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Are there any integer solutions to $2^x+1=3^y$ for $y>2$? for what values of $ x $ and $ y $ the equality holds
$2^x+1=3^y$
It is quiet obvious the equality holds for $x=1,y=1$ and $x=3,y=2$.
But further I cannot find why $x$ and $y$ cannot take higher values than this values.
| Note that $2\equiv -1$ mod $3$, which means that $x$ must be odd.
If $y$ is even, let $y=2z$ and we have $2^x=(3^z+1)(3^z-1)$ with some minor arranging. The two factors on the right-hand side differ by $2$, and must both be powers of $2$, so they must be $2$ and $4$ and $z=1$ which gives us the solution $x=3, y=2$.
It remains to deal with the case in which $x,y$ are both odd. We note that $3\equiv -1$ mod $4$ so if $x \ge 2$ and $y$ is odd the equation becomes $0+1\equiv -1$ mod $4$. So we must have $x\lt 2$ hence $x=1, y=1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/470568",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Complex Numbers: Finding solutions to $ (z^2-3z+1)^4 = 1 $ I have to find all solutions to
$$ (z^2-3z+1)^4 = 1 $$
What I thought could work was
$$z^2-3z+1= 1^{1/4} $$
Given that the 4 4th-roots of 1 are $1, i, -i, -1$ my idea was to look at each case separately. Starting with $1$ and with $z=a+bi \quad a,b \in R$:
$$z^2-3z+1= 1\\z^2-3z=0\\a^2+2abi-b^2=3a+3bi$$
From where you get
$$2ab=3b\ \to \ a=\frac 32\\a^2-b^2=3a\ \to \ \frac94-b^2=\frac92 \ \to \ b\notin R$$
So no possible solutions in this case.
I think the idea is okay but when I try to do the same with $i$:
$$z^2-3z+1= i\to (a^2-b^2-3a+1)+i(2ab-3b)=i $$
And I get:
$$a^2-b^2-3a+1 = 0\\2ab-3b=1 $$
Which can be solved but seems overly complicated...
Is what I've done correct? Any simpler ideas?
| Starting from $(z^2 - 3z + 1)^4 = 1$, we get
$$((z^2 - 3z + 1)^2 - 1)((z^2 - 3z + 1)^2 + 1) = 0.$$
For the first term we have that
$$(z^2 - 3z + 1)^2 - 1 = 0.$$
Hence, we get first $4$ roots, $z = 0, 1, 2, 3.$
Next, $$(z^2 - 3z + 1)^2 + 1 = 0$$
$$z^2 - 3z + 1 = i, -i.$$
Using the formula, we get the remaining roots, that is,
$$z = \frac{1}{2}( 3 \pm \sqrt{5 \pm 4i} ).$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Separating 18 people into 5 teams A teacher wants to divide her class of 18 students into 5 teams to work on projects,
with two teams of 3 students each and three teams of 4 students each.
a) In how many ways can she do this, if the teams are not numbered?
b) What is the probability that two of the students, Mia and Max, will be on the same team?
[This is not a homework problem.]
| Form each team one at a time.
$$\begin{align}
\text{First team: }&\binom{18}{3}\\
\text{Second team: }&\binom{15}{3}\\
\text{Third team: }&\binom{12}{4}\\
\text{Fourth team: }&\binom{8}{4}\\
\text{Fifth team: }&\binom{4}{4}
\end{align}$$
Then apply the rule of products.
The probability of any given student going into team one or two is $\frac{3}{18}$. The probability of a student going into team three, four, or five is $\frac{4}{18}$. So the probability of two students both going into team one is $\frac{3}{18}\frac{3}{18}$. Use the rule of sums to compute the total probability that they'll end up in the same team out of any of the five teams:
$$P=2\frac{3}{18}\frac{3}{18}+3\frac{4}{18}\frac{4}{18}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/472286",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Prove that: $ \cot7\frac12 ^\circ = \sqrt2 + \sqrt3 + \sqrt4 + \sqrt6$ How to prove the following trignometric identity?
$$ \cot7\frac12 ^\circ = \sqrt2 + \sqrt3 + \sqrt4 + \sqrt6$$
Using half angle formulas, I am getting a number for $\cot7\frac12 ^\circ $, but I don't know how to show it to equal the number $\sqrt2 + \sqrt3 + \sqrt4 + \sqrt6$.
I would however like to learn the technique of dealing with surds such as these, especially in trignometric problems as I have a lot of similar problems and I don't have a clue as to how to deal with those.
Hints please!
EDIT:
What I have done using half angles is this: (and please note, for convenience, I am dropping the degree symbols. The angles here are in degrees however).
I know that
$$ \cos 15 = \dfrac{\sqrt3+1}{2\sqrt2}$$
So,
$$\sin7.5 = \sqrt{\dfrac{1-\cos 15} {2}}$$
$$\cos7.5 = \sqrt{\dfrac{1+\cos 15} {2}} $$
$$\implies \cot 7.5 = \sqrt{\dfrac{2\sqrt2 + \sqrt3 + 1} {2\sqrt2 - \sqrt3 + 1}} $$
| \begin{align}
& \cot({7.5^\circ}) = \cot\left(\frac{45^\circ-30^\circ}{2}\right)
= \frac{\cos(45^\circ) + \cos(30^\circ)}{\sin(45^\circ)-\sin(30^\circ)} \\[10pt]
= {} & \frac{\frac{\sqrt{2}}{2} + \frac{\sqrt{3}}{2}}{\frac{\sqrt{2}}{2} - \frac{1}{2}} =\frac{ 2 + \sqrt{6} }{ 2 - \sqrt{2}}
=\sqrt{6} + \sqrt{3} + \sqrt{2} + 2
\end{align}
| {
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"timestamp": "2023-03-29T00:00:00",
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Find the equation which has 4 distinct roots
For which value of $k$ does the equation $x^4 - 4x^2 + x + k = 0 $ have four distinct real roots?
I found this question on a standardized test, and the answer presumably relies on a graphing calculator. Is there some method to solve this problem without the aid of a calculator?
| Discriminant of equation is
$$
\Delta(k) =
\begin{array}{|ccccccc|}
1 & 0 & -4 & 1 & k & 0 & 0 \\
0 & 1 & 0 & -4 & 1 & k & 0 \\
0 & 0 & 1 & 0 & -4 & 1 & k \\
4 & 0 & -8 & 1 & 0 & 0 & 0 \\
0 & 4 & 0 & -8 & 1 & 0 & 0 \\
0 & 0 & 4 & 0 & -8 & 1 & 0 \\
0 & 0 & 0 & 4 & 0 & -8 & 1
\end{array}
= 256 k^3 - 2048 k^2 + 3520 k + 229.
$$
If equation has distinct real roots, then discriminant is positive.
I hope we are talking about $k\in \mathbb{Z}$.
If $k\le -1$, then $\Delta(k)<0$.
$\Delta(0)=229>0$.
$\Delta(1)=1957>0$.
$\Delta(2)=1125>0$.
$\Delta(3)=-731<0$.
$\Delta(4)=-2075<0$.
$\Delta(5)=-1371<0$.
If $k\ge 6$, then $\Delta(k)>0$.
When $k\ge 6$, then $x^4 - 4x^2 + x + k>0$.
So, $k =0$, $k=1$, or $k=2$.
Links to see graphs/plots (and related info):
$x^4-4x^2+x+0$,
$x^4-4x^2+x+1$,
$x^4-4x^2+x+2$.
| {
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Divisibility of sequence Let the sequence $x_n$ be defined by $x_1=1,\,x_{n+1}=x_n+x_{[(n+1)/2]},$ where $[x]$ is the integer part of a real number $x$. This is A033485. How to prove or disprove that 4 is not a divisor of any its term? The problem belongs to math folklore. As far as I know it, M. Kontsevich authors that.
| Prove by induction that modulo $4$,
$y_n = (x_n, (x_{2n-1}, x_{2n}, x_{2n+1}))$ is always one of the following quadruplets :
$(1,(1,2,3)) ; (2,(1,3,1)) ; (2,(3,1,3)) ; (3,(3,2,1))$
($x_n$ and $x_{2n-1}$ determine the other two values so this is also saying that $(x_n,x_{2n-1})$ can never be $(1,2),(1,3),(2,2),(3,1)$ or $(3,2)$)
We check that $y_1 = (1,(1,2,3))$.
If $y_n = (1,(1,2,3))$ then since $1$ can only be followed by $2$ or $3$, either $x_{n+1} = 2$ and $y_{n+1} = (2,(3,1,3))$, either $x_{n+1} = 3$ and $y_{n+1} = (3,(3,2,1))$.
If $y_n = (3,(3,2,1))$ then since $3$ can only be followed by $1$ or $2$, either $x_{n+1} = 1$ and $y_{n+1} = (1,(1,2,3))$, either $x_{n+1} = 2$ and $y_{n+1} = (2,(1,3,1))$.
If $x_n = 2$ then $n = 2k$ and $y_k$ is $(1,(1,2,3))$ or $(3,(3,2,1))$.
In the first case we get $y_{n-1} = (1,(1,2,3)), y_n = (2,(3,1,3))$ and $y_{n+1} = (3,(3,2,1))$
In the other case we get $y_{n-1} = (3,(3,2,1)), y_n = (2,(1,3,1))$ and $y_{n+1} = (1,(1,2,3))$
Shape those quadruplet as a small triangle piece.
Here is the picture of the possible ways those pieces can interact : From the possible top and left pieces (determined by $x_n,x_{2n-1},x_{4n-3}$) we compute the other values and see that the next two pieces are still of the $4$ possible kind.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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I need to calculate $x^{50}$ $x=\begin{pmatrix}1&0&0\\1&0&1\\0&1&0\end{pmatrix}$, I need to calculate $x^{50}$
Could anyone tell me how to proceed?
Thank you.
| Evaluate the first few powers and guess
$\begin{pmatrix}1&0&0\\1&0&1\\0&1&0\end{pmatrix}^{2n}$ = $\begin{pmatrix}1&0&0\\n&1&0\\n&0&1\end{pmatrix}$, proof by induction. Then $x^{50}=\begin{pmatrix}1&0&0\\25&1&0\\25&0&1\end{pmatrix}$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Inequality for each $a, b, c, d$ being each area of four faces of a tetrahedron We know 'triangle inequality'. I'm interested in the generalization of this inequality.
Here is my question.
Question: How can we represent a necessary and sufficient condition for each positive number $a, b, c, d$ being each area of four faces of a tetrahedron?
I've tried to get some kind of inequality, but I'm facing difficulty.
| The Triangle Inequality is an aspect of the Law of Cosines.
$$\begin{align}
a \le b + c \quad b \le c + a \quad c \le a + b \quad &\implies \qquad |b-c| \le a \le b+c\\
&\implies b^2 + c^2 - 2 b c \le a^2 \le b^2 + c^2 + 2 b c \\
&\implies \exists\;\theta, \; 0 \leq \theta \leq \pi \quad \text{s.t.} \quad a^2 = b^2 + c^2 - 2 b c \cos\theta
\end{align}$$
where it turns out that "$\theta$" is exactly the angle that fits into the appropriate corner of the triangle. (Proof left to reader.)
For tetrahedra, we have this Law of Cosines involving face areas $W$, $X$, $Y$, $Z$ and dihedral angles $A$, $B$, $C$ (meeting at the vertex opposite face $W$) and $D$, $E$, $F$ (surrounding face $W$). For instance,
$$W^2 = X^2 + Y^2 + Z^2 - 2 Y Z \cos A - 2 Z X \cos B - 2 X Y \cos C$$
(with $A$ between faces $Y$ & $Z$, etc). Clearly, this gives the necessary condition
$$W \leq X + Y + Z$$
and its kin, although these are not sufficient.
Interestingly, when you've come to know tetrahedra like I know them, you realize that there are in fact seven faces to each of these things: the four familiar ("standard") ones, and three that I call "pseudo-faces". A pseudo-face is the projection of the tetrahedron into a plane parallel to a pair of opposite edges. I denote the areas of these $H$, $J$, $K$.
More-interestingly, there's a Law of Cosines involving pseudo-faces:
$$\begin{align}
Y^2 + Z^2 - 2 Y Z \cos A \quad &= H^2 = \quad W^2 + X^2 - 2 W X \cos D \\
Z^2 + X^2 - 2 Z X \cos B \quad &= J^2 = \quad W^2 + Y^2 - 2 W Y \cos E \\
X^2 + Y^2 - 2 X Y \cos C \quad &= K^2 = \quad W^2 + Z^2 - 2 W Z \cos F
\end{align}$$
which, together with the Law of Cosines above, proves this Sum-of-Squares identity:
$$W^2 + X^2 + Y^2 + Z^2 = H^2 + J^2 + K^2 \qquad(1)$$
Now, given seven ostensible areas (four standard and three pseudo), the Law of Cosines leads to Triangle-Inequality-like conditions, such as
$$|Y-Z| \leq H \leq Y+Z \qquad\qquad |W-X| \leq H \leq W + X \qquad (2)$$
(and likewise for $J$ and $K$). Of course, the areas must also satisfy the Sum-of-Squares identity $(1)$. But even this collection of conditions isn't sufficient to determine a tetrahedron. We need one more:
$$\begin{align}
0 \quad \leq \quad &2 W^2 X^2 Y^2 + 2 W^2 Y^2 Z^2 + 2 W^2 Z^2 X^2 + 2 X^2 Y^2 Z^2 + H^2 J^2 K^2 \\
&-H^2\left(W^2 X^2+Y^2 Z^2\right)
-J^2\left(W^2 Y^2+Z^2 X^2\right)
-K^2\left(W^2 Z^2+X^2 Y^2\right) \qquad (3)
\end{align}$$
When the right-hand side is in fact non-negative, it gives $81 V^4$, where $V$ is the volume of the tetrahedron.
Together, $(1)$, $(2)$, $(3)$ constitute my analogue of Menger's Theorem (which outlines conditions under which six edge-lengths can make a tetrahedron). For further information on this result, see my Bloog post "A Hedronometric Theorem of Menger".
FYI: The Bloog also has a number of other notes on "Hedronometry" ---my name for the dimensionally-enhanced trigonometry of tetrahedra--- both Euclidean and non-. (The earliest notes need some editing love. I was just using them for TeX practice waaaay-back-when. :)
So, one way to answer your question is this:
$W$, $X$, $Y$, $Z$ can be areas of faces of a tetrahedron if and only if there exist non-negative $H$, $J$, $K$ satisfying $(1)$, $(2)$, $(3)$.
| {
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"timestamp": "2023-03-29T00:00:00",
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If $ \cos x +2 \cos y+3 \cos z=0 , \sin x+2 \sin y+3 \sin z=0$ and $x+y+z=\pi$. Find $\sin 3x+8 \sin 3y+27 \sin 3z$
Problem : If $ \cos x +2 \cos y+3 \cos z=0 , \sin x+2 \sin y+3 \sin z=0$ and $x+y+z=\pi$. Find $\sin 3x+8 \sin 3y+27 \sin 3z$
Solution: Adding $ \cos x +2 \cos y+3 \cos z=0$ and $\sin x+2 \sin y+3 \sin z=0$,we get
$ (\cos x+\sin x) +2(\cos y+\sin y)+3(\cos z+\sin z) =0$
Am I doing right ?
| Here's another hint: $a^3 + b^3 + c^3 - 3abc = (a+b+c) (a^2 + b^2 +c^2 - ab-bc-ca)$.
We have $e^{ix} + 2e^{iy} + 3e^{iz} = 0 $.
Using the above identity, we get
$e^{i3x} + 8 e^{i3y} + 27e^{i3z} = 6e^{i(x+y+z)} = -6 $
Now compare real and imaginary parts.
| {
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$1^{k}+2^{k}+\cdots+n^{k}$ is not divisible by $n+2$ $k$ is odd number.
show that for arbitrary $n\in N$ , $1^{k}+2^{k}+\cdots+n^{k}$ is not divisible by $n+2$
| As $k$ is odd, $r^k+(n+2-r)^k$ is divisible by $r+n+2-r=n+2$
Putting $r=1,2,\cdots,n,n+1$ and adding we get,
$2\{1^k+2^k+\cdots +n^k+(n+1)^k\}$ is divisible by $n+2$
If $(n+2)$ divides $(1^k+2^k+\cdots +n^k),$
it will divide $2\{1^k+2^k+\cdots +n^k+(n+1)^k\}-2(1^k+2^k+\cdots +n^k)=2(n+1)^k$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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How to find this $\iint_{D}f(x)\,dxdy$ Let $x\in \left[0,\dfrac{\pi}{2}\right]$ and
$$f(x)=\int_{0}^{\cos^2{x}}\arccos{\sqrt{t}}\,dt-\int_{0}^{\sin^2{x}}\arcsin{\sqrt{t}}\,dt.$$
Find the value
$$\iint_{D}f(x)\, dxdy,\quad\text{where } D=\{(x,y)\mid x^2+y^2\le 1\}.$$
My idea:
$$f'(x)=-2\cos{x}\sin{x}\cdot x-2\sin{x}\cos{x}\cdot x=-2x\sin{2x}$$
so
$$f(x)=\int_{0}^{x}f(t)\,dt+f(0)=\cdots=\int_{0}^{1}\arccos{\sqrt{t}}\,dt+x\cos{2x}-\dfrac{1}{2}\sin{(2x)}$$
so
$$\iint_{D}f(x)\,dxdy=\cdots=\dfrac{\pi}{2}\int_{0}^{1}\arccos{\sqrt{t}}\,dt+2+\iint_{D}x\cos{(2x)}\,dxdy-1/2\iint_{D}\sin{2x}\,dxdy$$
But $$\iint_{D}x\cos{(2x)}\,dxdy$$ is very ugly, so my idea is not good? Thank you everyone have nice methods? Thank you
| You are off on the right foot; I agree that $f'(x)=-2 x \sin{2 x}$, and therefore $f(x) = x^2 \cos{2 x} - (x/2) \sin{2 x} + C$, where
$$\begin{align}C = f(0) &= \int_0^1 dt \, \arccos{\sqrt{t}}\\ &= 2 \int_0^1 du \, u \, \arccos{u}\\ &= [u^2 \arccos{u}]_0^1 + \int_0^1 du \frac{u^2}{\sqrt{1-u^2}} \\ &= 0 + \int_0^{\pi/2} d\theta \, \sin^2{\theta} \\ &= \frac{\pi}{4}\end{align}$$
The double integral may be simply expressed in Cartesians:
$$\begin{align}\iint_D dx dy \,f(x) &= \int_{-1}^1 dx \, \left (x^2 \cos{2 x} - \frac{x}{2} \sin{2 x} + \frac{\pi}{4} \right ) \int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}} dy \\ &= 2 \int_{-1}^1 dx \, \left (x^2 \cos{2 x} - \frac{x}{2} \sin{2 x} + \frac{\pi}{4} \right ) \sqrt{1-x^2}\end{align}$$
Now,
$$\frac{\pi}{2} \int_{-1}^1 dx \, \sqrt{1-x^2} = \frac{\pi^2}{4}$$
For the other integrals, consider the following integral:
$$J(a) = \int_{-1}^1 dx \, \sqrt{1-x^2} \, e^{i a x} = \frac{\pi}{a} J_1(a)$$
where $J_1$ is the Bessel function of the first kind of order $1$. Now,
$$J'(a) = i \int_{-1}^1 dx \, x \, \sqrt{1-x^2} \, e^{i a x} =\pi \frac{d}{da} \frac{J_1(a)}{a}$$
so that
$$\int_{-1}^1 dx \, x \, \sqrt{1-x^2} \, \sin{a x} = -\pi \left (\frac{J_1'(a)}{a} - \frac{J_1(a)}{a^2} \right ) = \frac{\pi}{2 a} \left [\frac{2 J_1(a)}{a} + J_2(a)-J_0(a) \right ] \\ =\frac{\pi}{a} J_2(a) $$
Evaluating $J''(a)$ in a similar fashion, we find that
$$\int_{-1}^1 dx \, x^2 \sqrt{1-x^2} \cos{a x} = \frac{\pi}{a} \left [J_1(a) - \frac{3 J_2(a)}{a} \right ]$$
Putting this all together, plugging in $a=2$, I get
$$\iint_D dx dy \,f(x) = \pi [ J_1(2)- 2 J_2(2)]+\frac{\pi^2}{4}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Why is L'Hôpital's rule giving the wrong answer? Given the following limit:
$$
\lim_{x \rightarrow -1} \frac{x^5+1}{x+1}
$$
The solution using L'Hôpital's rule:
$$
\lim_{x \rightarrow -1} \frac{x^5+1}{x+1} = \begin{pmatrix} \frac{0}{0} \end{pmatrix} \rightarrow \lim_{x \rightarrow -1} \frac{5x^4}{1} = 5 \cdot (-1)^4 = -5
$$
This is wrong. How come?
EDIT: Well, a perfect example of a huge blunder.
| As $\lim_{x\to-1}\frac{x^5+1}{x+1}$ is of the form $\frac00$
It's legal to apply L'Hospital's rule
$$\implies \lim_{x\to-1}\frac{x^5+1}{x+1}=\lim_{x\to-1}\frac{5x^4}{1}=5(-1)^4=5$$
Alternatively,
as $x^5+1=(x+1)(x^4-x^3+x^2-x+1)$
If $x+1\ne0,$ $\displaystyle\frac{x^5+1}{x+1}=x^4-x^3+x^2-x+1$
If $x\to-1,x\ne-1\iff x+1\ne0$
Alternatively,
we can put $x+1=h\implies h\to0$ as $x\to-1$
| {
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Number of divisors of $9!$ which are of the form $3m+2$ Total number of divisors of $9!$ which are is in the form of $3m+2$, where $m\in \mathbb{N}$
My Try: Let $ N = 9! = 1\times 2 \times 3 \times 2^2 \times 5 \times 2 \times 3 \times 7 \times 2^3 \times 3^2 = 2^7 \times 3^4 \times 5 \times 7$
Now If Here $N$ must be a mutiple of $3m+2$, means when $N$ is divided by $3$ It will gave a remainder $2$
But I did not understand how can i proceed further, thanks in advance
| There are $8 \times 2 \times 2=32$ divisors of $2^7 \times 5 \times 7$, and each is congruent to either $1 \pmod 3$ or $2 \pmod 3$.
If $2^i$ exactly divides the divisor $d$, for $i \in \{0,2,4,6\}$, we pair it up with the divisor $2d$. Each pair consists of one divisor congruent to $1 \pmod 3$ and one congruent to $2 \pmod 3$.
So exactly $16$ of the divisors of $2^7 \times 5 \times 7$ are of the form $2 \pmod 3$. These are the only divisors of $9!$ congruent to $2 \pmod 3$.
| {
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"answer_id": 1
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Prove that $\lim_{N \rightarrow\infty}{\sum_{n=1}^N{\frac{\lvert\mu(n)\rvert}{n}}}=\infty$ Prove that $\lim_{N \rightarrow\infty}{\sum_{n=1}^N{\frac{\lvert \mu(n)\rvert}{n}}}=\infty$. I try to write out a few terms and observe what will happen. I notice that the series will somehow be related to $\sum_{n=1}^N{\frac{1}{n}}$ but I don't know how to relate them. Can someone guide me?
| We make a very crude estimate of the proportion of square-free numbers in the interval $[2^N+1,2^{N+1}]$.
Note that $\le \frac{1}{4}$ of these numbers are divisible by $4$, and $\le \frac{1}{9}$ are divisible by $9$, and $\le \frac{1}{25}$ are divisible by $25$, and so on.
There is overlap, but even if we don't take account of that, the proportion of the numbers in the interval divisible by a square $\gt 1$ is
$$\le \frac{1}{4}+\frac{1}{9}+\frac{1}{25}+\frac{1}{49}\cdots.\tag{1}$$
We can find an upper bound for Sum (1), by using $\frac{1}{4}$ plus
$$\frac{1}{9}+\frac{1}{16}+\frac{1}{25}+\frac{1}{36}+\cdots,$$
which is less than
$$\frac{1}{2\cdot 3}+\frac{1}{3\cdot 4}+\frac{1}{4\cdot 5}+\frac{1}{5\cdot 6}+\cdots.\tag{2}$$
But Sum (2) is a telescoping series with sum $\frac{1}{2}$. That is because
$\frac{1}{2\cdot 3}=\frac{1}{2}-\frac{1}{3}$, and $\frac{1}{3\cdot 4}=\frac{1}{3}-\frac{1}{4}$, and so on.
When we add back the $\frac{1}{4}$ that we left out, we find that Sum (1) is $\le \frac{1}{4}+\frac{1}{2}=\frac{3}{4}$. So the proportion of numbers in $[2^N+1,2^{N+1}]$ that are square-free is at least $\frac{1}{4}$.
Thus, since there are $2^N$ numbers in the interval, and the reciprocal of each is $\ge \frac{1}{2^{N+1}}$, we have
$$\sum_{2^N+1}^{2^{N+1}}\frac{|\mu(n)|}{n}\ge 2^N\frac{1}{4}\frac{1}{2^{N+1}}=\frac{1}{8}.$$
Thus each of our infinitely many intervals makes a contribution of at least $\frac{1}{8}$ to the sum, and therefore our sum diverges.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/486307",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Compositeness of $n^4+4^n$ My coach said that for all positive integers $n$, $n^4+4^n$ is never a prime number.
So we memorized this for future use in math competition. But I don't understand why is it?
| In order to get a prime, we need $n$ odd. So $4^n=4\cdot 4^{2k}$ for some $k$, and therefore $4^n=4\cdot (2^k)^4$ where $k=\frac{n-1}{2}$.
Now use the factorization
$$x^4+4y^4=(x^2-2xy+2y^2)(x^2+2xy+2y^2),$$
with $x=n$ and $y=2^{(n-1)/2}$.
The case $n=1$ gives the lone prime. For all other $n$, we have $x^2-2xy+2y^2\gt 1$.
Remark: It is hard to judge whether the above factorization is "natural." Perhaps it will look more reasonable if we express $x^4+4y^4$ as a difference of squares:
$$x^4+4y^4=(x^2+2y^2)^2-4x^2y^2.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/489071",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 2,
"answer_id": 1
} |
How to compute the integral $ \int\frac{1}{x\sqrt{x^2 +3x}}dx$ Given a problem :
$$ \int\frac{1}{x\sqrt{x^2 +3x}}dx, $$
what is the best solution for this?
I am thinking about solving this problem by using :
$$ u = x+3;\qquad x = u-3; $$
So that we get :
$ \int\frac{1}{x\sqrt{x}\sqrt{x+3}} dx$, then $ \int\frac{1}{(u-3)^{3/2}(u)^{1/2}} du$,
then
$ \int (u)^{-1/2} (u-3)^{-3/2} du$.
Am I right so far? or is there a better method? Thanks.
| For $x>0$, $$\int\frac{1}{x\sqrt{x^2 +3x}}dx=\int\frac{1}{x^2\sqrt{1 +\frac{3}{x}}}dx = -\frac{1}{3}\int\frac{du}{\sqrt{u}}=-\frac{2}{3}\sqrt{u}=-\frac{2}{3}\sqrt{1+\frac{3}{x}} +C $$ where $$u=1+\frac{3}{x}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/489494",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
Evaluating $\int_a^b \frac12 r^2\ \mathrm d\theta$ to find the area of an ellipse I'm finding the area of an ellipse given by $\frac{x^2}{a^2}+\frac{y^2}{b^2} = 1$. I know the answer should be $\pi ab$ (e.g. by Green's theorem). Since we can parameterize the ellipse as $\vec{r}(\theta) = (a\cos{\theta}, b\sin{\theta})$, we can write the polar equation of the ellipse as $r = \sqrt{a^2 \cos^2{\theta}+ b^2\sin^2{\theta}}$. And we can find the area enclosed by a curve $r(\theta)$ by integrating
$$\int_{\theta_1}^{\theta_2} \frac12 r^2 \ \mathrm d\theta.$$
So we should be able to find the area of the ellipse by
$$\frac12 \int_0^{2\pi} a^2 \cos^2{\theta} + b^2 \sin^2{\theta} \ \mathrm d\theta$$
$$= \frac{a^2}{2} \int_0^{2\pi} \cos^2{\theta}\ \mathrm d\theta + \frac{b^2}{2} \int_0^{2\pi} \sin^2{\theta} \ \mathrm d\theta$$
$$= \frac{a^2}{4} \int_0^{2\pi} 1 + \cos{2\theta}\ \mathrm d\theta + \frac{b^2}{4} \int_0^{2\pi} 1- \cos{2\theta}\ \mathrm d\theta$$
$$= \frac{a^2 + b^2}{4} (2\pi) + \frac{a^2-b^2}{4} \underbrace{\int_0^{2\pi} \cos{2\theta} \ \mathrm d\theta}_{\text{This is $0$}}$$
$$=\pi\frac{a^2+b^2}{2}.$$
First of all, this is not the area of an ellipse. Second of all, when I plug in $a=1$, $b=2$, this is not even the right value of the integral, as Wolfram Alpha tells me.
What am I doing wrong?
| This is another way to do it when one know the area of a circle: Consider the area of a circle with radius 1 in coordinates $(\xi, \eta)$ this is:
$$
\int d\xi d \eta = \pi
$$
now if you define new coordinates in your ellipse equation $\xi = \frac{x}{a}, \quad \eta= \frac{y}{b}$ you obtain a circle of radius one: $\xi^2 + \eta^2 =1$
The area of the ellipse you want is $ \int dx dy = ab \int d\xi d\eta = \pi ab$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/493104",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 3
} |
Matrix Multiplication Problem I'm working on the following problem and I can't seem to come up with the right answer.
$$ \text{Let}: A^{-1} =
\begin{bmatrix}
1 & 2 & 1 \\
0 & 3 & 1 \\
4 & 1 & 2 \\
\end{bmatrix}
$$
Find a matrix such that:
$$ ACA =
\begin{bmatrix}
2 & 1 & 3 \\
-1 & 2 & 2 \\
2 & 1 & 4 \\
\end{bmatrix}
$$
Could someone point me in the right direction? Thanks!
| You have
$$
C=A^{-1}(ACA)A^{-1}.
$$
So
$$
C= \begin{bmatrix}
1 & 2 & 1 \\
0 & 3 & 1 \\
4 & 1 & 2 \\
\end{bmatrix}\,\cdot\,
\begin{bmatrix}
2 & 1 & 3 \\
-1 & 2 & 2 \\
2 & 1 & 4 \\
\end{bmatrix}\,\cdot\,
\begin{bmatrix}
1 & 2 & 1 \\
0 & 3 & 1 \\
4 & 1 & 2 \\
\end{bmatrix}.
$$
Now you can just perform the computation.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/493964",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Product polynomial in $\mathbb{F}_7$ I need to compute the product polynomial $$(x^3+3x^2+3x+1)(x^4+4x^3+6x^2+4x+1)$$
when the coefficients are regarded as elements of the field $\mathbb{F}_7$.
I just want someone to explain to me what does it mean when a cofficient (let us take 3 for example) is in $\mathbb{F}_7$ ? I know that $\mathbb{F}_7= \mathbb{Z}/7\mathbb{Z}$
| Well, you can multiply two polynomials in the usual way first, then you look at the coefficients and reduce them to modulo 7. For example:
$(3x^2+x+1)(x^2+3x+6) = 3x^2(x^2+3x+6)+x(x^2+3x+6)+1.(x^2+3x+6)=(3x^4+9x^3+18x^2)+(x^3+3x^2+6x)+(x^2+3x+6)=3x^4+10x^3+22x^2+9x+6$ which is the same as $3x^4+3x^3+x^2+2x-1$ in $\mathbb{Z}_7$ because $10 \equiv 3 \pmod{7}$, $22 \equiv 1 \pmod{7}$, $9 \equiv 2 \pmod{7}$ and $6 \equiv -1 \pmod{7}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/495418",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 2
} |
Prove $3|n(n+1)(n+2)$ by induction I tried proving inductively but I didn't really go anywhere. So I tried:
Let $3|n(n+1)(n+2)$.
Then $3|n^3 + 3n^2 + 2n \Longrightarrow 3|(n(n(n+3)) + 2)$
But then?
| Assume $n(n+1)(n+2) = 3p$.
$$(n+1)(n+2)(n+3) = \frac{n(n+1)(n+2)(n+3)}{n} =3p\cdot\frac{n+3}{n}=3p+3\cdot\frac{3p}{n}$$
Notice $\frac{3p}{n}$ is also an integer by assumption.
With this induction step, the base case has to be $n=1$. Prove the case for $n=0$ separately.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/497859",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 7,
"answer_id": 5
} |
Inductive Proof of Inequality $\textbf{Question:}$ Prove using induction that the following inequality holds for all positive integers $n$:
$$\dfrac{(1+a_1)(1+a_2)\cdots(1+a_n)}{1+a_1a_2\cdots a_n}\leq 2^{n-1},$$
where $a_1,a_2,\dots ,a_n\geq 1$.
$\textbf{Attempted (Incorrect) Solution:}$
Base Case: $\dfrac{(1+a_1)}{(1+a_1)}=1=2^0=2^{1-1}$.
Assume we have $\dfrac{(1+a_1)(1+a_2)\cdots(1+a_k)}{1+a_1a_2\cdots a_k}\leq 2^{k-1}$ for some positive integer $k$. Clearly, $\dfrac{(1+a_1)(1+a_2)\cdots(1+a_k)}{1+a_1a_2\cdots a_ka_{k+1}}\leq\dfrac{(1+a_1)(1+a_2)\cdots(1+a_k)}{1+a_1a_2\cdots a_k}\leq 2^{k-1}$, so $\dfrac{(1+a_1)(1+a_2)\cdots(1+a_k)(1+a_{k+1})}{1+a_1a_2\cdots a_ka_{k+1}}=\dfrac{(1+a_1)(1+a_2)\cdots(1+a_k)}{1+a_1a_2\cdots a_ka_{k+1}}+\dfrac{(1+a_1)(1+a_2)\cdots(1+a_k)a_{k+1}}{1+a_1a_2\cdots a_ka_{k+1}}\leq 2^{k-1}+\dfrac{(1+a_1)(1+a_2)\cdots(1+a_k)a_{k+1}}{1+a_1a_2\cdots a_ka_{k+1}}\leq 2^{k-1}+ \dfrac{(1+a_1)(1+a_2)\cdots(1+a_k)a_{k+1}}{a_1a_2\cdots a_ka_{k+1}}\leq 2^{k-1}+ \dfrac{2^k(a_1a_2\cdots a_k)a_{k+1}}{a_1a_2\cdots a_ka_{k+1}}=2^{k-1}+2^k=3\cdot2^{k-1}\not\leq 2^k.$
I've tried a couple of different approaches, but none of them have panned out. I feel like I might be missing some key insight/trick. Any hints/suggestions would be greatly appreciated. Thanks!
| Hint:
With $b_k = a_k a_{k+1} \ge 1, \quad c_k = a_k + a_{k+1}-1 \ge 1$,
$(1+a_k)(1+a_{k+1}) = 1+(c_k+1) + b_k \le (1 + b_k) + (1 + b_k c_k)$
and $b_kc_k \ge b_k = a_k a_{k+1}$
So split the LHS of the inductive step into two sums...
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/498390",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
How to prove $\cos 36^{\circ} = (1+ \sqrt 5)/4$? Given $4 \cos^2 x -2\cos x -1 = 0$.
Use this to show that $\cos 36^{\circ} = (1+ \sqrt 5)/4$, $\cos 72^{\circ} = (-1+\sqrt 5)/4$
Your help is greatly appreciated! Thanks
| Using $\cos5\theta = 16\cos^5 \theta - 20 \cos^3 \theta + 5 \cos \theta$;
$\cos 180 = 16\cos^5 36 - 20 \cos^3 36 + 5 \cos 36$
Let $\cos36 = x$:
$-1 = 16x^5 -20x^3 +5x$
Its solutions are ${-1,\frac{1- \sqrt{5}}{4}},\frac{1+ \sqrt{5}}{4}$
And as $\cos36 = x$, $\cos 36$ must be equal to one of them.
$\cos 36$ must be equal to $\frac{1+\sqrt{5}}{4}$, as $-1$ and $\frac{1- \sqrt{5}}{4}$ are negative.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/498600",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
If $x_12$, show that $(x_n)$ is convergent. If $x_1<x_2$ are arbitrary real numbers, and $x_n=\frac{1}{2}(x_{n-2}+x_{n-1})$ for $n>2$, show that $(x_n)$ is convergent. What is the limit?
The back of my textbook says that $\lim(x_n)=\frac{1}{3}x_1+\frac{2}{3}x_2$. I was thinking that if I show that the sequence is monotone increasing by induction, then I can "guess" (from the back of my textbook) that there is an upperbound of the limit and show by induction that all elements in $x_n$ are between $x_1$ and the upperbound and so it's bounded. Then by the monotone convergence theorem say it's convergent.
I'm not sure how to show that it is monotone, and then after showing convergence finding the limit, without magically guessing it.
| $\displaystyle{z \in {\mathbb C}\,,\quad \left\vert z\right\vert < 1}$.
$$
\sum_{n = 3}^{\infty}x_{n}\,z^{n}
=
{1 \over 2}\sum_{n = 3}^{\infty}x_{n - 2}\,z^{n}
+
{1 \over 2}\sum_{n = 3}^{\infty}x_{n - 1}\,z^{n}
$$
\begin{align}
------------&------------------\\
\sum_{n = 1}^{\infty}x_{n}\,z^{n} - x_{1}z - x_{2}z^{2}
&=
{1 \over 2}\,z^{2}\sum_{n = 1}^{\infty}x_{n}\,z^{n}
+
{1 \over 2}\,z\sum_{n = 2}^{\infty}x_{n}\,z^{n}
\\[3mm]&=
{1 \over 2}\,z^{2}\sum_{n = 1}^{\infty}x_{n}\,z^{n}
+
{1 \over 2}\,z\left(\sum_{n = 1}^{\infty}x_{n}\,z^{n} - x_{1}z\right)
\\------------&------------------
\end{align}
$$
\left(1 - {1 \over 2}\,z - {1 \over 2}\,z^{2}\right)
\sum_{n = 1}^{\infty}x_{n}\,z^{n}
=
x_{1}z + x_{2}z^{2} - {1 \over 2}\,x_{1}\,z^{2}
$$
\begin{align}
----&---------------------------\\
\sum_{n = 1}^{\infty}x_{n}\,z^{n}
&=
-\,
{2x_{1}z + \left(2x_{2} - x_{1}\right)z^{2}
\over
z^{2} + z - 2}
=
-\,
{2x_{1}z + \left(2x_{2} - x_{1}\right)z^{2}
\over
\left(z - 1\right)\left(z + 2\right)}
\\[3mm]&=
-\,{1 \over 3}\,z
\left[%
{2x_{1} + \left(2x_{2} - x_{1}\right)z \over z - 1}
-
{2x_{1} + \left(2x_{2} - x_{1}\right)z \over z + 2}\right]
\\[3mm]&=
-\,{1 \over 3}\,z
\left\{%
\left[%
2x_{2} - x_{1}
+
{x_{1} + 2x_{2} \over z - 1}
\right]
-
\left[%
2x_{2} - x_{1}
+
{4x_{1} - 4x_{2} \over z + 2}
\right]
\right\}
\\[3mm]&=
{1 \over 3}\left(x_{1} + 2x_{2}\right)z\,{1 \over 1 - z}
+
{2 \over 3}\left(x_{1} - x_{2}\right)z\,{1 \over 1 + z/2}
\\[3mm]&=
{1 \over 3}\left(x_{1} + 2x_{2}\right)\sum_{n = 1}^{\infty}z^{n}
+
{2 \over 3}\left(x_{1} - x_{2}\right)\sum_{n = 1}^{\infty}
\left(-1\right)^{n - 1}{z^{n} \over 2^{n - 1}}
\\[3mm]&=
\sum_{n = 1}^{\infty}\left[%
{1 \over 3}\left(x_{1} + 2x_{2}\right)
+
{2 \over 3}\left(x_{1} - x_{2}\right)\,{\left(-1\right)^{n - 1} \over 2^{n - 1}}
\right]z^{n}
\\----&---------------------------
\end{align}
$$
x_{n}
=
{1 \over 3}\left(x_{1} + 2x_{2}\right)
+
{2 \over 3}\left(x_{1} - x_{2}\right)\,
{\left(-1\right)^{n - 1} \over 2^{n - 1}}\,,
\qquad\qquad n \geq 1
$$
$$
\begin{array}{|c|}\hline\\
\color{#ff0000}{\large\quad%
\lim_{n \to \infty}x_{n}
\color{#000000}{\ =\ }
{1 \over 3}\,x_{1} + {2 \over 3}\,x_{2}
\quad}
\\ \\ \hline
\end{array}
$$
${\bf 'EASY\ WAY':}$
Look for a solution $x_{n} \propto \mu^{n}$. We get $\mu = 1$ and $\mu = -1/2$.
The general solution is
$x_{n} = A\ 1^{n} + B\left(-1/2\right)^{n} = A + B\left(-1\right)^{n}/2^{n}$.
With $x_{1} = A - B/2$ and $x_{2} = A + B/4$ we get $A = x_{1}/3 + 2x_{2}/3 =
\lim_{n \to \infty}x_{n}$. In addition, $B = 4\left(x1 - x_{2}\right)/3$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/502100",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 5,
"answer_id": 2
} |
Monotonic sequence and limit
For $x_n = \frac{1}{2+x_{n-1}}$ where $x_1 =1/2$, show that the
sequence is monotonic and find its limit.
What I first did was finding $x_{n+1}$, which equals $\frac{1}{2+x_n}$; then $x_{n+2}=\frac{1}{2+\frac{1}{2+x_n}}=\frac{x_n+2}{2x_n+5}$ thus it does eventually get smaller hence $x_n>x_{n+1}$. How can I finish this?
| We prove by induction that the odd subsequence is decreasing and that the even subsequence is increasing.
Proceed by observing that $x_{1} > x_{3}$ and $x_{2} < x_{4}$. Now assume that $$x_{2n-1} > x_{2n+1}, \quad x_{2n} < x_{2n+2}$$
Now we have
$$ x_{2n+3} = \frac{1}{2 + x_{2n+2}} < \frac{1}{2 + x_{2n}} = x_{2n+1}$$
With this, we have
$$ x_{2n+2} = \frac{1}{2 + x_{2n+1}} < \frac{1}{2 + x_{2n+3}} = x_{2n+4} $$
Thus, by induction we have our claim.
Also, prove by induction that $x_{2n} < x_{1}$ and $x_{2n-1} > x_{2}$ for all $n \in \mathbb{N}$.
Thus both the even and the odd subsequence converge. We need to show that they converge to the same value.
Now assume that the limit exists (justified below) and is equal to $L$. Then
$$L = \lim x_{n} = \lim\frac{1}{2 + x_{n-1}} = \frac{1}{2 + L}$$
from where we get that
$$ L = (\pm\sqrt{2} - 1)$$
But $L > 0$ (Why?) and hence $L = \sqrt{2} - 1$.
EDIT: For the existence of the limit, see the Edit in Did's proof.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/502244",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
Limit $ \sqrt{2\sqrt{2\sqrt{2 \cdots}}}$
Find the limit of the sequence $$\left\{\sqrt{2}, \sqrt{2\sqrt{2}}, \sqrt{2\sqrt{2\sqrt{2}}}, \dots\right\}$$
Another way to write this sequence is $$\left\{2^{\frac{1}{2}},\hspace{5 pt} 2^{\frac{1}{2}}2^{\frac{1}{4}},\hspace{5 pt} 2^{\frac{1}{2}}2^{\frac{1}{4}}2^{\frac{1}{8}},\hspace{5 pt} \dots,\hspace{5 pt}2^{\frac{1}{2^{n+1} - 2}}\right\}$$
So basically we have to find $\lim_{n\to\infty}2^{\frac{1}{2^{n+1} - 2}}$. This equates to $$\lim_{n\to\infty}2^{\frac{1}{2^{\infty+1} - 2}} \Longrightarrow 2^{\frac{1}{2^{\infty}}} \Longrightarrow 1$$
Is this correct? P.S. Finding that $S(n)$ was a pain!
| Hints:
$$x=\sqrt{2\sqrt{2\sqrt2\ldots}}\implies x=\sqrt{2x}\implies x^2=2x\implies\ldots$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/505270",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
} |
Find a closed form of $\sum_{i=1}^n i^3$ I'm trying to compute the general formula for $\sum_{i=1}^ni^3$. My math instructor said that we should do this by starting with a grid of $n^2$ squares like so:
$$
\begin{matrix}
1^2 & 2^2 & 3^2 & ... & (n-2)^2 & (n-1)^2 & n^2 \\
2^2 & 3^2 & 4^2 & ... & (n-1)^2 & n^2 & 1^2 \\
3^2 & 4^2 & 5^2 & ... & n^2 & 1^2 & 2^2 \\
\vdots &\vdots & \vdots&\ddots & \vdots & \vdots & \vdots \\
(n-2)^2 & (n-1)^2 & n^2 & ... & (n-5)^2 & (n-4)^2 & (n-3)^2 \\
(n-1)^2 & n^2 & 1^2 & ... & (n-4)^2 & (n-3)^2 & (n-2)^2 \\
n^2 & 1^2 & 2^2 & ... & (n-3)^2 & (n-2)^2 & (n-1)^2 \\
\end{matrix}
$$
And then sum the rows, and then sum all of the sums of rows. How would I then compute the general formula for $\sum_{i=1}^ni^3$?
| Perhaps your instructor wanted to show this.
The total sum must be
$$(1^3 + 2^3 + 3^3 + \dots +n^3) +\sum_{i=1}^{n-1} i (n-i)^2= n ( 1^2 + 2^2 + 3^2 + \dots + n^2)$$
Upon simplification i get
$$\sum_{i=1}^n i^3 = n^3 + \sum_{i=1}^{n-1} (n i^2 - i (n-i)^2) = n^3 + \sum_{i=1}^{n-1} -i \left(i^2-3 i n+n^2\right) \\
\implies 2 \sum_{i=1}^{n-1} i^3 + n^3 = n^3 + 3 n \sum_{i=1}^{n-1}i^2 - n^2 \sum_{i=1}^n i $$
Given that you know how to evaluate $\displaystyle \sum_{i=1}^n i^2$, you can calculate the value from above.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/506784",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Question about Fermat's Theorem I'm trying to find $2^{25} \mod 21 $. By Fermat's theorem, $2^{20} \cong_{21} 1 $. Therefore, $2^{25} = 2^{20}2^{5} \cong_{21} 2^5 = 32 \cong_{21} 11 $. However, the answer in my book is $2$! What am I doing wrong?
Also, I would like to ask what are the last two digits of $1 + 7^{162} + 5^{121} \times 3^{312} $
Thanks for your help.
| As $21$ is not prime we can use Carmichael function to show that $2^6\equiv1\pmod{21}$
In fact, $2^6=64\equiv1\pmod{21}$
$\displaystyle \implies 2^{25}=2\cdot(2^6)^4\equiv2\cdot1^4\pmod{21}$
For the last two digits of $1+7^{162}+(5^{121})(3^{312}),$
we need $1+7^{162}+(5^{121})(3^{312})\pmod{100}$
Now, observe that $\displaystyle7^2=49=50-1$
$\implies7^4=(50-1)^2=50^2-2\cdot50\cdot1+1^2\equiv1\pmod{100}$
$\displaystyle \implies 7^{162}=7^2\cdot(7^4)^{40}\equiv49\cdot1^{40}\pmod{100}\equiv49\ \ \ \ (1)$
Again, $5^{a+b}-5^a=5^a(5^b-1)\equiv0\pmod{100}$ if integer $a\ge2,b\ge0$
$\displaystyle \implies 5^{121}\equiv5^2\pmod{100}\equiv25=100c+25$(say) where $c$ is some integer
and $3\equiv-1\pmod4\implies 3^{312}\equiv(-1)^{312}\pmod4\equiv1=4d+1$ where $d$ is some integer
$\displaystyle \implies 5^{121}\cdot3^{312}=(100c+25)(4d+1)\equiv25\pmod{100}\ \ \ \ (2) $
Can you take it from here using $(1),(2)?$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Matrix Algebra Question (Linear Algebra) Find all values of $a$ such that $A^3 = 2A$, where
$$A = \begin{bmatrix} -2 & 2 \\ -1 & a \end{bmatrix}.$$
The matrix I got for $A^3$ at the end didn't match up, but I probably made a multiplication mistake somewhere.
| Any element of the set $\left\{\lambda\right\}$ of $A$ eigenvalues satisfy the equations
$$
\lambda^{2} = \left(a - 2\right)\lambda + 2\left(a - 1\right)\,,
\quad
\lambda^{3} = 2\lambda\,,
\quad
\sum_{\lambda}\lambda = a - 2
$$
Then, $\sum_{\lambda}\lambda^{3} = 2\left(a - 2\right)$. Also,
\begin{align}
\sum_{\lambda}\lambda^{3}
&=
\sum_{\lambda}\lambda\left[\left(a - 2\right)\lambda + 2\left(a - 1\right)\right]
=
\left(a - 2\right)\sum_{\lambda}\lambda^{2}
+
2\left(a - 1\right)\left(a - 2\right)
\\[3mm]&=
\left(a - 2\right)^{3} + 6\left(a - 1\right)\left(a - 2\right)
\end{align}
We get
$$
\left(a - 2\right)\left[\left(a - 2\right)^{2} + 6a - 8\right] = 0
$$
$$
\color{#ff0000}{\large%
a = 2\,,\qquad\qquad a = -1 \pm \sqrt{5\,}}
$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Basic Mathematical Induction I'm not quite sure how to approach this question.
I need to prove that for $$n\ge1$$
$$1^2+2^2+3^3+\dots+n^2=\frac16n(n+1)(2n+1)$$
Do I just plug $1$ and see if $$\frac16(1)((1)+1)(2(1)+1) = 1^2\text{ ?}$$
| Well what we want to do with induction is to show that for the set $S$ that satisfies our desired property, $S = \mathbb{N}$. We can do this on the naturals by a very special property. Every as part of the Peano axioms, we know that $\forall x \in \mathbb{N} $, $x$ has a successor also in $\mathbb{N}$ in the form $\text{s} (x) = x+1$. So if we can show that if the base natural $1$ holds and that our property holds for an arbitrary $x$ then it holds for its successor, we prove the property over the naturals. This works because if $1,x,\text{s}(x) \in S$ we can let $1$ be our $x$ and then we get that it holds for s$(1)$ so it holds for s$($s$(1))$ and so on.
So to the problem at hand. Let $S = \left\{ x \in \mathbb{N} : 1^2 + 2^2 + ... + x^2 = \frac{x(x+1)(2x+1)}{6} \right\}$. First we show the base case $1 \in S$. So, $$1^2 = 1 = \frac{1(1+1)(2*1+1)}{6} = \frac{6}{6} = 1$$
therefore $1 \in S$.
Next we form our inductive hypothesis and assume that some natural $k \in S$ which implies $1^2 + 2^2 + ... + k^2 = \frac{k(k+1)(2k+1)}{6}$. We will use this to make our inductive step and show $k+1 \in S$. So,
\begin{align*}
1^2 + 2^2 + ...+k^2 + (k+1)^2 =& \frac{k(k+1)(2k+1)}{6} + (k+1)^2\\
=&\frac{k(k+1)(2k+1)}{6} + \frac{6(k+1)^2}{6}\\
=&\frac{(k+1)[k(2k+1) + 6(k+1)]}{6}\\
=&\frac{(k+1)[2k^2+k + 6k+6)]}{6}\\
=&\frac{(k+1)[2k^2+7k+6)]}{6}\\
=&\frac{(k+1)[(k+2)(2k+3)]}{6}\\
=&\frac{(k+1)[(k+1+1)(2k+2+1)]}{6}\\
=&\frac{(k+1)((k+1)+1)(2(k+1)+1)}{6}\\
\end{align*}
Which is what we trying to show! $\blacksquare$
Side note: I should probably conclude that we showed that when $k \in S$ that $k+1 \in S$ which means that $S = \mathbb{N}$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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System of Linear Equations ($3\times 6$ matrix, parametric answer) Solve the system
\begin{array}{r@{}r@{}r@{}r@{}r@{}r@{}r@{}r}
x_1 & - 2 x_2 & - 2 x_3 & & + 5 x_5 & - 4 x_6 & = & -1 \cr
& & & - x_4 & + 4 x_5 & + 2 x_6 & = & 7 \cr
x_1 & - 2 x_2 & & & + 9 x_5 & - 6 x_6 & = & 3
\end{array}
\begin{bmatrix}
x1 \\ x2 \\ x3 \\ x4 \\ x5 \\ x6
\end{bmatrix}
= __ + t1 __ + t2 __ + t3 __ + t4 __ + t5 __
where t1, ..., t5 are all free variables.
Apologies, I'm still trying to figure out how to use LaTeX and don't know how to make t1,...,t5 italicized, or have the [x1,...,x5] inline.
Anyways, I've reduced the matrix to
\begin{Bmatrix}1 & -2 & 0 & 0 & 9 & -6 & 3 \\
0 & 0 & 1 & 0 & 2 & -5 & 2 \\
0 & 0 & 0 & 1 & -4 & -2 & -7 \\
\end{Bmatrix}
and I know 3 things that DO belong in the blanks.
$$
[x1,...,x6] = \langle 3,0,2,-7,0,0\rangle + t1\langle ,1,0,0,0,0\rangle + t2\langle -9,0,-2,4,1,0\rangle
$$
I'm pretty sure one of them should also be $\langle 6,0,5,2,0,1\rangle$, and then the rest are $\langle 0,0,0,0,0,0\rangle$, however that's incorrect and I'm stuck as to why.
Any help is appreciated, thank you
| It looks like it's simply a bug in the row reduction. I get a different reduced row echelon form:
$$
\begin{bmatrix} 1 & -2 & -2 & 0 & 5 & -4 & -1 \\ 0 & 0 & 0 & -1 & 4 & 2 & 7 \\ 1 & -2 & 0 & 0 &
9 & -6 & 3 \\ \end{bmatrix}
\xrightarrow{R_3 \gets R_3-R_1}
\begin{bmatrix} 1 & -2 & -2 & 0 & 5 & -4 & -1 \\ 0 & 0 & 0 & -1 & 4 & 2 & 7 \\ 0 & 0 & 2 & 0 &
4 & -2 & 4 \\ \end{bmatrix}
\xrightarrow{R_2 \leftrightarrow R_3}
\begin{bmatrix} 1 & -2 & -2 & 0 & 5 & -4 & -1 \\ 0 & 0 & 2 & 0 & 4 & -2 & 4 \\ 0 & 0 & 0 & -1 &
4 & 2 & 7 \\ \end{bmatrix}
\xrightarrow{R_1 \gets R_1+R_2}
\begin{bmatrix} 1 & -2 & 0 & 0 & 9 & -6 & 3 \\ 0 & 0 & 2 & 0 & 4 & -2 & 4 \\ 0 & 0 & 0 & -1 &
4 & 2 & 7 \\ \end{bmatrix}
\xrightarrow{R_2 \gets \tfrac{1}{2}R_2}
\begin{bmatrix} 1 & -2 & 0 & 0 & 9 & -6 & 3 \\ 0 & 0 & 1 & 0 & 2 & -1 & 2 \\ 0 & 0 & 0 & -1 &
4 & 2 & 7 \\ \end{bmatrix}
\xrightarrow{R_3 \gets -R_3}
\begin{bmatrix} 1 & -2 & 0 & 0 & 9 & -6 & 3 \\ 0 & 0 & 1 & 0 & 2 & \color{red}{-1} & 2 \\ 0 & 0 & 0 & 1 &
-4 & -2 & -7 \\ \end{bmatrix}
$$
We have free variables $x_2$, $x_5$, and $x_6$ (since their columns do not have leading entries). So let's set $x_2=t$, $x_5=u$ and $x_6=v$. The variables $x_1$, $x_3$, and $x_4$ can be solved in terms of the free variables via:
\begin{align*}
x_1-2t+9u-6v &= 3 \\
x_3+2u-v &= 2 \\
x_4-4u-2v &= -7.
\end{align*}
So the solutions are $$(x_1,x_2,x_3,x_4,x_5,x_6)=(3+2t-9u+6v,t,2-2u+v,-7+4u+2v,u,v)$$ for any $t,u,v$. We could alternatively write this as
$$(3,0,2,-7,0,0)+t(2,1,0,0,0,0)+u(-9,0,2,4,1,0)+v(6,0,1,2,0,1).$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove that $x^2 + xy + y^2 \ge 0$ by contradiction Using the fact that $x^2 + 2xy + y^2 = (x + y)^2 \ge 0$, show that the assumption $x^2 + xy + y^2 < 0$ leads to a contradiction...
So do I start off with...
"Assume that $x^2 + xy + y^2 <0$, then blah blah blah"?
It seems true...because then I go $(x^2 + 2xy + y^2) - (x^2 + xy + y^2) \ge 0$.
It becomes $2xy - xy \ge 0$, then $xy \ge 0$.
How is this a contradiction?
I think I'm missing some key point.
| Averaging the inequalities $(x+y)^2\ge0$ and $x^2+y^2\ge0$ we get $\frac12(x+y)^2+\frac12(x^2+y^2)\ge0$, that is $x^2+xy+y^2\ge0$, contradicting the assumption that $x^2+xy+y^2\lt0$.
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "6",
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Is $\prod \limits_{n=2}^{\infty}(1-\frac{1}{n^2})=1$ Question is to check if $\prod \limits_{n=2}^{\infty}(1-\frac{1}{n^2})=1$
we have $\prod \limits_{n=2}^{\infty}(1-\frac{1}{n^2})=\prod \limits_{n=2}^{\infty}(\frac{n^2-1}{n^2})=\prod \limits_{n=2}^{\infty}\frac{n+1}{n}\frac{n-1}{n}=(\frac{3}{2}.\frac{1}{2})(\frac{4}{3}.\frac{2}{3})(\frac{5}{4}.\frac{3}{4})...$
In above product we have for each term $\frac{a}{b}$ a term $\frac{b}{a}$ except for $\frac{1}{2}$.. So, all other terms gets cancelled and we left with $\frac{1}{2}$.
So, $\prod \limits_{n=2}^{\infty}(1-\frac{1}{n^2})=\frac{1}{2}$.
I would be thankful if some one can assure that this explanation is correct/wrong??
I am solving this kind of problems for the first time so, it would be helpful if some one can tell if there are any other ways to do this..
Thank you
| Your explanation as is is not sufficient. The part where you say that everything but $1/2$ gets cancelled needs more rigorous verification.
To demonstrate, let's "prove" that the infinite product $1\cdot1\cdot1\cdot\ldots$ equals $1/2$. Indeed,
$$
1\cdot 1 \cdot 1 \cdot \ldots =
\left(\frac{1}{2} \cdot 2\right) \cdot \left(\frac{1}{2} \cdot 2\right) \cdot \left(\frac{1}{2} \cdot 2\right)
\cdot \cdots.
$$
The $2$ in the first factor cancels the $\frac{1}{2}$ in the second, the $2$ in the second factor cancels the $\frac{1}{2}$ in the third, and so on. Everything but the very first $\frac{1}{2}$ gets canceled, so the infinite product equals $\frac{1}{2}$.
This is clearly wrong. For a correct explanation you should look at partial products, like in user17762's answer.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
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Infinite Series $\sum_{k=1}^{\infty}\frac{1}{(mk^2-n)^2}$ How can we prove the following formula?
$$\sum_{k=1}^{\infty}\frac{1}{(mk^2-n)^2}=\frac{-2m+\sqrt{mn}\pi\cot\left(\sqrt{\frac{n}{m}}\pi\right)+n\pi^2\csc^2\left(\sqrt{\frac{n}{m}}\pi\right)}{4mn^2}$$
What is the general method for finding sums of the form $\sum\limits_{k=1}^{\infty}\frac{1}{(mk^2-n)^\ell}, \ell\in\mathbb{N}$?
| Start with the infinite product expansion
$$\frac{\sin z}{z} = \prod_{k=1}^{\infty}\left(1 - \frac{z^2}{k^2\pi^2}\right)$$
Taking logarithm, substitute $z$ by $\pi\sqrt{x}$ and differentiate with respect to $x$, we find
$$
\sum_{k=1}^{\infty} \frac{1}{k^2 - x}
= -\frac{d}{dx} \left[ \sum_{k=1}^{\infty}\log\left(1 - \frac{x}{k^2}\right)\right]
= -\frac{d}{dx} \left[ \log\left(\frac{\sin(\pi \sqrt{x})}{\pi \sqrt{x}}\right) \right]
$$
Differentiate both sides with respect to $x$ for $\ell - 1 $ more times and then divide by $-(\ell-1)!$, we get in general:
$$
\sum_{k=1}^{\infty} \frac{1}{(k^2 - x)^\ell} = -\frac{1}{(\ell-1)!}
\frac{d^{\ell}}{dx^{\ell}} \left[ \log\left(\frac{\sin(\pi \sqrt{x})}{\pi \sqrt{x}}\right) \right]
$$
In the case $\ell = 2$, the RHS simplifies to
$$-\frac{1}{2x^2} + \frac{\pi}{4x}\left(
\frac{1}{\sqrt{x}}\cot(\pi\sqrt{x}) + \pi \csc(\pi\sqrt{x})^2
\right)$$
Substitute $x$ by $\frac{n}{m}$ will give you the formula you have for $\ell = 2$. Formula for other $\ell$ can be obtained by taking corresponding number of derivatives.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Do there exist an infinite number of $S_m=\frac{m(m+3)}{2}$ such that $S_m=S_p+S_q=S_r-S_t=S_uS_v$? Is the following true?
"Letting $S_n=\frac{n(n+3)}{2}$ for $n\in\mathbb N$, there exist an infinite number of $S_m$ such that
$$S_m=S_p+S_q=S_r-S_t=S_uS_v$$
where $S_i\gt 2\ (i=p,q,r,t,u,v).$"
Note that $S_i\gt 2\ (i=p,q,r,t,u,v)$. (otherwise this question is too easy.)
Motivation : I've known that there exist an infinite number of the above examples in the triangular numbers, which can be represented as $\frac{n(n+1)}{2}$. This got me interested in the above. This expectation seems true by using computer.
Examples :
$$S_{45}=S_{15}+S_{42}=S_{52}-S_{25}=S_{5}S_{9}$$
$$S_{60}=S_{42}+S_{42}=S_{63}-S_{18}=S_{7}S_{9}$$
$$S_{63}=S_{36}+S_{51}=S_{64}-S_{10}=S_{6}S_{11}$$
$$\vdots$$
$$S_{1845}=S_{330}+S_{1815}=S_{1858}-S_{218}=S_{41}S_{60}$$
| Claim 1) There exist an infinite number of triples $(a,b,c)$ of odd numbers such that: $$(\spadesuit)\;(a^2-9)(b^2-9)=8(c^2-9).$$
For any odd value of $b_0$, $(3,b_0,3)$ is clearly a solution of $(\spadesuit)$. This implies that the Pell equation
$$ \frac{b_0^2-9}{8}a^2-c^2 = D = 9\frac{b_0^2-17}{8}$$
has a solution $(a,c)=(3,3)$, hence it has an infinite number of solutions.
In particular, if $u,v$ satisfy
$$ u^2 - \frac{b_0^2-9}{8} v^2 = 1, $$
then
$$ \left(ua+vc,b_0,uc+va\frac{b_0^2-9}{8}\right) $$
is another solution of $(\spadesuit)$. For example, starting from $(a,b,c)=(3,5,3)$ we have that $(15,5,21)$ is another solution of $(\spadesuit)$.
However, an equivalent form of $(\spadesuit)$ is the following:
$$ S_{\frac{a-3}{2}} S_{\frac{b-3}{2}} = S_{\frac{c-3}{2}}, $$
so there exists an infinite number of "special numbers" $S_n$ that are product of two special numbers.
Claim 2) Every natural number $m\geq 3$ can be expressed as the difference of two special numbers.
This follows from the simple observation that $S_{n+1}-S_n = n+2$.
Claim 3) $m$ can be written as a sum of two special numbers iff $8m+18$ can be written as a sum of two squares both greater than $9$.
We have $8S_n=(2n+3)^2-9$, so $N=S_c+S_d$ implies $8N+18=(2c+3)^2+(2d+3)^2$. On the other hand, if $8N+18$ is the sum of two squares, it is the sum of two odd squares, since $8N+18\equiv 2\pmod{4}$, so $8N+18=C^2+D^2$ implies $N=S_{\frac{C-3}{2}}+S_{\frac{D-3}{2}}$.
Assume now that $(a,b,c)$ is a solution of $(\spadesuit)$. A simple inspection $\pmod{3}$ tells us that $3|c$, so $c=3d$, and $3$ divides at least one number between $a$ and $b$. Moreover:
$$8S_{\frac{c-3}{2}}+18 = c^2+9.$$
By choosing $b=5$, we have $c^2+9=2a^2=a^2+a^2$, so if $(a,5,c)$ is a solution of $(\spadesuit)$,
$$S_{\frac{c-3}{2}}$$
is the sum of two special numbers, and this is sufficient to prove the conjecture.
| {
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"timestamp": "2023-03-29T00:00:00",
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The relation between a real matrix and a complex matrix. Let $A$ and $B$ be $n\times n$ real matrices. Show that
$$\det\begin{pmatrix}A & -B \\ B & A \end{pmatrix}=|\det(A+iB)|^{2}.$$
| $$\begin{pmatrix}1/2 & -i/2 \\ 1/2 & i/2 \end{pmatrix}\begin{pmatrix}A & -B \\ B & A \end{pmatrix}\begin{pmatrix}1 & 1 \\ i & -i \end{pmatrix}=\begin{pmatrix}A-iB & 0 \\ 0 & A+iB \end{pmatrix}$$
The determinants are $i/2,\det\begin{pmatrix}A & -B \\ B & A \end{pmatrix},-2i$ and $|\det(A+iB)|^2$.
For a holomorphic transformation $\mathbb{C}^n\rightarrow\mathbb{C}^n$, the Jacobian is like $\begin{pmatrix}A & -B \\ B & A \end{pmatrix}$ due to the Cauchy-Riemann equation. On the other hand, if you complexify the (co)tangent bundle and make a coordinate change $dx_i,dy_i\rightarrow dz_i=dx_i+idy_i,d\bar{z}_i=dx_i-idy_i$, the Jacobian will be like $\begin{pmatrix}A-iB & 0 \\ 0 & \overline{A-iB} \end{pmatrix}$. And $\begin{pmatrix}1/2 & -i/2 \\ 1/2 & i/2 \end{pmatrix}$, $\begin{pmatrix}1 & 1 \\ i & -i \end{pmatrix}$ are just the transformation matrices.
| {
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"timestamp": "2023-03-29T00:00:00",
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What are the possible values of $a$ such that $f(x) = (x + a)(x + 1991) + 1$ has two integer roots? What are the possible values of $a$ such that $f(x) = (x + a)(x + 1991) + 1$ has two integer roots?
$(x + a)(x + 1991) + 1 = x^2 + (1991 + a)x + (1991a + 1)$
This is of the form $ax^2 + bx + c$. Applying the quadratic formula $\left(\frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\right)$, we get, that the rooots must equal:
$\frac{-(1991 + a) \pm \sqrt{(1991 + a)^2 - 4(1991a + 1)}}{2} = \frac{-(1991 + a) \pm \sqrt{1991^2 + a^2 + 2\times1991a - 4\times1991a -4}}{2} = \frac{-(1991 + a) \pm \sqrt{(1991 - a)^2 - 2^2}}{2} = \frac{-(1991 + a) \pm \sqrt{(1991 - a - 2)(1991-a+2)}}{2}$
For the formula to yield an integer, the discriminant must be a perfect square or $0$. Clearly, the discriminant will be zero if $a = 1993$ or $a = 1989$. The only thing that remains to be shown is that these are indeed the only possible solutions.
I couldn't think of any way to do so. Are there any alternate solutions, perhaps some that involve more number theory and less algebra?
| If $(1991-a)^2-2^2=b^2$, say, then $(1991-a)^2-b^2=4$, and you have two squares that differ by $4$. Of course $2^2$ and $0^2$ work, as do $(-2)^2$ and $0^2$; these are the solutions that you already have. And they’re the only ones: if $x$ and $y$ are integers, and $x^2-y^2=4$, then $(x-y)(x+y)=4$. The factors $x-y$ and $x+y$ have the same parity, so they must both be even, and therefore either both are $2$, or both are $-2$. In either case one of $x$ and $y$ is $0$ and the other is $2$, giving you the solutions that you already have.
| {
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centenes of $7^{999999}$
What is the value of the hundreds digit of the number $7^{999999}$?
Equivalent to finding the value of $a$ for the congruence $$7^{999999}\equiv a\pmod{1000}$$
| Using Carmichael function, $$\lambda(1000)=100\implies 7^{100}\equiv1\pmod{1000}$$
Alternatively,
we have $7^2=49=50-1,7^4=(50-1)^2=50^2-2\cdot50\cdot1+1^2=1+2400$
$7^{20}=(7^4)^5=(1+2400)^5=1+\binom512400+\binom52(2400)^2+\cdots\equiv1\pmod{1000}$
Now , $999999\equiv-1\pmod{100}\equiv-1\pmod{20}$
So in either case, $7^{999999}\equiv7^{-1}\pmod{1000}$
$$\text{Now, }\frac{1000}7=142+\frac67=142+\frac1{\frac76}=142+\frac1{1+\frac16}$$
So, the previous convergent of $\displaystyle\frac{1000}7$ is $\displaystyle142+\frac1{1+\frac11}=\frac{143}1$
Using convergent property (Theorem $3$ here) of continued fraction, $1000\cdot1-7\cdot143=-1$
$\displaystyle\implies 7^{-1}\equiv143\pmod{1000}$
| {
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probability density function, solve for c For the probability density function
$$ f(x)=c \frac{x}{x^2 + 1} \quad \rm{for} \quad 0\leq x \leq 2 $$
a) Find $c$.
b) find $E(X)$ and $VAR(X)$
To find $c$, you set the integral equal to $1$. So 1 = $c$ $\ln(x^2+1)\over 2$ from $0$ to $2$ and I solved for $c = 1.24$. Is this correct?
Can someone help on part b?
| For $\mathbb{E}(X)$ we have that
\begin{align}
\mathbb{E}(X) &= c\int_{0}^2\frac{x^2}{x^2+1}dx \\
&= c\int_{0}^2\frac{x^2+1-1}{x^2+1}dx \\
&= 2c- c\int_{0}^2\frac{1}{x^2+1}dx\\
&= 2c- c(\arctan(2)-\arctan(0)).
\end{align}
The variance is given by
$$
\text{var}(X) = \mathbb{E}(X^2)-\left(\mathbb{E}(X)\right)^2.
$$
Let us calculate $\mathbb{E}(X^2)$. We have that
\begin{align}
\mathbb{E}(X^2) &= c\int_{0}^2\frac{x^3}{x^2+1}dx.
\end{align}
Now,
\begin{align}
\int_{0}^2\frac{x^3}{x^2+1}dx&= \int_{0}^2x\frac{x^2-1+1}{x^2+1}dx\\
&=\int_{0}^2xdx-\int_0^2\frac{x}{x^2+1}dx,
\end{align}
and
\begin{align}
\int_0^2\frac{x}{x^2+1}dx = \frac{1}{2}\int_0^2\frac{1}{x^2+1}d(x^2) = \frac{1}{2}\left.\ln(x^2+1)\right|_0^2 = \frac{1}{2}\ln5.
\end{align}
From here, I believe that you will succeed to continue...
| {
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"timestamp": "2023-03-29T00:00:00",
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Using Exponential Generating Functions on Counting Problems Is it possible to use exponential generating functions to solve problems where repetition is wanted?
For example, if I wanted to solve the following problem which wants distinct possibilities...
How many different 5-letter words can be formed from the letters from the word ABRACADABRA if duplicated letters are allowed but no letter can be used more times than it occurs in the word ABRACADABRA ?
5 distinct letters, A repeated five times, B twice, C once, D once, R twice.
Then, the function would be
$$ F(x) = (1 + \frac {x}{1!} + \frac {x^2}{2!} + \frac {x^3}{3!} + \frac {x^4}{4!} + \frac {x^5}{5!}) (1 + \frac {x}{1!} + \frac {x^2}{2!})^2 (1 + \frac {x}{1!})^2 $$
$$ F(x) = \frac {x^{11}}{120} + \frac {3 x^{10}}{40} + \frac {2 x^9}{5} + \frac {19 x^8}{12}+ \frac{289 x^7}{60}+\frac{331 x^6}{30}+\frac{2221 x^5}{120}+\frac {545 x^4}{24} + \frac{121 x^3}{6}+\frac{25 x^2}{2}+5 x+1 $$
Therefore, the answer would be $5!(\frac {1271}{120})$
However, how would I be able to apply it to the following problem?
How many 5-card hands (from an ordinary deck) have at least one card of each suit?
4 suits, repeated 13 times
$ 52*51*50*49*48 = 311,875,200 $ total possibilities
$$ G(x) = (\frac {x^{13}}{13!} + \frac{x^{12}}{12!}+ \frac{x^{11}}{11!} + \frac{x^{10}}{10!} + \frac{x^9}{9!} + \frac{x^8}{8!} + \frac{x^7}{7!} + \frac{x^6}{6!} + \frac{x^5}{5!} + \frac{x^4}{4!} + \frac{x^3}{3!} + \frac{x^2}{2!} + \frac{x}{1!} + 1)^4 $$
Of course, this would just give me unique combinations, so it doesn't work. Is there an actual method to solve using exponential generating functions? Is it possible or even recommended?
| "How many 5-card hands (from an ordinary deck) have at least one card of each suit?"
First, as a sanity check, a non-GF solution. There must be two cards of one suit and one each of the remaining three suits. There are 4 ways to choose the special suit, $\binom{13}{2}$ ways to choose its two cards, and 13 ways to choose the card from each of the remaining three suits. So the number of hands is
$$4 \cdot \binom{13}{2} \cdot 13^3 = 685,464$$
Now let's try a solution using an ordinary power series generating function. Each suit can contribute 1 to 13 cards, and $n$ cards can be selected from a suit in $\binom{13}{n}$ ways, so the OPSGF is
$$f(x) = \left[ \binom{13}{1}x + \binom{13}{2} x^2 + \binom{13}{3}x^3 + \dots + \binom{13}{13} x^{13} \right]^4$$
The coefficient of $x^n$ is the number of n-card hands. If all we are interested in is the coefficient of $x^5$, we can write
$$f(x) = x^4 \left[ \binom{13}{1} + \binom{13}{2} x + O(x^2) \right]^4 = x^4 \left[ \binom{13}{1}^4 + \binom{4}{1} \binom{13}{1}^3 \binom{13}{2} x +O(x^2) \right]$$
so the coefficient of $x^5$ is $\binom{4}{1} \binom{13}{1}^3 \binom{13}{2}$, in agreement with our previous answer.
Now for the original question: the corresponding exponential generating function is
$$g(x) = \left[ \binom{13}{1}x + \frac{1}{2!} 2! \binom{13}{2} x^2 + \frac{1}{3!} 3! \binom{13}{3}x^3 + \dots + \frac{1}{13!} 13! \binom{13}{13} x^{13} \right]^4$$ which is the same as our OPSGF, because now we are considering the order of the cards in the hand to be significant, and $n$ cards can be selected from 13 in $n! \binom{13}{n}$ ways taking order into account.
The number of n-card hands, considering the order of the cards to be significant, is the coefficient of $\frac{1}{n!} x^n$. If you expand g(x) in the same way as we did for the OPSGF, the coefficient of $\frac{1}{5!} x^5$ is $5! \binom{4}{1} \binom{13}{1}^3 \binom{13}{2}$. But then if you do not want to consider the order of the cards to be significant, we have over-counted, and we must divide by 5! to compensate, leading us back to our original answer. The bottom line is that there is really not much difference between the two approaches, using an OPSGF or an EGF.
| {
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"timestamp": "2023-03-29T00:00:00",
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Find all couple $(x,y)$ for satisfy $\frac{x+iy}{x-iy}=(x-iy)$ I have a problem to solve this exercise, I hope someone help me.
Find all couple $(x,y)$ for satisfy $\frac{x+iy}{x-iy}=(x-iy)$
| $\frac{x+iy}{x-iy}=\frac{(x+iy)^2}{(x-iy)(x+iy)}=\frac{(x+iy)^2}{x^2+y^2}=\frac{x^2-y^2}{x^2+y^2}+i\frac{2xy}{x^2+y^2}$
Then you have to impose
$x=\frac{x^2-y^2}{x^2+y^2}$
$y=\frac{2xy}{x^2+y^2}$
and solve the system.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Use the Chinese remainder theorem to find the general solution of $x \equiv a \pmod {2^3}, \; x \equiv b \pmod {3^2}, \; x \equiv c \pmod {11}$ Help! Midterm exam is coming, but i still unable to solve this simple problem using the Chinese remainder theorem.
$$x \equiv a \pmod {2^3}, \quad x \equiv b \pmod {3^2}, \quad x \equiv c \pmod {11}.$$
| We know that $x$ has the form:
$$
x=a\cdot(s\cdot 3^2\cdot 11)+b\cdot(t\cdot 2^3\cdot 11)+c\cdot(r\cdot 2^3\cdot 3^2)
$$
where $s,t$ and $r$ have been found so that
$$
\begin{align}
s\cdot 3^2\cdot 11&\equiv 1\mbox{ mod }2^3\\
t\cdot 2^3\cdot 11&\equiv 1\mbox{ mod }3^2\\
r\cdot 2^3\cdot 3^2&\equiv 1\mbox{ mod }11
\end{align}
$$
Just try $s=1$, $s=2$ etc. until you find $s$ and similarly for $t$ and $r$. Or use Bezout's identity from the extended Eucledian algorithm to find $s,t$ and $r$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Calculation of all values of $a$ for which $3x^2+(4-2a)x-8-a^2\leq 0$. Calculation of all values of $a$ for which $3x^2+(4-2a)x-8-a^2\leq 0$.
Given that $x$ lies between $-3$ and $2$.
My Try:: Let $x = \alpha,\beta$ be the Roots of Given equation. where $-3<\alpha,\beta<2$
So $\displaystyle \alpha+\beta = \frac{2a-4}{3}$ and $\displaystyle \alpha.\beta = \frac{-(8+a^2)}{3}$
Now I Did not Understand How can I proceed
Help Required
Thanks
| The inequality can be written as:
$$(2x+1)^2 -9 \le (a+x)^2$$
Now if $-3 \le x \le 2$, the LHS takes values of $16$ at the end points. For the inequality to hold, we must then have $16 \le (a+2)^2$ and $16 \le(a-3)^2$. As the RHS is less convex, or is a flatter parabola, the intermediate points will fall in line.
Solving these, we have $a \le -6$ or $a \ge 2$ from the first condition
and $a \le -1$ or $a \ge 7$ from the second. Hence we can have any real $a \not \in (-6,7)$.
| {
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"timestamp": "2023-03-29T00:00:00",
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If $ac-bd=p$ and $ad+bc=0$, then $a^2+b^2\neq 1$ and $c^2+d^2\neq 1$? I'm trying to prove the following:
Let $a,b,c,d\in\Bbb{Z}$ and $p$ be a prime integer. If $ac-bd=p$ and $ad+bc=0$, prove that $a^2+b^2\neq 1$ and $c^2+d^2\neq 1$.
Actually I'm not even sure if this is correct. A proof or counter-example (in case this assertion is wrong) would be great.
I got the following 3 results:
*
*$p^2=(a^2+b^2)(c^2+d^2)$
*$b(c^2+d^2)=-pd$
*$a(c^2+d^2)=pc$
*$c(a^2+b^2)=pa$
*$d(a^2+b^2)=-pb$
I'm at a loss as to how to proceed beyond this. Assuming $c^2+d^2=1$ or $a^2+b^2=1$ does not seem to cause any contradictions.
Thanks in advance!
| The idea is that there aren't a lot of ways we can have $a^2 + b^2 = 1$ or $c^2 + d^2 = 1$ in the first place!
If $a^2 + b^2 = 1$, then one of $a, b$ is zero and the other has absolute value $1$. Let's say $a = 0$ and $b = \pm 1$. Then $ac - b d = -b d = p$, so $d$ is $\pm p$. But we also have $ad + b c = b c = 0$, so $c = 0$.
And now we've found a counterexample that shows the problem statement is false: take $a = 0, b = 1, c = 0, d = -p$. Notice we have $a^2 + b^2 = 1$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Solving system of equations with R Suppose I have the following function :
$f(x,y,z,a)= \cos(ax) + 12y^2 - 9az$
and I want to solve the following syste of equations.
$ f(x,y,z,1)= 10 $,
$f(x,y,z,5)= 7 $,
and
$f(x,y,z,-3)= 17 $.
These are equivalent to
$\cos(x) + 12 y^2 - 9 z(1) = 10$,
$ \cos(5x) + 12y^2 - 9 z(5) = 7 $,
and $ \cos(-3x) + 12y^2 +27z = 17 $.
How do I solve these equations for $x$, $y$, and $z$? I would like to solve these equations, if possible, using R or any other computer tools.
| $cos(x) + 12 y^2 - 9 az = 10$
$cos(5x) + 12 y^2 - 45 az = 7$
$cos(-3x) + 12 y^2 + 27 az = 17 = cos(3x) + …$
$12y^2$ is the easiest to eliminate, so this becomes
$12 y^2 = 10 + 9az - cos(x) = 7 + 45az - cos(5x) = 17 - 27az - cos(3x)$
Since each of these three functions now equal each other, we can subtract them from each other to get 3 pairs of $0$
$10 + 9az - cos(x) - 7 - 45az + cos(5x) = 3 - 36az - cos(x) + cos(5x) = 0$
$17 - 27az - cos(3x) - 10 - 9az + cos(x) = 7 - 36az + cos(x) - cos(3x) = 0$
$17 - 27az - cos(3x) - 7 - 45az + cos(5x) = 10 - 72az - cos(3x) + cos(5x) = 0$
We can now eliminate az by saying
$36az = 3 - cos(x) + cos(5x) = 7 + cos(x) - cos(3x)$
and then
$7 + cos(x) - cos(3x) - 3 + cos(x) - cos(5x) = 4 + 2cos(x) - cos(3x) = 0$
The cleanest identity for $cos(3x)$ is $4cos^3(x) - 3cos(x)$, which then gives us
$4 + 5cos(x) - 4cos^3(x) = 0$
You can then use cubic roots to calculate the possible values of cosx, which can then be plugged into the original formulae using the identities $cos(3x) = 4cos^3(x) - 3cos(x)$ and $cos(5x) = 16cos^5(x) - 20cos^3(x) + 5cos(x)$. After that, "y" and "z" can be calculated by subtracting the pairs of equations created by these new constraints.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove some divisibility results by induction Please hint me, I have two questions:
Prove by induction that:
1)
$$ {13}^n+7^n+19^n=39k,\,\, n\in\mathbb O$$
in which $\mathbb O$ is the set of odd natural numbers.
2)
$$ 5^{2n}+5^n+1=31t,~n\not=3k, $$ $n\in \mathbb N$
| With congruences instead of induction:
(1) You need to show that $13^n+7^n+19^n$ is divisible by both $3$ and $13$. This follows from
$$13^n+7^n+19^n\equiv 1^n+1^n+1^n=3\equiv 0 \ ({\rm mod\ } 3),$$
$$13^n+7^n+19^n\equiv 0^n+7^n+(-7)^n\ ({\rm mod\ }13),$$ where the last expression is $=0$ since $n$ is odd.
(2) Here, I'd like to rewrite $$5^{2n}+5^n+1=(5^n)^0+(5^n)^1+(5^n)^2={(5^n)^3-1\over 5^n-1}={5^{3n}-1\over 5^n-1},$$
using the formula for geometric series (assuming $n\neq 0$). Note that
$$5^{3n}-1=125^n-1\equiv 1^n-1=0\ ({\rm mod\ }31).$$
If $n$ is not divisible by $3$ then $n=3k+1$ or $n=3k+2$, and we have
$$5^{3k+1}-1=5\cdot 125^k-1\equiv 5\cdot 1^k-1=4\ ({\rm mod\ }31),$$
$$5^{3k+2}-1=25\cdot 125^k-1\equiv 25\cdot 1^k-1=24\ ({\rm mod\ }31)$$
So $31$ divides the product $(5^{2n}+5^n+1)(5^n-1)=5^{3n}-1$, but not the factor $5^n-1$. Since $31$ is a prime number, it has to divide the other factor $5^{2n}+5^n+1$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Problem finding in simple algebra It is given,
$$x= \sqrt{3}+\sqrt{2}$$
How to find out the value of $$x^4-\frac{1}{x^4}$$/
The answer is given $40 \sqrt{6}$ but my answer was not in a square-root form
I have done in thsi way:
$$x+ \frac{1}{x}= 2 \sqrt{3}$$
Then,
$$(x^2)^2-\left(\frac{1}{x^2}\right)^2= \left(x^2 + \frac{1}{x^2}\right)^2-2$$
But this way is not working. Where I am wrong?
| Oh, mistake $(x^2)^2+(\frac1{x^2})^2=(x^2+\frac1{x^2})^2-2$ !!
en
$$x+\frac1x=2\sqrt3$$
and
$$x-\frac1x=2\sqrt2$$
so
$$x^2+\frac1{x^2}=(x+\frac1x)^2-2=10$$
and
$$x^2-\frac1{x^2}=(x+\frac1x)(x-\frac1x)=4\sqrt6$$
It follows that
$$x^4-\frac1{x^4}=(x^2+\frac1{x^2})(x^2-\frac1{x^2})=40\sqrt6$$
| {
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"url": "https://math.stackexchange.com/questions/522712",
"timestamp": "2023-03-29T00:00:00",
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Solving $\frac{5}{t-3}-2=\frac{30}{t^2-9}$ I need help with solving this equation:
$$
\frac{5}{t-3}-2=\frac{30}{t^2-9}
$$
I tried to solve, but I always get false result. The result should be $-\frac{1}{2}$ but I always get $-\frac{1}{2}$ and $3$.
This is how I did it:
$$\begin{align}
\frac{5}{t-3}-2&=\frac{30}{t^2-9}\\
\frac{5-2\cdot(t-3)}{t-3}&=\frac{30}{t^2-9}\\
\frac{5-2t+6}{t-3}&=\frac{30}{t^2-9}\\
\frac{-2t+11}{t-3}&=\frac{30}{t^2-9}\\
\frac{-2t+11}{t-3}\cdot\left(t^2-9\right)&=30\\
\frac{(-2t+11)\cdot\left(t^2-9\right)}{t-3}&=30\\
\frac{(-2t+11)\cdot(t-3)\cdot(t+3)}{t-3}&=30\\
(-2t+11)\cdot(t+3)&=30\\
-2t^2-6t+11t+33&=30\\
-2t^2+5t+33&=30\\
-2t^2+5t+33-30&=0\\
-2t^2+5t+3&=0\\
\end{align}$$
Solving
$$
-2t²+5t+3=0
$$
I get:
$$
-\frac{1}{2}\text{ and }3
$$
Am I doing something wrong?
| The original equation is not defined for $t = 3$, hence, we need to omit that as a "solution" to the original equation.
Important: Always test possible solutions to see if they are defined for, and/or solve, the original equation. If not, we cannot include those as solutions to the original equation.
| {
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$a+b+c=0;a^2+b^2+c^2=1$ then $a^4+b^4+c^4$ is equal to what?
$a+b+c=0;a^2+b^2+c^2=1$ then $a^4+b^4+c^4$ is equal to what?
I tried to solve this problem, and I get $a^4+b^4+c^4 = 2(a^2b^2 + 1)$ but I'm not sure if it's correct
| Given a+b+c=0a+b+c=0 and (a2+b2+c2)=1(a2+b2+c2)=1, Now (a2+b2+c2)2=12=1(a2+b2+c2)2=12=1
(a4+b4+c4)+2(a2b2+b2c2+c2a2)=1.................(1)(a4+b4+c4)+2(a2b2+b2c2+c2a2)=1.................(1)
and from (a+b+c)2=0(a+b+c)2=0,
we get 1+2(ab+bc+ca)=0⇒(ab+bc+ca)=−121+2(ab+bc+ca)=0⇒(ab+bc+ca)=−12
again squaring both side , we get (ab+bc+ca)2=14(ab+bc+ca)2=14
(a2b2+b2c2+c2a2)+2abc(a+b+c)=14⇒(a2b2+b2c2+c2a2)=14(a2b2+b2c2+c2a2)+2abc(a+b+c)=14⇒(a2b2+b2c2+c2a2)=14
So put in eqn.... (1)(1) , we get
(a4+b4+c4)+2⋅14=1⇒(a4+b4+c4)=12
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/524335",
"timestamp": "2023-03-29T00:00:00",
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infinite summation formula help How do I find the following sum? $$\sum_{x=3}^\infty 1.536x\left(\frac{5}{8}\right)^x$$ It wouldn't be geometric because of the $x$ in front, right?
| Hint: You are correct that it's not a geometric series. What we can do is note that
$$\sum\limits_{n = 0}^{\infty} r^n = \frac{1}{1 - r}$$
Now differentiating termwise, we see that
$$\sum\limits_{n = 1}^{\infty} nr^{n - 1} = \frac{1}{(1 - r)^2}$$
This can be adapted for your series:
\begin{align}
\sum\limits_{x = 3}^{\infty} x \left(\frac 5 8 \right)^x &= \frac 5 8 \sum\limits_{x = 3}^{\infty} x \left(\frac 5 8\right)^{x - 1} \\
&= \frac 5 8 \left(\sum\limits_{x = 1}^{\infty} x \left(\frac 5 8 \right)^{x - 1} - 1 - 2 \cdot \frac 5 8\right) \\
&= \frac 5 8 \left(\frac{1}{(1 - (5/8))^2}-1 - \frac {10} 8\right)
\end{align}
| {
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"timestamp": "2023-03-29T00:00:00",
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How find the sum $\sum_{n=1}^{\infty}\frac{\binom{4n-4}{n-1}}{2^{4n-3}(3n-2)}$ today,I see a amazing math problem:
show that
$$\sum_{n=1}^{\infty}\dfrac{\binom{4n-4}{n-1}}{2^{4n-3}(3n-2)}=\dfrac{\sqrt[3]{17+3\sqrt{33}}}{3}-\dfrac{2}{3\sqrt[3]{17+3\sqrt{33}}}-\dfrac{1}{3}$$
This problem is from here.
But I consider sometimes,and I think it maybe use Taylor therom
$$\arcsin{x}=\sum_{n=0}^{\infty}\dfrac{(2n-1)!!}{(2n+1)(2n)!!}x^{2n+1}$$
Thank you,Now is 24:00 in beijing time,so I must go bed.I hope someone can help.Thank you
| The sum can be written as
$$\sum_{n=0}^\infty \binom{4n}{n}\frac{1}{(3n+1)2^{4n+1}}=\frac{1}{2}{\;}_3F_2 \left(\frac{1}{4},\frac{1}{2},\frac{3}{4}; \frac{2}{3},\frac{4}{3};\frac{16}{27} \right)$$
For the Hypergeometric Function, see equation (25) of http://mathworld.wolfram.com/HypergeometricFunction.html
$${\;}_3F_2 \left(\frac{1}{4},\frac{1}{2},\frac{3}{4}; \frac{2}{3},\frac{4}{3};y \right) = \frac{1}{1-x^3}$$
where $x$ and $y$ are related by
$$y=\left( \frac{4x(1-x^3)}{3}\sqrt[3]{4} \right)^3$$
Putting $y=16/27$, we get a polynomial equation in $x$.
$$1=16 x^3(1-x^3)^3$$
According to Mathematica, this equation has only two real roots which are
$$\begin{align*}
\alpha &= \frac{1}{\sqrt[3]{2}} \\
\beta &= \left\{\frac{1}{6} \left(5-\frac{4}{\left(19-3 \sqrt{33}\right)^{1/3}}-\left(19-3 \sqrt{33}\right)^{1/3}\right) \right\}^{1/3} \end{align*}$$
Numericaly, $\beta$ gives the right answer. So, the final result is
$$ \begin{align*}\sum_{n=0}^\infty \binom{4n}{n}\frac{1}{(3n+1)2^{4n+1}} &=\frac{1}{2\left(1-\beta^3 \right)} \\
&= \dfrac{\sqrt[3]{17+3\sqrt{33}}}{3}-\dfrac{2}{3\sqrt[3]{17+3\sqrt{33}}}-\dfrac{1}{3} \\ &\approx 0.543689\end{align*}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/526072",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
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Differentiation answer check $$f(x)=\arcsin \sqrt{\frac{x}{x+1}} + \arctan \sqrt{x} \mbox{.}$$
$$f'(x) = \frac{1}{\sqrt{1-\frac{x}{x+1}}} \cdot \frac{1}{2}\sqrt{\frac{x+1}{x}} \cdot -x^{-2} + \frac{1}{1+x} \cdot \frac{1}{2\sqrt{x}} = -\frac{\sqrt{x+1}}{2\sqrt{1-\frac{x}{x+1}}{\sqrt{x}}\frac{1}{x^2}} + \frac{1}{2\sqrt{x} + 2x}\mbox{.}$$
Is there anything else I can do with the answer?
| This is how I would submit. Nothing doing, sonny!
| {
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"timestamp": "2023-03-29T00:00:00",
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system of equations $\sqrt{x}+y = 11$ and $x+\sqrt{y} = 7$. If $x,y\in \mathbb{R}$ and $\sqrt{x}+y = 11\;$ and $x+\sqrt{y} = 7$. Then $(x,y) = $
$\underline{\bf{My\;\; Try::}}$ Let $x=a^2$ and $y=b^2$, Then equation is $a+b^2 = 11$ and $a^2+b = 7$.
$(a+b)+(a+b)^2-2ab = 18$ and Now Let $a+b=S$ and $ab=P$, we get $S+S^2-P=18$
Now I did not understand how can I solved it.
Help required
Thanks
| HINT:
Assuming $x,y$ are real we have $x=7-\sqrt y\le 7\implies a^2\le 7$
Putting $b=7-a^2,$
$$a+(7-a^2)^2=11\implies a^4-14a^2+a+38=0$$
Observe that $a=2$ is a solution
For the general case, we need to use the formula described here or here
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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More values of $a$ and $D $ on conditions set by me and a way to obtain more values. Here I define -: $\alpha=a+ \sqrt D$ and $\beta=a-\sqrt D$ Then find out values of $\alpha$ and $\beta$ satisfying- $$\alpha>1 \quad and \quad -1< \beta <1 $$
and both the variables are integers .I began finding some initial values-
\begin{array}{c|lcr}
& \text{a} & \text{D} & \text{} \\
\hline
& 1 & 2 \\
& 1 & 3 \\
& 2 & 2 \\
& 2 & 3 \\
& 2 & 5 \\
\end{array}
Can anyone make a very big list of such numbers and tell me how to generate the values of $\alpha \quad and \quad\beta$.
| Let's first consider that $a$ and $D$ are 2 real positive number. You have $3$ conditions on $a$:
\begin{equation}
\begin{array}{rcl}
a & > & 1 - \sqrt{D} \\
a & > & -1 + \sqrt{D} \\
a & < & 1 + \sqrt{D}
\end{array}
\end{equation}
Since $1 + \sqrt{D}$ is always greater than $1-\sqrt{D}$ and $-1+\sqrt{D}$, then you have the following:
$$\max(1-\sqrt{D}, -1 + \sqrt{D}) < a < 1 + \sqrt{D}$$
Now, we have to evaluate $\max(1-\sqrt{D}, -1 + \sqrt{D})$. It's easy to notice that:
\begin{equation}
\begin{array}{cc}
D=1 & \max(1-\sqrt{1}, -1 + \sqrt{1}) = 0 \\
D\geq 2 & \max(1-\sqrt{D}, -1 + \sqrt{D}) = ...
\end{array}
\end{equation}
Since $\sqrt{D} > 1$ when $D \geq 2$, then we can pose that $\sqrt{D} = 1 + \delta$, with $\delta > 0$. Then:
$$\max(1-\sqrt{D}, -1 + \sqrt{D}) = \max(1-1-\delta, -1 + 1+\delta) = \max(-\delta, \delta) = \delta = -1 + \sqrt{D}$$
Then, you can say that, for every $D \geq 1$ we have that $\alpha > 1$ and $-1 < \beta <1$ when
$$-1+\sqrt{D} < a < 1 + \sqrt{D}$$
Passing to integer when $D \geq 2$, then we have:
$$\lceil -1+\sqrt{D} \rceil \leq a \leq \lfloor 1 + \sqrt{D} \rfloor$$
or
$$ -1+\lceil \sqrt{D} \rceil \leq a \leq 1 + \lfloor \sqrt{D} \rfloor$$
Then:
if $D=1$, $0 < a < 2$, $a=1$
if $D=2$, $1 \leq a \leq 2$, $a=1, a=2$
if $D=3$, $1 \leq a \leq 2$, $a=1, a=2$
if $D=4$, $1 \leq a \leq 3$, $a=1, a=2, a=3$
and so on...
| {
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A pair of unbiased dice are rolled together till a sum of either 5 or 7 is obtained, find the probability that 5 comes before 7 Problem :
A pair of unaiased dice are rolled together till a sum of either 5 or 7 is obtained, find the probability that 5 comes before 7.
My approach :
Probability P(A) of getting 5 as sum on two dice is
$$P(A) = \frac{4}{36} = \frac{1}{9}$$
Let B the event that a sum of 7 occurs and C the event that neither a sum of 5 nor a sum of 7 occurs. We have :
$$P(B)= \frac{6}{36} = \frac{1}{6}$$
$$P(C) = \frac{26}{36} = \frac{13}{18}$$
Please suggest how to proceed further.
| You are interested that the game will end where you first get sum of $5$, and that it will happen before the first "sum of $7$". Hence, by noting that the first event has $4$ elementary outcomes while the second has $6$, and by using the law of total probability, you get
\begin{align}
P(\text{ 5 before 7}) &= \sum_{n=1}^{\infty}P(\text{ sum of $5$}|\text{$n-1$ times neither $5$ or $7$} )P(\text{$n-1$ times neither $5$ or $7$} )\\
&=\frac{4}{36} \frac{26^0}{36^0} + \frac{4}{36} \frac{26^1}{36^1}+\cdots+ \frac{4}{36} \frac{26^{n-1}}{36^{n-1}}\\
&=
\sum_{n=1}^{\infty} \left(\frac{26}{36}\right)^{n-1}\frac{4}{36}\\
& = \frac{1}{9}\frac{1}{1-13/18}\\
& = \frac{2}{5} .
\end{align}
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Find the remainder when $2^{47}$ is divided by $47$ So I am trying to work out how to find remainders in several different way.
I have a few very similar question,
1.) Find the remainder when $2^{47}$ is divided by 47
So i have a solution that says
$$2^{24} \equiv 2$$
$$2^{48} \equiv 4$$
$$2^{47} \equiv 2$$ Since $(2,47)=1$
and
2.)Find the remainder when $2^{32}$ is divided by $47$:
We have $$2^6 = 64 \equiv 17$$
$$2^{12} \equiv 17^2 = 289 \equiv 7$$
$$2^{24} \equiv 7^2=49 \equiv 2$$
$$2^8 \equiv 4*17=68 \equiv 21$$
$$2^{32}=2^{8}*2^{24} \equiv21*2 \equiv42$$
Okay so although i have some solutions here, Im not 100 percent sure how they derived this. For example on the second question, why have they started with $2^{6}$ before finding the remainder? Is there some steps missing or am I just missing the point?
Is this called something in particular? There are not many problems similar to and i do not know if it has a special name to search for.
P.S Please do not use fermat's little theorem, i want to understand this method, thanks :)
| 1) Find the remainder when $2^{47}$ is divided by $47$.
$2^1\equiv2\pmod{47}$
$2^2\equiv4\pmod{47}$
$2^4\equiv16\pmod{47}$
$2^8=(2^4)^2\equiv(16)^2=256\equiv21\pmod{47}$
$2^{16}=(2^8)^2\equiv(21)^2=441\equiv18\pmod{47}$
$2^{24}=2^8\cdot2^{16}\equiv21\cdot18=378\equiv2\pmod{47}$
$2^{48}\equiv2^2=4\pmod{47}$
$2*2^{47}\equiv2*2\pmod{47}$
Since $ca\equiv cb\pmod m$ and $(c,m)=1$ implies $a\equiv b\pmod m$, and since $(2,47)=1$, we can cancel the $2$:
$2^{47}\equiv2\pmod{47}$
2) Find the remainder when $2^{32}$ is divided by $47$.
From the previous solution:
$2^{24}\equiv2\pmod{47}$
$2^8\equiv21\pmod{47}$
$2^{32}=2^{24+8}=2^{24}\cdot2^8\equiv2\cdot21=42\pmod{47}$
| {
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"timestamp": "2023-03-29T00:00:00",
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Solution of $ 1+ 5 * 2^m =n^2$ Equation and some other question from this equation I want to find all integer Solutions of $ 1+ 5 * 2^m =n^2$ Equation . From this eqn I have to answer the following answer .
*
*I have to find an expression for $ n^2-1 $
*Are $ (n+1) $ and $ (n-1) $ both even or both odd or is one even and the other odd ?
*Let $ a = \frac{n-1}{2} $ . I have to find an expression for $ a(a+1) $ .
*If a is odd , is $ (a+1) $ even or odd ?
*From parts 3 and 4 , is it possible for $ a(a+1) =1 $ or $ a=1 $ ?
*I have to find the pnly possible values a can take and then have to find what m and n should be ?
I want to have the solution of the above equation . Then I will answer the above question .
| I am considering the case $m\ge 2$
Observe that $5\cdot2^m=n^2-1$ is even and so are $n\pm1$
and $(n-1,n+1)=(n-1,n+1-(n-1))=(n+1,2)=2$
alternatively, if integer $d>0$ divides both $n+1,n-1, d$ will divide $n+1-(n-1)=2$
$\implies (n-1,n+1)|2\implies (n-1,n+1)=2$ as $n\pm1$ are even
So, we have
$$5\cdot2^{m-2}=\frac{n+1}2\cdot \frac{n-1}2$$ with $\left(\frac{n+1}2,\frac{n-1}2\right)=1$
So, either $\displaystyle\frac{n+1}2=5; \frac{n-1}2=2^{m-2}$
or $\displaystyle\frac{n-1}2=5; \frac{n+1}2=2^{m-2}$
or $\displaystyle\frac{n-1}2=5\cdot2^{m-2} ;\frac{n+1}2=1$
or $\displaystyle\frac{n+1}2=5\cdot2^{m-2} ;\frac{n-1}2=1$
| {
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"timestamp": "2023-03-29T00:00:00",
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If $a^3 + b^3 +3ab = 1$, find $a+b$
Given that the real numbers $a,b$ satisfy $a^3 + b^3 +3ab = 1$, find $a+b$.
I tried to factorize it but unable to do it.
| There are a continuum of solutions to
$$
a^3+b^3+3ab=1
$$
Suppose that
$$
x=a+b
$$
then
$$
\begin{align}
1
&=a^3+(x-a)^3+3a(x-a)\\
&=x^3-3ax^2+3a^2x+3ax-3a^2
\end{align}
$$
which means
$$
\begin{align}
0
&=(x-1)\left(x^2+(1-3a)x+3a^2+1\right)
\end{align}
$$
So either $x=1$ irregardless of $a$, or
$$
x=\frac{3a-1\pm(a+1)\sqrt{-3}}{2}
$$
Thus, other than $x=1$, the only real $x$ is $-2$, which comes from $a=-1$.
That is, the only two real values of $a+b$ are $1$ and $-2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/540084",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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Evaluation of $\lim_{n\rightarrow \infty}\frac{1}{2n}\cdot \ln \binom{2n}{n}$
Evaluate
$$\lim_{n\rightarrow \infty}\frac{1}{2n}\cdot \ln \binom{2n}{n}.$$
$\underline{\bf{My\;\;Try}}::$ Let $\displaystyle y = \lim_{n\rightarrow \infty}\frac{1}{2n}\cdot \ln \binom{2n}{n} = \lim_{n\rightarrow \infty}\frac{1}{2n}\cdot \ln \left(\frac{(n+1)\cdot (n+2)\cdot (n+3)\cdots (n+n)}{(1)\cdot (2)\cdot (3)\cdots (n)}\right)$
$\displaystyle y = \lim_{n\rightarrow \infty}\frac{1}{2n}\cdot \left\{\ln \left(\frac{n+1}{1}\right)+\ln \left(\frac{n+2}{2}\right)+\ln \left(\frac{n+3}{3}\right)+\cdots+\ln \left(\frac{n+n}{n}\right)\right\}$
$\displaystyle y = \lim_{n\rightarrow \infty}\frac{1}{2n}\cdot \sum_{r=1}^{n}\ln \left(\frac{n+r}{r}\right) = \lim_{n\rightarrow \infty}\frac{1}{2n}\cdot \sum_{r=1}^{n}\ln \left(\frac{1+\frac{r}{n}}{\frac{r}{n}}\right)$
Now Using Reinman Sum
$\displaystyle y = \frac{1}{2}\int_{0}^{1}\ln \left(\frac{x+1}{x}\right)dx = \frac{1}{2}\int_{0}^{1}\ln (x+1)dx-\frac{1}{2}\int_{0}^{1}\ln (x)dx = \ln (2)$
My Question is , Is there is any method other then that like Striling Approximation OR Stolz–Cesàro theorem OR Ratio Test
If yes then please explain here
Thanks
| Notice $\displaystyle \sum_{k=0}^{2n}\binom{2n}{k} = 2^{2n}$ and $\displaystyle \binom{2n}{n} \ge \binom{2n}{k}$ for all $0 \le k \le 2n$, we have
$$
\frac{2^{2n}}{2n+1}\le \binom{2n}{n} \le 2^{2n}
\quad\implies\quad \log 2 - \frac{\log{(2n+1)}}{2n} \le \frac{1}{2n} \log \binom{2n}{n} \le \log 2$$
Since $\quad\displaystyle \lim_{n\to\infty} \frac{\log(2n+1)}{2n} = 0\quad$,
we get $\quad\displaystyle \lim_{n\to\infty} \frac{1}{2n}\log\binom{2n}{n} = \log 2$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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"answer_id": 1
} |
Maximum N that will hold this true Find the largest positive integer $N$ such that
$$\sqrt{64 + 32^{403} + 4^{N+3}}$$
is an integer
Is $N = 1003$?
| First note that
$$64 + 32^{403} + 4^{N+3} = 2^6 +2^{2015} + 2^{2N+6}.$$
It is easy to see that this cannot be a square of an integer if $2N+6 < 2015$, so now consider the way in which the expression can be a square:
$(2^3 + 2^{N+3})^2 = 2^6 + 2.2^3.2^{N+3} + 2^{2N+6} = 2^6 + 2^{N+7} + 2^{2N+6}$
Thus we need $N+7 = 2015$, so that $N=2008$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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absolute value inequalities When answer this kind of inequality
$|2x^2-5x+2| < |x+1|$
I am testing the four combinations when both side are +, one is + and the other is - and the opposite and when they are both -.
When I check the negative options, I need to flip the inequality sign?
Thanks
| if you take $x<0$
$-(2x^2-5x+2)<-(x+1)$
$-2x^2+6x-1<0$
you gets roots:
$x_1=\frac {1}{2}(3−7\sqrt7)≈2.82$
$x_2=\frac {1}{2}(3+\sqrt7)≈0.18$
Becouse we have x<0 (x is negative and our roots are pozitive) there are solutions:
$x \in { \emptyset}$
if $X>0$
$2x^2 - 5x + 2 < x +1$
$2x^2 - 6x + 1 < 0$
you gets roots:
$ x_1 = \frac {1}{2}( 3-\sqrt 7) \approx 0.18 $
$ x_2 = \frac {1}{2}( 3+\sqrt7) \approx 2.82$
so you get
$ \frac {1}{2}( 3-\sqrt7) \leq x \leq \frac {1}{2}( 3+\sqrt7) $
if we combine both results we get
$x \in {( \frac {1}{2}( 3-\sqrt 7), \frac {1}{2}( 3+\sqrt 7) )}$
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "3",
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Finite Sum $\sum\limits_{k=1}^{m-1}\frac{1}{\sin^2\frac{k\pi}{m}}$
Question : Is the following true for any $m\in\mathbb N$?
$$\begin{align}\sum_{k=1}^{m-1}\frac{1}{\sin^2\frac{k\pi}{m}}=\frac{m^2-1}{3}\qquad(\star)\end{align}$$
Motivation : I reached $(\star)$ by using computer. It seems true, but I can't prove it. Can anyone help?
By the way, I've been able to prove $\sum_{n=1}^{\infty}\frac{1}{n^2}=\frac{{\pi}^2}{6}$ by using $(\star)$.
Proof : Let
$$f(x)=\frac{1}{\sin^2x}-\frac{1}{x^2}=\frac{(x-\sin x)(x+\sin x)}{x^2\sin^2 x}.$$
We know that $f(x)\gt0$ if $0\lt x\le {\pi}/{2}$, and that $\lim_{x\to 0}f(x)=1/3$. Hence, letting $f(0)=1/3$, we know that $f(x)$ is continuous and positive at $x=0$. Hence, since $f(x)\ (0\le x\le {\pi}/2)$ is bounded, there exists a constant $C$ such that $0\lt f(x)\lt C$. Hence, substituting $x={(k\pi)}/{(2n+1)}$ for this, we get
$$0\lt \frac{1}{\frac{2n+1}{{\pi}^2}\sin^2\frac{k\pi}{2n+1}}-\frac{1}{k^2}\lt\frac{{\pi}^2C}{(2n+1)^2}.$$
Then, the sum of these from $1$ to $n$ satisfies
$$0\lt\frac{{\pi}^2\cdot 2n(n+1)}{(2n+1)^2\cdot 3}-\sum_{k=1}^{n}\frac{1}{k^2}\lt\frac{{\pi}^2Cn}{(2n+1)^2}.$$
Here, we used $(\star)$. Then, considering $n\to\infty$ leads what we desired.
| There not being a residue theory answer, I have copied excerpts from this answer:
$\frac{2n/z}{z^{2n}-1}$ has residue $1$ at $z=e^{\pi ik/n}$ and residue $-2n$ at $z=0$.
Thus,
$$
\begin{align}
\left(\frac{2i}{z-\frac1z}\right)^2\frac{2n/z}{z^{2n}-1}
&=\frac{-4z^2}{z^4-2z^2+1}\frac{2n/z}{z^{2n}-1}\\
&=\left(\frac1{(z+1)^2}-\frac1{(z-1)^2}\right)\frac{2n}{z^{2n}-1}
\end{align}
$$
has residue $\csc^2\left(\pi k/n\right)$ at $z=e^{\pi ik/n}$ except at $z=\pm1$. A bit of computation gives
$$
\begin{align}
\frac{2n}{z^{2n}-1}
&=\phantom{+}\frac1{z-1}-\frac{2n-1}2+\frac{(2n-1)(2n+1)}{12}(z-1)+O\!\left((z-1)^2\right)\\
&=-\frac1{z+1}-\frac{2n-1}2-\frac{(2n-1)(2n+1)}{12}(z+1)+O\!\left((z+1)^2\right)
\end{align}
$$
Therefore,
$$\newcommand{\Res}{\operatorname*{Res}}
\Res_{z=1}\left(\frac1{(z+1)^2}-\frac1{(z-1)^2}\right)\frac{2n}{z^{2n}-1}
=\frac14-\frac{4n^2-1}{12}=-\frac{n^2-1}3
$$
and
$$
\Res_{z=-1}\left(\frac1{(z+1)^2}-\frac1{(z-1)^2}\right)\frac{2n}{z^{2n}-1}
=\frac14-\frac{4n^2-1}{12}=-\frac{n^2-1}3
$$
Since the sum of the residues at all the singularities is $0$, we get that half the sum over the singularities except at $z=\pm1$ is
$$
\sum_{k=1}^{n-1}\csc^2\left(\frac{k}{n}\pi\right)=\frac{n^2-1}3
$$
| {
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"url": "https://math.stackexchange.com/questions/544228",
"timestamp": "2023-03-29T00:00:00",
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Prove $\left(\frac{a+b}{c} + \frac{b+c}{a} + \frac{c+a}{b}\right)\left(\frac{ab}{c^2+b^2} + \frac{bc}{a^2 + c^2} + \frac{ca}{a^2 + b^2}\right) \ge 9$ If $a,b,c \in \mathbb{R^+}$,then prove that the following inequality holds:
$$\left(\frac{a+b}{c} + \frac{b+c}{a} + \frac{c+a}{b}\right)\left(\frac{ab}{c^2+b^2} + \frac{bc}{a^2 + c^2} + \frac{ca}{a^2 + b^2}\right) \ge 9 \tag{1}$$
This is a inequality posted on the AoPS site and it's been answered by one user. The problem is that I can't quite understand the proof. The proof is:
After using AM-GM we need to prove:
$$(a^2c + a^2b + b^2c + b^2a + c^2a + c^2b)^3 \ge 27abc(a^2+b^2)(a^2+c^2)(b^2+c^2) \tag{2}$$
which holds from AM-GM.
The second part is clear to me, but I can't understand the part when he transforms $(1)$ into $(2)$. Because he said he used AM-GM I apply that inequality on both terms in $(1)$, but i end up with:
$$\sqrt[3]{\frac{(a+b)(b+c)(c+a)a^2b^2c^2}{abc(a^2+b^2)(a^2+c^2)(b^2+c^2)}} \ge 1$$
And I don't know how to continue. I used the fact that:$$(a+b)(b+c)(c+a) = a^2c + a^2b + b^2c + b^2a + c^2a + c^2b + 2abc$$
But the term $a^2b^2c^2$ in the numerator is making troubles.
From AM-GM and Nesbitt's Inequality we know that the first term on the LHS in $(1)$ is $\ge 6$ and i tried proving
$$\frac{ab}{c^2+b^2} + \frac{bc}{a^2 + c^2} + \frac{ca}{a^2 + b^2} \ge \frac 32$$
But again I end up with empty hands.
How can I prove it? Can someone explain me the first part in the AoPS answer?
| As njguliyev pointed out,
$$\left(\frac{a+b}{c} + \frac{b+c}{a} + \frac{c+a}{b}\right) = \frac{a^2c + a^2b + b^2c + b^2a + c^2a + c^2b}{abc}$$
$$\left(\frac{ab}{c^2+b^2} + \frac{bc}{a^2 + c^2} + \frac{ca}{a^2 + b^2}\right) \ge 3 \frac{\sqrt[3]{(abc)^2}}{\sqrt[3]{(a^2+b^2)(a^2+c^2)(b^2+c^2)}}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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A problem on limit involving various functions Find the value of $$\lim \limits_{x\to 0} \frac{\tan\sqrt[3]x\ln(1+3x)}{(\tan^{-1}\sqrt{x})^2(e^{5\large \sqrt[3]x}-1)}$$
Applying L'Hospital's rule does not seem to simplify the expression.
| Note that
$$\lim \limits_{y\to 0}\frac{\tan y}{y}=\lim_{y\to 0}\frac{\tan^{-1} y}{y}=\lim_{y\to 0}\frac{\ln (1+y)}{y}=\lim_{y\to 0}\frac{e^y-1}{y}=1.$$
Therefore,
$$\lim \limits_{x\to 0^+} \frac{\tan\sqrt[3]x\cdot \ln(1+3x)}{\left(\tan^{-1}\sqrt{x}\right)^2\cdot(e^{5 \sqrt[3]x}-1)}=\lim_{x\to 0^+}\frac{\sqrt[3]x\cdot 3x}{(\sqrt{x})^2\cdot 5 \sqrt[3]{x}}=\frac{3}{5}.$$
Edit: Some comments below suggest that the answer above may not be detailed enough. Let me explain a little more about the gap between the two lines of equations above. The limit we are concerned about can be written as
$$\lim \limits_{x\to 0^+}\left(\frac{\tan\sqrt[3]x}{\sqrt[3]x}\cdot \frac{\ln(1+3x)}{3x}\cdot \left(\frac{\sqrt{x}}{\tan^{-1}\sqrt{x}}\right)^2\cdot\frac{5 \sqrt[3]x}{e^{5 \sqrt[3]x}-1}\cdot \frac{\sqrt[3]x\cdot 3x}{(\sqrt{x})^2\cdot 5 \sqrt[3]{x}}\right). $$
On the one hand, as $x\to 0^+$, $\sqrt[3]x$, $3x$ $\sqrt x$ and $5\sqrt[3]x$ all approach to $0$, so from the first displayed line we know that each limit of first four terms in the product above exists and equals $1$. On the other hand, the limit of the last term exists and equals $\dfrac{3}{5}$. As a result, the original limit exists and equals $\dfrac{3}{5}$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Probability task (Find probability that the chosen ball is white.) I have this task in my book:
First box contains $10$ balls, from which $8$ are white. Second box contains $20$ from which $4$ are white. From each box one ball is chosen. Then from previously chosen two balls, one is chosen. Find probability that the chosen ball is white.
The answer is $0.5$. Again I get the different answer:
There are four possible outcomes when two balls are chosen:
$w$ -white, $a$ - for another color
$(a,a),(w,a),(a,w),(w,w)$.
Each outcome has probability:
$\frac{2}{10} \cdot \frac{16}{20}; \frac{8}{10} \cdot \frac{16}{20}; \frac{2}{10} \cdot \frac{4}{20}; \frac{8}{10} \cdot \frac{4}{20};$
In my opinion the probability that the one ball chosen at the end is white is equal to the sum of last three probabilities $\frac{8}{10} \cdot \frac{16}{20} + \frac{2}{10} \cdot \frac{4}{20} + \frac{8}{10} \cdot \frac{4}{20}=\frac{21}{25}$. Am I wrong or there is a mistake in the answer in the book?
| First box: $p(white) = \frac{8}{10}$
Second box: $p(white) = \frac{2}{10}$
One of the two is chosen. If you pick the first ball $(p=0.5)$, then $p(white)=0.8$. If you pick the second $(p=0.5)$, then $p(white) = 0.2$. Now notice that p(ball 1 is white) = p(ball 2 is nonwhite) and p(ball 2 is white) = p(ball 1 is white). This symmetry lets us conclude immediately that the probability of ending up with a white ball is $0.5$.
| {
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Prove that $ \int_{0}^{1}\frac{\sqrt{1-x^2}}{1-x^2\sin^2 \alpha}dx = \frac{\pi}{4\cos^2 \frac{\alpha}{2}}$. Prove that $\displaystyle \int_{0}^{1}\frac{\sqrt{1-x^2}}{1-x^2\sin^2 \alpha}dx = \frac{\pi}{4\cos^2 \frac{\alpha}{2}}$.
$\bf{My\; Try}::$ Let $x = \sin \theta$, Then $dx = \cos \theta d\theta$
$\displaystyle = \int _{0}^{1}\frac{\cos \theta }{1-\sin^2 \theta \cdot \sin^2 \alpha}\cdot \cos \theta d\theta = \int_{0}^{1}\frac{\cos ^2 \theta }{1-\sin^2 \theta \cdot \sin ^2 \alpha}d\theta$
$\displaystyle = \int_{0}^{1}\frac{\sec^2 \theta}{\sec^4 \theta -\tan^2 \theta \cdot \sec^2 \theta \cdot \sin^2 \alpha}d\theta = \int_{0}^{1}\frac{\sec^2 \theta }{\left(1+\tan ^2 \theta\right)^2-\tan^2 \theta \cdot \sec^2 \theta\cdot \sin^2 \alpha}d\theta$
Let $\tan \theta = t$ and $\sec^2 \theta d\theta = dt$
$\displaystyle \int_{0}^{1}\frac{1}{(1+t^2)^2-t^2 (1+t^2)\sin^2 \alpha}dt$
Now How can I solve after that
Help Required
Thanks
| Here is another method to solve the problem by using residue. Suppose $\alpha\in(0,\pi/2)$. Using $x=\sin(\theta)$ and $u=\tan(\theta), z=e^{i\theta}$, one has
\begin{eqnarray}
&&\int_0^1\frac{\sqrt{1-x^2}}{1-x^2\sin^2(\alpha)}\,\mathrm{d}x\\
&=&\int_0^{\pi/2}\frac{\cos^2(\theta)}{1-\sin^2(\theta)\sin^2(\alpha)}\,\mathrm{d}\theta\\
&=&\int_0^{\pi/2}\frac{\frac{1+\cos(2\theta)}{2}}{1-\frac{1-\cos(2\theta)}{2}\sin^2(\alpha)}\,\mathrm{d}\theta\\
&=&\frac14\int_0^{2\pi}\frac{1+\cos(\theta)}{2-(1-\cos(\theta))\sin^2(\alpha)}\,\mathrm{d}\theta\\
&=&\frac14\int_{|z|=1}\frac{1+\frac{z+z^{-1}}{2}}{2-(1-\frac{z+z^{-1}}{2})\sin^2(\alpha)}\frac{1}{iz}\,\mathrm{d}z\\
&=&\frac1{4i}\int_{|z|=1}\frac{(z+1)^2}{z[4+(z-1)^2\sin^2(\alpha)]}\,\mathrm{d}z\\
&=&\frac1{4i\sin^2(\alpha)}\int_{|z|=1}\frac{(z+1)^2}{z(z+\tan^2(\alpha/2))(z+\cot^2(\alpha/2))}\,\mathrm{d}z\\
&=&\frac1{4i\sin^2(\alpha)}2\pi i\bigg[\text{Res}(\frac{(z+1)^2}{(z+\tan^2(\alpha/2))(z+\cot^2(\alpha/2))},z=0)+\text{Res}(\frac{(z+1)^2}{z(z+\cot^2(\alpha/2))},z=-\tan^2(\alpha/2)\bigg]\\
&=&\frac{\pi}{4\cos^2(\alpha/2)}.
\end{eqnarray}
| {
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"timestamp": "2023-03-29T00:00:00",
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A closed form for $\int_0^\infty\frac{\ln(x+4)}{\sqrt{x\,(x+3)\,(x+4)}}dx$ I need to a evaluate the following integral
$$I=\int_0^\infty\frac{\ln(x+4)}{\sqrt{x\,(x+3)\,(x+4)}}dx.$$
Both Mathematica and Maple failed to evaluate it in a closed form, and lookups of the approximate numeric value $4.555919963334436...$ in ISC+ and WolframAlpha did not return plausible closed form candidates either. Does anybody by any chance have an idea if a closed form exists for this integral, and what it could be?
| This integral in more general form you can find in the tables of integrals by Prudnikov A. P., Brychkov Yu. A. and Marichev O. I.:
$$\int_0^{\infty}\!\!\!\!\frac{x^{\alpha-1}\ln^n(cx+d)}{(ax+b)^\sigma(cx+d)^\rho} \!\mathrm dx=(-1)^n\left(\!\frac{d}{c}\!\right)^\alpha\!\! b^{-\sigma}\!\frac{\partial^n}{\partial \rho^n}\!\!\left[d^{-\rho}\mathbf{B}(\alpha,\sigma+\rho-\alpha) \ _2F_1\!\!\left(\!\sigma,\alpha;\sigma+\rho;1\!-\!\frac{ad}{bc}\!\right)\!\right]$$
when $\Re\{\alpha\}>0; \ \Re\{\alpha-\sigma-\rho\}<0; \ |\arg(ax+b)|,|\arg(cx+d)|<\pi \ \forall x\in[0,\infty)$, where $\mathbf{B}(\cdot \ , \ \cdot)$ -is beta function and $_2F_1$ - is Gauss hypergeometric function.
Here we have $\alpha=\rho=\sigma=\frac{1}{2}, \ a=c=1, \ d=4, \ b=3$ so all the conditions are satisfied.
Then
$$
\begin{eqnarray}
I=\int_0^\infty\frac{\ln(x+4)}{\sqrt{x\,(x+3)\,(x+4)}}\mathrm dx&=&-\!\frac{2}{\sqrt{3}}\!\!\left(\!\frac{\partial}{\partial \rho}\!\!\left[\frac{\mathbf{B}\left(\frac{1}{2},\rho\right)}{4^{\rho}} \ _2F_1\!\left(\frac{1}{2},\frac{1}{2};\rho+\frac{1}{2};-\frac{1}{3}\right)\!\right]\right)_{\rho=\frac{1}{2}}=\\
&=& -\!\frac{2 \ _2F_1\!\left(\frac{1}{2},\frac{1}{2};1;-\frac{1}{3}\right)}{\sqrt{3}}\!\!\left(\!\frac{\partial}{\partial \rho}\!\!\left[\frac{\mathbf{B}\left(\frac{1}{2},\rho\right)}{4^{\rho}}\!\right]\right)_{\rho=\frac{1}{2}}-\\
&-&\!\frac{\mathbf{B}\left(\frac{1}{2},\frac{1}{2}\right)}{\sqrt{3}}\!\!\left(\!\frac{\partial}{\partial \rho}\!\!\left[\ _2F_1\!\left(\frac{1}{2},\frac{1}{2};\rho+\frac{1}{2};-\frac{1}{3}\right)\!\right]\right)_{\rho=\frac{1}{2}}
\end{eqnarray}
$$
Then we perform some simplifications (one can use Wolfram Mathematica ;) ) $$\left(\!\frac{\partial}{\partial \rho}\!\!\left[\frac{\mathbf{B}\left(\frac{1}{2},\rho\right)}{4^{\rho}}\!\right]\right)_{\rho=\frac{1}{2}}=-2\pi\ln 2, \\ \mathbf{B}\left(\frac{1}{2},\frac{1}{2}\right)=\pi, \qquad \ _2F_1\!\left(\frac{1}{2},\frac{1}{2};1;-\frac{1}{3}\right)=\frac{2}{\pi } \mathbf{K}\left(i\frac{1}{\sqrt{3}}\right)$$
where $\mathbf{K}(\cdot)$ is the complete elliptic integral of the first kind.
So $$I=\frac{8\ln 2}{\sqrt{3} }\mathbf{K}\left(i\frac{1}{\sqrt{3}}\right)-\frac{\pi}{\sqrt{3} }\left(\!\frac{\partial}{\partial \rho}\!\!\left[\ _2F_1\!\left(\frac{1}{2},\frac{1}{2};\rho+\frac{1}{2};-\frac{1}{3}\right)\!\right]\right)_{\rho=\frac{1}{2}}\approx 4.55592$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Limit of $\frac{1}{a} + \frac{2}{a^2} + \cdots + \frac{n}{a^n}$ What is the limit of this sequence $\frac{1}{a} + \frac{2}{a^2} + \cdots + \frac{n}{a^n}$?
Where $a$ is a constant and $n \to \infty$.
If answered with proofs, it will be best.
| Hint:
$\frac{1}{a} + \frac{2}{a^2} + \cdots + \frac{n}{a^n}$ = $(\frac{1}{a} + \frac{1}{a^2} + \cdots + \frac{1}{a^n}) + (\frac{1}{a^2} + \frac{1}{a^3} + \cdots + \frac{1}{a^n}) + ...$
Simplify each sum, factor out common terms and it will become more clear to solve. :)
| {
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Proving that $\sqrt{3} +\sqrt{7}$ is rational/irrational I took $\sqrt{3}+\sqrt{7}$ and squared it. This resulted in a new value of $10+2\sqrt{21}$.
Now, we can say that $10$ is rational because we can divide it with $1$ and as for $2\sqrt{21}$, we divide by $2$ and get $\sqrt{21}$.
How do I prove $\sqrt{21}$ to be rational/irrational?
Thanks
| Suppose $\sqrt{7}+\sqrt{3}={a\over b}$ rational number say
then ${4b\over a}={4\over \sqrt{7}+\sqrt{3}}=4 \cdot {\sqrt{7}-\sqrt{3}\over (\sqrt{7}+\sqrt{3})(\sqrt{7}-\sqrt{3})}=(\sqrt{7}-\sqrt{3})$
so $\sqrt{7}+\sqrt{3}+\sqrt{7}-\sqrt{3}={a\over b}+{4b\over a}$
so $\sqrt{7}={a\over 2b}+{2b\over a}$
so we are getting $\text{ an irrational number }\sqrt{7}=\text{ some rational number}(\Rightarrow\Leftarrow)$
| {
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"timestamp": "2023-03-29T00:00:00",
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A problem on sequence and series Suppose we have a decreasing sequence $\{x_n\}$ which converges to $0$.
Then is it true that the sum $$\sum_{n=1}^{\infty}\frac{x_n-x_{n-1}}{x_n}$$ diverges ?
| Here is a potentially useful trick. Since the $x_n$ are decreasing, you can take a chunk like:
$$\frac{x_{n+1} - x_n}{x_{n+1}} + \frac{x_{n+2} - x_{n+1}}{x_{n+2}} + \cdots + \frac{x_{m+n} - x_{m+n-1}}{x_{m+n}}$$
and bound each of the denominators by $x_n$, so
\begin{align*}&\frac{x_{n+1} - x_n}{x_{n+1}} + \frac{x_{n+2} - x_{n+1}}{x_{n+2}} + \cdots + \frac{x_{m+n} - x_{m+n-1}}{x_{m+n}} \\ \leq &\frac{x_{n+1} - x_n}{x_n} + \frac{x_{n+2} - x_{n+1}}{x_n} + \cdots + \frac{x_{m+n} - x_{m+n-1}}{x_n}\end{align*}
Note something nice happens in the numerators.
| {
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"timestamp": "2023-03-29T00:00:00",
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Judicious guess for the solution of differential equation $y''-2y'+5y=2(\cos t)^2 e^t$ I want to find the solutions of the differential equation: $y''-2y'+5y=2(\cos t)^2 e^t$.
I want to do this with the judicious guessing method and therefore I want to write the right part of the differential equations as the imaginary part of a something. How can I do this?
| We use the technique of the operator $D$. Note that $y^{\prime \prime} - 2y^{\prime} + 5y = (D^2 - 2D + 5)y$. Hence,
$$
(D^2 - 2D + 5)y = 2\cos^2(t) e^t = e^t[1 + \cos (2t)] = e^t + e^t\cos(2t) \quad \Rightarrow
$$
$$
y(t) = \dfrac{1}{D^2 -2D + 5}\cdot 0 + \dfrac{1}{D^2 -2D + 5}e^{1t} + \dfrac{1}{D^2 -2D + 5}e^t\cos(2t) \quad \Rightarrow
$$
$$
y(t) = \dfrac{1}{(D-1)^2 + 2^2}\cdot 0 + \dfrac{1}{1^2 -2\cdot 1 + 5}e^t + e^t\dfrac{1}{D^2 + 4}\cos(2t) \quad \Rightarrow
$$
$$
\quad (1) \quad y(t) = e^t[C_1\cos(2t) + C_2\sin(2t)] + \dfrac{1}{4}e^t + e^t\dfrac{1}{D^2 + 4}\cos(2t) \quad \Rightarrow
$$
But,
$$
\dfrac{1}{D^2 + 4}e^{2it} = e^{2it}\dfrac{1}{(D + 2i)^2 + 4}\cdot 1 = e^{2it}\dfrac{1}{(D + 4i)}\cdot \dfrac{1}{D}\cdot 1 = \dfrac{e^{2it}}{4i}\dfrac{1}{1 + \dfrac{D}{4i}}t
$$
$$
= \dfrac{e^{2it}}{4i}(1 - \dfrac{D}{4i} + \ldots)\cdot t = \dfrac{e^{2it}}{4i}(t - \dfrac{1}{4i}) = \dfrac{e^{2it}(1 - 4it)}{16}
$$
Thus,
$$
\dfrac{1}{D^2 + 4}\cos(2t) = Re\biggl[\dfrac{e^{2it} - 4ite^{2it}}{16}\biggr] = \dfrac{\cos(2t)}{16} + \dfrac{t\sin(2t)}{4} \quad (2)
$$
Replacing (2) in (1), we have
$$
y(t) = e^t[C_1\cos(2t) + C_2\sin(2t)] + \dfrac{1}{4}e^t + \dfrac{e^t\cos(2t)}{16} + \dfrac{te^t\sin(2t)}{4}
$$
| {
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Find the limit: $\lim \limits_{x \to 1} \left( {\frac{x}{{x - 1}} - \frac{1}{{\ln x}}} \right)$ Find:
$$\lim\limits_{x \to 1} \left( {\frac{x}{{x - 1}} - \frac{1}{{\ln x}}} \right)
$$
Without using L'Hospital or Taylor approximations
Thanks in advance
| Note that $$\log(x)=\int_1^x \frac{dt}{t}$$
and that for $1 < t < x$ $$2-t < \frac{1}{t} < \frac{\frac{1}{x}-1}{x-1}(t-1) +1.$$
Integration in $t$ from $1$ to $x$ results in $$0 < \frac{(3-x)(x-1)}{2} < \log(x) < \frac{(1+\frac{1}{x})(x-1)}{2}$$ for all $x\in(1,3)$. Therefore on the same interval
$$ \frac{2-x}{3-x}<\frac{x}{x-1}-\frac{1}{\log(x)}< \frac{x}{x+1} $$
and $$\lim_{x\downarrow 1}\left(\frac{x}{x-1}-\frac{1}{\log(x)}\right)=\frac{1}{2}.$$
For $x\in(0,1)$ all inequalities change direction and therefore also
$$\lim_{x\uparrow 1}\left(\frac{x}{x-1}-\frac{1}{\log(x)}\right)=\frac{1}{2}.$$
| {
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Proof: $2^{n-1}(a^n+b^n)>(a+b)^n$ If $n \in \mathbb{N}$ with $n \geq 2$ and $a,b \in \mathbb{R}$ with $a+b >0$ and $a \neq b$, then $$2^{n-1}(a^n+b^n)>(a+b)^n.$$
I tried to do it with induction. The induction basis was no problem but I got stuck in the induction step: $n \to n+1$
$2^n(a^{n+1}+b^{n+1})>(a+b)^{n+1} $
$ \Leftrightarrow 2^n(a\cdot a^n + b\cdot b^n)>(a+b)(a+b)^n$
$\Leftrightarrow a(2a)^n+ b(2b)^n>(a+b)(a+b)^n$
dont know what to do now :/
| Note that $a\not=b$ implies $(a-b)(a^n-b^n)\gt0$. Using this and the inductive hypothesis, we get
$$\begin{align}
(a+b)^{n+1}&=(a+b)^n(a+b)\cr
&\lt2^{n-1}(a^n+b^n)(a+b)\cr
&=2^{n-1}(a^{n+1}+ab^n+a^nb+b^{n+1})\cr
&=2^n(a^{n+1}+b^{n+1})-2^{n-1}(a^{n+1}-ab^n-a^nb+b^{n+1})\cr
&=2^n(a^{n+1}+b^{n+1})-2^{n-1}(a-b)(a^n-b^n)\cr
&\lt2^n(a^{n+1}+b^{n+1})
\end{align}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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The number of integral solutions for the equation $x-y = x^2 + y^2 - xy$ Find the number of integral solutions for the equation $x-y = x^2 + y^2 - xy$ and the equation of type $x+y = x^2 + y^2 - xy$
| For second equation:
$$x + y = x^2 + y^2 − xy$$
By dividing $xy$ on both sides
$$\frac{1}{x} + \frac{1}{y} = \frac{x}{y} + \frac{y}{x} -1 = y \text{ (say)}$$
Here for any real no. $a$
$$a + \frac{1}{a}\ge 2$$
So RHS will be $\ge1$.
But because only integer solutions are required:
LHS will be $\le 2$. (Assuming neither $x$ nor $y$ is zero).
So for this equality to be true $1\le y\le2$. Hence we need to consider cases only for $x =0,1$ and $y=0,1$
By substituting values 3 possible solutions are:
$x=0,y=0$;
$x=1,y=0$;
$x=0,y=1$;
| {
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How do I use Cauchy-Schwarz inequality? I wouldn't have posted if I hadn't searched in every site about Cauchy-Schwarz or bcs or Bunyakovsky inequality. But the only thing I can find is the statement and a proof.
In many answers in this SE I found referrals to CBS. but I cannot understand which vectors I can use to get the result. For example
$$\frac{a^4}{a^2+ab} + \frac{b^4}{b^2 + bc} + \frac{c^4}{c^2 + ac} \ge \frac{(a^2+b^2+c^2)^2}{a^2 + b^2 + c^2 + ab + ac + bc}$$
I have also noted that many times the inequality looks like
$$\frac{a^2}{b} + \frac{c^2}{d} + \frac{e^2}{f} \ge \frac{(a+c+e)^2}{b+d+f}$$
Is something like this valid? If yes, how can I end with this, using BCS (or any other inequality)?
| In case $b+d+f>0$, multiplying $(b+d+f)$ to both sides will give $(\frac{a^2}{b}+\frac{c^2}{d}+\frac{e^2}{f})(b+d+f)\geq (a+c+e)^2$, and it is a direct corollary from Cauchy-Schwarz inequality.
| {
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Test for convergence/divergence of $\sum_{k=1}^\infty \frac{k^2-1}{k^3+4}$ Given the series
$$\sum_{k=1}^\infty \frac{k^2-1}{k^3+4}.$$
I need to test for convergence/divergence. I can compare this to the series $\sum_{k=1}^\infty \frac{1}{k}$, which diverges.
To use the comparison test, won't I need to show that $\frac{k^2-1}{k^3+4}>\frac{k^3}{k^4}=\frac{1}{k}$, in order to state the original series diverges?
This doesn't seem to hold, I feel like I'm missing the obvious.
Any help is appreciated.
Thanks.
| When $k \geq 1$, we have $k^3+4 \leq k^3+4k^3 = 5k^3$ (Since $4 \leq 4k^3$ for $k \geq 1$). Also, when $k \geq 2$, we have $k^2-1 > \frac{k^2}{2}$ (This is again straightforward to verify, equality happens when $k = \sqrt{2}$ :)). Therefore,
$$\sum_{k=1}^\infty \frac{k^2-1}{k^3+4} = \sum_{k=2}^\infty \frac{k^2-1}{k^3+4} \geq \sum_{k=2}^\infty\frac{\frac{k^2}{2}}{5k^3} = \frac{1}{10}\sum_{k=2}^\infty\frac{1}{k}.$$
The lower bound is the harmonic series that clearly diverges. Hence the original series diverges.
| {
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Divisibility by $10^6$? Let $p_k$ be the $k^{th}$ prime number.
Find the least $n$ for which $(p_1^2+1)(p_2^2+1) \cdots (p_n^2+1)$ is divisible by $10^6$.
I have no idea where to start on this problem. Any help would be appreciated.
| For each prime above $2$, $p^2+1$ is even. For $2^2+1=5$ so we only need to look for the first $5$ odd primes so that $5\mid p^2+1$.
$p^2\equiv-1\pmod{5}\iff p\in\{2,3\}\pmod{5}$. Scanning the primes: $3,7,13,17,23$ are the first $5$ so that $p\in\{2,3\}\pmod{5}$. Therefore, the product
$$
(2^2+1)(3^2+1)(5^2+1)\dots(23^2+1)
$$
will be divisible by $10^6$. If some $p^2+1$ is divisible by $25$, you may need fewer terms. Indeed, $7^2+1=50$, so we only need
$$
(2^2+1)(3^2+1)(5^2+1)\dots(17^2+1)
$$
| {
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a mathematical problem on inequalities If $a,b,c,d$ are positive real numbers such that $a+b+c+d=1$,show that
$$ \frac{a^3}{b + c} + \frac{b^3}{c + d} + \frac{c^3}{d + a} + \frac{d^3}{a + b} > \frac 1 8$$
| By Cauchy-Schwarz Inequality, we have$$\left(\sum_{cyc} \frac{a^3}{b + c} \right)\cdot \sum_{cyc}(b+c) \ge (a^{3/2}+b^{3/2}+c^{3/2}+d^{3/2})^2$$
So it is sufficient to show that $a^{3/2}+b^{3/2}+c^{3/2}+d^{3/2} \ge \frac12$.
But we have from Power Means inequality,
$$\left(\frac{a^{3/2}+b^{3/2}+c^{3/2}+d^{3/2}}{4}\right)^{2/3}\ge \frac{a+b+c+d}{4}=\frac14 \implies a^{3/2}+b^{3/2}+c^{3/2}+d^{3/2} \ge \frac12$$
Equality is attained when $a=b=c=d=\frac14$.
| {
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How find this inequality minimum $\sum_{i=1}^{n}a^2_{i}-2\sum_{i=1}^{n-1}a_{i}a_{i+1}$
Let $n$ be a given positive integer, and let $a_{1},a_{2},\cdots,a_{n}\ge 0$ such that $a_{1}+a_{2}+\cdots+a_{n}=1$.
Find this minimum value
$$a^2_{1}+a^2_{2}+\cdots+a^2_{n}-2a_{1}a_{2}-2a_{2}a_{3}-2a_{3}a_{4}-\cdots-2a_{n-2}a_{n-1}-2a_{n-1}a_{n}.$$
I think maybe we can use this well known
$$a^2_{1}+a^2_{2}+\cdots+a^2_{n}\ge a_{1}a_{2}+a_{2}a_{3}+\cdots+a_{n-1}a_{n}+a_{n}a_{1}?$$
But this problem is only
$$-2a_{1}a_{2}-2a_{2}a_{3}-2a_{3}a_{4}-\cdots-2a_{n-2}a_{n-1}-2a_{n-1}a_{n}$$
so I can't it.Thank you for help.
By the way:
I don't know this problem have Someone research?if No,I think this is interesting problem.
| I show below that the minimum is $-\frac{1}{6}$ for $n \geq 4$, and $-\frac{1}{7}$ when $n = 3$.
If we put
$$
Q_n(a_1,a_2,\ldots ,a_n)=
a^2_{1}+a^2_{2}+\cdots+a^2_{n}-2a_{1}a_{2}-2a_{2}a_{3}-2a_{3}a_{4}-\cdots-2a_{n-2}a_{n-1}-2a_{n-1}a_{n}
$$
and
$$
R_n(a_1,a_2,\ldots ,a_n)=Q_n(a_1,a_2,\ldots,a_n)+\frac{(a_1+a_2+a_3+\ldots +a_n)^2}{6}
$$
then for $n\geq 4$, we have the identity
$$
R_n(a_1,a_2,a_3,\ldots,a_n)=
\frac{1}{42}\bigg(\sum_{k=1}^{n-2}a_k-5a_{n-1}+7a_n\bigg)^2
+\frac{1}{28}\bigg(\sum_{k=1}^{n-3}2a_k-5a_{n-2}+4a_{n-1}\bigg)^2
+\frac{1}{4}\bigg(\sum_{k=1}^{n-4}2a_k-2a_{n-3}+a_{n-2}\bigg)^2
$$
Since $Q_n(0,\ldots, 0,\frac{1}{6},\frac{1}{3},\frac{1}{3},\frac{1}{6})=-\frac{1}{6}$,
we see that the minimum is $-\frac{1}{6}$ for $n\geq 4$.
For $n=3$, the minimum is $-\frac{1}{7}$ attained at
$(\frac{2}{7},\frac{3}{7},\frac{2}{7})$ (thanks Macavity) because of
$$
Q_3(a_1,a_2,a_3)+\frac{(a_1+a_2+a_3)^2}{7}=
\frac{1}{56}\bigg(a_1-6a_2+8a_3\bigg)^2
+\frac{1}{8}\bigg(-3a_1+2a_2\bigg)^2
$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Critical points : $ z = \cos^2x + \cos^2y$ for $ y-x = \frac{\pi}{4}$ Find and classify the critic points (maximum, minimum or neither) of the function $$z=z(x,y) = \cos^2x + \cos^2y $$
if $y-x = \large\frac{\pi}{4}$
I've find $x = -\large\frac{\pi}{8} + \large\frac{k \pi}{2}$ and $y = \large\frac{\pi}{8} + \large\frac{k \pi}{2}$ ($k \in \Bbb{Z}$), but I sm having troubles to classify the points.
Thanks for your help!
| use
$$\cos^2{x}+\cos^2{y}=\dfrac{1}{2}(\cos{(2x)}+\cos{(2y)})+1=\cos{(x+y)}\cos{(x-y)}+1=\dfrac{\sqrt{2}}{2}\cos{(x+y)}+1$$
since
$$y=x+\dfrac{\pi}{4}$$
so
$$\cos^2{x}+\cos^2{y}=\dfrac{\sqrt{2}}{2}\cos{(2x+\dfrac{\pi}{4})}+1\in[1-\dfrac{\sqrt{2}}{2},1+\dfrac{\sqrt{2}}{2}]$$
if only if $2x+\dfrac{\pi}{4}=2k\pi$ and $2x+\dfrac{\pi}{4}=(2k-1)\pi$
| {
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$\sum_{i=1}^n \frac{1}{i(i+1)} = \frac{3}{4} - \frac{2n+3}{2(n+1)(n+2)}$ by induction. I am wondering if I can get some help with this question. I feel like this is false, as I have tried many ways even to get the base case working (for induction) and I can't seem to get it. Can anyone confirm that this is false? If I am wrong, I would really appreciate a hint.
$$\sum_{i=1}^n \frac{1}{i(i+1)} = \frac{3}{4} - \frac{2n+3}{2(n+1)(n+2)}$$
| For $n=1$, we have
$$\frac{1}{2} = \sum_{i=1}^1 \frac{1}{i(i+1)} = \sum_{i=1}^n \frac{1}{i(i+1)} \ne \frac{3}{4} - \frac{2n+3}{2(n+1)(n+2)} = \frac{3}{4} - \frac{2+3}{2(1+1)(1+2)} = \frac{1}{3}.$$
So, yes, I'd say you're right.
| {
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$1 + \frac{1}{1+2}+ \frac{1}{1+2+3}+ ... + \frac{1}{1+2+3+...+n} = ?$ How do I simplify the following series
$$1 + \frac{1}{1+2}+ \frac{1}{1+2+3}+ \frac{1}{1+2+3+4} + \frac{1}{1+2+3+4+5} + \cdot\cdot\cdot + \frac{1}{1+2+3+\cdot\cdot\cdot+n}$$
From what I see, each term is the inverse of the sum of $n$ natural numbers.
Assuming there are $N$ terms in the given series,
$$a_N = \frac{1}{\sum\limits_{n = 1}^{N} n} = \frac{2}{n(n+1)}$$
$$\Rightarrow \sum\limits_{n=1}^{N} a_N = \sum\limits_{n=1}^{N} \frac{2}{n(n+1)}$$
... and I'm stuck.
I've never actually done this kind of problem before (am new to sequences & series).
So, a clear and detailed explanation of how to go about it would be most appreciated.
PS- I do not know how to do a telescoping series!!
| Hint:
$$\frac{2}{n(n+1)} = 2\left(\frac{1}{n} - \frac{1}{n+1}\right)$$
NOTE:
$$\frac{1}{n(n+1)} = \frac{1}{n} - \frac{1}{n+1}$$
Hence:
$$\frac{m}{n(n+1)} = m\left(\frac{1}{n} - \frac{1}{n+1}\right)$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Proving that if these quadratics are equal for some $\alpha$, then their coefficients are equal Let
$$P_1(x) = ax^2 -bx - c \tag{1}$$
$$P_2(x) = bx^2 -cx -a \tag{2}$$
$$P_3(x) = cx^2 -ax - b \tag{3}$$
Suppose there exists a real $\alpha$ such that
$$P_1(\alpha) = P_2(\alpha) = P_3(\alpha)$$
Prove
$$a=b=c$$
Equating $P_1(\alpha)$ to $P_2(\alpha)$
$$\implies a\alpha^2 - b\alpha - c = b\alpha^2 - c\alpha - a$$
$$\implies (a-b)\alpha^2 + (c-b)\alpha + (a-c) = 0$$
Let
$$Q_1(x) = (a-b)x^2 + (c-b)x + (a-c)$$
This implies, $\alpha$ is a root of $Q_1(x)$.
Similarly, equating $P_2(\alpha)$ to $P_3(\alpha)$ and $P_3(\alpha)$ to $P_1(\alpha)$, and rearranging we obtain quadratics $Q_2(x)$ and $Q_3(x)$ with common root $\alpha$:
$$Q_2(x) = (b-c)x^2 + (a-c)x + (b-a)$$
$$Q_3(x) = (c-a)x^2 + (b-a)x + (c-b)$$
$$Q_1(\alpha) = Q_2(\alpha) = Q_3(\alpha) = 0$$
We have to prove that this is not possible for non-constant quadratics $Q_1(x), Q_2(x), Q_3(x)$.
EDIT:
I also noticed that for distinct $a, b, c \in \{1, 2, 3\}$:
$$Q_a(x) + Q_b(x) = -Q_c(x)$$
| If we assume that $ P_1(\alpha)=P_2(\alpha)=P_3(\alpha)=p $ :
\begin{align*}
p &= a\alpha^2 -b\alpha -c \qquad &&p= a\alpha^2 -b\alpha -c \\
p &= b\alpha^2 -c\alpha -a \qquad &&p= -a +b\alpha^2 -c\alpha \\
p &= c\alpha^2 -a\alpha -b \qquad &&p= -a\alpha -b-c\alpha^2\\
\end{align*}
then we get the Cramer system :
\begin{align*}
& p\begin{bmatrix}
1 \\[0.3em]
1 \\[0.3em]
1
\end{bmatrix} = \begin{bmatrix}
\alpha^2 & -\alpha & -1 \\[0.3em]
-1 & \alpha^2 & -\alpha \\[0.3em]
-\alpha & -1 & \alpha^2
\end{bmatrix}\begin{bmatrix}
a \\[0.3em]
b \\[0.3em]
c
\end{bmatrix}\\
&pI_{3}=A_{3,3}X_{3}
\end{align*}
if $ A_{3,3} $ is invertible then
\begin{equation*}
X_{3}=pA_{3,3}^{-1}I_{3}
\end{equation*}
Then the values of $ a $, $ b $ and $ c $ can be found as follows:
\begin{equation*}
a = p\frac { \begin{vmatrix} {\color{red}1} & -\alpha & -1 \\ {\color{red}1} & \alpha^2 & -\alpha \\ {\color{red}1} & -1 & \alpha^2 \end{vmatrix} } { \mid A_{3,3} \mid}, \quad b = p\frac { \begin{vmatrix} \alpha^2 & {\color{red}1} & -1 \\ -1 & {\color{red}1} & -\alpha \\ -\alpha & {\color{red}1} & \alpha^2 \end{vmatrix} } {\mid A_{3,3} \mid},\text{ and }c = p\frac { \begin{vmatrix} \alpha^2 & -\alpha & {\color{red}1} \\-1 & \alpha^2 & {\color{red}1} \\-\alpha & -1 & {\color{red}1}\end{vmatrix} } {\mid A_{3,3} \mid}
\end{equation*}
with $ \det(A_{3,3})=\mid A_{3,3} \mid=\alpha^{6}-3\alpha^{3}+\alpha^{2}-1 $ then we gwt the answer :
\begin{equation*}
a=b=c=p\left[ \dfrac{\alpha^{4}+\alpha^{3}+2\alpha^{2}-\alpha+1}{\alpha^{6}-3\alpha^{3}+\alpha^{2}-1 }\right]
\end{equation*}
| {
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circumradii of $\triangle{IAB}$ Let $I$ be the incenter of $\triangle{ABC}$. Let $R$ be the radius of the circle that circumscribes $\triangle{IAB}$. Find a formula for $R$ in term of other elements $a, b, c, A, B, C, r, R$ of $\triangle{ABC}$. I need this formula in order to prove a geometric inequality.
| Remember the extended law of sines:
$$R = \frac{a}{2\sin A} = \frac{b}{2\sin B} = \frac{c}{2\sin C}$$
The above gives the circumradius ($R$) for a triangle whose one angle is $A$ and the opposite side is $a$ and so on. Well, the labeling is a bit different here.
For our triangle $IAB$, the extended law of sines becomes: $$R = \frac{a}{2\sin\angle AIB}$$
So, we need to calculate $\angle AIB$, and we are done. This can be done by realizing that $IB$ is the angle bisector of $\angle A$ and $IA$ angle bisector of $\angle A$. So, $\angle AIB$ is:
$$180^{\circ} - (\angle IAB + \angle IBA) = 180^{\circ} - (\frac{\angle A}{2} + \frac{\angle B}{2}) = 180^{\circ} - (\frac{180^{\circ} - \angle C}{2}) = 90^{\circ} + \frac{C}{2} $$
Note that $\sin (90^{\circ} + \theta) = \cos\theta$. So we substitute back and our formula becomes:
$$R=\frac{AB}{2\cos\frac{C}{2}}$$
If you need proof of anything that I have used ask me. If you want to express $\cos\frac{C}{2}$ in terms of $\cos C$, you can do that by using the identity [and considering the positive root]:
$$\cos^2\frac{\theta}{2}=\frac{1+\cos\theta}{2}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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What is the last non-zero digit of $50!$? I'm looking for a fast method. I just multiplied all the numbers together modulo ten and divided by $5^{12}$ and $2^{12}$ modulo 10, which gave me $2$.
| You have correctly figured that $5^{12}$ is the highest power of $5$ that divides $50!$. Similarly, $\lfloor\frac{50}{2}\rfloor+\lfloor\frac{50}{4}\rfloor+\lfloor\frac{50}{8}\rfloor+\lfloor\frac{50}{16}\rfloor+\lfloor\frac{50}{32}\rfloor=25+12+6+3+1=47$, so $2^{47}$ is the highest power of $2$ that divides $50!$. We can similarly figure out the other highest powers of primes dividing $50!$: $\lfloor\frac{50}{3}\rfloor+\lfloor\frac{50}{9}\rfloor+\lfloor\frac{50}{27}\rfloor=16+5+1=22$, and $\lfloor\frac{50}{7}\rfloor+\lfloor\frac{50}{49}\rfloor=7+1=8$, and so forth, so in fact
$$50!=2^{47}\cdot3^{22}\cdot5^{12}\cdot7^{8}\cdot11^{4}\cdot13^{3}\cdot17^{2}\cdot19^2\cdot23^2\cdot29\cdot31\cdot37\cdot41\cdot43\cdot47.$$
Clearly, the number ends in exactly $12$ zeroes, so the desired digit is $50!/10^{12}\bmod10$, or
$$2^{35}\cdot3^{22}\cdot7^{8}\cdot1^{4}\cdot3^{3}\cdot7^{2}\cdot9^2\cdot3^2\cdot9\cdot1\cdot7\cdot1\cdot3\cdot7\bmod10,$$
which is not too hard to compute using identities like $3^4=9^2\equiv1\bmod10$, $3\cdot7\equiv1\bmod10$. (It might not be super elegant, but I don't think there is a much quicker method.)
| {
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"timestamp": "2023-03-29T00:00:00",
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Question on Summation and Geometric Progression Let $S_k$, $k=1,2,3,....,100$, denote the sum of the infinite geometric progression series whose first term is $\displaystyle\frac{k-1}{k!}$ and the common ratio is $\displaystyle\frac1k$. Then find the value of $$\frac{100^2}{100!}+\displaystyle\sum_{k=1}^{100}|(k^2-3k+1)S_k|$$
I could proceed as follows:
The given first term of the GP and the common ration gives the sum as $\displaystyle\frac{1}{(k-1)!}$. However, I am unable to solve the series after putting $S_k$ in the expression.
| Writing $k^2-3k+1 = (k^2-3k+2)-1 = (k-1)(k-2)-1$, we see that the term of the sum for $k \geqslant 3$ is
$$(k-1)(k-2)S_k - S_k = S_{k-2} - S_k,$$
so overall we get a telescoping sum
$$\begin{align}
\sum_{k=1}^{100} \lvert (k^2-3k+1)S_k\rvert &= S_1 + S_2 + \sum_{k=3}^{100} S_{k-2}-S_k\\
&= 2(S_1 + S_2) - S_{99} -S_{100}\\
&= 4 - \frac{1}{98!} - \frac{1}{99!}\\
&= 4 - \frac{100}{99!}\\
&= 4 - \frac{100^2}{100!}.
\end{align}$$
| {
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If 4n+1 and 3n+1 are both perfect sqares, then 56|n. How can I prove this?
Prove that if $n$ is a natural number and $(3n+1)$ & $(4n+1)$ are both perfect squares, then $56$ will divide $n$.
Clearly we have to show that $7$ and $8$ both will divide $n$.
I considered first $3n+1=a^2$ and $4n+1=b^2$. $4n+1$ is a odd perfect square.
- so we have $4n+1\equiv 1\pmod{8}$; from this $2|n$ so $3n+1$ is a odd perfect square.
- so $3n+1\equiv 1\pmod{8}$ so $8|n$ but I can't show $7|n$. How do I show this?
Thanks for the help.
| $3n+1=a^2$ and $4n+1=b^2$ gives $4a^2-3b^2=1$. With $x=2a$ and $y=b$ we get an instance of Pell's equation.
$$x^2-3y^2=1.$$
Trying small integers, it is obvious that the minimal solution is $4-3=1$, that is $x_1=2$ and $y_1=1$.
Therefore, all the solutions are of the form $(x_k,y_k)$ with $x_k+y_k\sqrt 3=(x_1+y_1\sqrt 3)^k$ or equivalently
$$ \begin{align}
x_{k+1}&=2x_k+3y_k\\
y_{k+1}&=x_k+2y_k,
\end{align} $$
(which already works starting from the trivial solution $x_0=1$, $y_0=0$).
This gives $3y_{k+1} =3x_k+6y_k$ which implies $$\begin{align}&x_{k+2}-2x_{k+1}=3x_k+2x_{k+1}-4x_k\\
\Leftrightarrow &x_{k+2}-4x_{k+1}+x_k=0
\end{align}$$
and $x_0=1$, $x_1=2$.
Clearly, the $x_i$ alternate in sign and we are only interested in the even ones which are those with odd index.
$$\begin{align}x_{2k+1} &= 4x_{2k}-x_{2k-1}\\&=4(4x_{2k-1}-x_{2k-2})-x_{2k-1}\\ &=15x_{2k-1}-x_{2k-1}-x_{2k-3}\\&=14x_{2k-1}-x_{2k-3}.\end{align}$$
Upon division by two, we obtain the recurrence for the solutions $a_k$ to $3n+1=a^2$ subject to the other condition.
$$a_k=14a_{k-1}-a_{k-2},$$
with $a_0=1$ and $a_1=13(=\dfrac{x_3}2)$.
Now, it is trivial to check that modulo $3$, all $a_k$ are $\equiv 1$ because $14\cdot 1 - 1 \equiv 1 \mod 3$. Which implies that $3 \mid a^2-1$.
Furthermore, modulo $7$, all $a_k$ are $\equiv \pm 1$ because $a_k \equiv - a_{k-2} \mod 7$ which implies that $7 \mid a^2-1$.
And finally, modulo $4$, all $a_k$ are $\equiv 1$ because $14-1 \equiv 1 \mod 4$ which implies that $8 \mid (a-1)(a+1)=a^2-1$.
So, in summary, $56 \mid \dfrac{a^2-1}3$ as desired.
| {
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Finding $\lim\limits_{x \to 0}\frac{(e^{2\tan(x)}-1) \cdot \ln(2-cos^2(x))}{\sqrt(1+x^3)-(cosx)}$ I have to find the $$\lim_{x \to 0}\frac{(e^{2\tan(x)}-1) \cdot \ln(2-\cos^2(x))}{\sqrt{1+x^3}-(\cos x)}$$ using only notable limits
I managed to solve it having as result $4$ using notable limits, it is quite simple resolve but the solution is not $4$...
I can't find any other way of solution, can somebody solve it using notable limits clearly? (I'm asking many questions regarding the calculus of limit because I'm preparing for a limit's test)
| $$\lim_{x \to 0}\frac{(e^{2\tan x}-1) \cdot \ln(2-\cos^2x)}{\sqrt{1+x^3}-\cos x}=\lim_{x \to 0}\frac{e^{2\tan x}-1}{2\tan x}\cdot\lim_{x \to 0}\frac{ \ln(1+\sin^2x)}{\sin^2x}\cdot\lim_{x \to 0}\frac{2\tan x\cdot\sin^2x}{\sqrt{1+x^3}-\cos x}=2\cdot1\cdot\lim_{x \to 0}\frac{\tan x}{x}\lim_{x \to 0}\cdot\frac{\sin^2 x}{x^2}\cdot\lim_{x \to 0}\frac{x^3}{\sqrt{1+x^3}-1+(1-\cos x)}=2\lim_{x \to 0}\frac{x^3}{\frac{x^3}{\sqrt{1+x^3}+1}+(1-\cos x)}= 2\lim_{x \to 0}\frac{1}{\frac{1}{\sqrt{1+x^3}+1}+\frac{1-\cos x}{x^3}}=\frac{2}{\frac{1}{2}+\lim_{x \to 0}\frac{1-\cos x}{x^2}\cdot\frac{1}{x}}=\frac{2}{\frac{1}{2}+\frac{1}{2}\cdot\lim_{x \to 0}\frac{1}{x}}.$$
If we see it as the lateral limit, then the limit is equal to $0.$
| {
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Proof of Heron's Formula for the area of a triangle
Let $a,b,c$ be the lengths of the sides of a triangle. The area is given by Heron's formula:
$$A = \sqrt{p(p-a)(p-b)(p-c)},$$
where $p$ is half the perimeter, or $p=\frac{a+b+c}{2}$.
Could you please provide the proof of this formula?
Thank you in advance.
| I noticed that many proofs for Herons formula use the properties of angles, which is why I wanted to show a proof that doesn't involve any, and seems to be simpler than the usual methods that are applied.
A triangle with side lengths $a, b, c,$ where $c$ is the greatest side length and an altitude($h$) that intercepts $c$ such that $c$ is the sum of two side lengths ,
$c = m + n$, using Pythagorean theorem and the area formula of a triangle, we can prove Herons formula.
First we find $a$ and $b$ in terms of $m$ and $n$ using Pythagorean theorem:
$a^2 - m^2 = h^2$
$b^2 - n^2 = h^2$
$a^2 - m^2 = b^2 - n^2$
$a^2 - b^2 = m^2 - n^2$
Then using $c = m + n$, eliminate $m$:
$a^2 - b^2 = (c-n)^2 - n^2$
$a^2 - b^2 = c^2 - 2cn$
$n = \frac{1}{2c}(c^2 + b^2 - a^2)$
We can eliminate $n$ from $b^2 - n^2 = h^2$ to find $h$:
$h = \sqrt{b^2 - (\frac{1}{2c}(c^2 + b^2 - a^2))^2}$
Plugging this into the area formula ($A = \frac{1}{2}ch$) gives:
$A = \frac{1}{2}c\sqrt{ b^2 - (\frac{1}{2c}(c^2 + b^2 - a^2))^2} $
$A = \frac{1}{2}c\sqrt{ \frac{1}{4c^2}(2a^2b^2+2a^2c^2-a^4-b^4+2b^2c^2-c^4)} $
$A = \sqrt{\frac{1}{16}(c^2 - (a - b)^2)(( a + b)^2 - c^2)} $
$A = \sqrt{\frac{1}{16}(a + b - c)( a + b + c)( b + c - a)(a + c - b)} $
$A = \sqrt{s(s - a)(s- b)(s- c)}$
Q.E.D.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/576831",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 5,
"answer_id": 3
} |
If $x^3+\frac{1}{x^3}=18\sqrt{3}$ then to prove $x=\sqrt{3}+\sqrt{2}$
If $x^3+\frac{1}{x^3}=18\sqrt{3}$ then we have to prove $x=\sqrt{3}+\sqrt{2}$
The question would have been simple if it asked us to prove the other way round.
We can multiply by $x^3$ and solve the quadratic to get $x^3$ but that would be unnecessarily complicated.Also, as $x^3$ has 2 solutions,I can't see how x can have only 1 value. But the problem seems to claim that x can take 1 value only.Nevertheless,is there any way to get the values of x without resorting to unnecessarily complicated means?
NOTE: This problem is from a textbook of mine.
| Set $a=x+\frac{1}{x}, b\sqrt3=a$. Then we want that
$$a^3-3a=18\sqrt3\iff b^3-b=6\iff(b-2)((b+1)^2+2)=0\iff b=2$$
So
$$2\sqrt3=x+\frac{1}{x}\iff x^2-2\sqrt{3}x+1=0\iff (x-\sqrt3)^2-2=0 \iff x=\sqrt3\pm\sqrt2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/583062",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 1
} |
Ratio of balls in a box A box contains some identical tennis balls. The ratio of the total volume of the tennis
balls to the volume of empty space surrounding them in the box is $1:k$, where $k$ is an
integer greater than one.
A prime number of balls is removed from the box. The ratio of the total volume of the
remaining tennis balls to the volume of empty space surrounding them in the box is 1:$k^2$.
Find the number of tennis balls that were originally in the box.
A few questions regarding this problem:
Does the shape of the box matter? I just let the volume of the box be a constant $V$. Also I noted that the ratio $\frac{1}{k^2} = \left( \frac{1}{k} \right)^2$. ie. New ratio is the old ratio squared. I also let the amount of balls in the box be $n$ and the amount of balls taken out be $p$ where $p$ is a prime, so the new amount of balls in the box is $n-p$.
This is about all I could do in this problem but I would like to be guided towards solving the problem (and I'm also interested in your thought processes and what ideas you initially think of so I can get a better idea of what to think of when doing problem solving) than just being given the solution.
Your help would be much appreciated.
| $V$ is the volume of the box and $s$ is the volume of each ball. Then we can write:
$\frac{ns}{V-ns}=\frac{1}{k} \to \frac{ns}{V}=\frac{1}{k+1}$
$\frac{(n-p)s}{V-(n-p)s}=\frac{1}{k^2} \to \frac{(n-p)s}{V}=\frac{1}{k^2+1}$
$\frac{n}{n-p}=\frac{k^2+1}{k+1} \to 1+\frac{p}{n-p}=k+1-\frac{2k}{k+1}$
$\frac{p}{n-p}=\frac{k^2-k}{k+1}$
$\frac{p}{n}=\frac{k^2-k}{k^2+1}$
if $gcd(n,p)=1$ then $k(k-1)n=p(k^2+1)$ so $k=ap$ or $k=ap+1$.
if $k=ap$ then $n=\frac{(a^2p^2+1)}{a^2p-a}=p+\frac{pa+1}{a^2p-a}$ so $\frac{pa+1}{a^2p-a} \geq 1$ so $a+1 \geq a^2p-pa$ so either $a=1$ or $\frac{a+1}{a^2-a}\geq p$ that is impossible for $a>1$ so $a=1$ and $n=\frac{p^2+1}{p-1}=p+1\frac{2}{p-1}$ and it is only natural when $p = 2$ so $n=5$ and $k=2$. or $p=3,n=5,k=3$
if $k=ap+1$ then $n=\frac{a^2p^2+2ap+2}{a^2 p+a}=p+\frac{ap+2}{a^2p+a}$ so $ap+2 \geq a^2p+a$ which is impossible for $a>1$, but for $a=1$,$p+\frac{p+2}{p+1}=p+1+\frac{1}{p+1}$ will not be a natural number so we should dismiss this case.
if $gcd(n,p)=p$ then $\frac{1}{t}=\frac{k^2-k}{k^2+1} \to t=\frac{k^2+1}{k^2-k}$ that cannot be a natural number for $k>1$, be cause it is strictly decreasing for $k>1$ and for $k=3$ it is less than 2 and for $t=2$ it is not integer.
Note that in the second case $n$ cannot be equal to $p$ because for no natural $k$, $k^2+1=k^2-k$.
So the final result is: $n=5,k=2,p=2$ or $n=5,k=3,p=3$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/583712",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 3
} |
De Moivre's formula to solve $\cos$ equation Use De Moivre’s formula to show that
$$
\cos\left(x\right) + \cos\left(3x\right) = 2\cos\left(x\right)\cos\left(2x\right)
$$
$$
\mbox{Show also that}\quad
\cos^{5}\left(x\right) = \frac{1}{16}\,\cos\left(5x\right) + \frac{5}{16}\,\cos\left(3x\right) + \frac{5}{8}\,\cos\left(x\right)
$$
Hence solve the equation $\cos\left(5x\right) = 16\cos^{5}\left(x\right)$ completely.
Express your answers as rational multiples of $\pi$.
| @Lauren, the approach comes down to the fact that $Re[(\text{cos}(x)+i\cdot \text{sin}(x))^3] = \text{cos}(3x)$.
So, $$\text{cos}(3x) \\ = \text{cos}^3(x) + 3\text{cos}(x)\cdot (i^2\cdot \text{sin}^2(x)) \\ = \text{cos}(x)\cdot (\text{cos}^2(x) - 3\text{sin}^2(x)) \\ = \text{cos}(x)\cdot (1-2\text{sin}^2(x)+ \text{cos}(2x)-1) \\ = \text{cos}(x)\cdot (2\text{cos}(2x) - 1) = \text{cos}(3x).$$
This proves part 1. Then the next part follows from the fact that $\text{cos}(3x) = -2\text{cos}(x)$. Why? I'll leave this as an exercise.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/583764",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Taylor Series for $\log(x)$ Does anyone know a closed form expression for the Taylor series of the function $f(x) = \log(x)$ where $\log(x)$ denotes the natural logarithm function?
| The Taylor series for $\ln$ centered at $1$ can be easily derived with the geometric series
$$\sum_{n=0}^\infty x^n=\frac{1}{1-x}$$
We start with the derivative of $\ln$, which is given by $1/x$ for every $x>0$. This derivative is equivalent to
$$\frac{1}{1-[-(x-1)]}$$
so if $x$ satisfies $|-(x-1)|=|x-1|<1$, this can be expanded as a geometric series.
\begin{align}
\frac{1}{1-[-(x-1)]} &= \sum_{n=0}^\infty [-(x-1)]^n\\
&=\sum_{n=0}^\infty (-1)^n(x-1)^n
\end{align}
Thus, $(\ln)'(x)=\sum_{n=0}^\infty (-1)^n(x-1)^n$ holds for every $x$ with $|x-1|<1$. We can get a series expression for $\ln\left(x_0\right)$ by integrating this identity from $1$ to $x_0$.
\begin{align}
\ln\left(x_0\right)-\ln(1) &= \sum_{n=0}^\infty (-1)^n\frac{\left(x_0-1\right)^{n+1}}{n+1}-\sum_{n=0}^\infty (-1)^n\frac{(1-1)^{n+1}}{n+1}\\
&= \sum_{n=1}^\infty (-1)^{n-1}\frac{\left(x_0-1\right)^n}{n}-\sum_{n=1}^\infty (-1)^{n-1}\frac{0^{n}}{n}\\
&= \sum_{n=1}^\infty \frac{(-1)^{n-1}}{n}\left(x_0-1\right)^n\\
&= \left(x_0-1\right)-\frac{\left(x_0-1\right)^2}{2}+\frac{\left(x_0-1\right)^3}{3}-\frac{\left(x_0-1\right)^4}{4}+\cdots
\end{align}
What if we want to find a series expansion centered at a point other than $1$, say at $a>0$? You can always do this directly by computing the derivatives of $\ln$ at $a$, but an easier method is to leverage the series we just derived and the identity
$$\ln\left(\frac{x}{y}\right)=\ln(x)-\ln(y)$$
To see this, evaluate $\ln$ at $x/a$, where $x$ is any positive real number. If $|x-a|<a$, we will have that $|x/a-1|<1$, so the Taylor series of $\ln$ centered at $1$ will converge to $\ln(x/a)$. We can then write
\begin{align}
\ln\left(\frac{x}{a}\right) &= \sum_{n=1}^\infty \frac{(-1)^{n-1}}{n}\left(\frac{x}{a}-1\right)^n\\
&= \sum_{n=1}^\infty \frac{(-1)^{n-1}}{n}\left(\frac{x-a}{a}\right)^n\\
&= \sum_{n=1}^\infty \frac{(-1)^{n-1}}{n}\frac{(x-a)^n}{a^n}\\&= \sum_{n=1}^\infty \frac{(-1)^{n-1}}{na^n}(x-a)^n\\
\end{align}
From $\ln(x/a)=\ln(x)-\ln(a)$, it immediately follows that
\begin{align}
\ln(x) &= \ln(a)+\sum_{n=1}^\infty \frac{(-1)^{n-1}}{na^n}(x-a)^n\\
&= \ln(a)+\frac{1}{a}(x-a)-\frac{1}{2a^2}(x-a)^2+\frac{1}{3a^3}(x-a)^3-\frac{1}{4a^4}(x-a)^4+\cdots
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/585154",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "20",
"answer_count": 4,
"answer_id": 0
} |
The probability of being dealt a 2 pair 5 card hand... Given a standard playing card deck of 52 cards, what is the probability of being dealt a 2 pair 5 card hand consisting exactly of one pair of face cards and one pair of NOT face cards is?
| First, I'll solve the problem assuming that it doesn't matter whether or not the fifth card matches either of the pairs.
There are $3 \cdot 4 = 12$ face cards in the deck, and $52 - 12 = 40$ non-face cards.
The number of choices for the positions of the pairs is
$$\binom{5}{2} \cdot \binom{3}{2}$$
The probability of getting the desired cards in a specific order is:
$$P_0 = \frac{12}{52} \cdot \frac{3}{51} \cdot \frac{40}{50} \cdot \frac{3}{49} \cdot \frac{48}{48}$$
$$P_0 = \frac{3}{13} \cdot \frac{1}{17} \cdot \frac{4}{5} \cdot \frac{3}{49}$$
$$P_0 = \frac{3 \cdot 2 \cdot 2 \cdot 3}{13 \cdot 17 \cdot 5 \cdot 7 \cdot 7}$$
$$P_0 = \frac{36}{54145}$$
The final probability is simply the product of this and the number of choices for positions of the pairs:
$$P = P_0 \cdot \binom{5}{2} \cdot \binom{3}{2} = \frac{216}{10829}$$
If it does matter that the fifth card does not match either of the pairs, then the last term in our calculation of $P_0$ becomes $44/48$ instead of $48/48$, which gives us a final answer instead of $$ P = \frac{198}{10829}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/586178",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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} |
How to show that $A_k=(-1)^k\binom nk$?
In the identity $$\frac{n!}{x(x+1)(x+2)\cdots(x+n)}=\sum_{k=0}^n\frac{A_k}{x+k},$$prove that $$A_k=(-1)^k\binom nk.$$
My try: The given identity implies $$\frac{1\cdot2\cdots n}{x(x+1)(x+2)\cdots(x+n)}=\frac{A_0}{x}+\frac{A_1}{x+1}+\dots+\frac{A_n}{x+n}.$$
Now putting $A_k=(-1)^k\binom nk,$$$\frac{1\cdot2\cdots n}{x(x+1)(x+2)\cdots(x+n)}=\frac1x-\frac{n}{x+1}+\dots+\frac{(-1)^n}{x+n}.$$ How to proceed further?
| Hint:
The number $A_k$ is the residue of the meromorphic function $$\frac{n!}{x(x+1) \cdots (x+n)} $$ at the point $x=-k.$ Since it's a simple pole the residue is $$A_k=\frac{n!}{\frac{d}{dx}[x(x+1) \cdots (x+n)]} \bigg|_{x=-k}=\frac{n!}{(-k)(-k+1) \cdots (-k+n)}, $$where in the final product the term $(-k+k)$ is obviously omitted.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/590934",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Solve limit with square root signs I have an homework question that I can't solve. Solve the following limit:
$$
\lim_{x \to \infty}\;(x^2+3)^\frac {1}{2}\ - (x^2 + x)^\frac {1}{2}\
$$
| For every $x\in D=(-\infty,-1]\cup[0,\infty)$ we have
$$
f(x):=\sqrt{x^2+3}-\sqrt{x^2+x}=\frac{(x^2+3)-(x^2+x)}{\sqrt{x^2+3}+\sqrt{x^2+x}}=\frac{3-x}{\sqrt{x^2+3}+\sqrt{x^2+x}}.
$$
It follows that
$$
\lim_{x\to\infty}f(x)=\lim_{x\to\infty}\frac{3-x}{\sqrt{x^2+3}+\sqrt{x^2+x}}=\lim_{x\to\infty}\frac{\frac{3}{x}-1}{\sqrt{1+\frac{3}{x}}+\sqrt{1+\frac{1}{x}}}=-\frac12.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/593024",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
How to prove $\sum_{n=0}^{\infty} \frac{n^2}{2^n} = 6$? I'd like to find out why
\begin{align}
\sum_{n=0}^{\infty} \frac{n^2}{2^n} = 6
\end{align}
I tried to rewrite it into a geometric series
\begin{align}
\sum_{n=0}^{\infty} \frac{n^2}{2^n} = \sum_{n=0}^{\infty} \Big(\frac{1}{2}\Big)^nn^2
\end{align}
But I don't know what to do with the $n^2$.
| There is an elementary and visually appealing solution, which is to think "matrix-like".
Knowing that $\sum_{n=0}^{\infty} 2^{-n}=2$, the sum $\sum_{n=0}^{\infty}n/2^n$ can be rewritten as:
$$
\begin{align}
&& 0/2^0 && + && 1/2^1 && + && 2/2^2 && + && 3/2^3 && + && \cdots && =
\\ = && && + && 2^{-1} && + && 2^{-2} && + && 2^{-3} && + && \cdots && +
\\ + && && && && + && 2^{-2} && + && 2^{-3} && + && \cdots && +
\\ + && && && && && && + && 2^{-3} && + && \cdots && +
\\ + && && && && && && && && + && \ddots
\end{align}
$$
summed first in the columns, then in the rows. It doesn't matter which direction you sum first, because all sums in both directions converge absolutely. So we can sum first in the rows, and then in the columns. Each row-sum is just a geometric series; the $m$-th row-sum is $r_m=\sum_{n=m}^{\infty} 2^{-n}=2^{1-m}$. Now summing up all rows, we have $\sum r_m = 2$.
What we just did is equivalent to:
$$
\sum_{n=0}^{\infty} \frac{n}{2^n} =
\sum_{n=0}^{\infty} \sum_{m=1}^{n} \frac{1}{2^n} =
\sum_{m=1}^{\infty} \sum_{n=m}^{\infty} 2^{-n}=
\sum_{m=1}^{\infty} 2^{1-m} =
2
$$
It's not hard to do it for a three-dimensional matrix, unless you try to draw it :)
$$
\sum_{n=0}^{\infty} \frac{n^2}{2^n} =
\sum_{n=0}^{\infty} \sum_{m=1}^{n} \frac{n}{2^n} =
\sum_{m=1}^{\infty} \sum_{n=m}^{\infty} \frac{n}{2^n} =
\sum_{m=1}^{\infty} \left\{
\sum_{n=0}^{\infty} \frac{n}{2^n} - \sum_{n=0}^{m-1} \frac{n}{2^n}
\right\}
$$
We can work out $\sum_{n=0}^{m-1} \frac{n}{2^n}$ in the same way we did for $\sum_{n=0}^{\infty} \frac{n}{2^n}$:
$$
\sum_{n=0}^{m-1} \frac{n}{2^n} =
\sum_{n=0}^{m-1} \sum_{l=1}^{n} \frac{1}{2^n} =
\sum_{l=1}^{m-1} \sum_{n=l}^{m-1} 2^{-n} =
\sum_{l=1}^{m-1} \left\{ 2^{1-l} - 2^{1-m} \right\} =
2 - \left(m+1\right)2^{1-m}
$$
Substituting:
$$
\sum_{n=0}^{\infty} \frac{n^2}{2^n} =
\sum_{m=1}^{\infty} \left\{
2 - \left[ 2 - \left(m+1\right)2^{1-m} \right]
\right\} =
\sum_{m=1}^{\infty} \left(m+1\right)2^{1-m} = \\
\sum_{m=1}^{\infty} \left\{
\left(m-1\right)2^{1-m} + 2\cdot 2^{1-m}
\right\} =
\sum_{m=0}^{\infty} m\cdot 2^{-m} + 4\sum_{m=1}^{\infty} 2^{-m} =
2 + 4 = 6
$$
Again, you can change the order of summations because all inner sums converge absolutely.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/593996",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "44",
"answer_count": 8,
"answer_id": 5
} |
Matrix and eigenvectors
$\quad$The matrix $\mathbf A=\frac19\begin{bmatrix}
7 & -2 & 0 \\
-2 & 6 & 3 \\
0 & 2 & 5 \\
\end{bmatrix}$ has eigenvalues $1$, $\frac23$ and $\frac13$n with the corresponding eigenvectors $ \mathbf v_1=\begin{bmatrix}
-2 \\
2 \\
1 \\
\end{bmatrix}$, $\mathbf{v}_2=\begin{bmatrix}
2 \\
1 \\
2 \\
\end{bmatrix}$, $\mathbf v_3=\begin{bmatrix}
1 \\
2 \\
-2 \\
\end{bmatrix}$.
$\text{(i)}$ Find the general solution of the equation $\mathbf x_{k+1}=\mathbf{Ax}_k$ if $\mathbf x_0=\begin{bmatrix}
1 \\
11 \\
-2 \\
\end{bmatrix}$.
For this question, does it mean I should find out what $x_k$ is in terms of $A$?
Can anyone help me please?
| You have
$$A=P^{-1}DP$$
With
$$P=\begin{bmatrix}
-2&2&1\\
2&1&2\\
1&2&-2\\
\end{bmatrix}$$
and
$$D=\begin{bmatrix}
1&0&0\\
0&2/3&0\\
0&0&1/3\\
\end{bmatrix}$$
Then
$$x_n=A^nx_0= P^{-1}DP \; P^{-1}DP \dots P^{-1}DP =P^{-1}D^nP \,x_0$$
With
$$D^n=\begin{bmatrix}
1&0&0\\
0&(2/3)^n&0\\
0&0&(1/3)^n\\
\end{bmatrix}$$
After simplification of the matrix product, and noticing $P^2=9I_3$, thus $P^{-1}=\frac19P$,
$$x_n=\frac1{3^n}\begin{bmatrix}
2^{n+1}-4\cdot3^{n}+3\\
4\cdot3^{n}+2^{n}+6\\
2\cdot\left(3^{n}+2^{n}-3\right)\\
\end{bmatrix}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/594071",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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showing any controllable system can be put in 'controller' form. I am looking at the proof of the following theorem: Let $\dot{x} =Ax + Bu$ be a controllable single input system, where $\Delta_A:= \det(\lambda I -A) = \lambda^n + a_1\lambda^{n-1} + \ldots + a_{n-1}\lambda + a_n$. Show this system is isomorphic to the system $\dot{x} = \bar{A}x + \bar{B}u$ where:
$$\bar{A} = \left( \begin{matrix} 0 & 1 & 0& \dots & 0 & 0\\
0 &0&1 &\ldots &0&0 & \\
\vdots &\vdots &\vdots & & \vdots &\vdots &\\
0&0&0& \ldots & 0& 1\\
-a_n& -a_{n-1}& -a_{n-2}& \ldots & -a_2& -a_1
\end{matrix}\right) ,\quad\bar{b}=\left( \begin{matrix} 0\\0\\ \vdots \\0\\0\\1 \end{matrix}\right)$$
This amounts to finding an invertible matrix $S$ such that $\bar{A}=S^{-1}AS$ and $\bar{B} = S^{-1}B$.
The proof however I really cannot follow. This theorem comes from a subject that I am taking for the second time. I distinctly remember begin quite impressed with this theorem last year when the teacher proved it with relative ease. However in the updated lecture notes we are given this year I cannot follow the proof at all.
It starts of with defining the required matrix $S$ via $S = \left(q_1\quad q_2 \quad \ldots \quad q_n\right)$ where
\begin{eqnarray}
q_n&=&B\\
q_{n-1} &=& AB + a_{1}B\\
q_{n-1} &=& A^2B + a_{1}AB +a_{2}B\\
&\vdots&\\
q_1 &=& A^{n-1}B +a_{1}A^{n-2}B + \ldots +a_{n-1}B
\end{eqnarray}
Can anyone show me why this choice of $S$ works? Thanks a lot in advance for any help!
| We need to show that $S \bar{A} = AS$. First look at $S \bar{A}$ (work it out).
$S \bar{A} = \begin{pmatrix} -a_n q_n & -a_{n-1} q_n + q_1 & -a_{n-2} q_n + q_2 & \dots & -a_1 q_n + q_{n-1} \end{pmatrix}$
Now $AS = \begin{pmatrix} A q_1 & A q_2 & \dots & A q_n \end{pmatrix}$. Let us look at each column seperately.
$A q_1 = A^n B + a_1 A^{n-1} B + \dots + a_{n-1} A B = -a_n B = -a_n q_n$.
It follows from the Cayley-Hamilton theorem, because $\Delta_A$ is the characteristic polynomial of $A$. Others will just follow from the definitions
$\begin{align}
A q_2 &= A^{n-1} B + a_1 A^{n-2} B + \dots + a_{n-2} A B = -a_{n-1} B + q_1 = -a_{n-1} q_n + q_1 \\
A q_3 &= A^{n-2} B + a_1 A^{n-3} B + \dots + a_{n-3} A B = -a_{n-2} B + q_2 = -a_{n-2} q_n + q_2 \\
&\vdots \\
A q_{n-1} &= A^2 B + a_1 AB = -a_2 B + q_{n-2} = -a_2 q_n + q_{n-2} \\
A q_n &= AB = -a_1 B + q_{n-1} = -a_1 q_n + q_{n-1} \\
\end{align}$
Hence, $S \bar{A} = AS$. Also, see that $S \bar{B} = q_n = B$.
Edit: Also note that $q_i$ are linearly dependant (ie. $S$ does not have an inverse) if the system is not controllable.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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How do you simplify this expression? $$\lim_{h\to0}(\frac{x}{h(x+h+1)} + \frac{1}{x+h+1} - \frac{x}{h(x+1)})$$
I know the answer is $$\frac{1}{(1+x)^2}$$
But I can't get there
| Take the lcm of the denominators and simplify: You will get
\begin{align}
\dfrac{x}{h(x+h+1)} + \dfrac{1}{x+h+1} - \dfrac{x}{h(x+1)} & = \dfrac{x(x+1) + h(x+1) - x(x+h+1)}{h(x+1)(x+h+1)}\\
& = \dfrac{x^2+x + hx + h - x^2 - hx - x}{h(x+1)(x+h+1)}\\
& = \dfrac{h}{h(x+1)(x+h+1)}\\
& = \dfrac1{(x+1)(x+h+1)}
\end{align}
Now let $h \to 0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/596253",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Is there a more conceptual proof of this fact? Equip ${\mathbb R}^3$ with the usual scalar product $(.|.)$. Let $A$ be the matrix
$$
A=
\left(\begin{matrix}
1 & 2 & 3 \\
-2 & 4 & 5 \\
-3 & -5 & 6 \\
\end{matrix}\right)
$$
With the help of a computer, I found an ugly, wholly computational proof of the following
fact :
Theorem There is no orthogonal basis $(b_1,b_2,b_3)$ such that
$(Ab_1|b_2)=(Ab_2|b_3)=(Ab_3|b_1)=0$.
(I argue as follows : there must be at least one index $i$ such that
$Ab_i$ is not a multiple of $b_i$, say $i=1$. Then there is a constant
$c$ such that $b_2=c(b_1 \wedge Ab_1)$, and we assume without loss
that $c=1$. Similarly we may assume $b_3=b_1 \wedge b_2$. We may express
everything in terms of the coordinates $x,y,z$ of $b_1$. Finally
we have a system of two polynomial equations in $x,y,z$ (coming
from $(Ab_2|b_3)=(Ab_3|b_1)=0$ ), which can be shown to have no real
solution by a formal computing system).
Does anyone know a more interesting and conceptual method ?
| You are essentially asking if there exists an orthogonal matrix $B$ such that
$$
B^TAB=\pmatrix{x&2a&0\\ 0&y&2b\\ 2c&0&z}\tag{1}
$$
for some $x,y,z,a,b,c\in\mathbb R$. By splitting into symmetric and skew-symmetric parts, we can rewrite the equation as
\begin{align*}
B^T\pmatrix{1\\ &4\\ &&6}B &= \pmatrix{x&a&c\\ a&y&b\\ c&b&z},\tag{2}\\
B^T\pmatrix{0&2&3\\ -2&0&5\\ -3&-5&0}B &= \pmatrix{0&a&-c\\ -a&0&b\\ c&-b&0}.\tag{3}
\end{align*}
Take the Frobenius norm on both sides of $(2)$ and $(3)$, we get
\begin{align*}
1^2+4^2+6^2 &= x^2+y^2+z^2+2(a^2+b^2+c^2),\\
2(2^2+3^2+5^2) &= 2(a^2+b^2+c^2).
\end{align*}
Therefore $x^2+y^2+z^2=-23$, which is insolvable. Hence the orthogonal basis in question does not exist.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/596681",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
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