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Linear Transformation in $\mathbb R_2[X]$ I'm learning linear algebra, specifically linear transformations, and need help with the following exercise:
Consider the linear transformation $f:\mathbb R_2[X] \rightarrow \mathbb R_2[X]$ satisfying the relations
\begin{align}
f(1+X+X^2)&=1+X+2X^2\\
f(X+X^2)&=1+3X+6X^2\\
f(X^2)&=X+4X^2
\end{align}
and let $Q(X) = 4+10X+\mu X^2, \mu \in \mathbb R$.
$(\alpha)$ Determine $f$.
$(\beta)$ Determine the value of $\mu$ for which $Q(X)$ belongs to $\operatorname{Im}f$. Also, find all the $P(X) \in \mathbb R_2[X]$ such that $f(P(x)) = Q(x)$.
Since I'm having difficulties for $(\beta)$, I'm going to share my thoughts for $(\alpha)$.
$(\alpha)$ We consider the standard basis $\mathscr{B}=\{1,X,X^2\}$ of $\mathbb R_2[X]$. Given that $f$ is a linear transformation, it can be represented by a matrix whose columns are the images of the standard basis vectors. The third column of the requested matrix is found immediately. It is given by
$$f(X^2)=X+4X^2.$$
By linearity, we determine the first two columns. One has
\begin{align}
f(1) &= f\left((1+X+X^2) - (X+X^2)\right) = f(1+X+X^2) - f(X+X^2)\\
&= (1+X+2X^2) - (1+3X+6X^2) = -2X - 4X^2
\end{align}
and
\begin{align}
f(X) &= f\left((X+X^2) - (X^2)\right) = f(X+X^2) - f(X^2)\\
&= (1+3X+6X^2) - (X+4X^2) = 1+2X+2X^2.
\end{align}
Hence, the matrix of $f$ relative to the basis $\mathscr{B}$ is
$$A = \begin{pmatrix}0 & 1 & 0\\-2 & 2 & 1\\-4 & 2 &4\end{pmatrix}$$
Is my work correct for $(\alpha)$? Since we are asked to determine $f$, I'm not sure if my answer is complete. I only determined the matrix representation of $f$. Can I explicitly express $f$ from here? Also, any help for $(\beta)$ would be appreciated. I can state the obvious by saying that $\operatorname{Im}f = \operatorname{Col}A$ but this does not help a lot.
|
I would say your solution for $\alpha$ is one of the equivalent ways of expressing $f$. Another would be to say that $$f(ax^2+bx+c) = a(4x^2+x) + b(2x^2+2x+1) + c(-4x^2-2x)$$
and then simplify that expression.
For $\beta$, you already have the matrix set up, so the question is for which value of $\mu$ does the equation $$\begin{pmatrix}0 & 1 & 0\\-2 & 2 & 1\\-4 & 2 &4\end{pmatrix}\begin{pmatrix}a\\b\\c\end{pmatrix} = \begin{pmatrix}4\\10\\\mu\end{pmatrix}$$
have a solution, and since the matrix has full rank, the answer should be simple.
|
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|
Find all complex numbers satisfying the equation $\bar{z}+1=iz^2+|z|^2$ Find all complex numbers z satisfying $$\bar{z}+1=iz^2+|z|^2$$
where $i=\sqrt{-1}$
I only know one way i.e. assuming $z=x+iy$ but that process is very cumbersome. I don't know how to proceed otherwise with a shorter approach.
|
The answer is$$z=i,i\omega,i\omega^2$$where $\omega,\omega^2$ are the two complex cube roots of unity.
Putting $z=x+iy$,we get
$$(x-y)^2=1+x$$and $$x^2-y^2=-y$$
Let $A=x+y$ and also let $B=x-y$
Substituting in $$(x-y)^2=1+x$$ we get $$B^2=\frac{A+B}{2}$$i.e.$$A=2B^2-B-2$$
Substituting in $$x^2-y^2=-y$$ we get $$AB=\frac{B-A}{2}$$ i.e.$$2AB=B-A$$ Thus $$2(2B^2-B-2)B=b-2B^2+B+2$$i.e $$2B^3-3B-1=0$$ Thus $$(B+1)(2B^2-2B-1)=0$$
Case I$$B=-1$$
Now $$A=2B^2-B-2=1$$
$$x=\frac{A+B}{2}=0;y=\frac{A-B}{2}=1$$Thus
$$z=i$$
Case II$$2B^2-B-1=0$$Now $$A=2B^2-B-2$$Thus $$A+B+2-2B-1=0$$i.e$$A-B=-1$$Thus $$y=-\frac{1}{2}$$
Also $$x^2-y^2=-y$$and therefore $$x^2=\frac{3}{4}$$
$$z=\pm\frac{\sqrt{3}}{2}-\frac{i}{2}=i\omega,i\omega^2$$
Finally one has three solutions $$z=i,i\omega,i\omega^2$$
where $\omega,\omega^2$ are the two complex cube roots of unity.
|
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|
Simplify $\frac {2^{n(1-n)}\cdot 2^{n-1}\cdot 4^n}{2\cdot 2^n\cdot 2^{(n-1)}}$ Simplify::
$$\frac {2^{n(1-n)}\cdot 2^{n-1}\cdot 4^n}{2\cdot 2^n\cdot 2^{(n-1)}}$$
My Attempt:
\begin{align}
&\frac {2^{n-n^2}\cdot 2^{n-1}\cdot 2^{2n}}{2\cdot 2^n\cdot 2^{n-1}}\\
&=\frac {2^{n-n^2+n-1+2n}}{2^{1+n+n-1}} \\
&=\frac {2^{4n-n^2-1}}{2^{2n}}
\end{align}
I could not move on. Please help me to continue.
|
$$\frac {2^{4n-n^2-1}}{2^{2n}}=2^{4n-n^2-1-2n}=2^{2n-n^2-1}=2^{-(n-1)^2}$$
where I have used $\frac{a^b}{a^c}=a^{b-c}$
|
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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.
|
Another solution(just for fun)
Let $F_j(x)= \sum_{x \in 0..n}\binom{n}{x} | x \mod 3 = j$
$F_0(x) + F_1(x) +F_2(x) =2^x$
$F_j(x)=F_j(x-1) + F_{(j-1) \mod 3}(x-1)$
Self-cast previous equation
$F_j(x) = F_j(x-2) + 2*F_{(j-1) \mod 3}(x-2) + F_{(j-2) \mod 3}(x-2)$ =
$2^{x-2} + F_{(j-1) \mod 3}(x-2)$
Another self-cast
$F_j(x) = 2^{x-2} + 2^{x-4} + 2^{x-6} + F_j(x-6)$
So, $F_2(28) = 2^{26} + 2^{24} + ... + 2^4 + F_2(4)$ = $2^{26} + 2^{24} + ... + 2^2 + F_1(2) = 2^{26} + 2^{24} + ... + 2^2 + \binom{2}{1} $ =
$2 + 4\sum_{i\in 0..12}4^i = 2 +4\frac{4^{13}-1}{3} = \frac{4^{14} - 4 + 6}{3}$ = $\frac{2^{28} +2}{3}$
|
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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
|
It seems that you've got $\frac{18}{23}$ as the minimum value incorrectly though the value itself is correct. (you have already corrected the error.)
Let $k=a^2+b^2$.
Then, we get
$$ab=\frac{7k-9}{9}\tag1$$
Also, from
$$k=(a+b)^2-2ab=(a+b)^2-2\times\frac{7k-9}{9}$$
we get
$$(a+b)^2=\frac{23k-18}{9}\tag2$$
Therefore, we have to have $$\frac{23k-18}{9}\ge 0\tag3$$
Under $(3)$, $a,b$ are the solutions of $$t^2\mp\sqrt{\frac{23k-18}{9}}\ t+\frac{7k-9}{9}=0$$
Since $a,b$ are real, we have to have
$$\left(\mp\sqrt{\frac{23k-18}{9}}\right)^2-4\cdot 1\cdot \frac{7k-9}{9}\ge 0\tag4$$
In order for real $a,b$ satisfying $(1)$ and $(2)$ to exist, it is necessary and sufficient that $(3)$ and $(4)$ hold, i.e.
$$\color{red}{\frac{18}{23}\le a^2+b^2\le\frac{18}{5}}$$
|
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|
Probability that my roll on a die will be higher than yours: Why divide by 6? I have to work out a question where on two fair, $6$ sided dice, what is the probability that the second die gives me a higher number than the first die.
So I broke the question down the long way and said "If you roll a $1$, I have a $\frac{5}{6}$ chance of beating you, if you roll a $2$, I have a $\frac{4}{6}$ chance of beating you, etc."
Then I added all these up but got the answer to be $\frac{5}{2}$.
More research showed that I was actually supposed to divide this number by $6$ to get my probability to be $\frac{15}{36} = \frac{5}{12}$, but I can't see why you divide it by $6$.
Can someone explain this to me please?
EDIT: The bit I am struggling with is the fact that why do we take the draw into consideration.
Thank you to everyone who has commented and answered, I now get that the division by 6 is because we include the probability of drawing. But in my specific question, there was nothing about a tie, I simply have to beat you, or I lose.
So why do we still include the possibility of a draw, when that's not part of the game?
|
Okay:
If the first die is a $1$ the probability of the second die being higher is $5/6$. But that is the probability only if the first die is $1$.
If the second die is $2$ the probability of the second die being higher is $2/3$. But again that is only the probability if the first die is $2$.
And so one. You have six different probabilities, form $5/6$ to $0$ depending on the what the first die is.
As the first die is equally likely to be any one of those, and as it must be one and only one and never more than one of those, then the total probability of the second die being higher than the first is THE AVERAGE of all the probabilities.
To find the average probability add up the probabilities and divide by the equally likely possibilities of the first die. i.e. $\frac 16[5/6 + 2/3 + 1/2 + 1/3 + 16 + 0] = 5/12$.
or..
(1,1)(1,2)(1,3)(1,4)(1,5)(1,6)
(2,1)(2,2)(2,3)(2,4)(2,5)(2,6)
(3,1)(3,2)(3,3)(3,4)(3,5)(3,6)
(4,1)(4,2)(4,3)(4,4)(4,5)(4,6)
(5,1)(5,2)(5,3)(5,4)(5,5)(5,6)
(6,1)(6,2)(6,3)(6,4)(6,5)(6,6)
Probability is: $P = 15/36$
$P = \frac {5/6+2/3+1/2 + 1/3 + 1/6 + 0}6$
$P = \frac {5+4+3+2+1+0}{36}$
$P = \frac 16*\frac 56 + \frac 16*\frac 23+\frac 16*\frac 12+\frac 16*\frac 13+\frac 16*\frac 16 + \frac 16*0$
They are all equal ways of thinking of it.
|
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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):
*
*$3,8,15,24,35$
*$\dfrac{1}{2},\dfrac{2}{3},\dfrac{3}{4},\dfrac{4}{5},\dfrac{5}{6}$
*$2, 4, 8, 16 \text{ and } 32$
*$-\dfrac{1}{6},\dfrac{1}{6},\dfrac{1}{2},\dfrac{5}{6},\dfrac{7}{6}$
*$25,-125,625,-3125,15625$
*$\dfrac{3}{2},\dfrac{9}{2},\dfrac{21}{2},21,\dfrac{75}{2}$
*$65, 93$
*$\dfrac{49}{128}$
*$729$
*$\dfrac{360}{23}$
*$3, 11, 35, 107, 323$; $3+11+35+107+323+...$
*$-1,\dfrac{-1}{2},\dfrac{-1}{6},\dfrac{-1}{24},\dfrac{-1}{120}$; $-1+(\dfrac{-1}{2})+(\dfrac{-1}{6})+(\dfrac{-1}{24})+(d\frac{-1}{120})+...$
*$2, 2, 1, 0, -1$; $2+2+1+0+(-1)+...$
*$1,2,\dfrac{3}{5},\dfrac{8}{5}$
|
in elementary school math the fraction $x\frac{y}{z}$ usually means $x+\frac{y}{z}$ and is called a mixed fraction.
However these are almost never used after junior high.
Most of the time when you see $x\frac{y}{z}$ the two terms should be multiplied, so it is equal to $\frac{xy}{z}$.
|
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|
Generating functions on recursive We have a recursive sequence, defined by $r_0 = r_1 = 0$, and
$$r_n = 7r_{n-1} + 4r_{n-2} + n + 1,\quad \text{for } n\geq 2.$$
Express the generating function of this sequence as a quotient of polynomials or products of polynomials.
I have written the sequence
$$R(x) = r_0(0)+r_1(x)+r_2(x^2)+.....+r_n(x^n)+...$$
$$2xR(x)= 2r(0)x+2r(1)x^2+....+2r(n-1)x^n$$
but then what? In other examples, the instructor solved for Fibonacci and he gets nearly the same sequence after summing up, but I couldn't figure out this case.
|
Indeed, begin by letting
$$R(x) = r_0 + r_1 x + r_2 x^2 + \dots = \sum _{n \ge 0} r_n x^n$$
be a formal series.
Notice that
$$R(x) = \underbrace{r_0} _{=0} + \underbrace{r_1} _{=0} x + \sum _{n \ge 2} \big( 7 r_{n-1} + 4 r_{n-2} + (n+1) \big) x^n = \\
\sum _{m \ge 1} 7 r_m x^{m+1} + \sum _{p \ge 0} 4 r_p x^{p+2} + \sum _{k \ge 3} k x^{k-1} = \\
7x \big( R(x) - \underbrace{r_0} _{=0} \big) + 4x^2 R(x) + \sum _{k \ge 3} (x^k)' = \\
7x R(x) + 4x^2 R(x) + \left( \sum _{k \ge 3} x^k \right) ' = \\
(7x + 4x^2) R(x) + \left( \frac 1 {1-x} - 1 - x - x^2 \right) ' = \\
(7x + 4x^2) R(x) + \frac 1 {(1-x)^2} - 1 - 2x ,$$
which implies that
$$(4x^2 + 7x - 1) R(x) = 1 + 2x - \frac 1 {(1-x)^2} = \frac {2x^3 - 3x^2} {(1-x)^2} , $$
whence one finally gets
$$R(x) = \frac {2x^3 - 3x^2} {(4x^2 + 7x - 1) (1-x)^2} .$$
|
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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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$(1-x)^n = 1/(1-x)^{-n}=1/(1+nx+n(n-1)x^2/2 +\cdots) < 1/(1+nx)$, assuming $ 0 < x < 1$.
|
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A system of Diophantine equations Is there any positive solution for the Diophantine system
$$\left\{\begin{array}{l}
x^2-4y=z^2 \\
y^2-4x=w^2
\end{array} \right. ,$$
except that $(x,y)=(4,4), (6,5)$ (up to symmetry)?
|
$x^2-4y=z^2$
$y^2-4x=w^2$
$x^2-y^2-4(y-x)=z^2-w^2$
$(x-y)(x+y+4)=(z-w)(z+w)$
Since factors in LHS and also in RHS of this relation are co-prime then we can write:
$x-y=z-w$
$x+y+4=z+w$
$⇒ 2x+4=2z$ $⇒z=x+2$
$2w=2y+4$ $⇒w=y+2$
Plugging z or w in 1st or 2nd equation we get:
$x+y=-1$
Therefore:
$z+w=x+y+4=-1+4=3$
Now we make following table for one set of possible solutions:
$z= 0, 1, 2, 3$
$w= 3, 2, 1, 0$
$x= -2, -1, 0, 1$
$y= 1, 0,-1, -2$
Now if $x-y=1$ then we get:
$x^2-y^2 +4=z^2-w^2$
or: $x^2+w^2 +4 = z^2+y^2$
Which gives: $x=6, w=1, z=4, y=5$
If $y-x=1$ then:
$y^2-x^2 +4=w^2-z^2$
Which gives: $y=6, z=1, w=4, x=5$
If $z=w=0$ then $x=y=4$
These are only possible sets of positive solutions.
|
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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 ?
|
If we solve $\quad x^4+2ax^3+x^2+2ax+1=0\quad$ for $a$ we get
\begin{equation}
a = -\frac{x^4 + x^2 + 1}{2 x^3 + x}\quad \text{for}\quad x^3 + x\ne0
\end{equation}
which suggests that $\quad a<0\quad $ but
$\quad x=1\implies a=-\frac{3}{4}\quad $ and $\quad x=-1\implies a=\frac{3}{4}$
If we simply try a range of values, for $F(x)$ above, we get
\begin{equation}
F(-5)=2.50384615384615\qquad
F(-4)=2.00735294117647\qquad
F(-3)=1.51666666666667\qquad
F(-2)=1.05\qquad
F(-1)=0.75\qquad
F(0)=???\qquad
F(1)=-0.75\qquad
F(2)=-1.05\qquad \\
F(3)=-1.51666666666667\qquad
F(4)=-2.00735294117647\qquad
F(5)=-2.50384615384615\qquad \end{equation}
fra
And this shows that $\qquad a>\frac{3}{4}\iff x\le -1$
|
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|
(floor function) sum of x: $\left\lfloor{\frac{x}{5}}\right\rfloor - \left\lfloor{\frac{x}{9}}\right\rfloor = \frac{x}{15}$ Find the sum of integers $x$ such that
$$\left\lfloor{\frac{x}{5}}\right\rfloor - \left\lfloor{\frac{x}{9}}\right\rfloor = \frac{x}{15}$$
I don't have much experience in solving problems with the floor function involved. I have rewritten the equation as
$$\frac{x}{45} + \left\{\frac{5x}{45}\right\} = \left\{\frac{9x}{45}\right\}$$
but don't know where to go from there.
[EDIT: Thanks, I understand how to solve it now.]
|
First, since the LHS is an integer, we know $x$ is a multiple of $15$. Thus, let $x = 15n$. Then our equation reduces to
$$n = 3n - \left\lfloor \frac{15n}{9}\right\rfloor \implies$$
$$2n = \left\lfloor \frac{15n}{9}\right\rfloor \leq \frac{15n}{9}$$
thus ruling out any positive solutions. Now note that
$$2n = \left\lfloor \frac{15n}{9}\right\rfloor \geq \frac{15n}{9}-1 \implies$$
$$n \geq -3$$
So any possible solutions for $n$ satisfy $-3 \leq n \leq 0$. Checking manually, we see that the only solutions are $n = 0, -1, -2$, i.e. $x = 0, -15, -30$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2080125",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Trignometric integration We have to solve the following integration.
$$\int\frac{\tan 2\theta}{\sqrt{\cos^6 \theta+\sin^6\theta}}\ d\theta$$
In this, I write the denominator as $\sqrt 1-3\sin^2\theta \cos^2\theta$
But now how to proceed?
|
hint: Put $t = \cos (2\theta)\implies 1 - 3\sin^2\theta\cos^2\theta = 1 - \dfrac{3}{4}\sin^2(2\theta)= \dfrac{1}{4} + \dfrac{3}{4}\cos^2(2\theta)\implies I = -\displaystyle \int\dfrac{dt}{t\sqrt{1+3t^2}}$. From this put again $\sqrt{3}t= \tan(\phi)$. Can you take it from here?
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2081519",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
How to find two variables $a,b \in {\bf Z}$ that the matrix $A$ is orthogonal I have to find two variables $a,b \in {\bf Z}$ that the given $n \times n$ matrix A becomes orthogonal.
\begin{equation*}
A =
\begin{pmatrix}
1&2 \\
a&b
\end{pmatrix}
\end{equation*}
I know that a $n \times n$ matrix is called orthogonal if $A^TA$ $=$ $id$ which means:
\begin{equation*}
\begin{pmatrix}
1&a \\
2&b
\end{pmatrix}
\cdot
\begin{pmatrix}
1&2 \\
a&b
\end{pmatrix}
=
\begin{pmatrix}
1+a^2&2+ab \\
2+ab&4+b^2
\end{pmatrix}
\stackrel{?}{=}
\begin{pmatrix}
1&0 \\
0& 1
\end{pmatrix}
\end{equation*}
|
There is no solution since the norm of the column vector of an orthogonal matrix must be $1$. (Note that the second column is $(2,b)^T$.)
|
{
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|
Factorization of quartic polynomial. I want to know other ways of factorization to get quadratic factors in this polynomial:
$$x^4+2x^3+3x^2+2x-3$$
Thanks in advance for your suggestions.
The original polynomial is $$x^6-2x^3-4x^2+8x-3$$ where the two found factors are $(x+1)$ and $(x-1)$ by synthetic division.
|
Splitting up the x^2 term works well for factorising quartic but one has to be smart in grouping terms
In this , write 3x^2 as x^2 + 2x^2 and group terms as
= x^2(x^2+2x+1) +2x^2+2x-3 =[x(x+1)]^2 +2x(x+1) -3
= [x(x+1) +3][x(x+1)-1]
|
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"timestamp": "2023-03-29T00:00:00",
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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.
|
We have $x- \left(2+\sqrt 5 \right)^{\frac{1}{3}} - \left(2-\sqrt 5 \right)^{\frac{1}{3}} = 0$
So $x^3 -(2-\sqrt 5)-(2+\sqrt 5) = 3x \left(2+\sqrt 5 \right)^{\frac{1}{3}}\left(2-\sqrt 5 \right)^{\frac{1}{3}}= -3x$
(invoking that $a^3+b^3+c^3 = 3abc$ when $a+b+c = 0$)
from which we see that $x$ is a root of $x^3+3x-4 =0$
Since the derivative is positive, this means it has only one real root, which by inspection is $x=1$
|
{
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|
Prove the Identity $\dfrac{\tan\theta}{\cos\theta-\sec\theta} = -\csc\theta$ Been trying this question for over an hour and would like to know how it's done. Thanks.
|
\begin{align}
& \frac{\frac{\sin\theta}{\cos\theta}}{\cos\theta - \frac{1}{\cos\theta}} \\[10pt]
= {} & \frac{\frac{\sin\theta}{\cos\theta}}{\frac{\cos^2\theta - 1}{\cos\theta}} \\[10pt]
= {} & \frac{\frac{\sin\theta}{\cos\theta}}{\frac{-(1 - \cos^2\theta)}{\cos\theta}} \\[10pt]
= {} & \frac{\frac{\sin\theta}{\cos\theta}}{\frac{-(\sin^2\theta)}{\cos\theta}} \\[10pt]
= {} & \frac{\sin\theta}{\cos\theta} \cdot \frac{\cos\theta}{-\sin^2\theta} \\[10pt]
= {} & \frac{-1}{\sin\theta} \\[10pt]
= {} & - \csc\theta
\end{align}
|
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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?
|
We do have that $$I =\int \frac {1}{a^2-x^2} dx =\frac {1}{2a}\ln \frac {x+a}{x-a} $$ But however we apply the absolute value function to arguments of logarithm functions in order to extend the antiderivative's domain giving us: $$I =\int \frac {1}{a^2-x^2} dx =\frac {1}{2a} \ln |\frac {x+a}{x-a}|$$Hope it helps.
|
{
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"timestamp": "2023-03-29T00:00:00",
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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?
|
HINT.- $$\begin{cases}\sin{x}+\sin{y}+\sin{z}=0\\\cos{x}+\cos{y}+\cos{z}=0\end{cases}\iff\begin{cases}2\sin\dfrac{x+y}{2}\cos\dfrac{x-y}{2}=-\sin z\\2\cos\dfrac{x+y}{2}\cos\dfrac{x-y}{2}=-\cos z\end{cases}\Rightarrow \tan\frac{x+y}{2}=\tan z$$
Hence, because of $x,y,z \in [-\pi,\pi]$, $$x+y=2z\iff x+y+z=3z\Rightarrow \color{red}{z=0 \text{ and }x+y=0}$$
NOTE.-Obviously instead of $z$ we can choose either $x$ and $y$ to be zero. Thus $$\color{red}{x^2+y^2+z^2=2t^2}$$ where $(x,y,z)=(t,-t,0)$ and with $t\in [-\pi,\pi]$.
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
Find an analytic function $w=u+iv$ such that $u=\frac{x(1-x^2-y^2)}{4x^2y^2+(1+y^2-x^2)^2}, w(0)=0$
Find an analytic function $w=u+iv$ such that
$$
u=\frac{x(1-x^2-y^2)}{4x^2y^2+(1+y^2-x^2)^2}, w(0)=0
$$
This is the assignment under b.) and on a.) the answer is given to use the substitutions:
$$x=\frac{z+\overline{z}}{2}; y=\frac{z-\overline{z}}{2i}.$$
the answer here is suppose to be:
$\frac{z}{1-z^2}$ and just cannot seem to get this. When simplifying the expression with these substitutions I get:
$$\frac{1}{2}\frac{(z+\overline{z})(1-z\overline{z})}{1-(z^2+{\overline{z}}^2)+\frac{(z^2+{\overline{z}}^2)^2}{4}-\frac{(z+\overline{z})^2(z-\overline{z})^2}{4}}$$
|
Simplifying the denominator gives
$$
u = \frac{(z + \bar{z})(z\bar{z}-1)}{2(z^2-1)(\bar{z}^2-1)}
$$
You can then separate the denominator to get
$$
u = \frac{1}{2}\left(\frac{z}{z^2-1} + \frac{\bar{z}}{\bar{z}^2-1}\right)
$$
This is clearly the real part of the analytic function $z/(z^2-1)$, as desired.
|
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|
Integrate $\int\sin^2(x)\cos^4(x)dx$ As in the title: how does one integrate $$ \int\sin^2(x)\cos^4(x)dx? $$
Any hints? Integration by parts doesn't seem to be a reasonable technique here.
|
Or you could separate it into two integrals right from the beginning:
$$
\int \sin^2 x \cos^4 x \ dx = \int \cos^4 x \ dx - \int \cos^6 x \ dx.
$$
Then, we have
$$
\cos^4 x = \frac{3 + \cos 4x + 4\cos 2x}{8}
$$
and
$$
\cos^6 x = \frac{10 + \cos 6x + 6 \cos 4x + 15 \cos 2x}{32}.
$$
On completing the integration, the answer should be:
$$
\int \sin^2 x \cos^4 x \ dx = \frac{x}{16} + \frac{\sin 2x}{64} - \frac{\sin 4x}{64} - \frac{\sin 6x}{192}+C.
$$
|
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|
Determine whether the following sequence is increasing or decreasing $\frac{n^2+2n+1}{3n^2+n}$ Determine whether the following sequence is increasing or decreasing:
$$\frac{n^2+2n+1}{3n^2+n}$$
I'm not sure whether my solution is correct:
$$\frac{n^2+2n+1}{3n^2+n}=\frac{n(n+2)+1}{n(3n+1)}=\frac{n+2}{3n+1}+\frac{1}{n(3n+1)}.$$
Let's prove $\frac{n+2}{3n+1}$ is a decreasing sequence.
$$a_n>a_{n+1} \Leftrightarrow \frac{n+2}{3n+1}>\frac{n+3}{3n+4}\Leftrightarrow(n+2)(3n+4)>(n+3)(3n+1)\Leftrightarrow3n^2+10n+8>3n^2+10n+3\Leftrightarrow 8>3$$
So $\frac{n+2}{3n+1}$ is a decreasing sequence and we know that $\frac{1}{n(3n+1)}$ is also decreasing so our given sequence is a decreasing sequence as a sum of $2$ decreasing sequences.
|
Let $$a_n=\frac{n^2+2n+1}{3n^2+n}$$ Then
$$a_{n+1}-a_n=\frac{-5n^2-11n-4}{(3n^2+7n+4)(3n^2+n)}<0$$
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
Finding $B,C$ such that $B\left[\begin{smallmatrix}1&2\\4&8\end{smallmatrix}\right]C=\left[\begin{smallmatrix}1&0\\0&0\end{smallmatrix}\right]$
State $B,C$, such that $B\begin{bmatrix} 1 & 2 \\ 4 & 8 \end{bmatrix}C=\begin{bmatrix}
1 & 0 \\
0 & 0
\end{bmatrix}$
I tried some things, but I ended up with non invertible matrices, so I stopped for the moment.
Is there a quick way to guesswork here? (it's an assignment so, maybe some "nice" values for the entries will do)
I don't even know how I multiply a matrix to get 0-Entries with invertible matrices.
So maybe let's start at that part?
|
$B,C$ are not unique.
$\begin {bmatrix} 1&2\\4&8 \end{bmatrix} = B^{-1}
\begin {bmatrix} 1&0\\0&0 \end{bmatrix}C^{-1}$
Suppose
$B^{-1} = \begin {bmatrix} b_{11}&b_{12}\\b_{21}&b_{22} \end{bmatrix}$ and
$C^{-1} = \begin {bmatrix} c_{11}&c_{12}\\c_{21}&c_{22} \end{bmatrix}$
$\begin {bmatrix} 1&2\\4&8 \end{bmatrix} = $$\begin {bmatrix} b_{11}&b_{12}\\b_{21}&b_{22} \end{bmatrix} \begin {bmatrix} 1&0\\0&0 \end{bmatrix}\begin {bmatrix} c_{11}&c_{12}\\c_{21}&c_{22}\end{bmatrix}\\
\begin {bmatrix} b_{11}&b_{12}\\b_{21}&b_{22} \end{bmatrix} \begin {bmatrix} c_{11}&c_{12}\\0&0 \end{bmatrix}
\\
\begin {bmatrix} b_{11}c_{11}&b_{11}c_{12}\\b_{21}c_{11}&b_{21}c_{12} \end{bmatrix}$
$b_{11}c_{11} = 1\\
b_{11}c_{12}= 2\\
c_{12} = 2c_{11}\\
b_{21}c_{11} = 4b_{11}\\
$
Choose the other elements to make your calculation easy. i.e. $\det B^{-1} = 1$
$B = \begin {bmatrix} 1&0\\-4&1 \end{bmatrix}$
$C = \begin {bmatrix} 1&-2\\0&1 \end{bmatrix}$
Should work.
Completely different approach.
Daigonlize $\begin{bmatrix} 1&2\\4&8 \end{bmatrix}$
$P^{-1}\begin{bmatrix} 1&2\\4&8 \end{bmatrix}P = \begin{bmatrix} 9&0\\0&0 \end{bmatrix}$
$\frac 1{81}\begin{bmatrix} 1&2\\-4&1\end{bmatrix}\begin{bmatrix} 1&2\\4&8 \end{bmatrix}\begin{bmatrix} 1&-2\\4&1\end{bmatrix} = \begin{bmatrix} 1&0\\0&0 \end{bmatrix}$
|
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|
Evaluate $f^{2016}(0)$ for the function $f(x)=x^2 \ln(x+1)$
Question: Evaluate $f^{2016}(0)$ for the function $f(x)=x^2 \ln(x+1)$
Background: I know we can use taylor/mclaurin/power series to solve this but these were not teached to me by my teacher and not part of the syllabus so we are required to find the $nth$ derivative and make a suitable observation and hence evaluate the function.
One thing I wasn't sure about was to make a new post or edit my old post asking the part of the question. So if someone could clarify that for me , it would be great for the future.
My attempt (for $n\geq 3$ I have used partial fractions decomposition but I have not showed the working):
$$f^{(0)}(x)=x^2 \ln(x+1)$$
$$f^{(1)}(x)= 2x\ln(x+1) + \frac{x^2}{x+1} $$
$$f^{(2)}(x)=2\ln(x+1) + \frac{4x}{x+1} - \frac{x^2}{(x+1)^2}$$
$$f^{(3)}(x)=\frac{2}{(x+1)}+\frac{2}{(x+1)^2}+\frac{2}{(x+1)^3}$$
$$f^{(4)}(x)=-\frac{2}{(x+1)^2}-\frac{6}{(x+1)^3}-\frac{8}{(x+1)^4} $$
$$f^{(5)}(x)=\frac{4}{(x+1)^3}+\frac{12}{(x+1)^4}+\frac{24}{(x+1)^5}$$
From here I can see that for $n\geq3$ we have:
$$ f^{(n)}(x)=(-1)^{n-1} \left[ \frac{2(n-3)!}{(x+1)^{n-2}} + \frac{2(n-2)!}{(x+1)^{n-1}} + \frac{(n-1)!}{(x+1)^n} \right]$$
(it doesn't work for $n <3$ since we can't have a negative factorial)
Now evaluating this at zero for the $2016$th derivative:
$$ f^{(2016)}(0)=(-1)^{2015} \left[ 2(2013)!+2(2014)!+2015! \right]$$
$$ \Longrightarrow f^{(2016)}(0)=(-1) \left[ 2(2013)!+2(2014)!+2015! \right]$$
Would this be correct?
|
It is okay to use partial fractions and derive the result, but you should still eventually prove that $n$-th derivative formula is correct. You claim it is for $n\geq 3$, so why not to use induction for this?
For $n=3$ you can simply verify this (and you did that).
For induction step, you assume
$$f^{(n)}(x)=(-1)^{n-1} \left[ \frac{2(n-3)!}{(x+1)^{n-2}} + \frac{2(n-2)!}{(x+1)^{n-1}} + \frac{(n-1)!}{(x+1)^n} \right]$$
and you want to show
$$f^{(n+1)}(x)=(-1)^{n+1-1} \left[ \frac{2(n+1-3)!}{(x+1)^{n+1-2}} + \frac{2(n+1-2)!}{(x+1)^{n+1-1}} + \frac{(n+1-1)!}{(x+1)^{n+1}} \right]=(-1)^{n} \left[ \frac{2(n-2)!}{(x+1)^{n-1}} + \frac{2(n-1)!}{(x+1)^{n}} + \frac{n!}{(x+1)^{n+1}} \right]$$
But you have
\begin{align}
f^{(n+1)}(x) &= \frac{d}{dx} f^{(n)}(x) \\
&= \frac{d}{dx} \left((-1)^{n-1} \left[ \frac{2(n-3)!}{(x+1)^{n-2}} + \frac{2(n-2)!}{(x+1)^{n-1}} + \frac{(n-1)!}{(x+1)^n} \right]\right) \\
&= \left((-1)^{n-1} \left[ \frac{d}{dx}\frac{2(n-3)!}{(x+1)^{n-2}} + \frac{d}{dx}\frac{2(n-2)!}{(x+1)^{n-1}} + \frac{d}{dx}\frac{(n-1)!}{(x+1)^n} \right]\right) \\
\end{align}
Now just derive the fractions and after simplification you will obtain the expected result for $f^{(n+1)}(x)$.
After you prove this you can safely use the formula to evaluate $f^{(2016)}(0)$ the way you suggested.
|
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|
Find all $n \in \mathbb{N}$ for which $(2^n + n) | (8^n + n)$.
Find all $n \in \mathbb{N}$ for which $(2^n + n) | (8^n + n)$.
$n = 1, 2, 6$ are some solutions. Also, if the above holds then
$$(2^n + n) | 2^n(2^n-1)(2^n+1)$$
and
$$(2^n + n)| n(2^n+1)(2^n-1)$$
I've tried using cases when $n$ is even or odd and have tried using modular arithmetic but am not able to proceed. Please help.
Thanks.
|
More generally,
if
$(a^n+n) |((a^3)^n+n)$,
then,
since
$(a^3)^n+n^3
=(a^n)^3+n^3
=(a^n+n)((a^2)^n-na^n+n^2)
$,
we have
$(a^3)^n+n
=(a^3)^n+n^3-n^3+n
=(a^n+n)((a^2)^n-na^n+n^2)-n^3+n
$
so
$(a^n+n)
| (n^3-n)
$.
Therefore
$a^n+n \le n^3-n$,
which bounds $n$
as a function of $a$.
In particular,
$a^n < n^3$,
so
$a < n^{3/n}$.
Since
$n^{3/n}
< e^{3/e}
< 3.1
$
for all $n$,
$4^{3/4}
< 3$,
and
$n^{3/n}
< 2$
for
$n \ge 10$,
we get these bounds on $n$:
If $a = 2$,
$n < 10$.
Trying
$\dfrac{8^n+n}{2^n+n}$,
the only integer values are,
according to Wolfy,
$(1, 3), (2, 11), (4, 205)$,
and
$(6, 3745)$.
If $a = 3$,
$n \le 3$.
Trying
$\dfrac{27^n+n}{3^n+n}$
we get,
again according to Wolfy,
the only solution is
$(1, 7)$.
There are no solutions for
$a \ge 4$.
|
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|
Show that $x-\sqrt{x^2-x+1}<\frac{1}{2} $ for every real number $x$, without using differentiation Let $f$ be a function defined by :
$$f(x)=x-\sqrt{x^2-x+1}$$
Show that:
$$\forall x\in\mathbb{R},\quad f(x)<\dfrac{1}{2} $$
without use notion of différentiable
let $x\in\mathbb{R}$
\begin{aligned}
f(x)-\dfrac{1}{2}&=x-\dfrac{1}{2}-\sqrt{x^2-x+1} \\
&=\dfrac{(x-\dfrac{1}{2})^2-|x^2-x+1|}{(x-\dfrac{1}{2})+\sqrt{x^2-x+1}}\\
&= \dfrac{-3/4}{(x-\dfrac{1}{2})+\sqrt{x^2-x+1}}
\end{aligned}
i don't know if $(x-\dfrac{1}{2})+\sqrt{x^2-x+1} $ positive or negative for all x in R
|
For $x\le 1/2$ is trivial. For $x > 1/2$:
$$x - 1/2 < \sqrt{x^2-x+1}\iff (x-1/2)^2< x^2-x+1\iff 1/4 < 1.$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
The sum of series $1\cdot 3\cdot 2^2+2\cdot 4\cdot 3^2+3\cdot 5\cdot 4^2+\cdots \cdots n$ terms The sum of series $1\cdot 3\cdot 2^2+2\cdot 4\cdot 3^2+3\cdot 5\cdot 4^2+\cdots \cdots n$ terms
i have calculate $a_{k} = k(k+2)(k+1)^2$
so $\displaystyle \sum^{n}_{k=1}a_{k} = \sum^{n}_{k=1}k(k+1)^2(k+2)$
i wan,t be able go further, could some help me with this, thanks
|
Hint :
Since we have
$$(k+3)(k+4)-(k-2)(k-1)=10k+10$$
$$\implies \frac{(k+3)(k+4)-(k-2)(k-1)}{10}=k+1$$
we get
$$\begin{align}k(k+2)(k+1)^2&=k(k+1)(k+2)\color{red}{(k+1)}\\\\&=k(k+1)(k+2)\cdot\color{red}{\frac{1}{10}((k+3)(k+4)-(k-2)(k-1))}\\\\&=\frac{1}{10}(k(k+1)(k+2)(k+3)(k+4)-(k-2)(k-1)k(k+1)(k+2))\\\\&=\frac1{10}(a_{k+4}-a_{k+2})\end{align}$$ where $$a_k=k(k-1)(k-2)(k-3)(k-4)$$
Now simplify your sum using the idea of telescoping series.
|
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|
Inverse Mellin transform of $f(s)= 2^{ \frac{s}{6} }\frac{\Gamma \left( \frac{s+1}{3/2} \right)}{ \Gamma \left( \frac{s+1}{2} \right)}$ Given the function
$$f(s)= 2^{ \frac{s}{6} }\frac{\Gamma \left( \frac{s+1}{3/2} \right)}{ \Gamma \left( \frac{s+1}{2} \right)},$$
can we find and inverse Mellin transform for $f(s)$? That is,
$$\frac{1}{2 \pi i}\int_{- i\infty}^{ i\infty} 2^{ \frac{s}{6} }\frac{\Gamma \left( \frac{s+1}{3/2} \right)}{ \Gamma \left( \frac{s+1}{2} \right)} x^{-s-1 } ds$$
for $x>0$.
I was wondering if the integral can be expressed in terms of hypergeometric functions? For example, this is very similar to the Mellin–Barnes integral
\begin{align}
{}_2F_1(a,b;c;z) =\frac{\Gamma(c)}{\Gamma(a)\Gamma(b)} \frac{1}{2\pi i} \int_{-i\infty}^{i\infty} \frac{\Gamma(a+s)\Gamma(b+s)\Gamma(-s)}{\Gamma(c+s)}(-z)^s\,ds
\end{align}
However, I am not sure how to connect it to my problem!
Thanks.
|
Let me offer a partial solution, which I might finish later. Repeatedly using the Gauss multiplication formula yields the following identity:
$$2^{ \frac{s}{6} }\frac{\Gamma \left( \frac{s+1}{3/2} \right)}{ \Gamma \left( \frac{s+1}{2} \right)}=\frac1{\sqrt{\pi } \sqrt[6]{2}}\frac{\Gamma \left(\frac{s}{6}+\frac{11}{12}\right) \Gamma \left(\frac{s}{6}+\frac{2}{3}\right) \Gamma \left(\frac{s}{6}+\frac{5}{12}\right)}{\Gamma\left(\frac{s}{6}+\frac{1}{2}\right) \Gamma \left(\frac{s}{6}+\frac{5}{6}\right)}\left(\sqrt{\frac38}\right)^{-s}$$
and now the inverse Mellin transform can be directly converted into a Meijer $G$-function, yielding
$$\frac{3\sqrt[6]{32}}{\sqrt{\pi}}G_{2,3}^{3,0}\left(\frac{27z^6}{512}\middle|
{{\frac12,\frac56}\atop{\frac{5}{12},\frac23,\frac{11}{12}}}\right)$$
This can then be further expanded into a sum of $3$ ${}_2 F_2$ hypergeometric functions, using this formula.
|
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Other way to derive/prove $a_n = \frac{r^{n+1} - 1}{r-1}$ for sum of geometric sequence? The way I know, for a sequence $a_1 = 1 + r + r^2 + ... + r^n$, is to create another sequence $a_2 = r \cdot a_1 = r + r^2 + ... + r^{n+1}$, then subtract $a_1$ from $a_2$, to end up with $$a_2 - a_1 = r^{n+1}-1 = ra_1-a_1 = a_1(r-1)$$
so that $$a_1 = \frac{r^{n+1} - 1}{r-1}$$
However, even though I don't have a problem "believing" the algebra, it isn't very intuitive to me.
Is there a different way to derive that formula without resorting to this algebraic trick? Or do you have a way to think about it that makes it more intuitive?
|
Do you remember fat fit identity ?$$(x-y)=(x-y).1\\
(x^2-y^2)=(x-y)(x+y)\\
(x^3-y^3)=(x-y)(x^2+xy+y^2)\\
(x^4-y^4)=(x^3+x^2y+xy^2+y^3)\\...\\
(x^n-y^n)=(x-y)(x^{n-1}+x^{n-2}y+x^{n-3}y^2+...x^{2}y^{n-3}+xy^{n-2}+y^{n-1})$$
so now :take $x=r ,y=1$
you will have
$$(x^n-y^n)=(x-y)(x^{n-1}+x^{n-2}y+x^{n-3}y^2+...x^{2}y^{n-3}+xy^{n-2}+y^{n-1})\\(r^n-1^n)=(r-1)(r^{n-1}+r^{n-2}1+r^{n-3}1^2+...r^{2}1^{n-3}+r1^{n-2}+1^{n-1})\\
(r^n-1)=(r-1)(r^{n-1}+r^{n-2}+...+r^2+r+1)\\
(r^{n-1}+r^{n-2}+...+r^2+r+1)=\frac{(r^n-1)}{r-1}$$
In the case of $|r|<1$ there is some visual proof too
like below
|
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|
If $n=2^{31}\cdot 3^{19}$ then how many divisors of $n^2$ are smaller than $n$ but they aren't divisor of $n$? If $n=2^{31}\cdot 3^{19}$ then how many divisors of $n^2$ are smaller than $n$ but they aren't divisor of $n$?
It is clear that the power of $2$ or $3$ should be bigger than how much there are in $n$.For example I found some cases:
Power of $2=32\Rightarrow$there are $19$ cases for the power of $3$
Power of $2=33\Rightarrow$there are $18$ cases for the power of $3$
Power of $2=34\Rightarrow$there are $18$ cases for the power of $3$
But it is hard to see how many numbers the power of $3$ or $2$ can be by increasing the power of the other.Any hints?
|
Notice that if $d|n^2$ and $d<n$ then $n^2/d>n$.
We conclude the number of divisors of $n^2$ less than $n$ are $\frac{\tau(n^2)-1}{2}$
So the answer is $\frac{\tau(n^2)-1}{2}-\tau(n)+1$
|
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|
Proving $\left(\frac{a}{a+b+c}\right)^2+\left(\frac{b}{b+c+d}\right)^2+\left(\frac{c}{c+d+a}\right)^2+\left(\frac{d}{d+a+b}\right)^2\ge\frac{4}{9}$ The inequality:
$$\left(\frac{a}{a+b+c}\right)^2+\left(\frac{b}{b+c+d}\right)^2+\left(\frac{c}{c+d+a}\right)^2+\left(\frac{d}{d+a+b}\right)^2\ge\frac{4}{9}$$
Conditions: $a,b,c,d \in \mathbb{R^+}$
I tried using the normal Cauchy-Scharwz, AM-RMS, and all such..
I think I can do it using some bash method like expanding the LHS and then maybe to go with homogeneity, normalization and then Muirhead, Jensen or something I dont know since I didn't go that way..
But can someone help me with a nice elegant solution. This is an olympiad question I was trying to solve, but couldn't manage an elegant solution.
|
Let $\{a,b,c,d\}=\{x,y,z,t\}$, where $x\geq y\geq z\geq t$, $x+y+z+t=4$, $x+t=2u$ and $y+z=2v$.
Hence, $u+v=2$ and by Rearrangement we obtain: $$\sum\limits_{cyc}\frac{a^2}{(a+b+c)^2}\geq\frac{x^2}{(x+y+z)^2}+\frac{y^2}{(x+y+t)^2}+\frac{z^2}{(x+z+t)^2}+\frac{t^2}{(y+z+t)^2}=$$
$$=\frac{x^2}{(4-t)^2}+\frac{t^2}{(4-x)^2}+\frac{y^2}{(4-z)^2}+\frac{z^2}{(4-y)^2}$$
and since $$\frac{x^2}{(4-t)^2}+\frac{t^2}{(4-x)^2}\geq\frac{-2(x+t)^2+16(x+t)-16}{(8-x-t)^2} \tag1$$ it's just
$$((x+t)(x^2+t^2)-12(x^2+xt+t^2)+48(x+t)-64)^2\geq0, \tag2$$
it remains to prove that $$\frac{-2u^2+8u-4}{(4-u)^2}+\frac{-2v^2+8v-4}{(4-v)^2}\geq\frac{4}{9},\tag3$$
where $u$ and $v$ are positive numbers such that $u+v=2$ or
$$\sum_{cyc}\left(\frac{-2u^2+8u-4}{(4-u)^2}-\frac{2}{9}\right)\geq0$$ or
$$\sum\limits_{cyc}\frac{(u-1)(17-5u)}{(4-u)^2}\geq0$$ or
$$\sum\limits_{cyc}\left(\frac{(u-1)(17-5u)}{(4-u)^2}-\frac{4}{3}(u-1)\right)\geq0$$ or
$$\sum_{cyc}\frac{(u-1)^2(13-4u)}{(4-u)^2}\geq0.$$
Done!
|
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How can I prove this inequality in a triangle: $\frac{1}{(b+c-a)} +\frac{1}{(c+a-b)} +\frac{1}{(a+b-c)} \gt \frac{1}{a} +\frac{1}{b} +\frac{1}{c}$?
If $a,b,c$ are the sides of a triangle then show that-
$\frac{1}{(b+c-a)} +\frac{1}{(c+a-b)} +\frac{1}{(a+b-c)} \gt \frac{1}{a} +\frac{1}{b} +\frac{1}{c}$
I got this problem from an old book on algebra. I've been trying for a long time but can't reach the answer.
My try on the question-
We know that in a triangle, the sum of two sides is always greater than the third side
$b+c\gt a$ $\implies$ $b+c-a \gt 0$
$c+a\gt b$ $\implies$ $c+a-b\gt 0$
$a+b \gt c$ $\implies$ $a+b-c\gt 0$
Also,
$\frac {1}{(b+c)} \lt \frac{1}{a}$
$\frac{1}{(c+a)} \lt \frac{1}{b}$
$\frac{1}{(a+b)} \lt \frac{1}{c}$
Adding,
$\frac{1}{(b+c)} + \frac{1}{(c+a)} +\frac{1}{(a+b)} \lt \frac{1}{a} + \frac{1}{b} + \frac{1}{c}$
How do I proceed further? All genuine answers are welcome :)
|
hint: use $\dfrac{1}{x}+\dfrac{1}{y} \ge \dfrac{4}{x+y}$ $ $ $3$ times. The first time for the first two terms with $x = a+b-c, y = b+c-a$.
|
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|
How to find the limit $z\rightarrow i$? How to calculate the limit of $\frac{z^{10}+1}{z^6+1}$ as $z\rightarrow i$?
I tried to take the limit but the denominator becomes zero. Does this mean that the limit is infinite?
|
To avoid L'Hopital rememeber that
$$\begin{align}x^{10}+1=&(x^2+1)(x^8-x^6+x^4-x^2+1)\\
x^6+1=&(x^2+1)(x^4-x^2+1)\end{align}$$
So the fraction simplifies to
$${x^8-x^6+x^4-x^2+1\over x^4-x^2+1}\to {5\over 3}$$
|
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Find all reals $x$,$y$ satisfying the following equation:
Find all positive reals $x,y \in \mathbb{R}^+$ satisfying:
$$\frac{x^9}{y} + \frac{y^9}{x} = 10-\frac{8}{xy}$$
Since this involves higher exponents I am unable to tackle this problem. Please help me.
|
Hint: Write it as:
$x^{10}+y^{10}=10xy-8$
Then use AM-GM:
$10xy-5=x^{10}+y^{10}+1+1+1\ge5\sqrt[5]{x^{10}y^{10}}=5x^2y^2$
Which gives $5(xy-1)^2\le0$, so $xy=1$
Then initial equation can be rewritten as $x^{10}+\frac{1}{x^{10}}=2$
Which by AM-GM again, (or by writing it as a $(x^{10}-1)^2=0$) has the solution, $x^{10}=1$, so $x=\pm1$
So $x=1,y=1$, or $x=-1,y=-1$
|
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Computing $ \lim \limits_{x \rightarrow \infty} \left(1+\frac{1}{x} \right)^{\sqrt{ax^2 +bx+c}}$ For $a \geq 0 $, the following limit
$$ L = \lim_{x \rightarrow \infty} \left(1+\dfrac{1}{x} \right)^{\sqrt{ax^2 +bx+c}}$$
can be computed by applying L'Hopital rule as follows
$$L = \exp \left(\lim_{x \rightarrow \infty} \frac{\log \left(1+\dfrac{1}{x} \right)}{\frac{1}{\sqrt{ax^2 +bx+c}}}\right) = \exp \left(\lim_{x \rightarrow \infty} \dfrac{\frac{-1}{x(x+1)}}{\frac{-(2ax+b)}{2(ax^2 +bx+c)^{3/2}}}\right) =$$
$$ = \exp \left(\lim_{x \rightarrow \infty} \dfrac{2(ax^2 +bx+c)^{3/2}}{x(x+1)(2ax+b)}\right) = \exp \left(\lim_{x \rightarrow \infty} \dfrac{2\left(a+ \frac{b}{x}+\frac{c}{x^2}\right)^{3/2}}{1\cdot\left(1+\frac{1}{x}\right)\left(2a+\frac{b}{x}\right)}\right) = \exp(\sqrt{a}).$$
I am wondering if exists another method to compute this limit.
Thanks for any hint!
|
I would start with
$$\left(1+\frac1x\right)^{\sqrt{ax^2+bx+c}} = \left(1+\frac1x\right)^{\sqrt a x \cdot\sqrt{1+\frac{b}{ax} + \frac{x}{ax^2}}} =\left(\left(1 + \frac1x\right)^x\right)^{\sqrt{a}\cdot \sqrt{1+\frac{b}{ax} + \frac{x}{ax^2}}}$$
Now, use the fact that, if $f$ and $g$ are continuous and all limits exist, $$\lim_{x\to\infty}f(x)^{g(x)} = \left(\lim_{x\to\infty}f(x)\right)^{\lim_{x\to\infty} g(x)}$$
|
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Show the recursive sequence is increasing
How do I show that the recursive sequence
$$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$$
Again I get stuck since I don't know how to handle the ceils and floors.
$\phantom{x}$
How do I go about showing the sequence is increasing? Maybe I'm making it harder than it actually is - is there by any chance an easier way?
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Show $a_{n+1}>a_n$ be induction. Note that $\lceil(n+1)/2\rceil $ is either $\lceil n/2\rceil$ or $\lceil n/2\rceil+1$. Hence by induction hypothesis (which is applicable if $\lceil n/2\rceil<n$, i.e., for $n\ge2$), we may use that $a_{\lceil(n+1)/2\rceil}$ is $>$ or $=$ to $a_{\lceil n/2\rceil}$. The same works for floor. So
$$\begin{align}a_{n+1}&=a_{\lfloor (n+1)/2\rfloor}+a_{\lceil (n+1)/2\rceil}+3(n+1)+1\\
&\ge a_{\lfloor n/2\rfloor}+a_{\lceil (n+1)/2\rceil}+3(n+1)+1\\
&\ge a_{\lfloor n/2\rfloor}+a_{\lceil n/2\rceil}+3(n+1)+1\\
&> a_{\lfloor n/2\rfloor}+a_{\lceil n/2\rceil}+3n+1\\
&=a_n.
\end{align}$$
But recall that this induction step works only for $n\ge 2$. Hence we need to show manually that $a_3>a_2>a_1$. We compute $a_1=3$, $a_2=10$, $a_3=26$, so all is fine.
|
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|
Express the $n$th smallest number in this set in terms of $n$
Suppose $n$ is a perfect square. Consider the set of all numbers which are the product of two numbers, not necessarily distinct, both of which are at least $n$. Express the $n$th smallest number in this set in terms of $n$.
If, for example, $n = 4$, then the numbers are $4^2, 4 \cdot 5, 4 \cdot 6, 5^2$. If $n = 9$, then the numbers are $9^2,9 \cdot 10, 9 \cdot 11, 10^2, 9 \cdot 12, 10 \cdot 11,9 \cdot 13, 10 \cdot 12, 11^2$. Thus we conjecture the answer is $(n+\sqrt{n}-1)^2$, but how do we prove this?
|
Clearly there $w=(n+\sqrt n -1)^2$ is at least the $n$'th number in the list.
We must only prove that $w<ab$ if $a+b>2n+2\sqrt{n}-2$, with both $a$ and $b\geq n$.
Clearly this product is minimized in the case $a=n$, Which leaves $b>n+2\sqrt{n}-2$, since $b$ is an integer we can take the minimal case $b=n+\sqrt{2n}-1$.
So we must only prove $n(n+2\sqrt{n}-1)\geq (n+\sqrt{n}-1)^2\iff -n>-2n-2\sqrt{n}+n+1=-n-2\sqrt{n}+1$.
So it is in fact true.
|
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|
Finding the matrix associated with a linear map
Find the matrix associated with the linear map $f:R^2 \rightarrow R^2$ defined by $f(x,y)=(3x-y,y-x)$ with respect to the ordered basis ${(1,0),(1,1)}$
Let the matrix be $A$ and let $f(x)=AX$ where
$$X=\begin{bmatrix}
x \\
y
\end{bmatrix}$$
and
$$A=\begin{bmatrix}
a & b \\
c & d
\end{bmatrix}$$
I tried solving for $a,b,c,d$ by using the basis vectors as $(x,y)$. That is, I took $(x,y)=(1,0)$ and $(x,y)=(1,1)$ to find $a,b,c,d$. But I am not getting the given answer.
|
You can set up the matrix with respect to the standard basis and then convert it to the matrix with respect to the given basis by multiplying with the appropriote change of bases matrices.
Altneratively, you can get the matrix of the linear map with respect to the given basis directly. In general: the $i$-th column of that matrix contains the (red) coordinates of the image of the $i$-th basis vector with respect to the (blue) basis in question; that means:
$$f \begin{pmatrix} 1 \\ 0 \end{pmatrix} = \begin{pmatrix} 3 \\ -1 \end{pmatrix}=\color{red}{4}\color{blue}{\begin{pmatrix} 1 \\ 0 \end{pmatrix}}\color{red}{-1}\color{blue}{\begin{pmatrix} 1 \\ 1 \end{pmatrix}} \implies C_1 =\color{red}{ \begin{pmatrix} 4 \\ -1 \end{pmatrix}}$$
and
$$f \begin{pmatrix} 1 \\ 1 \end{pmatrix} = \begin{pmatrix} 2 \\ 0 \end{pmatrix}=\color{red}{2}\color{blue}{\begin{pmatrix} 1 \\ 0 \end{pmatrix}}+\color{red}{0}\color{blue}{\begin{pmatrix} 1 \\ 1 \end{pmatrix}} \implies C_2 =\color{red}{ \begin{pmatrix} 2 \\ 0 \end{pmatrix}}$$
So the matrix $C$ of the linear map with respect to the given basis is:
$$C = \left( C_1 \vert C_2\right) = \begin{pmatrix} 4 & 2 \\ -1 & 0\end{pmatrix}$$
|
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|
If $2^{2017} + 2^{2014} + 2^n$ is a perfect square, find $n$. If $2^{2017} + 2^{2014} + 2^n$ is a perfect square, find $n$.
My first shot would be to assume the perfect square is $2^{2018}$, but how would I prove that? Even if it is, what is $n$? All help is appreciated.
|
If you want an answer then $(a+b)^2 = a^2 + 2ab + b^2$ should suggest an obvious answer by setting $a^2 = 2^{2014}; 2ab=2^{2017}; 2^n = c^2$.
i.e $(2^{1007} + 2^{\frac n2})^2 = 2^{2014} + 2*2^{1007}*2^{\frac n2} + 2^n= 2^{2014} + 2^{1 + 1007 + \frac n2} + 2^n$. We just have to solve for $1 + 1007 + \frac n2 = 2017$. So $n = 2018$.
But is it the only solution?
Bear with me.
Let $m$ be any positive integer. Let $m = \sum_{i=0}^k a_i2^i; a_i = \{0|1\}$ but its unique binary expansion.
Claim: If $m$ has 3 or more non-zero terms in its binary expansion then $m^2$ has more than 3 non-zero terms in its binary expansion.
Proof: Let $m = 2^a + 2^b + 2^c + \sum_{i= c+1}^k a_i 2^i; a < b < c$. ($a$ may equal $0$. and $a_i; i > c$ may all be $0$.)
Then $m = (1 + 2^{b'} + 2^{c'} + \sum_{i=c+1}^k a_i2^{i - a})2^a; c'= c-a;b'=b-a$
So $m^2 = [(1 + 2^{b'} + 2^{c'} + 2*2^{b'}2^{c'}+ 2^{2b'} + 2^{c'}) + 2(1 + 2^{b'} + 2^{c'})\sum_{i=c+1}^k a_i2^{i - a} + (\sum_{i=c+1}^k a_i2^{i - a})^2]2^{2a}$.
$= 2^a + 2^b + 2^c + 2^{1+b+c} +2^{2b} + 2^{2c} +..... $.
Note all the later terms, if they exist, are larger than $2^{2c}$ so $m^2$ has at least four non-zero terms in its binary expansion.
Okay...
So $2^{2017} + 2^{2014} + 2^n = m^2$ has at most three non-zero terms in its expansion. So $m$ has at most $2$ it its expansion.
So $m = 2^k$ or $m = 2^k + 2^j$.
$2^{2017} + 2^{2014} + 2^n = 2^{2k}$ is impossible.
And $2^{2017} + 2^{2014} + 2^n = (2^k + 2^j)^2 = 2^{2k} + 2*2^j*2^k + 2^{2k}$ has only $n = 2018$ for solution.
So $n = 2018$ is the only solution.
|
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|
Evaluating $2x^3+2x^2-7x+72$ where $x= \frac{3-5i}{2} $ So I have to find the value of :
$$2x^3+2x^2-7x+72$$ at $$x= \frac{3-5i}{2} $$
Where $i$ stands for $\sqrt{-1}$.
I know the obvious approach would be to substitute the value of $x$ in the equation, but the equation gets extremely messy in that process. Is there any easier or quicker approach for the sum.
P.s. The answer is $4$.
|
You have $\frac{3-5i}{2}$ is the solution of $2x^2-6x+17 = 0$.
Then, $$2x^3 + 2x^2-7x+72 = x(2x^2-6x+17) + 4(2x^2-6x+17) + 4 = 4.$$
|
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|
Find the determinant of a 5x5 matrix Find the determinant of the following matrix:
$$\begin{bmatrix}
1& 1& 1& 1& 1\\
3 & 3 &3 &3 &2\\
4& 4& 4& 3& 3\\
5& 5& 4& 4& 4\\
6& 5& 5& 5 &5\end{bmatrix}$$
Laplace doesn't seem like the best method here, can we somehow turn this into a triangular matrix so that the determinant is the product of the elements on the main diagonal?
I multiplied the first row by $(-3)$ and added it to he second one, then by $(-4)$ and added it to the third one, by $(-5)$ and added it to the fourth one, and by $(-6)$ and added it to last one.
$$\begin{vmatrix}
1& 1& 1& 1& 1\\
3 & 3 &3 &3 &2\\
4& 4& 4& 3& 3\\
5& 5& 4& 4& 4\\
6& 5& 5& 5 &5
\end{vmatrix}=\begin{vmatrix}
1& 1& 1& 1& 1\\
0& 0 &0 &0 &-1\\
0& 0& 0& -1& -1\\
0& 0& -1& -1& -1\\
0& -1& -1& -1 &-1
\end{vmatrix}$$
What should I do now?
|
Now you can expand by the first collumn, and you get
$$\begin{vmatrix} 0 & 0 &0 &-1 \\ 0 & 0 &-1 &-1 \\ 0 &-1 &-1 &-1 \\ -1 &-1&-1&-1 \end{vmatrix}$$
which is triangular, so the determinant equals $(-1)^4=1$.
|
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How can I evaluate this series? I have to evaluate the following sum:
$$\sum_{n=0}^{\infty} \frac{n^{2}-5n+2}{n!}$$
Could you recommend some guide for solving this? I have not found any information that I could use for this type of exercise.
|
Let us make the problem more general considering $$F(x)=\sum_{n=0}^{\infty} \frac{an^{2}+bn+c}{n!}x^n$$ First, write $$an^2+bn+c=a(n(n-1)+n)+bn+c=an(n-1)+(a+b)n+c$$ $$F(x)=a\sum_{n=0}^{\infty} \frac{n(n-1)}{n!}x^n+(a+b)\sum_{n=0}^{\infty} \frac{n}{n!}x^n+c\sum_{n=0}^{\infty} \frac{x^n}{n!}$$ $$F(x)=ax^2\sum_{n=0}^{\infty} \frac{n(n-1)}{n!}x^{n-2}+(a+b)x\sum_{n=0}^{\infty} \frac{n}{n!}x^{n-1}+c\sum_{n=0}^{\infty} \frac{x^n}{n!}$$ $$F(x)=ax^2\left(\sum_{n=0}^{\infty} \frac{x^n}{n!}\right)''+(a+b)x\left(\sum_{n=0}^{\infty} \frac{x^n}{n!}\right)'+c \left(\sum_{n=0}^{\infty} \frac{x^n}{n!}\right)$$ $$F(x)=ax^2e^x+(a+b)xe^x+c e^x$$ $$F(1)=(2a+b+c)\,e$$
If you had faced the problem of $$G(x)=\sum_{n=0}^{\infty} \frac{an^{3}+bn^2+cn+d}{n!}x^n$$ writing $n^3=n(n-1)(n-2)+3n(n-1)+n$ and applying the same approach, you would obtain $$G(1)= (5 a+2 b+c+d)\,e$$ and so on.
|
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|
Simplifying fraction with factorials: $\frac{(3(n+1))!}{(3n)!}$ I was trying to solve the limit:
$$\lim_{n \to \infty} \sqrt[n]{\frac{(3n)!}{(5n)^{3n}}}$$
By using the root's criterion for limits (which is valid in this case, since $b_n$ increases monotonically):
$$L= \lim_{n \to \infty} \frac{a_n}{b_n} = \lim_{n \to \infty} \frac{a_{n+1}-a_n}{b_{n+1}-b_n}$$
Now I realise using Sterling's formula would make everything easier, but my first approach was simplifying the factorial after applying the criterion I mentioned before. So, after a few failed attempts I looked it up on Mathematica and it said that $\frac{(3(n+1))!}{(3n)!}$ (which is one of the fractions you have to simplify) equals $3(n+1)(3n+1)(3n+2)$. Since I can't get there myself I want to know how you would do it.
Just so you can correct me, my reasoning was:
$$\frac{(3(n+1))!}{(3n)!} = \frac{3\cdot 2 \cdot 3 \cdot 3 \cdot 3 \cdot 4 \cdot (...) \cdot 3 \cdot (n+1)}{3 \cdot 1 \cdot 3 \cdot 2 \cdot 3 \cdot 3 \cdot (...) \cdot 3 \cdot n } = $$
$$= \frac{3^n(n+1)!}{3^{n}n!} = \frac{(n+1)!}{n!} = n+1$$
Which apparently isn't correct. I must have failed at something very silly. Thanks in advance!
|
By Stirling's inequality the answer is clearly $\left(\frac{3}{5e}\right)^3$. To prove it, you may notice that by setting
$$ a_n = \frac{(3n)!}{(5n)^{3n}} $$
you have:
$$ \frac{a_{n+1}}{a_n} = \frac{(3n+3)(3n+2)(3n+1)(5n)^{3n}}{(5n+5)^{3n+3}} = \frac{\frac{3n+3}{5n+5}\cdot\frac{3n+2}{5n+5}\cdot\frac{3n+1}{5n+5}}{\left(1+\frac{1}{n}\right)^{3n}}\to\frac{\left(\frac{3}{5}\right)^3}{e^3}$$
as $n\to +\infty$.
|
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|
Solve using generating functions the following recurrence $F_0 = F_1 = 0$
$F_2 = 1$
$F_n = F_{n-1} + F_{n-2} + F_{n-3} , n \ge 3 $
Hint: You will have to write the recurrence in relation to the Fibonacci sequence. I seem to be stuck in this method since I am at a point where $G(z) = \frac{z}{1-z-z^2-z^3}$ and I do not know how to proceed
|
The sequence
\begin{align*}
\left(F_n\right)_{n\geq 0}=(0,0,1,1,2,4,7,13,24,44,81,\ldots)
\end{align*}
has the corresponding generating function
\begin{align*}
G(z)=z^2+z^3+2z^4+4z^5+7z^6+\cdots
\end{align*}
Since the series expansion of $G(z)$ at $z=0$ starts with $z^2$ and the geometric series $\frac{1}{1-y}=1+y+y^2+\cdots$ starts with the constant term $1$ we use $z^2$ in the numerator of OPs generating function
\begin{align*}
G(z)&=\frac{z^2}{1-\left(z+z^2+z^3\right)}\tag{1}
\end{align*}
In the following it is convenient to use the coefficient of operator $[z^n]$ to denote the coefficient of $z^n$ in a series.
We apply the geometric series expansion of (1) and obtain for $n\geq 2$
\begin{align*}
[z^n]G(z)&=[z^n]z^2\sum_{k=0}^\infty \left(z+z^2+z^3\right)^k\\
&=[z^{n-2}]\sum_{k=0}^\infty z^k\left(1+z+z^2\right)^k\tag{2}\\
&=\sum_{k=0}^{n-2}[z^{n-2-k}]\left(1+z+z^2\right)^k\tag{3}\\
&=\sum_{k=0}^{n-2}[z^{n-2-k}]\sum_{l=0}^k\binom{k}{l}z^l\left(1+z\right)^l\tag{4}\\
&=\sum_{k=0}^{n-2}\sum_{l=0}^{n-2-k}[z^{n-2-k-l}]\binom{k}{l}\sum_{j=0}^l\binom{l}{j}z^j\\
&=\sum_{k=0}^{n-2}\sum_{l=0}^{n-2-k}\binom{k}{l}\binom{l}{n-2-k-l}\tag{5}\\
&=\sum_{k=0}^{n-2}\sum_{l=0}^{n-2-k}\binom{k}{n-2-k-l}\binom{t-2-k-l}{l}\tag{6}\\
&=\sum_{k=0}^{n-2}\sum_{l=0}^{k}\binom{n-2-k}{k-l}\binom{k-l}{l}\tag{7}\\
\end{align*}
Comment:
*
*In (2) we use the rule $[z^p]z^qA(z)=[z^{p-q}]A(z)$.
*In (3) we use the linearity of the coefficient of operator and use the rule from (2) again. Since the powers of $z$ are non-negative we restrict the upper limit of the left-most sum with $n-2$.
*In (4) we apply the binomial theorem and continue here and in the next line similarly as before.
*In (5) we select the coefficient of $z^{n-2-k-l}$. Next we transform the double sum to simplify the representation somewhat.
*In (6) we change the summation order of the inner sum by replacing the index $l$ with $n-2-k-l$.
*In (7) we change the summation order of the outer sum by replacing the index $k$ with $n-2-k$.
We conclude, the coefficients $F_n$ are
\begin{align*}
&F_0=F_1=0\\
&F_2=1\\
&F_n=\sum_{k=0}^{n-2}\sum_{l=0}^{k}\binom{n-2-k}{k-l}\binom{k-l}{l}\qquad\qquad n\geq 3
\end{align*}
Note: The coefficients $F_n$ are called Tribonacci numbers and are archived as OEIS/A000073.
|
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|
find the maximum and minimum value of the function $x^3+y^3-3x-12y+10$ find the maximum and minimum value of the function $x^3+y^3-3x-12y+10$? here the question is of $2$ variables and i am not able to solve that ,i know how to solve maximum and minimum question of 1 variable,so please help me to solve this
|
For positive variables by AM-GM we obtain:
$$x^3+y^3-3x-12y+10=x^3+1+1+y^3+8+8-3x-12y-8\geq$$
$$\geq3\sqrt[3]{x^3\cdot1\cdot1}+3\sqrt[3]{y^3\cdot8\cdot8}-3x-12y-6=-8.$$
The equality occurs for $x=1$ and $y=2$, which says that
$$\min_{x>0,y>0}(x^3+y^3-3x-12y+10)=-8$$
|
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|
if : $abc=8 $ then : $(a+1)(b+1)(c+1)≥27$ if : $$abc=8 :a,b,c\in \mathbb{R}_{> 0}$$
then :
$$(a+1)(b+1)(c+1)\ge 27.$$
My try :
$$(a+1)(b+1)(c+1)=1+(a+b+c)+(ab+ac+bc)+abc$$
$$(a+1)(b+1)(c+1)=1+(a+b+c)+(ab+ac+bc)+8, $$
then?
|
Using $\bf{A.M\geq G.M}$
$$\frac{a}{2}+\frac{a}{2}+1\geq 3\left(\frac{a^2}{4}\right)^{\frac{1}{3}}$$
Similarly $$\frac{b}{2}+\frac{b}{2}+1\geq 3\left(\frac{b^2}{4}\right)^{\frac{1}{3}}$$
$$\frac{c}{2}+\frac{c}{2}+1\geq 3\left(\frac{c^2}{4}\right)^{\frac{1}{3}}$$
So $$(a+1)(b+1)(c+1)\geq 27 \left(\frac{(abc)^2}{64}\right)^{\frac{1}{3}} = 27$$
equality hold when $\displaystyle a= b=c = 2$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Converting a power series to a function
$$\sum (n+1)^2x^n = {1+x\over (1-x)^3}$$
I tried to find a similar power series and start from there. I see $ x^n $ and it reminds me of geometric series, but $(n+1)^2$ confuses me. How do I prove this?
|
First use the fact that
$$\sum_{n=0}^\infty x^{n+1} = {x\over 1-x}$$
now take the derivative to get
$$\sum_{n=0}^\infty (n+1)x^n = {1\over (1-x)^2}$$
Now multiply by $x$ and take the derivative again:
$$\sum_{n=0}^\infty (n+1)^2x^n = {d\over dx}\left({x\over (1-x)^2}\right) = {(1-x)^2+2x(1-x)\over (1-x)^3}$$
$$={1-x^2\over (1-x)^4}={1+x\over (1-x)^3}$$
Another way: Note that $\displaystyle n^2= {n\choose 2}+{n+1\choose 2}$ is a sum of consecutive triangular numbers and that if $\displaystyle f(x) = \sum a_nx^n$ then $\displaystyle {f(x)\over 1-x}=\sum_k\left(\sum_{m\le k} a_m\right)x^k$. Applying this twice to $\displaystyle f(x) = {1\over 1-x}$ we see
$$G(x) = \sum_n {n+2\choose 2}x^n = {1\over (1-x)^3}$$
Which leads us to
$${1+x\over (1-x)^3}= G(x) + xG(x) = \sum {n+2\choose 2}x^n+{n+2\choose 2}x^{n+1}$$
$$=\sum\left({n+2\choose 2} + {n+1\choose 2}\right)x^n=\sum (n+1)^2x^n$$
|
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|
Is it possible to evaluate $\lim_{x\rightarrow 0}(-1+\cos x)^{\tan x}$? Let's say we have the following limit:
$$\lim_{x\rightarrow 0}(-1+\cos x)^{\tan x}$$
Would the following solution be correct?
The solution is incorrect, please see the correction of @YvesDaoust
\begin{align}
\lim_{x \rightarrow 0}(-1+\cos x)^{\tan x}
&= \lim_{x \rightarrow 0}\left((1-\cos x)^{\tan x}\cdot (-1)^{\tan x}\right) \\
&= \lim_{x \rightarrow 0}\left(1-\left(1-2\sin^2\left(\frac{x}{2}\right)\right)^{\tan x}\right) \cdot \lim_{x \rightarrow 0}(-1)^{\tan x} \\
&= \lim_{x \rightarrow 0}\left(2\sin^2 \left(\frac{x}{2}\right)\right)^{\tan x} \cdot 1\\
&= \lim_{x \rightarrow 0}2^{\tan x} \cdot \lim_{x \rightarrow 0}\sin \left(\frac{x}{2}\right)^{2\tan x} \\
&= 1 \cdot \lim_{x \rightarrow 0}\sin \left(\frac{x}{2}\right)^{2\tan x} \\
&= \lim_{x \rightarrow 0}\sin \left(\frac{x}{2}\right)^{2\frac{\sin x}{\cos x}} \\
&= \lim_{x \rightarrow 0}\sin \left(\frac{x}{2}\right)^{\frac{4\sin (\frac{x}{2})\cdot \cos (\frac{x}{2})}{\cos x}} \\
&= \lim_{x \rightarrow 0}\left(\sin \left(\frac{x}{2}\right)^{\sin \left(\frac{x}{2}\right)}\right)^{\frac{4\cos\frac{x}{2}}{\cos x}} \\
&= \left(\lim_{x \rightarrow 0}\sin \left(\frac{x}{2}\right)^{\sin \left(\frac{x}{2}\right)}\right)^{\lim_{x \rightarrow 0}\frac{4\cos\frac{x}{2}}{\cos x}} \\
&= \left(\lim_{u \rightarrow 0}u^u\right)^{4} \\
&= 1^4 \\
&= 1 \\
\end{align}
The result seems to be correct, but the way leading to it seems to be quite lengthy. Am I doing something redundant?
|
$$\cos x-1<0$$ in the neighborhood of $0$ so that the function cannot be evaluated (at best values closer and closer to $\pm1$ for rational exponents).
Hence the limit does not exist.
|
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|
System of two cubic equations: $x+y^2 = y^3$, $y+x^2=x^3$ I got stuck on this system of equations. Could you help and tell me how should I approach this problem?
\begin{align*}
x+y^2 &= y^3\\
y+x^2 &= x^3
\end{align*}
These are the solutions:
\begin{align*}
(0, 0); \\((1+\sqrt{5})/2, (1+\sqrt{5})/2); \\((1-\sqrt{5})/2, (1-\sqrt{5})/2);
\end{align*}
|
Here is a dynamical systems approach assuming that we only look for real solutions. Consider the map: $$f(x)=x^3-x^2=x^2(x-1)$$
Solving $f(x)=y$ and $f(y)=x$ is equivalent to solving $f(f(x))=x$. We want to show that a solution of the latter (a periodic orbit of period 2) is necessarily a fixed point, i.e. $f(x)=x$ which is trivially solvable.
So assume that $x=f(y), y=f(x)$. We may assume that $x\leq y$.
Now $f$ is strictly monotone increasing on $[1,+\infty)$ so if $1\leq x\leq y$ then $y=f(x)\leq x=f(y)$ implying $x=y$.
Similarly, as $f$ is strictly monotone on $(-\infty,0]$ then $x\leq y\leq 0$ implies $x=y$.
Finally, $f$ maps $(-\infty,1]$ into $(-\infty,0]$ so $x\in (0,1]$ implies $x=f(f(x))\leq 0$ (contradiction). We conclude that $x=y$ as wanted.
Remark: Incidently, the same argument shows that any real orbit of period $p$ must also be a fixed point. So e.g. $x+y^2=y^3$, $y+z^2=z^3$, $z+x^2=x^3$ has the same (real) solutions.
|
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Given positive integers $ x , y$ and that $3x^2 + x = 4y^2 +y$ Prove that $x-y$ is a perfect square This is some old Olympiad problem (I vaguely remember its Iranian )
I factorised it as
$ (x-y)(3x + 3y +1) = y^2 $
Then I fruitlessly tried to prove
$\gcd(x-y, 3x + 3y +1 ) = \gcd (6x+1,6y+1) = 1$
for proving that $x-y$ is a perfect square.
How do I continue my method or are there other ways to approach this intuitively?
|
Here is the method to approach this sort of problem. Note that $x>y$. Let $\gcd(x,y)=d$. Now, first note that $$x-y=4y^2-3x^2 \equiv 0 \pmod{d^2} \tag{1}$$
Now, $$4y^2-3x^2=4(y-x)(y+x)+x^2 \equiv 0 \pmod {x-y} $$
And $$4y^2-3x^2=3(y-x)(y+x)+y^2 \equiv 0 \pmod {x-y} $$
This gives us that $x-y$ is a common factor of $x^2$ and $y^2$, so it must divide their greatest common divisor, $d^2$. So we have $$d^2 \equiv 0 \pmod{x-y}$$
So $d^2$ and $x-y$ both divide each other. Since $x>y$, we have that $x-y=d^2$. Our proof is done.
|
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|
Find the integral of $\int \frac{x^4}{x-1}$ I was not happy with the responses so I went ahead and solved the integral to get feedback on the answer:
Dividing as the book shows us:
we get $x^4$ divided by $x-1$ which gives us the quotient plus the remainder over the divisor which is: $x^3 + \frac{x^3}{x-1}$
$\int x^3+\frac{x^3}{x-1} \to \frac{1}{4}x+x^3ln(\vert x-1 \vert)$
Instead of using shortcuts I want feedback on how it should be done with the process in the book.
|
Substitute $\text{u}=x-1$:
$$\int\frac{x^4}{x-1}\space\text{d}x=\int\frac{\left(1+\text{u}\right)^4}{\text{u}}\space\text{d}\text{u}=\int\left(\text{u}^3+4\text{u}^2+6\text{u}+\frac{1}{\text{u}}+4\right)\space\text{d}\text{u}=$$
$$\int\text{u}^3\space\text{d}\text{u}+4\int\text{u}^2\space\text{d}\text{u}+6\int\text{u}\space\text{d}\text{u}+\int\frac{1}{\text{u}}\space\text{d}\text{u}+4\int1\space\text{d}\text{u}=$$
$$\frac{\text{u}^4}{4}+4\cdot\frac{\text{u}^3}{3}+6\cdot\frac{\text{u}^2}{2}+\ln\left|\text{u}\right|+4\text{u}+\text{C}=$$
$$\frac{\left(x-1\right)^4}{4}+\frac{4\left(x-1\right)^3}{3}+3\left(x-1\right)^2+\ln\left|x-1\right|+4\left(x-1\right)+\text{C}$$
|
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$ \lim_{n \rightarrow \infty} n \left( \frac{1}{(n+1)^2} + \frac{1}{(n+2)^2} + \cdots + \frac{1}{(2n)^2} \right)$ as Riemann sum? I am trying to evaluate the limit $$
\lim_{n \rightarrow \infty} n \left( \frac{1}{(n+1)^2} + \frac{1}{(n+2)^2} + \cdots +
\frac{1}{(2n)^2} \right).$$
I have been trying to convert it to the Riemann sum of some integral, but have been unable to recongize what the integral should be. How should I go about solving this problem?
|
The polygamma functions are great to solve this kind of problems.
$$\sum_{k=1}^N \frac{1}{(x+k)^2}=\psi^{(1)}(x+1)-\psi^{(1)}(N+x+1)$$
$\psi^{(1)}(z)$ is the trigamma function, i.e.: the polygamma[1,z] function.
With $z=n$ and $N=n$ :
$$n\sum_{k=1}^n \frac{1}{(n+k)^2}=n\left(\psi^{(1)}(n+1)-\psi^{(1)}(2n+1) \right)$$
The asymptotic expansion of the trigamma function is : $\psi^{(1)}(z+1)=\frac{1}{z}-\frac{2}{z^2}+O\left(\frac{1}{z^3}\right)$
$$n\sum_{k=1}^n \frac{1}{(n+k)^2}=n\left(\frac{1}{n}-\frac{2}{n^2}-\frac{1}{2n}+\frac{2}{4n^2}+O\left(\frac{1}{n^3}\right) \right) = \frac{1}{2}+\frac{3}{2n}+O\left(\frac{1}{n^2}\right)$$
$$\lim_{n \rightarrow \infty} n\sum_{k=1}^n \frac{1}{(n+k)^2}=\frac{1}{2}$$
|
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How to prove that a Möbius image is a circle? Let $\mu(z) = \frac{z-ia}{z-ib}$, $z=x+iy$, where $x,y, a, b$ are real, with $a<b$. Define a line $l = x+ia$. We can see that the image of this line under $\mu$ is not a line:
$$\mu(l)=\frac{x}{x+i(a-b)}.$$
We also know that Möbius transformations map lines to lines or circles in $\mathbb{C}$. But the problem is - I can see that $\mu(l)$ is not a line, but can't see how it is a circle.
Here's what I do:
-> Take the modulus^2 of the equation above to get
$$ \mu^2 = \frac{x^2}{ x^2+(a-b)^2 }.$$
-> After some algebra I get
$$ (x\mu)^2 + (a-b)^2\mu^2 - x^2 =0 $$
But is this a circle? Actually, WolframAlpha doesn't think so. What is it that I'm not doing right?
|
The circle is formed by $\Re(\mu(x))$ and $\Im(\mu(x))$, with $x$ acting as a parameter. To avoid confusion I will call the parameter $t$, and use $c = a - b$.
Obtain real and imaginary parts as $$\frac{t}{t - ic} = \frac{t^2 + ict}{t^2 + c^2} = \frac{t^2}{t^2+c^2} + i\frac{ct}{t^2+c^2}$$ Now let $x = \frac{t^2}{t^2+c^2}$ and $y = \frac{ct}{t^2+c^2}$. Exclude $t$ to obtain and equation on $x, y$:
$xt^2 = t^2 + c^2$, so $t^2 = \frac{c^2}{1 - x}$ (and $t^2 + c^2 = xt^2 = \frac{c^2x}{1 - x}$).
Therefore, $y^2 = \frac{c^2t^2}{(t^2 + c^2)^2} = c^2t^2 (\frac{1-x}{c^2})^2 = c^2 \frac{c^2}{1-x} \frac{(1-x)^2}{c^4} = x(1-x)$, or
$$(x - \frac{1}{2})^2 + y^2 = \frac{1}{4}$$
|
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|
Proving $\text {limsup}_{x\to \infty}x\int_{x}^{x+1}\sin (t^2) \mathrm{d}t=1$ It is not difficult through the substitution $u=t^2$ to show that $$x\int_{x}^{x+1}\sin (t^2) \mathrm{d}t=\frac 12 \left(\cos(x^2)-\cos[(x+1)^2]\right)+r(x)$$
where $\lim_{x \to \infty} r(x) =0$, it seems intuitive that $\frac 12 \left(\cos(x^2)-\cos[(x+1)^2]\right)$ should always be able to reach $1$ no matter how far down along the $x$ axis we are. The solutions manual says that $\frac 12 \left(\cos(x^2)-\cos[(x+1)^2]\right)=\sin(s^2+\frac 14)\sin(s)$ where $s=x+\frac 12$ and then gives the following explanation
"Choose any integer $n >\frac {2-\epsilon}{8\epsilon}$ then the interval $I=\left(\frac 14 + \left(\left(2n +\frac 12 \right)\pi -\epsilon \right)^2,\frac 14 + \left(\left(2n +\frac 12 \right)\pi +\epsilon \right)^2\right)$ is longer than $2\pi$ and thus there is a $t\in I$ such that $\sin(t^2+\frac 14)=1$. But then $tf(t)>1-\epsilon$ (where $f(t)=\int_{t}^{t+1}\sin (u^2) \mathrm{d}u$) it follows that $\text {limsup}_{x\to \infty}x\int_{x}^{x+1}\sin (t^2) \mathrm{d}t=1$."
I understand why, given that $tf(t)>1-\epsilon$ for all $\epsilon$ would imply the statement, but what I don't get is why did the manual forget about the $\sin (t)$ that had come in the simplification. It is possible that $\sin (t)<1-\epsilon$ and then the argument does not work. I also understand the strategy of the proof, building a sequence of $t$'s that converge to $1$, but I cannot figure out where did the bounds for the interval $I$ come from. Is there no simpler, even more general strategy for proving this (or related) proposition?
|
Your initial idea of enforcing the substitution $t\to \sqrt{t}$ is a good one. Indeed we have
$$\int_x^{x+1} \sin(t^2)\,dt=\int_{x^2}^{(x+1)^2}\frac{\sin(t)}{2\sqrt{t}}\,dt$$
Now, let's integrate by parts with $u=\frac{1}{2\sqrt{t}}$ and $v=-\cos(t)$. Then, we have
$$\begin{align}
\int_x^{x+1} \sin(t^2)\,dt&=\frac{\cos(x^2)}{2x}-\frac{\cos((x+1)^2)}{2(x+1)}-\frac14 \int_{x^2}^{(x+1)^2}\frac{\cos(t)}{t^{3/2}}\,dt\\\\
&=\frac{1}{2x}\left(\cos(x^2)-\cos((x+1)^2)\right)+\underbrace{\frac{\cos((x+1)^2)}{x(x+1)}-\frac14 \int_{x^2}^{(x+1)^2}\frac{\cos(t)}{t^{3/2}}\,dt}_{=O\left(\frac1{x^2}\right)}\\\\
&=\frac1x \sin\left(\frac{(x+1)^2+x^2}{2}\right)\sin\left(\frac{(x+1)^2-x^2}{2}\right)+O\left(\frac1{x^2}\right)\\\\
&=\frac1x \sin\left((x+1/2)^2+1/4\right)\sin\left(x+1/2\right)+O\left(\frac1{x^2}\right)\\\\
\end{align}$$
The term $\displaystyle \sin(x+1/2)$
Now, let $1>\epsilon>0$ be given. Suppose that $x+1/2=(2n+1/2)\pi+\nu$, where $\nu\in [-\epsilon,\epsilon]$.
Then, $\sin(x+1/2)=\cos(\nu)$ and $\cos(\epsilon)\le \sin(x+1/2)\le 1$.
The term $\displaystyle \sin((x+1/2)^2+1/4)$
We now examine the argument $(x+1/2)^2+1/4$.
Note that for $x\in [(2n+1/2)\pi-\epsilon,(2n+1/2)\pi+\epsilon]$, the argument $(x+1/2)^2+1/4$ satisfies the inequalities
$$((2n+1/2)\pi -\epsilon)^2+1/4\le (x+1/2)^2+1/4\le ((2n+1/2)\pi -\epsilon)^2+1/4$$
If we select $n$ large enough such that the length of the interval $[((2n+1/2)\pi -\epsilon)^2+1/4,((2n+1/2)\pi +\epsilon)^2+1/4]$ is at least $2\pi$, then we can select $x\in [(2n+1/2)\pi - \epsilon,(2n+1/2)\pi+\epsilon]$ such that $\sin((x+1/2)^2+1/4)=1$.
Hence, we require that
$$((2n+1/2)\pi+\epsilon)^2+1/4-((2n+1/2)\pi-\epsilon)^2-1/4\ge 2\pi$$
This inequality is satisfied if $n\ge \frac{1-\epsilon}{4\epsilon}$. In that case, there exists an $x$ such that $\sin((x+1/2)^2+1/4)=1$.
Finally, for such an $x$ we have
$$x\int_x^{x+1}\sin(t^2)\,dt=\sin(x+1/2)+O\left(\frac1x\right)$$
Recall that $\cos(\nu)\le \sin(x+1/2)\le 1$, where $\nu\in[-\epsilon,\epsilon]$. Since $\epsilon>0$ is arbitrary, then we have
$$\limsup_{x\to \infty}x\int_x^{x+1}\sin(t^2)\,dt=1$$
as was to be shown!
|
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|
If $(x-a)^2+y^2=r^2$, prove that $r=\sqrt{(x_1-a)^2+y_1^2}=\sqrt{(x_2-a)^2+y_2^2}$ and $a=\frac{x_2^2+y_2^2-x_1^2-y_1^2}{2(x_2-x_1)}$. Consider this is taking place in the hyperbolic half plane model
If $x_1 \not= x_2$ then the line $\overleftrightarrow{AB}$ has the following equation $(x-a)^2+y^2=r^2$. Prove that $r=\sqrt{(x_1-a)^2+y_1^2}=\sqrt{(x_2-a)^2+y_2^2}$ and $$a=\frac{x_2^2+y_2^2-x_1^2-y_1^2}{2(x_2-x_1)}$$
Now I understand that $r$ comes from taking the square root of the equation and recognizing the radius is the same from both points. Now I'm not quite sure how to arrive at $a$ which seems to be the center of the half circle.
|
You have:
$$\sqrt{(x_1-a)^2+y_1^2}=\sqrt{(x_2-a)^2+y_2^2}$$
Squaring both the sides we get:
$$(x_1-a)^2+y_1^2=(x_2-a)^2+y_2^2$$
$$(x_1-a)^2-(x_2-a)^2=y_2^2-y_1^2$$
$$(x_1-a-x_2+a)(x_1-a+x_2-a)=y_2^2-y_1^2$$
$$(x_1-x_2)(x_1+x_2-2a)=y_2^2-y_1^2$$
$$x_1^2-x_2^2-2a(x_1-x_2)=y_2^2-y_1^2$$
$$2a(x_2-x_1)=x_2^2+y_2^2-x_1^2-y_1^2$$
$$a={x_2^2+y_2^2-x_1^2-y_1^2\over2(x_2-x_1)}$$
|
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|
Proving trigonometrical identity from given identity.
Given $$\dfrac{\sin^4 A}{a}+\dfrac{\cos^4 A}{b}=\dfrac{1}{a+b}$$ prove that:$$\dfrac{\sin^8 A}{a^3}+\dfrac{\cos^8 A}{b^3}=\dfrac{1}{(a+b)^3}$$
Using given, I proved: $b\sin^2 A=a\cos^2 A$ Help me proceeding after this.
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$\dfrac{\sin^4A}{a}+\dfrac{\left(1-sin^2A\right)^2}{b}-\dfrac{1}{a+b}=0$
$\sin^4A\left(\dfrac{1}{a}+\dfrac{1}{b}\right)-\dfrac{2\sin^2A}{b}+\dfrac{1}{b}-\dfrac{1}{a+b}=0$
$\left(\sin^2A-\dfrac{a}{a+b}\right)^2=0$
$\sin^2A=\dfrac{a}{a+b}$
$\cos^2A=\dfrac{b}{a+b}$
You can take it up from here.
|
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|
Fourier Coefficients for $\frac{3}{5-4\cos{2\theta}}$ I want to compute the (even) fourier coefficients for
$\dfrac 3 {5-4\cos 2\theta}$ on the interval $[0, 2\pi]$.
Namely, I want to compute the integral:
$$b_n = \int_0^{2\pi} \cos(n\theta) \frac 3 {5-4\cos 2\theta } \frac{d\theta}{2\pi}$$
Integrating this in mathematica, it seems like $b_n = 0$ if $n$ is odd and $b_n = \left(\frac 1 2 \right)^{n/2}$ for $n$ even.
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Complex analysis approach:
Let $$f(z)=\frac{3}{5-2(z^2+z^{-2})}=\frac{-3z^2}{2z^4-5z^2+2}=\frac{1}{2z^2-1}-\frac{2}{z^2-2}$$ Then $f(e^{i\theta})=\frac{3}{5-4\cos 2\theta}$.
Now, for $|z|<\sqrt{2}$, we have:
$$\frac{-2}{z^2-2}=\frac{1}{1-z^2/2}=\sum_{k=0}^{\infty} \frac{1}{2^k}z^{2k}$$
and for $|z|>\frac{1}{\sqrt{2}}$ you have:
$$\frac{1}{2z^2-1}=\frac{1}{2z^2}\frac{1}{1-\frac{1}{2z^2}}=\sum_{k=1}^{\infty}\frac{1}{2^k}z^{-2k}$$
Since $z=e^{i\theta}$ is in this range, this means that:
$$\frac{3}{5-4\cos 2\theta} = \sum_{k=\infty}^\infty \frac{1}{2^{|k|}}e^{i2k\theta}$$
Combining $k$ and $-k$ terms, this gives:
$$\frac{3}{5-4\cos2\theta} = 1 + \sum_{k=1}^{\infty} \frac{1}{2^{k-1}}\cos 2k\theta$$
I might have missed something in this computation, but this is close to your answer - the coefficients are $0$ if $n$ is odd and $\frac{1}{2^{k-1}}$ if $n=2k$.
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Find $A^n$ for all $n \in \mathbb{N}$. I'm learning linear algebra and need help with the following problem:
Let $A = \begin{pmatrix}-2 & -3 & -3\\-1 & 0 & -1\\5 & 5 & 6\end{pmatrix} \in M_{3x3}(\mathbb{R}).$ Find $A^n$ for all $n \in \mathbb{N}$.
My first thought was to compute $A^2$, $A^3$, $A^4$ and see if a pattern emerge. I used Mathematica to compute the power of $A$ to save me some time. The computations gave
$A^2 = \begin{pmatrix}-8 & -9 & -9\\-3 & -2 & -3\\15 & 15 & 16\end{pmatrix}, \quad A^3 = \begin{pmatrix}-20 & -21 & -21\\-7 & -6 & -7\\35 & 35 & 36\end{pmatrix}, \quad A^4 = \begin{pmatrix}-44 & -45 & -45\\-15 & -14 & -15\\75 & 75 & 76\end{pmatrix}$.
The above results are clearly showing a pattern relating the first four powers of $A$, for instance $(a_{31}) = (a_{32}) = (a_{33}) + 1$, but I'm unable to write the general form of a matrix $A^{k}$.
$(Q1)$ How do I find/write the general form of a matrix $A^{k}$?
$(Q2)$ I only compute the first four powers of $A$. How do I know that this pattern won't fail for some integer $k$?
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The matrix is similar to the diagonal matrix $D=diag(1,1,2)$. Hence we have
$$
A^n=(SDS^{-1})^n=SD^nS^{-1}
$$
with
$$
S=\begin{pmatrix} -5 & -4 & 3\cr -1 & -1 & 1\cr 6 & 5 & -5\end{pmatrix}.
$$
|
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|
if $z=2+4i$, find $\sqrt{z}$ when $\sqrt{z}=a+bi$ Question
if $z=2+4i$, find $\sqrt{z}$ when $\sqrt{z}=a+bi$
what I have so far
$\sqrt{z}=a+bi$, so I square both sides ->
$z=a^2+2abi-b^2$
we can substitute for z
$z=2+4i=a^2+2abi-b^2$
that means $a^2-b^2=2$ and $2abi=4i$
and ab=2 so then $b=\frac{a}{2}$
from then i dont know how to proceed from there
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You have
$$a^2-b^2=2\quad (1)\\
ab=2\to b=\frac{2}{a}\quad (2)$$
Plug $(2)$ in $(1)$
$$a^2-\frac{4}{a^2}=2\to a^4-2a^2-4=0$$
so
$$a^2=\frac{2\pm\sqrt{20}}{2}=1\pm\sqrt{5}$$
so,
$$a=\pm\sqrt{1+\sqrt{5}}\to b=\pm\frac{2}{\sqrt{1+\sqrt{5}}}=\pm \frac{\sqrt{\sqrt{5}-1}}{2}$$
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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...
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$U$ is the set of vectors $\vec x$, whose image under $f_A$ is $\vec 0$.
Since your matrix $A$ is rank 4 and $f_A:K^5 \rightarrow K^4$, it must have a $1$ dimensional Kernel.
So your answer should be subspace of $K^5$ spanned by 1 vector. What vector could that be from your reduced matrix?
|
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How many numbers smaller than $10^6$ contain exactly three '$9$'s and have an odd sum of digits? i came up with an idea that i choose
Even Even Even 9 9 9 - and i sort it in ways such that
all even are the same - $\frac{6!}{3!*3!}$
two are the same - $\frac{6!}{3!*2!}$
all are different - $\frac{6!}{3!}$
other one is that
Odd Odd Even 9 9 9
odd are the same - $\frac {6!}{3!*2!}$
all three are differnt - $\frac{6!} {3!}$
and my result is a sum of all this options.
Am i right?
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Assuming leading zeros allow us to assume all numbers have six digits you need.
A) 1 non-nine odd, two different evens, 3 nines
plus
B) 1 (non-nine odd), two same evens, 3 nines
plus
C) 3 non-nine odds, all different, 3 nines
plus
D) 3 non-nine odds, two the same, 3 nines
E) 3 non-nine odds, all the same, 3 nines.
There are $4$ odds to choose from and $5$ evens.
A) is $4*5*4*{3\choose 1}{6 \choose 3}$. That is, $4$ choices for the non-nine odd, $5$ for the first even, $4$ for the second even. There are ${3\choose 1}$ ways to arrange the odd digit among the two even digits. And there are ${6\choose 3}$ ways to place the three nines among six digits.
B) is $4*5{3\choose 1}{6 \choose 3}$. That is, $4$ choices for the first non-nine odd, $5$ choices for the two evens. There are ${3\choose 1}$ ways to arrange the odd digit among the two even digits. And there are ${6\choose 3}$ ways to place the three nines among six digits.
C) is $4*3*2{6\choose 3}$
D) is $4*3{3 \choose 2}{6\choose 3}/2$
E) is $4*{6 \choose 3}$
There is an alternative.
There is ${6 \choose 3}$ ways to arrange three $9$s is six positions. There are $9^3$ possible options for the remaining three digits. $5^3$ of them are all evens (as there are $5$ choices of non-nine evens). $4^3$ are all odd (as there are $4$ choices of non-nine odds). $4*5^2{3\choose 1}$ of those have $1$ odd and $2$ evens. $4^2*5{3\choose 1}$ of those have $2$ odds and $1$ even.
Detour: Verify the $9^3 = 5^3 + 5^2*4*{3\choose 1} + 5*4^2*{3\choose 1} + 4^3$
$9^3 = (5+4)^3 = 5^3 + 5^2*4*{3\choose 1} + 5*4^2*{3\choose 1} + 4^3$ is just the binomial theorem. Nothing to verify.
End of detour.
Of those $5^3$ (three evens) and $5*4^2*3$ (two odds, one even) have an even sum of digits. So when three $9$s are added the sum will be odd.
So there are ${6 \choose 3}(5^3 + 5*4^2*3)$ six digit numbers with exactly three $9$s and an even sum of digits.
|
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|
Number of ways to get $xy^2$ from $(x+y+5)^5$ I am trying to understand a proof of counting number of trees using the conclusion of Cayley's formula.
At the end they got a $(x+y+5)^5$ and they say that the number of ways to get $xy^2$ from this is $5 \cdot \binom{4}{2} \cdot 5^2 = 750 $ and I can't figure how. Can someone please explain?
thanks.
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$$(x+y+z)^n \to \dfrac{n!}{k_1!k_2!k_3!}x^{k_1}y^{k_2}z^{k_3} ,k_1+k_2+k_3=n\\$$
$$(x+y+5)^5 \to \dfrac{5!}{k_1!k_2!k_3!}x^{k_1}y^{k_2}5^{k_3} ,k_1+k_2+k_3=5\\ \to xy^2 \to ,1+2+k_3=5 \to k_3=2\\ \dfrac{5!}{1!2!2!}x^{1}y^{2}5^{2} \\25\dfrac{5!}{1!2!2!}x^{1}y^{2}=25\times 30 xy^2$$
|
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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.
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$$a^3+b^3+c^3-3abc=(a+b)^3-3ab(a+b)+c^3-3abc=\{(a+b)+c\}\{(a+b)^2-(a+b)c+c^2\}-3ab\{(a+b)+c\}$$
$$=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)$$
But $2(a^2+b^2+c^2-ab-bc-ca)=(a-b)^2+(b-c)^2+(c-a)^2$
Now if $a^3+b^3+c^3-3abc=0$
either $a+b+c=0$
But with $A+B+C=m\pi,\tan A+\tan B+\tan C=\tan A\tan B\tan C$
As $0<A,B,C<\pi; \tan A\tan B\tan C\ne0$
So, we need $(\tan A-\tan B)^2+(\tan B-\tan C)^2+(\tan C-\tan A)^2=0\ \ \ \ (1)$
As $\tan A,\tan B,\tan C$ are real, each of $(\tan A-\tan B)^2,(\tan B-\tan C)^2,(\tan C-\tan A)^2$ must be $\ge0$
$(1)\implies$ $$(\tan A-\tan B)^2=(\tan B-\tan C)^2=(\tan C-\tan A)^2=0$$
Observation : The proposition will hold if in the condition, $\tan$ is replaced with $\sin,\cos$
Can you prove them?
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|
Solve the differential equation $2r(1+s^2)dr + (1+r^4)ds = 0$ $2r(1+s^2)dr + (1+r^4)ds = 0$
$\Rightarrow \frac {2r}{1+r^4}dr + \frac {ds}{1+s^2} = 0$ $(1)$
Integrate both sides:
For integration of $\frac {2r}{1+r^4}dr$ $(2)$
I put $1+r^4=t\Rightarrow 4r^3dr=dt\Rightarrow 2rdr = \frac {dt}{2r^2} = \frac {dt}{2\sqrt{t-1}}$
Hence $(2)$ becomes:
*
*Integration of $\frac {dt}{2\sqrt{t-1}}$ = $\sqrt{t-1} = \sqrt{r^4} = r^2$
*Integration of $\frac {ds}{1+s^2} = arctan(s)$
Hence (1) becomes
$$r^2 + arctan(s) = c$$
But at the back, the answer written is $r^2 + s = c(1-r^2s)$
Can someone please tell me what did I do wrong.
|
In addition to @Did's hint, there is a relatively easy way to handle that integral.
We have $$ \frac {2r}{1+ r^4} \mathrm {d}r = - \frac {1}{1+ s^2 } \mathrm {d}s $$ $$\Rightarrow \int \frac {\mathrm {d}(r^2)}{1+(r^2)^2} = \int - \frac {1}{1+ s^2 } \mathrm {d}s$$ $$\Rightarrow \arctan r^2 = - \arctan s +c $$
Hope you can take it from here.
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|
Solving $\sin 2x + \cos 2x =-1$
$$\sin 2x + \cos 2x =-1$$
How would I go about solving this equation? Some hint on how to start, so I can try to figure it out on my own.
Thank you
|
Use sum to product for $\sin x + \sin y$ (it works also for $\cos x + \cos y$):
$$\begin{align}
\sin 2x + \cos 2x&=\sin 2x + \sin\left(\frac{\pi}{2}-2x\right)\\
&=2\sin\frac{2x+\pi/2-2x}{2}\cos\frac{2x-\pi/2+2x}{2}\\
&=2\sin\frac{\pi}{4}\cos\left(2x-\frac{\pi}{4}\right)\\
&=\sqrt 2\cos\left(2x-\frac{\pi}{4}\right)
\end{align}$$
The rest should be easy.
|
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|
Solve the inequality $\sin(x)\sin(3x) > \frac{1}{4}$
Find the range of possible values of $x$ which satisfy the inequation $$\sin(x)\sin(3x) > \frac{1}{4}$$
SOURCE : Inequalities (PDF)( Page Number 6; Question Number 306)
One simple observation is that both $x$ and $3x$ have to positive or negative simultaneously. I tried expanding $\sin(3x)$ by the regular indentity as :
$$\sin(x) \times \big(3\sin(x)-4\sin^3(x)\big) > \frac {1}{4}$$
$$\implies \sin^2(x)\times\big(3-4\sin^2(x)\big) >\frac{1}{4}$$
I do not find any way of proceeding. Wolfram Alpha gives 4 sets of answers. Do I have to observe this problem "case-by-case"? Can this question be solved without calculus ? Can anyone provide a hint to what should be done ?
Thanks in Advance ! :)
|
You are on the correct path.
Now you need to solve the following equation:
$$\sin^2(x)\cdot \big(3-4\sin^2(x)\big) =\frac{1}{4}$$
$$4\sin^2(x)\cdot\big(3-4\sin^2(x)\big)-1=0$$
$$12\sin^2x-16\sin^4x-1=0$$
$$16\sin^4x-12\sin^2x+1=0$$
So we get by solving, $$\sin^2x=\frac{12\pm \sqrt{144-4\cdot 16}}{2\cdot 16}=\frac{12\pm \sqrt{80}}{32}=\frac{6\pm 2\sqrt{5}}{16}=\left(\frac{\sqrt5 \pm 1}{4}\right)^2=(\sin 72^\circ)^2 \text{or} (\sin 18^\circ)^2$$
So the $4$ roots are $x=72^\circ$,$x=72^\circ$,$x=18^\circ$ and $x=18^\circ$.
Now observe that, to check the inequality, you have to check for three regions:
$$x<18^\circ$$ $$18^\circ<x<72^\circ$$ $$x>72^\circ$$
And see which region(s) satisfy the inequality.
|
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|
What is the z-transform of $\frac{1}{n}$ given that $n \geq 1$ and $x(n) = \frac{1}{n}$, how can I get the z-transform of $x(n)$?
Thanks
|
Good Question
$$\frac{u(n-1)}{n} \rightleftharpoons ??$$
We know that Z transform is defined as :
$$X(z) = \sum_{n=-\infty}^{\infty}x(n)z^{-n}$$
$$u(n) \rightleftharpoons \frac{z}{z-1}$$
By the use of time shifting property
$$x(n-1) \rightleftharpoons z^{-1} X(z)$$
$$u(n-1) \rightleftharpoons \frac{1}{z-1}$$
Differentiate the General formula of X(z) with respect to z
$$\frac{dX(z)}{dz} = -\sum_{n=-\infty}^{\infty}nx(n)z^{-n-1}$$
$$\frac{zdX(z)}{dz} = -\sum_{n=-\infty}^{\infty}nx(n)z^{-n}$$
$$nx(n) \rightleftharpoons -\frac{zdX(z)}{dz}$$
Now we want to find out the Z transform of $$\frac{u(n-1)}{n}$$
$$x(n) = \frac{u(n-1)}{n} \rightleftharpoons X(z)$$
$$n x(n) \rightleftharpoons -\frac{zdX(z)}{dz}$$
$$n [\frac{u(n-1)}{n}] \rightleftharpoons -\frac{zdX(z)}{dz}$$
$$u(n-1) \rightleftharpoons -\frac{zdX(z)}{dz}$$
Means
$$-\frac{zdX(z)}{dz} = \frac{1}{z-1}$$
$$\frac{dX(z)}{dz} = -\frac{1}{z(z-1)}$$
$$\frac{dX(z)}{dz} = [\frac{1}{z}] + [\frac{1}{z-1}]$$
$$X(z) = ln(\frac{z}{z-1})$$
Hence
$$\frac{u(n-1)}{n} \rightleftharpoons ln(\frac{z}{z-1})$$
Now we have some interesting series for natural log
$$\frac{1}{z} + \frac{1}{2z^2} + \frac{1}{3z^3} + ...... = ln(\frac{z}{z-1})$$
For z = 2
$$\frac{1}{2} + \frac{1}{2\times2^2} + \frac{1}{3\times2^3} + ...... = ln(2)$$
For $$z = i^2 = -1$$
$$\frac{1}{1} - \frac{1}{2} + \frac{1}{3} + ...... = ln(2)$$
|
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|
Need help understanding certain steps of the solution to a limit. I had trouble evaluating this limit and found this solution from another post of Stack Exchange. The point of this solution was trying to find the limit without using Taylor's series or L'Hospital's rule.
I just need help understanding certain steps:
*
*How did he go from step 3 to step 4.
*How did he go from step 4 to step 5.
His solution:
|
If we have:
$$4L=\lim_{x\to 0} \frac{\frac12\tan 2x-x}{x^3} \qquad L=\lim_{x\to 0} \frac{\tan x-x}{x^3}$$
substracting we would get
$$4L-L=3L=\lim_{x\to 0} \frac{\frac12\tan 2x-\tan x}{x^3}$$
Now, use that $\tan(2x)=\frac{2\tan x}{1-\tan^2 x}$ to go on:
\begin{equation*} \begin{split} 3L&=\lim_{x\to 0} \frac{\frac{\tan x}{1-\tan^2x}-\tan x}{x^3}=\lim_{x\to 0} \frac{\tan x}{x} \cdot \frac{\frac{1}{1-\tan^2 x}-1}{x^2}=\lim_{x\to 0} \frac{\tan x}{x}\cdot \frac{\tan^2 x}{x^2(1-\tan^2 x)}=\\=&\lim_{x\to 0} \frac{\tan x}{x}\cdot \frac{\tan^2 x}{x^2}\cdot\frac{1}{1-\tan^2 x}=\lim_{x\to 0} \frac{\tan^3 x}{x^3} \cdot \lim_{x\to 0} \frac{1}{1-\tan x^2}\end{split} \end{equation*}
Now, as $\lim_{x\to 0} \frac{1}{1-\tan^2 x}=1$, (because $\tan x\to 0$ when $x\to 0$) and $$\lim_{x\to 0} \frac{\tan^3 x}{x^3}=\left(\lim_{x\to 0} \frac{\tan x}{x}\right)^3=1^3=1$$ we get that $3L=1$.
|
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|
Coefficients of products of binomials What is the coefficient of the term $\prod_{k=1}^N{z_k^{N-1}}$ in the expansion of $\prod_{1\leq i<j\leq N}{(z_i+z_j)^2}$?
From the symmetry of the problem all that I can argue is that this coefficient shall be the largest of all the other coefficients.
|
This is more of a comment but slightly too large to post as such. If you are merely interested in how your coefficient grows with $N$, here are two crude bounds:
Let $C_N^*$ denote the coefficient of $\prod_{k=1}^N{z_k^{N-1}}$ in the expansion of $\prod_{1\leq i<j\leq N}{(z_i+z_j)^2}$. First note that
\begin{equation}
C_N^* \geq 2^{\frac{N (N-1)}{2}} \, .
\end{equation}
This can be easily seen by observing that from the expansion of
\begin{equation}
\prod_{1\leq i<j\leq N}{(z_i+z_j)^2} = \prod_{1\leq i<j\leq N}{(z_i^2 + 2 z_i z_j +z_j^2)}
\end{equation}
we can extract
\begin{equation}
\prod_{1\leq i<j\leq N}{2 z_i z_j} = 2^{\frac{N (N-1)}{2}} \prod_{k=1}^N{z_k^{N-1}} \, .
\end{equation}
An upper bound can be obtained by considering the expression $\prod_{1\leq i<j\leq N}{(2 z_i^2 + 2 z_i z_j + 2 z_j^2)}$; certainly any coefficient in the expansion of the preceding product will be larger than its counterpart for the original expression. An upper bound on $C_N^*$ is thus
\begin{equation}
C_N^* < 2^{\frac{N (N-1)}{2}} M_N\, .
\end{equation}
The variable $M_N$ is the amount of how many ways there are to combine the terms in the expansion of $\prod_{1\leq i<j\leq N}{(z_i^2 + z_i z_j + z_j^2)}$ to yield $\prod_{k=1}^N{z_k^{N-1}}$.
For the sake of argument, let us assume that $N$ is odd. An upper bound on $M_N$ can be obtained by noticing that this quantity certainly lies below the amount of ways to obtain any term involving $z_N^{N-1}$. To that end, we can pick $\frac{N-1}{2} - k$ out of $N-1$ terms to provide $z_N^2$. From the remaining $\frac{N-1}{2} - k$ terms involving $z_N$ we pick $2 k$ to provide $z_i z_N$ (for some $i$). This leaves us with $\frac{(N-1)^2}{2} - k$ terms to pick any of at most 3 items not involving $z_N$. Putting the above together we arrive at
\begin{align}
M_N &\leq \sum_{k=0}^{\frac{N-1}{2}} \binom{N-1}{\frac{N-1}{2}-k} \binom{\frac{N-1}{2}+k}{2k} 3^{(N-1)^2-k} \\
&\leq \sum_{k=0}^{\frac{N-1}{2}} 2^{N-1} 2^{\frac{N-1}{2}+k} 3^{(N-1)^2-k} \\
&< \sum_{k=0}^{\frac{N-1}{2}} 2^{N-1} 2^{\frac{N-1}{2}+k} 2^{2 (N-1)^2-2k} \\
&= 2^{\frac{3}{2}(N-1) + 2(N-1)^2} \sum_{k=0}^{\frac{N-1}{2}} 2^{-k} \\
&< 2^{\frac{3}{2}(N-1) + 2(N-1)^2 + 1} \, .
\end{align}
Thus
\begin{equation}
C_N^* < 2^{\frac{N (N-1)}{2} + \frac{3}{2}(N-1) + 2(N-1)^2 + 1} \, .
\end{equation}
The two bounds tell you that
\begin{equation}
\frac{1}{2} \leq \lim_{N \to \infty} \frac{1}{N^2} \log{C_N^*} < 2 + \frac{1}{2}
\end{equation}
|
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|
Is the product of two consecutive integers $1$ $\pmod n$? Is there a simple test for $n$ to determine if there exists an integer $x$ such that $x$($x+1$) $=$ $1$ $\pmod n$. For example, $n$ $=$ $3$ and $n$ $=$ $7$, there are no integers $x$ such that $x$($x+1$) $=$ $1$ $\pmod n$. But for $n$ $=$ $5$ and $n$ $=$ $11$, there are integers $x$ such that $x$($x+1$) $=$ $1$ $\pmod n$. There are namely,
$x = 2$,
$2*3$ $=$ $1$ $\pmod 5$
$x = 7$,
$7*8$ $=$ $1$ $\pmod {11}$
It is confusing to find out which integers $n$ have this property. Maybe there is a special (mathematical) property these numbers have. Help is appreciated. Thanks.
|
First we rewrite the congruence:
$$4x(x+1)=4\pmod{4n}\quad\Leftrightarrow\quad (2x+1)^2=5\pmod{4n}\ .$$
Now write $y=2x+1$ and consider various cases.
*
*If $n$ is even then the above implies $y^2=5\pmod8$, which has no solution.
*If $5^2\mid n$ then we get
$$\eqalign{y^2=5\pmod{25}\quad
&\Rightarrow\quad 5\mid y^2\cr
&\Rightarrow\quad 25\mid y^2\cr
&\Rightarrow\quad 25\mid 5\cr}$$
and again there is no solution.
*So for solutions to exist, $n$ is a product of $5$ (possibly) and prime powers $p^\alpha$ where $p$ is not $2$ or $5$. There is a solution iff
$$y^2=5\pmod{p^\alpha}$$
has a solution for every such prime power, which can be proved to be equivalent to
$$y^2=5\pmod p$$
having a solution for every $p\mid n$ (except $p=2,5$). Using the Legendre symbol one can show that this comes down to the following.
The congruence $x(x+1)=1\pmod n$ has a solution if and only if $n$ is a product of primes in which $5$ occurs only once (or not at all), and every other prime is congruent to $1$ or $4$ modulo $5$.
Here is a table of some low values of $n$. I have listed "ok" if the congruence has a solution, otherwise I have given a "bad" prime factor of $n$.
$$\def\ok{{\rm ok}}
\matrix{3&5&7&9&11&13&15&17&19&21&23&25&27&29&31&\cdots&55\cr
3&\ok&7&3&\ok&13&3&17&\ok&3&23&5&3&\ok&\ok&\cdots&\ok\cr}$$
Note that $p=5$ is "bad" for $n=25$ because it occurs twice, but it is "ok" for $n=5$ and $n=55$ because it only occurs once.
|
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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.
|
First observe that $(a+b)^2=c^4 \implies a+b = c^2 , c>0$ Then we can combine that with $a^2+b^2=c^3$ to get $(a-b)^2=2c^3-c^4$. As $a-b \in N, (2c^3-c^4)^{1/2}= c(2c-c^2)^{1/2} \in N$ Which implies $c = 1,2$ . Checking the two values, $c=0$ has no natural solutions and $c=1$ has one natural solution of $a=2,b=2,c=2$.
If we include zero in the set of natural numbers then for the cases $c=0,1$ we have $(0,1,1)$ and $(1,0,1)$ as solutions. For the case $c=0$ we also have a fourth solution of $(0,0,0)$
|
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|
finding ratio of two definite integration If $\displaystyle A = \int^{1}_{0}x^{\frac{7}{2}}(1-x)^{\frac{5}{2}}dx$ and $\displaystyle B = \int^{1}_{0}\frac{x^{\frac{3}{2}}(1-x)^{\frac{7}{2}}}{(x+3)^8}dx\;,$ then value of $AB^{-1} = $
Attempt: i have tried using gamma function $\displaystyle \int^{1}_{0}x^m(1-x)^ndx = \frac{(m-1)!\cdot (n-1)!}{(m+n-1)!}$
so $\displaystyle A=\int^{1}_{0}x^{\frac{7}{2}}(1-x)^{\frac{5}{2}}dx = \int^{1}_{0}x^{\frac{5}{2}}(1-x)^{\frac{7}{2}}dx = \frac{(\frac{3}{2})!\cdot (\frac{5}{2})!}{5!}$
wan,t be able to go further, could some help me, thanks
|
We first prove a result : $F(a,b,p)=\displaystyle\int_0^1 \frac{x^{a-1}(1-x)^{b -1}}{(x+p)^{a+b}}\,dx = \left(\frac{\beta(a,b)}{p^b(1+p)^a}\right) \\$
Proof :
We start with the integral $\displaystyle \beta(a,b) = \int\limits_0^1 x^{a-1}(1-x)^{b-1}\; dx \\$
The substitution $y = \dfrac{(p+1)x}{p+x}$ is pretty straight forward and gives the result,
$\displaystyle\int_0^1 \frac{x^{a-1}(1-x)^{b -1}}{(x+p)^{a+b}}\,dx = \left(\frac{\beta(a,b)}{p^b(1+p)^a}\right) \\$
Now the second integral is $\displaystyle \int\limits_0^1 \dfrac{x^\frac{3}{2}(1-x)^\frac{7}{2}}{(x+3)^8}\; dx \\ = \dfrac{1}{3}\left(\int\limits_0^1 \dfrac{x^\frac{3}{2}(1-x)^\frac{7}{2}}{(x+3)^8}\;(x+3-x)\; dx\right) \\= \dfrac{1}{3}\left(F(5/2,9/2,3)-F(7/2,9/2,3)\right)\\ = \dfrac{17\pi}{21233664\sqrt{3}}\\$
This is what we needed and you can carry on from here. In fact the ratio is $\dfrac{51840\sqrt{3}}{17}$
|
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|
How to deduce $\sin A+\sin B+\sin C=4\cos\frac A2\cos\frac B2\cos\frac C2$ from $A+B+C=\pi$?
If $A+B+C=\pi$,$$\sin A+\sin B+\sin C=4\cos\dfrac A2\cos\dfrac B2\cos\dfrac C2\tag1$$$$\sin A+\sin B-\sin C=4\sin\dfrac A2\sin\dfrac B2\cos \dfrac C2\tag2$$$$\cos A+\cos B+\cos C=4\sin\dfrac A2\sin\dfrac B2\sin\dfrac C2+1\tag3$$$$\cos A+\cos B-\cos C=4\cos\dfrac A2\cos\dfrac B2\sin\dfrac C2-1\tag4$$$$\tan A+\tan B+\tan C=\tan A\tan B\tan C\tag5$$$$\cot\dfrac A2+\cot\dfrac B2+\cot\dfrac C2=\cot\dfrac A2\cot\dfrac B2\cot\dfrac C2\tag6$$
Formulae $(1)$ through $(6)$ were given with the condition that $A+B+C=180^{\circ}$. I'm not sure how to arrive at them.
Question: How do you arrive at $(1)$ through $(6)$?
I need a place to start. I am well aware that$$\sin A+\sin B=2\sin\dfrac {A+B}2\cos\dfrac {A-B}2$$And$$\cos A+\cos B=2\cos\dfrac {A+B}{2}\cos\dfrac {A-B}2$$
However, I'm not sure how to get $\sin A\pm\sin B\pm\sin C$. I'm guessing it has something to do with the expansion of $\sin(A+B+C)$.
Note: In your answer, give a hint on where I can begin, then hide the rest of your answer.
|
$$4\cos\frac{A}{2}\cos\frac{B}{2}\cos\frac{C}{2}=2\cos\frac{A}{2}\left(\cos\frac{B+C}{2}+\cos\frac{B-C}{2}\right)=$$
$$=\cos\frac{A+B+C}{2}+\cos\frac{-A+B+C}{2}+\cos\frac{A+B-C}{2}+\cos\frac{A-B+C}{2}=$$
$$=\sin{A}+\sin{B}+\sin{C}$$
|
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|
Is this limit $\lim\limits_{x \to\, -8}\frac{\sqrt{1-x}-3}{2+\sqrt[3]{x}} = 0$? Is this limit $\lim\limits_{x \to\, -8}\frac{\sqrt{1-x}-3}{2+\sqrt[3]{x}} = 0$?
It's in indeterminant form, 0/0 when $x$ approaches $-8$. So I used LHopital's rule and got $$-\frac{3x^{\frac{2}{3}}}{2\sqrt{1-x}}$$ plug in $-8$ it is $-2(-1)^{2/3}$ which is imaginary. I used wolframalpha, the answer is $0$. So, which is correct?
|
Wolfram Alpha is not correct.
You're almost correct, except that we have
$$\left(-8\right)^{2/3}=(64)^{1/3}=4$$ or alternatively $$\left(-8\right)^{2/3}=(-2)^{2}=4$$
Hence, applying L'Hospital's Rule
$$\begin{align}
\lim_{x\to -8}\frac{\sqrt{1-x}-3}{2+\sqrt[3]{x}}&=\lim_{x\to -8}\left(-\frac{3x^{2/3}}{2\sqrt{1-x}}\right)\\\\
&=-\frac{12}{6}\\\\
&=-2
\end{align}$$
An alternative approach would be to rationalize the numerator and denominator such that
$$\begin{align}
\lim_{x\to -8}\frac{\sqrt{1-x}-3}{2+\sqrt[3]{x}}&=\lim_{x\to -8}\left(\left(\frac{\sqrt{1-x}-3}{2+\sqrt[3]{x}}\right)\left(\frac{\sqrt{1-x}+3}{\sqrt{1-x}+3}\right)\left(\frac{x^{2/3}-2x^{1/3}+4}{x^{2/3}-2x^{1/3}+4}\right)\right)\\\\
&=\lim_{x\to -8}\left(-\frac{x^{2/3}-2x^{1/3}+4}{\sqrt{1-x}+3}\right)\\\\
&=-2
\end{align}$$
as expected!
|
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|
$S=1+2i+3i^2+4i^3+\dots+(n+1)i^n$; $4\mid n$; closed form for $S$ $$S=1+2i+3i^2+4i^3+\dots+(n+1)i^n$$
where $4\mid n$. How can I simplify this exprerssion so as to obtain a general expression?
|
Consider functions $f(x) = 1 + 2x + 3x^2 +\ldots +(n+1)x^n$ and $F(x) = 1+ x + x^2 +\ldots + x^{n+1}$. We have $F'(x) = f(x)$. On the other hand, $F(x)(x-1) = x^{n+2} -1$ which implies $F(x) = \frac{x^{n+2}-1}{x-1}$ for $x\neq 1$. From this we have $$f(x) = F'(x) = \frac{(n+2)x^{n+1}(x-1)-(x^{n+2}-1)}{(x-1)^2}$$
All you have to do is evaluate at $i$.
|
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|
On the limit of $\sin^2 (\pi\sqrt{n^2+n})$ What is the limit of the sequence $\sin^2 (\pi\sqrt{n^2+n})$ as $n$ tends to infinity?
My Attemp: I replace the square root with $n+\frac 12$ (its equivalent) and the rest is routine:
$\lim \sin^2 (\pi\sqrt{n^2+n}) = \lim \sin^2 (\pi n + \frac{\pi}2) = 1$
Is this correct?
|
Hint. The sequence is convergent.
As $n$ tends to $+\infty$, we may write
$$
\begin{align}
u_n &:=\sin^2 \left( \pi \sqrt{n^2+n }\right)\\
&=\sin^2 \left( \pi n \:\sqrt{1+\frac{1}{n}}\right)\\
&=\sin^2 \left( \pi n \:\left(1+\frac{1}{2n}+\mathcal{O}\left(\frac{1}{n^2}\right)\right)\right)\\
&=\sin^2 \left( \pi n +\frac{\pi}{2}+\mathcal{O}\left(\frac{1}{n}\right)\right)\\
&=\left((-1)^n\sin \left(\frac{\pi}{2}+\mathcal{O}\left(\frac{1}{n}\right)\right)\right)^2
\end{align}
$$
Can you finish it?
|
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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.
|
The function $f(x)=x-3\sqrt{5\over x}$ increases from $-\infty$ at $x=0$ to $+\infty$ as $x\to\infty$, so the equation $x-3\sqrt{5\over x}=8$ has exactly one solution for $x\gt0$. Multiplying through by $x$ and moving everything to the left hand side, we have
$$x^2-8x-3\sqrt{5x}=0$$
Now let $u=x-\sqrt{5x}$. We can rewrite this equation as
$$x^2-11x+3u=0$$
On the other hand, moving the $-3\sqrt{5x}$ to the right hand side and squaring gives $x^4-16x^3+64x^2=45x$, or
$$(x^3-16x^2+64x-45)x=(x-5)(x^2-11x+9)x=0$$
But $x=0$ is obviously not a solution to $x-3\sqrt{5\over x}=8$. Nor is $x=5$. So the (unique positive) solution must satisfy $x^2-11x+9=0$. But since it also satisfies $x^2-11x+3u=0$, we see that $9=3u$, or $u=3$. We thus find that $x-\sqrt{5x}=3$.
|
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"timestamp": "2023-03-29T00:00:00",
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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?
$$2(ab + bc + ca)-a^2 + b^2 + c^2>0$$
$$2ab + 2bc + 2ca-a^2 + b^2 + c^2>0$$
$$2b(a+c) + 2ca-a^2 + b^2 + c^2>0$$
...?
|
Let $a=y+z$, $b=x+z$ and $c=x+y$.
Hence, $x$, $y$ and $z$ are positives and
$$\sum_{cyc}(2ab-a^2=\sum_{cyc}(2(x+z)(y+z)-(y+z)^2)=$$
$$=\sum_{cyc}(2x^2+6xy-2x^2-2xy)=4(xy+xz+yz)>0$$
|
{
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Solve $\lim_{x \rightarrow 0} \frac{e^x+e^{-x}-2}{x^2+2x}$ without using L'Hopital's rule I tried:
$$\lim_{x \rightarrow 0} \frac{e^x+e^{-x}-2}{x^2+2x} = \\
\frac{e^{x}(1+e^{-2x}-\frac{2}{e^x})}{x(x-2)} = \frac{e^x(1+e^{-2x})-2}{x(x-2)} = \frac{e^x(1+e^{-2x})}{x(x-2)} - \frac{2}{x(x-2)} = \\
\frac{1+e^{-2x}}{x} \cdot \frac{e^x}{x-2} - \frac{2}{x(x-2)} = ???$$
What do I do next?
|
You can use hyperbolics: $2\cosh x=e^x+e^{-x}$ and so the limit expression can be written as $$\frac{2}{x+2}\cdot\frac{\cosh x-1}{x-0}$$ and use the limit definition of the derivative on $y=\cosh x$ taken at $x=0$. Now give it a try from here.
|
{
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|
Definite integration with variable substitution gives wrong result I have the following definite integral:
$$\int^{5}_{-5}{\sqrt{25-x^2} dx}$$
I do a variable substitution:
$$y = 25 - x^2$$
$$x = \sqrt{25 - y}$$
$$dy = -2x~dx$$
$$dx = \frac{dy}{-2x} = -\frac{1}{2\sqrt{25-y}}$$
I get the new integral:
$$-\frac{1}{2}\int^{0}_{0}{\sqrt{\frac{y}{25-y}} dy}$$
From Integral table I get that this integral (24) may be solved as:
$$-\frac{1}{2}(-\sqrt{y(25-y)}-25\tan^{-1}\frac{\sqrt{y(25-y)}}{y-25})\biggr\rvert^0_{0}$$
Going back to $x$ I get:
$$-\frac{1}{2}(-\sqrt{(25-x^2)x^2}-25\tan^{-1}\frac{\sqrt{(25-x^2)x^2}}{-x^2})\biggr\rvert^5_{-5}$$
Then:
$$\frac{1}{2}x\sqrt{25-x^2}+\frac{25}{2}\tan^{-1}\frac{-\sqrt{25-x^2}}{x}\biggr\rvert^5_{-5}$$
But I get $0$ as result, while the right answer is $\frac{25\pi}{2}$.
Where is the mistake?
|
The "mistake" is $x=\sqrt{25 - y}$. What happens if $x < 0$? One needs to be careful when making non-injective substitutions.
|
{
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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 $7$
|
$(n^3-1)(n^3)(n^3+1) = n^9-n^3$
$n^6 \equiv 1 \pmod 7$ This is Fermat's little theorem. If $p$ prime $n^{p-1}\equiv 1\pmod p$
$n^9-n^3 \equiv n^3-n^3\equiv 0 \pmod 7$
$n^6 \equiv 1 \pmod 9$
If this is not obvious:
$n^2 \equiv 1 \pmod 3\\
n^2 = (3k+1)\\
n^6 = (3k+1)^3 = (27k^3 + 27k^2 + 9k + 1)$
$n^9-n^3 \equiv 0 \pmod 9$
|
{
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|
Finding $ E ( X Y \mid X + Y = w ) $ when $ X $ and $ Y $ are i.i.d $\mathcal N ( 0 , 1 ) $ variables $ X $ and $ Y $ are both independent and identically distributed random variables with normal distributions $ \mathcal N ( 0 , 1 ) $. What is $ E ( X Y \mid X + Y = w ) $?
I know this means that $ W=X+Y $ must be normally distributed as well with mean $ 0 $ and variance $ 2 $. I also know that $ E ( X Y ) = E ( X ) E ( Y ) $ because of independence. However, I am confused as to how we calculate the conditional expectation in this situation. Do we just take the integral of the normal pdf? What would the boundaries be?
|
Letting $W = X + Y$ then you can write out the joint Gaussian as
\begin{align*}
\begin{bmatrix}
X \\ Y \\ W
\end{bmatrix} \sim \mathcal{N} \left(
\begin{bmatrix}
0 \\ 0 \\0
\end{bmatrix},
\begin{bmatrix}
1 & 0 & 1 \\
0 & 1 & 1 \\
1 & 1 & 2
\end{bmatrix}
\right)
\end{align*}
without worrying too much that this covariance matrix is singular, therefore partitioning and conditioning we have
\begin{align*}
\begin{bmatrix} X \\ Y \end{bmatrix} \bigg| W = w \sim \mathcal{N}\left(
\begin{bmatrix}
\frac{w}{2} \\ \frac{w}{2} ,
\end{bmatrix}
\begin{bmatrix}
\frac{1}{2} & -\frac{1}{2} \\
-\frac{1}{2} & \frac{1}{2}
\end{bmatrix}
\right)
\end{align*}
Still singular but that's ok because of course this Gaussian is supported on the one dimensional space $X + Y = w$. Anyway we get
\begin{align*}
\mathbb{E}\left[ X Y | X + Y = w \right] &= \mbox{Cov}_{X + Y = w}(X,Y) + (\mathbb{E}[X|X+Y=w])^2 \\
&= \frac{w^2}{4} - \frac{1}{2}.
\end{align*}
|
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|
Evaluate $I_{n}=\int\frac{\cos nx}{5-4\cos x}=$ I stumbled across the following integral:-
$$I_{n}=\int\frac{\cos nx}{5-4\cos x}dx$$
where $n$ is a positive integer.
I have no idea how to proceed....I tried integration by parts and even writing
$\cos(x)=\frac{e^{ix}+e^{-ix}}{2}$
I couldn't make much headway.....
Any ideas on how to proceed would be appreciated.
EDIT:
This question is different from @amWhy has marked..I want to evaluate indefinite integral ..Not the definite one.
|
Hint:
It is just a piece of cake of creating the reduction formula:
$\int\dfrac{\cos nx}{5-4\cos x}~dx$
$=\int\dfrac{2\cos x\cos((n-1)x)}{5-4\cos x}~dx-\int\dfrac{\cos((n-2)x)}{5-4\cos x}~dx$ (according to http://mathworld.wolfram.com/Multiple-AngleFormulas.html)
$=\dfrac{1}{2}\int\dfrac{4\cos x\cos((n-1)x)}{5-4\cos x}~dx-\int\dfrac{\cos((n-2)x)}{5-4\cos x}~dx$
$=\dfrac{1}{2}\int\dfrac{(4\cos x-5+5)\cos((n-1)x)}{5-4\cos x}~dx-\int\dfrac{\cos((n-2)x)}{5-4\cos x}~dx$
$=-\dfrac{1}{2}\int\cos((n-1)x)~dx+\dfrac{5}{2}\int\dfrac{\cos((n-1)x)}{5-4\cos x}~dx-\int\dfrac{\cos((n-2)x)}{5-4\cos x}~dx$
$=-\dfrac{\sin((n-1)x)}{2(n-1)}+\dfrac{5}{2}\int\dfrac{\cos((n-1)x)}{5-4\cos x}~dx-\int\dfrac{\cos((n-2)x)}{5-4\cos x}~dx$
|
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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.
|
Start with
$$ \sum_{k=0}^{r+1} (-1)^k\binom{r+1}{k}4^{r+1-k} = 4^{r+1} + \sum_{k=1}^{r+1} (-1)^k\Bigg( \binom{r}{k} + \binom{r}{k-1}\Bigg)4^{r+1-k}$$
$$ = 4^{r+1} + \sum_{k=1}^{r+1} (-1)^k\binom{r}{k}4^{r+1-k} + \sum_{k=1}^{r+1} (-1)^k\binom{r}{k-1}4^{r+1-k}.$$
In the first sum the term $k=r+1$ is zero and you can add in the $k=0$ term back. In the second sum, reindex your $k$ to go from $0$ to $r$. So you'll have
$$ \sum_{k=0}^{r+1} (-1)^k\binom{r+1}{k}4^{r+1-k} = \sum_{k=0}^{r} (-1)^k\binom{r}{k}4^{r+1-k} + \sum_{k=0}^{r} (-1)^{k+1}\binom{r}{k}4^{r-k}. $$
Using your induction hypothesis,
$$ = 4*3^{r} - 3^{r} = 3^{r+1}.$$
However a much quicker proof would be to use the binomial theorem:
$$ (x+y)^r = \sum_{k=0}^{r} \binom{r}{k} x^k y^{r-k} $$
with $x=-1$ and $y=4$.
|
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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 $\frac{n+1}{(n+1)+1}$?
|
Well, I will tell you a method without having to use induction. We have, $$\frac {1}{2} = \frac {2-1}{2\times 1} = \frac {1}{1}-\frac {1}{2} $$ $$\frac {1}{6} = \frac {3-2}{3\times 2} = \frac {1}{2}-\frac {1}{3} $$ $$\vdots $$ $$\frac {1}{n (n+1)} = \frac {n+1-n}{n (n+1)} = \frac {1}{n}-\frac {1}{n+1} $$
Adding, we get, $$\frac {1}{2}+\frac {1}{6}+\cdots + \frac {1}{n(n+1)} = (\frac {1}{1}-\frac {1}{2})+(\frac {1}{2}-\frac {1}{3})+\cdots + ( \frac{1}{n}-\frac {1}{n+1}) = \frac {1}{1}-\frac {1}{n+1} = \frac {n}{n+1} $$
This is also called a telescoping sum. Hope it helps.
|
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|
Question from applications of derivatives. Prove that the least perimeter of an isoceles triangle in which a circle of radius $r$ can be inscribed is $6r\sqrt3$.
I have seen answer online on two sites. One is on meritnation but the problem is that answer is difficult and bad formatting. Other answer on topperlearning but that answer make uses of trigonometric functions. And I want to solve it without trignometric function.
So please can someone provide easy method.
|
Let $\Delta ABC$ be our triangle, where $AB=BC$ and $\measuredangle ABC=2x$.
Hence, $P_{\Delta ABC}=2r\left(\cot{x}+2\tan\left(\frac{\pi}{4}+\frac{x}{2}\right)\right)$ and we need to prove that $\min\limits_{\left(0,\frac{\pi}{2}\right)}f=3\sqrt3$, where
$$f(x)=\cot{x}+2\tan\left(\frac{\pi}{4}+\frac{x}{2}\right).$$
Indeed,
$$f'(x)=\frac{1}{\cos^2\left(\frac{\pi}{4}+\frac{x}{2}\right)}-\frac{1}{\sin^2x}=\frac{\left(\sin{x}-\cos\left(\frac{\pi}{4}+\frac{x}{2}\right)\right)\left(\sin{x}+\cos\left(\frac{\pi}{4}+\frac{x}{2}\right)\right)}{\cos^2\left(\frac{\pi}{4}+\frac{x}{2}\right)\sin^2x}.$$
Since $\sin{x}+\cos\left(\frac{\pi}{4}+\frac{x}{2}\right)>0$ for $0<x<\frac{\pi}{2}$,
we see that $f'(x)=0$ on $\left(0,\frac{\pi}{2}\right)$ for $x=\frac{\pi}{6}$ only
and since $f'(x)<0$ on $\left(0,\frac{\pi}{6}\right)$ and $f'(x)>0$ on $\left(\frac{\pi}{6},\frac{\pi}{2}\right)$,
we see that $f$ gets on $\left(0,\frac{\pi}{2}\right)$ a minimal value for $x=\frac{\pi}{6}$.
Since $f\left(\frac{\pi}{6}\right)=3\sqrt3$, we are done!
|
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|
Finding $\lim_{n\to\infty}(\frac{1}{\sqrt{n^2+1}} + \frac{1}{\sqrt{n^2+2}} + ... + \frac{1}{\sqrt{n^2+n}})$ I'm trying to find $\lim_{n\to\infty}(\frac{1}{\sqrt{n^2+1}} + \frac{1}{\sqrt{n^2+2}} + ... + \frac{1}{\sqrt{n^2+n}})$.
*
*I tried to use the squeeze theorem, failed.
*I tried to use a sequence defined recursively: $a_{n+1} = {a_n} + \frac{1}{\sqrt{(n+1)^2 +n+1}}$. It is a monotone growing sequence, for every $n$, $a_n > 0$. I also defined $f(x) = \frac{1}{\sqrt{(x+1)^2 +x+1}}$. So $a_{n+1} = a_n + f(a_n)$. But I'm stuck.
How can I calculate it?
|
It looks squeezable.
\begin{align}
\frac{n}{\sqrt{n^2+n}} \le \sum_{k=1}^n\frac{1}{\sqrt{n^2+k}} \le \frac{n}{\sqrt{n^2+1}}
\\
\\
\frac{1}{\sqrt{1+\frac{1}{n}}} \le \sum_{k=1}^n\frac{1}{\sqrt{n^2+k}} \le \frac{1}{\sqrt{1+\frac{1}{n^2}}}
\end{align}
|
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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?
|
Notice that $(a+b+c)^2=a^2+b^2+c^2+2(ab+bc+ac)=2c^2+2(ab+bc+ac)$, so the square of $a+b+c$ is even and thus $a+b+c$ is also even.
|
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|
$5x^2−10x+7$ in completed square form? I was studying quadratic equations and practicing to solve them using the technique of completing the squares.
My answer was as follows:
$$5x^2−10x=−7$$
$$5x^2−10x+25=18$$
$$x(x-5)^2−5(x-5)=18$$
$$\boldsymbol{(x-5)^2-18=0}$$
The answer in the book is:
$$\boldsymbol{5(x-1)^2+2=0}$$
I'm confused. I just started with these sums today. Could you please tell me where I've gone wrong?
|
We know that $ax^2+bx+c=0$ and we want to get this form $a(x+d)^2+e=0$.
This is a shortcut: $d=\frac{b}{2a}$ and $e=c-\frac{b^2}{4a}$.
Then we know that $5x^2-10x+7=0$, here we can see that $a=5,b=-10,$ and $c=7$.
So we simply substitute and we get $d=\frac{-10}{2(5)}$ and $e= 7-\frac{(-10)^2}{4(5)}$.
Now we simplify and we get $d=-1$ and $e=7-5 \rightarrow 2$. So our final answer is:
$5(x-1)^2+2=0$.
The other way is by leaving the equation as :
$5x^2-10x+( )=-7$ now factor 5 and we get $5(x^2-2x+( ))=-7$
Then we can use the following information that $(\frac{b}{2})^2$=$(\frac{2}{2})^2=1$
So we get $5(x^2-2x+1-1)=-7$ we take out the -1 since we know that $(x^2-2x+1)=(x-1)^2$
we get $5(x^2-2x+1)-5=-7 \rightarrow 5(x-1)^2+2=0$.
|
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|
How can I calculate the limit without using the L'Hopital's rule $$ \lim_{x \to 2} \frac{1-\sqrt{1-\sqrt{4x-8}}}{1-\sqrt{1-\sqrt{\frac{x-2}{x+2}}}} $$
I have tried to multiply with conjugates or use auxiliary variables but it did not arrive at all simple
|
$$\lim_{x \to 2} \frac{1-\sqrt{1-\sqrt{4x-8}}}{1-\sqrt{1-\sqrt{\frac{x-2}{x+2}}}}$$
Let's let $z = x-2$ (or $x = z + 2$). Then we get
$$\lim_{z \to 0} \frac{1-\sqrt{1-2\sqrt z}}{1-\sqrt{1-\sqrt{\frac{z}{z+4}}}}$$
\begin{align}
1-\sqrt{1-2\sqrt z}
&= \dfrac{1-(1-2\sqrt z)}{1+\sqrt{1-2\sqrt z}} \\
&= \dfrac{2\sqrt z}{1+\sqrt{1-2\sqrt z}}
\end{align}
\begin{align}
\frac{1}{1-\sqrt{1-\sqrt{\frac{z}{z+4}}}}
&=\frac{1+\sqrt{1-\sqrt{\frac{z}{z+4}}}}
{1-\left(1-\sqrt{\frac{z}{z+4}}\right)} \\
&=\frac{1+\sqrt{1-\sqrt{\frac{z}{z+4}}}}
{\sqrt{\frac{z}{z+4}}} \\
&=\frac{\sqrt{z+4}+\sqrt{z+4-\sqrt{z(z+4)}}}
{\sqrt{z}} \\
\end{align}
\begin{align}
\dfrac{1-\sqrt{1-\sqrt{4x-8}}}
{1-\sqrt{1-\sqrt{\frac{x-2}{x+2}}}}
&= \dfrac{2\sqrt z}
{1+\sqrt{1-2\sqrt z}}\cdot
\dfrac{\sqrt{z+4}+\sqrt{z+4-\sqrt{z(z+4)}}}
{\sqrt{z}}\\
&= 2\dfrac{\sqrt{z+4}+\sqrt{z+4-\sqrt{z(z+4)}}}
{1+\sqrt{1-2\sqrt z}}
\end{align}
If we now let $z=0$, we get
\begin{align}
\lim_{x \to 2} \frac{1-\sqrt{1-\sqrt{4x-8}}}
{1-\sqrt{1-\sqrt{\frac{x-2}{x+2}}}}
&= 2 \lim_{z \to 0} \dfrac{\sqrt{z+4}+\sqrt{z+4-\sqrt{z(z+4)}}}
{1+\sqrt{1-2\sqrt z}} \\
&= 2 \lim_{z \to 0} \dfrac{2+2}{1+1} \\
&= 4 \\
\end{align}
|
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|
Find a positive integer $a$ such that $4 \mid a$ and $9 \mid a+1$ and $25 \mid a+2$ I need to find a positive integer $a$ such that $4 \mid a$ and $9 \mid a+1$ and $25 \mid a+2$.
I tried converting this to a system of congruences, and I got the following.
$a \equiv 0$ mod $4$
$(a+1) \equiv 0$ mod $9$
$(a+2) \equiv 0$ mod $25$
And then I want to use Chinese Remainder Theorem. But, I think my congruences are wrong since they don't involve the variable $x$, but I can't imagine how it would fit in. It seems unnecessary but I think I am supposed to use this method.
What should my congruences look like?
|
$a \equiv 0 \pmod{4}$, so $a = 4k$, and $a+1 \equiv 0 \pmod{9}$, it means $4k+1 \equiv 0 \pmod{9}$, the minimum $k$(here $k \ge 0$) is $k = 2$. so, when $a \equiv 0 \pmod{4}$ and $a+1 \equiv 0 \pmod{9}$, the minimum $a$ is $a = 4k = 8$, all the $a$ is $8+(4 \cdot 9)m = 8+36m$.
then, we have $a+2 \equiv 0 \pmod{25}$, it means $8+36m+2 \equiv 0 \pmod{25}$, as $36m \equiv 15 \pmod{25}$. the minimum $m$(here $m \ge 0$) is $m = 15$.
so $a = 8+36m = 548$ is one solution.
|
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|
If $a^2+b^2 = c^2 + \sqrt{3}$, find value of $\angle C$.
In a triangle with sides $a$, $b$, $c$, consider this relationship:
$$a^2+b^2 = c^2 +\sqrt 3$$
Now find value of $\measuredangle C$ .
My try: I tried to use Laws of Cosines and Law of Sinee, but I didn't get any result.
|
Note: In what follows I am interpreting $a^2+b^2 = c^2 +\sqrt 3$ as meaning that $c^2$ is larger than what would be obtained using the Pythagorean theorem by $\sqrt 3$.
Since you must have $$a^2+b^2 = c^2 +\sqrt 3$$ the law of cosines states that in this case $$- 2\ a\ b\ cos (\measuredangle C) = \sqrt 3$$ Therefore the angle is determined by the formula $$\measuredangle C = \arccos(-\frac{\sqrt 3}{2\ a\ b})$$.
Note: if the interpretation should have been that $a^2 + b^2$ is smaller than $c^2$ (as obtained using the Pythagorean theorem) by $\sqrt 3$ then the above equation is the same but without the minus sign.
$\measuredangle C$ varies with the values of $a$ and $b$ therefore there are an infinite number of values for $\measuredangle C$. For example, in the figure below (showing both possible interpretations)
when $a$ and $b$ are both 1, then $\measuredangle C$ is $150 ^{\circ}$. When $a$ and $b$ are both $2$, then $\measuredangle C$ is $102.5^{\circ}$. In both cases $a^2 + b^2 = c^2 + \sqrt 3$ as required. Given any values of $a$ and $b$, the value of $\measuredangle C$ can be calculated to make $a^2 + b^2 = c^2 + \sqrt 3 $.
|
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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.
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Since $f(x)=\left(x+\frac{1}{x}\right)^2$ is a convex function, your inequality follows from Jensen:
$$LS\geq2\left(\frac{a+b}{2}+\frac{1}{\frac{a+b}{2}}\right)^2=\frac{25}{2}$$
|
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|
Is this elementary proof of FLT correct? Consider the classic FLT for $n=3$, $x^3+y^3=z^3 $
Without loss of generality,we can rewrite as $a^3+(b-a)^3=c^3…Equation (1)$
We can also assume there is no common factor between $a,b-a$ and $c …Assumption (1)$
Since $m+n$ is a factor of $m^3+n^3$, $b$ divides $c^3$.
Let $p$ be a common prime factor of $b$ and $c … Assumption (2)$
$p$ does not divide $a$ as if it did, $p$ divides $b-a$ and hence $p$ divides $a$ violating $Assumption (1)$
Rewriting Equation(1) we get, $a^3+b^3-a^3-3b^2 a+3ba^2=c^3…Equation (2)$
rearranging we get, $b^3-c^3=3b^2 a-3ba^2 …Equation (3)$
Note that $p^3$ divides the LHS of $Equation (3)$
Two cases:
Case I: $p \ne $n
On the RHS of $Equation (3)$, $p^2$ divides the first term $3b^2 a$ but not $3ba^2$
Case II: $p = $n
Now $p^3$ divides $3b^2a$ but not $3ba^2$.
In both cases, LHS of $Equation (3) \ne RHS $
The General case of FLT for any odd prime n:
Without loss of generality,we can rewrite FLT as $a^n+(b-a)^n=c^n…Equation (4)$
Expanding, we get
$a^n+b^n-nC_1 b^{n-1} a+nC_2 b^{n-2} a^2-⋯-nC_{n-2} b^2 a^{n-2}+nC_{n-1} ba^{n-1}-a^n=c^n…Equation (5)$
rearranging we get,
$b^n-c^n=nC_1 b^{n-1} a- nC_2 b^{n-2} a^2+ ⋯+nC_{n-2} b^2 a^{n-2}-nC_{n-1} ba^{n-1}…Equation (6)$
Therefore $p^n$ divides the LHS of $Equation (6)$
Two cases.Case I: $p \ne n$
On RHS of Equation (6), $p^2$ divides all the terms except the last term $nC_{n-1} ba^{n-1}$
Case II: $p=n$
Now $p^3$ divides all terms of RHS except the last term $nC_{n-1} ba^{n-1}$
In both cases, LHS of $Equation (6) \ne RHS $
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The proof is incorrect. The error is in the statement
$$``\text{$p^2$ divides the first term $3b^{2}a$ but not $3a^{2}b$}."$$
In case 1, when $p$ is different than $3$, you know that $p \nmid a$, $p \nmid 3$, and $p\mid b$. But, you do not know that $p^{2} \nmid b$. This is because you chose $p$ to be any prime factor of $b$ and $c$ (assumption 2), which does not exclude the possibility that $p^{2} \mid b$.
So, if $p^{2} \mid b$, then the claim that $p^{2} \nmid 3a^{2}b$ is false, and the conclusion no longer follows.
|
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|
Does there exist $a \in \mathbb{Q}$ such that $a^2 - a + 1$ is a square? Does there exist $a \in \mathbb{Q}$ such that $a^2 - a + 1$ is a square?
Context for this question: For pedagogical purposes I was trying to create an example of a cubic polynomial with both its critical values and its zeros all at integers. After fumbling around with it unsuccessfully for a while, I realized that the problem of creating such an example was equivalent to finding a rational number $a$ ($a \ne 0,1$) with the property that $a^2 - a + 1$ is a perfect square. (Equivalently, find two integers $m,n$ such that $m^2 - mn + n^2$ is a perfect square.) I haven't been able to find an example and strongly suspect that it's not possible, but can't see a simple proof of why that should be so.
Any proof or counterexample to the conjecture that no such $a$ exists?
EDIT: I may have found the beginning of an argument, within a few minutes of posting the question. I'd appreciate feedback.
Suppose $b$ is a positive integer such that $a^2-a+1=b^2$. Then $a-1 = a^2-b^2$, so $a-1= (a-b)(a+b)$. Now if $a-b > 1$ then we have
$$a-1 = (a-b)(a+b) > a+b$$ whence $b < -1$, contradicting the assumption that $b$ is a positive integer.
So if such a $b$ exists, it must be that $a-b \le 1$. Then.... (?)
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$$a^2-a+1=(a-\frac{1}{2})^2+\frac{3}{4}$$
So what you ask are there any rational numbers $u=a-\frac{1}{2}$ and $v$ such that,
$$u^2+\frac{3}{4}=v^2$$
$$v^2-u^2=\frac{3}{4}$$
$$(v-u)(v+u)=\frac{3}{4}$$
Let $x$ and $y$ be two real numbers that multiply to $\frac{3}{4}$, with the condition that $\frac{x+y}{2}$ and $\frac{y-x}{2}$ are both rational. Note that this condition equates to the condition that both $x$ and $y$ are rational.
Solving the system,
$$v-u=x$$
$$v+u=y$$
Yields the rational solutions $v=\frac{x+y}{2}$ and $u=\frac{y-x}{2}$. Which implies $a=\frac{y-x+1}{2}$.
If you restrict $v$ to be an integer. Then it must be that,
$$x+y=2v \in \mathbb{Z}$$
$$xy=\frac{3}{4}$$
Eliminating $y$ from this system gives,
$$x(2v-x)=\frac{3}{4}$$
Or,
$$4x^2-8vx+3=0$$
By the rational root theorem, the only rational solutions to the above equation are,
$$x=\pm \frac{3}{2}, \pm \frac{3}{4}, \pm 3, \pm \frac{1}{2}, \pm \frac{1}{4}, \pm 1$$
The only solutions which correspond to integer values of $v$ are $x=\pm \frac{3}{2}, \pm \frac{1}{2}$. These give $v=\pm 1$. So we have that,
$$x+y=\pm 2$$
$$xy=\frac{3}{4}$$
Solving for $x$ and $y$ leads to the conclusion that the only possible integer solutions for $a$ is $a=0$ or $a=1$.
|
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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.
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$\mathbb Q(\sqrt[3]{2} )$ is a field. Since $1, \sqrt[3]{2} , \sqrt[3]{4} $ form a basis for $\mathbb Q(\sqrt[3]{2} )$ over $\mathbb Q$, you know that
$$ (2-\sqrt[3]{2}+\sqrt[3]{4} )^{-1} = a + b\sqrt[3]{2}+ c\sqrt[3]{4} $$
for some $a, b, c \in \mathbb Q$.
One natural approach would be write the equation
$$ (2-\sqrt[3]{2}+\sqrt[3]{4} )( a + b\sqrt[3]{2}+ c\sqrt[3]{4}) = 1,$$
and expand the brackets.
If I've done this correctly, this gives
$$ (2a + 2b - 2c - 1) + (-a + 2b + 2c) \sqrt[3]{2}+ (a - b + 2c)\sqrt[3]{4} = 0.$$
But since $1, \sqrt[3]{2}$ and $ \sqrt[3]{4} $ are linearly independent over $\mathbb Q$, this means that the coefficients vanish individually:
$$ 2a + 2b - 2c = 1, \ \ \ -a + 2b + 2c = 0, \ \ \ a - b + 2c = 0.$$
Now you can go ahead and solve these simultaneous equations:
$$ a = \frac 3 {11}, \ \ \ b = \frac 2 {11}, \ \ \ c = - \frac 1 {22}, $$
So
$$ (2-\sqrt[3]{2}+\sqrt[3]{4} )^{-1} = \frac 3 {11} + \frac 2 {11}\sqrt[3]{2}+ - \frac 1 {22}\sqrt[3]{4} $$
|
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|
Infinite series of formula How do I compute this infinite series
$$ \sum_{k=0}^{\infty}{\frac{(7n+32)(3^n)}{n(n+2)(4^n)}}$$
I believe partial fraction decomposition is part of the solving method but am a little stuck because the $3^n$ and the $4^n$ make the decomposition method a bit strange:
${\frac{(7n+32)(3^n)}{n(n+2)(4^n)}} = {\frac{A}{n}} + {\frac{B}{n+2}} + {\frac{C}{4^n}}$
$(7n+32)(3^n) = A(n+2)(4^n) + B(n)(4^n) + C(n)(n+2)$
$7n(3^n) + 32(3^n) = An (4^n) + 2A(4^n) + B(n)(4^n) + C(n^2 + 2n)$
$7n(3^n) + 32(3^n) = (An +2A + Bn)(4^n) + C(n^2 + 2n)$
..Stuck on how to proceed.
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HINT:
$$\frac{(7n+32)(3^n)}{n(n+2)(4^n)}=\frac{(4^2(n+2)-3^2)(3^n)}{n(n+2)(4^n)}$$
$$=\dfrac{3^n}{n4^{n-2}}-\dfrac{3^{n+2}}{(n+2)4^n}$$
$$=16\left(\dfrac{(3/4)^n}n-\dfrac{(3/4)^{n+2}}{n+2}\right)$$
|
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Gain matrices of numerical schemes Introduction
Say I have a linear multiple degree of freedom system, as below:
\begin{equation*}
\frac{dx}{dt}=ax+by
\end{equation*}
\begin{equation*}
\frac{dy}{dt}=dx+ey
\end{equation*}
This can also be written as $\frac{dr}{dt}=Ar$ with $A$ given by
\begin{equation*}
A=
\begin{bmatrix}
a & b \\
c & d \\
\end{bmatrix}
\end{equation*}
Matrix $A$ can be used to write the numerical schemes (Euler, Heun and Runge-Kutta) using gain matrices.
For Euler is this
\begin{equation*}
r_{n+1}=r_{n}+\frac{dr}{dt}\Delta t=r_{n}+Ar_{n}\Delta t
\end{equation*}
and rewriting gives
\begin{equation*}
r_{n+1}=\left(I+A\Delta t\right)r_{n}.
\end{equation*}
Now we can multiply old vector $r_{n}$ with $G=(I+A\Delta t)$ to get the new vector $r_{n+1}$.
In a similar way, the gain matrices $(G)$ can be determined for the Heun and Runge-Kutta schemes.
For the Heun scheme follows
\begin{equation*}
\begin{split}
r_{n+1}&=r_{n}+\frac{1}{2}\left(\frac{dr}{dt}\rvert_{(r_{n},t_{n})}+\frac{dr}{dt}\rvert_{(r_{p},t_{n+1})}\right)\Delta t \\
&=r_{n}+\frac{1}{2}\left(Ar_{n}+Ar_{p}\right) \\
&=r_{n}+\frac{1}{2}Ar_{n}\Delta t+\frac{1}{2}Ar_{p}\Delta t \\
&=\left(I+\frac{1}{2}A\Delta t\right)r_{n}+\left(\frac{1}{2}A\Delta t\right)r_{p} \\
\end{split}
\end{equation*}
Question
Runge Kutta is defined as
\begin{equation*}
r_{n+1}=r_{n}+\frac{1}{6}(k_{1}+2k_{2}+2k_{3}+k_{4})\Delta t
\end{equation*}
with
\begin{align*}
k_{1}&=f\left(r_{n},t_{n}\right) \\
k_{2}&=f\left(r_{n}+\frac{1}{2}k_{1}\Delta t,t_{n}+\frac{1}{2}\Delta t\right) \\
k_{3}&=f\left(r_{n}+\frac{1}{2}k_{2}\Delta t,t_{n}+\frac{1}{2}\Delta t\right) \\
k_{4}&=f\left(r_{n}+k_{3}\Delta t,t_{n}+\Delta t\right)
\end{align*}
For the Runge-Kutta scheme I can't figure out how to get gain matrices. Somebody who knows how to do this?
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You just insert the $f$ of the homogeneous linear equation,
\begin{alignat}{3}
k_1&=f(y)&&&&=Ay
\\
k_2&=f(y+0.5Δt·k_1)&&=A(y+0.5Δt·k_1))&&=Ay+0.5Δt·A^2y
\\
k_3&=f(y+0.5Δt·k_3)&&=A(y+0.5Δt·k_3))&&=Ay+0.5Δt·A^2y+0.25Δt^2·A^3y
\\
k_4&=f(y+Δt·k_4)&&=A(y+Δt·k_4)&&=Ay+Δt·A^2y+0.5Δt^2·A^3y+0.25Δt^3·A^4y
\\\hline
&k_1+2k_2+2k_3+k_4&&&&=6Ay+3Δt·A^2y+Δt^2·A^3y+0.25Δt^3·A^4y
\end{alignat}
which ends up giving the exact 4th order Taylor polynomial for $e^{Δt⋅A}y$ in the RK4 step.
|
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|
Test convergence of the series and find its sum: $\sum_{n=0}^{\infty}\left(\frac{-1-2i}{2+3i}\right)^n$ Test convergence of the series and find its sum: $$\sum_{n=0}^{\infty}\left(\frac{-1-2i}{2+3i}\right)^n$$
My try:
This is simplified to $$\sum_{n=0}^{\infty}\left(\frac{-i-8}{13}\right)^n$$
Then how to proceed?
|
$\sum_{n=0}^{\infty}\left(\frac{-1-2i}{2+3i}\right)^n
$
Since
$\frac{-1-2i}{2+3i}
=\frac{-1-2i}{2+3i}\frac{2-3i}{2-3i}
=\frac{-8-i}{13}
$
and
$|\frac{-8-i}{13}|^2
=\frac{65}{169}
\lt 1
$,
the sum converges.
Applying the standard formula,
the sum is
$\begin{array}\\
\dfrac{1}{1-\frac{-8-i}{13}}
&=\dfrac{13}{13+(8+i)}\\
&=\dfrac{13}{21+i}\\
&=\dfrac{13}{21+i}\dfrac{21-i}{21-i}\\
&=\dfrac{13(21-i)}{442}\\
&=\dfrac{21-i}{34}\\
\end{array}
$
|
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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.
|
Using Holder's inequality we get that
$$(1+1+1+1)^{3/4}(x^4+y^4+1+1)^{1/4} \geq |x|+|y| +1+1 \geq4 $$
Therefore,
$$(x^4+y^4+1+1)^{1/4}\geq 4^{1/4}$$
We conclude,
$$x^4+y^4\geq 4-2=2$$
|
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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.