Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Jordan form of the "multiplicative table" matrix I have to find the Jordan form of the $(10\times10)-$matrix $A$ with the $n$th row formed by $n(1,2,3,4,5,6,7,8,9,10), \ \ 1 \leq n \leq 10$
I have calculated the determinant of $(A-xI)$ using Gaussian elimination, and I have that the result is $x^9(x-1)$, so there are ... | I want to thank @Hagen von Eitzen to make me see where was the mistake.
Given $A-xI$ the matrix defined upon, with Gaussian elimination of the type: $n$-th column $-$ $n$ times the first column, we end up with a matrix of the form:
$\left(\begin{matrix}1-x & 2x & 3x &... \\ 2 & -x & 0 &... \\ 3 & 0 & -x &... \\ . & ...... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1626282",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Why can we convert a base $9$ number to a base $3$ number by simply converting each base $9$ digit into two base $3$ digits?
Why can we convert a base $9$ number to a base $3$ number by simply
converting each base $9$ digit into two base $3$ digits ?
For example $813_9$ can be converted directly to base $3$ by notin... | Let's look at what you base $9$ number actually means. $$813_9=8\cdot 9^2+1\cdot 9^1+3\cdot 9^0$$
If we wish to write this as powers of $3$ with coefficients between $0$ and $2$, we can simply do
\begin{align}
3\cdot 9^0&=1\cdot 3^1+0\cdot 3^0\\
1\cdot 9^1&=0\cdot 3^3+1\cdot 3^2\\
8\cdot 9^2&=2\cdot 3^5+2\cdot 3^4\\
\e... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1627453",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 5,
"answer_id": 3
} |
Use the partial fraction to evaluate $\int \frac{\left(3y^2+3y+2\right)}{\left(y^2-1\right)\left(y+1\right)}dy$. $\int \frac{\left(3y^2+3y+2\right)}{\left(y^2-1\right)\left(y+1\right)}dy$
My try：
$\int \frac{\left(3y^2+3y+2\right)}{\left(y^2-1\right)\left(y+1\right)}dy$ ＝
$\int \frac{\left(3y^2+3y+2\right)}{\left(y-1\r... | The doubled factor $(y+1)^2$, does not contribute $\frac{B}{y+1}+\frac{C}{y+1}$. It contributes $\frac{B}{y+1}+\frac{C}{(y+1)^2}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1631643",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
On the divisors of $a^b + (\frac{a-b}2)^a + (\frac{b-a}2)^b+ b^a $ Let $a$ and $b$ be two odd natural numbers. Show that $\frac{a+b}{2}$ divides $$a^b + \left(\frac{a-b}2\right)^a + \left(\frac{b-a}2\right)^b+ b^a$$
| Well, that's simple. You need just one fact (easy to prove by induction): $x+y$ divides $x^n+y^n$ if $n$ is odd.
Now, $a^b+\left({b-a\over2}\right)^b$ is divisible by $a+{b-a\over2}={a+b\over2}$, and $b^a+\left({a-b\over2}\right)^a$ is divisible by $b+{a-b\over2}$ which also happens to be ${a+b\over2}$, hence so must b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1631792",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Evaluating the integral $\int{\frac{x^2-2x-1}{(x-1)^2(x^2+1)}}$ I successfully evaluated the following integral using partial fraction expansion, but am unsure of a few steps.
$$
\int{\frac{x^2-2x-1}{(x-1)^2(x^2+1)}} = \int\left( \frac{A}{x-1} + \frac{B}{(x-1)^2} + \frac{Cx+D}{x^2+1} \right) \\\ \\
\\
x^2-2x-1 = A(x-1)... | In general you factor the denominator (this can be hard), then
every linear factor $(x-a)^r$ gives you $r$ terms $\frac {A_1}{x-a}$, $\frac {A_2}{(x-a)^2}$, $\ldots$, $\frac {A_r}{(x-a)^r}$
every quadratic factor $(x^2+ bx+c)^s$ (it has complex roots, otherwise factor it to linear terms and apply the above) gives you $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1632009",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Is it possible to solve the following equation without using the Rational Root Theorem? Given $f(x)=x^4+2x^3+2x^2-2x-3$, where $x-1$ is a factor of $f(x)$, how is it possible to solve $f(x)$ without the Rational Root Theorem?
Here's my progress:
$$f(x)=x^4+2x^3+2x^2-2x-3$$
$$f(x)=(x-1)(x^3+3x^2-5x+3)$$
And there's wher... | $$\begin{aligned}
x^4+2x^3+2x^2-2x-3
&=(x^4+2x^2-3)+(2x^3-2x) \\
&=(x^2+3)(x^2-1)+2x(x^2-1) \\
&=(x^2-1)(x^2+3+2x) \\
&=(x^2-1)(x^2+2x+3)
\end{aligned}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1639613",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Show that for each $n \geq 2$, $\left(1 - \frac{1}{4}\right)\left(1 - \frac{1}{9}\right) \cdots \left(1 - \frac{1}{n^2}\right) = \frac{n + 1}{2n}$ Need to show that for each $n \in \mathbb{N}$, with $n \geq 2$,
$$\left(1 - \frac{1}{4}\right)\left(1 - \frac{1}{9}\right) \cdots \left(1 - \frac{1}{n^2}\right) = \frac{n + ... | Here is an inductive proof:
*
*For $n=2$ we get $$1-\frac{1}{2^{2}}=\frac{2+1}{2*2}=\frac{3}{4}$$
*Let $n=k>2$ and $$(1-\frac{1}{2^{2}})(1-\frac{1}{3^{2}})...(1-\frac{1}{k^{2}})=\frac{k+1}{2*k}$$
and now the induction step:
*Let $n=k+1$, then $$(1-\frac{1}{2^{2}})(1-\frac{1}{3^{2}})...(1-\frac{1}{k^{2}})(1-\frac{1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1640628",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
limit of $f(x) = \lim \limits_{x \to 0} (\frac{\sin x}{x})^{\frac 1x}$ Any ideas how to calculate this limit without using taylor?
$$f(x) = \lim \limits_{x \to 0} \left(\frac{\sin x}{x}\right)^{\frac1x}$$
| Here's a proof
that just uses
basic properties
of
$\sin, \cos$,
and
$\ln$.
Since
$-1 \le \cos(x) < 1$
and $\sin'(x) = \cos(x)$
and $\cos'(x) = -\sin(x)$,
$\sin(x)
=-\int_0^x \cos(t) dt
$
so
$|\sin(x)|
\le |x|
$.
Also,
since,
for $x > 0$,
$\ln(1+x)
=\int_1^{1+x} \frac{dt}{t}
=\int_0^{x} \frac{dt}{1+t}
$,
$\ln(1+x)
\le x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1642327",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 4
} |
finding real roots by way of complex I was given
$$x^4 + 1$$
and was told to find its real factors. I found the $((x^2 + i)((x^2 - i))$ complex factors but am lost as to how the problem should be approached.
My teacher first found 4 complex roots ( different than mine)
$$( x - \frac{\sqrt{2}}{2} - \frac{\sqrt{2}}{2} i... | $$x^4+1=0\Longleftrightarrow$$
$$x^4=-1\Longleftrightarrow$$
$$x^4=|-1|e^{\arg(-1)i}\Longleftrightarrow$$
$$x^4=e^{\pi i}\Longleftrightarrow$$
$$x=\left(e^{\left(2\pi k+\pi\right)i}\right)^{\frac{1}{4}}\Longleftrightarrow$$
$$x=e^{\frac{1}{4}\left(2\pi k+\pi\right)i}$$
With $k\in\mathbb{Z}$ and $k:0-3$
So, the solutio... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1646603",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
How to find $\dim W_1$, $\dim W_2$, $\dim W_1+W_2$, $\dim W_1\cap W_2$ for the following spans? Let $W_1=\{(1,1,2,1), (3,1,0,0)\}$ and $W_2=\{(-1,-2,0,1), (-4,-2,-2,-1)\}$
Apparently $\dim W_1=\dim W_2=2$.
For $\dim W_1\cap W_2$, since $(-4,-2,-2,-1)$ can be expressed as $-(1,1,2,1)-(3,1,0,0)$, $\dim W_1\cap W_2=1$. Bu... | Put the given vectors into the columns of matrices
\begin{align*}
w_1 &=
\left[\begin{array}{rr}
1 & 3 \\
1 & 1 \\
2 & 0 \\
1 & 0
\end{array}\right] &
w_2 &=
\left[\begin{array}{rr}
-1 & -4 \\
-2 & -2 \\
0 & -2 \\
1 & -1
\end{array}\right]
\end{align*}
The dimensions of $W_1$ and $W_2$ are the ranks of the matrices $w_... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1648036",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Is this identity about floor function true? When $a, b, c$ are positive integers, is this identity below is true for all $a, b, c$?
$$\left\lfloor \frac{\left\lfloor\frac ab \right\rfloor}c \right\rfloor =\left\lfloor \frac{\left\lfloor\frac ac \right\rfloor}b \right\rfloor $$
| Using division with remainder write
$a=qbc+r$ with $0\le r<bc$ and then write $r=q'b+r'$ with $0\le r'<b$, as well as $r=q''c+r''$ with $0\le r''<c$. So $a=qbc+q'b+r'=qbc+q''c+r''$.
Because $0\le r<bc$ we conclude that $0\le q'<c$ and $0\le q''<b$.
Then
$$ \left\lfloor\frac{\left\lfloor\frac ab\right\rfloor}c\right\r... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1648117",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Probability of an even number of sixes
We throw a fair die $n$ times, show that the probability that there are an even number of sixes is $\frac{1}{2}[1+(\frac{2}{3})^n]$. For the purpose of this question, 0 is even.
I tried doing this problem with induction, but I have problem with induction so I was wondering if my... | Let $E_n$ be the event that there are an even number of sixes rolled in $n$ rolls, and $O_n$ be the event that there are an odd number of sixes rolled in $n$ rolls. Let $X_i = 1$ if roll $i$ is a $6$, otherwise $X_i = 0$. Clearly, $$\Pr[E_n] + \Pr[O_n] = 1$$ for all $n$. Now, observe that $$\begin{align*} \Pr[E_n] &... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1648601",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Given M, can we find $2$ primes $a,b$ so that for all naturals $x,y$, $|a^x-b^y|>M$? For example, if $M = 2$, one can show that $3,17$ satisfy the above:
For any naturals $x,y$, $|3^x-17^y|>2$.
| Choose $a$ to be a prime larger than $M + 1$, and let $b$ be the first prime congruent to $1 \pmod{2 \cdot a}$, guaranteed to exist by Dirichlet's theorem. Now $a^x \equiv a \pmod{2 \cdot a}$ and $b^y \equiv 1 \pmod{2 \cdot a}$, so $\vert a^x - b^y \vert \ge a-1 \gt M$. Using this method, for $M=3$ we have $(a,b)=(5,... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1650598",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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The sum of the following infinite series $\frac{4}{20}+\frac{4\cdot 7}{20\cdot 30}+\frac{4\cdot 7\cdot 10}{20\cdot 30 \cdot 40}+\cdots$
The sum of the following infinite series $\displaystyle \frac{4}{20}+\frac{4\cdot 7}{20\cdot 30}+\frac{4\cdot 7\cdot 10}{20\cdot 30 \cdot 40}+\cdots$
$\bf{My\; Try::}$ We can write t... | The numerators suggest that you could make use of a power series involving exponents that are rational numbers with denominator $3$.
$$\begin{align}
\sum_{n=1}^{\infty}\frac{4\cdot7\cdot\cdots\cdot(3n+1)}{(n+1)!10^n}
&=\sum_{n=1}^{\infty}\frac{\frac43\cdot\frac73\cdot\cdots\cdot\frac{3n+1}3}{(n+1)!\left(10/3\right)^n}\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1652349",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 2,
"answer_id": 0
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Finding coefficient of polynomial? The coefficient of $x^{12}$ in $(x^3 + x^4 + x^5 + x^6 + …)^3$ is_______?
My Try:
Somewhere it explain as:
The expression can be re-written as: $(x^3 (1+ x + x^2 + x^3 + …))^3=x^9(1+(x+x^2+x^3))^3$
Expanding $(1+(x+x^2+x^3))^3$ using binomial expansion:
$(1+(x+x^2+x^3))^3 $
$= 1+3(x+... | $$(x^3+x^4+x^5+x^6+\cdots)^3=x^9(1+x+x^2+\cdots)^3=x^9\left(\dfrac1{1-x}\right)^3=x^9(1-x)^{-3}$$
Now, we need the coefficient of $x^3$ in $(1-x)^{-3}$
Now the $r+1,(r\ge0)$th term of $(1-x)^{-3}$ is $$\dfrac{-3(-4)(-5)\cdots(-r)(-r-1)(-r-2)}{1\cdot2\cdot3\cdot r}(-x)^r=\binom{r+2}2x^r$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1654126",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 4,
"answer_id": 0
} |
Find the equation of tangent at origin to the curve $y^2=x^2(1+x+x^2)$ How do I find the equation of tangent at $(0,0)$ to the curve $y^2=x^2(1+x+x^2)$ ?
Differentiating and putting the value of $x$ and $y$ gives an indeterminate form.
Can we trace the curve and geometrically make tangents and find their equation ?
| $$y^2=x^2+x^3+x^4$$
$$2y\frac{dy}{dx}=2x+3x^2+4x^3$$
$$\implies \frac{dy}{dx}=\frac{2x+3x^2+4x^3}{2y}$$
Now, equation of tangent line at (0,0) is (eqn 1):
$$y=\frac{dy}{dx}\big{|}_{(0,0)}(x)$$
Since $\dfrac{dy}{dx}$ yields 0/0 form, we use limits to evaluate the tangent:
$$\lim_{x,y\to0} \frac{2x+3x^2+4x^3}{2y} $$
$$ \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1654315",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Evaluation of $\int_{0}^{1}\frac{\arctan x}{1+x}dx$
Evaluation of $$\int_{0}^{1}\frac{\tan^{-1}(x)}{1+x}dx = \int_{0}^{1}\frac{\arctan x}{1+x}dx$$
$\bf{My\; Try::}$ Let $$I = \int_{0}^{1}\frac{\tan^{-1}(ax)}{1+x}dx$$
Then $$\frac{dI}{da} = \frac{d}{da}\left[\int_{0}^{1}\frac{\tan^{-1}(ax)}{1+x}\right]dx = \int_{0}^{1... | The substitution $x\mapsto \frac{1-x}{1+x}$ transforms the integral into
$$I=\int_0^1 \frac{\tan^{-1} x}{1+x}\,dx=\int_0^1 \frac{\frac{\pi}{4}-\tan^{-1} x}{1+x}\,dx.$$
Because $\displaystyle\,\,\,\, \tan^{-1} \frac{1-x}{1+x}=\tan^{-1}1-\tan^{-1} x=\frac{\pi}{4}-\tan^{-1} x.$
Taking the average of these two representat... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1654385",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
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Coordinates of a matrix So on the textbook, it gives an example:
If the basis of B matrix is{$\begin{bmatrix}1&0\\ 0&0\end{bmatrix}, \begin{bmatrix}0&1\\ 0&0\end{bmatrix},\begin{bmatrix}0&0\\ 1&1\end{bmatrix}, \begin{bmatrix}1&0\\ 1&0\end{bmatrix}$}
Then the B-coordinates of a matrix $\begin{bmatrix}a&b\\ c&d\end{bma... | $$(a-c+d)\begin{bmatrix}1&0\\ 0&0\end{bmatrix}+b \begin{bmatrix}0&1\\ 0&0\end{bmatrix}+d\begin{bmatrix}0&0\\ 1&1\end{bmatrix}+(c-d)\begin{bmatrix}1&0\\ 1&0\end{bmatrix}=\begin{bmatrix}a&b\\ c&d\end{bmatrix},$$
while
$$(a-d)\begin{bmatrix}1&0\\ 0&0\end{bmatrix}+b \begin{bmatrix}0&1\\ 0&0\end{bmatrix}+(c-d)\begin{bmatrix... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1658439",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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how we can calculate $ \frac {\sqrt {x^2} + \sqrt {y^2} }{2 \sqrt {xyz}}$? I teach math for Schools. How can Help me in the following past Olympiad question?
If $y,z$ be two negative distinct number and $x$ and $y$ be negate of each other, how we can calculate $ \displaystyle\frac {\sqrt {x^2} + \sqrt {y^2} }{2 \sqrt ... | $$\frac{\sqrt{x^2}+\sqrt{y^2}}{2\sqrt{xyz}}=\frac{2\sqrt{x^2}}{2\sqrt{-x^2z}}$$ using $y=-x$. This yields: $$\frac{2\sqrt{x^2}}{2\sqrt{x^2}\sqrt{-z}}=\frac{1}{\sqrt{-z}}$$. Note that $z<0$, so $-z>0$, so $\sqrt{-z}$ is well-defined.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1658764",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Evaluate $\lim_{x \to 0} \frac{\tan(\tan x)-\sin(\sin x)}{\tan x-\sin x}$ Evaluate $$\lim_{x \to 0}\frac{\tan(\tan x)-\sin(\sin x)}{\tan x-\sin x}$$
First I tried using L'Hopital's rule..but it's very lengthy
Next I have written the limits as
$$L=\lim_{x \to 0}\frac{\tan(\tan x)-\sin(\sin x)}{\tan x-\sin x}=\frac{\lim_... | \begin{align}
\lim_{x\to 0}\frac{\tan(\tan x)-\sin(\sin x)}{\tan x-\sin x}&= \lim_{x\to 0}\frac{\tan(\tan x)-\tan(\sin x)\cos(\sin x)}{\tan x - \sin x}\\
&=\lim_{x\to 0}\frac{\tan(\tan x)-\tan(\sin x)+\tan(\sin x)-\tan(\sin x)\cos(\sin x)}{\tan x -\sin x}\\
&=\lim_{x\to 0}\left(\frac{\tan(\tan x)-\tan(\sin x)}{\tan x-\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1659264",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Find $\int\frac{\sqrt{2-x-x^2}}{x^2}dx$ Find $\int\frac{\sqrt{2-x-x^2}}{x^2}dx$
Let $I=\int\frac{\sqrt{2-x-x^2}}{x^2}dx$
I took $\sqrt{2-x-x^2}$ as first function and $\frac{1}{x^2}$ as the second function and integrated it by parts,
$I=\int\frac{\sqrt{2-x-x^2}}{x^2}dx=\sqrt{2-x-x^2}\int\frac{1}{x^2}dx-\int\frac{-1-2x... | Let $$\displaystyle I = \int \frac{\sqrt{2-x-x^2}}{x^2}dx = \int\frac{\sqrt{(x+2)(1-x)}}{x^2}dx$$
Now Let $\displaystyle (x+2) = (1-x)t^2\;,$ Then $\displaystyle x = \frac{t^2-2}{t^2+1} = 1-\frac{3}{t^2+1}$
So $$\displaystyle dx = \frac{6t}{(t^2+1)^2}dt$$ and $$\displaystyle (1-x) = \frac{3}{t^2+1}$$
So Integral $$\dis... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1659444",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Integrate $\int\frac{\sin(2x)}{2+\cos x} dx$
$$\int\frac{\sin(2x)}{2+\cos x} dx $$
Attempt: let $u = 2 + \cos x$. Then $du = -\sin x$. As we can see, $u - 2 = \cos x$, and $\sin(2x) = 2\sin x\cos x$. Using the change of variables method, we see that
$$\int\frac{\sin(2x)}{2+\cos x} dx = -2 \int(u-2)\frac{1}{u}du = -2 ... | Consider integral
$$I=\int\frac{\sin(2x)}{2+\cos x}dx.$$ Through the linéarisation formula $\frac{\sin(2x)}{2}=\sin(x)\cos(x),$ we obtain
$$I=2\int\frac{\sin x\cos x}{2+\cos x}dx,$$ and by the change of variable $u=\cos x, \ x=\arccos x$ and $du=-\sin xdx,$ we arrive at
$$I=-2\int\frac{udu}{2+u}du=-2\int\frac{u+2-2}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1662503",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
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Solve the system of equation with answers as x=___+___s Solve the system:
\begin{cases}
-4x_1 + x_2=3 \\
8x_1-2x_2=-6
\end{cases}
I need the solution in the form of:
\begin{cases}
x_1=\underline{\hspace{1cm}}+\underline{\hspace{1cm}}\times s\\[0.3em]
x_2=\underline{\hspace{1cm}}+\underline{\hspace{1cm}}\times s\\[0.3em... | Reading off from your last matrix equation, you have
$$-4x_1 + x_2 = 3 \Rightarrow x_1 = 1/4 x_2 - 3/4$$
Now choose $x_2 = s$. Then $x_2 = 0 + 1s$ is equivalent to making that choice, and substituting $x_2 = s$ gives $x_1 = 1/4s - 3/4$ in the first equation.
$$x_2 = 0 + 1s$$
$$x_1 = - 3/4 + 1/4s$$
Now if $x$ is a vecto... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1663222",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Does $f_n(x) = \dfrac{x^2}{\sqrt{x^2+1/n}}$ converge uniformly to $|x|$ and why? It is easy to see that $f_n(x) = \dfrac{x^2}{\sqrt{x^2+1/n}}$ converge pointwise to $|x|$
Question: is the convergence uniform?
Attempt:
Want to show $\forall \epsilon > 0$, ... , $|f_n(x) - f| < \epsilon$
$|f_n(x) - f| = |\dfrac{x^2}{... | $$\left|\frac{x^2}{\sqrt{x^2+\frac{1}{n}}}-\frac{x^2}{\sqrt{x^2}}\right|=\left|\frac{x^2\left(\sqrt{x^2}-\sqrt{x^2+\frac{1}{n}}\right)}{\sqrt{x^2+\frac{1}{n}}\sqrt{x^2}}\right|\leq \left|\frac{x^2\left(\sqrt{x^2}-\sqrt{x^2+\frac{1}{n}}\right)}{\sqrt{x^2}\sqrt{x^2}}\right|$$
$$=\left|\sqrt{x^2}-\sqrt{x^2+\frac{1}{n}}\ri... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1664783",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
How many ways can you collect six dollars from $8$ people if $6$ people give $0$ or $1$ dollars and $2$ people each give $0$, $1$, or $5$ dollars? I must use a generating function to solve this question:
In how many ways can you collect six dollars from eight people if six people give either $0$ or $1$ dollars and the... | In general,
$$
(a+b+c)^2=a^2+b^2+c^2+2(ab+bc+ca).
$$
Thus
\begin{align}
(1+x+x^5)^2&=1+x^2+x^{10}+2(x+x^6+x^5)\\
&=1+2x+x^2+2x^5+2x^6+2x^{10}
\end{align}
and so the coefficient of $x^6$ is
\begin{align}
\binom{6}{6}+2\binom{6}{5}+\binom{6}{4}+2\binom{6}{1}+\binom{6}{0}&=1+2\cdot 6+15+ 2\cdot 6+1\\
&=41.
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1665107",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
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Proving that $1^n+2^n+3^n+4^n$ $(n\in \Bbb N)$ is divisible by 10 when $n$ is not divisible by 4 I was solving some math problems to prepare for math contests and came across this one:
Prove that $1^n+2^n+3^n+4^n$ $(n\in N)$ is divisible by 10 if and only if $n$ is not divisible by 4.
So, from what I understand, we hav... | To be slightly more concise:
Lemma
For any prime $p$, $x+x^2+\dots+x^{p-1} \equiv x(1+x+\dots+x^{p-2}) \equiv x(\frac{x^{p-1}-1}{x-1}) \equiv 0 \pmod p$ iff $x \not \equiv 1 \pmod p$
Note the following:
$$1^{n}+2^{n}+3^{n}+4^{n}\equiv 1+0+1+0 \equiv 0 \pmod 2$$
$$1^{n}+2^{n}+3^{n}+4^{n} \equiv 2^{n}+2^{2n}+2^{3n}+2^{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1665472",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Evaluation of $\lim_{x\rightarrow \infty}\left\{\left[(x+1)(x+2)(x+3)(x+4)(x+5)\right]^{\frac{1}{5}}-x\right\}$
Evaluation of $\displaystyle \lim_{x\rightarrow \infty}\left\{\left[(x+1)(x+2)(x+3)(x+4)(x+5)\right]^{\frac{1}{5}}-x\right\}$
$\bf{My\; Try::}$ Here $(x+1)\;,(x+2)\;,(x+3)\;,(x+4)\;,(x+5)>0\;,$ when $x\righ... | \begin{align}
&\lim_{x\to\infty}\left(\left((x+1)(x+2)(x+3)(x+4)(x+5)\right)^{\frac15}-x\right)\\
&=\lim_{x\to\infty}\cfrac{\left((1+\frac1x)(1+\frac2x)(1+\frac3x)(1+\frac4x)(1+\frac5x)\right)^{\frac15}-1}{\frac1x}\\
&=\lim_{h\to0}\cfrac{\left((1+h)(1+2h)(1+3h)(1+4h)(1+5h)\right)^{\frac15}-1}{h}\\
&=f'(0)\\
f(x)&=\left... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1666688",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 7,
"answer_id": 6
} |
Constant length of segment of tangent
Prove that the segment of the tangent to the curve $y=\frac{a}{2}\ln\frac{a+\sqrt{a^2-x^2}}{a-\sqrt{a^2-x^2}}-\sqrt{a^2-x^2}$ contained between the $y$-axis and the point of tangency has a constant length.
What I have done: First of all I found out slope $m$ of the tangent to the... | First, note that the given function $$y = f(x;a) = \frac{a}{2} \log \frac{a + \sqrt{a^2-x^2}}{a - \sqrt{a^2-x^2}} - \sqrt{a^2-x^2}$$ satisfies the relationship $$f(ax;a) = a f(x;1)$$ for $a > 0$; thus, $a$ is a scaling factor for positive reals, and it suffices to consider only the special case $a = 1$. This simplifie... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1669213",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Find the maximum value of $4x − 3y − 2z$ subject to $2x^2 + 3y^2 + 4z^2 = 1.$ Find the maximum value of $4x − 3y − 2z$ subject to $2x^2 + 3y^2 + 4z^2 = 1.$
My Attempt
let $S=4x − 3y − 2z$ and $ t=2x^2 + 3y^2 + 4z^2$. Then $t-s =2x^2 + 3y^2 + 4z^2 -(4x − 3y − 2z)= 2(x-1)^2 + 3(y+1/2)^2+4(z+1/4)^2 -3>=-3$
so $s<=t+3<=4$ ... | Hint: CS inequality $\implies$
$$(2x^2+3y^2+4z^2)(8+3+1) \ge (4x-3y-2z)^2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1669947",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
approximate the root of perturbed polynomial
Approximate the root of $$f(x)=(x-1)(x-1)(x-3)(x-4)-10^{-6}x^6$$ near $r=4$.
Do I have to use iterative method of finding the root, such as Bisection, Secant, etc? Is there other way?
| If
$f(x)
=(x-1)(x-2)(x-3)(x-4)-10^{-6}x^6
$,
if $c$ is small,
$\begin{array}\\
f(4+c)
&=(3+c)(2+c)(1+c)(c)-10^{-6}(4+c)^6\\
&=6c(1+c/2)(1+c/3)(1+c)-10^{-6}4^6(1+c/4)^6\\
&\approx 6c(1+c/2+c/3+c)-(2/5)^6(1+6c/4)\\
&=6c(1+11c/6)-(2/5)^6(1+3c/2)\\
&=c(6+(3/2)(2/5)^6) +11c^2-(2/5)^6\\
&\approx 6c -(2/5)^6\\
\end{array}
$
I... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1670812",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Calculating the determinant by upper triangular reduction - can you check if it's correct? Exercise:
Calculate $$\begin{vmatrix}
a_0 & a_1 & a_2 & \dots & a_n \\
-x & x & 0 & \dots & 0 \\
0 & -x & x & \dots & 0 \\
\dots & \dots & \dots & \dots & \dots \\
0 & 0 & 0 & \dots & x
\end{vmatrix}$$
My approach:
$$\begin{vm... | To verify this by other means, we denote the above determinant as $D_n$ and expand by cofactors along the last column. This yields $D_{n}=xD_{n-1}+x^n a_n$ with base case $D_0 =a_0$. (The $n$-by-$n$ submatrix in the corner yields $(-x)^n$, cancelling the $(-1)^n$ factor from the cofactor expansion.) It can then be read... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1671885",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How many integer pairs (x, y) satisfy $x^2 + 4y^2 − 2xy − 2x − 4y − 8 = 0$? How many integer pairs (x, y) satisfy $x^2 + 4y^2 − 2xy − 2x − 4y − 8 = 0$?
My Attempt
Let $f(x,y)=x^2 + 4y^2 − 2xy − 2x − 4y − 8$ . So $f(x,0)=x^2 − 2x − 8$ . $f(x,0)$ has two roots $x=4 , -2$ . So (4,0), (-2,0) are solution of the given ... | Rewrite the equation as
$$\frac14\left((x+2y-4)^2 + 3(x-2y)^2-48\right)=0.$$
Setting $a:=x+2y-4$ and $b:=x-2y$, we must have that $a$ and $b$ are integers such that
$$a^2+3b^2=48.\tag1$$
The only integer solutions to (1) satisfy $(|a|,|b|)=(0,4)$ and $(|a|,|b|)=(6,2)$. This gives six possibilities for $a$ and $b$, whic... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1672767",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
$ p \in Q[x] $ has as a root a fifth primitive root of unity, then every fifth primitive root of unity is a root of $p$. I'm extremely stuck. Can't figure it.
The conjugate is easy: let $w$ be a primitive root of unity, then $w^{-1}$ will also be a root, that's easy. But I'm missing $w^2$ and $w^3$. Why would they be ... | The fifth roots of $1$ form a cyclic group of order $5$, so any of them, different from $1$, is primitive. If $w$ is one of the primitive roots, the others are $\omega^2$, $\omega^3=\omega^{-2}$ and $\omega^4=\omega^{-1}=\bar{\omega}$.
It is clear that $(x-\omega)(x-\omega^2)(x-\omega^3)(x-\omega^4)=x^4+x^3+x^2+x+1$. T... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1673701",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 4
} |
Area of bounded between $y=x-1$ and $(y-1)^2=4 (x+1)$ Find the area enclosed by curves $y=x-1$ and $(y-1)^2=4 (x+1)$.
I found point of intersection as $(0,-1)$ and $(8,7)$ and set up integral as:
$|\int_{-1}^{7} [(y+1)-\frac{(y-1)^2}{4}-1]dy|$. Am I heading in right direction?
| Finding the points of intersection of the two curves: $(y-1)^2 = 4(y+2) \Rightarrow y^2-2y+1 - 4y-8 = 0\Rightarrow y^2-6y-7 = 0\Rightarrow (y+1)(y-7) = 0\Rightarrow y = -1, 7\Rightarrow A = \displaystyle \int_{-1}^7 \left(y+1- \left(\dfrac{(y-1)^2}{4}-1\right)\right)dy$. The difference between your answer and mine is t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1673912",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Proof for $\forall x\in \mathbb{Z}^+, \exists t \in \mathbb{Z}, 5 \nmid x \to ((x^2= 5t + 1) \vee (x^2 = 5t – 1))$ I am trying to write a proof for the following: If $x$ is a positive integer that is not divisible by $5$, then $x^2$ can be written either as $x^2 = 5t+1$ or $x^2 = 5t−1$ for some integer $t$. So far, I h... | If $x = 5n-1$, then
$$x^2 = (5n-1)(5n-1) = 5n(5n-1) - (5n-1) = 5n(5n-1) - 5n+1 $$
When you divide $x^2$ by $5$, the first two terms above has no remainders, and therefore, $x^2$ modulo $5$ is $1$. (note: $t$ in this case would be equal to $n(5n-1)-n$)
If $x = 5n-2$, then
$$x^2 = (5n-2)(5n-2) = 5n(5n-2) - 2(5n-2) = 5n(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1675055",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
Prove by induction that $n^2 < 2^n$ for all $n \geq 5$? So far I have this:
First consider $n = 5$. In this case $(5)^2 < 2^5$, or $25 < 32$. So the inequality holds for $n = 5$.
Next, suppose that $n^2 < 2^n$ and $n \geq 5$. Now I have to prove that $(n+1)^2 < 2^{(n+1)}$.
So I started with $(n+1)^2 = n^2 + 2n + 1$. Be... | There are, of course, many ways to force $2^{n + 1}$ to show up on the RHS. I like to bound the lower order terms as multiples of the leading term by repeatedly applying the given inequality $n \geq 5$ so that I can apply the inductive hypothesis in one shot. Indeed, observe that:
\begin{align*}
(n + 1)^2
&= n^2 + 2n +... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1675411",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 0
} |
Inverse of $4 \times 4 $ matrix Would you be so kind to provide me some examples of $4 \times 4 $ matrices that satisfy the relation $AB = I $
I am completely clueless to find randomly any two matrix $A$ & $B$. Do we have any rules or shortcuts to find them or do we have to do a trial and error method till we get the... | The standard "beginner's" method for matrix inversion is Gauss-Jordan elimination.
Here are some $4 \times 4$ examples that work out reasonably simply.
$$A = \left[ \begin {array}{cccc} 2&-2&-1&-3\cr 2&-3&0&-3
\cr -1&-1&2&0\cr -1&3&-2&-2
\end {array} \right],\ B = A^{-1} =
\left[ \begin {array}{cccc} -12&11&-9/2&3/... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1676779",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Probability that the roots of a quadratic equation are real
Roots of the quadratic equation $x^2+5x+3=0$ are $4\sin^2\alpha+a$ and $4\cos^2\alpha+a$. Another quadratic equation is $x^2+px+q=0$ where $p,q\in\mathbb{N}$ and $p,q\in[1,10]$. Find the probability that the roots of second quadratic equation are real and t... | As roots of the quadratic equation $x^2+5x+3=0$ are $x_1=4\cos^2\alpha+a$ and $x_2=4\sin^2\alpha+a$, so we get the magnitude of difference b/w the roots as
$$\left|x_1-x_2\right|=\left|4\cos^2\alpha-4\sin^2\alpha\right|=\left|\pm\sqrt{5^2-4(3)}\right|=\sqrt{13}\tag{1}$$
Now consider the quadratic equation $x^2+px+q=0$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1677360",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 1,
"answer_id": 0
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Using the usual notation for a triangle write $\sin^2A$ in terms of the sides a, b and c. This is an A-level trigonometric problem.
Using the usual notation for a triangle write $\sin^2A$ in terms of the sides $a$, $b$ and $c$.
Answer: $$\frac{(a+b-c)(a-b+c)(a+b+c)(-a+b+c)}{4b^2c^2}$$
The last step of the solution requ... | You can solve it in a simpler way: start from Al Kashi's formula:
$$a^2=b^2+c^2-2bc\cos A,$$
from which you deduce first $4b^2c^2\cos^2A= (b^2+c^2-a^2)^2$, and finally
\begin{align*}
4b^2c^2\sin^2A&=4b^2c^2- (b^2+c^2-a^2)^2\\
&=(2bc -b^2-c^2+a^2)(2bc +b^2+c^2-a^2)\\
&=[a^2-(b-c)^2][(b+c)^2-a^2]\\
&=[(a-b+c)(a+b-c)][(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1678338",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
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What's wrong with this transformation? I have the equation $\tan(x) = 2\sin(x)$ and I'd like to transform it in this way:
$$\tan(x) = 2\sin(x)
\Longleftrightarrow
\frac{\sin(x)}{\cos(x)} = 2\sin(x)
\Longleftrightarrow
\sin(x) = 2\sin(x)\cos(x)
\Longleftrightarrow
\sin(x) = \sin(2x)$$
But I'm getting a wrong result so I... | The solutions to $\sin(\frac{x}{2})=0$ or $\cos(\frac{3x}{2})=0$ are given by $$\frac{x}{2} = \pi k \text{ or } \frac{3x}{2} = \frac{\pi}{2}+\pi k, k \in \mathbb Z,$$ which can be rewritten as $$x = 2\pi k \text{ or } x = \frac{\pi}{3}+\frac{2\pi}{3}k, k \in \mathbb Z.$$
The solutions to $\cos(x) = \frac{1}{2}$ or $\si... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1678548",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 6,
"answer_id": 1
} |
If $x$, $\{x\}$, $\lfloor x\rfloor$ are in G.P, find $x$.
If $x$, $\{x\}$, $\lfloor x\rfloor$ are in Geometric Progression, find $x$; $x \neq 0$.
Here, $\{x\}=x-\lfloor x\rfloor$
Some properties are pretty evident:
$$0\leq \{x\} < 1 \tag{1}$$
$$\lfloor x \rfloor \leq x\tag{2}$$
But I can't seem to be able to use them... | Let $x=n+y$, $n=\lfloor x\rfloor$ and $y=\{x\}$ Since these numbers are in geometric progression, we have $x/y=y/n$, or
$$\frac {n+y}{y}=\frac yn$$
$$(n+y)n=y^2$$
$$y^2 - n^2 - ny = 0$$
$$y=\frac{n\pm n\sqrt{5}}2$$
$y$ is between 0 and 1, so $n(1\pm\sqrt{5})/2$ has to be between $0$ and $1$.
This is impossible for ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1680348",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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$\lim_{x \to -1} \frac{|x + 1|} { x^2 + 2x + 1}$? Is the solution: "no limit"
Because:
$$\frac{|x + 1| }{ (x^2 + 2x + 1)} = \frac{1}{(x + 1)}$$
Or is factoring different for $|x + 1|$?
| You can see that
$$
\frac{|x+1|}{x^2+2x+1}\ne\frac{1}{x+1}
$$
by just plugging in $x=-2$; the left-hand side is
$$
\frac{|-2+1|}{4-4+1}=1
$$
whereas the right-hand side is
$$
\frac{1}{-2+1}=-1
$$
Better, $x^2+2x+1=(x+1)^2=|x+1|^2$, so the correct simplification is
$$
\frac{|x+1|}{x^2+2x+1}=\frac{|x+1|}{|x+1|^2}=\frac{1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1683474",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Integrating $\int^b_a(x-a)^3(b-x)^4 \,dx$ I came across a question today...
The value of $\int^b_a(x-a)^3(b-x)^4 \,dx$ is
First I tried the property $\int^b_af(x)=\int^b_af(a+b-x)$. I got $\int^b_a(x-a)^4(b-x)^3 \,dx$, which can be simplified to: $\dfrac{b-a}{2}\int^b_a(x-a)^3(b-x)^3 \,dx$. Well now what? Do I have t... | By integration by parts,
\begin{align}
\int_a^b(x-a)^3 (x-b)^4 dx &= \left[\frac{1}{5}(x-a)^3(x-b)^5\right]_a^b-\int_a^b \frac{3}{5}(x-a)^2(x-b)^5 dx\\
&=-\frac{3}{5}\left(\left[\frac{1}{6}(x-a)^2(x-b)^6\right]_a^b-\int_a^b \frac{1}{3}(x-a)(x-b)^6dx\right)\\
&=\frac{1}{5}\left(\left[\frac{1}{7}(x-a)(x-b)^7\right]_a^b-\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1685217",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 2
} |
Find a mistake in integration So, the integral is:
$$I=\int \frac{3x+5}{x^2+4x+8}dx$$
and here is how I did it, but in the end, I got a wrong result:
$$x^2+4x+8=(x+2)^2+4=4\bigg[\bigg(\frac{x+2}{2}\bigg)^2+1\bigg]$$
$$I=\frac{1}{4} \int \frac{3x+5}{\big(\frac{x+2}{2}\big)^2+1}dx$$
substitution: $\frac{x+2}{2}=u$, $dx=2... | You're doing good. Maybe some passages can be done more easily (at least according to my tastes):
\begin{align}
\int\frac{3x+5}{x^2+4x+8}\,dx
&=\frac{1}{2}\int\frac{6x+10}{x^2+4x+8}\,dx\\[6px]
&=\frac{1}{2}\int\frac{6x+12-2}{x^2+4x+8}\,dx\\[6px]
&=\frac{3}{2}\int\frac{2x+4}{x^2+4x+8}\,dx-
\int\frac{1}{x^2+4x+8}\,dx
\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1685740",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 1,
"answer_id": 0
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Find a value of $n$ such that the coefficients of $x^7$ and $x^8$ are in the expansion of $\displaystyle \left(2+\frac{x}{3}\right)^{n}$ are equal
Question: Find a value of $n$ such that the coefficients of $x^7$ and $x^8$ are in the expansion of $\displaystyle \left(2+\frac{x}{3}\right)^{n}$ are equal.
My attempt... | $\displaystyle \left(2+\frac{x}{3}\right)^{n}=\sum_{i=0} ^n {n \choose k}({\frac{x}{3}})^i 2^{n-i}$
Now for $i =7\implies{n \choose 7}({\frac{x}{3}})^7 2^{n-7}$
and for $i =8\implies{n \choose 8}({\frac{x}{3}})^8 2^{n-8}$
now to get the coefficients of $x^8$ & $x^7$ equal. $\implies$
${n \choose 7}({\frac{1}{3}}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1685895",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 4
} |
General solution expressed in $a_0$ and $a_1$ of a Fibonacci-like sequence?
What is the general solution expressed in $a_0$ and $a_1$ of a Fibonacci-like sequence ?
I mean if $a_0,a_1$ are given and $a_{n+1}:=a_n+a_{n-1}$
$(\begin{array}{cc}a_n&a_{n-1}\end{array})=(\begin{array}{cc}a_{n-1}&a_{n-2}\end{array})\left(\b... | The general solution will be as follows:
$a_n = \dfrac{\varphi_2a_0 - a_1}{\varphi_2 - \varphi_1}\varphi_1^{n} + \dfrac{a_1 - \varphi_1a_0}{\varphi_2 - \varphi_1}\varphi_2^{n}$, where $\varphi_{1,2} = \dfrac{1 \pm \sqrt{5}}{2}$ -- roots of equation $\varphi^2-\varphi-1 = 0$. You can check it by substituting this formu... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1686998",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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proof - $(x,y) = (4,6)$ is the only solution for $x^3 + x^2 - 16 = 2^y$ I saw this question and on seeing the answers I believed it did not have to be so complicated. The pair $(x,y) = (4,6)$ only fits the equation.
Find all possible $(x, y)$ pairs for $x^3 + x^2 - 16 = 2^y$.
My efforts:
$$x^3 + x^2 - 16 = 2^y$$
$$x^... | $x^2(x+1) = 2^y + 16$
Since LHS is an integer, then we must have $y \ge 0$.
Since RHS is a positive integer, then we must have $x \ge 1$.
$x^2(x+1)$ is strictly increasing for $x \ge 0$.
$2^y + 16$ is strictly increasing for $y \ge 0$.
$$\begin{array}{n|c|c|}
\hline
n & n^2(n+1) & 2^n + 16 \\ \hline
0 & 0 & 17 \\
1 & ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1689962",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Solve the recurrence relation $a_n = 2a_{n-1} + 2^n$ with $a_0 = 1$ using generating functions Here is what I have so far, or what I know how to do, rather:
I am given this equation: $a_n = 2a_{n-1} + 2^n$ with $a_0 = 1$
So, with the $2a_{n-1}$, I know I can do the following.
We change the $a_n$ to $x^n$, so then we ha... | Let
$$f(x)=a_0+a_1x+a_2x^2+a_3x^3+\cdots$$
We know for a fact that $\frac{1}{1-x}=1+x+x^2+\cdots$ so we can compute
\begin{align}
2xf(x)+\frac{1}{1-2x}&=2a_0x+2a_1x^2+2a_2x^3+\cdots+1+(2x)+(2x)^2+\cdots\\
&=1+(2a_0+2)x+(2a_1+2^2)x^2+(2a_2+2^3)x^3+\cdots\\
&=a_0+a_1x+a_2x^2+a_3x^3+\cdots\\
&=f(x)
\end{align}
Now we get ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1690386",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Use of residues to find I=$\int_0^\infty \frac{\sin^2(x)}{1+x^4} dx$ I'm working on the problem $$I=\int_0^\infty \frac{\sin^2(x)}{1+x^4} dx$$
I found 4 singularities and i would like to use the singularities in the 1st and 2nd quadrants to solve this integral; i.e. $z_0=e^{i\frac{\pi}{4}}$ and $e^{i\frac{3\pi}{4}}$. M... | Let's start at the beginning because it appears that, although you have the correct formulation, you really do not seem to understand what is going on.(For example, why not use the singularities in the 3rd and 4th quadrants?)
Consider the contour integral:
$$\oint_C dz \frac{1-e^{i 2 z}}{1+z^4} $$
where $C$ is a semici... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1690986",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
How do I integrate $\int \frac{dx}{\sin^3 x + \cos^3 x}$? How do I integrate the following $$\int \frac{dx}{\sin^3 x + \cos^3 x}$$ ?
It appears that I am supposed to break this up into $(\sin x + \cos x)(1-\cos x \sin x)$, but the next thing to do is not apparent to me.
| $\;\;\displaystyle\frac{1}{(\sin x+\cos x)(1-\sin x\cos x)}=A\left(\frac{\sin x+\cos x}{1-\sin x\cos x}\right)+B\left(\frac{1}{\sin x+\cos x}\right)$
where $1=A(\sin x+\cos x)^2+B(1-\sin x\cos x)=(A+B)+(2A-B)\sin x\cos x$.
Then $A=\frac{1}{3}$ and $B=\frac{2}{3}$,
so $\displaystyle\int\frac{1}{\sin^3 x+\cos^3 x}dx=\in... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1694424",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 2
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Proof using strong induction a conjecture about $4^n$ Compute $4^1$, $4^2$, $4^3$, $4^4$, $4^5$, $4^6$, $4^7$, and $4^8$. Make a conjecture about the units digit of $4^n$ where $n$ is a positive integer. Use strong mathematical induction to prove your conjecture.
Solution:
$4^1$ = $4$
$4^2$ = $16$
$4^3$ = $64$
$4^4$ =... | To be proven: If $n$ is a positive integer, the units digit of $4^n$ is $4$ if $n$ is odd and $6$ if $n$ is even.
Proof by strong induction: (Base cases $n=1$ and $n=2$.)
Now assume that $k\ge 3$ and the result is true for all smaller positive values of $k$. The goal is to prove the theorem for $n=k$.
Let $\ell =k-2$; ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1696365",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
the system of diophantine equations: $x+y=a^3$; $xy=\dfrac{a^6-b^3}{3}$ has only trivial solutions. Without using Fermat's Last Theorem, how can one prove that the following system of diophantine equations has only trivial solutions:
$$x+y=a^3$$
$$xy=\dfrac{a^6-b^3}{3}$$
We suppose of course that $\gcd(x,y)=\gcd(a,x)=\... | $
\begin{cases}
1)&x = a^3 - y \\
2)&xy = \frac{a^6 - b^3}{3} \\
\end {cases}
$
First, subsitute $1)$ into $2)$
$(a^3 - y)y = \frac{a^6 - b^3}{3}$
$3y\left(a^3-y\right)=a^6-b^3$
$3a^3y-3y^2=a^6-b^3$
$-3y^2+3a^3y-a^6+b^3=0$
$y = \frac{-3a^3\pm \sqrt{-3a^6+12b^3}}{2\left(-3\right)}$
$y = \frac{3a^3\pm\sqrt{12b^3-3a^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1697652",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
} |
show this inequality with $a+b+c+d=1$ Let $a,b,c,d\ge 0$,and such $a+b+c+d=1$, show that
$$3(a^2+b^2+c^2+d^2)+64abcd\ge 1$$
use AM-GM
$$a^2+b^2+c^2+d^2\ge 4\sqrt{abcd}$$
it suffices to
$$4\sqrt{abcd}+64abcd\ge 1$$
| Here is a way using Schur and smoothing. Let $a$ be the minimum among $a, b, c, d$ and say $3p = b+c+d, \,q = bc+cd+db, \,r = bcd$. We define
$$F(a, b, c, d) = 3\sum_{cyc} a^2+64abcd $$
We will first show $F(a, b, c, d) \ge F(a, p, p, p)$.
$$\iff 3(b^2+c^2+d^2-3p^2) \ge 64a(p^3-r)$$
But $b^2+c^2+d^2-3p^2 = 2(3p^2-q)$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1698269",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 1
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The best way of integrating irrational functions Ok, so, here is the example integral:
$$I=\int\frac{x-2-\sqrt{-x^2-4x+4}}{x^2-\sqrt{-x^2-4x+4}}dx$$
I always solve these types of integrals using Euler's substitutions, but, recently, I came across some more difficult integrals like the one above which, after using Euler... | There is a mistake in your computation. I added my comments here since there is a limited number of characters I can put into the comment window.
\begin{equation}
x=\frac{4-2t}{1-t^2},\ \ \ dx=\frac{-2(t^2-4 t+1)}{(1-t^2)^2}dt
\end{equation}
Then after the substitution, the integral becomes:
\begin{equation}
I=\int \f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1698947",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
calculate $\sum_{1\le i\le r}\frac{i+1} { r+1}{2r-i\choose r-i}{s+i-2\choose i}+\frac{1}{r+1}{2r\choose r}$? prove the following equation :
$$\sum_{1\le i\le r}\frac{i+1} { r+1}{2r-i\choose r-i}{s+i-2\choose i}+\frac{1}{r+1}{2r\choose r}={s+2r-1\choose r} - {s+2r-1\choose r-1}$$
| Suppose we seek to verify that
$$\frac{1}{r+1} \sum_{q=0}^r (q+1) {2r-q\choose r-q}
{s-2+q\choose q} = {s+2r-1\choose r} - {s+2r-1\choose r-1}.$$
Introduce the integral representation
$${2r-q\choose r-q}
= \frac{1}{2\pi i}
\int_{|w|=\epsilon}
\frac{(1+w)^{2r-q}}{w^{r-q+1}}
\; dw$$
Note that this is zero when $q\gt r$ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1699118",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
find $\frac{3}{6}+\frac{3\cdot5}{6\cdot9}+\frac{3\cdot5\cdot7}{6\cdot9\cdot12}+\cdots$ find $\frac{3}{6}+\frac{3\cdot5}{6\cdot9}+\frac{3\cdot5\cdot7}{6\cdot9\cdot12}+\cdots$
I had $(1-x)^{-\frac{p}{q}}$ in mind.
$$S=\frac{3}{6}+\frac{3\cdot5}{6\cdot9}+\frac{3\cdot5\cdot7}{6\cdot9\cdot12}+\cdots$$
$$S+1=1+\frac{3}{6}+\f... | In the last step, you miss some multiple of $3$, and you miss one term of the expansion.
$$
S=\sum_{n\geq 1} \frac{(2n+1)!!}{(n+1)!3^n}=\sum_{n\geq 1} \binom{-\frac{1}{2}}{n+1}\frac{(-2)^{n+1}}{3^n}=3\sum_{n\geq 1} \binom{-\frac{1}{2}}{n+1}\left(\frac{-2}{3}\right)^{n+1}=3\left(\left(1-\frac{2}{3}\right)^{-\frac{1}{2}}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1699983",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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How was this factoring of $1-2x-x^2$ achieved? If I factor $1-2x-x^2$ using the quadratic formula I get
$$x=\frac{2\pm \sqrt{4-4(-1)(1)}}{2(-1)}$$
$$x=\frac{2\pm \sqrt{8}}{-2}$$
$$x=-1 \pm \sqrt{2}$$
Let $\alpha = -1 +\sqrt{2}$ and $\beta=-1-\sqrt{2}$.
So $1-2x-x^2=-(x-\alpha)(x-\beta)$.
In the image below, where did ... | Your $\alpha$ and $\beta$ are not the same as the $\alpha$ and $\beta$
in the image you copied.
In the image, they are factoring
$$1−2x−x^2 = -\left(\frac 1\alpha - x\right)\left(\frac 1\beta - x\right).$$
You can confirm that using their values
$\alpha = 1 + \sqrt2$ and $\beta = 1-\sqrt2$,
we have
\begin{align}
\frac ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1701147",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Solve $ 6\sin^2(x)+\sin(x)\cos(x)-\cos^2(x)=5 $
Question: Solve $$ 6\sin^2(x)+\sin(x)\cos(x)-\cos^2(x)=5 $$
$$ 0 \le x \le 360^{\circ} $$
My attempt:
$$ 6\sin^2(x)+\sin(x)\cos(x)-\cos^2(x)=5 $$
$$ 6(\frac{1}{2} - \frac{\cos(2x)}{2}) + \sin(x)\cos(x) -(\frac{1}{2} + \frac{\cos(2x)}{2}) = 5 $$
$$ 3 - 3\cos(2x)+ \s... | I'm not sure why you're using this complicated method. If you have an equation of the form
$$
a\sin^2x+b\sin x\cos x+c\cos^2x=d
$$
you can just observe that this is equivalent to
$$
a\sin^2x+b\sin x\cos x+c\cos^2x=d\sin^2x+d\cos^2x
$$
so it becomes
$$
(a-d)\sin^2x+b\sin x\cos x+(c-d)\cos^2x=0
$$
This is Roman83's solut... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1701363",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 9,
"answer_id": 6
} |
Let $a$ and $b$ be the coefficient of $x^3$ in $(1+x+2x^2+3x^3)^4$ and $(1+x+2x^2+3x^3+4x^4)^4$ respectively. Let $a$ and $b$ be the coefficient of $x^3$ in $(1+x+2x^2+3x^3)^4$ and $(1+x+2x^2+3x^3+4x^4)^4$ respectively.Find $(a-b).$
I tried to factorize $(1+x+2x^2+3x^3)$ and $(1+x+2x^2+3x^3+4x^4)$ into product of two ... | Let $p(x)=1+x+2x^2+3x^3$ and $q(x)=1+x+2x^2+3x^3+4x^4$. Then
$$(q(x))^4-(p(x))^4=(q(x)-p(x))A(x)$$
for some polynomial $A(x)$. But $q(x)-p(x)=4x^4$, so $(q(x))^4-(p(x))^4$ is divisible by $x^4$. It follows that all coefficients of $x^k$ for $k\le 3$ are $0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1704589",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
If $\csc \frac{\pi}{32}+\csc \frac{\pi}{16}+\csc \frac{\pi}{8}+\csc \frac{\pi}{4}+\csc \frac{\pi}{2}=\cot \frac{\pi}{k}$. If $\csc \frac{\pi}{32}+\csc \frac{\pi}{16}+\csc \frac{\pi}{8}+\csc \frac{\pi}{4}+\csc \frac{\pi}{2}=\cot \frac{\pi}{k}$.Find $k$
Let $\frac{\pi}{32}=\theta$
$\csc \frac{\pi}{32}+\csc \frac{\pi}{16... | We have the formula ie $\sin(2x)=2sinxcosx$ so we know value of $sin(45)$ substitute $45=2x$ so youll get $sin(22.5)=\pi/8$ . thus solving double angle identity many times you get all sines . then the work is almost done. To convert cos to sin use $sin^2(x)+cos^2(x)=1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1705618",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
If $A$ is a square matrix such that $A^{27}=A^{64}=I$ then $A=I$
If $A$ is a square matrix such that $A^{27}=A^{64}=I$ then $A=I$.
What I did is to subtract I from both sides of the equation:
$$A^{27}-I=A^{64}-I=0$$
then:
\begin{align*}
A^{27}-I &= (A-I)(A+A^2+A^3+\dots+A^{26})=0\\
A^{64}-I &= (A-I)(A+A^2+A^3+\dots+A... | \begin{align*}
& I = A^{64} = A^{2(27) + 10} = (A^{27})^2A^{10} = A^{10}\\
\implies & I = A^{27} = A^{2(10) + 7} = (A^{10})^2A^7 = A^7\\
\implies & I = A^{10} = A^{1(7)+3} = (A^7)^1A^3 = A^3\\
\implies & I = A^7 = A^{2(3) + 1} = (A^3)^2A = A.
\end{align*}
This is nothing more than the Euclidean algorithm applied to the... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1706061",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 2
} |
How many solutions in integers to the following equation
What is the number of positive integer solutions $(a, b)$ to $2016 + a^2 = b^2$?
We have,
$2016 = (b-a)(b+a) = 2^5 \cdot 3^2 \cdot 7$
$b - a = 2^{t_1} 3^{t_2} 7^{t_4}$ and
$a - b = 2^{t_5} 3^{t_6} 7^{t_7}$
Thus, we must have $t_1 + t_5 = 5$ and $t_2 + t_6 = 2$ ... | Just to close the problem out, since $t_1$ can not be either $0$ or $5$ (by parity), there are only $24$ possible choices of the $t_i$. Inspection shows that $b$ is always positive, but that we switch the sign of $a$ when we exchange the choices $\{t_1,t_2,t_4\},\;\{t_5,t_6,t_7\}$. Thus we can reject exactly half of ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1706193",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Simplify $2(\sin^6x + \cos^6x) - 3(\sin^4 x + \cos^4 x) + 1$ Here is the expression:
$$2(\sin^6x + \cos^6x) - 3(\sin^4 x + \cos^4 x) + 1$$
The exercise is to evaluate it.
In my text book the answer is $0$
I tried to factor the expression, but it got me nowhere.
| Let $t=\sin^2(x)$.
$$2(t^3+(1-t)^3)-3(t^2+(1-t)^2)+1=6t^2-6t+2-6t^2+6t-3+1=0.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1710182",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 2
} |
Is the intersection of $x^2 + y^2 + z^2 = 1$ and $x = \frac{1}{2}$ a manifold in $\mathbb{R}^3$? Is the intersection of $x^2 + y^2 + z^2 = 1$ and $x = \frac{1}{2}$ a manifold in $\mathbb{R}^3$?
I think that it is, because it can be parameterized by $f(x) = (\frac{1}{2},\sqrt{\frac{3}{4}} \cos x, \sqrt{\frac{3}{4}} \sin... | Let
$$f:\Bbb R^3\to\Bbb R^2,\quad f(x,y,z)=(x^2+y^2+z^2-1,x-1/2).$$
Then, your space is $f^{-1}(0)$. Now,
$$df=\begin{pmatrix}
2x & 2y & 2z \\
1 & 0 & 0
\end{pmatrix}
$$
which has full rank for all $(x,y,z)\in f^{-1}(0)$, so yes it is a manifold.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1714042",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
The Diophantine Equation: $x^3-3=k(x-3)$ I wish to know how to resolve the diophantine equation: $x^3-3=k(x-3)$ ? The problem is:
Find all integers $x\ne3$ such that $x-3\mid x^3-3$.
- From 250 Problem's in Elementary Number Theory, by W. Sierpinski
I have found the solutions to this problem by using rules in divisib... | You made a mistake at the beginning. $(x^3-3)-(x^3-3x^2)=3x^2-3$, not just $3x^2$. Thus, you have $(x-3) \mid (9x-3)$ by subtracting from $3x^2-9x$ and then $(x-3) \mid 24$ by subtracting from $9x-27$.
Here is the set of all of the integer divisors of 24:
$$\{-24, -12, -8, -6, -4, -3, -2, -1, 0, 1, 2, 3, 4, 6, 8, 12, ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1714433",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 1
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Find $\lim_{x\to0}\frac{(\tan(\frac{\pi}{4}+x))^{\frac{1}{x}}-e^2}{x^2}$ Find $\lim_{x\to0}\frac{(\tan(\frac{\pi}{4}+x))^{\frac{1}{x}}-e^2}{x^2}$
My attempt:
$\lim_{x\to0}(\tan(\frac{\pi}{4}+x))^{\frac{1}{x}}=e^2$
So the required limit is in $\frac{0}{0}$ form.
Then i used L hospital form.
$\lim_{x\to0}\frac{e^2(\frac... | You may observe that, by the Taylor expansion, you get, as $x \to 0$,
$$
\begin{align}
\tan x&=x+\frac{x^3}3+O(x^5) \tag1
\\ \frac{1+\tan x}{1-\tan x}&=1+2 x+2 x^2+\frac{8 x^3}{3}+O(x^5) \tag2
\end{align}
$$ giving
$$
\begin{align}
\log\left(\frac{1+\tan x}{1-\tan x}\right)&=2 x+\frac{4 x^3}{3}+O(x^5) \tag3
\\\\\frac1x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1715484",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Question Regarding Integration Within Summation
It expresses an integral within a summation procedure.
| $$\sum _{1}^{\infty}\log (\frac{k+a+1}{k+a})-\log(\frac{1+k+b}{k+b})$$
$$=\sum_1^\infty \log\frac{k+b(k+a+1)}{k+a(k+b+1)}$$
now sum it over
and expanding the expression gives
$$=\lim _{k \to \infty}\log \frac{(1+b)(2+a)}{(1+a)(2+b)}\frac {(2+b)(3+a)}{(2+a)(3+b)}.... \frac{(k+a+1)}{(k+b+1)}$$
as the consecutive expres... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1715625",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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Convergence of binomial series I know that $f(x)=\sum_{n=0}^{\infty}\binom{\alpha}{n}x^n$ converges for $|x|<1$
What I then have to show is that $(1+x)f'(x)=\alpha f(x)$ for $|x|<1$ and that any such $f$ is of the form $c(1+x)^\alpha$ for some constant c, and to use that fact to establish the binomial series.
I tried t... | $$\begin{align*}(1+x)f'(x)
&= (1+x) \sum_{n=1}^\infty \binom a n n x^{n-1}
= (1+x) \sum_{n=0}^\infty \binom a {n+1} (n+1) x^n \\
&= \sum_{n=0}^\infty \binom a {n+1} (n+1) x^n + \sum_{n=1}^\infty \binom a {n} n x^n \\
&= a + \sum_{n=1}^\infty \binom a {n+1} (n+1) x^n + \binom a {n} n x^n
= a + \sum_{n=1}^\infty a_n x^n
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1716587",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
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Does the sequence $\frac{n!}{1\cdot 3\cdot 5\cdot ... \cdot (2n-1)}$ converge? I'm trying to determine if this sequence converges as part of answering whether it's monotonic:
$$
\left\{\frac{n!}{1\cdot 3\cdot 5\cdot ... \cdot (2n-1)}\right\}
$$
First, I tried expanding it a bit to see if I could remove common factors i... | $a_{n+1} = a_n \frac{n+1}{2n+1}$ Each number in the sequence is slightly more that (1/2) as the one before it.
for $n>3, \dfrac{a_{n+1}}{a_n} < 0.6$
$a_n < 0.4 (0.6)^{n-2}$
The sequence converges to 0.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1722673",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 7,
"answer_id": 4
} |
Integral of quotients of $\sin$ function I am trying to calculate the definite integral
$$\int_0^\pi \frac{\sin(\frac{21}{2}x)}{\sin(\frac{1}{2}x)} dx.$$
Wolfram Alpha says here that the answer is $\pi$. I replaced 21 by other constants and think that in general, $\int_0^\pi \frac{\sin(\frac{n}{2}x)}{\sin(\frac{1}{2}x)... | This is exactly the closed form of the Dirichlet's Kernel when N = 10.
$\sum _{n=-N}^{N} $ $e^{-inx}$ = $\frac{sin(\frac{(2N+1)x}{2})}{sin(\frac{x}{2})}$
Integration the series term by term, the terms with positive values of n will cancel the terms with negative ones, and the value of the integral when $n=0$ is $ \int_... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1724930",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 1
} |
For any real number $x$, if $ x^3+2x+33\neq 0$, then $x+3 \neq0$ How to solve this type of sum using indirect proof. Appreciate if anyone can explain it step by step.
| Your statement is true. But we can even establish that the converse is true.
Note that $$x^3+2x+33=(x+3)(x^2-3x+11)$$
But, since the polynomial discriminant is negative, for any real number, $x^2-3x+11 \neq 0$.
So we have that if $x^3+2x+33=0 \Rightarrow x+3=0$.
Also, since $(-3)^3+2(-3)+33=0$, we have that $x+3=0 \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1726246",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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All 3 digit numbers the sum of whose digits is not greater than 16 My working: Let the digits of the numbers be $x,y,z$ where
\begin{align}1\leq x&\leq 9\\0\leq y,z&\leq 9\\x+y+z &\leq 16 \end{align}
I tried to solve it by making different cases here and using counting but it got really complicated and i couldnt get ve... | Let $h$ denote the hundreds digit, $t$ denote the tens digit, and $u$ denote the units digit. Since the digit sum is at most $16$,
$$h + t + u \leq 16 \tag{1}$$
where $1 \leq h \leq 9$, $0 \leq t \leq 9$, and $0 \leq u \leq 9$.
Let $h' = h - 1$. Then $h'$ is a non-negative integer satisfying the inequalities $0 \l... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1728950",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Question about an inequality which seems right but not easy to prove The origin problem is as follows:
let $a,b,c,d$ are positive real numbers,and $a+b+c+d=4$
prove:$$\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{d}+\frac{d^2}{a}\geq 4+\frac{1}{4}[(a−b)^2+(b−c)^2+(c−d)^2+(d−a)^2]$$
The solution is easy enough, which is to plu... | $\sum\limits_{cyc}\frac{a^2}{b}-4=\sum\limits_{cyc}\left(\frac{a^2}{b}-2a+b\right)=\sum\limits_{cyc}\frac{(a-b)^2}{b}\geq\frac{\left(\sum\limits_{cyc}|a-b|\right)^2}{4}\geq\frac{1}{4}\sum\limits_{cyc}(a-b)^2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1731535",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Show that $x^2 + y^2 + z^2 = x^3 + y^3 + z^3$ has infinitely many integer solutions.
Show that $x^2 + y^2 + z^2 = x^3 + y^3 + z^3$ has infinitely many integer solutions.
I am not able to find an idea on how to proceed with the above questions. I have found only the obvious solution $(1,1,1)$.
Could you please provide... | Let $y=1+a, z=1-a$. Then
$$x^3+2(1+3a^2)=2(1+a^2)+x^2$$
$$x^2-x^3=4a^2.$$
Let $1-x=4p^2$, then
$$x^2(1-x)=(4p^2-1)^24p^2=(2a)^2.$$
Let $a=p(4p^2-1)$. Then
$$(x,y,z)=(1-4p^2, 1+p(4p^2-1), 1-p(4p^2-1))$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1732049",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
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evaluate $\lim_{x\rightarrow 0}\frac{x}{\left | x-1 \right |-\left| x+1\right|}$ I just have a quick question about limits like this one: $$\lim_{x\rightarrow 0}\frac{x}{\left | x-1 \right |-\left| x+1\right|}$$
leaving it as is i get $$\lim_{x\rightarrow 0}\frac{x}{\left ( x-1 \right )-\left( x+1\right)}$$
$$\lim_{x\r... | another way: $\frac{x}{\sqrt{(x-1)^2}-\sqrt{(x+1)^2}} \\ =\frac{x}{\sqrt{(x-1)^2}-\sqrt{(x+1)^2}} \cdot \frac{\sqrt{(x-1)^2}+\sqrt{(x+1)^2}}{\sqrt{(x-1)^2}+\sqrt{(x+1)^2}} \\ =\frac{x [ \sqrt{(x-1)^2}+\sqrt{(x+1)^2}]}{(x-1)^2-(x+1)^2} \\ =\frac{x[ \sqrt{(x-1)^2}+\sqrt{(x+1)^2}]}{([x-1]-[x+1])([x-1]+[x+1])} \text{ n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1732851",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Bounding a sum involving a $\Re((z\zeta)^N)$ term This is a follow up to this question. Any help would be very much appreciated.
Let $k\in\mathbb{N}$ be odd and $N\in\mathbb{N}$. You may assume that $N>k^2/4$ or some other $N>ak^2$.
Let $\zeta:=\exp(2\pi i/k)$ and $\alpha_v:=\zeta^v+\zeta^{-v}+\zeta^{-1}$.
Here there a... | Consider first
$$\begin{align}
1-\sin^2\left(\frac{2\pi
v}{k}\right)\Re\left((\alpha_v\zeta)^N\right)+|\alpha_v|^{2N}&\leq 1+\sin^2\left(\frac{2\pi
v}{k}\right)\left|\Re\left((\alpha_v\zeta)^N\right)\right|+|\alpha_v|^{2N}
\\&\leq 1+\sin^2\left(\frac{2\pi
v}{k}\right)\left|(\alpha_v\zeta)^N\right|+|\alpha_v|^{2N}
\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1733216",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 1
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Finding the extrema on sphere Let $f(x,y,z) = xe^{yz}$ . Find the extrema on the sphere $x^2+y^2+z^2=1$ . We have $e^{yz}=2tx$ , $xze^{yz}= 2ty$ , $xye^{yz}=2tz$ solving by subtitutions i found that $z(x^4-2)=0$ implies that if $z=0$ then $y=0$ and $x= \pm1$ . if the other part is zero then i found that the solution ... | The unit sphere is a compact subset of $\mathbb{R}^3$ and $f$ is continuous, so the extrema exist and are reached.
Let $g=0$ be the constraint, with
\begin{align}
g(x,y,z)=x^2+y^2+z^2-1.
\end{align}
We are going to use the Lagrange multipliers theorem ; to do this, we first compute the differentials
\begin{align}
\math... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1734631",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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$a,b,c$ are the sides and $A,B,C$ are the angles of a triangle. If the roots of the equation $a(b-c)x^2+b(c-a)x+c(a-b)=0$ are equal then,
$a,b,c$ are the sides of a $\triangle ABC$ and $A,B,C$ are the respective angles. If the roots of the equation $a(b-c)x^2+b(c-a)x+c(a-b)=0$ are equal then $\sin^2 \bigl(\frac{A}{2}\... | As you noticed, the roots are,
$$x = 1\hspace{1 cm}\mathrm{ or }\hspace{1 cm}x = \frac{(a-b)c}{a(b-c)}$$
The roots are equal,$$1 = \frac{(a-b)c}{a(b-c)}.$$
Therefore, we can find
$$b = \frac{2ac}{a+c}.$$
Now let's talk about trigonometry. The half angle formula for sine is
$$\sin^2{\frac{W}{2}} = \frac{1 - \cos{W}}{2}$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1734726",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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what is the set of points in the complex plane which satisfies |z| = Re(z) + 2? what is the set of points in the complex plane which satisfies |z| = Re(z) + 2?
so $ \sqrt{x^2+y^2} = x + 2 $ this is not a circle or anything and it asks me to sketch it what should I do
| $$
\sqrt{x^2+y^2} = x = 2 \iff x^2+y^2 =(x+2)^2 \\
\iff x^2+y^2=x^2+4x+4 \\
\iff y^2=4x+4
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1735481",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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In a triangle $ABC,$if $(a+b+c)(a+b-c)(a-b+c)(b+c-a)=\frac{8a^2b^2c^2}{a^2+b^2+c^2}$ then the triangle is In a triangle $ABC,$if $(a+b+c)(a+b-c)(a-b+c)(b+c-a)=\frac{8a^2b^2c^2}{a^2+b^2+c^2}$ then the triangle is
$(A)$isosceles
$(B)$right angled
$(C)$equilateral
$(D)$ obtuse angled
$(a+b+c)(a+b-c)(a-b+c)(b+c-a)=\frac{... | Without loss of generality suppose $\cos C\ne 0 $.
Let the area of the triangle be T. The LHS is $16 T^2. $
The RHS is $8(\frac {1}{2}(a b \sin C)^2) (4\sin^2 C)c^2/(a^2+b^2+c^2)=32 T^2c^2/(a^2+b^2+c^2).$
And the denominator above is $a^2+b^2+c^2=(c^2+2 a b \cos C)+c^2=2(c^2+a b \cos C).$ So we have $$1=c^2/(\sin^2 c)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1736881",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find the sum of $-1^2-2^2+3^2+4^2-5^2-6^2+\cdots$
Find the sum of $$\sum_{k=1}^{4n}(-1)^{\frac{k(k+1)}{2}}k^2$$
By expanding the given summation,
$$\sum_{k=1}^{4n}(-1)^{\frac{k(k+1)}{2}}k^2=-1^2-2^2+3^2+4^2-5^2-6^2+\cdots+(4n-1)^2+(4n)^2$$
$$=(3^2-1^2)+(4^2-2^2)+(7^2-5^2)+(8^2-6^2)+\cdots+[(4n-1)^2-(4n-3)^2]+((4n)^2-... | We know that $$\begin{align*}1+2+3+...+(4n+1)&=\frac{(4n+1) \cdot (4n+2)}{2}\\&=(2n+1)(4n+1)\end{align*}$$
So, $$\begin{align*}&2+3+(2+2)+(3+2)+6+7+...+(4n-2)+(4n-1)+(4n-2+2)+(4n-1+2)=(2n+1)(4n+1)-1 \\&\implies 2(2+3+...(4n-1))+2n \cdot 2=(2n+1)(4n+1)-1 \\&\implies 2(4+6+6+7+...+(8n-2))=2((2n+1)(4n+1)-4n-1)\\&=16n^2+4n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1738560",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 6,
"answer_id": 3
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How to solve this question in more time efficient way?
Q) if$$x\sin a=y\cos a=\frac{2z\tan a}{1-\tan^2 a}$$ then find $4z^2(x^2+y^2)$a)$(x^2+y^2)^{3}$b)$(x^2-y^2)^3$c)$(x^2-y^2)^2$d)$(x^2+y^2)^2$
Ans:c
i solved this in a very long way:
$$x\sin a=y\cos a=\frac{2z\tan a}{1-\tan^2 a}=z\tan 2a$$$$\implies x= \frac{z\tan ... | $x\sin\alpha=y\cos\alpha\iff\tan\alpha=\dfrac yx$
$\dfrac{2\tan\alpha}{1-\tan^2\alpha}=\dfrac{2xy}{x^2-y^2}$
$$\implies\sin\alpha=\dfrac zx\cdot\dfrac{2xy}{x^2-y^2}=\dfrac{2yz}{x^2-y^2}$$
Use $$\dfrac1{(\sin\alpha)^2}-\dfrac1{(\tan\alpha)^2}=1$$
OR
similarly find $\cos\alpha=?$
Now find $\sin^2\alpha+\cos^2\alpha$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1738815",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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How to solve the integral $\int\limits_0^a {\frac{{\sqrt {{a^2} - {x^2}} }}{{b - x}}} \mathop{\mathrm{d}x}\\$? I have seen this integral:
$$\int\limits_0^a {\frac{{\sqrt {{a^2} - {x^2}} }}{{b - x}}} \mathop{\mathrm{d}x}\\$$
In this integral: $a$ and $b$ are constants.
I have try with two ways, but failed:
*
*$u = \... | First off I would let $x=-a\cos\theta$. Then
$$\begin{align}\int_0^a\frac{\sqrt{a^2-x^2}}{b-x}dx&=\int_{\frac{\pi}2}^{\pi}\frac{a^2\sin^2\theta}{b+a\cos\theta}d\theta\\
&=\int_{\frac{\pi}2}^{\pi}\frac{a^2(1-\cos^2\theta)}{b+a\cos\theta}d\theta\\
&=\int_{\frac{\pi}2}^{\pi}\left(-a\cos\theta+b+\frac{a^2-b^2}{b+a\cos\thet... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1742517",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
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Check: Radius of Convergence of the Sum of these Complex Taylor Series I just found the following Taylor series expansions around $z=0$ for the following functions:
*
*$\displaystyle \frac{1}{z^{2}-5z+6} = \frac{1}{(z-2)(z-3)} = \frac{-1}{(z-2)} + \frac{1}{(z-3)} = \sum_{n=0}^{\infty}\frac{z^{n}}{2^{n+1}} - \sum_{n=... |
From the representation
\begin{align*}
\frac{1}{z^2-5z+6}&=\frac{1}{(z-2)(z-3)}\tag{1}\\
\frac{1}{1-z-z^2}&=\frac{1}{\left(z-\left(\frac{1-\sqrt{5}}{-2}\right)\right)\left(z-\left(\frac{1+\sqrt{5}}{-2}\right)\right)}\tag{2}\\
\end{align*}
we can already deduce the radius of convergence.
Both expressions have two ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1742942",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Trigonometric limit $\lim_{x\to\pi/4}\frac{1-\tan x}{1-\sqrt{2}\sin x}$ The limit is
$$\lim_{x\to\pi/4}\frac{1-\tan x}{1-\sqrt{2}\sin x}$$ I was able to solve it using L'hopital and the answer that I got was $2$.
Can you please confirm if the answer is right and suggest some other way to evaluate the limit without usin... | Multiply both deminator and numerator to the $\left( 1+\sqrt { 2 } \sin { x } \right) $ and $\left( 1+\tan { x } \right) $ respectively
\begin{align*}
\lim_{x\to\frac{\pi}{4}}{\frac{1-\tan x}{1-\sqrt{2}\sin x}}={}&\lim_{x\to\frac{\pi}{4}}{\frac{\left(1-\tan^2\!x\right)\left(1+\sqrt2\sin x\right)}{\left(1-2\sin^2\!x\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1746581",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 0
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Finding the sum of series $\sum_{n=0}^∞ \frac{2^n + 3^n}{6^n}$ I am being asked to find the sum of the following convergent series :
$$\sum_{n=0}^∞ \frac{2^n}{6^n} + \frac{3^n}{6^n}$$
Attempting to generalize from partial sums yields nothing of interest:
$s_1 = \frac{5}{6}$
$s_2 = \frac{5}{6} + \frac{13}{36} = \frac{43... | Note that the sum $\sum_{n=0}^{\infty}r^{n}=\frac{1}{1-r}$ for $|r|<1$. According to this fact $\sum_{n=0}^{\infty}\frac{2^{n}+3^{n}}{6^{n}}=\sum_{n=0}^{\infty}\frac{2^{n}}{6^{n}}+\sum_{n=0}^{\infty}\frac{3^{n}}{6^{n}}=\sum_{n=0}^{\infty}(\frac{2}{6})^{n}+\sum_{n=0}^{\infty}(\frac{3}{6})^{n}=\frac{1}{1-1/3}+\frac{1}{1-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1753313",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Evaluation of limit at infinity: $\lim_{x\to\infty} x^2 \sin(\ln(\cos(\frac{\pi}{x})^{1/2}))$ $$\lim_{x\to\infty} x^2 \sin(\ln(\cos(\frac{\pi}{x})^{1/2}))$$
What I tried was writing $1/x=t$ and making the limit tend to zero and writing the cos term in the form of sin
| Using Taylor series for $\cos u$, $\sqrt{1+u}$, $\ln(1+u)$, and $\sin u$ when $u\to 0$. When $x\to \infty$, $\frac{1}{x} \to 0$, so
$$\cos\frac{\pi}{x} = 1- \frac{\pi^2}{2x^2} + o\left(\frac{1}{x^2}\right)$$
and
$$\sqrt{ \cos(\frac{\pi}{x}) } = \sqrt{ 1- \frac{\pi^2}{2x^2} + o\left(\frac{1}{x^2}\right) } = 1- \frac{\pi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1753645",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 7,
"answer_id": 0
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Find the limit of $\frac{n^4}{\binom{4n}{4}}$ as $n \rightarrow \infty$ $\frac{n^4}{\binom{4n}{4}}$
$= \frac{n^4 4! (4n-4)!}{(4n)!}$
$= \frac{24n^4}{(4n-1)(4n-2)(4n-3)}$
$\rightarrow \infty$ as $n \rightarrow \infty$
However, the answer key says that
$\frac{n^4}{\binom{4n}{4}}$
$= \frac{6n^3}{(4n-1)(4n-2)(4n-3)}$ this ... | It is worth generalizing the given limit: for positive integers $m$,
$$\begin{align*} a_n(m) &= \frac{n^m}{\binom{mn}{m}} \\ &= \frac{n^m m! \, (m(n-1))!}{(mn)!} \\ &= \frac{n^m m!}{(mn)(mn-1)\cdots(m(n-1)+1)} \\ &= \prod_{k=1}^m \frac{nk}{m(n-1) + k} \\ &= \prod_{k=1}^m \frac{k}{m + (k-m)/n}\end{align*}.$$ Consequen... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1754820",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Is the number of sequences with equal 0's and 1's small? I think that is n choose n/2. Though when I try to get a feel of what that number should be I get a much larger number than I expect:
i.e. we have
$$ n C \frac{n}{2} = \frac{n!}{\frac{n}{2}! \frac{n}{2}! }$$
using Stirling's?
Recall:
$$n! \sim \sqrt{2 \pi n} \lef... | Your derivation is correct, and does tell you that most sequences are unbalanced. The expression $\sqrt{2} \frac{2^n}{\sqrt{\pi} \sqrt{n}}$ is much smaller than $2^n$ when $n$ is large, by a factor of about $\sqrt{n}$. For instance, if $n=1000000$, your estimate says that only about $1/1000$ of all the binary sequenc... | {
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"url": "https://math.stackexchange.com/questions/1756171",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
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What is the probability of getting 2 same colour sweets and 1 different colour sweet? A little box contains $40$ smarties: $16$ yellow, $14$ red and $10$ orange.
You draw $3$ smarties at random (without replacement) from the box.
What is the probability (in percentage) that you get $2$ smarties of one color and another... | Your options are "Exactly two yellow smarties, exactly two red smarties, or exactly two orange smarties."
If $P$ represents your final probability, you need to add up the following probabilities:
$P($exactly two yellows$) \, + \, P($exactly two reds$) \, + \,P($exactly two orange$)$
The probability of getting exactly t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1757260",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Prove limit of a sequence in Newton's method Given the $ f(x)=x^3+x-1 $, I have shown so far that $ f$ has a unique root $r\in(0,1)$ and that for the sequence $(x_{n}), n>=0$ produced by Newton's method we have $$\lim_{n\to\infty} x_{n}=r$$ for every $x_{0}\in\mathbb{R}$.
How do I prove that $$\lim_{n\to\infty} \frac{x... | $$x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)} = x_n - \frac{x_n^3+x_n -1}{3x_n^2+1} = \frac{2x_n^3+1}{3x_n^2+1}$$
hence
$$x_{n+1}-r =\frac{2x_n^3+1}{3x_n^2+1} - r = \frac{2x_n^3 - 3x_n^2r+1-r}{3x_n^2+1} = \frac{2x_n^3 - 3x_n^2r+r^3}{3x_n^2+1} = \frac{(x_n-r)^2(2x_n+r)}{3x_n^2+1}$$
Therefore
$$\frac{x_{n+1}-r}{(x_n-r)^2} = ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1758154",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Sum of the series $\frac{1}{(1-x)(1-x^3)}+\frac{x^2}{(1-x^3)(1-x^5)}+\frac{x^4}{(1-x^5)(1-x^7)}+....$ If $|x|<1$, find the sum of infinite terms of following series:
$$\frac{1}{(1-x)(1-x^3)}+\frac{x^2}{(1-x^3)(1-x^5)}+\frac{x^4}{(1-x^5)(1-x^7)}+....$$
Could someone give me hint to solve this problem. I wrote $r_{th}$ t... | Hint: $$\frac{x^{2n}}{(1-x^{2n+1})(1-x^{2n+3})}=\frac{1}{x(1-x^2)}\left(\frac{1}{1-x^{2n+1}}-\frac{1}{1-x^{2n+3}}\right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1761826",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Proof of an inequality in $\mathbb{C}$ Let $z\in \mathbb{C}, n \geq 2$. Show this complex inequality
$$|z^n-1|^2\le |z-1|^2\left(1+|z|^2+\dfrac{2}{n-1}\Re{(z)}\right)^{n-1}$$
For $n=2$ the inequality is easy to prove:
$$|z^2-1|^2\le |z-1|^2\left(1+|z|^2+2\Re{(z)}\right)$$
$$\Longleftrightarrow |z+1|^2\le 1+|z|^2+2\Re{... | Firstly: $$\frac{z^n-1}{z-1}=z^{n-1}+z^{n-2}+\cdots1 \hspace{2cm}(1)$$
Let $z_k, \; k=0,\cdots,n-1$ be the roots of unity, i.e. the roots of $z^n-1=0$, and take $z_0=1$. Using Vieta's relations for the sum of the roots (https://en.wikipedia.org/wiki/Vieta's_formulas), one has $\sum\limits_{i=0}^{n-1} z_i=0$. Using that... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1762337",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 1,
"answer_id": 0
} |
You have to estimate $\binom{63}{19}$ in $2$ minutes to save your life. This is from the lecture notes in this course of discrete mathematics I am following.
The professor is writing about how fast binomial coefficients grow.
So, suppose you had 2 minutes to save your life and had to estimate, up to a factor of $100$... | With pen and paper, Stirling's approximation:
$$
\begin{align}
{63 \choose 19}
&= \frac{63!}{19! 44!} \\
&\doteq \frac{\sqrt{2\pi 63}}{\sqrt{2\pi 19}\sqrt{2\pi 44}}
\left( \frac{63}{e}\right)^{63}
\left( \frac{e}{19}\right)^{19}
\left( \frac{e}{44}\right)^{44} \\
&\doteq \sqrt{\frac{60}{2 \cdot ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1762693",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "44",
"answer_count": 6,
"answer_id": 2
} |
How do I determine if matrix A is diagonalizable? I am trying to figure out how to determine the diagonalizability of the following matrix:
$A=\begin{pmatrix}
1 &0 &0 &0 \\
2&1 & -3 & -2\\
3& 0 & 0 &-9 \\
-1& 0& -1& 0
\end{pmatrix}$
There are two distinct eigenvalues: $λ_1=λ_4=1$,$λ_2=3$ and $λ_3=-3$.
Can so... | The characteristic polynomial is
$$
\det
\begin{pmatrix}
1-X &0 &0 &0 \\
2&1-X & -3 & -2\\
3& 0 & 0-X &-9 \\
-1& 0& -1& 0-X
\end{pmatrix}
=(1-X)^2(3-X)(-3-X)
$$
You want to determine the geometric multiplicity of the eigenvalue $1$, that is the dimension of the eigenspace, which is the null space of $A-I$. L... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1770395",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
I do not understand how to solve this The two sequences of numbers { 1, 4, 16, 64, . . .} and { 3, 12, 48, 192, . . .} are mixed as follows: { 1, 3, 4, 12, 16, 48, 64, 192, . . .}. One of the numbers in the mixed series is 1048576. Then the number immediately preceding it is
| The second sequence is just 3 times the previous sequence. Note that no number in the first sequence is divisible by 3 while all numbers in the second are. The mixed sequence is in the form $\{4^0, 3\cdot 4^0, 4^1, 3\cdot 4^1, 4^2, 3 \cdot 4^2, 4^3, 3 \cdot 4^3, \ldots\}$. Seeing that 1048576 is not divisible by 3 it m... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1770796",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Evaluate the following limit: Find
$$\lim_{n \to \infty}\frac{1}{\sqrt{n}}\left[\frac{1}{\sqrt{2}+\sqrt{4}}+\frac{1}{\sqrt{4}+\sqrt{6}}+\cdots+\frac{1}{\sqrt{2n}+\sqrt{2n+2}}\right]$$
MY TRY:
$$
\begin{align}
\lim_{n \to \infty} &\frac{1}{\sqrt{n}} \biggl[
\frac{1}{\sqrt{2}+\sqrt{4}}
+ \frac{1}{\sqrt{4}+\sqrt{6}}
+ \... | If one attempts to use Cauchy's First Limit Theorem, then we start with
$$\begin{align}
\lim_{n\to \infty}\frac{1}{\sqrt{n}}\sum_{k=1}^n\frac{1}{\sqrt{2k}+\sqrt{2k+2}}&=\lim_{n\to \infty}\sqrt{n}\frac1n\sum_{k=1}^n\frac{1}{\sqrt{2k}+\sqrt{2k+2}} \tag 1\\\\
\end{align}$$
The Theorem reveals that since $\lim_{n\to \infty... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1773167",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
} |
Prove using factorials that ${n\choose k}+2{n\choose k+1}+{n\choose k+2}={n+2\choose k+2}$ Prove using factorials that ${n\choose k}+2{n\choose k+1}+{n\choose k+2}={n+2\choose k+2}$
I think I'm having a bit of algebra problem with this proof. Here is my work thus far:$$\frac{n!}{k!(n-k)!}+2\frac{n!}{(k+1)!(n-k-1)!}+\fr... | This result, and its generalizations, are made intuitively obvious by the relationship between Pascal's triangle and binomial coefficients; e.g., $$\begin{array}{ccccccc} \ldots & \binom{n}{k} & & \binom{n}{k+1} & & \binom{n}{k+2} & \ldots \\ & \ldots & \binom{n+1}{k+1} & & \binom{n+1}{k+2} & \ldots & \\ & & \ldots & \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1773699",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Only valid for Pythagoraean triples $\sqrt2+\frac{b}{\sqrt2+\frac{b}{\sqrt2+\frac{b}{\sqrt2\cdots}}}=\sqrt{c+a}$? $$\sqrt2+\frac{b}{\sqrt2+\frac{b}{\sqrt2+\frac{b}{\sqrt2\cdots}}}=\sqrt{c+a}$$
Where (a,b,c) are the Pythagoraean Triples and are satisfy by the Pythagoras theorem $a^2+b^2=c^2$
An example of Pythagoraean t... | Solving the Recurrence
Let
$$
x = \sqrt 2 + \frac{b}{x}
$$
Hence,
$$
x = \frac{\sqrt 2 x + b}{x} \implies x^2 = \sqrt 2 x + b \\
x^2 - x\sqrt2 - b = 0
$$
The roots of this equation are
$$
x_1, x_2 = \frac{\sqrt2 \pm \sqrt{ \sqrt{2}^2 + 4b}}{2} = \frac{\sqrt{2} \pm \sqrt{2 +4b}}{2}
$$
Proving that this works for all ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1773880",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Find all real numbers $a,b$ such that $|a|+|b|\geq\frac{2}{\sqrt{3}}$ and $|a\sin x+b\sin{2x}|\leq 1$ for all real $x$. Find all real numbers $a,b$ such that $|a|+|b|\geqslant\frac{2}{\sqrt{3}}$ and $|a\sin(x)+b\sin(2x)|\leqslant 1$ for all real $x$.
We could write the inequality as
$$
\left|\frac{a}{\sqrt{a^2+b^2}}\si... | Let $f(x)=a\sin x+b\sin 2x$. Note that $f(-x)=-f(x)$, so we can assume $a\ge 0$. Also $f(\pi-x)=a\sin x-b\sin 2x$, so we can also assume $b\ge 0$.
We look first at the maximum value of $k\sin x+\sin 2x$. Differentiating and putting $c=\cos x$ we get $kc+2(2c^2-1)=0$, so $4c^2+kc-2=0$, so $c=\frac{1}{8}(\sqrt{32+k^2}-k)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1775058",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 1
} |
Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.