Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
How to prove this trig. identity???(question 1 : solved)(question 2 : solved) question 1:
We know :
$$
\left\{
\begin{array}{ll}
\alpha \ + \beta\ +\theta\ = \pi\\
\alpha\ = \frac{\pi}{2} \\
\end{array}
\right.
$$
How to prove this :
$$
\sin(\alpha)\sin(\beta)\sin(\alpha\ -\ \beta) + \sin(\beta)\sin(\theta)\sin(\be... | Question 2 :
The relation $\alpha + \beta + \theta = \pi$ implies that $\cos (\alpha + \beta + \theta) = -1$. But
\begin{align}
\cos(\alpha + \beta + \theta) &= \cos\alpha \cos(\beta + \theta) - \sin \alpha\sin(\beta + \theta) \\
&= \cos(\alpha)(\cos\beta\cos\theta - \sin\beta \sin\theta) - \sin\alpha(\sin\beta \cos\t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1682586",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Solve the Integral $\int \frac{dx}{\left(x-2\right)^3\sqrt{3x^2-8x+5}}$ $$\int \frac{dx}{\left(x-2\right)^3\sqrt{3x^2-8x+5}}$$
I am pretty sure that a certain substitution should work.
I tried using $x-2=\frac{1}{t}$
And got the integral:
$$\int \:\frac{-\frac{dt}{t^2}}{\frac{1}{t^3}\sqrt{3\left(\frac{1}{t}+2\right)^2-... | From the point that you stopped:
$$\int \:\frac{tdt}{\sqrt{3\left(\frac{1}{t}+2\right)^2-8\left(\frac{1}{t}+2\right)^2+5}}$$
$$=\int\frac{tdt}{\sqrt{5-5\left(\frac 1 t+2\right)^2}}=\int\frac{tdt}{\sqrt{-\frac{5}{t^2}-\frac{20}{t}-15}}=-\frac{i}{\sqrt 5}\int\frac{t^2}{\sqrt{3t^2+4t+1}}dt$$
$$\overset{\text{complete th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1684228",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Symmetric matrix with nonnegative determinant Given positive reals $a,b$, and $c$, is it true that the matrix
$$
\begin{pmatrix} a^4+b^4+c^4 & a^3+b^3+c^3 & a^2+b^2+c^2 \\ a^3+b^3+c^3 & a^2+b^2+c^2 & a+b+c \\ a^2+b^2+c^2 & a+b+c & 3\\ \end{pmatrix}
$$
has nonnegative determinant?
I conjecture the answer is affirmativ... | Note that
$$
\begin{pmatrix} a^4+b^4+c^4 & a^3+b^3+c^3 & a^2+b^2+c^2 \\ a^3+b^3+c^3 & a^2+b^2+c^2 & a+b+c \\ a^2+b^2+c^2 & a+b+c & 3\\ \end{pmatrix}=\begin{pmatrix} a^2 & b^2 & c^2 \\ a & b & c \\ 1 & 1 & 1\\ \end{pmatrix}\begin{pmatrix} a^2 & a & 1 \\ b^2 & b & 1 \\ c^2 & c & 1\\ \end{pmatrix}
$$
The determinant t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1685498",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
$\lim_{x \to 0^+}\frac{\tan x \cdot \sqrt {\tan x}-\sin x \cdot \sqrt{\sin x}}{x^3 \sqrt{x}}$ Calculate:
$$\lim_{x \to 0^+}\frac{\tan x \cdot \sqrt {\tan x}-\sin x \cdot \sqrt{\sin x}}{x^3 \sqrt{x}}$$
I don't know how to use L'Hôpital's Rule.
I tried to make $\tan x =\frac{\sin x}{\cos x}$ for the term ${\sqrt{\tan x}}... | $$\begin{align}
\lim_{x\to0^{+}}\frac{\tan x \cdot \sqrt {\tan x}-\sin x \cdot \sqrt{\sin x}}{x^3 \sqrt{x}}
&=\lim_{x\to0^{+}}\frac{\tan^{3/2}x-\sin^{3/2} x}{x^3 \sqrt{x}}\\
&=\lim_{x\to0^{+}}\frac{\sin^{3/2}x}{x^{3/2}}\frac{\sec^{3/2}x-1}{x^2}\\
&=\lim_{x\to0^{+}}\left(\frac{\sin x}{x}\right)^{3/2}\cdot \lim_{n\to\inf... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1687677",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
$x^{2}+2x \cdot \cos b \cdot \cos c+\cos^2{b}+\cos^2{c}-1=0$ Solve this equation :
$$x^{2}+2x \cdot \cos b \cdot \cos c+\cos^2{b}+\cos^2{c}-1=0$$
Such that $a+b+c=\pi$
I don't have any idea. I can't try anything.
| "Completing the square" would be my choice.
$x^2+ (2cos(b)cos(c))x+ cos^2(b)cos^2(c)- cos^2(b)cos^2(c)+ cos^2(b)+ cos^2(c)- 1= 0$
$(x+ cos(b)cos(c))^2= 1- cos^2(b)- cos^2(c)+ cos^2(b)cos^2(c)$
You say that "$a+ b+ c= \pi$ but there is NO "a" in your equation.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1688739",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
} |
Solve in positive integers the equation $a^3+b^3=9ab$ Solve in positive integers the equation:
$$a^3+b^3=9ab$$
I try to:
$$\dfrac{a^2}{b}+\dfrac{b^2}{a}=9\Longrightarrow a^2<9b,b^2<9a$$
Of course, I can't solve it. Can anyone help?
| By the Arithmetic Mean Geometric Mean Inequality, we have
$$\frac{9}{2}=\frac{1}{2}\left(\frac{a^2}{b}+\frac{b^2}{a}\right)\ge \sqrt{ab}.$$
Now we only have a smallish number of possibilities to examine.
A simple congruential observation then dramatically cuts down the number of cases. For $a$ and $b$ must both be eve... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1691052",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
How to evaluate $\int_{0}^{1}\frac{\arctan x}{1+x^{2}}\ln\left ( \frac{1+x^{2}}{1+x} \right )\mathrm{d}x$ How to evaluate
$$\int_{0}^{1}\frac{\arctan x}{1+x^{2}}\ln\left ( \frac{1+x^{2}}{1+x} \right )\mathrm{d}x$$
I completely have no idea how to find the result.Mathematic gave me the following answer part of the integ... | Let $t=\arctan x$. Then
\begin{eqnarray}
&&\int_{0}^{1}\frac{\arctan x}{1+x^{2}}\ln\left ( \frac{1+x^{2}}{1+x} \right )\mathrm{d}x\\
&=&-\int_{0}^{\frac{\pi}{4}}t\ln[\cos t(\cos t+\sin t))]\mathrm{d}t\\
&=&-\frac{1}{2}\int_{0}^{\frac{\pi}{4}}t\ln[\cos t^2(\cos t+\sin t)^2]\mathrm{d}t\\
&=&-\frac{1}{2}\int_{0}^{\frac{\p... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1692452",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Prove that $1+\frac{1}{3}+\frac{1\cdot 3}{3\cdot 6}+\frac{1\cdot 3\cdot 5}{3\cdot 6 \cdot 9}+.........=\sqrt{3}$
Prove that $$1+\frac{1}{3}+\frac{1\cdot 3}{3\cdot 6}+\frac{1\cdot 3\cdot 5}{3\cdot 6 \cdot 9}+.........=\sqrt{3}$$
$\bf{My\; Try::}$ Using Binomial expansion of $$(1-x)^{-n} = 1+nx+\frac{n(n+1)}{2}x^2+\fr... | Here is a variation to obtain $\sqrt{3}$ based upon the generating function of the Central binomial coefficients
\begin{align*}
\sum_{n=0}^{\infty}\binom{2n}{n}z^n=\frac{1}{\sqrt{1-4z}}\qquad\qquad |z|<\frac{1}{4}
\end{align*}
We obtain
\begin{align*}
1&+\frac{1}{3}+\frac{1\cdot 3}{3\cdot 6}+\frac{1\cdot 3\cdot ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1692678",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
"answer_count": 1,
"answer_id": 0
} |
If $a^{2}+84a+2008=b^{2}$ what is $a+b$ let $a, b$ are two positive integer satisfy the condition $a^{2}+84a+2008=b^{2}$. Find out $a+b$
My Solution
$a^{2}+84a+2008=b^{2} \implies (a+42)^{2}+244=b^{2} \implies (b+a+42)(b-a-42)=2^{2}61$. Considering (244,1) , (122,2) ,(61,4) we observe only (244,1) give integer $(a,b)=(... | With the quadratic formula you get
$$a=\frac{-84\pm \sqrt{84^2-4(2008-b^2)}}{2}=-42\pm \sqrt{b^2-244}$$
Now with this, it is possible to obtain the root since $b^2-244=x^2$, and as $b$ and $a$ must be positive integers, that root must be too, so neccesarily $x$ too. This lets
$$(b+x)(b-x)=244$$
And that implies that:
$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1692763",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
If a family of straight lines is $\lambda^2 P+\lambda Q+R=0$ ,then the family of lines will be tangent to the curve $Q^2=4PR.$ I have read this theorem in my book but i do not know how to prove it.
If a family of straight lines can be represented by an equation $\lambda^2 P+\lambda Q+R=0$ where $\lambda $ is a paramete... | This is a pretty brute-force method but it works.
Test
To figure out what was going on, I tried a simple case: $P=x$, $Q=y$, $R=1$. The family of lines is $$\lambda^2x +\lambda y + 1 = 0$$ and the quadratic is $$y^2 = 4x$$ To say that the family of lines is tangent to the curve means that wherever they intersect, the... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1693078",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 1
} |
Find the Cartesian equation of the locus described by $\arg \left(\frac{z-2}{z+5} \right)= \frac{\pi}{4}$ My working:
$$ \frac{x + iy - 2}{x + iy + 5} $$
$$ \frac{(x - 2 + iy)(x+5-iy)}{(x + 5 + iy)(x+5-iy)} $$
$$ \frac{x^2+5x-ixy-2x-10+2iy+ixy+5iy+y^2}{x^2+5x-ixy+5x+25-5iy+ixy+5iy+y^2} $$
$$ \frac{x^2+3x-10+y^2+7iy}{x^... | Hint for another method:
Translate this problem into pure geometry: denote $A$ the point with affix $-5$, $B$ the point with affix $2$, $M$ the point with affix $z$. The condition translates into $\angle AMB=\frac\pi4$.
Now use the inscribed angle theorem: the locus of $M$ is a circular arc. The centre $O$ of this arc... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1693722",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
indefinite integral of $\cos(x) / x$? How can I evaluate ;
$$\int \frac{\cos(x)}{x} \, dx $$
I tried to do partial integration but it became to an infinite loop of partial integrations
| Recall the Taylor series expansion for $\cos x$
\begin{align*}\cos x &= 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + \cdots \\
\implies \frac{\cos x}{x} &= \frac{1}{x} - \frac{x}{2!} - \frac{x^3}{4!} - \frac{x^5}{6!} + \cdots \\
\therefore \int \frac{\cos x}{x} &= ln\ x - \frac{x^2}{2 \times 2!} + \frac{x^4}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1694748",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
If $f(x)=\lim_{n\to\infty}[2x+4x^3+\cdots+2nx^{2n-1}]$, $0
If $f(x)=\lim_{n\to\infty}[2x+4x^3+\cdots+2nx^{2n-1}]$, $0<x<1$, then find $\int f(x)\mathrm{d}x$
$$f(x)=\lim_{n\to\infty}2x[1+2x^2+\cdots+nx^{2n}]$$
$$S=\frac{f(x)}{2x}=1+2x^2+\cdots+nx^{2n}$$
$S$ is an AGP. I used the general method for finding $S$.
$$x^2S=x^... | You can turn that long expression into a infinite series.
$$ f(x) = \sum_{n=1}^\infty 2n x^{2n-1}
$$
Let $g = \int f(x)\,dx$. By termwise integration, (This is justified by differentiating $g$ and noticing that $g$ has the same radius of convergence as $f$. Then it follows from the uniqueness of antiderivative up to a ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1695309",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
System of inequalities $x^2+2x+\alpha\leq0$ and $x^2-4x-6\alpha\leq 0$ has unique solution.
Find all values of $\alpha$ for which the system of inequalities $x^2+2x+\alpha\leq0$
and $x^2-4x-6\alpha\leq 0$
has unique solution.
My Try: We can write it as $$\frac{x^2-4x}{6}\leq \alpha \leq -x^2-2x$$
So we get $$\frac{(x... | As both the Quadratic inequalities is less than zero, so we can say that their discriminant must be greater than or equal to Zero ,i.e., real roots must exists.
So, we will get $$4-4\alpha \ge 0$$ $$16 + 24\alpha \ge 0$$On solving these eqations, we will get $$\frac{-2}3 \le \alpha \le 1$$ Now, we can determine the roo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1697550",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Find a plane perpendicular to $yz$, passing by a point and making an angle with another plane The problem is to find the equation of a plane (let's call it $A$) that is perpendicular to the $yz$ plane, containing the point $P(2,1,1)$, and making an angle of $\cos^{-1} \frac{2}{3}$ with the plane $2x - y + 2z - 3 = 0$ (... | With the relationship you found between $b$ and $c$, as well as $x=0$, you have
$$
ax + by + cz + d = 0 \\
a(0) + by + (\frac{-3b}{4})z + d = 0 \\
4by - 3bz + 4d = 0 \\
$$
With the components of $(2,1,1)$, you then have
$$
4b(1) - 3b(1) + 4d = 0 \\
b = -4d \\
$$
With this relationship between $b$ and $d$, you finally h... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1698134",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
If a chord joining the points $P(a\sec\alpha,a\tan\alpha)$ and $Q(a\sec\beta,a\tan\beta)$ on the hyperbola $x^2-y^2=a^2$ is a normal to it at $P$,then If a chord joining the points $P(a\sec\alpha,a\tan\alpha)$ and $Q(a\sec\beta,a\tan\beta)$ on the hyperbola $x^2-y^2=a^2$ is a normal to it at $P$,then show that $\tan \... | As $(2)$ passes through $Q(a\sec\beta,a\tan\beta),$
$$\dfrac{\sec\beta}{\sec\alpha}=2-\dfrac{\tan\beta}{\tan\alpha}$$
Squaring and writing $\tan\alpha=p,\tan\beta=q,$
$$\dfrac{1+q^2}{1+p^2}=4+\dfrac{q^2}{p^2}-\dfrac{4q}p$$
$$\iff4-\dfrac{4q}p=\dfrac{1+q^2}{1+p^2}-\dfrac{q^2}{p^2}$$
$$\iff\dfrac{4(p-q)}p=\dfrac{p^2-q^2}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1698451",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Given $f(z)=(z^2-1)^{\frac{1}{2}}$ find the first three terms of the Laurent expansion Given
$$f(z)=(z^2-1)^{\frac{1}{2}}$$
which has a branch cut for $|z|<1$. Find the first three terms of the Laurent expansion.
I proceed by factoring out a $z^2$.
$$z(1-\frac{1}{z^2})^{\frac{1}{2}}$$
where I know the following expans... | This function has branch points at $z=+1$ and at $z=-1$ and a branch line connecting these two points. Because it is a square-root singularity, the branch line for $|z|>1$ cancels, as may be seen by considering the net phase change when both branch points are encircled:
\begin{align*}
\left.\arg\left(\sqrt{z^2-1}\right... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1699012",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
The number of real roots of $x^5 + 2x^3 + x^2 + 2 = 0 $ is The number of real roots of $x^5 + 2x^3 + x^2 + 2 = 0 $ is
A. 0;
B. 3;
C. 5;
D. 1.
I don't know how to solve this.
| Consider the polynomial $\mod{x^3 +1} $
$x^5 + 2x^3 + x^2 + 2 \equiv -x^2 -2 + x^2 + 2 \equiv 0$
Thus, the polynomial has a factor of $x^3+1$
The remainder of the question can be done as above.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1700615",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 8,
"answer_id": 7
} |
How do i find eigenvectors for a $3\times 3$-matrix when eigenvalues are mixed complex or real? My task is this;
Find the eigenvalues and eigenvectors for the matrix: $A = \begin{pmatrix}2 & 1 & -3\\4 & 2 & 3\\0 & 0 & 1\end{pmatrix}$.
My work so far is this:
Apply the lemma that states $\lambda$ is and eigenvalue for $... | Your eigenvalues are correct. Let's step through one eigenvector example and you can do the other two. We will use $\lambda = 0$. Note that there are other approaches to finding eigenvectors, this is just one approach.
To find the eigenvectors, we want to find $v$ such that $Av = \lambda v$ or $(A- \lambda I) v = 0$.
F... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1701336",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
solve $\int \frac{dx}{3+\sqrt{x+2}}$
$$\int \frac{dx}{3+\sqrt{x+2}}$$
How should I approach this? U-substituion does not seem to work
| Let $y = x + 2, dy = dx$.
$$ \int \frac{dx}{3+\sqrt{x+2}} = \int \frac{dy}{\sqrt{y} + 3}
$$
Let $z = \sqrt{y}, dz = dy/(2\sqrt{y})$.
$$ \int \frac{dy}{\sqrt{y} + 3} = \int \frac{2\sqrt{y} dz}{z+3} = 2\int \frac{z}{z+3} dz
$$
Then do partial fraction decomposition.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1701903",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Let $X = \dfrac{1}{25} \sum\limits_{i=1}^{25} X_i$ and $Y =\dfrac{5}{2}X - \dfrac{2}{5}$. What is $P(|Y| > 1)$?
Suppose that $X_1,X_2,\ldots,X_{25}$ are independent random variables from $\mathcal{N}(1, 4)$. Let $X = \dfrac{1}{25} \sum\limits_{i=1}^{25} X_i$ and $Y =\dfrac{5}{2}X - \dfrac{2}{5}$. What is the probabili... | Assuming "$4$" is the variance, ...
$X$ is a linear function of i.i.d. r.v.s. It's mean is $\frac{1}{25} ( 25 \cdot 1) = 1$. It's variance is $\left( \frac{1}{25} \right)^2 (25 \cdot 4) = 4/25$. (This is a standard result. There are several ways to get this. Google "sums of normally distributed variables" for a pl... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1702341",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
The graphs of $x^2+y^2+6x-24y+72=0$ and $x^2-y^2+6x+16y-46=0$ intersect at four points.Find the sum of the distances of these four points The graphs of $x^2+y^2+6x-24y+72=0$ and $x^2-y^2+6x+16y-46=0$ intersect at four points.Compute the sum of the distances of these four points from the point $(-3,2).$
$x^2+y^2+6x-24y... | note: for any point of 4, the distance to $(-3,2)$ is $\sqrt{(-3\pm\sqrt{36\pm 4\sqrt{41}}-(-3))^2+(10\pm \sqrt{41}-2)^2}=\sqrt{36\pm 4\sqrt{41}+(8\pm \sqrt{41})^2}=\sqrt{(10\pm\sqrt{41})^2}=10\pm\sqrt{41}$
so it is easy to see the final answer.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1702822",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
} |
How can I evaluate $\int_{0}^{1}\frac{(\arctan x)^2}{1+x^{2}}\ln\left ( 1+x^{2} \right )\mathrm{d}x$ How to calculate this relation?
$$I=\int_{0}^{1}\frac{(\arctan x)^2}{1+x^{2}}\ln\left ( 1+x^{2} \right )\mathrm{d}x=\frac{\pi^3}{96}\ln{2}-\frac{3\pi\zeta{(3)}}{128}-\frac{\pi^2G}{16}+\frac{\beta{(4)}}{2}$$
Where G is ... | Take $\arctan\left(x\right)=u,\,\frac{dx}{1+x^{2}}=du
$. Then $$I=\int_{0}^{1}\frac{\left(\arctan\left(x\right)\right)^{2}}{1+x^{2}}\log\left(1+x^{2}\right)dx=\int_{0}^{\pi/4}u^{2}\log\left(1+\tan^{2}\left(u\right)\right)du
$$ $$=-2\int_{0}^{\pi/4}u^{2}\log\left(\cos\left(u\right)\right)du
$$ and now using the Fouri... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1702974",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 1,
"answer_id": 0
} |
What is $\frac{{\partial f(\left| z \right|)}}{{\partial x}} = $? Let $f(z)$ is polynomial and $z=x+iy$ , $x,y\in R$.
What is $\frac{{\partial f(\left| z \right|)}}{{\partial x}} = $?
| $$\begin{align} & \frac{{\partial f(\left| z \right|)}}{{\partial x}}
\\ & =\frac{{\partial f(\sqrt{x^2+y^2})}}{{\partial x}}
\\ & =\frac{\partial f(\sqrt{x^2+y^2})}{\partial (\sqrt{x^2+y^2})}\cdot \frac{\partial (\sqrt{x^2+y^2})}{\partial x}
\\ & =\frac{d\{f(\sqrt{x^2+y^2})\}}{d(\sqrt{x^2+y^2})}\cdot \frac{\partial (\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1704264",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Series of squares of n integers - where is the mistake? Given the following two series:
$$1^3 + 2^3 + ... + n^3$$
$$0^3 + 1^3 + .... + (n-1)^3$$
I take the difference vertically of the two:
$$\left(1^3-0^3\right) + \left(2^3-1^3\right) + .... + \left(n^3-(n-1)^3\right)$$
This equals to $n^3$
If I now express this in su... | $$\sum_{i=1}^n i={n(n+1)\over2}\quad\text{ not }\quad {n(n-1)\over2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1704944",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 2
} |
Closed form for $\sum_{n=1}^\infty \int_0^1 \frac{x^{n-1}\ln^2(x)\ln(1-x)}{n^2} \,dx$ I am trying to get this to equal $\displaystyle-\frac {\pi^a}{b}$ for some positive integers $a$ and $b$ . My efforts so far give:
$\displaystyle \sum_{n=1}^\infty \int_0^1 \frac{x^{n-1}\ln^2(x)\ln(1-x)}{n^2} \,dx=A-B+C$
where $A=\d... | Hint. One may use some MZVs algebra.
For $n\geq1$, set
$$
I_n:=\frac1{n^2}\int_0^1 x^{n-1}\ln^2(x)\ln(1-x)\:dx,
$$ then, differentiating the Euler beta integral three times, one gets
$$
I_n=-\frac4{n^6}-2\frac{H_{n-1}}{n^5}+2\frac{\frac{\pi^2}6-H_{n-1,2}}{n^4}+2\frac{\zeta(3)-H_{n-1,3}}{n^3}
$$ summing with respect to... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1706158",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
Calculate the limit of: $x_n = \frac{\ln(1+\sqrt{n}+\sqrt[3]{n})}{\ln(1 + \sqrt[3]{n} + \sqrt[4]{n})}$, $n \rightarrow \infty$ Is it ok to solve the following problem this way? What I have done is to solve parts of the limit first (that converges to $0$), and then solve the remaining expression? Or is this flawed reaso... | With all due respect to the answers already posted, I'm not in favor of using "equivalents" without enough explanation, or writing $\lim_{n\to \infty}$ before the limit has been shown to exist. Instead, I would do this: The expression for $n>1$ is less than
$$\frac{\ln (3n^{1/2})}{\ln n^{1/3}} = \frac{\ln 3 + (1/2)\ln ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1706508",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
What is the Permutation Matrix in FFT DFT Factorization? Given:
$$F_N = \frac{1}{\sqrt{N}} \begin{bmatrix}
1&1&1&1&\cdots &1 \\
1&\omega&\omega^2&\omega^3&\cdots&\omega^{N-1} \\
1&\omega^2&\omega^4&\omega^6&\cdots&\omega^{2(N-1)}\\ 1&\omega^3&\omega^6&\omega^9&\cdots&\omega^{3(N-1)}\\
\vdots&\vdots&\vdots&\vdots&\ddot... | The permutation matrix P for FFT of $F_8$ is
$
P=\begin{bmatrix}
1&0&0&0&0&0&0&0\\
0&0&1&0&0&0&0&0\\
0&0&0&0&1&0&0&0\\
0&0&0&0&0&0&1&0\\
0&1&0&0&0&0&0&0\\
0&0&0&1&0&0&0&0\\
0&0&0&0&0&1&0&0\\
0&0&0&0&0&0&0&1
\end{bmatrix}
$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1709714",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Prove that $\lim_{x\to\infty} \frac{1}{x}\int_0^xf(t)dt$ exists and find it, where $f(t)$ is an alternating function. The problem: function $f$ alternates between $1$ and $-1$ on $[0^2; 1^2), [1^2; 2^2), ... [(n-1)^2; n^2)$, so on $[0^2; 1^2], \;f(x) = 1;$ on $[1^2; 2^2],\;f(x) = -1$ etc.
Prove that $\;\;\lim_{x\to\in... | Let $x \in \mathbf R$. Then there is a unique integer $n \in \mathbf N$ such that $n^2 \le x < (n+1)^2$. We have
\begin{align*}
\int_0^x f(t)\, dt &= \sum_{k=0}^{n-1} (-1)^k (2k+1) + (-1)^n(x-n^2)\\
&= 2\sum_{k=0}^{n-1} (-1)^k k + \frac 12\bigl(1 + (-1)^{n}\bigr) + (-1)^n(x-n^2)\\
&= \frac 12\bigl((-1)^{n-1}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1710142",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Induction for divisibility: $3\mid 12^n -7^n -4^n -1$ I must use mathematical induction to show that
$a_{n} = 12^n −7^n −4^n −1$ is divisible by 3 for all positive integers n.
Assume true for $n=k$
$a_{k} = 12^k -7^k -4^k -1$
Prove true for $n=k+1$
$a_{k} = 12^{k+1} -7^{k+1} -4^{k+1} -1$
$ = (12^k)(12) - (7^k)(7) - (... | Since $(x-12)(x-7)(x-4)(x-1)=x^4-24x^3+183x^2-496x+336$, we have that
$a_n=12^n-7^n-4^n-1$ satisfies
$$
a_n=24a_{n-1}-183a_{n-2}+496a_{n-3}-336a_{n-4}\tag{1}
$$
Since the first $4$ values
$$
a_1=0,a_2=78,a_3=1320,a_4=18078\tag{2}
$$
are all divisible by $6$, induction with $(1)$ and $(2)$ insure that $6\mid a_n$ for al... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1711256",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 4
} |
Perimeter and area of a regular n-gon. A friend of mine asked me how to derive the area and perimeter of a regular $n$-gon with a radius $r$ for a design project he is working on. I came up with this, but I want to make sure I didn't make any errors before giving it to him.
First, I assumed that the $n$-gon was inscrib... | Consider a regular polygon with side length $s$ inscribed in a circle with radius $r$. Let $\theta$ be the measure of a central angle subtended by a side of the regular polygon as shown in the figure below.
As you observed, since a full revolution is $2\pi$ radians, each central angle that subtends a side of an inscr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1712375",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 0
} |
Invert $\overline{x+1}$ in $\mathbb Q[x]/(x^2+x+1)$
Invert $\overline{x+1}$ in $\mathbb Q[x]/(x^2+x+1)$.
So i know that the coset representatives for $\mathbb Q[x]/(x^2+x+1) = \{a+bx : a,b \in \mathbb{Q}\}$. But I am unsure as to how to invert this. Any help?
would $\overline{x+1} = \overline{(x^2+x+1) - x^2} = \ov... | We need to find some polynomial $P(x)$ so that $P(x)\cdot (x+1) = 1$. In $\Bbb Q[x]/\langle x^2+x+1\rangle$ we have $x^2 = -x-1$, so there are no polynomials of degree 2 or higher. The polynomial $P$ that we seek therefore has the form $q_1x+q_0$. So we want $$(q_1x+q_0)(x+1) = 1\qquad \pmod{x^2+x+1}.$$
Expanding th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1712440",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Derive $\tan(3x)$ in terms of $\tan(x)$ using De Moivre's theorem
Derive the following identity: $$\tan(3x)=\frac{3\tan(x)-\tan^3(x)}{1-3\tan^2(x)}$$
The way I approached the questions is that I first derived $\sin(3x)$ and $\cos(3x)$ because $\tan(\theta)=\frac{\sin(\theta)}{\cos(\theta)}$. Then substituting:
$$\tan... | Wthout de Moivre:
$$\tan(2x+x)=\frac{\tan2x+\tan x}{1-\tan2x\tan x}=\frac{\frac{2\tan x}{1-\tan^2x}+\tan x}{1-\frac{2\tan x}{1-\tan^2x}\tan x}=$$
$$=\frac{3\tan x-\tan^3x}{1-3\tan^2x}=\tan x\frac{(\tan x-\sqrt3)(\tan x+\sqrt3)}{(\sqrt3\,\tan x-1)(\sqrt3\,\tan x+1)}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1713544",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Prove: $2^k$ is the sum of two perfect squares If $k$ is a nonnegative integer, prove that $2^k$ can be represented as a sum of two perfect squares in exactly one way. (For example, the unique representation of $10$ is $3^2+1^2$; we do not count $1^2+3^2$ as different.)
I understand that $2^{2n}=0+2^{2n}$ and $2^{2n+1}... | Use induction.
Note that if $$x^2+y^2=2^k\ (\text{where } k \geq 2)$$
If $x \equiv 1 \pmod 2$, then from $x^2+y^2 \equiv 0 \pmod 2$ we get $y \equiv 1 \pmod 2$.
However, then $$x^2+y^2 \equiv 1+1 \equiv 2 \pmod 4$$ A contradiction.
Thus, $x=2x_{1}, y=2y_{2}$. Thus $x_1^2+y_1^2 =2^{k-2}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1713733",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
Why $xy+yz+xz=1$ is two-sheeted hyperboloid? I can't see why $xy+yz+xz=1$ is two-sheeted hyperboloid.
I know that the equation for two-sheeted hyperboloid is: $\frac{x^2}{a^2}+\frac{y^2}{b^2}-\frac{z^2}{c^2}=-1$.
| You need to perform Gauß' reduction to see it:
\begin{align*}
xy+yz+zx&=(x+z)(y+z)-z^2=\frac14\Bigl((x+z+y+z)^2-(x+z-y-z)^2\Bigr)-z^2\\
&=\Bigl(\frac{x+y+2z}2\Bigr)^2-\Bigl(\frac{x-y}2\Bigr)^2-z^2.
\end{align*}
Hence the equation of the quadric can be written as
$$\Bigl(\frac{x-y}2\Bigr)^2+z^2-\Bigl(\frac{x+y+2z}2\Big... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1714267",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Plane $3x + y - z= 4$ touches the ellipsoid $2z^2 = \sqrt7(1 - 2x^2 -y^2)$ Show that the Plane $3x + y - z= 4$ touches the ellipsoid $2z^2 = \sqrt7(1 - 2x^2 -y^2)$
My attempt: First I tried to convert the equation of ellipsoid in general form and then further applying the condition of plane as tangent which is $$ \fra... | The equation of the plane in general form is
$$3 x + y - z - 4 = 0$$
and therefore its normal is $(3,1,-1)$. Its minimum distance $D$ to origin is
$$D = \frac{\lvert -4 \rvert}{\sqrt{3^2 + 1^2 + (-1)^2 }} = \frac{4}{\sqrt{11}} \approx 1.206$$
and the unit normal vector is $\hat n = (3/\sqrt{11},1/\sqrt{11},-1/\sqrt{11}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1714469",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Trigonometric substitution $\tan{\frac{x}{2}}=t$. What is $\cos{x}$ then? For example, the integral is:
$$\int \frac{\sin{x}}{3\sin{x}+4\cos{x}}dx$$
And we use the substitution: $\tan{\frac{x}{2}}=t$
Now, to get $\cos{x}$ in terms of $\tan\frac{x}{2}$, I first expressed $\cos^2\frac{x}{2}$ and $\sin^2\frac{x}{2}$ in te... | I'd rather have a more elegant way to do this, but this way at least works:
You've shown that $\cos x = \dfrac{1-t^2}{1+t^2}$, and then that $\sin^2 x = \dfrac{4t^2}{(1+t^2)^2}$.
Notice that if you start with $a=b=3$ and then square both sides of $a=b$ to get $a^2=b^2$, you can deduce that $a = \pm b$. That just means... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1719033",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
How to prove $1+\cos \left(\frac{2\pi}{7}\right)-4\cos^2 \left(\frac{2\pi}{7}\right)-8\cos^3 \left(\frac{2\pi}{7}\right) \neq 0$ The task is to prove the following non-equality by hand:
$$1+\cos \left(\frac{2\pi}{7}\right)-4\cos^2 \left(\frac{2\pi}{7}\right)-8\cos^3 \left(\frac{2\pi}{7}\right) \neq 0$$
Wolframalpha s... | We prove a closely related result, which in particular shows there is a typo in the given equation.
The number $e^{2\pi i/7}$ is a root of $x^7=1$, and therefore of $x^6+x^5+\cdots+x+1=0$, or equivalently of
$$(x^3+x^{-3})+(x^2+x^{-2})+(x+x^{-1})+1=0.\tag{1}$$
(We divided through by $x^3$.) Let $w=\frac{1}{2}(x+x^{-1})... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1720595",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 0
} |
Determine the first three non-zero terms in the Taylor polynomial approximation for the initial value problem: $y''+\sin(y)=0$ Having trouble understanding how to solve this problem. Did I at least set it up correctly?
$y''+\sin(y)=0,\;y(0)=1,\;y'(0)=0$
So assuming $y(x)=\sum_{n=0}^{\infty}a_nx^n$ then $y''(x)=\sum_{n... | $y = a_0 + a_1 x + a_2 x^2 + a_3 x^3$
$y(0) = 1, y'(0) = 0$
$y = 1 + a_2 x^2 + a_3 x^3$
$y'' = 2 a_2 + 6 a_3 x + 12 a_4 x^2$
$\sin y = y - y^3 / 6 + y^5/5!...$
$y^3 = 1 + 3 a_2 x^2 + 3 a_3 x^3 + 3 a_2x^4 + 3a_4 x^4...$
$y^5 = 1 + 5 a_2 x^2 + 5 a_3 x^3 + 10 a_2x^4 + 5a_4 x^4...$
$\sin y = (1-1/6+1/120...) + (1-1/2+1/24.... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1721013",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Solving the following trigonometric equation: $\sin x + \cos x = \frac{1}{3} $ I have to solve the following equation:
$$\sin x + \cos x = \dfrac{1}{3} $$
I use the following substitution:
$$\sin^2 x + \cos^2 x = 1 \longrightarrow \sin x = \sqrt{1-\cos^2 x}$$
And by operating, I obtain:
$$ \sqrt{(1-\cos^2 x)} = \dfrac... | Hint: I think the approach of @JanEerland is instructive. Here some thoughts how we could find this kind of substitution.
When looking at
\begin{align*}
\sin x+\cos x=\frac{1}{3}
\end{align*}
and we think of the trigonometric addition formulas we know that
\begin{align*}
\sin(x+a)=\sin x \cos a+\cos x\sin a
\en... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1722068",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 8,
"answer_id": 0
} |
Finding equation of sphere, $2|PB|=|PA|, A(-2,5,2)$ and $B(5,2,-1)$ I have question asking to find an equation for a sphere with points $P$ such that the distance from $P$ to $A$ is twice the distance from $P$ to $B$ with $A(-2,5,2)$ and $B(5,2,-1)$.
I have:
$$4[(x-5)^2+(y-2)^2+(z+1)^2] = [(x+2)^2+(y-5)^2+(z-2)^2]$$
$... | The first error is in the coefficient of $y$ :
$$3x^2-44x+3y^2\color{red}{-}6y+3z^2+12z=-87$$
The second error is in the coefficient of $x$ :
$$x^2-\color{red}{\frac{44}{3}}x+y^2-2y+z^2+4z=-29$$
Now, add $(-22/3)^2+(-1)^2+2^2$ to the both sides.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1722248",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
integrate $\int \frac{\tan^4x}{4}\cos^3x$
$$\int \frac{\tan^4x}{4}\cos^3x$$
$$\int \frac{\tan^4x}{4}\cos^3x=\frac{1}{4}\int \frac{\sin^4x}{\cos^4x}\cos^3x=\frac{1}{4}\int\frac{\sin^4x}{\cos x}=\frac{1}{4}\int\frac{\sin^2x\cdot\sin^2x}{\cos x}=\frac{1}{4}\int\frac{\sin^2x\cdot(1-\cos^2x)}{\cos x}=\frac{1}{4}\int \frac... | \begin{align}
\frac{1}{4}\int \tan^4 x \cos^3 x \, dx
&= \frac{1}{4} \int \frac{\sin^4 x}{\cos^4 x}\cos^3 x \, dx\\[0.3cm]
&= \frac{1}{4} \int \frac{\sin^4 x}{\cos^2 x}\cos x \, dx\\[0.3cm]
&= \frac{1}{4} \int \frac{\sin^4 x}{1-\sin^2x} \cos x \, dx
\end{align}
Now do a $u$-sub with $u = \sin x$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1723251",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
How to compute such a limit? Knowing $f(x,y) = 2x^2 +3y^2 -7x +15y$, one simply proves $$|f(x,y)|\leq 5(x^2+y^2)+22 \sqrt{x^2 + y^2}$$ How can I use this info to compute
$$ \lim_{(x,y)\to(0,0)} \frac{f(x,y) - 2(x^2+y^2)^{1/4}}{(x^2+y^2)^{1/4}}\;\;\; ?$$
Thanks!
| Approach limit with $y = mx$
then
$$
\lim_{(x,y)\to(0,0)} \frac{f(x,y) - 2(x^2+y^2)^{1/4}}{(x^2+y^2)^{1/4}} =
$$
$$
\lim_{x\to0} \frac{x(2x +3m^2x -7 +15m)}{x^{1/2}(1+m^2)^{1/4}} - 2 =
$$
$$
\lim_{x\to0} \frac{x^{1/2}( -7 +15m)}{(1+m^2)^{1/4}} - 2 = -2
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1723718",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
In $\triangle ABC$, if $\cos A\cos B\cos C=\frac{1}{3}$, then $\tan A\tan B+\tan B \tan C+\tan C\tan A =\text{???}$
In $\triangle ABC$, if
$$\cos A \cos B \cos C=\frac{1}{3}$$
then can we find value of
$$\tan A\tan B+\tan B \tan C+\tan C\tan A\ ?$$
Please give some hint. I am not sure if $\tan A \tan B+\tan ... | Let $$S=\tan A\tan B+\tan B\tan C+\tan C\tan A$$
Multiplying by $\cos A \cos B \cos C=\frac 13$, we get $$\frac 13S=\sin A\sin B\cos C+\cos A\sin B\sin C+\sin A\cos B \sin C$$
However,$$\cos(A+B+C)=-1=\cos A\cos B\cos C-\sin A\sin B\cos C-\sin A\cos B\sin C-\cos A\sin B\sin C$$
Therefore, $$\frac 13S=\cos A\cos B\cos C... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1724897",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 0
} |
Minimum point of the function $f (x) = 2 ^{- x-\sqrt {x ^ 2 + 1}} + 2 ^ {2x-2\sqrt {1-x ^ 2}}$ I would like to find without the use of derivatives of the minimum point of the function $$f (x) = 2 ^{- x-\sqrt {x ^ 2 + 1}} + 2 ^ {2x-2\sqrt {1-x ^ 2}}$$
(In fact, the point minimum is known: $x = 0$ and the minimum value ... | Let $a,b > 0$. Then for any positive $t$ inequality
$$at^2+\frac{b}t=at^2+\frac{b}{2t}+\frac{b}{2t}\ge3\sqrt[3]{\frac{ab^2}4}.$$
In our case we set $t=2^x$, $a=2^{-2\sqrt{1-x^2}}$, $b=2^{-\sqrt{1+x^2}}$. Then $ab^2=2^{-2(\sqrt{1-x^2}+\sqrt{1+x^2})}\ge\frac1{16}$ due to inequality $\sqrt{1-x^2}+\sqrt{1+x^2}\le2$, which ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1725925",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Using squeeze thorem find $ \lim_{n \to \infty}{\frac{1 \cdot 3 \cdot 5 \cdot ...\cdot (2n-1)}{2\cdot 4 \cdot 6 \cdot ...\cdot 2n}}$ It is already solved here at Math.stackexchange, but we haven't learned Stirling's approximation (at our school), so can it be solved using only squeeze theorem?
$$\lim_{n \to \infty}{\fr... | Hint multiply both numerator,denominator by $2.4.6....2n$ so you get $$\frac{(2n)!}{4^n(n!)^2}$$ where i get $4^n$ by taking $2$ common so we can write it as $$\frac{{2n\choose n}}{4^n}$$ thus we know $4^n\geq {2n\choose n}$. So limit is $0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1727158",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 0
} |
proof limit of $\frac{1}{(x^{2}-1)}$ = -1 using $\epsilon - \delta$ as $x_{0} \to 0$ Question: proof limit of $\frac{1}{(x^{2}-1)}$ = -1 using $\epsilon - \delta$ as $x_{0} \to 0$
Hi... I am stumped on an intro analysis problem.
so it is stated Show $lim \frac{1}{x^{2}-1}$ = -1 as $x_{0} \to 0$.
here is my work, and I... | Note that in our case,
$$|f(x)-L|=\frac{x^2}{|x^2-1|}.$$
If $|x|\lt 1/2$, then $|x^2-1|\gt \frac{3}{4}$ and therefore
$$|f(x)-L|\le \frac{4x^2}{3}.$$
Now it should not be hard to find an appropriate $\delta$. To make things even smoother, note that if $|x|\lt \frac{1}{2}$ then $x^2\le \frac{|x|}{2}$.
Added: We wanted ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1728376",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Prove that $\bigg\lfloor\frac{m}{p^k}\bigg\rfloor-\bigg\lfloor\frac{n}{p^k}\bigg\rfloor-\bigg\lfloor\frac{m-n}{p^k}\bigg\rfloor$ equal to zero or one
Prove that $\bigg\lfloor\frac{m}{p^k}\bigg\rfloor-\bigg\lfloor\frac{n}{p^k}\bigg\rfloor-\bigg\lfloor\frac{m-n}{p^k}\bigg\rfloor$ equal to zero or one for all $k,m,n\in \... | Since
$x = \left\lfloor x \right\rfloor + \left\{ x \right\}$
then we have
$$
\eqalign{
& \left\lfloor {x - y} \right\rfloor = - \left\lceil {y - x} \right\rceil = \cr
& = \left\lfloor x \right\rfloor - \left\lfloor y \right\rfloor + \left\lfloor {\left\{ x \right\} - \left\{ y \right\}} \right\rfloor =... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1728994",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Simpler Way to Write a sum of factorial I have a fairly simple question. I was wondering if there is a simpler way to write the following: $$\sum_{k=0}^{x-1}\frac{(n+k+1)!}{k!}$$
So for example, given $n=4$ and $x=3$, we would have $$1\cdot 2\cdot 3\cdot 4\cdot 5+2\cdot 3\cdot 4\cdot 5\cdot 6+3\cdot 4\cdot 5\cdot 6\cdo... | Note that we can write
$$\frac{(n+k+1)!}{k!}=\frac{1}{n+2}\left(\frac{(k+n+2)!}{k!}-\frac{(k+n+1)!}{(k-1)!}\right)$$
Therefore, evaluating the telescoping sum is trivial and yields
$$\begin{align}
\sum_{k=0}^{x-1}\frac{(n+k+1)!}{k!}&=(n+1)!+\sum_{k=1}^{x-1}\frac{(n+k+1)!}{k!}\\\\
&=(n+1)!+\frac{1}{n+2}\left(\frac{(x+n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1729598",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 2
} |
Prove the convergence and find the sum of series $\sum\limits_{n=1}^{\infty}\left(n^3\sin\frac{\pi}{3^n}\right)$. We know that $0<\sin\frac{\pi}{3^n}\le\frac{\sqrt 3}{2},\forall n\ge 1$. How to find the boundary for $n^3\sin\frac{\pi}{3^n}$ (how to use comparison test here)?
I tried using the ratio test, but the limit ... | You can evaluate the limit in this way
$$\begin{align}
\lim_{n\to\infty} \left( \frac{(n+1)^3}{n^3} \cdot \frac{\sin\frac{\pi}{3^{n+1}}}{\sin\frac{\pi}{3^{n}}}\right) &= \lim_{n \to \infty} \frac{(n+1)^3}{n^3} \lim_{n \to \infty} \frac{\sin\frac{\pi}{3^{n+1}}}{\sin\frac{\pi}{3^{n}}} \\
&= 1 \cdot \lim_{n \to \infty} \f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1730174",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 1
} |
Similarity of Linear Transformations in $\mathbb{R}^4$ Define $\mathbb{R}^4 \rightarrow \mathbb{R}^4$ by $T(x,y,z,w) = (x+z, y-z,z,w-z)$.
Let $$B = \{(1,2,3,4), (1,2,3,0), (1,2,0,0), (1,0,0,0)\}$$
$$B' = \{(1,0,-1,0), (0,2,3,0), (1,0,0,0), (1,1,1,1)\}$$
Find the matrix $[T]_B$ of the linear transformation $T$ with re... | You have calculated is the matrix $[T]_{B,E}$, where $E$ is the standard basis.
In general if $\mathscr{A}, \mathscr{B}$ are two bases then, the n th column of $[T]_{\mathscr{A}, \mathscr{B}}$ is $ \begin{bmatrix} \lambda_1\\ \vdots\\ \lambda_n \end{bmatrix}$,
where $T(a_n) = \Sigma_ i \lambda_i b_i $ and $a_i, \in \m... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1733600",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Why can't I solve for this second derivative? Here's the equation I have to find the second derivative point for.
$$f(x)=\frac{x+2}{x^{\frac{1}{2}}}$$
$$f'(x) = \frac{x-2}{2x^{\frac{3}{2}}}$$
From here I then calculate the second derivative and set it equal to 0.
But it doesn't work.. Take a look:
$$f''(x)=\frac{-x^\fr... | We proceed via the quotient rule:
$$f''(x)=\frac{2x^{3/2}-(x-2)3x^{1/2}}{4x^3}=\frac{2x^{3/2}-3x^{3/2}+6x^{1/2}}{4x^3}=x^{1/2}\left(\frac{6-x}{4x^3}\right)=\frac{6-x}{4x^{5/2}}$$
We can set $f''(x)=0$ and find that $f''(6)=0$. To find the points of inflection, we simply need to test a point less than and greater than $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1736589",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Finding series solution about zero $y''+x^2y'+4y=1-x^2$
To find a power series, one substitutes in $y= \sum_0^\infty a_nx^n$. So after substitution, I've gotten
$\sum_0^\infty (n+1)(n+2)a_{n+2}x^n + \sum_1^\infty (n-1)a_{n-1}x^n + 4\sum_0^\infty a_nx^n$
The recurrence relation would then be in terms of $a_{n-2}$ since... | For the equation
$$ y'' + x^2 \, y' + 4 \, y = 1 - x^2$$
use the solution
$$y = \sum_{n=0}^{\infty} a_{n} \, x^n$$
to obtain:
\begin{align}
\sum_{n} (n)(n-1) \, a_{n} \, x^{n-2} + \sum_{n} n \, a_{n} \, x^{n+1} + 4 \, \sum_{n} a_{n} &= 1 - x^2 \\
\sum_{n=0} [(n+1)(n+2) \, a_{n+2} + 4 \, a_{n}] \, x^{n} + \sum_{n=1} n ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1738404",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
$\forall x\in R,$find the range of the function $f(x)=\cos x(\sin x+\sqrt{\sin^2x+\sin^2\alpha});\alpha\in[0,\pi]$ $\forall x\in R,$find the range of the function $f(x)=\cos x(\sin x+\sqrt{\sin^2x+\sin^2\alpha});\alpha\in[0,\pi]$
$f(x)=\cos x(\sin x+\sqrt{\sin^2x+\sin^2\alpha});\alpha\in[0,\pi]$
$f'(x)=\cos x(\cos x+\... | You're derivative need work. $$f'(x)=-\sin(\sin(x)+\sqrt{\sin^2 x+\sin^2\alpha})\left(\cos x+\frac{2\sin x\cos x}{\sqrt{\sin^2 x+\sin^2\alpha}}\right).$$
Now determine when this is zero, which is precisely when $$\sin(\sin(x)+\sqrt{\sin^2 x+\sin^2\alpha})=0,$$ or $$\cos x+\frac{2\sin x\cos x}{\sqrt{\sin^2 x+\sin^2\alph... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1740147",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
curve of $(x^2+y^2)^2=2(x^2-y^2)$ The diagram shows the curve $(x^2+y^2)^2=2(x^2-y^2)$ and one of its maximum points $M$. Find the coordinates of $M$.
My attempt.
Differentiate the equation and I got
$\frac{dy}{dx}=-\frac{x(x^2+y^2-1)}{y(x^2+y^2+1)}$
How should I proceed?
| HINT.-You have a closed curve, symmetrical with respect to the two axes (a lemniscate indeed) and you have the alternative of take the explicit form for $y$ solving a quadratic equation; you have
$$y^4+2(x^2+1)y^2+(x^4-2x^2)=0$$ By symmetry you can take only the positive signs in solving so you have
$$y=\left(-x^2-1+\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1740632",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Find the first two terms in the perturbation expansion of the solution I want to find the $\mathcal{O}(1)$ and $\mathcal{O}(\epsilon)$ terms in the pedestrian expansion $y = y_0 + \epsilon y_1 + \epsilon ^2 y_2 + \dots$, where $y$ satisfies the following second order ODE:
$\frac{d^2y}{dx^2} + (1-\epsilon x)y = 0$,
wit... | You're right that $y_0 = \cos x$. To find $y_1$, we need to plug our expansion for $y$ into the ODE and collect the $\mathcal{O} (\epsilon)$ terms:
$\frac{d^2}{dx^2} (y_0 + \epsilon y_1 + \dots) + (1-\epsilon x)(y_0 + \epsilon y_1 + \dots) = 0$.
Taking order $\epsilon$ terms:
$\frac{d^2y_1}{dx^2} + y_1 - xy_0 = 0$.
Sub... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1741382",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Calculate the sum of the infinite series $ 1 + \frac{1+2}{2!} + \frac{1+2+3}{3!} .... $ Calculate the sum of the infinite series $ 1 + \frac{1+2}{2!} + \frac{1+2+3}{3!} .... $
My attempt : I recognised that this series can be decomposed into the taylor expansion of $ e $ around 0.
So I thought of writing the series as... | The general term is $$a_n=\frac{1}{n!}\sum_{i=1}^n i=\frac 12\frac{n (n+1)}{ n!}$$ So, now consider $$S=\frac 12\sum_{n=1}^\infty \frac{n (n+1)}{ n!}x^n=\frac 12\sum_{n=1}^\infty \frac{n (n-1)+2n}{ n!}x^n=\frac 12\sum_{n=1}^\infty \frac{n (n-1)}{ n!}x^n+\sum_{n=1}^\infty \frac{n}{ n!}x^n$$ $$S=\frac {x^2}2\sum_{n=1}^\i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1741937",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Show that there are no solutions to $x^2 + y^2 = 3z^2$ in $\Bbb{Z}$ I'm attempting to work through some of the questions in Whitelaw's "Introduction to Abstract Algebra" but am having some difficulty.
The question is as follows
Show that $\forall n \in \Bbb{Z},$ $n^2 \equiv 0\ \text{or}\ 1 \mod{3}.$ Show further tha... | Let's suppose that we have a solution $(a,b,c)$ where $c\neq 0$ is the smallest positive integer such that the equality holds.
$$a^2 + b^2= 3c^2 \Rightarrow 3|(a^2 + b^2)$$
But we know that this means $3|a$ and $3|b$. However, on squaring we get that $9$ divides the LHS, and so it must also divide the RHS:
$$a'^2 + b'^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1744456",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Find the sum to $n$ terms of the series $10+84+734+....$ Find the sum to n terms of the series $10+84+734+....$
*
*$\frac{9(9^n+1)}{10} + 1$
*$\frac{9(9^n-1)}{8} + 1 $
*$\frac{9(9^n-1)}{8} + n $
*$ \frac{9(9^n-1)}{8} + n^2$
My attempt:
I'm getting option $(4)$.
For $n=1$, all options are right,
For $n=2$, sum... | Note: I think formal issues are not that essential. But you have to reasonably show that you have checked all variants and derive the appropriate conclusion. Here is one variation of the theme.
We have four options each providing a sequence
\begin{align*}
(a_n)_{n\geq 1}=(a_1,a_2,a_3,\ldots)
\end{align*}
The sequen... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1745075",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 2
} |
Let $x \ge 0$. Determine a condtion on $|x-4|$ that'll assure $|\sqrt{x} - 2| < 10^{-2}$ I'm trying to understand the logic of this proof.
Let $x \ge 0$. Determine a condition on $|x-4|$ that'll assure $|\sqrt{x} - 2| < 10^{-2}$
Proof
$|\sqrt{x} - 2| = \frac{|(\sqrt{x} - 2)(\sqrt{x} + 2)|}{\sqrt{x} + 2} = \frac{x-4}{\... | When you say $$|\sqrt{x} - 2| = \frac{|(\sqrt{x} - 2)(\sqrt{x} + 2)|}{\sqrt{x} + 2} = \frac{|x-4|}{\sqrt{x} + 2} \le \frac{|x-4|}{2}$$ The final $\le$ comes from replacing $\sqrt x +2$ with $2$. As $\sqrt x \ge 0$, this is decreasing the denominator, which increases the fraction. The overall inequality
$$|\sqrt{x} - 2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1745557",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
How do I show that $\sum_{cyc} \frac {a^6}{b^2 + c^2} \ge \frac {abc(a + b + c)}2?$ Let $a, b, c$ be positive real numbers, show that
$$\frac {a^6}{b^2 + c^2} + \frac {b^6}{c^2 + a^2} + \frac {c^6}{a^2 + b^2} \ge \frac {abc(a + b + c)}2.$$
I think this is likely to turn out to be proved by Hölder, but I can't see how. ... | W.L.O.G., let $a\ge b\ge c$, then by the rearrangement inequality, we have
\begin{align*}
\frac{a^6}{b^2+c^2}+\frac{b^6}{a^2+c^2}+\frac{c^6}{a^2+b^2}
&\ge \frac{b^6}{b^2+c^2}+\frac{c^6}{a^2+c^2}+\frac{a^6}{a^2+b^2},\tag{1}\\
\frac{a^6}{b^2+c^2}+\frac{b^6}{a^2+c^2}+\frac{c^6}{a^2+b^2}
&\ge \frac{c^6}{b^2+c^2}+\frac{a^6}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1746046",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
In a $\triangle ABC,a^2+b^2+c^2=ac+ab\sqrt3$,then the triangle is In a $\triangle ABC,a^2+b^2+c^2=ac+ab\sqrt3$,then the triangle is
$(A)$equilateral
$(B)$isosceles
$(C)$right angled
$(D)$none of these
The given condition is $a^2+b^2+c^2=ac+ab\sqrt3$.
Using sine rule,
$a=2R\sin A,b=2R\sin B,c=2R\sin C$,we get
$\sin^2A+... | Since $0=a^2+b^2+c^2-ac-ab\sqrt{3}=\left(b-\frac{\sqrt{3}}2a\right)^2+\left(c-\frac12a\right)^2\,,$ we have $b=\frac{\sqrt{3}}2a$ and $c=\frac{1}{2}a$. Thus, $\angle A=\frac{\pi}{2}$, $\angle B=\frac{\pi}{3}$, and $\angle C=\frac{\pi}{6}$.
It can be shown that, if $\alpha,\beta,\gamma\in(0,\pi)$ and $\alpha+\beta+\gam... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1746491",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 0
} |
Obtaining the Rodrigues formula On $So(3)$ the algebra of a $3 \times 3$ skew symmetric matrices define Lie bracket $[A,B]=AB-BA$
Consider the exponential map $$EXP: So(3) \to So(3)$$.
We have the $So(3)$ matrix $$A=\begin{bmatrix} 0 & -c & b \\c & 0 & -a\\
-b & a & 0\end{bmatrix}$$
Upon letting $\theta=\sqrt{a^2 + b^... | "Rodrigues formula" will be easier to prove with $A/\theta=B$, thus the result we have to prove is under the form:
$$Exp(\theta B)=I_3 + (\sin \theta) B+ (1-\cos \theta) B^2 \ \ \ (0)$$
by taking this definition:
$$B:=\begin{bmatrix} 0 & -c & b \\c & 0 & -a\\
-b & a & 0\end{bmatrix} \ \ (1a) \ \ \text{with} \ \ \sqr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1747065",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
Limit of tan function This is a question from an old tutorial for a basic mathematical analysis module.
Show that $$\lim\limits_{n\to \infty}\tan^n(\frac{\pi}{4}+\frac{1}{n}) = e^2$$
My tutor has already gone through this in class but I am still confused. Is there anything wrong with the following reasoning?
Since $\... | Using the addition angle formula for the tangent function we can write
$$\tan^n\left(\frac{\pi}{4}+\frac1n\right)=\left(\frac{1+\tan\left(\frac1n\right)}{1-\tan\left(\frac1n\right)}\right)^n$$
Note that $\tan(x)=x+O(x^3)$. Then, we have
$$\begin{align}
\left(\frac{1+\tan\left(\frac1n\right)}{1-\tan\left(\frac1n\right... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1750591",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Pade approximant for the function $\sqrt{1+x}$ I'm doing the followiwng exercise:
The objective is to obtain an approximation for the square root of any given number using the expression
$$\sqrt{1+x}=f(x)\cdot\sqrt{1+g(x)}$$
where g(x) is an infinitesimal. If we choose $f(x)$ as an approximation of $\sqrt{1+x}$, then ... | Using the idea of continued fractions, observe that
$$
\sqrt{1+x}=1+(\sqrt{1+x}-1)=1+\frac{x}{2+(\sqrt{1+x}-1)}\\=1+\cfrac{x}{2+\cfrac{x}{2+(\sqrt{1+x}-1)}}=…
$$
continuing in an obvious repetitive pattern.
The computation of the (degree-balanced) partial fractions of this can be implemented (in Magma CAS) as
F<x>:=Rat... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1750968",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 1
} |
How to evaluate this limit using Taylor expansions? I am trying to evaluate this limit:
$\lim_{x \to 0} \dfrac{(\sin x -x)(\cos x -1)}{x(e^x-1)^4}$
I know that I need to use Taylor expansions for $\sin x -x$, $\cos x -1$ and $e^x-1$. I also realise that all of these are just their regular Taylor expansions with their f... | Using basic limits
\begin{eqnarray*}
\lim_{x\rightarrow 0}\left( \frac{\sin x-x}{x^{3}}\right) &=&-\frac{1}{6} \\
\lim_{x\rightarrow 0}\left( \frac{\cos x-1}{x^{2}}\right) &=&-\frac{1}{2} \\
\lim_{x\rightarrow 0}\left( \frac{x}{e^{x}-1}\right) &=&1
\end{eqnarray*}
which can be evaluated using L'HR, it follows that
\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1752176",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
Is the following solution correct? Question: $ \sqrt{x^2 + 1} + \frac{8}{\sqrt{x^2 + 1}} = \sqrt{x^2 + 9}$
My solution: $(x^2 + 1) + 8 = \sqrt{x^2 + 9} \sqrt{x^2 + 1}$
$=> (x^2 + 9) = \sqrt{x^2 + 9} \sqrt{x^2 + 1}$
$=> (x^2 + 9) - \sqrt{x^2 + 9} \sqrt{x^2 + 1} = 0$
$=> \sqrt{x^2 + 9} (\sqrt{x^2 + 9} - \sqrt{x^2 + 1}) =... | The equation says$$ \sqrt{x^2 + 1} + \frac{8}{\sqrt{x^2 + 1}}=\frac{x^2 + 9}{\sqrt{x^2 + 1}} = \sqrt{x^2 + 9}.$$
So either $x^2+9=0$ and $x=\pm3i$, or $\sqrt{x^2+9}=\sqrt{x^2+1}$, which is impossible (by squaring).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1752506",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 0
} |
Find the limit $\lim_{x\rightarrow 0} \frac{\sin x - x}{\tan x - x}$ Find the limit if it exists. (this exercise is taken from Calculus - The Classic Edition by Swokowski Chapter 10, section 1, no. 9, p.498)
$$\lim_{x\rightarrow 0} \frac{\sin x - x}{\tan x - x}$$
since the limit is $0/0$ therefore, we use L'Hopital's ... | $$
\lim_{x\to0}\frac{\sin x-x}{\tan x-x}\overset{L'H}{=}\lim_{x\to0}\frac{\cos x-1}{\frac{1}{\cos^2x}-1}=\lim_{x\to0}\frac{\frac{\cos x-1}{x^2}}{\frac{1-\cos^2 x}{x^2\cos^2x}}
$$
now,
$$
\lim_{x\to0}\frac{\cos x-1}{x^2}=-\frac{1}{2}
$$
and
$$
\lim_{x\to0}\frac{1-\cos^2 x}{x^2\cos^2x}=\lim_{x\to0}\left(\frac{\sin x}{x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1752715",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 7,
"answer_id": 5
} |
Convergence of a series implies convergence of another series Let $a_1,a_2,\cdots$ be a sequence of real numbers with $a_i\geq 0$. If $\sum_{n=1}^{\infty}\frac{1}{1+a_n}<\infty$ then show that $\sum_{n=1}^{\infty}\frac{1}{1+x_na_n}<\infty$ for each real sequence $x_i$ with $x_i\geq 0$ and $\lim \inf x_n >0$.
Suppose $x... | Convergence only depends on the tail of the series.
If $\liminf x_n \ge b > 0$, then there is some $N$ such that for $n > N$, $x_n > b/2$. Let $B = \max(1, 2/b)$.
Then for $n > N$,
$$\dfrac{1}{1+x_n a_n} \le \dfrac{1}{1 + (b/2) a_n} \le \dfrac{B}{1+a_n}$$
By a Comparison test, $\sum_n 1/(1+x_n a_n)$ converges.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1754551",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How do I solve for $m$ and $n$ While reading about nested radicals, I came across a theorem that said $\sqrt{m\sqrt[3]{4m-8n}+n\sqrt[3]{4m+n}}=\pm\frac {1}{3}\left(\sqrt[3]{(4m+n)^{2}}+\sqrt[3]{4(4m+n)(n-2n)}+\sqrt[3]{(n-2n)^{2}}\right)$
So I tried an easy example ($\sqrt{\sqrt[3]{5}-\sqrt[3]{4}}$) and got this System ... | Eliminate $m$ from the first system with
$$m=\frac{5-n^4}{4n^3},$$
giving
$$(5-n^4)^3(20-36n^4)=1024n^{12},$$
a quartic equation in $n^4$
$$9(n^4)^4-396(n^4)^3+750(n^4)^2-1500(n^4)+625=0.$$
Even Wolfram cannot find a closed-form solution !
With $-4$ in the RHS of the first equation, the equation is
$$9(n^4)^4+116(n^4... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1759721",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Multiplying binomials to come up with $ y^8 - 256 $ $$\ { (y^4 + 16) }{ (y^2 + 4) }{ (y + 2) }{ (y - 2) }$$
How do I multiply these to come up with $\ {y^8 - 256}$
| $$(a+b)(a-b)=a^2-b^2$$
$$\ { (y^4 + 16) }{ (y^2 + 4) }\color{red}{{ (y + 2) }{ (y - 2)} }=(y^4+16)(y^2+4)\color{red}{(y^2-4)}$$
$$(y^4+16)\color{red}{(y^2+4){(y^2-4)}}=(y^4+16)\color{red}{(y^4-16)}$$
$$\color{red}{(y^4+16)(y^4-16)}={(y^8-256)}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1761540",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
$\sqrt{x+938^2} - 938 + \sqrt{x + 140^2} - 140 = 38$ - I keep getting imaginary numbers $$\sqrt{x+938^2} - 938 + \sqrt{x + 140^2} - 140 = 38$$
My attempt
$\sqrt{x+938^2} + \sqrt{x + 140^2} = 1116$
$(\sqrt{x+938^2} + \sqrt{x + 140^2})^2 = (1116)^2$
$x+938^2 + 2*\sqrt{x+938^2}*\sqrt{x + 140^2} + x + 140^2 = 1116^2$
$2x... | For this sort of problem, sometimes it will be useful to replace
explicit numbers by variables to avoid distractions.
Let $a = 938$, $b = 140$, $c = 938 + 140 + 38 = 1116$.
Let $u = \sqrt{x+a^2}$ and $v = \sqrt{x+b^2}$, the equation at hand becomes
$$u + v = c\tag{*1}$$
Notice $u^2 - v^2 = (x+a^2) - (x+b^2) = a^2-b^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1764925",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
} |
So many logs with different bases
$ \large { 6 }^{ \log _{ 5 }{ x } }\log _{ 3 }( { x }^{ 5 } ) -{ 5 }^{ \log _{ 6 }{ 6x } }\log _{ 3 }{ \frac { x }{ 3 } } ={ 6 }^{ \log _{ 5 }{ 5x } }-{ 5 }^{ \log _{ 6 }{ x } }$
The sum of the solutions to the equation above can be expressed as $a^{b/c} + d$ find $abc+d$.
I ... | $$ \large { 6 }^{ \log _{ 5 }{ x } }\log _{ 3 }( { x }^{ 5 } ) -{ 5 }^{ \log _{ 6 }{ 6x } }\log _{ 3 }{ \frac { x }{ 3 } } ={ 6 }^{ \log _{ 5 }{ 5x } }-{ 5 }^{ \log _{ 6 }{ x } }$$
$$\large {6}^{\log _{5}{x} }\log _{3}({x}^{5})-{5}^{\log _{6}{6}+\log_6x}(\log_{3}x-\log_3 3)={6}^{\log _{ 5 }{5}+\log_5x}-{5}^{\log _... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1765122",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
$(x-1)(y-2)=5$ and $(x-1)^2+(y+2)^2=r^2$ intersect at four points $A,B,C,D$. Centroid of $\Delta ABC$ lies on $y=3x-4$, then the locus of $D$ $(x-1)(y-2)=5$ and $(x-1)^2+(y+2)^2=r^2$ intersect at four points $A,B,C,D$. If centroid of $\Delta ABC$ lies on $y=3x-4$, then what is the locus of $D$?
I did try a couple of th... | Point D is the intersection of line $y=3x$ (parallel to $y=3x-4$ and through origin) and curves: $(x-1)(3x-2)=5$ and $(x-1)^2+(y)^2=r^2$ .
$\begin{cases} y=3x \\(x-1)(y-2)=5 \\ (x-1)^2+(y+2)^2=r^2 \end{cases}$
$(x-1)(3x-2)=5$
$3x^{2}-5x-3=0$
$\begin{array}{} x_{D}=\frac{5+\sqrt{61}}{6} & y_{D}=\frac{5+\sqrt{61}}{2} \e... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1767166",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
solve $a_n=5a_{n-1}-4a_{n-2}+3\cdot2^n$ with initial conditions $a_0=1, a_1=10$ so i am pretty sure that i have solve the homogeneous solution correctly. $a_n^h = B\cdot 4^n+C\cdot1^n$ however I am not so confident on the particular solution. Here was my attempt.
Since $f(n)=3\cdot2^n$ I knew that the particular soluti... | As a way of checking, you could always fall back on the generating function route:
$$\begin{align*} f(z) &= \sum_{n=0}^\infty a_n z^n \\
&= a_0 + a_1 z + \sum_{n=2}^\infty (5a_{n-1} - 4a_{n-2} + 3(2^n))z^n \\
&= 1 + 10z + 5z(-1 + f(z)) - 4z^2 f(z) + 3\sum_{n=2}^\infty (2z)^n \\
&= 1 + 5z + (5z - 4z^2) f(z) + \frac{12z... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1768314",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
For positive $a$, $b$, $c$ with $a^4+b^4+c^4=3$, then $\sum_{cyc}\frac{a^2}{b^3+1}\geq \frac32$
If $a$, $b$, $c$ $\in (0, \infty)$, $a^4+b^4+c^4=3$, then: $$\sum_{cyc}\frac{a^2}{b^3+1}\geq \frac32$$
original problem image
I have been into inequalities lately and I am stuck with this. I used a famous inequality at fi... | Note that
$$\frac{1}{b^3+1} - \Big(1 - \frac{b^2+b}{4}\Big) = \frac{b(b^2+3b+1)(b-1)^2}{4(b+1)(b^2-b+1)}\ge 0.$$
It suffices to prove that
$$\sum_{\mathrm{cyc}} a^2\Big(1 - \frac{b^2+b}{4}\Big) \ge \frac{3}{2}$$
or
$$a^2+b^2+c^2 - \frac{1}{4}(a^2b^2+b^2c^2+c^2a^2) - \frac{1}{4}(a^2b+b^2c+c^2a)\ge \frac{3}{2}.$$
Note th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1770039",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Solving a radical equation for real $x$
Solve for $x \in \mathbb{R}$
$$\dfrac{\sqrt{x^2-x+2}}{1+\sqrt{-x^2+x+2}} - \dfrac{\sqrt{x^2+x}}{1+\sqrt{-x^2-x+4}} = x^2-1$$
I tried squaring the equation but it became a sixteen degree equation. I also tried substitutions, but that didn't help. There must be some elegant solut... | Firstly, we need to find what values of $x$ are acceptable. Solving the inequalities:
$$\begin{array}[t]{rl}
x^2-x+2 &\ge 0\\
x^2+x &\ge 0 \\
-x^2+x+2 &\ge 0\\
-x^2-x+4 &\ge 0
\end{array}
$$
yields
$$\left.\begin{array}{l}
x \in \mathbb R\\
x \in (-\infty,-1] \cup [0,+\infty)\\
x \in [-1,2] \\
x\in \left[ \frac{-1-\s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1770604",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
Solve $ 1 + \frac{\sqrt{x+3}}{1+\sqrt{1-x}} = x + \frac{\sqrt{2x+2}}{1+\sqrt{2-2x}} $
Solve for $x \in \mathbb{R}$
$$ 1 + \dfrac{\sqrt{x+3}}{1+\sqrt{1-x}} = x + \dfrac{\sqrt{2x+2}}{1+\sqrt{2-2x}} $$
I tried some substitutions and squaring but that didn't help. I also tried to use inequalities as done in my previous p... | I hope that my solution is simpler:
\begin{eqnarray}
&&1+\frac{\sqrt{x+3}}{1+\sqrt{1-x}}=x+\frac{\sqrt{2x+2}}{1+\sqrt{2-2x}}\\
&\Longleftrightarrow& x-1+\frac{\sqrt{2x+2}}{1+\sqrt{2-2x}}-\frac{\sqrt{x+3}}
{1+\sqrt{1-x}}=0\\
&\Longleftrightarrow& x-1+\frac{\sqrt{2x+2}+\sqrt{2x+2}\sqrt{1-x}-\sqrt{x+3}-\sqrt{x+3}\sqrt{2-2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1770721",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Use induction to prove that $2^n \gt n^3$ for every integer $n \ge 10$. Use induction to prove that $2^n \gt n^3$ for every integer $n \ge 10$.
My method:
If $n = 10$, $2^n \gt n^3$ where $2^{10} \gt 10^3$ which is equivalent to $1024 \gt 1000$, which holds for $n = 10$. $2^k \gt k^3$.
$2^{k + 1} \gt (k + 1)^3$
$2^{k ... | Hint: if $n\geq 10$, $$
\frac{(n+1)^3}{n^3}=\frac{n^3+3n^2+3n+1}{n^3}=1+\frac 3n+\frac 3{n^2}+\frac 1{n^3}\leq 1+\frac 3{10}+\frac 3{100}+\frac 1{1000}<2.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1772026",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
How many phone number with $8$ digits exist s.t divide $2,3,5$ and there is no repetitive digit in it? Here is my approach:
The last digit should be $0$ and the first digit does not $0$. Hence there are $9$ choices for the first digit, $8$ for second,...,$3$ for seventh.
So there are $9.8.7.6.5.4.3$ phone numbers tha... | Recall that a number is divisible by $3$ if and only if its digit sum is divisible by $3$. We will be choosing $7$ digits from $1,2,\dots,9$ whose digit sum is divisible by $3$.
The sum of the digits $1$ to $9$ is divisible by $3$, so the two digits we do not choose must have sum divisible by $3$. There are $12$ ways ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1773321",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Prove that $a^2b+b^2c+c^2a \leqslant 3$ for $a,b,c >0$ with $a^ab^bc^c=1$
Let $a,b,c >0$ and $a^ab^bc^c=1$. Prove that
$$a^2b+b^2c+c^2a \leqslant 3.$$
I don't even know what to do with the condition $a^ab^bc^c=1$. At first I think $x^x>1$, but I was wrong. This inequality is true, following by the verification from M... | The condition can be presented in form
$$a\ln a +b\ln b + c\ln c = 0.$$
Let
$$a\leq b\leq c,$$
then
$$\ln a \leq \ln b \leq \ln c,$$
and we can use Chebyshev sum inequality:
$$3(a\ln a +b\ln b + c\ln c) \geq (a+b+c)\ln abc,$$
$$abc\leq 1.$$
Evidently, the expression $a^2b+b^2c+c^2a$ achieves maximum when $abc=1.$
The s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1774794",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "17",
"answer_count": 5,
"answer_id": 4
} |
How to prove this trigonometric integral? $$ \displaystyle \int_{-\pi/4}^{\pi/4} {{\left(\dfrac{\cos x - \sin x}{\cos x + \sin x}\right)}^{\cos(2t)} \ dx} = \frac{\pi}{2 \sin(\pi \cos^2 t)}$$
I could simplify it to
$\displaystyle \int_0^1 {\left(t^n + \frac{1}{t^n}\right) \ \frac{dt}{1+t^2}}, \ n = \cos 2t $
From here... | \begin{align*}
\frac{\cos x-\sin x}{\cos x+\sin x} &=
\tan \left( \frac{\pi}{4}-x \right) \\
\int_{-\pi/4}^{\pi/4}
\left( \frac{\cos x-\sin x}{\cos x+\sin x} \right)^{n} dx &=
\int_{-\pi/4}^{\pi/4} \tan^{n} \left( \frac{\pi}{4}-x \right) dx \\
&= \int_{0}^{\pi/2} \tan^{n} u \, du \\
&= \frac{\pi}{2} \sec ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1775140",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
$a,b,c >0$, and $ab+bc+ca=3$, prove $(a^ab^bc^c)^{\frac{3}{a+b+c}} \geqslant \sqrt[3]{\frac{a^3+b^3+c^3}{3}}$ $a,b,c >0$, and $ab+bc+ca=3$, prove
$$(a^ab^bc^c)^{\frac{3}{a+b+c}} \geqslant \sqrt[3]{\frac{a^3+b^3+c^3}{3}}$$
I think the equality is only achieve when $a=b=c=1$. The condition $ab+bc+ca=3$ is necessary. I us... | It is not an answer, but rather some results which can be helpful for the further investigations.
*
*$abc \leq 1$
$\rhd$ $$\frac{ab +bc + ac}{3} \geq \sqrt[3]{a^2b^2c^2}\Rightarrow\sqrt[3]{a^2b^2c^2}\leq 1$$
$\lhd$
*One can prove that
$$\forall a_i > 0,\forall p_i>0: \sum\limits_{1}^{n}p_i = 1$$
the following ineq... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1775377",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 1,
"answer_id": 0
} |
Given two lines in Cartesian form, find the vector equation of a line which passes through the intersection of two lines. Given two lines in Cartesian form, find the vector equation of a line which passes through the intersection of two lines (and is perpendicular to both). No points given just the two equations. What ... | $x+1 = 2x+1 \Leftrightarrow x = 0$
$\frac{y}{3} = 2y + 1 \Leftrightarrow y = 6y + 3 \Leftrightarrow y = - \frac{3}{5} $
$ z - \frac{3}{2} = -z \Leftrightarrow 2z = \frac{3}{2} \Leftrightarrow z = \frac{3}{4} $
So the point $P$ of the intersection is : $P=(0,-\frac{3}{5},\frac{3}{4})$
The parallel vectors to the lines ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1776056",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Convert from Nested Square Roots to Sum of Square Roots I am looking for a way to easily discover how to go from a nested root to a sum of roots. For example,
$$\sqrt{10-2\sqrt{21}}=\sqrt{3}-\sqrt{7}$$
I know that if I set $\alpha=\sqrt{10-2\sqrt{21}}$, square both sides, I get
$$\alpha^2=10-2\sqrt{21}$$
Now I recogn... | Let $a$ and $b$ be non-negative rational numbers such that
$$\sqrt{10 - 2\sqrt{21}} = \sqrt{a} - \sqrt{b}$$
Squaring both sides of the equation yields
$$10 - 2\sqrt{21} = a - 2\sqrt{ab} + b$$
Matching rational and irrational parts yields the system of equations
\begin{align*}
a + b & = 10 \tag{1}\\
-2\sqrt{ab} & = -2\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1776159",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
If $x \geq C$, where $C > 0$ is a constant, then what is the least upper bound for $\dfrac{2x}{x + 1}$? The title says it all.
Since
$$f(x) = \dfrac{2x}{x + 1} = 2\left(1 - \dfrac{1}{x + 1}\right),$$
then because $x \geq C$ where $C > 0$, an upper bound is given by
$$\dfrac{2x}{x + 1} < 2.$$
Note that the lower bound i... | rewrite $f(x)$ in the form $$\frac{2}{1+\frac{1}{x}}$$ and consider the cases $$x>0$$ or $$x<0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1776908",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
How to prove that $(x^2+y)(y^2+z)(z^2+x)+2xyz \leqslant 10$, where $x,y,z \in \mathbb{R}^+$ and $x^2+y^2+z^2+xyz=4$ $x,y,z \in \mathbb{R}^+$ and $x^2+y^2+z^2+xyz=4$, prove
$$(x^2+y)(y^2+z)(z^2+x)+2xyz \leqslant 10$$
I try several trig substitutions but feel hopeless with the cyclic term here. The condition $x^2+y^2+z^2... | Let $x=\frac{2a}{\sqrt{(a+b)(a+c)}}$ and $y=\frac{2b}{\sqrt{(a+b)(b+c)}},$ where $a$, $b$ and $c$ are positives.
Thus, the condition gives $z=\frac{2c}{\sqrt{(a+c)(b+c)}}$ and we need to prove that
$$x^2y^2z^2+3xyz+\sum_{cyc}(x^3y^2+x^3z)\leq10$$ or
$$\frac{32a^2b^2c^2}{\prod\limits_{cyc}(a+b)^2}+\frac{12abc}{\prod\lim... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1777259",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
} |
Find $\lim_{n\to\infty}1+\sqrt[2]{2+\sqrt[3]{3+\sqrt[4]{4+\ldots+\sqrt[n]{n}}}}$ $$\lim_{n\to\infty}1+\sqrt[2]{2+\sqrt[3]{3+\sqrt[4]{4+\ldots+\sqrt[n]{n}}}}$$
I have no idea about this.
The equation can be written in its recursive form as:
$$f(n) = g(1,n)$$
Where
$$g(x,n) = [x\impliedby n]\cdot (x+ g(x+1,n))^{\frac 1x}... | Here I show that $1.9<Lim_{n→∞} a_n<2$ if:
$a_n=\sqrt{2+\sqrt [3]{3+\sqrt[4]{4+ . . . +\sqrt[n]{n}}}}$
We use following inequalities:
$\sqrt[k]{k+1}>1$ ; $\sqrt[k]k<\sqrt[k]{k+2}$. . (for $k>2$)
The first one is clear and the second one can be proved by induction. The series of numbers of $a_n $is increasing; using in... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1778930",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
Compute $\sum_{k=0}^{180} \cos^22k^\circ $ I'm trying to compute the following:
$$\sum_{k=0}^{180} \cos^22k^\circ $$
| Remark that:
$cos(x+90°) = sin(x)$
And that:
$cos^2(x) + sin^2(x) = 1$
So you can group your terms:
$$
\begin{align*}
\sum_{x=0}^{180} cos^2(x.2°) &= \sum_{x=0}^{44} cos^2(x.2°) \\
& + \sum_{x=45}^{89} cos^2(x.2°) \\
& + \sum_{x=90}^{134} cos^2(x.2°) \\
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1779371",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 2
} |
How to evaluate $\lim _{n\to \infty }\left(\frac{\sqrt{4n^3+3n}-2n\sqrt{n-2}}{\sqrt{2n+4}}\right)$? I have a problem with this limit, I don't know what method to use. I have no idea how to compute it. Is it possible to compute this limit with the McLaurin expansion? Can you explain the method and the steps used? Thanks... | To provide an alternative (may be a little expeditious but also need a little more background) to rationalization, we may use Taylor's expansion for $(1 + x)^\alpha, \alpha > 0$ at $0$ (which is also known as McLaurin expansion):
$$(1 + x)^\alpha = 1 + \alpha x + o(x). \tag{1}$$
In view of $(1)$, the expression of inte... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1780951",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
How to find the integral of a quotient of rational functions? How do I compute the following integral: $$\int \dfrac{x^4+1}{x^3+x^2}\,dx$$
My attempt:
We can write $$\dfrac{x^4+1}{x^3+x^2} = \dfrac{A}{x^2} + \dfrac{B}{x} + \dfrac{C}{x+1}$$
It is easy to find that
$A=1$,
$B=2$, and
$C=-1$.
Therefore
$$\frac{x^4+1}{x^3+... | Your $A,B,C$ are wrong. You can't write the expression in that form because the numerator won't have the correct degree.
If you combine the terms, you get
$$\frac{A(x+1)+Bx(x+1) +Cx^2}{x^3+x^2}$$
Note that the numerator is of degree at most $2$.
Instead, just perform polynomial division to get a quotient and a remaind... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1784930",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 6,
"answer_id": 0
} |
Closed form of $\int_{\frac{\pi}{4}}^{\frac{3\pi}{4}}\frac{\lfloor|\tan x|\rfloor}{|\tan x|}dx$? I am trying to find a closed form for the integral $$I=\int_{\frac{\pi}{4}}^{\frac{3\pi}{4}}\frac{\lfloor|\tan x|\rfloor}{|\tan x|}dx$$ So far, my reasoning is thus: write, by symmetry through $x=\pi/2$, $$I=2\sum_{n=1}^{\i... | Maybe we are lucky. We may notice that:
$$ 1+\frac{2n+1}{n^2(n+1)^2} = 1+\frac{1}{n^2}-\frac{1}{(n+1)^2} $$
and the roots of the polynomial $x^2(x+1)^2+2x+1$ are given by
$$ \alpha = \frac{1}{2}\left(-1-\sqrt{2}-\sqrt{2\sqrt{2}-1}\right), $$
$$ \beta = \frac{1}{2}\left(-1-\sqrt{2}+\sqrt{2\sqrt{2}-1}\right), $$
$$ \gamm... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1785414",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
How does one solve this recurrence relation? We have the following recursive system:
$$
\begin{cases}
& a_{n+1}=-2a_n -4b_n\\
& b_{n+1}=4a_n +6b_n\\
& a_0=1, b_0=0
\end{cases}
$$
and the 2005 mid-exam wants me to calculate answer of $ \frac{a_{20}}{a_{20}+b_{20}} $.
Do you have any idea how to solve this recursive equ... | Observe that
$$a_{n+1}+b_{n+1}=2a_n+2b_n=2(a_n+b_n)\;,$$
and $a_0+b_0=1$, so in general $a_n+b_n=2^n$.
Quickly calculating a few values, we see that the numbers $b_n$ are a little nicer than the numbers $a_n$:
$$\begin{array}{rcc}
n:&0&1&2&3&4\\
a_n:&1&-2&-12&-40&-112\\
b_n:&0&4&16&48&128\\
\end{array}$$
Concentrating... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1786760",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "17",
"answer_count": 4,
"answer_id": 1
} |
Determine all integral solutions of $a^2+b^2+c^2=a^2b^2$ I started like this :
$a^2+c^2=b^2(a^2-1)\\c^2 +1=(a^2-1)(b^2-1)$
but it's leads to nowhere.
can you help please ?
| This is a Pythagorean quadruple problem.there is a good way to prove that the only solution is (0,0,0)
assuming the equation is
$a^2+b^2+c^2=t^2$ ,∀ t=ab
all integer positive solution given by
$a=(l^2+m^2−n^2)/n , b=2l, c=2m$ ,and $t=(l^2+m^2+n^2)/n$
Then
$2l(l^2+m^2−n^2)=l^2+m^2+n^2$
one can find that $(l^2+m^2−n^2)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1787444",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Determining axis of rotation from the rotation matrix without using eigenvalues and eigenvectors
Consider the following rotation matrix:
$$R=\begin{pmatrix} -\frac{1}{\sqrt{2}}&0&\frac{1}{\sqrt{2}} \\
\sqrt{\frac{3}{8}} &\frac{1}{2}& \sqrt{\frac{3}{8}} \\
-\frac{1}{\sqrt{8}}&\sqrt{\frac{3}{4}}&-\frac{1}{\sqrt{8}}\en... | Consider Rodrigues' formula for a rotation matrix:
$$
R=I+\sin\varphi K+(1-\cos\varphi)K^2
$$
with
$$
K=\pmatrix{
0&-k_3&k_2\\
k_3&0&-k_1\\
-k_2&k_1&0
}\;,
$$
where $k$ is the (unit) rotation axis.
To my mind it seems easier to just take the antisymmetric part of $R$, which is $\sin\varphi K$, and normalise to get rid ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1788717",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Prove : $2(x^3+y^3+z^3)+3xyz \ge 3(x^2y+y^2z+z^2x)$ with $x,y,z \gt 0$ At first I tried to divide both side by $xyz$, the inequality became:
$$2\sum {\frac{x^2}{yz}}+3 \ge 3\sum{\frac xy}$$
Let $$\frac xy = a;\frac yz = b;\frac zx = c;$$
So all we have to prove is
$$2\sum \frac ab +3 \ge 3 \sum a $$ with $a,b,c$ being ... | By Schur inequality (whenever you get $xyz$ with positive coefficient on the higher side, worth trying Schur):
$$x^3+y^3+z^3 + 3xyz \geqslant (x^2y+y^2z+z^2x) + (xy^2+yz^2+zx^2)$$
So it is enough to show
$$x^3+y^3+z^3+(xy^2+yz^2+zx^2) \geqslant 2(x^2y+y^2z+z^2x)$$
which follows from three AM-GMs like $x^3+xy^2 \geqslan... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1791894",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Sketch the heart and indicate its orientation with arrows $ r = 1 - \cos(\theta)$. Find the area enclosed by the heart Hi all I am trying to figure out how to sketch the heart. Here is what I have tried so far:
$$r = 1 - \cos(\theta) \\
r(r = 1 - \cos(\theta)) \\
r^2 = r - r\cos(\theta) \\
$$
Use the fact that
$$r =\sq... | You can easily draw the shape by substituting convenient values for $\theta$. To find the area, use the formula $\displaystyle A=\frac{1}{2}\int_a^{b}r(\theta)^2d\theta$. The integration proceeds as follows:
$$2\cdot \frac{1}{2} \int_0^{\pi}(1-\cos \theta)^2d\theta$$
$$ \int_0^{\pi}(1-2\cos \theta+\cos^2\theta )d\theta... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1792115",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Prove $\sum\limits_{cyc}\left(\frac{a^4}{a^3+b^3}\right)^{\frac34} \geqslant \frac{a^{\frac34}+b^{\frac34}+c^{\frac34}}{2^{\frac34}}$ When $a,b,c > 0$, prove
$$\left(\frac{a^4}{a^3+b^3}\right)^{\frac34}+\left(\frac{b^4}{b^3+c^3}\right)^ {\frac34}+\left(\frac{c^4}{c^3+a^3}\right)^{\frac34} \geqslant \frac{a^{\frac34}+b^... | Version of 29.06.18
HINT
$$\mathbf{\color{brown}{Task\ transformations}}$$
Let
$$
\begin{cases}
&b^3+c^3= 2x^4\\
&c^3+a^3= 2y^4\\
&a^3+b^3= 2z^4
\end{cases}\Rightarrow
\begin{cases}
a^3=-x^4+y^4+z^4\\
b^3= x^4-y^4+z^4\\
c^3= x^4+y^4-z^4\tag1,
\end{cases}$$
then the issue equation transforms to
$$\rlap\bigcirc\!\sum\df... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1793680",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
"answer_count": 5,
"answer_id": 3
} |
The problem of surds and indices If $$ \frac 12 \times \left(\sqrt[3] a + \sqrt[3] b + \sqrt[3] c\right) = \frac 1{\sqrt[3] 3 - 1}\;\;\;(a>b>c)$$
then what is $a - 2b - 3c $ ?
| Use the factorization $x^3 - 1 = (x - 1)(x^2 + x + 1)$ with $x = \sqrt[3]{3}$ to get $2 = (\sqrt[3]{3}-1)(\sqrt[3]{9} + \sqrt[3]{3} + 1)$, or
$$\frac{1}{\sqrt[3]{3}-1} = \frac{1}{2}(\sqrt[3]{9}+\sqrt[3]{3} + \sqrt[3]{1})$$
So $a = 9$, $b = 3$, $c = 1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1794041",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
If $g(x) = xf(x^2)$ and $f(x)=\sum_{n=0}^\infty \sin(\frac{\pi}{n+2})x^n$, what is $f^{(20)}(0)$ and $g^{(35)}(0)$? My task is this:
(i)Let $$f(x)=\sum_{n=0}^\infty \sin\left(\frac{\pi}{n+2}\right)x^n.$$ Find $f^{(20)}(0)$ and $g^{(35)}(0)$ when $g(x) = xf(x^2)$.
(ii)Find $$\lim_{n\to\infty}\frac{f(x)-1-\frac{\sqrt(3)x... | Note that $$f'\left(x\right)=\sum_{n\geq0}n\sin\left(\frac{\pi}{n+2}\right)x^{n-1}=\sum_{n\geq1}n\sin\left(\frac{\pi}{n+2}\right)x^{n-1}
$$ since for $n=0
$ there is no addend. Furthermore $$f''\left(x\right)=\sum_{n\geq1}n\left(n-1\right)\sin\left(\frac{\pi}{n+2}\right)x^{n-2}=\sum_{n\geq2}n\left(n-1\right)\sin\left... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1796341",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"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.