Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Diophantine Equation: $xy+ax+by+c=0$ How to find integer solutions $x,y$ of $xy+ax+by+c=0$ for given $a,b,c \in \mathbb{Z}$?
Is there somewhere a treatise on this kind of equations?
| (I'm guessing that what was intended was $xy+ax+by+c=0$.)
\begin{align}
u & = x+y \\
v & = x-y \\[8pt]
\text{so that }x & = \frac{u+v}{2} \\[8pt]
y & = \frac{u-v}{2}
\end{align}
Then $xy = \dfrac{u^2-v^2}{4}$. So
$$
xy + ax + by + c =0
$$
becomes
$$
\frac{u^2-v^2}{4} + a\frac{u+v}{2} + b \frac{u-v}{2} + c =0,
$$
and then multiplying both sides by $4$, we get
$$
u^2 - v^2 + 2(a+b)u + 2(a-b)v + c = 0.
$$
Then one can complete the square, getting
$$
(u+a+b)^2 - (v-a+b)^2 = (a+b)^2+(a-b)^2 - c = 2a^2+2b^2-c.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/449845",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
} |
The evaluation of the infinite product $\prod_{k=2}^{\infty} \frac{k^{2}-1}{k^{2}+1}$ How does one show that$$ \prod_{k=2}^{\infty}\frac{k^{2}-1}{k^{2}+1} =\frac{\pi}{\sinh \pi} ?$$
My attempt: $$ \begin{align} \prod_{k=2}^{\infty}\frac{k^{2}-1}{k^{2}+1} &= \lim_{n \to \infty} \prod_{k=2}^{n}\frac{(k-1)(k+1)}{(k-i)(k+i)} \\ &= \lim_{n \to \infty} \frac{\Gamma(n) \Gamma(n+2) \Gamma(2-i) \Gamma(2+i)} {2 \Gamma(n+1-i) \Gamma(n+1+i)} \\ &= \lim_{n \to \infty} \frac{\Gamma(n)\Gamma(n+2) (1-i) \Gamma(1-i) (1+i)i \Gamma(i)}{2 \Gamma(n+1-i)\Gamma(n+1+i)} \\ &= \frac{\pi}{\sinh \pi}\lim_{n \to \infty} \frac{\Gamma(n)\Gamma(n+2)}{\Gamma(n+1-i) \Gamma(n+1+i)} \end{align}$$
I'm not sure how to go about showing that the limit evaluates to $1$.
EDIT:
To evaluate that limit we can use the fact that $ \displaystyle \frac{\Gamma(n)}{\Gamma(n+z)} \sim n^{-z}$ as $ n \to \infty$.
$$ \begin{align} \lim_{n \to \infty} \frac{\Gamma(n)\Gamma(n+2)}{\Gamma(n+1-i) \Gamma(n+1+i)} &= \lim_{n \to \infty} \frac{\Gamma(n) (n+1)n \Gamma(n)}{(n-i)\Gamma(n-i) (n+i)\Gamma(n+i)} \\ &= \lim_{n \to \infty} \frac{\Gamma(n)n^{-i}}{\Gamma(n-i)} \frac{\Gamma(n) n^{i}}{\Gamma(n+i)} \frac{n^2+n}{n^{2}+1} \\ &= (1)(1)(1) \\ &= 1 \end{align}$$
| If you know the product representation
$$\sin (\pi z) = \pi z \prod_{k = 1}^\infty \left(1 - \frac{z^2}{k^2}\right),\tag{1}$$
it is rather easy.
Setting $z = i$, we obtain
$$\frac{\sin (\pi i)}{\pi i} = \frac{\sinh \pi}{\pi} = \prod_{k=1}^\infty \left(1 - \frac{i^2}{k^2}\right) = \prod_{k=1}^\infty \left(1 + \frac{1}{k^2}\right) = 2 \prod_{k=2}^\infty \left(\frac{k^2+1}{k^2}\right).$$
On the other hand,
$$\prod_{k=2}^n \left(\frac{k^2-1}{k^2}\right) = \frac12\cdot \frac32\cdot \frac23\cdot \frac43 \dotsb \frac{n-1}{n}\cdot \frac{n+1}{n} = \frac12\cdot\frac{n+1}{n},$$
so
$$\prod_{k=2}^\infty \left(\frac{k^2-1}{k^2}\right) = \frac12.$$
Now divide.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/449914",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
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Sequence problem, find root the equation $x^3-5x+1=0$ has a root in $(0,1)$. Using a proper sequence for which $$|a(n+1)-a(n)|\le c|(a(n)-a(n-1)|$$ with $0<c<1$ , find the root with an approximation of $10^{-4}$.
| You can attempt this: define a sequence $x_n$ by
$$
x_{n+1} = \frac 15\left(1 + x_n^3\right).
$$
We know that if $x_1 \in (0, 1)$, then $x_n \in (0, 1)$ for all $n$. Next, we show that it satisfies the said condition:
\begin{align*}
x_{n+1} - x_n & = \frac 15(1 + x_n^3 - 5x_n)\\
& = \frac 15\left(1 + x_n^3 - (1 + x_{n-1}^3)\right)\\
& = \frac 15\left(x_n^3 - x_{n-1}^3\right) \\
& = \frac 15\left(x_n - x_{n-1}\right)\left(x_n^2 + x_nx_{n-1} + x_{n-1}^2\right) \\
\therefore
\left|x_{n+1} - x_n\right|
& = \frac 15\left|\left(x_n - x_{n-1}\right)\left(x_n^2 + x_nx_{n-1} + x_{n-1}^2\right)\right| \\
& \le \frac 15\left|x_n - x_{n-1}\right| \cdot 3 \\
& \le \frac 35\left|x_n - x_{n-1}\right|.
\end{align*}
From this, we see that the sequence is Cauchy, hence convergent. Let $x$ be the limit of the sequence. Take the limit $n \to \infty$ in the first equation (which defines $x_{n+1}$ from $x_n$) to get
$$
x = \frac 15(1 + x^3).
$$
This means $x$ is a root of $f(x) = x^3 - 5x + 1$.
To compute $x$ numerically within a given error tolerance $\epsilon$, we need to find $x_N$ such that $|x_N - x| \le \epsilon$. However, we do not know $x$, so instead, we can use the following criterion to find $N$: for all $n > N$, $|x_{n} - x_N| < \epsilon$.
This criterion is sufficient because if the inequality holds for all $n > N$,
it will follow that $\lim_{n \to \infty} |x_n - x_N| = |x - x_N| \le \epsilon$.
One convenient way to guarantee $|x_n - x_N| < \epsilon$ for all $n > N$ is via the triangle inequality:
$$
|x_n - x_N| \le \sum_{k=N}^{n-1} \left|x_{k+1} - x_k\right|
\le \sum_{k=N}^\infty \left|x_{k+1} - x_k\right|
\le \frac{\left(\frac 35\right)^N}{1 - \frac 35} = \frac 52\left(\frac 35\right)^N.
$$
So, if we can find $N$ such that $\frac 52\left(\frac 35\right)^N < \epsilon$, we will get $|x_N - x| \le \epsilon$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/450142",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
The map $p(x)\mapsto p(x+1)$ in vector space of polynomials I encountered this question in a previous year paper of an exam. I wish somebody could help me how to go by this question.
Choose the correct option
Let $N$ be the vector space of all real polynomials of degree at most $3$. Define
$$S:N \to N \ \text{by} \ (Sp)(x)=p(x+1), \ p \in N$$
Then the matrix of $S$ in the basis $\{1,x,x^2,x^3\}$, considered as column vectors, is given by
*
*\begin{bmatrix}1 &0 &0& 0\\
0& 2& 0& 0\\
0 &0& 3& 0\\
0& 0& 0& 4\end{bmatrix}
*\begin{bmatrix}1& 1& 1& 1\\
0& 1 &2 &3\\
0 &0& 1& 3\\
0& 0 &0 &1\end{bmatrix}
*\begin{bmatrix}1& 1& 2& 3\\
1 &1 &2& 3\\
2 &2 &2 &3\\
3 &3& 3 &3\end{bmatrix}
*\begin{bmatrix} 0& 0& 0& 0\\
1& 0 &0 &0\\
0& 1& 0& 0\\
0 &0 &1& 0\end{bmatrix}
How to tackle p(x+1) thing?
| So, you can begin computing the images of $1, x, x^2$ and $x^3$. For instance:
$$
S(1) = 1 \ ,
$$
because it doesn't matter wherever you evaluate the constant polynomial $1$, its value will always be $1$. Next, you compute also
$$
S(x) = x + 1 \ , \qquad S(x^2) = (x + 1)^2 = x^2 + 2x + 1
$$
and
$$
\qquad S(x^3) = (x+1)^3 = \dots
$$
Then, you think about the coordinates of these polynomials in the basis $1, x, x^2, x^3$. For instance,
$$
1 = 1\cdot 1 + 0\cdot x +0 \cdot x^2 + 0 \cdot x^3 \ .
$$
So the coordinates of $S(1) = 1$ are
$$
\begin{pmatrix}
1 \\
0 \\
0 \\
0 \\
\end{pmatrix}
$$
This way you've obtained the first column of your matrix. The second column will be the coordinates of $S(x) = x + 1$. That is,
$$
\begin{pmatrix}
1 \\
1 \\
0 \\
0 \\
\end{pmatrix}
$$
The third one, those of $S(x^2)$,
$$
\begin{pmatrix}
1 \\
2 \\
1 \\
0 \\
\end{pmatrix}
$$
And the last one?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/450870",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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} |
$n$ and $n^5$ have the same units digit? Studying GCD, I got a question that begs to show that $n$ and $n^5$ has the same units digit ...
What would be an idea to be able to initiate such a statement?
testing
$0$ and $0^5=0$
$1$ and $1^5=1$
$2$ and $2^5=32$
In my studies, I have not got "mod", please use other means, if possible of course.
I demonstrated in a previous period that
$$2|n^5-n$$because $$n^5-n=(n+1)n(5n^4+5n+5)$$,
and$$5|n^5-n$$By Fermat's Little Theorem
Only I do not understand what should happen to the units of the two numbers are equal ... What must occur?
| Without using any modular arithmetic:
$$n^5-n=n(n-1)(n+1)(n^2+1)=n(n-1)(n+1)(n^2-4+5)=n(n-1)(n+1)(n^2-4)+5n(n-1)(n+1)=$$
$$=(n-2)(n-1)n(n+1)(n+2)+5(n-1)n(n+1)$$
$(n-2)(n-1)n(n+1)(n+2)$ is the product of 5 consecutive integers thus divisible by 2 and 5.
$5n(n-1)(n+1)$ is multiple of $5$ and even.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/451927",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Finding a Liapunov function. Is my analysis correct? Consider the following system of differential equations $x'=f(x)$:
\begin{equation}
x_1' = -x_1 -x_2^3 \\
x_2' = x_1-x_2
\end{equation}
Here is the Liapunov function I wish to use: $V(x_1, x_2) = \alpha x_1^2 + \beta x_2^4$. So I need to find $\alpha$ and $\beta$ that satisfy the following 3 Liapunov conditions:
The conditions to be a Liapunov funtion w.r.t the origin $x_0$:
1) $V(0, 0) = 0$
Condition 1) is clearly true.
2) $V(x) > 0$ for all $x\neq x_0$ This is clearly true for $\alpha \geq 0$ and $\beta \geq 0$ but not both equal to $0$.
3) $\nabla V\cdot f \leq 0$ for all $x$:
\begin{align}
\nabla V\cdot f &= 2\alpha x_1(-x_1-x_2^3) + 4\beta x_2 (x_1-x_2)\\
&= -2\alpha x_1^2 -2\alpha x_1x_2^3 + 4\beta x_1x_2 - 4\beta x_2^2\\
&= -4\beta x_2^2 -2\alpha x_1^2 -2\alpha x_1x_2^3 +4\beta x_1x_2\\
&= -4\beta x_2^2 -2\alpha + x_1x_2(-2\alpha x_2^2+4\beta)
\end{align}
In order for condition 3) to hold, that is $\nabla V(x)\cdot f \leq 0$ for all $x$, then the third term must equal $0$. That is we need:
\begin{equation}
-2\alpha x_2^2+4\beta =0
\end{equation}
If this is the case, all solutions depend on $x_2$ unless both $\alpha$ and $\beta$ equal $0$ which cannot happen as condition 2) would not be satisfied.
This is where I am stuck. Please help!
| We are given
$$x_1' = -x_1 -x_2^3 \\
x_2' = x_1-x_2$$
with
$$V(x_1, x_2) = \alpha x_1^2 + \beta x_2^4$$
We have:
\begin{align}
V' &= 2 \alpha x_1 x'_1 + 4 \beta x_2^3 x'_2 \\
&=2 \alpha x_1(-x_1 -x_2^3) + 4 \beta x^3_2(x_1-x_2)\\
&=-2 \alpha x_1^2 -2 \alpha x_1 x^3_2 + 4 \beta x_1 x^3_2 - 4 \beta x^4_2 \\
&=-2 \alpha x^2_1 + x_1 x^3_2(4 \beta -2 \alpha) -4 \beta x^4_2
\end{align}
Now, what if we let $4 \beta - 2 \alpha = 0 \rightarrow 4 \beta = 2 \alpha$ to get rid of the $x_1 x^3_2$ term, we have:
$$V' = -2 \alpha x^2_1 - 4 \beta x^4_2 = -2 \alpha x^2_1 - 2 \alpha x^4_2 = -2 \alpha (x^2_1 + x^4_2)$$
Clearly this is negative $\forall (x_1,x_2)$, so we have (since $4 \beta = 2 \alpha \rightarrow \beta = \dfrac{\alpha}{2}$):
$$V(x_1, x_2) = \alpha x^2_1 + \beta x^4_2 = \alpha x^2_1 + \dfrac{\alpha}{2} x^4_2$$
Lets verify this:
\begin{align}
V' &= 2 \alpha x_1 x'_1 + 2 \alpha x^3_2 x'_2\\
&= 2 \alpha x_1(-x_1-x^3_2)+ 2 \alpha x^3_2(x_1-x_2) \\
&= -2 \alpha x^2_1 -2 \alpha x_1 x^3_2 + 2 \alpha x_1 x^3_2 - 2 \alpha x^4_2\\
&= -2 \alpha (x^2_1 + x^4_2)
\end{align}
That checks out and meets all the criteria for a Liapunov function.
Note: do you now see where you made a slight error with the derivative?
This is telling us there are no closed orbits and we can see this in a phase portrait plot as follows (we could have also found the Jacobian and critical points to determine this).
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
the strategy about a $0-1$ matrix game Given a $4\times4$ binary matrix as following,
$\left( \begin{matrix}
1 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 \\
0 & 0 & 1 & 0 \\
\end{matrix} \right)$
if you choose $a_{ij}$ in this matrix, then all the number in i th row and j th column will be changed( from 0 to 1 or from 1 to 0).
If all number in matrix turns to $0$, you win the game.
the steps may described like $a_{11}-a_{12}-a_{32}$.
Can any given $0-1$ matrix turn to a 0 matrix?
What is the strategy?
Thanks.
| This is yet another variant of the Lights Out puzzle. For each button (i,j) we define a toggle matrix $T_{ij}$ where the entry is 1 if the button in that location changes state, or if it doesn't.
For example, $T_{11}=\left(\begin{matrix}1&1&1&1&\\1&0&0&0&\\1&0&0&0\\1&0&0&0\end{matrix}\right)$
The initial matrix
$A=\left( \begin{matrix}
1 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 \\
0 & 0 & 1 & 0 \\
\end{matrix} \right)$
The problem turns to a solving a linear equations in {0,1} system.
$A+\sum{x_{ij}T_{ij}}=0(\mod2)$, $x_{ij}\in\{0,1\}$
which is equivalent to
$\sum{x_{ij}T_{ij}}=A(\mod2)$, $x_{ij}\in\{0,1\}$
then it becomes a linear equations as followings:
$\left(\begin{matrix}
1 & 1 & 1 & 1 & 0 & 0 & 1 & 0 & 0 \\
1 & 1 & 1 & 0 & 1 & 0 & 0 & 1 & 0 \\
1 & 1 & 1 & 0 & 0 & 1 & 0 & 0 & 1 \\
1 & 0 & 0 & 1 & 1 & 1 & 1 & 0 & 0 \\
0 & 1 & 0 & 1 & 1 & 1 & 0 & 1 & 0 \\
0 & 0 & 1 & 1 & 1 & 1 & 0 & 0 & 1 \\
1 & 0 & 0 & 1 & 0 & 0 & 1 & 1 & 1 \\
0 & 1 & 0 & 0 & 1 & 0 & 1 & 1 & 1 \\
0 & 0 & 1 & 0 & 0 & 1 & 1 & 1 & 1 \\
\end{matrix}\right)(x_{11},x_{12},x_{13},\cdots,x_{33})^T=(a_{11},a_{12},a_{13},\cdots,a_{33})^T$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Existence of linear mapping I am studying for an exam in linear algebra and I am having trouble solving the following:
Do linear mappings $\phi : \mathbb{R}^2 \rightarrow \mathbb{R}^2$ with the following properties exist?
$1)$ $\phi_1 \begin{pmatrix} 2 \\ 0 \end{pmatrix} = \begin{pmatrix} 0 \\ 1 \end{pmatrix} $, $\phi_1 \begin{pmatrix} 1 \\ 1 \end{pmatrix} = \begin{pmatrix} 5 \\ 2 \end{pmatrix} $, $\phi_1 \begin{pmatrix} 1 \\ 2 \end{pmatrix} = \begin{pmatrix} 2 \\3 \end{pmatrix} $
$2)$ $\phi_2 \begin{pmatrix} 2 \\ 3 \end{pmatrix} = \begin{pmatrix} 0 \\0\end{pmatrix} $, $\phi_2 \begin{pmatrix} 0 \\ 0 \end{pmatrix} = \begin{pmatrix} 2\\ 3 \end{pmatrix} $
$3)$ $\phi_3 \begin{pmatrix} 1 \\ -1 \end{pmatrix} = \begin{pmatrix} 1\\ 0 \end{pmatrix} $, $\phi_3 \begin{pmatrix} -2 \\ 5 \end{pmatrix} = \begin{pmatrix} -3 \\ 1 \end{pmatrix} $, $\phi_3 \begin{pmatrix} 2 \\ 1 \end{pmatrix} = \begin{pmatrix} 1 \\1 \end{pmatrix} $
I know that the following properties have to hold for a linear mapping:
*
*$f(\mathbf{x}+\mathbf{y}) = f(\mathbf{x})+f(\mathbf{y})$
*$f(\alpha \mathbf{x}) = \alpha f(\mathbf{x})$
*$f(0) = 0 $
I conclude that $2)$ is a not a linear mapping since $\phi(0)$ is not mapped to $0$.
But how shall I proceed with the others?
| You're quite right about $\phi_2$. For the others, you should see whether you can compute what one of the three values ought to be just using the two given values. So for instance $\begin{pmatrix}2\\1\end{pmatrix}=4\begin{pmatrix}1\\-1\end{pmatrix}+\begin{pmatrix}-2\\5\end{pmatrix}$. Can you conclude anything from this observation about whether the proposed value for $\phi_3\begin{pmatrix}2\\1\end{pmatrix}$ is consistent with the first two givens? If it is consistent, then you just need to argue that there's no inconsistency in the first two points. If it's not, then no such map can exist.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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closed form of $\int_{0}^{2\pi}\frac{dx}{(a^2\cos^2x+b^2\sin^2x)^n}$ closed form of $$\int_{0}^{2\pi}\frac{dx}{(a^2\cos^2x+b^2\sin^2x)^n}$$
for $a,b>0$
n=1 we get
$$\int_{0}^{2\pi}\frac{dx}{(a^2\cos^2x+b^2\sin^2x)^1}=\frac{2\pi}{ab}$$
n=2 we get
$$\int_{0}^{2\pi}\frac{dx}{(a^2\cos^2x+b^2\sin^2x)^2}=\frac{\pi(a^2+b^2)}{a^3b^3}$$
but what the integral for n ????
I hope to be there two different solution one by contour integration and the other by real analysis and thanks for all
| Here is a different solution using differentiation under integral sign.
We firstly look at the the case $n=1$, then
\begin{align*}
I_1(\alpha,\beta) & = \int_{0}^{2\pi}\frac{dx}{\alpha \cos^2x+\beta \sin^2x} \\
& = \int_{0}^{2\pi} \frac{\sec^2x}{\alpha + \beta \tan^2x} \,\mathrm{d}x\\
& = \frac{1}{\beta} \int_0^{2\pi} \frac{1}{\left(\sqrt{\frac{\alpha}{\beta}}\right)^2 + \tan^2 (x)}\; \mathrm{d}(\tan x)\,\\
& = \frac{1}{\sqrt{\alpha \beta}} \left[ \left(\tan^{-1}\left(\sqrt{\frac{\beta}{\alpha}}\tan (x)\right)\right) \right]_0^{2\pi} = \frac{2\pi}{\sqrt{\alpha\beta}}
\end{align*}
Where $a^2 = \alpha$ and $b^2 = \beta$.
Differentiating $I_n$ with respect to $\alpha$ and $\beta$ yields
\begin{align*}
\frac{\mathrm{d}}{\mathrm{d}\alpha} I_n & = - \int_0^{2\pi} \frac{n \cos^2x}{(\alpha \cos^2x + \beta \sin^2x)^{n+1}} \\
\frac{\mathrm{d}}{\mathrm{d}\beta} I_n & = - \int_0^{2\pi} \frac{n \sin^2x}{(\alpha \cos^2x + \beta \sin^2x)^{n+1}}
\end{align*}
Addition gives that
\begin{align}
\nabla I_n = \left( \frac{\mathrm{d}}{\mathrm{d}\alpha} +
\frac{\mathrm{d}}{\mathrm{d}\beta} \right) I_1
= -n \int_0^{2\pi} \frac{(\sin^2x + \cos^2x)\mathrm{d}x}{(\alpha \cos^2x + \beta \sin^2x)^2}
= -n I_{n+1}
\end{align}
Where $\nabla$ is the nabla-operator. This gives us the recurrence relation
\begin{align}
\nabla I_{n-1} = (1-n) I_n \tag{1}
\end{align}
Which can be used to evaluate each $n \in \mathbb{N} / \{ 1 \}$.
This is a fine answer, but if one wants an explicit expression it can
be shown by induction that
$$ \boxed{ \displaystyle
I_{n+1} = \frac{2\pi(-1)^{n}}{n!} \left( \frac{\partial }{\partial \alpha} + \frac{\partial}{\partial \beta} \right)^n \frac{1}{\sqrt{\alpha\beta}}
}
$$
Where the last part is the binomial expansion of the partial derivatives, and this holds for all $n \in \mathbb{N}$.
Example for $n=3$ we have
$$
\left( \frac{\partial }{\partial \alpha} + \frac{\partial}{\partial \beta} \right)^2 f(\alpha,\beta) = \left(\frac{\partial^2}{\partial \alpha^2}+2\frac{\partial^2}{\partial \alpha \, \partial \beta}+\frac{\partial^2}{\partial \beta^2}\right)f(\alpha,\beta)
$$
We can also express the partial derivatives as a sum like
$$
\left( \frac{\partial }{\partial \alpha} + \frac{\partial}{\partial \beta} \right)^n
=
\sum_{k=0}^b \binom{n}{k} \frac{\partial^{n-k} }{\partial \alpha^{n-k} }
\frac{\partial^{k} }{\partial \beta^{k} }
=
\sum_{k=0}^\infty \binom{n}{k} \frac{\partial^{n} }{\partial \alpha^{n-k} \,\partial^k \beta}
$$
without that doing us much good. For actually computing explicit values of $n$ the recurrence relation is more convenient.
EDIT: To give an intuition of the "closed" expression one can first rewrite
$(1)$ to
$$
I_n = - \frac{1}{n-1} \nabla I_{n-1}
$$
Direct computation of the first values gives
One can also show that the expression satisfies $(1)$ directly by insertion, and the use of the chain rule.
| {
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Summation and proof by induction question: $\sum_{j=1}^{n}j(j+1)(j+2)=\frac{n(n+1)(n+2)(n+3)}{4}$ I can't figure this out based on examples in textbooks, etc.
Show via induction that $\sum_{j=1}^{n}j(j+1)(j+2)=\frac{n(n+1)(n+2)(n+3)}{4}$
So far, I have:
(a) base case
$P(1)= 1(1+1)(1+2) = \frac{1(1+1)(1+2)(1+3)}{4} = 6 = 6$
(b) inductive step $P(k)\rightarrow P(k+1)$
If $P(k)$ is true, then
$P(k+1) = (2)(3)+(2)(3)(4)+(3)(4)(5)+\cdots+(k+1)(k+2)(k+3)= \frac{(k+1)(k+2)(k+3)(k+4)}{4}$
But this doesn't seem to be going anywhere. Any ideas?
| If $$\sum_{j=1}^{n}j(j+1)(j+2)=\frac{n(n+1)(n+2)(n+3)}{4}$$
then $$\sum_{j=1}^{n+1}j(j+1)(j+2)=\sum_{j=1}^{n}j(j+1)(j+2)+(n+1)(n+2)(n+3)$$
$$=\frac{n(n+1)(n+2)(n+3)}{4}+(n+1)(n+2)(n+3)=$$
$$=\frac{n(n+1)(n+2)(n+3)}{4}+\frac{4(n+1)(n+2)(n+3)}{4}=$$
$$=\frac{(n+1)(n+2)(n+3)(n+4)}{4}$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
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} |
Sequence and Series - If $a_n =\int^{\frac{\pi}{2}}_0 \frac{\sin^2nx}{\sin^2x}dx,$..... If $\displaystyle a_n =\int^{\frac{\pi}{2}}_0 \frac{\sin^2nx}{\sin^2x}dx, $ then find the value of
$$\begin{vmatrix}
a_1 & a_{51} & a_{101} \\ a_2 &a_{52} & a_{102}\\ a_3 & a_{53}&a_{103}\\
\end{vmatrix}.$$
My approach :
We know that $S_{n+1} - S_{n} = T_n$ where $S_{n+1} $ is sum of $n+1$ term and $S_n $ is sum of $n$ terms and $T_n $ is $n$th term.
Can we use this here somehow ..... as I used :
$\displaystyle \frac{\sin^2(n+1)x}{\sin^2x}- \frac{\sin^2nx}{\sin^2x} = \frac{\sin^2(n+1)x - \sin^2{nx}}{\sin^2x} $.... now what to do further... please suggest... thanks.
| HINT:
As $\displaystyle a_n =\int^{\frac\pi2}_0 \frac{\sin^2nx}{\sin^2x}dx, $
If $n=1, \displaystyle a_1 =\int^{\frac\pi2}_0 dx=\frac\pi2$
Else we are interested in $n>1$
$\displaystyle a_{n+1}-a_n=\int^{\frac\pi2}_0 \frac{\sin^2nx-\sin^2(n-1)x}{\sin^2x}dx$
Using $\sin^2A-\sin^2B=\sin(A-B)\sin(A+B),$
$\displaystyle a_{n+1}-a_n=\int^{\frac\pi2}_0 \frac{\sin (2n-1)x}{\sin x}dx=b_n$(say)
$\implies b_1=\displaystyle a_2-a_1=\int^{\frac\pi2}_0 \frac{\sin x}{\sin x}dx=\frac\pi2$
Now, $\displaystyle b_n-b_{n-1}=\int^{\frac\pi2}_0 \frac{\sin (2n-1)x-\sin(2n-3)x}{\sin x}dx$
Using $\sin C-\sin D=2\sin\frac{C-D}2\cos\frac{C+D}2, \displaystyle b_n-b_{n-1}=\int^{\frac\pi2}_02\cos(2n-2)dx$
As $n\ne1,b_n-b_{n-1}=\frac{2\sin(2n-2)}{2n-2}|_0^{\frac\pi2}=0$ as $\sin m\pi=0$ for integer $m$
$\implies b_n=b_{n-1}$
So if $n>1, b_n=\cdots=b_1=\frac\pi2$
$\implies$ if $n>1, a_{n+1}-a_n=b_n=\frac\pi2$ and we have $a_1=\frac\pi2$
So, $a_n=??$ for $n>1$
| {
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} |
$2+2 = 5$? error in proof $$\begin{align} 2+2 &= 4 - \frac92 +\frac92\\
&= \sqrt{\left(4-\frac92\right)^2} +\frac92\\
&= \sqrt{16 -2\times4\times\frac92 +\left(\frac92\right)^2} + \frac92\\
&= \sqrt{16 -36 + \left(\frac92\right)^2} +\frac92\\
&= \sqrt {-20 +\left(\frac92\right)^2} + \frac92\\
&= \sqrt{25-45 +\left(\frac92\right)^2} +\frac92\\
&= \sqrt {5^2 -2\times5\times\frac92 + \left(\frac92\right) ^2} + \frac92\\
&= \sqrt {\left(5-\frac92\right)^2} +\frac92\\
&= 5 + \frac92 - \frac92 \\
&= 5\end{align}$$
Where did I go wrong
| $\sqrt{\left(4 - \frac 9 2 \right)^2} = 4 - \frac 9 2 = -0.5.$
It's not true.
If $a \geq 0$, then $\sqrt{a} \geq 0$.
| {
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"answer_id": 3
} |
Prove with Induction for $n\in \mathbb{N}$ and $n$ is even for $1^2-3^2+5^2-7^2+\dots+(2n-3)^2-(2n-1)^2=-2n^2 $ I want to prove by indection, for $n\in\mathbb N$ even:
$$1^2-3^2+5^2-7^2+\dots+(2n-3)^2-(2n-1)^2=-2n^2 $$
what I did first is to check the numbers, so if $n$ is even
lets take $n=2$ so $(2\cdot 2-3)^2-(2\cdot 2-1)^2=-2\cdot 4$
lets take $n=4$ so $(2\cdot 4-3)^2-(2\cdot 4-1)^2\neq-2\cdot 16$
I did something wrong?
Thanks!
| For $n=4$ you need to show that $1^2-3^3+5^2-7^2 = -2\cdot 16$.
| {
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} |
Convert $\sin\theta/2$ using trig identities $$\sin\dfrac \theta2 = \sin^2θ+\cos^2θ-1$$
$$\sinθ = 2\sin^2θ+\cos^2θ-1$$
Am I on the right track?
| In the event that you are being asked to solve for $\theta$ given the equation $$\sin\left(\frac \theta2\right) = \sin^2 \theta + \cos^2 \theta - 1,$$
...note that by the Pythagorean Theorem, we know that $\sin^2 \theta + \cos^2 \theta = 1.\;$
$$\begin{align}\sin\left(\frac \theta2\right) & = \underbrace{(\sin^2 \theta + \cos^2\theta)}_{= 1} - 1 \\ \\ & = 1 - 1 \\ \\& = 0\end{align}$$
$$\implies \frac{\theta}{2} = \sin^{-1}(0) \implies \theta = 2 \sin^{-1}(0)$$
| {
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Solve the function equation $g^2(x)-g(x+1)-\dfrac{x^2+2x-6}{4}=0$ let $g(x)\in \Bbb R$ and for any $x\in \Bbb R$ such that
$$g^2(x)-g(x+1)-\dfrac{x^2+2x-6}{4}=0, g(0)=0$$
find $g(x)$
my idea
let $x\longrightarrow x+1$, then we have
$$g^2(x+1)-g(x+2)-\dfrac{(x+1)^2+2(x+1)-6}{4}=0$$
$$g^2(x+2)-g(x+3)-\dfrac{(x+2)^2+2(x+2)-6}{4}=0$$
$$\cdots\cdots $$
$$g^2(x+n)-g(x+n+1)-\dfrac{(x+n)^2+2(x+n)-6}{4}=0$$
following I can't work,Thank you
| Take $\,x=-1\;$:
$$g(-1)^2-g(0)-\frac{(-1)^2+2(-1)-6}4=0\implies g(-1)^2=-\frac74$$
which of course is impossible unless there are further conditions or we're working with complex numbers and not with the reals, as written, say.
| {
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$a_n$ given, then evaluate $\lim_{n\to \infty}a_n$. If $$a_n=\sqrt{4+\sqrt{4+3\sqrt{4+5\sqrt{4+7\sqrt{\cdots\sqrt{4+(2n-1)\sqrt{4+(2n+1)}}}}}}}$$for any natural number $n$,
then evaluate $\lim_{n\to \infty}a_n$.
Note that $a_1=\sqrt{4+\sqrt{4+3}}$ and $a_2=\sqrt{4+\sqrt{4+3\sqrt{4+5}}}$.
I don't have any good idea. I need your help.
| Define
$$
f(x)=\sqrt{4+x\sqrt{4+(x+2)\sqrt{4+(x+4)\sqrt{4+\dots}}}}
$$
then $f(x)^2=4+xf(x+2)$. This indicates we should look at $f(x)=x+2$.
Considering $f(x)=x+2$, we are lead to show inductively that
$$
x+2=\small\sqrt{4+x\sqrt{4+(x+2)\sqrt{4+\dots+\sqrt{4+(x+2k-4)\sqrt{4+(x+2k-2)(x+2k+2)}}}}}
$$
Define
$$
f_{k,x}(y)=\left\{\begin{array}{}
y&\text{if }k=0\\
\sqrt{4+xf_{k-1,x+2}(y)}&\text{if }k>0
\end{array}\right.
$$
unrolled, that is
$$
f_{k,x}(y)=\small \sqrt{4+x\sqrt{4+(x+2)\sqrt{4+\dots+\sqrt{4+(x+2k-4)\sqrt{4+(x+2k-2)y}}}}}
$$
Note that $f_{0,x}(x+2)=x+2$. Suppose that, for some $k\ge0$, $f_{k,x}(x+2k+2)=x+2$. Then
$$
\begin{align}
f_{k+1,x}(x+2k+4)
&=\sqrt{4+xf_{k,x+2}(x+2k+4)}\\
&=\sqrt{4+x(x+4)}\\
&=x+2
\end{align}
$$
Therefore, for all $k\ge0$, we have $f_{k,x}(x+2k+2)=x+2$.
We want to show that
$$
\lim_{k\to\infty}f_{k,x}(1)=x+2
$$
Note that
$$
\frac{f_{0,x+2k}(x+2k+2)}{f_{0,x+2k}(1)}=\frac{x+2k+2}{1}
$$
Suppose that, for some $j\ge0$,
$$
\frac{f_{j,x+2k-2j}(x+2k+2)}{f_{j,x+2k-2j}(1)}\le(x+2k+2)^{1/2^j}
$$
Then
$$
\begin{align}
\frac{f_{j+1,x+2k-2j-2}(x+2k+2)}{f_{j+1,x+2k-2j-2}(1)}
&=\sqrt{\frac{4+(x+2k-2j-2)f_{j,x+2k-2j}(x+2k+2)}
{4+(x+2k-2j-2)f_{j,x+2k-2j}(1)}}\\
&\le\sqrt{\frac{f_{j,x+2k-2j}(x+2k+2)}{f_{j,x+2k-2j}(1)}}\\
&\le(x+2k+2)^{1/2^{j+1}}
\end{align}
$$
Therefore, for all $j\ge0$,
$$
\frac{f_{j,x+2k-2j}(x+2k+2)}{f_{j,x+2k-2j}(1)}\le(x+2k+2)^{1/2^j}
$$
in particular, for $j=k$,
$$
\frac{f_{k,x}(x+2k+2)}{f_{k,x}(1)}\le(x+2k+2)^{1/2^k}
$$
Since $f_{k,x}(x+2k+2)=x+2$, we have
$$
(x+2)(x+2k+2)^{-1/2^k}\le f_{k,x}(1)\le (x+2)
$$
and by the squeeze theorem, we have
$$
\lim_{k\to\infty}f_{k,x}(1)=x+2
$$
as desired. Setting $x=1$, we get the answer to the question to be $3$.
| {
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Using partial fractions to find an antiderivative of $(x^2+2x)/(x+1)^2$
Evaluate $$
\int\frac{x^2+2x}{(x+1)^2}dx
$$
My solution
Let $u =x+1$,
$
du=dx
$.
Then
$
du(x^2+2x)=(x^2+2x)dx
$ and
$
x=u-1
$.
We get
$$
\int\frac{(u-1)^2+2(u-1)}{u^2}du
=
\int\frac{u^2-2u+1+2u-2}{u^2}du
=
\int\frac{u^2-1}{u^2}du
$$
which simplifies to
$$
\int(1-u^{-2})du
=u-\frac{u^{-1}}{-1} +C
=u+\frac{1}{u}+C
=x+1+\frac{1}{x+1}+C
$$
... But the answer is
$$
x+\frac{1}{x+1}+C
$$
What is wrong?
| That's perfectly fine. We can absorb the $1$ into the constant (for example, define $D=C+1$). Recall that general antiderivatives may differ by any arbitrary constant. To see this, try checking your work by taking the derivative of your answer (as well as the "official" answer). In either case, you'll end up with:
$$
\dfrac{x^2+2x}{(x+1)^2}
$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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what is the remainder when $(17^{3}+19^{3} + 21^{3}+23^{3})$ is divided by 83? what is the remainder when $(17^{3}+19^{3} + 21^{3}+23^{3})$ is divided by 83?
NOTE:$a^{3}+b^{3}=(a+b)(a^2-ab+b^2)$
| For every $a$, $(a+1)^3+(a-1)^3=2(a^3+3a)$ and $(a+3)^3+(a-3)^3=2(a^3+3\cdot9a)$ hence $s(a)=(a-3)^3+(a-1)^3+(a+1)^3+(a+3)^3$ is
$$
s(a)=4a^3+2\cdot3\cdot(1+9)a=4a(a^2+15).
$$
Since $83$ is prime, $83$ divides $s(a)$ if and only if $83$ divides $a$ or $83$ divides $a^2+15$. In your post, $a=20$, then $83$ does not divide $a$ but $a^2+15=415=5\cdot83$ hence $83$ divides $s(20)$.
| {
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How to find the sum of this : $\sqrt{1+\frac{1}{1^2}+\frac{1}{2^2}}+ \sqrt{1+\frac{1}{2^2}+\frac{1}{3^2}}+\sqrt{1+\frac{1}{3^2}+\frac{1}{4^2}}+.....$ How to find the sum of the following :
$\sqrt{1+\frac{1}{1^2}+\frac{1}{2^2}}+ \sqrt{1+\frac{1}{2^2}+\frac{1}{3^2}}+\sqrt{1+\frac{1}{3^2}+\frac{1}{4^2}}+.....+\sqrt{1+\frac{1}{1999^2}+\frac{1}{2000^2}}$
Please suggest as getting no clue on this... thanks..
| Hint: Try writing it as a series, then do a little bit of "algebra work".
| {
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Solve : $\frac{2012!}{2^{2010}}-\sum^{2010}_{k=1} \frac{k^2k!}{2^k}-\sum^{2010}_{k=1} \frac{k\cdot k!}{2^k}$ Solve : $$\frac{2012!}{2^{2010}}-\sum^{2010}_{k=1} \frac{k^2k!}{2^k}-\sum^{2010}_{k=1} \frac{k\cdot k!}{2^k}$$
Can we take like this :
Let us take (k+1)th term = $$\frac{(k+1)^2(k+1)!}{2^{k+1}} ; \frac{(k+1)(k+1)!}{2^{k+1}}$$ and (k-1)th term is : $$\frac{(k-1)^2(k-1)!}{2^{k-1}}; \frac{(k-1)(k-1)!}{2^{k-1}}$$
What can we do further to this series... please suggest ......thanks.
| The "2010" in
$\frac{2012!}{2^{2010}}-\sum^{2010}_{k=1} \frac{k^2k!}{2^k}-\sum^{2010}_{k=1} \frac{k\cdot k!}{2^k}$
is a MacGuffin.
If it is replaced by $n$,
this becomes
$
\frac{(n+2)!}{2^n}-\sum^{n}_{k=1} \frac{k^2k!}{2^k}-\sum^{n}_{k=1} \frac{k\cdot k!}{2^k}
$.
Adding the last two terms
(as Kunnysan did)
$\begin{align}
\sum^{n}_{k=1} \frac{k^2k!}{2^k}+\sum^{n}_{k=1} \frac{k\cdot k!}{2^k}
&=\sum^{n}_{k=1} \frac{k^2k!+k\cdot k!}{2^k}\\
&=\sum^{n}_{k=1} k!\frac{k^2+k}{2^k}\\
&=\sum^{n}_{k=1} k!\frac{k(k+1)}{2^k}\\
&=\sum^{n}_{k=1} (k+1)!\frac{k}{2^k}\\
&=\sum^{n}_{k=1} (k+1)!\frac{(k+2)-2}{2^k}
\quad \text{This is Kunnysan's very ingenious key step}\\
&=\sum^{n}_{k=1} \frac{(k+2)!}{2^k}
-\sum^{n}_{k=1} \frac{(k+1)!}{2^{k-1}}\\
&=\sum^{n+1}_{k=2} \frac{(k+1)!}{2^{k-1}}
-\sum^{n}_{k=1} \frac{(k+1)!}{2^{k-1}}\\
&=\frac{(n+2)!}{2^{n}}
-\frac{(2)!}{2^{0}}\\
&=\frac{(n+2)!}{2^{n}}
-2\\
\end{align}
$
so the result is $2$
independent of $n$.
| {
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"source": "stackexchange",
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Finding unit's digit in exponentiation Could someone please explain to me how to find the unit's digit in the following expression:
$$7^{95} - 3^{58}$$
| Note that $7^2$ "ends" in a $9$. So $7^4=7^2\cdot 7^2$ ends in a $1$, So $7^8=7^4\cdot 7^4$ ends in a $1$, and therefore $7^{12}=7^8\cdot 7^4$ ends in a $1$, and therefore $7^{16}$ ends in a $1$, and so on. In general, if $n$ is divisible by $4$, then $7^n$ ends in a $1$.
It follows that $7^{92}$ ends in a $1$. We are almost there. Just multiply by $7$ three more times. The result ends in a $3$.
Similarly, $3^2$ ends in a $9$, so $3^4$ ends in a $1$, Thus $7^{36}$ ends in a $1$, and therefore $7^{38}$ ends in a $9$.
Now we are subtracting a number that ends in $9$ from a number that ends in $3$. The result ends in $4$.
| {
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can someone explain this limit i have,
$$\lim_{x\to 0} \frac {\sqrt{x+49}-7}{3-\sqrt{x+9}}$$ the correct answer is $-\frac{3}{7}$ and in my case the result is $\frac{7}{3}$ i don't understand.
i tried this
$\lim_{x\to 0} \frac {\sqrt{x+49}-7}{3-\sqrt{x+9}}$=$\lim_{x\to 0} \frac {\sqrt{x+49}-7}{3-\sqrt{x+9}}.\frac {\sqrt{x+49}+7}{\sqrt{x+49}+7}.\frac {3+\sqrt{x+9}}{3+\sqrt{x+9}}$=
$\lim_{x\to 0} \frac {x+49-49}{9-x+9}.\frac {\sqrt{x+49}+7}{3+\sqrt{x+9}}= \frac {x(\sqrt{x+49}-7)}{x(3-\sqrt{x+9})}=\frac{14}{3}=\frac{7}{3}$ i don't know what is the error!
| You did the double multiplication together. That invites mistakes:
$$\lim_{x\to 0} \frac {\sqrt{x+49}-7}{3-\sqrt{x+9}}=\lim_{x\to 0} \frac {\sqrt{x+49}-7}{3-\sqrt{x+9}}\frac{\sqrt{x+49}+7}{\sqrt{x+49}+7}=$$
$$=\lim_{x\to 0}\frac x{\left(\sqrt{x+49}+7\right)\left(3-\sqrt{x+9}\right)}\frac{3+\sqrt{x+9}}{3+\sqrt{x+9}}=\lim_{x\to 0}\frac x{-x}\cdot\frac{3+\sqrt{x+9}}{\sqrt{x+49}+7}=$$
** because $\;(3-\sqrt{x+9})(3+\sqrt{x+9})=9-(x+9)=\color{red}{-x}\;$ **
$$=(-1)\frac{3+\sqrt9}{\sqrt{49}+7}=-\frac6{14}=-\frac37$$
| {
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Trace zero matrix that can't be written as $AB - BA$? According to this paper, every trace zero matrix over a field can be written in the form $AB - BA$. However, here's a basic counterexample:
Let $A = diag(a, -a)$ for some nonzero number a. Then $A = BC - CB$ if and only if a system of equations involving entries of $B,C,A$ is true. Then the entries of $A$ are:
$$\begin{align}
a = b_{1,1}c_{1,1} + b_{1,2}c_{2,1} - b_{1,1}c_{1,1} - b_{2,1}c_{1,2} \\
0 = b_{1,1}c_{1,2} + b_{1,2}c_{2,2} - b_{1,1}c_{2,1} - b_{2,1}c_{2,2} \\
-a = b_{2,1}c_{1,1} + b_{2,2}c_{2,1} - b_{1,2}c_{1,1} - b_{2,2}c_{1,2} \\
0 = b_{2,1}c_{1,2} + b_{2,2}c_{2,2} - b_{1,2}c_{2,1} - b_{2,2}c_{2,2}
\end{align}
$$
For $a$ nonzero the first and the last equations can't both be true. So where did I mess up?
| Consider
$$\begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}\begin{bmatrix} 0 & 0 \\ a & 0 \end{bmatrix} -\begin{bmatrix} 0 & 0 \\ a & 0 \end{bmatrix}\begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}$$
What made you conclude that the first and last of those equations can't be true?
| {
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Compute $ax^3+by^3$ given $ax^2+by^2$ and $ax+by$. Given are positive real numbers $A$ and $B$ and positive integers $a$ and $b$ such that
$$ \begin{aligned} ax+by &= A\\ ax^2+by^2 &= B.\end{aligned} \tag{*}$$
What are the possible values of
$$ ax^3 + by^3 = C?$$
Think of it as $a+b$ numbers, $a$ of them are $x$ and $b$ of them are $y$, such that the sum and the sum of their squares are given and the sum of their cubes is asked. I'm not sure if the answer can be expressed as a rational function of $A$, $B$, $a$ and $b$, but I hope so.
Note that one has the equation
$$ (A-ax)^2 = (by)^2 = b(B-ax^2),$$
thus
$$\begin{aligned} a(a+b)x^2 &= bB-A^2 + 2ax,\\ b(a+b)y^2 &= aB-A^2 + 2by.\end{aligned} \tag{1}$$
Multiplying with $x$ and adding gives
$$\begin{aligned} (a+b)C &= B(bx+ay)-A^2(x+y) + 2B \\ &= B(a+b)(x+y)- B(ax+by) - A^2(x+y)+2B \\&= \big(B(a+b)- A^2\big)\,(x+y) - AB+ 2B. \end{aligned}$$
So this boils down to determining possible values of $x+y$. Here I am a bit stuck. Note also that from (*) one can deduce (multiply the first equation with $x$, $y$, add and subtract the second):
$$ (a+b)xy = A(x+y) - B.$$
One could simply solve the quadratic equations (1), verify what combination of solutions works for the original problem and use this to calculate $C$. But actually I am interested in a solution that will work for higher order cases as well, (3 unknowns, sum of first, second and thirth powers is known and the sum of fourth powers must be determined) so I am looking for an approach avoids this. Maybe there is a quadratic equation with $x$ and $y$ as roots hidden somewhere?
| Well, you have 3 equations,
$$ax+by = A\tag{1}$$
$$ax^2+by^2 = B\tag{2}$$
$$ax^3+by^3 = C\tag{3}$$
If you want to express C in terms of $a,b,A,B$, you can eliminate $x,y$ between $(1), (2), (3)$ to have an equation purely in $a,b,A,B,C$ (easily done in Mathematica using the Resultant command).
However, we get a quadratic in C. But we can use (1) and (2) to simplify its discriminant D and I ended up with,
$$C = \frac{A(-2A^2+3B(a+b))\pm(a-b)(a b)D^3}{(a+b)^2}$$
where,
$$D= (x-y) = \sqrt{\frac{-A^2+(a+b)B}{ab}}\tag{4}$$
Hence, C is not a rational function of $a,b,A,B$, as you need to take square roots as shown by $(4)$.
| {
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"url": "https://math.stackexchange.com/questions/466978",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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what is the taylor expansion of $\ln(\sin x/x)$ what is the Taylor expansion of the function
$$f(x) = \ln \frac{\sin x}x$$
around the point $x=0$? Ignore powers of $x$ which are greater than $6$.
Here is my method:
$$\ln(1+x)=x-\frac{x^2}2 + \frac{x^3}3 -\frac{x^4}4,$$
so we should get the function
$$g(x)=\frac{\sin x}x$$
in the form of $1 + x$ to get the answer … or is there another way to consider the Taylor expansion of $\ln (u)$ by considering $f$ to be the answer of $$\frac{\sin x}x\cdots$$
Which way is the best?? What is the final answer???
| Your instincts are correct. Write
$$\frac{\sin{x}}{x} = 1 - \frac{x^2}{6} + \frac{x^4}{120} - \frac{x^6}{5040} + \cdots$$
We wish to maintain the series to $O(x^6)$. Then
$$\begin{align}\ln{\left ( \frac{\sin{x}}{x}\right)} &= \left (\frac{\sin{x}}{x} - 1\right ) - \frac12 \left (\frac{\sin{x}}{x} - 1\right )^2 + \frac13 \left (\frac{\sin{x}}{x} - 1\right )^3-\cdots \\ &= \left ( - \frac{x^2}{6} + \frac{x^4}{120} - \frac{x^6}{5040} + \cdots \right ) - \frac12 \left ( - \frac{x^2}{6} + \frac{x^4}{120} - \frac{x^6}{5040} + \cdots \right )^2+\\ &\frac13 \left ( - \frac{x^2}{6} + \frac{x^4}{120} - \frac{x^6}{5040} + \cdots \right )^3-\cdots \\ &= - \frac{x^2}{6} + \frac{x^4}{120} - \frac{x^6}{5040} + \cdots - \frac{x^4}{72} \left (1 - \frac{x^2}{10}+\cdots \right ) - \frac{x^6}{648}+\cdots \\ &= - \frac{x^2}{6} - \frac{x^4}{180}-\frac{x^6}{2835}+\cdots \end{align}$$
Note that
$$-\frac{1}{180} = \frac{1}{120} - \frac{1}{72}$$
$$-\frac{1}{2835} = -\frac{1}{5040}+\frac{1}{720}-\frac{1}{648}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/467986",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Compute $\int x^2 \cos \frac{x}{2} \mathrm{d}x$ I am trying to compute the following integral:
$$\int x^2 \cos \frac{x}{2} \mathrm{d}x$$
I know this requires integration by parts multiple times but I am having trouble figuring out what to do once you have integrated twice. This is what I have done:
Let $u = \cos \frac{x}{2}$ and $\mathrm{d}u = \frac{-\sin\left(\dfrac{x}{2}\right)}{2} \mathrm{d}x$ and $\mathrm{d}v = x^2$ and $v= \frac{x^3}{3}$.
\begin{align}
&\int x^2 \cos \frac{x}{2} \\
&\cos \frac{x}{2} \cdot \frac{x^3}{3} - \int \frac{x^3}{3} \cdot \frac{-\sin\left(\dfrac{x}{2}\right)}{2} \mathrm{d}x
\end{align}
So now I integrate $\int \frac{x^3}{3} \cdot \frac{-\sin\left(\dfrac{x}{2}\right)}{2} \mathrm{d}x$ to get:
\begin{align}
\frac{-\sin\left(\dfrac{x}{2}\right)}{2} \cdot \frac{x^4}{12} - \int \frac{x^4}{12} \cdot \frac{\cos x}{2}
\end{align}
Now, this is where I get stuck. I know if I continue, I will end up with $\frac{-\sin\left(\dfrac{x}{2}\right)}{2}$ again when I integrate $\cos \frac{x}{2}$. So, where do I go from here?
Thanks!
| $$\int x^2 \cos \frac{x}{2} \mathrm{d}x$$
Let $u = x^2$ and $\mathrm{d}u = 2x\,dx$ and let $dv = \cos\frac x2\,dx \implies v = 2 \sin \frac x2 $.
$$ 2x^2 \sin \frac x2 - \int 2x\cdot 2\sin \frac x2\,dx $$
You'll only need to do integration by parts one additional time. Let me know if you get stuck after that.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Finesse vs. brute force in solving quadratic equations In Higher Algebra by Hall and Knight, the following "artifice" for solving a certain type of equations is given:
Solve: $\sqrt{3x^2-4x+34} - \sqrt{3x^2-4x-11} = 9$
They make use of the fact that $(3x^2-4x+34) - (3x^2-4x-11) = 45$, and utilizing the formula $a^2 - b^2 = (a-b)(a+b)$, obtain a neat answer that $x = 3, -5/3$.
My attempt, however, results in something weird:
$\sqrt{3x^2-4x+34} - \sqrt{3x^2-4x+11} = 9$
Let $3x^2 - 4x = t$
$\Rightarrow \sqrt{t+34} + \sqrt{t+11}$ = 9
Squaring both sides,
$t + 34 + t + 11 + 2 \sqrt{(t+34)(t+11)} = 81$
$\Rightarrow 18-t = \sqrt{(t+34)(t+11)}$
Squaring again,
$t^2 - 36t + 324 = t^2 + 45t + 374$
$\Rightarrow t = -50/81$
$\Rightarrow 3x^2 - 4x = -50/81$
$\Rightarrow 243x^2 - 324x + 50 = 0$
The discriminant here is $\sqrt{324^2 - 200.243} = 56376$, which is not a perfect square and hence the roots are irrational.
Have I made a mistake somewhere or it is really the case that finesse leads to correct answers?
| Oops! Realized that I had written $-11$ as $11$. Sorry for the mistake! I'll try again and re-post if needed.
| {
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"timestamp": "2023-03-29T00:00:00",
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Conjugacy classes and orders of matrices. The following are prime decompositions in $\Bbb{Z}_7[x]$:
$x^8+1= (x^2-x-1)(x^2+x-1)(x^2+3x-1)(x^2+4x-1)$
$x^4+1= (x^2+3x+1)(x^2+4x+1)$
(a) Give representatives for the conjugacy classes of elements of order dividing 16 in $Gl_2(\Bbb{Z}_7)$ and give the order of each.
(b) Show that there are no elements of order 32 in $Gl_2(\Bbb{Z}_7)$.
(a) We can write the rational canonical forms as representatives.
$x^2-x-1 \rightarrow \begin{bmatrix} 0 & 1 \\1 & 1 \end{bmatrix}$
$x^2+x-1 \rightarrow \begin{bmatrix} 0 & 1 \\1 & -1 \end{bmatrix}$
$x^2+3x-1 \rightarrow \begin{bmatrix} 0 & 1 \\1 & -3 \end{bmatrix}$
$x^2+4x-1 \rightarrow \begin{bmatrix} 0 & 1 \\1 & -4 \end{bmatrix}$
$x^3+3x+1 \rightarrow \begin{bmatrix} 0 & -1 \\1 & -3 \end{bmatrix}$
$x^2+4x+1 \rightarrow \begin{bmatrix} 0 & -1 \\1 & -4 \end{bmatrix}$
$x^2+1 \rightarrow \begin{bmatrix} 0 & -1 \\1 & 0 \end{bmatrix}$.
The first four matrices have order 16, since they satisfy $x^8+1=0$ so $x^8=-1 \implies (x^8)^2 = 1$. The last two have order 8, since $x^4=-1 \implies x^8=1$.
(b)We don’t know the factorization of $x^{16}+1$, so we can’t check whether or not the matrices that satisfy it actually belong to $Gl_2(\Bbb{Z}_7)$. I wasn’t able to tell whether this was even reducible in $Z_7[x]$ to begin with, so I was kind of stuck here, and I was wondering if anybody could give me a hint...
Thanks in advance
| Part (a) Let $A\in\operatorname{GL}_2(\mathbb Z_7)$ be a matrix whose order divides $16$.
Then $A^{16} - I = 0$, so the minimal polynomial $f$ of $A$ is a divisor of $x^{16} - 1$.
We have
$$x^{16} - 1
= (x^8 + 1)(x^8 - 1)
= (x^8 + 1)(x^4 + 1)(x^2 + 1)(x+1)(x - 1)
$$
Since $A$ is $2\times 2$, the degree of $f$ is at most $2$.
Using the fact that $x^2 + 1$ is irreducible over $\mathbb Z_7$ (since $-1$ is not a square mod $7$) and the given factorizations of $x^4 + 1$ and $x^8+ 1$, we can enumerate all the possibilities for $f$.
Furthermore, in all the cases we can immediately give the exact order of $A$, which is given by the smallest exponent $k$ such that $f$ divides $x^k - 1$. Representatives for $A$ are given by the possible rational normal forms (which happen to be uniquely determined by $f$ in all cases). We get the following list:
$$
\begin{array}{ccc}
f & \operatorname{ord}(A) & \text{representatives} \\
\hline
x-1 & 1 & I \\
x+1 & 2 & -I \\
(x-1)(x+1) & 2 & \begin{pmatrix}1 & 0 \\ 0 & -1\end{pmatrix} \\
x^2 + 1 & 4 & \begin{pmatrix}0 & -1 \\ 1 & 0\end{pmatrix} \\
x^2 + 3x + 1 & 8 & \begin{pmatrix}0 & -1 \\ 1 & -3\end{pmatrix} \\
x^2 + 4x + 1 & 8 & \begin{pmatrix}0 & -1 \\ 1 & 3\end{pmatrix} \\
x^2 - x - 1 & 16 & \begin{pmatrix}0 & 1 \\ 1 & 1\end{pmatrix} \\
x^2 + x - 1 & 16 & \begin{pmatrix}0 & 1 \\ 1 & -1\end{pmatrix} \\
x^2 + 3x - 1 & 16 & \begin{pmatrix}0 & 1 \\ 1 & -3\end{pmatrix} \\
x^2 + 4x - 1 & 16 & \begin{pmatrix}0 & 1 \\ 1 & 3\end{pmatrix} \\
\end{array}
$$
Part (b)
Assume that that there is a matrix $A\in\operatorname{GL}_2(\mathbb Z_7)$ of order $32$.
Then the minimal polynomial $f$ of $A$ divides $x^{32} - 1$, but not $x^{16} - 1$ and is of degree $1$ or $2$.
In the algebraic closure of $\mathbb{Z}_7$ there is a zero $a$ of $f$ of multiplicative order $32$. (Otherwise, all zeros would have an order dividing $16$. Since $x^{32} - 1$ doesn't have multiple zeros, $f$ does not have multiple roots and so $f$ would divide $x^{16} - 1$.)
From $f(a) = 0$ we get $[\mathbb{Z}_7(a) : \mathbb{Z}_7] \leq \deg(f) = 2$.
So $\mathbb{Z}_7(a)$ is a field of size $7^1$ or $7^2$. Because up to isomorphisms, finite fields are uniquely determined by their order, and because $\mathbb{F}_7$ is a subfield of $\mathbb{F}_{49}$, we get $a\in\mathbb{F}_{49}^\times$.
Since $\mathbb{F}_{49}^\times \cong \mathbb{Z}_{48}$ does not contain an element of order $32$, this is a contradiction.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Quadratic equation - $\alpha$ and $\beta$ Roots If $\alpha$ and $\beta$ are the roots of the equation $x^2 + 8x - 5 = 0$, find the quadratic equation whose roots are $\frac{\alpha}{\beta}$ and $\frac{\beta}{\alpha}$.
My working out so far:
I know that $\alpha + \beta = -8$ and $\alpha \beta = -5$ (from the roots) and thenIi go on to work out that
$\alpha= -8-\beta$ and $\beta= -8-\alpha$, then I substitute into what the question asks me.
$\frac{-8-\beta}{-8-\alpha}$ and $\frac{-8-\alpha}{-8-\beta}$ however I do not know how to proceed further. I might be doing this completely wrong and my apologies for that.
Another solution came to me that if $\alpha$ and $\beta$ are roots of the other unknown equation. I can somehow manipulate that to find the answer. But I don't think that will work. All help is appreciated thank you.
| Since we want the zeroes of the new quadratic polynomial to be the mutual reciprocals $ \ \frac{\alpha}{\beta} \ $ and $ \ \frac{\beta}{\alpha} \ \ , $ it must be palindromic, that is, of the form $ \ Ax^2 + Bx + A \ \ $ (the sort of symmetry André Nicolas was referring to in his comments). Viete tells us that $ \ \alpha + \beta \ = \ -8 \ $ and that $ \ \alpha·\beta \ = \ -5 \ \ , $ as other posters have found. So we may conclude that the sum of the reciprocals of these roots is
$$ \frac{\alpha \ + \ \beta}{\alpha · \beta} \ \ = \ \ \frac{1}{\alpha} \ + \ \frac{1}{\beta} \ \ = \ \ \frac{-8}{-5} \ \ = \ \ \frac85 \ \ . $$
We may also divide the sum-of-roots equation through to find that
$$ \frac{\alpha \ + \ \beta}{\alpha} \ \ = \ \ 1 \ + \ \frac{ \beta}{\alpha} \ \ = \ \ \frac{-8}{\alpha} \ \ \ \text{and} \ \ \ \frac{\alpha \ + \ \beta}{\beta} \ \ = \ \ 1 \ + \ \frac{\alpha}{ \beta} \ \ = \ \ \frac{-8}{\beta} \ \ . $$
We thus obtain
$$ \left(1 \ + \ \frac{ \beta}{\alpha} \right) \ + \ \left(1 \ + \ \frac{\alpha}{ \beta} \right) \ = \ -8·\left(\frac{1}{\alpha} + \frac{1}{\beta} \right) \ \ = \ \ -\frac{64}{5} $$ $$ \Rightarrow \ \ \frac{\alpha}{ \beta} \ + \ \frac{ \beta}{\alpha} \ \ = \ \ -\frac{64}{5} \ - \ 2 \ \ = \ \ -\frac{74}{5} \ \ . $$
Our monic palindromic polynomial is therefore $ \ x^2 \ + \ \frac{74}{5} · x \ + \ 1 \ \ . $ [In answer to the question, a corresponding quadratic equation is $ \ A·\left(x^2 + \frac{74}{5} · x + 1 \right) \ = \ 0 \ \ , $ where any non-zero value for $ \ A \ $ will do.]
$$ \ \ $$
It isn't overly difficult to work with the roots themselves: the discriminant of the original quadratic polynomial is $ \ 8^2 + 4·5 \ = \ 84 \ \ , $ so the zeroes are $ \ \alpha \ , \ \beta \ = \ -4 \ \pm \ \sqrt{21} \ \ . $ From this, we have reasonably convenient calculation with "conjugate" numbers: $$ \frac{\alpha}{\beta} \ + \ \frac{\beta}{\alpha} \ \ = \ \ \frac{\alpha^2 \ + \ \beta^2}{\alpha·\beta} \ \ = \ \ \frac{(-4 \ + \ \sqrt{21})^2 \ + \ (-4 \ - \ \sqrt{21})^2}{(-4 \ + \ \sqrt{21})·(-4 \ - \ \sqrt{21})} \ \ = \ \ \frac{2·[ \ (-4)^2 \ + \ (\sqrt{21})^2 \ ]}{(-4)^2 \ - \ (\sqrt{21})^2} $$ $$ = \ \ \frac{2·( 16 \ + \ 21 )}{16 \ - \ 21} \ = \ -\frac{74}{5} \ \ . $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/471090",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
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Prove that if $x^n = y^n$ and $n$ is even, then $x = y$ or $x = -y$ This is a problem from Spivak's Calculus $3^{rd}$ ed., Chapter I, Problem $6$(d)
Prove that if $x^n = y^n$ and $n$ is even, then $x = y$ or $x = -y$.
Proof. Suppose $x^n = y^n$ and $n$ is even. We consider the following cases.
Case 1. $x \geq 0$ and $y \geq 0$. Now suppose, for the sake of contradiction, $x \neq y$. Then $x > y$ or $x < y$. If $x > y$, then $0 \leq y < x$, so $y^n < x^n$. But this contradicts our assumption that $x^n = y^n$. Similarly, $x < y$ leads to a contradiction. Thus $x = y$.
Case 2. $x \geq 0$ and $y < 0$. Now suppose, for the sake of contradiction, $x \neq -y$. Then $x > -y$ or $x < -y$. If $x > -y$, then $0 < -y < x$, so $(-y)^n < x^n$. Since $n$ is even, it follows that $y^n < x^n$. But this contradicts our assumption that $x^n = y^n$. Similarly, $x < -y$ leads to a contradiction. Thus $x = -y$.
Case 3. $x < 0$ and $y \geq 0$. Applying case 2 with $x$ and $y$ interchanged, we get $y = -x$. Therefore $x = -y$.
Case 4. $x <0$ and $y < 0$. Then $-x > 0$ and $-y > 0$. Applying case 1 to $-x$ and $-y$, we get $-x = -y$. Therefore $x = y$.
Is my proof correct? Is there a shorter way to prove this?
| In general
$$x^n=y^n\iff 0=x^n-y^n=(x-y)\left(x^{n-1}+x^{n-2}y+\ldots+xy^{n-2}+y^{n-1}\right)$$
If $\,x\neq y \;$ then we must have that the second factor to the right above equals zero.
But if $\,x,y >0\,$ ($\;x,y<0\;$), then that second factor is always positive (negative), so we can already focus on the case with different signs, and thus we can assume WLOG $\,x>0>y\;$ and the second factor's zero, thus:
$$x^{n-1}+x^{n-2}y+\ldots+xy^{n-2}+y^{n-1}=P+N\;,\;\;\text{with}$$
$$P:=x^{n-1}+x^{n-3}y^2+\ldots+xy^{n-2}\;,\;\;N:=x^{n-2}y+x^{n-4}y^3+\ldots y^{n-1}$$
Observe that $\,P>0\;,\;N<0\;$ (why? Remember $\,n\;$ is even...) , so we can put
$$x^{n-1}+x^{n-3}y^2+\ldots+xy^{n-2}=-\left(x^{n-2}y+x^{n-4}y^3+\ldots y^{n-1}\right)\iff$$
$$x\color{red}{\left(x^{n-2}+x^{n-4}y^2+\ldots+y^{n-2}\right)}=-y\color{red}{\left(x^{n-2}+x^{n-4}y^2+\ldots y^{n-2}\right)}$$
End now the proof that it must be $\,x=-y\;$ ...
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
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Ideal in the ring of upper triangular matrices I'm trying to describe an ideal of the ring $R=\left\{ \begin{pmatrix}a & b\\
0 & c
\end{pmatrix}:a,b,c \in \mathbb{R}\right\} $
It's easy to prove that $I=\left\{ \begin{pmatrix}0 & a\\
0 & 0
\end{pmatrix}:a\in\mathbb{R}\right\} $ and $J=\left\{ \begin{pmatrix}a & b\\
0 & 0
\end{pmatrix}:a,b\in\mathbb{R}\right\} $ are ideals of $R$
My question is: how can I find other ideals?
Any help would be appreciated.
Thanks.
| $\begin{pmatrix} a & b\\ 0 & c \end{pmatrix}
\begin{pmatrix} x & y\\ 0 & z \end{pmatrix} =
\begin{pmatrix} ax & ay+bz\\ 0 & cz \end{pmatrix}$.
If $a\neq 0$, you can vary $ax$ and $ay$ over $\mathbb{R}$, by varying $x$ and $y$. Similarly, if $b\neq 0$, you can vary $bz$ over $\mathbb{R}$, and if $c\neq 0$, you can vary $cz$ over $R$ by varying $z$. Thus, the different right ideals of $R$ are:
$\begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix}, \begin{pmatrix} 0 & 0 \\ 0 & \ast \end{pmatrix}, \begin{pmatrix} 0 & \ast \\ 0 & 0 \end{pmatrix}, \begin{pmatrix} 0 & \ast \\ 0 & \ast \end{pmatrix}, \begin{pmatrix} \ast & \ast \\ 0 & 0 \end{pmatrix}, \begin{pmatrix} \ast & \ast \\ 0 & \ast \end{pmatrix}$
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "9",
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Find the sum of the series $1^2-2^2+3^2-4^2+...-(2n)^2$ Find the sum of the series $$1^2-2^2+3^2-4^2+...-(2n)^2$$
I tried rewriting it as $$\sum_{r=1}^{2n}-1^{n+1}(r^2)$$ but it didn't help.
Also, looked at re-arranging as $$1^2+3^2+5^2+7^2+...+(2n-1)^2$$ and $$-2^2-4-6^2-8^2-...-(2n)^2$$
Still couldn't get to the given answer of $-n(2n+1)$
| Hint: $$1^2 - 2^2 + \ldots + (2n-1)^2 - (2n)^2 = 1^2 + 2^2 + \ldots + (2n)^2 - 8 (1^2 + 2^2 + \ldots + n^2)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/474980",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Any simplification or approximation of sqrt As you know, we could write $(a+b+c)^2$ as $a^2+b^2+c^2+2ab+2ac+2bc$.
what about $(a+b+c+\cdots)^{1/2}$? is there any expansion for $(a+b+c+\cdots)^{1/2}$?
Any simplification or approximation of $(a+b+c+\cdots)^{1/2}$ could help me.
Thanks in advance
| http://en.wikipedia.org/wiki/Binomial_theorem#Newton.27s_generalised_binomial_theorem
If $|a|>|b|$ then
\begin{align}
(a+b)^{1/2} & = a^{1/2} + \frac12 a^{1/2-1} b + \frac{(1/2)(1/2-1)}{2\cdot1} a^{1/2-2} b^2 \\[12pt]
& {}\quad{} + \frac{(1/2)(1/2-1)(1/2-2)}{3\cdot2\cdot1} a^{1/2-3} b^3 \\[12pt]
& {}\quad{} + \frac{(1/2)(1/2-1)(1/2-2)(1/2-3)}{4\cdot3\cdot2\cdot1} a^{1/2-4} b^4 + \cdots
\end{align}
But I'm not sure just how this should go with a sum of more than two terms, as in $(a+b+c)^{1/2}$. Notice that I said $|a|>|b|$, and notice the asymmetry in the expansion: $b$ is always raised only to integer powers; not so for $a$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Integral of $\sqrt{a-x^2}$ I know how to integrate stuff like $\sqrt{a-x^2}$, though I haven't done this in a while. So I wanted to check the answer after calculating. My calculations go like this ($\theta=sin^{-1}(\frac{x}{\sqrt{a}})$):
$$\int_{A}^{B}{\sqrt{a-x^2}}dx=\int_{A}^{B}{\sqrt{a-x^2}}\frac{1}{\frac{1}{\sqrt{a}}\frac{1}{\sqrt{1-\frac{x^2}{a}}}}d(sin^{-1}(\frac{x}{\sqrt{a}}))=\int_{sin^{-1}(\frac{A}{\sqrt{a}})}^{sin^{-1}(\frac{B}{\sqrt{a}})}{(a-x^2)d\theta}=\int_{sin^{-1}(\frac{A}{\sqrt{a}})}^{sin^{-1}(\frac{B}{\sqrt{a}})}{(a-((sin\theta)\sqrt{a})^2)d\theta}=[a\theta]_{sin^{-1}(\frac{A}{\sqrt{a}})}^{sin^{-1}(\frac{B}{\sqrt{a}})}-a\int_{sin^{-1}(\frac{A}{\sqrt{a}})}^{sin^{-1}(\frac{B}{\sqrt{a}})}{(\frac{1}{2}-\frac{1}{2}cos2\theta)d\theta}=[a\theta]_{sin^{-1}(\frac{A}{\sqrt{a}})}^{sin^{-1}(\frac{B}{\sqrt{a}})}-[a\frac{1}{2}\theta]_{sin^{-1}(\frac{A}{\sqrt{a}})}^{sin^{-1}(\frac{B}{\sqrt{a}})}+\frac{1}{2}a\int_{sin^{-1}(\frac{A}{\sqrt{a}})}^{sin^{-1}(\frac{B}{\sqrt{a}})}{cos2\theta\frac{1}{2}d(2\theta)}=[\frac{1}{2}a\theta]_{sin^{-1}(\frac{A}{\sqrt{a}})}^{sin^{-1}(\frac{B}{\sqrt{a}})}+\frac{1}{2}a[\frac{1}{2}sin2\theta]_{sin^{-1}(\frac{A}{\sqrt{a}})}^{sin^{-1}(\frac{B}{\sqrt{a}})}=[\frac{1}{2}a\theta]_{sin^{-1}(\frac{A}{\sqrt{a}})}^{sin^{-1}(\frac{B}{\sqrt{a}})}+[\frac{1}{2}a sin\theta cos\theta]_{sin^{-1}(\frac{A}{\sqrt{a}})}^{sin^{-1}(\frac{B}{\sqrt{a}})}=[\frac{1}{2}asin^{-1}(\frac{x}{\sqrt{a}})+\frac{1}{2}x\sqrt{a-x^2}]_{A}^{B}$$
However this doesn't seem to be right after I checked it multiple times with Wolfram Integrator and my calculator. Where did I go wrong?
| So you've got the integral:
$$\int_A^B \sqrt{a - x^2} \,dx = \int_A^B \sqrt{(\sqrt a)^2 - x^2}\,dx$$
Integrals in the form $\sqrt{a^2 - x^2}$ correspond nicely to using the substitution $x = a \sin \theta \implies \,dx = a \cos \theta d\theta$.
In your case, we have $\sqrt{(\sqrt{a})^2 - x^2}$, so our nice substitution would be $x = \sqrt a \sin\theta$. Such a substitution from the start greatly simplifies the calculation.
For a refresher on using trigonometric substitution, see the corresponding Wikipedia entry, which starts off integrands containing $a^2 - x^2$, and the use of the substitution $x = a \sin \theta$. There are a few other handy substitutions to use in some other choice situations posted in that same entry.
| {
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"timestamp": "2023-03-29T00:00:00",
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Ratio of two areas in a triangle Suppose you have the following triangle $ABC$:
with the following properties:
$$|AB|=4\cdot |AA'|,\>\>\>|AC|=4\cdot |CC'|\>\>\>|BC|=4\cdot |BB'|$$
I have to find the ratio of the total area of the triangle and the red area. I tried a lot of algebraic manipulations with the lengths of the sides, but couldnt solve it that way.
I really need hints for this one. Thanks!
| EDIT: Final solution.
First off, notice that $\triangle ABB',\triangle BCC', \triangle CAA'$ all have area equal to $\frac{1}{4}$ of the area of $\triangle ABC$. To see this, pick any side as the base when measuring $\triangle ABC$, say side $BC$. Then $BB'$ is a triangle with a common height and one fourth base, so its area is one fourth. This argument obviously holds for all three sides.
What might be more surprising is that all three of the small triangles at the corners have the same area. They all equal $\frac{1}{13}\times \frac{1}{4}$ of the area of $\triangle ABC$, or rather, area equal to one thirteenth of the three aforementioned triangles. To make this explicit, look at $\triangle ABB'$ and $\triangle AA'P$, where $P$ is the intersection of $A'C$ and $AB'$. We have the following relationships:
$$AB = 4AA' \text{, and }\angle{BAB'} = \angle{A'AP} = \theta$$
Next we have the following equations for the triangles' areas:
$$\triangle AA'P = \frac{1}{2}AP\times AA'\sin\theta \text{, and } \triangle ABB' = \frac{1}{2}AB\times AB'\sin\theta $$
Therefore, the ratio of their areas is $\dfrac{\triangle AA'P}{\triangle ABB'} = \dfrac{AP}{4AB'}$. Now as soon as I can prove that this last quantity equals $\frac{1}{13}$, the proof will be complete.
EDIT2:
Got it.
My solution requires Menelaus' Theorem. I will not prove it here.
By Menelaus' Theorem, we have that
$$\begin{align*} BC \times B'P \times AA' &= B'C \times AP \times BA' \\
BC \times B'P \times \frac{1}{4}AB &= \frac{3}{4}BC \times AP \times \frac{3}{4}AB \\
\dfrac{B'P}{AP} &= \dfrac{9}{4}\\
\dfrac{AB' - AP}{AP} &= \dfrac{9}{4}\\
\dfrac{AB'}{AP} &= \dfrac{13}{4} \implies \dfrac{AP}{AB'} = \dfrac{4}{13}\\
\end{align*}$$
Thus, as desired $\dfrac{AP}{4AB'} = \dfrac{1}{13}$. Thus, the area of the central triangle is $$\triangle ABC(1 - 3\times \frac{1}{4} + 3 \times \frac{1}{13}\times \frac{1}{4}) = \triangle ABC \times \frac{4}{13}$$
This is my first solution using trigonometry and the particular case of an equiangular triangle. According to Blue, due to affine equivalence of all triangles, this proof suffices for all cases.
Let $AB = BC = AC = 4k$, and hence $AA'=BB'=CC'=k$ and $\angle{ABC} = \angle{BCA} = \angle{BAC} = 60$.
I claim the central triangle is equiangular. Examine one of the three
small triangles at the corners of $\triangle ABC$. For convenience,
I'll focus on the top one. Call it $AA'P$, where $P$ is the
intersection of $CA'$ and $AB'$. Note the following $$ \angle{A'AP} = \angle{A'AB}\text{, and } \\ \angle{AA'P} = \angle{A'AC} = \angle{BB'A}$$ Furthermore, we know that $\angle{A'AB} + \angle{BB'A} = 120$, since the sum of the angles of any triangle equals 180. By this same principle, $\angle{A'PA} = 60$. This is true for all the
small triangles, and by the vertical angles theorem, the central
triangle is equiangular.
Notice that triangle $PA'A$ is similar to triangle $BB'A$. Thus we can
find the lengths of $PA$ and $PA'$. But first we must compute the
length of $B'A$. Via the law of cosines, $B'A = \sqrt{(4k)^2 +k^2 - 2(k)(4k)\cos(60)} = k\sqrt{13}$. Now using properties of similar
triangles, we have the following equality
$$\dfrac{PA}{AB}=\dfrac{A'A}{B'A} \implies PA=\dfrac{AB\times A'A}{B'A} = \dfrac{4k}{\sqrt{13}}$$
By a similar argument, we can calculate the length $PA' =
> \dfrac{k}{\sqrt{13}}$. So, the side lengths of the central triangle
all equal $k\sqrt{13} - \dfrac{5k}{\sqrt{13}} = \dfrac{8k}{\sqrt{13}}$. Finally, the ratio of the areas of two
equilateral triangles equals the ratio of the squares of their sides,
hence
$$\dfrac{\triangle Outer}{\triangle Inner} = \dfrac{16k^2}{\frac{64k^2}{13}} = \boxed{\dfrac{13}{4}}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "3",
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} |
$\tan A + \tan B + \tan C = \tan A\tan B\tan C\,$ in a triangle I want to prove this (where each angle may be negative or greater than $180^\circ$):
When $A+B+C = 180^\circ$ \begin{equation*} \tan A + \tan B + \tan C = \tan A\:\tan B\:\tan C. \end{equation*}
We know that
\begin{equation*}\tan(A+B) = \frac{\tan A+\tan B}{1-\tan A\tan B}\end{equation*} and that \begin{equation*}\text{and that}~A+B = 180^\circ-C.\end{equation*}
Therefore $\tan(A+B) = -\tan C.$
From here, I got stuck.
| \begin{eqnarray}
\tan A+\tan B+\tan C&=&\frac{\sin A\cos B+\sin B\cos A}{\cos A\cos B}+\tan(180^\circ-A-B)\\
&=&\frac{\sin(A+B)}{\cos A\cos B}-\frac{\sin(A+B)}{\cos(A+B)}\\
&=&\sin(A+B)\frac{\cos(A+B)-\cos A\cos B}{\cos A\cos B\cos(A+B)}\\
&=&-\frac{\sin A\sin B\sin(A+B)}{\cos A\cos B\cos(A+B)}\\
&=&-\tan A\tan B\tan(A+B)\\
&=&\tan A\tan B\tan C.
\end{eqnarray}
| {
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"source": "stackexchange",
"question_score": "15",
"answer_count": 11,
"answer_id": 7
} |
Prime Numbers And Perfect Squares Find all primes $p$ and $q$ such that $p^2 + 7pq + q^2$ is a perfect square.
One obvious solution is $p = q$ and under such a situation all primes $p$ and $q$ will satisfy.
Further if $p \neq q$ then we can assume without the loss of generality that $p > q$. Assuming this and that there exists at least one such perfect square I have tried to show some contradiction modulo $4$ as any odd perfect square leaves a remainder of $1$ when divided by $4$, but it is not working. However I firmly believe that $p = q$ is the only solution, but I have failed to prove this.
| COMMENT.- Just for fun, prove easily the impossibility for twin numbers and other couples of primes.
First, not possible for $p=2\lt q$. In fact $$q^2+14q+4=z^2\Rightarrow z^2\ge55\Rightarrow z\ge 8$$ Actually $\color{red}{z\ge 9}$ because $z$ must be odd. The discriminant of $q^2+14q+4-z^2=0$ is $ 45+z^2=t^2$ so making $t=z+h$ one has $45=2zh+h^2\ge18h+h^2$. It is clear how to finish.
$$***$$
Now with the twin numbers $(p,q)=(p,p+2)$ where $p\ge3$. One has
$$p^2+7p(p+2)+(p+2)^2=9p^2+18p+4=z^2$$
Because of $p\ge3$ we have $z\ge 12$ so $\color{red}{z\ge 13}$ because $z$ should be odd.
The equation in $3p$
$$(3p)^2+6(3p)+4-z^2=0$$ has discriminant $$9-4+z^2=t^2\iff5+z^2=t^2$$ This is impossible because $z$ being greater or equal than $13$ the minimal difference $t^2-z^2$ must be greater or equal than $ 14^2-13^2=27.$
$$***$$
There are much primes of the form $(p,q)=(p,p+4)$, for example $(3,7),(7,11),(13,17)$,…… For these we have $$p^2+7p(p+4)+(p+4)^2=z^2=9p^2+36p+16=z^2\Rightarrow z^2\ge 205\Rightarrow \color{red}{z\ge 15}$$ The equation
$$(3p)^2+12(3p)+16-z^2=0$$ has discriminant $36-16+z^2=t^2\iff 20=t^2-z^2$ but then we have $20\ge 16^2-15^2=31$, absurde.
Wants someone to try to see the case $(p,q)=(p+2h)\large ?$ This includes all the odd primes.
| {
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Prove that there are infinitely many perfect cubes of the form $p^2+3q^2$ Prove that there are infinitely many perfect cubes of the form $p^2+3q^2$ where $p$ and $q$ are integers.
Hint: one approach is to set $p^2+3q^2=(a^2+3b^2)^3$ and then find $(p,q)$ in terms of $a,b$.
Any different approach is very welcome!
| Use this from cubing Eusenstein integers, of the form $a+b\sqrt{\;-3}$.
For $a, b$ as integers, then
$$(a+b\sqrt{-3})^3 = (a^3 - 9ab^2) + (3a^2b - 3b^3)\sqrt{-3} $$
Whereapon, one sees that $(a^2+3b^2)^3 = (a^3-9ab^2)^2 + 3(3a^2b-3b^3)^2$ directly follow.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Quadratic substitution question: applying substitution $p=x+\frac1x$ to $2x^4+x^3-6x^2+x+2=0$ By using the substitution $p=x+\frac{1}{x}$, show that the equation $$2x^4+x^3-6x^2+x+2=0$$
reduces to $2p^2+p-10=0$.
I can't think of anything that produces a useful result, I tried writing p as $p=\frac{x^2+1}{x}$ and finding areas to substitute but have come with no progress. Could someone offer a slight hint on how to proceed?
Thanks
| $$p=x+x^{-1}$$
$$p^2=x^2+2+x^{-2}$$
$$2x^4+x^3-6x^2+x+2=0$$
$$2x^2+x-6+x^{-1}+2x^{-2}=0$$
$$2(x^2+2+x^{-2})+(x+x^{-1})-10=0$$
$$2p^2+p-10=0$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Middle school number theory Find at least three numbers that satisfy all three conditions:
(1) there is a remainder of $1$ when the number is divided by $2$;
(2) there is a remainder of $2$ when the number is divided by $3$;
(3) there is a remainder of $3$ when the number is divided by $4$.
Since the LCM($2, 3, 4$) is $12$, all of them can divide $12$ without giving a remainder.
So, $12 - 1=11$ gives
a remainder of $1$ when divided by $2$,
a remainder of $2$ when divided by $3$,
a remainder of $3$ when divided by $4$.
Similarly $12*2-1=23$
and $12*3-1=35$
How would I explain to middle school students the "pulling out of a hat" of
first the $LCM-1$,
and secondly the $LCM*(n)-1$ parts?
| The simple answer could just be write out the sets; i.e,
$S_1=\{...,-3,\color{red}{-1},1,3,5,7,9,\color{red}{11},13,15,17,19,21,\color{red}{23},25...\}$
$S_2=\{...,-4,\color{red}{-1},2,5,8,\color{red}{11},14,17,20,\color{red}{23},26,...\}$
$S_3=\{...,-5,\color{red}{-1},3,7,\color{red}{11},15,19,\color{red}{23},27,...\}$.
As noted in the other answer, immediately we see that $-1,11$, and $23$ satisfy the conditions. From there, we can note that perhaps every $12$th number also satisfies. Then note that $LCM(2,3,4)=12$. Finally construct your formula as $LCM\cdot{n}-1$.
I tried not to use any modular arithmetic since they are middle schoolers, so I hope this helps.
$\mathbf{EDIT}$ After thinking about it, the last thing i would do is verify that the number $12n-1$ leaves the appropriate remainders when divided by $2,3,$ and $4$. Convert $12n-1$ to $12n+11$ for arithmetic's sake and show
$$\frac{12n+11}{2}=6n+5+\frac{1}{2}$$
$$\frac{12n+11}{3}=4n+3+\frac{2}{3}$$
$$\frac{12n+11}{4}=3n+2+\frac{3}{4}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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How prove this inequality $(\sum a_{1}^{1.5})^2\ge \sum a_{1}\sum a_{1}a_{2}$ Now my question let $a_{1},a_{2},\cdots,a_{n}$ are positive numbers,and $a_{n+i}=a_{i},i=1,2,\cdots$,show that
$$(\sum a_{1}^{1.5})^2\ge \sum a_{1}\sum a_{1}a_{2}$$
my teacher (tian275461) have prove this
$$(a^{1.5}+b^{1.5}+c^{1.5})^2\ge (a+b+c)(ab+bc+ac)$$
He methods:let $a\longrightarrow a^2,b\longrightarrow b^2,c\longrightarrow c^2$
then
$$\Longleftrightarrow (a^3+b^3+c^3)^2\ge (a^2+b^2+c^2)(a^2b^2+c^2a^2+b^2c^2)$$
$$\Longleftrightarrow(\sum a^3)^2\ge \sum a^2\sum a^2b^2$$
note
$$(\sum a^3)^2=\sum a^2\sum a^4-\sum b^2c^2(b-c)^2$$
$$\Longleftrightarrow \sum a^2\left(\sum a^4-\sum a^2b^2\right)-\sum b^2c^2(b-c)^2\ge 0 $$
$$\Longleftrightarrow \dfrac{1}{2}\sum a^2\sum(b^2-c^2)^2-\sum b^2c^2(b-c)^2\ge 0$$
$$\Longleftrightarrow \dfrac{1}{2}(b-c)^2 \left(\sum a^2\sum (b+c)^2-2\sum b^2c^2\right) \ge 0$$
it suffices to show that
$$\sum b^2\sum (b+c)^2-2\sum b^2c^2\ge 0$$
and note that
$$\sum b^2\sum (b+c)^2-2\sum b^2c^2=2\sum a^4+2\sum a^3b+2\sum a^3c+2\sum a^2b^2+2\sum a^2bc\ge 0$$
for n=4,it only show that
$$(a^3+b^3+c^3+d^3)^2\ge (a^2+b^2+c^2+d^2)(a^2b^2+b^2c^2+c^2d^2+d^2a^2)$$
| Here is a proof in the case where the $\{a_i\}$ obey a condition for the variance about their mean.
More precisely, the formulation for the above condition is the following. Let $\bf a$ be the vector composed from the $\{a_i\}$, i.e. ${\bf a} = (a_1,a_2, \dots, a_{n-1}, a_n)$. Define the mean ${m} = \frac1n \sum a_i$. Define a vector ${\bf m} = (m,m, \dots, m)$. Let $\bf \bar a$ be the permutation of $\bf a$ which is constructed by shifting all components cyclically by one position, i.e. ${\bf \bar a} = (a_2,a_3, \dots, a_n, a_1)$. Then we require
$$
\cos^2({\bf a}, {\bf m})\geq \cos({\bf a}, {\bf \bar a})
$$
Writing this in components gives the equivalent
$$
\frac{m^2 (\sum_j a_j)^2}{n m^2 \sum_j a_j^2 } \geq \frac{\sum_j a_j a_{j+1}}{\sum_j a_j^2 }
$$
or $(\sum_j a_j)^2 \ge n \sum_j a_j a_{j+1}$. Alternatively, one can define the difference $d_j = a_j - m$. Then we have the condition $\sum_j d_j d_{j+1} < 0$.
As an example, consider ${\bf a} = (10,9,9,1)$. Then we have
$(\sum_j a_j)^2 = (29)^2 = 841 \ge n \sum_j a_j a_{j+1} = 4 \cdot(90 + 81 + 9 + 10) = 760$.
As a counterexample, consider ${\bf a} = (4,5,4,2,1,2)$. Then we have
$(\sum_j a_j)^2 = (18)^2 = 324 < n \sum_j a_j a_{j+1} = 6 \cdot(20 + 20 + 8 + 2 + 2 +8) = 360$.
(Note that, nevertheless, the original inequality holds in this case.)
Now it suffices to prove, instead of the original inequality, the tighter one:
$$
n (\sum_j a_j^{1.5})^2 \ge (\sum_j a_j)^3
$$
Setting $a_j = b_j^2$ gives
$$
(\frac1n \sum_j b_j^{3})^2 \ge (\frac1n \sum_j b_j^2)^3
$$
or
$$
\sqrt[3]{\frac1n \sum_j b_j^{3}} \ge \sqrt[2]{\frac1n \sum_j b_j^{2}}
$$
but this is just the power mean inequality.$\qquad \Box$
| {
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$x^2 +y^2 + z^2$ is irreducible in $\mathbb C [x,y,z]$
Is $x^2 +y^2 + z^2$ irreducible in $\mathbb C [x,y,z]$?
As $(x^2+y^2+z^2)= (x+y+z)^2- 2(xy+yz+zx)$,
$$(x^2+y^2+z^2)=\left(x+y+z+\sqrt{2(xy+yz+zx)}\right)\left(x+y+z-\sqrt{2(xy+yz+zx)}\right).$$
But how to show that none of these factors belong to $\mathbb C [x,y,z]$?
| We can consider $X^2+Y^2+Z^2$ as a polynomial in $R[X]$, where $R=\mathbb C[Y,Z]$. Now apply the Eisenstein criterion with the irreducible $Y+\mathrm iZ$ in $R$:
*
*It divides $Y^2+Z^2$;
*It does not divide $1$, the coefficient of $X^2$;
*It does not divide $(Y^2+Z^2)^2$.
Irreducibility follows.
| {
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"timestamp": "2023-03-29T00:00:00",
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Solutions to $\sqrt{x}+y=6,x^2+y^2=90$ $$\begin{gather}
\sqrt{x}+y=6 \tag{1} \\
x^2 + y^2 = 90 \tag{2}
\end{gather}$$
WE have to solve for $x$ and $y$(Note that 9 is an obvious value of x) My friend asked me this question earlier today, which he said he had made up himself. However, he himself has no idea how to solve for the two variables. I used the method of substitution to finally obtain $x(x+1)=6(21-2y)$. Then I figured that both sides must be non-negative, and hence the value of $y$ has to be less than $11$. Then we test for values less than $11$ which makes $6(21-2y)$ a product of two consecutive numbers. WE get $3$ as a value of $y$, and also $7$. But putting $7$ in (1) does not work.
But there is a flaw in my reasoning. When I get to the 'consecutive' part, I start assuming $x$ and $y$ are integers. Since the person who asked me is not familiar with complex numbers, WE can assume that the solutions are real. However, I am interested in extending the values of $x$ and $y$ beyond the reals, if there are any.
NOTE: From (2), we get the value of $y^2$ in terms of $x$. Then we figure out the value of $x$ in (1) and then substitute $y$ for $90-x^2$, which gives us $x(x+1)$.
| Another way to show there is only one real root...
$\sqrt{x} = 6 - y \implies y \le 6$
So we are looking for roots of $f(y) = (y-6)^4 + y^2 - 90$ where $y \le 6$. We already have $y=3$ as a root from observation.
Using $f(y) = (y-3)(y^3 -21y^2 +154y - 402)$, we need to check if the cubic factor has any real roots in the interval. We note,
If $g(y) = y^3 -21y^2 +154y - 402, g(6) = -18 < 0$ and $g'(y) = 3y^2 - 42y + 154 > 0$ so the cubic has no root with $y < 6$. Hence $(x, y) = (9, 3)$ is the only real solution.
| {
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"timestamp": "2023-03-29T00:00:00",
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Ellipse arcs. Draw a tangent line in the end point or make arc longer? I read this article:
link
It describes how to draw ellipse arcs at all from svg. Each ellipse is described with the following params (and I know them):
*
*x1, y1, x2, y2 - arc from point (x1, y1) to point (x2, y2);
*rx, ry - radiuses
*x-axis-rotation, which indicates how the ellipse as a whole is rotated relative to the current coordinate system
*large-arc-flag and sweep-flag, which are described on the following picture:
I need to draw a tangent line in the second point but I have no idea how to do this. I need coordinates of this line and the direction.
| I rewrote this answer completely, see its history for earlier versions.
Change the coordinate system in such a way that the ellipse becomes a circle of radius $1$. The following transformation achieves this:
$$ f: \begin{pmatrix}x\\y\end{pmatrix} \mapsto
\begin{pmatrix}\frac1{r_x}&0\\0&\frac1{r_y}\end{pmatrix}\cdot
\begin{pmatrix}
\cos\alpha & \sin\alpha \\
-\sin\alpha & \cos\alpha \\
\end{pmatrix}\cdot
\begin{pmatrix}x\\y\end{pmatrix}
= \begin{pmatrix}
\frac{x\cos\alpha+y\sin\alpha}{r_x} \\
\frac{y\cos\alpha-x\sin\alpha}{r_y}
\end{pmatrix}
$$
Apply this transformation to the end points to obtain new coordinates $(x_1',y_1')$ and $(x_2',y_2')$. The center of the ellipse corresponds to a point which has distance $1$ from these two points. So now you have to intersect two circles of radius $1$ around these. To do this, compute the difference vector
$$ v = \begin{pmatrix}v_x\\v_y\end{pmatrix} =
\begin{pmatrix}x_1'\\y_1'\end{pmatrix} -
\begin{pmatrix}x_2'\\y_2'\end{pmatrix}
$$
A circle of radius $1$ at the origin will intersect another circle of radius $1$ at point $v$ in two points which form a line that will pas through the midpoint between the centers, i.e. through the point $\frac12v$. A generic point on that line can be written as
$$ q = \begin{pmatrix}q_x\\q_y\end{pmatrix} =
\frac12\begin{pmatrix}v_x\\v_y\end{pmatrix} +
\lambda\begin{pmatrix}-v_y\\v_x\end{pmatrix} $$
To find those points where that line intersects the circle, compute the squared distance from the origin, and set that equal to one.
$$ \lVert q\rVert^2 = q_x^2 + q_y^2 = 1 $$
Solving this for $\lambda$ will result in two solutions, corresponding to the two points of intersection:
$$ \lambda = \pm\sqrt{\frac{1}{v_x^{2} + v_y^{2}} - \frac{1}{4}} $$
One of these will correspond to the case where the two flags agree, the other to that where they disagree. Try them both until you know which one is which. Once you have made that decision, you now have the point $q$ which corresponds to the center of your ellipse. The ellipse itself will be represented by a circle of radius $1$ around that point. The tangent to that circle through the origin will be the line
$$ q_xx + q_yy = 0 $$
You have to shift that line by $(x_2',y_2')$ and undo $f$ to obtain the equation of the tangent in your original coordinate system. To do that, you can combine all the coordinate transformations done so far into a single affine transformaton matrix.
\begin{align*}
M &=
\begin{pmatrix}
\frac1{r_x} & 0 & 0 \\
0 & \frac1{r_y} & 0 \\
0 & 0 & 1
\end{pmatrix} \cdot
\begin{pmatrix}
\cos\alpha & \sin\alpha & 0 \\
-\sin\alpha & \cos\alpha & 0\\
0 & 0 & 1
\end{pmatrix} \cdot
\begin{pmatrix}
1 & 0 & -x_2 \\
0 & 1 & -y_2 \\
0 & 0 & 1
\end{pmatrix} \\&=
\begin{pmatrix}
\frac{\cos\alpha}{r_x} &
\frac{\sin\alpha}{r_x} &
-\frac{x_2\cos\alpha + y_2\sin\alpha}{r_x} \\
-\frac{\sin\alpha}{r_y} &
\frac{\cos\alpha}{r_y} &
\frac{x_2\sin\alpha - y_2\cos\alpha}{r_y} \\
0 & 0 & 1
\end{pmatrix}
\end{align*}
Then you can take the parameters of the line and multiply them by the transpose of that matrix, or any multiple thereof.
$$ r_xr_yM^T\cdot\begin{pmatrix}q_x\\q_y\\0\end{pmatrix} = \begin{pmatrix}
-r_xq_y\sin\alpha + r_yq_x\cos\alpha \\
r_xq_y\cos\alpha + r_yq_x\sin\alpha \\
r_xx_2q_y\sin\alpha - r_xq_yy_2\cos\alpha -
r_yq_xx_2\cos\alpha - r_yq_xy_2\sin\alpha
\end{pmatrix}
$$
So the equation of your tangent in the original coordinate system will be
\begin{multline*}
\left( -r_xq_y\sin\alpha + r_yq_x\cos\alpha \right)x +
\left( r_xq_y\cos\alpha + r_yq_x\sin\alpha \right)y \\ = -
\left( r_xx_2q_y\sin\alpha - r_xq_yy_2\cos\alpha -
r_yq_xx_2\cos\alpha - r_yq_xy_2\sin\alpha \right)
\end{multline*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/488486",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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} |
Factorize : $x^6 − 10x^3 + 27$ I want to factorize $$x^6 − 10x^3 + 27$$
I tried two methods , first I let $y=x^3 $ and converted it into a quadratic but the solutions are not real .
The second method I tried was getting it to the form $a^3+b^3+c^3-3abc$ but I could only get close ( $(x^2)^3+(x)^3+3^3-3(x^2)(x)(3)$ but that is actually $x^6-8 x^3+27$ , close but not exact )
I'm run out of ideas , so any help is appreciated .
Note : I know that the answer is $$(x^2+2 x+3) (x^4-2 x^3+x^2-6 x+9)$$ , but I want to know how !
| Here is a backdoor way to get the quadratic factor.
Start with your factorization
$x^6-10x^3+27=(x^3-5+\sqrt{-2})(x^3-5-\sqrt{-2}).$
This is a factorization over $\mathbb Z[\sqrt{-2}]$, so seek a quadratic factor with the form
$x^2+2ax+(a^2+2b^2)=(x+a+b\sqrt{-2})(x+a-b\sqrt{-2}).$
The constant term in the quadratic polynomial has to divide $27$ and be positive; moreover $b=0$ cannot work since there are clearly no linear factors over the integers. Therefore $a^2+2b^2\in\{3,27\}$, from which $a=\pm1$ or $\pm5$ and $b=1$. So our candidates are limited to
$x^2+2x+3$
$x^2-2x+3$
$x^2+10x+27$
$x^2-10x+27.$
The third candidate fails because at $x=1$ it does not give divisors of $1^6-(10×1^3)+27=18$. The second and fourth candidates similarly fail at $x=2$. But the first candidate appears to give divisors of the sixth-degree polynomial for every absolutely small integer $x$, and when we try that factor by long division we find that it holds to give the quoted answer.
| {
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"answer_count": 5,
"answer_id": 4
} |
Help me solve this polynomial problem I have to find $a$,$b$ and $c$ so that this expression is a polynomial:
$$\frac{x^3+ax^2+bx+c}{x-1} + \frac{x^3+bx^2+cx+a}{x-2} + \frac{x^3+cx^2+ax+b}{x-3}$$
Please give me hints. Thanks!
| Divide with a remainder:
$$
\frac{x^3+ax^2+bx+c}{x-1}= x^2+b'x+c' +\frac{p}{x-1}
$$
and similarly for other summands. Then your question is reduced to the next one: when
$$
\frac{p}{x-1} +\frac{q}{x-2} +\frac{r}{x-3}
$$
is a polynomial? The answer: only for $p=q=r=0$, since the quadratic polynomial
$$
p(x-2)(x-3)+q(x-1)(x-3)+r(x-2)(x-1)
$$
cannot be divided by $(x-1)(x-2)(x-3)$. Hence $x^3+ax^2+bx+c$ is divided by $x-1$ etc.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Find all real matrices $A$ such that $A^2 = \mathrm{diag}(1,1,2,3,5,8,13)$ Let $A \in \mathcal{M}_{7 \times 7} (\mathbb{R})$ such that
$$A^2= \begin{pmatrix} 1&0&0&0&0&0&0\\0&1&0&0&0&0&0\\0&0&2&0&0&0&0 \\ 0&0&0&3&0&0&0 \\ 0&0&0&0&5&0&0 \\ 0&0&0&0&0&8&0 \\ 0&0&0&0&0&0&13\end{pmatrix} $$
How many matrices, which satisfy this condition could you find?
My friend told to me, that the correct answer is infinity. But I can find only one:
$$A^2= \begin{pmatrix} 1&0&0&0&0&0&0\\0&1&0&0&0&0&0\\0&0&\sqrt{2}&0&0&0&0 \\ 0&0&0&\sqrt{3}&0&0&0 \\ 0&0&0&0&\sqrt{5}&0&0 \\ 0&0&0&0&0&\sqrt{8}&0 \\ 0&0&0&0&0&0&\sqrt{13} \end{pmatrix} $$
I will grateful for hints, how find any other examples. I suppose that $A$ should depend from parametr which don't govern on product $A \cdot A$.
| We have $$\begin{pmatrix} \cos \varphi & \sin \varphi & 0 & \ldots & 0 \\ \sin \varphi & -\cos \varphi & 0 & \ldots & 0 \\ 0 & 0& 0& \ldots & 0\\ \vdots & \vdots& \vdots& \ddots & \vdots\\ 0 & 0& 0& \ldots & 0 \end{pmatrix}^2 = \begin{pmatrix} 1 & 0 & 0 & \ldots & 0 \\ 0 & 1 & 0 & \ldots & 0 \\ 0 & 0& 0& \ldots & 0\\ \vdots & \vdots& \vdots& \ddots & \vdots\\ 0 & 0& 0& \ldots & 0 \end{pmatrix}$$
Additionaly I noticed that
$$\begin{pmatrix} \cos \varphi & \sin \varphi & 0 \\ \sin \varphi & -\cos \varphi & 0 \\ 0 & 0& x \end{pmatrix} ^2 = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0& x^2 \end{pmatrix} $$
so I suppose that we can take
$$A= \begin{pmatrix} \cos \varphi&\sin \varphi &0&0&0&0&0\\\sin \varphi&-\cos \varphi&0&0&0&0&0\\0&0&\sqrt{2}&0&0&0&0 \\ 0&0&0&\sqrt{3}&0&0&0 \\ 0&0&0&0&\sqrt{5}&0&0 \\ 0&0&0&0&0&\sqrt{8}&0 \\ 0&0&0&0&0&0&\sqrt{13} \end{pmatrix} $$
Of course, $\varphi$ is any so we have infinity matrices.
| {
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"source": "stackexchange",
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} |
Limit of $s_n=\frac{1}{\sqrt{n}}\left(1+\frac{1}{\sqrt{2}}+\cdots +\frac{1}{\sqrt{n}}\right)$ \begin{align*}S_n=\frac{1}{\sqrt{n}}\left(1+\frac{1}{\sqrt{2}}+\cdots +\frac{1}{\sqrt{n}}\right)\end{align*}
how to calculate the limit $s_n$?
\begin{align*}\lim_{n\to \infty } \, S_n\end{align*}
| Consider the curve $y=\frac{1}{\sqrt{x}}$. We have
$$\int_1^{n+1}\frac{1}{\sqrt{x}}\,dx\lt 1+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+\cdots+\frac{1}{\sqrt{n}}\lt \int_0^n \frac{1}{\sqrt{x}}\,dx.$$
Evaluate the integrals. We get
$$2\sqrt{n+1}-2\lt 1+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+\cdots+\frac{1}{\sqrt{n}}\lt 2\sqrt{n}.$$
Divide everything by $\sqrt{n}$, and use Squeezing to conclude that our limit is $2$.
| {
"language": "en",
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"source": "stackexchange",
"question_score": "10",
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"answer_id": 2
} |
Prove that $(x^2-x^3)(x^4-x) = \sqrt{5}$, where $x= \cos(2\pi/5)+i\sin(2\pi/5)$
Prove $(x^2-x^3)(x^4-x) = \sqrt{5}$ if $x= \cos(2\pi/5)+i\sin(2\pi/5)$.
I have tried it by substituting $x = \exp(2i\pi/5)$
but it is getting complicated.
| Note $(x^2-x^3)(x^4-x)=-x^7+x^6+x^4-x^3$. The key is that $x$ is a root of unity, particularly one that gives $x^5=1$. This means $x^6=x$ and $x^7=x^2$ while $x^4=x^*$ and $x^3=(x^2)^*$ which means that our imaginary components fall out since $z+z^*=2a$ for $z=a+bi$. Squaring both sides to get rid of the radical reduces the problem nicely.
sorry for any errors as I've posted this from my phone.
| {
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"source": "stackexchange",
"question_score": "5",
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} |
Solve $z^4+1=0$ algebraically I know the result and how to solve it using trigonometry and De Moivre.
However, given that the complex number $z$ can be rewritten as $a+bi$, how can I solve it algebraically?
| $$ z^4 + 1 = 0 \Rightarrow z^4 + 2z^2 + 1 - 2z^2 = 0 $$
$$ \Rightarrow (z^2 + 1)^2 -(\sqrt{2}z)^2 = 0 $$
$$ \Rightarrow (z^2 - \sqrt{2} z + 1)( z^2 + \sqrt{2}z + 1 ) = 0 $$
use $$ ax^2 + bx + c = 0 \Rightarrow x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} $$
| {
"language": "en",
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"source": "stackexchange",
"question_score": "11",
"answer_count": 7,
"answer_id": 0
} |
How to solve $y' = \sqrt {x+y+1}$ How does one solve $y' = \sqrt{x+y+1}$? I try substituting $v=x+y+1$ and using substitution methods, but it turned out to be so messy.
| Another substitution is $x+y+1 = z^2$, so $1+y' = 2zz'$, from which $y' = 2zz'-1$:
$$
2zz'-1 = z \\
2zz' = z+1 \\
\frac {2zdz}{z+1} = dx \\
\int \left (1-\frac 1{z+1} \right ) dz = \frac 12\int dx \\
z - \ln \left |z+1 \right | = \frac x2 + C
$$
Starting this point, it makes sense to leave a solution as an implicit equation w.r.t $x,y$, so
$$
\sqrt{x+y+1} - \ln \left ( \sqrt{x+y+1}+1\right ) - \frac x2 = C
$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Finding Coefficients of Power Series Expansion Find the coefficients of $ a_{n}$ and $b_{n}$ for $ 0 ≤ n ≤ 4$ for the power series expansion of two linearly independent solutions of the ODE: $y'' -(e^{x}-1)y=0$.
This is what I've tried so far:
$\sum_{n=2}^{\infty} n(n-1)c_{n}x^{n-2} - \sum_{n=0}^{\infty}\frac{1}{n!}x^{n}\sum_{n=0}^{\infty}c_{n}x^{n} + \sum_{n=0}^{\infty}c_{n}x^{n} = 0$
$\sum_{n=0}^{\infty} (n+2)(n+1)c_{n+2}x^{n} - \sum_{n=0}^{\infty}\frac{1}{n!}x^{n}\sum_{n=0}^{\infty}c_{n}x^{n} + \sum_{n=0}^{\infty}c_{n}x^{n} = 0$
Don't know how to continue from here.
| Hint: $e^x = 1 + x + \dfrac{x^2}{2!}+\dfrac{x^3}{3!}+\dfrac{x^4}{4!}+\ldots$, $y_1(x) = x + a_2 x^2 + a_3 x^3 + a_4 x^4 + \ldots$, and $y_2(x) = 1 + b_2 x^2 + b_3 x^3 + b_4 x^4 + \ldots$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
calculate a trigonometric expression related to $\sin(\pi/5)$ I tried to solve a question which led to
$$4 \sin\frac{\pi}{5} \sin\frac{3\pi}{5} = \sqrt{5}$$
But, I got stuck to prove that.
Is there any easy solution?
| I will find the $\cos\frac{2\pi}{5}$ with basic complex number theory.
Let $z = \cos(\frac{2\pi}{5}) + i \sin(\frac{2\pi}{5})$, then $z^5=1$. (de Moivre's Theorem) Let's find the solution of this quintic equation.
This equation is factored into $(z-1)(z^4 + z^3 + z^2 + z +1 ) = 0$. $z=1$ is trivial solution.
Now let $u = z + z^4$, and $v= z^2 + z^3$. then $u + v = -1$, and $uv = z^3 + z^4 + z^6 + z^7 = z + z^2 + z^3 + z^4 = -1$. So $t^2 + t - 1 = 0$ has two solutions $u$ and $v$.
Solving the quadratic equation, we get $t = \frac{-1 \pm \sqrt{5}}{2}$.
Geometrically $u$ has positive real part, and $v$ has negative real part. so $z + z^4 = \frac{-1 + \sqrt{5}}{2}$. Since $z$ and $z^4$ are conjugate, the real part of $z$ is the half of it, i.e. $\cos\frac{2\pi}{5}=\frac{-1 + \sqrt{5}}{4}$.
Now, $\cos(\frac{2\pi}{5}) = 1-2\sin^2(\frac{\pi}{5})$, so you can find $\sin\frac{\pi}{5}$.
In this way you can find $\cos\frac{2\pi}{17}$, by decreasing the order of equation 16th to 8th, 4th, .... You need to take $u$ and $v$ in special way...
| {
"language": "en",
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Prove $ \frac{\sqrt{2}}{2} + \frac{\sqrt{3}}{4} + \frac{\sqrt{4}}{6} + \cdots + \frac{\sqrt{n+1}}{2n} > \frac{\sqrt{n}}{2} $ by induction
Prove by induction that for all $n > 0$,
$$
\frac{\sqrt{2}}{2} + \frac{\sqrt{3}}{4} + \frac{\sqrt{4}}{6} + \cdots
+ \frac{\sqrt{n+1}}{2n} > \frac{\sqrt{n}}{2}
$$
I have done the basis step, where $n = 1$ and showed that L.H.S is > R.H.S
For inductive step, I assume $n = k$ & L.H.S > R.H.S is true.
However, I am stuck at showing how for $k + 1$, it is also true for L.H.S > R.H.S
Any help will be much appreciated. Thanks!
| (Someone please check for error ... I'm not confident if I got this correct)
Since for the base case $n=1$: $\dfrac{\sqrt2}2>\dfrac{\sqrt 1}2$, you just have to prove that $\dfrac{\sqrt{n+1}}{2n}\geq\dfrac{\sqrt{n}}2-\dfrac{\sqrt{n-1}}2$ for $n>1$ since, if $a>b$ then $a+c>b+d$ for $c\geq d$.
\begin{align*}
\frac{\sqrt{n+1}}{2n}&\geq\frac{\sqrt n-\sqrt{n-1}}2\\ \\
\frac{\sqrt{n+1}}{n}&\geq\sqrt n-\sqrt{n-1}\\ \\
\frac{\sqrt{n+1}}{\sqrt n-\sqrt{n-1}}&\geq n\\ \\
\sqrt{n+1}(\sqrt n+\sqrt{n-1})&\geq n\\ \\
\sqrt{n^2+n}+\sqrt{n^2-1}&\geq n\\\square
\end{align*}
(note that $\sqrt{n^2+n}>n$ since $\sqrt{n^2}=n$ for $n\geq1$)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/502026",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
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Finding the determinant of a $4\times4$ matrix How does one find the determinant of a $4\times 4$ matrix? I am using Cramer's rule to solve a system of linear equations but don't know how to find the determinant of a $4\times 4$ matrix. Our matrix is with variables and not actual values so the answer will be in terms of the variables. Is there a rule/formula that I can use to get the determinant without using co-factor expansion?
The matrix given is
\begin{bmatrix}
1 & 1 & -2 & 3 \\
3 & 1 &-6 & 8 \\
0 & 7 & 3 & 0 \\
-1 & 8 & 7 & -4
\end{bmatrix}
, but then to use cramers rule and then to find the subsequent determinants the matrix will change? so was hoping for a general formula?
| A "quick" method to compute the determinant of a $4 \times 4$ matrix by hand, is by using cofactor expansion.
Let $ D =
\begin{pmatrix}
1 & 1 & -2 & 3 \\
3 & 1 &-6 & 8 \\
0 & 7 & 3 & 0 \\
-1 & 8 & 7 & -4
\end{pmatrix}$
Here we are going to choose which row/column contains the most $0$'s Clearly, row 3 contains the most. So we will exand along that row.
Start with the first term in row three, which is $0$ and multiply it to the $(n-1) \times (n-1)$ sub-matrix $M$, which you can obtain by covering up the column and rows that intersect at the term. Which should look like this:
$0 \cdot
\begin{pmatrix}
1 & -2 & 3 \\
1 &-6 & 8 \\
8 & 7 & -4
\end{pmatrix}$
As you can see, we can obtain a $3 \times 3$ matrix containing all original entries of our $4 \times 4$ matrix, excluding those from column 1 and row 2.
Now, we repeat this process for the remain terms in row three: $7,3,0$:
$\text{det}(D) =$
$ 0 \cdot
\begin{pmatrix}
1 & -2 & 3 \\
1 &-6 & 8 \\
8 & 7 & -4
\end{pmatrix} $
$- 7 \cdot
\begin{pmatrix}
1 & -2 & 3 \\
3 &-6 & 8 \\
-1 & 7 & -4
\end{pmatrix} +$
$ 3 \cdot
\begin{pmatrix}
1 & 1 & 3 \\
3 & 1 & 8 \\
-1 & 8 & -4
\end{pmatrix} - $
$0 \begin{pmatrix}
1 & 1 & -2 \\
3 & 1 &-6 \\
-1 & 8 & 7
\end{pmatrix}$
The matrices being mutliplied by $0$ can be discarded. So we are left with:
$\text{det}(D) = - 7 \cdot
\begin{pmatrix}
1 & -2 & 3 \\
3 &-6 & 8 \\
-1 & 7 & -4
\end{pmatrix}+$
$ 3 \cdot
\begin{pmatrix}
1 & 1 & 3 \\
3 & 1 & 8 \\
-1 & 8 & -4
\end{pmatrix}$
Expanding along row 1 in both resulting $3 \times 3$ matrices and we get:
$\text{det}(D) =$
$ -7 \cdot \{\begin{pmatrix}
-6 & 8 \\
7 & -4
\end{pmatrix} + $
$2 \cdot \begin{pmatrix}
3 & 8 \\
-1 & -4
\end{pmatrix} + $
$3 \cdot \begin{pmatrix}
3 &-6 \\
8 & 7
\end{pmatrix}\} + $
$
3 \cdot \{ \begin{pmatrix}
1 & 8 \\
8 & -4
\end{pmatrix} - $
$\begin{pmatrix}
3 & 8 \\
-1 & -4
\end{pmatrix} + $
$ 3 \cdot \begin{pmatrix}
3 & 1 \\
-1 & 8
\end{pmatrix}\}$
Now all you have left is to comupute six $2 \times 2$ matrices.
As far as computing this with variables, the same principles of cofactor expansion can be applied. Just determine which row/column contains the most $0$'s (if any) and expand along that row.
| {
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Rate of convergence for sequences I have numerically determined that the sequence $\{f_x\} = \frac{\sin(x^2)}{x^2}$ approaches $1$ (as $x$ approaches $0$) faster than the sequence $\{g_x\} = \frac{\sin^2(x)}{x^2}$. However, I am stuck for determing the rate of convergence of each sequence.
Should I use the small angle approximation $\sin(x)\approx x$ and the fact that $|\sin(x)|\leq 1$? Thanks.
| Let's start with the definition of rate of convergence of functions (There is also definition of rate of convergence of sequences):
Definition. Let $f$ be a function defined on the interval $(a,b)$ that
contains $x=0$, and suppose $\lim_{x\to 0}{f(x)=L}$. If there exists a
function $g$ for which $\lim_{x\to 0}{g(x)=0}$ and a positive constant
$K$ such that $$|f(x)-L| \le K|g(x)|$$ for all sufficiently small
values of $x$, then $f(x)$ is said to converge to $L$ with rate of
convergence $O(g(x))$.
Let $f(x)=\sin{x^2}/{x^2}$ and $g(x)=(\sin{x})^2/{x^2}$. From Taylor's Theorem, we know that
$$\sin{x}=x-\frac{1}{6}x^{3}\cos{\xi}$$
for some $\xi$ between $0$ and $x$. Hence,
$$f(x)=\frac{x^2-\frac{1}{6}x^{6}\cos{\xi}}{x^2}=1-\frac{1}{6}x^{4}\cos{\xi}$$
We know $\sin{x}=(1-\cos{2x})/{2}$ and
$$\cos{x}=1-\frac{1}{2}x^2+\frac{1}{24}x^4\cos{\xi}$$
for some $\xi$ between $0$ and $x$. Therefore,
$$g(x)=\frac{(1-1+2x^2-\frac{2}{3}x^4\cos{\xi})/2}{x^2}=\frac{x^2-\frac{1}{3}x^4\cos{\xi}}{x^2}=1-\frac{1}{3}x^2\cos{\xi}$$
Finally
$$|\frac{\sin{x^2}}{x^2}-1|=\frac{1}{6}|x^{4}\cos{\xi}|\le \frac{1}{6}|x^{4}|$$
and
$$|\frac{(\sin{x})^2}{x^2}-1|=\frac{1}{3}|x^2\cos{\xi}|\le \frac{1}{3}|x^2|$$
It follows that the rate of convergence of $f(x)$ is $O(x^4)$ and the rate of convergence of $g(x)$ is $O(x^2)$. It is common to express this in shorthand by writing
$$\frac{\sin{x^2}}{x^2}=1+O(x^4) \qquad and \qquad \frac{(\sin{x})^2}{x^2}=1+O(x^2)$$
That means an error term with rate of convergence $O(x^4)$ approaches zero faster than an error term with rate of convergence $O(x^2)$, as the value of $x$ approaches zero. Therefore $f(x)$ approaches one faster than $g(x)$ as $x$ approaches zero.
Source: A Friendly Introduction to Numerical Analysis by Brian Bradie.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Find $f'(0)$ for $f(x)=(2x+1)^3(3x+3)^2$. Find $f'(0)$ for $f(x)=(2x+1)^3(3x+3)^2$.
Do I use the chain rule for each or do I use the derivative product rule first
Please Help!!!
| $f(x) = (2x+1)^3\cdot (3x+3)^3$ and $f(0) = 27$
Taking $\log_{e}$ on both side
$\log_{e}f(x) = \log\left\{(2x+1)^3 \cdot (3x+3)^3\right\} = 3\cdot \log_{e}(2x+1)+3\cdot \log_{e}(3x+3)$
Now Diff. both side w.r.to $x$
$\displaystyle \frac{1}{f(x)}\cdot f^{'}(x) = \frac{3}{2x+1}\cdot 2+\frac{3}{3x+3}\cdot 3$
Put $x = 0$ on both side,
$\displaystyle \frac{f^{'}(0)}{27} = \frac{6}{1}+\frac{9}{3} = 9$
So $f^{'}(0) = 27 \times 9 = 243$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Integer $x$ for which $x^4+x^3+x^2+x+1$ is perfect square. Integer values of $x$ for which $\bf{x^4+x^3+x^2+x+1}$ is a Perfect Square.
$\underline{\bf{My\; Try}}$:: Let $\bf{x^4+x^3+x^2+x+1 = k^2}$, where $k\in \mathbb{Z}$
$4x^4+4x^3+4x^2+4x+1 = 4k^2 = (2k)^2$
Now How can I proceed after that
Help Required,
Thanks
| Hint: For all but finitely many integer values of $k$, we have
$$ (2k^2 + k)^2 < 4k^4 + 4k^3 + 4k^2 + 4k + 4 < (2k^2 + k + 2 ) ^ 2 $$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Damped Harmonic Oscillator and Response Function This is another one of those questions that I feel like I am almost there, but not quite, and it's the math that gets me. But here goes:
For a driven damped harmonic oscillator, show that the full width at half maximum of the response function $| R(\omega)|^2$ is $\gamma$. Where $\gamma $ is the damping factor. So we start off with: $$\ddot x + \gamma \dot x + \omega_0x = \frac{F_{ext}(t)}{m} = f(t)$$
And I move to solve this. I know that the solution will be $x_0 e^{i\omega t}$ and that I should get the following:
$$-\omega^2 x_0 e^{i\omega t} + (i\omega)x_0 e^{i\omega t}\gamma+ \omega_0^2x_0e^{i\omega t} = \frac{F_0 e^{i\omega t}}{m}$$
which turns into $$x_0 (\omega_0^2 -\omega^2 + (i\omega) \gamma ) = \frac{F_0}{m}= f_0$$ which then becomes $$x_0 = \frac{f_0}{\omega_0^2 -\omega^2 + (i\omega) \gamma }$$
The denominator should give me the response function $R(\omega)$. So I take the absolute value and square it and I get: $$|R(\omega^2)| = \frac{f_0}{\omega_0^2 -\omega^2 t + (i\omega) \gamma } \frac{f_0}{\omega_0^2 -\omega^2 t - (i\omega) \gamma } = \frac{f_0}{(\omega_0^2 -\omega^2)^2 + \gamma^2 \omega^2 }$$
So far so good. I want to find the maximum of this function and that means I want to know where the derivative of the denominator is zero. Taking that derivative: $$\frac{d}{d\omega}(\omega_0^2 -\omega^2)^2 + \gamma^2 \omega^2) = 2((\omega_0^2 -\omega^2)(2\omega) + 2\gamma^2 \omega = 0$$ and since it = 0 we can divide thru by $2\omega$ and we are left with $$-2(\omega_0^2 - \omega^2) + \gamma^2 = 0$$ and solving for $\omega$: $\omega = \sqrt{\omega_0^2 - \frac{\gamma^2}{2}} $. We will replace $\gamma$ with $2\beta$, leaving us $\omega = \sqrt{\omega_0^2 - 2 \beta^2} $
we know where the maximum is, but now we want to find the half-maximum. Plugging my $\omega$ back into my $|R(\omega)|^2$ expression, I have $$\frac{1}{(\omega_0^2 - \omega_0^2 + 2 \beta^2)^2 + 4\beta^4 (\omega_0^2 - 2\beta^2)}=\frac{1}{(2 \beta^2)^2 + 4\beta^4 (\omega_0^2 - 2\beta^2)}=\frac{1}{4 \beta^4 + 4\beta^2 \omega_0^2 - 8\beta^6}$$
which shows me the maximum of $|R(\omega)|^2$. At 1/2 that is the half maximum and I want to know what $\omega$ is at that point. So going back to my original equation I posit: $$|R(\omega^2)| = \frac{1}{(\omega_0^2 -\omega^2)^2 + 4\beta^2 \omega^2 }=\frac{1}{8 \beta^4 + 8\beta^2 \omega_0^2 - 16\beta^6}$$
But at this point I feel I have lost the plot. Going through this the whole thing struck me as more complicated than it needs to be. My text says showing this should be "an easy exercise." So I am turning to people here to see where I messed up.
Best to you all, and thanks.
| The problem is not so much the calculation, as that is all about quadratic functions, but to select suitable auxiliary variables so that the calculation gets a readable and thus verifiable organization.
I'll interpret the task, as did @Aurey, that the task asks for the lowest order terms of the answer assuming $0<γ=2β\ll ω_0$.
Set $x=ω^2−ω_0^2$, then the denominator is a quadratic form in $x$,
\begin{align}
(ω_0^2−ω^2)^2+4β^2ω^2
&=x^2+4β^2x+4β^2ω_0^2
\\
&=(x+2β^2)^2+4β^2(ω_0^2-β^2)
\\
&=D(x)
\end{align}
which shows the minimum value and thus the maximum of the frequency response at $x_{\min}=-2β^2$. Now we want to find the positions $x_\pm$ where the parabola has twice the minimum value, so that the fraction has half the maximum value.
\begin{align}
&&D(x_\pm)&=2·D(x_\min)
\\
\iff&&(x_\pm+2β^2)^2+4β^2(ω_0^2-β^2)&=2·4β^2(ω_0^2-β^2)
\\ \iff&&
(x_\pm+2β^2)^2&=4β^2(ω_0^2-β^2)
\\ \iff&&x_\pm&=-2β^2\pm2β\sqrt{ω_0^2-β^2}
\\
&&&=\pm2βω_0-2β^2+O(β^3)
\end{align}
or
$$
ω_\pm^2=ω_0^2\pm2βω_0-2β^2+O(β^3)=(ω_0\pm β)^2-3β^2+O(β^3)
$$
of which the positive square roots are, at the same order in $β$,
$$
ω_\pm=ω_0\pmβ-\frac{3β^2}{2ω_0}+O(β^3)=ω_0\pm\frac{γ}2-\frac{3γ^2}{8ω_0}+O(γ^3)
$$
Thus the distance between these points is not influenced by the asymmetry, as both roots are shifted, in the lowest terms, by the same amount. The searched for width is
$$
ω_+-ω_-=γ+O(γ^3).
$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Need to prove that $4\cdot 10^{2n} + 9\cdot 10^{2n-1} + 5$ is divisible by $99$ for all $n \in \mathbb{N} $, using induction. First, obviously, I figured out the base case. So I have $4\cdot 10^{2n} + 9\cdot 10^{2n-1} + 5 = 99k$ for some $k \in \mathbb{N} $. As for the inductive step, I was thinking about splitting it up into two parts; proving that the $4\cdot 10^{2n} + 9\cdot 10^{2n-1} + 5$ is divisible by $9$, and that it's divisible by $11$. Unfortunately, I factor it out and get $10(4\cdot10^{2n} + 9\cdot10^{2n-1} + 5)-45$, which is equal to $10(99k)-45$, but of $99$'s divisors, only $3$ divides $45$.
Is there a more ingenious way to factor this equation so that all of my problems are solved?
| Let $f_{n}\equiv4\cdot10^{2n}+9\cdot10^{2n-1}+5$. Then $f_{1}=4\cdot100+9\cdot10+5=495$,
and $99\mid495$ so $99\mid f_{1}$. Now, suppose that for some $n\in\mathbb{N}$,
$99\mid f_{n}$. Now, since $100\equiv1\text{ mod }99$,
\begin{align*}
f_{n+1} & \equiv 4\cdot10^{2\left(n+1\right)}+9\cdot10^{2\left(n+1\right)-1}+5\text{ (mod) }99\\
& \equiv 4\cdot10^{2n+2}+9\cdot10^{2n+1}+5\text{ (mod) }99\\
& \equiv 4\cdot10^{2n}\cdot10^{2}+9\cdot10^{2n-1}\cdot10^{2}+5\text{ (mod) }99\\
& \equiv4\cdot10^{2n}+9\cdot10^{2n-1}+5\text{ (mod) }99\\
& \equiv f_n \text{ (mod) } 99 \\
& \equiv 0 \text{ (mod) } 99
\end{align*}
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Implicit differentiation of a lemniscate at a point: So here's the problem:
Find the slope of the tangent line of :
$2(x^2 +y^2)^2 = 25(x^2 - y^2)$ at the point (3,1)
Cool:
So here's what I did:
Simplification step:
$2(x^2 +y^2)^2 = 25(x^2 - y^2) \Rightarrow 2(x^2+y^2)^2 = 25x^2 - 25y^2$
Differentiate both sides:
$4(x^2 +y^2)(2x(x') + 2y(y')) = 50x(x') - 50y(y')$
Simplify:
$(4x^2 + 4y^2)(2x + 2y(y'))= 50x - 50y(y')$
$(4x^2 + 4y^2)(2x + 2y(y'))= 50x - 50y(y')$
$8x^3 + 8x^2y(y') + 8xy^2 + 8y^3(y')= 50x - 50y(y')$
$8x^2y(y') + 8y^3(y') + 50y(y') = 50x - 8x^3 + 8xy^2$
$(y') (8x^2y + 8y^3 + 50y) = 50x - 8x^3 + 8xy^2$
$y' = \frac{50x - 8x^3 + 8xy^2}{8x^2y + 8y^3 + 50y}$
Cool, got y'...
Now, should I take the original equation and solve for y and substitute? I'm not even going to try putting down my steps on here.
The prompt is to find m=?
Someone please help?
| We start from
$$2(x^2+y^2)^2=25(x^2-y^2).$$
Don't hesitate, Differentiate. We get
$$4(x^2+y^2)(2x+2yy')=50x-50yy'.$$
Substitute $x=3$, $y=1$. We get
$$4(10)(6+2m)=150-50m,$$
and therefore $130m=-90$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/512941",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
} |
Distinct Mersenne numbers are coprime How can you prove that if $p$ and $q$ are distinct primes, then the following holds?:
$$(M_p,M_q)=1$$
Note: $M_n=2^n-1$, with $n$ prime number
| Let $z < x < y$ all be integers with $y = kx + z$ for some positive integer $k$.
I claim that $2^y - 1 \equiv 2^z - 1 \pmod{2^x - 1}$.
Proof:
$(2^y - 1) - 2^{y-x}(2^x - 1) = 2^y - 1 - 2^y + 2^{y-x} = 2^{y-x} - 1$, so
$2^y - 1 \equiv 2^{y-x} - 1 \pmod{2^x-1}$. By similar reasoning, it is also congruent to $2^{y-2x} - 1, 2^{y-3x} - 1, \ldots, 2^{y-kx} - 1 = 2^z - 1$.
Hence, $\gcd(2^y - 1, 2^x - 1) = \gcd(2^x - 1, 2^z - 1)$, and you can iterate this to decrease the exponents in a way similar to the numbers in the Euclidean algorithm. I trust you can take it from here.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
cubic equations which have exactly one real root Question is to check :
For any real number $c$, the polynomial $x^3+x+c$ has exactly one real root .
the way in which i have proceeded is :
let $a$ be one real root for $x^3+x+c$ i.e., we have $a^3+a+c=0$
i have seen that $(x-a)(x^2+ax+(a^2+1))=x^3+x-a^3-a$
But, $a^3+a+c=0$. So, $-a^3-a=c$.
so, $(x-a)(x^2+ax+(a^2+1))=x^3+x-a^3-a=x^3+x+c$
i.e, $(x-a)(x^2+ax+(a^2+1))=x^3+x+c$
Now, determinant for quadratic $x^2+ax+(a^2+1)$ is $a^2-4(a^2+1)=-3a^2-4 <0$ for any real number $a$
Thus, quadratic has no real root and so is the cubic $x^3+x+c$
I would like to know if this justification is sufficient and if this can be generalized.
I mean, can i say $x^3+ax^2+bx+c$ for $a,b,c\in \mathbb{Z}$ have exactly one real root. (Conditions apply)
can this be generalized to any odd degree polynomial (at least for some special cases)
| $(x^3+x+1)'=3x^2+1>0$ for all real $x$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Calculation of sum by using residue theory I am studying an example about the calculation of a summation by using residue theory. I understand how to calculate the sum in general -the frame of the solution way-, but I dont know how some parts of the solution to be found. I Will write the question and answer, please explain me these parts of the solution which I dont understand. Thank you:)
question:
$$\sum _{n=0}^{\infty}\binom{3n}{2n}\frac{1}{8n}=$$ $$ \frac{1}{2\pi i} \sum_{n=0}^{\infty}\int_{C_R}\frac{(1+z)^{3n}}{z^{2n+1}}\frac{1}{8^n}dz=\frac{1}{2\pi i}\int_{C_R}\sum_{n=0}^\infty [\frac{(1+z)^3}{8z^2}]^n\frac{1}{z}dz=\frac{1}{2\pi i} \int _{C_R}\frac{1}{1-\frac{(1-z)^3}{8z^2}}\frac{1}{z}dz =\frac{1}{2\pi i} \int_{C_R}\frac{8z^2}{8z^2-(1-z)^3}\frac{dz}{z}$$
$$=\frac{1}{2\pi i} \int_{C_R}\frac{8z}{8z^2-(1-z)^3}dz$$
I understand so far.
Now, I Try to find singularities, but I dont know how to find these singularities.
Its singularities are $-0.2$ and $4.2=2+\sqrt{5}$ (How to find?)
Now, sum is found, However I dont understand anything the following part
Please explain espacially here in a clear way
(i) $0\lt R\lt 2-\sqrt{5}$ then No residue insinde C, sum $=0$ (Not its answer)
(ii) $|2-\sqrt{5}|\lt R\lt 1$ only one residue insinde, sum$=\frac{8-4\sqrt{5}}{\sqrt{5}-5} =0.34$ (not its answer)
(iii) $1\lt R\lt 2+\sqrt{5}$ two residues sum $=\frac{2(1+\sqrt{5})}{5-\sqrt{5}}=2.34$ (its answer)
(iv) $R\gt 2+\sqrt{5}$ three residues sum $ =0$ (Not its answer)
| You have a cubic in the denominator:
$$8 z^2-(1-z)^3 = z^3+5 z^2+3 z-1$$
You should be able to see that $z=-1$ is a root of this cubic by inspection. By synthetic division, you will see that
$$z^3+5 z^2+3 z-1 = (z+1)(z^2+4 z-1)$$
So we see that the other poles are at $z_{\pm}=-2\pm\sqrt{5}$. Now use the fact that
$$\operatorname*{Res}_{z=z_0} \frac{8 z}{z^3+5 z^2+3 z-1} = \frac{8 z_0}{3 z_0^2+10 z_0+3}$$
at pole $z=z_0$. At each pole:
$$\operatorname*{Res}_{z=-1} \frac{8 z}{z^3+5 z^2+3 z-1} = 2$$
$$\operatorname*{Res}_{z=\sqrt{5}-2} \frac{8 z}{z^3+5 z^2+3 z-1} = \frac{3}{\sqrt{5}}-1$$
$$\operatorname*{Res}_{z=-\sqrt{5}-2} \frac{8 z}{z^3+5 z^2+3 z-1} = -\frac{3}{\sqrt{5}}-1$$
Whether these poles contribute to the integral, as you say, depends on $R$:
$$0 < R < \sqrt{5}-2 \implies 0$$
$$\sqrt{5}-2 < R < 1 \implies \frac{3}{\sqrt{5}}-1$$
$$1 < R < \sqrt{5}+2 \implies \frac{3}{\sqrt{5}}-1 + 2 = \frac{3}{\sqrt{5}}+1$$
$$R > \sqrt{5} + 2 \implies \frac{3}{\sqrt{5}}-1 + 2 - \frac{3}{\sqrt{5}}-1 = 0$$
This is essentially what you say.
| {
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"timestamp": "2023-03-29T00:00:00",
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Proof that $\mathbb Z[\sqrt{3}]$ is a Euclidean Domain Let $R_d$ be the ring defined as $R_d=\left \{ x+y\omega : x,y\in \mathbb{Z} \right\}$, where
$$\omega =
\begin{cases}
\sqrt{d}, & \text{if } \quad d \not \equiv 1\mod 4 \\
\frac{1+\sqrt{d}}{2}, & \text{if } \quad d\equiv 1\mod 4.
\end{cases}$$
It has been proven that $R_d$ is Euclidean for several positive values of $d$.
Does anyone know where I can find a proof that $R_d$ is Euclidean for $d=3$?
Thank you.
| Define the norm on $\mathbb Z[\sqrt 3]$ to be $N(a + b \sqrt 3) = \vert a^2 - 3 b^2 \vert$.
Let $\alpha, \beta \in \mathbb Z[\sqrt 3]$ with $\beta \neq 0$.
Say $\alpha = a + b \sqrt 3$ and $\beta = c + d \sqrt 3$.
Notice that
\begin{align*}
\frac\alpha\beta &= \frac{a + b \sqrt 3}{c + d \sqrt 3} \cdot \frac{c - d \sqrt 3}{c - d \sqrt 3} \\
&= \frac{ac - 3bd}{c^2 - 3d^2} + \frac{-ad + bc}{c^2 - 3d^2} \sqrt 3 \\
&= r + s\sqrt 3
\end{align*}
where $r = \displaystyle \frac{ac - 3bd}{c^2 - 3d^2}$ and $s = \displaystyle \frac{-ad + bc}{c^2 - 3d^2}$.
Let $p$ be the closest integer to $r$ and let $q$ be the closest integer to $s$. Notice that $\vert r - p \vert \leq 1/2$ and $\vert s - q \vert \leq 1/2$.
We want to show that $\alpha = (p + q\sqrt 3) \beta + \gamma$ for some $\gamma \in \mathbb Z[\sqrt 3]$ such that either $\gamma = 0$ or $N(\gamma) < N(\beta)$. (We'll show the latter holds always.)
Define $\theta := (r - p) + (s - q)\sqrt 3$ and define $\gamma = \beta \cdot \theta \in \mathbb Z[\sqrt 3]$ and observe that
\begin{align*}
\gamma &= \beta \cdot \theta\\
&= \beta ( (r - p) + (s - q)\sqrt 3)\\
&= \beta (r + s\sqrt 3) - \beta(p + q\sqrt 3) \\
&= \beta \cdot\frac\alpha\beta - \beta (p + q\sqrt 3) \\
&= \alpha - \beta (p + q\sqrt 3)
\end{align*}
Hence we have $\alpha = \beta(p + q\sqrt 3) + \gamma$.
Finally notice that
\begin{align*}
N(\gamma) &= N(\beta \cdot \theta) \\
&= N(\beta) \cdot N(\theta) \\
&= N(\beta) \cdot \vert (r - p)^2 - 3 (s - q)^2 \vert \\
&\leq N(\beta) \cdot \max\{ (r - p)^2, 3(s - q)^2\} \\
& \leq\frac34 N(\beta)\\
&< N(\beta)
\end{align*}
The key here was that $\vert (r - p)^2 - 3 (s - q)^2 \vert \leq \max\{ (r - p)^2, 3(s - q)^2\}$ since $(r - p)^2, 3(s - q)^2 \geq 0$ and then we use that $(r - p)^2 \leq 1/4$ and $3(s - q)^2 \leq 3/4$.
| {
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How is $\frac{\big(\frac{3}{2}\big)^{99}-1}{\big(\frac{3}{2}\big)^{100}-1}\approx\frac{1}{\big(\frac{3}{2}\big)}$ I read somewhere that $$\frac{\big(\frac{3}{2}\big)^{99}-1}{\big(\frac{3}{2}\big)^{100}-1}\approx\frac{1}{\big(\frac{3}{2}\big)}$$I don't know how to have it. Please let me know how this is approximated.
| Let $x>1$ and $q_n = \frac{x^n-1}{x^{n+1}-1}$.
We see that $q_n = \frac{1-\frac{1}{x^n}}{x-\frac{1}{x^n}}$, from which we have $\lim_n q_n = \frac{1}{x}$.
In this case we have $x= \frac{3}{2}$, hence $\lim_n \frac{(\frac{3}{2})^n-1}{(\frac{3}{2})^{n+1}-1} = \frac{1}{(\frac{3}{2})}$.
$n=99$ is sufficiently 'large' that the approximation works.
| {
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"timestamp": "2023-03-29T00:00:00",
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F ind the seven solutions to $x^{7} \equiv 1 \pmod{29}$
Use the fact that $2$ is a primitive root modulo 29 to find the seven solutions to $x^{7} \equiv 1 \pmod{29}$
As $2$ is primitive root modulo $29$ then $$2^{28} \equiv 1 \pmod{29} $$ $$2^{4*7} \equiv 1 \pmod{29} $$ $$16^{7} \equiv 1 \pmod{29} $$ where 16 is a solution to the equation. From this solution as I can get 6 missing solutions?
| We know that $2^{28} \equiv 1 \bmod 29$, i.e. $16^7 \equiv 1 \bmod 29$. However we can also conclude from this that:
$16^{14},16^{21},16^{28},16^{35},16^{42},16^{49} \equiv 1 \bmod 29$.
So $16^2, 16^3, 16^4, 16^5, 16^6, 16^7$ are the other $6$ solutions.
Alternatively: we know every $x\in\mathbb{Z}/29\mathbb{Z}$ can be written as $x=2^k$ for some $k$.
To satisfy $x^7 \equiv 1 \bmod 29$ we require $2^{7k} \equiv 1 \bmod 29$. But since $2$ is a primitive root mod $29$ this means that $7k \equiv 0 \bmod 28$, i.e. $k \equiv 0 \bmod 4$ so we may take $k=0,4,8,12,16,20,24$ (after this the powers of $2$ repeat).
| {
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Writing u as a linear combination of the vectors in S. Write vector
u = $$\left[\begin{array}{ccc|c}2 \\10 \\1\end{array}\right]$$
as a linear combination of the vectors in S. Use elementary row operations on an augmented matrix to find the necessary coefficients.
S = {
$v1$$\left[\begin{matrix}1\\2\\2\end{matrix}\right] , v2\left[\begin{matrix}4\\2\\1\end{matrix}\right],
v2\left[\begin{matrix}5\\4\\1\end{matrix}\right]
$ }. If it is not possible, explain why?
This is what i have so far:
S = {
$v1$$\left[\begin{matrix}1\\2\\2\end{matrix}\right] , v2\left[\begin{matrix}4\\2\\1\end{matrix}\right],
v3\left[\begin{matrix}5\\4\\1\end{matrix}\right].
v4\left[\begin{matrix}2\\10\\1\end{matrix}\right]
$ }
$c1$$\left[\begin{matrix}1\\2\\2\end{matrix}\right] , c2\left[\begin{matrix}4\\2\\1\end{matrix}\right],
c3\left[\begin{matrix}5\\4\\1\end{matrix}\right].
c4\left[\begin{matrix}2\\10\\1\end{matrix}\right]
=\left[\begin{matrix}0\\0\\0\end{matrix}\right]$
$c1$$\left[\begin{matrix}1\\2\\2\end{matrix}\right] , c2\left[\begin{matrix}4\\2\\1\end{matrix}\right],
c3\left[\begin{matrix}5\\4\\1\end{matrix}\right].
c4\left[\begin{matrix}2\\10\\1\end{matrix}\right]$
$
\begin{bmatrix}
1 & 4 & 5 & 2\\
2 & 2 & 4 & 10\\
2 & 1 & 1 & 1\\
\end{bmatrix}
$
Now i don't know how to do this. Help will greatly be appreciated.
Thanks
| You need to solve the system
$$ c_1\left[\begin{matrix}1\\2\\2\end{matrix}\right] + c2\left[\begin{matrix}4\\2\\1\end{matrix}\right]+
c3\left[\begin{matrix}5\\4\\1\end{matrix}\right]=\left[\begin{matrix}2\\10\\1\end{matrix}\right] .$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/521503",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Show that if a and b are positive integers, and $a^3 | b^2$, then a | b Show that if a and b are positive integers, and $a^3 | b^2$, then a | b.
If p is a prime divisor of a, and $p^r$ is the highest power of p dividing a. Then $p^{3r} | a^3$, and so $p^{3r} | b^2$. If $p^s$ is the highest power of p dividing b, then 3r ≤ 2s. And so we have, r ≤ (2/3)s < s.
But how come this implies that $p^r | b$, and eventually a | b.
| Let $a,b$ be elements of $Z^+$ (i.e., the positive integers) and suppose $a^3|b^2$.
Then, by definition $b^2=c*a^3$ for some $c$ such that $c$ is an element of $Z^+$.
Hence, $\sqrt{b^2} = b = \sqrt{c*a^3} = (\sqrt{a^2})*(\sqrt{c*a}) = a*\sqrt{c*a}$
This works for both roots of $b^2$ and $a^2$.
Consider case-by-case:
$\sqrt{b^2}$ = $b$ & $\sqrt{a^2}$ = $a$: $b = a*(\sqrt{c*a})$
$\sqrt{b^2}$ = -$b$ & $\sqrt{a^2}$ = $a$: -$b$ = $a*(\sqrt{c*a}) \Rightarrow b = -a*(\sqrt{c*a}) = a*(-\sqrt{c*a})$
$\sqrt{b^2}$ =$ b$ & $\sqrt{a^2} = -a: b = -a*(\sqrt{c*a}) = a*(-\sqrt{c*a})$
$\sqrt{b^2}$ = -$b$ & $\sqrt{a^2} = -a: -b = -a*(\sqrt{c*a}) \Rightarrow b = a*(\sqrt{c*a})$
Hence, $a|b$ by definition.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/522201",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
Find the value of : $\lim_{x\to\infty} \sqrt{x+\sqrt{x}}-\sqrt{x}$ I tried to multiply by the conjugate:
$\displaystyle\lim_{x\to\infty} \frac{\left(\sqrt{x+\sqrt{x}}-\sqrt{x}\right)\left(\sqrt{x+\sqrt{x}}+\sqrt{x}\right)}{\sqrt{x+\sqrt{x}}+\sqrt{x}}=\displaystyle\lim_{x\to\infty} \frac{x-x+\sqrt{x}}{\sqrt{x+\sqrt{x}}+\sqrt{x}}=\displaystyle\lim_{x\to\infty} \frac{\sqrt{x}}{\sqrt{x+\sqrt{x}}+\sqrt{x}}$
I don't even know if my rewriting has helped at all. How would I go about doing this?
| $\displaystyle \lim_{x\to\infty}\left(\sqrt{x+\sqrt x}-\sqrt x\right) = \lim_{x\to\infty}\sqrt{x}\left\{\left(1+\frac{1}{\sqrt{x}}\right)^{\frac{1}{2}}-1\right\}$
Now expand Using Binomialy
$\displaystyle = \lim_{x\to\infty}\sqrt{x}.\left\{\left(1+\frac{1}{2.\sqrt{x}}+\frac{1}{2}.\left(\frac{1}{2}-1\right).\frac{1}{(\sqrt{x})^2}+.........\right)-1\right\} = \lim_{x\to\infty}\sqrt{x}\left\{\frac{1}{2.\sqrt{x}}+\frac{1}{2}.\left(\frac{1}{2}-1\right).\frac{1}{(\sqrt{x})^2}+.........\right\}$
$\displaystyle =\frac{1}{2}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/524288",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 1
} |
Evaluate the limit $\lim_{x\to 2} \frac{x-2}{\sqrt{x^2+5}-3}$ I need to evaluate the following limit:
$\lim_{x\to 2} \frac{x-2}{\sqrt{x^2+5}-3}$
I have multiplied both sides by the conjugate $\sqrt{x^2+5}+3$ but am getting $x^2-4$ as the denominator. Is this the correct way to go about it?
| In
$\lim_{x\to 2} \frac{x-2}{\sqrt{x^2+5}-3}$
let $x = y+2$.
This becomes
$\begin{align}
\lim_{y\to 0} \frac{y}{\sqrt{(y+2)^2+5}-3}
&=\lim_{y\to 0} \frac{y}{\sqrt{y^2+4y+4+5}-3}\\
&=\lim_{y\to 0} \frac{y}{\sqrt{y^2+4y+9}-3}\\
&=\lim_{y\to 0} \frac{y}{\sqrt{y^2+4y+9}-3}\frac{\sqrt{y^2+4y+9}+3}{\sqrt{y^2+4y+9}+3}\\
&=\lim_{y\to 0} \frac{y(\sqrt{y^2+4y+9}+3)}{(\sqrt{y^2+4y+9}-3)(\sqrt{y^2+4y+9}+3)}\\
&=\lim_{y\to 0} \frac{y(\sqrt{y^2+4y+9}+3)}{(y^2+4y+9)-9}\\
&=\lim_{y\to 0} \frac{y(\sqrt{y^2+4y+9}+3)}{y^2+4y}\\
&=\lim_{y\to 0} \frac{\sqrt{y^2+4y+9}+3}{y+4}\\
&=\frac{\sqrt{9}+3}{4}\\
&=\frac{6}{4}\\
&=\frac{3}{2}\\
\end{align}
$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/524684",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
Prove that $\mathbb{Z}$ is isomorphic to $\mathbb{Z}_x$ Let $\mathbb{Z}_x$ denote the ring of integers with the operations $\odot, \oplus $ defined as $a \odot b$ = $ a+b -a b$ and $a \oplus b$ = $a+b -1$. Prove that $\mathbb{Z}$ is isomorphic to $\mathbb{Z}_x$.
Not sure how to start.
| So we want to find a bijection $f \colon \def\Z{\mathbb Z}\Z \to \Z$ such that
$$ f(a+b) = f(a) \oplus f(b) , \quad f(ab) = f(a) \odot f(b) $$
holds for all $a,b \in \Z$. Let's suppose we have such an $f$ and want to find $f(0)$, for all $a$ we must have
\begin{align*}
f(a) &= f(a+0)\\
&= f(a) \oplus f(0)\\
&= f(a) + f(0) - 1\\
\iff f(0) &= 1
\end{align*}
Using the same idea, we must have
\begin{align*}
f(a) &= f(a\cdot 1)\\
&= f(a) \odot f(1)\\
&= f(a) + f(1) - f(a)f(1)\\
\iff f(1) &= f(a)f(1)
\end{align*}
As $f$ is onto, we must have $f(1) = 0$. Now we can get an idea what to do for a general $n$:
\begin{align*}
f(n+1) &= f(n) \oplus f(1)\\
&= f(n) \oplus 0\\
&= f(n) - 1
\end{align*}
So if there is such an $f$, we must have: $f(1) = 0$, $f(n+1) = f(n)- 1$ for any $n$, so we must have $f(n) = 1-n$ for every $n$. Now check if this $f$ is the desired isomorphism.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/526300",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Find the area of the largest rectangle
A rectangle is formed by bending a length of wire of length $L$ around four pegs. Calculate the area of the largest rectangle which can be formed this way (as a function of $L$).
How am I supposed to do this? If I'm interpreting the question correctly, a square would have an area of $\dfrac{1}{16}L^2$. But I don't know how to find the maximum area. I'm guessing it involves finding the stationary point of some function of $L$, but which function that might be eludes me at the moment.
| You will get a rectangle of sides $a$ and $b$, whose area is $A=a\cdot b$, and perimeter $L=2a+2b$.
One approach is calculus: Let $x=a$, then $b=\frac L2-x$ and area is $A=x(\frac L2-x)=\frac L2x-x^2$, have the derivative equal $0$, and voila.
Second approach: $(a-b)^2\ge 0$ (equality only holds when $a-b=0$, so:
\begin{align}
a^2-2ab+b^2&\ge 0\\
4ab+a^2-2ab+b^2&\ge 4ab \\
a^2+2ab+b^2&\ge 4ab \\
(a+b)^2&\ge 4ab
\end{align}
Now, remeber $A=ab$ and $L=2a+2b$, hence $a+b=\frac L2$, and substituting:
\begin{align}
\left(\frac L2\right)^2 &\ge 4A \\
\frac{L^2}4&\ge4A \\
A&\le\frac{L^2}{16}
\end{align}
The equality only holds when $a-b=0$ which is when $a=b=\frac L4$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/528798",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
If $n$ divides $2^{2^{n} +1}+1$ $\to$ $n$ divides $2^{n}+1$? Find a counterexample to show that the following implication is not valid.
if $n$ divides $2^{2^{n} +1}+1$ $\to$ $n$ divides $2^{n}+1$
And show how to use it.
This question appeared on the topic Does $n \mid 2^{2^n+1}+1$ imply $n \mid 2^{2^{2^n+1}+1}+1$?
UPDATE: Sorry, I made a mistake when typing the implication. I typed if $n$ divides $2^{2^{n} +1}+1$ $\to$ $n$ divides $2^{n}+n$ but the correct is if $n$ divides $2^{2^{n} +1}+1$ $\to$ $n$ divides $2^{n}+1$
| The post you linked bounced around $n = 57$, and it happens to be a counterexample.
First, I'll show that $57\ |\ 2^{2^{57} + 1} + 1$. To do this, I'll show that it is divisble by 19 and 3.
$\phi(3) = 2$, so $2^{2^{57} + 1} \equiv 2^1 \bmod{3}$, and thus, we know that $2^{2^{57} + 1} + 1 \equiv 2^1 + 1 \equiv 0\bmod{3}$.
$\phi(19) = 18$ and $\phi(18) = 6$, so $2^{57} + 1 \equiv 2^3 + 1 = 9\bmod{18}$, and $2^{2^{57} + 1} + 1 \equiv 2^9 + 1 = 513 \bmod{19}$. $513 = 19 \times 27$, so $2^{2^{57} + 1} + 1$ is also divisible by 19.
From these, we can conclude that $57\ |\ 2^{2^{57} + 1} + 1$.
Now, $\phi(57) = 36$, so $2^{57} + 1 \equiv 2^{21} + 1 = 2097153 \bmod{57}$. Doing out the division, you can see that this actually leaves a remainder of 9 (For the lazy, $2097144 = 57 \times 36792$).
Hence, $57\ |\ 2^{2^{57} + 1} + 1$, but $57\not|\ 2^{57} + 1$.
(See Euler's theorem for how I reduced all those exponents.)
As for "how to use it", what specific use case did you want?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/528908",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Prove that in a sequence of numbers $49 , 4489 , 444889 , 44448889\ldots$ Prove that in a sequence of numbers $49 , 4489 , 444889 , 44448889\ldots$ in which every number is made by inserting $48$ in the middle of previous as indicated, each number is the square of an integer.
| $44...488...89$ has $n+1$ numbers "$4$", $n$ numbers "$8$", and the "$9$". So:
$$
44...488...89 = 4\cdot\frac{10^n-1}{9}\cdot 10^{n+1} + 8\cdot\frac{10^n-1}{9}\cdot 10 + 9
$$
Now, we say $10^n = y$ so
$$
\begin{align}
&\frac{4\cdot (10y-1)\cdot 10y + 8\cdot 10(y-1) + 81}{9}\\
&=\frac{400y^2 + 40y +1}{9}\\
&=\left(\frac{20y + 1}{3}\right)^2
\end{align}
$$
Note that as $y = (10^n)$, $3 | (20y +1)$, for any $n$ value.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/529597",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 4,
"answer_id": 0
} |
How to approximate this series? How to approximate this series, non-numerically?
$ S_n = \sum_{n=1}^{50} \sqrt{n}$
| Consider that:
$$2\left((n+1)\sqrt{n+1}-n\sqrt{n}\right)=3\sqrt{n+1}+\frac{n-1-\sqrt{n(n+1)}}{\sqrt{n}+\sqrt{n+1}}.$$
The last term is negative, but greater than $$-\frac{3}{2(\sqrt{n}+\sqrt{n+1})}$$
by the AM-GM inequality. This gives:
$$\sum_{n=1}^{50}\sqrt{n}\geq\frac{2}{3}\sum_{n=1}^{50}\left(n\sqrt{n}-(n-1)\sqrt{n-1}\right)=\frac{2}{3}50\sqrt{50},$$
$$-\frac{2}{3}50\sqrt{50}+\sum_{n=1}^{50}\sqrt{n}\leq\frac{1}{2}\sum_{n=1}^{50}\frac{1}{\sqrt{n-1}+\sqrt{n}}=\frac{1}{2}\sum_{n=1}^{50}\left(\sqrt{n}-\sqrt{n-1}\right)=\frac{1}{2}\sqrt{50}.$$
The same argument proves that the sum
$$\sum_{n=1}^{N}\sqrt{n}$$
is always between $$\frac{2}{3}N\sqrt{N}$$ and $$\frac{4N+3}{6}\sqrt{N}.$$
As pointed out in the previous comments, we can produce a tighter bound by considering that:
$$0\leq 2\left((n+1/2)\sqrt{n+1/2}-(n-1/2)\sqrt{n-1/2}\right)-3\sqrt{n}\leq \frac{1}{16}\left(\frac{1}{\sqrt{n+1/2}}-\frac{1}{\sqrt{n-1/2}}\right).$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/530432",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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How find this maximum $S_{\Delta ABC}$ in $\Delta ABC$,and $\angle ABC=60$,such that $PA=10,PB=6,PC=7$,
find the maximum $S_{\Delta ABC}$.
My try:let $AB=c,BC=a,AC=b$, then
$$b^2=a^2+c^2-2ac\cos{\angle ABC}=a^2+c^2-2ac$$
then
$$S_{ABC}=\dfrac{1}{2}ac\sin{60}=\dfrac{\sqrt{3}}{4}ac$$
Then I can't
| *
*$B(0,0)$
*$P=6\,e^{i\alpha}=6\,(\cos\alpha+i\sin\alpha)\iff P_x=6\cos\alpha$ , and $P_y=6\sin\alpha$ , with $\alpha\in(0,{\pi\over3})$
*$A=A_x+iA_y$ , with $(A_x-P_x)^2+(A_y-P_y)^2=PA^2=10^2=100$ ; $\frac{A_y}{A_x}=\tan\frac\pi3=\sqrt3$
*$C(C_x,0)$ , with $(C_x-P_x)^2+P_y^2=PC^2=7^2=49$
*Heron's Formula : $S_{ABC}=\sqrt{s(s-a)(s-b)(s-c)}\ $ , with $\ s = \frac{a\ +\ b\ +\ c}2$ , where :
*$a=BC=C_x$
*$b^2=AC^2=(A_x-C_x)^2+A_y^2$
*$c^2=AB^2=A_x^2+A_y^2$
*Good Luck ! :-)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/530887",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Largest box fitting inside an ellipsoid Find the volume of the largest box with sides parallel to the $xy$, $xz$, and $yz$ planes that can fit inside the ellipsoid $(x/a)^2 + (y/b)^2 + (z/c)^2 = 1$.
My answer: We want to maximize $f(x,y,z) = xyz$ subject to $(x/a)^2 + (y/b)^2 + (z/c)^2 \leq 1$. So $\nabla f $ parallel to $\nabla g$, i.e.
$$(yz, xz, xy) = \lambda \left( \frac{x}{a}, \frac{y}{b}, \frac{z}{c} \right)\\ \Rightarrow \frac{ayz}{x}=\frac{bxz}{y}= \frac{cxy}{z}\\ \Rightarrow ay^2 = bx^2 ; bz^2 = cy^2 ; az^2 = cx^2 \\ \Rightarrow (x,y,z) \text{ parallel to } \left( 1, \sqrt{\frac{a}{b}}, \sqrt{\frac{a}{c}}\right)$$
so we need to find $t$ such that $\left( t, t\sqrt{\frac{a}{b}}, t\sqrt{\frac{a}{c}}\right)$ is on the ellipsoid, i.e.
$$\left( \frac{t}{a\sqrt{a}}\right)^2 + \left( \frac{t}{b\sqrt{b}}\right)^2 + \left( \frac{t}{c\sqrt{c}}\right)^2 = 1.$$
So the volume is $$8t^3 \frac{1}{\sqrt{abc}}\\ = 8\frac{1}{\left(\sqrt{ \frac{1}{a^3} + \frac{1}{b^3} + \frac{1}{c^3} }\right)^3}\cdot \frac{1}{\sqrt{abc}}$$
But the answer which the T.A. has indicated is $\frac{8abc}{3\sqrt{3}}$ (of course there could be a mistake). Where is my error?
| It is not difficult to verify that the largest box in the unit sphere is the cube with sides $2/\sqrt3$, whence for such a sphere you get the volume $8/3\sqrt3$. Now notice that $f:\mathbb R^3\to\mathbb R^3$, $(x,y,z)\mapsto(ax,by,cz)$ maps unit sphere to your ellipsoid, and $V(f(X))=abc V(X)$ where $V$ is the volume and $X$ is any set. Moreover, $f$ brings your box to a box again. Therefore the maximum is necessarily attained for $f(X)$ where $X$ is the cube. And you get $V(f(X))=8abc/3\sqrt3$.
Your mistake is that $\partial_x (x/a)^2 = 2x/a^2$ and not $2x/a$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/531565",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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How to evaluate the series: $ \frac {20} {(5-4)(5^2-4^2)} + \frac {20^2} {(5^2-4^2)(5^3-4^3)} + \frac {20^3} {(5^3-4^3)(5^4-4^4)} + \ldots $ How to evaluate this infinte summation?
$$ \dfrac {20} {(5-4)(5^2-4^2)} + \dfrac {20^2} {(5^2-4^2)(5^3-4^3)} + \dfrac {20^3} {(5^3-4^3)(5^4-4^4)} + \ldots $$
Telescopic cancellation is the way to go, I feel, but I don't know how to get this into telescopic form.
Hints?
| We have:
$$
\frac{20}{(5−4)(5^2−4^2)}=\frac{4}{5-4}-\frac{4^2}{5^2-4^2}
$$
$$
\frac{20^2}{(5^2−4^2)(5^3−4^3)}=\frac{4^2}{5^2-4^2}-\frac{4^3}{5^3-4^3}
$$
$$
\frac{20^3}{(5^3−4^3)(5^4−4^4)}=\frac{4^3}{5^3-4^3}-\frac{4^4}{5^4-4^4}
$$
Means for gathering data through equations have
$$
\frac{20}{(5−4)(5^2−4^2)}+\frac{20^2}{(5^2−4^2)(5^3−4^3)}+\frac{20^3}{(5^3−4^3)(5^4−4^4)}+\cdots
$$
$$
=\frac{4}{5-4}-\frac{4^2}{5^2-4^2}+\frac{4^2}{5^2-4^2}-\frac{4^3}{5^3-4^3}+\frac{4^3}{5^3-4^3}-\frac{4^4}{5^4-4^4}+\cdots=\frac{4}{5-4}=4
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/532238",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Existence of positive integer k that are both squares Is there a positive integer k such that $4k+1$ and $9k+1$ are both squares?
| Let's find the $k$ for which $4k+1 = a^2$ is a square. $4k+1$ is odd, so $a$ is odd. Then we have
$$k = \frac{a^2-1}{4} = \frac{a-1}{2}\cdot\frac{a+1}{2} = \frac{a-1}{2}\left(\frac{a-1}{2}+1\right)$$
the product of two consecutive numbers. For convenience, let's write $k =m\cdot (m+1)$. So can $9k+1$ be a square then?
$$9k+1 = 9m(m+1)+1 = 9m^2 + 9m + 1 = (3m+1)^2 + 3m = (3m+2)^2 - 3(m+1)$$
lies strictly between two consecutive squares then, unless $m = 0$ or $m+1 = 0$, which means $k = 0$, but that is excluded since $k$ was supposed to be positive.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/534634",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How can we calculate $\int \:\frac{1}{\sqrt{x^2-1}}dx$
Solve the following integral: $$\int \:\frac{1}{\sqrt{x^2-1}}dx$$
I attempted to solve it intergradation by parts by doing a
$$\int \:1\:\frac{1}{\sqrt{x^2-1}} \, dx$$
and set $u$ be $\frac{1}{\sqrt{x^2-1}}$ and $dv/dx$ be $1$:
but as I start doing, it gets more complicated. What is the right direction to solve this?
| Here's an alternative way doing it. Rewrite the integral as:
$\int \frac{1}{\sqrt{x^2-1}} dx = \int \frac{1}{\sqrt{x^2-1}} \cdot \frac{x + \sqrt{x^2-1}}{x + \sqrt{x^2-1}} dx = \int \frac{1 + \frac{x}{\sqrt{x^2-1}} dx}{x + \sqrt{x^2-1}}$.
Now use a simple substitution $u=x+\sqrt{x^2-1}$, so that the integral becomes:
$\int \frac{du}{u} = \log u + C = \log{x+\sqrt{x^2-1}}+C.$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Solving two simultaneous equations Suppose that $x$, $y$ and $z$ are three integers (positive,negative or zero) such that we get the following relationships simultaneously
*
*$x + y = 1 - z$ and
*$x^3 + y^3= 1 - z^2$
Find all such $x$, $y$ and $z$.
I have tried to determine the values of $x$ and $y$ by fixing some values of $z$ but it is obvious that this method will not work. I also tried to determine the bounds of $x$ and $y$ and the related terms like $xy$ and ($x$$2$ +$y$$2$) but this this method failed as well. What then is the way out?
| To find all integer solutions, we begin by substituting $z=1-x-y$, into $z^2= 1-x^3-y^3$ to get
$$(1-x-y)^2=1-x^3-y^3.$$
Since $y=-x$ is already a clear solution, we will force the factor $(y+x)$ in what follows.
\begin{align*}
& (1-x-y)^2=1-x^3-y^3 \\
\Rightarrow & 1-2x+x^2-2y+2xy+y^2=1-x^3-y^3 \\
\Rightarrow & y^3+y^2+2xy-2y+x^3+x^2-2x=0 \\
\Rightarrow & y^3+y^2+y(2x-2)+(x^3+x^2-2x)=0.
\end{align*}
Divide the left by $(y+x)$ to see that
\begin{align*}
&\frac{y^3+y^2+y(2x-2)+(x^3+x^2-2x)}{y+x} \\
=&y^2+y(1-x)+(x^2+x-2) \\
=&y^2-xy+y+x^2+x-2,
\end{align*}
with a zero remainder. Hence
\begin{align*}
& y^3+y^2+y(2x-2)+(x^3+x^2-2x)=0 \\
\Rightarrow & (y+x)\left( y^2-xy+y+x^2+x-2 \right)=0 \\
\Rightarrow & (y=-x) \, \operatorname{or}\, \left(y^2-xy+y+x^2+x-2 =0 \right).
\end{align*}
Treating the latter case as a quadratic in $y$, and applying the quadratic formula, we get
\begin{align*}
& y^2-xy+y+x^2+x-2 =0 \\
\Rightarrow& y^2 +(1-x)y+\left( x^2+x-2 \right)=0 \\
\Rightarrow&y=\frac{-(1-x) \pm \sqrt{(1-x)^2-4(x^2+x-2)}}{2} \\
&\,\,\,\,=\frac{(x-1)\pm\sqrt{3(1-x)(x+3)}}{2}.
\end{align*}
Now
$$\sqrt{3(1-x)(x+3)}\geq 0 \Rightarrow -3 \leq x \leq 1.$$
Of these 5 integer test values for $x$, $x \in \{ -3,-2,0,1 \}$ produce integers $y$, and it produces them for both the $\pm$ on the solution $y$. This leaves the set of all integer solutions,
$$(x,-x,1),$$
from the $+$ branch on the solution $y$,
$$\{(-3, -2, 6), (-2, 0, 3), (0, 1, 0), (1, 0, 0)\},$$
and from the $-$ branch on the solution $y$, two more solutions arise not in that prior set,
$$\{(-2, -3, 6), (0, -2, 3)\}.$$
This constitutes all integer solutions.
| {
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How to prove $\cos ^6x+3\cos ^2x\space \sin ^2x+\sin ^6x=1$ Prove the following equation.
\begin{eqnarray}
\\\cos ^6x+3\cos ^2x\space \sin ^2x+\sin ^6x=1\\
\end{eqnarray}
I can't prove it by many methods I use.
Please give me some hints.
Thank you for your attention
| An alternate approach
From algebra we know
if $a+b+c=0$ then $ a^3+b^3+c^3 = 3.a.b.c$
$$Derivative \begin{cases}a+b+c = 0 & \text{(1)}\\ \Rightarrow a+b=-c \\ \Rightarrow (a+b)^3=(-c)^3 & \text{cubing both sides} \\ \Rightarrow a^3+b^3+3a^2b+3ab^2=(-c)^3 \\ \Rightarrow a^3+b^3+3ab(a+b)=(-c)^3 & \text{but }a+b = -c\text{ from (1)} \\ \Rightarrow a^3+b^3-3abc=-c^3 \\ \Rightarrow a^3+b^3+c^3=3abc & \text{ subtracting } c^3 \text{from both sides} \end{cases}$$
Now if $a=cos^2x$, $b=sin^2x$ and $c = -1$, we have
$$a+b+c=cos^2x + sin^2x - 1 = 0$$
thus we have
$$\Rightarrow (cos^2x)^3 + (sin^2x)^3 +(-1)^3 - 3 \cos ^2x .\sin ^2x . (-1) = 0 $$
$$\Rightarrow\cos ^6x+3\cos ^2x\space \sin ^2x+\sin ^6x=1$$
Hence Provded
| {
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$x^2+xy+y^2$ and $x^2-xy+y^2$ are not both perfect squares
Prove that $x^2+xy+y^2$ and $x^2-xy+y^2$ cannot be both perfect squares.
Surely $x$ and $y$ are natural numbers. If $x^2+xy+y^2 =a^2$ and $x^2-xy+y^2=b^2$ simultaneously then we have to show that there are no such integers $a$ and $b$.
I have tried that:
Suppose $x^2+xy+y^2=a^2$ and $x^2-xy+y^2=b^2$. Then $2xy=a^2-b^2=(a+b)(a-b)$, so one of $x$ and $y$ is even, but then I am stuck.
| Adding these equations gives $2(x^2+y^2) = a^2 + b^2$; unique factoriation of Gaussian integers shows that we essentially must have $a = x+y$ and $b = x-y$. Now combine this with what you already have.
| {
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What is the minimum value of $a+b+\frac{1}{ab}$ if $a^2 + b^2 = 1$? For the case when $a,b>0,$ I used AM-GM Inequality as follows that:
$\frac{(a+b+\frac{1}{ab})}{3} \geq (ab\frac{1}{ab})^\frac{1}{3}$
This implies that $(a+b+\frac{1}{ab})\geq 3$. Hence, the minimum value of $(a+b+\frac{1}{ab})$ is 3
But the answer is $2+\sqrt{2}$ ... how is it ?
| If $a,b>0$, one can make the substitution $u=ab,$ where $0<u\leq \frac 1 2$ (by AM-GM). Note that $$1=a^2+b^2=(a+b)^2-2ab,$$ hence $$a+b=\sqrt{1+2u}.$$ It follows that one needs to minimize $$f(u):=a+b+\frac 1{ab}=\sqrt{1+2u}+\frac 1 u, \qquad 0<u\leq \frac 1 2.$$
By Calculus, it is easy to see that $f$ has a global mimimum at $u_0\in (1,2)$ with $u_0^4-2u_0-1=0$, but $f$ is decreasing on $(0,\frac 1 2].$ It follows that $f$ achieves absolute minimum over $(0,\frac 1 2]$ at $u=\frac 1 2=ab,$ i.e. when $a=b=\frac{\sqrt{2}}2$ (the case of equality by AM-GM: $1=a^2+b^2\geq 2ab).$ Hence $$f(1/2)=2+\sqrt{2}$$ is the minimum of $a+b+\frac 1 {ab}.$ QED
| {
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Proving the second root of a quadratic equation If $\alpha$ is a root of the equation $4x^2+2x-1=0$, then prove that $4\alpha^3-3\alpha$ is the other root. How do I proceed? The sum of the roots, the product of the roots lead me nowhere. Should I find the roots of the equation and substitute in the given two expressions of $\alpha$ and check whether manipulating the first root gives me the second root (which seems much complicated), or is there any other way?
| Clearly, the absolute value of both roots are $<1$
$$0=(4x^2+2x-1)(x-1)=4x^3-3x-(2x^2-1)$$
If $x=\cos y,$ we have $$\cos3y=\cos2y$$
$\implies3y=360^\circ n\pm2y$ where $n$ is any integer.
'+'$\implies y=360^\circ n\implies x=1$
'-'$\implies y=72^\circ n$ where $n\equiv0,\pm1,\pm2\pmod5$
As $x=1$ does not satisfy $4x^2+2x-1=0,n\not\equiv0\pmod5$
So, the roots of $4x^2+2x-1=0$ are $\cos72^\circ,\cos144^\circ=-\cos36^\circ$
Now $4\cos^372^\circ-3\cos72^\circ=\cos3\cdot72^\circ=\cos216^\circ=\cdots=-\cos36^\circ$
?
Similarly, $\cos3\cdot(144^\circ)=\cdots=\cos72^\circ$
Can you take it from here
| {
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Volume of a sphere with three holes drilled in it. Suppose that the sphere $x^2+y^2+z^2=9$ has three holes of radius $1$ drilled through it. One down the $z$-axis, one along the $x$-axis, and one along the $y$-axis. What is the volume of the resulting solid? I can do it for two holes but I'm stuck on three.
| The 3 holes such drilled form an intersection of 3 cylinders in the center, plus 6 cylinder/cap pairs. I will treat each problem in turn.
Here is a picture of the situation:
Intersection of 3 cylinders
The problem is to find the volume of three orthogonal, intersecting cylinders:
$$\begin{align}x^2+y^2&=1\\x^2+z^2&=1\\ y^2+z^2&=1\end{align}$$
The intersection region is pictured below:
We note that there are two ways to bound the volume over $x$:
$$\begin{align}|x| &\le \sqrt{1-y^2}\\ |x| &\le \sqrt{1-z^2}\end{align}$$
Since we are computing the volume of the interior of the region defined by these bounds, it stands to reason that $|x|$ must be bounded by the smaller of these two bounds:
$$|x| \le \min{\left(\sqrt{1-y^2},\sqrt{1-z^2}\right)}$$
so that the volume integral takes the form
$$\int_{-1}^1 dz \: \int_{-\sqrt{1-z^2}}^{\sqrt{1-z^2}} dy \: \int_{-m(y,z)}^{m(y,z)} dx = 2 \int_{-1}^1 dz \: \int_{-\sqrt{1-z^2}}^{\sqrt{1-z^2}} dy \: m(y,z)$$
Below is a representation of the integration region for this integral:
The reason for the lines is because $\sqrt{1-y^2} \lt \sqrt{1-z^2}$ according to whether $|y| \gt |z|$. The integral is then symmetric over the regions bounded by the sloped lines; thus, we need only consider one such region and the others will yield the same result. Let's then consider the region surrounding the positive $y$ (horizontal) axis in the above figure. In this case, $m(y,z) = \sqrt{1-y^2}$; when we use polar coordinates, the integral becomes
$$\begin{align}8 \int_{-\pi/4}^{\pi/4} d\phi \: \int_0^1 d\rho \, \rho \sqrt{1-\rho^2 \cos^2{\phi}} &= 4 \frac{2}{3} \int_{-\pi/4}^{\pi/4} d\phi \: \left( 1- \left|\sin^3{\phi}\right|\right) \sec^2{\phi}\\ &= \frac{16}{3} - \frac{16}{3} \int_0^{\pi/4} d\phi \: \sin^3{\phi} \, \sec^2{\phi}\\ &= 8 \left (2 - \sqrt{2}\right ) \end{align}$$
Cylinder and cap
The cross-sectional geometry here corresponds to a rectangle of width $2$ inscribed symmetrically about a diameter of a circle of radius $3$, with the cap corresponding to the resulting circular segment. The distance from center of circle to short edge of rectangle is $\sqrt{3^2-1^2}=2 \sqrt{2}$, so the height of the cylinder outside the intersection above is $2 \sqrt{2}-1$. The volume of a cylinder is thus $\pi \cdot 1^2 \cdot (2 \sqrt{2}-1) = (2 \sqrt{2}-1) \pi$.
The volume of the cap is the difference between the volume of the spherical sector subtended by the solid angle defined by the hole and the corresponding cone. The volume of the spherical sector is given by $\frac13 (3)^3 \Omega$, where $\Omega$ is the solid angle. We find $\Omega$ by integrating over angle in spherical coordinates:
$$\Omega = \int_0^{\theta_0} d\theta \, \sin{\theta} \, \int_0^{2 \pi} d\phi = 2 \pi (1-\cos{\theta_0})$$
where $\sin{\theta_0} = \frac13 \implies \cos{\theta_0} = 2 \sqrt{2}/3$. Thus the volume of a sector is $6\pi (3-2 \sqrt{2})$.
The volume of a cone is $\frac13 \pi (1^2) 2 \sqrt{2} = 2 \sqrt{2} \pi/3$. Thus, the volume of a cap is
$$6\pi (3-2 \sqrt{2}) - \frac{2 \sqrt{2} \pi}{3} = \left (18 - \frac{38 \sqrt{2}}{3}\right )\pi$$
Putting it all together
The volume of the holes is $6$ times the sum of the volumes of the cylinder and cap, plus the volume of the intersection:
$$V_{\text{holes}} = 6 \pi (2 \sqrt{2}-1) + \left (108- 76 \sqrt{2}\right )\pi + 8 (2-\sqrt{2})= (102-64\sqrt{2})\pi + 8 (2-\sqrt{2})$$
The volume left over after drilling is therefore the volume of the sphere minus the volume of the holes, or
$$V = \frac{4 \pi}{3} (3)^3 - V_{\text{holes}} = (64 \sqrt{2}-66) \pi - 8 (2-\sqrt{2}) \approx 72.31$$
compared with the original volume of the sphere $36 \pi \approx 113.1$.
| {
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Use the ϵ-N definition of limit to prove that lim[(2n+1)/(5n-2)] = 2/5 as n goes to infinity Use the ϵ-N definition of limit to prove that lim[(2n+1)/(5n-2)] = 2/5 as n goes to infinity.
The way I do it is
Let ∊ > 0 be given. Notice N ∈ natural number (N) which satisfies {fill this box later}< N.
It follows that if n>=N, then n > {fill this box later}, so for such n, |(2n+1)/(5n-2)-2/5| = |9/(25n-10)| = 9/5|1/(5n-2)|
I am supposed to get to a something that is less than ∊
How to make this to less than ∊?
| Try setting your last expression to $\epsilon$ and work backwards. You'll find that the following will work:
Given any $\epsilon > 0$, let $N$ be any natural number that is greater than or equal to the real number:
$$\dfrac{\dfrac{9}{5\epsilon}+2}{5} + 1$$
(by the Archimedean Property, such an $N$ must exist). Now suppose that $n \geq N$ so that:
\begin{align*}
n \geq \dfrac{\dfrac{9}{5\epsilon}+2}{5} + 1
&\implies n > \dfrac{\dfrac{9}{5\epsilon}+2}{5} \\
&\implies 5n - 2 > \dfrac{9}{5\epsilon} \\
&\implies \epsilon > \frac{9}{5} \cdot \frac{1}{5n-2} \\
\end{align*}
Then observe that:
\begin{align*}
\left| \frac{2n + 1}{5n - 2} - \frac{2}{5} \right|
&= \frac{9}{5}\left| \frac{1}{5n - 2} \right| \\
&= \frac{9}{5} \cdot \frac{1}{5n - 2} \\
&< \epsilon \\
\end{align*}
as desired.
| {
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Volume of a cuboid whose diagonal and surface area is known The sum of length, breadth and depth of cuboid is $19$cm and its diagonal is $5\sqrt{5}$cm.
Its volume is:
a) 125
b) 236
c) 361
d) 486
Solution:
$$\ell^2 + b^2 +h^2 = 125\quad\text{ and }\quad \ell+b+h=19,$$
How can i find volume from this information?
| $l+b+h=19$
$l^2+b^2+h^2=125$
From these 2 equations, we get
$2(lb+bh+hl) = (l+b+h)^2-(l^2+b^2+h^2)
= 19^2-125 = 236 = S$
This is coincidentally the Surface area of the cuboid.
The Volume $lbh = \frac{S}{2(l^{-1}+b^{-1}+h^{-1})}$.
Assuming $l=b$, we can find $l$ and $h$ to be $\frac{19-\sqrt{7}}{3}$ and $\frac{19+2\sqrt{7}}{3}$ and the Volume is then $240.63$.
Max Volume is when $l=b=h$ and we get $254.037$, so answers $c$ and $d$ are our. All we can say is that answer $b$ ($236$) is possibly right as Volume. I don't think you can actually find the Volume.
| {
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Show that $\{ 1, 1-x , 1-2x + {1 \over 2} x^2\}$ a orthogonal system I have to show that the following set:
$$A = \left\{ 1, 1-x , 1-2x + {1 \over 2} x^2\right\}$$
is orthogonal system in relative to the inner product
$$\langle f, g\rangle = \int ^\infty_0 f(x)g(x)e^{-x}dx$$.
as far as I know, in order that $A$ will be a orthogonal system, for any $f,g \in A$,
$$\langle f, g\rangle = \begin{cases}0 & f \ne g \\ 1 & f = g \end{cases}$$
Lets take $f = 1, g = (1 - 2x + \frac{1}{2}x^2)$,
\begin{align}
\langle f, g\rangle = &\int^\infty_0 1\cdot(1 - 2x + \frac{x^2}{2})dx\\
= & \int^\infty_01-2x + \frac{x^2}{2} dx \\
= & \int^\infty_01 - 2x + \frac{1}{2}x^2 \\
= & \int^\infty_01 - \int^\infty_02x + \int^\infty_0 \frac {1}{2} x^2\\
= & x - \frac{x^2}{2} + \frac{x^3}{6}\bigg|^\infty_0 \\
= & 0 - (\infty - \frac{\infty^2}{2} + \frac{\infty^3}{6}) = \infty.
\end{align}
So how that set is orthogonal? thanks in advance.
| You've forgotten the $e^{-x}$ term in $\langle f,g\rangle$.
$$\langle f,g\rangle =\int_0^\infty 1\cdot(1-2x+\frac{x^2}{2})\cdot e^{-x} \mathrm{d}x=0$$
etc.
| {
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Prove by induction that $3\mid n^3 - n$
Prove by induction that $3\mid n^3 - n$.
I'm having an argument with my professor whether my exam was right or not. Before I sign a formal complain to get a review on my exam, I'd like to be sure it's correct. My answer:
Proof by induction:
Proposition: $\forall n \in\mathbb{ N}$ (belongs to natural numbers), $3 | (n^3 - n)$;
Basic Step: $P(0) = 0^3 - 0$ is divisible by 3.
Inductive Step:
$p(n+1) = (n+1)^3 - (n+1)$
$= n^3 + 3n^2 + 3n + 1 - n - 1$
$= n^3 + 3n^2 + 2n$
$= n^3 + 3n^2 + 2n - (n^3 - n)$
$= 3n^2 + 3n$
By induction, $p(n+1)$ is true if:
$3\mid n$ and $3\mid m$ and $3 | n - m$
to prove that, assume that $n$ is $3p$ and $m$ is $3q$, therefore $3 | 3p - 3q$.
My professor wanted:
$= (n^3 -n) + 3n^2 + (2n + n)$
| Note: the following "equality" in your argument is not valid (third to fourth line):
$${\bf n^3 + 3n^2 + 2n} = {\bf n^3 + 3n^2 + 2n} \color{red}{\bf - (n^3 - n)}$$
First, your missing the inductive hypothesis:
Assume that for $n = k$, we have $p(k) = k^3 - k$ is divisible by $3$.
Then, we show based on that assumption, that $p(k+1)$ is true.
Here is what your professor is arguing:
$$\begin{align} p(k+1) & = (k+1)^3 - (k+1) \\ \\
& = k^3 + 3k^2 + 3k + 1 - k - 1\\ \\
& = k^3 + 3k^2 + 2k \\ \\
& = k^3 - k + k + 3k^2 + 2k \\ \\
& = (\underbrace{k^3 - k}_{\large \text{IH}}) + 3(k^2 + k)\end{align}$$
By the inductive hypothesis (IH), we have that the first term is divisible by $3$, and we see that the second term is also divisible by three.
Hence $p(k + 1)$ is true.
Therefore, by induction, $n^3 - n$ is divisible by $3$.
| {
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Let $p=q+4a$. Prove that $\left( \frac{a}{p} \right) = \left( \frac{a}{q} \right)$. Here's a little number theory problem I'm wrestling with.
Let $p$ and $q$ be odd prime numbers with $p=q+4a$ for some $a \in \mathbb{Z}$. Prove that $$\left( \frac{a}{p} \right) = \left( \frac{a}{q} \right),$$
where $\left( \frac{a}{p} \right)$ is the Legendre symbol. I have been trying to use the law of quadratic reciprocity but to no avail. Can you help?
| Note that $p \equiv q \pmod{4}$, so $\frac{p-1}{2}\frac{q+1}{2} \equiv \frac{p-1}{2}\frac{p+1}{2} \equiv 0 \pmod{2}$.
\begin{align}
\left(\frac{a}{p}\right)=\left(\frac{4a}{p}\right)=\left(\frac{p-q}{p}\right)& =\left(\frac{-q}{p}\right) \\
& =\left(\frac{-1}{p}\right)\left(\frac{q}{p}\right) \\
&=(-1)^{\frac{p-1}{2}}\left(\frac{p}{q}\right)(-1)^{\frac{p-1}{2}\frac{q-1}{2}} \\
&=(-1)^{\frac{p-1}{2}\frac{q+1}{2}}\left(\frac{p-q}{q}\right) \\
&=\left(\frac{4a}{q}\right) \\
&=\left(\frac{a}{q}\right)
\end{align}
| {
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Derivative of following functions $f(x) = \cos^{-1}\left(1-2x^2\right)$ Derivative of following functions
(A) $f(x) = \cos^{-1}\left(1-2x^2\right)$
(B) $f(x) = \sin^{-1}(3x-4x^3)$
$\bf{My\; Try}::$ Given $f(x) = \cos^{-1}\left(1-2x^2\right)$ and function $f(x)$ is defined when $\left|1-2x^2\right|\leq 1\Rightarrow -1 \leq x\leq 1$
Now put $x=\sin \theta$, where $\displaystyle -\frac{\pi}{2} \leq \theta \leq \frac{\pi}{2}\Rightarrow -\pi\leq 2\theta \leq \pi$
we get $\displaystyle f(x) = \cos^{-1}(1-2\sin^2 \theta) = \cos^{-1}(\cos 2 \theta)$
$ = -2\theta \;\;\;, -\pi \leq 2\theta \leq 0$
$=+2\theta \;\;\;, 0 \leq 2\theta \leq \pi$
$ f(x) = -2\sin^{-1}(x) \;\;\;, -1 \leq x \leq 0$
$\;\;\; \;\;\;\; =+2\sin^{-1}(x) \;\;\;, 0 \leq x \leq 1$
So $\displaystyle f^{'}(x) = -\frac{2}{\sqrt{1-x^2}} \;\;\;, -1 \leq x \leq 0$
$\displaystyle \;\;\; \;\;\;\; =+\frac{2}{\sqrt{1-x^2}} \;\;\;, 0 \leq x \leq 1$
So function $f(x) = \cos^{-1}(1-2x^2)$ is not Differentiable at $x = 0$
Is my process is Right , If not Then How can we solve it
Help me
Thanks
| Your process seems to be fine.
Putting $x=\cos\theta,0\le \theta\le \pi$ as the principal value of inverse cosine function
I also reached at $$\cos^{-1}(1-2x^2)=\begin{cases} 2\cos^{-1}x-\pi &\mbox{if } \frac\pi2\le \theta\le \pi\implies0\ge x\ge -1 \\
\pi- 2\cos^{-1}x & \mbox{if } 0\le \theta\le \frac\pi2\implies1\ge x\ge0 \end{cases} $$
Observe that $$f'(x)=\frac{d\{\cos^{-1}(1-2x^2)\}}{dx}=\frac{d\{\cos^{-1}(1-2x^2)\}}{d(1-2x^2)}\cdot\frac{d(1-2x^2)}{dx}$$
$$=\frac{-1}{\sqrt{1-(1-2x^2)^2}}\cdot(-4x)=\frac{2x}{\sqrt{x^2-x^4}}$$
$$=\frac{2x}{|x|\sqrt{1-x^2}}\begin{cases} =\frac2{\sqrt{1-x^2}} &\mbox{if } |x|=x\iff x>0 \\ =-\frac2{\sqrt{1-x^2}} &\mbox{if } |x|=-x\iff x<0 &\\ \text{ is clearly undefined } &\mbox{if } x=0\end{cases} $$
| {
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"timestamp": "2023-03-29T00:00:00",
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How find this maximum $\frac{1}{a^2+1}+\frac{1}{b^2+1}+\frac{1}{1+c^2}$ let $a,b,c>0$ ,and such $a+b+c=3$.prove that
$$\dfrac{1}{a^2+1}+\dfrac{1}{b^2+1}+\dfrac{1}{1+c^2}<\dfrac{11}{5}$$
if this problem find this minimum,then
$$f(x)=\dfrac{1}{x^2+1}\ge ax+b$$
where $a=f'(1),a+b=f(1)$
since $$f'(x)=-\dfrac{2x}{(1+x^2)^2}\Longrightarrow f'(1)=-\dfrac{1}{2}$$
so
$$a=-\dfrac{1}{2},b=1$$
and
$$\dfrac{1}{1+x^2}+\dfrac{1}{2}x-1=\dfrac{x(x-1)^2}{(x^2+1)^2}\ge 0$$
then
$$\dfrac{1}{1+x^2}\ge-\dfrac{1}{2}x+1$$
so
$$\sum_{cyc}\dfrac{1}{1+a^2}\ge\sum_{cyc}-\dfrac{1}{2}a+1=-\dfrac{1}{2}(a+b+c)+3=\dfrac{3}{2}$$
But for maximum,I can't prove it.Thank you
| First show that $\dfrac{1}{1+x^2} < \dfrac{63 - 19x}{60}$ for $0 < x < 3$.
Then $\text{LHS} < \dfrac{3\cdot63 - 19(a + b + c)}{60} = \dfrac{189 - 57}{60} = \text{RHS}$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove that $\sum^n_{k=1} k^2 = \binom{n+1}{2} + 2\binom{n+1}{3}$ for $n\geq 2$ Prove, for all $n\geq 2$ that
$$\sum^n_{k=1} k^2 = \binom{n+1}{2} + 2\binom{n+1}{3}$$
Let us prove the inductive base for $n = 2$
$$\rm{LHS} = 1^2 + 2^2= 1 + 4 = 5$$
$$\rm{RHS} = \binom{3}{2} + 2\binom{3}{3} = 3 + 2\cdot 1 = 5$$
$$\rm{LHS} = \rm{RHS}$$
as desired.
Now, assume for some $k$,
$$1^2 + 2^2 + \dots + k^2 = \binom{k+1}{2} + 2\binom{k+1}{3}$$
To prove the inductive step, it is sufficient to prove that,
$$\binom{k+2}{2} + 2\binom{k+2}{3} - \binom{k+1}{2} - 2\binom{k+1}{3} = (k+1)^2$$
The LHS can be simplified as:
$$\frac{(k+2)!}{2!k!} + 2\left[\frac{(k+2)!}{3!(k-1)!}\right] - \frac{(k+1)!}{2!(k-1)!} - 2\left[\frac{(k+1)!}{3!(k-2)!}\right]$$
$$=\frac{3(k+2)! + 2k(k+2)! -3k(k-1)! - 2k(k-1)(k+1)!}{6k!}$$
$$=\frac{(k+1)!(3k + 6 - 2k^2 - 2k) + (k-1)!(2k^3 + 2k^2 - 3k)}{6k!}$$
$$=\frac{(k+1)!(-2k^2+k+6) + k!(2k^2+2k - 3)}{6k!}$$
$$=\frac{(k+1)(-2k^2 + k + 6) + 2k^2 +2k - 3}{6}$$
$$=\frac{-2k^3 + k^2 + 9k + 3}{6}$$
I'm pretty sure I'm making a mistake somewhere, but I can't figure it out. If someone could complete this inductive proof for me, I will be grateful.
Also, I don't feel satisfied with this ugly proof. Please add a combinatorial proof for this in your answer if possible.
| Perhaps this Wikipedia link might prove itself useful. The sums are computed through telescoping, each relying on the result of the previous one, which, in this case, is $\sum_1^n k=\frac{n(n+1)}2$, for instance like this.
| {
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Find $ S_{20} $ if $ \frac{ a_{n} +2}{2} = \sqrt{2S_{n}} $ for all integer n Define $ a_{n} $ is a sequence and all terms of $ a_{n} $ are positive. $ S_{n} $ is the summation of the first n terms. If $ \frac{ a_{n} +2}{2} = \sqrt{2S_{n}} $ for all integer n, then find $ S_{20} $
| $$\frac{a_1+2}{2}=\sqrt{2a_1} \Leftrightarrow (\sqrt{a_1}-\sqrt{2})^2=0$$
Thus
$$S_1=a_1=2$$
Next
$$\frac{ S_n-S_{n-1} +2}{2} = \sqrt{2S_n} \Leftrightarrow (\sqrt{S_n}-\sqrt{2})^2=S_{n-1}\Leftrightarrow \sqrt{S_n}=\sqrt{S_{n-1}}+\sqrt{2}$$
Let $T_n=\sqrt{S_n}$.
Then $T_1=\sqrt{2}$ and
$$T_n =T_{n-1}+\sqrt{2}$$
Can you find $T_{20}$?
| {
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"timestamp": "2023-03-29T00:00:00",
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Integer solutions of the equation $x^2+y^2+z^2 = 2xyz$
Calculate all integer solutions $(x,y,z)\in\mathbb{Z}^3$ of the equation $x^2+y^2+z^2 = 2xyz$.
My Attempt:
We will calculate for $x,y,z>0$. Then, using the AM-GM Inequality, we have
$$
\begin{cases}
x^2+y^2\geq 2xy\\
y^2+z^2\geq 2yz\\
z^2+x^2\geq 2zx\\
\end{cases}
$$
So $x^2+y^2+z^2\geq xy+yz+zx$. How can I solve for $(x,y,z)$ after this?
| Another method:
It's clear zero is a solution and moreover $xyz\ge0$. Let's prove there are no more solutions over the integers. By the generalized mean theorem:
$$\sqrt{\frac{x^2+y^2+z^2}{3}}>\sqrt[3]{xyz}$$
Which leads to:
$$x^2+y^2+z^2 > 3(xyz)^{2/3}$$
So now we know that whenever $3(xyz)^{2/3} > 2xyz$ there are no solutions. Simple algebra shows that this is true whenever $xyz< 27/8$, so the only possible integer solutions are:
$$(0,0,0), (0,a,b), (1, 1, 2), (1, 1, 3)$$
And their permutations / sign changes. Now verify these do not satisfy the equation.
$\blacksquare$
| {
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"timestamp": "2023-03-29T00:00:00",
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Proof: Divisible by 15 I have to proof that $16^m - 1$ is divisible by $15$. Is my following proof correct?
$$\begin{align}
16^m - 1=&\frac{16^{m+1}}{16}-1\\
=&\frac{16^{m+1}-16}{16} \\
=&(16^{m+1}-16)\cdot\frac{1}{16} \\
=&\underbrace{(16^{m+1}-16)}\cdot\frac{1}{15(1+1/15)} \\
=&\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,a\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\cdot\frac{1}{1+1/15}\cdot\frac{1}{15}\\
=&\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,a\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\cdot\frac{1}{16/15}\cdot\frac{1}{15}\\
=&\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,a\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\cdot\frac{1}{1}\cdot\frac{15}{16}\cdot\frac{1}{15}\\
=&\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\underbrace{a\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\cdot\frac{15}{16}}\cdot\frac{1}{15}\\
=&\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,b\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\cdot\frac{1}{15}\\
\end{align}$$
$$\therefore \boxed{16^m - 1=\frac{b}{15}}$$
Or is this the wrong way and I have to do it with mathematical induction?
| Just observe that $x^n-1$ factors as $(x-1)(x^{n-1} + x^{n-2} + \cdots + x + 1)$, and make the appropriate substitution.
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "1",
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How is $\dfrac1{(1-x)^5}=\sum_{n\geq0}{n+4\choose4}x^n$ Can someone please explain to me how is $$\dfrac1{(1-x)^5}=\sum_{n\geq0}{n+4\choose4}x^n$$
Thanks!
| Firstly$$\dfrac1{(1-x)^5}=(1-x)^{-5}$$
then from binomial theorem
$$(1-x)^{-5}=\sum_{n\geq0}(-1)^n\binom{-5}{n}x^n$$
because
$$\binom{-5}{n}=\frac{(-5)(-5-1)(-5-2)...(-5-(n-1))}{n(n-1)(n-2)..1}=$$
$$=(-1)^n\frac{(5)(5+1)(5+2)...(5+(n-1))}{n(n-1)(n-2)..1}=$$
$$=(-1)^n\frac{(n+4)(n+3)(n+2)...6\cdot5}{n(n-1)(n-2)..1}=$$
$$=(-1)^n\binom{n+4}{n}=(-1)^n\binom{n+4}{4}$$
finally
$$\dfrac1{(1-x)^5}=\sum_{n\geq0}(-1)^n\binom{-5}{n}x^n=\sum_{n\geq0}\binom{n+4}{4}x^n$$
| {
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"timestamp": "2023-03-29T00:00:00",
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How find this sum $\sum\limits_{n=0}^{\infty}\frac{1}{(3n+1)(3n+2)(3n+3)}$ Find this sum
$$I=\sum_{n=0}^{\infty}\dfrac{1}{(3n+1)(3n+2)(3n+3)}$$
My try: let
$$f(x)=\sum_{n=0}^{\infty}\dfrac{x^{3n+3}}{(3n+1)(3n+2)(3n+3)},|x|\le 1$$
then we have
$$f^{(3)}(x)=\sum_{n=0}^{\infty}x^{3n}=\dfrac{1}{1-x^3}$$
then we find the $f(x)$,Following is very ugly(can you someone can post your follow solution,) have other simple methods? Thank you very much.
| OK, so let's start with
$$\frac{d^3}{dx^3} f(x) = \frac{1}{1-x^3}$$
where $f(0)=f'(0)=f''(0)=0$. Note that we may use partial fractions and a little rearranging to find that
$$\frac{1}{1-x^3} = \frac13 \left [\frac{1}{1-x} + \frac{x+\frac12}{\left (x+\frac12\right)^2+\frac{3}{4}}+\frac{3}{2} \frac{1}{\left (x+\frac12\right)^2+\frac{3}{4}}\right ]$$
Integrating and using $f''(0)=0$ we get that
$$f''(x) = -\frac13 \log{(1-x)} + \frac16 \log{(1+x+x^2)} + \frac{1}{\sqrt{3}} \arctan{\frac{2 x+1}{\sqrt{3}}} - \frac{\pi}{6 \sqrt{3}}$$
From here the challenge will be managing the various terms rather than individual integrations, which are easily done through parts. Note that, for example,
$$\int dx \, \log{(1+x+x^2)} = x \log{(1+x+x^2)} - \int dx \frac{2 x^2+x}{x^2+x+1}$$
You may use partial fractions on the latter integral and determine that
$$\int dx \frac{2 x^2+x}{x^2+x+1} = 2 x - \frac12 \log{(1+x+x^2)} - \sqrt{3} \arctan{\frac{2 x+1}{\sqrt{3}}} +C$$
Using a substitution ($u=(2 x+1)/\sqrt{3}$) and integration by parts, we similarly find that
$$\int dx \, \arctan{\frac{2 x+1}{\sqrt{3}}} = \frac{\sqrt{3}}{2}\left [\frac{2 x+1}{\sqrt{3}} \arctan{\frac{2 x+1}{\sqrt{3}}} - \frac12 \log{(1+x+x^2)}\right ] + C$$
Omitting a lot of algebra, I find that, using $f'(0)=0$:
$$f'(x) = \frac13 (1-x) \log{(1-x)} - \frac16 (1-x) \log{(1+x+x^2)} + \frac{1+x}{\sqrt{3}} \arctan{\frac{2 x+1}{\sqrt{3}}} - \frac{\pi}{6 \sqrt{3}} (1+x)$$
Now we integrate once more. Again, the challenge is bookkeeping, as all the integrals may be done by parts. Using $f(0)=0$, I get that
$$f(x) = -\frac16(1-x)^2 \log{(1-x)} + \frac{1}{12} (x^2-2 x-2) \log{(1+x+x^2)} + \frac{(x^2+2 x)}{2 \sqrt{3}} \arctan{\frac{2 x+1}{\sqrt{3}}} - \frac{\pi}{12 \sqrt{3}} (x^2+2 x)$$
The sum is then
$$f(1) = \frac{\sqrt{3}}{12} \pi - \frac14 \log{3} \approx 0.178$$
which checks out with, e.g., WA.
| {
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"timestamp": "2023-03-29T00:00:00",
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