Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Solving an equation containing cube roots: $\sqrt[3]{5x+7}-\sqrt[3]{5x-12}=1$ I'm trying to figure out a way to solve this equation:
$$\sqrt[3]{5x+7}-\sqrt[3]{5x-12}=1.$$
I tried to cube both sides, but I ended up with an equation looking like this:
$$\sqrt[3]{(5x-12)(5x+7)}(\sqrt[3]{5x-1}-\sqrt[3]{5x+7})=-6.$$
At this... | Put $a = \sqrt[3]{5x+7}, b = \sqrt[3]{5x-12}\implies a - b = 1, a^3 - b^3 =19\implies a^2+ab+b^2 = \dfrac{a^3-b^3}{a-b} = \dfrac{19}{1} = 19\implies (a-b)^2+3ab = 19\implies 1^2+3ab = 19 \implies ab = 6\implies (b+1)b = 6\implies b^2+b-6 = 0\implies (b+3)(b-2)=0\implies b = 2, -3\implies 5x-12 = 2^3, (-3)^3 = 8,-27\imp... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2248632",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 6,
"answer_id": 0
} |
How i can prove that $\sum_{i=0}^{r+1}\binom{n-i}{r-i+1} = \binom{n}{r}$ I'm trying to prove this binomial coefficient identity.
$$\sum_{i=0}^{r+1}\binom{n-i}{r-i+1} = \binom{n}{r} $$
The textbook's hint is about using the identity
$$\binom{n}{r} =\binom{n}{n-r}$$
I've only reached this
$$\frac{1}{(n-1-r)!}\left(\frac... | You can use the Identity:
$$ \begin{pmatrix}
n \\
r
\end{pmatrix} =
\begin{pmatrix}
n - 1 \\
r
\end{pmatrix} + \begin{pmatrix}
n - 1 \\
r - 1
\end{pmatrix}$$
Recursively in the last term { $C(n - 1, r - 1)$ }. Doing this you will have:
$$ \begin{pmatrix} n \\r \end{pmatrix} = \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2250421",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
An object travelling on $x^2$ I have an object traveling along the curve $y=x^2$. $z$ is the distance from the origin and $dz/{dt}=1$ is the rate it's increasing per unit time. At what rate are my $x$ and $y$'s moving at the point (2,4)? In other words what is $dy/dt$ and $dx/dt$.
I've seen the geometry. Found $z$ as t... | I think this is simpler to understand. By definition
$$z=\sqrt{(x-0)^2+(y-0)^2}=\sqrt{x^2+y^2}$$
Since $y=x^2$
$$z=\sqrt{x^2+x^4}$$
Taking the derivative of this equation using the chain rule gives
$$z^{'}=\left({(x^2+x^4)}^{1/2}\right)^{'}\implies\frac{dz}{dt}=1/2(x^2+x^4)^{1/2-1}(x^2+x^4)^{'}\implies\frac{dz}{dt}=(1/... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2250641",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Find a 3 x 3 orthogonal matrix Find a 3x3 orthogonal matrix whose first column is $[\frac1{\sqrt{3}},{- \frac 1{\sqrt{3}}}, \frac1{\sqrt{3}}]$.
I was thinking this has to do with: ${Q}^T•Q=I_{n}$, but I still don't understand how to go about this.
| In this case we can also guess $3$ vectors without calling for Gram-Schmidt by noticing
$\begin{pmatrix} 1 \\ -1 \\ 1\end{pmatrix}\text{ , }\begin{pmatrix} 1 \\ 2 \\ 1\end{pmatrix}\text{ and }\begin{pmatrix} 1 \\ 0 \\ -1\end{pmatrix}\text{ are orthogonal.}$
By keeping first and third coordinate of $v_1,v_2$ equal, we w... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2252292",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Probability of picking at least x blue balls out of red and blue balls Given $n$ balls in total.
Each ball is either red or blue.
There are $ r $ red balls and $ b $ blue balls.
At random, and without replacement, we pick $ t $ balls out of the $ n $ balls.
How to calculate the probability for the event that at least $... | The probability of first getting $x$ blue balls, followed by $t-x$ red balls, equals:
$$\frac{b}{n} \cdot \frac{b-1}{n-1} \cdot \ldots \cdot \frac{b-x+1}{n-x+1} \cdot \frac{r}{n-x} \cdot \frac{r-1}{n-x-1} \cdot \ldots \cdot \frac{r-t+x+1}{n-t+1} = \frac{b!}{(b-x)!} \frac{r!}{(r-t+x)!} \frac{(n-t)!}{n!}$$
There are ${{t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2252950",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
How to find the angle between 2 intersection lines of four planes?
Given $L_1=\begin{cases}P_1: x-y+z=4\\ P_2: 2x+y-z=6 \end{cases}$
and $L_2=\begin{cases}P_1:x+y+z=4\\ P_2:2x+3y-z=6 \end{cases}$
$L_1$ and $L_2$ are the intersection lines of the given planes. Find the angle between the lines $L_1,L_2$
First in order... | Let $\vec{l_1}(a,b,c)$.
Thus, $a-b+c=0$ and $2a+b-c=0$, which gives $a=0$ and $b=c$ and $\vec{l_1}(0,1,1).$
Let $\vec{l_2}(a,b,c)$.
Thus, $a+b+c=0$ and $2a+3b-c=0$, which gives $b=-\frac{3}{4}a$, $c=-\frac{1}{4}a$
and $\vec{l_1}(4,-3,-1).$
Now,
$$\measuredangle\left(L_1,L_2\right)=\arccos\frac{|0\cdot4+1\cdot(-3)+1\cd... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2253072",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
How to find in a more formal way $\lfloor{(2+\sqrt3)^4}\rfloor$
Problem Statement:-
The largest integer which is less than or equal to $(2+\sqrt3)^4$ is
$\text{(A)}192\qquad\qquad \text{(B)}193\qquad\qquad \text{(C)}194\qquad\qquad\text{(C)}195\qquad\qquad$
My Solution:-
I started with the binomial expansion but as ... | Another way . . .
Let $a = 2 + \sqrt{3},\;\;b = 2 - \sqrt{3}$.
Then
$$a + b = 4$$
$$ab = 1$$
so
$$a^2 + b^2 = (a+b)^2 - 2ab = 4^2-2(1) = 14$$
hence
$$a^4 + b^4 = (a^2 + b^2)^2 - 2(ab)^2 = 14^2 - 2(1)^2 = 194\\[0pt]$$
Then
\begin{align*}
&a^4 + b^4 = 194\\[6pt]
\implies\; &193 < a^4 < 194\qquad\text{[since $0 < b < 1$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2254817",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
Given the square, calculate $\tan{\alpha}.$ Problem: Given the square $ABCD$, let $M$ be the midpoint on the side $|CD|$ and designate $\alpha=\angle AMB$. Calculate $\tan{\alpha}.$
Attempt: We can, without compromising generality, assume that the side of the square is equal to 1. So drawing a figure we get
I know th... | You've already found the lengths shown below.
Hence
\begin{align}
u^2 &= \left(\frac{\sqrt 5}{2}\right)^2 -\left(\frac{2}{\sqrt 5}\right)^2 \\
&= \frac 54 - \frac 45 \\
&= \frac{9}{20} \\
u &= \frac{3}{2\sqrt 5}
\end{align}
It follows that
$ \tan \alpha = \dfrac{\frac{2}{\sqrt 5}}{\frac{3}{2\sqrt... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2257339",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Given that $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{5}$ and $abc = 5$, solve for $a^3 + b^3 + c^3$ I recently encountered this question and have been stuck for a while. Any help would be appreciated!
Q: Given that
$$\frac{1}{a} + \frac{1}{b} + \frac{1}{c} = \frac{1}{5} \tag{1} \label{eq:1}$$
$$abc = 5 \tag{2} \la... | Let $k$ be an arbitrary real constant and $a$, $b$, $c$ be the roots of the cubic equation $x^3-kx^2+x-5=0$. Then we have $ab+bc+ca=1$ and $abc=5$. Also, we have $a+b+c=k$.
$$a^3+b^3+c^3=k(a^2+b^2+c^2)-(a+b+c)+15=k(a+b+c)^2-2k(ab+bc+ca)-(a+b+c)+15=k^3-2k-k+15=k^3-3k+15$$
Since there are infinitely many possible $k$ suc... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2258697",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
How can I solve $\int \frac{1}{\sqrt{x^2 + 7x} + 3}\, dx$ How can I solve the following integral?
$$\int \frac{1}{\sqrt{x^2 + 7x} + 3}\, dx$$
$$\begin{align}
\int \frac{1}{\sqrt{x^2 + 7x} + 3}\, dx&=\int \frac{1}{\sqrt{\left(x+\frac72\right)^2-\frac{49}{4}} + 3}\, dx
\end{align}$$
$$u = x+\frac{7}{2}, \quad a = \frac{... | Hint. By the change of variable
$$
x+\frac72=\frac72 \cdot \cosh u,\qquad dx=\frac72 \cdot \sinh u\:du,
$$ one gets
$$
\begin{align}
\int \frac{1}{\sqrt{x^2 + 7x} + 3}\, dx&=\int \frac{1}{\sqrt{\left(x+\frac72\right)^2-\frac{49}{4}} + 3}\, dx
\\\\&=\int \frac{\frac72\sinh u}{\frac72\sinh u + 3}\, du
\end{align}
$$ then... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2261818",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Is this the correct way to find the Laurent Expansion of the complex function $f(z) = \frac{z}{z^2 - 1}$? The function
$$
f(z) = \frac{z}{z^2 - 1}
$$
has singularities at $z = \pm 1$. I will expand about the point $z = 1$. Then, with the substitution $w = z+1$,
\begin{align*}
f(x) =& \frac{z}{z^2 - 1} \\[3mm]
=& \frac{... | At the penultimate line you have
$$\sum_{n=0}^\infty \frac{w^{n-1}}{2^{n+1}}
-\sum_{n=0}^\infty \frac{w^n}{2^{n+1}}.$$
This is
$$\frac1{2w}+\frac14+\frac w8+\frac{w^2}{16}+\cdots-\frac12-\frac w4-\frac{w^2} 8-\cdots=\frac1{2w}-\sum_{n=0}^\infty\frac{w^n}{2^{n+2}}.$$
Does this look more like a Laurent series?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2262582",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
If $\sin^4(x)=A+B\cos(2x)+C\cos(4x)$, then find $A$, $B$, and $C$.
If $\sin^4(x)=A+B\cos(2x)+C\cos(4x)$, then find $A, B$, and $C$.
Would I use half-angle identities for this problem?
| No, just take $x=0,\pi/4,\pi/2$, and you'll get three equations on $A,B,C$:
\begin{align}
0&=A+B+C\\ \ \\
\frac14&=A+0-C\\ \\
1&=A-B+C
\end{align}
Solving, we get $A=3/8$, $B=-1/2$, $C=1/8$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2264008",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Solving simultaneous equations with complex numbers and moduli Would anyone be able to give me a starting point as to how to approach this? A friend suggested squaring both sides, but I read something that said you can't square both sides of an equation(?).
Complex numbers and simultaneous equations question
If $x$ and... | Hint: let $z=x+iy$ and write the system as:
$$
\begin{cases}
\begin{align}
\lvert z \rvert &= 1 \\
\left| z - \frac{3}{2} \right| &= 2
\end{align}
\end{cases}
$$
Since $z \bar z = |z|^2=1$ the second equation is equivalent to:
$$
2^2 = \left| z - \frac{3}{2} \right|^2 = \left( z - \frac{3}{2} \right) \left( \bar z - \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2264553",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Show that $\sin(3x)$ is equivalent to $3\sin(x)\cos^2(x)-\sin^3(x)$
The expression $\sin(3x)$ is equivalent to:
A. ...
My book states the right answer is B which is $3\sin(x)\cos^2(x)-\sin^3(x)$.
I tried:
$$\sin(x)\cos(2x)+\cos(x)\sin(2x) = \\
\sin(x)(2\cos^2(x)-1)+\cos(x)\cdot2\sin(x)\cos(x) = \\
2\cdot\sin(x)\cos(x... | $$\sin(3x) = \sin(x+2x)$$
$$= \sin(x) \cos(2x) + \sin(2x) \cos(x)$$
$$ = \sin(x) ( \cos^2(x) - \sin^2 (x)) + \cos(x) (2 \sin(x) \cos(x)) $$
$$= 3 \sin(x) \cos^2 (x) - \sin^3 (x )$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2264820",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
Finding the value of $f(1)$ from a given functional equation
A function $f: \mathbb{Q}^+ \cup \{0\} \to \mathbb{Q}^+ \cup \{0\}$ is defined such that $$ f(x) + f(y) + 2xyf(xy) = \frac{f(xy)}{f(x+y)}$$
Then what is the value of $\left[f(1)\right]$ (where $[.]$ denotes the greatest integer function)?
I proceeded th... | *
*Assume $f(0) \neq 0$.
Then $f(0) + f(0) = \frac{f(0)}{f(0)} \rightarrow f(0) = \frac{1}{2}$,
and $f(x) + f(0) = \frac{f(0)}{f(x)} \rightarrow f(x)^2 + \frac{1}{2}f(x) -\frac{1}{2} = 0.$
*Assume $f(1) \neq 0$.
We also know that $f(1) + f(1) + 2f(1) = \frac{f(1)}{f(2)}$. Or $f(2) = \frac{1}{4}$.
However, $(\frac{1}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2265651",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Trouble calculating Fourier series of a function Calculate the Fourier series
$$
f(x) = \left\{\begin{aligned}
& 7\sin(x), && 0 \le x \le \pi\\
& 0, && \pi \le x \le 2\pi
\end{aligned}
\right.$$
I know that when $f(x+L) = f(x)$, for all real $x$, the Fourier series expansion is
$$f(x) = \dfrac{a_0}{2} + \sum_{n = 1}^... | Hint: The Fourier series of a function $f$ in $[a,a+2\ell]$ with period $T=2\ell$ is
$$f(x) = \dfrac{a_0}{2} + \sum_{i = 1}^{\infty} a_n \cos\left( \dfrac{n\pi x}{\ell} \right) + b_n\sin\left( \dfrac{n\pi x}{\ell} \right)$$
where
\begin{eqnarray}
a_0 &=& \dfrac{1}{\ell}\int_a^{a+2\ell}f(x) dx \\
a_n &=& \dfrac{1}{\ell... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2266598",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Find the $2\times 2$ matrix $D$ such that $P^{-1}DP=\begin{pmatrix} -4 & -15 \\ 2 & 7 \end{pmatrix}.$ Let $P = \begin{pmatrix} 2 & 5 \\ 1 & 3 \end{pmatrix}$. Find the $2 \times 2$ matrix $D$ such that
$$P^{-1} DP = \begin{pmatrix} -4 & -15 \\ 2 & 7 \end{pmatrix}.$$
I think matrix multiplication is not associative, so ... | Hint: The matrix $P$ is diagonalizable with distinct eigenvalues $1$ and $2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2267881",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Find the value of $\int ^{\infty}_{-\infty} \frac{\cos x}{e^x+e^{-x}} dx$ Show that $$\int ^{\infty}_{-\infty} \frac{\cos x}{e^x+e^{-x}}dx=\frac{\pi}{e^\frac{\pi}{2}+e^{-\frac{\pi}{2}}}.$$
How do we solve this? Since given function is even, it follows that
$$ \int ^{\infty}_{-\infty} \frac{\cos x}{e^x+e^{-x}}dx= 2\int... | $$
\begin{aligned}
\int_{-\infty}^{\infty} \frac{\cos x}{e^x+e^{-x}} d x
= & 2 \operatorname{Re} \int_0^{\infty} \frac{e^{-x} \cdot e^{i x}}{1+e^{-2 x}} d x \\
= & 2 \operatorname{Re} \sum_{n=0}^{\infty} \int_0^{\infty} e^{(i-1-2 n) x} d x \\
= & 2 \operatorname{Re} \sum_{n=0}^{\infty} \left(-1\right)^n\left[\frac{e^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2268181",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Rational numbers inequality proof - is this proof valid?
Knowing that $x^2+y^2 = 2 $ prove that $x+y \le 2$
I rewrite this as
$$x^2+y^2 \ge x+y$$
Now, multiply both sides by 2
$$x^2 -2x + 1 + y^2-2y + 1 -2 +x^2 + y^2 \ge 0$$
I substitue $2$ for $x^2+y^2$
$$(x-1)^2+(y-1)^2\ge0$$
Which is true
| Or (shortened, once we know how to poceed):
$$x+y\le x+y+\tfrac{ (x-1)^2+(y-1)^2}2=x+y+\tfrac{x^2-2x+1+y^2-2y+1}2=\frac{x^2+y^2+2}2=2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2270476",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
Solve this trigonometric inequality The values of $\lambda$ for which the equation $2\sin (x)-\sqrt \lambda \cos(x)=\sqrt2+\sqrt{2-\lambda}$ has solutions.
MY APPROOACH:
Now I know that max value of LHS of above expression can be $\sqrt{4+\lambda}$ and min value can be $\sqrt{4-\lambda}$. So $\sqrt2+\sqrt{2-\lambda}\in... | Obviously, we need $-4\le\lambda\le2$.
\begin{align*}
\sqrt{4+\lambda}&\ge\sqrt{2}+\sqrt{2-\lambda}\\
4+\lambda&\ge2+2\sqrt{2(2-\lambda)}+2-\lambda\\
\lambda&\ge \sqrt{4-2\lambda} \qquad\qquad (\text{Note: this implies that }\lambda\ge0)\\
\lambda^2&\ge4-2\lambda\\
(\lambda+1)^2&\ge5\\
\lambda+1&\ge\sqrt{5}\\
\lambda&\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2275022",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Multiple of $5$ If two natural nos. $x,y$ are selected at random then the probability that $x^2+y^2$ is a multiple of 5?
I think they should give a finite set for selecting natural nos. otherwise there are infinite ordered pairs of $x$ and $y$.
Is the question incomplete or am I missing some trick?
| Without loss of generality, you can assume that $x$ and $y$ are restricted to the set $\{0, 1, 2, 3, 4\}$, as pointed out by Lord Shark the Unknown in a comment. Now,
$$z \equiv 0 \pmod 5 \implies z^2 \equiv 0 \pmod 5$$
$$z \equiv 1 \pmod 5 \implies z^2 \equiv 1 \pmod 5$$
$$z \equiv 2 \pmod 5 \implies z^2 \equiv 4 \pm... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2278200",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Logarithmic-trigonomic inequality $\log_5 \sin(x) > \log_{25} \cos(\frac{x}{2}) $ $$\log_5 \sin(x) > \log_{25} \cos(\frac{x}{2})$$
I have simplified it to
$$\sin(\frac{x}{2}) \cdot \sin(\frac{x}{2}) \cdot \cos(\frac{x}{2}) > \frac{1}{4}$$
But what should I do next?
I have no idea about next steps!
Thanks!
| Note that
$$\log_5 \sin(x) = \frac{\log\sin(x)}{\log 5}$$
and
$$\log_{25} \cos(x/2) = \frac{\log\cos(x/2)}{\log 25}$$
so that your inequality reduces to
$$2\log\sin(x) > \log\cos(x/2).$$
This becomes
$$\log\sin^2(x) > \log\cos(x/2).$$
Since the logarithm is a monotonically increasing function, then
$$\sin^2(x) > \cos(x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2279592",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Evaluate $\int_0^1{\frac{u^{n+2}+u^n}{u+1}}\,du$ for $n\in\mathbb{N}$, $n\neq-1$ Background:
I got stuck in the question below because of a silly error.
Evaluate the integral using the substitution $u=\tan x$
$$\int^{\frac{1}{4}\pi}_0{(\tan^{n+2}x+\tan^{n}x)}\,dx$$
I realized that $du = \sec^2x\,dx$ and hence $du=\ma... | $$\frac{u^n}{u+a} = u^{n-1} - au^{n-2} + a^2u^{n-3} - \cdots + a^{n-2}u - a^{n-1} -\frac{a^n}{u+a} $$
So
\begin{align}
\int \frac{u^n}{u+a} du &= \int \left(u^{n-1} - au^{n-2} + a^2u^{n-3} - \cdots + a^{n-2}u - a^{n-1} -\frac{a^n}{u+a}\right) du\\
&= \frac{u^n}{n} - a \frac{u^{n-1}}{n-1} + a^2 \frac{u^{n-2}}{n-2} - \cd... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2280396",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Finding a generator of $\mathbb{F}_{49}^*$ I have a little doubt.
Let $\alpha\in\mathbb{F}_{49}$ such that $\alpha^2=3$ and $\mathbb{F}_{49}=\mathbb{F}_7(\alpha)$. Find a generator for the cyclic group $\mathbb{F}_{49}^*$ and find the minimal polynomial over $\mathbb{F}_7$
For the first part, it's enough to find a prim... | This are the powers of $1 + \alpha$ in $\mathbb{F}_{49}$:
$$\begin{array}{cccccc}
\color{blue}{1 + 1\alpha} &
4 + 2\alpha &
3 + 6\alpha &
0 + 2\alpha &
\color{blue}{6 + 2\alpha} &
5 + 1\alpha \\
\color{blue}{1 + 6\alpha} &
\color{red}{5 + 0\alpha} &
5 + 5\alpha &
6 + 3\alpha &
\color{blue}{1 + 2\alpha} &
0 + 3\alpha \\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2282918",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Prove an inequality involving powers of two Let $a,b,c$ be mutually distinct positive integers.
Prove
$$2^{a+b} + 2^{b+c} + 2^{a+c} < 2^{a+b+c} + 1$$
Any hints on how to start proving it?
| I assume the set of Natural Numbers to be $\mathbb N = \{1,2,3 \ldots \}$
Let $$\alpha =2^{-a} + 2^{-b} + 2^{-c}$$
And
$$\beta =1+2^{-(a+b+c)}$$
We can easily see that the maximum value of $\alpha$ will occur for minimum values of $a,b$, and $c$.
The minimum values of $a,b$ and $c$ , are $1,2$, and $3$ (All are distin... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2283150",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
expanding $\frac 1{\log\left( \frac{z+1}{z-1} \right) }$ at infinity Just as the title says, how do we expand such a function at infinity? I am told that the first term is $z/2$. My knowledge in complex analysis is very rusty, so if anyone can help me I'd be really thankful.
| The "quick and dirty" way to do this is to view this as a composition of series with well-known forms:
\begin{align*}
\left[ \ln \left( \frac{z + 1}{z - 1} \right) \right]^{-1} &= \left[ \ln \left( 1 + \frac{1}{z} \right) - \ln \left(1 - \frac{1}{z} \right) \right]^{-1} \\
&= \left[ \sum_{n=1}^{\infty} \frac{(-1)^{n+1}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2283946",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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How to solve this partial fraction decomposition? Please help me to solve the following partial fraction decomposition:
$$\frac{1-v^2}{v+v^3} = \frac{A}{v}+\frac{Bv+C}{1+v^2}$$
| The usual method would be to multiply both sides of this equation by $v+v^3$:
$$\begin{align}\frac{1-v^2}{v+v^3} &= \frac{A}{v}+\frac{Bv+C}{1+v^2}\\
1-v^2 &= A(1+v^2) + (Bv+C)(v)\\
1-v^2 &= Av^2 + A + Bv^2 + Cv\\
-v^2 + 0v + 1 &= (A+B)v^2 + Cv + A
\end{align}$$
At this point, you can equate coefficients on the two side... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2284041",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 0
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Find all postive integer $n$,such $(x+y+z)|(x^{2n+1}+y^{2n+1}+z^{2n+1})$
Let $0<x<y<z<p$,where $x,y,z$ are postive intgers,$p$ is prime number, and such
$$x^3\equiv y^3\equiv z^3\pmod p$$
Find all postive integer $n$,such $$(x+y+z)|(x^{n}+y^{n}+z^{n})$$
$n=1$ is clear
$n=2$
I have show that
$$p|x^2+xy+y^2, p|y^... | For the congruence $x^3\equiv a\pmod p$ to have three non-congruent solutions $x,y,z$ it is necessary that there exists a primitive cubic root $\omega\in\Bbb{Z}_p$. This is the case if and only if $p\equiv1\pmod3$. Otherwise cubing will be a bijection from $\Bbb{Z}_p$ to itself.
In that case we have $y\equiv\omega x$ a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2287402",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Find $\lim_{(x,y)\to(0,0)}\frac{1-\cos(x^2+y^2)}{(x^2+y^2)x^2y^2}$
$$\lim_{(x,y)\to(0,0)}\frac{1-\cos(x^2+y^2)}{(x^2+y^2)x^2y^2}$$
how should I approach this?
$$\lim_{(x,y)\to(0,0)}\frac{1-\cos(x^2+y^2)}{(x^2+y^2)}=0$$ but I still have $$\lim_{(x,y)\to(0,0)}\frac{1}{x^2y^2}$$
The answer says there is no limit
| Note that
$$\lim_{(x,y)\to(0,0)}\frac{1-\cos(x^2+y^2)}{(x^2+y^2)^2}=\frac{1}{2}.$$
So in there is a neighborhood $U$ of $(0,0)$ where $\frac{1-\cos(x^2+y^2)}{(x^2+y^2)^2}\geq \frac{1}{4}.$
Hence, for $x\not=0$, $y\not=0$ and $(x,y)\in U$,
$$\frac{1-\cos(x^2+y^2)}{(x^2+y^2)x^2y^2}=\frac{1-\cos(x^2+y^2)}{(x^2+y^2)^2}\cdo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2287961",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Number reduction before mod $$15\cdot25 \pmod {11}\equiv 4\cdot3 \pmod {11}$$
How does it work?
Full example
$$3^{(11-1)} \pmod{11} = 3\cdot27\cdot27\cdot27 \pmod{11}= 3\cdot5\cdot5\cdot5 \pmod{11} = 15\cdot25 \pmod{11}= 4\cdot3 \pmod{11} =1 \pmod{11}$$
| $a\equiv b \pmod{11}$ means that $11$ divides $a-b$. So if we want to see that $$(a+11)(b+11)\equiv ab \pmod{11}$$ we should check that $11$ is a divisor of $$(a+11)(b+11)-ab=ab+11a+11b+11*11-ab=11a+11b+121$$ This is clearly the case (in your example we have that $a=4$ and $b=14$). You can do this again for $b$ and you... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2290457",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Finding coefficient of $x^{15}$ $$(x+x^2+x^3+x^4+x^5)\cdot (x^2+x^3+x^4+…)^5$$
I have done a little :
$$x(1 + x+x^2+x^3+x^4)\cdot x^{10}(1 + x^2+x^3+…)^5$$
By generating functions:
$$\begin{align}&x^{11}\cdot\frac{1 - x^5}{1-x}\cdot\frac{1}{(1-x)^5}\\[1ex]
\implies &x^{11}(1 - x^5)\cdot\frac{1}{(1-x)^6}\\[1.5ex]
\impl... | You are most of the way home with what you have done already, the last step involves manipulating the summation by multiplying through by $x^{11}(1-x^5)=x^{11}-x^{16}$ like so
$$\begin{align}(x^{11}-x^{16})\sum_{n=0}^{\infty}\binom{n+5}{5}x^n&=\sum_{n=0}^{\infty}\binom{n+5}{5}(x^{n+11}-x^{n+16})\\&=\sum_{n=0}^{\infty}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2291167",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
If $a,b,c \geq 1$ then prove $\sqrt{a-1} + \sqrt{b-1} + \sqrt{c-1} < \sqrt{abc+c}$ If $a,b,c \geq 1$, then prove:
$$\sqrt{a-1} + \sqrt{b-1} + \sqrt{c-1} < \sqrt{abc+c}$$
My try:
$$A=\sqrt{a-1} + \sqrt{b-1} + \sqrt{c-1} \\ A^2 =(a-1)+(b-1)+(c-1)+2\sqrt{(a-1)(b-1)}+2\sqrt{(a-1)(c-1)}+2\sqrt{(b-1)(c-1)} $$
$$B=\sqrt{abc... | $$\sqrt{a-1}+\sqrt{b-1}=\sqrt{a+b-2+2\sqrt{(a-1)(b-1)}}=$$
$$=\sqrt{ab-(a-1)(b-1)+2\sqrt{(a-1)(b-1)}-1}=$$
$$=\sqrt{ab-\left(\sqrt{(a-1)(b-1)}-1\right)^2}\leq\sqrt{ab}.$$
Thus,
$$\sqrt{a-1}+\sqrt{b-1}+\sqrt{c-1}\leq\sqrt{ab}+\sqrt{c-1}=$$
$$=\sqrt{ab+1-1}+\sqrt{c-1}\leq\sqrt{(ab+1)c}=\sqrt{abc+c}.$$
Done!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2291796",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
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Prove that if $5^n$ begins with $1$, then $2^{n+1}$ also begins with $1$
Prove that if $5^n$ begins with $1$, then $2^{n+1}$ also begins with $1$ where $n$ is a positive integer.
In order for a positive integer $k$ to begin with a $1$, we need $10^m \leq k < 2 \cdot 10^m$ for some positive integer $m$. Thus since $5^... | $10^m < 5^n < 2*10^m$ [Note: equality can not hold as $5^n$ has no factor of $2$.]
$10^m < \frac {10^n}{2^n} < 2*10^m$
$1 < \frac {10^{n-m}}{2^n} < 2$
$2^n < 10^{n-m} < 2^{n+ 1}$
So $10^{k} < 2^{n+1}$ for $k = n-m$.
...
And $2^m5^m < 5^n < 2^{m+1} 5^m$
$2^{n+1}5^m < 2^{n-m + 1}5^n$
$2^{n+ 1} < 2^{n-m+1}5^{n-m}$
$2^{n+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2292419",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Computing $\sum_{n=0}^\infty \frac{(-1)^n}{2n+1} $
$$\sum_{n=0}^\infty \frac{(-1)^n}{2n+1} \space \text{ converges by the alternating series test, but what is its value?} $$
$1 - \sin(x) \space \text{ is an entire function} $, so by Weierstrass Factorization Theorem, it can be written as a product of its zeros, which... | While $f(x) = 1 - \sin x$ is indeed an entire function, the roots at $x = \frac{(4n+1)\pi}{2}$ and $\frac{-(4n+3)\pi}{2}$ are in fact double roots, which you can check since they are also roots of $f'(x)$. So, the factorisation you give isn't valid.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2292606",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 6,
"answer_id": 1
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How to find $\lim\limits_{x \to 8} \frac{\frac{1}{\sqrt{x +1}} - \frac 13}{x-8}$ I am trying to find the limit as $x\to 8$ of the following function. What follows is the function and then the work I've done on it.
$$ \lim_{x\to 8}\frac{\frac{1}{\sqrt{x +1}} - \frac{1}{3}} {x-8}$$
\begin{align}\frac{\frac{1}{\sqrt{x +... |
$$\frac{\frac{1}{x+1}-\frac{1}{9}}{(x-8)\left(\frac{1}{\sqrt{x +1}} + \frac{1}{3}\right)} = \frac{\color{red}{8-x}}{(x-8)\left(\frac{1}{\sqrt{x +1}} + \frac{1}{3}\right)}$$
Careful in the numerator:
$$\frac{1}{x+1}-\frac{1}{9} \ne 8-x$$
but rather:
$$\frac{1}{x+1}-\frac{1}{9}= \frac{9}{9(x+1)}-\frac{x+1}{9(x+1)} = \f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2294991",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Theory of Equations and co prime
If the range of quadratic polynomial $f(x)= 2x^{2}+3x+\frac{a}{b}$ is $[4, \infty)$ and $a$ and $b$ are co prime. Find $a+b$ .
Options are $50,46,49,58$.
What I did was I made a rough graph and the roots are complex. So I saw Discriminant is less than zero. Then another method I trie... | $$f(x)=2\left(x^2+\frac{3}{2}x\right)+\frac{a}{b}=2\left(x+\frac{3}{4}\right)^2-\frac{9}{8}+\frac{a}{b}$$
The minimum value is
$$\frac{-9}{8}+\frac{a}{b}=4$$
$$\frac{a}{b}=\frac{41}{8}$$
$a=41$ and $b=8$.
$a+b=49$.
Another method:
The minimum value of $f(x)$ is $4$. So the graph has the vertex with $y$-coordinate equa... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2295924",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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How to prove the given function is not differentiable analytically? Well the question presented to me is this. The given function is,
$$f\left( x \right) = \left\{ {\begin{array}{*{20}{c}}{\frac{1}{2}x + 2,\,\;\;\;x < 2}\\{\sqrt {2x} ,\;\;\;\;\;\;x \ge 2}\end{array}} \right.
$$
Now have to check whether the given func... | $$f (2^-)=\lim_{x\to 2^-} (\frac {x}{2}+2)=1+2=3$$
$$ f (2^+)=\lim_{2^+}\sqrt {2x}=2$$
$f $ is not continuous at $x=2$ thus it is not differentiable at $x=2$.
By definition, differentiable at $x=x_0$ means
$$\exists L\in \mathbb R \;\exists \eta>0 :\forall x\in (2-\eta,2+\eta) $$
$$f (x)=f (2)+(x-2)\Bigl (L+\epsilon (x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2296139",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
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I'm stuck at integrating an arc length over a circle + angle between curves Let $(\phi, \theta)$ be the usual spherical coordinates. And let $\gamma (t) = (\phi (t), \theta (t))$ be the curve given by
$$ \phi (t) = \ln\left(\cot\left(\frac{\pi}{4} - \frac{t}{2}\right)\right),$$ $$ \theta (t) = \frac{\pi}{2} - t.$$
$t \... | \begin{align}
& \int_0^1 \sqrt{ \left(\frac{\sin (\frac{\pi}{2} - t) }{2\cos(\frac{\pi}{4} - \frac t 2) \sin(\frac \pi 4 - \frac t 2)} \right)^2 + 1} \, dt \\[10pt]
\end{align}
As $$\sin^2(\frac \pi 4 - \frac t 2)cos^2(\frac \pi 4 - \frac t 2) = \frac{1 - \cos(\pi - 2t)}{8},$$ then
\begin{align}
& \int_0^1 \sqrt{ \left... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2296761",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Proof for sum of product of four consecutive integers I had to prove that
$(1)(2)(3)(4)+\cdots(n)(n+1)(n+2)(n+3)=\frac{n(n+1)(n+2)(n+3)(n+4)}{5}$
This is how I attempted to do the problem:
First I expanded the $n^{th}$ term:$n(n+1)(n+2)(n+3)=n^4+6n^3+11n^2+6n$.
So the series $(1)(2)(3)(4)+\cdots+(n)(n+1)(n+2)(n+3)$ wil... | On your last equation, putting $n=-2,-3,-4$ leads the equation equal to zero. Then $n=1$ and the value of $\frac{1*2*3*4*5}5$ leads $\frac{n(n+1)(n+2)(n+3)(n+4)}5$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2297845",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 6,
"answer_id": 5
} |
Find the extrema of $f(x,y) = \left( \frac{1}{2} - x^2 + y^2 \right) \cdot e^{1-x^2-y^2}$ I am asked to find the extrema of $$f(x,y) = \left( \frac{1}{2} - x^2 + y^2 \right) \cdot e^{1-x^2-y^2}$$
The partial derivatives could be easier, but that's not the issue:
\begin{align*}
f_x &= \left( e^{1-x^2-y^2} \right) \left(... | For $x=0$ but $y\ne 0$, we have
$$2y^2-1=0$$
$$y=\pm\sqrt{\frac{1}{2}}$$
For $y=0$ but $x\ne 0$, we have
$$2x^2-3=0$$
$$x=\pm\sqrt{\frac{3}{2}}$$
For $x,y\ne 0$, we have
$$2x^2=2y^2+3$$
$$-2y^2 -3+2y^2-1=0$$
$$-4=0$$
So no solution
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2298307",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Why can't the quadratic formula be simplified to $x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}=\frac{-b\pm(b-2)\sqrt{ac}}{2a}$? I am currently taking Algebra 1 (the school year's almost over ), and we just learned the quadratic formula, another method to solve quadratic equations:
$$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$$
However, this... | Because $\sqrt{b^2 - 4ac}\neq 2b\sqrt{ac}$. Say for example that $c = 0$ and $b\neq 0$. Then you have
\begin{align*}
\sqrt{b^2 - 4ac} &= \sqrt{b^2}\\
&= \left|b\right|\\
&\neq 0,
\end{align*}
so you can see that this simplification cannot be correct.
It seems that in your proposed simplification, you have completely di... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2300647",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 6,
"answer_id": 3
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Find $\lim_{x\to 0} \left\lfloor \frac{\tan 2x}{\sin x} \right\rfloor $
Find the limit
$$\lim_{x\to 0} \left\lfloor \dfrac{\tan 2x}{\sin x} \right\rfloor $$
My try:
$$ \tan 2x =\dfrac{\sin 2x}{\cos 2x}$$
$$\sin 2x =2\cos x\sin x$$
So:
$$\dfrac{\tan 2x}{\sin x}=\dfrac{2\cos x}{\cos 2x}$$
So:
$$\lim_{x\to 0} \left\lflo... | Let's consider
$$
f(x)=\frac{2\cos x}{\cos2x}
$$
over $(-\pi/2,\pi/2)$. Then
$$
f'(x)=2\frac{-\sin x\cos2x+2\cos x\sin2x}{\cos^22x}=
2\frac{\sin x(2\cos^2x+1)}{\cos^22x}
$$
Therefore $f$ has a local minimum at $0$ and $f(0)=2$.
In a suitable neighborhood of $0$ we have $2\le f(x)<3$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2303012",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 1
} |
Find the absolute max and min of f subject to the given constraint $f(x)=4x^4+y^4$ subject to the constraint $x^2+y^2=1$
My attempt at it was by finding partial derivitives:
$$f_x=16x^3 =0$$
$$f_y=4y^3=0$$
this means that we have an interior point $(0,0)$
Now Let $x=\cos(t)$, $y=\sin(t)$
$$f(\cos(t),\sin(t))=4\cos^4t+\... | Express $(x,y)$ in polar coordinates as $(\cos\theta,\sin\theta)$, and then let
$c= \cos^2 \theta$. Note that $\sin^2 \theta = 1-\cos^2 \theta$ so $x^4=c^2$ and $y = (1-c)^2$.
Then $$4x^4+y^4 = 4c^2 + (1-c)^2 = 5c^2 -2c+1
$$
Since $c$, being $\cos^2\theta$, is restricted by $0 \leq |c|\leq 1$,
this is maximized at ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2303236",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Having trouble integrating $ \ \ \int \frac{x^{3}+3x^{2}+2x+4}{x^{2}(x^{2}+2x+2)} dx $. I'm having trouble integrating $$\int \frac{x^{3}+3x^{2}+2x+4}{x^{2}(x^{2}+2x+2)} dx $$
My approach
*
*$ \frac{x^{3}+3x^{2}+2x+4}{x^{2}(x^{2}+2x+2)} =\frac{Ax+B}{x^{2}}+\frac{Cx+D}{x^{2}+2x+2} $ ,
*or, $ x^{3}+3x^{2}+2x+4=(Ax+B... | HINT: we have $$\frac{x^3+3x^2+2x+4}{x^2(x^2+2x+2)}=-\frac{1}{x}+\frac{2}{x^2}+\frac{2x+3}{x^2+2x+2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2304492",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Prove inequality $a<1Given $a,b,c$ are positive number satisfy $a<b<c;a+b+c=6;ab+bc+ca=9$. Prove that $a<1<b<3<c<4$
i think use quality:"The solutions of function $f(x)=(x-a)(x-b)(x-c)$ satisfy $a\le x_1\le b\le x_2\le c$ with $x_1,x_2\ (x_1<x_2)$ are extreme point of $f(x)$ ".
Help me
| Your idea is correct. $a, b, c$ are the zeros of
$$
f(x)=(x-a)(x-b)(x-c) = x^3 - 6 x^2 + 9x - abc
$$
Rolles's theorem states that $f'$ has a zero in each interval
$(a, b)$ and $(b, c)$. But
$$
f'(x) = 3x^2 - 12 x + 9 = 3(x-1)(x-3)
$$
has zeros $x_1=1$ and $x_2=3$. It follows that
$$
a < x_1 = 1 < b < x_2 = 3 < c \, ... | {
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"url": "https://math.stackexchange.com/questions/2305304",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Confusing differentials problem - Electrical resistance $ R = \frac{k}{r^2} $ with $ dr = 5\% $ I am struggling with a confusing differentials' problem. It seems like there is a key piece of information missing:
The problem:
The electrical resistance $ R $ of a copper wire is given by $ R = \frac{k}{r^2} $ where $ k ... | Just doing this by brute force:
$$R=\frac{k}{r^2}\quad\text{percentage error in $r$ is $\pm 5$ percent}$$
So $R_{min} =\frac{k}{(r+0.05r)^2} =\frac{k}{1.05^2r^2}=\frac{400}{441}\cdot\frac{k}{r^2}$
$R_{max} = \frac{k}{(0.95r)^2} =\frac{400}{361}\cdot \frac{k}{r^2}$
Then the percentage error is given by:
$$\frac{R_{max}-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2307021",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
minimizing vectors Could please check whether my solution is right?
Q. $$A=\begin{pmatrix}
1 & 1 \\
1 & 1 \\
0 & 0 \\
\end{pmatrix}$$Find the set of vectors $x$ that minimize the value $$||
A
x-
\begin{pmatrix}
1 \\
2 \\
3 \\
\end{pmatri... | I would use a more "basic" method.
Writing the vector x as $\begin{pmatrix}x \\ y \end{pmatrix}$, $Ax= \begin{pmatrix}1 & 1 \\ 1 & 1 \\ 0 & 0\end{pmatrix}\begin{pmatrix} x \\ y \end{pmatrix}= \begin{pmatrix} x+ y \\ x+ y \\ 0 \end{pmatrix}$ so that $Ax- \begin{pmatrix}1 \\ 2 \\ 3 \end{pmatrix}= \begin{pmatrix}x+ y- 1 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2308121",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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How many numbers in between $1$ and $250$ are not divisible by $2, 3, 5$ or $7$? How many numbers in between $1$ and $250$ are not divisible by $2, 3, 5$ or $7$? Using the Inclusion Exclusion Principle, I got the answer to be $57$ (i.e. $250-193$), but I'm worried I might have screwed up the arithmetic. Is there a quic... | I will first consider all integers up to $210$, as it is the L.C.M. of $2$, $3$, $5$ and $7$. The number of integers from $1$ to $210$ which are not divisible by $2$, $3$, $5$ or $7$ is
$$210-\frac{210}{2}-\frac{210}{3}-\frac{210}{5}-\frac{210}{7}+\frac{210}{6}+\frac{210}{10}+\frac{210}{14}+\frac{210}{15}+\frac{210}{2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2309178",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Find the value of $\frac{\sin 24x}{\cos18x \cos 6x}$. Given $\tan x = \sqrt{5} - 2 $, find the value of $$\frac{\sin 24x}{\cos 18x\cos 6x}.$$
Now I wrote numerator as $\sin (18 + 6)x$ and I used identity $\sin(A+B)$, after that I reduced problem to $\tan 18x + \tan 6x$
How do I proceed?
Thanks
| We have
\begin{align}
\tan 2x&=\frac{2\tan x}{1-\tan^2x}\\
&=\frac{2(\sqrt{5}-2)}{1-(\sqrt{5}-2)^2}\\
&=\frac{2(\sqrt{5}-2)}{1-5+4\sqrt{5}-4}\\
&=\frac{2(\sqrt{5}-2)}{4(\sqrt{5}-2)}\\
&=\frac{1}{2}
\end{align}
Then use
$$\tan 3A=\frac{3\tan A-\tan^3A}{1-3\tan^2A}$$
to deduce that
$$\tan 6x=\frac{11}{2}$$
Use it once a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2309459",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
If $\frac{3-\tan^2\frac\pi7}{1-\tan^2\frac\pi7}=k\cos\frac\pi7$, find $k$ I need help solving this question:
$$
\text{If }\frac{3-\tan^2\frac\pi7}{1-\tan^2\frac\pi7}=k\cos\frac\pi7\text{, find k.}
$$
I simplified this down to:
$$
\frac{4\cos^2\frac\pi7-1}{2\cos^2\frac\pi7-1}
$$
But am unable to proceed further. The val... | Set $z=e^{i\pi/7}$, so $z^7=e^{i\pi}=-1$ and $z^{-1}=\bar{z}$.
Then
$$
\cos\frac{\pi}{7}=\frac{z+\bar{z}}{2}=\frac{z^2+1}{2z},
\qquad
\sin\frac{\pi}{7}=\frac{z-\bar{z}}{2i}=\frac{z^2-1}{2iz}
$$
Therefore
$$
\tan\frac{\pi}{7}=-i\frac{z^2-1}{z^2+1}
$$
Plugging in the left hand side, we get
$$
\frac{3(z^2+1)^2+(z^2-1)^2}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2310116",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
Sum $\sum_{n=1}^{\infty}\frac{3n+7}{n(n+1)(n+2)}.$ How can we find the sum
$$\sum_{n=1}^{\infty}\frac{3n+7}{n(n+1)(n+2)}.$$
WolframAlpha says it's $13/4$ but how to compute it? In general how to approach this kind of sums of rational functions where denominator has distinct roots?
| $S=\sum_{n=1}^{+\infty}(\frac {3}{(n+1)(n+2)}+\frac {7}{n (n+1)(n+2)}) $
$$=3\sum_{n=1}^{+\infty}(\frac {1}{n+1}-\frac {1}{n+2}) $$
$$+7\sum_{n=1}^{+\infty}\frac{1}{n (n+1)(n+2)} $$
$$\implies S=\frac {3}{2}+7A$$
and
$$S=\sum_{n=1}^{+\infty}\frac {3 (n+2)+1}{n (n+1)(n+2)} $$
$$=3\sum_{n=1}^{+\infty}(\frac {1}{n}-\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2317136",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 2
} |
What are the area of a triangle with side lengths $\tan(x)$, $\cos(x)$ and $\sin(x)$?
Consider a nondegenerated right triangle with sides of length $\sin x$, $\cos x$, and $\tan x$ where $x$ is a real number.
Compute the possible values of the area of this triangle.
*
*I was thinking of more along the lines of ... | Since $\sin x$, $\cos x$ and $\tan x$ are all positive, we may assume that $\displaystyle x\in\left(0,\frac{\pi}{2}\right)$. Therefore, $\tan x>\sin x$.
If $\cos x>\tan x$, then $\cos x$ is the hypotenuse and
\begin{align}
\sin^2x+\tan^2x&=\cos^2x\\
\sin^2x(\cos^2x+1)&=\cos^4x\\
1-\cos^4x&=\cos^4x\\
\cos x&=\frac{1}{\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2318135",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
Solve the equation $\sqrt[3]{x+2}+\sqrt[3]{2x-1}+x=4$ Find $x\in R$ satisfy $$\sqrt[3]{x+2}+\sqrt[3]{2x-1}+x=4$$
The root of equation very bad
My try 1:
Let $\sqrt[3]{x+2}=a;\sqrt[3]{2x-1}=b\Rightarrow a^3-b^3=3-x$
Have: $a+b=4-x$
=>Root of a system of equation bad, too
My try 2:
Use quality $\sqrt[3]{a}\pm \sqrt[3]... | If we set $a = \sqrt[3]{x+2}$ and $b = \sqrt[3]{2x-1}$, since $a + b = 4 - x = 1 + (3 - x)$, we obtain that
\begin{cases}
a^{3} - b^{3} = a + b - 1\\
b^{3} = 2a^{3} - 5
\end{cases}
Hence we conclude that $b = 6 - a - a^{3}$, from whence we have the equation
\begin{align*}
(6 - a - a^{3})^{3} = 2a^{3} - 5
\end{align*}
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2319277",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
If $a^3+b^3+c^3=3$ so $2(a^2+b^2+c^2)-(ab+bc+ca) \geq 3$ I was looking at this question, and I derived this inequality from that
Let $a$, $b$ and $c$ be positive numbers such that $a^3+b^3+c^3=3.$ Prove that:
$$2(a^2+b^2+c^2)-(ab+bc+ca) \geq 3$$
I couldn't solve Rozenberg's inequality but I assumed it's true. So if ... | Alternative proof: the pqr method
Let $p = a + b + c$ and $q = ab + bc + ca$ and $r = abc$.
We will use $p^2 \ge 3q$ and $q^2 \ge 3pr$ which are well-known. See the remarks at the end.
Using $a^3 + b^3 + c^3 = p^3 - 3pq + 3r$, the condition is written as
$$p^3 - 3pq + 3r = 3. \tag{1}$$
Using (1) and $q^2 \ge 3pr$, we h... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2319488",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Find $\gcd (3 + \sqrt{13},\ 2 + 5\sqrt{13})$ in $\mathbb Z[(1+\sqrt{13})/2]$ My task is find $\gcd (3 + \sqrt{13},\ 2 + 5\sqrt{13})$ in $\mathbb Z[(1+\sqrt{13})/2]$.
Can you give me some advice?
Any help is highly appreciated.
| \begin{align*}
\text{Let}\;\;R &= \mathbb{Z}[(1+\sqrt{13})/2]\\[4pt]
a &=3 + \sqrt{13}\\[4pt]
b &=2 + 5\sqrt{13}\\[8pt]
\end{align*}
and suppose that $d \in R$ is a common divisor of $a,b$.
By definition,
\begin{align*}
& N(a) = a\bar{a} = \left(3 + \sqrt{13}\right)\left(3 - \sqrt{13}\right) = -4\\[4pt]
& N(b) = b\bar... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2322062",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Alternate method for solving missing area question I recently saw a puzzle in an advert for the website Brilliant.org, which went as follows:
What is the blue area?
Hint: Think outside the box
My answer:
I set the area to be found to $x$, the side length of the square to be $y$, and the sections to be $a,b,c,d$ as ... | The hint may assume two things: to draw extra lines 1) outside; 2) inside the box so that the solution is simple to understand.
Here is yet another method: draw the lines to divide the square into four rectangles indicated by $A, B, C$ and $3+3$:
$$\begin{cases}
A+B=8 \\
A+C=4 \\
CB=6A
\end{cases} \stackrel{(1)-(2)}\R... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2323550",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 12,
"answer_id": 0
} |
If $f (x) +f'(x) = x^3+5x^2+x+2$ then find $f (x)$
If $f (x) +f'(x) = x^3+5x^2+x+2$ then find $f(x)$.
$f'(x)$ is the first derivative of $f (x)$.
I have no idea about this question, please help me.
| We work with the general form of the equation $f(x)$, assuming it to be a cubic equation: $$f(x)=ax^3+bx^2+cx+d$$
We differentiate this to see that $$f'(x)=3ax^2+2bx+c$$
Now we can say that \begin{align}f(x)+f'(x)&=(ax^3+bx^2+cx+d)+(3ax^2+2bx+c)\\
x^3+5x^2+x+2&=ax^3+(b+3a)x^2+(c+2b)x+(c+d)\end{align}
We can compare coe... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2323688",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 7,
"answer_id": 6
} |
How to derive the Rotation Matrix from the Euler Formula I'm trying to understand how the two dimensional rotation matrix (i.e. $R \in \mathbb{R}^2$) can be derived from the Euler Formula ($e^{i\theta} = \cos \theta + i \sin \theta$). $R$ is given as:
$$
R(\theta) =
\begin{bmatrix}
\cos\theta & -\sin\theta\\
\sin\t... | The complex number $a+bi$ can be represented by the matrix $\displaystyle \begin{pmatrix} a & -b \\ b & a \end{pmatrix}$.
Note that $(a+bi)\pm(c+di)=(a\pm c)+(b\pm d)i$ and
$$ \begin{pmatrix} a & -b \\ b & a \end{pmatrix} \pm\begin{pmatrix} c & -d \\ d & c \end{pmatrix}=\begin{pmatrix} a\pm c & -(b\pm d) \\ b\pm d & ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2324847",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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probability or rolling 1 1 2 with six dice What is the probability of rolling at least two ones and at least one two (order not important) with six 6-sided die?
Equivalently, to win a game, I need two 1s and a 2, (1,1,2). What is the probability of my rolling at least this with 6 dice.
There are $6^6=46656$ possible r... | You can consider the possibility to find probability directly without inclusion-exclusion principle. This is a longer way, but there are those who like it. Consider all possibilities to roll at least two ones and at least one two with six 6-sided die. We can have exactly two $1$ and one $2$. Denote this event by $A_{21... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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Finding integer part of $(3+\sqrt{3})^4$ I need to find the integer part of $(3+\sqrt{3})^4$, ie the floor of it and I am unable to proceed.
I can find integer part of $(2+\sqrt{3})^4$ using the binomial theorem as $2-\sqrt{3}<1$.
Please give a hint. Thanks!
| One possibility is to look at $f(n) = (3+\sqrt{3})^n + (3-\sqrt{3})^n$. It's easy to see that $f(4) = 504$. You have
$$(3 + \sqrt{3})^4 = 3^4 + 4 \cdot 3^3 \cdot \sqrt{3} + 6 \cdot 3^2 \cdot 3 + 4 \cdot 3 \cdot 3 (\sqrt{3}) + 9 = 252 + 144 \sqrt{3}$$
and when you expand out $(3-\sqrt{3})^4$ similarly the terms with $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2328977",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
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Finding polynomial with roots $1/\alpha$, $1/\beta$, $1/\gamma$ for the roots of $x^3-4x^2+x+6=0$ If $\alpha$, $\beta$, $\gamma$ are the roots of $x^3-4x^2+x+6=0$, find the equation whose roots are
1) $1/\alpha$, $1/\beta$, $1/\gamma$
2) $\alpha^2$, $\beta^2$, $\gamma^2$
How to do these kind of questions. I am stuck.
I... | For the first one, set $x=1/y\implies$ $$\left(\dfrac1y\right)^3-4\left(\dfrac1y\right)^2+\left(\dfrac1y\right)+6=0$$
Multiply by $y^3$
For the second, $x^2=y,$
$$x(x^2+1)=4x^2-6\implies x(y+1)=4y-6$$
Square both sides and replace $x^2$ with $y$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2329973",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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Solving the functional equation $ f \left( \sqrt { x ^ 2 + y ^ 2 } \right) = f ( x ) f ( y ) $ I happened to run across this problem in Larson's Problem Solving through Problems and I wanted to ask how you would approach it.
A real-valued continuous function satisfies for all real $ x $ and $ y $ the functional equati... | Note that the constant zero function satisfies the equation. So here you must accept that $ 0 ^ x = 0 $, for all real numbers $ x $ ( which is problematic when $ x $ is close to $ 0 $). Otherwise your claim that $ f ( x ) = f ( 1 ) ^ { x ^ 2 } $ is not correct.
Letting $ x = y = 0 $ in
$$ f \bigg( \sqrt { x ^ 2 + y ^ 2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2330402",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
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How to find the limit of $\frac{\sqrt{x^2+9}-3}{x^2}$? as x approaches 0
How to find the limit of $\frac{\sqrt{x^2+9}-3}{x^2}$? as x approaches 0
I thought that this was going to be a simple problem but then it got more complicated than I have expected
So I first multiplied the expression with its conjugate on both t... | Note:
\begin{align*}
\frac{\sqrt{x^2+9}-3}{x^2}
= & \; \frac{\big(\sqrt{x^2+9}-3 \big) \big(\sqrt{x^2+9}+3 \big)}{x^2 \big( \sqrt{x^2+9}+3 \big)} \\
= & \; \frac{x^2+9-3^2}{x^2\big(\sqrt{x^2+9}+3 \big)} \\
= & \; \frac{1}{\sqrt{x^2+9} + 3}.
\end{align*}
So we have
\begin{align*}
\lim_{x \to 0} \frac{\sqrt{x^2+9}-3}{x^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2333129",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 6,
"answer_id": 2
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If $a,b,c,d>0$ prove $\frac{2a^2+bc}{a+b+c} +\frac{2b^2+cd}{b+c+d}+\frac{2c^2+da}{c+d+a}+\frac{2d^2+ab}{d+a+b} \geq a+b+c+d$ If $a,b,c,d>0$ prove
$$\frac{2a^2+bc}{a+b+c} +\frac{2b^2+cd}{b+c+d}+\frac{2c^2+da}{c+d+a}+\frac{2d^2+ab}{d+a+b} \geq a+b+c+d$$
I tried cauchy: $$\frac{2a^2+bc}{a+b+c} +\frac{2b^2+cd}{b+c+d}+\frac... | We need to prove that
$$\sum_{cyc}\frac{2a^2+bc}{a+b+c}\geq\sum_{cyc}a$$ or
$$\sum_{cyc}\left(\frac{2a^2+bc}{a+b+c}-a\right)\geq0$$ or
$$\sum_{cyc}\frac{(a-b)(a-c)}{a+b+c}\geq0.$$
Now, easy to use BW.
Indeed, let $a=\min\{a,b,c,d\}$, $b=a+u$, $c=a+v$ and $d=a+w$ and the rest for you.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2336234",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Chinese Remainder theorem polynomials Is there a program, small script, or easy method I can use to find the polynomial P(x) such that:
$P(x) = x+1$ $\pmod {x^2+1}$
$P(x) = x+2$ $\pmod {x^2+2}$
$P(x) = x+3$ $\pmod {x^2+3}$
$P(x) = x+4$ $\pmod {x^2+4}$
.........
This is easy to do with integers, but I can't get across t... | It didn't take too long to compute the following table of moduli.
\begin{array}{r|cccc}
& x^2 + 1 & x^2 + 2 & x^2 + 3 & x^2 + 4 \\
\hline
\dfrac 16(x^2+2)(x^2+3)(x^2+4) & 1 & 0 & 0 & 0 \\
-\dfrac 12(x^2+1)(x^2+3)(x^2+4) & 0 & 1 & 0 & 0 \\
\dfrac 12(x^2+1)(x^2+2)(x^2+4) & 0 & 0 & 1 & 0 \\
-\dfrac 16(x^2+1)(x^2+2)(x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2336499",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
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Find $[T_{c}]^{\alpha}_{\alpha}$ For $c \in R$, let $T_{c}: M_{2x2}(R) \rightarrow M_{2x2}(R)$ be the linear transformation defined by $T_{c}(A) = A + cA^{T}$ for every $A \in M_{2x2}$.
For the basis $\alpha = \left\{\begin{bmatrix} 1 & 0\\ 0 & 0\\\end{bmatrix} \begin{bmatrix} 0 & 0\\ 0 & 1\\\end{bmatrix} \begin{bmatri... | You are on the right way, but you made mistakes in your computation.
First:
$$
T_c\begin{pmatrix}0&1\\-1&0\end{pmatrix}=\begin{pmatrix}0&1\\-1&0\end{pmatrix}+c\begin{pmatrix}0&-1\\1&0\end{pmatrix}=\begin{pmatrix}0&1-c\\c-1&0\end{pmatrix}.
$$
Second: Since
$$
T_c\begin{pmatrix}1&0\\0&0\end{pmatrix}=(1+c)\begin{pmatrix}1... | {
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"url": "https://math.stackexchange.com/questions/2338518",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Rational values of trigonometric functions I am using extensively trigonometric functions when an angle is given in degrees.
Some of these functions like sine or cosine have rational values, for example, the well known example is that $\cos(\theta) =0.6 $ and $\sin(\theta) =0.8 $.
However besides the case of mul... | Let us take the Pythagorean theorem $a^2 + b^2 = c^2$ and divide both sides by $c^2$ to get $\frac{a^2}{c^2} + \frac{b^2}{c^2} = 1$. Now, we can use the well-known relationship between sine and cosine: $\sin^2(x) + \cos^2 (x) = 1$, and let $\sin(x) = \frac{a^2}{c^2}$, and $\cos(x) = \frac{b^2}{c^2}$.
Since the hypoten... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2339177",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
} |
How to find this minimum of the value
Let $x_{i}$, where $i\in\{1,2,\cdots,n\}$ be distinct real numbers. Find the minimum of the value of
$$\sum_{1\le i<j\le n}\left(\dfrac{1-x_{i}x_{j}}{x_{i}-x_{j}}\right)^2$$
It is clear when $n=2$ minimum of the value is $0$, when $x_{1}x_{2}=1$. But for $n\ge 3$,I can't fin... | We begin with a short proof for the case $n=3$ :
We make a substitution so we put :
$\frac{\lambda_1}{2}(\frac{1+(a-b)^2}{a-b})^2=(\frac{1-xy}{x-y})^2$
$\frac{\lambda_2}{2}(\frac{1+(b-c)^2}{c-b})^2=(\frac{1-zy}{z-y})^2$
$\frac{\lambda_3}{2}(\frac{1+(a-c)^2}{a-c})^2=(\frac{1-xz}{x-z})^2$
With $a,b,c$ and the condition $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2340586",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
"answer_count": 3,
"answer_id": 1
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Integration of secant $$\begin{align}
\int \sec x \, dx
&= \int \cos x \left( \frac{1}{\cos^2x} \right) \, dx \\
&= \int \cos x \left( \frac{1}{1-\sin^2x} \right) \, dx \\
& = \int\cos x\cdot\frac{1}{1-\frac{1-\cos2x}{2}} \, dx \\
&= \int \cos x \cdot\frac{2}{1+\cos2x} \, dx
\end{align}$$
I am stuck in here. Any h... | \begin{align*}\int\sec x\,\mathrm dx&=\int\frac1{\cos x}\,\mathrm dx\\&=\int\frac{\cos x}{\cos^2x}\,\mathrm dx\\&=\int\frac{\cos x}{1-\sin^2x}\,\mathrm dx.\end{align*}
Now, doing $\sin x=t$ and $\cos x\,\mathrm dx=\mathrm dt$, you get $\displaystyle\int\frac{\mathrm dt}{1-t^2}$. But\begin{align*}\int\frac{\mathrm dt}{1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2340800",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "18",
"answer_count": 5,
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Suppose that $P(x)$ is a polynomial of degree $n$ such that $P(k)=\frac{k^{2}}{k^{3}+1}$ for $k=0,1,\ldots,n$, find the value of $P(n+1)$ Suppose that $P(x)$ is a polynomial of degree $n$ such that $P(k)=\dfrac{k^{2}}{k^{3}+1}$ for $k=0,1,\ldots,n$. Find the value of $P(n+1)$
I tried by making a $f(x) = (x^3+1) P(x) - ... | If $p(x)$ is a polynomial with degree $\leq n$ and $\delta$ is the forward difference operator, $(\delta p)(x)=p(x+1)-p(x)$, we have that
$$ \left(\delta^{n+1}p\right)(x) = \sum_{k=0}^{n+1}\binom{n+1}{k}(-1)^k\,p(x+k) = 0\tag{1} $$
hence in our case
$$ P(n+1)=(-1)^n\sum_{k=0}^{n}\binom{n+1}{k}(-1)^k \frac{k^2}{k^3+1}.... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2343233",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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How to solve these functional equations $f(x) = 1 + f(x) + f(x^2) + \ldots $ and $f(x) = 1 + f(x) + \big(f(x^2)\big)^2 + \ldots $? The identity $\frac{1}{1-f(x)}=1+f(x)+\big(f(x)\big)^2+\big(f(x)\big)^3 + \ldots$ is well known.
Is there somewhere in the literature a method allowing us to solve the following functional ... | I think that there also isn't a solution for the second equation.
Plug in $x=0$. Then we get:
$
f(0) = 1 + f(0) + f(0)^2 + \cdots
$
So $f(0) = \frac{1}{1-f(0)}$ and $f(0)^2 - f(0) + 1 = 0$, and this has no roots over the reals. (Note that $f(0) \neq 1$ as then the equation for $f(0)$ won't converge).
Also note that for... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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If $a,b$ are roots of $3x^2+2x+1$ then find the value of an expression It is given that $a,b$ are roots of $3x^2+2x+1$ then find the value of:
$$\left(\dfrac{1-a}{1+a}\right)^3+\left(\dfrac{1-b}{1+b}\right)^3$$
I thought to proceed in this manner:
We know $a+b=\frac{-2}{3}$ and $ab=\frac{1}{3}$. Using this I tried to c... | The answer is $-10$.
Find the common denominator:
$$\left(\dfrac{1-a}{1+a}\right)^3+\left(\dfrac{1-b}{1+b}\right)^3=\frac{(1-ab-(a-b))^3+(1-ab+(a-b))^3}{(1+ab+(a+b))^3}=$$
$$\frac{2\cdot\left(\frac23\right)^3+2\cdot 3 \cdot\left(\frac23\right)\cdot (a-b)^2}{(\frac{2}{3})^3}=\frac{2\cdot\left(\frac23\right)^3+4\cdot ((a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2344931",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 2
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Trig equation $\sin(x)+\sin(3x)=0$, the answers are given in a factored form $\sin(x)+\sin(3x)=0$
So to solve this I tried the following:
-First I transformed this expression from sum to product because it equals zero
$\sin(x)+\sin(3x)=0 => 2\sin(\frac{x+3x}{2})\cdot\cos(\frac{x-3x}{2})=0$
-Second part I got to
$2\si... | \begin{align*}\sin (2x)=0 &\Longrightarrow 2x=k\pi\to x=\frac{k\pi}{2}\\
\cos x=0&\Longrightarrow x= \frac{\pi}{2}+k\pi=\frac{(2k+1)\pi}{2}\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2345345",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 1
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Prove that $(k^3)!$ is divisible by $(k!)^{k^2 + k + 1}$. Proof using properties of greatest integer function? I came across this problem in a regional Olympiad Exam. I'm giving the outline of my solution and a question related to it.
Problem
Let $k$ be any positive integer. Prove that $(k^3)!$ is divisible by $(k!)^{k... | In the interval $[1,k^3]$ there are $k^3-k^2$ numbers $\not\equiv 0\pmod{k}$, $k^2-k$ numbers that are $\equiv 0\pmod{k}$ but $\not\equiv 0\pmod{k^2}$, $k-1$ numbers that are $\equiv 0\pmod{k^2}$ but $\not\equiv 0\pmod{k^3}$ and just $1$ element that is a multiple of $k^3$. In particular $(k^3)!$ is for sure a multiple... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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How to evaluate $\int \cos^2x \ dx$ I am new to integration and I want to evaluate
$ \int \cos^2x \, dx $
What I done
Using simple chain rule,
$ \int(\cos x)^2 \\
= \int t^2 \,dt \ (t = \cos x) \\
= \frac{t^3}{3} \ + C $
By substitution of $t = \cos(x)$ I got with answer.
| You can make use of the double angle formula for cosine:
$$\begin{eqnarray}\cos(2x) & = & 2 \cos^2 (x) - 1 \\
\cos^2 (x) & = & \frac{1}{2}\left(\cos (2x) + 1\right)\\
\int \cos^2 x\ dx & = & \frac{1}{2} \int \left(\cos(2x) + 1 \right) dx \\
& = & \frac{1}{2}\left[\frac{1}{2}\sin(2x) + x + C \right] \\
& = & \frac{1}{4... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2349144",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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For real $a,b,c$ , if $a^2+b^2+c^2=ab+bc+ac$ ,then the value of $\frac{a+b} {c}$ is?
For real $a,b,c$ , if $a^2+b^2+c^2=ab+bc+ac$ ,then the value of
$\frac{a+b} {c}$ is how much?
Ans.
What I could gather:
from the identity,
$$ a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^3-ab+bc+ca)$$
We gather that RHS=0.
$$ =>a^3+b^3+c^3=3... | $a^2 + b^2 +c^2 = \frac {a^2 + b^2}{2} + \frac {b^2 + c^2}{2} + \frac {a^2 + c^2}{2}= ab + bc + ac\\
\frac {a^2 -2ab +b^2}{2} + \frac {b^2 -2bc +c^2}{2} + \frac {a^2 -2ac +c^2}{2}= 0\\
(a-b)^2 + (b-c)^2 + (c-a)^2 = 0\\
a = b = c$
$\frac {a+b}{c} = 2$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How many times does $k$ occur in the composition of $n$? How many times does the number $k$ occur in the composition of $n$?
Composition of Integer
In short, the difference between the partition of an integer and composition is the order of numbers. In partition, the order doesn't matter whereas, in composition, it do... | Suppose that $k$ (such that $0<k<n$) occurs at "position $p$" in a composition of $n$: $$\ldots+k+\ldots$$
by which I mean that everything to its left is a composition of $p$, and everything to its right is a composition of $n-p-k$.
If $p>0$ and $n-p-k>0$, there are $2^{p-1}$ compositions of $p$ and $2^{n-p-k-1}$ compo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2354225",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Integration by trig substitution - why is my answer wrong? Attempting integral:
$$-\int \frac{dx}{\sqrt{x^2-9}}$$
Let $x = 3\ sec\ \theta$ so that under the square root we have:
$$\sqrt{9\ sec^2\ \theta - 9}$$
$$\sqrt{9(sec^2\ \theta - 1)}$$
$$\frac{dx}{d\theta} =3\ sec\ \theta\ tan\ \theta$$
The $1/3$ and the $3$ canc... | $$\int\frac{dx}{\sqrt {x^2-9}}$$
By substitution
$$x=3\sec\theta $$
$$dx=3\sec\theta \tan \theta d\theta$$
$$\int \frac{3\sec\theta \tan \theta d\theta}{\sqrt{9\sec^2\theta-9}} $$
$$\int \frac{3\sec\theta \tan \theta d\theta}{3\tan\theta} $$
$$\int \sec\theta d\theta $$
$$ ln|\sec\theta+\tan\theta|$$
$$ ln| \frac{1+\s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2354524",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove using vector methods that the midpoints of the sides of a space quadrilateral form a parallelogram. Problem
Prove using vector methods that the midpoints of the sides of a space quadrilateral form a parallelogram.
My Solution
B (Conclusion): The midpoints of the sides of a space quadrilateral form a parallelogr... | Hint: If your four points are $a, b, c, d$, then the midpoints, in order around the quad, are
$$
p = \frac{1}{2}(a+b), q = \frac{1}{2}(b+c), r = \frac{1}{2}(c+d), s = \frac{1}{2}(d+a).
$$
For $pqrs$ to be a parallelogram, you need the edge from $p$ to $q$ to have the same direction vector as the edge from $s$ to $r$; ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2355280",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
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How to find the value of this expression Suppose I have a general quadratic equation $ ax^2 + bx + c = 0 $
And I have $ \alpha , \beta $ as roots, then what is the simplest way to find $ (a\alpha + b)^{-2} + (a\beta + b)^{-2} $ ?
I know the formula for finding the sum and product of roots, but that doesn't helped me i... | The roots of the equation $ax^2+bx+c=0$ are:
$$\alpha=\frac{-b-\sqrt{b^2-4ac}}{2a}; \beta=\frac{-b+\sqrt{b^2-4ac}}{2a}.$$
Now substitute these into the expression:
$$\frac{1}{(a\alpha+b)^2} + \frac{1}{(a\beta+b)^2}=$$
$$\frac{1}{(\frac{-b-\sqrt{b^2-4ac}}{2}+b)^2} + \frac{1}{(\frac{-b+\sqrt{b^2-4ac}}{2}+b)^2}=$$
$$\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2355718",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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prove $\sum_{cyc}\frac{1}{a(a+b)}\ge\frac{4}{ac+bd}$ Let $a,b,c,d$ be positives. Show that
$$\sum_{cyc}\dfrac{1}{a(a+b)}\ge\dfrac{4}{ac+bd}$$
Use Cauchy-Schwarz inequality we have
$$\sum_{cyc}\dfrac{1}{a(a+b)} \cdot\sum_{cyc}a(a+b)\ge (1+1+1+1)^2$$
it suffient
$$\dfrac{16}{\sum\limits_{cyc}(a^2+ab)}\ge\dfrac{4}{ac+bd}$... | By AM-GM $$\frac{2}{\sqrt{abcd}}\geq\frac{4}{ac+bd}.$$
Thus, it remains to prove that
$$\sum_{cyc}\frac{1}{a^2+ab}\geq\frac{2}{\sqrt{abcd}}.$$
Since the last inequality is homogeneous, we can assume that $abcd=1$.
Now, let $a=\frac{y}{x}$, $b=\frac{z}{y}$ and $c=\frac{t}{z}$, where $x$, $y$, $z$ and $t$ are positives.... | {
"language": "en",
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"source": "stackexchange",
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Find out |a+p+b+q| for the following question. Let$$f(x)=\begin{cases}
ax(x-1)+b&;x\lt1\\
x+2&;1\le x\le3\\
px^2+qx+2&;x\gt3
\end{cases}$$ is continuous for all x belongs to R except $x=1$ but $|f(x)|$ is differentiable everywhere and $f'(x)$ is continuous at x=3 and $|a+p+b+q|=k$, then k=?
|
$f′(x)$ is continuous at $x=3$
So derivatives of $x+2$ and $px^2 + qx + 2$ must be equal.
$1=2px + q$; $x=3$
$\boxed{1=6p + q}$
continuous for all x belongs to R except x=1
but |f(x)|
is differentiable everywhere
So it's graph should be like this (drawed very roughly, slopes should be equal at $x=1$)
If $x+2$... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Show that $y = \frac{2x}{x^2 +1}$ lies between $-1$ and $1$ inclusive. Prove, using an algebraic method,that $y=\frac{2x}{x^2 +1}$ lies between $-1$ and $1$ inclusive. Hence, determine the minimum and maximum points $y=\frac{2x}{x^2 +1}$ .
What I tried:
Firstly, I thought of using partial fractions but since $x^2 +1 =... |
For what values of $a$ does the equation
$$
\frac{2x}{1+x^2}=a
$$
admit a solution?
The equation can be rewritten
$$
ax^2-2x+a=0
$$
and its discriminant is
$$
4(1-a^2)
$$
which is $\ge0$ if and only if $-1\le a\le 1$. For $a=0$ it's not a degree $2$ equation, but of course $f(0)=0$, so also $a=0$ is a value attai... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2361415",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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"answer_id": 1
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Integral $\int \frac{x^2}{x^4+x^2+1}\ dx$ I have taken $(x^4+x^2+1)=u$ to differentiate it w.r.t. $x$ to get
$$\int \frac{x}{2u(2x^2+1)}\ du$$
| $$\dfrac{2x^2}{x^4+x^2+1}=\dfrac{x^2+1}{x^4+x^2+1}+\dfrac{x^2-1}{x^4+x^2+1}$$
$$=\dfrac{1+\dfrac1{x^2}}{x^2+1+\dfrac1{x^2}}+\dfrac{1-\dfrac1{x^2}}{x^2+1+\dfrac1{x^2}}$$
For the first integral:
As $\displaystyle\int\left(1+\dfrac1{x^2}\right)dx=x-\dfrac1x,$
write $x^2+1+\dfrac1{x^2}=1+\left(x-\dfrac1x\right)^2+2$
Can yo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2363639",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Value of a definite integral with irrational terms in the denominator Can someone help me to find the value of the following integration?
$$ \int_0^3\frac{dx}{\sqrt{x+1} + \sqrt{5x+1}} $$
I tried to rationalize the given expression, but that was not of much help.
I also tried to apply the following formula,
$$ \int_0^... | Rationalizing is probably the way to go. After rationalizing you can rewrite the integrand as,
$$ L =- \int_0^3 \frac{\sqrt{x+1} - \sqrt{5x+1}}{4x}\ dx $$
which can further be expanded into,
$$ L = -\frac 1 4 \int_0^3 \left( \frac{\sqrt{x+1}}{x} - \frac{\sqrt{5x+1}}{x} \right) \ dx \quad\quad \text{(1)}$$
which has an... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2366058",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Polynomial having rational coefficients and one root: $\sqrt{2}+\sqrt{3}-\sqrt{5}$
Form a polynomial of smallest degree having rational coefficients and one root as $\sqrt{2}+\sqrt{3}-\sqrt{5}$
Idea 1:
I thought that other roots would be just different combination of signs on the surds, ie
*
*$\sqrt{2}+\sqrt{3}+\... |
If $f(x)\in \mathbb{Q}[x]$ has $\sqrt{2}+\sqrt{3}-\sqrt{5}$ as a root,
then it has all eight numbers $\pm \sqrt{2} \pm \sqrt{3} \pm \sqrt{5}$
as roots. So the smallest degree of your polynomial is $8$.
The reason is that we have some automorphisms over $\mathbb{Q}(\sqrt{2},\sqrt{3},\sqrt{5})$. One is
$$\sigma(\s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2367821",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 1
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$\int_\gamma{(x-y)dx + (x+y)dy}, \quad \gamma : x^2 + 2y^2 = 1 , \quad 0 \leq y $ I'm asked to find
$$\int_\gamma{(x-y)dx + (x+y)dy}$$
where
$$\gamma : x^2 + 2y^2 = 1 , \quad 0 \leq y$$
(with positive direction) i.e the upper half of the ellipse $x^2 + 2y^2 = 1$.
My attempt
Let $\sigma = \gamma + \gamma_1$ where $\... | Here is an alternate approach using straight forward substitution.
Let $x=\cos\theta,\,y=\frac{\sqrt{2}}{2}\sin\theta,\,0\le\theta\le\pi$
\begin{eqnarray}
\int_\gamma{(x-y)dx + (x+y)dy}&=&\int_0^\pi-\left(\cos\theta+\frac{\sqrt{2}}{2}\sin\theta\right)\sin\theta+\left(\cos\theta+\frac{\sqrt{2}}{2}\sin\theta\right)\cdot\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2368362",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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Prove this trigonometric inequality about the angles of $\triangle ABC$
In $\Delta ABC$ show that
$$\cos{\frac{A}{2}}+\cos\frac{B}{2}+\cos\frac{C}{2}\ge \frac{\sqrt{3}}{2} \left(\cos\frac{B-C}{2}+\cos\frac{C-A}{2}+\cos\frac{A-B}{2}\right)$$
since
$$\frac{\sqrt{3}}{2}\left(\cos\frac{B-C}{2}+\cos\frac{C-A}{2}+\cos\df... | We can make also the following thing.
We'll prove that for all triangle the following inequality is true.
$$\sin\alpha+\sin\beta+\sin\gamma\geq\frac{\sqrt{3}}{2}(\cos(\alpha-\beta)+\cos(\alpha-\gamma)+\cos(\beta-\gamma)).$$
Indeed, let $R$ be a radius of circumcircle, $r$ be a radius of inscribed circle and $p$ be a s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2368630",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 2
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Quartic with 3 distinct roots Consider a quartic equation
$$x^4 – kx^3 + 11x^2 – kx + 1 = 0$$
The value of $k$ so that given equation has three real and distinct solutions can be equal to - i think that for any four degree to have 3 roots, its $f'(... | You already know a polynomial having one distinct root whose derivative has one root: $f(x) = x^2$. It is not automatic that the number of distinct roots decreases when you take the derivative of a polynomial -- the total number of roots (including multiplicities) decreases by one. So if it is one of the multiple roo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2369210",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 4
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Which area is larger, the blue area, or the white area? In the square below, two semicircles are overlapping in a symmetrical pattern. Which is greater: the area shaded blue or the area shaded white?
My Solution
Let the length of each side of the square be $2r$.
The area of the square is $4r^2$.
The two semi-circles h... | The area of the segment is “quarter of circle minus triangle”:
$$
\frac{1}{4}\pi r^2-\frac{1}{2}r^2=\frac{r^2}{4}(\pi-2)
$$
Thus half of the white area is “quarter of circle plus triangle minus segment”:
$$
\frac{1}{4}\pi r^2+\frac{1}{2}r^2-\frac{r^2}{4}(\pi-2)=r^2
$$
Therefore the white area is $2r^2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2369403",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "16",
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"answer_id": 1
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Integrate $\int t^{-\frac{1}{2}} e^{-\frac{c^2}{4t}} \, \mathrm{d}t$ I am trying to solve the following integral by hand, where $c$ is a constant:
$$\int t^{-\frac{1}{2}} e^{-\frac{c^2}{4t}} \, \mathrm{d}t$$
I have found a solution using Mathematica, but I was interested in how one might solve it manually.
At first I t... | Using the substitution suggested by Robert Israel quickly yields a solution:
Let $t=\cfrac{c^2}{4v^2}$.
Thus $\mathrm{d}t = -\cfrac{c^2}{2v^3}$ and $v=\cfrac{c}{2\sqrt{t}}$.
So,
\begin{align}
\int t^{-\frac{1}{2}} e^{-\frac{c^2}{4t}} \, \mathrm{d}t &= \int \left(\frac{c^2}{4}v^{-2}\right)^{-\frac{1}{2}} e^{-\frac{c^2}... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Limit $\lim_{x \to 2^{-}} \left ( \frac{1}{\sqrt[3]{x^{2} -3x+2}} + \frac{1}{\sqrt[3]{x^{2} -5x+6}} \right )$ $$\lim_{x \to 2^{-}} \left ( \dfrac{1}{\sqrt[3]{x^{2} -3x+2}} + \dfrac{1}{\sqrt[3]{x^{2} -5x+6}} \right )$$
I've tried using the $A^3-B^3$ identity, but that doesn't help. Also, I tried multiplying every fract... | Note that
\begin{align}
x^2-3x+2&=(2-x)(1-x)=(2-x)(2-x-1)
\\[4px]
x^2-5x+6&=(2-x)(3-x)=(2-x)(2-x+1)
\end{align}
Set $2-x=t^3$, so your expression becomes
$$
\frac{1}{\sqrt[3]{t^3(t^3-1)}}+\frac{1}{\sqrt[3]{t^3(t^3+1)}}
=\frac{1}{t}\frac{\sqrt[3]{t^3+1}+\sqrt[3]{t^3-1}}{\sqrt[3]{t^6-1}}
$$
and the limit is
$$
\lim_{t\to... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2370100",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
What is the probability that Cathy wins?
Alice, Bob and Cathy take turns (in that order) in rolling a six sided
die. If Alice ever rolls a 1, 2 or 3 she wins. If Bob rolls a 4 or a 5
he wins, and Cathy wins if she rolls a 6. They continue playing until
a player wins. What is the probability (as a fraction) that ... | C wins when:
1) A loses, B loses, C wins
or
2) A loses, B loses, C loses, A loses, B loses, C wins
or
3) A loses, B loses, C loses, A loses, B loses, C loses, A loses, B loses, C wins
or
..........
There are infinitely many possibilities, what leads to infinite series
$$
\sum_{i\ge 0}\left(\frac{3}{6}\cdot\frac{4}{6}\c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2374181",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Solve $\frac{\sqrt{5+2\sqrt{6}} + \sqrt{5-2\sqrt{6}}}{\sqrt{5+2\sqrt{6}} - \sqrt{5-2\sqrt{6}}} =\sqrt \frac{x}{2}$ The value of $x$ (considering only the positive root) satisfying the equation $$\frac{\sqrt{5+2\sqrt{6}} + \sqrt{5-2\sqrt{6}}}{\sqrt{5+2\sqrt{6}} - \sqrt{5-2\sqrt{6}}} = \sqrt{\frac{x}{2}}$$ is?
Please, ca... | Hint: Multiply the numerator and denominator of
$$\frac{\sqrt{5 + 2\sqrt{6}} + \sqrt{5 - 2\sqrt{6}}}{\sqrt{5 + 2\sqrt{6}} - \sqrt{5 - 2\sqrt{6}}}$$
by the conjugate of the denominator to obtain
$$\frac{\sqrt{5 + 2\sqrt{6}} + \sqrt{5 - 2\sqrt{6}}}{\sqrt{5 + 2\sqrt{6}} - \sqrt{5 - 2\sqrt{6}}} \cdot \frac{\sqrt{5 + 2\s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2375517",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
What did I do wrong in this proof? I have this answer marked partially correct. I'd like to get an idea of where I went wrong
Prove that for all integers $a,b \in \mathbb{Z}$, $(a+b)^5 \equiv a^5+b^5 \mod 5$.
My answer:
Suppose that $a,b \in \mathbb{Z}$. Then, by definition of congruence modulo, $5|(a+b)^5-(a^5+b^5)$ m... | Your idea is correct, but your proof looks backwards. I suggest the following modifications. Observe how now the proof looks like a tale of deduction rather than one of back-engineering, if you may.
Suppose that $a,b \in \mathbb{Z}$. Then, by definition of congruence
modulo, we must show that $5|(a+b)^5-(a^5+b^5)$. ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2380319",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
using Parseval's identity to estimate the value of $\Sigma_{n=1}^\infty \frac{1}{(2n-1)^2(2n+1)^2}$ There is a problem with two parts; The first part is asking to find Fourier series for $f(x)=|\sin(x)|$ on $[-\pi,\pi]$. And the second part wants to estimate the following using Parseval's identity:
$$
\sum_{n=1}^\infty... | The Fourier series of the function $f(x)$ is given by
$$
f(x) = \sum_{n=-\infty}^\infty c_n e^{i n x}
$$
with
$$
c_n = \frac{1}{2 \pi} \int_{-\pi}^\pi d x~f(x) e^{-\imath n x}
$$
In the case $f(x) = |\sin x |$, this results in
$$
c_n = 2 \frac{1}{2 \pi} \int_{0}^\pi d x~\sin x e^{-\imath n x} = \left\{ \begin{array}{l... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2384297",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Evaluating limits Evaluate the limit without L’Hôpital rule:
$$
\lim_{x \to 0}\frac{\sin^2{x}+2\ln\left(\cos{x}\right)}{x^4}
$$
My work is:
\begin{align}
L&=\lim_{x \to 0}\frac{\sin^2{x}-x^2}{x^4}+\lim_{x \to 0} \frac{x^2+2\ln\left(\cos{x}\right)}{x^4}\\
&=
\lim_{x \to 0}\frac{\sin{x}-x}{x^3}
\lim_{x \to 0}\frac{\sin{x... | Or we could've approached it differently from the beginning, if we observe that
$$2\ln(\cos x)=\ln(\cos^2 x)=\ln(1-\sin^2 x)$$
and use the series for
$$\ln(1-t)=-\sum_{n=1}^{\infty}\frac{t^n}{n}=-t-\frac{t^2}{2}-\frac{t^3}{3}-\cdots.$$
Then for the original limit:
$$\lim_{x\to0}\frac{\sin^2 x+2\ln(\cos x)}{x^4}=
\lim_{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2387185",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 1
} |
Show that the roots of the equation $x^2-4abx+(a^2+2b^2)^2=0$ are imaginary. Show that the roots of the equation $x^2-4abx+(a^2+2b^2)^2=0$ are imaginary.
My Attempt:
$$x^2-4abx+(a^2+2b^2)^2=0$$
Comparing above equation with $Ax^2+Bx+C=0$, we get
\begin{align}
A&=1 \\
B&=-4ab \\
C&=(a^2+2b^2)^2
\end{align}
Now,
\begin{a... | Suppose $ab\ne 0$.
Let $f (x) $ be your quadric.
$$\lim_{\infty}f (x)=+\infty $$
thus the roots will not be real if the minimum is $$f (c)>0$$ where $c $ is such that $$f'(c)=0.$$
$$f'(x)=2x-4ab \implies c=2ab $$
$$\implies f (c)=$$
$$4a^2b^2-8a^2b^2+(a^2+2b^2)^2=$$
$$(a^2+2b^2+2ab)(a^2+2b^2-2ab) =$$
$$((a+b)^2+b^2)((a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2387278",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Limit of $f(x)\cdot g(x)$ given that $\lim_{x\to 0}{\frac{f(x)}{\sin(2x)}}=2$ and $\lim_{x\to 0}{(\sqrt{x+4}-2)\cdot{g(x)}}=5$ Question:
If $$\lim_{x\to 0}{\left[\frac{f(x)}{\sin(2x)}\right]}=2$$ and $$\lim_{x\to 0}{\left[(\sqrt{x+4}-2)\cdot{g(x)}\right]}=5$$ then find $$\lim_{x\to 0}{[f(x)\cdot{g(x)}]}$$
But, from w... | Only one theorem applied.
Theorem. If $\lim\limits_{x\rightarrow c}f(x)=L$ and $\lim\limits_{x\rightarrow c}g(x)=M$, then $\lim\limits_{x\rightarrow c}[f(x)g(x)]=LM$
$$
\because \qquad \lim_{x\rightarrow0}\frac{f(x)}{\sin(2x)}=2\quad \text{and}\quad \lim_{x\rightarrow0}\left[(\sqrt{x+4}-2）\cdot g(x)\right]=5
$$
$$
\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2390040",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
On a determinant identity
Show that
$$\det \begin{bmatrix} -x & 1 & \dots & 1 \\ 1 & -x & \dots & 1 \\ \vdots & \vdots & \ddots & \vdots \\ 1 & 1 & \dots & -x \end{bmatrix} = (-1)^n(1+x-n)(1+x)^{n-1}$$
for all $n \in \mathbb{N}$.
I tried using induction but I didn't manage (using row/column swaps). Does anyone have ... | HINT
$$\begin{align}\begin{vmatrix} -x & 1 & \dots & 1 \\ 1 & -x & \dots & 1 \\ \vdots & \vdots & \ddots & \vdots \\ 1 & 1 & \dots & -x \end{vmatrix}&=\begin{vmatrix} -x+n-1& 1 & \dots & 1 \\ -x+n-1 & -x & \dots & 1 \\ \vdots & \vdots & \ddots & \vdots \\-x+n-1 & 1 & \dots & -x \end{vmatrix}\\&=(-x+n-1)\begin{vmatrix} ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2390545",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.