$
$$-\tan^2{x} \cdot \frac{2\tan{x}}{1-\tan^2{x}} \cdot \frac{1-\tan^2{x}}{2\tan{x}} = 1$$
$$\tan^2{x} = -1$$
This does obviously not compute. Why is my way wrong and how can I go about solving it?
| Your proof is fine up to
$-\tan^2{(x)}(\frac{2\tan{x}}{1-\tan^2{x}}) = \frac{2\tan{x}}{1-\tan^2{x}}$
Then you do $-\tan^2{x} \cdot (\frac{2\tan{x}}{1-\tan^2{x}}) \cdot (\frac{1-\tan^2{x}}{2\tan{x}}) = 1$.
But you can onlty do that if $\frac{2\tan{x}}{1-\tan^2{x}} \ne 0$.
So you need to say:
"Assume $\frac{2\tan{x}}{1-\tan^2{x}}\ne 0$ then
"$-\tan^2{x} \cdot (\frac{2\tan{x}}{1-\tan^2{x}}) \cdot (\frac{1-\tan^2{x}}{2\tan{x}}) = 1$
"$-\tan^2{x} = 1$.
"But this is impossible.
"So $\frac{2\tan{x}}{1-\tan^2{x}} = 0$"
And go on from there:
"So $\tan x = \sin x/\cos x = 0$.
"So $x = k\pi$".
=====
Or you can note $tan z = tan x \iff z = x + k\pi$.
So $3x = x + k\pi$ so $2x = k\pi$ so $x = \frac k 2 \pi$. But $\tan \frac k 2 \pi; k$ odd is undefined so $x = \frac k 2 \pi; k$ even or in other words $x = \frac n \pi$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1886474",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 2
} |
Find $\lim_{n\to\infty}\frac{n^2+5n^3}{2n^3+3\sqrt{4+n^6}}$ $$\frac{n^2+5n^3}{2n^3+3\sqrt{4+n^6}}$$
$$=\frac{\frac{1}{n}+5}{2+\frac{3}{n^3}\sqrt{4+n^6}}$$
$$=\frac{\frac{1}{n}+5}{2+3\sqrt{\frac{4+n^6}{n^9}}}$$
$$=\frac{0+5}{2+0}$$
$$=\frac{5}{2}$$
...but the given answer is $1$. Where did I make a mistake??
| In your answer, the term under the radical $\sqrt{4+n^6}$ does not play a real role anymore. Since this is potentially the "most complicated term", this situation is an hint (not a proof) that something has gone wrong with your proof. The exercise as lost most of its interest to test your skill.
If you factor your expression correctly with the term in $n$ with the highest power, you get:
$$\frac{n^2+5n^3}{2n^3+3\sqrt{4+n^6}} = \frac{5n^3 \left(1+\frac{1}{5n}\right)}{2n^3+3n^3\sqrt{4/n^6+1}} = \frac{5n^3 \left(1+\frac{1}{5n}\right)}{n^3 (2+3\sqrt{4/n^6+1})}\,.$$
As for the notations, it is discouraged (even wrong) to write something like $\frac{1/n+5}{2+3/n} = \frac{5}{2}$, because it is not true. At the last stages, you should reintroduce the notion of limits, instead of writing a series of equalities.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1887718",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Trying to find an explicit sum of an infinite series: $\sum_{n=1}^{\infty} \frac{ (-1)^n }{(2n+1) 3^n }$ I have the following series
$$\sum_{n=1}^{\infty}\frac{(-1)^n }{(2n+1)3^n}$$
I am trying to find an explicit sum. I know this looks like $\arctan x = \sum_{n=0}^{\infty} \frac{ (-1)^n x^{2n+1}}{2n+1} $. I do the following
$$ \sum_{n=1}^{\infty} \frac{ (-1)^n }{(2n+1) 3^n } = \sum_{n=0}^{\infty} \frac{ (-1)^n }{(2n+1) 3^n } - 1 =\sqrt{3} \sum_{n=1}^{\infty} \frac{ (-1)^n \sqrt{1/3}^{2n+1}}{(2n+1) } - 1$$
Since $(1/3)^n = (\sqrt{1/3})^2n = \sqrt{3} (\sqrt{1/3})^{2n+1}$. Thus, the sum is
$$ \sqrt{3} \sum_{n=1}^{\infty} \frac{ (-1)^n \sqrt{1/3}^{2n+1}}{(2n+1) } - 1 = \sqrt{3} \arctan(1/\sqrt{3}) = 1 = \boxed{\frac{ \sqrt{3} \pi }{6} - 1 }$$
Is this a correct solution? Do you guys a differenti method?
| As $\log\dfrac{1+x}{1-x}=2\sum_{r=0}^\infty\dfrac{x^{2r+1}}{2r+1}$
$$ \sum_{n=0}^{\infty} \frac{ (-1)^n }{(2n+1) 3^n }=\dfrac1{i/\sqrt3}\sum_{n=0}^{\infty}\dfrac{(i/\sqrt3)^{2n+1}}{2n+1}$$
Now $$2\sum_{n=0}^{\infty}\dfrac{(i/\sqrt3)^{2n+1}}{2n+1}=\log\dfrac{1+i/\sqrt3}{1-i/\sqrt3}=\log\dfrac{\sqrt3+i}{\sqrt3-i}=\log\dfrac{e^{i\pi/6}}{e^{-i\pi/6}}=\log(e^{i\pi/3})=\dfrac{i\pi}3$$
Considering principal values.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1888910",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
} |
Equations with Radicals Let $a$, $b$, $c$, and $d$ be distinct real numbers such that
\begin{align*}
a &= \sqrt{4 + \sqrt{5 + a}}, \\
b &= \sqrt{4 - \sqrt{5 + b}}, \\
c &= \sqrt{4 + \sqrt{5 - c}}, \\
d &= \sqrt{4 - \sqrt{5 - d}}.
\end{align*}
Compute $abcd$.
I squared it to
$a^4-8a^2-a+11=0,$
$b^4-8b^2-b+11=0,$
$c^4-8c^2+c+11=0,$
$d^4-8d^2+d+11=0,$
But I'm stuck here.
Could I get a full answer instead of a hint? Thanks!
| By your calculations, $a, b, -c, -d$ are exactly the roots of $x^4 -8x^2 - x + 11$. (We know they're distinct because $a, b, c, d$ are distinct and positive.) So we can use these formulas to find that the product of the roots of this polynomial is $11$ and hence $abcd = ab(-c)(-d) = 11$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1889197",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
infinite series integration $x^2 \tan^{-1} x^3$, why can't you directly integrate the infinite series representation? We have $$(\arctan(x^3))' = \frac{3x^2}{1+x^6} = \frac{3x^2}{1-(-x^6)} = 3x^2\sum_{n=0}^{\infty} (-x^6)^n.$$
Integrating this gives $$3\sum_{n=0}^{\infty} \frac{(-1)^n x^{6n+5}}{6n+3}.$$
Apparently it's correct except for the $\frac{1}{6n+3}$ which should be $(2n+1)$, if i remember correctly. But why? With many series you can directly use the integration rules you know, so why not here?
Why should I here write out the original infinite series and then integrate part by part and then make an infinite equation out of that?
| Well,
$$\begin{align}
\int 3x^2 \sum_{n=0}^\infty (-x^6)^n \,\textrm{d}x
& = \int 3x^2 \sum_{n=0}^\infty x^{12n} \,\textrm{d}x + \int -3x^8 \sum_{n=0}^\infty x^{12n} \,\textrm{d}x \\
& = 3 \int \sum_{n=0}^\infty x^{12n + 2} \,\textrm{d}x - 3\int \sum_{n=0}^\infty x^{12n + 8} \,\textrm{d}x \\
& = 3 \sum_{n=0}^\infty \int x^{12n+2} \,\textrm{d}x - 3 \sum_{n=0}^\infty \int x^{12n+8} \,\textrm{d}x \\
& = \sum_{n=0}^\infty \frac{3x^{12n+3}}{12n+3} - \sum_{n=0}^\infty \frac{3x^{12n+9}}{12n+9} = \textrm{arctan}(x^3)\end{align}.
$$
Or even shorter:
$$\begin{align}
\int 3x^2\sum_{n \ge 0} (-x^6)^n \,\textrm{d}x & = 3\sum_{n \ge 0} \int (-x^6)^n\cdot x^2 \,\textrm{d}x = \sum_{n \ge 0} \frac{3 x^3 (-x^6)^n}{6n+3} \\ & = \sum_{n \ge 0}\frac{(-1)^n (x^3)^{2n+1}}{2n+1} = \arctan (x^3).
\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1890061",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Closed form for this $\text{I}(u)=\int_u^\infty\frac{\sqrt{x}}{x^2-2x+2}\space\text{d}x$ integral? (spot my mistake) Solving this question, I came up with this (but mathematica gives me a different answer, also when I use a value for $u$) can someone spot my mistake?
My work:
$$\text{I}(u)=\int_u^\infty\frac{\sqrt{x}}{x^2-2x+2}\space\text{d}x=\lim_{n\to\infty}\int_u^n\frac{\sqrt{x}}{x^2-2x+2}\space\text{d}x=$$
Substitute $s=\sqrt{x}$ and $\text{d}s=\frac{1}{2\sqrt{x}}\space\text{d}x$.
This gives a new lower bound $s=\sqrt{u}$ and upper bound $s=\sqrt{n}$:
$$2\lim_{n\to\infty}\int_{\sqrt{u}}^{\sqrt{n}}\frac{s^2}{s^4-2s^2+2}\space\text{d}s=2\lim_{n\to\infty}\int_{\sqrt{u}}^{\sqrt{n}}\frac{s^2}{\left(s^2-(1+i)\right)\left(s^2-(1-i)\right)}\space\text{d}s=$$
Use partial fractions, I set:
*
*$$\text{z}=\frac{1+i}{2}\to\overline{\text{z}}=\frac{1-i}{2}$$
*$$\text{q}=1+i\to\overline{\text{q}}=1-i$$
$$2\lim_{n\to\infty}\int_{\sqrt{u}}^{\sqrt{n}}\left[\frac{\text{z}}{s^2-\overline{\text{q}}}+\frac{\overline{\text{z}}}{s^2-\text{q}}\right]\space\text{d}s=$$
$$2\lim_{n\to\infty}\left[\text{z}\cdot\int_{\sqrt{u}}^{\sqrt{n}}\frac{1}{s^2-\overline{\text{q}}}\space\text{d}s+\overline{\text{z}}\cdot\int_{\sqrt{u}}^{\sqrt{n}}\frac{1}{s^2-\text{q}}\space\text{d}s\right]=$$
Substitute, for the left integral $t=\frac{s}{\sqrt{\overline{\text{q}}}}$ and $\text{d}t=\frac{1}{\sqrt{\overline{\text{q}}}}\space\text{d}s$.
This gives a new lower bound $t=\frac{\sqrt{u}}{\sqrt{\overline{\text{q}}}}$ and upper bound $t=\frac{\sqrt{n}}{\sqrt{\overline{\text{q}}}}$:
Substitute, for the right integral $v=\frac{s}{\sqrt{\text{q}}}$ and $\text{d}v=\frac{1}{\sqrt{\text{q}}}\space\text{d}s$.
This gives a new lower bound $v=\frac{\sqrt{u}}{\sqrt{\text{q}}}$ and upper bound $v=\frac{\sqrt{n}}{\sqrt{\text{q}}}$:
$$-2\lim_{n\to\infty}\left[\frac{\text{z}}{\sqrt{\overline{\text{q}}}}\cdot\int_{\frac{\sqrt{u}}{\sqrt{\text{q}}}}^{\frac{\sqrt{n}}{\sqrt{\text{q}}}}\frac{1}{1-t^2}\space\text{d}t+\frac{\overline{\text{z}}}{\sqrt{\text{q}}}\cdot\int_{\frac{\sqrt{u}}{\sqrt{\text{q}}}}^{\frac{\sqrt{n}}{\sqrt{\text{q}}}}\frac{1}{1-v^2}\space\text{d}v\right]=$$
Now, use:
$$\int\frac{1}{1-x^2}\space\text{d}x=\frac{\ln\left|\frac{x+1}{x-1}\right|}{2}+\text{C}$$
$$-2\lim_{n\to\infty}\left[\frac{\text{z}}{\sqrt{\overline{\text{q}}}}\cdot\left[\frac{\ln\left|\frac{t+1}{t-1}\right|}{2}\right]_{\frac{\sqrt{u}}{\sqrt{\text{q}}}}^{\frac{\sqrt{n}}{\sqrt{\text{q}}}}+\frac{\overline{\text{z}}}{\sqrt{\text{q}}}\cdot\left[\frac{\ln\left|\frac{v+1}{v-1}\right|}{2}\right]_{\frac{\sqrt{u}}{\sqrt{\text{q}}}}^{\frac{\sqrt{n}}{\sqrt{\text{q}}}}\right]=$$
$$-2\lim_{n\to\infty}\left[\frac{\text{z}}{2\sqrt{\overline{\text{q}}}}\cdot\left[\ln\left|\frac{t+1}{t-1}\right|\right]_{\frac{\sqrt{u}}{\sqrt{\text{q}}}}^{\frac{\sqrt{n}}{\sqrt{\text{q}}}}+\frac{\overline{\text{z}}}{2\sqrt{\text{q}}}\cdot\left[\ln\left|\frac{v+1}{v-1}\right|\right]_{\frac{\sqrt{u}}{\sqrt{\text{q}}}}^{\frac{\sqrt{n}}{\sqrt{\text{q}}}}\right]=$$
$$-\lim_{n\to\infty}\left[\frac{\text{z}}{\sqrt{\overline{\text{q}}}}\cdot\left[\ln\left|\frac{t+1}{t-1}\right|\right]_{\frac{\sqrt{u}}{\sqrt{\text{q}}}}^{\frac{\sqrt{n}}{\sqrt{\text{q}}}}+\frac{\overline{\text{z}}}{\sqrt{\text{q}}}\cdot\left[\ln\left|\frac{v+1}{v-1}\right|\right]_{\frac{\sqrt{u}}{\sqrt{\text{q}}}}^{\frac{\sqrt{n}}{\sqrt{\text{q}}}}\right]=$$
$$-\lim_{n\to\infty}\left[\frac{\text{z}}{\sqrt{\overline{\text{q}}}}\cdot\left(\ln\left|\frac{\frac{\sqrt{n}}{\sqrt{\text{q}}}+1}{\frac{\sqrt{n}}{\sqrt{\text{q}}}-1}\right|-\ln\left|\frac{\frac{\sqrt{u}}{\sqrt{\text{q}}}+1}{\frac{\sqrt{u}}{\sqrt{\text{q}}}-1}\right|\right)+\frac{\overline{\text{z}}}{\sqrt{\text{q}}}\cdot\left(\ln\left|\frac{\frac{\sqrt{n}}{\sqrt{\text{q}}}+1}{\frac{\sqrt{n}}{\sqrt{\text{q}}}-1}\right|-\ln\left|\frac{\frac{\sqrt{u}}{\sqrt{\text{q}}}+1}{\frac{\sqrt{u}}{\sqrt{\text{q}}}-1}\right|\right)\right]=$$
Set:
$$\text{s}=\frac{\text{z}}{\sqrt{\overline{\text{q}}}}\to\overline{\text{s}}=\frac{\overline{\text{z}}}{\sqrt{\text{q}}}$$
$$-\lim_{n\to\infty}\left[\text{s}\cdot\left(\ln\left|\frac{\frac{\sqrt{n}}{\sqrt{\text{q}}}+1}{\frac{\sqrt{n}}{\sqrt{\text{q}}}-1}\right|-\ln\left|\frac{\frac{\sqrt{u}}{\sqrt{\text{q}}}+1}{\frac{\sqrt{u}}{\sqrt{\text{q}}}-1}\right|\right)+\overline{\text{s}}\cdot\left(\ln\left|\frac{\frac{\sqrt{n}}{\sqrt{\text{q}}}+1}{\frac{\sqrt{n}}{\sqrt{\text{q}}}-1}\right|-\ln\left|\frac{\frac{\sqrt{u}}{\sqrt{\text{q}}}+1}{\frac{\sqrt{u}}{\sqrt{\text{q}}}-1}\right|\right)\right]=$$
Assume $u\space\wedge\space n\in\mathbb{R}^+$:
*
*$$\text{A}=\left|\frac{\frac{\sqrt{u}}{\sqrt{\overline{\text{q}}}}+1}{\frac{\sqrt{u}}{\sqrt{\overline{\text{q}}}}-1}\right|=\left|\frac{\frac{\sqrt{u}}{\sqrt{\text{q}}}+1}{\frac{\sqrt{u}}{\sqrt{\text{q}}}-1}\right|=\frac{\sqrt{\sqrt{2}+\sqrt{u}\cdot\sqrt{2(1+\sqrt{2})}+u}}{\sqrt{\sqrt{2}-\sqrt{u}\cdot\sqrt{2(1+\sqrt{2})}+u}}$$
*$$\text{B}=\left|\frac{\frac{\sqrt{n}}{\sqrt{\text{q}}}+1}{\frac{\sqrt{n}}{\sqrt{\text{q}}}-1}\right|=\left|\frac{\frac{\sqrt{n}}{\sqrt{\text{q}}}+1}{\frac{\sqrt{n}}{\sqrt{\text{q}}}-1}\right|=\frac{\sqrt{\sqrt{2}+\sqrt{n}\cdot\sqrt{2(1+\sqrt{2})}+n}}{\sqrt{\sqrt{2}-\sqrt{n}\cdot\sqrt{2(1+\sqrt{2})}+n}}$$
$$-\lim_{n\to\infty}\left[\text{s}\cdot\left(\ln\left(\text{B}\right)-\ln\left(\text{A}\right)\right)+\overline{\text{s}}\cdot\left(\ln\left(\text{B}\right)-\ln\left(\text{A}\right)\right)\right]=-\left(\text{s}+\overline{\text{s}}\right)\lim_{n\to\infty}\left[\ln\left(\text{B}\right)-\ln\left(\text{A}\right)\right]=$$
Notice:
*
*$$\lim_{n\to\infty}\frac{\sqrt{\sqrt{2}+\sqrt{n}\cdot\sqrt{2(1+\sqrt{2})}+n}}{\sqrt{\sqrt{2}-\sqrt{n}\cdot\sqrt{2(1+\sqrt{2})}+n}}=1$$
*$$\text{s}+\overline{\text{s}}=2\Re[\text{s}]=\sqrt{\frac{1}{\sqrt{2}}-\frac{1}{2}}$$
$$\left(\text{s}+\overline{\text{s}}\right)\ln\left(\text{A}\right)=\sqrt{\frac{1}{\sqrt{2}}-\frac{1}{2}}\cdot\ln\left(\text{A}\right)$$
| the indefinitiv integral is given by $$-\frac{2 \tan ^{-1}\left(\frac{\sqrt{x}}{\sqrt{-1-i}}\right)}{(-1-i)^{3/2}}-\frac{2 \tan
^{-1}\left(\frac{\sqrt{x}}{\sqrt{-1+i}}\right)}{(-1+i)^{3/2}}$$
i hope this will help
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1891853",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
factors of $(x+y)^p-x^p-y^p$ Assume $x,y\in\mathbb{N}$ and $p$ - prime, $p\ge 5$.
$5xy(x+y)(x^2+xy+y^2)$ divides $(x+y)^5-x^5-y^5$ (in fact, there is equality).
$7xy(x+y)(x^2+xy+y^2)^2$ divides $(x+y)^7-x^7-y^7$ (there is equality also).
$11xy(x+y)(x^2+xy+y^2)$ divides $(x+y)^{11}-x^{11}-y^{11}$ but $11xy(x+y)(x^2+xy+y^2)^2$ does not.
It is easy to prove that $pxy(x+y)$ divides $(x+y)^p-x^p-y^p$.
Does $pxy(x+y)(x^2+xy+y^2)$ divide $(x+y)^p-x^p-y^p$ always?
Is there a characterisation of $p$'s for which $pxy(x+y)(x^2+xy+y^2)^2$ divides $(x+y)^p-x^p-y^p$ ?
Edit:
For $p=17$ there is no divisibility by $(x^2+xy+y^2)^2$, similarily as for $p=11$.
| Solving for $x$ in $x^2+xy+y^2=0$, we obtain
$$x = \frac{-y \pm \sqrt{y^2-4y^2}}{2}=\frac{-1 \pm i\sqrt{3}}{2}y$$
Call one solution (the one with positive imaginary part) $\xi y$, the other $\xi'y$, so that $\xi^3 = \xi'^3 = 1$. Then it's easy to show that $x = \xi y$ and $x = \xi'y$ are roots of $f(x,y)=(x+y)^p-x^p-y^p$, and so $(x-\xi y)(x-\xi'y)=x^2+xy+y^2$ divides $f$.
$(x^2+xy+y^2)^2$ will divide $f$ precisely when $x = \xi y, \xi' y$ are multiple roots of $f$, i.e. when they are roots of $\frac{\partial f}{\partial x} = p((x+y)^{p-1}-x^{p-1})$. Letting $x=\xi y$, we have
$$0 = (\xi y + y)^{p-1}-(\xi y)^{p-1}$$
$$0 = (1+\xi)^{p-1}-\xi^{p-1}$$
$$\xi^{p-1}=(1+\xi)^{p-1}$$
$$\xi = \zeta_{p-1}(1+\xi)$$
($\zeta_{p-1}$ is some $(p-1)^{th}$ root of unity, not necessarily primitive)
$$\zeta_{p-1} = \frac{\xi}{1+\xi} = \frac{1}{2}+\frac{\sqrt{3}}{2}i$$
That last number is a complex cube root of $-1$, and a primitive cube root of $6$. Had we used $\xi'$ here, we would have gotten the other complex cube root of $-1$. In order for $\zeta_{p-1}$ to be a primitive sixth roots of $1$, we need $p-1$ to be a multiple of six, and so $p \equiv 1$(mod $6$). This is also sufficient.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1892857",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Shortest distance between z axis and the line $x+y+2z=3, 2x+3y+4z+4=0$ z axis is : $x=0=y$
$x + B(y) = 0$ --> (1)
$x+y+2z-3+ A (2x+3y+4z+4 ) = 0$ --> (2)
These are two planes constructed through the two lines. We find the parallel planes so that it is easier to calculate the distance after that.
Therefore, for planes to be parallel :
$\frac{1+2A}{1} $ = $\frac{1+3A}{B}$ = $\frac{2+4A}{0}$
But now no values of A,B will satisfy this.
How to proceed further ?
| The intersection of the given planes is the line
\begin{align*}
\mathbf{r} = 13\mathbf{i} - 10\mathbf{j} + t(-2\mathbf{i}+\mathbf{k})
\end{align*}
The vector perpendicular to $\mathbf{k}$ (vector along $z$ axis) and the above line is $\mathbf{k} \times (-2\mathbf{i}+\mathbf{k}) = -2\mathbf{j}$. Hence the shortest distance is given by the projection of the line joining $(0,0,0)$ and $(13, -10, 0)$ on the vector $\mathbf{j}$ and this is 10.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1892958",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Is the choice for substitution correct for nonlinear second order differential equation? I am trying to understand why substitution destroys the information in original problem. I have the following second order nonlinear differential equation,
\begin{equation}
\frac{d^2 Y}{d x^2} + Y\sqrt{1 - Y^2} =0
\label{eq:xdef}
\end{equation}
for Y(0) = C and Y(0)' = 0
When I plot the above equation using numerical solver, I get a cosine function. I am trying to get rid of squareroot function in nonlinear equation. I am trying to avoid using trigonometric functions or exponential function in substitution. Therefore my attempt in substitution is to use:
\begin{equation}
z=\sqrt{1 - Y^2}
\label{}
\end{equation}
then the derivatives of the above substitution is,
\begin{equation}
\frac{dY}{dx}=-z\frac{dz}{dx}\frac{1}{Y}
\label{}
\end{equation}
\begin{equation}
\frac{d^2 Y}{d x^2}=(-(\frac{dY}{dx})^2-(\frac{dz}{dx})^2-z\frac{d^2 z}{d x^2})\frac{1}{Y}
\label{}
\end{equation}
then the nonlinear ODE simplifies to
\begin{equation}
\frac{d^2 z}{d x^2}+(\frac{dz}{dx})^2\frac{1}{z(1-z^2)}-(1-z^2)=0
\label{}
\end{equation}
\begin{equation}
z(0)=\sqrt{1 - C^2}
\label{}
\end{equation}
and z(0)' = 0
I think that the above results are correct. However when I try to plot it using numerical solver, instead a cosine curve shifted upward by constant 1 I get either error message (singularity) or some other curve (nonoscillating). Does that means that my substitution is invalid because information is lost? Or is my approach to plot it is incorrect? Is there a better substitute that will eliminate squareroot without destroying the information?
| You seem to have an error in your substitution formulas. Your first and second derivative formulas are correct. But the final equation in variable $z$ is not right. The right equation after substitution is
$$\frac{d^2z}{dx^2} + \frac{1}{z(1-z^2)}\left(\frac{dz}{dx}\right)^2 + z^2-1 = 0.$$
Is this substitution important or you need to find information on how to solve the equation? Because the original equation is a Hamiltonian system with one degree of freedom and as such is integrable, i.e. it is (more or less) explicitly solvable. One can write the system as a planar system (a system of two equations but involving only first derivatives)
$$\begin{align}
\frac{dY}{dx} &= V\\
\frac{dV}{dx} &= -Y\sqrt{1-Y^2}
\end{align}$$
Now, one way to go about this is to write the system as
$$\frac{dY}{V} = dx = - \frac{dV}{Y\sqrt{1-Y^2}}$$
$$\frac{dY}{V} = - \frac{dV}{Y\sqrt{1-Y^2}}$$ which after cross-multiplying leads to
$$Y\sqrt{1-Y^2}dY = -VdV$$
$$VdV + Y\sqrt{1-Y^2}dY = 0.$$
Integrating the last identity on both sides leads to
$$H(Y,V) = \frac{V^2}{2} + \int Y\sqrt{1-Y^2} dY = \frac{V^2}{2} - \frac{1}{3}(1-Y^2)^{\frac{3}{2}}=E_0$$
where $E_0$ is a constant (energy level). This means that a solution $(Y(x),V(x))$ to the system leaves the function $H$ invariant, i.e. $H(Y(x),V(x)) = E_0$ for all $x$. $H$ is the total energy of the system, where $\frac{V^2}{2}$ is the kinetic energy, while the therm $U(Y) = -\frac{1}{3}(1-Y^2)^{3/2}$ is the potential energy. To understand the trajectories of the system in the $(Y,V)$ plane, one can draw a graph of the potential $U(Y)$ and based on it one can reconstruct the dynamics of the system. On the other hand, since $V = \frac{dY}{dx}$, then the Hamiltonian gives us the equation
$$\left(\frac{dY}{dx}\right)^2 = 2E_0 + \frac{2}{3}(1-Y^2)^{\frac{3}{2}}$$ which turns into
$$\frac{dY}{dx} = \pm \sqrt{2E_0 + \frac{2}{3}(1-Y^2)^{\frac{3}{2}}}$$ and so you can write it as
$$\frac{dY}{ \sqrt{2E_0 + \frac{2}{3}(1-Y^2)^{\frac{3}{2}}}} = \pm dx$$ and by integrating both sides, one gets
$$\int_{Y_0}^{Y(x)} \frac{d\tilde{Y}}{ \sqrt{2E_0 + \frac{2}{3}(1-\tilde{Y}^2)^{\frac{3}{2}}}} = \pm(x-x_0),$$
so basically you can try to solve the integral
$$\int \frac{d{Y}}{ \sqrt{2E_0 + \frac{2}{3}(1-{Y}^2)^{\frac{3}{2}}}},$$ although the solutions are probably restricted to the real line holomorphic functions that uniformize an algebraic curve of genus 2 (a hyper-elliptic Riemann surface of genus 2), which is a bit tough.
Alternativly, you can try to run some simulations of the equation
$$\frac{dY}{dx} = \pm \sqrt{2E_0 + \frac{2}{3}(1-Y^2)^{\frac{3}{2}}},$$ having in mind that whenever the solution's derivative $Y'(x)$ becomes $0$, the sign of the equation switches (the sign $\pm$ in front of the square root). The behavior, for certain energy levels is oscillatory, kind of like sine and cosine, but definitely not exactly like them. And do not forget $-1 \leq Y \leq 1$.
Now if one performs the change $z=\sqrt{1-Y^2}$, then this is a type of canonical transformation of the Hamilton system, so the Hamiltonian of the old system, in terms of $(Y,V)$ turns into the Hamiltonian of the new system in terms of $(z,w)$ (which I wrote first), where the 2D transformation is
\begin{align}
z &= \sqrt{1-Y^2}\\
w &= - \frac{Y}{\sqrt{1-Y^2}}V \left(= \frac{dz}{dx}\right)
\end{align}
and the inverse is
\begin{align}
Y &= \sqrt{1-z^2}\\
V &= - \frac{z}{\sqrt{1-z^2}}w \left(= - \frac{z}{\sqrt{1-z^2}} \frac{dz}{dx}\right)
\end{align}
Consequently,
$$\frac{dY}{dx} = - \frac{z}{\sqrt{1-z^2}} \frac{dz}{dx}$$ and when we substitute it in the first-order equation (which would be equivalent to differentiating the substitution one more time, substituting it in the original equation and getting the equation I wrote first):
$$-\frac{z}{\sqrt{1-z^2}}\frac{dz}{dx} = \pm \sqrt{2E_0 + \frac{2}{3} z^3}.$$ If we square on both sides we get
$$\left(z\frac{dz}{dx}\right)^2 = \frac{2}{3}(1-z)(1+z)(3E_0 + z^3)$$ where $z=z(x), w = \frac{dz}{dx}(x)$ is a parametrization (after allowing $x \in \mathbb{C}$) of the hyper-elliptic Riemann surface
$$z^2 w^2 = \frac{2}{3}(1-z)(1+z)(3E_0 + z^3).$$
The general solution to the equation in terms of $z$ is then
$$ \int_{z_0}^{z(x)} \frac{\tilde{z} d\tilde{z}}{\sqrt{(1-\tilde{z})(1+\tilde{z})(3E_0 + \tilde{z}^3)}} = \pm \sqrt{\frac{2}{3}}(x-x_0).$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1893061",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
how to prove this combinatorial identity I accidentally find? Today when I solve a counting problem using different methods I find the following (seemingly correct) combinatorial identity, but I can't find it on the Internet and I can't prove its correctness neither. But I have verified its correctness with positive integer $n$ within $[0, 1000]$ using a simple computer program. Anyone can give a proof to this identity (or any link to its proof)?
$$\frac{(n+1)n}{2} \cdot n! = \sum\limits_{k=0}^{n} (-1)^k \cdot \frac{n!}{k!\cdot(n-k)!} \cdot (n-k)^{n+1}$$
And equivalently if you want,
$$\frac{(n+1)n}{2} = \sum\limits_{k=0}^{n} (-1)^k \cdot \frac{1}{k!\cdot(n-k)!} \cdot (n-k)^{n+1}$$
| Another way is to consider that the backward finite difference (backward Delta) is defined as
$$
\nabla _x \,f(x) = f(x) - f(x - 1)
$$
and we have that its $n$-th iteration is:
$$
\nabla _x ^n \,f(x) = \sum\limits_{\left( {0\, \leqslant } \right)\,k\,\left( { \leqslant \,n} \right)} {\left( { - 1} \right)^k \left( \begin{gathered}
n \\
k \\
\end{gathered} \right)\;f(x - k)}
$$
therefore the RHS is:
$$
\sum\limits_{\left( {0\, \leqslant } \right)\,k\,\left( { \leqslant \,n} \right)} {\left( { - 1} \right)^k \left( \begin{gathered}
n \\
k \\
\end{gathered} \right)\;\left( {n - k} \right)^{n + 1} } = \left. {\nabla _x ^n \,x^{n + 1} } \right|_{\,x = n}
$$
Now $x^{\,n + 1} $ is a polynomial of degree $n+1$ and we can express it in terms
of the Stirling Numbers of $2$nd kind and Falling Factorials of $x$ as
$$
x^{\,n + 1} = \sum\limits_{\left( {0\, \leqslant } \right)\,k\,\left( { \leqslant \,n} \right)} {\left\{ \begin{gathered}
n + 1 \\
k \\
\end{gathered} \right\}\;x^{\,\underline {\,k\;} } }
$$
The backward Delta of the falling factorial is given by:
$$
\begin{gathered}
\nabla _x \;x^{\,\underline {\,m\;} } = \left( {x\left( {x - 1} \right) \cdots \left( {x - m + 1} \right)} \right) - \left( {\left( {x - 1} \right)\left( {x - 2} \right) \cdots \left( {x - m} \right)} \right) = m\left( {x - 1} \right)^{\,\underline {\,m - 1\;} } \hfill \\
\nabla _x ^{\,n} \;x^{\,\underline {\,m\;} } = m^{\,\underline {\,n\;} } \left( {x - n} \right)^{\,\underline {\,m - n\;} } \quad \Rightarrow \quad \nabla _x ^{\,n} \;x^{\,\underline {\,m\;} } = 0\quad \left| {\;m < n} \right. \hfill \\
\end{gathered}
$$
therefore:
$$
\begin{gathered}
\nabla _x ^n \,x^{n + 1} = \nabla _x ^n \,\sum\limits_{\left( {0\, \leqslant } \right)\,k\,\left( { \leqslant \,n} \right)} {\left\{ \begin{gathered}
n + 1 \\
k \\
\end{gathered} \right\}\;x^{\,\underline {\,k\;} } } = \nabla _x ^n \,\left( {\left\{ \begin{gathered}
n + 1 \\
n + 1 \\
\end{gathered} \right\}\;x^{\,\underline {\,n + 1\;} } + \left\{ \begin{gathered}
n + 1 \\
n \\
\end{gathered} \right\}\;x^{\,\underline {\,n\;} } } \right) = \hfill \\
= \left( {\left\{ \begin{gathered}
n + 1 \\
n + 1 \\
\end{gathered} \right\}\;\left( {n + 1} \right)^{\,\underline {\,n\;} } \left( {x - n} \right)^{\,\underline {\,1\;} } + \left\{ \begin{gathered}
n + 1 \\
n \\
\end{gathered} \right\}n^{\,\underline {\,n\;} } \;\left( {x - n} \right)^{\,\underline {\,0\;} } } \right) = \hfill \\
= \left( {1\;\left( {n + 1} \right)^{\,\underline {\,n\;} } \left( {x - n} \right)^{\,\underline {\,1\;} } + \left( \begin{gathered}
n + 1 \\
n - 1 \\
\end{gathered} \right)n^{\,\underline {\,n\;} } \;\left( {x - n} \right)^{\,\underline {\,0\;} } } \right) = \hfill \\
= n!\left( {\;\left( \begin{gathered}
n + 1 \\
n \\
\end{gathered} \right)\left( {x - n} \right) + \left( \begin{gathered}
n + 1 \\
n - 1 \\
\end{gathered} \right)} \right) \hfill \\
\end{gathered}
$$
which, calculated at $x=n$ gives:
$$
\begin{gathered}
\sum\limits_{\left( {0\, \leqslant } \right)\,k\,\left( { \leqslant \,n} \right)} {\left( { - 1} \right)^k \left( \begin{gathered}
n \\
k \\
\end{gathered} \right)\;\left( {n - k} \right)^{n + 1} } = \left. {\nabla _x ^n \,x^{n + 1} } \right|_{\,x = n} = \hfill \\
= n!\left( \begin{gathered}
n + 1 \\
n - 1 \\
\end{gathered} \right) = n!\frac{{\left( {n + 1} \right)n}}
{2} \hfill \\
\end{gathered}
$$
thus proving your assertion, while generalizing it to other values of $x$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1895141",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 5,
"answer_id": 2
} |
Ellipse and hyperbola have the same foci I am trying to solve the following problem:
An ellipse and a hyperbola have the same foci, $A$ and $B$, and intersect at four points. The ellipse has major axis $50$, and minor axis $40$. The hyperbola has conjugate axis of length $20$. Let $P$ be a point on both the hyperbola and ellipse. What is $PA \times PB$?
So I say the center of the ellipse is at $(0,0)$ and the equation of the ellipse is $$\frac{x^2}{25^2}+\frac{y^2}{20^2}=1$$
I calculate that the foci of the ellipse are located at $(15,0)$ and $(-15,0)$.
The general equation of a hyperbola is: $$\frac{x^2}{a^2}-\frac{y^2}{b^2}=\pm 1 \quad \cdots \cdots (*)$$ since the length of the conjugate axis is $20$, we can say
$$2a = 20 \implies a = 10$$
Since $a$ is $10$ we search for $b$ with the condition that the hyperbola formed from $(*)$ will have foci at $(15,0)$ and $(-15,0)$.
I get $$b = 5 \sqrt{5}$$
Now plugging into $(*)$ the values I have for $a$ and $b$ and get,
$$\frac{y^2}{125}-\frac{x^2}{100}=-1$$
for the equation of the hyperbola. Now we need one intersection point and for that I used Mathematica and get $$P = \left ( \frac{50}{3}, \frac{20 \sqrt{5}}{3} \right )$$
The whole line of reasoning leads to the following diagram:
With $A$ and $B$ the foci and $P$ one of the points of intersection. I used the distance formula to get the length of $PA$ and $PB$ and got $15$ and $35$ as seen in the diagram. $$15 \times 35 = 525$$
Of course, this is not the answer given, which is $500$. Where did I go wrong?
Thanks to all for their nice solutions.
| Using elliptic coordinates:
\begin{align*}
z &= c\cosh (\alpha+\beta i) \\
(x,y) &= (c\cosh \alpha \cos \beta, c\sinh \alpha \sin \beta)
\end{align*}
Ellipse:
$$\frac{x^2}{c\cosh^2 \alpha}+\frac{y^2}{c\sinh^2 \alpha}=1$$
In this case, $$\
\begin{array}{rcccl}
\text{major axis} &=& 2c\cosh \alpha &=& 50 \\
\text{minor axis} &=& 2c\sinh \alpha &=& 40
\end{array}$$
Hyperbola:
$$\frac{x^2}{c\cos^2 \beta}-\frac{y^2}{c\sin^2 \beta}=1$$
In this case, $$\
\begin{array}{rcccl}
\text{transverse axis} &=& 2c\cos \beta &=& 2\sqrt{15^2-10^2} \\
\color{red}{\text{conjugate axis}} &=& 2c\sin \beta &=& 20
\end{array}$$
Now,
\begin{align*}
|z+c|+|z-c| &= 2c\cosh \alpha \\
|z+c|-|z-c| &= \pm 2c\cos \beta \\
4|z+c||z-c| &= (|z+c|+|z-c|)^2-(|z+c|-|z-c|)^2 \\
&= 4c^2(\cosh^2 \alpha-\cos^2 \beta) \\
|z+c||z-c| &= c^2(\cosh^2 \alpha-\cos^2 \beta) \\
&= c^2(\sinh^2 \alpha+\sin^2 \beta) \\
&= \left( \frac{40}{2} \right)^2+
\left( \frac{20}{2} \right)^2 \\
&= 500
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1896801",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
How can one find the factorization $a^4 + 2a^3 + 3a^2 + 2a + 1 = (a^2 + a + 1)^2$ from scratch? I want to prove $a^4 + 2a^3 + 3a^2 + 2a + 1$ always be perfect square for $a \in \mathbb{N}$. Using GeoGebra I found $a^4 + 2a^3 + 3a ² + 2a + 1 = (a^2 + a + 1)^2$ which is trivial to verify.
Since this equation has no real roots, it's not so easy to decompose it through division by $a-a_0$ where $a_0$ is a root. I don't want to use complex numbers if possible. How do find this factorization from scratch / without using a computer?
| Standard Solution:
Let $a^4+2a^3+3a^2+2a+1=(ba^2+ca+d)^2$ where $b,c,d \in \mathbb{R}$
Observe that
$$(ba^2+ca+d)^2=b^2a^4+2bca^3+(2bd+c^2)a^2+2cda+d^2$$
Comparing coefficient of like terms, we get
$$b^2=1 \Rightarrow b=\pm 1$$
$$bc=1 \Rightarrow c=\pm 1$$
$$ \pm2d+1=3 \Rightarrow d=\pm 1$$
Thus, $$a^4+2a^3+3a^2+2a+1=[\pm(a^2+a+1)]^2$$
Note: Although it appears that this method assumes that the quartic must be the square of a polynomials, but if you end up getting no answer, it means you arrived at a contradiction and the equation is not the square of a polynomial. Moreover, the if there are 2 answers, they are additive inverses of each other.
Alternate Solution:
Let $f(a)=a^4+2a^3+3a^2+2a+1$
Observe that $f(0)=1$ and that the coefficients follow this pattern: $1,2,3,2,1$
Dividing $f(a)$ by $a^2$ to get
$$g(a)=\frac{f(a)}{a^2}=a^2+2a+3+\frac{2}{a}+\frac{1}{a^2}=\left(a+\frac{1}{a}\right)^2+2\left(a+\frac{1}{a}\right)+1$$
Substitute $t=a+\dfrac{1}{a}$ to get
$t^2+2t+1=(t+1)^2$
Now, back-substitute to get
$$f(a)=a^2 \cdot \left(a+\frac{1}{a}+1\right)^2=(a^2+a+1)^2$$
But since $x^2=(-x)^2$, we get
$$\color{blue}{\boxed{\color{red}{a^4+2a^3+3a^2+2a+1=[\pm (a^2+a+1)]^2}}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1899346",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 1
} |
Boundedness of $\frac{1}{n}+\frac{1}{n-x}+\frac{1}{n-2x}+\dots+\frac{1}{n-(n-1)x}$ Let $0\leq x\le 1$, and $n$ a positive integer. Define the function
$$f(n,x)=\frac{1}{n}+\frac{1}{n-x}+\frac{1}{n-2x}+\dots+\frac{1}{n-(n-1)x}.$$
For which $x$ is the sum bounded by a constant for arbitrarily large $n$?
If $x=0$, this is $\frac{1}{n}+\frac{1}{n}+\dots+\frac{1}{n}=1$, a constant. If $x=1$, it is $\frac{1}{n}+\frac{1}{n-1}+\dots+\frac{1}{1}$, a divergent sum. The question is what happens in between.
As mfl wrote, the sum is bounded above by $\frac{1}{1-x}$. Is it also true that the sum converges to $\frac{1}{1-x}$?
| Hint
$$\frac{1}{n}+\frac{1}{n-x}+\frac{1}{n-2x}+\dots+\frac{1}{n-(n-1)x}\le \frac{n}{n-(n-1)x}\underbrace{\nearrow}_{n\to \infty} \frac{1}{1-x}$$ for $0<x<1.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1899815",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
The square of any odd integer is odd. Suppose that $n$ is an odd integer. Then $n = 2k + 1$ for some integer $k$.
Hence $n^2=(2k+1)(2k+1)=4k^2+4k+1=2(2k^2+2k)+1$.
Since $k$ is an integer, $2k^2 + 2k$ is an integer. Thus $n^2 = 2k' + 1$ for some integer $k'$.
Therefore $n^2$ is odd.
Is this correct?
| Using modular arithmetic:
$$n\equiv1\mod2\implies n^2\equiv1\mod2$$ because $n^2\bmod2=(n\bmod2)^2$ and $1^2=1$.
You can even handle the case of even and odd at the same time with
$$n\bmod2=(n\bmod2)^2=n^2\bmod2$$ because $0^2=0$ and $1^2=1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1900994",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 2
} |
How $99^{2} \mod 32$ is equal to $3^{2} \mod 32$? I know how to get $99^{2} \mod 32$ from the $(99\mod32)(99\mod32)\mod32$ and get the answer. But the solution guide I have, has converted $$99^{2}\mod32$$ to $$3^{2}\mod32$$. Any idea how?
| Just as a more concrete representation, note that
\begin{align}
99^2 & = (3 \times 32 + 3)^2 \\
& = (3 \times 32)^2 + 2 (3)(3 \times 32) + 3^2 \\
& = (3 \times 3 \times 32 \times 32) + (2 \times 3 \times 3 \times 32)
+ 3^2 \\
& = (288 \times 32) + (18 \times 32) + 3^2 \\
& = (306 \times 32) + 3^2
\end{align}
and hence $99^2 \bmod 32 = 3^2 \bmod 32 = 9 \bmod 32 = 9$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1904866",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Find perpendicular distance from point to line in 3D? I have a Line going through points B and C; how do I find the perpendicular distance to A?
$$A= (4,2,1)$$
$$B= (1,0,1)$$
$$C = (1,2,0)$$
| Another approach (a very low-tech solution). By the Pythagorean theorem we have
$$ AB=\sqrt{13},\qquad AC=\sqrt{10},\qquad BC=\sqrt{5}\tag{1}$$
hence $ABC$ is an acute triangle (since $AB^2<AC^2+BC^2$ and so on) and by calling $H_A$ the projection of $A$ on $BC$, we also have
$$ H_A B^2 - H_A C^2 = AB^2 - AC^2 = 3 \tag{2}$$
Since $H_A B + H_A C = BC = \sqrt{5}$, from $(2)$ it follows that $H_A B-H_A C = \frac{3}{\sqrt{5}}$, so:
$$ H_A B = \frac{4\sqrt{5}}{5}\qquad H_A C = \frac{\sqrt{5}}{5}\tag{3}$$
and:
$$\boxed{\phantom{\sum_{j=1}^{n^2}} H_A = \frac{4}{5}C+\frac{1}{5}B = \color{red}{\left(\frac{5}{5};\frac{8}{5};\frac{1}{5}\right)},\qquad AH_A^2 = AC^2-H_A C^2 = \color{red}{\frac{49}{5}}\phantom{\sum_{j=1}^{n^2}}}\tag{4} $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1905533",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "16",
"answer_count": 9,
"answer_id": 2
} |
Find any rational approximation of $a = \frac{\sin(\sqrt{2})}{\sqrt{2}}$ I need to find any ratinal approximation of $a = \frac{\sin(\sqrt{2})}{\sqrt{2}} $ with accuracy $d = \frac{1}{500}$.
I generally know how to do these type of tasks (using Taylor expansion), but I got problem with this one.
I defined function $f(x) = \begin{cases} 1 \ \mbox{dla} \ x=0 \\ \frac{\sin x}{x} \ \mbox{dla} \ x \neq 0 \end{cases}$, to handle problem with dividng by zero in $f(0)$.
I calculate first and second derivative:
$f'(x) = \frac{x^2 \cdot \cos x - \sin x}{x^2}$
$f''(x) = \frac{x^3 \cdot \sin x - 2x^2 \cdot \cos x + \sin x }{x^4}$
But well, what now? I have no idea how to use the Lagrange remainder theorem here.
| $\text{sinc}(x)$ is an entire function fulfilling
$$ \text{sinc}(x) = \sum_{n\geq 0}\frac{(-1)^n x^{2n}}{(2n+1)!} \tag{1}$$
hence
$$ \text{sinc}(\sqrt{2})=\sum_{n\geq 0}\frac{(-1)^n 2^n}{(2n+1)!}\tag{2}$$
The terms of the last series are decreasing in absolute value, hence
$$ \left|\,\text{sinc}(\sqrt{2})-\sum_{n=0}^{3}\frac{(-1)^n 2^n}{(2n+1)!}\right|\leq \frac{2^3}{(2\cdot 3+1)!}=\frac{1}{630}\tag{3} $$
and
$$ \sum_{n=0}^{3}\frac{(-1)^n 2^n}{(2n+1)!} = \color{red}{\frac{44}{63}}\tag{4} $$
is an approximation fulfilling the given constraints.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1906520",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Show that $9|2^{2n+1}+2^{4n+1}-4$
Show that $9|2^{2n+1}+2^{4n+1}-4$ ( n is a positive integer):
1-Using induction
2-Don't use induction
I have noticed that $2^m \equiv 2^{m+6} \pmod 9 $ but couldn't use it to solve the problem!
| $2x^2+2x-4=2(x^2+x-2)=2(x+2)(x-1)=2(x-1+3)(x-1)$ so, for $x=2^{2n}=4^n$,
$$
2^{2n+1}+2^{4n+1}-4=2(4^n-1+3)(4^n-1)
$$
and you just need to show that $3\mid 4^n-1$.
No induction
Since $4\equiv1\pmod{3}$, the statement is proved.
With induction
The base case is obvious. If $4^n-1=3k$ with $k$ integer,
$$
4^{n+1}-1=4\cdot 4^n-1=4(3k+1)=12k+4-1=3(4k+1)
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1907540",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 4
} |
Largest divisor of $P(n) = (n + 1)(n + 3)(n + 5)(n + 7)(n + 9)$
Let $P(n) = (n + 1)(n + 3)(n + 5)(n + 7)(n + 9)$. What is the largest integer that is a divisor of $P(n)$ for all positive even integers $n$?
I can see that obviously it is divisible by $3$ and $5$ and hence answer is $15$ but I can't prove it.
| $$P(10)=(11)(13)(15)(17)(19)=3.5.11.13.17.19$$
$$P(20)=(21)(23)(25)(27)(29)=3^4.5^2.7.23.29$$
we can see that
$$gcd(P(10),P(20))=15$$
$$P(n)=(n+1)(n+3)(n+5)(n+7)(n+9) \equiv (n+1)n(n+2)(n+1)n \equiv 0 \mod 3$$
since $n(n+1)(n+2) \equiv 0 \mod 3.$, that is product of 3 consecutive numbers must be divisible by $3$.
Also,
$$P(n)=(n+1)(n+3)(n+5)(n+7)(n+9) \equiv (n+1)(n+3)n(n+2)(n+4)\equiv 0 \mod 5$$
since product of 5 consecutive numbers is divisible by $5$.
Hence $15$ is the largest number that divides $P(n)$ for all even $n$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1908210",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 4
} |
Find $ ? = \sqrt[3] {1 + \sqrt[3] {1 + 2 \sqrt[3] {1 + 3 \sqrt[3] \cdots}}} $ I wonder about a closed form for
$ ? = \sqrt[3] {1 + \sqrt[3] {1 + 2 \sqrt[3] {1 + 3 \sqrt[3] {1 + 4 \sqrt[3] {1 + 5 \sqrt[3] \cdots}}}}} $
To be clear
$$? = \sqrt[3]{ 1 + \color{Red}{1}\sqrt[3]{ 1 + \color{Red}{2} \sqrt[3]{ 1 + \color{Red}{3} \sqrt[3]{\cdots}}}} $$
Where the red coefficients are just the natural numbers.
I tried solving the related equation $ f(x) ^3 = 1 + (x+y) f(x+1) $ for various fixed integer values $y,$ but I failed.
It appears
$$ ? = \sqrt[3] {1 + \sqrt[3] 5} $$
But I am not able to prove it.
See also https://en.m.wikipedia.org/wiki/Nested_radical#Ramanujan.27s_infinite_radicals
| Generalizing Ramanujan's Radical, you can get trivial results like this,
$$2=\sqrt[3]{4+1^2\sqrt[3]{10+3^{2}\sqrt[3]{16+5^{2}\sqrt[3]{22+7^{2}\sqrt[3]{28+9^{2}\sqrt[3]{...}}}}}}$$
$$3=\sqrt[3]{7+2^{2}\sqrt[3]{13+4^{2}\sqrt[3]{19+6^{2}\sqrt[3]{25+8^{2}\sqrt[3]{...}}}}}$$
Where the terms, $4,10,16,..$ are in an arithmetic progression with $d=6$. And so are the terms $7,13,19..$
While your stated question looks similar to Ramanujan's, it isn't. A closed form expression is highly unlikely.
Also, the above expressions are from;
$$n+1=\sqrt[3]{1+3n+n^{2}\sqrt[3]{1+3\left(n+2\right)+\left(n+2\right)^{2}\sqrt[3]{1+3\left(n+4\right)+\left(n+4\right)^{2}\sqrt[3]{...}}}}$$
Some approximations for the constant, one of them include;
$$\approx \sqrt{\frac{\sqrt{15}+\sqrt[3]{2}}{\sqrt{7}}}$$
At 6 decimal places, other approximations include;
$12^{\frac{2}{15}},\sqrt{\frac{97}{50}},\exp\left(\frac{1}{3}-\frac{1}{500}\right),\sinh^{-1}\left(\frac{17}{9}+.0001\right),\frac{39}{28}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1908336",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 1,
"answer_id": 0
} |
Integer partition of n into k parts recurrence I was learning integer partition of a number n into k parts(with minimum 1 in each part) and came across this recurrence :
part(n,k) = part(n-1,k-1) + part(n-k,k)
But, I cannot understand the logic behind this recurrence. Can someone please help me visualize this recurrence?
|
Let's consider an example: $n=8$ and $k=3$:
\begin{align*}
\\
P(n,k)&=P(n-1,k-1)+P(n-k,k)&\\
\\
P(8,3)&=P(7,2)+P(5,3)&\\
\\
5\quad&=\quad3\qquad+\quad2
\end{align*}
$$ $$
\begin{array}{rlrlrl}
&P(8,3)\qquad=&& P(7,2)\qquad\qquad+&&P(5,3)\\
\hline\\
8&=\color{red}{6+1}+1\qquad& 7&=6+1\\
&=\color{red}{5+2}+1\qquad&&=5+2\\
&=\color{red}{4+3}+1\qquad&&=4+3\\
&=\color{blue}{4+2+2}\qquad&&\qquad &\qquad5&=3+1+1\\
&=\color{blue}{3+3+2}\qquad&&\qquad&\qquad&=2+2+1
\end{array}
Note that the partitions of $P(8,3)$ that have smallest integer part equal to one correspond to the integer partitions of $P(7,2)$ whereas the partitions with smallest integer part $>1$ correspond to the partitions of $P(5,3)$:
\begin{align*}
4+2+2=(3+1+1)+3\\
3+3+2=(2+2+1)+3
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1908701",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 3,
"answer_id": 2
} |
Proof for an inequality problem using the rearrangement inequality I have found this problem in a book and would like someone to give me a good and elegant proof for it.
The problem is:
If $a_1,a_2,...,a_n$ are positive real numbers and $s=a_1+a_2+...+a_n$, then
$$\frac{a_1}{s-a_1}+\frac{a_2}{s-a_2}+...+\frac{a_n}{s-a_n} \ge \frac{n}{n-1}$$
The book does not provide solutions but only gives hints, which is that it can be solved like another problem using n variables
The problem is:
If $a,b,c$ are the lengths of the sides of a triangle, prove that
$$\frac{a}{b+c-a}+\frac{b}{c+a-b}+\frac{c}{a+b-c} \ge 3$$
Again only hints are given and this is the solution I am proposing:
let $(b_1,b_2,b_3)=\frac{a}{b+c-a},\frac{b}{c+a-b},\frac{c}{a+b-c}$
Let $(a_1,a_2,a_3)=a,b,c$
By the rearrangement inequality we have,
$$a_1b_1+a_2b_2+a_3b_3 \ge a_2b_1 +a_3b_2+a_1b_3$$
$$a_1b_1+a_2b_2+a_3b_3 \ge a_3b_1 +a_1b_2+a_2b_3$$
$$\frac{a}{b+c-a}+\frac{b}{c+a-b}+\frac{c}{a+b-c} \ge \frac{b}{b+c-a}+\frac{c}{c+a-b}+\frac{a}{a+b-c}$$
$$\frac{a}{b+c-a}+\frac{b}{c+a-b}+\frac{c}{a+b-c} \ge \frac{c}{b+c-a}+\frac{a}{c+a-b}+\frac{b}{a+b-c}$$
Let $$Q = \frac{a}{b+c-a}+\frac{b}{c+a-b}+\frac{c}{a+b-c}$$
Now by adding the two inequalities we have,
$$2Q \ge \frac{b}{b}+\frac{b}{c}-\frac{b}{a}+\frac{c}{c}+\frac{c}{a}-\frac{c}{b}+\frac{a}{a}+\frac{a}{b}-\frac{a}{c}+\frac{c}{b}+\frac{c}{c}-\frac{c}{a}+\frac{a}{c}+\frac{a}{a}-\frac{a}{b}+\frac{b}{a}+\frac{b}{b}-\frac{b}{c}=6+\frac{a-b-a+b}{a}+\frac{a-c-a+c}{b}+\frac{b-a-b+a}{c}=6$$
$$Q \ge 3$$
I do understand the general idea but what is needed in the problem above is a generalization of the proof I provided. However I fail at producing one each time I try. Can someone please provide me with one that is simple and easy to read. Thanks.
UPDATE:
Thanks for the answers but I am looking for a proof using the rearrangement inequality in a similar way to the example.
| Multiplying $s$ and $a_j$ by $n/s$ we can assume that $s=n$ and $a_1+\dotsm a_n=n$. Let us find the minimum of
$$
\frac{a_1}{n-a_1}+\dots+\frac{a_n}{n-a_n}
$$
over positive $a_j$ satisfying $a_1+\dotsm+a_n=n$. Using Lagrange multipliers, we take the function
$$
g(a_1,\dots,a_n,\lambda)=\frac{a_1}{n-a_1}+\dots+\frac{a_n}{n-a_n}+\lambda(a_1+\dotsm+a_n-n)
$$
and equate all of its derivatives with zero. We obtain
$$
\frac{n}{(n-a_j)^2}-\lambda=0,\quad j=1,\dots,n,\qquad
a_1+\dotsm+a_n=n.
$$
It is easily seen that the only solution is $a_1=a_2=\dotsm=1$, $\lambda=n/(n-1)^2$.
The minimum is attained at this point and is $n/(n-1)$. This proves the desired assertion.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1910308",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Prove the inequality $a^{2}+b^{2}+c^{2}+abc(a+b+c) \geq 2(ab+bc+ca)$, given $a^{2}b+b^{2}c+c^{2}a+a^{2}b^{2}c^{2}=4$
Let $a, b$ and $c$ be positive real numbers such that $a^{2}b+b^{2}c+c^{2}a+a^{2}b^{2}c^{2}=4$. Prove that
\begin{equation}
a^{2}+b^{2}+c^{2}+abc(a+b+c) \geq 2(ab+bc+ca)
\end{equation}
Let us consider that
\begin{equation*}
\begin{split}
\text{$(a+b+c)^{2}$} &=\text{$(a+b+c)(a+b+c)$} \\
&=\text{$a^{2}+ab+ac+ba+b^{2}+bc+ca+cb+c^{2}$} \\
&=\text{$a^{2}+b^{2}+c^{2}+2ac+2ab+2bc$} \\
&=\text{$a^{2}+b^{2}+c^{2}+2(ab+bc+ca)$}
\end{split}
\end{equation*}
and consider
\begin{equation*}
\begin{split}
\text{$a^{2}+b^{2}+c^{2}$} &=\text{$(a+b+c)^{2}-2(ab+bc+ca)$}
\end{split}
\end{equation*}
The L.H.S of the inequality (1)
\begin{equation}
\begin{split}
\text{$a^{2}+b^{2}+c^{2}+abc(a+b+c) $}
&=\text{$(a+b+c)^{2}-2(ab+bc+ca)+abc(a+b+c)$} \\
&=\text{$(a+b+c)^{2}+(a+b+c)abc-2(ab+bc+ca)$} \\
&=\text{$(a+b+c)\bigg[(a+b+c)+abc \bigg]-2(ab+bc+ca)$} \\
\end{split}
\end{equation}
I have not real proof anything yet. I have through out some work, but I have not think how would I use $a^{2}b+b^{2}c+c^{2}a+a^{2}b^{2}c^{2}=4$. Please If you any idea that would help, please write it down.
| We'll prove that $a+b+c\geq3$.
Let $a+b+c<3$, $a=kx$, $b=ky$ and $c=kz$, where $k>0$ and $x+y+z=3$.
Hence, $k<1$ and $4=a^2b+b^2c+c^2a+a^2b^2c^2=k^3(x^2y+y^2z+z^2x+k^3x^2y^2z^2)<x^2y+y^2z+z^2x+x^2y^2z^2$,
which is contradiction because we'll prove now that $x^2y+y^2z+z^2x+x^2y^2z^2\leq4$.
Indeed, we need to prove that $\frac{(x^2y+y^2z+z^2x)(x+y+z)^3}{27}+x^2y^2z^2\leq\frac{4(x+y+z)^6}{729}$.
But by Rearrangement $x^2y+y^2z+z^2x\leq\frac{4(x+y+z)^3}{27}-xyz$.
Hence, it remains to prove that $(x+y+z)^3\geq27xyz$, which is AM-GM.
Id est, $a+b+c\geq3$.
Hence, it's enough to prove that $9abc\geq(a+b+c)\sum\limits_{cyc}(2ab-a^2)$ or
$\sum\limits_{cyc}(a^3-a^2b-a^2c+abc)\geq0$, which is Schur.
Done!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1911655",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 1,
"answer_id": 0
} |
Is there a quicker proof to show that $2^{10^k} \equiv 7 \pmod{9}$ for all positive integers $k$? I noticed this pattern while playing with digit sums and noticed that the recursive digit sums (until you arrive at a single digit number) of numbers like, $2^{10}$, $2^{100}$, $2^{1000}$ and so on is always $7$. So, I decided to find out if it is true that for all positive integers $k$,
$$ 2^{10^k} \equiv 7 \pmod{9} $$
My proof is as follows:
Lemma 1. $\, 10^k \equiv 4 \pmod{6}$ for all integers $k \geq 1$
Proof. For all integers $k \geq 1$, the number $10^k + 2$ must be divisible by $6$ since it is even (implying divisibility by $2$), and its digit sum is $3$ (implying divisibility by $3$). Therefore, you can show that
\begin{align*}
10^k + 2 &\equiv 0 \pmod{6} \\
10^k &\equiv 4 \pmod{6} \,\, \text{ for all integers } k \geq 1
\end{align*}
Lemma 2. $\, 2^{4 + 6k} \equiv 7 \pmod{9}$, for all integers $k \geq 0$
Proof. If $a \equiv c \pmod{n}$ and $b \equiv d \pmod{n}$, then $a\,b \equiv c\,d \pmod{n}$.
And, by extension, $a\,b^k \equiv c\,d^k \pmod{n}$ (for integers $k \geq 0$). Therefore, with
\begin{align*}
2^4 \equiv 16 \equiv 7 \pmod{9}
\end{align*}
and
\begin{align*}
2^6 \equiv 64 \equiv 1 \pmod{9}
\end{align*}
we can show that,
\begin{align*}
2^{4 + 6k} \equiv 7 \cdot 1^k \equiv 7 \pmod{9} \,\, \text{ for all integers } k \geq 0
\end{align*}
Theorem. $\, 2^{10^k} \equiv 7 \pmod{9} $ for all integers $k \geq 1$
Lemma 1 implies that for all integers $k \geq 1$, $10^k = 4 + 6n$ where $n$ is some positive integer. Therefore,
\begin{align*}
2^{10^k} &= 2^{4 + 6n} \\
2^{10^k} \!\!\!\! \mod a &= 2^{4 + 6n} \!\!\!\! \mod a
\end{align*}
for any positive integer $a$. Using this result along with Lemma 2,
\begin{align*}
2^{10^k} \equiv 7 \pmod{9} \,\, \text{ for all integers } k \geq 1
\end{align*}
$$\tag*{$\blacksquare$}$$
I think the proof is correct, but I am not a fan of it. Mainly because, I think proving Lemma 2 is a much too long a way to prove this theorem - since it proves a generalization of the theorem first. Also, I discovered that lemma numerically, which feels a bit like cheating.
Either way, is there a quicker, more elegant proof which does not:
*
*Use Lemma 2, or prove some generalization of the theorem first.
*Uses Euler's Theorem. I feel that using Euler's Theorem is using a needlessly complicated theorem to prove something as simple as this.
I am sure some of the people here can come up with single line proofs. I am curious to see if there's such a proof.
| Induction? Base case when $k=1$ is clear, for inductive step we have: $2^{10^k}=(2^{10^{k-1}})^{10}\equiv 7^{10}\equiv 7\bmod 9$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1912681",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Inequality with $a+b+c=1$ Let $a,b,c\in\mathbb{R^+}$ such that $a+b+c=1$. Prove that $$\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}+3(ab+bc+ca)\geq\frac{11}{2}.$$
I am trying to resolve this problem but actually i found some issues:
$$\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}=3+\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\geq \frac{9}{2}$$
I only use Nesbitt's inequality. Then we only need that $3(ab+bc+ca)\geq 1$, but it is not correct, because $3(ab+bc+ca)\leq 1$.
Maybe some ideas will be more that grateful.
| Variational Approach
Let $x=a+b$, $y=b+c$, and $z=c+a$. Then $x+y+z=2$ and we are minimizing
$$
\frac1x+\frac1y+\frac1z+3\left(1-\frac{x^2+y^2+z^2}2\right)\tag{1}
$$
Furthermore, since $a+b+c=1$ and $a,b,c\ge0$, we have that $0\le x,y,z\le1$.
Since $x+y+z=2$, we get the restriction
$$
\delta x+\delta y+\delta z=0\tag{2}
$$
Taking the variation of $(1)$ gives
$$
\left(\frac1{x^2}+3x\right)\delta x+\left(\frac1{y^2}+3y\right)\delta y+\left(\frac1{z^2}+3z\right)\delta z=0\tag{3}
$$
Thus, to minimize $(1)$, we need to have $(3)$ whenever we have $(2)$. This means that
$$
\frac1{x^2}+3x=\frac1{y^2}+3y=\frac1{z^2}+3z\tag{4}
$$
Case $\boldsymbol{1}$: If $x\ne y$ and $y\ne z$ and $z\ne x$, then $(4)$ implies
$$
\frac{x+y}{x^2y^2}=\frac{y+z}{y^2z^2}=\frac{z+x}{z^2x^2}=3\tag{5}
$$
which implies that $x=y=z$.
Case $\boldsymbol{2}$: If $x=y\ne z$, then $2x+z=2$ and
$$
\frac{z+x}{z^2x^2}=3\implies12x^4-24x^3+12x^2+x-2=0\tag{6}
$$
Sturm's Theorem shows that there are no solutions to $(6)$ in $[0,1]$.
Therefore, $(1)$ is minimized when $x=y=z=\frac23$ and thus, $a=b=c=\frac13$. This means that
$$
\frac1{a+b}+\frac1{b+c}+\frac1{c+a}+3(ab+bc+ca)\ge\frac{11}2\tag{7}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1914966",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 5
} |
explicit formula for contraction of conus of homotopy equivalence Let $f: K^\bullet \to L^\bullet$ be homotopy equivalence. From theory of triangulated categories it follows that $C(f)^\bullet$ is contractible. But how can i produce explicit formula for homotopy of $Id_{C(f)^\bullet}$ and $0_{C(f)^\bullet}$?
| It is not that difficult! Let us write down carefully all the morphisms and identities that we have. We have morphisms of complexes
$$f^n\colon K^n \to L^n, \quad f^{n+1} \circ d^n_K = d_L^n\circ f^n$$
and
$$g^n\colon L^n \to K^n, \quad g^{n+1} \circ d^n_L = d_K^n\circ g^n,$$
and homotopies $h_K^n\colon K^n\to K^{n-1}$, $h_L^n\colon L^n\to L^{n-1}$ such that
$$\mathrm{id}_{K^{n+1}} - g^{n+1}\circ f^{n+1} = d_K^n\circ h_K^{n+1} + h_K^{n+2}\circ d_K^{n+1}$$
and
$$\mathrm{id}_{L^n} - f^n\circ g^n = d_L^{n-1}\circ h_L^n + h_L^{n+1}\circ d_L^n.$$
The cone $C (f)$ is formed by objects $C (f)^n \mathrel{\mathop:}= L^n\oplus K^{n+1}$ and differentials
$$d^n \mathrel{\mathop:}= \begin{pmatrix}
d^n_L & f^{n+1} \\
0 & -d^{n+1}_K
\end{pmatrix}.$$
We want to construct certain homotopy $h^n\colon C (f)^n \to C (f)^{n-1}$. In theory, it is possible to come up directly with a null-homotopy, i.e. find some $h^n$ such that
$$d^{n-1} \circ h^n + h^{n+1}\circ d^n \stackrel{???}{=} \mathrm{id}_{C (f)^n} = \begin{pmatrix}
\mathrm{id}_{L^n} & 0 \\
0 & \mathrm{id}_{K^{n+1}}
\end{pmatrix},$$
but instead, let us try a rather obvious candidate (which won't be a null-homotopy, but we'll see how to correct it)
$$h^n \mathrel{\mathop:}= \begin{pmatrix}
h^n_L & 0 \\
g^n & -h_K^{n+1}
\end{pmatrix}.$$
Multiplying matrices, we get
$$d^{n-1} \circ h^n + h^{n+1}\circ d^n =$$
$$\begin{pmatrix}
d^{n-1}_L & f^n \\
0 & -d^n_K
\end{pmatrix} \circ \begin{pmatrix}
h^n_L & 0 \\
g^n & -h_K^{n+1}
\end{pmatrix} + \begin{pmatrix}
h^{n+1}_L & 0 \\
g^{n+1} & -h_K^{n+2}
\end{pmatrix} \circ \begin{pmatrix}
d^n_L & f^{n+1} \\
0 & -d^{n+1}_K
\end{pmatrix} =$$
$$\begin{pmatrix}
d_L^{n-1}\circ h_L^n + f^n\circ g^n & -f^n\circ h_K^{n+1} \\
-d_K^n\circ g^n & d^n_K\circ h_K^{n+1}
\end{pmatrix} + \begin{pmatrix}
h_L^{n+1}\circ d_L^n & h_L^{n+1}\circ f^{n+1} \\
g^{n+1}\circ d^n_L & g^{n+1}\circ f^{n+1} + h^{n+2}_K\circ d^{n+1}_K
\end{pmatrix} =$$
$$\begin{pmatrix}
\mathrm{id}_{L^n} & - f^n\circ h_K^{n+1} + h_L^{n+1}\circ f^{n+1} \\
0 & \mathrm{id}_{K^{n+1}}
\end{pmatrix}.$$
Sadly, $k^n \mathrel{\mathop:}= - f^n\circ h_K^{n+1} + h_L^{n+1}\circ f^{n+1} \ne 0$, so $h^n$ is not a null-homotopy but a homotopy between the zero morphism and some morphism $\phi^n \mathrel{\mathop:}= \begin{pmatrix}
\mathrm{id}_{L^n} & k^n \\
0 & \mathrm{id}_{K^{n+1}}
\end{pmatrix}$. However, the latter is visibly an isomorphism of complexes, having $\psi^n \mathrel{\mathop:}= \begin{pmatrix}
\mathrm{id}_{L^n} & -k^n \\
0 & \mathrm{id}_{K^{n+1}}
\end{pmatrix}$ as its inverse. Hence the required null-homotopy is given by $h^n\circ \psi^n$:
$$d^{n-1}\circ h^n\circ \psi^n + h^{n+1}\circ \psi^{n+1}\circ d^n = d^{n-1}\circ h^n\circ \psi^n + h^{n+1}\circ d^n\circ \psi^n = \phi^n\circ \psi^n = \mathrm{id}_{C (f)^n}.$$
By the way, it is also true that if you have a chain contraction for the cone of $f^\bullet$, then $f^\bullet$ is a homotopy (proof: write the chain contraction $2\times 2$ matrix and look at the identity $d^{n-1} \circ h^n + h^{n+1}\circ d^n = \mathrm{id}_{C (f)^n}$).
And to make you laugh, here is this construction as described in a real GTM textbook (Rosenberg, "Algebraic K-Theory and Its Applications", p. 45):
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1917288",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
How to compare the values $\sqrt 2$ and $\ln(3).$ How to compare the values $\sqrt 2$ and $\ln(3)?$ I know only $\ln(x)<x$ and $\sqrt{x}<x$. Please help. Thanks.
| The trick is to look at $f(x) = \sqrt{x} - \ln(1+x)$.
Note that for $x = 1$,
$$1 - \ln(2) > 0$$
This is true since $\ln(x)$ is an increasing function and $1 < 2 < e$ implies
$$0 = \ln(1) < \ln(2) < \ln(e) = 1$$
Now look at the derivative
$$f'(x) = \frac{1}{2\sqrt{x}} - \frac{1}{x+1}$$
We would like to show that $f'(x) \ge 0$ for $x > 1$, which means $f(x)$ is increasing and must therefore be positive for $x > 1$ (since it is already positive at $x=1$ per above).
Consider the following equivalent inequalities (when $x > 1$):
$$\begin{eqnarray}
\frac{1}{2\sqrt{x}} - \frac{1}{x+1} &\ge& 0\\
\frac{1}{2\sqrt{x}} &\ge& \frac{1}{x+1}\\
x + 1 &\ge& 2\sqrt{x}
\end{eqnarray}$$
For the last inequality, $1+1 = 2 = 2\sqrt{1}$ and for $x > 1$, taking the derivatives agains shows that the left side grows at a rate of $1$ and the right side grows at the rate of $\frac{2}{2\sqrt{x}} = \frac{1}{\sqrt{x}} < 1$ when $x > 1$.
Thus the last equivalent inequality holds for $x > 1$, which implies $f'(x) \ge 0$ for $x > 1$ which implies $f(x)$ is positive for $x > 1$. In particular, $f(2) = \sqrt{2} - \ln(3) > 0$. Therefore $\sqrt{2} > \ln(3)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1917410",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 6,
"answer_id": 1
} |
How to prove $\sqrt{5}-\sqrt{3}$ is bigger than $\sqrt{15}-\sqrt{13}$ Although I can determine using a calculator that $\sqrt{5}-\sqrt{3}$ is larger than $\sqrt{15}-\sqrt{13}$, how would I go about proving that? My teacher gave us a hint which was to use the difference of two squares identity $(a^2-b^2) = (a-b)\cdot(a+b)$, but I don't see how to proceed.
Thanks in advance!
| \begin{align}
(\sqrt 5 - \sqrt 3)(\sqrt 5 + \sqrt 3)
&= (\sqrt{15} - \sqrt{13})(\sqrt{15} + \sqrt{13}) = 2
\\
\dfrac{\sqrt 5 - \sqrt 3}{\sqrt{15} - \sqrt{13}}
&= \dfrac{\sqrt{15} + \sqrt{13}}{\sqrt 5 + \sqrt 3} > 1
\\
\sqrt 5 - \sqrt 3 &> \sqrt{15} - \sqrt{13}
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1919604",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 7,
"answer_id": 6
} |
Determine the coefficient of $xy$ in the expansion of $(x+y+2)^7$ The problem: Determine the coefficient of $xy$ in the expansion of $(x+y+2)^7$
My approach:
We can rewrite the equation substituting $x+y =j$
$$(x+y+2)^7=(j+2)^7$$
This is simpler because we know the coefficients thanks to the formula:
$$(a+b)^{n}=\sum _{{k=0}}^{n}{n \choose k}a^{{n-k}}b^{{k}} $$
With $n=7$ we have $1,7,21,35,35,21,7,1$ as coefficients and the expansion looks like this:
$$j^7+14\cdot j^6+(21\cdot 2^2) j^5+(35\cdot 2^3) j^4+(35\cdot 2^4) j^3+(21\cdot 2^5) j^2+(7\cdot 26)j+2^7$$
Now the only time that $xy$ appears is in the expansion of $j^2$ therefore we have:
$$(21\cdot 2^5) j^2=(21\cdot 2^5) (x+y)^2= (21\cdot 2^5)(x^2+2xy+y^2)=21\cdot 2^5x^2+21\cdot 2^6xy+21\cdot 2^5y^2$$
The coefficient is $21\cdot 2^6$
Is this correct? Is there a simpler proof?
| If you know derivatives, denoting
$$f(x,y)=(x+y+2)^7=\sum_{i,j=0}^7 c_{ij}x^iy^j$$
then you are looking for $c_{11}$. It is easy to see that
$$c_{11}=\frac{\partial^2 f}{\partial x \partial y}(0,0)=7 \cdot 6 \cdot (0+0+2)^5$$
Intuitively: derivating with respect to $x$, and then setting $x=0$ eliminates all the terms which contain any power of $x$ besides the first power. You do then the same for $y$,
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1920638",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
} |
Transforming the equation $\cot x -\cos x = 0$ into the form $\cos x(1- \sin x) = 0$
I am required to write the equation $$\cot x - \cos x = 0$$
in the form
$$\cos x(1 - \sin x) = 0$$
What I reached is as follows,
\begin{align}
\cot x & = \frac{\cos x}{\sin x}\\[4pt]
\frac{\cos x}{\sin x} - \cos x & = 0\\[4pt]
\cos x\left(\frac{1}{\sin x} - 1\right) & = 0\\[4pt]
\cos x\left(\frac{1}{\sin x} - \frac{\sin x}{\sin x}\right) & = 0
\end{align}
How can I rewrite in the above format?
| $$
\cot(X)-\cos(X) = 0 \implies \frac{\cos X}{\sin X} - \cos X = 0 \implies \cos X\bigg(\frac{1}{\sin X} - 1\bigg) = 0
$$
$$
\implies \frac{\cos X(1 - \sin X)}{\sin X} = 0 \implies \cos X(1 - \sin X) = 0 \cdot \sin X = 0
$$
This applies when $\sin X \neq 0$. If it is zero, then $\cot X = \infty$ so the equation is not satisfied.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1921345",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
What is this number: $x = \sqrt{1+2^1\sqrt{1+2^2\sqrt{1+2^3\sqrt{1+...}}}}\approx 4.14031...$? I was watching this Mathologer video about Ramanujan's nested radical identity
$$
3 = \sqrt{1 + 2\sqrt{1+3\sqrt{1+4\sqrt{1+...}}}}\, ,
$$
and I decided to look at this for different sets of "radical coefficients". Using powers of $2$ yields the quantity
$$
x = \sqrt{1+2^1\sqrt{1+2^2\sqrt{1+2^3\sqrt{1+...}}}}\, .
$$
I can't figure out much about the nature of the number $x$. I can't even prove that it exists, let alone anything analytically about its value. Nevertheless, by looking at the truncation of this expression at different levels in Mathematica, it seems to be the case that:
1) The expression converges relatively quickly.
2) $x \approx 4.14031456214125981180937290...$
I tried feeding this number into an Inverse Symbolic Calculator... to no avail, although $x$ does match about 8 digits of
$$
\frac{5}{4}\, \frac{\sqrt{3} + \sqrt{5}\ln(5)}{\ln(5)}\, .
$$
Can anyone tell me anything else about this quantity?
Edited to add:
I note that the apparently simpler Nested Radical Constant
$$
\sqrt{1 + \sqrt{2 + \sqrt{3 + ...}}}
$$
has no closed form, so we clearly have no reason to expect my quantity $x$ to have a closed form either...
Edit 2:
There was an answer that someone posted, and then deleted for some reason, which gave a method for getting a lower bound on this quantity. A variation of that method leads to:
\begin{align}
x &= \sqrt{1+2^1\sqrt{1+2^2\sqrt{1+2^3\sqrt{1+...}}}}\\
&> \sqrt{1+2^1\sqrt{0+2^2\sqrt{0+2^3\sqrt{0+...}}}}\\
&= \sqrt{1+2^1 \times 2^{2/2}\times 2^{3/4}\times 2^{4/8}\times 2^{5/16}...}\\
&= \sqrt{1 + 2^{\sum_{m=0}^\infty \frac{m+1}{2^m}}}\\
&= \sqrt{1 + 2^4} = \sqrt{17} \approx 4.1231...
\end{align}
This is apparently a relatively tight lower bound.
| Interpreting the sequence as mvw, we have $x_1<\sqrt{2^1+2^1}=2$, $x_2<\sqrt{2^1\sqrt{1+2^2}+2^1\sqrt{1+2^2}}=2\sqrt[4]{1+2^2}<2^{1+3/4}$ and similarly $$x_n<2\sqrt[4]{2^2\sqrt{\cdots\sqrt{2^n+2^n}} +2^2\sqrt{\cdots\sqrt{2^n+2^n}}}=2^{2/2+3/2^2+\cdots+(n+1)/2^n}<2^3.$$ The modus operandi is to turn each $1$ into the term that stands at the right of it, which is certainly bigger than $1$. Since $x_n$ is also increasing, its limit exists. One might as well note that the analogous sequence with powers of any fixed $m>1$ does converge, and if $m=1$ it converges to the golden ratio.
I guess my method can be refined, or there is another one, to yield a better bound.
UPDATE
I thought it wouldn't be easy to majorize effectively and more sharply the sequence in a similar fashion, but I was wrong. It suffices to turn each $1$ into half the term that stands at the right of it. We thus get, for $n>1$, \begin{align} x_n &<\sqrt{(2^0+2^1)\sqrt{(2^1+2^2)\sqrt{\cdots\sqrt{2^{n-1}+2^n}}}} \\ &=2^{1/4+2/8+\cdots+(n-1)/2^n}\cdot3^{1/2+1/4+\cdots+1/2^n}<6. \end{align}
UPDATE 2 Finally nailed it! Let $P_m=\prod_{n=1}^\infty(1+2^{n+m})^{1/2^n}$. Then we find $$x<\sqrt{(1+2^1)\sqrt{(1+2^2)\sqrt{\cdots}}}=P_0<5.284,$$ $$x<\sqrt{1+2\sqrt{(1+2^2)\sqrt{(1+2^3)\sqrt{\cdots}}}}=\sqrt{1+2P_1}<4.429,$$ $$x<\sqrt{1+2\sqrt{1+2^2\sqrt{(1+2^3)\sqrt{(1+2^4)\sqrt{\cdots}}}}}=\sqrt{1+2\sqrt{1+4P_2}}<4.215 $$ etc.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1922740",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 2,
"answer_id": 1
} |
Complex Solutions to Polynomial Equations I have to find another complex solution to the polynomial equation $z^6 - 1 = 0$ given that $\frac{1}{2}−\frac{\sqrt{3}}{2}i$ is a solution.
I guess the question I have for this is that is there any other way other than expanding the brackets out completely?
| $$x^6-1=(x^3-1)(x^3+1)=(x-1)(x^2+x+1)(x+1)(x^2-x+1)$$
the roots are as follows
$$x_1=1$$
$$x_2=-1$$
roots of $x^2+x+1$ are
$$x_{3,4}=-\frac{1}{2}\pm \frac{\sqrt{3}i}{2}$$
roots of $x^2-x+1$ are
$$x_{5,6}=\frac{1}{2}\pm \frac{\sqrt{3}i}{2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1922994",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Finding $\lim_{x \to \infty} \left(\frac{a_{1}^{1/x}+a_{2}^{1/x}+\cdot\cdot\cdot{a_{n}}^{1/x}}{n}\right)^{nx}$ For non zero positive reals $a_{1},a_{2}\cdot\cdot\cdot a_{n}$ how to find
$$
\lim_{x \to \infty}
\left(\frac{a_1^{1/x} +a_2^{1/x}+\ldots +a_n^{1/x}}{n}\right)^{nx}?
$$
It becomes indeterminant form $1^{\infty}.$ But difficult to solve by L'Hospital's Rule. By using A.M.-G.M. inequality it comes that limit is $\geq a_{1}a_{2}\cdot\cdot\cdot a_{n}.$ I also tried by using Squeeze. Please help.Thanks.
| tl;dr
For $a_1,\dots, a_n$ positive numbers, and $x>0$, $$
\lim_{x\to\infty}\left(\frac{a_{1}^{\frac{1}{x}}+a_{2}^{\frac{1}{x}}+\cdots+{a_{n}}^{\frac{1}{x}}}{n}\right)^{nx} = \prod_{k=1}^na_{k}
$$
Proof. Detailed approach, where we use the (low-order) Taylor expansions $e^u=1+u+o(u)$ and $\ln(1+u)=u+o(u)$ when $u\to 0$.
You can rewrite
$$\begin{align}
\left(\frac{a_{1}^{\frac{1}{x}}+a_{2}^{\frac{1}{x}}+\cdots+{a_{n}}^{\frac{1}{x}}}{n}\right)^{nx}
&=
\exp\left(nx \ln \frac{a_{1}^{\frac{1}{x}}+a_{2}^{\frac{1}{x}}+\cdots+{a_{n}}^{\frac{1}{x}}}{n}\right)
= \exp\left(nx \ln \frac{1}{n}\sum_{k=1}^na_{k}^{\frac{1}{x}}\right)\\
&=
\exp\left(nx \ln \frac{1}{n}\sum_{k=1}^n e^{\frac{1}{x}\ln a_{k}}\right)
\end{align}$$
Since $n$ is a constant and $\frac{1}{x}\ln a_{k}\xrightarrow[x\to\infty]{} 0$ for each $k$, we have
$$
e^{\frac{1}{x}\ln a_{k}} = 1 + \frac{1}{x}\ln a_{k} + o\left(\frac{1}{x}\right)
$$
for each $k$ as $x\to\infty$, and therefore
$$\begin{align}
\left(\frac{a_{1}^{\frac{1}{x}}+a_{2}^{\frac{1}{x}}+\cdots+{a_{n}}^{\frac{1}{x}}}{n}\right)^{nx}
&=
\exp\left(nx \ln \frac{1}{n}\sum_{k=1}^n e^{\frac{1}{x}\ln a_{k}}\right)\\
&=
\exp\left(nx \ln\left( 1+ \frac{1}{nx}\sum_{k=1}^n\ln a_{k}+ o\left(\frac{1}{x}\right)\right)\right) \\
&= \exp\left(nx \left( \frac{1}{nx}\sum_{k=1}^n\ln a_{k}+ o\left(\frac{1}{x}\right)\right)\right) \\
&= \exp\left(\sum_{k=1}^n\ln a_{k}+ o\left(1\right)\right) \\
&= \exp\left(\ln \prod_{k=1}^na_{k}+ o\left(1\right)\right)
= e^{o(1)} \prod_{k=1}^na_{k} \\
& \xrightarrow[x\to\infty]{} \prod_{k=1}^na_{k}.
\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1924073",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Solve the equation: $x=\sqrt{a-\sqrt{a+x}},(a\geq 1)$
Solve the equation: $x=\sqrt{a-\sqrt{a+x}},(a\geq 1)$
We have: $x^2=a-\sqrt{a+x}$ , if we take both sides to the power of 2 we will get a 4th order equation which I don't know of!
Please help.
| We certainly have $x>0$. Then by squaring, the original equation can be written
$$a-x^2=\sqrt{x+a}.$$
The two members describe the parabolas of equations
$$\begin{cases}y=a-x^2,\\x=y^2-a\end{cases}$$ where $y>0$.
Adding these two equalities we get a pair of straight lines
$$y+x=y^2-x^2,$$ i.e. $$(y+x)(y-x-1)=0.$$
Hence, the requested solution lies in the first quadrant and is the positive root of
$$\begin{cases}y-x-1=0\\y=a-x^2\end{cases},$$
$$x^2+x+1-a=0,$$
$$x=\frac{-1+\sqrt{4a-3}}{2}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1926750",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 7,
"answer_id": 5
} |
show inequality $\frac{1}{128}\ge \sum_{1\le i\neq j\leq n}a_i^5a_j^3$ Let $a_1,...a_n$ be positive reals with sum 1. Show that
$$\frac{1}{128}\ge \sum_{1\le i\neq j\leq n}a_i^5a_j^3$$
It seems hard to show it .
| We'll prove the statement for nonnegative reals. Firstly, note that for any $a_i, a_j$ we have $(a_i-a_j)^2\ge 0\implies a_ia_j\le \frac{1}{4}(a_i+a_j)^2\le \frac{1}{4}$. Hence, the sum at question is at most $\frac{1}{16}\sum_{i\neq j}a_i^3a_j=\frac{1}{16}\sum_i a_i^3(1-a_i)$, so it suffices to check that $\sum_i a_i^3(1-a_i)\le \frac{1}{8}$ when $a_i$ sum to $1$. This we may do by induction on $n$. For the base case $n=2$, this is just $8a_1a_2(a_1^2+a_2^2)\le 1=(a_1+a_2)^4\Leftrightarrow (a_1-a_2)^4\ge 0$.
Now suppose the inequality holds for $n=k$. Suppose $a_1, \ldots, a_{k+1}$ have sum one, and WLOG suppose $a_1\le a_2\le \cdots \le a_{k+1}$ so that $a_1+a_2\le \frac{2}{n}\le \frac{3}{4}$. Then we claim that the sum in question increased under the transformation $(a_1, a_2)\to (0, a_1+a_2)$. Indeed, the difference of the sum after this transformation and before it is:
$$(a_1+a_2)^3-a_1^3-a_2^3-(a_1+a_2)^4+a_1^4+a_2^4=a_1a_2(3a_1+3a_2-4a_1^2-6a_1a_2-4a_2^2)\ge a_1a_2(4(a_1+a_2)^2-4a_1^2-6a_1a_2-4a_2^2)\ge 0$$
So, it suffices to check that $\sum_{i=1}^kb_i^3(1-b_i)\le \frac{1}{8}$, where $b_1=a_1+a_2, b_i=a_{i+1}, 2\le i\le k$. But this is covered by the inductive hypothesis (after reindexing the $b_i$), so the induction is complete and we're done.
Equality holds when two of the variables are $\frac{1}{2}$ and the rest are $0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1928070",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How to break $\ln(x+i \cdot y)$ into real and imaginary components? Wolfram shows that there are indeed Re and Im components without stating them: http://www.wolframalpha.com/input/?i=ln(x%2Bi*y)
How do I break $\ln(x+i \cdot y)$ into real and imaginary components?
| We want to express $x+iy$ as $re^{i\theta}$, for some value of $r$ and $\theta$.
First let's normalize $x+iy$ as $\sqrt{x^2+y^2}(\frac{x}{\sqrt{x^2+y^2}}+\frac{y}{\sqrt{x^2+y^2}}i)$.
We see that $r=\sqrt{x^2+y^2}$ and we see that $\tan(\theta)=\frac{y}{x}$. Therefore, we have $\theta=\arctan(\frac{y}{x})$. We can then write $x+iy=\sqrt{x^2+y^2}e^{i\arctan(\frac{y}{x})}$. So we have:
$$\ln(x+iy)=\ln(\sqrt{x^2+y^2}e^{i\arctan(\frac{y}{x})})$$
$$\ln(x+iy)=\frac{1}{2}\ln(x^2+y^2)+i\arctan(\frac{y}{x})$$
Our real part is $\frac{1}{2}\ln(x^2+y^2)$ and our imaginary part is $i\arctan(\frac{y}{x})$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1928333",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Why does this series expansion for $\frac{1}{\cos(x)}$ fail using $\frac{1}{1-x}$? I'm trying to do (2nd order) Taylor expansion for
$$\frac{1}{\cos(x)}$$ using $$\frac{1}{1-x}$$
What I do is write
$$\cos(x)=1-\frac{x^2}{2!}+\frac{x^4}{4!}+ \cdot\cdot\cdot, \text{ at } x_0=0$$
and
$$\frac{1}{1-x}=1+x+x^2+\cdot\cdot\cdot$$
so
$$\frac{1}{\cos(x)}=\frac{1}{1-(1-\cos(x))}=1+(1-\cos(x))+(1-\cos(x))^2$$
$$=1+(1-1+\frac{x^2}{2!})+(1-1+\frac{x^2}{2!})^2$$
$$=1+\frac{1}{2}x^2+\frac{1}{4}x^4$$
which is correct up to order 2 (or order 1?), but the coefficient of $x^4$ (is that 3rd or 4th order?) is wrong.
Why? How can I get the correct coefficients for higher orders?
And if I do want only 2nd order Taylor polynomial, then what is the term $\frac{1}{4}x^4$?
| Why do you make things more complex than what they are?
If you want the expansion of this function, say up to order $6$, simply write $\;\dfrac1{\cos x}=\dfrac1{1-\dfrac{x^2}2+\dfrac{x^4}{24}-\dfrac{x^6}{720}+o(x^7)}$, for instance, and set
$$u=\dfrac{x^2}2-\dfrac{x^4}{24}+\dfrac{x^6}{720}+o(x^7)$$
Now do the calculations for $\dfrac1{1-u}$ up to order $3$ (since the polynomial part of $u$ has order $2$), truncating the results for each power at order $6$:
\begin{alignat}{5}
1+u&=1+\frac{x^2}2&&-\frac{x^4}{24}&&+\frac{x^6}{720}&&+o(x^7)\\
{}+u^2&=&&+\frac{x^4}{4}&&-\frac{x^6}{24}&&+o(x^7)\\
{}+u^3&=&&&&+\frac{x^6}{8}&&+o(x^7)\\
\hline
\frac1{\cos x}&=1+\frac{x^2}2&&+\frac{5x^4}{24}&&+\frac{59x^6}{24}&&+o(x^7)
\end{alignat}
Alternatively, you can divide $1$ by $1-\dfrac{x^2}2+\dfrac{x^4}{24}-\dfrac{x^6}{720}$ by increasing powers (not the Euclidean division!) up to degree $6$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1930185",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Find the number of ways to select 2n balls from n identical blue balls, n identical red balls and n identical white balls, where n $\in$ $\mathbb N$ Q: Find the number of ways to select 2n balls from n identical blue balls, n identical red balls and n identical white balls, where n $\in$ $\mathbb N$.
My working:
$(x+x^2+x^3+...)^3$
$=x^3(1+x+x^2+...)^3$
$= x^3 \sum_{r=0}^\infty \begin{pmatrix} {r+3-1}\\{r}\end{pmatrix}x^r$
$= \sum_{r=0}^\infty \begin{pmatrix} {r+2}\\{r}\end{pmatrix}x^{3+r}$
Hence the number of ways is $\begin{pmatrix} {2n-1}\\{2n-3}\end{pmatrix}$=$\begin{pmatrix} {2n-1}\\{2}\end{pmatrix}$.
However, the actual answer is $\begin{pmatrix} {2n+2}\\{2}\end{pmatrix}$$-3$$\begin{pmatrix} {n+1}\\{2}\end{pmatrix}$. Did i go wrong somewhere in my proof? Or is it some conceptual understanding gone wrong? Thanks!
| This is same as the number of solutions to $$x+y+z = 2n$$ where $0 \leq x, y, z \leq n$. Thus we need the coefficient of $t^{2n}$ in
\begin{align*}
(1+t+t^2+ \cdots +t^n)^3 &= \left(\frac{1-t^{(n+1)}}{1-t}\right)^3\\
&=(1-3t^{(n+1)}+3t^{2(n+1)}-t^{3(n+1)})\left(1+\binom{3}{1}t + \binom{4}{2}t^2+ \cdots +\binom{3+k}{k+1}t^{k+1} + \cdots \right)
\end{align*}
Thus the required coefficient is
$$\binom{2n+2}{2n} - 3 \binom{n+1}{n-1} = \binom{2n+2}{2} - 3 \binom{n+1}{2} $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1932378",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
The roots of the cubic equation, $2x^3+px^2-(3+5i)x+q=0$ are $x\in \{ i , \frac{1}{k}, -1-k\} $ Find $p,q$ and $k$.
The roots of the cubic equation, $2x^3+px^2-(3+5i)x+q=0$ are $x\in \{ i , \frac{1}{k}, -1-k\} $
Use this information to find $p,q$ and $k$, where $k$ is real.
What I have done:
Consider $p(x)=2x^3+px^2+(-3-5i)x+q=0$
Since $x=i$ , $x=\frac{1}{k}$ and $x=-1-k$ are roots we will try subbing them in and seperating imaginary and real parts and solving simultaneously hopefully!
Starting of with $x=i$
$$\Longrightarrow p(i)=2(i)^3+p(i)^2+(-3-5i)i+q=0$$
$$ \Leftrightarrow -2i -p +5 - 3i +q=0$$
$$ \Leftrightarrow q-p+5-5i=0$$
$$ \Leftrightarrow (q-p+5)+i(-5)=0$$
So one equation we have is $q-p+5=0$
Next $x=\frac{1}{k}$
$$\Longrightarrow p(\frac{1}{k})=2(\frac{1}{k})^3+p(\frac{1}{k})^2+(-3-5i)\frac{1}{k}+q=0$$
$$\Leftrightarrow \frac{2}{k^3} + \frac{p}{k^2} - \frac{3}{k} - \frac{5i}{k}+q=0$$
$$ \Leftrightarrow (\frac{2}{k^3} + \frac{p}{k^2} - \frac{3}{k} +q) + i(\frac{-5}{k})=0$$
So another equation we have is $\frac{2}{k^3} + \frac{p}{k^2} - \frac{3}{k} +q=0$ since $\frac{-5}{k} \neq 0$
The last root we have is $x=-1-k$
$$\Longrightarrow p(-1-k)=2(-1-k)^3+p(-1-k)^2+(-3-5i)(-1-k)+q=0$$
$$ \Leftrightarrow -2(k+1)^3+p(k+1)^2+3+3k+5i+5ki+q=0 $$
$$ \Leftrightarrow (-2(k+1)^3+p(k+1)^2+3+3k+q)+i(5+5k) =0 $$
So 2 other equations we have are $-2(k+1)^3+p(k+1)^2+3+3k+q=0$ and $5+5k=0$
Hence the 4 equations we have are
$$q-p+5=0 $$
$$\frac{2}{k^3} + \frac{p}{k^2} - \frac{3}{k} +q=0$$
$$-2(k+1)^3+p(k+1)^2+3+3k+q=0$$
$$5+5k=0$$
For the last equation $k=-1$ and subbing this into the third equation I get $q=0$
and subbing that into the second equation I get $p=1$ but subbing $q$ and $p$ into $q-p+5$ does not make it true so where did I go wrong? I also do not think I have made any algebraic error.
| Via long division of the polynomial by $x-i$ we get
$$2x^3-px^2-(3+5i)x+q=(x-i)(2x^2+(p-2i)x+(-1+(p-5)i)$$
Immediately we run into an error in your working, since $q=i+(p-5)$, not $q-p+5=0$. Rewinding to the line where this false relation is introduced:
$$(q-p+5)+i(-5)=0$$
we see that going from this to $q-p+5=0$ implies that $-5i=0$, which is absurd.
The other relations you derive (with $x=\frac1k,-1-k$) are also flawed. The correct equations those incorrect relations stem from:
$$\left(\frac{2}{k^3} + \frac{p}{k^2} - \frac{3}{k} +q\right) + i\left(\frac{-5}{k}\right)=0$$
$$(-2(k+1)^3+p(k+1)^2+3+3k+q)+i(5+5k) =0$$
are of the form $a+bi=0$, but you cannot say that $a=0$ or $b=0$ in this case as you did because $a$ and $b$ can be complex here. They need to be restricted to the real numbers in order for you to conclude that $a=b=0$.
We shall solve for $p,q,k$ anyway. The remaining roots after the long division are all real, so the coefficients of the quotient polynomial must be too. $p-2i$ real implies $p=a+2i$ with $a$ real. $-1+(p-5)i=-3+(a-5)i$ is also real, so $a=5$, $p=5+2i$ and our remaining quadratic equation is
$$2x^2+5x-3=0$$
which turns out to have roots $\frac12$ and $-3$, so $k=2$. $q$ works out to be $i+(5+2i-5)=3i$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1932487",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
How is this property of real numbers proven?
Prove that $\dfrac{x^2}{(x − 1)^2} + \dfrac{y^2}{(y − 1)^2} + \dfrac{z^2}{(z − 1)^2} ≥ 1$ for all real numbers $x, y, z$, each different from $1$ and satisfying $xyz = 1$.
How do I prove this?
| it should be $$\frac{(xy+yz+zx-3)^2}{(x-1)^2(y-1)^2(z-1)^2}\geq 0$$
Hint: set $$x=a/b,y=b/c,z=c/a$$ in the given term
after the substituion we obtain
$${\frac {{a}^{2}}{{b}^{2}} \left( {\frac {a}{b}}-1 \right) ^{-2}}+{
\frac {{b}^{2}}{{c}^{2}} \left( {\frac {b}{c}}-1 \right) ^{-2}}+{
\frac {{c}^{2}}{{a}^{2}} \left( {\frac {c}{a}}-1 \right) ^{-2}}-1
$$
factoring all we get
$${\frac { \left( b{a}^{2}-3\,bca+{c}^{2}a+{b}^{2}c \right) ^{2}}{
\left( a-b \right) ^{2} \left( b-c \right) ^{2} \left( -c+a \right) ^
{2}}}
\geq 0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1932944",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
} |
Solve the equation $z^{3}=18+26i$, where $z=x+yi$ and $x,y$ are integers. $(x+yi)^{3}=(x^{3}\text{−}3xy^{2})+(3x^{2}y\text{−}y^{3})i=18+26i$
which leads to $$\begin{cases}
x^{3}\text{−}3xy^{2} & =18\\
3x^{2}y\text{−}y^{3} & =26
\end{cases}$$
The problem here is that I don't know to precede further.
I've been told to set $y=tx$ in the equatlity $18(3x^{2}y\text{−}y^{3})=26(x^{3}\text{−}3xy^{2})$ and then if $x\neq0$ and $y\neq0$ then $18(3t\text{−}t^{3})=26(1\text{−}3t^{2})$ which is equivalent to $$(3t−1)(3t^{2}-12t-13)=0$$. The only rational solution of this equation is $t=\frac{1}{3}$; hence, $x=3,y=1$, and $z=3+i$.
But I don't understand that solution (the substitution $y=tx$ to be specific) and how they ended up with $18(3t\text{−}t^{3})=26(1\text{−}3t^{2})$ from that substitution?
| $$x^{3}\text{−}3xy^{2} =18 \Rightarrow x(x^2-3y^2) = 18.$$
This means that $x$ is a divisor of $18$, i.e. $x \in \{1,2,3,6,9,18\}$.
Similarly:
$$3x^{2}y\text{−}y^{3} =26 \Rightarrow y(3x^2-y^2) = 26,$$
which means that $y$ is a divisor of $26$, i.e. $y \in \{1,2,13,26\}$.
Moreover, we know that:
$$|z^3|^2 = 18^2 + 26^2 = 1000.$$
Then, $|z|^2 = 1000^{\frac{1}{3}} = 10 = x^2 + y^2$
We can reduce the sets for $x$ and $y$, since $x^2 < 10$ and $y^2 < 10$. We get:
$$\begin{cases}x \in \{1,2,3\}\\
y \in \{1,2\}
\end{cases}$$
It is clear now that the only solution is $x= 3$ and $y=1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1934805",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Compute $\sum\limits_{n=1}^\infty \frac{5+4n-1}{3^{2n+1}}$ I have troubles finding the limit of the following series: $\sum_{n=1}^\infty \frac{5+4n-1}{3^{2n+1}}$
So far I figured it'd easier to split the sum into:
$\sum_{n=1}^\infty \frac{5}{3^{2n+1}} \sum_{n=1}^\infty \frac{4n-1}{3^{2n+1}}$
= $\sum_{n=1}^\infty 5 \cdot\frac{1}{3^{2n+1}} +\sum_{n=1}^\infty 4n-1 \cdot \frac{1}{3^{2n+1}}$
And with $\sum_{n=1}^\infty \frac{1}{w^n} = \frac{1}{w-1}$ you get the following terms:
$5\cdot \frac{1}{3^{2n+1}-1} + 4n-1\cdot \frac{1}{3^{2n+1}-1}$
which is bascially a sequence, but im stuck right here.. help is very appreciated!
| Hint. We have that for $|x|<1$,
$$\sum_{n=0}^{\infty}x^n=\frac{1}{1-x}\quad\mbox{and}\quad \frac{d}{dx}\left(\sum_{n=0}^{\infty} x^n\right)=\sum_{n=0}^{\infty}n x^{n-1}=\frac{1}{(1-x)^2}.$$
Hence
$$\sum_{n=1}^\infty \frac{5+4n-1}{3^{2n+1}}=\sum_{n=1}^\infty \frac{4+4n}{3\cdot 9^{n}}=\frac{4}{3}\sum_{n=1}^\infty (1/9)^{n}+\frac{4}{27}\sum_{n=1}^\infty n(1/9)^{n-1}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1935546",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Sum of a special type of series Is there a general method to derive the formula (without using mathematical induction) for the sum of the series of type $1*2+2*3+3*4+4*5+...+n(n+1)$ or perhaps $\frac{1}{(2*3*4)}+\frac{1}{(3*4*5)}+\frac{1}{(5*6*7)}+...+\frac{1}{(n*(n+1)*(n+2))}$?
I am a high school student so an elaborate explanation with example would help me best. I am looking for the formulas of those two series but I am more interested in the method of derivation of the formulas and also any general method by which such type of problems can be solved.
| Given that the rising and falling factorials are defined as
$$
\begin{array}{l}
n^{\,\overline {\,m\,} } = n\left( {n + 1} \right)\, \cdots \;\left( {n + m - 1} \right) \\
n^{\,\underline {\,m\,} } = n\left( {n - 1} \right)\, \cdots \;\left( {n - \left( {m + 1} \right)} \right) \\
n^{\,\underline {\, - \,m\,} } = \frac{1}{{\left( {n + m} \right)^{\,\underline {\,m\,} } }} = \frac{1}{{\left( {n + 1} \right)^{\,\overline {\,m\,} } }} \\
\end{array}
$$
where here we consider $n$ and $m$ integers,
and that the forward finite difference (integer $q$) is defined as:
$$
\begin{array}{l}
\Delta _{\,n} \;f(n) = f(n + 1) - f(n)\quad \quad \Delta _{\,n} ^q \;f(n) = \Delta _{\,n} \;\left( {\Delta _{\,n} ^{q - 1} \;f(n)} \right) \\
\Delta _{\,n} ^{ - 1} \;f(n) = \sum\limits_{k\, = \,a}^{n - 1} {f(k)} + c \\
\end{array}
$$
then for the falling and rise factorials we have in general
$$
\begin{array}{l}
\Delta _{\,n} ^q \,n^{\,\overline {\,m\,} } = m^{\,\underline {\,q\,} } \,\left( {n + q} \right)^{\,\overline {\,m - q\,} } \quad \quad \Delta _{\,n} ^q \,n^{\,\underline {\,m\,} } = m^{\,\underline {\,q\,} } \,n^{\,\underline {\,m - q\,} } \\
\sum\limits_{k\, = \,1}^{n - 1} {k^{\,\overline {\,m\,} } } \quad \left| {\; - 1 \ne m} \right.\,\quad = \frac{1}{{m + 1}}\,\left( {\left( {n - 1} \right)^{\,\overline {\,m + 1\,} } - 0^{\,\overline {\,m + 1\,} } } \right) = \frac{1}{{m + 1}}\,\left( {n - 1} \right)^{\,\overline {\,m + 1\,} } \\
\sum\limits_{k\, = \,1}^{n - 1} {k^{\,\underline {\, - m\,} } } \quad \left| {\;1 \ne m} \right.\,\quad = \sum\limits_{k\, = \,1}^{n - 1} {\frac{1}{{\left( {k + 1} \right)^{\,\overline {\,m\,} } }}} = \frac{1}{{1 - m}}\,\left( {n^{\,\underline {\,1 - m\,} } - 1^{\,\underline {\,1 - m\,} } } \right) = \\
= \frac{1}{{1 - m}}\,\left( {\frac{1}{{\left( {n + 1} \right)^{\,\overline {\,m - 1\,} } }} - \frac{1}{{2^{\,\overline {\,m - 1\,} } }}} \right) \\
\end{array}
$$
In the two particular cases you indicate, then
$$
\begin{array}{l}
\sum\limits_{k\, = \,1}^n {k^{\,\overline {\,2\,} } } = \sum\limits_{k\, = \,1}^n {k\left( {k + 1} \right)} = \frac{1}{3}\,n^{\,\overline {\,3\,} } \\
\sum\limits_{k\, = \,1}^{n - 1} {k^{\,\underline {\, - 3\,} } } = \sum\limits_{k\, = \,2}^n {\frac{1}{{k\left( {k + 1} \right)\left( {k + 2} \right)}}} = \\
= \frac{1}{{1 - 3}}\,\left( {\frac{1}{{\left( {n + 1} \right)^{\,\overline {\,2\,} } }} - \frac{1}{{2^{\,\overline {\,2} } }}} \right) = \frac{1}{2}\,\left( {\frac{1}{6} - \frac{1}{{\left( {n + 1} \right)\left( {n + 2} \right)}}} \right) \\
\end{array}
$$
Besides the above, you may also consider the following.
The first type, can be translated into
$$
n\left( {n + 1} \right) = n^{\,\overline {\,2\,} } = \sum\limits_{0\, \le \,k\, \le \,2} {\left[ \begin{array}{c}
2 \\
k \\
\end{array} \right]n^{\,k} }
$$
where $n^{\,\overline {\,2\,} }$ is the rising factorial and ${\left[ \begin{array}{c} 2\\ k\\ \end{array} \right]x^{\,k} } $
is the unsigned Stirling N. of 1st kind.
So the whole boils down into a combination of the sum of $n$ and $n^2$,
(and for exponents greater than $2$ will involve Bernoulli numbers).
Concerning the second, instead, it can be splitted by partial fractions into
$$
\frac{1}{{n\left( {n + 1} \right)\left( {n + 2} \right)}} = \frac{1}{2}\frac{1}{n} - \frac{1}{{n + 1}} + \frac{1}{2}\frac{1}{{n + 2}}
$$
and thus solved in terms of harmonic numbers in general, while in this simple case, the sum will turn out in cancelling many of terms as pointed out by Mariuslp in his comment.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1935715",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Prove $\sqrt{3} + \sqrt{5}$ is irrational If it is assumed that $\sqrt{3}$ is known to be irrational (not the case for $\sqrt{5}$), then prove that $\sqrt{3}+\sqrt{5}$ is irrational.
My approach:
Assume that $\sqrt{3}+\sqrt{5}$ is rational. Then there exist coprime integers $p$ and $q$ so that $\frac{p}{q}$ is rational and $\frac{p}{q}=\sqrt{3}+\sqrt{5}$. Thus $(\sqrt{3}+\sqrt{5})^2=2(4+\sqrt{15})=\frac{p^2}{q^2}$, which implies that $p^2$ is even, so $p$ is also even. Let $p:=2m$ for some integer $m$, then $p^2=4m^2$. Thus $q^2(4+\sqrt{15})=2m^2$, which implies that $q^2$ is even, and so $q$ is even. But this contradicts that $p$ and $q$ are coprime, and we arrive at a contradiction.
My way of proving this does not use the fact that $\sqrt{3}$ is
irrational. Please let me know if my proof is correct, and how to use
the above mentioned fact.
| Let $x=\sqrt 3 + \sqrt 5$. Then:
$$(x - \sqrt 3)^2 = 5$$
$$x^2 - 2 \sqrt 3 x + 3 = 5$$
$$x^2 - 2 = 2 \sqrt 3 x$$
$$(x^2 - 2)^2 = 12 x^2$$
$$x^4 -16 x^2 + 4 = 0$$
By the rational root theorem, if the last equation had rational roots then they would be integer divisors of $4$. But none of $\pm1, \pm2,\pm4$ are roots, so $x$ must be irrational.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1938023",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 1
} |
Sum of square roots........... $$
\mbox{If}\quad S =
1 + \,\sqrt{\,\frac{1}{2}\,}\, + \,\sqrt{\,\frac{1}{3}\,}\, +
\,\sqrt{\,\frac{1}{4}\,}\, +
\,\sqrt{\,\frac{1}{5}\,}\, + \cdots + \,\sqrt{\,\frac{1}{100}\,}\,\,,
$$
then what is the value of $\left\lfloor\,S\,\right\rfloor$ ?.
Here $\left\lfloor\,S\,\right\rfloor$ is the greatest integer function which is less than or equal to $S$.
| Hint:
$$S=\sum_{k=1}^{100}\frac{1}{\sqrt{k}}$$
$$\frac{1}{\sqrt{k+1}+\sqrt{k}}\lt \frac{1}{2\sqrt{k}}\lt \frac{1}{\sqrt{k+\frac{1}{2}}+\sqrt{k-\frac{1}{2}}}$$
$$\implies \sqrt{k+1} -\sqrt{k} \lt \frac{1}{2\sqrt{k}}\lt \sqrt{k+\frac12}-\sqrt{k-\frac{1}{2}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1939540",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
How to prove that $\sum\limits_{k=0}^{n} \binom{n+1}{k+1} = 2^{n+1} - 1$ using the Binomial Theorem? I have this proposition:
$$\sum_{k=0}^{n} \binom{n+1}{k+1} = 2^{n+1} - 1$$
How can I prove that? How to use the Binomial Theorem to solve that?
| Use inductive approach
For n = 0: $$\binom{1}{1} = 1 = 2^1 - 1$$ => TRUE
Assume that for n>=0, the following (*) is true $$\sum_{k=0}^{n} \binom{n+1}{k+1} = 2^{n+1} - 1$$
We need prove: $$\sum_{k=0}^{n+1} \binom{n+2}{k+1} = 2^{n+2} - 1$$
Remember: $$\binom{n}{k} = \binom{n-1}{k} + \binom{n-1}{k-1}$$
So: $$\binom{n+2}{k+1} = \binom{n+1}{k+1} + \binom{n+1}{k}$$
=> $$\sum_{k=0}^{n+1} \binom{n+2}{k+1} = \sum_{k=0}^{n+1}\binom{n+1}{k+1}+\sum_{k=0}^{n+1}\binom{n+1}{k}$$
$$=\binom{n+1}{n+2} + \sum_{k=0}^{n}\binom{n+1}{k+1} + \sum_{k=0}^{n}\binom{n+1}{k+1} - \binom{n+1}{0}$$
$$=2*\sum_{k=0}^{n}\binom{n+1}{k+1} - 1 = 2*2^{n+1} - 1 = 2^{n+2}-1$$
That is what to be proven.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1939729",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
Please help me understand the steps to simplify this radical Question: $$\frac{\frac{1}{\sqrt {x+h}}- \frac{1}{\sqrt x}}{h}$$
Solution given:
$$= \frac{1}{h} \cdot\frac{\sqrt x - \sqrt {x+h}} {\sqrt {x + h}\sqrt x} $$
$$= \frac{1}{h} \cdot\ \frac{x - (x+h)}{\sqrt{x + h} \sqrt x (\sqrt{x} + \sqrt{x + h})}$$
$$= \frac{1}{h} \cdot\ \frac{x - x - h}{x \sqrt{x + h} + (x + h) \sqrt x}$$
$$= \frac{1}{h} \cdot\ \frac{-h}{x \sqrt{x + h} + (x + h) \sqrt x}$$
$$= -\frac{1}{x \sqrt{x + h} + (x + h) \sqrt x}$$
I've studied and understood the material up to this point just fine. I get about rationalizing stuff, conjugate pairs etc, but I can't figure out of what the author has between each step to get to the next.
I can only comprehend the first and possibly the second step.
Source: http://www.themathpage.com/alg/multiply-radicals.htm
See problem 10, the last problem on the page.
| Step 1: Delete the chain break and multiply with the denominator of the other fraciton
Step 2: Use Binomial formula since $(a-b)(a+b)=a^2-b^2$
Step 3: Use Associative law of addition multiply everything out in the denominator
Step 4: $(x-x)=0$
Step 5: $\frac{-h}{h} = -1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1940187",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
In convex quadrilateral $ABCD,$ $\angle A \cong \angle C,$ $AB=CD=180,$ and $AD \ne BC.$ The perimeter of $ABCD$ is $640.$ Find $\cos A.$ In convex quadrilateral $ABCD,$ $\angle A \cong \angle C,$ $AB=CD=180,$ and $AD \ne BC.$ The perimeter of $ABCD$ is $640.$ Find $\cos A.$
I have no idea on where to start on this problem
| Hint: Let $AD = x$ and $BC = y$. Then $x+y+180+180 = 640$ so $x+y = 280$. By the Law of Cosines, we have
$$ BD^2 = AD^2+AB^2-2AD\cdot AB\cos\alpha = x^2+180^2-360x\cos\alpha $$
and
$$ BD^2 = BC^2+CD^2-2BC\cdot CD\cos\alpha = y^2+180^2-360y\cos\alpha. $$
Therefore
$$0 = x^2-y^2 - 360(x-y)\cos\alpha. $$
Can you take it from here?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1940303",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Number of perfect squares in $ 4^{6} \cdot 6^{7} \cdot 21^{8}$ Why can't we use this approach:
rewriting, $2^{19} \cdot 7^{8} \cdot 3^{15}$, and then
using $(3\cdot 2)\cdot(4^9)\cdot(49^4)\cdot(9^7)$.
Hence, number of perfect squares is $10\cdot 5\cdot 8$.
The answer seems to differ.
| I agree with you. If $n=4^{6} \cdot 6^{7} \cdot 21^{8}=2^{19} \cdot 3^{15}\cdot 7^{8}$ then the number of perfect squares which divide $n$ is
$$(\lfloor 19/2\rfloor+1)\cdot (\lfloor 15/2\rfloor+1)\cdot (\lfloor 8/2\rfloor+1)=10\cdot 8\cdot 5=400.$$
What is the given answer? It is not clear what "Number of perfect squares in" means.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1940521",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Find the locus of the vertices of the right circular cones that pass through the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1, z=0$
Prove that the locus of the vertices of the right circular cones that pass through the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1, z=0$ is $\frac{x^2}{a^2-b^2}-\frac{z^2}{b^2}=1, y=0$ or $\frac{y^2}{a^2-b^2}+\frac{z^2}{a^2}=-1, z=0$.
EDIT:
Here $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1, z=0$ is the base of the cone. Let the vertex of the cone be $(x_1,y_1,z_1)$. Let the generator be $$\frac{x-x_1}{l}=\frac{y-y_1}{m}=\frac{z-z_1}{n}.$$
Then $z=0$ implies any point on ellipse be $(x_1-lz_1/n, y_1-mz_1/n,0)$. It lies on the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ then the point satisfies the ellipse we get $$\frac{(nx_1-lz_1)^2}{a^2}+\frac{(ny_1-mz_1)^2}{a^2}=n^2$$
Eliminating $l,m,n$ we get the equation of the cone as $$\frac{1}{a^2}(zx_1-xz_1)^2+\frac{1}{b^2}(zy_1-yz_1)^2=(z-z_1)^2.$$ How to get the locus of vertex $(x_1,y_1,z_1)$?
Edit 2
How to get the locus of the vertices in the given two forms using purely mathematical way?
I not able to solve the problem. Please help.
| In this answer, I answer the question "What is the cone of the conic section?" by considering Dandelin Spheres, which are tangent to the cone and to the plane of a conic section; the key fact is that Dandelin Spheres are tangent to the plane of a conic section at the foci of that conic.
The figure below, taken from that answer, shows the view of the plane perpendicular to an ellipse through its major axis. The ellipse itself appears only as $\overline{PQ}$ (its major axis) and the Dandelin Spheres have become "Dandelin Circles" $\bigcirc{R}$ and $\bigcirc{R^\prime}$. Point $C$, the apex of the cone, is the intersection of lines through $P$ and $Q$ that are tangent to these circles.
With that introduction, identifying the locus of $C$ is straightforward. We simply observe that
$$|\overline{CS}| = |\overline{CT}| \quad\to\quad |\overline{CP}|+|\overline{PF}| = |\overline{CQ}| + |\overline{QF}| \tag{$\star$}$$
Writing $m$ and $c$ for the ellipse's major and "focal" radii, $(\star)$ says
$$|\overline{CP}| - |\overline{CQ}| = |\overline{QF}| - |\overline{PF}| = (m+c)-(m-c) = 2c$$
That is, the difference of $C$'s distances from $P$ and $Q$ is a constant, so that $C$ lies on a hyperbola with foci $P$ and $Q$ that passes through $F$ and $F^\prime$. This hyperbola has focal radius $m$, transverse radius $c$, and conjugate radius $n := \sqrt{m^2-c^2}$ (equal to the minor radius of the ellipse). Its equation is
$$\frac{w^2}{c^2} - \frac{z^2}{n^2} = 1 \qquad\to\qquad \frac{w^2}{m^2-n^2} - \frac{z^2}{n^2} = 1 \tag{$\star\star$}$$
where the $w$-axis is horizontal in the figure. That axis is either the $x$-axis or the $y$-axis, according to which of $a$ or $b$ is bigger in the given ellipse equation,
$$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$$
When $a$ is bigger, $\overline{PQ}$ aligns with the $x$-axis, so the figure shows the $xz$-plane; the locus of $C$ is
$$\frac{x^2}{a^2-b^2} - \frac{z^2}{b^2} = 1, \qquad y=0$$
When $b$ is bigger, $\overline{PQ}$ aligns with the $y$-axis, so the figure shows the $yz$-plane; the locus of $C$ is
$$\frac{y^2}{b^2-a^2} - \frac{z^2}{a^2} = 1, \qquad x=0$$
or, equivalently, with appropriate sign changes,
$$\frac{y^2}{a^2-b^2} + \frac{z^2}{a^2} = -1, \qquad x=0$$
Note. For the configuration shown in the figure, $C$ is on one semi-branch of the hyperbola, the "lower-right". Adjusting $\bigcirc{R}$ and $\bigcirc{R^\prime}$ to appear on the other sides of $\overline{PQ}$, and switching which circle is bigger, allows $C$ to travel throughout the entire hyperbola.)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1940885",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 0
} |
I conjecture this inequality $\sqrt[4]{\frac{(xy+yz+xz)(x^2+y^2+z^2)}{9}}\ge\sqrt[3]{\frac{(x+y)(y+z)(z+x)}{8}}$ Let $x,y,z>0$,prove or disprove
$$\sqrt[4]{\dfrac{(xy+yz+xz)(x^2+y^2+z^2)}{9}}\ge\sqrt[3]{\dfrac{(x+y)(y+z)(z+x)}{8}}$$
I tried many times ,use
$$x^2+y^2+z^2\ge\dfrac{1}{3}(x+y+z)^2$$
and kown $$9(x+y)(y+z)(z+x)\ge 8(x+y+z)(xy+yz+xz)$$
| Let $x+y+z=3u$, $y+xz+yz=3v^2$ and $xyz=w^3$.
Hence, it's obvious that our inequality is equivalent to $f(w^3)\geq0$, where $f(w^3)=w^3+A(u,v^2)$,
which says that it's enough to prove our inequality for the minimal value of $w^3$.
But $x$, $y$ and $z$ are positive roots of the equation
$$(X-x)(X-y)(X-z)=0$$ or
$$X^3-3uX^2+3v^2X-w^3=0$$ or
$$w^3=X^3-3uX^2+3v^2X$$
which says that a line $Y=w^3$ and a graph of $Y=X^3-3uX^2+3v^2X$ have three common points.
Thus, $w^3$ gets a minimal value, when a line $Y=w^3$ is a tangent line to the graph of $Y=X^3-3uX^2+3v^2X$, which happens for equality case of two variables.
Also we need to check the case $w^3\rightarrow0^+$.
Two these cases give a trivial inequalities:
*
*For $y=z=1$ we get $$(x-1)^2(2048x^7+6439x^6+18822x^5+29481x^4+33108x^3+24105x^2+9094x+1319)\geq0$$
*For $y=1$ and $z\rightarrow0^+$ we get
$$(4096x^6-729x^5+9372x^4-4374x^3+9372x^2-729x+4096)x^3\geq0$$
Done!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1942110",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
"answer_id": 1
} |
What is the largest number that divides $n^5-n$ whenever $n$ is odd? My attempt: Let $n=2k+1$, where $k$ is an Integer. Then $$n^5-n=(2k+1)(2k)2(k+1)2(2k^2+2k+1)=8(2k+1)(k)(k+1)(2k^2+2k+1)=16(2k+1)(P)(2k^2+2k+1)$$
where $P$ is an integer. So according to me the answer should be atleast $16$. How should I go about finding other factors from this expression. What other approaches can I use to solve problems of this kind?
Edit: This was a multiple choice question. The choices were:
*
*15
*30
*16
*240
*720
| With such problems, you should always factorize first and see if you can get anywhere from there. Also, plugging in specific values helps (do that with the factorized expressions). That already suffices here:
Okay, so let $N = n^5 - n = n·(n^4 - 1) = n·(n-1)·(n+1)·(n^2 + 1)$ for some $n ∈ ℕ$.
If $n$ is odd,
*
*all of $n-1$, $n+1$ and $n^2 + 1$ are even, one of $n-1$, $n+1$ even being divisible by $4$, so $N$ is divisible by $16$,
*one of $n$, $n-1$ and $n+1$ is divisible by $3$, so $N$ is divisible by $3$,
*either one of $n$, $n-1$ or $n+1$ is divisible by $5$ or else $n \equiv \pm 2 \bmod 5$, so $n^2 + 1 \equiv 5 \equiv 0 \bmod 5$, and $N$ is divisible by $5$.
So all numbers $N = n^5 - n$ with $n$ being odd are divisible by $16·3·5 = 240$.
For $n = 3$, $N = 3·2·4·10 = 240$, so there is no number larger than $240$ that divides all numbers $N = n^5 - n$ with $n$ being odd.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1942527",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
rational solutions of $x^4+x^2y^2+y^4=x^2$ I am trying to find rational solutions of $x^4+x^2y^2+y^4=x^2$, except for $(\pm 1,0), (0,0) $. I guess these three solutions are the only ones, but I failed to prove it. It seems that $x^4+x^2y^2+y^4=(x^2+xy+y^2)(x^2-xy+y^2) $ may be useful, but I do not know how to proceed.
| Start with the HINT by Piquito: $$(x^2+y^2)^2=x^4+x^2y^2+y^4+x^2y^2=x^2+x^2y^2$$ Rearranging yields $$(x^2+y^2)^2=x^2(y^2+1)$$ Since the LHS is a perfect square, and $x^2$ is a perfect square, it follows that $(y^2+1)$ must be a perfect square, and the only solution in the integers for that is $y=0$. The implications of $y=0$ have been elaborated in a previous answer.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1946362",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Compute limit of $\sum_{n=0}^\infty \frac{(-2)^n+2^{3n}}{3^n\cdot4^n}$
I need help to compute the limit of the following series: $\sum_{n=0}^\infty \frac{(-2)^n+2^{3n}}{3^n\cdot4^n}$
What I got so far:
It is probably easier to split the sum into the following:
$\sum_{n=0}^\infty \frac{(-2)^n+2^{3n}}{3^n\cdot4^n}$ = $\sum_{n=0}^\infty \frac{(-2)^n}{(3\cdot4)^n} + \sum_{n=0}^\infty \frac{8^{n}}{(3\cdot4)^n}=$
$\sum_{n=0}^\infty (\frac{-1}{6})^n + \sum_{n=0}^\infty (\frac{8}{12})^n= \sum_{n=0}^\infty (-1)^n (\frac{1}{6})^n + \sum_{n=0}^\infty (\frac{4}{6})^n = \sum_{n=0}^\infty (-1)^n (\frac{1}{6})^n + \frac{1}{1-\frac{4}{6}} = \sum_{n=0}^\infty (-1)^n (\frac{1}{6})^n + 3$
But here im stuck: I don't know how to deal with the alternating series that looks similar to another geometric series. Help to get further is very appreciated!
| use the geometric series
$${\frac {1}{1-x}}=\sum _{n=0}^{\infty }x^{n}\quad {\text{ for }}|x|<1\!$$
and let $x\rightarrow -x$ to get
$${\frac {1}{1+x}}=\sum _{n=0}^{\infty }(-1)^n(x)^{n}\quad {\text{ for }}|x|<1\!$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1948266",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.