Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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If $a,b,c \in \mathbb{R}$ and $a+b+c = 1$, prove that $(2a+b)(2b+c)(2c+a)+(1+a+2b)(1+b+2c)(1+c+2a) \leq 9$ If $a,b,c \in \mathbb{R}$ and $a+b+c = 1$, prove that
$(2a+b)(2b+c)(2c+a)+(1+a+2b)(1+b+2c)(1+c+2a) \leq 9$
| Hint :
As pointed out by user Albus Dumbledore we can just consider the case where all the variable are positive now try to show that with this constraint of positivity :
$$(2a+b)(2b+c)(2c+a)\leq 1$$
And :
$$(1+a+2b)(1+b+2c)(1+c+2a)\leq 8$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3912976",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Let E be the standard basis and S another basis for $\mathbb{R}^2$. Assume the matrix M is understood to be in terms of E. Find $[M]_S.$ The question: Let $E = \left\{\begin{pmatrix} 1\\ 0\end{pmatrix}, \begin{pmatrix} 0\\ 1\end{pmatrix} \right\}$ be the standard basis for $\mathbb{R}^2$. The set $S = \left\{\begin{pmatrix} 1\\ 2\end{pmatrix}, \begin{pmatrix} -3\\ 2\end{pmatrix} \right\}$ is another basis for $\mathbb{R}^2$, Assume the matrix $M =\begin{bmatrix} 3 & 5\\ -3 & 2 \end{bmatrix}$ is understood to be in terms of $E$. Find $[M]_S.$
I am really confused on what this is asking of me and how I should approach this. I think that it is saying that I have $[M]_E$, so then to find $M$. I have tried
$
\left[
\begin{array}{cc|cc}
3 & 5 & 1 & 0 \\
-3 & 2 & 0 & 1 \\
\end{array}
\right]
\rightarrow
\left[
\begin{array}{cc|cc}
1 & 0 & \frac{2}{21} & \frac{-5}{21}\\
0 & 1 & \frac{1}{7}& \frac{1}{7}\\
\end{array}
\right]
$, so that I am left with $M = \frac{1}{21}\begin{bmatrix}
2 & -5\\
3& 3\\
\end{bmatrix}$. I then used this new $M$ to solve for $[M]_S$ by doing $MS = \frac{1}{21}\begin{bmatrix}
2 & -5\\
3& 3\\
\end{bmatrix}
\begin{bmatrix}
1 & -3\\
2& 2\\
\end{bmatrix}=
\frac{1}{21}
\begin{bmatrix}
-8 & -16\\
9& -3\\
\end{bmatrix}
$.
But this seems extremely wrong. My professor didn't do a great job going over bases, so any advice would be appreciated.
| $$[M]_E =\begin{bmatrix} 3 & 5\\ -3 & 2 \end{bmatrix}$$
means that the linear map $M$ acts on the basis $E$ as
$$M\begin{bmatrix}1 \\ 0 \end{bmatrix} = 3\begin{bmatrix}1 \\ 0 \end{bmatrix}-3\begin{bmatrix}0 \\ 1 \end{bmatrix}, \quad M\begin{bmatrix}0 \\ 1 \end{bmatrix} = 5\begin{bmatrix}1 \\ 0 \end{bmatrix}+2\begin{bmatrix}0 \\ 1 \end{bmatrix}$$
or in general
$$M\begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 3 & 5\\ -3 & 2 \end{bmatrix}\begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix}3x+5y \\ -3x+2y \end{bmatrix}.$$
You have to find a matrix $$[M]_S = \begin{bmatrix} \alpha & \gamma \\ \beta & \delta\end{bmatrix}$$
such that
$$M\begin{bmatrix} 1 \\ 2\end{bmatrix} = \alpha \begin{bmatrix} 1 \\ 2\end{bmatrix} + \beta \begin{bmatrix} -3 \\ 2\end{bmatrix}, \quad M\begin{bmatrix} -3 \\ 2\end{bmatrix} = \gamma \begin{bmatrix} 1 \\ 2\end{bmatrix} + \delta \begin{bmatrix} -3 \\ 2\end{bmatrix}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3916118",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Show that $\sin\left(\frac{1}{n}\right)=\frac{1}{n}-\frac{1}{3!~n^3}+O\left(\frac{1}{n^5}\right)$
Show that $\sin\left(\frac{1}{n}\right)=\frac{1}{n}-\frac{1}{3!~n^3}+O\left(\frac{1}{n^5}\right)$.
In fact, this result is pretty obvious but when I did this in homework I basically got no points because I was to shallow in my reasoning. So here is my new attempt.
Proof:
We know that the Taylor series at point $0$ of $\sin(x)$ exists so the "tail" $\sum\limits_{k=2}^{\infty}\frac{(-1)^k}{(2k+1)!~~n^{2k+1}}$ of it also exists.
Basically, we now want to show that there exist a constant $C>0$ and an index $n_0$ such that for all $n>n_0$ it holds: $\Big|\sum\limits_{k=2}^{\infty}\frac{(-1)^k}{(2k+1)!~~n^{2k+1}}\Big|\leq C \cdot \frac{1}{n^5}$ which is equivalent to saying $\sum\limits_{k=2}^{\infty}\frac{(-1)^k}{(2k+1)!~~n^{2k+1}}=O\left(\frac{1}{n^5}\right)$.
For this we show that $\lim\limits_{n\to\infty}n^5\cdot\sum\limits_{k=2}^{\infty}\frac{(-1)^k}{(2k+1)!~~n^{2k+1}}$ converges.
Because $\sum\limits_{k=2}^{\infty}\frac{(-1)^k}{(2k+1)!~~n^{2k+1}}$ converges we are allowed to perform the multiplication with a fixed $n$: $n^5\cdot \sum\limits_{k=2}^{\infty}\frac{(-1)^k}{(2k+1)!~~n^{2k+1}}=\sum\limits_{k=2}^{\infty}\frac{(-1)^k}{(2k+1)!~~n^{2k-4}}$.
Now the question is what happens to $\sum\limits_{k=2}^{\infty}\frac{(-1)^k}{(2k+1)!~~n^{2k-4}}$ if $n\to\infty$? Again, this seems very obvious as $n$ only appears in the denominators but my tutor said that this is not a rigorous argument. So I guess that we must find a closed form expression for $\sum\limits_{k=2}^{\infty}\frac{(-1)^k}{(2k+1)!~~n^{2k-4}}$ or we must at least find an upper bound which attains a closed form expression. I would simply take a geometric series $\sum\limits_{k=1}^{\infty}\frac{1}{n^k}=\frac{1}{n-1}$ as $\frac{(-1)^k}{(2k+1)!~~n^{2k-4}}\leq\frac{1}{n^k}$ holds for all $n$ and $k$.
Hence, for all $n$ we have:
$$
0\leq n^5\cdot\sum\limits_{k=2}^{\infty}\frac{(-1)^k}{(2k+1)!~~n^{2k+1}}=\sum\limits_{k=2}^{\infty}\frac{(-1)^k}{(2k+1)!~~n^{2k-4}}\leq \frac{1}{n-1}\\\implies \lim\limits_{n\to\infty}n^5\cdot\sum\limits_{k=2}^{\infty}\frac{1}{(2k+1)!~~n^{2k+1}}=0.
$$
So for a constant $C>0$ we find an index $n_0$ which is large enough such that for all $n>n_0$:
$$\Big|n^5\cdot\sum\limits_{k=2}^{\infty}\frac{(-1)^k}{(2k+1)!~~n^{2k+1}}\Big|<C \\\implies \Big|\sum\limits_{k=2}^{\infty}\frac{(-1)^k}{(2k+1)!~~n^{2k+1}}\Big|<C \frac{1}{n^5}\\\implies
\sum\limits_{k=2}^{\infty}\frac{(-1)^k}{(2k+1)!~~n^{2k+1}}=O\left(\frac{1}{n^5}\right)\\\implies
\sin\left(\frac{1}{n}\right)=\frac{1}{n}-\frac{1}{3!~n^3}+O\left(\frac{1}{n^5}\right).
$$
Is this correct?
| Comment on the Approach in the Question
It would be more rigorous to cite Dominated Convergence as a reason that
$$
\lim_{n\to\infty}n^5\sum_{k=2}^\infty(-1)^k\frac1{(2k+1)!}\frac1{n^{2k+1}}=-\frac1{120}
$$
thereby bounding the remainder of the series.
One could also bound by a geometric series:
$$
\begin{align}
\left|\sum_{k=2}^\infty(-1)^k\frac1{(2k+1)!}\frac1{n^{2k+1}}\right|
&\le\sum_{k=2}^\infty\frac1{120n^{2k+1}}\\
&=\frac1{120\!\left(n^5-n^3\right)}
\end{align}
$$
However, both of these approaches require that you have either defined $\sin(x)$ as its Taylor series, or that you've shown that the Taylor series converges, which seems to be doing more than this question asks, so circularity is a concern.
Another Approach
Integrate by parts $4$ times:
$$
\begin{align}
\sin(x)
&=\int_0^x\cos(t)\,\mathrm{d}t\\
&=x-\int_0^x(x-t)\sin(t)\,\mathrm{d}t\\
&=x-\frac12\int_0^x(x-t)^2\cos(t)\,\mathrm{d}t\\
&=x-\frac16x^3+\frac16\int_0^x(x-t)^3\sin(t)\,\mathrm{d}t\\
&=x-\frac16x^3+\frac1{24}\int_0^x(x-t)^4\cos(t)\,\mathrm{d}t\\
&=x-\frac16x^3+O\!\left(x^5\right)
\end{align}
$$
Since
$$
\begin{align}
\left|\int_0^x(x-t)^4\cos(t)\,\mathrm{d}t\right|
&\le\left|\int_0^x(x-t)^4\,\mathrm{d}t\right|\\
&=\frac15|x|^5
\end{align}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3917184",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Prove recursion for sum One should show, that
$$S(n,x,y):= (y-n)\sum_{k=0}^n \binom{n}{k} (x+k)^{k+1} (y-k)^{n-k-1}$$
satisties the recursion
$$S(n,x,y)=x(x+y)^n +n S(n-1,x+1,y-1),\quad \text{for} ~n\geq1,\quad S(0,x,y)=x.$$
I know that we should use Abels' generalization of the binomial theorem, that is
$$\begin{equation} (x + y)^n = \sum_{k=0}^n \binom{n}{k} x (x-kz)^{k-1} (y + kz)^{n-k}\end{equation}$$
for any $x, y, z \in\mathbb{R}$.
STARTING SOLUTION:
If I start with the right side, I get
$$\begin{align*}x (x+y)^n &+ n S(n-1,x+1,y-1) \\
&= x (x+y)^n + n (y-n+1) \sum_{k=0}^{n-1} \binom{n-1}{k} (x+1+k)^{k+1} (y-1-k)^{n-k-2}\\
&= x\sum_{k=0}^n \binom{n}{k} x (x-kz)^{k-1} (y + kz)^{n-k} \\
&\hspace{2.1cm}+ n (y-n+1) \sum_{k=0}^{n-1} \binom{n-1}{k} (x+1+k)^{k+1} (y-1-k)^{n-k-2}\\
&=\cdots\\
&= (y-n) \sum_{k=0}^n \binom{n}{k} (x+k)^{k+1} (y-k)^{n-k-1} = S(n,x,y),\end{align*}$$
I do not really know how to continue from here, those anyone have a clue what could be inserted for z (and how one could shift the indices of the second sum to get the right term)?
| We start with the right-hand side and keep at first the focus on $nS(n-1,x+1,y+1)$.
We obtain
\begin{align*}
&\color{blue}{nS(n-1,x+1,y-1)}\\
&\qquad =n(y-n)\sum_{k=0}^{n-1}\binom{n-1}{k}(x+1+k)^{k+1}(y-1-k)^{n-k-2}\tag{1}\\
&\qquad= n(y-n)\sum_{k=1}^{n}\binom{n-1}{k-1}(x+k)^k(y-k)^{n-k-1}\tag{2}\\
&\qquad=(y-n)\sum_{k=0}^n\binom{n}{k}k(x+k)^k(y-k)^{n-k-1}\tag{3}\\
&\qquad\,\,\color{blue}{=S(n,x,y)-(y-n)\sum_{k=0}\binom{n}{k}x(x+k)^k(y-k)^{n-k-1}}\tag{4}
\end{align*}
Comment:
*
*In (1) we use the definition of $S(n,x,y)$.
*In (2) we shift the index $k$ by one.
*In (3) we use the binomial identity $\binom{n}{k}=\binom{n-1}{k-1}\frac{n}{k}$. We also start with index $k=0$ which doesn't change anything.
*In (4) we use $k=(x+k)-x$ and obtain a representation with $S(n,x,y)$.
Looking at (4) we observe the following is left to show:
\begin{align*}
(x+y)^n=\color{blue}{(y-n)}\sum_{k=0}\binom{n}{k}(x+k)^k(y-k)^{n-k-1}\tag{5}
\end{align*}
We use Abel's generalisation with $z=-1$ which gives
\begin{align*}
(x + y)^n = \color{blue}{x}\sum_{k=0}^n \binom{n}{k} (x+k)^{k-1} (y - k)^{n-k}\tag{6}
\end{align*}
and we see (6) is close to the wanted representation (5) indicating we could give the substitution $\color{blue}{x\to y-n}$ a try.
This substitution finally does the job.
We obtain
\begin{align*}
\color{blue}{(x+y)^n}&=((y-n)+(x+n))^n\tag{7}\\
&=\sum_{k=0}^n\binom{n}{k}(y-n)((y-n)+k)^{k-1}((x+n)-k)^{n-k}\tag{8}\\
&=\sum_{k=0}^n\binom{n}{k}(y-n)(y-(n-k))^{k-1}(x+(n-k))^{n-k}\\
&\,\,\color{blue}{=(y-n)\sum_{k=0}^n\binom{n}{k}(y-k)^{n-k-1}(x+k)^{k}}\tag{9}\\
\end{align*}
which is besides a factor $x$ the sum in (4) and the claim follows.
Comment:
*
*In (7) we use the substitution $x\to y-n$ and $y\to x+n$.
*In (8) we use Abel's generalised binomial theorem from (6).
*In (9) we change the order of summation $k\to n-k$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3919610",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
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} |
Given A and Find B such that $\frac{\lVert A_n-B\rVert_\infty}{\lVert A_n \rVert_\infty} - \frac{1}{K(A_n)}$ < $\frac{1}{n^2}$ $A_n=\begin{bmatrix}
1 & 2 \\
2 & 4+1/n^2\\
\end{bmatrix}$
How can we find singular B such
$\frac{\lVert A_n-B\rVert_\infty}{\lVert A_n \rVert_\infty} - \frac{1}{K(A_n)}$ < $\frac{1}{n^2}$
$A_n^{-1}=\begin{bmatrix}
4n^2+1 & -2n^2 \\
-2n^2 & n^2\\
\end{bmatrix}$
Then, we can find $K(A_n)$. But then I don't know what to do. (K refer condition number.)
| The condition number $\kappa(A) = \| A \| \| A^{-1}\| $ can take any norm. So I'll assume you're also using $\|\cdot \|_{\infty}$.
First we have $ \frac{\| A_{n} - B \|_{\infty} }{\| A_{n} \|_{\infty} } - \frac{1}{\kappa(A_{n})} = \frac{\| A_{n} - B \|_{\infty} }{\| A_{n} \|_{\infty} } - \frac{1}{\|A_{n}\|_{\infty} \| A_{n}^{-1} \|_{\infty}}$
Let's find $\| A_{n}\|_{\infty} $ and $\| A_{n}^{-1}\|_{\infty} $
$ \|A_{n}\|_{\infty} = \max \{3 , 6 + \frac{1}{n^{2}} \} = 6 + \frac{1}{n^{2}}$
Now, first $A^{-1} = \begin{bmatrix} 4n^{2} + 1 & -2n^{2} \\ -2n^{2} & n^{2} \end{bmatrix}$.
Now, we get $ \|A_{n}^{-1}\|_{\infty} = \max \{ 6n^{2}+1, 3n^2 \} = 6n^{2}+1 $.
Now, we have
$ \frac{\| A_{n} - B \|_{\infty} }{\| A_{n} \|_{\infty} } < \frac{1}{\kappa(A_{n})} + \frac{1}{n^2} $
Get a common denominator
$ \frac{\| A_{n} - B \|_{\infty} }{\| A_{n} \|_{\infty} } <\frac{n^{2}+\kappa(A_{n})}{n^{2} \kappa(A_{n})} $
Now multiply by $\kappa(A_{n})$
$\| A_{n} - B \|_{\infty} <\frac{\| A_{n} \|_{\infty} ( n^{2}+\kappa(A_{n})) }{n^{2} \kappa(A_{n})} $
Replace $\kappa(A_{n}) = \|A_{n}\|_{\infty} \| A_{n}^{-1} \|_{\infty}$
$\| A_{n} - B \|_{\infty} <\frac{\| A_{n} \|_{\infty} ( n^{2}+\kappa(A_{n})) }{n^{2} \|A_{n}\|_{\infty} \| A_{n}^{-1} \|_{\infty} } $
Now cancel $ \|A_{n} \|_{\infty}$
$\| A_{n} - B \|_{\infty} <\frac{n^{2}+\kappa(A_{n}) }{n^{2} \| A_{n}^{-1} \|_{\infty} } $
Replace $\kappa(A_{n}) = \|A_{n}\|_{\infty} \| A_{n}^{-1} \|_{\infty}$
$\| A_{n} - B \|_{\infty} <\frac{n^{2}+\|A_{n}\|_{\infty} \| A_{n}^{-1} \|_{\infty} }{n^{2} \| A_{n}^{-1} \|_{\infty} } = \frac{\| A_{n}\|_{\infty} }{n^{2}} + \frac{1}{\| A_{n}^{-1} \|_{\infty} }$
Now substitute in some things
$ = \frac{ 6 + \frac{1}{n^2} }{n^{2}} + \frac{1}{6n^{2}+1} =\frac{37n^{4}+12n^{2}+1}{n^{4}(6n^{2}+1)}$
And suppose that $ B = \begin{bmatrix} a & b \\ c & d\end{bmatrix} $ then we will find that $ \| A_{n} - B\|_{\infty} = \max \{ |1-a| + |2-b| , |2-c| + |4+\frac{1}{n^{2}} -d| \} $
and
$ \max \{ |1-a| + |2-b| , |2-c| + |4+\frac{1}{n^{2}} -d| \} < \frac{37n^{4}+12n^{2}+1}{n^{4}(6n^{2}+1)}$
Now
$ B = \begin{bmatrix} 1 & 2 \\ 2 & 4 \end{bmatrix}$
This matrix is singular
It gives us
$ \max \{ 0 , \frac{1}{n^{2}} \} = \frac{1}{n^{2}} < \frac{37n^{4}+12n^{2}+1}{n^{4}(6n^{2}+1)}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3920809",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Solve: $3\sin{2x}+4\cos{2x}-2\cos{x}+6\sin{x}-6=0$
Solve: $3\sin{2x}+4\cos{2x}-2\cos{x}+6\sin{x}-6=0$
My Try
$6\sin{x}\cos{x}+4(\cos^2{x}-\sin^2{x})-2\cos{x}+6\sin{x}-6=0$
I have expanded the equation, But I cannot proceed further, Any hint would be appreciated. Thank you!
| $x=2y$ we get after some manipulation $$-4\cos^4 y-36\cos^2y\sin^2 y+24\cos^3 y\sin y=0$$ $$-4\cos^2 y{(\cos y-3\sin y)}^2=0$$ can you end it??
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3921973",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 1
} |
Decompose $\frac{1}{a \sin x +b}$ into sum of trigonometric functions I am wondering if it is possible to write
$$f(x)=\frac{1}{a \sin x +b}$$
as a sum of functions that don't have trigonometric functions in the denominator? Or just pure trigonometric functions, without a constant added. So maybe something like this:
$$\frac{1}{a \sin x +b}=A\sin^3 x \cos^2x + B\sin x \cos ^4 x + C\frac{1}{\cos x}+ D\frac{\cos^2 x }{\sin x}+\dots $$
The reason I'm asking is that I need to integrate products of $f(x)$ with powers $\cos^n x$, and it would be really neat to get rid of the constant in the denominator, as integrals of the form $\int\frac{\cos^n x}{a \sin x +b}dx$ are extremely hard to solve, whereas integral tables contain solutions to $\int \sin^p x \cos^q x dx$ and similar integrals.
One potential starting point could be
$$\frac{1}{a \sin x +b}=\frac{a \sin x -b}{(a \sin x +b)(a \sin x -b)}=\frac{a \sin x -b}{(a^2 \sin^2 x -b^2)}=...$$
with the goal to eventually make the replacement $\sin^2 x +\cos^2 x =1$. Can such a method be formalized? Or is the decomposition I have in mind provable not possible? Thanks for your help!
| An idea for a solution just came to my mind, which I am going to post here hoping it might be useful for somebody in the future. Unfortunately, the condition doesn't work in my case, so that I might go with the Fourier series solution. But for $a\geq b$, we can write
$$\frac{1}{a \sin x +b}=\frac{1}{a}\frac{1}{\sin x +\frac{b}{a}}=\frac{1}{a}\frac{1}{\sin x +\sin\left(\arcsin\left(\frac{b}{a}\right)\right)}=\frac{1}{a}\frac{1}{2\sin\left(\frac{x+\arcsin\left(\frac{b}{a}\right)}{2}\right) \cos\left(\frac{x-\arcsin\left(\frac{b}{a}\right)}{2}\right)}.$$
Trigonometric expansion with Mathematica yields
$$\frac{1}{2\sin\left(\frac{x+\arcsin\left(\frac{b}{a}\right)}{2}\right) \cos\left(\frac{x-\arcsin\left(\frac{b}{a}\right)}{2}\right)}\\=\frac{1}{2} \sec \left(\frac{c}{2}\right) \sec (c) \sin \left(\frac{x}{2}\right) \sec \left(\frac{c}{2}-\frac{x}{2}\right)-\frac{1}{2} \csc \left(\frac{c}{2}\right) \sec (c) \sin \left(\frac{x}{2}\right) \csc \left(\frac{c}{2}+\frac{x}{2}\right)+\csc (c)$$
with $c=\arcsin\left(\frac{b}{a}\right)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3924529",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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Is $\mathbb{Z}[√13] $ a Unique factorization domain? I think it is not so as $12 $ can be written in two ways $12=2.6=(1+\sqrt{13})(-1+\sqrt{13})$. Are these two factorization unique upto irreducibles? Please help.
| $\mathbf{2}$ is Irreducible in $\boldsymbol{\mathbb{Z}\!\left[\sqrt{13}\right]}$
Suppose that
$$
\left(a+b\sqrt{13}\right)\left(c+d\sqrt{13}\right)=2\tag1
$$
where $2\nmid\left(a+b\sqrt{13}\right)$ and $2\nmid\left(c+d\sqrt{13}\right)$. Then
$$
ac+13bd=2\tag2
$$
and
$$
ad+bc=0\tag3
$$
If $2\mid a$, then since $2\nmid\left(a+b\sqrt{13}\right)$, we must have $2\nmid b$. Then $(2)$ says that $2\mid d$ and $(3)$ says that $2\mid c$. However, this means that $2\mid\left(c+d\sqrt{13}\right)$, which contradicts the hypotheses. Therefore, $2\nmid a$. Similarly, $2\nmid c$.
Multiplying $(2)$ by $ac$ and applying $(3)$ to eliminate $d$ gives
$$
a^2c^2-13b^2c^2=2ac\tag4
$$
and therefore,
$$
(ac-1)^2-1\equiv b^2c^2\pmod4\tag5
$$
However, since $a,c\equiv1\pmod2$, we must have $(ac-1)^2-1\equiv3\pmod4$. Since $3$ is not a quadratic residue mod $4$, $(5)$ can have no solutions.
Thus, $2$ is irreducible in $\mathbb{Z}\!\left[\sqrt{13}\right]$.
$\boldsymbol{\mathbb{Z}\!\left[\sqrt{13}\right]}$ is not a UFD
$$
\left(\sqrt{13}-3\right)\left(\sqrt{13}+3\right)=2\cdot2
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3924667",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 1
} |
Finite sum of combinations Suppose $a$ and $b$ are two positive integers with $a \leq b$. Do we have the following?
$$\sum_{k=0}^{a}\binom{a}{k}\binom{b}{b-k} = \binom{a+b}{a}$$
This is very intuitive to me: we want to choose $a$ items from a set of $a+b$ items. We can choose $k \leq a$ items from the first $a$ items and $b-k$ items from the last $b$ items.
| Note that
$$ (1+x)^{a+b}=(1+x)^a(1+x)^b. $$
By using
$$ (1+x)^n=\sum_{k=0}^n\binom{n}{k}x^k $$
one has
$$ (1+x)^{a+b}=\sum_{k=0}^n\binom{a+b}{k}x^k \tag1$$
and
$$ (1+x)^a(1+x)^b=\sum_{k=0}^a\binom{a}{k}x^k\sum_{k=0}^b\binom{b}{k}x^k=\sum_{j=0}^a\sum_{k=0}^b\binom{a}{j}\binom{b}{k}x^{j+k}. \tag2 $$
From (1), the coefficient of $x^a$ is
$$\binom{a+b}{a} $$
and from (2), the coefficient of $x^a$ is
$$ \sum_{j=0}^a\binom{a}{j}\binom{b}{a-j}. $$
So
$$ \sum_{j=0}^a\binom{a}{j}\binom{b}{a-j}=\binom{a+b}{a}. $$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Use of definition of limit to prove $\lim_{x→2} \dfrac{x}{x^2-2}=1$ I know that by definition I have to prove that $$\lim_{x→2}\dfrac{x}{x^2−2}=1⟺∀ϵ>0,∃δ>0,\, 0<|x−2|<δ⟹\left|\frac{x}{x^2−2}−1\right|<ϵ.$$
I have:
$\left|\dfrac{-x^2+x+2}{x^2-2}\right|= \left|\dfrac{-(x^2-x-2)}{x^2-2}\right|= \left|\dfrac{-(x^2-x+1-3)}{x^2-2}\right|=$
$\left|\dfrac{-(x-1)^2+3}{x^2-2}\right|$.
So, I don't know how to continue.
| Note that $-(x^2-x-2)=-(x-2)(x+1)$ and that therefore$$\frac x{x^2-2}-1=-(x-2)\frac{x+1}{x^2-2}.$$Besides, if $|x-2|<\frac12$, then $\frac32<x<\frac52$ and therefore $\frac14<x^2-2<\frac{17}4$; in particular, $|x^2-2|>\frac14$. On the other hand, $\frac52<x+1<\frac72$; in particular, $|x+1|<\frac72$. So,\begin{align}\left|-(x-2)\frac{x+1}{x^2-2}\right|&=|x-2|\frac{|x+1|}{|x^2-2|}\\&<|x-2|\frac{7/2}{1/4}\\&=14|x-2|.\end{align}Therefore, given $\varepsilon>0$, take $\delta=\min\left\{\frac\varepsilon{14},\frac12\right\}$, and then$$|x-2|<\delta\implies\left|\frac x{x^2-2}-1\right|<\varepsilon.$$
| {
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"source": "stackexchange",
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Is there any simple formula for $x \sum\limits_{k=1}^n \frac{p_{n-k}(x)}{k}$? Is there any simple formula?
$$
\begin{align}
p_0(x) &= 1 \\
p_n(x) &= x \sum\limits_{k=1}^n \frac{p_{n-k}(x)}{k}
\end{align}
$$
I am thankful for any help or hint.
Edit: The first few polynomials are
$$
\begin{align}
p_0(x) &= 1 \\
p_1(x) &= x \\
p_2(x) &= x^2 + \frac{1}{2}x \\
p_3(x) &= x^3 + x^2 + \frac{1}{3}x \\
p_4(x) &= x^4 + \frac{3}{2}x^3 + \frac{11}{12}x^2 + \frac{1}{4}x \\
p_5(x) &= x^5 + 2x^4 + \frac{7}{4}x^3 + \frac{5}{6}x^2 + \frac{1}{5}x
\end{align}
$$
I know that the first coefficient is always $1$, the second is $\frac{n-1}{2}$ and the last is $\frac{1}{n}$.
| This is not an answer!
Just too long for a comment.
A closed form could be quite complicated...
Using Mathematica FindSequenceFunction powerful feature, I found
$$\frac{x}{n}+\frac{ (2 (\psi ^{(0)}n+\gamma ))}{n}x^2+\frac{\left(6 \psi ^{(0)}n^2+12 \gamma \psi ^{(0)}n+6 \psi ^{(1)}n+6 \gamma ^2-\pi ^2\right)}{2 n}x^3+\ldots$$
where $\gamma$ is the Euler constant, and $\psi ^{(0)},\psi ^{(1)}$ are polygamma and its derivative.
Indeed we get
$$
\begin{array}{cr}
1 & x \\
2 & x^2+\frac{x}{2} \\
3 & x^3+x^2+\frac{x}{3} \\
4 & \ldots+\frac{3 x^3}{2}+\frac{11 x^2}{12}+\frac{x}{4} \\
5 & \ldots\frac{7 x^3}{4}+\frac{5 x^2}{6}+\frac{x}{5} \\
\end{array}
$$
| {
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"answer_count": 1,
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Prove that the function is Riemann-Integrable. This is an incorrect attempt! Look below for @Laithy 's advice and my second attempt.
I tried this problem but I'm not sure whether I am getting it right. I am having trouble with the very last step where I am supposed to show that $I^-(f) = I^+(f)$. Here is my attempt. Please point out any errors and help me justify why $I^-(f)$ and $I^+(f)$ are equal. Thank you!
$
f(x) =\begin{cases}
1 & 0\leq x < \frac{1}{2} \\
0 & x = \frac{1}{2} \\
1 & \frac{1}{2} < x \leq 1
\end{cases}
$
Let $P$ be the following partition $P = \{[0, \frac{1}{2}], [\frac{1}{2}, 1]\}$ of $[0, 1]$.
Consider $I_1 = [0, \frac{1}{2}]$. It follows that $m_1 = 0$ and $M_1 = 1$.
This implies $m_1(x_1 - x_0) = 0$ and $M_1(x_1 - x_0) = 1(\frac{1}{2} - 0) = \frac{1}{2}$.
Now consider $I_2 = [\frac{1}{2}, 1]$. It follows that $m_2 = 0$ and $M_2 = 1$.
This implies $m_2(x_2 - x_1) = 0$ and $M_2(x_2 - x_1) = 1(1 - \frac{1}{2}) = \frac{1}{2}$.
All this gives the following:
$S^- = 0$ and $S^+ = 1$.
Therefore...
$I^-(f) = \sup\{S^-, \sigma\in G[0, 1]\} = \sup\{0, \sigma\in G[0, 1]\}$
and
$I^+(f) = \inf\{s^+, \sigma\in G[0, 1]\} = \inf\{1, \sigma\in G[0, 1]\}$
How do I decide whether $I^-(f)$ equals $I^+(f)$?
| Okay, I tried this again and I hope @Laithy can help me figure out whether I got it correct this time. Feel free to chime in give your opinions, thanks!
$
f(x) =\begin{cases}
1 & 0\leq x < \frac{1}{2} \\
0 & x = \frac{1}{2} \\
1 & \frac{1}{2} < x \leq 1
\end{cases}
$
Let $\epsilon\in\mathbb{R}^+$, choose $\delta = \frac{\epsilon}{4}$ and let $P$ be the following partition of $[0, 1]$... $P = \{[0, \frac{1}{2}-\delta], [\frac{1}{2} - \delta, \frac{1}{2} + \delta], [\frac{1}{2} + \delta, 1]\}$.
Consider $[0, \frac{1}{2} - \delta]$. It follows that $m_1 = 1$ and $M_1 = 1$.
This implies: $m_1(x_1 - x_0) = \frac{1}{2} - \delta$ and $M_1(x_1 - x_0) = \frac{1}{2} - \delta$.
Now consider $[\frac{1}{2} - \delta, \frac{1}{2} + \delta]$. It follows that $m_2 = 0$ and $M_2 = 1$.
This implies: $m_2(x_2 - x_1) = 0$ and $M_2(x_2 - x_1) = 2\delta$.
Consider the final sub-division $[\frac{1}{2} + \delta, 1]$. It follows that $m_3 = 1$ and $M_3 = 1$.
This implies: $m_3(x_3 - x_2) = \frac{1}{2} - \delta$ and $M_3(x_3 - x_2) = \frac{1}{2} - \delta$.
All this gives the following results:
$U(f, P) = \frac{1}{2} - \delta + 2\delta + \frac{1}{2} - \delta = 1$
and
$L(f, P) = \frac{1}{2} - \delta + 0 + \frac{1}{2} - \delta = 1 - 2\delta$
Therefore...
$U(f, P) - L(f, P) = 1 - (1 - 2\delta) = 2\delta = 2\big(\frac{\epsilon}{4}\big) = \frac{\epsilon}{2} < \epsilon$
$f$ is Riemann-Integrable.
| {
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"source": "stackexchange",
"question_score": "2",
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How to prove that $50^{44}+30! \cdot44^{50}-24 $ is divisible by $31$? I am struggling with this example. This is a new topic for me. I would really appreciate some hints to understand it. I need part a) and b) to conclude the excercise.
I was thinking about Fermat, Euler or Wilson theorems but I am not sure if it is correct, of course I can use properties of congruences in number theory.
Part a): show that $50^{44}+30! \cdot 44^{50}-24 $ is divisible by $31$
Part b) : Conclude that $ (50^{44}+30!\cdot 44^{50})^{\varphi(31)} \equiv 1 \pmod{ 31}$
| \begin{align}
50^{44} + 30! \cdot 44^{50} &\equiv 50^{44} - 44^{50} \pmod{31}, \text{ By Wilson's Theorem } \\
&\equiv 2^{44} \cdot 5^{88} - 2^{100}\cdot 11^{50} \pmod{31} \\
&\equiv 2^{14}\cdot 5^{28} - 2^{10} \cdot 11^{20} \pmod{31}\text{, By Fermat's little Theorem } \\
&\equiv 2^{4} \cdot 5 - (-3)^{10} \pmod{31}, \\
&\equiv 2^4 \cdot 5 - 3^{10} \pmod{31} \\
&\equiv 2^4 \cdot 5 - 3^{3\times 3}\cdot 3 \pmod{31} \\
&\equiv 2^4 \cdot 5 - (-4)^3 \cdot 3 \pmod{31} \\
&\equiv 2^4 (5+12) \pmod{31} \\
&\equiv 2^4 (2^4+1) \pmod{31}
\\ &\equiv2^8 + 2^4 \pmod{31} \\
&\equiv 2^3 + 2^4 \pmod{31} \\
&\equiv 24 \pmod{31}
\end{align}
| {
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"source": "stackexchange",
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Let $f(x)=x^6-2x^3-8$ and $g(x)=x^2+2x+4$. Let $a_1$ through $a_6$ be its roots. Find the value of $\prod_{n=1}^{6} (g(a_n))$ Question:
Let $f(x)=x^6-2x^3-8$ and $g(x)=x^2+2x+4$. Let $a_1$ through $a_6$ be its roots. Find the value of $\displaystyle \prod_{n=1}^{6} (g(a_n))$
My process:
I first thought of setting $z=a_i^2+2a_i+4$ and then putting the value back into $f(z)$ and then from there finding the product using Vieta's. But the process turned incredibly long and tedious and I found it to be impossible to do by hand.
So I just went to to the brute force method, expanding the whole expression out. Writing all the terms would be really tedious so I would like to abbreviate it. Let $P= a_1 \cdot a_2 \cdots \cdot a_6$ and let $e_n$ denote the sum of roots taken $n$ at a time.
Here is the expression I got:
$$\prod_{i=1}^{6} (a_i^2+2i+4)=P^2+2e_5P+2^2e_4P+2^3e_3P+2^4e_2P+2^5e_1P+2^6P+2^5*4*e_5+2^4*4^2*e^4+2^3*4^3*e_3+2^2*4^4*e_2+2*4^5*e_1+4_6+4(\sum_{cyc}(a_1a_2a_3a_4a_5)^2))^+4^2(\sum_{cyc}(a_1a_2a_3a_4)^2)+4^3\sum_{cyc}(a_1a_2a_3)^2+4^4\sum_{cyc}(a_1a_2)^2+4^5\sum_{cyc}(a_1)^2$$
Which is a monster in its own right. However, I was extremely interested in the patterns that kept popping up. Like the $2e_5P+2^2e_4P+2^3e_3P+2^4e_2P+2^5e_1P+2^6P$ where the power of 2 and the index of $e$ keeps increasing and decreasing with each other. Therefore I wondered if there could be any general formula for any $f(x)$ and $g(x)$.
And that is my question, is there any way to find the answer efficiently without all this unnecessary calculation; and also can this process be generalised to any polynomials $f(x)$ and $g(x)$. And if it cannot be generalised, then is there any sort of algorithm or methodical way that one can take while solving this type of problem?
| The leading coefficient of $f$ is equal to one, so $\prod_{n=1}^{6} (g(a_n))$ is the resultant of $f$ and $g$, and that can be computed as a determinant whose entries are the coefficients of the polynomials:
$$
\prod_{n=1}^{6} (g(a_n)) = \operatorname{res}(f, g) =
\begin{vmatrix}
-8 & & 4 \\
0 & -8 & 2 & 4 \\
0 & 0 & 1 & 2 & 4 \\
-2 & 0 & & 1 & 2 & 4 \\
0 & -2 & & & 1 & 2 & 4 \\
0 & 0 & & & & 1 & 2 & 4 \\
1 & 0 & & & & & 1 & 2 \\
& 1 & & & & & & 1
\end{vmatrix}
= 1600
$$
where the empty entries are zero.
| {
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"answer_count": 4,
"answer_id": 2
} |
Proving that $\int_{0}^{1} \frac{dx}{1+x^a}>\frac{a}{1+a}, a>0$ Binding an integral from one or both sides is always interesting even if the integral is doable.
For instance, $$a>0, x \in (0,1) \implies 1>1-x^{2a} \implies \frac{1}{1+x^a} > 1-x^a.$$
Hence $$\int_{0}^{1} \frac{dx}{1+x^a}> \int_{0}^{1} (1-x^a)~ dx=\frac{a}{1+a}, a>0 ~~~(1)$$
The question is: How else (1) can be proved?
| We can know that
$$
\begin{align*}
1-\int_0^1{\frac{1}{1+x^a}\text{d}x}&=\int_0^1{\left( 1-\frac{1}{1+x^a} \right) \text{d}x}
\\
&=\int_0^1{\frac{x^a}{1+x^a}\text{d}x}
\\
&<\int_0^1{x^a\text{d}x}
\\
&=\frac{1}{a+1}
\end{align*}
$$
Therefore $\displaystyle \int_0^1{\frac{1}{1+x^a}\text{d}x}>\frac{a}{a+1}$.
| {
"language": "en",
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"source": "stackexchange",
"question_score": "1",
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Compute difficult integral $\int \frac{dx}{2 + x + \sqrt{1 - x^2}}$ To solve the integral
$$I = \int \frac{dx}{2 + x + \sqrt{1 - x^2}}$$
I have tried several things, such as $t = \arcsin x$, because $\cos(\arcsin x) = \sqrt{1 - x^2}$. If I am not wrong, we can conclude with this variable change
$$
I = \int \frac{\cos t\,dt}{2 + \sin t + \cos t}
$$
but if it were correct, how could I go on?
| Continue with $x=\sin t$
\begin{align}
&\int \frac{1}{2 + x + \sqrt{1 - x^2}} dx=\int \frac{\cos t}{2 + \sin t + \cos t}dt\\
= &\ \frac12\int \left(1+ \frac{\cos t-\sin t}{2 + \sin t + \cos t}-
\frac{2}{2 + \sin t + \cos t} \right)dt\\
=& \ \frac12t +\frac12 \ln(2 + \sin t + \cos t) -\sqrt2\tan^{-1} \frac{\tan\frac t2+1}{\sqrt2}\\
=&\ \frac12\sin^{-1}x+\frac12 \ln\left(2 + x+ \sqrt{1-x^2}\right) -\sqrt2\tan^{-1} \frac{1+x-\sqrt{1-x^2}}{x \sqrt2}
\end{align}
| {
"language": "en",
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"source": "stackexchange",
"question_score": "4",
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How to find the value of the constant $c$? Given $X,Y$ random variables continuously I need to find the Probability density function of $X$
$$f_{X,Y}(x,y) = \begin{cases} c e^{-y} \cdot e^{-\frac{ (y-1-x)^2 }{2}} & x >0 \\ 0 & x \leq 0
\end{cases}$$
my plan is to find $c$ and than I will be able to calculate $f_{\small {X}}$. to find $c$ I need to use the fact that $$1=\int _{-\infty }^{\infty }\int _0^{\infty }ce^{-y}e^{−\left(y−1−x\right)^2}\:dxdy$$
I am not sure if the limits of $y$ is true here, and I am stuck on calculate this integral.
assume that I found $c$ I just need to write
$$f_{\small {X}}=\int _{-\infty }^{\infty }ce^{-y}e^{−\left(y−1−x\right)^2}\:dy$$
Is my plan correct? how do I find $c$?
| To find $c$ we need to solve:
$$1=\int _{-\infty }^{\infty }\int _0^{\infty }ce^{-y}e^{−\left(y−1−x\right)^2}\:dxdy$$
this is a bit easier to solve is we first switch the order of integration and do a bit of rearranging
$$\frac{1}{c}=\int _{0}^{\infty }\int _{-\infty}^{\infty } \exp[-(y+(y-1-x)^2)] \:dydx$$
expanding and completing the square for the expression inside
$$y+(y-1-x)^2 = y^2-y(1+2x)+(1+x)^2$$
$$=\left(y-\frac{1+2x}{2}\right)^2-\left(\frac{1+2x}{2}\right)^2+(1+x)^2$$
$$=\left(y-\frac{1+2x}{2}\right)^2 +x +\frac{3}{4}$$
Returning to our original integral now
$$\frac{1}{c}=\int _{0}^{\infty }\int _{-\infty}^{\infty } \exp[-(y+(y-1-x)^2)] \:dydx$$
$$ = \int _{0}^{\infty }\int _{-\infty}^{\infty } \exp \left[-\left(y-\frac{1+2x}{2}\right)^2 - x -\frac{3}{4}\right] dydx$$
$$ = e^{-\frac{3}{4}} \int _{0}^{\infty } e^{ - x} \int _{-\infty}^{\infty } e^{-\left(y-\frac{1+2x}{2}\right)^2} dydx$$
Notice that the pdf of a normal distribution with mean $\left(\frac{1+2x}{2}\right)$ and var $\frac{1}{2}$ is
$$\frac{1}{\sqrt{\pi}}e^{-\left(y-\frac{1+2x}{2}\right)^2}\:$$
therefore the integral of the above from $-\infty$ to $\infty$ is $1$
$$ \frac{1}{c} = \sqrt{\pi} e^{-\frac{3}{4}} \int _{0}^{\infty } e^{ - x} \int _{-\infty}^{\infty } \frac{1}{\sqrt{\pi}}e^{-\left(y-\frac{1+2x}{2}\right)^2} dydx$$
$$ = \sqrt{\pi} e^{-\frac{3}{4}} \int _{0}^{\infty } e^{ - x} = \sqrt{\pi} e^{-\frac{3}{4}}$$
Thus $c = \frac{e^{\frac{3}{4}}}{\sqrt \pi}$
| {
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Alternate way to solve this algebra problem? Working on a problem that says given the absolute value of the difference of the roots of $ax^2 + bx + c$ as $2$, what is the absolute value of the difference of the roots of $ax^2 + 6bx + 36c$?
I reasoned that the difference of roots would just be $2\sqrt{b^2 - 4ac}$. Therefore, the value of $\sqrt{b^2 - 4ac}$ is $1$. Then, I used this information to answer the question:
$2\sqrt{36b^2 - 36(4ac)} =$ ?
$2\sqrt{36(b^2 - 4ac)}$
$12\sqrt{b^2 - 4ac}$
$=12$
The question hinted that I could use Vieta's formulas to solve the problem. Is my method correct, and is there a method involving Vieta's formulas?
| Note, for $ax^2+bx +c=0$
$$(x_1-x_2)^2=(x_1+x_2)^2-4x_1x_2=\frac{b^2}{a^2}-\frac{4c}a=2^2$$
Then, for $ay^2 + 6by + 36c=0$
$$(y_1-y_2)^2 = (y_1+y_2)^2-4y_1y_2=36\left(\frac{b^2}{a^2}-\frac{4c}a\right)=36\cdot 4
$$
Thus, $| y_1-y_2|= 12$.
| {
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"question_score": "2",
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Solving $\sqrt{x^2+ay^2} - \sqrt{x^2+y^2} = z$ for $x$ Suppose I have an expression as follows
$$\sqrt{x^2+ay^2} - \sqrt{x^2+y^2} = z$$
How would I go about rearranging this to deduce a value for x?
Thanks for the help.
| You have a system of two equations
$$\begin{cases}\sqrt{x^2+ay^2}-\sqrt{x^2+y^2} = z \\ \sqrt{x^2+ay^2}+\sqrt{x^2+y^2}
= \frac{(a-1)y^2}{z}\end{cases}$$
where the second one is a rationalization of the first. Square both equations and add them
$$4x^2 + 2(a+1)y^2 = z^2 + \frac{(a-1)^2y^4}{z^2} \implies x^2 = \frac{z^2}{4} - \frac{(a+1)y^2}{2}+\frac{(a-1)^2y^4}{4z^2}$$
You can take the positive or negative square root as needed.
| {
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Integral of $\exp(-\frac{c_1}{x}-\frac{c_2}{x^2})$ on positive real line Is it possible to find
$$
\int_{0}^{K}
\exp\left(-\,{c_{1} \over x} - {c_{2} \over x^{2}}\right)
\,\mathrm{d}x
$$
or at least a good approximation to it, where $c_{1}$ and $c_{2}$ are positive and $K$ is a very large number $?$.
I tried to consider the change of variables
$-u = -c_{1}/x - c_{2}/x^{2}$ but to no avail.
Note: Approximation can include known functions such as Erf, or Gamma, if necessary.
| As said, series expansion could be the solution. Writing
$$\exp\left(-\frac{a}{x}-\frac{b}{x^2}\right)=1+\sum_{n=1}^p \frac {c_n}{n! \,x^n}=f(x)$$ the $c_n$'s would be
$$\left(
\begin{array}{cc}
1 & -a \\
2 & a^2-2 b \\
3 & -a^3+6 a b \\
4 & a^4-12 b a^2+12 b^2 \\
5 & -a^5+20 b a^3-60 b^2 a \\
6 & a^6-30 b a^4+180 b^2 a^2-120 b^3 \\
7 & -a^7+42 b a^5-420 b^2 a^3+840 b^3 a \\
8 & a^8-56 b a^6+840 b^2 a^4-3360 b^3 a^2+1680 b^4 \\
9 & -a^9+72 b a^7-1512 b^2 a^5+10080 b^3 a^3-15120 b^4 a \\
10 & a^{10}-90 b a^8+2520 b^2 a^6-25200 b^3 a^4+75600 b^4 a^2-30240 b^5
\end{array}
\right)$$ and
$$\int\exp\left(-\frac{a}{x}-\frac{b}{x^2}\right)\,dx=1-a\log(x)-\sum_{n=2}^p \frac {c_n}{(n-1)\,n! \,x^{n-1}}+ C$$
For sure, there is a problem with a lower bound equal to $0$.
| {
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For which a the matrix is positive-definite We have $A=\begin{pmatrix}a&2\\ 2&a+1\end{pmatrix}$
And I want to know for which a the matrix is positive-definite.
So our requirement is $det(A)\ge0$
$(a*(a+1))-4\ge0$
$a^2+a-4\ge0$
I understand that in the points $\frac{-1+\sqrt{17}}{2},\frac{-1-\sqrt{17}}{2}$ the function are equal to zero.
Is it true to say that because the function is equal to zero in the points so $ ∀a>\frac{-1+\sqrt{17}}{2},\frac{-1-\sqrt{17}}{2}$
$a^2+a-4\ge0$?
| The last step is to note that since $a^2+a-4=(a-(-1+\sqrt{17})/2)(a-(-1-\sqrt{17})/2)$, we have that it is positive precisely on $(-\infty,(-1-\sqrt{17})/2)\cup((-1+\sqrt{17})/2,\infty)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3948898",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Number comparison: $5^{152}<2^{353}$ and $2^{1413}<3\cdot 5^{608}$ Is it possible to prove that $5^{152}<2^{353}$ and $2^{1413}<3\cdot 5^{608}$ without using a calculator or logarithms (middle school math only recommended)?
My idea for the first one was to use the obvious $5^3<2^7$ and then raise to the power of $50$ to get $5^{150}<2^{350}$. But since $5^2>2^3$, I couldn't use this approach to get the desired result.
Can you please help me find a relatively short proof for these inequalities? Thank you.
| As @Gottfried Helms pointed out, the binomial expansion works.
We have
\begin{align*}
&5^{152} < 2^{353} \\
\iff \quad & 10^{152} < 2^{505}\\
\iff \quad & 10^{152} < 1024^{50}\cdot 2^5\\
\iff \quad & 10^{152} < 10^{150}(1 + 3/125)^{50}\cdot 2^5\\
\iff \quad & \frac{25}{8} < (1 + 3/125)^{50} \\
\iff \quad & \frac{5}{2\sqrt{2}} < (1 + 3/125)^{25} \\
\Longleftarrow \quad & \frac{5}{2\sqrt{2}} < 1 + 25 \cdot \frac{3}{125}
+ \frac{25\cdot 24}{2}\frac{3^2}{125^2}\\
\iff \quad & \frac{5}{2\sqrt{2}} < \frac{1108}{625}
\end{align*}
which is true.
We are done.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3949007",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Find point farthest on $(x-5)^2+(y+1)^2+(z-7)^2=69$ from point $(2, 3, -6)$ So, I have the equation of a sphere $$(x-5)^2+(y+1)^2+(z-7)^2=69$$ and a point, $P(2, 3, -6)$. I am supposed to find the point on the sphere farthest away from $P$.
I tried to figure out how to use parametric equations to represent the sphere and maximize that using the distance formula. However, they are multiple variables in this equation, and this led me to nowhere.
| The center of the sphere is $C(5,-1,7)$. The line passing through $C$ and $P(2,3,-6)$ has parametric equations
$$
\begin{cases}
x=5+(2-5)t&x=5-3t\\
y=-1+(3-(-1))t&y=-1+4t\\
z=7+(-6-7)t&z=7-13t\\
\end{cases}
$$
Substitute in the equation of the sphere
$$(x-5)^2+(y+1)^2+(z-7)^2=69$$
$$(5-3t-5)^2+(-1+4t+1)^2+(7-13t-7)^2=69$$
$$194 t^2 = 69\to t=\pm\sqrt{\frac{69}{194}}$$
the point on the sphere farthest away from $P$ is:
$$\left(3 \sqrt{\frac{69}{194}}+5,-2 \sqrt{\frac{138}{97}}-1,13 \sqrt{\frac{69}{194}}+7 \right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3949141",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Can't find the inverse of Laplace Transform. Find $\mathcal{L}^{-1}{(F(s))}$, if given $F(s)=\dfrac{2}{s(s^2+4)}$.
I have tried as below.
To find inverse of Laplace transform, I want to make partial fraction as below.
\begin{align*}
\dfrac{2}{s(s^2+4)}=\dfrac{A}{s}+\dfrac{Bs+C}{s^2+4}=\dfrac{(A+B)s^2+Cs+4A}{s(s^2+4)}.
\end{align*}
After that, we have system of linear equation
\begin{align*}
A+B&=0\\
C&=0\\
4A&=2.
\end{align*}
Thus we have $A=2$, $B=-2$, and $C=0$.
Now, substituting $A, B, C$ and we have
\begin{align*}
\dfrac{2}{s(s^2+4)}=\dfrac{2}{s}+\dfrac{-2s}{s^2+4}.
\end{align*}
But the fact is
\begin{align*}
\dfrac{2}{s(s^2+4)}\neq \dfrac{2}{s}+\dfrac{-2s}{s^2+4} = \dfrac{8}{s^2+4}.
\end{align*}
I'm stuck here. I can't make a partial fraction for $F(s)$ and I can't find inverse of Laplace transform for $F(s)$.
Anyone can give me hint to give me hint for this problem?
| You made a mistake because $A \neq 2$
$4A=2$ then $A=\frac{1}{2}$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Solution of Wave Equation initial conditions A vibrating string fixed at $ x=0 $ and $ x=L $ undergoes oscillations described by the wave equation $ \frac{\partial^2u}{\partial x^2} = \frac{1}{c^2} \frac{\partial^2u}{\partial t^2} $ where $ u(x,t) $ represents the displacement from equilibrium of the string. Initially, the profile of the string is $ u(x,0) = \sin(\frac{4 \pi}{L}x) $ and its initial velocity is $ u_t(x,0)=8 \pi \sin( \frac{4 \pi}{L} x ) $.
My attempt at a solution:
Take $ u(x,t) = [ A \cos(kx) + B \sin(kx) ][ C \cos (kct) + D \sin(kct) ] $
$ u(x,0) = [A \cos(kx) + B \sin(kx)] \times C = \sin( \frac{4 \pi}{L} x) $.
So, $ A = 0; k = \frac{4 \pi}{L} $
Then, $ B \sin({ \frac{4 \pi}{L} x})[ C \cos (\frac{4 \pi}{L} ct) + D \sin( \frac{4 \pi}{L} ct )] $
$ u_t(x,t) = B \sin( \frac{4 \pi}{L} x)[ \frac{4 \pi}{L}c \times C \times -\sin( \frac{4 \pi}{L}ct) + \frac{4 \pi}{L} \times D \times \cos( \frac{4 \pi}{L} ct) ] $
$ u_t(x,0) = \frac{4 \pi}{L} c BD \sin( \frac{4 \pi}{L} x ) = 8 \pi \sin ( \frac{8 \pi}{L}x ) $
How do I solve this? Can I take $ BD = \frac{L}{c} \times 4 \cos( \frac{4 \pi}{L} x ) $
| You won't get $\sin(\frac{8\pi}{L}x)$. You are actually dealing with $\cos(\frac{4\pi}{L}c 0)=1$.
Update: I see that you may have a typo in the initial velocity condition which should have contained $\sin(\frac{8\pi}{L}x)$.
To handle that begin with the solutions as $\sum_n a_n e^{i\frac{2\pi n}{L}(x-ct)}+b_ne^{i\frac{2\pi n}{L}(x+ct)}$. With the initial conditions you can easily recognize the only coefficients left would be $a_2,b_2,a_4,b_4$. With the constraints of getting real solutions, you can see it is
$\frac12 \sin(\frac{4\pi}{L}(x-ct))+\frac12\sin(\frac{4\pi}{L}(x-ct) + \frac{L}{2c}\cos(\frac{8\pi}{L}(x-ct))-\frac{L}{2c}\cos(\frac{8\pi}{L}(x+ct)).$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3953005",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Prove relation about area of triangle
In the below picture, angle A is obtuse, AD is a median.
We are also given the relation $AB^2 = AF*AC$.
We want to prove that area of triangle $(ABC) = AB*AD$.
What I have tried:
Area of triangle is $A = \frac {1}{2}*AC*BF$.
Replacing AC from the given relation $AB^2 = AF*AC$, we get $A = \frac {1}{2}*\frac{AB^2*BF}{AF}$.
Also from Pythagoras, we replace $AB^2$ with $BF^2+AF^2$.
But I can't see how to involve AD and prove the required relation!
Any help please?
| As you have noted, the area $\cal A$ can be written as $AC\times BF/2$. Proving ${\cal A}=AB\times AD$ is equivalent to proving $$AB^2\times 4AD^2=AC^2\times BF^2\tag1$$ on squaring both sides.
The median $AD$ has the property that $4AD^2=2AC^2+2AB^2-BC^2$ as shown here. Using Pythagoras on $\triangle CFB$ yields $$BC^2=(AF+AC)^2+BF^2=AF^2+AC^2+2AF\times AC+BF^2.$$ Since $AF^2+BF^2=AB^2=AF\times AC$ we obtain $BC^2=3AB^2+AC^2$ so that $$4AD^2=AC^2-AB^2\tag2.$$ Finally, using the same identity, we have $BF^2=AF\times(AC-AF)$ so that $$AC^2\times BF^2=AF\times AC\times(AC^2-AF\times AC)=AB^2\times(AC^2-AB^2).$$ Substituting in $(2)$ gives $(1)$ so it follows that ${\cal A}=AB\times AD$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Please show that $f(x)=\sum \limits_{n=1}^\infty \frac{x^2}{x^2+n^2}$ is continuous on $\Bbb R$
Let $f(x)=\sum \limits_{n=1}^\infty \frac{x^2}{x^2+n^2}$ on $\Bbb R$. Please show that $f$ is continuous on $\Bbb R$.
I thought if I can prove that $f$ converges uniformly on $R$, then we are done. But I found that Show that $\sum_{n=1}^{\infty}\frac{x^{2}}{x^{2}+n^{2}}$ does not converge uniformly on $(-\infty,\infty)$., we have to find some other ways to prove it.
From Is $f(x)=\sum_{n=1}^\infty\frac{nx^2}{n^3+x^3}$ uniformly continuous on $[0,\infty)$? and Show $ f(x) = \sum_{n=1}^{\infty} \frac{nx}{n^3 + x^3}$ ,$\ g(x) = \sum_{n=1}^{\infty} \frac{x^4n}{(n^3 + x^3)^2}$ are bounded on $[0, \infty)$., I think I can use the same method to prove this question.
I think $f$ converges uniformly on any bounded intervals, hence $$f'(x)=2x\sum \limits_{n=1}^\infty \frac{1}{n^2+x^2}+x^2\sum \limits_{n=1}^\infty \frac{-2x}{(n^2+x^2)^2}$$
Let $g(x)=\sum \limits_{n=1}^\infty \frac{x}{n^2+x^2}$ and $h(x)=\sum \limits_{n=1}^\infty \frac{x^3}{(n^2+x^2)^2}$ on $\Bbb R$. We need to show that $g$ and $h$ are bounded on $\Bbb R$.
First, we discuss $g$.
For $x\in [-1,1]$, $$\left|\frac{x}{n^2+x^2} \right| \le \frac{1}{n^2}$$, so $\left|g(x) \right|\le \sum \limits_{n=1}^\infty \left|\frac{x}{n^2+x^2} \right|\le \sum \limits_{n=1}^\infty \frac{1}{n^2} < \infty$.
For $|x|>1$ be fixed and $m\in \Bbb N$, for any $n\in \Bbb N$ such that $$x(m-1)\le n < mx$$, we have $$\left|\frac{x}{n^2+x^2} \right| \le \left|\frac{x}{x^2(m-1)^2+x^2}\right|=\left|\frac{1}{x}\frac{1}{(m-1)^2+1}\right|$$ and there are at most $\lfloor x \rfloor +1$ of such $n$. $$\left|g(x) \right|\le \sum \limits_{n=1}^\infty \left|\frac{x}{n^2+x^2} \right|\le \sum \limits_{n=1}^\infty \left|\frac{\lfloor x \rfloor +1}{x}\right|\left|\frac{1}{(m-1)^2+1}\right|\le \sum \limits_{n=1}^\infty 2\frac{1}{(m-1)^2+1} < \infty $$
Hence, $g$ is bounded.
Now, consider $h$, if $x\in [-1,1]$, then $$\left|\frac{x^3}{(n^2+x^2)^2} \right|\le \frac{1}{n^4}$$
For $|x|>1$ and $m\in \Bbb N$ be fixed, for any $n\in \Bbb N$ such that $$x(m-1)\le n < mx$$, we have $$\left|\frac{x^3}{(n^2+x^2)^2} \right| \le \left|\frac{x^3}{(x^2+(m-1)^2x^2)^2} \right|=\left|\frac{1}{x}\frac{1}{(1+(m-1)^2)^2} \right|$$ and there are at most $\lfloor x \rfloor +1$ of such $n$.
$$\left|h(x) \right|\le \sum \limits_{n=1}^\infty \left|\frac{x^3}{(n^2+x^2)^2} \right|\le \sum \limits_{n=1}^\infty \left|\frac{\lfloor x \rfloor +1}{x}\right|\left|\frac{1}{((m-1)^2+1)^2}\right|\le \sum \limits_{n=1}^\infty 2\frac{1}{((m-1)^2+1)^2} < \infty$$ Hence, $h$ is bounded.
Finally, by mean value theorem, we can prove that $f$ is uniformly continuous on $\Bbb R$.
Appreciate any suggestion.
| A much shorter proof: continuity is local property. To prove that $f$ is continuous it is enough to prove that it is continuous at each point. Take any $x \in \mathbb R$. In $[x-1,x+1]$ the series converges uniformly by M-test since it is dominated by a constant times $\sum \frac 1 {n^{2}}$. Hence $f$ is continuous on $[x-1,x+1]$ and we are done.
| {
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"url": "https://math.stackexchange.com/questions/3960266",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Invert Trigonometric System of Equations In my research, I have a system of trig equations that I would like to invert. I know $\alpha$ and $\beta$ ($\theta$ is a constant) in the following system.
$$\cos \alpha = \cos \theta \cos x + \sin \theta \sin x \cos y$$
$$\tan \beta = \frac{\cos x +\cos \theta \cos \alpha}{\sin \theta \sin x \sin y} $$
I would like to find an inversion in the form:
$$\text{trig}(x) =\; ...$$
$$\text{trig}(y) = \;...$$
How can this be done?
| $$\cos(\alpha) = \cos (\theta) \cos( x) + \sin (\theta) \sin (x) \cos( y)\tag 1$$
$$\tan (\beta) = \frac{\cos (x) +\cos (\theta) \cos (\alpha)}{\sin (\theta) \sin (x) \sin (y)}\tag 2$$
From $(1)$
$$\cos(y)=\csc (\theta ) \csc (x) (\cos (\alpha )-\cos (\theta ) \cos (x))\tag 3$$
From $(2)$
$$\sin(y)=\cos (\alpha ) \cot (\beta ) \cot (\theta ) \csc (x)+\cot (\beta ) \csc (\theta )
\cot (x)\tag 4$$
Using $\cos^2(y)+\sin^2(y)=1$, we end with
$$1=(\cos (\alpha ) \cot (\beta ) \cot (\theta ) \csc (x)+\cot (\beta ) \csc (\theta )
\cot (x))^2+(\cot (\theta ) \cot (x)-\cos (\alpha ) \csc (\theta ) \csc (x))^2$$
Now, using the tangent half-angle substitution $x=2 \tan ^{-1}(t)$, we end with a quadratic in $t^2$
$$a + b t^2+c t^4 =0$$ where
$$a=16 \cos (\alpha ) \cos (2 \beta ) \cot (\theta ) \csc (\theta )+4 \sin ^2(\alpha )
\cos (2 \beta )+$$ $$\csc ^2(\theta ) (\cos (2 \alpha ) (\cos (2 \theta )+3)+9)+3
\cot ^2(\theta )-3$$
$$b=-2 \cos (2 (\alpha -\beta ))-2 \cos (2 (\alpha +\beta ))-16 \sin ^2(\alpha ) \csc
^2(\theta )-$$ $$4 \cos (2 \alpha )+4 \cos (2 \beta )-12$$
$$c=-16 \cos (\alpha ) \cos (2 \beta ) \cot (\theta ) \csc (\theta )+4 \sin ^2(\alpha )
\cos (2 \beta )+$$ $$\csc ^2(\theta ) (\cos (2 \alpha ) (\cos (2 \theta )+3)+9)+3
\cot ^2(\theta )-3$$
Applied to the case use by @Jean Marie $\alpha=\theta=\frac\pi 4 $ and $\beta=\tan ^{-1}(2))$, this leads to
$$132 t^4-152 t^2+36=0$$ giving the four roots
$$t=\pm \frac{1}{\sqrt{3}} \qquad \text{and} \qquad t=\pm \frac{3}{\sqrt{11}} $$ that is to say
$$x=\pm \frac \pi 3\approx \pm1.04720 \qquad \text{and} \qquad x=\pm2 \tan ^{-1}\left(\frac{3}{\sqrt{11}}\right)\approx 1.47063$$ and then the corresponding $y$'s using $(3)$ or $(4)$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Cyclic inequality with $a+b+c=1$
Let $a;b;c$ be positive real numbers such that $a+b+c=1$
Prove: $$a^2+b^2+c^2+a^2b+b^2c+c^2a+4abc \geq \dfrac{16}{27}$$
First, I tried with $abc \geq (a+b-c)(b+c-a)(c+a-b)$
But it gave nothing.
Then I tried to add: $a^2b+\dfrac{b}{9}+b^2c+\dfrac{c}{9}+c^2a+\dfrac{a}{9}-\dfrac{1}{9}$
Then AM-GM, but from it, I got: $\dfrac{2}{3}(a^2+b^2+c^2)+4abc \geq \dfrac{10}{27}$
I think it's wrong, I'm stuck here.
Can someone help? Thanks.
| Claim:$$27(a^2c+b^2a+c^2b+abc)\le 4{(a+b+c)}^3$$
Proof:WLOG $a=\min(a,b,c)$ then take $b=a+u,c=a+w$ where $u,w\ge 0$ then we have to prove $$-9au^2+9auw-9aw^2-4u^2+15u^2w-12uw^2-4w^3\le 0$$ $$-9a(u^2-uw+w^2)-{(u-2w)}^2(4u+w)\le 0$$ which is true
thus $$a^2c+b^2a+c^2b+abc\le \frac{4}{27}\tag 1$$
we have to prove $$a^2+b^2+c^2+4abc+a^2b+b^2c+c^2a\ge \frac{16}{27}$$ using (1) it suffices to prove:$$a^2+b^2+c^2+ab(a+b)+bc(b+c)+ac(a+c)+5abc\ge \frac{16}{27}+\frac{4}{27}$$ using $a+b+c=1$ and some mainipulation we have to prove $$ab+bc+ca-2abc\le \frac{7}{27}$$ WLOG $c=\min(a,b,c)$ then $c\le 1/3$ or $1-2c>0$ $$ab+bc+ca-2abc-\frac{7}{27}=\color{red}{ab}(1-2c)+c(1-c)-\frac{7}{27}$$ $$\le \color{red}{\frac{{(a+b)}^2}{4}}(1-2c)+c(1-c)-\frac{7}{27}\tag{AM-GM}$$ $$=\frac{{(1-c)}^2}{4}(1-2c)+c(1-c)-\frac{7}{27}$$ $$={(3c-1)}^2(6c+1)\frac{-1}{108}\le 0$$ which is true and proof is completed(phew!)
Acknowledgement:Earlier I only had a partial proof (I could prove only LHS$>15/27$) ,But tbanks to River-Li who proposed a stronger Claim(see comments) and made some corrections the proof is done!
| {
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"timestamp": "2023-03-29T00:00:00",
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Find the maximum and minimum value of $2x^2 + y^2 + z^2$ subject to $x + y + z = 10$. I am working on inequality problems for mathematical olympiads. I have come across this problem:
Find the maximum and minimum value of $2x^2 + y^2 + z^2$ subject to $x + y + z = 10$, with $x,y,z \ge 0 $
I found the minimum using Cauchy-Schwarz: $$100 = \left(\frac 1 {\sqrt 2} \left(\sqrt 2 x\right) + y + z\right)^2 \le \left(\left(1/\sqrt 2\right)^2 + 1^2 + 1^2\right) \left(\left(\sqrt 2 x\right)^2 + y^2 + z^2\right)=2.5\left(2x^2 + y^2 + z^2\right)$$
so $2x^2 + y^2 + z^2 \ge 40$ with equality if $x=2$ and $y=z=4$.
I am not sure how to find the maximum though. Could I have a hint? Ideally I would like to do it without using calculus, though if it is easy to do with calculus feel free to share the method.
| Substitute the value of $z$ in the expression: $$2x^2 +y^2 +(x+y-10)^2 $$ As increasing $x$ and $y$ increases the value of this expression, the maximum occurs when $x+y$ is maximum, i.e. $x+y=10$: $$2x^2 +(10-x)^2 = 3x^2 -20x +100$$
This is a parabola which attains its maximum value on $[0,10]$ at $x=10$, which is $200$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Proving a convoluted proof to an inequality: $x,y>0$ such that $x^2+y^2=1$. Prove that, $x^3+y^3 \geqslant \sqrt2 xy $ As the title described, I was trying to find an alternative proof to
If $x,y$ are positive numbers such that $x^2 + y^2=1$, prove that $x^3 + y^3 \geqslant \sqrt2 xy $.
Here's the proof that I've found (I'm sorry, I forgot where I got it):
Apply the Chebyshev's inequality on the tuplets $(x^2, y^2)$ and $\left( \frac1y, \frac1x\right)$, we have $$ \frac12 \left( \frac{x^2}y + \frac{y^2}x \right) \geqslant \frac{x^2 + y^2}2 \cdot \frac{1/y + 1/x}2 \quad \Rightarrow \quad \frac{x^2}y + \frac{y^2}x \geqslant \frac12 \left(\frac1x + \frac1y \right)$$ Apply AM-HM inequality on the tuplets $(x,y)$, we have $$ \frac{x+y}2 \geqslant \frac2{1/x + 1/y} \quad \Rightarrow \quad \frac1x + \frac1y \geqslant \frac4{x+y} $$ Apply Cauchy-Schwartz inequality on the tuplets $(x,y)$ and $(1,1)$, we have $$ x + y = x \cdot 1 + y\cdot 1 \leqslant \sqrt{x^2 + y^2} \sqrt2 = \sqrt2 $$ Combining these 3 inequalities above yield $$ \dfrac{x^2}y + \dfrac{y^2}x \geqslant \frac12 \cdot \frac4{x+y} \geqslant \frac12 \cdot \frac4{\sqrt2} = \sqrt2 $$ The result follows.
Now since I love to punish myself, I tried to find a harder proof as such:
We can let $(x,y) = (\cos\theta, \sin\theta) $, where $\theta \in (0, \tfrac\pi2) $. The inequality in question becomes $$ \begin{array} {l c l }
\cos^3 \theta + \sin^3 \theta &\geqslant &\sqrt2 \cos \theta \sin \theta \\
(\cos\theta + \sin\theta)(\cos^2 \theta - \sin \theta \cos \theta + \sin^2\theta) &\geqslant &\sqrt2 \cos \theta \sin \theta \\
(\cos\theta + \sin\theta)(1 - \sin \theta \cos \theta ) &\geqslant &\sqrt2 \cos \theta \sin \theta \\
\cos\theta + \sin\theta &\geqslant & \cos \theta \sin \theta ( \sqrt2 + \cos \theta + \sin \theta) \\
\dfrac1{\sin\theta \cos\theta} &\geqslant & \dfrac{\sqrt2}{\cos \theta \sin \theta} + 1 \\
\end{array}$$ Apply Weierstrass substitution ($t = \tan\frac\theta2$, where $0<t<1$) yields $$ \dfrac{(1+t^2)^2}{2t(1-t^2)} \geqslant \dfrac{\sqrt2 (1+t^2)}{(1-t^2) +2t} + 1
$$ which simplifies to $$ - \dfrac{t^6 - 2\sqrt2 t^5 - 3t^4 -8t^3 + 3t^2 + 2\sqrt2 t - 1}{ 2t(t-1)(t+1) (t^2 - 2t - 1)} \geqslant 0 $$ or $$ t^6 - 2\sqrt2 t^5 - 3t^4 -8t^3 + 3t^2 + 2\sqrt2 t - 1 \leqslant 0, \quad\quad\quad 0<t<1$$
Now how do I prove the sextic polynomial inequality above (which is true)?
| By CS-Engel you will get
$x^3+y^3 \ge \dfrac{(x^2+y^2)^2}{x+y} = \dfrac{(x^2+y^2)(x^2+y^2)}{x+y} = \dfrac{x^2+y^2}{x+y}$ ... (1)
By QM-AM inequality
$\dfrac{1}{\sqrt{2}} = \sqrt{\dfrac{x^2+y^2}{2}} \ge \dfrac{x+y}{2}$
or
$\dfrac{1}{x+y} \ge \dfrac{1}{\sqrt{2}}$ ...(2)
Combining (1) and (2) with AM-GM we get
$\dfrac{x^2+y^2}{x+y} \ge \dfrac{2xy}{\sqrt{2}} = \sqrt{2}xy$
as desired
| {
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Computing the definite integral $\int_0^1(-1)^{\left\lfloor\frac1x\right\rfloor}\,\mathrm dx$
$$\int_0^1(-1)^{\left\lfloor\frac1x\right\rfloor}\,\mathrm dx$$
For $x\in\left(\frac1{n+1},\frac1n\right)$, $\left\lfloor\frac1x\right\rfloor=n$, and $\frac1n-\frac1{n+1}=\frac1{n(n+1)}$. So the integral is equal to the Riemann sum
$$\int_0^1(-1)^{\left\lfloor\frac1x\right\rfloor}\,\mathrm dx=\lim_{n\to\infty}\sum_{k=1}^n\frac{(-1)^k}{k(k+1)}$$
Now, for $|x|<1$, we have
$$\frac x{1-x}=\sum_{k=1}^\infty x^k\\
\implies -x-\ln(1-x)=C_1+\sum_{k=1}^\infty\frac{x^{k+1}}{k+1}\\
\implies x-\frac{x^2}2+(1-x)\ln(1-x)=C_2+C_1x+\sum_{k=1}^\infty \frac{x^{k+2}}{(k+1)(k+2)}$$
If $x=0$, then we find $C_1=C_2=0$ so that
$$x-\frac{x^2}2+(1-x)\ln(1-x)=\sum_{k=1}^\infty \frac{x^{k+2}}{(k+1)(k+2)}\\
\implies 1-\frac x2+\frac{1-x}x\ln(1-x) = \sum_{k=2}^\infty \frac{x^k}{k(k+1)}\\
\implies 1+\frac{1-x}x\ln(1-x) = \sum_{k=1}^\infty \frac{x^k}{k(k+1)}$$
and when $x=-1$, we have
$$\int_0^1(-1)^{\left\lfloor\frac1x\right\rfloor}\,\mathrm dx=\sum_{k=1}^\infty\frac{(-1)^k}{k(k+1)}=\boxed{1-2\ln2}$$
Is this solution correct?
| According to this answer, we get:
$$\mathscr{S}_{-1}:=\int_0^1\left(-1\right)^{\left\lfloor\frac{1}{x}\right\rfloor}\space\text{d}x=\frac{-1+\ln\left(1-\left(-1\right)\right)-\left(-1\right)\ln\left(1-\left(-1\right)\right)}{-1}=$$
$$1-2 \ln (2)\approx-0.386294\tag1$$
| {
"language": "en",
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"source": "stackexchange",
"question_score": "3",
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"answer_id": 0
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How to solve $\lim_{n \to \infty}\frac{1}{\sqrt[3]{n^3+n+1}-\sqrt{n^2-n+2}}$ without L'Hopital? $\lim_{n \to \infty}\frac{1}{\sqrt[3]{n^3+n+1}-\sqrt{n^2-n+2}}$
$\lim_{n \to \infty}\frac{1}{\sqrt[6]{(n^3+n+1)^2}-\sqrt[6]{(n^2-n+2)^3}}$ but because this limit is still the type of $\frac{1}{\infty-\infty}$ I tried to do this:
$\lim_{n \to \infty}\frac{\sqrt[6]{(n^3+n+1)^2}+\sqrt[6]{(n^2-n+2)^3}}{(n^3+n+1)^2-(n^2-n+2)^3} = \lim_{n \to \infty}\frac{\sqrt[6]{(n^3+n+1)^2}+\sqrt[6]{(n^2-n+2)^3}}{3n^5-7n^4+15n^3-17n^2+14n-7}$ I'm totally stuck here. I would divide the fraction by $3n^5$ and then the solution is $0$. Not the correct answer.
Did I miss something?
| Always try to simplify algebraic calculations in a manner which reduces typing effort and visual clutter.
Clearly we can take $n$ common from both terms in denominator and hence denominator can be written as $n(a-b) $ where both $a, b$ tend to $1$. Further we can see that $a^3,b^2$ are radical free and hence we have $$n(a-b) =n(a-1-(b-1))=n\left((a^3-1)\cdot\frac{a-1}{a^3-1}-(b^2-1)\cdot\frac{b-1}{b^2-1}\right)\tag{1}$$ Just note that $$n(a^3-1)=n\left(\frac{1}{n^2}+\frac {1}{n^3}\right)\to 0$$ and $$n(b^2-1)=n\left(-\frac{1}{n}+\frac{2}{n^2}\right)\to - 1$$ It now follows from equation $(1)$ that denominator $n(a-b) $ tends to $$0\cdot\frac{1}{3}-(-1)\cdot\frac{1}{2}=\frac{1}{2}$$ and the expression under limit thus tends to $2$.
| {
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How would one solve the following: $\sin(\frac{\pi x}{30})+\sin(\frac{\pi x}{360})-\cos(\frac{\pi x}{30})-\cos(\frac{\pi x}{360})=0$ I want to use the idea of vectors to solve this problem: "How many times a day do the hour and minute hands of a clock face the opposite of each other?"
We know that the minute hand turns $6^{\circ}$ every minute. If I assume my starting point is 12 o'clock, then my angle (with respect to the positive x-axis, of course), $\theta_m(t)=90^{\circ}-6t$. Similarly, the angle between the hour-hand and the positive x-axis at any given time (in minutes), t, is $\theta_h= 90^{\circ}- \frac{1}{2}t$.
If we think of the hour and minute hands as two dimensional vectors whose tails are on the origin, then the vector of the minute hand can be specified as $\textbf{m}$, where:
\begin{align*}
\textbf{m} &= \begin{bmatrix}
\sin(6t) \\
\cos(6t) \\
\end{bmatrix}
\end{align*}
In the same way, the vector of the hour hand, $\textbf{h}$= \begin{bmatrix}
\sin( \frac{1}{2}t) \\
\cos( \frac{1}{2}t) \\
\end{bmatrix}
We want the instances where:
\begin{align*}
\textbf{m} &= -\textbf{h} \\
\textbf{m}+\textbf{h} &=\textbf{0} \\
\begin{bmatrix}
\sin(6t)+\sin\left( \frac{1}{2}t\right) \\
\cos(6t)+\cos\left( \frac{1}{2}t\right) \\
\end{bmatrix} &= \textbf{0}
\end{align*}
So we have:
$$\sin(6t)+\sin\left( \frac{1}{2}t\right) -\cos(6t)-\cos\left( \frac{1}{2}t\right)=0$$
In radians:
$$\sin\left( \frac{\pi t}{30}\right)+\sin\left( \frac{\pi t}{360}\right) -\cos\left( \frac{\pi t}{30}\right)-\cos\left( \frac{\pi t}{360}\right)=0$$
My question is, can I find out the roots of that by hand (for my puposes, $0<t \le 1440$)? I plotted it and the period seems to be changing.
Edit: So technically from the vector equation I get this:
$$sin(\frac{\pi t}{30})+sin(\frac{\pi t}{360})=0$$
$$cos(\frac{\pi t}{30})+cos(\frac{\pi t}{360})=0$$
It turns solving these get me the true answers, while solving the one in the title gives me unnecessary ones (I'm saying all of this by looking at the plots). So, how do I solve these 2 sets by hand?
| Ok, we have $$sin( \frac{\pi t}{30})+sin( \frac{\pi t}{360})=0$$
Let $ \frac{\pi t}{360}=y$. So we have: $$sin(12y)+sin(y)=0$$
As lan bhattacharjee pointed out, we could use Simpson's formulas here:
\begin{align*}
sin(12y)+sin(y) &= 2sin[ \frac{1}{2}(12y+y)]cos[ \frac{1}{2}(12y-y)] \\
&= 2sin( \frac{13y}{2})cos( \frac{11y}{2})
\end{align*}
So we have:
\begin{equation}
2sin( \frac{13y}{2})cos( \frac{11y}{2})=0
\end{equation}
From the main post it is also given:
$$cos( \frac{\pi t}{30})+cos( \frac{\pi t}{360})=0$$
From this we get:
\begin{equation}
2cos( \frac{13y}{2})cos( \frac{11y}{2})=0
\end{equation}
Comparing (1) & (2), it is apparent that $cos( \frac{11y}{2})$ has to be zero, because if it isn't zero, then we get $sin( \frac{13y}{2})=cos( \frac{13}{y})$. But we know that $sinx=cosx$ only when $sinx=cosx= \frac{1}{\sqrt{2}}$. Yet, we assumed that $cos( \frac{11y}{2})$ doesn't equal zero. That would leave us with $sin( \frac{13y}{2})cos( \frac{11y}{2}) \neq 0$, which is not true.
So the only way for equations 1 and 2 to be true is if:
\begin{align*}
cos( \frac{11y}{2}) &= 0 \\
\frac{11y}{2} &= \frac{\pi}{2}(2n+1), n \in \mathbb{Z} \\
y &= \frac{\pi}{11}(2n+1) \\
\frac{\pi t}{360} &= \frac{\pi}{11}(2n+1) \\
t &= \frac{360}{11}(2n+1)
\end{align*}
Since we're dealing with time, we want:
\begin{align*}
t &> 0 \\
... &> ... \\
n &\geq 0
\end{align*}
There are 1440 minutes in a single day:
\begin{align*}
t &\leq 1440 \\
... &\leq ... \\
n &\leq 21
\end{align*}
There are 22 integers between 0 and 22 (including 0 and 22). So, the hour and minute hands of a clock face opposite positions to each other 22 times a day.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Solving $\sqrt{a-\sqrt{a+x}} = x$ Solve the equation
$$\sqrt{a-\sqrt{a+x}} = x$$
My approach:
Tried shifting the variables into different options, but couldn't get anything out of it. So, please help.
| $$a-\sqrt{a+x}=x^2$$
$$a-x^2=\sqrt{a+x}$$
$$a+x=x^4-2ax^2+a^2$$
$$x^4-2ax^2-x+a^2-a=0$$
$$(x^2+x+1-a)(x^2-x-a)=0$$
You can factor the last expression by setting up $(x^2+px+q)(x^2+rx+s)=x^4-2ax^2-x+a^2-a$. And comparing the coefficient to find $p,q,r,s$.
| {
"language": "en",
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"source": "stackexchange",
"question_score": "1",
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Find $\int{x\sqrt{1-x^2}\arcsin{x}dx}$ I tried $$\int{x\sqrt{1-x^2}\arcsin{x}\ \mathrm{d}x}$$$$=\int{\arcsin{x}\ \mathrm{d}\left(\frac{2x^3(1-x^2)^\frac{3}{2}}{3}\right)}$$$$=\frac{2x^3(1-x^2)^\frac{3}{2}}{3}\arcsin{x}-\int{\frac{2x^3(1-x^2)^\frac{3}{2}}{3}\ \mathrm{d}\left(\arcsin{x}\right)}$$$$=\frac{2x^3(1-x^2)^\frac{3}{2}}{3}\arcsin{x}-\int{\frac{2x^3(1-x^2)^\frac{3}{2}}{3\sqrt{1-x^2}}\ \mathrm{d}x}$$$$=\frac{2x^3(1-x^2)^\frac{3}{2}}{3}\arcsin{x}-\frac{2}{3}\int{x^3\ \mathrm{d}x}+\frac{2}{3}\int{x^5\ \mathrm{d}x}$$$$=\cdots$$
This solution isn't true. Where am I wrong?
EDIT: Thank you all for the many different answers! If I could, I'd give you all the accepted answer.
| Let $x=\sin(\theta)$.* Then,
$$
\frac{dx}{d\theta}=\cos(\theta) \implies dx=\cos(\theta)d\theta \, .
$$
The integral becomes
$$
\int \sin(\theta) \left(\sqrt{1-\sin(\theta)^2}\right) \theta\cos(\theta)d\theta
$$
*To be more precise, we are making the substitution $\theta=\arcsin(x)$, so that $-\pi/2 \leq \theta \leq \pi/2$. This allows us to conclude that $\arcsin(\sin(\theta))=\sin(\theta)$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Why is $\sum\limits_{n=1}^{\infty} \frac{n^2-1}{n^3+1}$ divergent? I am stuck on this problem. I have checked on wolfram alpha, and the calculator says that this series is divergent. However, I can't find a way to prove it.
The problem:
$$\sum\limits_{n=1}^{\infty} \frac{n^2-1}{n^3+1}$$
So I tried doing this problem with the direct comparison test.
I have concluded that
$$
\frac{n^2}{n^3} >\frac{n^2-1}{n^3+1}
$$
$$
b_{n}> a_{n}
$$
$b_{n}$ is a p-series, where p=1. This means that the series $b_{n}$ is divergent, making the test direct comparison test inconclusive.
$$
\frac{n^2}{n^3} = \frac{1}{n}
$$
I appreciate any help... thanks! :)
| $$
\frac{n^2-1}{n^3+1} = \frac{n^2+1 -2}{n^3+1} = \frac{n^2+1}{n^3+1} -\frac{2}{n^3+1} \geq \frac{1}{n} - \frac{2}{n^3+1} \quad \forall\ n \in \mathbb{N}.
$$
Therefore,
\begin{align}\sum\limits_{n=1}^{\infty} \frac{n^2-1}{n^3+1} = \sum\limits_{n=1}^{\infty}\left(\frac{n^2+1}{n^3+1} -\frac{2}{n^3+1}\right)\\
\\
= \sum\limits_{n=1}^{\infty}\frac{n^2+1}{n^3+1}\ -\ \sum\limits_{n=1}^{\infty}\frac{2}{n^3+1} \\
\\
\geq \left(\sum\limits_{n=1}^{\infty} \frac{1}{n}\right)\ -\ L \\
\end{align}
for some $L\in \mathbb{R}.$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Prove inequality with $a;b;c \in R$ Let $a;b;c \in \mathbb{R} $ such that $a+b+c=0$
Prove that: $P=\dfrac{a-1}{a^2+8}+\dfrac{b-1}{b^2+8}+\dfrac{c-1}{c^2+8} \geq -\dfrac{3}{8}$
I tried to do this:
$\dfrac{8a-8}{a^2+8}+2+\dfrac{8b-8}{b^2+8}+2+\dfrac{8c-8}{c^2+8}+2 \geq 3$
$\Leftrightarrow \dfrac{(a+2)^2}{a^2+8}+\dfrac{(b+2)^2}{b^2+8}+\dfrac{(c+2)^2}{c^2+8} \geq \dfrac{3}{2}$
Then WLOG $c=max\{a;b;c\} \Rightarrow c \geq 0$
I hope I can prove: $\dfrac{(a+2)^2}{a^2+8}+\dfrac{(b+2)^2}{b^2+8} \geq f(c)$
So I only need to find minimum of $f(c)+\dfrac{(c+2)^2}{c^2+8}$
But I can't find $f(c)$, I stuck here.
| Your idea is very good and we will use that :)
As $a+b+c=0$ we can find a pair such that they have the same sign hence WLOG $ab\ge 0$. Now from your attempt we have to prove $$\Leftrightarrow \dfrac{(a+2)^2}{a^2+8}+\dfrac{(b+2)^2}{b^2+8}+\dfrac{(c+2)^2}{c^2+8} \geq \dfrac{3}{2}$$
by C-S $$\dfrac{(a+2)^2}{a^2+8}+\dfrac{(b+2)^2}{b^2+8}\ge \frac{{(a+b+4)}^2}{a^2+b^2+16}=\frac{{(4-c)}^2}{a^2+b^2+16}\ge \frac{{(4-c)}^2}{c^2+16} $$ here we used $a^2+b^2\le {(a+b)}^2=c^2\iff 2ab\ge 0$
it remains to prove $$\frac{{(4-c)}^2}{c^2+16}+\frac{{(c+2)}^2}{c^2+8}\ge \frac{3}{2}$$ $$\iff \frac{c^2{(c-4)}^2}{2(c^2+16)(c^2+8)}\ge 0$$ which is true
Equality when $a=b=c=0$,
$c=4,a=-2,b=-2$(upto permutations)
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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If $x_1, x_2$ are roots of $a\cos x+b\sin x+c=0$ for $x_1+x_2\ne 2k\pi$ show that $\sin(x_1+x_2)=\frac{2ab}{a^2+b^2}.$ If $x_1, x_2$ are roots of $a\cos x+b\sin x+c=0$ for $x_1+x_2≠2kπ$ show that $\sin(x_1+x_2)=\frac{2ab}{a^2+b^2}$
What I've done till now:
$$a\cos x_1+b\sin x+c-(a\cos x_2+b\sin x_2+c)=0$$
$a\cos x_1+b\sin x_1-a\cos x_2-b\sin x_2=0$
$a(\cos x_1-\cos x_2)+b(\sin x_1-\sin x_2)=0$
$a(-2\sin(\frac{x_1+x_2}{2})\sin(\frac{x_1-x_2}{2}))+b(2\sin(\frac{x_1-x_2}{2})\cos(\frac{x_1+x_2}{2}))=0$
$-a\sin(\frac{x_1+x_2}{2})+b(\frac{\cos x_1+x_2}{2})=0$
That's it. Does anyone know how to solve this?
| Just rewrite this equation in terms of complex exponentials:
$$\frac{a-bi}{2}e^{ix}+\frac{a+bi}{2}e^{-ix}+c=0$$
Setting $z=a+bi$ one can rewrite the equation as a quadratic in $w=e^{ix}$
$$\bar{z}w^2+cw+z=0$$
From Vieta's formulae we know that the product of the two roots of the equation is
$$w_1w_2=e^{i(x_1+x_2)}=\frac{z}{\bar{z}}$$
which, upon separating real and imaginary parts yields
$$\cos(x_1+x_2)=\frac{a^2-b^2}{a^2+b^2}~,~\sin(x_1+x_2)=\frac{2ab}{a^2+b^2}$$
| {
"language": "en",
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"source": "stackexchange",
"question_score": "4",
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Find constants $a$, $b$, $c$, $d$ and $e$ such that $\cos4x=a\sin^4x+b\sin^3x+c\sin^2x+d\sin x+e$ for all angles $x$ Basically, write $\cos4x$ as a polynomial in $\sin x$.
I've tried the double angles theorem and $\cos2x=\cos^2x-\sin^2x$. I'm still having trouble right now though.
Please help! Thanks!
| Double angle theorem once:
$\cos 4x = \cos^2 2x + \sin^2 2x$.
Double angle theorem twice:
$= (\cos^2 x -\sin^2 x)^2 + (2\sin x\cos x)^2$.
Expand chicken soup and rice:
$= \cos^4 x- 2\cos^2 x\sin^2 x + \sin^4 x +4\sin^2 x\cos^2 x$
Can you finish?
To get it entirely in terms of $\sin^k x$ replace $\cos^2 x$ with $1-\sin^2 x$ (keeping in mind $\cos^4 x = (\cos^2 x)^2$)
Can you finish?
$= (1 - \sin^2 x)^2 - 2(1-\sin^2 x) \sin^2 x + \sin^4 x +4\sin^2 x(1-\sin^2 x) $
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove that $(1+\frac{1}{n})^k < 1+ \frac{k}{n}+\frac{k^2}{n^2}$ Prove that $(1+\frac{1}{n})^k < 1+ \frac{k}{n}+\frac{k^2}{n^2}, \forall$ n, k nonnegative integers, $k\le n$.
I know that has something to do with Bernoulli's inequality $(1+\alpha)^x\ge 1+\alpha x, \alpha \ge -1, n\ge1$.
If I reconsider Bernoulli's inequality with $\alpha=\frac{1}{n}$ and $x=k$ it follows that
$(1+\frac{1}{n})^k\ge 1+\frac{k}{n}$, but I don't know how to continue.
I also tried to prove it with induction where I consider
$p(n):(1+\frac{1}{n})^k < 1+ \frac{k}{n}+\frac{k^2}{n^2}$ to be true and prove that $ p(n+1):(1+\frac{1}{n+1})^k < 1+ \frac{k}{n+1}+\frac{k^2}{(n+1)^2}$ to be also true, but it didn't work.
Thank you!
| Proof without induction/calculus:
It's easy to prove the inequality is true for $k=1,2$. If $k\ge 3$ we have
$$\left(1+\frac{1}{n}\right)^k - 1- \frac{k}{n}-\frac{k^2}{n^2} $$ $$= - \frac{k^2}{n^2} + \binom{k}{2} \frac{1}{n^2} + \sum_{i=3}^k \binom{k}{i} \frac{1}{n^i} = -\frac{k(k+1)}{2n^2}+\sum_{i=3}^k \binom{k}{i} \frac{1}{n^i} \tag1$$
Notice that
$$\frac{\binom{k}{i+1}/n^{i+1}}{\binom{k}{i}/n^i}=\frac{k-i}{n(i+1)} \le \frac{k-3}{4n}, \forall i \ge 3$$
Then $(1)$ is less then
$$-\frac{k(k+1)}{2n^2} + \frac{\binom{k}{3}/n^3}{1-\frac{k-3}{4n}}$$ and it suffices to prove
$$ \frac{\frac{k(k-1)(k-2)}{6n^3}}{1-\frac{k-3}{4n}} < \frac{k(k+1)}{2n^2}$$
Or equivalently
$$\frac{4(k-1)(k-2)}{4n-k+3} < 3(k+1) \iff 4(k-1)(k-2) < 3(k+1)(4n-k+3)\\
\stackrel{k\le n}{\Leftarrow} 4(k-1)(k-2) < 3(k+1)(4k-k+3) = 9(k+1)^2$$ which is trivially true.
| {
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"url": "https://math.stackexchange.com/questions/3991180",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Solve non-homogeneous recurrence $x_{n+2} - 2x_{n+1} + x_n = 8$ I'm trying to solve this recurrence $x_{n+2} - 2x_{n+1} + x_n = 8$, but the solution of the homogeneous is $a_n=b+cn$, and I don't know how to change it to find the non-homogeneous solution.
| $$x_{n+2} - 2x_{n+1} + x_n = 8\tag{1}$$
Set $n=n+1$
$$x_{n+3}-2x_{n+2}+x_{n+1}=8$$
Subtract the two equations
$$x_{n+3}-2x_{n+2}+x_{n+1}-x_{n+2} + 2x_{n+1} - x_n=0$$
$$x_{n+3}-3x_{n+2}+3x_{n+1}-x_n=0$$
Characteristic equation is
$$\lambda^3-3\lambda^2+3\lambda-1=0\to (\lambda-1)^3=0\to \lambda=1$$
and finally
$$x_n=an^2+bn+c\tag{2}$$
edit
from $(1)$ we get $x_0=8+2x_1-x_2$
from $(2)$ we have $x_0=c,\;x_1=a+b+c;\;x_2=4a+2b+c$
putting all together
$$\begin{cases}
c=8+2x_1-x_2\\
a+b+c=x_1\\
4a+2b+c=x_2\\
\end{cases}
$$
$$\begin{cases}
a+b+8+2x_1-x_2=x_1\to a+b=x_2-x_1-8\\
4a+2b+8+2x_1-x_2=x_2\to 4a+2b=2x_2-2x_1-8\\
\end{cases}
$$
$$\begin{cases}
a+b=x_2-x_1-8\\
2a+b=x_2-x_1-4\\
\end{cases}
$$
subtracting the two equations, we get $a=4$ so the $(2)$ becomes
$$x_n=4n^2+bn+c$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Given $\frac{A}{4B}+\frac{2a^3A}{B^3}-\frac{a^2A}{B^3}-\frac{2a^2A^2}{B^3}=0$, what does $\frac{a}{B}$ tend to as $\frac{A}{B}\to \infty$? In this problem, all of $A, a, B$ can vary.
My attempt was to let $\frac{A}{B}=C$ and $\frac{a}{B}=c$, which gives $ \frac{C}{4}+2Ac^3-c^2C-2Ac^2C=0$.
Now $C\to \infty $ means either $A\to \infty $ or $B\to 0 $. At this point I'm not sure at all how to proceed.
| You can rewrite the equation to get a function $c=f(C)$ and then calculate $\lim_{C\to\infty}f(C)$. It requires you to solve a cubic equation. However, there is an easier solution:
You rewrite the equation to this form:$$
C=\frac{-2Ac^3}{\frac14-c^2-2Ac^2}=\frac{2Ac^3}{(2A+1)c^2-\frac14}
$$
If $2A+1>0$ then:
$$
C=\frac{2Ac^3}{\left(c\sqrt{2A+1}-\frac12\right)\left(c\sqrt{2A+1}+\frac12\right)}
$$
You need to make something with $c$ to make $C$ go to infinity. Let's assume that $A$ is positive. Now what do you have to do to make $C=\infty$ if you can change only $c$? If $c$ approaches $\infty$, $\frac{1}{2\sqrt{2A+1}}$ from right or $-\frac{1}{2\sqrt{2A+1}}$ from right, $C$ goes to infinity. And vice-versa, if $C$ goes to infinity, $c$ approaches $\infty$, $\frac{1}{2\sqrt{2A+1}}$ from right or $-\frac{1}{2\sqrt{2A+1}}$ from right. And if $A$ is negative, we just flip the sign.
If $2A+1\leq0$ then $(2A+1)c^2-\frac14$ is always nonpositive. It also means that $A<0$. In this case, $c$ must go to $\infty$.
We have up to three results, so what now? It is because the equation gives up to three sollutions for each $C$. So the sollution varies on which of these solutions you pick.
To sum it up:
$$
\lim_{\frac{A}{B}\to\infty}=
\begin{cases}
\infty \text{ or} \pm\frac{1}{2\sqrt{2A+1}}, & \text{if $\liminf A>0$} \\
-\infty \text{ or} \pm\frac{1}{2\sqrt{2A+1}}, & \text{if $\liminf A>-\frac12$ and $\limsup A<0$} \\
\infty, & \text{if $\limsup A\leq-\frac12$}
\end{cases}
$$
Note that $\frac{A}{B}$ can't go to $\infty$ if $A=0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3993931",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Making vector subject of equation I have the following question
Solve the vector equation $\bf{a \times r}\ +\ \lambda\bf{r = c}$ for
r where $\lambda \ne 0$.
So far my approach has been
Dot both sides with c, which with some rearranging leads to $\bf{r}.(\bf{c \times a}\ +\ \lambda \bf{c}) = |{\bf{c}}|^2$. Leading me to believe that r is a point on a plane. I then converted this equation to Cartesian form and then to Parametric for so r is the subject.
However my answer looks very different to the provided answer and no working is provided.
The solution given was
$$\bf{r} = N^{-1} \left(\bf{(a.c)a} + \lambda^2\bf{c} - \lambda \bf{a} \times \bf{c}\right)\ \text{where}\ N = \lambda(k^2 + \lambda^2)$$
Is my answer equivalent to this and how was this solution reached?
| First off, I assume that in the provided solution, $\mathbf{k}$ should be $\mathbf{a}$, since the former doesn't appear in the problem statement, and after swapping it for $\mathbf{a}$ the answer coincides with mine.
The provided answer contains the term $\mathbf{a}\times\mathbf{c}$, which gives us a hint that we need to apply $\mathbf{a}\times$ to the equation. Indeed doing so gives us
$$(\mathbf{a}\cdot\mathbf{r})\mathbf{a} - (\mathbf{a}\cdot\mathbf{a})\mathbf{r} + \lambda\mathbf{a}\times\mathbf{r} = \mathbf{a}\times \mathbf{c}\ .$$
Now observe that we can use the original equation to substitute $\mathbf{a}\times\mathbf{r}$ with $\mathbf{c} -\lambda \mathbf{r}$, so that $(\mathbf{a}\cdot \mathbf{r})\mathbf{a} - |\mathbf{a}|^2 \mathbf{r} + \lambda\mathbf{c} - \lambda^2 \mathbf{r} = \mathbf{a}\times \mathbf{c}$, or
$$
(\mathbf{a}\cdot\mathbf{r})\mathbf{a}-(|\mathbf{a}|^2+\lambda^2)\mathbf{r} = \mathbf{a}\times\mathbf{c} - \lambda\mathbf{c}\ .
$$
At this point we still have a pesky $\mathbf{a}\cdot\mathbf{r}$ factor remaining, but since it's the coefficient of $\mathbf{a}$, it gets annihilated if we take the vector product with $\mathbf{a}$ again,
$$
-(|\mathbf{a}|^2 + \lambda^2)\mathbf{a}\times\mathbf{r} = (\mathbf{a}\cdot \mathbf{c})\mathbf{a} - (\mathbf{a}\cdot\mathbf{a})\mathbf{c} - \lambda\mathbf{a}\times \mathbf{c}\ .
$$
This might alarm you, because we seem to be reintroducing $\mathbf{a}\times \mathbf{r}$, but we already know how to transform it into something simpler. Making the same substitution for $\mathbf{a}\times\mathbf{r}$ as before, we get
$$
-(|\mathbf{a}|^2 + \lambda^2)(\mathbf{c} - \lambda\mathbf{r}) = (\mathbf{a}\cdot \mathbf{c})\mathbf{a} - |\mathbf{a}|^2\mathbf{c} - \lambda\mathbf{a}\times \mathbf{c}\ .
$$
Finally the $-|\mathbf{a}|^2\mathbf{c}$ terms will cancel, and rearranging the equation gives us the provided answer.
There are two main ideas behind the solution. The first is to use the properties of the cross product and vector triple product to get rid of factors that involve products of the unknown $\mathbf{r}$ with other vectors, so that we end up with an equation containing only a linear factor of $\mathbf{r}$ with a constant coefficient. The second is to use the original equation to substitute $\mathbf{a}\times\mathbf{r}$ repeatedly by an expression where $\mathbf{r}$ is only multiplied by a constant scalar.
Your attempt sounds correct in principle, but perhaps a little too "coordinate-y". In essence, reversing dot products is not a particularly easy thing to do. However, at least the expression $\mathbf{r}\cdot(\mathbf{c}\times\mathbf{a} + \lambda\mathbf{c}) = |\mathbf{c}|^2$ is consistent with the solution. To see this, take the dot product of the provided solution with $\mathbf{c}\times\mathbf{a} + \lambda\mathbf{c}$. In the denominator, all the triple products of $\mathbf{a}$, $\mathbf{a}$, $\mathbf{c}$ or of $\mathbf{a}$, $\mathbf{c}$, $\mathbf{c}$ vanish, so what remains is
$$
\lambda|\mathbf{a}\cdot\mathbf{c}|^2 + \lambda^3|\mathbf{c}|^2 + \lambda|\mathbf{a}\times\mathbf{c}|^2\ .
$$
Here the term $|\mathbf{a}\times\mathbf{c}|^2$ can be rewritten as a scalar triple product $(\mathbf{a}\times\mathbf{c})\cdot(\mathbf{a}\times\mathbf{c})$. We can cyclically permute it into $\mathbf{a}\cdot(\mathbf{c}\times(\mathbf{a}\times\mathbf{c}))$, and expand the result using the vector triple product relation, i.e. $\mathbf{a}\cdot[(\mathbf{c}\cdot\mathbf{c})\mathbf{a} - (\mathbf{c}\cdot\mathbf{a})\mathbf{c}] = |\mathbf{a}|^2|\mathbf{c}|^2 - |\mathbf{a}\cdot\mathbf{c}|^2$. Notice that second term here leads to a cancellation, so we end up with
$$
\lambda|\mathbf{a}|^2|\mathbf{c}|^2 + \lambda^3|\mathbf{c}|^2\ .
$$
After dividing by $N = \lambda(|\mathbf{a}|^2 + \lambda^2)$, you get $|\mathbf{c}|^2$, which is the right hand side of the equation you obtained.
| {
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"url": "https://math.stackexchange.com/questions/3994609",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find all solutions of congruence $3x^2−2x+9≡0\pmod {35}$ Find all solutions of congruence $3x^2 - 2x + 9 ≡ 0 \bmod 35$:
Attempt:
\begin{align}
3x^2 - 2x + 9 &\equiv 0 \bmod 35\tag{* 3} \\
9x^2-6x+27 &\equiv 0 \bmod 35 \tag{- 26} \\
(3x-1)^2 &\equiv -26 \bmod 35 \\
\\
-26 + 35 &= 9 = 3^2 \\
\\
\iff (3x-1)^2 &\equiv 3^2 \bmod 35 \\
\iff (3x-1-3)*(3x-1+3) &\equiv 0 \bmod 35 \\
\\
\implies \underbrace{3x - 4 \equiv 0 \bmod 35}_{(a)}
&\lor \underbrace{3x + 2 \equiv 0 \bmod 35}_{(b)}\\
\end{align}
Case $(a){:}\; 3x - 4 \equiv 0 \bmod 35$ $\Rightarrow 3x \equiv 4 \bmod 35$ $\Rightarrow 3x \equiv 39 \bmod 35$
$\quad\underset{ ( gcd(3,35) = 1)}{\implies} x \equiv 13 \bmod 35$
Case $(b){:}\; 3x +2 \equiv 0 \bmod 35$ $\Rightarrow 3x \equiv -2 \bmod 35$ $\Rightarrow 3x \equiv 33 \bmod 35$
$\quad\underset{ ( gcd(3,35) = 1)}{\implies} x \equiv 11 \bmod 35$
So $x = 13$ or $x = 11$.
Is it correct that way?
| At no point in your proof did you split the congruence into $\bmod5$ and $\bmod7$, so you missed two solutions as I will show below:
$$3x^2-2x+4=0\bmod5\implies x^2+x+3\equiv0\bmod5\implies x\equiv\{1,3\}\bmod5$$
$$3x^2-2x+2\equiv0\bmod7\implies x^2+4x+3\equiv0\bmod7\implies x\equiv\{-1,-3\}\bmod7$$
Combining with the Chinese remainder theorem this gives $x\equiv\{6,11,13,18\}\bmod35$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3995715",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Why combining two quadratic equations of a circle and a parabola creates extra solutions for $x$ We have a parabola and a circle with the following equations and their graph placed at the end of my question.
Parabola: $y^2 = 4x -4$
Circle: $(x-2)^2 + y^2 = 9$
My goal was to calculate their intersection points so I substituted $y^2$ from the parabola equation into the circle equation and I got
$(x-2)^2 + (4x-4)=9 \implies x^2 - 4x + 4 + (4x - 4) = 9 \implies x^2 = 9 \implies x = \pm3$
$x=3$ is the only correct solution but why is $x=-3$ produced as an extra invalid solution?
What is the exact mathematical explanation behind this? Why substituting one equation into the other has produced extra answers?
update
When I calculate $x$ from the parabola equation and substitute it in the circle equation, I don't get any extra answers for $y$:
$y^2=4x-4 \implies y^2 +4 = 4x \implies x = \frac{y^2}{4} + 1$
$(x-2)^2 +(4x-4)=9 \implies ((\frac{y^2}{4} + 1) - 2)^2 + (4x - 4)=9 \implies y^4 +8y - 128 = 0 \implies y^2=8,-16$
$y^2 = -16$ cannot be true so $y^2 = 8 \implies y=\pm 2\sqrt{2}$ and these are correct answers for $y$.
2nd update
I made a mistake in the calculation in the previous update although the final solutions where correct. I write the correct calculation:
$(x-2)^2 +y^2=9 \implies
((\frac{y^2}{4} + 1) - 2)^2 + y^2=9 \implies
(\frac{y^2}{4} - 1)^2 + y^2=9 \implies
(\frac{y^4}{16} - \frac{y^2}{2} + 1) + y^2=9 \implies
\frac{y^4}{16} + \frac{y^2}{2} + 1=9 \implies
(\frac{y^2}{4} + 1)^2=9 \implies
(\frac{y^2}{4} + 1)=\pm3 \implies
\frac{y^2}{4} =2,-4 \implies
y^2=8,-16$
| $$(x-2)^2 + y^2 = 9 \tag{A}$$
requires that $x \in [-1,5]$ and $y \in [-3,3]$.
$$y^2 = 4x-4 \tag{B}$$ requires that $x \in [1, \infty)$.
So, for $x$ to satisfy both (A) and (B), we must have
$x \in [1, \infty) \cap [-1,5] = [1,5]$.
Hence $x=-3$ will be an extraneous root.
| {
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"url": "https://math.stackexchange.com/questions/3998217",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 2
} |
Find $a_1+\frac{a_2}{2}+\frac{a_3}{2^2}+\cdots\infty$ A sequence $\left\{a_n\right\}$ is defined as $a_n=a_{n-1}+2a_{n-2}-a_{n-3}$ and $a_1=a_2=\frac{a_3}{3}=1$
Find the value of $$a_1+\frac{a_2}{2}+\frac{a_3}{2^2}+\cdots\infty$$
I actually tried this using difference equation method.Let the solution be of the form $a_n=\lambda^n$
$$\lambda^n=\lambda^{n-1}+2\lambda^{n-2}-\lambda^{n-3}$$ which gives the cubic equation $\lambda^3-\lambda^2-2\lambda+1=0$. But i am not able to find the roots manually.
| Let $$S=\sum_{n=1}^{\infty}\frac{a_n}{2^{n-1}}$$
We have,
$$S=\frac{1}{1}+\frac{1}{2}+\frac{3}{4}+P$$
Where $$P=\sum_{n=4}^{\infty}\frac{a_n}{2^{n-1}}$$
Using the given recurrence we have,
$$P=\sum_{n=4}^{\infty}\frac{a_{n-1}+2a_{n-2}-a_{n-3}}{2^{n-1}}$$
So we get,
$$P=\sum_{n=4}^{\infty}\frac{a_{n-1}}{2^{n-1}}+2\sum_{n=4}^{\infty}\frac{a_{n-2}}{2^{n-1}}-\sum_{n=4}^{\infty}\frac{a_{n-3}}{2^{n-1}}$$
By Change of variable for each summation we get
$$P=\sum_{k=3}^{\infty}\frac{a_k}{2^{k}}+2\sum_{k=2}^{\infty}\frac{a_k}{2^{k+1}}-\sum_{k=1}^{\infty}\frac{a_k}{2^{k+2}}$$
Which implies
$$P=\frac{1}{2}\sum_{k=3}^{\infty}\frac{a_k}{2^{k-1}}+\frac{2}{4}\sum_{k=2}^{\infty}\frac{a_k}{2^{k-1}}-\frac{1}{8}\sum_{k=1}^{\infty}\frac{a_k}{2^{k-1}}$$
$\implies$
$$P=\frac{1}{2}\left(\frac{a_3}{4}+P\right)+\frac{1}{2}\left(\frac{a_2}{2}+\frac{a_3}{4}+P\right)-\frac{S}{8}$$
Using $a_3=3,a_2=1$
$$P=\frac{3}{8}+\frac{P}{2}+\frac{1}{4}+\frac{3}{8}+\frac{P}{2}-\frac{S}{8}$$
Thus we get
$$\frac{S}{8}=\frac{3}{8}+\frac{1}{4}+\frac{3}{8}$$
$\implies$
$$S=8$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4000062",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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How does $b^2=a^2-ab$ lead to $\frac{b}{a}=\frac{-1+\sqrt 5}{2}$? So basically, I came across this particular solution of a problem that starts with
$$\frac {b}{a-b}=\frac ab \quad\implies\quad b^2=a^2-ab \quad\implies\quad\frac ba=\frac{-1+\sqrt 5}{2}$$
Can someone please explain how the $\frac ba=\frac{-1+\sqrt 5}{2}$ part came about?
The $\frac {b}{a-b}=\frac ab$ part is given as part of the problem.
| Dividing $b^2=a^2-ab$ by $a^2$ yields
$$
\left(\frac b a\right)^2 = 1-\frac b a.
$$
Let $z=\tfrac b a$ to get the quadratic equation $z^2=1-z$ or equivalently
$$
z^2+z-1 = 0.
$$
Solve this using your favorite method of solving a quadratic equation to obtain
$$
z = \frac { -1 \pm \sqrt{5} } 2.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4000170",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How do i prove this using precise definition of a limit? $(x^2 - 6x +11) = 6$
lim approaching 1
Given ϵ > 0
if |x−1|<δ then $|x^2 − 6x + 11 -6|$=$|x^2 - 6x +5|$<ϵ
this gets factored to:
$|x^2 - 6x +5|$ = (x-1)(x-5)<ϵ
This is where i am stuck, can someone explain to me what to do next?
So basically we start with $-1<x-1<1$ = $0<x<2$ for (x-1) and we find the equivalence for (x-5) $-5<x-5<-3$
is that correct?
Just to make sure but is it $-5<x-5<-3$ or $-5<x-5<3$? because if you add -5 to both sides it becomes -3.
| Here I propose the solution to the general case where $x$ approaches $a\in\mathbb{R}$ for the sake of curiosity.
In general, it can be proven that
\begin{align*}
\lim_{x\to a}(x^{2} - 6x + 11) = a^{2} - 6a + 11
\end{align*}
Indeed, let us consider that $|x - a| < \delta$. Then we have that
\begin{align*}
|(x^{2} - 6x + 11) - (a^{2} - 6a +11)| & = |x^{2} - a^{2} - 6(x - a)|\\\\
& = |(x-a)(x+a) - 6(x-a)|\\\\
& = |x-a||x + a - 6 |\\\\
& = |x-a||(x-a) + 2a - 6|\\\\
& \leq |x - a|(|x-a| + 2|a-3|)\\\\
& < \delta^{2} + 2\delta|a-3| = \varepsilon
\end{align*}
Consequently, for every $\varepsilon > 0$, there corresponds $\delta = -|a-3| + \sqrt{|a-3|^{2} + \varepsilon}$ such that
\begin{align*}
|x - a| < \delta \Rightarrow |f(x) - f(a)| < \varepsilon
\end{align*}
where $f(x) = x^{2} - 6x + 11$, and we are done.
Hopefully this helps!
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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From the system of equations prove $a+b+c=0$
$a,b,c$ are distinct real numbers and $x,y$ are also real numbers. we have these equations:
$${ \begin{cases}{a^3+ax+y=0} \\ {b^3+bx+y=0} \\
{c^3+cx+y=0}\end{cases} }$$ Prove $a+b+c=0$
I added all the equations together and get:
$$a^3+b^3+c^3+(a+b+c)x+3y=0$$
It is similar to Euler identity (because we have $a^3+b^3+c^3$). if $a+b+c=0$ then from Euler identity we can conclude $a^3+b^3+c^3=3abc$. and equation change to:
$$3abc+3y=0$$
But it seems that doesn't work.
| Hint: $a,b,c$ are the roots of cubic $$p(t)=t^3+tx+y$$
so by vieta $a+b+c=0$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
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Quadratic equation in trigonometric form $\sqrt 3 a\cos x + 2b\sin x = c$ Let a,b,c be three non-zero real number such that the equation, $\sqrt 3 a\cos x + 2b\sin x = c$, $x \in \left[ { - \frac{\pi }{2},\frac{\pi }{2}} \right]$, has two distict roots $\alpha$ and $\beta $ where $\alpha +\beta=\frac{\pi}{6}$. Then find the value of $\frac{b}{a}$.
My approach is as follow
$\sqrt 3 a\cos x + 2b\sin x = c$
$3{a^2}{\cos ^2}x = {c^2} + 4{b^2}{\sin ^2}x - 4bc\sin x$
$ {\sin ^2}x\left( {4{b^2} + 3{a^2}} \right) - 4bc\sin x + {c^2} - 3{a^2} = 0$
Roots are real when $16{b^2}{c^2} + 4\left( {4{b^2} + 3{a^2}} \right)\left( {{c^2} - 3{a^2}} \right) > 0$
$\sin \alpha + \sin \beta = \frac{{4bc}}{{4{b^2} + 3{a^2}}};\sin \alpha \sin \beta = \frac{{{c^2} - 3{a^2}}}{{4{b^2} + 3{a^2}}}$, how I will proceed from here
| You can see the equation as defining the intersections of the right half of the unit circle and a straight line. The bisector of the two points makes the angle $\dfrac{\alpha+\beta}2$ with the horizontal axis, and the line is perpendicular to it.
Hence
$$\dfrac{2b}{\sqrt3 a}=\tan\dfrac\pi{12}.$$
Analytical solution:
$$\cos\left(x-\arctan\dfrac{2b}{\sqrt3a}\right)=\frac c{\sqrt{3a^2+4b^2}}$$
and
$$x=\arctan\dfrac{2b}{\sqrt3a}\pm\arccos\frac c{\sqrt{3a^2+4b^2}}.$$
Obviously,
$$\alpha+\beta=2\arctan\dfrac{2b}{\sqrt3a}.$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Evaluating $\int_{0}^{\infty} \frac{1}{1 + x + x^2 + x^3 + x^4 + x^5}\mathrm{d}x$
How to evaluate $$\int_{0}^{\infty} \frac{1}{1 + x + x^2 + x^3 + x^4 + x^5} dx$$
My Attempt:
$$I = \int_{0}^{\infty} \frac{1}{1 + x + x^2 + x^3 + x^4 + x^5} dx$$
$$I = \int_{0}^{\infty} \frac{(1-x)}{(1 - x^6)} dx$$
Substituting $t = \frac{1}{x}$
$$I = \int_{0}^{\infty} \frac{(1-\frac{1}{x})}{(1 - \frac{1}{x^6})} \cdot \frac{1}{x^2} dx$$
$$I = \int_{0}^{\infty} \frac{(1-{x})(1 + x + x^2)}{(1-x^6 )} dx$$
This does not seem to work.
| Note that by replacing $x\to \frac{1}{x}$ $$I=\int_0^{\infty} \frac{1-x}{1-x^6}dx=\int_0^{\infty} \frac{(1-x)x^3}{1-x^6}dx$$Adding $$2I=\int_0^{\infty} \frac{(1-x)(1+x^3)}{1-x^6}= \int_0^{\infty} \frac{1-x}{1-x^3}=\int_0^{\infty} \frac{1}{x^2+x+1}$$ which should be easy..
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Looking for different approach to find $\left(x_1+\frac1{x_1}\right)\left(x_2+\frac1{x_2}\right)\left(x_3+\frac1{x_3}\right)$
If $x_1,x_2,x_3$ be roots of $x^3+3x+5=0$, find the value of
$$\left(x_1+\frac1{x_1}\right)\left(x_2+\frac1{x_2}\right)\left(x_3+\frac1{x_3}\right)$$
To solve this problem I expanded the expression and used Vieta's formula and got the answer $-\frac{29}5$
. but I wonder is there any other approach to solve this problem?
I tried this way as an alternative ansewr:
Consider $x_1,x_2,x_3$ be roots polynomial $P(x)=x^3+3x+5$. then $\frac1{x_1},\frac1{x_2},\frac1{x_3}$ are roots of $P(\frac1x):$
$$P(\frac1x)=0\quad \rightarrow \quad P(\frac1x)=(\frac1x)^3+\frac3x+5=0\rightarrow 5x^3+3x^2+1=0$$
But I don't know whether it helps or not.
| I'll expand on my comments above.
You want to find a cubic polynomial satisfied by $y=x+x^{-1}$ whenever $x$ is a root of $P$. Note that $x^3+3x+5=0$ gives $5x^{-1}=-(x^2+3)$ and $5x^{-3}=-(3x^{-2}+1)$. Therefore
\begin{align*}
-5x^{-1}&=x^2+3\\
25x^{-2}&=(x^2+3)^2=x^4+6x^2+9\\
&=x(-3x-5)+6x^2+9\\
&=3x^2-5x+9\\
-125x^{-3} & = 25(3x^{-2}+1)\\
&=9x^2-15x+52.
\end{align*}
Thus
\begin{align*}
5y &=5x+5x^{-1}=-x^2+5x-3\\
25 (y^2-2) &=25 x^2 +25 x^{-2} =28x^2-5x+9\\
125 (y^3-3y) &=125 x^3 + 125 x^{-3} =-9x^2-360x-677
\end{align*}
Eliminating $x^2$ gives
\begin{align*}
25 (y^2-2)+140y &=135x-75\\
125 (y^3-3y)-45y &=-405x-650
\end{align*}
or equivalently,
\begin{align*}
5y^2+28y &=27x-5\\
25y^3-84y &=-81x-130
\end{align*}
Now eliminate $x$ and we have
\begin{align*}
-145=-81x-130+3(27x-5)&=(25y^3-84y)+3(5y^2+28y)\\
&=25y^3+15y^2
\end{align*}
i.e.
$$
5y^3+3y^2+29=0
$$
so the product of roots $y$ is $-29/5$.
Remark: This is essentially the "standard method" for finding a polynomial equation for $y_i=q(x_i)$ where $q$ is a polynomial and $x_i$ are the roots of some polynomial $P$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4015863",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
What is the probability that there more rabbits than chickens in each of these three cages? Suppose we have $4$ chickens and $12$ rabbits, and put them into $3$ cages. What is the probability that there more rabbits than chickens in each of these three cages?
I was trying to solve it by using P(more rabbits than chicken) = 1-P(more chicken than rabbits). Since the number of chickens is far less than the number of rabbits and it would be easier to calculate.
Then, the cardinality of the sample space would be $3^{12+4}$. But I don't know how to continue.
| To Reader: As discussed in the comments, and as @Matthew Pilling explained in the comments, my solution probably have an issue (read comments provided by @Matthew Pilling). I cannot delete my answer since it has been already accepted.
==============================================================
Assumptions: Chickens are indistinguishable as well as Rabbits.
Note: For the case of distinguishable items, look at the other answer provided by @John L.
Let's say $c_{i}$ and $r_{i}$ are, respectively, the number of chickens and rabits in cage $i-th$. Therefore:
$$ c_{1} + c_{2} + c_{3} = 4 $$
$$ r_{1} + r_{2} + r_{3} = 12 $$
However, we want the number of rabbits to be more than the number of chickens in each cage. So,
$$ r_{1} = c_{1} + a_{1} + 1 $$
$$ r_{2} = c_{2} + a_{2} + 1 $$
$$ r_{3} = c_{3} + a_{3} + 1 $$
Where, $a_{i}$ is the auxiliary variable: $ a_{i} \ge 0 $
Finally, by substitution, we'll have:
$$ c_{1} + c_{2} + c_{3} = 4 $$
$$ a_{1} + a_{2} + a_{3} = 5 $$
So, we can first distribute the chickens and put them in cages, and then for each case, we can choose a different set of $a_{1}$, $a_{2}$, and $a_{3}$.
The final answer is:
$$ {4+3-1 \choose 3-1}{5+3-1 \choose 3-1} = {6 \choose 2}{7 \choose 2} = 315 $$
Now, you have the number of ways putting chickens and rabits to satisfy condition. Divide it by total cases and you will get the probability.
For total number of cases:
$$ c_{1} + c_{2} + c_{3} = 4 $$
$$ r_{1} + r_{2} + r_{3} = 12 $$
total number of cases (no condition) =
$$ {4+3-1 \choose 3-1}{12+3-1 \choose 3-1} = {6 \choose 2}{14 \choose 2} = 1365 $$
$$ probaility = \frac{315}{1365} $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4016006",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
A quite difficult limit: $\lim _{x\to 0}\frac{\exp(x^2)-\cos x-x\sin x}{\sinh^2x}$ I tried this limit with Taylor series. However, proceeding in this manner I obtained a series over another series. Furthermore, I do not know how to continue as long as we have $x$ tends to $0$. Were it to be $x$ tends to infinite, this limit would be simpler.
This is what I got
$$\begin{align}&
\lim _{x\to 0}\frac{\exp(x^2)-\cos x-x\sin x}{\sinh^2x} \\[6pt]
=\;&\frac{\frac12x^2+\frac{5}{8}x^4+\frac{23}{144}x^6+\frac{241}{5760}x^8+\frac{3359}{403200}x^{10}+\cdots}{{x^2+\frac{1}{3}x^4+\frac{2}{45}x^6+\frac{1}{315}x^8+\frac{2}{14175}x^{10}+\cdots}}
\end{align}$$
I can't continue.
| $$\frac{0.5x^2+\frac{5}{8}x^4+\frac{23}{144}x^6+\frac{241}{5760}x^8+\frac{3359}{403200}x^{10}+\ldots \:}{{x^2+\frac{1}{3}x^4+\frac{2}{45}x^6+\frac{1}{315}x^8+\frac{2}{14175}x^{10}+\ldots \:}}=\frac{0.5+\frac{5}{8}x^2+\frac{23}{144}x^4+\frac{241}{5760}x^6+\frac{3359}{403200}x^{8}+\ldots \:}{{1+\frac{1}{3}x^2+\frac{2}{45}x^4+\frac{1}{315}x^6+\frac{2}{14175}x^{8}+\ldots \:}} \to 1/2$$
as $ x \to 0.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4017441",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 1
} |
$\lim_{x\to 0} \frac{\cos \left(\pi \cdot \frac{1-\cos (ax)}{x^2}\right)}{x^2} $ I have to find for which values of $a \in \Bbb N, a \ne 0$ the following limit exists and it is finite:
$$\lim_{x\to 0} \frac{\cos \left(\pi \cdot \frac{1-\cos (ax)}{x^2}\right)}{x^2} $$
Applying L'Hôpital's rule:
$$\frac{1-\cos (ax)}{x^2}\sim \frac{\sin (ax) \cdot a}{2x}\sim \frac{a^2}{2}$$
Then $$ \frac{\cos (\pi \cdot \frac{1-\cos (ax)}{x^2})}{x^2} \sim \frac{\cos (\pi \cdot \frac{a^2}{2})}{x^2}.$$
$$\cos (\pi \cdot \frac{a^2}{2})=0 \implies a^2=1+2k, \quad k \in \Bbb N$$
and in this case the limit is $0$.
In the book the suggested solution is $a^2=1+2k$ for which the limit is $$ \frac{(-1)^k \cdot (2k+1)^2\cdot\pi}{24}$$
but I don't understand this solution.
Trying to solve the limit with $a^2=1+2k$:
$$ \frac{\cos \left(\pi \cdot \frac{1-\cos (ax)}{x^2}\right)}{x^2} \sim \frac{-\sin (\pi \cdot \frac{1-\cos ax}{x^2}) \cdot \pi \cdot \frac{\sin (ax) a x^2 - (1-\cos (ax)) 2x}{x^4}}{2x} $$$$ = \sin \left( \frac{\pi}{2}+k \pi\right) \cdot \pi \cdot \frac{(1-\cos(ax))2- \sin(ax) ax}{2 \cdot x^4}$$$$= (-1)^k \cdot \pi \cdot \frac{(1-\cos(ax))2- \sin(ax) ax}{2 \cdot x^4} $$
| You need to ensure that $$\dfrac{1-\cos ax} {x^2}\to\dfrac{2k+1}{2}$$ so that the limit in question exists. This clearly means that $a\neq 0$ and further $$a^2=2k+1$$ and thus $k\geq 0$.
Now we can write expression in question as $$(-1)^k\cdot\dfrac {\sin\left(\dfrac{(2k+1)\pi}{2}\left(1-\dfrac{2}{2k+1}\cdot \dfrac{1-\cos ax}{x^2}\right)\right)}{x^2}$$ which can be written as $(-1)^{k}(\sin f(x)) /x^2$ where $f(x) \to 0$. And we rewrite it as $$(-1)^k\cdot \frac{\sin f(x)}{f(x)} \cdot\frac{f(x)} {x^2}$$ Thus the desired limit equals the limit of $$(-1)^k\cdot\frac{1}{x^2}\cdot\frac{(2k+1)\pi}{2}\left(1-\frac{2}{2k+1}\cdot\frac{1-\cos ax} {x^2}\right)=(-1)^k\cdot \frac{\pi} {2}\frac{a^2x^2-4\sin^2(ax/2)}{x^4}$$ Putting $ax=2t$ we can see that the above equals $$(-1)^k\frac{a^{4}\pi}{8}\frac{t^2-\sin^2t}{t^4}$$ You can easily check via factoring that last factor tends to $1/3$ and hence the desired limit is $$(-1)^k\cdot\frac{(2k+1)^2\pi}{24}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4019497",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
Find the sum: $\sum_{n=0}^\infty \frac{(n!)^2}{(2n)!}x^n$
Find the sum:
$$\sum_{n=0}^\infty \frac{(n!)^2}{(2n)!}x^n$$
My try:
I played a bit with the coefficient to make it look easier/familiar:
First attempt:
$$\begin{align}
\sum_{n=0}^\infty \frac{(n!)^2}{(2n)!}x^n
&= \sum_{n=0}^\infty \frac{n!}{2^n(2n-1)!!}x^n \\
&= \sum_{n=0}^\infty \frac{n!}{(2n-1)!!}\left(\frac x2\right)^n
\end{align}$$
Second attempt:
$$\begin{align}
\sum_{n=0}^\infty \frac{(n!)^2}{(2n)!}x^n
&= \sum_{n=0}^\infty \frac{n!\cdot n!}{(n+n)!}x^n \\
&= \sum_{n=0}^\infty \frac{1}{{2n \choose n}}x^n
\end{align}$$
However, I could not proceed with any of them. Also, I have figured out that the convergence radius is $4$.
My research:
I have also found the same sum has been discussed at AoPS, which unfortunately uses Beta function that my course has not covered yet.
Entering the sum to Wolphram Alpha, I got the following output for the partial sum:
$$\sum_{n=0}^k\frac{(n!)^2x^n}{(2n)!}=\frac{4\sqrt{x}\left(\sin^{-1}\left(\frac{\sqrt{x}}{2}\right)-\frac{2^{2k}k!(k+1)!B_\frac{x}{4}\left(k+\frac{1}{2},\frac{3}{2}\right)}{(2k)!}\right)}{(4-x)^{3/2}}+\frac{4}{4-x}.$$
My background:
As I have already mentioned, I cannot use Gamma, Beta or similar functions. I only know about the convergence theorems on functional series and operations on them. So, I'm looking for some method that uses quite elementary tricks.
Thanks in advance.
| Suppose that $\displaystyle f(x) = \sum_{n\geqslant 0} a_n x^n$ with $a_0=1$. We shall assume that the series has non-zero radius of convergence. Then,
\begin{align*}
\frac{1}{2}\frac{d}{dx} \Big( x^2 f(x^2)\Big) = \sum_{n \geqslant 0} (n+1)a_n x^{2n+1} \quad \tag{1}\label{A}
\end{align*}
and
\begin{align*}
2x \frac{d}{dx} \Bigg( \frac{f(x^2)-1}{x} \Bigg)
&= \sum_{n \geqslant 1} 2(2n-1)a_n x^{2n-1} \\
&=\sum_{n \geqslant 0}2(2n+1) a_{n+1} x^{2n+1} \tag{2}\label{B}
\end{align*}
If $f$ now satisfies the differential equation,
\begin{align*}
2x \frac{d}{dx} \Bigg( \frac{f(x^2)-1}{x} \Bigg) = \frac{1}{2} \frac{d}{dx} \Big(x^2 f(x^2)\Big) \tag{3}\label{C}
\end{align*}
we can equate coefficients in the power series \eqref{A} and \eqref{B} to derive,
\begin{align*}
a_{n+1} = \frac{n+1}{2(2n+1)} a_n, \quad a_0 = 1
\end{align*}
which means
\begin{align*}
a_n = \frac{n!}{2^n(2n-1)(2n-3) \cdots 1} = \frac{(n!)^2}{(2n)!}.
\end{align*}
and
\begin{align*}
f(x) = \sum_{n \geqslant 0} \frac{(n!)^2}{(2n)!} x^n.
\end{align*}
Introduce $\phi(x) = (f(x^2)-1)/x$ so that the differential equation \eqref{C} becomes,
\begin{align*}
(4-x^2)\phi' -3x \phi = 2
\end{align*}
Multiply this by the integrating factor $(4-x^2)^{1/2}$ to obtain,
\begin{align*}
(4-x^2)^{3/2} \phi' - 3x (4-x^2)^{1/2}\phi=2(4-x^2)^{1/2}
\end{align*}
which is the same as
\begin{align*}
\frac{d}{dx} \Big( (4-x^2)^{3/2} \phi \Big) = 2(4-x^2)^{1/2} \tag{4}\label{D}.
\end{align*}
We note that by construction $\phi(0)=0$.
The right hand side may be integrated using the substitution $x=2\sin\theta$ to give,
\begin{align*}
\int_0^\xi 2(4-x^2)^{1/2} dx
&= 8\int_0^{\sin^{-1}(\xi/2)} \cos^2\theta d\theta \\
&=8 \Big[ \sin\theta\cos\theta \Big]_0^{\sin^{-1}(\xi/2)} + 8\int_0^{\sin^{-1}(\xi/2)} \sin^2\theta d\theta \\
&=8 (\xi/2) (1-\xi^2/4)^{1/2}+8\int_0^{\sin^{-1}(\xi/2)} 1-\cos^2\theta d\theta \\
&= 2 \xi (4-\xi^2)^{1/2} + 8\sin^{-1}(\xi/2) - 8\int_0^{\sin^{-1}(\xi/2)} \cos^2\theta d\theta
\end{align*}
from which we see,
\begin{align*}
\int_0^\xi 2(4-x^2)^{1/2} dx = \xi(4-\xi^2)^{1/2} + 4\sin^{-1}(\xi/2).
\end{align*}
Substituting this in \eqref{D}, and recalling $\phi(0)=0$, we get
\begin{align*}
\phi(x)(4-x^2)^{3/2} = x(4-x^2)^{1/2} + 4\sin^{-1}(x/2)
\end{align*}
yielding
\begin{align*}
f(x) &= \sqrt x \phi(\sqrt x) + 1 \\
&= 1 + \frac{x}{4-x}+\frac{4\sqrt x \sin^{-1}(\frac{\sqrt x }{2})}{(4-x)^{3/2}} \\
&=\frac{4}{4-x}\Bigg(1 + \frac{\sqrt{x} \sin^{-1}(\frac{\sqrt{x}}{2})}{\sqrt{4-x}} \Bigg)
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4020158",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 2,
"answer_id": 0
} |
How to prove this lower bound on $\log(1+x)$ for $x \geq 0$? How about $x > -1$? I want to show that for all real numbers $x \geq 0$:
$$
\log(1+x) \geq \frac{x(5x+6)}{2(x+3)(x+1)}.
$$
I'd like to understand each step necessary to prove this so that I can apply it to future problems. Amazingly, a popular online calculator immediately spits out that this inequality is true for all $x > -1$. Is there an easy way to prove this for $x \geq 0$? How about $x >-1$?
Attempt (and more info on how I'm stuck):
My guess is to use the Taylor series expansion of both sides at $x=0$ (but why is that sufficient for all $x \geq 0$?). Expanding both sides:
$$
\log(1+x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \ldots \geq x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{5x^4}{18} + \frac{7x^5}{27} - \ldots = \frac{x(5x+6)}{2(x+3)(x+1)}.
$$
I see that the early order terms match and then they don't. Not sure how to proceed from there. The signs are alternating on both sides, which I'm not sure how to deal with. Though, it seems the coefficients are converging to $0$ on both sides. It also seems the magnitude of the coefficients on the right-hand side is larger than the left-hand side (if you go term-wise), but how do I prove that? And do I need that fact to prove the inequality?
| Basically, the same proof as the great @AlbusDumbledore except (1) it more closely follows @RobertShore's suggestion and (2) it's not as nice. It might be easier to follow step-by-step though.
Let $f(x) = \log(1+x)$ and $g(x) = \frac{x(5x + 6)}{2(x+3)(x+1)}$. We want to show that $f(x) \geq g(x)$ for all $x \geq 0$.
Note that $f(0) = g(0) = 0$. Thus, it suffices to show that $g^\prime(x) \leq f^\prime(x)$ for $x \geq 0$, i.e., that $f$ grows faster than $g$. Also, note that $f^\prime(x) = \frac{1}{1+x}$ and $g^\prime(x) = \frac{7x^2 + 15x + 9}{(x+3)^2(1+x)^2}$. Then,
\begin{align}
&\frac{7x^2 + 15x + 9}{(x+3)^2(1+x)^2} \leq \frac{1}{1+x} \\
&\iff \frac{7x^2 + 15x + 9}{(x+3)^2} \leq 1+x \\
%&\iff \frac{7x^2 + 15x + 9}{(x+3)^2} -1 \leq x \\
&\iff \frac{3x}{(x+3)} \leq x \\
%&\iff \frac{3x}{(x+3)} - x \leq 0 \\
&\iff \frac{-x^2}{(x+3)} \leq 0 \\
&\iff \frac{x^2}{(x+3)} \geq 0,
\end{align}
which is clearly true for $x \geq 0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4021473",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
} |
Sum $\sum_{x}\sum_{y}xy\left(\frac{1}{2}\right)^{x-1}\cdot\left(\frac{1}{3}\right)^y$? $P(X=x,Y=y)=\left(\dfrac{1}{2}\right)^{x-1}\cdot\left(\dfrac{1}{3}\right)^y;x,y\ge1$. Find $E(XY)$ .
$$E(XY)=\sum_{x}\sum_{y}xy\left(\frac{1}{2}\right)^{x-1}\cdot\left(\frac{1}{3}\right)^y$$
$\sum_{y}y\left(\dfrac{1}{3}\right)^y\sum_{x}x\left(\dfrac{1}{2}\right)^{x-1}$
here $x,y\ge1$
I used this formula which is the sum of Arithmetic-Geometric progression.
$$a+(a+d)r+(a+2d)r^2\cdots = \frac{a}{1-r}+\frac{dr}{(1-r)^2}$$
$r$= common ratio(GP)
$d$= common difference(AP)
$a$= first term
$\bigg(1+2\cdot\dfrac{1}{2}+3\cdot\dfrac{1}{2^2}....\bigg)\cdot\bigg(1\cdot\dfrac{1}{3}+2\cdot\dfrac{1}{3^2}+3\cdot\dfrac{1}{3^3}....\bigg)=\bigg(\dfrac{1}{1-\frac{1}{2}}-\dfrac{1\cdot\frac{1}{2}}{(1-\frac{1}{2})^2}\bigg)\cdot\dfrac{1}{3}\bigg(\dfrac{1}{1-\frac{1}{3}}-\dfrac{1\cdot\frac{1}{3}}{(1-\frac{1}{3})^2}\bigg)=\bigg(4\cdot\dfrac{1}{3}\cdot\dfrac{3}{4}\bigg)=1$
But this is wrong. Answer lies between $2.50$ to $3.50$
| You have made a miscalculation. The formula has a plus sign where you have a minus sign: $$\frac{a}{1-r} \color{red}{+} \frac{dr}{(1-r)^2}.$$
The sum over $x$ is $$\frac{1}{1-1/2} + \frac{1/2}{(1-1/2)^2} = 4.$$ The sum over $y$ is $$\frac{1}{3}\left(\frac{1}{1-1/3} + \frac{1/3}{(1-1/3)^2}\right) = \frac{3}{4}.$$ Their product is $3$. You have an extra $1/3$ in your calculation because when you calculated the sum over $y$, you already included the $1/3$, but forgot that you did that so you multiplied again by $1/3$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4022020",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Finding $\frac{S_{\triangle BKC}}{S_{\triangle ABC}}$ Suppose circle $I$ is $\triangle ABC$’s incircle and $D, E,$ and $F$ are the tangent points of $BC, CA, AB$ and circle $I.$ Given that $DI$ intersects $EF$ at $K, BC = 10, AC = 8,$ and $AB = 7$ find $\frac{S_{\triangle BKC}}{S_{\triangle ABC}}.$
|
We use the identity $A = r \cdot s, $ where $s$ is sub-perimeter, $r$ is the inradius and $A$ is the area of the triangle.
So area of $\displaystyle \triangle ABC = \frac{10+8+7}{2} \cdot r = \frac{25r}{2}$
Area of $\triangle BIC = \frac{1}{2} BC \cdot DI = 5r$
Area of $\triangle BIK = \frac{1}{2} BD \cdot KI$, Area of $\triangle CIK = \frac{1}{2} CD \cdot KI$
So all we are left with is to find $KI$. For that, we draw internal angle bisector of $\angle A$ which meets $BC$ at $M$.
As $AF = AE, FI = IE, \angle AFI = \angle AEI = 90^0$, therefore $AI$ is perpendicular bisector of $FE$.
So $\angle IFK = \frac{\angle A}{2}$. Also, $\angle FIK = 180^0 - \angle FID = \angle B$.
That leads to $\triangle FIK \sim \triangle ABM$ (by A-A-A)
$\frac{KI}{FI} = \frac{BM}{AB} = \frac{14/3}{AB} \implies KI = \frac{2r}{3}$
(to calculate $BM$, we used the fact that $AM$ is the angle bisector and hence $BM:MC = AB:AC = 7:8$ and $BM + MC = 10$)
Getting to concluding steps, area of $\triangle BKC = 5r + \frac{1}{2} (BD+CD) \frac{2r}{3} = \frac{25r}{3}$
$\therefore \frac{S_{\triangle BKC}}{S_{\triangle ABC}} = \frac{2}{3}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4022276",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Inequality $2(a^2+b^2+c^2)^2 + 5(a^2+b^2+c^2) +30 abc \ge 3$ for $a+b+c=1$ Let $a,b,c$ be positive numbers with $a+b+c=1$. Show that
$$
2(a^2+b^2+c^2)^2 + 5(a^2+b^2+c^2) +30 abc \ge 3
$$
Equality holds for $a=b=c=\frac13$. One could try homogenizing:
$$
2(a^2+b^2+c^2)^2 + 5(a^2+b^2+c^2)(a+b+c)^2 +30 abc (a+b+c) - 3 (a+b+c)^4 \ge 0
$$
Expanding gives
$$
2 \sum_{cyc} a^4 - \sum_{cyc} a^3(b+c) - 2 \sum_{cyc} a^2 b^2 + 2 a b c \sum_{cyc} a \ge 0
$$
but I didn't manage to advance further.
| Another idea to deal with this type of inequality.
Assume $c=\min\{a,b,c\} \to 0< c \le \dfrac{1}{3}.$
Let $$f(a,b,c)=2 \left( {a}^{2}+{b}^{2}+{c}^{2} \right) ^{2}+5\left(a^2+b^2+c^2\right)+30abc$$
$$g(a,b,c)=2 \left[ \frac{1}{2} \left( a+b \right) ^{2}+c^2 \right] ^2+\frac{5}{2}
\left( a+b \right) ^{2}+5c^2+\dfrac{15}{2} \left( a+b \right) ^2c$$
We will so that
$$f(a,b,c)\ge g(a,b,c)\Leftrightarrow \left( a-b \right) ^{2} \left\{ 3a^2+2ab+3b^2+4{c}^{2}
+15\left(\frac{1}{3}-c\right) \right\}\ge 0\, \forall\,0<c\le \frac{1}{3}
$$
We only need to prove $$g(a,b,c)\ge 3\Leftrightarrow \frac{1}{2} c(c+1)(3c-1)^2\ge 0.$$
Done.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4023018",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
help faulhaber's formula summation problem I'm given that $\sum_{k=1}^{n}k=\frac{n(n+1)}{2}$
and told to prove $\sum_{k=1}^n (2k-1)=n^2$
$\sum_{k=1}^n (2k-1)=2\sum_{k=1}^n k- \sum_{k=1}^n 1=2\frac{n(n+1)}{2}-n$
$=2\frac{n(n+1)}{2}-n$
$=\frac{n\left(n+1\right)\cdot \:2}{2}-n$
$=n\left(n+1\right)-n$
$=n^2+n-n$
$=n^2$
$\sum_{k=1}^n (2k-1)=n^2$
next I need to write the summation that gives the sum
of the odd numbers found in the first $n$ rows of this sequence.
$1 = a_1$
$3 + 5 = a_2$
$7 + 9 + 11 = a_3$
$13 + 15 + 17 + 19 = a_4$
$21 + 23 + 25 + 27 + 29 = a_5$
$\vdots \vdots \vdots$
I'm givin that $n=3$ should yield the same result as $\sum_{k=1}^6 (2k-1)=36$
if you sub in $n^2$ to the formula for $\sum_{k=1}^{n}k=\frac{n(n+1)}{2}$ you get $\frac{n^{2}\left(n+1\right)^{2}}{2^{2}}$
at $n=3$; $\frac{3^{2}\left(3+1\right)^{2}}{2^{2}}=36$
I'm having trouble getting that formula into the correct form I believe this is the end result but I'm unsure how to prove the equality $\frac{n^{2}\left(n+1\right)^{2}}{2^{2}}=\sum_{k=1}^{n}k^{3}$
Then I need to replace that sum with an explicit formula of n
| Note that $a_i$ is a sum of $i$ terms. The first term is $(i^2 - i + 1)$ and the common difference of the terms is $2$.
Hence $$a_i = (i^2 - i + 1) + (i^2 - i + 3) + \ldots + (i^2 - i + 1 + 2(i-1))$$
Or, $$a_i = \sum_{j=1}^i \left(i^2 - i + (2j-1) \right)$$
Or, $$a_i = i^3 - i^2 + i^2 = i^3$$
Finally $$\sum_{i=1}^n a_i = \sum_{i=1}^n i^3 = \left(\frac{n(n+1)}{2}\right)^2$$
| {
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Given $a,b\in \mathbb N$, can $a^2+a b+b^2$ and $a^2- a b+b^2$ both be square numbers? The question arises if you search for two integer triangles ABC and AB'C with common vertices A and C, CA=b, CB=a, CB'=a+b and $\measuredangle ACB$ = $\measuredangle ACB'= 60°$. With c=AB and c'=AB' you get $c^2=a^2- a b+b^2$ and $c'^2=a^2+ a b+b^2$. I cannot find any $a,b\in \mathbb N$ such that $c,c'\in \mathbb N$. Is there an argument that shows that this is impossible?
| There are no positive integer solutions to these equations. One way of proving this is to consider the Diophantine equation
$$a^4+a^2b^2+b^4=(a^2+ab+b^2)(a^2-ab+b^2)=z^2.$$
We can tackle this by the method used in Does the equation $y^2=3x^4-3x^2+1$ have an elementary solution?
Theorem
There are no positive integer solutions of either of the equations
$$ x^4+x^2y^2+y^4=z^2\tag{1}$$ $$x^4-2x^2y^2-3y^4=z^2\tag{2}.$$
Proof
First note that, for either equation, we can suppose that the variables are pairwise coprime since a common factor of any pair of variables would be a factor of all and cancellation can occur.
An equation of form (1)
We can suppose that $x$ is odd. Then $z$ is odd and we have $y^2+y^4\equiv 0 \pmod 8$. Therefore $y$ is divisible by $4$ and we can write $y=2t$, where $t$ is even.
$ x^4+4x^2t^2+16t^4=z^2$ can be rewritten, using completing the square, as $$\left (\frac{x^2+2t^2-z}{2}\right )\left (\frac{x^2+2t^2+z}{2}\right)=-3t^4.$$
Since the two bracketed factors, $L$ and $M$ say, differ by the integer $z$ and have integer product, they are both integers. Furthermore, if $q$ is a prime common factor of $L$ and $M$, then $q$ would be a factor of both $z$ and $t$, a contradiction.
Therefore $\{L,M\}=\{au^4,cv^4\}$, where $ac=-3$ and $t=uv$, with $u$ and $v$ coprime and $uv$ even. Then $$au^4+cv^4=x^2+2t^2=x^2+2u^2v^2.$$
We can assume $v$ is even. Then $au^2\equiv 1\pmod 4$ and so $a=1$.
Therefore $u^4-2u^2v^2-3v^4=x^2$, an equation of form (2).
An equation of form (2)
Let $u,v,x$ be a pairwise coprime solution of $ u^4-2u^2v^2-3v^4=x^2$.
This can be rewritten, using completing the square, as $$\left (\frac{u^2-v^2-x}{2}\right )\left (\frac{u^2-v^2+x}{2}\right)=v^4.$$
The bracketed factors, $L$ and $M$, are again coprime integers.Therefore $\{L,M\}=\{aX^4,cY^4\}$, where $ac=1$ and $v=XY$, with $X$ and $Y$ coprime. Then $$X^4+X^2Y^2+Y^4=u^2,$$ an equation of form (1).
Fermat's infinite descent
We have seen that any positive integer solution $(x,y,z)$ of an equation of form (1) always leads to another positive integer solution $(X,Y,Z)$, where $Y=\frac{y}{2uX}<y$, which is impossible.
| {
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If for integer $x_1,x_2,x_3,x_4$ holds $x_i\cdot x_j = y_{ij}^2 - 13$ ($i\neq j$), is it possible that $x_1+x_2+x_3=10^6$? Let $a$, $b$, $c$, $d$ be integers. Given that a product of any two of them equals perfect square minus $13$, is it possible for the sum of some three of them to be equal to $10^6$?
My guess is that it is impossible.
My approach was to use the remainders modulo $5$, as $n^2\equiv0,\pm1\pmod{5}$.
If $a+b+c=10^6$, then $d(a+b+c)=k^2 + l^2 + m^2 - 39 = d \cdot 10^6$ for some $k,l,m$.
It is clear that (with the proper rearrangement of variables), $k^2\equiv l^2 \equiv 0,\; m^2 \equiv -1$ or $k^2\equiv l^2 \equiv -1,\; m^2 \equiv 1$.
One can elaborate a bit further on the possibilities for the remainders, but I can't arrive to any contradiction.
I have also tried solving this for smaller numbers in Wolfram, but it uses to much RAM.
| You are incredibly close. Consider the equation $k^2 + l^2 + m^2 -39 = d\cdot 10^6$ modulo $8$.
Note that $n^2 \equiv 0,1,4 \pmod 8$. Hence $k^2 + l^2 + m^2 \not\equiv 7 \pmod 8$. But $39 \equiv 7 \pmod 8$ and $d \cdot 10^6 \equiv 0 \pmod 8$. Hence there are no integer solutions to $k,l,m$.
| {
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Find all functions $f:\mathbb{R} \to \mathbb{R}$ such that $f(x^2 + y + f(y)) = 2y + (f(x))^2$
Find all functions $f:\mathbb{R} \to \mathbb{R}$ such that $$f(x^2 + y + f(y)) = 2y + (f(x))^2$$
I tried something like this but that inner $f(y)$ makes things worse to my opinion. Any help is appreciated.
| Here's my solution:
We claim that the only solution to the functional equation is $f(x)=x$, indeed it is easy to see that it is a solution. Now, we prove that it is the only one.
Let $P(x,y)$ denote the given assertion, by comparing $P(x,y)$ and $P(-x,y)$ we see that $f(x)^2=f(-x)^2$.
Then, we claim that $f$ is surjective, indeed, $$P(0,x): f(x+f(x))=2x+f(f(0))^2$$ and take any $k \in \mathbb{R}$, let $x=\frac{k-f(f(0))^2}{2}$, we get $f\left(\frac{k-f(f(0))^2}{2}+f\left(\frac{k-f(f(0))^2}{2}\right)\right)=k$.
Let $a\in \mathbb{R}$ such that $f(a)=a$, from $f(x)^2=f(-x)^2$, by letting $x=a$, we see that $f(-a)=0$. Moreover, we have $$P(0,a): 0=f(a+f(a))=2a+f(f(0))^2$$ $$P(0,-a): 0=f(-a+f(-a))=-2a+f(f(0))^2$$ subtracting both equation yields $a=0$. Therefore, we also have $f(x+f(x))=2x$ and $P(x,0) : f(x^2)=f(x)^2\ge 0$.
We now prove that $f$ is odd, notice that since $$P\left(x,-\frac{f(x)^2}{2}\right): f\left(x^2-\frac{f(x)^2}{2}+f\left(-\frac{f(x)^2}{2}\right)\right)=0$$ by the fact above that if $f(a)=0$, then $a=0$, we have $$-\frac{f(x)^2}{2}+f\left(-\frac{f(x)^2}{2}\right)=-x^2 \implies f\left(-\frac{f(x)^2}{2}+f\left(-\frac{f(x)^2}{2}\right)\right)=f(-x^2).$$ On the other hand, we have $$P\left(0,-\frac{f(x)^2}{2}\right): f\left(-\frac{f(x)^2}{2}+f\left(-\frac{f(x)^2}{2}\right)\right)=-f(x)^2$$ which means that $f(-x^2)=-f(x^2)=-f(x)^2 \le 0$. Therefore, we can see that $f(x)=-f(-x)$.
Now, there are two ways to proceed:
1st way: We show that $f$ is injective, denote $Q(x,y)$ by the new assertion $$P(\sqrt{x},y): f(x+y+f(y))=2y+f(\sqrt{x})^2=2y+f(x) \ \forall x\ge 0$$ if $x\le 0$, then use $P(\sqrt{-x},y)$. (Or just use the fact that $f$ is odd.) Then, by comparing $Q(x,y)$ and $Q(y,x)$, if $f(s)=f(t)$, then $s=t$. So, $f$ is bijective.
Next, since $f$ is bijective, so as $x+f(x)$. Thus, $f$ is additive! As $$f(x^2+y+f(y))=2y+f(x)^2=f(x^2)+f(y+f(y)).$$ And so, since $f$ is also positive on the positive reals, it is linear, by checking, $f\equiv x$.
2nd way: As we have shown that $f(x)=-f(-x)$, $$P(x,-x^2): -f(f(x)^2)=f(f(-x^2))=-2x^2+f(x)^2$$ also since $x^2-\frac{f(x)^2}{2}+f\left(-\frac{f(x)^2}{2}\right)=0 \implies -2x^2+f(x)^2=2f\left(-\frac{f(x)^2}{2}\right)=-2f\left(\frac{f(x)^2}{2}\right)$, we have $$f(f(x)^2)=2f\left(\frac{f(x)^2}{2}\right)\implies f(x)=2f\left(\frac{x}{2}\right) \ \forall x\ge 0$$ by surjectivity.
Next, as $f(x+f(x))=2x$, by replacing $x$ with $f(x)$ gives $$f(f(x)+f(f(x)))=2f(x)=f(2x) \implies f(x)+f(f(x))=2x \ \forall x\ge 0.$$ Since $f$ is positive on the positive reals, we use recurrence relation to solve for $f$, let $x_n=f^n(x)$, where $f$ is iterated $n$ times and define $x_0=x$ for a arbitrary real positive number, then this yields the following recurrence relation $$x_{n+1}+x_n-2x_{n-1}=0$$ which $x_n=\lambda_1\cdot (1)^n+\lambda_2 \cdot (-2)^n$ and $x_0=\lambda_1-\lambda_2$, $x_1=\lambda_1-2\lambda_2$ since $a_n\ge 0$, we must have $\lambda_2 =0$ otherwise $a_n$ would be negative for some arbitrary large, so we have $f(x_0)=x_1=\lambda_1=x_0$. Hence, as $x_0$ is arbitrary, we have $f(x)=x$ for all positive reals and since $f$ is odd $f\equiv x$.
| {
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Prove that $\frac{a^2}{a^2+bc}+\frac{b^2}{b^2+ac}+\frac{c^2}{c^2+ab}<2$ Prove that for every $a,b,c \in \mathbb{R}^{+}$ We have $$\frac{a^2}{a^2+bc}+\frac{b^2}{b^2+ac}+\frac{c^2}{c^2+ab}<2$$
Unfortunately i can just prove that :
$$\frac{a^2}{a^2+bc}+\frac{b^2}{b^2+ac}+\frac{b^2}{b^2+ac} <3$$
like this :
$$a^2+bc>a^2 \iff \frac{a^2}{a^2+bc}<1$$
and by the same method we have :
$$\frac{b^2}{b^2+ac}<1,\frac{b^2}{b^2+ac}<1$$
Adding them together will give us the desired inequality.
and please don't use any $\sum_{cyc}$ because I get confused with it.
| Suppose $a \geqslant b \geqslant c,$ we have
$$\frac{a^2}{a^2+bc} < 1,$$
and
$$\frac{b^2}{b^2+ca}+\frac{c^2}{c^2+ab} \leqslant \frac{b^2}{b^2+c^2}+\frac{c^2}{c^2+b^2} = 1.$$
Therefore
$$\frac{a^2}{a^2+bc}+\frac{b^2}{b^2+ca}+\frac{c^2}{c^2+ab} < 2 .$$
| {
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Modular quadratic equation question- Where did I go wrong? \begin{gather}
\frac{N^2+N}{2} \equiv 0 \pmod 4 \\
N^2+N\equiv 0\\
(N+\frac{1}{2})^2-\frac{1}{4}\equiv 0\\
4(N+\frac{1}{2})^2-1\equiv 0\\
2^2(N+\frac{1}{2})^2\equiv 1\\
(2N+1)^2\equiv 1 \\
2N+1 \equiv 1\\
2N=0\\
N=2,4,6,8...\\
2N+1 \equiv 3\\
2N=2\\
N=1,3,5,7...\\
\end{gather}
In other words I've shown $\forall N >0 \to \frac{N^2+N}{2} \equiv 0 \pmod 4$, which is false e.g. for $N=3$, $\frac{N^2+N}{2}$ has a remainder of 2
I'd like a correct solution as well. Thanks.
| What you did incorrectly is basically explained in Peter's answer, and several question comments. Note, though, the more general issue when manipulating congruences, since you are really dealing with integers, is that involving "fractions" of the form $\frac{p}{q}$ where $\gcd(p,q) = 1$, really means the division by $q$ is multiplying by its modulo multiplicative inverse, i.e., $q^{-1}$. However, only integers which are relatively prime to the modulus have inverses. Since $\gcd(4, 4) = 4 \neq 1$ and $\gcd(4, 2) = 2 \neq 1$, neither $2$ nor $4$ have an inverse in modulo $4$, so while working in that modulus you can't, in general, just "divide" by either value, e.g., $\frac{1}{2}$ and $\frac{1}{4}$ don't make sense.
As for how to correctly solve your congruence equation, note for any solution $N$, there's some integer $j$ where
$$\begin{equation}\begin{aligned}
& \frac{N^2 + N}{2} \equiv 0 \pmod{4} \iff \frac{N^2 + N}{2} = 4j \iff \\
& N^2 + N = 8j \iff N^2 + N \equiv 0 \pmod{8}
\end{aligned}\end{equation}\tag{1}\label{eq1A}$$
Due to $N$ and $N + 1$ being relatively prime to each other, with one being odd and one being even, then since $8 = 2^3$ has only prime factors of $2$, this then gives
$$\begin{equation}\begin{aligned}
& N(N + 1) \equiv 0 \pmod{8} \implies \\
& (N \equiv 0 \pmod{8}) \lor (N + 1 \equiv 0 \pmod{8}) \implies \\
& N \equiv 0, 7 \pmod{8}
\end{aligned}\end{equation}\tag{2}\label{eq2A}$$
This means the allowed values for $N$ are $N = 8k$ and $N = 8k + 7$ for any $k \in \mathbb{Z}$.
| {
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How to simplify the summation of a recurrence relation After solving the recurrence relation
$$T(n) = 3T(\frac{n}{3}) + n\log(n)$$
I get following equation
$$T(n)=3kT(\frac{n}{3k})+ n\log(n) + n\log(\frac{n}{3}) + n\log(\frac{n}{3^2})+\dots+n\log(\frac{n}{3^k})$$
I don't know how to simplify the summation and how to know the asymptotic function?
| Let $n=3^k$. We have,
$T(n)=3^kT(\frac{n}{3^k})+ n\log(n) + n\log(\frac{n}{3}) + n\log(\frac{n}{3^2})+\dots+n\log(\frac{n}{3^k})$
$=3^k.T(1)+n\log\left(n.\frac{n}{3^1}.\frac{n}{3^2}\ldots.\frac{n}{3^k}\right)$
$=3^k.1+n\log\left(n.\frac{n^k}{3^{1+2+\ldots+k}}\right)$ (since $T(1)=1$)
$=3^k+n\log\left(\frac{n^{k+1}}{3^{1+2+\ldots+k}}\right)$
$=3^k+n\log\left(\frac{(3^k)^{k+1}}{3^{k(k+1)/2}}\right)$
$=3^k+n\log\left(\frac{3^{k(k+1)}}{3^{k(k+1)/2}}\right)$
$=3^k+3^k\log\left(3^{k(k+1)/2}\right)$
$=3^k+3^k.k(k+1)/2.\log3$
$=3^k+\Theta(3^k.k^2)$
$=\Theta(3^k.k^2)$
$=\Theta(n.(\log n)^2)$ (since $n=3^k$)
Or use Master theorem:
$T(n) = 3T(\frac{n}{3}) + n\log(n)$, here $c_{crit}=\log_b a = \log_3 3 = 1$, $k=1$, hence, we have,
$T(n) = \Theta(n^{c_{crit}}\log^{k+1}n)=\Theta(n\log^2 n)$
| {
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Solve this limit $\lim\limits_{x\to 1}\left(\dfrac{2017}{1-x^{2017}} -\dfrac{2018}{1-x^{2018}}\right)$ \begin{align}
\lim\limits_{x\to 1}\left(\dfrac{2017}{1-x^{2017}} -\dfrac{2018}{1-x^{2018}}\right)
&= \lim\limits_{x\to 1}\dfrac{2017(x^{2017} + \dots + 1) - 2018(x^{2016} + \dots + 1)}{(1-x)(x^{2016} + \dots +1)(x^{2017} + \dots +1)}\\\\
&= \lim\limits_{x\to 1}\dfrac{2017x^{2017} - x^{2016} - \dots - 1}{(1-x)(x^{2016} + \dots +1)(x^{2017} + \dots +1)}
\end{align}
Did I do the right way with those step above?
I think next step is to separate:
$2017x^{2017} = \underbrace{x^{2017} + \dots + x^{2017}}_{\text{2017 addends}}$
Then combine with the rest and factorize $(1-x)$,
but it has a little confused after factorizing.
Please help me!!?
| The binomial theorem gives another way. Let $x=1+\varepsilon$ with $0<|\varepsilon|\ll1$. Then
$$x^n=1+n\varepsilon+\tfrac12n(n-1)\varepsilon^2+O(\varepsilon^3),\quad x^{n+1}=1+(n+1)\varepsilon+\tfrac12(n+1)n\varepsilon^2+O(\varepsilon^3).$$Hence
$$\frac n{1-x^n}-\frac{n+1}{1-x^{n+1}}=\frac1\varepsilon\left(\frac{-1}{1+\tfrac12(n-1)\varepsilon+O(\varepsilon^2)}-\frac{-1}{1+\tfrac12n\varepsilon+O(\varepsilon^2)}\right)$$
$$=-\frac1\varepsilon\frac{\tfrac12\varepsilon+O(\varepsilon^2)}{1+n\varepsilon+O(\varepsilon^2)}=-\frac{\tfrac12+O(\varepsilon)}{1+O(\varepsilon)}.$$
Therefore the limit as $\varepsilon\to0$ is $-\frac12$.
| {
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Prove $ (\frac{1}{cosA}-1)(\frac{1}{cosB}-1)(\frac{1}{cosC}-1) \ge 1$ Let $\triangle ABC$ be a acute triangle. Prove that: $$(\frac{1}{cosA}-1)(\frac{1}{cosB}-1)(\frac{1}{cosC}-1) \ge 1 $$
My attempt:
$$\Leftrightarrow (1-cosA)(1-cosB)(1-cosC)\ge cosA.cosB.cosC$$
$$\Leftrightarrow 1-2cosA.cosB.cosC + cosA.cosB + cosB.cosC+cosA.cosC \ge cosA+cosB+cosC $$
$$\Leftrightarrow cos^2A+cos^2B+cos^2C + cosA.cosB + cosB.cosC+cosA.cosC \ge cosA+cosB+cosC $$
$$ 0<cos A,cos B,cosC<1$$
$$cos^2A+cos^2B+cos^2C=1-2cosA.cosB.cosC\ge 3\sqrt[3]{cosA.cosB.cosC}$$
$$\Rightarrow cosA.cosB.cosC \le \frac{1}{8}$$
And I was stuck here. Could you help me ?
| Simple. Use AM-GM and Jensen's inequalities: $\cos A\cos B\cos C \le \dfrac{(\cos A+\cos B+\cos C)^3}{27}\le \dfrac{1}{27}\cdot (3\cos (\frac{A+B+C}{3}))^3=\left(\frac{1}{2}\right)^3 =\dfrac{1}{8}.$
| {
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trigonometric equations finding $x$, when $\cos(x)=0$ or some value I'm practicing trigonometric equations where we need to find the value of $x$, however sometimes I keep getting a completely different solution than the one in the book and I don't really understand where I am making a mistake (and it's always in finding the $x$ in the end). For example:
Problem: $$\cos x + \sqrt3\cos 2x + \cos3x=0$$ which gives: $$2\cos2x(\cos x + \frac{\sqrt3}{2})=0$$
and then: $2\cos2x=0$ , where I get that $x=\frac{\pi}{4}+\frac{k\pi}{2}, k\in \Bbb Z$ and: $\cos x= -\frac{\sqrt3}{2}$, where I get $x=\frac{5\pi}{6}+ 2k\pi, k\in \Bbb Z$ (which based on my book is the only solution I got right)
I believe I also have this: $\cos (2\pi - x)= -\frac{\sqrt3}{2} $, and the solution I got: $x=\frac{7\pi}{6}+ 2k\pi, k\in \Bbb Z$
I guess I'm getting something wrong since the solutions given are: $x=\frac{(2k+1)\pi}{2}$ ; $x=\frac{5\pi}{6}+2k\pi$; $x=-\frac{5\pi}{6}+2k\pi$, $k\in \Bbb Z$
| hint
If $$\cos(2x)(\cos(x)\frac{\sqrt{3}}{2})=0$$
then
$$\cos(2x)=0=\cos(\frac{\pi}{2})$$
or
$$\cos(x)=-\frac{\sqrt{3}}{2}=\cos(\pi-\frac{\pi}{6})$$
then
$$2x=\pm \frac{\pi}{2}+2k\pi=\frac{\pi}{2}+k\pi$$
or
$$x=\pm \frac{5\pi}{6}+2k\pi$$
The solutions are
$$\frac{\pi}{4}+k\frac{\pi}{2}\;,\;$$
$$ \frac{5\pi}{6}+2k\pi=-\frac{7\pi}{6}+2(k+1)\pi\;$$
and
$$\; \frac{-5\pi}{6}+2k\pi=\frac{7\pi}{6}+2(k-1)\pi$$
| {
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Find $\int \frac 1 {(x^2 +a^2)^2} dx $. I am trying to integrate $\int \frac 1 {(x^2+a^2)^2} \ dx$. The only thing that I can think to try is substitution, $u=x^2+a^2$ so that $\frac{du}{dx}=2x \Rightarrow du = 2x\ dx = 2\sqrt{u-a^2}\ du$ and then the integral becomes
$$ \int \frac{1}{u^2} (2\sqrt{u-a^2} \ du) $$
which does not seem productive.
I could try perhaps to separate with integration by parts, setting $u = \frac{1}{x^2 +a^2}$ and $dv = \frac{dx}{x^2+a^2}$. Then we obtain $\frac{du}{dx} = -\frac{1}{(x^2+a^2)^2}(2x)$ and $v = \frac 1 a \tan^{-1}(x/a)$. Then the integral becomes
$$ \begin{align*}
uv - \int v \ du &= \left(\frac{1}{x^2 +a^2}\right)\left(\frac 1 a \tan^{-1}(x/a)\right) - \int \left(\frac 1 a \tan^{-1}(x/a)\right) \left( -\frac{1}{(x^2+a^2)^2}(2x) \right) \ du
\end{align*}$$
but this also looks like it's headed nowhere good. Advice?
| If you do $x=ay$ and $\mathrm dx=a\,\mathrm dy$, your integral becomes$$\int\frac a{(a^2y+a^2)^2}\,\mathrm dy=\frac1{a^3}\int\frac1{(y^2+1)^2}\,\mathrm dy.$$On the other hand\begin{align}\int\frac1{(y^2+1)^2}\,\mathrm dy&=\int\frac1{y^2+1}\,\mathrm dy-\int\frac{y^2}{(y^2+1)^2}\,\mathrm dy\\&=\arctan(y)-\frac12\int\frac{2y}{(y^2+1)^2}y\,\mathrm dy\\&=\arctan(y)-\frac12\left(-\frac y{y^2+1}+\int\frac1{y^2+1}\,\mathrm dy\right)\\&=\frac12\arctan(y)+\frac12\frac y{y^2+1}\\&=\frac12\arctan\left(\frac xa\right)+\frac12\frac{ax}{x^2+a^2},\end{align}and therefore$$\int\frac1{(x^2+a^2)^2}\,\mathrm dx=\frac1{2a^3}\arctan\left(\frac xa\right)+\frac1{2a^2}\frac x{x^2+a^2}.$$
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Simplifying $z^2+i=0$ I need to simplify $z^2+i=0$ and find all solutions for $z$. I have seen that the solutions to $z=\sqrt{i}=\left(\frac{1}{\sqrt{2}}+i\frac{1}{\sqrt{2}}\right)$ and $\left(-\frac{1}{\sqrt{2}}-i\frac{1}{\sqrt{2}}\right)$. I was hoping to find a similar solution for $z=\sqrt{-i}\,$ but my attempt gives me $z=\pm i^{\frac{3}{2}}$
$$z=re^{i\theta} \,\,\& \,\, e^{i\pi}=-1 $$
then,
$$(re^{i\theta})^2=-i\\r^2e^{i2\theta}=ie^{i\pi+k(2\pi)}$$ where $k\in\mathbb{Z}$.
So, we have $\begin{cases} r^2=i \,\,\,\therefore r=\sqrt{i}\\ \theta=k\pi
\end{cases}$
Then, $$z_k=\sqrt{i} \, e^{i\left(\frac{\pi}{2}+k\pi\right)}$$
$$z_0=\sqrt{i}\left(\cos\frac{\pi}{2}+i\sin\frac{\pi}{2}\right)=i^\frac{3}{2}\\
z_1=\sqrt{i}\left(\cos(\frac{\pi}{2}+1)+i(\sin\frac{\pi}{2}+1)\right)=-i^\frac{3}{2}$$
I realized that this is literally the same as just solving $z=\sqrt{-i}=i\sqrt{i}=i^\frac{3}{2}$, however, I was hoping to find a solution of the form $x+iy$. I am not sure how to go about this problem a different way.
|
I was hoping to find a similar solution for z=−i−−√ but my attempt gives me z=±i32
I don't see why.
$z = x+yi$ and $z^2 = (x^2 -y^2) + 2xyi = -i$ so $x^2-y^2 = 1$ and $2xy = -1$ so $x^2 =y^2$ and $x = \pm |y|$ but $2xy =-1$ is negative os $x = -y$ and so $2xy=-1\implies -2y^2 = 1\implies y =\pm \frac 1{\sqrt 2}$ and $x = \mp \frac 1{\sqrt 2}$. and $z = \pm \frac 1{\sqrt 2} \mp \frac 1{\sqrt 2} i$.
Which shouldn't surprise us as if $(\pm \frac 1{\sqrt 2} \pm \frac 1{\sqrt 2} i)^2 = i$ then $i(\pm \frac 1{\sqrt 2} \pm \frac 1{\sqrt 2} i) = \pm \frac 1{\sqrt 2}i \mp \frac 1{\sqrt 2} =\mp \frac 1{\sqrt 2} \pm \frac 1{\sqrt 2}$ when squared should be $i^2*i = -i$.
===
Using polar coordidates $-i = 0 + (-1)i = \cos \frac {3\pi}2 +\sin(\frac {3\pi}2)i = e^{(\frac {3\pi}2 + 2k\pi)i}$ and so the square roots of $-i$ are
$e^{(\frac {3\pi}4 + k\pi)i} = \cos(\begin{cases}\frac {3\pi}4\\\frac {7\pi}4\end{cases})+\sin(\cos(\begin{cases}\frac {3\pi}4\\\frac {7\pi}4\end{cases}))i =\pm \frac 1{\sqrt 2} \mp \frac 1{\sqrt 2}i$
| {
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Show that no number of the form $2^a3^b$ is $3$-perfect. Show that no number of the form $2^a3^b$ is $3$-perfect.
A number is $3$-perfect if $\sigma(n)=3n$. I assumed that there exists a number $n=2^a3^b$ such that $\sigma(n)=\sigma(2^a3^b)=\sigma(2^a)\sigma(3^b)=3\cdot2^a3^b=3n$. Then I had $(1+2+\cdots +2^a)(1+3+\cdots+3^b)=2^a3^{b+1}$ and so $(2^{a+1}-1)({1\over 2}(3^{b+1}-1))=2^a3^{b+1}$. and through some algebra on the LHS I convinced myself that these two things are not equal to each other, contradiction.
I'm confident there is a better and more elementary way to do this problem. Any hints or solutions are greatly appreciated.
| Multiplying both sides of $(2^{a+1}-1)(\frac{3^{b+1}-1}{2})=2^a3^{b+1}$ by $2$, we get $(2^{a+1} - 1)(3^{b+1} - 1) = 2^{a+1} 3^{b+1}$.
But this is impossible, because $2^{a+1}-1 < 2^{a+1}$ and $3^{b+1}-1 < 3^{b+1}$.
| {
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Criteria for computing the integral of $e^{-\frac{x^{2}}{2 a} - \frac{y^{2}}{2 b} - k x y}$? In some problem the following integral is involved
\begin{equation}
\int_{-\infty}^{\infty} \mathrm{d} y \int_{-\infty}^{\infty} \mathrm{d} x \; e^{-\frac{x^{2}}{2 \alpha_{1}} - \frac{y^{2}}{2\alpha_{2}} - k x y}
\end{equation}
where $\alpha_{1}, \alpha_{2}$ and $k$ are positive constants. In the "limit for small $k$'' I attempt to solve this integral by rewriting the argument in the exponential as
\begin{equation}
-\frac{1}{2\alpha_{1}}\left(x + \alpha_{1}k y\right)^{2} - \frac{1}{2\alpha_{2}}\left(1 - \alpha_{1}\alpha_{2}k^{2}\right)y^{2}
\end{equation}
and integrating as if I had two Gaussian integrals (one for $x$ and one for $y$) which seems to yield the correct result (provided $1 - \alpha_{1}\alpha_{2}k^{2} > 0$). However, is this procedure valid? More specifically, when integrating over $x$, why can one integrate over a sum of variables just as if one were integrating over one variable?
| Express $\frac{x^2}{2a} + \frac{y^2}{2b} +kxy $ in a separable form with the variable changes
$$x=u\cos\theta -v \sin\theta, \>\>\> y=u\sin\theta +v \cos\theta$$
where, in the limit of small $k$ and $a\ne b$
$$\theta \approx \frac{ab}{b-a}k
$$
Then, the separable form reads
$$\frac{x^2}{2a} + \frac{y^2}{2b} +kxy
= \left( \frac1a - \frac{abk^2}{a-b}\right)\frac{u^2}2
+ \left( \frac1b - \frac{abk^2}{b-a}\right)\frac{v^2}2
$$
and
\begin{align}
& \int_{-\infty}^{\infty} \mathrm{d} y \int_{-\infty}^{\infty} \mathrm{d} x \; e^{-\frac{x^{2}}{2 a} - \frac{y^{2}}{2b} - k x y}\\
=& \int_{-\infty}^{\infty} \mathrm{d} v \int_{-\infty}^{\infty} \mathrm{d} u \; e^{-\left( \frac1a - \frac{abk^2}{a-b}\right)\frac{u^2}2- \left( \frac1b - \frac{abk^2}{b-a}\right)\frac{v^2}2}\\
=& \frac{2\pi}{\sqrt{\left(\frac1a - \frac{abk^2}{a-b}\right) \left( \frac1b - \frac{abk^2}{b-a}\right)}}\approx
\frac{2\pi\sqrt{ab}}{\sqrt{1-ab k^2}}
\end{align}
| {
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Probability St Petersburg Game I am reading a book on probability and there is an interesting chapter on the St Petersburg game - where a coin is flipped until a head is landed and the prize is £2 if there is a head on the first throw, £4 is there is a head on the second flip, £8 for a head on the third etc. The idea of this game is it introduces a paradox since someone should be prepared to pay any amount to play since the expected yield is infinite.
However in the book it introduces the notion of changing the rewards to 2,4,6,8.... etc rather than the doubling in the original game. The book contends that even though the possible payouts increase without bound the calculation of the expected value yields a sensible answer of £4. It mentions that this is simple to calculate so omits the calculations. I have pondering this and cant see how the £4 has been arrived at - a entry price of £4 would be a fair price for the game but why? Any help greatly appreciated. I am sure I'm missing something to do with summing to infinity perhaps
| If you are familiar with the series
$\frac12 + \frac14 + \frac18 + \frac1{16} + \cdots,$
you may recall that it adds up to $1$:
$$ 1 = \frac12 + \frac14 + \frac18 + \frac1{16} + \cdots, $$
which is just a number-heavy way to say that the probability is $1$ that the coin eventually will come up heads. (Probability $\frac12$ on the first throw, $\frac14$ that it takes exactly two throws, $\frac18$ that it takes exactly three throws, etc.)
If we multiply all the terms by $2$ we get twice as much for the sum:
\begin{align}
2 &= 1 + \frac12 + \frac14 + \frac18 + \cdots, \\
4 &= 2 + 1 + \frac12 + \frac14 + \cdots.
\end{align}
If we divide all the terms by $2$ we get half as much:
\begin{align}
\frac12 & = \frac14 + \frac18 + \frac1{16} + \frac1{32} + \cdots,\\
\frac14 & = \frac18 + \frac1{16} + \frac1{32} + \frac1{64} + \cdots,\\
\frac18 & = \frac1{16} + \frac1{32} + \frac1{64} + \frac1{128} + \cdots,\\
\end{align}
and so forth.
Rearranging terms in a series can alter the sum if some terms are positive and some are negative; but in the series you are looking at, all terms are positive, so we can write:
\begin{align}
2\cdot\frac12 + 4\cdot\frac14 &+ 6\cdot\frac18 + 8\cdot\frac1{16} + \cdots\\
&= 1 + 2\cdot\frac12 + 3\cdot\frac14 + 4\cdot\frac18 + \cdots\\
&= 1 + (1+1)\cdot\frac12 + (1+1+1)\cdot\frac14 + (1+1+1+1)\cdot\frac18 + \cdots\\
&= 1 + \frac12 + \frac14 + \frac18 + \cdots\\
&\phantom{{}= 1} + \frac12 + \frac14 + \frac18 + \cdots\\
&\phantom{{}= 1 + \frac12} + \frac14 + \frac18 + \cdots\\
&\phantom{{}= 1 + \frac12 + \frac14} + \frac18 + \cdots\\
&\phantom{{}= 1 + \frac12 + \frac14 + \frac18} + \cdots\\
&= 2 \\
&\phantom{{}= 2} + 1\\
&\phantom{{}= 2 + 1} + \frac12\\
&\phantom{{}= 2 + 1 + \frac12} + \frac14\\
&\phantom{{}= 2 + 1 + \frac12 + \frac14} + \cdots\\
&= 4.
\end{align}
| {
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"timestamp": "2023-03-29T00:00:00",
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For $0For $0<b \leq a$ prove that $$\frac{1}{8}\left(\frac{(a-b)^2}{a} \right) \leq \frac{a+b}{2}-\sqrt{ab} \leq \frac{1}{8}\left(\frac{(a-b)^2}{b} \right) $$
I was trying compare the following expressions
$$\frac{a+b}{2}-\sqrt{ab} =\frac{1}{2}(\sqrt{a}-\sqrt{b})^2$$
$$\frac{1}{8}\left(\frac{(a-b)^2}{a} \right)=\frac{1}{2a}\left(\frac{a-b}{2}\right)^2\leq\frac{1}{a}(a-b)^2$$
and use the fact that $\sqrt{a} \leq a$ when $a>0$ but I don´t get good results.
Any hint of how I should start?
| Hint:
$$
\frac{1}{2}(\sqrt a - \sqrt b )^2 = \frac{b}{{2(\sqrt a + \sqrt b )^2 }}\left( {\frac{{(a - b)^2 }}{b}} \right)
$$
and
$$
\frac{1}{2}(\sqrt a - \sqrt b )^2 = \frac{a}{{2(\sqrt a + \sqrt b )^2 }}\left( {\frac{{(a - b)^2 }}{a}} \right).
$$
Use the fact that $0<b\leq a$ to estimate the right-hand sides from above and below.
| {
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KVPY Question: Sequence and Series.
$S_n = \cfrac{7}{4 \cdot 1 \cdot 2} + \cfrac{10}{4^2 \cdot 2 \cdot 3} + \cfrac{13}{4^3 \cdot 3 \cdot 4} + ....$, then $S_{\infty}=$
I found the general term:
$$\sum_{n=1}^{n \rightarrow \infty} \cfrac{(3n+4)}{n(n+1)4^{n}} = $$
How to evaluate further? Please help.
I am at an elementary level, so this question might seem dumb, sorry.
| $$\frac{4n+3}{(n+1)4^n}=\frac 1{4^{n-1}\cdot n}-\frac 1{4^n\cdot (n+1)}$$
Now we give values $n=1 \rightarrow n$ and sum up and cancel similar terms :
$$\frac 1{4^0\cdot 1}-\frac 1{4\cdot 2}$$
$$\frac 1{4\cdot 2}-\frac 1{4^2\cdot 3}$$
$$\frac 1{4^2\cdot 3}-\frac 1{4^3\cdot 4}$$
$$\frac 1{4^3\cdot 4}-\frac 1{4^4\cdot 5}$$
$$\vdots$$
$$\frac 1{4^{n-2}\cdot(n-1) }-\frac 1{4^{n-1}\cdot n}$$
$$\frac 1{4^{n-1}\cdot n}-\frac 1{4^n\cdot (n+1)}$$
$$\therefore S_n=1-\frac 1{4^n(n+1)}$$
$$\Rightarrow \lim_{n\rightarrow \infty} S_n=1$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Determine if the following function is continuous on $ (0,0) $ Determine if the following function is continuous on $ (0,0) $
$$f(x,y)=\begin{cases}
(1+x^2y^2)^{-1/(x^2+y^2)},& \text{ if } (x,y)\neq (0,0) \\
1,& \text{ if } (x,y)=(0,0)
\end{cases}$$
Conjecturing that $ \lim _ {(x, y) \to (0,0)} f (x, y) = 1 $.
My attempt is,
$$\begin{align*}
0 &\leq | f (x) -L|\\
&=|(1+x^2y^2)^{-1/(x^2+y^2)}-1|\\ &\leq |(1+x^2y^2)^{-1/(x^2+y^2)}|+1\end{align*}$$
Hence I do not know what else to do because in another limit like this but with the euler function, you can limit that using that the function is strictly increasing, but here you cannot do that because $ (1 + x ^ 2y ^ 2) $ is not strictly increasing , so I don't know how to limit that exponent with the function $ (1 + x ^ 2y ^ 2) $.
| Alternatively, you can use Bernoulli's inequality to obtain the desired result. We have:
$\left(1+x^2y^2\right)^{\frac{1}{x^2+y^2}}\ge 1+\dfrac{x^2y^2}{x^2+y^2}=\dfrac{x^2+y^2+x^2y^2}{x^2+y^2}\implies \left(1+x^2y^2\right)^{-\frac{1}{x^2+y^2}}\le\dfrac{x^2+y^2}{x^2+y^2+x^2y^2}= h(x,y)$. Also: $x^2+y^2 \ge x^2 \implies \dfrac{1}{x^2+y^2} \le \dfrac{1}{x^2}\implies -\dfrac{1}{x^2+y^2} \ge -\dfrac{1}{x^2}\implies \left(1+x^2y^2\right)^{-\frac{1}{x^2+y^2}}\ge \left(1+x^2y^2\right)^{-\frac{1}{x^2}} = \left(\left(1+x^2y^2\right)^{-\frac{1}{x^2y^2}}\right)^{y^2}=g(x,y)$. So: $g(x,y) \le f(x,y) \le h(x,y)$. But both $h,g \to 1$ when $(x,y) \to (0,0)$ because $0 \le \dfrac{x^2y^2}{x^2+y^2} \le x^2$, and thus $\dfrac{x^2y^2}{x^2+y^2} \to 0$. Thus $f(x,y) \to 1 = f(0,0)$, proving continuity of $f$ at $(0,0)$. Note that $g(x,y) \to e^{-0} = 1$ when $(x,y) \to (0,0)$.
| {
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Limit of this sequence $ \lim_{n\to \infty}\sqrt[n]{\frac{2^n + 3^n + 4^n}{5^n + 6^n}}$? How to find the limit of this sequence. I tried using the squeeze theorem but no result yet.
$ \lim_{n\to \infty}\sqrt[n]{\frac{2^n + 3^n + 4^n}{5^n + 6^n}}$
| The limit is $\frac{4}{6} = \frac{2}{3}$. Prove that this is so.
HINT: Check that for each $\epsilon > 0$, there is an $n(\epsilon)$ such that for all $n>n(\epsilon)$, the denominator $2^n+3^n+4^n$ satisfies $$(1-\epsilon)4^n \le 2^n+3^n+4^n \le (1+\epsilon)4^n.$$ Likewise, there is an $n'(\epsilon)$ such that for all $n>n'(\epsilon)$, the numerator $5^n+6^n$ satisfies $$(1-\epsilon)6^n \le 5^n+6^n\le (1+\epsilon)6^n.$$
So for each $\epsilon > 0$ and for all $n > \max\{n(\epsilon), n'(\epsilon)\}$ the followign holds:
$$\sqrt[n]{\frac{1-\epsilon}{1+\epsilon}}\sqrt[n]{\frac{4^n}{6^n}} \le \sqrt[n]{\frac{5^n+6^n}{2^n+3^n+4^n}} \le \sqrt[n]{\frac{1+\epsilon}{1-\epsilon}}\sqrt[n]{\frac{4^n}{6^n}}$$
Can you finish from here.
| {
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For which values of $a$ does this system of equations $\mathbf{{not}}$ have a unique solution? Here's my system of linear equations:
$\begin{cases}
x + 2y + 2z = 1\\x + ay + 3z = 3\\x + 11y +az = 0\\
\end{cases}$
Thus I have the augmented matrix $\left[\begin{array}{ccc|c}1&2&2&1\\1&a&3&3\\1&11&a&0\end{array}\right]$
By row reduction, I obtain:
$\left[\begin{array}
{ccc|c}1&2&2&1\\0&a-2&1&2\\0&9&a-2&-1\end{array}\right]$
Unfortunately, I am stuck at this stage. I have tried swapping rows around but I didn't have much luck.
Update: I have managed to solve this with the use of the determinant.
Matrix of minors:
$\left[\begin{array}
{ccc}a^2-33&a-3&11-a\\2a-22&a-2&9\\6-2a&1&a-2\end{array}\right]$
Matrix of cofactors:
$\left[\begin{array}
{ccc}a^2-33&3-a&11-a\\22-a&9&a-2\\6-2a&-1&a-2\end{array}\right]$
Adjugate matrix:
$\left[\begin{array}
{ccc}a^2-33&22-2a&6-2a\\3-a&a-2&-1\\11-a&-9&a-2\end{array}\right]$
Det(A) = $1(a^2 - 33) + 2(3 - a) + 2(11 - a) = a^2 - 4a - 5$
$(a - 5)(a + 1) = 0$
Thank you all for your help!
| Let's rearrange the augmented matrix representing the system of linear equations, by swapping $R_2$ and $R_3$ (to avoid division by $a-2$ while row-reduction), to get
$\left[\begin{array}{ccc|c}1&2&2&1\\1&11&a&0\\1&a&3&3\end{array}\right]$ and a subsequent row-reduction will give you
$\left[\begin{array}{ccc|c}1&2&2&1\\0&9&a-2&-1\\0&0&1-\frac{(a-2)^2}{9}&2+\frac{a-2}{9}\end{array}\right]$
Unique solution:
in order for the system to have a unique solution, the coefficient and the augmented matrix both must be of full rank i.e., both must have rank $n=3$. To make that happen, we must have $1\neq\frac{(a-2)^2}{9} \implies a \notin \{-1,5\}$.
No solution:
for this to happen, we need to have the augment matrix rank higher than the coefficient matrix rank. This can happen if the coefficient matrix has rank < 3, e.g., if $1=\frac{(a-2)^2}{9} \implies a \in \{-1,5\}$ and augmented matrix has rank $n=3$ simultaneously, which means $2+\frac{a-2}{9} \neq 0 \implies a \neq -16$, i.e., for $a=-1,5$ the system will have no solution.
Infinitely many solutions:
it can happen iff both the coefficient and the augmented matrix are rank-deficient, i.e. for both of them the rank is $<3$. For this the last row of the augmented matrix needs to be zero. Which implies both $1-\frac{(a-2)^2}{9}=0 \implies a \in \{-1,5\}$ and $2+\frac{a-2}{9}=0 \implies a=-16$ simultaneously. Which is clearly impossible.
| {
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Surd Manipulations. If a=$-\sqrt {99}+\sqrt {999}+\sqrt {9999}$
b=$-\sqrt {99}-\sqrt {999}+\sqrt {9999}$
c=$-\sqrt {99}+\sqrt {999}-\sqrt {9999}$
Then $\displaystyle\sum_\limits{cyc} \frac{a^4}{(a-b)(a-c)}$ equals??
I first individually calculated (a-b), (b-c), (c-a) and tried to simplify the sum. However, I couldn't do it. Please help!
| Hint:
$$a^4(b-c)+b^4(c-a)+c^4(a-b)$$
$$=ab(a^3-b^3)-c(a^4-b^4)+c^4(a-b)$$
$$=(a-b)\{ab(a^2+ab+b^2)-c(a+b)(a^2+b^2)+c^4\}$$
$$=\cdots$$
$$=-(a-b)(b-c)(c-a)(a^2+b^2+c^2+ab+bc+ca)$$
Alternatively, if $f(a,b,c)=a^4(b-c)+b^4(c-a)+c^4(a-b)$
Clearly, $f(a,a,c)=0\implies a-b$ is a factor
Similarly, $(b-c)(c-a)$ are also factors
As $f(a,b,c)$ is symmetric about $a,b,c$
$$f(a,b,c)=(a-b)(b-c)(c-a)(p(a^2+b^2+c^2)+q(ab+bc+ca))$$
Compare the coefficients of $a^4$ to find $p=-1$
Can you find $q=-1?$
Now $a^2+b^2+c^2+ab+bc+ca=(a+b+c)^2-3(ab+bc+ca)$
| {
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Find the sum $\sum_{n=1}^\infty\frac{\sqrt{n}}{2^{n/2}}$ Find the sum $\sum_{n=1}^\infty\frac{\sqrt{n}}{2^{n/2}}$. Is it possible to find a closed form for this sum? I was trying to upper bound this sum by another series, but could not find one.
Can anyone help?
| The closed form using polylogarithms is given elsewhere. An alterantive for (upper and lower) bounds...
I note that for $n > 4$, $\displaystyle \frac{2}{2^{n/2}} < \frac{\sqrt{n}}{2^{n/2}} < \frac{1}{2^{n/4}}$.
The left inequality is easy: For $4 < n$, $2 < \sqrt{n}$, then divide through by $2^{n/2}$. Using the first three terms of the original series and then the lower bounding term subsequently, we get the lower bound $(1/4)(4 + 6\sqrt{2} + \sqrt{6} = 3.733{\dots}$.
For the right inequality, start with $n = 4$, observing $\displaystyle \sqrt{4} = 2 = 2^{4/4}$. Then $\frac{\mathrm{d}}{\mathrm{d}n}\sqrt{n} < 1/4$ for $n > 4$ and $\frac{\mathrm{d}}{\mathrm{d}n} 2^{n/4} > 1/4$ when $2^{n/4} \ln 2 > 1$, which is certainly the case for $n > 4$. Therfore,
$$ \frac{\sqrt{n}}{2^{n/4}} < 1 \qquad (n > 4) \text{.} $$
Now divide through by $2^{n/4}$. Splicing the three terms from the original series into the series of these upper bounding terms gives the upper bound $\sqrt{3/8} + \sqrt{2} + \frac{1}{2-2^{3/4}} = 5.169{\dots}$.
| {
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Can you solve $\int 1/(x^2+1)^2\, dx$ I know that the integral looks like like the anti-derivative of $\arctan$, but i don't know how to use this fact so I tried to use a fraction decomposition to: $$\frac{1}{(x^2+1)^2}$$
to solve the anti-derivative : $$\int \frac{1}{(x^2+1)^2}\, dx$$
so :$$\frac{1}{(x^2+1)^2}=\frac{A}{x^2+1}+\frac{B}{x^2+1}\iff \frac{1}{x^2+1}=A+B$$
$$1+0x^2=(A+B)x^2+(A+B)1$$
Thus I got this system of equations:
$$\cases{A+B=1 \\ A+B=0}$$ Which is impossible. I also tried to use substitution but it didn't help me.
| $I=\int \frac{dx}{(x^2+1)^2}=\frac{1}{2} \int \frac{1+x^2+1-x^2}{x^4+2x^2+1}.dx$
$\implies I= \frac{1}{2} \left ( \int \frac{1+x^2}{x^4+2x^2+1}.dx +\int \frac{1-x^2}{x^4+2x^2+1}.dx\right).$
$I_1=\int \frac{1+x^2}{x^4+2x^2+1}.dx=\int \frac{1+\frac{1}{x^2}}{x^2+2+\frac{1}{x^2}}.dx=\int \frac{1+\frac{1}{x^2}}{(x-\frac{1}{x})^2+4}.dx$
Now $t=x-\frac 1 x \implies dt=\left(1+\frac{1}{x^2} \right)dx$
$I_1=\int\frac{dt}{t^2+4}=\frac{1}{2}\tan^{-1}\left(\frac{t}{2}\right)$.
$I_2=\int \frac{1-x^2}{x^4+2x^2+1}.dx=-\int \frac{1-\frac{1}{x^2}}{x^2+2+\frac{1}{x^2}}.dx=-\int \frac{1-\frac{1}{x^2}}{(x+\frac{1}{x})^2+1}.dx$
Now $p=x+\frac 1 x \implies dp=\left(1-\frac{1}{x^2} \right)dx$
$I_2=\int\frac{dp}{p^2+1}=\tan^{-1}(p)$
$I=I_1+I_2=\frac{1}{2}\tan^{-1}\left(\frac{x-\frac 1 x}{2}\right)+\tan^{-1}(x+\frac 1 x).$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4065559",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 4
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Integration with complex constants So I saw a youtube video of the following integral, which can be solved by letting u=tanx, then you get
$\int \frac{1}{1+x^2} dx $ = ${tan}^{-1}x + C$
But it should also be solvable through partial fraction decomposition.
$\int \frac{1}{1+x^2}dx$ = $\int {\frac{A}{x-i} + \frac{B}{x+i}dx}$
$\frac{1}{1+x^2}$ = $\frac{A}{x-i}$ + $\frac{B}{x+i}$
$1$ = $A(x+i)$ + $B(x-i)$
$x=i$: $1$ = $2iA$
$A$=$\frac 1{2i}$ = -$\frac {i}{2}$
$x=-i$: $1 = -2iB$
$B$ = $\frac{i}2$
$\int \frac{1}{1+x^2}dx$ = $\int -\frac{i}{2(x-i)}$ + $ \frac{i}{2(x+i)}dx$ = $-\frac{i}2$ln$\lvert {x-i}\rvert$ + $\frac{i}2$$ln\lvert {x+i}\rvert$ + C = $\frac{i}2ln\lvert {x+i}\rvert - ln\lvert {x-i}\rvert$ + C
This should be equal to ${tan}^{-1}x + C$ right?
${tan}^{-1}({1})$ = $\frac\pi4$
$\frac{i}2(ln\lvert {1+i}\rvert - ln\lvert {1-i}\rvert)$ = $\frac{i}2(ln{\sqrt2}-ln{\sqrt2})$=O
So these are obviously not equal, then I thought why must i have absolute value sign in the logarithmic expressions, that rule I know only applies for integarting with real constants.
Thus,
$\int \frac{1}{1+x^2} dx = $$\frac{i}2(ln({x+i}) - ln({x-i}))$ + C = $\frac{i}2ln(\frac{({x+i})^2}{x^2+1}) + C$
Evaluated at x=1 gives us
$\frac{i}2ln(\frac{({1+i})^2}{1^2+1})$ = $\frac{i}2ln(i)$ = $\frac{i}2\cdot \frac{i\pi}{2}$ = $-\frac\pi4$
So, im quite close, but I must have missed something somewhere, and I cant find the mistake. But this triggers two other questions: How would I show that
$\frac{i}2ln(\frac{({x+i})^2}{x^2+1})$ = $\tan^{-1}x$ (or some other logarithmic expression if that expression is where the error lies)
and, why dont I need to take absolute value when integrating
$\int \frac{1}{x+a}dx$
with complex constants? Is the explanation as easy that I can evaluate all logarithmic expressions using complex numbers?
| $$e^{xi} = \cos{x}+i\sin{x}$$
$$e^{-xi} = \cos{x}-i\sin{x}$$
$$\sin{x} = \frac{ e^{xi}-e^{-xi} }{2i}$$
$$ \cos{x} = \frac{ e^{xi}+e^{-xi} }{2}$$
$$\tan{x} = \frac{\sin{x}}{\cos{x}} =\frac{1}{i} \frac{ e^{xi}-e^{-xi} }{e^{xi}+e^{-xi} }$$
$$\tan{x} = \frac{1}{i} \frac{ e^{xi}-e^{-xi} }{e^{xi}+e^{-xi} }$$
Say $ \frac{1}{i} \frac{ e^{xi}-e^{-xi} }{e^{xi}+e^{-xi} } = y$
$$ \tan{x} = y , x = \arctan{y}$$
$$ \frac{1}{i} \frac{ e^{xi}-e^{-xi} }{e^{xi}+e^{-xi} } = y$$
Say $x = i \log{z}$
$$ \frac{1}{i} \frac{ z^{-1}-z }{z^{-1}+z} = y$$
$$ z^{-1}-z = i y (z^{-1}+z) $$
$$1-z^2 = i y (1+z^2)$$
$$i y z^2+z^2 = 1-i y$$
$$ z = \sqrt{ \frac{ 1-i y}{1+i y} }$$
$$\arctan{y} = i \log{ \sqrt{ \frac{ 1-i y}{1+i y} }}$$
$$\arctan{y} = \frac{i}{2} \log{ \frac{ 1-i y}{1+i y }}$$
$$\arctan{y} = \frac{i}{2} \log{ \frac{ (1-i y)^2}{(1+i y )(1-i y)}}$$
$$\arctan{y} = \frac{i}{2} \log{ \frac{ 1-2y i -y^2}{1+y^2 }}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4068755",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Find maximize value of this expression Let $a;b;c \in [0;2]$
Find maximum value of: $P=ab(a^2-b^2)+bc(b^2-c^2)+ca(c^2-a^2)$
After some factoring, I found:
$P=(a-b)(b-c)(a-c)(a+b+c)$
Then square it: $P^2=(a-b)^2(b-c)^2(c-a)^2(a+b+c)^2$
Now $P^2$ be a symmetric expression. But it' still hard. Can someone help, please? Thank you.
| Let $P=f(a,b,c)$ and note that for $k\le \min\{2-a,2-b,2-c\}$
$$f(a+k,b+k,c+k)\ge f(a,b,c)$$
Thus, we can choose $a=2$ (WLOG). Now, we need to maximize
\begin{align}f(2,b,c)&=2b(4-b^2)+bc(b^2-c^2)+2c(c^2-4)
\\
&=2b(4-b^2)+c(b^3-8)+c^3(2-b)
\\
&=2b(4-b^2)+c(b-2)(b^2+2b+4-c^2)
\end{align}
Observe that $$b-2\le 0$$
$$b^2+2b+(4-c^2)\ge 0$$
Thus,
$$f(2,b,0)\ge f(2,b,c)$$
Now, we need to maximize
$$f(2,b,0)=2b(4-b^2)$$
Candidates are boundary values and values that make the first order equal to $0$. Checking all values gives us
$$\frac{32}{3\sqrt{3}}=f(2,\frac{2}{\sqrt{3}},0)\ge f(a,b,c)$$
Remember $f$ is symmetric.
Done!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4071059",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
} |
Expression of $\cos^{-1}\left(4x^3-3x\right)$ The Original question : (translated from French)
Let $f$ be the function
$$
f\left(x\right)=\cos^{-1}\left(4x^3-3x\right)
$$
*
*
*
*Find the definition domain of $f$.
*
*
*Compute $f'\left(x\right)$. Find an expression of $f\left(x\right)$ in terms of $\cos^{-1}\left(x\right)$.
My attempt :
$f$ is defined on $\left[-1;1\right]$ and
$$
f'\left(x\right)=\frac{3-12x^2}{\sqrt{1-\left(4x^3-3x\right)^2}}
$$
To answer the question I thought i could express it as
$$
f'\left(x\right)=\frac{3-12x^2}{\sqrt{-1+3x-4x^3}\sqrt{1-3x+4x^3}}
$$
and try a kind of decomposition to find it as
$$
f'\left(x\right)=a\frac{3-12x^2}{\sqrt{-1+3x-4x^3}}+b\frac{3-12x^2}{\sqrt{1-3x+4x^3}}
$$
What am I missing ?
| The problem with your approach is that $f$ is not differentiable at all points of $[-1,1]$.
Since domain is $[-1,1]$, let $x=\cos \theta$, where $\theta \in [0,\pi]$ so that $\cos $ function becomes a bijection and hence $\cos^{-1}x=\theta$. Note that $0\le 3\theta \le 3\pi$.
Now using trig. identities, it follows that:
$f(x)=\cos^{-1} (\cos 3\theta)$
Case 1: $0\le 3\theta \le \pi\implies 0\le \theta\le \pi/3$
$f(x)=3\theta =3 \cos^{-1}x$ for $\frac 12\le x \le 1$
Case 2: $\pi\lt 3\theta \le 2\pi \implies \pi/3\lt \theta \le 2\pi/3$
$\cos 3\theta=\cos (2\pi-(2\pi-3\theta))=\cos(-3\theta+2\pi)\implies f(x)=\cos^{-1}\cos 3\theta=2\pi-3\theta=2\pi-3\cos^{-1}x$ for $-\frac 12\le x\lt \frac 12$.
Case 3: $2\pi\lt 3\theta \le 3\pi\implies 2\pi/3\lt \theta \le\pi$
$f(x)=\cos^{-1}\cos 3\theta=\cos^{-1}\cos (3\theta-2\pi)=3\theta -2\pi=3\cos^{-1}x-2\pi$ for $-1\le x\lt -1/2$.
Can you find the derivative now?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4071853",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Is there any other approach available without trig substitution?
Let $f\colon [0, 1] \to \mathbb{R}$ be a continuous function such that for any $x, y \in [0, 1]$, $xf(y) + yf(x) \le 1$. Find
maximum value of $\int_0^1f(x)\,dx$.
Method given was using $x= \sin\theta$, we get $$\int_0^1 f(x)\, dx = \int_0^{\pi/2} f(\sin\theta)\cos\theta\, d\theta \tag{I}$$and similarly by $x= \cos\theta$, we would get $$\int_0^1 f(x)\,dx = \int_0^{\pi/2} f(\cos\theta)\sin\theta \,d\theta \tag{II},$$ adding $(\text{I})$ and $(\text{II})$ we get max value of the required integral to be $\pi/4$. Any other approach because the idea is very tricky and one may not realize for this type of substitution.
| I don't have an alternative approach to suggest. The first thing that occurred to me, which was to integrate over the two-dimensional region $[0,1]^2$, doesn't work:
\begin{align}
1=\int_0^1\int_0^1 1\,dx\,dy\ge\int_0^1\int_0^1(yf(x)+xf(y))\,dx\,dy&=2\int_0^1y\,dy\int_0^1 f(x)\,dx\\
&=\left[y^2\right]_0^1\int_0^1f(x)\,dx\\
&=\int_0^1f(x)\,dx.
\end{align}
My upper bound is not as tight as yours: $1>\pi/4\approx0.785$.
What you have done is to cleverly bound the integral by restricting the integration to a single contour, namely the arc of the unit circle that lies in the first quadrant. Your bound is, in fact, the best possible, as the example $f(x)=\sqrt{1-x^2}$ shows. First confirm that your maximum is attained:
\begin{align}
\int_0^1\sqrt{1-x^2}\,dx&=\int_0^{\pi/2}\sqrt{1-\sin^2\theta}\cos\theta\,d\theta\\
&=\int_0^{\pi/2}\left(\frac{1}{2}\cos2\theta+\frac{1}{2}\right)\,d\theta\\
&=\frac{\pi}{4}.
\end{align}
To verify that $f$ satisfies the required condition, observe that the function
$$
g(x,y)=y\sqrt{1-x^2}+x\sqrt{1-y^2}
$$
achieves a maximum of $1$ for $(x,y)\in[0,1]^2$ all along your contour. You can check this by observing that the equations
\begin{align}
0&=\frac{\partial}{\partial x}\left[
y\sqrt{1-x^2}+x\sqrt{1-y^2}\right]=\sqrt{1-y^2}-\frac{xy}{\sqrt{1-x^2}}\\
0&=\frac{\partial}{\partial y}\left[
y\sqrt{1-x^2}+x\sqrt{1-y^2}\right]=\sqrt{1-x^2}-\frac{xy}{\sqrt{1-y^2}}
\end{align}
are satisfied when $x^2+y^2=1$. To see that it is a maximum that is achieved at this locus of critical points, switch to polar coordinates,
$$
h(r,\theta)=g(r\cos\theta,r\sin\theta),
$$
and compute
\begin{align}
\frac{\partial}{\partial r}h(r,\theta)=&\left(\sin\theta\sqrt{1-r^2\cos^2\theta}+\cos\theta\sqrt{1-r^2\sin^2\theta}\right)\\
&\times\left(1-\frac{r^2\sin\theta\cos\theta}{\sqrt{1-r^2\cos^2\theta}\sqrt{1-r^2\sin^2\theta}}\right),\\
\frac{\partial}{\partial\theta}h(r,\theta)=&r\cos\theta\sqrt{1-r^2\cos^2\theta}+\frac{r^3\sin^2\theta\cos\theta}{\sqrt{1-r^2\cos^2\theta}}\\
&-r\sin\theta\sqrt{1-r^2\sin^2\theta}-\frac{r^3\cos^2\theta\sin\theta}{\sqrt{1-r^2\sin^2\theta}}.
\end{align}
One can check that, for fixed $\theta$, $\frac{\partial h(r,\theta)}{\partial r}$ is positive when $r<1$, zero when $r=1$, and negative when $r>1$, while, for $r$ fixed to $1$, $\frac{\partial h(r,\theta)}{\partial\theta}$ is zero, as expected.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4072354",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
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} |
Taylor series for $\sqrt[3]{x+1}$ I'm trying to find the Taylor series for $\sqrt[3]{x}$, but since the n-th derivative of the function $$f(x)=\sqrt[3]{x}$$
not definded at $x=0$, I've swithched to $f(x)=\sqrt[3]{x+1}$, This is my Attempt:
$$f(x)=\sum_{n=0}^\infty\frac{f^{(n)}(0)x^n}{n!}$$
$$\sqrt[3]{x+1}=1+\frac{x}{3}-\frac{2x^2}{9\cdot2!}+\frac{10x^3}{27\cdot 3!}-\frac{80x^4}{81\cdot4!}+...$$
but when i plug $x=125 \iff \sqrt[3]{x+1}=\sqrt[3]{126}$ i get something way bigger than the acttual value which is about $5.01...$, you can check it here.
But I can't find the mistake in the equation.
| For $x\gt 1$, expand in terms of $\frac1x<1$ to ensure convergence
$$\sqrt[3]{x+1}= \sqrt[3]{x} \sqrt[3]{1+\frac1x}
= \sqrt[3]{x} (1+\frac{1}{3x}-\frac{2}{9\cdot2!}\frac1{x^2}+\frac{10}{27\cdot 3!}\frac1{x^3}-\frac{80}{81\cdot4!}\frac1{x^4}+...)
$$
Then, set $ x=125$ to obtain the approximation
$$\sqrt[3]{126} =5(1+\frac1{3\cdot 125}-\cdots )=5.0133$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4073453",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 1
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Problem from Zorich's book Choose real numbers $a$ and $b$ so that the function $f(x)=\cos x - \dfrac{1+ax^2}{1+bx^2}$ is an infinitesimal of highest possible order as $x\to 0$.
My solution: Actually this problem is not so difficult and I guess that I've almost solved it. We'll split the solution into some cases:
*
*If $b=0$ and using Taylor's series one can write $f(x)=x^2(-1/2-a)+o(x^3)$ as $x\to 0$.
So we see that if $a=-1/2$ then $f(x)=o(x^3)$. If $a\neq -1/2$ then $f(x)=o(x)$ as $x\to 0$.
*Suppose that $b\neq 0$ then again using Taylor's series one can derive that $$f(x)=\cos x-(1+ax^2)(1+bx^2)^{-1}=x^2\left(-\frac{1}{2}-a+b\right)+x^4\left(\frac{1}{24}-b^2+ab\right)+x^6\left(-\frac{1}{720}-ab^2+b^3\right)+o(x^7).$$
One can show that if $a=-\dfrac{5}{12}$ and $b=\dfrac{1}{12}$ then $f(x)=o(x^5)$ as $x\to 0$.
Definition: If $f=o(g)$ and g is itself infinitesimal as $x\to 0$, we say that $f$ is an infinitesimal of higher order than $g$ as $x\to 0$.
Question: Can anyone explain in details which one has the highest possible order? I guess it should be in the case when $a=-\dfrac{5}{12}$ and $b=\dfrac{1}{12}$ but cannot explain it in a rigorous way. If someone can explain it in a great detail that would be awesome!
| Regardless of whether $b = 0$ or not, we can always write $$f(x) = \cos(x) - (1 + ax^2)(1 - bx^2 + b^2 x^4 - b^3 x^6 + \dots)$$ which leads to $$f(x) = (-a + b - 1/2)x^2 + (ab - b^2 + 1/24)x^4 + (-ab^2 + b^3 - 1/720)x^6 + \dots$$
Therefore for $a = -5/12$ and $b = 1/12$, we have $f(x) = 1/480 x^6 + \dots$ which has order of vanishing $6$.
On the other hand, if $f$ has order $\geq 6$, then we must have $−a+b−1/2 = 0$ and $ab−b^2+1/24 = 0$. It is easy to solve from these two equations that $a = -5/12$ and $b = 1/12$ is the only possibility.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4073904",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Rewriting function Given function
$$
f(x) = \frac{-2 + 2 \sqrt{x+1}}{x}
$$
and
$$
g(x) = \frac{2}{1+\sqrt{x+1}}
$$
Prove that $f(x) = g(x)$ where $x\neq 0$.
This question is from a 2018 examen in the Netherlands. I tried rewriting f(x) as follows:
$$
f(x) = \frac{-2 + 2 \sqrt{x+1} * \sqrt{x+1}}{x*\sqrt{x+1}}
$$
$$
f(x) = \frac{-2 + 2(x+1)}{x*\sqrt{x+1}}
$$
$$
f(x) = \frac{-2 + 2x + 2}{x*\sqrt{x+1}}
$$
$$
f(x) = \frac{2x}{x*\sqrt{x+1}}
$$
$$
f(x) = \frac{2}{\sqrt{x+1}}
$$
However, I can't find a way to get to the $h(x)$ function where the denominator has a 1 + still.
EDIT:
Just as I post this I notice that I made a mistake in the first step. By not multiplying against the whole numerator. I'll leave the question open anyways.
| Hint: consider the following identity
$$
(\sqrt{x+1} + 1)(\sqrt{x+1} - 1) = (\sqrt{x+1})^2 - 1^2 = x + 1 - 1 = x
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4075107",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Let $a, b, c$ be the three roots of $x^3-(k+1)x^2+kx+12=0$, where $k$ is a real number. If $(a-2)^3+(b-2)^3+(c-2)^3=-18$, find the value of $k$
Let $a, b, c$ be the three roots of the equation $x^3-(k+1)x^2+kx+12=0$, where $k$ is a real number. If $(a-2)^3+(b-2)^3+(c-2)^3=-18$, find the value of $k$.
Here is a problem from a Hong Kong competition that I found a way to solve, but I am interested in knowing other methods since my brute-force way doesn't seem to be the solution intended.
My attempt
Let $y=x-2$, the equation becomes
$$(2+y)^3-(k+1)(2+y)^2+k(2+6)+12=0$$
$$y^3+(5-k)y^2+(8-3k)y-2k+16=0$$
Let $y_1, y_2, y_3$ be the new roots of the equation. By Vieta's formula, we have
\begin{cases}
y_1+y_2+y_3=-5+k \\
y_1y_2+y_2y_3+y_1y_3=8-3k\\
y_1y_2y_3=2k-16
\end{cases}
And by $y_1^3+y_2^3+y_3^3 = (y_1+y_2+y_3)^3-3(y_1+y_2+y_3)(y_1y_2+y_2y_3+y_1y_3)+3y_1y_2y_3$, we have
\begin{align*}
y_1^3+y_2^3+y_3^3 &= (y_1+y_2+y_3)^3-3(y_1+y_2+y_3)(y_1y_2+y_2y_3+y_1y_3) +3y_1y_2y_3\\
&= (-5+k)^3-3(-5+k)(8-3k)+3(2k-16)\\
&=k^3-6k^2+12k-53
\end{align*}
By $y_1^3+y_2^3+y_3^3=-18$ given in the problem, we have
$$k^3-6k^2+12k-35=0$$
And $k=5$ is the only real solution.
| using the identity $$a^3+b^3+c^3=3abc+\frac{(a+b+c)}{2} ({(a-b)}^2+{(b-c)}^2+{(c-a)}^2)$$ $$-18=3(a-2)(b-2)(c-2)+\frac{a+b+c-6}{2}({(a-b)}^2+{(b-c)}^2+{(c-a)}^2)$$ Now $$x^3-(k+1)x^2+kx+12=(x-a)(x-b)(x-c)$$ $$\to (a-2)(b-2)(c-2)=2k-16$$ now use $${(a-b)}^2+{(b-c)}^2+{(c-a)}^2={(a+b+c)}^2-3(ab+bc+ca)={(k+1)}^2-3k$$ combining these we get the cubic $$k^3-6k^2+12k-35=0$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Find sum of all possible value of $(a+b)^2$
Let $P(x)$ be a quadratic with Complex coefficient whose coefficient of $x^2$ is $1$. Suppose the equation $P(P(x))=0$ has four distinct solution, $x=3,4,a,b$. Find sum of all possible value of $(a+b)^2$.
Solution: Either $P(3) = P(4)$ or not. We first see that if $P(3) = P(4)$ it's easy to obtain by Vieta's that $(a+b)^2 = 49$. Now, take $P(3) \neq P(4)$ and WLOG $P(3) = P(a), P(4) = P(b)$. Now, consider the parabola formed by the graph of $P$. It has vertex $\frac{3+a}{2}$. Now, say that $P(x) = x^2 - (3+a)x + c$. We note $P(3)P(4) = c = P(3)\left(4 - 4a + \frac{8a - 1}{2}\right) \implies a = \frac{7P(3) + 1}{8}$. Now, we note $P(4) = \frac{7}{2}$ by plugging in again. Now, it's easy to find that $a = -2.5, b = -3.5$, yielding a value of $36$. Finally, we add $49 + 36 = \boxed{085}$. ~awang11, charmander3333
Remark: We know that $c=\frac{8a-1}{2}$ from $P(3)+P(4)=3+a$.
My doubt how can we say that $P(3)(4)=c$ and $P(3)+P(4)=3+a$?.
| Since $3,4,a,b$ are roots of $P(P(x))$ and $P(3)$ and $P(4)$ are different.
$P(3)$ and $P(4)$ are distinct roots of $P(x)$.
As a result, from Vieta's formula, we know that we have $c=P(3)P(4)$ and $P(3)+P(4)=3+a$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4080046",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Eigenvalues of correlation matrix $\mathbf{\rho}$ with constants I'm tasked with finding the eigenvalues of $M = \begin{bmatrix} 1 & \rho & \rho \\ \rho & 1 & \rho \\ \rho & \rho & 1 \end{bmatrix}$. My difficulty here is in finding roots of the characteristic polynomial, since it's a cubic.
Attempt. Let's call the characteristic polynomail by $chp$, and let's use $x$ instead of $\lambda$ to have a more familiar-looking expression. We have
$$
\operatorname{chp}(x) = \det\left(\mathbf{\rho} - xI\right) = \det\begin{bmatrix} 1-x & \rho & \rho \\ \rho & 1-x & \rho \\ \rho & \rho & 1-x \end{bmatrix}
$$
Expanding on the first line, we obtain
$$
chp(x) = (1-x) \det \begin{bmatrix} 1-x & \rho \\ \rho & 1-x \end{bmatrix} - \rho \det \begin{bmatrix} \rho & \rho \\ \rho & 1-x \end{bmatrix} + \rho \det \begin{bmatrix} \rho & 1-x \\ \rho & \rho \end{bmatrix}.
$$
After simplification, you'd get
$$
chp(x) = 1 - 3\rho^2 + 2\rho^3 - 3x + 3\rho^2 x + 3x^2 - x^3.
$$
I couldn't solve $chp(x)=0$ analitically. Using Wolfram though, and coming back to the usual $\lambda$ notation, I got
$$
\lambda_1 = \lambda_2 = 1-\rho \qquad \text{and} \qquad \lambda_3 = 1+2\rho.
$$
Thanks for any help.
| There's another way that is in keeping with your method, but it'll be a lot easier to find the eigenvalues of this matrix in particular.
I'm taking it up from here:
$$\operatorname{chp}(x) = \det\left(\mathbf{\rho} - xI\right) = \det\begin{bmatrix} 1-x & \rho & \rho \\ \rho & 1-x & \rho \\ \rho & \rho & 1-x \end{bmatrix}$$
Now, to find the determinant of the above matrix, you can make use of some elementary row and column transformations before expanding the matrix along a row(or a column) to make the calculations a lot more simpler.
$$\begin{align}
\det\begin{bmatrix}
1-x & \rho & \rho\\
\rho & 1-x & \rho\\
\rho & \rho & 1-x
\end{bmatrix}&=
\det\begin{bmatrix}
1+2\rho-x & \rho & \rho\\
1+2\rho-x & 1-x & \rho\\
1+2\rho-x & \rho & 1-x
\end{bmatrix}&C_1\rightarrow C_1+C_2+C_3\\
&=
\left(1+2\rho-x\right)\det\begin{bmatrix}
1 & \rho & \rho\\
1 & 1-x & \rho\\
1 & \rho & 1-x
\end{bmatrix}&\text{Taking $1+2\rho-x$ out from $C_1$}\\
&=
\left(1+2\rho-x\right)\det\begin{bmatrix}
1 & \rho & \rho\\
0 & 1-x-\rho & 0\\
0 & 0 & 1-x-\rho
\end{bmatrix}& \begin{array}{1}
R_2\rightarrow R_2-R_1\\
R_3\rightarrow R_3-R_1
\end{array}
\end{align}$$
Now expanding by $C_1$, we get,
$$\det\begin{bmatrix}
1-x & \rho & \rho\\
\rho & 1-x & \rho\\
\rho & \rho & 1-x
\end{bmatrix}=\left(1+2\rho-x\right)\left(1-x-\rho\right)^2$$
I hope you can take it up from here and find the eigenvalues easily now.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4083700",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Limit $\underset{x\to 0}{\text{lim}}\frac{\sqrt[3]{a x+b} - 2}{x}$ equals to $\frac{5}{12}$ A friend of mine asked this question to me. It seems it's from Stewart.
Find the values of a and b such that
$\underset{x\to 0}{\text{lim}}\frac{\sqrt[3]{a x+b} - 2}{x} = \frac{5}{12}$
This is what I tried with better results.
For $b$:
$$\frac{\sqrt[3]{a x+b} - 2}{x} = \frac{5}{12} $$
$$\sqrt[3]{a x+b} - 2 = x\frac{5}{12} $$
$$\underset{x\to 0}{\text{lim}} \sqrt[3]{a x+b} - 2 = \underset{x\to 0}{\text{lim}} x\frac{5}{12} $$
$$\sqrt[3]{b} - 2 = 0 $$
$$\sqrt[3]{b} = 2 $$
$$ b = 8$$
For $a$:
$$a x+8 = (x\frac{5}{12} + 2)^3$$
$$a x+8 = \frac{125 x^3}{1728}+\frac{25 x^2}{24}+5 x+8$$
$$a x = \frac{125 x^3}{1728}+\frac{25 x^2}{24}+5 x$$
$$a = \frac{125 x^2}{1728}+\frac{25 x}{24}+5$$
$$\underset{x\to 0}{\text{lim}} a =\underset{x\to 0}{\text{lim}} \frac{125 x^2}{1728}+\frac{25 x}{24}+5$$
$$a = 5 $$
But the limit $$\underset{x\to 0}{\text{lim}}\frac{\sqrt[3]{5x+8} - 2}{x}$$ doens't go to $\frac {5}{12}$.
May someone help.
| As the denominator tends to $0$ the limit can't be finite unless the numerator tends to $0$ are well. So we must have $\sqrt[3]{a\cdot 0 + b} -2 = 0$ so evaluating the function at $x=0$ will yield indefinite form of $\frac 00$.
So $b=2^3 = 8$.
As the limit is indefinite form we can use L'hopital to calculate
$\lim_{x\to 0} \frac {(ax+8)^{\frac 13}}{x} = \lim_{x\to 0} \frac {\frac 13(ax+8)^{-\frac 23}a}1= \frac {5}{12}$
So as that isn't in indefinite form. we can achieve the limit by evalutation.
So $\frac 13(a\cdot 0+8)^{-\frac 23}a=\frac 5{12}$ and
$\frac 13(a\cdot 0+8)^{-\frac 23}a= \frac a{3\sqrt[3]8^2}= \frac a{12}$
So $\frac a{12} = \frac 5{12}$ and $a=5$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4086122",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 0
} |
Determine continuous function from piecewise derivative I'm having trouble solving this issue.
Determine a continuous function $f$ on the interval $[-2,2]$ whose derived function on $[-2,2] \setminus \{0\}$ is known to be the function:
$$
f(x)=
\left\lbrace
\begin{array}{lll}
\dfrac{x^2+4x+7}{2x^3-x^2+18x-9} & \text{ if} -2 \leq x < 0 \\
& \\
x^2 \sin^2 (x) & \text{ if } 0 < x \leq 2
\end{array}
\right.
$$
I tried to calculate the integral of the function in each definition interval, add a constant and impose conditions to determine the constant. I have the problem in the first interval. I can't integrate the function. How do I solve this problem? Thanks!
| You can use partial fractions. First, factorize the denominator as $$2x^3-x^2+18x-9 = (2x-1)(x^2+9) = (2x-1)(x+3i)(x-3i) $$ Now, set $$\frac{x^2+4x+7}{(2x-1)(x-3i)(x+3i)} = \frac{P}{2x-1} + \frac{Q}{x+3i} +\frac{R}{x-3i} \\ \implies x^2+4x+7 = P(x^2+9) +Q(2x-1)(x-3i) +R(2x-1)(x+3i)$$
Letting $x=\frac 12, 3i, -3i$ will give you $P=1, Q=\frac i3, R= -\frac i3 $. Hence, the integral is $$\int \frac{dx}{2x-1} +\frac i3 \int \frac{dx}{x+3i} -\frac i3 \int \frac{dx}{x-3i} \\ =\ln|2x-1| +\frac i3 \ln(x+3i) -\frac i3 \ln(x-3i) + C \\ =\ln(1-2x) +\frac i3 \left[ \ln \sqrt{x^2+9} +i(\pi +\tan^{-1} \frac 3x) \right] -\frac i3 \left[ \ln\sqrt{x^2+9} +i(-\pi -\tan^{-1}\frac 3x ) \right] + C \hspace{0.1 cm} (*) \\ = \ln(1-2x)-\frac 23 (\pi+\tan^{-1} \frac 3x ) + C \\ = \ln(1-2x)-\frac 23 \tan^{-1} \frac 3x +C’$$
$ \begin{align} (*) \ln(z) = \ln |z| +i\arg z \end{align}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4090039",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Finding Taylor Series from existing Series If the Taylor Series of $\ln(x)$ is known:
$$\ln(x) = (x-1) -\frac{1}{2}(x-1)^2+\frac{1}{3}(x-1)^3-\frac{1}{4}(x-1)^4+\frac{1}{5}(x-1)^5-...$$
Can one find the Taylor series of
$$f(x)= \frac{x}{1-x^2}$$
by manipulating the Taylor series of ln(x)?
| Notice that:
$$\int f(x)dx = \int \dfrac{x}{1-x^2}dx=-\frac{1}{2}\int \frac{du}{u}=-\frac{1}{2}\ln|1-x^2|+C$$
(via the $u$-substitution $u=1-x^2$)
From your note:
$$\ln(x)=(x-1)-\frac{1}{2}(x-1)^2+\frac{1}{3}(x-1)^3+\dots$$
So substitute $1-x^2$ into $x$ in the natural log taylor expansion, and take the derivative to obtain a series representation of $f(x).$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4094977",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 3
} |
$x^2+xy+xz+yz=6+2\sqrt{5}$. Find the minimum of $3x+y+2z$ Let $x,~y,~z>0$ satisfy $x^2+xy+xz+yz=6+2\sqrt{5}$. How to find the minimum of $3x+y+2z$? I can find it by Larange multiplier in calculus, but I wonder if there is a easy way using, say Cauchy-Schwartz inequality?
| Apply the AM-GM inequality
$$3x+y+2z = (x+y)+2(x+z) \ge 2\sqrt{2(x+y)(x+z)} = 2\sqrt{2(x^2+xy+xz+yz)}=2\sqrt{2}(\sqrt{5}+1)$$
The equality occurs iff $x+y = 2(x+z)= \sqrt{6+2\sqrt{5}}$ or
$$x+y = 2(x+z) = \sqrt{5}+1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4096980",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Generalizing Solutions for Trigonometric functions (Secant) I have been working on the problem: $3\sec^2(2x)-5=1$
Where you have to solve for x and then enter the generalized solutions in ascending order. This is using the Acrobatiq platform, but I am uncertain how to format the answer.
Thus far I performed the following:
*
*$3\sec^2(2x)-5=1$ (+5 both sides)
*$3\sec^2(2x)=6$
*$\sec^2(2x)=2$ (Divide 3 both sides)
*$\sqrt{\sec^2(2x)}=\sqrt2$ (Square both sides)
*$\sec(2x)=\pm \sqrt 2$
From $\pm \sqrt 2$ I get the points $\frac{\pi}{4}$ and $\frac{5 \pi}{4}$
Generalizing the solution I get $2x=\frac{\pi}{4}+2\pi k$ which I then obtain the solution: $\frac{9 \pi}{8}$ .
For $\frac{5 \pi}{4}$ I get $2x=\frac{5 \pi}{4}+2 \pi k$ = $\frac{13 \pi}{8}$
Dividing both solutions by 2 I get: ${\frac{9 \pi}{4}, \frac{13 \pi}{4}}$
(The question is stating that it wants the answer for $2\pi k$ though there are other solutions i.e. $\frac{\pi}{8}$)
When entering these answers for $x$ its incorrect.. am I performing something incorrect? Or is there specific formatting to the Acrobatiq platform? (If anyone is familiar)
| I think you should use the definition of $\sec$ and congruences to solve it in a simple way:
$$\sec^22x=2\iff \cos^2 2x=\frac12\iff \cos 2x=\pm\frac{\sqrt 2}2.$$
Now
*
*$\cos 2x=\frac{\sqrt 2}2\iff 2x\equiv \pm\frac\pi 4\pmod{2\pi}\iff x\equiv \pm\frac\pi 8\pmod{\pi}$
*$\cos 2x=-\frac{\sqrt 2}2\iff 2x\equiv \pm\frac{3\pi}4\pmod{2\pi}\iff x\equiv \pm\frac{3\pi}8\pmod{\pi}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4097126",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
How do you find delta in an epsilon-delta proof of continuity when the function is nonlinear and can't be factored? So we define $f:\mathbb{R}\to \mathbb{R}$ and $f(x)=x^2+3x-5$, and we're using the epsilon-delta definition to prove that it is continuous at 3.
So I know the general structure of an epsilon-delta proof, but we were never shown how to prove a nonlinear function. Like I have no idea what to choose delta as, and how to even figure out what we should choose delta as. I know we need to let epsilon be an arbitrary positive real number, and plug the equation so that $|x^2+3x-5-3|<\delta$ so that $|x^2+3x-8|<\delta$. I don't think you can factor this, so what do you do in order to choose delta?
| Let $\delta$ be the value we want and lets try to find what restrictions we need.
$|x - 3| < \delta$
$3 -\delta < x < 3 + \delta$. As $\delta$ can be arbitrarily small we can assume will always pick a $\delta \le 3$.
$0\le 9-6\delta + \delta^2 < x^2 < 9 +6\delta + \delta^2$
$(9 - 6\delta + \delta^2) + 3(3-\delta) -5 < x^2 + 3x -5 < (9 + 6\delta + \delta^2) + 3(3+\delta) - 5$
$13 - 9\delta + \delta^2 < x^2 +3x-5 < 13 +9\delta + \delta^2$
$-9\delta + \delta^2 < (x^2 + 3x - 5) - 13 < 9\delta + \delta^2$.
now as $\delta$ can be arbitrarily small we can assume we will always pick a $\delta \le 1$. And if $0 < \delta \le 1$ then $\delta^2 \le \delta$.
$9\delta + \delta^2 \le 9\delta + \delta = 10\delta$. And $-9\delta + \delta^2 >-9\delta > -10\delta$.
So
$-10\delta \le -9\delta + \delta^2 < (x^2 + 3x - 5)-13 <9\delta + \delta^2 < 10 \delta$.
SO
$-10\delta < (x^2 + 3x - 5)-13 < 10 \delta$
$|(x^2 + 3x - 5)-13| < 10\delta$.
$|(x^2 + 3x -5) - (3^3 + 3\cdot 3 - 5)| < 10\delta$
So if we choose a $\delta$ where $\delta \le \frac \epsilon {10}$ and $\delta \le 1$ and $\delta \le 3$ we will be good and proven our result.
So let $\delta = \min(\frac {\epsilon}{10}, 1)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4102286",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
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Fractions in Questions and Answers
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