Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Non-Linear System. Find the conditional expectation. I've had my test for this course and I think I failed it again. The hardest part for me is findig the correct distributions. This is a test exercise I couldn't figure out or at least, I probably failed the question. I hope I can get your help, so next time I won't fa... | There are two ways you could tackle the problem:
(a) Use "change of variables" from $X$ and $V$ to $X$ and $Y$ and then find the required conditional distribution and expectation.
(b) Use the "direct" method, which is what you've done.
For this particular problem, I think (a) is a slightly simpler but I will use (b) he... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1343778",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
$(x+y+z)^3-(y+z-x)^3-(z+x-y)^3-(x+y-z)^3=24xyz$? The question given is
Show that $(x+y+z)^3-(y+z-x)^3-(z+x-y)^3-(x+y-z)^3=24xyz$.
What I tried is suppose $a=(y+z-x),\ b=(z+x-y)$ and $c=(x+y-z)$ and then noted that $a+b+c=x+y+z$. So the question statement reduced to $(a+b+c)^3-(a^3+b^3+c^3)$. Then I tried to invoke t... | Pascal's (or Tartaglia's) Tetrahedron: the left outline is a binomial expansion of $(x+y)^3$, while the right outline is a binomial expansion of $(x+z)^3$ and the bottom outline is a binomial expansion of $(y+z)^3$. Each of them is highlighted in yellow for identification purposes. Furthermore, you may notice that the ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1344314",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 3
} |
Proving inequalities using Calculus In general how do you prove inequalities using calculus, I believe it is using maxima or minima right? For example
$$a^2b+b^2c+c^2a \le 3, \qquad a,b,c \ge 0,\quad a+b+c=3.$$
How would you use calculus, just a sketch, I am interested in the method itself.
| $f(a,b) = a^2b + b^2(3-a-b) + (3-a-b)^2a = a^2b + 3b^2-ab^2-b^3+9a+a^3+ab^2-6a^2-6ab+2a^2b=3a^2b+3b^2-b^3+9a+a^3-6a^2-6ab$. Thus: $f_a = 6ab+9+3a^2-12a-6b = 0 = f_b = 3a^2+6b-3b^2-6a\Rightarrow 3(a^2-b^2) = 6(a-b)\Rightarrow (a-b)(a+b-2)=0$. There are $2$ cases:
Case $1$: $a = b \Rightarrow 6a^2+9+3a^2-12a-6a = 0\Right... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1345008",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
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How can I calculate this limit? $\lim _{ m\rightarrow \infty }{ \left( \lim _{ n\rightarrow \infty }{ \cos ^{ 2n }{ \left( \pi m!x \right) } } \right) } $
Attempt : since $\cos ^{ 2 }{ x=\frac { 1+cos2x }{ 2 } } $ so we can write $$\lim _{ m\rightarrow \infty }{ \left( \lim _{ n\rightarrow \infty }{ \cos ^{ 2... | Let
\begin{equation*}
f_{m}(x)=\lim_{n\rightarrow \infty }(\cos m!\pi x)^{2n}.
\end{equation*}
Note that for $m$ fixed, $f_{m}(x)$ is the limit of geometric sequence, $%
r^{n}$ with
\begin{equation*}
r=(\cos m!\pi x)^{2}\in \lbrack 0,1].
\end{equation*}
If $r=1,$ the limit of the geometric sequence $(r^{n})$ is $1,$ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1345083",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Solving the differential equation $x^2y''+xy'-y=x^2$
$$x^2y''+xy'-y=x^2$$
My attempt:
Divided by $x^2$:
$$y''+\frac{y'}{x}-\frac{y}{x^2}=1$$
Now to solve the homogenous equation using Euler's method
$$y''+\frac{y'}{x}-\frac{y}{x^2}=0$$
To look for solution $y=x^r$
so $y'=rx^{r-1}$
$y''=r(r-1)x^{r-2}$
So:
$$r(r-1)x^{... | As per ChickenP's answer,
$y_c = Ax + \dfrac{B}{x}$
Now since the D.E. is linear, also account for a particular solution, i.e.
$y_p = s \cdot x^2$ so $y'' + \dfrac{y'}{x} - \dfrac{y}{x^2} = 2s + 2s -s = 3s = 1$ giving $s = \frac{1}{3}$, so $y_p = \frac{1}{3}x^2$
Then the full solution is
$y = Ax + \dfrac{B}{x} + \frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1346243",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Do more generalizations of Schur's inequality exist?
I meet this following problem
If $$n\ge 3,\sum_{i=1}^{n}\left(\prod_{j\neq i}(a_{i}-a_{j})\right)\ge 0$$
where $a_{i}$ are real numbers.
when $n=3$, it is Schur's inequality
so which $n$ such this inequality?
but for more generalization form of Schur'.s Ineq... | Now I have solve it, this inequality it is only true for $n=3,n=5$ holds.
For $n=3$ then inequality can write as
$$(a_{1}-a_{2})(a_{1}-a_{3})+(a_{2}-a_{1})(a_{2}-a_{3})+(a_{3}-a_{1})(a_{3}-a_{2})=a^2_{1}+a^2_{2}+a^2_{3}-a_{1}a_{2}-a_{1}a_{3}-a_{2}a_{3}\ge 0$$
it is clear true
for $n=5$,WLOG, we assume that
$$a_{1}\ge a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1346798",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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Solving a system of non-linear equations Let
$$(\star)\begin{cases}
\begin{vmatrix}
x&y\\
z&x\\
\end{vmatrix}=1, \\
\begin{vmatrix}
y&z\\
x&y\\
\end{vmatrix}=2, \\
\begin{vmatrix}
z&x\\
y&z\\
\end{vmatrix}=3.
\end{cases}$$
Solving the above system of three non-linear equations with three unknowns.
... | Given $x^2-yz = 1, \quad y^2-xz = 2, \quad z^2-xy = 3$, we can sum all of these to get
$$(x-y)^2+(y-z)^2+(z-x)^2 = 12 \tag{1}$$
OTOH, subtracting gives $(y^2-x^2)+z(y-x)=1 \implies (x+y+z)(y-x) = 1$
and similarly $(x+y+z)(z-y) = 1$, so we must have $y-x = z - y = a$, say. Using this in $(1)$,
$$a^2+a^2+4a^2=12 \impli... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1347813",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 4,
"answer_id": 0
} |
Find the area of triangle, given an angle and the length of the segments cut by the projection of the incenter on the opposite side.
In a triangle $ABC$, one of the angles (say $\widehat{C}$) equals $60^\circ$.
Given that the incircle touches the opposite side ($AB$) in a point that splits it in two segments having ... | An alternative approach: let we just find the inradius $r$. Assuming that $\widehat{C}=60^\circ$, $I_C$ is the projection of the incenter on the $AB$-side, $AI_C=a,BI_C=b$, we have:
$$\widehat{A}=2\arctan\frac{r}{a},\qquad \widehat{B}=2\arctan\frac{r}{b}\tag{1}$$
hence:
$$ \frac{\pi}{3} = \arctan\frac{r}{a}+\arctan\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1349445",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Help with Definite integral question Anyone please help with this question:
(a) Show that:
\begin{align}
\int_{0}^{a} f(x) dx = \int_{0}^{a} f(a-x) dx
\end{align}
(b) Hence show that:
\begin{align}
\int_{0}^{\frac{\pi}{4}} \frac{1-\sin(2x)}{1+\sin(2x)} dx = \int_{0}^{\frac{\pi}{4}} \tan^2{x} dx
\end{align}
And eva... | Using the double angle identities to express $\cos 2x$ in terms of $\cos x$ and $\sin x$, we get:
$$\frac{1-\cos 2x}{1+\cos 2x} = \frac{1- (1 - 2 \sin^2 x)}{1 + (2 \cos^2 x - 1)} = \frac{\sin^2 x}{\cos^2 x} = \tan^2 x$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1350001",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Evaluating a function at a point where $x =$ matrix. Given $A=\left(
\begin{array} {lcr}
1 & -1\\
2 & 3
\end{array}
\right)$
and $f(x)=x^2-3x+3$ calculate $f(A)$.
I tried to consider the constant $3$ as $3$ times the identity matrix ($3I$) but the answer is wrong. Appreciate any ideas.
| $f(A) = \begin{pmatrix}
1 & -1 \\ 2 & 3
\end{pmatrix}^2 - 3\begin{pmatrix} 1 & -1 \\ 2 & 3 \end{pmatrix} + 3\begin{pmatrix}
1 & 0 \\ 0 & 1
\end{pmatrix} $
$f(A) = \begin{pmatrix}
-1 & -4 \\ 8 & 7
\end{pmatrix} + \begin{pmatrix}
-3 & 3 \\ -6 & -9
\end{pmatrix} + \begin{pmatrix}
3 & 0 \\ 0 & 3
\end{pmatrix}$
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1350275",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
According to Stewart Calculus Early Transcendentals 5th Edition on page 140, in example 5, how does he simplify this problem? In Stewart's Calculus: Early Transcendentals 5th Edition on page 140, in example 5, how does
$$\lim\limits_{x \to \infty} \frac{\dfrac{1}{x}}{\dfrac{\sqrt{x^2 + 1} + x}{x}}$$
simplify to
$$\li... | Since you are calculating the limit at $x\to\infty$ you may assume that $x>0$. Then
$$\frac{\sqrt{x^2+1}+x}x=\frac{\sqrt{x^2+1}}{\sqrt{x^2}}+\frac xx=\sqrt{\frac{x^2+1}{x^2}}+1=\sqrt{1+\frac{1}{x^2}}+1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1350812",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
particular solution of a difference equation I am unable to find a particular solution of the following difference equation
$$
y[k-1]-5y[k]+6y[k+1]=-u[k-1]+4u[k]
$$
with $u[k]=\big(\frac{1}{2}\big)^k$.
This is what I tried so far. Because $u[k]=\big(\frac{1}{2}\big)^k$ we try $y_p[k]=\alpha \big(\frac{1}{2}\big)^k$. Su... | We change notation slightly. Note that $-u_{k-1}+4u_k=\frac{4}{2^k}-\frac{2}{2^k}=\frac{2}{2^k}$. We look for a solution of the shape $y_i=\alpha\frac{i}{2^i}$.
Substitute in $6y_{k+1}-5y_k+y_{k-1}$, forgetting about the $\alpha$ for a while. We get
$$\frac{6(k+1)}{2^{k+1}}-\frac{5k}{2^k}+\frac{k-1}{2^{k-1}}.$$
This s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1353075",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Calculus - Solving limits with square roots I am having trouble understanding how to solve this limit by rationalizing. I have the problem correct (I used Wolfram Alpha of course), but I still don't understand how it is completed. I was trying to solve this by multiplying both the numerator and the denominator by $\sqr... | A possible step-by-step solution: write $x=y+5$ (so that you are looking for a limit as $y\to 0$), and the denominator is $x-5=y$
$$\begin{align}
\sqrt{x^2+11} &= \sqrt{(y+5)^2+11} = \sqrt{y^2+10y + 36} = \sqrt{36}\sqrt{1+\frac{10}{36}y+\frac{y^2}{36}} \\
&= 6 \sqrt{1+\frac{5}{18}y+\frac{y^2}{36}}
\end{align}
$$
From t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1353339",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Finding bases such that the matrix representation is a block matrix where one submatrix is the identity matrix Problem: Let $L: \mathbb{R}^2 \rightarrow \mathbb{R}^3$ be a linear map with \begin{align*} [L]_{\alpha}^{\beta} = \begin{pmatrix} 2 & 3 \\ 4 & 6 \\ 6 & 9 \end{pmatrix} \end{align*} as the matrix representatio... | so as I mentioned in the comment, we first choose a proper basis $B$ in $\mathbb{R}^2$, we take
$$
B=\left\{\begin{pmatrix} 1 \\ 0 \end{pmatrix},\begin{pmatrix} -1.5 \\ 1 \end{pmatrix}\right\}
$$
and first keep the standard basis $A$ in $\mathbb{R}^3$, therefore our matrix looks like
$$
L^{B}_A=\begin{pmatrix} 2 & 0 \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1354357",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Question on proving quadratic inequality
Let $ax^2+bx+c$ = 0 be a quadratic equation and $\alpha$,$\beta$ are real roots. Condition for $\alpha < -1$ and $\beta > 1$. Show that $1 +\frac{c}{a}$ + $\left|\frac{b}{a}\right| < 0$.
I have tried but could not prove this inequality. I want to know how to solve this. Show... | Without loss of generality, let $a \gt 0$.
Now firstly, Since $\alpha$ is a root of $f(x)$, hence $a\alpha^2+b\alpha+c=0$.
$\Rightarrow$ $\alpha = \frac {-b - \sqrt{b^2-4ac}}{2a} \lt -1$ $\Rightarrow$ $-b - \sqrt{b^2-4ac} \lt -2a$ $\Rightarrow$ $\sqrt{b^2-4ac} \gt 2a-b$ $\Rightarrow$ $b^2-4ac \gt 4a^2-4ab+b^2$ $\Right... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1354848",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Arrangements of Chairs in a Circle
Ten chairs are arranged in a circle. Find the number of subsets of this set of chairs that contain at least three adjacent chairs.
Hints only please!
This is a confusing worded-problem.
We could break it into, $7$ cases but that would take very long?
Case 1: 3 chairs adjacent.
Wa... | The number of sets of exactly three adjacent chairs is $10$. Either they're chairs $1$, $2$, and $3$, or $2,3,4$, or $3,4,5$, etc. After $8,9,10$ comes $9,10,1$ and then $10,1,2$.
The number of subsets of the remaining set of seven chairs is $2^7=128$.
However, if one just picks $10\times128=1280$, then one is counti... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1356805",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 2
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How prove $\lim x_n = \sqrt{\frac{a}{b}}$ if $b \neq 0$ for $x_1=c$ and $x_{n+1}=\frac{x_n^2 + \frac{a}{b}}{2x_n}$? Let $x_1=c$ and $x_{n+1}=\frac{x_n^2 + \frac{a}{b}}{2x_n}$. How prove $\lim x_n = \sqrt{\frac{a}{b}}$ if $b \neq 0$ ?
| Let $r=\frac{a}{b}$ and assume that $r>0$ and $c>0$. To prove existence of the limit, rewrite the equation as follows:
$$
2x_{n+1}=x_n + \frac{r}{x_n}\quad (\star)
$$
Since $x_1=c>0$ and $r>0$, it follows by induction that all $x_n$ are positive. Applying the AM-GM Inequality, we see that $x_{n+1}\geq \sqrt{r}$ (for $n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1359098",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
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Evaluating a Summation with a binomial Problem:
Evaluate for $n=11$$$\begin{align} \sin^{4n}\left(\frac{\pi}{4n}\right) + \cos^{4n}\left( \frac{\pi}{4n}\right) = \frac{1}{4^{2n-1}} \left[ \sum_{r=0}^{n-1} \binom{4n}{2r} \cos\left(1 - \frac{r}{n} \right) \pi \, + \frac{1}{2} \binom{4n}{2n} \right]. \end{align} $$
Sorr... | With some trivial manipulation it straighforward to check that we just have to compute:
$$ \sum_{r=0}^{2n}\binom{4n}{2r}\cos\frac{4\pi r}{n}=\text{Re}\sum_{r=0}^{2n}\binom{4n}{2r}\exp\left(8r\cdot\frac{2\pi i}{4n}\right).\tag{1} $$
Since:
$$ \sum_{r=0}^{2n}\binom{4n}{2r} z^{2r} = \frac{1}{2}\sum_{k=0}^{4n}\binom{4n}{k}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1362141",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
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is there a general formula for cases like $\sqrt{2}$ = $\frac {2}{\sqrt{2}}$? I just noticed that $\sqrt{2}$ is equal to $\frac {2}{\sqrt{2}}$:
$\sqrt{2} = 1.414213562$
$\frac {2}{\sqrt{2}} = 1.414213562$
It is confirmed by a hand-calculator.
I tried to proof this as follows:
$\sqrt{2}$ = $\frac {2}{\sqrt{2}}$
$2^{\fra... | Yes. $$\frac{x^a}{x^b} = x^{a-b}$$ In this case we have $$\frac{2^1}{\sqrt{2}} = \frac{2^1}{2^{1/2}} = 2^{1- 1/2} = 2^{1/2} = \sqrt{2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1362633",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
"answer_id": 0
} |
Generalize multiples of $999...9$ using digits $(0,1,2)$ The smallest $n$ such that $9n$ uses only the three digits $(0,1,2)$ is $1358$, giving a product $12222$. For $99n$ this is $11335578$, giving $1122222222$. Similarly,
$999(111333555778)=111222222222222,$
$9999(1111333355557778) = 11112222222222222222, ...$
The... | It is readily verified that
$$\tag1(10^k-1)\cdot\frac{10^{4k}+2\cdot 10^{3k}+2\cdot 10^{2k}+2\cdot 10^k+2}{9} =\frac{10^{5k}-1}9+\frac{10^{4k}-1}{9}$$
and that the two fractions are in fact integers and the number on the right, being the sum of two repunits, is written with $1$ and $2$ only.
Remains to show that the n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1362933",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
} |
Euler's proof of sum of natural numbers I've to recheck Euler's proof of the sum of the natural numbers, but I don't know exactly what it is? It has something to do with the $\zeta(s)$?
| Euler was assuming the following things:
$$0+1+x+x^2+x^3+x^4+...=\sum_{k=0}^{\infty} x^k=\frac{1}{1-x},|x|<1$$
If we take the $\frac{d}{dx}$ (derivative) of that sum we get:
$$0+1+2x+3x^2+4x^3+5x^4+...=\sum_{k=0}^{\infty} (k+1)x^k=\frac{1}{(1-x)^2},|x|<1$$
Second thing he assumed was:
$$\sum_{k=0}^{\infty} (k+1)x^k=1-2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1363503",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Find lim$_{x \to 0}\left(\frac{1}{x} - \frac{\cos x}{\sin x}\right).$ $$\lim_{x \to 0} \left(\frac{1}{x} - \frac{\cos x}{\sin x}\right) = \frac{\sin x - x\cos x}{x\sin x}= \frac{0}{0}.$$
L'Hopital's: $$\lim_{x \to 0} \frac{f'(x)}{g'(x)} = -\frac{1}{x^2} + \frac{1}{\sin ^2x} = \frac{0}{0}.$$
Once again, using L'Hopital... | $$\begin{align}\lim_{x\to 0}\frac{\sin x-x\cos x}{x\sin x}&=\lim_{x\to 0}\frac{\cos x-(\cos x-x\sin x)}{\sin x+x\cos x}\\&=\lim_{x\to 0}\frac{x\sin x}{\sin x+x\cos x}\\&=\lim_{x\to 0}\frac{x}{1+\frac{x}{\sin x}\cdot \cos x}\\&=\frac{0}{1+1\cdot 1}\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1363719",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 7,
"answer_id": 0
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Solution verification for $y'$ when $y=\sin^{-1}(\frac{2x}{1+x^2})$ I was required to find $y'$ when $y=\sin^{-1}(\frac{2x}{1+x^2})$
This is my solution.
Above when I put $\sqrt{x^4-2x+1}=\sqrt{(1-x^2)^2}$ then I get the correct answer but when I put $\sqrt{x^4-2x+1}=\sqrt{(x^2-1)^2}$ then I get something else.
Now my... | HINT:
For real $a,$ $$\sqrt{a^2}=|a|$$
$=+a$ if $a\ge0$ and $=-a$ if $a<0$
Things will be clearer with the following :
Using my answer here: showing $\arctan(\frac{2}{3}) = \frac{1}{2} \arctan(\frac{12}{5})$ and $\arctan y=\arcsin\dfrac y{\sqrt{y^2+1}}$
$$2\arctan x=\begin{cases} \arcsin\frac{2x}{1+x^2} &\mbox{if } ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1367747",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Definite Integration with Trigonometric Substitution I'm working on a question that involves using trigonometric substitution on a definite integral that will later use u substitution but I am not sure how to go ahead with this.
$$\int_1^2\frac1{x^2\sqrt{4x^2+9}}dx$$
My first step was to use $\sqrt{a^2+x^2}$ as $x=a\ta... | $$\int_1^2\frac1{x^2\sqrt{4x^2+9}}dx = \int_a^b\frac{\frac{2}{3}\sec^2\theta}{\frac{9}{4}\tan^2\theta\sqrt{9\tan^2\theta+9}}d\theta$$
where $a = \tan^{-1}\frac{2}{3}$ and $b = \tan^{-1}\frac{4}{3}$
$$ \int_a^b\frac{2\sec\theta}{9\tan^2\theta}d\theta = \left(-\frac{2\text{cosec}\theta}{9} \right)_a^b$$
Hint: $\text{cose... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1369251",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Prove the inequality $\sqrt\frac{a}{a+8} + \sqrt\frac{b}{b+8} +\sqrt\frac{c}{c+8} \geq 1$ with the constraint $abc=1$ If $a,b,c$ are positive reals such that $abc=1$, then prove that $$\sqrt\frac{a}{a+8} + \sqrt\frac{b}{b+8} +\sqrt\frac{c}{c+8} \geq 1$$ I tried substituting $x/y,y/z,z/x$, but it didn't help(I got the r... | Let $a=\frac{x^2}{yz}$ and $b=\frac{y^2}{xz}$, where $x$, $y$ and $z$ are positives.
Hence, $c=\frac{z^2}{xy}$ and we need to prove that:
$$\sum_{cyc}\frac{x}{\sqrt{x^2+8yz}}\geq1.$$
Now, by Holder
$$\left(\sum_{cyc}\frac{x}{\sqrt{x^2+8yz}}\right)^2\sum_{cyc}x(x^2+8yz)\geq(x+y+z)^3.$$
Thus, it remains to prove that
$$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1369441",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "17",
"answer_count": 6,
"answer_id": 2
} |
Number theory with binary quadratic I found this questions from past year maths competition in my country, I've tried any possible way to find it, but it is just way too hard.
Given $$ \frac {x^2-y^2+2y-1}{y^2-x^2+2x-1} = 2$$ find $x-y$
I'm not sure if given choices is right... (A)2 (B)3 (C)4 (D)5 (E)6
I've tried to mo... | \begin{align} 2&= \frac {x^2-y^2+2y-1}{y^2-x^2+2x-1} \\
&=\frac {x^2-(y^2-2y+1)}{y^2-(x^2-2x+1)} \\
&= \frac {x^2-(y-1)^2}{y^2-(x-1)^2} \\
&= \frac {(x-y+1)(x+y-1)}{(y+x-1)(y-x+1)}\end{align}
$$\implies {x-y+1}=2y-2x+2\\
3(x-y)=1 $$
Thus, $x-y=\dfrac13$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1373243",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Polynomial of $11^{th}$ degree Let $f(x)$ be a polynomial of degree $11$ such that $f(x)=\frac{1}{x+1}$,for $x=0,1,2,3.......,11$.Then what is the value of $f(12)?$
My attempt at this is:
Let $f(x)=a_0+a_1x+a_2x^2+a_3x^3+......+a_{11}x^{11}$
$f(0)=\frac{1}{0+1}=1=a_0$
$f(1)=\frac{1}{1+1}=\frac{1}{2}=a_0+a_1+a_2+a_3+...... | Somewhat less vague...
$(x+1)f(x)-1$ is a polynomial of degree 12 with roots at every integer in $[0,11]$, so could be $$(x+1)f(x)-1 = A \prod_{c=0}^{11} x-c$$
for some/any (nonzero) constant $A$.
When $x=-1$, we have $(0)f(-1)-1 = A (-1)^{12} 12!$, or $-1 = A \, 12!$ and discover only $A = \frac{-1}{12!}$ is consisten... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1374128",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 1
} |
Solve $10x^4-7x^2(x^2+x+1)+(x^2+x+1)^2=0$ How to solve the following equation?
$$10x^4-7x^2(x^2+x+1)+(x^2+x+1)^2=0$$
My attempt:
$$ 10x^4 - (7x^2+1)(x^2+x+1)=0$$
Thats all i can
Update
Tried to open brakets and simplify:
$$(7x^2+1)(x^2+x+1) = 7x^4+7x^3+7x^2+x^2+x+1=7x^4+7x^3+8x^2+1 $$
$$10x^4 - (7x^2+1)(x^2+x+1)= 3x^4... | $10x^4-7x^2(x^2+x+1)+(x^2+x+1)^2=0$
Divide by $x^4$ on both sides,
$10-\frac{7x^2(x^2+x+1)}{x^4}+\frac{(x^2+x+1)^2}{x^4}=0$
$10-7(1+\frac{1}{x}+\frac{1}{x^2})+(1+\frac{1}{x}+\frac{1}{x^2})^2=0$
Put $(1+\frac{1}{x}+\frac{1}{x^2})=t$
$10-7t+t^2=0$,solving we get $t=2,5$
when $t=2$
$1+\frac{1}{x}+\frac{1}{x^2}=2$
simplify... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1374321",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 3
} |
Am I getting the right answer for the integral $I_n= \int_0^1 \frac{x^n}{\sqrt {x^3+1}}\, dx$?
Let $I_n= \int_0^1 \dfrac{x^n}{\sqrt {x^3+1}}\, dx$. Show that $(2n-1)I_n+2(n-2)I_{n-3}=2 \sqrt 2$ for all $n \ge 3$. Then compute $I_8$.
I get an answer for $I_8={{2 \sqrt 2} \over 135}(25-16 \sqrt 2)$, could somebody plea... | If we set
$$
I_n=\int_0^1\frac{x^n}{\sqrt{x^3+1}}\,\mathrm{d}x\tag{1}
$$
Then, integration by parts gives
$$
\begin{align}
I_{n+3}+I_n
&=\int_0^1\frac{x^n(x^3+1)}{\sqrt{x^3+1}}\,\mathrm{d}x\\
&=\int_0^1x^n\sqrt{x^3+1}\ \mathrm{d}x\\
&=\frac1{n+1}\int_0^1\sqrt{x^3+1}\ \mathrm{d}x^{n+1}\\
&=\frac1{n+1}\left[\sqrt2-\int_0... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1375090",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 3
} |
Common solutions of two inequations Find the real values of $a$ for which the inequations $x^2-4x-6a\leq 0$ and $x^2+2x+a\leq0$ have only one real solution common.
My attempt:
Let $\alpha$ be one real common root of two inequations.
then $ \alpha^2-4\alpha-6a\leq 0$.......(1)
$\alpha^2+2\alpha+a\leq 0$.......(2)
Subtra... | Hint:
$$x^2-4x-6a \le 0 \iff 2-\sqrt{4+6a} \le x \le 2+\sqrt{4+6a} \tag{A}$$
$$x^2+2x+a \le 0 \iff -1-\sqrt{1-a} \le x \le -1+\sqrt{1-a} \tag{B}$$
Now find the situations when $A \cap B$ is exactly one point, to get $a \in \{0, 1\}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1375193",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Derivation of Gradshteyn and Ryzhik integral 3.876.1 (in question) In the Gradshteyn and Ryzhik Table of Integrals, the following integral appears (3.876.1, page 486 in the 8th edition):
\begin{equation}
\int_0^{\infty} \frac{\sin (p \sqrt{x^2 + a^2})}{\sqrt{x^2 + a^2}} \cos (bx) dx =
\begin{cases}
\frac{\pi}{2} J_0 \l... | $$I(a,b,p)\equiv\int_0^{\infty}\frac{\sin\left(p\sqrt{x^2+a^2}\right)}{\sqrt{x^2+a^2}}\cos(bx)\,dx\\\\
$$
Enforcing the substitution $x\to a\sinh x$ and assuming that $a>0$ yields
$$\begin{align}
I(a,b,p)&=\int_0^{\infty}\sin\left(pa\cosh x\right)\cos(ba\sinh x)\,dx\\\\
&=\frac12\int_0^{\infty}\left(\sin\left(pa\cosh ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1376303",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 1,
"answer_id": 0
} |
Calculate the limit $\lim_{x\to 0} \left(\frac 1{x^2}-\cot^2x\right)$ The answer of the given limit is $2/3$, but I cannot reach it. I have tried to use the L'Hospital rule, but I couldn't drive it to the end. Please give a detailed solution!
$$\lim_{x\to 0} \left(\dfrac 1{x^2}-\cot^2x\right)$$
| Here's a non-series solution.
$$\lim_{x \rightarrow 0} \left(\frac{1}{x^2} - \cot^2(x)\right) = \lim_{x \rightarrow 0} \frac{\sin^2(x) - x^2\cos^2(x)}{x^2\sin^2(x)}$$
Instead of directly doing l'Hospital, we can make our life easier using $\lim_{x \rightarrow 0}\frac{\sin(x)}{x} = 1$ Or rather, $\lim_{x \rightarrow 0}\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1377444",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 5,
"answer_id": 3
} |
how can i prove this trigonometry equation I need help on proving the following:
$$\frac{\cos {7x} - \cos {x} + \sin {3x}}{ \sin {7x} + \sin {x} - \cos {3x} }= -\tan {3x}$$
So far I've only gotten to this step:
$$\frac{-2 \sin {4x} \sin {3x} + \sin {x}}{ 2 \sin {4x} \cos {3x} - \cos {x}}$$
Any help would be appreciated... | You are almost there (I think you have a typo in the expression you got) :
Using
$$\cos A-\cos B=-2\sin\frac{A+B}{2}\sin\frac{A-B}{2}$$
$$\sin A+\sin B=2\sin\frac{A+B}{2}\cos\frac{A-B}{2}$$
we have
$$\begin{align}\frac{\cos(7x)-\cos x+\sin(3x)}{\sin(7x)+\sin x-\cos(3x)}&=\frac{-2\sin(4x)\sin(3x)+\sin(3x)}{2\sin(4x)\co... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1377868",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Prove using mathematical induction that $x^{2n} - y^{2n}$ is divisible by $x+y$ Prove using mathematical induction that
$(x^{2n} - y^{2n})$ is divisible by $(x+y)$.
Step 1: Proving that the equation is true for $n=1 $
$(x^{2\cdot 1} - y^{2\cdot 1})$ is divisible by $(x+y)$
Step 2: Taking $n=k$
$(x^{2k} - y^{2k})$ is... | Hint: Rewrite:
$$x^{2k+2}-y^{2k+2}=(x^{2k+2}-y^{2k}x^2)+(y^{2k}x^2-y^{2k+2})=x^2(x^{2k}-y^{2k})+y^{2k}(x^2-y^2).$$
Added: Note it can be proved without induction:
$$x^{2n}-y^{2n}=(x^2)^n-(y^2)^n=(x^2-y^2)(x^{2(n-1)}+x^{2(n-2)}y^2+\dots+x^2y^{2(n-2)}+y^{2(n-1)}),$$
and the first factor is divisible by $x+y$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1377927",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 2
} |
How to integrate ${x^3}/(x^2+1)^{3/2}$? How to integrate $$\frac{x^3}{(x^2+1)^{3/2}}\ \text{?}$$
I tried substituting $x^2+1$ as t, but it's not working
| $x^2+1=t^2\Rightarrow x\,dx=t\,dt\;,\;x^2=t^2-1$
$$\int\frac{x^3}{(x^2+1)^{\frac{3}{2}}}dx=\int\frac{t^2-1}{t^2}dt=\int(1-t^{-2})dt=t+\frac{1}{t}+C$$
$t=\sqrt{1+x^2}\Rightarrow$ answer$=\sqrt{1+x^2}+\frac{1}{\sqrt{1+x^2}}+C=\frac{x^2+2}{\sqrt{x^2+1}}+C$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1378025",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 2
} |
Is $\left(45+29\sqrt{2}\right)^{1/3} + \left(45-29\sqrt{2}\right)^{1/3}$ an integer? The problem is the following:
Prove that this number
$$x = \left(45+29\sqrt{2}\right)^{1/3} + \left(45-29\sqrt{2}\right)^{1/3}$$
is an integer. Show which integer it is.
I thought that it has some relations with something like c... | Given the hint that $(45+29\sqrt{2})^{1/3} + (45-29\sqrt{2})^{1/3}$ is an integer.
One should first investigate whether
$$(45 + 29\sqrt{2})^{1/3} = a + b\sqrt{2}$$
for some integers $a, b$.
Notice $$(a+b\sqrt{2})^3 = a ( a^2 + 6 b^2 ) + (3a^2 + 2b^2)b\sqrt{2}$$
This suggest us to look for integer solutions for followi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1378183",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 2
} |
Find all values that solve the equation For which values a, the equation
$$ a\sin{x}+(a+1)\sin^2{\frac{x}{2}} + (a-1)\cos^2{\frac{x}{2}} =1 $$
has a solution?
My idea: I think it's possible to factorize equation or reduce equation to the form like: $a(\sin^2{\frac{x}{2}} + \cos^2{\frac{x}{2}}) =1 $
Let's go:
$$ 2a\sin{... | Note that your equation reduces to
\begin{align}
1
&= a\sin{x}+(a+1)\sin^2{\frac{x}{2}} + (a-1)\cos^2{\frac{x}{2}} \\
&= a \sin x +a \left(\sin^2{\frac{x}{2}} + \cos^2{\frac{x}{2}}\right) +
\sin^2{\frac{x}{2}}- \cos^2{\frac{x}{2}} \\
&= a \sin x +a + \sin^2{\frac{x}{2}}- \cos^2{\frac{x}{2}} \\
&= a \sin x +a - \cos(x) ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1378266",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
Find the equation of the circle. Find the equation of the circle whose radius is $5$ which touches the circle $x^2 + y^2 - 2x -4y - 20 = 0$ externally at the point $(5,5)$
| Express that the circle is through the point $(5,5)$
$$(5-x_c)^2+(5-y_c)^2=5^2,$$
and that the gradients are collinear a this point
$$(x_c-5)(2\cdot5-4)-(y_c-5)(2\cdot 5-2)=0.$$
From the second equation,
$$y_c=\frac{3x_c+5}4,$$
and plugging in the first,
$$x_c^2-10x_c+9=0.$$
The solutions are $(x_c,y_c)=(1,2)$ and $(9,... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1378916",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 3
} |
The value of the definite integral The value of the definite integral $\displaystyle\int\limits_0^\infty \frac{\ln x}{x^2+4} \, dx$ is
(A) $\dfrac{\pi \ln3}{2}$ (B) $\dfrac{\pi \ln2}{3}$ (C) $\dfrac{\pi \ln2}{4}$ (D) $\dfrac{\pi \ln4}{3}$
I tried using integration by parts,
\begin{align}
& \int_0^\infty \frac{\ln x... | Let $$\displaystyle I = \int_{0}^{\infty}\frac{\ln x}{x^2+4}dx = \int_{0}^{2}\frac{\ln x}{x^2+4}dx+\underbrace{\int_{2}^{\infty}\frac{\ln x}{x^2+4}dx}_{J}$$
Now for Calculation of $$\displaystyle J = \int_{2}^{\infty}\frac{\ln x}{x^2+4}dx\;,$$ Now put $\displaystyle x=\frac{4}{y}\;,$ Then $\displaystyle dx = -\frac{4}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1378974",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
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How to factor intricate polynomial $ ab^3 - a^3b + a^3c -ac^3 +bc^3 - b^3c $ I would like to know how to factor the following polynomial.
$$ ab^3 - a^3b + a^3c -ac^3 +bc^3 - b^3c $$
What is the method I should use to factor it? If anyone could help.. Thanks in advance.
| $ab^3 - a^3b + a^3c -ac^3 +bc^3 - b^3c =$
$ab^3 - a^3b + a^3c -b^3c -ac^3 + bc^3=$
$ab(b^2-a^2)+c(a^3-b^3)-c^3(a-b)=$
$ab(a+b)(b-a)+c(a-b)(a^2+ab+b^2)-c^3(a-b)=$
$(a-b)[-ab(a+b)+c(a^2+ab+b^2)-c^3]=$
$(a-b)(-a^2b-ab^2+ca^2+cab+cb^2-c^3)=$
$(a-b)(ca^2-a^2b+cab-ab^2+cb^2-c^3)=$
$(a-b)[a^2(c-b)+ab(c-b)+c(b^2-c^2)]= $
$(a-b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1379045",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Show by induction that $F_n \geq 2^{0.5 \cdot n}$, for $n \geq 6$ I have the following problem:
Show by induction that $F_n \geq 2^{0.5 \cdot n}$, for $n \geq 6$
Where $F_n$ is the $nth$ Fibonacci number.
Proof
Basis
$n = 6$.
$F_6 = 8 \geq 2^{0.5 \cdot 6} = 2^{\frac{6}{2}} = 2^3 = 8$
Induction hypothesis
Assume $F... | Suppose we knew for 2 values of n $F_n \ge 2^{\frac n2}$ i.e for n = 6 and n = 7.
We know this holds for n=6 and n=7.
We also know that $F_{n+1} = F_{n} + F_{n-1}$
So we assume for some k and k-1 (7 and 6) $F_{k-1} \ge 2^{\frac {k-1}{2}}$ and $F_{k+1} \ge 2^{\frac {k+1}{2}}$
We know $F_{k} \ge F_{n-1}$
so $F_{k+1} \ge ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1381545",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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Showing that $\frac{1}{2^n +1} + \frac{1}{2^n +2} + \cdots + \frac{1}{2^{n+1}}\geq \frac{1}{2}$ for all $n\geq 1$
Show that $$\frac{1}{2^n +1} + \frac{1}{2^n +2} + \cdots + \frac{1}{2^{n+1}}\geq \frac{1}{2}$$ for all $n\geq 1$
I need this in order to complete my proof that $1 + \frac{n}{2} \leq H_{2^n}$, but I don't ... | Hint for every $1\leq k\leq 2^n$
$$\frac{1}{2^n+k}\geq \frac{1}{2^{n+1}} $$
do every term in your sum is greater then $\frac{1}{2^{n+1}} $ and in the sum there is $2^n$ terms so your sum is greater then $2^n$ times $\frac{1}{2^{n+1}} $ $\cdots$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1381948",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 1
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How many divisors of the combination of numbers?
Find the number of positive integers that are divisors of at least one of $A=10^{10}, B=15^7, C=18^{11}$
Instead of the PIE formula, I would like to use intuition.
$10^{10}$ has $121$ divisors, and $15^7$ has $64$ divisors, and $18^{11}$ has $276$ divisors.
Number of... | Each divisor is of the form $2^i3^j5^k$, where the exponents are nonnegative.
If $i=0$ and $j=0$, then $0\le k \le 10$, which accounts for $11$ possibilities.
If $i=0$, $j>0$, and $k=0$, then $1\le j \le 22$, which accounts for $22$ possibilities.
If $i=0$, $j>0$, and $k>0$, then $1\le j,k \le 7$, which accounts for $4... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1382185",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
How to solve $\lfloor \sqrt{(k + 1)\cdot2009} \rfloor = \lfloor \sqrt{k\cdot2009} \rfloor$ Is there any way to solve this equation (or to tell how many solutions are there), other than checking all 2009 possibilities? $\lfloor \sqrt{(k + 1)\cdot2009} \rfloor = \lfloor \sqrt{k\cdot2009} \rfloor, 0 \le k \le 2009, k \in ... | As shown in N.S.'s answer, there are no solutions for $k \le 501$ since $\sqrt{2009(k+1)} > \sqrt{2009k}+1$ for all $k \le 501$. Also, for all $502 \le k \le 2009$, we have $\sqrt{2009(k+1)} < \sqrt{2009k}+1$.
Thus, for each $502 \le k \le 2009$, we have either $\lfloor\sqrt{2009(k+1)}\rfloor - \lfloor\sqrt{2009k}\rfl... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1383253",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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$p,q,r$ primes, $\sqrt{p}+\sqrt{q}+\sqrt{r}$ is irrational. I want to prove that for $p,q,r$ different primes, $\sqrt{p}+\sqrt{q}+\sqrt{r}$ is irrational.
Is the following proof correct?
If $\sqrt{p}+\sqrt{q}+\sqrt{r}$ is rational, then $(\sqrt{p}+\sqrt{q}+\sqrt{r})^2$ is rational, thus $p+q+r+2\sqrt{pq}+2\sqrt{pr}+2... | Suppose $\sqrt{p}+\sqrt{q}+\sqrt{r} = a \to (a-\sqrt{p})^2 = (\sqrt{q}+\sqrt{r})^2\to a^2-2a\sqrt{p} +p =q+r+2\sqrt{qr}\to a^2-q+p-r=2a\sqrt{p}+2\sqrt{qr}\to (a^2-q+p-r)^2=4a^2p+4qr+8a\sqrt{pqr}\to \sqrt{pqr} $ is a rational number, and this is not possible.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1384514",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 5,
"answer_id": 4
} |
Prove the series has positive integer coefficients How can I show that the Maclaurin series for
$$
\mu(x) = (x^4+12x^3+14x^2-12x+1)^{-1/4}
\\
= 1+3\,x+19\,{x}^{2}+147\,{x}^{3}+1251\,{x}^{4}+11193\,{x}^{5}+103279\,
{x}^{6}+973167\,{x}^{7}+9311071\,{x}^{8}+\cdots
$$
has positive integer coefficients? (I have others to d... | I can prove the integer property, but not (yet) positivity.
Lemma: The coefficients of Taylor series of $\mu(x)$ are integers.
Proof. We can write
$$x^4+12x^3+14x^2-12x+1=\left(1-x\right)^4\Bigl(1-8\nu(x)\Bigr),$$
where $\nu(x)$ is given by
$$\nu(x)=\frac{2}{1-x}-\frac{7}{\left(1-x\right)^2}+\frac{7}{\left(1-x\right)^3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1386819",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 4,
"answer_id": 0
} |
Find the value below
Let $a,b,c$ be the roots of the equation $$8x^{3}-4x^{2}-4x+1=0$$
Find $$\frac{1}{a^{3}}+\frac{1}{b^{3}}+\frac{1}{c^{3}}$$
It's just for sharing a new ideas, thanks:)
| A systematic way to compute sums of the form $a^{-k} + b^{-k} + c^{-k}$ is
using Newton's identities.
Notice
$$\begin{align}
a, b, c \text{ roots of } p(x) &= 8x^3 - 4x^2 - 4x + 1\\
\implies\quad
\frac{1}{a}, \frac{1}{b}, \frac{1}{c}
\text{ roots of } q(x) &= x^3 p(\frac{1}{x}) = x^3 - \color{red}{4} x^2 - \color{gree... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1387198",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
Proving that $1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\cdots+\frac{1}{2n-1}-\frac{1}{2n}=\frac{1}{n+1}+\frac{1}{n+2}+\cdots+\frac{1}{2n}$ I have written the left side of the equation as $$\left(1+\frac{1}{3}+\frac{1}{5}+\cdots+\frac{1}{2n-1}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+\cdots+\frac{1}{2n}\right).$$ I... | Try using $\Sigma$-notation to make your problem more manageable in terms of its algebraic expressions and the like. To this end, note that
$$
1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\cdots+\frac{1}{2n-1}-\frac{1}{2n}=\frac{1}{n+1}+\frac{1}{n+2}+\cdots+\frac{1}{2n}\tag{1}
$$
becomes
$$
\sum_{i=1}^n\frac{1}{2i-1}-\sum_{i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1387744",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 3,
"answer_id": 0
} |
How do I find the remainder of $4^0+4^1+4^2+4^3+ \cdots + 4^{40}$ divided by 17? Recently I came across a question,
Find the remainder of $4^0+4^1+4^2+4^3+ \cdots + 4^{40}$ divided by 17?
At first I applied sum of G.P. formula but ended up with the expression $1\cdot \dfrac{4^{41}-1}{4-1}$. I couldn't figure out how ... | HINT:
Observe that for any non-negative integer $a,$ $$1+4+4^2+4^3=(1+4)(1+4^2)\equiv0\pmod{17}$$
$$\implies\sum_{a=m}^n4^{4a}(1+4+4^2+4^3)\equiv0\pmod{17}$$
Here $m=0,n=9$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1388038",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 0
} |
Show that $\sqrt{4+2\sqrt{3}}-\sqrt{3}$ is rational using the rational zeros theorem What I've done so far:
Let $$r = \sqrt{4+2\sqrt{3}}-\sqrt{3}.$$
Thus, $$r^2 = 2\sqrt{3}-2\sqrt{3}\sqrt{4+2\sqrt{3}}+7$$
and $$r^4=52\sqrt{3}-28\sqrt{3}\sqrt{4+2\sqrt{3}}-24\sqrt{4+2\sqrt{3}}+109.$$
I did this because in a similar exam... | In terms of your method, I think the first thing I'd do is rewrite it as,
$$r + \sqrt{3} = \sqrt{4 + 2\sqrt{3}}.$$
If you square both sides,
$$r^2 + 3 + 2\sqrt{3}r = 4 + 2\sqrt{3}.$$
Isolating $\sqrt{3}$,
$$2\sqrt{3}(r - 1) = 1 - r^2.$$
Squaring again,
$$12(r-1)^2 = r^4 - 2r^2 + 1.$$
Expanding and rearranging,
$$(r-1)^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1388206",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 5,
"answer_id": 4
} |
Find the sum of binomial coefficients
Calculate the value of the sum
$$
\sum_{i = 1}^{100} i\binom{100}{i} = 1\binom{100}{1} +
2\binom{100}{2} +
3\binom{100}{3} +
\dotsb +
... | $$\begin{align}
\sum_{i=1}^ni\binom ni
&=\sum_{i=0}^{n-1}(i+1)\binom n{i+1}
\color{lightgray}{=\sum_{i=0}^{n-1}(i+1)\frac {n(n-1)^{\underline{i}}}{(i+1)i!}=\sum_{i=0}^{n-1}n\frac {(n-1)^{\underline{i}}}{i!}}\\
&=\sum_{i=0}^{n-1}n\binom {n-1}i\\
&=n\sum_{i=0}^{n-1}\binom {n-1}i\\
&=n(1+1)^{n-1}\\
&=n\cdot 2^{n-1}
\end{a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1388720",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
Card Draw question What are the ways to draw 13 cards from a pack of 52 cards such that
(a) the hand is void in at least one suit, (b) the hand is not void in any suit.(“void in a suit” means having no cards of that suit)
Current approach
I am thinking on following lines:
Case-1: All cards from exactly 1 suit $= {4 \c... | For (a), select the 3 suits represented out of 4, and then select 13 cards out of the remaining $52 - 13 = 39$, for a total of $\binom{4}{3} \cdot \binom{39}{13} = 32489701776$. For (b), it is all posibilities (taking 13 card out of 52) less the ones missing a suit. Part (a) uses at most 3 suits, by a similar reasoning... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1390748",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
$\int\limits_{1/2}^{2}\frac{1}{x}\sin (x-\frac{1}{x})dx$ $\int\limits_{1/2}^{2}\frac{1}{x}\sin (x-\frac{1}{x})dx$ has value equal to
$(A)0\hspace{1cm}(B)\frac{3}{4}\hspace{1cm}(C)\frac{3}{4}\hspace{1cm}(D)2 $
I tried to solve this question by putting $x-\frac{1}{x}=t$ and limits have changed to $\frac{-3}{2}$ to $\frac... | Let $$I(a) = \int^{a}_{\frac{1}{a}}\frac{1}{x}\sin \left(x-\frac{1}{x}\right)dx$$
Where $a>0$
Now $$I'(a) = \frac{1}{a}\sin \left(a-\frac{1}{a}\right)+\sin \left(\frac{1}{a}-a\right)\cdot a \cdot \frac{1}{a^2} = 0$$
So $$I'(a) = 0\Rightarrow I(a) = \mathcal{C}$$
Now put $a=1\;,$ in first equation, We get $I(1) = 0$
So ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1390926",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 4
} |
Coefficient of $x^9$ in $(1+x)\cdot (1+x^2)\cdot (1+x^3)\cdots(1+x^9)$
The Coefficient of $x^9$ in $(1+x)\cdot (1+x^2)\cdot (1+x^3)\cdots(1+x^9)$ is
My Try We Know that Coefficient of $x^9$ occur (which is $ = 1$) when we
multiply $x^0\cdot x^9$ and $x^1\cdot x^8$ and $x^2\cdot x^7$ and $x^3\cdot x^6$ and $x^4\cdo... | This is the same as the number of ways to express $9$ as the sum of distinct positive integers $\le 9$.
$9 = 9 = 8+1 = 7+2 = 6+3 = 6+2+1 = 5+4 = 5+3+1 = 4+3+2$
So there are eight ways.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1392725",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Finding $\lim_{(x,y)\rightarrow (0,0)}\frac{e^{-\frac{1}{x^2+y^2}}}{x^4+y^4}$ I want to find $\lim_{(x,y)\rightarrow 0}\frac{e^{-\frac{1}{x^2+y^2}}}{x^4+y^4}$.
What I tried:
Denote $x=r\cos(\theta)$ and $y=r\sin(\theta)$. So the limit is:
$$\lim_{r\rightarrow 0}\frac{e^{-\frac{1}{r^2(\cos^2(\theta)+\sin^2(\theta))}}}{r... | Let
$1/r^2
= z
$.
Then
$U(r, \theta)
=\frac{e^{-\frac{1}{r^2}}}{r^4(\cos^4(\theta)+\sin^4(\theta))}
=\frac{z^2e^{-z}}{\cos^4(\theta)+\sin^4(\theta)}
$.
Also,
$\begin{array}\\
\cos^4(\theta)+\sin^4(\theta)
&=\cos^4(\theta)+\sin^4(\theta)+2\cos^22(\theta)\sin^2(\theta)-2\cos^2(\theta)\sin^2(\theta)\\
&=(\cos^2(\theta)+\s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1393578",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
How to factor $4x^2 + 2x + 1$? I want to know how to factor $4x^2 + 2x + 1$? I found the roots using quadratic equation and got $-1 + \sqrt{-3}$ and $-1 - \sqrt{-3}$, so I thought the factors would be $(x - (-1 + \sqrt{-3}))$ and $(x - (-1 - \sqrt{-3}))$
However, according to MIT's course notes, the factors are $(1 - (... | The solution is $\frac{-1 \pm \sqrt {-3}}{4}$ (suppose we know what we mean here by $\sqrt{-3}$).
Then the expansion would be
$\begin{align*}
&4x^2+2x+1=4\left(x-\frac{-1+ \sqrt {-3}}{4}\right)\left(x-\frac{-1- \sqrt {-3}}{4}\right)\\
& = 4\left(x-\frac{1}{-1- \sqrt {-3}}\right)\left(x-\frac{1}{-1+ \sqrt {-3}}\right)\\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1393930",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 2
} |
Counting integral solutions Suppose $a + b + c = 15$
Using stars and bars method, number of non-negative integral solutions for the above equation can be found out as $15+3-1\choose15$ $ =$ $17\choose15$
How to extend this principle for finding number of positive integral solutions of
$a + b + 3c = 15$?
I tried to do i... | The generating function for the number of non-negative integral solutions to $a+b+3c=n$ is
$$
f(x)=\frac1{1-x}\frac1{1-x}\frac1{1-x^3}
$$
Since
$$
\begin{align}
(1-x)^{-2}
&=\sum_{k=0}^\infty\binom{-2}{k}(-x)^k\\
&=\sum_{k=0}^\infty\binom{k+1}{k}x^k\\
&=\sum_{k=0}^\infty(k+1)x^k
\end{align}
$$
and
$$
\begin{align}
\lef... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1394730",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
Find $\int \frac{1}{\sin\frac{x}{2}\sqrt{\cos^3\frac{x}{2}}}dx$ $\int \frac{1}{\sin\frac{x}{2}\sqrt{\cos^3\frac{x}{2}}}dx$
I tried to solve it.I put $\frac{x}{2}=t$ then $\int \frac{1}{\sin\frac{x}{2}\sqrt{\cos^3\frac{x}{2}}}dx=$
$\int \frac{2 dt}{\sin t\sqrt{\cos^3 t} }=\int \frac{2 dt}{\sin t \cos t\sqrt{\cos t} }$ a... | $$\int\frac{dx}{\sin\frac{x}{2}\sqrt{\cos^3\frac{x}{2}}}$$
$$=\int\frac{\sin\frac{x}{2}dx}{\sin^2\frac{x}{2}\sqrt{\cos^3\frac{x}{2}}}$$
$$=\int\frac{\sin\frac{x}{2}dx}{\left(1-\cos^2\frac{x}{2}\right)\sqrt{\cos^3\frac{x}{2}}}$$
Let $\cos\frac{x}{2}=t\implies -\frac{1}{2}\sin\frac{x}{2}=dt$
$$=-\frac{1}{2}\int\frac{dt}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1395518",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Number of non-decreasing functions?
Let $A = \{1,2,3,\dots,10\}$ and $B = \{1,2,3,\dots,20\}$.
Find the number of non-decreasing functions from $A$ to $B$.
What I tried:
Number of non-decreasing functions = (Total functions) - (Number of decreasing functions)
Total functions are $20^{10}$. And I think there are ${20... | Means Here function satisfy the following Condition.
$f(1),f(2),f(3),.......f(10)\in \left\{1,2,3,4,5,...,20\right\}$
$\; \bullet f(1)<f(2)<..........<f(10)\;,$ Total no. of possibilities $ = \displaystyle \binom{20}{10}$
$\;\bullet f(1)\leq f(2)<f(3)<.......<f(10)\;,$ Total no. of possibilities $\displaystyle \binom{2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1396896",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 3,
"answer_id": 0
} |
Prove that the equation $\ 5x^4 + x − 3 = 0\ $ has no rational solutions. I'm locked at $\ x\left(5x^3 + 1\right) = 3$. Not too sure where to go from there but I'm getting the feeling it's really really obvious..
| Let us suppose $x = a / b$ is a rational solution, where a, b are integers, mutually prime. Then:
$$
5(a/b)^4 + a/b -3 = 0
$$
and after multiplying by $b^4$:
$$
5a^4 + ab^3 -3b^4=0
$$
or equivalent
$$
a(5a^3 + b^3) = 3b^4 (*)
$$
The last equality implies that $a$ divides $3b^4$. Because $a$ and $b$ are coprime, then ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1397955",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 4
} |
Evaluate $\int e^x \sin^2 x \mathrm{d}x$ Is the following evaluation of correct?
\begin{align*} \int e^x \sin^2 x \mathrm{d}x &= e^x \sin^2 x -2\int e^x \sin x \cos x \mathrm{d}x \\ &= e^x \sin^2 x -2e^x \sin x \cos x + 2 \int e^x (\cos^2 x - \sin^2x) \mathrm{d}x \\ &= e^x \sin^2 x -2e^x \sin x \cos x + 2 \int e^x (1 -... | Notice
$$\int e^x\sin^2x\mathrm{d}x=$$ $$=\int e^x\left(\frac{1-\cos 2x}{2}\right)\mathrm{d}x$$
$$=\frac{1}{2}\int e^xdx-\frac{1}{2}\int e^x \cos 2x \mathrm{d}x$$
Using $\displaystyle \int e^{ax}\cos (bx) \mathrm{d}x=\frac{e^{ax}}{a^2+b^2}(a\cos bx+b\sin bx)$, we get
$$=\frac{1}{2}e^x-\frac{1}{2}\frac{e^x}{1^2+2^2}(\c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1398965",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 0
} |
How to get to this equality $\prod_{m=1}^{\infty} \frac{m+1}{m}\times\frac{m+x}{m+x+1}=x+1$?
How to get to this equality $$\prod_{m=1}^{\infty} \frac{m+1}{m}\times\frac{m+x}{m+x+1}=x+1?$$
I was studying the Euler Gamma function as it gave at the beginning of its history, and need to solve the following product operat... | $$\begin{align}
\prod_{n=1}^N\frac{m+1}{m}\frac{m+x}{m+x+1}&=\left(\prod_{n=1}^N\frac{m+1}{m}\right)\left(\prod_{n=1}^N\frac{m+x}{m+x+1}\right)\\\\
&=\left(\frac{2}{1}\frac{3}{2}\frac{4}{3}\cdots \frac{N}{N-1}\frac{N+1}{N}\right)\left(\frac{1+x}{2+x}\frac{2+x}{3+x}\frac{3+x}{4+x}\cdots \frac{N-1+X}{N+x}\frac{N+x}{N+1+X... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1400338",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
Find $\int_0^\frac{\pi}{2} \frac {\theta \cos \theta } { \sin \theta + \sin ^ 3 \theta }\:d\theta$ My Calc 2 teacher wasn't able to solve this:
$$\int_0^\frac{\pi}{2} \frac {\theta \cos \theta } { \sin \theta + \sin ^ 3 \theta }\:d\theta$$
Can someone help me solve this?
| You can transform the integral by parts to get
$$\int_0^{\pi/2} d\theta \frac{\theta \cos{\theta}}{\sin{\theta} (1+\sin^2{\theta})} = \int_0^{\pi/2} d(\sin{\theta}) \left (\frac1{\sin{\theta}}-\frac{\sin{\theta}}{1+\sin^2{\theta}} \right )\theta \\ = -\frac{\pi}{4} \log{2} - \int_0^{\pi/2} d\theta \left [\log{(\sin{\th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1400439",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 5,
"answer_id": 2
} |
Find Solution to an infinite Nested Radicals How do I find the solution to the following:
$$ \sqrt{ 7 - \sqrt{\frac{7}{2} + \sqrt{\frac{7}{4} - \sqrt{\frac{7}{16} + \sqrt{\frac{7}{256} - \ldots}}}}}$$
I first tried looking for a pattern for the denominators, but the $16$ seems to be throwing me off. Can we use calculus... | Hint 1:
$$X = \sqrt{7 - \dfrac{A}{\sqrt{2}}}$$
where $A = \displaystyle\sqrt{7 + \sqrt{7 - \sqrt{7 + \sqrt{7 - ....}}}}$
Hint 2:
$$(A^{2} - 7)^{2} = 7 - A$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1400523",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
LU-factorization: why can't I get a unit lower triangular matrix? I want to find an $LU$-factorization of the following matrix: \begin{align*} A = \begin{pmatrix} 3 & -6 & 9 \\ -2 & 7 & -2 \\ 0 & 1 & 5 \end{pmatrix} \end{align*} This matrix is invertible (the determinant is $33$), so I should be getting a $LU$ decompos... | In the framework of the Gaussian elimination procedure, you get $L$ with all $1$s on the diagonal by never truly rescaling rows. You rescale them in the intermediate step before you add them to another row, but then you "scale them back" so that you're only changing the row that you actually added to. You can always pe... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1400631",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Solve for $v$ - simplify as much as possible Solve for $v$. Simplify the answer.
$$-3 = -\frac{8}{v-1}$$
Here is what I tried:
$$-3 = \frac{-8}{v-1} $$
$$(-8) \cdot (-3) = \frac{-8}{v-1} \cdot (-8) $$
$$24 = v-1$$
$$25 = v$$
| It may be instructive to, as an exercise, try a few problems similar to this one without skipping any steps (as follows).
$$\begin{array}{lll}
-3&=&-\displaystyle\frac{8}{v-1}\\
-3&=&(-1)\cdot\displaystyle\frac{8}{v-1}\\
-3&=&\displaystyle\frac{-1}{1}\cdot\displaystyle\frac{8}{v-1}\\
-3&=&\displaystyle\frac{(-1)\cdot8... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1401865",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
Find $\int_{1}^{2}\frac{x^2-1}{x^3\sqrt{2x^4-2x^2+1}}dx$ $$\int_{1}^{2}\frac{x^2-1}{x^3\sqrt{2x^4-2x^2+1}}dx=\frac{u}{v}$$ where $u$ and $v$ are in their lowest form. Find the value of $\dfrac{1000u}{v}$
$$\int_{1}^{2}\frac{x^2-1}{x^3\sqrt{2x^4-2x^2+1}}dx=\int_{1}^{2}\frac{x^2-1}{x^3\sqrt{2x^2(x^2-1)+1}}dx$$ I put $x^2... | Let $$\displaystyle I = \int_{1}^{2}\frac{x^2-1}{x^3\sqrt{2x^4-2x^2+1}}dx = \int \frac{\left(1-x^{-2}\right)\cdot x^{-3}}{\sqrt{2-2x^{-2}+x^{-2}}}dx\;,$$
Now Let $x^{-2} = u\;,$ Then $-2x^{-3}dx = du$ and Changing Limit, We get
$$\displaystyle I = -\frac{1}{2}\int_{1}^{\frac{1}{4}}\frac{1-u}{\sqrt{u^2-2u+2}}du = -\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1403516",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
Triangle area inequalities; semiperimeter I've got stuck on this problem :
Proof that for every triangle of
sides $a$, $b$ and $c$ and area
$S$, the following
inequalities are true :
$4S \le a^2 + b^2$
$4S \le b^2 + c^2$
$4S \le a^2 + c^2$
$6S \le a^2 + b^2 + c^2$
The first thing that came to my mind was the inequal... | Those inequalities are very weak. It is trivial that
$$ 2S\leq ab\leq \left(\frac{a+b}{2}\right)^2\leq \frac{a^2+b^2}{2}$$
by the GM-AM-QM inequality, hence $S\leq\frac{a^2+b^2+c^2}{6}$ by adding the first three inequalities.
However, they cannot all hold as equalities, since a (Euclidean) triangle cannot have three ri... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1404335",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
Prove the trigonometric identity $\cos(x) + \sin(x)\tan(\frac{x}{2}) = 1$ While solving an equation i came up with the identity $\cos(x) + \sin(x)\tan(\frac{x}{2}) = 1$.
Prove whether this is really true or not. I can add that $$\tan\left(\frac{x}{2}\right) = \sqrt{\frac{1-\cos x}{1+\cos x}}$$
| Yup! We have $$\cos x + \sin x \tan \frac{x}{2} = (\cos ^2 \frac{x}{2} - \sin ^2 \frac{x}{2}) + (2 \sin \frac{x}{2} \cos \frac{x}{2}) \frac{\sin \frac{x}{2}}{\cos \frac{x}{2}} = \cos ^2 \frac{x}{2} + \sin ^2 \frac{x}{2} = 1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1407794",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 9,
"answer_id": 2
} |
If $ a + b + c \mid a^2 + b^2 + c^2$ then $ a + b + c \mid a^n + b^n + c^n$ for infinitely many $n$
Let $ a,b,c$ positive integer such that $ a + b + c \mid a^2 + b^2 + c^2$.
Show that $ a + b + c \mid a^n + b^n + c^n$ for infinitely many positive integer $ n$.
(problem composed by Laurentiu Panaitopol)
So far no i... | Here's a more intuitive way to get the idea of considering powers of $2$.
Added (below): in the same way we can prove that any $n=6k\pm1$ works.
Note that $a+b+c\mid(a+b+c)^2-(a^2+b^2+c^2)=2(ab+bc+ca)$. By The Fundamental Theorem of Symmetric Polynomials (FTSP), $a^n+b^n+c^n$ is an integer polynomial in $a+b+c$, $ab+bc... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1408323",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 6,
"answer_id": 4
} |
Find the least positive residue of $10^{515}\pmod 7$. I tried it, but being a big number unable to calculate it.
| $10^{6} \equiv 1 $ mod $7$ By Fermat's little theorem. Notice $515=6 \times 85+5$ so $10^{515} \equiv 10^{5}$ mod $7$. Notice $10 \equiv 3$ mod 7, so $10^{2} \equiv 3^{2} \equiv 2$ and $10^{3} \equiv 20 \equiv -1\equiv 6$ mod $7$, $10^{4} \equiv 60 \equiv 4$ mod $7$, so $10^{5} \equiv 40 \equiv 5$ mod $7$. So $10^{515... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1409132",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
Find min of $M=\frac{1}{2+\cos2A}+\frac{1}{2+\cos2B}+\frac{1}{2-\cos2C}$
Find min of $$M=\frac{1}{2+\cos2A}+\frac{1}{2+\cos2B}+\frac{1}{2-\cos2C}$$, where $A, B, C$ are three angle of triangle $ABC$
Using Cauchy-Schwarz, we obtain: \begin{align*}
M &= \frac{1}{2+\cos2A}+\frac{1}{2+\cos2B}+\frac{1}{2-\cos2C}\\
&\ge... | After using C-S we need to prove that \begin{align*}
&\displaystyle\frac{9}{6-2\cos\gamma\cos\left(\alpha-\beta\right)+1-2\cos^2\gamma}\geq\frac{6}{5} \\
\iff&\displaystyle4\cos^2\gamma+4\cos\gamma\cos\left(\alpha-\beta\right)+1\geq0 \\
\iff&\displaystyle\left(2\cos\gamma+\cos\left(\alpha-\beta\right)\right)^2+\s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1411021",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Vector Functions of One Variable Question
A particle moves along the curve of the intersection of the cylinders $y=-x^2$ and $z=x^2$ in the direction in which $x$ increases. (All distances are in cm.) At the instant when the particle is at the point $(1,\,-1,\,1)$ its speed is $9$ cm/s, and the speed is increasing at a... | As I mentioned in my comment above, you can figure out $\dfrac{d^2x}{dt^2}$ from your equation in $|\mathbf{v}|$, given that you know that the change of speed is 3cm/s/s.
\begin{eqnarray}
|\mathbf{v}| &=& (\sqrt{1+8x^2})\dfrac{dx}{dt}\\
\dfrac{d|\mathbf{v}|}{dt} &=& (\sqrt{1+8^2})\dfrac{d^2x}{dt^2} + \frac{8x}{\sqrt{1+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1412648",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Does the limit $\lim\limits_{x\to0}\left(\frac{1}{x\tan^{-1}x}-\frac{1}{x^2}\right)$ exist? Does the limit: $$\lim\limits_{x\to0}\frac{1}{x\tan^{-1}x}-\frac{1}{x^2}$$ exist?
| Using Why is $\arctan(x)=x-x^3/3+x^5/5-x^7/7+\dots$? on,
$$\lim_{x\to0}\dfrac{x-\tan^{-1}x}{x^2\tan^{-1}x}=\lim_{x\to0}\dfrac{\dfrac {x^3}3+O(x^5)}{x^2\tan^{-1}x} =\dfrac13\lim_{x\to0}\dfrac{x+O(x^3)}{\tan^{-1}x}$$ as $x\ne0$ as $x\to0$
Now set $\tan^{-1}x=y\implies x=\tan y,x\to0\implies y\to0$
$$\lim_{x\to0}\dfrac{x+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1412775",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 4
} |
Simulataneous equations Suppose you have the following system of linear congruence
$2x+5y$ is congruent to 1 (mod6)
$x+y$ is congruent to 5 (mod6)
where $x,y \in \mathbb{Z}$
How would you obtain a general solution for this system. Also is there a way to determine whether the system is solvable or not?
| The system
\begin{align*}
x+y & \equiv 5 \pmod{6}\\
2x+5y & \equiv 1 \pmod{6}
\end{align*}
can be rewritten as
$$
\begin{pmatrix}
1 & 1\\
2 & 5
\end{pmatrix}
\begin{pmatrix}
x\\y
\end{pmatrix}
\equiv \begin{pmatrix}
5\\1
\end{pmatrix} \pmod{6}
$$
Now consider using row-operations (modulo $6$ of course) to get
$$
\left[... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1414334",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
For $n\times n$ matrices, is it true that $AB=CD\implies AEB=CED$?
If $A,B,C,D,E$ are $n\times n$ matrices, does $AB=CD$ imply $AEB=CED$?
I only know that $AB=CD \implies ABE=CDE$, but I don't see how you can sandwhich $E$ within it.
Also, if $AB=CD=0$, does $\det(AB)=\det(CD)=0$?
I think this should be true becaus... | I don't think this works by sandwiching.....
$$A=\begin{pmatrix}
1 & 3 \\
0 & 1 \\
\end{pmatrix},
B=\begin{pmatrix}
1 & 2 \\
0 & 1 \\
\end{pmatrix},
C=\begin{pmatrix}
1 & 1 \\
0 & 1 \\
\end{pmatrix},
D=\begin{pmatrix}
1 & 4 \\
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1414795",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 4
} |
Proving that $\frac{1}{2}<\frac{2}{3}<\frac{3}{4}<$...$<\frac{n-1}{n}$ In an attempt to find a pattern, I did this:
Let a,b,c,d be non-zero consecutive numbers. Then we have:
$a=a$
$b=a+1$
$c=a+2$
$d=a+3$
This implies:
$\frac{a}{b}=\frac{a}{a+1}$
$\frac{b}{c}=\frac{a+1}{a+2}$
$\frac{c}{d}=\frac{a+2}{a+3}$
I don't know ... | Now that I understand better, another simple way to prove the inequality is:
$$\begin{align}& n^{2}-1<n^{2}\\
& \implies\dfrac{n^{2}-1}{n(n+1)}<\dfrac{n^2}{n(n+1)}\\
&\implies \dfrac{(n-1)(n+1)}{n(n+1)}<\dfrac{n^{2}}{n(n+1)}\\
&\implies \dfrac{n-1}{n}<\dfrac{n}{n+1}\\
\end {align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1416155",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 7,
"answer_id": 6
} |
Closed form for $\sum_{k=1}^\infty(\zeta(4k+1)-1)$ Wikipedia gives
$$\sum_{k=2}^\infty(\zeta(k)-1)=1,\quad\sum_{k=1}^\infty(\zeta(2k)-1)=\frac34,\quad\sum_{k=1}^\infty(\zeta(4k)-1)=\frac78-\frac\pi4\left(\frac{e^{2\pi}+1}{e^{2\pi}-1}\right)$$
from which we can easily find $\sum_{k=1}^\infty(\zeta(2k+1)-1)$ and $\sum_{k... | By expressing the infinite sum as a double sum and then switching the order of summation, we get
$$\begin{align} \sum_{n=1}^{\infty} \left[\zeta(4n+1) - 1 \right] &= \sum_{n=1}^{\infty} \sum_{m=2}^{\infty} \frac{1}{m^{4n+1}} = \sum_{m=2}^{\infty} \frac{1}{m}\sum_{n=1}^{\infty} \left( \frac{1}{m^{4}} \right)^{n} \\ &= \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1418568",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 3,
"answer_id": 1
} |
How to prove $\sum _{k=1}^{\infty} \frac{k-1}{2 k (1+k) (1+2 k)}=\log_e 8-2$? How to prove:
$$\sum _{k=1}^{\infty} \frac{k-1}{2 k (1+k) (1+2 k)}=\log_e 8-2$$
Is it possible to convert it into a finite integral?
| Let me try.
Note that the $$RHS = 3\ln 2 + 2 = 2 + 3 \sum_{k=1}^\infty \frac{(-1)^{k+1}}{k} = 2 + 3\sum_{k=1}^\infty \frac{1}{2k-1}-\frac{1}{2k}$$
We have $$\begin{eqnarray} \sum_{k=1}^\infty \frac{k-1}{2k(k+1)(2k+1)} &=& \sum_{k=1}^\infty \frac{2k - (k+1)}{2k(k+1)(2k+1)} \\ &=&\sum_{k=1}^\infty \left(\frac{1}{(k+1)(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1419166",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 4,
"answer_id": 0
} |
How to calculate the limit that seems very complex.. Someone gives me a limit about trigonometric function and combinatorial numbers.
$I=\displaystyle \lim_{n\to\infty}\left(\frac{\sin\frac{1}{n^2}+\binom{n}{1}\sin\frac{2}{n^2}+\binom{n}{2}\sin\frac{3}{n^2}\cdots\binom{n}{n}\sin\frac{n+1}{n^2}}{\cos\frac{1}{n^2}+\binom... | Consider $$\lim_{n\to\infty} n\frac{\displaystyle\sum\limits_{k=0}^n {n \choose k}\sin\frac{k+1}{n^2}}{\displaystyle\sum\limits_{k=0}^n {n \choose k}\cos\frac{k+1}{n^2}}.$$ If this limit equals $\alpha$, your limit equals $e^\alpha$. In order to simplify a bit, thanks to the Taylor expansions of $\sin$ and $\cos$ near ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1421920",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "21",
"answer_count": 3,
"answer_id": 1
} |
Find the value of $\sum_{n=1}^\infty \frac{1}{n(2n-1)}$ I will show two solutions of this problem.
First solution :
$$\sum_{n=1}^\infty \frac{1}{n(2n-1)}=\sum_{n=1}^\infty \left(\frac{2}{2n-1}-\frac{1}{n}\right)$$
$$=\sum_{n=1}^\infty \left(\int_0^1 (2x^{2n-2}-x^{n-1})dx\right)$$
$$=\int_0^1 \left( \sum_{n=1}^\infty(2x... | Here is an even more baffling calculation:
\begin{align*}
0 &= \sum_{n=1}^{\infty} \left( \frac{2}{2n} - \frac{1}{n} \right) \\
&= \sum_{n=1}^{\infty} \int_{0}^{1} (2 x^{2n-1} - x^{n-1}) \, dx \\
&= \int_{0}^{1} \sum_{n=1}^{\infty} (2 x^{2n-1} - x^{n-1}) \, dx \\
&= \int_{0}^{1} \left( \frac{2x}{1-x^2} - \frac{1}{1-x} ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1424359",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 1,
"answer_id": 0
} |
Solve $\vert x-2\vert+2\vert x-4\vert\leq \vert x+1\vert$ I was helping someone with abolute values and inequalities and found this question.
What is the easiest way to solve this?
The only thing I thought of is to add the L.H.S and graph it with the R.H.S to answer the questoin is there simpler way to deal with this?
... | You have:
$$
x-2\ge 0 \iff x\ge 2
$$
$$
x-4\ge 0 \iff x\ge 4
$$
$$
x+1\ge 0 \iff x\ge -1
$$
so we can split the inequality $|x-2|+2|x-4|-|x+1|\le 0$ in four systems:
$$
\begin{cases}
x<-1\\
2-x+2(4-x)+x+1\le 0
\end{cases}
$$
$$
\begin{cases}
-1\le x<2\\
2-x+2(4-x)-x-1\le 0
\end{cases}
$$
$$
\begin{cases}
2\le x<4\\
x-2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1428567",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
Calculating the $n^\text{th}$ derivative How do we calculate the $n^{\text{th}}$ derivative for
$$
\frac{x^3}{(x-a)(x-b)(x-c)}?
$$
How can I obtain the partial fraction for the given term?
| Using Partial fraction
Above Given fraction is a Improper fraction then by actual division, Since same degree in above and below.
I.e $\displaystyle \frac{x^3}{x\cdot x \cdot x}$
Then $$\displaystyle \frac{x^3}{(x-a)(x-b)(x-c)}= 1+\frac{f(x)}{(x-a)(x-b)(x-c)}$$
Where $f(x)$ is a degree of polynomial $2$
So $$\displays... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1429800",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Solve complex equation $z^4=a^{16}$ Let $a$ be some complex number. I have to solve equation
$$z^4=a^{16}$$
One is tempted to "simplify" it to $z=a^4$, so it is the solutions. But somebody told me, there are more solutions than that. Is it true and why? It seems counterintuitive.
| Note that every $a \in \mathbb{C}$ can be written as $a = re^{i\varphi}$, hence your equation delivers
$$z^4 = r^{16} e^{16i\varphi}$$
The complex roots are given by
$$z^n = re^{i\varphi} \implies z_k = r^{\frac{1}{n}}e^{i\left( \frac{\varphi}{n} + \frac{2\pi k}{n}\right)} \qquad k=0,\dots,n-1$$
So you have $n$ soluti... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1431097",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
All real solution of the equation $2^x+3^x+6^x = x^2$
Find all real solution of the equation $2^x+3^x+6^x = x^2$
$\bf{My\; Try::}$ Let $$f(x) = 2^x+3^x+6^x-x^2\;,$$ Now Using first Derivative
$$f'(x) = 2^x\cdot \ln 2+3^x\cdot \ln 3+6^x\cdot \ln 3-2x$$
Now for $x<0\;,$ We get $f'(x)>0,$ So function $f(x)$ is strictly... | The derivative $f'(x)$ is certainly greater than $h(x)=6^x \ln 6 - 2x.$ [I assume you had a typo when you wrote the term $6^x \ln 3,$ as the derivative of $6^x$ is $6^x \ln 6.$]
Now assume $x \ge 0$ and note $h(0)=\ln 6 >0.$ Also we have
$$h'(x)=6^x (\ln 6)^2 -2 > (3.2)\cdot 6^x -2\ge 1.2 >0,$$
using the underestimate ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1433802",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
Evaluating $\lim _{x\to 1}\left(\frac{\left(2x^2-1\right)^{\frac{1}{3}}-x^{\frac{1}{2}}}{x-1}\right)$ Without L'Hopital or Calculus? What is: $\lim _{x\to 1}\left(\frac{\left(2x^2-1\right)^{\frac{1}{3}}-x^{\frac{1}{2}}}{x-1}\right)$?
Thanks in advance
Much appreciated!
| It mostly depends on what you're allowed to use. In order to do it without recurring to derivatives, I'd split it into two parts:
\begin{align}
\lim _{x\to 1}\biggl(\frac{(2x^2-1)^{1/3}-x^{1/2}}{x-1}\biggr)
&=\lim _{x\to 1}\biggl(\frac{(2x^2-1)^{1/3}-1+1-x^{1/2}}{x-1}\biggr)
\\[6px]
&=\lim_{x\to 1}\left(\frac{(2x^2-1)^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1434258",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
How to find the equations whose roots are equal to the following numbers? How to find the equations whose roots are equal to the following numbers ?
(a) $\sin^2\frac{\pi}{2n+1}$,$\sin^2\frac{2\pi}{2n+1}$,$\sin^2\frac{3\pi}{2n+1}$,...,$\sin^2\frac{n\pi}{2n+1}$
(b)$\cot^2\frac{\pi}{2n+1}$,$\cot^2\frac{2\pi}{2n+1}$,$\cot^... | For $(b),$ see Trig sum: $\tan ^21^\circ+\tan ^22^\circ+...+\tan^2 89^\circ = ?$
For $(a),$
Method $\#1:$ $$\cot^2y=\dfrac{1-\sin^2y}{\sin^2y}$$
Method $\#2:$
Using De Moivre's formula,
$$\cos(2m+1)x+i\sin(2m+1)x=(\cos y+i\sin y)^{2m+1}=\cdots$$
Equating the imaginary parts (See this),
$$\sin(2m+1)x=(2n+1)\sin x-\cd... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1434767",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Why the maximum value of $|AB|+\sqrt{2}|CD|$ is $10$?
Assume a Convex quadrilateral $ABCD$ a,such $|AC|=1,|BC|=3\sqrt{2}$,and $|AD|=|BD|=\frac{\sqrt{2}}{2}|AB|$, find the maximum of the value $|AB|+\sqrt{2}|CD|$
The Book only have answer: $10$.
I wanted to solve this by using law of cosines in triangles $ABC$and $B... | Let $\angle ADC=x$, then $\angle BDC=90^\circ-x$. Apply the law of cosine to $\triangle ADC$ we have
$$2AD\cdot CD\cos x=AD^2+CD^2-1\tag{1}$$
and to $\triangle BCD$ we have
$$2BD\cdot CD \cos (90^\circ-x) = BD^2+CD^2-18\tag{2}.$$
Square both (1) and (2) and add them up we get
$$4AD^2\cdot CD^2 = (AD^2+CD^2-1)^2+(AD^2+C... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1435019",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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What is row reduced echelon form? How to row reduce this matrix? I'm not being able to grasp the concept of row reduced echelon form. Please, explain how to row reduce one of the the following matrices.
$A =
\begin{bmatrix}
1&3&4&5\\3&9&12&9\\1&3&4&1
\end{bmatrix}$
$B=
\begin{bmatrix}
1&2&1&2\\0&1&0&1\\-1&2&0&3... | You begin to put the matrix in echelon form, with the pivots equal to $1$, going downwards row after row. When that is done, all coefficients under a pivot (in the same column) are equal to $0$.Then you make the coefficients above the pivots equal to $0$, starting fom the last pivot and going upwards:
\begin{align*}
&\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1436191",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Maclaurin expansion of $y=\frac{1+x+x^2}{1-x+x^2}$ to $x^4$
Maclaurin expansion of $$\displaystyle y=\frac{1+x+x^2}{1-x+x^2}\,\,\text{to } x^4$$
I have tried by using Maclaurin expansion of $\frac1{1-x}=1+x+x^2+\cdots +x^n+o(x^n)$, but it seems not lead me to anything.
| The sortest way is division of numerator by the denominator along the increasing powers of $x$:
This division tell us that
\begin{align*}1+x+x^2&=(1-x+x^2)(1+2x+2x^2-2x^4)+x^5(-2+2x)\\
\text{so that}\quad\frac{1+x+x^2}{1-x+x^2}&=1+2x+2x^2-2x^4+\frac{x^5(-2+2x)}{1-x+x^2}\\&=1+2x+2x^2-2x^4+O(x^5).\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1439732",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 4
} |
How is it possible that $\int\frac{dy}{(1+y^2)(2+y)}$ = $\frac{1}{5}\int\frac{dy}{y+2}-\int\frac{ydy}{1+y^2}+\int\frac{2dy}{1+y^2} $? Suppose we have a fraction
$$I=\int\frac{dy}{(1+y^2)(2+y)}$$
How is it possible that
$$5I = \int\frac{dy}{y+2}-\int\frac{ydy}{1+y^2}+\int\frac{2dy}{1+y^2} ?$$
How are they using partial... | $$\begin{align*}
\frac1{y+2}-\frac{y}{1+y^2}+\frac2{1+y^2}&=\frac1{y+2}+\frac{2-y}{1+y^2}\\
&=\frac{(1+y^2)+(4-y^2)}{(y+2)(1+y^2)}\\
&=\frac5{(y+2)(1+y^2)}
\end{align*}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1442739",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Complex numbers - locus of a point Question:
If $z \neq 1$ and ${z^2} \over {z-1}$ is real, then find the locus of the point represented by the complex number $z$.
I'm not sure how to approach this question. I attempted to substitute $z = x + iy$, however, that didn't solve the problem. It's quite clear, by observat... | Let $\displaystyle z=x+iy$ in $\displaystyle \frac{z^2}{z-1} = \frac{(x+iy)^2}{x+iy-1} = \frac{x^2-y^2+2ixy}{(x-1)+iy}\times \frac{(x-1)-iy}{(x-1)-iy}$
So we get $\displaystyle \frac{(x^2-y^2)(x-1)+2xy^2+i\left[-y(x^2-y^2)++2xy(x-1)\right]}{(x-1)^2+y^2}$
$\displaystyle = \frac{(x^2-y^2)(x-1)+2xy^2}{(x-1)^2+y^2}+i\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1443122",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Solving $\int_0^{\frac{1}{2}} x^{\frac{3}{2}} (1-2x)^{\frac{3}{2}}dx$? How to solve the definite integral $\int_0^{\frac{1}{2}} x^{\frac{3}{2}} (1-2x)^{\frac{3}{2}}~dx$?
Funnily enough, this is actually a problem that showed up during my real analysis course. At first glance, the problem seemed solvable by using knowle... | $$\int_0^{\frac{1}{2}} x^{\frac{3}{2}} (1-2x)^{\frac{3}{2}}dx$$
Substitute $u = \sqrt{x}$
$$ = 2\int_0^{\frac{1}{\sqrt{2}}} u^4 (1-2u^2)^{\frac{3}{2}}dx$$
Substitute u = $\frac{\sin(s)}{\sqrt{2}}$, $du = \frac{\cos(s)}{\sqrt{2}}$ (this will be valid for $0 < s < \frac{\pi}{2}$)
$$ = \sqrt{2}\int_0^{\frac{\pi}{2}} \frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1443182",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
If $\frac{x^2+y^2}{x+y}=4$, then what are the possible values of $x-y$?
If $\frac{x^2+y^2}{x+y}=4$,then all possible values of $(x-y)$ are given by
$(A)\left[-2\sqrt2,2\sqrt2\right]\hspace{1cm}(B)\left\{-4,4\right\}\hspace{1cm}(C)\left[-4,4\right]\hspace{1cm}(D)\left[-2,2\right]$
I tried this question.
$\frac{x^2+y^2... | Given $$\displaystyle \frac{x^2+y^2}{x+y} = 4\Rightarrow x^2+y^2 = 4x+4y$$
So we get $$x^2-4x+4+y^2-4y+4 = 8\Rightarrow (x-2)^2+(y-2)^2 = (2\sqrt{2})^2$$
Now Put $$x-2 = 2\sqrt{2}\cos \phi\Rightarrow x = 2+2\sqrt{2}\cos \phi$$
and $$y-2 = 2\sqrt{2}\sin \phi\Rightarrow y = 2+2\sqrt{2}\sin \phi$$
So $$\displaystyle x-y =... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1443441",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 0
} |
Can't understand this pseudo-inverse relation. In the Answer to a different Question, a curious matrix relation came up:
M is symmetric and non-singular, G is non-symmetric and singular.
Theorem:
When $M$ is positive/negative definite, or more generally, if $G$ and $(G^\dagger G) M (G^\dagger G)$ have the same ranks th... | Primary matrices
The matrix
$$
\mathbf{M} =
\left[
\begin{array}{ccc}
1 & 2 & 3 \\
2 & 4 & 5 \\
3 & 5 & 6 \\
\end{array}
\right]
$$
has rank $\rho_{M} = 3$. The matrix
$$
\mathbf{G} =
\left[
\begin{array}{ccc}
1 & 0 & 0 \\
2 & 0 & 0 \\
0 & 0 & 0 \\
\end{array}
\right]
$$
has rank $\rho_{G} = 2$.
To make ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1444777",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
$\int\frac{x dx}{\sqrt{x^4+4x^3-6x^2+4x+1}}$ $\int\frac{x dx}{\sqrt{x^4+4x^3-6x^2+4x+1}}$
I was given this question by my senior.I tried to solve it but could not reach the answer.
Let $I= \int\frac{x dx}{\sqrt{x^4+4x^3-6x^2+4x+1}} $
$I=\int\frac{dx}{\sqrt{x^2+4x-6+\frac{4}{x}+\frac{1}{x^2}}}$
Then after repeated attem... | Note that:
$$(x+1)^4=x^4+4x^3+6x^2+4x+1$$
Then
$$(x+1)^4-12x^2=x^4+4x^3-6x^2+4x+1$$
$$(x+1)^4-12x^2=12x^2((\frac{(x+1)^2}{\sqrt{12}x})^2-1)$$
Let us define
$$f(x)=\frac{(x+1)^2}{\sqrt{12}x}$$
And derive it:
$$f'(x)=\frac{2(x+1)x-(x+1)^2}{12x^2}=\frac{x^2-1}{12x^2}$$
Now lets have a look at
$$[arcsin(f(x))]'=\frac{f'(x)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1445369",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Show that $5\cos^2x - 2\sqrt{3}\sin x\cos x + 3\sin^2x = \cos 2x - \sqrt{3}\sin2x + 4$ Show that $5\cos^2x - 2\sqrt{3}\sin x\cos x + 3\sin^2x = \cos 2x - \sqrt{3}\sin2x + 4$
Attempt.
I'm preparing for a trigonometry examination tomorrow. So I'm solving as many questions as possible. This is confusing. I've tried solvin... | Notice, the following formula
$$\cos^2A+\sin^2 A=1$$
$$2\sin A\cos A=\sin 2A$$
$$\cos^2 A-\sin^2 A=\cos 2A$$
Now, we have
$$LHS=5\cos^2 x-2\sqrt 3\sin x\cdot \cos x+3\sin^2x$$
$$=4\cos^2 x+\cos^2 x-\sqrt 3(2\sin x\cdot \cos x)+4\sin^2x-\sin^2 x$$
$$=4(\cos^2 x+\sin^2x)-\sqrt 3\sin 2x+(\cos^2 x-\sin^2 x)$$
$$=4(1)-\sqr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1456576",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Calculating $\lim_{x\to 0}\frac{x^2+x\cdot \sin x}{-1+\cos x}$ Without L'Hopital,
$$\lim_{x\to 0}\frac{x^2+x\cdot \sin x}{-1+\cos x}$$
That's
$$\frac{x^2+x\cdot \sin x}{-1+\left(1-2\sin^2\frac{x}{2}\right)} = \frac{x^2+x\cdot \sin x}{-2\sin^2\frac{x}{2}}$$
Let's split this
$$\frac{x\cdot x}{-2\cdot\sin\frac{x}{2}\cdot ... | You did this in a needlessly elaborate way.
$$\frac{x^{2}+x\sin x}{-1+\cos x}\approx\frac{x^{2}+x\left(x-x^{3}/6\right)}{-1+\left(1-x^{2}/2\right)}=\frac{2x^{2}-x^{4}/6}{-x^{2}/2}=-4$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1457217",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Hyperbolic Trig Inequality The following hyperbolic trig inequality came up.
$$0 \leq y \leq x \leq 2 \implies \sinh(x)-\sinh(y) \leq \sinh(x-y)\cdot e^{xy/2}.$$
I spent many hours trying to prove it. The first few terms of the Taylor series work out, but I couldn't get a general proof. I passes the usual sanity checks... | Completing Brevan Ellefsen's solution.
First, since
$\sinh x - \sinh y = 2 \sinh \frac{x-y}{2} \cosh \frac{x+y}{2}$,
and
$\sinh (x-y) = 2 \sinh \frac{x-y}{2} \cosh \frac{x-y}{2}$,
we have
$$
r\equiv\frac{\sinh x - \sinh y}{\sinh (x-y)}
= \frac{\cosh \frac{x+y}{2} } {\cosh \frac{x-y}{2}}
= \frac{1 + \tanh\frac x 2 \tanh... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1460482",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
guess the color of the next card, what is the payoff of this game? There are 26 red cards and 26 black cards on the table which are randomly shuffled and are facing down onto the table. The host turns up the cards one at a time. You can stop the game any time (even at the beginning of the game). Once you stops the gam... | [Thanks to Brian Scott for pointing out my original misinterpretation of this problem.]
The value of the game is zero.
Let $f(r,b)$ denote the value of the game if there are $r$ red cards and $b$ black cards remaining. If we stop at that point, the expected value is $(r-b)/(r+b)$; but as we also have the option of con... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1461915",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"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.