Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Find $\sum_{n=1}^{\infty} x^{\left\lfloor {n \over 2}\right\rfloor} y^{\left\lfloor {n + 1 \over 2}\right\rfloor}$ Let $x,y > 0, xy <1$. Find the sum
$$\sum_{n=1}^{\infty} x^{\left\lfloor {n \over 2}\right\rfloor} y^{\left\lfloor {n + 1 \over 2}\right\rfloor}$$
While I have some ideas how to test convergence, I don't q... | Note that for $n \in \mathbf N$ we have
$$ \def\fl#1{\left\lfloor#1\right\rfloor}\fl{\frac n2} + 1
= \fl{\frac{n+2}2}$$
and
$$ \fl{\frac{n+1}2}+1 = \fl{\frac{n+3}2} $$
So, if we call the sum $s := \sum_{n=1}^\infty x^{\fl{n/2}}y^{\fl{(n+1)/2}}$ and assume it converges we have,
\begin{align*}
xys &= \sum_{n=1}^\i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2034276",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Prove $4a^2+b^2+1\ge2ab+2a+b$ Prove $4a^2+b^2+1\ge2ab+2a+b$
$4a^2+b^2+1-2ab-2a-b\ge0$
$(2)^2(a)^2+(b)^2+1-2ab-2a-b\ge0$
Any help from here? I am not seeing how this can be factored
| Let $x=(2a,b,1)$ and $y=(1,2a,b)$. Then by, Cauchy-Schwarz,
$2ab+2a+b= <x,y> \le ||x||*||y||=\sqrt{4a^2+b^2+1}*\sqrt{4a^2+b^2+1}=4a^2+b^2+1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2035680",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 1
} |
If $1\leq a < b$ show that $\sqrt{b}-\sqrt{a}\leq \frac{1}{2}(b-a)$. Need help with proof. If $1\leq a < b$ show that $\sqrt{b}-\sqrt{a}\leq \frac{1}{2}(b-a)$.
Working:
$\frac{1}{2}(1-a)\leq 0<\frac{1}{2}(b-a)$ and $(1-\sqrt{a})\leq 0<\sqrt{b}-\sqrt{a}$
Trying to show $\frac{1}{2}(b-a)-(\sqrt{b}-\sqrt{a})$ is positive ... | One may observe that, for $b>a\ge1$,
$$
\begin{align}
(b-a)-2(\sqrt{b}-\sqrt{a})&=(\sqrt{b}-\sqrt{a})(\sqrt{b}+\sqrt{a})-2(\sqrt{b}-\sqrt{a})
\\\\&=(\sqrt{b}-\sqrt{a})\left[\sqrt{b}+\sqrt{a}-2 \right]
\\\\&=(\sqrt{b}-\sqrt{a})\left[(\sqrt{b}-1)+(\sqrt{a}-1) \right]
\\\\&>0
\end{align}
$$ by using that $x \mapsto \sqrt{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2036425",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
About fractions whose sum is a natural number Some days ago I found an old problem of an olympiad that I always found interesting. It asks to replace each $\boxed{}$ with the numbers $1,2\ldots 30$ without repeating any number, such that their sum is an integer number.
$$\frac{\boxed{}}{\boxed{}}+\frac{\boxed{}}{\boxed... | Here's what one can play around with (and supposedly, Ian Miller started similarly):
We can build as many integer fractions as possible, namely
$$\tag1\frac{30}{15}+\frac{28}{14}+ \frac{26}{13}+\frac{24}{12}+ \frac{22}{11}+
\frac{20}{10}+ \frac{18}{9}+ \frac{16}{8},$$
each summand equalling $2$.
As $14$ is already in... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2037453",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
} |
Convergence of $\sum_{n=1}^{\infty}\left(\sqrt{n^2+1} - \sqrt[3]{n^3+1}\right)$ I am trying to test the convergence of
$$\sum_{n=1}^{\infty}\left(\sqrt{n^2+1} - \sqrt[3]{n^3+1}\right)$$
The ratio test yields and inconclusive $1$. I am able to show $\sqrt{n^2+1} - \sqrt[3]{n^3+1} < 1$ but this is not enough for converge... | $$ \sqrt{n^2-1}-\sqrt[3]{n^3-1} = \left(n-\sqrt[3]{n^3-1}\right)-\left(n-\sqrt{n^2-1}\right)\tag{1} $$
so, due to $\frac{a^2-b^2}{a-b}=(a+b)$ and $\frac{a^3-b^3}{a-b}=(a^2+ab+b^2)$ we have:
$$\left(n-\sqrt[3]{n^3-1}\right)\leq \frac{1}{n^2},\qquad \left(n-\sqrt{n^2-1}\right)\geq \frac{1}{2n}\tag{2} $$
and these inequal... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2037637",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
If $a_{1}=1$ and $b_{1}=2$ show that $b_{5}-a_{5}<1/2^{45}$. Need help interpreting solution. If $a_{1}=1$ and $b_{1}=2$ show that $b_{5}-a_{5}<1/2^{45}$.
$a_{n+1}=\sqrt{a_{n}b_{n}}$ , and $b_{n+1}=\frac{1}{2}(a_{n}+b_{n})$.
I've managed to work out that $b_{n+1}-a_{n+1}<\frac{1}{8}(b_{n}-a_{n})^{2}$. It was the solut... | From
$$
b_{n+1}-a_{n+1}<\frac{1}{8}(b_{n}-a_{n})^{2},\quad n\ge1,
$$ one gets, with $n=1,2,3,4$,
$$
\begin{align}
b_{5}-a_{5}&<\frac{1}{8}(b_{4}-a_{4})^{2}
\\\\&<\frac{1}{8^3}(b_{3}-a_{3})^{2}
\\\\&<\frac{1}{8^7}(b_{2}-a_{2})^{2}
\\\\&<\frac{1}{8^{15}}(b_{1}-a_{1})^{2}.
\end{align}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2037814",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Show that $G_4(i)\neq 0$, and $G_6(\rho)\neq 0$, $\rho=e^{2\pi i /3}$ Let $G_k$ denote the Eisenstein series of weight $k$. I know that $G_k(i)=0$ if $k \not\equiv 0 \ (mod \ 4)$ and $G_k(\rho)=0$ if $k \not\equiv 0 \ (mod \ 6)$. However, I want to know how to show, that $G_4(i)\neq0$ and $G_6(\rho)\neq0$, without us... | Here's a direct proof that $G_4(i) \neq 0$ that uses only the series.
The point is simply to estimate the tail $|a|,|b| \geq N$ and then calculate the first few terms for $|a|,|b| < N$. In fact, using very simply bounds we can take $N=2$.
First note that $|a|+|b| \leq \sqrt{3}|a+bi|$ which follows from
$$ 2|a||b| \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2044686",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Number of occurrences of k consecutive 1's in a binary string of length n (containing only 1's and 0's) Say a sequence $\{X_1, X_2,\ldots ,X_n\}$ is given, where $X_p$ is either one or zero ($0 < p < n$). How can I determine the number of strings, which do contain at least one occurrence of consequent $1$'s of length $... | Here is an approach based upon generating functions. We start considering words with no consecutive equal characters at all.
These words are called Smirnov words or Carlitz words. (See example III.24 Smirnov words from Analytic Combinatorics by Philippe Flajolet and Robert Sedgewick for more information.)
A generatin... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2045496",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 6,
"answer_id": 2
} |
Non-linear recurrence with square root: $a_{n+2}=\sqrt{a_{n+1}\cdot a_{n}}$ How should I approach this problem:
$a_{n+2}=\sqrt{a_{n+1}\cdot a_{n}}$
where $a_0 = 2, a_1=8$
| An alternative approach to the one given by Oliver Oloa:
$$a_{n+2}=\sqrt{a_{n+1}\cdot a_{n}}$$
$$\implies a^2_{n+2}={a_{n+1}\cdot a_{n}}$$
$$\implies \frac{a_{n+2}}{a_{n+1}}=\frac{a_{n}}{a_{n+2}} \tag1$$
Similarly we can write that
$$\frac{a_{n+1}}{a_{n}}=\frac{a_{n-1}}{a_{n+1}} \tag2$$
$$\frac{a_{n}}{a_{n-1}}=\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2053822",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
How do I express x in terms of y for the following expression ? $y=2x+\frac{8}{x^2}-5$ How do I express x in terms of y for the following expression ?
$$y=2x+\frac{8}{x^2}-5$$
The reason I want to do this is to calculate the area between this curve, the y-axis, y=5 and y=1.
I found the answer to be 5 using a roundabout... | Basic algebra is correct.
From
$y=2x+\frac{8}{x^2}-5
$,
multiplying by $x^2$
we get
$yx^2=2x^3+8-5x^2
$,
or
$2x^3-(5-y)x^2+8
= 0
$.
This is,
unfortunately,
a cubic
which can be solved by the
traditionally messy formula.
Its derivative is
$6x^2-2(5-y)x
= 0
$
which has roots
$x=0$
and
$x=\dfrac{5-y}{3}
$.
Note that
at $x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2057584",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
How do you find the real solutions to these simultaneous equations? I am looking for all real $(a,b,c)$ that satisfy the following
\begin{equation}
\left\{
\begin{array}{l}2a + a^2b = b\\
2b + b^2c = c\\
2c + c^2a = a\\
\end{array}
\right.
\end{equation}
I know that $a=b=c = 0$ is the only real solution to the problem... | $b = \frac {2a}{1-a^2}$
$a = \tan t$
$b = \tan 2t\\
c = \tan 4t\\
a = \tan 8t$
$tan t = tan 8t\\
t + n\pi = 8t\\
7t = n\pi\\
t = \frac {n}{7} \pi$
$a,b,c = \tan \frac {n\pi}{7}, \tan \frac {2n\pi}{7},\tan \frac {4n\pi}{7}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2059206",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
show $\frac{y}{x^2+y^2} $ is harmonic except at $y=0,x=0$ Let $f(z)=u(x,y)+iv(x,y) $
where $$ f(z)=u(x,y)=\frac{y}{x^2+y^2}$$
show $u(x,y)$ is harmonic except at $z=0$
Attempt
$$ u=\frac{y}{x^2+y^2}=y(x^2+y^2)^{-1} $$
Partial derivatives with x
$$\begin{aligned}
u_x&= y *(x^2+y^2)^{-2}*-1*2x
\\ &= -y*2x(x... | $$
f(z) = \frac 1z = \frac{1}{x+iy} = \frac{x}{x^2+y^2} + i\frac{-y}{x^2+y^2}
$$
is holomorphic in $\Bbb C \setminus \{ 0 \}$. It follows that
$$
-\operatorname{Im} f(z) = \frac{y}{x^2+y^2}
$$
is harmonic in the same domain.
(It is a direct consequence of the Cauchy-Riemann
differential equations that real and ima... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2062523",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
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Prob. 3, Chap. 3 in Baby Rudin: If $s_1 = \sqrt{2}$, and $s_{n+1} = \sqrt{2 + \sqrt{s_n}}$, what is the limit of this sequence? Here's Prob. 3, Chap. 3 in the book Principles of Mathematical Analysis by Walter Rudin, 3rd edition:
If $s_1 = \sqrt{2}$, and $$s_{n+1} = \sqrt{2 + \sqrt{s_n}} \ \ (n = 1, 2, 3, \ldots),$$ ... | For the first part of your question, not answered above:
If $s_1=\sqrt{2}$ and
$$
s_{n+1}=\sqrt{2+\sqrt{s_n}}
$$
prove 1) that $\{ s_n \}$ converges and 2) that $s_n<2$ for any $n\in \mathbb{N}$..
First we show $\{ s_n\}$ is increasing by induction.
Base case:
$$
2>0\Rightarrow 2+\sqrt{2}>\sqrt{2}\Righta... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2066320",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 1
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Why is $(n+1)^{n-1}(n+2)^n>3^n(n!)^2$
Why is $(n+1)^{n-1}(n+2)^n>3^n(n!)^2$ for $n>1$
I can use $$(n+1)^n>(2n)!!=n!2^n$$ but in the my case, the exponent is always decreased by $1$, for the moment I don't care about it, I apply the same for $n+2$
$(n+2)^{n+1}>(2n+2)!!=(n+1)!2^{n+1}$
gathering everything together,
$... | Hint
Induction for the step $n+1$:
$$(n+2)^{n}(n+3)^{n+1}=\frac{(n+3)^{n+1}}{(n+1)^{n-1}}(n+1)^{n-1}(n+2)^{n}>3^n(n!)^2\frac{(n+3)^{n+1}}{(n+1)^{n-1}}$$
We may expect
$$\frac{(n+3)^{n+1}}{(n+1)^{n-1}}>3(n+1)^2 \quad (1)$$
in order to finish the induction.
Backing to $(1)$ we have an equivalent expression:
$$\left(\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2067511",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
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How to find the largest integer $n$ for which $n!$ can be expressed as the product of $n - 3$ consecutive integers I need to find the largest integer $n$ for which $n!$ can be expressed as the product of $n - 3$ consecutive integers. Example: $7! = 7 \cdot 8 \cdot 9 \cdot 10 $
| $n=23$ has an answer.
Given that $\binom{K}{n-3} = \frac{K(K-1)\cdots(K-(n-2))}{(n-3)!}$ this means you want a $K$:
$$\binom{K}{n-3} = n(n-1)(n-2)$$
Now, $\binom{K}{n-3}$ is an increasing function as $K$ increases.
And we have:
$\binom{n}{n-3} = \frac{n(n-1)(n-2)}{6} < n(n-1)(n-2)$ and $\binom{n+1}{n-3} = \frac{(n+1)n(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2067725",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
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Natural number which can be expressed as sum of two perfect squares in two different ways? Ramanujan's number is $1729$ which is the least natural number which can be expressed as the sum of two perfect cubes in two different ways. But can we find a number which can be expressed as the sum of two perfect squares in two... | Any Pythagorean triple $(A^2+B^2=C^2)$ provides a candidate where the $C^2$ can have $2$-or-more combinations of $A^2$ and $B^2$ that add up to it. These can be found by testing natural numbers of the form $(4n+1)$ with a range of $m$ values defined as shown below to see which, if any yield integers. We begin with Eucl... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2068122",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 13,
"answer_id": 12
} |
What is the minimal polynomial of $\sqrt{3} + \sqrt[3] {2}$ over $\mathbb{Q}$? What is the minimal polynomial of $\sqrt{3} + \sqrt[3]{2}$ over $\mathbb{Q}$?
I know the basic idea of what a minimal polynomial is--it is the lowest degree monic polynomial in $\mathbb{Q}[x]$ that has the above as a root. But how do you go... | Start with $x=\sqrt{3}+\sqrt[3]{2}$. Then
\begin{gather*}
x-\sqrt{3} = \sqrt[3]{2} \\
(x-\sqrt{3})^3 = 2 \\
x^3 - 3\sqrt{3}x^2 + 9x - 3\sqrt{3} = 2 \\
x^3 + 9x - 2 = 3\sqrt{3}(x^2+1) \\
(x^3+9x-2)^2 = 27(x^2+1)^2 \\
x^6 + 18x^4 - 4x^3 + 81x^2 - 36x + 4 = 27x^4 + 54x^2 + 27 \\
x^6 - 9x^4 - 4x^3 + 27x^2 - 3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2068608",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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Use the sum identity and double identity for sine to find $\sin 3x$. Q. Use the sum identity and double identity for sine to find $\sin 3x$.
$$
\begin{align}
\sin 3x &= \sin (2x + x)\\
&=\sin 2x \cos x + \cos 2x \sin x \\
&= (2\sin x \cos x) \cos x + (1 - 2\sin^2 x) \sin x\\
&=2\sin x \cos^2 x + \sin x - 2\sin^3 x \\
&... | You seem to have forgotten the first two terms inside the quotation marks.
$$2 \sin x - 2 \sin^3 x + \sin x - 2 \sin^3 x = 3 \sin x - 4 \sin^3 x$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2070117",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Simplify $\frac {2^{n(1-n)}\cdot 2^{n-1}\cdot 4^n}{2\cdot 2^n\cdot 2^{(n-1)}}$ Simplify::
$$\frac {2^{n(1-n)}\cdot 2^{n-1}\cdot 4^n}{2\cdot 2^n\cdot 2^{(n-1)}}$$
My Attempt:
\begin{align}
&\frac {2^{n-n^2}\cdot 2^{n-1}\cdot 2^{2n}}{2\cdot 2^n\cdot 2^{n-1}}\\
&=\frac {2^{n-n^2+n-1+2n}}{2^{1+n+n-1}} \\
&=\frac {2^{4n-n^2... | $\frac {2^{n(1-n)}.2^{n-1}.4^n}{2.2^n.2^{(n-1)}}$
First of all both N & D contains $2^{n-1}$ Eliminate them. Now you have,
$\frac {2^{n(1-n)}.4^n}{2.2^n}$
$\frac {2^n.2^{-n^2}.2^{2n}}{2.2^n}$
Now both N and D have $2^n$ eliminate it.
$\frac {2^{-n^2}.2^{2n}}{2}$
= $\frac {2^{2n}}{2.2^{n^2}}$
Edit -
According to your... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2071147",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
How to use the 'modulus' operator? This is a problem from BdMO $2012$ Dhaka region Question Paper:
The product of a number with itself is called its square. For example,
$2$ multiplied by $2$ is $4$, so $4$ is the square of $2$. If you take a square
number and multiply it with itself, what will be the largest poss... | So you can simply try out all $10$ possible options:
*
*$x\equiv0\pmod{10} \implies x^4\equiv0^4\equiv 0\equiv0\pmod{10}$
*$x\equiv1\pmod{10} \implies x^4\equiv1^4\equiv 1\equiv1\pmod{10}$
*$x\equiv2\pmod{10} \implies x^4\equiv2^4\equiv 16\equiv6\pmod{10}$
*$x\equiv3\pmod{10} \implies x^4\equiv3^4\equiv 81\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2071421",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 4,
"answer_id": 2
} |
Confusion in fraction notation $$a_n = n\dfrac{n^2 + 5}{4}$$
In the above fraction series, for $n=3$ I think the answer should be $26/4$, while the answer in the answer book is $21/2$ (or $42/4$). I think the difference stems from how we treat the first $n$. In my understanding, the first number is a complete part and ... | $3 \times \frac{3^2+5}{4} = 3 \times \frac{9+5}{4} = 3\times \frac{14}{4} = 3\times \frac{7}{2} = \frac{21}{2}$
Normally (even though in calculator this is often not true) the convention is that a number on the side of a fraction is multiplying that fraction. There should be a "times" either a cross or a dot, however o... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2074265",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "22",
"answer_count": 7,
"answer_id": 5
} |
Find the maximum power of $24$ in $(48!)^2$? Find the maximum power of $24$ in $(48!)^2$ ?
How to approach for such questions ?
| $24=\color\red{2^3}\cdot\color\green{3^1}$
The multiplicity of $\color\red{2}$ in $48!$ is $\sum\limits_{n=1}^{\log_{\color\red{2}}48}\Big\lfloor\frac{48}{\color\red{2}^n}\Big\rfloor=24+12+6+3+1=\color\red{46}$
The multiplicity of $\color\green{3}$ in $48!$ is $\sum\limits_{n=1}^{\log_{\color\green{3}}48}\Big\lfloor\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2075577",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 3,
"answer_id": 0
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If $ f(x) = \frac{\sin^2 x+\sin x-1}{\sin^2 x-\sin x+2},$ then value of $f(x)$ lies in the interval If $\displaystyle f(x) = \frac{\sin^2 x+\sin x-1}{\sin^2 x-\sin x+2},$ then value of $f(x)$ lies in the interval
assume $\sin x= t$ where $|\sin x|\leq 1$
let $\displaystyle y = \frac{t^2+t-1}{t^2-t+1}$
$\displaystyle yt... | This is an answer without using derivative.
As you did, let $t=\sin x$.
First of all, we have $y\not=1$ since for $y=1$ we get $t=\frac 32\gt 1$.
Since $y\not=1$, we have
$$\begin{align}y=\frac{t^2+t-1}{t^2-t+2}&\iff (y-1)t^2+(-y-1)t+2y+1=0\\&\iff t^2+\frac{-y-1}{y-1}t+\frac{2y+1}{y-1}=0\\&\iff \left(t-\frac{y+1}{2y-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2081524",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Use Mean value theorem to prove the following inequality A) Use the Mean value theorem to prove that
\begin{equation}
\sqrt{1+x} < 1 + \frac{1}{2}x \text{ if } x>0
\end{equation}
B) Use result in A) to prove that
\begin{equation}
\sqrt{1+x}>1+\frac{1}{2}x-\frac{1}{8}x^2 \text{ if } x>0
\end{equation}
Can someone give ... | OK, I'll show how I'd solve a:
We have
$$
f(x) = 1+ \frac{x}{2} - \sqrt{1+x}
$$
Clearly $f(0) = 0$. Let's look at the derivative of $f$. If we show that the derivative is strictly positive for $x>0$, the function is strictly increasing, hence positive (in combination with $f(0)$) for $x>0$. We have
$$
f'(x) = \frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2081769",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
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Prove that $(2+ \sqrt5)^{\frac13} + (2- \sqrt5)^{\frac13}$ is an integer When checked in calculator it is 1. But how to prove it?
Also it is not a normal addition like $x+ \frac1x$ which needs direct rationalization. So I just need something to proceed.
| An easy way is to simplify the nested radical, and then combine like terms.
An easy way to denest $\sqrt[m]{A+B\sqrt[n]{C}}$ is to assume the form $a+b\sqrt[n]{C}$ and expand both sides with the binomial theorem. $\sqrt[3]{2+\sqrt5}$ becomes$$2+\sqrt5=(a^3+15ab^2)+(3a^2b+5b^3)\sqrt5\tag1$$
From which we get a system of... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2082836",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 8,
"answer_id": 4
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Identifying $\mathbb{C} D_8$ as a product of matrix rings Let $G = D_8$ be the dihedral group of order $8$, i.e.,
\begin{align*}
G = \langle a,b \mid a^4 = 1 = b^2, ab = ba^{-1} \rangle.
\end{align*}
By standard results from ring/represention theory, the group algebra $\mathbb{C} G$ decomposes as a product of matrix ri... | I managed to work this out, so I'll provide my solution, which carries on from my work in the question.
The elements
\begin{align*}
e_{11} = \tfrac{1}{4}(1+ia-a^2-ia^3) \\
e_{22} = \tfrac{1}{4}(1-ia-a^2+ia^3)
\end{align*}
are orthogonal idempotents which are also orthogonal to each of the $e_i$. Observe that $e_{11}b =... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2083462",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Find recurrence relation in a counting problem In the book Galois Theory of Ian Stewart, problem 1.11 states: "Let $P(n)$ be the number of ways to arrange $n$ zeroes and ones in a row, given that ones occur in groups of three or more. Show that $P(n)=2P(n-1)-P(n-2)+P(n-4)$".
From the recurrence relation, I guess we ha... | $$
\begin{align}
P(n)
&=\overbrace{P(n-1)}^{\substack{\text{# of arrangements}\\\text{of $n-1$ digits}\\\text{prepending a $0$}}}
+\overbrace{P(n-1)}^{\substack{\text{# of arrangements}\\\text{of $n-1$ digits}\\\text{prepending a $1$}}}
-\overbrace{P(n-2)}^{\substack{\text{# of arrangements}\\\text{of $n-2$ digits}\\\t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2083571",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Proving $a\cos x+b\sin x=\sqrt{a^2+b^2}\cos(x-\alpha)$
Show that $a\cos x+b\sin x=\sqrt{a^2+b^2}\cos(x-\alpha)$ and find the correct phase angle $\alpha$.
This is my proof.
Let $x$ and $\alpha$ be the the angles in a right triangle with sides $a$, $b$ and $c$, as shown in the figure. Then, $c=\sqrt{a^2+b^2}$. The le... | You are largely correct. However you could prove the first part in a very simple way:
We can write $$P= a\cos x+b\sin x =\sqrt {a^2+b^2} \left[\frac {a}{\sqrt {a^2+b^2}}\cos x+\frac{b}{\sqrt {a^2+b^2}}\sin x \right] $$ Now we can take $\frac{a}{\sqrt {a^2+b^2}} $ as $\cos \alpha $ giving us $$P=\sqrt {a^2+b^2}[\cos \al... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2085687",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Provide a different method of proving:$\int_{-\infty}^{\infty}{1\over [\pi(x+e^{\pi})^2+\pi^{1/3}]^2}dx={1\over 2}$ Accidentally founded this particular integral producing a rational number
I can't be for sure it is correct, so can one provide a proof of it.
$$\int_{-\infty}^{\infty}{1\over [\pi(x+e^{\pi})^2+\pi^{1/3}... | Starting with,
$$\int_{-\infty}^{\infty}{1\over [A{u}^2+B]^2}du$$ you could let $u=\sqrt{\frac{B}{A}} \tan(\theta)$ so $du=\sqrt{\frac{B}{A}} \sec^2(\theta)d \theta.$ Using $1+\tan^2(x)=\sec^2(x)$, this gives, $$\sqrt{\frac{B}{A}}\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{\sec^2(\theta)}{(B\tan^2(\theta)+B)^2}d\theta... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2085788",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Determine whether the following sequence is increasing or decreasing $\frac{n^2+2n+1}{3n^2+n}$ Determine whether the following sequence is increasing or decreasing:
$$\frac{n^2+2n+1}{3n^2+n}$$
I'm not sure whether my solution is correct:
$$\frac{n^2+2n+1}{3n^2+n}=\frac{n(n+2)+1}{n(3n+1)}=\frac{n+2}{3n+1}+\frac{1}{n(3n... | You're solution is fine. Here is another, perhaps more efficient way forward.
We start with the decomposition in the OP as expressed by
$$\frac{n^2+2n+1}{3n^2+n}=\frac{n+2}{3n+1}+\frac{1}{n(3n+1)} \tag 1$$
Then, we simply note that the first term on the right-hand side of $(1)$ can be written as
$$\frac{n+2}{3n+1}=\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2086547",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Find the $\gcd[x+y+z; x^2+xy+z^2; y^2+yz+z^2; z^2+zx+x^2]$ What I have done:
There exists a non-zero integer $t$ such:
$$x+y+z=kt$$
$$x^2+xy+y^2=ut$$
$$y^2+yz+z^2=vt$$
$$z^2+zx+x^2=wt$$
$\implies$
$$(x-y)(x+y+z)=(u-v)t$$
$$(y-z)(x+y+z)=(v-w)t$$
$$(z-x)(x+y+z)=(w-u)t$$
$\implies$
$$\dfrac{x+y+z}{t}= \dfrac{u-v}{x-y}=\d... | I first noticed that $(x,y,z)$ are not interchangeable, and found only $gcd=1$ and $gdg=3$, with $(x,y,z)$ in the range $1$ to $999$ and co-prime in pairs.
I’ll use $(f_1,f_2,f_3,f_4)$ for $(x+y+z,x^2+xy+z^2,y^2+yz+z^2,z^2+zx+x^2)$
*
*When $x=3a+1,y=3b+1,z=3c+1$
$$f_1=3*(a+b+c+1)$$
$$f_2=3*(3a^2+3ab+3a+b+3c^2+2c+1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2088438",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Sum of $n$ terms of the given series. Find the sum of $n$ terms of the series:
$$\frac{1}{x+1}+\frac{2x}{(x+1)(x+2)}+\frac{3x^2}{(x+1)(x+2)(x+3)}+\frac{4x^3}{(x+1)(x+2)(x+3)(x+4)}+.......$$
Could someone give me slight hint to proceed in this question?
| Hint. One may observe that
$$
\frac{nx^{n-1}}{\prod _{k=1}^n (x+k)}=\left(1-\frac{x^n \Gamma(x+1)}{\Gamma(x+n+1)}\right)-\left(1-\frac{x^{n-1} \Gamma(x+1)}{\Gamma(x+n)}\right)
$$ giving
$$
\sum_{n=1}^N\frac{nx^{n-1}}{\prod _{k=1}^n (x+k)}=1-\frac{x^N \Gamma(x+1)}{\Gamma(x+N+1)}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2088797",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Exponent rules- basic algebra Can some one point out where I've gone wrong. The correct answer, aside from the 9, is on the left page. I tried it again, changing up what I used as the denominator and it looks even more incorrect. Was what I did on the right page wrong?
| We have: $\dfrac{\dfrac{x^{2}y^{-3}}{3z^{2}}-\dfrac{z^{-3}y^{-3}}{3x^{2}}}{\dfrac{x^{-4}y^{2}}{3z^{-2}}}$
$=\bigg(\dfrac{x^{2}y^{-3}}{3z^{2}}-\dfrac{z^{-3}y^{-3}}{3x^{2}}\bigg)\cdot\dfrac{3z^{-2}}{x^{-4}y^{2}}$
$=\bigg(\dfrac{(3x^{2})(x^{2}y^{-3})-(3z^{2})(z^{-3}y^{-3})}{(3z^{2})(3x^{2})}\bigg)\cdot\dfrac{3z^{-2}}{x^{-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2089304",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Drawing $\lvert z-2\rvert + \lvert z+2\rvert=5$ in complex plane If we substitute $z=x+iy$ to $\lvert z-2\rvert + \lvert z+2\rvert=5$ and solve for $iy$ we will get
$$ iy=-\frac{1}{2}(2x-5) $$
$$ iy=-\frac{1}{2}(2x+5) $$
Then we can draw like in the $\mathbb{R}^2$ plane. This way we get two parallel lines.
Do you think... | $1)$ If you want go through a geometric approach: by definition of ellipse one can conclude that $z$ is on an ellipse with foci $-2$ and $2$ and bigger axis equal to $5$.
$2)$ If you want go through an analytic approach you can write $z=x+iy$ and then
$$\sqrt{(x-2)^2+y^2}+\sqrt{(x+2)^2+y^2}=5$$
$$\sqrt{(x-2)^2+y^2}=5-\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2090645",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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$\forall n\in\mathbb{Z}$, find the $\gcd(n^2+1, (n+1)^2+1)$?
$\forall n\in\mathbb{Z}$, find the $\gcd(n^2+1, (n+1)^2+1)$.
I think this is a simple exercise, but I get this:
$(n+1)^2+1=n^2+2n+2$.
$n^2+2n+2 = (n^2+1)+(2n+1)$
then $\gcd(n^2+1, (n+1)^2+1)=\gcd(n^2+1, 2n+1)$
and $\displaystyle n^2+1 = \frac{n(2n+1)}{... | Note that $$(2n+3)(n^2+1) - (2n-1)((n+1)^2+1) = 5.$$
To get this equation, set $(n+a)(n^2+1)-(n+b)(n^2+2n+2) = c$ for some constants $a,b,c$, and equate coefficients. This gives $b = -\tfrac{1}{2}$, $a = \tfrac{3}{2}$, and $c = \tfrac{5}{2}$. Then, multiply both sides by $2$.
Hence, $\text{gcd}(n^2+1,(n+1)^2+1)$ divide... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2091252",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove that $\frac{x_1}{x_2}+\frac{x_2}{x_3}+\frac{x_3}{x_4}+\cdots +\frac{x_{n-1}}{x_n}+\frac{x_n}{x_1}<2n-1$ Let $x_1,x_2,x_3,\cdots ,x_n (n\ge2)$ be real numbers greater than $1.$ Suppose that $|x_i-x_{i+1}|<1$ for $i=1,2,3,\cdots,(n-1)$.
Prove that $$\frac{x_1}{x_2}+\frac{x_2}{x_3}+\frac{x_3}{x_4}+\cdots +\frac{x_... | Proof: I get this proof for two case
(1): if for all $k=1,2,\cdots,n-1$,have $a_{k}\le a_{k+1}$,then we have
$$a_{k}\le a_{k+1}<a_{k}+1$$
so we have
$$a_{i}<a_{i-1}+1<a_{i-2}+2<\cdots<a_{1}+(i-1)$$
then we have
$$\dfrac{a_{1}}{a_{2}}+\dfrac{a_{2}}{a_{3}}+\cdots+\dfrac{a_{n-1}}{a_{n}}+\dfrac{a_{n}}{a_{1}}<(n-1)+\dfrac{a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2091511",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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Find the Frenet Frame of a curve
Find the Frenet Frame of $\alpha(t)=(2t,t^2,\frac{t^3}{3})$ for $t\in\mathbb{R}$
First finding $\alpha'=(2,2t,t^2)$ and $\alpha''=(0,2,2t)$
$||\alpha'||=\sqrt{4+4t^2+t^4}=\sqrt{(t^2+2)^2}=t^2+2 \neq 1$ so not arc length parameterized
$T=\frac{\alpha'}{||\alpha'||}=(\frac{2}{t^2+2}, \f... | So using $B=\frac{\alpha' \times \alpha''}{||\alpha' \times \alpha''||}$
$\alpha' \times \alpha'' = (2t^2, -4t, 4)$
$||\alpha' \times \alpha''||=\sqrt{4t^4+16t^2+16}=2\sqrt{(t^2+2)^2}=2(t^2+2)$
$B=(\frac{t^2}{t^2+2}, \frac{-2t}{t^2+2}, \frac{2}{t^2+2})$
$N=-(T \times B)=(\frac{-2t^3-4t}{(t^2+2)^2}, \frac{-t^4+4}{(t^2+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2091839",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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then maximum and minimum value of $x+y$ if $x,y\in R$ and $x^3+y^3=2\;,$ then maximum and minimum value of $x+y$
using $\displaystyle \frac{x^3+y^3}{2}\geq \left(\frac{x+y}{2}\right)^3$
So $(x+y)^3\leq 2^3$ so $x+y\leq 2$
could some help me to find minimum value, thanks
| If $x,y$ are real numbers then there is no absolute minimum. In fact the value of $x+y$ can be made as close to zero as we like. We have $y = \sqrt[3]{2 - x^3}$
Obviously for very large $x$ we have that $y \approx -x$. Additionally the value must be positive, as $y > -x$, from the condition.
On the other side if we add... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2094891",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
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Irreducibles in $\mathbb{Q}$ and Inverses of Quadratics in $\mathbb{Q}(\alpha)$ Consider $f(x) = x^3 - 9x + 3 \in \mathbb{Q}[x],$ and let $\alpha$ be a root of $f(x).$ Prove that $f(x)$ is irreducible over $\mathbb{Q}.$ Furthermore, in the field $\mathbb{Q}(a),$ find $(3 \alpha^2 + 2 \alpha + 1)^{-1}$ in terms of the b... | I really like Lubin's approach. It is something I haven't seen before. The Euclidean Algorithm is tried and true, but as you will see, it is not fun in this case.
Using the division algorithm, we divide $x^3-9x+3$ by $3x^2 + 2x + 1$ to see that:
$$ x^3 - 9x + 3 = \left( \frac{1}{3}x-\frac{2}{9}\right) (3x^2+2x+1) + \le... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2096651",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Find the general solution to this Differential Equation. Given that $z=f(x,y)$ and $a\in \mathbb{R}$ is a constant, I have to solve the following differential equation:
$$ \frac{z \,dz+y \,dy}{y^2+z^2}=\frac{dx}{\sqrt{(x-a)^2+y^2+z^2}+(x-a)}.$$
I have not seen anything like this before so any ideas/hints would be much ... | Let rewrite equation:
$$\frac{z\,dz+y\,dy}{y^2+z^2}=\frac{dx}{\sqrt{(x-a)^2+y^2+z^2}+(x-a)}.$$
$$\frac{d(y^2+z^2)}{y^2+z^2}=\frac{2dx}{\sqrt{(x-a)^2+y^2+z^2}+(x-a)}.$$
With changing $y^2+z^2=t^2$ and $x-a=s$:
$$\frac{2tdt}{t^2}=\frac{2ds}{\sqrt{s^2+t^2}+s}.$$
or
$$\frac{\sqrt{s^2+t^2}+s}{t}dt=ds$$
This is a homogeneous... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2096961",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
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Evaluating $\quad\lim_{ x \to -\infty }\frac{\sqrt[n]{x+1}+\sqrt[n]{x+2}-2\sqrt[n]{x-3}}{\sqrt[n]{x-1}+\sqrt[n]{x-2}-2\sqrt[n]{x+3}}$ find limit :
$$\lim_{ x \to -\infty }\frac{\sqrt[n]{x+1}+\sqrt[n]{x+2}-2\sqrt[n]{x-3}}{\sqrt[n]{x-1}+\sqrt[n]{x-2}-2\sqrt[n]{x+3}}=\;\; ? \;\quad \text {given }\,n \in \mathbb{N}, n>2 ,\... | If we apply binomial expansion, we get
$$L=\lim_{x\to\pm\infty}\frac{\frac9{nx^{1-1/n}}+\mathcal O\left(\frac1{x^{2-1/n}}\right)}{-\frac9{nx^{1-1/n}}+\mathcal O\left(\frac1{x^{2-1/n}}\right)}\to-1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2097628",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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What are the values of $k$ when $kx^2+x+k$, has equal zeroes? Show that the quadratic equation $kx^2 + 2(x+1)=k$ has real roots for all the values of $k\in \mathbb{R}$
what i did
$kx^2+2x+2-k=0$
$4-4(2-k)(k)>0$
$4-8k+4k^2>0$
$(64±64)÷ 4(2)>0$
$128÷ 8>0$
$16>0$
please help me check
| We can just compute the roots using factorization:
$$
kx^2+2(x+1) = k \\
kx^2 +2x + 2 - k = 0 \\
kx^2 -x^2 + x^2 +2x + 1 + 1-k = 0 \\
(k-1)x^2 + (x+1)^2 + 1-k = 0 \\
(k-1)(x^2 - 1) + (x+1)^2 = 0 \\
(k-1)(x+1)(x - 1) + (x+1)^2 = 0 \\
(x+1)((k-1)(x - 1) + (x+1)) = 0 \\
(x+1)(kx - k - x + 1 + x+1) = 0 \\
(x+1)(kx - k ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2098312",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
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Power Series proofs $$\alpha(x) =\sum_{j=0}^\infty \frac{x^{3j}}{(3j)!}$$
$$\beta(x) = \sum_{j=0}^\infty \frac{x^{3j+2}}{(3j+2)!}$$
$$\gamma(x) = \sum_{j=0}^\infty \frac{x^{3j+1}}{(3j+1)!}$$
Show that $\alpha(x+y) = \alpha(x)α(y) + \beta(x)\gamma(y) + \beta(y)\gamma(x)$ for every $x, y \in\mathbb R$.
Show that $\alpha(... | For the latter, notice that \begin{align*} \alpha' &= \beta, \\ \beta' &= \gamma, \\ \gamma' &= \alpha. \end{align*} Then we see \begin{align*}\frac d {dx} (\alpha^3 + \beta^3 + \gamma^3 - 3\alpha \beta \gamma ) &= 3\alpha^2 \alpha'+ 3\beta^2 \beta' + 3\gamma^2 \gamma' - 3\alpha' \beta \gamma - 3\alpha \beta' \gamma -... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2098942",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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sum of series $(2^2-1)(6^2-1)+(4^2-1)(8^2-1)+\cdots \cdots +\cdots (100^2-1)(104^2-1)$ The sum of series $(2^2-1)(6^2-1)+(4^2-1)(8^2-1)+\cdots \cdots +\cdots (100^2-1)(104^2-1)$
Attempt Assume $\displaystyle S = \sum^{50}_{r=1}((2r)^2-1)((2r+4)^2-1) = \sum^{50}_{r=1}(4r^2-1)(4r^2+16r+15)$
$\displaystyle S = \sum^{50}_{... | You can find the sum using the idea of telescoping sum.
Since
$$((2r)^2-1)((2r+4)^2-1)=(2r-1)(2r+1)(2r+3)(2r+5)$$
we get
$$\small\begin{align}&\sum_{r=1}^{50}(2r-1)(2r+1)(2r+3)(2r+5)\\\\&=\frac{1}{10}\sum_{r=1}^{50}\color{red}{10}(2r-1)(2r+1)(2r+3)(2r+5)\\\\&=\frac{1}{10}\sum_{r=1}^{50}\color{red}{((2r+7)-(2r-3))}(2r-1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2099793",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
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Sum of given infinite series: $\frac14+\frac2{4 \cdot 7}+\frac3{4 \cdot 7 \cdot 10}+\frac4{4 \cdot 7 \cdot 10 \cdot 13 }+....$ Find the sum of infinite series
$$\frac{1}{4}+\frac{2}{4 \cdot 7}+\frac{3}{4 \cdot 7 \cdot 10}+\frac{4}{4 \cdot 7 \cdot 10 \cdot 13 }+....$$
Generally I do these questions by finding sum of $n$... | Notice that
$$\frac k{\prod_{m=1}^k(3m+1)}=\frac1{3\prod_{m = 1}^{k-1} (3m+1)}-\frac{1}{3\prod_{m = 1}^k (3m+1)}$$
Which gives us a telescoping series:$$S_N=\frac{1}{3} - \frac{1}{3\prod_{m = 1}^N (3m+1)}$$
which tends to $1/3$ as suspected.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2100279",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 2,
"answer_id": 0
} |
$\int\sin^2(x) \, dx$ - Which one is right? If I enter $\int\sin^2(x)\,dx$ into an integral calculator it returns:
$$\int\sin^2(x)\,dx=-\frac{\sin(2x)-2x}{4}+c$$
If I calculate it with Angular multiples I come to:
$$\int\sin^2(x)\,dx=\int\frac{1-\cos(2x)}2 \, dx=\frac{\sin(2x)-2x} 4 +c$$
Here is my calculation
If I cal... | With angular multiples,
$\int\sin^2(x)\,dx=\int\frac{1-\cos(2x)}2 \, dx=\frac{x}{2} - \frac{\sin(2x)}4 + c$
Now $\frac{x}{2} - \frac{\sin(2x)}4 + c$
$\frac{x}{2} - \frac{2\sin(x)\cos(x)}4 + c$
$\frac{x}{2} - \frac{\sin(x)\cos(x)}2 + c$
$\frac{- \sin(x)\cos(x) + x}2 + c$
That is with partial integration and substitution... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2100513",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Evaluating some limits $$f\left(x\right)\:=\:\frac{1}{1-e^{\frac{1}{x}}}, x \in (0, \infty )$$
As far as i know, $\lim _{x\to \infty }f\left(x\right)$ is $\frac{1}{+0}=\infty$. But, $\lim _{x\to \infty }\left(f\left(x\right)\:+\:x\right)=\frac{1}{2}$. How so ?
| As mentioned, $\frac{1}{1-e^{\frac{1}{x}}} \xrightarrow[x \to \infty]{} -\infty$. As for the limit of $f(x) + x$,
$\frac{1}{1-e^{\frac{1}{x}}} + x = \frac{1}{1-(1+\frac{1}{x} +\frac{1}{2x^2} +\mathcal{O}(\frac{1}{x^3}))} + x$
$= \frac{-x}{1+\frac{1}{2x} + \mathcal{O}(\frac{1}{x^2})} + x = -x(1-\frac{1}{2x} + \mathcal... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2100712",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
Calculate volume between two geometric figures I have a figure C that is defined as the intersection between the sphere $x^2+y^2+z^2 \le 1 $ and the cyllinder $x^2+y^2 \le \frac{1}{4}$.
How should i calculate the volume of this figure?
| Here is a sketch of the sphere and part of the infinite length cylinder:
(Large version)
It suggests to split the intersection volume into three volumes:
*
*the fully enclosed cylinder part $V_m$ in the middle and
*the two spherical caps, one on top $V_t$, one at the bottom $V_t$, where by symmetry $V_t = V_b$
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2103064",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 2
} |
If $2^{2017} + 2^{2014} + 2^n$ is a perfect square, find $n$. If $2^{2017} + 2^{2014} + 2^n$ is a perfect square, find $n$.
My first shot would be to assume the perfect square is $2^{2018}$, but how would I prove that? Even if it is, what is $n$? All help is appreciated.
| We are going to solve a more general question: For any given natural number $k$, find the integer $n$ satisfying that
$$
2^{2k}+2^{2k+3}+2^n=m^2,
$$
where $m$ is a natural number.
Simplifying the equation to $$9\cdot 2^{2k}+2^n=m^2 \Leftrightarrow 2^n=(m+p)(m-p), $$
where $p=3\cdot 2^{k}.$
Splitting the power of $2$ yi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2105224",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
"answer_id": 4
} |
Patterns in $\frac{80}{81}$ and $\frac{10}{81}$. The decimal form of $\frac{80}{81}$ is $0.987654320\ldots$ notice the expected $1$ is missing. The decimal form of $\frac{10}{81}$ is $0.12345679\ldots$ notice the expected $8$ is missing. Can someone expansion why the decimal form is the way they are? I think it has som... | This is because
$\dfrac1{(1-x)^2}
=\sum_{n=0}^{\infty} (n+1)x^n
$.
Putting $x=.1$,
this is
$\dfrac1{.9^2}
=\dfrac1{.81}
=\dfrac{100}{81}
=1+2/10+3/100+4/1000 + ...
$
so
$\dfrac{10}{81}
=1/10+2/100+3/1000+4/10000 + ...
=0.1234567...
$.
The next terms are
$8/10^8+9/10^9 +10/10^{10}+...
$,
but we get a carry here
(from th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2106973",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
} |
Given $AO:OA' + BO:OB' + CO:OC' = 92$ find the value of $AO:OA' \times BO:OB' \times CO:OC'$.
In $\triangle ABC$, points $A',B',C'$ are on sides $BC,AC,AB$ respectively. $AA', BB', CC'$ are concurrent at point $O$.
Given $AO:OA' + BO:OB' + CO:OC' = 92$ find the value of $AO:OA' \times BO:OB' \times CO:OC'$.
M... | Another way to solve this problem is to use the method of Mass Points. Assign the masses $a,b,c$ to the points $A, B, C$. This implies that we should assign the masses $(a+b), (b+c), (c+a), (a+b+c)$ to $C', A', B', O$. Then we have:
$$\frac{AO}{A'O} = \frac{b+c}{a} \quad \frac{BO}{B'O} = \frac{c+a}{b} \quad \frac{CO}{C... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2110362",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Prove that if $x^6+(x^2+y)^3$ is a perfect square, then $y$ is a multiple of $x^2$
Let $x$ and $y$ be integers with $x \neq 0$. Prove that if $x^6+(x^2+y)^3$ is a perfect square, then $y \equiv 0 \pmod{x^2}$.
We can expand the given expression as $$x^6+(x^2+y)^3 = 2x^6+3x^4y+3x^2y^2+y^3 = (2x^2+y)(x^4+x^2y+y^2)=k^2,$... | As far as I am aware, this is how one can prove it.
Let $a,b$ be integers that satisfy your condition.
So $$a^6+(a^2+b)^3=t^2$$
Then, dividing by $a^6$, we have $$1+\left(1+\frac{b}{a^2} \right)^3=\left(\frac{t}{a^3} \right)^2$$
However, the elliptic curve $1+x^3=y^2$ is an elliptic curve with rank $0$, , so it only ha... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2110577",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
} |
Ask about beautiful properties of $e$ One of students asked me about "some beautiful properties (or relation) of $e$". Then I list like below
\begin{align}
& e \equiv \lim_{x \to \infty} \left(1+\frac{1}{x} \right)^x\\[10pt]
& e = \sum_{k=0}^\infty \frac{1}{k!}\\[10pt]
& \frac{d}{dx} (e^x) = e^x\\[10pt]
& e^{ix} = \co... | Here is the (simple) continued fraction for $e-1.$ The pattern continues forever, 1,1,2,1,1,4,1,1,6,1,1,8,1,1,10,1,1,12,...
To get $e$ itself add one, this is the same as changing the first $1$ to a $2$
$$
\begin{array}{ccccccccccccccccccccccc}
& & 1 & & 1 & & 2 & & 1 & & 1 & & 4 & & 1 & & 1 & & 6 & \\
\frac{0}{1}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2110919",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 5,
"answer_id": 2
} |
Prove that $\frac{(n+2)^{n+1}}{(n+1)^n}-\frac{(n+1)^n}{n^{n-1}}For all $n\in\mathbb N$ prove that:
$$\frac{(n+2)^{n+1}}{(n+1)^n}-\frac{(n+1)^n}{n^{n-1}}<e$$
We can rewrite it in the following form
$$(n+2)\left(1+\frac{1}{n+1}\right)^{n}-(n+1)\left(1+\frac{1}{n}\right)^{n-1}<e$$ or
$$(n+1)\left(1+\frac{1}{n+1}\right)^{... | I do not know if this is a satisfactory answer for you.
Consider $$A=\frac{(n+2)^{n+1}}{(n+1)^n}\qquad , \qquad B=\frac{(n+1)^n}{n^{n-1}}$$ Take logarithms and expand as Taylor series for large values of $n$. This would lead to $$\log(A)=1+\log \left(n\right)+\frac{1}{2 n}+\frac{1}{3
n^2}-\frac{13}{12 n^3}+O\left(\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2111377",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Calculation error the length of an angle bisector How to find the length of an angle bisector ($BK$) in a triangle $A(1;4),~B(7;8),~C(9;2)$.
I use this formula:
$\frac{A_{1} \cdot x+B_{1} \cdot y+C_{1}}{\sqrt{A_{1}^{2}+ B_{1}^{2}}}=\frac{A_{2} \cdot x+B_{2} \cdot y+C_{2}}{\sqrt{A_{2}^{2}+ B_{2}^{2}}}$.
And, my result:... | Maybe you might find interesting this approach:
Once you have all tree points you have all sides ($AB,AC,BC$) and then you can use the angle bisector theorem:
$$\frac{AB}{AK}=\frac{BC}{KC}$$
and using that $AK+KC=AC$ you can find $AK$ and $KC$. After that you can use Stewart's theorem in order to find $BK$:
$$\frac{AB... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2111958",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Integration by parts: the variance of a standard normal I am calculating the variance of a standard normal, but I stuck with the following part (the answer is different from what I know). What is wrong with my calculation?
$$
\frac{1}{\sqrt{2 \pi}} \int_{-\infty}^{\infty} y^2 e^{- y^2 / 2} = \left[ y^2 \cdot \left( - \... | \begin{align}
\frac{1}{\sqrt{2 \pi}} \int_{-\infty}^{\infty} y^2 e^{- y^2 / 2}dy
&=\frac{2}{\sqrt{2 \pi}} \int_{0}^{\infty} y^2 e^{- y^2 / 2} dy\\
&\text{put $\frac{y^2}{2}=t$,we have $ydy=dt$}\\
&=\frac{2}{\sqrt{2 \pi}} \int_{0}^{\infty} e^{- t} \sqrt{2t}dt\\
&=\frac{2\sqrt 2}{\sqrt{2 \pi}} \int_{0}^{\infty} e^{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2112888",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Find length of intersection between 2 points and a sphere I have a sphere and 2 points. The points have (x,y,z) coordinates and the sphere is defined by its centre (0,0,0) and radius R. I am trying to find the length between the 2 points which intersects the sphere. How can I obtain the equation to describe this length... | The sphere is: $ x^2 + y^2 + z^2 =R^2 $ . The line is: $ x=x \\ y=x\cdot\frac{y_2-y_1}{x_2-x_1} -\frac{x_1(y_2-y_1)}{x_2-x_1} +y_1 =:r_yx+y_0 \\ z=x\cdot\frac{ z_2-z_1 }{x_2-x_1} -\frac{x_1(z_2-z_1)}{x_2-x_1} +z_1 =:r_zx+z_0$.
Sphere and line intersect when:
$ x^2 + (r_yx+y_0)^2 + (r_zx+z_0)^2 =R^2 \implies \\
x^2\cdo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2113001",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
if : $abc=8 $ then : $(a+1)(b+1)(c+1)≥27$ if : $$abc=8 :a,b,c\in \mathbb{R}_{> 0}$$
then :
$$(a+1)(b+1)(c+1)\ge 27.$$
My try :
$$(a+1)(b+1)(c+1)=1+(a+b+c)+(ab+ac+bc)+abc$$
$$(a+1)(b+1)(c+1)=1+(a+b+c)+(ab+ac+bc)+8, $$
then?
| By Holder $$\prod\limits_{cyc}(a+1)\geq\left(\sqrt[3]{abc}+1\right)^3=27$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2113791",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 3
} |
Help Determinant Binary Matrix I was messing around with some matrices and found the following result.
Let $A_n$ be the $(2n) \times (2n)$ matrix consisting of elements $$a_{ij} = \begin{cases} 1 & \text{if } (i,j) \leq (n,n) \text{ and } i \neq j \\ 1 & \text{if } (i,j) > (n,n) \text{ and } i \neq j \\ 0 & \text{oth... | As I commented : $detA = detB\cdot detc$, where $B = C$ with zero elements $a_{i,j}$ when $i = j$ and $1$ otherwise.
Now it's easy to see that $detB = n-1$, using definition of determinant. We have only one zero position in every row and only one $\sigma \in S_n$, which give us non-zero addition. So we get $det(B) = 1+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2113878",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
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How to solve this proportion Okay I think I'm just having a major brain block, but I need help solving this proportion for my physics class.
$$\frac {6.0\times 10^{-6}}{ x^2} = \frac {2.0\times 10^{-6}}{ (x-20)^2}$$
What's confusing me is the solution manual to this problem lists writing the proportion as,
$$\frac {(x... | The proportion they gave is correct. Simply divide both sides of the equation you started with by $6\cdot10^{-6}$ and multiply both sides by $(x-20)^2$.
So we have
$$
\frac{(x-20)^2}{x^2} =
\frac{2.0\cdot10^{-6}}{6.0\cdot10^{-6}}=\frac{1}{3}.$$
The left-hand side becomes
$$\frac{(x-20)^2}{x^2}=\frac{x^2-40x+400}{x^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2114285",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Question of computing a limit which tends to infinity $$f(x) = (x^3 + 2x^2 - 3x)/ (3x^2 + 3x - 6)$$
Question 1:
It is known that $\lim_{ x \rightarrow \infty} (f(x) - (mx +c))=0$
Find the values of $m$ and $c$,
Using the difference and sum rule and splitting out the values, simplifying $f(x)$,by dividing it with $x^2$,... | $\lim_{x \to \infty}((\frac {x^3+2x^2-3x}{3x^2+3x-6})-(mx+c))=0 \Rightarrow \lim_{x \to \infty} \frac {(x^3+2x^2-3x)-(mx+c)(3x^2+3x-6)}{3x^2+3x-6}=0 \Rightarrow \lim_{x \to \infty} \frac {x^3+2x^2-3x-[3mx^3+(3m+3c)x^2+(-6m+3c)x-6c]}{3x^2+3x-6}=0 \Rightarrow \lim_{x \to \infty} \frac {(1-3m)x^3+(2-3m-3c)x^2+(-3+6m-3c)x+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2116079",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
If $abc=1$ so $\sum\limits_{cyc}\sqrt{\frac{a}{4a+2b+3}}\leq1$. Let $a$, $b$ and $c$ be positive numbers such that $abc=1$. Prove that:
$$\sqrt{\frac{a}{4a+2b+3}}+\sqrt{\frac{b}{4b+2c+3}}+\sqrt{\frac{c}{4c+2a+3}}\leq1$$
The equality "occurs" also for $a\rightarrow+\infty$, $b\rightarrow+\infty$ and $a>>b$.
I tried AM-G... | We begin with a theorem :
Theorem :
Let $a,b,c,d,e,f$ be positive real number , with $a\geq b \geq c$ , $d\geq e \geq f $ under the three following conditions :
$a\geq d$ , $ab\geq de$ , $abc\geq def$ so we have :
$$a+b+c\geq d+e+f$$
Here we suppose that we have :
$a\geq b \geq 1 \geq c $
So to get the majorization we ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2116440",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 1,
"answer_id": 0
} |
Work Proof... Need help! I'm having trouble figuring out a proof at work, and to spare the business jargon I put it in terms of variables. Could someone help?
Prove:
$$\frac{A-B+Y}{1-C} \cdot C +\frac{A-B+X}{1-D} \cdot D = \frac{A-B}{1-C-D} \cdot(C+D)$$
where:
$$X=\frac{A-B+Y}{1-C} \cdot C$$
$$Y=\frac{A-B+X}{1-D} \cdo... | From the condition we get
$(1-C)X-CY=(A-B)C$ and $-DX+(1-D)Y=(A-B)D$ or
$$(1-C)(1-D)X-C(1-D)Y=(A-B)(1-D)C$$ and $$-DCX+C(1-D)Y=(A-B)CD$$ and after adding we get $((1-C)(1-D)-DC)X=(A-B)C$, which gives
$X=\frac{(A-B)C}{1-C-D}$ and $Y=\frac{(A-B)D}{1-C-D}$.
Thus, $$\frac{A-B+Y}{1-C} \cdot C +\frac{A-B+X}{1-D} \cdot D=\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2117272",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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If $\alpha $ and $\beta$ are roots of equation $a\tan\theta +b \sec\theta=c$. Prove that $\tan(\alpha + \beta)=\frac{2ac}{a^2-c^2}$ If $\alpha $ and $\beta$ are roots of equation $a\tan\theta +b \sec\theta=c$. Prove that $\tan(\alpha + \beta)=\frac{2ac}{a^2-c^2}$
i have converted tan to sin and cos and reached to $\si... | We have $a\tan\theta +b \sec\theta=c$. So,
\begin{align*}
c-a\tan \theta & = b \sec \theta\\
(c-a\tan \theta)^2 & = (b\sec \theta)^2\\
(a^2-b^2)\tan^2 \theta -2ac \tan \theta+(c^2-b^2) & = 0.
\end{align*}
This is a quadratic in $\tan \theta$. It has two solutions $\tan \alpha$ and $\tan \beta$. So
\begin{align*}
\ta... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2117468",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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How to simplify derivatives
The math problem asks to find the derivative of the function
$$y=(x+1)^4(x+5)^2$$
I get to the part $$(x+1)^4 \cdot 2(x+5) + (x+5)^2 \cdot 4(x+1)^3$$
How do they arrive at the answer
$$2(x+1)^3(x+5)(3x+11) ?$$
| $$(x+1)^4 \cdot 2(x+5) + (x+5)^2 \cdot 4(x+1)^3$$
$$(x+1)^3((x+1) \cdot 2(x+5) + (x+5)^2 \cdot 4)$$
$$(x+1)^3(x+5)( (x+1)\cdot 2 + (x+5) \cdot 4)$$
$$(x+1)^3(x+5)( 2)((x+1)+ (x+5)2)$$
$$2(x+1)^3(x+5)(x+1+2x+10)$$
$$2(x+1)^3(x+5)(3x + 11)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2118826",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Prove that; $8\cos^3 (\frac {\pi}{9})- 6\cos(\frac {\pi}{9})=1$ Prove that; $8\cos^3 (\frac {\pi}{9})- 6\cos(\frac {\pi}{9})=1$
My Attempt,
$$L.H.S=8\cos^3(\frac {\pi}{9}) - 6\cos(\frac {\pi}{9})$$
$$=2\cos(\frac {\pi}{9}) [4\cos^2(\frac {\pi}{9}) - 3]$$
$$=2\cos(\frac {\pi}{9}) [2+2\cos(\frac {2\pi}{9}) - 3]$$
$$=2\co... | $\cos 3x=4\cos^3x-3\cos x \Rightarrow 2\cos 3x=8\cos^3x-6\cos x$.
You can now see that if you put $x=\frac{\pi}{9}$, you get the result.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2118953",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
use of triple scalar product to prove an equality $a,b,c$ are unit vectors mutually inclined at an angle $\theta$.
Given that $$(a\times b) + (b\times c)=pa+qb+rc$$ where $p,q,r$ are constants and $a,b,c$ represent the above mentioned vectors. Prove that
$$r^2 + p^2 + \frac{q^2}{\cos\theta}=2.$$
| Let $\, \mathbf{v}=\mathbf{a} \times \mathbf{b}+\mathbf{b} \times \mathbf{c} =
p\mathbf{a}+q\mathbf{b}+r\mathbf{c}$
$\mathbf{a} \cdot \mathbf{v}$ gives
$$\mathbf{a} \cdot \mathbf{b} \times \mathbf{c} =
p+(q+r)\cos \theta \tag{1}$$
$\mathbf{b} \cdot \mathbf{v}$ gives
$$0 = q+(r+p)\cos \theta \tag{2}$$
$\mathbf... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2119392",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Prove that $ABCD$ is a rectangle. I am looking for a synthetic proof Let $ABCD$ a parallelogram and $ BE \perp AC, E \in (AC)$, $M$ is the middle of $[AE]$ and $N$ is the middle of $[CD]$. If $ BM \perp NM$ then $ABCD$ is a rectangle. I try to show that $MNCB$ is inscriptible. I am looking for a s... | $$\overrightarrow{BM}\cdot \overrightarrow{MN} = 0$$
$$ \Leftrightarrow (\overrightarrow{BA}\cdot \overrightarrow{BE})\cdot(\overrightarrow{AD} \cdot \overrightarrow{EC}) = 0$$
$$ \Leftrightarrow \color\red{\overrightarrow{BA}\cdot \overrightarrow{AD}} + \color\RoyalBlue{\overrightarrow{BA} \cdot \overrightarrow{EC}} +... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2119619",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Is $\cos(4x)+\sin(2x)$ periodic and how do I calculate the primitive period? My first attempt is under this, i can work out the primitive period of both of the $\cos(4x)$ and $\sin(2x)$ but how do I calculate the primitive period of $\cos(4x)+\sin(2x)$?
My attempt:
Let $u=4x$ then $x=\frac{u}{4}$ and $\cos(u)$ is $2\p... | Your attempt seems fine. Choose $T = \pi$, because that is a multiple of $T_1 = \pi /4$ and $T_2 = \pi$. Then, \begin{align} f(x + T) & = \cos (4(x+T)) + \sin (2(x+T)) \\ & = \cos (4x + 4T) + \sin (2x + 2T) \\ & = \cos(4x+4\pi) + \sin (2x + 2 \pi) \\ & = \cos (4x) + \sin (2x) \\ & = f(x). \end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2119925",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Probability of the second heart card If you pull a card and it is a heart, you turn it back and you pull another card. If it is not heart, you pull another card without returning the first card back. What is the probability that the second card is a heart.
| Case 1 -
First card is heart = $\frac {13}{52} = \frac {1}{4}$
Second card is heart = $\frac {13}{52} = \frac {1}{4}$
Probability = $\frac {1}{4} × \frac {1}{4} = \frac {1}{16}$
Case 2 -
First card not heart = $\frac {39}{52} = \frac {3}{4}$
Second card is heart = $\frac {13}{51}$
Probability = $\frac {3}{4} × \frac {1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2122681",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Find the determinant of order $100$ Find the determinant of order $100$:
$$D=\begin{vmatrix}
5 &5 &5 &\ldots &5 &5 &-1\\
5 &5 &5 &\ldots &5 &-1 &5\\
5 &5 &5 &\ldots &-1 &5 &5\\
\vdots &\vdots &\vdots &\ddots &\vdots &\vdots &\vdots\\
5 &5 &-1 &\ldots &5 &5 &5\\
5 &-1 &5 &\ldots &5 &5 &5\\
-1 &5 &5 &\ldots &5 &5 &5
\en... | Add all the rows to the last row and in the last row all entries will be $99 \cdot 5 - 1 = 494$. Now you can take $494$ outside of the determinant and now start subtracting $-5$ times the last row from each other row. Then the entries in the second diagonal will be $-6$, the last row will be $1$'s , while all other ent... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2122803",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 1
} |
An expression for $f(2x)$ as a rational function of $f(x)$ A real-valued function $f$ defined on the $\mathbb{R}\setminus\{1\}$ by
\begin{equation*}
f(x) = \frac{3x + 1}{x - 1} = 3 + \frac{4}{x - 1}
\end{equation*}
is invertible, and
\begin{equation*}
f^{-1}(x) = \frac{x + 1}{x - 3} = 1 + \frac{4}{x - 3} .
\end{equatio... | Here is the intended solution to the problem. This is a lot of work for a problem on a ten-minute competition. (There are six topics at each competition, and in each topic, there are three problems. The problem in the post comes from the topic called "Functions." I doubt anyone solved this problem during the competit... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2124248",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
Finding function when dx is integrated $\int (5-x^2+\frac{18}{x^4})dx$ I have a question on how to evaluate this integral:
$\int (5-x^2+\frac{18}{x^4})dx$
Is this correct?
$\int (5-x^2+\frac{18}{x^4})dx$
$ \int 5 dx = 5x + C$
$ \int -x^2 = \frac {-x^3}{3}= -\frac{1}{3}x^3 +C $
$ \int \frac {18}{x^4} = \int 18x^{-4} dx... | It is not quite correct. Note that
$$\int 18x^{-4}\,dx=18\,\left(\frac{x^{(-4+1)}}{(-4+1)}\right)+C=-6x^{-3}+C$$
On a side note, it is not good form to write "$5=5x$" as short hand notation for "$\int 5\,dx=5x$."
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2124833",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Calculate: $\lim_{x\to 0}{\frac{\tan x-\sin x}{x^3}}$
Calculate: $\lim{\frac{\tan x-\sin x}{x^3}}$ as $x\to 0$
I have solved it using a way that gives a wrong answer but I can't figure out why:
$$\lim{\frac{\tan x-\sin x}{x^3}}\\
=\lim{\frac{\tan x}{x^3}-\lim\frac{\sin x}{x^3}}\\
=\lim{\frac{1}{x^2}}\cdot\lim{\fra... | $\frac{\tan{x}-\sin{x}}{x^3}=\frac{2\sin{x}\sin^2\frac{x}{2}}{x^3\cos{x}}\rightarrow2\cdot\left(\frac{1}{2}\right)^2=\frac{1}{2}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2125706",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Cross-product properties from abstract definition Given two $3$D vectors $\mathbf{u}$ and $\mathbf{v}$ their cross-product $\mathbf{u} \times \mathbf{v}$ can be defined by the property that, for any vector $\mathbf{x}$ one has $\langle \mathbf{x} ; \mathbf{u} \times \mathbf{v} \rangle = {\rm det}(\mathbf{x}, \mathbf{u}... | I believe I've found an elegant proof.
*
*Assume that $ \mathbf{u}\times \mathbf{v} \ne 0$. For $\mathbf{x} = \mathbf{u}\times \mathbf{v}$ we have
\begin{align}
\|\mathbf{u}\times \mathbf{v}\|^4 &= \langle \mathbf{u}\times \mathbf{v},\mathbf{u}\times \mathbf{v}\rangle^2 \\
&=\det(\mathbf{u}\times \mathbf{v},\mathbf{u... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2127494",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 3
} |
How many 3-tuples satisfy $x_{1} + x_{2} + x_{3} = 11;$ $(x_{1} ,x_{2} ,x_{3}$ are nonnegative integers?) I know that the total number of choosing without constraint is
$\binom{3+11−1}{11}= \binom{13}{11}= \frac{13·12}{2} =78$
Then with x1 ≥ 1, x2 ≥ 2, and x3 ≥ 3.
the textbook has the following solution
$\binom{3+5−... | For the given equation: $$x_1+x_2+x_3 = 11, \;\text{ with } x_1, x_2, x_3 \;\text{ non-negative },$$ your solution is correct.
$$\binom{3+11−1}{11}= \binom{13}{11}= \frac{13!}{2!11!} = \frac{13·12}{2} =78$$
Your final answer is correct, (Now corrected: (but you failed to show that you need to divide $13\times 12$ by $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2129086",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 0
} |
Solve the system of non-homogeneous differential equations using the method of variation of parameters My exam question reads as follows:
Given the system:
$y'_1=y_1-y_2+\frac{1}{\cos \left(x\right)}$
$y'_2=2y_1-y_2$
Solve it using the method of variation of parameters, before that describe that method for solving syst... | I will map it out and you can fill in the details. We are given
$$y'= Ay + g = \begin{pmatrix}1 & -1 \\ 2 & -1\end{pmatrix}y+\begin{pmatrix}\dfrac{1}{\cos x} \\ 0\end{pmatrix}$$
Using eigenvalues / eigenvectors (or other approaches), we find the homogeneous solution
$$Y_h(x) = c_1\begin{pmatrix}\sin x + \cos x \\ 2\sin... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2130913",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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To evaluate $\frac{\cot25+\cot55}{\tan25+\tan55}+ \frac{\cot55+\cot100}{\tan55+\tan100}+\frac{\cot100+\cot25}{\tan100+\tan25}$ To evaluate $$\frac{\cot25^{\circ}+\cot55^{\circ}}{\tan25^{\circ}+\tan55^{\circ}}+ \frac{\cot55^{\circ}+\cot100^{\circ}}{\tan55^{\circ}+\tan100^{\circ}}+\frac{\cot100^{\circ}+\cot25^{\circ}}{\t... | We know that $$\cot A + \cot B = \frac {\cot A \cot B -1}{\cot (A+B)} $$
Now using this identity, we get, $$P =\cot 55^\circ [\cot 25^\circ + \cot 100^\circ] + \cot 25^\circ \cot 100^\circ $$ $$P =\cot 55^\circ [\frac {\cot 25^\circ \cot 100^\circ -1}{\cot 125^\circ}] + \cot 25^\circ \cot 100^\circ $$ $$P = \cot 25^\ci... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2133195",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
How do I find the point equidistant from three points $(x, y, z)$ and belonging to the plane $x-y+3z=0?$ I struggle to find the point ${P}$ equidistant from the points ${A(1,1,1), B(2,0,1), C(0,0,2)}$ and belonging to the plane ${x-y+3z=0}$.
Any tips?
| In $\Delta ABC$,
\begin{align*}
a &= BC \\
&= \sqrt{5} \\
b &= CA \\
&= \sqrt{3} \\
c &= AB \\
&= \sqrt{2} \\
a^2 &= b^2+c^2 \\
\angle A &= 90^{\circ} \\
O &= \frac{B+C}{2} \tag{circumcentre of $\Delta ABC$} \\
&= \left( 1,0,\frac{3}{2} \right) \\
\vec{AB} \times \vec{AC} &=
(1, -1, 0) \times (-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2133525",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
Convert rectangular equation $(x^2+y^2)^2 - 4(x^2-y^2) = 0$ to polar form So I am not sure how to do this problem : "Convert rectangular equation $(x^2+y^2)^2 - 4(x^2-y^2) = 0$ to polar form". Any help would be appreciated.
| $$0=(x^2+y^2)^2 - 4(x^2-y^2) $$
Recall that $x^2+y^2=r^2$ and $x=r\cos(\theta)$ and $y=r\sin(\theta)$
$$ 0=(r^2)^2-4(r^2 \cos^2(\theta)-r^2\sin^2(\theta))$$
$$ 0=r^4-4(r^2(\cos^2(\theta)-\sin^2(\theta))$$
$$ 0 = r^4-4r^2\cos(2\theta)$$
$$ 0 = r^2(r^2-4\cos(2\theta))$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2133985",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
How can you prove $\frac{n(n+1)(2n+1)}{6}+(n+1)^2= \frac{(n+1)(n+2)(2n+3)}{6}$ without much effort?
I will keep it short and take only an extract (most important part) of
the old task.
$$\frac{n(n+1)(2n+1)}{6}+(n+1)^2= \frac{(n+1)(n+2)(2n+3)}{6}$$
What I have done is a lot work and time consuming, I have "simply" s... | Oh goodness, you needn't so much work. All you need to do is this:
$$\begin{align}\frac{n(n+1)(2n+1)}6+(n+1)^2&=\frac{n(n+1)(2n+1)}6+\frac{6(n+1)^2}6\\&\tag 1=c\bigg(n(2n+1)+6(n+1)\bigg)\\&\tag2=c\bigg(2n^2+7n+6\bigg)\\&\tag3=c\bigg((n+2)(2n+3)\bigg)\end{align}$$
$(1)$ factor out $(n+1)/6$ and call it $c$.
$(2)$ expan... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2137292",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 5,
"answer_id": 3
} |
Prove that: $\sec^2 20^\circ +\sec^2 40^\circ +\sec^2 80^\circ = \textrm 36$ Prove that: $\sec^2 20^\circ +\sec^2 40^\circ +\sec^2 80^\circ = \textrm 36$
My Attempt:
$$L.H.S=\sec^2 20^\circ + \sec^2 40^\circ +\sec^2 80^\circ$$
$$=\dfrac {1}{\cos^2 20°} +\dfrac {1}{\cos^2 40°} +\dfrac {1}{\cos^2 80°}$$
$$=\dfrac {\cos^2... | Starting like dxiv,
let $a=\cos20^\circ,b=-\cos40^\circ,c=-\cos80^\circ,$
As $\cos(3\cdot20^\circ)=\dfrac12$
$\cos(3\cdot40^\circ)=-\dfrac12$
$\cos(3\cdot80^\circ)=-\dfrac12$
As $\cos3x=\cos60^\circ, 3x=360^\circ m\pm60^\circ$ where $m$ is any integer.
$\implies x=120^\circ m+20^\circ$ where $m\equiv-1,0,1\pmod3$
Now a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2137609",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
} |
$\int_{0}^{\frac{\pi}{4}}\frac{\tan^2 x}{1+x^2}\text{d}x$ on 2015 MIT Integration Bee So one of the question on the MIT Integration Bee has baffled me all day today $$\int_{0}^{\frac{\pi}{4}}\frac{\tan^2 x}{1+x^2}\text{d}x$$ I have tried a variety of things to do this, starting with
Integration By Parts Part 1
$$\frac{... | This is not the whole answer to this question but it is something new that wasn't mentioned here so I thought I would post it. Let us denote the unknown integral by $J$. Then substituting for $y= \tan(x)$ we get:
\begin{equation}
J = \int\limits_0^1 \frac{y^2}{1+\arctan(y)^2} \cdot \frac{1}{1+y^2} d y =
\int\limits_0^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2138866",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "64",
"answer_count": 7,
"answer_id": 0
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Minimize $\big(3+2a^2\big)\big(3+2b^2\big)\big(3+2c^2\big)$ if $a+b+c=3$ and $(a,b,c) > 0$
Minimize $\big(3+2a^2\big)\big(3+2b^2\big)\big(3+2c^2\big)$ if $a+b+c=3$ and $(a,b,c) > 0$.
I expanded the brackets and applied AM-GM on all of the eight terms to get :
$$\big(3+2a^2\big)\big(3+2b^2\big)\big(3+2c^2\big) \geq 3\... | By Lagrange multipliers, $(a,b,c)=(1,1,1)$ is the only stationary point inside the closed region $(a,b,c)\geq 0, a+b+c=3$. To prove it is a minimum is is enough to study the given function under the assumptions $c=0$ and $b=3-a$. The one variable function
$$ f(a) = 3(3+2a^2)(3+2(3-a)^2) $$
has absolute minima over the ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2141009",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 0
} |
Prove that $\prod \limits_{cyc}(a^3+a+1 ) \leq 27$
Let $a,b,c > 0$ such that $a^2 + b^2 + c^2 =3$. Prove that
$$(a^3+a+1)(b^3+b+1)(c^3+c+1) \leq 27$$
My attempt :
Let $\lambda = \prod \limits_{cyc} a^3+a+1$ .
Applying AM-GM on the set $\{(a^3+a+1),(b^3+b+1),(c^3+c+1)\}$ :
$$\lambda^{\frac{1}{3}}\leq\dfrac{\sum \limit... | From the expansion of $a^4-4a^3+6a^2-4a+1=(a-1)^{4} \ge 0$, we can simply the equation to get the following: $$4a^3+4a+4 \le a^4+6a^2+5 $$
So we have that
$$\prod_{cyc}(a^3+a+1)\leq\frac{1}{4^3}\prod_{cyc}(a^2+1)\prod_{cyc}(a^2+5) \le 27$$
From $\text{AM-GM}$ Inequality. We have the desired result.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2144093",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Proving functions are linearly independent I'm currently going through Harvard's Abstract Algebra using Michael Artin's book, and have no real way of verifying my proofs, and was hoping to make sure that my proof was right.
The question reads:
Let $V$ be the vector space of functions on the interval $[0, 1]$. Prove th... | Why not assume that $f(x) = a_1x^3+a_2 \sin x + a_3 \cos x$ is the zero function. Then note that $0 = f(0) = a_3$, so we may rewrite $f(x) = a_1x^3 + a_2 \sin x$. Evaluate this rewritten expression for $f$ at the point $\pi$ to show that $0 = f(\pi) = a_1 \pi^3 \Longrightarrow a_1 = 0$. Thus we can again rewrite $f(x) ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2145060",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 5,
"answer_id": 3
} |
Solutions to a Cubic Root I'm looking for a sanity check to my work.
We are supposed to find the solutions of the cubic polynomial, $x^3+3x+5=0$.
Let $x=w-\frac{1}{w}$ and substitute.
Then $(w-\frac{1}{w})^3+3(w-\frac{1}{w})+5=0$.
Then $w^3+5w^3-1=0$.
Finally $w^3=\frac{-5+\sqrt{29}}{2}$ and $\frac{-5-\sqrt{29}}{2}$.
S... | Here's a sanity check that shows something is wrong:
$$\sqrt[3]{-5+\sqrt{29}\over2}-\sqrt[3]{-5-\sqrt{29}\over2}=\sqrt[3]{-5+\sqrt{29}\over2}+\sqrt[3]{5+\sqrt{29}\over2}\gt0$$
since $\sqrt{29}\gt5$ (so that the first cube root is positive). But $x^3+3x+5\gt0$ for all $x\gt0$, so the real roots of $x^3+3x+5$ must be ne... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2146211",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Solve $\lim_{x \rightarrow +\infty}\frac{\sqrt{x}+x}{x+\sqrt[3]{x}}$ I tried first without L'Hôpital's rule:
$$\lim_{x \rightarrow +\infty}\frac{\sqrt{x}+x}{x+\sqrt[3]{x}} =
\frac{\sqrt{x}}{\sqrt[3]{x}} \cdot \frac{1+\frac{x}{\sqrt x}}{1+ \sqrt[3] x} = \frac{\sqrt x}{\sqrt[3] x} \cdot \frac{\frac{\sqrt x+x}{\sqrt x}}{... | In such cases, it is easier to perform a suitable transformation to obtain integer-valued powers: this suggests letting $x = u^6$, hence we have the equivalent limit $$\lim_{x \to \infty} \frac{x^{1/2} + x}{x+x^{1/3}} = \lim_{u \to \infty} \frac{u^3 + u^6}{u^6 + u^2} = \lim_{u \to \infty} \frac{u^{-3} + 1}{1 + u^{-4}}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2147487",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Geometric interpretation of complex number.
Let $z \in \mathbb{C}, z^2, z^3$ be the verticies of a right triangle. Find the geometric images of $z$.
I did not understand the question but I guess it want me to find the figure formed by $z$ under these constraints.
Let $z = x + iy$, then $z^2 = x^2 - y^2 +2xyi$ and $z... | If the vertex at the right angle is $z^2$, then $\dfrac{z^3-z^2}{z-z^2}$ must be purely imaginary. Thus $\dfrac{z^2-z}{1-z} = -z$ is purely imaginary. Thus $z = ri$ for some real $r$.
Let $z = re^{i\theta}$ be the vertex at the right angle. From the figure we have
\begin{align*}
AB^2 &= r^2 + r^4 -2r^3 \cos \theta \\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2148353",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Other Idea to show an inequality $\dfrac{1}{\sqrt 1}+\dfrac{1}{\sqrt 2}+\dfrac{1}{\sqrt 3}+\cdots+\dfrac{1}{\sqrt n}\geq \sqrt n$ $$\dfrac{1}{\sqrt 1}+\dfrac{1}{\sqrt 2}+\dfrac{1}{\sqrt 3}+\cdots+\dfrac{1}{\sqrt n}\geq \sqrt n$$
I want to prove this by Induction
$$n=1 \checkmark\\
n=k \to \dfrac{1}{\sqrt 1}+\dfrac{1}{... | Combining AM-HM $$\left(a_1+a_2+...+a_n\right)\left(\frac{1}{a_1}+\frac{1}{a_2}+...+\frac{1}{a_n}\right)\geq n^2$$
Thus
$$\left(\sqrt{1}+\sqrt{2}+...+\sqrt{n}\right)\left(\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+...+\frac{1}{\sqrt{n}}\right)\geq n^2$$
and
$$n\sqrt{n}\geq\left(\sqrt{1}+\sqrt{2}+...+\sqrt{n}\right)$$
so
$$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2149448",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 6,
"answer_id": 1
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A relation between two elements Let $G $ be a group and $ a, b\in G $ such that $a^{-1}b^2a=b^3$ and $b^{-1}a^2b=a^3$.
How can we show that $a=b$$\,$?
| $\bbox[5px,border:2px solid]{\begin{array}{c}b^2a=ab^3\\a^2b=ba^3\end{array}}\ $ these relations are equivalent to OP's since elements in a group are invertible.
If $G$ is commutative or if $a,b$ commute
We have $ab=ba$.
$b^3=a^{-1}b^2a=a^{-1}ab^2=b^2\Rightarrow b=1$
$a^3=b^{-1}a^2b=b^{-1}ba^2=a^2\Rightarrow a=1$
We ge... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2149551",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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$\lim \limits_{x\to 0} \frac {4^{\sin x}-2^{\tan 2x}}{3^{\sin 2x}-9^{\tan x}}=?$ find the limit :
$$\lim_{x\to 0} \frac {4^{\sin x}-2^{\tan 2x}}{3^{\sin 2x}-9^{\tan x}}=?$$
my try :
$$\lim_{x\to 0} \frac {2^{2\sin x}-2^{\tan 2x}}{3^{\sin 2x}-3^{2\tan x}}=$$
$$\lim_{x\to 0} \frac {2^{2\sin x}-2^{\frac{\sin 2x}{\cos2x}}}... | $$\lim_{x\to0}\frac {2^{\tan 2x}\left(2^{(2\sin x-\tan 2x)}-1\right)}{3^{(2\tan x} \left(3^{(\sin 2x-2\tan x)}-1\right)}$$
$$=\lim_{x\to0}\frac {2^{\tan 2x}}{3^{2\tan x}}\cdot\underbrace{\lim_{x\to0}\dfrac{2^{(2\sin x-\tan 2x)}-1}{2\sin x-\tan 2x}}_{(1)}\cdot\lim_{x\to0}\dfrac1{\underbrace{\dfrac{3^{(\sin 2x-2\tan x)}-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2149919",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Letters are arranged in boxes
Ways =(selecting one box from upper and lower rows)(selecting four boxes out of remaining)(number of ways arranging six letters in these boxes)
$$={3\choose 1}{3\choose 1}{6 \choose 4}\cdot6!$$
| [\begin{align}
& There\text{ are 3 cases to be considered : }\left( 1,1,1,3 \right)\text{ ,}\left( 2,1,1,2 \right)\text{ }and\text{ }\left( 3,1,1,1 \right) \\
& \underline{case\text{ 1}}\text{ : }top\text{ }row\text{ }1:6C1*3 \\
& \text{ }second\text{ }row\text{ }1:5C1 \\
& ~\text{ }th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2151402",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Maximum radius of intersection of a plane and an ellipsoid The question is to use Lagrange multipliers to show that the maximum value of $r = (x^2 + y^2 + x^2)^{1/2}$ subject to the following conditions
$$\frac{x^2}{a^2} + \frac{y^2}{b^2} + \frac{z^2}{c^2} = 1 \quad\quad\quad lx + my + nz = 0$$ satisfies $$\frac{l^2a^... | Multiply your first equation with $x$, second with $y$, third with $z$ and add them together. You will get $$(x^2+y^2+z^2)\frac{1}{r}+2\lambda\left(\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}\right)=r+2\lambda=0$$
Now use $2\lambda=-r$ to plug into your equations, find $x,y,z$ and plug it into the equation of the p... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2151854",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Natural parametrization of curves I want to find curvatures and torsions for the following curves but get stuck with their natural parametrizations ($s$ is natural if $|\dot{\gamma}(s)| = 1$). Can anyone help me?
(a) $e^t(\cos t,\sin t,1)$
(b) $(t^3+t,t^3-t,\sqrt{3}t^2)$
(c) $3x^2+15y^2=1, z=xy$
Update:
Here are my att... | $(a)$
\begin{align*}
\mathbf{\dot{r}}(t) &= e^{t}(\cos t-\sin t, \sin t+\cos t,1) \\
|\mathbf{\dot{r}}(t)| &= e^{t}\sqrt{(\cos t-\sin t)^2+(\sin t+\cos t)^2+1} \\
&= e^{t}\sqrt{3} \\
s &= \int_{0}^{t} |\mathbf{\dot{r}}(t)| \, dt \\
&= \sqrt{3}(e^{t}-1) \\
t &= \ln \left( 1+\frac{s}{\sqrt{3}} \right)
\end{al... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2153421",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Solve the equation : $\tan \theta + \tan 2\theta + \tan 3\theta = \tan \theta \tan 2\theta \tan 3\theta $ I've been having some trouble solving this equation. (The solution in my book is given as $ \frac {n \pi}{3}, n \in Z $)
Here is what I've done
$$\frac {\sin \theta}{\cos \theta} + \frac {\sin 2\theta} {\cos 2\thet... | You could have solved the problem using the tangent multiple angle formulae.
Using $t=\tan(\theta)$ $$\tan(2\theta)=\frac{2t}{1-t^2}\qquad \tan(3\theta)=\frac{3t-t^3}{1-3t^2}$$ This makes, after some minor simplifications
$$\tan \theta + \tan 2\theta + \tan 3\theta - \tan \theta \tan 2\theta \tan 3\theta=\frac{2 t \le... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2154221",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
Prove that $\frac{a}{\sqrt{a^2+b^2}}+\frac{b}{\sqrt{9a^2+b^2}}+\frac{2ab}{\sqrt{a^2+b^2}\times \sqrt{9a^2+b^2} }\leq \frac32.$
Prove that $$\dfrac{a}{\sqrt{a^2+b^2}}+\dfrac{b}{\sqrt{9a^2+b^2}}+\dfrac{2ab}{\sqrt{a^2+b^2}\times \sqrt{9a^2+b^2} }\leq \dfrac{3}{2}.$$
When is equality attained ?
My Attempt :
I could not t... | It's enough to prove our inequality for positives $a$ and $b$.
Let $a^2=\frac{x}{3}$, $b^2=y$ and $x^2+y^2=2kxy$.
Hence, $k\geq1$ and we need to prove that
$$\sqrt{\frac{x}{x+3y}}+\sqrt{\frac{y}{3x+y}}+2\sqrt{\frac{xy}{(3x+y)(x+3y)}}\leq\frac{3}{2}$$ or
$$\frac{\sqrt{x(3x+y)}+\sqrt{y(x+3y)}}{\sqrt{(3x+y)(x+3y)}}\leq\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2154453",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Solve $\lim_{x \rightarrow 0} \frac{e^x+e^{-x}-2}{x^2+2x}$ without using L'Hopital's rule I tried:
$$\lim_{x \rightarrow 0} \frac{e^x+e^{-x}-2}{x^2+2x} = \\
\frac{e^{x}(1+e^{-2x}-\frac{2}{e^x})}{x(x-2)} = \frac{e^x(1+e^{-2x})-2}{x(x-2)} = \frac{e^x(1+e^{-2x})}{x(x-2)} - \frac{2}{x(x-2)} = \\
\frac{1+e^{-2x}}{x} \cdot ... | Simply define $f(x) = e^x + e^{-x}.$ Then the expression equals
$$\frac{f(x) - f(0)}{x-0}\cdot \frac{x}{x^2+2x}.$$
As $x\to0,$ the first fraction $\to f'(0)$ by the definition of the derivative. The second fraction $\to 1/2.$ Since $f'(0)=0,$ the limit in question is $0\cdot (1/2) = 0.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2155353",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
If $f(n)$ be the integer closest to $\sqrt{n}.$ then value of $ \sum^{2016}_{k=1}\frac{1}{f(k)}$ If $f(n)$ be the integer closest to $\sqrt{n}.$ then value of $\displaystyle \sum^{2016}_{k=1}\frac{1}{f(k)}$
could some help me with ths, thanks
| Taking a cue from sequence A000194 in the OEIS, we have $$S = \sum_{k=1 }^{2016} \frac{1}{f (k)}$$ $$ = \frac {2}{1} + \frac {4}{2} + \frac {6}{3} + \cdots + \frac {2 \lfloor \sqrt {2016} \rfloor}{\lfloor \sqrt {2016} \rfloor} + \frac {36}{\lceil \sqrt {2016} \rceil}\,\,(\text{ why? })$$ $$=2+2+2\cdots +2 +\frac {36}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2157893",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Continued Fraction Identity Problem
Question: What went wrong in my work to$$\ln(1-x)=-\cfrac x{1-\cfrac x{2+x-\cfrac {2x}{3+2x-\cfrac {3x}{4+3x-\ddots}}}}\tag{1}$$
I started with the expansion$$\begin{align*}\ln(1-x) & =-x-\dfrac {x^2}2-\dfrac {x^3}3-\dfrac {x^4}4-\&\text{c}.\\ & =-x\left\{1+\left(\dfrac x2\right)+\... | You went wrong with conversion from the series to the Euler CF. It was almost right, but the numeric numerator factor should be squared. That is,
$$ \ln (1-x) = -\cfrac{x}{1 - \cfrac{1^2x}{2 + x - \cfrac{2^2x}{3 + 2x - \cfrac{3^2x}{\ddots}}}}. $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2158039",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"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.