Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Prove that the number is the product of two successive positive integers Prove that the number $\underbrace{11\cdots 1}_{100\text{ digits}}\underbrace{22\cdots 2}_{100\text{ digits}}$ is the product of two successive positive integers.
What is the general method for this class of proof (big integers)?
| Hint:
$$
\begin{align}
\underbrace{11\cdots 1}_{100\text{ digits}}\underbrace{22\cdots 2}_{100\text{ digits}} & = 1 \cdot 10^{199} + 1 \cdot 10^{198} + \cdots + 1 \cdot 10^{100} + 2 \cdot 10^{99} + 2 \cdot 10^{98} + \cdots + 2 \cdot 10^0 \\
& = 2 \sum_{k=0}^{99} 10^k + \sum_{k=100}^{199} 10^k \\
& = \frac{(10^{100}-1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2027522",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Basic geometry: find the angle Question: Find $\angle BDE$ when
$\angle A=20^\circ$, $\overline{AB}=\overline{AC}$, $\angle CBD=65^\circ$, and $\angle BCE=25^\circ$.
So far I figured out that $\angle BDE=\angle AED-15^\circ$ and $\overline{BD}\bot\overline{CE}$ that don't seem to be helping.
| This is a variation of the notorious Langley's problem.
We follow a trigonometric approach. Let $\angle BDE=x$.
i) Use the Law of sines in $\triangle EBC$,
$$\frac{\sin 25^\circ}{EB}=\frac{\sin75^\circ}{AB}.$$
ii) Use the Law of sines in $\triangle DBC$,
$$\frac{\sin 80^\circ}{BD}=\frac{\sin 35^\circ}{AB}.$$
iii) Us... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2029820",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
How can we factorize $a{b^2} + {a^2}b + {a^2}c + a{c^2} + {b^2}c + b{c^2} + 2abc$? How can we factorize $a{b^2} + {a^2}b + {a^2}c + a{c^2} + {b^2}c + b{c^2} + 2abc$?
| The factorization of $ab^2+a^2b+a^2c+ac^2+b^2c+bc^2+2abc$
is $(a + b) (a + c) (b + c)$.
First, take out the $(a+b)$ (because a and b are factors to exactly half of the polynomial):
$$((a^2b+a^2c+ac^2+abc)+(ab^2+b^2c+bc^2+abc))*(\frac{a+b}{a+b})$$Then we have:$$(a+b)(ab+ac+bc+c^2)$$Then take out $(a+c)$:$$(a+b)(a(b+c)+c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2030326",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
finding local behaviour as x tends to $0^+$ of particular solution of ode we have
$$ y' + xy = \cos x $$
I am asked to find the first three terms in local behaviour as $x \to 0^+$.
thought:
As $ x \to 0^+$, then $\cos x \sim 1 $ so
$$ y' + xy \sim 1 $$
which can be solved by using integrating factor $e^{x^2/2}$, thus
$... | \begin{equation}
y^\prime+xy=\cos x
\end{equation}
Let $y=\sum_{n=0}^{\infty}c_nx^n$. Then
\begin{eqnarray}
y^\prime+xy&=&\sum_{n=1}^{\infty}nc_nx^{n-1}+x\sum_{n=0}^{\infty}c_nx^n\\
&=&\sum_{n=1}^{\infty}nc_nx^{n-1}+\sum_{n=0}^{\infty}c_nx^{n+1}
\end{eqnarray}
Replacing each $n$ in the first sum with $n+1$ and each $n$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2032587",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
total number of positive and negative roots of the equation $ax^3+bx^2+cx+d=0$ Suppose $a,b,c,d$ are non zero real numbers and $ab>0,$ and
$\displaystyle \int^{1}_{0}(1+e^{x^2})(ax^3+bx^2+cx+d)dx = \int^{2}_{0}(1+e^{x^2})(ax^3+bx^2+cx+d)dx=0$
Then total number of positive and negative roots of the equation $ax^3+bx^2+... | We know $a,b$ have the same sign, so the number of sign changes in the cubic is either $0, 1$ or $2$. Thus by Descartes rule of signs, at most it can have only $2$ positive roots.
From the integrals given, it is clear there must be sign changes in both the intervals $(0,1)$ and $(1,2)$. So there are two positive roots... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2035617",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
find the value of given trigonometric equation If $\alpha$ and $\beta$ are two different values of $\theta$ which satisfy $$bc \cos{θ} \cos{\phi} + ac \sin{\theta} \sin{\phi} = ab,$$ then what is the value of $$(b^2 + c^2 - a^2) \cos{\alpha} \cos{\beta} + (c^2 + a^2 - b^2) \sin{\alpha} \sin{\beta}\,?$$
| Considering an ellipse
$$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$$
The equations of tangents at $[\alpha]$ and $[\beta]$ are
$$\frac{x\cos \alpha}{a}+\frac{y\sin \alpha}{b}=1$$
$$\frac{x\cos \beta}{a}+\frac{y\sin \beta}{b}=1$$
If the tangents meet at $(c\cos \phi, \, c\sin \phi)$, then
$$(c\cos \phi , \, c\sin \phi)=
\left(
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2040565",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
How many four letter words can be formed using the letterss of the word 'INEFFECTIVE'?
How many four letter words can be formed using the letterss of the word 'INEFFECTIVE'?
Please explain with details. Also, how can we clearly differentiate between permutations and combinations?
| There are $11$ letters in 'INEFFECTIVE'.
CASE $(1)$: There is only set of three same letters of the letter $E$. So $3$ same letters can be selected in one way. Out of the six remaining distinct letters we can select one in $\binom{6}{1}=6$ ways. Hence there are $6$ groups each of which contains the three same letter... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2041676",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Proving $\sum_{t=1}^j \sum_{r=1}^{t} (-1)^{r+t} \binom{j-1}{t-1} \binom{t-1}{r-1} f(r) = f(j)$ I've run into the following identity and trying to prove it:
Let $j \in \mathbb{N}$ and $f:\mathbb{N} \to \mathbb{R}$, then
$$
\sum_{t=1}^j \sum_{r=1}^{t} (-1)^{r+t} \binom{j-1}{t-1} \binom{t-1}{r-1} f(r) = f(j)
$$
I've s... | Note that it is:
$$
\begin{gathered}
f(n) = \sum\limits_{\left( {0\, \leqslant } \right)\,k\,\left( { \leqslant \,n} \right)} {\left( \begin{gathered}
n \\
k \\
\end{gathered} \right)g(k)} \quad \Leftrightarrow \quad g(n) = \sum\limits_{\left( {0\, \leqslant } \right)\,k\,\left( { \leqslant \,n} \right)} {\le... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2043204",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Find the curve of intersection of the surfaces $x^2+y^2=z$ and $x+y+z=1$
Find the curve of intersection of the surfaces $x^2+y^2=z$ and $x+y+z=1$.
So if we substitute $z=1-x-y$ into the equation $x^2+y^2=z$ we get $x^2+y^2=1-x-y$ which can be written in the form $\left(x+\frac{1}{2}\right)^2+\left(y+\frac{1}{2}\right... | There is an error in your algebra
$$\left(x+\frac{1}{2}\right)^2+\left(y+\frac{1}{2}\right)^2=\frac32$$
so parametrically continue from here.
$$ x = \sqrt{\frac32} \cos t -\frac12\,,y =\sqrt{\frac32} \sin t -\frac12 $$
Plug these into the second equation and find $z$ in terms of $t$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2044610",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
How to show $\left(1+\frac{1}{3}x+\frac{1}{24}x^2+\frac{1}{300}x^3+...\right)^2 = 1 +\frac{2}{3}x+\frac{7}{36}x^2+\frac{1}{30}x^3+...$? I tried to find the right handside of the equation by manipulating the series but I failed at getting the right handside of it.
$$\left(1+\frac{1}{3}x+\frac{1}{24}x^2+\frac{1}{300}x^3+... | Well, you can carefully multiply and don't forget any terms:
$$\left( 1 + \frac{1}{3}x+\frac{1}{24}x^2 + \cdots \right)^2 = \left( \color{blue}{1 + \frac{1}{3}x+\frac{1}{24}x^2 + \cdots }\right)\left(\color{red}{ 1 + \frac{1}{3}x+\frac{1}{24}x^2 + \cdots }\right)$$
Expanding and combining terms of the same degree:
*
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2044953",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Interesting limit involving gamma function $$\displaystyle \lim_{n \to \infty}{\dfrac{\Gamma\left(\frac{p}{m}+n+1\right) \Gamma\left(\frac{q}{m} + \frac{n}{2} + 1\right)}{\Gamma\left(\frac{q}{m}+n+1\right) \Gamma\left(\frac{p}{m} + \frac{n}{2} + 1\right)}}$$
This is from here.
If we add one more variable -
$$\displ... | From Abramowitz and Stegun we know
$$\ln \Gamma(x) = x \ln(x) - \frac{1}{2} \ln(x) - x + \frac{\ln(2\pi)}{2} + O(\tfrac{1}{x}) \qquad (x \to \infty)$$
Applying this formula here yields
$$\begin{align*}
\ln \Gamma\left(\frac{p}{m} + \frac{n}{t}\right) &= \left(\frac{p}{m} + \frac{n}{t}\right) \ln\left(\frac{p}{m} + \fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2047875",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Prove the inequality $\sum\limits_{cyc}\sqrt{x+yz}\ge\sqrt{xyz}+\sqrt{x}+\sqrt{y}+\sqrt{z}$
Let $x,y,z$ is positive real numbers, such that $\frac1x+\frac1y+\frac1z=1$. Prove the inequality
$$\sqrt{x+yz}+\sqrt{y+zx}+\sqrt{z+xy}\ge\sqrt{xyz}+\sqrt{x}+\sqrt{y}+\sqrt{z}$$
My work so far:
$\frac1x+\frac1y+\frac1z=1\Right... | An homogenization gives
$$\sum\limits_{cyc}\sqrt{x+yz\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)}\geq\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\sqrt{xyz}+\sqrt{x}+\sqrt{y}+\sqrt{z}$$ or
$$\sum\limits_{cyc}\sqrt{yz(x+y)(x+z)}\geq xy+xz+yz+\sum\limits_{cyc}x\sqrt{yz},$$ which is C-S:
$$\sqrt{yz(x+y)(x+z)}=\sq... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2051260",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
What is wrong with the "proof" for $\ln(2) =\frac{1}{2}\ln(2)$? I have got a question which is as follows:
Is $\ln(2)=\frac{1}{2}\ln(2)$??
The following argument seems suggesting that the answer is yes:
We have the series $1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\cdots$
which has a mathematically determined value $... | In your new series
$$\left(\frac 11 - \frac 12\right) - \frac 14 +
\left(\frac 13 - \frac 16\right) - \frac 18 +
\left(\frac 15 - \frac 1{10}\right) - \frac 1{12} +
\left(\frac 17 - \frac 1{14}\right) - \frac 1{16} \dots$$
Terms of the form $\frac{1}{2z - 1}$ occur 1/3 of the time instead of 1/2 the time like they do... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2051691",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 5,
"answer_id": 3
} |
Difficult coordinate geometry problem The base of a triangle passes through a fixed point $P(a,b)$ and its sides are respectively bisected at right angles by the lines $x+y=0$ and $x=9y$. If the locus of the third vertex is a circle, then find its equation.
Apart from the fact that the circumcentre of the given triangl... | Let $A(x,y)$ be the required locus.
First reflect $A(x,y)$ about $x+y=0$, we get
$$B=(-y,-x)$$
Now reflect $A(x,y)$ about $x-9y=0$, we get $C=(x',y')$
$$\frac{y-y'}{x-x'}=-9 \tag{1}$$
Also,
$$\frac{x-9y}{\sqrt{1+9^2}}=-\frac{x'-9y'}{\sqrt{1+9^2}}$$
$$x+x'=9(y+y') \tag{2}$$
Solving $(1)$ and $(2)$,
$$C(x',y')=\left( \fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2052268",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Why is $\log\left(\frac{2a+2b+3}{2a+2b+1}\right)>\frac{1}{a+b+1}$?
Why is $\log\left(\frac{2a+2b+3}{2a+2b+1}\right)>\frac{1}{a+b+1}$ ? with $a,b\ge 0$
If this is true (I don't know whether it is true, just inserted some values in wolfram alpha) then I can show that $\frac{1}{a+b+1}<\int_{a}^{a+1}\int_{b}^{b+1}\frac{1... | Set $a+b+1=1/c$, so
$$
\frac{2a+2b+3}{2a+2b+1}=
\frac{\frac{2}{c}+1}{\frac{2}{c}-1}=\frac{2+c}{2-c}
$$
and the inequality is
$$
\log\frac{2+c}{2-c}>c
$$
for $0<c\le1$.
Consider the function
$$
f(x)=\log(2+x)-\log(2-x)-x
$$
defined over $[0,1]$. Then $f(0)=0$ and
$$
f'(x)=\frac{1}{2+x}+\frac{1}{2-x}-1=\frac{x^2}{4-x^2}>... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2052645",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
How can I find the $n$'th derivative of $f(x)=x^2 \ln(x+1)$
How can I find the $n$'th derivative of $f(x)=x^2 \ln(x+1)$
I have tried this
$$f(x)=x^2\ln(x+1)$$
$$ f'(x) = 2x\ln(x+1) + \frac{x^2}{x+1}$$
$$ f''(x)=2\ln(x+1) + \frac{4x}{x+1} - \frac{x^2}{(x+1)^2}$$
However I do not see any pattern :(
| Taking one more derivative results in
$$
f^{(3)}(x) = \frac2{x+1} + \frac{(x+1)4-4x(1)}{(x+1)^2} - \frac{(x+1)^22x-x^22(x+1)}{(x+1)^4} = \frac{2 \left(x^2+3 x+3\right)}{(x+1)^3}.
$$
However, that's not the most convenient form for taking further derivatives: using partial fractions gives us
$$
f^{(3)}(x) = \frac{2}{x+1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2053469",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Find minimal value for product of the sums Find minimal value for product of the sums
$$\left( x+\frac{x^2}{2!}+\frac{x^3}{3!}+\dots \right)
\left( \frac{1}{x}+\frac{1}{2!x^2}+\frac{1}{3!x^3}+\dots \right)
$$
Here what we did about solution. We know these series equals Taylor expansion of $(e^x-1)(e^{\frac{1}{x}}-1)$. ... | Hint. Set, for $x>0$,
$$
f(x)=(e^x-1)(e^{\frac{1}{x}}-1)=e^{x+\frac{1}{x}}-e^x-e^{\frac{1}{x}}+1
$$ giving
$$
f'(x)=\left(1-\frac1{x^2} \right)e^{x+\frac{1}{x}}-e^x+\frac1{x^2}e^{\frac{1}{x}}=\frac{e^{\frac{1}{x}}-e^{\frac{1}{x}+x}-e^x x^2+e^{\frac{1}{x}+x} x^2}{x^2}
$$ then observe that $f'(1)=0$, $f'(x)<0$ for $0<x<1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2053603",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Limit of $a\sqrt{x+1}+b\sqrt{4x+1}+c\sqrt{9x+1}$ when $x\to\infty$, for given real numbers $a$, $b$, $c$
Find the limit of $a\sqrt{x+1}+b\sqrt{4x+1}+c\sqrt{9x+1}$ when $x\to\infty$, for given real numbers $a$, $b$, $c$.
I would like to see a solving method without l'Hopital or Taylor expansion.
| When $x$ is large, the leading behavior of
$\sqrt{x+1}$, $\sqrt{4x+1}$ and $\sqrt{9x+1}$ is $\sqrt{x}$, $2\sqrt{x}$ and $3\sqrt{x}$ respectively. One approach is isolate them from the sum and see
whether you can control the remainder. After you do this, you get
$$\begin{align}
&a\sqrt{x+1} + b \sqrt{4x+1} + c\sqrt{9... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2054190",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Integral of $\int{\tan^{5}(x)\sec^4(x)}dx$? Here's my attempt at the problem:
$\int{\tan^{5}(x)\sec^4(x)}dx=
\int{\frac{\sin^5(x)}{\cos^9(x)}}\,dx=
\int{\frac{\sin(x)(\sin^2(x))^2}{\cos^9(x)}}\,dx=
\int{\frac{\sin(x)(1-\cos^2(x))^2}{\cos^9(x)}}\,dx=
\int{\frac{(1-u^2)^2}{u^9}}\,du=
\int{\frac{u^4-2u^2+1}{u^9}}\,du=
\i... | Substituting $u=\tan(x),\,du=\sec^2(x)\,dx$
\begin{eqnarray}
\int\tan^5(x)\sec^4(x)\,dx&=&\int\tan^5(x)\sec^2(x)\sec^2(x)\,dx\\
&=&\int\tan^5(x)\left[\tan^2(x)+1\right]\sec^2(x)\,dx\\
&=&\int u^5\left(u^2+1\right)\,du\\
&=&\int u^7+u^5\,du\\
&=&\frac{1}{8}u^8+\frac{1}{6}u^6+c\\
&=&\frac{1}{8}\tan^8(x)+\frac{1}{6}\tan^6... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2054394",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
lagrange multiplier inequality question
Prove for every $x,y,z>0, that\
f(x,y,z)= \frac{x}{y+z}+\frac{y}{x+z}+\frac{z}{y+x}\geq \frac{3}{2} $
So since $f$ is a homogeneous function we can prove this on the following constraint $g(x,y,z)=x+y+z-1=0$.
However, I seem to be stuck now with this equation set:
$\frac{1}{y+z... | $$\frac{x}{y+z}=\frac{x+y+z}{y+z}-1$$
So we get $$LHS=\frac{2(x+y+z)(\frac{1}{x+y}+\frac{1}{y+z}+\frac{1}{z+x})}{2}-3 \ge \frac{9}{2}-3$$
Using AM-GM inequality,
$$LHS \ge \frac{9}{2}-3=\frac{3}{2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2056830",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Prove that $\sum\limits_{cyc}\sqrt{\frac{8ab+8ac+9bc}{(2b+c)(b+2c)}}\geq5$ Let $a$, $b$ and $c$ be non-negative numbers such that $ab+ac+bc\neq0$. Prove that:
$$\sqrt{\frac{8ab+8ac+9bc}{(2b+c)(b+2c)}}+\sqrt{\frac{8ab+8bc+9ac}{(2a+c)(a+2c)}}+\sqrt{\frac{8ac+8bc+9ab}{(2a+b)(a+2b)}}\geq5$$
I tried Holder, but without succ... | The Buffalo Way works. Due to symmetry, assume that $a\ge b\ge c$.
Let
$$X = \frac{9}{25}\frac{8ab + 9bc + 8ca}{(2b+c)(b+2c)}, \quad
Y = \frac{9}{25}\frac{8bc + 9ca + 8ab}{(2c+a)(c+2a)}, \quad
Z = \frac{9}{25}\frac{8ca + 9ab + 8bc}{(2a+b)(a+2b)}.$$
We need to prove that $\sqrt{X} + \sqrt{Y} + \sqrt{Z} \ge 3$.
We will ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2057006",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Another simple rule satisfied by the Fibonacci $n$-step constants? Given,
$$x^n(2-x)=1\tag1$$
for $n=2,3,4,\dots$
$$(x-1)(x^2-x-1)=0\\
(x-1)(x^3-x^2-x-1)=0\\
(x-1)(x^4-x^3-x^2-x-1)=0$$
the roots of which are the golden ratio, the tribonacci constant, the tetranacci constant, and so on.
Q: Is it true that for all inte... | For all integer $n$ define $a_n(x) := 1-2x^n+x^{n+1} = (x-1)(x^n - \dots -x^2 - x - 1).$
For $y=x^{-1}$ we have the two algebraic identities
$$ \frac{a_n(x)(1-x^{n+1})(1+2x+x^{n+1})}{x^{3n+3}} =
2y(1-y^{n-1})(1-y^{2n+2}) - (1-y^{n+1})^3,$$
$$ \frac{-a_n(x)(1 - x^{n + 1})^2(1 + 2 x^n + x^{n + 1})}{x^{5n+3}} =
(1-y^{n-1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2059800",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Given $ab+bc+ca=3abc$, prove $\sqrt{\frac{a+b}{c(a^2+b^2 )}}+\sqrt{\frac{b+c}{a(b^2+c^2)}}+\sqrt{\frac{c+a}{b(c^2+a^2 )}}\leq 3$ $a, b, c$ are positive real numbers such that $ab+bc+ca=3abc$
Prove∶
$$\sqrt{\frac{a+b}{c(a^2+b^2 )}}+\sqrt{\frac{b+c}{a(b^2+c^2)}}+\sqrt{\frac{c+a}{b(c^2+a^2 )}}\;\;\leq\; 3$$
| We need to prove that $$\sum\limits_{cyc}\sqrt{\frac{a+b}{c(a^2+b^2)}}\leq\frac{ab+ac+bc}{abc}$$ or
$$\sum\limits_{cyc}\sqrt{\frac{ab(a+b)}{a^2+b^2}}\leq\frac{ab+ac+bc}{\sqrt{abc}}$$
By C-S $\left(\sum\limits_{cyc}\sqrt{\frac{ab(a+b)}{a^2+b^2}}\right)^2\leq(ab+ac+bc)\sum\limits_{cyc}\frac{a+b}{a^2+b^2}$.
Thus, it remai... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2061465",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Using Divergence theorem to calculate flux
Let $W$ be the region bounded by the cylinder $x^2+y^2=4$, the plane $z=x+1$, and the $xy$-plane. Use the Divergence Theorem to compute the flux of $F = \langle z,x,y+z^2 \rangle$ through the boundary of $W$.
So far I've gotten to the point of computing div$(F)$ and integra... | If $x = r\cos\theta = -1$, then
\begin{align*}
r = -\frac{1}{\cos\theta} = -\sec\theta
\end{align*}
But since this means you will eventually have two compute two integrals, maybe polar coordinates is not the way to go.
\begin{align*}
\iint_D (x+1)^2\,dA &= \int_{-1}^2 \int_{-\sqrt{4-x^2}}^{\sqrt{4-x^2}}(x+1)^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2065449",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
"answer_count": 2,
"answer_id": 1
} |
continued fraction for $\sqrt{n^2 − 1}$ Question:
Find the continued fraction for $\sqrt{n^2 − 1}$, where $n \ge 2$ is
an integer.
My attempt:
$n - 1 = \sqrt{n^2} - 1 \lt \sqrt{n^2 − 1} \lt \sqrt{n^2}$
So far, $[n-1; ]$
$\sqrt{n^2 − 1} = n - 1 + \frac{1}{x}$
$\to \frac{1}{x} = \sqrt{n^2 - 1} - (n-1) \to \frac{1}{... | $$\begin{align*}\sqrt{n^2-1} &= (n-1) + \sqrt{n^2-1} - (n-1) \\& = (n-1) + \dfrac{1}{\dfrac{1}{\sqrt{n^2-1}-(n-1)}}\\ &=(n-1) + \dfrac{1}{\dfrac{\sqrt{n^2-1}+(n-1)}{2(n-1)}} \\ &= (n-1) + \dfrac{1}{1 + \dfrac{\sqrt{n^2-1}-(n-1)}{2(n-1)}} \\& = (n-1) + \dfrac{1}{1 + \dfrac{1}{\dfrac{2(n-1)}{\sqrt{n^2-1}-(n-1)}}} \\&=(n-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2066269",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
If $a+b+c+d=0$ and $\{a,b,c,d\}\subset[-1,1]$ so $\sum\limits_{cyc}\sqrt{1+a+b^2}\geq4$ Let $\{a,b,c,d\}\subset[-1,1]$ such that $a+b+c+d=0$. Prove that:
$$\sqrt{1+a+b^2}+\sqrt{1+b+c^2}+\sqrt{1+c+d^2}+\sqrt{1+d+a^2}\geq4$$
I tried Holder and more, but without success.
| Let $ {x,y,z}\in[-1, 1]$,and $x+y+z=0 $
Prove: $$\sqrt{1+x+y^2}+\sqrt{1+y+z^2}+\sqrt{1+z+x^2}\geq 3$$
Proof:
First we will show:
If $ab\geq 0, $ then:$\sqrt{1+a}+\sqrt{1+a}\geq 1+\sqrt{1+a+b}$
This is obvious！
Note that at least two of $ x+y^2, y+z^2 $ and $ z+x^2 $ have the same positive and negative values。Without lo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2066916",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 1
} |
Prove by induction that $n^3 < 3^n$ The question is prove by induction that $n^3 < 3^n$ for all $n\ge4$.
The way I have been presented a solution is to consider:
$$\frac{(d+1)^3}{d^3} = (1 + \frac{1}{d})^3 \ge (1.25)^3 = (\frac{5}{4})^3 = \frac{125}{64} <2 < 3$$
Then using this $$(d+1)^3 = d^3 \times \frac{(d+1)^3}{d^... | The best solution that I can produce using some of the ideas from above is:
Proving the base case :
For $n=4$, we have $P(4): 3^4 > 4^3 \Leftrightarrow 81 > 64$ which is true.
Assume that the inequality is true for some d:
Assume $P(d): 3^d > d^3$.
Then show that $P(d+1)$ is true:
$P(d+1): 3^{d+1} > (d+1)^3 $
Starting ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2068426",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 3
} |
prove $(a^3+1)(b^3+1)(c^3+1)\ge 8$ let $a,b,c\ge 0$ and such $a+b+c=3$ show that
$$(a^3+1)(b^3+1)(c^3+1)\ge 8$$
My research:It seem use Holder inequality,so
$$(a^3+1)(b^3+1)(c^3+1)\ge (abc+1)^3$$
Use AM-GM
$$abc\le\dfrac{(a+b+c)^3}{27}=1$$
what? then I think this method is wrong
| BY Jensen inequality with $\phi\left(t\right)=\ln(t^3+1)$
$$\phi\left(\frac13\sum_{i=1}^{3}x_i\right)\le \frac13\sum_{i=1}^{3}\phi\left(x_i\right) $$
since $a+b+c= 3$ we get
$$\ln(2)=\phi\left(1\right)\le \frac13\left(\ln(a^3+1)+\ln(b^3+1)+\ln(c^3+1)\right) $$
hence $$\ln8= 3\ln(2)\le \ln \left((a^3+1)(b^3+1)(c^3+1)\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2069201",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
What is the remainder left after dividing $1! + 2! + 3! + ... + 100!$ by $5$? I have this question as a homework.
What is the remainder left after dividing $1! + 2! + 3! + \cdots + 100!$ by $5$?
I tried this: I noticed that every $n! \equiv 0 \pmod{5}$ for every $n\geq 5$.
For $n < 5$:
$$\begin{align*}
1! &\equiv 1 ... | You are right.
Notice that the unit digit of the sum is $3$, because all the terms in the sum end up with a zero from $5!$ and hence to find the unit digit, we just need to find the unit digit of the sum $1!+2!+3!+4!=33$ which is $3$.
So, when the sum $1! + 2! + 3! + \cdots + 100!$ is divided by $5$, it leaves a remai... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2069717",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 3,
"answer_id": 2
} |
Prove that if $10 \mid (a^2+ab+b^2)$, then $1000 \mid (a^3-b^3)$.
Let $a$ and $b$ be integers. Prove that if $10 \mid (a^2+ab+b^2)$, then $1000 \mid (a^3-b^3)$.
I saw that $a^3-b^3 = (a-b)(a^2+ab+b^2)$ and $(a^2+ab+b^2) = (a+b)^2-ab$. How can we use the fact that $10 \mid (a^2+ab+b^2)$ to solve this question?
| Here's a more general result:
Let $\Psi_n(x,y)$ denote the $n$th homogenized cyclotomic polynomial. If $x,y\in\mathbb Z$, every prime divisor of $\Psi_n(x,y)$ is either:
*
*$1\pmod n$
*a prime divisor of $n$
*a prime divisor of both $x$ and $y$
This is a simple corollary of the same result for the u... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2070405",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
A fair coin is tossed repeatedly and independently until two consecutive heads or two consecutive tails appear? A fair coin is tossed repeatedly and independently until two consecutive heads or two consecutive tails appear . What is the probability of the number of tosses ?
I tried it as :
For success, either we end u... | While other answers are correct, they don't explain the solution.
I am assuming your question is about the expected number of coin tosses. Let $X$ be the discreet random variable that is the number of throws until two of the same coin are observed. Your question is asking about the expected value of $X$ i.e. $E(X)$.
I... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2071174",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
The sum of $\binom{n}{0}\binom{3n}{2n}-\binom{n}{1}\binom{3n-3}{2n-3}+\binom{n}{2}\binom{3n-6}{2n-6}+\cdots \cdots $
The sum of $$\binom{n}{0}\binom{3n}{2n}-\binom{n}{1}\binom{3n-3}{2n-3}+\binom{n}{2}\binom{3n-6}{2n-6}-\cdots \cdots \cdots $$
$\bf{My\; Try::}$ We can write above sum as $$\sum^{n}_{k=0}(-1)^k\binom{n... | \begin{align*}
&\binom{n}{0}\binom{3n}{2n}-\binom{n}{1}\binom{3n-3}{2n-3}+\binom{n}{2}\binom{3n-6}{2n-6}-\cdots \\
&=\binom{n}{0}\binom{3n}{n}-\binom{n}{1}\binom{3n-3}{n}+\binom{n}{2}\binom{3n-6}{n}-\cdots
\end{align*}
The right hand side is the coefficient of $x^n$ in
\begin{align*}
&\binom{n}{0}(1+x)^{3n} - \binom{n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2071532",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Choosing one type of ball without replacement. Suppose I have $9$ balls, among which $3$ are green and $6$ are red. What is the probability that a ball randomly chosen is green?
It is $\dfrac{3}{9}=\dfrac{1}{3}$.
If three balls are randomly chosen without replacement, then what is the probability that the three balls ... | a) Your first answer is correct.
b) All 3 are green without replacement
= $\dfrac39 \cdot \dfrac28 \cdot \dfrac17$ = $\dfrac1{84}$
c) 2 balls are green out of 3 without replacement.
= $ \left(\dfrac{3}{9} \cdot \dfrac{2}{8} \cdot \dfrac{6}{7} \right ) + \left(\dfrac{3}{9} \cdot \dfrac{6}{8} \cdot \dfrac{2}{7} \ri... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2075588",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Why is the solution to $x-\sqrt 4=0$ not $x=\pm 2$? If the equation is $x-\sqrt 4=0$, then $x=2$.
If the equation is $x^2-4=0$, then $x=\pm 2$.
Why is it not $x=\pm 2$ in the first equation?
| $x^2 - 2^2 = 0$
$(x+2)(x-2)=0$
$x+2=0$ or $x-2=0$
$\sqrt4 = 2$
$x-\sqrt4=0$
$x=2$
Why can't the square root of $4$ be $-2$ instead of $2$, if $-2$ times $-2$ also equals $4$?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2075745",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 13,
"answer_id": 10
} |
System of $3$ nonlinear equations
Find positive solutions to the following:
\begin{align*}
x^2+y^2+xy=&1\\
y^2+z^2+yz=&3\\
z^2+x^2+xz=&4
\end{align*}
I simplified and got $x+y+z=\sqrt{7}$ and $x^2+y^2+xy=1$. How do I continue?
| All positive real solutions are given by
$$
(x,y,z)=\frac{1}{\sqrt{7}}(2,1,4).
$$
I obtained the solutions by finding one linear equation (note that $x+y+z= 7$ does not hold), namely $2y + z - 3x=0$.
Then replacing $z=3x-2y$ yields two quadratic equations, which are easy to solve with the formula for quadratic equation... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2076499",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Show that $\sqrt{4 + 2\sqrt{3}} - \sqrt{3}$ is rational.
Show that $\sqrt{4 + 2\sqrt{3}} - \sqrt{3}$ is rational.
I've tried to attempt algebra on this problem. I noticed that there is some kind of nesting effect when trying to solve this. Please help me to understand how to attempt to denest this number.
Any help wo... | Many questions with
sum or difference of square roots
can be solved with
conjugating.
So,
if $s = \sqrt{4 + 2\sqrt{3}} - \sqrt{3}$,
and $t = \sqrt{4 + 2\sqrt{3}} + \sqrt{3}$,
$\begin{array}\\
st
&=(\sqrt{4 + 2\sqrt{3}} - \sqrt{3})(\sqrt{4 + 2\sqrt{3}} + \sqrt{3})\\
&=4 + 2\sqrt{3}-3\\
&=1 + 2\sqrt{3}\\
\end{array}
$
Si... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2076737",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 6,
"answer_id": 5
} |
Unexpected Proofs Using Generating Functions I recently came across this beautiful proof by Erdős that uses generating functions in a unique way:
Let $S = \{a_1, \cdots, a_n \}$ be a finite set of positive integers such that no two subsets of $S$ have the same sum. Prove that $$\sum_{i=1}^n \frac{1}{a_i} < 2.$$
Quest... |
Sicherman dice are the only pair of 6-sided dice that are not normal dice, bear only positive integers, and have the same probability distribution for the sum as normal dice.
The faces on the dice are numbered 1, 2, 2, 3, 3, 4 and 1, 3, 4, 5, 6, 8.
(Source: Wikipedia article on Sicherman dice)
We can prove this fact... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2077997",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "38",
"answer_count": 5,
"answer_id": 4
} |
Simple limit. Used L'Hôpital's rule. Didn't work. $\lim_{x\rightarrow 0}f(x)$
$f(x)=\frac{\exp (\arcsin \left (x \right ))-\exp (\sin \left (x \right ))}{\exp (\arctan \left (x \right ))-\exp (\tan \left (x \right ))}$
I tried the L'Hôpital's rule as mentioned in the title and replaced ($\arcsin x$,$\arctan x$,$\s... | We can proceed as follows
\begin{align}
L &= \lim_{x \to 0}\frac{e^{\arcsin x} - e^{\sin x}}{e^{\arctan x} - e^{\tan x}}\notag\\
&= \lim_{x \to 0}e^{\sin x - \tan x}\cdot\frac{e^{\arcsin x - \sin x} - 1}{e^{\arctan x - \tan x} - 1}\notag\\
&= \lim_{x \to 0}\frac{e^{\arcsin x - \sin x} - 1}{\arcsin x - \sin x}\cdot\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2078291",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
Linear transformation matrix of vector space $\mathbb R^{2x2}$
Select a basis B of a vector space $\mathbb R^{2x2}$ and for linear transformation $f:\mathbb R^{2x2}→\mathbb R^{2x2}$ given by $f(X) = \begin{pmatrix}1 & 0 \\ 1 & 0 \end{pmatrix}X+\begin{pmatrix}1 & 1 \\ 1 & 1 \end{pmatrix}X^T$ compute the matrix relative... | Speedding: your calculations are correct. Another way. If $X=\begin{pmatrix}{x_1}&{x_2}\\{x_3}&{x_4}\end{pmatrix}\in\mathbb{R}^{2\times 2}$ then, $$f(X)=\begin{pmatrix}1 & 0 \\ 1 & 0 \end{pmatrix}X+\begin{pmatrix}1 & 1 \\ 1 & 1 \end{pmatrix}X^T=\ldots=\begin{pmatrix}{2x_1+x_2}&{x_2+x_3+x_4}\\{2x_1+x_2}&{x_2+x_3+x_4}\en... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2078951",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Want to show $e_1 (x) := \lim_{N\rightarrow \infty} \sum_{\mu=-N}^{N} \frac{1}{x+\mu} = \frac{1}{x} - \sum_{m=1}^{\infty} \gamma_m x^{2m-1}$ I want to prove
\begin{align}
e_1 (x) := \lim_{N\rightarrow \infty} \sum_{\mu=-N}^{N} \frac{1}{x+\mu}
\end{align}
for $\mu\neq 0$, $|x|<1$,
\begin{align}
e_1(x) = \frac{1}{x} -... | My trial is as follows
\begin{align}
e_1 (x) &:= \lim_{N\rightarrow \infty} \sum_{\mu=-N}^{N} \frac{1}{x+\mu} \\
&= \frac{1}{x} + \lim_{N\rightarrow \infty} \sum_{\mu=1}^N \frac{2x}{x^2-\mu^2} \\
& = \frac{1}{x} - \lim_{N\rightarrow \infty} \sum_{\mu=1}^N \frac{2x}{\mu^2-x^2} \\
&= \frac{1}{x} - \lim_{N\rightarrow ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2079220",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
For which $n$ this inequality is always true $a^5+1\ge a^3+a^n$ for every $a>0$? For which $n$ this inequality is always true $a^5+1\ge a^3+a^n$ for every $a>0$?
1.$1$
2.$2$
3.$3$
4.$4$
5.$5$
My attempt:It's clear that it is true for $n=2$.Because:
$(a^3-1)(a^2-1) \ge 0$
Is true because $a^3-1$ and $a^2-1$ are both n... | Note that equality always holds for $a = 1$.
$
a^5 + 1 \geq a^3 + a^n \\
\implies a^5 - a^3 \geq a^n - 1 \\
\implies a^3(a + 1)(a - 1) \geq a^n - 1
$
Case 1: $a > 1$. Then,
$$
a^4 + a^3 \geq 1 + a + a^2 + \cdots + a^{n-1}
$$
Case 2: $0 < a < 1$. Then,
$$
a^4 + a^3 \leq 1 + a + a^2 + \cdots + a^{n-1}
$$
Now, both cases ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2080262",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
Find the number of three elements sets of positive integers $\{a,b,c\}$ such that $a\times b\times c=2310$. Find the number of three elements sets of positive integers $\{a,b,c\}$ such that $a\times b\times c=2310$.
My attempt:
Now, $2310=2\times 3\times 5\times 7\times 11=a\times b\times c$
$a=2^{x_{1}}3^{y_{1}}5^{z_{... | $5=3+1+1, 1+2+2.$ So making sets without integer $1,$ first possibility, $10$ ways. Second possibility, $15$ ways. If integer $1$ is taken once $, 5=3+2, 4+1.$
First possibility $10$ ways, second possibility $5$ ways. So total $40$ ways.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2080757",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 5,
"answer_id": 3
} |
Identifying the digits of $37 \cdot aaaa\ldots a$. With a calculator, I have noticed that the integer $37$ multiplied with some particular numbers yields numbers with some structures.
For instance, let $aaaa\ldots a$ be a natural number of $n$ identical digits. Then, $ 37 \cdot aaaa\ldots a$ is a number with $n+1$ or ... | Observe the patterns in the written calculation of these products
$$a=1\to\begin{matrix}3&7\\&3&7\\&&3&7\\&&&3&7\end{matrix}$$
$$a=2\to\begin{matrix}7&4\\&7&4\\&&7&4\\&&&7&4\end{matrix}$$
$$a=3\to\begin{matrix}1&1&1\\&1&1&1\\&&1&1&1\\&&&1&1&1\end{matrix}$$
$$a=4\to\begin{matrix}1&4&8\\&1&4&8\\&&1&4&8\\&&&1&4&8\end{matr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2083110",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
When is the fraction $\frac{n+7}{2n+7}$ the square of a rational number?
Find all positive integers $n$ such that the fraction $\dfrac{n+7}{2n+7}$ is the square of a rational number.
The answer says $n = 9,57,477$, but how do we prove this? I wrote $\dfrac{n+7}{2n+7} = \dfrac{a^2}{b^2}$ for some $a,b \in \mathbb{Z^+}... | $$\gcd(n+7,2n+7)=\gcd(n+7,n)=\gcd(7,n)\in\{1,7\}$$
If the $\gcd=1$, then $n+7,2n+7$ are both squares:
$$n+7=a^2,2n+7=b^2\implies 2a^2-b^2=7$$
If the $\gcd=7$, then $\frac{n+7}{7},\frac{2n+7}{7}$ are both integer squares. Write $n=7m$:
$$m+1=a^2,2m+1=b^2\implies2a^2-b^2=1$$
Finding exact solutions from here requires stu... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2083696",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
} |
Integral of $2x^2 \sec^2{x} \tan{x}$ I've been trying for a while to find $\int{( 2x^2 \sec^2{x} \tan{x} )} dx$, using integration by parts.
I always end up getting a more complicated integral in the second part of the equation.
For example:
$$ \int{( 2x^2 \sec^2{x} \tan{x} )} dx =
\\ 2x^2 \tan^2x - \int{\tan{x} \cdot ... | We have,
$\int \sec ^{2}x\tan {x}dx$
Put v = $\sec{x}$
Then dv = 2$\sec{x}\tan{x}$
So $\int (v) dv=\frac{1}{2}v^{2}=\frac{1}{2}\sec ^{2}x.$
Now use this result in question,
2$\int{(x^2 \sec^2{x} \tan{x} )} dx$
Integrating by parts, integrate $\sec^2x \tan{x}$ and differentiate $x^2$
We have,
2$\left[\frac{1}{2}x^{2}\se... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2084215",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
Integer solutions to nonlinear system of equations $(x+1)^2+y^2 = (x+2)^2+z^2$ and $(x+2)^2+z^2 = (x+3)^2+w^2$
Do there exist integers $x,y,z,w$ that satisfy \begin{align*}(x+1)^2+y^2 &= (x+2)^2+z^2\\(x+2)^2+z^2 &= (x+3)^2+w^2?\end{align*}
I was thinking about trying to show by contradiction that no such integers exi... | We need to find (or prove that it does not exist) an odd number $a$ that is the difference of two squares:
$$y^2-z^2=a$$
This is quite easy. We only need to write $a$ as the product of two different numbers: $a=mn$. If $m>n$, say, then the easy possbility is that $m=y+z$ and $n=y-z$. We only need to solve for $y$ and $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2085253",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 1
} |
Prove that there do not exist integers that satisfy the system
Prove that there do not exist integers $x,y,z,w,t,$ that satisfy \begin{align*}(x+1)^2+y^2 &= (x+2)^2+z^2\\(x+2)^2+z^2 &= (x+3)^2+w^2\\(x+3)^2+w^2 &= (x+4)^2+t^2.\end{align*}
I thought about using a modular arithmetic argument. The given system is equival... | My solution is similar, in spirit, to Fimpellizieri's, but organized a little differently ...
Suppose $x,y,z,w,t$ are integers such that
\begin{align}
2x + 3 &= y^2 - z^2 \tag{eq1}\\
2x + 5 &= z^2 - w^2 \tag{eq2}\\
2x + 7 &= w^2 - t^2 \tag{eq3}
\end{align}
\begin{align}
\text{Equation } (1) &\implies y,z \text{ have ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2085316",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
How to calculate $w = F(v)$ direct and also over the relation $c_{B}(w) = A\cdot c_{A}(v)$ We have linear transformation $F = \mathbb{P_{2}} \rightarrow \mathbb{R^{2}} $, where $ F(p(t)) = \begin{pmatrix} p(0) \\ P(1) \end{pmatrix},$
$ A = \{1,t,t^{2}\},$
$B=\left \{ \begin{pmatrix} 1\\ 0\end{pmatrix},\begin{pmatrix} 0... | By definition, we have
$$ F(v) = \begin{pmatrix} v(0) \\ v(1) \end{pmatrix} = \begin{pmatrix} a \\ a + b + c \end{pmatrix}. $$
Since $v = a \cdot 1 + b \cdot t + c \cdot t^2$, we also have
$$ c_A(v) = \begin{pmatrix} a \\ b \\ c \end{pmatrix} $$
and so
$$ c_B(F(v)) = M^A_B c_A(v) = \begin{pmatrix} 1 & 0 & 0 \\ 1 & 1 & ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2087749",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
About an integration technique My professor wrote this :
$\displaystyle \int x^2\log(\sqrt{1-x^2}) \,\mathrm{d}x = \frac{1}{3}\int \log(\sqrt{1-x^2}) \, \mathrm dx^3$
Can someone explain me what is going on with the integration variable ?
| $$\int x^2 \log\sqrt{1-x^2} dx = \int x^2 \log\sqrt{1-(x^3)^{2/3}} dx$$
Let $u = x^3$, so $du = 3x^2 dx$ or
$$dx = \frac{du}{3x^2}$$
so the integral becomes
$$\int \frac{x^2}{3x^2} \log\sqrt{1-u^{2/3}} du = \frac{1}{3} \int \log\sqrt{1-u^{2/3}} du \ .$$
Resubstituting $u = x^3$ yields the rather confusing expression o... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2089164",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Show that if $2+2\sqrt{28n^2+1}$ is an integer then it must be perfect square. As written in title, I want to prove that
If $n$ is an integer, show that if $2+2\sqrt{28n^2+1}$ is an integer than it must be perfect square.
I m struggling in making a start . Please help.
| With suggestions of @barak manos and @ Alqatrkapa I Noted that I can improve my post.
Notice that $2+2\sqrt{28n^2+1}$ is an even integer. Also, $28n^2+1$ is a perfect square of an odd integer say $m$ (Because $28n^2+1$ is odd itself).
Now, $$28n^2=m^2-1=(m+1)(m-1)\implies 7n^2=(\frac{m+1}{2})(\frac{m-1}{2})$$
Hence, ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2090665",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 3,
"answer_id": 2
} |
Solving quadratic congruences mod powers of a prime. The problem statement is to prove that there are exactly two solutions in the 5-adic numbers to $x^2+1 =0$.
My investigation suggests that there should be solutions, one based on each solution to $x^2+1\equiv 0 \pmod 5$, which are 2 and 3 (the additive inverse of 2)... | Suppose $x^2\equiv -1 \bmod 5^k$, so that $x^2\equiv d\cdot5^k-1 \bmod 5^{k+1}$ for some $d \in \{0,1,2,3,4\}$. Now consider:
$x^2 \bmod 5^{k+1}$
$(x+5^k)^2 \bmod 5^{k+1}$
$(x+2\cdot5^k)^2 \bmod 5^{k+1}$
$(x+3\cdot5^k)^2 \bmod 5^{k+1}$
$(x+4\cdot5^k)^2 \bmod 5^{k+1}$
Multiplying these out, we get (mod $5^{k+1}$):
$d\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2090786",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
A simple proof by induction $P(n) = \frac{6^{2n} - 3^n}{11} \in \mathbb{N}$ Verify by induction that $P(n) = \frac{6^{2n} - 3^n}{11} \in \mathbb{N} \quad \forall n \ge 1 \in \mathbb{N}$
Basis: $P(1) \Rightarrow \frac{33}{11} \in \mathbb{N}$.
Induction: if the statement holds for some $n$ $\Rightarrow$ holds for $n+1$
I... | So the statement $P(n)$ could also be formulated that $11$ divides $6^{2n} -3^n$. So $P(n+1)$ would consider
\begin{align}
6^{2(n+1)} - 3^{n+1} & = \\
6^{2n} \times 36 - 3 \times 3^n & = \\
3(12 \times 6^{2n} - 3^n )&= \\
3(11\times 6^{2n} + 6^{2n} - 3^{n}) & = \ldots
\end{align} Can you use $P(n)$ in order to prove ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2091861",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 6,
"answer_id": 3
} |
How did he simplify this problem like this? $$\left(\frac{x}{12}+\frac{x}{18}\right)t=x$$
$$\left(\frac{1}{12}+\frac{1}{18}\right)t=1$$
I want a solid rule, how did he simplify the problem in the image like this!
| If
$$\left(\frac{x}{12}+\frac{x}{18}\right)t=x\tag{1}$$
then in general it is not true that
$$\left(\frac{1}{12}+\frac{1}{18}\right)t=1\tag{2}$$
Counterexample: if $t=36$ and $x=0$, then you see that $(1)$ is equivalent to $0=0$ which is true, but $(2)$ is equivalent to $5=1$ which is not.
However, if you already know ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2092894",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Find the bound of the following sequences Find the bound of the following sequences:
$$a)\ \frac{n^2}{n^2+1}$$
$$b)\ \frac{n^3}{n+1}$$
My attempt:
$a)$$$0<\frac{n^2}{n^2+1}<1, \forall n\in \Bbb N $$
$$\frac{n^2}{n^2+1}>0 \Leftrightarrow (n^2>0) \land (n^2+1)>0$$
which is true $\forall n\in \Bbb N$.
$$\frac{n^2}{n... | $a)$ is correct, and so are the bounds.
For $b)$, it is true that $\forall n \in \mathbb N$, both $n^3 \geq 0$ and $n+1 \geq 0$ (infact, $n+1 \geq 1$) hold true. Therefore, their quotient is also positive i.e. $\frac {n^3}{n+1} \geq 0$ is also true.
However, to analyse the other direction, we can use the following tr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2093029",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
For $x+y=1$, show that $x^4+y^4\ge \frac{1}{8}$ As in the title. Let $x,y$ be two real numbers such that $x+y=1$. Prove that $x^4+y^4\ge \frac{1}{8}$.
Any hints? Basically, the only method I am aware of is plugging $y=1-x$ into the inequality and investigating the extrema of the function, but I don't think it's the bes... | If $a,b\in \mathbb{R}$, then $(a+b)^2\leq 2(a^2+b^2)$ since $2ab\leq a^2+b^2$.
Applying this with $x$ and $y$ we get
$$ 1=(x+y)^2\leq 2(x^2+y^2)$$
so $x^2+y^2\geq \frac{1}{2}$, and then applying the inequality again with $x^2$ and $y^2$ we get
$$ \frac{1}{4}\leq (x^2+y^2)^2\leq 2(x^4+y^4)$$
which is the desired result.... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2095390",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 10,
"answer_id": 2
} |
Write a product of two numbers in non decimal base If $A=(b-1) (b-1) (b-1)$ and $B=(b-1) (b-1)$ are written in base $b$, what is $A\times B$ in base $b$?
I've tried developing $A \times B$ in decimal base but couldn't get this product back in base $b$.
Beside this, is there a general procedure to compute product in n... | Your two numbers as the largest possible three digit and largest possible two digit number written in base $b$. I.e. they are $1000_b-1_b$ and $100_b-1_b$. Algebraically these are equal to $b^3-1$ and $b^2-1$ respectively. As such their product will be given by:
$$(b^3-1)(b^2-1)=b^5-b^3-b^2+1$$
This however has negativ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2096179",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Decomposition of this partial function I came across this
$$\int \frac{dx}{x(x^2+1)^2}$$
in "Method of partial functions" in my Calculus I book.
The thing is that, however, decomposing $\frac{1}{x(x^2+1)^2}$ would take long in this way:
$$\frac{1}{x(x^2+1)^2}= \frac{A}{x}+\frac{Bx+C}{x^2+1}+\frac{Dx+E}{(x^2+1)^2}$$
... | One way to simplify this a bit, is to multiply numerator and denominator of your integrand by $x$:
$$\int \frac{x}{x^2(x^2+1)^2}dx$$
Now substituting $y = x^2$; $dy = 2x dx$, you get the integral
$$\frac{1}{2}\int \frac{1}{y(y+1)^2}dy$$
Splitting this in partial fractions is more straightforward than what you had befo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2096391",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 1
} |
Differentiate by first principle,$f(x)=\frac{x^2}{\sin x}$ Differentiate by first principle,$f(x)=\frac{x^2}{\sin x}$
$$f'(x)=\lim_{h\to0}\frac{f(x+h)-f(x)}{h}$$
$$f'(x)=\lim_{h\to0}\frac{\frac{(x+h)^2}{\sin(x+h)}-\frac{x^2}{\sin x}}{h}$$
$$f'(x)=\lim_{h\to0}\frac{\frac{(x+h)^2\sin x-x^2\sin(x+h)}{\sin(x+h)\sin x}}{h}... | \begin{eqnarray}
\dfrac{f(x+h)-f(x)}{h}&=&\dfrac{1}{h}\left[\dfrac{(x+h)^2}{\sin(x+h)}-\dfrac{x^2}{\sin x}\right]\\
&=&\dfrac{(x+h)^2\sin x-x^2\sin(x+h)}{h\sin x\sin(x+h)}\\
&=&\dfrac{[(x+h)^2-x^2]\sin x+x^2\sin x-x^2\sin(x+h)}{h\sin x\sin(x+h)}\\
&=&\dfrac{[(x+h)^2-x^2]\sin x-x^2[\sin(x+h)-\sin x]}{h\sin x\sin(x+h)}\\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2097066",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Compute $5^{10,000}\equiv \mod 52$ I'm unsure how to solve questions like these, we are given an example below but I'm still confused about how they got from $5^{100\times10+0}$ to $5^{0}$.
ie) Compute $5^{1000} \bmod 77$
$$
5^{1000}=5^{(166⋅6+4)}=5^4=25^2=4^2=16=2 \pmod 7\\
5^{1000}=5^{(100⋅10+0)}=5^0=1 \pmod{11}\\
5^... | One uses the effective Chinese remainder theorem:
Let $7u+11v=1$ a Bézout's relation between $7$ and $11$. Then $$(x\equiv a\mod 7 \;\text{ and }\; x\equiv b\mod 11)\iff (x\equiv 7bu+11av\mod 7\times11).$$
One ingredient is Lil' Fermat: as $7$ and $11$ are prime, $x^6\equiv1\mod 7$ for all $x\not \equiv 0\mod 7$ and... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2097894",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
What's the name of the following method for dividing polynomials? It's not long-division nor synthetic division I saw this method in some random PDF and am intrigued of the exact method used. I can't find any page of this method on the web because I'm not sure what you'd call this method.
Here is the method:
For solvi... | What you have is literally polynomial long division written out without division signs. Indeed, what is written out here is the essence of polynomial long division, which is all about finding the coefficient of the factor that returns the highest degree term in the original polynomial.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2099461",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 1,
"answer_id": 0
} |
Proof that $\frac{a}{a+3b+3c}+\frac{b}{b+3a+3c}+\frac{c}{c+3a+3b} \ge \frac{3}{7}$ for all $a,b,c > 0$ So I am trtying to proof that $\frac{a}{a+3b+3c}+\frac{b}{b+3a+3c}+\frac{c}{c+3a+3b} \ge \frac{3}{7}$ for all $a,b,c > 0$. First I tried with Cauchy–Schwarz inequality but got nowhere.
Now I am trying to find a conve... | WLOG $a\geq b\geq c.$ $$\text {Let } x=1/(a+3b+3c),\; y=1/(b+3c+3a),\; z=1/(c+3a+3b).$$ We have $a+3b+3c\leq b+3c+3a\leq c+3b+3a.$ Therefore $$(1).\quad x\geq y\geq z>0.$$ Let $f(a,b,c)=ax+by+cz.$
Differentiating $f$ by $c,$ keeping $a$ and $b$ constant, we have $$\partial f/\partial c=-3ax^2-3by^2+3(a+b)z^2=3(a(z^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2100641",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 7,
"answer_id": 4
} |
Fibonacci numbers divisible by p for some prime p? This came up in a series of notes on number theory I'm reading. The question is:
Prove that the $(p-1)$th or the $(p+1)$th Fibonacci number is divisible by $p$ for some prime $p$
I'm very much lost on this question, so I thought I'd see if I could find a proof here.
| Fix a prime $p\ge 7$. We have by Cassini's identity that
$$
F_{p-1}F_{p+1} = F_p^2-(-1)^{p-1} \,\,\,\,(=\, F_p^2-1).
$$
Hence, it is enough to prove that
$$
F_p^2 \equiv 1\pmod{p}.
$$
But we have
\begin{align*}
F_p^2&=\left(\frac{1}{\sqrt{5}}\left(\frac{1+\sqrt{5}}{2}\right)^p-\frac{1}{\sqrt{5}}\left(\frac{1-\sqrt{5}}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2100728",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Finding out the range of values of the argument of a complex number Let $z,z^2,z^3,z^4$ be the four complex numbers.If these taken in order form a cyclic quadrilateral then the question is to find out the range of values of $\theta$ where $\theta$=$arg(z)$ and $\theta €(0,2\pi)$.
Since it forms a cyclic quadrilateral ... | You've already got $|z|=1$ correctly.
Note here that we have to have
$$z\not=z^2,z\not=z^3,z\not=z^4,z^2\not=z^3,z^2\not=z^4,z^3\not=z^4$$
giving
$$\theta\not=\frac{2}{3}\pi,\pi,\frac{4}{3}\pi$$
Also, we have to consider the expression "these taken in order".
For $0\lt\theta\le\frac{\pi}{2}$, since we have $0\lt\theta... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2102434",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Two fair dice are rolled. What is the conditional probability that at least one lands on 6 given that the dice land on different numbers? $F =$ The probability that the dice land on different numbers
$F = {(1,2),(1,3),(1,4),(1,5),(1,6),(2,1),(2,3),(2,4),(2,5),(2,6),(3,1),(3,2),(3,4),(3,5),(3,6),(4,1),(4,2),(4,3),(4,5),... | There are only 10 cases where the dice have different numbers: (1, 6), (2, 6), ..., (5, 6), (6, 5), (6, 4), ..., (6, 1).
So the answer is then $\frac{\frac{10}{36}}{\frac{5}{6}}=\frac{1}{3}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2102523",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Determine limit: $\lim_{x \to -\infty} (\sqrt{x^2 +2x} - \sqrt{x^2 - 2x}) $ Determine the limit of:
$$\lim_{x \to -\infty} \left(\sqrt{x^2 +2x} - \sqrt{x^2 - 2x}\right) $$
I've tried a few times, most notably the following two versions. I'm looking for a comment on both, since both amount to a wrong answer.
First attem... | The second is almost good, to mistake from the fact that if $x<0$, $\sqrt{x^2}=-x$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2103112",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
How to compute the sum $\sum_{n=0}^\infty \tfrac{n^2}{2^n}$?
How to find this sum :
$\sum_{n=0}^\infty \dfrac{n^2}{2^n}$
$\sum_{n=0}^\infty \dfrac{n^2}{2^n}=\dfrac{1}{2}+\dfrac{4}{4}+\dfrac{9}{8}+\dfrac{16}{16}+\dfrac{25}{32}+\dfrac{36}{64}+\dfrac{49}{128}+\dots$
Now $\sum_{n=0}^\infty \dfrac{n}{2^n}\leqslant \sum_{n... | Hint: For $|x|<1$
\begin{align*}
\sum_{n=0}^{\infty }n^{2}x^{n}&=\sum_{n=0}^{\infty }n\left ( n+1 \right )x^{n}-\sum_{n=0}^{\infty }nx^{n} \\
&=x\sum_{n=0}^{\infty }n\left ( n+1 \right )x^{n-1}-x\sum_{n=0}^{\infty }nx^{n-1}\\
&=x\left ( \sum_{n=0}^{\infty }x^{n+1} \right )''-x\left ( \sum_{n=0}^{\infty }x^{n} \right )'... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2104107",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Calculate velocity vector along a curve (as a function of time) I have a curve defined by an equation: $y = f(x)$
In my case, the equation is a polynomial $y = ax^3 + bx^2 + cx + d$.
I also have $2$ boundary conditions. Point $A$ and Point $B$. I know the velocity and position of $A$ and $B$ and I know the time taken t... | Using cubic spline:
\begin{align*}
p(x) &=
\frac{x-b}{a-b}f(a)+\frac{x-a}{b-a}f(b) \\
& \quad +
(x-a)(x-b)\left \{
\frac{x-b}{(a-b)^{2}} \left[ f'(a)-\frac{f(a)-f(b)}{a-b} \right]+
\frac{x-a}{(b-a)^{2}} \left[ f'(b)-\frac{f(b)-f(a)}{b-a} \right]
\right \} \\[5pt]
p'(x)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2104250",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Finding two possible values of $z^2 + z + 1$ given that $z$ is one of the three cube roots of unity. Factorise $z^3 - 1 $.
If $z$ is one of the three cube roots of unity, find the two possible values of $z^2 + z + 1$.
Factorising gives you :
$(z - 1)(z^2 + z + 1) = 0$ since $z$ is one of the three cube roots of unity.
... | $$
0 = z^2 + z + 1 = (z+1/2)^2 - (1/2)^2 + 1 = (z+1/2)^2 + 3/4 \iff \\
z + \frac{1}{2} = \pm i \frac{\sqrt{3}}{2} \iff \\
z = \frac{-1 \pm i \sqrt{3}}{2}
$$
By the way:
$$
1 + z + z^2 = \frac{z^3-1}{z-1}
$$
as a finite term geometric series.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2105017",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Probability of cards drawn from a deck
The cards are drawn from a well shuffled deck of 52 cards one after
the other without replacement. The probability of first card being a
spade and the second a black king is ?
Here, is my approach,
Upon first draw we got a black spade king
$$P(\text{first card is spade}) = \... | Upon first draw we got a black spade king
$$P(\text{first card is $K \spadesuit$}) = \frac{1}{52}$$
$$ P(\text{second black king i.e $K\clubsuit$}) = \frac{1}{51} $$
Upon first draw we don't get a black spade king
$$P(\text{first card is $\spadesuit$ but not $K\spadesuit$}) = \frac{12}{52}$$
$$ P(\text{second black $K$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2105956",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Evaluate the series $\sum_{n=0}^{\infty} \frac{n^2-1}{n!}\frac{x^n}{n-1}$ I would like to show that the following sum converges $\forall x \in \mathbb{R}$ as well as calculate the sum:
$\sum_{n=0}^{\infty} \frac{n^2-1}{n!}\frac{x^n}{n-1}$
First for the coefficient:
$\frac{n^2-1}{n!(n-1)}=\frac{n+1}{n!}$
Then, what I ... | I think it might be easier to not incur the $(n+1)^2$, but to simply break up the sum into
$$
\sum_{n=0}^\infty \frac{n+1}{n!} x^n \;\; =\;\; \sum_{n=0}^\infty \frac{n}{n!} x^n + \sum_{n=0}^\infty \frac{x^n}{n!}.
$$
Clearly the second sum goes to $e^x$, but we can notice that
$$
\sum_{n=0}^\infty \frac{nx^n}{n!} \;\; ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2106152",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Find a minimum of $x^2+y^2$ under the condition $x^3+3xy+y^3=1$ As in the title, I've tried to find a maximum and mininum of $x^2+y^2$ when $x^3+3xy+y^3=1$ holds. It is not too hard to show that $x^2+y^2$ has no maximum, but I can't find a minimum.
Lagrange multiplier gives a dirty calculation so I can't handle it. Is ... | Noting that
$$x^3+y^3=(x+y)^3-3xy(x+y)$$
and
$$(x+y)^3-1=(x+y-1)\left[(x+y)^2+(x+y)+1\right]$$
gives
$$x^3+3xy+y^3=1 \\\\ \implies\ (x+y)^3-1-3xy(x+y)+3xy=0 \\\\ \implies\ (x+y-1)\left[(x+y)^2+(x+y)+1-3xy\right]=0 \\\\ \implies\ (x+y-1)(x^2+y^2-xy+x+y+1)=0$$
So either $x+y=1$ or $x^2+y^2-xy+x+y+1=0$.
Treating the latte... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2107048",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 0
} |
computation of minimum value of a function
How will we calculate the minimum value ilof the equation with the given condition.
| Using Holder,s Inequality
$\displaystyle (x^2+8y^2+27z^2)\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\geq \bigg[\bigg(x^2\cdot \frac{1}{x}\cdot \frac{1}{x}\bigg)^{\frac{1}{3}}+\bigg(8y^2\cdot \frac{1}{y}\cdot \frac{1}{y}\bigg)^{\frac{1}{3}}+\bigg(27z^2\cdot \frac{1}{z... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2108077",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
How to find $\lim_{x\to1}\frac{\sqrt[3]{x+7}+2\sqrt{3x+1}-6}{\sqrt[4]{x}-1}$ without using L'Hospital Rule? Compute the following limit. I've tried using l'Hospital. And it work the result was $\dfrac{7}{3}$. But how can I do this without this rule? I am trying to help I friend who hasn't done derivatives yet.
$$\lim_{... | Let $x=(u+1)^4$:
$$\frac{\sqrt[3]{(u+1)^4+7}+2\sqrt{3(u+1)^4+1}-6}u\\=\frac{\sqrt[3]{u^4+4u^3+6u^2+4u+8}+2\sqrt{3u^4+12u^3+18u^2+12u+4}-6}u$$
then some binomial expansions:
$$\sqrt[3]{8+(u^4+4u^3+6u^2+4u)}=2+\frac1{12}(u^4+4u^3+6u^2+4u)+\mathcal O(u^2)$$
$$\sqrt{4+(3u^4+12u^3+18u^2+12u)}=2+\frac14(3u^4+12u^3+18u^2+12u)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2110708",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
How do I find the limit of the following sequence? $a_0$ to $a_5$ are set independently, all positive.
Then let us define the sequence as $a_n=\frac{\sum_{i=1}^6a_{n-i}}{6},\;\forall n\geq6$.
How do I find $\lim_{n\rightarrow\infty}a_n$?
Or, will it converge at all? Or will it just be exactly the limit value after s... | Hint:
Note that if
$$x_n = \begin{pmatrix}
a_{n-5}\\
a_{n-4}\\
a_{n-3}\\
a_{n-2}\\
a_{n-1}\\
a_{n}
\end{pmatrix} ; \, \, \, \, \, A = \begin{pmatrix}
0 & 1 & 0 & 0 & 0 & 0 \\
0 & 0 & 1 & 0 & 0 & 0 \\
0 & 0 & 0 & 1& 0 & 0 \\
0 & 0 & 0 & 0 & 1& 0 \\
0 & 0 & 0 & 0 & 0 & 1\\
1/6 & 1/6 &1/6 &1/6 &1/6 &1/6
\end{pmatrix},
$$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2116109",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Find the limit: $\lim_{x\to 0} \frac{8^x-7^x}{6^x-5^x}$ Find the limit:
$$\lim_{x\to 0} \frac{8^x-7^x}{6^x-5^x}$$
I really have no idea what to do here, since I obviously can't plug in $x=0$ nor divide by highest power...Any help is appreciated!
| $$\lim_{x\to0} \frac{8^x-7^x}{6^x-5^x} = \lim_{x\to0} \frac{8^x-7^x}{6^x-5^x} \cdot \frac{x}{x} = \lim_{x\to0} \frac{8^x-1-(7^x-1)}{6^x-1-(5^x-1)}\cdot\frac{x}{x} = \lim_{x\to0} \frac{8^x-1-(7^x-1)}{x}\cdot\lim_{x\to0}\frac{x}{6^x-1-(5^x-1)} = \bigg(\lim_{x\to0}\frac{8^x-1}{x}-\frac{7^x-1}{x}\bigg)\cdot\lim_{x\to0} \fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2118307",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
In right triangle $ABC$ ( $BC$ is hypotenuse),$D$ is a point on $BC$.Calculate $AB$ given that...
In right triangle $ABC$ ( $BC$ is hypotenuse),$D$ is a point on $BC$.Calculate $AB$ given that: $AD$=5,$BD=3$,$CD=6$.
| You also can use Stewart's Theorem:
$$\frac{AC^2}{CD\cdot BC}-\frac{AD^2}{CD\cdot BD}+\frac{AB^2}{BD\cdot BC}=1\\ \frac{81-AB^2}{6\cdot 9}-\frac{5^2}{6\cdot 3}+\frac{AB^2}{3\cdot 9}=1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2120648",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Difficult trigonometry problem to find the minimum value Find the minimum value of $5\cos A + 12\sin A + 12$.
I don't know how to approach this problem. I need help.
I'll show you how much I got...
$$5\cos A +12\sin A + 12 = 13(5/13\cos A +12/13\sin A) + 12$$
| $(p\cos A+q\sin A)^2+(p\sin A-q\cos A)^2=p^2+q^2$ See Brahmagupta-Fibonacci Identity
$\implies(p\cos A+q\sin A)^2\le p^2+q^2$ the equality occurs if $p\sin A-q\cos A=0$
$\iff-\sqrt{p^2+q^2}\le p\cos A+q\sin A\le \sqrt{p^2+q^2}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2122332",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 7,
"answer_id": 2
} |
If $abc=1$ so $\sum\limits_{cyc}\frac{a^3c}{(a+c)(b+c)}\geq\frac{3}{4}$ Let $a$, $b$ and $c$ be positive numbers such that $abc=1$. Prove that:
$$\frac{a^3c}{(a+c)(b+c)}+\frac{b^3a}{(b+a)(c+a)}+\frac{c^3b}{(c+b)(a+b)}\geq\frac{3}{4}$$
I tried C-S and BW. It does not help.
| Let $a = \frac{x}{y}, \ b = \frac{y}{z}, \ c = \frac{z}{x}; \ x, y, z > 0$.
It suffices to prove that $f(x, y, z)\ge 0$ where
\begin{align*}
f(x,y, z) &= 4\, x^7\, z^5 - 3\, x^6\, y^3\, z^3 + 4\, x^6\, y^2\, z^4 + 4\, x^5\, y^7 - 3\, x^5\, y^5\, z^2 - 3\, x^5\, y^2\, z^5 + 4\, x^4\, y^6\, z^2 - 6\, x^4\, y^4\, z^4\\
&... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2123675",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
How did Descartes come up with the spoof odd perfect number $198585576189$? We call $n$ a spoof odd perfect number if $n$ is odd and and $n=km$ for two integers $k, m > 1$ such that $\sigma(k)(m + 1) = 2n$, where $\sigma$ is the sum-of-divisors function.
In a letter to Mersenne dated November $15$, $1638$, Descartes sh... | If $22021$ were prime, we would have $$\sigma(d)=\sigma(3^2\cdot 7^2\cdot 11^2\cdot 13^2)\cdot 22022=(1+3+3^2)\cdot(1+7+7^2)\cdot(1+11+11^2)\cdot(1+13+13^2)\cdot 22022=2d$$ which can be verified by multiplication
I guess Descartes calculated $\sigma(3^2\cdot 7^2\cdot 11^2\cdot 13^2)=3^2\cdot 7\cdot 13\cdot 19^2\cdot 61... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2124167",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 1,
"answer_id": 0
} |
How do I solve $(x+1)^5 +36x+36 = 13(x+1)^3$? I tried $$(x+1)^5 + 36 x + 36 = 13 (x +1)^3\\
(x+1)^5 + 36(x+1) = 13 (x +1)^3\\
(x+1)^4 +36 = 13 (x+1)^2
$$
But, don't understand how to solve further. Can somebody show step by step please. Thanks!
| Assuming that you don't see the direct substitution $t=x+1$ for whatever reason, note that $(x+1)^5 +36x+36 - 13(x+1)^3$ is a monic polynomial with the free term $1+36-13=24\,$. By the rational root theorem, try the divisors of $24$ for possible rational roots, and enjoy your luck.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2125300",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 4
} |
How does this simplify to 1? I am working on this differetiation problem:
$ \frac{d}{dx}x(1-\frac{2}{x})$
and I am currently stuck at this point:
$1\cdot \left(1-\frac{2}{x}\right)+\frac{2}{x^2}x$
Symbolab tells me this simplifies to $1$ but I do not understand how. I am under the impression that;
$1\cdot \left(1-\frac... | $1\cdot \left(1-\frac{2}{x}\right)+\frac{2}{x^2}x $
= $1\cdot 1- 1 \cdot \frac{2}{x} + \frac{2}{x}$
= $1- \frac{2}{x} + \frac{2}{x}$
=$1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2127110",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Find the polynomials which satisfy the condition $f(x)\mid f(x^2)$
I want find the polynomials which satisfy the condition
$$f(x)\mid f(x^2).$$
I want to find such polynomials with integer coefficients, real number coefficients and complex number coefficients.
For example, $x$ and $x-1$ are the linear polynomials ... | I don't have "the answer" per se, but I think I can provide a new perspective on it.
Let's work over the complex numbers $\mathbb{C}$. That way, we can write every polynomial as the product of its roots, i.e. in the form $f(x) = (x - r_1)^{e_1}...(x - r_n)^{e_n}$ for distinct $r_1, ... , r_n \in \mathbb{C}$.
Let's fi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2128306",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
Prove $a_n = n 3^{n-1}$, recurrence relation. Let $a_0 = 0, a_1 = 1$, and let $a_{n+2} = 6a_{n+1} - 9a_n$ for $n\geq 0.$
Prove $a_n = n 3^{n-1}$ for $n\geq 0.$
attempt: Suppose $a_0 = 0, a_1 = 1$, and let $a_{n+2} = 6a_{n+1} - 9a_n$.
Then notice when $n = 0$, we have $a_2 = 6_{1} - 9a_0 = 6-0 = 6$.
And $a_2 = 2*(3^{2-... | Induction.
True for $n=0,n=1$.
We will assume true up to $n$, will show holds for $n+1$:
$$a_{n+1} = 6 a_n -9 a_{n-1} = 6 n 3^{n-1} - 9 (n-1) 3^{n-2}=(2n -(n-1))3^n=(n+1)3^{(n+1)-1}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2134345",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Find the maximum and minimum value of $\cos^2x-6\sin(x)\cos(x)+3\sin^2x+2$ Find the maximum and minimum value of $$\cos^2x-6\sin(x)\cos(x)+3\sin^2x+2$$.
i simplified and reach to expression as follows :
$5 + 2\sin(x)[\sin(x)-3\cos(x)]$. How do i go from here?
Thanks
| HINT:
Let $y=\cos^2x-6\sin(x)\cos(x)+3\sin^2x+2$
Divide both sides by $\cos^2x$ and use $\sec^2x=1+\tan^2x$ to form a Quadratic Equation in $\tan x$
Now as $\tan x$ is real, the discriminant must be $\not<0$
Or divide both sides by $\sin^2x$ to form a Quadratic Equation in $\cot x$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2134658",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Find the GCD of ... Find the GCD of:
$y^2-10y+24+6x-9x^2$, $2y^4-18x^2y^2-48xy^2-32y^2$.
My Attempt:
$$
\begin{split}
1^{st} \text{expression} &= y^2-10y+24+6x-9x^2\\
&={(y)^2-2\cdot(y)\cdot5+(5)^2}-(5)^2+24+6x-9x^2\\
&=(y-5)^2-{25-24-6x+9x^2}\\
&=(y-5)^2-{(1)^2-2\cdot(1)\cdot(3x)+(3x)^2}\\
&=(y-5)^2-(1-3x)^2\\
&=(y-5+... | The problem is that sometimes you can't find a factorization for the algebric expressions. If you can is just to use Mark's approach.
If not, a more general way is use the same approach that we use to find the $\gcd$ between integer numbers. Which is based on Euclides's division algorithm.
Step $1)$: Divide $$a(y)=2y... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2135188",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
How to find this integral... $$\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}e^{-\left(3x^2+2\sqrt2xy+3y^2\right)}dxdy$$
I have no idea how to integrate this function. If the middle $xy$ term would not have been present it would have been easy. But the $xy$ term is causing a problem.
| $$I=\int_{-\infty}^{+\infty}\int_{-\infty}^{+\infty}e^{-(3x^2+2\sqrt2xy+3y^2)}dxdy$$
Let $x=r\cos\theta$, $y=r\sin\theta$.
Converting to polar form,
$$\begin{align}I&=\int_{0}^{2\pi}\int_{0}^{\infty}e^{-(3r^2\cos^2\theta+2\sqrt2r^2\sin\theta\cos\theta+3r^2\sin^2\theta)}rdrd\theta\\
&=\int_{0}^{2\pi}\int_{0}^{\infty}e^{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2135511",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Find all points where $f$ is continuous $f:\mathbb{R}^2 \rightarrow \mathbb{R}$
$f(x,y)=\begin{cases}
\frac{\sin{y}+2yx^2}{y}, & y \gt0 \\
\frac{y+3x}{1-y}, & y\le 0
\end{cases}$
I have to check
a) What points is $f$ continuous at?
b) Is $f$ class $C^1$ at $(1,0)$
My idea for a)
Function $f$ is continuous at all point... | As you say, the function is continuous for all $y \neq 0$, and so we need to consider $y=0$.
$$\lim_{y \to 0^+} \left(\frac{\sin y +2x^2y}{y}\right) \ = \ \lim_{y\to 0^+}\left(\frac{\sin y}{y} + \frac{2x^2y}{y}\right) \ = \ 1+\lim_{y \to 0^+}\left(\frac{2x^2y}{y}\right) \ = \ 1+2x^2$$
$$\lim_{y\to 0^-}\left(\frac{3x+y}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2138648",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How to integrate ${dx}/{\sqrt{\frac{2K}{m}\left(\frac{1}{x}-\frac{1}{R}\right)}}$
If
$\displaystyle\frac{dx}{dt}=-\sqrt{\frac{2K}{m}\left(\frac{1}{x}-\frac{1}{R}\right)}$
and $x(0)=R$, find $t$ when $x=0$.
I have no idea how to integrate $\displaystyle\frac{dx}{\sqrt{\frac{2K}{m}\left(\frac{1}{x}-\frac{1}{R}\righ... | Starting with
$$\int\frac{dx}{\sqrt{\frac{2K}{m}\left(\frac{1}{x}-\frac{1}{R}\right)}}$$
We will replace all the constant with simpler ones. We let $\frac{2K}{m} = a^2$ and we let $R=b^2$
$$\int\frac{dx}{a\sqrt{\left(\frac{1}{x}-\frac{1}{b^2}\right)}}$$
The obvious thing now is to let $x \to b^2 x$
$$\frac{b}{a}\int\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2138883",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
interval of convergence of the power series $\sum_{n=1}^\infty(1+\frac{1}{n})^{n^2}x^n$ Find the radius of convergence and the interval of convergence of the power series
$\sum_{n=1}^\infty(1+\frac{1}{n})^{n^2}x^n$
My attempt: The root test brings us to
$\lim_{n\rightarrow\infty}|(1+\frac{1}{n})^nx|$,
but I'm not sure ... |
Note: Both tests, root test as well as ratio test can be used to analyze the convergence behavior. But, the first one is in this case considerably simpler to use.
We recall the validity of the limit
\begin{align*}
\lim_{n\rightarrow\infty}\left(1+\frac{1}{n}\right)^n=e
\end{align*}
Root test:
We obtain
\begin{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2139683",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Limit of function with natural logarithm I need help solving this problem. I tried L'hospital and rearranging but nothing worked.
$$
\lim_\limits{x→∞} x^2\left(\ln\left(1 + \frac{1}{x}\right)- \frac{1}{x+1}\right)
$$
| Using L'Hospital rule we get $$\\ \lim _{ x\rightarrow \infty }{ \frac { \ln { \left( 1+\frac { 1 }{ x } \right) -\frac { 1 }{ x+1 } } }{ \frac { 1 }{ { x }^{ 2 } } } } =\lim _{ x\rightarrow \infty }{ \frac { \frac { x }{ x+1 } \cdot \left( -\frac { 1 }{ { x }^{ 2 } } \right) +\frac { 1 }{ { \left( x+1 \right) ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2139810",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
Is there a quicker way to evaluate $\int_{0}^{\infty} \frac{1-x^2}{x^4+3x^2+1}\ dx$? The integral is: $$\int_{0}^{\infty} \frac{1-x^2}{x^4+3x^2+1}\ dx$$
My procedure:
$$4\int_{0}^{\infty} \frac{1-x^2}{(2x^2+3)^2-5}\ dx=4\int_{0}^{\infty} \frac{1-x^2}{(2x^2+3-\sqrt5)(2x^2+3+\sqrt5)}\ dx$$
$$$$Using partial fractions to ... | We have
\begin{align*}
\int \frac{1-x^2}{x^4 + 3x^2 + 1} \, dx
&= - \int \frac{1 - x^{-2}}{(x + x^{-1})^2 + 1} \, dx \\
&= - \arctan\left(x + \frac{1}{x}\right) + C.
\end{align*}
Here, we utilized the substitution $u = x + x^{-1}$ when we move to the second line. For the definite integral, we have
$$ \int_{0}^{\infty} ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2143667",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
Solving and proving inequalities? If $a,b$ and $c$ are positive real numbers, how do I prove that:
$$\frac{a^3}{b^2-bc+c^2}+
\frac{b^3}{c^2-ca+a^2} + \frac{c^3}{a^2-ab+b^2} \geq 3 \cdot
\frac{ab+bc+ca}{a+b+c}.$$
and when is equality?
Are there general techniques to solve these symmetric and or cyclic inequalities?
| By C-S $$\sum_{cyc}\frac{a^3}{b^2-bc+c^2}=\sum_{cyc}\frac{a^4}{b^2a+c^2a-abc}\geq\frac{(a^2+b^2+c^2)^2}{\sum\limits_{cyc}(a^2b+a^2c-abc)}.$$
Thus, it remains to prove that
$$(a^2+b^2+c^2)^2(a+b+c)\geq3(ab+ac+bc)\sum\limits_{cyc}(a^2b+a^2c-abc)$$ or
$$\sum_{cyc}(a^5+a^4b+a^4c-a^3b^2-a^3c^2-6a^3bc+5a^2b^2c)\geq0.$$
Sinc... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2144113",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Evaluating $\frac{z_2-\sqrt{3}i}{z_3-\sqrt{3}i}$ Let $A(z_1),B(z_2),C(z_3)$ are complex numbers satisfying $|z-\sqrt{3}i|=1$ and $3z_1+\sqrt{3}i=2z_2+2z_3$. The question asks to find the value of $$\frac{z_2-\sqrt{3}i}{z_3-\sqrt{3}i}$$ I tried to shift the origin to $\sqrt{3}i$ and transformed the original condition in... | We can rewrite the equation as $3(z_1 - \sqrt {3} i) = 2(z_2-\sqrt {3}i)+2(z_3-\sqrt {3}i)$
i.e. $3\vec{OZ_1} = 2\vec{OZ_2}+2 \vec{OZ_3}$
So if $\theta = \pm \arg \frac{z_2-\sqrt{3}i}{z_3-\sqrt{3}i}$ is the angle between $\vec{OZ_2}$ and $\vec{OZ_3}$, we have from the above equation that
$9 = 2^2+2^2+8 \cos \theta \Ri... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2145406",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Find the remainder of $\frac{3^{11}-1}{2}$ divided by $9$ I have the following question:
Find the remainder of $$\frac{3^{11}-1}{2}$$ divided by $9$
I tried to reformat the question:
$$\frac{3^{11} -1 } {2} \times \frac{1}{9} = \frac{3^{11}-1}{18}$$
Since $3^2 = 9$
$$\frac{3^2(3^9) -1}{3^2 \times2}$$
I don't know whe... | The tool you need is modular arithmetic.
$3^{11}$ is congruent to $0$ mod $9$
$3^{11} - 1$ is congruent to $8$ mod $9$
The only hard part is the division. But, fortunately, you can uniquely divide by $2$ mod $9$:
$0*2 = 0,
1*2 = 2,
2*2 = 4,
3*2 = 6,
4*2 = 8,
5*2 = 1,
6*2 = 3,
7*2 = 5,$ and
$8*2 = 7$
so $8/2 = 4$ modulo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2146798",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 2
} |
An interesting limit: $\lim_{n\rightarrow \infty}\frac{1}{n^{2}+1}+\frac{2}{n^{2}+2}+\cdots+\frac{n}{n^{2}+n}$ $$\lim_{n\rightarrow \infty}\frac{1}{n^{2}＋1}＋\frac{2}{n^{2}＋2}＋…＋\frac{n}{n^{2}＋n}$$
Thanks in advance
| HINT:
$$\frac{1}{n^2+n}\sum_{k=1}^n(k)\le \sum_{k=1}^n\frac{k}{n^2+k}\le \frac{1}{n^2+1}\sum_{k=1}^n (k)$$
SPOILER ALERT: Scroll over the highlighted area to reveal the solution
We can write the limit as$$\lim_{n\to\infty}\sum_{k=1}^n\frac{k}{n^2+k}$$Then, using $n^2+n\ge n^2+k\ge n^2+1$ we have $$\frac{1}{n^2+n}\sum... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2148266",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Find $N$ given a sequence $a_n$
In the sequence {$a_n$}, $$a_n=\frac{a_1+a_2+...+a_{n-1}}{n-1} \, , \quad n \ge 3$$
If $a_1+a_2 \ne 0$ and the sum of the first $N$ terms is $12(a_1+a_2)$, find $N$.
Kind of lost on where to start with this one. My initial thought was, $$\sum^{N}_{n=3}\frac{a_1+a_2+...+a_{n-1}}{n-1}=... | \begin{align*}
S_2 &= a_1+a_2 \\
S_n &= a_1+a_2+\ldots+a_n \\
a_{n+1} &= \frac{a_1+\ldots+a_{n}}{n} \\
&= \frac{S_n}{n} \\
S_{n+1} &= S_{n}+\frac{S_n}{n} \\
&= \frac{n+1}{n} S_{n} \\
&= \frac{n+1}{n} \cdot \frac{n}{n-1} \cdot \ldots \cdot \frac{4}{3} \cdot \frac{3}{2} S_{2} \\
&= \frac{n+1}{2} S_2 \\
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2149290",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Find the greatest integer not exceeding $A$ if $A = \sqrt{x^2+\sqrt{4x^2+\sqrt{16x^2+\sqrt{100x^2+39x+\sqrt{3}}}}}$ and $x=20062007$.
Find the greatest integer not exceeding $A$ if $$A = \sqrt{x^2+\sqrt{4x^2+\sqrt{16x^2+\sqrt{100x^2+39x+\sqrt{3}}}}}$$ and $x=20062007$.
The nested square roots are very tricky and I am... | A).
Show that $A>20062008$:
$$E:=100x^2+39x+\sqrt{3}>64x^2+16x+1 = (8x+1)^2;$$
$$D:=16x^2 + \sqrt{E} > 16x^2 + (8x+1) = (4x+1)^2;$$
$$C:=4x^2+ \sqrt{D} > 4x^2+(4x+1) = (2x+1)^2;$$
$$B:=x^2+\sqrt{C} > x^2+(2x+1) = (x+1)^2;$$
$$A=\sqrt{B}>x+1.$$
B). Show that $A<20062009$. More rough estimation:
$$E<(10x+2)^2;$$
$$D<16x^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2150083",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Proving that $c$ is the mean proportional between $a$ and $b$.
If $a≠b$ and $a:b$ is the duplicate ratio of $(a+c):(b+c)$, prove that $c$ is the mean proportional between $a$ and $b$.
My attempt:
I have assumed that $c$ is the mean proportional between $a$ and $b$, expressed $a$ and $c$ in terms of $b$ and then subst... | Hint:
$$
\require{cancel}
\begin{align}
\frac{a}{b}=\frac{(a+c)^2}{(b+c)^2} \quad & \implies \quad a(b+c)^2=b(a+c)^2 \\
& \iff \quad a(b^2+2bc+c^2)=b(a^2+2ac+c^2) \\
& \iff \quad ab^2 +\cancel{2abc}+ac^2-a^2b-\cancel{2abc}-bc^2=0 \\
& \iff \quad ab(b-a)+(a-b)c^2=0 \\
& \iff \quad (a-b)(c^2-ab)=0
\end{align}
$$
All ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2150803",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How come rearrangement of a convergent series may not converge to the same value / or does not converge at all By looking at Riemann's Rearrangement theorem I wonder, How come a convergent series particular re-arrengement may not converge to the same value of the original series.
Isn't the below true?
Let $\phi : N \to... | Here's a concrete example where rearrangement of a convergent series results in a series that converges to a different value.
Consider
$$
1 - 1 + \frac{1}{2} - \frac{1}{2} + \frac{1}{3} - \frac{1}{3}
+ \frac{1}{4} - \frac{1}{4} + \frac{1}{5} - \frac{1}{5} + \ldots
$$
The partial sums of this series alternate 0 between... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2151423",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 6,
"answer_id": 4
} |
Proving G.Bauer's Pi Formula
Question: How do you prove this $\pi$ formula found by G.Bauer in $1859$$$\dfrac 2\pi=1-5\left(\dfrac {1}{2}\right)^3+9\left(\dfrac {1\cdot3}{2\cdot4}\right)^3-13\left(\dfrac {1\cdot3\cdot5}{2\cdot4\cdot6}\right)^3+\&\text{c}.$$
I'm not sure what to do in this case. I found it in a wikipe... | This is what I have so far:
Start with the Hypergeometrical Series$$_4F_3\left[\begin{array}{c c}\frac 12n+1,n,-x,-y\\\frac 12n,x+n+1,y+n+1\end{array};-1\right]=\dfrac {\Gamma(x+n+1)\Gamma(y+n+1)}{\Gamma(n+1)\Gamma(x+y+n+1)}\tag1$$
And let $n=-x=-y=\frac 12$. Therefore, $(1)$ takes the form
$$\begin{align*}_4F_3\left[... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2153349",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.