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If $f(x)\mod (x-1)=5$, $f(x)\mod (x+1)=3$, $f(x)\mod (x+2)=5$ then find the remainder when $f(x)$ is divided by $x^3+2x^2-x-2$
If $f(x)\mod (x-1)=5$, $f(x)\mod (x+1)=3$, $f(x)\mod (x+2)=5$ then find the remainder when $f(x)$ is divided by $x^3+2x^2-x-2$
I don't have any idea as to how to proceed with this question. I have figured out that $x^3+2x^2-x-2=(x-1)(x+1)(x+2)$.
But I don't know how to use this information to find the remainder $x^3+2x^2-x-2$ would give. A small hint as to how to proceed would be enough.
|
The final condition implies that the divisor is of order $3$, so the remainder is of order $2$. We can write $$f(x)=(x-1)(x+1)(x+2)Q(x) +ax^2+bx+c$$ We have $f(1) =5$, so $$a+b+c=5...(1)$$ and similarly $f(-1)=3$ and $f(-2)=5$, so , $$a-b+c=3...(2)$$ and $$4a-2b+c=5$$ Solving for $a,b,c$, we get $a=1, b=1, c=3$. Hence our remainder is $$ax^2+bx+c = x^2+x+3.$$
|
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|
Mean and mode of a Beta random variable A continuous random variable is said to have a $Beta(a,b)$ distribution if its density is given by
$f(x) = (1 / \text{B}(a,b))x^{a-1} (1-x)^{b-1}$ if $0 < x < 1$
Find mean , var, mode if $a = 3, b = 5.$
This is throwing me off with the beta distribution. I'm not sure if it changes the way i solve for the mean, var , mode.
My approach is that since mean of a continuous random variable which is basically the Expected value of X (EX), so we can just do $\int x * f(x) dx$. We can find $ f(x) = (105) x^{2} (1-x)^{4} $ and then $\int x *f(x) dx = 1/105 \int_{0}^{1} x^3 * (1-x)^4 dx = 105 * \beta(4,5) = 105 * ((6*24 )/ 40320)$
|
\begin{equation}
\begin{array}{rcl} f^\prime(x)=\dfrac{df(x)}{dx}& = & \dfrac{1}{B(a,b)}\dfrac{d}{dx}x^{a-1}(1-x)^{b-1}\\
& = & \dfrac{1}{B(a,b)}\left[(a-1)x^{a-2}(1-x)^{b-1}-x^{a-1}(b-1)(1-x)^{b-2}\right]\\
& = & \dfrac{1}{B(a,b)}x^{a-2}(1-x)^{b-2}\left[(a-1)(1-x)-(b-1)x\right]\\
\end{array}
\end{equation}
\begin{equation}
\begin{array}{rcl}
f^\prime(x)=0 & \Rightarrow & (a-1)(1-x)-(b-1)x=0 \Rightarrow a-1-x(a-1+b-1)=0\\
& \Rightarrow & x=\dfrac{a-1}{a+b-2} \quad \text{exists when } \;\; a>1\;\; \& \;\;b>1
\end{array}
\end{equation}
Mode exists when $a>1$ and $b>1$ and is equals to $\dfrac{a-1}{a+b-2}$
|
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|
Find two integer numbers given the LCM and the difference between them I started a Computer Science degree this year and I have a subject called 'Discrete Mathematics' which this unit (divisibility) is annoying me a lot, because I can't find the key to solve the problems as like as other units and subjects.
The problem that I can't solve is this:
Find two integer numbers 'a' and 'b' given that their difference is 1080 and their LCM is 3900
Thanks
|
In the search of a more elegant solution rather than brute force, we know that as $\text{lcm}(a,b)=2^2\cdot 3\cdot 5^2\cdot 13$ that if both $a$ and $b$ had a factor of $13$ that $a-b$ must also be divisible by $13$. Similarly if both $a$ and $b$ had a factor of $5^2$ that $a-b$ must be divisible by $5^2$.
Since $a-b=1080$ is not divisible by $13$ nor divisible by $25$, we know that exactly one of the two is divisible by $13$ and that exactly one of the two is divisible by $25$.
Also, since $a-b=1080=2^3\cdot 3^3\cdot 5$, one can also reason that both $a$ and $b$ must be divisible by $2^2\cdot 3\cdot 5=120$.
(They must both be divisible by four because otherwise one will be divisible by four and the other is not, implying their difference would not have been divisible by four which is a contradiction. similarly for the other factors. In general, try proving that $\gcd(a-b,\text{lcm}(a,b))$ divides both $a$ and $b$)
This brings our candidate answers to either $|a|=2^2\cdot 3\cdot 5^2\cdot 13=3900$ and $|b|=2^2\cdot 3\cdot 5=120$ or $|a|=2^2\cdot 3\cdot 5\cdot 13=780$ and $|b|=2^2\cdot 3\cdot 5^2=300$
It is clear that since $|3900\pm 120|>3780> 1080$ that it could not be the first result.
The second result on the other hand does work with $780-(-300)=1080$, yielding $a=780$ and $b=-300$. We already know that $\text{lcm}(780,-300)=3900$ by construction.
|
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|
Mathematical Induction with Exponents: $1 + \frac12 + \frac14 + \dots + \frac1{2^{n}} = 2 - \frac1{2^{n}}$ Prove $1 + \frac{1}{2} + \frac{1}{4} + ... + \frac{1}{2^{n}} = 2 - \frac{1}{2^{n}}$ for all positive integers $n$.
My approach was to add $\frac{1}{2^{n + 1}}$ to both sides for the induction step. However, I got lost in the algebra and could not figure out the rest of the proof. Any help would be greatly appreciated.
Thank you in advance.
|
Let $S(n)$ be the statement: $1+\dfrac{1}{2}+\dfrac{1}{4}+\cdots+\dfrac{1}{2^{n}}=2-\dfrac{1}{2^{n}}$
First do basis step:
$S(1):\hspace{5 mm}$ $1+\dfrac{1}{2^{1}}=\dfrac{3}{2}$
and$\hspace{9 mm}$ $2-\dfrac{1}{2^{1}}=\dfrac{3}{2}$
Then do inductive step:
Assume $S(k)$ is true, or $1+\dfrac{1}{2}+\dfrac{1}{4}+\cdots+\dfrac{1}{2^{k}}=2-\dfrac{1}{2^{k}}$
$S(k+1):\hspace{5 mm}$ $1+\dfrac{1}{2}+\dfrac{1}{4}+\cdots+\dfrac{1}{2^{k}}+\dfrac{1}{2^{k+1}}$
$\hspace{18 mm}=2-\dfrac{1}{2^{k}}+\dfrac{1}{2^{k+1}}$
$\hspace{18 mm}=2+\dfrac{1}{2^{k+1}}-\dfrac{1}{2^{k}}$
$\hspace{18 mm}=2+\dfrac{2^{k}-2^{k+1}}{2^{k+1}\cdot{2^{k}}}$
$\hspace{18 mm}=2+\dfrac{2^{k}\big(1-2\big)}{2^{k+1}\cdot{2^{k}}}$
$\hspace{18 mm}=2-\dfrac{1}{2^{k+1}}$
So $S(k+1)$ is true whenever $S(k)$ is true.
Therefore, $1+\dfrac{1}{2}+\dfrac{1}{4}+\cdots+\dfrac{1}{2^{n}}=2-\dfrac{1}{2^{n}}$.
|
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Double Integrals - Finding the volume of a unit disk under a function For a Unit Disk
$x^2 + y^2 \ge 1$
And for a function of $x$ and $y$
$f(x, y) = 3 + y - x^2$
I want to find the volume underneath the function bound by the unit disk.
At first I integrated the function with respect to $y$, and made the upper bound $\sqrt{1 - x^2}$ and made the lower bound $-\sqrt{1 - x^2}$. I did this using Pythagoras' theorem to express the upper and lower bounds of the circle in terms of $x$. The resulting expression is $6\sqrt{1-x^2}$
Then I have to integrate the expression I just found with respect to $x$, and the upper and lower bounds are $1$ and $-1$ respectively.
However, the expression given is as follows:
I do not understand why the expression has changed.
The expression that I worked out $6\sqrt{1-x^2}$ represents the area of a slice with a constant $x$ value, and surely I just integrate this expression with respect to x between the bounds of $1$ and $-1$ to find the total volume.
|
The volume is, by definition:
$$\int_{-1}^1\int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}}(3+y-x^2)dydx\stackrel{(**)}=\int_{-1}^1(3-x^2)\left(2\sqrt{1-x^2}\right)dx=\int_{-1}^1\left(6-2x^2\right)\sqrt{1-x^2}dx$$
(**) Explanation: we have
$$\int_{-1}^1\int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}}(3+y-x^2)dydx=\int_{-1}^1\left(\left.(3-x^2)y+\frac12y^2\right)\right|_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}}dx= etc.$$
|
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|
Prove that $(\frac{bc+ac+ab}{a+b+c})^{a+b+c} \geq \sqrt{(bc)^a(ac)^b(ab)^c}$
Prove that $$\left( \frac{bc+ac+ab}{a+b+c} \right)^{a+b+c} \ge \sqrt{(bc)^a(ac)^b(ab)^c}$$
where $a,b,c>0$
My attempt:
I couldn't proceed after that. Please help me in this regard, thanks!
|
If you want to use the weighted AM-GM inequality, considering
\begin{align*}
\frac{bc+ca+ab}{a+b+c} &=
\frac{2(bc+ca+ab)}{2(a+b+c)} \\
&= \frac{(b+c)\color{red}{\boldsymbol{a}}+
(c+a)\color{green}{\boldsymbol{b}}+
(a+b)\color{blue}{\boldsymbol{c}}}{(b+c)+(c+a)+(a+b)} \\
\left( \frac{bc+ca+ab}{a+b+c} \right)^{2(a+b+c)}
&\ge
\color{red}{\boldsymbol{a}}^{b+c}
\color{green}{\boldsymbol{b}}^{c+a}
\color{blue}{\boldsymbol{c}}^{a+b} \\
\left( \frac{bc+ca+ab}{a+b+c} \right)^{a+b+c}
&\ge \sqrt{a^{b+c}b^{c+a}c^{a+b}} \\
&= \sqrt{(bc)^{a}(ca)^{b}(ab)^{c}}
\end{align*}
|
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|
Show that ${(F_n^2+F_{n+1}^2+F_{n+2}^2)^2\over F_{n}^4+F_{n+1}^4+F_{n+2}^4}=2$
If $F_n$ is the $n$-th Fibonacci number ($1,1,2,3,5,8,\dots$), show that
$${(F_n^2+F_{n+1}^2+F_{n+2}^2)^2\over F_{n}^4+F_{n+1}^4+F_{n+2}^4}=2$$
I have tested with a lot of Fibonacci numbers and it seem to obey the ruse, but I don't know how simplify it to 2.
I try:
Let $a=F_n$, $b=F_{n+1}$ and $c=F_{n+2}$
$a^4+b^4+c^4+2(ab)^2+2(ac)^2+2(bc)^2=2a^4+2b^4+2c^4$
$2(ab)^2+2(ac)^2+2(bc)^2=a^4+b^4+c^4$
I am not sure, what to do next. Can anyone help by completing the prove?
|
replace $c$ with $a+b$ and it should simplify.
$a^4+b^4+c^4 = a^4+b^4+(a+b)^4 = 2a^4+4a^3b+6a^2b^2+4ab^3+2b^4$
$2(ab)^2+2(ac)^2+2(bc)^2 = 2a^2b^2 + 2a^2(a^2+2ab+b^2) + 2b^2(a^2+2ab+b^2) \\
= 2a^4+4a^3b+6a^2b^2+4ab^3+2b^4$
|
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|
Is it possible to have three real numbers that have both their sum and product equal to $1$? I have to solve $ x+y+z=1$ and $xyz=1$ for a set of $(x, y, z)$. Are there any such real numbers?
Edit : What if $x+y+z=xyz=r$, $r$ being an arbitrary real number. Will it still be possible to find real $x$, $y$, $z$?
|
Suppose there actually is such a triple, $(x,y,z)$, of real numbers. Let $u$ be a variable and let's force $x$, $y$, and $z$ to be the roots of a polynomial. We would consider \begin{align*}
0 &= (u-x)(u-y)(u-z) \\
&= u^3 - (x+y+z)u^2 + (xy + xz + yz) u - (x y z) \\
&= u^3 - u^2 + (xy + xz + yz) u - 1 \\
&= u^3 - u^2 + A u - 1 \text{,}
\end{align*} where we have used the abbreviation $A = xy + xz + yz$. If we let $A$ range over the complex numbers, we get three (generically complex) roots that satisfy your equation. Clearly, if $x$, $y$, and $z$ are real, $A$ is real, so we should restrict $A$ to the reals, but this isn't quite strong enough to make $x$, $y$, and $z$ real. (In fact, if $A$ is greater than about $-2.6$, two of the roots are complex. The exact bound for $A$ is the one real root of $4A^3 - A^2 - 18 A + 31$, the discriminant of the polynomial in $u$.)
The above says we could take $A = -3$ and then $x$, $y$, and $z$ are \begin{align*}
-1, 1-\sqrt{2}, \text{ and } 1+\sqrt{2} \text{.}
\end{align*} The sum of the two conjugates is $2$ and adding the first, the sum of all three is $1$. The product of the conjugates is $1 - 2 = -1$, since this is an example of a difference of two squares, and so the product of all three is $1$.
|
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Show that $(1+x)^{1-x}(1-x)^{1+x}<1$ If $x$ is a positive proper fraction. Prove that $$(1+x)^{1-x}(1-x)^{1+x}<1$$
Actually this question has two parts I can't do the $1^{st}$ part but the $2^{nd}$ part was quite easy with respect to the $1^{st}$ one. The $2^{nd}$ was to show that $$a^bb^a<(\frac{a+b}{2})^{a+b} $$ I observe that if $1^{st}$ is true then if I will put $$x=\frac{a-b}{a+b}$$ the equation of part $1^{st}$ will take the form $$a^bb^a<(\frac{a+b}{2})^{a+b} $$ and hence proved but I am unable to prove that $(1+x)^{1-x}(1-x)^{1+x}<1$
|
Since $\log$ is concave,
$$
\begin{align}
\frac{1-x}2\log(1+x)+\frac{1+x}2\log(1-x)
&\le\log\left(\frac{1-x}2(1+x)+\frac{1+x}2(1-x)\right)\\
&=\log\left(1-x^2\right)
\end{align}
$$
Therefore,
$$
\color{#090}{(1+x)^{1-x}(1-x)^{1+x}}\le\color{#C00}{\left(1-x^2\right)^2}
$$
|
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|
Simplifying the second derivative I need to simplify the below with respect to $y''$. I'm given: $$y''=-\frac{3x^2y^3-3x^3y^2\left(-\frac{x^3}{y^3}\right)}{y^6}$$
The final result should look like:
$$-\frac{3x^2(y^4+x^4)}{y^7}$$
Here are my steps:
*
*Combine term: $-3x^3y^2\left(-\frac{x^3}{y^3}\right)$.
$$-3x^3y^2\left(-\frac{x^3}{y^3}\right)=\frac{3x^6y^2}{y^3}$$
Our expression is now:
$$y''=-\frac{3x^2y^3+\frac{3x^6y^2}{y^3}}{y^6}$$
*Multiply the final term in the numerator, $\frac{3x^6y^2}{y^3}$, by the it's reciprocal in the denominator, $\frac{y^6}{1}$, to eliminate the complex fraction.
$$\frac{3x^6y^2}{y^3} * \frac{y^6}{1}=\frac{3x^6y^8}{y^3}$$
Our expression is now:
$$y''=-\frac{3x^2y^3}{y^6}+\frac{3x^6y^8}{y^3}$$
*Multiply term $\frac{3x^6y^8}{y^3}$ by $\frac{y^3}{y^3}$ to make common denominators.
$$\frac{3x^6y^8}{y^3}*\frac{y^3}{y^3}=\frac{3x^6y^{11}}{y^6}$$
Our separate terms can now be added, and our expression will look like:
$$y''=-\frac{3x^2y^3+3x^6y^{11}}{y^6}$$
*Now factor the numerator, our expression will become:
$$y''=-\frac{3x^2y^3(1+x^4y^{8})}{y^6}$$
Have I made a mistake? I'm not sure where to go from here. I need to achieve this expression as stated at the top my question:
$$y''=-\frac{3x^2(y^4+x^4)}{y^7}$$
|
$$y''=-\frac{3x^2y^3-3x^3y^2\left(-\frac{x^3}{y^3}\right)}{y^6}=-\frac{3x^2y^3+3x^3\left(\frac{x^3}{y}\right)}{y^6}=y''=-\frac{\left(\frac{3x^2y^4+3x^6}{y}\right)}{y^6}=-\frac{3x^2y^4+3x^6}{y^7}=-\frac{3x^2(y^4+x^4)}{y^7}$$
|
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|
Is $\int_0^\infty \vert \sin x \vert^{x^2} \ dx$ convergent? To study the question, I'm looking at the convergence of the series
$$u_k = \int_{k \pi}^{(k+1) \pi}\vert \sin x \vert^{x^2} \ dx,$$ using the inequalities
$$2 \int_0^{\frac{\pi}{2}} \sin^{(k+1)^2\pi^2} x \ dx \le u_k \le 2 \int_0^{\frac{\pi}{2}} \sin^{k^2 \pi^2} x \ dx.$$
But I'm not able to get a good approximate of
$$v_k = \int_0^{\frac{\pi}{2}} \sin^{k^2 \pi^2} x \ dx$$
Any good idea?
|
I finally found another solution based on more basic technics.
$$\begin{align}
v_k&= \int_0^{\frac{\pi}{2}} \sin^{k^2 \pi^2} x \ dx = \int_0^{\frac{\pi}{2}} \cos^{k^2 \pi^2} x \ dx\\
&\ge \int_0^{\frac{\sqrt{2}}{k \pi}} \cos^{k^2 \pi^2} x \ dx\\
&\ge \int_0^{\frac{\sqrt{2}}{k \pi}} (1-\frac{x^2}{2})^{k^2 \pi^2} \ dx\\
&\ge \frac{\sqrt{2}}{k\pi} \left(1-\frac{1}{k^2 \pi^2}\right)^{k^2\pi^2} \sim\frac{\sqrt{2}}{e k \pi}
\end{align}$$
Based on the inequality
$$\cos x\ge 1-\frac{x^2}{2} \text{ for } x \ge 0$$
Hence the integral diverges as the harmonic series diverges.
|
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|
$\binom{28}{2}+\binom{28}{5}+\binom{28}{8}+\cdots+\binom{28}{26}=\frac{2}{3}(2^{27}+1)$ I need to prove the following identity
$$
\binom{28}{2}+\binom{28}{5}+\binom{28}{8}+\cdots+\binom{28}{26} = \frac{2}{3}(2^{27}+1).
$$
I have tried to use the fact that $\binom{m}{n}=\binom{m}{m-n}$ but it doesn't help.
|
Hint. Let $w=\exp(2\pi i/3)$. Then $w^3=1$ and $1+w+w^2=0$, and hence
$$
(1+1)^{28}+w(1+w)^{28}+w^2(1+w^2)^{28}=3 \left(\binom{28}{2}+\binom{28}{5}+\cdots+\binom{28}{26}\right),
$$
since the sum above is a real number.
Next
$$
1+w=-w^2\quad\Rightarrow\quad(1+w)^{28}=w^{54}=w^2\\
1+w^2=-w\quad\Rightarrow\quad(1+w^2)^{28}=w^{28}=w
$$
Finally
$$
(1+1)^{28}+w(1+w)^{28}+w^2(1+w^2)^{28}=2^{28}+2.
$$
|
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|
find range of $a^2+b^2$ without trigonometric substution given $7a^2-9ab+7b^2=9$ and $a,b$ are real no. then find range of $a^2+b^2$ without trigonometric substution
from $7a^2-9ab+7b^2=9$
$\displaystyle ab = \frac{7(a^2+b^2)-9}{9}$
put into inequality $\displaystyle a^2+b^2 \geq 2ab$
$\displaystyle a^2+b^2 \geq \frac{14(a^2+b^2)-18}{9}$
$\displaystyle a^2+b^2 \leq \frac{18}{5}$
i wan,t be able to find minimum,could some help me with this
|
For the maximum, note that:
$$9=7a^2-9ab+7b^2 = \frac{5}{2}(a^2+b^2) + \frac{9}{2}(a-b)^2 \ge \frac{5}{2}(a^2+b^2)$$
[ EDIT ] For the minimum, note that:
$$9=7a^2-9ab+7b^2 = \frac{23}{2}(a^2+b^2) - \frac{9}{2}(a+b)^2 \le \frac{23}{2}(a^2+b^2)$$
|
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Confusion in fraction notation $$a_n = n\dfrac{n^2 + 5}{4}$$
In the above fraction series, for $n=3$ I think the answer should be $26/4$, while the answer in the answer book is $21/2$ (or $42/4$). I think the difference stems from how we treat the first $n$. In my understanding, the first number is a complete part and should be added to fraction, while the book treats it as part of fraction itself, thus multiplying it with $n^2+5$.
So, I just want to understand which convention is correct.
This is from problem 6 in exercise 9.1 on page 180 of the book Sequences and Series.
Here is the answer sheet from the book (answer 6, 3rd element):
*
*$3,8,15,24,35$
*$\dfrac{1}{2},\dfrac{2}{3},\dfrac{3}{4},\dfrac{4}{5},\dfrac{5}{6}$
*$2, 4, 8, 16 \text{ and } 32$
*$-\dfrac{1}{6},\dfrac{1}{6},\dfrac{1}{2},\dfrac{5}{6},\dfrac{7}{6}$
*$25,-125,625,-3125,15625$
*$\dfrac{3}{2},\dfrac{9}{2},\dfrac{21}{2},21,\dfrac{75}{2}$
*$65, 93$
*$\dfrac{49}{128}$
*$729$
*$\dfrac{360}{23}$
*$3, 11, 35, 107, 323$; $3+11+35+107+323+...$
*$-1,\dfrac{-1}{2},\dfrac{-1}{6},\dfrac{-1}{24},\dfrac{-1}{120}$; $-1+(\dfrac{-1}{2})+(\dfrac{-1}{6})+(\dfrac{-1}{24})+(d\frac{-1}{120})+...$
*$2, 2, 1, 0, -1$; $2+2+1+0+(-1)+...$
*$1,2,\dfrac{3}{5},\dfrac{8}{5}$
|
I don't think there can ever be a mixed fraction of the form $n\frac{n^2+5}{4}$ if $n$ $\in$ $\mathbb{N}$. Please note that if it were a mixed fraction then $n^2+5$ would denote the remainder while $4$ is the divisor and this would never be possible as for $n$ $\in$ $\mathbb{N}$, $n^2+5 \gt 4$ always.
Hence, this expression would definitely denote $n\times$$\frac{n^2+5}{4}$.
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|
if $xP(x)=x^t+P(x-1)$ find the $t\equiv ?\pmod 3$ Let polynomials $P(x)\in Z[x]$,and such $\deg{(P(x))}=t-1$,and such for any real $x$ have
$$xP(x)=x^t+P(x-1)$$
Find the $t\equiv ?\pmod 3$?
I try Let $P(x)=x^{t-1}+a_{t-2}x^{t-2}+\cdots+a_{1}x+a_{0}$ where $a_{i}\in Z$
so we have
$$a_{t-2}x^{t-1}+a_{t-3}x^{t-2}+\cdots+a_{1}x^2+a_{0}x=(x-1)^{t-1}+a_{t-2}(x-1)^{t-2}+\cdots+a_{1}(x-1)+a_{0}$$
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Let $P(x) = \displaystyle\sum_{k = 0}^{t-1}a_kx^k$ where $a_k \in \mathbb{Z}$. Also, for $r = 0,1,\ldots,5$ define $S_r = \displaystyle\sum_{\substack{k = 0\\k \equiv r \pmod{6}}}^{t-1}a_k$.
Note that for $x = e^{i\tfrac{\pi}{3}} = \tfrac{1}{2}+\tfrac{\sqrt{3}}{2}i$ we have $x-1 = -\tfrac{1}{2}+\tfrac{\sqrt{3}}{2}i = e^{i\tfrac{2\pi}{3}}$.
By plugging $x = e^{i\tfrac{\pi}{3}}$ into the identity $xP(x) - P(x-1) = x^t$ and using the fact that $e^{i\tfrac{\pi}{3}k}$ and $e^{i\tfrac{2\pi}{3}k}$ are $6$-periodic, we get:
$$e^{i\tfrac{\pi}{3}}P\left(e^{i\tfrac{\pi}{3}}\right) - P\left(e^{i\tfrac{2\pi}{3}}\right) = e^{i\tfrac{\pi}{3}t}$$
$$\sum_{k = 0}^{t-1}a_ke^{i\tfrac{\pi}{3}(k+1)} - \sum_{k = 0}^{t-1}a_ke^{i\tfrac{2\pi}{3}k} = e^{i\tfrac{\pi}{3}t}$$
$$\sum_{k = 0}^{t-1}a_k\left(e^{i\tfrac{\pi}{3}(k+1)} - e^{i\tfrac{2\pi}{3}k}\right) = e^{i\tfrac{\pi}{3}t}$$
$$\left[-\tfrac{1}{2}+\tfrac{\sqrt{3}}{2}i\right]S_0 + \left[-\tfrac{1}{2}+\tfrac{\sqrt{3}}{2}i\right]S_2 + \left[-\tfrac{3}{2}-\tfrac{\sqrt{3}}{2}i\right]S_3 + \left[1-\sqrt{3}i\right]S_4 + \left[\tfrac{3}{2}+\tfrac{\sqrt{3}}{2}i\right]S_5 = e^{i\tfrac{\pi}{3}t}$$
Equating real and imaginary parts gives:
$$\tfrac{1}{2}\left[-S_0-S_2-3S_3+2S_4+3S_5\right] = \cos\left(\tfrac{\pi}{3}t\right)$$
$$\tfrac{\sqrt{3}}{2}\left[S_0+S_2-S_3-2S_4+S_5\right] = \sin\left(\tfrac{\pi}{3}t\right)$$
Then, $2$ times the real part equation plus $\tfrac{2}{\sqrt{3}}$ times the imaginary part equation gives:
$$4(S_5-S_3) = 2\cos\left(\tfrac{\pi}{3}t\right)+\tfrac{2}{\sqrt{3}}\sin\left(\tfrac{\pi}{3}t\right)$$
Since the $a_k$'s are integers, the $S_r$'s are also integers. So, $4(S_5-S_3)$ must be a multiple of $4$. But, $$2\cos\left(\tfrac{\pi}{3}t\right)+\tfrac{2}{\sqrt{3}}\sin\left(\tfrac{\pi}{3}t\right) = \begin{cases}2 & \text{if} \ t \equiv 0 \ \text{or} \ 1 \pmod{6} \\ 0 & \text{if} \ t \equiv 2 \ \text{or} \ 5 \pmod{6} \\ -2 & \text{if} \ t \equiv 3 \ \text{or} \ 4\pmod{6}\end{cases}.$$
Therefore, $t \equiv 2 \ \text{or} \ 5 \pmod{6}$, and thus, $\boxed{t \equiv 2 \pmod{3}}$.
Note: It is easy to check that $t = 2$ and $P(x) = x+1$ is a solution. There may be other solutions for larger $t \equiv 2\pmod{3}$.
Further Analysis: By substituting the sum for $P(x)$ into $xP(x) = x^t+P(x-1)$, we have:
$$\sum_{k = 0}^{t-1}a_kx^{k+1} = x^t + \sum_{n = 0}^{t-1}a_n(x-1)^n$$
$$\sum_{k = 1}^{t}a_{k-1}x^k = x^t + \sum_{n = 0}^{t-1}a_n\left(\sum_{k = 0}^{n}\dbinom{n}{k}(-1)^{n-k}x^k\right)$$
$$a_{t-1}x^t + \sum_{k = 1}^{t-1}a_{k-1}x^k = x^t + \sum_{k = 0}^{t-1}\left(\sum_{n = k}^{t-1}a_n\dbinom{n}{k}(-1)^{n-k}\right)x^k$$
By equating corresponding coefficients, we get:
$$a_{t-1} = 1$$
$$a_{k-1} = \sum_{n = k}^{t-1}a_n\dbinom{n}{k}(-1)^{n-k} \ \text{for} \ k = 1,\ldots,t-1$$
$$0 = \sum_{n = 0}^{t-1}a_n(-1)^{n}$$
The first two equations uniqely determine the coefficients $a_k$ for $k = 0,\ldots,t-1$. Then, the polynomial $P(x) = \displaystyle\sum_{k = 0}^{t-1}a_kx^k$ satisfies $xP(x) = x^t+P(x-1)$ iff the third equation is satisfied.
I tested this recursion for $1 \le t \le 56$ using a computer program. It appears that $t = 2$ was the only value of $t$ in that range for which $P(x)$ satisfies $xP(x) = x^t+P(x-1)$.
|
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|
in a triangle $ABC,$ if $2c^2=a^2+b^2,$ then largest possible degree measure of angle $C$ is In a triangle $ABC,$ if $2c^2=a^2+b^2,$ then largest possible degree measure of angle $C$ is
cosine formula $\displaystyle \cos C = \frac{a^2+b^2-c^2}{2ab} = \frac{c^2}{2ab}<\frac{(a+b)^2}{2ab}=\frac{a^2+b^2}{2ab}+1$
wan,t be able to process after that, could some help me
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We know that $$\cos C =\frac {a^2+b^2-c^2 }{2ab} =\frac {c^2}{2ab} =\frac {a^2+b^2}{4ab} $$ Now applying the AM-GM inequality, we get, $$\frac {a^2+b^2}{4ab}\geq \frac {1}{2} \Rightarrow C\leq \frac {\pi}{3} $$ Hope it helps.
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|
A simpler proof of $(1-x)^n<\frac{1}{1+nx}$ I proved the following inequality:
Let $x\in\mathbb{R}, 0<x<1, n\in\mathbb{N}$, then $(1-x)^n<\frac{1}{1+nx}$
however, judging from the context in the exercise book, I feel like there is a much simpler way to prove it, but I can't see it. So I'm asking for that simpler alternative proof, or at least a hint to it. My proof as follows:
Bernoulli inequality states that for $-1<x, x\neq 0, n\in \mathbb{N},n>1$ the following is true:$(1+x)^n>1+nx$.
Thus, $\frac{1}{(1+x)^n}<\frac{1}{1+nx}$ is also true.
Then I need to show that $(1-x)^n<\frac{1}{(1+x)^n}$, which is equivalent to $\frac{1}{(1-x)^n}>(1+x)^n$, which I prove by induction:
Basecase: $n=1$ $1=\frac{1-x}{1-x}\Leftrightarrow 1=\frac{1}{1-x}-\frac{x}{1-x}\Leftrightarrow 1+\frac{x}{1-x}=\frac{1}{1-x}$
Let $a,b\in \mathbb{R_{>0}}$ and $0>b>1$ Then $\left[ a>ab \right]\Leftrightarrow \left[a<\frac{a}{b}\right]$. Thus $\left[ 0<x<1\right] \Rightarrow \left[ x<\frac{x}{1-x}\right]$
Thus $1+\frac{x}{1-x}=\frac{1}{1-x} \Rightarrow 1+x<\frac{1}{1-x} \square$
Inductive step: Assume $(1+x)^n<\frac{1}{(1-x)^n}$. Need to show $(1+x)^{n+1}<\frac{1}{(1-x)^{n+1}}$.
$(1+x)^{n+1}<\frac{1}{(1-x)^{n+1}}\Leftrightarrow (1+x)^n\cdot(1+x)<\frac{1}{(1-x)^n} \cdot \frac{1}{1-x}$
Let $a,b,c,d \in \mathbb{R_{>0}}$. Then $[a>c]\wedge[b>d] \Rightarrow [ab>cd]$. $(1+x)^n<\frac{1}{(1-x)^n}$ was the assumption and $1+x>\frac{1}{1-x}$ was the basecase, therefore $(1+x)^n\cdot(1+x)<\frac{1}{(1-x)^n} \cdot \frac{1}{1-x} \Leftrightarrow (1+x)^{n+1}<\frac{1}{(1-x)^{n+1}} \square$
Thus $\frac{1}{(1+x)^n}<\frac{1}{1+nx}$ and $(1-x)^n<\frac{1}{(1+x)^n}$ are both true, which implies the original statement $(1-x)^n<\frac{1}{1+nx} \square$ If I were to count the proof of the Bernoulli inequality by induction, it would mean that I used induction twice in order to prove something that basic, which to me doesn't seem to be a sensible thing to do.
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First note:
$$\frac{1+nx}{1-x}=\frac{(1+nx)(1+x)}{(1-x)(1+x)}=\frac{1+(n+1)x+nx^2}{1-x^2}\ge\frac{1+(n+1)x}{1}$$
So:
$$\frac{1+nx}{1+(n+1)x}\ge1-x$$
Take a product of these expressions for $n=0,1,\cdots,N-1$ to see
$$\frac{1}{1+Nx}\ge(1-x)^N$$
(this can easily be recast as an induction)
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|
Deriving $\cosh^{-1}{x}=\ln\left(x+\sqrt{x^2-1}\right)$ Let $y=\cosh^{-1}{x}$. Then, $x=\cosh{y}=\frac{1}{2}\left(e^y+e^{-y}\right)$. Multiplying by $2e^y$, we get $2xe^y=e^{2y}+1$. Solving $e^{2y}-2xe^y+1=0$ by the quadratic formula, we have $e^y=\frac{2x\pm\sqrt{4x^2-4}}{2}=x\pm\sqrt{x^2-1}$. We find that both roots are possible. Thus, $y=\ln\left(x\pm\sqrt{x^2-1}\right)$. Why did we fail to eliminate the minus?
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As $y\ge 0$, $\;\mathrm e^y\ge 1$. Now $\mathrm e^y$ is a root of the quadratic polynomial $\;P(t)=t^2-2xt+1$. Just observe that, as $P(1)=2(1-x) < 0$ for $x>1$, $1$ separates the roots. Hence $\mathrm e^y$ is the greatest root — with a + sign.
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|
Prob. 17, Chap. 3 in Baby Rudin: For $\alpha > 1$, how to obtain these inequalities from this recurrence relation? Here's Prob. 17, Chap. 3 in the book Principles of Mathematical Analysis by Walter Rudin, 3rd edition:
Fix $\alpha > 1$. Take $x_1 > \sqrt{\alpha}$, and define $$x_{n+1} = \frac{\alpha + x_n}{1+x_n} = x_n + \frac{\alpha - x_n^2}{1+x_n}.$$
(a) Prove that $x_1 > x_3 > x_5 > \cdots$.
(b) Prove that $x_2 < x_4 < x_6 < \cdots$.
(c) Prove that $\lim x_n = \sqrt{\alpha}$.
My effort:
From the recursion formula, we can obtain
$$
\begin{align}
x_{n+1} &= \frac{ \alpha + x_n}{1+ x_n} \\
&= \frac{ \alpha + \frac{\alpha + x_{n-1}}{1+x_{n-1}} }{ 1 + \frac{\alpha + x_{n-1}}{1+x_{n-1}} } \\
&= \frac{ (\alpha + 1) x_{n-1} + 2 \alpha }{ 2x_{n-1} + ( 1 + \alpha ) } \\
&= \frac{\alpha+1}{2} + \frac{2 \alpha - \frac{(\alpha+1)^2}{2} }{2x_{n-1} + ( 1 + \alpha ) } \\
&= \frac{\alpha+1}{2} + \frac{ \alpha - \frac{\alpha^2+1 }{2} }{2x_{n-1} + ( 1 + \alpha ) }.
\end{align}
$$
What next?
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$$\text {Let } a_n=(\sqrt \alpha)\tan (d_n+\pi /4) \text { with } d_n\in (0,\pi /4).$$ Using $\tan (x +\pi /4)=(1+\tan x)/(1-\tan x)$ when $\tan x \ne 1,$ we arrive, after some calculation, that $$\tan d_{n+1}=k \tan d_n$$ $$ \text { where } k=\frac {1-\sqrt {\alpha} }{1+\sqrt {\alpha}}.$$ Observe that $0>k>-1$ because $\alpha >1.$
Remark. I tried this substitution by heuristic analogy with a substitution for Heron's method: If $\sqrt {\alpha} \ne b_1>0$ and $b_{n+1}=(b_n+\alpha /b_n)/2,$ then $b_n>\sqrt {\alpha}$ for $n\geq 2.$ Let $b_n=\sqrt {\alpha} \coth c_n$ for $n\geq 2$, with $c_2>0.$ Then $c_n=2^{n-2}c_2$ for $n\geq 2.$ In this Q I tried $\tan$ because $\tan d_n$ alternates $\pm. $
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Why is $\lim\limits_{x\to -\infty} \frac{1}{\sqrt{x^2+2x}-x}$ equal to $0$? So I made one exercise, which was $\lim_{x\to +\infty} \frac{1}{\sqrt{x^2-2x}-x}$ I solved this one by:
$\lim \limits_{x\to +\infty} \frac{1}{\sqrt{x^2-2x}-x} \frac{\sqrt{x^2-2x}+x}{\sqrt{x^2-2x}+x} $
$\lim \limits_{x\to +\infty} \frac{\sqrt{x^2}\sqrt{1-\frac{2}{x}}+x}{-2x}$
$\lim \limits_{x\to +\infty} \frac{\sqrt{1-\frac{2}{x}}+1}{-2} = -1 $
The next exercise is $\lim \limits_{x\to -\infty} \frac{1}{\sqrt{x^2+2x}-x}$
I thought that I could solve this one in the same way by:
$\lim \limits_{x\to -\infty} \frac{1}{\sqrt{x^2+2x}-x} \frac{\sqrt{x^2+2x}+x}{\sqrt{x^2+2x}+x} $
$\lim \limits_{x\to -\infty} \frac{\sqrt{x^2}\sqrt{1+\frac{2}{x}}+x}{2x}$
$\lim \limits_{x\to -\infty} \frac{\sqrt{1+\frac{2}{x}}+1}{2} = 1 $
But apparently, the answer is $0$...
Why can't the second one be solved in the same way as the first?
And why is the answer $0$?
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Let $-1/x=h$
$x^2+2x=\dfrac{1-2h}{h^2}\implies\sqrt{x^2+2x}=\dfrac{\sqrt{1-2h}}{\sqrt{h^2}}=\dfrac{\sqrt{1-2h}}{|h|}$
As $h\to0^+, h>0,|h|=+h$
$$\lim_{x\to-\infty}\dfrac1{\sqrt{x^2+2x}-x}=\lim_{h\to0^+}\dfrac h{\sqrt{1-2h}+1}=?$$
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|
How to evaluate integral: $\int x^2\sqrt{x^2+1}\;dx.$ I want to solve the integral:$$\int x^2\sqrt{x^2+1}\;dx.$$
I did $x = \tan t$, then it is equal to:$$\int\frac{\tan^2 t}{\cos^3 t}\;dx.$$
Or:$$\int\frac{\sin^2 t}{\cos^5 t}\;dx.$$
I stuck there. Any help will be much appreciated...
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Going the hyperbolic instead of the trigonometric route, we can let $x:=\sinh t$, then $dx=\cosh tdt$.
\begin{align}
\int x^2\sqrt{x^2+1}dx &= \int\sinh^2t\sqrt{\sinh^2t+1}\cosh tdt\\
&=\int\sinh^2t\cosh^2tdt\\
&=\int\left(\sinh t\cosh t\right)^2dt\\
&=\frac{1}{4}\int\sinh^2(2t)dt\\
&=\frac{1}{8}\int\left(\cosh(4t)-1\right)dt\\
&=\frac{1}{32}\left(\sinh(4t)-4t\right) + C
\end{align}
Since $t=\text{arsinh }x$ by inversion of our initial substitution, we have
\begin{align}
\int x^2\sqrt{x^2+1}dx &= \frac{1}{32}\left[\sinh(4\text{arsinh }x)-4\text{arsinh }x\right]+C\\
&=\frac{1}{8}\left((x+2x^3)\sqrt{x^2+1}-\text{arsinh }x\right) + C
\end{align}
If you want, you can write out $\text{arsinh }x$ using logarithms.
|
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Show that that $a>\frac{3}{4}$ Let $x^4+2ax^3+x^2+2ax+1=0$ .
If this equation has at least two real negative roots then show that that $a>\frac{3}{4}$.
I tried to solve the equation first by using the sub $t=x+ \frac{1}{x}$.
Then I got $t^2+2at-1=0$.
Thereafter how can I proceed to get the required inequality ?
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Another approach to determine the behaviour of $f(x)=x^4 + 2ax^3 + x^2 + 2ax + 1$ is by quartic equation analysis.
Define $a,b,c,d,e$ as the coefficients of $f(x)$.
$$a=1,b=2k,c=1,d=2k,e=1$$
where you want to prove that $f(x)$ has two real negative roots when $k>\frac{3}{4}$
Define the discriminant which will allow us to determine the nature of the roots by analyzing its sign: $$\Delta_{f_a} = 256 a^3e^3 - 192 a^2bde^2 - 128a^2c^2e^2 + 144a^2cd^2e - 27a^2d^4 + 144ab^2ce^2 - 6ab^2d^2e - 80abc^2de + 18 abcd^3 + 16ac^4e - 4ac^3d^2 - 27b^4 e^2 + 18b^3cde - 4b^3d^3 - 4b^2c^3e + b^2 c^2 d^2$$
Define $P,Q,D$ which will be useful to evaluate if $f(x)$ will have 4 complex-roots or two negative real roots + two complex conjugate roots.
$$P=8ac - 3 b ^ 2$$ $$Q= b^3 + 8da^2 - 4abc$$ $$D=64 a^3e - 16a^2c^2 + 16 ab^2c - 16a^2bd - 3 b^4$$
Now let's test when $k<\frac{3}{4}$ and $k>\frac{3}{4}$
$$k=0.74<\frac{3}{4}$$
$$P=1.4288, Q=9.16179, D=33.6064, \Delta_{f_{0.74}}=9.13574$$
Since $\Delta_{f_{0.74}}>0, P>0, D>0$ there are two pair of complex roots.
$$k=0.76>\frac{3}{4}$$
$$\Delta_{f_{0.76}}=-9.62079$$
Since $\Delta_{f_{0.76}}<0$ $f(x)$ has two distinct real roots and two complex conjugate root. By Descarte's rule of signs we determine that $f(x)$ has two real negative roots.
Thus when $k>\frac{3}{4}$ $f(x)$ will have two real negative roots.
|
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What is the value of $a+b+c+d$?
Four integers $a,b,c,d$ make all the statements below true. What is the value of $a+b+c+d$?
(i) $10 \leq a,b,c,d \leq 20$
(ii) $ab-cd = 58$
(iii) $ad-bc = 110$
Adding the two equations together gives $ab+ad-cd-bc = 168$ and so $$a(b+d)-c(b+d) = (a-c)(b+d) = 168 = 2^3 \cdot 3 \cdot 7.$$ Then since $20 \leq b+d \leq 40$, we see that $b+d = 21,24,28$. In the first case, $b+d = 21$ and $a-c = 8$. Thus, $(a,c) = (18,10),(19,11)$ and $(b,d) = (10,11),(11,10)$. How do we deal with the other cases?
|
Consider $(iii)-(ii)$, we have
$$(a+c)(d-b) = (ad-bc) - (ab-cd) = 110 - 58 = 52$$
Since $10 \le a, c \le 20 \implies 20 \le a+c \le 40$ and the only divisor
of $52$ between $20$ and $40$ is $26$, we find
$$a+c = 26, d - b = 2$$
Consider $(ii)+(iii)$, we have
$$(a-c)(b+d) = (ab-cd) + (ad - bc) = 58+160 = 168$$
Since $10 \le b, d \le 20 \implies 20 \le b+d \le 40$ and the divisors of
$168$ between $20$ and $40$ are $21, 24, 28$, we have 3 possibilities.
$$
\begin{cases}
b+d = 21,a-c = 8\\
b+d = 24,a-c = 7\\
b+d = 28,a-c = 6
\end{cases}$$
Notice
$$\begin{cases}
a+c = 26 &\implies a-c = 26 - 2c \equiv 0\pmod 2\\
d-b = 2 &\implies b+d = 2 + 2b \equiv 0 \pmod 2
\end{cases}
$$ This rules out the first and second case. This leaves us with only
one and only one possibility:
$$b+d = 28\quad\implies\quad a+b+c+d = 26 + 28 = 54$$
|
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|
Prove that $(2+ \sqrt5)^{\frac13} + (2- \sqrt5)^{\frac13}$ is an integer When checked in calculator it is 1. But how to prove it?
Also it is not a normal addition like $x+ \frac1x$ which needs direct rationalization. So I just need something to proceed.
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Use the general formula $x^3+y^3=(x+y)\left((x+y)^2-3xy\right)$.
Here $x=\sqrt[3]{2+\sqrt{5}}$ and $y=\sqrt[3]{2-\sqrt{5}}$.
Let $a=\sqrt[3]{2+\sqrt{5}}+\sqrt[3]{2-\sqrt{5}}$ and
$$b=\sqrt[3]{2+\sqrt{5}}\sqrt[3]{2-\sqrt{5}}=\sqrt[3]{2^2-(\sqrt{5})^2}=-1$$
$$(2+\sqrt{5})+(2-\sqrt{5})=4$$
$$=a\left(a^2-3b\right)=a^3+3a$$
$$a^3+3a-4=0$$
You can apply Rational Root Theorem/Test, polynomial division, fundamental theorem of algebra, etc.
$$(a-1)\left(a^2+a+4\right)=0$$
$a^2+a+4=0$ has no real solutions, but clearly $a$ is real, so $a=1$.
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|
Which integration formula for $\frac{1}{a^2-x^2}$ is correct? In my book integration formula for $\frac{1}{x^2-a^2}$ is given as $\frac{1}{2a}\ln(\frac{x-a}{x+a})$.
From the above formula we can write the formula for integration of $\frac{1}{a^2-x^2}$ as $-\frac{1}{2a}\ln(\frac{x-a}{x+a}) = \frac{1}{2a}\ln(\frac{x+a}{x-a}) $ [as we know $\ln(\frac{1}{x})=-\ln(x)$]
But, in my book the integration of $\frac{1}{a^2-x^2}$ is written as $\frac{1}{2a}\ln(\frac{x+a}{{a-x}}) $.
We know that the integration in the second case cannot be $\frac{1}{2a}\ln(\frac{x+a}{x-a}) $ and $\frac{1}{2a}\ln(\frac{x+a}{{a-x}}) $ at the same time. Which formula is the correct one and why?
|
Neither is correct.
It's not restrictive to assume $a>0$ (for $a=0$ the antiderivative has a different form; for $a<0$ change the intervals below accordingly).
The function $\frac{1}{x^2-a^2}$ is defined for $x\ne\pm a$ and is negative over $(-a,a)$, positive over $(-\infty,-a)$ and $(a,\infty)$.
The sign of $\frac{x-a}{x+a}$ is the same as the sign of $x^2-a^2$ (for $x\ne\pm a$, of course), so it's clear that
$$
\frac{1}{2a}\ln\frac{x-a}{x+a}
$$
cannot be an antiderivative over $(-a,a)$, because it is undefined there.
There's a simple way out. An antiderivative of $1/x$ is $\ln|x|$ (up to an arbitrary constant over $(-\infty,0)$ and an arbitrary constant on $(0,\infty)$, with no connection between each other).
The partial fraction decomposition is
$$
\frac{1}{x^2-a^2}=\frac{1}{2a}\frac{1}{x-a}-\frac{1}{2a}\frac{1}{x+a}
$$
so we can write
$$
\int\frac{1}{x^2-a^2}\,dx=
\frac{1}{2a}(\ln|x-a|-\ln|x+a|)=
\frac{1}{2a}\ln\left|\frac{x-a}{x+a}\right|
$$
up to arbitrary constants in the intervals $(-\infty,-a)$, $(-a,a)$ and $(a,\infty)$.
Since $1/(a^2-x^2)=-1/(x^2-a^2)$, we have
$$
\int\frac{1}{a^2-x^2}\,dx=
-\frac{1}{2a}\ln\left|\frac{x-a}{x+a}\right|=
\frac{1}{2a}\ln\left|\frac{x-a}{x+a}\right|^{-1}=
\frac{1}{2a}\ln\left|\frac{x+a}{x-a}\right|
$$
again up to constants as before.
|
{
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"url": "https://math.stackexchange.com/questions/2083201",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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|
Find $x^2+y^2+z^2$ if $x+y+z=0$ and $\sin{x}+\sin{y}+\sin{z}=0$ and $\cos{x}+\cos{y}+\cos{z}=0$ for any $x,y,z \in [-\pi,\pi]$ Find $x^2+y^2+z^2$ if $x+y+z=0$ and $\sin{x}+\sin{y}+\sin{z}=0$ and $\cos{x}+\cos{y}+\cos{z}=0$ for any $x,y,z \in [-\pi,\pi]$.
My attempt:I found one answer $x=0,y=\frac{2\pi}{3},z=-\frac{2\pi}{3}$ which gives the answer $x^2+y^2+z^2=\frac{8{\pi}^2}{9}$.But I want a way to find the answer using equations.ANy hints?
|
Although this is same to answer of @schrodingersCat Let $x\geq y\geq0\geq z$
$$\cos(x+y)^2=1-\sin(x+y)^2$$
$$⇔1-(\sin{x}+\sin{y})^2=(\cos{x}+\cos{y})^2$$
$$⇔1+2(\cos{x}\cos{y}+\sin{x}\sin{y})=0$$
$$⇔\cos(x-y)=-1/2 $$
$$x-y=2\pi/3$$
and
$$\sin{x}+\sin{y}=-\sin{z}$$
$$⇔\sin(y+2\pi/3)+\sin{y}=\sqrt3/2$$
$$⇔(1/2)\sin{y}+\sqrt3/2\cos{y}=\sqrt3/2$$
$$⇔\sin(y+\pi/3)=\sqrt3/2$$
then we get $$x=2\pi/3 ,y=0,z=-2\pi/3$$, if angle(x,y,z) are points of unit circle, they consist of the regular triangle. The answer of $x^2+y^2+z^2$ is unique and,
$$\displaystyle x^2+y^2+z^2=\frac{8\pi^2}9 $$
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/2083271",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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"answer_id": 2
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|
Show that $x-\sqrt{x^2-x+1}<\frac{1}{2} $ for every real number $x$, without using differentiation Let $f$ be a function defined by :
$$f(x)=x-\sqrt{x^2-x+1}$$
Show that:
$$\forall x\in\mathbb{R},\quad f(x)<\dfrac{1}{2} $$
without use notion of différentiable
let $x\in\mathbb{R}$
\begin{aligned}
f(x)-\dfrac{1}{2}&=x-\dfrac{1}{2}-\sqrt{x^2-x+1} \\
&=\dfrac{(x-\dfrac{1}{2})^2-|x^2-x+1|}{(x-\dfrac{1}{2})+\sqrt{x^2-x+1}}\\
&= \dfrac{-3/4}{(x-\dfrac{1}{2})+\sqrt{x^2-x+1}}
\end{aligned}
i don't know if $(x-\dfrac{1}{2})+\sqrt{x^2-x+1} $ positive or negative for all x in R
|
Notice that
$$x-\sqrt{x^2-x+1}=x-\sqrt{(x-\frac{1}{2})^2+\frac{3}{4}}\\\sqrt{(x-\frac{1}{2})^2+\frac{3}{4}}> (x-\frac{1}{2})
$$
Now consider if $x<\frac{1}{2}$ the inequality is trivial,consider then $x\geq \frac{1}{2}$ then you're allowed to square the inequality and since $\frac{3}{4}>0$ the inequality is always true.Hence
$$ x-\sqrt{x^2-x+1}< \frac{1}{2}$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
The sum of series $1\cdot 3\cdot 2^2+2\cdot 4\cdot 3^2+3\cdot 5\cdot 4^2+\cdots \cdots n$ terms The sum of series $1\cdot 3\cdot 2^2+2\cdot 4\cdot 3^2+3\cdot 5\cdot 4^2+\cdots \cdots n$ terms
i have calculate $a_{k} = k(k+2)(k+1)^2$
so $\displaystyle \sum^{n}_{k=1}a_{k} = \sum^{n}_{k=1}k(k+1)^2(k+2)$
i wan,t be able go further, could some help me with this, thanks
|
Hint:
Observe that
\begin{align*}
a_k&=k(k+2)(k+1)^2\\
&=(k+1-1)(k+1+1)(k+1)^2\\
&=\left[(k+1)^2-1\right](k+1)^2\\
&=(k+1)^4-(k+1)^2
\end{align*}
Now, here is something that can be useful in order to compute the sum of the fourth powers and squares
Geometric interpretation for sum of fourth powers
|
{
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 0
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|
Find the inverse of a polynomial in a quotient ring Consider $\mathbb Z_5[x]/I$ with $I$ as ideal generated by $b=x^3+3x+2$. If $(x+2) + I$ is element of $\mathbb Z_5[x]/I$ that has an inverse. Find the inverse of $(x+2) + I$.
I stuck to get the inverse because the gcd of that is not $1$.
|
Perform the Euclidean division by Horner's scheme (in $\mathbf F_5$):
$$\begin{matrix}\\ \\\times -2\quad\end{matrix}
\begin{matrix}
\hline
1&0&-2&2\\
\downarrow&-2&-1&1\\
\hline
\ 1&-2&2&-2
\end{matrix}
$$
Thus $0=x^3+3x+2\;(=x^3-2x+2)=(x+2)(x^2-2x+2)-2$, whence, multiplying both sides by $-2$,
$$(x+2)(x^2-2x+2)=2\implies(x+2)(-2x^2-x+1)=2\cdot (-2)=1$$
so that $\quad(x+2)^{-1}=-2x^2-x+1$.
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
Help solve simultaneous equations I need to solve the system of equations
$$6x+y=3\tag 1 $$
$$x^2+y^2=16\tag 2$$
So first, I rearrange $(1)$ to $y=3-6x$ and substitute that into $(2)$.
I get $$x^2+(3-6x)^2 = 16$$
which is $$x^2 + 36x^2-36x+9-16=0$$
which is $$37x^2-36x-7=0$$.
This I need to solve by completing the square.
|
$ 37 x^2 - 36 x - 7 = 0 $
Factor out the $37$ from the first two terms
$ 37 ( x^2 - \dfrac{36}{37} x ) - 7 = 0$
Take half of $\dfrac{36}{37}$ which is $\dfrac{18}{37} $ and add it with its sign to $ x $ then square both as follows
$ 37 ( x - \dfrac{18}{37} )^2 - 7 = 37 ( \dfrac{18}{37} )^2$
Multiply through by $37^2$
$ 37 ( 37 x - 18)^2 - 37^2 (7) = 37 (18)^2 $
Divide through by $37$
$ (37 x - 18 )^2 - 7(37) = (18)^2 = 324 $
So,
$ (37 x - 18)^2 = 324 + 7(37) = 583 $
Therefore,
$ x = \dfrac{1}{37} \left ( 18 \pm \sqrt{ 583 } \right)$
|
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|
Given the $ x+y+z =3$ Prove that $ x^3+y^3+z^3+6xyz \geq 27/4$ Also, $x$,$y$ and $z$ are positive.
I have figured out that it would be enough to prove that LHS is bigger than $(27-\text{LHS})/3$, then by substituting 27 with $ (x+y+z)^3 $, it's enough to prove that
$$x^3+y^3+z^3+6xyz \geq x^2(3-x)+y^2(3-y)+z^2(3-z).$$
And I'm stuck. I feel like this has to be provable somehow with rearrangement, since the right hand side is conveniently ordered (if $x>y$, then $3-x<3-y$). But the equality only holds when one of the variables is $0$ and the other two are $1.5$, which isn't the case for rearrangement.
|
Without loss of generality, we assume $z\ge \frac{x+y+z}{3}=1$. We temporarily fix $z$ and check how to arrange $x$ and $y$ to achieve a minimum. Now $y=3-x-z$.
So $f(x)=x^3+(3-x-z)^3+z^3+6x(3-x-z)z$, and the derivative of $f$ is
$$f'(x) = 3x^2-3(3-x-z)^2 + 6(3-x-z) z-6xz=x(-3*2*(z-3) -6z-6z) + (-3(z-3)^2+6(3-z)z) = x(-18z+18) + 3(3-z)(9z-9) $$
and $f''(x)=-18z+18 \le 0$, which means $f(x)$ assumes its minimum at the boundary of $[0, 3-z]$, then either $x=0$ or $y=0$. Without loss of generality, we assume $y=0$, and now we want to prove $x^3+z^3\ge 27/4$ given $x+z=3$ and $x\ge 0$ and $z\ge 0$, which is much easier than the original problem.
|
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"timestamp": "2023-03-29T00:00:00",
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|
solving inhomogenous $2$ x $2$ ODE?
Let $y'=\begin{pmatrix}1 & 2 \\ 3 & 6\end{pmatrix}y+\begin{pmatrix}x \\ sin(x)\end{pmatrix}$ with $y_0=\begin{pmatrix}0 \\ 0\end{pmatrix}$
Now I want to solve it, but don't know how to continue:
$1)$ solve the homogeous ODE $y'=Ay$
$\Rightarrow y_{hom}(x)=e^{0x} \cdot c_1\begin{pmatrix}-2 \\ 1\end{pmatrix}+c_2 \cdot e^{7x}\begin{pmatrix}1 \\ 3\end{pmatrix}=c_1\begin{pmatrix}-2 \\ 1\end{pmatrix}+c_2 \cdot e^{7x}\begin{pmatrix}1 \\ 3\end{pmatrix}$
$2)$ solve particular ODE $y'=Ay+b$ with variation of parameters:
$y_{par}=Y_{hom}\cdot c(x)=\frac{1}{7}\begin{pmatrix}e^{7x}+6 & 3(e^{7x}-1)\\2(e^{7x}-1) & 6e^{7x}+1\end{pmatrix}\begin{pmatrix}\dot c_1(x) \\ \dot c_2(x)\end{pmatrix}=\begin{pmatrix}\ x \\ \ sin(x)\end{pmatrix}$
Now I solved for $c_1$ and $c_2$, but I guess it's wrong because I couldn't solve it without calculator and than I checked it , it was wrong :/ $c_1=\dfrac{\mathrm{e}^{-7x}\left(-1029\sin\left(x\right)+\mathrm{e}^{7x}\left(7350\cos\left(x\right)+1225x^2\right)-147\cos\left(x\right)+\left(2100x-300\right)\mathrm{e}^{14x}\right)}{17150}+C$
$c_2=\dfrac{\mathrm{e}^{-7x}\left(1029\sin\left(x\right)+\mathrm{e}^{7x}\left(1225\cos\left(x\right)-1225x^2\right)+147\cos\left(x\right)+\left(350x-50\right)\mathrm{e}^{14x}\right)}{8575}+C$
Is this way correct or is there another one ?
Thanks in advance
|
We are given
$$y'= Ay + g = \begin{pmatrix}1 & 2 \\ 3 & 6\end{pmatrix}y+\begin{pmatrix}x \\ \sin x\end{pmatrix}, y_0=\begin{pmatrix}0 \\ 0\end{pmatrix}$$
You correctly found the homogeneous solution
$$y_h(x) = c_1\begin{pmatrix}-2 \\ 1\end{pmatrix}+c_2 \cdot e^{7x}\begin{pmatrix}1 \\ 3\end{pmatrix}$$
Since you want to use Variation of Parameters, we will follow Example 2 (I assume you understand the theory, but it is also on the site).
$$Y = \begin{pmatrix}-2 & e^{7 x}\\ 1 & 3 e^{7x}\end{pmatrix} \implies Y^{-1} =
\begin{pmatrix}
-\dfrac{3}{7} & \dfrac{1}{7} \\
\dfrac{e^{-7 x}}{7} & \dfrac{2 e^{-7 x}}{7} \\
\end{pmatrix}$$
We now form
$$Y^{-1} g = \begin{pmatrix}
-\dfrac{3}{7} & \dfrac{1}{7} \\
\dfrac{e^{-7 x}}{7} & \dfrac{2 e^{-7 x}}{7} \\
\end{pmatrix}. \begin{pmatrix}
x \\
\sin x \\
\end{pmatrix} =
\begin{pmatrix}
\dfrac{\sin x}{7}-\dfrac{3 x}{7} \\
\dfrac{1}{7}x e^{-7 x} +\dfrac{2}{7} e^{-7 x} \sin x \\
\end{pmatrix}$$
Next we integrate
$$\displaystyle \int_0^x Y^{-1}g ~dx =
\begin{pmatrix}
\dfrac{1}{14} \left(-3 x^2 -2 \cos x +2\right) \\
\dfrac{74-e^{-7 x} (175 x + 49 \cos x + 343 \sin x + 25)}{8575} \\
\end{pmatrix}$$
We can now write the particular solution $y_p(x) = Y \displaystyle \int_0^x Y^{-1}g~ dx$
$$y_p(x) = \begin{pmatrix} \dfrac{3675 x^2-175 x+74 e^{7 x}-343 \sin x+2401 \cos x-2475}{8575} \\ \dfrac{-3675 x^2-1050 x+444 e^{7 x}-2058 \sin x-2744 \cos x+2300}{17150} \end{pmatrix}$$
|
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|
How to evaluate $\lim _{x\to \infty }\left(\frac{x+3}{\sqrt{x^2-5x}}\right)^{x^2\sin\left(1/x\right)}$?
How to evaluate $\lim _{x\to \infty }\left(\frac{x+3}{\sqrt{x^2-5x}}\right)^{x^2\sin\left(1/x\right)}$?
My Try:
$$\lim _{x\to \infty }\left(x^2\sin\left(\frac{1}{x}\right)\ln\left(\frac{x+3}{\sqrt{x^2-5x}}\right)\right) = \lim _{t\to 0 }\left(\frac{1}{t^2}\sin\left(t\right)\ln\left(\frac{\frac{1}{t}+3}{\sqrt{\frac{1}{t^2}-\frac{5}{t}}}\right)\right)$$
Now $\sin(x) \approx x, x \rightarrow 0$ so:
$$\approx \lim _{t\to 0 }\left(\frac{1}{t}ln\left(\frac{\left(3t+1\right)\sqrt{-5t+1}}{1-5t}\right)\right)$$
At this point i used the rule of the de l'Hôpital so:
$$\lim _{t\to 0 }\left(\frac{1}{t}ln\left(\frac{\left(3t+1\right)\sqrt{-5t+1}}{1-5t}\right)\right) = \lim _{t\to 0}\left(\frac{\frac{-15t+11}{2\left(-5t+1\right)\left(3t+1\right)}}{1}\right) = \frac{11}{2}$$
So:
$$\lim _{x\to \infty }\left(\left(\frac{x+3}{\sqrt{x^2-5x}}\right)^{x^2\sin\left(\frac{1}{x}\right)}\right) = \color{red}{e^\frac{11}{2}}$$
Which it is the exact result of the proposed limit.
My question is, there is another method, different from mine to get the same result? (Preferably without resorting to de l'Hôpital rule).
|
Notice that
$$\frac{x+3}{\sqrt{x^2-5x}} = \left(\dfrac{x^2+6x+9}{x^2-5x}\right)^{1/2} = \left(1+ \dfrac{11x+9}{x^2-5x}\right)^{1/2}.$$
By setting $y = \dfrac{x^2-5x}{11x+9}$ and since $x \rightarrow \infty \implies y \rightarrow \infty$, we obtain
$$ x = \dfrac{5+11y + \sqrt{121y^2+146y+25}}{2}.$$
Thus, $$\lim_{x \rightarrow \infty} \left(\dfrac{x+3}{\sqrt{x^2-5x}}\right)^{x/2} = \lim_{y \rightarrow \infty} \left( 1+\dfrac{1}{y}\right)^{\left(\dfrac{5+11y + \sqrt{121y^2+146y+25}}{4}\right)} = $$
$$= \lim_{y \rightarrow \infty} \left( 1+\dfrac{1}{y}\right)^{5/4}\cdot \left( 1+\dfrac{1}{y}\right)^{11y/4}\cdot \left( 1+\dfrac{1}{y}\right)^{11y/4\cdot \left(\sqrt{1+146/(121y)+25/(121y^2)}\right)} = e^{11/2},$$
Since $g(y) = \sqrt{1+ \frac{146}{121y}+\frac{25}{121y^2}}$ is continuous and $\lim_{y \rightarrow \infty} g(y)$ exists.
Yet, $$\lim_{x \rightarrow \infty}\dfrac{\sin(1/x)}{1/x} = 1.$$
Therefore, we have
$$\lim_{x \rightarrow \infty} \left(\dfrac{x+3}{\sqrt{x^2-5x}}\right)^{x^2\sin\left(1/x\right)} = e^{11/2}.$$
|
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|
Show the recursive sequence is increasing
How do I show that the recursive sequence
$$a_n = a_{\lfloor n/2 \rfloor} + a_{\lceil n/2 \rceil} +3n+1, \quad
n\geq 2, \phantom{x} a_1 = 3$$
is an increasing sequence?
1. attempt:
If I can show that $a_{n+1}-a_n>0$, I would be able to show it is increasing.
\begin{align*}
a_{n+1} - a_n & = a_{\left\lfloor \frac{n+1}{2} \right\rfloor} + a_{\left\lceil \frac{n+1}{2} \right\rceil} +3(n+1)+1 - ( a_{\lfloor n/2 \rfloor} + a_{\lceil n/2 \rceil} +3n+1) \\
& = a_{\left\lfloor \frac{n+1}{2} \right\rfloor} + a_{\left\lceil \frac{n+1}{2} \right\rceil}+3 - a_{\lfloor n/2 \rfloor} - a_{\lceil n/2 \rceil}
\end{align*}
I can't put $a$ together because the indexes are different (because of the ceils and floors).
2. attempt:
Proof by induction
Base case: $n=2$ then $a_2 = a_1 + a_1 + 3\cdot 2 + 1 = 3+3+7=13$ so $a_1<a_2$.
Testing with more gives: $a_1 < a_2 < a_3 =26 < a_4 = 39<... $
Assume $a_n<a_{n+1}$.
Now I want to show that $a_{n+1} < a_{n+2}$
$$ a_{n+2} = a_{\lfloor (n+2)/2 \rfloor} + a_{\lceil (n+2)/2 \rceil} +3(n+2)+1$$
Again I get stuck since I don't know how to handle the ceils and floors.
$\phantom{x}$
How do I go about showing the sequence is increasing? Maybe I'm making it harder than it actually is - is there by any chance an easier way?
|
Suppose for all $m\le n-1$, $a_{m+1}\ge a_m$.
Case 1: $n=2k$. Note that
$$ \lfloor \frac{2k+1}2 \rfloor=k,\lceil\frac{2k+1}2 \rceil=k+1. $$
Then
\begin{eqnarray}
a_{n+1} - a_n & =&a_{2k+1}-a_{2k}\\
&=& a_{\left\lfloor \frac{2k+1}{2} \right\rfloor} + a_{\left\lceil \frac{2k+1}{2} \right\rceil} +3(2k+1)+1 - ( a_{\lfloor 2k/2 \rfloor} + a_{\lceil 2k/2 \rceil} +3\cdot2k+1) \\
& = &a_{k+1} - a_{k}+3\\
&\ge&a_{k+1} - a_{k}\\
&\ge&0
\end{eqnarray}
Do the same for $n=2k-1$.
|
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|
Find the value after Remainder Theorem The expression $4x^3+21x-6$ has the same remainder when divided by $x-a$ or by $x+b$, where $a$ not equals to $b$. Find the value of $a^2+b^2-ab$. I have the answer but I don't know how to work through the question.
|
Given same remainder, by Polynomial Remainder Theorem we have,
$\displaystyle\ \ \ \ \ \ f(a)=f(-b)$
$\displaystyle\Rightarrow 4a^3+21a-6=-4b^3-21b-6$
$\displaystyle\Rightarrow 4(a^3+b^3)+21(a+b)=0$
$\displaystyle\Rightarrow \frac{(a^3+b^3)}{(a+b)}=a^2+b^2-ab=\frac{-21}{4}$
|
{
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|
What is the minimum value of $x^2+y^2$ subject to $x^3+y^3+xy=1$?
What is the minimum value of $x^2+y^2$ under the constraint $x^3+y^3+xy=1$? Please do not use partial differentials (multivariable calculus) or Lagrange multipliers. You can use elementary algebra or single variable calculus.
I plotted the graph of $x^3+y^3+xy=1$. It seems the minima occurs when $x=y$ but I don't know why that will be true. Any ideas?
|
The cubic $x^3+y^3+xy+1$ is symmetric in $y=x$. So lets rotate it by $\frac{\pi}{4}$. Let $x=\frac{x`}{\sqrt{2}}+\frac{y`}{\sqrt{2}}$ and $y=\frac{x`}{\sqrt{2}}-\frac{y`}{\sqrt{2}}$. Note that the function to minimize becomes $x`^2+y`^2$.
$$\left(\frac{x`}{\sqrt{2}}+\frac{y`}{\sqrt{2}}\right)^3+\left(\frac{x`}{\sqrt{2}}-\frac{y`}{\sqrt{2}}\right)^3+\left(\frac{x`}{\sqrt{2}}+\frac{y`}{\sqrt{2}}\right)\left(\frac{x`}{\sqrt{2}}-\frac{y`}{\sqrt{2}}\right)-1=0$$
$$\frac{3x`y`^2}{\sqrt{2}}+\frac{x`^3}{\sqrt{2}}-\frac{y`^2}{2}+\frac{x`^2}{2}-1=0$$
We can solve for $y`^2$.
$$y`^2=\frac{-\sqrt{2}x`^3-x`^2+2}{3\sqrt{2}x`-1}$$
As this is positive solutions will only exist for $-\sqrt{2}x`^3-x`^2+2\ge0$. Note the root of $-\sqrt{2}x`^3-x`^2+2=0$ can be found by solving: $-2\left(2\left(\frac{x`}{\sqrt{2}}\right)^3+\left(\frac{x`}{\sqrt{2}}\right)^2-1\right)=0$. Also note that $y`^2$ is strictly decreasing on this domain.
Using calculus to find the minimum of $x`^2+y`^2$ gives a value outside this domain so the minimum must be at the end of the domain, i.e. at the real root of $2\left(\frac{x`}{\sqrt{2}}\right)^3+\left(\frac{x`}{\sqrt{2}}\right)^2-1=0$. And the corresponding $y`$ is $0$. So rotating back to $x$ and $y$ shows that the minimum is on the line $x=y$ and is the real root of $2x^3+x^2-1$.
|
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|
$a+b+c+ab+bc+ca+abc=1000$ find the minimum value of $a+b+c$ Given $a+b+c+ab+bc+ca+abc=1000$.
Find the minimum value of $a+b+c$.
Now we are considering $a$, $b$, $c$ to be integers and here in lies the pertinent problem as I could not get an answer in integers but in fractions.
|
Without loss of generality, suppose that $a\leq b\leq c$. Also, write $N:=1001$. Since $$(a+1)(b+1)(c+1)=N\,,$$ we have that $$|a+1|\,|b+1|\leq N\,.$$ This clearly means
$$|a+1|+|b+1|\leq N+1\,.$$
(If $x$ and $y$ are positive integers with $xy\leq N$, then $(x-1)(y-1)\geq 0$, which implies that $x+y\leq xy+1\leq N+1$.)
Thus,
$$-(a+1)-(b+1)\leq |a+1|+|b+1|\leq N+1\text{ or }a+b\geq -N-3\,.$$
Since $c+1$ is clearly positive, we have $c+1\geq 1$, which means $c\geq 0$. That is,
$$a+b+c\geq -N-3\,.$$
The equality holds iff $(a,b,c)=(-N-1,-2,0)$.
If $a$, $b$, and $c$ are required to be nonnegative, then by the AM-GM Inequality, we have
$$\frac{(a+1)+(b+1)+(c+1)}{3}\geq \sqrt[3]{(a+1)(b+1)(c+1)}=\sqrt[3]{1001}>10\,.$$
That is, $a+b+c>27$, or $a+b+c\geq 28$. As $(a,b,c)=(6,10,12)$ works, the minimum value of $a+b+c$ is indeed $28$. (Unlike the first part of this answer, $1001$ can't be replaced by an arbitrary positive integer $N$. For each $N$, a different analysis is required.)
|
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|
Sum $\frac {1}{9} + \frac {1}{18} +\frac {1}{30} +\frac {1}{45} +\frac {1}{63} \ldots$ What is the sum of the following series:
$\frac {1}{9} + \frac {1}{18} +\frac {1}{30} +\frac {1}{45} +\frac {1}{63} \ldots$
I tried to do it through telescope method but it didn't work.
|
I believe we have,
$$\frac{2}{3} \sum_{n=2}^{\infty} \frac{1}{n(n+1)}$$
Clearly telescoping will work.
How I got it,
Let's look at denominators.
$$9,18,30,45,63,...$$
Take difference of consecutive two.
$$9,12,15,18,..$$
This is clearly given by $3(x-1)+9$ because the difference between every two consecutive numbers is $3$.
So,
$$f(x+1)-f(x)=3x+6$$
Actually shifting everything, the terms, to the right $1$ so that the terms start at $x=2$ gives,
$$f(x+1)-f(x)=3(x-1)+6=3x+3$$
This is convenient to do because it factors,
$$f(x+1)-f(x)=3(x+1)$$
Now sum both sides of the top equation from $x=2$ to $n-1$ and something magical will happen. Keep in mind $f(2)=9$.
|
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|
Area of the shaded part.
My Attempt,
Area of square $=10^2=100 cm^2$.
Area of circle $=\pi r^2=25\pi cm^2$.
What should I do further?
|
$\begin{align} \color{darkorange}{\textsf{Shaded}} &= (\color{red}{\textsf{Triangle}} + \color{green}{\textsf{Polygon}})-\color{blue}{\textsf{CircleSegment}}\\&= \color{red}{\left(\frac{1\times2}{2}\right)} + \color{green}{\left(3 + \frac{3\times 4}{2}\right)}-\color{blue}{\left(\frac{\arctan{(3/4)}}{2\pi}\pi \times 5^2\right)} \\&= \color{red}{1} + \color{green}{9}-\color{blue}{\frac{5^2}{2}\arctan{\frac 3 4}} \\ &= 10 -\color{blue}{12.5\arctan{\frac 3 4)}}\\ &\approx \color{darkorange}{1.95624...}\end{align}$
|
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|
Area between geodesic and arc of constant lattitude A geodesic is a line representing the shortest route between two points on a sphere, for example on the Earth treated here as a perfect sphere.
Two points on Earth having the same latitude can be also connected with the line being a part of a circle for selected constant latitude.
Differences between these two lines can be visualized with the use of
this Academo program presenting the situation in the context of the map of Earth.
Question:
*
*How to calculate the area between these two lines?
(Assume for example that the starting point is $(\alpha, \beta_1)=(45^\circ, -120^\circ)$ and the destination $(\alpha, \beta_2)=(45^\circ, 0^\circ))$ - the arc of constant latitude $45^\circ$ has length $120^\circ$.
In wikipedia a formula for an area of a spherical polygon is presented, but the polygon is limited in this case with parts of geodesics. Is it possible somehow transform these formulas of spherical geometry into the case of finding the area between geodesic and arc of constant latitude?
|
Summary: If $A$ and $B$ lie on the latitude line at angle $0 < \alpha < \pi/2$ north of the equator on a sphere of unit radius, and at an angular separation $0 < \theta = \beta_{2} - \beta_{1} < \pi$, then the "digon" bounded by the latitude and the great circle arc $AB$ (in blue) has area
\begin{align*}
\pi - \theta \sin\alpha - 2\psi
&= \text{sum of interior angles} - \theta \sin\alpha \\
&= \pi - \theta \sin\alpha - 2\arccos\left(\frac{\sin\alpha(1 - \cos\theta)}{\sqrt{\sin^{2}\theta + \sin^{2}\alpha(1 - \cos\theta)^{2}}}\right).
\end{align*}
If $A$ and $B$ have longitude-latitude coordinates $(0, \alpha)$ and $(\theta, \alpha)$, their Cartesian coordinates (on the unit sphere) are
$$
A = (\cos\alpha, 0, \sin\alpha),\quad
B = (\cos\theta\cos\alpha, \sin\theta \cos\alpha, \sin\alpha).
$$
Let $C = (0, 0, 1)$ be the north pole, $G$ the "gore" (shaded) bounded by the spherical arcs $AC$, $BC$, and the latitude through $A$ and $B$, and $T$ the geodesic triangle with vertices $A$, $B$, and $C$.
Lemma 1: The area of $G$ is $\theta(1 - \sin\alpha)$.
Proof: The spherical zone bounded by the latitude through $A$ and $B$ and containing the north pole has height $h = 1 - \sin\alpha$ along the diameter through the north and south poles. By a theorem of Archimedes, this zone has area $2\pi h = 2\pi(1 - \sin\alpha)$. The area of the gore $G$, which subtends an angle $\theta$ at the north pole, is
$$
(\theta/2\pi)2\pi(1 - \sin\alpha) = \theta(1 - \sin\alpha).
$$
Lemma 2: The area of $T$ is $\theta - \pi + 2\arccos\dfrac{\sin\alpha(1 - \cos\theta)}{\sqrt{\sin^{2}\theta + \sin^{2}\alpha(1 - \cos\theta)^{2}}}$.
Proof: If $\psi$ denotes the interior angle of $T$ at either $A$ or $B$, the area of $T$ is the angular defect, $\theta + 2\psi - \pi$. To calculate $\psi$, note that the unit vector $n_{1} = \frac{A \times C}{\|A \times C\|} = (0, -1, 0)$ is orthogonal to the great circle $AC$, the unit vector
$$
n_{2} = \frac{A \times B}{\|A \times B\|}
= \frac{(-\sin\theta \sin\alpha, \sin\alpha(\cos\theta - 1), \cos\alpha \sin\theta)}{\sqrt{\sin^{2}\theta + \sin^{2}\alpha(1 - \cos\theta)^{2}}}
$$
is orthogonal to the great circle $AB$, and
$$
\cos\psi = n_{1} \cdot n_{2}
= \frac{\sin\alpha(1 - \cos\theta)}{\sqrt{\sin^{2}\theta + \sin^{2}\alpha(1 - \cos\theta)^{2}}}.
$$
This completes the proof of Lemma 2.
The area of the digon is the difference,
\begin{align*}
A &= \theta(1 - \sin\alpha) - (\theta + 2\psi - \pi)
= \pi - \theta \sin\alpha - 2\psi \\
&= \pi - \theta \sin\alpha - 2\arccos\left(\frac{\sin\alpha(1 - \cos\theta)}{\sqrt{\sin^{2}\theta + \sin^{2}\alpha(1 - \cos\theta)^{2}}}\right).
\end{align*}
When $\alpha = 0$, the area vanishes for $0 < \theta < \pi$ (because the latitude through $A$ and $B$ coincides with the great circle arc), while if $\alpha$ is small and positive, the area is close to $\pi$ when $\theta = \pi$ (because $A$ and $B$ are nearly antipodal and the great circle arc passes through the north pole).
|
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|
Series shown up in Quantum Mechanics Here is the series, by some concepts in quantum mechanics, it should be 1, however, I don't know how to prove it. Could someone show me the proof? Thanks.
$$\lim_{k\to \infty}\sum_{n=1}^{k} \frac{4}{n^2\pi^2}(1-\cos\frac{n\pi}{2})^2
$$
|
Let $S$ be defined as
$$
S=\frac{4}{\pi^2}\sum_{k\geq 1}\frac{1}{k^2}\left(1-\cos\left(\frac{\pi k}{2}\right)\right)^2
$$
Decompose $S$ according to the remainder of $n$ modulo $4$.
Since $\cos(\pi n/2)=1$ for $n=0 \mod 4$ one part of the sum exactly cancels. Furthermore $\color{blue}{\cos(n\pi/2)=0}$ for $\color{blue}{n=1,3\mod 4}$ and $\cos(n\pi/2)=-1$ for $n=2\mod 4$ which means that we can equally write
$$
S=\frac{4}{\pi^2}\sum_{k\geq 0}\left(\color{blue}{\frac{1}{(4k+1)^2}+\frac{1}{(4k+3)^2}}+\frac{2^2}{(4k+2)^2}\right)=\frac{4}{\pi^2}\sum_{k\geq 0}\left(\color{blue}{\frac{1}{(2k+1)^2}}+\frac{2^2}{(4k+2)^2}\right)=\\\frac{4}{\pi^2}\sum_{k\geq 0}\left(\color{blue}{\frac{1}{(2k+1)^2}}+\frac{1}{(2k+1)^2}\right)=\frac{8}{\pi^2}{\sum_{k\geq 0}\frac{1}{(2k+1)^2}}
$$
or
$$
S=1
$$
Since
$$
{\sum_{k\geq 0}\frac{1}{(2k+1)^2}}=\sum_{k\geq 1}\frac{1}{k^2}-\sum_{k\geq 1}\frac{1}{(2k)^2}=\frac{\pi^2}{6}-\frac{1}{4}\frac{\pi^2}{6}={\frac{\pi^2}{8}}
$$
From a number theoretical view we can identify the sum as
$$
S=\frac{4}{\pi^2}L(\chi_1,2)=\frac{4}{\pi^2}\sum_{n\geq1}\frac{(1-\chi_1(n))^2}{n^2}
$$
where $\chi_1(n)$ is the non-trivial Dirichlet character associcated with the map $Z/4Z\rightarrow \mathbb{S_1}$
|
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|
Find integer solution of a system of equations Find all such natural $n$ such that
$
\dfrac{12n-21143}{25}
$
and
$
\dfrac{2n-3403}{25}
$
both are squares of prime numbers.
So far I have found by computer only solution: $ 2014.$
How to prove taht there are no more solutions?
I was tried first to solve the system
\begin{cases}
12n=21143 \mod 25,\\
2n=3403 \mod 25,
\end{cases}
and find that $n=14 \mod 25$. Then I substitute $n=14+25 \cdot k$ into the system
\begin{cases}
12(14+25 \cdot k)-21143= 25 p^2,\\
2(14+25 \cdot k)-3403= 25 q^2,
\end{cases}
and get
\begin{cases}
12k= 839+ p^2,\\
2k= 135+ q^2,
\end{cases}
or $10k=704 +p^2-q^2.$ It follows that $p^2-q^2=6 \mod 10$. But how to find now all $k?.$
Edit. If we rewrite it in the way $p^2-q^2=16=4^2 \mod 10$ then we get the $p=5 \mod 10$ and $q=3 \mod 10.$
Sorry, one solutions.
|
Your equations are
\begin{cases}
12k= 839+ p^2\\
2k= 135+ q^2
\end{cases}
and we observe that $$p^2-q^2 = 10k - 704 = 5(2k-141)+1$$
so $\,p^2 - q^2 = 1 \pmod{5}$ which tells us that $p^2 = 1 \mod 5$ and $q = 0 \mod 5$, or $p=0 \mod 5$ and $q^2 = -1 \mod 5$. We consider the first case here, and the second later. Since $q$ is prime and divisible by $5$, $q=5$. Then
$$
2k = 135+25 = 160 \implies k=80
$$
and
$$
12k = 960 =- 839+p^2\\
P^2 = 121\\
p=11
$$
Finally,
$$
12 n - 21143 = 25p^2= 3025\\
12n = 24168\\
n=2014
$$
or
$$
2 n - 3403= 25q^2= 625\\
2n = 4028\\
n=2014
$$
So that is the one solution.
Now consider the second case: $p=0 \mod 5$ and $q^2 = -1 \mod 5$.
$$12k=839+p^2 = 144 \\
k=72$$
And the other equation becomes
$$2\cdot 72 = 135 + q^2\\
q=3$$
So that will give a second solution, with
$$
n = 14 + 25*k = 14 + 25\cdot 72=1814 $$
Kudos to Leox for noting the second case, which I had overlooked in my original answer.
|
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|
What is the coefficient of $x^2$ in the polynomial.. What is coefficient of $x^2$ in polynomial $(1-4x)^6(1+3x)^8$ ?
I know how to do it for $(1-4x)^6$ for example, but how to do it for the product of two polynomials?
|
Let $f(x) = (1-4x)^6(1+3x)^8 = uv$, where $u = (1-4x)^6$ and $v=(1+3x)^8$
\begin{align*}
f''(x) &= u'' v + 2 u'v' + v''\\
&= 6 \cdot 5 \cdot (-4)^2 (1-4x)^4 (1+3x)^8 + 2 \cdot 6 \cdot (-4) \cdot 8 \cdot 3 (1-4x)^5 (1+3x)^7 \\
& \qquad + 8 \cdot 7 \cdot (3)^2 (1-4x)^6 (1+3x)^6
\end{align*}
Hence coefficient of $x^2$ is $\frac{1}{2!}f''(0) = (480 - 1152 + 504)/2 = -84 $
|
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|
Solve $y'=\frac{2}{x^2}-y^{2}$
$$y'=\frac{2}{x^2}-y^{2}$$
$y=\frac{k}{x}$ is a solution, find the general solution
So we have riccati differential equation.
$y=z+\frac{k}{x}\iff z=y-\frac{k}{x}$
$y'=z'-\frac{k}{x^2}$
Plugin it to the ode we get:
$$z'-\frac{k}{x^2}=\frac{2}{x^2}-(z+\frac{k}{x})^2$$
$$z'=\frac{2-k^2+k}{x^2}-\frac{2kz}{x}-z^2$$
Which is again a riccati differential equation do I need to guess a particular $k$? for example $k=-1$?
|
Hint: if $y=\frac{k}{x}$ is a solution then pluging in the equation we get $$\\ -\frac { k }{ { x }^{ 2 } } =\frac { 2 }{ { x }^{ 2 } } -\frac { { k }^{ 2 } }{ { x }^{ 2 } } \\ { k }^{ 2 }-k-2=0\\ \left( k-2 \right) \left( k+1 \right) =0\\ $$
|
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|
System of two cubic equations: $x+y^2 = y^3$, $y+x^2=x^3$ I got stuck on this system of equations. Could you help and tell me how should I approach this problem?
\begin{align*}
x+y^2 &= y^3\\
y+x^2 &= x^3
\end{align*}
These are the solutions:
\begin{align*}
(0, 0); \\((1+\sqrt{5})/2, (1+\sqrt{5})/2); \\((1-\sqrt{5})/2, (1-\sqrt{5})/2);
\end{align*}
|
Since $y=x^3-x^2$, we get $x+(x^3-x^2)^2=(x^3-x^2)^3$ or
$$x(x^2-x-1)(x^6-2x^5+2x^4-2x^3+2x^2-x+1)=0$$
and since $x^6-2x^5+2x^4-2x^3+2x^2-x+1=(x^6-2x^5+x^4)+(x^4-2x^3+x^2)+(x^2-x+1)>0$,
we get your answer.
|
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|
Find the sum of the series $\frac{1}{1\cdot 2\cdot 3}+\frac{1}{4\cdot 5\cdot 6}+\frac{1}{7\cdot 8\cdot 9}+...=$ Given : $$\frac{1}{1\cdot 2\cdot 3}+\frac{1}{4\cdot 5\cdot 6}+\frac{1}{7\cdot 8\cdot 9}+ ...$$
What is the sum of this series, how can one rewrite it to look simpler ?
EDIT :
Actually I found how to rewrite it :
$$\sum _{n=1}^{\infty }\:\frac{1}{\left(3n-2\right)\left(3n-1\right)3n}$$
|
Let's find the sum of the geometric series: $$\sum_{k=0}^\infty x^{3k}=\frac{1}{1-x^{3}}$$
For $0<x<1$. Now let's integrate this series by term three times:
$$\sum_{k=0}^\infty \frac{x^{3k+1}}{3k+1}=\int_0^x \frac{dt}{1-t^{3}}$$
$$\sum_{k=0}^\infty \frac{x^{3k+2}}{(3k+1)(3k+2)}=\int_0^x \int_0^y \frac{dt~ dy}{1-t^{3}}$$
$$\sum_{k=0}^\infty \frac{x^{3k+3}}{(3k+1)(3k+2)(3k+3)}=\int_0^x \int_0^y \int_0^z \frac{dt~ dy~ dz}{1-t^{3}}$$
Changing $k+1 \to k$ we have:
$$\sum_{k=1}^\infty \frac{x^{3k}}{(3k-2)(3k-1)3k}=\int_0^x \int_0^y \int_0^z \frac{dt~ dy~ dz}{1-t^{3}}$$
For $x=1$ we have:
$$\sum_{k=1}^\infty \frac{1}{(3k-2)(3k-1)3k}=\int_0^1 \int_0^y \int_0^z \frac{dt~ dy~ dz}{1-t^{3}}$$
Looks complicated, but if we use Cauchy formula for repeated integration, we simply get:
$$\int_0^1 \int_0^y \int_0^z \frac{dt~ dy~ dz}{1-t^{3}}=\frac{1}{2} \int_0^1 \frac{(1-t)^2dt}{1-t^{3}}=\frac{1}{2} \int_0^1 \frac{(1-t)dt}{1+t+t^2}$$
So the integral form of this series is:
$$\sum_{k=1}^\infty \frac{1}{(3k-2)(3k-1)3k}=\frac{1}{2} \int_0^1 \frac{(1-t)dt}{1+t+t^2}=\frac{1}{12} \left(\sqrt{3} \pi -3 \log (3)\right)$$
To solve the integral we just need to modify the functions so it looks like the definitions of the arctangent and the logarithm. It's not that hard, but I will add the full solution if the OP needs it.
A little hint:
$$1+t+t^2=\frac{3}{4}+\left(\frac{1}{2}+t \right)^2=\frac{3}{4} \left(1+ \frac{4}{3}\left(\frac{1}{2}+t \right)^2 \right)$$
Thus, we need a change of variable: $$u=\frac{2}{\sqrt{3}} \left(\frac{1}{2}+t \right)$$
|
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|
Evaluate: $\frac{1}{2^1}-\frac{2}{2^2}+\frac{3}{2^2}-\frac{4}{2^4}+\ldots-\frac{100}{2^{100}}$ The problem is to evaluate the following sum:
$$\frac{1}{2^1}-\frac{2}{2^2}+\frac{3}{2^3}-\frac{4}{2^4}+\ldots-\frac{100}{2^{100}}$$
My approach was to find the common denominator ($2^{100}$), then the series becomes:
$$ \frac{1\cdot 2^{99}-2\cdot 2^{98}+3\cdot 2^{97}-4\cdot 2^{96}+\ldots -100\cdot 2^0}{2^{100}}$$
And split it this way:
$$ \frac{(1\cdot 2^{99}+3\cdot 2^{97}+\ldots+99\cdot 2^{1})-(2\cdot 2^{98}+4\cdot 2^{96}+\ldots+100\cdot 2^{0})}{2^{100}}$$
But it seems that cacelling out the terms is not a promising approach. Any hints?
|
$$I=\frac{1}{2^1}-\frac{2}{2^2}+\frac{3}{2^3}-\frac{4}{2^4}+\ldots-\frac{100}{2^{100}}$$
thus
$$2I=1-\frac{2}{2^1}+\frac{3}{2^2}-\frac{4}{2^3}+\ldots-\frac{100}{2^{99}}$$
as a result
$$2I+I=1-\left(\frac{2}{2^1}-\frac{1}{2^1}\right)+\left(\frac{3}{2^2}-\frac{2}{2^2}\right)-\left(\frac{4}{2^3}-\frac{3}{2^3}\right)+\cdots-\left(\frac{100}{2^{99}}-\frac{99}{2^{99}}\right)-\frac{100}{2^{100}}$$
therefore
$$3I=\underbrace{\left(1-\frac 12+\frac 14-\frac 18+\cdots-\frac{1}{2^{99}}\right)}_{\frac{2^{101}-2}{3\times 2^{100}}}-\frac{100}{2^{100}}=\frac{2^{101}-302}{3\times 2^{100}}$$
|
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|
Fourier Coefficients for $\frac{3}{5-4\cos{2\theta}}$ I want to compute the (even) fourier coefficients for
$\dfrac 3 {5-4\cos 2\theta}$ on the interval $[0, 2\pi]$.
Namely, I want to compute the integral:
$$b_n = \int_0^{2\pi} \cos(n\theta) \frac 3 {5-4\cos 2\theta } \frac{d\theta}{2\pi}$$
Integrating this in mathematica, it seems like $b_n = 0$ if $n$ is odd and $b_n = \left(\frac 1 2 \right)^{n/2}$ for $n$ even.
|
Hint: Observe
\begin{align}
\frac{3}{5-4\cos 2\theta} = \frac{3/5}{1-\frac{4}{5}\cos 2\theta} = \frac{3}{5}\sum^\infty_{n=0} \left(\frac{4}{5}\right)^n \cos^n 2\theta
\end{align}
and
\begin{align}
\cos^n 2\theta = \left(\frac{e^{2i\theta}+e^{-2i\theta}}{2}\right)^n = \frac{1}{2^n}\sum^n_{k=0}\binom{n}{k} e^{2ik\theta}e^{-2i(n-k)\theta}.
\end{align}
Thus
\begin{align}
\int^{2\pi}_0 e^{2ik\theta}e^{-2i(n-k)\theta} \frac{e^{in\theta}+e^{-in\theta}}{2}\ dx = \text{ something }.
\end{align}
Of course, you could have just consider the integral
\begin{align}
\int^{2\pi}_0 \cos(n\theta) \cos^{n} 2\theta\ \frac{d\theta}{2\pi}.
\end{align}
|
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|
Finding basis of subspace from homogenous equations (checking) The question;
$U = \{x |Ax = 0\}$ If $ A = \begin{bmatrix}1 & 2 & 1 & 0 & -2\\ 2 & 1 & 2 & 1 & 2\\1 & 1 & 0 & -1 & -2\\ 0 & 0 & 2 & 0 & 4\end{bmatrix}$
Find a basis for $U$.
To make it linearly independent, I reduce the rows of $A$;
$\begin{bmatrix}1 & 0 & 0 & 0 & 0\\0 & 1 & 0 & 0 & -2\\0 & 0 & 1 & 0 & 2\\0 & 0 & 0 & 1 & 0\end{bmatrix}\begin{bmatrix}x_1 & x_2 & x_3 & x_4\end{bmatrix} = 0$
So I figure I just have to list the rows out as vectors;
Basis vectors = $\begin{pmatrix}1\\0\\0\\0\\0\end{pmatrix} , \begin{pmatrix}0\\1\\0\\0\\-2\end{pmatrix} , \begin{pmatrix}0\\0\\1\\0\\2\end{pmatrix} , \begin{pmatrix}0\\0\\0\\1\\0\end{pmatrix}$
(can't figure out how to surround it all with curly brackets in MathJax).
Am I doing this properly? It feels too simple...
|
Observe that row operations doesn't change the solution for the system of linear equations. From the last column of the reduced row echelon form (RREF) of $A$,
$$\begin{bmatrix}1 & 0 & 0 & 0 & 0\\0 & 1 & 0 & 0 & -2\\0 & 0 & 1 & 0 & 2\\0 & 0 & 0 & 1 & 0\end{bmatrix}\begin{bmatrix}x_1 \\ x_2 \\ x_3 \\ x_4 \\ x_5\end{bmatrix} = 0,\tag{*}\label{*}$$
we see that if $x = (x_1,\dots,x_5)^T \in U$, $Ax = 0$, so $\sum\limits_{j=1}^5 a_i x_i = 0$, where $a_i$ denotes the $i$-th column of $A$. From \eqref{*}, we observe that $(x_1,x_2,x_3,x_4,x_5) = (0,-2,2,0,-1)$ is a possible solution. By the rank-nullity theorem, since $\mathop{\mathrm{rank}}(A) = 4$, so $\dim U = \dim\mathsf{N}(A) = 5 - 4 = 1$. Hence, we conclude that $x = (0,-2,2,0,-1)^T$ is a basis for $U$.
|
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|
Evaluating $\sum_{n \in \mathbb{N}} \frac{1}{n^2}$ with the Poisson summation formula So, there is the following exercise at the end of a lecture script
By using the function $f(x)=e^{-a|x|}$ ($x \in \mathbb{R}, a>0$) evaluate
$$\sum_{n \in \mathbb{N}} \frac{1}{n^2+a^2}$$
and by letting $a \to 0$ find $\sum_{n \in \mathbb{N}} \frac{1}{n^2}$.
Okay, this task wants me to use the Poisson summation formula:
$$\sum_{k \in \mathbb{Z}} \hat{f}(k)=2\pi \sum_{k \in \mathbb{Z}} f(2\pi k)$$
As hinted I computed $\hat{f}(k)=\frac{2a}{a^2+k^2}$. Therefore
$$\sum_{n \in \mathbb{N}}\frac{1}{n^2+a^2}=\frac{1}{2a}\sum_{n \in \mathbb{N}} \hat{f}(n) $$
Further I should decompose the sum since I don't know if Poisson's formula is valid if the sum only runs over the natural numbers. We have $$\sum_{n \in \mathbb{Z}}\frac{1}{n^2+a^2}=2\sum_{n \in \mathbb{N}}\frac{1}{n^2+a^2}+\frac{1}{a^2}$$
All in all I get
$$\begin{align}\sum_{n \in \mathbb{N}}\frac{1}{n^2+a^2}&=-\frac{1}{2a^2}+\frac{1}{2} \sum_{n \in \mathbb{Z}}\frac{1}{n^2+a^2}\\ &=-\frac{1}{2a^2}+\frac{1}{4a} \sum_{n \in \mathbb{Z}} \hat{f}(n)\\ &=-\frac{1}{2a^2}+\frac{\pi}{2a} \sum_{n \in \mathbb{Z}} f(2\pi n) \\ &=-\frac{1}{2a^2}+\frac{\pi}{2a} \sum_{n \in \mathbb{Z}} e^{-2\pi a |n|}\end{align}$$
But I am not sure how to apply the limit $a \to 0$ now. I should get $\frac{\pi^2}{6}$ I guess.
EDIT: As hinted I continue
$$\begin{align}\sum_{n \in \mathbb{N}}\frac{1}{n^2+a^2}&=-\frac{1}{2a^2}+\frac{\pi}{2a} \left(-1+2\sum_{n \in \mathbb{N_0}} e^{-2\pi a n} \right) \\
&=-\frac{1}{2a^2}+\frac{\pi}{2a} \left(-1+ \frac{2}{1-e^{-2\pi a}} \right) \\ &=-\frac{1}{2a^2}+\frac{a\pi}{2a^2} \left(-\frac{1-e^{-2\pi a}}{1-e^{-2\pi a}}+ \frac{2}{1-e^{-2\pi a}} \right) \\
&=-\frac{1}{2a^2}+\frac{a\pi}{2a^2} \frac{1+e^{-2\pi a}}{1-e^{-2\pi a}} \\&=-\frac{1-e^{-2\pi a}}{2a^2 (1-e^{-2\pi a})}+\frac{a\pi}{2a^2} \frac{1+e^{-2\pi a}}{1-e^{-2\pi a}}
\\&=\frac{-1+e^{-2\pi a}+a\pi+a\pi e^{-2\pi a}}{2a^2-2a^2e^{-2\pi a}}\end{align}$$
Now, using a few times L'Hospital I indeed got $\frac{\pi^2}{6}$. Surely not the most elegant way. Thanks a lot for all your help and hints.
|
The Poisson summation formula leads to:
$$ \sum_{n\geq 0}\frac{1}{n^2+a^2}=\frac{1+\pi a \coth(\pi a)}{2a^2}\tag{1}$$
and $\lim_{a\to 0^+}\sum_{n\geq\color{red}{ 1}}\frac{1}{n^2+a^2}$ can be computed through de l'Hospital theorem or just by recalling that in a neighbourhood of the origin
$$ z\coth(z)=1 + \frac{z^2}{3}+O\left(z^4\right).\tag{2}$$
|
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|
Find that solution $ϕ$ which satisfies $ϕ(0) = 0, ϕ(1) = 3$ for $y'' = 3x + 1$. Find the particular solution to the following second order linear ordinary differential equation that satisfies the $ϕ$
$y'' = 3x + 1$, $ϕ(0) = 0, ϕ(1) = 3$
My solution:
$y' = \frac{3x^2}{2} + x + c_1$
$y = \frac{x^3}{2} + \frac{x^2}{2} + c_1x + c_2$
$ϕ(0) = 0 => 0 + 0 + 0 + c_2 = 0 => c_2 = 0$
$ϕ(1) = 3 => \frac{1}{2} + \frac{1}{2} + c_1 + c_2 = 3$
I am in trouble in step 4 where I need to find the value of $c_1$. How do I go about finding this? Should I do this: $\frac{1}{2} + \frac{1}{2} + c_1 + c_2 = 3 => 1 + 2c = 3 => c_1 = 1$? Please help me!
|
You already have c2=0 substitute it and find c1.
|
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|
While I am solve $(4+t^2)\frac{dy}{dt} + 2ty = 4t$, I don't know why my answer is wrong.. $(4+t^2)\frac{dy}{dt} + 2ty = 4t$
$(4+t^2)\frac{dy}{dt}= 4t -2ty$
$(4+t^2)dy = 4t dt -2tydt$
$\int (4+t^2)dy = \int( 4t -2ty )dt$
$4y+t^2y = 2t^2 -t^2y+ c$
$y(4+2t^2)= 2t^2 +c $
$y = \frac{2t^2 + c}{4+2t^2}$
This is my answer, but the text book answer is $y = \frac{2t^2 + c}{4+t^2}$
This is text book.... answer and solution
I don't know why my answer is wrong.
so... to solve this kind of question... can't I solve like what I did ? ( multiply both side with dt)
|
The basic error was integrating before the variables were separated.
\begin{eqnarray}
(4+t^2)\frac{dy}{dt} + 2ty &=& 4t\\
(4+t^2)\,dy+2ty\,dt&=&4t\,dt\\
(4+t^2)\,dy+(2ty-4t)\,dt&=&0\\
(4+t^2)\,dy+2t(y-2)\,dt\\
\frac{1}{y-2}\,dy+\frac{2t}{4+t^2}\,dt&=&=0\\
\ln\vert y-2\vert+\ln\vert 4+t^2\vert&=&\ln\vert c\vert\\
\ln\vert(y-2)(4+t^2)\vert&=&\ln\vert c\vert\\
(y-2)(4+t^2)&=&c\\
y-2&=&\frac{c}{4+t^2}\\
y&=&\frac{c}{4+t^2}+2\\
y&=&\frac{c+8+2t^2}{4+t^2}\\
y&=&\frac{2t^2+C}{4+t^2}
\end{eqnarray}
|
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|
Prove decimal number pattern for its n-bit 2's compliment If we let $B = b_{n−1}b_{n−2} · · · b_1b_0$ be an $n$-bit $2$’s complement integer.
How can we show that the decimal value of $B$ is $−b_{n−1}\cdot2^{n−1} + b_{n−2} \cdot 2^{n−2} + b_{n−3}\cdot 2^{n−3} + \cdots + b_1\cdot 2 + b_0$.
|
Hint: let $\;A=-B\;$ then, by the definition of $2$'s complement, $A$ is obtained from $B$ by flipping all bits and adding $1\,$:
$$
A = (1-b_{n-1})\cdot 2^{n-1} + (1-b_{n-2})\cdot 2^{n-2} + \cdots + (1-b_1) \cdot 2 + (1-b_0) + 1
$$
Let $B\,'$ be the given expression $B\,' = −b_{n−1}\cdot2^{n−1} + b_{n−2} \cdot 2^{n−2} + \cdots + b_1\cdot 2 + b_0\,$. Then:
$$
\require{cancel}
\begin{align}
A + B\,' & = - 2 b_{n-1}\cdot 2^{n-1} + (2^{n-1}+2^{n-2}+\cdots+2+1) + 1 = \\
& = -b_{n-1} \cdot 2^{n} + (2^n - \cancel{1}) + \cancel{1} \\
& \equiv \;0 \pmod{2^n}
\end{align}
$$
Since $2$'s complement arithmetic is $\bmod 2^n\;$ it follows that $A+B\,'=0$ $\iff$ $B\,'=-A=B$.
|
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|
If $\tan^3A+\tan^3B+\tan^3C=3\tan(A)\tan(B)\tan(C)$, prove triangle ABC is equilateral triangle If $\tan^3A+\tan^3B+\tan^3C=3\tan(A)\tan(B)\tan(C)$, prove triangle ABC is equilateral triangle
Now i remember a identity which was like if $a+b+c=0$,then $a^3+b^3+c^3=3abc$. So i have $\sum_{}^{} \tan(A)=0$. How do i proceed?
Thanks.
|
Note that if $A,B,C$ are angles of a triangle, we have that $$\tan A+\tan B +\tan C=\tan A \tan B \tan C$$ As seen here. If $$0=\tan A+\tan B +\tan C=\tan A \tan B \tan C$$
Then we have that for at least one angle among $A,B,C$ is $0$ or $\pi$. Thus, it is a contradiction. So $$\tan A+\tan B +\tan C \neq 0$$
Now exploit the fact $$a^3+b^3+c^3-3abc=0$$ is true only if $a=b=c$ given that $a+b+c \neq 0$. Thus $\tan A=\tan B= \tan C$. Thus $\triangle {ABC}$ is equilateral.
|
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|
Question on Linear Combinations and Vectors This is a very simple question I just wanted to make sure I was doing correctly.
Express the vector
$$
\underline{v} = \left(\matrix{2\\-1\\5\\-3\\6}\right)
$$as a linear combination of $\mathbf{e_1}, \mathbf{e_2}, \mathbf{e_3}, \mathbf{e_4}$, and $\mathbf{e_5}$ in $\mathbb{R}^5$.
So would I just write it out as
$$
2\mathbf{e_1}-\mathbf{e_2}+5\mathbf{e_3}-3\mathbf{e_4}+6\mathbf{e_5}
$$
or is there more involved here?
|
An easy way to see this is to put the standard basis vectors into a matrix. This may be overkill, but this question is currently unanswered.
Note that taking each standard basis vector in $\mathbb{R}^5$ as columns into a matrix simply yields the identity matrix $I_5$ or some permutation of it.
$
\begin{align*}
I_5 = [e_1, e_2, e_3, e_4, e_5] =
\begin{bmatrix}
1 & 0 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 & 0 \\
0 & 0 & 1 & 0 & 0 \\
0 & 0 & 0 & 1 & 0 \\
0 & 0 & 0 & 0 & 1 \\
\end{bmatrix}
\end{align*}
$
Now a linear combination can be found by simply multiplying:
$
\begin{align*}
\small
I_5\underline{v} =
\begin{bmatrix}
1 & 0 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 & 0 \\
0 & 0 & 1 & 0 & 0 \\
0 & 0 & 0 & 1 & 0 \\
0 & 0 & 0 & 0 & 1 \\
\end{bmatrix}
\begin{bmatrix}
2 \\
-1 \\
5 \\
-3 \\
6 \\
\end{bmatrix}
&=
2\begin{bmatrix}
1 \\
0 \\
0 \\
0 \\
0 \\
\end{bmatrix}
+
(-1)\begin{bmatrix}
0 \\
1 \\
0 \\
0 \\
0 \\
\end{bmatrix}
+
5\begin{bmatrix}
0 \\
0 \\
1 \\
0 \\
0 \\
\end{bmatrix}
+
(-3)\begin{bmatrix}
0 \\
0 \\
0 \\
1 \\
0 \\
\end{bmatrix}
+
6\begin{bmatrix}
0 \\
0 \\
0 \\
0 \\
1 \\
\end{bmatrix} \\
\\
&=
2e_1 -e_2 +5e_3 - 3e_4 + 6e_5
\end{align*}
$
|
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|
Find the minimum of $\overline{AP}+\overline{AQ}$. My niece asked me a math question: There is a point $A(-3,2)$ on the $x$-$y$ plane. A line $L$ is perpendicular to the line $x=y$. Line $L$ and $y=f(x)=2^x$ has an intersection point $P$. Line $L$ and $y=g(x)=\log_2 x$ has an intersection point $Q$. Find the minimum of $\overline{AP}+\overline{AQ}$.
Note: My niece is a freshman in a senior high school. She have not learned calculus. Therefore, I can't use calculus to solve this question.
The following is my try. First, let the coordinates of $P$ be $(a,2^a)$. Because $f(x)$ and $g(x)$ are inverse functions of each other, the coordinates of $Q$ is $(2^a,a)$. Then
$$
\overline{AP}+\overline{AQ}
= \sqrt{(-3-a)^2+(2-2^a)^2} + \sqrt{(-3-2^a)^2+(2-a)^2}. \tag{1}
$$
Because the two terms in $(1)$ are both positive numbers, we can use the inequality of arithmetic and geometric means (AM-GM inequality) to obtain
$$
\begin{align}
\overline{AP}+\overline{AQ} & \geq 2\sqrt{ \sqrt{(-3-a)^2+(2-2^a)^2} \cdot \sqrt{(-3-2^a)^2+(2-a)^2}} \\
& = 2 \{ [ (-3-a)^2+(2-2^a)^2][(-3-2^a)^2+(2-a)^2]\}^{1/4} \\
& = 2 [(a^2+6a+9+4-4\cdot 2^a+2^{2a})(2^{2a}+6\cdot 2^a+9+a^2-4a+4)]^{1/4} \\
& = 2 [(2^{2a}-2^{a+2}+a^2+6a+13)(a^{2a}+3\cdot 2^{a+1}+a^2-4a+13)]^{1/4}. \tag{2}
\end{align}
$$
I don't know how to continue.
|
Sorry, but AM-GM is unable to be applied in this case as $\overline{AP} \neq \overline {AQ}$ for all $P$, $Q$.
However, the triangle inequality can be applied nicely in this case. First, we must reformulate the points.
Let us name the points $A(-3,2)$, $B(2,-3)$, $P(a,2^a)$, $Q(2^a,a)$,
Note that $$ \overline{AP}+\overline{AQ}=\sqrt{(2^a-2)^2+(a+3)^2} + \sqrt{(2^a+3)^2+(a-2)^2}=\overline{BP}+\overline{AP} $$
As seen in equation $(1)$ in your question. So the question becomes equivalent to minimizing the value of $\overline{BP}+\overline{AP}$. However, we have that $$\overline{BP}+\overline{AP} \ge \overline{AB}=5 \sqrt{2}$$
From the triangle inequality. Note the equality is true when $P$ is the intersection of $x+y=-1$ and $y=2^{x}$. The existance of this intersection can be confirmed through IVT.
As @amd has guessed, this is when $A, P, Q$ are colinear.
|
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|
Find the Derivative of $f(x)=\frac{7}{\sqrt {x}}$ using the definition. I get that $$\frac{d}{dx}\left(7\times\dfrac{1}{\sqrt{x}}\right)=\frac{d}{dx}(7x^{.5})=\dfrac{7}{2}x^{-.5}$$ is the derivative, but I can't ever use $\lim_{h \to 0} \dfrac{f(x+h)-f(x)}{h}$.
If someone or anyone could go step by step and do the problem, I would be eternally grateful.
|
$$f(x)=\dfrac{7}{\sqrt{x}}\\f'(x)=\lim\limits_{h \to0}\dfrac{f(x+h)-f(x)}{h}\\f'(x)=\lim\limits_{h \to0}\dfrac{\frac{7}{\sqrt{x+h}}-\frac{7}{\sqrt{x}}}{h}\\f'(x)=\lim\limits_{h \to0}\dfrac{\bigg(\frac{7}{\sqrt{x+h}}-\frac{7}{\sqrt{x}}\bigg)\bigg(\frac{7}{\sqrt{x+h}}+\frac{7}{\sqrt{x}}\bigg)}{h\bigg(\frac{7}{\sqrt{x+h}}+\frac{7}{\sqrt{x}}\bigg)}\\f'(x)=\lim\limits_{h \to0}\dfrac{\frac{49}{x+h}-\frac{7}{\sqrt{x}}}{h\bigg(\frac{7}{\sqrt{x+h}}+\frac{7}{\sqrt{x}}\bigg)}\\f'(x)=\lim\limits_{h \to0}\dfrac{49x-49x-49h}{h\bigg(\frac{7}{\sqrt{x+h}}+\frac{7}{\sqrt{x}}\bigg)(x(x+h))}\\f'(x)=\lim\limits_{h \to0}\dfrac{-49}{\bigg(\frac{7}{\sqrt{x+h}}+\frac{7}{\sqrt{x}}\bigg)(x(x+h))}\\f'(x)=\dfrac{-49}{(\frac{14}{\sqrt{x}})x^2}\\f'(x)=\dfrac{-7}{2x^{\frac{3}{2}}}\\f'(x)=-\dfrac{7}{2}x^{-\frac{3}{2}}$$
|
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|
Integrate $\int \frac{dx}{\sqrt {7-6x}-x^2}$ $$\int \frac{dx}{\sqrt {7-6x}-x^2}$$
Have no idea how to start with this. How do i make a perfect square of the variable?
|
I assume you are trying to solve the integral
$$I:=\int \frac{\mathrm d x}{\sqrt{7-6x-x^2}}$$
because you mentioned to make a perfect square.
Since $$7-6x-x^2= 16-(x+3)^2= \frac{1}{16}\left(1- \left( \frac{x+3}{4}\right)^2\right)$$
the integral becomes
$$I= \int \frac{\mathrm d x}{\frac{1}{4}\sqrt{1- \left( \frac{x+3}{4}\right)^2 }} = \arcsin \left( \frac{x+3}{4}\right).$$
In the last step you need to substitute $t= \frac{x+3}{2}$.
|
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|
solving a Diophantine system of two equations Find all triples $(a,b,c) \in \mathbb{N}$ satisfing the following equations:-
\begin{align}
a^2 + b^2 & = c^3 \\
(a + b)^2 & = c^4
\end{align}
Thank you for your help.
|
Hint: Since $(a^2+b^2)c=c^4=(a+b)^2$ we have that $a^2+b^2$ divides $(a+b)^2$. This implies $a=b$ for positive integers. Then we have $2a^2=c^3$, so that $(a,b,c)=(2,2,2)$.
|
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Alternate methods to solve this equation The question is to find the roots of the equation:
$2(x^2+ \frac{1}{x^2}) - 3(x+ \frac{1}{x}) - 1 =0$
My attempt:
$2(x^2+ \frac{1}{x^2} + 2) - 3(x+ \frac{1}{x}) - 1 - 4 = 0$
⇒$2(x+ \frac{1}{x})^2 - 3(x+ \frac{1}{x}) - 1 - 4 = 0$
⇒$2(x+ \frac{1}{x})^2 - 3(x+ \frac{1}{x}) - 1 - 4 = 0$
Let $(x+ \frac{1}{x})$ be $'a'$
Therefore the given equation becomes,
$2a^2 -3a - 5 = 0$
⇒$2a^2 -a(5-2) - 5 = 0$
⇒$2a^2 -5a + 2a - 5 = 0$
⇒$a(2a - 5) + 1(2a - 5) = 0$
⇒$(2a - 5)(a + 1) = 0$
⇒$(2a - 5) = 0$ or $(a + 1) = 0$
⇒$a = 5/2$ or $a = -1$
When $a = 5/2$
⇒$(x+ \frac{1}{x}) = 5/2$
⇒$2(x^2 + 1) = 5x$
⇒$2x^2 - 5x + 2 = 0$
⇒$2x^2 -x(4 + 1) + 2 = 0$
⇒$2x^2 - 4x - x + 2 = 0$
⇒$2x(x - 2) - 1(x - 2) = 0$
⇒$(x - 2)(2x - 1) = 0$
⇒$(x - 2) = 0$ or $(2x - 1) = 0$
⇒$x = 2$ or $x = 1/2$
When $a = -1$
$(x+ \frac{1}{x}) = -1$
⇒$x^2 + 1 = -x$
⇒ $x^2 + x + 1 = 0$
⇒$x = \frac {-1 \pm \sqrt{1^2 - 4.1.1}}{2.1}$
⇒$x = \frac {-1 + \sqrt{1^2 - 4.1.1}}{2.1}$ or $\frac {-1 - \sqrt{1^2 - 4.1.1}}{2.1}$
⇒$x = \frac {-1 + \sqrt{3}i}{2}$ or $x = \frac {-1 - \sqrt{3}i}{2}$
My question is: Is there any elegant technique to solve this solve because i find my method lengthy and cumbersome.
|
Hint -
$(x+ \frac{1}{x}) = t$
Also,
$(x^2+ \frac{1}{x^2}) = t^2 - 2$
Hope it helps.
|
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|
Evaluate the given limit Given a function $f : R → R$ for which $|f(x) − 3| ≤ x^2$. Find
$$\lim_{ x\to0}\frac{f(x) - \sqrt{x^2 + 9}}{x}$$
Can the function $f(x)$ be considered as $x^2 + 3$ and go about evaluating the limit using the Limit laws?
|
Apply law |a+b| <= |a| + |b|
Let $L = \frac{f(x)-\sqrt{x^2+9}}{x}$
$$
|L| = \lvert\frac{f(x) - 3 + 3 - \sqrt{x^2+9}}{x}\rvert \le |\frac{f(x)-3}{x}| + |\frac{3 - \sqrt{x^2+9}}{x}| \\
\le |\frac{x^2}{x}| + |\frac{(3-\sqrt{x^2+9})(3+\sqrt{x^2+9})}{x(3+\sqrt{x^2+9})}| = |x| + |\frac{x}{3+\sqrt{x^2+9}}|
$$
Hence,
$$
\lim_{x \rightarrow 0}{|L|} \le \lim_{x \rightarrow 0}{(|x| + |\frac{x}{3+\sqrt{x^2+9}}|)} = 0 \quad (1)
$$
Since $|L| \ge 0 \quad \forall x$, we also have $\lim_{x \rightarrow 0}{|L|} \ge 0 \quad (2)$
From (1) and (2) we have $\lim_{x \rightarrow 0}{|L|} = 0$, or $\quad \lim_{x \rightarrow 0}L = 0$
|
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|
Solve $\lim_{x \rightarrow 1}\frac{3\ln(x) - x^2+1}{x-1}$ without using L'Hôpital's I tried:
$$\lim_{x \rightarrow 1}\frac{3\ln(x) - x^2+1}{x-1} = \\
\frac{3\ln(x) + (1-x^2)}{-1(1-x)} = \\
\frac{3\ln(x)+ (1-x)(1+x)}{-(1-x)} = \\
\frac{3\ln(x)}{x-1} + \frac{1+x}{-1} = \\
\frac{\ln{x^3}}{x-1} - 1-x = \\
???$$
What do I do next? Remember, I can't use L'Hôpital.
|
$$\lim_{x\to1}\frac{\ln x}{x-1}=1$$ is a basic limit from high school, and expresses the logarithm is differentiable at $x=1$ and has derivative equal to $1$.
|
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|
Equality using floor function Let $n\in \mathbb{N}$. How we can show this :
$$\lfloor \sqrt{n}+\sqrt{n+1}+\sqrt{n+2}+\sqrt{n+3}\rfloor =\lfloor \sqrt{16n+20}\rfloor$$
by using the concavity of $x\longmapsto \sqrt{x}$.
I read an article a few years ago about this but I can not find it on the internet.
|
Let's first show the $\leq$ direction: By concavity,
$$\frac{\sqrt{n} + \sqrt{n+1}}{2} < \sqrt{n + \frac{1}{2}}$$
so
$$\sqrt{n} + \sqrt{n+1} < \sqrt{4n + 2}\,\,\,\,\, (1)$$
Similarly
$$\sqrt{n+2} + \sqrt{n+3} < \sqrt{4n + 10} \,\,\,\,\, (2)$$
And using the same principle again:
$$\sqrt{4n + 2} + \sqrt{4n + 10} < \sqrt{16n + 24}$$ So, adding $(1)$ and $(2)$:
$$ \sqrt{n} + \sqrt{n+1} + \sqrt{n+2} + \sqrt{n+3} < \sqrt{16n + 24} \,\,\,\,\, (\ast)$$
To prove the $\leq$ direction, it's enough to show that there is no integer between $\sqrt{16n+20}$ (excluded) and $\sqrt{16n + 24}$ (included). The four critical numbers,
i) $8(2n+3)$,
ii) $16n + 23$,
iii) $ 2(8n + 11)$ and
iv) $16n + 21$
are never squares:
i) and iii) because they have an odd multiplicity of prime factor 2, ii) because it's $\equiv 3 \text{ mod } 4$ and iv) because it's $\equiv 5 \text{ mod } 16$.
Therefore $\lfloor \sqrt{16n + 20} \rfloor = \lfloor \sqrt{16n + 24} \rfloor $ and by $(\ast)$, the inequality with $\leq$ holds.
Now let's show the $\geq$: By the arithmetic-geometric mean inequality:
$$ \frac{\sqrt{n} + \sqrt{n + 1} +\sqrt{n+2} + \sqrt{n+3}}{4} > (n(n+1)(n+2)(n+3))^{1/8} \,\,\, (\ast \ast) $$
Note that $(n(n+1)(n+2)(n+3))=((n+\frac{3}{2})^2 - \frac{1}{4})((n+\frac{3}{2})^2 - \frac{9}{4}) > (n+\frac{5}{4})^4$ for all $n > 3$ since $n^2 +3n > n^2 + \frac{5}{2}n + \frac{25}{16}$ if $ n > 3$ (so it remains at the end to verify the identiy for 1,2,3 by hand).
Now substituting this back into $(\ast \ast)$ gives
$$ \frac{\sqrt{n} + \sqrt{n + 1} +\sqrt{n+2} + \sqrt{n+3}}{4} > \sqrt{n+\frac{5}{4}}$$
for all $n>3$
and multiplying through by $4$ finally
$$ \sqrt{n} + \sqrt{n + 1} +\sqrt{n+2} + \sqrt{n+3} > \sqrt{16n+20}$$ completing the proof for $n>3$. For $n = 1,2,3$, the original identity is easy to check by hand.
|
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|
How to solve $x - 3\sqrt{\frac{5}{x}} = 8$ for $x - \sqrt{5x}$? I have a problem from my textbook. By using $x - 3\sqrt{\frac{5}{x}} = 8$ how can we find the value of $x - \sqrt{5x}$. I have derived the equation that's given so much, but i couldn't find the answer. Solvings or hints are appreciated.
|
we can write $$x-8=3\sqrt{\frac{5}{x}}$$ after squaring this equation we get
$$x^2-16x+64=9\cdot \frac{5}{x}$$ multiplying by $$x\ne 0$$ we obtain
$$x^3-16x^2+64x-45=0$$ factorizing this equation we get
$$(x-5)(x^2-11x+9)=0$$ can you finish now?
|
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|
Epsilon-Delta: Prove $\frac{1}{x} \rightarrow 7$ as $x \rightarrow \frac{1}{7}$
Prove that $\displaystyle\frac{1}{x} \rightarrow 7$ as $\displaystyle x \rightarrow \frac{1}{7}$.
I need to show this with an $\epsilon-\delta$ argument. Still figuring these types of proofs out though, so I could use some tips/critiques of my proof, if it is correct at all.
It might not be so clear, but I use the fact that $\displaystyle\left|x - \frac{1}{7}\right| < \delta$ several times in the proof.
For $\varepsilon > 0$, let $\displaystyle\delta = \min\left\{\frac{1}{14}, \frac{\varepsilon}{98}\right\}$. Then $\displaystyle \left|x - \frac{1}{7}\right| < \delta$ implies:
$$\left|\frac{1}{7}\right| = \left|\left(-x + \frac{1}{7}\right) + x\right| \leq \left|x - \frac{1}{7}\right| + \left|x\right| < \frac{1}{14} + |x|,$$ and so $\displaystyle |x| > \frac{1}{14}$.
Also, $\displaystyle \left|x - \frac{1}{7}\right| < \delta$ implies:
$$\left|\frac{1}{x} - 7\right| = \left|\frac{1-7x}{x}\right| = 7\frac{\left|x - \frac{1}{7}\right|}{|x|} < 98\left|x - \frac{1}{7}\right| < \frac{98\varepsilon}{98} = \varepsilon.$$
Thus for $\varepsilon > 0$, $\displaystyle\left|\frac{1}{x} - 7\right| < \varepsilon$ if $\displaystyle\left|x - \frac{1}{7}\right| < \delta$, for $\displaystyle \delta = \min\left\{\frac{1}{14}, \frac{\varepsilon}{98}\right\}$.
|
for $\beta \gt 0$ write:
$$
x = \frac{1+\beta}7
$$
then
$$
\begin{align}
7-\frac1x\ & \le 7\bigg(1-\frac1{1+\beta}\bigg) \\[14pt]
& = \frac{7\beta}{\beta+1}\\[14pt]
& \le 7\beta
\end{align}
$$
thus:
$$
x-\frac1{7} \le \frac{\beta}7 \Rightarrow 7 -\frac1x\le 7\beta
$$
with this established, the $\epsilon-\delta$ argument is straightforward. a minor adjustment provides the argument for the case $x=\frac{1-\beta}7$
|
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|
If a matrix commutes with a set of other matrices, what conclusions can be drawn? I have a very specific example from a book on quantum mechanics by Schwabl, in which he states that an object which commutes with all four gamma matrices,
$$
\begin{pmatrix}
1 & 0 & 0 & 0\\
0 & 1 & 0 & 0\\
0 & 0 & -1 & 0\\
0 & 0 & 0 & -1\\
\end{pmatrix}
\begin{pmatrix}
0 & 0 & 0 & 1\\
0 & 0 & 1 & 0\\
0 & -1 & 0 & 0\\
-1 & 0 & 0 & 0\\
\end{pmatrix}
\begin{pmatrix}
0 & 0 & 0 & -i\\
0 & 0 & i & 0\\
0 & i & 0 & 0\\
-i & 0 & 0 & 0\\
\end{pmatrix}
\begin{pmatrix}
0 & 0 & 1 & 0\\
0 & 0 & 0 & -1\\
-1 & 0 & 0 & 0\\
0 & 1 & 0 & 0\\
\end{pmatrix},
$$
must be a multiple times the unit matrix. These matrices don't seem to span all $4 \times 4$ matrices so why would this be the case? I have asked around but no one seems to know the answer.
|
Call your four matrices $A,B,C,D$ respectively. While they indeed don't span $M_4(\mathbb C)$, the point is that the algebra they generate is the whole matrix space. So, any matrix that commutes with $A,B,C,D$ must in turn commute with all members of $M_4(\mathbb C)$. In fact, if we put $X=\frac{B\,(AC-C)\,A}{2i}$ and $Y=\frac{B\,(AC+C)\,A}{2i}$, the canonical basis of $M_4(\mathbb C)$ can be obtained as polynomials in $A,B,C,D$:
\begin{align*}
E_{11}&=\frac12(X^2+X),&E_{14}&=E_{11}B,&E_{13}&=E_{11}D,\\
E_{22}&=\frac12(X^2-X),&E_{23}&=E_{22}B,&E_{24}&=-E_{22}D,\\
E_{33}&=\frac12(Y^2+Y),&E_{32}&=-E_{33}B,&E_{31}&=-E_{33}D,\\
E_{44}&=\frac12(Y^2-Y),&E_{41}&=-E_{44}B,&E_{42}&=E_{44}D,\\
E_{12}&=E_{13}E_{32},\\
E_{21}&=E_{24}E_{41},\\
E_{34}&=E_{31}E_{14},\\
E_{43}&=E_{42}E_{23}.
\end{align*}
|
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|
Prove that $a^2 + b^2 + c^2 \lt 2(ab + bc + ca)$ Given that $a, b$ and $c$ are the sides of a triangle.
How to prove that $a^2 + b^2 + c^2 \lt 2(ab + bc + ca)$?
Maybe any hint? Am I going to wrong direction?
$$2(ab + bc + ca)-a^2 + b^2 + c^2>0$$
$$2ab + 2bc + 2ca-a^2 + b^2 + c^2>0$$
$$2b(a+c) + 2ca-a^2 + b^2 + c^2>0$$
...?
|
Alternatively, by cosine law
\begin{align*}
b^2+c^2-a^2 &= 2bc\cos A \\
c^2+a^2-b^2 &= 2ca\cos B \\
a^2+b^2-c^2 &= 2ab\cos C \\
a^2+b^2+c^2 &= 2(bc\cos A+ca\cos B+ab\cos C) \\
& \le 2(bc+ca+ab)
\end{align*}
|
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|
Prove that $\left(\frac{3+\sqrt{17}}{2}\right)^n + \left(\frac{3-\sqrt{17}}{2}\right)^n$ is always odd for any natural $n$.
Prove that $$\left(\frac{3+\sqrt{17}}{2}\right)^n + \left(\frac{3-\sqrt{17}}{2}\right)^n$$
is always odd for any natural $n$.
I attempted to write the binomial expansion and sum it so the root numbers cancel out, and wanted to factorise it but didn't know how. I also attempted to use induction but was not sure how to proceed.
|
HINT:
Say $\left(\frac{3+\sqrt{17}}{2}\right)=a$ and $\left(\frac{3-\sqrt{17}}{2}\right)=b$.
Now observe that:
$$\left(\frac{3+\sqrt{17}}{2}\right)^n + \left(\frac{3-\sqrt{17}}{2}\right)^n$$
$$=a^n+b^n$$
$$=(a+b)(a^{n-1}+b^{n-1})-ab(a^{n-2}+b^{n-2})$$
$$=\color{red}{3\cdot\left[\left(\frac{3+\sqrt{17}}{2}\right)^{n-1}+ \left(\frac{3-\sqrt{17}}{2}\right)^{n-1}\right]+ 2\cdot \left[\left(\frac{3+\sqrt{17}}{2}\right)^{n-2} + \left(\frac{3-\sqrt{17}}{2}\right)^{n-2}\right]}$$
Now use strong induction and see what you can do.
P.S. $3 \times \mathrm{odd} + 2\times \mathrm{odd} = \mathrm{odd + even} = \mathrm{odd}$
Hope this helps you.
|
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|
Prove that $(n^3 - 1)n^3(n^3 + 1)$ is divisible by $504$ How to prove that $(n^3 - 1)n^3(n^3 + 1)$ is divisible by $504$? Factoring $504$ yields $2^3 \cdot 3^2 \cdot 7$. Divisibility by $8$ can be easily seen, because if $n$ is even then $8 | n^3$, else $(n^3 - 1)$ and $(n^3 + 1)$ are both even and one of these is divisible by $4$ so $8|(n^3 - 1)(n^3 + 1)$. I'm stuck at proving divisibility by $9$ and $7$
|
For 9: if $n$ is divisible by 3, then the conclusion is obvious. If $n \equiv 1 \bmod 3$, then $n^3 \equiv 1 \bmod 9$ (proof: write $n^3 = (3k+1)^3 = 27k^3 + 27k^2 + 9k + 1$; all terms but the constant are killed modulo 9). If $n \equiv -1 \bmod 3$, then $n^3 \equiv -1 \bmod 9$ (proof is similar).
For 7, the conclusion follows from arberavdullahu's comment about cubic residues. If you'd rather it be more explicit: let $m \in \{0, \ldots, 6\}$ be such that $m \equiv n \bmod 7$; then $n^3 = (7k+m)^3 = 343k^3 + 147 k^2 m + 21km^2 + m^3$; all terms but $m^3$ die modulo 7, so you just have to check that all of $0^3, \ldots, 6^3$ are congruent to one of $-1, 0, 1$. In general, though, $n^p \equiv m^p \mod r$ as long as $n \equiv m \mod r$, so this step is easy (the mod-9 step, where we had to move from mod-3 considerations to mod-9, was a bit harder).
|
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|
How can I find all the matrices that commute with this matrix? I would like to find all the matrices that commute with the following matrix
$$A = \begin{pmatrix}2&0&0\\ \:0&2&0\\ \:0&0&3\end{pmatrix}$$
I set $AX = XA$, but still can't find the solutions from the equations.
|
If
$B:= \left( \begin{matrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 1 \end{matrix} \right)$,
then your matrix becomes $A = 2I + B$. Thus a matrix $C$ will commute with $A$ if and only if $C$ commutes with $B$. But note
$BC = \left( \begin{matrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 1 \end{matrix} \right)\left( \begin{matrix} a & b & c \\ d & e & f \\ g & h & i \end{matrix} \right) = \left( \begin{matrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ g & h & i \end{matrix} \right)$
and
$CB = \left( \begin{matrix} a & b & c \\ d & e & f \\ g & h & i \end{matrix} \right) \left( \begin{matrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 1 \end{matrix} \right) = \left( \begin{matrix} 0 & 0 & c \\ 0 & 0 & f \\ 0 & 0 & i \end{matrix} \right)$.
It follows that $BC = CB$ if and only if $c=f = g = h = 0$
|
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|
Showing $3^n=\sum_{k=0}^n{(-1)^k\binom{n}{k}4^{n-k}}$. This is Exercise 1.9.3 of F. M. Goodman's "Algebra: Abstract and Concrete".
Show $$3^n=\sum_{k=0}^n{(-1)^k\binom{n}{k}4^{n-k}}.$$
My Attempt:
I tried induction on $n$ as follows.
Base: Try when $n=1$. Then $LHS=3^1=3$ and
$$\begin{align}
RHS&=\sum_{k=0}^1{(-1)^k\binom{1}{k}4^{1-k}} \\
&=(-1)^0\binom{1}{0}4^1+(-1)^1\binom{1}{1}4^0 \\
&=4-1 \\
&=3
\end{align}$$ so the result holds for $n=1$.
Assume the result for $n=r\in\mathbb{N}$. Then $$3^r=\sum_{k=0}^r{(-1)^k\binom{r}{k}4^{r-k}}.$$
Consider when $n=r+1$: I have $3^{r+1}=3\cdot 3^r=3\sum_{k=0}^r{(-1)^k\binom{r}{k}4^{r-k}}$ and have considered
$$\sum_{k=0}^{r+1}{(-1)^k\binom{r+1}{k}4^{r+1-k}}$$ by writing $\binom{r+1}{k}=\binom{r}{k}+\binom{r}{k-1}$ but to no avail.
|
One option is to notice the right side is a binomial expansion. But your solution should work. For the induction step, we have
$$\begin{align*}
& \sum_{k=0}^{r+1}(-1)^k\binom{r+1}{k}4^{r+1-k} \\
=& 4^{r+1}+\sum_{k=1}^{r+1} (-1)^k\binom{r+1}{k}4^{r+1-k} \\
=& 4^{r+1}+\sum_{k=0}^r (-1)^{k+1}\binom{r+1}{k+1} 4^{r-k} \\
=& 4^{r+1}+\sum_{k=0}^{r-1} (-1)^{k+1}\left( \binom{r}{k+1}+\binom{r}{k}\right)4^{r-k} + (-1)^{r-1}\\
=& 4^{r+1}+\sum_{k=0}^{r-1}(-1)^{k+1}\binom{r}{k+1}4^{r-k}+\sum_{k=0}^{r-1}(-1)^{k+1}\binom{r}{k}4^{r-k}+(-1)^{r-1} \\
=& 4^{r+1}+\sum_{k=1}^{r}(-1)^k\binom{r}{k}4^{r-k+1}-\sum_{k=0}^{r}(-1)^k\binom{r}{k}4^{r-k}+(-1)^{r}+(-1)^{r-1} \\
=& 4^{r+1}+\sum_{k=0}^{r}(-1)^k\binom{r}{k}4^{r-k+1}-4^{r+1}-3^r \\
=& 4\sum_{k=0}^{r}(-1)^k\binom{r}{k}4^{r-k}-3^r \\
=& 4 \cdot 3^r - 3^r \\
=& 3\cdot 3^r \\
=& 3^{r+1}
\end{align*}$$
|
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|
Prove $\frac12+\frac16+...+\frac{1}{n(n+1)}=\frac{n}{n+1}$ for $n \in \mathbb{N}$ I am using Induction:
Base Case $n=1$ holds ; $\frac12$= $\frac{1}{(1)+1}$
Assume $\frac{n}{n+1}$ is true from some $n \in \mathbb{N}$.
Then $\frac12+\frac16+...+\frac{1}{n(n+1)}+ \frac{1}{(n+1)((n+1)+1)}=\frac{n}{n+1}+ \frac{1}{(n+1)((n+1)+1)}$. By the Inductive Hypothesis
from here do I simplfy the RHS showing that it equals $\frac{n+1}{(n+1)+1}$?
|
For completeness sake, we evaluate the sum without induction: Note that $$ \frac{1}{k(k+1)} = \frac{(k+1) - k}{k(k+1)} = \frac{k+1}{k(k+1)} - \frac{k}{k(k+1)} = \frac{1}{k} - \frac{1}{k+1}.$$ Therefore, $$\sum_{k=1}^n \frac{1}{k(k+1)} = \sum_{k=1}^n \frac{1}{k} - \frac{1}{k+1} = \sum_{k=1}^n \frac{1}{k} - \sum_{k=2}^{n+1} \frac{1}{k} = \frac{1}{1} - \frac{1}{n+1} = \frac{n}{n+1}.$$
|
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|
Prove that the sum of pythagorean triples is always even Problem: Given $a^2 + b^2 = c^2$ show $a + b + c$ is always even
My Attempt, Case by case analysis:
Case 1: a is odd, b is odd. From the first equation,
$odd^2 + odd^2 = c^2$
$odd + odd = c^2 \implies c^2 = even$
Squaring a number does not change its congruence mod 2.
Therefore c is even
$ a + b + c = odd + odd + even = even$
Case 2: a is even, b is even. Similar to above
$even^2 + even^2 = c^2 \implies c$ is even
$a + b + c = even + even + even = even$
Case 3:
One of a and b is odd, the other is even
Without loss of generality, we label a as odd, and b as even
$odd^2 + even^2 = c^2 \implies odd + even = c^2 = odd$
Therefore c is odd
$a + b + c = odd + even + odd = even$
We have exhausted every possible case, and each shows $a + b + c$ is even. QED
Follow Up:
Is there a proof that doesn't rely on case by case analysis?
Can the above be written in a simpler way?
|
Hint
Write $a+b+c=k$, so
$$a^2+b^2=(a+b)^2-2ab= (k-c)^2-2ab=c^2 → k^2-2(kc+ab)=0→k^2=2(kc+ab)$$
|
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|
Solving Linear Equation with non-invertible matrix? Find X when AX = B.
A= \begin{bmatrix}1&2&1\\2&5&4\end{bmatrix}
B = \begin{bmatrix}3&1\\5&4\end{bmatrix}
I know how to solve for x when A is invertible. I know it would just be $A^{-1}$B.
I'm confused with how to do this when A is non-invertible, I was trying to use the free variables. I can give the actual matrices if needed.
Thanks
|
reducing, saving a little typing
$$
\left(
\begin{array}{rrr|rr}
1 & 2 & 1 & 3 & 1 \\
2 & 5 & 4 & 5 & 4
\end{array}
\right)
$$
$$
\left(
\begin{array}{rrr|rr}
1 & 2 & 1 & 3 & 1 \\
0 & 1 & 2 & -1 & 2
\end{array}
\right)
$$
$$
\left(
\begin{array}{rrr|rr}
1 & 0 & -3 & 5 & -3 \\
0 & 1 & 2 & -1 & 2
\end{array}
\right)
$$
As Robert said from the beginning, you now have familiar augmented systems for the two columns of $X,$
$$
\left(
\begin{array}{rrr|r}
1 & 0 & -3 & 5 \\
0 & 1 & 2 & -1
\end{array}
\right)
$$
$$
\left(
\begin{array}{rrr|r}
1 & 0 & -3 & -3 \\
0 & 1 & 2 & 2
\end{array}
\right)
$$
|
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Prove that if LCMs are equal, then the numbers are equal too. For $a,b \in \mathbb{N}$, how do I prove:
$$lcm(a,a+5)=lcm(b,b+5) \implies a=b$$
|
As Batominovski pointed out in his comment, $a$ and $b$ have to be positive integers, otherwise there are counterexamples to the statement.
We're going to use the following identities:
i) $ab=[a,b](a,b)$ where $[a,b]$ and $(a,b)$ denotes the $\text{lcm}$ and the $\gcd$ of $a$ and $b$, respectively.
ii) $[ca,cb]=c[a,b]$.
We have $(a,a+5)\mid (a+5)-a=5$, so $(a,a+5)=1$ or $(a,a+5)=5$. Analogously, $(b,b+5)=1$ or $(b,b+5)=5$. So we have four cases to analyze:
I) $(a,a+5)=(b,b+5)=1:$ By the first identity we deduce that $$a(a+5)=[a,a+5]=[b,b+5]=b(b+5).$$ Now, if $a>b$, then $a(a+5)>b(b+5)$ and if $a<b$, then $a(a+5)<b(b+5)$, therefore it must be $a=b$.
II) $(a,a+5)=1$ and $(b,b+5)=5$. Let's write $b=5b'$, then $b+5=5(b'+1)$. On the other hand, by the first identity we have $$a(a+5)=[a,a+5]=[b,b+5]=[5b',5(b'+1)]\overset{\text{by (ii)}}=5[b',b'+1]=5b'(b'+1).$$
Thus, $5\mid a(a+5)$, so $5\mid a$ and then $5\mid a+5$, contradiction; or $5\mid a+5$ and then $5\mid a$, contradiction again.
III) $(a,a+5)=5$ and $(b,b+5)=1$. By the symmetry of $a$ and $b$ this case is analogous to the previous one.
IV) $(a,a+5)=(b,b+5)=5$. Using the first identity we deduce that $$a(a+5)=5[a,a+5]=5[b,b+5]=b(b+5).$$
So, if $a>b$ or $a<b$ we will have $a(a+5)>b(b+5)$ or $a(a+5)<b(b+5)$, respectively. Hence, it must be $a=b$.
|
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|
Find the difference of two pronic numbers whose sum is 240. How do I do this without Excel/brute force, and how do I explain this to an 11 year old please? There's apparently supposed to be a trick to this because it's in a standardised exam.
Well I noticed that the difference between two consecutive numbers in the sequence of pronic numbers is an element of $\{4,6,8,10,12,...\}$.
Answer: I guess the pronic numbers are 30 and 210 and hence the difference is 180.
See here for a long solution. I lost my notes, but I think I can use quadratic formula here (or I'm confusing this with another question), but the 11 year old doesn't know quadratic. I think curriculum varies a lot from where she is as opposed to from where I am.
|
if $a(a+1)+b(b+1)=240$, then must one of them less than $120$ or equal $120$. we can let $a < b$. as $11 \cdot 12 > 120 > 10 \cdot 11$, let us try $A$ less than $10 \cdot 11$, B great than $11 \cdot 12$, here $A=a(a+1), B=b(b+1)$.
the minimum $A$ is $A = 1 \cdot 2$, if $A=1 \cdot 2$, $B$ is the maximum, but if $B = 15 \cdot 16$, we have $A+B=242>240$, so $B < 15 \cdot 16$, the maximum $B$ is $14 \cdot 15$.
so we have $11 \cdot 12 \le B \le 14 \cdot 15$
when $A = 10 \cdot 11$, then if $B = 11 \cdot 12, 12 \cdot 13, 13 \cdot 14, 14 \cdot 15$, $A+B \neq 240$. note, when check $A+B \neq 240$, we can try the last digit of $A+B$ $=$ or $\neq$ zero, the same check below.
when $A=9 \cdot 10$, then check $B = 11 \cdot 12, 12 \cdot 13, 13 \cdot 14, 14 \cdot 15$ by last digit, only need check $B=14 \cdot 15$, but $A+B = 9 \cdot 10+14 \cdot 15 \neq 240$, no solution.
when $A=8 \cdot 9$, then check $B = 11 \cdot 12, 12 \cdot 13, 13 \cdot 14, 14 \cdot 15$ by last digit, no solution.
when $A=7 \cdot 8$, then check $B = 11 \cdot 12, 12 \cdot 13, 13 \cdot 14, 14 \cdot 15$ by last digit, no solution.
when $A = 6 \cdot 7$, then check $B = 11 \cdot 12, 12 \cdot 13, 13 \cdot 14, 14 \cdot 15$ by last digit, no solution.
when $A = 5 \cdot 6$, then we have your solution.
|
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|
Proving: $\left(a+\frac1a\right)^2+\left(b+\frac1b\right)^2\ge\frac{25}2$
For $a>0,b>0$, $a+b=1$ prove:$$\bigg(a+\dfrac{1}{a}\bigg)^2+\bigg(b+\dfrac{1}{b}\bigg)^2\ge\dfrac{25}{2}$$
My try don't do much, tough
$a+b=1\implies\dfrac{1}{a}+\dfrac{1}{b}=\dfrac{1}{ab}$
Expanding: $\bigg(a+\dfrac{1}{a}\bigg)^2+\bigg(b+\dfrac{1}{b}\bigg)^2\\=a^2+b^2+\dfrac{1}{a^2}+\dfrac{1}{b^2}+4\\=(a+b)^2-2ab+\bigg(\dfrac{1}{a}+\dfrac{1}{b}\bigg)^2-\dfrac{2}{ab}+4\\=4-2ab-\dfrac{2}{ab}+1+\dfrac{1}{a^2b^2}\\=4-2\bigg(\dfrac{a^2b^2+1}{ab}\bigg)+\dfrac{a^2b^2+1}{a^2b^2}\\=4-\bigg(\dfrac{a^2b^2+1}{ab}\bigg)\bigg(2-\dfrac{1}{ab}\bigg)\\=4-\bigg(ab+\dfrac{1}{ab}\bigg)\bigg(2-\dfrac{1}{ab}\bigg)$
Seems to using Cauchy-Schwartz. Please help.
|
A shortcut that I found (Correct me if I am wrong in spotting):
Using AM-GM Inequality, we get
$$\frac{a+b}{2}\ge \frac{2}{\frac{1}{a}+\frac{1}{b}}$$
$$\frac{1}{a}+\frac{1}{b}\ge \frac{4}{a+b}$$
$$\frac{1}{a}+\frac{1}{b}\ge \frac{4}{1}$$
Using Cauchy-Schwarz Inequality,
$$\bigg(a+\dfrac{1}{a}\bigg)^2+\bigg(b+\dfrac{1}{b}\bigg)^2$$
$$\ge \bigg(a+\dfrac{1}{a}+b+\dfrac{1}{b}\bigg)^2$$
$$\ge (1+4)^2$$
$$\ge 25$$
$$\ge \frac{25}{2}$$
Hope this helps.
|
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|
Prove that if $x_{n+2}=\frac{2+x_{n+1}}{2+x_n},$ then $x_n$ converges
Let $x_0 > 0$, $x_1 > 0$ and $$x_{n+2} = \dfrac{2+x_{n+1}}{2+x_n}$$ for $n \in \{0,1,\dots\}$. Prove that $x_n$ converges.
My attempt:
If $x_n$ converges and $\lim\limits_{n\rightarrow\infty}x_n=a$ so $a=\frac{2+a}{2+a}$, which gives $a=1$. Now,
$$x_{n+2}-1=\frac{x_{n+1}-x_n}{2+x_n}$$
and what is the rest? Thank you!
|
Clearly every $x_n$ is positive. With this we know that $$x_{n+2} = \frac{2 + x_{n+1}}{2 + x_n} = \frac{2}{2+x_n} + \frac{x_{n+1}}{2+x_n} < 1 + \frac{x_{n+1}}{2}.$$
Applying this recursively we get $$x_n < 1 + \frac{x_{n-1}}{2} < 1 + \frac{1 + \frac{x_{n-2}}{2}}{2} = 1 + \frac{1}{2} + \frac{x_{n-2}}{4} < $$
$$< 1 + \frac{1}{2} + \frac{1 + \frac{x_{n-3}}{2}}{4} = 1 + \frac{1}{2} + \frac{1}{4} + \frac{x_{n-3}}{8} < \frac{x_{n-k}}{2^k} + \sum_{i=0}^{k-1} \frac{1}{2^i}.$$
This bound is valid for any $k = 1 \ldots n$. For $k = n$ we get $$x_n < \frac{x_0}{2^n} + \sum_{i=0}^{n-1} \frac{1}{2^i} < \frac{x_0}{2^n} + 2. $$
Edit: This proves the sequence is bounded. But to get convergence we still need to prove something else, like it is monotone.
|
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|
Abstract Algebra (Multiplicative Inverse using the Euclidean Algorithm) The question I am trying to solve is to find the multiplicative inverse of
$2-\sqrt[3]{2}+\sqrt[3]{4}$.
So far I am come up with the following.
Let $\alpha = 2-\sqrt[3]{2}+\sqrt[3]{4}$.
Then $s(x)=2-x+x^2$ and $x=\sqrt[3]{2}.$
Then $t(x)=x^3-2$.
The $gcd(s(x),t(x))=1$.
This implies there exists $\alpha(x),\beta(x)\in Q[x]$ such that $\alpha(x)s(x)+\beta(x)t(x)=1$.
Then $\frac{t(x)}{s(x)}=(x+1)+(\frac{-x-4}{2-x+x^2})$.
Then $(x^3-2)=(x+1)(2-x+x^2)+(-x-4)$ and $r_1=-x-4$.
Then $\frac{s(x)}{r_1(x)}=\frac{2-x+x^2}{-x-4}=(-x+5)+(\frac{22}{-x-4})$.
Then $2-x+x^2=(-x+5)(-x-4)+22$.
That is far as I have gone and I'm not sure if I am correct or where to go next.
|
The following Maple command
gcdex(x^3-2,2-x+x^2,x,'s','t'); s; t;
will produce the gcd (1) and the factors $s(x),t(x)$ such that
$$
s(x)(x^3-2) +t(x) (2-x+x^2)=1.
$$
You'll see that
$$
s(x)=- 5/22 + 1/22 x
$$
and
$$
t(x)= 3/11 + 2/11 x - 1/22 x^2.
$$
Accordingly,
$$
t(\sqrt[3]2) \cdot (2-\sqrt[3]2+\sqrt[3]4)=1,
$$
or
$$
(3/11 + 2/11 \sqrt[3]2 - 1/22 \sqrt[3]4) \cdot (2-\sqrt[3]2+\sqrt[3]4)=1.
$$
|
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|
Prove that the following determinant is equal to $\sin(2(x+y))$
Prove that
$$\left|\begin {array} c
\cos(x+y) & \sin(x+y) & -\cos(x+y)\\
\sin(x-y) & \cos(x-y) & \sin(x-y) \\
\sin{2x} & 0 & \sin (2y)
\end {array}\right|
=\sin(2(x+y))$$
It is my question. I have tried several types of operations such as $C_1=C_1 + C_3$ and many others but I failed. Somebody please help me.
|
Posing $X=x+y$ and $Y=x-y$, the determinant becomes:
$$\Delta=\left|\begin {array} c
\cos(X) & \sin(X) & -\cos(X)\\
\sin(Y) & \cos(Y) & \sin(Y) \\
\sin(X+Y) & 0 & \sin (X-Y)
\end {array}\right|$$
Because of the zero at $(3,2)$, Sarrus's rule yields an expression of the determinant as a sum of four terms (instead of six):
$$\Delta=\cos(X)\cos(Y)\sin(X-Y) + \sin(X)\sin(Y)\sin(X+Y) -\sin(X)\sin(Y)\sin(X-Y)\\+\cos(X)\cos(Y)\sin(X+Y)$$
which, after simplifications, turns out to be $\sin(2X)=\sin(2x+2y)$ because:
\begin{align}\Delta &= \sin(X+Y)\big[\cos(X)\cos(Y)+\sin(X)\sin(Y)\big] \\&\qquad\qquad + \sin(X-Y)\big[\cos(X)\cos(Y) -\sin(X)\sin(Y)\big]\\ &= \sin(X+Y)\cos(X-Y) +\sin(X-Y)\cos(X+Y)\\ &= \sin(2X)
\end{align}
|
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|
Can the cosine function be written in "simpler" expressions? I recently found that the constants $\pi$ and $\phi=\frac{1+\sqrt{5}}{2} $ can be related by the identity
$$
2\cos \frac{\pi}{5} = \phi.
$$
Is there some way to write the cosine function by "simpler" mathematical expressions?
|
Gauss introduced a method for cyclotomic polynomials that quickly gives such expressions. Let $n$ be odd and $\omega$ a primitive $n$-th root of unity. Then one of th values indicated by the expression $\omega + (1/\omega)$ is, indeed, $ x =2 \cos (2 \pi / n).$ These satisfy monic polynomials with integer coefficients. If you prefer $ x =2 \cos ( \pi / n)$ you need half angle formulas.
Oh, given the way these are constructed, all the roots are real and given by $2 \cos 2k\pi/n$
Fairly consistent for prime $n.$
$n=5$
$$ x^2 + x - 1 $$
$n=7$
$$ x^3 + x^2 -2x-1 $$
$n=11$
$$ x^5 + x^4 - 4 x^3 - 3 x^2 + 3x +1 $$
$n=13$
$$ x^6 + x^5 - 5 x^4 - 4 x^3 + 6 x^2 + 3 x - 1 $$
$n=17$
$$ x^8 + x^7 - 7 x^6 - 6 x^5 + 15 x^4 + 10 x^3 - 10 x^2 - 4 x + 1 $$
$n=19$
$$ x^9 + x^8 -8 x^7 - 7 x^6 +21 x^5 + 15 x^4 -20 x^3 - 10 x^2 +5 x + 1 $$
These are all in pages 3-23 of Reuschle (1875). I learned the method in Galois Theory by David A. Cox, and wrote working programs for fixed polynomial degrees 3, 5, 7.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Is $\frac1p(\sin^px+\cos^px)-\frac1q(\sin^qx+\cos^qx)$ constant for some reals $p$ and $q$. We know for
$$f(x)=\dfrac14(\sin^4x+\cos^4x)~~~;~~~g(x)=\dfrac16(\sin^6x+\cos^6x)$$
have $f(x)-g(x)=\dfrac{1}{12}$. My question is
Are there other real $p$ and $q$ such that
$$f(x)=\dfrac1p(\sin^px+\cos^px)~~~;~~~g(x)=\dfrac1q(\sin^qx+\cos^qx)$$
give us $f(x)-g(x)=C$ for a real constant $C$?
I had some idea but they were not useful. Thanks.
|
Let $p,q$ be integers. If $p=q$, one has nothing to do. Suppose $p<q$. Let
$$F(x)=\dfrac1p(\sin^px+\cos^px)-\dfrac1q(\sin^qx+\cos^qx)$$
and then
\begin{eqnarray}
F'(x)&=&\sin^{p-1}x\cos x-\cos^{p-1}x\sin x-\sin^{q-1}x\cos x+\cos^{q-1}x\sin x\\
&=&\sin x\cos x(\sin^{p-2}x-\cos^{p-1}x-\sin^{q-2}x-\cos^{q-1}x).
\end{eqnarray}
If $f(x)-g(x)$ is constant, then $F'(x)\equiv0$ and hence
$$ \sin^{p-2}x-\cos^{p-1}x-\sin^{q-2}x-\cos^{q-1}x\equiv0.$$
Let
$$ h(x)=\sin^{p-2}x-\cos^{p-1}x-\sin^{q-2}x-\cos^{q-1}x $$
and then
\begin{eqnarray}
h'(x)&=&(p-2)\sin^{p-3}x\cos x+(p-2)\cos^{p-3}x\sin x-(q-2)\sin^{q-3}x\cos x-(q-2)\cos^{q-3}x\sin x\\
&=&\sin x\cos x\big\{(p-2)\big[\sin^{p-4}x+\cos^{p-4}x\big]-(q-2)\big[\sin^{q-4}x+\cos^{q-4}x\big]\big].
\end{eqnarray}
Since $h(x)\equiv0$, $h'(x)\equiv0$. Clearly if $p=2$, then $q=2$. Suppose $p>2$. Let $x=\frac{\pi}{4}$ and then one has
$$ (\frac{\sqrt{2}}{2})^{p}=\frac{q-2}{p-2}(\frac{\sqrt{2}}{2})^{q}$$
or
$$ 2^{\frac12(q-p)}=\frac{q-2}{p-2}. $$
Let $p-2=2m,q-2=2n$ ($m<n$) and then one has
$$ 2^{n-m}=\frac{n}{m} $$
from which one has $n=m2^r$ ($r>0$). So
$$ 2^{m(2^r-1)}=2^r$$
or $$ m(2^r-1)=r. $$
Noting if $r>1$, $2^r-1>r$ and hence $m(2^r-1)>r$, one must have $r=1$ and hence $m=1,n=2$. Thus $p=4,q=6$.
|
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|
Prove the inequality, power. $\{ x,y \in\Bbb R\ \}$ If $x+y = 2$ then prove the inequality:
$x^4 + y^4 \ge 2$
How I started
*
*$(x+y)^2 = 4$
*$x^2 + y^2 = 4 - 2xy$
*$(x^2+y^2)^2 - 2(xy)^2 \ge 2$
*$(4-2xy)^2 - 2(xy)^2 \ge 2$
*$16-16xy + 4(xy)^2 -2(xy)^2 - 2 \ge 0$
*$2(xy)^2 - 16xy + 14 \ge 0$
*for $t=xy$
*$2t^2 - 16t + 14 \ge 0$
It isn't always true, I think that I should have a assumption for $t$, but I don't know how should I do this.
|
$x+y = 2\\
(x+y)^2 = 4\\
(x-y)^2 \ge 0$
add them together
$2x^2 + 2y^2 \ge 4\\
x^2 + y^2 \ge 2$
repeat:
$x^4 + y^4 \ge 2$
|
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|
Rank of the $n \times n$ matrix with ones on the main diagonal and $a$ off the main diagonal I want to find the rank of this $n\times n$ matrix
\begin{pmatrix}
1 & a & a & \cdots & \cdots & a \\
a & 1 & a & \cdots & \cdots & a \\
a & a & 1 & a & \cdots & a \\
\vdots & \vdots & a& \ddots & & \vdots\\
\vdots & \vdots & \vdots & & \ddots & \vdots \\
a & a & a & \cdots &\cdots & 1
\end{pmatrix}
that is, the matrix whose diagonals are $1's$ and $a$ otherwise, where $a$ is any real number.
My first observation is when $a=0$ the rank is $n$ and when $a=1$ the rank is $1.$ Then I can assume $a\neq 0, 1$ and proceed row reduction to find its pivot rows. I obtain
\begin{pmatrix}
1 & a & a & \cdots & a \\
0 & 1+a & a & \cdots & a \\
0 & a & 1+a & \cdots & a \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
0 & a & a & \cdots & 1+a
\end{pmatrix}
by subtracting the first row multiplied $a$ for each row below the first, and then divides the factor $(1-a)$,
and stuck there. Any hints/helps?
|
If you haven't figured it out yet here's the solution: if $a=1$ $\mathrm{Rank}(A)=1$ otherwise $\mathrm{Rank}(A)=n$ where
$$A:= \begin{pmatrix}
1&a&a&\cdots&a\\
a&1&a&\cdots&a\\
a&a&1&\cdots&a\\
\vdots&\vdots&\vdots&\ddots&\vdots\\
a&a&1&\cdots&1
\end{pmatrix}$$
You've already shown that you know if $a\in\{0,1\}$. So as for the rest, your close to a solution. The next set of Row equations are as follows $R_i-R_1 \to R_i$ such that $1<i\leq n$. This gives the matrix,
$$A_2= \begin{pmatrix}
1&a&a&\cdots&a\\
-1&1&0&\cdots&0\\
-1&0&1&\cdots&0\\
\vdots&\vdots&\vdots&\ddots&\vdots\\
-1&0&0&\cdots&1
\end{pmatrix}$$
The next Row equation is $\frac{R_1-\sum_{i=2}^n aR_i}{1+a(n-1)}\to R_1$. This gives the matrix
$$A_3= \begin{pmatrix}
1&0&0&\cdots&0\\
-1&1&0&\cdots&0\\
-1&0&1&\cdots&0\\
\vdots&\vdots&\vdots&\ddots&\vdots\\
-1&0&0&\cdots&1
\end{pmatrix}$$
The next Row equations are $R_i+R_1\to R_i$ where $1<i\leq n$. This gives
$$A_4= \begin{pmatrix}
1&0&0&\cdots&0\\
0&1&0&\cdots&0\\
0&0&1&\cdots&0\\
\vdots&\vdots&\vdots&\ddots&\vdots\\
0&0&0&\cdots&1
\end{pmatrix}=I$$
Note that $\mathrm{Rank}(A)=\mathrm{Rank}(I)=n$ as desired.
Karma made note that if $a=\frac{1}{1-n}$ then
$$A_3= \begin{pmatrix}
0&0&0&\cdots&0\\
-1&1&0&\cdots&0\\
-1&0&1&\cdots&0\\
\vdots&\vdots&\vdots&\ddots&\vdots\\
-1&0&0&\cdots&1
\end{pmatrix}$$
This implies that for $a=\frac{1}{1-n}$ the $\mathrm{Rank}(A)=n-1$.
|
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|
For $2x+2y-3z\geq0$ and similar prove that $\sum\limits_{cyc}(7z-3x-3y)(x-y)^2\geq0$
Let $x$, $y$ and $z$ be non-negative numbers such that
$2x+2y-3z\geq0$, $2x+2z-3y\geq0$ and $2y+2z-3x\geq0$. Prove that:
$$(7z-3x-3y)(x-y)^2+(7y-3x-3z)(x-z)^2+(7x-3y-3z)(y-z)^2\geq0$$
I have a proof for the following weaker inequality.
Let $x$, $y$ and $z$ be non-negative numbers such that
$2x+2y-3z\geq0$, $2x+2z-3y\geq0$ and $2y+2z-3x\geq0$. Prove that:
$$(11z-4x-4y)(x-y)^2+(11y-4x-4z)(x-z)^2+(11x-4y-4z)(y-z)^2\geq0.$$
For the proof we can use the following lemma.
Let $a$, $b$, $c$, $x$, $y$ and $z$ be real numbers such that
$x+y+z\geq0$ and $xy+xz+yz\geq0$. Prove that:
$$(a-b)^2z+(a-c)^2y+(b-c)^2x\geq0.$$
Proof.
Since $x+y+z\geq0$, we see that $x+y$ or $x+z$ or $y+z$ is non-negative because if
$x+y<0$, $x+z<0$ and $y+z<0$ so $x+y+z<0$, which is contradiction.
Let $x+y\geq0$.
If $x+y=0$ so $xy+xz+yz=-x^2\geq0$, which gives $x=y=0$ and since $x+y+z\geq0$,
we obtain $z\geq0$, which gives $(a-b)^2z+(a-c)^2y+(b-c)^2x\geq0.$
Id est, we can assume $x+y\geq0$.
Now, $$(a-b)^2z+(a-c)^2y+(b-c)^2x=(a-b)^2z+(a-b+b-c)^2y+(b-c)^2x=$$
$$=(x+y)(b-c)^2+2(a-b)(b-c)y+(y+z)(a-b)^2$$
and since $x+y>0$, it's enough to prove that
$$y^2-(x+y)(y+z)\leq0,$$
which is
$xy+xz+yz\geq0,$
which ends a proof of the lemma.
Now we can prove a weaker problem.
From the condition we have:
$$\sum_{cyc}(2x+2y-3z)(2x+2z-3y)=\sum_{cyc}(9xy-8x^2)\geq0$$
and we see that $\sum\limits_{cyc}(11z-3x-3y)=5(x+y+z)\geq0$ and
$\sum\limits_{cyc}(11z-3x-3y)(11y-3x-3z)=9\sum\limits_{cyc}(9xy-8x^2)\geq0$
and by the lemma we are done!
This way does not help for the starting inequality.
Any hint please. Thank you!
|
Proof:
Because both the conditions and the conclusion are homogeneous, we may assume that $x+y+z=1$.
Geometrically,
$$\Delta=\{(x,y,z)\in \mathbb{R}^3 |\ 2x+2y−3z≥0, 2x+2z−3y≥0, \\
2y+2z−3x≥0, x+y+z=1\}$$
is a triangle . By solveing three linear equations
$$2x+2y−3z=0,\ 2x+2z−3y=0,\ x+y+z=1;$$
$$2x+2z−3y=0,\ 2y+2z−3x=0,\ x+y+z=1;$$
$$2x+2y−3z=0,\ 2y+2z−3x=0,\ x+y+z=1;$$
we get three vertices of the triangle $\Delta$. They are $(\frac{1}{5}, \frac{2}{5}, \frac{2}{5})$, $(\frac{2}{5}, \frac{1}{5}, \frac{2}{5})$ and $(\frac{2}{5}, \frac{2}{5}, \frac{1}{5})$.
Then $\Delta=u(\frac{1}{5}, \frac{2}{5}, \frac{2}{5})+v(\frac{2}{5}, \frac{1}{5}, \frac{2}{5})+w(\frac{2}{5}, \frac{2}{5}, \frac{1}{5}),\ u+v+w=1, u\geq 0,v\geq 0,w\geq 0$.
Let $x=\frac{1}{5}u+\frac{2}{5}v+\frac{2}{5}w,\ y=\frac{2}{5}u+\frac{1}{5}v+\frac{2}{5}w,
\ z=\frac{2}{5}u+\frac{2}{5}v+\frac{1}{5}w$. We may write the inequality
$$(7z−3x−3y)(x−y)^2 +(7y−3x−3z)(x−z)^2 +(7x−3y−3z)(y−z)^2\geq 0$$
as
$$\dfrac{1}{25}[(u+v-w)(u-v)^2+(v+w-u)(w-v)^2+(u+w-v)(u-w)^2]\geq 0.$$
Obvioursly, it is schur's inequality. Equality holds for $(x,y,z)=(1,1,1)$, and for
$(x,y,z)=(4,3,3)$ or any cyclic permutation.
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/2181568",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "2",
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|
Incorrectly solving the determinant of a matrix Compute $det(B^4)$, where $B =
\begin{bmatrix}
1 & 0 & 1 \\
1 & 1 & 2 \\
1 & 2 & 1
\end{bmatrix}
$
I created
$C=\begin{bmatrix}
1 & 2 \\
1 & 1 \\
\end{bmatrix}
$ and
$
D= \begin{bmatrix}
1 & 1 \\
1 & 2 \\
\end{bmatrix}
$
$det(B) = -det(C) -2det(D) = -(1-2) - 2(2-1) = -(-1)-2(1)=1-2=1, 1^4=1$
However, the correct answer is 16. I'm confused on where I made my wrong turn
|
You have calculated det$(B) $ wrong - you just got the order of signs mixed up, when calculating the determinant by expanding into $ 2 \times 2 $ matrices.
It is det $(B) = 1 \begin{vmatrix} 1 & 2 \\ 2 & 1 \\ \end{vmatrix} - 0 \begin{vmatrix} 1 & 2 \\ 1 & 1 \\ \end{vmatrix} + 1 \begin{vmatrix} 1 & 1 \\ 1 & 2 \\ \end{vmatrix} $.
Hence det$(B) = 1 ( 1 - 4 ) - 0(1- 2 ) + 1 ( 2-1 ) = -3 + 1 = -2 $.
And it seems you used the formula det$(A^n) = ($det$(A)) ^ n $ which is correct and hence det$(B^4) = ($det$(B))^4 = ( -2 )^4 = 16 . $
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2184156",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Using L'Hopital Rule, evaluate $\lim_{x \to 0} {\left( \frac {1} {x^2}-\frac {\cot x} {x} \right)}$ Using L'Hopital Rule, evaluate $$ \lim_{x \to 0} {\left( \frac {1} {x^2}-\frac {\cot x} {x} \right)}$$
I find this question weired.
If we just combine the two terms into one single fraction, we get$$\lim_{x \to 0} {\frac {1-x\cot x} {x^2}}=\frac10=\infty$$
If we follow L'Hopital Rule, this is $\infty-\infty$ form. We follow the following process to convert it into $\frac00$form.$$\infty_1 -\infty_2=\frac 1{\frac 1{\infty_1}}-\frac 1{\frac 1{\infty_2}}=\frac {\frac 1{\infty_2}-\frac 1{\infty_1}}{{\frac 1{\infty_1}}{\frac 1{\infty_2}}}$$
So we will get $$\lim_{x \to 0} {\left( \frac {1} {x^2}-\frac {\cot x} {x} \right)}=\lim_{x \to 0} {\frac {x\tan x-x^2}{x^3\tan x}}$$
If you keep differentiating using the rule you will get rid of the form of $\frac00$ in the third step of differentiation, which give you the answer $1 \over 3$. This method is very tedious. Trust me, you don't want to try.
I am wondering is there a smarter way of solving this question?
Thanks.
|
Note that
$$
\lim_{x\to0}x\cot x=\lim_{x\to0}\frac{x}{\sin x}\cos x=1
$$
so $\dfrac{1-x\cot x}{x^2}$ is an indeterminate form $0/0$ at $0$.
You can certainly use l’Hôpital:
$$
\lim_{x\to0}\dfrac{1-x\cot x}{x^2}
=
\lim_{x\to0}\frac{-\cot x+\frac{x}{\sin^2x}}{2x}=
\lim_{x\to0}\frac{x-\sin x\cos x}{2x\sin^2x}
$$
However, this doesn't seem very inviting, but not hard at all. Observe that you can compute instead
$$
\lim_{x\to0}\frac{x-\sin x\cos x}{2x^3}=
\lim_{x\to0}\frac{1-\cos^2x+\sin^2x}{6x^2}=
\lim_{x\to0}\frac{2\sin^2x}{6x^2}
$$
Alternatively you can do
$$
\frac{1}{x^2}-\frac{\cot x}{x}=
\frac{1}{x^2}-\frac{\cos x}{x\sin x}=
\frac{\sin x-x\cos x}{x^2\sin x}
$$
which is much nicer:
$$
\lim_{x\to0}\frac{\sin x-x\cos x}{x^2\sin x}=
\lim_{x\to0}\frac{\sin x-x\cos x}{x^3}\frac{x}{\sin x}
$$
Since the limit of the second fraction is $1$, we can just compute
$$
\lim_{x\to0}\frac{\sin x-x\cos x}{x^3}=
\lim_{x\to0}\frac{x\sin x}{3x^2}
$$
which is fairly easy.
|
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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|
Extrema under constraints
Find the critical points of the function $f(x_1, x_2)=x_1x_2$ under the constraint $2x_1+x_2=b$.
Using the method of Lagrange multipliers I got the following:
\begin{equation*}L(x_1,x_2,\lambda )=x_1x_2-\lambda \cdot \left (2x_1+x_2-b\right )\end{equation*}
\begin{align*}&L_{x_1}(x_1,x_2,\lambda)=0 \Rightarrow x_2-2\lambda=0 \\ & L_{x_2}(x_1,x_2,\lambda)=0 \Rightarrow x_1-\lambda=0 \\ & L_{\lambda}(x_1,x_2,\lambda)=0 \Rightarrow -\left (2x_1+x_2-b\right )=0\end{align*}
Solving this system we get the critical point $\left (\frac{b}{4}, \frac{b}{2}\right )$.
To check what extrema (if there exists) it is, we do the following:
$$f_{x_1} =x_2 , \ f_{x_2}=x_1 , \ f_{x_1x_1}=0 . \ f_{x_1x_2}=1 , \ f_{x_2x_2}=0$$
Then: \begin{equation*}f_{x_1x_2}\left (\frac{b}{4}, \frac{b}{2}\right )=1>0 \ \text{ and } \ f_{x_1x_1}\left (\frac{b}{4}, \frac{b}{2}\right )f_{x_2x_2}\left (\frac{b}{4}, \frac{b}{2}\right )-\left (f_{x_1x_2}\left (\frac{b}{4}, \frac{b}{2}\right )\right )^2=0\cdot 0-1=-1<0\end{equation*}
Therefore, $\left (\frac{b}{4}, \frac{b}{2}\right )$ is a saddle point. Is this correct? Because at Wolfram there are some maxima.
$$$$
Then I want to check if there are extrema if we have an other constraint, $\{(x_1, x_2)\in \mathbb{R}^2 \mid x_1\geq 0, x_2\geq 0\}$.
A critical point is \begin{equation*}\nabla f=\begin{pmatrix}0 \\ 0\end{pmatrix}\Rightarrow \begin{pmatrix}x_2 \\ x_1\end{pmatrix}=\begin{pmatrix}0 \\ 0\end{pmatrix} \Rightarrow x_1=0 \ \text{ und } \ x_2=0\end{equation*}
But this point is again a saddle point, right?
Then I wan to check if there are extrema if we have an other constraint, $\{(x_1, x_2)\in \mathbb{R}^2 \mid x_1\geq 0, x_2\geq 0\}$.
A critival point is \begin{equation*}\nabla f=\begin{pmatrix}0 \\ 0\end{pmatrix}\Rightarrow \begin{pmatrix}x_2 \\ x_1\end{pmatrix}=\begin{pmatrix}0 \\ 0\end{pmatrix} \Rightarrow x_1=0 \ \text{ and } \ x_2=0\end{equation*}
But this point is again a saddle point, right?
|
$$\begin{array}{ll} \text{maximize} & x y\\ \text{subject to} & 2x + y = b\end{array}$$
From the equality constraint, we have $y = b - 2 x$. Let
$$g (x) := x (b - 2 x)$$
The derivative of $g$ vanishes at $\frac b4$. Hence, the maximizer is $(\bar x, \bar y) := \left(\frac b4, \frac b2\right)$ and the maximum is $\frac{b^2}{8}$. For example, if $b = 4$, we have
|
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|
Cubic Root verification I'm working on a cubic root question and goes like this...
Let, r,s, and t be the roots of the equation $x^3+ax^2+bx+c = 0$. Re-write the expressions in terms of a,b, and c.
I'm given $r^2+s^2+t^2$. I know that, given 3 roots, I can factor the cubic equation as $(x-r)(x-s)(x-t)$, but don't know what to replace $x$ by in order to get $r^2+s^2+t^2$.
|
We haven't studied Vieta's formula in my class just yet.
Pretending to not know Vieta's formulas, note that if:
$$P(x)=x^3+ax^2+bx+c = (x-r)(x-s)(x-t)$$
then:
$$P(-x)=-x^3+ax^2-bx+c = -(x+r)(x+s)(x+t)$$
It follows that:
$$
\begin{align}
P(x)P(-x) & = (x^3+ax^2+bx+c)(-x^3+ax^2-bx+c) \\
& = (ax^2+c)^2-(x^3+bx)^2 \\
& = -x^6 +x^4(a^2-2b)+\cdots
\end{align}
$$
and also:
$$
\begin{align}
P(x)P(-x) & = -(x^2-r^2)(x^2-s^2)(x^2-t^2) \\
& =-x^6+x^4(r^2+s^2+t^2)+\cdots
\end{align}
$$
Now equate the coefficient of $x^4$ between the two expressions.
|
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|
Find $k$ such that $\frac{1}{(1+x)^k}+\frac{1}{(1+y)^k}+\frac{1}{(1+z)^k}\geq\frac{3}{2^k}$ for $xyz=1$.
Find all $k\in\mathbb{R}^+$ such that for all $xyz=1$, $x,y,z\in\mathbb{R}^+$, we
have
$$\frac{1}{(1+x)^k}+\frac{1}{(1+y)^k}+\frac{1}{(1+z)^k}\geq\frac{3}{2^k}.$$
If we set $$x=\frac{a}{b},\,y=\frac{b}{c},\,z=\frac{c}{a},$$ where $a,b,c\in\mathbb{R}^+$, then the inequality is equivalent to $$\frac{a^k}{(a+b)^k}+\frac{b^k}{(b+c)^k}+\frac{c^k}{(c+a)^k}\geq\frac{3}{2^k}.$$ For $k=1$ this is obviously false, as we take $b\to0$ and $c\to\infty$. In an olympiad test I saw that the case $k=\sqrt3$ is true (though I do not know how to prove that). I then thought that maybe this is true for all $k>1$. Am I correct and how do I prove it?
|
I think the answer is $k_{min}=\log_23$.
Let $x=e^a$, $y=e^b$, $z=e^c$ and $f(x)=\frac{1}{(1+e^x)^k}$.
Hence, $a+b+c=0$ and we need to prove that
$$\sum_{cyc}f(a)\geq3f\left(\frac{a+b+c}{3}\right)$$
But $f''(x)=\frac{2ke^x(ke^x-1)}{(1+e^x)^{k+2}}\geq0$ for all $k\geq1$ and $x\geq0$.
Thus, by Vasc's RCF Theorem for all $k\geq1$ it's enough to prove our inequality
for $y=x$ and $z=\frac{1}{x^2}$, which gives $\frac{2}{(1+x)^k}+\frac{1}{\left(1+\frac{1}{x^2}\right)^k}\geq\frac{3}{2^k}$
and for $x\rightarrow+\infty$ we get $k\geq\log_23$.
|
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|
Does the sequence $x_{n+1}=x_n+\frac{1}{x_n}$ converge or diverge? Set $x_1=a$, where $a > 0$ and let $x_{n+1}= x_n + \frac{1}{x_n}$. Determine if the sequence $ \lbrace X_n \rbrace $ converges or diverges.
I think this sequence diverges since Let $x_1= a > 0$ and $x_{n+1}= x_n + (1/x_n)$ be given. Let $x_1=2$ since $a >0$. Then...
$x_2 = 2 + \frac{1}{2} = 2.5$
$x_3 = 2.5 + \frac{1}{2.5} = 2.9$
$x_4 = 2.9 + \frac{1}{2.9} = 3.2448$
$x_5 = 3.2448 + \frac{1}{3.2448} = 3.5530$
$x_6 = 3.5530 + \frac{1}{3.5530} = 3.8344$
$x_7 = 3.8344 + \frac{1}{3.8344} = 4.0952$
$x_8 = 4.0952 + \frac{1}{4.0952} = 4.3394$
$x_9 = 4.3394 + \frac{1}{4.3394} = 4.5698$
$x_{10} = 4.5698 + \frac{1}{4.5698} = 4.7887$
but others have said it converges so I'm confused on whether it converges or diverges? Can someone please explain.
|
Assume the sequence converges i.e. $\displaystyle\lim_{n\to \infty} x_n = C< \infty.$
Then, as $n\to\infty$ we have $\,x_{n}\approx x_{n+1}$.
Taking limit of $x_{n+1}=x_n+\frac{1}{x_n}$ we get
$$
\lim_{n\to\infty} x_{n}= \lim_{n\to\infty} \left(x_n+\frac{1}{x_n} \right)\implies C = C + \frac{1}{C} \iff \frac{1}{C} = 0 \quad \text{ – } \text{ contradiction}
$$
The last expression results in contradiction even if $\,C\sim 0$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Find alternating series that converges to $ \int_0^{1/2}x\log(1+x^3)dx $ I need to find the alternating series that converges to $ \int_0^{1/2}x\log(1+x^3)\,dx $
Here's what I did:
$$
\frac{d}{dx}[\log(1+x^3)]=\frac{1}{1+x^3}=\frac{1}{1-(-x)^2}=\sum_{n=1}^\infty(-x^3)^{n-1}=1-x^3+x^6-x^9+-...
$$
$$\begin{align}
f(x)&=x\log(1+x^3)=x\int(1-x^3+x^6-x^9+-...)\\\\
&=x\left[x-\frac{x^4}{4}+\frac{x^7}{7}-\frac{x^{10}}{10}+-...+ C\right]
\end{align}$$
$$
f(0)=0; C=0
$$
$$
f(x)=x^2-\frac{x^5}{4}+\frac{x^8}{7}-\frac{x^{11}}{10}+-...
$$
$$\begin{align}
\int_0^{1/2}x\log(1+x^3)dx&=\int_0^{1/2}(x^2-\frac{x^5}{4}+\frac{x^8}{7}-\frac{x^{11}}{10}+-...)\,dx\\\\
&=\left.\left[\frac{x^3}{3}-\frac{x^6}{6*4}+\frac{x^9}{7*9}-\frac{x^{12}}{10*12}+-...\right]\right|_0^{1/2}\\\\
&=\frac{1}{2^3(3)}-\frac{1}{2^6(6)(4)}+\frac{1}{2^9(7)(9)}-\frac{1}{2^{12}(10)(12)}+-...
\end{align}$$
Is my method correct?
|
I would do it this way:
Using
$\ln(1+z)
=\sum_{k=1}^{\infty} \dfrac{(-1)^{k-1}x^k}{k}
$,
which converges for your range,
$\begin{array}\\
\int_0^{1/2}x\ln(1+x^3)dx
&=\int_0^{1/2}x\left(\sum_{k=1}^{\infty} (-1)^{k-1}\dfrac{x^{3k}}{k}\right)dx\\
&=\sum_{k=1}^{\infty} \dfrac{(-1)^{k-1}}{k}\int_0^{1/2}x^{3k+1}dx\\
&=\sum_{k=1}^{\infty} \dfrac{(-1)^{k-1}}{k}\dfrac{x^{3k+2}}{3k+2}\big|_0^{1/2}\\
&=\sum_{k=1}^{\infty} \dfrac{(-1)^{k-1}}{k(3k+2)2^{3k+2}}\\
&=\dfrac1{5\cdot 2^5}-\dfrac1{2\cdot8\cdot 2^8}+...\\
\end{array}
$
More generally,
$\begin{array}\\
\int x^a\ln(1+x^b)dx
&=\int x^a\left(\sum_{k=1}^{\infty} (-1)^{k-1}\dfrac{x^{bk}}{k}\right)dx\\
&=\sum_{k=1}^{\infty} \dfrac{(-1)^{k-1}}{k}\int x^{a+bk}dx\\
&=\sum_{k=1}^{\infty} \dfrac{-1)^{k-1}x^{a+1+bk}}{k(a+1+bk)}\\
\end{array}
$
|
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"timestamp": "2023-03-29T00:00:00",
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|
If $(a+b)(a+c)(b+c)=8$ then $\prod\limits_{cyc}(2a+bc)\leq27$
Let $a$, $b$ and $c$ be non-negative numbers such that $(a+b)(a+c)(b+c)=8$. Prove that:
$$(2a+bc)(2b+ac)(2c+ab)\leq27$$
My trying:
We need to prove that
$$8abc+a^2b^2c^2+\sum_{cyc}(4a^2b^2+2a^3bc)\leq27$$ or
$$abc\prod_{cyc}(a+b)+a^2b^2c^2+\sum_{cyc}a^2b^2\sqrt[3]{\prod_{cyc}(a+b)^2}+abc\sum_{cyc}a^2\sqrt[3]{\prod_{cyc}(a+b)}\leq\frac{27}{64}\prod_{cyc}(a+b)^2.$$
We made a homogenization! But what is the rest?
Thank you!
|
let : $p= a+b+c, \ q= ab+bc+ca, \ r=abc$
then : $pq=8+r \ (\ p\ge 3, \ 0\le q\le 3)$
$$q^2\ge 3pr \Rightarrow q \le \dfrac{3p^2-\sqrt{9p^4-96p}}{2}$$
$$by \ Shur : q^3+9r \ge 4pq \Rightarrow q \ge \dfrac{72-p^3}{5p}$$
$$f(p,q) = (p-2)^2q^2+2(p^2-4)(p-4)q+16p(4-p) \le 27$$
So we only need to consider the inequality in 2 cases :
*
*$f\left(p, \dfrac{3p^2-\sqrt{9p^4-96p}}{2}\right) \le 27, \ p\ge 3 \Leftrightarrow $
$$(p-3)(243p^5-465p^4-559p^3-3109p^2+8849p+243) \ge 0 $$
*$f\left(p, \dfrac{72-p^3}{5p}\right)\le 27, \ 3\le p \le 4 \Leftrightarrow $
$$(p-3)(p^7-11p^6+11p^5-71p^4+523p^3-287p^2+768p-6912) \le 0 , \ 3\le p \le 4$$
|
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|
Using the substitution method for a simple integral I have been playing with the substitution rule in order to test some ideas with computational graphs. One of the things I'm doing is applying the substitution to well known, and easy, integrals. For example, let's use that method to find the indefinite integral for
$$f(x) = x^2$$
Using the rule $\int x^n dx = \frac{x^{n+1}}{n+1} + C$, we get
$$F(x) = \frac{x^3}{3} + C$$
So, let's do the following substitution:
$$u = x^2$$
$$\frac{du}{dx} = 2x \Leftrightarrow dx = \frac{1}{2x} du$$
So, performing the substitution in the integral of $f(x)$ gives us
$$\int x^2 dx = \int u \frac{1}{2x} du = \frac{1}{2x} \frac{u^2}{2} + C = \frac{(x^2)^2}{4x} + C = \frac{x^3}{4} + C$$
Have I done anything wrong with the substitutions??
Thanks in advance!
|
$\int \frac{1}{2x}u du$ = $\int \frac{1}{2\sqrt{u}}\cdot u du$ = $ \int \frac{\sqrt{u}}{2}du$ = $u^{\frac{3}{2}}\cdot\frac{2}{3}\cdot \frac{1}{2} + C$
$= \frac{u^{\frac 3 2}}{3} + C = \frac{x^3}{3} + C$ as required.
|
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|
proof by induction that $3^{2n-1} + 2^{2n-1}$ is divisible by $5$ I am supposed to proof by induction that $3^{2n-1} + 2^{2n-1}$ is divisible by $5$ for $n$ being a natural number (integer, $n > 0$).
What I have so far:
Basis: $n = 1$
\begin{align}
3^{2 \cdot 1-1} + 2^{2 \cdot 1-1} & = 3^1 + 2^1\\
& = 5
\end{align}
Assumption: $3^{2n-1} + 2^{2n-1}$ is divisible by $5$ for $n = k \in \mathbb{N}$.
$5 \mid (3^{2n-1} + 2^{2n-1}) \implies 3^{2n-1} + 2^{2n-1} = 5m$, $m \in \mathbb{Z}$
Proof: Let $n = k + 1$
\begin{align}
3^{2 \cdot (k+1)-1} + 2^{2 \cdot (k+1)-1} & = 3^{2k+2-1} + 2^{2k+2-1}\\
& = 3^{2k+1} + 2^{2k+1}\\
& = 3^{2k} \cdot 3^1 + 2^{2k} \cdot 2^1\\
& = 3^{2k} \cdot 3 + 2^{2k} \cdot 2
\end{align}
And here I got stuck. I don't know how to get from the last line to the Assumption. Either I am overlooking a remodeling rule or I have used a wrong approach.
Anyway, I am stuck and would be thankful for any help.
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Let's use, from hypothesis, that $$3^{2k-1}+2^{2k-1}=5p\to 3^{2k-1}=5p-2^{2k-1}$$ so
$$3^{2k+1}+2^{2k+1}=9\cdot3^{2k-1}+4\cdot2^{2k-1}=9\cdot(5p-2^{2k-1})+4\cdot2^{2k-1}=\\
45p-5\cdot2^{2k-1}=5(9p-2^{2k-1})$$
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"language": "en",
"url": "https://math.stackexchange.com/questions/2195460",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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|
Solve $x^2 + 12 = y^4$ in the integers.
Solve $x^2 + 12 = y^4$ in the integers.
My thought was to factor $x^2 - y^4$ = $(x - y^2)(x + y^2) = -12$. I'm not sure if this is the right approach, and I'm not sure where to go from here.
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Rearrange as
$$y^4-x^2=12$$
Because they are squares, any positive value of $y$ or $x$ has a corresponding negative value, so it’s sufficient to find the positive values.
$$(y^2-x)(y^2+x)=12$$
So $[(y^2-x),(y^2+x)]$ is $(1,12)$ or $(2,6)$ or $(3,4)$
(Note: because of the symmetry, we can ignore $(12,1)$ or $(6,2)$ and $(4,3)$.)
Put $a=(y^2-x)$ and $b=(y^2+x)$ then $y^2=(a+b)/2$ and $x=(b-a)/2$
Clearly only $(2,6)$ can fit, as the others give fractions.
So $y^2=(2+6)/2=4$ and $x=(6-2)/2=2$
The positive answer is $(x,y)=(2,2)$ and the full answer is
$$(x,y)=(\pm2,\pm2)$$
Please let me know if you need any more details.
|
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"url": "https://math.stackexchange.com/questions/2198056",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "2",
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|
Solve the inequality $(x^2-2\sqrt{a}\cdot x+1)(2^x+\lg a)<0$
Solve the inequality $$(x^2-2\sqrt{a}\cdot x+1)(2^x+\lg a)<0$$
My work so far:
1) $a>0$
2) Let $f(x)=(x^2-2\sqrt{a}\cdot x+1)(2^x+\lg a)$.
$f(x)=0$ if and only if $x^2-2\sqrt{a}\cdot x+1=0$ or $2^x+\lg a=0$
$$x^2-2\sqrt{a}\cdot x+1=0$$
$$x_{1,2}=\sqrt a \pm \sqrt{a-1}$$
and $$2^x+\lg a=0$$
$$x=\log_2\lg\frac1a$$
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you must do case work for $a$:
1) $$a>1$$ then we have $$2^x+\lg(a)>0$$ and we have to solve $$x^2-2\sqrt{a}x+1<0$$ this is equivalent to
$$(x-\sqrt{a})^2<a-1$$ and further
$$|x-\sqrt{a}|<\sqrt{a-1}$$
can you proceed?
2)$$a=1$$ then we have $$(x^2-2x+1)<0$$ which is impossible.
3) the case $$0<a<1$$ is for you!
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"url": "https://math.stackexchange.com/questions/2199811",
"timestamp": "2023-03-29T00:00:00",
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|
Evaluate $\int_0^{\frac{\pi}{2}} \frac{\sin^3(x)}{\cos^3(x)+\sin^3(x)} dx$
Evaluate $\int_0^{\frac{\pi}{2}} \frac{\sin^3(x)}{\cos^3(x)+\sin^3(x)} dx$
$$\int_0^{\frac{\pi}{2}} \frac{\sin^3(x)}{\cos^3(x)+\sin^3(x)} dx=\int_0^{\frac{\pi}{2}} 1-\frac{\cos^3(x)}{\cos^3(x)+\sin^3(x)} dx=\int_0^{\frac{\pi}{2}} 1-\frac{1}{1+\tan^3(x)} dx$$
Then to evaluate $\int \frac{1}{1+\tan^3(x)}dx$, I let $\tan(x)=t$ and hence $dt = \sec^2(x) dx$, which implies $dx=\frac{1}{\sec^2(x)}dt$.
Then I don't know how to continue.
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I entered this into Wolfram alpha and this is what it gives me.
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{
"language": "en",
"url": "https://math.stackexchange.com/questions/2200716",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 3
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|
If $\dfrac {1}{a+b} +\dfrac {1}{b+c}=\dfrac {3}{a+b+c}$, prove that $\angle B=60^\circ $ If $\dfrac {1}{a+b}+\dfrac {1}{b+c}=\dfrac {3}{a+b+c}$, prove that $\angle B=60^\circ$.
My Attempt
$$\dfrac {1}{a+b}+\dfrac {1}{b+c}=\dfrac {3}{a+b+c}$$
$$\dfrac {a+2b+c}{(a+b)(b+c)}=\dfrac {3}{a+b+c}$$
$$a^2-ac-b^2+c^2=0$$.
How to prove further?
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You're almost there. You had
$$ \frac{a+2b+c}{(a+b)(b+c)} = \frac{3}{a+b+c} $$
which implies
$$ a^2 + ab + ac + 2ab + 2b^2 + 2bc + ac + bc + c^2 = 3ab + 3b^2 + 3ac + 3bc $$
so
$$ a^2 + c^2 - ac = b^2 $$
Now recall the cosine rule $b^2 = a^2 + c^2 - 2ac\cos(\angle B)$. This means that $\angle B = \arccos(\frac{1}{2}) = 60^\circ$.
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{
"language": "en",
"url": "https://math.stackexchange.com/questions/2201496",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 2
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System of Equation. (Complicated) Given that $x,y \in \mathbb{R}$. Solve the system of equation.
$$5x(1+\frac{1}{x^2+y^2})=12$$
$$5y(1-\frac{1}{x^2+y^2})=4$$
My attempt,
I made them become $x+\frac{x}{x^2+y^2}=\frac{12}{5}$ and $y-\frac{y}{x^2+y^2}=\frac{4}{5}$.
I don't know how to continue from here. Hope anyone would provide some different solutions. Thanks in advance.
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First of all get rid of $x^2+y^2$
$$ \frac{12}{4x} +\frac{4}{5y} = 2,\quad 2x+6y = 5xy $$
It is a rectangular hyperbola passing through origin with asymptotes parallel to axes reminding one of $ x= t, y= 1/t $ when axes are asymptotes. So solve for $x,y$
$$ x= \frac{6y}{5y-2},\, y= \frac{2x}{5x-6}$$
The asymptotes are
$$ x= \frac65 ,\, y= \frac{2}{5} $$
We take hyperbola in the form
$$ ( x-\frac65 ) (y- \frac25) = k $$
Since it goes through the origin, $k$ must be the product $k=\dfrac65 \cdot \dfrac25 = \dfrac{12}{25} $ so as not to leave behind any constant term.
We can now find $x,y$ upto a parameter $t$
$$ ( x-\frac65 ) (y- \frac25) = \frac{12}{25}$$
$$\rightarrow x= \frac65 + t \sqrt{12}/5,\, y= \frac25 + \sqrt{12}/{5t},\, $$
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{
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"url": "https://math.stackexchange.com/questions/2203287",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 4
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|
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.