Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
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 ... | \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 $z$; $$
x^3+y^3+x^3=3xyz \implies x+y+z=0$$ Notic... | {
"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 the... | 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 $$C=\begin{bmatrix}0&\frac12&-1\\\frac12&0&-\frac12\\-1&-\frac12&5\end{bmatrix}.$$ Tan... | {
"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 div... | $$32\cdot2^{5k+1}+5\cdot5^{k+2}=5(2^{5k+1}+5^{k+2})+27(2^{5k+1})$$ 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
$$f(a... | I used $QR$ factorization rather than attempting to solve the normal equations. We have
$$\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{\sq... | {
"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
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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 ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 3
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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 $$(a,b,c,d)=(1+6z^3,\,1-6z^3,\,-6z^2,\,1)$$ ... | 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:
$$(a+b)^3\equiv0\p... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2647249",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
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Prove that $(x-y)(y-z)(z-x) \leq \frac{1}{\sqrt{2}}$
If $x,y,z$ are real and $x^2+y^2+z^2=1$, prove that$$(x-y)(y-z)(z-x) \leq \frac{1}{\sqrt{2}}.$$
Equality is achieved in some strange cases: For example, if $x = -\dfrac{1}{\sqrt{2}}$, $y = 0$ and $z = \dfrac{1}{\sqrt{2}}$, then $(x-y)(y-z)(z-x)=\dfrac{1}{\sqrt{2}}$... | Here is a proof similar to the solution of IMO 2006 problem
3
(well... at least to the solution I found on the IMO):
As Michael Rozenberg noticed, it suffices to prove the inequality
\begin{equation}
\left( x^2 +y^2 +z^2 \right) ^3 \geq2\left( x-y\right) ^2 \left(
y-z\right) ^2 \left( z-x\right) ^2 .
\label{dari... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2648520",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
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How to solve the equation $x^2+2=4\sqrt{x^3+1}$? From the Leningrad Mathematical Olympiad, 1975:
Solve $x^2+2=4\sqrt{x^3+1}$.
In answer sheet only written $x=4+2\sqrt{3}\pm \sqrt{34+20\sqrt{3}}$.
How to solve this?
| Let $u = x^2-x+1$, $v = x+1$ and $\lambda = \sqrt{\frac{u}{v}}$, we have
$$\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... | {
"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
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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}... | {
"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
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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... | 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 $|k|\geq 2$.... | {
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"url": "https://math.stackexchange.com/questions/2650244",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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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 ... | Maple expresses it as
$$ {\frac { \left( i\sqrt {3}+1 \right) \sqrt {4+2\,{x}^{2}-2\,i{x}
^{2}\sqrt {3}}\sqrt {4+2\,{x}^{2}+2\,i{x}^{2}\sqrt {3}} \left( i{\it
EllipticF} \left( \left( i\sqrt {3}+1 \right) x/2,(1+i\sqrt {3})/2
\right) \sqrt {3}+3\,{\it EllipticF} \left( \left( i\sqrt {3}+1
\right) x/2,(1+i\sqrt {3})... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2651254",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
How many sub-square matrices does a square matrix have and is there a simple formula for it? Consider an $n \times n$ matrix $M$. I want to find the determinant for ALL sub-square matrices of $M$. There may be a better way but my method is to find all sub-square matrices and check them individually.
*
*How many s... | If you admit the zero-by-zero matrix as a square submatrix, then the number
of square submatrices is
$$\sum_{k=0}^n\binom nk^2.$$
This can be written as
$$\sum_{k=0}^n\binom nk\binom n{n-k}$$
which can be recognised as the coefficient of $t^n$ in $(1+t)^n(1+t)^n$
etc.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2652008",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
prove that radius is a root of $(3-2\sqrt{2})t^2-2\sqrt{2}t+2=0$ a circle is described to pass through the origin and to touch the lines $x=1,x+y=2$ then prove that the radius of the circle is a root of $(3-2\sqrt{2})t^2-2\sqrt{2}t+2=0$
solution i try
length of perpendicular from $(1-r,2r-1)$ is $\displaystyle \frac{1... | let $$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$$
can you finish?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2655225",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find the values of the parameters for which the matrices commute
I have two matrices:
$A = \left( \begin{array}{cc}
1 & a \\
b & 6
\end{array} \right)
%
\ \ \ \ B = \left( \begin{array}{cc}
4 & c \\
d & 2
\end{array} \right)$.
How can I find the values of the parameters $a, b, c, d$ for which the matrices will com... | We have
$$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}.$$
Hence
\begin{eqn... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2655369",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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Expected number of turns in dice throwing I generated a transition probability matrix for a scenario where I throw five dice and set aside those dice that are sixes. Then, I throw the remaining dice and again set aside the sixes - then I repeat this procedure until I get all the sixes. $X_n$ here represents the number ... | An other method
Each turn you set aside any die which shows a six.
Let $F_n$ be the expected number of turns until you set aside at least one of $n$ die. Let $E_n$ be the expected number of turns until you set aside all $n$ die. Let $p_n(k)$ be the (conditional)probability of setting aside $k$ die in a turn when gi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2656232",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
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Finding the length of one side of a triangle In $\triangle\,$ABC if BC is unity and $\sin\frac{A}{2} = x_1, \sin\frac{B}{2} = x_2, \cos\frac{A}{2} = x_3, \cos\frac{B}{2} = x_4$ with $$ \left(\frac{x_1}{x_2}\right)^ {2007} - \left(\frac{x_3}{x_4}\right)^{2006} = 0 $$ then the length of AC is equal to $\ldots$.
I tried u... | In a given $\triangle$ both $\displaystyle \frac{A}{2},\frac{B}{2}\in\bigg(0,\frac{\pi}{2}\bigg).$
For $x\in(0,\dfrac{\pi}{2})$, $\sin x$ is a strictly increasing function whereas $\cos x$ is a strictly decreasing function.
Now if $\displaystyle \frac{A}{2}>\frac{B}{2}$ then $\displaystyle \sin\frac{A}{2}>\sin\frac{B}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2656705",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Creating a basis for U Consider the subspace
$$
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... | 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 & ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2658900",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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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: $$\left(\frac{1}{2}+\cos \frac{\pi}{20}\right)\left(\frac{1}{2}+\cos \frac{3\pi}{... | Put $$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)$$
Then you have to find $f({1\over 2})/({1\over 2}+{\sqrt{2}\over2})$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2659008",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
"answer_id": 3
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Basis for Matrix A
Find a basis for all $2\times2$ matrices $A$ for which
$A\begin{bmatrix} 1 & 1 \\ 1 & 1 \end{bmatrix}$ = $\begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}$.
Maybe I'm dumb-- but isn't $A$ just the $0$ matrix? In which case, the base is simply the $0$ matrix as well?
| If you let $A = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$ then your equation says,
$$ \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{pma... | {
"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
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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 $$\frac{a+c}{b+c}-\frac{f(a,b,d)+f(c,b,d)}{f(b,a,d)+f(c,a,d)}=-{\frac { \left( 3\,c+d \right) \left( a-b \right) \left( abc+abd+ca
d+cbd \right) \left( c+a \right) }{ \left( 3\,{a}^{2}{b}^{2}c+3\,{a}^
{2}{b}^{2}d+3\,{a}^{2}b{c}^{2}+4\,{a}^{2}bcd+2\,{a}^{2}b{d}^{2}+3\,{a}
^{2}{c}^{2}d+2\,{a}^{2}c{d}^{2}+6\,a{b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2659411",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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How to find $\lim_\limits{n\to \infty }(1-\frac{5}{n^4})^{(2018n+1)^4}$
How to find $$\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... | 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_... | {
"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
$$ x = \sqrt[... | Since the degree is 3, if you have a solution, you can make a factorisation and have a second degree equation to solve :
$$\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) \qua... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2661654",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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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 $$\frac14=x^4... | You can write $z=e^{it}=\cos t+i\sin t$. Then $z^2+\overline z^2=2\cos 2t$.
So we need to solve $\cos2t=\pm\frac12$. Thus $2t=\pm\frac\pi3+2k\pi$
or $2t=\pm\frac{2\pi}3+2k\pi$, that is $t=\pm\frac\pi6+k\pi$ or
$t=\pm\frac\pi3+k\pi$, so that $z=e^{it}=\frac12(\pm\sqrt3\pm i)$
or $\frac12(\pm1\pm i\sqrt3)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2662090",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 3
} |
Integral of a shifted Gaussian distribution with an error function In the course of computing a convolution of two functions, I have simplified it to a single variable integral of the form
$$\int_0^\infty xe^{-ax^2+bx}\mathrm{erf}(cx+d) dx$$
where $\mathrm{erf}$ is the error function defined as $$\ \mathrm{erf}(x) = \f... | 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{\... | {
"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 ... | 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 ... | {
"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 solv... | \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{\dfra... | {
"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(\fr... | 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
$$A=\frac{3^x+2^x-2}{4^x+2^x-2} =\frac{ (\log (2)+\log (3))x+\frac{1}{2} \left(\log ^2(2)+\log
^2(3)\right)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2664317",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Solving this system of equations without a CAS: $4=a+c$, $5=d+ac+b$, $4=ad+bc$, $4=bd$ Solving this system of equations :
$$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}$$ The third system with two unknowns gi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2664582",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Is the sequence $a_{n}= {{n}\choose{[n/2]}}^{1/n}$ increasing? Let the sequence $a_{n}= {{n}\choose{[n/2]}}$$^{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 deri... | 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 assumptio... | {
"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 $(... | {
"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 h... | 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 inte... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2665211",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 0
} |
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 co... | 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)}}\\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2665530",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
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}... | 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:
$$ \int_{0}^{z}\frac{x^{-1/3}}{\sqrt{1-x^2}}\,dx = z^{2/3}\sum_{n\geq 0}\frac{3\binom{2n}{n}}{(6n+2)4^n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2665993",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Is a proof using modular arithmetic in a question like this valid? It's been two years or so since I've finished my math undergrad (and I'm doing something non-math related now, unfortunately), so I apologize if what is to follow isn't a very good question!
Prove that for all Integers $n$, $n(n + 1)(2n + 1)$ will alway... | $$n(n+1)(2n+1)=n(n+1)(2n+4-3)=2\underbrace{n(n+1)(n+2)}-3n(n+1)$$
Use The product of n consecutive integers is divisible by n factorial
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2668673",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "27",
"answer_count": 4,
"answer_id": 2
} |
Finding the non singular matrix. $(1)$ Suppose $A$ and $B$ satisfy the relation $B^2+AB+2I=0$. Are the following four matrices necessarily non singular?
(a)$A$ ( b)$B$ (c) $A+2I$ (d) $B+2I$
My attempt:- $$B^2+AB+2I=0$$
$$\Rightarrow B^2+AB= -2I$$
$$\Rightarrow B(A+B)= -2I$$
... | You are correct that $B$ has to be invertible. Let's construct an example which shows the matrices $A$, $A + 2I$ and $B + 2I$ can be singular.
$B$ is invertible so we can express $A$ from $B^2 + AB + 2I = 0$ as $$A = B + 2B^{-1}$$
Using the Spectral theorem, we conclude that the eigenvalues of $A$ are of the form $x + ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2669090",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
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When two digit numbers in base $5$ are multiplied the result is $4103_5$. What are the numbers in base $5$? When two digit numbers in base $5$ are multiplied the result is $4103_5$. What are the numbers in base $5$?
Well given by two digit numbers in base $5$ I tried out the multiplication and tried to simplify.
$(ab_5... | So $n_1 = 5a + b; n_2 = 5c + d$ and $(5a+b)(5c+d) = 4103_5 = 4*5^3 + 1*5^2 + 3$
$(5a + b)(5c+d) = ac*5^2 + 5(bc + ad) + bd = 4*5^3 + 1*5^2 + 3 = 21*5^2 + 0 + 3$
So $ac \approx 4*5$ which is a very large value considering $a,c \le 4$.
Note $40_5*100_5 = 4000_5 < 4103_5$ so as $(5a+c) < 100_5$ we now $(5b+d) > 40_5$ and ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2669326",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
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Roots of $x^3+5x-18=0$ using Cardano's method Given that $x^3+5x-18=0$. We have to solve it using Cardano's method.
Using trial $x=2$ will be a root. Dividing the equation by $x-2$ we shall get the other quadratic equation and solving that one, we shall obtain all the roots.
But when I am trying to solve the equation... | Cardono's formula says that one root is$$\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}.$$A... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2670354",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
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Showing $x^4-x^3+x^2-x+1>\frac{1}{2}$ for all $x \in \mathbb R$
Show that
$$x^4-x^3+x^2-x+1>\frac{1}{2}. \quad \forall x \in \mathbb{R}$$
Let $x \in \mathbb{R}$,
\begin{align*}
&\mathrel{\phantom{=}}x^4-x^3+x^2-x+1-\frac{1}{2}=x^4-x^3+x^2-x+\dfrac{1}{2}\\
&=x^2(x^2-x)+(x^2-x)+\dfrac{1}{2}=(x^2-x)(x^2+1)+\dfrac{1}{2... | Factorise and complete the square of the first bit to understand function:
$$\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*}$$
Then $(x^2+1)\left[(x-\frac12)^2-\frac14\right]\geq-\frac14$, a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2670433",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 7,
"answer_id": 4
} |
Evaluate the limit $\lim_{n\to \infty}\sqrt{n^2 +2} - \sqrt{n^2 +1}$ I know that
$$\lim_{n\to \infty}(\sqrt{n^2+2} - \sqrt{n^2+1})=0.$$
But how can I prove this?
I only know that $(n^2+2)^{0.5} - \sqrt{n^2}$ is smaller than $\sqrt{n^2+2} - \sqrt{n^2}$ = $\sqrt{n^2+2} - n$.
Edit:
Thank Y'all for the nice and fast... | Hint: $\displaystyle \sqrt{n^2+2}-\sqrt{n^2+1}=\frac{1}{\sqrt{n^2+2}+\sqrt{n^2+1}}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2670779",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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Find the value of a 5th-root expression. Simplify and find the value of the expression:
$$\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)]$$
We know that $(a+b)=x$.
Simlifying $a×b$ that is
$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2673142",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
how to find the equation of a tangent line to a circle, given its slope and the eq. of the circle? So I have a circle: $(x-2)^2 + (y-2)^2 = 25$ and I have a tangent line to this circle, with a slope of $m= -3/4$.
I have to find the equation of the tangent line, so I know the radius of the circle is $r = 5$ and I wrote... | Here is the derivation for the equation of the line with slope $m$ that is tangent to a circle of radius $r$ centered on the origin. (If the center of the circle was not centered on the origin, we can simply shift the equation we found.)
Let the required tangent line be $y=mx+b$ and the point of tangency be $(x_0, y_0)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2676066",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 5
} |
A set of matrix with 2 properties Let $A \subset M_2(R)$ a set with the properties:
1)If $X,Y \in A$, then $X+Y \in A$;
2)If $X \in M_2(R)$ with $tr(X^tX)=1$, then $X \in A$.
Prove that $A=M_2(R)$.
I wrote that for a matrix $X \in M_2(R)$ with the elements $a,b,c,d$ we have by the second propertie that if $a^2+b^2+c^2+... | The matrices $E_{i, j}, (i,j)\in \{1,2\}^2,$ such that $$e_{k,l} = \begin{cases} 1, & \text{ if } (k,l) = (i,j) \\0,& \text{ otherwise} \end{cases}$$ are in $A$.
Now let's show that $\pmatrix{\lambda & 0 \\ 0 & 0} \in A$. Since $\lambda = \lfloor \lambda \rfloor + (\lambda - \lfloor \lambda \rfloor$) we can assume $\l... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2677752",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
$\sum_{i=0}^n \frac{1}{(i+3)(i+4)} = \frac{n}{4(n+4)}$ (prove by induction) I'm having some difficulty proving by induction the following statement.
$$\sum_{i=0}^n \frac{1}{(i+3)(i+4)} = \frac{n}{4(n+4)}$$
I have shown that $\sum_{i=0}^n \frac{1}{(i+3)(i+4)} = \frac{n}{4(n+4)}$ holds for $n=1$ (equals $\frac{1}{20}$) ... | $$\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)}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2679614",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Find the sum to $n$ terms of the series: $1^2.1+2^2.3+3^2.5+....$ Find the sum to $n$ terms of the series:
$$1^2.1+2^2.3+3^2.5+.....$$
My Attempt:
Here, $n^{th}$ term of $1,2,3,....=n$
$n^{th}$ term of $1^2,2^2,3^2,....=n^2$
Also, $n^{th}$ term of $1,3,5,....=2n-1$
Hence, $n^{th}$ term of the given series is $t_n=n^2(2... | Hint:
$$\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}$$$$\sum_{i=1}^{n}i^2=\frac{n(n+1)(2n+1)}{6}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2680816",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 2
} |
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... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2683178",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 2
} |
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:
... | {
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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 str... | 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 $\textrm{A.M.}\ge\textrm{H.M.}$
\... | {
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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 $n=k$
$$P(k):1^2-2^2+3^2-.....+(-1)^{k-1}k^2=(-1)^{k-1}\frac{k.(k+1)}{2}$$.
$$P(k+1):1^2-2^2+3^2-.....+(-1)^{k-1}k^... | L.H.S. of $P(k+1)$ should be $1^2-2^2+3^2-\cdots+(-1)^{k-1}k^2+(-1)^k(k+1)^2$
which is equal to
$\displaystyle \frac{(-1)^{k-1}k(k+1)}{2}+(-1)^k(k+1)^2=\frac{(-1)^{k}[-k(k+1)+2(k+1)^2]}{2}=\frac{(-1)^{k}(k+1)(k+2)}{2}$
| {
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Let $x_n=\frac{1}{3}+(\frac{2}{5})^2+(\frac{3}{7})^3+....(\frac{n}{2n+1})^2$ Is $(x_n) $cauchy sequence? Let $$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(\... | 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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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)
$$\... | We need to prove that $$\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.$$
Now, let $a+b+c$ be constant and $a^3+b^3+c^3... | {
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Prove that $\left|\frac{1}{z^4-4z^2+3}\right|\leq \frac{1}{3}$, if $|z|=2$
Prove that $$\left|\dfrac{1}{z^4-4z^2+3}\right|\leq \dfrac{1}{3},\quad\text{ if}\quad |z|=2$$
I know that I can complete squares on the denominator ($(z^2-2)^2-1$). I know the triangular inequality and some other few things... but I can't work... | For $|z|=2$ the triangle inequality gives
$$
|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 -... | {
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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 f... | 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... | {
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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 quadrati... | (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 =... | {
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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... | 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 sinc... | {
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"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... | 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... | {
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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... | 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... | {
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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 ... | 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. Contra... | {
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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>|... | 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 dele... | {
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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,
$$\int\frac{1}{bx^2(\frac{a}{b}-\frac{1}{x})} \, dx = \fr... | Note that
$$\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$$
| {
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"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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Generating function of a sequence with a sum I am new to generating functions and I know some of the basics how to create them. But now I am not sure how to create a generating function of this sequence:
$$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... | 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{... | {
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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)... | 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... | {
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"url": "https://math.stackexchange.com/questions/2709465",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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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{... | Why not write $3+\sqrt3=\sqrt3(1+\sqrt3)$?
| {
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"source": "stackexchange",
"question_score": "8",
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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 t... | 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 chi... | {
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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\alp... | 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
| {
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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 fo... | 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$$\frac{\frac{4^{n+1}}{\sqrt{3n+4}}}{\frac{4^n}{\sqrt{3n+1}}}=4\sqrt{\frac{3n+1}{3n+4}}.$$Ca... | {
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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}]+... | 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
$$x=y=z$$ a... | {
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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,
$$ \zeta(3) \leq \frac{5}{2}\sum_{n=1}^{3}\frac{(-1)^{n+1}}{n^3 \binom{2n}{n... | {
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If $b-a=mc$ then prove that $\cot (\dfrac {B-A}{2})=\dfrac {1+m.\cos (B)}{m.sin (B)}$ If $b-a=mc$ then prove that $\cot (\dfrac {B-A}{2})=\dfrac {1+m.\cos (B)}{m.sin (B)}$
My Attempt:
We know,
$$\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... | 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... | {
"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 th... | 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:
$$\tan (\theta) + \tan (2\theta) = \tan (3\theta)$$
$$\dfrac {\sin (\theta)}{\cos (\theta)}+ \dfrac {\sin (2\theta)}{\cos (2\theta)}=\dfrac {\sin (3\theta)}{\cos (3\t... | \begin{align}
\tan (\theta) + \tan (2\theta) &= \tan(3\theta) \\
\tan (\theta) + \tan (2\theta) &= \tan(\theta + 2\theta) \\
\dfrac{\tan (\theta) + \tan (2\theta)}{1}
&= \dfrac{\tan(\theta) + \tan(2\theta)}
{1-\tan (\theta) \tan (2\theta)} \\
\end{align}
So, either $\tan (\theta) + \tan (2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2725922",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Simple approaches to prove that $\lim\limits_{x\to 0}\left(\frac{1}{x^2}-\frac{1}{\sin^2 x}\right)=-\frac13\ $?
Find $\lim_{x\to 0}(\frac{1}{x^2}-\frac{1}{\sin^2 x})$
My attempt:
$\lim_{x\to 0}(\frac{1}{x^2}-\frac{1}{\sin^2 x})$
=$\lim_{x\to 0}(\frac{\sin^2 x -x^2}{x^2 \sin^2 x})$ ($\frac{0}{0}$ form)
Applying L'Ho... | $$\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}$$
\begin{align*}
\lim_{x\to 0}\left(\frac{1}{x^2}-\frac{1}{\sin^2 x}\right)&=\lim_{x\to 0}\frac{\sin^2x-x^2}{x^2\sin^2x}\\
&=\left(\lim_{x\to 0}\frac{\sin x-x}{x^3}\right)\left(\lim_{x\to 0}\frac{x\sin x+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2731206",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
The integer part of $ \sum_{k=2}^{9999} \frac{1}{\sqrt{k}} $ For $k>0 $
$ \frac{1}{2\cdot \sqrt{k+1}}<\sqrt{k+1}-\sqrt{k}<\frac{1}{2\cdot \sqrt{k}} $,
the integer part of $ \sum_{k=2}^{9999} \frac{1}{\sqrt{k}} $
My attempt here was not complying with any of the options given as my attempt was to find the min. value fo... | for $t\in [k,k+1]$ we have,
$$
\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}}
$$
and we get
\begin{eqnarray}
\sum_{k=1}^{10^4-1}\frac{1}{\sqrt{k+1}}&\leq &\sum_{k=1}^{10^4-1}\int_k^{k+1}\frac{dt}{\sqrt{t}}&\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2732776",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
$\lim_{n\to+\infty} \frac{1}{n\log(n)}\sum_{k=4}^{n}\frac{2k}{k^2-2k-3}$
Given the limit:
$$\lim_{n\to+\infty} \frac{1}{n\log(n)}\sum_{k=4}^{n}\frac{2k}{k^2-2k-3} = \alpha$$
Find the value of $\alpha$
I know the series does not converge (it is equivalent to the harmonic series. Correct me, please, if I am wrong).
D... | From the $\lim\limits_{n\to+\infty} \frac{1}{n\log(n)}\sum\limits_{k=4}^{n}\frac{2k}{k^2-2k-3}$ first dealing with the sum equals to the followings:
$\sum\limits_{k=4}^{n}\frac{2k}{k^2-2k-3}=\sum\limits_{k=4}^{n}\frac{2k}{(k+1)(k-3)}=\frac{1}{2}\sum\limits_{k=4}^{n}\big(\frac{k}{k-3}-\frac{k}{k+1}\big)=\frac{1}{2}\sum... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2735194",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Why can't I cancel $2x-3$ from $(2x-3)(x+5)=9(2x-3)$? Why are these simplifications wrong?
$$\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 mea... | {
"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
} |
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 d... | 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 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2737652",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
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... | 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 $1=\dfrac{2n+18}{2n+18}$.
$$\begin{align} \frac{2n-18}{n^2-8n+8}\times\frac{2n+18}{2n+18}&=\frac{4n^2-324}{2n^3 - 16n^2 + 16n + 18n^2 -144n + 144} \\ \\ &=\frac{2n^2-162}{n^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2739291",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
Inconsistency for solving $x' = x^{1/2}$ The proposed system $x' = x^{1/2}$ can be solved easily to obtain $x(t) = \frac{1}{4} (t^2 + t c + c^2)$, where $c$ is the integration constant.
However, differentiate the newly-found $x(t)$, one gets: $x(t)' = \frac{1}{2}t+\frac{1}{4}c$. This implies that $x^{1/2} = \frac{1}{2}... | The solution of
$$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$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2741103",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Product of sums which equal to sum of product We can be sure that
$$\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}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2741897",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
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:
$$
\begin{align}
\text{At Point P(-2,4):} \hspace{30pt} y &= x^2 \\
\frac{dy}{d... | Notice in your first lines of reasoning at the point $P(-2,4)$ that is when $x=-2$, the slope of the tangent line at the point is
$$ \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) $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2744119",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
Evaluate $\lim_{n\to \infty} n^{-n^2}\left( \prod_{r=0}^{n-1} \left(n+\frac {1}{3^r}\right) \right) ^n$
Evaluate $$\lim_{n\to \infty} n^{-n^2}\left( \prod_{r=0}^{n-1} \left(n+\frac {1}{3^r}\right) \right) ^n$$
Since on substituting $n=\infty$ we get a indeterminate form of $1^{\infty}$. Hence we can write the same li... | In your attempt, you rewrote the expression in an incorrect way (you forgot the logarithm, when rewriting $x = e^{\log x}$). How did you get that second expression?
You have
$$
\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 ... | {
"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,
$$ P \le... | Your mistake is in $$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\})$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2748446",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
If $\frac{\cos^4 \alpha}{x}+\frac{\sin^4 \alpha}{y}=\frac{1}{x+y}$,prove that $\frac{dy}{dx}=\tan^2\alpha$ If $\frac{\cos^4 \alpha}{x}+\frac{\sin^4 \alpha}{y}=\frac{1}{x+y}$,prove that $\frac{dy}{dx}=\tan^2\alpha$
It is very long to direct differentiate it.Can someone help me?
| Using cauchy schwarz inequality $$\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)$$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2748907",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
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How to show that $a_n=1+1/\sqrt{2}+\cdots+(1/\sqrt{n-1})-2\sqrt{n}$ has an upper bound. Let $a_n=1+1/\sqrt{2}+\cdots+(1/\sqrt{n-1})-2\sqrt{n}$ while $a_1=-2,n\ge2 $ ,
I need to prove that $a_n$ converges.
I proved that it is monotonically increasing and tried to prove that it is upper-bounded by induction but failed ... | Note: To prove something converges you don't have to figure out what it converges to. And to prove something is bounded above you don't have to find the least upper bound; it's enough just to find any upper bound (and prove it is an upper bound).
So
$a_{n+1} - a_n = \frac 1{\sqrt{n}} - 2\sqrt{n+1} + 2\sqrt{n}$.
Clai... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2748996",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
Finding solutions for $\sin(z)=\sin(2)$ having some difficulty wrapping my head around how to methodically do this style of question:
Find all solutions $z\in ℂ$ for $\sin(z)=\sin(2)$
I attempted to solve this by using the identity,
$\sin(x)\cosh(y)+i\cos(x)\sinh(y)=\sin(2)$,
so that,
$\sin(x)\cosh(y)=\sin(2)$ and $\c... | $$\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 \rig... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2751593",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
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 th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2755323",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
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:
$$L(x)=\left(\frac{8}{x}+1\right)\sqrt{x^2+9}, \\
L'(x)=\left(-\frac{8}{x^2}\right)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2757174",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Finding the Laurent series of the following function I need to find the laurent series and the residue of the following complex function
$$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}}=... |
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 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2758338",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
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:
$$I_n=\frac {x^{n+1}}{(n+1)\sqrt{... | $$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).
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2758516",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
} |
Find $x, y\in\mathbb Z$ such that $\frac{x}{2} + \frac{x}{y} - \frac{3}{2} = \frac{10}{y}$ Find $x, y\in\mathbb Z$ such that $\dfrac{x}{2} + \dfrac{x}{y} - \dfrac{3}{2} = \dfrac{10}{y}$
My try
$$\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\ne... | 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.
$$ (y+2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2761984",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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Generating function to find the sum of digits How many integers between $30,000$ and $100,000$ have a sum of $15$ or less?
I was approaching this problem with a generating function:
$$g(x) = (x^3+x^4+x^5+x^6+x^7+x^8+x^9)(1+x+...+x^9)^4$$
First I'm going to pull out $x^3$ then I would need the coefficients of $x$ to t... | We want the number of solutions in integers to
$$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$$
with the additional constraint $0 \le x_6$. More generally, we seek the generating function of... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2762449",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Finding an appropriate upper bound for $\frac{x^2-4}{x-1}$ at $x=0$ and $x=2$ I am investigating the continuity of the function $\frac{x^2-4}{x-1}$ at $x=0$ and $x=2$.
Here's a part of my proof;
$$\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$$
... |
$$\lvert f(x)-f(0)\rvert =\frac{\left\lvert x\right\rvert\left\lvert x+4\right\rvert}{\left\lvert x-1\right\rvert}$$
The challenge I have is to find an appropriate upper bound for $\frac{1}{x-1}$ at $x=0$ and $x=2$. Please, can anyone help out?
You want to stay close enough to $x=0$ to ensure $\tfrac{1}{x-1}$ stays b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2764567",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
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Calculating a determinant. $D_n$=\begin{vmatrix}
a & 0 & 0 & \cdots &0&0& n-1 \\
0 & a & 0 & \cdots &0&0& n-2\\
0 & 0 & a & \ddots &0&0& n-3 \\
\vdots & \vdots & \ddots & \ddots & \ddots&\vdots&\vdots \\
\vdots & \vdots & \vdots & \ddots & \ddots& 0&2 \\
0 & \cdots & \cdots & \cdots &0&a&1 \\
n-1 & n-2 & n-3 & \cdots... | Expand with respect to the first line: the term obtained with $a$ is $aD_{n-1}$. For the second one, we get $(-1)^{n+1}$ times a determinant that can be expanded with respect to the first column. This lead to the recurrence relation
$$
D_n=aD_{n-1}-\left(n-1\right)^2a^{n-2}.
$$
Letting $b_n:=a^{-n}D_n$ for $a\neq 0$ a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2765475",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
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Find vector $x \not= 0$ that satisfies the equation $Ax = x$. Given a matrix A:
$$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 & \f... | 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
$$B=A-I= \begin{bmatrix}
-1 & 0 & 0 & \frac{1}{2} & \frac{1}{2} \\
\frac{1}{3} & -1 & 1 & 0 & 0 \\
\frac{1}{3} & 0 & -1 & 0 & \frac{1}{2} \\
\frac{1}{3} & \frac{1}{2} & ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2766684",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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The shortest distance from a point to the graph of the function To compute the distance from the point (5,5) to the graph of xy=4. I choose an arbitrary point (u,v) on the graph of $xy=4$.
I get $d(u,v)=\sqrt{(u-5)^2+(v-5)^2}$ again $(u,v)$ satisfies equation of hyperbola so that $uv=4$. Now what shall i do next?
| I offer this as an alternative solution.
Looking at the graph of $xy=4$ and the point $P=(5,5)$, clearly we need only consider the branch of $xy=4$ that is in the first quadrant. So we need only consider the function $y = \dfrac 4x$ where $x > 0$.
Consider the particular point $Q=\left(\xi, \dfrac{4}{\xi}\right)$ on t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2767570",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Sum of series $\sum^{10}_{i=1}i\bigg(\frac{1^2}{1+i}+\frac{2^2}{2+i}+\cdots \cdots +\frac{10^2}{10+i}\bigg)$
The Sum of series $$\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... | 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 expre... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2773345",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
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 $a = 3$ ,$b=3$
$$\sqrt{a^2+b^2}=\sqrt{3^2+3^2} = \sqrt{18}=3\sqrt{2}$$
$$\theta =\arctan\biggr(\frac{b}{a}\biggr) =\arctan\bigg... | It might be useful to simplify before converting:
$$\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}$$$$= - \frac{243}{8}ie^{-i\frac{5}{3}\pi} = -\frac{243}{8}e^{i\frac{5}{6... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2776335",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Use residues to evaluate the improper integral $\int_0^\infty\frac{x^3\sin(x)}{(x^2+4)(x^2+16)}dx$ I am trying to solve the below problem using residues,
$$\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... | *
*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 $I =Im(\int_{-\infty}^\infty\frac{z^3}{(z^2+4)(z^2+16)}e^{iz}dz)$.
$$I = 2\pi i (Res_{2i} \frac{z^3}{(z^2+4)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2777478",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Prime factor $p$ of $10^{2^k}+ 1$ is $p \equiv 1 \pmod{2^{k+1}}$ I am asked to show that for when $n$ is a power of 2, any prime factor $p$ of $10^n + 1$ is $p \equiv 1 \pmod { 2n}$.
For $n = 1$, we have $p = 10^1 +1 = 11 \equiv 1 \pmod 2$
For $n > 1$ we now have that $10^n \equiv -1 \pmod{p}$ so $-1$ is a quadratic r... | $10^{2^k} \equiv -1 \bmod p$ implies that the order of $10$ mod $p$ is $2^{k+1}$.
By Fermat, $10^{2^{k+1}} \equiv 1 \bmod p$, and so $2^{k+1}$ divides $p-1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2779489",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Solving $\cos(2x)-\sin(x)= 0$ within the domain $[0,2\pi]$. Why am I missing some solutions? I‘m trying to figure this one out:
Solve for $x$ within the domain $[0,2\pi]$:
$$\cos(2x)-\sin(x)= 0$$
I figured this is $\sin(x)=-1$ and $\sin(x) = \frac 12$, which, when taking the $\arcsin$, gives $x=\frac{\pi}6$ and $x=... | \begin{align*}
\cos(2x) - \sin x & = 0\\
1 - 2\sin^2x - \sin x & = 0\\
1 - \sin x - 2\sin^2x & = 0\\
1 - 2\sin x + \sin x - 2\sin^2x & = 0\\
1(1 - 2\sin x) + \sin x(1 - 2\sin x) & = 0\\
(1 + \sin x)(1 - 2\sin x) & = 0
\end{align*}
\begin{align*}
1 + \sin x & = 0 & 1 - 2\sin x & = 0\\
\sin x & = -1 & -2\sin x & = -1\\
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2780080",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
Computing the integral of a rational function I need to compute the integral $\int \dfrac{2x}{(x^2+x+1)^2} \cdot dx$. I tried using the integration of a rational function technique, with $\frac{Ax+B}{x^2+x+1}+\frac{Cx+D}{(x^2+x+1)^2}$, but this simply returned $C=2$ and $A,B,D = 0$, so it doesn't really change anythin... | $$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)... | {
"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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Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.