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How prove binomial cofficients $\sum_{k=0}^{[\frac{n}{3}]}(-1)^k\binom{n+1}{k}\binom{2n-3k}{n}=\sum_{k=[\frac{n}{2}]}^n\binom{n+1}{k}\binom{k}{n-k}$ How prove this $$\sum_{k=0}^{[\frac{n}{3}]}(-1)^k\binom{n+1}{k}\binom{2n-3k}{n}=\sum_{k=[\frac{n}{2}]}^n\binom{n+1}{k}\binom{k}{n-k}$$
This equation How prove it? Thank you
I want take this $$f(x)=(1-x)^{n+1}?$$
But I can't deal this $[n/3]$,
Thank you for you help
|
Start from the RHS. We are counting the number of ways to choice $k$ elements among $n+1$, then choice $n-k$ elements among the $k$ elements previously chosen. If we imagine to assign a $+1$ weight in the first step, then increase the weight in the second step, we are counting the number of ways to assign a $+1$ or $+2$ weigth to the elements of a subset of $\{1,\ldots,n+1\}$ in such a way that the sum of the weights is just $n$. Rephrasing in the analytic combinatorics framework:
$$RHS=\sum_{k=\lfloor n/2\rfloor}^{n}\binom{n+1}{k}\binom{k}{n-k}=[x^n]\,\left((1+x+x^2)^{n+1}\right).\tag{1}$$
Now $1+x+x^2 =(1+x)^2-x$, so, for istance:
$$[x^n](1+x+x^2)^{n+1} = [x^n]\sum_{j=0}^{n+1}\binom{n+1}{j}(-1)^j x^j(x+1)^{2n+2-2j}=\sum_{j=0}^{n+1}(-1)^j\binom{n+1}{j}\binom{2n+2-2j}{n-j},$$
just like $1+x+x^2=\frac{1-x^3}{1-x}$ and
$$\frac{1}{(1-x)^m}=\sum_{j=0}^{+\infty}\binom{m+j-1}{j}x^j,$$
give:
$$[x^n](1+x+x^2)^{n+1}=[x^n]\left(\sum_{j=0}^{n+1}\binom{n+1}{j}(-1)^j x^{3j}\right)\cdot\left(\sum_{j=0}^{+\infty}\binom{n+j}{j}x^j\right).\tag{2}$$
Regarding $(2)$ as a Cauchy product gives $RHS=LHS$, QED.
The saddle-point method gives also:
$$[x^n](1+x+x^2)^{n+1}=\frac{3^{n+2}}{\sqrt{\pi(12n+30)}}\left(1+O\left(\frac{1}{\sqrt{n}}\right)\right).$$
|
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|
Show if $a^2+b^2 \le 2$ then $a+b \le 2$ If $a^2+b^2 \le 2$ then show that $a+b \le2$
I tried to transform the first inequality to $(a+b)^2\le 2+2ab$ then $\frac{a+b}{2} \le \sqrt{1+ab}$ and I thought about applying $AM-GM$ here but without result
|
Let $a=\sqrt{2}\cos\theta$, $y=\sqrt{2}\sin\theta$. Then $a^2+b^2=2$, and $a+b=\sqrt{2}(\cos\theta+\sin\theta)$, which is a maximum at $\theta=\frac{\pi}{4}$, at which case $a+b=2$. So $a+b\le 2$.
|
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|
IMO problem 4, $1998$ Determine all pairs $(a, b)$ of positive integers such that $ab^{2} + b + 7$ divides
$a^{2}b + a + b$. I really have no idea where to start with this. This is the first IMO problem that I attempted, because it looked to be the easiest, but I honestly have no idea where to begin. Any help or hints would be appreciated.
|
I've got a different solution to the one offered here already.
If $ab^2+b+7 \mid a^2b+a+b$, then $ab^2+b+7 \mid b(a^2b+a+b) - a(ab^2+b+7) = b^2-7a$
Now there are three cases left to check:
*
*$b^2-7a = 0$. This yields the infinite set of solutions $(a,b) = (7k^2, 7k)$, where $k\ge1$
*$b^2-7a > 0$. Hence, we have $ab^2+b+7 \le b^2-7a <b^2< ab^2+b+7$, a contradiction, so no solutions.
*$b^2-7a < 0$. Hence, we have $ab^2+b+7 \le 7a-b^2 \Rightarrow a(b^2-7) \le -b^2-b-7$. The RHS is negative, and since $a > 0$ we must have $b^2-7 <0 \Rightarrow b=1$ or $2$. Checking these cases individually we get two more solutions: $(a, b) = (11, 1), (49, 1)$.
Therefore, the solutions are $(a, b) = (11, 1), (49, 1), (7k^2, 7k)$
|
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|
Max. and Min. value of $f(x,y) = \frac{x-y}{x^4+y^4+6}$ [1] Calculation of Max. and Min. value of $\displaystyle f(x) = \sqrt{x^3-6x^2+21x+18}$, where $\displaystyle -\frac{1}{2}\leq x\leq 1$
[2] Calculation of Max. and Min. value of $\displaystyle f(x,y) = \frac{x-y}{x^4+y^4+6}\;,$ where $x,y\in \mathbb{R}$
$\bf{My\; Try::(1)}$Let $g(x) = x^3-6x^2+21x+18$, Now Using Derivative Test.
$g^{'}(x)=3x^2-12x+21=3(x^2-4x+7)=3\left\{(x-2)^2+3\right\}>0\;\forall x\in \mathbb{R}$
So $g(x)$ is Strictly Increasing function.
So $g(x)$ is Min. when $\displaystyle x = -\frac{1}{2}$ and $g(x)$ is Max. when $x=1$
$\bf{My\; Try::(2)}$ Let $\displaystyle f(x,y) = z=\frac{x-y}{x^4+y^4+6}.$
Using $\bf{A.M\geq G.M},\;\; $
$x^4+1\geq 2x^2$ and $y^4+1\geq 2y^2\;,$ we get $x^4+y^4+2\geq 2(x^2+y^2)\Rightarrow x^4+y^4+6\geq (x^2+y^2)+4$
Now I did not understand How can I solve it.
Help Required
Thanks.
|
Hint: Have a look at this for the second problem
Second partial derivative test
|
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|
When is $3^n + n$ a power of 2? For what $n \in \mathbb{N}$ is $3^n + n$ a power of $2$?
|
This is not a complete solution, but shows that any nontrival solutions must be mindbogglingly big if it exists at all.
We have the trivial solutions $3^0+0=2^0$ and $3^1+1=2^2$.
Assume $3^n+n=2^m$ with $n\ge 2$. Then clearly $m>n$.
We find $1+n\equiv 0\pmod 2$, hene $n$ is odd. Then $3^n\equiv 3\pmod 8$ and hence $n\equiv 5\pmod 8$, esp. $m\ge 6$.
From $3^8\equiv 1\pmod{32}$ we have $3^n\equiv 19\pmod{32}$, hence $n\equiv 13\pmod {32}$. A pattern emerges.
Lemma: For $k\in\mathbb N$ we have $3^{2^k}\equiv 1\pmod {2^{k+2}}$
Proof: This is true for $k=1$ and from
$$3^{2^{k+1}}-1=(3^{2^k}-1)(3^{2^k}+1) $$
the claim follows by indcution because $3^{2^k}+1$ is even. $_\square$
Propositio: Assume for some $1\le k< a<2^k$ we have that for all nontrivial solutions of $3^n+n=2^m$ we have $n\equiv a\pmod{2^k}$.
Then for all nontrivial solutions of $3^n+n=2^m$ we have $n\equiv -3^a\pmod{2^{k+2}}$.
Proof: Let $3^n+n=2^m$ be a nontrivial solution.
Then with $n=2^kb+a$ for some $b\in\mathbb N_0$ and by the lemma
$$ 3^n=3^a\cdot (3^{2^k})^b\equiv 3^a\pmod {2^{k+2}}.$$
As $m>n\ge a>k$ we find $3^a+n\equiv 0\pmod{2^{k+2}}$. $_\square$
Using the proprosition we can start with $(k,a)=(3,5)$ and repeatedly replace this with $(k+2, -3^a\bmod {2^{k+2}})$. The process either ends with a pair $(k,a)$ with $a\le k$ (and then necessarily $a=\in\{k,k-1\}$) or it never ends. In the latter case we conclude that no nontrivial solution exists, in the former case we may have found a solution, and if we give upprematurely, we at least obtain an estimate and modular condition for all nontrivial solutions. The sequence starts
$$ (3,5), (5,13), (7,45), (9,173), (11,685), (13,685), (15,25261)$$
and after a few more steps one reaches
$$(k,a)=(201,864075976670532385554180581999784042802808809920656868008621)$$
Especially, $m>n>8.64\cdot 10^{59}$. Also we can continue at leat until $k\approx 8.64\cdot 10^{59}$ and expect $a$ to grow accordinglyet.c sothat the sequence never end and presumably no solution exists.
By taking logarithms, we also find that $\frac{m}{n}$ is an extremely good approximation to $\frac{\ln 3}{\ln 2}$, which is also a hint towards non-existence of a solution.
|
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How can you find the cubed roots of $i$? I am trying to figure out what the three possibilities of $z$ are such that
$$
z^3=i
$$
but I am stuck on how to proceed. I tried algebraically but ran into rather tedious polynomials. Could you solve this geometrically? Any help would be greatly appreciated.
|
$z^3=i$ $⟹ z=\sqrt [3] {i}$
We know,
$\sqrt [3] {i} =a+bi$
$⟹ i=(a+bi)^3 = a^3-3ab^2+3a^2bi-b^3i= a^3-3ab^2+(3a^2b-b^3)i$
From this we can say,
$a^3-3ab^2=0$________________(i)
$3a^2b-b^3=1$_______________(ii)
From (i) we get,
$a(a^2-3b^2)=0$
So, $a=0$
Or,
$a^2-3b^2=0$ $⟹ a=±b\sqrt {3}$
By setting the value of $a$ in (ii) we get,
If $a=0$, $ b=-1$
If $a=±b\sqrt {3}$, $ b=1/2$
Then, $a=±b\sqrt {3}=±(1/2)\sqrt {3}=±\sqrt {3}/2$
So, $(a,b)=(0,-1),(±\sqrt {3}/2,1/2)$
Now, there are 3 values of $z$.
(1) $\sqrt [3] {i} =a+bi= 0+(-1)i= -i$
(2) $\sqrt [3] {i} =a+bi= \dfrac {\sqrt {3}}{2}+\dfrac {i}{2}$
(3) $\sqrt [3] {i} =a+bi= -\dfrac {\sqrt {3}}{2}+\dfrac {i}{2}$
|
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|
Real number multiplicative inverses expressed in another form I've been asked to express the multiplicative inverse of $3 + \sqrt{5}$ in the form $c + d\sqrt{5}$, where $c,d$ are rational numbers.
I understand that for some rational numbers $c,d$ we must have:
$$1 = (3 + \sqrt{5})(c + d\sqrt{5}).$$
I was able to answer for the multiplicative inverse of $2 +\sqrt{3}$. We find that $1 = (2 +\sqrt{3})(c + d\sqrt{3})$ where $c = 2$ and $d = -1$. However this seems to be related to the original $2 +\sqrt{3}$; and in the problem at hand this is not the case. So I'm rather confused.
How could i go about solving this? I'd like steps without the answer IF possible; moreover, how could i prepare for more general questions in this form?
|
Let's generalise this: you want to write $\dfrac{1}{x+y\sqrt{z}}$ in the form $b+c\sqrt{z}$:
$$\dfrac{1}{x+y\sqrt{z}} = \dfrac{(x-y\sqrt{z})}{(x+y\sqrt{z})(x-y\sqrt{z})} = \dfrac{x-y\sqrt{z}}{x^2-y^2z} = \dfrac{x}{x^2-y^2z} + \dfrac{-y}{x^2-y^2z} \sqrt{z}.$$
So $b= \frac{x}{x^2-y^2z}$ and $c= \frac{-y}{x^2-y^2z}$.
It works for complex numbers too.
|
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|
How Find this $T_{20}-S_{20}$
Let $\{a_{n}\},\{b_{n}\}$ be sequences such that
$$\sqrt{\dfrac{a_{n}+1}{b_{n}}}=\dfrac{1}{n}, \quad S_{n}=\sqrt{T_{n-1}} \quad(n>1)$$
where $\displaystyle S_{n}=\sum_{i=1}^{n}a_{i}$, $\displaystyle T_{n}=\sum_{i=1}^{n}b_{i}$, and $b_{1}=1$.
Find $T_{20}-S_{20}$.
(a) $15980$
(b) $35910$
(c) $43910$
(d) $43920$
My try: Since
$$n^2(a_{n}+1)=b_{n},S^2_{n}=T_{n-1}$$
so
$$S^2_{n+1}-S^2_{n}=a_{n+1}[S_{n+1}+S_{n}]=T_{n}-T_{n-1}=b_{n}.$$
Then I don't know how to continue. Thank you.
I guess this
$$1^3+2^3+\cdots+n^3=(1+2+\cdots+n)^2$$
maybe usefull,so I guess
$$a_{n}=n-1,b_{n}=n^3$$
and after I find this is true,But I can't prove it( without Mathematical induction)
|
We'll prove that $a_n=(n-1)$ and $b_n=n^3$ by complete induction.
For $n=1$, we have $b_1=1$ straight from problem statement and $a_1=0$ follows from $\sqrt{\frac{a_1+1}{b_1}}=1$.
For any higher $n$, we have
$$S_n=\sum_{k=1}^n a_k=\left(0+1+\ldots+(n-2)\right)+a_n=\frac{(n-1)(n-2)}{2}+a_n$$ and
$$T_{n-1}=\sum_{k=1}^{n-1} b_k = \left(1^3 + 2^3 + \ldots + (n-1)^3\right) = \left(\frac{n(n-1)}{2}\right)^2$$
Equating $S_n^2 = T_{n-1}$ gives us $$a_n^2 + 2\frac{(n-1)(n-2)}{2}a_n + \left(\frac{(n-1)(n-2)}{2}\right)^2 = \left(\frac{n(n-1)}{2}\right)^2$$
which, after simplification, yields
$$\left(a_n - (n-1)\right)\left(a_n + (n-1)^2\right) = 0$$
This gives us two possibilities: $a_n = (n-1)$ or $a_n = -(n-1)^2$. The latter would yield negative $S_n$, though, and that is impossible due to $S_n$ being equal to a square root; leaving the former as the only option.
$b_n$ can be obtained directly from $\sqrt{\frac{a_n+1}{b_n}}=\frac{1}{n}$; giving us $b_n=n^3$ as expected.
Now that we know the expressions for $a_n$ and $b_n$, it is straightforward to calculate $S_n=\frac{1}{2}n(n-1)$ and $T_n=\frac{1}{4}\left(n(n+1)\right)^2$ and $T_n-S_n=\frac{1}{4}n(n^3 + 2n^2 - n + 2)$ so $T_{20}-S_{20}=43910$.
|
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Two-variable limit question $\lim\limits_{(x,y)\rightarrow (0,0)} \dfrac{x^2y^2}{(x^2+y^4)\sqrt{x^2+y^2}}$
How to solve this two-variable limit? Thanks :D
|
This might also be an approach:
$$0 \le \frac{x^2y^2}{(x^2 + y^4)\sqrt{x^2 + y^2}} \le \frac{x^2y^2}{x^2\sqrt{x^2+y^2}} \le \frac{y^2}{\sqrt{y^2}} \le |y|$$
So, since $(x,y) \to (0,0)$ then $|y| \to 0$ and passing to a limit we obtain that the initial expression goes to 0.
|
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Find a closed form for the sum $∑(x^3 - 2x)$ from $x=1$ to any number $n$ Find a closed form for the sum $∑(x^3 - 2x)$ from $x=1$ to any number $n$.
Can someone explain to me what a closed form is and how to approach this problem?
|
A closed form is an expression that doesn't involve any looping/repeatition/recursion (like sums, products, sequences, etc.). It is not exactly clear, for example some consider binomials $\binom{n}{k}$ a closed form expression, while others do not. However, in case of sums of polynomials, it is quite clear: it is another polynomial. For example $\sum_{k=1}^{n}k = \frac{n(n+1)}{2} = \frac{1}{2}n^2+\frac{1}{2}n$.
In your case we will try to calculate
$$F(n) = \sum_{k=1}^{n} k^3-2k.$$
Observe, that $P(k) = k^3-2k$ is a polynomial of degree $3$, and so, its sum will be of degree $4$ (this is closely related to the fact that the integral of polynomial of degree $d \in \mathbb{N}$ is of degree $d+1$).
Now, take an arbitrary polynomial of degree $4$:
$$F(n) = a_4n^4+a_3n^3+a_2n^2+a_1n+a_0$$
Note, that it has $5$ unknowns, namely $a_4,a_3,\ldots,a_0$. This allows us to take some $5$ values of $F$ calculated by hand and obtain a linear set of equations (here I started with $-2$ to make calculations easier, but starting with $0$ or any other number is ok too):
\begin{align*}
\begin{cases}
F(-2) &= -1 \\
F(-1) &= 0 \\
F(0) &= 0 \\
F(1) &= -1 \\
F(2) &= 3 \\
\end{cases}
\end{align*}
We could solve this right away, but it would be tedious. To make it shorter, note the two zeros at $0$ and $-1$, this means $F(n) = n(n+1)G(n)$ where $G$ is of degree $2$. Some simple manipulations (i.e. $G(n) = \frac{F(n)}{n(n+1)}$) get us
\begin{align*}
\begin{cases}
G(-2) &= \dfrac{-1}{(-2)\cdot(-1)} &= -\dfrac{1}{2} \\
G(1) &= \dfrac{-1}{1\cdot2} &= -\dfrac{1}{2} \\
G(2) &= \dfrac{3}{2\cdot3} &= \dfrac{1}{2} \\
\end{cases}
\end{align*}
Solving the above is easy, we get $G(n) = \frac{n^2}{4}+\frac{n}{4}-1$, and so $$F(n) = n(n+1)\left(\frac{n^2}{4}+\frac{n}{4}-1\right).$$
I hope this helps $\ddot\smile$
|
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How to solve the following system of equations. $2x^2-3xy+2y^2=2\frac{3}{4}\\x^2-4xy+y^2+\frac{1}{2}=0$
I tried all the methods that I know, but I could't isolate $x$ or $y$ to form one equation.
|
By multiplying the second by $-2$ you get:
$$2x^2-3xy+2y^2=\frac{6}{4}$$
$$-2x^2+8xy-2y^2=1$$
Then adding them you get:
$$5xy=\frac{10}{4} \Rightarrow xy=\frac{1}{2} \Rightarrow x=\frac{1}{2y}$$
Substituting this in the second equation you get:
$$\frac{1}{4y^2}-2+y^2+\frac{1}{2}=0 \Rightarrow \frac{1}{4y^2}+y^2=\frac{3}{2}$$
Multiplying this equation by $4y^2$ you get:
$$1+4y^4=6y^2 \Rightarrow 4y^4-6y^2+1=0$$
Let $u=y^2$:$$4u^2-6u+1=0$$
Then continue by solving this quadratic polynomial...
|
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How find this integral $I=\int_{-\infty}^{+\infty}\frac{x^3\sin{x}}{x^4+x^2+1}dx$ Find this integral
$$I=\int_{-\infty}^{+\infty}\dfrac{x^3\sin{x}}{x^4+x^2+1}dx$$
my idea:
$$I=2\int_{0}^{+\infty}\dfrac{x^3\sin{x}}{x^4+x^2+1}dx$$
because
$$\dfrac{x^3\sin{x}}{x^4+x^2+1}\approx\dfrac{\sin{x}}{x},x\to\infty$$
so
$$I=\int_{-\infty}^{+\infty}\dfrac{x^3\sin{x}}{x^4+x^2+1}dx$$
converges
then I can't,Thank you
|
May I confess that I sawer nicer integrals ?
Anyway, being patient and using Mhenni Benghorbal's suggestions plus a series of integrations by parts as well as a few changes of variables, the antiderivative is
$$\frac{1}{24} e^{-\sqrt[6]{-1}} \left(e^{\sqrt{3}} \left(\left(\sqrt{3}-3 i\right)
\left(\text{Ei}\left(i x+(-1)^{5/6}\right)-\text{Ei}\left(\frac{1}{2} \left(-2 i
x-\sqrt{3}+i\right)\right)\right)+\left(\sqrt{3}+3 i\right) e^i
\left(\text{Ei}\left(\frac{1}{2} \left(-2 i
x-\sqrt{3}-i\right)\right)-\text{Ei}\left(\frac{1}{2} \left(2 i
x-\sqrt{3}-i\right)\right)\right)\right)-\left(\sqrt{3}-3 i\right) e^i
\left(\text{Ei}\left(\frac{1}{2} \left(-2 i
x+\sqrt{3}-i\right)\right)-\text{Ei}\left(\frac{1}{2} \left(2 i
x+\sqrt{3}-i\right)\right)\right)+\left(\sqrt{3}+3 i\right)
\left(\text{Ei}\left(\frac{1}{2} \left(-2 i
x+\sqrt{3}+i\right)\right)-\text{Ei}\left(\frac{1}{2} \left(2 i
x+\sqrt{3}+i\right)\right)\right)\right)$$ Once integrated between $-\infty$ and $\infty$, again after some patience, I arrived to $$\frac{\left(-i+\sqrt{3}+\left(\sqrt{3}+i\right) e^i\right) e^{-\sqrt[6]{-1}} \pi }{2
\sqrt{3}}$$ that is to say $$-\frac{1}{3} e^{-\frac{\sqrt{3}}{2}} \pi \left(\sqrt{3} \sin
\left(\frac{1}{2}\right)-3 \cos \left(\frac{1}{2}\right)\right)$$ which reduces to $$\frac{2}{\sqrt{3}} e^{-\sqrt{3}/2} \pi \cos \frac{\pi+3}{6}$$ which corresponds to heropup's result.
Thanks to Mhenni Benghorbal and heropup for their suggestions and ideas.
|
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Factoring a hard polynomial This might seem like a basic question but I want a systematic way to factor the following polynomial:
$$n^4+6n^3+11n^2+6n+1.$$
I know the answer but I am having a difficult time factoring this polynomial properly. (It should be $(n^2 + 3n + 1)^2$). Thank you and have a great day!
|
Exploit the symmetry of the palindromic polynomial to factorise as follows,
$$\begin{align*}
n^4 + 6n^3 + 11n^2 + 6n + 1
&= n^2\left( n^2 + \frac{1}{n^2} + 6n + \frac{6}{n} + 11 \right)\\
&= n^2\left( (n + 1/n)^2 + 6(n + 1/n) + 9 \vphantom{\frac{1}{n^2}} \right)\\
&= n^2\left( n + 1/n + 3 \right)^2\\
&= \left( n^2 + 3n + 1 \right)^2,\\
\end{align*}$$
having essentially substituted $k = n + \frac{1}{n}$ and used $k^2 - 2 = n^2 + \frac{1}{n^2}$.
|
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|
QR-factorization of a tridiagonal matrix super diagonals question I understand it is possible to QR-factorize a tridiagonal matrix A by performing Given's plane rotations:
$$ J(n-1,n)J(n-2,n-1)... J(1,2) A =R$$ where $R$ is upper triangular. I have read that in this case the first two super-diagonals of R will be non-zero. I am having a hard time visualizing this. I can see the given product gives an upper-triangular easily, but i struggle to see what happens above the diagonal. Any help would be greatly appreciated.
EDIT:
I have found now that $J(k-1,k) $ only affects rows $k-1$ and $k \space \space$, so if we can show $$J(2,3)J(1,2)A$$ has nonzero etc then we are done
EDIT 2: I have shown it this way. If anyone still wishes to answer then go ahead.
|
When working with Givens rotations on structured matrices, it is always instructive to draw a picture and of course to notice that pre-multiplying with $J(k,k+1)$ acts only on rows $k$ and $k+1$ (and post-multiplying affects the columns $k$ and $k+1$). In the pictures, $\times$ denotes nonzero entries, $\color{red}\times$ indicates the entries affected by the rotation, $\color{blue}\otimes$ marks the entry eliminated by the rotation, and $\color{red}\otimes$ shows where a new entry will be created:
$$
\begin{split}
\underbrace{\begin{bmatrix}
\times & \times & & & & \\
\times & \times & & & & \\
& & 1 & & & \\
& & & 1 & & \\
& & & & 1 & \\
& & & & & 1
\end{bmatrix}}_{J(1,2)}
\underbrace{\begin{bmatrix}
\color{red}\times & \color{red}\times & \color{red}\otimes & & & \\
\color{blue}\otimes & \color{red}\times & \color{red}\times & & & \\
& \times & \times & \times & & \\
& & \times & \times & \times & \\
& & & \times & \times & \times \\
& & & & \times & \times
\end{bmatrix}}_{A}
&=
\underbrace{\begin{bmatrix}
\times & \times & \times & & & \\
& \times & \times & & & \\
& \times & \times & \times & & \\
& & \times & \times & \times & \\
& & & \times & \times & \times \\
& & & & \times & \times
\end{bmatrix}}_{J(1,2)A}
\\
\underbrace{\begin{bmatrix}
1 & & & & & \\
& \times & \times & & & \\
& \times & \times & & & \\
& & & 1 & & \\
& & & & 1 & \\
& & & & & 1
\end{bmatrix}}_{J(2,3)}
\underbrace{\begin{bmatrix}
\times & \times & \times & & & \\
& \color{red}\times & \color{red}\times & \color{red}\otimes & & \\
& \color{blue}\otimes & \color{red}\times & \color{red}\times & & \\
& & \times & \times & \times & \\
& & & \times & \times & \times \\
& & & & \times & \times
\end{bmatrix}}_{J(1,2)A}
&=
\underbrace{\begin{bmatrix}
\times & \times & \times & & & \\
& \times & \times & \times & & \\
& & \times & \times & & \\
& & \times & \times & \times & \\
& & & \times & \times & \times \\
& & & & \times & \times
\end{bmatrix}}_{J(2,3)J(1,2)A}
\\
\underbrace{\begin{bmatrix}
1 & & & & & \\
& 1 & & & & \\
& & \times & \times & & \\
& & \times & \times & & \\
& & & & 1 & \\
& & & & & 1
\end{bmatrix}}_{J(3,4)}
\underbrace{\begin{bmatrix}
\times & \times & \times & & & \\
& \times & \times & \times & & \\
& & \color{red}\times & \color{red}\times & \color{red}\otimes & \\
& & \color{blue}\otimes & \color{red}\times & \color{red}\times & \\
& & & \times & \times & \times \\
& & & & \times & \times
\end{bmatrix}}_{J(2,3)J(1,2)A}
&=
\underbrace{\begin{bmatrix}
\times & \times & \times & & & \\
& \times & \times & \times & & \\
& & \times & \times & \times & \\
& & & \times & \times & \\
& & & \times & \times & \times \\
& & & & \times & \times
\end{bmatrix}}_{J(3,4)J(2,3)J(1,2)A}
\\
\underbrace{\begin{bmatrix}
1 & & & & & \\
& 1 & & & & \\
& & 1 & & & \\
& & & \times & \times & \\
& & & \times & \times & \\
& & & & & 1
\end{bmatrix}}_{J(4,5)}
\underbrace{\begin{bmatrix}
\times & \times & \times & & & \\
& \times & \times & \times & & \\
& & \times & \times & \times & \\
& & & \color{red}\times & \color{red}\times & \color{red}\otimes \\
& & & \color{blue}\otimes & \color{red}\times & \color{red}\times \\
& & & & \times & \times
\end{bmatrix}}_{J(3,4)J(2,3)J(1,2)A}
&=
\underbrace{\begin{bmatrix}
\times & \times & \times & & & \\
& \times & \times & \times & & \\
& & \times & \times & \times & \\
& & & \times & \times & \times \\
& & & & \times & \times \\
& & & & \times & \times
\end{bmatrix}}_{J(4,5)J(3,4)J(2,3)J(1,2)A}
\\
\underbrace{\begin{bmatrix}
1 & & & & & \\
& 1 & & & & \\
& & 1 & & & \\
& & & 1 & & \\
& & & & \times & \times \\
& & & & \times & \times
\end{bmatrix}}_{J(5,6)}
\underbrace{\begin{bmatrix}
\times & \times & \times & & & \\
& \times & \times & \times & & \\
& & \times & \times & \times & \\
& & & \times & \times & \times \\
& & & & \color{red}\times & \color{red}\times \\
& & & & \color{blue}\otimes & \color{red}\times
\end{bmatrix}}_{J(4,5)J(3,4)J(2,3)J(1,2)A}
&=
\underbrace{\begin{bmatrix}
\times & \times & \times & & & \\
& \times & \times & \times & & \\
& & \times & \times & \times & \\
& & & \times & \times & \times \\
& & & & \times & \times \\
& & & & & \times
\end{bmatrix}}_{J(5,6)J(4,5)J(3,4)J(2,3)J(1,2)A=R}
\end{split}
$$
|
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|
$\frac{1}{1-e^{\frac{ik\pi}{n+1}}}+\frac{1}{1-e^{-\frac{ik\pi}{n+1}}}=1$? I'm working on an assignment where part of it is showing that $S_k=0$ for even $k$ and $S_k=1$ for odd $k$, where
$$S_k:=\sum_{j=0}^{n}\cos(k\pi x_j)= \frac{1}{2}\sum_{j=0}^{n}(e^{ik\pi x_{j}}+e^{-ik\pi x_{j}}) $$
Here $x_j=j/(n+1)$.
So, working through the algebra:
$$\frac{1}{2}\sum_{j=0}^{n}(e^{ik\pi x_{j}} +e^{-ik\pi x_{j}}) =\dots
=\frac{1}{2}\cdot\frac{1-e^{ik\pi}}{1-e^{\frac{ik\pi}{n+1}}}+\frac{1}{2}\cdot\frac{1-e^{-ik\pi}}{1-e^{-\frac{ik\pi}{n+1}}}
$$
Obviously $S_k=0$ for even $k$'s, since $e^{i\pi\cdot\text{even integer}}=1$. But when $k$ is odd we get $$\frac{1}{1-e^{\frac{ik\pi}{n+1}}}+\frac{1}{1-e^{-\frac{ik\pi}{n+1}}}$$
which isn't obviously one to me, at least. Wolfram alpha confirms it equals 1.
My question: How does one see that it equals 1?
|
To see that the given expression is equal to $1$, set
$\omega = e^{\frac{ik\pi}{n + 1}}, \tag{1}$
then the expression becomes
$\dfrac{1}{1 - \omega} + \dfrac{1}{1 - \bar{\omega}} = \dfrac{1 - \bar{\omega} + 1 - \omega}{(1 - \omega)(1 - \omega)} = \dfrac{2 - (\omega + \bar{\omega})}{(1 - \omega)(1 - \bar{\omega})}$
$= \dfrac{2 - (\omega + \bar{\omega})}{(1 + \omega \bar{\omega}) - (\omega + \bar{\omega})} = \dfrac{2 - (\omega + \bar{\omega})}{2 - (\omega + \bar{\omega})} = 1, \tag{2}$
since $\omega \bar{\omega} = 1$.
Hope this helps. Cheerio,
and as always,
Fiat Lux!!!
|
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|
Calculating $ x= \sqrt{4+\sqrt{4-\sqrt{ 4+\dots}}} $ If $ x= \sqrt{4+\sqrt{4-\sqrt{ 4+\sqrt{4-\sqrt{ 4+\sqrt{4-\dots}}}}}} $
then find value of 2x-1
I tried the usual strategy of squaring and substituting the rest of series by x again but could not solve.
|
Solution without complex factorization:
Let,
$ x= \sqrt{4+\sqrt{4-\sqrt{ 4+\dots}}} $
$ y= \sqrt{4-\sqrt{4+\sqrt{ 4-\dots}}} $
Clearly,
$x = \sqrt{4 + y} \implies x^2 - y = 4$ (i)
$y = \sqrt{4 - x} \implies y^2 + x = 4$ (ii)
from (i) and (ii) we get,
$x^2 - y = y^2 + x$
$\implies (x+y)(x-y) = x+y$
Since, $x,y \ge 0$, we have $x+y \ne 0$
So, $x = y + 1$ (iii)
Putting (iii) in (i) or (ii) we get:
$x = \frac{\sqrt{13}+1}{2}$.
|
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|
Difference of consecutive cubes never divisible by 5. This is homework from my number theory course.
Since $(x+1)^3-x^3=3x^2+3x+1$ and $x^3-(x+1)^3=-3x^2-3x-1$, to say that the difference of two cubes is divisible by 5 is the same as saying that $3x^2+3x+1\equiv 0\mod 5$ or $-3x^2-3x-1\equiv 0\mod 5$. Both of these statements imply that $x(x+1)\equiv 3\mod 5$. Thus I can finish this by showing that there are no such integers which satisfy $x(x+1)\equiv 3\mod 5$.
I want to say that it is sufficient to check by hand for the values 0,1,2,3, and 4 (for which it is not true), but other than following this by a messy induction I was wondering if there is an easier way to show that there are no integers such that $x(x+1)\equiv 3\mod 5$?
|
$x = 5n + k$, then $x^2 + x - 3 = (5n + k)^2 + 5n + k - 3 = k^2 + k - 3\pmod 5$ now check $k = 0, 1, 2, 3$, and $4$ you don't get $0\pmod 5$.
|
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|
Product of inverse matrices $ (AB)^{-1}$ I am unsure how to go about doing this inverse product problem:
The question says to find the value of each matrix expression where A and B are the invertible 3 x 3 matrices such that
$$A^{-1} = \left(\begin{array}{ccc}1& 2& 3\\ 2& 0& 1\\ 1& 1& -1\end{array}\right)
$$ and
$$B^{-1}=\left(\begin{array}{ccc}2 &-1 &3\\ 0& 0 &4\\ 3& -2 & 1\end{array}\right)
$$
The actual question is to find $ (AB)^{-1}$.
$ (AB)^{-1}$ is just $ A^{-1}B^{-1}$ and we already know matrices $ A^{-1}$ and $ B^{-1}$ so taking the product should give us the matrix
$$\left(\begin{array}{ccc}11 &-7 &14\\ 7& -4 &7\\ -1& 1 & 6\end{array}\right)
$$
yet the answer is
$$
\left(\begin{array}{ccc} 3 &7 &2 \\ 4& 4 &-4\\ 0 & 7 & 6 \end{array}\right)
$$
What am I not understanding about the problem or what am I doing wrong? Isn't this just matrix multiplication?
|
$(A^{-1}B^{-1})$ is not always equal to $(AB)^{-1}$.
Analogous to matrix transpose $(AB)^T = B^TA^T$, we have $(AB)^{-1} = B^{-1}A^{-1}$.
Further, matrix multiplication is not commutative. Here is a proof to show this, but we can see this fact from a simple counterexample involving two square matrices $A$ and $B$.
$A = \begin{pmatrix}1 & 2\\\ -1 & 0\end{pmatrix}$ $B = \begin{pmatrix}1 & -1\\\ 0 & 1\end{pmatrix}$
$AB = \begin{pmatrix}1 & 1\\\ -1 & 1\end{pmatrix}$, $BA = \begin{pmatrix}2 & 2\\\ -1 & 0\end{pmatrix}$
$AB \neq BA$
|
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|
Factoring a difference of 2 cubes I am trying to factorize the expression $(a - 2)^3 - (a + 1)^3$ and obviously I would want to put it in the form of $(a - b)(a^2 + ab + b^2)$
So I start off with the first $(a - b)$ and I get $(a - 2) - (a + 1)$ which I simplify from $(a^2 + a -2a -2)$ to $(a^2 -3a -2)$
Now I'm up to $(a^2 + ab + b^2)$ and I $a^2$ would equal to $(a^2 - 4)$, $ab$ would be $(a - 2) * (a + 1)$ which is $(a^2 + a -2a -2)$ and $b^2$ would be $(a^2 + 1)$..
Then we get $(a^2 - 3a - 2)((a^2 - 4) (a^2 + a - 4a)(a^2 + 1)) $
At this point I get confused because I'm not sure if I did anything correct, and I don't know how to continue this. Any help is much appreciated
Regards,
|
Also in the second parenthesis $a$ squared will be $(a-2)$ squared i.e. $(a-2)(a-2) = a^2 -4a + 4$,
and similarly the $b$ squared will be $(a+1)$ squared i.e. $(a+1)(a+1)$ i.e. $a^2 + 2a + 1$.
I think having $a$ in the expression you want to factorize and in the identity you are trying to use is confusing you. Why not instead factorize $(x-2)^3 - (x+1)^3$?
Then in your identity $a$ is $(x-2)$ and $b$ is $(x+1)$ and you get
$( (x-2)-(x+1) )( (x-2)^2 + (x-2)(x+1) + (x+1)^2)$
which finally simplifies down to $-3(3x^2 - 3x +3)$ which gives
$-9(1 - x +x^2)$.
Then turn it back to $a$ at the end to get
$-9(1 - a +a^2)$
and as a check, try $a = 0$:
$(0-2)^3 -(0+1)^3 = -8 - 1 = -9$
$-9(1 - 0 + 0^2) = -9(1) = -9$
|
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|
Find the arc length of the curve $x = 1/6*(y^2+ 3)^{3/2}$ from $y = 0$ to $y = 1$ I am trying to find the arclength of the curve $$x = 1/6\cdot\left(y^2 + 3\right)^{3/2},\;\; 0\leq y\leq 1$$ I got this far and now I am stuck and don't know what to do next. Any help please?
$$\begin{align} dx & = \left(1/6\right)\cdot\left(3/2\right)\cdot\left(y^2 + 3\right)^{1/2}\cdot\left(2y\right)\\ \\
& = (1/4)(2y)\cdot\left(y^2 + 3\right)^{1/2}\\ \\
& = (y/2)\cdot\left(y^{2}+3\right)^{1/2}\end{align}$$
$$\begin{align}L & = \int_0^1 \sqrt{1 + \left(y/2\left(y^2 + 3\right)^{1/2}\right)^{2}}\\ \\
&= \int_0^1 \sqrt{1 + \left(y^{2}/4\right)\left(y^2 + 3\right)}\\ \\
&= \int_0^1 \sqrt{1 + \left(y^{4} + 3y^{2}\right)/4}\end{align}$$
|
Simpson's Rule with n=2 already gives a result correct to four decimal places.
$$\frac16\left(1+4\sqrt{1+\left(\frac{1}{2^4}+3\left(\frac{1}{2^2}\right)\right)/4}+\sqrt{2}\right)\approx1.1336\ldots$$
(A CAS says that the integral is $1.13359\ldots$)
|
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|
How to solve simple trigonometry equation. So we are learning trigonometry in school and I would like to ask for a little help with these. I would really appreciate if somebody can explain me how I can solve such equations :)
*
*$\sin 3x \cdot \cos 3x = \sin 2x$
*$2( 1 + \sin^6 x + \cos^6 x ) - 3(\sin^4 x + \cos^4 x) - \cos x = 0$
*$3 \sin^2 x - 4 \sin x \cdot \cos x + 5 \cos^2 x = 2$
*$\sin^2 x - \sin^4 x + \cos^4 x = 1$
In our students book they're poorly explained with 2 pages, I tried to find a solution on the web, but still couldn't find similar examples. All we got from our teacher was paper with few formulas and we basically have no idea when to use them. I would show what I've tried, but the problem is that I have no idea to even start solving such equations.
|
For the third one
$$3\sin^2x-4\sin x\cos x+5\cos^2x=2\\
\Rightarrow\sin^2x-4\sin x\cos x+4\cos^2x=2-2\sin^2x-\cos^2x\\
\Rightarrow(\sin x-2\cos x)^2=2(1-\sin^2x)-\cos^2x\\
\Rightarrow(\sin x-2\cos x)^2=2\cos^2x-\cos^2x\\
\Rightarrow(\sin x-2\cos x)^2=cos^2x\\
\Rightarrow(\sin x-2\cos x)^2-\cos^2x=0\\
\Rightarrow(\sin x-2\cos x+\cos x)(\sin x-2\cos x-\cos x)=0\\
\Rightarrow(\sin x-\cos x)(\sin x-3\cos x)=0$$
Then
$$\sin x-\cos x=0\Rightarrow\tan x=1$$ and $$\sin x-3\cos x=0\Rightarrow\tan x=3$$
For the first one I don't know how far your book is covering. I have used the following two rules for that $$\sin 2x=2\sin x\cos x$$ and $$\sin 3x=3\sin x-4\sin^3x$$ So it goes like this
$$\sin 3x .\cos 3x=\sin2x\\
\Rightarrow 2\sin 3x .\cos 3x=2\sin2x\\
\Rightarrow\sin6x=2\sin2x\\
\Rightarrow\sin3(2x)=2\sin2x\\
\Rightarrow 3\sin2x-4\sin^32x=2\sin2x\\
\Rightarrow 3\sin2x-4\sin^32x-2\sin2x=0\\
\Rightarrow \sin2x-4\sin^32x=0\\
\Rightarrow \sin2x(1-4\sin^22x)=0\\
$$
Then $$\sin 2x=0$$ and $$\sin2x=\pm \frac{1}{2}$$
Hope it helps.
|
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|
Definite Integral $\int_0^1\frac{\ln(x^2-x+1)}{x^2-x}\,\mathrm{d}x$ $$\int_0^1\frac{\ln(x^2-x+1)}{x^2-x}\,\mathrm{d}x$$
WA gives $\pi^2/9$
|
Mhenni has struck first with the approach I have taken, but I would like to elaborate. Again, the integrand may be Taylor expanded:
$$\begin{align}-\int_0^1 dx \frac{\log{[1-(x-x^2)]}}{x-x^2} &= \sum_{n=0}^{\infty} \frac1{n+1} \int_0^1 dx \, x^n (1-x)^n\\ &= \sum_{n=0}^{\infty} \frac1{n+1} \frac{n!^2}{(2 n+1)!}\\ &=2 \sum_{n=0}^{\infty} \frac1{(2 n+2) (2 n+1) \binom{2 n}{n}} \end{align}$$
It turns out that
$$\frac{\arcsin{x}}{\sqrt{1-x^2}} = \sum_{n=0}^{\infty} \frac{2^{2 n} x^{2 n+1}}{(2 n+1) \binom{2 n}{n}} $$
So the sum in question is simply
$$4 \int_0^1 dx \frac{\arcsin{(x/2)}}{\sqrt{1-x^2/4}} = 8 \int_0^{\pi/6} d\theta \, \theta = \frac{\pi^2}{9}$$
as was to be shown.
|
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|
Quadratic Congruence for every prime Let $p$ be a prime number. Prove that exists $a,b \in \mathbb{Z}$ such that $p|a^2+b^2+1$. What I've tried:
If $p \equiv 1 \mod 4$, we put $b=0$ and the condition is simply $a^2 \equiv -1 \mod p$ which has a solution by basic quadratic reprocity.
If $p \equiv 3 \mod 4$, I did not see anything simple that might work and some examples don't show any apparent pattern ($7|3^2+2^2+1,$ $19|6^2+1^2+1$, $23 | 6^2+3^2+1$, etc.)
Thank you.
|
The result is obvious for $p=2$, and as you point out it is straightforward for primes of the form $4k+1$. But we give a proof that works for all odd primes.
Let $A$ be the set $\left\{0^2,1^2,2^2,\dots, \left(\frac{p-1}{2}\right)^2\right\}$.
Let $B$ be the set $\left\{-1-0^2, -1-1^2, -1-2^2, \dots, -1-\left(\frac{p-1}{2}\right)^2\right\}$.
It is easy to show that any two elements of $A$ are incongruent modulo $p$, as are any two elements of $B$.
But the number of elements of $A$ is $\frac{p-1}{2}+1$, as is the number of elements of $B$.
The sum of the sizes of $A$ and $B$ is $p+1\gt p$. Thus by the Pigeonhole Principle there exist $x\in A$, $y\in B$ such that $x\equiv y\pmod{p}$.
It follows that there is an $a$ and a $b$ such that $a^2\equiv -1-b^2\pmod{p}$. This is what we wanted to prove.
|
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|
Pre Calculus Expression The questions is:
$$\dfrac{3(x+2)^2(x-3)^2 - 2(x+2)^3(x-3)}{(x-3)^4}$$
My answer is: $$\dfrac{3(x+2)^2 + 6x^2-4}{(x-3)^2}$$
Am I right? If not, where have I failed?
|
$$\dfrac{3(x+2)^2(x-3)^2 - 2(x+2)^3(x-3)}{(x-3)^4}$$
Let's factor the $(x-3)$ in the numerator and denominator.
$$\dfrac{(x-3)\left[3(x+2)^2(x-3) - 2(x+2)^3\right]}{(x-3)(x-3)^3}$$
Now we can cancel out $(x-3)$ in the numerator and the denominator. That gives us:
$$\dfrac{3(x+2)^2(x-3) - 2(x+2)^3}{(x-3)^3}$$
Let us expand the equation.
$$\dfrac{3(x+2)^2(x-3) - 2(x+2)^3}{(x-3)^3}$$
$$=\dfrac{3(x^2+4x+4)(x-3)+2(x^3-6x^2+12x-8)}{x^3-9x^2+27x-27}$$
$$=\dfrac{3x^3-3x^2+4x^2-12x+4x-12+2x^3-12x^2+24x+16}{x^3-9x^2+27x-27}$$
Combine like terms:
$$\dfrac{3x^3-3x^2+4x^2-12x+4x-12+2x^3-12x^2+24x+16}{x^3-9x^2+27x-27}$$
$$=\dfrac{5x^3-11x^2-14x+4}{x^3-9x^2+27x-27}$$
We cannot factor the numerator nor simplify it any further. So, the answer to your question is:
$$\displaystyle \boxed{\dfrac{3(x+2)^2(x-3)^2 - 2(x+2)^3(x-3)}{(x-3)^4}=\dfrac{5x^3-11x^2-14x+4}{x^3-9x^2+27x-27}}$$
|
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|
How do I get from $\frac{-x+1}{-x+2}$ to $1 + \frac{1}{x-2}$ wolframalpha tells me it's the same but I can not follow how to get from one to another.
$$\frac{-x+1}{-x+2} = \frac{1-x}{2-x} = \>? \dots$$
I don't get any further, always end up where I started.
|
$$\begin{align} \dfrac{1-x}{2-x} & = \dfrac{-(x - 1)}{-(x-2)} \\ \\ &= \dfrac{x-1}{x-2} \\ \\ &= \dfrac {x-1 \color{blue}{\bf - 1 + 1}}{x-2} \\ \\ & = \dfrac{(x-2) +1}{x - 2}\\ \\ & = \dfrac{x-2}{x-2} + \dfrac 1{x-2}\\ \\ & = 1 + \frac{1}{x - 2}\end{align}$$
|
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|
How do I solve delta epsilon proofs for quadratic equations? For the $\lim\limits_{x\rightarrow 2} (x^2 + 5x - 2) = 12$ I need to show how to find a $\delta$ such that $|f(x) - L| < \varepsilon$ for all $x$ satisfying $0 < |x - a| < \delta$.
Help is appreciated.
|
We have show that for all $\epsilon>0$ there is a $\delta >0$ such that
$$
|x-x_0|<\epsilon \implies | (x^2+5x-2) - 12|<\epsilon .
$$
We will manipulate the expression $| (x^2+5x-2) - 12|<\epsilon$ to find a candidate to $\delta$.
\begin{align}
| (x^2+5x-2) - 12|<\epsilon \implies & -\epsilon <x^2+5x-14<\epsilon \\
\implies& -\epsilon <\left(x-\frac{5}{2}\right)^2 -\frac{25}{4}-14<\epsilon \\
\implies& \frac{81}{4}-\epsilon <\left(x-\frac{5}{2}\right)^2<\frac{81}{4}+\epsilon \\
\implies& \sqrt{\frac{81}{4}-\epsilon} <\left|x-\frac{5}{2}\right|<\sqrt{\frac{81}{4}+\epsilon} \\
\implies& \left|x-\frac{5}{2}\right|<\sqrt{\frac{81}{4}+\epsilon} \\
\implies& - \sqrt{\frac{81}{4}+\epsilon}<x-\frac{5}{2}<\sqrt{\frac{81}{4}+\epsilon} \\
\implies& \frac{1}{2}- \sqrt{\frac{81}{4}+\epsilon}<x-2<\sqrt{\frac{81}{4}+\epsilon}+\frac{1}{2} \\
\implies& |x-2|<\max\left\{\left|-\sqrt{\frac{81}{4}+\epsilon}+\frac{1}{2}\right|,\left| +\sqrt{\frac{81}{4}+\epsilon}+\frac{1}{2}\right| \right\} \\
\end{align}
Set $\delta=\max\left\{\left|-\sqrt{\frac{81}{4}+\epsilon}+\frac{1}{2}\right|,\left| +\sqrt{\frac{81}{4}+\epsilon}+\frac{1}{2}\right| \right\}$. In the above inequalities there is no loss of generality if we fix $\epsilon < \frac{81}{4}$. Then
\begin{align}
0<|x-2|<\delta \implies & -\delta < x-2 <+\delta \\
\implies& \frac{1}{2}- \sqrt{\frac{81}{4}+\epsilon}<x-2<\sqrt{\frac{81}{4}+\epsilon}+\frac{1}{2} \\
\vdots\quad & \\
& \\
\vdots \quad & \\
\implies & | (x^2+5x-2) - 12|<\epsilon
\end{align}
|
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|
Prove an equation about binomial coefficients Could we prove:
$ \sum_{k} \binom{2k}{k}\binom{n+k}{m+2k} \frac{(-1)^k}{k+1} = \binom{n-1}{m-1}$ when $m,n \in N$
|
Here is a similar approach using basic complex variables.
Suppose we seek to evaluate
$$\sum_{k\ge 0} {2k\choose k} {n+k\choose m+2k}
\frac{(-1)^k}{k+1}$$
where $n\ge m.$
Introduce the integral representation
$${n+k\choose m+2k}
= \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^{m+2k+1}} (1+z)^{n+k} \; dz.$$
This yields the following integral for the sum:
$$\frac{1}{2\pi i}
\int_{|z|=\epsilon}
\sum_{k\ge 0} {2k\choose k} \frac{(-1)^k}{k+1}
\frac{1}{z^{m+2k+1}} (1+z)^{n+k} \; dz
\\= \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{(1+z)^n}{z^{m+1}}
\sum_{k\ge 0} {2k\choose k} \frac{(-1)^k}{k+1}
\frac{(1+z)^k}{z^{2k}} \; dz.$$
Now the OGF of the Catalan numbers is
$$\sum_{k\ge 0} {2k\choose k} \frac{1}{k+1} w^k
= \frac{1-\sqrt{1-4w}}{2w}.$$
Substituting this closed form that we computed
into the integral yields
$$\frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{(1+z)^n}{z^{m+1}}
\frac{1-\sqrt{1+4(1+z)/z^2}}{(-1)\times 2(1+z)/z^2} \; dz
\\= - \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{(1+z)^n}{z^{m-1}}
\frac{1-\sqrt{1+4(1+z)/z^2}}{2(1+z)} \; dz
\\= - \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{(1+z)^n}{z^m}
\frac{z-\sqrt{z^2+4(1+z)}}{2(1+z)} \; dz.$$
Observe that $z^2+4(1+z) = (z+2)^2$ so the integral becomes
$$- \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{(1+z)^n}{z^m}
\frac{z-(z+2)}{2(1+z)} \; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{(1+z)^{n-1}}{z^m} \; dz.$$
This last integral can be evaluated by inspection and yields
$$[z^{m-1}] (1+z)^{n-1} =
{n-1\choose m-1}.$$
Note that for the convergence of the Catalan GF series we require $|(1+z)/z^2|\lt 1/4$ (distance to the nearest singularity). Now $|(1+z)/z^2|\le (1+\epsilon)/\epsilon^2$ so we may take $\epsilon = N \ge 5.$
Remark: the same functions appear here as in @robjohn's answer,
who was first.
A trace as to when this method appeared on MSE and by whom starts at this
MSE link.
|
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|
Proof of $ e^x-x^2 \gt 1 $ when $ x \gt 0$ and $x$ is a real number . I want to Prove $ e^x-x^2 \gt 1 $ when $ x \gt 0$ and $x$ is a real number . For this purpose , my trying is as the following :
$ e^x-x^2 = \{1+x+\dfrac{x^2}{2!} + +\dfrac{x^3}{3!}++\dfrac{x^4}{4!}+ ....... \}-x^2$
$= 1+\{x-\dfrac{x^2}{2!} + +\dfrac{x^3}{3!}++\dfrac{x^4}{4!}+ ....... \}$
So $ e^x-x^2 \gt 1 $ is true if $\{x-\dfrac{x^2}{2!} + +\dfrac{x^3}{3!}++\dfrac{x^4}{4!}+ ....... \} \ge $ Please help me finding the convergence of this term .
|
The second derivative of $e^x-x^2-1$ gives you the inflection point at $x = ln(2)$.
The problem is equivalent to $e^{ln(2)}-len(2)^2>1$ or $1>ln(2)$, wich is true
|
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|
Find complex solutions Find all solutions of the equation $4\sin(z) + 5 = 0$.
So $4\sin(z) + 5 = 4\sin(x+iy) + 5 = 0$. Hence $\sin(x+iy) = \frac{-5}{4}$. So $x+iy = \sin^{-1}(\frac{-5}{4})$ which gives us $x+iy =$ ...
I am not sure what is really being asked here or how to achieve it. Help?
|
$\displaystyle\sin z=-\frac54$
$\displaystyle\cos z=\pm\sqrt{1-\frac{25}{16}}=\pm\frac{3i}4$
Using Euler formula, $$e^{iz}=\cos z+i\sin z=\pm\frac{3i}4+i\left(-\frac54\right)=-\frac{(5\mp3)i}4=-2i\text{ or }-\frac i2$$
|
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|
EXERCISE VERIFICATION: Find where $f(x):=|x|+|x+1|$ is differentiable and calculate its derivative Could someone verify my exercise?
a) $f(x):=|x|+|x+1|$
First, analyse the roots of each absolute value, where they go to zero:
$$|x|:=\left\{\begin{matrix}
& x& x>0 \\
& x- & x<0\\
& 0 & x=0
\end{matrix}\right.$$
$$|x+1|:=\left\{\begin{matrix}
& x+1& x>-1 \\
& -(x+1) & x<-1\\
& 0 & x=-1
\end{matrix}\right.$$
So the principal function will be:
$$f(x):=|x|+|x+1|:=\left\{\begin{matrix}
& x+x+1 &=2x+1& x\ge0 \\
& -x+x+1&=1 & -1\le x<0\\
& -x-x-1 &=-(2x+1) & x<-1
\end{matrix}\right.$$
So, the derivatives wil be:
for $x\ge0$:
$$\lim_{x\to c} \dfrac{f(x)-f(c)}{x-c}=\lim_{x\to c} \dfrac{2x+1-2c-1}{x-c}=\lim_{x\to c} \dfrac{2(x-c)}{x-c}=\lim_{x\to c}2= 2$$
for $-1\le x<0$:
$$\lim_{x\to c} \dfrac{f(x)-f(c)}{x-c}=\lim_{x\to c} \dfrac{1-1}{x-c}=\lim_{x\to c} \dfrac{0}{x-c}=\infty$$ so, $f(x)$ isn´t differentiable at $.-1\le x<0$
for $x<-10$:
$$\lim_{x\to c} \dfrac{f(x)-f(c)}{x-c}=\lim_{x\to c} \dfrac{-(2x+1)+(2c+1)}{x-c}=\lim_{x\to c} \dfrac{-2(x-c)}{x-c}=\lim_{x\to c}-2= -2$$
So, $f(x)$ isn´t differentiable at $-1\le x<0$
|
It is a good idea to draw the function first:
From this we guess that the function is differentiable except at $x=-1$ and $x=0$.
Your formula for $f$ above is correct, and from this we see that for $x \in \mathbb{R} \setminus \{-1,0\}$, the function $f$ is differentiable.
For $x=-1$, we see that ${f(-1+h)-f(-1) \over h} = 0$ for $h \in (0,1)$ and
${f(-1+h)-f(-1) \over h} = 2$ for $h \in (-1,0)$, hence it cannot be differentiable at $x=-1$ (since the limit as $h \to 0$ doesn't exist). The same
sort of analysis shows that $f$ is not differentiable at $x=0$ as well.
|
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|
$\epsilon$-$\delta$ limit proof that $\lim_{n\to \infty}\frac{n^2-n+2}{3n^2+2n-4}=\frac{1}{3}$ I need to prove that $$\lim_{n\to \infty}\frac{n^2-n+2}{3n^2+2n-4}=\frac{1}{3}$$ using the epsilon definition.
I'm having specific trouble understanding how to make it less than epsilon once I've simplified the equation.
|
$$
\left|\frac{n^2-n+2}{3n^2+2n-4} - \frac 1 3\right| = \left|\frac{3(n^2-n+2)}{3(3n^2+2n-4)} - \frac{3n^2+2n-4}{3(3n^2+2n-4)}\right|
$$
$$
= \left|\frac{-5n+10}{3(3n^2+2n-4)}\right| \le \frac{\text{some constant}}{n}
$$
Massage the whole thing a bit to figure out what "constant" can serve in this role, and then the whole thing is ${}<\varepsilon$ if $n$ is big enough.
PS:
We have
$$
\left|\frac{-5n+10}{3(3n^2+2n-4)}\right|.
$$
$$
3n^2 + 2n - 4.
$$
Let's try to show that this is $>n^2$ if $n$ is big enough. We want
\begin{align}
3n^2+2n - 4 & > n^2 \\[6pt]
2n^2 + 2n - 4 & > 0 \\[6pt]
2\left(n^2+n+\frac 1 4 \right) - 4 - \frac 1 2 & > 0 \tag{completing the square} \\[6pt]
\left(n+\frac 1 2 \right)^2 & > \frac 9 4 \\[6pt]
n & > 1.
\end{align}
So we have
$$
\left|\frac{-5n+10}{3(3n^2+2n-4)}\right| < \frac{|-5n+10|}{n^2}\text{ if } n \ge 2,
$$
and
$$
|-5n+10| \le |-5n|+|10| = 5n+10 = 5(n+1).
$$
Can we prove $\dfrac{5(n+1)}{n^2} \le \dfrac 6 n \text{ if }n>\text{something?}$
We would need $5(n+1)\le 6n$. That yields $n\ge 6$.
We need $n\ge 2$ and $n\ge 6$.
So if $n\ge 6$ then the fraction above is $<\dfrac 6 n$.
(You should check details and adjust them as needed . . .)
Then if $n>\dfrac 6 \varepsilon$, you've got it.
|
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|
How do I find the maximum and minimum of a sinusoidal function? I understand basic $\sin x$ and $\cos x$ min/max, but I am having a problem solving the minimum and maximum of the following:
$f(x) = \sin^2 x - \sin x$
Oh, and the range is $0 \le x \le \frac{3\pi}{2}$
|
$$\begin{align*} f(x) &= \sin^2 x - \sin x \\ &= \sin^2 x - 2 \cdot \tfrac{1}{2} \sin x + (\tfrac{1}{2})^2 - \tfrac{1}{4} \\ &= \left( \sin x - \tfrac{1}{2} \right)^2 - \tfrac{1}{4}\end{align*}.$$ Because the square of a real number is nonnegative, $f$ attains a minimum if $\sin x = \frac{1}{2}$, and the consequences are straightforward.
To determine the maximum value, observe that $|\sin x| \le 1$; consequently, $f$ is maximized if $(\sin x - \tfrac{1}{2})^2$ is made as large as possible. By inspection, this occurs if $\sin x = -1$.
|
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|
For any Fermat number $F_n=2^{2^n}+1$ with $n>0$, establish that $F_n\equiv5\,\text{or}\,8\pmod9$ For any Fermat number $F_n=2^{2^n}+1$ with $n>0$, establish that $F_n\equiv5\,\text{or}\,8\pmod9$ according as $n$ is odd or even.
Since for any $n\ge3$,$2^{2^n}\equiv2^{2^{n-2}}\pmod9$, if $n=2k+3$
then
$$2^{2^n}=2^{2^{2k+3}}\equiv2^{2^{2k+1}}=(2^{2^{2k}})^2\pmod9?$$
if $n=2k+4$
then
$$2^{2^n}=2^{2^{2k+4}}\equiv2^{2^{2k+2}}=(2^{2^{2k}})^4\pmod9?$$
but I don't know how continue?
|
The proof is by induction on $n$. We have $F_1\equiv 5\pmod{9}$.
Note that $F_{n+1}-1=(F_n-1)^2$. This is because $2^{2^{n+1}}=(2^{2^n})^2$.
Rewrite as
$$F_{n+1}=(F_n-1)^2+1.$$
Now we do the induction step.
If $F_k\equiv 5\pmod{9}$, then $(F_k-1)^2\equiv 16\equiv 7\pmod{9}$, and therefore $F_{k+1}\equiv 8\pmod{9}$.
Similarly, if $F_k\equiv 8\pmod{9}$, then $(F_k-1)^2\equiv 4\pmod{5}$, and therefore $F_{k+1}\equiv 5\pmod{9}$.
|
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|
Finding an algebraic proof for $r{n \choose r} = n{n-1 \choose r-1}$ I can't seem figure this proof out.
How are both sides equal.
$$r{n \choose r} = n{n-1 \choose r-1}$$
|
Here are a couple of approaches in addition to the approach used by Dror and TMM.
Generating Functions:
$$
\frac{\mathrm{d}}{\mathrm{d}x}(1+x)^n=\sum_{r=1}^n r\binom{n}{r}x^{r-1}
$$
$$
\begin{align}
n(1+x)^{n-1}
&=\sum_{r=0}^{n-1}n\binom{n-1}{r}x^r\\
&=\sum_{r=1}^nn\binom{n-1}{r-1}x^{r-1}\\
\end{align}
$$
Compare the coefficients of $x^{r-1}$
Induction: Suppose it is true for row $n-1$ in Pascal's Triangle, then
$$
\begin{align}
r\binom{n}{r}
&=r\left[\binom{n-1}{r}+\color{#C00000}{\binom{n-1}{r-1}}\right]\\
&=r\binom{n-1}{r}+\color{#C00000}{(r-1)\binom{n-1}{r-1}+\binom{n-1}{r-1}}\\
&=\color{#00A000}{(n-1)\binom{n-2}{r-1}+(n-1)\binom{n-2}{r-2}}+\binom{n-1}{r-1}\\
&=\color{#00A000}{(n-1)\binom{n-1}{r-1}}+\binom{n-1}{r-1}\\
&=n\binom{n-1}{r-1}
\end{align}
$$
|
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|
Solving an equality in 2 variables I need to prove that
$$\left(a + \frac{1}{a}\right)^2 +\left(b + \frac{1}{b}\right)^2 \gt \frac{25}{2}$$
if $a+b = 1$ and $a b \le 1/4$
I'd like a hint. Solve the equality first to $a$ or $b$, or stay in a and b as to get
$a b \le 4$ in the inequality ?
|
Apply Cauchy-Schwarz Inequality to argue that,
$(a+\frac{1}{a}+b+\frac{1}{b})^2 \le (1+1)((a+\frac{1}{a})^2+(b+\frac{1}{b})^2)$,
And further, $\frac{1}{a}+\frac{1}{b} = \frac{a+b}{ab}=\frac{1}{ab}\ge \frac{4}{(a+b)^2}=4$.
So the first expression becomes,
$25=(1+4)^2 \le (a+\frac{1}{a}+b+\frac{1}{b})^2 \le (1+1)((a+\frac{1}{a})^2+(b+\frac{1}{b})^2)$
That is $(a+\frac{1}{a})^2+(b+\frac{1}{b})^2 \ge 25/2$.
and Equality holds iff $a=b=\frac12$.
|
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|
How can I solve this question? Compute the value of the following improper integral. If it is divergent, type "Diverges" or "D".
$$\int_0^2 \frac{dx}{\sqrt{4-x^2}}$$
Do I make $u= 4-x^2$ then $du= -2x \, dx$
Not exactly sure..
|
Consider the substitution $x = 2\sin\theta$. From this we have $dx = 2\cos\theta\cdot d\theta$. Substitute these in and see what happens:
$$\int\frac{2\cos\theta \cdot d\theta}{\sqrt{4 - 4\sin^2\theta}}\\\\
= \int\frac{2\cos\theta \cdot d\theta}{2\cos\theta}\\
= \int d\theta\\
= \theta + C\\
= \sin^{-1}{\frac{x}{2}} + C$$
Hence,
$$\int_0^2\frac{dx}{\sqrt{4 - x^2}} = \left[\sin^{-1}{\frac{x}{2}}\right]_0^2\\
= \frac{\pi}{2}$$
|
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|
Solve two equations for $a$ and $b$ \begin{cases}
c_2=\dfrac{c_1}{a} \left( \left(\dfrac{c_3}{b}\right)^3 - 1 \right) \\[2ex]
b^2 = a^2 + c_3^2 + 2(a)\, (c_3)\, (c_4) \\
\end{cases}
I am stuck at this point. Not sure on how to move forward. ( A small change made)
|
$a=(\frac{c_1}{c_2})((\frac{c_3}{b})^3-1)$
$a^2=(\frac{c_1}{c_2})^2((\frac{c_3}{b})^3-1)^2$
substitute $a^2$ and $a$ in the second equation
$b^2=(\frac{c_1}{c_2})^2((\frac{c_3}{b})^3-1)^2+c_3^2+2c_3c_4(\frac{c_1}{c_2})((\frac{c_3}{b})^3-1)$
$b^2c_2^2=c_1^2(c_3^3-b^3)^2+c_3^2c_2^2b^3+2c_3c_4c_1c_2(c_3^3-b^3)$
$0=c_1^2c_3^6-2c_1^2c_3^3b^3+b^6c_1^2+c_3^2c_2^2b^3+2c_3c_4c_1c_2c_3^3-2c_3c_4c_1c_2b^3-b^2c_2^2$
$0=b^6c_1^2+b^3(c_3^2c_2^2-2c_1^2c_3^3-2c_3c_4c_1c_2)-b^2c_2^2+c_1^2c_3^6+2c_3c_4c_1c_2c_3^3$
$b^3c_3^2c_2^2-2c_1^2c_3^3b^3-2c_3c_4c_1c_2b^3=b^2c_2^2-b^6c_1^2-c_1^2c_3^6-2c_3c_4c_1c_2c_3^3$
here we see that $b=c_3$ will make the expresion valid(LHS=RHS),the possibility of other roots is not obvious,and is questioneable
|
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|
evaluation of $\int \cos (2x)\cdot \ln \left(\frac{\cos x+\sin x}{\cos x-\sin x}\right)dx$
Compute the indefinite integral
$$
\int \cos (2x)\cdot \ln \left(\frac{\cos x+\sin x}{\cos x-\sin x}\right)\,dx
$$
My Attempt:
First, convert
$$
\frac{\cos x+\sin x}{\cos x-\sin x} = \frac{1+\tan x}{1-\tan x} = \tan \left(\frac{\pi}{4}+x\right)
$$
This changes the integral to
$$
\int \cos (2x)\cdot \ln \left(\tan \left(\frac{\pi}{4}+x\right)\right)\,dx
$$
Now let $t=\left(\frac{\pi}{4}+x\right)$ such that $dx = dt$. Then the integral with changed variables becomes
$$
\begin{align}
\int \cos \left(2t-\frac{\pi}{2}\right)\cdot \ln (\tan t)dt &= \int \sin (2t)\cdot \ln (\tan t)dt\\
&= -\ln(\tan t)\cdot \frac{\cos (2t)}{2}+\frac{1}{2}\int \frac{\sec^2(t)}{\tan t}\cdot \cos (2t)\\
&= -\ln(\tan t)\cdot \frac{\cos (2t)}{2}+\frac{1}{2}\int \cot (2t)dt\\
&= -\ln(\tan t)\cdot \frac{\cos (2t)}{2}+\frac{1}{2}\ln \left|\sin (2t)\right|
\end{align}
$$
where $t=\displaystyle \left(\frac{\pi}{4}+x\right)$.
Is this solution correct? Is there another method for finding the solution?
|
Let's recall the definition of the Inverse Hyperbolic Tangent:
$$\begin{cases}\operatorname{arctanh}(x)=\frac{1}{2}\ln\left(\frac{1+x}{1-x}\right)\\\frac{d}{dx}\operatorname{arctanh}(x)=\frac{1}{1-x^2}\end{cases}; x\in(-1,1)$$
So, the integral can be rewritten as:
$$\require{cancel}\begin{align}\int\cos(2x)\log\left(\frac{\cos(x)+\sin(x)}{\cos(x)-\sin(x)}\right)dx &=\color{red}{2}\int\cos(2x)\color{red}{\frac{1}{2}}\log\left(\frac{\cancel{\cos(x)}}{\cancel{\cos(x)}}\left(\frac{1+\frac{\sin(x)}{\cos(x)}}{1-\frac{\sin(x)}{\cos(x)}}\right)\right)dx\\&=2\int\cos(2x)\operatorname{arctanh}\left(\tan(x)\right)dx\\&= \sin(2x)\operatorname{arctanh}\left(\tan(x)\right)-\int\frac{\sin(2x)}{1-\tan^2(x)}\sec^2(x)dx\\&= \sin(2x)\operatorname{arctanh}\left(\tan(x)\right)-\int\frac{\sin(2x)\cancel{\cos^2(x)}\cancel{\sec^2(x)}}{\underbrace{\cos^2(x)-\sin^2(x)}_{\cos(2x)}}dx\\&= \sin(2x)\operatorname{arctanh}\left(\tan(x)\right)+\frac{\log\left|\cos(2x)\right|}{2}+C\end{align}$$
|
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|
Help with proof using induction: $1 + \frac{1}{4} + \frac{1}{9}+\cdots+\frac{1}{n^2}\leq 2-\frac{1}{n}$ I am having trouble with the following proof:
For every positive integer $n$: $$1 + \frac{1}{4} + \frac{1}{9}+\cdots+\frac{1}{n^2}\leq 2-\frac{1}{n}$$
My work: I have tried to add $\frac{1}{(k+1)^2}$ to $2-\frac{1}{k}$ in the inductive step and reduce it down to $2-\frac{1}{k+1}$ but cannot do so. I am beginning to think that my entire approach is wrong.
|
Inductive Hypothesis: suppose $\sum_{k=1}^n \frac{1}{k^2} \leq 2 - \frac{1}{n}$.
Inductive Step: then suppose $\sum_{k=1}^{n+1} \frac{1}{k^2} \leq 2 - \frac{1}{n} + \frac{1}{(1+n)^2}. $
So it suffices to show that $ - \frac{1}{n} + \frac{1}{(1+n)^2} \leq - \frac{1}{n+1}$.
But this, by simple algebra, is equivalent to $ n \leq n+1 $, which is obviously true $\forall n \in \mathbb{N}$.
EDIT Proof that $ - \frac{1}{n} + \frac{1}{(1+n)^2} \leq - \frac{1}{n+1}$ is equivalent to $n \leq n+1$.
Add $\frac{1}{n}$ to both sides to se that what we need to show is $\frac{1}{(1+n)^2} \leq \frac{1}{n} - \frac{1}{n+1} = \frac{1}{n(n+1)}.$ Multiply both sides by $(n+1)$. Then take reciprocals and remember to swap the inequality sign.
|
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|
If a prime can be expressed as sum of square of two integers, then prove that the representation is unique. If a prime can be expressed as sum of two squares, then prove that the representation is unique.
My attempt:
If $a^2+b^2=p$, then it is obvious that $a,b$ of different parity.
Now, I assume the contraposition that the representation is not unique, $p=a^2+b^2=c^2+d^2$. Again, $c,d$ are of different parity.
Now, let $b,d$ be even and $a,c$ be odd.
So, $a^2+b^2=c^2+d^2 \implies a^2-c^2=d^2-b^2 \implies (a+c)(a-c)=(d-b)(d+b)$.
I cannot proceed any further. Please help.
|
Here's an answer without Algebraic Number Theory. I found it in Shanks, Solved and Unsolved Problems in Number Theory.
Assume $$p=a^2+b^2=c^2+d^2\tag1$$ with all variables positive integers. Then $$p^2=(a^2+b^2)(c^2+d^2)=a^2c^2+a^2d^2+b^2c^2+b^2d^2$$ and you can verify by just multiplying everything out that $$p^2=(ac+bd)^2+(ad-bc)^2\tag2$$ and $$p^2=(ac-bd)^2+(ad+bc)^2\tag3$$ By (1), we have $$(p-a^2)d^2=(p-c^2)b^2$$ which implies $$p(d^2-b^2)=(ad-bc)(ad+bc)\tag4$$ From (4), $p$ divides $ad-bc$, or $p$ divides $ad+bc$. If $p$ divides $ad-bc$, then from (2) we get $ad-bc=0$, so $d^2-b^2=0$, so $b=d$. If $p$ divides $ad+bc$, then from (3) we get $ac=bd$. Now $a$ and $b$ are relatively prime, so $a$ divides $d$, and $b$ divides $c$. Then by (1) we have $a=d$, and we have proved that the two representations of $p$ are the same.
This is probably something like what @Konstantinos was getting at in his answer.
|
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|
Why is $ \lim_{x \to \infty} \ x^{2/x} = 1$ Why is $\displaystyle \lim_{x \to \infty} \ x^{2/x} = 1$ since this is an indeterminate form $\infty^{0}$ and I can't see any manipulation that would suggest this result?
|
$\lim \limits_{x \to \infty} x^{\frac{1}{x}} = 1$
Proof using AM-GM and Sandwich Theorem
$\frac{1 + 1 + 1 + \dots + \sqrt{x} + \sqrt{x}}{x} \geq \sqrt[x]{x} \geq 1$
$\frac{x - 2 + 2\sqrt{x}}{x} \geq \sqrt[x]{x} \geq 1$
$1 - \frac{2}{x} + \frac{2}{\sqrt{x}} \geq \sqrt[x]{x} \geq 1$
$\lim \limits_{x \to \infty} 1 - \frac{2}{x} + \frac{2}{\sqrt{x}} = 1$
$\therefore \lim \limits_{x \to \infty} x^{\frac{1}{x}} = 1 \implies \lim \limits_{x \to \infty} x^\frac{2}{x} = 1$
|
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|
Could someone please help me evaluate this integral? The integral is
$$
\int_{0}^{1/\sqrt{\vphantom{\large A}2\,}}
{\arccos\left(x\right) \over \sqrt{\vphantom{\large A}1 -x^2\,}\,}\;{\rm d}x
$$
I was just wondering if I could use substitution to solve this problem or if I had to solve it a different way.
|
We have to evaluate the definite integral:
$$\int_0^{1/\sqrt{2}}\frac{\arccos x}{\sqrt{1-x^2}} \ dx$$
Just use u-substitution. Let:
$$u=\arccos x \implies du=-dx$$
$$\int_0^{1/\sqrt{2}}u \ -du=-\int_0^{1/\sqrt{2}}u \ du=\left.-\frac{u^2}{2}\right|_0^{1/\sqrt{2}}$$
Substituting $\arccos x$ back for $u$:
$$\int_0^{1/\sqrt{2}}\frac{\arccos x}{\sqrt{1-x^2}} \ dx=\left.-\frac{1}{2}(\arccos x)^2+C \ \right|_0^{1/\sqrt{2}}$$
Now we just evaluate the definite integral.
$$\left.-\frac{1}{2}(\arccos x)^2+C \ \right|_0^{1/\sqrt{2}}=\left[-\frac{1}{2}\arccos \left(\frac{1}{\sqrt{2}}\right)^2+C\right]-\left[-\frac{1}{2}(\arccos 0 )^2+ C\right]$$
You should know that $\arccos \left(\frac{1}{\sqrt{2}}\right)=\dfrac{\pi}{4}$ and $\arccos 0=\dfrac{\pi}{2}$.
$$-\frac{1}{2}\cdot\left(\frac{\pi}{4}\right)^2+C+\frac{1}{2}\cdot\left(\frac{\pi}{2}\right)^2-C$$
$$=-\frac{1}{2}\cdot\frac{\pi^2}{16}+\frac{1}{2}\cdot\frac{\pi^2}{4}$$
$$=\frac{\pi^2}{8}-\frac{\pi^2}{32}$$
$$=\frac{4\pi^2}{32}-\frac{\pi^2}{32}$$
$$=\frac{3\pi^2}{32}$$
$$\displaystyle \color{green}{\therefore \int_0^{1/\sqrt{2}}\dfrac{\arccos x}{\sqrt{1-x^2}} \ dx=\dfrac{3\pi^2}{32}}$$
Hope I helped!
|
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Maximum of $(1-q_1)(1-q_2)\ldots(1-q_n)$ I'm trying to find the maximum of $(1-q_1)(1-q_2)\ldots(1-q_n)$ where $n\ge 2$, on a the set $\{(q_1,\ldots , q_n) :q_1^2+q_2^2+\ldots+q_n^2=1 \ q_i\ge 0 \}$ (With the condition $q_i\ge0$ this is just the upper half of the sphere). This appeared to be a simple Lagrange multipliers question but calculating the derivative is becoming a problem. How is this done?
I was thinking of trying the fact that the product is the volume of a box having $(1,1\ldots 1)$ and $(q_1,\ldots ,q_n)$ at its diagonal but I can't figure out how to continue.(I'm guessing this reduces the problem to something geometrical)
Updates: Empirical evidence suggests that the maximum is when the $q_i$ are equal as pointed out by @Sabyasachi , I let WolframAlpha solve the Laplace multiplier equations for $n=2,3,4$ and got the same result(that they are equal). One solution which does not turn up in WA's solutions is that $q_3=0$ and $q_1=q_2=\frac {1} {\sqrt{2}}$ which is on the sphere and returns a larger value.
|
Addendum: I've included the full set of solutions Mathematica finds for the case $n=3$ at the end of this answer.
The comments and answers so far don't show how Lagrange multipliers can solve the problem (because of the non-negativity condition). Here's how that technique can be used.
To maximize a function $f(q_1,\dots, q_n)$ with the requirement that $q_i\ge0$ and an additional constraint $c(q_1,\dots, q_n)=0$, it suffices to maximize the function $f(s_1^2,\dots, s_n^2)$ subject to the constraint $c(s_1^2,\dots, s_n^2)=0$, then choose those solutions with all $s_i$ real that give the largest value of $f$. For these, $(s_1^2,\dots, s_n^2)$ is a non-negative solution to the original constrained maximization question.
For this particular question, then, maximize $\prod(1-s_i^2)$ subject to $\sum s_i^4=1$.
Here’s the three-variable case, where we need to maximize $(1-a)(1-b)(1-c)$ for $a,b,c\ge0$, subject to $a^2+b^2+c^2=1$.
Maximize $(1-x^2)(1-y^2)(1-z^2)$ subject to $x^4+y^4+z^4=1$. Using Lagrange multipliers, let $g(x,y,z,\lambda)=(1-x^2)(1-y^2)(1-z^2)-\lambda(x^4+y^4+z^4-1)$, and solve the following system of equations over the real numbers:
$$\begin{align}
0 = \frac{\partial f}{\partial x} = & -2x(1-y^2)(1-z^2)+4x^3\lambda\\
0 = \frac{\partial f}{\partial y} = & -2y(1-x^2)(1-z^2)+4y^3\lambda\\
0 = \frac{\partial f}{\partial z} = & -2z(1-x^2)(1-y^2)+4z^3\lambda\\
0 = \frac{\partial f}{\partial \lambda} = & x^4+y^4+z^4-1\\
\end{align}$$
When a solution $(x,y,z,\lambda)$ of this system gives a maximum among all real solutions, $(x^2,y^2, z^2)$ is a solution to the original question.
The system is no fun to solve, and I didn't try to generalize to see if it gives a nice answer the original question in general. In theory it will, but whether it's any simpler than the other answers, I don't know.
Here are all of Mathematica’s real solutions to the system above, written in terms of $a=x^2,b=y^2,c=z^2$.
$
\begin{array}{llll}
(a,b,c) & (1-a)(1-b)(1-c)\\ \hline\\
\left(\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}, 0\right) & \left(1-\frac{1}{\sqrt{2}}\right)^2 \\
\left(\frac{1}{\sqrt{2}} , 0 , \frac{1}{\sqrt{2}}\right) & \left(1-\frac{1}{\sqrt{2}}\right)^2 \\
\left(0 , \frac{1}{\sqrt{2}} , \frac{1}{\sqrt{2}} \right)& \left(1-\frac{1}{\sqrt{2}}\right)^2 \\
\left(\frac{1}{\sqrt{3}} , \frac{1}{\sqrt{3}} , \frac{1}{\sqrt{3}} \right)& \left(1-\frac{1}{\sqrt{3}}\right) ^3 \\
\left(\frac{2}{3}, \frac{2}{3} , \frac{1}{3} \right)& \frac{2}{27} \\
\left(\frac{2}{3} , \frac{1}{3} , \frac{2}{3} \right)& \frac{2}{27} \\
\left(\frac{1}{3} , \frac{2}{3} , \frac{2}{3} \right)& \frac{2}{27} \\
(1 , 0 , 0) & 0 \\
(0 , 1 , 0) & 0 \\
(0 , 0 , 1) & 0 \\
\end{array}
$
|
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|
Does $7$ divide $2 x^2 - 4y^2$ for all $x,y$? Does $7$ divide $2 x^2 - 4y^2$ for all $x,y \in \mathbb{Z}$?
|
Let $x = 7a + 1$ and $y = 7b + 1$ for integers $a,b$. Then,
$$\begin{align}2a^2 - 4b^2 &\equiv 2\cdot1^2 - 4\cdot1^2\pmod 7\\
&= 5\pmod7\\
&\equiv 5 \pmod 7\end{align}$$
So it is not generally true, because we can always generate integer pairs (though not all of them) that give a remainder of $5$ when divided by $7$ as shown above. Using the above result, we see that $(x,y) = (1, 1), (8, 1), (1, 8), (8,8)$ are all counter examples.
In fact, you can play around with the constant term in $x$ and $y$ to see how it affects the remainder.
However, if $x = 7a + 2$ and $y = 7b + 4$, then we see that
$$\begin{align}2x^2 - 4y^2\equiv2\cdot2^2-4\cdot4^2\pmod7\\
= -56\pmod7\\
\equiv 0 \pmod7\end{align}$$
Hence, the result will hold true if $x = 7a + 2$ and $y = 7b + 4$, for integers $a,b$.
|
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|
Proving inequality $x^xy^y \geq (\frac{x+y}{2})^{x+y}$ Prove that for all $x,y>0$ the following inequality $x^xy^y \geq (\frac{x+y}{2})^{x+y}$ is true.
It smells like Jensen inequality, but all I can get is that $\frac{x+y}{2}ln(x) + \frac{x+y}{2} ln(y) \geq xln(\frac{x+y}{2})+yln(\frac{x+y}{2})$
|
Let $f:(0, \infty) \rightarrow \mathbb{R}$ be given by $x\ln(x)$. $f''(x) = \frac{1}{x} > 0$, therefore $f$ is convex. By Jensen inequality,
$$f\left(\frac{x + y}{2}\right) \leq \frac{1}{2}f(x) + \frac{1}{2}f(y)$$
That is,
$$\left(\frac{x+y}{2}\right)\ln\left(\frac{x + y}{2}\right) \leq \frac{1}{2} x\ln(x) + \frac{1}{2}y\ln(y)$$
That is,
$$\left(\frac{x + y}{2}\right)^{\frac{x + y}{2}} \leq x^{\frac{x}{2}} y^{\frac{y}{2}}$$
Squaring both sides gives the desired result. QED.
Note that in a very similar way, you can show,
$$x^xy^yz^z \geq \left(\frac{x + y + z}{3}\right)^{x+y+z}$$
And even generalize the result further.
|
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|
Interference beats for a more general trigonometric sum Suppose I have three frequencies $\alpha,\beta,\gamma$ that are all close in value, and I consider the sum $\sin(\alpha x) +\sin(\beta x) +\sin(\gamma x)$ If there were only summands I could find a million explanations online that shows me how to rewrite this as a sinusoid amplitude-modulated by another sinusoid. But what about the case of three summands?
Can we write this as $A(x)\sin(\frac{\alpha+\beta+\gamma}{3}x)$ where $A(x)$ is some reasonably simple function?
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Let us expand
$$\sin\left(\dfrac{\alpha x}{3}+\dfrac{(\beta x+\gamma x)}{3}\right)$$
We get
$$\sin \dfrac{\alpha x}{3} \cos \dfrac{(\beta x+\gamma x)}{3} + \cos \dfrac{\alpha x}{3} \sin \dfrac{(\beta x+\gamma x)}{3}$$
Let us now expand
$$\cos \dfrac{(\beta x+\gamma x)}{3}= \cos \dfrac{\beta x}{3} \cos \dfrac{\gamma x}{3} - \sin \dfrac{\beta x}{3} \sin \dfrac{\gamma x}{3}$$
And
$$\sin \dfrac{(\beta x+\gamma x)}{3}= \sin \dfrac{\beta x}{3} \cos \dfrac{\gamma x}{3} + \cos \dfrac{\beta x}{3} \sin \dfrac{\gamma x}{3}$$
Substitution simply gives
$$\sin\left(\dfrac{\alpha x}{3}+\dfrac{(\beta x+\gamma x)}{3}\right)=\\ \sin \dfrac{\alpha x}{3} \cos \dfrac{\beta x}{3} \cos \dfrac{\gamma x}{3} - \sin \dfrac{\alpha x}{3}\sin \dfrac{\beta x}{3} \sin \dfrac{\gamma x}{3} \\
+ \cos \dfrac{\alpha x}{3} \sin \dfrac{\beta x}{3} \cos \dfrac{\gamma x}{3} + \cos \dfrac{\alpha x}{3}\cos \dfrac{\beta x}{3} \sin \dfrac{\gamma x}{3}$$
I do not know what you mean by "reasonably simple", however you can expand as follows:
$$\sin(\dfrac{x}{3})=\sum^\infty_{k=0}\dfrac{(-1)^k\left(\dfrac{x}{3}-\dfrac{\pi}{2}\right)^{2k}}{(2k)!}$$
Maybe this will give you idea on the function $A(x)$, on which I am willing to bet is not "simple", satisfying
$$\sin(\alpha x)+\sin(\beta x)+\sin(\gamma x)=A(x)\sin\left(\dfrac{\alpha+\beta+\gamma}{3}x\right)$$
|
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Solve $5\sin^2(x) + \sin(2x) - \cos^2(x) = 1$ I tried
$$5\sin^2(x) + 2\sin(x)\cos(x)- (1-\sin^2(x)) = 1.$$
Simplifying,
$$6\sin^2(x) + 2\sin(x)\cos(x) -2 = 0$$
Then I'm stuck!
|
Using that $$-2 = -2\sin^2(x) - 2\cos^2(x)$$ from $$6\sin^2(x) + 2\cos(x)\sin(x) - 2 = 0$$ we get $$4\sin^2(x) + 2\cos(x)\sin(x) - 2\cos^2(x) = 0$$ and dividing by $\cos^2(x)$ we have $$4\tan^2(x) + 2\tan(x) - 2 = 0$$
Can you continue from here? Note that when dividing by $\cos^2(x)$ we assume that it is not $0$. What if it is $0$?
|
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|
homogeneous coordinates I am to use homogeneous coordinates to calculate a standard matrix for a projection onto the line $4x-2y=6$ from the point $(3,10)$.
I'm not sure what homogeneous coordinates are and neither how to use them to calculate my problem? Please believe me, I've been looking for information on various sources!
|
In 2D points in homogeneous coordinates have the form $P = (x,y,1)$ and lines $L=(a,b,c)$ such that the equation for the line can be found by
$$ L \cdot P = 0 \} a x + b y + c = 0 $$
So in your case, the homogeneous coordinates for the point is $P=(3,10,1)$ and the line $L=(4,-2,-6)$.
The trick with homogeneous coordinates is that if you multiply them with a scalar value, it does not change the underlying geometry. This is powerful be because the coordinates of a point $P=(a,b,c)$ are $(x,y) = (\frac{a}{c},\frac{b}{c})$. To test if a point belongs to a line simply do the dot product $P\cdot L=0$ and check for zero. The coordinates of a line joining two points $P$, $Q$ are
$$\begin{pmatrix} p_x\\p_y\\1 \end{pmatrix} \times \begin{pmatrix} q_x\\q_y\\1 \end{pmatrix} = \begin{pmatrix} p_y-q_y \\ q_x-p_x \\ p_x q_y - p_y q_x \end{pmatrix} $$
and similarly the point where two lines $L=(a,b,c)$ and $M=(u,v,w)$ meet is
$$\begin{pmatrix} a\\b\\c \end{pmatrix} \times \begin{pmatrix} u\\v\\w \end{pmatrix} = \begin{pmatrix} b w - c v \\ c u-a w \\ a v - b u \end{pmatrix} $$
The above has coordinates $(x,y) = \left( \frac{b w -c v}{a v - b u} , \frac{c u-a w}{a v - b u} \right)$ which if you do the calculations with vectors you will end up with the same value.
Futhermore, the minimum distance of the line $L=(a,b,c)$ to the origin is $d = -\frac{c}{\sqrt{a^2+b^2}}$ and the point closest to the origin has coordinates $P=(-a c,-b c, a^2+b^2) \equiv \left(-\frac{a c}{a^2+b^2},-\frac{b c}{a^2+b^2}\right)$
In addition, the minimum distance of the line $L=(a,b,c)$ to the point $P=(u,v,w)$ is
$$ {\rm dist}(L,P) = \frac{a u + b v + c w}{w \sqrt{a^2+b^2}} $$
and the projected point on the line has coordinates
$$ {\rm proj}(L,P) = \left( b( b u-a v)-a c, -a ( b u - a v)- b c, w (a^2+b^2) \right) $$
An affine 3×3 transfromation can be defined for translations and rotations.
$$ \begin{bmatrix} 1 & 0 & t_x \\ 0 & 1 & t_y \\ 0 & 0 & 1 \end{bmatrix} \begin{pmatrix} p_x\\p_y\\1 \end{pmatrix} = \begin{pmatrix} p_x+t_x\\p_y+t_y\\1 \end{pmatrix}$$
$$ \begin{bmatrix} \cos\theta & -\sin\theta & 0 \\ \sin\theta & \cos\theta & 0 \\ 0 & 0 & 1 \end{bmatrix} \begin{pmatrix} p_x\\p_y\\1 \end{pmatrix} = \begin{pmatrix} p_x \cos\theta-p_y \sin \theta\\p_x \sin\theta + p_y \cos\theta\\1 \end{pmatrix}$$
With these basics you can built what you need from here.
|
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Integer solutions of $800000007 = x^2+y^2+z^2$ Prove that the equation, $800000007 = x^2+y^2+z^2$ has no solutions in integers.(That is $8$ followed by $7$ zeroes, with a $7$ at the end).
I tried checking modulo $3$, $5$, $7$, and $10$, but couldn't reach any conclusions.
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Hint: have a look at modulo $8$.
details:
If $x^2+y^2+z^2 = A $
then $A\neq 7\mod 8$:
$$
x^2\in \{0,1,4\}\mod 8\\
x^2+y^2+z^2 \in \{0,1,4\}+\{0,1,4\}+\{0,1,4\} =
\{0,1,2,3,4,5,6\}\mod 8.
$$
NB: the general equation $x^2+y^2+z^2 = A $ has integer solutions iff $A$ has not the form
$$
4^N(8k+7).
$$
Let us prove that if $A$ has this form there is no solution:
Assume it is not true.
Let $N$ be the smallest integer for which the equation has a solution of the form
$4^N(8k+7)$. $N>0$ because of the preceding proof. Then,
$$
x^2+y^2+z^2 =0\mod 4.
$$
As the squares are $0,1$ then $2$ divides $x,y,z$.
Then
$$
\left(\frac x2\right)
^2+\left(\frac y2\right)^2+
\left(\frac z2\right)^2 = 4^{N-1}(8k+7);
$$
this is impossible from the definition of $N$.
|
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|
a manipulation of Fibonacci recurrence Let $F_n$ be the Fibonacci number, and we know
$F_{n+2} = F_{n+1} + F_{n} $ with $F_0 =1,F_1 = 1$
And this can be manipulated to
$F_{n+6} = 4F_{n+3} + F_n$
if we let n be a multiple of 3, we can write $A_r = F_{3r}$, and we get
$A_{r+2} = 4A_{r+1} + A_r$ with $A_0=1, A_1=1$
So I am wondering, can we prove the relation:
$F_{r(k+2)} = AF_{r(k+1)} + BF_{rk}$, $k \in N$
exist for every $r \geq 1$ ? And is the pair of $(A,B)$ unique for every value of $r$?
|
Recall the theory of recurrence relations: The characteristic equation of the fibonacci sequence is $X^2 - X - 1 = 0 $.
This tells us that $F_n = a \alpha^n + b \beta ^n $ (for some constants $a$ and $b$), where $ \alpha + \beta = -1 $ and $ \alpha \beta = -1$.
Setting $A_n = F_{rn}$, we get that $$A_n = a(\alpha^r)^n + b (\beta^r)^n.$$
Hence, $A_n$ must satisfy the recurrence relation
$$ A_n = A A_{n-1} + B A_{n-2},$$
which has the characteristic equation $X^2 - AX - B = 0$, which gives us $A= \alpha^r + \beta^r $ and $ -B= \alpha^r \beta^r = (-1)^r $.
Note: $A$ can be determined using Newton's polynomials. As Nate remarks, $ A = \alpha^r + \beta^r = F_{r} + F_{r-2}$.
|
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|
How find this $\frac{3x^3+125y^3}{x-y}$ minimum value
let $x>y>0$,and such $xy=1$, find follow minimum of the value
$$\dfrac{3x^3+125y^3}{x-y}$$
My idea: let $x=y+t,t>0$
then
$$\dfrac{3x^3+125y^3}{x-y}=\dfrac{3(y+t)^3+125y^3}{t}=3t^2+3yt+3y^2+\dfrac{128y^3}{t}$$
and $$(y+t)y=1$$
I think this can use AM-GM inequality.But I can't.
Thank you very much
|
First note that $x>1$ and simplify like this:
$$\frac{3x^3+125y^3}{x-y}=\frac{3x^6+125x^3y^3}{x^4-x^3y}=\frac{3x^6+125}{x^4-x^2}$$
Set $t=x^2$. Now we want to find $a\ge0$ such that
$$\frac{3t^3+125}{t^2-t}\ge a\Longleftrightarrow3t^3+at+125\ge at^2$$
for $t>1$ where the equality is possible.
Note that we could use AM-GM inequality on $t^3+t^3+t^3+at+5^3$, but that equality is possible only if all terms are equal, i.e. $t^3=at=5^3\Longrightarrow t=5, a=5^2$.
Luckily this gives us exactly what we wanted:
$$t^3+t^3+t^3+5^2t+5^3\ge5\sqrt[5]{t^{10}5^5}=5^2t^2$$
Therefore $a=25$ is the minimum and it's possible only for $t=5\Longleftrightarrow x=\sqrt5, y=\dfrac1{\sqrt5}$.
|
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|
Equation in rational numbers? Is it true that this equation $6=\frac{x^2}{y^2+1}$ has no solutions in rational numbers?
If so, why?
It is quite evident that it has no solutions in integers (because $y^2+1$ never divides $3$).
|
Equivalently, we are looking for a solution of $6(a^2+b^2)=c^2$ in integers not all $0$.
Suppose to the contrary that there is such a solution. Let $3^k$ be the highest power of $3$ that divides both $a$ and $b$. Let $a=3^ks$ and $b=3^k t$.
Then $3$ does not divide $s^2+t^2$, so the highest power of $3$ that divides $6(a^2+b^2)$ is $3^{2k+1}$. It follows that $6(a^2+b^2)$ cannot be a perfect square.
|
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|
Question with Stokes theorem Show with Stokes that $\oint_C (y\mathbf{i}+z\mathbf{j}+x\mathbf{k})\bullet d\mathbf{r}=\sqrt{3}\pi a^2$ when $C$ is intersection of $x^2+y^2+z^2=a^2$ and $x+y+z=0$.
My work:
$$z=g(x,y)=-x-y$$
$$\mathbf{N}=\frac{-\frac{\partial g}{\partial x}\mathbf{i}-\frac{\partial g}{\partial y}\mathbf{j}+\mathbf{k}}{\sqrt{1+\left(\frac{\partial g}{\partial x}\right)^2+\left(\frac{\partial g}{\partial y}\right)^2}}=\pm\frac{1}{\sqrt 3}(\mathbf{i}+\mathbf{j}+\mathbf{k})$$
$$dS=\sqrt{1+\left(\frac{\partial g}{\partial x}\right)^2+\left(\frac{\partial g}{\partial y}\right)^2}dA=\pm\sqrt 3 dA$$
So stokes says that:
$$\oint_C (y\mathbf{i}+z\mathbf{j}+x\mathbf{k})\bullet d\mathbf{r}=\iint_R \nabla \times \mathbf{F}\bullet\mathbf{N}dS$$
$$\begin{align} \iint_R \nabla \times \mathbf{F}\bullet\mathbf{N}dS&=\iint_R (-\mathbf{i}-\mathbf{j}-\mathbf{k})\bullet(\pm\frac{1}{\sqrt 3}(\mathbf{i}+\mathbf{j}+\mathbf{k}))\pm\sqrt{3}dA \\
&=\pm\sqrt{3} \iint_R \pm\frac{3}{\sqrt{3}} dA \\
&=3 \iint_R 1 dA \\
&=3\int_0^{2\pi}d\phi\int_0^{a}rdr=3\pi r^2\end{align}$$
Soo... I got it wrong. Can someone explain what my mistake was?
|
Your double integral is over a circular region, but the intersection of the sphere and plane given is actually $x^2+xy+y^2=\frac{a^2}{2}$, an ellipse.
Edited for more detail:
Note that $\iint_R\,dA$ is exactly the area enclosed by this ellipse, so we just need to find this area and multiply it by 3.
One way to do this would be continue using calculus and convert to polar coordinates:
$$r^2+r^2\cos\theta\sin\theta=\frac{a^2}{2}$$
$$r^2=\frac{a^2}{2(1+\cos\theta\sin\theta)}$$
This leads to a pretty ugly integral which I don't really want to think about right now...
Another way would be to use some linear algebra. Note that $2x^2+2xy+2y^2$ corresponds to the symmetric matrix $\begin{pmatrix} 2 & 1 \\ 1 & 2 \end{pmatrix}$. If you orthogonally diagonalize this matrix, you should get eigenvalues of 1 and 3 and corresponding orthonormal eigenvectors of $\begin{pmatrix} \frac{1}{\sqrt 2} \\ -\frac{1}{\sqrt 2}\end{pmatrix}$ and $\begin{pmatrix}\frac{1}{\sqrt 2} \\ \frac{1}{\sqrt 2}\end{pmatrix}$.
What does this tell us? We can rewrite the equation of the ellipse as
$$\frac{\left(\frac{1}{\sqrt 2}x-\frac{1}{\sqrt 2}y\right)^2}{a^2}+\frac{\left(\frac{1}{\sqrt 2}x+\frac{1}{\sqrt 2}y\right)^2}{\frac{a^2}{3}}=1$$
So this ellipse has semimajor axis of length $a$ and semiminor axis of length $\frac{a}{\sqrt 3}$, and it encloses an area of $\frac{\pi a^2}{\sqrt 3}$.
Multiplying this by the 3 from above, we get the desired result.
|
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|
How to solve $x$ in $(x+1)^4+(x-1)^4=16$? I'm trying my hand on these types of expressions. How to solve $x$ in $(x+1)^4+(x-1)^4=16$? Please write any idea you have, and try to keep it simple. Thanks.
|
We have our equation to solve for:
$$(x+1)^4+(x-1)^4=16$$
First, expand the terms in the left hand side.
$$x^4+4x^3+6x^2+4x+1+x^4-4x^3+6x^2-4x+1=16$$
Simplify the left hand side.
$$2x^4+12x^2+2=16$$
Factor $2$ out from both sides.
$$x^4+6x^2+1=8$$
Move $8$ to the left hand side.
$$x^4+6x^2-7=0$$
Let $x^2=a$. Now our equation becomes:
$$a^2+6a-7=0$$
Factor it.
$$(a-1)(a+7)=0$$
Reverse the substitution.
$$(x^2-1)(x^2+7)=0$$
Factor some more (yes even the sum of squares)
$$(x+1)(x-1)(x+\sqrt7i)(x-\sqrt7i)=0$$
The four roots are $x=\pm 1$ and $x=\pm \sqrt{7}i$. But if you do not want complex roots, then the two real roots are $x=1$, $x=-1$.
Hope I helped!
|
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|
Jordan similar matrix I have matrix
$B = \begin{bmatrix}1 & 1 & -2 & 0\\2 & 1 & 0 & 2 \\ 1 & 0 & 1 & 1 \\ 0 & -1 & 2 & 1\end{bmatrix}$.
I found the characteristic polynomial $(1-x)^4$ and was able to get my Jordan Matrix $J = \begin{bmatrix}1 & 1 & 0 & 0\\0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 1 \\ 0 & 0 & 0 & 1\end{bmatrix}$. I need to find $P$ and $P^{-1}$ such that $B=PJP^{-1}$, but I got completely stuck after finding the eigenvectors.
Would really appreciate the help.
Thanks.
|
Note: It is important to learn to type problems using MathJax for many reasons, see this wonderful MathJax Basic Tutorial and Quick Reference.
We have the matrix:
$$B = \begin{bmatrix}1 & 1 & -2 & 0\\2 & 1 & 0 & 2 \\ 1 & 0 & 1 & 1 \\ 0 & -1 & 2 & 1\end{bmatrix}$$
The characteristic polynomial is:
$$(\lambda - 1)^4 = 0 \implies \lambda_{1,2,3,4} = 1$$
To find the four linearly independent eigenvectors, we try solving $[A - \lambda_i I]v_i = 0$, which yields only two independent eigenvectors as:
$$v_1 = \begin{bmatrix}-1\\ 0 \\ 0 \\1 \end{bmatrix}, v_3 = \begin{bmatrix}0\\ 2 \\ 1 \\0 \end{bmatrix}$$
Note, I could have called these any two eigenvectors I wanted, but I am matching your $J$, so chose them as shown above.
Now, we need to find two generalized eigenvectors, we set up and solve $[A-I]v_2 = v_1$. After RREF, we have:
$$v_2 = \begin{bmatrix}0\\-1\\ 0 \\ 0 \end{bmatrix}$$
We repeat this process using $[A-I]v_4 = v_3$, yielding:
$$v_4 = \begin{bmatrix}1\\0\\ 0 \\ 0 \end{bmatrix}$$
We can now write $P$ as a linear combination of those column eigenvectors, yielding:
$$P = [v_1 | v_2 | v_3 | v_4] = \begin{bmatrix}-1 & 0 & 0 & 1 \\ 0 & -1 & 2 & 0 \\ 0 & 0 & 1 & 0 \\ 1 & 0 & 0 & 0 \end{bmatrix}$$
It is a simple matter to find $P^{-1}$ and now we have:
$$J = P^{-1}AP = \begin{bmatrix}1 & 1 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 1 \\ 0 & 0 & 0 & 1\end{bmatrix}$$
|
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If $AD=BD$, $\angle ADC=3\angle CAB$, $AB=\sqrt{2}$, $BC=\sqrt{17}$, $CD=\sqrt{10}$. Find $AC$ In quadrilateral $ABCD$, we have
$$AD=BD,\angle ADC=3\angle CAB,AB=\sqrt{2},$$ $$BC=\sqrt{17},CD=\sqrt{10}$$
Find the $AC=?$
My idea: let $$\angle CAB=x.\angle ADC=3x,\angle ADB=y,$$
then we have
$$\angle CAD=90-\dfrac{y}{2}-x,\angle ACD=\dfrac{y}{2}+90-2x$$then we have
$$\dfrac{\dfrac{\sqrt{2}}{2}}{\sin{\dfrac{y}{2}}}=BD$$
and
$$\dfrac{BD}{\sin{BCD}}=\dfrac{DC}{\sin{DBC}}=\dfrac{BC}{\sin{BDC}}$$
then
$$\dfrac{\sqrt{2}}{\sin{\dfrac{y}{2}}\sin{BCA}}=\dfrac{\sqrt{10}}{\sin{DBC}}=\dfrac{\sqrt{17}}{\sin{(3x-y)}}$$
and in $\Delta ABC$,we have
$$\dfrac{\sqrt{17}}{\sin{x}}=\dfrac{\sqrt{2}}{\sin{ACB}}=\dfrac{AC}{\sin{ABC}}$$
in $\Delta ADC$,we have
$$\dfrac{AC}{\sin{3x}}=\dfrac{\sqrt{10}}{\sin{(90-y/2-x)}}=\dfrac{AD}{\sin{(90+y/2-2x)}}$$then I fell very ugly,and I can't.maybe have other idea. Thank you
|
$ABCD$ on a square lattice">
With three distances given, each as the square root of an integer which is the sum of two squares, this problem looks as if the points are meant to have integer $x$ and $y$ coordinates. So treat the problem as one in coordinate geometry.
We are given:
\begin{align}
AD&=BD\tag{1},\\
\angle ADC&=3\angle CAB\tag{2},\\
AB&=\sqrt2\tag{3},\\
BC&=\sqrt{17}\tag{4},\\
CD&=\sqrt{10}.\tag{5}
\end{align}
To satisfy (3), define cartesian coordinates so that $A=(0, 1)$ and $B=(1, 0)$. By (1),
$D$ lies on the line $l$ with equation $y=x$.
Next, locate $C$ by trying each lattice point satisfying (4),
then seeing if there is a suitable location for $D=(x, x)$ satisfying (5).
If $C=(0, 4)$ or $(2, 4)$ then $\angle CAB>90^\circ$ so (2) cannot be satisfied with the angles' values correctly specified.
If $C=(5, -1)$ then no $D$ on $l$ is near enough $C$.
If $C=(5, 1)$ then $D=(2, 2)$ or $(4, 4)$. $\angle CAB=45^\circ$ so $\angle ADC=135^\circ$ by (2). $D=(4, 4)$ makes $\angle ADC$ too small. However, $D=(2, 2)$ makes $\angle ADC=135^\circ$ as required, because, where $E=(3, 0)$, $\angle ADE=90^\circ$ and $\angle EDC=45^\circ$ as it is one of the acute angles of an isosceles right-angled triangle. In this case $AC=5$.
$C=(0, -4)$ and $C=(2, -4)$ yield reflections in $AB$ of the situation when $C=(5, 1)$ and $C=(5, -1)$ respectively, so these need not be further checked.
Thus $AC=5$.
|
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|
Riemann's zeta as a continued fraction over prime numbers. Riemann's zeta function is a function with many faces, I mean representations. I recently derived this one, bellow, as a continued fraction over prime numbers.
$$
\zeta(s)=1
+\cfrac{\frac{1}{2^{s}}}{1-\frac{1}{2^{s}}
-\cfrac{\frac{2^{s}-1}{3^{s}}}{1+\frac{2^{s}-1}{3^{s}}
-\cfrac{\frac{3^{s}-1}{5^{s}}}{1+\frac{3^{s}-1}{5^{s}}
-\cfrac{\frac{5^{s}-1}{7^{s}}}{1+\frac{5^{s}-1}{7^{s}}
-\cfrac{\frac{7^{s}-1}{11^{s}}}{1+\frac{7^{s}-1}{11^{s}}
-\ddots}}}}}
$$
... and I'd like to know if this is known in the literature and if so I'd appreciate to have references about it.
Thanks.
|
The continued fraction representation above had its origins on another problem I was working on sometime ago.
It's based on a very simple way of looking at the Euler's product representation of $\frac{1}{\zeta(s)}$. Interestingly it applies to every infinite product.
And this is as follows
$$
\frac{1}{\zeta(s)}=\left(1-\frac{1}{2^s}\right)-\left(1-\frac{1}{2^s}\right)\frac{1}{3^s}-\left(1-\frac{1}{2^s}\right)\left(1-\frac{1}{3^s}\right)\frac{1}{5^s}-\cdots
$$
From here its easy to derive the above continued fraction using Euler's continued fraction formula.
And thats it, It's nice and eventually a new thing.
EDIT
Just to make it clear, note that
$$
\begin{align*}
\frac{1}{\zeta(s)}&=\left(1-\frac{1}{2^s}\right)\left[\left(1-\frac{1}{3^s}\right)-\left(1-\frac{1}{3^s}\right)\frac{1}{5^s}-\left(1-\frac{1}{3^s}\right)\left(1-\frac{1}{5^s}\right)\frac{1}{7^s}-\cdots\right]\\
&=\left(1-\frac{1}{2^s}\right)\left(1-\frac{1}{3^s}\right)\left[\left(1-\frac{1}{5^s}\right)-\left(1-\frac{1}{5^s}\right)\frac{1}{7^s}-\cdots\right]\\
&\vdots\\
&=\prod_{p\in\mathbb{P}}\left(1-\frac{1}{p^{s}}\right)
\end{align*}
$$
where $\mathbb{P}$ is the set of the prime numbers.
EDIT
To derive the continued faction just put $\frac{1}{\zeta(s)}$ in the form
$$
\frac{1}{\zeta(s)}=1-\frac{1}{2^s}\left(1+\frac{2^s-1}{3^s}\left(1+\frac{3^s-1}{5^s}\left(1+\frac{5^s-1}{7^s}\left(1+\frac{7^s-1}{11^s}\left(1+\ddots\right ) \right ) \right ) \right ) \right)
$$
and then just apply the Euler continued fraction formula.
So we can write this as
$$
\frac{1}{\zeta(s)}=1-\frac{1}{2^s}-\frac{1}{2^s}\frac{2^s-1}{3^s}-\frac{1}{2^s}\frac{2^s-1}{3^s}\frac{3^s-1}{5^s}-\frac{1}{2^s}\frac{2^s-1}{3^s}\frac{3^s-1}{5^s}\frac{5^s-1}{7^s}-\cdots
$$
Now, let $a_1=-\frac{1}{2^s};a_2=\frac{2^s-1}{3^s};a_3=\frac{3^s-1}{5^s};a_4=\frac{5^s-1}{7^s}\cdots$ and we'll get the Euler continued fraction formula.
|
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How find this positive integer $(a,b)$ such $\left(\frac{a}{b}-\left[\frac{a}{b}\right]\right)\left[\frac{a}{b}\right]=2013$ let $a,b$ is positive integer numbers,and such
$$\begin{cases}
\gcd(a,b)=1\\
b\le 100\\
\left(\dfrac{a}{b}-\left[\dfrac{a}{b}\right]\right)\left[\dfrac{a}{b}\right]=2013
\end{cases}$$
Find the pairs $(a,b)$
where $[x]$ is the largest integer not greater than $x$.
my idea:
since $$2013=3\times 11\times 61$$
and let $a=kb+r,0<r<b$,nad we note
$$\left[\dfrac{a}{b}\right]\in N,0\le \dfrac{a}{b}-\left[\dfrac{a}{b}\right]<1$$.and then I can't.Thank you
|
Let $a=kb+r$, where $k, r \in \mathbb{Z}$ and $0 \leq r<b$. Since $\gcd(a, b)=1$, we have $\gcd(r, b)=1$.
Now the third equation becomes $\frac{r}{b}k=2013$, so $b \mid rk$. Since $\gcd(r,b)=1$, we get $b \mid k$. Let $k=lb, l \in \mathbb{Z}$, so $rl=2013$.
Now $r \mid 2013$ and $r<b \leq 100$, so $r=1, 3, 11, 33, 61$.
At this point, note that each choice of $r, b$ such that $r<b \leq 100, r \mid 2013$ and $\gcd(r, b)=1$ corresponds to exactly one solution for $(a, b)$. (Since $a=kb+r=lb^2+r=\frac{2013}{r}b^2+r$)
If $r=1$, we have $1<b \leq 100, \gcd(b, 1)=1$, giving $99$ solutions for $b$.
If $r=3$, we have $3<b \leq 100, \gcd(b, 3)=1$, giving $97-32=65$ solutions for $b$.
If $r=11$, we have $11<b \leq 100, \gcd(b, 11)=1$, giving $89-8=81$ solutions for $b$.
If $r=33$, we have $33<b \leq 100, \gcd(b, 33)=1$, giving $67-22-6+2=41$ solutions for $b$.
If $r=61$, we have $61<b \leq 100, \gcd(b, 61)=1$, giving $39$ solutions for $b$.
This gives a total of $325$ solutions for $(r,b)$ and hence for $(a, b)$.
|
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|
Find all triangles of which perimeter and area are numerically equal Find all triangles of which perimeter and area are numerically equal. I have got solution for right angle triangles but not of others
|
The area is $A = \sqrt{s(s-a)(s-b)(s-c)}$, where $s$ is the semiperimeter. Thus we get
$$(a+b+c)^2 = \frac{a+b+c}{2}(\frac{a+b+c}{2} - a)(\frac{a+b+c}{2} - b)(\frac{a+b+c}{2} - c).$$
We can further simplify this to
$$16(a+b+c) = (-a+b+c)(a-b+c)(a+b-c).$$
Let $u = -a+b+c$, $v = a-b+c$, $w = a+b-c$. Then
$$16(u+v+w) = u v w.$$
In particular any $u,v$ such that $uv > 16$ give a solution for $w$:
$$w = \frac{16(u+v)}{uv-16}.$$
Now for such $u,v,w$ we have that $a = \frac{v+w}{2}$, $b = \frac{w+u}{2}$ and $c = \frac{u+v}{2}$ are the sides of a triangle whose area is equal to its perimeter.
|
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|
How prove this $\cos{x}+\cos{y}+\cos{z}=1$ Question:
let $x,y,z\in R$ and such $x+y+z=\pi$,and such
$$\tan{\dfrac{y+z-x}{4}}+\tan{\dfrac{x+z-y}{4}}+\tan{\dfrac{x+y-z}{4}}=1$$
show that
$$\cos{x}+\cos{y}+\cos{z}=1$$
My idea: let $$x+y-z=a,x+z-y=b,y+z-x=c$$
then
$$a+b+c=\pi$$
and
$$\tan{\dfrac{a}{4}}+\tan{\dfrac{b}{4}}+\tan{\dfrac{c}{4}}=1$$
we only prove
$$\cos{\dfrac{b+c}{2}}+\cos{\dfrac{a+c}{2}}+\cos{\dfrac{a+b}{2}}=1$$
Use
$$\cos{\dfrac{\pi-x}{2}}=\sin{\dfrac{x}{2}}$$
$$\Longleftrightarrow \sin{\dfrac{a}{2}}+\sin{\dfrac{b}{2}}+\sin{\dfrac{c}{2}}=1$$
let
$$\tan{\dfrac{a}{4}}=A,\tan{\dfrac{b}{4}}=B,\tan{\dfrac{\pi}{4}}=C$$
then
$$A+B+C=1$$
and use $$\sin{2x}=\dfrac{2\tan{x}}{1+\tan^2{x}}$$
so we only prove
$$\dfrac{2A}{1+A^2}+\dfrac{2B}{1+B^2}+\dfrac{2C}{1+C^2}=1$$
other idea:let
$$\dfrac{y+z-x}{4}=a,\dfrac{x+z-y}{4}=b,\dfrac{x+y-z}{4}=c$$
then we have
$$a+b+c=\dfrac{\pi}{4},\tan{a}+\tan{b}+\tan{c}=1$$
we only prove
$$\cos{(2(b+c)}+\cos{2(a+c)}+\cos{2(a+b)}=\sin{(2a)}+\sin{(2b)}+\sin{(2c)}=1$$
then I fell very ugly, can you some can help?
Thank you very much!
|
Now,I have solution this problem:let $x,y,z\in R$ and such $x+y+z=\pi$,and
$$\tan{\dfrac{y+z-x}{4}}+\tan{\dfrac{x+z-y}{4}}+\tan{\dfrac{x+y-z}{4}}=1$$
show that
$$\cos{x}+\cos{y}+\cos{z}=1$$
$$\dfrac{y+z-x}{4}=a,\dfrac{x+z-y}{4}=b,\dfrac{x+y-z}{4}=c$$
we have
$$a+b+c=\dfrac{\pi}{4},\tan{a}+\tan{b}+\tan{c}=1$$
we only prove following
$$\cos{(2(b+c)}+\cos{2(a+c)}+\cos{2(a+b)}=\sin{(2a)}+\sin{(2b)}+\sin{(2c)}=1$$
since $$\tan{a}+\tan{b}+\tan{(\dfrac{\pi}{4}-a-b)}=1\Longrightarrow
\tan{a}+\tan{b}+\dfrac{1-\tan{(a+b)}}{1+\tan{(a+b)}}=1$$
$$\tan{a}+\tan{b}=\dfrac{2\tan{(a+b)}}{1+\tan{(a+b)}}$$
$$\Longrightarrow 1=\tan{a}+\tan{b}-\tan{a}\tan{b}$$
then
$$\sin{(a+b)}=\cos{(a-b)}$$
other hand we have
\begin{align*}\sin{(2x)}+\sin{(2y)}+\sin{(2z)}&=2\sin{(x+y)}\cos{(x-y)}+1-2\sin^2{(x+y)}\\
&=2\sin{(x+y)}[\cos{(x-y)}-\sin{(x+y)}]+1\\
&=1
\end{align*}
|
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|
Finding the limit of a sequence by diagonalising a matrix Consider the sequence described by:
$\frac11 , \frac32 , \frac75 , ... ,\frac {a_{n}}{b_{n}}$
where $ a_{n+1} = a_n +2b_n $ and $b_{n+1} = a_n+b_n$
Find a matrix $A$ such that
$$\begin{bmatrix} a_{n+1} \\b_{n+1} \end{bmatrix} = A \begin{bmatrix} a_{n} \\b_{n} \end{bmatrix}$$
By diagonalising $A$ find explicit formulae for $a_n \text{ and } b_n $ and hence show that
$$\lim_{n \to \infty} \frac {a_n} {b_n} = \sqrt{2}$$
My progress so far:
$A= \begin{bmatrix} 1 & 2 \\ 1 & 1 \end{bmatrix}$
with eigenvalues $ \lambda_1 = 1 + \sqrt{2} \text{ and } \lambda_2 = 1 -\sqrt{2}$ that correspond to eigenvectors $(\sqrt{2},1) , (-\sqrt{2},1)$.
So the matrix $P=\begin{bmatrix} \sqrt{2} & -\sqrt{2} \\ 1 & 1 \end{bmatrix}$ is such that $P^{-1}A P =\begin{bmatrix} 1+ \sqrt{2} & 0 \\ 0 & 1 - \sqrt{2} \end{bmatrix}$
I am unsure how this helps find explicit formulae for $ a_n \text{ and } b_n$.
Update:
The Solution I was given defined a coordinate system ,such that $\begin{bmatrix} 1 \\1 \end{bmatrix}$ becomes $P^{-1} \begin{bmatrix} 1 \\1 \end{bmatrix} = \begin{bmatrix} p_1 \\p_2 \end{bmatrix}$ where $$p_1 = \frac {\sqrt{2}+1} {2 \sqrt{2}} , p_2 = \frac {\sqrt{2}-
1} {2 \sqrt{2}} $$
then
$a_n = p_1 \lambda_1^{n-1} \sqrt{2} - p_2 \lambda_2^{n-1} \sqrt{2}$ and $b_n = p_1 \lambda_1^{n-1} + p_2 \lambda_2^{n-1} $
therefore $$\frac {a_n} {b_n} = \frac { p_1 \lambda_1^{n-1} \sqrt{2} - p_2 \lambda_2^{n-1} \sqrt{2}} {p_1 \lambda_1^{n-1} + p_2 \lambda_2^{n-1}}$$
$$ = \frac {\sqrt{2} -(p_2/p_1)(\lambda_2/ \lambda_1)^{n-1} \sqrt{2}} {1+(p_2/p_1)(\lambda_2/ \lambda_1)^{n-1}}$$
$$= \sqrt{2} ( \frac {1-(p_2/p_1)(\lambda_2/ \lambda_1)^{n-1} } {1 + (p_2/p_1)(\lambda_2/ \lambda_1)^{n-1}})$$
Since $ 0 <|\lambda_2 /\lambda_1| <1 $ we deduce that
$$\lim_{n \to \infty} \frac {a_n} {b_n} = \sqrt{2}$$
My Solution:
Upon following the advice of Git Gud and Mark Bennet, I was able to find:
$$ a_n = \frac12 (1 +\sqrt{2}) \lambda_1^{n-1}- \frac12 (\sqrt{2} -1) \lambda_2^{n-1} $$
and
$$ b_n = \frac {(1 +\sqrt{2}) \lambda_1^{n-1} - (\sqrt{2}-1 ) \lambda_2^{n-1}}{2\sqrt{2}}$$
so $ \frac {a_n} {b_n}$ simplifies to:
$$ \frac {a_n} {b_n} = \frac {\sqrt2 ((1 + \sqrt2)\lambda_1^{n-1} - (\sqrt2 -1 )\lambda_2^{n-1})} { (1 +\sqrt{2}) \lambda_1^{n-1} - (\sqrt{2}-1 ) \lambda_2^{n-1}} $$
*
*How and why are they able to define such a coordinate system?
*Why does this coordinate system simplify the question?
*Is there any discrepancy between my solution and their solution?
|
Define generating functions $A(z) = \sum_{n \ge 0} a_n z^n$ and similarly $B(z)$; multiply the recurrences by $z^n$ and sum over $n \ge 0$. Recognize some sums to get:
\begin{align}
\frac{A(z) - a_0}{z} &= A(z) + 2 B(z) \\
\frac{B(z) - b-0}{z} &= A(z) + B(z)
\end{align}
Solve the resulting linear system:
\begin{align}
A(z) &= \frac{1 + z}{1 - 2 z - z^2} \\
B(z) &= \frac{1}{1 - 2 z - z^2}
\end{align}
The zeros of the denominator are $-1 \pm \sqrt{2}$, the traditional dance with partial fractions gets a bit ugly:
\begin{align}
A(z) &= \frac{2 + \sqrt{2}}{2^{3/2}} \cdot \frac{1}{1 - (1 + \sqrt{2}) z}
- \frac{2 - \sqrt{2}}{2^{3/2}} \cdot \frac{1}{1 - (1 - \sqrt{2}) z} \\
B(z) &= \frac{1 + \sqrt{2}}{2^{3/2}} \cdot \frac{1}{1 - (1 + \sqrt{2}) z}
- \frac{1 - \sqrt{2}}{2^{3/2}} \cdot \frac{1}{1 - (1 - \sqrt{2}) z}
\end{align}
This is just a pair of geometric series for each, the first term dominates in each case. As you want:
$$
\lim_{n \to \infty} \frac{a_n}{b_n}
= \frac{2 + \sqrt{2}}{1 + \sqrt{2}}
= \frac{3 \sqrt{2} + 4}{3}
$$
|
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|
proof of $\sin(x)$ infinite series I can't seem to find a proof for this
$$\sin( x )= x - \frac{x^3}{3!} + \frac{x^5}{5!} – \frac{x^7}{7!}+\cdots$$
Which euler used to proof the basel problem.
Would be great if it anyone show us how to this equality
Thanks
|
You can use the fact that:
$$\sin(x) = \dfrac{e^{ix}-e^{-ix}}{2i}$$ and that:
$$e^{ix} = 1+(ix)+\frac{(ix)^2}{2!}+\frac{(ix)^3}{3!}+\cdots$$
$$e^{-ix} = 1+(-ix)+\frac{(-ix)^2}{2!}+\frac{(-ix)^3}{3!}+\cdots$$
So:
$$\begin{align*}\sin(x) &= \dfrac{e^{ix}-e^{-ix}}{2i} \\ &= (1-1)+ \frac{(ix)-(-ix)}{2i}+\frac{1}{2i}\bigg(\frac{-x^2}{2}-\frac{x^2}{2}\bigg)+\frac{1}{2i}\bigg(\frac{-ix^3}{3!}-\frac{ix^3}{3!}\bigg)+\cdots \\
&= 0+\frac{2ix}{2i}+0+\frac{-2ix^3}{2i3!}+\cdots \\
&= x-\frac{x^3}{3!}+\frac{x^5}{5!}-\cdots\end{align*}$$
Also using the fact that $i^2 = -1, i^3 = -i, i^4 = i, i^5 = -1,$ etc.
|
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|
Area of a five pointed star A 5 pointed star is inscribed in a circle of radius $r$.
Prove that the area of the star is
$$
\frac{10 \tan\left(\tfrac{\pi}{10}\right)}{3-\tan^2\left(\tfrac{\pi}{10}\right)} r^2
$$
|
Consider the following diagram of the 5-pointed star:
Clearly the area of the 5-star is $10$ times the area of the orange-shaded triangle $\triangle OAP$ which, in turn, equals half the height times the base:
$$
A\left(\triangle OAP\right) = \tfrac{1}{2} \bar{AC} \cdot \bar{OP} = \tfrac{1}{2} \bar{AC} \cdot r
$$
To find $\bar{AC}$ notice that it is a joint side of two right triangles, $\triangle OAC$ and $\triangle PAC$, hence
$$
\bar{OC} = \frac{\bar{AC}}{\tan\left(\angle AOC\right)} \qquad \bar{PC} = \frac{\bar{AC}}{\tan\left(\angle APC\right)} = r - \bar{OC} \tag{1}
$$
By the symmetry $\angle EOF = \tfrac{2}{5} \pi$, and $\angle AOB = \tfrac{2}{5} \pi$, and hence $\angle AOC = \tfrac{1}{2} \angle AOB = \tfrac{1}{5} \pi$. By the inscribed angle theorem applied to angle $\angle EPF$,
$$
\angle APF \equiv \angle EPF = \frac{1}{2} \angle EOF = \frac{\pi}{5} \quad \therefore \quad \angle APC = \tfrac{1}{2} \angle APF = \frac{\pi}{10}
$$
Using these angles in eq. $(1)$:
$$
r = \frac{\bar{AC}}{\tan\left(\tfrac{1}{5} \pi \right)} + \frac{\bar{AC}}{\tan\left(\tfrac{1}{10} \pi \right)}
$$
and solving for $\bar{AC}$ we get
$$
A\left(\triangle OAP\right) = r^2 \cdot \frac{1}{2} \frac{1}{\frac{1}{\tan\left(\tfrac{1}{5} \pi \right)} + \frac{1}{\tan\left(\tfrac{1}{10} \pi \right)}} = r^2 \frac{ \tan\left(\frac{\pi}{5}\right) \tan\left(\frac{\pi}{10}\right)}{2 \left(\tan\left(\frac{\pi}{5}\right) + \left(\frac{\pi}{10}\right)\right)} \tag{2}
$$
Furthermore, using angle doubling formula for tangent in the previous eq. $(2)$:
$$
\tan\left(\frac{\pi}{5}\right) = \frac{2 \tan\left(\frac{\pi}{10}\right)}{1 - \tan^2\left(\frac{\pi}{10}\right)}
$$
we have
$$
A\left(\triangle OAP\right) = r^2 \frac{\tan\left(\frac{\pi}{10}\right)}{3- \tan^2\left(\frac{\pi}{10}\right)}
$$
We finish by recalling the area of the 5-star equals 10 times the area of $\triangle AOP$:
$$
A\left({\huge \star}\right) = \frac{10 \cdot \tan\left(\frac{\pi}{10}\right)}{3- \tan^2\left(\frac{\pi}{10}\right)} r^2
$$
|
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|
Trig Identity Proofs I'm having a really hard time understanding how to do these. The directions are to verify that each of the following is an identity:
$$\dfrac{\csc x}{\cot x+\tan x}=\cos x$$
I have to get the left side to equal the right.
|
$$ \frac{ \csc x }{ \cot x + \tan x} = \frac{\frac{1}{\sin x}}{\frac{\cos x}{\sin x} + \frac{ \sin x}{ \cos x}} = \frac{\frac{1}{\sin x}}{\frac{ \cos^2 x + \sin^2 x}{\sin x \cos x}} = \frac{ \frac{1}{\sin x}}{\frac{1}{\cos x \sin x}} = \frac{ \cos x \sin x}{\sin x} = \cos x$$
I have used the identiy $\sin^2 x + \cos^2 x = 1 $
|
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|
Solve the System of Equations in $x$ and $y$ \begin{equation}
x+\frac{3\,x-y}{x^2+y^2}=3 \tag{1}
\end{equation}
\begin{equation}
y=\frac{x+3\,y}{x^2+y^2} \tag{2}
\end{equation}
|
We have:
\begin{equation}
x\left(1+\frac{3}{x^2+y^2}\right)=3+\frac{y}{x^2+y^2} \tag{3}
\end{equation}
\begin{equation}
y\left(1-\frac{3}{x^2+y^2}\right)=\frac{x}{x^2+y^2} \tag{4}
\end{equation}
Multiplying Eqn $3$ with $y$ and Eqn $4$ with $x$ we get
\begin{equation}
x\,y\left(1+\frac{3}{x^2+y^2}\right)=3\,y+\frac{y^2}{x^2+y^2} \tag{5}
\end{equation}
\begin{equation}
x\,y\left(1-\frac{3}{x^2+y^2}\right)=\frac{x^2}{x^2+y^2} \tag{6}
\end{equation} Adding Eqns $5$ and $6$ we get
$$y=\frac{1}{2\,x-3}$$ Using $y=\frac{1}{2\,x-3}$ in Eqn $2$ we get Bi Quadratic in $x$ as:
$$4x^4-24x^3+57x^2-63x+26=0$$
Can i get a Hint without involvement of BiQuadratic...
|
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|
Finding $a$ and $b$ from $a^3+b^3$ and $a^2+b^2$ Question 1
Two numbers are such that the sum of their cubes is 14 and the sum of their squares is 6. Find the sum of the two numbers.
I did
$a^2+b^2=6$ and $a^3+b^3=14$ Find $a$ and $b$, two numbers. but got lost when trying to algebraicly solve it.
Thank you, Any help is appreciated
|
There are lots of ways of solving this. One is to note that $a$ and $b$ are the roots of the quadratic equation $$0=(x-a)(x-b)=x^2-(a+b)x+ab=x^2-px+q$$ Where we use $p=a+b$ and $q=ab$ for the unknown coefficients.
Then we have $a^2-pa+q=b^2-pb+q=0$ and adding the equations we obtain $$(a^2+b^2)-p(a+b)+2q=0$$ which becomes$$6-p^2+2q=0$$
We also have (after multiplying the quadratic equation by $x$ and substituting $a,b$ and adding):
$$(a^3+b^3)-p(a^2+b^2)+q(a+b)=0$$which becomes $$14-6p+pq=0=28-12p+2pq=28-12p+p(p^2-6)$$ (substituting for $2q$ from the first equation)
Which gives the cubic for $p$ $$p^3-18p+28=0$$ This can be solved using standard methods. Pretty much any method will come down to solving an equivalent cubic.
Note that $p=a+b$ is what you are asked to find.
|
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|
Find value range of $2^x+2^y$ Assume $x,y \in \Bbb{R}$ satisfy $$4^x+4^y = 2^{x+1} + 2^{y+1}$$, Find the value range of $$2^x+2^y$$
I know $x=y=1$ is a solution of $4^x+4^y = 2^{x+1} + 2^{y+1}$ , but I can't go further more. I can only find one solution pair of $4^x+4^y = 2^{x+1} + 2^{y+1}$. It seems very far from solve this question...
|
Set $a=2^x$, $b=2^y$, then the problem is equivalent to finding the range of $a+b$ where $a,b>0$ and
$$a^2+b^2=2a+2b$$
Without loss of generality $a\ge b$ and we can make the substitution $c=a+b$, $d=a-b$, so now $c>d\ge0$, and we require
\begin{align*}\left(\frac{c+d}2\right)^2+\left(\frac{c-d}2\right)^2&=(c+d)+(c-d)\\
c^2+2cd+d^2+c^2-2cd+d^2&=8c\\
c^2+d^2&=4c\end{align*}
Observe that
$$c^2\le c^2+d^2=4c\Longleftrightarrow c(c-4)\le0\Longleftrightarrow c\le4$$
and
$$2c^2>c^2+d^2=4c\Longleftrightarrow c(c-2)>0\Longleftrightarrow c>2$$
It's easy to see all such values can be attained and the answer is the interval $(2,4]$.
|
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|
How many ways move n pies to m distances? A table size $1\times (m+n)$ squares. Give $n$ pies on the $n$ first squares. Now, I want move $n$ pies to the end of table by $m.n$ steps ($m$ steps for each pie), satify conditions one pie only move to an empty-adjacent square. Define $S(n,m)$ is number of all ways move to finished.
Prove that $$ S(n,m)=\dfrac{(n-1)^!(m-1)^!(m.n)!}{(n+m-1)^!} $$
There are symbol $x^!=1!2!...x!$
Example: $n=2; \; m=3$
$S(2,3)=\dfrac{(2-1)^!(3-1)^!(2.3)!}{(2+3-1)^!}=\dfrac{1!1!2!6!}{1!2!3!4!}=5$
$\begin{matrix}\begin{array}{|c|c|c|c|c|} \hline \circ&\circ&\;&\;&\;\\ \hline\end{array} & &\\ \downarrow & &\\ \begin{array}{|c|c|c|c|c|} \hline \circ&\;&\circ&\;&\;\\ \hline\end{array} & &\\ \downarrow & \searrow & \\ \begin{array}{|c|c|c|c|c|} \hline \circ&\;&\;&\circ&\;\\ \hline\end{array} & &\begin{array}{|c|c|c|c|c|} \hline \;&\circ&\circ&\;&\;\\ \hline\end{array}\\ \downarrow & \searrow & \downarrow\\ \begin{array}{|c|c|c|c|c|} \hline \circ&\;&\;&\;&\circ\\ \hline\end{array} & &\begin{array}{|c|c|c|c|c|} \hline \;&\circ&\;&\circ&\;\\ \hline\end{array}\\ \downarrow & \swarrow & \downarrow \\ \begin{array}{|c|c|c|c|c|} \hline \;&\circ&\;&\;&\circ\\ \hline\end{array} & &\begin{array}{|c|c|c|c|c|} \hline \;&\;&\circ&\circ&\;\\ \hline\end{array}\\ \downarrow & \swarrow & \\ \begin{array}{|c|c|c|c|c|} \hline \;&\;&\circ&\;&\circ\\ \hline\end{array} & &\\ \downarrow & &\\ \begin{array}{|c|c|c|c|c|} \hline \;&\;&\;&\circ&\circ\\ \hline\end{array} & &\end{matrix}$
|
Please take a look at the following articles:
Article One
Article Two
|
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|
Higher-Order Approximation of Catalan-Numbers I have a question considering the higher-order approximations of the Catalan-Numbers, following the book Analytic Combinatorics by Flajolet and Sedgewick. First we set
$$
C_n = \frac{1}{n+1} \binom{2n}{n} = \frac{1}{n+1} [z^n] (1-4z)^{-1/2} = \frac{4^n}{n+1} [z^n] (1-z)^{-1/2},
$$
where we use that the generating function of the central binomial coefficient is $(1-4z)^{-1/2}$.
Using Theorem VI.1 from the quoted book, we know that
$$
(1-z)^{-\alpha} \sim \frac{n^{\alpha-1}}{\Gamma(\alpha)} \left( 1 + \frac{\alpha(\alpha -1) }{2n} + \frac{\alpha(\alpha -1) (\alpha -2) (3 \alpha -1)}{24 n^2} + O\left( \frac{1}{n^3}\right) \right)
$$
(the full approximation can be found in the book, page 382).
So we get
$$
C_n = \frac{4^n}{n+1} \frac{n^{-1/2}}{\sqrt{\pi}}\left(1-\frac{1}{8n}+\frac{1}{128n^2}+O\left( \frac{1}{n^3}\right) \right).
$$
Yet the book states (page 384)
$$
C_n = \frac{4^n}{\sqrt{\pi n^3}}\left(1-\frac{9}{8n}+\frac{145}{128n^2}+O\left( \frac{1}{n^3}\right) \right).
$$
I can see that we want to add a bit to the numerators, since we slash a bit of the denominators. But I don't understand the particular choice of the factors.
Why do we add the sum of the prior fractions (we add $1$ to the $n^{-1}$ term, $9/8$ to the $n^{-2}$ term etc) to each fraction?
|
Note that $$\dfrac{n}{n+1} = 1 - \dfrac{1}{n} + \dfrac{1}{n^2} + \ldots $$
and
$$ \left(1 - \dfrac{1}{n} + \dfrac{1}{n^2} + \ldots\right)
\left( 1 - \dfrac{1}{8n} + \dfrac{1}{128 n^2} + \ldots\right)
= 1 - \dfrac{9}{8n} + \dfrac{145}{128 n^2} + \ldots $$
|
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|
counterexamples to $ \det \Big(A^2+B^2\Big)\ge \det(AB-BA) $ $n\geq3$. A and B are two $n\times n$ reals matrices. For $n\times n$, Could one give counterexamples to show that
$$ \det \Big(A^2+B^2\Big)\ge \det(AB-BA) \tag{$*$}$$
is not necessarily true?
Well, I won't do more than $3\times3$: the following $A,B$ show (*) is not right
$$A=\left(\begin{array}{ccc}0&2&0\\0&0&1\\0&0&0\end{array}\right), B=\left(\begin{array}{ccc}0&0&0\\1&0&0\\0&1&0\end{array}\right)$$
What is the counterexample for $n\times n$? Thanks a lot!
|
Let \begin{equation}A=\begin{bmatrix}I_{n-2} & \mathbf{0} & \mathbf{0} \\ \mathbf{0} & 0 & -1 \\ \mathbf{0} & 1 & 0\end{bmatrix},\end{equation} then \begin{equation}A^2=\begin{bmatrix} I_{n-2} & \mathbf{0} & \mathbf{0} \\\mathbf{0}&-1 & 0 \\ \mathbf{0}& 0 & -1\end{bmatrix}.\end{equation} Now make $B$ a matrix filled with zeros, except for the very last diagonal entry, make this entry $\sqrt{2}$. Then $B^2$ has all zeros except for the last entry on the diagonal which is $2$, and so we have \begin{equation}A^2+B^2=\begin{bmatrix} I_{n-2} & \mathbf{0} & \mathbf{0} \\\mathbf{0}&-1 & 0 \\ \mathbf{0}& 0 & 1\end{bmatrix}.\end{equation}
We then also have \begin{equation}AB-BA=\begin{bmatrix} \mathbf{0} & \mathbf{0} & \mathbf{0} \\\mathbf{0}&0 & -\sqrt{2} \\ \mathbf{0}& -\sqrt{2} & 0\end{bmatrix},\end{equation} and so then det$(A^2+B^2)=-1$ and det$(AB-BA)=0$ for all $n \geq 3$.
|
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|
If $\sin A = \cfrac{3}{5}$ with $A$ in QII, find $\sec2A$. If $\sin A = \cfrac{3}{5}$ with $A$ in QII, find $\sec2A$.
I'm getting $\sec2A=\cfrac{25}{7}$. Is that correct?
|
Since $\sin A = \frac{3}{5} $, then $\cos A = \frac{4}{5} $. Therefore,
$$ \sec (2A) = \frac{1}{\cos(2A)} = \frac{1}{\cos^2 A - \sin^2 A} = \frac{1}{\frac{16}{25} - \frac{9}{25}} = \frac{1}{\frac{7}{25}} = \frac{25}{7}$$
|
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|
No. of real solutions of the equation $2 \cos (\frac{x^2 + x}{6}) = 2^x + 2^{-x} $ How many real solutions are there of the equation $2 \cos (\frac{x^2 + x}{6}) = 2^x + 2^{-x} $?
Please illustrate.
|
Let $y=2^x$, then
$$
\begin{align}
\ln y&=\ln2^x\\
\ln y&=x\ln 2\\
y&=e^{x\ln 2}.
\end{align}
$$
Consequently, $2^{-x}=e^{-x\ln 2}$ and
$$
\begin{align}
2\cos\left(\frac{x^2+x}{6}\right)&=e^{x\ln 2}+e^{-x\ln 2}\\
\cos\left(\frac{x^2+x}{6}\right)&=\frac{e^{x\ln 2}+e^{-x\ln 2}}{2}
\end{align}
$$
Now, let $x=i\theta$, then
$$
\begin{align}
\cos\left(\frac{(i\theta)^2+i\theta}{6}\right)&=\frac{e^{i\theta\ln 2}+e^{-i\theta\ln 2}}{2}\\
\cos\left(\frac{-\theta^2+i\theta}{6}\right)&=\cos(\theta\ln 2)\\
\frac{-\theta^2+i\theta}{6}&=\theta\ln 2\\
\theta^2+(6\ln2-i)\theta&=0\\
\theta(\theta+6\ln2-i)&=0\\
\theta_1=0&\text{ or }\ \theta_2=i-6\ln2.
\end{align}
$$
Thus, $\large x_1=0$ and $\large x_2=-(1+6i\ln2)$. The real solution is only $\large\color{blue}{x=0}$.
|
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|
Non-Homogenous System answer not matching Find the Genereal Solution of $\vec{x}^{'}=\begin{pmatrix}2&-1\\3&-2 \end{pmatrix}\vec{x}+\begin{pmatrix}1\\-1 \end{pmatrix}e^t$
I found the eigenvalues to be $\lambda=\pm 1$
Therefore the eigenvectors are $e_1=\begin{pmatrix}1\\1 \end{pmatrix}$ and $e_2=\begin{pmatrix} 1\\3 \end{pmatrix}$
Therefore $\phi(t)=\begin{pmatrix} e^t & e^{-t} \\ e^{t} & 3e^{-t} \end{pmatrix}$ and $\phi^{-1}(t)=\dfrac{1}{3e^{-t}e^t-e^te^{-t}}\begin{pmatrix} 3e^{-t}& -e^{-t}\\-e^{t}&e^{t} \end{pmatrix}=\dfrac{1}{2}\begin{pmatrix} 3e^{-t}& -e^{-t}\\-e^{t}&e^{t} \end{pmatrix}$
Next I multiply by $g(t)$ $$\dfrac{1}{2}\begin{pmatrix} 3e^{-t}& -e^{-t}\\-e^{t}&e^{t} \end{pmatrix}\begin{pmatrix}e^t\\-e^t \end{pmatrix}=\begin{pmatrix} 3e^{-t}e^{t}+e^{-t}e^{t}\\-e^{t}e^{t}-e^{t}e^{t} \end{pmatrix}=\begin{pmatrix}4\\-2e^{2t} \end{pmatrix}$$
Taking the integral yields $$\begin{pmatrix} 4t\\-e^{2t} \end{pmatrix}$$
Next I multiply by $\phi(t)$ $$\dfrac{1}{2}\begin{pmatrix} e^t & e^{-t} \\ e^{t} & 3e^{-t} \end{pmatrix}\begin{pmatrix} 4t\\-e^{2t} \end{pmatrix}=\dfrac{1}{2}\begin{pmatrix}4te^t-e^t\\4te^t-3e^t \end{pmatrix}$$
Now I write $x_h + 2\begin{pmatrix}1\\1 \end{pmatrix}te^{t}-\dfrac{1}{2}\begin{pmatrix} 1\\3 \end{pmatrix}e^{t}$
The book got $x_h + 2\begin{pmatrix}1\\1 \end{pmatrix}te^{t}+\begin{pmatrix} 1\\0 \end{pmatrix}e^{t}$
Where are they getting the $\begin{pmatrix}1\\0 \end{pmatrix} $ from?
|
Hint:
How about that $\dfrac{1}{2}$ term when you multiply by $g(t)$?
Update
I agree with your solution.
|
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|
Number theory - rational number Are there any $x, y$ that fit in below
$\sqrt{4y^2-3x^2}$ such that an rational number is yielded.
Appreciate if explanation is given.
|
You want $4y^2-3x^2=n^2 \iff (2y-n)(2y+n) = 3x^2$
So for e.g. we can set $2y-n = 3, 2y+n = x^2 \implies y= \dfrac{x^2+3}4, n = \dfrac{x^2-3}2$ and both will be integer if $x \equiv 1 \pmod 2$. Thus $x = 2k+1, y = k^2+k+1$ should work for any $k \in \mathbb{Z}$, and we find
$$4y^2-3x^2 = 4(k^2+k+1)^2-3(2k+1)^2 = (2k^2+2k-1)^2$$
giving you at least one infinite family of solutions. You could try other ways of factoring $3x^2$ if you need to find other solutions.
|
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|
Guess the closed form on the following sequence? any help would be appreciated, have no idea where to start
$u_1 = 2/3$ and $u_{k+1}$ such that:
$$u_k + \frac{1}{(k+2)(k+3)}$$ for all, k are natural numbers
guess a general formula (i.e the closed form) of the sequence
|
Hint: So $u_1=\frac{2}{3}$ and $u_2=\frac{2}{3}+\frac{1}{(3)(4)}$ and $u_3=\frac{2}{3}+\frac{1}{(3)(4)}+\frac{1}{(4)(5)}$ and so on.
Note that $\frac{1}{(3)(4)}=\frac{1}{3}-\frac{1}{4}$ and $\frac{1}{(4)(5)}=\frac{1}{4}-\frac{1}{5}$.
Do a couple more terms and notice the beautiful cancellations (telescoping). In general $\frac{1}{(k+2)(k+3)}=\frac{1}{k+2}-\frac{1}{k+3}$.
|
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|
Maximum value of $abc$ for $a, b, c > 0$ and $ab + bc + ca = 12$ $a,b,c$ are three positive real numbers such that $ab+bc+ca=12$.
Then find the maximum value of $abc$
|
I do not understand that. Same for any number can make this combination.
equation: $XY+XZ+YZ=N$
Solutions in integers can be written by expanding the number of factorization: $N=ab$
And vospolzovavschis solutions of Pell's equation: $p^2-(4k^2+1)s^2=1$
$k$ - what some integer number given by us.
Solutions can be written:
$X=ap^2+2(ak+b+a)ps+(2(a-2b)k+2b+a)s^2$
$Y=2(ak-b)ps+2(2ak^2+(a+2b)k+b)s^2$
$Z=bp^2-2(2b+a)kps+(4bk^2-2ak-b)s^2$
And more:
$X=-2bp^2+2(k(4b+a)+b)ps-2((4b+2a)k^2+(2b-a)k)s^2$
$Y=-(2b+a)p^2+2(k(4b+a)-b-a)ps-(8bk^2-(4b+2a)k+a)s^2$
$Z=bp^2-2(2b+a)kps+(4bk^2-2ak-b)s^2$
Perhaps these formulas for someone too complicated.
Then equation: $XY+XZ+YZ=N$
If we ask what ever number: $p$
That the following sum can always be factored: $p^2+N=ks$
Solutions can be written.
$X=p$
$Y=s-p$
$Z=k-p$
8 - really is the maximum number of turns. But the smallest one can hardly get.
|
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|
Integral: $\int \frac{dx}{\sqrt{x^{2}-x+1}}$ How do I integrate this? $$\int \frac{dx}{\sqrt{x^{2}-x+1}}$$
I tried solving it, and I came up with $\ln\left | \frac{2\sqrt{x^{2}-x+1}+2x-1}{\sqrt{3}} \right |+C$. But the answer key says that the answer should be $\sinh^{-1}\left ( \frac{2x-1}{\sqrt{3}} \right )+C$. Any answer is very appreciated.
|
We have, $$∫\frac{dx}{\sqrt{x^2-x+1}}=\text{?}$$
Now, observe the expression
$$x^2-x+1=\left(x^2-x\right)+1=\left(x-\frac{1}{2}\right)^2+1-\frac{1}{4}=\left(x-\frac{1}{2}\right)^2+\left(\frac{\sqrt{3}}{2}\right)^{2}$$
$$∴ ∫\frac{dx}{\sqrt{x^2-x+1}}=∫\frac{dx}{\sqrt{\left(x-\frac{1}{2}\right)^2+\left(\frac{3}{2}\right)^2}} $$
Now, I use standard formula $$4∫\frac{dx}{\sqrt{x^2+a^2}} =\ln\left(x+\sqrt{x^2+a^2 }\right)+C, \quad C\in\Bbb R$$
$$ ∫\frac{dx}{\sqrt{\left(x-\frac{1}{2}\right)^2+\left(\sqrt{\frac{3}{2}}\right)^2}}=\ln\left(\left(x-\frac{1}{2}\right)+\sqrt{\left(x-\frac{1}{2}\right)^2+\left(\frac{\sqrt{3}}{2}\right)^2}\right)$$
$$=\ln\left(\frac{2x-1}{2}+\sqrt{\left(\frac{2x-1}{2}\right)^2+\left(\frac{\sqrt{3}}{2}\right)^2}\right)$$
$$=\ln\left(\frac{√3}{2} \left(\frac{2x-1}{\sqrt{3}}+\sqrt{\left(\frac{2x-1}{\sqrt{3}}\right)^2+1}\right)\right) $$
take out $\frac{\sqrt 3}{2}$
$$=\ln\left(\frac{2x-1}{√3}+\sqrt{\left(\frac{2x-1}{\sqrt{3}}\right)^2+1}\right)+\ln \frac{\sqrt{3}}{2}$$
$$=\sinh^{-1} \left(\frac{2x-1}{√3}\right)+C'$$
|
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|
Finding an explicit formula for $a_n$ defined recursively by The sequence $a_n$ is defined recursively by $a_0=0$, $a_n=4a_{n-1}+1$. I must use generating functions to solve this. $n\geq1$.
I have found a pattern:
$$\sum_{n=1}^\infty(4a_{n-1}+1)x^n = x+5x^2+21x^3+85x^4+341x^5+\ldots$$
If we subtract 1 from each term, respectively (ignoring the first term) we would have:
$$x+4x^2+20x^3+84x^4+340x^5+\ldots$$
As you can see, the pattern is the second term increases by 4, the third term increases by 4^2, the fourth term increases by 4^3, and the fifth term increases by 4^4. I wanted to use the formula for the geometric series, but the common ratio isn't 4, but $4^n$. Is there any way I can use this pattern to help me find the explicit formula, using generating functions?
Using $$\sum_{n=0}^\infty 4^nx^n = \frac{1}{1-4x}$$ Isn't helping me too much. But I have a feeling I can manipulate this generating function to solve this.
However, this might seem helpful:
$$\sum_{n=0}^\infty4^nx^{n+1} = 4^0x+4^1x^2+4^2x^3+4^3x^4+\ldots = x+4x^2+16x^3+64x^4+\ldots$$
After some work I have gotten this:
$$A(x)=\frac{x^2}{(1-x)(x-4)}$$ Where $A(x)$ is our unknown generating function (which we have been looking for). But I don't think this is correct because it does not match the answer down below.
Edit: I have finally figured it out, I will be back with my LaTex code to present it.
Final edit:
Find an explicit formula for $a_n$ using the generation function technique.
\newline We will begin by multiplying each side by $x^n$ and summing over $n\geq1$. We will also let $A(x)=\sum_{n=0}^\infty a_nx^n$ be the unknown generating function which we are looking for.
$$\begin{align*}
\implies \sum_{n=1}^\infty a_nx^n&=\sum_{n=1}^\infty (4a_{n-1}+1)x^n
\end{align*}
$$
Because
$a_0=0$ is given, we can rewrite $A(x)$:
$$
\begin{align*}
A(x)=a_0+\sum_{n=1}^\infty a_nx^n=\sum_{n=1}^\infty a_nx^n
\end{align*}
$$
Now, we continue to manipulate our equality.
$$
\begin{align*}
\implies \sum_{n=1}^\infty a_nx^n&=\sum_{n=1}^\infty (4a_{n-1}+1)x^n \\[2mm]
&=\sum_{n=1}^\infty 4a_{n-1}x^n + x^n \\[2mm]
&=\sum_{n=1}^\infty 4a_{n-1}x^n+\sum_{n=1}^\infty x^n \\[2mm]
&=\sum_{n=1}^\infty 4a_{n-1}x^n+ \frac{x}{1-x} \\[2mm]
&=\sum_{n=0}^\infty 4a_{n}x^{n+1}+ \frac{x}{1-x} \\[2mm]
&=a_0+\sum_{n=1}^\infty 4a_{n}x^{n+1}+ \frac{x}{1-x} \\[2mm]
&=\sum_{n=1}^\infty 4a_{n}x^{n+1}+ \frac{x}{1-x} \\[2mm]
&=4x\sum_{n=1}^\infty a_{n}x^{n}+ \frac{x}{1-x}. \\[2mm]
\end{align*}$$
It is clear now, that we have a multiple of our previously stated $A(x)$ on the left and right hand side.
$$
\begin{align*}
\implies&\sum_{n=1}^\infty a_{n}x^{n}=4x\sum_{n=1}^\infty a_{n}x^{n}+ \frac{x}{1-x} \\[2mm]
\implies&A(x)=4xA(x)+\frac{x}{1-x} \\[2mm]
\implies&A(x)-4xA(x)=\frac{x}{1-x}\\[2mm]
\implies& A(x)(1-4x)=\frac{x}{1-x}\\[2mm]
\implies& A(x)=\frac{x}{(1-x)(1-4x)}=\frac{x}{4x^2-5x+1} \\[2mm]
\end{align*}$$
Thus, the explicit formula for $a_n$ is given by $A(x)=\dfrac{x}{4x^2-5x+1}$ through the use of generating functions.
Finally, it follows that we have $$A(x)=\sum_{n=0}^\infty\frac{4^n-1}{3}x^n=\frac{x}{4x^2-5x+1}.$$
|
The characteristic equation for the homogenous part is: $x^2 - 4x = 0$ gives: $x = 0$ and $x = 4$. So the solution to the equation is: $a_n = c\cdot 4^n + b$. Now $a_0 = 0$ and $a_1 = 1$ gives: $0 = c + b$, and $1 = 4c + b$. So $b = -c$, and $c = \dfrac{1}{3}$, and $b = -\dfrac{1}{3}$. So: $a_n = \dfrac{4^n - 1}{3}$
|
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|
given $\cosh u = x$ find $\sinh u$ I'm asked to show that:$\newcommand{\arcosh}{\operatorname{arcosh}}$
$\int{x \arcosh x}dx = \frac{1}{4}(2x^2 -1)\arcosh x - \frac{1}{4}x\sqrt{x^2 -1} + C$
If I integrate by parts:
let $u = \arcosh x \Rightarrow \dfrac{du}{dx} = \dfrac{1}{\sqrt{x^2 -1}}$
and let $\dfrac{dv}{dx} = x \Rightarrow v = \frac{1}{2}x^2$
using the formula for integration by parts and rearranging gives:
$I = \frac{1}{2}x^2 \arcosh x - \frac{1}{2}\int{\sqrt{x^2 - 1} + \dfrac{1}{\sqrt{x^2 -1}}}dx$
If I use the substitution:
let $x = \cosh u$
Using this substitution and rearranging gives:
$I = \frac{1}{2}x^2 \arcosh x - \frac{1}{2}\int{\cosh^2 u} du$
$\Rightarrow I = \frac{1}{2}x^2 \arcosh x - \frac{1}{4}[\sinh u \cosh u + u] + C$
I want to find this in terms of $x$. To eliminate $u$, I will find $\sinh u$ in terms of $x$.
I know that $\cosh u = x$
$\Rightarrow \cosh^2 u = x^2$
$\Rightarrow \sinh^2 u + 1 = x^2$
$\Rightarrow \sinh^2 u = x^2 - 1$
$\Rightarrow \sinh u = \pm \sqrt{x^2 - 1}$
I can therefore say:
$I = \frac{1}{2}x^2 \arcosh x - \frac{1}{4}[\pm x\sqrt{x^2 - 1} + \arcosh x] + C$
$\Rightarrow I = \frac{1}{4}(2x^2 - 1)\arcosh x \pm \frac{1}{4}x\sqrt{x^2 - 1} + C$
Have I misunderstood something about $\sinh u$?
Could someone please explain why I must ignore the negative option when square rooting $\sinh^2 u$?
Thank you.
|
$\sinh u$ is negative when $u$ is negative and positive otherwise whereas $\cosh u$ is always positive. Therefore, the same is true for $\cosh u\sinh u$. To transfer this property to $x \sqrt{x^2 -1}$, we take the square root to be positive and let $x$ do the job.
|
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|
$p$ is an odd prime number where $p=3k+1\Longleftrightarrow\exists a,b\in\Bbb Z^+$ such that $p=a^2+ab+b^2$ Let $p$ be odd prime number,show that:
$$p=3k+1\Longleftrightarrow \exists a,b\in\Bbb Z^+ \textrm{ such that } p=a^2+ab+b^2$$
I guess this is true because I find
when: $p=7,k=2$,and $$7=2^2+2\cdot 1+1^2$$
(2) when $p=13,k=4$,and
$$13=1^2+1\cdot 3+3^2$$
and so on.
How do I prove this ?
|
HINT:
For odd prime $p$,
$$a^2+ab+b^2=p\implies(2a+b)^2\equiv-3b^2\mod p$$
$$\iff\left(\frac{2a+b}b\right)^2=-3\mod p$$
Check when $-3$ is a Quadratic Residue $\pmod p$
Now, use this
Alternatively, $$\left(\frac{-3}p\right)=\left(\frac{-1}p\right)\cdot\left(\frac3p\right)$$
$-1$ is a quadratic residue modulo $p$ if and only if $p\equiv 1\pmod{4}$
Using Quadratic Reciprocity Theorem
$$\left(\frac3p\right)\left(\frac p3\right)=(-1)^{\frac{(p-1)(3-1)}4}=(-1)^{\frac{p-1}2}$$
$$a\equiv\pm1\pmod 3\implies a^2\equiv1\pmod3$$
Can you take it home from here?
|
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How did Euler realize $x^4-4x^3+2x^2+4x+4=(x^2-(2+\alpha)x+1+\sqrt{7}+\alpha)(x^2-(2-\alpha)x+1+\sqrt{7}-\alpha)$? How did Euler find this factorization?
$$\small x^4 − 4x^3 + 2x^2 + 4x + 4=(x^2-(2+\alpha)x+1+\sqrt{7}+\alpha)(x^2-(2-\alpha)x+1+\sqrt{7}-\alpha)$$
where $\alpha = \sqrt{4+2\sqrt{7}}$
I know that he had some super powers, like he was sent to us from a super intelligent alien universe just to humiliate our intelligence, but how the hell did he do that three centuries ago? :|
|
Hint:As @illysial, @Mehdi and others indicated, the substitution $x = t+1$ transforms the expression $x^4 − 4x^3 + 2x^2 + 4x + 4$ into $t^4-4 t^2+7$.
|
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Show that $\sum_{n=0}^{\infty}\frac{1}{4n^4+1}=\frac{1}{2}+\frac{\pi}{4}\tanh\left(\frac{\pi}{2}\right)$ Show that
$$\sum_{n=0}^{\infty}\frac{1}{4n^4+1}=\frac{1}{2}+\frac{\pi}{4}\tanh\left(\frac{\pi}{2}\right)$$.
I am thinking of using Fourier series and Parseval's identity to tackle this, I tried $x^4$, $4x^4+1$, however these did not work quite well.
|
It can be shown using Poisson summation formula that
$$\sum_{n\in\mathbb{Z}}\frac{1}{n^2+b^2}=\frac{\pi}{b}\tanh \pi b.$$
Setting $b_{\pm}=e^{\pm i\pi/4}c$ in this formula, we get
\begin{align}
\sum_{n\in\mathbb{Z}}\frac{1}{n^4+c^4}&=\frac{1}{b_+^2-b_-^2}\left(\sum_{n\in\mathbb{Z}}\frac{1}{n^2+b_-^2}-\sum_{n\in\mathbb{Z}}\frac{1}{n^2+b_+^2}\right)=\\ &=\frac{\pi}{2ic^2}\left[\frac{\tanh \pi b_-}{b_-}-\frac{\tanh \pi b_+}{b_+}\right].\tag{1}
\end{align}
On the other hand
$$\tanh\frac{\pi e^{\pm i\pi/4}}{\sqrt{2}}=\tanh\frac{\pi(1\pm i)}{2}=\coth\frac{\pi}{2}.$$
Hence, setting $c=1/\sqrt2$ in (1), we get
$$\sum_{n\in\mathbb{Z}}\frac{1}{n^4+\frac14}=2\pi\coth\frac{\pi}{2}.$$
From this one easily finds
$$\boxed{\displaystyle\sum_{n=0}^{\infty}\frac{1}{4n^4+1}=\frac14\sum_{n=0}^{\infty}\frac{1}{n^4+\frac14}=\frac{\pi}{4}\coth\frac{\pi}{2}+\frac12}$$
(Note the misprint $\coth\rightarrow\tanh$ in your question).
|
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Convergent or divergent? $\sum_{n=1}^{\infty}{\frac{n^n}{(n+1)^{n+1}}}$ Does the series
$$\sum_{n=1}^{\infty}{\frac{n^n}{(n+1)^{n+1}}}$$
converge?
|
$$\begin{align*}
\frac{n^n}{(n+1)^{n+1}} &= \frac{1}{n+1}\left(\frac{n}{n+1}\right)^n = \frac{1}{n+1}\left(\frac{1}{1+\frac{1}{n}}\right)^n
= \frac{1}{n+1}e^{-n\ln\left(1+\frac{1}{n}\right)} \\
&= \frac{1}{n+1}e^{-n\left(\frac{1}{n}+o\left(\frac{1}{n}\right)\right)}
= \frac{1}{n+1}e^{-1+o\left(1\right)} \operatorname*{\sim}_{n\to\infty} \frac{e^{-1}}{n}
\end{align*}$$
As the series $\sum_{n=1}^\infty \frac{e^{-1}}{n}$ diverges, so does this one.
|
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Trying to prove M is a manifold Let $M$ be the set of all points $(x, y, z) \in \mathbb{R^3}$ satisfying both of the equations
$x^3 + y^3 + z^3 = 1$ and $x + y + z = 1$. Prove that M is a manifold, except
perhaps near the points $(x; y; z) = (-1; 1; 1) ; (1;-1; 1) ; (1; 1;-1)$. What is
the dimension of $M$?
I have calculated my $D_m=(3x^2 + 3y^2 + 3z^2)$ but I am not sure where to go from here.
|
Consider $\Phi = x^3 + y^3 + z^3 - 1$, and $\Psi = x + y + z - 1$, and $\Lambda = (\Phi, \Psi)$, then $M = \Lambda^{-1}(0,0)$.
Consider the matrix $\begin{pmatrix} \Phi_x & \Phi_y & \Phi_z \\ \Psi_x & \Psi_y & \Psi_z \end{pmatrix} = \begin{pmatrix} 3x^2 & 3y^2 & 3z^2 \\ 1 & 1 & 1 \end{pmatrix}$
Consider the matrices : $D_1 = \begin{pmatrix} 3x^2 & 3y^2 \\ 1 & 1 \end{pmatrix}$.
$D_2 = \begin{pmatrix} 3x^2 & 3z^2 \\ 1 & 1 \end{pmatrix}$
$D_3 = \begin{pmatrix} 3y^2 & 3z^2 \\ 1 & 1 \end{pmatrix}$
We have: $\det(D_1) = 3(x^2 - y^2)$
$\det(D_2) = 3(x^2 - z^2)$
$\det(D_3) = 3(y^2 - z^2)$.
To prove that $M$ is a manifold we only need to prove that it cannot occur at the same time that: $\det(D_1) = \det(D_2) = \det(D_3) = 0$. So suppose it can. Then this means:
$x^2 = y^2 = z^2$.
So if $x = y = z$, then $x = y = z = \dfrac{1}{3}$, and $x^3 = y^3 = z^3 = \dfrac{1}{3}$, contradiction.
So if $x = -y$ then $x^3 + y^3 = 0 = x + y$. So $z = z^3 = 1$, but then $y^2 = z^2 = 1$, so $y = 1$ or $-1$. Then $x = 1$ or $-1$. But these points are excluded in the problem so it cannot be the case that all three determinants be zero. Thus one of them must be non-zero, which means the matrix has full rank = $2$ which is also the dimension of $M$.
|
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|
Is there an elegant way to simplify $\frac{\tan(x+20^{\circ })-\sin(x+20^{\circ })}{\tan(x+20^{\circ })+\sin(x+20^{\circ })}$ I wonder how to solve this equation: $$\frac{\tan(x+20^{\circ })-\sin(x+20^{\circ })}{\tan(x+20^{\circ })+\sin(x+20^{\circ })}=4\sin^{2}\left(\frac{x}{2}+10^{\circ }\right)$$ in an elegant/shorter way.
My way:
$$\frac{\sin(x+20^{\circ })-\sin(x+20^{\circ })\cdot cos(x+20^{\circ})}{\sin(x+20^{\circ })+\sin(x+20^{\circ })\cdot \cos(x+20^{\circ})}=2[1-(\cos(x+20^{\circ}))]$$
so
$$1-\cos(x+20^{\circ})=2[1-\cos(x+20^{\circ})][1+\cos(x+20^{\circ})]$$
which gives
$$\cos(x+20^{\circ})=1\Rightarrow x=-20^{\circ}+360^{\circ}k\\\cos(x+20^{\circ})=-0.5\Rightarrow x=140^{\circ}+360^{\circ}k,\, \, x=-100^{\circ}+360^{\circ}k$$
Thanks.
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Setting $\displaystyle\frac x2+10^\circ=y\iff x+20^\circ=2y$
$$\frac{\tan2y-\sin2y}{\tan2y+\sin2y}=\frac{\sin2y\cdot\dfrac{1-\cos2y}{\cos2y}}{\sin2y\cdot\dfrac{1+\cos2y}{\cos2y}}=\frac{\cos2y\sin2y(1-\cos2y)}{\cos2y\sin2y(1+\cos2y)}=\tan^2y$$ if $\cos2y\sin2y\ne0$
Now $\cos2y\sin2y=0\iff\sin4y=2\sin2y\cos2y=0\implies4y=n\cdot180^\circ$ where $n$ is any integer
So, we need $\displaystyle y\ne n\cdot45^\circ\ \ \ \ (1)$ and we have $\displaystyle\tan^2y=4\sin^2y,$
$\displaystyle\implies\sin^2y=4\sin^2y\cos^2y\iff\sin^2y(1-4\cos^2y)=0$
If $\displaystyle\sin^2y=0,\sin y=0\implies y=m\cdot180^\circ$ which is inadmissible by $(1)$
So, $\displaystyle1-4\cos^2y=0\iff\cos^2y=\frac14\iff\cos2y=2\cos^2y-1=-\frac12=\cos120^\circ$
$\displaystyle\implies2y=360^\circ\cdot r\pm120^\circ\iff y=180^\circ\cdot r\pm60^\circ$ where $r$ is any integer
So, $x=2y-20=\cdots$
|
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|
Finding eigenvectors of a matrix I want to find all eigenvalues and eigenvectors of the matrix $\begin{bmatrix}0&1&0\\0&0&1\\-1&0&0\end{bmatrix}$.
Here is how I find eigenvalues:
$$\begin{align*}
\det(A - \lambda I) &= \det \Bigg(\begin{bmatrix}0&1&0\\0&0&1\\-1&0&0\end{bmatrix} - \begin{bmatrix}\lambda&0&0\\0&\lambda&0\\0&0&\lambda \end{bmatrix} \Bigg)\\
&= \det \Bigg(\begin{bmatrix} -\lambda&1&0 \\ 0&-\lambda&1 \\ -1&0&-\lambda \end{bmatrix} \Bigg)\\
&= -\lambda^3 - 1\\
\therefore \lambda =& -1
\end{align*}$$
Using eigenvalue that I found ($-1$), I want to find eigenvectors:
$$\begin{align*}
(A - \lambda I)\vec{V} =& 0\\
\Bigg(\begin{bmatrix}0&1&0\\0&0&1\\-1&0&0\end{bmatrix} - \begin{bmatrix}-1&0&0\\0&-1&0\\0&0&-1\end{bmatrix}\Bigg) \begin{bmatrix}x\\y\\z \end{bmatrix} =& \begin{bmatrix}0\\0\\0\end{bmatrix}\\
\begin{bmatrix}1&1&0\\0&1&1\\-1&0&1\end{bmatrix} \begin{bmatrix} x\\y\\z \end{bmatrix} = & \begin{bmatrix}0\\0\\0\end{bmatrix}\\
\begin{bmatrix} x+y \\ y+z \\ -x+z \end{bmatrix} = & \begin{bmatrix}0\\0\\0\end{bmatrix}\\
\end{align*}$$
But what I should do from now? What is really the eigenvectors? Does this means that I have unlimited eigenvectors and any number that satisfies three equations can be eigenvectors?
|
We could put the solution in a cleaner form:
Let $\lambda =-1$ be our eigenvalue. Then we have
$$
\begin{align}
M&=A-\lambda I\\&= \begin{bmatrix}1&1&0\\0&1&1\\-1&0&1\end{bmatrix}
\end{align}
$$
Let the corresponding eigenvector be $\overrightarrow{v}$:
$$
\overrightarrow{v}=\begin{bmatrix}v_1\\v_2\\v_3\end{bmatrix}
$$
such that $M\overrightarrow{v}=\overrightarrow{0},\ \overrightarrow{v}\ne0$. Then we have
$$
v_1+v_2=0\\ v_2+v_3=0\\-v_1+v_3=0
$$
Solving the system of linear equations, we have:
$$
ker(M)=\left\{\;t\begin{pmatrix}1\\-1\\1\end{pmatrix},\; t\in\mathbb{R}\backslash \left\{0\right\} \right\}
$$
Hence the corresponding eigenvector is
$$
\overrightarrow{v}=\begin{bmatrix}1\\-1\\1\end{bmatrix}
$$
|
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|
Find the minimum value of $P=\frac{1}{4(x-y)^2}+\frac{1}{(x+z)^2}+\frac{1}{(y+z)^2}$ Let $x,y,z$ be real numbers such that $x>y>0, z>0$ and $xy+(x+y)z+z^2=1$.
Find the minimum value of $$P=\frac{1}{4(x-y)^2}+\frac{1}{(x+z)^2}+\frac{1}{(y+z)^2}$$
I tried using some ways, but failed. Please give me an idea. Thank you.
|
notice that: $x - y = (x+z) - (y+z)$. So let $a = x+z$, and $b = y+z$, then:
$ab = 1$, and minimize: $P(a,b) = \dfrac{1}{4(a-b)^2} + \dfrac{1}{a^2} + \dfrac{1}{b^2} = a^2 + \dfrac{1}{a^2} + \dfrac{1}{4(a^2 - 2 + b^2)} = a^2 + \dfrac{1}{a^2} + \dfrac{1}{4\left(a^2 + \dfrac{1}{a^2} - 2\right)}= t + \dfrac{1}{4t - 8}$, with $t = a^2 + \dfrac{1}{a^2} \geq 2$.
So now consider $P(a,b) = f(t) = t + \dfrac{1}{4t-8}$ on $2 \leq t < \infty$.
We have: $f'(t) = 1 -\dfrac{4}{(4t-8)^2} = 1 - \dfrac{1}{4t^2 - 16t + 16} = 0 \iff 4t^2 - 16t + 16 = 1 \iff (2t-4)^2 = 1 \iff 2t - 4 = 1 \iff t = \dfrac{5}{2} \iff a = \sqrt{2}$, and then $b = \dfrac{1}{\sqrt{2}}$.
So: $f_{min} = f(\frac{5}{2}) = 3 = P_{min}$
|
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|
$(a+b) \cdot (a-b)=a^2-b^2, \forall a,b \in R \text{ iff R is commutative }$ Let $R$ a ring. Show that $(a+b) \cdot (a-b)=a^2-b^2, \forall a,b \in R \text{ iff R is commutative }$.
That's what I have tried:
$$(a+b) \cdot (a-b)=a^2-b^2 \Rightarrow a \cdot a+a(-b)+ba+b(-b)=a^2-b^2 \Rightarrow a^2-ab+ba-b^2= a^2-b^2 \Rightarrow -ab+ba=0_R \Rightarrow ba=ab, \text{ so R is commutative}$$
Could you tell me if it is right??
|
If $\;R\;$ is commutative then
$$(a-b)(a+b)=a^2+\overbrace{ab-ba}^{=0\;, \text{ by comm.!}}-b^2=a^2-b^2$$
|
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|
How prove $\frac{a^2}{a+2b^2}+\frac{b^2}{b+2c^2}+\frac{c^2}{c+2a^2}\ge 1$
let $a,b,c$ be postive real numbers ,and such $$ab+bc+ac=3$$
show that
$$\dfrac{a^2}{a+2b^2}+\dfrac{b^2}{b+2c^2}+\dfrac{c^2}{c+2a^2}\ge 1$$
This problem is from Secrets In Inequalities volume 1 page 30,example 1.24. the comment.the author this case is a bit more diffcult,But this author can't post solution.Thank you
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By AM-GM (with all sums being cyclic),
$$\sum \frac{a^2}{a+2b^2} = \sum a - \sum \frac{2ab^2}{a+2b^2}\ge \sum a - \sum \frac{2ab^2}{3\sqrt[3]{ab^4}}= \sum a - \frac23\sum (ab)^{2/3}$$
By Power Mean Inequality, $$1 = \frac{ab+bc+ca}3 \ge \sqrt[\frac23]{\frac{\sum (ab)^{2/3}}3} \implies \sum (ab)^{2/3} \le 3$$
It remains to show that $\sum a \ge 3$ which follows from the well known
$$\left(\sum a\right)^2 \ge 3\sum ab = 9$$
|
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|
Help with definition of derivative My textbook says the definition is this:
$$\lim\limits_{\textbf{x} \rightarrow \textbf{x}_0} \frac{\|f(\textbf{x}) - f(\textbf{x}_0) - \textbf{T}(\textbf{x} - \textbf{x}_0)\|}{\|\textbf{x} - \textbf{x}_0\|} = 0$$
I am trying to learn how to use it on, say for example, the function $f(x,y) = xy$ at the point $(1, 1)$.
So we have:
$$\lim\limits_{(x,y) \rightarrow (1,1)} \frac{\|xy - 1 - \textbf{T}(\textbf{x} - \textbf{x}_0)\|}{\|(x,y) - (1,1)\|} = 0$$
$$\lim\limits_{(x,y) \rightarrow (1,1)} \frac{\|xy - 1 - \textbf{T}(\textbf{x} - \textbf{x}_0)\|}{\sqrt{(x - 1)^2 + (y - 1)^2}} = 0$$
$$\lim\limits_{(x,y) \rightarrow (1,1)} \frac{\|xy - 1 - [y(x - 1) + x(y - 1)]\|}{\sqrt{(x - 1)^2 + (y - 1)^2}} = 0$$
$$\lim\limits_{(x,y) \rightarrow (1,1)} \frac{\|xy - 1 - xy + y - xy + x\|}{\sqrt{(x - 1)^2 + (y - 1)^2}} = 0$$
$$\lim\limits_{(x,y) \rightarrow (1,1)} \frac{\|- 1 + y - xy + x\|}{\sqrt{(x - 1)^2 + (y - 1)^2}} = 0$$
But if I plug in $(1,1)$, I get an indeterminate form $\frac{0}{0}$. Is everything correct so far?
|
Notice taking limit $x \neq 1$, and $y \neq 1$, but they get very close to $(1,1)$, and this is enough to get through the problem...
So starting at your last line: $|| - 1 + y - xy + x|| = |x - 1|\cdot |y - 1|$, so:
$0 \leq \dfrac{||-1 + y - xy + x||}{\sqrt{(x - 1)^2 + (y - 1)^2}} = \dfrac{|x - 1|\cdot |y - 1|}{\sqrt{(x - 1)^2 + (y - 1)^2}} \leq |x - 1|$. So:
$|x - 1| \to 0$ when $x \to 1$, and by squeeze theorem the middle term $\to 0$.
|
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|
convergence of a sum of two geometric series I would like to use the ratio and root test on the following series:
s = 1/2 + 1/3 + (1/2)^2 + (1/3)^2 + .. = a1 + a2 + a3 + ...
where a2 is (1/2)^2 + (1/3)^2 for example
I know we have a sum of two geometric series so the sum will be convergent but I'd like to find the following results after applications of the root and ratio test:
$lim_{n \rightarrow \infty} Inf\left(\frac{a_{n+1}}{a_n}\right) = 0 $
$lim_{n \rightarrow \infty} Sup\left(\frac{a_{n+1}}{a_n}\right) = +\infty $
$lim_{n \rightarrow \infty} Inf \sqrt[n]{a_n} = \frac{1}{\sqrt{3}} $
$lim_{n \rightarrow \infty} Sup \sqrt[n]{a_n} = \frac{1}{\sqrt{2}} $
I know that $\frac{(a_{n+1})}{(a_n)} = \frac{(3^{(n+1)} + 2^{(n+1)})}{(6.(3^n+2^n))}$ and $(a_n) = \frac{3^n + 2^n}{3^n . 2^n}$. But what to do afterwards ?
How to get these calculations ?
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So I believe you have decided to describe this series by, $$ \sum_{n = 1}^{\infty} \left({\frac{1}{2}}\right)^n + \left({\frac{1}{3}}\right)^n $$
Root Test:
Now by the expansion of the Binomial Theorem for each $n \in \Bbb N$,
$$ 0 \lt \sqrt[n]{ \left({\frac{1}{2}}\right)^n + \left({\frac{1}{3}}\right)^n } \lt \sqrt[n]{\left({\frac{1}{2} + \frac{1}{3}}\right)^n } = \left({\frac{1}{2} + \frac{1}{3}}\right) = \frac 5 6 $$
$$ 0 \lt \lim \sqrt[n]{ \left({\frac{1}{2}}\right)^n + \left({\frac{1}{3}}\right)^n } \lt 1 $$
See this. From this it follows that our series converges.
Ratio Test:
First note that $ \left({\frac{1}{2}}\right)^{n + 1} + \left({\frac{1}{3}}\right)^{n + 1} = \frac{1}{2}\left({\frac{1}{2}}\right)^{n} + \frac{1}{3}\left({\frac{1}{3}}\right)^{n} \lt \frac{1}{2}\left({\frac{1}{2}}\right)^{n} + \frac{1}{2}\left({\frac{1}{3}}\right)^{n} $ for each $n \in \Bbb N$
Therefore,
$$ 0 \lt \frac{ \left({\frac{1}{2}}\right)^{n + 1} + \left({\frac{1}{3}}\right)^{n + 1} }{ \left({\frac{1}{2}}\right)^n + \left({\frac{1}{3}}\right)^n} \lt \frac{1}{2}\frac{ \left({\frac{1}{2}}\right)^n + \left({\frac{1}{3}}\right)^n }{ \left({\frac{1}{2}}\right)^n + \left({\frac{1}{3}}\right)^n } = \frac{1}{2} $$
whence it follows that
$$ 0 \lt \lim \frac{ \left({\frac{1}{2}}\right)^{n + 1} + \left({\frac{1}{3}}\right)^{n + 1} }{ \left({\frac{1}{2}}\right)^n + \left({\frac{1}{3}}\right)^n} \lt \frac 1 2 \lt 1 $$
Clearly, our series again converges.
It is also probably worth mentioning that you can easily prove that if $\sum a_n$ and $\sum b_n$ are convergent series then $\sum a_n + b_n$ will converge too. And the sum will be the sum of the two convergent series. Therefore the convergence of the geometric series $\sum (\frac{1}{2})^n$ and $\sum (\frac{1}{3})^n$ imply the convergence of yours.
|
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|
Can be rational? Can $z=\sqrt{1-4y^x}$ be a rational number, where $x>2$ is an integer and $y$ is a rational number and $y>0$?
I have tried with $x=2$ and it has rational solutions, for example $y=12/25$ and then $z=7/25$. But I haven't found any rational solution with $x>2$
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Too long for a comment:
$\sqrt{1-4y^x}=z\quad=>\quad1-4y^x=z^2\iff1-4~\bigg(\dfrac mn\bigg)^x=\bigg(\dfrac ab\bigg)^2,\quad\begin{cases}\gcd\big(~a~,b\big)=1.\\\gcd\big(m,n\big)=1.\end{cases}$
$\iff\underbrace{n^x\overbrace{(b^2-a^2)}^{b~>~a}=\overbrace{4b^2m^x}^{>~0}}_{\gcd(a,b){\large=}1,\quad\gcd(m,n){\large=}1.}~=>~n^x$ can only be given by $4b^2$, and $b^2-a^2$ can only come from
$4m^x$. Now, $4$ can be broken up into a product of two factors in only two ways: as $1\cdot4$, or as $2\cdot2$,
giving us the three possible cases $~\begin{cases}~n^x=4b^2\\m^x=b^2-a^2\end{cases}\quad,~\begin{cases}~~~n^x=2b^2\\2m^x=b^2-a^2\end{cases}\quad,~\begin{cases}~~~n^x=b^2\\4m^x=b^2-a^2\end{cases}$
|
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For which $n$ is $ \int_0^{\pi/2} \frac{\mathrm{d}x}{2+\sin nx}= \int_0^{\pi/2} \frac{\mathrm{d}x}{2+\sin x}=\frac{\pi}{3\sqrt{3\,}\,}$? I have been trying to figure out for which $n$ is $$ \int_0^{\pi/2} \frac{\mathrm{d}x}{2+\sin nx} = \int_0^{\pi/2} \frac{\mathrm{d}x}{2+\sin x}=\frac{\pi}{3\sqrt{3\,}\,}$$
Using maple I got the following values below 100
$$
n = 1
,\,2
,\,12
,\,13
,\,14
,\,24
,\,25
,\,26
,\,36
,\,37
,\,38
,\,48
,\,49
,\,50
,\,60
,\,61
,\,62
,\,73
,\,74
,\,85
,\,86
,\,98
,\,110
$$
But I am having a hard time seeing any pattern. I tried calculating the integral directly.
The integral seems to converge for all $n$, but a closed form seems hard. Any help would be appreciated =)
EDIT: I ran a few more tests and can not find any more integer values that work. Are the list above exhaustive?
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Sub $y=n x$ and use a Weierstrass substitution $t=\tan{(y/2)}$ and get that the integral is
$$\frac1{n} \int_0^{\tan{(\pi n/4)}} \frac{dt}{1+t+t^2} $$
Evaluate this by expressing the denominator as $(t+1/2)^2+3/4$ to get an expression for the integral and the stated equation:
$$\frac{2}{\sqrt{3} n} \arctan{ \left [\frac{\sqrt{3}}{2 \cot{\left (\frac{\pi}{4} n \right )+1}} \right ] } = \frac{\pi}{3 \sqrt{3}} $$
A quick scan shows this works for $n=1$ and $n=2$. I guess you would have to define the branch of the arctangent according to the value of $n$. For example, add $2 \pi$ to each multiple of $8$ for $n$. For example,
$$\arctan{ \left [\frac{\sqrt{3}}{2 \cot{\left (\frac{\pi}{4} \right )+1}} \right ] } = \frac{\pi}{6} $$
$$\arctan{ \left [\frac{\sqrt{3}}{2 \cot{\left (\frac{9 \pi}{4} \right )+1}} \right ] } = \frac{\pi}{6} + 2\pi$$
$$\arctan{ \left [\frac{\sqrt{3}}{2 \cot{\left (\frac{17 \pi}{4} \right )+1}} \right ] } = \frac{\pi}{6} + 4\pi$$
and so on. Thus, for $n=12$, we have
$$\frac{2}{12 \sqrt{3}} \arctan{ \left [\frac{\sqrt{3}}{2 \cot{\left (\frac{12 \pi}{4} \right )+1}} \right ] } = \frac{1}{6 \sqrt{3}} 2 \pi = \frac{\pi}{3 \sqrt{3}} $$
For $n=13$, we have
$$\frac{2}{13\sqrt{3}} \arctan{ \left [\frac{\sqrt{3}}{2 \cot{\left (\frac{13 \pi}{4} \right )+1}} \right ] } = \frac{2 }{13 \sqrt{3}} \left ( 2 \pi+\frac{\pi}{6}\right ) = \frac{\pi}{3 \sqrt{3}} $$
|
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|
Inequality $\sum\frac{x}{(x + n^2)^2}<\frac{1}{2} \sum \frac{1}{x + n^2} $
$x\geq0$, then, we have
$$\sum_{n=1}^{\infty}\frac{x}{(x + n^2)^2}<\frac{1}{2} \sum_{n=1}^{\infty} \frac{1}{x + n^2} $$
The problem is not easy, even $x=1$. Any help will be appreciated
|
From a purely algebraic point of view, this problem is quite interesting if we first notice that $$\frac {d}{dx} \Big( \frac {1}{x+n^2} \Big)=-\frac{1}{\left(x+n^2\right)^2}$$ The second point is to recognize that
$$ \sum_{n=1}^{\infty} \frac{1}{x + n^2}=\frac{\pi \sqrt{x} \coth \left(\pi \sqrt{x}\right)-1}{2 x}$$ So $$\sum_{n=1}^{\infty}\frac{x}{(x + n^2)^2}-\frac{1}{2} \sum_{n=1}^{\infty} \frac{1}{x + n^2}=\frac{\pi ^2 x \text{csch}^2\left(\pi \sqrt{x}\right)-1}{4 x}$$ which is always negative.
|
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Number of Rolls of Fair Dice to get '6' and '5' A Fair Dice is Thrown Repeatedly. Let $X$ be number of Throws required to get a '$6$' and $Y$ be number of throws required to get a '$5$'. Find $$E(X|Y=5)$$
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This is my Approach, Let me know if i am on right track...
$$E(X|Y=5)=\sum_{x\in \mathbb{X}}xP(X=x|Y=5)=\sum_{x \in \mathbb{X}}\frac{xP(X=x,Y=5)}{P(Y=5)}$$
The set $\mathbb{X}$ is $\left\{1,2,3,4,6,7,8,\cdots \infty \right\}$ So
$$E(X|Y=5)=\sum_{k=1,2,3,\cdots,\infty_{k\ne 5}}\frac{k\,P(X=k,Y=5)}{P(Y=5)}$$ We have
$$P(X=1,Y=5)=\left(\frac{1}{6}\right)^2\left(\frac{2}{3}\right)^3$$
$$P(X=2,Y=5)=\left(\frac{1}{6}\right)^2\left(\frac{2}{3}\right)^3$$
$$P(X=3,Y=5)=\left(\frac{1}{6}\right)^2\left(\frac{2}{3}\right)^3$$
$$P(X=4,Y=5)=\left(\frac{1}{6}\right)^2\left(\frac{2}{3}\right)^3$$
$$P(X=6,Y=5)=\left(\frac{1}{6}\right)^2\left(\frac{2}{3}\right)^4$$
$$P(X=7,Y=5)=\left(\frac{1}{6}\right)^2\left(\frac{2}{3}\right)^5$$
$$P(X=8,Y=5)=\left(\frac{1}{6}\right)^2\left(\frac{2}{3}\right)^6$$ and so on $\cdots \cdots$
Thus
$$\sum_{k=1,2,3..}k\,P(X=k,Y=5)=\left(\frac{1}{36}\right)\left(\frac{8}{27}\right)\left(10+6\left(\frac{2}{3}\right)+7\left(\frac{2}{3}\right)^2+8\left(\frac{2}{3}\right)^3+9\left(\frac{2}{3}\right)^4+\cdots\right)=\left(\frac{1}{36}\right)\left(\frac{8}{27}\right)\left(5+5+6\left(\frac{2}{3}\right)+7\left(\frac{2}{3}\right)^2+8\left(\frac{2}{3}\right)^3+9\left(\frac{2}{3}\right)^4+\cdots\right)$$ Now for $|x|<1$
$$ \left(1-x\right)^{-2}=1+2x+3x^2+4x^3+5x^4+6x^5+7x^6+\cdots+\infty $$ $\implies$
$$g(x)=5+6x+7x^2+8x^3+\cdots+\infty=\frac{\left(1-x\right)^{-2}}{x^4}-\left(\frac{1+2x+3x^2+4x^3}{x^4}\right)$$ So
$$g\left(\frac{2}{3}\right)=21$$ So
$$\sum_{k=1,2,3..}k\,P(X=k,Y=5)=\left(\frac{1}{36}\right)\left(\frac{8}{27}\right)(26)=\frac{52}{243}$$
Also $$P(Y=5)=\frac{5^4}{6^5}$$ So
$$E(X|Y=5)=\left(\frac{52}{243}\right)\left(\frac{6^5}{5^4}\right)=\frac{1664}{625}$$
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/804414",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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|
Integral $\int_0^1\log(1+x)\frac{1+x^2}{(1+x)^4}dx=-\frac{\log 2}{3}+\frac{23}{72}$ EDIT: Small Typo Fixed now, Thanks to Sir Chen Wang!
Hi I am trying to prove this result without using a series approach
$$
\int_0^1\log(1+x)\frac{1+x^2}{(1+x)^4}dx=-\frac{\log 2}{3}+\frac{23}{72}.
$$I know we can just solve it by writing
$$
\sum_{n=1}^\infty \frac{(-1)^{n+1}}{n}\left( \int_0^1\frac{x^n}{(1+x)^4}dx+\int_0^1 \frac{x^{2+n}}{(1+x)^4}dx\right),
$$
which leads to summing a bunch of Harmonic numbers which is not so easy.
This method is brute force but I often have trouble summing harmonic numbers, can we prove it another way?
Thanks
|
$$
\begin{align*}
I&=\int^1_0\log(1+x)\frac{1+x^2}{(1+x)^4}dx\\
&=-\frac13\int^1_0\log(1+x)d\left(\frac{2+3x+3x^2}{(1+x)^3}\right)\\
&=-\frac13\left(\left.\log(1+x)\frac{2+3x+3x^2}{(1+x)^3}\right|^1_0-\int^1_0\frac{2+3x+3x^2}{(1+x)^3}d\,\log(1+x)\right)\\
&=-\frac{\log 2}{3}+\frac13\int^1_0\frac{2+3x+3x^2}{(1+x)^4}dx\\
&=-\frac{\log 2}{3}+\frac13\int^1_0\left(\frac{2}{(1+x)^4}-\frac{3}{(1+x)^3}+\frac{3}{(1+x)^2}\right)dx\\
&=-\frac{\log 2}{3}+\frac13\left(\frac{7}{12}-\frac{9}{8}+\frac{3}{2}\right)\\
&=-\frac{\log 2}{3}+\frac{23}{72}.
\end{align*}
$$
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/804673",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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|
Proof of basic properties Can anyone provide proof of properties such as:
$$a(b+c) = ab+ac$$
$$(a+b)^2 = a^2+2ab+b^2$$
And exponent rules:
$$a^n \cdot a^m = a^{n+m}$$
$$(a^n)^m = a^{n \cdot m}$$
For $a, b, c \in \mathbb{R}$
I'm quite sure that the proof is the same or very similar for the first two and the last two, which is why I pick so 'many' examples.
Thanks.
|
The first equation is an axiom of $\mathbb{R}$
The second equation is an application of the first one and uses another axiom of $\mathbb{R}$, the commutative property :
$$(a+b)^2 = (a+b)\cdot (a+b) = a\cdot a + a\cdot b + b\cdot a + b\cdot b = a^2+2ab+b^2$$
The two next proves are made for integers. We can extand these results saying
$$\underbrace{a \cdot a \cdot a \cdot\ldots \cdot a}_{\lfloor m\rfloor \textrm{ times}}\cdot a^{m-\lfloor m\rfloor} = \underbrace{a \cdot a \cdot a \cdot\ldots \cdot a}_{m \textrm{ times}}$$
The third one uses the commutative and associative properties :
$$\underbrace{a \cdot a \cdot a \cdot\ldots \cdot a}_{m \textrm{ times}}\cdot \underbrace{a \cdot a \cdot a \cdot\ldots \cdot a}_{n \textrm{ times}}= \underbrace{a \cdot a \cdot a \cdot\ldots \cdot a}_{m+n \textrm{ times}}$$
The last one is the same of the third one :
$${\underbrace{\underbrace{a \cdot a \cdot a \cdot\ldots \cdot a}_{m \textrm{ times}}\cdot \underbrace{a \cdot a \cdot a \cdot\ldots \cdot a}_{m \textrm{ times}}\cdot \cdot\ldots \cdot\underbrace{a \cdot a \cdot a \cdot\ldots \cdot a}_{m \textrm{ times}}}_{n \textrm{ times}}} = \underbrace{a \cdot a \cdot a \cdot\ldots \cdot a}_{m\cdot n \textrm{ times}}$$
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/807566",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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|
Linear Algebra matrix notation My question is referring to the following $4 \times 6$ matrix:
$$\begin{bmatrix}
0 & 1 & 2 & 0 & 0 & 2 \\
0 & 0 & 0 & 1 & 0 & 3 \\
0 & 0 & 0 & 0 & 1 & 4 \\
0 & 0 & 0 & 0 & 0 & 0 \\
\end{bmatrix}$$
Then the book states "It is useful to denote the column numbers in which the leading entries $1$ occur, by $c_1, c_2, ...c_r$, with remaining columns being $c_{r+1} ... c_n$ where $r$ is the number of non-zero rows. For this matrix is gives the example of $r=3$, $c_1 = 2$, $c_2=4$, $c_3=5$, $c_4=1$, $c_5=3$, $c_6=6$.
I don't understand this, what does it mean? How did they get all those numbers for each $c$?
Thanks everyone for your time.
|
I haven't seen this notation before, and I don't see the purpose of it, which is probably why it is so hard to understand. This seems to be some sort of intermediate (and complicated...) way of describing reduced row echelon forms.
The definition, stated a little less haphazardly, is:
*
*(1) $r$ is the number of nonzero rows in the matrix
*(2) $c_j$, for $j \le r$ : look at the $j^{th}$ nonzero row, $c_j$ is the column of the first nonzero value
*(3) $c_j$, for $j > r$ : look at the $j^\text{th}$ column that doesn't contain a pivot. Suppose the $j^{th}$ column without a pivot is the $n^\text{th}$ column in the matrix. $c_j = n$.
Example:
Here is your example matrix, with boxes around the pivots (the first non zero elements per row):
$$\begin{bmatrix}
0 & \boxed{1} & 2 & 0 & 0 & 2 \\
0 & 0 & 0 & \boxed{1} & 0 & 3 \\
0 & 0 & 0 & 0 & \boxed{1} & 4 \\
0 & 0 & 0 & 0 & 0 & 0 \\
\end{bmatrix}$$
This matrix has $3$ nonzero rows, so $r=3$:
$${
\boxed{\begin{array} {c c c c c c}
0 & 1 & 2 & 0 & 0 & 2 \\
0 & 0 & 0 & 1 & 0 & 3 \\
0 & 0 & 0 & 0 & 1 & 4 \\
\end{array}}\\
\begin{array} {c c c c c c}
0 & 0 & 0 & 0 & 0 & 0 \\
\end{array}
}$$
So now we want to find $c_1, c_2, c_3$. To find $c_1$, look at the first nonzero row, and find out what column the pivot is in:
$$\begin{bmatrix}
0 & \boxed{1} & 2 & 0 & 0 & 2 \\
0 & 0 & 0 & 1 & 0 & 3 \\
0 & 0 & 0 & 0 & 1 & 4 \\
0 & 0 & 0 & 0 & 0 & 0 \\
\end{bmatrix} \text{First nonzero row, pivot in column 2}$$
So $c_1 = 2$. For $c_2$:
$$\begin{bmatrix}
0 & 1 & 2 & 0 & 0 & 2 \\
0 & 0 & 0 & \boxed{1} & 0 & 3 \\
0 & 0 & 0 & 0 & 1 & 4 \\
0 & 0 & 0 & 0 & 0 & 0 \\
\end{bmatrix} \text{Second nonzero row, pivot in column 4}$$
So $c_2 = 4$. For $c_3$:
$$\begin{bmatrix}
0 & 1 & 2 & 0 & 0 & 2 \\
0 & 0 & 0 & 1 & 0 & 3 \\
0 & 0 & 0 & 0 & \boxed{1} & 4 \\
0 & 0 & 0 & 0 & 0 & 0 \\
\end{bmatrix} \text{Third nonzero row, pivot in column 5}$$
So $c_3 = 5$. Now we want to find the rest of the $c$ values, look at the columns without pivots:
$$
\boxed{\begin{array} {c} 0 \\ 0 \\ 0 \\ 0 \end{array}}
\begin{array} {c} 1 \\ 0 \\ 0 \\ 0 \end{array}
\boxed{\begin{array} {c} 2 \\ 0 \\ 0 \\ 0 \end{array}}
\begin{array} {c} 0 \\ 1 \\ 0 \\ 0 \end{array}
\begin{array} {c} 0 \\ 0 \\ 1 \\ 0 \end{array}
\boxed{\begin{array} {c} 2 \\ 3 \\ 4 \\ 0 \end{array}}
$$
That's columns $1, 3$, and $6$. So $c_4 = 1$, $c_5 = 3$, and $c_6 = 6$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
linear transformation $\Bbb R^3 \to \Bbb R^2$ For linear transform $T ( 1,1,1 ) = (1 , 2),\,\, T ( 1, 1, 0) = (2 , 3)$ and $\,\,T ( 1,0,0 ) = ( 3 , 4)$,from $\mathbb{R}^3$ to $\mathbb{R}^2$, the function used was : Choose one :
a. $T ( x, y, z) = (x - y - 3z , 4x - y - 2z )$
b. $T ( x, y, z) = (3x + y + z, 4x + y + z)$
c. $T ( x, y, z) = (3x - y - z, 4x - y -z )$
d. $T ( x, y, z) = (x - 3y - z, 4x - y -z )$
e. $T ( x, y, z) = (3x + y - z, 4x - y + z)$
|
The answer is (c). One can rewrite the given equations in matrix form as follows:
$$ T \cdot\begin{bmatrix} 1 & 1 & 1 \\ 1 & 1 & 0 \\ 1 & 0 & 0 \end{bmatrix} = \begin{bmatrix} 1 & 2 & 3 \\ 2 & 3 & 4 \end{bmatrix}.$$
Hence, we have
$$ T = \begin{bmatrix} 1 & 2 & 3 \\ 2 & 3 & 4 \end{bmatrix} \cdot\begin{bmatrix} 1 & 1 & 1 \\ 1 & 1 & 0 \\ 1 & 0 & 0 \end{bmatrix}^{-1} = \begin{bmatrix} 1 & 2 & 3 \\ 2 & 3 & 4 \end{bmatrix} \cdot \begin{bmatrix} 0 & 0 & 1 \\ 0 & 1 & -1 \\ 1 & -1 & 0 \end{bmatrix} = \begin{bmatrix} 3 & -1 & -1 \\ 4 & -1 & -1 \end{bmatrix}.$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/810141",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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|
I want to prove $k(x,t)=\frac{1}{\sqrt{4\pi t} } e^{\frac{-x^2}{4t}} $ I have this integral $$u(x,t)=\int _{-\infty}^{\infty} f(\eta)\left[\frac{1}{2\pi}\int _{-\infty}^{\infty}e^{iw(x-\eta)-w^2t}\ dw\right]\ d\eta=\int _{-\infty}^{\infty}k(x-\eta,t)f(\eta)\ d\eta$$
I want to prove $$k(x,t)=\frac{1}{\sqrt{4\pi t} }\ e^{\Large \frac{-x^2}{4t}}
$$
Thanks for helping me out.
|
Completing square gives
\begin{align*}
&= \int_{-\infty}^\infty \exp ( i\omega (x-\eta ) - \omega^2t)d\omega \\
&= \int_{-\infty}^\infty \exp \left( - \left( i (x-\eta ) \cdot \frac{1}{2\sqrt t}\right )^2+2 \omega \sqrt{t} i (x-\eta ) \cdot \frac{1}{2\sqrt t} - \omega^2t + \left( i (x-\eta ) \cdot \frac{1}{2\sqrt t}\right )^2\right)d\omega \\
&= \exp\left(- \frac{(x-\eta)^2}{4t} \right )\int_{-\infty}^\infty \exp \left(- \left(\omega \sqrt t - \frac {i (x-\eta)} {2 \sqrt t} \right )^2 \right )d\omega\\
&= \exp\left(- \frac{(x-\eta)^2}{4t} \right) \frac 1 {\sqrt t} \int_{-\infty}^\infty \exp \left(- \left(\omega \sqrt t - \frac {i (x-\eta)} {2 \sqrt t} \right )^2 \right )d(\omega \sqrt t) \\
&= \exp\left(- \frac{(x-\eta)^2}{4t} \right) \sqrt{\frac \pi t}
\end{align*}
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/814708",
"timestamp": "2023-03-29T00:00:00",
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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.