Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
proving the limit of a function using the $\epsilon , \delta $ definition Use the $\epsilon , \delta $ definition of a limit to verify that $$\lim \limits_{x \to 1} \frac{x^2 +x}{3x-2}=2$$
Let, $$\lim \limits_{x \to 1} \frac{x^2 +x}{3x-2}-2 = \frac{x^2+x -6x+4}{3x-2}= \frac{x^2-5x+4}{3x-2} = \frac{(x-4)(x-1)}{3x-2}$$
... | Continuing your analysis: If $|x-1|<{5\over 6}$ then
$$
\left|{x^2+x\over 3x-2}-2\right|=\left|{(x-4)(x-1)\over 3x-2}\right|\le {23/6\over 3/2}\cdot|x-1|={23\over 9}|x-1|,
$$
because $|x-4|<23/6$ and $|3x-2|>3/2$ for $x\in(1/6,11/6)$. Now make your choice of $\delta>0$ (for a given $\epsilon>0$) taking care that it is ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1481704",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Proving some identities in the set of natural numbers without using induction... I'm not sure how to prove some of the identities without using induction, for example:
$$1+2+3+...+n=\frac{n(n+1)}{2}$$
$$1^2+2^2+...+n^2=\frac{n(n+1)(2n+1)}{6}$$
$$1^3+2^3+...+n^3=(\frac{n(n-1)}{2})^2$$
What my teacher suggested and did f... | For the first one there is a nice way to find it is sum let
$S = 1 + 2 + 3 \cdots + n $ then
$S = n + (n-1) + (n-2) + \cdots + 2 + 1 $ add them by terms (first term with the first second with the second we get
$2S = (n+1) + (n+1) + (n+1) \cdots + (n+1) $ but we have $n$ of $(n+1)$ so
$2S = n(n+1) \Rightarrow S =... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1490653",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Continuity of $\frac{x^5-4x^3y^2-xy^4}{(x^2+y^2)^2}$ at (0, 0) I am having trouble proving that $\dfrac{x^5-4x^3y^2-xy^4}{(x^2+y^2)^2}$ is continuous at $(0, 0)$ if we set the value at $(0, 0)$ to be $0$.
I don't see a way to prove this as I cannot factor this into partial fractions.
| As an another approach:
$$\bigg|\frac{x^5-4x^3y^2-xy^4}{(x^2+y^2)^2}\bigg|=|x|\bigg|\frac{x^4-4x^2y^2-y^4}{(x^2+y^2)^2}\bigg|=|x|\bigg|\frac{(x^2+y^2)^2-6x^2y^2-2y^4}{(x^2+y^2)^2}\bigg| \\ \le |x|\bigg(1+\bigg|\frac{6x^2y^2+2y^4}{(x^2+y^2)^2}\bigg|\bigg)\le |x|\bigg(1+\bigg|\frac{6x^2y^2+6y^4}{(x^2+y^2)^2}\bigg|\bigg) ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1491859",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Verify $\int\sec x\ dx=\frac12 \ln \left\lvert\frac{1+\sin x}{1-\sin x}\right\rvert + C$ Question says it all, how can I verify the following?
$$\int\sec x\ dx=\frac12 \ln \left|\frac{1+\sin x}{1-\sin x}\right| + C$$
| We have
$$\begin{array}{lll}
\sec^2 x&=& \tan^2 x + 1\\
\sec^2 - \tan^2x&=& 1\\
(\sec x + \tan x)(\sec x - \tan x)&=&1\\
(\sec x + \tan x)^2(\sec x - \tan x)&=&\sec x + \tan x\\
(\sec x + \tan x)^2&=&\frac{\sec x + \tan x}{\sec x - \tan x}\\
\mid\sec x + \tan x\mid&=&\bigg(\frac{\sec x + \tan x}{\sec x - \tan x}\bigg)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1492942",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 4
} |
A polynomial problem related to lx^2 + nx + n If the roots of $lx^2 + nx + n = 0$ are in the ratio $p:q$, find the value of $\sqrt{\frac{p}{q}}$ + $\sqrt{\frac{q}{p}}$ + $\sqrt{\frac{n}{l}}$.
How to go about this problem?
| Notice,
let $kp$ & $kq$ be the roots of $lx^2+nx+n=0$ hence satisfying the equation we get $$lp^2k^2+npk+n=0\tag 1$$
$$lq^2k^2+nqk+n=0\tag 2$$
solving (1) & (2) for $k$, as follows $$\frac{k^2}{n^2p-n^2q}=\frac{k}{nlq^2-nlp^2}=\frac{1}{nlp^2q-nlpq^2}$$
$$\implies k^2=\frac{n^2p-n^2q}{nlp^2q-nlpq^2}=\frac{n^2(p-q)}{nlp... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1493495",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
how to find this limit : lim x to infinity how can I find this limit which is become infinite
$\lim _{x\to \infty }\left(x(\sqrt{x^2+1}-x)\right)$
can I use conjugate method ?
that what I'm doing until now
$= x\left(\frac{\sqrt{x^2+1}-x}{1}\cdot \:\frac{\sqrt{x^2+1}+x}{\sqrt{x^2+1}+x}\right)\:$
$= x\left(\frac{1}{\sqrt... | Yes, you can use conjugates for $\left(\sqrt{x^2+1}-x\right)$ as follows
Notice,
$$\lim_{x\to \infty}x\left(\sqrt{x^2+1}-x\right)$$
$$=\lim_{x\to \infty}\frac{x\left(\sqrt{x^2+1}-x\right)\left(\sqrt{x^2+1}+x\right)}{\left(\sqrt{x^2+1}+x\right)}$$
$$=\lim_{x\to \infty}\frac{x(x^2+1-x^2)}{\sqrt{x^2+1}+x}$$
$$=\lim_{x\t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1494166",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Find all the solutions of diophantine eq: $x^3-2xy^2+y^3-s^2=0$ Given $x,y,s$ are natural numbers:
$$x^3-2xy^2+y^3-s^2=0$$
I found the solutions using wolfram alpha
$$(x,y,s) = (1,2,1), (6,10,4), (4,8,8)$$
But how do I prove these are the only solutions? Any tips or reference to papers that study this diophantine equat... | There is a beautiful parameterization by Lagrange and Legendre:
$$x^3 + a x^2y + b x y^2 + c y^3 = z^2\tag1$$
where,
$$x = u^4 - 2b u^2 v^2 - 8c u v^3 + (b^2 - 4a c)v^4$$
$$y = 4v(u^3 + a u^2v + b u v^2 + c v^3)$$
and $z$ is a $6$-deg polynomial.
In your case, with $a,b,c = 0,-2,1$, we have,
$$x^3 - 2x y^2 + y^3 = z^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1494816",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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How to prove $\frac xy + \frac yx \ge 2$ I am practicing some homework and I'm stumped.
The question asks you to prove that
$x \in Z^+, y \in Z^+$
$\frac xy + \frac yx \ge 2$
So I started by proving that this is true when x and y have the same parity, but I'm not sure how to proceed when x and y have opposite partiy
Th... | If $x=y,$ then it is true.
Suppose $x>y$ then there is exists positive integer $n$ such that $x=y+n.$
This implies $\frac{x}{y}=1+\frac{n}{y}$ and $\frac{y}{x}=1-\frac{n}{x}$.
Adding these equations, we obtain
\begin{align}
\frac{x}{y}+ \frac{y}{x}=2+n\left(\frac{1}{y}-\frac{1}{x}\right)=2+n\frac{x-y}{xy}> 2\quad as\qu... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1494887",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 10,
"answer_id": 4
} |
What would be the fastest way of solving the following inequality $\frac{(x+1)}{(x-1)(x+2)}>\frac{(x)}{(x-1)(x-3)}$ What would be the fastest way of solving the following inequality:
$\frac{(x+1)}{(x-1)(x+2)}>\frac{(x)}{(x-1)(x-3)}$
| Reducing to the same denominator, we get
$$\frac{(x+1)}{(x-1)(x+2)}>\frac{(x)}{(x-1)(x-3)}\iff -\frac{4x+3}{(x+2)(x-1)(x-3)}>0 $$
Now the sign of $-\dfrac{4x+3}{(x+2)(x-1)(x-3)}$, when defined, is the sign of the product $p(x)=-(4x+3)(x+2)(x-1)(x-3)$, which by Bolzano's theorem can change sign only at $-2, -\dfrac34,1$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1495145",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 4
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Solve using AM GM Inequality if possible Let x, y, z be non-zero real numbers such that $\dfrac{x}{y} + \dfrac{y}{z} + \dfrac{z}{x} = 7$ and $\dfrac{y}{x} + \dfrac{z}{y} + \dfrac{x}{z} = 9$, then $\dfrac{x^3}{y^3} + \dfrac{y^3}{z^3} + \dfrac{z^3}{x^3}$ is equal to?
I don't really know how to solve this. any methods wou... | This method doesn't use the $AM-GM$ inequality.
$a^3 +b^3 +c^3 = (a+b+c)^3 -3(a+b+c)(ab+ac+bc)+3abc$
Applying it to your question, we have
$\dfrac{x^3}{y^3} + \dfrac{y^3}{z^3} + \dfrac{z^3}{x^3}$
= $(\dfrac{x}{y} + \dfrac{y}{z} + \dfrac{z}{x})^3 -3(\dfrac{x}{y} + \dfrac{y}{z} + \dfrac{z}{x})(\dfrac{y}{x} + \dfrac{z}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1495357",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
Distributing multiplication of rational functions I am having trouble distributing with fractions. This
$$ \left(\frac{1}{(x + 3)} + \frac{(x + 3)}{(x - 3)}\right)\, (9 - x^2) = -\frac{(x^2-3)}{9-x^2}(9-x^2) $$
has the answer $\left\{\left[x = - \frac{9}{7}\right]\right\}$.
I start solving by cancelling $(9-x^2)$ on th... | $\left(\frac{1}{x+3} + \frac{x+3}{x-3} \right) ( 9 -x^2) -x^2 -3$
It will be
$\frac{9-x^2}{x+3} + \frac{(x+3)(9-x^2)}{x-3} = -x^2 - 3 $
after that
$\frac{(9-x^2)(x-3)}{(x+3)(x-3)} + \frac{(x+3)(x+3)(9-x^2)}{(x-3)(x+3)} = -x^2 -3 $
Note that $9-x^2 = -(x^2 - 9) = -(x+3)(x-3)$, i.e., you can cancel it with the denom... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1496225",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
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What is the number of solution to $x^2\equiv 1\pmod{p^k}$? We learned about quadratic reciprocity and how it relates to the number of solution to $x^2 \equiv 1 \pmod{p}$.
I was wondering, is there a general formula for the
number of solutions modulo $p^k$ to $x^2 \equiv 1 \pmod{p^k}$?
Thank you very much!
| If $p$ is odd, then $x^2\equiv 1\pmod{p}$ has exactly two solutions, because if $p\mid x^2-1=(x+1)(x-1)$, then either $p\mid x+1$ or $p\mid x-1$ (by Euclid's Lemma). The solutions are $x\equiv \pm 1\pmod{p}$.
$x^2\equiv 1\pmod{p^k},\, k\ge 2$ for an odd prime $p$ too has exactly two solutions $x\equiv \pm 1\pmod{p^k}$.... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1497078",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Expanding brackets problem: $(z - 1)(1 + z + z^2 + z^3)$ I have:
$(z - 1)(1 + z + z^2 + z^3)$
As, I have tried my own methods and enlisted the help on online software, but as well as them not all arriving at the same solution, I can't follow their reasoning.
I tried to gather all the like terms:
$(z - 1)(z^6+1)$
And th... | I like to do factorizations like this by writing
$$
z^7-1=(z-1)(\cdots)
$$
and try to figure out what should replace the dots. First, we'd like to have a factor of $z^7$ in the end result, so the $z$ on the right side should be multiplied by $z^6$ to get this. Then, we write
$$
z^7-1=(z-1)(z^6+\cdots).
$$
At this po... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1500084",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Solve $[\frac{x^2-x+1}{2}]=\frac{x-1}{3}$ How can one solve the equation :
$[\frac{x^2-x+1}{2}]=\frac{x-1}{3}$ ?
Such that $[x]$ is the integer part of $x$.
By definition :
$[\frac{x^2-x+1}{2}]<=\frac{x^2-x+1}{2}<[\frac{x^2-x+1}{2}]+1$.
So :
$\frac{x-1}{3}<=\frac{x^2-x+1}{2}<\frac{x-1}{3}+1$.
1)
$\frac{x-1}{3}<=\frac{... | Write the original equation as:
$$\left\lfloor\frac{x^2-x+1}{2}\right\rfloor=\frac{x-1}{3} \tag{1}$$
By definition of the floor function, both sides of the equation must be integers, i.e. $x\in\mathbb{Z}$. So then $x^2-x+1$ is an odd integer, whether $x$ is even or odd. Therefore
$$\left\lfloor\frac{x^2-x+1}{2}\right\r... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1501997",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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How to find the determinant of this $5 \times 5$ matrix? How can I find the determinant of this matrix?
I know in matrix $3 \times 3$
$$A= 1(5\cdot 9-8\cdot 6)-2 (4\cdot 9-7\cdot 6)+3(4\cdot 8-7\cdot 5) $$
but how to work with a $5\times 5$ matrix?
| Once the matrix starts getting large, it can be easier to use row- or column-reduction to find the determinant, especially if there aren’t many sparse rows or columns to take advantage of in iterated Laplace expansions. This technique makes use of the facts that swapping two rows/columns changes the sign of the determi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1502099",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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$A+B+C+7$ is perfect square where $A,B,C$ numbers with repeating digits. Let $A=\underbrace{444...4}_{2m},$ $B=\underbrace{222...2}_{m+1}$ and $C=\underbrace{888...8}_{m}$.
Prove that $A+B+C+7$ is perfect square.
How to prove that problem?
I have checked it for small values of $m$ and it's true but I Can't prove it in... | Proof:
$A=\underbrace{444...4}_{2m}=\dfrac{4}{9}\cdot \underbrace{999...9}_{2m}=\dfrac{4}{9}(10^{2m}-1).$
$B=\underbrace{222...2}_{m+1}=\dfrac{2}{9}\cdot \underbrace{999...9}_{m+1}=\dfrac{2}{9}(10^{m+1}-1).$
$C=\underbrace{888...8}_{m}=\dfrac{8}{9}\cdot \underbrace{999...9}_{m}=\dfrac{8}{9}(10^{m}-1).$
Hence $$A+B+C+7=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1503670",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
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In the real number system,the equation $\sqrt{x+3-4\sqrt{x-1}}+\sqrt{x+8-6\sqrt{x-1}}=1$ has how many solutions? In the real number system,the equation $\sqrt{x+3-4\sqrt{x-1}}+\sqrt{x+8-6\sqrt{x-1}}=1$ has how many solutions?
I tried shifting the second term to the rhs and squaring.Even after that i'm left with square ... | Hint:
Notice that $$x+3-4\sqrt{x-1}=x-1-4\sqrt{x-1}+4=(\sqrt{x-1}-2)^2$$
and
$$x+8-6\sqrt{x-1}=x-1-6\sqrt{x-1}+9=(\sqrt{x-1}-3)^2=(3-\sqrt{x-1})^2$$
After, you can try by cases.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1503958",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 0
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Can a pre-calculus student prove this?
a and b are rational numbers satisfying the equation $a^3 + 4a^2b = 4a^2 + b^4$
Prove $\sqrt a - 1$ is a rational square
So I saw this posted online somewhere, and I kind of understand what the question is saying. I'm interesting in doing higher order mathematics but don't quite... | As randomgirl and Michael point out, there are counterexamples, such as $a=b=0$. However, it is true when $a \ge 1$. In principle it can be proved with just pre-calculus mathematics, but only an exceptional pre-calculus student could prove it.
Suppose $a^3 + 4a^2b = 4a^2+b^4$, where $a$ and $b$ are rational numbers wi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1505661",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "33",
"answer_count": 8,
"answer_id": 3
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$\epsilon-N$ for $\lim_{n\to\infty} \frac{3n+1}{2n+5}=\frac32$. I want to prove using $\epsilon-N$ that $\lim_{n\to\infty} \frac{3n+1}{2n+5}=\frac32$.
Firstly we will find a sufficently large $n\in \Bbb N$:
\begin{align*}
\epsilon\quad\gt&\quad\left|\frac{3n+1}{2n+5} - \frac32 \right|\\
=&\quad \left|\frac{3n+1 -3n-7.5... | What is the problem?
$\large\mid{3n+1\over2n+5}-{3\over2}\mid={13\over4n+10}$
$\large\mid{3N+1\over2N+5}-{3\over2}\mid={13\over4N+10}$
$\large\mid{3(n+1)+1\over2(n+1)+5}-{3\over2}\mid={13\over4(n+1)+10}$
So for $n>N$ the second one is obviously larger than the first one and the first one is obviously larger than the th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1509081",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
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Sum of $(a+\frac{1}{a})^2$ and $(b+\frac{1}{b})^2$ Prove that:
$$
\left(a+\frac{1}{a}\right)^2+\left(b+\frac{1}{b}\right)^2\ge\frac{25}{2}
$$
if $a,b$ are positive real numbers such that $a+b=1$.
I have tried expanding the squares and rewriting them such that $a+b$ is a term/part of a term but what I get is completely... | Without loss of generality we can choose $a=sin^2x$ and $b=cos^2x$ for $x \in \left(0 \: \frac{\pi}{2}\right)$
Now $$a^2+b^2=sin^4x+cos^4x=1-2sin^2xcos^2x=1-\frac{sin^22x}{2}=\frac{3+cos4x}{4} \ge \frac{1}{2}$$ with Equality at $x=\frac{\pi}{4}$
Also $$2sinxcosx \le 1$$ $\implies$
$$\frac{1}{sin^4xcos^4x} \ge 16$$
No... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1510416",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Complex numbers lies on ellipse $ z $ is a variable complex number such that $ |z|=2$. Show that the point $ z+ \frac{1}{z} $ lies on an ellipse of eccentricity $\frac{4}{5}$ in the complex plane.
| As you know $|z|=2$, you can write $z$ as a function of its argument $$z = 2\cos\varphi + 2i\sin\varphi $$
and then $$\begin{align}\frac 1z & = \frac 12\cos(-\varphi) + \frac i2\sin(-\varphi) \\ & = \frac 12\cos\varphi - \frac i2\sin\varphi\end{align}$$
so $$\begin{align}z+\frac 1z & = \left(2+\frac 12\right)\cos\varph... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1514099",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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a conjecture of two equivalent q-continued fractions related to the reciprocal of the Göllnitz-Gordon continued fraction A111374-OEIS Given the square of the nome $q=e^{2i\pi\tau}$ and ramanujan theta function $f(a,b)=\sum_{k=-\infty}^{\infty}a^{k(k+1)/2}b^{k(k-1)/2}$ with $|q|\lt1$, define,
$$\begin{aligned}M(q)=\cfra... | For $n\in\mathbb{N}$, let $q_n=\exp\frac{2\pi\mathrm{i}\tau}{n}$, so $q_n^n=q$.
Use formula $(***)$ from
that post,
but with $q$ replaced with $q^2$, so it reads
$$\small\cfrac{1}{1-q^2+\cfrac{(a+bq^2)(aq^2+b)}
{1-q^6+\cfrac{q^2(a+bq^4)(aq^4+b)}
{1-q^{10}+\cfrac{q^4(a+bq^6)(aq^6+b)}{1-q^{14}+\cdots}}}}
= \frac{(-a^2q^6... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1514471",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Coefficient of $x^{5}$ in expansion of determinant How to find the coefficient of $x^{5}$ in expansion of determinant $\begin{vmatrix}
7 & 2 & 1 & 3 & x & 7 & 2\\
2 & 8 & 7 & x & 5 & 2 & 8\\
1 & 7 & x & 8 & 4 & 1 & 7\\
3 & x & 8 & 10 & 6 & 3 & 9\\
x & 5 & 4 & 6 & 2 & 10 & x\\
7 & 2 & 1 & 3 & 10 & x & 2\\
2... | Here is one approach which should work:
Let
$$A(X) =\begin{vmatrix}
7 & 2 & 1 & 3 & x & 7 & 2\\
2 & 8 & 7 & x & 5 & 2 & 8\\
1 & 7 & x & 8 & 4 & 1 & 7\\
3 & x & 8 & 10 & 6 & 3 & 9\\
x & 5 & 4 & 6 & 2 & 10 & x\\
7 & 2 & 1 & 3 & 10 & x & 2\\
2 & 8 & 7 & 9 & x & 2 & 8
\end{vmatrix}$$
Then, you are looking for... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1514794",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Number of ordered positive rationals (x,y,z) satisfying following conditions. How many ordered triples $(x,y,z)$ of positive rational numbers satisfy the conditions: $x+y+z$, $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}$, and $xyz$ are all integers.
| Then $yz+xz+xy$ is also an integer. It follows from the Vieta relations that $x$, $y$, and $z$ are the roots of a monic cubic with integer coefficients, so they are integers.
Thus we want all triples $(x,y,z)$ of positive integers such that $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}$ is an integer.
Now it is a short search,... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1516989",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Finding the value of integers of a logarithmic function. Expressing $\frac{2x^2-9x-6}{x(x^2-x-6)}$ in partial fraction would give you: $\frac{16}{5(x+2)}$-$\frac{1}{5(x-3)}$-$\frac{1}{x}$
Given that $\int_{4}^6\frac{2x^2-9x-6}{x(x^2-x-6)}dx=ln\frac{m}{n},$ determine the values for the integers m and n.
| It turns out the partial fraction decomposition is incorrect.
It can be verified that instead,
$$\begin{align}\int_4^6 \frac{2x^2 - 9x - 6}{x(x^2 - x - 6)} dx &= \int_4^6\frac{1}{x} + \frac{2}{x + 2} - \frac{1}{x-3}dx\\
&=\left[\ln x + 2 \ln (x + 2) - \ln (x - 3)\right]_4^6\\
&= \ln\frac{6}{4} + 2\ln \frac{8}{6} - \ln\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1518677",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
how to solve the following mordell equation:$ y^2 = x^3 - 3$ i just started solving mordell's equations and get a little bit stuck. For example:
$ y^2 = x^3 - 3$.
I know that $x$ must be odd, for if $x$ is even $y^2 \equiv 5 \pmod{8}$. So $x \equiv \{1,3\} \pmod{4}$. Now note the following if we add 4 at both sides:
$y... | Yes, I think this is the best way. Keith Conrad has answered this with many examples in his article Examples of Mordell's equation; e.g., compare with Theorem $2.2$ and $2.3$, which is very similar, for $y^2=x^3-5$ and $y^2=x^3-6$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1520668",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Evaluate the Integral: $\int^\pi_0\cos^6\theta\ d\theta$ $\int^\pi_0\cos^6\theta\ d\theta$
So I split the trig value into:
$\int^\pi_o\cos^5\theta\ cos\theta\ d\theta$
Then I utilized the Pythagorean theorem for $cos^5\theta$
$\int^\pi_o(1-sin^5\theta)\ cos\theta$
I utilized u-substitution:
$u=sin\ \theta$
$du=cos\ \th... | Answer (using partial integration):
$$\int\cos^6(x)\space\space\text{d}x=$$
$$\frac{1}{6}\sin(x)\cos^5(x)+\frac{5}{6}\int\cos^4(x)\space\space\text{d}x=$$
$$\frac{5}{24}\sin(x)\cos^3(x)+\frac{1}{6}\sin(x)\cos^5(x)+\frac{5}{8}\int\cos^2(x)\space\space\text{d}x=$$
$$\frac{5}{24}\sin(x)\cos^3(x)+\frac{1}{6}\sin(x)\cos^5(x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1521105",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
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Prove $ \forall x >0, \quad \sqrt{x +2} - \sqrt{x +1} \neq \sqrt{x +1}-\sqrt{x}$
I would like to prove
$$ \forall x >0, \quad \sqrt{x +2} - \sqrt{x +1} \neq \sqrt{x +1}-\sqrt{x}$$
*
*I'm interested in more ways of proving it
My thoughts:
\begin{align}
\sqrt{x+2}-\sqrt{x+1}\neq \sqrt{x+1}-\sqrt{x}\\
\fr... | Hint #1:
Assume that $\sqrt{x + 2} - \sqrt{x + 1} = \sqrt{x + 1} - \sqrt{x}$ for some $x > 0$.
Hint #2:
Derive a contradiction.
Hint #3:
This proof (by contradiction) results to some changes in the notation you used in your proof.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1523179",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 7,
"answer_id": 2
} |
Prove that $\lim\limits_{n\to \infty}\frac{n^2+n-1}{3n^2+1}=\frac{1}{3}$ using the definition
Prove $$\lim_{n\to \infty}\frac{n^2+n-1}{3n^2+1}=\frac{1}{3}$$ using the definition.
Let there be $\varepsilon>0$ we need to find $N<n$ such that $\Big|\frac{n^2+n-1}{3n^2+1}-\frac{1}{3}$$
\Big|<\varepsilon$
$$\left|\frac{n^... | $|\frac{3n-4}{9n^2+3}|<\frac{3n}{9n^2}=\frac{1}{3n}\leq\frac{1}{3N}<\epsilon$ whenever $n\geq N>\frac{1}{3\epsilon}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1525794",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Associative Property What I learn as a basic theory of associative is that $$(a \times b) \times c = a \times (b \times c) \text{ and } (a+b)+c = a+(b+c).$$
However when doing my exercises on this topic, I came across questions with only 2 variables. Eg. $a^2+b^2$. Is this associative and why?
The binary operation is g... | For $a\ast b$ to be associative it should be
$$(a\ast b)\ast c=a\ast (b\ast c),\text{ for all }a,b,c\in \Bbb R,\tag{1}$$
since I assume we are discussing a sum of squares of real numbers. This is, it should be
$$(a^2+b^2)^2+c^2=a^2+(b^2+c^2)^2,\ \forall a,b,c\in \Bbb R.$$
But now it's not difficult to find a counterexa... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1526327",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 5
} |
Convergence of the series $\sum\limits_{n = 4}^\infty {\frac{{n + 1}}{{(n + 5)(n + 4)(n - 3)}}}$? To analyze the convergence of the
$$\sum\limits_{n = 4}^\infty {\frac{{n + 1}}{{(n + 5)(n + 4)(n - 3)}}}$$
series I used the criterion of integral $$\displaystyle\int_4^\infty {\frac{{x + 1}}{{(x + 5)(x + 4)(x - 3)}}dx}... | Assume that the series begins at $n=4$. Then, we have
$$n+5\ge n$$
$$n+4\ge n$$
$$n-3\ge \frac14 n$$
$$n+1\le 2n$$
Therefore, have
$$\frac{n+1}{(n+5)(n+4)(n-3)}\le \frac{2n}{\frac14 n^3}=8\frac1{n^2}$$
Finally, using the result $\sum_{n=1}^\infty \frac1{n^2}=\frac{\pi^2}{6}$ reveals
$$\sum_{n=4}^{\infty}\frac{n+1}{(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1526806",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
Finding limit of $\left(\frac{n^2 + n}{n^2 + n + 2}\right)^n$ Please, help me to find limit of this sequence:
$\lim_{n\to \infty} \left(\frac{n^2 + n}{n^2 + n + 2}\right)^n$
| $$\lim_{n\to\infty}\left(\frac{n^2+n}{n^2+n+2}\right)^n=$$
$$\lim_{n\to\infty}\exp\left(\ln\left(\left(\frac{n^2+n}{n^2+n+2}\right)^n\right)\right)=$$
$$\lim_{n\to\infty}\exp\left(n\ln\left(\frac{n^2+n}{n^2+n+2}\right)\right)=$$
$$\exp\left(\lim_{n\to\infty}n\ln\left(\frac{n^2+n}{n^2+n+2}\right)\right)=$$
$$\exp\left(\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1528471",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Integrating $\frac{1}{(x^2+1)^2}$ How do I integrate $\frac{1}{(x^2+1)^2}$?
I've tried to use the fact that $\int \frac{1}{(x^2+1)}=Arctan(x)$ but I don't know how to. I think it's by parts. Tried using $u'=(x^2+1)^{-1}$ and $v=(x^2+1)^{-1}$ in the formula $\int u'v=vu-\int uv'$, but it doesn't seem to help...
| First consider
$$I = \int \frac{dx}{(x^2+1)^{2}}$$
and the substitution $x=\tan\theta$. This leads to $dx = sec^{2}\theta \, d\theta$ and $x^2 + 1 = sec^{2}\theta$ and
\begin{align}
I &= \int \frac{sec^{2}\theta}{sec^{4}\theta} \, d\theta \\
&= \int \frac{d\theta}{sec^{2}\theta} \\
&= \int \cos^{2}\theta \, d\theta \\
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1529122",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
Finding points of a function's graph that are closest to a given point A question from my calculus book states,
Which points on the graph $y=4-x^2$ are the closest to the point (0,2)?
Using some of my notes, I have a formula as follows (not sure what it's actually called):
$$d=\sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$$
So ... | This is a classic optimization problem where you are trying to minimize the distance from the point $(0, 2)$. Given the function $$f(x) = 4 - x^2$$, we can represent all of the points on that curve as, $(x, 4 - x^2)$. Now, we can use the distance formula: $$d = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$$. When we plug in $(x, 4 -... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1531154",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Finding $\int _0^a\sqrt{1+\frac{1}{4x}}dx$ to calculate arclength So I'm trying to find the arclength of $x^{0.5}$ and its tougher than I thought. Tried substitutions like $\dfrac{\cot^2x}{4}$ and some other trig subs but they got me nowhere. Any tips?
$$\int _0^a\sqrt{1+\frac{1}{4x}}dx$$
Edit:
This is what I got so fa... | You need to make a first substitution:
let u= 1+1/4x
yhen you will have to integrate: sqrt(1+u^2)du which is easy.
at the end you will get a formula with ln, tan inverse, and sec.
Hope it works
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1531672",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
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Solving the recurrence $a_{n+2} = 3a_{n+1} - 2a_n, a_0 = 1, a_1 = 3$ using generating functions
Solve the following recurrence using generating functions: $a_{n+2} = 3a_{n+1} - 2a_n, a_0 = 1, a_1 = 3$.
My partial solution:
We can rewrite $a_{n+2} = 3a_{n+1} - 2a_n$, as $a_{n+2} - 3a_{n+1} + 2a_n = 0$, and we let $A(z... | \begin{align}
\frac{C}{1-z}+\frac{D}{1-2z} &= \frac{C(1-2z)}{(1-z)(1-2z)} + \frac{(1-z)D}{(1-z)(1-2z)} \\\\ &= \frac{(C+D)+(-2C-D)z}{(1-z)(1-2z)} \\\\ &= \frac{(a_0)+(a_1-3a_0)z}{(1-z)(1-2z)}
\end{align}
You now have 2 linear equations and easily solve for $C$ and $D$ in terms of $a_0$ and $a_1$.
What is the power seri... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1532413",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Construct generator matrix given generator polynomial? How would I take a generator polynomial and construct a generator matrix out of it for a cyclic code?
For example, I have a cyclic code in:
$R_{15}=GF(2)[x] / \langle x^{15} + 1\rangle$
This is given by the generator polynomial:
$g(x) = x^8 + x^4 + x^2 + x + 1$
So,... | This is correct. The rows of $G$ correspond to the generator polynomial (expressed as a binary $n-$tuple) with all its (non cyclical) $k$ shifts, which correspond to the "canonical" input messages $u_0=(1,0,0,0 \cdots) \equiv 1$, $u_1=(0,1,0,0 \cdots) \equiv x $, $u_2=(0,0,1,0 \cdots) \equiv x^2 $, and so on.
This is t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1532750",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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Given 3 equally likely outcomes - H, T, and B) - why is P(exactly 1 H in 3 tries) = 0.4444.... and not 0.381? I have a three-sided dice, with three letters: $H$, $T$ and $B$. I want to figure out the probability that, out of 3 rolls, I get exactly 1 H. I figured I'd start by finding the probability that the 3 rolls do ... | These are the probabilities:
*
*$P_0=P(H=\color\red0,T=\color\green0,B=\color\orange3)=\dfrac{3!}{\color\red0!\cdot\color\green0!\cdot\color\orange3!}\cdot\dfrac{1}{3^3}=\dfrac{1}{27}$
*$P_1=P(H=\color\red0,T=\color\green1,B=\color\orange2)=\dfrac{3!}{\color\red0!\cdot\color\green1!\cdot\color\orange2!}\cdot\dfrac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1534579",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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if $a ≡ b\pmod {2n}$ then prove $a^2 ≡ b^2 \pmod {4n}$
Let $n$ be positive number, if $a \equiv b \pmod{2n}$, prove that
$a^2 \equiv b^2 \pmod{4n}$.
By the congruence in hypothesis, we have $a-b = 2nk$ where $k$ is an integer.
Then $a = b+2nk$ and $a^2 = b^2+4n^2k^2+4knb$. From this we get $a^2-b^2 = 4kn(kn+b)$.
No... | Suppose
$a \equiv b \bmod 2kn
$.
($k=1$ in this case.)
Then
$a
=b+2jkn
$
for some $j$,
so
$a^2
=b^2+4bkjn+4j^2k^2n^2
=b^2+4kn(bj+j^2kn)
$
so
$a^2
\equiv b^2 \bmod 4kn
$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1537761",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Evaluating the limit $\mathop {\lim }\limits_{x \to 0} \frac{{x(1 - 0.5\cos x) - 0.5\sin x}}{{{x^3}}}$ For evaluating the limit $\lim\limits_{x \to 0} \frac{x(1 - 0.5\cos x) - 0.5\sin x}{x^3}$, I proceeded as follows:
$$\lim_{x \to 0} \left(\frac{x(1 - 0.5\cos x)}{x^3} - \frac{0.5}{x^2}\left(\frac{\sin x}{x}\right)\rig... | Use Taylor's development at order $3$:
*
*$\cos x= 1-\dfrac{x^2}2+o(x^2)$, hence $\;x\bigl(1-\frac12\cos x\big)=\dfrac x2+\dfrac{x^3}4+o(x^3)$,
*$\sin x=x-\dfrac{x^3}6+o(x^3)$.
Thus the numerator is
$$\dfrac x2+\dfrac{x^3}4-\frac x2+\dfrac{x^3}{12}+o(x^3)=\dfrac{x^3}3+o(x^3)$$
and finally
$$\frac{x\bigl(1-\frac1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1541165",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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What is $\frac{x^{10} + x^8 + x^2 + 1}{x^{10} + x^6 + x^4 + 1}$ given $x^2 + x - 1 = 0$?
Given that $x^2 + x - 1 = 0$, what is $$V \equiv \frac{x^{10} + x^8 + x^2 + 1}{x^{10} + x^6 + x^4 + 1} = \; ?$$
I have reduced $V$ to $\dfrac{x^8 + 1}{(x^4 + 1) (x^4 - x^2 + 1)}$, if you would like to know.
| $$x^2=1-x$$
$$x^4=(1-x)^2=x^2-2x+1=1-x-2x+1=2-3x$$
$$x^6=(1-x)(2-3x)=3x^2-5x+2=3(1-x)-5x+2=5-8x$$
$$x^8=(2-3x)^2=9x^2-12x+4=9(1-x)-12x+4=13-21x$$
$$x^{10}=(1-x)(13-21x)=21x^2-34x+13=21(1-x)-34x+13=34-55x$$
Therefore,
$$x^{10}+x^8+x^2+1=34-55x+13-21x+1-x+1=49-77x$$
$$x^{10}+x^6+x^4+1=34-55x+5-8x+2-3x+1=42-66x$$
Your equ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1542378",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
"answer_id": 0
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sum of solutions of $\{(x,y,z)\mid x+y+z=k\}$, $k = 1,\ldots,N$ What is the sum of non-negative integer solutions of $\{(x,y,z)\mid x+y+z=k\}$, $k = 1,\ldots,N$?
I know that $\{(x,y,z)\mid x+y+z=k\}$ has $\binom{k+3-1}{3-1}=\binom{k+2}{2}$ non-negative integer solutions. Thus, the answer to the question above is $$\su... | The set of solutions of the set of equations
$$x + y + z = k \qquad 1 \leq k \leq N \tag{1}$$
in the non-negative integers is the set of solutions of the inequality
$$x + y + z \leq N \tag{2}$$
in the non-negative integers with the exception of $(0, 0, 0)$. The number of solutions of the inequality in the non-negati... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1543498",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
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Find an integer having the remainders $2,3,4,5$ when divided by $3,4,5,6$, respectively.
Find an integer having the remainders $2,3,4,5$ when divided by $3,4,5,6$, respectively.
My work:
We consider the congruences $x \equiv 2 \pmod 3$, $x \equiv 3 \pmod 4$, $x \equiv 4 \pmod 5$, $x \equiv 5 \pmod 6$. We can reduce t... | I too found the answer to be: x ≡ -1 (mod 60).
My solution:
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1544093",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Finding expected value of random variable You and a friend go to a wine bar. The wine bar offers the following wine styles for ordering:
*
*2 of the wines offered are a Rosé style wine
*3 of the wines offered are a White style wine
*and 5 of the wines offered are a Red style wine
All in all there are 10 unique w... | We first do it more or less in the way you attempted. There are $\binom{10}{7}$ equally likely ways to choose $7$ wines.
First we find the number of ways to choose $0$ Pinks. There is only one way, so $\Pr(Y=0)=\frac{1}{\binom{10}{7}}$. For the future, note that this is $\frac{\binom{3}{0}\binom{7}{7}}{\binom{10}{7}}$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1545271",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Indefinite integral with substitution For my engineering math course I got a couple of exercises about indefinite integrals. I ran trought all of them but stumbled upon the following problem.
$$\int \frac{1-x}{\sqrt{1+x-2x^2}}\,dx $$
We can write $1+x-2x^2$ as $(1-x)(2x+1)$
So I got:
$$
\int \frac{1-x}{\sqrt{1+x-2x^2... | Hint: When you arrived at $~\displaystyle\int\sqrt{\frac{1-x}{2x+1}}~dx,~$ you should have immediately substituted
$\dfrac{1-x}{2x+1}=u^2.~$ Then the entire integrand would have been reduced to a rational function.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1549067",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 3
} |
$c$ is the value of $x^3+3x-14$ where $x=\sqrt[3]{7+5\sqrt2}-\frac{1}{\sqrt[3]{7+5\sqrt2}}$.Find the value of $a+b+c$ $a=\sqrt{57+40\sqrt2}-\sqrt{57-40\sqrt2}$ and $b=\sqrt{25^{\frac{1}{\log_85}}+49^{\frac{1}{\log_67}}}$ and $c$ is the value of $x^3+3x-14$ where $x=\sqrt[3]{7+5\sqrt2}-\frac{1}{\sqrt[3]{7+5\sqrt2}}$.Fin... | (1) $a=(5+4\sqrt{2})-(4\sqrt{2}-5)=10$
(2) $b=\sqrt{25^{\log_5{8}}+49^{\log_7{6}}}=\sqrt{8^2+6^2}=10$
(3) $x=\sqrt[3]{7+5\sqrt2}-\frac{1}{\sqrt[3]{7+5\sqrt2}}=(\sqrt{2}+1)-\frac{1}{\sqrt{2}+1}=(\sqrt{2}+1)-(\sqrt{2}-1)=2$, $c=8+6-14=0$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1552710",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
How to prove $\tanh ^{-1} (\sin \theta)=\cosh^{-1} (\sec \theta)$ As the question says
How to prove
$$\tanh ^{-1} (\sin \theta)=\cosh^{-1} (\sec \theta)$$
I have tried to solve it
The end result that got for RHS
$$=\log \frac{1+\tan\frac{\theta}{2}}{1-\tan \frac{\theta}{2}}$$
I am stuck here Please help
| We know by formula,$$\tanh^{−1}x=\frac{1}{2}\log\left(\frac{1+x}{1-x}\right)$$ and $$\cosh^{-1} x=\log (x+\sqrt{x^2-1})$$
Now putting $x=\sin \theta$ in the formula for $\tanh ^{-1}x$, we have that
$$\tanh^{−1}(\sin \theta)=\frac{1}{2}\log\left(\frac{1+\sin \theta}{1-\sin \theta}\right)$$
$$=\frac{1}{2}\log\left(\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1552782",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
Question on indefinite integrals I have to integrate:
$$I_2 = \int \frac{e^{2x} - e^{x} + 1}{(e^x\cos(x) - \sin(x))\cdot \left(e^x\sin(x) + \cos(x)\right)} \text{d}x$$
I simply can't understand from where to begin with. Please help me in solving this problem.
| I have a better solution to your problem
Here it goes:
\begin{aligned}
& \int \frac{e^{2 x}-e^{x}+1}{\left(e^{x} \sin x+\cos x\right)\left(e^{x} \cos x-\sin x\right)} dx \\
=& \int \frac{e^{2 x}-e^{x}+1}{\left(e^{2 x}+1\right) \sin \left(x+\tan ^{-1} \frac{1}{e^{x}}\right) \cdot \cos \left(x+\tan ^{-1} \frac{1}{e^{x}}\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1552994",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 3
} |
How do I determine if a matrix is in a span of two other matrices? If \begin{align}
A =\begin{bmatrix}
1 & 1 \\
-1 & 1 \\
\end{bmatrix}
\end{align}
and
\begin{align}B = \begin{bmatrix}
1 & -1 \\
1 & 0 \\
\end{bmatrix}
\end{align}
So, how can I check if $C$ is in span $(A,... | Hint: If you can find $κ, λ$ such that $$κ\begin{pmatrix}1 & 1\\-1&1 \end{pmatrix}+λ\begin{pmatrix}1 & -1\\1&0 \end{pmatrix}=\begin{pmatrix}1 & 2\\3&4 \end{pmatrix}$$ then, the answer is yes. You have $4$ equations with $2$ unknowns, the first of the equations being $κ\cdot1+λ\cdot1=1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1554664",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Zero divided by zero must be equal to zero What is wrong with the following argument (if you don't involve ring theory)?
Proposition 1: $\frac{0}{0} = 0$
Proof: Suppose that $\frac{0}{0}$ is not equal to $0$
$\frac{0}{0}$ is not equal to $0 \Rightarrow \frac{0}{0} = x$ , some $x$ not equal to $0$ $\Rightarrow$ $2(\frac... | The problem is that you assume $\frac00$ is = some $x$ belong to the real number group.$\frac00$ is just not defined for you to assume it to be $=x$.
Basically its one of the seven indeterminate forms of mathematics.Look here for the complete list-https://en.wikipedia.org/wiki/Indeterminate_form#List_of_indeterminate_... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1554929",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "52",
"answer_count": 16,
"answer_id": 12
} |
Find the area of the region enclosed by the inner loop of the curve $r = 4 + 8 \sin \theta$ The loop is generated for $\theta \in \left[\frac76\pi, \frac{11}{6}\pi\right]$
(this is from setting $r = 0$).
So, $$A = \int_{\frac76\pi}^{\frac{11}{6}\pi} \frac12(4 + 8 \sin \theta)^2 \,d\theta=\\
= \int_{\frac76\pi}^{\frac{... | As much as Wolfram|Alpha is impressive, here is the answer again using the SageMath free/libre software:
sage: integral((1/2)*(4 + 8 * sin(x))^2, x, 7*pi/6, 11*pi/6).simplify()
16*pi - 24*sqrt(3)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1555516",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Combinatorics : Choosing r elements from a multiset of n elements, given some elements in multiset are identical to each other Given a multiset = {a, a, a, b, b, b, b, c, c}. The multiset has:
number of a's = 3
number of b's = 4
number of c's = 2
In how many ways we can select any 3 elements from this multiset?
My sol... | Here the numbers are so small that it’s not hard simply to write out the terms of degree $3$ and count them:
$$\begin{align*}
&x^3\cdot1\cdot1\\
&x^2\cdot1\cdot x\\
&x^2\cdot x\cdot 1\\
&x\cdot 1\cdot x^2\\
&x\cdot x\cdot x\\
&x\cdot x^2\cdot 1\\
&1\cdot x\cdot x^2\\
&1\cdot x^2\cdot x\\
&1\cdot x^3\cdot 1
\end{align*}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1556194",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Proving of this trigonometric identity $$\frac{\cot \beta}{\csc \beta - 1} + \frac{\cot \beta}{\csc \beta + 1} = 2 \sec \beta$$
What I've done:
$$\frac{\frac{\cos \beta}{\sin \beta}} {\frac{1}{\sin \beta} +1} +
\frac{\frac{\cos \beta}{\sin \beta}} {\frac{1}{\sin \beta} -1}=\\
=\frac{\frac{\cos \beta}{\sin \beta}} {\fra... | The mistake is in the third line of your derivation: the denominators should be $(1-\sin\beta)(1+\sin\beta)$, or $1-\sin^2\beta$, instead of $\sin^2\beta$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1559131",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
If $ c(a+b)\cos \frac{B}{2}=b(a+c)\cos \frac{C}{2}\;,$ prove that it is Isosceles Triangle
In a $\triangle ABC\;,$ If $\displaystyle c(a+b)\cos \frac{B}{2}=b(a+c)\cos \frac{C}{2}\;,$ Then how can we prove that $\triangle ABC$
is an Isoceles $\triangle.$
$\bf{My\; Try::}$ Using $\displaystyle \frac{a}{\sin A}=\frac{b... | $$\sin C\cos\dfrac{A-B}2=\sin\dfrac C2\cdot2\sin\dfrac{A+B}2\cos\dfrac{A-B}2 =\sin\dfrac C2\left(\sin\dfrac A2+\sin\dfrac B2\right)$$
So,
$$\sin C\cdot \cos \left(\frac{A-B}{2}\right)=\sin B\cdot \cos \left(\frac{A-C}{2}\right)$$
$$\iff\sin\dfrac C2\left(\sin\dfrac A2+\sin\dfrac B2\right)=\sin\dfrac B2\left(\sin\dfrac ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1560934",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 3
} |
On the complete solution to $x^2+y^2=z^k$ for odd $k$? While trying to answer this question, I was looking at a computer output of solutions to $x^2+y^2 = z^k$ for odd $k$ and noticed certain patterns. For example, for $k=5$ we have $x,y,z$,
$$10, 55, 5\\25, 50, 5\\38, 41, 5\\117, 598, 13\\122, 597, 13\\338, 507, 13\\7... | If $x^2+y^2=z^k$, then $(x+iy)(x-iy)=z^k$. The Gaussians are a UFD, so if $\gcd(x+iy,x-iy)=1$ then this forces $x+iy=(r+si)^k$ for some integers $r$ and $s$. I think this is your form 1.
If there exists $d$ dividing both $x+iy$ and $x-iy$, then $d$ must divide their sum, $2x$, and their difference, $2iy$. So if $x$ an... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1562228",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Find all incongruent solutions of $x^8\equiv3\pmod{13}$. Find all incongruent solutions of $x^8\equiv3\pmod{13}$.
I know that $2$ is a primitive root of $13$ and that $2^4\equiv3\pmod{13}$, so we want to solve $x^8\equiv2^4\pmod{13}$.
Now, $\gcd(8,\phi(13))=\gcd(8,12)=4$ divides the exponent of $2$, which is $4$, so $x... | We have that $Z/13Z$ it's a field. In $Z/13Z$ the inverse of $3$ is $9$ so if you multiply by $9$ will have $9x^8$ is congruent by 1 modulo 9. In Z/13Z, $x^{12}$ is congruent by 1 modulo 13. So $9x^8-x^{12}$ is congruent by 0 modulo 13. So $x^8(x^4-9)$ is congruent with 0 modulo 13. Result that x is congruent by 0 mod... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1565487",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
Find the value of the following $n \times n$ determinantes Find the value of the following $n \times n$ determinantes
*
*$$\begin{vmatrix}
a_1+x & x & x & \ldots & x \\
x & a_2+x & x & \ldots & x \\
x & x & a_3+x & \ldots & x \\
\vdots & \vdots& &\ddots& \vdots\\
x & x & x & \ldots & a_n+x \\
\end{vmatrix}$$
*$... | Let
\begin{align}
T_n &=
\begin{vmatrix}
a_1+x & x & x & \cdots & x \\
x & a_2+x & x & \cdots & x \\
x & x & a_3+x & \cdots & x \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
x & x & x & \cdots & a_n+x
\end{vmatrix} \\
&=
\begin{vmatrix}
a_1+x & x & x & \cdots & x \\
-a_1 & a_2 & 0 & \cdots & 0 \\
-a_1 & 0 & a_3 & \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1567508",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
prove inequality $2(x+z)^3<(3x+z)(x+3z)$ Let $0<x<1,0<z<1$. Then
$$
2(x+z)^{3}<(3x+z)(x+3z)
$$
This checks out numerically, but I don't know why.
| In the proof I'll use the fact that : $x^3<x^2$, $z^3<z^2$ ,$x^2<x$, $z^2<z$ and $2xz<x^2+z^2$. We have
\begin{align}
2(z+x)^3=2x^3+2z^3+6x^2z+6z^2x&< 2x^2+2z^2+12xz\\ &=2x^2+2z^2+10xz+2xz\\
&<2x^2+2z^2+10xz+z^2+x^2\\
&=(3x+z)(x+3z)
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1569772",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
Find all positive integers $(a,b,c,n)$ such that $2^n=a!+b!+c!$ I have found the solutions by a little calculation $(2,3,5,7)$ and $(2,3,4,5)$. But I don't know if there's any other solutions or not?
| Assume $1\le a\le b\le c$. Then $2^n=a!(1+b\cdots(a+1)+c\cdots(a+1)$.
It follows that either $a=2$ or $a=1$.
Consider the case $a=2$. Then one of the numbers $b\cdots(a+1)$ and $c\cdots(a+1)$ must be even, and the other odd, and since $b\cdots(a+1)$ divides $c\cdots(a+1)$ it follows that $b\cdots(a+1)$ must be odd, an... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1571706",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Find the domain of $\sec^{-1}(\frac{2+x}{1-x})^{\frac{1}{x}}$ Find the domain of $\sec^{-1}(\frac{2+x}{1-x})^{\frac{1}{x}}$
Since the domain of definition of $\sec^{-1} x$ is $x\leq-1$ or $x\geq 1$.
So $(\frac{2+x}{1-x})^{\frac{1}{x}}\leq -1$ or $(\frac{2+x}{1-x})^{\frac{1}{x}}\geq 1$
The domain of $(\frac{2+x}{1-x})^... | $\Longrightarrow$The Range of $(\frac{2+x}{1-x})^{\frac{1}{x}}$ =
[0,$\infty$)
The domain of Sec$^{-1}x$ = $\mathbb{R}$- (-1,1)
$\Longrightarrow$1$\leq$$\left(\frac{2+x}{1-x}\right)^{\frac{1}{x}}$$\Longrightarrow$
x$\geq$ -$\frac{1}{2}$$\Longrightarrow$ the domain of $\sec^{-1}(\frac{2+x}{1-x})^{\frac{1}{x}}$
is [$\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1571797",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Infinite trigonometric summation Can the following summation be written in a finite number of terms:
$$\sum_{r=1}^{\infty}\frac{\tan(\theta/2^r)}{2^{r-1}\cos(\theta/2^{r-1})}$$
I tried to simplify the expression using trigonometric identities and then converting the infinite summation into a definite integral. I couldn... | Let $a_n = \dfrac{\tan(\frac{x}{2^n})}{2^{n-1}\cos(x/2^{n-1})}$. We want to find $\sum_{n=1}^{\infty} a_n$.
Let's write out the first term:
$$a_1 = \frac{\tan(x/2)}{\cos(x)}$$
and invoke the identity $\tan(\frac{x}{2}) = \csc(x)-\cot(x)$:
$$a_1 = \frac{\csc(x)-\cot(x)}{\cos(x)} \\ = \frac{1}{\sin(x)\cos(x)} - \frac{\c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1573701",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Existence of primitive Pythagorean triples Let $(a,b,c)$ be a Pythagorean triple, which means $c^2=a^2+b^2$.
If $c$ is odd and $a$ & $b$ are relatively prime, then there exist integers $m$ and $n$ such that $c=m^2+n^2, ~a=m^2-n^2, ~b=2mn$.
One can easily check the above by proving $\gcd(\frac{a+c}{2},\frac{a-c}{2})=1$.... | $$\text{For any C, if } n=\frac{-1\pm \sqrt{2C-1}}{2}\text { yields an integer for }n, \text{then you have a primitive triple.}$$
$$A=2n^2+1\quad B=2n^2+2n\quad C=2n^2+2n+1 \text{ where }C-B=1$$
$$\text{If }n=\sqrt{\frac{C-1}{4}}\text{ yields an integer for n, you have a primitive triple.}$$
$$A=4n\quad B=4n^2-1\quad C... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1573809",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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About equality of nested radicals. Allow me, please, reformulate this problem.
The equal numbers $$a=\sqrt{13}+\sqrt{10+2\sqrt{13}}$$
$$b=\sqrt{5+2\sqrt3}+\sqrt{18-2\sqrt3+2\sqrt{65-26\sqrt3}}$$ have the same minimal polynomial (over $\mathbb Z$) $$f(x)=x^4-46x^2-104x-43$$ in which it appears just one sign change so t... | The two examples are of the following form:
Take $a,b,c\in \mathbb{R}$ such that $a=b^2-4c$, and suppose that all the square roots below are well defined. Then if
$$\alpha=\sqrt{a}+\sqrt{2b+2\sqrt{a}}$$
and
$$\beta=\sqrt{b+2\sqrt{c}}+\sqrt{a+b-2\sqrt{c}+2\sqrt{ab-2a\sqrt{c}}}$$
we have $\alpha=\beta$.
We have only to ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1575825",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Finding the points on the graph of the function $\frac{1}{3}x^3 + x^2 - x - 1$ at which the slope is $-1$; $2$ Please help me. Find the points on the graph of $\frac{1}{3}x^3 + x^2 - x - 1$ at which the slope is
(a) $-1$; (b) $2$.
I don't know where to begin.
| The slope of the graph is equal to the derivative $$\frac{d}{dx}\left(\frac13x^3+x^2-x-1\right)=x^2+2x-1$$ Now solve for (a) $$x^2+2x-1=-1 \iff x^2+2x=0 \iff x(x+2)=0$$ and for (b) $$x^2+2x-1=2 \iff x^2+2x-3=0 \iff (x-1)(x+3)=0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1576515",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Can you simplify this term? $$X=\frac{\frac{c}{r^2}+\frac{1-c}{(1+r)^{T+1}}}{\frac{c}{r}+\frac{1-c}{(1+r)^T}-1}$$
| $X=\frac{\frac{c}{r^2}+\frac{1-c}{(1+r)^{T+1}}}{\frac{c}{r}+\frac{1-c}{(1+r)^{T}}-1}$
$====================$
$X=\frac{\frac{c}{r^2}+\frac{1-c}{(1+r)^{T+1}}}{\frac{c-r}{r}+\frac{1-c}{(1+r)^{T}}}$
$====================$
$X=\frac{\frac{c(1+r)^{T+1}+(1-c)r^2}{r^2(1+r)^{T+1}}}{\frac{(c-r)(1+r)^{T}+(1-c)r}{r(1+r)^{T}}}$
$===... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1578116",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Circle and a line that passes through it Given a line with equation: $y=ax-3$ that passes through a circle with equation $(x-1)^2+(y-1)^2= 1$. Find the range of values of $a$.
I tried graphing and got: $0<x<2$ and $0<y<2$.
I also tried finding $a$ by substituting $x$ and $y$ into $y=ax-3$
which really confuses me.
Cou... | we substitute
$y=ax-3$
in
$(x-1)^2+(y-1)^2= 1$
$(x-1)^2+(ax-3-1)^2= 1$
$(x-1)^2+(ax-4)^2= 1$
$x^2-2x+1+a^2x^2-8ax+16=1$
we have this second degree equation
$x^2(1+a^2)+x(-2-8a)+16=0$
with
$\Delta =(-2-8a)^2-4\cdot(1+a^2)\cdot 16$
if $\Delta <0$
the equation have no solution and the line not intersect the circle
if $\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1578561",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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Prove $\frac{x}{\sqrt{1+x^2}} \lt \arctan x$ for every $x \gt 0$
Prove $$\frac{x}{\sqrt{1+x^2}} \lt \arctan x$$ for every $x \gt 0$.
I proved half of it with lagrange rule but that I got stuck. Any ideas?
I can upload my work if you want.
| let be
$\displaystyle f:[0,\infty) \rightarrow \mathbb{R} $
$f(x)=\frac{x}{\sqrt{1+x^2}} - \arctan x$
$\displaystyle f'(x)=\frac{x' \cdot \sqrt{1+x^2} - x \cdot (\sqrt{1+x^2})'}{(\sqrt{1+x^2})^2} -(\arctan x)'$
$\displaystyle f'(x)=\frac{1 \cdot \sqrt{1+x^2} - x \cdot \frac{1}{2 \sqrt{1+x^2}}\cdot 2 \cdot x}{(\sqrt{1+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1578798",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 3
} |
How can one show the inequality Let $a,b,n$ be natural numbers (in $\mathbb{N}^*$) such that
$a>b$ and $n^2+1=ab$
How can one show that $a-b\geq\sqrt{4n-3}$, and for what values of $n$ equality holds?
I tried this:
We suppose that $a-b\geq\sqrt{4n-3}$ so
$$a^2-2ab+b^2\geq 4n-3\,.$$
And we have $ab=n^2+1$.
So $a^2-2(n^2... | Let $x:=a-n$ and $y:=b+x-n$ (that is, $a=n+x$ and $b=n-x+y$). It can be easily seen that $x$ and $y$ are positive integers. We want to find $a-b=2x-y$. Note that $n^2+1=ab$ implies that $x^2-xy=ny-1$, or equivalently, $(2x-y)^2=y^2+4(ny-1)$. As $y\geq 1$, we have $$(a-b)^2=(2x-y)^2\geq 1+4(n-1)=4n-3\,$$ Therefore,... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1581101",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
I would like to calculate $\lim_{x \to \frac{\pi}{6}} \frac{2 \sin^2{x}+\sin{x}-1}{2 \sin^2{x}-3 \sin{x}+1}$ I want to calculate the following limit: $$\lim_{x \to \frac{\pi}{6}} \frac{2 \sin^2{x}+\sin{x}-1}{2 \sin^2{x}-3 \sin{x}+1}$$
or prove that it does not exist. Now I know the result is $-3$, but I am having troub... | Notice, $$\lim_{x\to \pi/6}\frac{2\sin^2 x+\sin x-1}{2\sin^2 x-3\sin x+1}$$
$$=\lim_{x\to \pi/6}\frac{\sin x-(1-2\sin^2 x)}{2-3\sin x-(1-2\sin^2 x)}$$
$$=\lim_{x\to \pi/6}\frac{\sin x-\cos 2x}{2-3\sin x-\cos 2x}$$
Apply L'hospital's rule for $\frac00$ form
$$=\lim_{x\to \pi/6}\frac{\cos x+2\sin 2x}{-3\cos x+2\sin2x}$$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1582062",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
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If $\arcsin x+\arcsin y+\arcsin z=\pi$,then prove that $(x,y,z>0)x\sqrt{1-x^2}+y\sqrt{1-y^2}+z\sqrt{1-z^2}=2xyz$ If $\arcsin x+\arcsin y+\arcsin z=\pi$,then prove that $(x,y,z>0)$
$x\sqrt{1-x^2}+y\sqrt{1-y^2}+z\sqrt{1-z^2}=2xyz$
$\arcsin x+\arcsin y+\arcsin z=\pi$,
$\arcsin x+\arcsin y=\pi-\arcsin z$
$\arcsin(x\sqrt{1... | Let $\sin^{-1}(x) = A\Rightarrow x=\sin A$ and $\sin^{-1}(y)=B\Rightarrow y=\sin (B)$
and $\sin^{-1}(z)=C\Rightarrow z=\sin C$
So above we have given $A+B+C = \pi$
Now we have to prove $\sin 2A+\sin 2B+\sin 2C=4\sin A\sin B\sin C$
So for $\bf{L.H.S}$
$$\sin 2A+\sin 2B+\sin 2C = 2\sin (A+B)\cdot \sin (A-B)+2\sin C\cos C... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1585113",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
If $(-1) \cdot (-1) = +1$ shouldn't $(+1) \cdot (+1) = -1$? The common multiplication rules are
$$(-1) \cdot (-1) = +1 \\
(+1) \cdot (+1) = +1$$
But these rules seem asymmetric. Because of these rules it is not possible e.g. to solve the equation
$$x^2 = -1 $$
without introducing complex numbers. There would be two "... | Math would certainly break down, here's why. Assume that every number is equal to itself, addition is associative, multiplication distributes over addition , $1$ and $-1$ are additive inverses, $1$ is the multiplicative identity, and $0$ is the additive identity. Then it follows that:
$$\begin{array}{lll}
\bigg(1\cdot ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1586477",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 4
} |
Prove This Inequality ${\pi \over 2} \le \sum_{n=0}^{\infty} {1 \over {1+n^2}} \le {\pi \over 2} + 1$ $${\pi \over 2} \le \sum_{n=0}^{\infty} {1 \over {1+n^2}} \le {\pi \over 2} + 1$$
I see I should use Riemann sum, and that
$$\int_0^{\infty} {dx \over {1+x^2}} \le \sum_{n=0}^{\infty} {1 \over {1+n^2}}$$
But how do I e... | Hint. For each $n=0,1,2,\cdots$, you have, for all $x \in [n,n+1]$,
$$
n^2+1\leq 1+x^2\leq (n+1)^2+1
$$ giving
$$
\frac1{(n+1)^2+1}\leq \frac1{x^2+1}\leq \frac1{n^2+1}
$$ then integrating with respect to $x$ from $n$ to $n+1$,
$$
\int_n^{n+1}\frac1{(n+1)^2+1}\:dx\leq \int_n^{n+1}\frac1{x^2+1}\:dx\leq \int_n^{n+1}\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1588542",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 1
} |
Prove $1^2-2^2+3^2-4^2+......+(-1)^{k-1}k^2 = (-1)^{k-1}\cdot \frac{k(k+1)}{2}$ I'm trying to solve this problem from Skiena book, "Algorithm design manual".
I don't know the answer but it seems like the entity on the R.H.S is the summation for series $1+2+3+..$. However the sequence on left hand side is squared series... | Hint: In order to show
\begin{align*}
\sum_{j=0}^k(-1)^{j-1}j^2=(-1)^{k-1}\frac{k(k+1)}{2}\qquad\qquad k\geq 0\tag{1}
\end{align*}
we consider sequences $(a_k)_{k\geq 0}$ and the corresponding generating functions $\sum_{k=0}^{\infty}a_kx^k$ as building blocks to generate the left hand sum in (1). This enables us to ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1588818",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 7,
"answer_id": 2
} |
Remainder of the numerator of a harmonic sum modulo 13 Let $a$ be the integer determined by
$$\frac{1}{1}+\frac{1}{2}+...+\frac{1}{23}=\frac{a}{23!}.$$
Determine the remainder of $a$ when divided by 13.
Can anyone help me with this, or just give me any hint?
| HINT:
Take the equation:
$\frac{1}{1}+\frac{1}{2}+...+\frac{1}{23}=\frac{a}{23!}$
Multiply each side by $23!$:
$\frac{23!}{1}+\frac{23!}{2}+...+\frac{23!}{23}=a$
Divide each side by $13$:
$\frac{23!}{1\times13}+\frac{23!}{2\times13}+...+\frac{23!}{23\times13}=\frac{a}{13}$
There is only one non-integer element in the... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1589591",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 1
} |
Prove that given $a,b,c > 0$, it is possible to construct a triangle with sides of length $a,b,c$ if and only if $pa^2+qb^2 > pqc^2$
Prove that given $a,b,c > 0$, it is possible to construct a triangle with sides of length $a,b,c$ if and only if $pa^2+qb^2 > pqc^2$ for any $p,q$ with $p+q = 1$.
Should I prove this us... | Below is a complete solution. Don't look at it if you want to solve this problem by yourself. Also, yes, the Triangle Inequality is definitely required in this problem.
Assume that $a,b,c>0$ satisfy the condition that $pa^2+qb^2>pqc^2$ for any $p,q \in\mathbb{R}$ such that $p+q=1$. Choosing $p$ and $q$ to be $\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1589759",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Calculation of $\max$ and $\min$ value of $f(x) = \frac{x(x^2-1)}{x^4-x^2+1}.$
Calculation of $\max$ and $\min$ value of $$f(x) = \frac{x(x^2-1)}{x^4-x^2+1}$$
My try: We can write $$f(x) = \frac{\left(x-\frac{1}{x}\right)}{\left(x^2+\frac{1}{x^2}\right)-1} = \frac{\left(x-\frac{1}{x}\right)}{\left(x-\frac{1}{x}\right... | take you solution,
$$f(t)=\dfrac{t}{t^2+1}$$
since
$$f(t)=-f(-t)$$
so we only consider $t\ge 0$ Use AM-GM inequality we have
$$f(t)=\dfrac{t}{t^2+1}\le\dfrac{t}{2t}=\dfrac{1}{2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1590463",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
Volume Integral of this set I'm not sure about this exercise. Be:
$$E=\left\{(x,y,z) \in \mathbb{R^3} : x\geq 0, y\geq 0, 0\leq z\leq \frac{1}{\sqrt{x^2+y^2}}-1\right\}$$
Find:
$$\int_{E} z\, \max\{x,y\}\: dx \, dy \, dz$$
My idea is that since $x,y\geq 0$, one of the two variables will be greater than the other in a ... | Partial answer:
The upper limit of integration for the $x$ variable doesn't actually go to infinity. Note that the bounds for $z$ imply an additional constraint on the range of $x$ and $y$:
$$\begin{align}
0\le z\le\frac{1}{\sqrt{x^2+y^2}}-1
&\implies0\le\frac{1}{\sqrt{x^2+y^2}}-1\\
&\implies1\le\frac{1}{\sqrt{x^2+y^2}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1592065",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Uniform convergence of $ \sum_{n=1}^\infty \frac{x\cdot\sin\sqrt {x/n}} {n + x} $ I need to check on the uniform convergence.
$$ \sum_{n=1}^\infty \frac{x\cdot\sin\sqrt {\frac x n}} {n + x} $$
on the interval:
a) $\; (0, 1) $
b) $\; (1, +\infty) $
I think I need to use Cauchy ratio, but I cannot understand how
| For part $(a)$:
$$\begin{align}
\left|\sum _{n=1}^\infty \frac{x\sin {\sqrt {\frac{x}{n}}}}{n+x}\right| &\le \sum _{n=1}^\infty \left|\frac{x\sin {\sqrt {\frac{x}{n}}}}{n+x}\right|\\\\
&\le \sum_{n=1}^\infty \frac{\left|x\,\sqrt{\frac{x}{n}}\right|}{n+x}\,\,\text{since} |\sin x|\le |x|\\\\
&\le \sum_{n=1}^\infty \left... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1592791",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Calculate $\int_0^{1}\frac{x^{-1 - x}\,\,\,\left(1 - x\right)^{x - 2}}{\mathrm{B}(1 - x\,, \,x)}\,\mathrm{d}x$ How does one calculate
$\displaystyle{\int_0^{1}\frac{x^{-1 - x}\,\,\,\left(1 - x\right)^{x - 2}}
{\mathrm{B}(1 - x\,,\,x)}\,\mathrm{d}x}$ ?.
The observation
$\displaystyle{\int_0^{1}\frac{x^{-1 - x}\,\,\,\lef... | For the proof, we will be using the following fact $$(1-\alpha)^{\left(\frac{1}{\alpha}-1\right)} = \int_0^1 \frac{\sin\left(\pi x \right)}{\pi (1-\alpha x)} \left[x^x\left(1-x\right)^{1-x}\right]^{-1}\ dx.$$
Taking your original sum, we have
\begin{align}
\int_0^1 \frac{x^{-1-x}(1-x)^{x-2}}{\Gamma(x)\Gamma(1-x)} dx &... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1593476",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 3,
"answer_id": 2
} |
Finding conditional expectation on the basis of conditional probabilities Knowing that
*
*$P(Y=1\mid X=5)=1/3$,
*$P(Y=5\mid X=5)=2/3$.
Calculate $E(Y\mid X=5)$ and $E(XY^2\mid X=5)$. How to solve this question? I have no idea whatsoever.
| Since the conditional probabilities sum up to $1$ then that is all for $Y\mid X=5$. So $$E[Y\mid X=5]=1\cdot P(Y=1\mid X=5)+5\cdot P(Y=5\mid X=5)=1\cdot\frac13+5\cdot\frac23=\frac{11}3$$ and
\begin{align}E[XY^2\mid X=5]&=E[5Y^2\mid X=5]=5E[Y^2\mid X=5]\\[0.2cm]&=5\left((1)^2\cdot P(Y=1\mid X=5)+(5)^2\cdot P(Y=5\mid X=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1593717",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Problem in a Pell equation If you take the continued fraction out of $\sqrt7$, you get [2;1, 1, 1, 4...] which yields:
$$2+\frac1{1+\frac1{\frac11+1}}=\frac8{3}$$
and indeed, $$8^2-7⋅3^2=1$$
However, if you take the continued fraction out of $\sqrt13$ you get [3;1, 1, 1, 1, 6...] which yields: $$3+\frac1{1+\frac1{1+\fr... | If you take a prime $p \equiv 1 \pmod 4$ you are guaranteed to get $-1$ first, but get back to $1$ if you repeat the periodic part of the CF.
Note $$18^2 + 13 \cdot 5^2 = 649,$$
$$ 2 \cdot 18 \cdot 5 = 180, $$
$$ 649^2 - 13 \cdot 180^2 = 1. $$
The continued fraction for $\sqrt {13}$ in the display I like. For a conver... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1594284",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Solve integral $\int{\frac{x^2 + 4}{x^2 + 6x +10}dx}$ Please help me with this integral:
$$\int{\frac{x^2 + 4}{x^2 + 6x +10}}\,dx .$$
I know I must solve it by substitution, but I don't know how exactly.
| Hint:
$$\int{\frac{(x^2 + 4)dx}{x^2 + 6x +10}}=\int{\frac{x^2 + 6x + 10-6x-6}{x^2 + 6x +10}}dx=x-6\int{\frac{x+1}{x^2 + 6x +10}}dx$$
Next hint:
$$6\frac{x+1}{x^2 + 6x +10}=3\frac{\color{blue}{2x+6}-4}{x^2 + 6x +10}$$
Next hint:
$$x^2 + 6x +10 = (x+3)^2+1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1594342",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 0
} |
Solving functional equation $f(4x)-f(3x)=2x$ Given that $f(4x)-f(3x)=2x$ and that $f:\mathbb{R}\rightarrow\mathbb{R}$ is an increasing function, find $f(x)$. My thoughts so far: subtituting $\frac{3}{4}x$, $\left(\frac{3}{4}\right)^2x$, $\left(\frac{3}{4}\right)^3x$, $\ldots$, we get that:
$$f(4x)-f(3x)=2x$$
$$ f\left(... | It's almost correct. You are correct that we can deduce that
$$f(4x)=f(3x)+2x$$
and by repeatedly applying this we get
$$f(4x)=f\left(4\left(\frac{3}4\right)^kx\right)+\sum_{n=0}^{k-1}2\left(\frac{3}4\right)^{n}x$$
You make an error on the next step, however. You mean to take a limit as $k$ goes to infinity, but you to... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1594409",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 1,
"answer_id": 0
} |
Use the identity $\cos 3\theta = 4 \cos^3\theta- 3 \cos \theta$ to solve the cubic equation $t^3 + pt + q = 0$ when $p, q \in \mathbb{R}$. I'm self studying Ian Stewart's Galois Theory and this is Exercise 1.8 from his Third Edition:
Use the identity $\cos 3\theta = 4 \cos^3\theta- 3 \cos \theta$ to
solve the cubic... | Note that, since $27q^2+4p^3<0$ and since $27q^2\geqslant0$, $p<0$. So, it makes sense to define $u=\sqrt{-\frac43p}$. Consider the substitution $t=u\cos\theta$. Then $t^3+pt+q$ becomes $u^3\cos^3\theta+pu\cos\theta+q$ and\begin{align}u^3\cos^3\theta+pu\cos\theta+q=0&\iff\frac{u^3\cos^3\theta+pu\cos\theta+q}{u^3/4}=0\\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1594680",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 1,
"answer_id": 0
} |
Is there a relationship between Rotors and the Rodrigues' rotation formula I am trying to understand quaternion in general, and it seems like the path to making sense of how they actually work is to first understand rotors and other techniques related to rotations. By looking at the equations and just reading briefly a... | We could "derive" the Rodrigues formula by "vectorizing" the quaternion sandwitch product. First, let us recall the sadwitch product:
$v' = Q \ v \ Q^*$
Where $v$ is a pure quaternion (its real part equal to zero) and $Q$ is a unit quaternion and $Q^*$ is its conjugate:
$Q^{*} = q_0 - i q_1 - j q_2 - k q_3$
Aplying th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1594957",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Obtaining an estimate for the Lagrangian $L=H^*$ from the Hamiltonian $H$ This is from C. Evans' PDE book, page 130. The convex function $H:\mathbb{R}^n\to\mathbb{R}$ is $C^2$ and satisfies $$ H\big(\frac{p_1+p_2}{2}\big) \leq \frac{1}{2}H(p_1) + \frac{1}{2}H(p_2) - \frac{\theta}{8}|p_1-p_2|^2. $$ The lagrangian $L$ is... | As your hint says, for every $q_1, q_2 \in \mathbb R^n$ there exist $p_1, p_2 \in \mathbb R^n$ such that $L(q_j) = p_j \cdot q_j - H(p_j)$ for $j \in \{1,2\}$. Combining with the inequality for $H$,
\begin{align}
\frac{1}{2} L(q_1) + \frac{1}{2} L(q_2)
&= \frac{p_1 \cdot q_1 + p_2 \cdot q_2}{2} - \frac{1}{2} H(p_1) - \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1596497",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Proof by induction that $5|11^n-6$ for all positive integers $n$
Prove by induction that $5|11^n-6$ for all positive integers $n$
Let $p(n) = 11^n-6.$ We have $p(1) = 5$, thus it holds for $p(1)$. Assume it holds for $p(k)$. We will prove that it's true for $p(k+1)$. We have $p(k+1) = 11^{k+1}-6$. But $6 = 11^k-f(k).... | Suppose $5\mid 11^n-6\implies 11^n-6=5k$ for some $k\in\mathbb{Z}$.
\begin{align*}
11(11^n-6)&=11\cdot5k \\
11^{n+1}-11\cdot 6+60&=11\cdot 5k+60 \\
11^{n+1}-6&=11\cdot 5k+60 \\
11^{n+1}-6&=5(11k+12)
\end{align*}
$$11^{n+1}-6=5(11k+12)\implies 5\mid 11^{n+1}-6$$
So we have shown
$$5\mid 11^n-6\implies 5\mid 11^{n+1}-6$$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1596657",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
solving $\int \frac{dx}{\sqrt{-x^2-12x+28}}$
$$\int \frac{dx}{\sqrt{-x^2-12x+28}}$$
First we need to use completing the square $-(x^2+12x-28)=-(x+6)^2+64$
So we have $\int \frac{dx}{\sqrt{-(x+6)^2+64}}$ I know that it is a general form of $\arcsin(\frac{x+6}{8})$ but how can I solve it using substitution?
| Its not really necessary to go for substitution. Use basic differentiation of inverse trig functions.
$\begin{align}
\int\frac{dx}{\sqrt{a^2-x^2}}&=
\sin^{-1}(\frac{x}{a})+C\\
&=\int\frac{dx}{\sqrt{64-(x+6)^2}}\\
&=\sin^{-1}\frac{(x+6)}{8}+C
\end{align}
$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1599953",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Has this equation appeared before? I want to know if the following equation has appeared in mathematical literature before, or if it has any important significance.
$$\sqrt{\frac{a+b+x}{c}}+\sqrt{\frac{b+c+x}{a}}+\sqrt{\frac{c+a+x}{b}}=\sqrt{\frac{a+b+c}{x}},$$
where $a,b,c$ are any three fixed positive real and $x$ is... | I edit this answer to explicit the degree of the resulting polynomial having $x$ as a root (I wonder about the exact origin of this strange unknown x. Explanation is tedious and it will be presented abbreviated as possible, just to have the degree of the polynomial without using algebraic number theory which is not so ... | {
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"url": "https://math.stackexchange.com/questions/1600124",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
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prove that $(x-y)^3+(y-z)^3+(z-x)^3 > 0.$
If $x<y<z$ are real numbers, prove that $$(x-y)^3+(y-z)^3+(z-x)^3 > 0.$$
Attempt
We have that $(x-y)^3+(y-z)^3+(z-x)^3=-3 x^2 y+3 x^2 z+3 x y^2-3 x z^2-3 y^2 z+3 y z^2$. Grouping yields $3(x^2z+xy^2+yz^2)-3(x^2y+xz^2+y^2z)$. Then I was thinking of using rearrangement but then... | Hint, $(z-x)^3=\left((z-y)+(y-x)\right)^3$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1601604",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 5,
"answer_id": 4
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Prove that $\frac{a}{\sqrt{a^2+1}}+\frac{b}{\sqrt{b^2+1}}+\frac{c}{\sqrt{c^2+1}} \leq \frac{3}{2}$
Let $a,b,$ and $c$ be positive real numbers with $ab+bc+ca = 1$. Prove that $$\dfrac{a}{\sqrt{a^2+1}}+\dfrac{b}{\sqrt{b^2+1}}+\dfrac{c}{\sqrt{c^2+1}} \leq \dfrac{3}{2}$$
Attempt
The $ab+bc+ca = 1$ condition reminds of ... | Set $a=\cot A$ etc.
$\implies\sum\cot A\cot B=1$
$\iff\tan A+\tan B+\tan C=\tan A\tan B\tan C$
$\implies A+B+C=n\pi$ where $n$ is any integer
Now use In $ \triangle ABC$ show that $ 1 \lt \cos A + \cos B + \cos C \le \frac 32$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1603415",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
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System and triangle If $a,b$ and $c$ are the lengths of sides of the triangle, find $x,y$ and $z$ such that
$$\begin{array}{l}
x^2y^2+x^2z^2=axyz\\
y^2x^2+y^2z^2=bxyz\\
z^2x^2+z^2y^2=cxyz
\end{array}$$
I see that $x=y=z=0$ is a solution. Moreover, $x=y=0$ and $z \in \mathbb{R}$ is solution, or $x=z=0$ and $y \in \mat... | If $a, b, c$ are the side length of a triangle then
$$
u = \frac 12(b+c-a) \, , \quad v = \frac 12(c+a-b) \, ,\quad w = \frac 12(a+b-c)
$$
are positive real numbers such that
$$
a = v + w \, , \quad b = w + u \, ,\quad c = u + v \, .
$$
(This is the so-called "Ravi substitution", see for example
Ravi Substitutio... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to integrate $\int (\tan x)^{1/ 6} \,\text{d}x$? How do I compute the following integral
$$
I=\int (\tan x)^{1/ 6} \,\text{d}x
$$
| HINT (the integral will become very very big):
$$\int\left(\tan(x)\right)^{\frac{1}{6}}\space\text{d}x=\int\sqrt[6]{\tan(x)}\space\text{d}x=$$
Substitute $u=\tan(x)$ and $\frac{\text{d}u}{\text{d}x}=\sec^2(x)$:
$$\int\frac{\sqrt[6]{u}}{1+u^2}\space\text{d}u=$$
Substitute $s=\sqrt[6]{u}$ and $\frac{\text{d}s}{\text{d... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1609654",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 2
} |
In $\triangle ABC,a,b,c$ are the sides of triangle satisfying $a^4+b^4+c^4-4abc+1=0$.Find $\frac{a^2+b^2+c^2}{S}$ In $\triangle ABC,a,b,c$ are the sides of triangle satisfying $a^4+b^4+c^4-4abc+1=0$
Find the value of $\frac{a^2+b^2+c^2}{S}$,where $S$ is area of the triangle $ABC$and find the value of $1+\frac{R}{r}$ wh... | HINT:
Using AM-GM inequality
$$\dfrac{a^4+b^4+c^4+1}4\ge\sqrt[4]{a^4b^4c^4\cdot1}$$
Can you take it from here?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1612232",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Prove that $ax+by+cz+2\sqrt{(xy+yz+xz)(ab+bc+ca)}\le{a+b+c}$ Let $a,b,c,x,y,z$ be positive real numbers such that $x+y+z=1$. Prove that
$$ax+by+cz+2\sqrt{(xy+yz+xz)(ab+bc+ca)}\le{a+b+c}$$.
my try:
$2\sqrt{(xy+yz+xz)(ab+bc+ca)}\le{\frac{2(a+b+c)}{3}}$
But this is not the right choice because
$ax+by+cz\le{\frac{a+b+c}... | WLOG: $$a+ b+ c= 1$$
AM-GM:
$$ax+ by+ cz+ 2\sqrt{\left ( xy+ yz+ zx \right )\left ( ab+ bc+ ca \right )}$$
$$\leq ax+ by+ cz+ xy+ yz+ zx+ ab+ bc+ ca$$
$$\Leftarrow xy+ yz+ zx+ ab+ bc+ ca= \frac{1- x^{2}- y^{2}- z^{2}}{2}+ \frac{1- a^{2}- b^{2}- c^{2}}{2}$$
$$\leq 1- ax- by- cz$$
$$\Leftrightarrow \left ( x- a \rig... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1613047",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
Finding Laurent's series of a function I am trying express the function $$f(z)=\frac{z^3+2}{(z-1)(z-2)}$$ like a Laurent's series in each ring centering in $0$, but I do not now how could I express it, in first I said that $$f(z)=(z^3+2)\left[\frac{1}{z-2}-\frac{1}{z-1} \right] $$
Ok, now, I see two posibilities: $A\eq... | Executing the polynomial division we have
$$
f(z)=\frac{z^2+3}{(z-1)(z-2)}=1+\frac{3z+1}{(z-1)(z-2)},
$$
and with partial fractions
$$
f(z)=\frac{z^2+3}{(z-1)(z-2)}=1-\frac{4}{z-1}+\frac{7}{z-2}.
$$
This can be rewritten as
$$
f(z)=1+4\cdot\frac{1}{1-z}-\frac{7}{2}\cdot\frac{1}{1-\frac{z}{2}}.
$$
Using the geometric se... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1614506",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Calculate $\lim_{x \to 0} \frac{\sin(\sin x)-x(1-x^2)^\frac{1}{3}}{x^5}$ by Taylor's theorem. I have to calculate : $$\lim_{x \to 0} \frac{\sin(\sin x)-x(1-x^2)^\frac{1}{3}}{x^5}$$ by using Taylor's theorem.
I know that :$$\sin(x) = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \ldots$$
But I don't know how ... | Expand $\sin(\sin x)$ similar to $\sin x=x-\frac{x^3}{3!}+\frac{x^5}{5!}-...$, i.e. $$\sin(\sin x)=\sin x-\frac{\sin^3x}3!+\frac{\sin^5 x}{5!}-...$$ near $x=0$.
And $(1-x^2)^{1/3}$ similar to $(1+x)^n=1+nx+\frac{(n-1)n}{2!}x^2+\frac{(n-2)(n-1)n}{3!}x^3+...$
Now, expand each $\sin x$ using taylor then forming combinatio... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1614626",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
If $(x,y)$ satisfies $x^2+y^2-4x+2y+1=0$ then the expression $x^2+y^2-10x-6y+34$ CANNOT be equal to If $(x,y)$ satisfies $x^2+y^2-4x+2y+1=0$ then the expression $x^2+y^2-10x-6y+34$ CANNOT be equal to
$(A)\frac{1}{2}\hspace{1cm}(B)8\hspace{1cm}(C)2\hspace{1cm}(D)3$
$(x,y)$ satisfies $x^2+y^2-4x+2y+1=0$
$(x,y)$ satisfi... | As you note, the restriction is
$$(x-2)^2+(y+1)^2=2^2$$
That means the distance from point $P(x,y)$ to the point $A(2,-1)$, $AP$, is $2$.
As you also note, the function is
$$f(x,y)=(x-5)^2+(y-3)^2$$
which is the square of the distance of point $P(x,y)$ to point $B(5,3)$, $PB$. Note that the distance from point $A(2,-1)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1617077",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 1
} |
Not getting the same solution when using the rule sin(x)\x=1 on a limit There is a rule in limits that when $x$ approaches zero:
$$\frac{\sin\left(x\right)}{x}=1$$
So I used this rule on the following exercise:
Evaluate
$$
\lim _{x\to 0}\:\frac{x-\sin\left(x\right)}{\sin\left(2x\right)-\tan\left(2x\right)}
$$
I... | The problem you encountered was that both $\sin(x)-\tan(x)=O\!\left(x^3\right)$ and $x-\sin(x)=O\!\left(x^3\right)$, this means that substituting $x$ for $\sin(x)$ may substantively change the limit.
To be precise,
$$
\begin{align}
\sin(x)-\tan(x)
&=\left(x-\frac{x^3}6+O\!\left(x^5\right)\right)-\left(x+\frac{x^3}3+O\!... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1618711",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 2
} |
Find the minimum value of $\sin^{2} \theta +\cos^{2} \theta+\sec^{2} \theta+\csc^{2} \theta+\tan^{2} \theta+\cot^{2} \theta$
Find the minimum value of
$\sin^{2} \theta +\cos^{2} \theta+\sec^{2} \theta+\csc^{2} \theta+\tan^{2} \theta+\cot^{2} \theta$
$a.)\ 1 \ \ \ \ \ \ \ \ \ \ \ \ b.)\ 3 \\
c.)\ 5 \ \ \ \ \ \ \ \ \ \... | Just using algebra$$y=\sin^{2} (\theta) +\cos^{2} (\theta)+\sec^{2} (\theta)+\csc^{2} (\theta)+\tan^{2} (\theta)+\cot^{2} (\theta)=2 \csc ^2(\theta )+2 \sec ^2(\theta )-1$$ Taking derivatives and simplifying $$y'=4 \tan (\theta) \sec ^2(\theta)-4 \cot (\theta) \csc ^2(\theta)=-32 \cot (2 \theta) \csc ^2(2 \theta)$$ $$y... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1620239",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 6,
"answer_id": 4
} |
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