Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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$f(x)=x^3-ax^2-ax+1,\;\;|x_1|+|x_2|+|x_3|=3$
Find all $a\in \mathbb R$ such that the zero-points of the function
$$f(x)=x^3-ax^2-ax+1$$ Satisfy the equation: $$|x_1|+|x_2|+|x_3|=3$$
My work:
$$
\begin{split}
f(x) &= x^3-ax^2-ax+1 \\
&= x^3+1-ax(x+1)\\
&= (x+1)(x^2-x+1-ax) \\
&=(x+1)(x^2-(a+1)x+1) \\
... | $ x_1 = -1 $, we have
$$ |x_2| + |x_3| = 2 $$
now, if $ x_2, x_3 $ real Number
can be four solution.
*
*$ x_2 > 0, x_3 > 0 $
$x_2 + x_3 = a+1$
$a =1$
*
*$ x_2 < 0, x_3 < 0 $
$x_2 + x_3 = -(a+1)$
$a =-3$
*
*$ x_2 < 0, x_3 > 0 $ or $ x_2 > 0, x_3 < 0 $
$ x_2 - x_3 = +- \sqrt{(a+1)^2 - 4} $
If $x_2 = x_3$
So,
$ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3486826",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Let $(x-1)^3$ divides $p(x)+1$ and $(x+1)^3$ divides $p(x)-1$. Find a polynomial $p(x)$ of degree 5. Let $(x-1)^3$ divides $p(x)+1$ and $(x+1)^3$ divides $p(x)-1$. Find a polynomial $p(x)$ of degree 5.
Here's what I have tried:—
As $(x-1)^3$ divides $p(x)+1$,
$p(1)+1=0$, $p(1)=-1$
$p(-1)=1$
Letting $p(x)=a_5x^{5} + a_... | Let $p(x)$ be of degree 5. Since $(x-1)^3|(p(x)+1)$ and $(x+1)^3|(p(x)-1)$ there exists two polynomials $f(x)=ax^2+bx+c$ and $g(x)=a_1 x^2+b_1 x +c_1$ such that
$$ (x-1)^3f(x)-1 = p(x) = (x+1)^3g(x)+1$$
$$ (x-1)^3f(x)- (x+1)^3g(x)-2 = 0$$
$$(x-1)^3(ax^2+bx+c)- (x+1)^3(a_1 x^2+b_1 x +c_1)-2 =0 \enspace \cdots(\ast) $$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3487745",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Determinant of the matrix $(\omega^i \theta^j)_{i,j = 0,1,2}$ Given that $\omega=\frac{-1+\sqrt{-3}}{2}$ is a complex cube root of 1, $\theta$ is a real cube root of 2 and $\omega^{2}+\omega+1=0$, I am trying to maipulate the matrix
\begin{pmatrix}
1 & \theta &\theta^{2} \\
\omega & \omega \theta & \omega \theta^{2} \... | $$\begin{vmatrix}
1 & \theta &\theta^{2} \\
\omega & \omega \theta & \omega \theta^{2} \\
\omega^{2} & \omega^{2}\theta & \omega^{2}\theta^{2}
\end{vmatrix}
=\omega\,\omega^2\theta\,\theta^2\begin{vmatrix}
1 & 1&1\\
1&1& 1 \\
1 & 1 & 1
\end{vmatrix}
$$
The values of $\theta$ and $\omega$ don't matter.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3488010",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Determine if function $f(x)=\ln \left ( \frac{x+1}{x-1} \right )$ is odd or even. Determine if function $f(x)=\ln \left ( \frac{x+1}{x-1} \right )$ is odd or even.
My solution:
$$
\begin{align}
f(-x)&=\ln\left ( \frac{-x+1}{-x-1} \right )\\
&=\ln \left ( \frac{-(x-1)}{-(x+1)}\right )\\
&=\ln\left ( \frac{x-1}{x+1} \rig... | While your conclusion is fine, there are still some problems with your computations, where some cautions is needed.
By definition, an even or odd function must have a symmetric domain of definition, so the first thing is to make this domain explicit: $\mathcal{D} := (-\infty, 1) \cup (1, +\infty)$.
Then we want to prov... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3491295",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
If $\alpha, \beta, \gamma, \delta$ are distinct roots of equation $x^4 + x^2 + 1 = 0$ then $\alpha^6 + \beta^6 + \gamma^6 + \delta^6$ is I tried to find discriminant first and got two roots $$\frac{-1 + \sqrt{3}i}{2}$$ and $$\frac{-1 - \sqrt{3}i}{2}$$
I tried taking $x^2 = t$ and solving equation for root $w$ and $w^2$... | Hint: Recall that $(a-b)(a^2+ab+b^2) = a^3-b^3$
Multiplying $$x^4 + x^2 + 1 = 0$$ with $x^2-1$ we get $$x^6-1=0$$ so $x^6=1$ and now is easy to finish...
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3494035",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 7,
"answer_id": 0
} |
limit and summation I have to calculate $$ \lim_{m\rightarrow\infty}\sum_{n=0}^{\infty}\frac{1}{m^n} \sum_{k=0}^{\infty} {2k \choose k } {2n \choose k}\left( \frac{-1}{2} \right)^k$$
We know that $$ \sum_{k=0}^{2n} {2k \choose k } {2n \choose k}\left( \frac{-1}{2} \right)^k=\frac{1}{4^n}\binom{2n}{n}=\frac1\pi\int_... | For third limit, you could also have used that, by the Wallis product,
$$\binom{2 n}{n} \sim \frac {4^n} {\sqrt{\pi n}}$$ making
$$\frac{1}{4^n}\binom{2n}{n}\sim \frac {1} {\sqrt{\pi n}}$$
Remember also that there exist tight bounds for the central binomial coefficient
$$\frac {4^n} {\sqrt{4 n}} \leq \binom{2 n}{n} \l... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3494444",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Why does the beta distribution become U shaped when $\alpha$ and $\beta$ <1? In the Beta distribution (used to model Bernoulli probabilities), the $\alpha$ and $\beta$ parameters can be interpreted as the number of heads$+1$ and the number of tails$+1$ seen. So, if they were both $2$, it would lean towards the coin bei... | The Beta distribution is
$\dfrac{x^{a-1}(1-x)^{b-1}}{B(a, b)}
$
so the shape depends only on
$f(x)
=x^{a-1}(1-x)^{b-1}
$.
$f'(x)
=(a - 1) x^{a - 2} (1 - x)^{b - 1} - (b - 1) x^{a - 1} (1 - x)^{b - 2}\\
=x^{a - 2} (1 - x)^{b - 2}((a - 1) (1 - x) - (b - 1) x)
$
so $f'(x) = 0$
when
$(a - 1) (1 - x)
= (b - 1) x
$
or
$a-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3494530",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Evaluate $\iint_{[0,1]^2}\frac{10x^2-1}{\sqrt{x^3+y^3}}dxdy$ I'm trying to prove $$I=\iint_{[0,1]^2}\frac{10x^2-1}{\sqrt{x^3+y^3}}dxdy=\frac83(\sqrt2-1)$$ in an elementary way.
What I've tried
I'm not satisfied with my following method.
Integrating w.r.t. $x$:
$$I=\int_0^1\left[-\frac{20}3 \left(y^{3/2}-\sqrt{y^3+1}\... | $$ I=\int_0^1\int_0^1 \frac{10x^2-1}{\sqrt{x^3+y^3}}dxdy=\int_0^1\int_0^1 \frac{10y^2-1}{\sqrt{x^3+y^3}}dxdy$$
$$\Rightarrow I=5\int_0^1\int_0^1 \frac{x^2+y^2}{\sqrt{x^3+y^3}}dxdy-\int_0^1\int_0^1 \frac{1}{\sqrt{x^3+y^3}}dxdy$$
$$=10\int_0^1 \int_0^y \frac{x^2+y^2}{\sqrt{x^3+y^3}}dxdy-2\int_0^1\int_0^y \frac{1}{\sqrt{x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3495004",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Find coefficient of $x^{10}$ in $x^2(x^2-3x^3-1)^6$ I'm trying to solve the following problem (final exams preparation):
Find coefficient of $x^{10}$ in $x^2(x^2-3x^3-1)^6$.
The steps I have taken so far:
$x^2(x^2-3x^3-1)^6 = $
$=x^2\sum \binom{6}{n1,n2,n3}(x)^{2n_1}(-3)^{n_2}(x)^{3n_2}(-1)^{n_3}$
$n_1+n_2+n_3=6$
$2n... | I'll translate the work you did into English.
So the coefficient of the $x^{10}$ term in $x^2(x^2-3x^3-1)^6$ is the same as the coefficient of the $x^8$ term in $(x^2-3x^3-1)^6$. To figure out which terms in that trinomial power contribute that that term, we effectively want to find the non-negative solutions to $2a+3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3496297",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
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If $z = \frac{(\sqrt{3} + i)^n}{(\sqrt{3}-i)^m}$, find the relation between $m$ and $n$ such that $z$ is a real number. I am given the following number $z$:
$$z = \dfrac{(\sqrt{3} + i)^n}{(\sqrt{3} - i)^m}$$
with $n, m \in \mathbb{N}$. I have to find a relation between the natural numbers $n$ and $m$ such that the numb... | You may continue as follows,
$$z = 2^{n - m} \cdot \dfrac{\cos \dfrac{n \pi}{6} + i \sin \dfrac{n \pi}{6}}
{\cos \dfrac{11 m \pi}{6} + i \sin \dfrac{11 m \pi}{6}}$$
$$=2^{n - m} \cdot \dfrac{\left(\cos \dfrac{n \pi}{6} + i \sin \dfrac{n \pi}{6}\right)
\left(\cos \dfrac{11m \pi}{6} - i \sin \dfrac{11m \pi}{6}\right)}
{\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3498784",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 4
} |
Numbers 'poetics' 1 and 2019
One number is called 'poetic' when it can be represented of only one form how $2^a+2^b+2^c$ such that $a\ge b \ge c\ge 0.$ How many poetic numbers are between $1$ and $2019?$
Attempt: I separate in cases, I find that every $2^k$ with $k\ge2 , 2^k + 1$ with $k\ge1$ and every case with $a>... | We have $a\ge b\ge c\ge0$. But there is also a requirement that the representation is unique. So for example: 6 does not count because it is $4+1+1$ and also $2+2+2$.
Indeed, we have
(1): $2^a+2^a+2^b=2^{a+1}+2^{b-1}+2^{b-1}$ for $b\ge1$ and
(2): $2^a+2^b+2^b=2^{a-1}+2^{a-1}+2^{b+1}$ for $a>b+1$. [For $a=b+1$ the two ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3500587",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Evaluating $\lim_{x \to 1}\frac{\sin(x^2-x)}{x^2-1}$ and $\lim_{x \to 1}\frac{\ln(x^2+x-1)}{\ln(x^3+x-1)}$ I found these limits and I was unable to solve them due to the occurring indeterminations
$$\lim_{x \to 1}\frac{\sin(x^2-x)}{x^2-1}$$
$$\lim_{x \to 1}\frac{\ln(x^2+x-1)}{\ln(x^3+x-1)}$$
Can someone help me, please... | For the second one $$\frac{\log(x^2+x-1)}{\log(x^3+x-1)}$$ using what Lord Shark the Unknown suggested, it becomes
$$\frac{\log \left(1+3t+t^2\right)}{\log \left(1+4t+3 t^2+t^3\right)}$$ Use the Taylor expansion of $\log(1+\epsilon)$ with $\epsilon=(3t+t^2)$ for the numerator and $\epsilon=(4t+3 t^2+t^3)$ for the denom... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3500775",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 3
} |
How many solutions does the expression $3^x(3^x+1)+2=|3^x-1|+|3^x+2|$ have? I can solve this by opening every single modulus, doing alternate combinations of $\pm$ and checking whether every $3^x$ is $>0$ or not. But that consumes a lot of time. Is there is a shorter way to solve such questions.
The answer is 1
| $\because\forall x\in\mathbb R,3^x>0,\therefore 3^x+2>0$,so
$$\begin{align}
3^x(3^x+1)+2 & =|3^x-1|+|3^x+2|\\
\Longleftrightarrow \quad\,\,\,\, 3^{2x}+3^x+2 & =|3^x-1|+3^x+2\\
\Longleftrightarrow \qquad\qquad\quad 3^{2x} & =|3^x-1|
\end{align}$$
And when $3^x>1$,it’s
$$3^{2x}>3^{2x}-1=(3^x+1)(3^x-1)>3^x-1$$
so it haven... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3503998",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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Taking the gradient of $f(\mathbf{x}) = \frac{1}{2}\|\mathbf{A} \mathbf{x} - \mathbf{b}\|_2^2$ Section 4.5 of the textbook Deep Learning by Goodfellow, Bengio, and Courville, says that the gradient of
$$f(\mathbf{x}) = \dfrac{1}{2}\|\mathbf{A} \mathbf{x} - \mathbf{b}\|_2^2$$
is
$$\nabla_{\mathbf{x}} f(\mathbf{x}) = \m... | We must take the derivative with finesse, and that means we use the chain rule. Note that $f = g \circ h$, where $h(x) = Ax-b$ and $g(u) = (1/2) \|u\|^2$. The derivatives of $h$ and $g$ are $h'(x) = A$ and $g'(u) = u^T$. So by the chain rule
$$
f'(x) = g'(h(x)) h'(x) = (Ax-b)^T A.
$$
The gradient of $f$ is
$$
\nabla f(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3508373",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
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Why is this the nearest integer? I was trying to prove that $(N+\sqrt{N^2-1})^k$, where k is a positive integer, differs from the integer nearest to it by less than $(2N-\frac{1}{2})^{-k}$. Note: N is an integer greater than 1.
So, I tried to look for the answer of the question, which I have taken it from an exam pape... | By induction on $k$, we have
$$ (N\pm \sqrt{N^2-1})^k=a\pm b\sqrt{N^2-1}$$
with $a,b\in\Bbb Z$.
And of course
$(N-\frac12)^2=N^2-N+\frac14<N^2-1 $ implies
$$ 2N-\frac12<N+\sqrt{N^2-1}<2N.$$
Now from
$$ (N+\sqrt{N^2-1})^k(N-\sqrt{N^2-1})^k=((N+\sqrt{N^2-1})(N-\sqrt{N^2-1}))^k=1^k=1,$$
we conclude that $(N+\sqrt{N^2-1})... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3508730",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Suppose that $z\in\mathbb C$ with $|z^2+1|\le 1$. How to prove $|z+1|\geq\frac12$. Let $z\in\mathbb C$ with $|z^2+1|\le 1$. I want to prove $|z+1|\geq\frac12$.
I noticed that $|z^2+1|\le 1$ means that $z$ lies in a cassini oval. I tried with the substitution $z=r\exp(i\theta)$ where $r\geq0, \theta\in[0,2\pi]$. The co... | It suffices to prove that
$$|z+1| < \tfrac{1}{2} \quad \Longrightarrow \quad |z^2+1| > 1.$$
To this end, let $z = a + \mathrm{i}b$.
From $|z+1| < \frac{1}{2}$, we have
$(a+1)^2 + b^2 < \frac{1}{4}$ which results in
$a^2 - b^2 > a^2 + (a+1)^2 - \frac{1}{4} = 2(a+\frac{1}{2})^2 + \frac{1}{4} > 0$.
Then, we have
$|z^2 + ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3511920",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Solve inverse trigonometric equation $\frac{\pi}{6}=\tan^{-1} \frac{11}{x} -\tan^{-1} \frac{1}{x}$ How do I go about solving for $x$ when I have:
$\frac{\pi}{6}=\tan^{-1} \left( \frac{11}{x} \right)-\tan^{-1}\left( \frac{1}{x} \right)$.
| The formula for the tangent of a difference says
$$
\tan(a-b) = \frac{\tan a-\tan b}{1+\tan a \tan b}.
$$
If $p = \tan a$ and $q=\tan b$ and $a = \tan^{-1}p$ and $b=\tan^{-1} q$ then we have
$$
\tan(a-b) = \frac{p-q}{1+pq}
$$
so
$$
a-b = \tan^{-1} \frac{p-q}{1+pq}
$$
or in other words
$$
\tan^{-1} p - \tan^{-1} q = \ta... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3513047",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 4
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Gaussian integral over a surface I have to solve the following integral:
$$
I(s)=\int_{S(s)} \frac{1}{2\pi^{3/2}}e^{-\frac{x^2+y^2+2z^2}{2}}dxdydz
$$
where $S(s)$ is the surface defined by $s=\sqrt{(x-y)^2+4z^2}$.
I parametrised $S(s$) using:
\begin{cases}
x = \frac{\xi+s\cos(\theta)}{2} \\
y = \frac{\xi-s\cos(\theta)... | Your first attempt is entirely correct and the value of the integral is indeed
\begin{align}
I(s) &= \frac{1}{2\pi^{3/2}}\frac{s}{2\sqrt{2}}\int \limits_{-\infty}^\infty \mathrm{e}^{-\frac{\xi^2+s^2}{4}} \mathrm{d} \xi \int \limits_0^{2\pi}\sqrt{2-\cos^2(\theta)} \, \mathrm{d} \theta = \frac{s \mathrm{e}^{-s^2 /4}}{2\p... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3513552",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Find area of ellipse $5x^2 -6xy +5y^2=8$
Find the area of ellipse whose equation in the $xy$- plane is given by $5x^2 -6xy +5y^2=8$
My attempt : I know that area of ellipse $ = \pi a b$ ,where $a$ is semi-major axis and $b$ is semi minor axis
Now if we make matrix $\begin{bmatrix} 5 & -3 \\-3& 5\end{bmat... | Given $$5x^2-6xy+5y^2=8$$
Curve represent ellipse whose center is at $(0,0)$ and it is symmetrical about $y=x$ and $y=-x$ line
Put $y=x$ in $5x^2-6xy+5y^2=8$, getting $=\pm \sqrt{2}.$
So point of intersection of line and ellipse is $(\sqrt{2},\sqrt{2})$ or $(-\sqrt{2},-\sqrt{2})$
Which is notning but distance length o... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3514276",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 2
} |
Inverse with Fermat, Modulo I wonder about an equation given as a solution to the following task:
Calculate the multiplicative inverse $$5^{−1} \pmod {13}$$
The solution ends in this equation:
$$
5^{11} \equiv 5^{10} \cdot 5 \equiv −1\cdot 5 \equiv 8 \pmod {13}
$$
and the check:
$$
5 \cdot 8 \equiv 40 \equiv... | Question 2: Why is $5^{10}\equiv - \pmod {13}$.
Well....Because it is.
$5^2 \equiv 25\equiv -1\pmod {13}$ so $5^{10}\equiv (-1)^5\equiv -1 \pmod {13}$.
The real question is why are we trying to find $5^{10}$?
$5^{11}*5 \equiv 5^{12} \equiv 1 \pmod {13}$ so $5^{-1}\equiv 5^{11}$ and..... well, I guess the author just... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3515204",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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Prove inequality of series sum by induction How should i prove by induction that:
$ 1 + 2^{(-2)} + 3^{-2} + ...+n^{(-2)} \le 2-n^{-1} $
I haven't had inequalities with sum of series at the class about induction so i can't really wrap my head around it.
| Induction is induction.
Base case: $1^{-2} = 1 \le 2- 1^{-1} = 1$.
Induction case:
If $1 + 2^{-2}+ 3^{-2} + ....... + n^{-2} \le 2-n^{-1}$ then
$1 + 2^{-2}+ 3^{-2} + ....... + n^{-2} + (n+1)^{-2} \le 2-n^{-1} + (n+1)^{-2}$.
So all we have to do is prove $ 2-n^{-1} + (n+1)^{-2} \le 2- (n+1)^{-1}$.
Can we do that?
All ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3519186",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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Represent $f(x)=\ln x$ as power series, in powers of $(x-4)$ I was requested to represent $f(x)=\ln x$ as a power series, in powers of $(x-4)$. I only started studying power series today, and though I was able to represent different functions as power series I can't seem to work this one out, specially with the conditi... | There are a few ways to go about this.
The most direct way.
$f(x) = f(x-a) + f'(x-a) (x-a) + \frac {f''(x-a)}{2!} (x-a)^2 + \cdots + \frac {f^{(n)}(x-a)}{n!} (x-a)^n + \cdots$
I hope that this looks familiar to you.
We set $a = 4$
$f(4) = \ln 4\\
f'(x) = \frac {1}{x}, f'(4) = \frac {1}{4}\\
f''(x) = -\frac {1}{x^2}, f'... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3524982",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
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In how many ways can we choose 4 different numbers from the set ${1,2,3,...,8,9,10}$ so that no two numbers are next to each other? I did this question using PIE and I'm confused as to why I'm not getting the right answer.
My approach:
Use complementary counting. There are $\binom{10}{4}$ ways to choose 4 different num... | It looks like you applied the Inclusion-Exclusion Principle to cases with two consecutive, three consecutive, and four consecutive numbers. However, you should instead apply the Inclusion-Exclusion Principle to pairs of consecutive numbers.
There are $\binom{10}{4}$ ways to choose four numbers from the set $\{1, 2, 3,... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3525172",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Derivative of $\dfrac{\sqrt{3-x^2}}{3+x}$ I am trying to find the derivative of this function
$f(x)=\dfrac{\sqrt{3-x^2}}{3+x}$
$f'(x)=\dfrac{\dfrac{1}{2}(3-x^2)^{-\frac{1}{2}}\frac{d}{dx}(3-x)(3+x)-\sqrt{3-x^2}}{(3+x)^2}$
$=\dfrac{\dfrac{-2x(3+x)}{2\sqrt{3-x^2}}-\sqrt{3-x^2}}{(3+x)^2}$
$=\dfrac{\dfrac{-x(3+x)}{\sqrt{3-... | Yet another method in case you're interested: first square both sides
$$y^2=\frac{3-x^2}{(3+x)^2}$$
Then differentiate both sides:
$$2y\frac{dy}{dx}=\frac{(3+x)^2(-2x)-(3-x^2)2(3+x)}{(3+x)^4}$$
$$=-\frac{6(x+1)}{(3+x)^3}$$
Then divide both sides by $2y$ to get the result.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3525571",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 7,
"answer_id": 3
} |
Minimum value of $\frac{ax^2+by^2}{\sqrt{a^2x^2+b^2y^2}}$ If $$x^2+y^2=1$$
Prove that Minimum value of $$f(x,y)=\frac{ax^2+by^2}{\sqrt{a^2x^2+b^2y^2}}$$ is
$$\frac{2\sqrt{ab}}{a+b}$$
My try:
I used basic Trigonometry:
Let $x=\cos t$ and $y=\sin t$
Then we get $$f(x,y)=g(t)=\frac{a\cos^2 t+b\sin^2 t}{\sqrt{a^2\cos^2 t+b... | Note that
\begin{eqnarray*}
f(x,y) = \frac{ax^2+by^2}{\sqrt{a^2x^2+b^2y^2}} \\
\end{eqnarray*}
\begin{eqnarray*}
= \frac{ \sqrt{ab}}{a+b} \left( \sqrt{ \frac{a^2x^2+b^2y^2}{ab}} + \sqrt{ \frac{ab}{a^2x^2+b^2y^2}} \right)
\end{eqnarray*}
\begin{eqnarray*}= \frac{ \sqrt{ab}}{a+b} \left( \left( \sqrt[4]{ \frac{a^2x^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3526498",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
Value of $f(f(x) – 2y) = 2x – 3y + f(f(y) – x)$ Let $f : \mathbb{R} \rightarrow \mathbb{R} $ be a polynomial function satisfying
$$f(f(x) – 2y) = 2x – 3y + f(f(y) – x),$$
where $x, y \in \mathbb{R}$. Find the value of $$f(21) – f(14).$$
| You are given
$$f(f(x) – 2y) = 2x – 3y + f(f(y) – x) \tag{1}\label{eq1A}$$
Since $x,y \in \mathbb{R}$, you can let $x = \frac{3y}{2}$ in \eqref{eq1A} to get
$$\begin{equation}\begin{aligned}
f\left(f\left(\frac{3y}{2}\right) - 2y\right) & = 2\left(\frac{3y}{2}\right) - 3y + f\left(f\left(y\right) - \frac{3y}{2}\right) ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3529999",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Infinity Summation problem (ANSWERED) Show that $$\sum_{n=3}^∞ 96n^4+156n^3+33n^2+54n+3 = \frac{100-π^3}{87}$$
Only way I know how to solve this is to put it in my graphing calculator, but that isn't much help in this situation.
This is a question I am solving on my own that I have discovered. If you send the answer an... | (Big) hint: you can rewrite
$$
\frac{1}{4n^4+16n^3+23n^2+14n+3}= 2\left( \frac{1}{2n+1}-\frac{1}{2n+3}\right) - \frac{1}{(n+1)^2}
$$
and now, you have the sum of a telescoping series and a convergent series (whose sum is known).
That is, we can compute the sum of the telescopic series as
$$
\sum_{n=1}^N 2\left( \frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3531051",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Find all primes $p \geq 5$ such that $6^p \cdot (p - 4)! + 10^{3p}$ is divisible by $p$
Find all primes $p \geq 5$ such that $6^p \cdot (p - 4)! + 10^{3p}$ is divisible by $p$
I've tried this : First check $(p - 4)!:$
\begin{align*}
(p - 1)! &\equiv -1 \text{(mod p)}\tag{by Wilson's Theorem} \\
(p - 1)(p - 2)(p - 3)(... | HINT.-After you have $6^p(p-4)!\equiv 1\pmod p$ you need
$$1+(10^p)^3\equiv1+10^3=1001=7\cdot11\cdot13\equiv0$$ then you have three solutions $7,11$ and $13$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3531874",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
prove $x^2 - x + 1$ divides $x^{10} - x^7 + x^4 + ax + b$ for some $a, b$ in an arbitrary field Let $F$ be an arbitrary field, I need to prove that $x^2 - x + 1$ divides $x^{10} - x^7 + x^4 + ax + b$ for some $a, b \in F$
The difficulty that I am currently facing is that since $F$ is an arbitrary field, $x^2 - x + 1$ m... | $$f(x)=x^4(x^6-x^3+1)+(ax+b)$$
Now reduce $f(x)$ by $x^2=x-1$ to have remainder $R(x)$ as
$$\implies R(x)=(x-1)^2[(x-1)^3-x(x-1)+1]+(ax+b)$$
$$\implies R(x)=(x-1-2x+1)[(x-1)(x^2-2x+1-x)+1])+ax+b$$
$$\implies R(x)=(x-1)^2[(x-1)(-2x)+1]+ax+b$$
$$\implies R(x)=(x-1)^2[-2(x-1)+2x+1]+ax+b$$
$$\implies R)x)=3(x-1)^2+ax+b \im... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3532539",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
ordered pair of $(p,q)$ in $20!$
A rational number given in the form
$\displaystyle \frac{p}{q},\;\;p,q\in \mathbb{Z}^{+}\;,\frac{p}{q}\in(0,1)$ and $p,q$ are coprime to each other.If $pq=20!.$ Then number of ordered pair of $p,q$ are
what i try
$20!=2^{18}\cdot 3^{8}\cdot 5^4\cdot 7^2\cdot 11\cdot 13\cdot 17\cdot 19... | That $\gcd(p,q)=1$ means each prime power factor of $20!$ must either be wholly in $p$ or wholly in $q$. The fraction $\frac pq$ can never equal $1$, and if it is greater than $1$ then $\frac qp$ is less than $1$ and vice versa.
There are $2^8$ ways to assign the prime powers to either $p$ or $q$, so the number of frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3532841",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Prove the following inequality: $\int_{0}^{\frac{\pi }{2}}\frac{sin(x)}{\sqrt{9-sin^{4}(x)}}dx\geq \frac{1}{3}$
Prove the following inequality: $$ \int_{0}^{\frac{\pi}{2}}\frac{\sin
x}{\sqrt{9-\sin^{4}x}}\ \mathrm dx\geq\frac{1}{3}. $$
I am thinking of replacing the equation with $\int_{0}^{\frac{\pi }{2}}\frac{\si... | \begin{align*}\int_{0}^{\pi/2} \frac{\sin{x}}{\sqrt{9-\sin^4{x}}}dx &\geq \int_{0}^{\pi/2}\frac{\sin{x}}{\sqrt{9}}dx\\& = \frac{1}{3}\int_{0}^{\pi/2} \sin{x}dx \\&=\frac{1}{3}[-\cos{\pi/2}-(-\cos{0})]\\&=\frac{1}{3}[0+1]\\&=\frac{1}{3}
\end{align*}
The inequality is because $\sqrt{9-\sin^4{x}}\leq \sqrt{9}$ due to posi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3535654",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Limit of $\lim_{x\to \infty} \sqrt[3]{x^3+2x}-\sqrt{x^2-2x}$ $$\lim_{x\to \infty} \sqrt[3]{x^3+2x}-\sqrt{x^2-2x}$$
I tried to used $(a^3-b^3)=(a-b)(a^2+ab+b^2)$ but it did not worked out so I tried to use the squeeze theorem.
$$0=\sqrt[3]{x^3}-\sqrt{x^2}\leq \sqrt[3]{x^3+2x}-\sqrt{x^2-2x}\leq\sqrt[3]{8x^3}-\sqrt{4x^2}=... | Tips:
Try to find and sum up the two limits:
$$\lim_{x\to\infty} \sqrt[3]{x^3-2x}-x \text{ and }\lim_{x\to\infty}x-\sqrt{x^2-2x}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3536477",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
"answer_id": 3
} |
Laplace transform problem help I'm stuck with this Laplace transform problem.
Using Laplace transform, solve initial problem:
$$
\ y''' - y = 1
\\ y(0) = y'(0) = 1
\\ y''(0) = 0
$$
And from that we would get:
$$
\ s^3Y - s^2 - s - Y = \frac{1}{s}
$$
I've got it until the point I get:
$$
\ Y = \frac{s^3 + s^2 + 1}{s(s... | Are you familiar with the general partial fraction method? With a quadratic (order 2) polynomial in the denominator, you should write a general linear (order 1) polynomial in the numerator. Then putting things over a common denominator,
$$\frac{A}{s} + \frac{B}{s-1} + \frac{Cs + D}{s^2+s+1} = \frac{(A+B+C)s^3+(B-C+D)s^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3537954",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Integral $\int_0^{\pi/2}x\arctan\left(\tfrac{1}{\sqrt3}+\tfrac{2}{\sqrt3}\tan x\right)dx$
Evaluate the integral $$P=\int_0^{\pi/2}x\arctan\left(\tfrac{1}{\sqrt3}+\tfrac{2}{\sqrt3}\tan x\right)dx.$$
Context:
I started trying to evaluate the integral $$J=\int_0^\infty \frac{\arctan(x)^2}{x^2+x+1}dx,$$
and the integral ... | $$J=\int_0^\infty \frac{\arctan^2 x}{1+x+x^2}dx\overset{x=\tan t}=\int_0^\frac{\pi}{2}\frac{t^2}{1+\sin t\cos t}dt\overset{2t=\frac{\pi}{2}-x}=\frac14\int_{-\frac{\pi}{2}}^\frac{\pi}{2}\frac{\left(\frac{\pi}{2}-x\right)^2}{2+\cos x}dx$$
$$=\frac12\int_0^\frac{\pi}{2}\frac{\frac{\pi^2}{4}+x^2}{2+\cos x}dx=\frac{\pi^3}{2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3538345",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
} |
$a_n=\sqrt{2-a_{n-1}},a_1=\sqrt{2}$ Given
$$a_n=\sqrt{2-a_{n-1}},a_1=\sqrt{2}$$
I calculated $a_1$ to $a_5$
$$\sqrt{2},
\sqrt{2-\sqrt{2}},
\sqrt{2-\sqrt{2-\sqrt{2}}}, \\
\sqrt{2-\sqrt{2-\sqrt{2-\sqrt{2}}}}, \\
\sqrt{2-\sqrt{2-\sqrt{2-\sqrt{2-\sqrt{2}}}}}$$
which made me think of $\sin/\cos$. So I divided each by $2$, c... | Let $x_n$ for $n\in\Bbb N$ be the sequence defined by
\begin{align}
x_0&=\frac\pi 2&
x_n&=\frac{\pi-x_{n-1}}2
\end{align}
and $a_n=2\cos(x_n)$.
Then $0\leq x_n\leq\frac\pi 2$, hence $0\leq a_n\leq 2$.
Moreover:
\begin{align}
a_n^2
&=4\cos^2(x_n)\\
&=2\cos(2x_n)+2\\
&=2\cos(\pi-x_{n-1})+2\\
&=-2\cos(x_{n-1})+2\\
&=-a_{n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3538786",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Integral $\int{\frac{\sin^3(x)}{2\cos^2(x)-3\sin^2(x)}}dx$ I want to integrate the following expression:
$$\int{\frac{\sin^3(x)}{2\cos^2(x)-3\sin^2(x)}}dx$$
I tried using a $t=\tan(\frac{x}{2})$ substitution but the terms were not cancelled. Then I tried using $1=\cos^2(x)+\sin^2(x)$ but couldn't get anywhere. I hope s... | Hint:
Recall that $$\sin^2(x) = 1 - \cos^2(x)$$
Therefore,
$$\int\dfrac{\sin^3(x)}{2\cos^2(x) - 3\sin^2(x)}\,\mathrm dx\equiv\int\dfrac{1 - \cos^2(x)}{5\cos^2(x) - 3}\sin(x)\,\mathrm dx$$
Let $u = \cos(x)\implies\mathrm du = -\sin(x)\mathrm dx$. So,
$$\int\dfrac{1 - \cos^2(x)}{5\cos^2(x) - 3}\sin(x)\,\mathrm dx\equiv\i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3540625",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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Prove that $\sum\limits_{n=0}^\infty\binom{2n}n\frac n{4^n(n+1)^2}=\ln(16)-2$
Prove that
$$\sum\limits_{n=0}^\infty\binom{2n}n\frac n{4^n(n+1)^2}=\ln(16)-2$$
According to Wolfram the above holds. Could someone show me the steps for this?
| HINT
First note that
$$\sum_{n\geq0}\binom{2n}n\frac n{4^n(n+1)^2}=\sum_{n\geq0}\binom{2n}n\frac{(n+1)-1}{4^n(n+1)^2}=\sum_{n\geq0}\binom{2n}n\frac1{4^n(n+1)}-\sum_{n\geq0}\binom{2n}n\frac1{4^n(n+1)^2}$$
Now recall that
$$\sum_{n\geq0}\binom{2n}n\frac{x^n}{4^n}=\frac1{\sqrt{1-x}},~~~\text{for }|x|<1$$
The task boils do... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3543054",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Is there any simpler way to find a remainder in multiple divisions? I got a question as follows.
$3x-5$ is the remainder when unknown $f(x)$ is divided by $x^2-x+1$
that has relatively complicated roots. Find the remainder when $f(x)$
is divided by $(x^2-x+1)(x-1)$. Express your answer in terms of
unknown $f(1)... | A possible way is as follows:
*
*$f(x) = (x^2-x+1)q(x) + 3x-5$
*$q(x) = (x-1)r(x) + c$
Hence,
$$f(x) = (x^2-x+1)((x-1)r(x) + c) + 3x -5$$ $$= (x^2-x+1)(x-1)r(x) + c(x^2-x+1) + 3x-5$$
Now you can find $c$:
$$f(1) =c -2 \Leftrightarrow c=2+f(1)$$
So, you get
$$f(x) = (x^2-x+1)(x-1)r(x) + \color{blue}{(2+f(1))(x^2-x+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3543784",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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integral ordered pair $(x,y)$ in $x^2+y^2=2x+2y+xy$
Find all possible integral ordered pair $(x,y)$ in $x^2+y^2=2x+2y+xy$
what i try $y^2-(x+2)y+x^2-2x=0$
$$y=\frac{x+2\pm\sqrt{(x+2)^2-4(x^2-2x)}}{2}$$
$$y=\frac{x+2\pm \sqrt{-3x^2+12x+4}}{2}$$
How do i solve it Help me please
| Hint:
If $x\le-1$ or $x\ge5$ then $-3x^2+12+4<0$, so its square root is not real;
that limits the possibilities to a few you could check.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3543962",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
How to calculate $\lim\limits_{ x \to \infty } \frac {7^{2x} +7^{-2x} }{ 3(7^{2x} - 7^{-2x})}$ $$\lim\limits_{ x \to \infty } \frac {7^{2x} +7^{-2x} }{ 3(7^{2x} - 7^{-2x})}$$
someone please help i’m not sure how to compute this. i’ve tried to do it this way:$$\lim\limits_{ x \to \infty } \frac {7^{2x} +7^{-2x} }{ 3(7^{... | Put the constant outside of the limit:
$$
\begin {align}
l=&\lim\limits_{ x \to \infty }\color{blue}{\frac 1 3} \frac {7^{2x} +7^{-2x} }{ (7^{2x} - 7^{-2x})} \\
l=&\color{blue}{ \frac 1 3}\lim\limits_{ x \to \infty } \frac {7^{2x} +7^{-2x} }{ 7^{2x} - 7^{-2x}} \\
l =& \frac 1 3\lim\limits_{ x \to \infty } \frac {\color... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3544652",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
No positive real such that $\left\lfloor\frac{25}{x}+\frac{49}{a}\right\rfloor=\left\lfloor\frac{144}{x+a}-1\right\rfloor$
Let $a>0$. Prove $\nexists x\in\mathbb R^+$ s.t.
$$\left\lfloor\frac{25}{x}+\frac{49}{a}\right\rfloor=\left\lfloor\frac{144}{x+a}-1\right\rfloor$$
I know that $$k\in\mathbb Z\implies\left(\forall... | Hint: The floor function, $\lfloor x\rfloor$, is bounded by $x-1< \lfloor x\rfloor\le x$ so you have $\text{RHS}\le \left(\frac{144}{x+a}-1\right)$ and $\left(\frac{25}{x}+\frac{49}{a}\right)-1<\text{LHS}$. By considering their (single) intersection, consider how you can form an inequality with $\left(\frac{144}{x+a}-1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3545287",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Changing index in summation This question seems simple but it's been eating me for 20 mins. on a book I seen something like this :
$$
\sum_{k=8}^{\infty}\left(\frac{5}{6}\right)^{k-1}\frac{1}{6} = \frac{1}{6}\left(\frac{5}{6}\right)^{7}\sum_{j=0}^{\infty}\left(\frac{5}{6}\right)^{j}
$$
Could anyone tell me why the abov... | If in doubt, write the terms out explicitly ... lets leave the $1/6$ out
\begin{eqnarray*}
\sum_{k=8}^{\infty}\left(\frac{5}{6}\right)^{k-1} &=& \left(\frac{5}{6}\right)^{7} + \left(\frac{5}{6}\right)^{8} + \left(\frac{5}{6}\right)^{9} + \cdots \\
&=& \left(\frac{5}{6}\right)^{7} \left(1 + \frac{5}{6} + \left(\frac{5... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3546071",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Proof of $n\ge 6\implies 2n−8\le n^2−8n+ 16$ using induction.
I have to do a proof by induction for this theorem:
For each $n\in \mathbb N$ such that $n\ge 6$ we have $$2n−8\le n^2−8n+ 16$$
Is this possible, or should I do a different type of proof. I am confused how to prove that $1\in S$ when $n\ge 6$, so I am thi... | Base case: Let $n=6$. Then the LHS is $4$ whereas the RHS is $(6-4)(6-4)=4$, so $2n-8\le n^2-8n+16=(n-4)^2$ is true for $n=6$.
Induction Hypothesis: Assume that for some fixed $k\ge 6$ we have $$2k-8\le (k-4)^2.\tag{$I$}$$
When $n=k+1$: Suppose $n=k+1$. Then
$$\begin{align}
2(k+1)-8&=2k-6\\
&=(2k-8)+2 \\
&\le (k-4)^2+2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3549404",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Proving $(\sin^2 \alpha+\sin\alpha \cos \alpha)^{\sin \alpha}(\cos^2 \alpha+\sin \alpha \cos \alpha)^{\cos \alpha}\leq 1$
If $\alpha \in \left(0, \frac{\pi}{2}\right)$, prove that:
$$(\sin^2 \alpha+\sin\alpha \cos \alpha)^{\sin \alpha}(\cos^2 \alpha+\sin \alpha \cos \alpha)^{\cos \alpha}\leq 1$$
I know $\sin \alpha$ ... | The inequality is equivalent to: $$(1-\cos^2 \alpha+\sin\alpha \cos \alpha)^{\sin \alpha}(1-\sin^2 \alpha+\sin \alpha \cos \alpha)^{\cos \alpha}\leq 1$$
By Bernoulli Inequality, $$(1-\cos^2 \alpha+\sin\alpha \cos \alpha)^{\sin \alpha}(1-\sin^2 \alpha+\sin \alpha \cos \alpha)^{\cos \alpha}\\\ \\ \leq \left(1+(-\cos^2 \a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3551745",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
For $\triangle ABC$, prove $\sum_{cyc}\frac{1}{4+\cos A\cos(B-C)}\geq\frac23$
For the triangle $\triangle ABC$, prove the inquality,
$$\frac1{4+\cos A\cos(B-C)}+\frac1{4+\cos B\cos(C-A)}+\frac1{4+\cos C\cos(A-B)}\ge \frac23$$
where $A$, $B$ and $C$ are the vertex angles.
I had trouble tackling it. Have treated i... | First, we use the trig identity $ \cos A \cos B = \frac{1}{2} (\cos(A+B) + \cos(A-B)) $:
$$\frac1{4 + \cos A \cos(B-C)}+\frac1{4 +\cos B \cos(C-A)} + \frac1{4 + \cos C \cos(A-B)} \ge \frac23$$
$$ \Leftrightarrow \frac1{4 + \frac{1}{2} (\cos(A+B-C) + \cos(A-B+C))} + \frac1{4 + \frac{1}{2} (\cos(B+C-A) + \cos(B-C+A))} ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3553471",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Finding the limit of $\sqrt{4x^2+x+7}+2x$ I've been working on this problem for a while now, but I can't solve it
$$\lim\limits_{x\to-\infty}{\sqrt{4x^2+x+7}+2x}$$
I've tried multiplying by
$$\frac{\sqrt{4x^2+x+7}-2x}{\sqrt{4x^2+x+7}-2x}$$
but I didn't get it. Am I missing something really obvious? Can someone hel... | $$
\begin{aligned}
\lim_{x\to-\infty}\left(\sqrt{4x^2+x+7}+2x\right) &= \lim_{x\to-\infty}\frac{4x^2+x+7-4x^2}{\sqrt{4x^2+x+7}-2x}\\
& =\lim_{x\to-\infty}\frac{x+7}{\sqrt{4x^2+x+7}-2x}\\
&=\lim_{x\to-\infty}\frac{-1-\frac{7}{x}}{\sqrt{4+\frac{1}{x}+\frac{7}{x^2}}+2}\\
&=-\frac{1}{4}
\end{aligned}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3553757",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 0
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A random sphere containing the center of the unit cube Inspired by a Putnam problem, I came up with the following question:
A point in randomly chosen in the unit cube, a sphere is then created using the random point as the center such that the sphere must be contained inside the cube (In other words, the largest spher... | Here's the solution, confirming the previous answers, using symmetry as I suggested. I'll reorder my variables to conform with previous discussions. So we're going to consider the cube $[-1,1]^3$ and restrict to centers of the sphere in the pyramid $0\le x\le y\le z\le 1$. This means that the face $z=1$ will be the clo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3563395",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "17",
"answer_count": 2,
"answer_id": 1
} |
Show that $n \ln \left(1+\frac{1}{n}\right) \geq \frac{2 n}{2 n+1}$
Hence or otherwise show that for all positive integers $n$
$$
n \ln \left(1+\frac{1}{n}\right) \geq \frac{2 n}{2 n+1}
$$
This is related to another part of this question where I proved $\ln \left(\frac{4-t}{t}\right) \geq 2-t$ for $0<t \leq 2$ by int... | Divide by $n$ both sides, you need to prove: $\ln\left(1+x\right) > \dfrac{2x}{2+x}= 2-\dfrac{4}{2+x}, 0 < x \le 1, x = \dfrac{1}{n}$. Consider $f(x) = \ln(1+x) +\dfrac{4}{2+x}-2, 0 < x \le 1\implies f'(x) = \dfrac{1}{1+x} - \dfrac{4}{(2+x)^2} = \dfrac{(x+2)^2-4(x+1)}{(1+x)(2+x)^2}= \dfrac{x^2+4x+4-4x-4}{(1+x)(2+x)^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3564966",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
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Evaluate the indefinite integral $\int \frac{\sin^3\frac\theta 2}{\cos\frac\theta2 \sqrt{\cos^3\theta + \cos^2\theta + \cos\theta}} d\theta$ $$\int \frac{\sin^3\frac\theta 2}{\cos\frac\theta2 \sqrt{\cos^3\theta + \cos^2\theta + \cos\theta}} d\theta$$
My Attempt:
$$I = \int \frac{\sin^2\frac\theta2\cdot\sin\frac\theta2\... | Set $\sqrt t=y,t=y^2,dt=2y\ dy$
$$I =\int\dfrac{(y^2-1)2y}{(y^2+1)y\sqrt{y^2(y^4+y^2+1)}}dy$$
$$=2\int\dfrac{1-1/y^2}{(y+1/y)\sqrt{y^2+1+1/y^2}}dy$$
Set $\int(1-1/y^2)dy =z$
Then $\sqrt{z^2+1}=u\implies z^2+1=u^2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3567411",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 3,
"answer_id": 0
} |
Is there another method ? of this summation $\sum_{k=0}^{n}{n \choose k}\cos kx$ Calculate or simplify $$\sum_{k=0}^{n}{n \choose k}\cos (kx)$$
by using complex method.
Let $A=\sum_{k=0}^{n}{n \choose k}\cos(kx)$ and $B=\sum_{k=0}^{n}{n \choose k}\ i \sin(kx)$ , then
$$\begin{align*}
A+iB &= \sum_{k=0}^{n}{n \choose k}... | Here's a induction proof. I think the proof using complex numbers is better, but the next proof illustrates a useful method.
We'll prove the following in parallel
$$\sum_k\binom{n}{k}\cos(k\ x) = 2^n\cos^n\left(\frac{x}{2}\right)\cos\left(\frac{n\ x}{2}\right)$$
$$\sum_k\binom{n}{k}\sin(k\ x) = 2^n\cos^n\left(\frac{x}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3568164",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Finding the MacLaurin series of $\frac{x+3}{2-x}$ I did:
$$\frac{x+3}{2-x} = \frac{x}{2-x}+\frac{3}{2-x} = x(\frac{1}{2-x})+3(\frac{1}{2-x}) = \\
= \frac{x}{2}(\frac{1}{1-\frac{x}{2}})+\frac{3}{2}(\frac{1}{1-\frac{x}{2}}) = \\
= \frac{x+3}{2}\sum(\frac{x}{2})^n = ...?$$
The answer my professor got was $\frac{(3 + x)}{(... | $$S=\frac{x+3}{2-x} = \frac{x-2+5}{2-x}=-1+\frac{5}{2-x}$$
$$S=-1+\frac 52\frac{1}{1-\dfrac x2}$$
$$S=-1+\frac 52\sum_{n=0}^{\infty}\left (\dfrac x 2\right )^n$$
Change the indice of the sum:
$$S=-1+\frac 52+\frac 52\sum_{n=1}^{\infty}\left (\dfrac x 2\right )^n$$
Finally:
$$S=\frac 32+\frac 52\sum_{n=1}^{\infty}\left ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3570936",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
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Prove $\frac{(a+b+c)^2}{(a^2+b^2+c^2)}=\frac{\cot(A/2)+\cot(B/2)+\cot(C/2)}{\cot(A)+\cot(B)+\cot(C)}$ $\frac{(a+b+c)^2}{(a^2+b^2+c^2)}=\frac{\cot(A/2)+\cot(B/2)+\cot(C/2)}{\cot(A)+\cot(B)+\cot(C)}$
I need to solve this trigonometric identity for a trianlge.
I'm not allowed to use the formula $a+b+c=s$ where 's' is per... | For a triangle ABCz, we have denominator of LHS as
$$D=(a^2+b^2+c^2)=[(b^2+c^2-a^2)+(c^2+a^2-b^2)+(a^2+b^2-c^2)]$$ $$=[2bc \cos A+2ca \cos B+ 2ab \cos C]=2abc[\cos A/a+\cos B/b+\cos C/c]=\frac{abc}{R}[\cot A+\cot B+\cot C].~~~~(1)$$ In the last step we have used $ a/\sin A=b/\sin B=c/\sin C=2R.$
Next take the numerator... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3573793",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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If $x^4+12x-5$ has roots $x_1,x_2,x_3,x_4$ find polynomial with roots $x_1+x_2,x_1+x_3,x_1+x_4,x_2+x_3,x_2+x_4,x_3+x_4$ I have the polynomial $x^4+12x-5$ with the roots $x_1,x_2,x_3,x_4$ and I want to find the polynomial whose roots are $x_1+x_2,x_1+x_3,x_1+x_4,x_2+x_3,x_2+x_4,x_3+x_4$.
I found the roots $x_1=-1+\sqrt{... | By Vieta's formulae, we have $x_1 + x_2 + x_3 + x_4 = 0$, $x_1x_2 + x_1x_3 + x_1x_4 + x_2x_3 + x_2x_4 + x_3x_4 = 0$, $x_1x_2x_3 + x_1x_2x_4 + x_1x_3x_4 + x_2x_3x_4 = -12$, and $x_1x_2x_3x_4 = 5$. We can now calculate \begin{align*}(x_1+x_2)+(x_1+x_3)+(x_1+x_4)+(x_2+x_3)+(x_2+x_4)+(x_3+x_4) &= 3(x_1+x_2+x_3+x_4) \\ &= 0... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3574530",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 1
} |
D-operator-methods Solve the following differential equation:
$$(D^2-2D+1)y=x^2e^{3x}$$
I found the $C.F.=(c_1+c_2x)e^x$
$$\begin{align}
P.I. & =\frac{x^2e^{3x}}{(D^2-2D+1)}\\
& =e^{3x}\frac{x^2}{(D+3)^2-2(D+3)+1}\\
& =e^{3x}\frac{x^2}{D^2+6D+9-2D-6+1}\\
& =e^{3x}\frac{x^2}{D^2+4D+4}\\
& =\frac{e^{3x}}{4}x^2\left(1+\fr... | Your solution is missing a second order derivative term at this step
$$\frac{e^{3x}}{4}x^2\left(1-\frac{D^2+4D}{4}+........\right)$$
repeating the steps where $\text{C.F.}$ represents the homogeneous solution and $\text{P.I.}$ the particular solution
$$\begin{align}
\text{P.I.} & =\frac{x^2e^{3x}}{(D^2-2D+1)}\\
& =e^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3575097",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Evaluating integral using residue calculus Evaluating the following using residue calculus:
$$I = \int_0^{2\pi} {sin^4\theta}d\theta$$
I have simplified to:
$$\oint_{\lvert z \rvert = 1} \frac{(z^2-1)^4}{16iz^5}$$
z= 0 is the isolated singular point here.
Let
$$g(z) = (z^2 -1)^4 $$ and
$$h(z) = 16iz^5$$
$h(z)$ has a z... | $\oint_{|z| = 1} \frac {(z^2-1)^4}{16i z^5} \ dz$
Rather than differentiating (which is legit, but not necessary) find the Laurent series. Which just means expanding the binomial.
$\frac {1}{16i z^5} - \frac {4}{16i z^3} + \frac {6}{16i z} - \frac {4}{16i} z +\frac {1}{16i} z^3$
$\frac {6}{16i z}$ is the only term ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3575578",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove $\cos x +\cos y = 2\cos(\frac{x+y}{2})\cos(\frac{x-y}{2})$ Prove that $$\cos(x) + \cos(y) = 2\cos(\frac{x+y}{2})\cos(\frac{x-y}{2})$$ holds true for any $x, y \in \mathbb{R}$.
Even though I managed to prove its brother $\sin(x) + \sin(y)$, I haven't been able to tackle this one.
Important identities needed for th... | Hint: $ x \rightarrow (x+y)/2 $ and $y \rightarrow (x-y)/2$ in the second & third equations. Now add them together.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3576912",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
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Confused about positive and negative signs: Find the value of $\frac{(\sqrt5 +2)^6 - (\sqrt5 - 2)^6}{8\sqrt5}$. Without tables or a calculator, find the value of $\displaystyle\frac{(\sqrt5 +2)^6 - (\sqrt5 - 2)^6}{8\sqrt5}$.
I do not understand how the positive/negative signs are obtained as shown in the book; is there... | Alternatively, let $a=9+4\sqrt5$ and $b=9-4\sqrt5$. Then, $a+b=18$, $ab=1$ and,
$$\displaystyle\frac{(\sqrt5 +2)^6 - (\sqrt5 - 2)^6}{8\sqrt5}
=\frac{a^3 - b^3}{8\sqrt5}=\frac{(a-b)(a^2 +ab+b^2)}{8\sqrt5}$$
$$=\frac{\sqrt{(a+b)^2-4ab}\>[(a+b)^2 -ab)]}{8\sqrt5}
=\frac{\sqrt{18^2-4}\>(18^2 -1)}{8\sqrt5}=323$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3578191",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 4
} |
Calculate the variance for the probability density function $f(x) = x/4$ for $1 \le x \le 3$ I am trying to use this formula:
$$V(X) = E(x^2)-E^2(x)$$
I already calculated $E(X) = \frac{13}{6}$.
I tried to calculate $E(x^2)$:
$$E(x) = \int_1^3 xf(x) dx \Leftrightarrow E(x^2) = \int_1^3x^2f(x^2) = ...$$ and so on. The f... | I think you have made a mistake at the calculation. The variance is $$Var(X)=E(X^2)-[E(X)]^2=\int_1^3 x^2\cdot f(x)\, dx-\left(\int_1^3 x\cdot f(x) \, dx\right)^2$$
$$\int_1^3 x^2\cdot \frac{x}4\, dx-\left(\int_1^3 x\cdot \frac{x}4 \, dx\right)^2=\int_1^3 \frac{x^3}4\, dx-\left(\int_1^3 \frac{x^2}4 \, dx\right)^2$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3578989",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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Comparing $2$ infinite continued fractions
$A = 1 +\dfrac{1}{1 + \frac{1}{1 + \frac{1}{\ddots}}} \\
B = 2 +\dfrac{1}{2 + \frac{1}{2 + \frac{1}{\ddots}}}$
Given the two infinite continued fractions $A$ and $B$ above, which is larger, $2A$ or $B?$
I used the golden ratio on the $2$ and came up with:
$A = 1 + \dfrac{1}{... | An alternative approach
$$A = 1 +\dfrac{1}{1 + \frac{1}{1 + \frac{1}{\ddots}}} < 1 +\dfrac{1}{1} =2$$
$$B = 2 +\dfrac{1}{2 + \frac{1}{2 + \frac{1}{\ddots}}} > 2$$
showing $$A < B$$
Thus
$$2A = 2 +\dfrac{2}{A} > 2 +\dfrac{2}{2} =3$$
$$B = 2 +\dfrac{1}{B} < 2 + \dfrac12 $$
showing $$2A > B$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3584000",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 0
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Evaluate: $S=\sum_{j=1}^{\infty}\prod_{k=1}^{j}\frac{2k}{j+k+1}$ Evaluate of this sum:
$$S=\sum_{j=1}^{\infty}\prod_{k=1}^{j}\frac{2k}{j+k+1}$$
Expand out the sum:
$$S=\prod_{k=1}^{1}\frac{2k}{k+2}+\prod_{k=1}^{2}\frac{2k}{k+3}+\prod_{k=1}^{3}\frac{2k}{k+4}+\cdots$$
$$S=\frac{2}{3}+\frac{2}{4}\cdot\frac{4}{5}+\frac{2}{... | The product can be simplified to:
$$\sum_{j=1}^{\infty}\prod_{k=1}^{j}\frac{2k}{j+k+1}=
\sum_{j=1}^{\infty} \frac{j!(j+1)! 2^j}{(2j+1)!}=
\sum_{j=1}^{\infty} \frac{2^j}{\binom{2j+1}{j}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3585115",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
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Exact expression of a trigonometric integral Let $a>2$ be a real number and consider the following integral
$$
I(a)=\int_0^\pi\int_0^\pi \frac{\sin^2(x)\sin^2(y)}{a+\cos(x)+\cos(y)} \mathrm{d}x\,\mathrm{d}y
$$
My question. Does there exist a closed-form expression of $I(a)$?
Some comments. Since $a-2<a+\cos(x)+\cos(y... | With CAS help:
$$\int _0^{\pi }\int _0^{\pi }\frac{\sin ^2(x) \sin ^2(y)}{a+\cos (x)+\cos (y)}dydx=\\\mathcal{L}_q\left[\int _0^{\pi }\int _0^{\pi }\mathcal{L}_a^{-1}\left[\frac{\sin ^2(x) \sin ^2(y)}{a+\cos (x)+\cos
(y)}\right](q)dydx\right](a)=\\\mathcal{L}_q\left[\int_0^{\pi } \left(\int_0^{\pi } e^{-q (\cos (x)+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3590145",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 3,
"answer_id": 0
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Integration through partial fractions with complex roots In integrating the following: $\frac{1}{(x^2+2x+3)^2}$ I am trying to use partial fraction decomposition as follows:
$\frac{1}{(x^2+2x+3)^2} = \frac{Ax + B}{x^2+2x+3} + \frac{Cx + D}{(x^2+2x+3)^2}$
Which gives me:
$1 = (Ax + B)(x^2 +2x +3) + (Bx + C)$
And that ge... | Hint:
Notice that $\left(x^2 + 2x + 3\right)^2\equiv\left((x + 1)^2 + 2\right)^2$. That is,
$$\int\frac1{\left(x^2 + 2x + 3\right)^2}\,\mathrm dx\equiv\int\frac1{\left((x + 1)^2 + 2\right)^2}\,\mathrm dx.$$
Let $u = x + 1\implies\mathrm du = \mathrm dx$. So,
$$\int\frac1{\left((x + 1)^2 + 2\right)^2}\,\mathrm dx\equiv\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3590515",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Let $a,b,c\in R$. If $f(x)=ax^2+bx+c$ is such that $a+b+c=3$ and $f(x+y)=f(x)+f(y)+xy$ for all $x,y\in R$...
Let $a,b,c\in R$. If $f(x)=ax^2+bx+c$ is such that $a+b+c=3$ and $f(x+y)=f(x)+f(y)+xy$ for all $x,y\in R$. Then $\sum_{n=1}^{10} f(n)$ is
From the first equation
$$a(x^2+y^2+2xy)+b(x+y)+c=ax^2+bx+c+ay^2+by+c+... | From $a+b+c=3$, we have $$f(1)=a+b+c=3$$
and from $f(x+y)=f(x)+f(y)+xy$, we have
$$f(x+1)=f(x)+f(1)+x=f(x)+x+3$$
therefore
$$\sum_{n=1}^{10}f(n+1)=\sum_{n=1}^{10} (f(n)+n+3)$$
or $$\sum_{n=2}^{11}f(n)=\sum_{n=1}^{10}f(n)+\sum_{n=1}^{10}(n+3)$$
can you proceed?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3592681",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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What is the general $n \times n$ form of the divisibility matrix? Background + Motivation
I had the following idea of using digits as vectors. Let us have say I have a three digit number $a$ in the basis $\lambda$:
$$ a = a_0 + a_1 \lambda + a_2 \lambda^2$$
Now, we want to convert $\lambda$ coefficients to $\lambda+1$... | We make use of the following:
$$a = a_0 + a_1 \lambda + a_2 \lambda^2 + \dots $$
Or with $\lambda + 1$ coefficients:
$$ a = b_0 + b_1 (\lambda +1) + b_2 (\lambda +1)^2 + \dots$$
We define $\lambda +1 = \beta $ and combine the above $2$ equations:
$$ a_0 + a_1(\beta -1) + a_2 (\beta -1)^2 + \dots = b_0 + b_1 \beta + b_... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3593305",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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Show that $(n+1)^{2/3} -n^{2/3} <\frac{2}{3} n^{-1/3}$ for all positive integers
I am trying to prove
$$(n+1)^{2/3} -n^{2/3} <\frac{2}{3} n^{-1/3}$$
for all positive integers.
My attempts so far have been to Taylor expand the left hand side:
$$(n+1)^{2/3} -n^{2/3}\\
=n^{2/3}\big((1+1/n)^{2/3} -1\big)\\
=n^{2/3}... | For $f(x) = x^{\frac{2}{3}}$, then by the Mean Value Theorem,
$(n+1)^{\frac{2}{3}}-n^{\frac{2}{3}}=\frac{f(n+1)-f(n)}{(n+1)-n} = f'(c) = \frac{2}{3}c^{-\frac{1}{3}}$ for some $c \in (n,n+1)$.
Since $f'(x)=\frac{2}{3}x^{-\frac{1}{3}}=\frac{2}{3(^3\sqrt{x})}$ is decreasing, then $c > n \implies f'(c) < f'(n) = \frac{2}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3593439",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 0
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Roots of the equation $(x – 1)(x – 2)(x – 3) = 24$ The equation $(x – 1)(x – 2)(x – 3) = 24$ has the real root equal to 'a' and the complex roots 'b' and 'c'. Then find the value of $\frac{bc}{a}$
My approach is as follow $y=f(x)=(x – 1)(x – 2)(x – 3) - 24=0$
$y'=3x^2-12x+11=0$
Solving we get $x=2\pm\sqrt{\frac{1}{3}}$... | Rewrite
$$(x – 1)(x – 2)(x – 3) = 24$$
as
$$x^3-6x^2+11x-30=0$$
which factorizes as
$$(x-5)(x^2-x+6)=0$$
Thus,
$$\frac{bc}a= \frac 65$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3594574",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Identity for Catalan numbers Assume $C_{n}=\frac{1}{n+1} \binom{2n}{n}$ the Catalan numbers. I want to prove the following identity with generating functions.
$$C_{n+1}=\sum\limits_{k=0}^{\lfloor \frac{n}{2}\rfloor} \binom{n}{2k}2^{n-2k}C_k$$
I know that the generating function for the Catalan numbers is $C(x)=\frac{1-... | We may also use
$$C_n = {2n\choose n} - {2n\choose n+1}.$$
getting for the RHS of the sum
$$\sum_{k=0}^{\lfloor n/2 \rfloor}
{n\choose 2k} 2^{n-2k} {2k\choose k}
- \sum_{k=0}^{\lfloor n/2 \rfloor}
{n\choose 2k} 2^{n-2k} {2k\choose k+1}.$$
Now for the first piece we have
$${n\choose 2k} {2k\choose k} =
\frac{n!}{(n-2k... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Define $I_n=\int_0^1\frac{x^n}{\sqrt{x^2+1}}dx$ for every $n\in\mathbb{N}$. Prove that $\lim_{n\to\infty}nI_n=\frac{1}{\sqrt 2}$.
Question: Define $I_n=\int_0^1\frac{x^n}{\sqrt{x^2+1}}dx$ for every $n\in\mathbb{N}$. Prove that $$\lim_{n\to\infty}nI_n=\frac{1}{\sqrt 2}$$.
My approach: Given that $I_n=\int_0^1\frac{x^... | you can use the Binomial series
$${{\left( 1+{{x}^{2}} \right)}^{-1/2}}=\sum\nolimits_{k=0}^{\infty }{\left( \begin{align}
& -1/2 \\
& \ \ k \\
\end{align} \right){{x}^{2k}}}$$
$${n{I}_{n}}=n\int_{0}^{1}{{{x}^{n}}{{\left( 1+{{x}^{2}} \right)}^{-1/2}}dx}=n\sum\nolimits_{k=0}^{\infty }{\left\{ \left( \begin{align}
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3598920",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 5,
"answer_id": 2
} |
Summation of binomial coefficients with indices having difference of 4
Let $S$ $=$ $\binom{n}3 $ $+$ $\binom{n}7$ $+$ $\binom{n}{11} $ $.....$
Find $S$.
This question can be solved by taking the difference of the binomial series of $\frac{-(1+x)^n - (1-x)^n}{2}$ and $\frac{(1+ix)^n - (1-ix)^n}{2i}$ , where $i^2$ $=$... | Let $$S=\sum_{k=0}^{n} {n \choose 4k+3}$$
$$(1+x)^n=1+{n \choose 1} x+ {n \choose 2} x^2+ {n \choose 3}x^3+ {n \choose 4} x^4+...+{n \choose n}x^n~~~(1)$$
Let $z^4=1$ its roots are $1,a,a^2,a^3 (1,i,-1,-i)$, with $1+a+a^2+a^3=0$
Put $x=w$ in (1), then
putting $x=1,a,a^2,a^3(1,i,-1,-i)$ in (1) we get
$$2^n=\sum_{k=0}^{n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3604752",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Prove that : $m_{a}m_{b}m_{c}\leq\frac{Rs^{2}}{2}$ Let $m_{a},m_{b},m_{c}$ be the lengths of the medians and $a,b,c$ be the lengths of the sides of a given triangle , Prove the inequality :
$$m_{a}m_{b}m_{c}\leq\frac{Rs^{2}}{2}$$
Where :
$s : \operatorname{Semiperimeter}$
$R : \operatorname{circumradius}$
I know the... |
Note that the triangles ABD and EDC are similar. Then,
$$\frac{AD}{BD}=\frac{CD}{ED}\implies \frac{m_a}{\frac a2}=\frac{\frac a2}{AE-m_a}
\implies m_a^2 -AE\cdot m_a + \frac {a^2}4=0$$
which, since $AE \le 2R$ and $a=2R\sin A$, leads to
$$m_a =\frac12(AE+\sqrt{AE^2-a^2})\le \frac12\left[2R+\sqrt{(2R)^2-(2R\sin A)^2}\r... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3605636",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Parameters of the Mandart inellipse What are the parameters of the Mandart ellipse
in terms of its associated triangle?
It looks that information on the Mandart ellipse
is less known than its Steiner counterpart.
Easily found references
wiki,
MathWorld
look incomplete,
so the provided answer collects the missing part... |
The Mandart inellipse of a triangle
is an ellipse inscribed within the triangle,
tangent to its sides
at the contact points
$X_a,X_b,X_c$
of its excircles.
Its parameters
are expressed in terms of the coordinates of vertices $A,B,C$,
side lengths $a,b,c$,
semiperimeter $\rho=\tfrac12\,(a+b+c)$,
inradius $r$ and
circum... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3608055",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Suggestions for $ \lim_{(x,y)\to (0,0)} \frac{x-\sqrt{xy}}{x^2-y^2} $? I'm trying to evaluate $$
\lim_{(x,y)\to (0,0)} \frac{x-\sqrt{xy}}{x^2-y^2}
$$
over the domain $x>0$, $y>0$.
============
My attempt:
$f(x,x^2)\to +\infty$; so if the limit exists it must be $+\infty$.
I tried to evaluate the limits "near" $(x,... | If the limit exists, then it must be equal to the limit along any line, for example, $y=x/4$.
In that case,
$$ \lim_{(x,y)\to (0,0)} \frac{x-\sqrt{xy}}{x^2-y^2} =\lim_{x\downarrow 0} \frac{8}{15}\frac{x}{x^2}=\frac{8}{15}\lim_{x\downarrow 0}\frac{1}{x}=\infty.$$
| {
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"url": "https://math.stackexchange.com/questions/3608328",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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If the equation $(x^2-4)^3(x^3+1)^n(x^2-5x+6)^m=0$ has 18 roots, find m+n. If the equation $(x^2-4)^3(x^3+1)^n(x^2-5x+6)^m=0$ has 18 roots, find $m+n$.
I did and I got
$$(x+2)^3(x-2)^{3+m}(x+1)^n(x^2-x+1)^n(x-3)^m=0$$, so I find $3+3+m+n+2n+m=18\implies 2m+3n=12$, the answer is m+n=5. What I have to do now?
| In the first factor the multiplicity of the solution $x=2\vee x=-2$ is $3$, so here we have $6$ solutions.
In the second case the solution $x=-1$ has $n$ multiplicity, while in third factor there are solutions: $x=-2 \vee x=-3$ with multiplicity $m$.
We have to have:
$$3n+2m=12$$
So, every $(m,n)\in N$ such that $3n+2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3610858",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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Solving $u''(x)-\frac{u'(x)^2}{u(x)}+\frac{u'(x)}{x}=u(x)^2$ I am looking to solve the following ODE:
$$u''(x)-\frac{u'(x)^2}{u(x)}+\frac{u'(x)}{x}=u(x)^2.$$
As far as I can see, none of the common methods to solve ODEs (separation of variables, integrating factor, Laplace transform, etc.) yield anything useful.
Can I ... | $\def\d{\mathrm{d}}$From the original equation $\dfrac{\d^2 u}{\d x^2} - \dfrac{1}{u} \left( \dfrac{\d u}{\d x} \right)^2 + \dfrac{1}{x} \dfrac{\d u}{\d x} = u^2$, letting $v = \dfrac{1}{ux^2}$ as @ClaudeLeibovici does yields$$
x^2 v \frac{\d^2 v}{\d x^2} - x^2 \left( \frac{\d v}{\d x} \right)^2 + xv \frac{\d v}{\d x} ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3611555",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
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Find all units of $\Bbb Q[x] / (x^2-1)$ Find all units of $\Bbb Q[x] / (x^2-1)$.
I just tried to find them by simply putting $(ax+b)(cx+d)=1$. But it was too complicated to find the undetermined coefficients. I cannot come up with anything related in case the ideal is not irreducible
Could you give me a few hints...?
| First, $b+(x^2-1)$ is a unit for all $0\ne b\in\mathbb{Q}$.
Also note that if for $a\ne0$, $ax+b+(x^2-1)$ is a unit, then so is $x+\frac{b}{a}+(x^2-1)$. So we need to check elements of the form $x+b+(x^2-1)$.
Now:
$$(x+b)(cx+d)=cx^2+dx+bcx+bd=c+bd+(d+bc)x=1$$
i.e. $d=-bc$ and $c+bd=1$ which gives
$$c-b^2c=1\implies c=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3611957",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Given a polynomial with roots $a, b, c, d, e$, find the polynomial whose roots are $abc, abd, abe, ...$ Let $p(x)=x^5-4x^4+3x^3-2x^2+5x+1$ and say $a, b, c, d, e$ are the roots of $p$. Find the polynomial whose roots are $abc, abd, abe, acd, ace, ade, bcd, bce, bde, cde$.
By Viete's theorem we just need to find the val... | $p(x) = x^5-4x^4+3x^3-2x^2+5x+1$
$p$ has $5$ roots donated by $a$, $b$, $c$, $d$ and $e$
The elementary symmetric functions of the roots are
$a+b+c+d+e = 4$
$de+ce+be+ae+cd+bd+ad+bc+ac+ab = 3$
$cde+bde+ade+bce+ace+abe+bcd+acd+abd+abc = 2$
$bcde+acde+abde+abce+abcd = 5$
$abcde = -1$
Let $z = abc$, Computing the elemen... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3612245",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Seeking alternative methods for $\int _0^1\frac{\ln \left(x^2-x+1\right)}{x\left(1-x\right)}\:dx$ I've solved this one by first tackling,
$$\int _0^{\infty }\frac{\ln \left(x^2-x+1\right)}{x\left(1-x\right)}\:dx$$
But i'd like to know other ways to solve it since the way i did it was a bit lengthy and not that straight... | On the path of Dennis Orton...
\begin{align}J&=\int _0^1\frac{\ln \left(x^2-x+1\right)}{x\left(1-x\right)}\:dx\\
&=\int _0^1\frac{\ln \left(x^2-x+1\right)}{x}\:dx+\int _0^1\frac{\ln \left(t^2-t+1\right)}{1-t}\:dt\\
&\overset{x=1-t}=2\int _0^1\frac{\ln \left(x^2-x+1\right)}{x}\:dx\\
&=2\left(\int _0^1\frac{\ln \left(\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3613591",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 7,
"answer_id": 2
} |
Finding a closed form for $\sum_{k=0}^n (k^3-k-3)$ using generating functions I have been given a few hints. first of all I've been told to use the formula (for general series $a_n,b_n$):
$$\sum_{n=0}^\infty\left(\sum_{k=0}^n a_k b_{n-k}\right)x^n=\left(\sum_{i=0}^\infty a_i x^i\right)\left(\sum_{i=0}^\infty b_i x^i \r... | \begin{align}
&\frac{3x^3-3x^2+9x-3}{(1-x)^5}\\
&=(3x^3-3x^2+9x-3)\sum_{n=0}^\infty \binom{n+4}{4}x^n\\
&=\sum_{n=0}^\infty (3x^3-3x^2+9x-3) \binom{n+4}{4}x^n\\
&= 3 \sum_{n=0}^\infty \binom{n+4}{4}x^{n+3}-3\sum_{n=0}^\infty \binom{n+4}{4}x^{n+2}+9 \sum_{n=0}^\infty \binom{n+4}{4}x^{n+1}-3\sum_{n=0}^\infty \binom{n+4}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3614895",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
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Expressions for system of equations in a neighborhood of the origin, $x' = y+y^2 - 2xy + x^2$, $y'=x+y^2 - 2xy + x^2.$ Do you guys agree with my solution to the following problem? Please provide feedback if possible, thanks!
Find expressions for the local stable and local unstable manifolds for the following system of ... | The stable/unstable manifolds $W^s$/$W^u$ of the equilibrium point $0$ are the manifolds tangent to the stable/unstable eigenspace $E^s$/$E^u$ at $0$ (with the same dimension as the corresponding eigenspace) s.t. the trajectories starting in $W^s$/$W^u$ converge to $0$ when $t \to \pm \infty$.
After the change of varia... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3616630",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
All integer values of $\frac{a^2+b^2+1}{ab-1}$ Determine all possible values of $\frac{a^2+b^2+1}{ab-1}$ where $a,b$ are positive integers.
I am quite certain one should use a Vieta jumping argument, but I cannot complete it.
Let $\frac{a^2+b^2+1}{ab-1} = k \in \mathbb{Z}$ with $a+b$ being minimal. Clearly $k > 0$ and ... | If the ratio is $n,$ then we have $x^2 - nxy + y^2 + (n+1) = 0,$ with both $x,y$ positive integers.
Now, if $2y > nx,$ then we have $nx - y < y,$ which means that "jumping" the $y$ leads us to a smaller point in terms of $x+y.$ Same thing, if $2x > ny,$ jumping $x$ decreases $x+y.$ So, Hurwitz defined a "ground" solut... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3616943",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
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Finding Max/Min of a function in three variables, subject to two constraints
Find the maximizers or minimizers of
$$f(x_1,x_2,x_3) = \frac{1}{x^2_1 + x^2_2 + x^2_3}$$
with the given constraints
$$h_1(x_1,x_2,x_3) = 1 - x^2_1 - 2x^2_2 - 3x^2_3 = 0$$
and
$$h_2(x_1,x_2,x_3) = x_1 + 2x_2 + x_3 = 0.$$
My solutio... | Hint.
You found
$$
\begin{cases}
\dfrac{-2x_1}{(x^2_1 + x^2_2 + x^2_3)^2} = -2x_1 \lambda_1 + \lambda_2 \\
\dfrac{-2x_2}{(x^2_1 + x^2_2 + x^2_3)^2} = -4x_2 \lambda_1 + 2\lambda_2 \\
\dfrac{-2x_3}{(x^2_1 + x^2_2 + x^2_3)^2} = -6x_3 \lambda_1 + \lambda_2
\end{cases}
$$
Calling
$$
\mu_1 = \lambda_1(x^2_1 + x^2_2 + x^2_3)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3620740",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Is it valid to use operations on both sides before inequality is proven? As part of a bigger proof, I am trying to prove the inequality:
$$\frac{ab}{a^2 + b^2}< \frac{1}{2}$$
Is the following proving method correct?
$$\frac{ab}{(a^2)+(b^2)} <\frac{1}{2}$$
$$2ab < (a^2 + b^2)$$
$$0 <(a-b)^2$$
$a \neq b$ thus $a - b \neq... | What you've shown is $$\frac{ab}{a^2 + b^2} < \frac{1}{2} \implies 0 < (a-b)^2$$
But what you are (apparently) required to show is $$0 < (a-b)^2 \implies \frac{ab}{a^2 + b^2} < \frac{1}{2}$$
So the correct proof is basically the reverse of what you've written, namely:
Suppose $a ≠ b$. Then
$$\begin{align} 0 < (a-b)^2 ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
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Find $f^{(80)}(27)$ where $f(x)=(x+3)^{\frac{1}{3}}\cdot (x-27)$
Suppose that $f(x)=(x+3)^{\frac{1}{3}}\cdot (x-27)$. Use a Taylor series expansion to find $f^{(80)}(27)$.
I tried the following:
\begin{align}
f'(x) &= (x+3)^{\frac{1}{3}}\cdot 1+(x-27)\cdot \frac{1}{3}(x+3)^{\frac{-2}{3}}\\
%
f''(x)
&= \frac{1}{3}(x... | It is clear that,
$$(x+3)^{\frac{1}{3}} = 30^{\frac{1}{3}}[1+\frac{(x-27)}{30}]^{\frac{1}{3}}.$$
Therefore,
$$(x+3)^{\frac{1}{3}} = 30^{\frac{1}{3}}\sum_{k=0}^{\infty}\frac{\Gamma(\frac{4}{3})}{\Gamma(k+1)\Gamma(\frac{4}{3}-k)}\frac{(x-27)^{k}}{30^{k}},$$
and,
$$f(x) = 30^{\frac{1}{3}}\sum_{k=0}^{\infty}\frac{\Gamma(\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3630448",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 1
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Integral $\int \frac{2x^5-2x^4+2x^3+3}{2x^4-2x^3-x^2+1}dx$ as partial fraction solved using matrix equation in order to solve the integral $$\int \frac{2x^5-2x^4+2x^3+3}{2x^4-2x^3-x^2+1}\mathrm dx,$$
the expression inside the integral can be expressed as
$$(2x^5-2x^4+2x^3+3/2x^4-2x^3-x^2+1)= x+(A/(x-1))+(B/(x-1)^2)+(Cx... | Hint. This is an exercise in simplifying the integrand.
First divide to obtain $$x+\frac{3x^3-x+3}{2x^4-2x^3-x^2+1}.$$ We now focus on the second part, and we immediately recognise that the denominator vanishes when $x=1.$ Hence $x-1$ is a factor of the denominator. Dividing out, we can express the fraction as $$\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3631398",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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How many non-negative integer solutions are there for $a+b+c+d=25$ if $a\geq 1, b\geq 2,c\leq 6,d\leq 14$
How many non-negative integer solutions are there for $a+b+c+d=25$ if $a\geq 1, b\geq 2,c\leq 6,d\leq 14$
So first I let $x= a-1$, $y=b-2$ and get:
$x+y+c+d=22$
And if all are non-negative I get that there are ${... | Your solution is almost correct.
You subtract the case $"c\geq 7\; and\;d\geq 15"$ once in the case $c\geq 7$ and once in the case $d\geq 15$. So, you need to add the $1$ at the end.
Here for reconfirmation of the result the same calculation using generating functions:
$$[x^{22}]\frac{1}{1-x}\cdot\frac{1}{1-x}\cdot\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3635856",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Synthetic Proof for a Geometry problem A while back, this question was asked on MSE:
Find the length of $CE$
In fact, allow me to phrase the problem in a slightly different manner:
In quadrilateral $ABCD$, $AB=6$, $\angle{ABC}=90°$, $\angle{BCD}=45°$ and $\angle{CAD}=2\angle{ACB}$. If $DE$ is perpendicular to $AC$ wit... | Let $F$ on $CD$ so that $AF\parallel BC$ and $H$ on $AC$ so that $FH\perp CD$.
Since $AF\parallel BC$, it follows that
$$\angle CAF = \angle ACB = \frac{\angle CAD}{2}.$$
Thus $\angle CAF = \angle DAF$. Also since $AF\parallel BC$,
$$\angle AFD = \angle BCD = 45^\circ.$$
Therefore $\angle AFD = \angle AFH$. It follow... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3641917",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
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Find general solution for $\tan{x} +\tan{2x} +\tan{3x}= 0$ My approach :-
$\tan{(x+2x)}= \tan{3x}$
$\tan{x}+\tan{2x} = \tan{3x}(1-\tan{x}.\tan{2x})$
Putting this in my equation ,
$\tan{3x}.(2-\tan{x}tan{2x}) =0$
Now either
$\tan{3x}= 0$ or $\tan{x}\tan{2x}=2$
Hence we get either
$x=\frac{n\pi}{3}$ or $x= n\pi ±... | Remember
$$\tan (2 x)=\frac{2 \tan (x)}{1-\tan ^2(x)}$$
and
$$\tan(3x)=\frac{\tan (x) \left(\tan ^2(x)-3\right)}{3 \tan ^2(x)-1}$$
So the given equation can be written as
$$\frac{2 \tan (x)}{1-\tan ^2(x)}+\frac{\left(\tan ^2(x)-3\right) \tan (x)}{3 \tan ^2(x)-1}+\tan (x)=0$$
Adding the fractions
$$\frac{2 \tan (x) \lef... | {
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"url": "https://math.stackexchange.com/questions/3642067",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
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Domain of $f(x,y) = \ln((16-x^2-y^2)(x^2+y^2-4))$ $f(x,y) = \ln((16-x^2-y^2)(x^2+y^2-4))$
I'm stuck in this one because this can be rewritten as:
$$f(x,y) = \ln(16-x^2-y^2) + \ln(x^2+y^2-4)$$
Yet, the domain of the given function is $\{(16-x^2-y^2>0)\land(x^2+y^2-4>0)\} \lor \{(16-x^2-y^2<0)\land (x^2+y^2-4<0)\}$. But ... | In real analysis $\ln(ab)=ln(a)+ln(b)$ is true iff $a>0$ and $b>0$, because in other case the logarithms don't make sense.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3643553",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Prove that for any integers $a,b,c,$ there exists a positive integer $n$ such that the number $n^3+an^2+bn+c$ is not a perfect square.
Question: Prove that for any integers $a,b,c,$ there exists a positive integer $n$ such that the number $n^3+an^2+bn+c$ is not a perfect square.
Solution: Let $f:\mathbb{N}\to\mathbb... | Let's put:
$$f(n)=n^3+an^2+bn+c$$
Suppose the opposite: that for some $a,b,c$ function $f(n)$ is always a perfect square for every $n$.
It means that, for example:
$$f(n-1)=(n-1)^3+a(n-1)^2+b(n-1)+c=p^2\tag{1}$$
$$f(n+1)=(n+1)^3+a(n+1)^2+b(n+1)+c=q^2\tag{2}$$
...with $p,q$ being integers. Now subtract (1) from (2) and ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3644427",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Consider the polynomial $x^3+2x^2-5x+1$ with roots $\alpha$ and $\alpha^2+2\alpha-4$. Find the third root in terms of $\alpha$ Consider the polynomial $x^3+2x^2-5x+1$ with roots $\alpha$ and $\alpha^2+2\alpha-4$. Find the third root $\beta$ in terms of $\alpha$
I have that $\alpha^3+2\alpha^2-5\alpha+1 = 0$, so $\alpha... | Call the roots $\alpha, \alpha', \alpha''$ with $\alpha' = \alpha^2 + 2\alpha - 4$.
$$p(x) = x^3 + 2x^2 - 5x + 1 = (x - \alpha)(x - \alpha')(x - \alpha'') = q(x)(x - \alpha'')$$
with $$q(x) = x^2 + (-\alpha - 3 \alpha + 4) x + (\alpha^3 + 2 \alpha^2 - 4\alpha) = x^2 + (-\alpha - 3 \alpha + 4) x + (\alpha - 1)$$
(replac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3646675",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 3
} |
Integer within certain interval. I have to show that there is a unique integer within the interval $[a,b]$ where
$a = -\frac{3}{2} + \sqrt{\frac{9}{4} + 2(n+1)}$
and
$b = -\frac{1}{2} + \sqrt{\frac{1}{4} + 2(n+1)}$
as well as $n \in \mathbb{N}$.
Now the uniqueness is easy by showing that there can be one integer within... | Assume the contrary - $\nexists$ an integer lying in $[a,b]$. Then, $\exists\ m \in \mathbb{N}$ such that $[a,b] \subset [m,m+1]$.
So,
\begin{align}
m &< a \\
\implies m+\frac{3}{2} &< \sqrt{\frac{9}{4} + 2(n+1)} \\
\implies m^2 + 3m &< 2n+ 2
\end{align}
and
\begin{align}
b &< m+1 \\
\implies \sqrt{\frac{1}{4} + 2(n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3648185",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Proving two binomial identities I would like to show that
\begin{align}
&\sum_{j=n-k}^n\binom nj(1-x)^{n-j-1}x^{j-1}(j-nx)\\
&\qquad=\binom n{n-k}(n-k)(1-x)^kx^{n-k-1}\sum_{k=0}^{n-1}\frac{(-1)^k}n\binom{n-1}k\binom n{n-k}(n-k)(1-x)^kx^{n-k-1}\\
&\qquad=(-1)^{n-1}\sum_{k=0}^{n-1}\binom{n-1}k\binom{n+k-1}k(-x)^k
\end{al... | $\def\peq{\mathrel{\phantom{=}}{}}$For the first identity, because for $0 \leqslant j \leqslant n - 1$,\begin{align*}
&\peq \binom{n}{j} j (1 - x)^{n - j} x^{j - 1} - \binom{n}{j + 1} (j + 1) (1 - x)^{n - j - 1} x^j\\
&= \frac{n!}{(j - 1)!\, (n - j)!} (1 - x)^{n - j} x^{j - 1} - \frac{n!}{j!\, (n - j - 1)!} (1 - x)^{n ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3650275",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 2
} |
Rational function integration (not sure wich technique to use) I have the integral $$\int \frac{x^2 - 11x -2}{(x-2)(x+2)(2x+1)} dx $$
From what I have learned in class, it seems that I need to use polynomial long division
is that the correct attitude for that problem? and if yes when I try to devide I have problem beca... | Basically, it can be solved by using partial fraction decomposition :
\begin{equation}
\begin{array}{c}
=\int\left(\frac{71}{21(2 x+11)}-\frac{6}{7(x+2)}-\frac{1}{3(x-2)}\right) \mathrm{d} x \\
=\frac{71}{21} \int \frac{1}{2 x+11} \mathrm{d} x-\frac{6}{7} \int \frac{1}{x+2} \mathrm{d} x-\frac{1}{3} \int \frac{1}{x-2} ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3651938",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Evaluate $ I=\iint\limits_S (x^5+z)\ dy\ dz\ \ \text{where}\ S\ \text{is an inner side of a hemisphere}\ x^2+y^2+z^2=R^2,\ z\leqslant 0 $ Evaluate $I$:
$$
I=\iint\limits_S (x^5+z)\ dy\ dz\ \ \text{where}\ S\ \text{is an inner side of a hemisphere}\ x^2+y^2+z^2=R^2,\ z\leqslant 0
$$
My attempt:
$$
\begin{aligned}
&z=-\s... | You forgot to multiply by another factor of $r$. The integral should be
$$
\int_0^R \frac{r^7}{\sqrt{R^2 - r^2}} dr.
$$
Take $r = R\sin u$ then you obtain
$$
\int_0^R \frac{r^7}{\sqrt{R^2 - r^2}} dr \;\; =\;\; \int_0^{\frac{\pi}{2}} \frac{R^7\sin^7u}{R\sqrt{1-\sin^2u}}R\cos udu \;\; =\;\; \int_0^{\frac{\pi}{2}} R^7\si... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3652335",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
n-th power of a matrix using the division of polynomials. Consider the matrix
$$
A=\begin{pmatrix}
0 & 0 & 0\\
-2 & 1 & -1\\
2 & 0 & 2
\end{pmatrix}
$$
*
*Calculate $A^3-3A^2+2A$.
*What is the remainder of the division of the polynomial $X^n$ by the polynomial $X^3-3X^2+2X$.
*Calculate $A^n$ for every natural num... | Write
$$x^n = q(x)(x^3-3x^2+2x) + r(x) = q(x)x(x-1)(x-2)+r(x)$$
for polynomials $q,r \in \Bbb{R}[x]$ where $\deg r \le 2$. Plugging in $x = 0,1,2$ gives
$$r(0) = 0, \quad r(1)=1, \quad r(2)=2^n$$
so $r(x) = (2^{n-1}-1)x^2+(-2^{n-1}+2)x$. Now we get
$$A^n = q(A)(A^3-3A^2+2A) + r(A) = r(A) = (2^{n-1}-1)A^2+(-2^{n-1}+2)A$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3652565",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Pairs $(a,b)\in F^2$ such that $a^6+b^6=1$ where $F$ is a finite field of 25 elements Let $F$ be a field of 25 elements, and consider the group $G$ of all $2\times 2$ matrices $A$ with entries in $F$ satisfying $A_5A=I$, where $I$ is the identity matrix and
$A_5=\begin{pmatrix} a^5 & c^5 \\ b^5 & d^5 \end{pmatrix}... | One needs to recognise that $a^6$ is the norm map from $F=\Bbb F_{25}$ to the field
$\Bbb F_5$ of order $5$, so $a^6\in\{0,1,2,3,4\}$. $a^6=0$ only has the solution $0$
but for $c\in\{1,2,3,4\}$ $a^6=c$ has six solutions in $F$. We need to count
the solutions of $(a^6,b^6)=(c,d)$ where $(c,d)$ is one of the pairs $(0,1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3652929",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
$\lim_n \frac{1}{n}\int_{1/n^2}^{+\infty} \frac{\tanh(\frac{\sqrt{x}}{n})}{x^2\sqrt{(n+1)x}}\ dx$ Let
$$f(x)=\frac{\tanh(\frac{\sqrt{x}}{n})}{x^2\sqrt{(n+1)x}} \ \ \ \text{and} \ \ f_n=f\chi_{[ 1/n^2,+\infty)}$$
I want to compute $$\lim_n \frac{1}{n}\int_{1/n^2}^{+\infty}f \ dx$$
My solution is the following:
Since $$... | Change variables with $$ y = \frac{\sqrt{x}}{n} \Rightarrow dy = \frac{dx}{2n\sqrt{x}}. $$
Also, $$ \sqrt{x} = ny \Rightarrow x^2 = n^4y^4. $$
When $ x = \frac{1}{n} $ we get that $ y = \sqrt{1 + \frac{1}{n^2}} \to 1.$
Finally, observe that $$ \int_1^{\infty} \frac{tanh(y)}{y^5} < + \infty$$ hence $$ \lim_{n \to \infty... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3656913",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Find the closed path of given system of ordinary differential equation Consider the system of ordinary differential equations:
$$\begin{cases}\frac{dx}{dt}=4x^3y^2-x^5y^4\\
\frac{dy}{dt}=x^4y^5+2x^2y^3\end{cases}$$
Then for this system there exist
$1).$ A closed path in $\left \{(x,y) \in \mathbb{R^2}|x^2+y^2 \leq 5 ... | Starting from this line in your answer:
$$-(x^2y^2+2)ydx+(4-x^2y^2)xdy=0$$
$$-x^2y^2(ydx+xdy)-2ydx+4xdy=0$$
$$-x^2y^2d(xy)-2ydx+4xdy=0$$
Divide by $xy$:
$$-xyd(xy)-2\dfrac {dx}x+4\dfrac {dy}y=0$$
$$-(xy)^2-4\ln x+8\ln y=0$$
$$-(xy)^2+4\ln \dfrac {y^2}{x}=C$$
Or if you prefer:
$$(xy)^2+4\ln \dfrac x{y^2}=C_1$$
Our answe... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3661361",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
Taylor series approximation of Gaussian $Q$ function , $Q(x) = \frac{1}{{\sqrt {2\pi } }}\int_x^\infty {{e^{ - \frac{{{v^2}}}{2}}}dv}$ $ \require{newcommand}\newcommand{\Erfc}{\operatorname{Erfc}}$
I am trying to find a Taylor series expansion for the Gaussian $Q$ function. I have seen that error function $\Erfc(x)$ is... | You can obtain a Taylor series as follows:
\begin{align*}
Q(x) & = \frac{1}{{\sqrt {2\pi } }}\int_0^{ + \infty } {e^{ - t^2 /2} dt} - \frac{1}{{\sqrt {2\pi } }}\int_0^x {e^{ - t^2 /2} dt} = \frac{1}{2} - \frac{1}{{\sqrt {2\pi } }}\int_0^x {e^{ - t^2 /2} dt}
\\ &
= \frac{1}{2} - \frac{1}{{\sqrt {2\pi } }}\int_0^x {\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3672516",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How to prove this algebraic version of the sine law? How to solve the following problem from Hall and Knight's Higher Algebra?
Suppose that
\begin{align}
a&=zb+yc,\tag{1}\\
b&=xc+za,\tag{2}\\
c&=ya+xb.\tag{3}
\end{align}
Prove that
$$\frac{a^2}{1-x^2}=\frac{b^2}{1-y^2}=\frac{c^2}{1-z^2}.\tag{4}$$
(I suppose tha... | By (1) and (3), $a=ay^2 + bxy +bz.$ Thus, $a(1-y^2)=b(xy+z)$ so that $$a^2(1-y^2)=ab(xy+z).$$
In a similar way, we derive from (2) and (3) that
$$b^2(1-x^2)=ab(xy+z).$$
Thus, the left sides of the two displayed equations are equal, yielding the first equality in (4). By symmetry, we're done.
IOW replace $(a,c)$ by $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3674276",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "26",
"answer_count": 4,
"answer_id": 2
} |
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.