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How to solve $x^3 \equiv 1 \pmod{37}$ We are asked to solve $x^3 \equiv 1 \pmod{37}$. I know that the answer is $10$ since $27\cdot37 = 999$ and $10^3 = 1000$ but how do I show this rigorously? If it helps, we are given the primitive roots of $37$ which are $2, 5, 13, 15, 17, 18, 19, 20, 22, 24, 32$, and $35$. But I am not sure how this is useful.
|
Just for fun another way.
Because of $\mathbb F_{37}$ is a field the equation in $\mathbb F_{37}[x]$ $$x^3=1\iff (x-1)(x^2+x+1)=0$$
must have three roots so, since $1$ is clearly root, one can find out the other two ones solving $$x^2+x+1=0$$ We have
$$x+y=-1=36\\xy=1$$ Exploring $$35+1\\34+2\\33+3\\32+4\\31+5\\30+6\\29+7\\28+8\\27+9\\26+10$$ We stop because $26\cdot10=260\equiv 1\pmod{37}$, thus the roots are $1,10$ and $26$.
|
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|
Proving distributive law of natural numbers Is my proof correct?
If we define multiplication for natural numbers as
$a \times S(b) = (a \times b) + a$
$a \times 0 = 0$
And addition as
$a + 0 = a$
$a + S(b) = S(a+b)$
Where $S(n)$ is the successor function of $n$ (and assume we've already proven commutative property, etc).
Then I wish to prove that $a \times (b + c) = (a \times b) + (a \times c)$ via induction on $c$.
Base case: Let $c=0$. Then $a \times (b + 0) = a \times b$ and $(a \times b) + (a \times 0) = (a \times b) + 0 = a \times b$. Therefore $a \times (b + 0) = (a \times b) + (a \times 0)$.
Inductive Step: Suppose $a \times (b + c) = (a \times b) + (a \times c)$ for some $c$. We wish to show that $a \times (b + S(c)) = (a \times b) + (a \times S(c))$.
$a \times (b + S(c)) = a \times (S(b + c)) = a \times (b+c) + a = (a \times b) + (a \times c) + a$
and
$(a \times b) + (a \times S(c)) = (a \times b) + (a \times c) + a$
Since both sides can be shown to equal the same quantity $(a \times b) + (a \times c) + a$, we close the induction.
|
Perfect ... though you may want to note that you need associativity of addition, since:
$a \times (b + S(c)) = a \times (S(b + c)) = (a \times (b+c)) + a = \color{red}{(}(a \times b) + (a \times c)\color{red}) + a$
while:
$(a \times b) + (a \times S(c)) = (a \times b) + \color{red}((a \times c) + a\color{red})$
|
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|
Solve an equation with integer variables I am trying to find all the integers $(n,m)$ such that $$2^{2n+1} + 2^n + 1 = m^2$$.
With a simple python program, I find that $n=0$, $n=4$ are the only solutions less than $50$. However with the precision of number I can not get accurate results when $n > 50$. Does any one have an idea which could help me to solve this equation ?
|
Suppose $n > 2$. We have $m^2 -1 = (m-1)(m+1)\equiv 0 \mod 2^n$, so one of $m-1$ and $m+1$ is divisible by $2^{n-1}$ (the other is divisible by $2$ but no higher power of $2$). By symmetry $m \to -m$, we may assume it is $m-1$. Thus
$m = 1 + k 2^{n-1}$. With this substitution the equation becomes
$$ 2^{2n} \left(2 - \frac{k^2}{4}\right) + (1-k) 2^n = 0$$
or
$$ 2^{n-2} = \frac{k-1}{8-k^2}$$
Now $\gcd(k-1,8-k^2) = \gcd(k-1,8-k) = \gcd(k-1,7)$ is either $1$ or $7$.
$k-1$ must be even, so $k$ is odd and $8-k^2$ is odd (and can only be $\pm 1$ or $\pm 7$). Only for $k=-3$ do we get $(k-1)/(8-k^2) = 4 $ a power of $2$. This corresponds to the solution $n = 4$, $m = 1 - 3 \cdot 2^3 = -23$.
|
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|
solving $x\,dx + xy^2\,dx + y\,dy + yx^2\,dy=0$ I have to solve $x\,dx + xy^2\,dx + y\,dy + yx^2\,dy=0$
Dividing by $dx$ we have
$x + xy^2 + yy' + yy'x^2=0$
From where,
$$\frac{yy'}{1+y^2}+\frac{x}{1+x^2}=\frac{y\,dy}{1+y^2}+\frac{x\,dx}{1+x^2}=\\ =\frac{d(y^2+1)}{1+y^2}+\frac{d(x^2+1)}{1+x^2}= \frac{1}{2}d\ln(1+y^2)+\frac{1}{2} d\ln(1+x^2)=\frac{1}{2}d\ln(1+y^2)(1+x^2)=0$$
Let $c=(1+y^2)(1+x^2)$, so our equation becomes:
$$
d\ln c=0
$$
So what should I do here, should I integrate, or should I divide by $dx$?
If I divide by dx I get the expression $2x+2yy'+2xy^2+2x^2yy'=0$ which has $x$, $y$ and $y'$ and doesn't help me get anywhere.
Thanks in advance.
|
it is $$-\frac{y'(x)}{\frac{1+y(x)^2}{y(x)}}=\frac{x}{1+x^2}$$
|
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|
Triangle inequality with the exradii $r_{a}$, $r_{b}$, $r_{c}$, the medians $m_{a}$, $m_{b}$, $m_{c}$ Given a triangle with the exradii $r_{a}$, $r_{b}$, $r_{c}$, the medians $m_{a}$, $m_{b}$, $m_{c}$. Show that
$$r_{a}^{2}+ r_{b}^{2}+ r_{c}^{2}\geq 3\sqrt{3}. S+ \left ( m_{a}- m_{b} \right )^{2}+ \left ( m_{b}- m_{c} \right )^{2}+ \left ( m_{c}- m_{a} \right )^{2}$$
I can only show the weaker inequality
$$r_{a}^{2}+ r_{b}^{2}+ r_{c}^{2}\geq 3\sqrt{3}. S$$
I need to the help. Thanks!
|
Let $a=y+z$, $b=x+z$ and $c=x+y$.
Hence, $x$, $y$ and $z$ they are positives and we need to prove that
$$\sum_{cyc}\frac{4S^2}{(b+c-a)^2}\geq3\sqrt{3xyz(x+y+z)}+2\sum_{cyc}(m_a^2-m_am_b)$$ or
$$\sum_{cyc}\frac{xyz(x+y+z)}{x^2}+\frac{1}{2}\sum_{cyc}\sqrt{\left(4x(x+y+z)+(y-z)^2\right)\left(4y(x+y+z)+(x-z)^2\right)}\geq3\sqrt{3xyz(x+y+z)}+3\sum_{cyc}(x^2+xy).$$
Now, $$\sqrt{3xyz(x+y+z)}\leq xy+xz+yz$$ and by C-S and AM-GM
\begin{align}
&\sum_{cyc}\sqrt{\left(4x(x+y+z)+(y-z)^2\right)\left(4y(x+y+z)+(x-z)^2\right)} \\
=&\sum_{cyc}\sqrt{(4x^2+y^2+z^2+4xy+4xz-2yz)(x^2+4y^2+4xy+4yz-2xz)} \\
=&\sum_{cyc}\sqrt{((2x+y-z)^2+8xz)((2y+x-z)^2+8yz)} \\
\geq&\sum_{cyc}((2x+y-z)(2y+x-z)+8z\sqrt{xy}) \\
\geq&\sum_{cyc}(2x+y-z)(2y+x-z)+24\sqrt[3]{x^2y^2z^2} \\
=&\sum_{cyc}(5x^2-xy)+24\sqrt[3]{x^2y^2z^2}.
\end{align}
Thus, it's enough to prove that
$$\sum_{cyc}\frac{yz(x+y+z)}{x}+\frac{1}{2}\sum_{cyc}(5x^2-xy)+12\sqrt[3]{x^2y^2z^2}\geq3(xy+xz+yz)+3\sum_{cyc}(x^2+xy)$$ or
$$\frac{2(x^2y^2+x^2z^2+y^2z^2)(x+y+z)}{xyz}+24\sqrt[3]{x^2y^2z^2}\geq\sum_{cyc}(x^2+13xy).$$
Now, let $x+y+z=3u$, $xy+xz+yz=3v^2$ and $xyz=w^3$.
Thus, we need to prove that
$$\frac{2(9v^4-6uw^3)3u}{w^3}+24w^2\geq9u^2+33v^2$$ or $f(v^2)\geq0,$ where
$$f(v^2)=\frac{2u(9v^4-6uw^3)}{w^3}+8w^2-3u^2-11v^2.$$
But, $$f'(v^2)=\frac{36uv^2}{w^3}-11>0,$$ which says that it's enough to prove that $f(v^2)\geq0$ for the minimal value of $v^2$,
which happens for equality case of two variables.
Since the last inequality is homogeneous, then after replacing $x$ on $x^3$ we need to prove that
$$\frac{2(2x^6+1)(x^3+2)}{x^3}+24x^2\geq x^6+2+26x^3+13$$ or
$$(x-1)^2(3x^7+6x^6+9x^5-6x^4+3x^3+12x^2+8x+4)\geq0,$$ which is obvious.
|
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|
Multiplicative group of integers modulo $p$ $\mathbb{Z}/7\mathbb{Z}=\{1,2,3,4,5,6\}$.
$6\times 6=1~{\rm mod}~ 7$ implies $6$ is an element of order $2$; however, we know that $\mathbb{Z}/7\mathbb{Z}\cong C_7$, not containing an element of order $2$.
I found it incredibly confusing, what have I missed? Any help will be appreciated.
|
The multiplicative group modulo a prime $n$ and a group opperator of multiplication and the element $0$ removed is equivalent to an additive group modulo $n-1$ with zero in.
$A=\mathbb Z/7\mathbb Z-\{0\}^{\times}\cong B=\mathbb Z/6\mathbb Z^{+}$
$1 \to 0$ because $1\times a = a \to 0 + a = a$.
$5 \to 1$ because then $A = <5> = \{5^0,5^1,5^2,5^3,5^4,5^5\} = \{1,5,4,6,2,3\}$ $\to$ $B = <1> = \{0,1,1+1,1+1+1,1+1+1+1,1+1+1+1+1\} = \{0,1,2,3,4,5,\}$.
$4 \to 2$ because $5\times 5 = 4 \to 1+1 = 2$.
$6 \to 3$ beacause $5\times 4 = 6\to 1+2 = 3$ and because $6\times 6=1\to 3+3 = 0$.
$2\to 4$ because $5\times 6 = 4\times 4 = 2 \to 1+3 = 2+2 = 4$ and because $2\times 4 = 0 \to 4+2 = 0$.
$3\to 5$ because $5\times 2=4\times 6 = 3 \to 1+4=2+3 = 5$ and because $3\times 5 = 1 \to 5 + 1 = 6$.
|
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|
Factoring $x^6+x^5+x^4+x^3+x^2+x+1$ into two cubics On page 587 of Dummit and Foote, the authors demonstrate producing irreducible polynomials over $\mathbb{F}_p$ recursively. This is after Proposition 18 which states "The polynomial $x^{p^n}-x$ is precisely the product of all the distinct irreducible polynomials in $\mathbb{F}_p[x]$ of degree $d$ where $d$ runs through all the divisors of $n$."
So, to find the irreducible cubics over $\mathbb{F}_2$, we consider the divisors of $\frac{x^8-x}{x(x-1)}=x^6+x^5+x^4+x^3+x^2+x+1$.
This apparently factors into the two cubics $x^3+x+1$ and $x^3+x^2+1$, but I am having trouble seeing why this is so. How do I factor the polynomial above as such?
I'd really like to know how to go about factoring the polynomial $x^6+x^5+x^4+x^3+x^2+x+1$ assuming I don't know it factors as the two cubics above.
|
We want to find $(x^3+ax^2+bx+c)(x^3+dx^2+ex+f)$. We have little choice for $c$ and $f$, as they must be $1$. Also, changing $x$ into $1/x$ and multiplying by $x^6$ should leave the thing fixed: since we get
$$
(x^3+bx^2+ax+1)(x^3+ex^2+dx+1)
$$
we need either $a=b$ and $d=e$ or $a=e$ and $b=d$. This is because of unique factorization, of course.
The first case is easily dismissed:
$$
(x^3+ax^2+ax+1)(x^3+dx^2+dx+1)
$$
would give, for the degree $5$ term, $a+d=1$, so, without loss of generality, $a=0$; however $x^3+1$ is not irreducible.
We remain with
$$
(x^3+ax^2+bx+1)(x^3+bx^2+ax+1)
$$
and we easily get $a=1$ and $b=0$ (or conversely).
|
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|
Find all integer solutions of: $\;\frac{1}{m}+\frac{1}{n}-\frac{1}{mn^2}=\frac{3}{4}$ I found the following problem from the 10th Iranian Mathematical Olympiad in Crux Magazine.
Find all integer solutions of $$\frac{1}{m}+\frac{1}{n}-\frac{1}{mn^2}=\frac{3}{4}$$
Initially it looked like a typical quadratic problem, however I hit a dead end each time I solve it.
My methodology is as follows,
$$\frac{n^2+mn-1}{mn^2}=\frac{3}{4}$$
$$\implies 4n^2+4mn-4 = 3mn^2$$
$$\implies (4-3m)n^2+4m \cdot n-4=0$$
I used the quadratic formula, and got,
$$n = \frac{-4m \pm \sqrt{(4m)^2-4\cdot(4-3m)\cdot(-4)}}{2(4-3m)}$$
I do the usual algebraic manipulations and drop at,
$$n = \frac{-2m \pm 2\sqrt{m^2-3m+4}}{4-3m}$$
I am unsure how I go ahead from this.
Some help would be much appreciated.
Cheers!
|
The other answers may be more elegant, but here is a way to continue from the point you reached. Assume $m$ and $n$ are integers. From
$$n = \frac{-2m \pm 2\sqrt{m^2-3m+4}}{4-3m}$$
you know that $-2m \pm 2\sqrt{m^2-3m+4}$ must be a multiple of
the integer $4-3m$; in particular, it must be an integer.
Therefore $2\sqrt{m^2-3m+4}$ is an integer
(since it is a difference of the integer $-2m$ and another integer),
which implies that $4(m^2-3m+4)$ is the square of an integer.
Suppose $4(m^2-3m+4) = k^2$ for some integer $k.$
Complete the square:
\begin{align}
k^2 &= 4(m^2-3m+4)\\ &= 4m^2 - 12m + 16 \\
&= (2m - 3)^2 + 7
\end{align}
and therefore the difference between the two squares $(2m - 3)^2$
and $k^2$ is $7.$
This is possible only if the two squares are $9$ and $16.$
Therefore $k^2 = 16$ and $$(2m - 3)^2 = 9.$$
This quadratic equation in $m$ has two roots, $m = 0$ and $m = 3,$
but $m = 0$ cannot be true for any solution to the original problem
(since we require $1/m$ to be defined); therefore
$$ m = 3.$$
Plug this into your equation for $n$ and confirm that the only possible integer value of $n$ is $n = 2.$
Alternative method:
Suppose that you solved for $m$ as in several other answers,
so that you found the equation
$$ m = \frac{4(n^2 - 1)}{3n^2 - 4n}. $$
If you know a few facts about rational functions (polynomials divided by polynomials), you can sketch $m$ as a function of $n$ free-hand
(without the assistance of a graphing calculator or software)
and solve the problem graphically.
First, plot the functions $m = 4(n^2 - 1)$ and $m = 3n^2 - 4n$
as shown below.
Because $3n^2 - 4n = 0$ at $n=0$ and $n = \frac43,$ we know that the rational function has vertical asymptotes at the lines $n=0$ and $n = \frac43.$
By looking at the highest-order coefficients of $4(n^2 - 1)$ and $3n^2 - 4n,$
we know the rational function has a horizontal asymptote at $m = \frac43.$
We know the rational function has the same zeros as $4(n^2 - 1),$
namely $n = \pm1.$
And by carefully accounting for the direction in which $3n^2 - 4n$ crosses the $n$-axis and the sign of $4(n^2 - 1)$ at each crossing,
we can find out how the rational function approaches each of its asymptotes.
The result is something like the following graph:
Assuming we sketched this free-hand, we do not yet know how close the curve is to various points with integer coordinates.
But since the middle branch has asymptotes $n=0$ and $n = \frac43,$ its only possible integer solution is at $n=1$ (which indeed occurs at $(1,0)$).
Since the left branch passes through $(-1,0),$ its only possible other
integer solution is at $m=1$ (at which we can solve for $n,$ discovering that $n$ is not an integer in that case).
We can then chip away at the right branch, either by finding $m$ when
$n = 2$ (the smallest possible integer value of $n$ on that branch)
or by solving for $n$ when $m = 2$
(the smallest possible integer value of $m$ on that branch).
Either way, we quickly find out that $(2,3)$ is a solution,
that the only other possible integer solution would be at $m = 2,$
and that in fact the curve crosses $m = 2$ at a non-integer value of $n$;
therefore there are no other solutions.
We eliminate $(1,0)$ and $(-1,0)$ by referring to the original problem statement, leaving only $n = 2, m = 3.$
|
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Is this proof of the irrationality of the square root of $3$ valid? Assume that $\sqrt{3}=\frac{a}{b}$ where $\frac{a}{b}$ is in its simplest possible form. So $3=\frac{a^2}{b^2}$, hence $3b^2=a^2$. If $b$ is even then $a^2$ and $a$ are also even. That means they have a common factor of $2$. But this is impossible since $\frac{a}{b}$ is in its simplest form. So if $\frac{a}{b}$ does exist, $a$ and $b$ must both be odd. If $a=2k+1$ and $b=2m+1$ then $3=\frac{4k^2+4k+1}{4m^2+4m+1}$. After a few steps we get to $4(3m^2+3m)−4(k^2+k)=2$. If $3m^2+3m=c$ and $k^2+k=d$ then $4c−4d=2$ and so $4(c−d)=2$. This has no whole number solutions and that means that $a$ and $b$ are not even and not odd and that cannot be. So $\frac{a}{b}$ does not exist.
|
Short version:
$$c^2\bmod4=c\bmod2$$
so that with $a,b$ not both even,
$$(3a^2-b^2)\bmod4\ne0.$$
|
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Define an isomorphism from $\mathbf{R}^2 $to $S$, where $S=\operatorname{span} .... $(image) Not a clue what to do with this question. We only briefly touched on isomorphisms in class,and I can't find any example online that uses transformations on two basis vectors, as in this question. Help a plebe out?
|
For part $a)$ You can define a Linear map $T:\mathbf{R^2}\to S$ as follows
$$T\left(c_1\begin{pmatrix}1\\0\\\end{pmatrix}+c_2\begin{pmatrix}0\\1\\\end{pmatrix}\right) = c_1\begin{pmatrix}3\\1\\4\\\end{pmatrix}+c_2\begin{pmatrix}2\\-7\\1\\\end{pmatrix}$$
The remaining task is then to show that with the above definition $T$ is an isomorphism. You can show this by proving that $\operatorname{null}T = \{0\}$ and $\operatorname{range}T =S$. The first of these two claims follows from the linear independence of $\begin{pmatrix}3\\1\\4\\\end{pmatrix}$ and $\begin{pmatrix}2\\-7\\1\\\end{pmatrix}$ and the latter claim is evident if you consider that $S$ by definition is the span of these two vectors.
For $b)$ Note that $v = \begin{pmatrix}10\\11\\15\\\end{pmatrix} = 4\cdot\begin{pmatrix}3\\1\\4\\\end{pmatrix}-1\cdot\begin{pmatrix}2\\-7\\1\\\end{pmatrix}$ consequently with the above choice of $T$
$$T^{-1}(v) = 4\cdot\begin{pmatrix}1\\0\end{pmatrix}-1\cdot\begin{pmatrix}0\\1\end{pmatrix}= \begin{pmatrix}4\\-1\end{pmatrix}$$
|
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Horizontal asymptotes of function defined as an inequation I was given as a homework:
Suppose $f$ is a function of $x$ and, $\forall x\in \mathbb{R}$, it holds
$$
\left| f(x)-\frac{1}{x}\right| \le \frac{2x^2+x|x|+2}{x^2+1}
$$
Find the horizontal asymptotes of $f(x)$.
I tried to use the Squeeze Theorem, rewritting the expression as:
$$
-\frac{2x^2+x|x|+2}{x^2+1} \le f(x)-\frac{1}{x} \le \frac{2x^2+x|x|+2}{x^2+1}
$$
Which, rearranging, gives
$$
\frac{-2x^3-x^2|x|-2x+x^2+1}{x^3+x} \le f(x) \le \frac{2x^3+x^2|x|+2x+x^2+1}{x^3+x}\\
g(x) \le f(x) \le h(x)
$$
The limits, however, are not equal:
$$
\lim_{x\rightarrow\infty}g(x)=-3\\
\lim_{x\rightarrow\infty}h(x)=3\\
\lim_{x\rightarrow-\infty}g(x)=-1\\
\lim_{x\rightarrow-\infty}h(x)=1\\
$$
And I can't use the Squeeze Theorem to find the horizontal asymptotes. Any hints on how to solve it?
|
There will be no hints on how to solve it because the condition you gave is not enough to completely determine horizontal asymptotes. For $x < 0$, the bounding expression becomes $$\frac{2x^2+x|x|+2}{x^2+1} = \frac {x^2 + 2}{x^2+1} = 1 + \frac 1{x^2+1}$$
which converges to $1$ as $x \to -\infty$, while for $x > 0$, it becomes
$$\frac{2x^2+x|x|+2}{x^2+1} = \frac {3x^2 + 2}{x^2+1} = 3 - \frac 1{x^2+1}$$
which converges to $3$ as $x \to \infty$. Thus choosing $f(x) = \frac1x + a$ will satisfy the inequality for sufficiently large values of $|x|$ for any $a$ with $|a| \le 1$.
Thus we can find functions $f(x)$ satisfying the condition and having any horizontal asymptote between $-1$ and $1$. Further, there is no reason a function cannot have different horizontal asymptotes to the left and to the right. For example, $\tan^{-1}x$ has a left asymptote of $-\frac \pi 2$ and a right asymptote of $\frac \pi 2$. Your condition requires the left asymptote $f(x)$ to be between $-1$ and $1$, but the right asymptote can be anywhere between $-3$ and $3$.
But that is assuming that the asymptotes exist at all. Your condition is not sufficient to require even that: $f(x) = \frac 1x + a\sin x$ satisfies the inequality for large $x$, but has no horizontal asymptotes.
(I haven't bothered to determine if these functions satisfy the inequality everywhere because it doesn't matter - if it doesn't, we can replace $a$ with a function having $a$ as horizontal asymptote, and dropping to $0$ anywhere necessary to satisfy the inequality.)
|
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|
Proving the trig identity $\frac{1-\cos\theta}{\sin\theta} = \frac{\sin\theta}{1+\cos\theta}$ without cross-multiplying I need to prove the following identity.
$$\frac{1-\cos\theta}{\sin\theta} = \frac{\sin\theta}{1+\cos\theta}$$
I want to prove it by deduction rather than cross multiplying.
|
You could use the half angle formulae .
$\large\frac{1-\cos\theta}{\sin\theta}$
$=\large\frac{2\sin^2\frac\theta2}{2\sin\frac\theta2.cos\frac\theta2}$
$= \large\frac{2\sin\frac\theta2}{2\cos\frac\theta2}$
multiply both numerator and denominator by $\cos\frac\theta2$
$=\large\frac{ 2\sin\frac\theta2\cos\frac\theta2}{2\cos^2\frac\theta2}$
$= \large\frac{\sin\theta}{\cos\theta+1}$
|
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|
Verifying Isomorphism between finite fields On page 587 in Dummit and Foote, we are given that $f_1(x)=x^4+x^3+1$ and $f_2(x)=x^4+x+1$ are two irreducible quartics over $\mathbb{F}_2$.
The authors then claim that a "simple calculation" verifies that $\alpha (x)=x^3+x^2$ is a root of $f_2(x)$ in $\mathbb{F}_{16}=\mathbb{F}_2[x]/(x^4+x^3+1)$, and we may use this fact to verify that $\mathbb{F}_2[x]/(x^4+x+1)\cong\mathbb{F}_2[x]/(x^4+x^3+1)$ with $x\mapsto x^3+x^2$.
However, I am having difficulty verifying this calculation. We have that $f_2(\alpha(x))=(x^3+x^2)^4+(x^3+x^2)+1=x^{12}+x^8+x^3+x^2+1$, yet I don't see why this should equal $0$ in $\mathbb{F}_2[x]/(x^4+x^3+1)$. I know that I will have to use the fact that $x^4+x^3+1=0$ in this field, but I'm not sure how. What am I missing?
|
The answer is easy. We have
$$
x^{12}+x^8+x^3+x^2+1=
(x^8 + x^7 + x^6 + x^5 + x^4 + x^2 + 1)(x^4 + x^3 + 1)=0
$$
because $x^4+x^3+1=0$.
|
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|
Proving Schwarz inequlity. Is the following proof correct?
The Schwarz inequality states:
$$
x_1 \cdot y_1 + x_2 \cdot y_2 \leq \sqrt{(x_1^2+x_2^2)\cdot(y_1^2+y_2^2)}
$$
From
$$
(x_1^2+x_2^2)\cdot(y_1^2+y_2^2)
$$
We can deduce:
$$
(x_1^2+x_2^2)\cdot(y_1^2+y_2^2) = (x_1y_1+x_2y_2)^2 + (x_1y_2-x_2y_1)^2
$$
We can rewrite it in form:
$$
(x_1y_2-x_2y_1) = \sqrt{(x_1^2+x_2^2)\cdot(y_1^2+y_2^2) - (x_1y_1+x_2y_2)^2}
$$
Meaning that
$$
(x_1^2+x_2^2)\cdot(y_1^2+y_2^2) - (x_1y_1+x_2y_2)^2 \geq 0
$$
From which the Schwarz inequality can be easily deduced.
$$
\sqrt{(x_1^2+x_2^2)\cdot(y_1^2+y_2^2)} \geq x_1y_1+x_2y_2
$$
|
Yes it is correct even if from here
$$(x_1^2+x_2^2)\cdot(y_1^2+y_2^2) = (x_1y_1+x_2y_2)^2 + (x_1y_2-x_2y_1)^2$$
we can deduce directly that
$$(x_1^2+x_2^2)\cdot(y_1^2+y_2^2) - (x_1y_1+x_2y_2)^2=(x_1y_2-x_2y_1)^2 \geq 0$$
This proof is nice but it is limited to n=2, for a general proof we can start from
$$\sum x_iy_i\le \sqrt{\sum x_i^2\sum y_i^2}$$
and observe that WLOG for homogenity we can rescale $x$ and $y$ such that $\sum x_i^2=\sum y_i^2=1$ and thus we need to prove that
$$\sum x_iy_i\le 1$$
then observe that
$x_iy_i\le \frac12(x_i^2+y_i^2)\iff (x_i-y_i)^2\ge 0$
then
$$\sum x_iy_i\le \frac12 \sum (x_i^2+y_i^2) =\frac12 (\sum x_i^2+\sum y_i^2)=1$$
|
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|
Elementary geometry problem about two squares.
Consider the picture above, where $ABCD$ and $EFGC$ are squares with areas respectively $A$ and $B$. Find the shaded area.
Well, I observerd the following: If the intersection between $BG$ and $EF$ is the midpoint of $EF$, then the area will be $\dfrac{A}{2} + \dfrac{3B}{4}$, since the square $EFGC$ will be divided into a rectangle plus a triangle with half the area of the rectangle, leaving out an identical triangle. But then when I change the square $EFGC$, the intersection will change, and the expression will change. Is there a more general formula that includes this case? I can't see how that's possible.
A colleague of mine did the following:
"Area of triangle $DBC$ is $\dfrac{A}{2}$. Area of triangle $GCB$ is $\dfrac{\sqrt{A}\sqrt{B}}{2}$. If $H$ is $BG \cap EF$, triangle $BEH$ has height $\sqrt{A}-\sqrt{B}$ and is similar to $GFH$, with ratio $\dfrac{\sqrt{A}-\sqrt{B}}{\sqrt{B}} = \dfrac{\sqrt{A}}{\sqrt{B}}-1$, then base of $BEH$ is $\dfrac{A}{\sqrt{B}} - \sqrt{A}$. Then the trapezoid $EHGC$ has area $\dfrac{\Bigg(\dfrac{A}{\sqrt{B}} - \sqrt{A}+\sqrt{B}\Bigg)\sqrt{B}}{2} = \dfrac{A - \sqrt{AB} + B}{2}$, then the shaded area is equal to $A + \dfrac{B-\sqrt{AB}}{2} $."
But I don't understand this part: "then base of $BEH$ is $\dfrac{A}{\sqrt{B}} - \sqrt{A}$".
I think his formula is not correct because I tried an example with $A=2$,$B=1$ and $H$ being midpoint of $EF$, and it didn't work, but maybe I'm doing something wrong.
Can someone please explain what's happening? Thanks.
|
$$ \triangle EHB \sim \triangle FHG
\\ \implies \frac{FH}{EH} = \frac{FG}{EB} = \frac{\sqrt B}{\sqrt A - \sqrt B}$$
So $$ EH = FH ( \frac {\sqrt A -\sqrt B} {\sqrt B} ) $$
But we also know that
$$ FH + EH = EF = \sqrt B $$
So
$$FH(1+ \frac {\sqrt A -\sqrt B} {\sqrt B} ) = \sqrt B $$
$$ \implies FH(\sqrt B + \sqrt A - \sqrt B ) = B $$
$$\implies FH = \frac { B }{\sqrt A} $$
$$ A_{TOT}= \frac 12A + B - \frac 12 FH\cdot FG
\\ = \frac 12A + B - \frac 12 \frac { B }{\sqrt A} \cdot \sqrt B
\\ = \frac 12A + \left (1- \frac 12 \sqrt \frac BA \right )B
$$
|
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|
Can I apply regular derivative instead of implicit describing rectangle diagonal?
The sides of this rectangle increase in such a way that dz/dt=1 and
dx/dt=3dy/dt. At the instant when x=4 and y=3, what is the value of
dx/dt?
Rectangle with diagonal z
I solve it like this:
$\dfrac{dz}{dy}=(\sqrt{x^2+y^2})'=\dfrac{1}{2\sqrt{x^2+y^2}}2y$
$\dfrac{dz}{dy}=\dfrac{dz}{dt}*\dfrac{dt}{dy}$
$\dfrac{y}{\sqrt{x^2+y^2}}=1*\dfrac{dt}{dy}$, so $\dfrac{dy}{dt}=\dfrac{\sqrt{x^2+y^2}}{y}$
Given that $\dfrac{dx}{dt}=3\dfrac{dy}{dt}$, conclude $\dfrac{dx}{dt}=3\dfrac{\sqrt{x^2+y^2}}{y}$
Substituting $x=4, y=3$ the answer is 5.
The answer in textbook uses implicit derivative:
$z^2=x^2+y^2$, which implies $2z\dfrac{dz}{dt}=2x\dfrac{dx}{dt}+2y\dfrac{dy}{dt}=\dfrac{dx}{dt}(2x+\dfrac{2}{3}y)$
Substituting values for x,y and $\dfrac{dz}{dt}=1$ we get that answer is 1.
Where did I get wrong? My suspicions are:
*
*Maybe it's incorrect to use $\dfrac{dz}{dy}$ when z depends on x too
*Working with z instead of $z^2$ is prohibited
|
The problem with your answer is that your first derivative only takes the derivative with respect to $y$. $x$ is essentially ignored. $\frac{dz}{dy}$ is not the same as $\frac{\partial z}{\partial y}$.
I would, indeed, start with $z^2 = x^2 + y^2$ because it is just a much simpler problem. Starting the way you did isn't problematic, it just becomes a much harder equation.
$$
z = \sqrt{x^2 + y^2} \\
d(z) = d\left(\sqrt{x^2 + y^2}\right) \\
dz = \frac{1}{2}\frac{1}{\sqrt{x^2 + y^2}} d(x^2 + y^2) \\
dz = \frac{1}{2}\frac{1}{\sqrt{x^2 + y^2}} (2xdx + 2ydy)\\
dz = \frac{x}{\sqrt{x^2 + y^2}}dx + \frac{y}{\sqrt{x^2 + y^2}}dy \\
$$
We can divide both sides by $dt$ to get
$$
\frac{dz}{dt} = \frac{x}{\sqrt{x^2 + y^2}}\frac{dx}{dt} + \frac{y}{\sqrt{x^2 + y^2}}\frac{dy}{dt} \\
$$
Then, putting in our substitutions, we get
$$
1 = \frac{x}{\sqrt{x^2 + y^2}}\frac{dx}{dt} + \frac{1}{3}\frac{y}{\sqrt{x^2 + y^2}}\frac{dx}{dt} \\
1 = \frac{x + \frac{1}{3}y}{\sqrt{x^2 + y^2}}\frac{dx}{dt} \\
\frac{dx}{dt} = \frac{\sqrt{x^2 + y^2}}{x + \frac{1}{3}y} \\
\frac{dx}{dt} = \frac{\sqrt{4^2 + 3^2}}{4 + \frac{1}{3}3} \\
\frac{dx}{dt} = \frac{5}{5} = 1
$$
Also note that you could have used $\frac{dz}{dy}$ just fine as well, but you MUST apply the derivative with respect to $y$ as well for that to work. So, your first step would result in:
$$
\frac{dz}{dy} = \frac{1}{2}\frac{1}{\sqrt{x^2 + y^2}} (2x\frac{dx}{dy} + 2y\frac{dy}{dy})\\
$$
And the $\frac{dy}{dy}$ would reduce to $1$, but I wanted to be sure that you saw it.
|
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|
How to factor this polynomial over $\mathbb{F}_2$ On page 587 in Dummit and Foote, the authors say the polynomial $\frac{x^{16}-x}{x(x-1)(x^2+x+1)}$ can be factored into quartics over $\mathbb{F}_2$ as $(x^4+x^3+x^2+x+1)(x^4+x^3+1)(x^4+x+1)$.
I am having trouble seeing this. When I divide the polynomial $x^{16}-x$ by $x(x-1)(x^2+x+1)$ using long division, I get $x^{12}+x^9+x^6+x^3+1$. However, I am not sure how to factor this polynomial into quartics. How do I do this?
|
Over $\mathbb Z$ we have
$$
\begin{align}
x^{16}-x &= x(x^{15}-1)
\\&= x \Phi_1(x) \Phi_3(x) \Phi_5(x) \Phi_{15}(x)
\\&= x (x - 1) (x^2 + x + 1) (x^4 + x^3 + x^2 + x + 1) (x^8 - x^7 + x^5 - x^4 + x^3 - x + 1)
\end{align}
$$
and so your polynomial is $(x^4 + x^3 + x^2 + x + 1) (x^8 - x^7 + x^5 - x^4 + x^3 - x + 1)$.
You only need to factor $(x^8 - x^7 + x^5 - x^4 + x^3 - x + 1)$ mod $2$.
|
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|
Finding unit digit of $f(10)$
If we define $$f(x)=\left\lfloor \frac {x^{2x^4}}{x^{x^2}+3}\right\rfloor$$ and we have to find unit digit of $f(10)$
I had tried approximation, factorization and substitutions like $x^2=u$ but it proved of no use. Moreover the sequential powers are feeling the hell out of me. Can someone please provide me with some hints
|
First substitute in $10$ for $x$
$$f(10)=\left\lfloor \frac {10^{2\cdot 10^4}}{10^{10^2}+3}\right\rfloor\\
=\left\lfloor \frac {10^{20000}}{10^{100}+3}\right\rfloor$$
Now ask Alpha
or ask Python
and the answer is $3$
Added in response to the comment: You can do long division in base $10^{100}$. Unfortunately the numerator still has $200$ digits, so it will be a long haul. The denominator is a simple $13$. I suspect you are intended to write
$$f(10)=\left\lfloor \frac {10^{2\cdot 10^4}}{10^{10^2}+3}\right\rfloor\\
=\left\lfloor 10^{19900}\frac {10^{100}}{10^{100}+3}\right\rfloor\\
=\left\lfloor 10^{19900}\frac {1}{1+3\cdot 10^{-100}}\right\rfloor\\
=\left\lfloor 10^{19900}(1-3\cdot 10^{-100}+(3\cdot 10^{-100})^2-(3\cdot 10^{-100})^3+\ldots )\right\rfloor$$
and note that all the terms with exponents less than $199$ get too many zeros from the $10^{19900}$ to matter, then evaluate the term with exponent $199$. Then note that the term with exponent $200$ doesn't carry and is positive, so you only care about the term with exponent $199$. We have $-(3^{199})\equiv -7 \equiv 3 \pmod {10}$ so the answer is $3$.
|
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|
In any triangle, if $\frac {\cos A+2\cos C}{\cos A+2\cos B}=\frac {\sin B}{\sin C}$, then the triangle is either isosceles or right-angled In any triangle, if $\dfrac {\cos A+2\cos C}{\cos A+2\cos B}=\dfrac {\sin B}{\sin C}$, prove that the triangle is either isosceles or right angled.
My Attempt:
Given:
$$\dfrac {\cos A+2\cos C}{\cos A+2\cos B}=\dfrac {\sin B}{\sin C}$$
$$\dfrac {\dfrac {b^2+c^2-a^2}{2bc}+2\dfrac {a^2+b^2-c^2}{2ab}}{\dfrac {b^2+c^2-a^2}{2bc}+2\dfrac {a^2+c^2-b^2}{2ac}}=\dfrac {b}{c}$$
On simplification,
$$\dfrac {ab^2+ac^2-a^3+2a^2c+2b^2c-2c^3}{ab^2+ac^2-a^3+2a^2b+2bc^2-2b^3}=\dfrac {b}{c}$$
|
Alternate approach:
Cross multiply
$\cos A \sin C + \sin 2C = \sin B \cos A + \sin 2B$
$\cos A (\sin C - \sin B) - (\sin 2B - \sin 2C)=0$
$\cos A * 2* \sin \frac{C-B}{2} \cos \frac{B+C}{2} - 2* \sin (B-C) \cos(B+C)=0$
$\cos A * 2* \sin \frac{C-B}{2} \cos \frac{B+C}{2} + 2* \sin (B-C) \cos A=0$
$2\cos A [\sin \frac{C-B}{2} \cos \frac{B+C}{2} + 2 \sin \frac{C-B}{2} \cos \frac{B-C}{2}] = 0$
$2\cos A *\sin \frac{B-C}{2}[\cos \frac{B+C}{2} - 2\cos\frac{B-C}{2}]=0$
IF:
$\cos \frac{B+C}{2} - 2\cos\frac{B-C}{2} =0 \implies \tan \frac B2 \tan \frac C2 = -\frac 13$ and $\frac B2<90 $ and $\frac C2 < 90$
|
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|
Integrate $\int\frac{1}{x^3+1}dx$ The problem is, as stated:
$$\int\frac{1}{x^3+1}dx$$
I tried using substitution: $t^3 = x^3 + 1$ but didn't get far with that. I also tried setting: $t = x^3 + 1$, with no luck again.
I tried partial decomposition but I didn't know how to integrate $$\int\frac{1}{x^2-x+1}$$ and I kept getting that term when expanding $x^3 + 1$
Any help would be much appreciated.
|
Method 1
$$\int \frac {dx}{x^2-x+1}=\frac {4}{3}\int \frac {dx}{1+\frac {4(x-1/2)^2}{3}}$$
Put $u=x-1/2$
Hence $$\frac 43\int \frac {dx}{1+\frac {4(x-1/2)^2}{3}} =\frac 43\int \frac {du}{1+\frac {4u^2}{3}}$$
And note that $$\int \frac {dx}{1+x^2}=\arctan x$$
I hope you can take it from here
|
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|
Show that there are no rational solutions on the ellipse $2x^2 + 3y^2 = 1$ Show that there are no rational solutions on the ellipse $2x^2 + 3y^2 = 1$.
Want to make sure that my proof is correct.
Suppose there are rational solutions.
Note $2x^2 + 3y^2 = 1 \iff 2x^2 + 3y^2 = z^2\ |\ x,y.z \in \mathbb{Z}$
Since taking $(\frac{x}{z}, \frac{y}{z})$ as our rational solution, $2(\frac{x}{z})^2 + 3(\frac{y}{z})^2 = 1 \implies 2x^2 + 3y^2 = z^2$.
Now reducing $mod\ 3$, we have $2x^2 + 3y^2 \equiv 1 \ mod\ 3 \implies 2x^2 \equiv 1 \ mod\ 3 \implies -1x^2 \equiv 1\ mod\ 3 \implies x^2 \equiv -1\ mod\ 3 $, which is impossible.
|
You correctly arrived at the following: If there are rational points on this ellipse then the equation$$2x^2+3y^2=z^2\tag{1}$$
has integer solutions $\ne(0,0,0)$. But starting from here the argument is more intricate than you assumed.
If there are integer solutions of $(1)$ then after dividing them by a suitable power of $3$ we may assume that not all three of $x$, $y$, $z$ are divisible by $3$. Modulo $3$ we can rewrite $(1)$ as
$$z^2+x^2\equiv0\qquad{\rm mod}\ 3\ .$$
Since all squares are $=0$ or $=1$ modulo $3$ this is only possible if both $x$ and $z$ are $\equiv0$ modulo $3$. But then $(1)$ would imply that $3y^2=z^2-2x^2$ is divisible by $9$, hence $y$ would be divisible by $3$ as well, which we have excluded.
|
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|
In any $\triangle ABC$, prove that: $\frac {\cos B-\cos C}{\cos A +1}=\frac {c-b}{a}$ In any $\triangle ABC$, prove that: $\dfrac {\cos B-\cos C}{\cos A +1}=\dfrac {c-b}{a}$
My Attempt:
$$\begin{align}
\text{R.H.S.}&=\dfrac {c-b}{a} \\[4pt]
&=\frac {a\cos B+b\cos A-a\cos C-c\cos A}{b\cos C+c\cos B} \\[4pt]
&=\dfrac {a(\cos B-\cos C)+(b-c)\cos A}{b\cos C+c\cos B}
\end{align}$$
|
Obvious if $B=C$
Otherwise using this $$\dfrac{\cos B-\cos C}{\sin C-\sin B}=\cot\dfrac A2$$
$$\dfrac{1+\cos A}{\sin A}=?$$
|
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|
Proving determinant with variables I have a problem that asks:
Prove that det$\begin{pmatrix}
1 && 1 && 1 \\
a && b && c \\
a^2 && b^2 && c^2 \end{pmatrix} = (c-a)(c-b)(b-a)$
I was thinking of solving it like normal and see if I can end up with $(c-a)(c-b)(b-a)$
What I did:
det$\begin{pmatrix}
b && c \\
b^2 && c^2 \end{pmatrix}$ -
det$\begin{pmatrix}
a && c \\
a^2 && c^2 \end{pmatrix}$ +
det$\begin{pmatrix}
a && b \\
a^2 && b^2 \end{pmatrix}$
which gives
$(bc^2 - cb^2)-(ac^2-ca^2)+(ab^2-ba^2)$
which simplifies to
$(b-a)c^2 + (a-c)b^2 + (c-b)a^2$
and this is where I got stuck. Am I even approaching this problem correctly? Any help would be appreciated.
|
which simplifies to $\;\; (b-a)c^2 + (a-c)b^2 + (c-b)a^2$
Hint: $\;(b-a)c^2 + (a-c)b^2 + (c\color{red}{-a+a}-b)a^2\,=\,\ldots$
|
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|
Taking the derivative of $x^4\sin(x)\cos(x)$, which step is wrong? I'm trying to take the derivative of $x^4\sin(x)\cos(x)$ and I keep getting the wrong answer.
My steps:
$$\frac {d}{dx}[x^4\sin(x)\cos(x)]$$
Apply product rule:
$$\frac {d}{dx}[x^4](\sin(x)\cos(x)+x^4\frac {d}{dx}[\sin(x)\cos(x)]$$
Simplify first part:
$$4x^3\sin(x)\cos(x)+x^4\frac {d}{dx}[\sin(x)\cos(x)]$$
Apply product rule to second part:
$$\cos(x)\cos(x)+(-\sin(x))$$
Add them all together: $$4x^3\sin(x)\cos(x)+x^4\cos^2(x)-\sin(x)$$
So something is wrong as the correct answer is $$-x^4\sin^2(x)+x^4\cos^2(x)+4x^3\cos(x)\sin(x)$$
Got the biggest headache from this one, would really appreciate help! Thanks!
|
$$\dfrac{d}{dx}(x^4\cdot \sin(x)\cdot cos(x))$$
Let $u = x^4$ and $v = \sin(x) \cdot \cos(x)$. From the well known formula we know
$$
\begin{align}
\dfrac{d}{dx}(u \cdot v) & = u \dfrac{dv}{dx} + v \dfrac{du}{dx} \\
& = x^4 \dfrac{d}{dx}(\sin(x) \cdot \cos(x)) + \sin(x) \cdot \cos(x) \cdot \dfrac{d}{dx}(x^4) \\
& = x^4 \dfrac{d}{dx}(\sin(x) \cdot \cos(x)) + 4 \cdot x^3 \sin(x) \cdot \cos(x)
\end{align}
$$
Expression simplified but we still need to solve $\dfrac{d}{dx}(\sin(x) \cdot \cos(x))$
let $u = \sin(x)$ and $v = cos(x)$
$$
\begin{align}
\dfrac{d}{dx}(u \cdot v) & = u \dfrac{dv}{dx} + v \dfrac{du}{dx} \\
& = \sin(x) \dfrac{d}{dx}(\cos(x)) + \cos(x) \cdot \cos(x) \cdot \dfrac{d}{dx}(\sin(x)) \\
& = \sin(x) \cdot(- \sin(x)) + cos(x) \cdot \cos(x) \\
& = (\cos(x))^2 - (\sin(x))^2
\end{align}
$$
All we need is to substitute this back.
$$
\begin{align}
\dfrac{d}{dx}(x^4\cdot \sin(x)\cdot cos(x)) & = x^4 \dfrac{d}{dx}(\sin(x) \cdot \cos(x)) + 4 \cdot x^3 \sin(x) \cdot \\
& = x^4 ((\cos(x))^2 - (\sin(x))^2) + 4 \cdot x^3 \sin(x)
\end{align}
$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Probability Problem Inquiry In an examination, the probability of getting a credit is 1/3. If four students are selected at random, what is the probability that at least one of them got a credit?
From the book:
$P = 1 - F$
$P$ = 1 - (2/3)(2/3)(2/3)(2/3) = 65/81
The book solved the problem by considering the complement (opposite) of the term "at least one" from the formula: 1 = P(A) + P(A'). I want to solve the problem by normal means, that is by following the favorable outcome of at least 1. Sadly, I can't seem to get the answer.
My analysis:
$P$(getting a credit) = $1$
$P$(at least one of them get a credit) = ?
$P$(Not getting a credit) = $1$ - $(\frac{1}{3}) = \frac{2}{3}$
n(objects) = 4
The problem asked us to get the probability of at least one of them got a credit; so it means that out of the 4, the possible outcomes are:
(One of them got a credit and the other 3 didn't) OR (two of them got a credit and the other 2 didn't) OR (three of them got a credit and 1 didn't) OR (All four of them got a credit)
So if we put it into mathematical expression:
$(\frac{1}{3})(\frac{2}{3})^3$ + $(\frac{1}{3})^2(\frac{2}{3})^2$ + $(\frac{1}{3})^3(\frac{2}{3})$ + $(\frac{1}{3})^4(\frac{2}{3})^0$ = 5/27
Question: From my analysis above, I don't know where I got wrong.. I think i got the possible outcomes right. So what is wrong with my analysis? Where did I go wrong? How can I avoid it next time? If you get the right solution, would you mind explaining it clearly?
Any help would be appreciated. Thank you in advance.
|
Since the probability for each student receiving credit is the same, the Binomial distribution applies. The probability of exactly $k$ successes in $n$ trials, each of which has probability $p$ of success is
$$\Pr(X = k) = \binom{n}{k}p^k(1 - p)^{n - k}$$
where $p^k$ is the probability of $k$ successes, $(1 - p)^{n - k}$ is the probability of $n - k$ failures, and $\binom{n}{k}$ is the number of ways of selecting which $k$ of the $n$ trials result in a success.
In your attempt, you did not take into account the way the successes could be distributed among the students.
Since you are interested in the probability that at least one student got a credit, we need to find
$$\Pr(X \geq 1) = \Pr(X = 1) + \Pr(X = 2) + \Pr(X = 3) + \Pr(X = 4)$$
Using the formula above with $n = 4$ and $p = 1/3$ gives
\begin{align*}
\Pr(X \geq 1) & = \sum_{k = 1}^{4} \binom{4}{k}\left(\frac{1}{3}\right)^k\left(\frac{2}{3}\right)^{4 - k}\\
& = \binom{4}{1}\left(\frac{1}{3}\right)\left(\frac{2}{3}\right)^3 + \binom{4}{2}\left(\frac{1}{3}\right)^2\left(\frac{2}{3}\right)^2 + \binom{4}{3}\left(\frac{1}{3}\right)^3\left(\frac{2}{3}\right)^1 + \binom{4}{4}\left(\frac{1}{3}\right)^4\left(\frac{2}{3}\right)^0\\
& = \frac{32}{81} + \frac{24}{81} + \frac{8}{81} + \frac{1}{81}\\
& = \frac{65}{81}
\end{align*}
as you found by subtracting the probability that no student received credit from $1$, which is easier. The factor $\binom{4}{k}$ that you are missing in your calculations accounts for which $k$ of the four students receives credit.
|
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|
For which values of m is the improper integral convergent For which values of m is the improper integral convergent:
$\int_0^\infty \frac{1}{(x^2+x)^mln^2(1+2x)}tg(\frac{x^4}{x^7+x^4+1})dx$
I've been at this one for a while, here's what I've got so far:
$\int_0^\infty \frac{1}{(x^2+x)^mln^2(1+2x)}tg(\frac{x^4}{x^7+x^4+1})dx$ =
$\int_0^1 \frac{1}{(x^2+x)^mln^2(1+2x)}tg(\frac{x^4}{x^7+x^4+1})dx$ + $\int_1^\infty \frac{1}{(x^2+x)^mln^2(1+2x)}tg(\frac{x^4}{x^7+x^4+1})dx$
$(x^2+x)^m$ ~ $x^m$ as $x->0$
$ln^2(1+2x)$ ~ $(2x)^2$ as $x->0$
$tg(\frac{x^4}{x^7+x^4+1})$ ~ $\frac{x^4}{x^7+x^4+1}$ as $x->0$
$(x^2+x)^m$ ~ $(x^2)^m$ as $x->\infty$
$ln^2(1+2x)$ ~ $ln^2x$ as $x->\infty$
$tg(\frac{x^4}{x^7+x^4+1})$ ~ $\frac{x^4}{x^7+x^4+1}$ as $x->\infty$
I get two integrals:
$\int_0^1 \frac{x^4}{x^m4x^2(x^7+x^4+1)}dx$ = $1/4\int_0^1 \frac{1}{\frac{x^m}{x^2}(x^7+x^4+1)}dx$
$\int_1^\infty \frac{x^4}{(x^2)^pln^2x(x^7+x^4+1)}dx$
And that's about as far as I've figured it out so far, I would be grateful if someone could help out.
|
Note that
$$\int_0^\infty \frac{1}{(x^2+x)^m \ln^2(1+2x)}\tan\left(\frac{x^4}{x^7+x^4+1}\right)dx
\\=\int_0^1 \frac{1}{(x^2+x)^m \ln^2(1+2x)}\tan\left(\frac{x^4}{x^7+x^4+1}\right)dx\\+\\+\int_1^\infty \frac{1}{(x^2+x)^m \ln^2(1+2x)}\tan\left(\frac{x^4}{x^7+x^4+1}\right)dx$$
and for $x\to 0^+$
$$\frac{1}{(x^2+x)^m \ln^2(1+2x)}\tan\left(\frac{x^4}{x^7+x^4+1}\right)\sim \frac{x^4}{4x^{2+m}}\sim \frac{1}{4x^{m-2}}$$
and thus for $m<3$ the first integral converges by limit comparison test with $\int_0^1 \frac{1}{4x^{m-2}}$ and for $x\to \infty$
$$\frac{1}{(x^2+x)^m \ln^2(1+2x)}\tan\left(\frac{x^4}{x^7+x^4+1}\right)\sim \frac{1}{x^{2m+3}\ln^2 x}$$
and thus for $2m+3\ge1\implies m\ge -1$ the second integral converges by limit comparison test with $\int_1^\infty \frac{1}{x^{2m+3}\ln^2 x}$ .
Therefore the given integral converges for $-1\le m<3$.
|
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|
How do I find $f(x)$ when provided $g(x)$ and $(f\circ g)(x)$? Let $g(x) = 2x^2−2x−1$.
Find a function $f$ such that $(f\circ g)(x) = −8x^6 + 24x^5 −8x^4 − 24x^3 + 6x^2 + 10x + 6.$
$f(x) =$
|
Let's suppose that $f(x)=Ax^3+Bx^2+Cx+D$.
Because $2^3=8$, we can see that $A=-1$.
So let's substract $-(2x^2-2x-1)^3=-8x^6+24x^5-12x^4-16x^3+6x^2+6x+1$ from $−8x^6 + 24x^5 −8x^4 − 24x^3 + 6x^2 + 10x + 6$. The result is $4x^4-8x^3+4x+5$.
$2^2=4$, so $B=1$. Now substract $(2x^2-2x-1)^2$ from $4x^4-8x^3+4x+5$. The result is $4$. So $C=0$ and $D=4$.
So the function you are looking for is: $$f(x)=-x^3+x^2+4$$
|
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|
$y^2+5xy+6x^2-9x-4y=0$ where $x$ and $y$ are integers Need some help solving this for integers $x$ and $y$:
$$
y^2+5xy+6x^2-9x-4y=0
$$
I managed to make something like this:
$$
(y+3x-4)(y+2x)=x\\
(y+3x)(y+2x-3)=y
$$
Find integers for $x$ and $y$ that satisfy the equations above.
But, what do I do next, or is this a bad approach?
|
Hint: written as a quadratic in $\,y\,$, the discriminant of $\;y^2+(5x-4)y+3x(2x-3)\,$ must be a perfect square: $\;\Delta_y=(5x-4)^2-12x(2x-3)=x^2 - 4 x + 16=(x-2)^2+12\,$.
|
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|
How to prove that $x'Bx=0$ for all $n\times 1$ vectors $x$ if and only if $B'=-B$?
Theorem. $x^TBx=0$ for all $n \times 1$ vectors $x$ if and only if $B^T=-B$.
Then how to prove this?
|
Suppose
$B^T = -B; \tag 1$
then
$(x^TBx)^T = x^TB^Tx = -x^TBx; \tag 2$
since $x^TBx$ is a scalar, that is, a $1 \times 1$ matrix, we have
$x^TBx = (x^TBx)^T, \tag 3$
whence (2) yields
$x^TBx = -x^TBx \Longrightarrow x^TBx = 0; \tag 4$
likewise, if
$x^TBx = 0 \tag 5$
for all $x$, then
$y^TBx + x^TBy = x^TBx + y^TBx + x^TBy + y^TBy = (x + y)^TB(x + y) = 0, \tag 6$
so
$y^TBx = -x^TBy; \tag 7$
now again, the quantities on each side of this equation are scalars, so in particular
$y^TB^Tx = (x^TBy)^T = x^TBy, \tag 8$
so (7) may be written
$y^TBx = -y^TB^Tx, \tag 9$
for all $x$ and $y$; thus
$Bx = -B^Tx, \tag{10}$
implying
$B = -B^T. \tag{11}$
|
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|
How to solve this equation for c? I want to solve the equation
$$
-\frac{a}{2}\left(c+\sqrt{c^2+4}\right)=-\frac{a-1}{2}\left(c-\sqrt{c^2+4}\right)
$$
for $c$, where $a$ is just a constant.
What I get is
$$
\frac{c-\sqrt{c^2+4}}{c+\sqrt{c^2+4}}=\frac{a}{a-1}.
$$
I think there now is some "trick" to solve this for $c$.
|
By distributing,
$$-\frac{ac}{2}-\frac{a\sqrt{c^2+4}}{2}=-\frac{ac}{2}+\frac{a\sqrt{c^2+4}}{2}+\frac{c}{2}-\frac{\sqrt{c^2+4}}{2}$$
Well,
$$0=(a-\frac{1}{2})\sqrt{c^2+4}+\frac{c}{2}$$
$$\frac{1}{1-2a}=\frac{\sqrt{c^2+4}}{c}$$ with assumption that $c \ne0.$
Squaring, $$\frac{1}{(1-2a)^2}=\frac{c^2+4}{c^2}=1+\frac{4}{c^2}$$
$$\frac{4}{c^2}=\frac{4a-4a^2}{(1-2a)^2}$$
By fliping, and multiplying both sides by $4$,
$$c^2=\frac{(1-2a)^2}{a-a^2}$$
You now have $c$ without rationalising surds.
|
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|
The biggest $\gcd(11n+4, 7n+2)$? I am having problems solving this problem with greatest common divisors:
What is the greatest common divisor of $11n+4$ and $7n +2$?
I tried Euclidean algorithm, and I tried to deduce the answer and I tried to incorporate
$$\gcd(a,b)\cdot\mathrm{lcm}(a,b)=|ab|$$
|
\begin{align}
(11n+4) &= 1(7n+2) + (4n+2) \\
(7n+2) &= 2(4n+2) + (-n-2) \\
(4n+2) &= -4(-n-2) - 6
\end{align}
This suggests that the biggest gcd must be a divisor of $6$.
Such a value should occur when $n=-2$.
\begin{align}
11(-2)+4 &= -18 \\
7(-2)+2 &= -12 \\
\gcd(-18, -12) &= 6
\end{align}
|
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"timestamp": "2023-03-29T00:00:00",
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|
If $\frac{b+c}{2k-1}=\frac{c+a}{2k}=\frac{a+b}{2k+1}$, then $\frac{\sin A}{k+1}=\frac{\sin B}{k}=\frac{\sin C}{k-1}$
If $$\frac{b+c}{2k-1}=\frac{c+a}{2k}=\frac{a+b}{2k+1}$$
show that
$$\frac{\sin A}{k+1}=\frac{\sin B}{k}=\frac{\sin C}{k-1}$$
where $k$ is an integer such that $2 < k \neq 4$, and where $a$, $b$, $c$ are sides of $\triangle ABC$.
Actually, I don't have any idea. Please someone help. It would be great if someone who answers include how he/she found where to begin.
|
Let $m$ be the common value of the initial fractions, so that
$$\begin{align}
a + b &= ( 2 k + 1 ) m \tag{1} \\
b + c &= ( 2 k - 1 ) m \tag{2}\\
c + a &= ( 2 k \phantom{+1\;\,}) m \tag{3}
\end{align}$$
We can combine these equations to get
$$\begin{align}
\phantom{-}(1)-(2)+(3):&\quad 2 a = 2(k+1)m \\
(1)+(2)-(3):&\quad 2b= 2(k\phantom{+1\;\,})m \\
-(1)+(2)+(3):&\quad 2c= 2(k-1)m
\end{align} \tag{4}$$
By the Law of Sines,
$$\sin A : \sin B : \sin C = a : b : c = k + 1 : k : k - 1 \tag{5}$$
and the result follows. $\square$
|
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|
attempt to solve a Bernoulli equation I tried solving the Bernoulli equation $y'-y\tan x = y^4 \cos x $ by equating the left hand side to 0 and finding the homogeneous solution, however the equation turned out to be too complex and without the ability to isolate x. Any hints on how to solve this?
|
$y' - y\tan(x) = y^4\cos(x)$
divide throughout by $y^4$
$\frac{y'}{y^4} - \frac1{y^3}\tan(x) = \cos(x)$
let $z= \frac{1}{y^3}\implies z' = \frac{-3}{y^4}y'$
$\frac{-z'}{3} -z\tan(x) = \cos(x)$
$z' + 3\tan(x)\,z = -3\cos(x)$
it is now a Linear differential equation
Integrating factor , $I=e^{\int3\tan(x)}=e^{\ln(\sec^3(x))} = \sec^3(x)$
the solution is given by ;
$z\cdot I =\int-3\cos(x)\cdot I\,dx$
integrate and sub back for $z$
Can you proceed further?Ask if you need help.
EDIT: on how we got $z\cdot I =\int-3\cos(x)\cdot I\,dx$;
we have $z' + 3\tan(x)\,z = -3\cos(x)$
multiply throughout by $\sec^3(x)$
we get ; $z'\cdot\sec^3(x) +3\cdot\tan(x)\cdot\sec^3(x)\cdot z = -3\cos(x)\cdot\sec^3(x)$
recognize that the LHS is a product rule derivative of $z\cdot\sec^3(x)$
ie $d(z\cdot\sec^3(x)) =(z'\cdot\sec^3(x) +3\cdot\tan(x)\cdot\sec^3(x)\cdot)\,dx z$
therefore the equation becomes ;
$\big(z\cdot\sec^3(x)\big)' = 3\cos(x)\cdot\sec^3(x)$
integrating on both sides gives us;
$z\cdot I =\int-3\cos(x)\cdot I\,dx$; $\quad$ where $I= \sec^3(x)$
|
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|
Beads and necklaces We have a several number of beads in a box and want to make 3 necklaces.
We firstly take some out of the box, put $3/4$ of them to the 1st necklace and equally share the rest $1/4$ to the 2nd and 3rd.
Then we take some more out of the box, put $1/4$ of them for the 2nd necklace and equally share the rest $3/4$ to the 1st and 3rd.
Finally we take the remaining beads, put $1/12$ to the 3rd necklace and equally share the rest 11/12 to the 1st and 2nd.
We then count the beads of the 3 necklaces and find them to be in ratio $3:2:1$. What is the least number of beads that we initially had in the box?
$A$ = number of beads initially in the box
We firstly take $x$, put $3x/4$ in the 1st necklace and equally share the rest $x/4$.
So the 1st necklace has $3x/4$
2nd has $x/8$
3rd has $x/8$
Then we take $y$, put $y/4$ in the 2nd necklace and equally share the rest $3y/4$
So 1st now has $3x/4+3y/8$
2nd has $x/8+y/4$
3rd has $x/8+3y/8$
Then we take the remaining, put $(Α-x-y)/12$ in the 3rd necklace and equally share the rest $11(A-x-y)/12$ in the other two.
So the 1st necklace has \begin{align}3x/4+3y/8 + 11(Α-x-y)/24 =&\; 3x/4+3y/8-11x/24-11y/24+11Α/24\\ =&\; 7x/24-2y/24+11Α/24\end{align}
2nd has $$x/8+y/4+ 11(Α-x-y)/24 = 11Α/24-8x/24-5y/24$$
3rd has $$x/8+3y/8 +(Α-x-y)/12 = Α/12+x/24+7y/24$$
And we know that $$(7x/24-2y/24+11Α/24) : (11Α/24-8x/24-5y/24) : (Α/12+x/24+7y/24) = 3:2:1$$
Any ideas on how to continue?
|
The last result implies two equations:
\begin{align}
\frac{\frac{11}{24}Α+\frac{7}{24}x-\frac{2}{24}y}{\frac{11}{24}Α-\frac{8}{24}x-\frac{5}{24}y}=&\; \frac32,\\[2ex]
\frac{\frac{11}{24}Α-\frac{8}{24}x-\frac{5}{24}y}{\frac{2}{24}Α+\frac{1}{24}x+\frac{7}{24}y}=&\; \frac21.
\end{align}
Solving this system for $x$ and $y$ implies that
$$x=\frac{11}{51}A,\quad\text{and}\quad y=\frac{13}{51}A.$$
We need the following numbers to be integers:
$$A,\;\; x,\;\; y,\;\; \frac{11}{24}Α,\;\;\frac{7}{24}x,\;\;\frac{2}{24}y,\;\;\frac{11}{24}Α-\frac{8}{24}x,\;\;\frac{5}{24}y,\;\;\frac{2}{24}Α,\;\;\frac{1}{24}x,\;\;\text{and}\;\;\frac{7}{24}y.$$
Notice that $51=3*17$, and $24=2^3*3$. Since $11$ is prime, for $x$ to be an integer $A$ must be divisible by $51$. For all other fractions above to be integers it is necessary and sufficient that $A$ also be divisible by $24$.
Therefore, $$A=\rlap{\underbrace{\phantom{2^3*3}}_{24}}2^3*
\overbrace{3*17}^{51}=408$$ is the lowest possible value for $A$.
|
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|
Largest cone that can be inscribed in a sphere
Prove that the volume of the largest cone that can be inscribed in a sphere of radius R is $\frac{8}{27}$ of the volume of the sphere.
My Attempt
r : radius of the base
h : height of the cone
$$
x=\sqrt{R^2-r^2}\implies h=R+x=R+\sqrt{R^2-r^2}\\
V(r)=\frac{1}{3}\pi r^2h=\frac{1}{3}\pi r^2(R+\sqrt{R^2-r^2})\\
V'(r)=\frac{\pi}{3}\big( 2r.(R+\sqrt{R^2-r^2})+r^2\frac{-r}{\sqrt{R^2-r^2}} \big)\\=\frac{\pi}{3}r\big( 2R+2\sqrt{R^2-r^2})-\frac{r^2}{\sqrt{R^2-r^2}} \big)\\=\frac{\pi}{3}r\big( \frac{2R\sqrt{R^2-r^2}+2R^2-2r^2-r^2}{\sqrt{R^2-r^2}} \big)\\=\frac{-\pi}{3}r\big( \frac{2r^2-2R\sqrt{R^2-r^2}-2R^2}{\sqrt{R^2-r^2}} \big)
$$
How do I proceed further and find the points where $V'(r)=0$ and prove the above statement ?
Note: I would like to stick with $r$ as the variable.
|
I would write
$$V=\frac{1}{3}\pi r^2h$$ then we get
$$r^2+x^2=R^2$$ and $$x+R=h$$ and so $$V(x)=\frac{1}{3}\pi(R^2-x^2)(x+R)$$
Or you can write: $$h=x+R$$ and $$x=\sqrt{R^2-r^2}$$ then we have
$$V=\frac{1}{3}\pi r^2(\sqrt{R^2-r^2}+R)$$
$$V'(r)=1/3\,{\frac {\pi\,r \left( 2\,{R}^{2}-3\,{r}^{2}+2\,\sqrt { \left( R-r
\right) \left( R+r \right) }R \right) }{\sqrt { \left( R-r \right)
\left( R+r \right) }}}
$$
Solving the equation $$V'(r)=0$$ we get $$r_{opt}=\frac{2\sqrt{2}R}{3}$$
|
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|
Arrangement of red and white balls with no adjacent red balls. I have $5$ white balls and $3$ red balls. Putting these $8$ balls in a row with no adjacent red balls, what is the number of arrangements?
My approach:
I tried the inclusive-exclusive method, by getting all the possible arrangements and then subtracting from the number of times the red ones would be together:
Total: $$\frac{8!}{5! 3!} = 56$$
Then, I ''glued'' together all the red ones, making it $1$ red ball and $5$ white balls:
$$\frac{6!}{5! 1!} = 6$$
And finally, I glued one pair of red balls and left one alone, making it:
$$\frac{7!}{5!1!1!} = 42$$
Subtracting the above from the total would give $8$ arrangements, but actually the answer is $20$. I would appreciate knowing if my approach is correct, or else, what I'm doing wrong to solve this question.
|
Method 1: The approach @JMoravitz wrote in the comments is the easiest way to handle this problem. Place the white balls in a row. This creates six spaces, four between successive white balls and two at the ends of the row.
$$\square w \square w \square w \square w \square w \square$$
To ensure that the red balls are separated, we must choose three of these six spaces in which to place a red ball, which can be done in
$$\binom{6}{3} = 20$$
ways.
Method 2: We subtract those arrangements with two or more consecutive red balls from the total number of arrangements.
As you realized, the total number of arrangements is the number of ways we can select three of the eight positions for the red balls, which is $\binom{8}{3}$.
Arrangements with exactly two red balls together: Line up the five white balls in a row, creating six spaces, as shown above. Choose one of these spaces in which to place two adjacent red balls and one of the remaining five spaces in which to place the other red ball. Thus, there are $6 \cdot 5 = 30$ such arrangements.
Arrangements with all three red balls together: Line up the five white balls in a row, creating six spaces. Choose one of these six spaces in which to place the three red balls. There are $6$ such arrangements.
Hence, there are
$$\binom{8}{3} - 6 \cdot 5 - 6 = 56 - 30 - 6 = 20$$
arrangements in which no two of the red balls are together.
Method 3: We use the Inclusion-Exclusion Principle.
As noted above, there are $\binom{8}{3}$ arrangements of five white and three red balls. From these, we must subtract those arrangements in which at least one pair of red balls is adjacent.
A pair of red balls is adjacent: We have seven objects to arrange, a block of two red balls, a single red ball, and five white balls. We choose five of the seven positions for the white balls, one of the remaining two for the single red ball, and place the block of two red balls in the final open position, which can be done in
$$\binom{7}{5}\binom{2}{1}\binom{1}{1} = \frac{7!}{5!2!} \cdot \frac{2!}{1!1} \cdot \frac{1!}{1!0!} = \frac{7!}{5!} = 7 \cdot 6 = 42$$
such arrangements, as you found.
However, we have subtracted those arrangements in which there are two pairs of consecutive red balls (those arrangements with three consecutive red balls) twice, once when we designated the first two red balls as the pair of consecutive red balls and once when we designated the last two red balls as the pair of consecutive red balls. Therefore, we need to add arrangements with two pairs of consecutive red balls to the total. This is where you made your mistake.
Two pairs of consecutive red balls: Since there are only three red balls, this can only occur if all three red balls are consecutive. We showed above that there are $\binom{6}{1}$ such arrangements.
By the Inclusion-Exclusion Principle, the number of arrangements of five white and three red balls in which no two red balls are consecutive is
$$\binom{8}{3} - \binom{7}{5}\binom{2}{1}\binom{1}{1} + \binom{6}{1} = 56 - 42 + 6 = 20$$
Method 4: @Lifeforbetter asked how to do this problem by first placing the red balls, then inserting the white balls.
Let $x_1$ be the number of white balls to the left of the first red ball; let $x_2$ be the number of white balls between the first and second red balls; let $x_3$ be the number of white balls between the second and third red balls; let $x_4$ be the number of white balls to the right of the third red ball. Since there are five white balls,
$$x_1 + x_2 + x_3 + x_4 = 5 \tag{1}$$
Since no two of the red balls are adjacent, $x_2, x_3 \geq 1$, while $x_1, x_4 \geq 0$. Let $x_2' = x_2 - 1$; let $x_3' = x_3 - 1$. Then $x_2'$ and $x_3'$ are nonnegative integers. Substituting $x_2' + 1$ for $x_2$ and $x_3' + 1$ for $x_3$ in equation 1 yields
\begin{align*}
x_1 + x_2' + 1 + x_3' + 1 + x_4 & = 5\\
x_1 + x_2' + x_3' + x_4 & = 3 \tag{2}
\end{align*}
Equation 2 is an equation in the nonnegative integers. A particular solution corresponds to the placement of $4 - 1 = 3$ addition signs in a row of $3$ ones. For instance,
$$1 + + 1 1 +$$
corresponds to the solution $x_1 = 1$, $x_2' = 0$, $x_3' = 2$, $x_4 = 0$ ($x_1 = 1$, $x_2 = 1$, $x_3 = 3$, $x_4 = 0$). The number of such solutions is
$$\binom{3 + 4 - 1}{4 - 1} = \binom{6}{3}$$
since we must choose which three of the six positions required for three ones and three addition signs will be filled with addition signs.
|
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|
Figure out all primes p and q such that Figure out all primes p and q such that
$p^3$ + 19$q^3$ + 2018 is the cube of a prime.
|
Let r be the prime such that
$p^3 + 19q^3 + 2018 = r^3$
Since 2 is the only even number prime but r does not equal to 2 (trivial), by observing the even-odd of the LHS and RHS, either $p=2$ or $q=2$ .
If $p = 2$, $2^3 + 19q^3 + 2018 = r^3$
Let $s^3 = 19q^3$
Since 19 is not a perfect cube, s or $s^2$ cannot be a rational number, hence $p=2$ cannot be a correct answer
If $q = 2$, $p^3 + 19*2^3 + 2018 = r^3$
$p^3+ 19*8 + 2018 = r^3$
$r^3 – p^3 = 2170$
$(r-p)(r^2+rp+p^2) = 2170$
2170 is not prime, but for the product of the 2 factors $(r-p)$ and $(r^2+rp+p^2)$ ends in 0 as last digit, one of the factor is divisible by 5.
EITHER
$(r – p)$ is divisible by 5………………….(A)
Since 5 is the only prime divisible by 5, and both r or p are odd numbers, $(r – p)$ must be divisible by 2 as well – hence, $(r – p)$ is divisible by both 5 and 2 – and is divisible by 10
let $r – p = 10k$ where k is an integer
$10k ((r^2+rp+p^2) = 2170$
$k(r^2+rp+p^2) = 217$ (a prime), so $k=217$ (incorrect) or
$r^2+rp+p^2 = 217$ and $k=1$
since $√217 = 14.73$, we have $r,p<14$
By trial and error we get $(r,p) = (3,13)$ or $(13,3)$ –
but since $r-p = 10$, $r=13, p=3$ is the correct answer
OR
$(r^2+rp+p^2)$ is divisible by 5………………..(B)
By checking the last digit of $r^2+rp+p^2$ when p and q are prime numbers not equal to 5, the last digit can only be 1,3,7,9 for all combinations of non-2 primes (r,p). The only way $r^2+rp+p^2$ can be divisible by 5 is unless (r,p) = (5,5) which is not a correct answer. So there is no solution in (B)
Therefore, $(p,q,r) = (3,2,13)$
Check – $3^3 + 19 * 2^3 + 2018 = 2197 = 13^3$
.
|
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|
Prove the following determinant Prove the following:
$$\left|
\begin{matrix}
(b+c)&a&a \\
b&(c+a)&b \\
c&c&(a+b) \\
\end{matrix}\right|=4abc$$
My Attempt:
$$\left|
\begin{matrix}
(b+c)&a&a \\
b&(c+a)&b \\
c&c&(a+b) \\
\end{matrix}\right|$$
Using $R_1\to R_1+R_2+R_3$
$$\left |
\begin{matrix}
2(b+c)&2(a+c)&2(a+b) \\
b&(c+a)&b \\
c&c&(a+b) \\
\end{matrix}\right|$$
Taking common $2$ from $R_1$
$$2\left|
\begin{matrix}
(b+c)&(a+c)&(a+b) \\
b&(c+a)&b \\
c&c&(a+b) \\
\end{matrix}\right|$$
How do I proceed further?
|
Let me give a solution without using the rule of Sarrus for fun. $\require{cancel}$
\begin{align}
& \begin{vmatrix}
b+c&a&a \\
b&c+a&b \\
c&c&a+b \\
\end{vmatrix} \\
&= \frac{1}{abc}\, \begin{vmatrix}
a(b+c)&ab&ca \\
ab&b(c+a)&bc \\
ca&bc&c(a+b) \\
\end{vmatrix} \\
&= \frac{1}{abc}\, \begin{vmatrix}
0&-2bc&-2bc \\
ab&b(c+a)&bc \\
ca&bc&c(a+b) \\
\end{vmatrix} \tag{$R_1 \to R_1 - R_2 - R_3$} \\
&= \frac{1}{abc}\, \begin{vmatrix}
0&-2bc&0 \\
ab&b(c+a)&-2ab \\
ca&bc&0 \\
\end{vmatrix} \tag{$C_3 \to C_3 - C_1 - C_2$} \\
&= \frac{1}{a\cancel{bc}}\, [-(-2\cancel{bc})]\, \begin{vmatrix}
ab&-2ab \\
ca&0 \\
\end{vmatrix} \\
&= \frac{2}{\cancel{a}} \, [-(-2ab)(c\cancel{a})] \\
&= 4abc
\end{align}
|
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|
$f(x)=2x+\cot^{–1}x+\log (\sqrt{1+x^2}−x)$ increasing or decreasing in $\mathscr{R}$.
Show that $f(x)=2x+\cot^{–1}x+\log(\sqrt{1+x^2}−x)$ is increasing in $\mathscr{R}$.
My Attempt
$$
f'(x)=2-\frac{1}{1+x^2}+\frac{\frac{x}{\sqrt{1+x^2}}-1}{\sqrt{1+x^2}−x}\\
=2-\frac{1}{1+x^2}+\frac{\frac{x-\sqrt{1+x^2}}{\sqrt{1+x^2}}}{\sqrt{1+x^2}−x}\\
=2-\frac{1}{1+x^2}-\frac{1}{\sqrt{1+x^2}}\\
=\frac{2(1+x^2)-1-\sqrt{1+x^2}}{1+x^2}\\
=\frac{1+2x^2-\sqrt{1+x^2}}{1+x^2}
$$
How do I show that $1+2x^2\geq\sqrt{1+x^2}\implies f'(x)\geq 0$ and thus the given function is increasing in $\mathscr{R}$ ?
|
HINT:
For $x\neq0$, $$1+2x^2>\sqrt{1+x^2}\impliedby1+4x^2+4x^4>1+x^2\impliedby 4x^4+3x^2>0$$
|
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|
Find $f_{10}$ of the sequence defined by $f_{n+1}=\frac{8f_n}{5}+\frac{6 \sqrt{4^n-f_n^2}}{5}$ Find $f_{10}$ of the sequence defined by $$f_{n+1}=\frac{8f_n}{5}+\frac{6 \sqrt{4^n-f_n^2}}{5}$$ given $f_0=0$
My Approach:
Letting $f_n=2^n b_n$ we get
$$b_{n+1}=\frac{4b_n}{5}+\frac{3 \sqrt{1-b_n^2}}{5}$$
Now letting $b_n=\cos(x_n)$ we get
$$\cos(x_{n+1})=\cos(x_n-\theta)$$ where $\cos(\theta)=\frac{4}{5}$
Now Since $f_0=0$ we have $b_0=0$ and $x_0=\frac{\pi}{2}$
Now we have
$$x_{n+1}=x_n-\theta$$
Putting $n=0,1,2,3 \cdots 10$ and adding all we get
$$x_{10}=\frac{\pi}{2}-10\theta$$
Hence
$$b_{10}=\cos\left(\frac{\pi}{2}-10\theta\right)=\sin(10\theta)=\sin\left(10\arcsin\left(\frac{3}{5}\right)\right)$$
How to proceed further?
|
Simply writing it out by hand quickly yields
$$f_1=\frac{6}{5},\qquad f_2=\frac{96}{25},\qquad f_3=\frac{936}{125},\qquad f_4=\frac{384}{25},$$
and the rest should only take a few minutes. Here's a few more, it's not that hard:
$$f_5=\frac{3744}{125},\qquad f_6=\frac{1536}{25},\qquad f_7=\frac{14976}{125},\qquad f_8=\frac{6144}{25}.$$
I must say that $f_8$ and $f_{10}$ took slightly longer, but all in all this is no more than ten minutes work:
$$f_9=\frac{59904}{125},\qquad f_{10}=\frac{24576}{25}.$$
|
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|
What is the solution of this definite integral: $\int_{0}^{2\pi} \ln\big(C-D\cos(f)\big)\cos(nf)df$ I encountered this integral in my calculations:
$$\int_{0}^{2\pi} \ln\big(C-D\cos(f)\big)\cos(nf)df$$
Here, $n$ is a natural number, $C, D$ are constants, such that $C\gt D$.
I tried to find solution in Table of integrals,series and products and by using Wolfram mathematica, but I have not managed to find it.
|
For $r<1$, we have
$$ \log{(1+r^2-2r\cos{\theta})} = \log{(1-re^{i\theta})(1-re^{-i\theta})} \\
= \log{(1-re^{i\theta})}+\log{(1-re^{-i\theta})} \\
= -\sum_{k=1}^{\infty} \frac{1}{k}r^k (e^{ik\theta}+e^{-ik\theta}) \\
= -2\sum_{k=1}^{\infty} \frac{r^k}{k} \cos{k\theta}, $$
provided that we choose all the logarithms to have value $0$ when their argument is $1$, i.e. $r=0$. Thus
$$ \int_0^{2\pi} \cos{n\theta}\log{(1+r^2-2r\cos{\theta})} \, d\theta = -2\sum_{k=1}^{\infty} \frac{r^k}{k} \int_0^{2\pi} \cos{k\theta}\cos{n\theta} \, d\theta = -2\pi\frac{r^n}{n}, $$
using that the series is uniformly convergent.
How does this relate to your integral? We can write the integrand as $\log{(1+r^2)}+\log{(1-\frac{2r}{1+r^2} \cos{\theta})}$ without affecting the result for $n \neq 0$. So if we split $\log{(C-D\cos{\theta})}=\log{C}+\log{(1-(D/C)\cos{\theta})}$, we can take $C/D = 2r/(1+r^2) $, and solving this gives
$$ r = \frac{1-\sqrt{1-(D/C)^2}}{D/C}, $$
since we need $r<1$. Thus
$$ \log{(C-D\cos{\theta})} = \log{C} + \log{1-(D/C)\cos{\theta}} = \log{C}-\log{(1+r^2)} +\log{(1+r^2-2r\cos{\theta})} \\
= \log{\left( \frac{C(D/C)^2}{2(1-\sqrt{1-(D/C)^2})} \right)} + \log{(1+r^2-2r\cos{\theta})}, $$
which we can integrate. We therefore conclude that
$$ \int_0^{2\pi} \cos{n\theta}\log{(C-D\cos{\theta})} \, d\theta = \begin{cases}
2\pi \log{\left( \frac{C(D/C)^2}{2(1-\sqrt{1-(D/C)^2})} \right)} & n=0 \\
-\frac{2\pi}{n} \left( \frac{1-\sqrt{1-(D/C)^2}}{D/C} \right)^n& n \in \mathbb{Z} \setminus \{0\}.
\end{cases} $$
|
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|
Partial fractions integral. Compute $\int_{0}^{1} \frac {x^4+1}{x^6+1}dx$
Evaluate $\int_{0}^{1} \frac {x^4+1}{x^6+1}dx$.
I directly approached this with partial fractions and rewritten $x^6+1=(x^2)^3+1=(x^2+1)(x^4+x^2+1)$.
Therefore the integral is:
$$\int_{0}^{1} \frac {x^4+1}{(x^2+1)(x^4+x^2+1)}dx=-2\int_{0}^{1} \frac {x}{1+x^2}dx+2\int_{0}^{1}\frac 1{1+x^2}dx-\int_{0}^1\frac {(x-1)^2}{(x^2+1)^2+(\frac {\sqrt{3}}2)^2}dx$$
But I don't know how to solve the last term.
I answered the question... I realized my silly mistake, sorry for bothering...
|
Set $x^2=y$ $$\dfrac{y^2+1}{y^3+1}=\dfrac A{y+1}+\dfrac{By+C}{y^2-y+1}$$
$$y^2=y^2(A+B)+y(B+C-A)+A+C$$
$\implies A+C=1\iff C=1-A, B+C=A\implies B=A-C=2A-1,$
$1=A+B=3A-1\iff A=\dfrac23, B=\dfrac13,C=\dfrac13$
$$\implies\dfrac{x^4+1}{x^6+1}=\dfrac1{3(x^2+1)}+\dfrac{x^2+1}{3(x^4-x^2+1)}$$
Now $\dfrac{x^2+1}{x^4-x^2+1}=\dfrac{1+\dfrac1{x^2}}{\left(x-\dfrac1x\right)^2+1}$
set $x-\dfrac1x=u$
|
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|
Integrating $\int_0^{2\pi} \frac1{a\cos(t)+b\sin(t)+c}dt$ with $\sqrt{a^2+b^2}=1Let $a,b,c\in\mathbb{R}$ such that $\sqrt{a^2+b^2}=1<c$.
Find $I:=\int_{0}^{2\pi} \frac{1}{a\cos(t)+b\sin(t)+c}dt$.
So, I set $z=e^{it}$ and then $dz=ie^{it}dt$ and so $\frac{dz}{iz}=dt$. Therefore,
$\begin{align*} I & =\frac{1}{i}\int_{|z|=1} \frac{dz/z}{a\cdot\frac{z+z^{-1}}{2}+b\cdot\frac{z-z^{-1}}{2i}+c} \newline &=\frac{2}{i}\int_{|z|=1} \frac{dz}{a\cdot(z^2+1)-ib\cdot(z^2-1)+2cz} \newline &=\frac{2}{i}\int_{|z|=1} \frac{dz}{(a-ib)z^2+2cz+a+ib}\newline &= \frac{2}{i}\int_{|z|=1} \frac{dz}{(z-z_+)(z-z_-)}\end{align*}$
Where $z_{\pm}=\frac{-2c\ \pm\ \sqrt{4c^2-4(a+ib)(a-ib)}}{2(a-ib)}=\frac{-c\ \pm\ \sqrt{c^2-(a^2+b^2)}}{(a-ib)}=\frac{-c\ \pm\ \sqrt{c^2-1}}{(a-ib)}$.
Therefore, if we define $f(z)=\frac{1}{z-z_-}$, we get that $f$ is analytic in $\{z:|z|\leq 1\}$ since
$$|z_-|=\Bigg|\frac{-c-\sqrt{c^2-1}}{a-ib}\Bigg|=\bigg|c+\sqrt{c^2-1}\bigg|>|c|>1$$
Finally from Cauchy's Integral Formula:
$$I=\frac{2}{i}\cdot 2\pi i\cdot f(z_+)=4\pi \frac{1}{z_+-z_-}=\frac{2\pi(a-ib)}{\sqrt{c^2-1}}$$
Which is not always real... So my question is what is the problem.
Moreover, sometimes it is correct, e.g. $a=1,b=0,c=2$ gives the correct answer $\frac{2\pi}{\sqrt{3}}$.
|
Note, we have
\begin{align}
\frac{a}{\sqrt{a^2+b^2}}\cos(t)+\frac{b}{\sqrt{a^2+b^2}}\sin(t) = \cos(t-\delta)
\end{align}
where
\begin{align}
\delta=\cos^{-1}\frac{a}{\sqrt{a^2+b^2}}.
\end{align}
Hence it follows
\begin{align}
\int^{2\pi}_0 \frac{dt}{c+a\cos(t)+b\sin(t)} = \int^{2\pi}_0 \frac{dt}{c+\cos(t-\delta)} =\int^{2\pi-\delta}_{-\delta}\frac{dt}{c+\cos(t)}.
\end{align}
However, since $(c+\cos(t))^{-1}$ is $2\pi$ period, then the integral is equivalent to
\begin{align}
\int^{2\pi}_0 \frac{dt}{c+\cos(t)} =\frac{2\pi}{\sqrt{c^2-1}}
\end{align}
where the evaluation is done with contour integral.
|
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|
Rewriting the nth term of a series in the form $ar^{n-1}$ In the formula below, where does the $\frac{4}{3}$ come from and what happened to the $3$? How did they get the far right answer? Taken from Stewart Early Transcendentals Calculus textbook.
$$\sum^\infty_{n=1} 2^{2n}3^{1-n}=\sum^\infty_{n=1}(2^2)^{n}3^{-(n-1)}=\sum^\infty_{n=1}\frac{4^n}{3^{n-1}}=\sum_{n=1}^\infty4\left(\frac{4}{3}\right)^{n-1}$$
|
$$\frac{4^n}{3^{n-1}}=\frac{4^{1+(n-1)}}{3^{n-1}}$$
$$=\frac{4^1\cdot 4^{n-1}}{3^{n-1}}$$
$$=4\cdot \frac{4^{n-1}}{3^{n-1}}$$
$$=4\cdot\big(\frac{4}{3}\big)^{n-1}$$
|
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|
To find the smallest value of 'k' for the following equation
Let $\mathrm a,b$ are positive real numbers such that for $\mathrm a - b = 10$, then the smallest value of the constant $\mathrm k$ for which $\mathrm {\sqrt {x^2 + ax}} - {\sqrt{x^2 + bx}} < k$ for all $\mathrm x>0$, is?
I don't get how to approach this problem. Any help would be appreciated.
|
An elementary way ($x>0$):
$$f(x) = \sqrt{x^2 + ax} - \sqrt{x^2+bx} = \frac{x^2 + ax - (x^2+bx)}{\sqrt{x^2 + ax} + \sqrt{x^2+bx}} =\frac{(a - b)x}{x\sqrt{1 + \frac{a}{x}} + x\sqrt{1 + \frac{b}{x}}}= \frac{(a - b)}{\sqrt{1 + \frac{a}{x}} + \sqrt{1 + \frac{b}{x}}} < \frac{a-b}{2}= 5$$
|
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|
Find the integral $\int\frac{dx}{\sqrt{x^2-a^2}}$
Evaluate the integral of $\frac{1}{\sqrt{x^2-a^2}}$
Put $x=a\sec\theta\implies dx=a\sec\theta\tan\theta d\theta$
$$
\begin{align}
& \ \ \ \int \frac{dx}{\sqrt{x^2-a^2}} \\
&=\int \frac{a\sec\theta\tan\theta d\theta}{\sqrt{a^2\sec^2\theta-a^2}} \\
&=\int \frac{a\sec\theta\tan\theta d\theta}{a\sqrt{\tan^2\theta}}\\
&= \int\frac{a\sec\theta\tan\theta d\theta}{a\color{red}{\tan\theta}}\\
&=\int\sec\theta d\theta \\
&=\log \lvert \sec\theta+\tan\theta \rvert|+C_0 \\
& \mbox{where $C_0$ is an arbitrary constant of integration } \\
&=\log \left\lvert \frac{x}{a}+\sqrt{\frac{x^2}{a^2}-1} \right\rvert + C_0 \\
&= \log \left\lvert \frac{x+\sqrt{x^2-a^2}}{a} \right\rvert +C_0 \\
&= \log \left\lvert x+\sqrt{x^2-a^2} \right\rvert -\log|a|+C_0 \\
&= \log|x+\sqrt{x^2-a^2}|+ C,
& \mbox{ where $C = \log \lvert a \rvert + C_0$}.
\end{align}
$$
This is how it is solve in my reference. But, $\sqrt{\tan^2\theta}=|\tan\theta|$ right ?
Then, does that imply
$$
\int\frac{dx}{\sqrt{x^2-a^2}}=\int\frac{a\sec\theta\tan\theta d\theta}{a\color{red}{|\tan\theta|}}=\color{red}{\pm}\int\sec\theta d\theta
$$
Why am I getting this confusion and is the first solution complete ?
|
This is an elementary integral, using an inverse hyperbolic function.
By the formula for inverse functions, $$\cosh'(x)=\sinh(x)=\sqrt{\cosh^2(x)-1}$$ lets you establish
$$\text{arcosh}'(x)=\frac1{\sqrt{x^2-1}}.$$
It is known that
$$\text{arcosh}(x)=\log(x+\sqrt{x^2-1})$$ though the form $\text{arcosh}$ is fine.
|
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|
Find all integral ordered pairs $(n,k)$ such that $\left\lfloor\frac{n^2+18n+10}{2}\right\rfloor = k^2$. I had two problems that I want to solve. The first one was easy, but the second one... not so much:
First Problem:
Find all values of $n$ such that$$\frac{(n+1)(n+9)+8n+1}{2} = n^2.$$
My Attempt:
$$(n+1)(n+9) = n^2 + 10n + 9.$$ Therefore, the problem is asking me to find all values of $n$ such that $$\begin{align} \frac{n^2 + 18n + 10}{2} &= n^2 \\ \\ \Leftrightarrow n^2 + 18n + 10 &= 2n^2 \\ \\ \Leftrightarrow 18n + 10 &= n^2 \\ \\ \Leftrightarrow -n^2 + 18n + 10 &= 0.\end{align}$$ By the Fundamental Theorem of Algebra (FTOA), I know that there must exist only two distinct solutions for $n$. Then, using the quadratic formula, I obtain that
$$\begin{align} n &=\frac{-18\pm\sqrt{18^2 - (-4\cdot 10)}}{2} \\ \\ &= \frac{18}{2}\pm\frac{\sqrt{324 + 40}}{2} \\ \\ &= 9\pm\sqrt{364}.\end{align}$$ These are the only solutions, and since $364$ is not a square number, then these solutions are also irrational.
Second Problem:
Find all ordered pairs $(n,k)\in\mathbb{Z}^2$ such that $$\left\lfloor\frac{(n+1)(n+9)+8n+1}{2}\right\rfloor=k^2$$ for which $n\neq k$ and we denote by $\left\lfloor x\right\rfloor$ rounding $x$ to the lowest integer.
How must I approach this problem? Because we have the $\left\lfloor\ldots\right\rfloor$ function, I assume that in at least one case, $(n+1)(n+9)+8n+1$ is odd. Therefore, in at least one case, $n$ must be odd. By trial and error, it seems like there exist only three solutions: $(2, 5), (3, 6), (4, 7)$.
My Attempt:
Executed Trial and Error and expanded the numerator, but it did not help.
Thank you in advance.
Edit (Another Attempt):
If $n$ is even, then $\exists a \in\mathbb{Z}$, the set of all integers, such that $n = 2a$. Simplifying results in $2a^2 + 24a$ $+ \, 5 = k^2$. Since the quadratic is congruent to $1$ modulo $4$ (or $a$ is a quadratic residue modulo $4$), then it definitely can be a square.
When $k=0$ then $a=-6\pm\sqrt{536}\notin\mathbb{Z}$ so $k\neq 0$.
When $k = 1$ then $a=-6\pm\sqrt{544}\notin\mathbb{Z}$ so $k\neq 1$.
When $k=2$ then $a=-6\pm\sqrt{568}\notin\mathbb{Z}$ so $k\neq 2$.
Now I go back to trial and error....
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The floor function seems annoying so lets get rid of it, letting $n=2m$ we get:
$\frac{(2m+1)(2m+9)+16m +1}{2}=k^2$.
So $4m^2+36m+10=2k^2$ or $(2m+9)^2 = 2k^2+71$
letting $n=2m+1$ we get
$\frac{(2m+2)(2m+10)+16m+9-1}{2}=k^2$
So $4m^2+40m+28 = 2k^2$ or $(2m+10)^2=2k^2+72$
Hence the problem reduces to finding all values of $a^2-2k^2=71$ or $a^2-2k^2=72$ which can be done with Pell equations.
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|
How can I solve this Combination with indistinguishable-objects problem? This question is not about how to solve the problem, but is about why doesn't my solution work.
A bowl has $2$ red, $2$ green, and $2$ blue marbles. How many combinations are possible if we take $3$ random marbles at a time?
I know that the answer should be, $7$.
Coz,
*
*BBG
*BBR
*BGG
*BGR
*BRR
*GGR
*GRR
But, when I want to calculate this using a combination formula, nothing comes out which is near $7$:
$\frac{^6C_3}{2! \cdot 2! \cdot 2!}$
$ = \frac{6!}{3! \cdot (6-3)!} \cdot \frac{1}{2! \cdot 2! \cdot 2!}$
$ = \frac{6!}{3! \cdot 3!} \cdot \frac{1}{2! \cdot 2! \cdot 2!}$
$ = \frac{6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1}{3 \cdot 2 \cdot 1 \cdot 3 \cdot 2 \cdot 1} \cdot \frac{1}{2! \cdot 2! \cdot 2!}$
$ = \frac{6 \cdot 5 \cdot 4 }{3 \cdot 2 \cdot 1} \cdot \frac{1}{2! \cdot 2! \cdot 2!}$
$ = \frac{5 \cdot 4 }{2! \cdot 2! \cdot 2!}$
$ = \frac{5}{2!} $
$ = \frac{5}{2}$
So, what am I doing wrong?
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You can essentially break the possibilities down into 3 cases:
*
*all the colors are the same; that cannot happen so it yields 0 possibilities
*two marbles have the same color and the last one is different, as much possibilities as ordered couples of colors (6 possibilities).
*all marbles are different, 1 possibility
In the end, you get 7 possibilities.
|
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|
Finding the value of $6b-3a$
Consider $f(x)= [x]^2 - [x+6]$ and $g(x)= 3kx^2+ 2x + 4(1-3k)$ where
$[x]$ denotes the floor function.
Let $A= \{x ~|~ f(x)= 0\}$ and $k \in [a,b]$ for which every element
of set A satisfies the inequality $g(x)\ge 0$.
1) The set $A$ is equal to ?
2) The value of $6b- 3a =$?
The first part is easy. It's just the solution of the quadratic; $[x]^2- [x]- 6 = 0 $
$\therefore \mathbb A = [-2, -1) \cup [3,4)$ which is the right answer.
But I am stuck on the second part of the question. I am unable to use the constraint on the values of $x$ to find k.
using the condition I have got:
$k(12- 3x^2) \ge -{2(x+2)} $
For $x=2$ this is true,
For $x\ne 2$:
$\implies3k(2-x) \ge -2 $
What do I do next?
The answer is :
$2$
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You need for $k$ to satify:
$$k (3x^2-12) =3k(x-2)(x+2)\geq -2(x+2)$$ for all $x \in \Bbb A= [-2, -1) \cup [3,4)$.
You can then consider the three cases:
*
*$x=-2$ then the inequality is verified.
*For all $x \in (-2, -1)$, as $x-2 <0$ and $x+2>0$ the inequality is equivalent to:
$$k \leq \frac{2}{3(2-x)}$$
and as it must be true for any $x$ and $\inf_{x \in(-2,-1)} \frac{2}{3(2-x)}=\frac{2}{3(2-(-2))}=\frac{1}{6}$ it is the same as:
$$k \leq \frac{1}{6}$$
*For all $x \in (3, 4)$, as $x-2 >0$ and $2+x>0$ the inequality is equivalent to:
$$k \geq -\frac{2}{3(x-2)}$$
and once again as $\sup_{x \in(3,4)} \frac{-2}{3(x-2)}= \frac{-2}{3(4-2)}=\frac{-1}{3}$ it is true iff:
$$k \geq -\frac{1}{3}$$
Thus $g(x) \geq 0$ for all $x \in \{-2\} \cup (-2, -1) \cup [3,4)$ iff:
$$-\frac{1}{3} \leq k \leq \frac{1}{6}$$
i.e:
$$a=-\frac{1}{3}$$
$$b=\frac{1}{6}$$
and:
$$6b-3a=1-(-1)=2$$
|
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Find a corresponding eigenvector for each eigenvalue Let A =$\begin{pmatrix} 6 & 1 \\ 2 & 7 \end{pmatrix}$
(a) Find the eigenvalues of A.
(b) Find a corresponding eigenvector for each eigenvalue in part (a).
My attempt
a) Eigenvalues:
$$\begin{vmatrix} 6-\lambda& 1 \\ 2 & 7-\lambda \end{vmatrix}=0 \Rightarrow \lambda^2-13\lambda +40=0 \Rightarrow \lambda_1=5, \lambda_2=8.$$
b) If $\lambda=5$, then
$\begin{pmatrix} 6-5 & 1 \\ 2 & 7-5 \end{pmatrix}$ = $\begin{pmatrix} 1 & 1 \\ 2 & 2 \end{pmatrix}$ $\Rightarrow $ Assuming this as B.
Then $ B\bar { x } =\bar { 0 } $,
$\begin{pmatrix} 1 & 1 \\ 2 & 2 \end{pmatrix}$ $\begin{pmatrix}X_1\\X_2\end{pmatrix}=0$
$\begin{pmatrix} 1 & 1 & 0 \\ 2 & 2 & 0 \end{pmatrix}$ By doing row reduction $=>$ $R_2->R_2-2R_1$
$\begin{pmatrix} 1 & 1 & 0 \\ 0 & 0 & 0 \end{pmatrix}$ By doing row reduction $
$X_1+X_2=0$
$X_1=-X_2$
Let $X_2=1$, then $X_1=-1$
$\begin{pmatrix}X_1\\X_2\end{pmatrix}=$ $\begin{pmatrix}-1\\1\end{pmatrix}$
But, the answer is $\begin{pmatrix}X_1\\X_2\end{pmatrix}=$ $\begin{pmatrix}1\\-1\end{pmatrix}$..
I verified many times but I ended up getting the answer as $\begin{pmatrix}X_1\\X_2\end{pmatrix}=$ $\begin{pmatrix}-1\\1\end{pmatrix}$
Can anyone please verify which is the correct answer.
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If $v$ is an eigenvector to your matrix, so is $\xi \cdot v$ for any $\xi \in \mathbb{R} \setminus \left\{ 0\right\}$. This can be extended to a complex setting.
Moreover if $u$ and $v$ are eigenvectors to the same eigenvalue, all their linear combinations (except $0$) are as well eigenvectors to that eigenvalue. This is why there is the concept of "eigenspaces".
In conclusion: Never mind the sign of an eigenvector or be startled when your calculation yields a scalar multiple.
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|
Inequality with $abc=1$, Given 3 positive numbers $a, b, c$ satisfying $abc= 1$
Prove: $$\frac{1}{a^{5}+ b^{5}+ c^{2}}+ \frac{1}{b^{5}+ c^{5}+ a^{2}}+ \frac{1}{c^{5}+ a^{5}+ b^{2}}\leq 1$$
My opinion:
Let: $x= \frac{a}{b}, y= \frac{b}{c}, z= \frac{c}{a}$
We have to prove:
$$\sum \frac{x^{2}y^{5}z^{5}}{x^{7}z^{5}+x^{2}y^{10}+y^{5}z^{7}}\leq 1$$
I want to prove:
$$\frac{x^2y^5z^5}{x^7z^5+x^2y^{10} +y^5z^7} \leq \frac{\frac{x^2}{z^2}}{2\sqrt{\frac{x^9}{z^9}}+1}$$
Help me, please! Thanks!
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By Muirhead we obtain:
$$\sum_{cyc}\frac{1}{a^5+b^5+c^2}\leq\sum_{cyc}\frac{1}{a^4b+ab^4+c^3ab}=\sum_{cyc}\frac{1}{ab(a^3+b^3+c^3)}=\frac{a+b+c}{a^3+b^3+c^3}\leq1.$$
For the proof of the last inequality we can use also AM-GM:
$$a^3+b^3+c^3=\frac{1}{9}\sum_{cyc}(5a^3+2b^3+2c^3)\geq\frac{1}{9}\sum_{cyc}9\sqrt[9]{\left(a^3\right)^5\left(b^3\right)^2\left(c^3\right)^2}=$$
$$=\sum_{cyc}\sqrt[3]{a^5b^2c^2}=\sum_{cyc}\sqrt[3]{a^3}=a+b+c.$$
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What will be the possible permutations of set S={1,2,3,4}. And how many are of those odd and even? I have solved it as per my knowledge and understanding.
Since there are 4 elements so no. of permutations will be
$$4! = 24$$
$$(1,2,3,4), (1,2,4,3), (1,3,2,4), (1,3,4,2), (1,4,2,3), (1,4,3,2), (2,1,3,4), (2,1,4,3), (2,3,1,4), (2,3,4,1), (2,4,1,3), (2,4,3,1), (3,2,1,4), (3,2,4,1), (3,1,2,4), (3,1,4,2), (3,4,2,1), (3,4,1,2), (4,2,3,1), (4,2,1,3), (4,3,2,1), (4,3,1,2), (4,1,2,3), (4,1,3,2)$$
Is it correct?
• And for even and odd-
Even: product of even no of transpositions. e.g., $(1,2,3) = (1,2)(1,3)$ is even
Odd: product of odd no of transpositions. e.g., $(1,2,3,4) = (1,2)(1,3)(1,4)$ is odd.
• But if I apply this theory in this question then all permutations will be odd.
I am confused here, is my solution incorrect? May be there will be more permutations with less elements like $(1,2,3) , (1,2,4)$?
• Also there is a swapping logic! If the numbers are swapped odd times then it is odd and even otherwise
So in this case $(1,2,3,4)$ is even (no swaps)
$(3,2,1,4)$ = Swap 1<->3 is odd
Also $(4,2,1,3)$ = Swap 3<->4 then swap 1<->4 = even permutation
What will be the answer to my question?
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There are twenty four elements in $\mathcal{S}_4$, namely $$\{\operatorname{id}, (1\ 2), (1\ 3), (1\ 4), (2\ 3),\ (2\ 4), (3\ 4), (1\ 2\ 3),\ (1\ 2\ 4), (1\ 3\ 4), (2\ 3\ 4),\ (1\ 3\ 2),\ (1\ 4\ 2),\ (1\ 4\ 3), (2\ 4\ 3),\ (1\ 2\ 3\ 4), (1\ 3\ 4\ 2), (1\ 4\ 2\ 3), (4\ 3\ 2\ 1), (2\ 4\ 3\ 1), (3\ 2\ 4\ 1), (1\ 2)(3\ 4), (1\ 3)(2 \ 4),\ (1\ 4)(2\ 3)\}.$$ In general; there are $n!$ elements in $\mathcal{S}_n.$
Now, list the cycle types of $\mathcal{S}_4$ $$\{(1, 1, 1, 1), (2,1,1),(2,2),(3,1),(4)\}.$$ List how many cycles each cycle type takes, let's call this quantity $\gamma(a)$ where $a$ corresponds to each cycle type. Then $$\gamma(1,1,1,1)=4,\quad\gamma(2,1,1)=3,\quad\gamma(2,2)=2,\quad\gamma(3,1)=2,\quad\gamma(4)=1.$$ Then, a permutation is even if $n-\gamma(a)$ is even. So, the even permutations are the ones of cycle type $(1,1,1,1)$, $(2,2)$, $(3,1)$. This gives us a total of twelve even permutations, namely $$\{\operatorname{id}, (1\ 2)(3\ 4), (1\ 3)(2 \ 4),\ (1\ 4)(2\ 3),(1\ 2\ 3),\ (1\ 2\ 4), (1\ 3\ 4), (2\ 3\ 4),\ (1\ 3\ 2),\ (1\ 4\ 2),\ (1\ 4\ 3), (2\ 4\ 3)\},$$ with the other twelve permutations being odd.
In general, there are $n!/2$ even and odd permutations in $\mathcal{S}_n$.
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What is the simplest way to obtain asymptotics of $\sum_{k=0}^n \binom{n}{k} \frac{1}{(1+k)^2}$? In the middle of a proof, I have had to analyze the asymptototic behavior of
$$
\mathbb{E}\left[\frac{1}{(1+X)^2}\right] = \frac{1}{2^n}\sum_{k=0}^n \binom{n}{k}\frac{1}{(1+k)^2}\tag{1}
$$
where $X$ is Binomially distributed with parameters $n$ and $1/2$. (I also had to handle $\mathbb{E}\left[\frac{1}{(1+X)^4}\right]$, but let's start with the square).
Now, it is easy to compute
$\mathbb{E}\left[\frac{1}{1+X}\right], \mathbb{E}\left[\frac{1}{(1+X)(2+X)}\right]$, but (1) does not have any closed form (the hypergeometric function is not considered by me as a closed form).
I know that, as $n\to\infty$,
$$
\mathbb{E}\left[\frac{1}{(1+X)^2}\right] = \frac{4}{n^2} - \frac{4}{n^3} + O\left(\frac{1}{n^4}\right) \tag{2}
$$
(see e.g. [1], which implies this but deals with general probability $p$ and power $r$ instead of $p=1/2$ and $r=2$), but that seems overkill.
What is the simplest and most elegant way to derive (2)?
*
*Via a simple argument (concentration of the Binomial around $n/2+O(\sqrt{n})$) is it easy to show that is it $\Theta(1/n^2)$. Applying Jensen also shows a lower bound of $\frac{4}{n^2}+\Theta\left(\frac{1}{n^3}\right)$.
*via a still simple argument, involving comparing it to the (explicitly computable) $\mathbb{E}\left[\frac{1}{(1+X)(2+X)}\right]$ and bounding the difference, it is not hard to show it is $\frac{4}{n^2} + \Theta\left(\frac{1}{n^3}\right)$.
But that's not necessarily elegant, and also doesn't quite lead to (2) (I reckon the second approach can be made work, but it'll get messy, and will definitely not end up being a "proof from the book")
[1] Francisco Cribari-Neto, Nancy Lopes Garcia, and Klaus LP Vasconcellos. A note on inverse moments of binomial variates. Brazilian Review of Econometrics, 20(2):269–
277, 2000.
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New Answer. If $p(x)$ is a non-constant polynomial of degree $d$ having no zero on $\mathbb{N}_0 = \{0,1,\cdots\}$ and $m \geq d$, then we may expand
$$ \frac{1}{p(x)} = \sum_{k=d}^{m} \frac{c_k}{(x+1)\cdots(x+k)} + r(x),$$
where $c_k$'s are constants and $r(x)$ has no pole on $\mathbb{N}_0$ and satisfies $r(x) = \mathcal{O}(\lvert x\rvert^{-m-1})$ near $\infty$. Constants $c_k$'s may be determined by comparing Laurent expansions of both sides at $\infty$, or by clever algebraic manipulation in nice situations. Then it follows that
$$ \mathbf{E}\bigg[\frac{1}{p(X)}\bigg]
= \sum_{k=d}^{m} \frac{c_k 2^k}{(n+1)\cdots(n+k)} + \mathcal{O}\left(\frac{1}{n^{m+1}}\right). $$
Example 1. For each $m\geq2$, we have
$$ \frac{1}{(x+1)^2} = \sum_{k=2}^{m} \frac{(k-2)!}{(x+1)\cdots(x+k)} + \frac{(m-1)!}{(x+1)^2(x+2)\cdots(x+m)} $$
So
$$ \mathbf{E}\bigg[\frac{1}{(x+1)^2}\bigg] = \sum_{k=2}^{m} \frac{(k-2)!2^k}{(n+1)\cdots(n+k)} + \mathcal{O}\left(\frac{1}{n^{m+1}}\right) $$
Similarly,
Example 2. Comparing the Laurent expansion near $x=\infty$, we find that
$$ \frac{1}{(x+1)^4} = \frac{1}{(x+1)\cdots(x+4)} + \frac{6}{(x+1)\cdots(x+5)} + r(x)$$
where $r(x) = \frac{109 + 120 x + 35 x^2}{(x+1)^4(x+2)\cdots(x+5)}$ is $\mathcal{O}(x^{-6})$ as $x\to\infty$. This gives
$$ \mathbf{E}\bigg[\frac{1}{(x+1)^4}\bigg] = \frac{16}{(n+1)\cdots(n+4)} + \frac{192}{(n+1)\cdots(n+5)} + \mathcal{O}\left(\frac{1}{n^6}\right). $$
Old Answer. Write $X = Y+Z$, where $Y$ and $Z$ are independent and $Y\sim\operatorname{Bin}\big(n-1,\frac{1}{2}\big)$ and $Z\sim\operatorname{Ber}\big(\frac{1}{2}\big)$. Then by conditioning on $Y$,
\begin{align*}
\mathbf{E}\bigg[ \frac{1}{(1+X)^2} \bigg]
&= \frac{1}{2}\mathbf{E}\bigg[ \frac{1}{(1+Y)^2} + \frac{1}{(2+Y)^2} \bigg] \\
&= \mathbf{E}\bigg[ \frac{1}{(1+Y)(2+Y)} \bigg] + \frac{1}{2}\mathbf{E}\bigg[ \bigg( \frac{1}{1+Y} - \frac{1}{2+Y} \bigg)^2 \bigg].
\end{align*}
Now using the probability generating function $f(z)=\mathbf{E}[z^Y] = \left(\frac{1+z}{2}\right)^{n-1}$, for each $k = 1, 2, \cdots$ we check that
\begin{align*}
\mathbf{E}\bigg[ \frac{1}{(1+Y)\cdots(k+Y)} \bigg]
&= \int_{0}^{1} \frac{(1-z)^{k-1}}{(k-1)!} f(z) \, dz \\
&= \int_{\frac{1}{2}}^{1} \frac{2^k}{(k-1)!} (1-w)^{k-1}w^{n-1}\,dw \\
&= \int_{0}^{1} \frac{2^k}{(k-1)!} (1-w)^{k-1}w^{n-1}\,dw + \mathcal{O}_k(2^{-n}) \\
&= \frac{2^k (n-1)!}{(n+k-1)!} + \mathcal{O}_k(2^{-n}).
\end{align*}
Here, the implicit bounds depend only on $k$. In particular,
*
*$ \mathbf{E}\Big[ \frac{1}{(1+Y)(2+Y)} \Big] = \frac{4}{n(n+1)} + \mathcal{O}(2^{-n}) $,
*$ \mathbf{E}\Big[ \Big( \frac{1}{1+Y} - \frac{1}{2+Y} \Big)^2 \Big]
= \mathbf{E}\Big[ \frac{1}{(1+Y)^2(2+Y)^2} \Big]
\leq \mathbf{E}\Big[ \frac{6}{(1+Y)(2+Y)(3+Y)(4+Y)} \Big] = \mathcal{O}\Big(\frac{1}{n^4}\Big). $ Here, we utilized the simple inequality that if $x \geq 0$ and $0 < a < b$ then $\frac{a}{a+x} \leq \frac{b}{b+x}$.
Therefore it follows that
$$ \mathbf{E}\bigg[ \frac{1}{(1+X)^2} \bigg]
= \frac{4}{n(n+1)} + \mathcal{O}\left(\frac{1}{n^4}\right)
= \frac{4}{n^2} - \frac{4}{n^3} + \mathcal{O}\left(\frac{1}{n^4}\right). $$
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|
A limit involving $\cos x$ and $x^2$ The question is finding the value of $L = \lim_{x \to 0}\frac{1 - \cos x \cos 2x \cos 3x}{x^2}$ if it exists . I've found the answer using taylor's formula but I'm looking for other solutions maybe using trigonometric identities .
|
Alternatively:
$$\cos 3x=\cos (2x+x)=\cos 2x\cos x-\sin 2x\sin x.$$
$$L = \lim_{x \to 0}\frac{1 - \cos x \cos 2x \cos 3x}{x^2}=\lim_{x \to 0}\frac{1 - \cos x \cos 2x (\cos 2x\cos x-\sin 2x\sin x)}{x^2}=\\
\lim_{x \to 0}\frac{1 - \cos^2 x(1-\sin ^22x)+\cos x\cos 2x\sin 2x\sin x}{x^2}=\\
\lim_{x \to 0}\frac{\sin^2x + \cos^2x\sin ^22x+\cos x\cos 2x\sin 2x\sin x}{x^2}=\\
\lim_{x\to 0} \frac{\sin^2x}{x^2}+\lim_{x\to 0} \frac{4\sin^22x}{(2x)^2}+\lim_{x\to 0} \frac{2\sin 2x\sin x}{2x^2}=1+4+2=7.$$
|
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|
Solve differential equation $(1+x^2) \frac{dy}{dx} - 2xy = x$ Solve differential equation $$(1+x^2) \frac{dy}{dx} - 2xy = x$$
I simplified it to $$\frac{1}{1+2y} dy = \frac{x}{1+x^2} dx$$
$$ \int \frac{1}{1+2y} dy =\int \frac{x}{1+x^2} dx $$
$$\frac{1}{2} \int \frac{2}{1+2y} dy = \frac{1}{2} \int \frac{2x}{1+x^2} dx$$
$$\frac{1}{2} \ln | 1 + 2y | = \frac{1}{2} \ln | 1+x^2 | + C $$
From here, I got stuck. I have to remove y from here to solve it.
The answer in the textbook gave $ y= (k(1+x^2) - 1)/{2}$ I believe $k$ is the integration constant. How do I remove the $\ln$ from both sides?
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You got : $\quad\frac{1}{2} \ln | 1 + 2y | = \frac{1}{2} \ln | 1+x^2 | + C $
$$\frac{1}{2} \ln | 1 + 2y | - \frac{1}{2} \ln | 1+x^2 | = C $$
$$\ln | 1 + 2y | - \ln | 1+x^2 | = 2C $$
$$\ln \frac{| 1 + 2y |}{ | 1+x^2 |} = 2C $$
Thus $\quad\frac{ 1 + 2y }{ 1+x^2 }=$constant.
$$\frac{ 1 + 2y }{ 1+x^2 }=c$$
$$1+2y=c(1+x^2)$$
$$y=\frac12\left(c(1+x^2)-1\right)$$
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|
Solve the differential equation: $\frac{dy}{dx} = \sqrt{ \frac{1-y^2}{1-x^2} } $ $$\frac{dy}{dx} = \sqrt{ \frac{1-y^2}{1-x^2} }$$
Simplified it to: $$ \int \frac{1}{\sqrt{1-y^2}} dy = \int \frac{1}{\sqrt{1-x^2}} dx$$
$$\implies\sin^{-1} y = ( \sin^{-1} x + C)$$
$$y = (\sin (\sin^{-1} x) \cos C_1) + \cos(\sin^{-1} x) (\sin C_1)$$
How to I simplify $\cos(\sin^{-1} x)$ to get the final answer ?
|
$\cos^2(\arcsin(x))=1-\sin^2(\arcsin(x))=1-x^2$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2769521",
"timestamp": "2023-03-29T00:00:00",
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|
Sequence $x_n$ such that $x_{n}\equiv 0\pmod {27}\Longleftrightarrow n\equiv 3\pmod 9$ Let $x_{n}$ be a sequence such that $x_{0}=9,x_{1}=89$,and such
$$x_{n+2}=10x_{n+1}-x_{n}\forall n\ge 0$$
Show that
$$x_{n}\equiv 0\pmod {27}\Longleftrightarrow n\equiv 3\pmod 9$$
Is there a very simple way to prove that a sequence has a period of 9 in the sense of mod 27?
|
Note the corresponding characteristic equation is
$$ \lambda^2-10\lambda+1=0 $$
which has two roots $r_1=5+2\sqrt6,r_2=5-2\sqrt6$. Therefore $x_n$ can be expressed as
$$ x_n=C_1r_1^n+C_2r_2^n. $$
Using $x_0=9,x_1=89$, it is easy to see $C_1=\frac92+\frac{11}{\sqrt6},C_2=\frac92-\frac{11}{\sqrt6}.$ So
\begin{eqnarray}
x_n&=&\bigg(\frac92+\frac{11}{\sqrt6}\bigg)(5+2\sqrt6)^n+\bigg(\frac92-\frac{11}{\sqrt6}\bigg)(5-2\sqrt6)^n\\
&=&\frac92\bigg[(5+2\sqrt6)^n+(5-2\sqrt6)^n\bigg]+\frac{11}{\sqrt6}\bigg[(5+2\sqrt6)^n-(5-2\sqrt6)^n\bigg]\\
&=&9\sum_{2k\le n}\binom{n}{2k}5^{n-2k}6^k+22\sum_{2k-1\le n}\binom{n}{2k-1}5^{n-2k+1}6^{k-1}\\
&\equiv&9\cdot5^n+22\bigg[\binom{n}{1}5^{n-1}+\binom{n}{3}5^{n-3}6\bigg]\mod27\\
&\equiv&5^{n-3}\bigg[9\cdot5^3+22n\cdot5^2+22n(n-1)(n-2)\bigg]\mod27\\
&\equiv&-9+10n-5n(n-1)(n-2)\mod27
\end{eqnarray}
Clearly if $x_n\equiv0\mod27$, then $3|n$ since $3|n(n-1)(n-2)$. Let $n=3k$. Then
\begin{eqnarray}
x_n&\equiv&-9+10n-5n(n-1)(n-2)\mod27\\
&\equiv&-9+30k-15k(3k-1)(3k-2)\mod27\\
&\equiv&3\bigg[-3-15k(9k^2-9k+4)\bigg]\mod27\\
&\equiv&3(-3-60k)\mod27\\
&=&-9(20k+1)\mod27\\
\end{eqnarray}
Thus if $x_n\equiv0\mod27$, then $3|(20k+1)$ which implies that $k=3l+1$ or
$$ n\equiv3\mod 9$$
Conversely if $ n\equiv3\mod 9$, it is not hard to show $x_n\equiv0\mod27$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2772217",
"timestamp": "2023-03-29T00:00:00",
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|
The ellipse of the maximum area contained between the curves $\frac{\pm 1}{x^2+c} $
The ellipse of the maximum area contained between the curves $\frac{\pm 1}{x^2+c}, c > 0 $
It seems to me that the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2} = 1$ of the maximum area when $ a = c; b = \frac{1}{c} $
That is, it looks like this:
The problem is that I'm not sure that an ellipse here can be a finite area
I will be glad to any hint or solution
|
Consider the case of the curves $y= \pm\frac{1}{x^2+1}$ (the general case can reduced to this case). The ellipse of largest area is not $2 x^2 + y^2=1$ but $ \frac{x^2}{2} + 2 y^2 =1$, tangent to the graphs at points $(\pm 1, \pm \frac{1}{2})$.
There is a family of ellipses tangent to the graph
$$\frac{x^2}{u}+\frac{y^2}{v}=1$$ where $v= \frac{27 u}{4(u+1)^3}$. The maximum of $uv = \frac{27 u^2}{4(u+1)^3}$ occurs at $u=2$, with $v=\frac{1}{2}$.
So, take your ellipse, double the horizontal diameter, and divide the vertical diameter by $\sqrt{2}$ to get the maximal ellipse. Its area is $\pi$, $1/2$ the area between the curves. Note that the common tangent to the ellipse and the curve passes through the vertex of the curve.
|
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Why am I getting inconsistent limit results? My objective is to solve the following limit $$\lim_{x \to \infty} \frac{a^x - b^x}{a^{x+1} - b^{x+1}},$$ where $a$ and $b$ are real constants, such that $a \gt b \gt 0$.
I (apparently) managed to solve it in two different ways, leading to two distinct answers, which of course, means something is wrong.
Method 1:
$$\lim_{x \to \infty} \frac{a^x - b^x}{a^{x+1} - b^{x+1}}\frac{a^{-x}}{a^{-x}}=\lim_{x \to \infty} \frac{1 - (\frac{b}{a})^x}{a - a(\frac{b}{a})^{x+1}}$$
using the L´Hospital rule, we get
$$\lim_{x \to \infty} \frac{-(\log{\frac{b}{a}})(\frac{b}{a})^x}{-a(\log{\frac{b}{a}})(\frac{b}{a})^{x+1}} = \frac{1}{a}(\frac{b}{a})^{-1} = \frac{1}{b}.$$
Method 2:
$$\lim_{x \to \infty} \frac{a^x - b^x}{a^{x+1} - b^{x+1}}\frac{b^{-x}}{b^{-x}}=\lim_{x \to \infty} \frac{(\frac{a}{b})^x - 1}{b(\frac{a}{b})^{x+1} - b}$$
using the L´Hospital rule, we get
$$\lim_{x \to \infty} \frac{(\log{\frac{a}{b}})(\frac{a}{b})^x}{b(\log{\frac{a}{b}})(\frac{a}{b})^{x+1}} = \frac{1}{b}(\frac{a}{b})^{-1} = \frac{1}{a}.$$
What am I doing wrong?
Edit: added constraints for $a$ and $b$.
|
Note that after this step
$$\lim_{x \to \infty} \frac{a^x - b^x}{a^{x+1} - b^{x+1}}\frac{a^{-x}}{a^{-x}}=\lim_{x \to \infty} \frac{1 - (\frac{b}{a})^x}{a - a(\frac{b}{a})^{x+1}}$$
we need some condition to apply l'Hopital (i.e. indeterminate form $\frac{\infty}{\infty}$ or $\frac{0}{0}$) and similarly for Method 2. That's not the correct way to solve the limit.
To solve note that for $x$ real we need $a,b>0$ and we can distinguish the two cases $a>b$ or $b>a$. For $x$ integer we can also consider $a,b <0$.
Notably for $x$ real and $a,b>0$
*
*for $a>b\implies \lim_{x \to \infty} \frac{a^x - b^x}{a^{x+1} - b^{x+1}}\frac{a^{-x}}{a^{-x}}=\lim_{x \to \infty} \frac{1 - (\frac{b}{a})^x}{a - a(\frac{b}{a})^{x+1}}=\frac1a$
*for $b>a\implies \lim_{x \to \infty} \frac{a^x - b^x}{a^{x+1} - b^{x+1}}\frac{b^{-x}}{b^{-x}}=\lim_{x \to \infty} \frac{(\frac{a}{b})^x - 1}{b(\frac{a}{b})^{x+1} - b}=\frac1b$
For $x$ integer with $a\cdot b<0$
*
*for $|a|>|b|\implies \lim_{x \to \infty} \frac{a^x - b^x}{a^{x+1} - b^{x+1}}=\lim_{x \to \infty} \frac{a^{x}}{a^{x+1}}\frac{1 - (\frac{b}{a})^x}{1 - (\frac{b}{a})^{x+1}}=\frac1a$
*for $|b|>|a|\implies \lim_{x \to \infty} \frac{a^x - b^x}{a^{x+1} - b^{x+1}}=\lim_{x \to \infty} \frac{b^{x}}{b^{x+1}}\frac{(\frac{a}{b})^x - 1}{(\frac{a}{b})^{x+1} - 1}=\frac1b$
|
{
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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|
When does $\frac{a+b}{2}$ and $\sqrt{ab}$ have inversed tens digits and ones digits? Let $a$ and $b$ be natural numbers, and
$$A = \frac{a+b}{2}$$
$$B = \sqrt{ab}$$
It's given that $A$ and $B$ are two-digit numbers such that the tens digit of $A$ is the same as the ones digit of $B$, and the tens digit of $B$ is the same as the ones digit of $A$.
So $A = 10x + y\;\,$and$\;B = 10y + x$.
Also given is $A\ne B$.
What is $a$ and $b$?
My teacher gave us the answer without explaining it as: $a = 98$ and $b = 32$, which makes $A = 65$ and $B=56$.
My question is: How do you prove this? I know $98 = 2\cdot 7^2$ and $32 = 2^5$, but I don't understand how to find this specific answer.
|
Since $a+b=2A$, and $ab=B^2$, it follows that $a,b$ are roots of the quadratic equation
$$t^2-(2A)t+B^2=0$$
so the discriminant
$$D=4(A^2-B^2)$$
must be a perfect square, hence $A^2-B^2$ must be a perfect square.
Also, since $A\ne B$, we get that $A^2-B^2$ is a nonzero perfect square.
\begin{align*}
\text{Then}\;\;A^2-B^2&=(A+B)(A-B)\\[4pt]
&=\bigl((10x+y)+(10y+x)\bigr)\bigl((10x+y)-(10y+x)\bigr)\\[4pt]
&=\bigl(11(x+y)\bigr)\bigl(9(x-y)\bigr)\\[4pt]
\end{align*}
hence $11(x+y)(x-y)$ must be a nonzero perfect square.
So we must have $x > y$, and at least one of the factors $(x+y),\;(x-y)$ must be a multiple of $11$.
But $x-y$ can't be a multiple of $11$, since $1 \le x-y \le 8$.
Then since $3 \le x+y \le 17$, we get that $x+y=11$, and $x-y$ is a perfect square.
Of the pairs $(x,y)$ of digits
$$(x,y) = (9,2),\;\;(x,y) = (8,3),\;\;(x,y) = (7,4),\;\;(x,y) = (6,5)$$
with $x > y$ and $x+y=11$, the pair $(x,y)=(6,5)$ is the only one for which $x-y$ is a perfect square.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Limitations of Cramer's rule? I'm trying to find the line about which this matrix reflects a vector
$\frac{1}{2}\left(\begin{array}{cc}
-1 & -\sqrt{3} \\
-\sqrt{3} & 1 \\
\end{array}\right)$ $\left(\begin{array}{c}
x \\
y \\
\end{array}\right)$ = $\left(\begin{array}{c}
x \\
y \\
\end{array}\right)$
using cramer's rule i.e
$x = \frac{(\frac{1}{2})^2 \left|\begin{array}{cc}
x & -\sqrt{3} \\
y & 1 \\
\end{array}\right|}{(\frac{1}{2})^2 \left|\begin{array}{cc}
-1 & -\sqrt{3} \\
-\sqrt{3} & 1 \\
\end{array}\right|}$ and $y = \frac{(\frac{1}{2})^2 \left|\begin{array}{cc}
x & -\sqrt{3} \\
y & 1 \\
\end{array}\right|}{(\frac{1}{2})^2 \left|\begin{array}{cc}
-1 & -\sqrt{3} \\
-\sqrt{3} & 1 \\
\end{array}\right|}$
But for some reason this gives me this $y=\frac{-5}{\sqrt{3}}x$ (incorrect) when I use the x-component and $y=-\sqrt{3}x$ when I use the y-component. Is there some obvious assumption I'm missing? Can Cramer's rule not be used like this? Any help would be deeply appreciated.
|
Note, that the matrix has a factor $\frac{1}{2}$ in front.
So, you made a small mistake when substituting $\begin{pmatrix} x \\ y \end{pmatrix}$ into the determinant:
$$x = \frac{(\frac{1}{2})^2 \left|\begin{array}{cc}
\color{blue}{2}x & -\sqrt{3} \\
\color{blue}{2}y & 1 \\
\end{array}\right|}{(\frac{1}{2})^2 \left|\begin{array}{cc}
-1 & -\sqrt{3} \\
-\sqrt{3} & 1 \\
\end{array}\right|}$$
Btw., it is much easier to solve
$$\left[ \frac{1}{2}\left(\begin{array}{cc}
-1 & -\sqrt{3} \\
-\sqrt{3} & 1 \\
\end{array}\right) - \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \right]\begin{pmatrix} x \\ y \end{pmatrix} = \frac{1}{2}\left(\begin{array}{cc}
-3 & -\sqrt{3} \\
-\sqrt{3} & -1 \\
\end{array}\right)\begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2775159",
"timestamp": "2023-03-29T00:00:00",
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|
An integral involving a Gaussian and an Owen's T function. Let $a,b,c \ge 0$.
In the course of solving gaussian integral of power of cdf : $\int_{-\infty}^{+\infty} \Phi(x)^n \cdot \phi(a+bx) \cdot dx$ we came across a following integral:
\begin{eqnarray}
T(a,b,c):= \int\limits_a^\infty \phi(\xi) T(\xi b,c) d\xi
\end{eqnarray}
where $T(.,.)$ is the Owen's T function https://en.wikipedia.org/wiki/Owen%27s_T_function. As the Wikipedia page tells us this function is related to likelihoods of a bivariate Gaussian distribution and furthermore the integral itself is related to likelihoods of a triple-variate Gaussian distribution.
Now, by going to spherical coordinates we managed to find our integral in case $a=0$. We have:
\begin{equation}
T(0,b,c) = \frac{1}{4\pi} \arcsin\left(
\frac{c}{\sqrt{(1+b^2)(1+c^2)}}\right)
\end{equation}
My question is therefore what is the value of the integral for generic $a>0$ ?
|
It turns out that the answer is actually quite simple and can be obtained using elementary methods. We have:
\begin{eqnarray}
T(a,b,c)= \int\limits_{{\mathbb R}^3} 1_{\xi_0 > a} 1_{\xi_1 > \xi_0 b} 1_{c \xi_1 > \xi_2 >0} \prod\limits_{p=0}^2 \phi(\xi_p) d \xi_p
\end{eqnarray}
Now we go to spherical coordinates as follows:
\begin{eqnarray}
\xi_0&=&r \sin(\theta) \cos(\phi)\\
\xi_1&=&r \sin(\theta) \sin(\phi)\\
\xi_2&=& r \cos(\theta)
\end{eqnarray}
where $\theta \in[0,\pi/2)$ and $\phi\in[-\pi/2,\pi/2)$ which follows from the fact that all three coordinates are positive.
Now we carefuly analyze the inequality conditions:
\begin{eqnarray}
1_{\xi_0 > a} &:& r \sin(\theta) \cos(\phi) > a \quad \Longrightarrow \quad \theta > \frac{a}{r \cos(\phi)} \quad \Longrightarrow \quad \theta > \arcsin(\frac{a}{r \cos(\phi)})\\
1_{\xi_1>\xi_0 b} &:& r \sin(\theta) \sin(\phi) > b r \sin(\theta) \cos(\phi) \quad \Longrightarrow \quad \tan(\phi) > b\\
1_{c \xi_1> \xi_2 >0} &:& c r \sin(\theta) \sin(\phi) > r \cos(\theta) \quad \Longrightarrow \quad c \sin(\phi) > \cot(\theta) \quad \Longrightarrow \quad \theta > arccot(c \sin(\phi))
\end{eqnarray}
Now, in the last equation on the right in the top line above we must have $r > a/\cos(\phi) = a \sqrt{1+\tan(\phi)^2} > a \sqrt{1+b^2}$.
Therefore we write down the integral in spherical coordinates as follows:
\begin{eqnarray}
(2\pi)^{3/2}T(a,b,c)&=& \int\limits_{a\sqrt{1+b^2}}^\infty e^{-1/2 r^2} r^2 \int\limits_{\arctan(b)}^{\arccos(a/r)} \int\limits_{\arcsin(\frac{a}{(r \cos(\phi))})\vee arccot(c \sin(\phi))}^{\pi/2} \sin(\theta) d\theta \quad d\phi \quad dr\\
&=& \int\limits_{a\sqrt{1+b^2}}^\infty e^{-1/2 r^2} r^2 \int\limits_{\arctan(b)}^{\arccos(a/r)}
%
\left[\cos(\arcsin(\frac{a}{r \cos(\phi)})) \wedge
\cos(arccot(c \sin(\phi)))
\right]
%
\quad d\phi \quad dr\\
&=&
\int\limits_{a\sqrt{1+b^2}}^\infty e^{-1/2 r^2} r^2 \int\limits_{\arctan(b)}^{\arccos(a/r)}
%
\left[
\frac{c \sin(\phi)}{\sqrt{1+c^2 \sin(\phi)^2}} \wedge
\frac{\sqrt{r^2 \cos(\phi)^2-a^2}}{r \cos(\phi)}
\right]
%
\quad d\phi \quad dr \\
&=&
\int\limits_{a\sqrt{1+b^2}}^{a\sqrt{1+b^2(1+c^2)}} e^{-1/2 r^2} r^2
\int\limits_{\arctan(b)}^{\arccos(a/r)}
\frac{\sqrt{r^2 \cos(\phi)^2-a^2}}{r \cos(\phi)} d\phi dr +\\
&&
\int\limits_{a\sqrt{1+b^2(1+c^2)}}^\infty e^{-1/2 r^2} r^2
\left(
\int\limits_{\arctan(b)}^{\arccos(\frac{a \sqrt{1+c^2}}{\sqrt{a^2 c^2 + r^2}})}
\frac{c \sin(\phi)}{\sqrt{1+c^2 \sin(\phi)^2}}
d\phi +
%
\int\limits_{\arccos(\frac{a \sqrt{1+c^2}}{\sqrt{a^2 c^2 + r^2}})}^{\arccos(a/r)}
\frac{\sqrt{r^2 \cos(\phi)^2-a^2}}{r \cos(\phi)} d\phi
\right) =\\
&=&
\int\limits_{a\sqrt{1+b^2}}^{a\sqrt{1+b^2(1+c^2)}} e^{-1/2 r^2} r^2
\left(
\frac{a}{r} \arctan[\frac{a b}{\sqrt{r^2-a^2(1+b^2)}}]-
\arctan[\frac{r b}{\sqrt{r^2-a^2(1+b^2)}}]
\right)dr +\\
&& \frac{\pi ^{3/2}}{2 \sqrt{2}} \left(\sqrt{\frac{2}{\pi }} a \left(\sqrt{b^2+1}-1\right) e^{-\frac{1}{2} a^2 \left(b^2+1\right)}-\frac{\sqrt{2} a e^{-\frac{1}{2} a^2 \Delta^2} \left(\Delta \left(\pi -2 \arctan\left(\frac{c}{\Delta}\right)\right)-2 \arctan\left(\frac{1}{c}\right)\right)}{\pi ^{3/2}}-\text{erf}\left(\frac{a \sqrt{b^2+1}}{\sqrt{2}}\right)+\frac{\text{erf}\left(\frac{a \Delta}{\sqrt{2}}\right)
\left(\pi -2 \arctan\left(\frac{c}{\Delta}\right)\right)}{\pi }+\frac{2 \arctan\left(\frac{c}{\Delta}\right)}{\pi }\right)
\end{eqnarray}
where $\Delta:=\sqrt{1+b^2(1+c^2)}$.
As a sanity check we look at the case $a=0$. In here only the very last term survives and we have:
\begin{eqnarray}
(2\pi)^{3/2} T(0,b,c)&=& \sqrt{\frac{\pi}{2}} \arctan(\frac{c}{\Delta})\\
\Longrightarrow\\
T(0,b,c)&=& \frac{1}{4\pi} \arctan(\frac{c}{\Delta})= \frac{1}{4\pi} \arcsin(\frac{c}{\sqrt{(1+b^2)(1+c^2)}})
\end{eqnarray}
as it should be.
|
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|
Prove $\sin(\alpha -\beta)+\sin(\alpha-\gamma)+\sin(\beta-\gamma)=4\cos\frac{\alpha-\beta}{2}\sin\frac{\alpha-\gamma}{2}\cos\frac{\beta-\gamma}{2}$ Here is a problem from Gelfand's Trigonometry:
Let $\alpha, \beta, \gamma$ be any angle, show that $$\sin(\alpha -\beta)+\sin(\alpha-\gamma)+\sin(\beta-\gamma)=4\cos\left(\frac{\alpha-\beta}{2}\right)\sin\left(\frac{\alpha-\gamma}{2}\right)\cos\left(\frac{\beta-\gamma}{2}\right).$$
I have tried to worked through this problem but cannot complete it. If I let $A= \alpha -\beta$, $B=\beta-\gamma$ and $C= \beta-\gamma$, and $A+B+C=\pi$ (now $A$, $B$ and $C$ are angles of a triangle), then I could prove the equality. But without this condition, I am stuck.
Could you show me how to complete this exercise?
|
Use factoring and algebra. Let $\;X:=e^{ix},\;Y:=e^{iy},\;Z:=e^{iz}.\;$
Then the following equations hold
$$\;\sin(x) = \frac{X-X^{-1}}{2i}, \; \cos(x)=\frac{X+X^{-1}}2,\;
\sin(x-y) = \frac{X^2-Y^2}{2iXY},\; \cos(x-y)=\frac{X^2+Y^2}{2XY}.$$
Summing and factoring these equations using a Computer Algebra System gives the equation
$$ \sin(x\!-\!y) \!+\! \sin(x\!-\!z) \!+\! \sin(y\!-\!z) \!=\!
\frac{X^2\!-\!Y^2}{2iXY} \!+\! \frac{X^2\!-\!Z^2}{2iXZ} \!+\! \frac{Y^2\!-\!Z^2}{2iYZ} \!=\! \frac{(X\!+\!Y)(X\!-\!Z)(Y\!+\!Z)}{2iXYZ}. $$
$$\textrm{Also now we have}
\quad\cos\frac{x-y}2 = \frac{X+Y}{2\sqrt{XY}},\quad
\sin\frac{x-z}2 = \frac{X-Z}{2i\sqrt{XZ}},\quad
\cos\frac{y-z}2 = \frac{Y+Z}{2\sqrt{YZ}},$$
and the result follows. Of course, trigonometric identities can and have also been used to prove it.
With a slight sign change this more symmetrical trigonometric
identity is also true
$$ \sin(x\!-\!y) \!+\! \sin(y\!-\!z) \!+\! \sin(z\!-\!x) \!=\!
-4 \sin\frac{x-y}2 \sin\frac{y-z}2 \sin\frac{z-x}2 . $$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Exact value of a convergent series: $\sum_{n=1}^\infty \frac{1}{n^3+6n^2+8n}$ I have a series
$$\sum_{n=1}^\infty \frac{1}{n^3+6n^2+8n}.$$
I know the series converges because
$$\frac{1}{n^3+6n^2+8n}\le \frac{1}{n^3}, $$
since $p=3>1$, I know that $\sum 1/n^3$ converges. But I am not sure how to figure out what it converges to.
|
Just another way using partial sums.
Using Mohammad Riazi-Kermani's answer, use partial fraction decomposition to get
$$a_n=\frac 1{n^3+6n^2+8n}=\frac {1}{n(n+4)(n+2)}=-\frac{1}{4 (n+2)}+\frac{1}{8 (n+4)}+\frac{1}{8 n}$$ and consider
$$S_p=\sum_{n=1}^p a_n$$ Using harmonic numbers
$$\sum_{n=1}^p \frac 1n=H_p\qquad \sum_{n=1}^p \frac 1{n+2}=H_{p+2}-\frac{3}{2}\qquad \sum_{n=1}^p \frac 1{n+4}=H_{p+4}-\frac{25}{12}$$
Now, use the asymptotics
$$H_q=\gamma +\log \left({q}\right)+\frac{1}{2 q}-\frac{1}{12
q^2}+O\left(\frac{1}{q^3}\right)$$ Replace and continue with Taylor expansions for large $p$ to get
$$S_p=\frac{11}{96}-\frac{1}{2 p^2}+O\left(\frac{1}{p^3}\right)$$ which shows the limit and also how it is approached.
|
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"url": "https://math.stackexchange.com/questions/2776091",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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|
Power series of $f(x)=\ln (x^2+4)$ I am supposed to find a power series representation of
$$f(x)=\ln\left(x^{2}+4\right).$$
Then, I am to graph it and observe what happens as $n$ increases. My attempt at a solution:
$$\ln\left(x^2+4\right) = \int \frac{1}{x^2+4}\,dx = \frac{1}{4}\int \frac{1}{\frac{x^2}{4}+1}\,dx = \frac{1}{4}\int \frac{1}{1-\left(-\frac{x^2}{4}\right)}\,dx.$$
Now that it is in the $\frac{1}{1-x}$ format, the power series representation is $\sum_{n=0}^{\infty} \int \frac{1}{4}\left(-\frac{x^2}{4}\right)^n$. The first terms are
$$\frac{1}{4} \left(x-\frac{x^5}{80}+\frac{x^7}{448}-\frac{x^9}{2304}+\cdots\right).$$
But as I graph these, they look nothing like the graph of $f(x)=\ln(x^2 +4)$. I am not sure if I turned the revised formula into a sum correctly.
|
Your integration is not correct $$\ln\left(x^2+4\right) \ne \int \frac{1}{x^2+4}\,dx $$ You need to start with $$\ln\left(x+4\right) = \int \frac{1}{4+x}dx$$ and go from there.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2781612",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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|
Integrate $\sin^{-1}\frac{2x}{1+x^2}$
Integrate $\sin^{-1}\frac{2x}{1+x^2}$
The solution is given in my reference as: $2x\tan^{-1}x-\log(1+x^2)+C$.
But, is it a complete solution ?
My Attempt
$$
\int 2\tan^{-1}x \, dx=\int \tan^{-1}x \cdot 2\,dx=\tan^{-1}x\int2\,dx-\int\frac{1}{1+x^2}\int2\,dx\cdot dx\\
=\tan^{-1}x \cdot2x-\int\frac{2x}{1+x^2}\,dx=2x\tan^{-1}x-\log(1+x^2)+C
$$
$$
2\tan^{-1}x=\begin{cases}\sin^{-1}\frac{2x}{1+x^2}\text{ if }|x|\leq{1}\\
\pi-\sin^{-1}\frac{2x}{1+x^2}\text{ if }|x|>{1}\text{ and }x>0\\
-\pi-\sin^{-1}\frac{2x}{1+x^2}\text{ if }|x|>{1}\text{ and }x<0
\end{cases}\\
\sin^{-1}\frac{2x}{1+x^2}=\begin{cases}2\tan^{-1}x\text{ if }|x|\leq{1}\\
\pi-2\tan^{-1}x\text{ if }|x|>{1}\text{ and }x>0\\
-\pi-2\tan^{-1}x\text{ if }|x|>{1}\text{ and }x<0
\end{cases}
$$
$$
\int\sin^{-1}\frac{2x}{1+x^2}\,dx=\begin{cases}\int2\tan^{-1}x\,dx&\text{ if } |x|\leq{1}\\\int\pi\, dx-\int2\tan^{-1}x\,dx&\text{ if }|x|>{1}&\text{ and } x>0\\-\int\pi \,dx-\int2\tan^{-1}x\,dx&\text{ if }|x|>{1}&\text{ and } x<0\end{cases}=\begin{cases}\color{red}{2x\tan^{-1}x-\log(1+x^2)+C\text{ if } |x|\leq{1}}\\\pi x-2x\tan^{-1}x+\log(1+x^2)+C&\text{ if }|x|>{1}&\text{ and } x>0\\-\pi x-2x\tan^{-1}x+\log(1+x^2)+C&\text{ if }|x|>{1}&\text{ and } x<0\end{cases}$$
So don't we have two more cases for our solution rather than that is given in my reference, right ?
|
We know that $\tan(\theta/2) = \frac{\sin(\theta)}{1+\cos(\theta)}$ and $\cos(\theta) = \sqrt{1-\sin^2(\theta)}$.
So, letting $x = \sin(\theta)$ $(\implies \theta = \arcsin(x))$ we have:
$\tan(\theta/2) = \frac{x}{1+\sqrt{1-x^2}} \implies \frac{\theta}{2} = \arctan\left(\frac{x}{1+\sqrt{1-x^2}}\right) \implies \arcsin(x) = 2\arctan\left(\frac{x}{1+\sqrt{1-x^2}}\right)$.
With that in hands, we just have to calculate $\arcsin\left(\frac{2x}{1+x^2}\right) = 2\arctan\left(\frac{\frac{2x}{1+x^2}}{1+\sqrt{1-\frac{4x^2}{1+2x^2+x^4}}}\right)$.
Now observe that $\frac{\frac{2x}{1+x^2}}{1+\sqrt{1-\frac{4x^2}{1+2x^2+x^4}}} = \frac{\frac{2x}{1+x^2}}{1+\sqrt{\frac{1-2x^2+x^4}{1+2x^2+x^4}}} = \frac{\frac{2x}{1+x^2}}{1+\frac{1-x^2}{1+x^2}} = \frac{\frac{2x}{1+x^2}}{\frac{2}{1+x²}} = x$. I thik this solve the equality you were struggling with.
|
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"url": "https://math.stackexchange.com/questions/2781904",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
degree of $\sqrt{2} + \sqrt[3]{3}$ over $\mathbb Q$ How can i find the degree of the minimal polynomial $P \in \mathbb Q[x]$ such that $P(\sqrt{2} + \sqrt[3]{3}) = 0$ ?
Recently i proved that $\mathbb Q[\sqrt{2} + \sqrt{3}] = \mathbb Q[\sqrt{2}, \sqrt{3}]$ using $(\sqrt{2} + \sqrt{3})^{-1} = \sqrt{3} - \sqrt{2}$, so $2\sqrt{3} = (\sqrt{3} - \sqrt{2}) + (\sqrt{2} + \sqrt{3})$ etc.
But how can i express $\sqrt{2}$ or $\sqrt[3]{3}$ with $\sqrt{2} + \sqrt[3]{3}$?
Is $\mathbb Q[\sqrt{2} + \sqrt[3]{3}]$ equal to $\mathbb Q[\sqrt{2}, \sqrt[3]{3}]$?
Thanks
|
Hint:
Check which automorphisms of ${\mathbf Q}[\sqrt 2,\sqrt[3]{3},e^{2\pi i/3}]$ fix ${\mathbf Q}[\sqrt2+\sqrt[3]{3}]$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2783025",
"timestamp": "2023-03-29T00:00:00",
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|
Solve in prime numbers the equation $p^q+q^r=r^p$
Find all triples of prime numbers $(p,q,r)$ such that $$p^q+q^r=r^p.$$
I proved that when $r=2$, the equation becomes $$p^q+q^2=2^p.$$ Then I tried to use reciprocity laws and Fermat's little theorem. I could prove that $p\equiv 7\pmod 8$ and that $p>q$.
The equation appeared in some olympiad . They asked to prove that $r=2$. So I am trying to find at least one triple.
|
It is clear that precisely one of $p$, $q$ and $r$ must equal $2$. We'll first show that in fact $r=2$:
Observation 1: $r=2$.
If $q=2$ then reducing mod $3$ shows that
$$p^2+2\equiv r^p\pmod{3},$$
so either $p=3$ or $r=3$. Both are easily verified to be impossible.
If $p=2$ then
$$2^q+q^r=r^2,$$
but for all primes $q,r>2$ we have $q^r>r^2$, a contradiction.
This leaves us with finding odd primes $p$ andd $q$ satisfying
$$p^q+q^2=2^p.$$
We first make a few more simple observations.
Observation 2: $q^2\equiv2\pmod{p}$.
By Fermat's little theorem
$$q^2\equiv p^q+q^2\equiv2^p\equiv2\pmod{p}.$$
Observation 3: $q<p$.
As $p>2$ we have $2^q<p^q+q^2=2^p$, so $q<p$.
The problem can be rephrased in the number ring $\Bbb{Z}[\sqrt{2}]$. This is a unique factorization domain with unit group $\{\pm1\}$. Setting $m:=\frac{p-1}{2}$ we can rewrite the equation above as
$$p^q=-(q+2^m\sqrt{2})(q-2^m\sqrt{2}).$$
The greatest common divisor of the two factors on the right hand side divides both $2q$ and $p^q$. Because $p$ and $q$ are distinct, these two factors are coprime and so both are $q$-th powers in $\Bbb{Z}[\sqrt{2}]$. Let $a,b\in\Bbb{Z}$ be such that
$$q+2^m\sqrt{2}=(a+b\sqrt{2})^q.$$
Then $q-2^m\sqrt{2}=(a-b\sqrt{2})^q$ and hence
$$p^q=-(a+b\sqrt{2})^q(a-b\sqrt{2})^q=(2b^2-a^2)^q,$$
which shows that $p=2b^2-a^2$.
Observation 4: $a=\pm q$ and $b\mid2^m$. (Thanks to barto's comment)
Set $\alpha:=a+b\sqrt{2}$ and $\bar{\alpha}:=a-b\sqrt{2}$. Because
$$(\alpha+\bar{\alpha})\mid(\alpha^2+\bar{\alpha}^q)\qquad\text{ and }
\qquad(\alpha-\bar{\alpha})\mid(\alpha^2-\bar{\alpha}^q),$$
we see that $2a\mid2q$ and $2b\sqrt{2}\mid2^{m+1}\sqrt{2}$. The binomial expansion
$$q+2^m\sqrt{2}=(a+b\sqrt{2})^q,$$
shows that $q\mid a^q$, and hence that $q\mid a$ so $a=\pm q$.
Observation 5: $b=\pm1$.
By observation 3 we have $a^2=q^2<p^2$, and so
$$p=2b^2-a^2>2b^2-p^2.$$
Because $p>q>2$ we have $p\geq5$ and so $b^2<\frac{p^2+p}{2}\leq(p-1)^2$.
Also, by observation 2
$$p=2b^2-a^2=2b^2-q^2\equiv2b^2-2\pmod{p},$$
and so $b^2\equiv1\pmod{p}$. This means $b=\pm1$.
We have $a=\pm q$ and $b=\pm1$ and so
$$p=2b^2-a^2=2-q^2<0,$$
a contradiction.
|
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"timestamp": "2023-03-29T00:00:00",
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|
ODE power series solution $y'+xy=1+x$ So i have been told to find the power series solution to the following ode $$y'+yx=1-x$$
Using the substitution $y=\sum_{n=0}^{\infty}a_nx^n$, i can rewrite the equation as the following;
$$\sum_{n=1}^{\infty}na_nx^{n-1}+\sum_{n=0}^{\infty}a_nx^{n+1}=1+x$$
That is
$$\sum_{n=0}^{\infty}(n+1)a_{n+1}x^{n}+\sum_{n=1}^{\infty}a_{n-1}x^{n}=1+x$$
I then combined the series and the non homogeneous terms to obtain
$$(a_1-1)+x(2a_2+a_0-1)+\sum_{n=2}^{\infty}[(n+1)a_{n+1}+a_{n-1}]x^n=0$$
Setting $a_1=1$ and $2a_2+a_0-1=0$, I can say that $(n+1)a_{n+1}+a_{n-1}=0$, meaning i can say $a_n=\frac{-a_{n-2}}{n}$ for $n=3,4,5...$
I began subbing in the values for n to try and get a relation, i got
$a_3=-\frac{1}{3}$, $a_4=\frac{a_0-1}{4\cdot2}$, $a_5=\frac{1}{5\cdot3}$, $a_6=-\frac{a_0-1}{6\cdot4\cdot2}$...
This is where I am stumped, I said that $$a_{2k}=\frac{(-1)^k(a_0-1)}{2k!!}, k=2,3,4...$$
And
$$a_{2k+1}=\frac{(-1)^n}{(2k-1)!!},k=1,2,3..$$
Am i correct in saying this? And if so how to i get to an answer from these statements? Any help would be greatly appriciated.
|
$$ 2\cdot4\cdot6\cdots2k=2^kk! $$
$$ 1\cdot3\cdot5\cdots(2k-1)=\frac{(2k)!}{2^kk!}$$
As for your solution, you will have to separate into two series of the form $$ y=a_0y_0+y_1 $$
ADDENDUM:
Note that your textbook did not bother to find the general terms for $3\cdot5\cdot7\cdot(2n+1)$. I worked it out using the general terms, but separated the odd and even terms in the sum.
Starting with
$$\sum_{n=1}^{\infty}na_nx^{n-1}+\sum_{n=0}^{\infty}a_nx^{n+1}=1+x$$
we adjust the indices to coordinate the exponents, getting
\begin{equation}
\sum_{n=0}^{\infty}(n+1)a_{n+1}x^n+\sum_{n=1}^{\infty}a_{n-1}x^{n}=1+x
\end{equation}
Next, we cast out the first term of the first sum to bring the two summations into alignment, and subtract the terms on the right to the left side.
\begin{equation}
a_1-1-x+\sum_{n=1}^{\infty}(n+1)a_{n+1}x^n+\sum_{n=1}^{\infty}a_{n-1}x^{n}=0
\end{equation}
We can now combine the summations.
\begin{equation}
a_1-1-x+\sum_{n=1}^{\infty}\left[(n+1)a_{n+1}+a_{n-1}\right]x^{n}=0
\end{equation}
Next we cast out the $x$ to the power $1$ term from the summation.
\begin{equation}
a_1-1+(2a_2+a_0-1)x+\sum_{n=2}^{\infty}\left[(n+1)a_{n+1}+a_{n-1}\right]x^{n}=0
\end{equation}
Setting all coefficients to $0$ gives
\begin{eqnarray}
a_1&=&1\\
a_2&=&\frac{1-a_0}{2}\\
a_{n+1}&=&-\frac{a_{n-1}}{n+1}\text{ for }n\ge2
\end{eqnarray}
Clearly the even and odd coefficients beginning with the $x^2$ term should be separated, giving the relations
\begin{eqnarray}
a_{2k}&=&-\frac{a_{2k-2}}{2k}\text{ for }k>1\\
a_{2k+1}&=&-\frac{a_{2k}}{2k+1}\text{ for }k\ge1
\end{eqnarray}
So
\begin{eqnarray}
a_2&=&\frac{1-a_0}{2}\\
a_4&=&-\frac{a_2}{4}=(-1)\frac{1}{2\cdot4}\cdot(1-a_0)\\
a_6&=&-\frac{a_4}{6}=(-1)^2\frac{1}{2\cdot4\cdot6}\cdot(1-a_0)\\
&\vdots&\\
a_{2k}&=&\frac{(-1)^{k-1}}{2^kk!}\cdot(1-a_0)\text{ for }k\ge2
\end{eqnarray}
For the odd power terms we get
\begin{eqnarray}
a_3&=&-\frac{a_1}{3}=-\frac{1}{3}\\
a_5&=&\frac{(-1)^2}{3\cdot5}\\
a_7&=&\frac{(-1)^3}{3\cdot5\cdot6}\\
&\vdots&\\
a_{2k+1}&=&(-1)^k\frac{2^kk!}{(2k+1)!}\text{ for }k\ge1
\end{eqnarray}
This gives a general solution of
\begin{equation}
y=a_0+x+\sum_{k=1}^\infty(-1)^k\frac{2^kk!}{(2k+1)!}x^{2k+1}+(1-a_0)\sum_{k=1}^\infty\frac{(-1)^{k-1}}{2^kk!}x^{2k}
\end{equation}
|
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"timestamp": "2023-03-29T00:00:00",
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|
Integrate $\int \frac {dx}{1+\sin x}$ Integrate $\int \dfrac {dx}{1+\sin x}$
My attempt:
$$\begin{align}&=\int \dfrac {dx}{1+\sin x}\\
&=\int{\dfrac{dx}{1+\dfrac{2\tan \left( \dfrac{x}{2} \right)}{1+\tan ^2\left( \dfrac{x}{2} \right)}}}\\
&=\int{\dfrac{1+\tan ^2\left( \dfrac{x}{2} \right)}{\left( 1+\tan \left( \dfrac{x}{2} \right) \right) ^2}}\,dx\end{align}$$
|
See Tangent_half-angle_substitution
If $z = \tan \big(\frac x2 \big)$, then $dx = \dfrac{2 dz}{1+z^2}$
and $\sin(x) = \dfrac{2z}{1+z^2}$
\begin{align}
\int\frac{dx}{1+\sin x}
&= \int\frac{1}{1+\sin x} \cdot dx \\
&= \int \dfrac{1}{\left(1+\dfrac{2z}{(1+z^2)}\right)}
\cdot \dfrac{2 dz}{1+z^2} \\
&= \int \dfrac{2 \, dz}{(1+z)^2} \\
&= -\dfrac{2}{1+z} + C \\
&= -\dfrac{2}{1 + \tan(\frac x2)} \\
&= \dfrac{-2}{1 + \dfrac{\sin x}{1+\cos x}} \\
&= -2\dfrac{1+\cos x}{1 + \cos x + \sin x}
\end{align}
|
{
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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|
How do I evaluate this integral ? How do you evaluate this integral $\int \frac{x+1}{(x^2-x+1)^3}dx$ ?
I found out:
$\int \frac{x+1}{(x^2-x+1)^3}dx = \frac{1}{2}\int \frac{2x-1}{(x^2-x+1)^3}dx + \int \frac{3}{(x^2-x+1)^3}dx
$
First integral is nicely solvable now.
I have rearrange the second one:
$
\int \frac{3}{(x^2-x+1)^3}dx = \int \frac{3}{((x-\frac{1}{2})^2+\frac{3}{4})^3}dx=\frac{1}{64}\int \frac{3}{((2x-1)^2+3)^3}dx
$
Let $y=2x-1$, then I have:
$\frac{1}{128}\int \frac{3}{(y^2+3)^3}dy$
What is this integral equeal to ?
|
At last I have solved it myself.
$
arctg(y)=\int \frac{1}{y^2+1}dy =
$
Using per parters
$
=\frac{1}{y^2+1}dy-\int\frac{-2y^2}{(y^2+1)^2}dy=\frac{1}{y^2+1}+2\int\frac{y^2+1-1}{(y^2+1)^2}dy=\frac{1}{y^2+1}+2\int\frac{1}{y^2+1}dy-2\int\frac{1}{(y^2+1)2}dy
$
Now I know what is the $\int\frac{1}{(y^2+1)^2}$ equal to.
By proceeding same steps as I did above I will get what I need.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2792057",
"timestamp": "2023-03-29T00:00:00",
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|
Computing a divergent integral $\iiint r^5sin\theta cos\theta$ Problem
Evaluate
$$I = \iiint_V z \sqrt{x^2+y^2+z^2}dxdydz$$
Where V is :
$$\sqrt{x^2+y^2} \leq z\leq \sqrt{1-x^2-y^2}$$
Solution:
I must use spherical coordinates so:
$$x=r\cos\psi \sin{\theta}$$
$$y=r \sin{\psi} \sin{\theta}$$
$$z = r \cos \theta$$
$$r \geq 0; 0 \leq \psi \leq 2 \pi; 0 \leq \theta \leq \pi$$
Jacobian matrix J= $r^2sin\theta$
The new V looks like this:
$$\sqrt{r^2\cos^2 \psi \sin^{2} \theta + r^2 \sin^{2} \psi \sin^{2} \theta} \leq r\cos \theta \leq \sqrt{1-(r^2\cos^2 \psi \sin^2 \theta + r^{2}\sin^2\psi \sin^{2}\theta)}$$
From these we conclude that:
$$\sin \theta \leq \cos \theta$$
$$r \leq \frac{1}{\sqrt{2}\sin \theta}; r \leq 1$$
And so the new V is:
$$ 0 \leq r \leq \frac{1}{\sqrt{2}\sin \theta}$$
$$ 0 \leq \theta \leq \pi/4$$
$$0 \leq \psi \leq 2\pi$$
And so our integral is:
$$\int \limits_0^{2\pi }\int \limits_0^{\frac{\pi \:}{4}}\int \limits^{\frac{1}{\sqrt{2}\sin\left(\theta \right)}}_0r^4\cos\left(\theta \right)\sin \left(\theta \right)drd\theta \:d\psi $$
Question here somewhere I a may be making a mistake, because later on the integral is diverging
\begin{align}
\int \limits_0^{2\pi }\int \limits_0^{\frac{\pi \:}{4}}\int \limits^{\frac{1}{\sqrt{2}\sin\left(\theta \right)}}_0r^4\cos\left(\theta \right)\sin \left(\theta \right)drd\theta \:d\psi &= \frac{1}{5} \int \int r^{5}\cos\theta \sin{\theta} \in[0, \frac{1}{\sqrt{2}}\sin\theta]\\
&=\frac{1}{20\sqrt{2}} \int \int \frac{d \sin\theta}{\sin^{4}\theta}\\
&= \frac{-3}{20\sqrt{2}} \int_{0}^{2\pi} \frac{1}{\sin^{2}\theta} |^{\pi/4}_{0} d\psi
\end{align}
and here in $\sin(0)$ the integral is diverging
Question:
What can I do about this problem? Am I making a mistake? What Should I fix? What is the solution to this Integral?
Update
$$\sqrt{x^2+y^2} \leq z\leq \sqrt{1-x^2-y^2}$$
to spherical coordinates
$$\sqrt{r^2\cos^2 \psi \sin^{2} \theta + r^2 \sin^{2} \psi \sin^{2} \theta} \leq r\cos \theta \leq \sqrt{1-(r^2\cos^2 \psi \sin^2 \theta + r^{2}\sin^2\psi \sin^{2}\theta)}$$
$$r \sin{\theta} \leq r \cos{\theta} \leq \sqrt{1-r^{2} \sin^{2}{\theta}}$$
And From here:
$$\sin{\theta} \leq \cos{\theta}$$
$$r \leq 1 $$
$$r^{2} \sin^{2}{\theta} \leq 1 - r^{2} \sin^{2}{\theta} <=> r^{2} \sin^{2}{\theta} +r^{2} \sin^{2}{\theta} \leq 1 <=> r^{2} \leq \frac{1}{2 sin^{2} \theta} <=> r \leq \frac{1}{\sqrt{2} sin{\theta}} $$
|
There is not the slightest reason why the integral of a nice function over a compact domain should be divergent.
I'm using geographical longitude $\phi$ and geographical latitude $-{\pi\over 2}\leq\theta\leq{\pi\over2}$. The condition $z\geq\sqrt{x^2+y^2}$ then translates into $r\sin\theta\geq r\cos\theta$, or $\theta\geq{\pi\over4}$, and the condition $z\leq\sqrt{1-x^2-y^2}$ translates into $r\leq 1$. Since the Jacobian is $r^2\cos\theta$ we obtain
$$I=2\pi\int_0^1\int_{\pi/4}^{\pi/2} r\sin\theta\ r\ r^2\cos\theta\>d\theta\>dr=2\pi\int_0^1r^4\>dr\int_{\pi/4}^{\pi/2}\sin\theta\cos\theta\>d\theta={\pi\over10}\ .$$
|
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|
Verifying a Stubborn Inequality I have an inequality which arose in an unrelated problem, and it's been proving to get the better of me. I essentially know it has to be true, but cannot verify it. For integers $n_{1}, n_{2}, \ell_{1}, \ell_{2}$, I want to show that
$$8 n_{1} + 8n_{2} -\ell_{1}^{2} - \ell_{2}^{2} - 2\ell_{1} \ell_{2} > 0$$
Given only the constraints $4n_{1} - \ell_{1}^{2} \geq -1$, $4n_{2} - \ell_{2}^{2} >0$, $n_{1} \geq 0$, and $n_{2} > 0$. I am also happy to assume $\ell_{1} + \ell_{2} >0$, and that $\ell_{1} \geq \ell_{2} >0$. In my particular problem, this case will suffice. Does anyone have any tips on proving this?
I realize (given the form of the constraints) it's tempting to rewrite the lefthand side as
$$(4n_{1} -\ell_{1}^{2})+(4n_{2} - \ell_{2}^{2}) + 4n_{1} + 4n_{2} - 2\ell_{1} \ell_{2},$$
but the term $-2 \ell_{1} \ell_{2}$ is troubling, as it is negative by my assumptions.
|
With less complicated symbols, you want to prove that $8a+8b-x^2-y^2-2xy>0$ provided $4a-x^2\ge-1$ and $4b-y^2>0$.
First notice that $4b-y^2>1$, because $4b-y^2=1$ implies $y^2\equiv 3\pmod{4}$, which is impossible. Now rewrite the expression as
\begin{align}
2(4a-x^2)+2(4b-y^2)+x^2+y^2-2xy
&=2(4a-x^2)+2(4b-y^2)+(x-y)^2\\
&>-2+2+0=0
\end{align}
|
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}
|
Evaluating $\int_0^1 \frac{x-x^2}{\sin \pi x} dx = \frac{7 \zeta(3)}{\pi^3}.$ I tried to use the series for $\sin \pi x$ and maybe find something related to $\zeta(3)$, but didn't work. I'm guessing this integral needs more than the little calculus that I know.
\begin{equation}
\int_0^1 \frac{x-x^2}{\sin \pi x} dx = \frac{7 \zeta(3)}{\pi^3}.
\end{equation}
|
First, denote the integral below as $I$$$I=\int\limits_0^1dx\space\frac {x(1-x)}{\sin\pi x}$$and through integration by parts on $u=x-x^2$, then we have
$$\begin{align*}I & =-\frac 1{\pi}(x-x^2)\log\cot\left(\frac {\pi x}2\right)\,\Biggr\rvert_0^1+\frac 1{\pi}\int\limits_0^1dx\, (1-2x)\log\cot\left(\frac {\pi x}2\right)\\ & =\frac 1{\pi}\int\limits_0^1dx\,\log\cot\left(\frac {\pi x}2\right)-\frac 2{\pi}\int\limits_0^1dx\, x\log\cot\left(\frac {\pi x}2\right)\\ & =-\frac 8{\pi^3}\int\limits_0^{\pi/2}dx\, x\log\cot x\tag1\end{align*}$$
where equation ($1$) comes from making the substitution $x\mapsto\frac {\pi x}2$. The latter integral can be evaluated by splitting up the natural log into two separate integrals and using the Fourier series for $\log\sin x$ and $\log\cos x$, which I have included below
$$\begin{align*}\log\cos x & =\sum\limits_{k\geq1}(-1)^{k-1}\frac {\cos2kx}{k}-\log 2\tag2\\\log\sin x & =-\sum\limits_{k\geq1}\frac {\cos 2kx}k-\log 2\tag3\end{align*}$$
Expanding ($1$) gives
$$I=-\frac 8{\pi^3}\underbrace{\int\limits_0^{\pi/2}dx\, x\log\cos x}_{I_1}+\frac 8{\pi^3}\underbrace{\int\limits_0^{\pi/2}dx\, x\log\sin x}_{I_2}\tag4$$
Call the first and second integrals $I_1$ and $I_2$ respectively. Using ($2$) and ($3$) gives the following two identities
$$\begin{align*}I_1 & =\int\limits_0^{\pi/2}dx\,\left(\sum\limits_{k\geq1}\frac {(-1)^{k-1}\cos 2kx}k-x\log 2\right)\\ & =\sum\limits_{k\geq1}\frac {(-1)^{k-1}}k\left[\frac {\pi}{4k^2}\sin\pi k+\frac 1{4k^3}\cos\pi k-\frac 1{4k^2}\right]-\frac {\pi^2}8\log2\\ & =\frac 14\sum\limits_{k\geq1}\frac {(-1)^{k-1}}{k^3}\cos\pi k-\frac 14\sum\limits_{k\geq1}\frac {(-1)^{k-1}}{k^3}-\frac {\pi^2}8\log 2\\ & \color{blue}{=-\frac 14\zeta(3)-\frac 3{16}\zeta(3)-\frac {\pi^2}8\log 2}\tag5\end{align*}$$
As a side note, the infinite sum with $\sin\pi k$ vanishes because $\sin\pi k=0$ for $k\in\mathbb{Z}$. In a similar manner, $I_2$ can be integrated as follows
$$\begin{align*}I_2 & =-\int\limits_0^{\pi/2}dx\,\left(\sum\limits_{k\geq1}\frac {\cos 2kx}k+x\log 2\right)\\ & =-\sum\limits_{k\geq1}\frac 1k\left[\frac {\pi}{4k}\sin\pi k+\frac 1{4k^2}\cos\pi k-\frac 1{4k^2}\right]-\frac {\pi^2}8\log 2\\ & =-\frac 14\sum\limits_{k\geq1}\frac {\cos\pi k}{k^3}+\frac 14\sum\limits_{k\geq1}\frac 1{k^3}-\frac {\pi^2}8\log 2\\ & \color{red}{=\frac 14\zeta(3)+\frac 3{16}\zeta(3)-\frac {\pi^2}8\log 2}\tag6\end{align*}$$
Substituting the results for ($5$) and ($6$) into ($4$) leaves us with
$$\begin{align*}I & =-\frac 8{\pi^3}\left[\color{blue}{-\frac 14\zeta(3)-\frac 3{16}\zeta(3)}\color{red}{-\frac 14\zeta(3)-\frac 3{16}\zeta(3)}\right]\\ & =\frac 7{\pi^3}\zeta(3)\end{align*}$$
Multiply by $-1$ to get the integral under question
$$\int\limits_0^1dx\space\frac {x^2-x}{\sin\pi x}\color{brown}{=-\frac 7{\pi^3}\zeta(3)}$$
|
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Sum of $\sum_{n=1}^{\infty} \frac {x^n}{(n-1)!} $ the task is to find sum of $\sum_{n=1}^{\infty} \frac {x^n}{(n-1)!} $. I've tried to factor out $x$ to get $x\sum_{n=1}^{\infty} \frac {{x}^{n-1}}{(n-1)!}$ and integrate the sum, which yields $x\sum_{n=1}^{\infty} \frac {x^n}{n!} $ which I can sum as $xe^x$. Then I need to derivate it which yields $e^x+xe^x$. The correct answer is supposed to be $xe^x$. What step am I doing wrong?
Thanks for your help.
|
Note that you have to integrate the complete expression $x\sum_{n=1}^\infty \frac{x^{n-1}}{(n-1)!}$ if you want to use integration. We then obtain by using integration by parts
\begin{align*}
\int \underbrace{x}_{u}\underbrace{\sum_{n=1}^\infty \frac{x^{n-1}}{(n-1)!}}_{v^\prime}\,dx
&= x\sum_{n=1}^\infty \frac{x^{n}}{n!}-\int \sum_{n=1}^\infty \frac{x^{n}}{n!}\,dx\\
&=x\left(e^x-1\right)-\int \left(e^x-1\right)\,dx\\
&=x\left(e^x-1\right)-e^x+x+C\\
&=xe^x-e^x+C\tag{1}
\end{align*}
and differentiation of (1) gives
\begin{align*}
\frac{d}{dx}\left(xe^x-e^x+C\right)&=\left(e^x+xe^x\right)-e^x\\
&=xe^x
\end{align*}
as wanted.
More convenient is reindexing the series. This way we obtain
\begin{align*}
\color{blue}{\sum_{n=1}^\infty\frac{x^n}{(n-1)!}}&=\sum_{n=0}^\infty \frac{x^{n+1}}{n!}=x\sum_{n=0}^\infty\frac{x^n}{n!}
\color{blue}{=xe^x}
\end{align*}
|
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"timestamp": "2023-03-29T00:00:00",
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|
Inequality in number theory Prove that:
(i)$5<5^{\frac{1}{2}}+5^{\frac{1}{3}}+5^{\frac{1}{4}}$
(ii)$8>8^{\frac{1}{2}}+8^{\frac{1}{3}}+8^{\frac{1}{4}}$
(iii)$n>n^{\frac{1}{2}}+n^{\frac{1}{3}}+n^{\frac{1}{4}}$ for all integer $n\geq9$
Can raising both sides to exponent $12$ help
|
HINT: For (ii) we have to prove that
$$6>2\sqrt{2}+\sqrt[4]{8}$$
This can be written as
$$6-2\sqrt{2}>\sqrt[4]{8}$$
Raise to the power $4$
we have
$$3080>2112\sqrt{2}$$
Squaring we get
$$565312>0$$
For (i) we get by $$AM-GM$$
$$\frac{5^{1/2}+5^{1/3}+5^{1/4}}{3}\geq \sqrt[3]{5^{13/12}}$$ so we have to Show that
$$3\sqrt[3]{5^{13/12}}>5$$ this is equivalent to $$27^{12}>5^{23}$$ and this is
$$150094635296999121>11920928955078125$$
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/2796837",
"timestamp": "2023-03-29T00:00:00",
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|
Finding factors of second order complex polynomial. It concerns with finding roots of complex polynomial:
$x^2+(i-1)x+(2+i)$. One way is to find roots by factoring, as:$(x-i)(x -(1-2i))=(x-i)(x-1+2i)=(x^2 -x +ix +i +2)=x^2 +x(i-1)+(i+2)$
But, this is a guess game for me, with no formal process to get in this case.
The logical process is to take roots of the quadratic equation, but I hope the quadratic formula does not work for complex coefficients.
Update -- Based on comments by @WillJagy, the square root is $\pm1\pm3i$, hence the equation is having roots as :$\frac{-(i-1)\pm\sqrt{-6i-8}}2\implies \frac{(1-i)\pm(\pm 1\pm 3i)}2$ can have possible values: (i) for $1+3i$, get: $x_{11} = 1+i, x_{12} = -2i$
(ii) for $1-3i$, get: $x_{21} = 1-2i, x_{22} = i$
(iii) for $-1+3i$, get: $x_{11} = i, x_{21} = 1-2i$
(iv) for $-1-3i$, get: $x_{11} = -2i, x_{21} = 1+i$
So, the $4$ root pairs should be tried one by one:
(i) $1+3i =>(x-1-i)(x+2i) = x^2 +ix -x -2i +2$, mismatch
(ii) $1-3i => (x-i)(x-1+2i) = x^2 -x +ix +i +2$, matches original equation
(iii) same as for (ii),
(iv) same as for (i)
Why only two root pairs match, and the other don't?
Some answers on mse : 1
Update 2 -- To add to the answer by @Skip, where the fourth quadrant angle was transformed to the first one by taking out $\sqrt{-1} = i$ as common factor. For mixed sign discriminant, let $D = -6 +8i$, then angle cannot be taken\manipulated to be in the first quadrant; & can lie in the 2nd or 3rd quadrant.
In 2nd quad., $\sin$ of negative angle is positive, while in 3rd quad. both $\sin, \cos$ are negative.
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Let $z=a+bi, a,b, \in \mathbb{R}$, where $$z^2 = -8-6i$$
$$(a+bi)^2=(-8-6i)$$
$$a^2-b^2+2abi=-8-6i$$
that is we have
$$a^2-b^2=-8\tag{1}$$
$$ab=-3\tag{2}$$
Substituting $(2)$ into $(1)$,
$$a^2-\frac{9}{a^2}=-8$$
$$a^4+8a^2-9=0$$
$$(a^2-9)(a^2+1)=0$$
$$a=\pm3$$
when $a=3$, from $ab=-3$,$b=-1$.
when $a=-3$, from $ab=-3$, $b=1$
There are only two solutions for $z^2=-8-6i$, $z_1 = 3-i$ and $z_2 = -3+i$ and we have $z_2 = -z_1$. That is the solutions to $z^2 = -8-6i$ are $\pm(3-i)$
The solution to the original problem is
$$x=\frac{(1-i) \pm z_1}{2}$$
You might have solved for $a$ and substitute it into $(1)$ and thought that for each $a$, there are $2$ possible values of $b$'s. However, remember that you have to check all the constraitns, constraint $(2)$ tells us that for each non-zero $a$, there is exactly one $b$.
You shold be expecting $2$ roots counting multiplicity for a quadratic equation from fundamental theorem of algebra.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Proof that continued fractions are finite for rationals? How does one prove that the continued fraction representations of rational numbers are finite?
For every $x\in\mathbb{R}$, the (simple) continued fraction representation of $x$ is:
$$
x = [a_0; a_1, a_2, ...] = a_0 + \frac{1}{a_1 + \frac{1}{a_2 + \frac{1}{...}}}
$$
where $a_0\in\mathbb{Z}$ and $a_k\in\mathbb{N}$ for $k\geq 1$, which are themselves obtained as follows:
$$
\begin{align}
r_0 &= x \\
\forall k \geq 0,\quad a_k &= \lfloor r_k \rfloor \\
\forall k \geq 0,\quad r_{k+1} &=
\begin{cases}
1 / (r_k-a_k) & \text{if } r_k > a_k \\
0 & \text{otherwise}
\end{cases}
\end{align}
$$
and if there exists $n$ for which $r_n > r_{n+1} = 0$, then we correct $a_n\mapsto
a_n-1$.
Clearly if the sequence $a_k$ converges to 0, then $x$ is rational. But the converse does not seem trivial at all; why does this recursion necessarily terminate if $x = p/q$? Contraposition does not seem evident to me here either. Is there another way to think about this?
|
it is the Euclidean Algorithm, that is all. Lots of people use "back-substitution" to finish the Extended Algorithm and find the Bezout combination, I prefer to write this as a continued fraction.
$$ \gcd( 12345, 1601 ) = ??? $$
$$ \frac{ 12345 }{ 1601 } = 7 + \frac{ 1138 }{ 1601 } $$
$$ \frac{ 1601 }{ 1138 } = 1 + \frac{ 463 }{ 1138 } $$
$$ \frac{ 1138 }{ 463 } = 2 + \frac{ 212 }{ 463 } $$
$$ \frac{ 463 }{ 212 } = 2 + \frac{ 39 }{ 212 } $$
$$ \frac{ 212 }{ 39 } = 5 + \frac{ 17 }{ 39 } $$
$$ \frac{ 39 }{ 17 } = 2 + \frac{ 5 }{ 17 } $$
$$ \frac{ 17 }{ 5 } = 3 + \frac{ 2 }{ 5 } $$
$$ \frac{ 5 }{ 2 } = 2 + \frac{ 1 }{ 2 } $$
$$ \frac{ 2 }{ 1 } = 2 + \frac{ 0 }{ 1 } $$
Simple continued fraction tableau:
$$
\begin{array}{cccccccccccccccccccc}
& & 7 & & 1 & & 2 & & 2 & & 5 & & 2 & & 3 & & 2 & & 2 & \\
\frac{ 0 }{ 1 } & \frac{ 1 }{ 0 } & & \frac{ 7 }{ 1 } & & \frac{ 8 }{ 1 } & & \frac{ 23 }{ 3 } & & \frac{ 54 }{ 7 } & & \frac{ 293 }{ 38 } & & \frac{ 640 }{ 83 } & & \frac{ 2213 }{ 287 } & & \frac{ 5066 }{ 657 } & & \frac{ 12345 }{ 1601 }
\end{array}
$$
$$ $$
$$ 12345 \cdot 657 - 1601 \cdot 5066 = -1 $$
|
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"url": "https://math.stackexchange.com/questions/2804921",
"timestamp": "2023-03-29T00:00:00",
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|
Solving for the determinant only given one column of values. Given
$$ \det
\begin{bmatrix}
a & 1 & d \\
b & 1 & e \\
c & 1 & f \\
\end{bmatrix}
= 4
$$
and
$$ \det
\begin{bmatrix}
a & 1 & d \\
b & 2 & e \\
c & 3 & f \\
\end{bmatrix}
= -2
$$
I am asked to find
$$ \det
\begin{bmatrix}
a & 8 & d \\
b & 8 & e \\
c & 8 & f \\
\end{bmatrix}
$$
along with,
$$ \det
\begin{bmatrix}
a & 4 & d \\
b & 5 & e \\
c & 6 & f \\
\end{bmatrix}
$$
How would I go about doing this, I understand that the first one would just be 32 since when any row (or column) is multiplied by a scalar the determinant is multiplied by the same value. How do I find the determinant of the second matrix?
|
You're right about the first question. For the second one, note that\begin{align}\det\begin{bmatrix} a & 4 & d \\ b & 5 & e \\ c & 6 & f \\ \end{bmatrix}&=\det\begin{bmatrix} a & 3+1 & d \\ b & 3+2 & e \\ c & 3+3 & f \\ \end{bmatrix}\\&=\det\begin{bmatrix} a & 3 & d \\ b & 3 & e \\ c & 3 & f \\ \end{bmatrix}+\det\begin{bmatrix} a & 1 & d \\ b & 2 & e \\ c & 3 & f \\ \end{bmatrix}.\end{align}
|
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"timestamp": "2023-03-29T00:00:00",
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|
Which values for $x$ make $x(x + 180)$ a square?
Which values for $x$ make $x(x + 180)$ a square?
$x = 12, 16, 60$ are some values.
Then maybe solving $x(x + 180) = y^2 $ would give other values?
Tried using the general formula, but that would only give 2 values each time I find a suitable $y$, not all values. And I am stuck in the process:
$x(x + 180) = y^2 \Rightarrow x^2 + 180x - y^2 = 0.$
$x = \frac{-180 \pm \sqrt{180^2 +4y^2}}{2} = \frac{-180 \pm \sqrt{8100 +y^2}}{2}$ And then I beleive that I should find a sum of squares that is a square, and that gives a positive $x.$
|
Note that $$y^2=x(x+180)=(x+90-90)(x+90+90)=(x+90)^2-90^2$$
and then $$90^2=(x+90)^2-y^2=(x+90+y)(x+90-y) $$
As the factors on th right have the same parity (differ by $2y$) and the left is even, we are looking for factorizations of $90^2$ into even factors. Any such factorization $90^2=(2u)(2v)$ (or simply $45^2=uv$) with wlog $u\ge v$ leads to valid solution $y=u-v$ and $x=u+v-90$. (Don't forget to allow negative $u,v$)
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
Find $x+y$, if integers $x$ and $y$ satisfy the equation $y+1/x=25/3$ Find $x+y$, if $x$ and $y$ are integers and satisfy the equation $y+1/x=25/3$
so I got to the answer by placing $3$ in the $x$ cause it looked like a fraction and $8$ was left for $y$,
My question is: Is there any other/better way to solve this algebraically?
|
We have: $y = 8+\dfrac{x-3}{3x}\implies \dfrac{x-3}{3x}\in \mathbb{Z}\implies x-3=3nx\implies x-3nx=3\implies(3n-1)x=-3\implies x\mid 3 \implies x = \pm 1, \pm 3$. Since $y$ is an integer, $x = 3$. Thus $y = 8$, and $x+y = 3+8 = 11$.
|
{
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"url": "https://math.stackexchange.com/questions/2810517",
"timestamp": "2023-03-29T00:00:00",
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|
Cool way of finding $\cos\left(\frac{\pi}{5}\right)$ while trying to solve a problem, I stumbled upon a way of finding $\cos\left(\frac{\pi}{5}\right)$ using identities and the cubic formula. Is it possible to find other values of sine or cosine in a similar way ?
Consider
$$\cos\left(\frac{\pi}{5}\right) - \cos\left(\frac{2\pi}{5}\right).$$
Using the difference of cosines identity, we have
$$\cos\left(\frac{\pi}{5}\right) - \cos\left(\frac{2\pi}{5}\right) = -2\sin\left(\frac{3\pi}{10}\right)\sin\left(-\frac{\pi}{10}\right).$$
Now we change the RHS using the identity $\sin(x) = \cos\left(\frac{\pi}{2}-x\right)$ and the fact that $\sin(x)$ is odd.
$$-2\sin\left(\frac{3\pi}{10}\right)\sin\left(-\frac{\pi}{10}\right) = 2\cos\left(\frac{\pi}{5}\right) \cos\left(\frac{2\pi}{5}\right)$$
So,
$$\cos\left(\frac{\pi}{5}\right) - \cos\left(\frac{2\pi}{5}\right) = 2\cos\left(\frac{\pi}{5}\right) \cos\left(\frac{2\pi}{5}\right)$$
Now we make use of the identity $\cos(2x)=2\cos^2(x)-1$.
$$\cos\left(\frac{\pi}{5}\right) - 2\cos^2\left(\frac{\pi}{5}\right)+1 = 2\cos\left(\frac{\pi}{5}\right) \left(2\cos^2\left(\frac{\pi}{5}\right)-1\right)$$
Let $y=\cos\left(\frac{\pi}{5}\right)$ and we have
$$y-y^2+1=2y(2y^2-1)$$
$$4y^3+2y^2-3y-1=0$$
which has the correct solution
$$ y=\frac{\sqrt{5}+1}{4} =\cos\left(\dfrac{\pi}{5}\right) $$
One of the roots is also $\sin\left(\dfrac{\pi}{10}\right)$ which I'm guessing is because you end up with the same cubic if you apply the above to sin too.
|
Everything begins with finding $\sin\left(\frac{\pi}{10}\right)$
Okay... Lets say $x=\frac{\pi}{10}$
$$5x=\frac{\pi}{2}$$
$$2x=\frac{\pi}{2}-3x$$
$$\sin(2x)=\sin\left(\frac{\pi}{2}-3x\right)$$
$$\sin(2x)=\cos(3x)$$
$$2\sin(x)\cos(x)=4\cos^3(x)-3\cos(x)$$
$$2\sin(x)\cos(x)-4\cos^3(x)+3\cos(x)=0$$
$$\cos(x)\left(2\sin(x)-4\cos^2(x)+3\right)=0$$
$$2\sin(x)-4\cos^2(x)+3=0$$
$$2\sin(x)-4(1-sin^2(x))+3=0$$
$$2\sin(x)-4+4\sin^2(x)+3=0$$
$$4\sin^2(x)+2\sin(x)-1=0$$
Now use the formula $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$
$$a=4,b=2,c=-1$$
$$=\frac{-2\pm \sqrt{2^2-4(4)(-1)}}{2(4)}$$
$$=\frac{-2\pm \sqrt{20}}{8}$$
$$\sin\left(\frac{\pi}{10}\right)=\frac{-1\pm \sqrt{5}}{4}$$
One another way of finding the value of $\cos\left(\frac{\pi}{5}\right)$
$$\cos\left(\frac{\pi}{5}\right)=\cos\left(2\left(\frac{\pi}{10}\right)\right)$$
$$=1-2\left(\sin\frac{\pi}{10}\right)^2$$
Since $\sin\left(\frac{\pi}{10}\right)=\frac{\sqrt{5}-1}{4}$
$$=1-2\left(\frac{\sqrt{5}-1}{4}\right)^2$$
$$=1-\frac{(\sqrt{5}-1)^2}{8}$$
$$=\frac{8-5-1+2\sqrt{5}}{8}$$
$$=\frac{2+2\sqrt{5}}{8}$$
$$\cos\left(\frac{\pi}{5}\right)=\frac{1+\sqrt{5}}{4}$$
Now lets find $\sin\left(\frac{\pi}{5}\right)$
$$\sin\left(\frac{\pi}{5}\right)=\sqrt{1-\cos^2\left(\frac{\pi}{5}\right)}$$
$$=\sqrt{1-\left(\frac{1+\sqrt{5}}{4}\right)^2}$$
$$=\sqrt{1-\frac{1+5+2\sqrt{5}}{16}}$$
$$=\sqrt{\frac{10-2\sqrt{5}}{16}}$$
$$\sin\left(\frac{\pi}{5}\right)=\frac{\sqrt{10-2\sqrt{5}}}{4}$$
|
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|
Splitting up an infinite sum I am playing with the following example trying to determine if $\frac{1}{n}-\frac{1}{\sqrt{n}}$ is in $\ell^2$.
Here is what I have,
\begin{align}
\sum\limits_{n=1}^\infty \left( \frac{1}{n}-\frac{1}{\sqrt{n}}\right)^2 &= \sum\limits_{n=1}^\infty \left(\frac{\sqrt{n}-n}{n\sqrt{n}}\right)^2\\
&=\sum\limits_{n=1}^\infty\frac{n-2n\sqrt{n}+n^2}{n^3}\\
&=\sum\limits_{n=1}^\infty\frac{1}{n^2}-2\sum\limits_{n=1}^\infty\frac{1}{n\sqrt{n}}+\sum\limits_{n=1}^\infty\frac{1}{n}
\end{align}
The first summand is in $\ell^2$, the second summand is in $\ell^2$, but the third summand is not. Thus the sequence $\frac{1}{n}-\frac{1}{\sqrt{n}}$ is not in $\ell^2$. I think I may be making an error when I split the sums. Allowing such splitting would result in things like,
\begin{align}
\sum\limits_{n=1}^\infty\frac{1}{n^2} &= \sum\limits_{n=1}^\infty\frac{1-n^2}{n^2}+\sum\limits_{n=1}^\infty\frac{n^2}{n^2}
\end{align}
What am I missing here?
|
Put $a_n=\frac{1}{n}-\frac{1}{\sqrt{n}}, b_n =\frac{1}{n}$ and $c_n=\frac{1}{\sqrt{n}}$.
Suppose that $(a_n) \in \ell^2$. Since $(b_n) \in \ell^2$ and since $\ell^2$ is a vector space, we get that $(c_n)=(b_n)-(a_n) \in \ell^2$, a contradiction. Hence $(a_n) \notin \ell^2$.
|
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|
Prove that $\frac{dy}{dx}+\sec^2(\frac{\pi}{4}-x)=0$
If $y=\sqrt{\frac{1-\sin 2x}{1+\sin 2x}}$, prove that $\frac{dy}{dx}+\sec^2(\frac{\pi}{4}-x)=0$
My attempts:
Attempt 1:
$y=\sqrt{\frac{1-\sin 2x}{1+\sin 2x}}$
$\implies y=\sqrt{\frac{\sin^2 x+\cos^2 x-2\sin x \cos x}{\sin^2 x+\cos^2 x+2\sin x \cos x}}$
$\implies y=\sqrt{\frac{(\sin x -\cos x)^2}{(\sin x+\cos x)^2}}$
$\implies y=\frac{\sin x -\cos x}{\sin x +\cos x}$
$\therefore \frac{dy}{dx}=\frac{(\sin x +\cos x)(\cos x+\sin x)-(\sin x -\cos x)(\cos x -\sin x)}{(\sin x +\cos x)^2}$
$\implies \frac{dy}{dx}=\frac{(\sin x+\cos x)^2+(\sin x -\cos x)^2}{(\sin x+\cos x)^2}$
$\implies \frac{dy}{dx}=\frac{2(\sin^2 x+\cos^2 x)}{(\cos(\frac{\pi}{2}-x)+\cos x)^2}$
$\implies \frac{dy}{dx}=\frac{2}{(2\cos(\frac{\pi}{4})\cos(\frac{\pi}{4}-x))^2}$
$\implies \frac{dy}{dx}=\frac{2}{4\cos^2(\frac{\pi}{4})\cos^2(\frac{\pi}{4}-x)}$
$\implies \frac{dy}{dx}=\frac{2}{4\times\frac{1}{2}\cos^2(\frac{\pi}{4}-x)}$
$\implies \frac{dy}{dx}=\sec^2(\frac{\pi}{4}-x)$
$\implies \frac{dy}{dx}-\sec^2(\frac{\pi}{4}-x)=0$
But i have to prove $\frac{dy}{dx}+\sec^2(\frac{\pi}{4}-x)=0$
Attempt 2:
$y=\sqrt{\frac{1-\sin 2x}{1+\sin 2x}}$
$\implies y=\sqrt{\frac{\cos^2 x+\sin^2 x-2\cos x \sin x}{\cos^2 x+\sin^2 x+2\cos x \sin x}}$
$\implies y=\sqrt{\frac{(\cos x -\sin x)^2}{(\cos x+\sin x)^2}}$
$\implies y=\frac{\cos x -\sin x}{\cos x +\sin x}$
$\therefore \frac{dy}{dx}=\frac{(\cos x +\sin x)(-\sin x-\cos x)-(\cos x -\sin x)(-\sin x +\cos x)}{(\cos x +\sin x)^2}$
$\implies \frac{dy}{dx}=\frac{-(\cos x+\sin x)^2+(\cos x -\sin x)^2}{(\cos x+\sin x)^2}$
$\implies \frac{dy}{dx}=\frac{(\cos x -\sin x)^2-(\cos x+\sin x)^2}{(\cos x+\sin x)^2}$
$\implies \frac{dy}{dx}=\frac{-4\sin x\cos x}{(\cos x+\cos(\frac{\pi}{2}-x))^2}$
$\implies \frac{dy}{dx}=\frac{-4\sin x\cos x}{(2\cos(\frac{\pi}{4})\cos(x-\frac{\pi}{4}))^2}$
$\implies \frac{dy}{dx}=\frac{-4\sin x\cos x}{4\cos^2(\frac{\pi}{4})\cos^2(x-\frac{\pi}{4})}$
$\implies \frac{dy}{dx}=\frac{-4\sin x\cos x}{4\times \frac{1}{2}\cos^2(x-\frac{\pi}{4})}$
$\implies \frac{dy}{dx}=\frac{-2\sin x\cos x}{\cos^2(x-\frac{\pi}{4})}$
Attempt 3:
$y=\sqrt{\frac{1-\sin 2x}{1+\sin 2x}}$
$\implies y=\sqrt{\frac{(1-\sin 2x)(1-\sin 2x)}{(1+\sin 2x)(1-\sin 2x)}}$
$\implies y=\frac{1-\sin 2x}{\cos 2x}$
$\therefore \frac{dy}{dx}=\frac{\cos 2x(-2\cos 2x)-(1-\sin 2x)(-2\sin 2x)}{\cos^2 2x}$
$\implies \frac{dy}{dx}=\frac{-2\cos^2 2x+2\sin 2x(1-\sin 2x)}{\cos^2 2x}$
My questions:
(i) Why does Attempt 2 give a result different from that given by Attempt 1.
(ii) How do i prove the result?
|
In fact, the result you want is not true: it depends on the values of $x$.
Having said this, with a small modification, any of your methods should work. I'll use your third method. Note that you have
$$y=\sqrt{\frac{(1-\sin2x)^2}{\cos^22x}}\ ;$$
this gives
$$y=\left|\frac{1-\sin2x}{\cos2x}\right|\ ,$$
and the omission of the absolute values is your main problem.
First, suppose that $-\pi/4<x<\pi/4$. Then $1-\sin2x>0$ and $\cos2x>0$; so
$$y=\frac{1-\sin2x}{\cos2x}\ .$$
Your differentiation is correct, and you can continue
$$\eqalign{\frac{dy}{dx}
&=\frac{-2\cos^2 2x+2\sin 2x(1-\sin 2x)}{\cos^2 2x}\cr
&=\frac{2\sin2x-2}{\cos^22x}\cr
&=-2\,\frac{1-\sin2x}{1-\sin^22x}\cr
&=-\frac2{1+\sin2x}\cr
&=-\frac2{1+\cos(\frac\pi2-2x)}\cr
&=-\frac1{\cos^2(\frac\pi4-x)}\cr
&=-\sec^2\Bigl(\frac\pi4-x\Bigr)\cr}$$
which is what you want.
If, however $\pi/4<x<3\pi/4$, then $1-\sin2x>0$ and $\cos2x<0$; so
$$y=-\frac{1-\sin2x}{\cos2x}$$
and exactly the same working will give
$$\frac{dy}{dx}=\sec^2\Bigl(\frac\pi4-x\Bigr)\ .$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
What is the dimension of the vector space consisting of all 3 by 3 symmetric matrices? What is a basis for it? Could you please explain the following and give me a meticulous solution:"What is the dimension of the vector space consisting of all 3 by 3 symmetric matrices? What is a basis for it?"
Solution till now:
Let Matrix $$A=\begin{pmatrix}a&b&c\\d&e&f\\h&i&j\end{pmatrix}$$
Since it's symmetric $c=h$, $d=b$, $f=i$
$$A=\begin{pmatrix}a&b&c\\b&e&i\\c&i&j\end{pmatrix}$$
$A$ could be written as $$A=a \times\begin{pmatrix}1&0&0\\0&0&0\\0&0&0\end{pmatrix} +b\times\begin{pmatrix}0&1&0\\1&0&0\\0&0&0\end{pmatrix}+c\times\begin{pmatrix}0&0&1\\0&0&0\\1&0&0\end{pmatrix}+e\times\begin{pmatrix}0&0&0\\0&1&0\\0&0&0\end{pmatrix}+i\times\begin{pmatrix}0&0&0\\0&0&1\\0&1&0\end{pmatrix}+j\times\begin{pmatrix}0&0&0\\0&0&0\\0&0&1\end{pmatrix}$$
What's the conclusion?
Is it six dimentional matrix?
|
Hint:
Any symmetric matrix $\mathbf{A}\in M_{3,3}(\mathbb{R})$ can be written as
$$\mathbf{A}=\begin{pmatrix}a&b&c\\b&d&e\\c&e&f\end{pmatrix}$$
Can you write $\mathbf{A}$ as a sum of some matrices multiplied by some scalars?
You have found that any symmetric matrix may be written $$A=a\begin{pmatrix}1&0&0\\0&0&0\\0&0&0\end{pmatrix} +b\begin{pmatrix}0&1&0\\1&0&0\\0&0&0\end{pmatrix}+c\begin{pmatrix}0&0&1\\0&0&0\\1&0&0\end{pmatrix}+e\begin{pmatrix}0&0&0\\0&1&0\\0&0&0\end{pmatrix}+i\begin{pmatrix}0&0&0\\0&0&1\\0&1&0\end{pmatrix}+j\begin{pmatrix}0&0&0\\0&0&0\\0&0&1\end{pmatrix}\tag{1}$$ The trick is to realize that the set of all symmetric matrices forms a vector space, and that you may consider each of the $6$ matrices as vectors. To emphasize this, $(1)$ may be written as $$A=aM_1+bM_2+cM_3+eM_4+iM_5+jM_6$$ Next thing is to note that $M_1,M_2,M_3,M_4,M_5,M_6$ are all linearly independent, because the only way to have $A$ be the zero-matrix is precisely when $a=b=c=e=i=j=0$. It therefore follows that $\{M_1,M_2,M_3,M_4,M_5,M_6\}$ is a basis for your vector space, and the dimension is $6$.
|
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|
integral $\int \frac{1-a\cos x}{1+a^2-2a\cos x}dx$ I'm trying to solve this integral: $$\int \frac{1-a\cos x}{1+a^2-2a\cos x}dx.$$
My trouble: Using software, i found that if $|a|>1$, the integral is equal to $$\arctan \left(\frac{x}{2}\right)+\frac{1-a^2}{|a-1||a+1|}\arctan \left(\frac{|a+1|}{|a-1|}\tan \left( \frac{x}{2} \right) \right )+C.$$
If $|a|<1$, then the antiderivate is $$x-\arctan \left(\frac{x}{2}\right)+\frac{1-a^2}{|a-1||a+1|}\arctan \left(\frac{|a+1|}{|a-1|}\tan \left( \frac{x}{2} \right) \right )+C.$$
Integrating over the interval $[0,2\pi]$, the integral is equal to $2\pi$ if $|a|<1$ and $0$ if $|a|>1$. Using some skills, like this:
$$\int \frac{1-a\cos x}{1+a^2-2a\cos x}dx =\frac{1}{2}\int \frac{2-2a\cos x+a^2-a^2}{1+a^2-2a\cos x}dx $$
$$=\frac{1}{2}\int dx+\frac{1}{2}\int \frac{1-a^2}{1+a^2-2a\cos x}dx$$
I found the answer $$\frac{x}{2}+\frac{1-a^2}{|a-1||a+1|}\arctan \left(\frac{|a+1|}{|a-1|}\tan \left( \frac{x}{2} \right) \right )+C.$$
Integrating this over the same interval as before, the result is not correct. So, my question is: Using only real analysis, is it possible to calculate this integral? If the answer is yes, how? What makes $x-\arctan (x/2)$ show up instead of $x/2$ (and the same for only $\arctan (x/2)$)?
|
Your antiderivative is correct for $\frac{|x|}2\le \pi/2$. Thus,
$$\begin{align}
\int_0^{2\pi}\frac{1-a\cos(x)}{a^2+1-2a\cos(x)}\,dx&=\int_{-\pi}^\pi \frac{1-a\cos(x)}{a^2+1-2a\cos(x)}\,dx\\\\
&=\left.\left(\frac x2+\text{sgn}(1+a)\text{sgn}(1-a)\arctan\left(\left|\frac{a+1}{a-1}\right|\tan(x/2)\right)\right)\right|_{-\pi}^\pi\\\\
&=\pi\left(1+\text{sgn}(1+a)\text{sgn}(1-a)\right)\\\\
&=\begin{cases}2\pi&,|a|<1\\\\0&,|a|>1\end{cases}
\end{align}$$
as expected!
|
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|
Simplify: $\frac{1+\cos2x}{1-\cos x}-\frac{4\cos^2(\frac{x}{2})}{\tan^2x}$ Simplify:
$$\frac{1+\cos2x}{1-\cos x}-\frac{4\cos^2(\frac{x}{2})}{\tan^2x}$$
So what I've tried was:
$$\frac{1+\cos2x}{1-\cos x}-\frac{4\cos^2(\frac{x}{2})}{\tan^2x}=\frac{1+\cos^2x-\sin^2x}{1-\cos x}-\frac{4\cos^2(\frac{x}{2})}{\tan^2x}=$$
$$=\frac{2\cos^2x}{1-\cos x}-\frac{4\cos^2(\frac{x}{2})}{\tan^2x}=\frac{2\cos^2x\,\tan^2x-4\cos^2(\frac{x}{2})(1-\cos x)}{(1-\cos x)(\tan^2x)}=$$
$$=\frac{2\sin^2x-4\cos^2(\frac{x}{2})+4\cos^2(\frac{x}{2})\cos x}{(1-\cos x)(\tan^2x)}=\frac{2\cdot2\sin^2(\frac{x}{2})\cos^2(\frac{x}{2})-4\cos^2(\frac{x}{2})+4\cos^2(\frac{x}{2})\cos x}{(1-\cos x)(\tan^2x)}=$$
$$=\frac{4\cos^2(\frac{x}{2})(\sin^2(\frac{x}{2})-1+\cos x)}{(1-\cos x)(\tan^2x)}=\frac{4\cos^2(\frac{x}{2})(\sin^2(\frac{x}{2})-1+\cos^2(\frac{x}{2})-\sin^2(\frac{x}{2}))}{(1-\cos x)(\tan^2x)}=$$
$$=\frac{4\cos^2(\frac{x}{2})(-1+\cos^2(\frac{x}{2}))}{(1-\cos x)(\tan^2x)}=\frac{4\cos^2(\frac{x}{2})(-\sin^2(\frac{x}{2}))}{(1-\cos x)(\tan^2x)}=\frac{-2\sin^2(x)}{(1-\cos x)(\tan^2x)}$$
This is farthest I got but the answer is supposed be: $0$,
according to the solutions, I might've missed something on the process even though I checked it a couple of time..
I'm preparing for the Tel-Aviv univeristy math entry test and I took this exercise from math-exercises.com to practise (that is where the solution is from).
|
$$u=\frac{1+cos2x}{1-cosx}-\frac{4cos^2(\frac{x}{2})}{tan^2x}$$
Using $\tan(x/2)=t$, $\sin x=\dfrac{2t}{1+t^2}$ and $\cos x=\dfrac{1-t^2}{1+t^2}$
$$U=\dfrac{2\cos^2 x}{1-cosx}-2\dfrac{(1+\cos^2x)}{\tan^2x}$$
Then
$$U=2\dfrac{(1-t^2)^2(1+t^2)}{(1+t^2)^22t^2}-2\dfrac{\frac{2}{1+t^2}}{4t^2}(1-t^2)^2$$
Eventually:
$$U=\dfrac{(1-t^2)^2}{(1+t^2)t^2}-\dfrac{(1-t^2)^2}{(1+t^2)t^2}=0$$
|
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|
Hw to find the two second order linear recursive common items? Given two sequences
$$a(n)=4a(n-1)-a(n-2), \quad a(1)=1，\; a(2)=5;$$
$$b(n)=10b(n-1)-b(n-2), \quad b(1)=1，\; b(2)=11.$$
If some $a(n)=b(m)$, $m,n\in\mathbb{N}$, exist?(Except a(1)=b(1) ) If there are no common items, how to prove it?
The Two solution equation follows from wolfram alpha.
Thanks a lot, I love you master!
|
Linear difference equations are much like differential equations
To solve
$$
a_n = 4 a_{n-1}-a_{n-2},\: a_1=1,\; a_2 = 5
$$
we assume that $a_n = \alpha^n$ and substituting results in
$$
(1-4\alpha+\alpha^2)\alpha^n = 0
$$
now solving
$$
1-4\alpha+\alpha^2 = 0
$$
gives
$$
a_n = c_1(-2-\sqrt5)^n+c_1(-2+\sqrt5)^n
$$
Analogously for
$$
b_n = c_3(5-2\sqrt6)^2+c_4(5+\sqrt6)^n
$$
Considering the initial conditions we have
$$
a_n =-\frac{\left(1+\sqrt{3}\right) \left(2 \left(2-\sqrt{3}\right)^n+\sqrt{3} \left(2-\sqrt{3}\right)^n-\left(2+\sqrt{3}\right)^n\right)}{2
\left(2+\sqrt{3}\right)} = \frac{1}{2} \left(1+\sqrt{3}\right) \left(\left(2+\sqrt{3}\right)^{n-1}-\left(2-\sqrt{3}\right)^n\right)
$$
and
$$
b_n = -\frac{\left(2+\sqrt{6}\right) \left(5 \left(5-2 \sqrt{6}\right)^n+2 \sqrt{6} \left(5-2 \sqrt{6}\right)^n-\left(5+2 \sqrt{6}\right)^n\right)}{4
\left(5+2 \sqrt{6}\right)} = \frac{1}{4} \left(2+\sqrt{6}\right) \left(\left(5+2 \sqrt{6}\right)^{n-1}-\left(5-2 \sqrt{6}\right)^n\right)
$$
I hope this helps.
NOTE
$$
a_n = \eta_n + \gamma_n\sqrt 3\\
b_n =\xi_n+\mu_n\sqrt6
$$
|
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|
Number of ways to distribute 10 things among 6 people given that the number of things given to two people doesn't exceed 4? Here is a more specific question:
Find the number of ways of giving $10$ identical gift boxes to 6
people : $A$, $B$, $C$, $D$, $E$, $F$ in such a way that total number
of boxes given to $A$ and $B$ together does not exceed $4$.
I am currently learning about problems which are related to Combinations with Repetitions, for the most part of it I am able to solve the basic questions, but this one stumped me.
Can you suggest how to approach this problem?
|
Let $x_A$, $x_B$, $x_C$, $x_D$, $x_E$, and $x_F$ represent the number of gift boxes given to persons $A$, $B$, $C$, $D$, $E$, and $F$, respectively. Since a total of ten boxes are distributed to these six people,
$$x_A + x_B + x_C + x_D + x_E + x_F = 10 \tag{1}$$
Since $A$ and $B$ together receive at most four of these gifts, we must solve equation 1 in the nonnegative integers subject to the restriction that $x_A + x_B \leq 4$.
This can be solved by casework. If
$$x_A + x_B = k \tag{2}$$
then for equation 1 to be satisfied, we must have
$$x_C + x_D + x_E + x_F = 10 - k \tag{3}$$
for $0 \leq k \leq 4$.
A particular solution of the equation
$$x_1 + x_2 + \cdots + x_n = m \tag{4}$$
in the nonnegative integers corresponds to the placement of $n - 1$ addition signs in a row of $m$ ones. For instance, if $m = 10$ and $n = 5$,
$$1 + + 1 1 1 + 1 1 + 1 1 1 1$$
corresponds to the solution $x_1 = 1$, $x_2 = 0$, $x_3 = 3$, $x_4 = 2$, $x_5 = 4$. Consequently, the number of solutions of equation 4 in the nonnegative integers is
$$\binom{m + n - 1}{n - 1}$$
since we must choose which $n - 1$ of the $m + n - 1$ positions required for $m$ ones and $n - 1$ addition signs will be filled with addition signs.
Thus, equation 2 has
$$\binom{k + 2 - 1}{2 - 1} = \binom{k + 1}{1} = k + 1$$
solutions and equation 3 has
$$\binom{10 - k + 4 - 1}{4 - 1} = \binom{13 - k}{3}$$
solutions in the nonnegative integers. Thus, when $x_A + x_B = k$ and $x_C + x_D + x_E + x_F = 10 - k$, the number of solutions of equation 1 is
$$(k + 1)\binom{13 - k}{3}$$
Since $0 \leq k \leq 4$, the number of ways of distributing ten gifts to persons $A$, $B$, $C$, $D$, $E$, and $F$ so that $A$ and $B$ together receive at most $4$ of those gifts is
$$\sum_{k = 0}^{4} (k + 1)\binom{13 - k}{3} = \binom{13}{3} + 2\binom{12}{3} + 3\binom{11}{3} + 4\binom{10}{3} + 5\binom{9}{3}$$
|
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|
Probability that all three shirts were worn Mr. Jones has three shirts: Red, Green and White. Each day he picks randomly a red shirt with probability of $\frac{1}{2}$, green with probability of $\frac{1}{3}$ and white with probability of $\frac{1}{6}$.
What is the probability that he wears all three shirts after 6 days?
My try:
$$\binom{6}{1}\binom{5}{1}\binom{4}{1}\frac{1}{2}\cdot\frac{1}{3}\cdot\frac{1}{6}\cdot[\binom{3}{2}(\frac{1}{2})^2\cdot[\frac{1}{3}+\frac{1}{6}]+[\binom{3}{2}(\frac{1}{3})^2\cdot[\frac{1}{2}+\frac{1}{6}]+[\binom{3}{2}(\frac{1}{6})^2\cdot[\frac{1}{2}+\frac{1}{3}]+\binom{3}{1}\binom{2}{1}\frac{1}{2}\cdot\frac{1}{3}\cdot\frac{1}{6}]$$
First, we pick possessions for red, green and white. Then, we multiply it by the other combinations possible of the other three possessions left, and summing it up (I didn't count all possibilities). The problem is the answer is bigger than 1, so it must be wrong. Also, I guess there is more elegant way.
Thanks in advance.
|
Let $R$ denote the event that he wears a red shirt in the $6$ days.
Let $G$ denote the event that he wears a green shirt in the $6$ days.
Let $W$ denote the event that he wears a white shirt in the $6$ days.
To be found is: $$P(R\cap G\cap W)=1-P(R^{\complement}\cup G^{\complement}\cup W^{\complement})=$$$$1-P(R^{\complement})-P(G^{\complement})-P(W^{\complement})+P(R^{\complement}\cap G^{\complement})+P(R^{\complement}\cap W^{\complement})+P(G^{\complement}\cap W^{\complement})=$$$$1-\left(\frac36\right)^6-\left(\frac46\right)^6-\left(\frac56\right)^6+\left(\frac16\right)^6+\left(\frac26\right)^6+\left(\frac36\right)^6$$
|
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|
Is the function $f(x,y) = \frac{xy^3}{x^4+y^2}$ a $C^1$-function? Consider the function
$$
f(x,y) = \frac{xy^3}{x^4+y^2}.
$$
with $f(0,0) = 0$.
Is this a $C^1$-function?
Firstly we compute:
\begin{align*}
& D_1f(x,y) = \frac{y^3(x^4+y^2)-xy^3(4x^3)}{(x^4+y^2)^2} = \frac{y^3(y^2-3x^4)}{(x^4+y^2)^2}, \\
& D_2f(x,y) = \frac{3xy^2(x^4+y^2)-xy^3(2y)}{(x^4+y^2)^2} = \frac{xy^2(3x^4+y^2)}{(x^4+y^2)^2},
\end{align*}
and $D_1f(0,0)= D_2f(0,0) = 0$. The function is differentiable at $(0,0)$ with $f'(\vec{0}) = [0,0]$, since we can verify that:
\begin{align*}
\frac{|f(x,y)|}{\|\vec{x}\|} & = \frac{|x||y|^3}{x^4+y^2} \cdot \frac{1}{\sqrt{x^2+y^2}} = \frac{|x||y|}{\frac{x^4}{y^2} + 1} \cdot \frac{1}{\sqrt{x^2+y^2}} = \frac{|x||y|}{\left( \frac{x^2}{y} \right)^2 + 1} \cdot \frac{1}{\|\vec{x}\|} \\
& \leq |x||y| \cdot \frac{1}{\|\vec{x}\|} \leq \|\vec{x}\|^2 \cdot \frac{1}{\|\vec{x}\|} = \|\vec{x}\| \to 0 \textrm{ als } \vec{x} \to \vec{0}.
\end{align*}
In order to determine if it is $C^1$, we need to determine whether the partial derivatives are continuous on the entire domain. I am trying to determine this for the point $(0,0)$, and so I need to determine
$$
\lim_{(x,y) \to (0,0)} \frac{y^3(y^2-3x^4)}{(x^4+y^2)^2},
$$
and
$$
\lim_{(x,y) \to (0,0)} \frac{xy^2(3x^4+y^2)}{(x^4+y^2)^2}.
$$
I'm stuck in this step, I don't know how to compute these limits.
|
You have
$$\left\vert \frac{y^3(y^2-3x^4)}{(x^4+y^2)^2} \right\vert \le \vert y \vert^3\frac{(y^2+3x^4)}{(x^4+y^2)^2} \le 3 \vert y \vert^3 \frac{(y^2+x^4)}{(x^4+y^2)^2}= 3 \frac{\vert y \vert^3}{x^4+y^2} \le 3\vert y \vert$$ hence
$$\lim_{(x,y) \to (0,0)} \frac{y^3(y^2-3x^4)}{(x^4+y^2)^2} = 0.$$
Using a similar argument you also get
$$\lim_{(x,y) \to (0,0)} \frac{xy^2(3x^4+y^2)}{(x^4+y^2)^2}=0.$$
This proves that the partial derivatives are continuous at the origin and that $f$ is $\mathcal C^1$.
|
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|
What is $x$ if $\sqrt{x+4+2\sqrt{x+3}}=\frac{x+8}{3}$? I need to find $x$, given that
$$\sqrt{x+4+2\sqrt{x+3}}=\frac{x+8}{3}$$
I simplified this to $x^4+14x^3+105x^2+68x-188=0$. According to Symbolab, that is not correct. That's why I'm goint to write out my attempt so you can point out where my mistake is.
My attempt
$$\sqrt{x+4+2\sqrt{x+3}}=\frac{x+8}{3}$$
$$\left(\sqrt{x+4+2\sqrt{x+3}}\right)^2=\left(\frac{x+8}{3}\right)^2$$
$$x+4+2\sqrt{x+3}=\frac{x^2+16x+64}{9}$$
$$9x+36+18\sqrt{x+3}=x^2+16x+64$$
$$-x^2-7x-28=-18\sqrt{x+3}$$
$$x^2+7x+28=18\sqrt{x+3}$$
$$(x^2+7x+28)^2=(18\sqrt{x+3})^2$$
$$x^4+7x^3+28x^2+7x^3+49x^2+196x+28x^2+196x+784=324x+972$$
$$x^4+14x^3+105x^2+68x-188=0$$
Where is my mistake? Even if this were true, I still wouldn't be able to solve it without a calculator (I can't use Rational Root Theorem on such a big numbers!).
By the way, the solution should be (again, according to Symbolab) $x \in \{1,-2\}$.
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It's $$\sqrt{1+2\sqrt{x+3}+x+3}=\frac{x+8}{3}$$ or
$$\sqrt{\left(1+\sqrt{x+3}\right)^2}=\frac{x+8}{3}$$
$$1+\sqrt{x+3}=\frac{x+8}{3}$$ or
$$\sqrt{x+3}=\frac{x+5}{3}$$ and since $x\geq-3$, it's
$$9(x+3)=(x+5)^2,$$
which gives $x=1$ or $x=-2.$.
|
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|
Limits applications in geometric problem We have the following situation
The goal is to find $$\lim_{a \to b } \frac{a-b}{c-d} $$
Thought
As $a $ tends to $b$ then we see we are gonna have a rectangle which means that $2 \alpha $ is gonna tend to $ \frac{\pi}{2}$. In other words, $\alpha $ is gonna tend to $\frac{ \pi }{4} $. Now notice
$$ a = \frac{ x }{ \tan \alpha }, b = \frac{ y }{ \tan \alpha} $$
and
$$ d = \frac{ y}{\cos \alpha}, c = \frac{x}{\sin \alpha } $$
now
$$ \lim_{a-b } \frac{ a-b}{c-d} = \lim_{\alpha \to \frac{\pi}{4} } = \frac{\frac{ x }{ \tan \alpha }- \frac{ y }{ \tan \alpha}}{\frac{ y}{\cos \alpha}- \frac{x}{\sin \alpha }} = \frac{x-y}{\frac{2}{\sqrt{2}}x - \frac{2}{\sqrt{2}} y } = \frac{ \sqrt{2} }{2} $$
Is this a correct solution? My
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It helps to first spell out what the construction is:
*
*start with the orthogonal "axes" through $O$ and pick point $A$ on the vertical axis;
*pick an angle $\alpha \lt \pi/4$ and draw $AB$ such that $B$ is on the horizontal axis and $\angle BAO = \alpha$;
*the (other) line through $A$ at angle $\alpha$ with $AB$ intersects the perpendicular in $B$ on $AB$ at $C$;
*the vertical through $C$ intersects the horizontal axis at $D$.
It follows from the above that the entire construction is uniquely defined by $a$ and $\alpha$, so the next step is to express the relevant quantities in terms of $a$ and $\alpha$ alone:
*
*$c = \dfrac{a}{\cos \alpha}\;\;(1)\;$ from right triangle $\triangle OBA$;
*$d = c \tan \alpha = \dfrac{a \sin \alpha}{\cos^2 \alpha} \;\;(2)\;$ from right triangle $\triangle BCA$ and $(1)$;
*$b = d \sin \alpha = \dfrac{a \sin^2 \alpha}{\cos^2 \alpha} = a \tan^2 \alpha \;\;(3)\;$ from right triangle $\triangle DCB$ and $(2)$;
Therefore:
$$\require{cancel}
\frac{a-b}{c-d} = \frac{\cancel{a} \left(1 - \tan^2 \alpha\right)}{\dfrac{\cancel{a}}{\cos \alpha}\left(1 - \tan \alpha\right)} = \frac{\cos \alpha\,\cancel{(1-\tan \alpha)}(1+\tan \alpha)}{\cancel{1 - \tan \alpha}} = \sin \alpha + \cos \alpha
$$
As the OP well noted $\,a \to b\,$ corresponds to $\,\alpha \to \dfrac{\pi}{4}\,$, so the limit is $\,\sin \dfrac{\pi}{4} + \cos \dfrac{\pi}{4} = \sqrt{2}\,$.
What went wrong in OP's proof started here...
$$ a = \frac{ x }{ \tan \alpha }, \color{red}{b = \frac{ y }{ \tan \alpha}} $$
The second part should rather be $\,\color{red}{b = y \cdot \tan \alpha}\,$.
Using $(3)$ above, in fact $\,y=\dfrac{b}{\tan \alpha}=\dfrac{a \tan^2 \alpha}{\tan \alpha}=a \tan \alpha = x\,$.
|
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|
Explain this convergence among Pythagorean triplets Why do the ratios of successive values of integers $a$ and $c$, where $a^{2}+(a+1)^{2}=c^{2}$, appear to converge to
$$\frac{a_{n+1}}{a_{n}},\frac{c_{n+1}}{c_{n}}\rightarrow3+2\sqrt{2}$$
I rigorously generated all the {a,c} pairs where $a^{2}+(a+1)^{2}=c^{2}$ for $a<10^6$:
{3, 5}, {20, 29}, {119, 169}, {696, 985}, {4059, 5741}, {23660, 33461}, {137903, 195025}, {803760, 1136689}...
We can see that the fit to the ratio just gets better and better...
$$\frac{c_{8}}{c_{7}}=\frac{1136689}{195025}\approx5.82842712473...$$
$$3+2\sqrt{2}\approx5.82842712475... $$
Why does this ratio emerge? And with no other pairs between that satisfy the condition?
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$$\begin{align}
a^{2}+(a+1)^{2}&=c^{2} \\
2a^2+2a+1&=c^2 \\
2\left(a+\frac{1}{2}\right)^2 +\frac{1}{2}&=c^2 \\
(2a+1)^2+1&=2c^2 \\
2c^2-(2a+1)^2&=1 \\
2c^2-d^2&=1 \quad | \quad d=2a+1
\end{align}$$
The above pell equation: $2c^2-d^2=1$ factors into $(c\sqrt 2-d)(c\sqrt 2 +d)=1$. With the initial solution being $(c_0,d_0)=(1,1)$, we have that $(\sqrt 2-1)(\sqrt 2 +1)=1$. Since we are permitted to multiply any equation by a constant we choose $(\sqrt 2-1)^2(\sqrt 2 +1)^2=(3-2\sqrt2)(3+2\sqrt 2)=1^2=1$. Doing this results in
$$\begin{align}
(c\sqrt 2-d)(3-2\sqrt 2)(c\sqrt 2 +d)(3+2\sqrt 2)&=1 \\
&\implies \\
2(3c+2d)^2-(4c+3d)^2&=1
\end{align}$$
This allows you to show iterative solutions:
$$\begin{cases}
c_{k+1}=3c_k+2d_k \\
d_{k+1}=4c_k+3d_k
\end{cases}$$
Which can be solved by the method I outlined here to produce:
$$c_k=\frac{2+\sqrt 2}{4} \left(3+2\sqrt 2\right)^k+\frac{2-\sqrt 2}{4} \left(3-2\sqrt 2\right)^k$$
note that the effect of $(3-2\sqrt 2)^k$ diminishes with increasing k, explaining your observation
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|
Find all possible relationships between $a$ and $b$
Find all possible relationships between $a\in \mathbb{R}$ and $b \in \mathbb{R}$ such that
$$\frac{a^2b^2(a+b)^2}{4}=a^2+ab+b^2+2$$
I can see that if $ab=2$, then we have equality. However, this was through observation only. Are there any other ways of determining such relationships?
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To simplify things, lets first substitute $x=ab$ and $y=a+b$, to get
$$\frac{x^2 y^2}{4} = y^2 - x + 2\\
x^2 y^2 = 4y^2 - 4x + 8 \\
(x^2-4) y^2 + 4(x - 2) = 0\\
(x-2)((x+2) y^2 + 4) = 0$$
So either $x=2$, or $(x+2)y^2+4=0$.
So $ab=2$ or $(ab+2)(a+b)^2+4=0$.
The latter case can be rearranged and factored further, using the temporary substitution $z=ab+2$:
$$(ab+2)(a+b)^2+4=0\\
z(a+b)^2+4=0\\
za^2 + zb^2 + 2abz + 4=0\\
za^2 + zb^2 + 2(z-2)z + 4=0\\
za^2 + zb^2 + z^2 + (z-2)^2=0\\
za^2 + zb^2 + z^2 + a^2b^2=0\\
(a^2+z)(b^2+z)=0\\
(a^2+ab+2)(b^2+ab+2)=0$$
This means that the original equation was actually factorisable as
$$(ab-2)(a^2+ab+2)(b^2+ab+2)=0$$
And we actually have exactly three possible relationships: $ab=2$; $a^2+ab+2=0$; and $b^2+ab+2=0$.
|
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|
Is it possible to construct three cevians so that three of the parts are equivalent? Recently I have started wondering if it is possible to construct two cevians in a triangle so that at least three of four parts which the cevians divide the triangle in are equivalent. I think the answer is yes, but do not know how to construct them.
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Yes you can! In the triangle above, our cevians are $BE$ and $CF$. The three areas that are equal are $[BFX]$, $[AFXE]$, and $[CEX]$.
The point $X$ is defined with barycentric coordinates $(1/2, 1/4, 1/4)$. In other words, we have the ratios:
$$\frac{DX}{DA} = \frac12$$
$$\frac{EX}{EB} = \frac14$$
$$\frac{FX}{FC} = \frac14$$
There are actually two other solutions, and they involve some other regions. If you're looking for just one example, then this is the one.
First, I will show that the triangle above satisfies your condition, with three of these areas being equal. Since $\frac{EX}{EB} = \frac{FX}{FC}$, we see that $\triangle XEF \sim \triangle XBC$, so $EF || BC$. This implies that the heights of $\triangle FBC$ and $\triangle EBC$ are the same, hence their areas are equal. This implies that $[FBX] = [ECX]$.
Now, let both of these areas be $A$. We see that $[BCX] = 3A$ by base ratios, and hence, $[BDX] = [CDX] = 1.5A$. Since $[BXD] = [BXA]$, we must have $[AFX] = .5A$, and similarly, $[AEX] = .5A$, so $[AFXE] = A$, as desired. This result can be derived using easier methods, like barycentric coordinates and mass points.
Now I'll derive the other two possibilities. First, I prove that the quadrilateral $AFXE$ must be one of the three equal area regions. Assume that it is not. Then $[BFX] = [BXC] = [CEX]$, and thus, $FX = XC$ and $EX = XB$. But the triangles $BFX$ and $CEX$ are directly congruent and $BF||CE$, which is absurd.
Now, let $X$ have homogenized barycentric coordinates $(x, y, z)$ so we have $x+y+z=1$. We then have $F(x:y:0) = (\frac{x}{x+y}, \frac{y}{x+y}, 0)$, and trivially, $A(1,0,0)$. If we assume without loss of generality that $[ABC]=1$, then the area of $\triangle AFX$ can be evaluated with the following determinant:
$$[AFX] = \begin{vmatrix}
1&0&0\\
\frac{x}{x+y}&\frac{y}{x+y}&0\\
x&y&z\\
\end{vmatrix}=\frac{1}{x}\cdot\frac{1}{x+y}\cdot\begin{vmatrix}
x&0&0\\
x&y&0\\
x&y&z\\
\end{vmatrix}=\frac{xyz}{x(x+y)} = \frac{yz}{x+y}$$
We slew the determinant by factoring out common factors in rows, and then using row reduction to make the computation trivial.
We use the same method to derive the following areas:
$$[AFXE] =\frac{yz}{x+y}+\frac{yz}{x+z}$$
$$[BFX] =\frac{xz}{x+y}$$
$$[CEX] =\frac{xy}{x+z}$$
$$[BXC] = x$$
The last area can be derived from interpreting barycentric coordinates as areal coordinates.
With this new algebraic foundation, we will now completely eliminate $[BXC]$ as one of the possible areas. We know that $[AFXE]$ is one of the areas. First, without loss of generality, let $[BFX]$ be another one of the possible areas. Then:
$$\frac{yz}{x+y}+\frac{yz}{x+z} = \frac{xz}{x+y}$$
This expands and simplifies to:
\begin{equation}
(*)\ \ \ y^2 + 2xy + yz = x^2 + xz
\end{equation}
Now assume that $[BXC]$ is one of the areas. Then:
$$[BFX] = [BXC] \implies \frac{xz}{x+y} = x \implies x+y=z$$
Since $x+y+z=1$, We derive that $x+y=z=\frac12$. Hence $y = \frac12-x$, so we can eliminate $y$ and $z$ in $(*)$:
$$(\frac12-x)^2 + 2x(\frac12-x) + (\frac12-x)(\frac12) = x^2+\frac12x$$
This has the unique positive root $x = \frac{\sqrt{5}-1}{4}$. Hence we have the solution:
$$x = \frac{\sqrt{5}-1}{4}$$
$$y = \frac{3-\sqrt{5}}{4}$$
$$z = \frac12$$
Similarly, if we chose $[CEX]$ to instead be our second area, we would have the solution:
$$x = \frac{\sqrt{5}-1}{4}$$
$$y = \frac12$$
$$z = \frac{3-\sqrt{5}}{4}$$
This wraps up the case of $[CEX]$ being a possible area. Our last case to consider $[AFXE] = [BFX] = [CEX]$, which will result in the solution I presented first.
$(*)$ still holds true, and now we have:
$$\frac{xz}{x+y} = \frac{xy}{x+z}$$
This expands to:
$$xy+y^2 = xz + z^2$$
Now if we restate $(*)$ for the reader's convenience:
$$y^2 + 2xy + yz = x^2 + xz$$
We see that the equations can be subtracted to obtain:
$$xy + yz = x^2 - z^2$$
Factoring:
$$y(x+z) = (x+z)(x-z) \implies x = y+z$$
Once again, from $x+y+z=1$, this implies $y+z=x=\frac12$. Substitute $z = \frac12 - y$ into $(*)$ to solve for $y$, so we derive the following solution:
$$x = \frac12$$
$$y = \frac14$$
$$z = \frac14$$
Which is precisely the original solution.
Ok this might have been slightly overkill.
|
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|
Showing that $\sin(\sqrt{4 \pi^{2}n^{2} + x})$ converges uniformly on $[0,1]$
Suppose we are considering the sequence of functions $f_{n}(x)=\sin(\sqrt{4 \pi^{2}n^{2} + x})$ and I am having trouble showing that that $f_{n}$ converges uniformly on the interval $[0,1]$.
An idea, I've tried is to consider the Taylor series:
$$\sin(\sqrt{4 \pi^{2}n^{2} + x}) = (\sqrt{4 \pi^{2}n^{2} + x})- \frac{(\sqrt{4 \pi^{2}n^{2} + x})^{3}}{6} + O((\sqrt{4 \pi^{2}n^{2} + x})^{5})$$
but I haven't gotten anything useful as of yet.
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Your sequence converges uniformly to $0$.
\begin{align}
\left|\sin\left(\sqrt{4\pi^2n^2+x}\right)\right| &= \left|\sin\left(\sqrt{4\pi^2n^2+x}\right)-\sin(2\pi n)\right| \\
&= \left|2\sin\left(\frac{\sqrt{4\pi^2n^2+x} - 2\pi n}2\right)\cos\left(\frac{\sqrt{4\pi^2n^2+x} + 2\pi n}2\right)\right|\\
&\le 2\left|\sin\left(\frac{\sqrt{4\pi^2n^2+x} - 2\pi n}2\right)\right|\\
&\le 2\left|\frac{\sqrt{4\pi^2n^2+x} - 2\pi n}2\right|\\
&= \sqrt{4\pi^2n^2+x} - 2\pi n\\
&= \frac{4\pi^2n^2+x - 4\pi^2n^2}{\sqrt{4\pi^2n^2+x} + 2\pi n}\\
&= \frac{x}{\sqrt{4\pi^2n^2+x} + 2\pi n}\\
&\le \frac{1}{4\pi n} \xrightarrow{n\to\infty} 0
\end{align}
uniformly in $x \in [0,1]$.
|
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|
Sum of infinite series $ {1+ \frac{2}{6} + \frac{2\cdot5}{6\cdot12} + \frac{2\cdot5\cdot8}{6\cdot12\cdot18} + \cdots}$. Prove that $1+ \frac{2}{6} + \frac{2\cdot5}{6\cdot12} + \frac{2\cdot5\cdot8}{6\cdot12\cdot18} +\cdots=4^{\frac13}$
I tried it in the backward method... I rewrote $4^{\frac13}$ in this way...
$(1+3)^{\frac13}$ and expanded it in the binomial expansion method, but it doesn't help in any way.
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using binomial expansion $(1-x)^{-\frac{2}{3}} =1-\frac{2}{3}(-x)+\frac{\frac{-2}{3}\times\frac{-5}{3}}{2!}(-x)^2+\frac{\frac{-2}{3} \cdot \frac{-5}{3} \cdot \frac{-8}{3}}{3!} \cdot (-x)^3 +\ldots$
when $x=\frac{1}{2}$ we get:
$(1-\frac{1}{2})^{-\frac{2}{3}}=(\frac{1}{2})^{-\frac{2}{3}}=(2)^\frac{2}{3}=(2^2)^\frac{1}{3}=4^{\frac{1}{3}}$
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|
Evaluate: $\lim_{x \to \pi/4}\frac{8-\sqrt{2}(\sin x+\cos x)^5}{1-\sin 2x}$ without L'Hopitals I have to find the limit of $\lim\limits_{x \to \pi/4}\frac{8-\sqrt{2}(\sin x+\cos x)^5}{1-\sin 2x}$
here is my try $\lim\limits_{x \to \pi/4}\frac{8-\sqrt{2}(\sin x+\cos x)^5}{1-\sin 2x}=\lim\limits_{x \to \pi/4}\frac{4(1-\cos^5(\frac{\pi}{4}- x))}{\sin^2 (\frac{\pi}{4}- x)}$
now observe that for $\frac{\pi}{6} \leq x \leq \frac{\pi}{3}$,
$2.5-(1-\cos^5(\frac{\pi}{4}- x))\leq\frac{1-\cos^5(\frac{\pi}{4}- x)}{\sin^2 (\frac{\pi}{4}- x)}\leq 2.5-(\sin^2 (\frac{\pi}{4}- x))$
then
$\lim\limits_{x \to \pi/4}(2.5-(1-\cos^5(\frac{\pi}{4}- x)))\leq \lim\limits_{x \to \pi/4}\frac{1-\cos^5(\frac{\pi}{4}- x)}{\sin^2 (\frac{\pi}{4}- x)}\leq \lim\limits_{x \to \pi/4} (2.5-(\sin^2 (\frac{\pi}{4}- x)))$
$2.5\leq \lim\limits_{x \to \pi/4}\frac{1-\cos^5(\frac{\pi}{4}- x)}{\sin^2 (\frac{\pi}{4}- x)}\leq 2.5$
this implies $4\lim\limits_{x \to \pi/4}\frac{(1-\cos^5(\frac{\pi}{4}- x))}{\sin^2 (\frac{\pi}{4}- x)}=4\cdot 2.5=10$
is my answer correct? wolfram gave the same answer. is there any other simpler method?
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$\lim\limits_{x \to \pi/4}\frac{8-\sqrt{2}(\sin x+\cos x)^5}{1-\sin 2x}$
$= \lim_{x \to \frac{\pi}4}\dfrac{\sqrt2(\sqrt2^5 - (\sin x + \cos x)^5)}{1- \sin 2x} $
Using,
$\color{blue}{\lim _{x \to a}\frac{x^n -a^n}{x-a}=na^{n-1}}$
$= \lim _{x \to \pi/4}\dfrac{\sqrt 2 (5.\sqrt2^{4})(\sqrt 2 -(\sin x +\cos x))}{1- \sin 2x}$
Using,
$\color{blue}{a-b= \dfrac{a^2 - b^2 }{a+b}}$
$= \lim_{x \to \pi/4}\dfrac{\sqrt 2. 20 ( 2-(\sin ^2 x + \cos^2 x+\sin 2x) )}{(\sin x + \cos x + \sqrt 2)(1- \sin 2x)}$
$= \lim _{x \to \pi/4}\dfrac{\sqrt 2 .20}{2\sqrt 2}$
$= 10$
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|
Solve the equation $X^2+X=\text{a given matrix}$ I want to solve the quadratic matrix equation
$$X^2+X=\begin{pmatrix}1&1\\1&1\end{pmatrix}$$
If I put $X$ in the form
$$X=\begin{pmatrix}a&b\\c&d\end{pmatrix}$$
then I find complicated equations. Is there a simple way to tackle the problem without using diagonalization?
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Your method also works, though cumbersome (not asap):
$$X^2+X=1 \Rightarrow \begin{pmatrix}a&b\\c&d\end{pmatrix}^2+\begin{pmatrix}a&b\\c&d\end{pmatrix}=\begin{pmatrix}1&1\\1&1\end{pmatrix} \Rightarrow\begin{cases}a^2+a+bc=1\\ ab+bd+b=1\\ ac+c+cd=1\\ bc+d^2+d=1\end{cases}.$$
Subtract the last from the first:
$$a^2-d^2+a-d=0 \Rightarrow (a-d)(\underbrace{a+d+1}_{\ne 0 \ \text{from (2)}})=0 \Rightarrow d=a.$$
Plug $d=a$ in the second and third equations and subtract:
$$(2ab+b)-(2ac+c)=0 \Rightarrow (b-c)(\underbrace{2a+1}_{\ne 0 \ \text{from (2)}})=0 \Rightarrow c=b.$$
Plug $c=b$ into the first and second equations and add them:
$$a^2+a+b^2+2ab+b=2 \Rightarrow (a+b)^2+(a+b)-2=0 \Rightarrow \\
a+b=-2;1 \Rightarrow 1) \ b=-a-2; \ 2) \ b=-a+1.$$
$1$) Plug $b=-a-2$ into the first equation:
$$a^2+a+(-a-2)^2=1 \Rightarrow 2a^2+5a+3=0 \Rightarrow a=-\frac32; -1 \Rightarrow \\
X_1=\begin{pmatrix}a&b\\c&d\end{pmatrix}=\begin{pmatrix}-\frac32&-\frac12\\-\frac12&-\frac32\end{pmatrix};\\
X_2=\begin{pmatrix}a&b\\c&d\end{pmatrix}=\begin{pmatrix}-1&-1\\-1&-1\end{pmatrix}.$$
$2$) Plug $b=-a+1$ into the first equation:
$$a^2+a+(-a+1)^2=1 \Rightarrow 2a^2-a=0 \Rightarrow a=0; \frac12 \Rightarrow \\
X_3=\begin{pmatrix}a&b\\c&d\end{pmatrix}=\begin{pmatrix}0&1\\1&0\end{pmatrix};\\
X_4=\begin{pmatrix}a&b\\c&d\end{pmatrix}=\begin{pmatrix}\frac12&\frac12\\\frac12&\frac12\end{pmatrix}.$$
|
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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.