Q
stringlengths 70
13.7k
| A
stringlengths 28
13.2k
| meta
dict |
|---|---|---|
Calculating conditional expectation and variance of multivariate normal
Suppose $\mathbf{Y} = \begin{bmatrix} Y_1 \\ Y_2 \\ Y_3 \end{bmatrix}
\sim N(\mu, \Sigma)$ where $\mu = \begin{bmatrix} 1 \\ 2 \\ 3
\end{bmatrix}$ and $\Sigma = \begin{bmatrix} 2 & -1 & 1 \\ -1 & 5 & 1
\\ 1 & 1 & 3 \end{bmatrix}$. Calculate $\mathbb{E}[Y_3 | Y_1 = y_1,
Y_2 = y_2]$ and $\mathbb{V}[Y_3 | Y_1 = y_1, Y_2 = y_2]$.
I know the general formula in the case when: $\mathbf{Y} = \begin{bmatrix} \mathbf{Y}_1 \\ \mathbf{Y}_2 \end{bmatrix} \sim N\left(\begin{bmatrix} \mu_1 \\ \mu_2 \end{bmatrix}, \begin{bmatrix} \Sigma_{11} & \Sigma_{12} \\ \Sigma_{21} & \Sigma_{22} \end{bmatrix} \right)$ then $\mathbf{Y}_2 | \mathbf{Y}_1 = \mathbf{y}_1 \sim N(\mu_2 + \Sigma_{21} \Sigma_{11}^{-1} ( \mathbf{y}_1 - \mu_1), \Sigma_{22} - \Sigma_{21} \Sigma_{11}^{-1} \Sigma_{12})$. However, I am unsure of how to apply it here, could someone please show me?
|
Partition the multivariate normal random vector
$\mathbf{Y}$
consisting of two subvectors as
$\mathbf{Y}=\left(\begin{array}{c}
\mathbf{Y^{(1)}}\\
\mathbf{Y^{(2)}}\\
\end{array} \right)$
where
$\mathbf{Y^{(1)}}=\left(\begin{array}{c}
Y_{1}\\
Y_{2}\\
\end{array} \right) $
and
$\mathbf{Y^{(2)}}=\left(\begin{array}{c}
Y_{3}
\end{array} \right)$.
Accordingly partition the mean vector as
$\mathbf{\mu}=\left(\begin{array}{c}
\mathbf{\mu^{(1)}}\\
\mathbf{\mu^{(2)}}\\
\end{array} \right)$
where
$\mathbf{\mu^{(1)}}=\left(\begin{array}{c}
\mu_{1}\\
\mu_{2}\\
\end{array} \right) =\left(\begin{array}{c}
1\\
2\\
\end{array} \right)$
and
$\mathbf{\mu^{(2)}}=\left(\begin{array}{c}
\mu_{3}
\end{array} \right) = 3$.
and variance-covariance matrix as
$\Sigma = \left(\begin{array}{ccc}
2 & -1 & 1\\
-1 & 5 & 1\\
1 & 1 & 3\\
\end{array} \right) = \left(\begin{array}{cc|c}
2 & -1 & 1\\
-1 & 5 & 1\\
\hline
1 & 1 & 3\\
\end{array} \right) = \left(\begin{array}{cc}
\Sigma_{11}&\Sigma_{12}\\
\Sigma_{21}&\Sigma_{22}\\
\end{array} \right)$
Plug-in the values in the expressions for mean and variance you have stated above to get the solution.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2485500",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
}
|
Use induction to prove following sum identity Using induction, find an explicit formula for the sum $\displaystyle\sum_{i=1}^n\frac{3}{i^2+3i}.$
My attempt: After writing the the first few partial sums, I was able to tell that the denominators of the sum will have the form of $(i+3)!/6$ but I'm not sure how to proceed or find a nice pattern for the numerators: 3,21,146,1022,...
Any help would be appreciated.
Edit: Use Induction, not telescoping. I know how to telescope, what I am looking for is the explicit formula of the sum without using telescoping.
|
It looks like a telescoping series to me. $\frac{3}{i(i+3)}=\frac{1}{i}-\frac{1}{i+3}$, so write out some terms and see what cancels. It looks like you are left with $$1 + \frac{1}{2}+ \frac{1}{3} - \frac{1}{n+1}- \frac{1}{n+2}- \frac{1}{n+3}$$
To do induction, claim $S_N=1 + \frac{1}{2}+ \frac{1}{3} - \frac{1}{n+1}- \frac{1}{n+2}- \frac{1}{n+3}$. Show that it works when $N=1$ (base case).
Now prove that $$S_{N+1}=\left(\sum_{i=1}^{N}\frac{3}{i^2+3i}\right) + \frac{3}{(N+1)^2+3(N+1)}$$
And this is $$\left(1 + \frac{1}{2}+ \frac{1}{3} - \frac{1}{n+1}- \frac{1}{n+2}- \frac{1}{n+3}\right)+\frac{3}{(N+1)^2+3(N+1)}$$
using the induction hypothesis.
Show this adds up to $$S_{N+1}= 1 + \frac{1}{2}+ \frac{1}{3} - \frac{1}{N+2}- \frac{1}{N+3}- \frac{1}{N+4}$$ which is your formula you are trying to prove.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2486511",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
}
|
What is the name of this theorem of Jakob Steiner's, and why is it true? In The Secrets of Triangles a remarkable theorem is attributed to Jakob Steiner.
Each side of a triangle is cut into two segments by an altitude. Build squares on each of those segments, and the alternating squares sum to each other.
The book doesn't include a proof, and I'm not sure how to start.
Does this theorem have a name? How could one go about proving this beautiful relationship?
|
Label the squares' side lengths $a, b, c, d, e, f $ (clockwise from $A$). The claim is that $$a^2+c^2+e^2=b^2+d^2+f^2$$
Let $x$ be the altitude from $A$.
Let $y$ be the altitude from $B$.
Let $z$ be the altitude from $C$.
By the Pythagorean theorem applied to the two right triangles that include the altitude from $A$, we have:
$$x^2+c^2=(a+b)^2$$
$$x^2+d^2=(e+f)^2$$
By the Pythagorean theorem applied to the two right triangles that include the altitude from $B$, we have:
$$y^2+a^2=(e+f)^2$$
$$y^2+b^2=(c+d)^2$$
By the Pythagorean theorem applied to the two right triangles that include the altitude from $C$, we have:
$$z^2+e^2=(c+d)^2$$
$$z^2+f^2=(a+b)^2$$
Labeling the six Pythagorean equations above $(1)$ through $(6)$, we can add $(1)$, $(3)$, and $(5)$ to get:
$$ x^2+y^2+z^2 +a^2+c^2+e^2=(a+b)^2+ (c+d)^2 + (e+f)^2$$
Add $(2)$, $(4)$, and $(6)$:
$$ x^2+y^2+z^2 +b^2+d^2+f^2=(a+b)^2+ (c+d)^2 + (e+f)^2$$
Notice that the right sides of the above two equations are equal, so we may equate the left sides:
$$ x^2+y^2+z^2+a^2+c^2+e^2= x^2+y^2+z^2+b^2+d^2+f^2 $$
Now subtract $x^2+y^2+z^2$ from both sides, and we are done.
$$a^2+c^2+e^2=b^2+d^2+f^2$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2486718",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "46",
"answer_count": 7,
"answer_id": 3
}
|
How to calculate the limit $\lim_{x \to 1} \frac{x + x^2 +\cdots+ x^n - n}{x - 1}$? How can I find the limit of the following function:
$$\begin{equation*}
\lim_{x \to 1}
\frac{x + x^2 +\cdots+ x^n - n}{x - 1}
\end{equation*}$$
Any help will be appreciated.
Thanks!
|
Remember that $$\sum_{i=1}^n x^i=\frac{x \left(x^n-1\right)}{x-1}$$ Let $x=y+1$ and consider
$$S_n=\frac {-n+\sum_{i=1}^n x^i }{x-1}=\frac{\frac{(y+1) \left((y+1)^n-1\right)}{y}-n}{y}$$ Now, using the binomial theorem or Taylor series $$(y+1)^n=1+n y+\frac{1}{2} (n-1) n y^2+\frac{1}{6} (n-2) (n-1) n y^3+O\left(y^4\right)$$
$$(y+1)^n-1=n y+\frac{1}{2} (n-1) n y^2+\frac{1}{6} (n-2) (n-1) n y^3+O\left(y^4\right)$$
$$\frac{(y+1) ^n-1}{y}=n +\frac{1}{2} (n-1) n y+\frac{1}{6} (n-2) (n-1) n y^2+O\left(y^3\right)$$
$$(y+1)\frac{(y+1) ^n-1}{y}=n+\frac{1}{2} n (n+1) y+\frac{1}{6} n \left(n^2-1\right) y^2+O\left(y^3\right)$$
$$(y+1)\frac{(y+1) ^n-1}{y}-n=\frac{1}{2} n (n+1) y+\frac{1}{6} n \left(n^2-1\right) y^2+O\left(y^3\right)$$ making
$$S_n=\frac{1}{2} n (n+1) +\frac{1}{6} n \left(n^2-1\right) y+O\left(y^2\right)$$ which shows the limit and how it is approached when $y\to 0$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2487214",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 2
}
|
Different ways to tackle the integral $\int_0^1\sqrt\frac x{1-x}\,dx$ $$\int_0^1\sqrt\frac x{1-x}\,dx$$
I saw in my book that the solution is $x=\cos^2u$ and $dx=-2\cos u\sin u\ du$.
I would like to see different approaches, can you provide them?
|
Here is another way that involves rationalising the numerator first.
For $0 \leqslant x < 1$ we can write
\begin{align*}
\int_0^1 \sqrt{\frac{x}{1 - x}} \, dx &= \int_0^1 \sqrt{\frac{x}{1 - x}} \cdot \frac{\sqrt{x}}{\sqrt{x}} \, dx\\
&= \int^1_0 \frac{x}{\sqrt{x - x^2}} \, dx
\end{align*}
Now rewriting the numerator as the derivative of the denominator we have
\begin{align*}
\int_0^1 \sqrt{\frac{x}{1 - x}} \, dx &= -\frac{1}{2} \int^1_0 \frac{(1 - 2x) - 1}{\sqrt{x - x^2}} \, dx\\
&= -\frac{1}{2} \int^1_0 \frac{1 - 2x}{\sqrt{x - x^2}} \, dx + \frac{1}{2} \int^1_0 \frac{dx}{\sqrt{x - x^2}} \, dx\\
&= I_1 + I_2
\end{align*}
The first of these integrals can be found using a substitution of $x = u + 1/2$. The result is
$$I_1 = \int^{1/2}_{-1/2} \frac{u}{\sqrt{1/4 - u^2}} \, du = 0,$$
as the integrand is odd between symmetric limits.
The second integral can be found by first completing the square. As
$$x - x^2 = \frac{1}{4} - \left (x - \frac{1}{2} \right )^2,$$
we have
$$I_2 = \frac{1}{2} \int^1_0 \frac{dx}{\sqrt{\frac{1}{2^2} - \left (x - \frac{1}{2} \right )^2}} \, dx = \frac{1}{2} \sin^{-1} (2x - 1) \Big{|}^1_0 = \frac{\pi}{2}.$$
Thus
$$\int_0^1 \sqrt{\frac{x}{1 - x}} \, dx = \frac{\pi}{2}.$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2490654",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 9,
"answer_id": 1
}
|
How to prove $|z|\cdot |w|=|z \cdot w|$ form complex numbers? I am trying to prove $|z|\cdot |w|=|z \cdot w|$
$|z\cdot w|=\sqrt{(ac-bd)^{2}+(ad+bc)^{2}}$
and
$|z|\cdot|w|=\sqrt{a^{2}+b^{2}}\cdot\sqrt{c^{2}+d^{2}}=\sqrt{(a^{2}+b^{2})\cdot(c^{2}+d^{2})}=\sqrt{(a^{2}c^{2}+a^{2}d^{2}+b^{2}c^{2}+b^{2}d^{2})}$
And I'm stuck
Help?
|
Now, $$(ac-bd)^2+(ad+bc)^2=a^2c^2-2abcd+b^2d^2+a^2d^2+2abcd+b^2c^2=$$
$$=a^2c^2+a^2d^2+b^2c^2+b^2d^2=a^2(c^2+d^2)+b^2(c^2+d^2)=(a^2+b^2)(c^2+d^2)$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2496326",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
}
|
Invariant factors So I was asked the following question: Let $ G = \{ 1 + a_1x + a_2x^2 +a_3x^3 : a_i \in \mathbb{Z}/2\mathbb{Z} \} $, and define a binary operation on $ G $ by $ p(x) * q(x) = p(x)q(x) \bmod{x^4} $. This makes $ G $ a $ \mathbb{Z} $-module. Find the invariant factors of $ G $.
I understand that the operation makes $ G $ a finite abelian group ($ \mathbb{Z} $-module) hence we can apply the classification theorem. If I can find a set of generators and all possible relations between them then I know how to find the invariant factors. But I'm not sure how to do this. Any hints?
|
The elements we have are:
$$\{1,1+x,1+x^2,1+x+x^2,1+x^3,1+x+x^3,1+x^2+x^3,1+x+x^2+x^3\}$$
It can be useful to see what the subgroups $\langle g\rangle$ look like for all $g\in G$, as this gives us a good idea what to pick as generators.
Additionally, this is an abelian group of order $8$, so it's isomorphic to one of the following: $$\mathbb{Z}_8,\mathbb{Z}_4\times\mathbb{Z}_2,\mathbb{Z}_2\times\mathbb{Z}_2\times\mathbb{Z}_2$$
To distinguish these cases, it can be useful to find the order of some elements:
\begin{array}{cccc}
g & g^2 & g^3 & g^4\\\hline
1+x & 1+x^2 & 1+x+x^2+x^3 & 1 \\
1+x^2 & 1 &&\\
1+x+x^2 & 1+x^2 & 1+x+x^3 & 1 \\
1+x^3 & 1 \\
1+x+x^3 & 1+x^2 & 1+x+x^2 & 1 \\
1+x^2+x^3 & 1 \\
1+x+x^2+x^3 & 1+x^2 & 1+x & 1
\end{array}
This gives us a decent amount of information.
The big thing it tells us is all of the inverses in this group, and all of the orders.
We can see that $1+x$ and $1+x+x^2$ are two elements of order $4$ that aren't powers of a single generator.
So, one them is the other times the element of order $2$.
Now, note that:
$$(1+x)(1+x^3) = 1+x+x^3$$
So, if we fix $1+x = g$, and $1+x^3 = s$, we get that $|g| = 4$, $|s| = 2$, and that:
\begin{array}{cccccccc}
1 & 1+x & 1+x^2 & 1+x+x^2 & 1+x^3 & 1+x+x^3 & 1+x^2 +x^3 & 1+x+x^2+x^3 \\
1 & g & g^2 & g^3s & s & gs & g^2s & g^{3}
\end{array}
Is how each element is expressed in terms of generators.
From here, it should be easy to establish the isomorphism with $\mathbb{Z}/4\mathbb{Z}\times\mathbb{Z}/ 2\mathbb{Z}$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2497473",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
Summation over roots of unity Find the value of $\displaystyle\sum_{r=1}^{4} \frac{1}{2-\alpha^r} $
where $ \alpha^k (k=0,1,2,3,4,5) $ are fifth roots of unity.
My approach:-
As we know that $ \alpha^k (k=0,1,2,3,4,5) $ are fifth roots of unity, then $ \alpha^k - 1$ should be equal to zero. Therefore the final answer to the summation $\displaystyle\sum_{r=1}^{4} \frac{1}{2-\alpha^r} $ should be $\displaystyle\sum_{r=1}^{4} 1 = 4 $
But the answer given is $ \dfrac{ 49}{31} $
Any help or hint will be much appreciated!
|
If $\alpha^5 = 1$, $\frac{1}{2-\alpha} = c_0 + c_1 \alpha + \ldots + c_4 \alpha^4$ where
$$\eqalign{ 2 c_0 - c_4 &= 1\cr
2 c_1 - c_0 &= 0\cr
2 c_2 - c_1 &= 0\cr
2 c_3 - c_2 &= 0\cr
2 c_4 - c_3 &= 0\cr}$$
The solution of this is $$c_0 = 16/31,\; c_1 = 8/31,\; c_2 = 4/31,\; c_3 = 2/31,\; c_4 = 1/31$$
Then $$ \eqalign{\sum_{r=1}^4 \frac{1}{2-\alpha^r} &= \frac{16}{31}\sum_{r=1}^4 1 +
\frac{8}{31}\sum_{r=1}^4 \alpha^r + \frac{4}{31}\sum_{r=1}^4 \alpha^{2r} +
\frac{2}{31}\sum_{r=1}^4 \alpha^{3r} + \frac{1}{31}\sum_{r=1}^4 \alpha^{4r}\cr &= \frac{16}{31} \cdot 4 - \frac{8}{31} - \frac{4}{31} - \frac{2}{31} - \frac{1}{31} = \frac{49}{31}}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2497816",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 4,
"answer_id": 1
}
|
Probability a bit in a bit string is 1 after swapping Stuck on a homework question, so I could use all the help I could get.
Let $x = x(1), \dots , x(n)$ be a bit string containing exactly $m$ occurrences of 1. Consider the following operation on $x$: we choose a random pair of indices $(i,j),$ and we swap $x(i)$ and $x(j)$ so that $x'(i) = x(j),$ $x'(j) = x(i),$ while $x'(k) = x(k)$ if $k \neq i$ and $k \neq j.$ (If $i = j,$ therefore, then we change nothing.) Let $X_1 = x,$ and let $X_2, \dots, X_N$ be obtained by such a sequence of operations (always swapping a new random pair) that so $X_{r+1} = X_r$. The number of 1s remains $m$ in each iteration. Show for each $i$, we have $P(X_N (i) = 1) \rightarrow \frac{m}{n}$ as $N \rightarrow \infty$.
We're given this hint: Consider the last time $i$ was swapped.
I've gathered that the probability that $i$ is swapped on any given iteration is $1-(1- \frac{1}{n})^2$. I've also figured out that the probability $i$ is 1 after a swap is $\frac{m}{n}$, as there are m choices for i to change to (including itself) after being selected for a swap, but I'm not sure how to apply this.
|
Let's start with this:
\begin{align*}
P(X_N(i)=1|X_{N-1}(i)=0) &= \frac{2m}{n^2}
\end{align*}
Because if $X_{N-1}(i)=0$, then we get $X_N(i)=1$ if the first index chosen is $i$ and the second index is one of the $m$ out of $n$ spots that have a $1$, or if the first index is one of the $m$ out of $n$ spots that have a $1$ and the second index chosen is $i$.
Then similarly, we can reason out $P(X_N(i)=1|X_{N-1}(i)=1)$ by saying that it is what happens when we do not {choose $i$ and one of the $n-m$ indexes that have $0$s, in either order}. That is:
\begin{align*}
P(X_N(i)=1|X_{N-1}(i)=1) &= 1 - \left( \frac{2(n-m)}{n^2} \right) \\
&= \frac{n^2 - 2n + 2m}{n^2}
\end{align*}
Therefore:
\begin{align*}
P(X_N(i)=1) = \big(P(X_{N-1}(i)=1) \cdot (n^2 - 2n + 2m) + P(X_{N-1}(i)=0) \cdot 2m \big)/n^2
\end{align*}
But since $X_{N-1}$ can be only $0$ or $1$, $P(X_{N-1}(i)=0) + P(X_{N-1}(i)=1) = 1$, and:
\begin{align*}
P(X_N(i)=1) = \big(P(X_{N-1}(i)=1) \cdot (n^2 - 2n + 2m) + (1-P(X_{N-1}(i)=1)) \cdot 2m \big)/n^2
\end{align*}
Let's define $t_N$ as $P(X_N(i)=1)$, making this:
\begin{align*}
t_N &= \big(t_{N-1} \cdot (n^2 - 2n) + 2m \big)/n^2 \\
&= t_{N-1} \left(1 - \frac{2}{n}\right) + \frac{2m}{n^2}
\end{align*}
It's now straightforward to show that $t_N$ converges to $\frac{m}{n}$:
Start by defining $s_N$ as $t_N - \frac{m}{n}$, so that $t_N = s_N + \frac{m}{n}$. Then we have:
\begin{align*}
s_N + \frac{m}{n} &= \left(s_{N-1} + \frac{m}{n}\right) \left(1 - \frac{2}{n}\right) + \frac{2m}{n^2} \\
s_N + \frac{m}{n} &= s_{N-1} \left(1 - \frac{2}{n}\right) + \frac{m}{n} - \frac{2m}{n^2} + \frac{2m}{n^2} \\
s_N &= s_{N-1} \left(1 - \frac{2}{n}\right)
\end{align*}
So obviously
$$
s_N = s_1 \left(1 - \frac{2}{n}\right)^{N-1}
$$
and
$$
\lim_{N \to \infty} s_N = 0
$$
which gives us what we want about $t_N$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2499066",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
}
|
Solve $\lfloor \sqrt x +\sqrt{x+1}+\sqrt{x+2}\rfloor=x$ $$\lfloor \sqrt x +\sqrt{x+1}+\sqrt{x+2}\rfloor=x$$
I tried to solve this equation.
First thing is $\lfloor \sqrt x +\sqrt{x+1}+\sqrt{x+2}\rfloor \in \mathbb{Z} $ so $x \in \mathbb{Z}$
second $$\sqrt x +\sqrt{x+1}+\sqrt{x+2} \geq \sqrt 0 +\sqrt{0+1}+\sqrt{0+2} \\\to x \in \mathbb{N}$$ so we can check $x=1,2,3,4,5,6,7,8,9,\ldots$ by a MATLAB program. I checked the natural numbers to find solution. I found $x=8,9$ worked here.
Now my question is about somehow an analytical solving of the equation, or another idea. Can any one help me? Thanks in advance.
|
You can use inequalities to simplify your problem.
Since $\lfloor x \rfloor \le x$. Therefore we've
\begin{align}
x&= \lfloor \sqrt x+\sqrt {x+1}+\sqrt{x+2} \rfloor \\
&\le \sqrt x+\sqrt {x+1}+\sqrt{x+2} \\
&\le 3\sqrt {x+2}\\
\end{align}
$$\implies x^2 \le 9(x+2) \; ; x \in \mathbb Z$$
This gives us the range $x \in [-1,10] \tag1$.
Also $\lfloor x \rfloor \ge x-1$. Therefore we've
\begin{align}
x&= \lfloor \sqrt x+\sqrt {x+1}+\sqrt{x+2} \rfloor \\
&\ge \sqrt x+\sqrt {x+1}+\sqrt{x+2} \color{red}{-1}\\
&\ge 3\sqrt {x}-1\\
\end{align}
$$\implies x+1 \ge 3\sqrt x$$
$$\implies x^2-7x+1\ge 0 \, ; x \in \mathbb Z$$
This gives us $x \in (-\infty, 0]\cup [7,\infty)\tag2$
Taking intersection of $(1)$ and $(2)$, and taking care of domain, I.e. $x\ge 0$, we get
$$\color{blue}{x \in \{0,7,8,9,10\}}$$
Now you can check for $x=0,7,8,9,10$, which is quite easy now.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2499746",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 7,
"answer_id": 1
}
|
$\iint_A\frac{y}{x^2+y^2}dxdy, \quad A: [\space {(x,y): \space x^2+y^2\ge 4, \quad x^2+y^2\le2 \sqrt{x^2+y^2}+2x}]$ I've been given a "fun" integral and I just want to know if I'm doing it good or I missed something.
$$\iint_A\frac{y}{x^2+y^2}dxdy, \quad A: [\space {(x,y): \space x^2+y^2\ge 4, \quad x^2+y^2\le2 \sqrt{x^2+y^2}+2x}]$$
And this is my solution:
$$
\begin{matrix}
x=r \cos \varphi \\
y=r \sin \varphi \\
\end{matrix}
$$
$$r\ge2, \quad \begin{matrix}
r^2 \le 2r+2r \cos \varphi \\
r \le 2(1+ \cos \varphi) \\
\end{matrix}$$
$$x=2 \Rightarrow r \cos \varphi=2 \Rightarrow r=\frac{2}{\cos \varphi}$$
Now I evaluate the integral:
$$
\begin{split}
I
&= \iint_A \frac{y}{x^2+y^2} dxdy \\
&= 2\int_0^{\pi/2} d\varphi
\left(\int_2^{2(1+\cos \varphi)}\frac{\sin \varphi}{r}dr
+\int_{\frac{2}{\cos \varphi}}^{2(1+\cos \varphi)}
\frac{\sin \varphi}{r}dr\right)\\
&= 2\int_0^{\pi/2}
\sin \varphi
\left(\ln|2+2\cos \varphi|
- \ln2
+ \ln|2+2 \cos|2+2 \cos \varphi|
- \ln \left| \frac{2}{\cos \varphi} \right|
\right) d \varphi\\
&= 2\int_0^{\pi/2}
\sin \varphi
(2 \ln|1+\cos \varphi|+\ln|\cos \varphi|)d\varphi\\
&\text{substitute } \cos \varphi =t \\
&= -2\int_1^0(2\ln(1+t)+\ln(t))dt=8\ln2-2
\end{split}
$$
Of course, I skipped many trivial calculation steps.
Can anyone check it or maybe comment if I'm doing it wrong?
|
The integrand is odd function respect to $y$ and the area is symmetric about $x$-axis, so the integral is $0$. On the other hand, you can write
$$\iint_A\frac{y}{x^2+y^2}dxdy=\int_{-\pi/2}^{\pi/2}\int_2^{2(1+\cos\theta)}\dfrac{r\sin\theta}{r^2}r\,dr\,d\theta=0$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2501896",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
}
|
Prove by induction that $\sum_{i=1}^{n} i \cdot 2^i = (n-1) \cdot 2^{n+1} +2$. Help finding my mistake
Prove by induction:
$$\sum_{i=1}^{n} i \cdot 2^i = (n-1) \cdot 2^{n+1} +2$$
Basis: let $p(n)$ be the predicate.
Let $n=1$ this gives $(1-1) \cdot 2^1+1+2 = 2$ and $1 \cdot 2^1 = 2$ so its true for $p(1)$
Induction: assume $n=k$ thus $$\sum_{i=1}^{k} i \cdot 2^i = (k-1) \cdot 2^{k+1} +2$$
when $n=k+1$:
$$\sum_{i=1}^{k+1} i \cdot 2^i = \sum_{i=1}^{k} i \cdot 2^i + ((k+1)-1) \cdot 2^{(k+1)+1} +2$$
$$ =(k-1) \cdot 2^{k+1} +2 + ((k+1)-1) \cdot 2^{k+2} +2$$
$$= 2^{k+1}k - 2^{k+1} +2 +2^{k+2}(k+1) - 2^{k+2}+2$$
$$=\frac{2^{k+1}k-2^{k+1}+2 \cdot 2^{k+1}(k+1)-2 \cdot 2^{k+1} +4}{2}$$
$$=2^{k+1}k-2^{k+1}+2^{k+1}(k+1)-2^{k+1}+2$$
$$= (k-1) \cdot 2^{k+1}+((k+1)-1) \cdot 2^{k+1} +2$$
I've been messing around with the arithmetic part of this for a while now and it's getting frustrating, I'm thinking that I've possibly made a mistake at the beginning of the induction step and that's why I cant get this to work, or maybe I'm just missing something in the algebra, either way can someone please help.
|
Assuming $P(n)$ is true, we have
$$\begin{align*}
\sum_{i=1}^{n+1} i \cdot 2^i
&= \sum_{i=1}^n i\cdot2^i + (n+1)\cdot2^{n+1} \\\\
&= (n-1)\cdot2^{n+1}+ 2 +(n+1)\cdot2^{n+1} \\\\
&= 2^{n+1}\cdot((n-1)+(n+1))+2 \\\\
&=2^{n+1}\cdot(2n)+2 \\\\
&=n\cdot2^{n+2}+2 \\\\
&=((n+1)-1)\cdot2^{((n+1)+1)}+2
\end{align*}$$ as desired.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2502457",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
}
|
Finding the limit of a function . How can I calculate the following limit:
\begin{equation*}
\lim_{x \rightarrow a}
\frac{\sqrt{x} - \sqrt{a} + \sqrt{x-a} }{\sqrt{x^2 - a^2}}
\end{equation*}
I feel that I should multiply by the conjugate, but which conjugate?
|
Hint: Write $y=\sqrt{x}$ and $b=\sqrt{a}$. Then you get:$${y-b+\sqrt{y^2-b^2}\over \sqrt{y^4-b^4}}={\sqrt{y-b}+\sqrt{y+b}\over \sqrt{(y+b)(y^2+b^2)}}$$ So the limit is $1\over \sqrt{2a}$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2503609",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
}
|
Counting the number of words containing a specific subword, and not another subword Consider the $10$ letter word
$$PRRAAAATTM$$
and all the words formed by rearranging its letters. How many of these words contain the subword $RAT$ but do not contain the subword $MAP$?
This is what I tried:
Let $R$ be the set of words containing the subword $RAT$.
Let $M$ be the set of words containing the subword $MAP$.
Then I want to compute $|R-M| = |R| - |R\cap M$|.
To count $|R|$, we count the number of words containing at least one $RAT$, and subtract the number of words containing $RAT$ at least twice + number of words containing $RAT$ at least three times - number of words containining $RAT$ at least four times - ... and so on.
Since only two $RAT$'s are possible, then
$|R|$ = number of words containing at least one $RAT$ - number of words containing two $RAT$'s.
Number of words containing at least one $RAT$ is
$$\binom{8}1 \times \frac{7!}{3}$$
and number of words containing two $RAT$'s is
$$\binom{6}2 \times \frac{4!}{2!}.$$
So
$$|R| = \binom{8}1\times\frac{7!}{3!} - \binom{6}2\times\frac{4!}{2!}
$$
Now with similar reasoning,
$$|M\cap R| = \binom{6}2\times\frac{4!}{2!} - \binom{4}3
$$
So
$$|R-M| = \binom{8}1\times\frac{7!}{3!} - 4!\times\binom{6}2 + \binom{4}3$$
|
@mathlove has provided you with a nice answer. Here is an alternative approach:
Number of words containing RAT: We have eight objects to arrange: RAT, A, A, A, M, P, R, T. We can choose three positions for the As, then arrange the remaining five distinct objects in the remaining five places in
$$\binom{8}{3}5!$$
ways.
Number of words containing two RATs: We have six objects to arrange: RAT, RAT, A, A, M, P, R. We choose two of the six positions for the RATs, two of the remaining four positions for the As, then arrange the remaining two distinct objects in the remaining two places in
$$\binom{6}{2}\binom{4}{2}2!$$
ways.
Hence,
$$|R| = \binom{8}{3}5! - \binom{6}{2}\binom{4}{2}2!$$
Number of words containing RAT and MAP: We have six objects to arrange: MAP, RAT, A, A, R, T. We can choose two of the six positions for the As, then arrange the remaining four distinct objects in the remaining four positions in
$$\binom{6}{2}4!$$
ways.
Number of words containing two RATs and MAP: We have four objects to arrange: MAP, RAT, RAT, A. We can choose two of the four positions for the RATs, then arrange the remaining two distinct objects in the remaining two positions in
$$\binom{4}{2}2!$$
ways.
Hence,
$$|M \cap R| = \binom{6}{2}4! - \binom{4}{2}2!$$
Therefore,
$$|R - M| = |R| - |M \cap R| = \binom{8}{3}5! - \binom{6}{2}\binom{4}{2}2! - \left[\binom{6}{2}4! - \binom{4}{2}2!\right]$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2513410",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
}
|
Find the number of real roots for $x+\sqrt{a^2-x^2}=b$, $a>0$, $b>0$, as a function of $a$ and $b$
Given: (1) $x+\sqrt{a^2-x^2}=b$, $(a,b,x)\subset \mathbb R$, $a>0$, $b>0$.
Find: number of roots for (1), given possible values for $a$ and $b$.
This is a question from a book for the preparation for math contests.
It states as final answer: (a) 1 root if $b<a$; and (b) 2 roots if $a<b<a\sqrt{2}$.
I'm having difficulties on finding this answer. I don't know whether it is correct or perhaps I'm not finding the right approach.
I started moving $x$ in (1) to the left, to get
$$\sqrt{a^2-x^2}=b-x$$
Before proceeding with squaring both sides, I saved 2 needed conditions for checking the final solution (c1) $a^2-x^2\ge 0$ and (c2) $b-x\ge 0$. Then squaring both sides, we get:
$$a^2-x^2=b^2+x^2-2bx\Leftrightarrow 2x^2-2bx+(b^2-a^2)=0$$
with discriminant $\triangle$ defined by:
$$\triangle=4(2a^2-b^2)$$
From this it is easy to see that a condition for 2 roots is (c3) $\sqrt{2}a>b,$ and for 1 root is (c4) $\sqrt{2}a=b,$ as $a>0$ and $b>0$. Then I find the roots as $$x=\frac{2b\pm \sqrt{\triangle}}{4}=\frac{b\pm \sqrt{2a^2-b^2}}{2}$$
From this point, I can't see a way to reach the stated answer, if it is right.
Full solutions or helpful hints are welcome. Sorry if it is a duplicate.
|
it must be $$a\geq |x|$$ and $$b\geq x$$
after squaring we get the equation
$$0=2x^2-2bx+b^2-a^2=0$$
solving this equation we get
$$x_1=\frac{b}{2}+\frac{1}{2}\sqrt{2a^2-b^2}$$
$$x_2=\frac{b}{2}-\frac{1}{2}\sqrt{2a^2-b^2}$$
so we get the following solution set
$$x=b$$ and $a=b$
$$x=\frac{1}{2}(b\pm i\sqrt{b^2-2a^2}$$ and $$0<a<\frac{b}{\sqrt{2}}$$
$$x=\frac{b}{2}$$ and $$a=\frac{b}{\sqrt{2}}$$
$$x=\frac{1}{2}(b\pm \sqrt{2a^2-b^2})$$ and $$\frac{b}{\sqrt{2}}<a<b$$
$$x=\frac{1}{2}(b-\sqrt{2a^2-b^2})$$ and $$a\geq b$$
Observe that i need time nto type all these formulas
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2514902",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 6,
"answer_id": 5
}
|
When $f(x)=\frac{-b^2m-ba+ax}{-mx-bm-a}$ is an integer $a,b,m,x$ are positive integers.
For which $x>0$ is $f(x)$ an integer?
$$f(x)=\frac{-b^2m-ba+ax}{-mx-bm-a}$$
I been trying to play with it, I changed it to:
$$\frac{b^2m-a\left(b+x\right)}{a+m\left(b+x\right)}$$
And then I been trying to say:
$$a+m\left(b+x\right)| b^2m-a\left(b+x\right) $$
$$a+m\left(b+x\right)| (a+m\left(b+x\right))(b+x)$$
So
$$a+m\left(b+x\right)| b^2m-a\left(b+x\right)+ (a+m\left(b+x\right))(b+x)$$
$$a+m\left(b+x\right)| m\left(\left(b+x\right)^2+b^2\right)$$
But I don't see how it helps, so please help me.
|
Here is a partial answer: Note that
\begin{eqnarray*}
f(x)&=&\frac{b^2m+ba-ax}{bm+a+mx}=\frac{b^2m+ba+bmx-bmx-ax-mx^2+mx^2}{bm+a+mx}\\
&=&\frac{(b-x)(bm+a+mx)+mx^2}{bm+a+mx}=b-x+\frac{mx^2}{bm+a+mx}.
\end{eqnarray*}
So $f(x)$ is an integer if and only if $\frac{mx^2}{bm+a+mx}$ is. This means there is a positive integer $k$ such that
$$mx^2=k(bm+a+mx)=kbm+ka+kmx.$$
In particular $ka=mc$ for some positive integer $c$, so we get
$$mx^2-kmx-kbm+mc=0\qquad\text{ and hence }\qquad x^2-kx-(kb+c)=0.$$
So by the quadratic formula
$$x=\frac{k}{2}\pm\frac{1}{2}\sqrt{k^2+4kb+4c},$$
is an integer. This is an integer if and only if $k^2+4kb+4c$ is a square. So the question is then:
Given positive integers $b$ and $c$, for which positive integers $k$ is
$k^2+4kb+4c$ a square?
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2518818",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
}
|
Limit of $f(x)$ when $x$ goes to zero Let $f(x) = \frac{1 + \tan x + \sin x - \cos x}{\sin^2 x + x^3}$ . Find value of $\lim_{x \to 0} f(x)$ if it exists . I can solve it using L'Hospital's Rule and Taylor series but I'm looking for another way suing trigonometric identities .
|
As $x\to 0$ we have
$\tan x\sim x;\;\sin x \sim x;\;1-\cos x\sim \dfrac{x^2}{2}$
So the limit becomes
$$\lim_{x\to 0} \frac{1 + \tan x + \sin x - \cos x}{\sin^2 x + x^3}=\lim_{x\to 0}\frac{x+x+\frac{x^2}{2}}{x^2+x^3}=\lim_{x\to 0}\frac{4x+x^2}{2(x^2+x^3)}=\lim_{x\to 0}\frac{x(4+x)}{2x^2(1+x)}=$$
$$=\lim_{x\to 0}\frac{4+x}{2x(1+x)}=\infty$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2520013",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
}
|
GCD of $n^3+3n+1$ and $7n^3+18n^2-n-2$ is always $1$ A problem I found yesterday says to prove $\gcd(n^3+3n+1, 7n^3+18n^2-n-2)=1$ for all integers $n\ge 1$. To begin, I used the Euclidean algorithm to observe that $7n^3+18n^2-n-2=7\left(n^3+3n+1\right)+\left(18n^2-22n-9\right)$, so $$\gcd(n^3+3n+1, 7n^3+18n^2-n-2)=\gcd(n^3+3n+1, 18n^2-22n-9).$$ From here, I got stuck however, since the next step of polynomial division involves rational numbers. I observed that the GCD cannot be a multiple of $3$, since if $3$ divides the term on the right, then $3\mid n$, however, then $3\nmid n^3+3n+1$. Similarly, by a parity argument, both terms are odd. Any suggestions for how to finish the problem from here?
|
It appears the problem fails for $n=309$. WolframAlpha says so here. See discussion on AoPS here on how to find $n=309$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2520554",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
}
|
How would one calculate absolute error under root operations? I know absolutely nothing about error analysis and I was wondering if someone could help me out. I want to approximate the value of $t$ that satisfies this equation: $$\sqrt{(\frac{1}{200}e^{2t}-\frac{1}{200}e^{-2t})^{2}+(\frac{1}{50}e^{2t})^{2}}=1.$$ I want to drop the term with the negative power, to get the following:
\begin{align}
\sqrt{(\frac{1}{200}e^{2t})^{2}+(\frac{1}{50}e^{2t})^{2}}&=1 \\
e^{2t}\sqrt{(\frac{1}{200})^{2}+(\frac{1}{50})^{2}}&=1 \\
2t+\frac{1}{2}\ln{\left((\frac{1}{200})^{2}+(\frac{1}{50})^{2}\right)
}&=0 \\
t&=-\frac{1}{4}\ln\left((\frac{1}{200})^{2}+(\frac{1}{50})^{2}\right)
\end{align}
However, I don't know by how much this approximates my actual value of t. How would one go about calculating the absolute error? Thanks in advance.
|
I'll not answer the error propagation part, but show you how to calculate the $t$ value explicitly: By squaring the inital equation and setting $X = e^{2t}$, you obtain
$$\frac{1}{200^2}(X-\frac{1}{X})^2 + \frac{1}{50^2}X^2 = 1$$
$$\frac{1}{200}(X^2 - 2 + \frac{1}{X^2}) + \frac{4}{50}X^2 = 200$$
$$\frac{1}{200}(X^4 - 2X^2 + 1) + \frac{4}{50}X^4 = 200 X^2$$
$$(X^4 - 2X^2 +1) + 16 X^4 = 40000 X^2$$
$$17X^4 - 40002X^2 + 1 = 0$$
Now substitute $Y = X^2$, solve the quadratic equation, take the square root to obtain $X$ and then the logarithm to obtain $t$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2522939",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
Graph the set of all complex numbers $z_n=\frac{2n+1}{n-i},n\in\mathbb R$ I'm a bit rusty on my complex numbers, how would you solve the following problem on paper?
Determine and sketch (graph) the set of all complex numbers of form:
$$z_n=\frac{2n+1}{n-i},n\in\mathbb R$$
Rationalizing yields $$S=\left\{z_n\in\mathbb C : z_n=\frac{n + 2 n^2}{1 + n^2}+\frac{1 + 2 n}{1 + n^2}i\right\}$$
How do I proceed to sketch (graph) this now on paper? (Wolframalpha yields a circle)
I assume I need to find the center and the radius of that circle which would be enough to sketch the graph, but I can't quite proceed from this point on.
|
I'll use $x$ instead of $n$.
Note that$$\frac{2x^2+x}{x^2+1}=1+\frac{x^2+x-1}{x^2+1}=1+\frac{2x^2+2x-2}{2(x^2+1)}$$and that$$\frac{2x+1}{x^2+1}=\frac12+\frac{-x^2+4x+1}{2(x^2+1)}.$$So$$\frac{2x+1}{x-i}=1+\frac i2+\frac{2x^2+2x-2}{2(x^2+1)}+\frac{-x^2+4x+1}{2(x^2+1)}i.$$Now, observe that$$\left(\frac{2x^2+2x-2}{2(x^2+1)}\right)^2+\left(\frac{-x^2+4x+1}{2(x^2+1)}\right)^2=\frac54.$$Therefore, your set is contained in the circle with center $1+\frac i2$ and radius $\frac{\sqrt5}2$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2523581",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
}
|
Proof explanation: finding the coefficient of $(r+1)$th term in the expansion of $\left(1-6x\right)^{-\frac{1}{2}}$? Here is the answer of the math..
$\displaystyle\left(1-6x\right)^{-\frac{1}{2}}$
$\displaystyle=\frac{\left(-\frac{1}{2}\right)\left(-\frac{1}{2}-1\right)....\left(-\frac{1}{2}-r+1_{ }\right)}{r!}\left(-6x\right)^r$
$\displaystyle=\frac{\left(-1\right)^r\left(\frac{1}{2}\right)\left(\frac{1}{2}+1\right)\left(\frac{1}{2}+2\right)....\left(r-1+\frac{1}{2}\right)}{r!}\left(-1\right)^r.\:x^r.\:2^r.3^r$
$\displaystyle=\left(-1\right)^{2r}\:\frac{1.3.5.7.....\left(2r-1\right)}{2^r\:.\:r!}\:\:x^r.\:2^r.3^r$
$\displaystyle=\frac{\left\{1.3.5.7.....\left(2r-1\right)\right\}\left(2.4.6....2r\right)}{r!\:\left(2.4.6.....2r\right)}\:\:x^r.\:3^r$
$\displaystyle=\frac{1.2.3.4.....2r}{r!\:2^r\:\left(1.2.3.\:....r\right)}\:\:x^r.\:3^r$
$\displaystyle=\frac{\left(2r\right)!\:\:.\:3^r}{2^r\:.\:\left(r!\right)^2}\:\:x^r$
$\displaystyle=\left(\frac{3}{2}\right)^r\:\frac{\left(2r\right)!\:}{\:\left(r!\right)^2}\:.\:\:x^r$
I can't understand the 4th and 6th line of this math . Please explain me that lines..
|
In line 4, $(n+1/2) = (2n+1)/2$ for all n running from 0 to r-1 and in denominator there become 2's r times which give $2^r$ in denominator. In next line they have multiplied $1.2...r$ and they distribute 2's for each n running from 1 to r which becomes $2.4....2r$ ($2^r$ is in numerator too). In next line they take back 2's which becomes $2^r$ in denominator and in numerator they rearrange each term which become $(2r)!$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2524472",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
}
|
recurrence relation, all terms of the sequence positive Let $a_1=a$, $a_2=\frac{1}{a}-a$, $a_{n+1}=\frac{n}{a_n}-a_n-a_{n-1}$ for $n=2,3,4,...$.
Find all $a$ such that $(a_n)$ is a sequence of positive reals.
My attempt was to look at $a_3=\frac{3a^2-1}{a-a^3}$, $a_4=\frac{8a^3-4a}{3a^4-4a^2+1}$ and a few more, $a_1>0$ gives $a>0$, $a_2>0$ gives $a\in(0,1)$, $a_3>0$ gives $a\in(\frac{1}{\sqrt{3}},1)$, but this probably doesn't give important information and further terms are nasty.
|
Some observations...Posting it as answer since it is too long as comment. With $a_1 = a, a_2 = \frac{1}{a} - a$ and $a_n(a_{n+1} + a_n + a_{n-1}) = n$, we get
$$S = a_2a_3 + (a_3 + a_4)^2+(a_4 + a_5)^2 + ... + (a_{n-1} + a_n)^2 -
(a_4^2 + a_5^2 + ...a_{n-1}^2) + a_na_{n+1}$$, where $S = \frac{n(n+1)}{2} - 3$.
So, $(a_4 + a_5)^2 + ... + (a_{n-1} + a_n)^2 = S - (a_2a_3 + a_na_{n+1}) + (a_4^2 + a_5^2 + ...a_{n-1}^2)$. Since LHS is $> 0$,
$S - (a_2a_3 + a_na_{n+1})$ is either $ > 0$ or It must be $S - (a_2a_3 + a_na_{n+1}) < (a_4^2 + a_5^2 + ...a_{n-1}^2)$. The lowest $n$ where this condition is applies is $n=5$ and the $a$ value we calculate must be applicable for $n>5$ I guess...
One more thing to try is to do alternate summation to cancel out products, like
$a_3^2 + a_3a_2 + a_3a_4 - a_4a_5 -a_4^2 -a_3a_4 + a_5a_6 + a_5^2 +a_4a_5 -a_6a_7 - a_6^2 - a_5a_6 ... = 3 - 4 + 5 -6 + 7 = \sum_{1}^{n}{n}{(-1)}^{n-1}$ and see whether we can get anywhere simpler...
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2525989",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "22",
"answer_count": 4,
"answer_id": 3
}
|
A positive real number $x$ with the property $x^3=3$ is irrational. I have the following problems:
1) There exists a positive real number $x$ such that $x^3=3$.
2) A positive real number $x$ with the property $x^3=3$ is irrational.
My Idea for 1) would be (there might be a few mistakes here):
Let $S = \{ x \geq 0 | x^3 \leq 3\}$, with $1\ge0$ and $1^3=1$ we have $1 \in S$, so $S \neq \emptyset$. With that we have $\forall x \in S, x < 3$, so S is bounded above. With the upper bound property of real numbers, $S$ has a least upper bound $s$:
$s=sup(S)$.
Since $1$ is in $S $, we know that $s>1$.
Now $s$ either is the solution, or one of the follwing two cases are true:
I) $s^3<3$
Let: $\varepsilon = \frac{3-s^3}{3s+1}$. By assumption $0<\varepsilon<1$, so that:
$(s+\varepsilon)^3=s^3+3s^2\varepsilon+3s\varepsilon^2+\varepsilon^3 \le s^3+3s^2\varepsilon+3s\varepsilon^2+\varepsilon^2=s^3+\frac{3-s^3}{3s+1}(3s+1)=3$.
Hence, $s + \varepsilon$ is also in $S$, in which case $s$ can not be an upper bound for $S$. This is a contradiction, so this case is not possible.
II) $s^3>3$
Let: $\varepsilon = \frac{s^3-3}{3s}$. Again $\varepsilon>0$, so that:
$(s-\varepsilon)^3=s^3-3s^2\varepsilon+3s\varepsilon^2-\varepsilon^3\ge s^3-3s^2\varepsilon+3s\varepsilon^2=s^3-3s\frac{s^3-3}{3s}=3.$
Hence, $s -\varepsilon$ is another upper bound for $S$, so that $s$ is not the least upper bound for $S$. This is a contradiction, so that this case is not possible.
Having eliminated these two cases, we are left with $s^3 = 3$, which is what we wanted to prove.
2) However I don't know how to proof that a positive real number with the property $x^3=3$ is irrational. It would be really nice if someone could help!
Edit: Made a correction regarding $(s+\varepsilon)^3$ and $(s-\varepsilon)^3$ (Hope this is correct)
|
To prove it's irrational, proceed just like in the proof that $\sqrt{2}$ is irrational. Assume there are integers, in lowest terms, such that $\frac{a^3}{b^3} = 3$. So, $a^3 = 3b^3$. Show that $3$ must divide both $a$ and $b$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2526743",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
}
|
Given $ \frac{1}{r}+\frac{1}{s}=a; \frac{1}{r}\times\frac{1}{s}=b; a+b=r; a\times b=s$, find $a$. (Brazilian Math Olympics, 2016)
Given: $\{a,b,r,s\}\subset \mathbb R$, $a>0$,
$\frac{1}{r}$, $\frac{1}{s}$ are roots for $x^2-ax+b=0$, and
$a$,$~b$ are roots for $x^2-rx+s$.
Find: the numeric value of $a$.
This is question 3, level 2, phase 3, Brazilian Math Olympics (OBM, 2016). No answer provided.
The first step is easy, using Girard relations, we can get the system
$$
\left\{
\begin{array}{c}
\frac{1}{r}+\frac{1}{s}=a \\
\frac{1}{r}\times\frac{1}{s}=b\\
a+b=r\\
a\times b=s
\end{array}
\right.
$$
I'm having a hard time on solving this system of equations. All tricks I know seems to lead to nowhere. It was considered a hard question in the contest (level 2 in OBM is for students up to 9th grade).
Hints and solutions are appreciated. Sorry if this is a duplicate.
|
$
\left\{
\begin{array}{l}
\frac{1}{r}+\frac{1}{s}=a \\
\frac{1}{r}\cdot\frac{1}{s}=b\\
a+b=r\\
a b=s
\end{array}
\right.
$
$
\left\{
\begin{array}{l}
\frac{1}{a+b}+\frac{1}{a b}=a \\
\frac{1}{(a+b) (a b)}=b
\end{array}
\right.
$
$$a^5+a^4-2 a^3-2 a^2-2 a-1=0\to \left(a^2-a-1\right) \left(a^3+2 a^2+a+1\right)=0$$
The only positive solution is $\color{red}{a=\dfrac{1+\sqrt{5}}{2}}$
Because $a^3+2 a^2+a+1=0$ has only one real solution which is negative
Indeed $P(a)=a^3+2 a^2+a+1$
has $P(-2)=-1$ and $P(-1)=1$ which means that there is a zero in $(-2,-1)$
Furthermore $P'(a)=3 a^2+4 a+1=(a+1) (3 a+1)$
$P'(a)=0$ for $a=-1;\;a=-\frac13$
$P''(a)=6a+4$
$P''(-1)=-2$ means that $(-1,1)$ is a maximum
$P''(-1/3)=2$ means that $\left(-\frac13,\frac{23}{27}\right)$ is a minimum
Anyway the function doesn't have any other real zeros.
As you can see in the picture below.
Hope this can help. It required some work to factor correctly the equation in the unknown $a$ because at first I tried with $b$ which leads to more complicated algebraic equations. Nice problem anyway, thank you.
$$...$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2527552",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
}
|
Number of Partitions of $n$ No Part Appears Exactly Once Equals to $n$ Partitioned into 0,2 ,3 or 4 (Mod 6 ) I want to prove that the number of partitions of $n$ in which no part appears exactly once is equal to the numbers of partitions of $n$ into parts that are congruent to one of $0,2,3,$ or $4$ mod $6$.
My approach
Intuitively, no part appearing not just once dually requires each part appears $3,5,7... <n$ number of times which means odd times just without 1.
Now I want to develop bijection between the first combinatorics and the second combinatorics. However, how could I make a function that relates each $3,5,7...<n $, which the number of cases depends on the size of $n$ into only 4 basic cases which rendered by $6$?
I think I am pretty bit lost. Any guidance to proceed further?
|
Let
\begin{align}
&A(n):= \text{no. of partitions of $n$ in which no part appears exactly once,}
\\&B(n):= \text{no. of partitions of $n$ into parts that are congruent to one of 0, 2, 3, or 4 mod 6.}
\end{align}
Now, for some indeterminate $q$, we have
\begin{align}
\sum_{n=0}^\infty A(n)q^n&=(1+q^2+q^3+\cdots)(1+q^4+q^6+\cdots)\cdots(1+q^{2k}+q^{3k}+\cdots)\cdots
\\&=\frac{(1-q^2)(1+q^2+q^3+\cdots)(1-q^4)(1+q^4+q^6+\cdots)\cdots(1-q^{2k})(1+q^{2k}+q^{3k}+\cdots)\cdots}{(1-q^2)(1-q^4)\cdots(1-q^{2k})\cdots}
\\&=\frac{(1+q^3)(1+q^6)\cdots(1+q^{3k})\cdots}{(1-q^2)(1-q^4)\cdots(1-q^{2k})\cdots}
\\&=\frac{(1-q^3)(1+q^3)(1-q^6)(1+q^6)\cdots(1-q^{3k})(1+q^{3k})\cdots}{((1-q^2)(1-q^4)\cdots(1-q^{2k})\cdots)((1-q^3)(1-q^6)\cdots(1-q^{3k})\cdots)}
\\&=\frac{1}{(1-q^2)(1-q^3)(1-q^4)(1-q^6)\cdots(1-q^{6k-4})(1-q^{6k-3})(1-q^{6k-2})(1-q^{6k})\cdots}
\\&=\sum_{n=0}^\infty B(n)q^n
\end{align}
Comparing coefficients on both sides, we obtain $$A(n)=B(n).$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2536114",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
}
|
How to find the Laplace inversion of $\frac{p}{p^4+4}$? How to calculate $$\mathscr{L}^{-1}\left\{\frac{p}{p^4+4}\right\}$$ where $\mathscr{L}$ is the Laplace transform operator? I thing need to apply some partial fraction first but am unable to work it out. Any idea or further help will be very good to me.
|
$p^4+4$ has 4 roots: $p=\pm 1 \pm i$, so it can be written as: $(p-1-i)(p-1+i)(p+1-i)(p+1+i)=(p^2-2p+2)(p+2p+2)$, so:
$$\frac{p}{p^4+4}=\frac{p}{(p^2-2p+2)(p+2p+2)}=\frac{Ap+B}{p^2-2p+2}+\frac{Cp+D}{p^2+2p+2}$$
The solution is $A=0, B=\frac{1}{4}, C=0, D=-\frac{1}{4}$ (I can show more detail of the computation if you need it), so:
$$\frac{p}{p^4+4}=\frac{\frac{1}{4}}{p^2-2p+2}-\frac{\frac{1}{4}}{p^2+2p+2}=\frac{\frac{1}{4}}{(p-1)^2+1}-\frac{\frac{1}{4}}{(p+1)^2+1}$$
We know that $\mathscr{L}\{e^{ax}f(x)\}=F(p-a), \mathscr{L}\{\sin{(ax)}\}=\frac{a}{p^2+a^2}, \mathscr{L}\{\cos{(ax)}\}=\frac{p}{p^2+a^2}$, so:
$$\mathscr{L}^{-1}\left\{\frac{\frac{1}{4}}{(p-1)^2+1} \right\}=\frac{1}{4}e^{x}\sin{x}$$
$$\mathscr{L}^{-1}\left\{\frac{\frac{1}{4}}{(p+1)^2+1} \right\}=\frac{1}{4}e^{-x}\sin{x}$$
Finaly:
$$\mathscr{L}^{-1}\left\{\frac{p}{p^4+4}\right\}=\frac{1}{4}e^{x}\sin{x}-\frac{1}{4}e^{-x}\sin{x}$$
$$\mathscr{L}^{-1}\left\{\frac{p}{p^4+4}\right\}=\frac{1}{4}\sin{x}(e^{x}-e^{-x})$$
$$\mathscr{L}^{-1}\left\{\frac{p}{p^4+4}\right\}=\frac{1}{4}\sin{x}(2\sinh{x})$$
$$\mathscr{L}^{-1}\left\{\frac{p}{p^4+4}\right\}=\frac{1}{2}\sin{x}\sinh{x}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2542569",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
}
|
General formula for the partial sums I m having trouble figuring out how to find the general formula for partial sums of the following two series.
*
*$\sum_{i=2}^n \frac{1}{i^2-1} = \frac{3}{4}-\frac{1}{2n}-\frac{1}{2(n+1)}$
*$\lim_{n\to\infty} \frac{4n^2 -n^3}{10+2n^3} = -\frac{1}{2} $
Both are in convergence and divergence section in the link at example-2 and example-5 respectively.
Let me know how those are being derived
thanks
|
Make it telescopic. For the first one, notice that:$$\frac{1}{i^2-1}=\frac{1}{2}(\frac{1}{i-1}-\frac{1}{i}+\frac{1}{i}-\frac{1}{i+1})$$
This is obtained by noting that $i^2-1$ in the denominator can be factored into two linear terms, namely $i+1$ and $i-1$. Then we assume we have a factorization in the form of:
$$\frac{1}{i^2-1}=\frac{A}{i-1}+\frac{B}{i+1}$$
This gives:
$$A(i+1)+B(i-1)=1$$
which has to hold for any value of $i$. Plug in $i=-1$ and $i=+1$ to see that $A=\frac{1}{2}$ and $B=-\frac{1}{2}$.
This gives
$$\frac{1}{i^2-1}=\frac{1}{2}\left(\frac{1}{i-1}-\frac{1}{i+1}\right)$$
Now this looks very close to our goal. We are just one step away. Add and subtract $\frac{1}{i}$ and you will obtain the desired telescopic form.
For the second one, factor out the highest power and use what you know about limits when $n$ goes to infinity. This is how it's done:
$$\lim_{n\to \infty}\frac{4n^2-n^3}{10+2n^3}=\lim_{n\to \infty}\frac{n^3(\frac{4}{n}-1)}{n^3(\frac{10}{n^3}+2)}=\frac{-1}{2}$$
Note that $n^3$ in the numerator and the denominator are cancelled and you are left with something that goes to zero except for the constant terms.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2543894",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
uncomplicated basis for $\mathbb Q(d)$ over $\mathbb Q$ where $d$ is a root of $x^4-14x^2+9$
Find an uncomplicated basis for $\mathbb Q(d)$ over $\mathbb Q$ where $d$ is a root of $x^4-14x^2+9$. Then a basis for $\mathbb Q(\sqrt{7+2\sqrt{10}})$ over $\mathbb Q$.
I thought the way to make a basis was to take the root of $d$ (which is $\pm\sqrt{7\pm2\sqrt{10}}$) and then these are the elements of the basis along with 1. So
{$1, \sqrt{7+2\sqrt{10}}, -\sqrt{7+2\sqrt{10}}, \sqrt{7-2\sqrt{10}}, -\sqrt{7-2\sqrt{10}}$} is our basis for $\mathbb Q(d)$.
I don't know how to find a basis of $\sqrt{7+2\sqrt{10}}$ then because I can't make any linear combinations.
|
You should note that, by denesting (justified by $7^2-40=3^2$),
$$
\sqrt{7+2\sqrt{10}}=\sqrt{\frac{7+3}{2}}+\sqrt{\frac{7-3}{2}}=\sqrt{5}+\sqrt{2}
$$
Since $(\sqrt{5}+\sqrt{2})^2=7+2\sqrt{10}$ and
$$
(\sqrt{5}+\sqrt{2})^3=
5\sqrt{5}+15\sqrt{2}+6\sqrt{5}+2\sqrt{2}=11\sqrt{5}+17\sqrt{2}
$$
we see that $\{1,\sqrt{2},\sqrt{5},\sqrt{10}\}$ is a basis, because these numbers are linear combinations of $\{1,d,d^2,d^3\}$ and conversely.
Note that, if $d$ is a root of an irreducible polynomial $f(x)$, then $\mathbb{Q}(d)$ is not, in general, generated by the roots of $f(x)$. Consider, for example, $f(x)=x^3-2$.
What's true is that, if $d$ is a root and $f(x)$ has degree $n$, then a basis is $\{1,d,d^2,\dots,d^{n-1}\}$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2548181",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
}
|
Finding the degree of minimal polynomial of a $10 \times 10$ matrix with entries $a_{ij}=1-(-1)^{i+j}$?
Q. What is the degree of minimal polynomial of a $10 \times 10$ matrix with entries $a_{ij}=1-(-1)^{i+j}$?
My approach : Let the matrix be denoted by $A$. Then $$A=
\left(\begin{matrix}
0 & 2 & 0 & 2 & 0 & 2 & 0 & 2 & 0 & 2 \\
2 & 0 & 2 & 0 & 2 & 0 & 2 & 0 & 2 & 0 \\
0 & 2 & 0 & 2 & 0 & 2 & 0 & 2 & 0 & 2 \\
2 & 0 & 2 & 0 & 2 & 0 & 2 & 0 & 2 & 0 \\
0 & 2 & 0 & 2 & 0 & 2 & 0 & 2 & 0 & 2 \\
2 & 0 & 2 & 0 & 2 & 0 & 2 & 0 & 2 & 0 \\
0 & 2 & 0 & 2 & 0 & 2 & 0 & 2 & 0 & 2 \\
2 & 0 & 2 & 0 & 2 & 0 & 2 & 0 & 2 & 0 \\
0 & 2 & 0 & 2 & 0 & 2 & 0 & 2 & 0 & 2 \\
2 & 0 & 2 & 0 & 2 & 0 & 2 & 0 & 2 & 0 \\
\end{matrix}\right)
$$
Some observations follow : $\text {Rank} A=2$, $\det A = 0$ and $\text {tr} A=0$.
Since $\det A=0$, there exists an eigenvalue which is $0$.
Since $\text {Rank} A=2$, number of non-zero eigenvalues can not exceed $2$.
Since $\text {tr} A=0$, number of non-zero eigenvalues are either $2$ or $0$.
Case 1. Number of non-zero eigenvalues is $2$: In this case they must be $\lambda$ and $-\lambda$ for trace should be zero. Thus the matrix is forced to be diagonalizable. This implies that the minimal polynomial is $P(x)=x(x-\lambda)(x+\lambda).$
Case 2. Number of non-zero eigenvalues is $0$: Here all diagonal entries of the Jordan form are $0$. We have either one $3\times 3$ jordan block or two $2\times 2$ Jordan blocks in order to preserve rank of $A$. Thus the minimal polynomial is $P(x)=x^3$ or $P(x)=x^2$ respectively.
However I checked that for even order matrices with order $\gt 2$ we have eigenvalues of the type we found in Case 1 above. That means for $\text {even $\times$ even}$ matrices, eigenvalues are always $0,\lambda,-\lambda$ for some $\lambda \neq 0$. How do I show this in order to discard the case 2 entirely?
|
Hint: Apply your matrix to the vector
$$\begin{pmatrix}1 \\ -1 \\ 1 \\ -1 \\ 1\\-1\\1\\-1\\1\\-1\end{pmatrix}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2549191",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
}
|
Find the number of solutions of $a_0+a_1+a_2=17$ if $2\le a_0\le 5$, $3\le a_1\le 6$, $4\le a_2\le 7$.
Find the number of solutions of $a_0+a_1+a_2=17$ if $2\le a_0\le 5$,
$3\le a_1\le 6$, $4\le a_2\le 7$.
This is an exmaple given by my professor, and his solution is:
The number of solutions to this equation satisfying the given
constraints is equal to the coefficient of $x^{21}$ in the expression
$$\left(\sum_{n=2}^5x^n\right)\left(\sum_{n=3}^6x^n\right)\left(\sum_{n=4}^7x^n\right)=x^9\left(\sum_{n=0}^3x^n\right)\left(\sum_{n=0}^3x^n\right)\left(\sum_{n=0}^3x^n\right).$$
This must be 3, since the coefficient of $x^8$ in $(1+x+x^2+x^3)^3$ is
3.
I understand actually nothing here. The expression of summation just pops up, and then comes the answer. What I only know that the idea of generating function is used here.
Can someone help explain how the steps in the solution come?
|
generating function.
This is another method that you may be interested.
$2\le a_0\le 5$, $3\le a_1\le 6$, $4\le a_2\le 7$.
$0\le a_0-2\le 5-2$, $0\le a_1-3\le 6-3$, $0\le a_2-4\le 7-4$.
$0\le b_0\le 3$, $0\le b_1\le 3$, $0\le b_2\le 3$.
$ a_0-2 = b_0, a_1-3 = b_1, a_2-4 = b_1$
$ a_0 =b_0+2, a_1 = b_1+3, a_2 = b_2+4$
$b_0+b_1+b_2+2+3+4 = 17$
$b_0+b_1+b_2 = 17-9 = 8$
Now $y_0 = 3-b_0, y_1 = 3-b_1, y_2= b_2 = 3-b_2, y_i's \gt 0$
$9-y_0+y_1+y_2 = 8$
$ y_0+y_1+y_2 = 1$
The number of solutions $= {(1+3-1)\choose(3-1)} = {3\choose2} = 3$
What I have given you is another solution using stars and bars.
$(x^2+x^3+x^4+x^5)(x^3+x^4+x^5+x^6)(x^4+x^5+x^6+x^7)$ is what your professor has shortened with summation signs.
How did you get this, the first expresssion in the bracket corresponds to values that $a_0$ can take. The way it is represented in generating function is $a_0=0$ is 1, $a_0=1$ is $x$, $a_0=2$ is $x^2$ and so on. $a_0$ can take values of {2,3,4,5} hence the first four terms of $x$ in the first bracket. Next a_1=3 is $x^3$, $a_1=4$ is $x^4$. $a_1=5$, is $x^5$ and $a_1=6$ is $x^6$. and hence the first four terms of x in the second bracket and now you can figure out how the thrid bracket of x terms are derived.
Now take $x^2$, $x^3$ and $x^4$ from all three bracketed expresssions what you then have is
$$x^2(1+x+x^2+x^3)\cdot x^3(1+x+x^2+x^3)\cdot x^4(1+x+x^2+x^3)=x^9(1+x+x^2+x^3)^3$$
What you are looking for is the coefficient of $x^{17}$, since we already took out $x^9$, we are looking for the coefficient of $x^8$ which is second term of the expansion of the expression in the brackets. if you expand $(1+x+x^2+x^3)^3$ using the binomial theorem, the equation becomes
$$(1+x+x^2+x^3)^3=\sum_{n=0}^3\binom{n}{3}1^n(x+x^2+x^3)^{3-n}$$
There can't be $x^8$ in $(x+x^2+x^3)^0$, $(x+x^2+x^3)$, and $(x+x^2+x^3)^2$, so $x^8$ appears only in $(x+x^2+x^3)^3$. Note that
$$(x+x^2+x^3)^3=(x+x^2+x^3)^2\cdot (x+x^2+x^3)=(x^6+2x^5+\dots)(x+x^2+x^3)$$
Hence, we obtain that the coefficient of $x^8$ is 3.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2550082",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
}
|
solution using synthetic geometry I managed to solve this problem only using complex numbers but I'd like to solve it using synthetic geometry and I can't.
Can someone help me to solve this problem using synthetic geometry?
Let $ABC$ an acute triangle with $AB > AC$ . Let $O$ its circumcenter and let $D$ the midpoint of $BC$. The circle of diameter $AD$ intersects again $AB$ and $AC$ in $E$ and in $F$, respectively. Let $M$ the midpoint of $EF$.
Prove that $MD$ is parallel to $AO$.
This is my solution. But, as I wrote above, I'd like to solve it using synthetic geometry and I can't.
Setting the origin of the plane in O and the points A, B and C on the circumference of unit radius, we have that
$ a \bar{a} = 1 \text{;} \ b \bar{b} = 1 \text{;} \ c \bar{c} = 1 $.
Because $ D $ is the midpoint of $ BC $, we can write
1) $ d = \dfrac{b + c}{2} $
and because $ AD $ is a diamtere of the new circle then said $ Q $ his midpoint we have
2) $ q = \dfrac{a + d}{2} = \dfrac{2a + b + c}{4}$
Said $ M_{1} $ the projection of $ Q $ on $ AB $, then
$ m_{1} = \frac{1}{2} \left[ \left( \dfrac{\bar{q} - \bar{a}}{\bar{b} - \bar{a}}\right) (b - a) + a + q \right] $
but
$\dfrac{1}{\bar{b} - \bar{a}} = \dfrac{1}{\dfrac{1}{b} - \dfrac{1}{a}} = \dfrac{ab}{a – b} $
and so
$ m_{1} = \frac{1}{2} \left[ \bar{q} ab (-1) - \dfrac{ab}{a} (- 1) + a + q \right] \Rightarrow $
$m_{1} = \frac{1}{2} \left( a + b + q - ab\bar{q} \right) $
In the same way, said $ M_{2} $ the projection of $ Q $ on $ AC $ we have
$ m_{2} = \frac{1}{2} \left[ \dfrac{\bar{q} - \bar{a}}{\bar{c} - \bar{a}} (c - a) + a + q \right] \Rightarrow $
$ m_{2} = \frac{1}{2} \left( a + c + q - ac\bar{q} \right) $
Because $ Q $ is the center of the new circle passing through $ A \text{,} D \text{,} E \text{,} F $ we have that $ M_{1}Q $ is axes of $ AE $ and that $ M_{2}Q $ in axes of $ AF $.
So we have that $ M_{1} $ is midpoint of $ AE $ and that $ M_{2} $ is midpoint of $ AF $.
So we can write
$ m_{1} = \dfrac{a + e}{2} \ \Rightarrow \ e = 2m_{1} – a $
and also
$ m_{2} = \dfrac{a + f}{2} \ \Rightarrow \ f = 2m_{2} - a$
The point $ M $ is defined as midpoint of $ EF $ so we have
$ m = \dfrac{e + f}{2} = \dfrac{2m_{1} + 2m_{2} - 2a}{2} = m_{1} + m_{2} - a$
therefore replacing $ m_{1} $ and $ m_{2} $ we have
$m = \frac{1}{2} a + \frac{1}{2} c + \frac{1}{2} q + \frac{1}{2} a + \frac{1}{2} b + \frac{1}{2} q - \frac{ab\bar{q}}{2} - \frac{ac\bar{q}}{2} - a \ \Rightarrow $
$m = \dfrac{b + c}{2} + q - a\bar{q} \dfrac{b + c}{2}$
We know, furthemore, that $ \bar{q} $ is:
3) $\bar{q} = \dfrac{2\bar{a} + \bar{b} + \bar{c}}{4} = \frac{1}{4} \left( \frac{2}{a} + \frac{1}{b} + \frac{1}{c} \right) = \dfrac{2bc + ab + ac}{4abc}$
Now we write the equation of the parallel line through $D$ parallel to $AO$:
let $z$ be a generic point of this line it is possible to write
4) $ \dfrac{z - d}{a - 0} = \dfrac{\bar{z} - \bar{d}}{\bar{a} – 0} $
Replacing $d$ with the expression of 1) and taking into account that $ \bar {a} = \frac{1}{a} $, we get
5) $z - \dfrac{b + c}{2} = a^{2} \left( \bar{z} - \dfrac{\bar{b} + \bar{c}}{2} \right) = a^{2}\bar{z} - \dfrac{a^{2} (\bar{b} + \bar{c})}{2}$
If the $ M $ point belongs to this line, $ m $ must satisfy equation 5). Substituting the value of $ m $ we have
$\dfrac{b + c}{2} + q - a \bar{q} \cdot \left( \dfrac{b + c}{2} \right) - \dfrac{b + c}{2} = a^{2}\left( \dfrac{\bar{b} + \bar{c}}{2} + \bar{q} - \bar{a}q \cdot \dfrac{\bar{b} + \bar{c}}{2} \right) - \dfrac{a^{2} (\bar{b} + \bar{c})}{2} \Rightarrow $
$ q - a \bar{q} \cdot \left( \dfrac{b + c}{2} \right) = a^{2}\bar{q} - \dfrac{aq( b + c)}{2bc} \ \Rightarrow $
$ q \left( 1 + \dfrac{ab + ac}{2bc} \right) = \bar{q}a \left( a + \dfrac{b + c}{2} \right)$
Substituting the values of $ q $ in 2) and the value of $ \bar{q} $ in 3) we have
$ \dfrac{(2a + b + c)(2bc + ab + ac)}{8bc} = \dfrac{a (2bc + ab + ac) (2a + b + c)}{a \cdot 8bc} $
which is, obviously, an identity, so $ MD $ is parallel to $ AO $, as we wanted to prove.
|
Considering two cases where only one of the two conditions is satisfied,
first let $ABC$ be an acute triangle but with $AB=AC$, as in the figure above. Since $MD$ and $AO$ are collinear, they are parallel in that they do not intersect.
Next, suppose $AB>AC$, but that the angle at $C$ is right. Then since $AD$ is the diameter of the lesser circle, $\angle AFD$ is right. But $\angle ACB$ is also right. Therefore $F$ and $C$ coincide, and in the triangle on base $EB$, which lies on $AO$ extended,$$\frac{FM}{ME}=\frac{CD}{DB}$$ making $MD$ again parallel to $AO$.
Alternatively, keeping $O$ within $\triangle ABC$ so that $\angle ACB$
is acute, and with $J$ as the midpoint of $AD$, if we move $C$ toward $A$ until $\angle CAB$ is right, then $CB$ is a diameter, so that $D$ coincides with $O$, and $M$ with $J$ and $H$. And since triangles $AME$ and $ADB$ are isosceles, then$$\angle DME=2\angle MAE$$and$$\angle CDM=2\angle DBA$$But $\angle DBA=\angle MAE$.
Therefore $$\angle CDM=\angle DME$$and $FE$ is parallel to $CB$.
Note that in all of these cases the angle between $CB$ and $FE$ equals $\angle OAD$ between the diameters. This is clearly $0^o$ in the first case, and in the second case (second figure) $\angle EAD$ and $\angle ECD$ at the circumference stand on common arc $ED$ and hence are equal. The alternative second case (third figure) is like the first case.
In cases where both conditions are satisfied, that is where $$\angle ABC<\angle ACB<90^o$$$EF$, $BC$ intersect at some point $G$, with $\angle BGE$ again equal to the angle of the diameters.
In the fourth figure, triangles $GDH$ and $DMH$ have the angle at $H$ in common.
And $$\angle GDH=\angle DMH$$ This was true in the first case above, where $\angle CDM=\angle DME$.
And in the second case, second figure, since$$\angle DCE=\angle DAE=\angle MDA$$then triangles $CDH$ and $DMH$ are similar, making $\angle CDH=\angle DMH$.
In the alternative second case (third figure) again $\angle CDM=\angle DME$.
Since in the general case (fourth figure) $\angle GDH=\angle DMH$, and the angle at $H$ is shared, then $$\angle DGH=\angle MDH$$
But $\angle DGH$ is equal to the angle of the diameters.
Therefore $$\angle MDH=\angle OAD$$ and $MD$ is parallel to $AO$.
[The foregoing is the nearest I get to formal synthetic proof so far.
Not sure about the validity here of using borderline cases to argue the general case. Suggestions welcome.]
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2550308",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
}
|
Let $\angle BAC =90 ^{\circ} AB=15 ,CD=10 ,AD=5$ Then $OA=?$
Let $\angle BAC =90 ^{\circ} AB=15 ,CD=10 ,AD=5$ Then $OA=?$
|
$A = (0,0)\\
B = (15,0)\\
C = (0,15)
D = (0,5)\\
O = (x,y)\\
(x,y)\cdot(x-15,y) = 0\\
x^2 + y^2 = 15x\\
(x,y-5)\cdot(x,y-15) = 0\\
x^2 + y^2 - 20y + 75 = 0\\
20y = 15 x + 75\\
y= \frac 34 x + \frac {15}{4}\\
x^2 + (\frac 34 x + \frac {15}{4})^2 = 15x\\
16x^2 + (3 x + 15)^2 = 240x\\
25x^2 + (90-240) x + 225 = 0\\
x^2 -6x + 9 = 0\\
(x-3)^2 = 0\\
x = 3\\
y = \frac 94 + \frac {15}{4} = 6\\
\|OA\| = \sqrt {3^2 + 6^2} = 3\sqrt 5$
If you have not learned about Euclidean inner products, we can do it with the Pythagorean theorem
$\|AO\|$ is the length of $AO$
$\|AO\| = \sqrt {x^2 + y^2}\\
\|OB\| = \sqrt {(15-x)^2 + y^2}\\
\|AO\|^2 + \|OB\|^2 = \|AB\|\\
x^2 + y^2 + x^2 - 30 x + 15^2 + y^2 = 15^2\\
x^2 + y^2 = 15 x$
and similarly
$\|DO\|^2 + \|OC\|^2 = \|DC\|\\
x^2 + (y-5)^2 + x^2 + (y-15)^2 = 10^2\\
2x^2 + y^2 - 10y + 25 y^2-30 y + 225 = 100\\
2x^2 + 2y^2 - 40y + 150 = 0\\
x^2 + y^2 - 20y + 75 = 0$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2551133",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
}
|
Find the real solutions for the system: $x^2+4xy-2y^2=5(x+y)$, $ 5x^2-xy-y^2=7(x+y)$
Find all real solutions for the system:
$$\left\{
\begin{array}{l}
x^2+4xy-2y^2=5(x+y)\\
5x^2-xy-y^2=7(x+y)\\
\end{array}
\right.
$$
This is from a math olympiad training book. No answer provided.
It is easy to spot that $(x,y)=(0,0)$ is a solution, by inspection. But I'm not being able to find other solutions, if they exist. I tried all tricks I know but was not getting to anything useful.
Hints and answers are appreciated. Sorry if this is a duplicate.
|
Note that $$\begin{align*}5\times\text{Eq.(2)}-7\times\text{Eq.(1)} \equiv 18x^2-33xy+9y^2=0 \\\equiv (y-3x)(3y-2x)=0\\\implies y=\frac{2x}{3} \text{ and } y=3x\end{align*}$$
Substituting $y=3x$ into $\text{Eq.(1)}$, we have, $$x^2+12x^2-2(9x^2)=5(4x) \implies x^2+4x=0 \\\implies \boxed{(x=0, y=0) \text{ and } (x=-4, y= -12)}$$
Similarly, $y=\frac{2x}{3}$ gives us, $$x^2+\frac{8x^2}{3}-\frac{8x^2}{9}=5(\frac{5x}{3}) \implies x^2-3x=0 \\\implies \boxed{(x=0,y=0) \text{ and } (x=3, y=2)}$$
Hope it helps.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2552252",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
}
|
How to find matrix exponential $e^A$ I have the matrix $$A =\begin{pmatrix} 0 & 1 \\ - 1 & 0 \end{pmatrix}$$
and I have to find $e^A$
I've found two complex-conjugate eigenvalues $\lambda_{1,2} = \pm i$
so substracting $\lambda_1 = i$ from the matrix's diagonal I got:
$$A_1 = \begin{pmatrix} -i & 1 \\-1 & i \end{pmatrix}$$
and therefore. to find eigenvector I have to solve the system:
$$A_1 = \begin{pmatrix} -i & 1 \\-1 & i \end{pmatrix} \begin{pmatrix} x_1 \\ x_2 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix}$$
so the first eigenvector is $h_1 = \begin{pmatrix}1 \\ i\end{pmatrix}$
and the second one is $h_2 = \begin{pmatrix}1 \\ -i\end{pmatrix}$
so the general solution is $$x(t) = C_1e^{it}\begin{pmatrix} 1 \\i\end{pmatrix} + C_2e^{-it}\begin{pmatrix} 1 \\-i\end{pmatrix}$$
I know that now I have to solve two Cauchy's problems for the standard basis $\mathbb{R}^2$ with vectors $v_1 = (1, 0)$ and $v_2 = (0,1)$ But I do not know how to approach it for complex numbers
|
Like here For any $a \in \mathbb{R}$ evaluate $ \lim\limits_{n \to \infty}\left(\begin{smallmatrix} 1&\frac{a}{n}\\\frac{-a}{n}&1\end{smallmatrix}\right)^{n}.$ Employing the Identification $$ 1 \equiv \left(\begin{matrix} 1& 0\\0&1 \end{matrix}\right)~~~\text{and}~~~~ i \equiv \left(\begin{matrix} 0& 1\\-1&0 \end{matrix}\right)=A.$$
we get, $$\color{red}{A^n = i^n ~~~and~~~~ e^A =\sum_{n=0}^{\infty} \frac{i^n}{n!} = e^i = \cos 1+i \sin 1 = \begin{pmatrix}\cos 1&\sin 1\\-\sin1&\cos1\end{pmatrix}.}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2552613",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
}
|
Let $a^{4} + b^{4} + a^{2} b^{2} = 60 \ \ a,b \in \mathbb{R}$ Then prove that : Let $a^{4} + b^{4} + a^{2} b^{2} = 60 \ \ a,b \in \mathbb{R}$ Then prove that :
$$4a^{2} + 4b^{2} - ab \geq 30$$
My attempt: :
$$4a^{2} + 4b^{2} - ab \geq 30 \\ 4(a^2+b^2)-ab \geq30 \\4(60-a^2b^2)-ab\geq30\\ 240-30\geq4(ab)^2+ab\\ 4(ab)^2+ab \leq210$$
Now what ?
|
we have to prove that $$a^2+b^2\geq \frac{15}{2}+\frac{ab}{4}$$
if $$\frac{15}{2}+\frac{ab}{4}<0$$ then is nothing to prove, in the other case we have
$$a^4+b^4+2a^2b^2\geq \frac{225}{4}+\frac{a^2b^2}{16}+\frac{15}{4}ab$$ for $$a^4+b^4$$ we Substitute $60-a^2b^2$ and we have to prove
$$60-a^2b^2+2a^2b^2\geq \frac{225}{4}+\frac{a^2b^2}{16}+\frac{15}{4}ab$$
rearranging and simplifying we get
$$a^2b^2-4ab+4\geq 0$$ which is $$(ab-2)^2\geq 0$$ which is true.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2552804",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 0
}
|
Evaluate $\int_0^a \sqrt{\frac{x^3}{a^3-x^3}} dx$. Evaluate following in terms of Gamma function:
$$\int_0^a \sqrt{\frac{x^3}{a^3-x^3}} dx$$
I don't know how to proceed. So, please tell the intuition behind the solution.
|
Assuming $a>0$,
$$\int_{0}^{a}\sqrt{\frac{x^3}{a^3-x^3}}\,dx \stackrel{x \mapsto az}{=} a\int_{0}^{1}z^{3/2}(1-z^3)^{-1/2}\,dz \stackrel{z \mapsto u^{1/3}}{=}\frac{a}{3}\int_{0}^{1}u^{-1/6}(1-u)^{-1/2}\,du $$
equals
$$\tfrac{a}{3}\,B\left(\tfrac{5}{6},\tfrac{1}{2}\right)=a \cdot\frac{\Gamma\left(\tfrac{5}{6}\right)\sqrt{\pi}}{\Gamma\left(\tfrac{1}{3}\right)}=\frac{2^{4/3}\pi a}{\sqrt{3}\,\Gamma\left(\tfrac{1}{3}\right)^3}=\frac{\text{AGM}(2,\sqrt{2+\sqrt{3}})}{2\pi\cdot 3^{1/4}}\cdot a $$
by Legendre duplication formula and the reflection formula. The last equality follows from the relation between $\Gamma\left(\frac{1}{3}\right)$ and an elliptic integral and allows a very efficient numerical evaluation of the LHS through the arithmetic-geometric mean.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2553744",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
}
|
Prove $5^n + 2 \cdot 3^n - 3$ is divisible by 8 $\forall n\in \mathbb{N}$ (using induction)
Prove $5^n + 2 \cdot 3^n - 3$ is divisible by 8 $\forall n\in \mathbb{N}$
Base case $n = 1\to 5 + 6 - 3 = 8 \to 8 \mid 8 $
Assume that for some $n \in \mathbb{N}\to 8 \mid 5^n + 2 \cdot 3^n - 3$
Showing $8 \mid 5^{n+1} + 2 \cdot 3^{n+1} - 3$
$$5^{n+1} + 2 \cdot 3^{n+1} - 3$$ $$5\cdot 5^n + 2\cdot 3\cdot 3^n - 3$$ $$ (5^n + 2\cdot 3^n - 3) + 4\cdot 5^n + 2\cdot 2\cdot3^n $$ $$ 5\cdot(5^n + 2\cdot 3^n - 3) + 4\cdot 5^n + 2\cdot 2\cdot3^n - 4\cdot(5^n + 2\cdot 3^n - 3)$$
$$ [5\cdot(5^n + 2\cdot 3^n - 3)] - [4\cdot 3^n - 12]$$
The first term divides by 8 but I am not sure how to get the second term to divide by 8.
|
If $f(m)=5^m+2\cdot3^m-3,$
Let use eliminate either of $5^n$ or $3^n$
Method$\#1:$
$$f(n+1)-5f(n)=3^n(6-10)-(3-3\cdot5)=4(3-3^n)$$
Now as $3^n$ is odd, for integer $n\ge0,3-3^n$ is even
$\implies f(n+1)-5f(n)$ is divisible by $8$
$\implies8|f(n+1)\iff8|f(n)$ as $8\nmid5$
Now establish the base case $f(0)$
Method$\#2:$
$$f(n+1)-3f(n)=?$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2554949",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 5
}
|
Finding: $\lim\limits_{x\to -\infty} \frac{6x^2+5\cos{\pi x}}{\sqrt{x^4+\sin{5\pi x}}}$
I'm running into problems with this limit:
$$\lim_{x\to -\infty} \frac{6x^2+5\cos{\pi x}}{\sqrt{x^4+\sin{5\pi x}}}$$
I've tried using l'Hospitals rule, however we will alway keep the $\cos(\pi x)$ expression, as well for $\sin(5\pi x)$. Also, terms do not cancel out with $\sin$ and $\pi$ since the product with $5$.
Can anyone give me a hint solving this?
Thanks in advance
Kind regards,
|
Hint: The functions sine and cosine are bounded. Therefore, even though they can oscillate wildly at infinity, they can be controlled.
This implies that in limit operations, we have the following rules of thumb that are correct if used appropriately:
$$1.\lim_{x\to \infty}(0 \times \sin(x)) = 0$$
$$2.\lim_{x\to \infty}(0 \times \cos(x)) = 0$$
$$3.\lim_{x\to \infty}(\frac{\sin(x)}{\infty}) = 0$$
$$4.\lim_{x\to \infty}(\frac{\cos(x)}{\infty}) = 0$$
In all of the above equalities, $0$ in the parentheses is meant to be taken as something infinitesimal. Note that $3.$ and $4.$ can be thought as special cases of $1.$ and $2.$ respectively.
Now factor out $x^2$ in the numerator and $x^4$ under the square root in the denominator and proceed.
If that hint is not enough, hover your mouse over the orange area:
$$\lim_{x \to -\infty}\frac{6x^2+5\cos\pi x}{\sqrt{x^4+5\sin 5\pi x}} = \lim_{x \to -\infty}\frac{x^2\left(6 + \frac{5\cos \pi x}{x^2}\right)}{\sqrt{x^4(1 + \frac{5\sin 5\pi x}{x^4})}}=\lim_{x \to -\infty}\frac{6 + \frac{5\cos \pi x}{x^2}}{\sqrt{1 + \frac{5\sin 5\pi x}{x^4}}}=6$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2555213",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
}
|
Why is it the case that if $a$ is a primitive root of $x^p=1$, then $\frac{x^p-1}{x-1}=(x-a^2)(x-a^4)...(x-a^{2(p-1)})=1+x+x^2+...+x^{p-1}$? If $p$ is an odd prime. Why is it the case that if $a$ is a primitive root of $x^p=1$, then $\frac{x^p-1}{x-1}=(x-a^2)(x-a^4)...(x-a^{2(p-1)})=1+x+x^2+...+x^{p-1}$? I can see why $\frac{x^p-1}{x-1}=1+x+x^2+...+x^{p-1}$, but I’m unsure why the even powers of the primitive root are the roots of such things.
|
Note that as $ a^p= 1$. So in the factorisation after $ p-1/2$ terms you have $ a^{p+1} ,a^{p+3} .... a^{2p-2} $ which are $ a^3,a^5,...a^{p-2} $ and thus the expression on R.H.S turns out to be $$ \prod_{i=1}^{p-1} ( x -a^i) $$
which then gives you the factorisation that comes from fact that it is a primitve root
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2565024",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
Sequence : $a_{n+1}=2a_n-a_{n-1}+2$ Let $c$ be a positive integer. The sequence $a_1, a_2, \ldots$ is defined by $a_1=1, a_2=c$ and
$a_{n+1}=2a_n-a_{n-1}+2$ for all $n \geq 2$.
Prove that for each $n \in \mathbb{N}$ there exists $k \in \mathbb{N}$ such that $a_na_{n+1} = a_k$.
My attempt :
Trying with small numbers, I see that $a_n=(n-1)c+(n-2)^2$ and will prove by induction.
$a_1 = 1 , a_2 = c+0 = c$ is true.
Suppose that $a_k=(k-1)c+(k-2)^2$ and $a_{k+1}=kc+(k-1)^2$ are true.
$a_{k+2} = 2a_{k+1}+a_k+2 = 2[kc+(k-1)^2]-[(k-1)c+(k-2)^2]+2=(k+1)c+k^2$
so $a_n=(n-1)c+(n-2)^2$ is true.
$a_na_{n+1} = [(n-1)c+(n-2)^2][nc+(n-1)^2] $
$= n(n-1)c^2 + (n-1)^3c + n(n-2)^2c + (n^2-3n+2)^2 = (k-1)c+(k-2)^2$
Please suggest how to solve this equation.
|
@Ian, the solutions to the quadratic are:
$-cn-n^2+3n$ and $cn+n^2-c-3n+4$ which are both integers. The second solution gives, for $n=1,2,\ldots$, the assignment $k(n)= 2, c+2, 2c+4$, which are all positive indices.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2568850",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
}
|
Wrong Wolfram Alpha result for $\lim_{(x,y)\to(0,0)}\frac{xy^4}{x^4+x^2+y^4}$?
I'm trying to solve this limit:
$$
\lim_{(x,y)\to(0,0)}\frac{xy^4}{x^4+x^2+y^4}
$$
Here's my attempt:
$$0 \le |\frac{xy^4}{x^4+x^2+y^4} - 0| = \frac{|x|y^4}{x^4+x^2+y^4},$$ and since $x^4+x^2 \ge0$ then $\frac{y^4}{x^4+x^2+y^4} \le 1$ so
$$
\frac{|x|y^4}{x^4+x^2+y^4} \le |x|,$$ so
$$
0 \le \lim_{(x,y)\to(0,0)}|\frac{xy^4}{x^4+x^2+y^4} - 0| \le \lim_{(x,y)\to(0,0)} |x| = 0,
$$ and using the squeeze theorem the limit is $0$.
But if I input the limit in wolfram alpha, it says that the limit doesn't exist. Here is the link to the limit in Wolfram Alpha.
|
You can also use polar coordinate by letting $x = r\cos \theta$ and $y = r\sin\theta$ and the limit becomes $$\lim_{r\to 0} \frac{r^5 \cos \theta \sin^4\theta}{r^4\cos^4\theta + r^2\cos^2\theta + r^4\sin^4\theta} = \lim_{r\to 0} r^3\left(\frac{\cos\theta\sin^4\theta}{r^2\cos^4\theta+\cos^2\theta + r^2\sin^4\theta}\right) = 0$$
Today's lesson is : Don't trust Wolframalpha entirely
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2571132",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
}
|
Simplify $\frac{7^{ \log_{5} 15 }+3^{2+\log_{5}7}}{7^{\log_{5}3}}$ I know that the result of this expression is 16 but how do I get to that result?
$$\frac{7^{ \log_{5} 15 }+3^{2+\log_{5}7}}{7^{\log_{5}3}}$$
|
We simplify the sum in two parts. First notice that $\displaystyle{\frac{7^{\log_5 15}}{7^{\log_5 5}} = 7^{\log_5 15 - \log_5 3} = 7^{\log_5 5} = 7}$. Next rewrite $7^{\log_5 3} = 3^{\log_3 7 \log_5 3}$, so the second part becomes $3^{2 + \log_5 7 - \log_3 7 \log_5 3}$, but $\log_3 7 \log_5 3 = \log_5 3^{\log_3 7} = \log_5 7$ so the second part is equivalent to $3^{2 + \log_5 7 - \log_5 7} = 9$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2571364",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 2
}
|
Why is function domain of fractions inside radicals not defined for lower values than those found by searching for domain of denominator in fraction? Consider function $y = \sqrt{\frac{1-2x}{2x+3}}$. To find the domain of this function we first find the domain of denominator in fraction:
$2x+3 \neq 0$
$2x \neq -3$
$x \neq - \frac{3}{2}$
So, the domain of $x$ (for fraction to be valid) is $x \in \left(- \infty, - \frac{3}{2}\right) \cup \left(- \frac{3}{2}, + \infty\right)$.
Then we find the domain for whole fraction:
$\frac{1-2x}{2x+3} \ge 0$
$1-2x \ge 0$
$-2x \ge -1$
$x \le \frac{1}{2}$
My textbook says that the (real) domain of the whole $y$ function is $x \in \left(- \frac{3}{2}, \frac{1}{2}\right]$.
I understand why the function is not defined in values larger than $\frac{1}{2}$ (because condition is $x \le \frac{1}{2}$), but I don't understand why it can't be have values less than $- \frac{3}{2}$ (because condition says only $x \neq - \frac{3}{2}$).
I checked the domain of this function and the domain given in the textbook is correct. Function has imaginary values for $x$ values less than $- \frac{3}{2}$ or bigger than $\frac{1}{2}$. It is undefined in $- \frac{3}{2}$.
Real values only in $\left(- \frac{3}{2}, \frac{1}{2}\right]$ domain.
|
Note that $\frac{1-2x}{2x+3} \geq 0$ implies $1-2x \geq 0$ only if $2x+3>0$.
You will get $1-2x\leq 0$ for $2x+3<0$, because what you're doing basically is multiplying $\frac{1-2x}{2x+3} \geq 0$ by a negative number.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2572000",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 4
}
|
How many even numbers of four distinct digits greater than 5000 are possible How many even numbers of four distinct digits greater than 5000 are possible? Please help me
The only thousand digit that are possible 5,6,7,8,9.
The only hundred digit that are possible are 1, 2, 3, 4, 5, 6, 7, 8, 9
The only ten digit that are possible are 1, 2, 3, 4, 5, 6, 7, 8, 9
The only unit digit that are possible are 2, 4, 6, 8
5x9x9x4=1620
|
THe first digit can be $5,6,7,8,9$. Those are four posibilities.
The second digit can be any of the ten $0,1,2..., 9$ (for some inexplicable reason you didn't include $0$) but the second digit must be different from the first. So there are $9$ options.
The third digit must be different from the first two so there are $8$ options.
The fourth must be different for the first three so so there are $7$ options.
The last digit must be even so it is $0,2,4,6,8$ and it must be different that the first four and ... we have no idea how many of the first four are even or not. So we are found in the Alps. Dang.
Start over.
Do two cases. Either the first number is even $6,8$ (2 options) or it is $5,7,9$ (3 options).
That last digit must be even so if the first is even then the second must be different so there are $4$ options because it must be different. If the first is not even there are $5$ options.
The second digit is different that the first or last so there are $8$ options.
The third digit must be different than the other three so there are $7$ options.
So if the first digit is even there are $2*4*8*7$. And if the first digit is not even there are $3*5*8*7$.
So there are $2*4*8*7 + 3*5*8*7 = (2*4 + 3*5)*8*7 = (8+15)*56 = 23*56= 1288$ such numbers.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2572290",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
}
|
Find an angle of a triangle on a larger triangle which cut through its midpoint
In triangle $\triangle BAC$ with $\angle ABC = 30\deg$. $D$ is the midpoint of $BC$. We join $A$ and $D$ and $\angle CDA = 45 \deg$. Find $\angle BAC$.
On applying Sine rule,
$$\frac{2x}{\sin {(15+\theta)}}=\frac{AC}{\sin 30}$$
and also
$$\frac{x}{\sin \theta}=\frac{AC}{\sin 45}$$
Where $x$ is $CD$ or $DB$ and $\theta$ is $\angle CAD$.
But solving this gives $$\frac{\sin {(15+\theta)}}{\sin \theta}=\sqrt 2$$
Is this correct?
|
Your reasoning looks good to me. Using your second equation, $$AC=\frac{x}{\sqrt{2}\sin{\theta}}$$ Now substituting $AC$ in the first equation, $$\frac{2x}{\sin{(15+\theta)}}=\frac{2x}{\sqrt{2}\sin{\theta}}$$ or $$\sin{(15+\theta)}=\sqrt{2}\sin{\theta}$$
Using trig identity, $$\cos15\sin\theta+\sin15\cos\theta=\sqrt{2}\sin{\theta}$$ Dividing by $\sin\theta$ we get $$\cot\theta=\frac{\sqrt{2}-\cos15}{\sin15}$$ Knowing that $\sin15=\frac{\sqrt{3}-1}{2\sqrt{2}}$, $\cos15=\frac{\sqrt{3}+1}{2\sqrt{2}}$ we get $\cot\theta=\sqrt{3}$, $\theta=30°$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2574815",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
}
|
Find all $\tau$ in $S_4$ such that $\tau (12)(34) \tau^{-1} = (12)(34)$ I know there should be 8 but I have only found these 6 so far, so I am missing 2:
$(1)$
$(1 2)$
$(3 4)$
$(12)(34)$
$(13)(24)$
$(14)(23)$
I know that if $\sigma$ is a $k$-cycle and $\tau\sigma\tau^{-1} = \sigma$ then $\tau$ is a power of $\sigma$. But what can we say about $\tau$ when sigma is a product of disjoint cycles?
|
In extended notation, the permutations $\tau\in S^4$ such that
$$\tau(1,2)(3,4)\tau^{-1} = (\tau(1),\tau(2))(\tau(3),\tau(4))=(1,2)(3,4)$$ clearly are
$$ \left(\begin{smallmatrix}1 & 2 & 3 & 4 \\ 1 & 2 & 3 & 4\end{smallmatrix}\right),\left(\begin{smallmatrix}1 & 2 & 3 & 4 \\ 1 & 2 & 4 & 3\end{smallmatrix}\right),\left(\begin{smallmatrix}1 & 2 & 3 & 4 \\ 2 & 1 & 3 & 4\end{smallmatrix}\right),\left(\begin{smallmatrix}1 & 2 & 3 & 4 \\ 2 & 1 & 4 & 3\end{smallmatrix}\right)$$
$$ \left(\begin{smallmatrix}1 & 2 & 3 & 4 \\ 3 & 4 & 1 & 2\end{smallmatrix}\right),\left(\begin{smallmatrix}1 & 2 & 3 & 4 \\ 3 & 4 & 2 & 1\end{smallmatrix}\right),\left(\begin{smallmatrix}1 & 2 & 3 & 4 \\ 4 & 3 & 1 & 2\end{smallmatrix}\right),\left(\begin{smallmatrix}1 & 2 & 3 & 4 \\ 4 & 3 & 2 & 1\end{smallmatrix}\right)$$
one just have to decide if the sets $\{1,2\}$,$\{3,4\}$ get switched or not, then if the action of the permutation on $\{1,2\}$ and $\{3,4\}$ is order-preserving or not. Three independent binary choices, hence eight chances.
The eight elements above form a non-abelian subgroup of $S_4$ with $8$ elements and exactly two elements of order $4$, thus a group isomorphic to the dihedral group $D_8$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2575058",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
}
|
How to prove non negativeness of $f(x)=1-\cos x-\frac{x^2}{20}$ I'm trying to prove that $f(x)=1-\cos x -\frac{x^2}{20}$, defined on $[-\pi, \pi]$, is a non negative function.
How do I prove that $f(x)\ge 0 $ for all $x \in [-\pi, \pi]$?
Hints?
|
HINT
Since
$$1-\frac{x^2}{2}\le \cos(x)\le1-\frac{x^2}{2}+\frac{x^4}{24}$$
Thus
$$f(x)=1-\cos x -\frac{x^2}{20}\geq 1-1+\frac{x^2}{2}-\frac{x^4}{24}-\frac{x^2}{20}=\frac{9x^2}{20}-\frac{x^4}{24}\geq 0 \quad x\in[-\pi,\pi]$$
Indeed
$$\frac{9x^2}{20}-\frac{x^4}{24}=x^2\left(\frac{9}{20}-\frac{x^2}{24}\right)\geq0 \iff x^2 \leq \frac{216}{20}=\frac{54}{5} \iff \\x\in \left[-\sqrt{\frac{54}{5}},\sqrt{\frac{54}{5}}\right]\implies x\in\left[-\pi,\pi\right]$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2575383",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
}
|
Solving a Matrix Equation with Maple I am attempting to use Maple to solve a matrix equation of the form $aB^2 + bB + cI = B^{-2}$, where $B$ is a $3 \times 3$ matrix, and I is the $3 \times 3$ identity matrix (i.e. to find the values a, b and c which satisfy the equation).
My idea was to use a piece of code like
A:=Matrix([[7,4,-2],[4,7,5],[2,-3,8]]);
d:=Vector([[8],[5],[2]]);
for matrix inversion to solve a system of three linear equations but that would assume that the right-hand side of the equation is a column vector when it is actually another $3 \times 3$ matrix. Is there a simple way to look at this which I am missing?
|
using new letters, there is the characteristic polynomial for $B.$ As $B$ is 3 by 3 and invertible, we have $r \neq 0$ in
$$ B^3 + p B^2 + q B + r I = 0. $$
When needed, this gives us the useful $$ B^3 = -p B^2 - q B - r I . $$
Next, we get
$$ B^4 = -p B^3 - q B^2 - r B = -p (-p B^2 - q B - r I ) - q B^2 - r B , $$
$$ B^4 = (p^2 - q) B^2 + (pq -r)B + prI. $$
Looks right. For you, these coefficients will be integers.
Also, you should confirm with the actual numbers that show up.
Back to $ B^3 + p B^2 + q B + r I = 0, $ we have $r \neq 0$ in
$$ B^3 + p B^2 + q B = -rI, $$
$$ B^2 + p B + q I = -r B^{-1}. $$ So
$$ B^{-1} = - \frac{1}{r} B^2 - \frac{p}{r} B - \frac{q}{r} I. $$
Nice rational coefficients, no guesswork. Finally, we get $B^{-2} = \left(B^{-1} \right)^2,$ which involves those coefficients, and will have $B^4$ and $B^3.$ All you do is substitute the expressions we found for $B^4$ and $B^3,$ add like terms, you get $B^{-2}$ in terms of $B^2, B,I.$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2576224",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
using a base 3 decimal to express as a base 10 fraction using geometric series
Express $0.\overline{21}_3$ as a base 10 fraction in reduced form.
So I was able to solve it by setting $x=\overline{.21}$, but the solution also briefly mentioned another way using the geometric series:
A quick way to get the answer by using the geometric series is: $(0.212121 \ldots)_3 = \frac{7}{9} + \frac{7}{81} + \frac{7}{729} + \dots = \frac{7}{8}.$
However, I'm having a hard time understanding how to actually use the geometric series (the above answer is not clear to me).
|
If you are not sure on how to compute $(0.2121\ldots)_{3}$, we have: $$\begin{align} E = (0.\color{red}{2}\color{green}{1}\color{red}{2}\color{green}{1}\ldots)_3 = \frac{1}{3}\times \color{red}{2} + \frac{1}{3^2}\times \color{green}{1} + \frac{1}{3^3}\times \color{red}{2} + \frac{1}{3^4}\times \color{green}{1} + \ldots \\ = \frac{2}{3} + \frac{1}{9} + \frac{2}{27} + \frac{1}{81} + \ldots \\ = \frac{7}{9}+ \frac{7}{81}+\ldots \end{align}$$
Now note that: $$\begin{align} E = \frac{7}{9}+\frac{7}{81}+\frac{7}{729}+\ldots \\ =\frac{7}{9}+\frac{7}{9}\times \frac{1}{9} + \frac{7}{9}\times [\frac{1}{9}]^2\ldots \\ =\frac79(1+\frac19 + [\frac19]^2+\ldots)\end{align}$$
Now, this is an infinite geometric progression. Can you compute the required sum?
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2576843",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
}
|
Solve $x^3 +y^3 + z^3 =57$ How can we solve $x^3 + y^3 + z^3 =57$ efficiently in a shorter way. $x$ $y$ and $z$ are integers. Given that modulus of $x$ $y$ and $z$ is less than or equal to five. We can of course do by hit and trial but what is the method of solving such questions. I actually stumbled upon this equation while solving a determinant. How to proceed. Pls help
|
Since $57$ is odd, we need $x+y+z$ odd.
Since $3\mid 57$ and $n\equiv n^3\bmod 3$, we need $3\mid x+y+z$.
Given that $-5\le x,y,z\le5$, it's relatively quick to eliminate $x+y+z=\pm 9$ and identify that we need $x+y+z=\pm 3$.
Then taking $x\ge y\ge z$ we must have $x\ge 3$ initially and then after considering a couple of cases we get $x\ge 4$, and we can also quickly eliminate $x=5$ leaving only $x=4$ to explore, which quickly leads to $y=1, z=-2$.
If all permutations are required then there are $3!=6$ arrangements of these values, of course.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2576947",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
}
|
Property of Medians and Cirumcircle Let $ABC$ be a non-isosceles triangle. Medians of $\triangle ABC$ intersect the circumcircle in points $L,M,N$. If $L$ lies on the median of $BC$ and $LM=LN$, then prove that $2a^2=b^2+c^2$.
My Attempt:
Let $G$ be the centroid of $\triangle ABC$ and $D$ be the mid-point of $BC$.
Since $LM=LN$, therefore $LM$ and $LN$ will subtend equal angles on the circumference of circumcircle
$\Rightarrow \angle GBL=\angle LCG$
$GL=GL$
Altitude of $\triangle BGL$ from $B$ to $GL$=Altitude of $\triangle LGC$ from $C$ to $GL$
Area($\triangle BGL$)=Area($\triangle LGC$)
Now, based on this ,can it be said that $\square GBLC$ is a parallelogram. If yes then
$$DL=GD=\frac{m_{a}}{3}$$
$\Rightarrow AD.DG=BD.DC$
$$m_{a}.\frac{m_{a}}{3}=\frac{a^2}{4}$$
$\Rightarrow 4m^2_{a}=3a^2$
$\Rightarrow 2b^2+2c^2-a^2=3a^2$
$\Rightarrow b^2+c^2=2a^2$
I am not sure whether this justification is sufficient(the one that I have written in bold). What more can be added to seal the issue
|
Let $\measuredangle LAC=\alpha_1$, $\measuredangle LAB=\alpha_2$, $\measuredangle MBA=\beta_1$, $\measuredangle MBC=\beta_2$, $\measuredangle NCB=\gamma_1$ and $\measuredangle NCA=\gamma_2$.
Thus,
$$\gamma_1+\alpha_2=\measuredangle NBM+\measuredangle BML=\measuredangle NML=\measuredangle MNL=\measuredangle MNC+\measuredangle LNC=\beta_2+\alpha_1,$$
which gives
$$\cos(\gamma_1+\alpha_2)=\cos(\beta_2+\alpha_1)$$ or
$$\cos\gamma_1\cos\alpha_2-\sin\gamma_1\sin\alpha_2=\cos\beta_2\cos\alpha_1-\sin\beta_2\sin\alpha_1$$ or by law of cosines and by law of sines
$$\frac{a^2+m_c^2-\frac{c^2}{4}}{2am_c}\cdot\frac{c^2+m_a^2-\frac{a^2}{4}}{2cm_a}-\frac{\frac{c}{2}\sin\beta}{m_c}\cdot\frac{\frac{a}{2}\sin\beta}{m_a}=$$
$$=\frac{a^2+m_b^2-\frac{b^2}{4}}{2am_b}\cdot\frac{b^2+m_a^2-\frac{a^2}{4}}{2bm_a}-\frac{\frac{b}{2}\sin\gamma}{m_b}\cdot\frac{\frac{a}{2}\sin\gamma}{m_a}$$ or
$$\frac{\left(a^2+\frac{1}{4}(2a^2+2b^2-c^2)-\frac{c^2}{4}\right)\left(c^2+\frac{1}{4}(2b^2+2c^2-a^2)-\frac{a^2}{4}\right)}{acm_c}-\frac{ac\sin^2\beta}{m_c}=$$
$$=\frac{\left(a^2+\frac{1}{4}(2a^2+2c^2-b^2)-\frac{b^2}{4}\right)\left(b^2+\frac{1}{4}(2b^2+2c^2-a^2)-\frac{a^2}{4}\right)}{abm_b}-\frac{ab\sin^2\gamma}{m_b}$$ or
$$\frac{(3a^2+b^2-c^2)(3c^2+b^2-a^2)}{4acm_c}-\frac{ac\left(\frac{2S}{ac}\right)^2}{m_c}=$$
$$=\frac{(3a^2+c^2-b^2)(3b^2+c^2-a^2)}{4abm_b}-\frac{ab\left(\frac{2S}{ab}\right)^2}{m_b}$$ or
$$\frac{(3a^2+b^2-c^2)(3c^2+b^2-a^2)-\sum\limits_{cyc}(2a^2b^2-a^4)}{cm_c}=$$
$$=\frac{(3a^2+c^2-b^2)(3b^2+c^2-a^2)-\sum\limits_{cyc}(2a^2b^2-a^4)}{bm_b}$$ or
$$\frac{a^4+c^4-b^4-4a^2c^2}{c\sqrt{2a^2+2b^2-c^2}}=\frac{a^4+b^4-c^4-4a^2b^2}{b\sqrt{2a^2+2c^2-b^2}}$$ or
$$(a^2+b^2+c^2)^2(b^2+c^2-2a^2)(b^2-c^2)(a+b+c)(a+b-c)(a+c-b)(b+c-a)=0$$ and we are done!
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2577452",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
}
|
Two fair coins are tossed until both turn up heads A penny and a dime are tossed together until both turn up heads, after which no more tosses are made. Find the expected number of times the penny comes up heads.
What I've tried:
Let $X$ and $Y$ be the number of times the penny and dime come up heads, respectively. Then, for $x = 1,2,3,\ldots$
$$P(X = x) = \sum_{y=1}^\infty P(X = x, Y = y) \\ = \sum_{y=1}^x P(X=x,Y=y) + \sum_{y=x+1}^\infty P(X=x, Y=y) \\ = \sum_{y=1}^x \sum_{n=x}^\infty \left( \frac{1}{4} \right) \binom{n-1}{x-1} \left( \frac{1}{2} \right)^{n-1} \binom{n-1}{y-1} \left( \frac{1}{2} \right)^{n-1} + \\ \sum_{y=x+1}^\infty \sum_{n=y}^\infty \left( \frac{1}{4} \right) \binom{n-1}{x-1} \left( \frac{1}{2} \right)^{n-1} \binom{n-1}{y-1} \left( \frac{1}{2} \right)^{n-1} \\ = \sum_{y=1}^x \sum_{n=x}^\infty \left( \frac{1}{4} \right)^n \binom{n-1}{x-1} \binom{n-1}{y-1} + \sum_{y=x+1}^\infty \sum_{n=y}^\infty \left( \frac{1}{4} \right)^n \binom{n-1}{x-1} \binom{n-1}{y-1}$$
Is there a way to simplify the above expression? Or is there an easier approach that I'm missing?
|
Imagine a clock that "ticks" only when the penny comes up heads. We need he expected number of ticks $T$ until the dime comes up heads. This follows a geometric distribution with success probability $\dfrac{1}{2}$. So $\mathbb{E}[T] = 2$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2577576",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
}
|
Why is my alternate method of calculating scalar products not working? The exercise is such: Given that $|\vec{a}| = 3$, $|\vec{b}| = 2$ and $\varphi = 60^{\circ}$ (the angle between vectors $\vec{a}$ and $\vec{b}$), calcluate scalar product $(\vec{a}+2\vec{b}) \cdot (2\vec{a} - \vec{b})$.
My initial thought was to solve the requested product by "sticking" separately calculated fragments together and then follow the definition of scalar product, which is
$$\vec{a} \cdot \vec{b} = |\vec{a}| \cdot |\vec{b}| \cdot \cos\varphi.$$
Via observation
$$\vec{a} \cdot \vec{a} = |\vec{a}| \cdot |\vec{a}| \cdot \cos0^{\circ} = |\vec{a}|^2 \Longrightarrow |\vec{a}| = \sqrt{\vec{a} \cdot \vec{a}}$$
we can get values of $|\vec{a}+2\vec{b}|$ and $|2\vec{a} - \vec{b}|$ (mind the notation: $a^2$ in this case represents $a^2 = |\vec{a}|^2$, and $b^2 = |\vec{b}|^2$):
\begin{align*}
|\vec{a}+2\vec{b}| &= \sqrt{(\vec{a}+2\vec{b}) \cdot (\vec{a}+2\vec{b})} =\sqrt{a^2 + 4|\vec{a}| \cdot |\vec{b}| \cdot \cos 60^{\circ} + b^2} = \\
&= \sqrt{3^2 + 4 \cdot 3 \cdot 2 \cdot \frac{1}{2}+2^2} = 5.
\end{align*}
Similarly,
\begin{align*}
|2\vec{a} - \vec{b}| &= \sqrt{(2\vec{a} - \vec{b}) \cdot (2\vec{a} - \vec{b})} = \sqrt{4a^2 - 4|\vec{a}| \cdot |\vec{b}| \cdot \cos 60^{\circ} + b^2} = \\
&=\sqrt{4 \cdot 3^2 - 4 \cdot 3 \cdot 2 \cdot \frac{1}{2} + 2^2} = 2\sqrt{7}.
\end{align*}
Now we plug both results in:
$$(\vec{a} + 2\vec{b}) \cdot (2\vec{a} - \vec{b}) = |\vec{a} + 2\vec{b}| \cdot |2\vec{a} - \vec{b}| \cdot \cos60^{\circ} = 5 \cdot 2\sqrt{7} \cdot \frac{1}{2} = 5\sqrt{7}.$$
However, this is not the right solution according to my textbook. The correct result is $19$. I thought about it for a bit and took a different route:
\begin{align*}
(\vec{a} + 2\vec{b}) \cdot (2\vec{a} - \vec{b}) &= \vec{a} \cdot 2\vec{a} - \vec{a} \cdot \vec{b} + 4 \vec{a} \cdot \vec{b} - 2\vec{b} \cdot \vec{b} = \\
&=2a^2 - |\vec{a}| \cdot |\vec{b}| \cdot \cos 60^{\circ} + 4 |\vec{a}| \cdot |\vec{b}| \cdot \cos 60^{\circ} - 2b^2 = \\
&= 2 \cdot 3^2 - 3 \cdot 2 \cdot \frac{1}{2} + 4 \cdot 3 \cdot 2 \cdot \frac{1}{2} - 2 \cdot 2^2 = 19.
\end{align*}
The second method clearly worked, while the first one failed miserably. But my question is why did my first approach fail? Did I get the wrong results when calculating $|\vec{a}+2\vec{b}|$ and $|2\vec{a} - \vec{b}|$? I have no idea. Please, help me understand my mistakes. Thank you in advance.
|
Note $$2\vec {b}\cdot 2\vec {b}=4b^2$$
Compare this with what you have in your first approach when calculating
$$
|\vec {a}+2\vec {b}|
$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2578792",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 0
}
|
Show that $f(x,y,z)=6x^2+4y^2+2z^2+4xz-4yz \geq 0$ for all $x, y, z$ except for $x=y=z=0$ When I try to factor the quadratic form, I end up with
$$6x^2+4y^2+2z^2+4xz-4yz = 2((x+z)^2+2x^2+(y-z)^2-z^2)$$
which does not ensure that $f(x,y,z) \geq 0$ for all $x, y, z \geq 0$ since the $z$ term is negative. How should these kinds of problems be tackled?
|
$$6x^2+4y^2+2z^2+4xz−4yz=(4x^2+z^2+4xz)+(4y^2+z^2−4yz)+2x^2=(2x+z)^2+(2y-z)^2+2x^2\ge 0$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2579074",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 2
}
|
Factoring $x^7+3x^6+9x^5+27x^4+81x^3+243x^2+729x+2187$
Question: How would you factor$$P(x)=x^7+3x^6+9x^5+27x^4+81x^3+243x^2+729x+2187$$
I thought for a while and realized that the coefficients are in powers of $3$, so $x=-3$ is a factor. Taking that factor out, we see that the septic is equal to$$P=(x+3)(x^2+9)(x^4+81)$$I'm wondering, however, if there is a quicker way to factor it because the original method was pretty tedious.
|
$$P(X)=x^7+3x^6+3^2x^5+3^3x^4+3^4x^3+3^5x^2+3^6x+3^7=\frac{x^8-3^8}{x-3}\\=\frac{(x^4-3^4)(x^4+3^4)}{x-3}=\frac{(x^2-3^2)(x^2+3^2)(x^4+3^4)}{x-3}=\frac{(x-3)(x+3)(x^2+3^2)(x^4+3^4)}{x-3}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2580330",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 1
}
|
How can I find the inverse of this infinite triangular matrix? I want to find the inverse of the following matrix
\begin{bmatrix}
1&0&0&0&0&\cdots&0&0&\cdots\\
0&1&0&0&0&\cdots&0&0&\cdots\\
\binom{2}{0}&0&1&0&0&\cdots&0&0&\cdots\\
\binom{4}{1}&\binom{2}{0}&0&1&0&\cdots&0&0&\cdots\\
\binom{6}{2}&\binom{4}{1}&\binom{2}{0}&0&1&\cdots&0&0&\cdots\\
\vdots&\vdots&\vdots&\vdots&\vdots&\ddots&\vdots&\vdots&\ddots\\
\binom{2(n-1)-2}{(n-1)-2}&\binom{2(n-1)-4}{(n-1)-3}&\binom{2(n-1)-6}{(n-1)-4}&\binom{2(n-1)-8}{(n-1)-5}&\binom{2(n-1)-10}{(n-1)-6}&\cdots&1&0&\cdots\\
\binom{2n-2}{n-2}&\binom{2n-4}{n-3}&\binom{2n-6}{n-4}&\binom{2n-8}{n-5}&\binom{2n-10}{n-6}&\cdots&0&1&\cdots\\
\vdots&\vdots&\vdots&\vdots&\vdots&\ddots&\vdots&\vdots&\ddots&
\end{bmatrix}
in order to get a different approach to the question
How can I find the general term of this recursive sequence?
|
Hmm, I'm not sure where your problem is.
Let's consider $L$ as the empty-matrix with all entries in the first principal subdiagonal set to $1$.
Then your matrix $M$ can be written as evaluation of the power series
$$ f(x)= x^0 + 0 \cdot x + \sum_{k=2}^\infty \binom{2k-2}{k-2}x^k \tag {1.1}$$
writing
$$ M= f(L) \tag {1.2} $$
The reciprocal of $f(x)$ can be found by
$$g(x)={1\over f(x)}= 1- (1 x^2+4 x^3 + 14 x^4 + 48 x^5 + 165 x^6 + 572 x^7 + ... +c_k x^k + ...) \tag {2.1}
$$
By this the inverse of $M$ should be definable by $g(x)$ leading to
$$ M^{-1}= g(L) = L^0 - (1 L^2 + 4 L^3 + 14 L^4 + 48 L^5 + 165 L^6 + 572 L^7 + ... +c_k L^k + ...) \tag {2.2}
$$
and I think the coefficients $c_k$ have been discussed already in your earlier question.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2581354",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
}
|
Is 2018 special because of these properties? I discovered that: $$2018=(6^2)^2+(5^2)^2+(3^2)^2+(2^2)^2$$
We also have: $$13^2+43^2=2018$$
And we have:
$$2018=44^2+9^2+1^2$$
I somehow tend to believe that there could be a finite number of these numbers that are sum of two squares, three squares and four fourth powers.
So we have a system of three Diophantine equations:
$$n=a^2+b^2$$
and $$n=c^2+d^2+e^2$$
and $$n=f^4+g^4+h^4+i^4$$
where, $n,a,b,c,d,e,f,g,h,i \in \mathbb N$.
Is there a finite number of these numbers?
Edit : Also, it is $$2018=35^2+26^2+8^2+7^2+2^2$$ a sum of five squares.
And of $$2018=11^3+7^3+7^3+1^3$$ four cubes.
|
Sum of 4 distinct fourth powers:
$$2018 =2^4 + 3^4 + 5^4 + 6^4$$
as well as sum of 12 consecutive squares:
$$2018 = 7^2 + 8^2 + 9^2 + 10^2 + 11^2 + 12^2 + 13^2 + 14^2 + 15^2 + 16^2 + 17^2 + 18^2$$
and "smallest number equal to the product of two primes which is also equal to the sum of 33 distinct primes": http://oeis.org/A102238
and the third least k > 0 such that the nextprime(k$\times$primorial(n)) - k$\times$primorial(n) is composite: http://oeis.org/A071771
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2586660",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 3,
"answer_id": 2
}
|
Evaluate lim$_{n\rightarrow\infty}$ $\frac{1-2+3-4+5-...............+\left(-2n\right)}{\sqrt{n^{2}+1}+\sqrt{n^{2}-1}}$
Evaluate $$\lim_{n\rightarrow\infty}\frac{1-2+3-4+5-\cdots+\left(-2n\right)}{\sqrt{n^{2}+1}+\sqrt{n^{2}-1}}.$$
My Approach
Let $A_{n}=1+3+5+\cdots\left(2n-1\right)$ and $B_{n}=2+4+6+\cdots\left(2n\right)$, then $A_n=\frac{n}{2}\left(1+2n-1\right)=n^{2}$
and
$B_{n}=\frac{n}{2}\left(2+2n\right)=n+n^{2}$.
Therefore,
$$\lim_{n\rightarrow\infty}\frac{A_{n}-B_{n}}{\sqrt{n^{2}+1}+\sqrt{n^{2}-1}}=\lim_{n\rightarrow\infty}\frac{-n}{\sqrt{n^{2}+1}+\sqrt{n^{2}-1}}=-\frac{1}{2}.$$
But the book mentions that the answer is $0$.
|
One may write
$$
1-2+3-4+5-\cdots+\left(-2n\right)=\underbrace{-\left(2-1\right)}_{\color{red}{-1}}\:\underbrace{-\left(4-3\right)}_{\color{red}{-1}}-\cdots\underbrace{-\left(2n-(2n-1)\right)}_{\color{red}{-1}}=\color{red}{-n}
$$ giving
$$
\frac{1-2+3-4+5-\cdots+\left(-2n\right)}{\sqrt{n^{2}+1}+\sqrt{n^{2}-1}}=\frac{-n}{n\sqrt{1+1/n^{2}}+n\sqrt{1-1/n^{2}}} \to -\frac12
$$ as $n\to \infty$.
Your result is correct to me.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2586747",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
}
|
A Question About A Calculation Of A Determinant.
Calculate $\det(A)$.
$$A =\begin{bmatrix}
a&b&c&d\\
-b&a&-d&c\\
-c&d&a&-b\\
-d&-c&b&a\\
\end{bmatrix}.$$
This is an answer on a book:
$$A A^T = (a^2+b^2+c^2+d^2) I.$$
$$\det(A) = \det(A^T).$$
$$\det(A)^2 = (a^2+b^2+c^2+d^2)^4.$$
The coefficient of $a^4$ in $\det(A)$ is $1$.
Therefore,
$$\det(A) = (a^2+b^2+c^2+d^2)^2.$$
I cannot understand the above answer.
It is obvious that $\det(A)$ is a multivariable polynomial $p(a,b,c,d)$.
Maybe, the following is true:
$p(a,b,c,d) = (a^2+b^2+c^2+d^2)^2$ for $(a,b,c,d) \in A$,
$p(a,b,c,d) = -(a^2+b^2+c^2+d^2)^2$ for $(a,b,c,d) \in B$,
where, $A \cup B = \mathbb{R}^4, A \cap B = \emptyset$.
Please prove this is impossible.
Thank you very much, Mr. Kavi Rama Murthy.
Proof:
$p(a,b,c,d)$ is a continuous function.
$(a,b,c,d)=(0,0,0,0)$ is the only solution for $p(a,b,c,d)^2 = (a^2+b^2+c^2+d^2)^4 = 0$.
so,
$(a,b,c,d)=(0,0,0,0)$ is the only solution for $p(a,b,c,d) = 0$.
$p(1,1,0,0) = \begin{vmatrix}
1&1&0&0\\
-1&1&0&0\\
0&0&1&-1\\
0&0&1&1\\
\end{vmatrix} = \begin{vmatrix}
1&1&0&0\\
0&2&0&0\\
0&0&1&-1\\
0&0&0&2\\
\end{vmatrix}
= 4 > 0$.
Let $(w_0, x_0, y_0, z_0) \in \mathbb{R}^4 - (0,0,0,0)$.
We can assume $w_0 \ne 0$ without loss of generality.
$f(x) := p(x, 1, 0, 0)$ is a one variable continuous function.
By the intermediate-value theorem, $f(w_0) = p(w_0, 1, 0, 0) > 0$.
$g(x, y, z) := p(w_0, x,y,z)$ is a three variable continuous funtion.
Again, by the intermediate-value theorem, $g(x_0, y_0, z_0) = p(w_0,x_0,y_0,z_0) > 0$.
Hence, $p(w, x, y, z) \geq 0$ for all $(w,x,y,z) \in \mathbb{R}^4$.
So, $p(w, x, y, z) = (w^2+x^2+y^2+z^2)^2$ for all $(w,x,y,z) \in \mathbb{R}^4$.
|
The book tells you that $A A^\top = x I$ with $x = (a^2+b^2+c^2+ d^2)$. It follows that
$$\det(A A^\top)= \det(x I) = x^4$$
But one has
$$\det(A A^\top) = \det(A)\det(A^\top) = (\det(A))^2 $$
Hence $\det(A)=\pm x^2$. The coefficient of $a^4$ shows that it is $+x^2$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2587255",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
}
|
Parametric equation with a trigonometric function in exponent I'm having trouble thinking of any solution or idea to solve this math problem. Any help will be appreciated.
$(a^2 - 1)*2^{-\sin^2 x} = a^2 - 4a + 3 $,
$a = ?$
The equation should have real solutions.
|
For $a=1$ the equation is satisfied for every $x$. Let's look for solutions different from $1$, so we can divide both sides by $a-1$, getting
$$
(a+1)2^{-{\sin^2x}}=a-3
$$
so
$$
2^{-{\sin^2x}}=\frac{a-3}{a+1} \tag{*}
$$
(note that $a=-1$ would lead to a contradiction).
Since $-1\le-{\sin^2x}\le0$, we have $1/2\le 2^{-{\sin^2x}} \le 1$.
Thus we need
$$
\frac{1}{2}\le \frac{a-3}{a+1}\le 1
$$
that is satisfied for $a\ge7$. Conversely, the equation (*) has solutions for every $a\ge7$.
How to find $a\ge7$? We need
$$
\frac{1}{2}\le \frac{a+1-4}{a+1}\le 1
$$
that is
$$
\frac{1}{2}\le 1-\frac{4}{a+1}\le 1
$$
that is
$$
0\le\frac{4}{a+1}\le\frac{1}{2}
$$
Therefore $a+1>0$ and $8\le a+1$, that is, $a\ge7$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2591702",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
}
|
Roots of $6z^5+5z^4+4z^3+3z^2+2z+1$ By All the zeroes of $p(z)$ lie inside the unit disk, I know there all roots are inside the unit disk. How can I remove the boundary. Wolfram alpha tells me there all there are no solutions on the unit circle. Furthermore, how can I prove the e.g. all roots are inside the circle with radius 0.9?
(Basically, this is a transition function of a system, I want to prove this system is stable.)
|
Multiply the given polynomial by $(1-z)^2$ to get $f(z)=1-7z^6+6z^7$. This is the characteristic polynomial of the matrix
$$P=\begin{pmatrix}0&0&0&0&0&0&-\tfrac{1}{6}\\
1&0&0&0&0&0&0\\
0&1&0&0&0&0&0\\
0&0&1&0&0&0&0\\
0&0&0&1&0&0&0\\
0&0&0&0&1&0&0\\
0&0&0&0&0&1&\tfrac{7}{6}\\\end{pmatrix}$$
Let $M$ be the maximum modulus of the roots of your polynomial. We have
$$ 5M^n\geq \left|\text{Tr}(P^n)-2\right| $$
so $M\geq \frac{2}{3}$ by considering $n=13$. Let us consider $g(z)=z^7-7z+6$, characteristic polynomial of the matrix
$$Q=\begin{pmatrix}0&0&0&0&0&0&-6\\
1&0&0&0&0&0&7\\
0&1&0&0&0&0&0\\
0&0&1&0&0&0&0\\
0&0&0&1&0&0&0\\
0&0&0&0&1&0&0\\
0&0&0&0&0&1&0\\\end{pmatrix}$$
and let $N$ be the maximum modulus of the roots of $g(z)$. By a similar argument, $N\geq\frac{4}{3}$, so all the roots of the given polynomial lie in the annulus $$\frac{2}{3}\leq |z|\leq\frac{3}{4}.$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2592267",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
}
|
Seems Simple, forgetting some fundamentals - $\frac{1.6}{3.01} = \frac{x}{1000+x}$ I am having trouble remembering/finding simple source to review a few fundamentals.
I know I need to factor and try getting x alone. Though not able to recreate answer. Seems it would have multiple roots. Would be great to see the proper steps to solve.
$$\frac{1.6}{3.01} = \frac{x}{1000+x}$$
|
It's much easier than you think.
$\frac ab = \frac cd \iff ad = bc; b\ne 0; d\ne 0$.
So
$\frac{1.6}{3.01} = \frac{x}{1000+x}\implies$
$(1000 + x)\frac{1.6}{3.01} = (1000 + x) \frac{x}{1000+x}\implies$
$(1000+x)\frac{1.6}{3.01}= x \implies$
$(1000+x)\frac{1.6}{3.01}*3.01 = x*3.01 \implies$
$1.6(1000 + x) = 3.01 x$
Just solve that.
(i.e.
$1600 + 1.6 x = 3.01 x$
$1600 = 3.01 x - 1.6 x = 1.41 x$
$x = \frac {1600}{1.41} $
=====
If you want to be pervese:
$\frac {1.6}{3.01} = \frac x{1000 + x}$
$\frac {160}{301} = \frac 1{\frac {1000}x + 1}$
$\frac {301}{160} = \frac {1000}x + 1$
$1\frac {141}{160}-1 = \frac {1000}x$
$\frac {141}{160} = \frac {1000}x$
$\frac {160}{141} = \frac x{1000}$
$\frac {160000}{141} = x$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2593793",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
}
|
Find the range of $f(x)=11\cos^2x+3\sin^2x+6\sin x\cos x+5$ I'm trying to solve this problem.
Find the range of $f(x)=11\cos^2x+3\sin^2x+6\sin x\cos x+5$
I have simplified this problem to $$f(x)= 8\cos^2x+6\sin x\cos x+8$$ and tried working with $g(x)= 8\cos^2x+6\sin x\cos x$.
I factored out the $2\cos x$ and rewrote the other factor as a linear combination of cosine.
It reduces down to $$g(x)=10\cos x\cos\left(x-\tan^{-1}\frac{3}{4}\right)$$
But then I'm stuck here. Please help me. Perhaps there's a different way to approach this?
|
Rewrite as following
$$\begin{align}
f(x) & = 8 + 6 \sin(x ) \cos(x ) + 8\cos^2(x) - 4 + 4 \\
&= 12 + 3 \sin(2x) + 4 \cos(2x) \\
&= 12 + 5 \left(\frac{3}{5} \sin(2x) + \frac{4}{5} \cos(2x)\right) \\
&= 12 + 5 \sin(2x + \arctan{\tfrac{4}{3}})
\end{align}$$
Now its easy since $\sin(...)$ always lies in $[-1,1]$, max/min values are $12 \pm 5$.
So maximum value is $17$ and minimum is $7$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2594442",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 0
}
|
How did they get this equation comparing three ratios? I was reading from an old maths textbook. It was giving some examples on how to solve ratios. I stumbled upon this example and felt perplexed after reading only part of it.
We're given this equation.
$$\frac{x}{l(mb+nc-la)} = \frac{y}{m(nc+la-mb)} = \frac{z}{n(la + mb - nc)}$$
And asked to prove that
$$\frac{l}{x(by + cz - ax)} = \frac{m}{y(cz+ax-by)} = \frac{n}{z(ax + by -cz)}$$
He starts by doing this:
$$\frac{\frac{x}{l}}{mb + nc - la} = \frac{\frac{y}{m}}{nc + la - mb} = \frac{\frac{z}{n}}{la + mb -nc}$$
Which I understand. Then, he goes on to say this:
We have $$\frac{\frac{x}{l}}{mb + nc - la} = \frac{\frac{y}{m}}{nc + la - mb} = \frac{\frac{z}{n}}{la + mb -nc}$$
$$= \frac{\frac{y}{m} + \frac{z}{n}}{2la}$$
These are similar expressions.
$$\therefore \frac{ny + mz}{a} = \frac{lz + nx}{b} = \frac{mx + ly}{c}$$
This is the portion of the proof that I don't understand. How did he go from
$= \frac{\frac{y}{m} + \frac{z}{n}}{2la}$ to $\frac{ny + mz}{a} = \frac{lz + nx}{b} = \frac{mx + ly}{c}$? And, also, what does he mean by these are "similar expressions."
The textbook I'm reading is called Higher Algebra a Sequel to Elementary Algebra for Schools by Henry Sinclair and Samuel Ratcliff Knight.
Thanks for the help.
|
The trick is that
$$\frac{a}{b}=\frac{c}{d} \implies \frac{a}{b}=\frac{c}{d}=\frac{a+c}{b+d}$$
$$A=\frac{\frac{x}{l}}{mb + nc - la} = \frac{\frac{y}{m}}{nc + la - mb} = \frac{\frac{z}{n}}{la + mb -nc}$$
Use each two terms
$$A = \frac{\frac{x}{l} + \frac{y}{m}}{2nc} = \frac{\frac{x}{l} + \frac{z}{n}}{2mb} = \frac{\frac{y}{m} + \frac{z}{n}}{2la}$$
Then you arrive naturally at the end.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2594588",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 3,
"answer_id": 2
}
|
Solve $y^{(4)}-2y^{(3)}+2y'-y=xe^x.$ Solve $y^{(4)}-2y^{(3)}+2y'-y=xe^x.$
The characteristic equation is $(r-1)^3(r+1)\Rightarrow y_h=(C_1+C_2x+C_3x^2)e^x+C_4e^{-x}.$
The problem is the particular equation. Why doesn't it work with the ansatz
$$y=(ax+b)e^x?$$
I get
\begin{array}{lcl}
y & = & e^x(ax+b) \\
y' & = & e^x(a(x+1)+b)\\
y'' & = & e^x(a(x+2)+b) \\
y^{(3)} & = & e^x(a(x+3)+b) \\
y^{(4)} & = & e^x(a(x+4)+b)
\end{array}
Setting these in i get $$e^x[((a(x+4)+b))-2((a(x+3)+b))+2(e^x(a(x+1)+b))-(e^x(ax+b))] = e^x\cdot 0=0.$$
So this doesn't work. Is it because I already have corresponding powers of $x$
in the homogenous solution? How can i fix my ansatz?
|
The particular solution is $$y*=\frac{1}{(D-1)^3(D+1)}xe^x=e^x\frac{1}{D^3(D+2)}x=e^x\frac{1}{2D^3(1+\frac{D}{2})}x=e^x\frac{1}{2D^3}(1-\frac{D}{2}+...)x$$
$$=e^x\frac{1}{2D^3}(x-\frac{1}{2})=e^x\frac{1}{2}\int{\int{\int{(x-\frac{1}{2})dx}dx}dx}=\frac{e^x}{2}(\frac{x^4}{24}-\frac{x^3}{12})=e^x(\frac{x^4}{48}-\frac{x^3}{24}) $$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2594875",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 1
}
|
Geometric proof for $\tan{(3x)}=\tan{(x)}\tan{\left(\frac{\pi}{3}-x\right)}\tan{\left(\frac{\pi}{3}+x\right)}$ Are there geometric proofs for the identitity
$$\tan{(3x)}=\tan{(x)}\tan{\left(\frac{\pi}{3}-x\right)}\tan{\left(\frac{\pi}{3}+x\right)}$$
My try:
Thank in advances.
|
I have another simple proof.
Assume $\angle AFD = \frac{\pi}{3}$, $\angle FAD = 4x$, and $\angle FDA = 4y$. Obviously, $y = \frac{\pi}{6}-x$, $\textit{i.e.}$, $x+y = \frac{\pi}{6}$.
$AQ$ bisects $\angle FAD$, and $AC$ bisects $\angle FAQ$.
Also, $DP$ bisects $\angle FDA$, and $DB$ bisects $\angle FDP$.
And $R = AQ \cap DP$, $H = AQ \cap DB$, $G= AC \cap DP$, $E = AC \cap DB$.
Then, $\angle AED = \angle AFD + \angle FAE + \angle FDE =
\frac{\pi}{3} + x + y = \frac{\pi}{2}$. Similarly, $\angle ARD = \frac{2\pi}{3}$.
$\text{area of } \triangle ADP =\frac{1}{2} AR \cdot DP \sin \frac{2\pi}{3}
= \frac{1}{2} AP \cdot DF \sin \frac{\pi}{3}$.
Thus, $AR\cdot DP = AP \cdot DF$.
It means that $AR:AP = DF:DP$.
We know that $AR:AP=RG:GP$, and $DF:DP=FB:BP$. Thus $RG:GP = FB:BP$. Therefore $RF \parallel GB$.
Because $R$ is the incenter of $\triangle AFD$, $\angle RFP = \frac{\pi}{6}$.
Hence, $\angle GBP = \frac{\pi}{6}$.
Thus, $\angle BGE = \angle BAG + \angle ABG = \frac{\pi}{6} + x$, and
$\angle EBC = \angle EBG = \frac{\pi}{2} - \angle BGE = \frac{\pi}{3} - x$.
Additionally, $\angle ECD = \frac{\pi}{2} - y = \frac{\pi}{2} - \left(\frac{\pi}{6} -x \right) = \frac{\pi}{3} + x$. Therefore, $\square ABCD$ satisfies the condition of problem, and we showed that $\angle EAD = 3x$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2596954",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
}
|
Product of length of segments in Ellipse.
If the normal at any point $P$ on the Ellipse $\dfrac{x^2}{a^2}+\dfrac{y^2} {b^2}=1$ meets the axis in $G$ and $g$ respectively, then find $PG\cdot Pg$, in terms of $a$ and $b$.
I tried considering the parametric point as $(a\cos\theta,b\sin\theta)$ on the ellipse, then constructed both tangent and normal at that point then found the coordinated where normal cut the axes, then found the length of the perpendicular from these coordinates on the tangent (at $(a\cos\theta,b\sin\theta)$), but I end up with following: $$PG\cdot Pg=a^2b^2(a^2\sin^2\theta+b^2\cos^2\theta)$$
please help.
|
Let $P(u,v)$.
Thus, $$\frac{xu}{a^2}+\frac{yv}{b^2}=1$$ is an equation of the tangent to ellipse.
Thus the slope of the normal it's $\frac{va^2}{b^2u}$ and the equation of the normal it's
$$y-v=\frac{va^2}{b^2u}(x-u),$$ which gives
$$g\left(0,v\left(1-\frac{a^2}{b^2}\right)\right)$$ and
$$G\left(\left(1-\frac{b^2}{a^2}\right)u,0\right),$$
which gives $$PG=\sqrt{\frac{b^4u^2}{a^4}+v^2}$$ and $$Pg=\sqrt{u^2+\frac{a^4v^2}{b^4}}.$$
Thus, $$PG\cdot Pg=\frac{a^4v^2+b^4u^2}{a^2b^2}$$ and since $$\frac{u^2}{a^2}+\frac{v^2}{b^2}=1$$ only, we see that $PG\cdot Pg$ depend on $P$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2597836",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
}
|
When are these eigenvalues non-negative? I'm trying to find a pair of real numbers $(a,b)$ which ensure that some matrix is strictly positive semi-definite.The eigenvalues of this matrix are $$\lambda=1 + a \pm \sqrt{(c+b)^2+2(x'-ax)^2}$$ and $$\lambda=1 - a \pm \sqrt{(c-b)^2+2(x'-ax)^2}.$$
I therefore need one of the eigenvalues to be zero and the rest non-negative.
For a fixed $c,x,x' \in \mathbb R$, what is the best way to determine some $a,b$ exist? If not, are there conditions on $x,x',c$ so that $a,b$ exist? Here I know that $0\leq x,x' \leq 1/\sqrt2$ and $0 \leq c \leq 1$.
|
I will use the notations $X=\sqrt{2}x$ and $X^\prime=\sqrt{2}x^\prime$ as it makes every expression simpler.
I will prove the following:
If $c^2 +{X^\prime}^2\leq 1$, then:
*
*Pick any $b$ in the interval $\left[0, \sqrt{1-{X^\prime}^2}-c\right]$
*If $X\neq 1$ put $a=\dfrac{-\left(1+XX^\prime\right) +\sqrt{\left(X+X^\prime\right)^2+\left(1-X^2\right)\left(c+b\right)^2}}{1-X^2}$
*If $X=1$ put $a=\dfrac{(c+b)^2+{X^\prime}^2-1}{2\left(1+X^\prime\right)}$
(a, b) is a solution and the vanishing eigenvalue is $1 + a - \sqrt{(c+b)^2+(X'-aX)^2}$
Note that the smallest of the four eigenvalues is either $$\lambda_1=1 + a - \sqrt{(c+b)^2+(X'-aX)^2}\qquad or\qquad \lambda_2=1 - a - \sqrt{(c-b)^2+(X'-aX)^2}$$
Your problem is equivalent to finding $a,b$ such that one of these situations occurs:
*
*$0=\lambda_1\leq\lambda_2$
*$0=\lambda_2\leq\lambda_1$.
I'll focus on the first situation.
First, $0=\lambda_1 $ is equivalent to
$$a+1\geq 0 \qquad and \qquad a^2 \left(1-X^2\right)+2a\left(1+XX^\prime\right)+\left(1-{X^\prime}^2-(c+b)^2\right)=0$$
When $X=1$ (the largest allowed value) this is a degree $1$ equation whose unique solution $$a=\frac{(c+b)^2+{X^\prime}^2-1}{2\left(1+X^\prime\right)}$$
is easily seen to satisfy $a+1\geq 0$ as required.
When $X\neq 1$, you need $\left(1+XX^\prime\right)^2-\left(1-X^2\right)\left(1-{X^\prime}^2-(c+b)^2\right)$ to be non-negative (we assume this for now and will look at it closer when we will be looking for $b$), there are then two solutions for $a$:
$$a=\dfrac{-\left(1+XX^\prime\right) \pm\sqrt{\left(1+XX^\prime\right)^2-\left(1-X^2\right)\left(1-{X^\prime}^2-(c+b)^2\right)}}{1-X^2}$$
but since $a+1$ needs to be nonnegative, there is hope only for the solution with "$+$" and the condition $a+1\geq 0$ becomes
$$\left(1+XX^\prime\right)^2-\left(1-X^2\right)\left(1-{X^\prime}^2-(c+b)^2\right)\geq \left(X^2+XX^\prime\right)^2 $$
since $u\leq v$ and $u^2\leq v^2$ are equivalent if $u$ and $v$ are known to be non-negative.
After crossing out the nonnegative common factor $\left(1-X^2\right)$ this simplifies into $$(X+X^\prime)^2+(c+b)^2\geq 0$$ which is always true.
So far, we have shown that $0=\lambda_1$ is equivalent to:
*
*When $X=1$, $$a=\frac{(c+b)^2+{X^\prime}^2-1}{2\left(1+X^\prime\right)}$$
*When $X\neq 1$, $$a=\dfrac{-\left(1+XX^\prime\right) +\sqrt{\left(X+X^\prime\right)^2+\left(1-X^2\right)\left(c+b\right)^2}}{1-X^2}$$
Now let's have a look at $\lambda_1\leq\lambda_2$. Because of all the square roots, it would be extremely painful to solve exactly. However there are some ranges of parameters that allow us to make this easy.
Note that $\lambda_1\leq\lambda_2$ has the form $$2a\leq \text{(some difference of square roots)}$$
I will solve $$2a\leq 0\leq \text{(that difference of square roots)}$$
First, $a\leq 0$ is equivalent to $$b\leq \sqrt{1-{X^\prime}^2}-c$$
It is easy to check in the case $X=0$, and not too difficult in the case $X\neq 0$.
Second, the difference of square roots on the right is $$\sqrt{(c+b)^2+(X'-aX)^2} - \sqrt{(c-b)^2+(X'-aX)^2}$$ and since $c$ is nonnegative, this difference is nonnegative if and only if $b\geq 0$. Finding $b$ such that $0\leq b\leq \sqrt{1-{X^\prime}^2}-c$ is possible if and only if $c\leq \sqrt{1-{X^\prime}^2}$.
The announced result follows.
The result was double-checked by computer.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2599156",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
}
|
Proving a formula for distributing $n$ objects into $r$ non-empty boxes / formula for number of onto functions Distribution of $n$ distinct objects into $r$ different boxes if empty boxes are not allowed or in each box at least one object is put is:
$$r^n - \dbinom{r}{1}(r-1)^n+ \dbinom{r}{2}(r-2)^n- \dbinom{r}{3}(r-3)^n + \cdots + (-1)^{r-1}\dbinom{r}{r-1}·1$$
It is a really daunting formula. I do not know how to prove it. It is given that this formula can be proved using principle of inclusions and exclusion and using set theory, but how?
Edit: This formula is also used to find the number of onto functions from a $A$ with $|A|=n$ to $B$ with $|B|=r$.
I would like to see the proof using elementary combinatorics and principle of inclusion and exclusion.
|
$\def\peq{\mathrel{\phantom{=}}{}}$For each $1 \leqslant k \leqslant r$, denote by $A_k$ the set of scenarios in which the $k$-th box is empty. The quantity to be found is$$
|\overline{A_1} \cap \cdots \cap \overline{A_r}| = |S| - |A_1 \cup \cdots \cup A_r|,
$$
where $S$ is the set of all scenarios.
Now, by inclusion and exclusion principle,\begin{align*}
|A_1 \cup \cdots \cup A_r| &= \sum_{1 \leqslant k \leqslant r} |A_k| - \sum_{k_1 < k_2} |A_{k_1} \cap A_{k_2}| + \cdots + (-1)^{j - 1} \sum_{k_1 < \cdots < k_j} |A_{k_1} \cap \cdots \cap A_{k_j}|\\
&\peq + \cdots + (-1)^{r - 1} |A_1 \cap \cdots \cap A_r|. \tag{1}
\end{align*}
Because for any $k_1 < \cdots < k_j$, the set $A_{k_1} \cap \cdots \cap A_{k_j}$ contains exactly all the scenarios in which the $k_1$-th, …, $k_j$-th boxes are empty, then$$
|A_{k_1} \cap \cdots \cap A_{k_j}| = (r - j)^n.
$$
Also, there are $\displaystyle\binom{r}{j}$ ways to select $j$ boxes from these $r$ boxes, thus\begin{align*}
(1) &= \sum_{1 \leqslant k \leqslant r} (r - 1)^n - \sum_{k_1 < k_2} (r - 2)^n + \cdots + (-1)^{j - 1} \sum_{k_1 < \cdots < k_j} (r - j)^n + \cdots + (-1)^{r - 1} · 0^n\\
&= (r - 1)^n \sum_{1 \leqslant k \leqslant r} 1 - (r - 2)^n \sum_{k_1 < k_2} 1 + \cdots + (-1)^{j - 1} (r - j)^n \sum_{k_1 < \cdots < k_j} 1 + \cdots + (-1)^{r - 1} · 0^n\\
&= (r - 1)^n \binom{r}{1} - (r - 2)^n \binom{r}{2} + \cdots + (-1)^{j - 1} (r - j)^n \binom{r}{j} + \cdots + (-1)^{r - 1} · 0^n,
\end{align*}
and\begin{align*}
&\peq |\overline{A_1} \cap \cdots \cap \overline{A_r}| = |S| - |A_1 \cup \cdots \cup A_r|\\
&= r^n - \left( (r - 1)^n \binom{r}{1} - (r - 2)^n \binom{r}{2} + \cdots + (-1)^{j - 1} (r - j)^n \binom{r}{j} + \cdots + (-1)^{r - 1} · 0^n \right)\\
&= r^n - (r - 1)^n \binom{r}{1} + (r - 2)^n \binom{r}{2} + \cdots + (-1)^j (r - j)^n \binom{r}{j} + \cdots + (-1)^r · 0^n\\
&= r^n - (r - 1)^n \binom{r}{1} + (r - 2)^n \binom{r}{2} + \cdots + (-1)^j (r - j)^n \binom{r}{j} + \cdots + (-1)^{r - 1} · 1^n.
\end{align*}
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2605383",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
}
|
Can't find error at completing the square I am desperatly looking for the mistake I did when completing the square.
I have a function $f(x)=-4.905x^2+5x+6$
Nothing special. So when I was trying to find the peak of the curve I ran into a problem and couldn't figure out why this happens, since I have repeated the task about 5 times.
I used the form: $(x+\frac{b}{2a})^2=(\frac{b}{2a})^2-\frac{c}{a}$
When I solve left hand side to get the value of -0.509683996 which seems to fit for the x value of the peak. When I put in this value for x into the original equation I also receive the correct value of about y=7.53 for the peak.
But when I try to read of the the peak by using the RHS my equation seems to crash: I received after serial trials always something that is not equal to the y values of the vertex :
https://www.symbolab.com/solver/step-by-step/%5Cfrac%7B5%5E%7B2%7D%7D%7B-9.81%5E%7B2%7D%7D-%5Cfrac%7B6%7D%7B-4.905%7D
what happened? Why is my right hand side not equal to 7.53?
I must do something extremely wrong when trying to calculate right hand side and receive the y value of the vertex.
|
It seems that you are confusing finding the solutions of $ax^2+bx+c=0$ where $a\not=0$ with finding the vertex of the parabola $y=ax^2+bx+c$.
In order to find the solutions, we have
$$\begin{align}ax^2+bx+c=0&\implies x^2+\frac bax=-\frac ca\\\\&\implies x^2+\frac bax+\frac{b^2}{4a^2}=\frac{b^2}{4a^2}-\frac ca\\\\&\implies \left(x+\frac{b}{2a}\right)^2=\frac{b^2-4ac}{4a^2}\\\\&\implies x+\frac{b}{2a}=\pm\sqrt{\frac{b^2-4ac}{4a^2}}\\\\&\implies x=-\frac{b}{2a}\pm\sqrt{\frac{b^2-4ac}{4a^2}}\end{align}$$
In order to find the vertex of the parabola $y=ax^2+bx+c$, we have
$$\begin{align}y&=ax^2+bx+c\\\\&=a\left(x^2+\frac bax\right)+c\\\\&=a\left(x^2+\frac bax+\frac{b^2}{4a^2}-\frac{b^2}{4a^2}\right)+c\\\\&=a\left(\left(x+\frac{b}{2a}\right)^2-\frac{b^2}{4a^2}\right)+c\\\\&=a\left(x+\frac{b}{2a}\right)^2+c-\frac{b^2}{4a}\end{align}$$
So, the vertex is
$$\left(-\frac{b}{2a},c-\frac{b^2}{4a}\right)$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2607657",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
}
|
Show that if $x+y+z=x^2+y^2+z^2=x^3+y^3+z^3=1$ then $xyz=0$ For any real numbers $x,y,z$ such that $x+y+z = x^2+y^2+z^2 = x^3+y^3+z^3 =1\\$ show that $x \cdot y \cdot z=0$.
I think that because $x \cdot y \cdot z = 0$ at least one of the three number should be equal to zero, but I'm stuck into relating the other things.
|
We have $$(x+y+z)(x^2-xy+y^2+z^2)=x^3+y^3+z^3+xz^2+yz^2+x^2z-xyz+y^2z$$ so $$x^2-xy+y^2+z^2=1+xz^2+yz^2+x^2z-xyz+y^2z$$ so $$-xy=xz(x+z)+yz(y+z)-xyz$$ giving $$xyz=xy+xz-xyz+yz-xyz\implies 3xyz=xy+xz+yz$$ Now $$1=(x+y+z)^2=x^2+y^2+z^2+2(xy+xz+yz)\implies xy=xz+yz=0$$ Hence $$\boxed{3xyz=0\implies xyz=0.}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2612380",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 8,
"answer_id": 6
}
|
Find two different latin squares of order $5$ In the following question I am trying to find two different Latin squares of order $5$
Latin Square #1
\begin{array}
& &1 &2 &3 &4 &5 \\
&5 &1 &2 &3 &4 \\
&4 &5 &1 &2 &3 \\
&3 &4 &5 &1 &2 \\
&2 &3 &4 &5 &1 \\
\end{array}
This Latin square was made by filling in the first row by $1,2,3,4,5$ and then filling in the second row by shifting cyclically one position to the right, and so on...
Latin Square #2
\begin{array}
& &1 &2 &3 &4 &5 \\
&2 &3 &4 &5 &1 \\
&3 &4 &5 &1 &2 \\
&4 &5 &1 &2 &3 \\
&5 &1 &2 &3 &4 \\
\end{array}
This Latin square was made by filling in the first row by $1,2,3,4,5$ and then filling in the second row by shifting cyclically one position to the left, and so on...
So my questions are
1) Are these 2 different Latin squares of order $5$
2) Are my explanations for forming each Latin square correct?
|
They are different, in the sense that they're unequal.
They are the same, in the sense that they are isotopic, meaning that there is a way to permute the rows, columns, and symbols of one to obtain the other. In this case, we can permute the rows of the first one to obtain the second one.
The Cayley table of $\mathbb{Z}_5$ is the second example (with different symbols). It has no $2 \times 2$ subsquares ($2 \times 2$ submatrices which are Latin squares) since, in a group table, that would imply a subgroup of order $2$, violating Lagrange's Theorem. So we can get an inequivalent Latin square by creating one with a $2 \times 2$ subsquare, i.e., by completing
$$
\begin{bmatrix}
1 & 2 & \cdot & \cdot & \cdot \\
2 & 1 & \cdot & \cdot & \cdot \\
\cdot & \cdot & \cdot & \cdot & \cdot \\
\cdot & \cdot & \cdot & \cdot & \cdot \\
\cdot & \cdot & \cdot & \cdot & \cdot \\
\end{bmatrix}
$$
to a Latin square.
(Your explanations for how your Latin squares were generated make sense to me.)
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2615005",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
How many strings of digits of length $n$ contains a substring of $k$ identical consecutive digits? Let $f(n, k)$ be the number of strings of digits of length $n$ that contain a substring of $k$ identical consecutive digits. What would be a recursive or closed-form way to compute $f(n, k)$?
For example, $f(n, 2)=10^n - 10\times 9^n$ by using complementary counting. There are $10$ choices for each digit ignoring the restriction, and there are $9$ choices for each digit after the first if we can't repeat any two consecutive digits.
|
We consider the alphabet $V=\{0,1,\ldots,9\}$. We are looking for the number $g(n,k)$ of strings of length $n$ having runs at most length $k-1$. The wanted number is $$f(n,k)=10^n-g(n,k)$$
Strings with no consecutive equal characters at all are called Smirnov or Carlitz words. See example III.24 Smirnov words from Analytic Combinatorics by Philippe Flajolet and Robert Sedgewick for more information.
A generating function for the number of Smirnov words over a q-ary alphabet is given by
\begin{align*}
\left(1-\frac{qz}{1+z}\right)^{-1}
\end{align*}
Replacing occurrences of $0$ in a Smirnov word by one up to $k-1$ zeros generates words having runs of $0$ with length less than $k$.
\begin{align*}\
z\longrightarrow z+z^2+\cdots+z^{k-1}=\frac{z\left(1-z^{k-1}\right)}{1-z}
\end{align*}
The same can be done for the other digits.
The resulting generating function is for $q=10$
\begin{align*}
\left(1- \frac{10\cdot \frac{z\left(1-z^{k-1}\right)}{1-z}}{1+\frac{z\left(1-z^{k-1}\right)}{1-z}}\right)^{-1}
&=\frac{1-z^k}{1-qz+(q-1)z^{k}}
\end{align*}
Denoting with $[z^n]$ the coefficient of $z^n$ in a series we obtain the number of wanted words of length $n$ as
\begin{align*}
\color{blue}{f(n,k)}&=10^n-g(n,k)\\
&\color{blue}{=[z^n]\left(\frac{1}{1-10z}-\frac{1-z^k}{1-10z+9z^{k}}\right)}
\end{align*}
Example: Let's look at an example. We take $k=3$. We obtain with some help of Wolfram Alpha
\begin{align*}
\frac{1}{1-10z}-\frac{1-z^3}{1-10z+9z^{k}}=10 z^3 + \color{blue}{190} z^4 + 2800 z^5 +\cdots
\end{align*}
The blue colored coefficient of $z^4$ shows there are $\color{blue}{190}$ words of length $4$ built from characters $\{0,1,2,\ldots 9\}$ and runs of a digit with length at least $k=3$.
The number $190$ is easily to check. Strings of length $n=4$ and runs with length $3$ have the form
\begin{align*}
abbb\qquad\text{or}\qquad bbba
\end{align*}
with $a,b\in\{0,1,\ldots,9\}$ given $2\cdot 10\cdot 10=200$ different strings. Since $abbb=bbba$ iff $a=b$ we have $10$ strings counted twice and to subtract from $200$ giving a total of
\begin{align*}
\color{blue}{f(4,3)=190}
\end{align*}
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2615086",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
}
|
Proving in different ways that $n^{n-1}-1$ is divisible by $(n-1)^2$. I have this amazing exercise which explicitly says prove in at least six different way that
$n^{n-1}-1$ is divisible by $(n-1)^2$ where $n$ is an integer.
So far I have only prove it as follows: By geometric sum we have
$$\sum_{k=0}^{n-2}n^k=\frac{n^{n-1}-1}{n-1}$$
However, since $n\equiv 1\mod (n-1)$ we have
$$\frac{n^{n-1}-1}{n-1}=\sum_{k=0}^{n-2}n^k\equiv \sum_{k=0}^{n-2}1^k\mod(n-1)$$
that is $$\frac{n^{n-1}-1}{n-1}=\sum_{k=0}^{n-2}n^k\equiv n-1\mod(n-1)\Longleftrightarrow \frac{n^{n-1}-1}{n-1} \equiv 0\mod(n-1)$$
and this prove that $(n-1)^2$ divides $n^{n-1}-1$.
Does anyone have another approach different from mine.?
Note that I do necessary need one to give all six approaches as was asked in the exercise one or two is plainly enough for me.
|
$$
\begin{align}
n^{n-1}-1 &= (n-1) ( n^{n-2} \color{red}{- 1} + n^{n-3} \color{red}{- 1} +\ldots + n^2 \color{red}{- 1} + n \color{red}{- 1} + 1 \color{red}{+n-2}) \\
&= (n-1) \big(\,(n-1)(\ldots) + (n-1)(\ldots)+ \ldots + (n-1)(n+1)+ (n-1)+ (n-1))\,\big)
\end{align}
$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2616895",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 2
}
|
Find the range of $x$ for the convergence of the series $\sum_{n=1}^{\infty} \frac{1}{\left(-3\right)^{n+2}} \frac{\left(4x-12\right)^{n}}{n^{2}+1}$
Question: Find the range of $x$ for the convergence of the series$$\sum_{n=1}^{\infty} \frac{1}{\left(-3\right)^{n+2}} \frac{\left(4x-12\right)^{n}}{n^{2}+1}$$
MY Approach: By $n$th term divergence test,
$$\lim_{n\rightarrow\infty} \frac{1}{\left(-3\right)^{n+2}} \frac{\left(4x-12\right)^{n}}{n^{2}+1}=0 \Longleftrightarrow |4x-12|<1\Longleftrightarrow\frac{11}{4}<x<\frac{13}{4},$$
not in the options.
Edit
|
$$\sum_{n=1}^{\infty}\frac{1}{\left(-3\right)^{n+2}}\frac{\left(4x-12\right)^{n}}{n^{2}+1}=\sum_{n=1}^{\infty}\frac{1}{\left(9\right)}\frac{\left(4-\frac{4x}{3}\right)^{n}}{n^{2}+1}$$
according to Cauchy root test , we get
$$|4-\frac{4x}{3}|\leqslant 1$$
so
$$\frac{9}{4}\leqslant x\leqslant \frac{15}{4}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2618920",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
}
|
How can one prove that this polynomial is non-negative? How one can prove the following inequality?
$$58x^{10}-42x^9+11x^8+42x^7+53x^6-160x^5+118x^4+22x^3-56x^2-20x+74\geq 0$$
I plotted the graph on Wolfram Alpha and found that the inequality seems to hold. I was unable to represent the polynomial as a sum of squares.
It looks quite boring to approximate the derivative to be zero and use some numerical methods to show that values near local minimums proves that the inequality really holds everywhere.
|
For $x<0$ it's obvious.
But for $x\geq0$ we obtain:
$$58x^{10}-42x^9+11x^8+42x^7+53x^6-160x^5+118x^4+22x^3-56x^2-20x+74=$$
$$=(x^3-x^2-x+1)(58x^7+16x^6+85x^5+85x^4+207x^3+47x^2)+$$
$$+287x^4-138x^3-103x^2-20x+74>0,$$
where $$287x^4-138x^3-103x^2-20x+74=$$
$$=(16x^2-4x-5)^2+(31x^4-10x^3+x^2)+(40x^2-60x+49)>0.$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2619131",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
}
|
System of equations involving complex numbers I'm getting confused to figure this work out. The only thing that came into my head was using AM-GM inequality, but i just get stuck. Here's the problem:
Let $a,b,c,$ be the complex numbers such that $abc=1$. and
\begin{cases}
a^{20}+b^{20} + c^{20} &= \frac{1}{a^{20}} + \frac{1}{b^{20}} + \frac{1}{c^{20}} \\
a^{17}+b^{17}+c^{17} &= \frac{1}{a^{17}} + \frac{1}{b^{17}} + \frac{1}{c^{17}}\\
a^{2017} + b^{2017}+c^{2017} &= \frac{1}{a^{2017}} + \frac{1}{b^{2017}} + \frac{1}{c^{2017}}\\
\end{cases}
show that $1 \in \{a,b,c\}$.
Please help me to solve this question, any thought would be helpful.
|
Hint. If $abc=1$ then, for any integer $n$, the equation
$$a^{n}+b^{n} + c^{n}= \frac{1}{a^{n}} + \frac{1}{b^{n}} + \frac{1}{c^{n}}$$
is equivalent to
$$(1-a^n)(1-b^n)(1-c^n)=0.$$
Moreover note that $\gcd(20,17)=1$, $\gcd(2017,17)=1$, and $\gcd(2017,20)=1$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2620257",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
}
|
Computing the definite integral $\int _0^a \:x \sqrt{x^2+a^2} \,\mathrm d x$
Compute the following definite integral $$\int _0^a \:x \sqrt{x^2+a^2} \,\mathrm d x$$
This is what I did:
$u = x^2 + a^2 $
$du/dx = 2x$
$du = 2xdx$
$1/2 du = x dx$
$\int _0^a\:\frac{1}{2}\sqrt{u}du = \frac{1}{2}\cdot \frac{u^{\frac{3}{2}}}{\left(\frac{3}{2}\right)}$ from $0$ to $a$.
$\frac{1}{3}\cdot \left(x^2+a^2\right)^{\frac{3}{2}}$ from $0$ to $a$.
I eventually got:
$\frac{1}{3}\left(81+a^2\right)^{\frac{3}{2}}-\frac{1}{3}\left(a^2\right)^{\frac{3}{2}}$
but this was incorrect.
The correct answer was:
$\frac{1}{3}\left(2\sqrt{2}-1\right)a^3$
Any help?
|
You calculated the antiderivative correctly.
$$\frac{1}{3}\cdot \left(x^2+a^2\right)^{\frac{3}{2}}$$ from $0$ to $a$ is $$\frac{1}{3}\cdot \left(a^2+a^2\right)^{\frac{3}{2}}-\frac{1}{3}\cdot \left(0^2+a^2\right)^{\frac{3}{2}}=\\\frac{1}{3}\cdot((2a^2)^\frac{3}{2}-(a^2)^\frac{3}{2})=\frac{1}{3}\cdot(2^\frac{3}{2}a^3-a^3)=\frac{1}{3}(2^\frac{3}{2}-1)a^3$$
Which explains the answer in your book. I don't know how you got $$\frac{1}{3}\left(81+a^2\right)^{\frac{3}{2}}-\frac{1}{3}\left(a^2\right)^{\frac{3}{2}}$$ from the previous step.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2621750",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
}
|
The number of ways in which one can choose three distinct numbers from the set so that the product of the chosen numbers is divisible by $9$
Consider the set $A=\{1,2,3,...,30\}$
. The number of ways in which one can choose three
distinct numbers from $A$ so that the product of the chosen numbers is divisible by $9$ is $X$
Find $X$
This question is from KVPY SA 2017.
My working:
Nos in set divisible by $3$(but not by $9$)-$3,6,12,15,21,24,30$-Total $7$
Nos in set divisible by $9$-$9,18,27$-Total $3$
Therefore, the ans should be $7.7.20+7.7.7+7.7.3+7.3.3+3.3.3+3.20.20+3.3.20+3.7.20$
Where $.$ is multiplication
Is this correct?
|
We solve the opposite: let $Y$ denote the number of ways one can choose three distinct numbers for which their multiple not divisible by 9 i.e. divisible by 3 and not 9 or not divisible by 3 at all. First note that there are $10$ multiples of 3 among those $3$ are also a multiple of 9 in $\{1,2,...,30\}$ so the number of ways of choosing 3 numbers in case 1 is to choose exactly one multiple of 3 not of 9 to $7$ different ways and the others are chosen from the rest of the set providing $7\times 20\times 19$ different cases. The 2nd case also provides $20\times 19\times 18$ different cases and choosing three arbitrary number is possible to $30\times 29\times 28$ cases. So:$$X=30\times 29\times 28-20\times 19\times 18-7\times 20\times 19=14860$$If we don't mind ordering the answer would then be:$$X=\binom{30}{3}-\binom{20}{3}-\binom{7}{1}\binom{20}{2}=1590$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2622034",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
}
|
Partial Fraction problem solution deviates from the Rule Question:
Compute $\displaystyle \int\frac{x^2+1}{(x^2+2)(x+1)} \, dx$
My Approach:
As per my knowledge this integral can be divided in partial Fraction of form $\dfrac{Ax+B}{x^2+px+q}$ and then do the following as per to integrate it.
Solution:
Taking $\dfrac{x^2+1}{(x^2+2)(x+1)}=\dfrac{Ax^2+Bx+C}{x^2+2}+\dfrac{D}{x+1}$
Rule given: Denominator g(x) contains quadratic tractor (may not be factorisable).
To each non-repeated quadratic factor of the form $x^2+px+q$(or $x^2+q$, $q$ not equal to $0$), there
Should be a partial Fraction of the form $Ax+B/(x^2+px+q)$.
My problem:
I can't understand why the solution provided deviates from the rule that I have studied to solve these kind of problems.
Book:
ISC MATHEMATICS XII
Publishers:
Kalyani
|
you must write $$\frac{x^2+1}{(x^2+2)(x+1)}=\frac{Ax+B}{x^2+2}+\frac{C}{x+1}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2622263",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
}
|
The Arithmetic Mean (A.M) between two numbers exceeds their Geometric Mean (G.M.) The Arithmetic Mean (A.M) between two numbers exceeds their Geometric Mean (G.M.) by $2$ and the GM exceeds the Harmonic Mean (H.M) by $1.6$. Find the numbers.
My Attempt:
Let the numbers be $a$ and $b$. Then,
$$A.M=\dfrac {a+b}{2}$$
$$G.M=\sqrt {ab}$$
$$H.M=\dfrac {2ab}{a+b}$$
According to question:
$$\dfrac {a+b}{2} =\sqrt {ab}+2$$
$$\dfrac {a+b}{2}-2=\sqrt {ab}$$
$$a+b-4=2\sqrt {ab}$$
Also,
$$\sqrt {ab}=\dfrac {2ab}{a+b} + 1.6$$
Then,
$$a+b-4=2(\dfrac {2ab}{a+b} + 1.6)$$
$$a+b-4=\dfrac {4ab+3.2(a+b)}{a+b}$$
$$(a+b-4)(a+b)=4ab+3.2(a+b)$$
$$(a+b)^2-4(a+b)=4ab+3.2a+3.2b$$
How do I solve further?
|
$$AM=GM+2$$
$$GM=HM+1.6$$
Since $$GM^2=AM\cdot HM,$$
$$GM^2=(GM+2)(GM-1.6)$$
$$GM^2=GM^2+0.4GM-3.2$$
$$GM=8$$
$$AM=10$$
$$\sqrt{ab}=8, \frac{a+b}{2}=10$$
$$ab=64, a+b = 20$$
The numbers are $16$ and $4$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2625855",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
}
|
Subgroups of General Linear Group I have a question about the order of the subgroups of $GL_2(\mathbb{R})$ generated by the following matrix:
$\begin{pmatrix}
1&-1 \\
-1&0
\end{pmatrix}$
So I computed several powers of this matrix and realized that the order of the subgroup generated by this matrix is infinity. However, I do not know how to prove it in general that:
${\begin{pmatrix}
1&-1 \\
-1&0
\end{pmatrix}}^n\neq \begin{pmatrix}
1&0 \\
0&1
\end{pmatrix} \forall n\in\mathbb{Z}.$
Is there any way to prove this in general? I might have did not remember something from Linear Algebra to prove this. Any suggestion is really appreciated.
Here are what I got for my calculations:
${\begin{pmatrix}
1&-1 \\
-1&0
\end{pmatrix}}^2= \begin{pmatrix}
2&-1 \\
-1&1
\end{pmatrix} $
${\begin{pmatrix}
1&-1 \\
-1&0
\end{pmatrix}}^3= \begin{pmatrix}
3&-2 \\
-2&1
\end{pmatrix} $
${\begin{pmatrix}
1&-1 \\
-1&0
\end{pmatrix}}^4= \begin{pmatrix}
5&-3 \\
-3&2
\end{pmatrix} $
${\begin{pmatrix}
1&-1 \\
-1&0
\end{pmatrix}}^5= \begin{pmatrix}
8&-5 \\
-5&3
\end{pmatrix} $
${\begin{pmatrix}
1&-1 \\
-1&0
\end{pmatrix}}^6= \begin{pmatrix}
13&-8 \\
-8&5
\end{pmatrix} $
|
Hint: When you diagonalize this matrix, you get
$
S\begin{pmatrix} \frac{-1 + \sqrt{5}}{2} & 0 \\ 0 & \frac{1-\sqrt{5}}{2} \end{pmatrix}S^{-1}$
for some ugly matrices $S$ and $S^{-1}$. These numbers are larger in norm than $1$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2626045",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
}
|
Equation System with 4 real variables I need to solve the next equation system:
Find all real numbers $a,b,c,d$ such that:
$$
\left\{
\begin{array}{c}
a+b+c+d=20 \\
ab+ac+ad+bc+bd+cd=150 \\
\end{array}
\right.
$$
I tried something like this:
$b+c+d=20-a$
And i put the second equation like this
$a(b+c+d) + bc+bd+cd=150$
Getting
$20a-a^2 +bc+bd+cd=150$
But i see that this is useless, so i don't know how to start this problem.
|
By $(a+b+c+d)^2=a^2+b^2+c^2+d^2+2(ab+ac+ad+bc+bd+cd)$, we find $$ a^2+b^2+c^2+d^2 = 100 $$
Also by Cauchy-Schwarz inequality, $(a+b+c+d)^2 \leq (1^2 + 1^2 + 1^2 +1^2)(a^2+b^2+c^2+d^2)$ and therefore we yields: $$ 400 \leq 400 $$
Then, the equality condition occurs if and only if $a=b=c=d=5$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2627063",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
}
|
Solve $\gcd(a,b)=2$, $3a+b^2 =3388$, $a>0$, $b>49$ I have this problem and I can't do it.
$$\begin{cases}
\gcd(a,b)=2 \\
3a+b^2 =3388 \\
a>0,b>49
\end{cases}
$$
I've tried writing $a=2a'$ and $b=2b'$, but then I have $3a'+2b'^2=1694$ and I don´t know what to do
|
Assuming $a$ needs to be positive, the equation $3a + b^2 = 3388$ shows that $b \leq 58$. Since $gcd(b,a) =2 $ implies that $b$ is divisible by $2$, so the only options are $b = 50,52,54,56,58$. Note that $4$ divides $3388$, so if $4$ divides $b$, then $4$ divides $3a$ and hence $4$ divides $a$. This is not acceptable, since $gcd(b,a)$ is required to be $2$. Thus we find $b \neq 52,56$.
Examining $3a + b^2 = 3388$ modulo $3$, we see that $b^2 \equiv 1 \mod 3$. Thus $b \equiv 1, 2 \mod 3$ are the only possibilities. Hence $b$ is not $54$.
Now the only remaining possibilities are $b = 50, 58$, which you can check are both solutions.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2628361",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
}
|
equations of triangle's sides, given equations of two bisectors and one point of triangle? The equations two angles bisectors are $x-3y-6=0$ and $x+y-2=0$. We also know that one point of the triangle is $A(2,-4)$. Clearly, this point doesn't satisfy those two equations, and by finding the intersection of the bisectors and A we can write the third bisector's equation. This is all I can come up with. Any hints?
|
Let $D$ be the incentre. $D=(3,-1)$.
The slope of $AD$ is $3$.
Let $\angle BAC=2a$, $\angle ABC=2b$ and $\angle ACB=2c$. Then $a+b+c=\frac{\pi}{2}$.
Note that $\angle BDC=\pi-b-c=\frac{\pi}{2}+a$.
If $\theta$ is the acute angle between the two given angle bisectors, then
$$\tan\theta=\frac{\frac{1}{3}-(-1)}{1+(-1)(\frac{1}{3})}=2$$
As $\frac{\pi}{2}+a$ is obtuse, $\frac{\pi}{2}+a=\pi-\theta$ and hence $a=\frac{\pi}{2}-\theta$. $\tan a=\frac{1}{2}$.
Let $m$ be the slope of a line making an angle $a$ with $AD$.
\begin{align*}
\tan a&=\left|\frac{m-3}{1+3m}\right|\\
\frac{1}{2}&=\left|\frac{m-3}{1+3m}\right|\\
1+3m&=\pm2(m-3)\\
m&=1\quad\textrm{or}\quad-7
\end{align*}
The equations of two of the sides are $y=-7x+10$ and $y=x-6$.
We can find $B$ and $C$ by solving these two lines with the angle bisectors.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2629104",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
}
|
Formatting Functions To Avoid Loss Of Significant
Rewrite the following to avoid loss of significant
*
*$\ln(x+1)-\ln(x)$ where $x>>1$
*$\cos^2(x)-\sin^2(x)$ where $x\approx \frac{\pi}{4}$
*$\sqrt{x^2+1}-x$ where $x>>1$
*$\sqrt{\frac{1+\cos x}{2}}$
*
*Using taylor expansion we get $$x-\frac{x^2}{2}+\frac{x^3}{3}-[(x-1)-\frac{(x-1)^2}{2}+\frac{(x-1)^3}{3}]=1+\frac{(x-1)^2-x^2}{2}+\frac{x^3-(x-1)^3}{3}$$
*Using taylor expansion we get
$$(1-\frac{x^2}{2!}+\frac{x^4}{4!})^2-(x-\frac{x^3}{3!}+\frac{x^5}{5!})^2$$
*$$\sqrt{x^2+1}-x=(\sqrt{x^2+1}-x)\cdot (\frac{\sqrt{x^2+1}+x}{\sqrt{x^2+1}+x})=\frac{1}{\sqrt{x^2+1}+x}$$
*$$\sqrt{\frac{1+\cos x}{x}}\approx \sqrt{\frac{1+1-\frac{x^2}{2!}+\frac{x^4}{4!}}{2}}$$
Is this valid?
|
You shouldn't use Taylor series for these.
*
*$\log(x+1)-\log(x) = \log(1+\frac{1}{x})$
*$\cos^2 x - \sin^2 x = \cos 2x$
*$\sqrt{x^2+1} - x = (\sqrt{x^2+1} - x) \frac{\sqrt{x^2+1}+x}{\sqrt{x^2+1}+x}$
and simplify.
*$1+\cos x = 2\cos^2 \frac{x}{2}$. This identity should help.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2630552",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
}
|
What's the expected number of times I have to roll two die until they both sum $7$? Here is my guess: the probability of summing $7$ on two rolls is $\frac 16$. This means if I repeat the experiment many times I'll roll $7$ one sixth of them (approximately). Hence,
$$N \cdot \bigg(\cfrac 16\bigg) \cdot 7 = 7$$
where $N$ is the total number of rolls. That gives me a total number of $6$ rolls on average to sum $7$.
I'm not quite sure so I'm all open to suggestions! Thanks in advance.
|
The probability of doing it after one roll is $1/6$, in two is $5/6 \times 1/6$, in three $(5/6)^2 1/6$ and so on ... we get
\begin{eqnarray*}
E(7)=1 \times \frac{1}{6} + 2 \times \frac{5}{6} \times\frac{1}{6} + 3 \times \left(\frac{5}{6}\right)^2 \times\frac{1}{6}+\cdots = \frac{1}{6} \sum_{i=1}^{\infty} i \left( \frac{5}{6} \right)^i \\
\end{eqnarray*}
Now recall that
\begin{eqnarray*}
\sum_{i=1}^{\infty} i x^{i-1} =\frac{1}{(1-x)^2}.
\end{eqnarray*}
So
\begin{eqnarray*}
E(7)= \frac{1}{6} \sum_{i=1}^{\infty} i \left( \frac{5}{6} \right)^{i-1} =\frac{1}{6} \frac{1}{(1-\frac{5}{6})^2}=6
\end{eqnarray*}
So the expected value is $\color{red}{6}$ as expected.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2633273",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 3,
"answer_id": 2
}
|
Help proving $\forall x \geq 2: \cos(2 \pi/x) \leq 1 - 1/x^2$ I want to prove the following inequality:
$$\forall x \geq 2: \cos(2 \pi/x) \leq 1 - 1/x^2
$$
I believe it holds: numerical evidence.
I tried to prove it using taylor series, but it didn't work so well. Any ideas?
|
Note that by Taylor's series $\forall x$
$$\cos x\le 1-\frac{x^2}2+\frac{x^4}{24}$$
thus
$$\cos \frac{2\pi}{x}\le 1-\frac{2\pi^2}{x^2}+\frac{16\pi^4}{24x^4}\le 1-\frac{1}{x^2}$$
indeed
$$1-\frac{2\pi^2}{x^2}+\frac{2\pi^4}{3x^4}\le 1-\frac{1}{x^2}\iff\frac{1}{x^2}-\frac{2\pi^2}{x^2}+\frac{2\pi^4}{3x^4}\le0$$
$$\iff 3x^2-6\pi^2x^2+2\pi^4\le0\iff x^2(6\pi^2-3)\ge2\pi^4\iff x^2\ge \frac{2\pi^4}{6\pi^2-3}\approx3.465$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2637857",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
}
|
Determine the whether the sequence $a_{n+1} = 3 - \frac{1}{a_n} \text{ for n > 1}$ is convergent or divergent. Consider the recursively defined sequence $a_n = 1$
$$a_{n+1} = 3 - \frac{1}{a_n} \text{ for n > 1}$$
Is the sequence convergent?
This is my attempt:
First, we prove that the sequence is positive, and monotonically increasing, using induction.
$\textbf{Base Case:}$ $a_1 = 1$ and $a_2 = 2$. $a_1 < a_2$ and $a_1,a_2 > 0$.
$\textbf{Inductive Hypothesis:}$ $a_n < a_{n+1}$ where $a_n, a_{n+1}$.
$\textbf{Inductive Step:}$ We prove that $a_n < a_{n+1} \Rightarrow a_{n+1} < a_{n+2}$.
By our induction hypothesis:
$$a_n < a_{n+1}$$
$$-a_n > -a_{n+1}$$
$$-\frac{1}{a_n} < -\frac{1}{a_{n+1}}$$ (True by our IH since $a_n, a_{n+1} > 0$ and thus, $-a_n, -a_{n+1}$ share the same sign).
$$3-\frac{1}{a_n} < 3-\frac{1}{a_{n+1}}$$
$$a_{n+1} < a_{n+2}$$.
Note that $a_{n+1} > 0$ by our IH, so $a_{n+2} > 0$.
Now we prove that the sequence is bounded.
Observe that $a_n$ is monotonically increasing, which means that $ - \frac{1}{a_n}$ is monotonically increasing as well and it is upped bounded by $0$. Thus, $3-\frac{1}{a_n}$ is upper bounded by $3$.
We have a sequence that is monotone and bounded. Hence, by the Monotone Convergence Theorem, This sequence converges.
I was wondering if this method is correct.
|
In your proof of $a_{n+1}>a_n$ you used that $a_n>0$, but it is not proven.
I like the following reasoning.
$$a_{n+1}-\frac{3-\sqrt5}{2}=3-\frac{3-\sqrt5}{2}-\frac{1}{a_n}=\frac{3+\sqrt{5}}{2}-\frac{1}{a_n}=\frac{1}{\frac{3-\sqrt5}{2}}-\frac{1}{a_n}>0$$ by induction because $a_1=1>\frac{3-\sqrt5}{2}.$
$$a_{n+1}-\frac{3+\sqrt5}{2}=3-\frac{3+\sqrt5}{2}-\frac{1}{a_n}=\frac{3-\sqrt{5}}{2}-\frac{1}{a_n}=\frac{1}{\frac{3+\sqrt5}{2}}-\frac{1}{a_n}<0$$ by induction because $a_1=1<\frac{3+\sqrt5}{2}.$
Thus, for all natural $n$ we got:
$$\frac{3-\sqrt5}{2}<a_n<\frac{3+\sqrt5}{2}.$$
In another hand, $$a_{n+1}-a_n=3-a_n-\frac{1}{a_n}=\frac{\left(a_n-\frac{3-\sqrt5}{2}\right)\left(\frac{3+\sqrt5}{2}-a_n\right)}{a_n}>0,$$
which says that $a$ converges.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2639499",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
}
|
Want to understand how a fraction is simplified The fraction is used to determine the sum of a telescopic series.
$$\sum_{k=1}^\infty\frac{1}{(k+1)\sqrt{k}+k\sqrt{k+1}}$$
This is the solved fraction.
$$\frac{1}{(k+1)\sqrt{k}+k\sqrt{k+1}} = \frac{(k+1)\sqrt{k}-k\sqrt{k+1}}{(k+1)^2k-k^2(k+1)} = \frac{(k+1)\sqrt{k}-k\sqrt{k+1}}{k^3+2k^2+k-k^3-k^2} = \frac{(k+1)\sqrt{k}-k\sqrt{k+1}}{k(k+1)} = \frac{\sqrt{k}}{k}-\frac{\sqrt{k+1}}{k+1} = \frac{1}{\sqrt{k}}-\frac{1}{\sqrt{k+1}}$$
I want to understand what is being done on every step,mainly the last three.
Thank you.
|
*
*First equal: multiply and divide by $(k+1)\sqrt k-k\sqrt{k+1}$
*Second equal: Expand the denominator
*Third equal: cancel obvious terms in the denominator and write $k^2+k=k(k+1)$.
*Fourth equal: distribute along the minus in the numerator, and cancel the obvious factors $(k+1)$ in the first summand, and $k$ in the second one:
$$
\frac{(k+1)\sqrt{k}-k\sqrt{k+1}}{k(k+1)}
=\frac{(k+1)\sqrt{k}}{k(k+1)}-\frac{k\sqrt{k+1}}{k(k+1)}
=\frac{\sqrt k}k-\frac{\sqrt{k+1}}{k+1}.
$$
*Fifth equal: $\frac{\sqrt k}{k}=\frac1{\sqrt k}$, and similarly for $k+1$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2640521",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
}
|
Expected number and probability of a series of 1s in a bit string. Suppose the sequence 1001111011011 has a total of $4$ "blocks" of one, because number of contiguous sequences made of ones (1, 1111,11 and 11) count to 4. Given a random bit array of length $N$, what's the expected value of the "number of 1 blocks"? Related to that, what is the probability there are $M$ blocks in a bit a array of length $N$?
|
Using $z$ for zeros and $w$ for ones and $u$ for runs we get the
generating function
$$(1+z+z^2+\cdots)
\\ \times \sum_{q\ge 0} u^{q} (w+w^2+\cdots)^q (z+z^2+\cdots)^q
\\ \times (1+uw+uw^2+\cdots)$$
which is
$$\frac{1}{1-z}
\\ \times \sum_{q\ge 0} u^{q} \frac{w^q}{(1-w)^q} \frac{z^q}{(1-z)^q}
\\ \times \left(1+u\frac{w}{1-w}\right)$$
or
$$\frac{1}{1-z}
\frac{1}{1-uwz/(1-z)/(1-w)}
\frac{1-w+uw}{1-w}
\\ = \frac{1-w+uw}{(1-z)(1-w)-uwz}.$$
As a sanity check we put $u=1$ and $w=z$ to obtain
$$\frac{1}{(1-z)^2 - z^2} = \frac{1}{1-2z}$$
and we see that we have accounted for all strings of length $N.$ We no
longer need the distinction between zeros and ones here so we obtain
$$\frac{1-z+uz}{(1-z)^2-uz^2}
= \frac{1-z+uz}{(1-z)^2}\frac{1}{1-uz^2/(1-z)^2}
\\ = \left(\frac{1}{1-z} + \frac{uz}{(1-z)^2}\right)
\frac{1}{1-uz^2/(1-z)^2}.$$
Extract the coefficient on $[u^M]$ to get
$$\frac{z^{2M}}{(1-z)^{2M+1}}
+ \frac{z^{2M-1}}{(1-z)^{2M}}
\\ = \left(\frac{z}{1-z} + 1\right) \frac{z^{2M-1}}{(1-z)^{2M}}
= \frac{z^{2M-1}}{(1-z)^{2M+1}}.$$
Continue with the coefficient on $[z^N]$ to obtain
$$[z^N] \frac{z^{2M-1}}{(1-z)^{2M+1}}
= [z^{N+1-2M}] \frac{1}{(1-z)^{2M+1}}
\\ = {N+1-2M+2M\choose 2M} = {N+1\choose 2M}$$
for a probability of
$$\bbox[5px,border:2px solid #00A000]{
\frac{1}{2^N} {N+1\choose 2M}.}$$
We get for the expectation
$$\sum_{M=0, \; 2M\le N+1} M {N+1\choose 2M}
= \sum_{M=1, \; 2M\le N+1} M {N+1\choose 2M}
\\ = \frac{N+1}{2} \sum_{M=1, \; 2M\le N+1} {N\choose 2M-1}
= \frac{N+1}{2} \sum_{q=0}^N {N\choose q} \frac{1-(-1)^q}{2}
\\ = \frac{N+1}{4} 2^N.$$
Dividing by $2^N$ then yields
$$\bbox[5px,border:2px solid #00A000]{
\frac{N+1}{4}.}$$
This may also be obtained by linearity of expectation. We place a zero
value at the front of the string and may then count runs of ones by
the number of times a zero was followed by a one in the modified
string, which gives $1/2 + (N-1)/4,$ the same answer ($N-1$ places
where we transition from one bit to the next).
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2643788",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
}
|
How many trees on $\{1,2,3,4,5,6,7\}$ have a vertex of degree 2?
How many trees on $\{1,2,3,4,5,6,7\}$ have a vertex of degree 2 ?
Attempt -
It feels like an inclusion exclusion problem (using kailey's code) , let's define $|A_i| \Rightarrow $ vertex $i$ is of degree $1$.
$|Ai\, \cup A_j$| - vertices $i,j$ are of degree $1$ and so on.
So following my calculation we might get -
$7 \cdot 5\cdot 6^4 - \binom{7}{2}\binom{5}{2}\cdot2\cdot5^3+ \binom{7}{3}\binom{5}{3}\cdot3!\cdot4^2 - \binom{7}{4}\binom{5}{4}\cdot4!\cdot3 + \binom{7}{5}5!$
What do you think? thank you !
|
I think that your formula is correct. It gives $16380$. Below there is an alternative approach with the same final result.
For a labeled tree with $7$ vertices we have that
$$\sum_{v \in \{1,2,3,4,5,6,7\}} \deg v = 2(7-1)=12.$$
So if a tree has NOT vertices of degree $2$ and it has $m$ leaves of degree $1$ then
$$12\geq l+3(7-l)$$
which means that $l$ can be $5$ or $6$ (there is at least one vertex which is not a leaf). Hence the degree sequence is
$$1,1,1,1,1,3,4\quad\text{or}\quad 1,1,1,1,1,1,6.$$
By Cayley's formula there are $7^{7−2}$ labeled trees with $7$ vertices and the number of trees with vertices $1,2,3,4,5,6,7$ of degrees $d_1, d_2,\dots, d_7$ is
$$\binom{7-2}{d_1-1,d_2-1,\dots, d_7-1}.$$
Hence the number of trees with vertices $1,2,3,4,5,6,7$ have at least a vertex of degree $2$ is
$$7^{5}-7\cdot 6\cdot\frac{5!}{2!3!}-7\cdot \frac{5!}{5!}=16380.$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2644294",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
}
|
Closed algebraic form of $\cos(\frac{\pi}{7})$ We all know that $\cos(\frac{\pi}{6})=\frac{\sqrt 3}{2}$, $\cos(\frac{\pi}{4})=\frac{\sqrt 2}{2}$ and $\cos(\frac{\pi}{3})=\frac{1}{2}$. One can also prove that $\cos(\frac{\pi}{5})=\frac{\sqrt 5+1}{4}$. But it seems that $\cos(\frac{\pi}{7})$ cannot be put in a closed form algebraic expression. Is there a proof of such a claim? I feel like this has to do with the Abel–Ruffini theorem.
The same for $\cos(\frac{\pi}{8})$ and $\cos(\frac{\pi}{9})$. But $\cos(\frac{\pi}{10})=\sqrt{\frac{5}{8}+\frac{\sqrt{5}}{8}}$.
|
For any $n\geq 3$, $\cos\left(\frac{2\pi}{n}\right)$ is al algebraic number over $\mathbb{Q}$ with degree $\frac{\varphi(n)}{2}$, since its minimal polynomial can be computed from $\Phi_n(x)$ in a straightforward way. Our case is given by $n=14$, where the minimal polynomial of $\cos\left(\frac{\pi}{7}\right)$ is given by $q(x)=8x^3-4x^2-4x+1$. By letting $\omega=\frac{-1+\sqrt{-27}}{2}$, through the cubic formula we get
$$ \cos\left(\frac{\pi}{7}\right) = \frac{1}{6}\left(1+\frac{7^{2/3}}{\omega^{1/3}}+7^{1/3}\omega^{1/3}\right). $$
It might be interesting to notice that the LHS is very close to $\frac{9}{10}$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2644775",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
}
|
When is $(p^2-1)/8$ even? I am trying to find for what values of p $\frac{p^2-1}{8}$ will be even number $\frac{p^2-1}{8}=2m\implies (p-1)(p+1)=8(2m)$ then can I write $p\equiv 1\pmod8$ or $p\equiv -1\pmod8$ ?
Also trying to find for what values of p it will be an odd number
$\frac{p^2-1}{8}=(2m+1)\\\implies (p-1)(p+1)=8(2m+1)$$\text{ will it be } p\equiv 1\pmod8$ or $p\equiv -1\pmod8$ again? p is an odd prime
|
I think you want to look at congruences modulo $16$, not $8$. After all, $\frac{p^2-1}8=2m$ if and only if $p^2-1=16m$, that is $p^2\equiv1\pmod{16}$. Similarly $\frac{p^2-1}8$ is odd if and only if $p^2-1=8+16m$, that is $p^2\equiv9\pmod{16}$.
By writing down all eight possibilities, you see that $n^2\equiv1\pmod{16}$ if and only if $n$ is congruent to one of $1,7,9,15$ modulo $16$. The other four possibilities, $n=3,5,11,13$ square to numbers $\equiv9\pmod{16}$. And I think that’s the whole story.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2645398",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
}
|
A sum of series problem: $\frac{3}{1!+2!+3!} + \frac{4}{2!+3!+4!} + \cdots + \frac{2008}{2006!+2007!+2008!}$ I have a question regarding the sum of this series:
$$\frac{3}{1!+2!+3!} + \frac{4}{2!+3!+4!} + \cdots + \frac{2008}{2006!+2007!+2008!}$$
My approach:
I found that this sum is equal to:
$$\sum_{n=3}^{2008}\frac{n}{(n-2)!+(n-1)!+(n)!}$$
I reduced it to :
$$\sum_{n=3}^{2008}\frac{1}{n(n-2)!}$$
Please suggest how to proceed further.
|
What we want is $$\sum_{n=1}^{N} \dfrac{n+2}{n! + (n+1)! +(n+2)!}$$
\begin{align}
\dfrac{n+2}{n! + (n+1)! +(n+2)!} & = \dfrac{n+2}{n! \left( 1 + (n+1) + (n+1)(n+2) \right)}\\
& = \dfrac{n+2}{n! \left( n^2 + 4n + 4 \right)}\\
& = \dfrac1{n! \left( n+2 \right)}\\
& = \dfrac{n+1}{(n+2)!}\\
& = \dfrac{n+2}{(n+2)!} - \dfrac1{(n+2)!}\\
& = \dfrac1{(n+1)!} - \dfrac1{(n+2)!}
\end{align}
Can you finish it off from here?
Move your mouse over the gray area below for the complete answer.
\begin{align}\sum_{n=1}^{N} \dfrac{n+2}{n! + (n+1)! +(n+2)!} & = \sum_{n=1}^{N} \left( \dfrac1{(n+1)!} - \dfrac1{(n+2)!}\right)\\ & = \left( \dfrac1{2!} - \dfrac1{3!} + \dfrac1{3!} - \dfrac1{4!} + \dfrac1{4!} - \dfrac1{5!} + \cdots + \dfrac1{(N+1)!} - \dfrac1{(N+2)!}\right)\\ & = \dfrac1{2!} - \dfrac1{(N+2)!}\end{align} Set $N=2006$ to get the answer to your question.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2647385",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 1
}
|
Inverse trigonometic functions Question:
Prove that $\tan^{-1}(\frac{1}{2} \tan2A)+\tan^{-1}(\cot A)+\tan^{-1}(\cot^{3} A)=0$
MyProblem
We can just use the formula of $\tan^{-1} A +\tan^{-1} B$ but I think it would be a waste of time. Is there any other shorter and simpler method to solve it.
|
The hint:
Prove that
$$\left(\tan^{-1}(\frac{1}{2} \tan2A)+\tan^{-1}(\cot A)+\tan^{-1}(\cot^{3} A)\right)'=0$$ and check $A=\frac{\pi}{2}.$
Indeed, $$\left(\arctan\left(\frac{1}{2}\tan2x\right)+\arctan\cot{x}+\arctan\cot^3x\right)'=$$
$$=\frac{\frac{1}{\cos^22x}}{1+\frac{1}{4}\tan^22x}+\frac{-\frac{1}{\sin^2x}}{1+\cot^2x}+\frac{3\cot^2x\cdot\left(-\frac{1}{\sin^2x}\right)}{1+\cot^6x}=$$
$$=\frac{4}{4\cos^22x+\sin^22x}-1-\frac{3\sin^2x\cos^2x}{\sin^6x+\cos^6x}=$$
$$=\frac{4}{4-3\sin^22x}-1-\frac{3\sin^2x\cos^2x}{1-3\sin^2x\cos^2x}=$$
$$=\frac{1}{1-3\sin^2x\cos^2x}-1-\frac{3\sin^2x\cos^2x}{1-3\sin^2x\cos^2x}=0.$$
Thus, our expression is a constant on all interval of the domain:
$$(0,\pi)\setminus\left\{\frac{\pi}{4},\frac{3\pi}{4}\right\},$$
for which it's enough to check $$x\in\left\{\frac{\pi}{8},\frac{\pi}{2},\frac{7\pi}{8}\right\},$$ which gives $\pi$ on $\left(0,\frac{\pi}{4}\right)$, $0$ on $\left(\frac{\pi}{4},\frac{3\pi}{4}\right)$ and $-\pi$ on $\left(\frac{3\pi}{4},\pi\right).$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2649051",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
}
|
Prove by induction that $1+2^{2^n}+2^{2^{n+1}}$ is divisible by $7$ As said in the title you have to show by induction that 1+$2^{2^n}+2^{2^{n+1}}$ is divisible by 7. So you start with n=0, that gives 1+2+4=7. So the start is shown.
Let 1+$2^{2^n}+2^{2^{n+1}}$ be divisble by 7 for a n.
I've tried several attempts but I ended up in a mess.
For example the latest attempt:
We have $1+2^{2^{n+1}}+2^{2^{n+2}}$
Adding $2^{2^n}-2^{2^n}$ gives you
$1+2^{2^{n+1}}+2^{2^n}+2^{2^{n+2}}-2^{2^n}$
As the first three summands resemble our Assumption, only $2^{2^{n+2}}-2^{2^n}$ needs to be proved as a multiple of seven.
But I am stuck at trying to show this.
As stated that might be a incorrect approach by myself, so I'm not really sure whether or not that is going in the correct direction. In the end I'd really appreciate some help on this question!
|
Starting with $n = 0$, we have $1+2^{2^0}+2^{2^{0+1}} = 7$, which is clearly divisible by $7$. Now suppose inductively that $n \ge 1$ and argument holds for all $n$. Then for $n+1$, we have
$$1+2^{2^{n+1}}+2^{2^{n+2}} = 1+2^{2\cdot2^n}+2^{2\cdot2^{n+1}} = 1+(2^{2^n})^2+(2^{2^{n+1}})^2$$
Now and and subtract $(2^{2^n})^2$, we have
$$1+2(2^{2^n})^2+(2^{2^{n+1}})^2-(2^{2^n})^2 = [1+(2^{2^n})^2]^2-(2^{2^n})^2$$ $$ = [1+(2^{2^n})^2+2^{2^n}]\cdot[1+(2^{2^n})^2-2^{2^n}]$$
$$ = [1+2^{2^{n+1}}+2^{2^n}]\cdot[1+2^{2^{n+1}}-2^{2^n}]$$
Here, notice that by inductive hyphothesis, $7|(1+2^{2^{n+1}}+2^{2^n})$. Therefore $7$ divides the whole expression and argument holds for $n+1$. Therefore by induction, argument holds for all $n$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2649803",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
}
|
Three summations
Given
$$S_n=\sum^{n-1}_{i=0}\sum^{i-1}_{j=0}\sum^{j-1}_{k=0} (i+j+k), $$
there are positive integers $A$ and $B$ such that
$$\frac{1}{S_3}+\frac{1}{S_4}+\frac{1}{S_5}+\dots=A-\frac{2\pi^2}{B}$$
Find $A+B$.
MyApproach:
I need to solve the innermost summation first and then proceed to the last one. But I did not proceed. The summation for $N$ natural numbers are $n(n+1)/2$.
Any help?
|
Expanding the triple summation:
$$\begin{align} S_n=&\sum^{n-1}_{i=0}\sum^{i-1}_{j=0}\sum^{j-1}_{k=0} (i+j+k)=\\ &\sum^{n-1}_{i=0}\sum^{i-1}_{j=0} \left(ij+j^2+\frac{j(j-1)}{2}\right)=\\ &\sum^{n-1}_{i=0}\left(i\cdot \frac{i(i-1)}{2}+\frac{3}{2}\cdot \frac{(i-1)i(2i-1)}{6}-\frac{1}{2}\cdot \frac{(j-1)j}{2}\right)=\\
\frac14&\sum^{n-1}_{i=0}\left(4i^3-6i^2+2i\right)=\\
\frac14&\left(4\cdot \frac{(n-1)^2n^2}{4}-6\cdot\frac{(n-1)n(2n-1)}{6}+2\cdot\frac{(n-1)n}{2}\right)=\\
\frac14&n(n-1)^2(n-2).\end{align}$$
Hence:
$$\begin{align}\frac{1}{S_n}=\frac{4}{n(n-1)^2(n-2)}=&4\left(\frac{1}{n(n-2)}-\frac{1}{(n-1)^2}\right)=\\
&4\left(\frac{1}{2(n-2)}-\frac{1}{2n}-\frac{1}{(n-1)^2}\right)=\\ &\frac{2}{n-2}-\frac{2}{n}-\frac{4}{(n-1)^2}.\end{align}$$
So:
$$\sum_{n=3}^{\infty}\frac{1}{S_n}=2\sum_{n=3}^{\infty} \left(\frac{1}{n-2}-\frac{1}{n}\right)-4\sum_{n=3}^{\infty} \frac{1}{(n-1)^2}=\\
2\cdot \frac32-4\cdot \frac{\pi^2}{6}=3-\frac{2\pi^2}{3}.$$
Finally, $A+B=3+3=6$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2652769",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
}
|
Given $x = \sqrt[4]{x^{3}+6x^{2}}$, determine whether the sum of all possible roots of $x$ is equal to $1$. Given $$x = \sqrt[4]{x^{3}+6x^{2}}$$
Quantity $A$: Sum of all possible roots of $x$
Quantity $B$: $1$
My solution: I've taken fourth power on both sides of the equation to get $$x^{4} = x^{3} + 6x^{2}.$$
Rearranging and factorizing $x^{2}$ gives $$x^{2}(x^{2} - x - 6) = 0.$$
Further factorizing gives $$x^{2}(x-3)(x+2) = 0.$$
So the roots of $x$ are $0$,$-2$,and $3$.
So the sum of the roots is $1$. Thus, my answer is $C$ $(A=B)$.
But the answer in the practice test was $A$. Why is this so?
|
$-2$ is not a solution of $x = \sqrt[4]{x^{3}+6x^{2}}$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2652883",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
}
|
How to calculate the integral $\int\frac{1}{\sqrt{(x^2+8)^3}}dx$? I need to solve something like this
$$\int\frac{1}{\sqrt{(x^2+8)^3}}dx$$
Wolfram alpha says the solution is $$\frac{x}{8\sqrt{x^2+8}} + c$$
The problem is that the integrand is obtained by the quotient rule:
$$\bigg(\frac{g(x)}{h(x)}\bigg)'=\frac{g'(x)h(x)-g(x)h'(x)}{h^2(x)}$$
$$\bigg(\frac{x}{8\sqrt{x^2+8}}\bigg)'=\frac{1}{8}\cdot\frac{\sqrt{x^2+8}-\frac{x^2}{\sqrt{x^2+8}}}{x^2+8}=\frac{1}{8}\cdot\frac{\frac{x^2+8-x^2}{\sqrt{x^2+8}}}{x^2+8}=\frac{1}{8}\cdot\frac{8}{\sqrt{(x^2+8)^3}}=\frac{1}{\sqrt{(x^2+8)^3}}$$
It there a way to extract the solution from these types of integrals which argument is born from the easy quotient rule?
|
I don't know of any general formula for the quotient rule.
But for an integral of the form
$$\int \frac{dx}{\left(ax^n+b\right)^{k}}$$
You can try the following substitutions:
*
*If $n=1$, put $u=ax+b$. ($n=0$ is trivial.)
*If $n=2$, put $x=\left(\frac{b}{a}\right)^\frac{1}{n} tan u$.
*If $n\ge 3$, put $x=\left(\frac{b}{a}\right)^\frac{1}{n} u$ and then factorise the polynomial in the denominator into linear and quadratic terms and then, after using the method of partial fractions to separate the terms, apply 1 and 2 as above.
This is just a general procedure, and may not be applicable always. Some problems will possess solutions obtained only by special methods.
Hope this helps.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2653216",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 2
}
|
Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.