Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Find $p$ for which all solutions of system/equation are real There is system of $5$ equations
$$
a+b+c+d+e = p; \\
a^2+b^2+c^2+d^2+e^2 = p; \\
a^3+b^3+c^3+d^3+e^3 = p; \\
a^4+b^4+c^4+d^4+e^4 = p; \\
a^5+b^5+c^5+d^5+e^5 = p, \\
\tag{1}
$$
where $p\in\mathbb{R}$.
One can prove (like here) that $a,b,c,d,e$ are roots of eq... | One can also use inequalities to give some restrictions on the solution.
We have $$ \sum a^2(a-1)^2 = \sum a^2 - 2 \sum a^3 + \sum a^4 = p - 2p + p = 0 $$ any solution $(a,b,c,d,e)$. Since $x^2(x-1)^2 \ge 0$ for all real $x$, each of $a, b, c, d, e$ must be either $0$ or $1$. Obviously, they cannot be all distinct.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1207678",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Finding a recurrence relation, first few terms of power series solution to differential equation I'm attempting to find a recurrence relation and the first few terms of a power series solution for the differential equation:
$$(1-x^2)y'' - 2xy' + \lambda y = 0$$
Where $\lambda$ is some integer.
So I've assumed a soluti... | For the differential equation $(1-x^{2}) y'' - 2 x y' + \lambda y = 0$ it is seen that if $y$ is of the form $\sum a_{n} x^{n}$ the following holds.
\begin{align}
y(x) &= \sum_{n=0}^{\infty} a_{n} \, x^{n} \\
y'(x) &= \sum_{n=0}^{\infty} n \, a_{n} \, x^{n-1} \\
y''(x) &= \sum_{n=0}^{\infty} n(n-1) \, a_{n} \, x^{n-2}
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1212451",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Problem in permutation groups involving conjugates I have to find a permutation $a$ satisfying $ a xa^{-1}=y$ where $ x=(12) (34)$ and $y=(56) (13)$
My attempt in solving the problem was-
$$ a(12)(34)a^{-1}= a(12)(a^{-1}a)(34)a^{-1}= (a(12)a^{-1})(a(34)a^{-1})=(56)(13)$$
$$\implies (a(1) \ a(2)) (a(3) \ a(4))= (56)(1... | Your solution is correct! One way to confirm this would be to write out
$$
a^{-1}=
\begin{pmatrix}
5 & 6 & 1 & 3& 4& 2\\1 & 2 & 3 & 4 & 5 & 6
\end{pmatrix}
$$
and
$$
x=(12)(34)
$$
and
$$
a=
\begin{pmatrix}
1 & 2 & 3 & 4 & 5 & 6\\5 & 6 & 1 & 3& 4& 2
\end{pmatrix}
$$
and simply follow the "route" of each element:
$$
5\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1214267",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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Proving $(n+1)!>2^{n+3}$ for all $n\geq 5$ by induction I am stuck writing the body a PMI I have been working on for quite some time.
Theorem: $∀n∈N ≥ X$, $(n+1)!>2^{n+3}$
I will first verify that the hypothesis is true for at least one value of $n∈N$.
Consider $n=3$: (not valid)
$$(3+1)!>2^{3+3} \implies 4!>2^{6} \imp... | Your first two steps are correct. You have established that:$$(n+1)!\gt2^{n+3}$$is first true when $n=5$. You then correctly assumed that it holds for some $n=k\ge5$, giving you:$$(k+1)!\gt2^{k+3}\tag{1}$$And, finally you also proceeded in the right direction by then saying that you now need to prove that this implies ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1214914",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 0
} |
Find minimum of $P=\frac{\sqrt{3(2x^2+2x+1)}}{3}+\frac{1}{\sqrt{2x^2+(3-\sqrt{3})x +3}}+\frac{1}{\sqrt{2x^2+(3+\sqrt{3})x +3}}$ For $x\in\mathbb{R}$ find minimum of $P$.
$P=\dfrac{\sqrt{3(2x^2+2x+1)}}{3}+\dfrac{1}{\sqrt{2x^2+(3-\sqrt{3})x +3}}+\dfrac{1}{\sqrt{2x^2+(3+\sqrt{3})x +3}}$
Source : Viet Nam national test for... | note when $x=0$, the three items are equal which hints the min is $\sqrt{3}$.
first take $\dfrac{1}{\sqrt{a}}+\dfrac{1}{\sqrt{b}} \ge \sqrt{\dfrac{8}{a+b}}$ which is easy to prove by AM-GM.
$P \ge \dfrac{\sqrt{3(2x^2+2x+1)}}{3}+\sqrt{\dfrac{8}{4x^2+6x+6}} \ge \dfrac{\sqrt{3(2x^2+2x+1)}}{3}+\sqrt{\dfrac{8}{6x^2+6x+6}} $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1215784",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Residue at essential singularity I need a little help with the following problem. I've tried many ways, but i didnt succeed. I think there needs to be a trick or something, some transformation. The task is to find the residue of the function at its singularity e.g. z=-3
\begin{equation}
f(z)=\cos\left(\frac{z^2+4z-1}{z... | $$
\begin{align}
&\cos\left(\frac{z^2+4z-1}{z+3}\right)\\
&=\cos\left((z+3)-2-\frac4{z+3}\right)\\
&=\cos\left((z+3)-\frac4{z+3}\right)\cos(2)+\sin\left((z+3)-\frac4{z+3}\right)\sin(2)\tag{1}
\end{align}
$$
Since $\cos\left((z+3)-\frac4{z+3}\right)$ is an even function of $z+3$, its residue at $z=-3$ is $0$.
The $(z+3)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1217126",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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limit of $\frac{xy-2y}{x^2+y^2-4x+4}$ as $(x,y)$ tends to $(2,0)$ Am I able to substitute $x$ by ($k$+$2$) with $k$ tending to $0$, then using polar coordinates to deduce its limit?!
\begin{equation*}
\lim_{(x,y)\to(2,
0)}{xy-2y\over x^2+y^2-4x+4}.
\end{equation*}
| Set $z=x-2$ then $z\to 0$ as $x\to 2$ and
$\lim\limits_{(x,y)\to (2,0)}\frac{(x-2)y}{(x-2)^2+y^2}=\lim\limits_{(z,y)\to (0,0)}\frac{zy}{z^2+y^2}=\lim\limits_{(my,y)\to(0,0)}\frac{my^2}{(1+m^2)y^2}=\frac{m}{(1+m^2)}$
Thus there is no limit.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1218520",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
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Proving $ 1+\frac{1}{4}+\frac{1}{9}+\cdots+\frac{1}{n^2}\leq 2-\frac{1}{n}$ for all $n\geq 2$ by induction Question:
Let $P(n)$ be the statement that $1+\dfrac{1}{4}+\dfrac{1}{9}+\cdots +\dfrac{1}{n^2} <2- \dfrac{1}{n}$. Prove by mathematical induction. Use $P(2)$ for base case.
Attempt at solution:
So I plugged in $... | Hint: You assume that the statement is true for $n=k$. In other words,
$$
1+\frac{1}{4}+\frac{1}{9}+\cdots+\frac{1}{k^2}<2-\frac{1}{k}.
$$
Now, add $\frac{1}{(k+1)^2}$ to both sides to get
$$
1+\frac{1}{4}+\frac{1}{9}+\cdots+\frac{1}{k^2}+\frac{1}{(k+1)^2}<2-\frac{1}{k}+\frac{1}{(k+1)^2}.
$$
What you would really like... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1220203",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 3
} |
Norm of the sum of two vectors This problem has two parts.
Part a): $x$ and $y$ are vectors. If $||x|| = 7, ||y|| = 11$, what is the smallest value possible for $||x+y||$? (Note: the || || denotes the norm of a vector).
This is what I have tried so far: I put vector $x$ equal to $\begin{pmatrix} a \\ b \end{pmatrix}$... | for the second part it is best to use vectors
$|\vec{x} + \vec{y}|^2 = (\vec{x} + \vec{y})\cdot ( \vec{x} + \vec{y})=
|\vec{x}|^2 + |\vec{y}|^2 +2\vec{x} \cdot \vec{y}$
solve the equation
$|7|^2 =
|5|^2 + |4|^2 +2(\vec{x} \cdot \vec{y})$
for $\vec{x} \cdot \vec{y}$
which you can use to evaluate
$|2\vec{x} -3 \vec{y}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1221766",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 2
} |
$ax^2+bx+c=0$ has roots $x_1,x_2$. what are the roots of $cx^2+bx+a=0$. Given solution: Dividing the first equation by $x^2$ we get
$c(\frac{1}{x^2})+b(\frac{1}{x})+a=0$
so $(\frac{1}{x_1}),(\frac{1}{x_2})$ are the roots of $cx^2+bx+a=0$.{How?It is not obvious to me.}
The answers so far are proving retrospectively that... | Another way to see it is to look at sum and product of roots. $x_1$ and $x_2$ are roots of $ax^2+bx+c=0$ so $x_1+x_2=-\frac{b}{a}$ and $x_1\cdot x_2=\frac{c}{a}$.
$$\frac{a}{c}=\frac{1}{x_1}\cdot \frac{1}{x_2}$$
$$\frac{1}{x_1}+\frac{1}{x_2}=\frac{x_1+x_2}{x_1x_2}=-\frac{b}{c}$$
and this shows that $\frac{1}{x_1}$ and ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1222799",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove that if $k \in \mathbb{N}$, then $k^4+2k^3+k^2$ is divisble by $4$ I am trying to solve by induction and have established the base case (that the statement holds for $k=1$).
For the inductive step, I tried showing that the statement holds for $k+1$ by expanding $(k+1)^4+2\cdot(k+1)^3+(k+1)^2$, but this equals $16... | $\underline{\text{Proof by induction:}}$
First, show that this is true for $k=1$:
$1^4+2\cdot1^3+1^2=4$
Second, assume that this is true for $k$:
$k^4+2k^3+k^2=4n$
Third, prove that this is true for $k+1$:
$(k+1)^4+2(k+1)^3+(k+1)^2=$
$4(k+1)^3+\color{red}{k^4+2k^3+k^2}=$
$4(k+1)^3+\color{red}{4n}=$
$4[(k+1)^3+n]$
Pleas... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1228643",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 3
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Maximum value of $ x^2 + y^2 $ given $4 x^4 + 9 y^4 = 64$ It is given that $4 x^4 + 9 y^4 = 64$.
Then what will be the maximum value of $x^2 + y^2$?
I have done it using the sides of a right-angled triangle be $2x , 3y $ and hypotenuse as 8 .
| Since you tagged Lagrange Multipliers, I will use that method.
Allow that $F(x,y) = 4x^4+9y^4-64$ and $G(x,y)=x^2+y^2-C$, where $C$ is treated as a constant and is the maximum you wish to find.
We allow that $\nabla F(x,y) = \lambda \nabla G(x,y)$ and find two new equations:
$16x^3 = \lambda 2x$ and
$36y^3 = \lambda 2y... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1234320",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 3
} |
Difficulty turning a quadratic equation to "vertex"-form I'm having difficulty reducing a quadratic equation to its "vertex-form" by following my textbook and nearly every tutorial I can find online.
The starting equation is:
$$f(x) = -2x^2 + 16x - 24$$
Next, I divided each term by $-2$ (to leave the first term as $x^2... | You made your error when you divided by $-2$. When you divided $f(x) = -2x^2 + 16x - 24$ by $-2$, you obtained $\color{red}{f(x) = x^2 - 8x + 12}$. The expression in red is not equal to the original function since the $y$-intercept of the original function is $-24$, while the $y$-intercept of the function in red is $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1234579",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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linear mapping question If $f$ is a linear mapping from $R^3$ $\rightarrow$ $R^3$, and $fof$=$2$$f$.
Suppose that $v$ $\in$ $Imf$
Can i say that $fof$($v$)=$2$$f$($v$) $\Rightarrow$ $fof$($v$)=$f$($2$$v$) $\Rightarrow$ $f^{-1}$($f$($f$($v$))=$f^{-1}$($f$($2$$v$)) $\Rightarrow$ $f$($v$)=$2$$v$ ? i know the other way by ... | Take any basis in $\Bbb R^3$ and write the matrix $A$ of the map $f$ in the basis. YOu initial condition essentially says that $$A^2=2A.$$This implies that the eigenvalues of $A$ lie in the set $\{0,2\}$.
Without losing generality we can suppose that $A$ is already in its Jordan normal form. There are several possibili... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1235908",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove (or disprove): $a\times b=c\times d$, the solutions for $x$ in the equation $\frac {a^x+b^x}{c^x+d^x} = \frac{a+b}{c+d}$ is only $\pm 1$. Prove (or disprove): If $a,b,c,d$ are positive real numbers with $a\times b=c\times d$, then the only solutions for $x$ in the equation $$\frac {a^x+b^x}{c^x+d^x} = \frac{a+b}{... | Clearly, if $a = c, b = d$ or $a = d, b = c$, then all $x$ satisfy the equation. Hence the statement is false. However, one can show that this is the only exception, namely, if $a,b,c,d > 0$ such that $ab = cd$ and $\{a,b\}\neq\{c,d\}$, then $x = \pm 1$ is the only solution to $$\frac{a^x+b^x}{a+b} = \frac{c^x+d^x}{c+d... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1237382",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 1,
"answer_id": 0
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Find $\lim\limits_{x \to 0} \frac{(1+3x)^{1/3}-\sin(x)-1}{1-\cos(x)}$ I would like to find using Taylor series :
$$\lim\limits_{x \to 0} \frac{(1+3x)^{1/3}-\sin(x)-1}{1-\cos(x)}$$
So I compute the taylor series of the terms at the order $1$ :
$(1+3x)^{1/3}=1+x+o(x)$ and $-\sin x -1=-1-x+o(x)$ and $1-\cos(x)$ does not h... | I will show another way to compute this limit. Let us transform the original expression as follows:
\begin{eqnarray*}
\frac{(1+3x)^{\frac{1}{3}}-1-(\sin x)}{1-(\cos x)} &=&\frac{\left( (1+3x)^{%
\frac{1}{3}}-1-x\right) -\left( \sin x-x\right) }{1-(\cos x)} \\
&=&\frac{9\times \left( \frac{(1+3x)^{\frac{1}{3}}-1-\frac{1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1237934",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
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solving given linear equation So before you guys judge me, I honestly am so clueless with this so please bare with my dumb questions.
I have been at this equation that I've been going for an hour now,
$\frac{(3x-1)}{2} -2 = 5-\frac{4x}{3+1}$ and I'm trying to solve for $x$.
So I go with
$6\cdot(3x-1)/2-2=(5-4x)/3 +1$ ... | Your Equation looks like:
$$\frac {3x+1} {2} - 2 = 5-\frac{4x}{3+1}$$
Adding in the denominator on the right.
$$\frac {3x+1} {2} - 2 = 5-\frac{4x}{4}$$
Which simplifies to
$$\frac {3x+1} {2} - 2 = 5-x$$
We want to eliminate the denominator so we multiply by $2$.
$$ 2\frac {3x+1} {2} - (2)2 = (2)5-(2)x$$
Simplifying
$$3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1238142",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Prove by mathematical induction: $\frac{1}{n}+\frac{1}{n+1}+\dots+\frac{1}{n^2}>1$ Could anybody help me by checking this solution and maybe giving me a cleaner one.
Prove by mathematical induction:
$$\frac{1}{n}+\frac{1}{n+1}+\dots+\frac{1}{n^2}>1; n\geq2$$.
So after I check special cases for $n=2,3$, I have to prove... | Here's a "cleaner solution" that's an improvement of the result, though it doesn't use Induction.
Strengthened claim:
$$ \frac{1}{n} + \frac{1}{n+1} + \ldots + \frac{1}{3n-2} \geq 1.$$
Proof: Since $ \frac{ 1}{ 2n-1 - k} + \frac{1}{2n-1 +k} = \frac{2(2n-1)}{(2n-1 - k)(2n-1+k)} \geq \frac{2(2n-1)}{(2n-1)^2} = \frac{2}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1239518",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 6,
"answer_id": 5
} |
Simultaneous Quadratic Equations: $x^2 + y ^ 2 - 2 x + 6y - 35 = 0$ and $2x + 3y = 5$ I've been given the task to simultaneously solve:
$$x^2 + y ^ 2 - 2 x + 6y - 35 = 0$$
$$2x + 3y = 5$$
I've tried applying the substitution method by reordering the second equation to both $x$ and $y$, but both times I have got a surd.... | From the second equation, we have that $x=\frac{5-3y}{2}$. Plugging this into the first equation,
$$\left(\frac{5-3y}{2}\right)^2+y^2-(5-3y)+6y-35=0\\
\frac{9y^2-30y+25}{4}+y^2+9y-40=0$$
Multiplying by $4$,
$$9y^2-30y+25+4y^2+36y-160=0\\
13y^2+6y-135=0$$
Applying the quadratic formula,
$$y=\frac{-6\pm\sqrt{6^2-4(13)(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1242861",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
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Method to integrate $\cos^4(x)$ Here my attempts for integrating $\cos^4(x)$ in few methods.
1st method.
$(\cos^2x)^2=(\frac{1}{2})^2(1+\cos2x)^2$
$=\frac{1}{4}(1+2\cos2x+\cos^22x)=\frac{1}{4}(1+2\cos2x)+\frac{1}{4}(\cos^22x)$
$=\frac{1}{4}(1+2\cos2x)+\frac{1}{4}(\frac{1}{2}(1+\cos4x))$
$=\frac{3}{8}+\frac{1}{2}\cos2... | I think you gave the most straightforward methods, including the sure-to-success replacement by $e^{ix}+e^{-ix}$. Basically any other method is going to be a more obfuscated form of this one.
You could also give a shot to the rational parametrization of the unit circle, by using $\cos x = \frac{1-t^2}{1+t^2}$ and $dx =... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1247913",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Evaluating a double integral with Fubini's theorem We want to evaluate the double integral $$ \int \int_{R} f(x,y) ~ \text{d}x \text{d}y $$ over the rectangle $R$ given by $0 \leq x \leq 2, 0 \leq y \leq 1$, with the function $f$ defined by $$ f(x,y) = \dfrac{xy(x^2 - y^2)}{(x^2 + y^2)^3} \quad \text{ for } (x, y) \neq... | using polar coordinates f(x,y)=f(r,a)=(sin4a)/r^2 if a is not zero ans 0 otherwise.If we take the ray a=pi/8 we can see that f is not bounded near the origen and so it is not continuous: f(0,0)=0
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1248105",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Is this a good proof of the binomial identity? Prove that the binomial identity ${n\choose k} = {n-1\choose k-1} + {n-1\choose k}$ is true using the following expression: $(1+x)^n = (1+x)(1+x)^{n−1}$ and the binomial theorem.
What I have:
We know from the binomial theorem that:
$$(x+1)^n= {n\choose 0} x^0 + {n\choose 1... | Your proof can be slightly modified as follows:
Note that
$x(x+1)^{n-1}=x^n+{n-1 \choose 1}x^{n-1}+{n-1 \choose 2}x^{n-2}+...+{n-1 \choose n-2}x^2+x$
$(x+1)^{n-1}=x^{n-1}+{n-1 \choose 1}x^{n-2}+{n-1 \choose 2}x^{n-3}+...+{n-1 \choose n-2}x+1$
Add the 2 equations and look at the addition of like terms.
$(x+1)^n=x^n + (... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1248296",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 2
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How to Find the Function of a Given Power Series? (Please see edit below; I originally asked how to find a power series expansion of a given function, but I now wanted to know how to do the reverse case.)
Can someone please explain how to find the power series expansion of the function in the title? Also, how would you... | $$\frac{1}{1-x}=1+x+x^2+x^3+x^4+...$$
$$x(\frac{1}{1-(x+x^3)})=x(1+(x+x^3)+(x+x^3)^2+(x+x^3)^3+.....)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1249107",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 0
} |
Could you explain the expansion of $(1+\frac{dx}{x})^{-2}$? Could you explain the expansion of $(1+\frac{dx}{x})^{-2}$?
Source: calculus made easy by S. Thompson.
I have looked up the formula for binomial theorem with negative exponents but it is confusing. The expansion stated in the text is:
$$\left[1-\frac{2\,dx}{x}... | Consider the expression $$A=(1+a)^{-2}=\frac 1{(1+a)^2}=\frac 1 {1+2a+a^2}$$ and, hoping you already know it, perform the long division. Limiting to first terms, you will arrive to $$A=1-2 a+3 a^2-4 a^3+5 a^4+\cdots$$ Now, replace in this last expression $a$ by $\frac{dx}x$; this will lead to $$A=1 - 2\left(\frac{dx}{x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1250593",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Precalculus unit circle with imaginary axis. (a) Suppose $p$ and $q$ are points on the unit circle such that the line through $p$ and $q$ intersects the real axis. Show that if $z$ is the point where this line intersects the real axis, then $z = \dfrac{p+q}{pq+1}$.
(b) Let $P_1 P_2 \dotsb P_{18}$ be a regular 18-gon. S... | There probably is an easy, intuitive way to solve part (a), but apparently neither of us has found it. Therefore we try the straightforward but tedious ways: analytic geometry for the intersection of the line and the $x$-axis, and real- and imaginary-parts of the complex numbers for the expression.
Let $p=a+bi$ and $q=... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
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Proving that $\frac{1}{n}+\frac{1}{n+1}+\cdots+\frac{1}{2n}>\frac{13}{24}$ by induction. Where am I going wrong? I have to prove that
$$\frac{1}{n}+\frac{1}{n+1}+\dots+\frac{1}{2n}>\frac{13}{24}$$ for every positive integer $n$.
After I check the special cases $n=1,2$, I have to prove that the given inequality holds fo... | Let
$$ A_n = \frac{1}{n}+\frac{1}{n+1}+\ldots+\frac{1}{2n}=H_{2n}-H_{n-1}.\tag{1}$$
Then:
$$ A_n-A_{n+1} = \frac{1}{n}-\frac{1}{2n+1}-\frac{1}{2n+2}>0 \tag{2} $$
hence the sequence $\{A_n\}$ is decreasing and to prove our claim it is enough to show that:
$$ \lim_{n\to +\infty}A_n >\frac{13}{24}=0.541666\ldots.\tag{3}$$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1252576",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 4,
"answer_id": 2
} |
Evaluate the integral $\int \frac{x}{(x^2 + 4)^5} \mathrm{d}x$
Evaluate the integral $$\int \frac{x}{(x^2 + 4)^5} \mathrm{d}x.$$
If I transfer $(x^2 + 4)^5$ to the numerator, how do I integrate?
| with $u = x^2 + 4$:
$$ \int (x^2 + 4)^{-5 } {x dx}
= \frac 12 \int u^{-5} {du}
= - \frac 1{8} u^{-4}
= - \frac 1{8} (x^2 + 4)^{-4}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1252652",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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Prove that $f : [-1, 1] \rightarrow \mathbb{R}$, $x \mapsto x^2 + 3x + 2$ is strictly increasing.
Prove that $f : [-1, 1] \rightarrow \mathbb{R}$, $x \mapsto x^2 + 3x + 2$ is strictly increasing.
I do not have use derivatives, so I decided to apply the definition of being a strictly increasing function, which should ... | The proof given looks fine to me, modulo my comment (see above).
Here's another way to see it, sans calculus:
Choose $x_1, x_2 \in [-1, 1]$ with $x_1 > x_2$. Then
$f(x_1) - f(x_2) = x_1^2 + 3x_1 + 2 - x_2^2 - 3x_2 - 2 = x_1^2 - x_2^2 + 3(x_1 - x_2)$
$= (x_1 + x_2)(x_1 - x_2) + 3(x_1 - x_2) = (x_1 + x_2 + 3)(x_1 - x_2)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1253120",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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"answer_id": 1
} |
Using implicit differentiation with a fraction How do I solve this? What steps? I have been beating my head into the wall all evening.
$$ x^2 + y^2 = \frac{x}{y} + 4 $$
| We can multiply both sides of the equation
$$x^2 + y^2 = \frac{x}{y} + 4 \tag{1}$$
by $y$ to obtain
$$x^2y + y^3 = x + 4y \tag{2}$$
Differentiating equation 2 implicitly with respect to $x$ yields
\begin{align*}
2xy + x^2y' + 3y^2y' & = 1 + 4y'\\
(x^2 + 3y^2 - 4)y' & = 1 - 2xy\\
y' & = \frac{1 - 2xy}{x^2 + 3y^2 - 4} \t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1255368",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Finding the sum of the trigonometric serie: There are two series:
$$1) 1+\dfrac{\cos{x}}{p}+\dfrac{\cos{2x}}{p^2}+...+\dfrac{\cos{nx}}{p^n}=\sum_{k=0}^{n}{\dfrac{\cos{kx}}{p^k}}$$
$$2) \dfrac{\sin{x}}{p}+\dfrac{\sin{2x}}{p^2}+...+\dfrac{\sin{nx}}{p^n}=\sum_{k=0}^{n}{\dfrac{\sin{kx}}{p^k}}$$
Where $p$ its a real constan... | Without complex numbers:
$$C_n=\sum_{k=0}^n\frac{\cos(kx)}{p^k}=\sum_{k=1}^{n+1}\frac{\cos(k-1)x}{p^{k-1}}=\sum_{k=1}^n\frac{\cos(k-1)x}{p^{k-1}}+\frac{\cos nx}{p^n}\\
=p\sum_{k=1}^{n}\frac{\cos kx\cos x+\sin kx\sin x}{p^k}+\frac{\cos nx}{p^n}\\
=p(C_n-1)\cos x+pS_n\sin x+\frac{\cos nx}{p^n}.$$
$$S_n=\sum_{k=0}^n\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1255455",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove that $n^2(n^2+1)(n^2-1)$ is a multiple of $5$ for any integer $n$. Prove that $n^2(n^2+1)(n^2-1)$ is a multiple of $5$ for any integer $n$.
I was thinking of using induction, but wasn't really sure how to do it.
| If $n$ is divisible by five, then $n^2(n^2+1)(n^2-1)$ is divisible by $5$, since $n^2$ is a factor.
Now suppose that $n$ is not divisble by $5$. Then $n=5m+r$ for some $r=1,2,3,4$. Note that $(n^2 + 1)(n^2-1) = n^4-1$.
$$n^4 = (5m+r)^4 = (5m)^4 + 4 (5m)^3r + 6 (5m)^2 r^2 + 4 (5m) r^3 + r^4$$ by the binomial theorem. We... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1257632",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Compare $\sum_{k=1}^n \left\lfloor \frac{k}{\varphi}\right\rfloor$ ... Given two integer sequences
\begin{equation*}
\displaystyle A_n=\sum_{k=1}^n \left\lfloor \frac{k}{\varphi}\right\rfloor ,
\end{equation*}
\begin{equation*}
B_n=\left\lfloor\dfrac{n^2}{2\varphi}\right\rfloor-\left\lfloor \dfrac{n}{2\varphi^2}\right\... | The difference $A_n-B_n$ can be proven to have no fixed upper or lower bound.
Let $S_n = \sum_{k=1}^n \lfloor k\varphi \rfloor$
To calculate $S_n$ let $n'=\lfloor n\varphi\rfloor$ and $T_n = \sum_{k=1}^n \lfloor k\varphi^2 \rfloor$
$\lfloor k\varphi\rfloor$ and $\lfloor k\varphi^2 \rfloor$ are complementary Beatty sequ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1259142",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
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"answer_id": 1
} |
Prove the Apollonius' theorem. Let in a $\Delta ABC$, D is the midpoint of $BC$.Prove that:
$AB^2+AC^2=2CD^2+2AD^2$
MY ATTEMPT :
Given that $BD=DC$ and we construct $E \ such \ that\ AE=EC\implies AC=\frac{EC}{2} \ and \ DE||AB \implies DE=\frac{AB}{2}$
For the $\Delta DEC$ we have $DC^2=DE^2+EC^2 \implies 4DC^2=AC^2+... | Use the cosine rule twice:
$$\cos B=\dfrac {c^2+\frac {a^2}{4}-m^2}{2\times c\times \frac {a}{2}}\implies ac\cos B=c^2+\dfrac {a^2}{4}-m^2$$
$$\cos C=\dfrac {b^2+\frac {a^2}{4}-m^2}{2\times b\times \frac {a}{2}}\implies ab\cos C=b^2+\dfrac {a^2}{4}-m^2$$
Adding the above two and using $a=b\cos C+c\cos B$, we get $$a^2=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1259332",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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What is the value of $\frac{a^3}{a^6+a^5+a^4+a^3+a^2+a+1}$ If $\frac{a}{a^{2}+1} = \frac{1}{3}$ Then find the value of $$\frac{a^3}{a^6+a^5+a^4+a^3+a^2+a+1}$$
Any hints to help me?
| Using $a^2=3a-1$ we can compute $a^3=8a-3,\; a^4=21a-8, ... , a^6=144a-55$ by iteration. Then your term equals
$$
\frac{8a-3}{29(8a-3)}=\frac{1}{29}.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1260592",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Proof that $(n^7-n^3)(n^5+n^3)+n^{21}-n^{13}$ is a multiple of $3$. I proved that $$(n^7-n^3)(n^5+n^3)+n^{21}-n^{13}$$ is a multiple of $3$ through the use of Little Fermat's theorem but i want to know if there exist other proofs(maybe for induction). How can I demonstrate it?
This my proof:
$$n^3(n^4-1)(n^5+n^3)+n^{13... | I will prove the induction method here for $n \geq 0$
Base case : $n=0,1$
Clearly $0$ works.
now $(1^7-1^3)(1^5 + 1^3) + 1^{21} - 1^13 = 0$ which is also a multiple of $3$
Now assume it works for a value $k \geq 0$ that is $$(k^7-k^3)(k^5 + k^3) + k^{21} - k^{13}$$ is a mulitple of $3$
That means that $3m = (k^7-k^3)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1261045",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 4
} |
Find the least nonnegative residue of $3^{1442}$ mod 700 So I have that $700=7\cdot2^2\cdot5^2$ and I got that $3^2\equiv1\pmod2$ so then $3^{1442}\equiv1\pmod2$ also $3^2\equiv1\pmod{2^2}$ so $3^{1442}\equiv1\pmod{2^2}$ which covers one of the divisors of $700$. Im not sure if I'm supposed to use $2$ or $2^2$ and I wa... | Go $\pmod4$, $\pmod7$ and $\pmod{25}$. We have
\begin{align}
3^2 \equiv 1\pmod4\\
3^6 \equiv 1\pmod7\\
3^{20} \equiv 1\pmod{25}
\end{align}
This gives us that
\begin{align}
3^{60} \equiv 1\pmod4\\
3^{60} \equiv 1\pmod7\\
3^{60} \equiv 1\pmod{25}
\end{align}
This means
$$3^{60} \equiv 1\pmod{700}$$
Note that $3^{1442} =... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1263865",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
What is the connection between the discriminant of a quadratic and the distance formula? The $x$-coordinate of the center of a parabola $ax^2 + bx + c$ is $$-\frac{b}{2a}$$
If we look at the quadratic formula
$$\frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$
we can see that it specifies two points at a certain offset from the cen... | For the equation $ax^2 + bx + c = 0$ the second coefficients represents the average of the two roots:
$$ \frac{1}{2}(r_1 + r_2) = -\frac{b}{2a} $$
Instead of using the quadratic formula directly we can just observe that:
$$ (x - r_1)(x - r_2) = x^2 - (r_1 + r_2) + r_1 r_2$$
This also recovers the product mean of the ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1264091",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "19",
"answer_count": 8,
"answer_id": 3
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Ways to place 3 red, 4 blue and 5 green wagons such that no 2 blue wagons were standing next to each other As the title says I need to find the number of ways to to place 3 red, 4 blue and 5 green wagons such that no 2 blue wagons were standing next to each other. The wagons of the same color are considered completely ... | Here is a different approach.
We first line up the three red and five green wagons (leaving enough space between each pair of wagons so that another wagon can be parked between them). This can be done by choosing which three of the eight positions will be reserved for the red wagons, which can be done in $\binom{8}{3}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1264407",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Solving an equation with LambertW function I have function $f(x)$ which has a maximum at $x^*$ ($\frac{d^2f(x)}{dx^2}<0$ can be shown):
$$f(x)=\frac{1}{2} (1-x) \log \left(\frac{2 d g x}{1-x}+1\right)$$
In the research article, the solution for $x^*$ is given as
$$x^*=\frac{(2 d g-1)-Q(2)}{(2 d g-1) \left(Q(2)+1\right... | Start with
$$\exp\left(\frac{2 d g}{2 d g x-x+1}\right)= \left(\frac{2 d g x-x+1}{1-x}\right)$$
Next, divide both sides by $e$ to obtain
$$\exp\left(\frac{(2 d g-1)(1-x)}{2 d g x-x+1}\right)= e^{-1}\left(\frac{2 d g x-x+1}{1-x}\right)$$
Then, multiply both sides by $2dg-1$ to reveal
$$ze^z=(2dg-1)/e$$
where $z=\frac{(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1267582",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
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How to prove a Fibonacci inequality using Strong Induction? Using strong induction I am trying to prove that
$$F_n \geq \left(\frac{1+\sqrt{5}}{2}\right)^{n-2} \text{ for all } n \geq 2$$
for the Fibonacci Sequence defined by: $F_0 = 0$, $F_1 = 1$, and $F_n = F_{n-1} + F_{n-2}$ for $n \geq 2$.
I know how to do strong ... | \begin{align}
F_n & = F_{n-1}+F_{n-2} \\[8pt]
& \ge \left(\frac{1+\sqrt{5}}{2}\right)^{n-3} + \left( \frac{1+\sqrt{5}}{2} \right)^{n-4} & & \text{(by the induction hypothesis)} \\[8pt]
& = \left(\frac{1+\sqrt{5}}{2}\right)^{n-4} \left( \frac{1+\sqrt{5}}{2} + 1 \right) & & \text{(Here we just pulled out a common factor.... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1269082",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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Solvability of $a_{1}^{2} = a_{2}^{n} + a_{3}^{n} + a_{4}^{n}$ when $n \geq 5$ is prime? Aware of a Darmon-Merel theorem that asserts that if $n \geq 5$ is prime then the equation $a_{1}^{2} = a_{2}^{n} + a_{3}^{n}$ has no solution in relatively prime integers $a_{1}, a_{2}, a_{3},$ I wonder if this is also true, under... | For $0<a\leq b\leq c\leq 100$ with $\gcd(a,b,c)=1$, we have the sums including the one by user R. Israel,
$$3^5+ \color{brown}{49}^5+ 69^5 = 42971^2$$
$$20^5+ 68^5+ \color{brown}{81}^5 = 70313^2$$
$$19^5+ 80^5+ 97^5 = 108934^2$$
$$ \color{brown}{16}^5+ 83^5+ 97^5 = 111926^2$$
While your question was on prime $n\geq 5$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1269338",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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Finding this weird limit involving periodic functions with periods 5 and 10. If $f(x)$ and $g(x)$ are two periodic functions with periods 5 and 10 respectively, such that:
$$\lim_{x\to0}\frac{f(x)}x=\lim_{x\to0}\frac{g(x)}x=k;\quad k>0$$
then for $n\in\mathbb N$, the value of :
$$\lim_{n\to\infty}\frac{f(5(4+\sqrt{15})... | The trick is that the conjugates [Galois conjugates, not complex conjugates] of $4+\sqrt{15}$ and $4+\sqrt{14}$ have modulus less than $1$, and if $a,b\in \mathbb{Z}$ then
$$(a+\sqrt{b})^n + (a-\sqrt{b})^n \in \mathbb{Z}$$
for all $n\in \mathbb{N}$.
So we have
\begin{align}
f\bigl(5(4+\sqrt{15})^n\bigr) &= f\Bigl(5\big... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1269729",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
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To find jordan canonical form Which of the following matrices have Jordan canonical form of equal to the $3\times 3$ matrix
$$
\begin{pmatrix}
0 & 1 & 0 \\
0 & 0 & 0 \\
0 & 0 & 0
\end{pmatrix}$$
a)$
\begin{pmatrix}
0 & 0 & 1 \\
0 & 0 & 0 \\
0 & 0 & 0
\end{pmatrix}$
b)$
\begin{pmatrix}
0 & 0 & 1 \\
0 & 0 & 1 \\
0 & 0 ... | Finding the eigenvalues of any of these matrices is not particularly difficult; they are all upper-triangular.
Hint: Suppose that $A$ is a $3 \times 3$ matrix with zero as its only (complex) eigenvalue. Note that $A$ has the desired J-C form if and only if $A \neq 0$ but $A^2 = 0$ (why?).
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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"answer_id": 0
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The limit of this function when x goes to minus infinity? I'm looking for the limit of $$\mathop {\lim }\limits_{x \to - \infty } \left( {\sqrt {1 + x + {x^2}} - \sqrt {1 - x + {x^2}} } \right)$$
I know it should be -1, but for some reason I always get to 1.
I'm not sure where the difference between $$- \infty$$ and ... | The problem is when you move $1/x$ inside the square roots! $1/x$ is negative but pulling it into the square roots like that makes it positive.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1272182",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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How does $-\frac{1}{x-2} + \frac{1}{x-3}$ become $\frac{1}{2-x} - \frac{1}{3-x}$ I'm following a solution that is using a partial fraction decomposition, and I get stuck at the point where $-\frac{1}{x-2} + \frac{1}{x-3}$ becomes $\frac{1}{2-x} - \frac{1}{3-x}$
The equations are obviously equal, but some algebraic mani... | I am a grade 8 student, so I may not be able to explain really well.
First, I need to prove that $-\frac {1} {x-2}=\frac {1} {2-x}$
To prove, let's assume that "$x$" can be any number, for instance, I take $x$=8.
So by substituting,
\begin{align}
-\frac {1} {x-2} & = -\frac {1} {8-2}\\
& = -\frac {1} {6}
\end{align}
An... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1275071",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 7,
"answer_id": 1
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How to compute $\sum_{n=0}^{\infty}{\frac{3^n(n + \frac{1}{2})}{n!}}$? I have to compute the series $\displaystyle\sum_{n=0}^{\infty}{\frac{3^n(n + \frac{1}{2})}{n!}}$.
$$\displaystyle\sum_{n=0}^{\infty}{\frac{3^n(n + \frac{1}{2})}{n!}} = \sum_{n=0}^{\infty}{\frac{3^n\frac{1}{2}}{n!}} + \sum_{n=0}^{\infty}{\frac{3^nn}... | $$\begin{align}\sum\limits_{n=0}^\infty \frac{n\cdot 3^n}{n!}&=0+\sum\limits_{n=1}^\infty \frac{3^n}{(n-1)!}\\&=3\sum\limits_{n=0}^\infty \frac{3^n}{n!}\\&=3\cdot e^3\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1275769",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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$5(a^2+b^2)$ covers all numbers $=a_2^2+a_2^2=b_1^2+b_2^2$? I start by noting that 4a*2b=2a*4b
I write 4a*2b as $((2a+b)+(2a-b))*((2a+b)-(2a-b)) = (2a+b)^2-(2a-b)^2$
I follow a similar principle for 2a*4b which I write as $(a+2b)^2-(a-2b)^2$
to arrive at
$(2a+b)^2-(2a-b)^2= (a+2b)^2-(a-2b)^2$ or
$(2a+b)^2+(a-2b)^2= (... | $$ 13 (a^2 + b^2) = (3a+2b)^2 + (2a-3b)^2 = (3a-2b)^2 + (2a+3b)^2$$
$$ 17 (a^2 + b^2) = (4a+b)^2 + (a-4b)^2 = (4a-b)^2 + (a+4b)^2$$
$$ 29 (a^2 + b^2) = (5a+2b)^2 + (2a-5b)^2 = (5a-2b)^2 + (2a+5b)^2$$
$$ 37 (a^2 + b^2) = (6a+b)^2 + (a-6b)^2 = (6a-b)^2 + (a+6b)^2$$
$$ 41 (a^2 + b^2) = (5a+4b)^2 + (4a-5b)^2 = (5a-4b)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1276217",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Prove series convergence The problem states:
a) If $\sum_1^\infty {a_n}$ converges and ${b_n}=n^\frac{1}{n}{a_n}$, then $ \sum_1^\infty{b_n}$ converges, and
b) If $\sum_1^\infty {a_n}$ converges and ${b_n}=\frac{{a_n}}{(1+|{a_n}|)}$, then $\sum_1^\infty {b_n}$ converges.
For a) We have that ${c_n}=n^\frac{1}{n}$ is a s... | In (a) it's correct with one detail: since $c_n$ is decreasing, you should check whether it is bounded from below.
Problem (b) is a bit funny since it isn't true if we don't assume anything else, e.g. that $\displaystyle \sum_{n=1}^{\infty} a_n$ converges absolutely. Here's a counterexample: $a_n$ is the sequence of al... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1279942",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
} |
An interesting point of a triangle. (Help needed to prove a statement.) Consider a triangle whose sides are segments of $\color{red}{\text{line}}$, $\color{blue}{\text{line}}$, $\color{green}{\text{line}}$ falling in the circum-circle $c$. Let $\color{red}{\text{P}}$,$\color{green}{\text{P}}$, $\color{blue}{\text{P}}$ ... | Any two ideal triangles are congruent, or in other words, any non-degenerate triangle inscribed into the unit circle can be mapped to any other by a projective transformation which fixes the unit circle. So without loss of generality you can restrict your considerations to one special case.
I like coordinates, so I'd s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1280198",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
Minimum value of trigonometric equation Find the minimum value of the expression $$y=\frac{16-8\sin^{2} 2x +8\cos^{4} x}{\sin^{2} 2x} .$$ When I convert the expression completely into $2x$, cross multiply and make the discriminant of the quadratic equation greater than $0$, I get the minimum value $-\infty$. I know it ... | $$8\cos^4x = 8\left(\dfrac{1+\cos (2x)}{2}\right)^2 = 8\left(\dfrac{1+2\cos (2x)+ \cos^2(2x)}{4}\right)=2+4\cos (2x)+2\cos^2(2x) = 2+4t+2t^2, t = \cos (2x) \Rightarrow y = \dfrac{16-8(1-t^2)+2+4t+2t^2}{1-t^2} = \dfrac{10+10t^2+4t}{1-t^2}=f(t), -1 \leq t \leq 1$$. Can you take it from here?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1280639",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Computing a limit similar to the exponential function I want to show the following limit:
$$
\lim_{n \to \infty}
n
\left[
\left( 1 - \frac{1}{n} \right)^{2n}
- \left( 1 - \frac{2}{n} \right)^{n}
\right]
= \frac{1}{e^{2}}.
$$
I got the answer using WolframAlpha, and it seems to be correct numerically, but I ... | We can proceed in the following manner
\begin{align}
L &= \lim_{n \to \infty}n\left\{\left(1 - \frac{1}{n}\right)^{2n} - \left(1 - \frac{2}{n}\right)^{n}\right\}\notag\\
&= \lim_{n \to \infty}n\left(1 - \frac{2}{n}\right)^{n}\left\{\dfrac{\left(1 - \dfrac{1}{n}\right)^{2n}}{\left(1 - \dfrac{2}{n}\right)^{n}} - 1\right\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1282610",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 1
} |
Diagonal-free Sudoku grid I have a Sudoku grid with the property that diagonally adjacent elements are distinct (it is also a torus under the same property).
My question is up to isomorphism, is the grid unique?
Here's the grid:
$$\begin{array}{|c|c|c|c|c|c|c|c|c|}
\hline
6& 5& 7& 3& 4& 2& 1& 9& 8\\ \hline
9& 8& 1& 5&... | In the general case, a grid will be in an isomorphism equivalence class of $9!\cdot 3!^4 \cdot 2$ corresponding to permutations of the symbols, permutations of the rows and columns, and the symmetry of the square which is not already covered by permutation of the rows and columns. I'm only counting $3!^2$ permutations ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1286014",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 1,
"answer_id": 0
} |
Cannot understand an Integral $$\displaystyle \int _{ \pi /6 }^{ \pi /3 }{ \frac { dx }{ \sec x+\csc x } } $$
I had to solve the integral and get it in this form.
My attempt:
$$\int _{ \pi /6 }^{ \pi /3 }{ \frac { dx }{ \sec x+\csc x } } $$
$$=\int _{\frac{\pi}{6}}^{ \frac{\pi}{3}} \dfrac{\sin x \cos x }{ \sin x+\cos ... | In general if you have an integral
$$\int_a^b f(x)dx$$
Upon making the substitution $t=g(x)$ the limits of the integral become $g(a)$ and $g(b)$. In your case you made the substitution $t=\tan \frac x2$ so the lower limit becomes $\tan \frac{\pi}{12}$ and the upper limit becomes $\tan \frac{\pi}{6}$. Thus the integral ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1288034",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
Proving $x\ln(\frac{x}{a})+y\ln(\frac{y}{b})\geq (x+y)\ln(\frac{x+y}{a+b})$
Let a,b,x,y be positive reals. Prove $x\ln(\frac{x}{a})+y\ln(\frac{y}{b})\geq (x+y)\ln(\frac{x+y}{a+b})$
I don't have any olympic background, so I may be missing some standard trick.
The inequality looks closely related to the concavity of $\... | use Cauchy-Schwarz inequality we have
$$\left(\dfrac{x^2}{a}+\dfrac{y^2}{b}\right)(a+b)\ge (x+y)^2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1289020",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Demonstrate that the two formulae for a scalar product are equivalent. In the figure below, three vectors are joined together to form a triangle.
The name of each vector is a single letter in boldface,
each vector is specified by three lengths in an $xyz$ coordinate system,
and vectors $\boldsymbol{a}$ and $\boldsymbol... | Vector $\boldsymbol{a}$ is the sum of vectors $\boldsymbol{b}$ and $\boldsymbol{c}$.
I.e., $\boldsymbol{a} = \boldsymbol{b} + \boldsymbol{c}$. Solving for $\boldsymbol{c}$ we have
\begin{eqnarray*}
\boldsymbol{c} &=& \boldsymbol{a} - \boldsymbol{b} \\
&=& (a_x, a_y, a_z) - (b_x, b_y, b_z) \\
&=& \b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1292089",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Find the generating function of this sequence I need to find the generating function of the sequence $c_n = (a_0, a_1, a_2, \ldots)$, where:
$$a_n =
\begin{cases}
2^{n/2} & \text{if $n$ is even,} \\
1 & \text{if $n$ is odd.}
\end{cases}$$
I have written out the first few terms of the sequence:
$$(1, 1, 2, 1, 4, 1, 8,... | Let us set $A(X)$ the generating function whose coefficients are $(a_n)$. I claim that :
$$A(X)-\frac{1}{1-X}=\sum_{n=1}^{\infty}(2^n-1)X^{2n}=\sum_{n=1}^{\infty}\sum_{k=0}^{n-1}2^kX^{2n} $$
$$\sum_{n=1}^{\infty}\sum_{k=0}^{n-1}2^kX^{2n}=X^2\sum_{n=1}^{\infty}\sum_{k=0}^{n-1}2^kX^{2(n-1)}=X^2\sum_{n=0}^{\infty}\sum_{k=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1292391",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
} |
Proving that $\frac{1}{a^3(b+c)}+\frac{1}{b^3(c+a)}+\frac{1}{c^3(a+b)}\ge\frac32$ using derivatives
Let $a,b,c\in\mathbb{R}^+$ and $abc=1$. Prove that
$$\frac{1}{a^3(b+c)}+\frac{1}{b^3(c+a)}+\frac{1}{c^3(a+b)}\ge\frac32$$
This isn't hard problem. I have already solved it in following way:
Let $x=\frac1a,y=\frac1b,z... | You can use this way to do. Your inequality $L\ge \frac32$ is equivalent to
$$ 2[x^2(x+y)(x+z)+y^2(x+y)(y+z)+z^2(x+z)(y+z)]\ge 3(x+y)(x+z)(y+z). $$
Let
$$ F(x,y,z)=2[x^2(x+y)(x+z)+y^2(x+y)(y+z)+z^2(x+z)(y+z)]-3(x+y)(x+z)(y+z)-\lambda(xyz-1). $$
Then set
$$ \frac{\partial F}{\partial x}=0, \frac{\partial F}{\partial y}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1292759",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
"answer_id": 0
} |
Verifying $\frac{\cos{2\theta}}{1 + \sin{2\theta}} = \frac{\cot{\theta} - 1}{\cot{\theta} + 1}$ My math teacher gave us an equality involving trigonometric functions and told us to "verify" them. I tried making the two sides equal something simple such as "1 = 1" but kept getting stuck. I would highly appreciate if som... | Use
$\cos(2x)=\cos^2(x) - \sin^2(x)$ and
$1+\sin(2x) = 1+2\sin(x)\cos(x) = \sin^2(x)+\cos^2(x)+2\sin(x)\cos(x)=(\sin(x)+\cos(x))^2$
So we have
$$\frac{\cos(2x)}{1+\sin(2x)}=\frac{\cos^2(x) - \sin^2(x)}{(\sin(x)+\cos(x))^2}=\frac{\cos(x) - \sin(x)}{\cos(x) + \sin(x)}=\frac{\cot(x)-1}{\cot(x)+1}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1292982",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 3
} |
About right identity which is not left identity in a ring Let $S$ be the subset of $M_2(\mathbb{R})$ consisting of all matrices of the form
$\begin{pmatrix}
a & a \\
b & b
\end{pmatrix}$
The matrix $\begin{pmatrix}
x & x \\
y & y
\end{pmatrix}$ is right identity in $S$ if and only if $x+y=1$. Fine, I can see that.
B... | Your result:
$\begin{pmatrix}
x & x \\
y & y
\end{pmatrix}\begin{pmatrix}
a & a \\
b & b
\end{pmatrix}=\begin{pmatrix}
x(a+b) & x(a+b) \\
y(a+b) & y(a+b)
\end{pmatrix}$
shows that in $S$ does not exists a left identity.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1293167",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
what is the decomposition of $(x^2+y^2+z^2)(x+y+z)(x+y-z)(-x+y+z)(x-y+z)-8x^2y^2z^2$? I want to have a decomposition of this :
$$(x^2+y^2+z^2)(x+y+z)(x+y-z)(-x+y+z)(x-y+z)-8x^2y^2z^2$$
I have tried all possible calculation which came to my mind,I will describe one of it which is better but no result:
I put $(x+y-z)=a$ ... | Maple does the job by
factor(expand((x^2+y^2+z^2)*(x+y+z)*(x+y-z)*(-x+y+z)*(x-y+z)-8*x^2*y^2*z^2));
$$ - \left( {x}^{2}-{y}^{2}-{z}^{2} \right) \left( {x}^{2}+{y}^{2}-{z}^{
2} \right) \left( {x}^{2}-{y}^{2}+{z}^{2} \right)
.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1293271",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Find the length of chord $BC$.
On a semicircle with diameter $AD$. Chord $BC$ is parallel to the diameter.Further each of the chords $AB$ and $CD$ has length of $2$ cm while $AD$ has the length $8$ cm.Find the length of $BC$.
$a.)7.5\quad cm\\
\color{green}{b.)7\quad cm}\\
c.)7.75\quad cm\\
d.)\text{cannot be determi... | Join the points B & D to obtain right $\Delta ABD$. Thus, we get $$BD=\sqrt{(AD)^2-(AB)^2}=\sqrt{8^2-2^2}=2\sqrt{15}$$ Now, draw a perpendicular say BM from the point B to the hypotenuse AD in right $\Delta ABD$. Then the normal distance between the parallel chords BC & AD is equal to BM i.e. length of perpendicular d... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1293801",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Is this true that $(\cos^2A+\cos^2B+\cos^2C+2\cos A\cos B\cos C=1 \implies A+B+C=\pi)$? Assume that $A,B,C$ are positive real numbers and $A,B,C \in (0,\frac{\pi}{2}]$ and we have
$$\cos^2A+\cos^2B+\cos^2C+2\cos A\cos B\cos C = 1 $$
prove or disprove that $$A+B+C=\pi$$
| With an obvious shorthand notation, we can solve the equation for $\cos C$:
$$c^2+2abc+a^2+b^2-1=0,$$
$$c=-ab\pm\sqrt{a^2b^2-a^2-b^2+1}=-ab\pm\sqrt{(1-a^2)(1-b^2)}.$$
Then
$$\cos C=-\cos A\cos B\pm \sin A\sin B=-\cos(A\pm B)$$
and
$$\pm C=\pi\pm A\pm B.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1294092",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Solving the Diophantine equation $x^n-y^n=1001$ For all $n \in \mathbb{N}$, solve the Diophantine equation $x^n-y^n=1001$, where $x,y \in \mathbb{N}$.
The cases $n=1,2$ are trivial ones. But for $n>2$ I can't find any solutions. How could I prove that there are no integer solutions for $n>2$?
| Since $x,y \in \mathbb{N}$ and $x^4$ - $y^4$= $(y + h)^4$ - $y^4 = 4y^3h + 6y^2h^2 + 4yh^3 + h^4$
we have x - y = 1 for all n≥4 because $7^4$ = 2041 > 1001 then 7 and greater than 7 candidates (11, 13, 77,…) must be discarded. Therefore ${n\choose1}y^{n-1} +{n\choose2}y^{n-2} + … + {n\choose1}y + 1$ = 1001; hence
y[$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1294357",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
how to parameterize the ellipse $x^2 + xy + 3y^2 = 1$ with $\sin \theta$ and $\cos \theta$ I am trying draw the ellipse $x^2 + xy + 3y^2 = 1$ so I can draw it. Starting from the matrix:
$$ \left[ \begin{array}{cc} 1 & \frac{1}{2} \\ \frac{1}{2} & 3 \end{array}\right]$$
I computed the eigenvalues $2 \pm \frac{1}{2}\sq... | let us define $\theta$ by $$\cos\theta = \frac1{\sqrt{10+4\sqrt 5}}, \sin \theta=\frac{2+\sqrt 5}{\sqrt{10+4\sqrt 5}}.$$ then you can verify that
$$\pmatrix{1&1/2\\1/2&3}\pmatrix{\cos \theta&-\sin \theta\\\sin \theta&\cos \theta} = \pmatrix{\cos \theta&-\sin \theta\\\sin \theta&\cos \theta}\pmatrix{2+\sqrt5/2&0\\0&2-\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1294442",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Finding the real root of the polynomial $2x^3-3x^2+2 $ I want to get exactly roots of this equation...
$2x^3-3x^2+2 = 0$
I try to solve it but can not find the solution.
wolframealpha just give me aproximation..
I know the real root is $-1< root <-1/2$.
| Here are the three exact solutions, which you can find by cubic root methods such as Cardano's:
$\left\{\frac{1}{2} \left(1-\frac{1}{\sqrt[3]{3-2 \sqrt{2}}}-\sqrt[3]{3-2
\sqrt{2}}\right),\frac{1}{2}+\frac{1}{4} \sqrt[3]{3-2 \sqrt{2}} \left(1-i
\sqrt{3}\right)+\frac{1+i \sqrt{3}}{4 \sqrt[3]{3-2 \sqrt{2}}},\frac{1}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1296026",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
The lines $x+2y+3=0$ , $x+2y-7=0$ and $2x-y+4=0$ are sides of a square. Equation of the remaining side is? I found out the area between parallel lines as $ \frac{10}{\sqrt{5}} $ and then I used
$ \frac{|\lambda - 4|}{\sqrt{5}} = \frac{10}{\sqrt{5}} $ to get the values as $-6$ and $14$ . I am getting the final equatio... | It is clear that the lines: $x+2y+3=0$ & $x+2y-7$ are parallel hence the forth line (side of square) must be parallel to third line: $2x-y+4=0$ Hence, let the forth line: $2x-y+c=0$ having slope $2$.
Let each side of square $a$ then it is equal to the distance between the first two (first & second) parallel lines calcu... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1296690",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
What mistake have I made when trying to evaluate the limit $\lim \limits _ {n \to \infty}n - \sqrt{n+a} \sqrt{n+b}$? Suppose $a$ and $b$ are positive constants.
$$\lim \limits _ {n \to \infty}n - \sqrt{n+a} \sqrt{n+b} = ?$$
What I did first:
I rearranged $\sqrt{n+a} \sqrt{n+b} = n \sqrt{1+ \frac{a}{n}} \sqrt{1+ \frac{b... | Your error is in this equality:
\begin{equation*}
\lim\limits_{n\rightarrow \infty }n-n\sqrt{1+\frac{a}{n}}\sqrt{1+\frac{b}{n}}%
=0\ \ \ \ \ \ (Here\ it\ is).
\end{equation*}
Explanation. You known that
\begin{eqnarray*}
\lim_{n\rightarrow \infty }n &=&+\infty ,\ \ \ and \\
\lim_{n\rightarrow \infty }n\sqrt{1+\frac{a}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1297035",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 7,
"answer_id": 4
} |
Axis angle and length of ellipse For an ellipse defined by
$$x = a \cos(t + \alpha)$$
$$y = b \cos(t + \beta)$$
What are the angles and lengths of each axis?
I've tried to work backwards from the expression for a rotated ellipse but I can't seem to equate the trigonometric expressions.
| You can write the coordinates at time $t$
\begin{eqnarray}
\left( \begin{array}{c}
x(t) \\
y(t)
\end{array}
\right) =
\left( \begin{array}{cc}
a_{11} & a_{12} \\
a_{21} & a_{22}
\end{array}
\right)\cdot
\left( \begin{array}{c}
\cos t \\
\sin t
\end{array}
\right)
\end{eqnarray}
Consider a singular value decomposition... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1297438",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
How to solve certain types of integrals I'm asking for a walk through of integrals in the form:
$$\int \frac{a(x)}{b(x)}\,dx$$
where both $a(x)$ and $b(x)$ are polynomials in their lowest terms. For instance $$\int \frac{x^3+2x}{x^2+1}\,dx$$
Is there a trick to doing these? or will I have to integrate by a clever subs... | The first step is generally to divide out the integrand so as to get a polynomial plus a rational function whose numerator has lower degree than its denominator. Here you get
$$\frac{x^3+2x}{1+x^2}=x+\frac{x}{x^2+1}\;.$$
Integrating the $x$ term (and, more generally, the polynomial quotient) is easy, so we’ve reduced t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1297995",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 1
} |
Proof of sum in an inequality I was having hard time solving this one, any help will be greatly appreciated.
prove that:
$$
{39\over e^2}\le\sum_{n=1}^\infty {4n^2-1\over e^n}-{3\over e}\le{54\over e^2}
$$
| We know $\sum_{n=0}^\infty x^n=\frac{1}{1-x}$. Taking derivatives $\sum_{n=1}^\infty nx^{n-1}=\frac1{(1-x)^2}$. Multiply both sides by $x$ and take another derivative to get $\sum_{n=1}^\infty n^2x^{n-1}=\frac{1-x^2}{(1-x)^4}$. Thus $\sum_{n=1}^\infty 4n^2x^n=\frac{4x(1-x^2)}{(1-x)^4}$. Thus $\sum_{n=1}^\infty(4n^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1298668",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
how to determine the conditions that 2 vectors parallel? Given 2 vectors,$u=(3,5)$,$v=(s,s^2)$,in what situations do u and v parallel?$(s≠0)$
In order to be parallel,$u$ must be proportional to $v$,vice verse.Let $k$ be a scalar $neq 0$,then $ku=(3k,5k)=v=(s,s^2)$,which gives $3k=s$;$5k=s^2 \rightarrow 5k=9k^2 (k \neq ... | There's nothing wrong. Your first value of $k = \frac{5}{9}$ means that $\frac{5}{9} u = v$. Your second value of $k = \frac{9}{5}$ is for k multiplying the other vector, and means that $\frac{9}{5} v = u$. But that's exactly the same you got before.
$$\frac{5}{9} u = v \implies u = \frac{9}{5} v$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1303263",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Prove that $\sin (\theta) + \cos(\theta) \ge 1$
Let $\theta$ be an arbitrary acute angle. Prove that $\sin (\theta) + \cos(\theta) \ge 1$.
$$\big(\sin (\theta) + \cos (\theta)\big)^2 = 1 + 2 \sin(\theta)\cos(\theta)\ge 0$$
so, \begin{align*}\big(\sin(\theta)) + \cos(\theta)\big)^2 &> 1\\
\big(\sin(\theta)+ \cos(\th... | This is a brute force approach.
There is an $R$ and $\alpha$ such that $\sin \theta + \cos \theta = R\sin(\theta + \alpha)$.
First, $R\sin(\theta + \alpha) = R\cos\alpha\sin\theta + R\sin\alpha\cos\theta$.
So $R\sin\alpha = R\cos\alpha = 1$. Since $\tan\alpha = 1$, $\alpha = 45^\circ$. Adding the squares of $R\sin\alph... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1304594",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 3
} |
General solution of the system of equations It seems like an easy question to solve but could not figure it out:
If it is known that the following system of equations have a solution $x = x_1(t)$, where $x_1(t)$ is a second order polynomial and $y= y_1(t)$ where $y_1(t)$ is a first order polynomial.
$$x' = (t-t^2)x + (... | $$x' = (t-t^2)x + (t^3-t^2+t+1)y$$
$$y' = (1-t)x + (t^2-t+1)y$$
Letting $x = at^2+bt+c$ and $y = mt+n$, we get that
$$2at+b = (t-t^2)(at^2+bt+c) + (t^3-t^2+t+1)(mt+n) =$$
$$(m-a)t^4+(a-b-m+n)t^3+(b-c+m-n)t^2+(c+m+n)t+n$$
and
$$m = (1-t)(at^2+bt+c) + (t^2-t+1)(mt+n)$$
From the first equation, we get that $m-a=0$, $a-b-m... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1305934",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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How to evaluate $\lim\limits_{n\to\infty}\frac{1}{n}((x+\frac{a}{n})^2+(x+\frac{2a}{n})^2+...+(x+\frac{(n-1)a}{n})^2)$ I don't know how to transform the expression $\frac{1}{n}((x+\frac{a}{n})^2+(x+\frac{2a}{n})^2+...+(x+\frac{(n-1)a}{n})^2)$
The solution, after transformation is
$\frac{n-1}{n}x^2+2\frac{1+...+(n-1)}{... | This is just the limit of the Riemann sum of $\displaystyle\int_{0}^1(x+a)^2da$, which is $\displaystyle x^2+x+\frac{1}{3}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1306361",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Both ways seems right to me The question: $$\int {2e^x \over e^{2x}-1}dx $$
Solution 1 (we solv it this way in the exam today)
$\int {2e^x \over (e^x-1)(e^x+1)}dx = \int {(\frac{A}{e^x-1} + \frac{B}{e^x+1})dx} \qquad $from this, $A=1$ and $B=1$. Then,
$\int {\frac{1}{e^x-1} }dx + \int {\frac{1}{e^x+1}}dx = \ln(e^x-1)+\... | You can re-write you integrand as $$\frac{2}{e^{x}-e^{-x}}=\text{cosech} x.$$ Thus
\begin{eqnarray}
\int\text{cosech}xdx &=& \ln \text{tanh} \frac{x}{2}+c \\
&=& \ln \left(\frac{e^{\frac{x}{2}}-e^{\frac{x}{2}}}{e^{\frac{x}{2}}+e^{\frac{x}{2}}}\right)+c
\end{eqnarray}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1306637",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 4
} |
find the minimum value of $\sqrt{x^2+4} + \sqrt{y^2+9}$ Question:
Given $x + y = 12$, find the minimum value of $\sqrt{x^2+4} + \sqrt{y^2+9}$?
Key:
I use $y = 12 - x$ and substitute into the equation, and derivative it.
which I got this
$$\frac{x}{\sqrt{x^2+4}} + \frac{x-12}{\sqrt{x^2-24x+153}} = f'(x).$$
However, aft... | Just another way:
$$\sqrt{x^2+4}+\sqrt{y^2+9} \ge \sqrt{(x+y)^2+(2+3)^2} = 13$$
where we have used the triangle inequality.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1307398",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 0
} |
Is this correct $(\cot x)(\sin x)=2(\cot x)^2$ $0≤x≤2\pi$ $x= \pi/2 , 3\pi/2$ $1.4371774 , 5.139467567$ Steps I took:
$$\cot x \sin x=2\cot^2 x$$
$$\cot x \sin x-2\cot^2 x=0$$
$$\cot x (\sin x-2\cot x)=0$$
$$\cot x \left(\frac{\sin^2 x}{\sin x} -2\frac{\cos x}{\sin x} \right)=0$$
$$\cot x \left(\frac{\sin^2 x-2\cos x}{... | i will use $$x = \cos t, y = \sin t , x^2 + y^2 = 1.$$ you have $$\cot t\sin t = 2 \cot^2 t \to \frac x y y=2\frac {x^2}{y^2} \to 0=x(2x-y^2)=x(-x^2 + 2x + 1)$$ the solutions are $$x = 0, x = \frac{-2 \pm 2\sqrt 2}{-2} =1+\sqrt 2, \sqrt 2 - 1$$ since $x < 1,$ we have $$ x= 0, x = \sqrt 2 - 1$$
and the corresponding ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1310760",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Roots of unity question. Question
Let $\omega=\cos\dfrac{4\pi}{7}+i\sin\dfrac{4\pi}{7}$. Show that
$\omega-1=2\sin\dfrac{2\pi}{7}\left(\cos\dfrac{11\pi}{14}+i\sin\dfrac{11\pi}{14}\right)$.
My attempt
Observe that $\omega$ is a seventh root of unity. Label the roots $1, \nu, \nu^2,\ldots,\nu^6$. Then $\omega=\nu^2$.
We... | Using $$\cos(2\theta)-1=-2\sin^2\theta,\ \ \sin(2\theta)=2\sin\theta\cos\theta$$
$$-\sin\theta=\cos\left(\frac{\pi}{2}+\theta\right),\ \ \cos\theta=\sin\left(\frac{\pi}{2}+\theta\right)$$gives you$$\begin{align}\omega-1&=\cos\frac{4\pi}{7}+i\sin\frac{4\pi}{7}-1\\&=\cos\frac{4\pi}{7}-1+i\sin\frac{4\pi}{7}\\&=-2\sin^2\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1311046",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
For which values of a parameter an equation has one Real root The following equation is given $$\log_{x-1}(x^2+2ax) - \log_{x-1}(8x-6a-3)=0$$
And I am trying to find for which values of $a$ it has only one root, which is real.
It is obvious that $$x-1>0 \Rightarrow x>1, x\neq2 $$
Also $$x^2+2ax>0$$ $$8x-6a-3>0 \Rightar... | For $D=0$, $a=1$ is the only solution.
For $D=(a-1)(a-13)\gt 0$, note that $x_2\not=2$ and that $x_1=2$ for $a=\frac{9}{10}$ (and $a=\frac{9}{10}$ is sufficient). In the following, let us separate it into cases.
*
*For $a\lt -\frac 12$, we have $\frac{6a+3}{8}\lt 0\lt 1\lt -2a.$ So, the condition we need is $x_2\gt ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1311145",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
What is the most unusual proof you know that $\sqrt{2}$ is irrational? What is the most unusual proof you know that $\sqrt{2}$ is irrational?
Here is my favorite:
Theorem: $\sqrt{2}$ is irrational.
Proof:
$3^2-2\cdot 2^2 = 1$.
(That's it)
That is a corollary of
this result:
Theorem:
If $n$ is a positive integer... | Using the rational root theorem on $x^2 - 2 = 0$ is a very simple and elegant way of proving the irrationality of $ \sqrt{2} $. Peersonally, I like it beacuse it can be explained easily to high school students
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1311228",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "114",
"answer_count": 19,
"answer_id": 9
} |
Find probability of exactly one $6$ in first ten rolls of die, given two $6$s in twenty rolls I am trying to calculate the probability that, when rolling a fair die twenty times, I roll exactly one $6$ in the first ten rolls, given that I roll two $6$s in the twenty rolls.
My thoughts
Let $A = \{\text {Exactly one 6 in... | ((10C1)(1/6)(5/6)^5)(10C1)(1/6)(5/6)^5
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1312058",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 2
} |
Prove a combinatorial identity: $ \sum_{n_1+\dots+n_m=n} \prod_{i=1}^m \frac{1}{n_i}\binom{2n_i}{n_i-1}=\frac{m}{n}\binom{2n}{n-m}$ Prove the combinatorial identity
$$
\sum_{n_1+\ldots+n_m=n} \;\;
\prod_{i=1}^m \frac{1}{n_i}\binom{2n_i}{n_i-1}=\frac{m}{n}\binom{2n}{n-m}, \enspace n_i>0,i=1,\ldots,m
$$
I "discove... | Using the generating function of the Catalan numbers the left is
$$[z^n] \left(-1 + \frac{1-\sqrt{1-4z}}{2z}\right)^m.$$
This is
$$\frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{n+1}}
\left(-1 + \frac{1-\sqrt{1-4z}}{2z}\right)^m \; dz.$$
Using Lagrange inversion put $1-4z=w^2$ so that $1/4-z=1/4 \times w^2$
or $z=1/4... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1312226",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "17",
"answer_count": 3,
"answer_id": 0
} |
Show that $a^{25} \pmod{88}$ is congruent with $ a^{5} \pmod {88}.$
Show that $a^{25} \pmod{88}$ is congruent with $ a^{5} \pmod {88}.$
I have proved it in the case that $\gcd(88,a)=1$, but in the other case , I don't know it. Any ideas?
| Note that $88=11 \cdot 8$, so $\Bbb Z _{88} \simeq \Bbb Z _8 \times \Bbb Z _{11}$. Therefore, you must show that $a^{25} = a^5$ in both $\Bbb Z _8$ and $\Bbb Z _{11}$. We shall use Euler's theorem: in $\Bbb Z _n$, we have $a^{\varphi (n)} = 1$, where $\varphi$ is Euler's function. Note that $\varphi(8)=4$ and $\varphi ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1315604",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 1
} |
solving difficult complex number proving if $z= x+iy$ where $y \neq 0$ and $1+z^2 \neq 0$, show that the number $w= z/(1+z^2)$ is real only if $|z|=1$
solution :
$$1+z^2 = 1+ x^2 - y^2 +2xyi$$
$$(1+ x^2 - y^2 +2xyi)(1+ x^2 - y^2 -2xyi)=(1+ x^2 - y^2)^2 - (2xyi)^2$$
real component
$$(1+ x^2 - y^2)x - yi(2xyi) = x + x^3... | A another approach is to utilise (as @mann and myself concluded) the fact that
$$
w = \bar{w}
$$
where bar represents the complex conjugate.
$$
\frac{z}{1+z^2} = \frac{\bar{z}}{1+\bar{z}^2}
$$
re-arrange to yeild
$$
z + z\bar{z}^2 = \bar{z} + \bar{z}z^2 \implies z-\bar{z} = \bar{z}z^2-z\bar{z}^2=z\bar{z}\left(z-\bar{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1316782",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 1
} |
How to integrate $\int_{-1}^1 \tan^{-1}\left(\frac{1}{\sqrt{1-x^2}}\right )\,dx$? Evaluate
$$\int_{-1}^1 \tan^{-1}\bigg (\dfrac{1}{\sqrt{1-x^2}}\bigg ) dx $$
Could somebody please help integrate this without using Differentiation under the Integral Sign?
| $$
\begin{aligned}
I &=\int_{-1}^{1} \tan ^{-1}\left(\frac{1}{\sqrt{1-x^{2}}}\right) d x \\
&=2 \int_{0}^{1} \tan ^{-1}\left(\frac{1}{\sqrt{1-x^{2}}}\right) d x \\
& \stackrel{IBP}{=} 2\left[\tan ^{-1}\left(\frac{1}{\sqrt{1-x^{2}}}\right)\right]_{0}^{1}-2 \int_{0}^{1} x \cdot\frac{1}{1+\left(\frac{1}{\sqrt{1-x^{2}}}\r... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1318120",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Group isomorphism and matrices Let $\mathbb{F}$ be a field. Consider the following three groups-
$$G=\left\{\begin {pmatrix}
1&a&b\\
0&1&c\\
0&0&1\\
\end{pmatrix} : a,b,c\in \mathbb{F}\right\} $$
$$H=\left\{\begin {pmatrix}
1&0&d\\
0&1&0\\
0&0&1\\
\end{pmatrix} : d\in \mathbb{F}\right\} $$
$$T=\left\{\begin {pmatrix}
x... | Do the multiplication
$$
\begin{pmatrix}
1&a&b\\
0&1&c\\
0&0&1
\end{pmatrix}
\begin{pmatrix}
1&a'&b'\\
0&1&c'\\
0&0&1
\end{pmatrix}
=\begin{pmatrix}
1&a'+a&b'+ac'+b\\
0&1&c'+c\\
0&0&1
\end{pmatrix}
$$
which means that the map
$$
f\colon G\to T,\qquad
\begin{pmatrix}
1&a&b\\
0&1&c\\
0&0&1
\end{pmatrix}\mapsto\begin{pmat... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1321962",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Show if $\sum_{n=1}^{\infty} \frac{(-1)^n}{1+\sqrt{n}}$ is absolute convergent or divergent Show if $\sum_{n=1}^{\infty} \frac{(-1)^n}{1+\sqrt{n}}$ is absolute convergent or divergent
First i subbed numbers in
$$\lim_{n \to \infty} \frac{(-1)^n}{1+\sqrt{n}} = \frac{-1}{1+\sqrt{1}} + \frac{1}{1+\sqrt{2}} - \frac{-1}{1+\... | $$\sum_{n=1}^{\infty}\left|\frac{(-1)^n}{1+\sqrt{n}}\right|=\sum_{n=1}^{\infty} \frac1{1+\sqrt{n}}>\frac12\sum_{n=1}^{\infty} \frac1{\sqrt{n}}$$ is divergent.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1322459",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 3
} |
How to determine the Laurent expansion of $\tan{z}$ around $z=0$ that is convergent in $z=\pi$ I want to determine the Laurent expansion of $\tan{z}$ around $z=0$ that is convergent in $z=\pi$ (only the first couple of terms). Now I know that if $\sum_{n=-\infty}^{\infty}c_nz^n$ then
$$c_n=\frac{1}{2\pi i}\int_K \frac{... | Oke so I decided to just brute-force this for the residu in $z=\frac{\pi}{2}$ ($z=-\frac{\pi}{2}$ is similar)
$$\begin{align}
(z-\frac{\pi}{2})\tan{z-\frac{\pi}{2}}&=(z-\frac{\pi}{2})\frac{\sin{z-\frac{\pi}{2}}}{\cos{z-\frac{\pi}{2}}}\\
&=(z-\frac{\pi}{2})\frac{1-\frac{1}{2!}(z-\frac{\pi}{2})^2+\frac{1}{4!}(z-\frac{\pi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1323844",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Mistake with Integration with Beta, Gamma, Digamma Fuctions Problem: Evaluate:
$$I=\int_0^{\pi/2} \ln(\sin(x))\tan(x)dx$$
I tried to attempt it by using the Beta, Gamma and Digamma Functions. My approach was as follows:
$$$$
Consider $$I(a,b)=\int_0^{\pi/2} \sin^a(x)\sin^b(x)\cos^{-b}(x)dx$$
$$$$
$$=\dfrac{1}{2}\beta... | OK, I realize this does not really answer your question (but look at the update below), in the sense it does not point at your error. Anyways, I think that you should be flexible with your methods doing integrals, so here is how it can be done with the change of variables I suggested:
You end up with the integral
$$
I... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1324719",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Use Newton’s method to find the solution of $2x^3+5x-6=0$ by using the following steps Use Newton’s method to find the solution of $$2x^3+5x-6=0$$
by using the following steps
a) Find, $f(0)$ where $$f(x)=2x^3+5x-6$$
b) Find, $f(1)$ where $$f(x)=2x^3+5x-6$$
c) Choose a guess point $$x_0$$
d) Perform the iterati... | As you wrote, Newton method updates the initial guess $x_0$ according to $$x_{n+1} = x_n -\frac{f(x_n )}{f'(x_n )}$$ So, if $f(x)=2 x^3+5 x-6$, $f'(x)=6 x^2+5$, this gives, after simplification $$x_{n+1} =\frac{4 x_n^3+6}{6 x_n^2+5}$$ As AjmalW answered, choose $x_0=1$ and start repeating the iterations. You will find... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1325636",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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High School Trigonometry ( Law of cosine and sine) I am preparing for faculty entrance exam and this was the question for which I couldn't find the way to solve (answer is 0). I guess they ask me to solve this by using the rule of sine and cosine:
Let $\alpha$, $\beta$ and $\gamma$ be the angles of arbitrary triangle... | The Law of Cosines is equivalent to (and is often proven via) the statements
$$a = b \cos\gamma + c\cos\beta \qquad b = c \cos\alpha + a \cos\gamma \qquad c = a \cos\beta + b \cos\alpha$$
so, your sum becomes
$$\frac{c \cos\alpha - a \cos\gamma}{a\sin\gamma} + \frac{a \cos\beta - b \cos\alpha}{b\sin\alpha} + \frac{b \c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1326863",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 8,
"answer_id": 0
} |
Probability that on three rolls of dice, there will be at least one 6 showing up? What is the probability that on three rolls of dice, there will be at least one 6 showing up?
Attempt:
Since there can be one six or two sixes or three sixes on three rolls, I considered separate cases and added them up.
So $(1/6)(5/6)(5... | You have forgotten to include th fact that any six can occur in any of the rolls, i.e. you have multiply the probabilities for exactly one and two sixs by three, giving
$$P=3\cdot\frac{1}{6}\cdot\frac{5}{6}\cdot\frac{5}{6}+3\cdot\frac{1}{6}\cdot\frac{1}{6}\cdot\frac{5}{6}+\frac{5}{6}\cdot\frac{5}{6}\cdot\frac{5}{6}=\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1327201",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 1
} |
Given $x^2 + 4x + 6$ as factor of $x^4 + ax^2 + b$, then $a + b$ is I got this task two days ago, quite difficult for me, since I have not done applications of Vieta's formulas and Bezout's Theorem for a while. If can someone solve this and add exactly how I am supposed to use these two theorem's on this task, I would ... | The first solution I came up with involves actually finding the roots of the quadratic, then employing Factor/Remainder theorem, so while it is a tad more tedious, it's also very direct. I will follow with an alternative solution that is more elegant.
By the quadratic formula, roots of $x^2 + 4x + 6$ are the complex co... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1328210",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 3
} |
Differential equation $y'' \cdot y^3 = 1$ I use these substitutions $y'=p(y)$ and $y'' = p' \cdot p$ to solve the equation, thus I have the consequence of the solution's steps:
$$ p'py^3 = 1 \implies p'p = \frac{1}{y^3} \implies \frac {dp}{dy} p = \frac {1}{y^3} \implies \int p dp = \int \frac{1}{y^3} dy \implies \\ p ... | Multiply both sides by $2y'/y^3$. Then
$$ 2y'y'' = \frac{y'}{y^3}, $$
and we can recognise both sides as derivatives. Integrating once,
$$ y'^2 = A^2-\frac{1}{2y^2} = \frac{A^2 y^2-1}{y^2}, $$
and then you can rearrange to
$$ \frac{y^2y'^2}{A^2 y^2-1} = 1, $$
and then take a square root and integrate:
$$ x_0 \pm x = \i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1330225",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
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Find the cubic equation of $x=\sqrt[3]{2-\sqrt{3}}+\sqrt[3]{2+\sqrt{3}}$ Find the cubic equation which has a root $$x=\sqrt[3]{2-\sqrt{3}}+\sqrt[3]{2+\sqrt{3}}$$
My attempt is
$$x^3=2-\sqrt{3}+3\left(\sqrt[3]{(2-\sqrt{3})^2}\right)\left(\sqrt[3]{(2+\sqrt{3})}\right)+3\left(\sqrt[3]{(2-\sqrt{3})}\right)\left(\sqrt[3]{(... | Identifying with Cardano's formula
$$x=\sqrt[3]{-\frac q2-\sqrt{\frac{q^2}4+\frac{p^3}{27}}}+\sqrt[3]{-\frac q2+\sqrt{\frac{q^2}4+\frac{p^3}{27}}},$$
we find $q=-4$, then $p=-3$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1331417",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 8,
"answer_id": 1
} |
Maximum value of trigonometric expression If
$r=3+\tan c \tan a, \quad
q=5+\tan b \tan c, \quad
p=7+\tan a \tan b$
Provided $a,b,c$ are positive and
$a+b+c=\dfrac{\pi}2$
Find the maximum value of
$\sqrt p + \sqrt q + \sqrt r$ .
| Let $\sqrt p=P$ etc.
$$R^2-3+Q^2-5+P^2-7=1\iff P^2+Q^2+R^2=16$$
Let $P=4\cos A,Q=4\sin A\cos B,R=4\sin A\sin B$
$P+Q+R=4[\cos A+\sin A(\cos B+\sin B)]\le4[\cos A+\sqrt2\sin A]$ if $\sin A\ge0$
Now $\cos A+\sqrt2\sin A=\sqrt3\cos\left(A-\arccos\dfrac1{\sqrt3}\right)\le\sqrt3$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1332293",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Finding flux over a surface If $F = 6z\mathbf i + (2x+y)\mathbf j -x\mathbf k$, evaluate $\int_S \mathbf F \cdot \mathbf n ds$ over the surface bounded by the cylinder $x^2 + z^2 = 9. x=0, y=0, z=0 $ and $ y= 6$
Okay, so I know this is a quarter cylinder in the first octant. Now assuming the surface they're talking abo... | For a direct computation, note that your surface is the union of the five surfaces parameterized by
\begin{align*}
\mathbf r(x,y)&=\left\langle x,y,\sqrt{9-x^2}\right\rangle & (x,y)&\in[0,3]\times[0,6] \\
\mathbf s(x,y) &= \langle x,y,0\rangle & (x,y)&\in[0,3]\times[0,6] \\
\mathbf t(y,z) &=\langle0,y,z\rangle & ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1334124",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Complex Integration with trignometric function
Verify that
$\int_0^{\frac{\pi}{2}}\frac{d\theta}{a+\sin^2\theta}=\frac{\pi}{2[(a(a+1)]^\frac{1}{2}}$
I know that $\sin\theta=\frac{e^{i\theta}-e^{-i\theta}}{2}$ then I did
$$\int_0^{\frac{\pi}{2}}\frac{d\theta}{a+\sin^2\theta}=\int_0^{\frac{\pi}{2}}\frac{d\theta}{a+\f... | NOTE 1:
First note that then integral diverges if $-1\le a\le 0$. So, we will assume that either $a>0$ or $a<-1$.
NOTE 2:
We will simplify things quite a bit by first invoking the identity
$$\sin^2x=\frac{1-\cos 2x}{2} \tag 1$$
By simplifying first, we will reduce the denominator of a contour integral from a ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1334389",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Compute definite integral Question: Compute
$$\int_0^1 \frac{\sqrt{x-x^2}}{x+2}dx.$$
Attempt: I've tried various substitutions with no success. Looked for a possible contour integration by converting this into a rational function of $\sin\theta$ and $\cos \theta$.
| Let us make the problem a bit more general considering $$I=\int \frac{\sqrt{x-x^2}}{x+a}dx$$ and let us apply the same approach Mario G proposed. We then arrive to $$\begin{align*}
I=\int\frac{\sqrt{x-x^2}}{x+2}dx
&=\frac{1}{2}\int\frac{\cos^2( t)}{\sin t+(2a+1)}\,dt\\
\end{align*}$$ Now, let use the tangent half-angle... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1340612",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 6,
"answer_id": 3
} |
Arithmetc Progression containing odd terms Here is the question:
If for an AP of odd number of terms,the sum of all the terms is $\frac{15}{8}$ times the sum of the terms in odd places then find the number of terms in the AP.
my try:First of all i thought that in an odd AP there will be (2n+1) terms becuase (2n) term... | First consider a constant progression, $t_n=1$. Then the ratio of the sums is $\dfrac{2n}{n+1}=\dfrac{15}8$ so that $\color{green}{n=15}$.
$$\frac{1+1+1+1+1+1+1+1+1+1+1+1+1+1+1}{1+1+1+1+1+1+1+1}=\frac{15}8.$$
This is compatible with a linear progression $t_n=n$, as
$$\frac{1+2+3+4+5+6+7+8+9+10+11+12+13+14+15}{1+3+5+7+9... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1342978",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.