Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Sum of first $n$ triangle numbers, without induction Background
I wish to calculate $$ S= \sum_{i = 1}^{n}\frac{k(k+1)}{2}$$
I know what the answer is going to be, since this is essentially the sum of the first $n$ triangle numbers.
I.e. $S = (1) + (1+2) + (1+2+3) + \cdots + (1+2+3+\cdots+n)$
All solutions I've seen se... | Another way to deal with your sum, and in general
with the sum of the binomial coefficient in the upper index,
is to consider that
$$
\Delta _{\,n} \left( \begin{gathered}
n \\
m \\
\end{gathered} \right) = \left( \begin{gathered}
n + 1 \\
m \\
\end{gathered} \right) - \left( \begin{gathered}
n \\
m ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1906224",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 4
} |
Prime numbers of the form $(1\times11\times111\times1111\times...)-(1+11+111+1111+...)$ Let
$$R(1) = 1-1,$$
$$R(2) = (1\times11) - (1+11),$$
$$R(3) = (1\times11\times111) - (1+11+111),$$ and so on...
$$R(4)=1355297\quad\text{(a prime number!)}$$
$R(4)$ is the only prime I found of such form up to $R(200)$. Are there an... | *
*Just a very basic observation, not a complete answer. Let's note by
$$a_n=1111...1$$
with $n$ of $1's$. So we have
$$a_n \equiv 1 \pmod{10}$$
And
$$\prod_{k=1}^{n} a_k \equiv 1 \pmod{10}$$
And
$$\sum_{k=1}^{n} a_k \equiv n \pmod{10}$$
And
$$R(n)=\prod_{k=1}^{n} a_k - \sum_{k=1}^{n} a_k \equiv 1 - n \pmod{10}$$
So, ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1907229",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 4,
"answer_id": 2
} |
Find the value of constants $c_1, c_2, c_3, c_4$ for which function $f: \mathbb{R} \rightarrow \mathbb{R}$ is differentiable The Problem:
Find the value of constants $c_1, c_2, c_3, c_4$ for which function $f: \mathbb{R} \rightarrow \mathbb{R} $ (written below) is differentiable:
$$
f(x) = \begin{cases}
c_1 \arctan... | $\lim_\limits{x\to 1^+} f'(x)$
We don't need to find the exact derivative of $c_3(x−1)(x−2)⋯(x−2016)+c_4(x+1)$
When we differentiate we get $ c_3 (x-2)\cdots(x-2016) + c_3(x-1) [(x-3)\cdots(x-2016) + (x-2)(x-4)\cdots]+c_4$
And as $x$ approaches $1, c_3(x-1)[\text{everything inside the brackets}] = 0$
leaving $c_3 (2015... | {
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"url": "https://math.stackexchange.com/questions/1908841",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Does not algebraic multiplicity= geometric multiplicity $\Rightarrow$ the matrix is diagonalizable?
let $A=\left(\begin{array}{ccc} -5 & -1 & 6 \\ -2 & -5 & 8 \\ -1 & -1 & 1 \end{array}\right)$ and $B=\left(\begin{array}{ccc} -9 & 3 & -3 \\ -14 & 4 & -7 \\ -2 & 1 & -4 \end{array}\right)$ matrices above $\mathbb{C}$ ar... | Indeed they are not similar. But $A$ is in fact not diagolizable.
It has a size 3 Jordan block:
$(A+3E)^2 \neq 0$ (whereas $(B+3E)^2=0$). This is also what you got using the $m_\lambda$ formulation.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1909422",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Aptitude simplification question Given $$p+q+r=1\;\;\;\;\&\;\;\;p^2+q^2+r^2=2\;\;\;\;\&\;\;\;p^3+q^3+r^3=3$$
find the value of $$p^4+q^4+r^4$$
| More generally, we have that for all $a,b,c\in\mathbb R$, $n\in\mathbb Z$, $n\ge 0$:
$$S_{n+3}=S_{n+2}S_1-S_{n+1}(ab+bc+ca)+S_{n}(abc),$$
where $S_n=a^n+b^n+c^n$. It can be proved by simply expanding.
Also notice $ab+bc+ca=\frac{(a+b+c)^2-\left(a^2+b^2+c^2\right)}{2}$.
Now apply this for your problem. Let $(a,b,c)=(p,q... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1910144",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Evaluate $(\sqrt{5}+\sqrt{6}+\sqrt{7})(\sqrt{5}+\sqrt{6}-\sqrt{7})(\sqrt{5}-\sqrt{6}+\sqrt{7})(-\sqrt{5}+\sqrt{6}+\sqrt{7})$ Evaluate $(\sqrt{5}+\sqrt{6}+\sqrt{7})(\sqrt{5}+\sqrt{6}-\sqrt{7})(\sqrt{5}-\sqrt{6}+\sqrt{7})(-\sqrt{5}+\sqrt{6}+\sqrt{7})$
I find out the answer $104$ but my friend find the answer $96$ using c... | Hint:
let $x=\sqrt(5)+\sqrt(6)$ and $y=\sqrt(5)-\sqrt(6)$
so
$$(x+\sqrt{7})(x-\sqrt{7})(y+\sqrt{7})(-y+\sqrt{7})$$
then you can use the fact of
$$(a-b)(a+b)=a^2-b^2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1910720",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
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How to show $\frac{1}{xy}$ = $\frac{1}{x}\frac{1}{y}$? I'm trying to prove that $\frac{1}{xy}$ = $\frac{1}{x}*\frac{1}{y},\forall x,y\neq 0$ using the field axioms of addition, multiplication, and the distributive law: $(x+y)z = xz+yz$ but am having a hard time doing so.
Things I've tried:
I've tried letting $\frac{1}{... | Remember the fields axioms (explained below). First you should think about which one to use in the statement of the desired identity. Certainly the demonstration through the axiom (M3). Second, you must realize that the secret of proof is to use the uniqueness of the axiom (M3).
Fixed real numbers $a$, $b$ there are r... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1911462",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
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Suppose that $q=\frac{2^n+1}{3}$ is prime then $q$ is the largest factor of $\binom{2^n}{2}-1$ I am curious for a starting point to prove the following claim
Suppose that $q={2^n+1 \above 1.5pt 3}$ is prime then $q$ is the largest factor of $\binom{2^n}{2}-1$
For example take $n=61$. Then $q ={2^{61}+1 \above 1.5pt ... | First of all, we have to show that $\frac{2^n+1}{3}$ is a factor of $$\binom{2^n}{2}-1=\frac{2^n(2^n-1)}{2}-1=\frac{2^{2n}-2^n-2}{2}$$
To show this, we consider
$$(2^n+1)^2-3(2^n+1)=2^{2n}+2\cdot 2^n+1-3\cdot2^n-3=2^{2n}-2^n-2$$
So, $\binom{2^n}{2}-1$ is divisble by $2^n+1$ and therefore by $\frac{2^n+1}{3}$.
But $(\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1912522",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 2
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Prove that every number of the sequence $49,4489,444889,\ldots$ is a perfect square
Prove that every number of the sequence $49,4489,444889,\ldots$ is a perfect square.
We can observe that $49=7^2, 4489=67^2, 444889=667^2, \ldots$
I have tried expanding terms of the sequence, and to express it as a whole square. But... | The $n$th term is $T_{n}=9+8.10+8.10^2+...8.10^{n-1}+4.10^n+...+4.10^{2n-1}$:
$$
\begin{align}
T_n &= 9+8.10(1+10+...10^{n-2})+4.10^n(1+...+10^{n-1}) \\
&= 9+8.10(\frac{10^{n-1}-1}{9})+4.10^n (\frac{10^{n}-1}{9}) \\
&=9+8.\frac{10^{n}-10}{9}+4.\frac{10^{2n}-10^n}{9} \\
&=\frac{81+8.10^n-80+4.10^{2n}-4.10^n}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1912893",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 0
} |
Solve quadratic fraction I would like to simplify the fraction
$$\frac{x^2-2x}{x^2+x-6}$$
I know from Mathematica that it should equal $\frac3{3x}$ but how do I get there?
| $x^2+x-6 = (x-2)(x+3)\rightarrow \frac{x(x-2)}{(x+3)(x-2)}=\frac{x}{x+3}$, so your result is incorrect, as the next counterexample shows:
$x=1; \frac{x^2-2x}{x^2+x-6}=\frac{1}{4}\neq 1=\frac{3}{3x}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1917872",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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prove that $a^2(p-q)(p-r)+b^2(q-p)(q-r)+c^2(r-p)(r-q)\ge 0$ If $a,b,c$
are the sides of a triangle and $p,q,r$ are positive real numbers then prove that $a^2(p-q)(p-r)+b^2(q-p)(q-r)+c^2(r-p)(r-q)\ge 0$
After modification. I have to prove $(a^2 p^2+b^2 q^2 + c^2 r^2) \ge pr(a^2+c^2-b^2)+qr(b^2+c^2-a^2)+pq(a^2+b^2-c^2)$
| We'll prove that your inequality is true for all reals $p$, $q$ and $r$
and $a$, $b$ and $c$ are lengths-sides of triangle.
Indeed,
$$a^2(p-q)(p-r)+b^2(q-p)(q-r)+c^2(r-p)(r-q)=$$
$$=\frac{1}{2}\left((p-q)^2(a^2+b^2-c^2)+(p-r)^2(a^2+c^2-b^2)+(q-r)^2(b^2+c^2-a^2)\right)\geq0$$
because $\sum\limits_{cyc}(a^2+b^2-c^2)=a^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1918278",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Defining the foci of "slanted" ellipse equation How to define the foci ($F_1,F_2$) coordinates of the slanted ellipse $x^2+4xy+9y^2=9$?
| Your ellipse is centered at the origin, hence in order to find its foci it is enough to find its vertices, or the direction of its axis. To do that, we may compute the stationary points of the quadratic form $q(x,y)=x^2+4xy+9y^2$ under the constraint $x^2+y^2=1$ through the method of Lagrange's multipliers. $\nabla q =... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1921399",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
Expanding $(2^b-1)\cdot(1+2^b+2^{2b}+ \cdots + 2^{(a-1)b})$ Can someone please explain to me how the following works (primarily interested in an explanation of the second step when $2^b$ is expanded?
I understand how each series cancels out to equal $2^n-1$ at the end.
$$\begin{align*}
xy&=(2^b-1) \cdot (1 + 2^b + 2^{2... | $$\begin{align*}
xy&=(2^b-1) \cdot (1 + 2^b + 2^{2b} + \cdots + 2^{(a-1)b})\\
&=2^b \cdot (1 + 2^b + 2^{2b} + \cdots + 2^{(a-1)b}) - (1 + 2^b + 2^{2b} + \cdots + 2^{(a-1)b})\\
&=(2^b + 2^{2b} + 2^{3b} + \cdots + 2^{ab})-(1+2^b+2^{2b}+ \cdots + 2^{(a-1)b})\\
&= \left[2^{ab} - 1\right] + \left[(2^b + 2^{2b} + \cdots + 2^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1922044",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
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Evaluate $\lim_{x\to0}\frac{e-(1+x)^\frac1x}{x}$ Somebody asked this and I think it's quite interesting as I couldn't figure out how to evaluate this but the Wolfram Alpha says its limit is $\frac e2$.
$$\lim_{x\to0}\frac{e-(1+x)^\frac1x}{x}$$
Could someone help here?
| For $0<x<1$: We have $$\frac {1}{x}\ln (1+x)=1-x/2+(x^2/3-x^3/4)+(x^4/5-x^5/6)+...=$$ $$=1-x/2+x^2/3-(x^3-x^4/5)-(x^5-x^6/7)-...$$ Therefore $$1- x/2<\frac {1}{x}\ln (1+x)<1-x/2+x^2/3.$$ Therefore $$e(1-e^{-x/2+x^2/3})<e-(1+x)^{\frac {1}{x}} <e(1-e^{-x/2}).$$
Both $(-x/2+x^2/3)$ and $(-x/2)$ belong to the interval $(-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1922138",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 6,
"answer_id": 4
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Angle between edges in a square pyramid
How do you calculate the missing angle? I couldn't figure it out. This is not homework, just a math practice question that I made up to test myself.
| First proof by analytical geometry.
Let
*
*$A(1,1,0),B(1,-1,0),C(-1,-1,0),D(-1,1,0)$ be the basis' vertices,
*$S(0,0,z)$ be the summit, with unknown $z$ and
*$a=SA=SB=SC=SD$.
*$\alpha$ be the unknown angle between $\vec{SA}$ and $\vec{SC}$.
Then $\tag{1} \cases{\vec{SA}=(1,1,0)-(0,0,z)=(1,1,-z)\\ \vec{SB}=(1,-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1922673",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
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Find the vlues of the trigonometric functions from the given information
$\tan t = \dfrac{1}{4}$, terminal point of $t$ is in the third quadrant.
So $\tan t = \dfrac{-\sqrt{1-\cos^2 t}}{\cos t} = \dfrac{1}{4} $, because it's in the third quadrant. After solving for $\cos^2 t$ get as an answer that $\cos ^2 t = \dfrac... | If
$$\tan t = \frac{1}{4}$$
then
$$t = \arctan \left(\frac{1}{4}\right) = 14.03°$$
or $0.244$ radians.
Beware
$$\tan t = \frac{\sin t}{\cos t} = \frac{\sqrt{1 - \cos^2t}}{\cos t}$$
In your formula, there is a minus sign which is wrong.
Hence
$$\frac{\sqrt{1 - \cos^2t}}{\cos t} = \frac{1}{4}$$
$$4\sqrt{1 - \cos^2 t} = ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1923043",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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How to calculate $\lim_{x \to 0} \left( \cos(\sin x) + \frac{x^2}{2} \right)^{\frac{1}{(e^{x^2} -1) \left( 1 + 2x - \sqrt{1 + 4x + 2x^2}\right)}}$? I need some help with computing this limit:
$$\lim_{x \to 0} \left( \cos(\sin x) + \frac{x^2}{2} \right)^{\frac{1}{(e^{x^2} -1) \left( 1 + 2x - \sqrt{1 + 4x + 2x^2}\right)}... | This answer is a codicil to that of Paramanand Singh. I give here a different calculation of $$\lim\limits_{x\to 0} \frac{\cos(\sin x)-1+\frac{x^2}{2}}{x^4}$$
Frequent use of the replacement of $\sin y$ by $\frac{\sin y}{y}y$ is made.
We have
$$ \frac{\cos(\sin x)-1+\frac{x^2}{2}}{x^4}= \frac{\cos(\sin x)-\cos x}{x^4... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1925856",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
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Finding the smallest value of the sum
The multiplication of three natural numbers $a$, $b$ and $c$ equals $2016$. What is the smallest value of $a + b + c$?
I started by a prime factorization to find that $2016 = 2^5 \cdot 3^2 \cdot 7$.
What should I do next?
| If we consider the fact that $7=2\cdot3\cdot\frac76$, then $2016=2^6\cdot3^3\cdot\frac76$, and it becomes easy to represent this number as a product of $\color\red3$ natural numbers which are as close to each other as possible:
*
*$a=2^{6/\color\red3}\cdot3^{3/\color\red3}=12$
*$b=2^{6/\color\red3}\cdot3^{3/\color\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1928473",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
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Closed form of $k$-th term in the recurrence $D_j = 1 + z^2\left(1-\frac{1}{D_{j-1}}\right)$, where $D_0 = 1+z^2$ I have $D_0=1+z^2$ and $D_{j}=1+z^2(1-1/D_{j-1})$ for $j>0$. I want to write an expression for arbitrary $D_k$, and what I have is below.
$D_{k}=1+z^2\Bigg(1-\bigg(1+z^2\big(1-(1+z^2(1-(\cdots(1+z^2)^{-1}\c... | If you work out the first few $D_j$'s explicitly, you should see a pattern:
$$D_1=1+z^2\left(1-{1\over1+z^2}\right)=1+z^2\left(z^2\over1+z^2\right)={1+z^2+z^4\over1+z^2}$$
$$D_2=1+z^2\left(1-{1+z^2\over1+z^2+z^4}\right)=1+z^2\left(z^4\over1+z^2+z^4\right)={1+z^2+z^4+z^6\over1+z^2+z^4}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1929532",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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The integral $\int \frac{x^2(1-\ln(x))}{(\ln(x))^4-x^4} dx$? $$\int \frac{x^2(1-\ln(x))}{\ln^4(x)-x^4} dx$$
I tried to factorize the denominator so that I could apply integration by parts but that didn't help me at all.
| HINT:
$$\mathcal{I}(x)=\int\frac{x^2\left(1-\ln(x)\right)}{\ln^4(x)-x^4}\space\text{d}x=$$
$$\frac{1}{2}\int\frac{\ln(x)-1}{x^2+\ln^2(x)}\space\text{d}x-\frac{1}{4}\int\frac{1+x}{x\left(x+\ln(x)\right)}\space\text{d}x+\frac{1}{4}\int\frac{1-x}{x\left(\ln(x)-x\right)}\space\text{d}x$$
*
*Now, for $\int\frac{1-x}{x\le... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1931376",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Prove or disprove $\sum\limits_{cyc}\frac{x^4+y^4}{x+y}\le 3\frac{x^4+y^4+z^4}{x+y+z}$
Let $x,y,z\ge 0$. Prove or disprove
$$\dfrac{x^4+y^4}{x+y}+\dfrac{z^4+y^4}{z+y}+\dfrac{z^4+x^4}{x+z}\le 3\dfrac{x^4+y^4+z^4}{x+y+z}$$
This is what I tried. Without loss of generality, let $x+y+z=1$, then
$$\Longleftrightarrow \su... | we have to show that $$\frac{3(x^4+y^4+z^4)}{x+y+z}-\frac{x^4+y^4}{x+y}-\frac{z^4+y^4}{z+y}-\frac{z^4+x^4}{x+z}\geq 0$$
assuming we have $$x=\min(x,y,z)$$ then we set
$$y=x+u,z=x+u+v$$ and we have after some algebra
$$\left( 12\,{u}^{2}+12\,uv+12\,{v}^{2} \right) {x}^{5}+ \left( 30\,{u}
^{3}+45\,{u}^{2}v+75\,u{v}^{2}+3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1932583",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Given $abc=1$ and $0< c \leq b \leq1\leq a$, prove that $8(a+b+c)^2\le9(1+a^2)(1+b^2)(1+c^2)$ I can't make progress with proving this inequality.
I have tried opening the brackets and using $abc=1$ in order to obtain the following:
$$a^2+b^2+ \frac 1{a^2b^2}+18+9a^2b^2+\frac 9{a^2}+\frac9{b^2}\ge 16 \left (ab+\frac 1a+... | Since the right side of your inequality does not depend on substitution $a\rightarrow-a$,
it's enough to prove it for positive variables.
We'll prove that your inequality is true for all positives $a$, $b$ and $c$ such that $abc=1$.
Indeed, let $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.
Hence, our inequality is equiva... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1933815",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to find the square roots of $z = 5-12i$ I am asked the following question:
Find the square roots of $z = 5-12i$
I know that this problem can be easily solved by doing the following:
$$
z_k^2 = 5-12i\\
(a+bi)^2 = 5-12i\\
(a^2-b^2) + i(2ab) = 5-12i\\
\\
\begin{cases}
a^2 - b^2 = 5\\
2ab = -12
\end{cases} \quad \Rig... | If you let $\theta$ be the angle in the first case, then using the fact that $\tan \theta = -12/5$, you can find $\cos\theta$ and $\cos\frac{\theta}{2}$ indirectly. One of the roots would be:
$\sqrt{13}(-\frac{3}{\sqrt{13}} + i \frac{2}{\sqrt{13}}) = z_1.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1934734",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
Show that $1^n+2^n+3^n+4^n$ is divisible by 5 if and only if n is not divisible by 4 Show that $$1^n+2^n+3^n+4^n$$ is
divisible by 5 if and only if n is not divisible by 4.
I don't find the relation why must divisible by 5 anyone can give a hint or give some part of proof?
| If $4 \mid n \implies n = 4k $
$1^{4k} + 2^{4k} + 3^{4k} + 4^{4k} \equiv 1^k + 1^k + 1^k + 1^k \equiv 1 + 1+ 1 + 1 \equiv 4 \space \text{(mod 5)}$
If $4 \nmid n \implies n = 4k + 1 \vee n = 4k + 2 \vee 4k + 3$
For, $n = 4k + 1$ you get:
$1^{4k+1} + 2^{4k+1} + 3^{4k+1} + 4^{4k+1} \equiv 1^k \times 1 + 1^k \times 2 + ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1935352",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
This sequence $\lfloor \sqrt{2003}\cdot n\rfloor $ contains an infinite number of square numbers
Show that: the sequence $\lfloor \sqrt{2003}\cdot n\rfloor $ contains an infinite number of square numbers.
Maybe consider the Pell equation to solve this problem. But how to do this? Thanks
| CW again. Comment on the pretty idea in the other answers: Suppose we have
$$ x^2 - d y^2 = -k, $$
with small integer $k > 0.$
$$ ( x - y \sqrt d) (x + y \sqrt d) = -k, $$
$$ x - y \sqrt d = \frac{-k}{x + y \sqrt d}, $$
$$ x = y \sqrt d - \frac{k}{x + y \sqrt d}, $$
$$ x^2 = xy \sqrt d - \frac{kx}{x + y \sqrt d}, $$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1937001",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 3,
"answer_id": 2
} |
Differentiation involving implicit and parametric. I was asked to differentiate the term,
$$X^3 + XY^2 - Y^3$$
For such I reached,
$$3X^2 + 2XY\dfrac{dy}{dx} - 3Y^2\dfrac{dy}{dx}$$
The apparent answer is,
$$3X^2 + Y^2 + 2XY\dfrac{dy}{dx} - 3Y^2\dfrac{dy}{dx}$$
How exactly this is reached as I am new to the whole imp... | You forgot about the product rule. Assuming $y=y(x),$
\begin{align*}
\frac{d}{dx} \left( x^3 +xy^2 - y^3 \right) &= 3x^2 + \frac{d}{dx} \left(xy^2 \right) -3y^2 \frac{dy}{dx} \\
&= 3x^2 + 2xy \frac{dy}{dx} + y^2 - 3y^2 \frac{dy}{dx}
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1937573",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Compute $\int^{\pi/2}_0 \frac{dx}{(a^2\cos^2 x + b^2 \sin ^2 x)^2}$ I have tried solving this for about an hour and will probably resort to head banging in some time:
$$\int ^{\frac{\pi}{2}}_{0} \dfrac{dx}{(a^2\cos^2 x + b^2 \sin ^2 x)^2}$$
I first divided by $\cos^4 x$ and then subsequently put $\tan x = t$, to get:
$... | For a solution that doesn't involve complex analysis: Try the substitution $t = \frac{a}{b} \tan u$. Then $du = \frac{ab}{a^2 + b^2 t^2} \,dt$ and
your integral becomes
$$\frac{1}{a^3 b^3} \int_0^{\pi/2} (b^2 \cos^2 u + a^2 \sin^2 u) \,du$$
which is more doable.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1939026",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
"answer_id": 4
} |
Expressing $\sqrt[3]{7+5\sqrt{2}}$ in the form $x+y\sqrt{2}$ Express $\sqrt[3]{(7+5\sqrt{2})}$ in the form $x+y\sqrt{2}$ with $x$ and $y$ rational numbers.
I.e. Show that it is $1+\sqrt{2}$.
| You can assume that the nested radical can be expressed in $a+b\sqrt{2}$ form. More specifically, we have $$\sqrt[m]{A+B\sqrt[n]{C}}=a+b\sqrt[n]{C}\tag{1}$$
With your question, we have $$\sqrt[3]{7+5\sqrt{2}}=a+b\sqrt{2}\tag{2}$$
Cubing both sides, we get $$7+5\sqrt{2}=(a^3+6ab^2)+(3a^2b+2b^3)\sqrt{2}\tag{3}$$
And equa... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1940784",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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Closed form solution for infinite summation Thomas, Bruckner & Bruckner, Elementary Real Analysis.
Prove that for all r > 1,
$$\frac{1}{r - 1} = \frac{1}{r+1} + \frac{2}{r^2 + 1} + \frac{4}{r^4 + 1} + \frac{8}{r^8 + 1} + \cdots$$
So far I have $$ \frac{1}{r-1} -\frac{1}{r+1} = \frac{2}{r^2 -1} $$
$$\sum_{n=1}^\infty ... | For any $x\in(0,1)$ we have
$$\frac{1}{1-x}= (1+x)(1+x^2)(1+x^4)(1+x^8)\cdots \tag{1} $$
that can be read as every positive integer has a unique representation in base-$2$.
By taking $\frac{d}{dx}\log(\cdot)$ of both sides, we get:
$$ \frac{1}{1-x}=\frac{1}{1+x}+\frac{2x}{1+x^2}+\frac{4x^3}{1+x^4}+\frac{8x^7}{1+x^8}+\c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1942285",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
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Solving the equation $(x^2-3x+1)^2=4x^2-12x+9$ I need to solve the following equation:
$$(x^2-3x+1)^2=4x^2-12x+9.$$
I think I need to bring everything to one side but I don't know anything else.
| $$(x^2-3x+1)^2=4x^2-12x+9$$
$$(x^2-3x+1)^2=4(x^2-3x+1)+5$$
$$(x^2-3x+1)^2-4(x^2-3x+1)-5=0$$
now let $u=(x^2-3x+1)^2$
so
$$u^2-4u-5=0$$
$$(u+1)(u-5)=0$$
then complete the solution
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1943179",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 6,
"answer_id": 3
} |
Total Expectation problem to solve in 2 ways - regular and by the law of total expectancy Im trying to solve the question in the way of total expectancy (meaning $E[X]=E[E[X|Y]]$).
the question: "A fair cube with 4 sides is thrown twice. each side has a number from 1 to 4. Let's mark Y - maximum of two results. Evaluat... | Who knows if anyone is looking anymore, never mind OP, but this is a concept I'm trying to grasp as well. I think I understand Ami's dissatisfaction with her own solution -- I too started by computing each conditional probability and wondered what I was supposed to have learned -- and so I've come up with my own attem... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1943854",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
The limit of $\frac{1^{1/3} + 2^{1/3} + \cdots + n^{1/3}}{n^{4/3}} $ I want to evaluate the following quotient limit:
$$\lim_{n \to \infty}\frac{1^{1/3} + 2^{1/3} + \cdots + n^{1/3}}{n^{4/3}} $$
I know an exact same answer is already in here.
I want to avoid using the fact that the integral is a limit of Riemann Sum, i... | $$\lim_{n \to \infty} \frac{1^{1/3} + 2^{1/3} + \cdots + n^{1/3}}{n^{4/3}} = \lim_{n \to \infty} \frac{1^{1/3} + 2^{1/3} + \cdots + n^{1/3}}{\frac{3}{4} (n^{\frac{4}{3}} - 1)}\cdot\frac{\frac{3}{4} (n^{\frac{4}{3}} - 1)}{n^{4/3}}$$
First term goes to $1$ by your $\frac{f(1) + f(2) + ... + f(n)}{F(n)}\to1$ result, secon... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1944831",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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What is $\lim_{x\to0}\frac{|x+3|(\sqrt{ax+b}-2)}x$ when it exists?
For what values of $a$ and $b$ does the following limit exist and what is the limit in those cases?
$$\lim_{x\to0}\frac{|x+3|(\sqrt{ax+b}-2)}x$$
Actually this is an assignment which I need to do, but I have no idea where to start, please help!
| A more brute-force approach:
$$|x+3| = \begin{cases}
x + 3, & x \geq -3 \\
-(x+3) = -x - 3, & x < -3
\end{cases}$$
so
$$\dfrac{|x+3|(\sqrt {ax+b}-2)}{x} = \begin{cases}
\dfrac{(x+3)(\sqrt {ax+b}-2)}{x}, & x \geq -3 \\
\dfrac{(-x-3)(\sqrt {ax+b}-2)}{x}, & x < -3 \text{.}
\end{cases}$$
For $x$ within a "small" neighbo... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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Prove this 3x3 determinant using properties of determinant.
Prove this $3\times 3$ determinant using properties of determinant.
$$\begin{vmatrix}y+z & x & x \\ y & z+x & y \\ z & z & x+y\end{vmatrix} = 4xyz$$
I have been trying to solve this one for over an hour now. I really can't even get started with it. The que... | The determinant
$$
\det(A) = \det(a_1, a_2, a_3)
$$
is an alternating multi linear form, this means
$$
\det(a_1+ \alpha a_2, a_2, a_3) = \det(a_1, a_2, a_3) + \alpha \det(a_2, a_2, a_3)
$$
because $\det$ is linear in the first argument as multi linear form.
Because it is an alternating form, if we exchange the first t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1951553",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Simplifying $\frac{\;\frac{x}{1-x}+\frac{1+x}{x}\;}{\frac{1-x}{x}+\frac{x}{1+x}}$
$$\frac{\;\;\dfrac{x}{1-x}+\dfrac{1+x}{x}\;\;}{\dfrac{1-x}{x}+\dfrac{x}{1+x}}$$
I can simplify this using the slow way ---adding the numerator and denominator, and then dividing--- but my teacher told me there is another way. Any help?
| Let me try.
$$\frac{\frac{x}{1-x}+\frac{1+x}{x}}{\frac{1-x}{x}+\frac{x}{1+x}} = \frac{\frac{1}{1-x}-1 + \frac{1}{x} + 1}{\frac{1}{x}-1 + 1-\frac{1}{1+x}} = \frac{\frac{1}{1-x} + \frac{1}{x}}{\frac{1}{x} - \frac{1}{1+x}} = \frac{\frac{1}{x(1-x)}}{\frac{1}{x(1+x)}} = \frac{1+x}{1-x}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1954425",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 3
} |
Calculus Integral $\int \frac{dt}{2^{t} + 4}$
I try resolve this integral, with $u = 2^t + 4 $, but can´t ... plz
$$\int \frac{dt}{2^t + 4} = \frac{1}{\ln 2}\int\frac{du}{u(u-4)} = \text{??} $$
| Hint:
$$\frac{1}{2^x+4}=\frac{2^x-2^x+4}{4(2^x+4)}=\frac{1}{4} \left(1-\frac{2^x}{2^x+4}\right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1955721",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 2
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Is $x^2+x+1$ ever a perfect power? Using completing the square and factoring method I could show that the equation $x^2+x+1=y^n$, where $x,y$ are positive odd and $n$ is positive even integers, does not have solution, but I could not show that for positive odd $x,y$ and odd $n>1$ the equation does (does not) have solu... | Your equation can be rewritten as
$$\frac{x^{3}-1}{x-1} = y^{N}.$$
The Diophantine equation
$$ \frac{x^{n} − 1}{x-1} = y^{q} \quad x > 1, \quad y>1, \quad n>2 \quad q \geq 2 \quad \tag1$$
was the subject matter of a couple of papers of T. Nagell from the $1920's.$ Some twenty-odd years later, W. Ljunggren clarified som... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1956093",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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"answer_id": 0
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Lower bound for cyclic expression in four variables We are given four positive real numbers $a,b,c,d$. I would like to prove that
$$\frac ab+\frac bc+\frac cd+\frac da\ge4$$
To start solving this I assumed
$$a\ge b\ge c\ge d$$
Therefore
$$\frac ab,\frac bc,\frac cd\ge 1,\ \frac da\le 1$$
and the last one is causing the... | As lab bhattacharjee commented, You can use the AM-GM inequality. By applying it here, we can obtain:
$$
\begin{align*}
\frac{\frac{a}{b} + \frac{b}{c} + \frac{c}{d} + \frac{d}{a}}{4} & \ge \sqrt[4]{\frac{abcd}{bcda}}\\
& = \sqrt[4]{1}\\
& = 1
\end{align*}
$$
Therefore:
$$
\frac{a}{b} + \frac{b}{c} + \frac{c}{d} + \fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1956583",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Sum of three numbers and the sum of their squares: $x^2+y^2+z^2 \geq \frac{1}{3}$ for $x+y+z=1$ We have $x, y, z \in \mathbb R$ such that $x+y+z=1$. Prove that $x^2+y^2+z^2 \geq \frac{1}{3}$. I am able to do this using the relationship between the power and arithmetic means. Is there a way to not use this relationship?... | For any real numbers $x,y,z$, we have
$$\begin{align}
0\le(x-y)^2+(y-z)^2+(z-x)^2&\implies2(xy+yz+zx)\le2(x^2+y^2+z^2)\\
&\implies(x+y+z)^2\le3(x^2+y^2+z^2)
\end{align}$$
So if $x+y+z=1$, then ${1\over3}\le x^2+y^2+z^2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1957020",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove or disprove $\sqrt[3]{\frac{(ab+bc+ac)(a^2+b^2+c^2)}{9}}\ge\sqrt[4]{\frac{(a^2b^2+b^2c^2+c^2a^2)}{3}}$ Let $a,b,c>0$ prove or disprove
$$\sqrt[3]{\dfrac{(a+b+c)(a^2+b^2+c^2)}{9}}\ge\sqrt[4]{\dfrac{(a^2b^2+b^2c^2+c^2a^2)}{3}}$$
since
$$(a^2+b^2+c^2)^2\ge 3(a^2b^2+b^2c^2+a^2c^2)\tag{1}$$
other
$$(a+b+c)^2\le 3(a^2... | After a quick look:
$$\sqrt[3]{\dfrac{(a+b+c)(a^2+b^2+c^2)}{9}}\ge \sqrt[3]{\dfrac{(2s+c)(2s^2+c^2)}{9}} \ge \sqrt[4]{\dfrac{(s^4+2s^2c^2)}{3}} \ge \sqrt[4]{\dfrac{(a^2b^2+b^2c^2+c^2a^2)}{3}}$$
where $s=\frac{a+b}{2}$ and we assume $c=\min(a,b,c)$.
The left inequality follows from $2(a^2+b^2)\ge (a+b)^2$.
The right ine... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1957863",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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cubic equation has $3$ distinct roots $\alpha,\beta,\gamma$ equation $x^3-9x^2+24x+c=0$ has $3$ distinct roots $\alpha,\beta,\gamma$, then $\lfloor \alpha \rfloor+\lfloor \beta \rfloor +\lfloor \gamma \rfloor =$
and $\lfloor \alpha \rfloor = \alpha-\{\alpha\},\lfloor \beta \rfloor = \beta-\{\beta\},\lfloor \gamma \rflo... | Direct from vieta's relation, we get $\alpha^{2}+\beta^{2}+\gamma^{2}=-b/a=9$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1959455",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
The ideal generated by $X^{2}-Y^{3}$ and $Y^{2}-X^{3}$ is not prime
I tried to show that the ideal generated by $X^{2}-Y^{3}$ and $Y^{2}-X^{3}$ is not prime in $\mathbb{C}[X,Y]$.
I noticed that $X^{2}-Y^{3}-(Y^{2}-X^{3})=(X-Y)(X^{2}+XY+Y^{2}+X+Y)$. Now, if $X-Y$ is in $(X^{2}-Y^{3},Y^{2}-X^{3})$ there exists $f(X,Y... | $X(X^2-Y^3)+(Y^2-X^3)=-XY^3+Y^2=Y^2(-XY+1)$. $-XY+1, Y^2$
Suppose that $-XY+1=f(X,Y)(X^2-Y^3)+g(X,Y)(Y^2-X^3)$, we deduce that $1=f(0,Y)(-Y^3)+g(0,Y)Y^2=Y^2(-f(0,Y)Y+g(0,Y))$ impossible.
Suppose that $Y^2=f(X,Y)(X^2-Y^3)+g(X,Y)(Y^2-X^3)$. Set $X=Y=1$, we have $1=f(1,1)(1^2-1^3)+g(1,1)(1^2-1^3)=0$ impossible, so $Y^2$ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1959819",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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To find number of distinct terms of binomial expansion I have binomial expansion as
$(x+\frac{1}{x} + x^2 + \frac{1}{x^2})^{15}$.
How do i find number of distinct terms in it. Distinct in sense means terms having different powers of $x$?
I have simplified this as $\frac{1}{x^{30}}(x^3+x+x^4+1)^{15}$. How do i proceed... | We have that
$$(x^4+x^3+x+1)^n=(x+1)^n\cdot(x^3+1)^n.$$
Now the powers of $x$ in $(1+x)^n$ are the integers in the interval $[0,n]$.
Moreover for $n\geq 2$, the multiplication of $(x^3+1)(x+1)^n$ gives a sum of powers of $x$ which cover the range $[0,n]\cup [3,n+3]=[0,n+3]$.
Hence, for $(x+1)^n\cdot(x^3+1)^n$ the powe... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1960353",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 1
} |
distinguishable balls distinguishable boxes where each box contains at least 2 balls Consider $m$ distinguishable balls and $n$ distinguishable boxes where $m > n$ (the boxes and balls are already distinguishable, say they come with preassigned distinct labels).
How many ways are there to distribute the balls into boxe... | We solve the case of indistinguishable boxes and distinguishable
boxes. The combinatorial species in the first case is
$$\mathfrak{P}_{=n}(\mathfrak{P}_{\ge 2}(\mathcal{Z}))$$
which gives the EGF
$$G(z) = \frac{(\exp(z)-z-1)^n}{n!}.$$
Extracting coefficients we get
$$m! [z^m] G(z)
= m! [z^m] \frac{(\exp(z)-z-1)^n}{n!}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1963468",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
How to prove the limit of a sequence by definition: $\lim((n^2 + 1)^{1/8} - n^{1/4}) = 0$? I have to prove the limit of a sequence using $\epsilon - N$ definition: $$\lim_{n\to\infty}((n^2 + 1)^{1/8} - n^{1/4}) = 0$$.
Attempt:
We want to show: $\forall \epsilon>0$ $\exists N \in \mathbb{N}$, s.t. if $n \ge N$, then $|... | Just keep (fearlessly) your argument going:
$$(n^2 + 1)^{1/8} - n^{1/4} = \frac{(n^2 + 1)^{1/4} - n^{1/2}}{(n^2 + 1)^{1/8} + n^{1/4}}$$
consequently
\begin{align*}(n^2 + 1)^{1/8} - n^{1/4} &= \frac{(n^2 + 1)^{1/4} - n^{1/2}}{(n^2 + 1)^{1/8} + n^{1/4}}\cdot \frac{(n^2 + 1)^{1/4} + n^{1/2}}{(n^2 + 1)^{1/4} + n^{1/2}} \\ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1964070",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Permutation or combination for repeated chosen items? I have this problem:
Choosing 5 vertices from the picture at random, what are the odds of at least 4 of them belonging to the pyramid?
My book says:
Because we are asked the probability of at least 4 vertices belonging
to the pyramid, we must consider two possib... | The book's solution is correct. Your approach is viable, but you made an error in its execution.
The author of your text solved the problem by considering which five-element subsets of the vertices contain at least four vertices of the pyramid. That is, the author is making an unordered selection of the vertices. Si... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1966036",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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Find the range of $a$ such $|f(x)|\le\frac{1}{4},$ $|f(x+2)|\le\frac{1}{4}$ for some $x$.
Let $$f(x)=x^2-ax+1.$$
Find the range of all possible $a$ so that there exist $x$ with
$$|f(x)|\le\dfrac{1}{4},\quad |f(x+2)|\le\dfrac{1}{4}.$$
A sketch of my thoughts: I write
$$f(x)=\left(x-\dfrac{a}{2}\right)^2+1-\dfrac... | My Work: I will approach in a very short way.we have been given
two characteristic equations of a function and their range .
$$f(x)=x^2-ax+1$$
$$\text{and, we know}$$
$$\lvert x^2-ax+1 \rvert \le \dfrac{1}{4}$$
$$\lvert (x+2)^2-a(x+2)+1 \rvert \le \dfrac{1}{4}$$
from the absolute value,we can write-
$$-\dfrac{1}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1967726",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 10,
"answer_id": 8
} |
How do I solve differentil equation if the begining values are given? How do I solve differentil equation if the begining values are given?
$\frac{dx}{dt}=x^2+5x$ with x(0)=-3. I need to find x(t).
$\int \frac{dx}{x^2+5x}=\int dt$
So when I put it on Symbollab I get that left side is $-\frac{2}{5}arctanh(\frac{2}{5}(x+... | In your calculation the initial step
$$
\int \frac{dx}{x^2+5x}
\ne \int \frac{dx}{x^2(1+x/5)}
= \int \frac{dx}{x^2 + x^3/5}
$$
is wrong.
You can decompose the fraction using polynomial factorization and then look for a sum of the fractions with the factors as denominator
$$
\frac{1}{x^2+5x}
= \frac{1}{x(x+5)}
= \frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1969278",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Trigonometry and integrals related picture
In my book it says $ \frac{x}{R} = \tan\theta$
ok, that is pretty obvious, but then it says that it implies that
$$ \frac{dx}{r^2} = \frac{R\,d\theta}{r^2 \cos^2 \theta} = \frac{d\theta}{R}$$
I really cannot understand how $d\theta$ got involved at all. Can anyone please tr... | You have $\dfrac x R = \tan\theta,$ so $x = R\tan\theta.$ If you know that $$\frac d {d\theta} \tan\theta = \sec^2\theta = \frac 1 {\cos^2\theta},$$ then, assuming $R$ remains constant as $x$ and $\theta$ change, this yields
$$
\frac{dx}{d\theta} = \frac d {d\theta}(R\tan\theta) = R\sec^2\theta = \frac{R}{\cos^2\theta}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1969999",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
To find the range of $\sqrt{x-1} + \sqrt{5-x}$ To find the range of $\sqrt{x-1} + \sqrt{5-x}$.
I do not know how to start?
Thanks
| Clearly $1\le x\le5$
If $y=\sqrt{x-1}+\sqrt{5-x}>0, y^2=4+2\sqrt{(x-1)(5-x)}$
Now $\sqrt{(x-1)(5-x)}\ge0$
and using AM-GM inequality, $\sqrt{(x-1)(5-x)}\le\dfrac{x-1+5-x}2=?$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1975431",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 1
} |
Find the equation solution $\left[ x\right] +\left[ \dfrac{9}{x}\right]=6$ solve this following equation
$$\left[ x\right] +\left[ \dfrac{9}{x}\right]=6$$
the equation obvious have a root $x=3$. But how to solve other roots?
| *
*for $x\le\lfloor{\frac{9}{6}}\rfloor$ there is no answer since $\lfloor{\frac{9}{x}}\rfloor$ is at least $6$ and when $\lfloor{x}\rfloor$ becomes zero, the other term is $7$.
*for $\frac{9}{6}<x \le \frac{9}{5}$, we have $\lfloor{\frac{9}{x}}\rfloor=5$ and $\lfloor{x}\rfloor=1$, hence all this range is a valid ans... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1978119",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
If two consecutive numbers are removed from the series $1+2+3+\ldots+n$ the average becomes $99/4$. Find the two numbers. The initial average will be $\frac{n+1}{2}$. If the two numbers are $k$ and $k+1$ then the new average will be $\frac{n(n+1)/2-(2k+1)}{n-2}$. I couldn't figure further even though I got the relation... | This approach finds that the new average lies within $\pm 1$ of the original average. This significantly narrows down possibilities, and the solution can then be found easily by elimination.
After removing the two numbers, the new average, $a$, is given by
$$\begin{align}
\frac {99}4=a&=\frac{\frac{n(n+1)}2-(2k+1)}{n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1978490",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 1
} |
How do I find an irrational root of a quartic function? For example, the equation $2x^4 - 3x^3 - 1 = 0$ has no rational roots, but it does have an irrational root between x = 1 and x = 2. If it had a rational root, I would be able to find it, but I am a bit confused on how to find an irrational root.
| In this case consider:
\begin{align}
2 \, x^4 - 3 \, x^3 -1 &= (a \, x^2 + b \, x + 1)(x^2 + c \, x -1) \\
&= a \, x^4 + (ac + b) \, x^3 + (-a + bc + 1) \, x^2 + (-b + c) \, x - 1
\end{align}
This leads to $b = c$, $a=2$, $1 - a + bc = -1 + b^2 = 0$ or $b = \pm 1$, and
$ac + b = 3b = -3$ which narrows $b$ to $b = -1$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1981617",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Rolling three dice - sum of two results equals third one Is my answer correct?
Three six-sided dice are thrown, what is the probability that the sum of two of them equals the third one?
If order does not matter, there are 9 such possibilities:
*
*(1, 1, 2)
*(1, 2, 3)
*(1, 3, 4)
*(1, 4, 5)
*(1, 5, 6)
*(2, 2, 4... | If you imagine there are three distinguishable dices (either different colors, or you throw them one after the other), then there are $6^3=216$ possible cases.
Now your enumeration counts down to :
$(1,1,2)$ : $3$ cases
$(1, 2, 3)$ : $6$ cases
$(1, 3, 4)$ : $6$ cases
$(1, 4, 5)$ : $6$ cases
$(1, 5, 6)$ : $6$ cases
$(2,... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1983517",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Find best-fit parabola to the given data
Find the parabola $At^2+Bt+C$ that best approximates the data set $t= -1,0,1,2,3$ and $b(t) = 5,2,1,2,5$.
Would I be using least squares such that $x = (A^TA)^{-1}A^Tb$? Thank you in advance.
| Problem statement
Given a sequence of $m=5$ data points $\left\{ x_{k}, y_{k} \right\}_{k=1}^{m}$, and the trial function
$$
y(x) = a_{0} + a_{1} x + a_{2} x^{2}.
$$
The linear system is
$$
\begin{align}
\mathbf{A} a &= y\\
%
\left[
\begin{array}{rrr}
1 & -1 & 1 \\
1 & 0 & 0 \\
1 & 1 & 1 \\
1 & 2 & 4 \\
1 & 3 &... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1984449",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
"answer_id": 3
} |
Find the following limit with a sum I am trying to solve the following limit, which includes a sum:
$\lim_{n\to\infty} [\frac{1}{n^2}(2+\frac{3^2}{2}+\cdots+\frac{(n+1)^n}{n^{n-1}})]$
So far, the only thing I have been able to do is to bound the limit between 0 and e:
*
*If $(2+\frac{3^2}{2}+\cdots+\frac{(n+1)^n}{n^... | The expression equals
$$\frac{\sum_{k=1}^{n} (k+1)^k/k^{k-1}}{n^2}.$$
Stolz-Cesaro says we should look at
$$\frac{(n+2)^{n+1}/(n+1)^n}{(n+1)^2 - n^2}=\frac{(1+1/(n+1))^n(n+2)}{2n+2}.$$
The last expresion $\to \frac{e}{2}.$ So by S-C, the limit is $e/2.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1985151",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Find $24 \cot^2 x$
Suppose $x$ is in the interval $[0,\pi/2]$ and $\log_{24 \sin x} (24 \cos x) = \frac{3}{2}$. Find $24 \cot^2 x$.
From the given equation we have $24 \cos{x} = (24 \sin{x})^{\frac{3}{2}}$ and so $24\cot^2{x} = 24^2\sin{x}$. How do we continue?
| A Solution
Since
$$
(24\cos(x))^2=(24\sin(x))^3\tag{1}
$$
we have
$$
24\sin^3(x)+\sin^2(x)-1=0\tag{2}
$$
We can try the rational root theorem to get
$$
\sin(x)=\frac13\tag{2}
$$
Now use
$$
24\cot^2(x)=24^2\sin(x)=192\tag{4}
$$
Uniqueness of the Solution
Since $\frac{\mathrm{d}}{\mathrm{d}u}\left(24u^3+u^2-1\right)=72u... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1986002",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Choose an infinite GP from given terms for a particular sum Is it possible to choose an infinite GP from amongst the terms 1, 1/2, 1/4, 1/8, 1/16 ... with a sum a) 1/5? b) 1/7 ?
My approach was simply choosing terms and probable ratios to match the sum...I would be grateful if you could reveal any trick for this quest... | Since all terms are of the form $\frac{1}{2^n}$, any geometric progression will be with ratio $\frac{1}{2^k}$
The summation of a geometric series is:
$$\sum\limits_{k=0}^\infty ar^k = \frac{a}{1-r}$$
The question then is what choice of $a$ (initial term) and what choice of $r$ (ratio) will give the desired results.
Sin... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1988749",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Is $x+1$ a factor of $x^{2016}-1$? $$x^{2016}-1=(x-1)(1+x+x^2+x^3+\dots+x^{2015})$$
If $x+1$ is a factor of $x^{2016}-1$, then $(1+x+x^2+x^3+\dots+x^{2015})=(x+1)G(x)$, where $G(x)$ is some polynomial.
What is $G(x)$ if $x+1$ is also a factor?
| \begin{align}
x^{2016}-1=&(x^2)^{1008}-1=(x^2-1)((x^2)^{1007}+\dots+x^2+1)\\
&=(x+1)(x-1)(x^{2014}+\dots+x^2+1).
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1991409",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Is it inequality true Please help me prove the inequality:
$$\sqrt[3]{3+\sqrt[3] 3}+\sqrt[3]{3-\sqrt[3] 3}<2\sqrt[3] 3$$
Thanky for your help and your attention.
| Let be
$$a=\sqrt[3]{3+\sqrt[3] 3}, b=\sqrt[3]{3-\sqrt[3] 3}$$
now we have
$$a^3+b^3=\left(\sqrt[3]{3+\sqrt[3] 3}\right)^3+\left(\sqrt[3]{3-\sqrt[3] 3}\right)^3=3+\sqrt[3]3+3-\sqrt[3]3=6.$$
Becouese,
$$a^3+b^3>a^2b+b^2a$$
we have
$$ab(a+b)<a^3+b^3=6$$
Now
$$3ab(a+b)<18$$
or
$$3a^2b+3ab^2<18$$
Add in the both sides equti... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1991759",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Pythagorean triples $(a,b,c)$ and divisibility of $a$ or $b$ by $3$. Call a triple of integers $(a, b, c)$ a Pythagorean triple if $a^2 + b^2 = c^2$
, i.e., if $a, b, c \in \mathbb{N^*}$ are the (measures of) sides of a right triangle. Examples of Pythagorean triples are (3, 4, 5), (5, 12, 13), (8, 15, 17) and (3312, ... | A number that is not divisible by $3$ may be written either as $3m+1$ or as $3m+2$ for some integer $m$. The square of such a number is then
$$
(3m+1)^2 = 9m^2 + 6m + 1 = 3(3m^2 + 2m) + 1\\
(3m + 2)^2 = 9m^2 + 12m + 4 = 3(3m^2 + 4m + 1) + 1
$$
and we see that the square of such a number is always of the form $3n+1$ for... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1995091",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Show series $1 - 1/2^2 + 1/3 - 1/4^2 + 1/5 - 1/6^2$ ... does not converge I was wondering if my proof is correct, and if there are any better alternative proofs. Or maybe proof that use nice tricks i might need in the future.
$$1 - \frac{1}{2^2} + \frac{1}{3} - \frac{1}{4^2} + \frac{1}{5} - \frac{1}{6^2} \ldots = \sum_... | First, there seems to be quite a bit of confusion in the comments concerning "grouping". Let's take a look at it, essentially:
\begin{align}
1 - \frac{1}{2^2} + \frac{1}{3} - \frac{1}{4^2} + \frac{1}{5} - \frac{1}{6^2} +\cdots &\stackrel{?}{=}
\left(1 - \frac{1}{2^2}\right) + \left(\frac{1}{3} - \frac{1}{4^2}\right) + ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1996702",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
Prove that $x_{n+1}=x_n+0.8x_n(1-x_n)-0.072$ converges Assume that $x_1$ is a real number and for $n \gt 1$ :
$x_{n+1}=x_n+0.8x_n(1-x_n)-0.072$
a) Assume that $x_1=0.6 $
Prove that $\{x_n\}$ converges and find its limit.
b) Find a set of real numbers like $A$ such that if $x_1 \in A$ then $\{x_n\}$ is convergent. ... | Let
\begin{align}
f(x) &= x+0.8x(1-x)-0.072 \\
&= x+\frac{4}{5}x(1-x) - \frac{9}{125} \\
&= \frac{9}{10}+\frac{9}{25}\left(x-\frac{9}{10}\right)-\frac{4}{5} \left(x-\frac{9}{10}\right)^2
\end{align}
I obtained that last expression by expanding $f$ in a series about $9/10$. You can expand out the last two expression... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1997689",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Find the value of $b-d$ If it's given that ->
$a^{3} = b^{2}$ ;
$c^{5} = d^{4}$ ;
$a - c = 19$
Find $b-d$
PS- I have given 4 hours of my life to this question and even then I am NOT able to get even the SLIGHTEST hint . So , i request you all to explain the solution of this question . Thanks in Advance !!
| Put $a=x^2, b=x^3, c=y^4, d=y^5$:
$$\begin{align}
a^3=b^2\;&\Rightarrow (x^2)^3=(x^3)^2\\
c^5=d^4\; &\Rightarrow (y^4)^5=(y^5)^4\\
\text{Given} \qquad\qquad a-c&=19\\
x^2-y^4&=19\\
(x-y^2)(x+y^2)&=19\end{align}
$$
As $19$ is prime and $y^2>0$, and also given that $x,y$ are integers, possible combinations are
$$
x-y^2=1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1997779",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Prove that $(a \cdot b) \text{ mod } n = ((a \text{ mod } n) \cdot (b \text{ mod } n)) \text{ mod } n$
Prove for all $n \in \mathbb{N}, n >1$ and $a,b \in \mathbb{Z}$ that
$$(a \cdot b) \text{ mod } n = ((a \text{ mod } n) \cdot (b \text{ mod
} n)) \text{ mod } n$$
I'm not sure if this is a proof but I tried anothe... | Suppose that:
$$\begin{cases}
a \mod n = r_a\\
b \mod n = r_b\\
(a \cdot b) \mod n = r\\
\end{cases}$$
or equivalently
$$\begin{cases}
a = q_a \cdot n + r_a\\
b = q_b \cdot n + r_b\\
a \cdot b = q \cdot n + r\\
\end{cases}.$$
Then:
$$r_a \cdot r_b = (a-q_a \cdot n)\cdot (b-q_b \cdot n) = \\
= a \cdot b - a \cdot q_b \c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1998233",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
How to show $\sum^{n}_{k=0}{a_kb_k=A_nb_{n+1}+\sum^{n}_{k=1}{A_k(b_k-b_{k+1})}}$ by induction?
Let $n\in\mathbb{N}$ and $a_1,a_2,\cdots,a_n,b_1,b_2,\cdots,b_{n+1}\in\mathbb{R}$ and $A_k=\sum^{n}_{k+1}{a_j}$ with $1\leq k\leq n$.
Show that $\sum^{n}_{k=1}{a_kb_k=A_nb_{n+1}+\sum^{n}_{k=1}{A_k(b_k-b_{k+1})}}$ by inductio... | Posting my answer from chat:
$$
\begin{align}
\sum^{n+1}_{k=1}{a_kb_k} &= a_{n+1}b_{n+1}+\sum^{n}_{k=1}{a_kb_k} \\
&= a_{n+1}b_{n+1}+A_nb_{n+1}+\sum^{n}_{k=1}{A_k(b_k-b_{k+1})} \\
&= a_{n+1}b_{n+1}+A_nb_{n+1} - A_{n+1} (b_{n+1}-b_{n+2})+\sum^{n+1}_{k=1}{A_k(b_k-b_{k+1})} \\
&= A_{n+1}b_{n+2} + a_{n+1}b_{n+1} - (A_{n+1}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1999135",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Determinant with variables What is the following determinant?
$$\begin{vmatrix}1+a & b & c & d \\a & 1+b & c & d \\a & b & 1+c & d \\a & b & c & 1+d \end{vmatrix}$$
I calculated it as $0$ but I do not think it is right. Thanks in advance.
| If you are not familiar with the concepts used in Thomas Andrews' (elegant) answer; you can try to compute the determinant by using properties to create zeroes.
For example; subtract the second row from the other rows and then add the first column to the second to get:
$$\begin{vmatrix}1+a & b & c & d \\a & 1+b & c & d... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2000558",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 2
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If $\sin \alpha +\cos \alpha =1.2$, then what is $\sin^3\alpha + \cos^3\alpha$? If $\sin \alpha +\cos \alpha =1.2$, then what is $\sin^3\alpha + \cos^3\alpha$?
All I know is that $\sin^{3}a+\cos^{3}a$ is equal to
$$(\sin a + \cos a)(\sin^{2} a - \sin a \cos a + \cos^{2} a)= \dfrac{6}{5}(\sin^{2} a - \sin a \cos a + \... | $$\sin^3 \alpha+\cos^3 \alpha=(\sin \alpha+\cos \alpha)(\sin^2 \alpha-\sin \alpha\cos \alpha+\cos^2 \alpha)$$
$$1.2^3=\sin^3\alpha+\cos^3 \alpha+3\sin \alpha\cos \alpha(\sin \alpha+\cos \alpha)$$ It follows $$\sin \alpha\cos \alpha=0.22$$ and because $$\sin \alpha+\cos \alpha=1.2$$ one has sum and product of two unknow... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2001078",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 3
} |
Find the solutions of the differential equation $y\frac{dy}{dx}+\sqrt{x-xy^4}=0$ Find the solutions of the differential equation $y\frac{dy}{dx}+\sqrt{x-xy^4}=0$
Attempt: If $u=x-xy^4$ then $\frac{du}{dx}=1-(y^4+4xy^3\frac{dy}{dx})$ But I couldn't solve it from here. Do you have any suggestion? Thanks
| $$ y\frac{dy}{dx} + \sqrt x \sqrt{1-y^4} = 0 \\ y \frac{dy}{dx}=-\sqrt x \sqrt{1-y^4}\\ \int\frac{y}{\sqrt{1-y^4}}dy = - \int \sqrt x dx = -\frac23 x \sqrt x + C \\ \int \frac12\frac{d}{dy} \arcsin(y^2)dy = -\frac{2}{3}x\sqrt x + C \\ \arcsin(y^2) = -\frac43 x\sqrt x + C \\ y = \pm \sqrt{\sin\left(-\frac43 x\sqrt x + ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2002353",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Solving higher-degree trigonometric equation Is there any method to find value of $\sin (A)$ from $\sin (9A)$ having known value of $\sin (9A) = \sin (30 degree) = 0.5$?
In $\sin (9A)$, being a nine-degree equation, there will be nine-roots. Can we find all nine roots from $\sin (9A) = 0.5$?
| Using trigoniometric idendities (and not the way that @EmilioNovati is using):
*
*$$\sin\left(9\text{A}\right)=\sin\left(\text{A}\right)\left(1+2\cos\left(2\text{A}\right)\right)\left(1+2\cos\left(6\text{A}\right)\right)$$
*$$\cos\left(2\text{A}\right)=\cos^2\left(\text{A}\right)-\sin^2\left(\text{A}\right)$$
*$$\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2004875",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
$\int_{0}^{2\pi} \frac{\cos{3\theta}}{5-4\cos{\theta}}\,d{\theta}$ by using complex/real integration How to evaluate $\int_{0}^{2\pi} \frac{\cos{3\theta}}{5-4\cos{\theta}}\,d{\theta}$
by using complex integration?
I assume $z=e^{i{\theta}}$, $\frac{1}{iz}dz=d{\theta}$,
$$\cos{\theta}=\frac{z+z^{-1}}{2}
\quad\mbox{and}... | Complex Method
We better use $\cos 3x=Re(e^{3xi})=Re(z^3)$ to make the life easier.
Let $z=e^{xi}$, then the integral is transformed to a contour integration with a unit circle in anti-clockwise direction.
\begin{aligned}
I&=\operatorname{Re} \oint_{|z |=1}\frac{z^3}{5-\frac{4}{2}\left(z+\frac{1}{z}\right)} \cdot \frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2004999",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Prove by induction, that $ \sum_{i=1}^n \frac{i}{(i+1)!}= \frac{1}{(n+1)(n-1)!}$ If I'm not wrong,
$$\sum_{i=1}^n \frac{i}{(i+1)!}= \frac{1}{(n+1)(n-1)!},$$
but I am having trouble proving it by induction...
If $n=1$ the formulas coincide.
Sup $n=k$ is valid: $$\sum_{i=1}^k \frac{i}{(i+1)!}= \frac{1}{(k+1)(k-1)!}.$$
Th... | For $n=2$ we have
$$
\frac{1}{2!}+\frac{2}{3!}=\frac{1}{2}+\frac{1}{3}=\frac{5}{6}
$$
On the other hand,
$$
\frac{1}{(2+1)(2-1)!}=\frac{1}{3}
$$
so your conjecture is wrong.
For $n=3$ we have
$$
\frac{1}{2!}+\frac{2}{3!}+\frac{3}{4!}=
\frac{12}{4!}+\frac{8}{4!}+\frac{3}{4!}=
\frac{23}{24}
$$
and you can make a better c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2008704",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Calculate the multiplicative inverse of $7$ in $\mathbb{Z}_{12}$ I'm not sure how to do this but I gave it a try:
I know that $7$ must have an inverse because $\text{ gcd }(12,7)=1$
Now, on this, use euclidean algorithm. We have $$7x \equiv 1 \text{ mod } 12$$
$$x \equiv 7^{-1}(\text{ mod } 12)$$
So
$$12 = 7 \cdot 1+5$... | You have
\begin{align}
\color{red}{12}&=\color{red}{7}\cdot1+\color{red}{5}\\
\color{red}{7} &=\color{red}{5}\cdot1+\color{red}{2}\\
\color{red}{5} &=\color{red}{2}\cdot2+\color{green}{1}
\end{align}
Then
$$
1=\color{red}{5}-\color{red}{2}\cdot2=
\color{red}{5}-2(\color{red}{7}-\color{red}{5})=
-2\cdot\color{red}{7}+3\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2009475",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Unique pair of positive integers $(p,n)$ satisfying $p^3-p=n^7-n^3$ where $p$ is prime
Q. Find all pairs $(p,n)$ of positive integers where $p$ is prime and $p^3-p=n^7-n^3$.
Rewriting the given equation as $p(p+1)(p-1)=n^3(n^2+1)(n+1)(n-1)$, we see that $p$ must divide one of the factors $n,n+1,n-1,n^2+1$ on the $\te... | To eliminate the case $p\lt n^2$, note that $p\lt n^2$ implies $p+1\lt n^2+1$ and $p-1\lt n^2-1$, and this gives $p^3-p=p(p+1)(p-1)\lt n^2(n^2+1)(n^2-1)\le n^3(n^2+1)(n^2-1)=n^7-n^3$.
Remark: The paragraph that argues that either $p\gt n^2$ or $p\lt n^2$ isn't really necessary. It's obvious that $p\not=n^2$, since pr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2010543",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
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Solve for x if $1 + x + x^2 + x^3 .... = 2$ I've gone as far as rewriting it as
$$\sum_{n=1}^\infty x^{n-1} = 2$$
Not sure where to go from here.
| $$|x|<1 \to \\1+x+x^2+x^3+x^4+...=2\\ 1+x(1+x+x^2+x^3+x^4+...)=2\\1+x(2)=2\\x=\frac{1}{2}$$
or
$$\\1+x+x^2+x^3+x^4+...=2\\ 1+x+x^2(1+x+x^2+x^3+x^4+...)=2\\1+x+x^2(2)=2\\2x^2+x-1=0\\x=\frac{1}{2} \color{red}{\checkmark}\\x=-1 ,\color{red}{|x|<1}$$
so $x=-1 $ is not acceptable
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2012289",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
convergence of $\sum_{n=1}^{\infty}(n^2+1)\log^\alpha \left ( \frac{ \sinh \frac{1}{n}}{\sin\frac{1}{n}} \right )$ $ \sum_{n=1}^{\infty}(n^2+1)\log^\alpha \left ( \frac{\sinh\frac{1}{n}}{\sin\frac{1}{n}} \right ) $
Find the value of $\alpha$ for which the series converges
please, I have no idea to approach the solution... | Hint. By using Taylor series expansions, as $n \to \infty$, we have
$$
\frac{\sinh \frac1n}{\sin \frac1n} = 1+\frac{1}{3 n^2}+O\left(\frac{1}{n^4}\right)
$$ and
$$
(n^2+1)\log^\alpha \left(\frac{\sinh\frac{1}{n}}{\sin\frac{1}{n}}\right)=\frac{1}{3^a n^{2\alpha-2}}+O\left(\frac{1}{n^{2\alpha}}\right)
$$ the given series... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2012452",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Inconclusive second derivative test rigorous proof I need to find and identify stationary points of the following function:
$f(x,y) = x^4 + 2x^2y^2 - y^4 - 2x^2 + 3$
Second derivative test appears to be inconclusive (Hessian equal to zero) at the point $(0,0)$. By plotting this appears to be a maximum. How does one pr... | Suppose $(x,y)$ lies in the unit disc centered at the origin (i.e. $x^2+y^2 \le 1$).
\begin{align}f(x,y) - f(0,0) &= x^4+2x^2y^2 - y^4-2x^2\\
&\le x^4 + 2x^2y^2 - y^4 - 2x^4 & x^4 \le x^2\\
&=-(x^2-y^2)^2\\
&\le 0.
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2018823",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Solve the equation $\sqrt{3x^2-7x-30}-\sqrt{2x^2-7x-5}=x-5$
Problem Statement:-
Solve the equation $\sqrt{3x^2-7x-30}-\sqrt{2x^2-7x-5}=x-5$
Attempt at solution:-
Let $\alpha=\sqrt{3x^2-7x-30}\;\;\; \text{&} \;\;\;\beta=\sqrt{2x^2-7x-5}$.
Then, we have $$\alpha-\beta=x-5\tag{1}$$
And, we have $$\alpha^2-\beta^2=x^2-2... | rewriting as $$\sqrt{3x^2-7x-30}=x-5+\sqrt{2x^2-7x-5}$$
after squaring we get
$$3x^2-7x-30-(x-5)^2-2x^2+7x+5=2(x-5)\sqrt{2x^2-7x-5}$$
simplifying
$$10x-50=2(x-5)\sqrt{2x^2-7x-5}$$
$$5x-25=(x-5)\sqrt{2x^2-7x-5}$$
squaring again
$$- \left( 2\,x+5 \right) \left( x-6 \right) \left( x-5 \right) ^{2}=0$$
from here you wil... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2019991",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Factorise $f(x) = x^3+4x^2 + 3x$ Not sure if this belongs here, but I'm slowly trudging through my studies for Math 1 and wondered if y'all could give feedback and/or corrections on the following factorisation question:
$$ \text{Factorise}: f(x) = x^3+4x^2+3x $$
Firstly, the GCD of the above is $x$:
$$x(x^2+4x+3)$$
Now... | Avoid the 'box method' which will only work for quadratics with nice integer roots and doesn't really tell you anything about what's going on.
Suppose we have a quadratic $p(x)$ that factorizes as:
$$
p(x) = (x + s)(x + t)
$$
where we don't know what $s$ and $t$ are yet. If we multiply out the brackets, then we get:... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2020321",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 7,
"answer_id": 6
} |
How can I derive what is $1\cdot 2\cdot 3\cdot 4 + 2\cdot 3\cdot 4\cdot 5+ 3\cdot 4\cdot 5\cdot 6+\cdots + (n-3)(n-2)(n-1)(n)$ ?? I'd like to evaluate the series $$1\cdot 2\cdot 3\cdot 4 + 2\cdot 3\cdot 4\cdot 5+ 3\cdot 4\cdot 5\cdot 6+\cdots + (n-3)(n-2)(n-1)(n)$$
Since I am a high school student, I only know how to p... | In general, we have
$$\sum_{k=1}^nk(k+1)(k+2)\dots(k+p)=\frac{n(n+1)(n+2)\dots(n+p+1)}{p+2}$$
Prove by induction,
$$\sum_{k=1}^{n+1}k(k+1)(k+2)\dots(k+p)\\=(n+1)(n+2)\dots(n+p+1)+\sum_{k=1}^nk(k+1)(k+2)\dots(k+p)\\=(n+1)(n+2)\dots(n+p+1)+\frac{n(n+1)(n+2)\dots(n+p+1)}{p+2}\\=\frac{\color{#034da3}{(p+2)}\color{#ee8844}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2021231",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 10,
"answer_id": 0
} |
How can we show $\cos^6x+\sin^6x=1-3\sin^2x \cos^2x$? How can we simplify $\cos^6x+\sin^6x$ to $1−3\sin^2x\cos^2x$?
One reasonable approach seems to be using $\left(\cos^2x+\sin^2x\right)^3=1$, since it contains the terms $\cos^6x$ and $\sin^6x$. Another possibility would be replacing all occurrences of $\sin^2x$ by $1... | Factoring from $u^3+v^3=(u+v)(u^2-uv+v^2)$:
\begin{align}
\cos^6x+\sin^6x&=(\cos^2x+\sin^2x)(\cos^4x-\cos^2x\sin^2x+\sin^4x)\\
&=(\cos^2x+\sin^2x)^2-2\cos^2x\sin^2x-\cos^2x\sin^2x\\
&=1-3\cos^2x\sin^2x
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2023931",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 11,
"answer_id": 1
} |
Prove that: $\sum\limits_{cyc}\frac{1}{a}\sum\limits_{cyc}\frac{1}{1+a^2}\geq\frac{16}{1+abcd}$ Let $a$, $b$, $c$ and $d$ be positive numbers. Prove that:
$$\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}\right)\left(\frac{1}{1+a^2}+\frac{1}{1+b^2}+\frac{1}{1+c^2}+\frac{1}{1+d^2}\right)\geq\frac{16}{1+abcd}$$
I t... | My second solution
By Cauchy-Bunyakovsky-Schwarz inequality, we have
\begin{align}
&\frac{1}{1+a^2} + \frac{1}{1+b^2} + \frac{1}{1+c^2} + \frac{1}{1+d^2}\\
=\ & \frac{(1/a)^2}{1+(1/a)^2} + \frac{(1/b)^2}{1+(1/b)^2} + \frac{(1/c)^2}{1+(1/c)^2} + \frac{(1/d)^2}{1+(1/d)^2}\\
\ge\ & \frac{(1/a + 1/b + 1/c + 1/d)^2}{4 + (1/... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2025871",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
If $S=\frac{1}{1+1^2+1^4}+\frac{2}{1+2^2+2^4}+\cdots+\frac{n}{1+n^2+n^4}$, then calculate $14S$.
If $$S=\frac{1}{1+1^2+1^4}+\frac{2}{1+2^2+2^4}+\cdots+\frac{n}{1+n^2+n^4}\,$$ find the value of $14S$.
The question can be simplified to:
Find $S=\sum\limits_{k=1}^n\,t_k$ if $t_n=\dfrac{n}{1+n^2+n^4}$.
As $1+n^2+n^4$ f... | HINT:
As $n^4+n^2+1=(n^2+1)^2-n^2=(n^2-n+1)(n^2+n+1)$
$$\dfrac{2n}{n^4+n^2+1}=\dfrac{(n^2+n+1)-(n^2-n+1)}{(n^2-n+1)(n^2+n+1)}=?$$
If $f(m)=\dfrac1{m^2-m+1}, f(m+1)=?$
See Telescoping series
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2029773",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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How to solve these limits using a formula for logarithm limit(without applying L'Hopitale rule) How to solve these limits using a formula for logarithm limit(without applying L'Hopitale rule)
$$ \lim_{x \to 0} \frac{\sqrt{1 + \tan(x)} - \sqrt{1 + \sin(x)}}{x^3} $$
$$ \lim_{x \to 0} \frac{\arctan 2x}{\sin[2 \pi(x+10)]}$... | Let us consider $$A= \frac{\sqrt{1 + \tan(x)} - \sqrt{1 + \sin(x)}}{x^3}$$ and use Taylor series around $x=0$ $$\sin(x)=x-\frac{x^3}{6}+O\left(x^5\right)$$ $$1+\sin(x)=1+x-\frac{x^3}{6}+O\left(x^5\right)$$ $$\tan(x)=x+\frac{x^3}{3}+O\left(x^5\right)$$ $$1+\tan(x)=1+x+\frac{x^3}{3}+O\left(x^5\right)$$ Now, using the gen... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2031082",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 1
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Sum first $n$ squares I am trying to find the formula for the sum of square number but I am struggle
with it. I now that it isn't complex but I am young in maths.
What I've done:
Using odd numbers to find the pattern
Using triangle number and the relative
formula:
$$\frac{n(n+1)}{2}$$
1 $\rig... | What you can do is follow Euler's method :
\begin{equation*}
\sum_{k=1}^n ((k+1)^3 - k^3) = (n+1)^3 - 1
\end{equation*}
But also :
\begin{equation*}
\sum_{k=1}^n ((k+1)^3 - k^3) = \sum_{k=1}^n (3k^2 + 3k +1)
\end{equation*}
Thus, knowing $\sum_{k=1}^n k = \frac{n(n+1)}{2}$, we get :
\begin{equation*}
3\sum_{k=1}^n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2031811",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Quadratic equation 4 need some guidance with a quadratic equation.
Suppose $x^2+20x-4000=0$
Here is what I have done so far;
Using the Quadratic Equation $x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$ where from the above equation, $a=1$, $b=20$, $c=-4000$, we find
$$\begin{align*}x&=\frac{-20+\sqrt{(20)^2-(4)(1)(-4000)}}{2\times ... | Did you figure it out?? I think everyone pointed out where you went wrong, but just to be sure:
$x=\frac{-20+\sqrt{20^2-(4)(1) (-4000)}}{2*\ 1}$
$\frac{1}{2} (-20-20 \sqrt{41})$
= $54.03124237$
And similarly
$x=\frac{-20-\sqrt{20^2-(4)(1) (-4000)}}{2*\ 1}$
=$\frac{1}{2} (-20+20 \sqrt{41})$
$=-74.03124237$
I used Mathe... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2033253",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Prove for all $x$, $x^8+x^6-4x^4+x^2+1\ge0$ Prove for all $x$
$x^8+x^6-4x^4+x^2+1\ge0$
By completing the square you get
$(x^4-2)^2+(x^3)^2+(x)^2-3\ge0$
I'm stuck about the $-3$
| Let $$P(x)=x^8+x^6-4x^4+x^2+1.$$
then
$$P'(x)=8x^7+6x^5-16x^3+2x.$$
but
$$P(\pm 1)=P'(\pm 1)=0$$
$$\implies P(x)=(x^2-1)^2(x^4+3x^2+1)$$
$$\implies \forall x\in\mathbb R \;\;P(x)\geq 0.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2035806",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 3
} |
Prove that $\sum\limits_{cyc}\sqrt{a^2+10bc}\geq\frac{1}{2}\sum\limits_{cyc}\sqrt{22a(b+c)}$ Let $a$, $b$ and $c$ be non-negative numbers. Prove that:
$$\sum\limits_{cyc}\sqrt{a^2+10bc}\geq\frac{1}{2}\sum\limits_{cyc}\sqrt{22a(b+c)}$$
I tried SOS, C-S, Holder, Mixing Variables and more, but without success.
| The Buffalo Way works although it is an ugly solution.
WLOG, assume that $a+b+c=3$. Write the inequality as
$$\sum_{\mathrm{cyc}} \sqrt{\frac{a^2+10bc}{11}} \ge \sum_{\mathrm{cyc}} \sqrt{\frac{a(b+c)}{2}}.$$
Let $X = \frac{a^2+10bc}{11}, \ Y = \frac{b^2+10ca}{11}, \ Z = \frac{c^2+10ab}{11}$
and $U = \frac{a(b+c)}{2}, \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2036890",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Prove a geometry question about angles and radii in five collinear circles? Our teacher give us this question and I worked on it but I couldn't find a way to prove that. Is it possible to help me to prove that?
Thanks.
| Assumptions
I'm assuming that the following are true:
*
*$A$, $B$, $G$, $H$, $I$, $K$ are collinear
*$B$, $G$, $H$, $I$, $K$ are the centres of their respective circles
*$A$, $C$, $D$, $E$, $F$, $J$ are collinear and the line on which they lie is tangent to all the circles.
Definitions
We know these angles:
*
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2038799",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 5
} |
Evaluate exactly the following series that includes fibonacci How can i find the exact value of this?
$$\sum_{n=1}^∞ \frac{f_n}{100^n}$$
Assuming that $ f_1=1$ and $f_2=1$ and $f_n=f_{n-1}+f_{n-2}$
I can approximate it to 0.01010203050813.........
But what is the exact value? Should i use $\lim_{n\to ∞}$ ? And how to s... | Following the hint of @lulu above, we let
$$
F(x) = \sum_{n = 1}^\infty f_nx^n
$$
We are after $F(\frac1{100})$, and to do that, we need a good expression for $F(x)$. The Fibonacci recurrence relation gives
$$
F(x) = \sum_{n = 1}^\infty f_nx^n = x + x^2 + \sum_{n = 3}^\infty f_nx^n\\
= x + x^2 + \sum_{n = 3}^\infty(f_... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2039202",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Inverse Nasty Integral Intrigued by the original question on a nasty integral, one wonders what functions $f(x)$ exist such that
$$\int_0^\infty f(x)\; dx=\frac {\pi}2$$
Something to do with the area of a half-circle with unit radius perhaps?
Edited To Add
It should be specified that $f(x)$ should not contain $\pi$... | Using Glasser's Master theorem, it is very easy to cook up very nasty looking
integral which evaluates to $\frac{\pi}{2}$.
For example, start from the integral
$$\int_{-\infty}^\infty \frac{1}{u^2+1} du = \pi\tag{*1}$$
If one replace the $u$ in integrand by $\displaystyle\;x - \frac{1}{x} - \frac{1}{x+1} - \frac{1}{x+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2040121",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Solve for all real values of $x$ and $y$ Solve the system of equations for all real values of $x$ and $y$
$$5x(1 + {\frac {1}{x^2 +y^2}})=12$$
$$5y(1 - {\frac {1}{x^2 +y^2}})=4$$
I know that $0<x<{\frac {12}{5}}$ which is quite obvious from the first equation.
I also know that $y \in \mathbb R$ $\sim${$y:{\frac {-4}{5... | for $$x,y\ne 0$$ we obtain
$$1+\frac{1}{x^2+y^2}=\frac{12}{5x}$$
$$1-\frac{1}{x^2+y^2}=\frac{4}{5y}$$
adding both we get
$$5=\frac{6}{x}+\frac{2}{y}$$
from here we obtain
$$y=\frac{2x}{5x-6}$$
can you proceed?
after substitution and factorizing we get this equation:
$$- \left( x-2 \right) \left( 5\,x-2 \right) \left(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2041484",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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Find $\sin \theta$ if $\tan \theta +\sec \theta =1.5 $ Find $\sin \theta$ if $\tan \theta +\sec \theta =1.5 $
$$\tan \theta +\sec \theta =1.5 $$
$$2\tan \theta +2\sec \theta =3 $$
$$2\sec \theta =3-2\tan \theta$$
$$4\sec^2 \theta =(3-2\tan \theta)^2$$
$$4+4\tan^2 \theta =9-12\tan \theta+4\tan^2 \theta$$
So I get $$\tan... | In the second solution, when $\sin \theta = -1$, what does $\cos \theta$ equal?
That's right: $0$. And you multiplied through by that. Doing so introduced a new (but wrong) solution.
To be more correct, the SQUARING introduced a second solution, as it often does; that solution happened to be invalid because it corres... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2042019",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
Prove the following trignometric inequality for all real $x$ Prove the following trignometric inequality for all $x \in \Bbb R$
$$x^2 \sin(x) + x \cos(x) + x^2 + {\frac 12} >0$$
take $x$ in the form of radians.
This particular question is the seventh question of the 1995 Indian RMO.
| Put $t=\tan(\frac{x}{2})$. Now substitute it in the original expression. We have $$ x^{2}\sin x + x\cos x + x^{2} + \frac{1}{2}$$ $$= (1+\sin x)x^{2} + x\cos x+\frac{1}{2}$$ $$=(1 + \frac{2t}{1+t^{2}})x^{2} + x(\frac{1-t^{2}}{1+t^{2}}) + \frac{1}{2}$$ $$= \frac{(1+t)^{2}}{1+t^{2}}[x^{2} - x\frac{1-t^{2}}{(1+t)^{2}} + \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2043013",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
} |
Finding volume of a solid over region E, bounded below by cone $z = \sqrt{x^2 + y^2}$ and above by sphere $ z^2 = 1- x^2 - y^2$ I set up the triple integral as follows:
$v=\int^1_0\int^{\sqrt{x-1}}_{-\sqrt{x}}\int^{\sqrt{1-x^2-y^2}}_{\sqrt{x^2+y^2}}(1)dzdydx$ Then went to polar coords getting $$\int^{2\pi}_0\int_0^1(\s... | Note that the solid is of revolution, formed by rotating the circular sector between $z=\sqrt {1-x^2}$ and $z=x$ (about the $z$-axis).
From cylinder method, we have that
\begin{align}V&=2\pi \int_{0}^{\sqrt2 \over 2}x|\sqrt {1-x^2}-x|dx\\
&=2\pi({1\over3} - {\sqrt2\over6})
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2044672",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Finding $ \int^1_0 \frac{\ln(1+x)}{x}dx$ There is supposed to be a clean solution to the integral below, maybe involving some symmetry
$$ \int^1_0 \frac{\ln(1+x)}{x}dx$$
I have tried integration by parts as followed:
$\ln(x+1)=u$ ,$\frac{1}{x+1} dx = du$ and also $\frac{1}{x}dx=dv$, $\ln(x)=v$. Then, the integral becom... | HINT:
use the expansion $$ln(1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+\ldots$$
$$\frac{ln(1+x)}{x}=1-\frac{x}{2}+\frac{x^2}{3}-\frac{x^3}{4}+\ldots$$
$$\int \frac{ln(1+x)}{x}\ dx=x-\frac{x^2}{2^2}+\frac{x^3}{3^2}-\frac{x^4}{4^2}+\ldots$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2046219",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 7,
"answer_id": 4
} |
Solve $\frac{5}{x+3}+\frac{2}{2x+6}=4$ I want to solve $\frac{5}{x+3}+\frac{2}{2x+6}=4$ for $x$
I try this;
$5(2x+6)+2(x+3)=4(x+3)(2x+6)$
$12x+36 = 4(2x^2+12x+18)$
$8x^2+36x+36=0$
Where I would then go to the formula, however the answer says -1.5, what am I doing wrong.
| Three answers:
i) $\frac n0$ is undefined. If you are ever given an expression $\frac 5{x+3}$ you know from the very beginning that it is impossible for these to be $\frac 50$ so it is impossible for $x +3 =0$. $x +3 \ne 0$ and $x \ne -3$.
So whatever you do, you know $x = -3$ will not be an acceptable solution.
So t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2046493",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 7,
"answer_id": 5
} |
Prove equilateral triangle in another triangle We got triangle $ABC$ ($AC = BC$). ∠$ACB = 120^\circ$. $AM = MN = NB = \frac 13AB$.
Prove that CNM is equilateral triangle.
Any ideas how to do that? I'd be grateful if someone helps me!
P.S.I am sorry that the drawing is not very accurate!
| Let us use complex numbers. There is no loss of generality in assuming $C$ as the origin, $X$ axis along $CA$ and $A$ is the point $(1,0)$. Then $B$ is represented by the complex number $e^{\frac{2\pi i }{3}} = -\frac{1}{2} + i \frac{\sqrt{3}}{2}$. $M$ and $N$ are respectively
$$m = \frac{1 \cdot e^{\frac{2\pi i }{3}} ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2046651",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 3
} |
Calculating $\int_0^{\pi/3}\cos^2x+\dfrac{1}{\cos^2x}\mathrm{d}\,x$ I've been given the following exercise:
Show that the exact value of $$\int_0^{\pi/3}\cos^2x+\frac{1}{\cos^2x}\,\mathrm{d}x = \frac{\pi}{6}+\frac{9}{8}\sqrt{3}$$
Can someone help me with this?
| We can separate the integral into $$\int_{0}^{\frac{\pi}{3}} \cos^{2}(x) + \sec^{2}(x) dx = \int_{0}^{\frac{\pi}{3}}\cos^{2}(x)dx +\int_{0}^{\frac{\pi}{3}} \sec^{2}(x) dx =I_1 + I_2$$. For the first integral we have $$ I_1 = \int_{0}^{\frac{\pi}{3}} \cos^{2}(x) dx =\int_{0}^{\frac{\pi}{3}} \frac{1+\cos 2x}{2} dx = \fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2047646",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Solving $(1+x^2)y' - 2xy = (1+x^2)\arctan(x)$ I'm asked to solve the differential equation: $$(1+x^2)y' - 2xy = (1+x^2)\arctan(x).$$
I rewrite it:
$$y' - \frac{2x}{1+x^2}y = \arctan(x).$$
The integrating factor is:
$$e^{-\int{\frac{2x}{1+x^2}}dx} = e^{-\ln{1+x^2}} = (e^{ln{1+x^2}})^{-1} = \frac{1}{1+x^2}.$$
$$\frac{... | Take into account that
$$
\frac{d}{dx}\arctan(x) = \frac{1}{1+x^2}
$$
so that
$$
\int dx\; \frac{\arctan(x)}{1+ x^2} = \int dx\; \arctan(x) \frac{d}{dx}\arctan(x) = \int dx\; \frac{d}{dx}\left(\frac{1}{2}\arctan^2(x)\right) = \frac{1}{2}\arctan^2(x) + c
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2048151",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
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Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.