Q
stringlengths 70
13.7k
| A
stringlengths 28
13.2k
| meta
dict |
|---|---|---|
Inequality $\max\left(\frac{1-a^2}{3-2a},\frac{1-b^2}{3-2b}\right)\geq\frac{12}{19}\cdot\frac{3-a^2-2b^2}{6-a-3b}$ Let $0\leq a,b\leq 1$. Prove that
$$\max\left(\frac{1-a^2}{3-2a},\frac{1-b^2}{3-2b}\right)\geq\frac{12}{19}\cdot\frac{3-a^2-2b^2}{6-a-3b}.$$
Equality holds when $a=0$ and $b=2/3$. This seems to be the only case of equality.
The left-hand side is symmetric in $a$ and $b$ but the right-hand side isn't, so we can't simply assume $a\geq b$ or vice-versa.
A natural thing to try would be to divide into two cases according to which term in the max is greater, but the inequality $$\frac{1-a^2}{3-2a}\geq \frac{1-b^2}{3-2b}$$ doesn't translate easily to nice form of $a$ in terms of $b$.
|
hint: $$\frac{1-a^2}{3-2a}-\frac{1-b^2}{3-2b}=\frac{(a-b)(2-3a-3b+2ab)}{(2a-3)(2b-3)}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1892819",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
}
|
By using a geometric series and a factorisation, compute the first three terms of this given Taylor expansion By using the geometric series
$$\frac{1}{1-x} = \sum_{n=0}^\infty x^n$$
and the factorisation
$$\frac{1}{1-3x+2x^2}= \left(\frac{1}{1-2x}\right) \left(\frac{1}{1-x}\right)$$
compute the first three terms of the Taylor expansion of
$$\frac{1}{1-3x+2x^2} \text{ around } x=0$$
My theory of how to do this question is input the summation to the factorisation equation and solve so the summation equals
$$\sum_{n=0}^\infty x^n = \frac{1-2x}{1-3x+2x^2}$$
and solve by starting with $x=0$ and going to $x=2$.
But I feel this may be too simple to be true. Are my workings correct?
Thanks in advance
|
That could work, but an easier way is the following. Note that since
$$
\frac{1}{1-x} = \sum_{n=0}^\infty x^n
$$
it follows that
$$
\frac{1}{1-2x} = \sum_{n=0}^\infty (2x)^n = \sum_{n=0}^\infty 2^nx^n = 1 + 2x + 4x^2 +8x^3 + \cdots
$$
Now, since
$$
\frac{1}{1 - 3x + 2x^2} = \frac{1}{1 - 2x}\frac{1}{1-x}
$$
we can expand the right-hand side to be
$$
\frac{1}{1 - 2x}\frac{1}{1-x} = (1 + 2x + 4x^2 + \cdots)(1 + x + x^2 + x^3 + \cdots)
$$
from which point we can collect terms appropriately to get the answer you're looking for.
The point is, you can multiply convergent (or even non-convergent, if you feel like it!) series to obtain the series for the product. Since you know the first two series for this one...
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1893913",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
}
|
Let $z$ be a complex number $\ne 0$. What is the absolute value of $z\sqrt{z}$? $\color{red}{\mathbf{EDIT}}$ The question was misinterpreted - it was actually: 'what is the absolute value of $z/\bar{z}$?'; I'am grateful for the answers given on the original problem though and will keep this up as is in case someone else has a similar issue.
Exercise 3, page 379 of "Basic Mathematics" by S.Lang.
Problem: Let $z$ be a complex number $\ne 0$. What is the absolute value of $z\sqrt{z}$ $?
My approach to the question goes as follows:
Let $z = x + iy$ for real numbers $x, y$ different than $0$.
We have
\begin{align}
z\sqrt{z} & = (x + iy)\sqrt{z + iy} \\
& = x\sqrt{x+iy} + iy\sqrt{x+iy} &&\text{by distributivity}
\end{align}
We have
\begin{align}
|z\sqrt{z}| & = \sqrt{(x\sqrt{x + iy})^2 + (y\sqrt{x + iy})^2} &&\text{by definition} \\
& = \sqrt{x^2(x + iy)^2 + y^2(x + iy)^2} \\
& = \sqrt{(x^2 + y^2)(x + iy)} &&\text{by factoring} \\
& = \sqrt{(x^2 + y^2)} \sqrt{(x + iy)} \\
& = |z| \sqrt{z} &&\text{by definition} \\
\end{align}
The author's solution is $1$.
Thank you.
|
Using polar coordinates, you write $z=re^{i\theta}$, from which $\bar z=re^{-i\theta}$, thus
$$
z/\bar z=\frac{re^{i\theta}}{re^{-i\theta}}=e^{2i\theta}
$$
whose absolute value is $1$ as requested.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1894307",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
}
|
Minimum and maximum with $1+\frac{a+x}{b+x}+\frac{a+x}{c+x}$ Let $a,b,c\geq 1$ and $x,y,z\geq 0$. What are the minimum and maximum of
$$f(a,b,c,x,y,z)=\frac{1}{1+\frac{a+x}{b+x}+\frac{a+x}{c+x}}+\frac{1}{1+\frac{b+y}{a+y}+\frac{b+y}{c+y}}+\frac{1}{1+\frac{c+z}{a+z}+\frac{c+z}{b+z}}?$$
Since each term is no more than $1$, the maximum is not more than $3$. When $a=b=c$, $f(a,b,c,x,y,z)=1$.
|
Without loss of generality, take $a\ge b\ge c$.
By mediant inquality,
$$1 \le \frac{a+x}{b+x} \le \frac{a}{b}$$
$$1 \le \frac{a+x}{c+x} \le \frac{a}{c}$$
$$3 \le 1+\frac{a+x}{c+x}+\frac{a+x}{c+x} \le
1+\frac{a}{b}+\frac{a}{c}$$
$$\frac{b}{a} \le \frac{b+y}{a+y} \le 1$$
$$1 \le \frac{b+y}{c+y} \le \frac{b}{c}$$
$$2+\frac{b}{a} \le 1+\frac{b+y}{a+y}+\frac{b+y}{c+y} \le 2+\frac{b}{c}$$
By mediant inquality,
$$\frac{c}{a} \le \frac{c+z}{a+z} \le 1$$
$$\frac{c}{b} \le \frac{c+z}{b+z} \le 1$$
$$1+\frac{c}{a}+\frac{c}{b} \le 1+\frac{c+z}{a+z}+\frac{c+z}{b+z} \le 3$$
For fixed $a\ge b\ge c$,
$$\frac{1}{1+\frac{a}{b}+\frac{a}{c}}+\frac{1}{2+\frac{b}{c}}+\frac{1}{3}
\le f(a,b,c,x,y,z) \le \frac{1}{3}+\frac{1}{2+\frac{b}{a}}+\frac{1}{1+\frac{c}{a}+\frac{c}{b}}
$$
Take $\frac{a}{b}, \frac{a}{c}, \frac{b}{c} \to \infty$,
$$\frac{1}{3} \le f(a,b,c,x,y,z) \le \frac{11}{6}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1895187",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
}
|
Is the series $\sqrt{1} - \sqrt{2} + \sqrt{3} - \sqrt{4} + \dots$ summable? Is the series $\sqrt{1} - \sqrt{2} + \sqrt{3} - \sqrt{4} + \dots$ summable? I think it diverges although:
$$ \sqrt{n+1} - \sqrt{n} \approx \frac{1}{2\sqrt{n}}$$
for example by the Mean Value Theorem $f(x+1)-f(x) \approx f'(x)$ and then I might argue:
$$ \sum_{n \geq 1} (-1)^{n+1} \sqrt{n} = \frac{1}{2}\sum_{m \geq 1} \frac{1}{\sqrt{2m}} = \infty $$
Are these Cesaro summable? For an even number of terms:
$$\sqrt{1} - \sqrt{2} + \sqrt{3} - \sqrt{4} + \dots - \sqrt{2n}
\approx - \frac{1}{2\sqrt{2}}\left( \frac{1}{\sqrt{1}} +
\frac{1}{\sqrt{2}} + \dots + \frac{1}{\sqrt{n}} \right)
\approx \sqrt{\frac{n}{2}}$$
so the Cesaro means tend to infinity. Does any more creative summation method work?
The result is from paper called "The Second Theorem of Consistency for Summable Series" in Vol 6 of the Collected Works of GH Hardy
the series $1 - 1 +1 - 1 \dots$ is summable $(1,k)$ for any $k$ but not summable $(e^n, k)$ for any value of $k$.
The series $\sqrt{1} - \sqrt{2} + \sqrt{3} - \sqrt{4} + \dots$ is summable $(n,1)$ but not $(e^{\sqrt{n}},1)$ and so on...
Here things like $(1,k), (n,1)$ refer to certain averaging procedures, IDK
|
How about an Abel sum?
$$
\sum_{n=1}^\infty (-1)^{n+1}\sqrt{n}\;x^n = -\mathrm{Li}_{-1/2}(-x)
$$
for $|x|<1$ and converges as $x \to 1^-$ to the value
$$
-\mathrm{Li}_{-1/2}(-1) \approx 0.3801048
$$
So we call that value the Abel sum of the divergent series $\sum_{n=1}^\infty (-1)^{n+1}\sqrt{n}$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1896464",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 5,
"answer_id": 1
}
|
Find the second smallest integer such that its square's last two digits are $ 44 $ Given that the last two digits of $ 12^2 = 144 $ are $ 44, $ find the next integer that have this property.
My approach is two solve the equation $ n^2 \equiv 44 \pmod{100}, $ but I do not know how to proceed to solve that equation.
I try a different path by letting $ n = 10x + y $ for some integers $ x, y, $ where $ 0 \le y \le9. $ Then $ n^2 \equiv 44 \; \pmod{100} $ can be reduced to $ 20xy + y^2 \equiv 44 \pmod{100}. $ At this point I let $ x $ run from $ 0, 1, 2, \dots $ and find the integer $ y \in \mathbb{Z}_{100} $ such that $ y^2 + 20xy - 44 = 0. $
My question is is there an alternative way to tackle this problem without having to try each $ x $ and $ y? $ Maybe try to solve the initial congruence equation $ n^2 \equiv 44 \pmod{100}. $
|
$n^2 \equiv 44 \pmod {100}$
By Chinese remainder theorem, $n^2 \equiv 0 \pmod 4$ and $n^2 \equiv 19 \pmod {25}$ is an solution for the formula above.
$$n^2 \equiv 0 \pmod 4 \tag{1}$$
$$n^2 \equiv 19 \pmod {25} \tag{2}$$
By (1), $n = 2k$. Substitude into (2),
$$k^2 \equiv 19 \times 4^{-1} \equiv 19 \times 19 \equiv 11 \pmod {25} \tag{3}$$
By (3), $k \equiv 6 \pmod {25}$ or $k \equiv 19 \pmod {25}$,
when $k = 25w +6$, $n = 50w + 12$.
when $k = 25w +19$, $n = 50w + 38$.
Addendum: The procedure to get the inverse 4 under module 25 is same to get quadratic residue for 11 under module 25.
Here is the manifestation for the latter one.
Since $n^2 \equiv 11 \equiv 1 \pmod 5$, by numerating, $n \equiv 1,4 \pmod 5$.
When $n = 5k+1$, $n \equiv 1, 6, 11, 16, 21 \pmod {25}$
Check one after another and eliminate, gets $n \equiv 6 \pmod{25}$.
When $n = 5k+4$, same procedure plays.
So $k \equiv 6,19 \pmod{25}$ such that $k^2 \equiv 11 \pmod{25}$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1898485",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
}
|
Limit of $\mathbf{tan}^2(\theta)[1-\mathbf{sin}(\theta)]$ Evaulate the limit of $\mathbf{tan}^2(\theta)[1-\mathbf{sin}(\theta)]$.
$$\lim \limits_{\theta \to \frac{\pi}{2}} \mathbf{tan}^2(\theta)[1-\mathbf{sin}(\theta)] $$
I have attempted the problem by direct substitution and get:
$$\mathbf{tan}^2(\frac{\pi}2)[1-\mathbf{sin}(\frac{\pi}2)] $$
$$\infty^2 \cdot [0]$$
$$\infty \cdot 0$$
I do not know whether to state that it is zero, but I graphed it and got $\frac{1}{2}$. The limit has to be evaluated algebraically.
|
$$\lim _{ \theta \to \frac { \pi }{ 2 } }{ \tan } ^{ 2 }(\theta )[1-{ \sin }(\theta )]=\lim _{ \theta \to \frac { \pi }{ 2 } }{ \tan } ^{ 2 }(\theta )\frac { 1-\sin ^{ 2 }{ (\theta ) } }{ 1+{ \sin }(\theta ) } =\lim _{ \theta \to \frac { \pi }{ 2 } }{ \tan } ^{ 2 }(\theta )\frac { \cos ^{ 2 }{ (\theta ) } }{ 1+{ \sin }(\theta ) } =\\ =\lim _{ \theta \to \frac { \pi }{ 2 } }{ \frac { \sin ^{ 2 }{ (\theta ) } }{ 1+{ \sin }(\theta ) } } =\frac { 1 }{ 2 } $$
Another way is to apply L'hospital's rule
$$\lim _{ \theta \to \frac { \pi }{ 2 } }{ \tan } ^{ 2 }(\theta )\left[ 1-\sin { \left( \theta \right) } \right] =\lim _{ \theta \to \frac { \pi }{ 2 } }{ \frac { 1-\sin { \left( \theta \right) } }{ \cot ^{ 2 }{ \left( \theta \right) } } } =\lim _{ \theta \to \frac { \pi }{ 2 } }{ \frac { -\cos { \left( \theta \right) } }{ 2\cot { \left( \theta \right) \left( -\frac { 1 }{ \sin ^{ 2 }{ \left( \theta \right) } } \right) } } } =\frac { 1 }{ 2 } $$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1899409",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 0
}
|
Show $\lim\left ( 1+ \frac{1}{n} \right )^n = e$ if $e$ is defined by $\int_1^e \frac{1}{x} dx = 1$ I have managed to construct the following bound for $e$, which is defined as the unique positive number such that $\int_1^e \frac{dx}x = 1$.
$$\left ( 1+\frac{1}{n} \right )^n \leq e \leq \left (\frac{n}{n-1} \right )^n$$
From here, there must surely be a way to deduce the well-known equality
$$\lim_{n \rightarrow \infty} \left ( 1+ \frac{1}{n} \right )^n = e$$
I have come up with the following, but I am not absolutely certain if this is correct or not.
PROPOSED SOLUTION:
The lower bound is fine as it is, so we shall leave it alone. Note that
$$\begin{align*}
\left ( \frac{n}{n-1} \right )^n &= \left ( 1+\frac{1}{n-1} \right )^{n} \\ &=
\left ( 1+\frac{1}{n-1} \right )^{n-1} \left ( 1+\frac{1}{n-1} \right )
\end{align*}$$
So using the fact that the limit distributes over multiplication, we have
$$\lim_{n \rightarrow \infty} \left ( \frac{n}{n-1} \right )^n = \lim_{n \rightarrow \infty} \left ( 1+\frac{1}{n-1} \right )^{n-1} \lim_{n \rightarrow \infty} \left ( 1+\frac{1}{n-1} \right ) $$
Since $$\lim_{n \rightarrow \infty} \left ( 1+\frac{1}{n-1} \right ) = 1 $$
and
$$\lim_{n \rightarrow \infty} \left ( 1+\frac{1}{n-1} \right )^{n-1} = \lim_{m \rightarrow \infty} \left ( 1+\frac{1}{m} \right )^m = e $$
We then have the required result
$$\lim_{n \rightarrow \infty} \left ( 1+ \frac{1}{n} \right )^n = e$$
|
"$ ( 1+\frac{1}{n} )^n \leq e \leq (\frac{n}{n-1})^n$"$
I'm assuming you've shown $( 1+\frac{1}{n} )^n < ( 1+\frac{1}{n+1} )^{n+1}$ so we can say $\lim_{n\rightarrow \infty}( 1+\frac{1}{n} )^n = c \le e$ and likewise $\lim_{n\rightarrow \infty}(\frac{n}{n-1} )^n = d \ge e$.
So $\frac cd = \lim_{n\rightarrow \infty}\frac{( 1+\frac{1}{n} )^n}{(\frac{n}{n-1} )^n}= \lim_{n\rightarrow \infty}(\frac{(n+1)(n-1)}{n^2})^n=$
$\lim_{n\rightarrow \infty}(1- \frac{1}{n^2})^n = \lim 1 - \frac{n}{n^2} = 1$
So $c = d$ and $c = e = d$.
So $c = \lim ( 1+\frac{1}{n} )^n = e$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1900042",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 6,
"answer_id": 5
}
|
Help proving the product of any four consecutive integers is one less than a perfect square Apparently this is a true statement, but I cannot figure out how to prove this. I have tried setting
$$(m)(m + 1)(m + 2)(m + 3) = (m + 4)^2 - 1 $$
but to no avail. Could someone point me in the right direction?
|
I will show that
$16m(m+1)(m+2)(m+3) = 16(m^2 + 3m + 1)^2 - 16$
It will follow that
$$m(m+1)(m+2)(m+3) = (m^2 + 3m + 1)^2 - 1$$
Proof:
\begin{align}
16 m(m+1)(m+2)(m+3)
&=(2m)(2m+2)(2m+4)(2m+6)\\
& ---\text{let $2m = n-3$}---\\
&=(n-3)(n-1)(n+1)(n+3)\\
&=(n-3)(n+3) \cdot (n-1)(n+1) \\
&=(n^2 - 9)(n^2 - 1) \\
&=n^4 - 10n^2 + 9 \\
&=(n^2 - 5)^2 - 16 \\
& ---\text{Note $n = 2m+3$}---\\
&=(4m^2 + 12m + 4)^2 - 16\\
&=16(m^2 + 3m + 1)^2 - 16\\
\end{align}
addendum
I just noticed this proof.
\begin{align}
m(m+1)(m+2)(m+3)
&= m(m+3) \cdot (m+1)(m+2) \\
&= (m^2 + 3m)(m^2 + 3m + 2) \\
&= ((m^2 + 3m + 1) - 1) \cdot ((m^2 + 3m + 1) + 1)\\
&= (m^2+3m + 1)^2 - 1
\end{align}
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1900365",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 8,
"answer_id": 7
}
|
Find the minimnum of the $\max{(2x_{1}+x_{2},2x_{2}+x_{3},\cdots,2x_{n-1}+x_{n})}$ Let $n$ is give postive integers,and for $x_{i}\ge 0$,such
$$x_{1}+x_{2}+\cdots+x_{n}=1$$
find $I$ minimum of the value
$$I=\max{(2x_{1}+x_{2},2x_{2}+x_{3},\cdots,2x_{n-1}+x_{n})}$$
I try to find the
$$2x_{1}+x_{2}\le I$$
$$2x_{2}+x_{3}\le I$$
$$\cdots$$
$$2x_{n-1}+x_{n}\le I$$
add all
$$(n-1)I\ge 3(x_{1}+\cdots+x_{n})-x_{1}-2x_{n}=3-x_{1}-2x_{n}$$
but $x_{1}+x_{n}$ maximum is?
|
Assume that $n\geq 2$. Observe that, as $x_1,x_2,\ldots,x_n\geq0$ (in particular, $x_{n-1}\geq 0$), we have
$$\begin{align}
S&:=\sum_{i=1}^{n-2}\,\left(\frac{2^i-(-1)^i}{3}\right)\,2^{n-2-i}\,\left(2x_{i}+x_{i+1}\right)+2^{n-2}\left(2x_{n-1}+x_n\right)
\\
&=\left(\frac{2^n-(-1)^n}{3}\right)\,x_{n-1}+2^{n-2}\,\sum_{i=1}^{n}\,x_i\geq 2^{n-2}\,\sum_{i=1}^n\,x_i=2^{n-2}\,.
\end{align}$$
On the other hand,
$$\begin{align}
S&\leq I\left(\sum_{i=1}^{n-2}\,\left(\frac{2^i-(-1)^i}{3}\right)\,2^{n-2-i}+2^{n-2}\right)
\\&= I\,\left(2^{n-2}+(n-2)\frac{2^{n-2}}{3}-\frac{(-1)^n}{3}\,\sum_{i=1}^{n-2}(-2)^{n-2-i}\right)
\\
&=\frac{I}{3}\left((n+1)2^{n-2}-(-1)^n\,\sum_{i=0}^{n-3}\,(-2)^i\right)
\\
&=\frac{I}{3}\Biggl((n+1)2^{n-2}-(-1)^n\,\left(\frac{1-(-2)^{n-2}}{3}\right)\Biggr)
\\
&=\frac{I}{3^2\cdot 2^2}\big((3n+3)\,2^n-4\cdot(-1)^n+2^n\big)
\\
&=\frac{I}{3^2\cdot 2^2}\big((3n+4)\, 2^n-4(-1)^n\big)\,.
\end{align}\,.$$
Ergo,
$$I\geq \frac{3^2\cdot 2^2\cdot S}{(3n+4)\, 2^n-4(-1)^n}\geq \frac{9\cdot 2^n}{(3n+4)\, 2^n-4(-1)^n}\,.$$
Therefore,
$$I\geq \frac{9\cdot 2^n}{(3n+4)\, 2^n-4(-1)^n}=:m_n\,.$$
The equality holds if and only if $2x_i+x_{i+1}=m_n$ for all $i=1,2,\ldots,n-1$ and $x_{n-1}=0$, which then yields the unique minimizing point:
$$\left(x_1,x_2,\ldots,x_n\right)=\frac{m_n}{3}\,\Biggl(1-\frac{1}{(-2)^{n-2}},1-\frac{1}{(-2)^{n-3}},1-\frac{1}{(-2)^{n-4}},\ldots,0,3\Biggr)\,.$$
For instances, $m_2=1$, $m_3=\frac{2}{3}$, $m_4=\frac{4}{7}$, and $m_5=\frac{8}{17}$ with the following respective minimizing points: $(0,1)$, $\left(\frac{1}{3},0,\frac{2}{3}\right)$, $\left(\frac{1}{7},\frac{2}{7},0,\frac{4}{7}\right)$, and $\left(\frac{3}{17},\frac{2}{17},\frac{4}{17},0,\frac{8}{17}\right)$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1902192",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
}
|
Prove that $\left(\frac{n}{2}\right)^n \gt n! \gt \left(\frac{n}{3}\right)^n \qquad n\ge 6 $
$$\left(\frac{n}{2}\right)^n \gt n! \gt \left(\frac{n}{3}\right)^n \qquad n\ge 6 $$
I tried it prove it by mathematical induction but failed .
For $n=6$
$$(3)^6 \gt 6!$$
Now for $n=k$
$$\left(\frac{k}{2}\right)^k \gt k!$$
Now for $n=k+1$
$$\left(\frac{(k+1)}{2}\right)^{k+1} \gt (k+1)!$$
$$\left(\frac{(k+1)}{2}\right)^{k} \gt 2(k!)$$
|
METHODOLOGY $1$: Proof by Induction
PRIMER:
In THIS ANSWER, I showed using Bernoulli's Inequality that $\left(1+\frac1n\right)^n$ is a monotonically increasing sequence. Therefore, its minimum is $2$ when $n=1$. It is clearly bounded by $3$ as can be shown applying the binomial theorem and applying simple estimates.
$$\begin{align}\left(1+\frac1n\right)^n&=1+1\\\\&+\frac1{2!}\left(1-\frac1n\right)+\frac{1}{3!}\left(1-\frac1n\right)\left(1-\frac2n\right)+\cdots +\frac{1}{n!}\left(1-\frac1n\right)\cdots\left(1-\frac{n-1}n\right)\\\\&\le 1+1+\frac12+\frac1{2^2}+\cdots +\frac{1}{2^{n-1}}\\\\&\le 3\end{align}$$
Now, assume for some $n>6$ that $\left(\frac{n}{2}\right)^n\ge n!\ge \left(\frac{n}{3}\right)^n$. Then, for $n+1$ we have
$$\begin{align}
\left(\frac{n+1}{2}\right)^{n+1}&=\left(\frac{n+1}{2}\right)\,\left(\frac{n}{2}\right)^n\,\left(1+\frac1n\right)^n\\\\
&\ge \left(\frac{n+1}{2}\right)\,n!\,\left(1+\frac1n\right)^n\\\\
&=(n+1)!\,\frac12 \,\left(1+\frac1n\right)^n\\\\
&\ge (n+1)!
\end{align}$$
as was to be shown, and
$$\begin{align}
\left(\frac{n+1}{3}\right)^{n+1}&=\left(\frac{n+1}{3}\right)\,\left(\frac{n}{3}\right)^n\,\left(1+\frac1n\right)^n\\\\
&\le \left(\frac{n+1}{3}\right)\,n!\,\left(1+\frac1n\right)^n\\\\
&=(n+1)!\,\frac13 \,\left(1+\frac1n\right)^n\\\\
&\le (n+1)!
\end{align}$$
as was to be shown!
METHODOLOGY $2$: Proof Invoking Bounds of Stirling Formula
We can use the bounds found from development of Stirling's Formula (See "Advanced Calculus, John Olmsted, 1961, pp. 490-491.)
$$\left(1+\frac{1}{12(n+1)}\right)\sqrt{2\pi n}\left(\frac{n}{e}\right)^n\le n!\le \left(1+\frac{1}{12(n-2)}\right)\sqrt{2\pi n}\left(\frac{n}{e}\right)^n \tag 1$$
for $n\ge 3$.
For the left-hand side of $(1)$, we write
$$\begin{align}
n!&\ge \left(1+\frac{1}{12(n+1)}\right)\sqrt{2\pi n}\left(\frac{n}{e}\right)^n\\\\
&=\color{blue}{\left(1+\frac{1}{12(n+1)}\right)}\color{green}{\sqrt{2\pi n}\left(\frac{3}{e}\right)^n}\left(\frac{n}{3}\right)^n\\\\
&\ge \color{blue}{(1)}\,\color{green}{(1)}\,\left(\frac{n}{3}\right)^n\\\\
&=\left(\frac{n}{3}\right)^n
\end{align}$$
For the right-hand side of $(1)$, we write
$$\begin{align}
n!&\le \left(1+\frac{1}{12(n-2)}\right)\sqrt{2\pi n}\left(\frac{n}{e}\right)^n\\\\
&=\color{blue}{\left(1+\frac{1}{12(n-2)}\right)}\color{green}{\sqrt{2\pi n}\,\left(\frac{2}{n}\right)^n}
\,\left(\frac{n}{2}\right)^n
\end{align}$$
The function $f(x)=\sqrt{2\pi x}\left(\frac{3}{e}\right)^x$ can easily be shown to monotonically decrease for $x>\frac{1}{2\log(e/2)}$. Therefore, for $x=n\ge 6$, the function is less than $f(6)<\frac{48}{49}$. And inasmuch as $ 1+\frac{1}{12(n-2)}\le \frac{49}{48}$ for $n\ge 6$, we have
$$\begin{align}
n!&\le \left(1+\frac{1}{12(n-2)}\right)\sqrt{2\pi n}\left(\frac{n}{e}\right)^n\\\\
&=\color{blue}{\left(1+\frac{1}{12(n-2)}\right)}\color{green}{\sqrt{2\pi n}\,\left(\frac{2}{n}\right)^n}
\,\left(\frac{n}{2}\right)^n\\\\
&\le \color{blue}{\left(\frac{48}{49}\right)}\,\color{green}{\left(\frac{49}{48}\right)}\,\left(\frac{n}{2}\right)^n
\end{align}$$
And we are done!
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1902324",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
}
|
Calculate the sum of $\sum_{n=1}^\infty (-1)^{n-1} \frac{n+1}{2^\frac{n}{2}}$ So, we have $$\sum_{n=1}^\infty (-1)^{n-1} \frac{n+1}{2^\frac{n}{2}}$$
I immediately noticed that $ln(1+x) = \sum_{n=1}^\infty (-1)^{n-1} \frac{x^n}{n}$ , for $|x|\lt 1$
It reminds of this sum, if we switch nominator and denominator and multiply it by $2^{n/2+1}$, we would get something similar.
Thats all i got so far.
|
One may start with the standard finite evaluation:
$$
1+x+x^2+...+x^n=\frac{1-x^{n+1}}{1-x}, \quad |x|<1. \tag1
$$ Then by differentiating $(1)$ we get
$$
1+2x+3x^2+...+nx^{n-1}=\frac{1-x^{n+1}}{(1-x)^2}+\frac{-(n+1)x^{n}}{1-x}, \quad |x|<1, \tag2
$$ by making $n \to +\infty$ in $(2)$, using $|x|<1$, gives
$$
\sum_{n=0}^\infty(n+1)x^n=\frac{1}{(1-x)^2}. \tag3
$$
Now just insert $x:=-\dfrac1{\sqrt{2}}$ into $(3)$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1902487",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
}
|
Finding the value of theta using the $6$ trigonometric functions on an $xy$ plane. Here's a link of what I am trying to learn about.
http://www.webpages.uidaho.edu/learn/math/lessons/lesson03/3_05.htm
Now I have one question. How will I find the value for theta in each function if the hypotenuse has no exact square root and is now a decimal.
Let $(14,5)$ be a point on the terminal side.
$x = 14$
$y = 5$
\begin{align*}
r & = \sqrt{14^2+5^2}\\
r & = \sqrt{196+25}\\
r & = \sqrt{221}\\
r & = 14.86
\end{align*}
What do I need to do with the calculator?
|
Based on a now deleted comment, it is my understanding that the terminal side of the angle passes through the point $(14, 5)$. You used the Pythagorean Theorem to conclude correctly that $r = \sqrt{221}$.
Using the trigonometric formulas
\begin{align*}
\sin\theta & = \frac{y}{r} & \csc\theta & = \frac{r}{y}\\
\cos\theta & = \frac{x}{r} & \sec\theta & = \frac{r}{x}\\
\tan\theta & = \frac{y}{x} & \tan\theta & = \frac{x}{y}
\end{align*}
with the values $x = 14$, $y = 5$, and $r = \sqrt{221}$ yields the exact values
\begin{align*}
\sin\theta & = \frac{5}{\sqrt{221}} & \csc\theta & = \frac{\sqrt{221}}{5}\\
\cos\theta & = \frac{14}{\sqrt{221}} & \sec\theta & = \frac{\sqrt{221}}{14}\\
\tan\theta & = \frac{5}{14} & \tan\theta & = \frac{14}{5}
\end{align*}
Now that we have the exact values, we can plug them into the calculator to obtain the approximations
\begin{align*}
\sin\theta & \approx 0.34 & \csc\theta & \approx 2.97\\
\cos\theta & \approx 0.94 & \sec\theta & \approx 1.06\\
\tan\theta & \approx 0.36 & \cot\theta & = 2.8
\end{align*}
where I have rounded to the nearest hundredth except for $\cot\theta$. The value for $\cot\theta$ is exact.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1903325",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
If $a$, $b$, and $c$ are sides of a triangle, then $\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}<2$.
Let $a,b,c$ be the lengths of the sides of a triangle. Prove that
$$\sum_{\text{cyc}}\frac{a}{b+c}=\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}<2\,.$$
Attempt. By clearing the denominators, the required inequality is equivalent to
$$a^2(b+c)+b^2(c+a)+c^2(a+b)>a^3+b^3+c^3\,.$$
Since $b+c>a$, $c+a>b$, and $a+b>c$, the inequality above is true. Is there a better, non-bruteforce way?
|
Hint: Without loss of generality, suppose $c$ is the largest side. Hence, $c<a+b$. Also,
$$\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\leq\frac{a}{a+b}+\frac{b}{a+b}+\frac{c}{a+b}\,.$$
Note that the bound is sharp. In the limit $a\to 0$ and $b\to c$, we have the sum goes to $2$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1903775",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 3
}
|
Finding the matrix representation of a linear mapping
Let $V=\left \{ A\in M_2\left(\mathbb{C} \right )\mid \text{tr}(A)=0 \right \}$ and $T:V\rightarrow V$ defined by $T\left(A\right)=\left(3\,\mathrm{i}-2\right)A+\left(4-6\,\mathrm{i}\right)A^t$
Find the matrix representation of $T$ with regard to the basis $B=\left\{\left(\begin{matrix}1&0\\0&-1\end{matrix}\right),\left(\begin{matrix}0&1\\0&0\end{matrix}\right),\left(\begin{matrix}0&0\\1&0\end{matrix}\right)\right\}$
What I have done is to plug the basis vectors (in this case matrices) to the transformation:
$$T(\begin{pmatrix}1&0\\0&-1\end{pmatrix})=\begin{pmatrix}3i-2&0\\0&2-3i\end{pmatrix}+\begin{pmatrix}4-6i&0\\0&-4+6i\end{pmatrix}=\begin{pmatrix}2-3i&0\\0&-2+3i\end{pmatrix}$$
$$T(\begin{pmatrix}0&1\\0&0\end{pmatrix})=\begin{pmatrix}0&3i-2\\0&0\end{pmatrix}+\begin{pmatrix}0&0\\4-6i&0\end{pmatrix}=\begin{pmatrix}0&3i-2\\4-6i&0\end{pmatrix}$$
$$T(\begin{pmatrix}0&0\\1&0\end{pmatrix})=\begin{pmatrix}0&0\\3i-2&0\end{pmatrix}+\begin{pmatrix}0&4-6i\\0&0\end{pmatrix}=\begin{pmatrix}0&4-6i\\3i-2&0\end{pmatrix}$$
So the matrix is the sum?
$$\begin{pmatrix}2-3i&2-3i\\2-3i&3i-2\end{pmatrix}$$
Or should I put them as as vector columns?
|
No the matrix is the matrix whose columns are the images. You need to find the images, and rewrite them as coordinate vectors with respect to $B$.
$$T(B_1) = (2-3i)B_1 + 0B_2 + 0B_3$$
$$T(B_2) = 0B_1 + (-2+3i)B_2 + (4-6i)B_3$$
$$T(B_3) = 0B_1 + (4-6i)B_2 + (-2+3i)B_3$$
Therefore
$$[T]_B = \begin{pmatrix} 2-3i & 0 & 0\\ 0 & -2+3i & 4-6i\\ 0 & 4-6i & -2+3i\end{pmatrix}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1904341",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
Prove inequality $\ln \left( \frac{e-e^x}{1-x} \right) \leq \sqrt{\frac{1+x+x^2}{3}}$ for $0The first function could be called 'exponential mean' of $y$ and $x$:
$$f(y,x)=\ln \left( \frac{e^y-e^x}{y-x} \right)$$
We can obtain it by Cauchy mean value theorem.
What is interesting, it appears numerically that:
$$\ln \left( \frac{e-e^x}{1-x} \right) \leq \sqrt{\frac{1+x+x^2}{3}}, \qquad 0<x \leq 1$$
$$\ln \left( \frac{e-e^x}{1-x} \right) \geq \sqrt{\frac{1+x+x^2}{3}}, \qquad x \geq 1$$
How would you prove both of these inequalities?
Around $1$ both functions are very close. See the plot below:
This is related to my recent question.
The series expansion doesn't seem to be the best way.
The first function can be represented:
$$\ln \left( \frac{e-e^x}{1-x} \right)=1+\ln(1-e^{-(1-x)})-\ln(1-x)$$
But that's not any better.
|
A little demonstration (not a full solution, but I'll try to expand it later):
If we were allowed to use CAS, such as Mathematica (or spent a little time differentiating), we could just square both sides and expand the left side into series aroud $x=1$:
$$\ln^2 \left( \frac{e-e^x}{1-x} \right) =1+(x-1)+\frac{1}{3} (x-1)^2+\frac{1}{24} (x-1)^3+\frac{1}{960} (x-1)^4+O((x-1)^5)$$
Now we expand and simplify the first three terms:
$$\ln^2 \left( \frac{e-e^x}{1-x} \right) =\frac{1}{3}+\frac{x}{3}+\frac{x^2}{3}\color{blue}{+\frac{1}{24} (x-1)^3+\frac{1}{960} (x-1)^4}+O((x-1)^5)$$
Comparing with the square of the right-hand side we immediately see:
$$\ln^2 \left( \frac{e-e^x}{1-x} \right)- \frac{1+x+x^2}{3}=\frac{1}{24} (x-1)^3+\frac{1}{960} (x-1)^4+O((x-1)^5)$$
Thus, around $x=1$ we have:
$$\ln^2 \left( \frac{e-e^x}{1-x} \right)- \frac{1+x+x^2}{3}=-\frac{1}{24} (1-x)^3+\frac{1}{960} (1-x)^4-O((1-x)^5) \leq 0, $$ $$x \to 1^-$$
$$\ln^2 \left( \frac{e-e^x}{1-x} \right)- \frac{1+x+x^2}{3}=\frac{1}{24} (x-1)^3+\frac{1}{960} (x-1)^4+O((x-1)^5) \geq 0, $$ $$ x \to 1^+$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1904473",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
}
|
Proving $\sum_{x=0}^{n-1} \cos\left(k +x{2\pi\over n}\right) =\sum_{x=0}^{n-1} \sin\left(k +x{2\pi\over n}\right) =0. $ Is there anybody who can help me show the following?
$$
\sum_{x=0}^{n-1} \cos\left(k +x{2\pi\over n}\right) =0
\qquad\hbox{and}\qquad
\sum_{x=0}^{n-1} \sin\left(k +x{2\pi\over n}\right) =0
$$
I can only prove this claim for the sine expression when $k=0$, by grouping the output values such that they all cancel out (for both $n$ even and odd). I have thought about using complex numbers in exponential form to prove it, but I can't do it ... :P
I would really appreciate an algerbraic proof.
|
Let $k$ be an arbitrary real number. Consider the trigonometric identity
$$2 \cos A\, \sin B = \sin(A + B) - \sin(A - B)$$
Setting $A = k + x(\frac{2\pi}{n})$ and $B = \frac{\pi}{n}$, we get
$$2\cos\left(k + x\frac{2\pi}{n}\right) \sin\left(\frac{\pi}{n}\right) = \sin\left[k + \left(x + \frac{1}{2}\right)\frac{2\pi}{n}\right] - \sin\left[k + \left(x - \frac{1}{2}\right)\frac{2\pi}{n}\right]$$
Summing from $x = 0$ to $x = n-1$ and letting $u_x = k + \left(x - \frac{1}{2}\right)\frac{2\pi}{n}$, we obtain
\begin{align}&\sum_{x = 0}^{n-1} 2\cos\left(k + x\frac{2\pi}{n}\right)\sin\left(\frac{\pi}{n}\right) \\
&= \sum_{x = 0}^{n-1}\, [\sin(u_{x+1}) - \sin(u_x)] \\
&= (\sin(u_1) - \sin(u_{0})) + (\sin(u_2) - \sin(u_1)) +\cdots + (\sin(u_n) - \sin(u_{n-1}))\\
&= \sin(u_n) - \sin(u_0)\\
&= 0\end{align}
Since $\sin(\frac{\pi}{n}) \neq 0$, we deduce that
$$\sum_{x = 0}^{n-1} \cos\left(k + x\frac{2\pi}{n}\right) = 0.\tag{*}\label{eq1}$$
If we take equation \eqref{eq1} and replace $k$ with $k - \pi/2$, we get
$$\sum_{x = 0}^{n-1}\sin\left(k + x\frac{2\pi}{n}\right) = 0.$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1905189",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
}
|
Trouble understanding proof of the inequality - $(\frac{1}{a}+1)(\frac{1}{b}+1)(\frac{1}{c}+1) \ge 64 $, for $a,b,c > 0$ and $a+b+c = 1$ I was looking into this problem in a book discussing inequalities, However I found the proof quite hard to understand.The problem is as follows:
Let $a,b,c$ be positive numbers with $a+b+c=1$, prove that
$$\left(\frac{1}{a}+1\right)\left(\frac{1}{b}+1\right)\left(\frac{1}{c}+1\right) \ge 64$$
and the proof provided was the following:
Note that $$ abc \le (\frac{a+b+c}{3})^3 = \frac{1}{27} \tag{1}$$
by AM-GM inequality. Then
$$(\frac{1}{a}+1)(\frac{1}{b}+1)(\frac{1}{c}+1)=1+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}+\frac{1}{abc} \tag{2}$$
$$\ge 1+\frac{3}{\sqrt[3]{abc}}+\frac{3}{\sqrt[3]{(abc)^2}} +\frac
{1}{abc} \tag{3}$$
$$=(1+\frac{1}{\sqrt[3]{abc}})^3 \ge 4^3 \tag{4}$$
Steps 1 and 2 are easy for me to understand, but if someone could help me with steps 3 and 4 ,I would be very thankful.
|
I have any solution.
Note that
$a+1=a+a+b+c\geq 4\sqrt[3]{a^2bc} \tag{1}$
by AM-GM
Analog we get that
$b+1=b+a+b+c\geq 4\sqrt[3]{ab^2c} \tag{2}$
and
$c+1=c+a+b+c\geq 4\sqrt[3]{abc^2} \tag{3}$
So
$$(a+1)(b+1)(c+1)\geq 64abc\Longleftrightarrow (\dfrac{a+1}{a})(\dfrac{b+1}{b})(\dfrac{c+1}{c})\geq 64$$
The equality will hold by $a=b=c=1$.
Proved
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1905278",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
}
|
Discontinuous at infinitely many points While doing a worksheet on real analysis I came across the following problem.
$Q$. Let $f$ be a function defined on $[0,1]$ with the following property.
For every $y \in R$, either there is no $x$ in $[0,1]$ for which $f(x)=y$ or there are exactly two values of $x$ in $[0,1]$ for which $f(x)=y$.
(a) Prove that $f$ cannot be continuous on $[0,1]$.
(b) Construct a function $f$ which has the above property.
(c) Prove that any such function with this property has infinitely many discontinuous on $[0,1]$.
I really have absolutely no idea how to solve the problem. Even constructing the function is proving pretty difficult. Any help would be appreciated asap.
|
This example shows that the image of $f$ can be the whole of $\mathbb{R}$:
\begin{align*}
f(0) & = 0, \\
f(x) & = \begin{cases}
\frac{3}{2^n} - 8x &
\left( \frac{1}{2^{n + 1}} \leqslant x < \frac{5}{2^{n + 3}} \right) \\
\frac{1}{2^{n - 2}} - 8x &
\left( \frac{5}{2^{n + 3}} \leqslant x < \frac{3}{2^{n + 2}} \right) \\
8x - \frac{5}{2^n} &
\left( \frac{3}{2^{n + 2}} \leqslant x < \frac{7}{2^{n + 3}} \right) \\
8x - \frac{3}{2^{n - 1}} &
\left( \frac{7}{2^{n + 3}} \leqslant x < \frac{1}{2^n} \right),
\end{cases}
\\
f(x) & = \begin{cases}
\frac{1}{8x - 5} &
\left( \frac{1}{2} \leqslant x < \frac{5}{8} \right) \\
\frac{1}{8x - 6} &
\left( \frac{5}{8} \leqslant x < \frac{3}{4} \right) \\
\frac{1}{7 - 8x} &
\left( \frac{3}{4} \leqslant x < \frac{7}{8} \right) \\
\frac{1}{8 - 8x} &
\left( \frac{7}{8} \leqslant x < 1 \right),
\end{cases} \\
f(1) & = 0.
\end{align*}
(The variable $n$ takes all positive integral values.)
\begin{alignat*}{2}
f\left(\left[\frac{1}{2^{n + 1}}, \frac{5}{2^{n + 3}}\right)\right) & =
f\left(\left[\frac{5}{2^{n + 3}}, \frac{3}{2^{n + 2}}\right)\right) && =
\left(-\frac{1}{2^{n - 1}}, -\frac{1}{2^n}\right], \\
f\left(\left[\frac{3}{2^{n + 2}}, \frac{7}{2^{n + 3}}\right)\right) & =
f\left(\left[\frac{7}{2^{n + 3}}, \frac{1}{2^n}\right)\right) && =
\left[\frac{1}{2^n}, \frac{1}{2^{n - 1}}\right),
\end{alignat*}
\begin{alignat*}{2}
f\left(\left[\frac{1}{2}, \frac{5}{8}\right)\right) & =
f\left(\left[\frac{5}{8}, \frac{3}{4}\right)\right) && =
\left(-\infty, -1\right], \\ \\
f\left(\left[\frac{3}{4}, \frac{7}{8}\right)\right) & =
f\left(\left[\frac{7}{8}, 1\right)\right) && =
\left[1, \infty\right).
\end{alignat*}
Amusingly, $f$ is continuous at $0$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1905560",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
"answer_id": 4
}
|
prove $abc \ge 8$ for $\frac{1}{1+a}+\frac{1}{1+b}+\frac{1}{1+c}=1$ I am reading this book about inequalities and the chapter about AM-GM inequalities includes this problem:
Let $a,b,c$ be positive numbers for which $\frac{1}{1+a}+\frac{1}{1+b}+\frac{1}{1+c}=1$, prove that
$$abc \ge 8$$
The book does not provide full solutions but only hints, and the one for this question is that it is similar to this problem:
Let $a,b,c$ be positive numbers with $a+b+c=1$, prove that
$$(\frac{1}{a}-1)(\frac{1}{b}-1)(\frac{1}{c}-1) \ge 8$$
again there's no solution provided but what I came up with is the following (you can correct me if it is wrong):
$$a+b=1-c$$
$$a+c=1-b$$
$$b+c=1-a$$
$$(\frac{1}{a}-1)(\frac{1}{b}-1)(\frac{1}{c}-1)=(\frac{1-a}{a})(\frac{1-b}{b})(\frac{1-c}{c})=(\frac{b+c}{a})(\frac{a+c}{b})(\frac{a+b}{c})=(\frac{b}{a}+\frac{c}{a})(\frac{a}{b}+\frac{c}{b})(\frac{a}{c}+\frac{b}{c})$$
By AM-GM we have for each term:
$$(\frac{b}{a}+\frac{c}{a}) \ge 2\sqrt{\frac{bc}{a^2}}$$
$$(\frac{a}{b}+\frac{c}{b}) \ge 2\sqrt{\frac{ac}{b^2}}$$
$$(\frac{a}{c}+\frac{b}{c}) \ge 2\sqrt{\frac{ab}{c^2}}$$
By multiplying the three inequalities we have
$$(\frac{b}{a}+\frac{c}{a})(\frac{a}{b}+\frac{c}{b})(\frac{a}{c}+\frac{b}{c}) \ge 8 \sqrt{\frac{a^2b^2c^2}{a^2b^2c^2}}=8$$
as desired.
Can someone please provide me a proof for the first problem as I cannot find any way around it, Thanks.
|
If the problem states that a, b, and c are positive integers, this is pretty easy. Find a common denominator on the LHS, multiply both sides by the denominator and simplify. Then you get: $$3+2(a+b+c)+ab+ac+bc=1+(a+b+c)+ab+ac+bc+abc$$ Simplify this and you get $$2+a+b+c=abc$$ The lowest possible non-zero integers are 1, 2, and 3, therefore the lowest value abc can have is 8.
If, however, a, b, and c are the lowest possible real numbers, this is a very different problem.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1906492",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
}
|
Prove the inequality $\left(\frac1a+\frac1b+\frac1c\right)\left(\frac1{1+a}+\frac1{1+b}+\frac1{1+c}\right)\ge\frac9{1+abc}$
Let $a,b,c>0$. Prove the inequality
$$\left(\frac1a+\frac1b+\frac1c\right)\left(\frac1{1+a}+\frac1{1+b}+\frac1{1+c}\right)\ge\frac9{1+abc}$$
My work so far:
Use AM-GM:
$$\frac1{1+a}+\frac1{1+b}+\frac1{1+c}=$$
$$=\frac{\frac1a}{1+\frac1a}+\frac{\frac1b}{1+\frac1b}+\frac{\frac1c}{1+\frac1c}\ge$$
$$\ge3\sqrt[3]{\frac{\frac1a\cdot\frac1b\cdot\frac1c}{\left(1+\frac1a\right)\left(1+\frac1b\right)\left(1+\frac1c\right)}}=$$
$$\ge\frac3{\sqrt[3]{abc}}\sqrt[3]{\frac{1}{\left(1+\frac1a\right)\left(1+\frac1b\right)\left(1+\frac1c\right)}}$$
|
By Rearrangement
$\sum\limits_{cyc}\frac{1}{a}\sum\limits_{cyc}\frac{1}{1+a}=\sum\limits_{cyc}\frac{1}{a(1+a)}+\sum\limits_{cyc}\frac{1}{b(1+a)}+\sum\limits_{cyc}\frac{1}{c(1+a)}\geq\sum\limits_{cyc}\frac{2}{b(1+a)}+\sum\limits_{cyc}\frac{1}{c(1+a)}$.
Now by AM-GM
$(1+abc)\sum\limits_{cyc}\frac{1}{b(1+a)}=\sum\limits_{cyc}\frac{1+abc+b+ab}{b(1+a)}-3=\sum\limits_{cyc}\left(\frac{1+b}{b(1+a)}+\frac{a(1+c)}{1+a}\right)-3\geq$
$\geq2\sum\limits_{cyc}\sqrt{\frac{a(1+b)(1+c)}{b(1+a)^2}}-3\geq6-3=3$.
By the same way $(1+abc)\sum\limits_{cyc}\frac{1}{c(1+a)}\geq3$ and we are done!
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1908753",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
}
|
How can it be proven that $\frac{x}{y}+\frac{y}{x}\geq2$, with $x$ and $y$ positive? So I realized that I have to prove it with the fact that $(x-y)^2+2xy=x^2+y^2$
So $\frac{(x+y)^2}{xy}+2=\frac{x}{y}+\frac{y}{x}$ $\Leftrightarrow$ $\frac{(x+y)^2}{xy}=\frac{x}{y}+\frac{y}{x}-2$
Due to the fact that $(x+y)^2$ is a square, it will be positive
$x>0$ and $y>0$ so $xy>0$
So $\frac{(x+y)^2}{xy}>0$
So $\frac{x}{y}+\frac{y}{x}>2$
But the problem is that I have proven that $\frac{x}{y}+\frac{y}{x}>2$, but in the case of $x=y$, it is equal, so it has to be $\geq$...
Could someone help?
|
You made a few mistakes here.
First, actually $(x+y)^2=x^2+2xy+y^2$, what you need is $(x-y)^2+2xy=x^2+y^2$. Now, you get $xy>0$ but $(x-y)^2 \geq 0$, so $\frac{(x-y)^2}{xy} \geq 0$, and this gives you $$\frac{x^2-2xy+y^2}{xy} \geq 0$$
which gives $\frac{x}{y}+\frac{y}{x} \geq 2$, which is the desired inequality.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1911037",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
}
|
Why does $f_A(X,X)>0$ for every $X$ (Question 8 of section 8.2 of Hoffman and Kunze's Linear Algebra) I'm studying Hoffman and Kunze's linear algebra book and I'm having troubles how to prove this exercise on page 276:
The only part of this exercise I couldn't prove was $f_A(X,X)>0$ for every real column matrix $X$.
|
Let's compute:
$$
X^t A X = \begin{pmatrix} x & y \end{pmatrix}
\begin{pmatrix} a & c \\ c & b \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = x(ax+cy)+y(cx+by) = ax^2+2cxy + by^2
$$
Now use the additional hypothesis that $a, b>0$ and $ab > c^2$ (which comes from $\det A > 0$). So we get
$$
ax^2+2cxy+by^2 \geq ax^2 + 2cxy+\frac{c^2}{a} y^2
= \frac{1}{a}(a^2x^2+2acxy+c^2 y^2) = \frac{1}{a}(ax+cy)^2
$$
which will be positive except when $ax=-cy$. Similarly,
$$
ax^2+2cxy+by^2 \geq \frac{c^2}{b} x^2 + 2cxy+ b y^2
= \frac{1}{b}(c^2x^2+2bcxy+b^2 y^2) = \frac{1}{a}(cx+by)^2
$$
which will be positive except when $cx=-by$.
Now in case when $ax=-cy$ and $cx=-by$ both hold, we can multiply them to get $abxy = c^2xy$. Since $ab>c^2$, this forces $xy=0$ $\Rightarrow$ either $x=0$ or $y=0$. However, if $x=0$ but $y\neq 0$, then $cx=-by$ cannot hold (since $b>0$). Similarly, if $y=0$ but $x\neq 0$, then $ax=-cy$ cannot hold (since $a>0$). We conclude that if the equality in both cases above hold, then $x=y=0$, that is $X$ is the zero vector.
So the contrapositive says that if $X$ is a non-zero vector, then the quantity $ax^2+2cxy+by^2$ is positive.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1916249",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
}
|
Integral of exponential of trigonometric functions I'm looking for a proof of the result
$$
J_0 \left( x \sqrt{a^2 + b^2}\right) = \frac{1}{2\pi} \int_0^{2\pi} e^{ix \, (a \cos \theta \, + \, b \sin \theta)} \, \text{d} \theta
$$
$J_0$ is the Bessel function of order $0$.
|
We have that
$$a\cos\theta + b\sin\theta = \sqrt{a^2+b^2}\cos(\theta-\psi),\qquad \psi=\arctan\frac{b}{a}\tag{1}$$
hence
$$\frac{1}{2\pi}\int_{0}^{2\pi}e^{ix(a\cos\theta+b\sin\theta)}\,d\theta = \frac{1}{2\pi}\int_{0}^{2\pi}e^{ix\sqrt{a^2+b^2}\cos\theta}\,d\theta \tag{2}$$
and the claim follows by expanding $e^{ix\sqrt{a^2+b^2}z}$ as a Taylor series centered at $z=0$ and performing termwise integration, since:
$$ \int_{0}^{2\pi}\cos^{2n}(\theta)\,d\theta = \frac{2\pi}{4^n}\binom{2n}{n}.\tag{3}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1917245",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
Unable to justify solution for a problem with exponential constant $e$ involved. For the question,
$e^{2x} + e^x - 2 = 0.$
I was asked to solve for $x$.
What I performed,
$e^{2x} + e^x = 2.$
$e^x(e^x + 1) = 2.$
For the 2 solutions involved,
\begin{align}
e^x &= 2\\
\ln2 &= x\\
.69 &= x
\end{align}
OR
\begin{align}
e^x + 1 &= 2\\
e^x &= 1\\
\ln1 &= x\\
0 &= x
\end{align}
My textbook states a single solution which is $0$.
|
Let $y=e^{x}$. Then, your equation becomes
$$
y^{2}+y-2=0\implies (y+2)(y-1)=0.
$$
Thus, $y=-2$ or $y=1$. Hence, $e^{x}=-2$ or $e^{x}=1$. Since the first is impossible, it follows that $e^{x}=1\implies x=0$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1918064",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
}
|
Finding values to make a set a basis Question:
For which values of $a\in \Bbb R$ is the set below a basis for $M_{2x2}(\Bbb R)$ as a vector space over $\Bbb R$?
$${(a,2a,2,3a), (1,2,2a,3), (1,2a,a+1,a+2), (1,a+1,2,2a+1)}$$ (these vectors should be matrices but I do not know how to do them in MathJax)
My thinking is that in order to be a basis the set must be linearly independent. Thus the sum of the set of vectors and the set of scalar coefficients must equal zero where all scalar coefficients are equal to zero. If this is the case then any number could be $a$ without affecting the linear independence.
Is my thinking totally off? Can someone explain this problem if it is.
|
In order for the matrices presented to form a basis they need to be linearly independent. This means that for scalars $x_i$, the equation:
$$
x_1 \begin{pmatrix} a & 2a \\ 2 & 3a \end{pmatrix}
+ x_2 \begin{pmatrix} 1 & 2 \\ 2a & 3 \end{pmatrix}
+ x_3 \begin{pmatrix} 1 & 2a \\ a+1 & a+2 \end{pmatrix}
+ x_4 \begin{pmatrix} 1 & a+1 \\ 2 & 2a+1 \end{pmatrix}
= \begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix}
$$
should have only the trivial solution. To solve this, we need to solve the system:
$$
\begin{align}
ax_1 + x_2 + x_3 + x_4 &= 0 \\
2ax_1 + 2x_2 +2ax_3 +(a+1)x_4 &= 0 \\
2x_1 + 2ax_2 + (a+1)x_3 + 2x_4 &= 0 \\
3ax_1 +3x_2 +(a+2)x_3 + (2a+1)x_4 &= 0
\end{align}
$$
Ensuring that this system has only the trivial solution amounts to making sure that $\det{A} \neq 0$ where:
$$
A = \begin{pmatrix} a & 1 & 1 & 1 \\ 2a & 2 & 2a & a+1 \\ 2 & 2a & a+1 & 2 \\ 3a & 3 & a+2 &2a+1 \end{pmatrix}
$$
After finding the determinant of $A$ you can find which $a$ values result in $\det{A} \neq 0$ It is precisely these values of $a$ which result in the set being a basis.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1918599",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
}
|
Prove inequality $ (a+b+c)(a^7+b^7+c^7)\ge(a^5+b^5+c^5)(a^3+b^3+c^3)$ for $a,b,c\ge 0$ Prove that:
$$ (a+b+c)(a^7+b^7+c^7)\ge(a^5+b^5+c^5)(a^3+b^3+c^3)$$
I already know that this can be proven using Cauchy Schwarz, but I don't really see how to apply it here. I'm looking for hints.
|
Using Holder's inequality, we have that
\begin{align*}
a^5+b^5+c^5 &= a^{\frac{1}{3}}a^{\frac{14}{3}} + b^{\frac{1}{3}}b^{\frac{14}{3}}+c^{\frac{1}{3}}c^{\frac{14}{3}}\\
&\le (a+b+c)^{\frac{1}{3}}(a^7+b^7+c^7)^{\frac{2}{3}},
\end{align*}
and
\begin{align*}
a^3+b^3+c^3 &= a^{\frac{2}{3}}a^{\frac{7}{3}} + b^{\frac{2}{3}}a^{\frac{7}{3}}+c^{\frac{2}{3}}a^{\frac{7}{3}}\\
&\le (a+b+c)^{\frac{2}{3}}(a^7+b^7+c^7)^{\frac{1}{3}}.
\end{align*}
Multiplying them together, we are done.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1919468",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 2
}
|
Mistake in basic algebra, I think? Problem prove $(2n+1)+(2n+3)+(2n+5)+...+(4n-1) =3n^2$
Induction proof: base case $n=1$ assume true for all $n$ prove for $n+1$.
The $n$th or last term becomes $(4(n+1)-1)=4n+3$.
We also sub $n+1$ in for all $n$ the $n-1$ term is $(4n-1)$ and the first term is $2(n+1)+1=2n+3$
The right side is $3(n+1)^2 = 3(n^2 + 2n +1 ) $
Next thing is I appear to be missing the first term in the sum $(2n+1)$ on the left.
Adding $(2n+1)$ to both sides and subtracting $4n+3$ from both sides we get the $n$ case that equals $ 3n^2 $ on the left and
$$3n^2 + 6n +3 + 2n+1 -4n -3= 3n^2 + 4n+1$$
Which leaves me with $(2n+1)+(2n+3)+(2n+5)+...+(4n-1) =3n^2 +4n +1$
Which is approximately $4n+1$ on the right bigger than what I started with and is exactly the same on the left algebraically I must of done something wrong.
Which leads me to my question, what was it?
|
$\sum_\limits{k=1}^{n} (2(n+k)-1) = 3n^2$
Base case: $n = 1$
$(2(1+1) - 1) = 3$
Suppose:
$\sum_\limits{k=1}^{n} (2(n+k)-1) = 3n^2$
We will show that
$\sum_\limits{k=1}^{n+1} (2((n+1)+k)-1) = 3(n+1)^2$
based on the inductive hypothesis
$\sum_\limits{k=1}^{n+1} (2((n+1)+k)-1)\\
\sum_\limits{k=1}^{n} (2(n+1)+k)-1) + 4n+3\\
\big(\sum_\limits{k=1}^{n} (2(n+k)-1)\big)+\big(\sum_\limits{k=1}^{n}2\big)+ 4n+3\\
3n^2 + 2n + 4n+3\\
3(n+1)^2$
QED
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1919985",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
}
|
Prove: $\cos^3{A} + \cos^3{(120Β°+A)} + \cos^3{(240Β°+A)}=\frac {3}{4} \cos{3A}$ Prove that:
$$\cos^3{A} + \cos^3{(120Β°+A)} + \cos^3{(240Β°+A)}=\frac {3}{4} \cos{3A}$$
My Approach:
$$\mathrm{R.H.S.}=\frac {3}{4} \cos{3A}$$
$$=\frac {3}{4} (4 \cos^3{A}-3\cos{A})$$
$$=\frac {12\cos^3{A} - 9\cos{A}}{4}$$
Now, please help me to continue from here.
|
If $\cos3x=\cos3A$
$3x=n360^\circ\pm3A$ where $n$ is any integer
$x=n120^\circ+ A$ where $n=0,1,2$
As $\cos3x=4\cos^3x-3\cos x,$ the roots of $4c^3-3c-\cos3A=0$ are
$c_{n+1}=\cos(n120^\circ+ A)$ where $n=0,1,2$
Using Vieta's formula, $$c_1+c_2+c_3=0, c_1c_2c_3=\dfrac{\cos3A}4$$ $$\text{ and }c_1c_2+c_2c_3+c_3c_1=-\dfrac34$$
We need $c_1^3+c_2^3+c_3^3$
Now use $$a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1921191",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 8,
"answer_id": 7
}
|
Stars and bars to find "how many $x$ digit numbers are there with sum of digits $y$"? This question poses a seemingly very simple method to solve problems of the sort "how many $x$ digit numbers are there with sum of digits $y$?", but I don't understand it. Why are the "bad" solutions in correspondence to the solutions of $y_1 + 10 + x_2 + x_3 + x_4 + x_5 = 23$? What's the idea?
For instance, what "correct solution" corresponds to taking $y_1=13$? How to solve this problem?
|
The referred question asks for the number of non-negative integer solutions of
\begin{align*}
x_1+x_2+x_3+x_4+x_5=23
\end{align*}
with only one additional restriction $0\leq x_1\leq 9$. In this case we do not need the inclusion-exclusion principle.
In order to determine all integer solutions with
\begin{align*}
x_1+x_2+x_3+x_4+x_5=23\qquad\qquad0\leq x_1\leq 9,0\leq x_2,x_3,x_4,x_5\tag{1}
\end{align*}
we look at all integer solutions of
\begin{align*}
x_1+x_2+x_3+x_4+x_5=23\qquad\qquad\qquad\qquad 0\leq x_1,x_2,x_3,x_4,x_5\tag{2}
\end{align*}
and subtract all solutions of
\begin{align*}
x_1+x_2+x_3+x_4+x_5=23\qquad\qquad\qquad x_1\geq 10, 0\leq x_2,x_3,x_4,x_5\tag{3}
\end{align*}
The solutions of (3) are the so-called bad solutions in the referred question, meaning the invalid solutions which are to subtract when determining the solutions of (2).
In order to calculate (2) with the nice range $x_1,x_2,x_3,x_4,x_5\geq 0$ we can use the stars-and-bars
technique and obtain
\begin{align*}
\binom{23+4}{4}=\binom{27}{4}=17550
\end{align*}
In order to calculate (3) we transform the range by a proper substitution which enables us to apply the stars-and-bars technique again. Instead of $x_1\geq 10$, we substitute $x_1=y_1+10$ and the range $x_1\geq 10$ can then be transformed to $y_1+10\geq 10$ or equivalently $y_1\geq 0$.
This way we obtain from (3)
\begin{align*}
x_1+x_2+x_3+x_4+x_5&=23\qquad\qquad\qquad x_1\geq 10, 0\leq x_2,x_3,x_4,x_5\\
(y_1+10)+x_2+x_3+x_4+x_5&=23\qquad\qquad\qquad 0\leq y_1,x_2,x_3,x_4,x_5\\
y_1+x_2+x_3+x_4+x_5&=13
\end{align*}
We are now in the same situation as in (2) and can apply the stars-and-bars technique again.
\begin{align*}
\binom{13+4}{4}=\binom{17}{4}=2380
\end{align*}
$$ $$
We finally obtain the number of wanted solutions as the total number of solutions of (2) minus the number of solutions of (3)
\begin{align*}
\binom{27}{4}-\binom{17}{4}=17550-2380=15170
\end{align*}
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1922623",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
}
|
Proving this identity: $\frac{\sin^2\theta}{\cos^2\theta} + \frac{\cos^2\theta}{\sin^2\theta} = \sec^2\theta \csc^2\theta - 2$ I have tried solving this trig. identity, but I get stuck when it comes to the $-2$ part. Any suggestions?
$$\frac{\sin^2\theta}{\cos^2\theta} + \frac{\cos^2\theta}{\sin^2\theta} = \sec^2\theta \csc^2\theta - 2$$
|
hint: $\dfrac{a^2}{b^2} + \dfrac{b^2}{a^2} = \dfrac{a^4+b^4}{a^2b^2}= \dfrac{(a^2+b^2)^2}{a^2b^2}-2$. Apply this identity for $a = \cos \theta, b = \sin \theta$ to get the desired identity.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1923555",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 1
}
|
Prove that $\binom{k}{p} \equiv \left\lfloor \frac{k}{p} \right\rfloor \pmod p$ for all odd prime number $p$ and $k\ge 3$ I don't really know how to start this exercise. Do I have to use p-adic valuation ?
If it's the case it will give $\nu_p(\binom{k}{p})= \sum \limits_{r=1}^{\infty}\left(\left\lfloor \frac{k}{p^r} \right\rfloor-\left\lfloor \frac{p}{p^r} \right\rfloor -\left\lfloor \frac{k-p}{p^r} \right\rfloor \right)$.
Thanks in advance !
|
Let us define $jp$, with $j$ positive integer, the greatest multiple of $p$ lower than $k$. Now let us consider the definition of $\binom{k}{p}$ as $$\frac {k (k-1)(k-2)(k-3)...[k-(p-1)]}{p!}$$ The numerator is formed by $p$ consecutive numbers and can be expressed as a product of factors that are $\equiv i \pmod p$, where $i$ assumes all integer values between $0$ (this is the case of $jp$) and $p-1$. Because $1 \cdot 2 \cdot 3....\cdot (p-1) \equiv -1 \pmod p$, the whole numerator can be written as the product of $jp$ and a number $\equiv -1 \pmod p$. As a result, dividing the numerator by $p $ we get a number that is $\equiv -j \pmod p$.
On the other hand, the denominator $p!=p(p-1)!$ is the product of $p$ and a quantity $\equiv -1 \pmod p$, so that dividing it by $p$ we get a number that is $\equiv -1 \pmod p$.
Based on the considerations above, $\binom{k}{p}$ is given by a fraction where, simplifying both numerator and denominator by $p$, we are left with a ratio between a number that is $ \equiv -j \pmod p $ and another one that is $ \equiv -1 \pmod p $. It follows that the result of the fraction is $ \equiv j \pmod p $. Because $j=\lfloor k/p \rfloor $, we finally obtain
$$\binom{k}{p} \equiv \left\lfloor \frac{k}{p} \right\rfloor \pmod p$$
EDIT: since a comment asked to better explain the calculations for the numerator, I added an example below.
Let us set, for example, $k=7$ and $p=3$. So we have $jp=6$ and $j=2$. The binomial coefficient is given by the ratio
$$\frac {7 (7-1)(7-2)}{3!} $$
The numerator is formed by $3$ consecutive numbers and can be expressed as a product of factors that are $\equiv i \pmod 3$, where $i$ assumes all integer values between $0$ and $p-1=2$. In fact, the numerator can be written as $6(6+1)(3+2) $. The product of these factors, not considering $jp=6$, is $$(p-1)! \mod p = (3-1)! \pmod 3=2 \pmod 3$$
(note that we can also consider this last number as $-1 \pmod p $, although it is not necessary). Then, the numerator can be written as the product of $jp=6$ and a number $\equiv 2 \pmod 3$, i.e. it is $\equiv jp (p-1)! \pmod p=\equiv (6 \cdot 2) \pmod 3$.
The denominator is $3! $ and then can be written as the product of $p=3$ and $(p-1)!=2$, where this last number is $\equiv (p-1)! \pmod p = \equiv 2 \pmod 3$.
Then the initial binomial coefficient is the ratio of a quantity $ \equiv (6 \cdot 2) \pmod 3$ and another quantity that is $ \equiv (3 \cdot 2) \pmod 3$. Their ratio is then a number $\equiv 2 \pmod 3$, where the number $2$ simply reflects the initial value of $j$, which in turn is equal to $ \left\lfloor \frac{7}{3} \right\rfloor$. We then conclude that
$$\binom{7}{3} \equiv \left\lfloor \frac{7}{3} \right\rfloor \pmod 3$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1924309",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
}
|
How do I prove that for $ x,y,z > 0$ $\frac{x}{y+z} + \frac{y}{x+z} + \frac{z}{x+y} > 3/2$? I'm solving a graduate entrance examination problem.
We are required to establish the inequality using the following result:
for $x,y > 0$, $\frac{x}{y} + \frac{y}{x} > 2$ (1), which is easy to prove as it is equivalent to $(x - y)^2 > 0$.
But when it comes to an inequality combining $x, y, z$, I got stuck as I've tried to develop the expression into one single fraction and obtain something irreducible.
Any hints ? My intuition tells me that for $x,y,z >0$, any fraction of the form $\frac{x}{y+z}$ is greater than 1/2. As there are three fractions of this kind with mute variables playing symmetrical roles, we get: $1/2 + 1/2 + 1/2 = 3/2$.
I just don't figure out how to play with the result (1).
|
Apply your formula three times:
$$\frac{x+y}{x+z}+\frac{x+z}{x+y}>2$$
$$\frac{y+x}{y+z}+\frac{y+z}{y+x}>2$$
$$\frac{z+x}{z+y}+\frac{z+y}{z+x}>2$$
Combining gives:
$$\frac{x+y}{x+z}+\frac{x+z}{x+y}+\frac{y+x}{y+z}+\frac{y+z}{y+x}+\frac{z+x}{z+y}+\frac{z+y}{z+x}>6$$
$$\frac{x+y+2z}{x+y}+\frac{x+2y+z}{x+z}+\frac{2x+y+z}{y+z}>6$$
$$1+\frac{2z}{x+y}+1+\frac{2y}{x+z}+1+\frac{2x}{y+z}>6$$
$$2\left(\frac{z}{x+y}+\frac{y}{x+z}+\frac{x}{y+z}\right)>3$$
$$\frac{z}{x+y}+\frac{y}{x+z}+\frac{x}{y+z}>\frac{3}{2}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1928892",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 0
}
|
Approximation of square root I have a square root in a problem which needs to be approximated. I'm not entirely sure how to do this algebraically.
$$ \sqrt{10^2-(6.9\times 10^{-2})^2}$$
The answer the problem is proposing as the approximation is $$ 10[1-\frac 1 2(6.9)^2\times10^{-6}]$$
This hasn't exactly been the most reputable textbook, however, so it could be wrong.
My attempt:
It should be able to approximated by the square-root of 100, because the other number is so small.
$$ 10 -(6.9)^2\times 10^{-4} $$
I can see they factored a 10 out, so I go ahead and do that.
$$ 10[1- \frac 1 {10}(6.9)^2\times 10^{-4}] $$
$$ 10[1-(6.9)^2\times 10^{-5}] $$
I'm not sure where exactly I'm slipping up, or what I'm comprehending incorrectly, any help would be greatly appreciated.
|
Use the Taylor series for $(1+x)^{1/2}$.
First we rewrite $(A+B)^{\alpha}$ to be in a form similar to this :
$(A+B)^{\alpha} = A^{\alpha}(1 + \frac{B}{A})^{\alpha}$.
With $A \gg B$ the ratio $\frac{B}{A}$ is "small" and we do a Taylor series for $(1+x)^{\alpha}$
$$(1+x)^{\alpha} = 1 + \frac{\alpha}{1!}x + \frac{\alpha(\alpha-1)}{2!}x^{2} + \ldots$$
and now replace each $x$ with $\frac{B}{A}$ to get
$$(A+B)^{\alpha} = A^{\alpha}\left( 1 + \frac{\alpha}{1!} \frac{B}{A} + \frac{\alpha(\alpha-1)}{2!}\left( \frac{B}{A} \right)^{2} + \ldots \right)$$
Putting in $\alpha = \frac{1}{2}$ gives
$$(A+B)^{\frac{1}{2}} = A^{\frac{1}{2}}\left( 1 + \frac{1}{2} \frac{B}{A} - \frac{1}{8}\left( \frac{B}{A} \right)^{2} + \ldots \right)$$
In your case $A = 10^{2}$ and $B = -(0.69)^{2}$ so $A^{\frac{1}{2}} = 10$ and $\frac{B}{A} = -(0.069)^{2} = -(6.9 \times 10^{-3})^{2} = -0.00004761...$
Therefore
$$\begin{align}
(A+B)^{\frac{1}{2}} &= \sqrt{10^{2}}\Big( 1 + \frac{1}{2}\frac{-(0.69)^{2}}{10^{2}} + \ldots \Big) = 10\left( 1 - \frac{0.00004761}{2} + \ldots \right)
\\&\approx 10*(1-0.000023805) = 9.99976195
\end{align}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1929331",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
}
|
Show that $\frac 1x \ge 3 - 2\sqrt{x}$ for all positive real numbers $x$. Describe when we have equality. Show that $\frac 1x \ge 3 - 2\sqrt{x}$ for all positive real numbers $x$. Describe when we have equality.
I tried manipulating the equation but nothing helps. Any answer is greatly appreciated.
|
$$
\frac{1}{x} \geq 3 - 2\sqrt{x} \iff \frac{1+2x\sqrt{x} -3x}{x} \geq 0 \iff \frac{(\sqrt{x}-1)^2(\sqrt{x} + \frac{1}{2})}{x} \geq 0
$$
We can cancel off the $(\sqrt{x}-1)^2$, and the $x$ on the bottom, since $x$ must be positive, and squares are always positive. This leaves us with $\sqrt{x} + \frac{1}{2} \geq 0$, which is true for all positive $x$. Hence this inequality also applies for all positive $x$.
Equality will apply when $x=1$, since $\sqrt{x}+\frac{1}{2}$ is strictly positive, and $\frac{1}x$ is strictly positive for $x \neq 0$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1929829",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
}
|
$ax^3+8x^2+bx+6$ is exactly divisible by $x^2-2x-3$, find the values of $a$ and $b$ Find the values of $a$ and $b$ for which the polynomial $ax^3+8x^2+bx+6$ is divisible by $x^2-2x-3$.
|
by using the long division we get
$$\frac{ax^3+8x^2+bx+6}{x^2-2x-3}=ax+(8+2a)+\frac{(7a+b+16)x+6a+30}{x^2-2x-3}$$
now the reminder should be zero
$$7a+b+16=0\tag 1$$
$$6a+30=0\tag2$$
$$a=-5, b=19$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1930266",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 9,
"answer_id": 0
}
|
Proof of the inequality $2^{n} < \binom{2n}{n} < 2^{2n}$? As review for a midterm I am asked to prove the inequality:
$2^{n} < \binom{2n}{n} < 2^{2n}, n > 1.$
What I have is a two-part inductive proof. It is not hard to show for $2^{n} < \binom{2n}{n}$:
Base step:
Let $n=2$:
$2^{2} < \frac{(2n)!}{2!2!} < 2^{2*2}$
$4 < 6 < 16$
Inductive Step:
Show $2^{k+1} < \frac{[2(k+1)]!}{(k+1)!(k+1)!}.$
We have
$2^{k+1} = 2^{k}*2,$
so
$2^{k+1} < 2 * \frac{2k!}{k!k!}.$
Since
$\frac{(2k+2)!}{(k+1)*k!*(k+1)*k!} = \frac{2(k+1)*(2k+1)*2k!}{(k+1)k!(k+1)k!} = (2k+1) * \frac{2k!}{k!k!}$
and
$(2k+1) > 2, k \geq 2,$
we can conclude the first part of the inequality. However, I can't make the second part work. By similar algebra I arrive at:
$(2k+1) * \frac{2k}{k!k!} < 2^{2(k+1)} = 2^{2k}*2^{2},$
but
$ (2k+1) \nless 4, k \geq 2.$
What have I done wrong?
|
You may avoid induction. We have
$$ \binom{2n}{n}=\frac{2n}{n}\cdot\frac{2n-1}{n-1}\cdot\ldots\cdot\frac{n+1}{2}>2^n $$
since every factor of the middle term, except the very first one, is greater than $2$.
On the other hand,
$$ \binom{2n}{n}<\sum_{k=0}^{2n}\binom{2n}{k} = 2^{2n} = 4^n $$
by the binomial theorem.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1932085",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
}
|
If $\cos A=\tan B$, $\cos B=\tan C$ β¦ If $\cos A=\tan B$, $\cos B=\tan C$ and $\cos C=\tan A$, prove that $\sin A=\sin B=\sin C$.
My Attempt.
Let us consider $x$, $y$ and $z$ as:.
$$x = \tan^2A$$
$$y = \tan^2B$$
$$z = \tan^2C$$
$$\cos^2A = \tan^2B$$
$$\frac {1}{\sec^2A}= \tan^2B$$
$$\frac {1}{1 + \tan^2A} = \tan^2B$$
$$\frac {1}{1 + x} = y$$
$$(1 + x)y = 1\tag{1}$$
Similarly,
$$(1 + y)z = 1\tag{2}$$
$$(1 + z)x = 1\tag{3}$$
Please help me to continue from here.
|
$$\tan^2C=\cos^2B=\dfrac1{1+\tan^2B}=\dfrac1{1+\cos^2A}=\dfrac1{2-\sin^2A}$$
$$\cos^2C=\tan^2A$$
$$\implies\tan^2C=\sec^2C-1=\dfrac{1-\cos^2C}{\cos^2C}=\dfrac{1-\tan^2A}{\tan^2A}=\dfrac{1-2\sin^2A}{\sin^2A}$$
Equating the values of $\tan^2C$ and writing $\sin^2A=x$
$$\dfrac1{2-x}=\dfrac{1-2x}x\implies x^2-3x+1=0\implies x=\dfrac{3-\sqrt5}2$$
As $0\le x\le1$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1935885",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
}
|
$\lim_{z\to 0} \frac{z^2-\overline{z}}{|z|}$ $$\lim_{z\to 0} \frac{z^2-\overline{z}^2}{|z|}$$
If we suppose $z = x+iy$, then the limit becomes:
$$\lim_{(x,y)\to (0,0)}\frac{x^2-y^2+2xyi-(x-yi)^2}{\sqrt{x^2+y^2}} = \lim_{(x,y)\to(0,0)}\frac{x^2-y^2+2xyi-x^2+2xyi+y^2}{\sqrt{x^2+y^2}} = \lim_{(x,y)\to(0,0)}\frac{4ixy}{\sqrt{x^2+y^2}} = \infty$$
Am I right?
|
HINT:
$$\left|\frac{xy}{\sqrt{x^2+y^2}}\right|\le \frac12\sqrt{x^2+y^2}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1936086",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
0.1999... and 0.2000 are the same? From Sipser's Introduction to the Theory of Computation:
A slight problem arises because certain numbers, such as 0.1999... and 0.2000..., are equal even though their decimal representations are different.
Can someone explain? Because that's news to me.
|
We can write
$$\begin{align}
0.1999 \cdots &= 0.1+0.09+0.099+ 0.999\cdots \\
&= \dfrac{1}{10}+\dfrac{9}{100}+\dfrac{9}{1000}+\dfrac{9}{10000} + \cdots\\
&= \dfrac{1}{10}+\dfrac{9}{10^2}+\dfrac{9}{10^3}+\cdots \\
&= \dfrac{1}{10}+9\left(\dfrac{1}{10^2}+\dfrac{1}{10^3}+\cdots\right) \\
&= \dfrac{1}{10}+\dfrac{9}{10}\left(\dfrac{1}{10}+\dfrac{1}{10^2}+\cdots\right) \\
&= \dfrac{1}{10}+\dfrac{9}{10}\left[\dfrac{1/10}{1-1/10} \right]
\end{align}$$
since
$$\dfrac{1}{10}+\dfrac{1}{10^2}+\cdots = \sum_{k=1}^{\infty}\left(\dfrac{1}{10}\right)^k = \dfrac{1/10}{1-1/10}$$
as given here (see equation (9)). The denominator "cancels out" the $9/10$ since $1 - 1/10 = 9/10$, and so we end up with $$\dfrac{1}{10}+\dfrac{1}{10}=\dfrac{2}{10}=0.2\text{.}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1936159",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
}
|
Series and probability The task at hand:
Suppose there are 3 players (A,B,C), who have 3 different dices with the following chances (per roll) to win:
Player A has a dice with a $\frac{1}{3}$ chance to win
Player B has a dice with a $\frac{2}{5}$ chance to win
Player C has a dice with a $\frac{4}{7}$ chance to win
Player A starts rolling the dice, followed by Player B, then C and then starting with A all over again until one player wins.
Calculate the chance of player A to win.
My question is wether the following approach is valid or has a logical error:
$\sum_{n=0}^\infty (\frac{2}{3}\frac{3}{5}\frac{3}{7})^n(\frac{1}{3}) = \frac{1}{3}\sum_{n=0}^\infty (\frac{2\cdot3\cdot3}{3\cdot5\cdot7})^n= \frac{1}{3}\sum_{n=0}^\infty (\frac{18}{105})^n= \frac{1}{3}\frac{1}{1-(\frac{18}{105})}= \frac{1}{3}\frac{1}{(\frac{87}{105})} =\frac{105}{261} \sim 0.4023$
|
Let $x$ be the probability that A wins. So $x$ equals P(A rolls a win) + P(B does not roll a win and C does not roll a win)*$x$. In other words:$$x = \frac{1}{3} + \frac{2}{3}\frac{3}{5}\frac{3}{7}x$$
and so
$$x = \frac{35}{3*29} \approx 0.4023.$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1937593",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
}
|
How to prove $\sin 10^\circ = \frac{-1+\sqrt{9-8\sin 50^\circ}}{4}$? How to prove this identity?
$$\sin 10^\circ = \frac{-1+\sqrt{9-8\sin 50^\circ}}{4}$$
Is this a particular case of a more general identity? Also, is it possible to give a geometric proof of this equality?
|
From the expression given for $\sin 10^{\circ}$ it is obvious that it is a root of a quadratic equation. Comparing $$\frac{-1 + \sqrt{9 - 8\sin 50^{\circ}}}{4}$$ with $$\frac{-b + \sqrt{b^{2} - 4ac}}{2a}$$ we can see that it almost fits with $a = 2, b = 3, c = \sin 50^{\circ}$ and we have then $$\frac{-1 + \sqrt{9 - 8 \sin 50^{\circ}}}{4} = \frac{1}{2} + \frac{-b + \sqrt{b^{2} - 4ac}}{2a}$$ Our job is now complete if we can show that $\alpha = \sin 10^{\circ} - (1/2) = \beta - (1/2)$ is a root of the equation $$ax^{2} + bx + c = 0$$ We have then $$a(\beta - 1/2)^{2} + b(\beta - 1/2) + c = 0$$ or $$4a\beta^{2} + 4(b - a)\beta + a - 2b + 4c = 0$$ or $$2\sin^{2}10^{\circ} + \sin 10^{\circ} + \sin 50^{\circ} = 1$$ or $$2\sin^{2}10^{\circ} + 2\sin 30^{\circ}\cos 20^{\circ} = 1$$ or $$2\sin^{2}10^{\circ} = 1 - \cos 20^{\circ}$$ which is true via the identity $1 - \cos A = 2\sin^{2}(A/2)$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1939066",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
}
|
Calculate $\int{\frac{x^\alpha}{(x^2+a^2)^{\frac{3}{2}}}} \ \mathrm{d}x$ for $\alpha = 1,2,4,6$? What is the most straightforward way to calculate the following indefinite integral? $$\int{\frac{x^\alpha}{(x^2+a^2)^{\frac{3}{2}}}}\ \mathrm{d}x,$$
for $\alpha = 1, 2, 4,5,6$?
|
HINTS:
*
*When $\alpha=1$, substitute $u=x^2+\text{a}^2$ and $\text{d}u=2x\space\text{d}x$:
$$\mathcal{I}_1(\text{a},x)=\int\frac{x}{\left(x^2+\text{a}^2\right)^{\frac{3}{2}}}\space\text{d}x=\frac{1}{2}\int\frac{1}{u^{\frac{3}{2}}}\space\text{d}u=-\frac{1}{\sqrt{u}}+\text{C}=\text{C}-\frac{1}{\sqrt{x^2+\text{a}^2}}$$
*When $\alpha=2$, substitute $u=x^2+\text{a}^2$ and $\text{d}u=2x\space\text{d}x$:
$$\mathcal{I}_2(\text{a},x)=\int\frac{x^2}{\left(x^2+\text{a}^2\right)^{\frac{3}{2}}}\space\text{d}x=\int\frac{1}{\sqrt{x^2+\text{a}^2}}\space\text{d}x-\text{a}^2\int\frac{1}{\left(x^2+\text{a}^2\right)^{\frac{3}{2}}}\space\text{d}x$$
Use:
*
*Assume that $\text{a}\ne0$:
$$\int\frac{1}{\sqrt{x^2+\text{a}^2}}\space\text{d}x=\ln\left|x+\sqrt{x^2+\text{a}^2}\right|+\text{C}$$
*$$\int\frac{1}{\left(x^2+\text{a}^2\right)^{\frac{3}{2}}}\space\text{d}x=\frac{x}{\text{a}\sqrt{1+\frac{x^2}{\text{a}^2}}}+\text{C}$$
So:
$$\mathcal{I}_2(\text{a},x)=\ln\left|x+\sqrt{x^2+\text{a}^2}\right|+\frac{x}{\text{a}\sqrt{1+\frac{x^2}{\text{a}^2}}}+\text{C}$$
3. When $\alpha=4$, substitute $x=\text{a}\tan(u)$ and $\text{d}x=\text{a}\sec^2(u)\space\text{d}u$, after that substitute $v=\sin(u)$ and $\text{d}v=\cos(u)\space\text{d}u$:
$$\mathcal{I}_4(\text{a},x)=\int\frac{x^4}{\left(x^2+\text{a}^2\right)^{\frac{3}{2}}}\space\text{d}x=\text{a}^2\int\frac{\tan^2(u)}{\sec(u)}\space\text{d}u=\text{a}^2\int\frac{v^4}{\left(v^2-1\right)^2}\space\text{d}v$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1944931",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
}
|
Finding dy/dx by implicit differentiation with the quotient rule I've been stuck on a certain implicit differentiation problem that I've tried several times now.
$$
\frac{x^2}{x+y} = y^2+6
$$
I know to take the derivatives of both sides and got:
$$
\frac{(x+y)2x-\left(1-\frac{dy}{dx}\right)x^2}{(x+y)^2} = 2y\frac{dy}{dx}
$$
I reduced that to get:
$$2x^2 +2xy-x^2-x^2*dy/dx=(2y*dy/dx)(x+y)^2$$
I then divided both sides by (2y*dy/dx) and multiplied each side by the reciprocals of the first three terms of the left side. Then I factored dy/dx out of the left side and multiplied by the reciprocal of what was left to get dy/dx by itself. I ended up with:
$$dy/dx=(2y(x+y)^2)/(4x^7y)$$
but this answer was wrong. I only have one more attempt on my online homework and I can't figure out where I went wrong. Please help!
|
An idea to avoid the cumbersome and annoying quotient rule: multiply by the common denominator
$$\frac{x^2}{x+y}=y^2+6\implies xy^2+6x+y^3+6y-x^2=0\implies$$
$$y^2+2xyy'+6+3y^2y'+6y'-2x=0\implies(2xy+3y^2+6)y'=2x-y^2-6\implies$$
$$y'=\frac{2x-y^2-6}{2xy+3y^2+6}$$
If nevertheless you want to use the quotient rule:
$$\frac{2x(x+y)-x^2}{(x+y)^2}-\frac{x^2}{(x+y)^2}y'=2yy'\implies$$
$$(2y(x+y)^2+x^2)y'=x^2+2xy\implies y'=\frac{x^2+2xy}{2y(x+y)^2+x^2}$$
Now, how come both expressions we got are equal?. Well, for one we can use, for example, that $\;\cfrac{x^2}{x+y} = y^2+6\;$ , so
$$\frac{x^2+2xy}{2y(x+y)^2+x^2}=\frac{2x-y^2-6}{2xy+3y^2+6}=\frac{2x-\frac{x^2}{x+y}}{2xy+2y^2+\frac{x^2}{x+y}}\iff$$
$$\frac{x^2+2xy}{2y(x+y)^2+x^2}=\frac{x^2+2xy}{2x^2y+4xy^2+2y^3+x^2}\iff$$
$$\frac{x^2+2xy}{2x^2y+4xy^2+2y^3+x^2}=\frac{x^2+2xy}{2x^2y+4xy^2+2y^3+x^2}\;\;\color{green}\checkmark$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1945594",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
Solving this $4$ variable diophantine equation Is there a way to solve this Diophantine equation in $a,b,c,d$? $$19a^3-33a^2b+3a^2c+30a^2d+21ab^2+24abc-12abd-15ac^2-54acd-30ad^2+ $$ $$2b^3-12b^2c-6b^2d+42bc^2+108bcd+60bd^2-7c^3-51c^2d-99cd^2-56d^3=0$$
Wolfram Alpha unfortunately cannot understand my input whenever I input this equation, and I know no strategy to solve these kinds of equations. So basically, I'm stuck...
I want just all possible values for $a,b,c,d$ just like what Wolfram Alpha does.
Note: $a\neq b\neq c\neq d\neq 0$
|
Your equation is homogeneous of degree $3$, so if $(a,b,c,d)$ is a solution so is $(ta,tb,tc,td)$ for any $t$. Thus it makes sense to look for primitive solutions, which are solutions with greatest common divisor $1$. I found some $288$ primitive solutions with $a,b,c \in [-20\ldots -1, 1\ldots 20]$ and $d \in [1 \ldots 20]$. I don't see an obvious pattern.
Here are some of those solutions:
$$ \matrix{ a & b & c & d\cr
-4 & -1 & -8 & 1\cr
-4 & 1 & -8 & 3\cr
-3 & 1 & -7 & 3\cr
-3 & 1 & 3 & 1\cr
-3 & 5 & -1 & 4\cr
-2 & 1 & -6 & 3\cr
-2 & 1 & -4 & 2\cr
-2 & 2 & -4 & 3\cr
-2 & 3 & -4 & 4\cr
-2 & 3 & 2 & 1\cr
-2 & 4 & -4 & 5\cr
-2 & 5 & -4 & 6\cr
-1 & -1 & -5 & 4\cr
-1 & 1 & -5 & 3\cr
-1 & 2 & -3 & 3\cr
-1 & 5 & 1 & 1\cr
1 & -2 & -1 & 2\cr
1 & -1 & -5 & 2\cr
1 & 1 & -3 & 3\cr
1 & 2 & -7 & 6\cr
2 & 1 & -2 & 3\cr
2 & 2 & 4 & 1\cr
2 & 3 & 4 & 2\cr
2 & 4 & 4 & 3\cr
2 & 5 & 4 & 4\cr
2 & 6 & 4 & 5\cr
3 & -4 & -7 & 1\cr
3 & 1 & -1 & 3\cr
3 & 2 & -5 & 6\cr
3 & 4 & -3 & 4\cr
4 & 3 & 8 & 1\cr
5 & 1 & 1 & 3\cr
5 & 2 & -3 & 6\cr
6 & 1 & 2 & 3\cr
7 & 1 & 3 & 3\cr
7 & 2 & -1 & 6\cr
8 & 1 & 4 & 3\cr
}$$
Did those coefficients come from somewhere in particular or are they just arbitrary?
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1945775",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
}
|
Find limit of sequence sum Let us have sum of sequence (I'm not sure how this properly called in English): $$X(n) = \frac{1}{2} + \frac{3}{4}+\frac{5}{8}+...+\frac{2n-1}{2^n}$$
We need
$$\lim_{n \to\infty }X(n)$$
I have a solution, but was unable to find right answer or solution on the internet.
My idea:
This can be represented as $$ \frac{1}{2} + \frac{1}{4} + \frac{2}{4} + \frac{3}{8}+\frac{2}{8}+\frac{5}{16} + \frac{2}{16} ... + ...$$
Which is basically
$\frac{1}{2} + \frac{1}{2} +\frac{1}{4}+\frac{1}{8}$ - 1/2 + geometric progression + our initial sum divided by 2.
And then I thought: hey, so I can figure out one part of this sum, and second is twice smaller, and then it forms a cycle! (I suppose).
So it would be $B_1 = 1/2 + b_1/(1-1/2) = 3/2$
$$lim_{n \to\infty }X(n) = B_1/(1-1/2) = 3$$
Is this correct?
|
Let
\begin{align*}
f(x) &= \sum_{n=1}^{\infty} \frac{x^{2n-1}}{2^n}\\
&=\frac{x}{2}\sum_{n=0}^{\infty}\left(\frac{x^2}{2}\right)^n\\
&=\frac{x}{2-x^2}.
\end{align*}
Then
\begin{align*}
f'(x) = \frac{2+x^2}{(2-x^2)^2}.
\end{align*}
Therefore,
\begin{align*}
\lim_{n\rightarrow \infty}X(n) = f'(1)=3.
\end{align*}
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1947082",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
}
|
The system of equations: $\begin{cases} 2x^2=4y^2+3z^2+2; \\ 13x=4y+3z+29 \end{cases}$ Solve in positive integers the system of equations:
$$\begin{cases}
2x^2=4y^2+3z^2+2;
\\
13x=4y+3z+29
\end{cases}$$
My work so far:
I used wolframalpha: $x=3,y=1,z=2$.
|
$$2x^2=4y^2+3z^2+2\tag1$$
$$13x=4y+3z+29\tag2$$
From $(1)(2)$,
$$13^2(4y^2+3z^2+2)=2(13x)^2=2(4y+3z+29)^2,$$
i.e.
$$644y^2+(-464-48z)y+489z^2-348z-1344=0$$
See this as a quadratic equation on $y$.
Since the discriminant has to be larger than or equal to $0$, we have to have
$$(-464-48z)^2-4\cdot 644(489z^2-348z-1344)\ge 0,$$
i.e.
$$\small -1.5\approx -\frac{236}{155}=\frac{2(87-49\sqrt{81})}{465}\le \frac{2(87-49\sqrt{69})}{465}\le z\le \frac{2(87+49\sqrt{69})}{465}\le \frac{2(87+49\sqrt{81})}{465}=\frac{352}{155}\approx 2.3$$
giving $$z=1,2$$
I think that you can continue from here.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1953480",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
}
|
Find all functions $f:\mathbb Z \rightarrow \mathbb Z$ such that $f(0)=2$ and $f\left(x+f(x+2y)\right)=f(2x)+f(2y)$
Find all functions $f:\mathbb Z \rightarrow \mathbb Z$ such that $f(0)=2$ and
$$f\left(x+f(x+2y)\right)=f(2x)+f(2y)$$
for all $x \in \mathbb Z$ and $y \in \mathbb Z$
My work so far:
1) $x=0$ $$f\left(f(2y)\right)=f(2y)+2$$
2) $y=0$ $$f\left(x+f(x)\right)=f(2x)+2$$
3) Let $n\ge 0$. Use induction we have
If $f(2n)=2n+2$ then $$f(f(2n))=f(2n+2)=2n+2+2=2n+4$$
Hence, if $k=2m\ge0$ then $f(k)=k+2$
4) $n<0$
I need help here
|
Supposing that we already know that $f(2n) = 2n + 2$ for $n \geq 0$, we can proceed as follows.
For some positive integer $k$, let $x = 2k$ and $y = -k$ in the functional equation. We get that
$$
f(2k + 2) = f( 2k + f(2k - 2k) ) = f(4k) + f(-2k).
$$
Now $2k + 2$ and $4k$ are positive even integers, and so this tells us that
$$
2k + 4 = 4k + 2 + f(-2k),
$$
and hence
$$
f(-2k) = -2k + 2.
$$
Thus $f(2n) = 2n + 2$ holds for all integers $n$.
We know show that $f$ sends odd integers to odd integers. Suppose to the contrary that there are some integers $a$ and $b$ such that
$$
f(2a - 1) = 2b.
$$
Take $x = -1$ in the functional equation to get that for any integer $y$, that
$$
f(f(2y - 1) - 1) = f(-2) + f(2y) = 2y + 2.
$$
This gives us that
$$
\begin{align*}
f(2a - 1) & = 2b & f(2b - 1) & = f(f(2a - 1) - 1) = 2a + 2 \\
f(2a + 1) & = f(f(2b - 1) - 1) = 2b + 2 & f(2b + 1) & = f(f(2(a+1)-1)-1) = 2a + 4 \\
f(2a + 3) & = f(f(2(b+1)-1)-1) = 2b + 4 & f(2b + 3) & = f(f(2(a+2)-1)-1) = 2a + 6 \\
& \vdots & & \vdots
\end{align*}
$$
and inductively, we obtain that for every non-negative integer $k$, we have that
$$
f(2a + 2k - 1) = 2b + 2k \quad\text{ and }\quad f(2b + 2k - 1) = 2a + 2k + 2.
$$
Now suppose that $b \leq a$. Then there is some $k \geq 0$ such that $a = b + k$, which gives us that
$$
2b = f(2a - 1) = f(2b + 2k - 1) = 2a + 2k + 2 = 2b + 4k + 2,
$$
which implies that $4k + 2 = 0$, a contradiction. Similarly, if $b > a$, then there is some $k \geq 0$ such that $b = a + k$. We then obtain that
$$
2a + 2 = f(2b - 1) = f(2a + 2k - 1) = 2b + 2k = 2a + 4k,
$$
which gives us that $4k = 2$, again a contradiction.
We conclude that $f(2y - 1)$ is odd for every integer $y$. Thus for every integer $y$, we have that $f(2y - 1) - 1$ is even, and hence we have that
$$
2y + 2 = f(f(2y - 1) - 1) = f(2y - 1) + 1
$$
and hence that
$$
f(2y - 1) = 2y + 1
$$
for every integer $y$.
We see that $f(n) = n + 2$ for every integer $n$, and can verify that this does indeed solve the functional equation.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1954840",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 0
}
|
Sum of series $1+\frac{1}{4}+\frac{1\times3}{4\times8}+\frac{1\times3\times5}{4\times8 \times12}+\cdots$
Sum of series :
$$1+\frac{1}{4}+\frac{1\times3}{4\times8}+\frac{1\times3\times5}{4\times8 \times12}+\cdots$$
$n$-th term of series is
$$a_{n} =\frac{1\times3 \times5\times\cdots \times(2n-1)}{4\times8\times12\times \cdots \times4n} = \prod^{n}_{k=1}{2k-1\over4k}$$
I can not go further,
|
Note, the general term $a_n, n\geq 1$ is
\begin{align*}
a_n=\frac{1\cdot 3\cdot 5\cdots (2n-1)}{4\cdot8\cdot 12\cdots (4n)}
=\frac{1\cdot 3\cdot 5\cdots (2n-1)}{(1\cdot 2\cdot 3\cdots n)\cdot 4^n}
=\frac{(2n-1)!!}{n!}\cdot \frac{1}{4^n}\\
\end{align*}
with $$(2n-1)!!=(2n-1)\cdot(2n-3)\cdots 5\cdot 3\cdot 1$$ double factorials.
We obtain
\begin{align*}
1+\sum_{n=1}^\infty\frac{(2n-1)!!}{n!}\cdot\frac{1}{4^n}
&=\sum_{n=0}^\infty\frac{(2n)!}{n!(2n)!!}\cdot\frac{1}{4^n}\tag{1}\\
&=\sum_{n=0}^\infty\frac{(2n)!}{n!n!2^n}\cdot\frac{1}{4^n}\tag{2}\\
&=\sum_{n=0}^\infty\binom{2n}{n}\frac{1}{8^n}\tag{3}\\
&=\left.\frac{1}{\sqrt{1-4z}}\right|_{z=\frac{1}{8}}\tag{4}\\
&=\frac{1}{\sqrt{1-\frac{1}{2}}}\\
&=\sqrt{2}
\end{align*}
Comment:
*
*In (1) we apply $(2n)!=(2n)!!\cdot (2n-1)!!$.
*In (2) we use $(2n)!!=n!2^n$.
*In (3) we use $\binom{2n}{n}=\frac{(2n)!}{n!n!}$.
*In (4) we use the generating function of the central binomial coefficient.
Add-on [2017-03-12] according to a comment from @navinstudent.
Using the binomial series expansion we obtain
\begin{align*}
\frac{1}{\sqrt{1-4z}}&=(1-4z)^{-\frac{1}{2}}=\sum_{n=0}^\infty \binom{-\frac{1}{2}}{n}(-4z)^n\qquad\qquad |z|<\frac{1}{4}
\end{align*}
Since
\begin{align*}
\binom{-\frac{1}{2}}{n}&=\frac{\left(-\frac{1}{2}\right)\left(-\frac{1}{2}-1\right)\cdots\left(-\frac{1}{2}-n+1\right)}{n!}\\
&=\frac{1}{n!}\left(-\frac{1}{2}\right)\left(-\frac{3}{2}\right)\cdots\left(-\frac{2n-1}{2}\right)\\
&=\frac{(-1)^n}{2^nn!}\cdot(2n-1)!!
=\frac{(-1)^n}{2^nn!}\cdot\frac{(2n)!}{(2n)!!}
=\frac{(-1)^n}{2^nn!}\cdot\frac{(2n)!}{2^nn!}\\
&=\frac{(-1)^n}{4^n}\binom{2n}{n}
\end{align*}
we get $$\sum_{n=0}^\infty \binom{-\frac{1}{2}}{n}(-4z)^n=\sum_{n=0}^\infty \binom{2n}{n}z^n$$ and step (4) follows.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1959377",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
}
|
Tricky Limits Problem () Problem:
If $\lim_{x \to 0}{\sin2x\over x^3}+a+{b\over x^2}=0$ then find the value of $3a+b$.
My attempt: $\lim_{x \to 0}{\sin2x\over x^3}+a+{b\over x^2}=\lim_{x \to 0}{\sin2x\over 2x}({2\over x^2})+a+{b\over x^2}={2+b+ax^2\over x^2}$.From this we can conclude that $a=0$ and $b=-2$, hence $3a+b=-2$. However the answer is $2$. Where am I going wrong?
|
we have
$sin(X)=X-\frac{X^3}{6}+X^3\epsilon(X)$
with $\displaystyle{ \lim_{X\to 0} \epsilon(X)=0}$.
thus
$sin(2x)=2x-\frac{4x^3}{3}+x^3\epsilon(x)$.
so
$\frac{sin(2x)}{x^3}+a+\frac{b}{x^2}$
$=\frac{2}{x^2}-\frac{4}{3}+a+\frac{b}{x^2}+\epsilon(x)$.
the limit is $0$ if
$a=\frac{4}{3}$ and $b=-2$.
finally
we will have
$3a+b=2$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1962646",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
}
|
Finding other vectors that form a basis The question I'm working on is:
For the given vectors $v, w$ in $\mathbb R^4$, find other vectors $u_1, ..., u_n$ such that {${v, w, u_1,...,u_n}$} form a basis of $\mathbb R^4$.
$v = \begin{bmatrix} 1 \\ 2 \\ 3 \\ 4 \end{bmatrix}$ and $w = \begin{bmatrix} 4 \\ 3 \\ 2 \\ 1 \end{bmatrix}$
I'm not sure how I'd find other vectors that form a basis for a subspace.
I know the first step is to get the matrix given into Reduced Row Echelon Form, and that all the vectors in a basis are linearly independent. However, putting these two together and reducing leaves:
$\begin{bmatrix} 1 & 0 \\ 0 & 1 \\ 0 & 0 \\ 0 & 0\end{bmatrix}$
Which I'm not sure what to do with to find a basis.
|
You want the determinant
$$
\begin{vmatrix}
1&4&a&w\\
2&3&b&x\\
3&2&c&y\\
4&1&d&z
\end{vmatrix}\ne0.
$$
Row reduction: If we subtract the previous row from each row, we get
$$
\begin{vmatrix}
1&4&a&w\\
1&-1&b-a&x-w\\
1&-1&c-b&y-x\\
1&-1&d-c&z-y
\end{vmatrix}.
$$
Repeat, and we get
$$
\begin{vmatrix}
1&4&a&w\\
0&-5&b-2a&x-2w\\
0&0&c-2b&y-2x\\
0&0&d-2c&z-2y
\end{vmatrix}
=
-5\,\begin{vmatrix}
c-2b&y-2x\\ d-2c&z-2y
\end{vmatrix}.
$$
We need, then, this last $2\times 2$ determinant to be nonzero. If we take $b=x=c=y=0$, we have
$$
(c-2b)(z-2y)-(d-2c)(y-2x)=-dy.
$$
It is then enough to take $d=y=1$. So,
$$
\begin{bmatrix}0\\0\\0\\1\end{bmatrix},\ \ \ \begin{bmatrix}0\\1\\0\\0\end{bmatrix}
$$
foot the bill.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1963125",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
Find the derivative of $\sin { \left( 3{ x }^{ 2 }+x \right) } $ How do you find the derivative of
$$\sin { \left( 3{ x }^{ 2 }+x \right) } $$
using the derivative definition and not the chain rule.
This is how far I was able to get
Any help would be appreciated. Thanks!
|
\begin{eqnarray*}\lim _{ h\rightarrow0 }&{ \frac { \sin { \left( 3{ \left( x+h \right) }^{ 2 }+x+h \right) } -\sin { \left( 3{ x }^{ 2 }+x \right) } }{ h } } \\=&\lim _{ h\rightarrow0 }{ \frac { 2\sin { \frac { \left( 3{ \left( x+h \right) }^{ 2 }+x+h \right) -\left( 3{ x }^{ 2 }+x \right) }{ 2 } \cos { \frac { \left( 3{ \left( x+h \right) }^{ 2 }+x+h \right) +\left( 3{ x }^{ 2 }+x \right) }{ 2 } } } }{ h } } \\
=&\lim _{ h\rightarrow 0 }{ \frac { 2\sin { \frac { h\left( 6x+3{ h }+1 \right) }{ 2 } \cos { \frac { 6{ x }^{ 2 }+6xh+3{ h }^{ 2 }+2x+h }{ 2 } } } }{ h } }\\
=&\lim _{ h\rightarrow 0 }{ \frac { \sin { \frac { h\left( 6x+3{ h }+1 \right) }{ 2 } \cos { \frac { 6{ x }^{ 2 }+6xh+3{ h }^{ 2 }+2x+h }{ 2 } } } }{ \frac { h\left( 6x+3{ h }+1 \right) }{ 2 } } } \left( 6x+3{ h }+1 \right) \\
=&\lim _{ h\rightarrow 0 }{ \cos { \frac { 6{ x }^{ 2 }+6xh+3{ h }^{ 2 }+2x+h }{ 2 } } } \left( 6x+3{ h }+1 \right) \\
=&\cos { \left( \frac { 6{ x }^{ 2 }+2x }{ 2 } \right) \left( 6x+1 \right) }
\end{eqnarray*}
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1963332",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
}
|
How prove this equation has only one solution $\cos{(2x)}+\cos{x}\cdot\cos{(\sqrt{(\pi-3x)(\pi+x)}})=0$
Let $x\in (0,\dfrac{\pi}{3}]$.
Show that this equation
$$\cos{(2x)}+\cos{x}\cdot\cos{(\sqrt{(\pi-3x)(\pi+x)}})=0$$
has a unique solution $x=\dfrac{\pi}{3}$
I try to the constructor $$f(x)=\cos{(2x)}+\cos{x}\cdot\cos{(\sqrt{(\pi-3x)(\pi+x)}})=0\, , \quad\quad f(\dfrac{\pi}{3})=0$$but I use found this function is not a monotonic function
see wolframpha
Now the key How to prove this function $f(x)$ in $(0,\frac{\pi}{3})$ has no solution, since
$$f\left(\frac{\pi}{6}\right)=\dfrac{1}{2}+\dfrac{1}{\sqrt{3}}\cos{\left(\dfrac{1}{2}\sqrt{\dfrac{7}{3}}\pi\right)}=-0.138\cdots<0$$
in other words, how to prove that
$$f(x)<0,\forall x\in(0,\dfrac{\pi}{3}) \, .$$
|
Not a solution, only a change of the question.
$\displaystyle a:=\sqrt{\frac{\pi-x}{2}+\sqrt{(\frac{\pi-x}{2})^2-x^2}}$ , $\enspace \displaystyle b:=\sqrt{\frac{\pi-x}{2}-\sqrt{(\frac{\pi-x}{2})^2-x^2}}$
=> $\enspace ab=x$ , $a^2+b^2=\pi-x$ , $a^2-b^2=\sqrt{(\pi-3x)(\pi+x)}$
The problem $f(x)=0$ with $\displaystyle 0<x<\frac{\pi}{3}$ changes to
$\displaystyle \cos(2ab)=\cos(a^2+b^2)\cos(a^2-b^2)=\frac{1}{2}(\cos(2a^2)+\cos(2b^2)) \enspace$ or converted
$\cos(2a^2)+\cos(2b^2)-2\cos(2ab)=0$ .
With $\enspace \alpha:=2a^2$ , $\beta:=2b^2\enspace $ and because of $\enspace 2ab=2\pi-\alpha-\beta\enspace $ one gets the condition
$\cos(\alpha)+\cos(\beta)-2\cos(\alpha+\beta)=0 \enspace $ and $\enspace \alpha\beta=(2\pi-\alpha-\beta)^2\enspace$ with $\enspace\displaystyle 0<\beta<\frac{2\pi}{3}<\alpha<2\pi$ . E.g. :
$\displaystyle (\alpha;\beta)=(\pi;\arctan\frac{1}{3})$ is a solution but we get
$3.65<(\pi-\beta)^2<3,7<\pi\beta<3.87\enspace$ with it.
Generally we have to show $\enspace \alpha\beta\neq (2\pi-\alpha-\beta)^2\enspace$ for all solutions of
$\cos(\alpha)+\cos(\beta)-2\cos(\alpha+\beta)=0 \enspace$ in the given value range.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1963425",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 2,
"answer_id": 0
}
|
How do you find the area of a parallelogram with the vertices? How do you find the area of a parallelogram with the following vertices; $A(4,2)$, $B(8,4)$, $C(9,6)$ and $D(13,8)$.
|
The absolute value of the cross product of two vectors $\vec{a}, \vec{b} \in \mathbb{R}^3$ spanning the parallelogram is its area:
$$A_\text{parallelogram}= \left|\vec{a}\times\vec{b}\right|$$
So in your case we have to write the points in $\mathbb{R}^2$ as vectors in $\mathbb{R}^3$ and apply the formula:
$\vec{AB}Β = \begin{pmatrix}8\\4\\0\end{pmatrix} -\begin{pmatrix}4\\2\\0\end{pmatrix} =\begin{pmatrix}4\\2\\0\end{pmatrix}$
$\vec{AD}Β = \begin{pmatrix}13\\8\\0\end{pmatrix} -\begin{pmatrix}4\\2\\0\end{pmatrix} =\begin{pmatrix}9\\6\\0\end{pmatrix}$
$A_\text{parallelogram}= \left|\vec{AB}\times\vec{AD}\right| = \left| \begin{pmatrix}4\\2\\0\end{pmatrix} \times \begin{pmatrix}9\\6\\0\end{pmatrix} \right| = \left|\begin{pmatrix}0\\0\\6\end{pmatrix} \right| = 6$
You might have noticed that this simplifies to
$$A_\text{parallelogram}= (b_1 - a_1)(d_2-a_2)-(b_2-a_2)(d_1-a_1)$$
$$= (8 - 4)(8-2)-(4-2)(13-4)=-24-(-18)=6$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1966062",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 6,
"answer_id": 1
}
|
If 'z' and 'w' are two complex numbers, prove that: |z+w|<_|z|+|w|. If 'z' and 'w' are two complex numbers, prove that: |z+w|<_|z|+|w|.
My Attempt;
If z is given as x + iy, and w = u + iv,
then |z| = β(x^2 + y^2), = r, say;
and |w| = β(u^2 + v^2), = p, say
Then we can also view z as r cis ΞΈ,
and w as p cis Ο (i.e., p (cos Ο + i sin Ο) in
polar form; so that ΞΈ and Ο can be
determined from x, y, u, and v -- to within
2Ο.)
Now if z + w = q, say,
consider the triangle formed by z, w, and
q.
The angle opposite q is Ο - |Ο - ΞΈ| .
So we have |q|^2 = |z|^2 + |w|^2 + 2 |
Is this correct. Please somebody check it. And if there is any other method... then plaese do it.
z| |w| cos (ΞΈ + Ο),
which is < |z + w|^2, since cos x
|
Perhaps writing the expression explicitly it can be clearer:
$$\begin{cases}z=a+bi\\{}\\w=x+iy\end{cases}\implies |z+w|=\sqrt{(a+x)^2+(b+y)^2}\stackrel?\le\sqrt{a^2+b^2}+\sqrt{x^2+y^2}$$
Well, now square both sides:
$$a^2+2ax+x^2+b^2+2by+y^2\stackrel?\le a^2+b^2+x^2+y^2+2\sqrt{(a^2+b^2)(x^2+y^2)}\iff$$
$$\iff2(ax+by)\stackrel?\le2\sqrt{(a^2+b^2)(x^2+y^2)}\iff$$
$$\iff a^2x^2+2abxy+b^2y^2\le a^2x^2+a^2y^2+b^2x^2+b^2y^2\iff$$
$$2abxy\le a^2y^2+b^2x^2\iff a^2y^2+b^2x^2-2abxy\ge0\iff$$
$$\iff (ay-bx)^2\ge0\;\;\;\color{green}\checkmark$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1968175",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
Prove that $\log_{2}(n) = O(n^{1-\epsilon})$ for any $0 \leq \epsilon < 1$ We assume that $n>0$. I would like to get some feedback on my proof of this statement.
When we want to show that $f(n) = O(g(n))$, we need to find some positive constants $c, n_{0}$ such that for every $n \geq n_{0}$ the following holds: $f(n) \leq cg(n)$. Here $f(n) = \log_{2}(n)$ and $g(n) = n^{1-\epsilon}$, so we want to find some positive constants $c,n_{0}$ such that $\log_{2}n \leq cn^{1-\epsilon}$ for every $n\geq n_{0}$.
For $\log_{2}n \leq cn^{1-\epsilon}$ to be true, then $\log_{2}(\log_{2}n) \leq \log_{2}(cn^{1-\epsilon})$ must be true. Doing the math from there we have:
$\log_{2}(\log_{2}n) \leq \log_{2}(c) + \log_{2}(n^{1-\epsilon}) = \log_{2}(c) + (1-\epsilon)\log_{2}(n)$
because $n$ is positive, $\log_{2}(n)$ will also be positive so we have
$\log_{2}(\log_{2}n) \leq (\log_{2}(c) + (1-\epsilon))\log_{2}(n)$
now at this point, someone could suggest that all we need to do is find a constant $c$ such that $(\log_{2}(c) + (1-\epsilon)) \geq 1$. This happens when $c = 2^{\epsilon}$.
However shouldn't we also prove that $\log_{2}(\log_{2}n) \leq \log_{2}(n)$ for every $n>n_{0}$? I have seen many proofs where they just pick an $n_{0}$ without formally going over the argument. My claim is that we can pick $n_{0} = 4$. So now we need to prove that $\log_{2}(\log_{2}n) \leq \log_{2}(n)$ for every $n>4$ and together with the constant that we picked above this will complete the proof of the original statement.
Let $h(n) = \log_{2}(\log_{2}n) - \log_{2}(n)$. We have that $\frac{\partial h}{\partial n} = \frac{1-\ln(n)}{n\ln(2)\ln(n)}$. This derivative becomes 0 when $n = e$. Since $h(4) = -1 < 0$ this means that $h(n) < 0$ for every $n>=4$. So $\log_{2}(\log_{2}n) \leq \log_{2}(n)$ for every $n>=4$, which means that we can pick $n_{0} = 4$.
|
Note: Here we analyse at some length OPs proof, identify some problems and propose an alternative while following OPs approach.
OPs proposed solution: $c=2^{\varepsilon}, n_0=4$
At first we check the proposed solution. OPs result $c=2^\varepsilon$ and $n_0=4$ gives
\begin{align*}
\log_2(n)\leq 2^{\varepsilon}n^{1-\varepsilon}\qquad\qquad n\geq 4
\end{align*}
We obtain by letting
*
*$n=4$:
\begin{align*}
&LHS:\qquad\log_2(4)=\log_2\left(2^2\right)=2\\
&RHS:\qquad2^{\varepsilon}\cdot 4^{1-\varepsilon}=2^{2-\varepsilon}>2\qquad\qquad 0\leq \varepsilon<1
\end{align*}
$n=4$ looks fine, but one more plausibility check with another power of $2$:
*
*$n=2^3$:
\begin{align*}
&LHS:\qquad\log_2(8)=\log_2\left(2^3\right)=3\\
&RHS:\qquad2^{\varepsilon}\cdot 8^{1-\varepsilon}=2^{3-2\varepsilon}
\end{align*}
Since
\begin{align*}
3\leq 2^{3-2\varepsilon}
\,\,\Longleftrightarrow\,\, \log_2(3)\leq 3-2\varepsilon\\
\end{align*}
we obtain
\begin{align*}
\varepsilon\leq \frac{3}{2}-\frac{1}{2}\log_2(3)\leq 0.7075\ldots
\end{align*}
We observe the last inequality is not valid for all $\varepsilon<1$ and this contradiction shows the proposed solution is not correct.
We will soon identify the problem, but first let's go on with OPs approach which is promising.
Taking logarithms:
It's a good idea to take logarithms in order to make the inequality better accessible.
\begin{align*}
\log_2(n)\leq c\cdot n^{1-\varepsilon}\tag{1}
\end{align*}
Since the logarithm $\log_2$ is a strictly increasing function we obtain from (1)
\begin{align*}
\log_2\left(\log_2(n)\right)&\leq \log_2(c\cdot n^{1-\varepsilon})\\
&=\log_2(c)+(1-\varepsilon)\log_2(n)\tag{2}
\end{align*}
$$ $$
OPs simplified inequality:
In order to estimate values for $c$ and $n_0$ OP tries to simplify the inequality and argues as follows:
(OP:) because $n$ is positive, $\log_2(n)$ will also be positive, so we have
\begin{align*}
\log_2\left(\log_2(n)\right)\leq(\log_{2}(c) + (1-\epsilon))\log_{2}(n)
\end{align*}
*
*At first the minor deficiency: From $n$ being positive it does not follow that $\log_2(n)$ is positive, since with $n=1$ we have $\log_2(1)=0$. The next minor is that from being positive we cannot deduce the validity of the inequality. The value has to be $\geq 1$ to deduce the validity of the inequality.
*The major deficiency is that this is an estimation at the wrong side of the inequality, since we actually obtain the following inequality chain
\begin{align*}
\log_2\left(\log_2(n)\right)\leq\log_2(c)+(1-\varepsilon)\log_2(n)\leq(\log_{2}(c) + (1-\epsilon))\log_{2}(n)
\end{align*}
Note, the inequality at the RHS is not helpful since determining $c$ and $n_0$ so that
\begin{align*}
\log_2\left(\log_2(n)\right)\leq(\log_{2}(c) + (1-\epsilon))\log_{2}(n) \qquad\qquad n\geq n_0
\end{align*}
is valid does not imply that (2) is valid for $n\geq n_0$.
We need instead an expression $A(n,\varepsilon)$ which is in between:
\begin{align*}
\log_2\left(\log_2(n)\right)\leq A(n,\varepsilon)\leq\log_{2}(c) + (1-\epsilon)\log_{2}(n) \qquad\quad n\geq n_0
\end{align*}
If we can now find $c$ and $n_0$ with
\begin{align*}\log_2\left(\log_2(n)\right)\leq A(n,\varepsilon)\qquad\qquad n\geq n_0\tag{3}
\end{align*}
then we may also deduce the validity of (2) from which the validity of (1) follows.
Note: It is this estimation at the wrong side which is the reason for the incorrect values of $c$ and $n_0$ stated by OP.
Another simplified inequality: $A(n,\varepsilon)$
We have to find $c$ and $n_0$ so that (2) is true for all $n\geq n_0$.
Since $\log_2\left(\log_2(n)\right)$ increases at a much slower rate than $\log_2(n)$ it is the factor $(1-\varepsilon)$ which can invalidate the inquality for small $n$. The factor is effective especially when $1-\varepsilon$ is very close to zero, which means $\varepsilon$ is very close to $1$.
So, we estimate $\varepsilon$ by a smaller value, which can be properly used for calculation.
Let $k=k(\varepsilon)\in\mathbb{N}$ with
\begin{align*}
2^{-k}<1-\varepsilon
\end{align*}
then we are looking for $c$ and $n_0$ so that the following inequality chain is valid for $n\geq n_0$:
\begin{align*}
\log_2\left(\log_2(n)\right)\leq \log_2(c)+2^{-k}\log_2(n)\leq \log_2(c)+(1-\varepsilon)\log_2(n)
\end{align*}
We simplify the inequality chain even more by setting $c=1$. We obtain
\begin{align*}
\log_2\left(\log_2(n)\right)\leq 2^{-k}\log_2(n)\leq(1-\varepsilon)\log_2(n)\tag{4}
\end{align*}
We obtain by setting $n=2{^{2^k}}$ in (4)
\begin{align*}
k\leq 2^{-k}2^k=1
\end{align*}
This is not a satifying solution but it indicates a more promising approach.
We obtain by setting $n=2{^{2^{2k}}}$ in (4)
\begin{align*}
2k\leq 2^{-k}2^{2k}=2^k
\end{align*}
which is true for all positive integers $k$,
resulting finally in the values
\begin{align*}
c=1\qquad\qquad\text{and}\qquad\qquad n_0=2{^{2^{2k}}}
\end{align*}
and we obtain according to (4)
\begin{align*}
\log_2(\log_2(n))&\leq 2^{-k}\log_2(n)\\
&\leq (1-\varepsilon)\log_2(n)\qquad\qquad n\geq n_0:=2^{2^{2k}}
\end{align*}
$$ $$
Conclusion:
If we set $k=k(\varepsilon)\in\mathbb{N}$ so that $2^{-k}<1-\varepsilon$ and $c=1$ then
\begin{align*}
\log_2(n)\leq n^{1-\varepsilon}\qquad\qquad n\geq n_0:=2^{2^{2k}}
\end{align*}
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1968257",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
}
|
How do I prove that $x^n-y^n=(x-y)(x^{n-1}+x^{n-2}y+\dotsb+xy^{n-2}+y^{n-1})$ I am unsuccessfully attempting a problem from Spivak's popular book 'Calculus' 3rd edition. The problem requires proof for the following equation:
$$x^n-y^n=(x-y)(x^{n-1}+x^{n-2}+\dotsb+xy^{n-2}+y^{n-1})$$
The solution to the problem, contained in the answer book, is as follows:
\begin{align*}
x^n-y^n &= (x-y)(x^{n-1}+x^{n-2}y+\dotsb+xy^{n-2}+y^{n-1})\\
&= x(x^{n-1}+x^{n-2}y+\dotsb+xy^{n-2}+y^{n-1})\\
&\qquad -[y(x^{n-1}+x^{n-2}y+{...}+xy^{n-2}+y^{n-1})]&\Rightarrow \mathbf{Equation1}\\
&=x^n+x^{n-1}y+\dotsb+x^2y^{n-2}+xy^{n-1}\\
&\qquad -[x^{n-1}y+x^{n-2}y^2+xy^{n-1}+y^n]&\Rightarrow \mathbf{Equation2}\\
&=x^n-y^n
\end{align*}
While I believe that the distributive law was used to arrive at Equation 1, I do not understand how Equation 2 was arrived at.
I have tried to solve this independently to no avail. I cannot seem to understand how $x^n$ and $y^n$ came about in Equation 2 for example.
To summarise the question, what principle was Equation 2 based upon? And how was this applied in the above problem.
|
I like using $\sum$ like this:
$\begin{array}\\
(x-y)(x^{n-1}+x^{n-2}y+\dotsb+xy^{n-2}+y^{n-1})
&=(x-y)\sum_{k=0}^{n-1} x^{n-1-k} y^k\\
&=x\sum_{k=0}^{n-1} x^{n-1-k} y^k-y\sum_{k=0}^{n-1} x^{n-1-k} y^k\\
&=\sum_{k=0}^{n-1} x^{n-k} y^k-\sum_{k=0}^{n-1} x^{n-1-k} y^{k+1}\\
&=\sum_{k=0}^{n-1} x^{n-k} y^k-\sum_{k=1}^{n} x^{n-k} y^{k}\\
&=x^n+\sum_{k=1}^{n-1} x^{n-k} y^k-\left(y^n+\sum_{k=1}^{n-1} x^{n-k} y^{k}\right)\\
&=x^n-y^n+\sum_{k=1}^{n-1} x^{n-k} y^k-\sum_{k=1}^{n-1} x^{n-k} y^{k}\\
&=x^n-y^n\\
\end{array}
$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1968874",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 0
}
|
Calculate limit involving $\sin$ function Calculate the following limit:
$$\lim_{x \rightarrow 0} \frac{x-\overbrace{\sin (\sin (...(\sin x)...))}^{150\ \text{times}\ \sin}}{x^3}$$
I tried applying L'Hospital's rule, but it got too messy.
Thank you in advance!
|
Use short form of series expansion:
$$
\sin(x)=x-\frac{x^3}{6}+O(x^5)\\
\sin(\sin(x))=\sin(x)-\frac{\sin^3(x)}{6}+O(\sin^5(x))\to\\
\sin(\sin(x))=x-\frac{x^3}{6}+O(x^5)-\frac{(x-\frac{x^3}{6}+O(x^5))^3}{6}+O(x^5)\to\\
\sin(\sin(x))=x-\frac{x^3}{3}+O(x^5)
$$
Important part here is to notice that each additional $\sin(...)$ leads to the expression of the same form, while adding $-\frac{x^3}{6}$.
Then,
$$\overbrace{\sin (\sin (...(\sin x)...))}^{150\ \text{times}\ \sin}=x-150\frac{x^3}{6}+O(x^5)$$
Hence,
$$\lim_{x \rightarrow 0} \frac{x-\overbrace{\sin (\sin (...(\sin x)...))}^{150\ \text{times}\ \sin}}{x^3}=\lim_{x \rightarrow 0} \frac{x-x+150\frac{x^3}{6}+O(x^5)}{x^3}=\lim_{x \rightarrow 0}(25+O(x^2))=25$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1969399",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 0
}
|
How can we solve $\lim\limits_{n \to \infty} \left (1 + \frac{1}{n} \right)^n$? Taken from Wikipedia:
The number $e$ is the limit $$e = \lim_{n \to \infty} \left (1 + \frac{1}{n} \right)^n$$
Graph of $f(x) = \left (1 + \dfrac{1}{x} \right)^x$ taken from here.
Its evident from the graph that the limit actually approaches $e$ as $x$ approaches $\infty$. So I tried approaching the value algebraically. My attempt:
$$\lim_{n \to \infty} \left (1 + \frac{1}{n} \right)^n$$
$$= \lim_{n \to \infty} \left(\frac{n + 1}{n}\right)^n$$
$$= \left(\lim_{n \to \infty} \left(\frac{n + 1}{n}\right) \right)^n$$
$$= 1^\infty$$
which is an indeterminate form. I cannot think of any other algebraic manipulation. My question is that how can I solve this limit algebraically?
|
An algebraic way is Binomial Expansion, which is given by
$$\begin{eqnarray*}
\left(1+\frac{1}{n}\right)^{\!n} &=& 1+n\left(\frac{1}{n}\right)+\frac{n(n-1)}{2!}\left(\frac{1}{n}\right)^{\!2}+\frac{n(n-1)(n-2)}{3!}\left(\frac{1}{n}\right)^{\!3}+\cdots \\ \\
&=& 1+1+\frac{n(n-1)}{n^2}\cdot\frac{1}{2!}+\frac{n(n-1)(n-2)}{n^3}\cdot \frac{1}{3!}+\cdots \\ \\
\end{eqnarray*}$$
Now every term in front there goes to $1$ for $n\to \infty$, so that the limit gives
$$
\lim_{n \to \infty}\left(1+\frac{1}{n}\right)^{\!n} = 1 + 1 + \frac{1}{2!} + \frac{1}{3!} + \cdots = \mathrm{e}.
$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1969934",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
}
|
How to evaluate $\lim\limits_{x\to 0} \frac{\arctan x - \arcsin x}{\tan x - \sin x}$ I have a stuck on the problem of L'Hospital's Rule,
$\lim\limits_{x\to 0} \frac{\arctan x - \arcsin x}{\tan x - \sin x}$ which is in I.F. $\frac{0}{0}$
If we use the rule, we will have
$\lim\limits_{x\to 0} \frac{\frac{1}{1+x^2}-\frac{1}{\sqrt{1-x^2}}}{\sec^2x-\cos x}$.
So, I think that I approach this problem in the wrong way.
Have you guy any idea?
|
The first step is OK, but you should try and simplify things before going on.
The numerator is
$$
\frac{1}{1+x^2}-\frac{1}{\sqrt{1-x^2}}=
\frac{\sqrt{1-x^2}-1-x^2}{(1+x^2)\sqrt{1-x^2}}=
\frac{-x^2(3+x^2)}{(1+x^2)\sqrt{1-x^2}(\sqrt{1-x^2}+1+x^2)}
$$
The denominator is
$$
\frac{1}{\cos^2x}-\cos x=\frac{1-\cos^3x}{\cos^2x}=
\frac{(1-\cos x)(1+\cos x+\cos^2x)}{\cos^2x}
$$
so you can rewrite your limit as
$$
\lim_{x\to0}
\frac{x^2}{1-\cos x}
\frac{-(3+x^2)\cos^2x}{(1+x^2)\sqrt{1-x^2}(\sqrt{1-x^2}+1+x^2)(1+\cos x+\cos^2x)}
$$
which is easy to compute (the second fraction is not βindeterminateβ).
Of course, Taylor expansion is easier.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1971171",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 3
}
|
functional equation of type $f(x)-f(y) = f\left(\frac{x-y}{1-xy}\right)$ If $f(x)-f(y) = f\left(\frac{x-y}{1-xy}\right)$ and $f$ has domain equal to $(-1,1)$, then which of the following is the function satisfying the given functional equation?
Options:
(a) $2-\ln\left(\frac{1+x}{1-x}\right)$
(b) $\ln\left(\frac{1-x}{1+x}\right)$
(c) $\frac{2x}{1-x^2}$
(d) $\tan^{-1}\left(\frac{1+x}{1-x}\right)$
From the options given we can easily get $f(x) = \ln\frac{1-x}{1+x}$,
but how can I solve using analytical method? Help me please.
|
Taking $y=0$ gives $f(0) = 0$. Using this we can rewrite the functional equation as
$$\frac{f(x)-f(y)}{x-y} = \frac{1}{1-xy}\frac{f\left(\frac{x-y}{1-xy}\right) - f(0)}{\left(\frac{x-y}{1-xy}\right)}$$
Now assuming $f$ is differentiable at $x=0$ we can take the limit $y\to x$ above to obtain the ODE $$f'(x) = \frac{1}{1-x^2}f'(0) = \frac{f'(0)}{2}\left[\frac{1}{1-x} + \frac{1}{1+x}\right]$$
which can be integrated with the initial condition $f(0) = 0$ to get the general solution $f(x) = C\log\left(\frac{1-x}{1+x}\right)$ where $C = -\frac{f'(0)}{2}$ is a free constant.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1974085",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
}
|
Equation which has to be solved with logarithms I need some help how to solve these equations for $x$. I think I have to use logarithms but still not sure how to do it and would be really grateful if someone could explain me.
$x^2 \cdot 2^{x + 1} +2 ^{\lvert x - 3\rvert + 2}
= x^2 \cdot 2^{\lvert x - 3\rvert + 4} + 2^{x - 1}$
$(x^2 - 7x + 5)^{x^2-2x-15} = 1$
|
$$x^2 \cdot 2^{x + 1} +2 ^{\lvert x - 3\rvert + 2}
= x^2 \cdot 2^{\lvert x - 3\rvert + 4} + 2^{x - 1}$$
Let us distinguish two cases, $x\ge3$ and $x\le3$, to get rid of the absolute value.
*
*$x\ge3$:
$$x^2 \cdot 2^{x + 1} +2 ^{x-1}
= x^2 \cdot 2^{x+1} + 2^{x - 1},$$ which is an identity !
*$x\le3$:
$$x^2 \cdot 2^{x + 1} +2 ^{5-x}
= x^2 \cdot 2^{7-x} + 2^{x - 1}$$
which we rewrite
$$\left(2x^2-\frac12\right)2^x=2^6\left(2x^2-\frac12\right)2^{-x},$$
so that $$x=\pm\frac12\text{ or }x=3.$$
$$(x^2 - 7x + 5)^{x^2-2x-15} = 1$$
$a^b=1$ when $a=1$ or $a=-1\land\text{even}(b)$ or $b=0$, so
$$a=0\to x=\frac{7\pm\sqrt33}2,$$
$$a=-1\to x=1\text{ or 6},$$ where $6$ must be rejected as it yields an odd exponent, and
$$b=0\to x=-3,5.$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1976125",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 2
}
|
$x+\dfrac{1}{y}=y+\dfrac{1}{z}=z+\dfrac{1}{x}$, prove that $xyz=1$ Let $x>0$, $y>0$ and $z>0$ such that $x\neq y $ , $y\neq z$ and $x\neq z$. If $x+\dfrac{1}{y}=y+\dfrac{1}{z}=z+\dfrac{1}{x}$, prove that $xyz=1$
|
Hint: Notice that
\begin{align} x-y &= \frac{y-z}{yz} \\
y-z &= \frac{z-x}{xz} \\
z-x &= \frac{x-y}{xy}.
\end{align}
Now multiply.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1976348",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
}
|
Finding characteristic features of a parabola known by two points and their tangents This is from a math competition so it must not be something really long
If a parabola touches the lines $y=x$ and $y=-x$ at $A(3,3) $ and $b(1,-1)$ respectively, then
(A) equation of axis of parabola is $2x+y=0$
(B)slope of tangent at vertex is $1/2$
(C) Focus is $(6/5,-3/5)$
(D) Directrix passes through $(1,-2)$
I thought the axis would be the angle bisector of the tangents passing through the focus but it turns out that is not the case in a parabola so how can I find anything..
|
Conditions (A),(B),(C),(D) are not needed to determine the tilted parabola ( $xy$ term non-zero). Because out of five constants needed to determine a conic, one can be reduced as zero determinant for parabola.You have given two points and two slopes which are quite sufficient.
$$ (x, y, y^{ \prime} ) = (3,3,1), (1,-1,-1) $$
Parabola involving $xy,x^2, y^2,x,y $ terms can be solved for y in a quadratic as:
$$ y^2 - 2y ( ax+b) + (ax+b)^2 - (cx+d) =0 $$
$$ y = (ax +b) \pm \sqrt{ c x +d } $$
$\pm$ symbol separates regions on either side of vertical tangent. Plug in given point coordinates:
$$ 3 a + b+ \sqrt{ 3c+d} =3 , \quad a + b+ \sqrt{ c+d} = -1 ,\quad a + c \ /( 2 \sqrt{ c+d} ) = 1 ,\quad a - c \ /( 2 \sqrt{ c+d} ) = -1 $$
Using a CAS to reduce tedium we get $ (a,b,c,d) = ( 1/2, -3/4, 9/4,-27/16) $
The parabola is plotted to verify everything .It checks out (C), (D) but (A),(B) are inconsistent with first inputs and are clearly incorrect.
EDIT1
Some calculations needed (in progress)at focus and tangent orthogonal intersection on directrix
For parabola $ 4 a y = x^2 $
Points of tangency
$$ (2at, a t^2)\quad (2a/t, a/t^2) $$
Point of intersection of polar chord tangents on directrix which happens at right angles at a point D
$$ a [( t-1/t), -1] $$
Length of tangents $T_1,T_2$ given by
$$ (T_1/a)^2 = t^4+ 3 t^2 + 1/t^2 +3 = 2, \quad (T_2/a)^2 = 1/t^4+ 3/ t^2 + t^2 +3 = 18 $$
$$\rightarrow t= \frac13 $$
$OF$ is perpendicular on hypotenuse AB
$$ \frac{1}{OF^2} =\frac{1}{2} + \frac{1}{ 9 \cdot 2} $$
$$ OF = \frac{3}{\sqrt 5} ... $$
$$ \cos \beta= \frac{OF}{OB} = \frac{ 3}{\sqrt10},\quad \sin \beta= = \frac{ 1}{\sqrt10}\quad $$
to be continued
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1977306",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
}
|
Given $P(x)=x^{4}-4x^{3}+12x^{2}-24x+24,$ then $P(x)=|P(x)|$ for all real $x$ $$P(x)=x^{4}-4x^{3}+12x^{2}-24x+24$$
I'd like to prove that $P(x)=|P(x)|$. I don't know where to begin. What would be the first step?
|
By completing the square for $x^4+12x^2\cdots$,
$$P(x)=(x^2+6)^2-4x^3-24x-12=(x^2+6)^2-4x(x^2+6)-12\\
=(x^2+6)(x^2-4x+6)-12.$$
The minimum value of the first factor is $6$ and that of the second is $2$.
Or completing the square for $x^2(x^2-4x\cdots)$,
$$P(x)=x^2(x-2)^2+8x^2-24x+24=x^2(x-2)^2+8(x^2-3x+3).$$
As you can easily check, the trinomial on the right has no real root.
Or completing the fourth power for $x^4-4x^3\cdots$,
$$P(x)=(x-1)^4+P(x)-(x^4-4x^3+6x^2-4x+1)\\=(x-1)^4+6x^2-20x+23$$ and the trinomial on the right has no real root.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1981110",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 10,
"answer_id": 0
}
|
Transform recursive sequence to direct. I am taking the GRE General Exam in a few weeks and there are some problems about sequences that I have found a bit difficult, e.g given a sequence in recursive form like $S_{n} = S_{n-1} - 10$ and some value for this sequence $S_{3}=0$ what is the value $S_{25}$?
I know that the sequence in the direct form is $S_{n} = -10n + 30 $ but how this turns out?
for example can someone tell me step by step whats $S_{n} = 2S_{n-1}-4$ direct formula given that $S_{1}=6$
Thanks
|
This is easy - given a recursively defined sequence $S_n$ you:
(1) transform it to a form s.t. there are no "exceptions" assuming that $S_n$ is zero for negative $n$. E.g. if $S_n=2S_{n-1} - 4$ and $S_1=6$, the equation $S_n=2S_{n-1} - 4$ (assuming that $S_{-1}$=0) would output $S_0=2*0-4=-4$, so we have to add +4*[n=0] and for the same reason for $S_1$ we have to add + 10 *[n=1] (where [blabla(n)] is a characteristic function of blabla - equal 1 if blabla(n), and 0 otherwise).
So:
$S_n=2S_{n-1} - 4 + 10*[n=1] +4*[n=0]$
(2) Now you define a function $S(x) = S_0x^0 + S_1x^1 + S_2x^2+... = \sum_0^{\infty} S_nx^n$
On the other hand, due to our recursive equation:
$S(x) = \sum_{n=0}^{\infty} S_nx^n =\\
\sum_{n=0}^{\infty} (2S_{n-1} - 4 + 10*[n=1] + 4*[n=0])x^n =\\
\sum_{n=0}^{\infty} (2S_{n-1} - 4)x^n + 10x +4 =\\
2\sum_{n=0}^{\infty} S_{n-1}x^n - 4 \sum_n^{\infty} x^n +10x +4=\\
2x\sum_{n=0}^{\infty}S_{n-1}x^{n-1} - 4 (\frac{1}{1-x}) +10x +4=\\
2x\sum_{n=0}^{\infty}S_{n}x^{n} + \frac{4}{x-1} + 10x +4=\\
2x S(x) +\frac{10x^2 - 6x}{x-1}$
So:
$S(x) = \frac{10x^2-6x}{(1-2x)(x-1)} = \frac{1}{1-2x} + \frac{4}{1-x} -5$
(3) So now we may expand it back to a series:
$S(x) = \sum_{n=0}^{\infty}(2x)^n + 4\sum_{n=0}^{\infty}x^n -5 =\\
\sum_{n=0}^{\infty} (2^n +4)x^n -5$
And as we defined $S(x)$ to be $\sum_{n=0}^{\infty} S_nx^n$ we may conclude that $S_n = 2^n+4 -5[n=0]$.
Checking for $n=0$: 1+4-5=0 ok, for $n=1$: 2+4=6 ok, for $n=2$ 4+4=8 ok!... :)
This method is pretty universal.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1984274",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
}
|
If $A \in M_3(\mathbb C)$ and $\operatorname{tr}(A)=\operatorname{tr}(A^{-1})=0$ and $\det(A)=1$, then $A^3=I_3$ Assume that we have a $3\times 3$ matrix like $A$ in which the coordinates come from the set of complex numbers ($\mathbb C$).
We know that $\operatorname{tr}(A)=\operatorname{tr}(A^{-1})=0$ and $\det(A)=1$ .
Prove that $A^3=I_3$ .
I don't know what's the relationship between $A^3$ and the trace !
|
Solving the set of equations:
$$\begin{cases}
\lambda_1+\lambda_2+\lambda_3=0\ \\
\lambda_1^{-1}+\lambda_2^{-1}+\lambda_3^{-1}=0 \\
\lambda_1\lambda_2\lambda_3=1
\end{cases}
$$
you get $6$ solutions, but we can work just on one of these WLOG, that is:
$$\lambda_1 = 1, \lambda_2 = e^{-i\frac{2\pi}{3}} ~ \text{and} ~ \lambda_3 = e^{i\frac{2\pi}{3}}.$$
Indeed, each of the $6$ solution is a permutation of the previous.
Using Cayley-Hamilton theorem, we know that:
$$A^3 - tr(A)A^2 + \frac{1}{2}\left((tr(A))^2 - tr(A^2)\right)A - \det(A)I_3 = 0.$$
By substitution, it is obvious that:
$$A^3 - \frac{1}{2}tr(A^2)A - I_3 = 0.$$
It is clear that $tr(A^2) = \lambda_1^2+\lambda_2^2+\lambda_3^2.$
Then:
$$tr(A^2) = 1^2 + \left(e^{-i\frac{2\pi}{3}}\right)^2 + \left(e^{i\frac{2\pi}{3}}\right)^2 = \\
= 1 + e^{-i\frac{4\pi}{3}} + e^{i\frac{4\pi}{3}} = 1 + 2\cos\left(\frac{4\pi}{3}\right) = \\ = 1 + 2\left(-\frac{1}{2}\right) =
1 - 1 = 0.$$
Finally, you can state that:
$$A^3 - I_3 = 0 \Rightarrow A^3 = I_3.$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1986427",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 2
}
|
Trigonometric limit I couldn't find this limit , can someone help me?
$$\lim_{x \to \infty} \frac{\arctan(x+1) - \arctan(x)}{\sin\left(\frac{1}{x+1}\right) - \sin\left( \frac 1x\right)}$$
|
Hint. By using Taylor series expansions, as $u \to 0$, one has
$$
\begin{align}
\arctan u&=u-\frac{u^3}{3}+o(u^3)
\\\sin u&=u-\frac{u^3}{6}+o(u^3)
\end{align}
$$ giving, as $x \to \infty$ $\big($with $u=\frac1{x}$ or $u=\frac1{x+1}$$\big)$,
$$
\begin{align}
\arctan x&=\frac{\pi}2-\arctan \frac1x =\frac{\pi}2-\frac1x+O\left(\frac1{x^3} \right)
\\\arctan (x+1)&=\frac{\pi}2-\arctan \frac1{x+1} =\frac{\pi}2-\frac1x+\frac1{x^2} +O\left(\frac1{x^3} \right)
\\\sin \frac1x&=\frac1x+O\left(\frac1{x^3} \right)
\\\sin \frac1{x+1}&=\frac1x-\frac1{x^2} +O\left(\frac1{x^3} \right)
\end{align}
$$ then, as $x \to \infty$,
$$
\frac{\arctan(x+1) - \arctan(x)}{\sin\left(\frac{1}{x+1}\right) - \sin\left( \frac 1x\right)}=\frac{\frac1{x^2}+O\left(\frac1{x^3} \right)}{-\frac1{x^2}+O\left(\frac1{x^3} \right)} \to -1.
$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1991898",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
}
|
$n$-derivative of function $f(x)=e^x \sin x$ at $x=0$ I have function $f(x) = e^x \sin{x}$ and must found $f^{(n)}(0)$
$f'(x) = e^x(\sin{x} + \cos{x}) $
$f''(x) = 2 e^x \cos{x}$
$f'''(x) = 2 e^x (\cos{x} - \sin{x})$
$f''''(x) = -4 e^x \sin{x}$
$f'''''(x) = -4 e^x (\sin{x} + \cos{x})$
I think $f^{(n)}(0) = \alpha (-1)^n x^{2n+1}$ but I can't find $\alpha$
|
I think the most appropriate approach is already given by @JackyChong.
Slightly more elaborated we have
\begin{align*}
\color{blue}{\left.\frac{d^n}{dx^n}\left(e^x\sin x\right)\right|_{x=0}}
&=\left.\frac{d^n}{dx^n}\left(\Im e^{(1+i)x}\right)\right|_{x=0}\\
&=\Im \left.\left(\frac{d^n}{dx^n}e^{(1+i)x}\right)\right|_{x=0}\\
&=\Im\left.\left((1+i)^ne^{(1+i)x}\right)\right|_{x=0}\\
&=\Im(1+i)^n\\
&=\Im\left(\sqrt{2}\left(\cos \frac{\pi}{4}+i\sin\frac{\pi}{4}\right)^n\right)\\
&=2^{\frac{n}{2}}\Im\left(\cos\frac{n\pi}{4}+i\sin\frac{n\pi}{4}\right)\\
&\color{blue}{=2^{\frac{n}{2}}\sin\frac{n\pi}{4}}
\end{align*}
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1992070",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
"answer_id": 2
}
|
Does ($\frac{p-1}{2})$! satisty $x^2$ = 1modp? I'm trying to show that for p prime = 3 mod4, ($\frac{p-1}{2})!^2$ = 1 modp.
I assume I use Wilson's for the factorial and Euler's Totient Function for the (p-1), but I'm not sure how to string them together.
|
It is the case that, for prime $p \equiv 3 \pmod{4}$
$$
\frac{p-1}{2}!^2 \equiv 1 \pmod{p}
$$
To see why, first notice $k(p-k) \equiv -k^2 \equiv -(p-k)^2 \pmod{p}$. Begin with Wilson's Theorem
$$
(p-1)! = 1 \cdot 2 \cdot 3 \cdot 4\cdots (p-4)(p-3)(p-2)(p-1) \equiv -1 \pmod{p}
$$
Now rearrange this product in a clever way
$$
(p-1)! = 1 (p-1) 2 (p-2) 3 (p-3) 4 (p-4) \cdots \frac{p-1}{2} \frac{p+1}{2} \equiv -1 \pmod{p}
$$
Using our above fact, we obtain
$$
-1^2 \cdot -2^2 \cdot -3^2 \cdot -4^2 \cdots -(\frac{p-1}{2})^2 \equiv -1 \pmod{p}
$$
So we have
$$
(-1)^{\frac{p-1}{2}} 1^2\cdot2^2\cdot3^2\cdot4^2\cdots(\frac{p-1}{2})^2 \equiv -1 \pmod{p}
$$
If $p \equiv 3 \pmod{4}$, we have $(-1)^{\frac{p-1}{2}} = -1$ so
$$
1^2\cdot2^2\cdot3^2\cdot4^2\cdots(\frac{p-1}{2})^2 = \left(\frac{p-1}{2}\right)!^2 \equiv 1 \pmod{p}
$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1992547",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
If $A$ is a real $n \times n$ matrix satisfying $A^3 = I$ then Trace of $A$ is always If $A$ is a real $3 \times 3$ matrix satisfying $A^3$ = I such that $ A \neq I $ .Then, Trace of A is always
*
*$0$
*$1$
*$-1$
*$3$
I proceed as follows: from given,
$\min(x)=x-1$ or $x^2+x+1$ or $(x-1)(x^2+x+1)$
$\min(x)\ne x-1$ as $A\ne I$
so, $\min(x)=x^2+x+1$ or $(x-1)(x^2+x+1)$
now, how to proceed after this step
any help would be appreciated
|
Here are two matrices $A$ with $A^3 = A$:
$$
A = \begin{pmatrix}1 & 0 & 0\\ 0 & 1 & 0 \\ 0 & 0 & -1\end{pmatrix}\qquad A = \begin{pmatrix}-1 & 0 & 0\\ 0 & -1 & 0 \\ 0 & 0 & -1\end{pmatrix}
$$
Since these have diffferent traces, the question does not have a definite answer.
Edit: The matrix
$$
A = \begin{pmatrix} 0 & 1 & 0\\ 0 & 0 & 1 \\ 1 & 0 & 0\end{pmatrix}
$$
has the desired property (for the edited question) and has trace 0. Hence, if the question is well-posed, the answer must be 0.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1994406",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
}
|
If $f(x)= \lim_ {n \to \infty} \sum_{r=1}^{n} \frac{n}{n^2+x^2r^2}$, find the required value We have
$$f(x)= \lim_ {n \to \infty} \sum_{r=1}^{n} \frac{n}{n^2+x^2r^2}$$, then find the value of $\sum_{k=1}^{3} k f(k)$.
Could someone give me slight hint as how to find $f(x)$ here.
|
Since the terms of the sum are non-negative and monotonically decreasing in $r$, we can write
$$\int_1^{n+1} \frac{n}{n^2+x^2y^2}\,dy \le \sum_{r=1}^n\frac{n}{n^2+x^2r^2}\le \int_0^n \frac{n}{n^2+x^2y^2}\,dy \tag 1$$
Evaluation of the integrals in $(1)$ is straightforward and reveals
$$\frac{\arctan\left(\frac{x}{1+(n+1)x^2/n^2}\right)}{x}\le \sum_{r=1}^n\frac{n}{n^2+x^2r^2}\le \frac{\arctan(x)}{x}$$
whereupon application of the squeeze theorem yields the coveted result
$$\lim_{n\to \infty}\sum_{r=1}^n\frac{n}{n^2+x^2r^2}=\frac{\arctan(x)}{x}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1999572",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
}
|
Find all $\alpha$ such that the series converges Find all values of $\alpha$ such that series $$\sum^\infty_{n=1} \left( \frac{1}{n \cdot \sin(1/n)} - \cos\left(\frac{1}{n}\right) \right)^\alpha$$ converges.
I used Maclaurin for $\sin$ and $\cos$ and got:
$$a_n = \left( \frac{1}{1 - \dfrac{1}{3!n^2} + \ldots} - 1 + \frac{1}{2!n^2} - \frac{1}{4!n^4} + \ldots \right) ^ \alpha$$
Put it together in one fraction seems to be a hard thing to do.
|
HINT:
You are on the right track. Note that
$$\frac{1}{1-\frac1{6n^2}+O\left(\frac1{n^4}\right)}=1+\frac1{6n^2}+O\left(\frac1{n^4}\right)$$
Hence, we see that
$$\frac{1}{n\sin(1/n)}-\cos(1/n)=\frac2{3n^2}+O\left(\frac1{n^4}\right)$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1999706",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
}
|
Compute the sum of a series given by non constant coefficient recurrence relation $a_k=(\frac{1}{2})^k-\frac{k}{n-k+1}a_{k-1}$ and $a_1=\frac{1}{2}+\frac{1}{n}(\frac{1}{2})^n-\frac{1}{n}$.($n\ge2$).I want to compute $\lim_{n\to\infty}\sum_{k=1}^{n-1}a_k$.
My first attempt is to find out a general form for $a_k$, and I tried to cast the recurrence relation into the form $f(k)a_k+f(k-1)a_{k-1}=g(k)$ and then the recurrence relation will become linear constant-coefficient one by taking $b_k=f(k)a_k$. However, I failed to convert it.
Also, since $\lim_{n\to\infty}\sum_{k=1}^{n-1}a_k\le\sum_{k=1}^{\infty}(\frac{1}{2})^k=1$, I tried to see if we can evaluate $\lim_{n\to\infty}\sum_{k=1}^{n-1}\frac{k}{n-k+1}a_{k-1}$.
Thanks for any hint.
|
Preliminary Result
Using the Beta Function Integral, we get
$$
\begin{align}
\sum_{k=0}^m\frac{(-1)^k}{\binom{n}{k}}
&=(n+1)\sum_{k=0}^m\int_0^1(-1)^kt^k(1-t)^{n-k}\,\mathrm{d}t\\
&=(n+1)\sum_{k=0}^m\int_0^1(1-t)^n\left(\frac t{t-1}\right)^k\,\mathrm{d}t\\
&=(n+1)\int_0^1(1-t)^{n+1}\left(1-\left(\frac t{t-1}\right)^{m+1}\right)\mathrm{d}t\\
&=(n+1)\left(\frac1{n+2}+\frac{(-1)^m}{n+2}\frac{(n-m)!(m+1)!}{(n+1)!}\right)\\
&=\frac{n+1}{n+2}\left[1+\frac{(-1)^m}{\binom{n+1}{m+1}}\right]\tag{1}
\end{align}
$$
Now that we have this result, we could verify it more easily by induction.
Apply To The Question
Start with
$$
a_k=\frac1{2^k}-\frac{k}{n-k+1}a_{k-1}\tag{2}
$$
and
$$
a_1=\frac12+\frac1n\frac1{2^n}-\frac1n\tag{3}
$$
Multiply $(2)$ by $(-1)^k\binom{n}{k}$ and rearrange
$$
\underbrace{(-1)^k\binom{n}{k}a_k}_{\large u_k}-\underbrace{(-1)^{k-1}\binom{n}{k-1}a_{k-1}}_{\large u_{k-1}}=(-1)^k\binom{n}{k}\frac1{2^k}\tag{4}
$$
Thus, we can solve for $u_k=(-1)^k\binom{n}{k}a_k$
$$
\begin{align}
\overbrace{(-1)^k\binom{n}{k}a_k}^{\large u_k}
&=\overbrace{\ -na_1\ \ \vphantom{\binom{n}{j}}}^{\large u_1}+\sum_{j=2}^k\overbrace{(-1)^j\binom{n}{j}\frac1{2^j}}^{\large u_j-u_{j-1}}\\
&=-\frac1{2^n}+\sum_{j=0}^k(-1)^j\binom{n}{j}\frac1{2^j}\tag{5}
\end{align}
$$
Therefore,
$$
\begin{align}
\sum_{k=0}^na_k
&=-\frac1{2^n}\sum_{k=0}^n\frac{(-1)^k}{\binom{n}{k}}+\sum_{j=0}^n\sum_{k=j}^n\frac{(-1)^{j+k}}{2^j}\frac{\binom{n}{j}}{\binom{n}{k}}\tag{6a}\\
&=-\frac1{2^n}\frac{n+1}{n+2}\left(1+(-1)^n\right)+\frac{n+1}{n+2}\sum_{j=0}^n\left(-\frac12\right)^j\binom{n}{j}\left[(-1)^n+\frac{(-1)^j}{\binom{n+1}{j}}\right]\tag{6b}\\
&=-\frac1{2^n}\frac{n+1}{n+2}\left(1+(-1)^n\right)+\left(-\frac12\right)^n\frac{n+1}{n+2}\\
&+\frac{n+1}{n+2}\sum_{j=0}^n\frac1{2^j}\left(1-\frac j{n+1}\right)\tag{6c}\\
&=-\frac1{2^n}\frac{n+1}{n+2}\left(1+(-1)^n\right)+\left(-\frac12\right)^n\frac{n+1}{n+2}\\
&+\frac{n+1}{n+2}\left(2-\frac1{2^n}\right)-\frac1{n+2}\left(2-\frac{n+2}{2^n}\right)\tag{6d}\\
&=\frac{n}{n+2}\left(2-\frac1{2^n}\right)\tag{6e}
\end{align}
$$
Explanation:
$\text{(6a)}$: multiply $(5)$ by $\left.(-1)^k\middle/\binom{n}{k}\right.$ and sum in $k$, then change the order of summation
$\text{(6b)}$: apply $(1)$
$\text{(6c)}$: Binomial Theorem and $\left.\binom{n}{j}\middle/\binom{n+1}{j}\right.=1-\frac j{n+1}$
$\text{(6d)}$: $\sum\limits_{j=0}^n\frac1{2^j}=2-\frac1{2^n}$ and $\sum\limits_{j=0}^n\frac j{2^j}=2-\frac{n+2}{2^n}$
$\text{(6e)}$: combine terms
Subtracting $a_0=1-\frac1{2^n}$ and $a_n=0$ from $\text{(6e)}$, we get
$$
\bbox[5px,border:2px solid #C0A000]{\sum_{k=1}^{n-1}a_k
=\frac{n-2}{n+2}+\frac2{n+2}\frac1{2^n}}\tag{7}
$$
and finally,
$$
\bbox[5px,border:2px solid #C0A000]{\lim_{n\to\infty}\sum_{k=1}^{n-1}a_k=1}\tag{8}
$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1999943",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
}
|
An Identity for Pell-numbers The Pell-numbers are defined recursively by:
$P_0 = 0,
P_1 = 1$ and $
P_{n+2} = 2P_{n+1} + P_n$
I am stuck trying to prove the identity:
$P_{2n+1}^2 - 1 = 4P_n P_{n+1}(P_{n+1}^2 - P_n^2)$
A proof would be great, otherwise a list of known Pell-identies would also help
|
There is a well-know Pell-identity: $P_{2n+1} = P_n^2 + P_{n+1}^2$ which can be found in this paper. It can also be shown directly.
Using this, we just need to show: $(P_n^2 + P_{n+1}^2)^2 - 1 = 4P_n P_{n+1}(P_{n+1}^2 - P_n^2)$
This is done by induction: The statement clearly holds for $n = 0$, therefore suppose it holds for any $n$ and consider the following argument:
$(P_{n+1}^2 + P_{n+2}^2)^2 - 1 = (P_{n+1}^24P_{n+1}^2 + 4P_nP_{n+1} + P_n^2)^2 - 1$
$= 16P_{n+1}^2(P_n + P_{n+1})^2 + 8P_{n+1}(P_n + P_{n+1})(P_n^2 + P_{n+1}^2) + (P_n^2 + P_{n+1}^2)^2 -1$
$= 16P_{n+1}^2(P_n + P_{n+1})^2 + 8P_{n+1}(P_n + P_{n+1})(P_n^2 + P_{n+1}^2) + 4P_n P_{n+1}(P_{n+1}^2 - P_n^2)$
$= 4P_{n+1}(P_n + P_{n+1})(4 P_{n+1}(P_n + P_{n+1}) + 2 P_n^2 + 2 P_{n+1}^2 + P_n(P_{n+1} - P_n))$
$= 4P_{n+1}(P_n + P_{n+1})(P_n^2 + 5P_{n+1}P_n + 6P_{n+1})$
$= 4P_{n+1}(P_n + P_{n+1})(P_n + 2P_{n+1})(P_n + 3P_{n+1})$
At this point, we can just apply the fact that $P_n + P_{n+1} = P_{n+2}$, and
$P_n + P_{n+1} = P_{n+2} - P_{n+1}$ and substitute that into the above equation:
$4P_{n+1}P_{n+2}(P_{n+2} - P_{n+1})((P_{n+2} + P_{n+1}) = 4P_{n+1}P_{n+2}(P_{n+2}^2 - P_{n+1}^2)$
which completes the induction.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2000382",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
}
|
Is there anything like upper tridiagonal matrix? How to find the determinant of such a matrix? I want to find the determinant of the following matrix.
$$\left[\begin{matrix} -\alpha_1 & \beta_2 & -\gamma_3 & 0 & 0 & 0 & \cdots & 0&0
\\ 0 & -\alpha_2 & \beta_3 & -\gamma_4 & 0 & 0 & \cdots & 0 & 0
\\0 & 0 & -\alpha_3 & \beta_4 & -\gamma_5&0&\cdots&0 & 0
\\\vdots &\vdots&\ddots &&&&&\vdots&\vdots
\\0&0&0&0&0&0&\cdots&\beta_{N-2}&-\gamma_{N-1}
\\0&0&0&0&0&0&\cdots&-\alpha_{N-2}&\beta_{N-1}\end{matrix}\right]$$
Is there any closed form formula for the determinant of such a matrix?
Or can I establish some recurrence relation for the determinant, like the tridiagonal matrix?
|
The matrix that you put in the question is not square since it has size $(N-2)\times (N-1)$ (you can check by counting the $\alpha's$ in each row and column).
If your matrix is supposed to have the form
$$T_n=\begin{bmatrix}a_1 & b_1 & c_1 & 0 & \cdots &0\\ 0 & a_2 & b_2 & 0 & \cdots&0\\ 0 & 0 & a_3 & b_3 &\cdots & 0\\ \vdots & \vdots &\vdots &\ddots & \cdots & 0\\ 0 & 0 & \cdots & & a_{n-1} & b_{n-1}\\ 0 & 0 &\cdots & & 0 & a_n\end{bmatrix}$$
then it is simply an upper triangular matrix and the determinant is $\displaystyle{\det(T_n)=\prod_{i=1}^n a_i}$.
If your matrix has the form:
$$A_n=\begin{bmatrix}a_1 & b_1 & 0 & 0 & \cdots &0\\ c_1 & a_2 & b_2 & 0 & \cdots&0\\ 0 & c_2 & a_3 & b_3 &\cdots & 0\\ \vdots & \vdots &\vdots &\ddots & \cdots & 0\\ 0 & 0 & \cdots & & a_{n-1} & b_{n-1}\\ 0 & 0 &\cdots & & c_{n-1} & a_n\end{bmatrix}$$
Then I can provide a recursive relation. To start, notice that \begin{align*}
\det(A_1)&=a_1, \text{ and } \\
\det(A_2)&=a_1a_2-b_1c_1 \text{ as expected.}\\
&\\
\det(A_3)&=a_3\cdot \det(A_2)-a_1b_2c_2\\
&=a_1a_2a_3-b_1c_1a_3-a_1b_2c_2\\
&=a_3\cdot \det(A_2)-b_2c_2\cdot \det(A_1)
\end{align*}
Now, for the matrix $A_{n+1}$ we have
\begin{align*}
\det(A_{n+1})&=\begin{vmatrix}a_1 & b_1 & 0 & 0 & \cdots &0 & 0\\ c_1 & a_2 & b_2 & 0 & \cdots&0 & 0\\ 0 & c_2 & a_3 & b_3 &\cdots & 0 & 0\\ 0 & 0 & c_3 & \ddots & \ddots & 0 & 0 \\ \vdots & \vdots &\vdots &\ddots & a_{n-1} & b_{n-1} & 0 \\ 0 & 0 & \cdots & 0 & c_{n-1} & a_{n} & b_{n}\\ 0 & 0 &\cdots & 0 & 0 & c_{n} & a_{n+1}\end{vmatrix}\\
&=(-1)^{(n+1)+(n+1)}\cdot a_{n+1}\cdot \begin{vmatrix}a_1 & b_1 & 0 & 0 & \cdots &0\\ c_1 & a_2 & b_2 & 0 & \cdots&0\\ 0 & c_2 & a_3 & b_3 &\cdots & 0\\ 0 & 0 & c_3 & \ddots & \ddots & 0 \\ \vdots & \vdots &\vdots &\ddots & a_{n-1} & b_{n-1}\\ 0 & 0 & \cdots & 0 & c_{n-1} & a_{n}\end{vmatrix}\\
&\\
& + (-1)^{(n+1)+n}\cdot c_{n-1}\cdot \begin{vmatrix}a_1 & b_1 & 0 & 0 & \cdots & 0\\ c_1 & a_2 & b_2 & 0 & \cdots & 0\\ 0 & c_2 & a_3 & b_3 &\cdots & 0\\ 0 & 0 & c_3 & \ddots & \ddots & 0 \\ \vdots & \vdots &\vdots &\ddots & a_{n-1} & 0 \\ 0 & 0 & \cdots & 0 & c_{n-1} & b_{n-1}\end{vmatrix}\\
&\\
&=a_{n+1}\cdot \det(A_n)-c_{n}b_{n}\det(A_{n-1}).
\end{align*}
So, the sequence of determinants $\{x_n=\det(A_n):n\in\mathbb{N}\}$ can be defined recursively by
$$\begin{cases}x_0=1\\ x_1=a_1\\ x_{n+1}=a_{n+1}x_n-b_nc_n\cdot x_{n-1}\end{cases}$$
I asked in this question if some closed form can be obtained. So far, I can say that in general we have
$$\begin{bmatrix}x_{n+1}\\ x_n\end{bmatrix}=\left(\begin{bmatrix}a_{n+1} & -b_{n}c_{n}\\ 1 & 0\end{bmatrix}\cdots \begin{bmatrix}a_3 & -b_2c_2\\ 1 & 0\end{bmatrix}\cdot \begin{bmatrix}a_2 & -b_1c_1\\ 1 & 0\end{bmatrix}\right)\cdot \begin{pmatrix}a_1\\ 1\end{pmatrix}$$
but it feels kind of ironic to me that to find the determinant we "only" have to multiply these $n$ matrices
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2003709",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
Evaluate $\lim_\limits{x \to +\infty}\left(x\ln (1+x)-x\ln x + \arctan\frac{1}{2x}\right)^{x^2\arctan x}$ $$\lim_{x \to +\infty}\left(x\ln (1+x)-x\ln x + \arctan\frac{1}{2x}\right)^{x^2\arctan x}$$
My attempt
\begin{align*}
&=\exp \lim_\limits{x \to +\infty} x^2\arctan x \cdot\ln\left[x\ln (1+x)-x\ln x + \arctan\frac{1}{2x}\right]\\
&=\exp \lim_{x \to +\infty} \frac{\pi}{2}x^2 \ln\left[\frac{\ln x}{\frac{1}{x}} + \frac{\ln \left(1+\frac{1}{x}\right)}{\frac{1}{x}}-\frac{\ln x}{\frac{1}{x}} + \arctan\frac{1}{2x}\right]\\
&=\exp \lim_{x \to +\infty} \frac{\pi}{2}x^2\ln\left[1 + \arctan \frac{1}{2x}\right]
\end{align*}
Here I don't know what to do. One thing I did was to multiply and divide by $\arctan \frac{1}{2x}$ in order to get rid of the logarithm. But I ended up with
$$\frac{\pi}{2}x^2\arctan \frac{1}{2x}$$
$\arctan \frac{1}{2x} \to \pi/2$ as $x \to +\infty$ but that doesn't lead to the right answer (I'd get a limit of $+\infty$).
|
$"=^{*}"$ denotes equality by L'Hospital's Rule.
Define $$\displaystyle f(x) = x\ln\Big(\frac{x+1}{x}\Big)\text{ and }g(x) = f(x)+\arctan\Big(\displaystyle \frac{1}{2x}\Big),\text { for } x > 0.$$ $$\displaystyle 1 = \lim_{x \to \infty}\frac{x}{x+1}=^{*}\lim_{x \to \infty}\frac{\ln\Big(\displaystyle \frac{x+1}{x}\Big)}{x^{-1}} = \lim_{x \to \infty}f(x)= \lim_{x\to\infty}g(x).$$An easy calculation shows that $\displaystyle \lim_{x\to\infty}g'(x)=0.$ $$\displaystyle {1}/{3} = \lim_{x \to \infty}\frac{x^3(32x^3+8x^2-1)}{6(x+1)^2(4x^{2}+1)^2}=\lim_{x \to \infty}\frac{x^4}{6}\Big(\frac{16x^2}{(4x^2+1)^2}-\frac{1}{x(x+1)^2}\Big)= \lim_{x\to\infty}\frac{g''(x)}{6x^{-4}}=^{*}\lim_{x\to\infty}\frac{g'(x)}{-2x^{-3}}=\lim_{x\to\infty}\Big(\frac{1}{g(x)}\Big(\frac{g'(x)}{-2x^{-3}}\Big)\Big)=^{*}\lim_{x\to\infty}\frac{\ln(g(x))}{x^{-2}}=\lim_{x\to\infty}\ln(g(x)^{\displaystyle x^2}).$$
Hence, $\displaystyle \lim_{x\to\infty}(g(x)^{\displaystyle x^2})= e^{\frac{1}{3}}$.Thus, $\displaystyle \lim_{x\to\infty}((g(x)^{\displaystyle x^{2}})^ {\displaystyle\arctan(x)})= e^{\frac{1}{3}\displaystyle\lim_{x\to\infty}\arctan(x)}=e^\frac{\pi}{6}.$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2009310",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
}
|
Showing $\sum\limits_{cyc}ab\le\frac14\left(\sum\limits_{cyc}a\right)^2$
$a,b,c,d>0$ with $a+b+c+d=4$ then $\sum\limits_{cyc}ab\le\frac14\left(\sum\limits_{cyc}a\right)^2$
How can I apply AM-GM here, is there a generalization, say we have $n$ variables summing up to $n$ then,
$\sum\limits_{cyc}a_1\cdot a_2\dots a_k\le\frac1{n^{k-1}}\left(\sum\limits_{cyc}a\right)^k$
because the following inequality should be also true
$\sum\limits_{cyc}abc\le\frac1{16}\left(\sum\limits_{cyc}a\right)^3$
|
Do we need that $a+b+c+d=4$ beacuse if we full expand we get that
$$a^2+b^2+c^2+d^2+2ac+2bd\geq 2ab+2bc+2cd+2da=2(a+c)(b+d)$$
or
$$(a+c)^2+(b+d)^2\geq 2(a+c)(b+d)$$
Which is true always, or maybe I did a mistake?
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2009517",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
}
|
Solve the integral $\int{\frac{x^2dx}{\sqrt[n]{x^2(3-x)}}} $ My attempt: $$\int{\frac{x^2dx}{\sqrt[n]{x^2(3-x)}}}=\int{\frac{x^2dx}{\sqrt[n]{\left( \frac{3-x}{x} \right)x^3 }}}$$
Let
$$ t^n = \frac{3-x}{x} ~~\rightarrow~~ nt^{n-1}dt=-3\frac{dx}{x^2} $$
or
$$ x = \frac{3}{t^n+1} ~~ \rightarrow~~dx=-\frac{3nt^{n-1}}{(t^n+1)^2}dt $$
But this substitution is not helping me.
|
Let $x=3z$
\begin{align}
\int \frac{x^{2-2/n}}{(3-x)^{1/n}} dx &= 3^{-1/n} \int \frac{x^{2-2/n}}{(1-x/3)^{1/n}} dx \\
&= 3^{3-3/n} \int z^{2-2/n} (1-z)^{-1/n} dz \\
&= 3^{3-3/n} \mathrm{B}_{z} \left( 3-\frac{2}{n}, 1-\frac{1}{n} \right) \\
&= 3^{3-3/n} \frac{n}{3n-2} z^{3-2/n} {}_{2}\mathrm{F}_{1}\left(3-\frac{2}{n},\frac{1}{n};4-\frac{2}{n};z \right) \\
&= \frac{1}{3^{1/n}} \frac{n}{3n-2} x^{3-2/n} {}_{2}\mathrm{F}_{1}\left(3-\frac{2}{n},\frac{1}{n};4-\frac{2}{n};\frac{x}{3} \right)
\end{align}
Note:
\begin{align}
\mathrm{B}_{z}(p,q) &= \int_{0}^{z} t^{p-1} (1-t)^{q-1} \mathrm{d}t \\
&= \frac{z^{p}}{p} \,{}_{2}\mathrm{F}_{1}(p,1-q;p+1;z)
\end{align}
The incomplete beta function and hypergeometric function.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2011861",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
}
|
Prime factor of $A=14^7+14^2+1$ Find a prime factor of $A=14^7+14^2+1$. Obviously without just computing it.
|
Same solution as Jyrki's, written differently
$$A=14^7+14^2+1=14^7-14+(14^2+14+1)\\
=14(14^6-1)+(14^2+14+1)=14(14-1)(14^2+14+1)(14^3+1)+(14^2+14+1)$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2012344",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
}
|
If the matrices $A^3 = 0$, $B^3=0$ and $AB=BA$ then show this: The question: If $A$ and $B$ are square matrices of the same type such that $A^3=0$, $B^3=0$ and $AB=BA$. Show that
$$\left(I+A+\frac{A^2}{2!}\right)\left(I+B+\frac{B^2}{2!}\right)=I+(A+B)+\frac{(A+B)^2}{2!}+\frac{(A+B)^3}{3!}+\frac{(A+B)^4}{4!}$$
This is how I started.
$$\left(I+A+\frac{A^2}{2!}\right)\left(I+B+\frac{B^2}{2!}\right)=I+(A+B)+\frac{A^2+B^2}{2!}+AB+\frac{AB^2}{2!}+\frac{A^2B}{2!}+\frac{A^2B^2}{2!2!}$$
I tried to get
$$\frac{A^2+B^2}{2!}+AB+\frac{AB^2}{2!}+\frac{A^2B}{2!}+\frac{A^2B^2}{2!2!}$$
to be equal to $$\frac{(A+B)^2}{2!}+\frac{(A+B)^3}{3!}+\frac{(A+B)^4}{4!}$$ But I'm not getting there.
|
Try to expand the right-hand side instead. You have
\begin{align}
\frac{1}{2}(A + B)^2 &= \frac{1}{2}A^2 + AB + \frac{1}{2}B^2\\[1mm]
\frac{1}{6}(A + B)^3 &= \frac{1}{2}A^2B + \frac{1}{2}AB^2\\[1mm]
\frac{1}{24}(A + B)^4 &= \frac{1}{4}A^2B^2,
\end{align}
where I have used that $A$ and $B$ commute as well as $A^k = B^k = 0$ for all integers $k \geq 3$. Summing everything up gives you the desired result.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2012593",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
}
|
find upper bound M taylor inequality Let $f(x) = \frac{1}{1-x}$
Find an upper bound M for $|f^{(n+1)}(x)|$ on the interval $[-1/2,1/2]$
I found derivative of the function and plugged in the points to see which one gives the largest value.
$f^{'}(x)=\frac{1}{(1-x)^2}$
$f^{'}(1/2)=4$
The solution was: $2^{n+2}(n+1)!$
I know the formula $|R_n(x)|= \frac{M|x-a|^{n+1}}{(n+1)!}$ but not sure how they got the M. Any help?
|
That is because you need to find the largest value of the $n$th derivative of $f$ over the interval [-0.5,0.5]. Given $f(x) = \dfrac{1}{1-x}$,
\begin{align*}
f'(x) & = \frac{1}{(1-x)^2}.\\
f''(x) & = \frac{1\cdot 2}{(1-x)^3}.\\
f'''(x) & = \frac{1\cdot 2\cdot 3}{(1-x)^4}.\\
\vdots & \qquad \vdots \\
f^{(n)}(x) & = \frac{1\cdot 2 \cdot\ldots\cdot n}{(1-x)^{n+1}} = \frac{n!}{(1-x)^{n+1}}.
\end{align*}
Thus, the maximum of $f^{(n)}(x)$ over the interval $[-0.5,0.5]$ is given by
$$ \max_{x\in [-0.5,0.5]} f^{(n)}(x) = \max_{x\in [-0.5,0.5]} \frac{n!}{(1-x)^{n+1}} = \frac{n!}{(1-0.5)^{n+1}} = 2^{n+1}n!.$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2013844",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
}
|
Find all solutions to $m^2 β n^2 = 105$, for which both m and n are integers Can I get a little help for this question?
Find all the integer solutions to $m^2 β n^2 = 105$.
|
$(m+n)(m-n) = 105$
$(m+n)(m-n) = 3\cdot 5\cdot 7$
4 Solutions :
Solve them by yourself.
$m-n = 1$, $m+n = 3\cdot 5\cdot 7$.
$m-n = 3$, $m+n = 5\cdot 7$-
$m-n = 5$, $m+n = 3\cdot 7$.
$m-n = 7$, $m+n = 3\cdot 5$.
But notice those $m^2 - n^2$, it's clear that $m$ and/or $n$ could be negative - it would not change a thing, thus $4\times 4=16$ solutions in total.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2013934",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
Prove using induction that $2^{4^n}+5$ is divisible by 21 I have to show, using induction, that $2^{4^n}+5$ is divisible by $21$. It is supposed to be a standard exercise, but no matter what I try, I get to a point where I have to use two more inductions.
For example, here is one of the things I tried:
Assuming that $21 |2^{4^k}+5$, we have to show that $21 |2^{4^{k+1}}+5$.
Now, $2^{4^{k+1}}+5=2^{4\cdot 4^k}+5=2^{4^k+3\cdot 4^k}+5=2^{4^k}2^{3\cdot 4^k}+5=2^{4^k}2^{3\cdot 4^k}+5+5\cdot 2^{3\cdot 4^k}-5\cdot 2^{3\cdot 4^k}=2^{3\cdot 4^k}(2^{4^k}+5)+5(1-2^{3\cdot 4^k})$.
At this point, the only way out (as I see it) is to prove (using another induction) that $21|5(1-2^{3\cdot 4^k})$. But when I do that, I get another term of this sort, and another induction.
I also tried proving separately that $3 |2^{4^k}+5$ and $7 |2^{4^k}+5$. The former is OK, but the latter is again a double induction.
Is there an easier way of doing this?
Thank you!
EDIT
By an "easier way" I still mean a way using induction, but only once (or at most twice). Maybe add and subtract something different than what I did?...
Just to put it all in a context: a daughter of a friend got this exercise in her very first HW assignment, after a lecture about induction which included only the most basic examples. I tried helping her, but I can't think of a solution suitable for this stage of the course. That's why I thought that there should be a trick I am missing...
|
$2^{4^k}+5 \equiv (-1)^{4^k}+5 \equiv 1+5 \equiv 0 \pmod 3$
Proof by induction that $2^{4^k}+5 \equiv 0 \pmod 7$ for all non negative ingeters $k$.
If $k=0$, then $2^{4^k}+5 \equiv 2+5 \equiv 0 \pmod 7$
Suppose that $2^{4^m}+5 \equiv 0$ for some non negative integer $m$.
Then
$2^{4^{m+1}}+5 \equiv
2^{4\cdot 4^m}+5 \equiv
(2^4)^{4^m}+5 \equiv
16^{4^m}+5 \equiv
2^{4^m}+5 \equiv 0 \pmod 7$
Hence, by mathematical induction $2^{4^k}+5 \equiv 0 \pmod 7$ for all non negative integers $k$.
It follows that $2^{4^k}+5 \equiv 0 \pmod{21}$ for all non negative integers $k$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2018239",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 3
}
|
Let $p$ be a prime of the form $3k+2$ that divides $a^2+ab+b^2$. Prove that $a,b$ are both divisible by $p$. [Solution verification] Problem:Let $p$ be a prime of the form $3k+2$ that divides $a^2+ab+b^2$ for some $a,b\in \mathbb{Z}$. Prove that $a,b$ are both divisible by $p$.
My Attempt: $a^2+ab+b^2\equiv 0 \pmod p\Rightarrow a^3\equiv b^3\pmod p\Rightarrow a^{3k}\equiv b^{3k}\pmod p.$
Next, observe that due to FLT we have $a^{3k+1}\equiv b^{3k+1}\pmod p.$ Now if $p\not|a$ and $p\not|b$, then $\gcd(a,p)=\gcd(b,p)=1.$ Therefore we can use can conclude that $$a^{\gcd(3k,3k+1)}\equiv b^{\gcd(3k,3k+1)}\pmod p\Rightarrow a\equiv b\pmod p.$$ Therefore $$a^2+ab+b^2\equiv 0 \pmod p\Rightarrow 3b^2\equiv 0\pmod p \text{ and } 3a^2\equiv 0\pmod p.$$ Which implies that $p|3$ which is a contradiction. Hence Proved.
I would like to know whether this proof is correct or not. I am unsure about the use of $\gcd$ in the exponent. Moreover, I acknowledge that this question has been asked before, but I've not seen any answer using this fact explicitly. The fact being: Let $\gcd(a,m)=\gcd(b,m)=1$, then if $a^{x}\equiv b^x\pmod m$ and $a^y\equiv b^y\pmod m\Rightarrow a^{\gcd(x,y)}\equiv b^{\gcd(x,y)}\pmod m.$
|
Start with $a^3\equiv b^3\bmod p$.
If $p$ divides $b$, then $p$ divides $a$.
If $p$ does not divide $b$, there is $c$ such that $bc \equiv 1 \bmod p$. Then $(ac)^3 \equiv 1 \bmod p$.
If $ac \not\equiv 1 \bmod p$, then the order of $ac \bmod p$ is $3$ and so $3$ divides $p-1$, which contradicts $p=3k+2$.
If $ac \equiv 1 \bmod p$, then $a \equiv b$ and $a^2+ab+b^2 \equiv 3b^2$. If this is a multiple of $p$, then $p=3$ because $p$ does not divide $b$, which contradicts $p=3k+2$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2018745",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
}
|
A Lagrange Mulipliers Problem My problem is this: Find the min and max values of $f(x,y)=x^2+3xy+y^2$ on the domain $(x-1)^2+y^2=1$.
I used lagrange multipliers to find that the $y$ coordinate satisfies $f\left(x\right)=8y+(46y^2)/3-16y^3-24y^4=0$ for any critical point $(x,y)$, and $x=2y^2+(2/3)y$
Using Wolfram Alpha, I found that this has roots (aprox.),
$y=-0.97110$
$y=-0.45311$
$y=0.75755$
$y=0$
Wolfram alpha tells me that the function is optimized at the first and third points. However, it is unable to get them in exact form that is not horrendous(multiple radicals involving i). But my professor is expecting an exact answer-and certainly not the exact answer I have.
Am I missing something?
|
$f(x,y,\lambda) = x^2 + y^2 + xy + \lambda (x^2 - 2x + y^2)\\
\frac {\partial f}{\partial x} = 2(1+\lambda) x + y - 2\lambda = 0\\
\frac {\partial f}{\partial y} = x + 2(1+\lambda) y = 0\\
\frac {\partial f}{\partial \lambda} = x^2 - 2x + y^2 = 0\\
$
$x,y,\lambda = 0$ is one solution (and is the min)
$2\lambda = -\frac {x+2y}{y}\\
2(1+\lambda) = -\frac{x}{y}$
$-\frac {x^2}{y} + y + \frac {x+2y}{y} = 0\\
-x^2 + y^2 + x+2y = 0\\
x^2 + y^2 - 2x = 0\\
2y^2 -x + 2y = 0\\
x = \sqrt {1-y^2}+1\\
2y^2 +2y-\sqrt{1-y^2} -1 = 0\\
2y^2 +2y-1 = \sqrt{1-y^2}\\
4y^4 + 8y^3 - 4y + 1 = 1-y^2\\
4y^4 + 8y^3 +y^2 - 4y = 0\\
y(4y^3 + 8y^2 + y - 4)= 0\\
$
And now we need the cubic formula:
$y = \sqrt[3]{\frac {31}{108}- \sqrt{(\frac {31}{108})^2-(\frac {13}{36})^3}}+\sqrt[3]{\frac {31}{108}+ \sqrt{(\frac {31}{108})^2-(\frac {13}{36})^3}}-\frac 23\approx 0.57638\\
x = \sqrt{1-y^2}+1 \approx 1.81718\\
f(x,y) = 4.68175$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2020489",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
}
|
if a,b,c,d are integers such that 5 divides ($a^4+b^4+c^4+d^4$) then 5 also divides (a+b+c+d) i want to show that
if a,b,c,d are integers such that 5 divides ($a^4+b^4+c^4+d^4$) then 5 also divides (a+b+c+d) but im not quite sure how to aproach this problem.
|
Hint
$$a^4-1=(a^2-1)(a^2+1)=(a-1)(a+1)(a^2-4+5)=(a-2)(a-1)(a+1)(a+2)+5(a^2-1)$$
Now, if $5 \nmid a$ one of $(a-2),(a-1),(a+1),(a+2)$ is divisible by $5$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2020821",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
}
|
Construction of multiplication and addition tables for GF(4) with Modulo ($x^2 + x + 1$) I am still trying to understand polynomial arithmetic with Galois Field.
Can someone explain to me what this question is asking to be constructed. I have searched all over google for GF(4) with modulo and I am not getting anywhere.
I have read over a few of the other posts here on the forums but none seem to apply to my specific situation. Can anyone help me understand GF with modulo of a polynomial?
Thanks in advance.
|
What you're asking saying $GF(4)$ modulo $x^2+x+1$ is equivalent to do:
$$GF(2)\cong \frac{\mathbb{F}_2[x]}{x^2+x+1}$$
So the sum of $GF(2)$ is the same as in $\mathbb{F}_2[x]$ and then doing the modulo $x^2+x+1$: Since all your elements of degree $2$ or greater can be reduced considering $x^n=x^2x^{n-2}=(x+1)x^{n-2}$, you only will have elements of degree $1$ or $0$, and this will be the table:
$$
\begin{array}{ | c | c | c |c| c|}
\hline
& \color{green}{0} & \color{green}{1} & \color{green}{x} & \color{green}{x+1}\\ \hline
\color{green}{0} & 0 & 1 & x & x+1 \\ \hline
\color{green}{1} & 1 & 0 & x+1 & x \\ \hline
\color{green}{x} & x & x & 0 & 1\\ \hline
\color{green}{x+1} & x+1 & x & 1 & 0\\ \hline
\hline
\end{array}$$
For the case of the multiplication, you can do it
$$
\begin{array}{ | c | c | c |c| c|}
\hline
& \color{green}{0} & \color{green}{1} & \color{green}{x} & \color{green}{x+1}\\ \hline
\color{green}{0} & 0 & 0 & 0 & 0 \\ \hline
\color{green}{1} & 0 & 1 & x & x+1 \\ \hline
\color{green}{x} & 0 & x & x+1 & 1\\ \hline
\color{green}{x+1} & 0 & x+1 & 1 & x\\ \hline
\hline
\end{array}$$
Why? It's clear that $1\cdot a=a$ for any element or why $0\cdot a=0$, so let's see the case with $x+1$. Look that $(x+1)x=x^2+x$, but by your quotient, $x^2+x+1\equiv 0$, so $x^2+x\equiv 1$ (we're modulo 2), and by the same reason, $(x+1)^2=x^2+2x+1\equiv x^2+1\equiv x$ And that's the way the multiplication happens.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2020956",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
If $ \sin \alpha + \sin \beta = a $ and $ \cos \alpha + \cos \beta = b $ , then show that $\sin(\alpha + \beta) = \frac {2ab } { a^2 + b^2} $ I've been able to do this, but I had to calculate $ \cos (\alpha + \beta) $ first. Is there a way to do this WITHOUT calculating $\cos(\alpha+\beta)$ first ?
Here's how I did it by calculating $\cos(\alpha+\beta)$ first
$ a^2 + b^2 = \sin ^2 \alpha + \sin ^2 \beta + 2 \sin \alpha \sin \beta + \cos ^2 \alpha + \cos ^2 \beta + 2 \cos \alpha \cos \beta $
$a^2 + b^2 = (\sin^2\alpha + \cos^2\alpha) + (\sin ^2 \beta + \cos^2 \beta) + 2(\cos\alpha\cos\beta + \sin\alpha\sin\beta)$
$a^2 + b^2 = 2 (1 + \cos(\alpha-\beta))$
$ \frac{a^2 + b^2}{2} = (1 + \cos(\alpha - \beta))$
$ b^2 - a^2 = (\cos ^2\alpha - \sin^2\alpha) + (\cos^2 \beta - \sin^2\beta) + 2\cos\alpha\cos\beta - 2\sin\alpha\sin\beta$
$b^2 - a^2 = (\cos^2\alpha - (1 - cos^2\alpha)) +(1-\sin^2\beta) - \sin^2\beta)) + 2(\cos\alpha\cos\beta - \sin\alpha\sin\beta) $
$b^2 - a^2 = 2 (\cos^2\alpha - \sin^2\beta + \cos(\alpha+\beta))$
$b^2 - a^2 = 2(\cos(\alpha+\beta)\cos(\alpha-\beta)+\cos(\alpha+\beta))$
$\frac{b^2 - a^2}{2} = \cos(\alpha+\beta)\{\cos(\alpha-\beta) + 1 \}$
$\frac{b^2 - a^2}{2} = \cos(\alpha+\beta)\{\frac{b^2+a^2}{2}\}$
$\cos(\alpha+\beta) = \frac {a^2 + b^2 } {a^2 - b^2}$
Then I just calculated $\sin(\alpha + \beta)$ by $1 - \cos^2(\alpha+\beta)$
|
$\sin \alpha+\sin \beta=2\sin(\frac{\alpha+\beta}{2})\cos(\frac{\alpha-\beta}{2})=a$ (1)
$\cos \alpha+\cos \beta=2\cos(\frac{\alpha+\beta}2)\cos(\frac{\alpha-\beta}2)=b$ (2)
Divide (1) and (2)
We get
$$\tan(\frac{\alpha+\beta}2) =\frac ab$$
We have the formula
$$\sin(\alpha+\beta) =\frac{2\tan(\frac{\alpha+\beta}2)}{1+\tan^2(\frac{\alpha +\beta}2)}$$
therefore $$\sin(\alpha+\beta)=\frac{2\frac ab}{1+\frac {a^2}{b^2}}=\frac{2ab}{a^2+b^2}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2021356",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
}
|
Compute $\int_{-\infty}^{+\infty}(1+\frac{1}{v^2})\exp(-\frac{u^2}{2\sigma_1^2\sigma_2^2}(\sigma_2v+\frac{\sigma_1}{v})^2)\,dv$
Compute $$\int_{-\infty}^{+\infty}(1+\frac{1}{v^2})e^{-\frac{u^2}{2\sigma_1^2\sigma_2^2}(\sigma_2v+\frac{\sigma_1}{v})^2}\,dv$$
Does this integral has a close form solution? What if $\sigma_1=\sigma_2=1$, $u=\sqrt2$, i.e, $$\int_{-\infty}^{+\infty}(1+\frac{1}{v^2})e^{-(v+\frac{1}{v})^2}dv$$
|
First we analyze the special case in which $u=\sqrt2$, and $\sigma_1=\sigma_2=1$. Noting that $\left(x+\frac 1x\right)^2=\left(x-\frac 1x\right)^2 +4$, we have
$$\begin{align}
I&=\int_{-\infty}^\infty\left(1+\frac1{x^2}\right)e^{-\left(x+\frac 1x\right)^2}\,dx\\\\
&=e^{-4}\int_{-\infty}^\infty\left(1+\frac1{x^2}\right)e^{-\left(x-\frac 1x\right)^2}\,dx \tag 1\\\\
&=2e^{-4}\int_{-\infty}^\infty e^{-u^2}\,du \tag 2\\\\
&=2\sqrt{\pi}e^{-4}
\end{align}$$
CAUTION:
In going from $(1)$ to $(2)$, we first split the integral as
$$\int_{-\infty}^\infty\left(1+\frac1{x^2}\right)e^{-\left(x-\frac 1x\right)^2}\,dx=\int_{-\infty}^0 \left(1+\frac1{x^2}\right)e^{-\left(x-\frac 1x\right)^2}\,dx+\int_{0}^\infty\left(1+\frac1{x^2}\right)e^{-\left(x-\frac 1x\right)^2}\,dx \tag 3$$
Then, we separately enforce the substitution $u=x-\frac1x$ in each of the integrals on the right-hand side of $(3)$.
For the general case, we enforce the substitution $x\to \sqrt{\frac{\sigma_1}{\sigma_2}}x$ to obtain
$$\begin{align}
I&=\sqrt{\frac{\sigma_2}{\sigma_1}}\int_{-\infty}^\infty \left(\frac{\sigma_1}{\sigma_2}+\frac1{x^2}\right)e^{-\frac{u^2}{2\sigma_1\sigma_1}\left(x+\frac 1x\right)^2}\,dx\\\\
&=\sqrt{\frac{\sigma_2}{\sigma_1}}\int_{-\infty}^\infty \left(1+\frac1{x^2}\right)e^{-\frac{u^2}{2\sigma_1\sigma_1}\left(x+\frac 1x\right)^2}\,dx+\sqrt{\frac{\sigma_2}{\sigma_1}}\left(\frac{\sigma_1}{\sigma_2}-1\right)\int_{-\infty}^\infty e^{-\frac{u^2}{2\sigma_1\sigma_1}\left(x+\frac 1x\right)^2}\,dx\\\\
&=2\sqrt{\frac{\sigma_2}{\sigma_1}}e^{-2\frac{u^2}{\sigma_1\sigma_1}}\int_{-\infty}^\infty e^{-\frac{u^2}{2\sigma_1\sigma_1}\left(x-\frac 1x\right)^2}\,d\left(x-\frac1x\right)\\\\
&+\sqrt{\frac{\sigma_2}{\sigma_1}}\left(\frac{\sigma_1}{\sigma_2}-1\right)e^{-2\frac{u^2}{\sigma_1\sigma_1}}\int_{-\infty}^\infty e^{-\frac{u^2}{2\sigma_1\sigma_1}\left(x-\frac 1x\right)^2}\,dx\\\\
&=2\sigma_2e^{-2\frac{u^2}{\sigma_1\sigma_1}}\sqrt{\frac{2\pi}{u^2}}+\sqrt{\frac{\sigma_2}{\sigma_1}}\left(\frac{\sigma_1}{\sigma_2}-1\right)e^{-\frac{u^2}{2\sigma_1\sigma_1}}\int_{-\infty}^\infty e^{-\frac{u^2}{2\sigma_1\sigma_1}\left(x-\frac 1x\right)^2}\,dx \tag4
\end{align}$$
Making the substitution $x\to 1/x$ in the integral on the right-hand side of $(4)$ over the intervals $0$ to $\infty$ and $-\infty$ to $0$ separately, reveals
$$\begin{align}
\int_{-\infty}^\infty e^{-\frac{u^2}{2\sigma_1\sigma_1}\left(x-\frac 1x\right)^2}\,dx&=\int_{-\infty}^\infty e^{-\frac{u^2}{2\sigma_1\sigma_1}\left(x-\frac 1x\right)^2}\frac{1}{x^2}\,dx\\\\
&= \frac12\int_{-\infty}^\infty \left(1+\frac1{x^2}\right)e^{-\frac{u^2}{2\sigma_1\sigma_1}\left(x-\frac 1x\right)^2}\,dx\\\\
&=\int_{-\infty}^\infty e^{-\frac{u^2}{2\sigma_1\sigma_1}\left(x-\frac 1x\right)^2}\,d\left(x-\frac1x\right)\\\\
&=\sqrt{\frac{2\pi \sigma_1\sigma_2}{u^2}}\tag 5
\end{align}$$
Substituting $(5)$ into $(4)$ yields
$$\bbox[5px,border:2px solid #C0A000]{I=\left(\sigma_1+\sigma_2\right)\sqrt{\frac{2\pi}{u^2}}e^{-2\frac{u^2}{\sigma_1\sigma_1}}}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2023685",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
}
|
Real values of $a$ for $x^4+(a-1)x^3+x^2+(a-1)x+1=0$ to have at least two negative roots
Problem Statement:-
Find all the values of $a$ for which the equation
$$x^4+(a-1)x^3+x^2+(a-1)x+1=0$$
possess at least two negative roots.
I know that there has been a post regarding this same problem here, but I have a different issue with the problem than that posted in the OP in the above post.
So, I don't have an issue of finding a solution and coincidentally I had the same approach as the one posted by RicardoCruz.
But still if you have a solution than you are more than welcome to post one, well come on who doesn't want some good reputation change ;P.
My Attempt at a solution:-
For those who wanna skip to my main issue may start to read from the next quote.
We can see very clearly that $x=0$ isn't a root of the equation $$x^4 +(a-1)x^3 +x^2 +(a-1)x+1=0$$
hence it is safe to divide the equation throughout by $x^2$ for arriving at the condition which the problem wants.
So, we get
$$\left(x^2+\dfrac{1}{x^2}+2\right)+(a-1)(x+\dfrac{1}{x})-1=0\\
\implies \left(x+\dfrac{1}{x}\right)^2+(a-1)(x+\dfrac{1}{x})-1=0$$.
Now, let $z=x+\dfrac{1}{x}$ then the equation becomes
$$z^2+(a-1)z-1=0\\
\implies z=\dfrac{-(a-1)\pm\sqrt{\left(a-1\right)^2+4}}{2}$$
Now, its necessary to note that $\sqrt{\left(a-1\right)^2+4}\gt(a-1)$, hence for the roots to be negative $$\left(x+\dfrac{1}{x}\right)=\dfrac{-(a-1)-\sqrt{\left(a-1\right)^2+4}}{2}$$
As, we know that $\text{A.M. $\ge$ G.M.}$ for non negative real numbers, and as we are dealing with negative roots of the equation, so we will be applying $\text{A.M. $\ge$ G.M.}$ to $-x,-\dfrac{1}{x}$, we get
$$\dfrac{-\left(x+\dfrac{1}{x}\right)}{2}\ge1\implies x+\dfrac{1}{x}\le-2\\
\implies \left(x+\dfrac{1}{x}\right)=\dfrac{-(a-1)-\sqrt{\left(a-1\right)^2+4}}{2}\le -2$$
But as the roots are supposed to be distinct hence, we have to exclude the case of equality, because it occurs when $x=\dfrac{1}{x}=-1$, which results in $x+\dfrac{1}{x}=-2\implies (x-1)^2=0$, hence $x$ has a repeated root. So, we get
$$\dfrac{-(a-1)-\sqrt{\left(a-1\right)^2+4}}{2}\lt -2\implies \sqrt{\left(a-1\right)^2+4}\gt 5-a$$
Now, this is where I am having problem, that's right solving the "inequality". I always had thought of this before but never tried to test it before, so bear with it and follow me on to a trip to solve inequalities in a weird way.
I did a case study for the inequality, which is as follows:-
Case-1:-
If $(5-a)\gt 0\implies a\lt 5$, then there is no problem in squaring the inequality as both sides are positive.
$$\left(a-1\right)^2+4\gt a^2+25-10a\implies a\gt \dfrac{5}{2}$$
$$\therefore a\in\left(\dfrac{5}{2},5\right)$$
Case-2:-
If $5-a\lt0$, then we might be faced with a problem cause, there might be a case where $|5-a|\gt\sqrt{\left(a-1\right)^2+4}$, which would make the inequality $\sqrt{\left(a-1\right)^2+4}\gt 5-a$ on squaring change the sign of inequality to be in accordance with the condition $|5-a|\gt\sqrt{\left(a-1\right)^2+4}$.
So, on solving the inequality we get,
$$\sqrt{\left(a-1\right)^2+4}\gt 5-a\implies (a-1)^2+4\lt(5-a)^2$$
Now, why did I change the sign of inequality was due to the fact that I was finding the interval where(as per the condition I stated before) $|5-a|\gt\sqrt{\left(a-1\right)^2+4}$. So, just like $5\gt-6$ but on squaring the inequality we get $25\lt36$. So that's what I did.
This results in $a\lt\dfrac{5}{2}\cup a\gt5\implies a\in\emptyset$
So, on combining both the cases we get $a\in\left(\dfrac{5}{2},5\right)$
But from RobertCruz's solution and the books answer it seems that $ a\in\left(\dfrac{5}{2},\infty\right)$ is the correct solution, so what am I doing wrong here.
|
From the place where you have $x+1/x\; (=z)$ in terms of $a,$ write $x+1/x=r$ where $$r=(-(a-1)-\sqrt {(a-1)^2+4})/2 <0.$$ So $x^2-rx+1=0,$ giving $$x=(r\pm \sqrt {r^2-4}\;)/2<0.$$ This yields two negative roots iff $r^2-4>0,$ that is, iff $|r|>2$ because $r^2-4<0$ yields no real negative $x$, and $r^2-4=0$ yields exactly one negative root. In terms of $a,$ this is $$2<|(-(a-1)-\sqrt {(a-1)^2+4}\;)/2|=|(\;a-1+\sqrt {(a-1)^2+4}\;)/2| .$$
(Because we must have $a>1,$ else we have $ x^4+(a-1)x^3+x^2+(a-1)x+1>0$ whenever $x<0$.)
For $a>1$ this is equivalent to $a-1>3/2,$ that is, $a>5/2.$
Another method would be: Let $f(x)=x^4+(a-1)x^3+x^2+(a-1)x+1$ with $a>1.$ Then $$(f(x)<0\land x<0)\iff$$ $$ a-1=-\frac {x^4+x^2+1}{-x^3-x}=$$ $$=\frac {(x+1/x)^2x^2-x^2}{-x^2(x+1/x)}=$$ $$=\frac {u^2-1}{u}$$ where $u=|x|+|1/x|=-x-1/x.$
The AGM inequality implies that $|x|+|1/x|\geq 2.$ Now show that $\frac {u^2-1}{u}\geq \frac {3}{2}$ for $u\geq 2.$ And I leave the rest to you.
Remark. For $u\geq 2$ let $u=2+b.$ Then $\frac {u^2-1}{u}-\frac {3}{2}=\frac {8b^2+2b}{2b+4}.$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2023856",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
}
|
How can $\frac{x^3-4x^2+4x}{x^2-4}$ be both $0$ and "undefined" when $x = 2$? Suppose I have a function defined as $$F(x)= \frac{x^3-4x^2+4x}{x^2-4}$$
Now I want to find the value of $F(2)$. I can do it in 2 ways:
*
*Put $x=2$ and solve the function. It will give:
$$F(2)=\frac{0}{0}$$ which is not defined.
*Solve $F(x)$ first and then put $x=2$.
$$F(x)= \frac{x(x-2)^2}{(x-2)(x+2)}=\frac{x(x-2)}{x+2}$$
It will give $${F(2)=\frac{0}{4}}$$ which is zero.
How can zero equal not defined?
|
The reason is simply that
$$ \frac{x^3-4x^2+4x}{x^2-4}$$
and
$$\frac{x(x-2)}{x+2}$$
are two different expressions.
Their values indeed coincide for $x\ne2$ (and they are both undefined for $x=-2$), but they are not "mandated" to be equal at $x=2$.
This symptom reflects the difference between
$$\frac{x-2}{x-2}$$ and $$1.$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2027810",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 2
}
|
Find the locus. let $A$ & $B$ be two points on the ellipse $\displaystyle \frac{x^2}{a^2}+\frac{y^2}{b^2} = 1$ such that they subtend $90^{0}$
on the centre of the ellipse. what is the locus of points of intersection of tangents at $A$ & $B$
|
Let $\mathcal{E}$ be the ellipse $\left\{\; (u,v) \in \mathbb{R}^2 \;: \;\frac{u^2}{a^2} + \frac{v^2}{b^2} = 1\; \right\}$.
For any point $P = (x,y)$ outside $\mathcal{E}$, there are two tangents of $\mathcal{E}$ passing through $P$.
Let $A = (u_1,v_1)$ and $B = (u_2,v_2)$ be the intersections of these two tangents with $\mathcal{E}$.
For simplicity of argument, let us first assume $u_1, u_2 \ne 0$.
Let $Q = (u,v)$ be either $A$ or $B$ and let $\displaystyle\;t = \frac{v}{u}\;$.
In order for $QP$ to be a tangent line of $\mathcal{E}$, $P$ and $Q$ satisfies
$$\frac{xu}{a^2} + \frac{yv}{b^2} = 1\iff u\left(\frac{x}{a^2}+\frac{yt}{b^2}\right) = 1 \implies u^2\left(\frac{x}{a^2}+\frac{yt}{b^2}\right)^2 = 1$$
Since $Q$ lies on $\mathcal{E}$,
$$\frac{u^2}{a^2} + \frac{v^2}{b^2} = 1 \iff u^2\left(\frac{1}{a^2} + \frac{t^2}{b^2}\right) = 1$$
This implies $t$ satisfies a quadratic equation.
$$\left(\frac{x}{a^2}+\frac{yt}{b^2}\right)^2 = \frac{1}{a^2} + \frac{t^2}{b^2}
\quad\iff\quad
\frac{t^2}{b^2}\left(1-\frac{y^2}{b^2}\right) - 2t\frac{xy}{a^2b^2} + \frac{1}{a^2}\left(1 - \frac{x^2}{a^2}\right) = 0
$$
It is clear $t_1 = \frac{v_1}{u_1}$ and $t_2 = \frac{v_2}{u_2}$ are the two roots of this polynomial. In order for $A$, $B$ to subtend a right angle at center of $\mathcal{E}$, we need $$u_1u_2 + v_1v_2 = 0 \iff u_1u_2(1+t_1t_2) = 0 \iff t_1t_2 = -1$$
Apply Vieta's formula to
above quadratic polynomial, the condition becomes
$$\frac{\frac{1}{a^2}\left(1 - \frac{x^2}{a^2}\right)}{\frac{1}{b^2}\left(1-\frac{y^2}{b^2}\right)} = t_1t_2 = -1
\quad\iff\quad \frac{x^2}{a^4} + \frac{y^2}{b^4} = \frac{1}{a^2}+\frac{1}{b^2}$$
This is the equation of another ellipse:
$$\mathcal{E}' = \left\{ \; (x,y) \in \mathbb{R}^2 \;:\; \frac{b^2x^2}{a^2(a^2+b^2)} + \frac{a^2y^2}{b^2(a^2+b^2)} = 1\; \right\}$$
When $u_1$ or $u_2$ is zero, it is easy to see the tangent lines through $A$, $B$ intersects at one of the 4 points $(\pm a,\pm b)$. Since $\mathcal{E}'$ also contains these 4 points, $\mathcal{E}'$ is the locus we seek.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2028467",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
}
|
On the asymptotic behaviour of the Cauchy product of harmonic numbers $\sum_{k=1}^n H_k H_{n-k+1}$ In this post we take in our hands two simple tools. The first is the generating function of the harmonic numbers, you can see it in this Wikipedia, section 3, that holds for $|z|<1$. The second is the Cauchy product formula $$ \left( \sum_{n=1}^\infty a_n \right) \left( \sum_{n=1}^\infty b_n \right) = \sum_{n=1}^\infty a_k b_{n-k+1},$$ where the convergence is assumed, there are theorems that tell you when is convergent.
In our case our factors are convergents because are the same, and this is well defined: we consider the square of the generating function and after we take the integral $\int_0^{1/2}$ to get if there are no mistakes $$\sum_{n=1}^\infty\frac{1}{(n+2)2^{n+2}}\sum_{k=1}^{n}H_k H_{n-k+1}=\int_0^{1/2}\left(\sum_{n=1}^\infty H_n z^n\right)^2 dz=\int_0^{1/2}\left(\frac{\log(1-z)}{1-z}\right)^2 dz.$$ Notice that is required use the Cauchy product and swap the sign of the series and integral. The integral in RHS is computed in a closed form as $2(\log 2-1)^2$.
I don't know if is well known from the literature this sequence $$\sum_{k=1}^n H_k H_{n-k+1}.$$ The sequence starts as $1,3, \frac{71}{12}, \frac{29}{3}, \frac{638}{45}, \frac{349}{18}, \frac{14139}{560}, \frac{79913}{2520}\ldots$ And I would like to know what's about its asymptotic behaviour. One know that the plot of this arithmetic function is smooth (with a well defined slope), see how Wolfram Alpha can show us the plot of partial sums if you type the code
sum HarmonicNumber[k]HarmonicNumber[1000-k+1], from k=1 to 1000
in its online calculator.
Question. What's about the asymptotic behaviour of $$\sum_{k=1}^{n}H_k H_{n-k+1}$$
as $n\to\infty$? I am saying a big oh, or small oh statement or your answer as an asymptotic equivalence. You can provide me hints to get it with summation (I don't know if it is easy), or have you another idea? Thanks in advance.
Thus feel free to add hints, references if you need it, or a more detailed answer.
|
Lemma For $n \ge 2$,
$$\begin{align}
\rho_n \stackrel{def}{=}\sum_{k=1}^{n-1} H_k H_{n-k}
&= (n+1)\left[\psi'(n+2)-\psi'(2) + (\psi(n+2)-\psi(2))^2 \right]\\
&= (n+1)\left[\psi'(n+2)-\left(\frac{\pi^2}{6}-1\right) + (H_{n+1} - 1)^2\right]
\end{align}
$$
where $\psi(x)$ is the digamma function.
In following expansion
$$
\rho_n = \sum_{k=1}^{n-1} H_k H_{n-k}
= \sum_{k=1}^{n-1}\left(\sum\limits_{p=1}^k \frac{1}{p}\right)\left(\sum_{q=1}^{n-k}\frac{1}{q}\right)
$$
the term $\displaystyle\;\frac{1}{pq}$ appear $(n+1) - (p+q)$ times. This leads to
$$\begin{align}
\rho_n
&= (n+1)\sum_{\substack{1 \le p, q\\p+q \le n}}\frac{1}{pq} - 2\sum_{\substack{1 \le p, q\\p+q \le n}}\frac{1}{p}
= (n+1)\sum_{\substack{1 \le p, q\\p+q \le n}}\frac{1}{pq} - 2\sum_{p=1}^{n-1}\frac{n-p}{p}\\
&= (n+1)\sum_{\substack{1 \le p, q\\p+q \le n}}\frac{1}{pq} - 2n(H_{n} - 1)\\
\implies\quad \frac{\rho_n}{n+1} - \frac{\rho_{n-1}}{n} &=
\sum_{\substack{1 \le p, q\\p+q = n}}\frac{1}{pq} - \frac{2n}{n+1}(H_{n} - 1) + \frac{2(n-1)}{n}(H_{n-1}-1)\\
&= \frac{1}{n}\sum_{\substack{1 \le p, q\\p+q = n}}\left(\frac{1}{p}+\frac{1}{q}\right) + \frac{2}{n+1}(H_n-1) + \frac{2}{n} H_{n-1} = \frac{2}{n+1}(H_{n} - 1)
\end{align}
$$
Notice last expression can be rewritten as
$$\frac{2}{n+1}(H_{n} - 1) = 2(H_{n+1}-H_n)(H_n-1) = (H_{n+1}-1)^2 - (H_n-1)^2 - \frac{1}{(n+1)^2}\\
= (\psi'(n+2) + (\psi(n+2)-\psi(2))^2)
- (\psi'(n+1) + (\psi(n+1)-\psi(2))^2)$$
The expression
$\displaystyle\;\frac{\rho_n}{n+1} - (\psi'(n+2) + (\psi(n+2)-\psi(2))^2)\;$
is a constant independent of $n$.
Evaluate this expression $n = 2$, we find the constant equal to $-\psi'(2)$ and our lemma follows.
For large $x$, we have following asymptotic expansion of $\psi$ and $\psi'$.
$$
\psi(x) \asymp
\log(x) - \frac{1}{2x} - \sum_{k=1}^\infty \frac{B_{2k}}{2k\;x^{2k}}
\quad\text{ and }\quad
\psi'(x) \asymp \frac{1}{x} + \frac{1}{2x^2} + \sum_{k=1}^\infty \frac{B_{2k}}{x^{2k+1}}
$$
where $B_k$ is the $k^{th}$ Bernoulli number. As a result,
$$\rho_{n-2} \asymp (n-1)\left[\small
\frac{1}{n} + \frac{1}{2n^2} + \sum_{k=1}^\infty \frac{B_{2k}}{n^{2k+1}} -
\left(\frac{\pi^2}{6} - 1\right)
+ \left(
\log(n) - \frac{1}{2n} - \sum_{k=1}^\infty \frac{B_{2k}}{2k\;n^{2k}} - (1-\gamma)
\right)^2
\right]
$$
For large $n$, the leading behavior of $\rho_n$ is $n\log(n)^2$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2029378",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
}
|
Find $\lim \limits_{n \to \infty}\left( 1 + \sqrt{2} + \sqrt[3]{3} + \dots \sqrt[n]{n} \right) \ln{2n+1 \over n}$ I am looking for
$$\lim \limits_{n \to \infty}\left( 1 + \sqrt{2} + \sqrt[3]{3} + \dots \sqrt[n]{n} \right) \ln{2n+1 \over n}$$
We notice $\ln{2n+1 \over n} = \ln\left({1 + {n+1 \over n}}\right)$. We also know that ${x \over 1 + x} \le \ln(1+x)$. From this, we get
$${n+1 \over 2n+1} \le \ln\left({1 + {n+1 \over n}}\right)$$
Then we notice $n \le 1 + \sqrt{2} + \sqrt[3]{3} + \dots \sqrt[n]{n}$, so
$$n{n+1 \over 2n+1} \le \left( 1 + \sqrt{2} + \sqrt[3]{3} + \dots \sqrt[n]{n} \right) \ln{2n+1 \over n}$$
$n{n+1 \over 2n+1} \to + \infty$, so
$$\lim \limits_{n \to \infty}\left( 1 + \sqrt{2} + \sqrt[3]{3} + \dots \sqrt[n]{n} \right) \ln{2n+1 \over n} = + \infty$$
Is this reasoning correct?
|
Your reasoning is correct. But you could say that all terms in bracket (the first part) are greater than 1, and second part goes to $ln(2)$, which is non zero. So the total limit goes to Infinity.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2029922",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
}
|
Finding height and area of non-right triangle - Heron's Formula? I would like to calculate the area for a triangle such that $a^2+b^2-c^2=1$ (an almost Pythagorean triple).
I know that the triangle is non-right, so I would like to use $\text{Area}=\frac{1}{2}ab\sin C$... but I do not know how to represent $\sin C$ since I don't have any actual values.
I know about Heron's formula where $S=\frac{a+b+c}{2}$ and $\text{Area}=\sqrt{s(s-a)(s-b)(s-c)}$, but I feel like that gets too lengthy with our side lengths?
Edit to add: For $Area = \frac{1}{4}\sqrt{4a^2b^2-1}$ as shown by @zipirovich, can this area ever be an integer if $a,b,c$ are positive integers and $a,b >1$? Or, is this impossible?
|
Your equation can be rewritten as $c^2=a^2+b^2-1$. Comparing it with the Law of Cosines $c^2=a^2+b^2-2ab\cos C$, we can see that $2ab\cos C=1$ or $\cos C=\frac{1}{2ab}$. Then
$$\sin C=\sqrt{1-\cos^2C}=\sqrt{1-\left(\frac{1}{2ab}\right)^2}=\sqrt{1-\frac{1}{4a^2b^2}},$$
and the area is
$$\text{Area}=\frac{1}{2}ab\sin C=\frac{1}{2}ab\sqrt{1-\frac{1}{4a^2b^2}}=\frac{1}{2}\sqrt{a^2b^2-\frac{1}{4}}=\frac{1}{4}\sqrt{4a^2b^2-1}.$$
So there's no single answer because there are many such triangles, but here's an answer in the sense of having a formula for it. Going back to where we found that $\cos C=\frac{1}{2ab}$: as long as $2ab>1$, we can always find the angle $C$ and build such a triangle.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2031032",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
}
|
Solve system of equation using matrices (4 variables) The Question:
Solve using matrices.
$$2w-2x-2y+2z=10\\w+x+y+z=-5\\3w+x-y+4z=-2\\w+3x-2y+2z=-6$$
My work:
$$
\begin{bmatrix}
2&-2&-2&2&10\\
1&1&1&1&-5\\
3&1&-1&4&-2\\
1&3&-2&2&-6\\
\end{bmatrix}
\rightarrow
\begin{bmatrix}
1&-1&-1&1&5\\
0&2&2&0&-10\\
0&4&-1&1&-11\\
0&-2&3&-1&1\\
\end{bmatrix}
\rightarrow\\
\begin{bmatrix}
1&-1&-1&1&5\\
0&2&2&0&-10\\
0&0&-5&0&9\\
0&0&5&-1&-9\\
\end{bmatrix}
\rightarrow
\begin{bmatrix}
1&-1&-1&1&5\\
0&2&2&0&-10\\
0&0&-5&0&9\\
0&0&0&-1&0\\
\end{bmatrix}\\[6ex]
z=0, \;y=\frac{-9}5,\; x=\frac{34}5,\; w=10
$$
The correct answer is $z=-1, \;y=-2,\; x=-3,\; w=1$. What did I do wrong?
|
The second matrix should be:
$$\begin{bmatrix}1&-1&-1&1&5\\0&2&2&0&-10\\0&4&2&1&-17\\0&4&-1&1&-11\end{bmatrix}$$
Note the row operations were:
$1)\; -R_1+R_2\to R_2\\
2) \;-3R_1+R_3\to R_3\\
3) \;-R_1+R_4\to R_4$
Also, in your "correct" answer, $y$ should be $-2$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2036797",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
}
|
How many three-digit numbers can be generated from 1, 2, 3, 4, 5, 6, 7, 8, 9, such that the digits are in ascending order? Solution for the same is given below
Numbers starting with 12 β 7 numbers
Numbers starting with 13 β 6 numbers;
14 β 5, 15 β 4, 16 β 3, 17 β 2, 18 β 1.
Thus total number of
numbers starting from 1 is given by the sum of 1 to 7 = 28.
Number of numbers starting from 2- would be given by the sum of 1 to 6 = 21
Number of numbers starting from 3- sum of 1 to 5 = 15
Number of numbers starting from 4 β sum of 1 to 4 = 10
Number of numbers starting from 5 β sum of 1 to 3 = 6
Number of numbers starting from 6 = 1 + 2 = 3
Number of numbers starting from 7 = 1
Thus a total of: 28 + 21 + 15 + 10 + 6 + 3 + 1 = 84 such numbers.
suggest some trick to approach without thinking about this gibberish given above?
|
You need to choose three digits from nine. Once the digits have been selected, the order in which they should be placed is determined. Hence $\binom{9}{3}$ numbers are possible.
You are seeing binomial coefficients in your method too: once the first digit is fixed, there are $\binom{n}{2}$ ways to choose the remaining two digits, where $n$ is the number of digits greater than the first digit. Equating our answers gives a binomial coefficient identity,
$$
\binom{9}{3}=\binom{8}{2}+\binom{7}{2}+\binom{6}{2}+\binom{5}{2}+\binom{4}{2}+\binom{3}{2}+\binom{2}{2},
$$
which you can think about.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2037138",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
}
|
How do I calculate surface integral? For the vector field $a = [-z^2β2z,-2xz+2y^2,-2xz-2z^2]^T$ and the area $F$ on the cylinder $x^2 + z^2 = 4$ , which is above the ground plane $z = 0$ , in front of the plane $x = 0$ and between the cross plane $y = 0$ and lies to the their parallel plane $y = 2$ , calculate the following integral:
$\int_{F}^{} \! a\cdot dn \, = ?$
So I use that:
$x=2cos(u)$
$y=v$
$z=2sin(u)$
and than I calculate normal vector.
I get integral
$\int_{0}^{2}\int_{0}^{\Pi/2}\begin{pmatrix}-z^2β2z\\-2xz+2y^2\\-2xz-2z^2\end{pmatrix}\cdot \begin{pmatrix}2sin(u)\\ 0\\ 2sin(u)\end{pmatrix}dudv $
$\int_{0}^{2}\int_{0}^{\Pi/2}\begin{pmatrix}-(2sin(u))^2β4sin(u)\\-2(2cos(u))(2sin(u))+2v^2\\-2(2cos(u))(2sin(u))-2(2sin(u))^2\end{pmatrix}\cdot \begin{pmatrix}2sin(u)\\ 0\\ 2sin(u)\end{pmatrix}dudv $
$\int_{0}^{2}\int_{0}^{\Pi/2}\begin{pmatrix}-8sin^3(u)β8sin^2(u)\\0\\-16cos(u)sin^2(u))-16sin^2(u)\end{pmatrix} $
I was to lazy to replace x,y and z, but at the end I get wrong solution and I need a lot of time to calculate, is there any better method?
|
If you evaluate the dot product correctly, you get
\begin{align*}
\int_0^2\int_0^{\pi/2} &\left(-8\sin^3 u -8 \sin^2 u - 16 \cos u \sin^2 u - 16 \sin^2 u\right) \,du\,dv\\
&=(-8)\int_0^2 1 \,dv \cdot \int_0^{\pi/2} \left(\sin^3 u + \sin^2u + 2 \cos u \sin^2 u + 2 \sin^2 u\right)\,du \\
&=(-16)\int_0^{\pi/2} \left((1-\cos^2 u)\sin u + 2 \cos u \sin^2 u + 3 \sin^2 u\right)\,du \\
&=(-16)\int_0^{\pi/2} \left(\sin u -\cos^2 u \sin u + 2 \cos u \sin^2 u + 3 \sin^2 u\right)\,du \\
\end{align*}
Now
\begin{align*}
\int_0^{\pi/2} \sin u \,du &= \left.-\cos u\right|^{\pi/2}_0 = 1 \\
\int_0^{\pi/2} \cos^2 u \,\sin u \,du
&= \int_0^1 y^2\,dy = \frac{1}{3}\quad \text{(substitute $y=\cos u$)}\\
\int_0^{\pi/2} \sin^2u \cos u\,du &= \frac{1}{3}\quad \text{(substitute $y=\sin u$)}\\
\int_0^{\pi/2} \sin^2u\,du
&=\int_0^{\pi/2} \left(\frac{1-\cos 2u}{2}\right)\,du
=\left[\frac{u}{2} - \frac{1}{4}\sin 2u\right]^{\pi/2}_0 = \frac{\pi}{4}
\end{align*}
So the integral is
$$
(-16)\left[1 -\frac{1}{3} + \frac{2}{3} + 3 \cdot \frac{\pi}{4}\right]
= -\frac{64}{3} - 12\pi
$$
(verified with Wolfram Alpha.)
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2045013",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
}
|
How to find $ker(A)$ How can I find $ker(A)$ for
$A=
\begin{pmatrix}
2 & 1 & 1 \\
4 & 2 & 2 \\
3 & 0 & 1
\end{pmatrix}
$?
What does this represent?
Edit: OK, using the Hints i get
$2x+y+z=0$
$3x+z=0$
$y-x=0$
$y=x$
And now?
|
Straight from wikipedia:
$$\ker(A) = \left\{\mathbf{v}\in V : A\cdot \mathbf{x} = \mathbf{0}\right\}$$
In your case $V=\mathbb{R}^3$ so you want to find all vectors in the usual 3 space which make this true:
$$A\bf{x}=
\begin{pmatrix}
2 & 1 & 1 \\
4 & 2 & 2 \\
3 & 0 & 1
\end{pmatrix}\begin{pmatrix}x\\y\\z\end{pmatrix}=\begin{pmatrix}0\\0\\0\end{pmatrix}$$
The science here is getting $A$ to reduced row echelon form. If that reduced matrix is $B$, then $\ker(A) = \ker(B)$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2045504",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
}
|
Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.