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Solve $a^3 -b^3 -c^3=3abc, a^2=2(b+c)$ in natural numbers.
Solve $a^3 -b^3 -c^3=3abc, a^2=2(b+c)$ in natural numbers.
Substituting $a=\sqrt{2(b+c)}$ in the cubic equation, we get:
$2\sqrt{2}(b+c)^{\frac{3}{2}} - b^3 -c^3 = 3\sqrt{2(b+c)}bc$
$2\sqrt{2}(b+c)\sqrt{b+c} - b^3 -c^3 = 3\sqrt{2(b+c)}bc$
Not able to proceed further, although I think that binomial expansion of $\sqrt{b+c}$ can yield some more steps. But, am confused about applying that too.
| \begin{align}x^3+y^3+z^3-3xyz & = x^3+y^3+3x^2y+3xy^2+z^3-3xyz-3x^2y-3xy^2 \\
& =(x+y)^3+z^3-3xy(x+y+z)\\
&= (x+y+z)((x+y)^2+z^2-(x+y)z)-3xy(x+y+z) \\ & =(x+y+z)(x^2+2xy+y^2+z^2-xy-xz-3xy) \\ & =(x+y+z)(x^2+y^2+z^2-xy-yz-zx)\end{align}
Therefore, for distinct $x,y $ and $z$; $$
x^3+y^3+x^3=3xyz \implies x+y+z=0$$ Notice that your equation can rewritten as $$(a)^3+(-b)^3+(-c)^3=3 (a) (-b)(-c)$$
$$\implies a-b-c=0$$
Substitute $a=b+c$ in your second equation,
$$
a^2=2(b+c) \implies (b+c)^2=2(b+c) \implies b+c=2$$
So finally , we've $a=2$ and $b+c=2$.
So the only natural solution to these equations is $\color{blue}{(a,b,c)=(2,1,1)}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2637276",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Finding the distance between two parallel tangents of a rational function A Curve C has equation $y=\frac{2x-5}{x-1}$
a) A line y = mx+c is tangent to the curve. Find a condition for c in terms of m.
This can be solved relatively easily
$mx + c = \frac{2x-5}{x-1}$
$mx^2 + (c-m-2)x -c+5=0$
take discriminant as 0, as there is only one point at which the tangent intersects the curve
$c^2 +(2m-4)c +m^2 -16m+4=0$
By completing the square for c and m separately
$(c+m-2)^2-(m-2)^2+(m-8)^2-64+4=0$
$(c+m-2)^2 = 12m$
$c = 2-m\pm\sqrt{12m}$
b) Two parallel lines are each tangents to the curve, touching at points P(p,q) and R(r,s) with p<1
I have plotted this graph using desmos and attached an image of it
P can be taken to be a point in the top-left part of the graph and R a point in the bottom-right section of the graph.
If the distance $PR$ is $d$, find $d^2$ in terms of $m$, the gradient of each of the lines.
No calculus is required for this part.
$d^2=(p-r)^2 +(q-s)^2$
the tangents intersecting P and Q are given by:
$y = mx - mp + q$
$y = mx - mr +s$
$PR$ is perpendicular to the tangents
Therefore $grad_{PR} = -\frac{1}{m}$
Using the original equation:
$q = \frac{2p-5}{p-1}$
and $s = \frac{2r-5}{r-1}$
$q-s = \frac{3(p-r)}{(p-1)(r-1)}$
also take the leftmost tangent as $y_1$ and the other as $y_2$
$y_1 = mx - mp + \frac{2p-5}{p-1}$
$y_2 = mx - mr + \frac{2r-5}{r-1}$
$grad_{PR}=-\frac{1}{m} =\frac{p-r}{q-s} = \frac{(p-1)(r-1)}{3}$
$m = -\frac{3}{(p-1)(r-1)}$
$(q-s)^2 = \frac{9(p-r)^2}{(p-1)^2(r-1)^2}$
$(p-r)^2=\frac{(p-r)^2(p-1)^2(r-1)^2}{(p-1)^2(r-1)^2}$
$d^2 = \frac{(p-r)^2[9+(p-1)^2(r-1)^2]}{(p-1)^2(r-1)^2}$
$(p-1)^2(r-1)^2 = \frac{9}{m^2}$
$d^2 = (p-r)^2(m^2+1)$
using the conclusion from part a
$y_1 = mx + 2 - m - \sqrt{12m}$
$y_2 = mx+ 2- m + \sqrt{12m}$
Hence $ 2 - m - \sqrt{12m} = - mp + \frac{2p-5}{p-1}$
and $ 2- m + \sqrt{12m} =- mr + \frac{2r-5}{r-1} $
I still need to find $p-r$ in order to express $d^2$ in terms of $m$
I've been working on this problem for ages so any help would be much appreciated
The answer is
$d^2 = 4(\frac{3}{m}+3m)$
however, this is not particularly useful without a solution to it
c) By differentiating $d^2$ with respect to $m$, or otherwise, find the shortest distance between the two branches of the curve.
This part evidently does require calculus.
Any suggestions as to how I should answer this question would also be helpful.
| Here’s a somewhat different approach using homogeneous coordinates.
The graph of this rational function is a hyperbola, so rewrite the equation as $$xy-2x-y+5 = 0$$ and then put it in the matrix form $\mathbf x^TC\mathbf x=0$, with $$C=\begin{bmatrix}0&\frac12&-1\\\frac12&0&-\frac12\\-1&-\frac12&5\end{bmatrix}.$$ Tangent lines $\mathbf l$ to this conic satisfy the dual conic equation $\mathbf l^TC^{-1}\mathbf l=0$. Setting $\mathbf l=(m,-1,c)^T$, corresponding to the generic equation $mx-y+c=0$ of a line, gives the equation $$\begin{bmatrix}m&-1&c\end{bmatrix} \begin{bmatrix}\frac13&\frac83&\frac13\\\frac83&\frac43&\frac23\\\frac13&\frac23&\frac13\end{bmatrix} \begin{bmatrix}m\\-1\\c\end{bmatrix} = \frac13\left(c^2+(2m-4)c+(m^2-16m-4)\right) = 0$$ with solutions $$c = 2-m\pm2\sqrt{3m}.$$ The point of tangency of a tangent line $\mathbf l$ is its pole $C^{-1}\mathbf l$: $$\begin{bmatrix}\frac13&\frac83&\frac13\\\frac83&\frac43&\frac23\\\frac13&\frac23&\frac13\end{bmatrix} \begin{bmatrix}m \\ -1 \\ 2-m\pm2\sqrt{3m} \end{bmatrix} = \begin{bmatrix} -2\pm2\sqrt{\frac m3} \\ 2m\pm4\sqrt{\frac m3} \\ \pm2\sqrt{\frac m3}\end{bmatrix}$$ which when converted back into inhomogeneous Cartesian coordinates is $$\left( 1\mp\sqrt{\frac3m}, 2\pm\sqrt{3m} \right).$$ The first choice of signs gives the point $P$, the other $R$.
Since the problem you’re working on says, “or otherwise,” here’s an otherwise. The curve is a hyperbola, so the minimum distance between its branches occurs at the vertices, and this distance is just the hyperbola’s transverse axis length. This can be computed directly from the matrix $C$: it is equal to $$2\sqrt{{-\det C \over \lambda_+^2\lambda_-}},$$ where $\lambda_+$ and $\lambda_-$ are the positive and negative eigenvalues of the upper-left $2\times2$ submatrix of $C$. This mysterious formula is the end-result of converting the general conic equation into standard form. The eigenvalues are easily seen to be $\pm\frac12$, and a simple computation then gives a minimal distance of $2\sqrt6$.
You could also find this distance by finding the hyperbola’s vertices, which are the intersections of the hyperbola with its transverse axis. The hyperbola’s center lies at the intersection of its asymptotes, here $(1,2)$, and its transverse axis is a bisector of the asymptotes, so the equation of its transverse axis is $(x-1)+(y-2)=0$, or $x+y=3$. Solving the resulting system of equations produces the points $(1\pm\sqrt3,2\mp\sqrt3)$ and the distance between them is again $2\sqrt6$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2642137",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Prove that $2^{5n + 1} + 5^{n + 2} $ is divisible by 27 for any positive integer My question is related to using mathematical induction to prove that $2^{5n + 1} + 5^{n + 2} $ is divisible by 27 for any positive integer.
Work so far:
(1) For n = 1:
$2^{5(1) + 1} + 5^{(1) + 2} = 26 + 53 = 64 + 125 = 189$
Check if divisible by $27$: $189$ mod $27$ = $0$
As no remainder is left, the base case is divisible by $27$.
(2) Assume $n = k$, then $2^{5k + 1} + 5^{k + 2} = 27k$
(3) Prove that this is true for n = k + 1:
$$2^{5(k + 1) + 1} + 5^{(k + 1) + 2} $$
$$= 2^{5k + 5 + 1} + 5^{k + 1 + 2} $$
$$ = 32 * 2^{5k + 1} + 5 * 5^{k + 2}$$
$$= ? $$
I know I am supposed to factor out 27 somehow, I just cant seem to figure it out. Any help would be appreciated.
| $$32\cdot2^{5k+1}+5\cdot5^{k+2}=5(2^{5k+1}+5^{k+2})+27(2^{5k+1})$$ Both terms now have factors in them divisible by 27:
*
*First has $2^{5k+1}+5^{k+2}$
*Second has $27$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2642314",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 4
} |
Exponential least squares $a+be^x$ is this correct? I am trying to find a least squares approximation for $a,b$ for the following data set
$$g(x)=a+be^x,\quad \begin{array}{|l|l|l|l|}
\hline
i & 0 & 1 & 2 \\ \hline
x & 1 & 2 & 3 \\ \hline
y & 0 & 2 & 2 \\ \hline
\end{array}$$
We want to find a stationary point of
$$f(a,b)=\sum_i(a+be^{x_i}-y_i)^2$$ so we need
$$f'(a,b)=\begin{pmatrix}\sum _{i}\:2\left(a+be^{x_i}-y_i\right)\\ \sum _i\left
(a+be^{x_i}-y_i\right)2e^{x_i}\end{pmatrix}$$
Now we have to solve the the non linear system $f'(a,b)=0$. The jacobi matrix is $$J_{f'}=\begin{pmatrix}6&\sum _i2e^{x_i}\\ \sum \:_i2e^{x_i}&\sum \:_i2e^{2x_i}\end{pmatrix}=\begin{pmatrix}6&2\left(e+e^2+e^3\right)\\ 2\left(e+e^2+e^3\right)&2\left(e+e^4+e^6\right)\end{pmatrix}$$
Now the newton-raphson iteration? Choose the starting $x^0=(a_0,b_0)=(0,0)$ and apply
$$x^1=x^0-\begin{pmatrix}6&2\left(e+e^2+e^3\right)\\ \:2\left(e+e^2+e^3\right)&2\left(e+e^4+e^6\right)\end{pmatrix}^{-1}\begin{pmatrix}-4\\ \:-4e^2-4e^3\end{pmatrix}=\begin{pmatrix}\frac{16e^5+8e^4+8e^3-8e}{8e^6-8e^5-8e^3-4e^2+12e}\\ \frac{-16e^3-16e^2+8e}{8e^6-8e^5-8e^3-4e^2+12e}\end{pmatrix}$$
now reaching this horrifying expression makes me believe I did something horribly wrong. Is this way of solving this problem correct?
| I used $QR$ factorization rather than attempting to solve the normal equations. We have
$$\begin{bmatrix}1&e\\1&e^2\\1&e^3\end{bmatrix}\begin{bmatrix}a\\b\end{bmatrix}\approx\begin{bmatrix}0\\2\\2\end{bmatrix}$$
After $QR$ factorization we have
$$\begin{bmatrix}\frac1{\sqrt3}&\frac{-e-2}{\sqrt{6(e^2+e+1)}}\\
\frac1{\sqrt3}&\frac{-e+1}{\sqrt{6(e^2+e+1)}}\\
\frac1{\sqrt3}&\frac{2e+1}{\sqrt{6(e^2+e+1)}}\end{bmatrix}\begin{bmatrix}\sqrt3&\frac e{\sqrt3}(e^2+e+1)\\
0&\frac{e(e-1)}3\sqrt{6(e^2+e+1)}\end{bmatrix}\begin{bmatrix}a\\b\end{bmatrix}=QR\begin{bmatrix}a\\b\end{bmatrix}\approx\begin{bmatrix}0\\2\\2\end{bmatrix}$$
Multiplying both sides on the left by $Q^T$ we arrive at
$$\begin{bmatrix}\sqrt3&\frac e{\sqrt3}(e^2+e+1)\\
0&\frac{e(e-1)}3\sqrt{6(e^2+e+1)}\end{bmatrix}\begin{bmatrix}a\\b\end{bmatrix}=\begin{bmatrix}\frac4{\sqrt3}\\
\frac{2(e+2)}{\sqrt{6(e^2+e+1)}}\end{bmatrix}$$
Dividing both sides on the left by
$$\begin{bmatrix}\sqrt3&0\\
0&\frac{e(e-1)}3\sqrt{6(e^2+e+1)}\end{bmatrix}$$
We have
$$\begin{bmatrix}1&\frac e3(e^2+e+1)\\
0&1\end{bmatrix}\begin{bmatrix}a\\b\end{bmatrix}=\begin{bmatrix}\frac43\\
\frac{e+2}{e(e-1)(e^2+e+1)}\end{bmatrix}$$
Finally, subtracting $\frac e3(e^2+e+1)$ times the second row from the first our solution is
$$\begin{bmatrix}a\\b\end{bmatrix}=\begin{bmatrix}\frac{e-2}{e-1}\\
\frac{e+2}{e(e-1)(e^2+e+1)}\end{bmatrix}$$
Comparing numerically with the solution originally obtained from solving the normal equations, we have agreement as best we can tell, so we conclude that the solution in the original question should have worked in one step because the system of equations was in fact linear! So what went wrong? The problem was that the matrix
$$\begin{bmatrix}-4\\-4e^2-4e^3\end{bmatrix}$$
Should have been
$$\begin{bmatrix}-8\\-4e^2-4e^3\end{bmatrix}$$
Fix that and simplify and you should get the solution @Moo and I agree on.
EDIT: Not to mention that also you should have
$$J_{f^{\prime}22}=2(e^2+e^4+e^6)$$
Rather than what you have there.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2645896",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Proving that $\frac {1}{3x^2+1}+\frac {1}{3y^2+1}+\frac {1}{3z^2+1}\geq \frac {3}{16 } $
Let $x,y,z\geq 1$ and $x+y+z=6$. Then $$\frac {1}{3x^2+1}+\frac {1}{3y^2+1}+\frac {1}{3z^2+1}\geq \frac {3}{16 }. $$
I tried to use Cauchy- Schwartz inequality but it doesn't work.
| The problem is symmetric in $x,y$ and $z$ ; and the "feasible" region is convex.. . $\therefore $ the minimum will occur when $x=y=z$, or else at a boundary point...
but on the boundary at least one of $x,y,z$ is $1$, and we easily get a lower bound of $\frac14$, coming from that term...
Finally if $x=y=z=2$
we get $$\frac1{3\cdot2^2+1}+\frac1{3\cdot 2^2+1}+\frac1{3\cdot 2^2+1}=\frac3{13}\ge \frac3{16}$$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2646467",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
On integer solutions to $a^3+b^3+c^3+d^3=3$ I was challenged to prove that there are infinitely many solutions to the equation$$a^3+b^3+c^3+d^3=3\ \ \text{ with }(a,b,c,d)\in\mathbb Z^4$$
That was easy: elementary algebra is enough to prove that $\forall z\in \mathbb Z$ then $$(a,b,c,d)=(1+6z^3,\,1-6z^3,\,-6z^2,\,1)$$ is a solution. Call these solutions and their permutations trivial.
Define a nonsum as a value of $k\in\mathbb Z$ for which there is no solution with $a+b=k$.
What about proving (or disproving/perfecting) the following conjectures?
*
*If polynomials $A$, $B$, $C$, $D$ with integer coefficients are such that $\forall z\in \mathbb Z,\ (A(z),B(z),C(z),D(z))$ is a solution, and one such solution is non-trivial, then all the polynomials are constant.
*There are nevertheless infinitely many non-trivial solutions.
*Any $k\equiv0\pmod3$ is a nonsum.
*$k=7$ is a nonsum.
*There are infinitely many nonsums with $k\not\equiv0\pmod3$
*If $k$ is a nonsum, then $-k$ is a nonsum.
*If $k$ is not a nonsum (that is, if there exists a solution with $a+b=k$), then there are infinitely many solutions with $a+b=k$.
| Conjecture 3 is true.
Note first that for any integer $n$, $n^3 \equiv -1,0$, or $1\pmod 9$. Hence any solution of $a^3+b^3+c^3+d^3=3$ must contain three cubes congruent to $1\pmod 9$ and one cube congruent to $0\pmod9$.
Now suppose $a+b = k\equiv0\pmod3$. We have:
$$a^3+b^3=(a+b)^3-3ab(a+b)$$
where:
$$(a+b)^3\equiv0\pmod{3^3}\quad \text{and}\ \textit{a fortiori}\quad \equiv0\pmod9$$
$$3ab(a+b)\equiv0\pmod9$$
Hence $a^3+b^3\equiv0\pmod9$ which is inconsistent with the condition stated above, since it implies either both $a^3$ and $b^3$ congruent to $0\pmod9$ or one of them congruent to $-1$ and the other to $1\pmod9$.
Similar but not identical reasoning shows that conjecture 4 is true, and more generally any $k\equiv0\pmod7$ is a nonsum.
For any integer $n$, $n^3 \equiv -1,0$, or $1\pmod 7$. Hence any solution of $a^3+b^3+c^3+d^3=3$ must contain either (Case 1) three cubes congruent to $1\pmod 7$ and one cube congruent to $0\pmod7$ or (Case 2) four cubes congruent to $-1\pmod7$
Now suppose $a+b = k\equiv0\pmod7$. We have:
$$a^3+b^3=(a+b)(a^2-ab+b^2)\equiv0\pmod7$$
But this is inconsistent with both Case 1 and Case 2 above, since it implies either both $a^3$ and $b^3$ congruent to $0\pmod7$ or one of them congruent to $-1$ and the other to $1\pmod7$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2647249",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
} |
Prove that $(x-y)(y-z)(z-x) \leq \frac{1}{\sqrt{2}}$
If $x,y,z$ are real and $x^2+y^2+z^2=1$, prove that$$(x-y)(y-z)(z-x) \leq \frac{1}{\sqrt{2}}.$$
Equality is achieved in some strange cases: For example, if $x = -\dfrac{1}{\sqrt{2}}$, $y = 0$ and $z = \dfrac{1}{\sqrt{2}}$, then $(x-y)(y-z)(z-x)=\dfrac{1}{\sqrt{2}}$.
Note that the claim is obvious in the case when $x \geq y \geq z$ (since $(x-y)(y-z)(z-x)\leq 0$ in this case). But the inequality is not symmetric in $x, y, z$ (only cyclic). Thus, you cannot WLOG assume that $x \geq y \geq z$. (But you can WLOG assume that $x \leq y \leq z$ because of the previous observation.)
| Here is a proof similar to the solution of IMO 2006 problem
3
(well... at least to the solution I found on the IMO):
As Michael Rozenberg noticed, it suffices to prove the inequality
\begin{equation}
\left( x^2 +y^2 +z^2 \right) ^3 \geq2\left( x-y\right) ^2 \left(
y-z\right) ^2 \left( z-x\right) ^2 .
\label{darij1.eq.goal1}
\tag{1}
\end{equation}
Indeed, if this inequality is proved, then we can take square roots on both
sides, and obtain
\begin{align*}
\sqrt{\left( x^2 +y^2 +z^2 \right) ^3 } & \geq\sqrt{2\left(
x-y\right) ^2 \left( y-z\right) ^2 \left( z-x\right) ^2 }\\
& =\left\vert \sqrt{2}\left( x-y\right) \left( y-z\right) \left(
z-x\right) \right\vert \geq\sqrt{2}\left( x-y\right) \left( y-z\right)
\left( z-x\right) ,
\end{align*}
which is the claim of the original post in a homogenized form.
Here is the idea of the proof of \eqref{darij1.eq.goal1} (see below for the
implementation): We notice that if we replace $x,y,z$ by $x+p,y+p,z+p$ for a
fixed $p\in\mathbb{R}$, then the right hand side of \eqref{darij1.eq.goal1}
does not change, but the left hand side may become smaller if we pick $p$
appropriate. The trick is to pick $p$ such that the left hand side becomes as
small as possible. This will make the inequality \eqref{darij1.eq.goal1}
sharper and easier to prove (spoiler: the two sides will differ by a single square).
Picking the right $p$ is easy: The sum $\left( x+p\right) ^2 +\left(
y+p\right) ^2 +\left( z+p\right) ^2 $ is a quadratic polynomial in $p$,
and is easily seen to be minimized for $p=-\dfrac{x+y+z}{3}$.
Let us actually do this. So let us set $p=-\dfrac{x+y+z}{3}$. Then,
\begin{equation}
x^2 +y^2 +z^2 \geq\left( x+p\right) ^2 +\left( y+p\right) ^2 +\left(
z+p\right) ^2 ,
\label{darij1.eq.side1}
\tag{2}
\end{equation}
because a straightforward computation shows that
\begin{align*}
& \left( x^2 +y^2 +z^2 \right) -\left( \left( x+p\right) ^2 +\left(
y+p\right) ^2 +\left( z+p\right) ^2 \right) \\
& =\dfrac{1}{3}\left( x+y+z\right) ^2 \geq0\qquad\left( \text{since
squares are nonnegative}\right) .
\end{align*}
Both sides of the inequality \eqref{darij1.eq.side1} are nonnegative reals;
thus, we can take them to the $3$-rd power and obtain
\begin{equation}
\left( x^2 +y^2 +z^2 \right) ^3 \geq\left( \left( x+p\right)
^2 +\left( y+p\right) ^2 +\left( z+p\right) ^2 \right) ^3
.
\label{darij1.eq.side13}
\tag{3}
\end{equation}
But a straightforward (if ugly) computation shows that
\begin{align*}
& \left( \left( x+p\right) ^2 +\left( y+p\right) ^2 +\left(
z+p\right) ^2 \right) ^3 -2\left( x-y\right) ^2 \left( y-z\right)
^2 \left( z-x\right) ^2 \\
& =\dfrac{2}{27}\left( 2x-y-z\right) ^2 \left( 2y-z-x\right) ^2 \left(
2z-x-y\right) ^2 \geq0
\end{align*}
(since squares are nonnegative). Hence,
\begin{equation}
\left( \left( x+p\right) ^2 +\left( y+p\right) ^2 +\left( z+p\right)
^2 \right) ^3 \geq2\left( x-y\right) ^2 \left( y-z\right) ^2 \left(
z-x\right) ^2 .
\end{equation}
Combining this with
\eqref{darij1.eq.side13}, we find $\left( x^2 +y^2 +z^2 \right) ^3
\geq2\left( x-y\right) ^2 \left( y-z\right) ^2 \left( z-x\right) ^2
$. This proves \eqref{darij1.eq.goal1}.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2648520",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
How to solve the equation $x^2+2=4\sqrt{x^3+1}$? From the Leningrad Mathematical Olympiad, 1975:
Solve $x^2+2=4\sqrt{x^3+1}$.
In answer sheet only written $x=4+2\sqrt{3}\pm \sqrt{34+20\sqrt{3}}$.
How to solve this?
| Let $u = x^2-x+1$, $v = x+1$ and $\lambda = \sqrt{\frac{u}{v}}$, we have
$$\begin{align}u + v & = (x^2-x+1) + (x + 1) = x^2 + 2 \\ &= 4\sqrt{x^3+1} = 4\sqrt{(x+1)(x^2-x+1)}\\ &= 4\sqrt{uv}\end{align}$$
This leads to
$$\lambda^2 + 1 = 4\lambda \implies \lambda = 2 \color{red}{\pm} \sqrt{3}$$
As a result,
$$\begin{align} &x^2 - x + 1 = \lambda^2 (x+1) = (7\color{red}{\pm} 4\sqrt{3})(x+1)\\
\iff & x^2 - (8 \color{red}{\pm} 4\sqrt{3})x = (6\color{red}{\pm} 4\sqrt{3})\\
\iff & (x - (4 \color{red}{\pm} 2\sqrt{3}))^2 = (4 \color{red}{\pm} 2\sqrt{3})^2 + (6 \color{red}{\pm} 4\sqrt{3})
= 34 \color{red}{\pm} 20\sqrt{3}\\
\implies & x = 4 \color{red}{\pm} 2\sqrt{3} \color{blue}{\pm} \sqrt{34 \color{red}{\pm} 20\sqrt{3}}
\end{align}$$
Notice $34-20\sqrt{3} \approx -0.651 < 0$. For real $x$,
the red $\color{red}{\pm}$ sign need to be $+$. This leaves us with
$$x = 4 + 2\sqrt{3} \color{blue}{\pm} \sqrt{34 + 20\sqrt{3}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2649304",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
} |
What is the limit of $3^{1/n}$ when n approaches infinity Graphically, I see that $\lim_{n->\infty}3^{1/n}$ approaches $1$. However, how to show $\lim_{n->\infty}3^{1/n} = 1$ step by step?
| Binomial theorem:
For $x\ge 0:$
$(1+x)^n \ge 1 + nx + n(n-1)\dfrac{x^2}{2!} \ge $
$n^2\dfrac{x^2}{4}$ for $n\ge 2.$
Set $x=\dfrac{2√3}{n}:$
$(1+\dfrac{2√3}{n})^n \ge n^2\dfrac{(4)(3)}{4n^2}= 3;$
$ (1+\dfrac{2√3}{n})^{n} \ge 3$ , or
$1 +\dfrac{2√3}{n} \ge 3^{1/n} .$
With lower bound $1$:
$1\lt 3^{1/n} \le 1+ \dfrac{2√3}{n}.$
Limit $n \rightarrow \infty$ is?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2649381",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 7,
"answer_id": 4
} |
range of $k$ in quadratic equation the exhaustive set of values of $k$ for which $|kx-2|=2x^2+kx+4$ has at least one positive root
solution i try $|kx-2|$ is either $(kx-2)$ or $-(kx-2)$
$kx-2=2x^2+kx+4\Rightarrow 2x^2=-6$ no real solution
$-kx+2=2x^2+kx+4\Rightarrow x^2+kx+4=0$
$\displaystyle x=\frac{-k\pm \sqrt{k^2-16}}{4}$
$|k|\geq 4$ i did not understand how i use condition at least one positive root
| If $-kx + 2 = 2x^2 + kx + 4$ then $2 = 2x^2 + 2kx + 4$ thus $0 = 2x^2 + 2kx - 2$. $$\therefore x^2 + kx - 1 = 0\tag1$$ and not $$x^2 + kx + 4 = 0.\tag2$$ Equation $(1)$ is correct, but Eq. $(2)$ is false. Therefore, you should have that $$x = \frac{-k\pm\sqrt{k^2 - 4}}{2}$$ which brings us to the fact that $|k|\geq 2$. Since we need a positive root $x$, then for some $n\in\mathbb{Z}$, we have $k^2 - 4 = n^2$. Since $4 = 2^2$ then we have that $k = 2$. $$\therefore x = \frac{-2\pm 0}{2} = -1.$$ This means that $k\neq 2$ because in this case, $x<0$. But, remember that $2^2 = (-2)^2$ so that means we can let $k = -2$. $$\therefore x = \frac{2\pm 0}{2} = 1.$$ It follows, then, that $(k, x) = (-2, 1)$ and $x$ has at least one positve solution.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2650244",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Finding $\int\frac{x^2-1}{\sqrt{x^4+x^2+1}}$ Finding
$$\int\frac{x^2-1}{\sqrt{x^4+x^2+1}}$$
Try: I have tried it to convert it into $\displaystyle \left(x+\frac{1}{x}\right)=t$ and $\displaystyle \left(1-\frac{1}{x^2}\right)$ but nothing happen.
I felt that it must be in Elliptical Integral of first kind, second kind or in third kind.
Can someone explain me how to write in elliptical form
| Maple expresses it as
$$ {\frac { \left( i\sqrt {3}+1 \right) \sqrt {4+2\,{x}^{2}-2\,i{x}
^{2}\sqrt {3}}\sqrt {4+2\,{x}^{2}+2\,i{x}^{2}\sqrt {3}} \left( i{\it
EllipticF} \left( \left( i\sqrt {3}+1 \right) x/2,(1+i\sqrt {3})/2
\right) \sqrt {3}+3\,{\it EllipticF} \left( \left( i\sqrt {3}+1
\right) x/2,(1+i\sqrt {3})/2 \right) -2\,{\it EllipticE} \left(
\left( i\sqrt {3}+1 \right) x/2,(1+i\sqrt {3})/2 \right) \right) }{16\;
\sqrt {x^4+x^2+1}}}
$$
So yes, it is expressed in terms of elliptic integrals of the first and second kinds (EllipticF is first kind, EllipticE is second kind).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2651254",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
How many sub-square matrices does a square matrix have and is there a simple formula for it? Consider an $n \times n$ matrix $M$. I want to find the determinant for ALL sub-square matrices of $M$. There may be a better way but my method is to find all sub-square matrices and check them individually.
*
*How many sub-square matrices does a square matrix have and is there a simple formula for it?
The other problem is, after checking several videos I am not sure what counts as a sub-square matrix. I thought I could use the formula for the sum of squares, i.e.
$$S(n) = \frac{n(n+1)(2n+1)}{6}$$
But this gives sixteen $(1 \times 1)$ matrices, nine $(2 \times 2)$ matrices, four $(3 \times 3)$ matrices and one $(4 \times 4)$ matrices, i.e. a total of $30$ sub-square matrices.
\begin{pmatrix}
2 & 3 & 1 & 1 \\
1 & 2 & 3 & 1 \\
1 & 1 & 2 & 3 \\
3 & 1 & 1 & 2
\end{pmatrix}
But, for example, I can find thirty-six $(2 \times 2)$ matrices by observation, i.e. six for any pair of rows. For example, the first two rows give:
\begin{equation}
\begin{pmatrix}
2 & 3 \\
1 & 2
\end{pmatrix}
\begin{pmatrix}
2 & 1 \\
1 & 3
\end{pmatrix}
\begin{pmatrix}
2 & 1 \\
1 & 1
\end{pmatrix}
\begin{pmatrix}
3 & 1 \\
2 & 3
\end{pmatrix}
\begin{pmatrix}
3 & 1 \\
2 & 1
\end{pmatrix}
\begin{pmatrix}
1 & 1 \\
3 & 1
\end{pmatrix}
\end{equation}
If I do the same for rows $1$ and $3$, $1$ and $4$, $2$ and $3$, $2$ and $4$, and $3$ and $4$, I get thirty-six sub-square matrices by considering only $(2 \times 2)$ matrices. So now I am wondering if my way of counting sub-square matrices is wrong.
Any ideas?
| If you admit the zero-by-zero matrix as a square submatrix, then the number
of square submatrices is
$$\sum_{k=0}^n\binom nk^2.$$
This can be written as
$$\sum_{k=0}^n\binom nk\binom n{n-k}$$
which can be recognised as the coefficient of $t^n$ in $(1+t)^n(1+t)^n$
etc.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2652008",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
prove that radius is a root of $(3-2\sqrt{2})t^2-2\sqrt{2}t+2=0$ a circle is described to pass through the origin and to touch the lines $x=1,x+y=2$ then prove that the radius of the circle is a root of $(3-2\sqrt{2})t^2-2\sqrt{2}t+2=0$
solution i try
length of perpendicular from $(1-r,2r-1)$ is $\displaystyle \frac{1-r+2r-1-2}{\sqrt{2}}=r^2$
$r=2(\sqrt{2}-1)$ which not satisfy $(3-2\sqrt{2})t^2-2\sqrt{2}t+2=0$
How to find correct radious in that question
| let $$x_M,y_M$$ the coordinates of the centre of the circle, and $r$ the searched radius, then we get
$$\left|\frac{x_M+y_M-2}{\sqrt{2}}\right|=r$$
$$|x_M-1|=r$$
$$x_M^2+y_M^2=r^2$$
can you finish?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2655225",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Find the values of the parameters for which the matrices commute
I have two matrices:
$A = \left( \begin{array}{cc}
1 & a \\
b & 6
\end{array} \right)
%
\ \ \ \ B = \left( \begin{array}{cc}
4 & c \\
d & 2
\end{array} \right)$.
How can I find the values of the parameters $a, b, c, d$ for which the matrices will commute?
I tried calculate $AB$ and $BA$ and compare the matrices elements, but that is not enough to get the result.
I think this have simple solution, but I have difficulties with it.
| We have
$$AB=\begin{pmatrix}
1&a\\
b&6\\
\end{pmatrix}\begin{pmatrix}
4&c\\
d&2\\
\end{pmatrix}=\begin{pmatrix}
4+ad&c+2a\\
4b+6d&bc+12
\end{pmatrix}$$
and
$$BA=\begin{pmatrix}
4&c\\
d&2\\
\end{pmatrix}\begin{pmatrix}
1&a\\
b&6\\
\end{pmatrix}=\begin{pmatrix}
4+bc&4a+6c\\
d+2b&ad+12
\end{pmatrix}.$$
Hence
\begin{eqnarray}
ad&=&bc\\
0&=&2a+5c\\
2b+5d&=&0\\
\end{eqnarray}
Thus $a=\frac{-5}{2}c$ and $b=\frac{-5}{2}d$ and thus $\frac{-5}{2}cd=\frac{-5}{2}dc$ is automatic. In other words for all $c,d$ and $a=\frac{-5}{2}c$ and $b=\frac{-5}{2}d$ the matrices commute.
Although the equations you get are not linear, a solution to the linear second and third equation automatically gives a solution to the non-linear equation! That makes this easy.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2655369",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Expected number of turns in dice throwing I generated a transition probability matrix for a scenario where I throw five dice and set aside those dice that are sixes. Then, I throw the remaining dice and again set aside the sixes - then I repeat this procedure until I get all the sixes. $X_n$ here represents the number of dices that are sixes after n rolls.
$$\begin{pmatrix}\frac{5^5}{6^5} & \frac{3125}{6^5} & \frac{1250}{6^5} & \frac{250}{6^5} & \frac{25}{6^5} & \frac{1}{6^5}\\\ 0 & \frac{625}{6^4} & \frac{500}{6^4} & \frac{150}{6^4} & \frac{20}{6^4} & \frac{1}{6^4} \\\ 0& 0 & \frac{125}{6^3}& \frac{75}{6^3}& \frac{15}{6^3} & \frac{1}{6^3} \\\ 0 & 0& 0& \frac{25}{6^2}& \frac{10}{6^2}& \frac{1}{6^2}& \\ 0 & 0 & 0 & 0 & \frac{5}{6} & \frac{1}{6} \end{pmatrix}$$
I want to figure out how many turns it takes for me on average to get all sixes.
I'm not even sure where to start with this problem. Is it a right approach to write a program where I calculate $P^n$ and see when the 6th column all equals to 1?
Any pointers would be greatly appreciated.
| An other method
Each turn you set aside any die which shows a six.
Let $F_n$ be the expected number of turns until you set aside at least one of $n$ die. Let $E_n$ be the expected number of turns until you set aside all $n$ die. Let $p_n(k)$ be the (conditional)probability of setting aside $k$ die in a turn when given that you set aside at least one.
$$\begin{align}p_n(k) &=\binom nk\dfrac{ 5^{n-k}}{(6^n-5^n)}\mathbf 1_{k\in \{1,..,n\}}
\\ F_n &= 6^n/(6^n-5^n)
\\ E_1 &= F_1\\ &=6
\\ E_2 &= F_2+p_2(1)E_1 \\ &= 36/11+2(5/11)\cdot 6\\ & = 96/11
\\ E_3 &= F_3+p_3(1)E_2+p_3(2)E_1
\\ E_4 &= F_4+p_4(1)E_3+p_4(2)E_2+p_4(3)E_1
\\ E_5 &= F_5+p_5(1)E_4+p_5(2)E_3+p_5(3)E_2+p_5(4)E_1
\\ E_6 &= F_6+p_6(1)E_5+p_6(2)E_4+p_6(3)E_3+p_6(4)E_2+p_6(5)E_1
\end{align}$$
$$$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2656232",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
Finding the length of one side of a triangle In $\triangle\,$ABC if BC is unity and $\sin\frac{A}{2} = x_1, \sin\frac{B}{2} = x_2, \cos\frac{A}{2} = x_3, \cos\frac{B}{2} = x_4$ with $$ \left(\frac{x_1}{x_2}\right)^ {2007} - \left(\frac{x_3}{x_4}\right)^{2006} = 0 $$ then the length of AC is equal to $\ldots$.
I tried using sine rule and arrive at an expression for b but have not been able to do so.
Any help would be appreciated. Thanks in advance.
| In a given $\triangle$ both $\displaystyle \frac{A}{2},\frac{B}{2}\in\bigg(0,\frac{\pi}{2}\bigg).$
For $x\in(0,\dfrac{\pi}{2})$, $\sin x$ is a strictly increasing function whereas $\cos x$ is a strictly decreasing function.
Now if $\displaystyle \frac{A}{2}>\frac{B}{2}$ then $\displaystyle \sin\frac{A}{2}>\sin\frac{B}{2}$ and $\displaystyle \cos\frac{A}{2}<\cos\frac{B}{2}$. So $x_{1}>x_{2}$ and $x_{3}<x_{4}$.
So $x^{2007}_{1}x^{2006}_{4}=x^{2007}_{2}x^{2006}_{3}$ not possible.
Similarly if $\displaystyle \frac{A}{2}<\frac{B}{2}$ then $\displaystyle \sin\frac{A}{2}<\sin\frac{B}{2}$ and $\displaystyle \cos\frac{A}{2}>\cos\frac{B}{2}$. So $x_{1}<x_{2}$ and $x_{3}>x_{4}$.
So $x^{2007}_{1}x^{2006}_{4}=x^{2007}_{2}x^{2006}_{3}$ not possible.
Therefore the only possibility is $\displaystyle \frac{A}{2}=\frac{B}{2}.$
So $x_{1}=x_{2}$ and $x_{3}=x_{4}.$
In that case $\triangle ABC$ is isosceles with $\angle ABC=\angle CAB.$ So $AC=BC=1$ unit.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2656705",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Creating a basis for U Consider the subspace
$$
U=\text{span} \lbrace \left[\begin{array}{c}
1\cr
5\cr
0\cr
4
\end{array}\right], \left[\begin{array}{c}
5\cr
1\cr
7\cr
-1
\end{array}\right], \left[\begin{array}{c}
3\cr
7\cr
0\cr
12
\end{array}\right] \rbrace
$$
of $\mathbb{R}^4$. Create a basis
$$
\lbrace \left[\begin{array}{c}
-1\cr
3\cr
-2\cr
2
\end{array}\right],\left[\begin{array}{c}
2\cr
2\cr
2\cr
2
\end{array}\right],x \rbrace
$$
for U.
$$
x =
\left[\Rule{0pt}{4.8em}{0pt}\right.
?
\left]\Rule{0pt}{4.8em}{0pt}\right.
$$
I have no clue where to start for this linear algebra problem. I know that the bases for U must be linear independent, however I do not know how to create the last basis of U to be this way. I think it has something to do with the third row since it has 0 zeroes in the spanning set.
| Consider the matrix
$$
A=\begin{bmatrix}
-1 & 2 & 1 & 5 & 3 \\
3 & 2 & 5 & 1 & 7 \\
-2 & 2 & 0 & 7 & 0 \\
2 & 2 & 4 & -1 & 12
\end{bmatrix}
$$
Perform Gaussian elimination and consider the columns of $A$ corresponding to the pivot columns in the reduced form.
\begin{align}
\begin{bmatrix}
-1 & 2 & 1 & 5 & 3 \\
3 & 2 & 5 & 1 & 7 \\
-2 & 2 & 0 & 7 & 0 \\
2 & 2 & 4 & -1 & 12
\end{bmatrix}
&\to
\begin{bmatrix}
1 & -2 & -1 & -5 & -3 \\
0 & 8 & 8 & 16 & 16 \\
0 & -2 & -2 & -3 & 6 \\
0 & 6 & 6 & 9 & 18
\end{bmatrix}
&&\begin{aligned}
R_1&\gets -R_1\\
R_2&\gets R_2-3R_1\\
R_3&\gets R_3+2R_1\\
R_4&\gets R_4-2R_1
\end{aligned}
\\[6px]
&\to
\begin{bmatrix}
1 & -2 & -1 & -5 & -3 \\
0 & 1 & 1 & 2 & 2 \\
0 & 0 & 0 & 1 & 2 \\
0 & 0 & 0 & -3 & -6
\end{bmatrix}
&&\begin{aligned}
R_2&\gets \tfrac{1}{8}R_2\\
R_3&\gets R_3+2R_2\\
R_4&\gets R_4-6R_2
\end{aligned}
\\[6px]
&\to
\begin{bmatrix}
1 & -2 & -1 & -5 & -3 \\
0 & 1 & 1 & 2 & 2 \\
0 & 0 & 0 & 1 & 2 \\
0 & 0 & 0 & 0 & 0
\end{bmatrix}
&&R_4\gets R_4+3R_3
\end{align}
This shows
*
*The rank of the matrix is $3$, so indeed the two new vectors you're given belong to $U$
*The first, second and fourth columns of $A$ are linearly independent.
Thus you can choose
$$
x=\begin{bmatrix} 5 \\ 1 \\ 7 \\ -1 \end{bmatrix}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2658900",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Finding value of product of Cosines
Finding $$\left(\frac{1}{2}+\cos \frac{\pi}{20}\right)\left(\frac{1}{2}+\cos \frac{3\pi}{20}\right)\left(\frac{1}{2}+\cos \frac{9\pi}{20}\right)\left(\frac{1}{2}+\cos \frac{27\pi}{20}\right)$$
My Try: $$\left(\frac{1}{2}+\cos \frac{\pi}{20}\right)\left(\frac{1}{2}+\cos \frac{3\pi}{20}\right)\left(\frac{1}{2}+\sin \frac{\pi}{20}\right)\left(\frac{1}{2}-\sin\frac{3\pi}{20}\right)$$
So we have $$\left(\frac{1}{2}+\cos \frac{\pi}{20}\right)\left(\frac{1}{2}+\sin\frac{\pi}{20}\right)\left(\frac{1}{2}+\cos \frac{3\pi}{20}\right)\left(\frac{1}{2}-\sin\frac{3\pi}{20}\right)$$
Could some help me to solve it, Thanks in Advanced
| Put $$f(x)=\left(x+\cos \frac{\pi}{20}\right)\left(x+\cos \frac{3\pi}{20}\right)\left(x+\cos \frac{5\pi}{20}\right)\left(x+\cos \frac{7\pi}{20}\right)\left(x+\cos \frac{9\pi}{20}\right)$$
Then you have to find $f({1\over 2})/({1\over 2}+{\sqrt{2}\over2})$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2659008",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
"answer_id": 3
} |
Basis for Matrix A
Find a basis for all $2\times2$ matrices $A$ for which
$A\begin{bmatrix} 1 & 1 \\ 1 & 1 \end{bmatrix}$ = $\begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}$.
Maybe I'm dumb-- but isn't $A$ just the $0$ matrix? In which case, the base is simply the $0$ matrix as well?
| If you let $A = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$ then your equation says,
$$ \begin{pmatrix} a + b & a + b \\ c+d & c+d \end{pmatrix} = \begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix}$$
i.e $a = -b$ and $c = -d$ which means we can rewrite $A$ as,
$$\begin{pmatrix} a & -a \\ -d & d \end{pmatrix} = a \begin{pmatrix} 1 & -1 \\ 0 & 0 \end{pmatrix} + d \begin{pmatrix} 0 & 0 \\ -1 & 1 \end{pmatrix}$$
And so you are interested in the subspace $S = \textbf{span}\left\{\begin{pmatrix} 1 & -1 \\ 0 & 0 \end{pmatrix},\begin{pmatrix} 0 & 0 \\ -1 & 1 \end{pmatrix}\right\} $
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2659300",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
Inequality with positive reals For each $x,y,z>0$, define
$$
f(x,y,z)=\frac{3x^2(y+z)+2xyz}{(x+y)(y+z)(z+x)}.
$$
Fix also $a,b,c,d>0$ with $a\le b$. How can we prove "reasonably" the following inequality?
$$
\frac{f(a,b,d)+f(c,b,d)}{f(b,a,d)+f(c,a,d)} \le \frac{a+c}{b+c}.
$$
| we get $$\frac{a+c}{b+c}-\frac{f(a,b,d)+f(c,b,d)}{f(b,a,d)+f(c,a,d)}=-{\frac { \left( 3\,c+d \right) \left( a-b \right) \left( abc+abd+ca
d+cbd \right) \left( c+a \right) }{ \left( 3\,{a}^{2}{b}^{2}c+3\,{a}^
{2}{b}^{2}d+3\,{a}^{2}b{c}^{2}+4\,{a}^{2}bcd+2\,{a}^{2}b{d}^{2}+3\,{a}
^{2}{c}^{2}d+2\,{a}^{2}c{d}^{2}+6\,a{b}^{2}{c}^{2}+8\,a{b}^{2}cd+3\,a{
b}^{2}{d}^{2}+8\,ab{c}^{2}d+4\,abc{d}^{2}+3\,a{c}^{2}{d}^{2}+6\,{b}^{2
}{c}^{2}d+3\,{b}^{2}c{d}^{2}+3\,b{c}^{2}{d}^{2} \right) \left( c+b
\right) }}
$$
and this is positive since we have $$a\le b$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2659411",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
How to find $\lim_\limits{n\to \infty }(1-\frac{5}{n^4})^{(2018n+1)^4}$
How to find $$\lim_{n\to \infty }(1-\frac{5}{n^4})^{(2018n+1)^4}$$
My attempt:
$$\lim _{n\to \infty }(1-\frac{5}{n^4})^{(2018n+1)^4}=\lim _{n\to \infty }\left(1-\frac{5}{n^4}\right)^{\left(4\cdot 2018n+4\right)}=\lim _{n\to \infty }\left(\left(1-\frac{5}{n^4}\right)^{\left(4\cdot 2018n\right)}+\left(1-\frac{5}{n^4}\right)^4\right)=\lim _{n\to \infty }\left(\left(\left(1-\frac{5}{n^4}\right)^{n}\right)^{4*2018}+\left(1-\frac{5}{n^4}\right)^4\right)$$
so we get :
*
*$\lim _{n\to \infty }\left(1-\frac{5}{n^4}\right)^4=1$
let:
$b_n=\frac{-5}{n^3}$
$\lim _{n\to \infty }\frac{-5}{n^3}=0$
*
*$\lim _{n\to \infty }\left(\left(1-\frac{5}{n^4}\right)^n\right)^{4\cdot 2018}=\lim \:_{n\to \:\infty \:}\left(\left(1+\frac{b_n}{n}\right)^n\right)^{4\cdot \:2018}=\left(e^0\right)^{4\cdot 2018}=1$
so the answer is :
$$\lim _{n\to \infty }(1-\frac{5}{n^4})^{(2018n+1)^4}=1+1=2$$
Is this answer correct if not how can I find the limit ?
thanks.
| You can also write
$$(1-5/n^4)^{(2018n +1)^4} = [(1-5/n^4)^{n^4}]^{(2018n +1)^4/n^4}.$$
Inside the brackets we have the limit $e^{-5},$ while the outer exponent $\to 2018^4.$ Thus the desired limit is $e^{-5\cdot 2018^4}.$ (We have used the result: If $a_n \to a\in (0,\infty)$ and $b_n\to b\in \mathbb R,$ then $a_n^{b_n} \to a^b.$ This is worthwhile proving for yourself, as it comes up often.)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2661461",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
I can't find the other two solutions to this equation. As an exercise, I have to use Cardano's formula
$$
x^3 = px + q$$
$$x = \sqrt[3]{\frac{q}{2}+\sqrt{\frac{q^2}{4} - \frac{p^3}{27}}} + \sqrt[3]{\frac{q}{2}-\sqrt{\frac{q^2}{4} - \frac{p^3}{27}}}
$$
to solve the equation
$
x^3 = 15x+4.
$
I finally get
$$ x = \sqrt[3]{2 + 11i} + \sqrt[3]{2 - 11i}$$
$$ x = 4 $$
but I have no idea how to find the other two solutions to the equation, and the procedures I've found googling are not employing Cardano's formula.
Thank you for your help.
| Since the degree is 3, if you have a solution, you can make a factorisation and have a second degree equation to solve :
$$\begin{align*}
x^3&=15x+4 \tag{1}\\
4^3&=15\times 4 +4\tag{2}
\end{align*}
$$
with (1)-(2) :
$$x^3-4^3=15x-15\times 4 $$
so (with $a^3-b^3=(a-b)(a^2+ab+b^2)$ )
$$ (x-4)(x^2 + 4x +16)= 15(x-4) \quad \Leftrightarrow \quad (x-4)(x^2 + 4x +16-15)= 0$$
and you just have to solve then
$$ x^2 +4x +1$$
to find the others solutions :
$$ x=-2\pm \sqrt{3}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2661654",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Find all $z$ s.t $|z|=1$ and $|z^2 + \overline{z}^2|=1$ Here's my attempt. Let $z=x+i\ y$, then $$z^2=x^2-y^2+i\ 2xy$$ and $$\bar z^2=x^2-y^2-i\ 2xy$$
Then, $$z^2+\bar z^2=2x^2-2y^2$$ so $$1=|z^2+\bar z^2|=\sqrt{(2x^2)^2+(-2y^2)^2}$$
Simpliflying the expression above, we get $$1=4x^4+4y^4$$
which gives us $$\frac14=x^4+y^4$$. I am stuck here. Is it wrong approach? is there an easier one?
| You can write $z=e^{it}=\cos t+i\sin t$. Then $z^2+\overline z^2=2\cos 2t$.
So we need to solve $\cos2t=\pm\frac12$. Thus $2t=\pm\frac\pi3+2k\pi$
or $2t=\pm\frac{2\pi}3+2k\pi$, that is $t=\pm\frac\pi6+k\pi$ or
$t=\pm\frac\pi3+k\pi$, so that $z=e^{it}=\frac12(\pm\sqrt3\pm i)$
or $\frac12(\pm1\pm i\sqrt3)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2662090",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 3
} |
Integral of a shifted Gaussian distribution with an error function In the course of computing a convolution of two functions, I have simplified it to a single variable integral of the form
$$\int_0^\infty xe^{-ax^2+bx}\mathrm{erf}(cx+d) dx$$
where $\mathrm{erf}$ is the error function defined as $$\ \mathrm{erf}(x) = \frac{2}{\sqrt{\pi}} \int_0^x e^{-t^2} dt.$$
I've looked through A Table of Integrals of the Error Functions, especially the formulas on pages 8 and 9. There are a lot of integrals that are similar, but I couldn't find a way to simplify this integral into something that was in that table. I also have tried differentiating under the integral sign using the constants $c$ and $d$ as parameters, but that only seemed to complicate the integration on the parameter after computing the integral with respect to $x$.
Is there any way to find a closed form of this integral?
| The result reads:
\begin{eqnarray}
&&\int\limits_0^\infty x e^{-a x^2-b x} \text{erf}(c x+d) dx= \frac{e^{\frac{b^2}{4 a}}}{4 a} \left( \right.\\
&& \frac{2 b \left(\arctan\left(\frac{\sqrt{a} (b+2 c d)}{2 a d-b c}\right)+\arctan\left(\frac{2 \sqrt{a} d}{b}\right)-\arctan\left(\frac{c}{\sqrt{a}}\right)\right)}{\sqrt{\pi } \sqrt{a}} +\\
&& 2 e^{-\frac{b^2}{4 a}} \text{erf}(d)-\frac{\sqrt{\pi } b \text{erf}\left(\frac{2 a d-b c}{2 \sqrt{a \left(a+c^2\right)}}\right)}{\sqrt{a}}+\frac{2 c e^{-\frac{(b c-2 a d)^2}{4 a \left(a+c^2\right)}}
\text{erfc}\left(\frac{b+2 c d}{2 \sqrt{a+c^2}}\right)}{\sqrt{a+c^2}} +\\
&& -\frac{4 \sqrt{\pi } b \left(T\left(\frac{2 a d-b c}{\sqrt{2} \sqrt{a \left(a+c^2\right)}},\frac{\sqrt{a} (b+2 c d)}{2 a d-b c}\right)+T\left(\frac{b}{\sqrt{2} \sqrt{a}},\frac{2 \sqrt{a}
d}{b}\right)\right)}{\sqrt{a}}\\
&&\left.\right)
\end{eqnarray}
where $T(\cdot,\cdot)$ is the Owen's t function https://en.wikipedia.org/wiki/Owen%27s_T_function .
For[count = 1, count <= 200, count++,
{a, b, c} = RandomReal[{0, 5}, 3, WorkingPrecision -> 50];
d = RandomReal[{-5, 5}, WorkingPrecision -> 50];
I1 = NIntegrate[x Exp[-a x^2 - b x] Erf[c x + d], {x, 0, Infinity},
WorkingPrecision -> 20];
1/a NIntegrate[
x Exp[-x^2 - b/Sqrt[a] x] Erf[c/Sqrt[a] x + d], {x, 0, Infinity}];
1/a Exp[
b^2/(4 a)] NIntegrate[(x - b/(2 Sqrt[a])) Exp[-x^2 ] Erf[
c/Sqrt[a] ( x - b/(2 Sqrt[a])) + d], {x, b/(2 Sqrt[a]),
Infinity}];
1/a Exp[
b^2/(4 a)] (NIntegrate[(x) Exp[-x^2 ] Erf[
c/Sqrt[a] ( x - b/(2 Sqrt[a])) + d], {x, b/(2 Sqrt[a]),
Infinity}] -
b Sqrt[Pi]/(2 Sqrt[a]) 2 T[b/Sqrt[2 a],
c/Sqrt[a], (-((b c)/(2 a)) + d) Sqrt[2]]);
I2 = E^(b^2/(4 a))/(
4 a) ((2 b)/(
Sqrt[a] Sqrt[\[Pi]]) (-ArcTan[c/Sqrt[a]] +
ArcTan[(2 Sqrt[a] d)/b] +
ArcTan[(Sqrt[a] (b + 2 c d))/(-b c + 2 a d)]) +
2 E^(-(b^2/(4 a))) Erf[d] - (b Sqrt[\[Pi]])/Sqrt[a]
Erf[(-b c + 2 a d)/(2 Sqrt[a (a + c^2)])] + (2 c)/Sqrt[
a + c^2] E^(-((b c - 2 a d)^2/(4 a (a + c^2))))
Erfc[(b + 2 c d)/(2 Sqrt[a + c^2])] - (4 b Sqrt[\[Pi]])/Sqrt[
a] (OwenT[b/(Sqrt[2] Sqrt[a]), (2 Sqrt[a] d)/b] +
OwenT[(-b c + 2 a d)/(Sqrt[2] Sqrt[a (a + c^2)]), (
Sqrt[a] (b + 2 c d))/(-b c + 2 a d)]));
If[Abs[I2/I1 - 1] > 10^(-3),
Print["results do not match", {a1, a2, c, d, {I1, I2}}]; Break[]];
If[Mod[count, 10] == 0, PrintTemporary[count]];
];
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2662259",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
How to find the number of elements in least common multiple problem? The problem is as follows:
Over a table there is a certain number of muffins, if we count them by
multiples of four the remainder is three. If we count them by
multiples of six the remainder is five but if we count them by
multiples of ten the remainder is nine. What is the least number of
muffins over the table?
What I tried to do is to build up the equations to get the supposed number as follows:
$\textrm{n is defined as the number of muffins}$
$4k+3=n$
$6k+5=n$
$10k+9=n$
Then i equated those to get the value of $k$.
$4k+3=6k+5$
$2k=-2$
$k=-1$
$10k+9=6k+5$
$4k=-4$
$k=-1$
$4k+3=10k+9$
$6k=-6$
$k=-1$
However in all cases I reach to $-1$ and it does not seem to be reasonable to use it in any of those equations. I'm lost in this one can somebody help me to address this situation the easiest way possible?.
| The reason you are getting a nonsensical answer is because there is no reason why the values of $k$ in your three equations should be the same. All you really know is that there are integers $a,b,c$ such that
$$\begin{align}
n &= 3 + 4a, \\
n &= 5 + 6b, \\
n &= 9 + 10c.
\end{align}$$
To solve this, you can iteratively substitute. From the first equation $n = 3 + 4a$, we see that
$$ 3 + 4a = 5 + 6b,$$
or rather that $2a = 1 + 3b = 4 + 3(b-1) = 4 + 3b'$ for some integer $b'$. Note that $b'$ must be even so that the right hand side is even (which must occur as the left hand side is even). So we can write $b' = 2b''$ for some integer $b''$.
Then from $2a = 4 + 6b''$, we have that $a = 2 + 3b''$. Substituting into $n = 3 + 4a$, we find that $n = 3 + 4(2 + 3b'') = 11 + 12b''$ for some integer $b''$. It's annoying to write $b''$, so I'll write $d$ instead of $b''$.
The point is that the information that
$$\begin{align}
n &= 3 + 4a \\
n &= 5 + 6b
\end{align}$$
is contained within the single equation $n = 11 + 12d$. Indeed, if $n = 11 + 12d$ for some integer $d$, then its remainder upon dividing by $4$ is $3$, and its remainder upon dividing by $6$ is $5$. We showed the converse above. Thus from two equations, we have one.
Now we repeat, this time with
$$\begin{align}
n &= 11 + 12 d \\
n &= 9 + 10 c.
\end{align}$$
Equating, we see that $9 + 10c = 11 + 12d$, or rather that $5c = 1 + 6d$. As above, we massage the right hand side a bit and write $5c = 25 + 6(d - 4) = 25 + 6d'$. In this form, we see that $d'$ must be divisible by $5$ to make the right hand side divisible by $5$ (as the left hand side is divisible by $5$). So we can write $d' = 5 d''$ for some integer $d''$. Thus $5c = 25 + 30d''$, or rather $c = 5 + 6d''$. Writing $d''$ is annoying, so I use $e$ instead.
Then we have that $c = 5 + 6e$. Substituting into $n = 9 + 10c$, we have that
$$ n = 9 + 10(5 + 6e) = 59 + 60e.$$
Notice that if $n = 59 + 60e$, then dividing by $4$ leaves remainder $3$, dividing by $6$ leaves remainder $5$, and dividing by $10$ leaves remainder $9$. And to get $n = 59 + 60e$, we just massaged the equations above.
Thus the set of integers that leave the three necessary remainders are exactly those integers of the form $59 + 60e$. The smallest (positive) integer of this form is $59$. (Note that $-1$ is also of this form, and you found $-1$ by essentially guessing a value for $e$ at the start).
So the answer is $59$.
Addendum: the computations I performed were elementary, but they can seem a bit uninspired. But I knew to do this because I was modifying a proof of the Chinese Remainder Theorem, which would apply naively if $\gcd(4,6,10) = 1$ (which it's not). The computations feel more natural when considered in congruences. To see this, let's redo the first pair reduction using congruences.
So $n = 3 + 4a$ is the same as $n \equiv 3 \mod 4$, and $n = 5 + 6b$ is the same as $n \equiv 5 \mod 6$. So we are trying to solve the simultaneous congruence
$$\begin{align}
n &\equiv 3 \mod 4, \\
n &\equiv 5 \mod 6.
\end{align}$$
To do this we write $n = 3 + 4a$ and substitute into the second congruence, getting
$$ 3 + 4a \equiv 5 \pmod 6 \implies 2a \equiv 1 \pmod 3 \implies a \equiv 2 \pmod 3.$$
And thus $a = 2 + 3b$ for some $b$, so that $n = 3 + 4(2 + 3b) = 11 + 12 b$ for some $b$. Or rather $a \equiv 11 \pmod {12}$.
This is exactly as above, except without the various variable shenanigans I employed previously. The rest of the argument works the same.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2662332",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Find $\lim_{x\to \infty} \left[f\!\left(\sqrt{\frac{2}{x}}\,\right)\right]^x$. Let $f:\mathbb{R} \to \mathbb{R}$ be such that $f''$ is continuous on $\mathbb{R}$ and $f(0)=1$ ,$f'(0)=0$ and $f''(0)=-1$ .
Then what is $\displaystyle\lim_{x\to \infty} \left[f\!\left(\sqrt{\frac{2}{x}}\,\right)\right]^x?$
When I was solving this problem, I supposed $f(x)$ to be a polynomial of degree two (because $f''$ is continuous) i.e. $f(x)=ax^2+bx+c$ and found coefficients with the help of given values . I got $f(x)=\frac{-x^2}{2} +1$. After solving , I found limit to be $e^{-1}$. I know this is a particular case.
Questions
$1$ : Will the limit be same for all functions with these properties ?
$2$ : Please give me some method which works for all such $f(x)$.
$3$: I want to practice more questions of this kind, please give me some references i.e. books, problem books, any online source.
Any kind of help will be highly appreciated. Thanks!
| \begin{align*}
&\lim_{x\rightarrow\infty}\log\left(f\left(\sqrt{\dfrac{2}{x}}\right)\right)^{x}\\
&=\lim_{x\rightarrow\infty}x\log\left(f\left(\sqrt{\dfrac{2}{x}}\right)\right)\\
&=\lim_{x\rightarrow\infty}\dfrac{\log\left(f\left(\sqrt{\dfrac{2}{x}}\right)\right)}{\dfrac{1}{x}}\\
&=\lim_{x\rightarrow\infty}\dfrac{\dfrac{1}{f\left(\sqrt{\dfrac{2}{x}}\right)}\cdot f'\left(\sqrt{\dfrac{2}{x}}\right)\cdot\dfrac{1}{\sqrt{\dfrac{2}{x}}}\cdot-\dfrac{1}{x^{2}}}{-\dfrac{1}{x^{2}}}\\
&=\lim_{x\rightarrow\infty}\dfrac{\dfrac{1}{f\left(\sqrt{\dfrac{2}{x}}\right)}\cdot f'\left(\sqrt{\dfrac{2}{x}}\right)}{\sqrt{\dfrac{2}{x}}}\\
&=\lim_{x\rightarrow\infty}\dfrac{f'\left(\sqrt{\dfrac{2}{x}}\right)}{f\left(\sqrt{\dfrac{2}{x}}\right)\sqrt{\dfrac{2}{x}}}\\
&=\lim_{x\rightarrow\infty}\dfrac{f''\left(\sqrt{\dfrac{2}{x}}\right)\cdot\dfrac{1}{\sqrt{\dfrac{2}{x}}}\cdot-\dfrac{1}{x^{2}}}{\dfrac{1}{\sqrt{\dfrac{2}{x}}}\cdot-\dfrac{1}{x^{2}}\cdot f\left(\sqrt{\dfrac{2}{x}}\right)+\sqrt{\dfrac{2}{x}}\cdot f'\left(\sqrt{\dfrac{2}{x}}\right)\cdot\dfrac{1}{\sqrt{\dfrac{2}{x}}}\cdot-\dfrac{1}{x^{2}}}\\
&=\dfrac{-1}{1+0}\\
&=-1.
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2663374",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 5,
"answer_id": 0
} |
Compute the limit $\lim_{x\to 0} \frac{3^x+2^x-2}{4^x+2^x-2}$
Compute the limit $\lim_{x\to 0} \frac{3^x+2^x-2}{4^x+2^x-2}$
Here is what I have done so far:
\begin{align}
\lim_{x\to 0} \frac{3^x+2^x-2}{4^x+2^x-2}
&=
\lim_{x\to 0} \left(\frac{3^x-1}{4^x+2^x-2}+\frac{2^x-1}{4^x+2^x-2}\right)\\
&=\lim_{x\to 0} \left(\frac{3^x-1}{x}\frac{x}{4^x+2^x-2}+\frac{2^x-1}{x}\frac{x}{4^x+2^x-2}\right)\\
&=\ln 3\lim_{x\to 0} \frac{x}{4^x+2^x-2}+\ln 2\lim_{x\to 0} \frac{x}{4^x+2^x-2}
\end{align}
| Just added for your curiosity since you already received good answers.
Since $$a^x=e^{x \log(a)}$$ using Taylor series around $x=0$
$$a^x=1+ \log (a)x+\frac{1}{2} \log ^2(a) x^2+O\left(x^3\right)$$ This makes
$$A=\frac{3^x+2^x-2}{4^x+2^x-2} =\frac{ (\log (2)+\log (3))x+\frac{1}{2} \left(\log ^2(2)+\log
^2(3)\right)x^2+O\left(x^3\right) } {(\log (2)+\log (4))x+\frac{1}{2} \left(\log ^2(2)+\log
^2(4)\right)x^2+O\left(x^3\right) }$$ Divide top and bottom by $x$ and make the long division to get, after simplifications,
$$A=\frac{\log (6)}{\log (8)}-\frac{ \log \left(\frac{4}{3}\right) \log
(54)}{18 \log (2)}x+O\left(x^2\right)$$ which shows the limit and how it is approached.
But you can use it for approximation. For illustration, use $x=0.1$. This would give $A\approx 0.852457$ while the exact value would be $A\approx 0.852248$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2664317",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Solving this system of equations without a CAS: $4=a+c$, $5=d+ac+b$, $4=ad+bc$, $4=bd$ Solving this system of equations :
$$4=a+c$$ $$5=d+ac+b$$ $$4=ad+bc$$ $$4=bd$$
Anyone knows how to solve this without a CAS?
| A particular solution is $(a,b,c,d)=(4,4,0,1)$. This becomes from the following:
$$\begin{cases}a+c=4\\ac+b+d=5\\ad+bc=4\\bd=4\end{cases}\Rightarrow\begin{cases}c(4-c)+b+d=5\\d(4-c)+bc=4\\bd=4\end{cases}\Rightarrow\begin{cases}4c-c^2+\dfrac 4d+d=5\\4d-cd+\frac{4c}{d}=4\end{cases}$$ The third system with two unknowns give as resultant $$d^6-5d^5+12d^4-40d^3+48d^2-80d+64=(d-1)(d-4)(d^2+4)^2=0$$ which gives the solution above and others whose calculation is left for the OP.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2664582",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Is the sequence $a_{n}= {{n}\choose{[n/2]}}^{1/n}$ increasing? Let the sequence $a_{n}= {{n}\choose{[n/2]}}$$^{1/n}$ for every $n\in\mathbb{N}$, where $[x]=\max\{k \in \mathbb{Z} | k\leq x\}$ the floor function. How can I show that this sequence is increasing? I have tried using logarithms and gamma functions (for derivative) but I didn't get far. Any ideas?
| You can use the identity $k\cdot \binom{m}{k} = m\cdot \binom{m-1}{k-1}$ and apply induction over n by showing $a_{n+2} \ge a_n$.
For $n=2 $ & $ n=3:$$$ a_1 \le a_2 \le a_3 \Leftrightarrow \binom{1}{0} \le \sqrt{\binom{2}{1}} \le \binom{3}{1}^{1/3} \Leftrightarrow 1 \le \sqrt{2} \le 3^{1/3}$$
Given that the assumption is proven for n, we now need to show $a_{n+2} \ge a_n$. First reshape $a_{n+2}$:
$$ a_{n+2} = \binom{n+2}{[\frac{n}{2}+2]}^{\frac{1}{n+2}} = \bigg(\frac{[n/2]+1}{[n/2]+1}\binom{n+2}{[\frac{n}{2}]+1}\bigg)^{\frac{1}{n+2}} = \bigg((n+2) \binom{n+1}{[\frac{n}{2}]}\bigg)^{\frac{1}{n+2}} =
\bigg((n+2)\frac{(n+1)}{(n+1) - [n/2])} \binom{n}{[\frac{n}{2}]}\bigg)^{\frac{1}{n+2}}
$$
Now check whether the inequality holds:
$$
a_{n+2} \ge a_n\\
\Leftrightarrow \bigg((n+2)\frac{(n+1)}{(n+1) - [n/2])} \binom{n}{[\frac{n}{2}]}\bigg)^{\frac{1}{n+2}} \ge \binom{n}{[n/2]}^\frac{1}{n}
\\\Leftrightarrow
\bigg(\bigg((n+2)\frac{(n+1)}{(n+1) - [n/2])} \binom{n}{[\frac{n}{2}]}\bigg)^{\frac{1}{n+2}} \bigg)^{n+2}
\ge
\bigg(\binom{n}{[n/2]}^\frac{1}{n}\bigg)^{n+2} \\
\Leftrightarrow
(n+2)\frac{(n+1)}{(n+1) - [n/2])} \binom{n}{[\frac{n}{2}]}
\ge
\binom{n}{[n/2]} \cdot \binom{n}{[n/2]}^\frac{2}{n} \\
\Leftrightarrow
(n+2)\frac{(n+1)}{(n+1) - [n/2])} \ge \binom{n}{[n/2]}^\frac{2}{n}
$$
In this last inequality, the right hand side is smaller than 4, because
$$
a_n \le \bigg(\sum_{i=0}^n \binom{n}{i}\bigg)^{1/n} = (2^n-1)^{1/n} \le (2^n)^{1/n} = 2
$$
And because in the left hand side in the inequality $(n+2) \ge 4$ for $n\ge 2$, and because the other factor is $\ge 1$, the statement is proofen.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2664807",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
What is the last nonzero digit at the end of 10!? What is the last nonzero digit at the end of $10!$ ? What is the last nonzero digit at the end
of $100!$ ? What is the last nonzero digit at the end of $1,000,000!$?
Is there a formula to find out the pattern?
| For $10!$:
The last non-zero digit is formed as the product of the digits in the unit places (note that $2\cdot 5$ has to be left out):
$$3 \cdot 4 \cdot 6 \cdot 7 \cdot 8 \cdot 9 \equiv 8 \mbox{ mod } 10 \rightarrow \mbox{last non-zero digit: }8$$
For $100!$:
Here the last non-zero digit is formed by $10$ blocks of $(3 \cdot 4 \cdot 6 \cdot 7 \cdot 8 \cdot 9)$:
$$(3 \cdot 4 \cdot 6 \cdot 7 \cdot 8 \cdot 9)^{10} \equiv 8^{10} \equiv 4 \mbox{ mod } 10 \rightarrow \mbox{last non-zero digit: }4$$
For $1000000!$:
Here you have $10^5$ blocks of last digits $(3 \cdot 4 \cdot 6 \cdot 7 \cdot 8 \cdot 9)$:
$$(3 \cdot 4 \cdot 6 \cdot 7 \cdot 8 \cdot 9)^{10^5} \equiv 8^{10 \cdot 10 \cdot 10 \cdot 10 \cdot 10 } \equiv 4^{10 \cdot 10 \cdot 10 \cdot 10} \equiv \ldots \equiv 6 \mbox{ mod } 10 \rightarrow \mbox{last non-zero digit: }6$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2665033",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
integral $ \int_0^{\frac{\pi}{3}}\mathrm{ln}\left(\frac{\mathrm{sin}(x)}{\mathrm{sin}(x+\frac{\pi}{3})}\right)\ \mathrm{d}x$ We want to evaluate
$ \displaystyle \int_0^{\frac{\pi}{3}}\mathrm{ln}\left(\frac{\mathrm{sin}(x)}{\mathrm{sin}(x+\frac{\pi}{3})}\right)\ \mathrm{d}x$.
We tried contour integration which was not helpful. Then we started trying to use symmetries, this also doesn't work.
| Here is an approach that uses real methods only.
\begin{align*}
I &= \int_0^{\pi/3} \ln \left (\frac{\sin x}{\sin \left (x + \frac{\pi}{3} \right )} \right ) \, dx\\
&= \int_0^{\pi/3} \ln (\sin x) \, dx - \int_0^{\pi/3} \ln \left (\sin \left (x + \frac{\pi}{3} \right ) \right ) \, dx.
\end{align*}
If in the second integral appearing on the right a substitution of $x \mapsto x - \frac{\pi}{3}$ is enforced, one has
\begin{align*}
I &= \int_0^{\pi/3} \ln (\sin x) \, dx - \int_{\pi/3}^{2\pi/3} \ln (\sin x) \, dx\\
&= 2 \int_0^{\pi/3} \ln (\sin x) \, dx - \int_0^{2\pi/3} \ln (\sin x) \, dx.\tag1
\end{align*}
Now consider the integral
$$I(\alpha) = \int_0^\alpha \ln (\sin x) \, dx, \quad 0 < \alpha < \pi.\tag2$$
Taking advantage of the well-known identity
$$\ln (\sin x) = -\ln 2 - \sum_{k = 1}^\infty \frac{\cos (2kx)}{k}, \quad 0 < x < \pi,$$
substituting this result into (2), after interchanging the summation with the integration before integrating one finds
$$I(\alpha) = -\alpha \ln 2 - \frac{1}{2} \sum_{k = 1}^\infty \frac{\sin (2k \alpha)}{k^2} = -\alpha \ln 2 - \frac{1}{2} \text{Cl}_2 (\alpha).$$
Here $\text{Cl}_2 (\varphi)$ denotes the Clausen function of order two.
In terms of the Clausen function of order two the integral in (1) can be written as
$$I = -\text{Cl}_2 \left (\frac{2\pi}{3} \right ) + \frac{1}{2} \text{Cl}_2 \left (\frac{4\pi}{3} \right ).\tag3$$
From the duplication formula for the Clausen function of order two, namely
$$\text{Cl}_2 (2\theta) = 2 \text{Cl}_2 (\theta) - 2 \text{Cl}_2 (\pi - \theta), \quad 0 < \theta < \pi,$$
if we set $\theta = 2\pi/3$ in the above duplication formula, as
$$\text{Cl}_2 \left (\frac{4\pi}{3} \right ) = 2 \text{Cl}_2 \left (\frac{2\pi}{3} \right ) - 2 \text{Cl}_2 \left (\frac{\pi}{3} \right ),$$
the expression for our integral in (3) can be expressed more simply as
$$\int_0^{\pi/3} \ln \left (\frac{\sin x}{\sin \left (x + \frac{\pi}{3} \right )} \right ) \, dx = -\text{Cl}_2 \left (\frac{\pi}{3} \right ).$$
| {
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"source": "stackexchange",
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Evaluating a Difference Operator of a Product of Factorial Polynomials I'm looking for a verification on a Difference Operator question.
Evaluate the following difference; $$\Delta x^{(4)}x^{(-3)}$$
I actually did it a couple of different ways, but as there is no answer key, I would just like to know that it is correct. For the problem, we need that
$$x^{(n)}=x(x-1)...(x-n+1)$$
$$x^{(-n)}=\frac{1}{(x+n)^{(n)}}$$
and
$$\Delta{f(x)}=f(x+1)-f(x)$$
Now, for the solution;
$$\begin{eqnarray*}\Delta x^{(4)}x^{(-3)}&=&(x+1)^{(4)}(x+1)^{(-3)}-x^{(4)}x^{(-3)}\\&=&\frac{(x+1)^{(4)}}{(x+4)^{(3)}}-\frac{x^{(4)}}{(x+3)^{(3)}}\\&=&\frac{(x+1)^{(4)}(x+1)-x^{(4)}(x+4)}{(x+4)^{(4)}}\\&=&\frac{x^{(3)}(x+1)(x+1)-x^{(3)}(x+4)(x-3)}{(x+4)^{(4)}}&=&\frac{x^{(3)}[(x^2+2x+1)-(x^2+x-12)]}{(x+4)^{(4)}}\\&=&x^{(3)}x^{(-4)}(x+13)
\end{eqnarray*}$$
While I got the same answer using a Difference operator product identity
$$\Delta[f(x)g(x)]=g(x)\Delta f(x)+f(x+1)\Delta g(x)$$
the final factor seems out of place, so I just want to verify its existence.
| Everything looks fine.
We obtain on the one hand
\begin{align*}
\color{blue}{\Delta\left(x^{(m)}x^{(-n)}\right)}&=(x+1)^{(m)}(x+1)^{(-n)}-x^{(m)}x^{(-n)}\\
&=\frac{(x+1)^{(m)}}{(x+n+1)^{(n)}}-\frac{x^{(m)}}{(x+n)^{(n)}}\\
&=\frac{(x+1)x^{(m-1)}(x+1)}{(x+n+1)^{(n+1)}}-\frac{x^{(m-1)}(x-m+1)(x+n+1)}{(x+n+1)^{(n+1)}}\\
&=x^{(m-1)}x^{(-n-1)}\left[(x+1)^2-(x+1-m)(x+1+n)\right]\\
&\,\,\color{blue}{=x^{(m-1)}x^{(-n-1)}\left[(m-n)x+mn+m-n\right]}
\end{align*}
$$ $$
On the other hand we obtain
\begin{align*}
\color{blue}{x^{(-n)}}&\color{blue}{\Delta x^{(m)}+(x+1)^{(m)}\Delta x^{(-n)}}\\
&=x^{(-n)}\left[(x+1)^{(m)}-x^{(m)}\right]+(x+1)^{(m)}\left[(x+1)^{(-n)}-x^{(-n)}\right]\\
&=\frac{(x+1)^{(m)}-x^{(m)}}{(x+n)^{(n)}}+(x+1)^{(m)}\left[\frac{1}{(x+n+1)^{(n)}}-\frac{1}{(x+n)^{(n)}}\right]\\
&=\frac{x^{(m-1)}}{(x+n+1)^{(n+1)}}\left[(x+1)-(x-m+1)\right](x+n+1)\\
&\qquad+\frac{x^{(m-1)}}{(x+n+1)^{(n+1)}}(x+1)\left[(x+1)-(x+n+1)\right]\\
&=x^{(m-1)}x^{(-n-1)}m\left(x+n+1\right)-x^{(m-1)}x^{(-n-1)}n(x+1)\\
&\,\,\color{blue}{=x^{(m-1)}x^{(-n-1)}\left[(m-n)x+mn+m-n\right]}
\end{align*}
Both equality chains coincide and putting $m=4$ and $n=3$ we finally get
\begin{align*}
\Delta\left(x^{(4)}x^{(-3)}\right)=x^{(3)}x^{(-4)}\left(x+13\right)
\end{align*}
in accordance with OP's result.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to determine $ \int \sec^{\frac{1}{3}}(x) dx $? How to determine $$ \int \sec^{\frac{1}{3}}(x) dx ?$$
An attempt:
$$ \int \frac{1}{\cos^{1/3}(x)}dx = \int \frac{\sin^{2}(x) + \cos^{2}(x)}{\cos^{1/3}(x)}dx$$
$$ = \int \frac{\sin^{2}(x)}{\cos^{1/3}(x)}dx + \int \cos^{2/3}(x) \cos(x) dx $$
Let $ A=\int \frac{\sin^{2}(x)}{\cos^{1/3}(x)}dx $, we focus on $\int \cos^{2/3}(x) \cos(x) dx$.
$$\int \underbrace{\cos^{2/3}(x)}_{u} \underbrace{\cos(x) dx}_{dv} = \sin(x) \cos^{2/3}(x) + (2/3) \underbrace{\int \frac{\sin^{2}(x)}{\cos^{1/3}(x)}dx }_{A} $$
We encounter $A$ once more.
How to continue? Or is there a better approach? Thanks.
The hint says use Trig. substitution two times
| Your question is equivalent to finding a primitive for $\sin(x)^{-1/3}$, or to finding a primitive for $\frac{x^{-1/3}}{\sqrt{1-x^2}}$. The latter is of course a hypergeometric function related to elliptic integrals:
$$ \int_{0}^{z}\frac{x^{-1/3}}{\sqrt{1-x^2}}\,dx = z^{2/3}\sum_{n\geq 0}\frac{3\binom{2n}{n}}{(6n+2)4^n}z^{2n}=\frac{3}{2}z^{2/3}\cdot{}_2 F_1\left(\tfrac{1}{3},\tfrac{1}{2};\tfrac{4}{3};z^2\right).$$
Some special values can be found through Euler's Beta function. For instance:
$$\begin{eqnarray*} \int_{0}^{\pi/2}\cos(x)^{-1/3}\,dx &=& \int_{0}^{\pi/2}\sin(x)^{-1/3}\,dx = \int_{0}^{1}\frac{dz}{z^{1/3}\sqrt{1-z^2}}=\frac{1}{2}\int_{0}^{1}u^{-2/3}(1-u)^{-1/2}\,du\\&=&\frac{1}{2}B\left(\tfrac{1}{3},\tfrac{1}{2}\right)=\frac{\Gamma\left(\tfrac{1}{3}\right)\,\Gamma\left(\tfrac{1}{2}\right)}{2\,\Gamma\left(\tfrac{5}{6}\right)}=\sqrt{\frac{\Gamma\left(\tfrac{1}{6}\right)^3}{2^{11/3}\sqrt{3\pi}}}=\frac{2^{1/2}3^{1/4}\pi}{\text{AGM}(1+\sqrt{3},\sqrt{8})}.\end{eqnarray*} $$
The representation through the AGM mean gives this tight inequality, for instance:
$$ \int_{0}^{\pi/2}\cos(x)^{-1/3}\,dx \leq \pi\sqrt{\frac{\sqrt{3}}{\sqrt{2}+\sqrt{6}}}.$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Is a proof using modular arithmetic in a question like this valid? It's been two years or so since I've finished my math undergrad (and I'm doing something non-math related now, unfortunately), so I apologize if what is to follow isn't a very good question!
Prove that for all Integers $n$, $n(n + 1)(2n + 1)$ will always be divisible by 6.
I can do that using induction, but I wanted to try a different way. Does it work to use modular arithmetic in the following way?
Let $f(n) = n(n+1)(2n+1) = 2n^3 + 3n^2 + n$. All we need to show is that $f(n)$ is divisible by both $2$ and $3$ for any choice of $n$.
Evaluate mod $2$.
$f(n) = n(n+1)(2n+1) = n(n+1)(0 + 1)$ mod $2 = n(n+1)$ mod $2$. Two consecutive numbers; one of them must be even, and so $f(n)$ is divisible by $2$.
Evaluate mod $3$
There are three possible residues for n modulo $3$: $0, 1,$ or $2$.
If the residue is $0$, then $f(n)$ is divisible by $3$.
If the residue is $1$, then $f(n) = n(n+1)(2n+1) = 1(1+1)(2*1+1) = 1(2)(3) = 0$ mod $3$.
If the residue is $2$, then $f(n) = 2(2+1)(2*2+1) = 2(3)(2) = 0$ mod $3$.
In any case, $f(n)$ is divisible by $3$.
Since $f(n)$ is divisible by $2$ and by $3$, it is divisible by $6$. The result follows.
Thank you!
| $$n(n+1)(2n+1)=n(n+1)(2n+4-3)=2\underbrace{n(n+1)(n+2)}-3n(n+1)$$
Use The product of n consecutive integers is divisible by n factorial
| {
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Finding the non singular matrix. $(1)$ Suppose $A$ and $B$ satisfy the relation $B^2+AB+2I=0$. Are the following four matrices necessarily non singular?
(a)$A$ ( b)$B$ (c) $A+2I$ (d) $B+2I$
My attempt:- $$B^2+AB+2I=0$$
$$\Rightarrow B^2+AB= -2I$$
$$\Rightarrow B(A+B)= -2I$$
$$\Rightarrow |B(A+B)|= |-2I|$$
$$\Rightarrow |B||A+B|= (-2)^3|I|=-8$$
So, $$|B|\not= 0$$
Reqd. Answer is $B$.
Am I Correct?
| You are correct that $B$ has to be invertible. Let's construct an example which shows the matrices $A$, $A + 2I$ and $B + 2I$ can be singular.
$B$ is invertible so we can express $A$ from $B^2 + AB + 2I = 0$ as $$A = B + 2B^{-1}$$
Using the Spectral theorem, we conclude that the eigenvalues of $A$ are of the form $x + \frac2x$, where $x \in \sigma(B)$.
$A$ is singular if and only if $0\in \sigma(A)$ so $$0 = x + \frac2x \iff x = \pm i\sqrt{2}$$
Meaning $\pm i\sqrt{2} \in \sigma (B)$.
On the other hand, $A + 2I$ is singular if and only if $-2 \in \sigma(A)$, which happens when
$$-2 = x + \frac2x \iff x = -1 \pm i$$
So it has to be $-1 \pm i \in \sigma(B)$.
Now, setting $B = \begin{pmatrix} \sqrt{2}i & 0 & 0\\ 0 & -2 & 0\\0 & 0 & -1 + i \end{pmatrix}$ yields $$A = B + 2B^{-1} = \begin{pmatrix} 0 & 0 & 0\\ 0 & -3 & 0\\0 & 0 & -2 \end{pmatrix}$$
so $A, A + 2I$ and $B + 2I$ are all singular.
Therefore, $(b)$ is the only correct option.
Interestingly, the above discussion also shows that $A, A + 2I$ and $B + 2I$ cannot all simultaneously be singular if the matrices are $2 \times 2$ since then $\sigma(B)$ can have at most two elements.
| {
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"timestamp": "2023-03-29T00:00:00",
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When two digit numbers in base $5$ are multiplied the result is $4103_5$. What are the numbers in base $5$? When two digit numbers in base $5$ are multiplied the result is $4103_5$. What are the numbers in base $5$?
Well given by two digit numbers in base $5$ I tried out the multiplication and tried to simplify.
$(ab_5)(cd_5)=4103_5$ or $(5a+b)(5c+d)=4*5^3+1*5^2+0*5^1+3$
Then get that $5^2ac+5ad+5cb+bd=5^3*4+5^2+3$
I notice that $bd=3$ so b=1,3 and $d=1,3$. Now this is where I'm stuck I don't see a clear way to get what the numbers should be without a lot of trial and error.
| So $n_1 = 5a + b; n_2 = 5c + d$ and $(5a+b)(5c+d) = 4103_5 = 4*5^3 + 1*5^2 + 3$
$(5a + b)(5c+d) = ac*5^2 + 5(bc + ad) + bd = 4*5^3 + 1*5^2 + 3 = 21*5^2 + 0 + 3$
So $ac \approx 4*5$ which is a very large value considering $a,c \le 4$.
Note $40_5*100_5 = 4000_5 < 4103_5$ so as $(5a+c) < 100_5$ we now $(5b+d) > 40_5$ and vice versa. So $a = c = 4$.
$16*5^2 + 5(4b + 4d) + db = 4*5^3 + 1*5^2 + 3 = 21*5^2 +3$
$20(b+d) + db = 5^3 + 3$
So $20(b+d) \approx 125$ so $b + d \approx 6$.
$bd \equiv 3\mod 5$ and $b,d \le 4$ so $bd \le 13$. so $bd=3,8,13$.
$13$ is prime so that is out. $bd = 3\implies b,d = 1,3$ and $b + d = 4\not \approx 6$. $bd = 8\implies b,d =2,4$ and $b + d = 6$.
So if an such number exist they are $42_5$ and $44_5$.
And $(4*5 + 2)(4*5+4) = 16*5^2 + 5(2*4 + 4*4) + 8 =$
$3*5^3 + 5^2 + 5(24) + 5 + 3 = $
$3*5^3 + 5^2 + 5(25 - 1) + 5 + 3=$
$3*5^3 + 5^2 + 5^3 +3 =$
$4*5^3 + 5^2 + 3 = 4103_5$.
And $42_5*44_5 = 4103_5$ or in base 10 $22*24 = 528$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Roots of $x^3+5x-18=0$ using Cardano's method Given that $x^3+5x-18=0$. We have to solve it using Cardano's method.
Using trial $x=2$ will be a root. Dividing the equation by $x-2$ we shall get the other quadratic equation and solving that one, we shall obtain all the roots.
But when I am trying to solve the equation using Cardan\rq s method, the calculation is becoming very difficult. I don\rq t know why. Please help.
Here is how did I proceed.
Let $x=u+v$. Then $x^3=u^3+v^3+3uvx$ i.e. $x^3-3uvx-(u^3+v^3)=0$. So $-3uv=5$ and $u^3+v^3=18$. Clearly $u^3, v^3$ are the roots of
\begin{align}
&t^2-(u^3+v^3)t+(uv)^3=0\\
\Rightarrow &t^2-18t-\frac{125}{27}=0\\
\Rightarrow &27t^2-(27\times 18)t-125=0
\end{align}
and from here when we are getting the roots of $t$, they are very complicated. Hence I do not know how to simplify them so that $x=2$ finally be achieved along with the other two roots.
Please help me
| Cardono's formula says that one root is$$\sqrt[3]{9+\sqrt{81+\frac{125}{27}}}+\sqrt[3]{9-\sqrt{81+\frac{125}{27}}}=\sqrt[3]{9+\frac{34}3\sqrt{\frac23}}+\sqrt[3]{9-\frac{34}3\sqrt{\frac23}}.$$But$$\sqrt[3]{9+\frac{34}3\sqrt{\frac23}}=1+2\sqrt{\frac23}\text{ and }\sqrt[3]{9-\frac{34}3\sqrt{\frac23}}=1-2\sqrt{\frac23}.$$Adding these two numbers, you'll get $2$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Showing $x^4-x^3+x^2-x+1>\frac{1}{2}$ for all $x \in \mathbb R$
Show that
$$x^4-x^3+x^2-x+1>\frac{1}{2}. \quad \forall x \in \mathbb{R}$$
Let $x \in \mathbb{R}$,
\begin{align*}
&\mathrel{\phantom{=}}x^4-x^3+x^2-x+1-\frac{1}{2}=x^4-x^3+x^2-x+\dfrac{1}{2}\\
&=x^2(x^2-x)+(x^2-x)+\dfrac{1}{2}=(x^2-x)(x^2+1)+\dfrac{1}{2}\\
&=x(x-1)(x^2+1)+\dfrac{1}{2}.
\end{align*}
Is there any way to solve this question?
| Factorise and complete the square of the first bit to understand function:
$$\begin{align*}
x^4-x^3+x^2-x+0.5&=x^4+x^2-(x^3+x)+\frac12\\
&=x(x^2+1)(x-1)+\frac12\\
&=(x^2+1)(x^2-1)+\frac12\\
&=(x^2+1)\left[(x-\frac12)^2-\frac14\right]+\frac12
\end{align*}$$
Then $(x^2+1)\left[(x-\frac12)^2-\frac14\right]\geq-\frac14$, and $\frac12-\frac14>0$ as required.
I will leave you to work out why $(x^2+1)\left[(x-\frac12)^2-\frac14\right]\geq-\frac14$ holds - it just requires a little thought :-)
| {
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Evaluate the limit $\lim_{n\to \infty}\sqrt{n^2 +2} - \sqrt{n^2 +1}$ I know that
$$\lim_{n\to \infty}(\sqrt{n^2+2} - \sqrt{n^2+1})=0.$$
But how can I prove this?
I only know that $(n^2+2)^{0.5} - \sqrt{n^2}$ is smaller than $\sqrt{n^2+2} - \sqrt{n^2}$ = $\sqrt{n^2+2} - n$.
Edit:
Thank Y'all for the nice and fast answers!
| Hint: $\displaystyle \sqrt{n^2+2}-\sqrt{n^2+1}=\frac{1}{\sqrt{n^2+2}+\sqrt{n^2+1}}$
| {
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Find the value of a 5th-root expression. Simplify and find the value of the expression:
$$\sqrt[5]{\frac{123+\sqrt{15125}}{2}}+\sqrt[5]{\frac{123-\sqrt{15125}}{2}}.$$
I tried to rationalise it.
It was of no use..
| Let
$$\sqrt[5]{\frac{123+\sqrt{15125}}{2}}+\sqrt[5]{\frac{123-\sqrt{15125}}{2}}=x$$
Raise to the power of 5 on both sides,
$$(\sqrt[5]{\frac{123+\sqrt{15125}}{2}}+\sqrt[5]{\frac{123-\sqrt{15125}}{2}})^5=x^5$$
Use the identity:
$$(a+b)^5=a^5+b^5+5ab[(a+b)^3-ab(a+b)]$$
We know that $(a+b)=x$.
Simlifying $a×b$ that is
$\sqrt[5]{\frac{123+\sqrt{15125}}{2}}×\sqrt[5]{\frac{123-\sqrt{15125}}{2}},$ we get $1$.
By substituting, the expression simplifies to the polynomial $$x^5=123+5[x^3-x]$$
$$x^5-5x^3+5x-123=0$$
Solving the degree 5 polynomial, we get
$$x=3$$..$Ans..$
| {
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how to find the equation of a tangent line to a circle, given its slope and the eq. of the circle? So I have a circle: $(x-2)^2 + (y-2)^2 = 25$ and I have a tangent line to this circle, with a slope of $m= -3/4$.
I have to find the equation of the tangent line, so I know the radius of the circle is $r = 5$ and I wrote the equation of the tangent line as:
$$y = -3/4x + h$$
So now I have to find $y, x$ and $h$, but I don't know if I can just replace $x$ and $y$ with the center points? Or do I have to find the point-line distance (and why?)
| Here is the derivation for the equation of the line with slope $m$ that is tangent to a circle of radius $r$ centered on the origin. (If the center of the circle was not centered on the origin, we can simply shift the equation we found.)
Let the required tangent line be $y=mx+b$ and the point of tangency be $(x_0, y_0)$. We need to find $b$ in terms of $m$ and $r$.
The slope of the tangent line $m$ must equal to the slope of the circle at $(x_0, y_0)$. So let's find the equation for slope of the circle at $(x_0, y_0)$.
The equation of the circle is
$$x^2 + y^2 = r^2.$$
Implicitly differentiating, we have
$$2x + 2y \frac{dy}{dx} = 0.$$
Rearranging we have
$$\frac{dy}{dx} = -\frac{x}{y}.$$
So we have $$m = -\frac{x_0}{y_0}.$$
Therefore, we have these three equations:
*
*$y_0 = m x_0 + b$
*$m = -\frac{x_0}{y_0}$
*${x_0}^2 + {y_0}^2 = r^2$
We need to find $b$ in terms of $m$ and $r$.
Solving the second equation for $x_0$ we have $x_0 = -m y_0$. Substituting this to the first equation we have
$$b = y_0 - m x_0 = y_0 - m (-m y_0) = y_0 (1+m)^2.$$
Substituting $x_0 = -m y_0$ to the third equation we have
$$m^2 {y_0}^2 + {y_0}^2 = r^2.$$
Solving for ${y_0}^2$ we have
$$y_0^2 = \frac{r^2}{m^2+1}.$$
Taking the square root we get
$$y_0 = \pm \frac{r}{\sqrt{m^2+1}}.$$
Therefore
$$b = \pm \frac{r}{\sqrt{m^2+1}} \cdot (1+m^2) = \pm r \sqrt{m^2+1}.$$
Note that we have two values for $b$, so there are actually two tangent lines with the same slope.
Hence, the tangent lines with slope $m$ for a circle of radius $r$ centered on the origin are given by
$$y = mx \pm r \sqrt{m^2+1}.$$
Finally, if the center of the circle was instead $(h, k)$, the equation of the tangent lines are
$$y-k = m(x-h) \pm r \sqrt{m^2+1}.$$
In your case, $(h, k) = (2,2)$, $r=5$, and $m = -3/4$. So the tangent lines are
$$y-2 = -\frac{3}{4}(x-2) \pm 5\sqrt{\left(-\frac{3}{4}\right)^2 + 1}.$$
Which can be simplified to
$$y = -\frac{3}{4} x + \frac{39}{4}$$
and
$$y = -\frac{3}{4} x - \frac{11}{4}.$$
| {
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A set of matrix with 2 properties Let $A \subset M_2(R)$ a set with the properties:
1)If $X,Y \in A$, then $X+Y \in A$;
2)If $X \in M_2(R)$ with $tr(X^tX)=1$, then $X \in A$.
Prove that $A=M_2(R)$.
I wrote that for a matrix $X \in M_2(R)$ with the elements $a,b,c,d$ we have by the second propertie that if $a^2+b^2+c^2+d^2=1$, then $X \in A$. Using this, for a random matrix $X$ with the elements $a,b,c,d$, we that $\frac{1}{\sqrt{a^2+b^2+c^2+d^2}}X \in A$ and if $\sqrt{a^2+b^2+c^2+d^2}$ is an integer, then $X\in A$. I don't know how to continue. Please help!
| The matrices $E_{i, j}, (i,j)\in \{1,2\}^2,$ such that $$e_{k,l} = \begin{cases} 1, & \text{ if } (k,l) = (i,j) \\0,& \text{ otherwise} \end{cases}$$ are in $A$.
Now let's show that $\pmatrix{\lambda & 0 \\ 0 & 0} \in A$. Since $\lambda = \lfloor \lambda \rfloor + (\lambda - \lfloor \lambda \rfloor$) we can assume $\lambda \in (0,1)$.
So we're looking for $X =\pmatrix{a & b \\ c & d}$ and $Y = \pmatrix{\lambda - a & -b \\ -c & -d}$ such that $$a^2+b^2+c^2+d^2= (\lambda-a)^2+b^2+c^2+d^2 = 1$$
so that $X, Y \in A$, implying $X+Y = \pmatrix{\lambda & 0 \\ 0 & 0} \in A$.
From the equation above, we get $a = \frac{\lambda}{2}$ ; now let's set $c, d=0$ and $b= \frac{\sqrt{4-\lambda^2}}{2}$, and we're done.
Likewise $\lambda E_{i,j} \in A$ for all $(i,j)$. Since $\pmatrix{a & b \\ c & d} = aE_{1,1} + bE_{1,2} + cE_{2,1} + dE_{2,2}$, eventually $A = \mathcal{M}_2(\Bbb R)$.
| {
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$\sum_{i=0}^n \frac{1}{(i+3)(i+4)} = \frac{n}{4(n+4)}$ (prove by induction) I'm having some difficulty proving by induction the following statement.
$$\sum_{i=0}^n \frac{1}{(i+3)(i+4)} = \frac{n}{4(n+4)}$$
I have shown that $\sum_{i=0}^n \frac{1}{(i+3)(i+4)} = \frac{n}{4(n+4)}$ holds for $n=1$ (equals $\frac{1}{20}$) , but I am getting stuck on the induction step.
As far as I know I have to show $$\sum_{i=0}^n \frac{1}{(i+3)(i+4)} = \frac{n}{4(n+4)}$$
implies
$$\sum_{i=0}^{n+1} \frac{1}{(i+3)(i+4)} = \frac{n+1}{4(n+5)}$$
To do this I think I should add the number $\frac{1}{(n+4)(n+5)}$ to $\frac{n}{4(n+4)}$ and see if it gives $\frac{n+1}{4(n+5)}$ , if I am not mistaken.
When trying to do that however I get stuck. I have:
$$\frac{n}{4(n+4)} +\frac{1}{(n+4)(n+5)} = \frac{n(n+4)(n+5)}{4(n+4)^2(n+5)} + \frac{4(n+4)}{4(n+4)^2(n+5)} = \frac{n(n+4)(n+5)+4(n+4)}{4(n+4)^2(n+5)} = \frac{n(n+5)+4}{4(n+4)(n+5)}$$
However beyond this point I don't know how to reach $\frac{n+1}{4(n+5)}$ I always just end up at the starting point of that calculation.
So I think that either my approach must be wrong or I am missing some trick how to simplify $$\frac{(n(n+5)+4}{4(n+4)(n+5)}$$
I would be very grateful for any help, as this is a task on a preparation sheet for the next exam and I don't know anyone, that has a correct solution.
| $$\frac{n}{4(n+4)} +\frac{1}{(n+4)(n+5)} = \frac{n(n+4)(n+5)}{4(n+4)^2(n+5)} + \frac{4(n+4)}{4(n+4)^2(n+5)} = \frac{n(n+4)(n+5)+4(n+4)}{4(n+4)^2(n+5)} = \frac{(n(n+5)+4}{4(n+4)(n+5)}$$
Last expression can be simplified by $n+4$
$$\frac{n(n+5)+4}{4(n+4)(n+5)}=\frac{(n+1)}{4(n+5)}=\frac{(n+1)}{4((n+1)+4)}$$
| {
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Find the sum to $n$ terms of the series: $1^2.1+2^2.3+3^2.5+....$ Find the sum to $n$ terms of the series:
$$1^2.1+2^2.3+3^2.5+.....$$
My Attempt:
Here, $n^{th}$ term of $1,2,3,....=n$
$n^{th}$ term of $1^2,2^2,3^2,....=n^2$
Also, $n^{th}$ term of $1,3,5,....=2n-1$
Hence, $n^{th}$ term of the given series is $t_n=n^2(2n-1)$
| Hint:
$$\sum_{i=1}^{n}(2i^3-i^2)=2\sum_{i=1}^{n}i^3-\sum_{i=1}^{n}i^2$$
$$\sum_{i=1}^{n}i^3=\frac{n^2(n+1)^2}{4}$$$$\sum_{i=1}^{n}i^2=\frac{n(n+1)(2n+1)}{6}$$
| {
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Given that $P(x)$ is a polynomial such that $P(x^2+1) = x^4+5x^2+3$, what is $P(x^2-1)$? How would I go about solving this? I can't find a clear relation between $x^2+1$ and $x^4+5x^2+3$ to solve $P(x^2-1)$.
| Here is essentially
division by $x^2+1$
that is just
plug and grind.
$\begin{array}\\
x^4+5x^2+3
&=x^4+x^2+4x^2+3\\
&=x^2(x^2+1)+4x^2+4-1\\
&=x^2(x^2+1)+4(x^2+1)-1\\
&=(x^2+4)(x^2+1)-1\\
&=(x^2+1+3)(x^2+1)-1\\
&=(x^2+1)^2+3(x^2+1)-1\\
\end{array}
$
so,
as
José Carlos Santos got,
$P(x)=x^2+3x-1$.
Then
$P(x^2-1)
=(x^2-1)^2+3(x^2-1)-1
=x^4-2x^2+1+3x^2-3-1
=x^4+x^2-3
$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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proving that $n^n \le (n!)^2$ I want to prove that $n^n \le (n!)^2$.
Now I tried by induction:
for $n=1$ ,$1=1$ and $P(1)$ is true
I suppose that $P(n)$ is true and I have to demonstrate that $P(n+1)$ is true
$$((n+1)!)^2=(n+1)^2*(n!)^2 \le (n+1)^2*n^n=(n+1)(n+1)*n^n=(n+1)^n$$
But I'm not sure about the last passage
| Notice: $(n!) = (1*2*...... *n)^2 = (1*2*...... *n)*(1*2*...... *n)=$
$(1*n) * (2*(n-1)) * ....... * (n*1)$.
If we can prove that $k*(n-k+1) \ge n$ we will have
$(1*n) * (2*(n-1)) * ....... * (n*1) \ge n*n*n*....* = n^n$, and that will be it.
We can probably prove $k*(n-k+1) \ge n$ by induction but I'd rather note:
$k \le n$ and $k\ge 1$ so $k(n-k + 1) = kn - k^2 + k = n + (k-1)n -k^2 + k \ge n + (k-1)k - k^2 +k = n$.
.....
Intuitively this is also a result of AM-GM. $k$ and $n-k + 1$ average to $\frac {n+1}2$ so if $m = \frac {n+1}2$ and $k = m - d$ and $n-k+1 = m + d$ we have $k(n-k + 1) = (m - d)(m+d) = m^2 - d^2 \ge m^2 = \frac {(n+1)^2}4$. If $n\ge 3$ then $\frac {(n+1)^2}4 \ge \frac {(n+1)^2}{n+1} = n+ 1> n$. And for $n=1,2,3$ then $\frac {(n+1)^2}4 = 1, \frac 94, 4 \ge n$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2684033",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Is there a direct proof of the inequality $cyclic \sum\frac {a^2+1}{b+c}\ge 3$ Let $a,b,c>0$ reals.
Prove that
$$cyclic \sum\frac {a^2+1}{b+c}\ge 3$$
I proved it using Nesbitt inequality
$$cyclic \sum \frac {a}{b+c} \ge \frac {3}{2} $$
and the fact that
$$a+\frac {1}{a}\ge 2$$
But i would like to know if there is a straight proof without Nesbitt inequality.
| Let $k=a+b+c$. Then
$$\frac{a^2+1}{b+c}=\frac{a^2-k^2+k^2+1}{k-a}=-a-k+\frac{k^2+1}{k-a}$$
So,
\begin{align*}
cyclic\sum\frac{a^2+1}{b+c}&=-a-b-c-3k+(k^2+1)\left(\frac{1}{k-a}+\frac{1}{k-b}+\frac{1}{k-c}\right)\\
&\ge -4k+3(k^2+1)\left(\frac{3}{(k-a)+(k-b)+(k-c)}\right)
\end{align*}
as $\textrm{A.M.}\ge\textrm{H.M.}$
\begin{align*}
-4k+3(k^2+1)\left(\frac{3}{(k-a)+(k-b)+(k-c)}\right)&=-4k+\frac{9(k^2+1)}{2k}\\
&=\frac{k}{2}+\frac{9}{2k}\\
&\ge3
\end{align*}
as $\textrm{A.M.}\ge\textrm{G.M.}$
| {
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How to deal with $(-1)^{k-1}$ It's a problem on mathematical induction.
$$1^2-2^2+3^2-.....+(-1)^{n-1}n^2=(-1)^{n-1}\frac{n.(n+1)}{2}$$
I have proved it for values of $n=1,2$.
Now I assume for $n=k$
$$P(k):1^2-2^2+3^2-.....+(-1)^{k-1}k^2=(-1)^{k-1}\frac{k.(k+1)}{2}$$.
$$P(k+1):1^2-2^2+3^2-.....+(-1)^{k-1}k^2+(k+1)^2=(-1)^{k-1}\frac{k.(k+1)}{2}+(k+1)^2\\=\frac{(k+1)}{2} [(-1)^{k-1}.k+2k+2]$$
I need suggestion to deal with the $(-1)^{k-1}$ so that I can prove the whole. Any help is appreciated.
| L.H.S. of $P(k+1)$ should be $1^2-2^2+3^2-\cdots+(-1)^{k-1}k^2+(-1)^k(k+1)^2$
which is equal to
$\displaystyle \frac{(-1)^{k-1}k(k+1)}{2}+(-1)^k(k+1)^2=\frac{(-1)^{k}[-k(k+1)+2(k+1)^2]}{2}=\frac{(-1)^{k}(k+1)(k+2)}{2}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2688026",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Let $x_n=\frac{1}{3}+(\frac{2}{5})^2+(\frac{3}{7})^3+....(\frac{n}{2n+1})^2$ Is $(x_n) $cauchy sequence? Let $$x_n=\frac{1}{3}+\left(\frac{2}{5}\right)^2+\left(\frac{3}{7}\right)^3+\cdots+\left(\frac{n}{2n+1}\right)^n$$ Is $(x_n)$ Cauchy sequence ?
My work
for $n>m:$
$$\begin{align}|x_n-x_m|&=\left|\frac{1}{3}+\left(\frac{2}{5}\right)^2+\cdots+\left(\frac{n}{2n+1}\right)^n-\left(\frac{1}{3}+\left(\frac{2}{5}\right)^2+\cdots+\left(\frac{m}{2m+1}\right)^m\right)\right|\\
&=\left|\left(\frac{m+1}{2(m+1)+1}\right)^{m+1}+\cdots+\left(\frac{n}{2n+!}\right)^n\right|\end{align}$$ how to proceed from here
| Hint. Note that for $k\geq 1$
$$0<\left(\frac{k}{2k+1}\right)^k=(2+1/k)^{-k}<2^{-k}.$$
Therefore the for $n>m\geq 1$,
$$|x_n-x_m|=\sum_{k=m+1}^{n} \left(\frac{k}{2k+1}\right)^k\leq \sum_{k=m+1}^{\infty} 2^{-k}=\frac{1}{2^{m}}.$$
| {
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"source": "stackexchange",
"question_score": "3",
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Inequality $\frac{a^3+b^3+c^3}{3}\geq\sqrt{\frac{a^4+b^4+c^4}{3}}$ Let $a$, $b$ and $c$ be non-negative numbers such that $(a+b)(a+c)(b+c)=8$. Prove that:
$$\frac{a^3+b^3+c^3}{3}\geq\sqrt{\frac{a^4+b^4+c^4}{3}}$$
Some attempts:
*
*From the condition follows $a^3+b^3+c^3 = (a+b+c)^3 -24$
*It is known (see here)
$$\frac{a+b+c}{3}\geq\sqrt[27]{\frac{a^3+b^3+c^3}{3}}$$
*Setting $2x=a+b$, $2y = b+c$, $2z = a+c$, we can express $a =x+z-y$ etc. The condition then becomes $xyz = 1$ which can be parametrized with free variables $0\leq q \leq 2 \pi /3 $ and arbitrary $r$ by
$$
x = \exp(r \cos q) \; ; \; y = \exp(r \cos (q + 2 \pi /3)) \; ; \; z = \exp(r \cos (q - 2 \pi /3))
$$
Using that, the condition can be removed and then calculus may be used.
*The question may be interpreted geometrically. Expressions such as $a^3+b^3+c^3 = $const. and $a^4+b^4+c^4 =$ const. can be interpreted as hypersurfaces of what has been called an N(3)-dimensional ball in p-norm, see here. A nice visualization is given in here. Then properties such as extrema, convexity etc. of these surfaces can be used.
I couldn't put the pieces together.
| We need to prove that $$\left(\frac{a^3+b^3+c^3}{3}\right)^6\geq\left(\frac{a^4+b^4+c^4}{3}\right)^3\left(\frac{(a+b)(a+c)(b+c)}{8}\right)^2$$ or
$$\left(\frac{a^3+b^3+c^3}{3}\right)^6\geq\left(\frac{a^4+b^4+c^4}{3}\right)^3\left(\frac{(a+b+c)^3-a^3-b^3-c^3}{24}\right)^2.$$
Now, let $a+b+c$ be constant and $a^3+b^3+c^3$ be constant.
Thus, by the Vasc's EV Method:
https://www.emis.de/journals/JIPAM/images/059_06_JIPAM/059_06_www.pdf
Corollary 1.8, case 3(a)
$a^4+b^4+c^4$ gets a maximal value for an equality case of two variables.
Since the last inequality is homogeneous,
it's enough to check what happens for $b=c=1$, which gives
$$\left(\frac{a^3+2}{3}\right)^6\geq\left(\frac{a^4+2}{3}\right)^3\left(\frac{(a+1)^2}{4}\right)^2$$ or $f(a)\geq0,$ where
$$f(a)=6\ln(a^3+2)-3\ln(a^4+2)-4\ln(a+1)-3\ln3+4\ln2.$$
Now, $$f'(a)=\frac{2(a-1)(a^6+4a^5+4a^4-12a^3-10a^2+8a+8)}{(a^3+2)(a^4+2)(a+1)}$$
and since $$a^6+4a^5+4a^4-12a^3-10a^2+8a+8>0,$$ we are done!
By the same way we can prove your stronger inequality:
$$\sqrt[27]{\frac{a^3+b^3+c^3}{3}}\geq\sqrt[53]{\frac{a^4+b^4+c^4}{3}}$$ and my old
$$\frac{a+b+c}{3}\geq\sqrt[53]{\frac{a^4+b^4+c^4}{3}}$$ is proven!
| {
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Prove that $\left|\frac{1}{z^4-4z^2+3}\right|\leq \frac{1}{3}$, if $|z|=2$
Prove that $$\left|\dfrac{1}{z^4-4z^2+3}\right|\leq \dfrac{1}{3},\quad\text{ if}\quad |z|=2$$
I know that I can complete squares on the denominator ($(z^2-2)^2-1$). I know the triangular inequality and some other few things... but I can't work so well with absolute value on $\mathbb{C}$, if anyone could help me...
| For $|z|=2$ the triangle inequality gives
$$
|z^2 -3| \ge |z^2| - 3 = |z|^2 - 3 = 1 \, ,\\
|z^2 -1| \ge |z^2| - 1 = |z|^2 - 1 = 3 \, ,
$$
and therefore
$$
|z^4 - 4z^2 +3| = |(z^2-3)(z^2-1)| = |z^2-3 | \cdot | z^2-1 | \ge 1 \cdot 3 = 3 \, .
$$
Alternatively: You already found that
$$
z^4 - 4z^2 +3 = (z^2 -2)^2 -1
$$
and for $|z| = 2$ the triangle inequality gives
$$
|z^2 -2| \ge |z|^2 - 2 = 2^2 - 2 = 2 \\
\Longrightarrow
|z^4 - 4z^2 +3| \ge | z^2 -2|^2 -1 \ge 2^2 - 1 = 3 \, .
$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Show that the equation $y^2 = x^3 +xz^4$ has no solutions in nonzero integers $x,y,z$.
Show that the equation $y^2 = x^3 +xz^4$ has no solutions in nonzero integers $x,y,z$.
There is given hint: Suppose that there is a solution. First show that it can be reduced to a solution satisfying $\gcd(x,z)=1$. Then use the fact that $x^3+ xz^4= x(x^2+z^4)$ is a perfect square to show that there are no solutions other than $x=y=0$.
The hint tells to prove by contradiction approach. So, $y\mid x,z\implies y\mid p$, where $p=(x,z)$. Hence, $\exists X,Z\in \mathbb{Z},\,x=Xp, z=Zp\implies y^2= X^3p^3 +XpZ^4p^4$$\implies y^2= Xp^3(X^2+Z^4p^2)$.
Is it valid to conclude from the last line, any of the below two separate conclusions :
1. The set of common divisors of $y^2, x^3, xz^4$ are the same; hence let $gcd(y^2, x^3, xz^4)=p'$
2.$\,\,p^3\mid y^2\implies \exists y=p^2Y, \,\,\,\,p^2\mid y$
| For starers, we must find $x,y,z$ all positive integers.
There are two possibilities : either $x$ is a square, say $x=A^2$ or otherwise $x=a^2b$ in which b is a square free integer.
Case 1. $y^2=A^2(A^4+z^4) \implies (y/A)^2 = Y^2 = A^4+z^4$ which is impossible due to the Fermat's LT (strong version for $n=4$).
Case 2. $y^2=a^2b(a^4b^2+z^4) \implies (y^2/a^2) = b(a^4b^2+z^4) \implies b|(a^4b^2+z^4) \implies b|z^4 \implies b|z$. So, $(y/ab)^2 = b(a^4+z_1^4b^2)$ in which $z=z_1b$. Again, $b|a^4$ which implies $b|a$. Thus, $(y/ab^2)^2 = (a_1^4b^3+z_1^4b) = b(a_1^4b^2+z_1^4)$ in which $a=a_1b$. So every time we find a new positive integer satisfying in the non-ending sequence $y>y_1=y/a>y_2=y/(ab)> \dots$ unless $a=1=b$ thus $x=1$. But if $x=1$ $y^2=1+z^4$ has positive integer solutions contradicting the mentioned Fermat's Theorem.
| {
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"timestamp": "2023-03-29T00:00:00",
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Ho do you approximate $4 \epsilon x + x^2 -1=0 $ using $x=x_0+x_1 \epsilon + x_2 \epsilon^2 $? Consider the quadratic $4 \epsilon x + x^2 -1=0 $ and suppose we want to approximate the solution to the equation using
$$x=x_0+x_1 \epsilon + x_2 \epsilon^2 $$
Why does substituting the line directly above into the quadratic yield $x_0^2 -1=0$?
I know that after substituting into the quadratic the term independent of epsilon is $x_0^2 -1$ but why must it be zero? thanks
| (Variables renamed because lazy.)
$x=u+vc+ wc^2
$
in
$ 4cx + x^2 -1=0
$
gives,
assuming that
$c$ is small,
$\begin{array}\\
0
&=4cx + x^2 -1\\
&=4c(u+vc+ wc^2) + (u+vc+ wc^2)^2 -1\\
&=4cu+4vc^2+ 4wc^3 + u^2+2uvc+v^2c^2+2uwc^2+O(c^3) -1\\
&=u^2-1+c(4u+2uv)+c^2(4v+v^2+2uw)+O(c^3)\\
\end{array}
$
For this to hold,
$u^2 = 1$,
$4u+2uv = 0$,
and
$4v+v^2+2uw = 0$.
From the first,
$u = \pm 1$.
From the second,
$0
= 4u+2uv
=2u(2+v)
$.
Since
$u \ne 0$,
$v = -2$.
From the third,
$0
= -8+4\pm 2w
$
so
$w
=\mp 2
$.
Therefore
$x
=\pm 1 -2c\mp 2c^2
$.
| {
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Solve $\sqrt {x^2-3}=x-3$ in $\mathbb R$
Solve $\sqrt {x^2-3}=x-3$ in $\mathbb R$
My attempt:
$|x^2-3|=(x-3)^2$
So $-(x^2-3)=(x-3)^2$ or $(x^2-3)=(x-3)^2$
If $-(x^2-3)=(x-3)^2=x^2+9-6x$
So no solutions in $\mathbb R$
And if $(x^2-3)=(x-3)^2$
So $x^2-3=x^2+9-6x$
Now, can I delete $x^2$ with $x^2$ ? Like this
$x^2-x^2-3-9+6x=0$
$6x=12$
$x=2$
But $f(2)$ isn’t equal to $0$?
| When you solve an equation by squaring both sides, it's possible that extraneous solutions are introduced. An extraneous solution is a value that you get at the end (like $x=2$ here) that doesn't actually solve the original equation. This is exactly what happened here. Since $x=2$ is the only value you got, and since it doesn't solve the original equation, that means the equation has no real solutions.
This can also be verified by graphing $y = \sqrt{x^2-3}$ and $y = x-3$ on the same axes and noticing that they never intersect.
Note also that $\left(\sqrt{x^2-3}\right)^2$ is just $x^2 - 3$, not $|x^2-3|$. So considering $-(x^2 - 3) = (x-3)^2$ is technically incorrect, but you got lucky here in that it also has no real solutions. Still, you should remove it completely from your work.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2704837",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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} |
Finding value of $\frac{49b^2-33bc+9c^2}{a^2}$
If $a,b,c$ are positive real numbers such that
$a^2+ab+b^2=9, b^2+bc+c^2=52,$
$c^2+ac+a^2=49$. Then finding $\displaystyle \frac{49b^2-33bc+9c^2}{a^2}$ is
Try: let $O$ be a point inside the Triangle $ABC$ such that angle $OAB$ and angle $OBC$ and $OCA$ is $120^\circ$. So we have
$OA=a,OB=b,OC=c$ and using cosine formula we have $AB=3,BC=7,CA=\sqrt{52}$.
Now using Area of triangle is $\displaystyle ab+bc+ca=\frac{42}{\sqrt{3}}$.
Could some help me to solve it , Thanks
| Applying cosine rule on $\triangle OBC$, we have
$$BC=\sqrt{b^2+bc+c^2}=\sqrt{52}$$
Similarly,
$$CA=7$$
By applying cosine rule on $\triangle ABC$
$$BC^2=AB^2+AC^2-2AB\cdot AC \cos \angle CAB$$
$$52=9+49-2(3)(7)\cos \angle CAB$$
$$\cos \angle CAB =\frac{1}{7} $$
Let's apply inversion at point $A$ with radius $1$, and let $O'$ be the image of $O$, $B'$ be the image of $B$ and $C'$ be the image of $C$.
Then $\triangle ABC \sim \triangle AC'B'$ , $\triangle AOB \sim \triangle AB'O'$, and $\triangle AOC \sim \triangle AC'O'$.
Hence
$$O'B'=\frac{BO \cdot AB' }{AO}=\frac{b\cdot AB'\cdot AB}{AO\cdot AB}=\frac{b}{3a}$$
$$O'C'=\frac{CO \cdot AC' }{AO}=\frac{c\cdot AC'\cdot AC}{AO\cdot AC}=\frac{c}{7a}$$
$$B'C'=\frac{CB \cdot AC' }{AB}=\frac{CB\cdot AC'\cdot AC}{AC\cdot AB}=\frac{CB}{7\cdot 3}=\frac{\sqrt{52}}{21}$$
Let's compute $\cos \angle B'O'C'$, from quadrilateral $AO'B'C'$.
$$\angle AB'O'+ \angle B'O'C'+\angle O'C'A + \angle C'AB'= 2\pi $$
$$\angle AOB+ \angle B'O'C'+\angle COA + \angle CAB= 2\pi $$
$$\frac{2\pi}3+ \angle B'O'C'+ \frac{2\pi}3+ \angle CAB= 2\pi $$
$$ \angle B'O'C'= \frac{2\pi}3 - \angle CAB$$
\begin{align}\cos \angle B'O'C' &= -\frac12 \cos \angle CAB +\frac{\sqrt3}2 \sin \angle CAB\\
&=-\frac1{14} +\frac{\sqrt3}2 \frac{\sqrt{48}}{7}\\
&= \frac{11}{14}\end{align}
Applying cosine rule on $\triangle O'B'C'$,
$$B'C'^2=O'B'^2+O'C'^2-2\cdot O'B'\cdot O'C' \cos \angle B'O'C'$$
$$\frac{52}{21^2}=\left(\frac{b}{3a} \right)^2+\left(\frac{c}{7a} \right)^2-2\cdot \left(\frac{b}{3a} \right)\cdot \left(\frac{c}{7a} \right) \cdot \frac{11}{14}$$
$$\frac{52}{7^2}=\left(\frac{b}{a} \right)^2+\left(\frac{3c}{7a} \right)^2-2\cdot \left(\frac{b}{a} \right)\cdot \left(\frac{c}{7a} \right) \cdot \frac{33}{14}$$
$$\frac{52}{7^2}=\left(\frac{7b}{7a} \right)^2+\left(\frac{3c}{7a} \right)^2-\frac{33bc}{(7a)^2}$$
$$\frac{(7b)^2+(3c)^2-33bc}{a^2}=52$$
Remark:
In general, if $$a^2+ab+b^2=\gamma$$
$$b^2+bc+c^2=\alpha$$
$$c^2+ac+a^2=\beta$$
then
$$a^2\alpha =\beta b^2 + \gamma c^2 - Dbc $$
where $$D=\left[ - \frac{ \beta + \gamma - \alpha}{2} \right] + \frac{\sqrt3}{2} \sqrt{4 \beta \gamma - (\alpha+\beta-\gamma)^2}$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Prove that $∀a, b, c ∈ \mathbb Z$ with $b \neq 0$, if $b|c$, then $\gcd(a, b) = \gcd(a + c, b)$.
Prove that $∀a, b, c ∈ \mathbb Z$ with $b \neq 0$, if $b|c$, then $\gcd(a, b) = \gcd(a + c, b)$.
I've proved it in the following way:
$\gcd(c,b) = b.$
Let $\gcd(a,b) = x$, $x|a$ and $x|b$
Since $b$ is the $\gcd$ of $c$,
if $x|b$ then $x|c$
Thus, $\gcd(a+c, b) = x = \gcd(a, b)$.
Was wondering if this is a solid proof. If not, is there anyway I can improve on this? Thanks for helping out
| The conclusion $\gcd(a+c, b) = x = \gcd(a, b)$ is not that clear from your previous parts.
One way to show the claim $\gcd(a, b) = \gcd(a + c, b)$ is to split the task into two parts:
$\qquad\qquad\gcd(a, b)|\gcd(a + c, b)\qquad$ and $\qquad\gcd(a+c, b)|\gcd(a, b)$
from which equality follows.
We obtain
\begin{align*}
\begin{array}{rlr}
\gcd(a,b)|b&\qquad\qquad(\text{by definition of }\gcd)&\qquad (1)\\
b|c&\qquad\qquad(\text{by assumption})\\
\Longrightarrow \gcd(a,b)|c&\qquad\qquad(\text{by transitivity of }|)&\qquad(2)\\
\\
\gcd(a,b)|a&\qquad\qquad(\text{by definition of }\gcd)&\qquad(3)\\
\\
\Longrightarrow \gcd(a,b)|(a+c)&\qquad\qquad(\text{by(2) and (3)})&\qquad(4)\\
\\
\color{blue}{\Longrightarrow \gcd(a,b)|\gcd(a+c,b)}&\qquad\qquad(\text{by(1) and (4)})&\qquad\color{blue}{(\text{I})}\\
\end{array}
\end{align*}
We further obtain
\begin{align*}
\begin{array}{rlr}
\gcd(a+c,b)|b&\qquad\qquad(\text{by definition of }\gcd)&\qquad(5)\\
b|c&\qquad\qquad(\text{by assumption})\\
\Longrightarrow \gcd(a+c,b)|c&\qquad\qquad(\text{by transitivity of }|)&\qquad(6)\\
\\
\gcd(a+c,b)|(a+c)&\qquad\qquad(\text{by definition of }\gcd)&\qquad(7)\\
\\
\Longrightarrow \gcd(a+c,b)|a&\qquad\qquad(\text{by(6) and (7)})&\qquad(8)\\
\\
\color{blue}{\Longrightarrow \gcd(a+c,b)|\gcd(a,b)}&\qquad\qquad(\text{by(5) and (8)})&\qquad\color{blue}{(\text{II})}\\
\end{array}
\end{align*}
We conclude from (I) and (II)
\begin{align*}
\color{blue}{\gcd(a,b)=\gcd(a+c,b)}
\end{align*}
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Spivak's Calculus: Chapter 1, Problem 18b (Quadratic determinant less than zero) The problem in question is as follows:
18b Suppose that $b^2 -4c \lt 0$. Show that there are no numbers $x$ that satisfy $x^2 + bx + c = 0$; in fact, $x^2 + bx + c \gt 0$ for all $x$. Hint: complete the square.
Trying to apply the hint, I began by constructing $b^2 - 4c < 0 \therefore (b-\frac{2c}{b})^2 - \frac{4c^2}{b^2} \lt 0$, but manipulating this ultimately just leads you to $b^2 \lt 4c$ which you didn't need to complete the square to get anyway.
The only other idea I had was that one could construct the quadratic equation beginning from the assumption that $x^2 + bx + c = 0$ and then go for proof by contradiction e.g.
$x^2 + bx + c =0$
$x^2 + bx = -c$
$x^2 + bx + (\frac{b}{2})^2 = -c + (\frac{b}{2})^2$
$(x + \frac{b}{2})^2 = \frac{b^2 - 4c}{4}$
$\therefore$ Given that for all real values of $x$ and $b$, $(x + \frac{b}{2})^2 \gt 0$, by transitivity of equality, $\frac{b^2 - 4c}{4} \gt 0$
$\therefore 4(\frac{b^2 - 4c}{4}) \gt 4(0)$
$\therefore b^2 - 4c \gt 0$ for all x such that $x^2 + bx + c = 0$
But that still leaves the statement "in fact, $x^2 + bx + c \gt 0$ for all $x$" unproven, unless it's supposed to obviously follow, in which case I'm not seeing how.
| Suppose for the sake of contradiction that $x^2 + bx + c = 0$ for the given hypothesis of $4c > b^2$.
Multiplying the given equation by $4$ and making use of our hypothesis yields:
$$0 = 4x^2 + 4bx + 4c > 4x^2 + 4bx + b^2 = (2x + b)^2$$
But, we cannot have that $0$ is strictly greater than a real number squared. Contradiction. QED.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2706487",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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} |
Delta Epsilon Proof for Ratio of Polynomials
Prove that $f$ is continuous at $x_0$ using the $\delta$ and $\epsilon$ criteria for:
$f(x)=\frac{x^3-2}{x+3}$, $x_0=1$.
I've made it halfway there, about; since $$|\frac{x^3-2}{x+3}+\frac{1}{4}|=|x-1||\frac{4x^2+4x+5}{4(x+3)}|$$
So then the issue is that we need $\delta>|x-1|$ but also bigger than the other expression. The only issue is, I don't really know how to find such a $\delta$. How can you proceed from here?
| This is a bit long but I wanted to go into detail so that you could apply this approach more easily to other examples.
Unless the function has a vertical tangent at $x_0$ there will be some positive bound $B$ satisfying
\begin{equation}
\left\vert \frac{f(x)-L}{x-x_0}\right\vert\le B \tag{1}
\end{equation}
on some deleted $c$-neighborhood $(x_0-c,x_0)\cup(x_0,x_0+c)$. Then if we let
\begin{equation}
\delta=\min\left\{c,\frac{\epsilon}{B}\right\} \tag{2}
\end{equation}
it follows that
\begin{eqnarray}
\left\vert \frac{f(x)-L}{x-x_0}\right\vert\cdot\left\vert x-x_0 \right\vert<B\cdot\frac{\epsilon}{B}\\
\left\vert f(x)-L\right\vert<\epsilon
\end{eqnarray}
For this particular exercise we wish to find a $B$ satisfying
\begin{equation}
\left\vert\frac{\frac{x^3-2}{x+3}+\frac{1}{4}}{x-1} \right\vert\le B
\end{equation}
For $x\ne1$ this simplifies to
\begin{equation}
\left\vert (x-2)+\frac{29}{4(x+3)}\right\vert\le B \tag{3}
\end{equation}
So let us investigate whether there is such a bound $B$ if we choose $c=1$ from equation (2) above.
\begin{equation}
\vert x-1\vert<1
\end{equation}
Then begin with
\begin{equation}
-1<x-1<1
\end{equation}
and try to work our way to an expression that looks like inequality $(3)$.
\begin{eqnarray}
-1 &<x-1<1\\
-2 &<x-2<0
\end{eqnarray}
Now we must add in the $\frac{29}{4(x+3)}$ term. We know its graph is an hyperbola which is decreasing to the right of its vertical asymptote at $x=-3$ so its largest value on $(0,1)
\cup(1,2)$ is $\frac{29}{12}$ at $x=0$ and its smallest value is $\frac{29}{20}$ at $x=2$. Therefore we continue
\begin{array}
\phantom{ }-2&<&x-2&<0\\
\tfrac{29}{20}&<&\frac{29}{4(x+3)}&<\frac{29}{12}\\
-2+\tfrac{29}{20}&<&x-2+\frac{29}{4(x+3)}&<\frac{29}{12}\\
-\tfrac{11}{20}&<&x-2+\frac{29}{x+3}&<\frac{29}{12}\\
&&\left\vert4(x-2)+\frac{29}{4(x+3)} \right\vert&<\frac{29}{12}<3
\end{array}
So let $\delta=\min\left\{1,\frac{\epsilon}{3}\right\}$. Then $\vert x-1\vert<1$ and $\vert x-1\vert<\frac{\epsilon}{3}$. Therefore
\begin{eqnarray}
\left\vert(x-2)+\frac{29}{4(x+3)} \right\vert\cdot\vert x-1\vert&<3\cdot\frac{\epsilon}{3}\\
\left\vert\frac{x^3-2}{x+3}+\frac{1}{4} \right\vert&<\epsilon
\end{eqnarray}
Therefore $\lim_{x\to1}f(x)=-\frac{1}{4}=f(1)$ so the function is continuous at $x=1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2706668",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
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} |
Evaluate $\int \frac{1}{ax^2-bx}\,dx$ with substitution Evaluate $\int\frac{1}{ax^2-bx}\,dx$
First try:
$$\int\frac{1}{ax^2-bx}\,dx = \int\frac{1}{bx^2(\frac{a}{b}-\frac{1}{x})}\,dx$$
By substituting $u=\frac{a}{b}-\frac{1}{x}$ so $du=\frac{1}{x^2}\,dx$ we have,
$$\int\frac{1}{bx^2(\frac{a}{b}-\frac{1}{x})} \, dx = \frac{1}{b} \int\frac{1}{u} \, du = \frac{\ln|u|}{b}+C = \frac{\ln|\frac{a}{b}-\frac{1}{x}|}{b}+C=\frac{\ln|\frac{ax-b}{bx}|}{b}+C$$
Second try:
$$\int\frac{1}{ax^2-bx} \, dx = \frac{1}{b}\int\frac{b}{x^2(a-\frac{b}{x})} \, dx$$
By substituting $u=a-\frac{b}{x}$ so $du=\frac{b}{x^2}dx$ we have,
$$\frac{1}{b}\int\frac{b}{x^2(a-\frac{b}{x})}dx = \frac{1}{b}\int\frac{1}{u}\,du = \frac{\ln|u|}{b}+C = \frac{\ln|a-\frac{b}{x}|}{b}+C=\frac{\ln|\frac{ax-b}{x}|}{b} + C$$
It seems to me that $\frac{\ln|\frac{ax-b}{bx}|}{b}+C\neq\frac{\ln|\frac{ax-b}{x}|}{b}+C$, so, whats the matter ?!
| Note that
$$\frac{\ln|\frac{ax-b}{bx}|}{b}+C=\frac{\ln|\frac{ax-b}{x}|}{b} + \frac{\ln |\frac{1}{b}|}{b} + C=\frac{\ln|\frac{ax-b}{x}|}{b}+C_2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2707839",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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Generating function of a sequence with a sum I am new to generating functions and I know some of the basics how to create them. But now I am not sure how to create a generating function of this sequence:
$$a_0 = 0$$
$$ a_n = n + c (\sum_{k=0}^{n-1} a_k)$$ for n >=1
where c is a real non-zero constant.
I guess I need to add $ \frac {1}{(1-x)^2}$ (which is generating function of the n (sequence 1, 2, 3, 4, ...)) to the rest. But how do I evaluate the rest?
Thank you for any tips.
| We denote the generating function of $a_n$ with $A(x)=\sum_{n=0}^\infty a_n x^n$.
Part 1: Generating function of $\sum_{k=0}^{n-1}a_k$:
A generating function of the sum of the first $n+1$ elements $\sum_{k=0}^n a_k$ can be derived as Cauchy product of $A(x)$ with $\frac{1}{1-x}$.
We obtain
\begin{align*}
A(x)\color{blue}{\cdot\frac{1}{1-x}}=\left(\sum_{k=0}^\infty a_kx^k\right)\left(\sum_{l=0}^\infty x^l\right)
=\sum_{n=0}^\infty\left(\sum_{{k+l=0}\atop{k,l\geq 0}}a_k\right)x^n
=\sum_{n=0}^\infty\color{blue}{\left(\sum_{k=0}^na_k\right)} x^n
\end{align*}
It is convenient to use the coefficient of operator $[x^n]$ to denote the coefficient of $x^n$ of a series. This way we can write e.g.
\begin{align*}
\color{blue}{[x^n]A(x)}=[x^n]\sum_{k=0}^\infty a_kx^k\color{blue}{=a_n}
\end{align*}
We get for $n\geq 1$
\begin{align*}
\sum_{k=0}^{n-1}a_k=[x^{n-1}]\frac{A(x)}{1-x}=[x^n]\frac{xA(x)}{1-x}\tag{1}
\end{align*}
Part 2: Generating function of $n$: Using the binomial series expansion we obtain for $n\geq 0$
\begin{align*}
\frac{1}{(1-x)^2}&=\sum_{n=0}^\infty\binom{-2}{n}(-x)^n=\sum_{n=0}^\infty\binom{n+1}{n}x^n\\
&=\sum_{n=0}^\infty (n+1)x^n=\sum_{n=0}^\infty nx^n+\frac{1}{1-x}
\end{align*}
Here we use the binomial identities $\binom{-p}{q}=\binom{p+q-1}{q}(-1)^q$ and $\binom{p}{p-1}=\binom{p}{1}=p$. It follows
\begin{align*}
\sum_{n=0}^\infty nx^n=\frac{1}{(1-x)^2}-\frac{1}{1-x}=\frac{x}{(1-x)^2}\\
\end{align*}
or equivalently
\begin{align*}
n=[x^n]\frac{x}{(1-x)^2}\qquad\qquad n\geq 0\tag{2}
\end{align*}
Putting all together: We consider
\begin{align*}
a_n&=n+c\sum_{k=0}^{n-1}a_k\\
a_0&=0
\end{align*}
and obtain from (1) and (2)
\begin{align*}
a_n&=[x^n]A(x)=[x^n]\frac{x}{(1-x)^2}+c[x^n]\frac{xA(x)}{1-x}\qquad\qquad n\geq 0\\
\end{align*}
respectively
\begin{align*}
A(x)&=\frac{x}{(1-x)^2}+c\cdot\frac{xA(x)}{1-x}\\
\end{align*}
from which
\begin{align*}
\color{blue}{A(x)=\frac{x}{(1-(c+1)x)(1-x)}}
\end{align*}
follows.
| {
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"url": "https://math.stackexchange.com/questions/2708045",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Evaluate the Legendre's Symbol $\frac{2017}{5003}$ I got 1, this is the way I approached this. Any suggestion to get the answer faster or less messy is appreciated.
$\frac{2017}{5003} = (-1)^{(2016)(5002)/4}(\frac{5003}{2017})$
= $\frac{969}{2017} = (-1)^{(968)(2016)/4}(\frac{2017}{969})$
= $\frac{79}{969} = (-1)^{(78)(968)/4}(\frac{969}{79})$
= $\frac{21}{79} = (-1)^{(20)(78)/4}(\frac{79}{21})$
= $\frac{16}{21} = (-1)^{(15)(20)/4}(\frac{21}{16})$
= $\frac{16}{21} \rightarrow \frac{4}{7} \times \frac{4}{3} \rightarrow \frac{2^2}{7} \times \frac{2^2}{4} \rightarrow 1 \times 1 =1.$
| You forgot all remainders are not primes, and the Legendre's symbol is multiplicative. Thus, since $969=3\cdot 17\cdot 19$, we have that
\begin{align}
\Bigl(\frac{969}{2017}\Bigr)&=\Bigl(\frac{3}{2017}\Bigr)\Bigl(\frac{17}{2017}\Bigr)\Bigl(\frac{19}{2017}\Bigr)=\Bigl(\frac{2017}{3}\Bigr)\Bigl(\frac{2017}{17}\Bigr)\Bigl(\frac{2017}{19}\Bigr) \\
&=\Bigl(\frac{1}{3}\Bigr)\Bigl(\frac{11}{17}\Bigr)\Bigl(\frac{3}{19}\Bigr)=\Bigl(\frac{11}{17}\Bigr)\Bigl(\frac{3}{19}\Bigr)\\
&=\Bigl(\frac{17}{11}\Bigr)\cdot (-1)\Bigl(\frac{19}{3}\Bigr)=-\Bigl(\frac{6}{11}\Bigr)\Bigl(\frac{1}{3}\Bigr)=-\Bigl(\frac{6}{11}\Bigr)=+1.
\end{align}
| {
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"timestamp": "2023-03-29T00:00:00",
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Showing $(3 + \sqrt{3})$ is not a prime ideal in $\mathbb{Z}[\sqrt{3}]$ Let $I = (3+\sqrt{3})$
Looking at the field norm we note that $N(3 + \sqrt{3}) = 6$. We also know that $\mathbb{Z}[\sqrt{3}]$ is a Euclidean Domain.
We want to find some $\alpha, \beta \in \mathbb{Z}[\sqrt{3}]$ s.t. $\alpha \cdot \beta = 3 + \sqrt{3}$. This requires $N(\alpha)\cdot N(\beta) = 6$.
So $N(\alpha) \in \{\pm 2\}$ and $N(\beta) \in \{\pm 3\}$.
$N(a+b\sqrt{3}) = a^2 - 3b^2 = -2$ when $\alpha = 1 - \sqrt{3}$ which is a non-unit since $N(\alpha) \neq 1$.
Then we have $N(c + d\sqrt{3}) = c^2 - 3d^2 = -3$ when $\beta = -3 - 2\sqrt{3}$ which is also a non-unit.
Then we see $\alpha\cdot\beta=(1-\sqrt{3})(-3-2\sqrt{3}) = 3 + \sqrt{3}$
Since this is a non trivial factorization of $3+\sqrt{3}$, then we see that $(1-\sqrt{3})\cdot(-3-2\sqrt{3})\in I$.
It remains to show that neither $\alpha$ or $\beta$ are in $I$.
Taking $\frac{1-\sqrt{3}}{3+\sqrt{3}} = \frac{(1-\sqrt{3})(3-\sqrt{3})}{6} = \frac{(3-\sqrt{3} - 3\sqrt{3} +3)}{6} = \frac{-4\sqrt{3}}{6}$ which is not in $\mathbb{Z}[\sqrt{3}]$. So there will be some remainder implying that $\alpha$ is not in $I$.
Doing the same thing we calculate $\frac{-3-2\sqrt{3}}{3+\sqrt{3}} = \frac{-3-\sqrt(3)}{6}$. Again, it will yield a remainder so we conclude that both $\alpha$ and $\beta$ are not in $I$, yet $\alpha \cdot \beta \in I$.
Thus $(3+\sqrt{3})$ is not prime.
Is this attempt correct? Is there a shorter way to go about this?
| Why not write $3+\sqrt3=\sqrt3(1+\sqrt3)$?
| {
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In how many ways can $3$ girls and $5$ boys be placed in three distinct cars that each hold at most three children?
There are $3$ cars of different makes available to transport $3$ girls and $5$ boys on a field trip. Each car can hold up to $3$ children. Find the number of ways in which they can be accommodated, if the internal arrangement of children inside the car is considered to be immaterial.
I considered three cases:
*
*$3$ G in car 1, $3$ B in car 2, and $2$ B in car 3.This can be done in $\binom{3}{3}\binom{5}{3}\binom{2}{2}3!=60$ ways;
*$2$ G, $1$ B in car 1; $1$ G, $2$ B in car 2; and $2$ B in car 3. This can be done in $\binom{3}{2}\binom{5}{1}\binom{1}{1}\binom{4}{2}\binom{2}{2}3!=540$ ways;
*$1$ G, $2$ B in car 1; $1$ G, $2$ B in car 2; and $1$ G, $1$ B in car 3. This can be done in $\binom{3}{1}\binom{5}{2}\binom{2}{1}\binom{3}{2}\binom{1}{1}\binom{1}{1}\frac{3!}{2!}=540$ ways.
The total number of ways thus equals $540+540+60=1140,$ but the answer given is $1680$. Where did I go wrong?
| There is no need to distinguish between boys and girls. Since there are $8$ children for nine seats available in the three cars I suggest a $9^{\rm th}$ child that will withdraw at the last moment.
There are ${9\choose3}$ ways to choose the $3$ children for the first car, then ${6\choose3}$ ways to choose the three children for the second car, and the remaining $3$ children will be put in the third car. It follows that there are
$${9\cdot8\cdot7\cdot6\cdot5\cdot 4\over(3!)^2}=1680$$
ways to accommodate the children.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2713476",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
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Derivative of the function $\sin^{-1}\Big(\frac{2^{x+1}}{1+4^x}\Big)$
Find y' if $y=\sin^{-1}\Big(\frac{2^{x+1}}{1+4^x}\Big)$
In my reference $y'$ is given as $\frac{2^{x+1}\log2}{(1+4^x)}$. But is it a complete solution ?
Attempt 1
Let $2^x=\tan\alpha$
$$
\begin{align}
y=\sin^{-1}\Big(\frac{2\tan\alpha}{1+\tan^2\alpha}\Big)=\sin^{-1}(\sin2\alpha)&\implies \sin y=\sin2\alpha=\sin\big(2\tan^{-1}2^x\big)\\
&\implies y=n\pi+(-1)^n(2\tan^{-1}2^x)
\end{align}
$$
$$
\begin{align}
y'&=\pm\frac{2.2^x.\log2}{1+4^x}=\pm\frac{2^{x+1}.\log2}{1+4^x}\\
&=\color{blue}{\begin{cases}
\frac{2^{x+1}.\log2}{1+4^x}\text{ if }-n\pi-\frac{\pi}{2}\leq2\tan^{-1}2^x\leq -n\pi+\frac{\pi}{2}\\
-\frac{2^{x+1}.\log2}{1+4^x}\text{ if }n\pi-\frac{\pi}{2}\leq2\tan^{-1}2^x\leq n\pi+\frac{\pi}{2}
\end{cases}}
\end{align}
$$
Attempt 2
$$
\begin{align}
y'&=\frac{1}{\sqrt{1-\frac{(2^{x+1})^2}{(1+4^x)^2}}}.\frac{d}{dx}\frac{2^{x+1}}{1+4^x}\\
&=\frac{1+4^x}{\sqrt{1+4^{2x}+2.4^x-4^x.4}}.\frac{(1+4^x)\frac{d}{dx}2^{x+1}-2^{x+1}\frac{d}{dx}(1+4^x)}{(1+4^x)^2}\\
&=\frac{(1+4^x).2^{x+1}.\log2-2^{x+1}.4^x.\log2.2}{\sqrt{1+4^{2x}-2.4^x}.(1+4^x)}\\&=\frac{2^{x+1}\log2\big[1+4^x-2.4^x\big]}{\sqrt{(1-4^x)^2}.(1+4^x)}\\&=\frac{2^{x+1}\log2\big[1-4^x\big]}{|{(1-4^x)}|.(1+4^x)}\\
&=\color{blue}{\begin{cases}\frac{2^{x+1}\log2}{(1+4^x)}\text{ if }1>4^x>0\\
-\frac{2^{x+1}\log2}{(1+4^x)}\text{ if }1<4^x
\end{cases}}
\end{align}
$$
In both my attempts i am getting both +ve and -ve solutions. Is it the right way to find the derivative?
And how do I connect the domain for each cases in attempt 1 and attempt 2 ?
| The trickiest part is the ^2 part let me know if you do not follow that step. I do not know how to use the fancy formatting so I had to write it out.
Solution
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Bounds on $\frac{2n!}{(n!)^2}$ Here is my problem:
Use induction to show that for $n\ge1$, $$\frac{4^n}{\sqrt{4n}} \le \frac{(2n)!}{(n!)^2}\le \frac{4^n}{\sqrt{3n+1}}.$$
What I have so far:
For $n=1$ it is $2≤2≤2$, and I tried to originally do this problem by assuming that the case holds for $n\ge 1$, and if it hold for n+1, than it should hold for all $n\ge 1$.
I started with $n+1$ case: $$\frac{4^{n+1}}{\sqrt{4(n+1)}}\le\frac{(2(n+1))!}{((n+1)!)^2}\le\frac{4^{n+1}}{\sqrt{3(n+1)+1}},$$ but this did not really lead me anywhere.
| Note that$$\frac{\frac{\bigl(2(n+1)\bigr)!}{(n+1)!^2}}{\frac{(2n)!}{n!^2}}=\frac{(2n+2)(2n+1)}{(n+1)^2}=2\frac{2n+1}{n+1}.$$On the other hand$$\frac{\frac{4^{n+1}}{\sqrt{4(n+1)}}}{\frac{4^n}{\sqrt{4n}}}=2\sqrt{\frac n{n+1}}$$and$$\frac{\frac{4^{n+1}}{\sqrt{3n+4}}}{\frac{4^n}{\sqrt{3n+1}}}=4\sqrt{\frac{3n+1}{3n+4}}.$$Can you take it from here?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2714383",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Find the number of real solutions to the system of equations $x=\frac{2z^2}{1+z^2},y=\frac{2x^2}{1+x^2},z=\frac{2y^2}{1+y^2}$ My approach is naive:
Given
$x=\frac{2z^2}{1+z^2},y=\frac{2x^2}{1+x^2},z=\frac{2y^2}{1+y^2}$,
$[\frac{1}{x}+\frac{1}{y}+\frac{1}{z}]=\frac{1}{2}\cdot[\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}]+3$
What to do next?
Tried it using trigonometry by replacing $z^2$ by $tan^2\theta$ but could not get promising results.
Is there any trick to such genre of problems?
| It is easy to show that $$x,y,z>0$$ Clearly that $$x=y=z=0$$ is one solution. Now we assume $$x\geq y\geq z$$ this is equivalent to
$$\frac{z^2}{1+z^2}\geq \frac{2x^2}{1+x^2}\geq \frac{2y^2}{1+y^2}$$ From here we get
$$\frac{z^2}{1+z^2}\geq \frac{x^2}{1+x^2}$$ this implies $$z\geq x$$ and so on, and we get
$$x=y=z$$ and $$1+x^2=2x$$ this means $$(x-1)^2=0$$ or $$x=y=z=1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2722309",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Prove that $1+\frac{1}{2^3}+\frac{1}{3^3}+\cdots+\frac{1}{2018^3}<\frac{5}{4}$ Question: Prove that $1+\frac{1}{2^3}+\frac{1}{3^3}+\cdots+\frac{1}{2018^3}<\frac{5}{4}$
Attempt: When I tried solving this problem, I paired up $1$ with $\frac{1}{2018^3}$, $\frac{1}{2^3}$ with $\frac{1}{2017^3}$ etc. but it wasn't useful.
| By creative telescoping / Euler's acceleration method
$$ \zeta(3)=\sum_{n\geq 1}\frac{1}{n^3} = \frac{5}{2}\sum_{n\geq 1}\frac{(-1)^{n+1}}{n^3 \binom{2n}{n}}\tag{1} $$
and since the last series is rapidly convergent and with alternating signs,
$$ \zeta(3) \leq \frac{5}{2}\sum_{n=1}^{3}\frac{(-1)^{n+1}}{n^3 \binom{2n}{n}} = \frac{1039}{864}=[1;4,1,14,1,10]\leq [1;4]=\frac{5}{4}.\tag{2}$$
$(1)$ has been historically crucial in Apery's proof of $\zeta(3)\not\in\mathbb{Q}$, for instance.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2723338",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
If $b-a=mc$ then prove that $\cot (\dfrac {B-A}{2})=\dfrac {1+m.\cos (B)}{m.sin (B)}$ If $b-a=mc$ then prove that $\cot (\dfrac {B-A}{2})=\dfrac {1+m.\cos (B)}{m.sin (B)}$
My Attempt:
We know,
$$\tan (\dfrac {B-A}{2} )=\dfrac {b-a}{b+a}.\cot (\dfrac {C}{2})$$.
Then
$$b-a=mc$$
$$(b+a).\tan (\dfrac {B-A}{2}).\cot (\dfrac {C}{2})=mc$$
| Since (b-a) = mc
m = (b-a)/c
L.H.S = $\frac{1+\frac{\left(b-a\right)}{c}CosB}{\frac{(b-a)}{c}SinB}$
= $\frac{c+\left(b-a\right)CosB}{\left(b-a\right)SinB}$
$=\frac{SinC+\left(SinB-SinA\right)CosB}{ \begin{array}{c}
\left(SinB-SinA\right)SinB \\
\end{array}
}
$
$ =\frac{{\mathrm{sin} \left(A+B\right)\ }-SinA.CosB+SinB.CosB}{ \begin{array}{c}
{Sin}^2B-SinA.SinB \\
\end{array}
}$
$ =\frac{CosASinB+SinBCosB}{ \begin{array}{c}
{Sin}^2B-SinA.SinB \\
\end{array}
}$
$ =\frac{CosA+CosB}{SinB-SinA}$
=$cot\left(\frac{B-A}{2}\right)$ = R.H.S
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2724383",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Solve system of equations: $2x^2-4y^2-\frac{3}{2}x+y=0 \land 3x^2-6y^2-2x+2y=\frac{1}{2}$ $$
\begin{cases}
2x^2-4y^2-\frac{3}{2}x+y=0 \\
3x^2-6y^2-2x+2y=\frac{1}{2}
\end{cases}
$$
I multiplied the first with $-6$ and the second with $4$ and get two easier equations:
$9x-6y=0 \land -8x+8y=2 $ and out of them I get that $x=\frac{1}{2}$ and that $y=\frac{3}{4}$ but when I put it back into the original systems equation I dont get the right answer. Can somebody explain why?
| Multiplying the fírst equation by$-6$ and the second by $4$ and adding both we get
$$x+2y=2$$ or $$x=2-2y$$ plugging this in the first equation we have
$$2(2-2y)^2-4y^2-\frac{3}{2}(2-2y)+y=0$$
Can you solve this?
Simplifying and factorizing we get $$(2 y-5) (2 y-1)=0$$ so $$y=\frac{5}{2}$$ or $$y=\frac{1}{2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2725579",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Solve $\tan (\theta) + \tan (2\theta) = \tan (3\theta)$
Find the general solution of:
$$\tan (\theta) + \tan (2\theta) = \tan (3\theta)$$
My Attempt:
$$\tan (\theta) + \tan (2\theta) = \tan (3\theta)$$
$$\dfrac {\sin (\theta)}{\cos (\theta)}+ \dfrac {\sin (2\theta)}{\cos (2\theta)}=\dfrac {\sin (3\theta)}{\cos (3\theta)}$$
$$\dfrac {\sin (\theta+2\theta)}{\cos (\theta) \cos (2\theta)}=\dfrac {\sin (3\theta)}{\cos (3\theta)}$$
| \begin{align}
\tan (\theta) + \tan (2\theta) &= \tan(3\theta) \\
\tan (\theta) + \tan (2\theta) &= \tan(\theta + 2\theta) \\
\dfrac{\tan (\theta) + \tan (2\theta)}{1}
&= \dfrac{\tan(\theta) + \tan(2\theta)}
{1-\tan (\theta) \tan (2\theta)} \\
\end{align}
So, either $\tan (\theta) + \tan (2\theta)=0$, or
$\tan (\theta) = 0$, or $\tan (2\theta)=0$
\begin{align}
\tan(\theta) + \tan(2\theta) &= 0 \\
\tan(\theta) + \dfrac{2 \tan(\theta)}{1 - \tan^2(\theta)} &= 0 \\
3 \tan(\theta) - \tan^3(\theta) &= 0 \\
\tan(\theta) &\in \{0, \pm \sqrt 3\}
\end{align}
So $\theta \in \left\{
n\pi, \pm\frac 13\pi + n\pi, \pm\frac 14\pi + n\pi
: n \in \mathbb Z
\right\}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2725922",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Simple approaches to prove that $\lim\limits_{x\to 0}\left(\frac{1}{x^2}-\frac{1}{\sin^2 x}\right)=-\frac13\ $?
Find $\lim_{x\to 0}(\frac{1}{x^2}-\frac{1}{\sin^2 x})$
My attempt:
$\lim_{x\to 0}(\frac{1}{x^2}-\frac{1}{\sin^2 x})$
=$\lim_{x\to 0}(\frac{\sin^2 x -x^2}{x^2 \sin^2 x})$ ($\frac{0}{0}$ form)
Applying L'Hospital's Rule we get,
=$\lim_{x\to 0}(\frac{2\sin x \cos x -2x}{2x \sin^2 x+ 2x^2\sin x \cos x})$
=$\lim_{x\to 0}(\frac{\sin 2x -2}{2x \sin^2 x+ x^2\sin 2x})$ ($\frac{0}{0}$ form)
Applying L'Hospital's Rule we get,
=$\lim_{x\to 0}(\frac{2\cos 2x}{2(\sin^2 x+2x\sin x \cos x)+ (2x\sin 2x+2x^2\cos 2x)})$
=$\lim_{x\to 0}(\frac{2\cos 2x}{2(\sin^2 x+x\sin 2x)+ (2x\sin 2x+2x^2\cos 2x)})$
=$\lim_{x\to 0}(\frac{2\cos 2x}{2(\sin^2 x+x\sin 2x)+ 2(x\sin 2x+x^2\cos 2x)})$
=$\lim_{x\to 0}(\frac{2\cos 2x}{2(\sin^2 x+x\sin 2x+x\sin 2x+x^2\cos 2x)})$
=$\lim_{x\to 0}(\frac{\cos 2x}{\sin^2 x+x\sin 2x+ x\sin 2x+x^2\cos 2x})$
=$\lim_{x\to 0}(\frac{\cos 2x}{\sin^2 x+2x\sin 2x+x^2\cos 2x})$ ($\frac{0}{0}$ form)
Applying L'Hospital's Rule we get,
=$\lim_{x\to 0}(\frac{-2\sin 2x}{2\sin x \cos x+2(\sin 2x+ 2x\cos 2x)+(2x\cos 2x-2x^2\sin 2x})$
=$\lim_{x\to 0}(\frac{-2\sin 2x}{3\sin 2x+6x\cos 2x-2x^2\sin 2x})$ ($\frac{0}{0}$ form)
Applying L'Hospital's Rule we get,
=$\lim_{x\to 0}(\frac{-4\cos 2x}{6\cos 2x+6(\cos 2x-2x\sin 2x)-2(2x\sin 2x+2x^2\cos 2x)})$
=$\lim_{x\to 0}(\frac{-4\cos 2x}{12\cos 2x-12x\sin 2x-4x\sin 2x-4x^2\cos 2x)})$
=$\frac{-4\cos 0}{12 \cos 0-0-0-0}$
=$\frac{-4}{12}$
=$\frac{-1}{3}$
My problem: This method is very lengthy and involves a lot of calculations. Is there is any other (better and efficient) method to evaluate this limit?
| $$\lim_{x\to 0}\frac{\sin x-x}{x^3}=\lim_{x\to 0}\frac{\cos x-1}{3x^2}=\lim_{x\to 0}\frac{-\sin x}{6x}=-\frac{1}{6}$$
\begin{align*}
\lim_{x\to 0}\left(\frac{1}{x^2}-\frac{1}{\sin^2 x}\right)&=\lim_{x\to 0}\frac{\sin^2x-x^2}{x^2\sin^2x}\\
&=\left(\lim_{x\to 0}\frac{\sin x-x}{x^3}\right)\left(\lim_{x\to 0}\frac{x\sin x+x^2}{\sin^2x}\right)\\
&=\left(\lim_{x\to 0}\frac{\sin x-x}{x^3}\right)\left(\lim_{x\to 0}\frac{x}{\sin x}+\lim_{x\to 0}\frac{x^2}{\sin^2x}\right)\\
&=\left(-\frac{1}{6}\right)(1+1)\\
&=-\frac{1}{3}
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2731206",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
The integer part of $ \sum_{k=2}^{9999} \frac{1}{\sqrt{k}} $ For $k>0 $
$ \frac{1}{2\cdot \sqrt{k+1}}<\sqrt{k+1}-\sqrt{k}<\frac{1}{2\cdot \sqrt{k}} $,
the integer part of $ \sum_{k=2}^{9999} \frac{1}{\sqrt{k}} $
My attempt here was not complying with any of the options given as my attempt was to find the min. value for the series which the question does not ask. What to do?
Options are:
A. 198
B. 197
C. 196
D. 195
I know $ \sum_{k=1}^{n} \frac{1}{k} \approx log n $ But this does not help in this case.
Also $ \sum_{k=2}^{9999} \frac{1}{\sqrt{k}} < 2(\sqrt{10000}-\sqrt{2}) $
This also does not help.
| for $t\in [k,k+1]$ we have,
$$
\frac{1}{\sqrt{k+1}}\leq \frac{1}{\sqrt{t}}\leq \frac{1}{\sqrt{k}}
$$
then
$$
\frac{1}{\sqrt{k+1}}\leq \int_k^{k+1}\frac{dt}{\sqrt{t}}\leq \frac{1}{\sqrt{k}}
$$
and we get
\begin{eqnarray}
\sum_{k=1}^{10^4-1}\frac{1}{\sqrt{k+1}}&\leq &\sum_{k=1}^{10^4-1}\int_k^{k+1}\frac{dt}{\sqrt{t}}&\leq& \sum_{k=1}^{10^4-1}\frac{1}{\sqrt{k}}\\
\sum_{k=2}^{10^4}\frac{1}{\sqrt{k}}&\leq &\int_1^{10^4}\frac{dt}{\sqrt{t}}&\leq& \sum_{k=1}^{10^4-1}\frac{1}{\sqrt{k}}\\
\frac{1}{100}+\sum_{k=2}^{10^4-1}\frac{1}{\sqrt{k}}&\leq &\int_1^{10^4}\frac{dt}{\sqrt{t}}&\leq& 1+\sum_{k=2}^{10^4-1}\frac{1}{\sqrt{k}}\\
\frac{1}{100}+\sum_{k=2}^{10^4-1}\frac{1}{\sqrt{k}}&\leq &2(100-1)&\leq& 1+\sum_{k=2}^{10^4-1}\frac{1}{\sqrt{k}}\\
\end{eqnarray}
we can conclude that
$$
197\leq \sum_{k=2}^{10^4-1}\frac{1}{\sqrt{k}}\leq 198-\frac{1}{100}<198
$$
so
$$
E\left(\sum_{k=2}^{10^4-1}\frac{1}{\sqrt{k}}\right)=197
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2732776",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
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$\lim_{n\to+\infty} \frac{1}{n\log(n)}\sum_{k=4}^{n}\frac{2k}{k^2-2k-3}$
Given the limit:
$$\lim_{n\to+\infty} \frac{1}{n\log(n)}\sum_{k=4}^{n}\frac{2k}{k^2-2k-3} = \alpha$$
Find the value of $\alpha$
I know the series does not converge (it is equivalent to the harmonic series. Correct me, please, if I am wrong).
Does it let me apply the following:
$$\lim_{n\to+\infty} \frac{1}{n\log(n)}\sum_{k=4}^{n}\frac{2k}{k^2-2k-3} = \lim_{n\to+\infty}\frac{2n}{n\log(n)(n^2-2n-3)}$$
Or not?
Thank you in advance.
| From the $\lim\limits_{n\to+\infty} \frac{1}{n\log(n)}\sum\limits_{k=4}^{n}\frac{2k}{k^2-2k-3}$ first dealing with the sum equals to the followings:
$\sum\limits_{k=4}^{n}\frac{2k}{k^2-2k-3}=\sum\limits_{k=4}^{n}\frac{2k}{(k+1)(k-3)}=\frac{1}{2}\sum\limits_{k=4}^{n}\big(\frac{k}{k-3}-\frac{k}{k+1}\big)=\frac{1}{2}\sum\limits_{k=1}^{n}\big(\frac{k+3}{k}-\frac{k+3}{k+4}\big)=\frac{1}{2}\sum\limits_{k=1}^{n}\big(\frac{3}{k}-\frac{1}{k+4}\big)$
Forming further:
$\frac{1}{2}\sum\limits_{k=1}^{n}\big(\frac{3}{k}-\frac{1}{k+4}\big)=\sum\limits_{k=1}^{n}\frac{k+4+2}{k(k+4)}=\sum\limits_{k=1}^{n}\frac{1}{k}+\sum\limits_{k=1}^{n}\frac{2}{k(k+4)}=\sum\limits_{k=1}^{n}\frac{1}{k}+\frac{1}{2}\sum\limits_{k=1}^{n}\big(\frac{1}{k}-\frac{1}{k+4}\big)$
$\sum\limits_{k=1}^{n}\big(\frac{1}{k}-\frac{1}{k+4}\big)$ is a harmonic series equal to $\frac{50}{24}-\frac{1}{n+1}-\frac{1}{n+2}-\frac{1}{n+3}-\frac{1}{n+4}$
Finally
$\lim\limits_{n\to+\infty} \frac{1}{n\ln(n)}\big(\sum\limits_{k=1}^{n}\frac{1}{k}+\frac{1}{2}\sum\limits_{k=1}^{n}\big(\frac{1}{k}-\frac{1}{k+4}\big)\big)=\lim\limits_{n\to+\infty} \frac{1}{n\ln(n)}\big(\gamma+\ln(n)+\frac{25}{24}+\frac{1}{2}\big(-\frac{1}{n+1}-\frac{1}{n+2}-\frac{1}{n+3}-\frac{1}{n+4}\big)\big)$
where $\gamma$ is the Euler-Macheroni constant.
$\lim\limits_{n\to+\infty} \frac{1}{n}+\frac{\gamma+\frac{25}{24}}{n\ln(n)}+\frac{\big(-\frac{1}{n+1}-\frac{1}{n+2}-\frac{1}{n+3}-\frac{1}{n+4}\big)}{n\ln(n)}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2735194",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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Why can't I cancel $2x-3$ from $(2x-3)(x+5)=9(2x-3)$? Why are these simplifications wrong?
$$\begin{align}
(2x-3)(x+5)=9(2x-3) &\quad\to\quad \frac{(2x-3)(x+5)}{2x-3} = \frac{9(2x-3)}{2x-3} \quad\to\quad x+5 = 9\\[4pt]
x(x+2)=x(-x+3) &\quad\to\quad \frac{x(x+2)}{x} = \frac{x(-x+3)}{x} \quad\to\quad x+2=-x+3
\end{align}$$
| Remember the one rule when it comes to division:
Never divide by zero
If you want to remove $(2x-3)$ you must check that it is not equal to $0$ that is, you have to assume that $x\neq \frac{3}{2}$. Same goes with division by $x$.
In your first example, if you are looking for $x$, dividing by $(2x-3)$ without notice means that you lost a solution $\left(x=\frac{3}{2}\right)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2735413",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
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Probability of being dealt a pair in a poker hand
In poker, which is more likely to being dealt? A pair or two-pairs?
Solve using probability.
Attempt:
First, lets find the probability of a pair. The size of sample space is ${52 \choose 5}$. Now as for the ways to pick a pair, First we select the ranks, this can be done in ${13 \choose 1}$ ways and we want 2 suits from the 4 so this we do in ${4 \choose 2 }$ ways. Now, for the remaining three cards since one rank already been chosen we only have ${12 \choose 3}$ ways to pick the ranks of remaining three and each can be chosen in ${4 \choose 1 }$ ways (1 from each suit). Thus,
$$ P( pair) = \frac{ {13 \choose 1 }{4 \choose 2 }{12 \choose 3 } { 4 \choose 1 } }{ {52 \choose 5 }}$$
Now, for two pairs, like previous part, first pick rank ${13 \choose 1 }$ and ${4 \choose 2 }$ suits. Now, for the second pair, we now pick ${12 \choose 1}$ and ${4 \choose 2 }$ suits. Finally, the last card can be chosen in ${11 \choose 1 }4 $ ways. Thus,
$$ P( 2 \; pairs) = \frac{ {13 \choose 1 } {4 \choose 2 } { 12 \choose 1 } {4 \choose 2 } {11 \choose 1 } {4 \choose 1} }{ { 52 \choose 5} } $$
Hence,
$$ \frac{ P(pair ) }{P(2 pairs) } = \frac{ {12 \choose 3 } }{ {12 \choose 1 } {4 \choose 2 }{ 11 \choose 1 } } \approx 20/3$$
So, getting a pair is 20/3 more likely than two pairs. IS this correct argument?
| For one pair, the count should be
$$\binom{13}{1}\binom{4}{2}\binom{12}{3}\binom{4}{1}^3$$
Explanation:
*
*Choose the rank for the pair:$\;\binom{13}{1}$ choices.
*Choose the two cards for that rank:$\;\binom{4}{2}$ choices.
*Choose the three non-paired ranks:$\;\binom{12}{3}$ choices.
*Choose the cards for each of those ranks:$\;\binom{4}{1}^3$ choices.
Thus, you forgot to apply the factor $\binom{4}{1}$ for each of the $3$ non-paired ranks.
For two pairs, the count should be
$$\binom{13}{2}\binom{4}{2}^2\binom{11}{1}\binom{4}{1}$$
Explanation:
*
*Choose the ranks for the two pairs:$\;\binom{13}{2}$ choices.
*Choose the two cards for each of those ranks:$\;\binom{4}{2}^2$ choices.
*Choose the non-paired rank:$\;\binom{11}{1}$ choices.
*Choose the card for that rank:$\;\binom{4}{1}$ choices.
But in your count
$$\binom{13}{1}\binom{4}{2}\binom{12}{1}\binom{4}{2}\binom{11}{1}\binom{4}{1}$$
the second pair could just as well have been the first, so you are counting each two-pair hand twice. Thus, your count is twice the correct count.
Using the correct counts, the ratio works out to ${\large{\frac{80}{9}}}\approx 8.9$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2737652",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove for every natural number $n$, with $n ≥ 7$, that $\frac{(2n−18)}{(n^2−8n+ 8)} < 1$. Using induction, Prove for every natural number $n$, with $n ≥ 7$, that $\frac{(2n−18)}{(n^2−8n+ 8)} < 1$. I cannot get the original equation to match the k+1 equation, and im not sure what I am messing up. I would post my work but I would rather see it done out completely from scratch. Any help appreciated!
| Partial Answer as Method of Induction is Not Used.
For curiosity, I have provided an answer that does not use induction.
Multiply the fraction by $1=\dfrac{2n+18}{2n+18}$.
$$\begin{align} \frac{2n-18}{n^2-8n+8}\times\frac{2n+18}{2n+18}&=\frac{4n^2-324}{2n^3 - 16n^2 + 16n + 18n^2 -144n + 144} \\ \\ &=\frac{2n^2-162}{n^3-8n^2+8n+9n^2-72n+72} \\ \\ &= \frac{2n^2-162}{n^3+n^2-64n+72}.\end{align}$$ So now, given that $n\in\mathbb{N}_{\geqslant 7}$, we want to show that $$\begin{align} 2n^2 - 162 &< n^3 +n^2-64n+72 \\ \\ \Leftrightarrow n^2-234&<n^3-64n \\ \\ \Leftrightarrow n^2+64n-234&<n^3.\end{align}$$ So, since $n\geqslant 7$ then there will always exist a constant $c_n$ such that $$n^2 + 64n - (234+n^3-c_n)=0$$ iff (if and only if) $$\begin{align} n&=\frac{-64\pm\sqrt{4096+4(234+n^3-c_n)}}{2} \\ \\ &= 2\left(-16\pm\sqrt{1258+n^3-c_n}\right).\end{align}$$ But since $n>0$ then $$n=2\left(\sqrt{1258+n^3-c_n}-16\right).$$ Of course, to prove the inequality, we need to show that $c_n>0$. Of course $c_n\neq0$ because the constant must always be changing, depending on the value of $n$. Also, given the strict inequality, $c_n\not<0$, thus $c_n>0$.
This proves the inequality.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2739291",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
Inconsistency for solving $x' = x^{1/2}$ The proposed system $x' = x^{1/2}$ can be solved easily to obtain $x(t) = \frac{1}{4} (t^2 + t c + c^2)$, where $c$ is the integration constant.
However, differentiate the newly-found $x(t)$, one gets: $x(t)' = \frac{1}{2}t+\frac{1}{4}c$. This implies that $x^{1/2} = \frac{1}{2}t+\frac{1}{4}c$. However,
$$(x^{1/2})^2 = \left(\frac{1}{2}t+\frac{1}{4}c\right)^2 = \frac{1}{4}\left(t^2 + \frac{1}{2}tc + c^2\right) \neq \frac{1}{4} \left(t^2 + t c + c^2\right) = x(t)$$
Does anyone know why the inconsistency occurs? I understand there is another solution, but it is also inconsistent.
| The solution of
$$x'=\sqrt x$$ is given by $$x=\frac{1}{4} \left(t+c\right)^2\implies x'=\color{red}{2 \times}\frac{1}{4} \left(t+c\right) =\frac{1}{2} \left(t+c\right)=\sqrt x$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2741103",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Product of sums which equal to sum of product We can be sure that
$$\left(\sum\limits_{k=0}^{n}\frac{1}{k+1}\right)\left(\sum\limits_{k=0}^{n}\binom{n}{k}\frac{(-1)^k}{k+1}\right)= \sum\limits_{k=0}^{n}\binom{n}{k}\frac{(-1)^k}{(k+1)^2}$$
Is there any similar identities or some types of generalization to find them?
| A computer search will produce the identity for $p\ge 1$ a positive
integer:
$$\bbox[5px,border:2px solid #00A000]{
\sum_{k=0}^n \frac{1}{k+p}
\sum_{k=0}^n {n\choose k} \frac{(-1)^k}{k+p}
= \sum_{k=0}^n {n\choose k} \frac{(-1)^k}{(k+p)^2}.}$$
This is
$$(H_{n+p} - H_{p-1})
\sum_{k=0}^n {n\choose k} \frac{(-1)^k}{k+p}
= \sum_{k=0}^n {n\choose k} \frac{(-1)^k}{(k+p)^2}.$$
To evaluate the sum on the LHS we introduce
$$f(z) = (-1)^n \frac{n!}{z+p} \prod_{q=0}^n \frac{1}{z-q}.$$
We then obtain
$$\mathrm{Res}_{z=k} f(z)
= (-1)^n \frac{n!}{k+p}
\prod_{q=0}^{k-1} \frac{1}{k-q} \prod_{q=k+1}^n \frac{1}{k-q}
\\ = (-1)^n \frac{n!}{k+p} \frac{1}{k!} \frac{(-1)^{n-k}}{(n-k)!}
\\ = (-1)^k \frac{1}{k+p} {n\choose k}.$$
With residues adding to zero we find
$$\sum_{k=0}^n {n\choose k} \frac{(-1)^k}{k+p}
= - \mathrm{Res}_{z=-p} f(z) - \mathrm{Res}_{z=\infty} f(z).$$
Now for the residue at infinity we get formally that
$$\mathrm{Res}_{z=\infty} f(z) =
- \mathrm{Res}_{z=0} \frac{1}{z^2} f(1/z)
\\ = (-1)^{n+1} n! \times \mathrm{Res}_{z=0} \frac{1}{z^2}
\frac{1}{1/z+p} \prod_{q=0}^n \frac{1}{1/z-q}
\\ = (-1)^{n+1} n! \times \mathrm{Res}_{z=0} \frac{1}{z^2}
\frac{z}{1+pz} \prod_{q=0}^n \frac{z}{1-qz}
\\ = (-1)^{n+1} n! \times \mathrm{Res}_{z=0} z^n
\frac{1}{1+pz} \prod_{q=0}^n \frac{1}{1-qz} = 0.$$
We also have
$$- \mathrm{Res}_{z=-p} f(z) =
(-1)^{n+1} n! \prod_{q=0}^n \frac{1}{-p-q}
\\ = n! \prod_{q=0}^n \frac{1}{p+q}
= n! \frac{(p-1)!}{(n+p)!} = \frac{1}{p} {n+p\choose p}^{-1}.$$
We thus obtain for the LHS the closed form
$$\frac{1}{p} {n+p\choose p}^{-1} (H_{n+p} - H_{p-1}).$$
We use
$$g(z) = (-1)^n \frac{n!}{(z+p)^2} \prod_{q=0}^n \frac{1}{z-q}.$$
for the RHS. Observe that the residue at infinity is certainly zero
because we are dividing the bound on $|z|=R$ of $|f(z)|$ by an extra
factor $R-p.$ This leaves the residue at $z=-p$ and we get
$$- \mathrm{Res}_{z=-p} g(z) =
(-1)^{n+1} n!
\left.\left(\prod_{q=0}^n \frac{1}{z-q}\right)'\right|_{z=-p}
\\ = (-1)^{n} n!
\left.\prod_{q=0}^n \frac{1}{z-q}
\sum_{q=0}^n \frac{1}{z-q}
\right|_{z=-p}
\\ = (-1)^{n+1} n!
\left.\prod_{q=0}^n \frac{1}{z-q}
\right|_{z=-p}
\sum_{q=0}^n \frac{1}{p+q}
\\ = \frac{1}{p} {n+p\choose p}^{-1} (H_{n+p} - H_{p-1}).$$
This concludes the proof. With this curious identity the first
impression is that we have not understood the rules of basic
arithmetic involving sums and products, yet it reveals itself to be
true.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2741897",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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How to find the equation of the tangent to the parabola $y = x^2$ at the point (-2, 4)? This question is from George Simmons' Calc with Analytic Geometry. This is how I solved it, but I can't find the two points that satisfy this equation:
$$
\begin{align}
\text{At Point P(-2,4):} \hspace{30pt} y &= x^2 \\
\frac{dy}{dx} &= 2x^{2-1} \\
&= 2x = \text{Slope at P.}
\end{align}
$$
Now, the equation for any straight line is also satisfied for the tangent:
$$
\begin{align}
y - y_0 &= m(x - x_0) \\
\implies y - y_0 &= 2x (x - x_0) \\
\text{For point P, } x_0 &= -2 \text{ and } y_0 = 4 \\
\implies y - 4 &= 2x(x+2)\\
\implies y - 4 &= 2x^2 + 4x\\
\implies y &= 2x^2 + 4x +4\\
\end{align}
$$
This is where the problem occurs. If I were to try to solve for $y$ using:
$$
y = ax^2+bx+c \implies y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
$$
I'd get:
$$
\begin{align}
y &= 2x^2+4x+4 \text{ and, at x-intercept: }\\
x &= \frac{-4 \pm \sqrt{4^2 - (4\times2\times4)}}{2\times2} \\
x &= \frac{-4 \pm \sqrt{16 - 32}}{4} \\
x &= \frac{-4 \pm 4i}{4} \\
x &= -1 \pm i
\end{align}
$$
Is this the correct direction, or did I do something wrong?
| Notice in your first lines of reasoning at the point $P(-2,4)$ that is when $x=-2$, the slope of the tangent line at the point is
$$ \frac{d}{dx} (x^2) \bigg|_{x=-2} = 2x \bigg|_{x=-2} = -4 $$
Thus, we have $m=-4$ and given the point one has
$$ y - 4 = -4(x+2) $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2744119",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
Evaluate $\lim_{n\to \infty} n^{-n^2}\left( \prod_{r=0}^{n-1} \left(n+\frac {1}{3^r}\right) \right) ^n$
Evaluate $$\lim_{n\to \infty} n^{-n^2}\left( \prod_{r=0}^{n-1} \left(n+\frac {1}{3^r}\right) \right) ^n$$
Since on substituting $n=\infty$ we get a indeterminate form of $1^{\infty}$. Hence we can write the same limit as
$$\exp\left({\lim_{n\to \infty} \left(\frac {\prod_{r=0}^{n-1} \left(n+\frac {1}{3^r}\right)}{n^{n-1} }-n \right)} \right)$$
Which evaluates to $e^{3/2}$ . Is it correct? I would also like to know if there are any other methods for this problem
| In your attempt, you rewrote the expression in an incorrect way (you forgot the logarithm, when rewriting $x = e^{\log x}$). How did you get that second expression?
You have
$$
\left( \prod_{r=0}^{n-1} \left(1+\frac {1}{3^r}\right) \right)^n
= \exp\left( n \sum_{r=0}^{n-1} \log(n+3^{-r})\right)
= \exp\left( n^2\log n + n\sum_{r=0}^{n-1} \log(1+\frac{3^{-r}}{n})\right) \tag{1}
$$
so that
$$
n^{-n^2} \left( \prod_{r=0}^{n-1} \left(1+\frac {1}{3^r}\right) \right)^n
= \exp\left( n\sum_{r=0}^{n-1} \log(1+\frac{1}{n3^r})\right)\,. \tag{2}
$$
Now, you have $\log(1+u) = u+O(u^2)$ when $u\to 0$, from which
$$\begin{align}
n^{-n^2} \left( \prod_{r=0}^{n-1} \left(1+\frac {1}{3^r}\right) \right)^n
&= \exp\left( n\sum_{r=0}^{n-1} \left(\frac{1}{n3^r}+ O\left(\frac{1}{n^23^r}\right)\right)\right) \\
&= \exp\left( \sum_{r=0}^{n-1} \left(\frac{1}{3^r}+ O\left(\frac{1}{n3^r}\right)\right)\right) \\
&= \exp \left( \frac{3}{2}(1+o(1)\right) \\
&\xrightarrow[n\to\infty]{} \boxed{e^{3/2}}
\end{align}$$
indeed.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2744809",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Finding $P(X + 10/X > 7 )$ where $X$ is uniform
Let $X$ be a continuous r.v. with a continuous uniform distribution on
$[0,10]$. What is $P \left( X + \frac{10}{X} > 7 \right)$?
Attempt
We notice that $X+ \frac{10}{X} > 7$ can be rewritten as $X^2 + 10 - 7X > 0$ which is equivalent to $(X-5)(X-2) > 0$. Thus,
$$ P \left( X + \frac{10}{X} > 7 \right) = P[(X-5)(X-2)>0] = P[ \{X>2 \} \cup \{ X < 5 \} ] $$
Since we have two cases: either $X>5$ and $X>2$ OR $X<5$ and $X<2$. now, using inclusion-exclusion we get
$$ P(X>2) + P(X<5) - P(2<X<5) = 1 - F(2) + F(5) - F(5) + F(2) = 1$$
What is wrong with my solution? According to my solution key, I should get $\frac{7}{10}$. What am I missing here?
| Your mistake is in $$P[(X-5)(X-2)>0]=P(\{X>2\} \cup \{X<5\})$$ This equality is not correct. Instead $$P[(X-5)(X-2)>0]=P(\{X>5\} \cup \{X<2\})$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2748446",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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If $\frac{\cos^4 \alpha}{x}+\frac{\sin^4 \alpha}{y}=\frac{1}{x+y}$,prove that $\frac{dy}{dx}=\tan^2\alpha$ If $\frac{\cos^4 \alpha}{x}+\frac{\sin^4 \alpha}{y}=\frac{1}{x+y}$,prove that $\frac{dy}{dx}=\tan^2\alpha$
It is very long to direct differentiate it.Can someone help me?
| Using cauchy schwarz inequality $$\frac{(\cos^2\alpha)^2}{x}+\frac{(\sin^2 \alpha)^2}{y}\geq \frac{\cos^2 x+\sin^2 x}{x+y}=\frac{1}{x+y}$$.
And equality holds when $$\frac{\cos^2 \alpha}{x}=\frac{\sin^2 x}{y}$$
So $$y=\tan^2 (\alpha)\cdot x\Rightarrow \frac{dy}{dx}=\tan^2(\alpha)$$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2748907",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How to show that $a_n=1+1/\sqrt{2}+\cdots+(1/\sqrt{n-1})-2\sqrt{n}$ has an upper bound. Let $a_n=1+1/\sqrt{2}+\cdots+(1/\sqrt{n-1})-2\sqrt{n}$ while $a_1=-2,n\ge2 $ ,
I need to prove that $a_n$ converges.
I proved that it is monotonically increasing and tried to prove that it is upper-bounded by induction but failed to.
Also, it was told that $a_n$ converges to $-2<L<-1$ so I tried to show by induction that $a_n$ is bounded by $-1$, but I'm always stuck with $a_{n+1} \le -1 + 1/\sqrt{n}$ or something like that.
How can I prove that $a_n$ is bounded from above?
| Note: To prove something converges you don't have to figure out what it converges to. And to prove something is bounded above you don't have to find the least upper bound; it's enough just to find any upper bound (and prove it is an upper bound).
So
$a_{n+1} - a_n = \frac 1{\sqrt{n}} - 2\sqrt{n+1} + 2\sqrt{n}$.
Claim: $2\sqrt n + \frac 1{\sqrt n} > 2\sqrt{n+1}$ for all natural $n$.
Pf: As all terms are positive and greater than zero...
$2\sqrt n + \frac 1{\sqrt n} > 2\sqrt{n+1} \iff$
$(2\sqrt n + \frac 1{\sqrt n})^2 > (2\sqrt{n+1})^2 \iff $
$4 n + 4 + \frac 1n > 4(n+1) \iff$
$\frac 1n > 0$.
So it it is true.
And so $a_{n+1} - a_n = 2\sqrt{n} + \frac 1{\sqrt n} - 2\sqrt{n+1} > 0$ so $a_{n+1} > a_n$.
So $\{a_i\}$ is monotonically increasing.
Claim: $a_{n} \le -1 + \frac 1{\sqrt{n-1}} \le 0$ for all $n\ge 2$.
Base case: $a_2 = 1 - 2\sqrt{2} \le 1-2 = -1 < -1 + \frac 1{\sqrt{1}} \le 0$.
Inductive step:
If $a_n \le -1 +\frac 1{\sqrt{n-1}}$ then
$a_{n+1} = a_n + \frac 1{\sqrt{n}} - 2\sqrt{n} + 2\sqrt{n-1}$
$\le -1 +\frac 1{\sqrt{n-1}}+ \frac 1{\sqrt{n}} - 2\sqrt{n} + 2\sqrt{n-1}$
We proved above that $\frac 1{\sqrt{n-1}}- 2\sqrt{n} + 2\sqrt{n-1} > 0$ so
$a_{n+1} \le -1 +\frac 1{\sqrt{n-1}}+ \frac 1{\sqrt{n}} - 2\sqrt{n} + 2\sqrt{n-1} < -1 +\frac 1{\sqrt{n}} < -1 + 1 = 0$.
So the claim is true.
$a_n \le -1 + \frac 1{\sqrt{n-1}} \le 0$ so $\{a_i\}$ is bounded above by $0$.
Claim 2: $\{a_i\}$ is bounded above by $-1$.
Claim: Let $\epsilon > 0$. Then $\frac 1{\sqrt{n}} < \epsilon \iff n > \frac 1{\epsilon^2}$.
So for all $n + 1 > \frac 1{\epsilon^2}$ then $a_{n+1} \le -1 + \frac 1{\sqrt{n}} < - 1 + \epsilon$. But as $\{a_i\}$ is monotonically increasing. If $m \le n \le \frac 1{\epsilon^2} -1$ then $a_m < a_{n+1} < -1 + \epsilon$ and if $n \ge n+1 > \frac 1{\epsilon^2}$ then $a_m \le -1 + \frac 1{\sqrt{m-1}} < -1 + \epsilon$.
So all $a_i < 1 + \epsilon$ for all $\epsilon > 0$.
So $a_i \le -1$.
And $\{a_i\}$ is monotonically increasing.
So $\{a_i\}$ converges.
That's it. Now, I don't have any idea what it converges to. My proof was crude and ham-fisted. But it was a legitimate proof.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Finding solutions for $\sin(z)=\sin(2)$ having some difficulty wrapping my head around how to methodically do this style of question:
Find all solutions $z\in ℂ$ for $\sin(z)=\sin(2)$
I attempted to solve this by using the identity,
$\sin(x)\cosh(y)+i\cos(x)\sinh(y)=\sin(2)$,
so that,
$\sin(x)\cosh(y)=\sin(2)$ and $\cos(x)\sinh(y)=0$,
but this seemed to lead to a dead end, so I'm unsure as to what the next step should be...
Any help is much appreciated
| $$\begin{align}
& \text{This problem still can be solved following your procedure}: \\
& \cos \left( x \right)\sinh \left( y \right)=0\Rightarrow \cos \left( x \right)=0\ OR\ \sinh \left( y \right)=0 \\
& \\
& If\sinh \left( y \right)=0,\ then\ y=0\ ,and\ hence\ \\
& \sin \left( x \right)\cosh \left( y \right)=\sin \left( x \right)\cosh \left( 0 \right) \\
& \quad \quad \quad \quad \quad \quad =\sin \left( x \right)\times 1 \\
& \quad \quad \quad \quad \quad \quad =\sin \left( x \right) \\
& \quad \quad \quad \quad \quad \quad \Rightarrow \sin \left( x \right)=\sin \left( 2 \right)\Rightarrow x=2+2n\pi . \\
& \\
& If\ \cos \left( x \right)=0,\ then\ \sin \left( x \right)=\pm \sqrt{1-\cos {{\left( x \right)}^{2}}}=\pm \sqrt{1-{{0}^{2}}}=\pm 1,and\ hence \\
& \ \sin \left( x \right)\cosh \left( y \right)=\pm \cosh \left( y \right)=\sin \left( 2 \right) \\
& \quad \quad \quad \quad \quad \quad \,\,\Rightarrow \ \cosh \left( y \right)=\pm \sin \left( 2 \right)\ \\
& \quad \quad \quad \quad \quad \quad \,\,\Rightarrow \ \cosh \left( y \right)=\sin \left( 2 \right)\ since\ \cosh \left( y \right),\sin \left( 2 \right)>0.\ \\
& \quad \quad \quad \quad \quad \quad \,\,\Rightarrow y={{\cosh }^{-1}}\left( \sin \left( 2 \right) \right). \\
& \\
& Finally\ z=\left( 2+2n\pi \right)+i{{\cosh }^{-1}}\left( \sin \left( 2 \right) \right) \\
\end{align}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2751593",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Preimage of two-dimensional function We have the function $f:(\mathbb{R}^2,\|\cdot\|_2)\rightarrow (\mathbb{R},|\cdot |)$ with \begin{equation*}f(x,y)=\begin{cases}y-x & y\geq x^2 \\ 0 & y<x^2\end{cases}\end{equation*}
How could we draw the preimage $f^{-1}((1,\infty))$ ? Could you give me a hint?
| Since we want the pre-image of $(1,\infty)$, we can ignore all regions where the function is equal to $0$. That is, focus on $y \ge x^2$. In particular, we have $f(x,y) = 1 \iff y - x = 1 \iff y = x + 1$. Hence, $f(x,y) > 1 \iff y > x + 1$ and $y \ge x^2$.
At this point, we could stop here and say that you can draw the graphs of $y = x + 1$ and $y = x^2$, and shade the regions that are above both lines (including only the line of the second graph).
To be even more explicit, notice that $x^2 = x + 1 \Rightarrow x = \frac{1 \pm \sqrt{5}}{2}$. A quick test shows that $x^2 > x + 1$ for $x < \frac{1 - \sqrt{5}}{2}$ or $x > \frac{1 + \sqrt{5}}{2}$, and $x^2 < x + 1$ for $\frac{1 - \sqrt{5}}{2} < x < \frac{1 + \sqrt{5}}{2}$. Hence,
$$f^{-1}((1,\infty)) = \{(x,y) \mid y \ge x^2 \text{ and } x < \frac{1 - \sqrt{5}}{2} \text{ or } x > \frac{1 + \sqrt{5}}{2}\} \cup \{(x,y) \mid y > x + 1 \text{ and } \frac{1 - \sqrt{5}}{2} \le x \le \frac{1 + \sqrt{5}}{2}\}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2755323",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Determining the minimal length of the hypotenuse How can I solve the following problem?
The hint I received is that the minimal length is not given by $L=15$. I feel like I need more information to solve this problem because I can't get any further. Any hints, tips, solutions are appreciated!
Thank you.
| Alternatively, denote the hypotenuse of small right-angle triangle by $z$. Then:
$$\frac{L}{z}=\frac{8+x}{x} \ \ \text{and} \ \ z^2=x^2+3^2 \Rightarrow\\
L=\left(\frac{8}{x}+1\right)z \to min, \ \ s.t. \ \ z=\sqrt{x^2+9}.$$
Subsituting:
$$L(x)=\left(\frac{8}{x}+1\right)\sqrt{x^2+9}, \\
L'(x)=\left(-\frac{8}{x^2}\right)\sqrt{x^2+9}+\left(\frac 8x+1\right)\frac{x}{\sqrt{x^2+9}}=0 \Rightarrow \\
\frac{-8(x^2+9)+(8+x)x^2}{x^2\sqrt{x^2+9}}=0 \Rightarrow x^3=72 \Rightarrow x=2\sqrt[3]{9}.$$
Hence:
$$\tan \alpha = \frac 3x \Rightarrow \alpha = \arctan \frac{3}{2\sqrt[3]{9}}=0.624763... \ \text{rad}=35.8^\circ.\\
L(2\sqrt[3]{9})=14.99216...$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Finding the Laurent series of the following function I need to find the laurent series and the residue of the following complex function
$$f(z)=(z+1)^2e^{3/z^2}$$
at $z=0$.
Since $e^z=\sum z^n/n!$, then
$$e^{3/z^2}=\sum_{n=0}^\infty \frac{3^n/n!}{z^{2n}}$$
thus
$$f(z)=(z^2+2z+1)\sum_{n=0}^\infty \frac{3^n/n!}{z^{2n}}=\sum_{n=0}^\infty \frac{3^n/n!}{z^{2(n-1)}}+\sum_{n=0}^\infty \frac{2\cdot3^n/n!}{z^{2n-1}}+\sum_{n=0}^\infty \frac{3^n/n!}{z^{2n}}$$
which, with a shift of index and expansion of positive powers, can be expressed as
$$f(z)=z^2+3+\sum_{n=1}^\infty\left(\frac{3}{n+1}+1\right)\frac{3^n/n!}{z^{2n}}+\sum_{n=1}^\infty\frac{2\cdot3^n/n!}{z^{2n-1}}$$
so the residue is given by evaluating the numerator of the second series at $n=1$, so its value is $6$. I tried using WolframAlpha and Mathematica to check my answer, but both would not return a value. Would this be correct? Also, is there a way to put the two sums together (one gives the coefficients of even powers, while the other of the odd) so I can have the principal part of the laurent series expressed only with one sum?
|
Using the coefficient of operator $[z^n]$ to denote the coefficient of $z^n$ of a series, we obtain
\begin{align*}
\color{blue}{\mathrm{res}_{z=0}f(z)}&=[z^{-1}](z+1)^2e^{3/z^2}\\
&=\left([z^{-3}]+2[z^{-2}]+[z^{-1}]\right)\sum_{j=0}^\infty 3^jz^{-2j}\\
&=0+2\cdot 3^1+0\\
&\,\,\color{blue}{=6}
\end{align*}
and see the residue of $f$ at $z=0$ is $6$.
Expanding $f$ into a Laurent series at $z=0$ we obtain
\begin{align*}
\color{blue}{f(z)}&\color{blue}{=(z+1)^2e^{3/z^2}}\\
&=(z^2+2z+1)\sum_{n=0}^\infty \frac{3^n}{n!}\frac{1}{z^{2n}}\\
&=\sum_{n=0}^\infty \frac{3^{n}}{n!}\frac{1}{z^{2n-2}}
+2\sum_{n=0}^\infty \frac{3^n}{n!}\frac{1}{z^{2n-1}}
+\sum_{n=0}^\infty \frac{3^n}{n!}\frac{1}{z^{2n}}\\
&=\sum_{n=-1}^\infty \frac{3^{n+1}}{(n+1)!}\frac{1}{z^{2n}}
+2\sum_{n=-1}^\infty \frac{3^{(n+1)}}{(n+1)!}\frac{1}{z^{2n+1}}
+\sum_{n=0}^\infty \frac{3^n}{n!}\frac{1}{z^{2n}}\\
&\color{blue}{=z^2+2z+\sum_{n=0}^\infty\frac{3^n}{n!}\left(\frac{3}{n+1}+1\right)\frac{1}{z^{2n}}
+2\sum_{n=0}^\infty \frac{3^{(n+1)}}{(n+1)!}\frac{1}{z^{2n+1}}}\tag{1}\\
&=z^2+2z+\sum_{m=0}^\infty\frac{3^{\frac{m}{2}}}{(\frac{m}{2})!}\left(\frac{3}{\frac{m}{2}+1}+1\right)\frac{1+(-1)^m}{2}\frac{1}{z^{m}}\\
&\qquad+2\sum_{m=0}^\infty \frac{3^{(\frac{m-1}{2}+1)}}{(\frac{m-1}{2}+1)!}\cdot\frac{1-(-1)^m}{2}\frac{1}{z^{m}}\\
&=z^2+2z+\sum_{m=0}^\infty\left(\frac{3^{\frac{m}{2}}}{2(\frac{m}{2})!}\left(\frac{m+8}{m+2}\right)\left(1+(-1)^m\right)\right.\\
&\qquad\qquad\qquad\qquad\quad\left.+ \frac{3^{\frac{m+1}{2}}}{\left(\frac{m+1}{2}\right)!}\left(1-(-1)^m\right)\right)\frac{1}{z^{m}}\tag{2}
\end{align*}
We observe that putting all terms together as in (2) is rather cumbersome and the representation in (1) seems to be more convenient. It clearly shows the terms for odd and even powers and we can also easily derive from (1) the residue by inspecting the right-most sum in (1) with $n=0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2758338",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
Find recurrence for $I_n$. Let $I_n=\int_{0}^{1/2} \frac {x^n}{\sqrt{1-x^2}}dx.$ I must find a recurrence for this so I just started using interation by parts:
Let $$f'(x)=x^n\to f(x)=\frac{x^{n+1}}{n+1}$$ and
$$g(x)=\frac 1{\sqrt{1-x^2}}\to g'(x)=\frac x{\sqrt{(1-x^2)^3}}$$
Therefore:
$$I_n=\frac {x^{n+1}}{(n+1)\sqrt{1-x^2}} - \int_{0}^{\frac 12}\frac{x^{n+2}}{(1-x^2)\sqrt{1-x^2}}dx$$
And I can't continue from here..
I tried rewriting $x^{n+2}=x^2x^n=(1-x^2+1)x^n$ but it's no good.
| $$I_n=\int_0^{1/2}\frac{x^n}{\sqrt{1-x^2}}dx\\
=-x^{n-1}\sqrt{1-x^2}\,\Big\vert_0^{1/2}+(n-1)\int_0^{1/2}x^{n-2}\sqrt{1-x^2}dx\\
-(1/2)^{n-1}(3/4)^{1/2}+(n-1)\int_0^{1/2}\frac{x^{n-2}-x^n}{\sqrt{1-x^2}}dx\\
=-(1/2)^{n-1}(3/4)^{1/2}+(n-1)(I_{n-2}-I_n).
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2758516",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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} |
Find $x, y\in\mathbb Z$ such that $\frac{x}{2} + \frac{x}{y} - \frac{3}{2} = \frac{10}{y}$ Find $x, y\in\mathbb Z$ such that $\dfrac{x}{2} + \dfrac{x}{y} - \dfrac{3}{2} = \dfrac{10}{y}$
My try
$$\dfrac{x}{2} + \dfrac{x}{y} - \dfrac{3}{2} = \dfrac{10}{y} \iff \dfrac{y}{2} = \dfrac{-x + 10}{x-3} \ \ (x\neq 3,\ 10, \ y\neq 0, \ x, \ y\in\mathbb Z)$$
Let $t = \dfrac{y}{2} = \dfrac{-x + 10}{x-3}.$ We have $y = 2t\neq 0,$ and $x = \dfrac{3t + 10}{t + 1} = 3 + \dfrac{7}{t+1}.$
Then, $x\in\mathbb Z \Leftrightarrow (t+1)|7 \Leftrightarrow t\in \{-8, -2, 6\}.$
Hence, $$x\in \{-4, 2, 4\}, \mbox { and } y\in \{-16, -4, 12\}.$$
| Note that listing possible values as sets is not concise enough as it suggests that, for example, x=4 and y=-16 work as a solution. Instead you should pair your answers.
Also, the method you used does give correct solutions, but it is not exhaustive. Rather, try a more standard Diophantine decomposition method.
$$ (y+2)(x-3) = 14 $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2761984",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Generating function to find the sum of digits How many integers between $30,000$ and $100,000$ have a sum of $15$ or less?
I was approaching this problem with a generating function:
$$g(x) = (x^3+x^4+x^5+x^6+x^7+x^8+x^9)(1+x+...+x^9)^4$$
First I'm going to pull out $x^3$ then I would need the coefficients of $x$ to the $15$th, $14$th, $13$th etc correct? If so, can someone help me through finding the coefficients?
Also finding the coefficient of x to the 15th would be the same as finding the coefficient of x to the 12 is I pull out x$^3$
| We want the number of solutions in integers to
$$x_1+x_2+x_3+x_4+x_5 \le 15$$
where $3 \le x_1 \le 9$ and $0 \le x_i \le 9$ for $2 \le i \le 5$. This is the same as the number of solutions to
$$x_1+x_2+x_3+x_4+x_5+x_6 = 15$$
with the additional constraint $0 \le x_6$. More generally, we seek the generating function of the number of solutions $a_r$ to
$$x_1+x_2+x_3+x_4+x_5+x_6 = r$$
with the above constraints.
The generating function is
$$\begin{align}
f(x) &= (x^3 + x^4 + x^5 + \dots +x^9) (1+x+x^2+\dots +x^9)^4(1+x+x^2+\dots) \\
&=x^3 \cdot \frac{1-x^7}{1-x} \cdot \left( \frac{1-x^{10}}{1-x} \right)^4 \cdot \frac{1}{1-x} \\
&=x^3 (1-x^7)(1-x^{10})^4 (1-x)^{-6} \\
&= x^3 (1-x^7) \cdot \sum_{i=0}^4 (-1)^i \binom{4}{i} x^{10i} \cdot \sum_{j=0}^{\infty} \binom{6+j-1}{j} x^j
\end{align}$$
From the final equation above we can extract the coefficient of $x^{15}$:
$$a_{15}=\binom{6+12-1}{12}-\binom{4}{1} \binom{6+2-1}{2}-\binom{6+5-1}{5} = \boxed{5852}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2762449",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Finding an appropriate upper bound for $\frac{x^2-4}{x-1}$ at $x=0$ and $x=2$ I am investigating the continuity of the function $\frac{x^2-4}{x-1}$ at $x=0$ and $x=2$.
Here's a part of my proof;
$$\lvert f(x)-f(0)\rvert =\left\lvert\frac{x^2-4}{x-1}-4\right\rvert$$
$$=\left\lvert \frac{x(x+4)}{x-1}\right\rvert$$
$$=\left\lvert \frac{x(x+4)}{x-1}\right\rvert$$
$$=\frac{\left\lvert x\right\rvert\left\lvert x+4\right\rvert}{\left\lvert x-1\right\rvert}$$
The challenge I have is to find an appropriate upper bound for $\frac{1}{x-1}$ at $x=0$ and $x=2$. Please, can anyone help out?
|
$$\lvert f(x)-f(0)\rvert =\frac{\left\lvert x\right\rvert\left\lvert x+4\right\rvert}{\left\lvert x-1\right\rvert}$$
The challenge I have is to find an appropriate upper bound for $\frac{1}{x-1}$ at $x=0$ and $x=2$. Please, can anyone help out?
You want to stay close enough to $x=0$ to ensure $\tfrac{1}{x-1}$ stays bounded.
You obviously don't want to go near $x=1$, since $\tfrac{1}{x-1}$ then becomes (arbitrary) large. Limit yourself to a distance of e.g. $\tfrac{1}{2}$ from $x=0$, then:
$$|x|<\tfrac{1}{2} \iff -\tfrac{1}{2} < x < \tfrac{1}{2}\iff -\tfrac{3}{2} < x-1 < -\tfrac{1}{2}$$
What does this tell you about $|x-1|$ and thus about $\tfrac{1}{|x-1|}$?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2764567",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Calculating a determinant. $D_n$=\begin{vmatrix}
a & 0 & 0 & \cdots &0&0& n-1 \\
0 & a & 0 & \cdots &0&0& n-2\\
0 & 0 & a & \ddots &0&0& n-3 \\
\vdots & \vdots & \ddots & \ddots & \ddots&\vdots&\vdots \\
\vdots & \vdots & \vdots & \ddots & \ddots& 0&2 \\
0 & \cdots & \cdots & \cdots &0&a&1 \\
n-1 & n-2 & n-3 & \cdots & 2 & 1& a\\
\end{vmatrix}
I tried getting the eigenvalues for A =
\begin{vmatrix}
0 & 0 & 0 & \cdots &0&0& n-1 \\
0 & 0 & 0 & \cdots &0&0& n-2\\
0 & 0 & 0 & \ddots &0&0& n-3 \\
\vdots & \vdots & \ddots & \ddots & \ddots&\vdots&\vdots \\
\vdots & \vdots & \vdots & \ddots & \ddots& 0&2 \\
0 & \cdots & \cdots & \cdots &0&0&1 \\
n-1 & n-2 & n-3 & \cdots & 2 & 1& 0\\
\end{vmatrix}
For $a=0$ , the rank of the matrix is $2$ , hence $\dim(\ker(A)) = n-2 $
$m(0)>=n-2$
However, I was not able to determine the other eigenvalues.
Testing for different values of n :
for $n=2$ :
$D_2 = a^2-1$
for $n=3$ :
$D_3 = a^3 -5a$
$D_n$ seems to be equal to $a^n - a^{n-2}\sum_{i=1}^{n-1}i^2$ .
However I'm aware that testing for different values of $n$ is not enough to generalize the formula.
Thanks in advance.
| Expand with respect to the first line: the term obtained with $a$ is $aD_{n-1}$. For the second one, we get $(-1)^{n+1}$ times a determinant that can be expanded with respect to the first column. This lead to the recurrence relation
$$
D_n=aD_{n-1}-\left(n-1\right)^2a^{n-2}.
$$
Letting $b_n:=a^{-n}D_n$ for $a\neq 0$ allows to derive and easier recurrence relation whose resolution shows that the formula mentioned in the opening post, namely,
$$
D_n=a^n - a^{n-2}\sum_{i=1}^{n-1}i^2,
$$
is correct.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2765475",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
"answer_id": 0
} |
Find vector $x \not= 0$ that satisfies the equation $Ax = x$. Given a matrix A:
$$A = \begin{bmatrix}
0 & 0 & 0 & \frac{1}{2} & \frac{1}{2} \\
\frac{1}{3} & 0 & 1 & 0 & 0 \\
\frac{1}{3} & 0 & 0 & 0 & \frac{1}{2} \\
\frac{1}{3} & \frac{1}{2} & 0 & 0 & 0 \\
0 & \frac{1}{2} & 0 & \frac{1}{2} & 0 \\
\end{bmatrix}$$
I have to find a vector $x \not= 0$ that satisfies the equation $Ax = x$.
I have no idea how to approach this. Any hints would be appreciated.
Thanks in advance.
| HINT
Note that
$$Ax = x\iff Ax-Ix=0\iff (A-I)x=0$$
thus the original problem is equivalent to find the null space for
$$B=A-I= \begin{bmatrix}
-1 & 0 & 0 & \frac{1}{2} & \frac{1}{2} \\
\frac{1}{3} & -1 & 1 & 0 & 0 \\
\frac{1}{3} & 0 & -1 & 0 & \frac{1}{2} \\
\frac{1}{3} & \frac{1}{2} & 0 & -1 & 0 \\
0 & \frac{1}{2} & 0 & \frac{1}{2} & -1 \\
\end{bmatrix}
$$
| {
"language": "en",
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"source": "stackexchange",
"question_score": "2",
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The shortest distance from a point to the graph of the function To compute the distance from the point (5,5) to the graph of xy=4. I choose an arbitrary point (u,v) on the graph of $xy=4$.
I get $d(u,v)=\sqrt{(u-5)^2+(v-5)^2}$ again $(u,v)$ satisfies equation of hyperbola so that $uv=4$. Now what shall i do next?
| I offer this as an alternative solution.
Looking at the graph of $xy=4$ and the point $P=(5,5)$, clearly we need only consider the branch of $xy=4$ that is in the first quadrant. So we need only consider the function $y = \dfrac 4x$ where $x > 0$.
Consider the particular point $Q=\left(\xi, \dfrac{4}{\xi}\right)$ on the curve $y = \dfrac 4x$. The slope of the tangent to the curve at the point $Q$ is
$y'=-\dfrac{4}{\xi^2}$.
So the slope of the normal line to the curve at that point is $\dfrac 14 \xi^2$.
The equation of the normal line to the curve at point $P$ is therefore
$$y-\dfrac{4}{\xi} = \dfrac 14\xi^2(x- \xi)$$
$Q$ will be the closests point on the curve to the point $P=(5,5)$ when the normal line contains the point $Q$. So
\begin{align}
5-\dfrac{4}{\xi} &= \dfrac 14\xi^2(5- \xi) \\
20\xi -16 &= \xi^3(5-\xi) \\
20\xi -16 &= 5\xi^3-\xi^4 \\
\xi^4 - 5\xi^3+20\xi-16 &= 0 \\
\xi^4 - 5\xi^3 + 4\xi^2 - 4\xi^2 + 20\xi - 16 &= 0 \\
\xi^2(\xi^2 - 5\xi + 4) - 4(\xi^2 - 5\xi + 4) &= 0 \\
(\xi^2 - 4)(\xi^2 - 5\xi + 4) &= 0 \\
(\xi + 2)(\xi - 2)(\xi - 1)(\xi - 4) &= 0 \\
\xi &\in \{ 1,2,4 \}
\end{align}
\begin{array}{c}
\xi & Q & d(P,Q) \\
\hline
1 & (1,4) &\sqrt{17} \\
2 & (2,2) & \sqrt{18} \\
4 & (4,1) & \sqrt{17} \\
\hline
\end{array}
So the points on the curve $xy=4$ that are closest to the point $(5,5)$ are $(1,4)$ and $(4,1)$ and the distance is $\sqrt{17}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2767570",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Sum of series $\sum^{10}_{i=1}i\bigg(\frac{1^2}{1+i}+\frac{2^2}{2+i}+\cdots \cdots +\frac{10^2}{10+i}\bigg)$
The Sum of series $$\sum^{10}_{i=1}i\bigg(\frac{1^2}{1+i}+\frac{2^2}{2+i}+\cdots \cdots +\frac{10^2}{10+i}\bigg)$$
Try: Let $$S=\sum^{10}_{i=1}\frac{i}{1+i}+2^2\sum^{10}_{i=1}\frac{i}{2+i}+\cdots \cdots \cdots +10^2\sum^{10}_{i=1}\frac{i}{10+i}$$
$$S=\sum^{10}_{i=1}\sum^{10}_{j=1}\bigg[\frac{i}{i+j}-1\bigg]-100$$
Could some help me How to solve it, Thanks in advanced
| HINT:
Write
$$\begin{align}S&=\sum^{10}_{i=1}\frac{i}{1+i}+2^2\sum^{10}_{i=1}\frac{i}{2+i}+\cdots+10^2\sum^{10}_{i=1}\frac{i}{10+i}\\&=\sum^{10}_{i=1}\frac{i}{1+i}+4\sum^{11}_{i=2}\left(\frac{i}{1+i}-\frac1{1+i}\right)+\cdots+100\sum^{19}_{i=10}\left(\frac{i}{1+i}-\frac9{1+i}\right)\end{align}$$ and combine like expressions.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2773345",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Converting $\frac{(3+3i)^5 (-2+2i)^3}{(\sqrt 3 +i)^{10}}$ to trigonometric form
Convert complex to trig.
$$\frac{(3+3i)^5 (-2+2i)^3}{(\sqrt 3 +i)^{10}}$$
Let us consider
$$(3+3i)^5$$
Here $a = 3$ ,$b=3$
$$\sqrt{a^2+b^2}=\sqrt{3^2+3^2} = \sqrt{18}=3\sqrt{2}$$
$$\theta =\arctan\biggr(\frac{b}{a}\biggr) =\arctan\biggr(\frac{3}{3}\biggr) = \arctan (1)$$
$$\theta = \frac{\pi}{4}$$
Thus in trigonometric form we get
$$Z_1 = \biggr [3\sqrt2 \bigg(\cos \frac{\pi}{4}+i\sin \frac{\pi}{4}\bigg)\bigg]^5$$
$$Z_1 =(3\sqrt 2)^5 \biggr [\bigg(\cos \frac{5\pi}{4}+i\sin \frac{5\pi}{4}\bigg)\bigg]$$
Let us consider
$$(-2+2i)^3$$
$a = -2$, $b=2$
$$\sqrt{a^2+b^2}=\sqrt{(-2)^2+2^2} =2\sqrt{2}$$
$$\theta =\arctan\biggr(\frac{b}{a}\biggr) =\arctan\biggr(\frac{2}{-2}\biggr) = \arctan (1)$$
$$\theta = \frac{3\pi}{4}$$
$$Z_2 =(2\sqrt 2)^3 \biggr [\bigg(\cos \frac{9\pi}{4}+i\sin \frac{9\pi}{4}\bigg)\bigg]$$
Multiplying $Z_1 \cdot Z_2$ we get
$$Z_1 \cdot Z_2 = (3\sqrt 2)^5 \cdot (2\sqrt2)^3 \biggr [\bigg(\cos \frac{5\pi}{4}+\cos \frac{9\pi}{4}\bigg)+i\sin \bigg(\frac{5\pi}{4}+i\sin\frac{9\pi}{4}\bigg)\bigg]$$
$$Z_1 \cdot Z_2 = 31104\biggr [\bigg(\cos \frac{7\pi}{2}+i\sin \frac{7\pi}{2}\bigg)\bigg]$$
Let us consider
$$(\sqrt 3+i)^{10}$$
Here $a = \sqrt 3$, $b =1$
$$\sqrt{a^2+b^2}=\sqrt{(\sqrt3)^2+1^2} = \sqrt{4}=2$$
$$\theta = \arctan \biggr(\frac{1}{\sqrt 3}\biggr) = \frac{\pi}{6}$$
$$Z_3 = 2^{10} \biggr [ \cos \biggr(\frac{5\pi}{3}\biggr)+i\sin \biggr(\frac{5\pi}{3}\biggr)\biggr]$$
Now we have
$$\frac{Z_1 \cdot Z_2}{Z_3} = \frac{31104\biggr [\bigg(cos \biggr(\frac{7\pi}{2}\biggr) +i\sin \biggr(\frac{7\pi}{2}\biggr) \bigg]}{2^{10} \biggr [ \cos \biggr(\frac{5\pi}{3}\biggr)+i\sin \biggr(\frac{5\pi}{3}\biggr)\biggr]} = \boxed {30.375 \biggr [ \cos \biggr(\frac{11\pi}{6}\biggr)+i\sin \biggr(\frac{11\pi}{6}\biggr)\biggr]}$$
Is my assumption correct?
| It might be useful to simplify before converting:
$$\frac{(3+3i)^5 (-2+2i)^3}{(\sqrt 3 +i)^{10}} = \frac{3^5\cdot 2^3\cdot ((1+i)(-1+i))^3(1+i)^2}{4^{10}}(\sqrt 3 - i)^{10}= \frac{3^5\cdot 2^3\cdot (-2)^3\cdot2i}{4^{10}}2^{10}e^{-i\frac{5}{3}\pi}$$$$= - \frac{243}{8}ie^{-i\frac{5}{3}\pi} = -\frac{243}{8}e^{i\frac{5}{6}\pi}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2776335",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Use residues to evaluate the improper integral $\int_0^\infty\frac{x^3\sin(x)}{(x^2+4)(x^2+16)}dx$ I am trying to solve the below problem using residues,
$$\int_0^\infty\frac{x^3\sin(x)}{(x^2+4)(x^2+16)}dx$$
This is what I have so far:
Firstly, Change the equation to $z$ as follows
$$f(z)=\frac{z^3}{(z^2+4)(z^2+16)}dz$$
Then, Identify the singularities in the upper half plane - these being $2i$ and $4i$ where $R>4$.
This brings us to the 2 equations:
$$\operatorname*{Res}_{z=2i} [f(z)e^{iz}]=\frac{z^3e^{iz}}{(z+2i)(z^2+16)}\biggr]_{z=2i}$$
and
$$\operatorname*{Res}_{z=4i} [f(z)e^{iz}]=\frac{z^3e^{iz}}{(z^2+4)(z+4i)}\biggr]_{z=4i}$$
Now when I try to solve each of these equations I am going wrong somewhere and ending up with some ridiculous answer... any help with this would be greatly appreciated!
| *
*Note that the residue method refers to the improper integral $\int_{\color{blue}{-\infty}}^\infty\frac{x^3sin(x)}{(x^2+4)(x^2+16)}dx = 2\int_0^\infty\frac{x^3sin(x)}{(x^2+4)(x^2+16)}dx$.
*You then calculate $I =Im(\int_{-\infty}^\infty\frac{z^3}{(z^2+4)(z^2+16)}e^{iz}dz)$.
$$I = 2\pi i (Res_{2i} \frac{z^3}{(z^2+4)(z^2+16)}e^{iz} + Res_{4i} \frac{z^3}{(z^2+4)(z^2+16)}e^{iz})$$
*Note, that the poles at $z=2i$ and $z=4i$ have order 1.
$$Res_{2i} \frac{z^3}{(z^2+4)(z^2+16)}e^{iz} = \lim_{z\rightarrow 2i} \frac{(z-2i)z^3}{(z-2i)(z+2i)(z^2+16)}e^{iz}= \frac{(2i)^3}{4i((2i)^2+16)}e^{-2} =-\frac{1}{6e^2}$$
$$Res_{4i} \frac{z^3}{(z^2+4)(z^2+16)}e^{iz} = \lim_{z\rightarrow 4i} \frac{(z-4i)z^3}{(z^2+4)(z-4i)(z+4i)}e^{iz}= \frac{(4i)^3}{((4i)^2+4)(8i)}e^{-4} =\frac{2}{3e^4}$$
So,
$$\int_0^\infty\frac{x^3sin(x)}{(x^2+4)(x^2+16)}dx = Im(\frac{1}{2}2\pi i (-\frac{1}{6e^2} + \frac{2}{3e^4})) = \boxed{\frac{\pi}{3e^2}(\frac{2}{e^2}-\frac{1}{2})}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2777478",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Prime factor $p$ of $10^{2^k}+ 1$ is $p \equiv 1 \pmod{2^{k+1}}$ I am asked to show that for when $n$ is a power of 2, any prime factor $p$ of $10^n + 1$ is $p \equiv 1 \pmod { 2n}$.
For $n = 1$, we have $p = 10^1 +1 = 11 \equiv 1 \pmod 2$
For $n > 1$ we now have that $10^n \equiv -1 \pmod{p}$ so $-1$ is a quadratic residue $\pmod p$ hence $ p \equiv 1 \pmod 4 \Rightarrow p \equiv 1 + 4l \pmod{2n}$ for some $0 \leq l < \frac{n}{2}$
Additionally, we have that $2n \mid 10^n \Rightarrow 10^n + 1 \equiv 1 \pmod{2n}$
I don't quite know how to conclude now that $p \equiv 1 \pmod{2n}$, I can't quite see why that would be true.
| $10^{2^k} \equiv -1 \bmod p$ implies that the order of $10$ mod $p$ is $2^{k+1}$.
By Fermat, $10^{2^{k+1}} \equiv 1 \bmod p$, and so $2^{k+1}$ divides $p-1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2779489",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Solving $\cos(2x)-\sin(x)= 0$ within the domain $[0,2\pi]$. Why am I missing some solutions? I‘m trying to figure this one out:
Solve for $x$ within the domain $[0,2\pi]$:
$$\cos(2x)-\sin(x)= 0$$
I figured this is $\sin(x)=-1$ and $\sin(x) = \frac 12$, which, when taking the $\arcsin$, gives $x=\frac{\pi}6$ and $x=-\frac{\pi}2$.
Now my question is:
I‘m supposed to get $\dfrac{\pi}6$, $\dfrac{5\pi}6$, $\dfrac{3\pi}2$, but adding $2\pi$ gets me only $\dfrac{\pi}6$, $\dfrac{3\pi}2$, $\dfrac{7\pi}2$ within the $[0,2\pi]$ domain.
Any idea what I‘m doing wrong here?
| \begin{align*}
\cos(2x) - \sin x & = 0\\
1 - 2\sin^2x - \sin x & = 0\\
1 - \sin x - 2\sin^2x & = 0\\
1 - 2\sin x + \sin x - 2\sin^2x & = 0\\
1(1 - 2\sin x) + \sin x(1 - 2\sin x) & = 0\\
(1 + \sin x)(1 - 2\sin x) & = 0
\end{align*}
\begin{align*}
1 + \sin x & = 0 & 1 - 2\sin x & = 0\\
\sin x & = -1 & -2\sin x & = -1\\
& & \sin x & = \frac{1}{2}
\end{align*}
Consider the diagram below.
The sine of an angle in standard position (vertex at the origin and initial side on the positive $x$-axis) is the $y$-coordinate of the point where the terminal side of the angle intersects the unit circle. Two angles have the same sine if their terminal sides intersect the unit circle at points with the same $y$-coordinate. Such angles are symmetric with respect to the $y$-axis. Thus, $\sin\theta = \sin\varphi$ if $\varphi = \pi - \theta$. Any two coterminal angles will also have the same sine. Hence, $\sin\theta = \sin\varphi$ implies
$$\varphi = \theta + 2k\pi, k \in \mathbb{Z}$$
or
$$\varphi = \pi - \theta + 2k\pi, k \in \mathbb{Z}$$
You found that a particular solution of the equation $\sin x = -1$ is
$$x = \arcsin(-1) = -\frac{\pi}{2}$$
Hence, all solutions of that equation have the form
\begin{align*}
\theta & = -\frac{\pi}{2} + 2k\pi, k \in \mathbb{Z} & \theta & = \pi - \left(-\frac{\pi}{2}\right) + 2k\pi\\
& & & = \frac{3\pi}{2} + 2k\pi
\end{align*}
If you list the solutions, you will notice that both equations yield the same set of solutions. The only one within the interval $[0, 2\pi]$ is $\dfrac{3\pi}{2}$.
You found that one solution of the equation $\sin x = \dfrac{1}{2}$ is $$x = \arcsin\left(\frac{1}{2}\right) = \frac{\pi}{6}$$ Hence, the general solution is
\begin{align*}
x & = \frac{\pi}{6} + 2k\pi, k \in \mathbb{Z} & x & = \pi - \frac{\pi}{6} + 2k\pi, k \in \mathbb{Z}\\
& & & = \frac{5\pi}{6} + 2k\pi, k \in \mathbb{Z}
\end{align*}
Of these angles, the only ones in the interval $[0, 2\pi]$ are $\dfrac{\pi}{6}$ and $\dfrac{5\pi}{6}$.
By only considering the arcsine of the angle, you missed the angle with the same sine that is obtained by reflection in the $y$-axis. There is only one such angle within each period if the terminal side of the angle lies on the $y$-axis (as is the case for $\theta = -\frac{\pi}{2}$) and two such angles otherwise (as is the case for $\theta = \frac{\pi}{6}$).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2780080",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
Computing the integral of a rational function I need to compute the integral $\int \dfrac{2x}{(x^2+x+1)^2} \cdot dx$. I tried using the integration of a rational function technique, with $\frac{Ax+B}{x^2+x+1}+\frac{Cx+D}{(x^2+x+1)^2}$, but this simply returned $C=2$ and $A,B,D = 0$, so it doesn't really change anything.
I also tried using a $u$ substitution, setting $u=x^2+x+1$. This made the numerator $2x=\frac{du}{dx}-1$, but I'm not really sure if I can do that/how to solve an integral with a derivative as a part of it.
How would I go about solving this?
Thanks for your time.
| $$2\int { \frac { x }{ (x^ 2+x+1)^ 2 } } $$
$$2\int { \frac { x }{ ((x+\frac { 1 }{ 2 } )^ 2+\frac { 3 }{ 4 } )^ 2 } } $$
Apply u-substitution: $u=x+\frac12$
$$2\int { \frac { 8(2u-1) }{ (4u^ 2+3)^ 2 } } du$$
$$2(8)\int { \frac { 2u-1 }{ (4u^ 2+3)^ 2 } } du$$
Apply the Sum Rule
$$2(8)(\int { \frac { 2u }{ (4u^ 2+3)^ 2 } } du-\int { \frac { 1 }{ (4u^ 2+3)^ 2 } du } )$$
Now, $$\int { \frac { 2u }{ (4u^ 2+3)^ 2 } } =- { \frac { 1 }{ 4(4u^ 2+3) } } $$
Now, $$\int { \frac { 1 }{ (4u^ 2+3)^ 2 } du=\frac { 1 }{ 12\sqrt { 3 } } (arctan(\frac { 2 }{ \sqrt { 3 } } u)+\frac { 1 }{ 2 } sin(2arctan(\frac { 2 }{ \sqrt { 3 } } u))) } $$
$$=2(8)(- { \frac { 1 }{ 4(4u^ 2+3) } }- { \frac { 1 }{ 12\sqrt { 3 } } (arctan(\frac { 2 }{ \sqrt { 3 } } u)+\frac { 1 }{ 2 } sin(2arctan(\frac { 2 }{ \sqrt { 3 } } u))) } $$
After doing small calculations and substituting $u=x+\frac12$,
$$\int\frac{2x}{x^2+x+1}dx=16(-\frac{1}{4(4x^2+4x+4)}-\frac{1}{24\sqrt{3}}(2arctan(\frac{2x+1}{\sqrt{3}})+sin(2arctan(\frac{2x+1}{\sqrt{3}}))))+C$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2781665",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
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