Q
stringlengths 70
13.7k
| A
stringlengths 28
13.2k
| meta
dict |
|---|---|---|
Prove $\Sigma_{cyc}(\frac{a}{b-c}-3)^4\ge193$ The inequality is expected original question of this MSE question. The exact statement is "If $a$, $b$ and $c$ are positive real numbers and none of them are equal pairwise, prove the following inequality."
$$\Sigma_{cyc}\left(\frac{a}{b-c}-3\right)^4\ge193$$
Full expanding gives 12-degree polynomial with about 90 terms. It starts with $\Sigma_{cyc}(a^{12}-16a^{11}b+8a^{11}c)$ and it does not look good for Muirhead or Schur.
Also I tried substitution of $\frac{a}{b-c}=x$, $\frac{b}{c-a}=y$ and $\frac{c}{a-b}=z$. Then by $uvw$, it suffices to show when $x=y$ (See answer to linked question for details). That is, $\frac{a}{b-c}=\frac{b}{c-a}$ or $c=\frac{a^2+b^2}{a+b}$, therefore either $a<c<b$ or $b<c<a$.
Given the constraints, it is clear that $x>0$ and substituting $y=x$, $z=-\frac{1+x^2}{2x}$ gives nonnegative polynomial for $0<x$ (which is not nonnegative polynomial for all $x$).
However, it looks like I cannot deduce $x>0$ from the fact it is enough to consider $x=y$.
How can I prove it? Thank you!
|
It remains to make two steps only.
*
*For $$\frac{a}{b-c}=\frac{b}{c-a}$$ or
$$c=\frac{a^2+b^2}{a+b}$$ it's enough to prove that
$$2\left(\frac{a}{b-\frac{a^2+b^2}{a+b}}-3\right)^2+\left(\frac{a^2+b^2}{a^2-b^2}-3\right)^4\geq193.$$
Now, let $a=tb$.
Thus, we need to prove that
$$2\left(\frac{t}{b-\frac{t^2+1}{t+1}}-3\right)^2+\left(\frac{t^2+1}{t^2-1}-3\right)^4\geq193$$ or
$$335t^8+1024t^7+338t^6-1280t^5-742t^4+640t^3+196t^2-128t+95\geq0,$$ which is obviously true for $t>0$.
*For $w^3\rightarrow0$ let $\frac{c}{a-b}\rightarrow0$.
Thus, we need to prove that
$$\left(\frac{a}{b}-3\right)^4+\left(\frac{b}{-a}-3\right)^4+81\geq193$$ or
$$(a^2-4ab-b^2)^2(a^4-4a^3b+8a^2b^2+4ab^3+b^4)\geq0,$$ which is obvious again.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3073351",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
}
|
Help on solutions of the congruence $f(x)=x^3+4x+8 \equiv 0 \pmod {15}$ I'm doing a little exercise, solve the congruence $f(x)=x^3+4x+8 \equiv 0 \pmod {15}$.
I know that $15=3 \times 5$ and they are relatively prime, so I can split the congruence into:
a) $f(x) \equiv 0 \pmod {3}$
b) $f(x) \equiv 0 \pmod {5}$
I proceed now with the individual solutions.
a) I can reduce the function because I know $x^3 \equiv x \pmod 3$, so it became $5x \equiv -8 \pmod 3 \Rightarrow x \equiv 2 \pmod 3$ and this is one solution
b) I can't reduce because the degree of the monic polynomial $f(x)$ is less than $x^5 \equiv x \pmod 5$. If I try all the values of $x \in [0,1,2,3,4]$ into $f(x)$, I can't obtain any zero $(x,f(x))\rightarrow (0,8),(1,13),(2,24),(3,47),(4,88)$.
So seems like there is no solution.
At this point my question is: the only solution is that from the point a)? Or maybe I'm missing something or maybe I'm wrong!
|
As other answrers note, there are no solutions because the equation fails $\bmod 5$.
Here an alternate method for solving cubic equations $\bmod 5$ is explored. Suppose the equation has the form
$ax^2+bx^2+cx+d\equiv 0\bmod 5$ Eq. 1
If $d\equiv 0$ then $x\equiv 0$ is a root and the quadratic equation $ax^2+bx+c=0$ may be solved for the other root bystanders techniques for quadratic equations.
Otherwise multiply by $ax-b$ to obtain a quartic equation
$a^2x^4+(ac-b^2)x^2+(ad-bc)x-bd\equiv 0$
and then, since $d\not\equiv 0$ forces $x\not\equiv 0$, we must have $x^4\equiv 1$. Thereby
$(ac-b^2)x^2+(ad-bc)x+(a^2-bd)\equiv 0$ Eq. 2
Or perhaps easier to remember:
$a(cx^2+dx+a)-b(bx^2+cx+d)\equiv 0$ Eq. 2a
Note that any zero root to this equation must be rejected as inconsistent with the cubic equation having $d\not\equiv 0$, and a root satisfying $ax=b$ must be checked against the cubic equation because multiplying the original cubic by $ax-b$. Apart from these checks Eq. 2 or 2a will be equivalent to the original cubic.
For the problem at hand we have $x^3+4x+8\equiv 0$. Then from Eq. 2a we see immediately that $4x^2+8x+1=0$. Multiplying by $4$ to render the equation monic and completing the square gives $(x+1)^2\equiv 2$, which has no solution as $2$ is not a quadratic residue $\bmod 5$. So the cubic equation also fails.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3078065",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
}
|
How do I solve for $a$ and $b$ in $\lim\limits_{x \to ∞} x \left(2 +(3+x) \ln \left( \dfrac{x+a}{x+b} \right) \right) = 2$? Given $\lim\limits_{x \to ∞} x \left(2 +(3+x) \ln \left( \dfrac{x+a}{x+b} \right) \right) = 2$
I need to solve for $a$ and $b$, so here we go,
$\lim\limits_{x \to ∞} x \left(2 +(3+x) \ln \left( \dfrac{x+a}{x+b} \right) \right)$
$= \lim\limits_{x \to ∞} x \left(2 +(3+x) \left(\ln (1+\frac{a}{x}) - \ln(1+\frac{b}{x}) \right) \right)$
$=\lim\limits_{x \to ∞} x \left(2 +(3+x)\left( \dfrac{a-b}{x} \right) \right)$
$=\lim\limits_{x \to ∞} \left(2x +(3+x)\left( a-b \right) \right)$
$=\lim\limits_{x \to ∞} \left(2x + 3(a-b) + x(a-b) \right) $
Since the limit exists, $2x$ and $x(a-b)$ must cancel out so $a-b = 2$
But if I do this, I'm left with $3(a-b) = 2$ which gives me a contradiction since $a-b$ is supposed to be $2$.
What do I do now? And where exactly have I gone wrong?
Thank you!
|
Hint: you have to expand $\log (1+y)$ up to the term in $y^{2}$ to answer this question. $\log (1+y)=y-\frac {y^{2}} 2+0(y^{2})$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3079016",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 1
}
|
Evaluating $\lim_{x\to0^+}\frac{2x(\sin x)^2+\frac{2x^7+x^8}{3x^2+x^4}-\arctan(2x^3)}{\ln(\frac{1+x^2}{1-x^2})-2x^2+xe^{-{1\over x}}}$ To start with, the term $xe^{-{1\over x}}$ can be ignored.Then splitting the term $\ln(\frac{1+x^2}{1-x^2})=\ln(1+x^2)-\ln(1-x^2)$ and expanding both of them up to order 3 we have in the denominator $2x^2+{2x^6\over 3}-2x^2={2x^6\over3}$
As for the numerator $\arctan(2x^3)\sim2x^3$ and ${2x^7+x^8\over 3x^2+x^4}\sim{2\over3}x^5$.So if we expand $\sin x$ up to order 2,then take the square (ignoring higher order terms) and multiply by $2x$ we have $2x(\sin x)^2\sim2x^3-{2\over3}x^5$ which cancels out with the other terms,giving us an indeterminate form.So I think we should have $x^6$ term in the numerator.
But if I expand $\sin x$ up to order 3 and take square, I fail to obtain $x^5$ term(so that when multiplied by $2x$ gives me $x^6$).How can this be done?
|
The Taylor series approach suggested by @Matteo and @WillJagy is the way to go. Here is that approach made a bit more explicit:
*
*Divide the numerator and denominator by x^6.
*Construct the resulting first order Taylor series for both numerator and denominator.
*Observe what the limits must be.
$$\lim_{x\to 0^+} \, \frac{-\tan ^{-1}\left(2 x^3\right)+\frac{(x+2) x^5}{x^2+3}+2 x \sin ^2(x)}{x \exp \left(-\frac{1}{x}\right)-2 x^2+\log \left(\frac{x^2+1}{1-x^2}\right)}$$
$$\lim_{x\to 0^+} \,{ {(\frac{1}{3}-\frac{2 x}{15}+o(x^2) )}\over{(\frac{2}{3}+e^{-1/x}(\frac{1}{x^5}+o(x^2)))}}=\frac{1}{2}$$
Here is a plot of the function:
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3079733",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
}
|
Prove that for all integers $n$ if $3 \mid n^2$, then $3 \mid n$ Prove that for all integers $n$ if $3$ | $n^2$, then $3$ | $n$.
I figured using contrapostive was the best method by using the definition "an integer $k$ is not divisible by 3 if and only if there exists an integer $k$ such that $n=3k+1$ or $n=3k+2$. Also using the definition $a$ divides $b$ written $a$|$b$ if $b=ac$ for some $c$ in integers.
Here it goes:
$$3\nmid n$$
$$3k+1=3n$$
$$\frac{3k+1}{3}=n$$
|
Presumably you have had the division theorem.
For $n$ and integer there exist integer $k, r$ so that $n = 3k + r$ where $0 \le r < 3$. So $r = 0, 1$ or $2$.
Can you accept that?
If $r=0$ then $3|n$.
If $r = 1$ then $n^2 = (3k + 1)^2 = 9k^2 + 6k + 1 = 3(3k^2 + 2k) + 1$ and $3\not \mid n^2$ and that's a contradiction.
If $r = 2$ then $n^2= (3k + 2)^2 = 9k^2 + 12k + 4 = 3(k^2 + 4k+ 1) + 1$ and $3\not \mid n^2$ and that's a contradiction.
So if $3|n^2$ then the only possibility is $n = 3k$ for some $k$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3080079",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
}
|
$a+b=c+d$ , $a^3+b^3=c^3+d^3$ , prove that $a^{2009}+b^{2009}=c^{2009}+d^{2009}$
Let $a, b, c, d$ be four numbers such that $a + b = c + d$ and $a^3 + b^3 = c^3 + d^3$. Prove that $a^{2009}+b^{2009}=c^{2009}+d^{2009}$.
I've got $a^2+b^2-ab=c^2+d^2-cd$.
I tried squaring or cubing it repeatedly but I didn't get what I wanted.
Now how do I proceed?
|
If $a=-b$ so $c=-d$ and we are done.
Let $a+b\neq0.$
Thus, $$(a+b)^3-3ab(a+b)=(c+d)^3-3cd(c+d)$$ or
$$ab(a+b)=cd(c+d)$$ or $$ab=cd$$ and the rest is smooth.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3080393",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 7,
"answer_id": 4
}
|
Find the numerical value of this expression If $x$ is a complex number such that $x^2+x+1=0$, then the numerical value of $(x+\frac{1}{x})^2+(x^2+\frac{1}{x^2})^2+(x^3+\frac{1}{x^3})^2+\ldots+(x^{27}+\frac{1}{x^{27}})^{2}$ is equal to?
A) 52 . B) 56 . C) 54. D)58 . E)None of these
Where is this question from? I'm pretty sure it comes from one high school math contest, does anyone one know which math contest and of course i can't solve, I have an answer key but I don't know the solution.
|
Multiply both sides by $x-1\ne 0$:
$$x^2+x+1=0 \Rightarrow (x-1)(x^2+x+1)=0 \Rightarrow x^3-1=0 \Rightarrow x^3=1\Rightarrow \\
x^{54}=1;x^{-54}=1 \quad (1)$$
Expand:
$$(x+\frac{1}{x})^2+(x^2+\frac{1}{x^2})^2+(x^3+\frac{1}{x^3})^2+\ldots+(x^{27}+\frac{1}{x^{27}})^{2}=\\
[x^2+x^4+\cdots+x^{54}]+[x^{-2}+x^{-4}+\cdots+x^{-54}]+2\cdot 27=\\
\frac{x^2(x^{54}-1)}{x^2-1}+\frac{x^{-2}(x^{-54}-1)}{x^2-1}+54\stackrel{(1)}=54.\\
$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3082719",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 1
}
|
$\int\limits_0^\infty {x^4 \over (x^4-x^2+1)^4}\ dx$ I want to calculate
$$\int\limits_0^\infty \frac{x^4}{(x^4-x^2+1)^4}dx$$
I have searched with keywords "\frac{x^4}{(x^4-x^2+1)^4}" and "x^4/(x^4-x^2+1)^4". But there are no results
|
Here is another way to get to the same point as what @Sangchul Lee gives.
Let
$$I = \int_0^\infty \frac{x^4}{(x^4 - x^2 + 1)^4} \, dx.$$
Then
$$I = \int_0^1 \frac{x^4}{(x^4 - x^2 + 1)^4} \, dx + \int_1^\infty \frac{x^4}{(x^4 - x^2 + 1)^4} \, dx.$$
Enforcing a substitution of $x \mapsto 1/x$ in the right most integral leads to
\begin{align}
I &= \int_0^1 \frac{x^4 (1 + x^6)}{(x^4 - x^2 + 1)^4} \, dx\\
&= \int_0^1 \frac{x^4 (1 + x^2)(x^4 - x^2 + 1)}{(x^4 - x^2 + 1)^4} \, dx\\
&= \int_0^1 \frac{x^4 (1 + x^2)}{(x^4 - x^2 + 1)^3} \, dx\\
&= \int_0^1 \frac{1 + 1/x^2}{\left [\left (x - \frac{1}{x} \right )^2 + 1 \right ]^3} \, dx.
\end{align}
On setting $-u = x - 1/x$, $-du = (1 + 1/x^2) \, dx$ one has
\begin{align}
I &= \int_0^\infty \frac{du}{(u^2 + 1)^3}\\
&= \int_0^{\frac{\pi}{2}} \cos^4 \theta \, d\theta\\
&= \int_0^{\frac{\pi}{2}} \left (\frac{1}{2} \cos 2\theta + \frac{1}{8} \cos 4\theta + \frac{3}{8} \right ) \, d\theta\\
&= \frac{3\pi}{16},
\end{align}
as expected.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3082789",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
}
|
Is it possible to reach the initial arrangement?
We have a stack of $n$ books piled on each other, and labeled by $1, 2, ..., n$. In each round we make $n$ moves in the following manner: In the $i$-th move of each turn, we turn over the $i$ books at the top, as a single book. After each round we start a new round similar to the previous one. Show that after some moves, we will reach the initial arrangement.
Say $n=4$ and initial arrangement of books $(a,b,c,d)$. First we act on it with identical transformation $\pi_1=id$ which leaves everything as it was. Then we act on it with
$$\pi_2 = \left(
\begin{array}\\
1 & 2 & 3 & 4 \\
2 & 1 & 3 & 4
\end{array}\right)$$
and we get $(b,a,c,d)$, then we act on this one with $$\pi_3 = \left(
\begin{array}\\
1 & 2 & 3 & 4 \\
3 & 2 & 1 & 4
\end{array}\right)$$ and we get $(c,a,b,d)$ and then $$\pi_4 = \left(
\begin{array}\\
1 & 2 & 3 & 4 \\
4 & 3 & 2 & 1
\end{array}\right)$$ and we get $(d,b,a,c)$ and then we repeat acting with $$\pi_1,\pi_2,\pi_3,\pi_4,\pi_1,\pi_2,...$$
Now what do we get if we repeat enough time $$\sigma = \pi_4\circ \pi_3\circ \pi_2\circ \pi_1$$ on starting $(a,b,c,d)$? If we repeat this $\sigma $ exactly $24$ times (which is the order of symmetric group $S_4$) we shoud get initialy arrangment. Clearly this can be easly generalized for arbitrary $n$.
Is this correct?
Edit: As suggested in comment by Lord Shark the Unknown, It shoud be considered also the first and last front of a book. So I should observe 8-couple $(a1,a2,b1,b2,c1,c2,d1,d2)$, where $x1$ is first front and $x2$ last one, instead of 4-couple $(a,b,c,d)$ and act on it with $S_8$?
|
For clear reference, here is a complete cycle of moves. Negative number represents book face down.
1 2 3 4
-1 2 3 4
-2 1 3 4
-3 -1 2 4
-4 -2 1 3
4 -2 1 3
2 -4 1 3
-1 4 -2 3
-3 2 -4 1
3 2 -4 1
-2 -3 -4 1
4 3 2 1
-1 -2 -3 -4
1 -2 -3 -4
2 -1 -3 -4
3 1 -2 -4
4 2 -1 -3
-4 2 -1 -3
-2 4 -1 -3
1 -4 2 -3
3 -2 4 -1
-3 -2 4 -1
2 3 4 -1
-4 -3 -2 -1
1 2 3 4 << return to initial state
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3083542",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
}
|
Integral of $\int{\frac{1}{\sqrt{x(1+x^2)}}dx}$ I was trying to solve the following question:
Evaluate: $$\int{\frac{1}{\sqrt{x(1+x^2)}}dx}$$
This is an unsolved question in my sample papers book and so I believe it should have an elementary primitive.
But, I don't know how to start this. I want a hint to get started with this. Thanks!
|
We can also express the solution in terms of the Beta Function and the Incomplete Beta Function as I cover here:
\begin{align}
I&= \int \frac{1}{\sqrt{x\left(1 + x^2\right)}}\:dx = \int_0^x \frac{t^{-\frac{1}{2}}}{\left(t^2 + 1 \right)^{\frac{1}{2}}} \:dt\\
&=\frac{1}{2} \left[ B\left(\frac{1}{2} - \frac{-\frac{1}{2} + 1}{2} , \frac{-\frac{1}{2} + 1}{2} \right) - B\left(\frac{1}{1 + x^2};\frac{1}{2} - \frac{-\frac{1}{2} + 1}{2} , \frac{-\frac{1}{2} + 1}{2} \right) \right] \\
&=\frac{1}{2} \left[ B\left( \frac{1}{4} , \frac{1}{4} \right) - B\left(\frac{1}{1 + x^2};\frac{1}{4} , \frac{1}{4} \right) \right]
\end{align}
Using the relationship between the Beta function and the Gamma Function we find that:
\begin{equation}
B\left( \frac{1}{4} , \frac{1}{4} \right) = \frac{\Gamma\left(\frac{1}{4}\right)\Gamma\left(\frac{1}{4}\right)}{\Gamma\left(\frac{1}{4} + \frac{1}{4}\right)} = \frac{\Gamma\left(\frac{1}{4}\right)^2}{\Gamma\left(\frac{1}{2}\right)} = \frac{\Gamma\left(\frac{1}{4}\right)^2}{\sqrt{\pi}}
\end{equation}
Thus our integral $I$ becomes:
\begin{equation}
I = \int \frac{1}{\sqrt{x\left(1 + x^2\right)}}\:dx = \int_0^x \frac{t^{-\frac{1}{2}}}{\left(t^2 + 1 \right)^{\frac{1}{2}}} \:dt = \frac{1}{2} \left[ \frac{\Gamma\left(\frac{1}{4}\right)^2}{\sqrt{\pi}} - B\left(\frac{1}{1 + x^2};\frac{1}{4} , \frac{1}{4} \right) \right]
\end{equation}
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3085740",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
}
|
Find $n$ such that polynomial is divisible Find $n \in N$, such that:
$$(x^2+x+1)^2 | x^n+(x+1)^n+1 = P(x)$$
If I let $Q(x) = x^2 + x + 1$, I have that $x^3 \equiv 1$ mod$Q(x)$, so I can work out the cases for remainder of $n$ when divided by $3$ ..
But what to do next, because I need $Q^2(x)$ dividing $P(x)$ ?
I suppose this could be done with complex zeroes, but is there a faster way ?
|
Since $(x^2+x+1)^2$ divides $P(x):=x^n+(x+1)^n+1$, we get that $Q(x):=x^2+x+1$ divides its derivative $P'(x)=n(x^{n-1}+(x+1)^{n-1})$.
Let us see when $n(x^{n-1}+(x+1)^{n-1})$ is divisible by $Q$ by working modulo $Q(x)$. Since $x^2+x+1=0$ we get $x^2=-(x+1)$ and
$$x^3=x^2x=-(x+1)x=-x^2-x=(x+1)-x=1,$$
hence
$$x^{3k}=1, x^{3k+1}=x, x^{3k+2}=x^2=-(x+1)$$
for all $k\in\mathbb{N}$, and similarly
$$(x+1)^k=(-1)^kx^{2k}.$$
Then we get
$$n(x^{3k}+(x+1)^{3k})=n(1+(-1)^{3k}),$$
so it is $0$ iff $k$ is odd, i.e., if $3k=6k'+3$,
and $$n(x^{3k+1}+(x+1)^{3k+1})=n(x+(-1)^{3k+1}x^2),$$
$$n(x^{3k+2}+(x+1)^{3k+2})=n(x^2+(-1)^{3k+2}x),$$
which can never be zero.
Hence $Q(x)^2$ divides $P(x)$ iff $n\equiv 4\pmod{6}$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3085998",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
}
|
How to solve $\lim \left(\frac{n^3+n+4}{n^3+2n^2}\right)^{n^2}$ I can't seem to find a way to solve:
$$\lim \left(\dfrac{n^3+n+4}{n^3+2n^2}\right)^{n^2}$$
I've tried applying an exponential and logaritmic to take the $n^2$ out of the exponent, I've tried dividing the expression, but I don't get anywhere that brings light to the solution.
Any ideas?
|
$$
\begin{align*}
L &= \lim_{n\to\infty}\left(\frac{n^3+n+4}{n^3+2n^2}\right)^{n^2} \\
&= \lim_{n\to\infty}\left(\frac{n^3 +2n^2 - 2n^2+n+4}{n^3+2n^2}\right)^{n^2}\tag1 \\
&= \lim_{n\to\infty}\left(1 + \frac{- 2n^2+n+4}{n^3+2n^2}\right)^{n^2} \tag2 \\
&= \lim_{n\to\infty}\left(1 + \frac{- 2n^2+n+4}{n^3+2n^2}\right)^{\frac{n^2(n^3+2n^2)(n-2n^2+4)}{(n^3+2n^2)(n-2n^2+4)}} \tag3 \\
&= \lim_{n\to\infty}e^{n^2(n-2n^2+4)\over(n^3+2n^2)} \tag4
\end{align*}
$$
Now consider:
$$
\lim_{n\to\infty}{n^2(n-2n^2+4)\over(n^3+2n^2)} = -\infty
$$
Hence your limit is:
$$
e^{-\infty} = 0
$$
Description of steps:
*
*$(1)$ add and subtract $2n^2$
*$(2)$ perform division
*$(3)$ multiply the power by the reciprocal of the fraction inside parentheses
*$(4)$ use the limit for $(1 + {1\over x^n})^{x_n}$ when $x_n \to\ \infty$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3087012",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
}
|
$\sum\limits_{n=1}^{\infty}\arctan{\frac{2}{n^2+n+4}}$ $$\sum\limits_{n=1}^{\infty}\arctan{\frac{2}{n^2+n+4}}$$
We know that : $\arctan{x} - \arctan{y} = \arctan{\frac{x-y}{1+xy}}$ for every $ xy > 1 $
I need to find two numbers which satisfy: $ab = n^2+2n+3 $ and $ a-b =2$ in order to telescope.
Edit: I am very sorry on the denominator it should have been $n^2+n+4$ not $n^2+2n+4$
Similarly here :
$$\sum\limits_{n=1}^{\infty}\arctan{\frac{8n}{n^4-2n^2+5}}$$
The result should be $ \arctan 2 $ on the first one and $ \pi/2 + \arctan2 $ on the second one.
|
$$\sum\limits_{n=1}^{\infty}\arctan{\frac{2}{n^2+n+4}} = \sum\limits_{n=1}^{\infty}\arctan{\frac{2(n+1)-2n}{n(n+1)(1+\frac{4}{n(n+1)})}} =
\sum\limits_{n=1}^{\infty}\arctan{\frac{\frac{2}{n}-\frac{2}{n+1}}{1+\frac{2}
{n}\cdot\frac{2}{n+1}}} =\sum\limits_{n=1}^{\infty}\arctan{\frac{2}{n}-\arctan{\frac{2}{n+1}}} =\arctan2$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3090636",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
}
|
How to prove that $\csc^2x=\sum_{-\infty}^{\infty}\frac{1}{(x-n\pi)^2}$ I was reading the book The Princeton Companion to Mathematics
On page $293$, there is a statement
$$\csc^2x=\sum_{n=-\infty}^{\infty}\frac{1}{(x-n\pi)^2}\tag{1}$$
Here is one method from the book:
Observe first that
$$h(x)= \sum_{n=-\infty}^{\infty}\frac{1}{(x-n\pi)^2}$$
is well-defined for each real $x$ that is not a multiple of $\pi$. Note also that $h(x+\pi)=h(x)$ and $h(\frac{1}{2}\pi-x) = h(\frac{1}{2}\pi+x)$. Set $f(x) = h(x) - \csc^2(\pi x)$. By showing that there are constans $K_1$ and $K_2$ such that
$$0<\sum_{n=1}^{\infty}\frac{1}{(x-n\pi)^2}<K_1$$
and
$$0<\csc^2x-\frac{1}{x^2}<K_2$$
for all $0<x\leq\frac{1}{2}\pi$, we deduce that there is a constant $K$ such that $|f(x)|<K$ for all $0<x<\pi$. Simple calculations show that
$$f(x)=\frac{1}{4}\left(f\left(\frac{1}{2}x\right)+f\left(\frac{1}{2}(x+\pi)\right)\right). \tag{2}$$
A single application of $(2)$ shows that $|f(x)|<\frac{1}{2}K$ for all $0<x<\pi$, and repeated applications show that $f(x) = 0$. Thus
$$\csc^2x=\sum_{n=-\infty}^{\infty}\frac{1}{(x-n\pi)^2}$$
for all real noninteger $x$.
Are here other methods to prove (1)?
|
METHODOLOGY $1$: FOURIER SERIES
We begin by writing the Fourier series,
$$\cos(xy)=a_0/2+\sum_{n=1}^\infty a_n\cos(nx) \tag1$$
for $x\in [-\pi/\pi]$. The Fourier coefficients are given by
$$\begin{align}
a_n&=\frac{2}{\pi}\int_0^\pi \cos(xy)\cos(nx)\,dx\\\\
&=\frac1\pi (-1)^n \sin(\pi y)\left(\frac{1}{y +n}+\frac{1}{y -n}\right)\tag2
\end{align}$$
Substituting $2$ into $1$, dividing by $\sin(\pi y)$ reveals
$$\begin{align}
\pi \cos(xy)/\sin(\pi y)&=\frac1y +\sum_{n=1}^\infty (-1)^n\left(\frac{1}{y -n}+\frac{1}{y +n}\right)\cos(nx)\\\\
&=\sum_{n=-\infty}^\infty \frac{(-1)^n\cos(nx)}{y-n}\tag3
\end{align}$$
Now differentiate with respect to $y$, set $x=\pi$, then put $y=x/\pi$, and divide by $\pi^2$. Can you finish?
METHODOLOGY $2$: PRODUCT REPRESENTATOIN OF THE SINE FUNCTION
As another way forward, we note that
$$\sin(\pi x)=\pi x\prod_{n=1}^\infty\left(1-\frac{x^2}{n^2}\right)$$
Taking the logarithm, and differentiating reveals
$$\begin{align}
\pi \cot(\pi x)&=\frac1x +\sum_{n=1}^\infty \frac{2x}{x^2-n^2}\\\\
&=\sum_{n=-\infty}^\infty \frac{1}{x-n}
\end{align}$$
Differentiating again and enforcing the substitution $x\to x/\pi$ and dividing by $\pi^2$ yields
$$\csc^2(x)=\sum_{n=-\infty}^\infty \frac{1}{(x-n\pi)^2}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3091415",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
}
|
The greatest value of $|z|$ if $\Big|z+\frac{1}{z}\Big|=3$ where $z\in\mathbb{C}$
$\bigg|z+\dfrac{1}{z}\bigg|=3$ then the greatest value of $|z|$ is ___________
My Attempt
$$
\bigg|z+\frac{1}{z}\bigg|=\bigg|\dfrac{z^2+1}{z}\bigg|=\frac{|z^2+1|}{|z|}=3\\
\bigg|z+\frac{1}{z}\bigg|=3\leq|z|+\frac{1}{|z|}\implies |z|^2-3|z|+1\ge0\\
|z|=\frac{3\pm\sqrt{5}}{2}\implies\color{red}{|z|\in(-\infty,\frac{3-\sqrt{5}}{2}]\cup[\frac{3+\sqrt{5}}{2},+\infty)}
$$
$$
\bigg|z+\frac{1}{z}\bigg|=3\geq\bigg||z|-\frac{1}{|z|}\bigg|\\
|z|^2-3|z|-1\leq0\text{ or }|z|^2+3|z|-1\geq0\\
|z|=\frac{3\pm\sqrt{13}}{2}\text{ or }|z|=\frac{-3\pm\sqrt{13}}{2}\\
\color{red}{|z|\in[\frac{3-\sqrt{13}}{2},\frac{3+\sqrt{13}}{2}]}\text{ or }\color{red}{|z|\in(-\infty,\frac{-3-\sqrt{13}}{2}]\cup[\frac{-3+\sqrt{13}}{2},+\infty)}
$$
The solution given in my reference is $\dfrac{3+\sqrt{13}}{2}$, why am I not able to find it in my attempt ?
Note: A similar question has been asked before, If $∣z+\frac{1}{z}∣=a$ then find the minimum and maximum value of $|z|$, but it does not have any answer which address why we choose a specific version of the triangle inequality.
|
The Joukowsky transform $z\to z+1/z$ transforms circles of radius $r>1$ into ellipses of semi-axes $r-1/r$ and $r+1/r$. The max value of $|z|$ corresponds to the radius of the circle such that the smallest semi-axis $r-1/r=3$. Solving this equation $r^2-3r-1=0$ for $r>0$ you get the result $r=(3+\sqrt{13})/2$ which agrees with yours.
The minor semi axes correspond to the purely imaginary values that jmerry found.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3091795",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
}
|
How do I prove this relationship between positive terms of a G.P.?
$a$, $b$, $c$, and $d$ are positive terms of a G.P. This is the relationship I'm trying to prove:
$$\frac1{ab} + \frac1{cd} > 2 \left(\frac1{bd} + \frac1{ac} - \frac1{ad}\right)$$
This question was listed under the section on Arithmetic, Geometric, and Harmonic Means. So, I tried using those.
$$A = \frac1{ab}$$
$$B = \frac1{cd}$$
$$A.M. = \frac12(A + B) = \frac12\left(\frac1{ab}+\frac1{cd}\right)$$
This gives the correct term on the left side as well as the $2$ on the right side.
But I am completely blank from here. How do I proceed from here to prove this relationship using means?
|
So $b=ax, c=ax^2$, and $d=ax^3$, and you have to prove: $$\frac1{a^2x} + \frac1{a^2x^5} > 2 (\frac1{a^2x^4} + \frac1{a^2x^2} - \frac1{a^2x^3})$$
or $$1 + \frac1{x^4} > 2 (\frac1{x^3} + \frac1{x} - \frac1{x^2})$$
or $$\boxed{x^4+1>2(x+x^3-x^2)}$$
or
$$x^4-2x^3+2x^2-2x+1>0\iff (x^2-x)^2+(x-1)^2>0,$$ which is obviously true.
You can try to prove the boxed inequality also like this, using AM-GM inequality: $$x^4+x^2\geq 2\sqrt{x^4\cdot x^2} = 2x^3$$ and$$x^2+1\geq 2\sqrt{x^2\cdot 1} = 2x$$
so $$x^4+2x^2+x = (x^4+x^2)+(x^2+1)> 2x^3+2x$$ and we are done again.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3092180",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
}
|
Find the remainder of the division of $x^n+5$ with $x^3+10x^2+25x$ Find the remainder of the division of $x^n+5$ with $x^3+10x^2+25x$ over $\mathbb{Q}$
What I tried to do is to write $x^n+5=p(x)(x^3+10x^2+25x)+Ax^2+Bx+C$, where $p(x)$ is a polynomial of degree $n-3$. If I set $x=0$ I obtain that $C=5$.
Now $x^3+10x^2+25x=x(x+5)^2$ and by setting $x=-5$, I get that
$5A-B=(-1)^n5^{n-1}$
But I need one more equation to be able to find the coefficients of the remainder, and I can't get one. What should I do?
|
Hint $\ $ Subtracting $\, C = 5\,$ then cancelling $x$ yields
$\qquad f(x) := x^{\large n-1} = B + Ax + (x+5)^{\large 2} p(x) $
Thus $\, B + Ax = f(-5) + f'(-5)(x+5)\,$ by Taylor expansion at $\,x = -5\,$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3092642",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
}
|
How can I simplify this fraction problem? I have the problem $\frac{x^2}{x^2-4} - \frac{x+1}{x+2}$ which should simplify to $\frac{1}{x-2}$
I have simplified $x^2-4$, which becomes:
$\frac{x^2}{(x-2)(x+2)} - \frac{x+1}{x+2}$
However, if I combine the fractions I get, $x^2-x-1$ for the numerator, which can't be factored. That's where I get stuck.
How can I get $\frac{1}{x-2}$ out of this problem?
|
Write $$\frac{x^2}{(x-2)(x+2)}-\frac{x+1}{x+2}=\frac{x^2}{(x-2)(x+2)}-\frac{(x+1)(x-2)}{(x+2)(x-2)}=…$$
Note that it must be $$x\ne 2,-2$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3093945",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
}
|
For the equation $y = 4x^2 + 8x + 5$, what are the integer values of x such that y/13 is an integer? For the equation $y = 4x^2 + 8x + 5$, what are the integer values of x such that y/13 is an integer?
For example, if x = 3, $y = 4(3^2) + 8(3) + 5$ = 65 which is divisible by 13
if x = 8, y = 325 which is divisible by 13
if x = 16, y = 1157 which is divisible by 13
if x = 21, y = 1937 which is divisible by 13
I am guessing that values of x = 13i + 3 or x = 13i + 8 where i is an integer will result in a value of y that is evenly divisible by 13.
How do you prove that x = 13i + 3 or x = 13i + 8 will result in a value of y that is evenly divisible by 13?
Is there a general proof to find values of x that will result in a value of y that is evenly divisible by an odd integer p?
|
Hint $\bmod 13\!:\,\ 0\equiv -3(4x^2\!+\!8x+5)\equiv x^2\! +\!2x\!-\!15\equiv (x\!+\!5)(x\!-\!3)\ $ so $\,x\equiv -5,3$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3095147",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 2
}
|
How to prove $\frac{(x+y)(y+z)(z+x)}{4xyz}≥\frac{x+z}{y+z}+\frac{y+z}{x+z}$ for $x,y,z>0$?
Prove that for $x,y,z$ positive numbers:
$$
\frac{(x+y)(y+z)(z+x)}{4xyz}≥\frac{x+z}{y+z}+\frac{y+z}{x+z}
$$
I tried to apply MA-MG inequality:
$x+y≥2\sqrt{xy}$ and the others and multiply them but it becomes
$\frac{(x+y)(y+z)(z+x)}{4xyz}≥2$ but $\frac{x+z}{y+z}+\frac{y+z}{x+z}≥2$ so it doesn't work like that.
I don't know what to apply here.
|
We need to prove that
$$\frac{(x+y)(x+z)(y+z)}{4xyz}-2\geq\frac{x+z}{y+z}+\frac{y+z}{x+z}-2$$ or
$$\sum_{cyc}\frac{(x-y)^2}{xy}\geq\frac{4(x-y)^2}{(y+z)(x+z)},$$ which is true by C-S:
$$\sum_{cyc}\frac{(x-y)^2}{xy}\geq\frac{(x-y+x-z+z-y)^2}{xy+xz+yz}=$$
$$=\frac{4(x-y)^2}{xy+xz+yz}\geq\frac{4(x-y)^2}{z^2+xy+xz+yz}=\frac{4(x-y)^2}{(y+z)(x+z)}.$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3096406",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
}
|
Calculate limit with squares $\lim_{n \to \infty}\left(\sqrt[3]{n^3+2n-1}-\sqrt[3]{n^3+2n-3}\right)^6 \cdot (1+3n+2n^3)^4$
$$\lim_{n \to \infty}\left(\sqrt[3]{n^3+2n-1}-\sqrt[3]{n^3+2n-3}\right)^6 \cdot (1+3n+2n^3)^4 $$
What I did was to multiply it and I got $\frac{1}{2}$ as the final result. Could someone confirm if it's correct?
|
The hint:
Since $$a-b=\frac{a^3-b^3}{a^2+ab+b^2},$$ we obtain:
$$\left(\sqrt[3]{n^3+2n-1}-\sqrt{n^3+2n-3}\right)^6(1+3n+2n^3)^4=$$
$$=\left(\tfrac{n^3+2n-1-(n^3+2n-3)}{\sqrt[3]{(n^3+2n-1)^2}+\sqrt[3]{(n^3+2n-1)(n^3+2n-3)}+\sqrt[3]{n^3+2n-3)^2}}\right)^6(1+3n+2n^3)^4=$$
$$=\frac{64(1+3n+2n^3)^4}{\left(\sqrt[3]{(n^3+2n-1)^2}+\sqrt[3]{(n^3+2n-1)(n^3+2n-3)}+\sqrt[3]{n^3+2n-3)^2}\right)^6}=$$
$$=\frac{64\left(\frac{1}{n^3}+\frac{3}{n^2}+2\right)^4}{\left(\sqrt[3]{\left(1+\frac{2}{n^2}-\frac{1}{n^3}\right)^2}+\sqrt[3]{\left(1+\frac{2}{n^2}-\frac{1}{n^3}\right)\left(1+\frac{2}{n^2}-\frac{3}{n^3}\right)}+\sqrt[3]{\left(1+\frac{2}{n^2}-\frac{3}{n^3}\right)^2}\right)^6}.$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3106079",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
}
|
How do we factor out $x^2 - x -2$ in this expression? Suppose we're given that
$$x^4 - 2x^3 +x-2$$
How do we factor out $x^2 - x -2$ in this expression?
$$(x^4 -x^3- 2x^2)-(x^3-x^2-2x)+(x^2-x-2) = x^2(x^2 -x-2)-x(x^2-x-2)+(x^2-x-2) = (x^2-x-2)(x^2-x+1)$$
This satisfies with what we want to get. However, I do not seem to have understood what is done there. Could I get your assistance in order to understand it? Perhaps there's better way of factoring.
Regards
|
If $x^2 - x -2$ is indeed a factor of $x^4 - 2x^3 + x -2$, that means I can write:
$$ x^4 - 2x^3 + x - 2 = P(x)(x^2 -x - 2)$$
In order to reproduce the $x^4$ term, that means that the leading term of $P(x)$ is $x^2$. Now
$$ x^2(x^2 -x -2) = x^4 -x^3 -2x^2 $$
which is not the same as what we started with. Let's try to salvage the situation by adding/subtracting whatever we need to both sides of this equation in order to make the RHS agree with our original polynomial. We need another $-x^3$, we need to get rid of the $-2x^2$ and we need to add $x-2$:
$$ \begin{align}
x^2(x^2-x-2) -x^3 +2x^2 + x -2 & = x^4 - x^3 -2x^2 - x^3 + 2x^2 + x -2 \\
& = x^4 -2x^3 +x -2
\end{align}$$
We continue the same way, starting with the $-x^3$ term on the LHS. We will try to write $$-x^3 + 2x^2 + x - 2 = Q(x)(x^2 - x -2)$$ We see that the leading term of $Q(x)$ must be $-x$ in order to reproduce the $-x^3$ term.
$$ (-x)(x^2 -x -2) = -x^3 +x^2 + 2x$$
which is not the same as $-x^3 + 2x^2 + x - 2$, so we'll add/subtract whateve we need to both sides: we need an extra $x^2$, we need to take away $x$ and add $-2$:
$$ (-x)(x^2 -x -2) +x^2 - x - 2= -x^3 +x^2 + 2x +x^2 - x - 2 = -x^3 + 2x^2 + x - 2$$
We continue on with what's left in the LHS: we want to write
$$x^2 - x - 2= R(x)(x^2 - x - 2)$$
Obviously R(x) is the constant polynomial $R(x) = 1$. So gathering the pieces together we can say:
$$ \begin{align}
x^4 - 2x^3 + x - 2 &= x^2(x^2-x-2) -x^3 +2x^2 + x -2 \\
&= x^2(x^2-x-2) -x(x^2-x-2) + x^2-x-2 \\
&= x^2(x^2-x-2) -x(x^2-x-2) + 1(x^2-x-2) \\
&= (x^2 - x + 1)(x^2-x-2)
\end{align}$$
Maybe this makes it clearer where the original manipulations came from, but what I did above is pretty much long division (of polynomials) without saying so. In long division, you would arrange as in long division of numbers but instead of calculating successive digits of the (numerical) quotient, you calculate successive terms of the (polynomial) quotient. It looks like this:
$$
\require{enclose}
\begin{array}{r}
{\color{red}{x^2}} {\color{blue}{-x}} + {\color{green}1} \\
x^2 - x - 2 \enclose{longdiv}{x^4 - 2x^3 + \phantom{2x^2 +} x - 2} \\
\underline{\color{red}{x^4 - \phantom{2}x^3 - 2x^2}\phantom{+x - 2}} \\
\phantom{x^4 }\phantom{- 2}-x^3 + 2x^2 + x - 2 \\
\underline{\color{blue}{-x^3 + x^2 + 2x}\phantom{- 2}} \\
\phantom{x^4 + -x^3}{x^2 -x -2} \\
\underline{\color{green}{x^2 -x -2 }}\\
\underline{0}
\end{array}
$$
Hope this helps.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3106729",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 4
}
|
How to solve this recurrence relation using generating functions: $a_n = 10 a_{n-1}-25 a_{n-2} + 5^n\binom{n+2}2$? How can we solve the following recurrence relation using GF?
$a_n = 10 a_{n-1}-25 a_{n-2} + 5^n {n+2 \choose 2}$ , for each $n>2, a_0 = 1, a_1 = 15$
I think that most of it is pretty straightforward. What really concerns me is this part
$5^n{n+2 \choose 2}$
Problem further analyzed
After trying to create generating functions in the equation we end up in this
$\sum_{n=2}^\infty a_n x^n = 10 \sum_{n=2}^\infty {a_{n-1}} \sum_{n=2}^\infty a_{n-2} x^n + \sum_{n=2}^\infty {n+2 \choose 2}5^nx^n$
Lets take this part:
$ \sum_{n=2}^\infty {n+2 \choose 2}5^nx^n = \sum_{n=2}^\infty \frac{(n+1)(n+2)}{2} 5^nx^n $
then?
|
Hint for finding the GF.
Note that $(n+1)(n+2)x^n=\frac{d^2}{dx^2}\left(x^{n+2}\right)$
and therefore
$$\sum_{n=2}^\infty {n+2 \choose 2}5^nx^n = \frac{1}{2\cdot 5^2}\frac{d^2}{dx^2}\left(
\sum_{n=2}^\infty (5x)^{n+2}
\right).$$
Then recall that $\sum_{n=0}^{\infty}z^n=\frac{1}{1-z}$.
Moreover in your attempt it should be
$$\sum_{n=2}^\infty a_n x^n = 10 \sum_{n=2}^\infty {a_{n-1}}x^n-25 \sum_{n=2}^\infty a_{n-2} x^n + \sum_{n=2}^\infty {n+2 \choose 2}5^nx^n.$$
that is
$$\sum_{n=2}^\infty a_n x^n = 10x \sum_{n=1}^\infty {a_{n}}x^n-25x^2 \sum_{n=0}^\infty a_{n} x^n + \sum_{n=2}^\infty {n+2 \choose 2}5^nx^n.$$
Can you take it from here and find $f(x) =\sum_{n=0}^\infty a_n x^n$?
Alternative way for finding $a_n$ without the GF.
The given recurrence is a non-homogeneous linear recurrence relation with constant coefficients with characteristic equation
$$z^2-10z+25=(z-5)^2=0$$
and non-homogeneous term $5^n {n+2 \choose 2}$ which is a second degree polynomial multiplied by a power of $5$, which is the root of multiplicity $2$ of the characteristic equation.
Hence the general term of the recurrence with $a_0 = 1$, $a_1 = 15$ has the form
$$a_n= 5^n(An^4+Bn^3+Cn+D)$$
where $A,B,C,D$ are real constant to be determined.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3108550",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
}
|
Minimum point of $x^2+y^2$ given that $x+y=10$ How do you I approach the following question:
Find the smallest possible value of $x^2 + y^2$ given that $x + y = 10$.
I can use my common sense and deduce that the minimum value is $5^2 + 5^2 = 50$. But how do you approach this mathematically?
Thanks in advance.
|
To minimize $f(x,y) = x^2+y^2$; given that $y=10-x$.
Then:
$$f(x,10-x)=x^2+(x-10)^2$$
$f$ is minimized for $f'=0$, so:
$$f'=2x+2(x-10)=0$$
Gives:
$$x=5\land y=5$$
Thus we have:
$$\min (x^2+y^2)=5^2+5^2=50$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3108814",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 9,
"answer_id": 7
}
|
Studying the convergence of the series $\sqrt{2}+\sqrt{2+\sqrt{2}}+\sqrt{2+\sqrt{2+\sqrt{2}}}+ \cdots$ Studying the convergence of the series $$\sqrt{2}+\sqrt{2+\sqrt{2}}+\sqrt{2+\sqrt{2+\sqrt{2}}}+ \cdots$$
I saw this problem and I tried to do it my own way but I don't know what I'm doing wrong because I'm getting a divergent series while in the solution it says that it's convergent.
This is what I tried to do
I can write $\sqrt{2}=2\cos\frac{\pi}{4}=2\cos\frac{\pi}{2^2}$
$\sqrt{2+\sqrt{2}}=\sqrt{2+2\cos\frac{\pi}{4}}=2\sqrt{\frac{1+\cos2\frac{\pi}{8}}{2}}=2\cos\frac{\pi}{8}=2\cos\frac{\pi}{2^3}$
Continuing like this I get $$\underbrace{\sqrt{2+\sqrt{2+\cdots+\sqrt{2}}}}_{n\text{ times}}=2\cos\frac{\pi}{2^{n+1}}$$
So I can write the series $$\sqrt{2}+\sqrt{2+\sqrt{2}}+\sqrt{2+\sqrt{2+\sqrt{2}}}+ \cdots=2\sum_{n=1}^\infty\cos\frac{\pi}{2^{n+1}}$$
Since $$\lim_{n\to\infty}a_{n}=\lim_{n\to\infty}\cos\frac{\pi}{2^{n+1}}=1\ne0$$ the series diverges
|
I think what is meant is that the sequence $a_0 = \sqrt{2}, a_{n+1} = \sqrt{2 + a_n}$ converges. This can be shown by proving that for all $n \in \mathbb{N}$ we have $a_n \leq 2$ by induction. It is clear that the sequence is monotonically increasing. Since the sequence is monotonically increasing and has upper bound $2$ it follows that the sequence converges.
Induction step is: $a_{n+1} = \sqrt{2 + a_n} \leq \sqrt{2+2} = 2$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3109128",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
}
|
Proving inequality involving radical I have tried many examples for the following inequality in Mathematica. It is likely true. I need some help proving it.
For $x_1, x_2, y_1, y_2 \geq 0$,
\begin{align*}
6(x_1^5y_1 + x_1y_1^5) + 4(x_2^3y_2 + x_2y_2^3) \leq 6(x_1^6 + x_2^4)^{\frac{5}{6}}(y_1^6 + y_2^4)^{\frac{1}{6}} + 6(x_1^6 + x_2^4)^{\frac{1}{6}}(y_1^6 + y_2^4)^{\frac{5}{6}}.
\end{align*}
I guess the difficulty of this is from $x_2$ and $y_2$ parts because the highest and lowest degrees of them from LHS and RHS are not the same.
|
Note by Hölder's inequality
$$(x_1^6+x_2^4)^{5/6} (y_1^6+y_2^4)^{1/6} \geqslant x_1^5y_1+x_2^{10/3}y_2^{2/3}$$
and
$$(x_1^6+x_2^4)^{1/6} (y_1^6+y_2^4)^{5/6} \geqslant x_1y_1^5+x_2^{2/3}y_2^{10/3}$$
Using this, it is left to show
$$3(x_2^{10/3}y_2^{2/3}+x_2^{2/3}y_2^{10/3}) \geqslant 2(x_2^3y_2 + x_2y_2^3)$$
In fact as $(\frac{10}3, \frac23) \succ (3, 1)$, by Muirhead (or you can use AM-GM) the stronger inequality holds:
$$x_2^{10/3}y_2^{2/3}+x_2^{2/3}y_2^{10/3} \geqslant x_2^3y_2 + x_2y_2^3$$
Hence:
$$6(x_1^5y_1 + x_1y_1^5) + \color{red}{6}(x_2^3y_2 + x_2y_2^3) \leq 6(x_1^6 + x_2^4)^{\frac{5}{6}}(y_1^6 + y_2^4)^{\frac{1}{6}} + 6(x_1^6 + x_2^4)^{\frac{1}{6}}(y_1^6 + y_2^4)^{\frac{5}{6}}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3112749",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
}
|
Finite double sum $\sum_{k=0}^N\sum_{l=0}^M\left\lfloor\frac{k+l}{c}\right\rfloor$; any advanced summation technique?
Let $M,N,c$ be positive integer. It was astonishing when trying to solve $\sum_{k=0}^N\sum_{l=0}^M\left\lfloor\frac{k+l}{c}\right\rfloor$ to obtain this rather complex looking result
\begin{align*}
\color{blue}{\sum_{k=0}^N}&\color{blue}{\sum_{l=0}^M\left\lfloor\frac{k+l}{c}\right\rfloor}\tag{1}\\
&=-\frac{1}{2}c(1-c)\left\lfloor\frac{M}{c}\right\rfloor\left\lfloor\frac{N}{c}\right\rfloor\left\lfloor\frac{2c-2}{c}\right\rfloor\\
&\quad+\frac{1}{2}m(m+1)\left\lfloor\frac{N}{c}\right\rfloor\left\lfloor\frac{c+m-1}{c}\right\rfloor
+\frac{1}{2}n(n+1)\left\lfloor\frac{M}{c}\right\rfloor\left\lfloor\frac{c+n-1}{c}\right\rfloor\\
&\quad+\left(m+1-\frac{c}{2}\right)(n+1)\left\lfloor\frac{M}{c}\right\rfloor+\left(n+1-\frac{c}{2}\right)(m+1)\left\lfloor\frac{N}{c}\right\rfloor\\
&\quad+\frac{1}{2}(n+1)c\left\lfloor\frac{M}{c}\right\rfloor^2+\frac{1}{2}(m+1)c\left\lfloor\frac{N}{c}\right\rfloor^2\\
&\quad+c(m+n+2-c)\left\lfloor\frac{M}{c}\right\rfloor\left\lfloor\frac{N}{c}\right\rfloor\\
&\quad+\frac{1}{2}c^2\left\lfloor\frac{M}{c}\right\rfloor\left\lfloor\frac{M}{c}\right\rfloor
\left(\left\lfloor\frac{M}{c}\right\rfloor+\left\lfloor\frac{N}{c}\right\rfloor\right)\\
&\quad+\left(\frac{1}{2}((m+n)^2+3(m+n)+2)-\left(m+n+\frac{3}{2}\right)c+\frac{1}{2}c^2\right)\left\lfloor\frac{m+n}{c}\right\rfloor
\end{align*}
Moreover what really baffled me was the long and cumbersome road to find this identity. In fact I needed to derive closed formulas for the three sums ($n,a,c$ integer $n\geq 0, c>0, 0\leq a <c$):
\begin{align*}
&\sum_{k=0}^n\left\lfloor\frac{k+a}{c}\right\rfloor,\qquad\sum_{k=0}^n k\left\lfloor \frac{k+a}{c}\right\rfloor,\qquad\sum_{k=0}^n\left\lfloor\frac{k+a}{c}\right\rfloor^2\\
\end{align*}
and used these intermediate results to calculate (1) with tedious, lengthy calculations. You might have a look at this answer to see some steps.
Question: Do we have some other, maybe more advanced techniques to tackle (1) more efficiently than I did in the referred answer?
|
Let $$S_n=\sum_{k=0}^{n-1}\left\lfloor \frac kc\right\rfloor$$
and
$$ T_n=\sum_{k=0}^{n-1}S_k.$$
Then we are looking for
$$ \begin{align}\sum_{k=0}^N\sum_{l=0}^M\left\lfloor\frac{k+l}{c}\right\rfloor
&=\sum_{k=0}^N(S_{M+k+1}-S_k)\\
&=T_{N+M+2}-T_{M+1}-T_{N+1}\end{align}$$
Note that for $n=qc+r$ with $0\le r<c$, we have
$$S_n=c\sum_{k=0}^{q-1}k+rq=\frac c2{q(q-1)}+rq.$$
By the same trick,
$$\begin{align}T_n&=\sum_{k=0}^{q-1}\left(\frac c2{k(k-1)}+\frac{k(k-1)}2k\right)+r\cdot\frac c2{q(q-1)} +\frac{r(r-1)}2q\\
&=\frac {cq(q-1)(q-2)}6+\frac{q(q-1)(q-2)(3q-1)}{12}+\frac{rq(cq-c+r-1)}2\end{align}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3114262",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
}
|
Proving that $\lim_{v\rightarrow \infty}\left[\frac{v^2}{v^3 + 1} + \frac{v^2}{v^3 + 2} +\cdots + \frac{v^2}{v^3+v}\right]= 1 $ I wonder if my solution that $\lim_{v\to\infty}\left[\frac{v^2}{v^3 + 1} + \frac{v^2}{v^3 + 2} + \cdots + \frac{v^2}{v^3+v}\right]= 1 $ is correct.
$$\frac{v^2}{v^3 + 1} + \frac{v^2}{v^3 + 2} + ... + \frac{v^2}{v^3+v} = \frac{\frac{1}{v}}{1 + \frac{1}{v^3}} + \frac{\frac{1}{v}}{1 + \frac{2}{v^3}} + ... + \frac{\frac{1}{v}}{1+\frac{1}{v^2}} \rightarrow 0 $$
So, the limit is not equal to 1.
Where is my mistake ?
|
I think that its good idea to use the sandwich theorem.
Let $a_v $ the sequence that you posted. Then
$$a_v \leq \frac{v^2}{v^3 + v} + \cdots + \frac{v^2}{v^3 +v} = b_v $$
And
$$b_v = v\left(\frac{v^2}{v^3+v}\right) \to 1$$
On the other hand you must use
$$c_v = \frac{v^2}{v^3 +1} + \cdots + \frac{v^2}{v^3 +1} \leq a_v $$
And..
$$c_v = v\left(\frac{v^2}{v^3+1}\right) \to 1$$
And
$$c_v \leq a_v \leq b_v $$
With $$\lim c_v = \lim b_v = 1$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3114947",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
}
|
Prove that $a^2+b^2+c^2\geqslant\frac{1}{3}$ given that $a\gt0, b\gt0, c\gt0$ and $a+b+c=1$, using existing AM GM inequality Using the AM and GM inequality, given that
$a\gt0, b\gt0, c\gt0$ and $a+b+c=1$ prove that
$$a^2+b^2+c^2\geqslant\frac{1}{3}$$
|
HINT: You can use your idea of squaring $a+b+c$, but also note that $\color{blue}{ab+bc+ca \le a^2 + b^2 + c^2}$, which you can prove with the help of AM-GM. (Hint for proving this: the AM-GM inequality tells us what about $a^2 + b^2, b^2+c^2$ and $c^2+a^2$?)
One more hint (based on a suggestion from user qsmy): let $x = a^2+b^2+c^2$ and $y = ab+bc+ca$. Squaring both sides of $a+b+c=1$ gives $x+2y=1$, and the blue inequality is $x\geq y$. Can you see it now?
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3115752",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
}
|
Evaluate this question based on series and limits.
For $a \in \mathbb R,a≠-1$
$$\lim_{n\to\infty}\frac{1^a+2^a+\cdots +n^a}{(n+1)^{a-1}[(na+1)+(na+2)+(na+3)+\cdots+(na+n)]}=\frac{1}{60}$$
Then find the values of $a $.
I tried to solve this problem using approximation but I got the the value of $a$ as $\dfrac{-3}{2}$. The source of this question is JEE advanced $2013$ of India.
|
For $a>-1$ and $a\neq -\frac 12$ we have
\begin{align}
\frac{1^a+2^a+\cdots +n^a}{(n+1)^{a-1}[(na+1)+(na+2)+(na+3)+\cdots+(na+n)]}
&\sim\frac {n^{a+1}/(a+1)}{n^{a-1}(a+1/2)n^2}\\
\to\frac 2{(a+1)(2a+1)}
\end{align}
from which $a=7$.
For $a<-1$ the numerator converges to positive value, hence the limit is $-\infty $.
Finally, for $a=-\frac 12$ we have
\begin{align}
\frac{1^a+2^a+\cdots +n^a}{(n+1)^{a-1}[(na+1)+(na+2)+(na+3)+\cdots+(na+n)]}
&\sim\frac {2\sqrt n}{n^{-3/2}n/2}\\
\sim 4n\to +\infty
\end{align}
The answer $a=7$ is confirmed:
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3115904",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
}
|
Combining simultaneous linear recurrence relations I was studying the sequence A249665, which I will call $a_n$, and came up with a second sequence $b_n$ for which I could prove that the following recurrence holds.
\begin{align}
a_n&=a_{n-1}+a_{n-4}+a_{n-5}+b_{n-2}+b_{n-3}
\\b_n&=a_{n-1}+a_{n-2}+a_{n-3}+a_{n-4}+b_{n-2}+b_{n-3}+b_{n-4}
\end{align}
Here are some initial values.
\begin{align}
(a_0,...)&=(0,1,1,1,2,6,14,28,56,118,254,541,1140,...)
\\(b_0,...)&=(0,0,1,2,4,8,17,37,79,166,349,738,1563,...)
\end{align}
On the OEIS they give a linear recurrence in three sequence, so when I saw that I felt good about myself for finding one in only two sequences. However a bit further on, they give a recurrence in $a_n$ only, which caught me by surprise.
$$a_n=2a_{n-1}-a_{n-2}+2a_{n-3}+a_{n-4}+a_{n-5}-a_{n-7}-a_{n-8}$$
The companion matrix of this recurrence has the same dimension as the one for my recurrence, which leads me to believe there should be some method to find this recurrence out of mine.
So is there a general method to combine two simultaneous linear recurrences into one? Or can you at least find a way to deduce the one from OEIS from the simultaneous one I found?
|
I think I figured it out. At least for this specific case. I think the idea applies pretty generally, but can become much more complicated. This case is just a bit of algebraic manipulation.
Note that from the first recurrence formula we get $b_n+b_{n-1}=a_{n+2}-a_{n+1}-a_{n-2}-a_{n-3}$. We find
\begin{align}
a_n
&=a_{n-1}+a_{n-4}+a_{n-5}+b_{n-2}+b_{n-3}
\\&=a_{n-1}+a_{n-4}+a_{n-5}+b_{n-3}
\\&+(a_{n-3}+a_{n-4}+a_{n-5}+a_{n-6}+b_{n-4}+b_{n-5}+b_{n-6})
\\&=a_{n-1}+a_{n-3}+2a_{n-4}+2a_{n-5}+a_{n-6}
\\&+(b_{n-3}+b_{n-4})+(b_{n-5}+b_{n-6})
\\&=a_{n-1}+a_{n-3}+2a_{n-4}+2a_{n-5}+a_{n-6}
\\&+(a_{n-1}-a_{n-2}-a_{n-5}-a_{n-6})+(a_{n-3}-a_{n-4}-a_{n-7}-a_{n-8})
\\&=2a_{n-1}-a_{n-2}+2a_{n-3}+a_{n-4}+a_{n-5}-a_{n-7}-a_{n-8}.
\end{align}
To get an idea of the general approach, consider the problem of finding a recurrence for $b_n$. By applying the first recurrence formula on every $a$ term in the second, we obtain
\begin{align}
b_n
&=b_{n-2}+b_{n-3}+b_{n-4}
\\&+(a_{n-2}+a_{n-5}+a_{n-6}+b_{n-3}+b_{n-4})
\\&+(a_{n-3}+a_{n-6}+a_{n-7}+b_{n-4}+b_{n-5})
\\&+(a_{n-4}+a_{n-7}+a_{n-8}+b_{n-5}+b_{n-6})
\\&+(a_{n-5}+a_{n-8}+a_{n-9}+b_{n-6}+b_{n-7}).
\end{align}
By the second recurrence formula we get $a_n+a_{n-1}+a_{n-2}+a_{n-3}=b_{n+1}-b_{n-1}-b_{n-2}-b_{n-3}$, giving
\begin{align}
b_n
&=b_{n-2}+2b_{n-3}+3b_{n-4}+2b_{n-5}+2b_{n-6}+b_{n-7}
\\&+(b_{n-1}-b_{n-3}-b_{n-4}-b_{n-5})
\\&+(b_{n-4}-b_{n-6}-b_{n-7}-b_{n-8})
\\&+(b_{n-5}-b_{n-7}-b_{n-8}-b_{n-9})
\\&=b_{n-1}+b_{n-2}+b_{n-3}+3b_{n-4}+2b_{n-5}+b_{n-6}-b_{n-7}-2b_{n-8}-b_{n-9}.
\end{align}
If you want to practice this youself, I advise to try and come up with a recurrence formula for the partial sums of fibonacci numbers, so give a recurrence formula for $s_n$ when $f_n=f_{n-1}+f_{n-2}$ and $s_n=s_{n-1}+f_n$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3118542",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
}
|
Is my integral in fully reduced form?
I have to integrate this:
$$\int_0^1 \frac{x-4}{x^2-5x+6}\,dx$$
Now
$$\int_0^1 \frac{x-4}{(x-3)(x-2)}\, dx$$
and by using partial fractions we get
$$\frac{x-4}{(x-3)(x-2)} = \frac{A}{x-3} + \frac{B}{x-2}$$
$$x-4 = A(x-2) + B(x-3)$$
$$= Ax - 2A + Bx - 3B$$
$$x-4 = (A+B)x - 2A - 3B$$
so
$$A+B=1$$ or $$2X+2B = 2$$
$-B = -2$ and $B = 2$ and $A = -1$
Then
$$\int_0^1 \frac{x-4}{x^2-5x+6} = \int_0^1 \frac{-1}{x-3}dx + \int_0^1 \frac{2}{x-2} dx$$
usub using $u = x-3$ and $du = dx$
so $$ -1 \ln | x-3 | \rbrack_0^1 + 2 \ln |x-2| \rbrack_0^1$$
$$-1 ( \ln 2 - \ln 3) + 2 (\ln 1 - \ln 2)$$
$$-\ln 2 + \ln 3 + 2\ln 1 - 2\ln 2 = -3 * \ln 2 + \ln 3 + 2\ln 1$$
Is there anyway to simplify this further? Wolfram has the answer at $-\ln(8/3)$ and I'm not sure how to simplify to here? How does my work look?
|
It is fine, except that I don't see where that $3$ at the last line comes from. Anyway,\begin{align}-\bigl(\log(2)-\log(3)\bigr)+2\bigl(\log(1)-\log(2)\bigr)&=\log(3)-3\log(2)\\&=\log(3)-\log(2^3)\\&=\log\left(\frac38\right)\\&=-\log\left(\frac83\right).\end{align}
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3118767",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
}
|
What is the number of pairs of $(a,b)$ of positive real numbers satisfying $a^4+b^4<1$ and $a^2+b^2>1$?
The number of pairs of $(a,b)$ of positive real numbers satisfying
$a^4+b^4<1$ and $a^2+b^2>1$ is -
$(i)$0
$(ii)$1
$(iii)$2
$(iv)$ more than 2
Solution:We have $a,b>0$,
According to the given situation,$0<a^4+b^4<1<a^2+b^2\implies a^4+b^4<a^2+b^2\implies 0<a^2(a^2-1)<b^2(1-b^2)\implies b^2(1-b)(1+b)<0 ;a^2(a-1)(a+1)>0\implies a\in(-\infty,-1)\cup (0,1);b\in(-1,0)\cup(1,\infty)$
But, in particular, if we choose $a=-1/2,b=1/2$,then $a^4+b^4=1/16+1/16=1/8<1$ and $a^2+b^2=1/4+1/4=1/2<1$.Which contradicts the given condition $a^2+b^2>1$
Where is the mistake in my approach? How should I approach this problem(means how to think ?). Can Triangle inequality be used here?
Please provide some hint...
|
Since the inequalities are strict, options (ii) and (iii) are ruled out by general notions of continuity: As soon as you have a single solution $(a,b)$ in the first quadrant, you have an entire open ball of solutions of some (possibly very small) radius. So it suffices to exhibit a single solution. The values $a=b=0.8$ do the trick: $0.8^2=0.64\gt0.5$, so $a^2+b^2\gt1$ but $0.8^4=0.64^2\lt0.7^2=0.49\lt0.5$, so $a^4+b^4\lt1$.
If the argument invoking continuity seems too advanced, a little extra work shows that $(a,b)=(0.8,0.8)$, $(0.8,0.7)$, and $(0.7,0.8)$ already give more than two solutions.
As for the OP's question, "Where is the mistake in my approach?," there are multiples errors in it. Each of the following implications is incorrect:
$$\begin{align}
a^4+b^4\lt a^2+b^2&\implies 0\lt a^2(a^2-1)\\
0\lt b^2(1-b^2)&\implies b^2(1-b)(1+b)\lt0\\
b^2(1-b)(1+b)\lt0&\implies b\in(-1,0)\cup(1\infty)\\
a^2(a-1)(a+1)\lt0&\implies a\in(-\infty,-1)\cup(0,1)
\end{align}$$
But even if these implications were correct (or were replaced with something that is correct), the "contradiction" the OP obtains with a particular choice of values at the end of the string of implications is not a contradiction, because the implications only go in one direction (as pointed out in a comment by Hagen von Eitzen). I.e., $"\implies"$ is not the same as $"\iff"$. (Note also that the problem calls for pairs of positive real numbers, so it's inappropriate to $a=-1/2$ anyway.)
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3122623",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 3
}
|
Finding powers of 2 and 3 in modular arithmetic
Find all the powers of $2$ and $3$ modulo $17$.
How would you solve this question and explain the steps please!
|
Recall that:
$$2^m\equiv k \pmod{17} \implies 2^{m+1}\equiv 2k \pmod {17}$$
We can then list, remembering that if our $k$ is greater than $17$, we subtract $17$ from it.
$$1\equiv 1 \pmod {17}$$
$$2\equiv 2 \pmod {17}$$
$$4\equiv 4 \pmod {17}$$
$$8\equiv 8 \pmod {17}$$
$$16\equiv 16 \pmod {17}$$
$$32\equiv 15 \pmod {17}$$
(Because $16\cdot 2= 32$, but $32>17$ so we apply $32-17=15$)
$$64\equiv 13 \pmod {17}$$
(Because $15\cdot 2 = 30$, but $30>17$, so we apply $30-17=13$)
$$128\equiv 9 \pmod {17}$$
(same as above)
$$256\equiv 1 \pmod {17}$$
And now we have a cycle, so $2^k\equiv 1,2,4,8,9,13,15,16\pmod{17}$
Now do the same for the powers of $3$. As a check, you should get $3^k\not\equiv 0 \pmod{17}$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3123193",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
}
|
Can you help me solve this algebra problem? Hi I need to solve this problem and I don’t know how. I’d appreciate your help.
If $x - \frac{ayz}{x^2} = y - \frac{azx}{y^2} = z - \frac{axy}{z^2}$ and $x\neq y\neq z$, then $$x - \frac{ayz}{x^2} = y - \frac{azx}{y^2} = z - \frac{axy}{z^2} = x + y + z - a$$
I think I need to
$x^3 - ayz = x^2k$
$y^3 - azx = y^2k,$
$z^3 - axy = z^2k$
then to multiply both sides of each equality by some quantity, add them all together and factor but I don’t know how to find that quantity.
|
Denote $k=xyz$ and $b$ the common value of $x-\frac{ayz}{x^2}=y-\frac{azx}{y^2}=z-\frac{axy}{z^2}$. We can see that the equation
$$
t-\frac{ak}{t^3}=b\tag{*}
$$ is satisfied by $t=x,y,z$. Hence $x,y,z$ are distinct, non-zero roots of $(*)$. Note that $(*)$ can be written as a polynomial equation of degree 4
$$
t^4-b t^3-ak=0.
$$ By Vieta's formula, the other root $w$ satisfies
$$
x+y+z+w=b,\quad xyzw=-ak.
$$ Since $k=xyz\ne 0$ it follows that
$$
w=-a=b-x-y-z,
$$ hence we get
$$
x-\frac{ayz}{x^2}= y-\frac{azx}{y^2}=z-\frac{axy}{z^2}=b=x+y+z-a.
$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3123743",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 0
}
|
Stuck on a Geometry Problem
$ABCD$ is a square, $E$ is a midpoint of side $BC$, points $F$ and $G$ are on the diagonal $AC$ so that $|AF|=3\ \text{cm}$, $|GC|=4\ \text{cm}$ and $\angle{FEG}=45 ^{\circ}$. Determine the length of the segment $FG$.
How can I approach this problem, preferably without trigonometry?
|
Let $H$ be the midpoint of $AC$ and $\angle EIC= 90^{\circ}$. We can observe that $$FH+3=HG+4,\quad FH+HG=x.$$ So we obtain $HG=\frac{x-1}2$. Since two corresponding angles are congruent; $\angle FEG =\angle EHG=45^{\circ}$ and $\angle EGF=\angle HGE$, we have that $\triangle FEG$ and $\triangle EHG$ are similar to each other. This gives $$FG:EG=EG:HG\implies EG^2 = FG\cdot HG=\frac{x(x-1)}2.$$ Now, note that $EI=\frac14 AC=\frac{x+7}4$ and $IG=IH-GH=\frac{x+7}{4}-\frac{x-1}2=\frac{9-x}{4}$. Since $\triangle EIG$ is a right triangle, by Pythagorean theorem, we find that
$$
EG^2=\frac{x(x-1)}{2}=EI^2+IG^2=\frac{(x+7)^2}{16}+\frac{(9-x)^2}{16},
$$ which implies $x=5 $ or $x=-\frac{13}3$. Since $x>0$, we get $x=5$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3126169",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "25",
"answer_count": 12,
"answer_id": 4
}
|
integrate sin(x)cos(x) using trig identity. Book tells me the answer is:
$$ \int \sin(x)\cos(x) dx = \frac{1}{2} \sin^{2}(x) + C $$
however, I get the result:
$$
\sin(A)\cos(B) = \frac{1}{2} \sin(A-B)+\frac{1}{2}\sin(A+B)
$$
$$
\begin{split}
\int \sin(x)\cos(x) dx
&= \int \left(\frac{1}{2}\sin(x-x) + \frac{1}{2}\sin(x+x)\right) dx \\
&= \int \left(\frac{1}{2}\sin(0) + \frac{1}{2}\sin(2x)\right) dx \\
&= \int \left(\frac{1}{2}\sin(2x)\right) dx \\
&= -\frac{1}{2} \frac{1}{2}\cos(2x) +C\\
&= -\frac{1}{4} \cos(2x) +C
\end{split}
$$
How did the book arrive at the answer $\frac{1}{2}\sin^2(x)$?
|
Because
$$\frac{1}{2} \sin^2(x) = \frac{1-\cos(2x)}{4}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3126227",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 2
}
|
How to deduce the following complex number problem I am stuck with the following problem that says:
Using the result $$x^n-1=(x^2-1)\prod_{k=1}^{(n-2)/2}[x^2-2x\cos \frac{2k\pi}{n}+1],$$ if $n$ be an even positive integer, deduce that $$\sin \frac{\pi}{32}\sin \frac{2\pi}{32}\sin \frac{3\pi}{32}.........\sin \frac{15\pi}{32}=\frac{1}{2^{13}}$$
Can someone point me in the right direction with some explanation? Thanks in advance for your time.
|
Fixing $n=32$, we can rewrite the equation like this:
$$\frac{x^{32}-1}{x^2-1} = \prod_{k=1}^{15}\left(x^2+1-2x\cos\frac{k\pi}{16}\right)$$
for all $x\neq\pm1$.
We will take the limit as $x\to 1$ of both sides of the equality.
First, we have (using L'Hôpital's rule)
$$\lim\limits_{x\to 1} \frac{x^{32}-1}{x^2-1} = \lim\limits_{x\to 1}\frac{32x^{31}}{2x} = 16$$
Then, plugging in $x=1$, we have
$$x^2+1-2x\cos\frac{k\pi}{16} = 2-2\cos\frac{k\pi}{16} = 4\cdot\frac{1-\cos(k\pi/16)}{2} = 4\sin^2\frac{k\pi}{32}$$
The product becomes
$$\prod_{k=1}^{15}4\sin^2\frac{k\pi}{32}= 4^{15}\left(\prod_{k=1}^{15}\sin\frac{k\pi}{32}\right)^2$$
Note that the limit as $x\to 1$ of the product is equal to its value at $x=1$ because the expression is continuous (it is the product of continuous functions).
Now, equating the limits we get
$$16 = 4^{15}\left(\prod_{k=1}^{15}\sin\frac{k\pi}{32}\right)^2$$
As all the factors in the product are positive, we can take the square root of both sides and we get
$$\sin \frac{\pi}{32}\sin \frac{2\pi}{32}\sin \frac{3\pi}{32}\cdots\sin \frac{15\pi}{32}= \prod_{k=1}^{15}\sin\frac{k\pi}{32} =\frac{1}{2^{13}}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3129039",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
}
|
Find $\lim\limits_{t \to \infty} \int_{0}^{t} \frac{\mathrm dx}{(x^2+a^2)(x^2+b^2)(x^2+c^2)}$
Find $$\lim_{t \to \infty} \int_{0}^{t} \frac{1}{(x^2+a^2)(x^2+b^2)(x^2+c^2)}\mathrm dx$$ where $a,b,c$ are strict positive and dinstinct real numbers.
I know it should be something with arctangent but I don't know how to get there. Can somebody give me some tips, please?
|
Using partial fraction decomposition, we have that
$$\begin{align*}
\frac1{(x^2+a^2)(x^2+b^2)(x^2+c^2)}&=\frac1{c^2-a^2}\left(\frac1{(x^2+a^2)(x^2+b^2)}-\frac1{(x^2+b^2)(x^2+c^2)}\right)\\&=\frac1{c^2-a^2}\left(\frac{1}{b^2-a^2}\left(\frac1{x^2+a^2}-\frac1{x^2+b^2}\right)\right)\\&-\frac1{c^2-a^2}\left(\frac{1}{c^2-b^2}\left(\frac1{x^2+b^2}-\frac1{x^2+c^2}\right)\right).
\end{align*}$$
Using $\lim_{t\to\infty}\int_0^t \frac{1}{x^2+\alpha^2}\mathrm dx=\frac{\pi}{2\alpha}$, $\ \alpha>0$, we can obtain the integral:
$$\begin{align*}
I&=\frac1{c^2-a^2}\frac1{b^2-a^2}\left(\frac1{a}-\frac1b\right)\frac{\pi}2-\frac1{c^2-a^2}\frac1{c^2-b^2}\left(\frac1{b}-\frac1c\right)\frac{\pi}2
\\&=\frac1{c^2-a^2}\frac{\pi}{2ab(a+b)}-\frac1{c^2-a^2}\frac{\pi}{2bc(b+c)}\\
&=\frac{\pi(a+b+c)}{2abc(a+b)(b+c)(c+a)}.
\end{align*}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3131197",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
}
|
Calculate the sum of fractionals
Let $n \gt 1$ an integer. Calculate the sum: $$\sum_{1 \le p \lt q \le
n} \frac 1 {pq} $$ where $p, q$ are co-prime such that $p + q > n$.
Calculating the sum for several small $n$ value I found out that the sum is always $\frac 1 2$.
Now, I'm trying to prove the sum is $\frac 1 2$ using induction by $n$. Suppose it's true for all values less or equals to $n$, trying to prove it for $n + 1$.
$$\sum_{1 \le p \lt q \le n +1,p+q>n+1} \frac 1 {pq} = \sum_{1 \le p \lt q \le n} \frac 1 {pq} + \sum_{1 \le p \lt q = n +1} \frac 1 {pq} = \sum_{1 \le p \lt q \le n} \frac 1 {pq} + \frac 1 {n+1} \sum_{1 \le p \lt n +1} \frac 1 {p} \tag1$$
In the second sum, $p$ and $n+1$ are coprime.
$$\sum_{1 \le p \lt q \le n,p+q>n+1} \frac 1 {pq} = \sum_{1 \le p \lt q \le n, p+ q>n} \frac 1 {pq} - \sum_{1 \le p \lt q \le n, p+ q=n+1} \frac 1 {pq} = \frac 1 2 - \sum_{1 \le p \lt q \le n, p+ q=n+1} \frac 1 {pq} \tag 2$$
From (1) and (2) I have to prove that $$\frac 1 {n+1} \sum_{1 \le p \lt n +1} \frac 1 {p} = \sum_{1 \le p \lt q \le n, p+ q=n+1} \frac 1 {pq} \tag3$$
where $p,q$ are co-prime and I'm stuck here.
|
If $p+q=n+1$ and $(p,n+1)=1$, then it follows from Euclidean Algorithm that $(p,q)=1$. Now, we are to prove:
$$\frac{1}{n+1} \cdot \sum_{1 \leqslant p<n+1}{\frac{1}{p}}=\sum_{1 \leqslant p<n+1}{\frac{1}{pq}}=\frac{1}{2}\sum{\frac{1}{p(n+1-p)}}$$
This can be seen easily by writing: $$\frac{1}{p(n+1-p)}=\frac{1}{n+1}\bigg(\frac{1}{p}+\frac{1}{(n+1-p)}\bigg)$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3133340",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
}
|
Find the number of ordered triplets satisfying $5\left(x+\frac{1}{x}\right)=12\left(y+\frac{1}{y}\right)=13\left(z+\frac{1}{z}\right)$ Find the number of ordered triplets $(x,y,z)$ of real numbers satisfying $$5\left(x+\frac{1}{x}\right)=12\left(y+\frac{1}{y}\right)=13\left(z+\frac{1}{z}\right)$$
and
$$xy+yz+zx=1$$
My try:
Letting:
$$\left(x+\frac{1}{x}\right)=\frac{k}{5}\tag{1}$$
$$\left(y+\frac{1}{y}\right)=\frac{k}{12}\tag{2}$$
$$\left(z+\frac{1}{z}\right)=\frac{k}{13}\tag{3}$$
By $AM-GM$
$$\left(z+\frac{1}{z}\right)^2 \ge 4$$
So $$k^2 \ge 676$$
Adding $(1),(2),(3)$ we get:
$$x+y+z+\frac{xy+yz+zx}{xyz}=\frac{281k}{780}$$
$$x+y+z+\frac{1}{xyz}=\frac{281k}{780}$$
Any clue from here?
|
From the $xy+yz+zx=1$ we get $x = \frac{1-yz}{y+z}$. Substitute that into $5 (x+1/x)-12 (y+1/y)$. Taking the numerator of this and the numerator of $12 (y + 1/y) - 13 (z + 1/z)$, you can eliminate $y$ and get $z (z^4-1) = 0$. Since $z=0$ is not allowed, $z = \pm 1$. From that you can find the $y$ and $x$ values.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3133577",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
}
|
Confusion in this algebraic limit approaching infinity
Question : Evaluate $\lim_{x \to \infty}{\frac{1}{1+x^2} + \frac{2}{2+x^2} + \frac{3}{3+x^2}+...+\frac{x}{x + x^2}}$.
My working:
$$\lim_{x \to \infty}{\frac{1}{1+x^2} + \frac{2}{2+x^2} + \frac{3}{3+x^2}+...+\frac{x}{x + x^2}}$$
Dividing and transforming each fraction by $x^2$in the numerator and denominator.
$$=\lim_{x \to \infty}{\frac{\frac{1}{x^2}}{\frac{1}{x^2} +1} +...+\frac{\frac{1}{x}}{\frac{1}{x} + 1}}$$
Using algebra of limits we get,
$$0+0+...+0 = 0$$
But when solving this using Sandwich Theorem I get,
$$\text{Let }\lim_{x \to \infty}{\frac{1}{1+x^2} + \frac{2}{2+x^2} + \frac{3}{3+x^2}+...+\frac{x}{x + x^2}} = g(x)$$
$$g(x) < \frac{1}{1+x^2} + \frac{2}{2 + x^2}+...+\frac{x}{1+x^2}$$
$$g(x) < \frac{x(x+1)}{2(1 + x^2)}$$
$$g(x) > \frac{1}{x+x^2} + \frac{2}{x+x^2} +...+\frac{x}{x+x^2}$$
$$g(x) > \frac{x(x+1)}{2(x+x^2)} \to g(x) > \frac{1}{2} $$
$$\text{So, we get } \frac{1}{2}<g(x)<\frac{x(x+1)}{2(1 + x^2)}$$
Applying limits,
$$\lim_{x \to \infty}{\frac{x^2+x}{2(1+x^2)}} = 1/2$$
$$\text{Finally, we get } \frac{1}{2}<g(x)<\frac{1}{2}$$
By Sandwich theorem, $\lim_{x \to \infty}{\frac{1}{1+x^2} + \frac{2}{2+x^2} + \frac{3}{3+x^2}+...+\frac{x}{x + x^2}} =\frac{1}{2}$
So, my question is Why am I getting two different answers when worked out differently.
If there is any error or misconception in my working, please correct me.
|
Two different answers with different methods in solving limits.
here you know the two important things are there one is $x\rightarrow\infty$ and and also the terms are variying as 1,2,3.....
you mean by individualy taking them as constant even when they are going to $\infty$
for example this below
similar misleading question by your point of view
here also the sum might tend to something different.
but the $k$ as a variable makes difference.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3135748",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
}
|
Help finding the derivative of $f(x) = \cos{\left(\sqrt{e^{x^5} \sin{x}}\right)}$ I am trying to find the derivative of $f(x) = \cos(\sqrt{(e^{x^5} \sin(x)})$.
I keep getting the wrong answer, and I'm not sure what I'm doing wrong.
$$\frac{d}{dx} e^{x^5} = e^{x^5} \cdot 5x^4$$
$$\frac{d}{dx} \sin(x) = \cos(x)$$
$$\frac{d}{dx} (e^{x ^5} \cdot \sin(x)) = [(e^{x^5} \cdot 5x^4) \cdot \sin(x)] + [\cos(x) \cdot e^{x^5}]$$
$$\frac{d}{dx} \sqrt{x} = \frac{1}{2\sqrt{x}}$$
$$\frac{d}{dx} \sqrt{(e^{x^5} \sin(x))} = \frac{[(e^{x^5} \cdot 5x^4) \cdot \sin(x)] + [\cos(x) \cdot e^{x^5}]}{2 \sqrt{e^{x^5} \sin(x)}}$$
Therefore, since $\frac{d}{dx} \cos(x) = -\sin(x)$, I have
$$ f'(x) = -\sin(\sqrt{e^{x^5} \sin(x)}) \cdot \frac{[(e^{x^5} \cdot 5x^4) \cdot \sin(x)] + [\cos(x) \cdot e^{x^5}]}{2 \sqrt{e^{x^5} \sin(x)}}$$
However, the website I'm using, "WeBWorK", says this is incorrect.
|
It may be a bit overkill but you can actually integrate your result and show it is equivalent to what you started with
$$ -\frac{1}{2}\int \left(\sin(\sqrt{e^{x^5} \sin(x)} \right) \frac{5x^4e^{x^5} \sin(x) + \cos(x) e^{x^5}}{\sqrt{e^{x^5} \sin(x)}} dx,$$
let
$$u(x) = \sqrt{e^{x^5} \sin(x)}, \qquad du = \frac{5x^4e^{x^5} \sin(x) + \cos(x) e^{x^5}}{\sqrt{e^{x^5} \sin(x)}} dx, $$
so the integral is simply
$$ - \int \sin(u)du, $$
which is what you started with.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3138417",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 2
}
|
How many numbers of the form $m^2 + \sqrt{2} n^2$ are between $1 \times 10^6$ and $2 \times 10^6$? I have a purely computational question today. How many numbers of the form $m^2 + \sqrt{2} n^2$ with $m,n \in \mathbb{Z}$ are between $1 \times 10^6$ and $2 \times 10^6$ ?
$$ \# \big\{ (m,n): 1 \times 10^6 < m^2 + \sqrt{2} \, n^2 < 2 \times 10^6 \big\} =\; ? $$
The asymptotic answer for this sequence of numbers can be found using Weyl's law, for exmaple that:
$$ \# \big\{ (m,n) : 0 < m^2 + \sqrt{2} \, n^2 < X \big\} \sim \frac{\pi}{4 \sqrt{2}} \, X$$
This result is saing that numbers of the form $m^2 + \sqrt{2} \, n^2$ are roughtly linearly distributed on the real line $\mathbb{R}$.
The Weyl's law estimate gives $\frac{\pi}{4\sqrt{2}} \times 10^6 \approx 555,630 + 0.36726\dots$ Perhaps with a computer it's possible to obtain an exact answer? E.g. using Python or sage.
|
COMMENT.- Without calculator the problem is
tedious but not difficult.
The required points are enclosed between the ellipses of equations $m^2 + \sqrt{2} n^2=10^6$ and $m^2 + \sqrt{2} n^2=2\cdot10^6$. Define the following sets in the first quadrant:
$$A=\{(0,y)\text { with } y\gt0 \}\\B=\{(x,0)\text { with } x\gt0 \}\\C=\{(x,y)\text { with } x\gt0,y\gt0 \}$$ By symmetry we have for the total number $N$ of points
$$N=2A+2B+4C$$ $A$ has $573$ points and $B$ has $413$.
One has for the calculation of the set $C$ $$\frac{\sqrt{10^6-m^2}}{\sqrt[4]2}\lt y\lt\sqrt[4]2\sqrt{10^6-m^2}\text { where } 1\le m\le1000$$ Assuming that @rogerl's comment above is true, there are $4\cdot660074+1972=\color{red}{2\space642\space268}$ points in total.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3139082",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
}
|
Find the maximum integer $n$ such that $e^{x-1}\ln x+\frac{e^x+2x-1}{x^2} > 2\sqrt{n}-1$
If $$e^{x-1}\ln x+\frac{e^x+2x-1}{x^2} > 2\sqrt{n}-1 \quad (n \in \mathbb{N}^+),$$
find the maximum of $n$.
I can prove that the inequality holds when $n=4$.
My proof:
When $n=4$:
$$e^{x-1}\ln x+\frac{e^x+2x-1}{x^2} > 3\quad(n \in \Bbb N^+)$$
Case I. $0<x<1.$
$$e^{x-1}\ln x>(x+1)\left(1-\frac{1}{x}\right)=x-\frac{1}{x}$$
Hence
$$e^{x-1}\ln x+\frac{e^x+2x-1}{x^2} > 3\quad(n \in \Bbb N^+)$$
$$\iff e^x>1+x+\frac{x^2}{2}$$
Case II. $x\geq 1.$
$$e^{x-1}\ln x>x\left(1-\frac{1}{x}\right)$$
Hence
$$e^{x-1}\ln x+\frac{e^x+2x-1}{x^2} > 3\quad(n \in \Bbb N^+)$$
$$\iff e^x>1+x+\frac{x^2}{2}+\frac{x^3}{6}$$
Q.E.D.
But I am not sure whether the inequality holds when $n=5$.
Any ideas?
|
We will prove that $\max n=4$.
Let $$f(x)=1+e^{x-1}\ln x+\frac{e^x+2x-1}{x^2}\tag1.$$ We wish to determine $\max n$ such that $$\min f(x)=2\sqrt{\max n}\implies \max n=\frac14[\min f(x)]^2\tag2.$$ Then $$f'(x)=\frac{e^x\ln x}e+\frac{e^x}{ex}+\frac{(e^x+2)x^2-2x(e^x+2x-1)}{x^4}=0\tag3$$ for critical points, which can be rearranged to give $$x^2e^x(x\ln x+1)+e((x-2)e^x-2x+2)=0\tag4.$$ Denote the LHS of $(3)$ as $g(x)$. Then \begin{align}g'(x)&=x(x+2)e^x(x\ln x+1)+x^2e^x(\ln x+1)+e((x-1)e^x-2)\\&=xe^x((x^2+3x)\ln x+2x+2)+e((x-1)e^x-2)\tag5\end{align} and $g'(x)$ is monotonically increasing for all $x>0$ since $$g''(x)=(x^3+6x^2+6x)\ln x+(3x^2+(9+e)x+2)\tag6$$ which is clearly positive. Since $g'(0.5)<0$ and $g'(1)>0$, there exists a unique real number $k\in(0.5,1)$ such that $g(k)$ is a critical point. It can be shown that $g(k)$ is a minimum value since $g(x)<0$ for $0<x<1$.
Now $g(x)$ is monotonically increasing for all $x>k$ and numerical experimentation on a simple scientific calculator reveals that $g(1.3109)<0$ and $g(1.3110)>0$. Between these two values, $$f(1.3109)\approx4.471809644,\,\quad f(1.3110)\approx4.471809635\tag7.$$ From $(2)$, since $n\in\Bbb N^+$, $$\boxed{\max n<\frac14\cdot4.471809644^2=4.999270\cdots<5\implies \max n=4.}\tag8$$ The bound is extremely tight.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3140248",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
}
|
Multi variable calculus, equation tangent to plane question So i think i may have it right but now sure...please check and help me to see if i got it right thanks!
Question:
$f(x,y) = 1 + x^2 + y^2$, find vector $v$ tangent to plane of graph at $f(1,1,3)$.
Answer:
$f(x,y) = 1 + x^2 + y^2$ and $g(x,y,z) = 1 + x^2 + y^2 – z$
$g(x,y,z) = f(x,y) – z$
$0 = f(x,y) – z$
$z = f(x,y) $
$z = 1 + x^2 + y^2 $
$z = 1 + 1 + 1$ (sub in 2 points i know already $f(1,1,3)$ for $x$ and $y$).
$z = 3 $
$g(x,y,z) = 1 + x^2 + y^2 – 3z$
$g(x,y,z) = 1 + x^2 + y^2 –3z$
$\text{grad} g(x,y,z) = 1 + x^2 + y^2 – 3z$
$\text{grad} g(x,y,z) = (2x, 2y, –3) $
$(x,y,z)\cdot \text{grad} g(1,1,3) = (1,1,3))\cdot \text{grad} g(1,1,3) $
$(x,y,z)\cdot (2,2,-3) = (1,1,3))\cdot (2,2,-3) $
$2x+ 2y -3z = 2 + 2 – 9 $
$2x+ 2y -3z = -5 $
$-3z = -5 - 2x -2y $
$3z = 2x + 2y +5 $
Is this right? Thanks!
|
No, your gradient is not correct (and, in your statement, $f(1,1,3)$ is misleading since $f$ is a function of TWO variables).
Let $f(x,y) = 1 + x^2 + y^2$ then
the gradient of
$$g(x,y,z):=f(x,y)-z=1 + x^2 + y^2 – z$$
at $(1,1,f(1,1))=(1,1,3)$, is
$$\left(\frac{\partial g}{\partial x},\frac{\partial g}{\partial y},\frac{\partial g}{\partial z}\right)_{(1,1,3)}=
(2x,2y,-1)_{(1,1,3)}=(2,2,-1).$$
Now choose a vector ${\bf v}$ such that the scalar product ${\bf v}\cdot (2,2,-1)$ is zero.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3140889",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
minimum value of $2\cos \alpha\sin \beta+3\sin \alpha\sin \beta+4\cos \beta$ Let $\alpha,\beta$ be real numbers ; find the minimum value of
$2\cos \alpha\sin \beta+3\sin \alpha\sin \beta+4\cos \beta$
What I tried :
$\bigg|4\cos \beta+(2\cos \alpha+3\sin \alpha)\sin \beta\bigg|^2 \leq 4^2+(2\cos \alpha+3\sin \alpha)^2$
How do I solve it ? Help me please
|
Property to note : $a\cos x + b\sin x = \pm\sqrt{a^2 +b^2}$, So,
$$2\cos\alpha + 3\sin\alpha = \pm\sqrt{13}$$
Taking the minimum value of the expression,
$$-\sqrt{13}\sin\beta +4\cos\beta = \pm\sqrt{29}$$
Therefore, the minimum value of the expression is $-\sqrt{29}$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3143555",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
}
|
A simpler method to prove $\log_{b}{a} \times \log_{c}{b} \times \log_{a}{c} = 1?$ This is the way I think about it:
$
1 = \log_aa = \log_bb = \log_cc \\~\\
\textbf{Using the ‘change of base rule':} \\
\log_{a}{b} = \frac{\log_{b}{b}}{\log_{b}{a}}, \hspace{0.3cm} \log_{b}{c} = \frac{\log_{c}{c}}{\log_{c}{b}}, \hspace{0.3cm} \log_{c}{a} = \frac{\log_{a}{a}}{\log_{a}{c}}\\~\\
\rightarrow \log_{b}{a} = \frac{\log_{b}{b}}{\log_{a}{b}}, \hspace{0.3cm} \log_{c}{b} = \frac{\log_{c}{c}}{\log_{b}{c}}, \hspace{0.3cm} \log_{a}{c} = \frac{\log_{a}{a}}{\log_{c}{a}}\\~\\
\rightarrow \large\log_{b}{a} \times \log_{c}{b} \times \log_{a}{c}\rightarrow\\~\\
\rightarrow\frac{\log_{b}{b}}{\log_{a}{b}} \times \frac{\log_{c}{c}}{\log_{b}{c}} \times \frac{\log_{a}{a}}{\log_{c}{a}} \overset? = 1\\
\rightarrow \frac{1}{\log_{a}{b} \times \log_{b}{c} \times \log_{c}{a}} \overset? = 1\\
\rightarrow \log_{a}{b} \times \log_{b}{c} \times \log_{c}{a} = 1\\
\rightarrow \log_{a}{b} \times \log_{b}{c} \times \log_{c}{a} \overset? = \log_{b}{a} \times \log_{c}{b} \times \log_{a}{c}\\
\rightarrow \frac{\log_{a}{b} \times \log_{b}{c} \times \log_{c}{a}}{\log_{b}{a} \times \log_{c}{b} \times \log_{a}{c}} = 1 \\
\rightarrow \frac{\log_{a}{b}}{\log_{b}{a}} \times \frac{\log_{b}{c}}{\log_{c}{b}} \times \frac{\log_{c}{a}}{\log_{a}{c}} = 1\\~\\
\small \text{------ Using the ‘change of base rule' again ------} \\~\\
\Large \therefore \hspace{0.2cm}\log_{b}{a} \times \log_{c}{b} \times \log_{a}{c} = 1
$
Any other methods?
Thanks in advance.
|
You can generalize this result with logarithm rule:
$\log_{b} a = \dfrac{\log_{e}a}{\log_{e}b} = \dfrac{\ln a}{\ln b}$. That is, the identity is valid for all $a_i$ positives reals numbers and differents of $1$:
$$
\log_{a_2}a_1 \cdot \log_{a_3}a_2 \cdot \log_{a_4}a_3 \cdots \log_{a_n}a_{n-1}\cdot \log_{a_1}a_n = 1.
$$
Firt all, we have for $n=2$, implies $\log_{a_2}a_1 \cdot \log_{a_1}{a_2} =
\dfrac{\ln a_1}{\ln a_2}\cdot \dfrac{\ln a_2}{\ln a_1} = 1.$
Your case is for $n=3$, that is
$$
\log_{a_2}a_1 \cdot \log_{a_3}{a_2} \cdot \log_{a_1}{a_3} =
\dfrac{\ln a_1}{\ln a_2}\cdot
\dfrac{\ln a_2}{\ln a_3} \cdot
\dfrac{\ln a_3}{\ln a_1} = 1.
$$
In general, suposse that the identity is valid for some natural number $n$, and we need to prove the valid for $n+1$. Then
$$
\log_{a_2}a_1 \cdot \log_{a_3}a_2 \cdot
\log_{a_4}a_3 \cdots \log_{a_n}a_{n-1}\cdot \log_{a_{n+1}}a_n \cdot \log_{a_1}{a_{n+1}} =\\
\dfrac{\ln a_1}{\ln a_2}\cdot
\dfrac{\ln a_2}{\ln a_3} \cdot
\dfrac{\ln a_3}{\ln a_4} \cdots
\dfrac{\ln{a_{n-1}}}{\ln{a_n}} \cdot
\dfrac{\ln{a_{n}}}{\ln{a_{n+1}}} \cdot
\dfrac{\ln{a_{n+1}}}{\ln{a_1}} =1.
$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3143783",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
}
|
How is $\frac{dx }{z(x+y) } = \frac{dy}{z(x-y) } = \frac{dz }{x^2 + y^2 } $ equivalent to $\frac{ y dx + xdy - zdz}{0}=\frac{ xdx - ydy -zdz}{0}$? In the book of PDE by Kumar, it is given that
However, I couldn't figure out how is
$$\frac{dx }{z(x+y) } = \frac{dy}{z(x-y) } = \frac{dz }{x^2 + y^2 } $$
is equivalent to
$$\frac{ y dx + xdy - zdz}{0 } = \frac{ x dx - y dy -z dz}{ 0} .$$
|
Some notes about solving PDE
$$z(x+y)z_x+z(x-y)z_y=x^2+y^2\qquad (1)$$
*
*see
https://www.math24.net/first-integrals-page-2/
Example 4
*In our case two independent first integrals is:
$$y^2+2xy-y^2=C_1, \quad z^2-2xy=C_2$$
*General solution of PDE $(1)$ is
$$z^2=2xy+f(y^2+2xy-x^2)$$
*Let $u=z^2$. Then from $(1)$ we get linear PDE
$$(x+y)u_x+(x-y)u_y=2x^2+2y^2\qquad (2)$$
with general solution
$$u=2xy+f(y^2+2xy-x^2)$$
*$u_p=2xy$ is particular solution of $(2)$
*$y^2+2xy-y^2=C_1$ is general solution of ODE
$$\frac{dx }{x+y} = \frac{dy}{x-y}.$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3144760",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
}
|
How to derive this sequence: $1^3+5^3+3^3=153,16^3+50^3+33^3=165033,166^3+500^3+333^3=166500333,\cdots$? I found it is interesting but I don't know how the R.H.S is coming from the L.H.S,
i.e, how to derive this sequence?
The sequence is as follows:
$$1^3+5^3+3^3=153$$
$$16^3+50^3+33^3=165033$$
$$166^3+500^3+333^3=166500333$$
$$1666^3+5000^3+3333^3=166650003333$$
$$.$$
$$.$$
$$.$$
$$\text{and so on}$$
So,any help please...
|
First some notations, let express the repunits $r_n=11\cdots 1=\dfrac{(10^n-1)}9$ in term of $k=10^n$
$\begin{cases}
a_n&=16\cdots 6&=10^{n-1}+6\ r_{n-1}&=\frac{k-4}6\\
b_n&=50\cdots 0&=5\times 10^{n-1}&=\frac{k}2\\
c_n&=33\cdots 3&=3\,r_n&=\frac{k-1}3\end{cases}$
The sum of cubes
$f(n)=a_n^3+b_n^3+c_n^3=\left(\frac{k-4}6\right)^3+\left(\frac k2\right)^3+\left(\frac{k-1}3\right)^3=\dfrac{k^3-k^2+2k-2}6=\dfrac{(k-1)(k^2+2)}6$
Can be expressed like this:
$$\boxed{f(n)=\dfrac{(10^n-1)(10^{2n}+2)}{6}}$$
But we are interested in the non-factored expression:
$\begin{align}f(n)&=\dfrac{k^3-k^2+2k-2}6\\&=\dfrac{k^3-4k^2+3k^2+2k-2}6\\&=\left(\dfrac{k^2-4k^2}6\right)+\left(\dfrac {3k^2}6\right)+\left(\dfrac{2k-2}6\right)=k^2\,a_n+k\,b_n+c_n=\overline{a_nb_nc_n}\end{align}$
Where
$$\overline{a_nb_nc_n}=\underbrace{16\cdots 6}_{a_n\times 10^{2n}}\;\underbrace{50\cdots 0}_{b_n\times 10^n}\;\underbrace{33\cdots 3}_{c_n}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3151306",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
}
|
What is wrong with this proof of $3\arcsin x$? We know that
\begin{align}
2\arcsin x&= \arcsin \left(2x\sqrt{1-x^2}\right) \tag{1}\\
\arcsin x + \arcsin y &= \arcsin \left[x\sqrt{1-y^2}+y\sqrt{1-x^2}\right] \tag{2}\\
3\arcsin x &= \arcsin x + 2\arcsin x \tag{3}
\end{align}
Thus $x=x, y=2x\sqrt{1-x^2}$
using ($1$), ($2$) and ($3$):
\begin{align}
3\arcsin x &= \arcsin \left[x\sqrt{1-2x\sqrt{1-x^2}^2}+ 2x(1-x^2)\right]\\
&= \arcsin \left[x\sqrt{1-4x^2(1-x^2)}+ 2x(1-x^2)\right]\\
&= \arcsin \left[x\sqrt{1-2(2x^2)+(2x^2)^2}+ 2x(1-x^2)\right]\\
&= \arcsin \left[x\sqrt{(2x^2-1)^2}+ 2x(1-x^2)\right]\\
&= \arcsin \left[x|2x^2-1|+ 2x(1-x^2)\right]
\end{align}
If $2x^2-1$ is positive, then $|2x^2-1|$ is $2x^2 -1$.
If $2x^2-1$ is negative, then $|2x^2-1|$ is $-2x^2+1$.
Range of $x$ is $-1\leq x \leq 1 \implies 0\leq x^2 \leq 1 \implies 0\leq 2x^2 \leq 2$.
For $x\in\left(\frac{-1}{\sqrt2}, \frac{+1}{\sqrt2}\right)$, then $2x^2-1$ is negative.
For $x\in\left(-1, \frac{-1}{\sqrt2}\right) \cup \left(\frac{+1}{\sqrt2}\ , 1\right)$, then $2x^2-1$ is positive.
Thus for $x\in\left( \frac{-1}{\sqrt2}, \frac{+1}{\sqrt2}\right)$
\begin{align}
3\arcsin x &= \arcsin [x|2x^2-1|+ 2x(1-x^2)]\\
&= \arcsin \left[-2x^3 +x+ 2x(1-x^2)\right]\\
&= \arcsin [3x - 4x^3]
\end{align}
Thus for $x\in\left(-1, \frac{-1}{\sqrt2}\right) \cup \left(\frac{+1}{\sqrt2}, 1\right)$
\begin{align}
3\arcsin x &= \arcsin [x|2x^2-1|+ 2x(1-x^2)]\\
&= \arcsin [2x^3- x+2x-2x^3]\\
&= \arcsin[x]
\end{align}
But clearly, $3\arcsin x = \arcsin[3x-4x^3]$.
So what is wrong whith this proof?
|
A geometric point of view might be illuminating.
Suppose $0\leq x \leq \frac{1}{\sqrt 2}$. Consider the figure below, where we start from a right-angled triangle $\triangle ABC$ with sides $\overline{AB} = \sqrt{1-x^2}$ and $\overline{BC} = x$, and hypotenuse $\overline{AC}=1$. The choice of $x$ we have made guarantees that
$$ \alpha = \angle BAC$$
is in the range $\left[0, \frac{\pi}{4}\right].$
By definition is
$$ \alpha = \arcsin x.$$
Extend first $CB$ to a segment $BD \cong BC$, then draw from $D$ the line perpendicular to $AD$ that intersects the extension of $AC$ in $E$. Finally draw from $C$ the perpendicular to $BC$ that meets $ED$ in $F$.
Define
\begin{equation}\beta = \angle CAD = 2\alpha.\tag{1}\label{eq:1}\end{equation}
We have, by definition,
\begin{equation}\beta = \arcsin \left(\frac{\overline{ED}}{\overline{AE}}\right).\tag{2}\label{eq:2}\end{equation}
From $\triangle DCF \sim \triangle ABC$ find
$$\overline{CF} = \frac{2x^2}{\sqrt{1-x^2}}$$
and
$$\overline{DF} = \frac{2x}{\sqrt{1-x^2}}.$$
By Pythagorean Theorem on $\triangle ADE$, and by $\triangle CEF \sim \triangle CED$
\begin{equation}
\begin{cases}
1+\overline{ED}^2 = (1+ \overline{EC})^2\\
\overline{EC} = \frac{x}{\sqrt{1-x^2}}\overline{ED}.
\end{cases}
\end{equation}
Solving the system yields
$$\overline{ED} = \frac{2x\sqrt{1-x^2}}{1-2x^2}$$
and
$$\overline{AC} = 1 + \overline{EC} = \frac{1}{1-2x^2}.$$
Using these results in \eqref{eq:2} and then cosidering the identity \eqref{eq:1} leads to
$$2\arcsin x = \arcsin \left(2x\sqrt{1-x^2}\right).$$
For $\frac{1}{\sqrt 2} \leq x \leq 1$, I would consider the triangle below, where again $\overline{BC} = x$, $\overline{AC} = 1$ and $D$ is the point symmetrical to $C$ with respect to line $AB$. Draw then from $C$ the perpendicular to $AD$ that meets its extension in $E$. Define then $\alpha$ as before and
$$\beta = \angle CAE = \pi - \angle CAD = \pi -2\alpha.$$
Use then the fact that $\sin \beta = \sin 2\alpha$.
Finally, for negative $x$ just define $\overline{BC} = -x$ and proceed as above, taking advantage of the sine odd symmetry.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3151681",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
}
|
Finding $\lim_{n \to \infty} \int_0^n \frac{dx}{1+n^2\cos^2x}$ Find $$\lim_{n \to \infty} \int_0^n \frac{dx}{1+n^2\cos^2x}$$
I tried:
*
*mean value theorem.
*variable change with $ \tan x = t $ but I need to avoid the points which are not in the domain of $\tan$ and it's complicated.
|
Using the substitution $u=\tan{(x)}\Rightarrow du=\sec^2{(x)}dx$ turns the integral into
$$\int_0^\infty \frac{du}{u^2+n^2+1}+\bigg\lfloor\frac{n-\frac\pi2}{\pi}\bigg\rfloor\int_{-\infty}^\infty \frac{du}{u^2+n^2+1}+\int_{-\infty}^{\tan{(n)}}\frac{du}{u^2+n^2+1}$$
$$=\bigg[\frac1{\sqrt{n^2+1}}\arctan{\bigg(\frac{u}{\sqrt{n^2+1}}\bigg)}\bigg]_0^\infty+\bigg\lfloor\frac{n-\frac\pi2}{\pi}\bigg\rfloor\bigg[\frac1{\sqrt{n^2+1}}\arctan{\bigg(\frac{u}{\sqrt{n^2+1}}\bigg)}\bigg]_{-\infty}^\infty+\bigg[\frac1{\sqrt{n^2+1}}\arctan{\bigg(\frac{u}{\sqrt{n^2+1}}\bigg)}\bigg]_{-\infty}^{\tan{(n)}}$$
$$=\frac\pi{\sqrt{n^2+1}}+\frac\pi{\sqrt{n^2+1}}\bigg\lfloor\frac{n-\frac\pi2}{\pi}\bigg\rfloor+\frac1{\sqrt{n^2+1}}\arctan{\bigg(\frac{\tan{(n)}}{\sqrt{n^2+1}}\bigg)}$$
$$=\frac\pi{\sqrt{n^2+1}}\bigg\lfloor\frac{n+\frac\pi2}{\pi}\bigg\rfloor+\frac1{\sqrt{n^2+1}}\arctan{\bigg(\frac{\tan{(n)}}{\sqrt{n^2+1}}\bigg)}$$
$$=\frac{\pi\big\lfloor\frac{n}{\pi}+\frac12\big\rfloor+\arctan{\big(\frac{\tan{(n)}}{\sqrt{n^2+1}}\big)}}{\sqrt{n^2+1}}$$
Taking the limit as $n\to\infty$ we have
$$\lim_{n\to\infty}\frac{\pi\big\lfloor\frac{n}{\pi}+\frac12\big\rfloor+\arctan{\big(\frac{\tan{(n)}}{\sqrt{n^2+1}}\big)}}{\sqrt{n^2+1}}=\lim_{n\to\infty}\frac{n+\arctan{\big(\frac{\tan{(n)}}{n}\big)}}{n}=\lim_{n\to\infty}\bigg(1+\frac{\arctan{\big(\frac{\tan{(n)}}{n}\big)}}{n}\bigg)=1$$
The final limit of $\frac{\arctan{\big(\frac{\tan{(n)}}{n}\big)}}{n}$ is equal to zero because $\frac{-\pi}2\lt \arctan{(x)} \lt \frac\pi2$ so $\frac{-\pi}{2n}\lt \frac{\arctan{\big(\frac{\tan{(n)}}{n}\big)}}{n} \lt \frac\pi{2n}$ and both bounds tend to zero in the limit.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3152960",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
}
|
Summing up the series $\sum_{n=0}^\infty C_{2n}^n r^{2n} $ Here $|r|<1/2$, so that the series converge.
I can do it by using contour inregration.
$$ S =\sum_n r^{2n} \frac{1}{2\pi i } \oint_C \frac{1}{z}(z+ \frac{1}{z})^{2n} dz
\\
= \frac{1}{2\pi i } \oint_C \frac{1}{z} r^{2n} (z+ \frac{1}{z})^{2n} dz \\
= \frac{1}{2\pi i } \oint_C \frac{1}{z} \frac{1}{1- r^2 (z+1/z)^2} dz . $$
Here $C$ is the unit circle in the complex plane.
It is not so tedious to get the final result, which is $1/\sqrt{1-4r^2 }$.
However, can anyone give a more direct solution?
|
We can use the binomial series expansion. In order to do so we recall the binomial identity
\begin{align*}
\binom{2n}{n}=(-4)^n\binom{-\frac{1}{2}}{n}\tag{1}
\end{align*}
and obtain
\begin{align*}
\color{blue}{\sum_{n=0}^\infty \binom{2n}{n}r^{2n}}&=\sum_{n=0}^\infty (-4)^n\binom{-\frac{1}{2}}{n}r^{2n}\\
&=\sum_{n=0}^\infty\binom{-\frac{1}{2}}{n}(-4r^2)^n\\
&=\left(1-4r^2\right)^{-\frac{1}{2}}\\
&\,\,\color{blue}{=\frac{1}{\sqrt{1-4r^2}}}
\end{align*}
The identity (1) is valid since we have
\begin{align*}
\color{blue}{(-4)^n\binom{-\frac{1}{2}}{n}}&=(-4)^n\frac{1}{n!}\left(-\frac{1}{2}\right)\left(-\frac{3}{2}\right)\cdots\left(-\frac{1}{2}-(n-1)\right)\\
&=\frac{2^n}{n!}(2n-1)!!\\
&=\frac{2^n}{n!}\frac{(2n)!}{(2n)!!}\\
&=\frac{2^n}{n!}\frac{(2n)!}{2^nn!}\\
&=\frac{(2n)!}{n!n!}\\
&\,\,\color{blue}{=\binom{2n}{n}}
\end{align*}
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3153633",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
}
|
Find a set of points in the given complex plane Here's the Question: Find a set of points in the complex plane that satisfies:
$$|z-i|+|z+i| = 1$$
Now from triangle inequality I found:
$$|z+z+i-i|=|2z|\geq1 $$
Which refers that there's no soluton and the set should be empty.
But if I substitute z=x+iy and simplify then:
$$ \Rightarrow \sqrt{ x^2 +(y-1)^2}+\sqrt {x^2 +(y+1)^2} = 1 $$ $$\Rightarrow \sqrt{ x^2 +(y-1)^2} = 1- \sqrt {x^2 +(y+1)^2}$$ Squaring both sides, $$\Rightarrow x^2 +(y-1)^2 = 1-2\sqrt{x^2 +(y+1)^2}+ x^2+(y+1)^2$$ Simplify and square again, $$\Rightarrow (4y+1)^2 =4(x^2+(y+1)^2)$$ $$\Rightarrow 16y^2+8y+1=4x^2+4y^2+8y+4$$ $$ \Rightarrow 12y^2-4x^2=3$$
Which implies a hyperbola. Now should I consider the points on the hyperbola as the expected points or there should be no points to satisfy the equation?
|
There is no such point because for all $z \in \mathbb{C}$ you have
$$|z-i|+|z+i| \geq |z-i - (z+i)| =|2i| = 2 > 1$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3155410",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
}
|
Automorphism of a matrices ring
Let $R$ be the ring of $3 \times 3$ matrices with coefficients in $\Bbb Z_5$.
For every $g \in GL_3(\Bbb Z_5)$ prove that the function $$f\colon R \rightarrow R$$ defined as $$x \mapsto g^{-1}xg$$ is an automorphism of $R$.
If I choose the matrix $$g = \begin{pmatrix}
0 & 0 & 1 \\
0 & 1 & 0 \\
1 & 0 & 0
\end{pmatrix},$$
how many matrices $x\in R$ such that $f(x)=x$ are there?
I have no idea how to solve this. Any help?
|
This is much more general: suppose $R$ is a ring and $g$ is invertible in $R$. Then the map $f\colon R\to R$, $f(x)=gxg^{-1}$ is an automorphism of $R$.
Indeed, $f(x+y)=g(x+y)g^{-1}=gxg^{-1}+gyg^{-1}=f(x)+f(y)$ and
$$
f(xy)=gxyg^{-1}=gxg^{-1}gyg^{-1}=f(x)f(y)
$$
Obviously, $f(1)=1$.
Since the map $x\mapsto g^{-1}xg$ is the inverse of $f$, we are done.
For the second part, you want $gxg^{-1}=x$, so $gx=xg$. If
$$
x=\begin{pmatrix}
a_{11} & a_{12} & a_{13} \\
a_{21} & a_{22} & a_{23} \\
a_{31} & a_{32} & a_{33}
\end{pmatrix}
$$
then
$$
gx=\begin{pmatrix}
a_{31} & a_{32} & a_{33} \\
a_{21} & a_{22} & a_{23} \\
a_{11} & a_{12} & a_{13} \\
\end{pmatrix}
\qquad
xg=\begin{pmatrix}
a_{13} & a_{12} & a_{11} \\
a_{23} & a_{22} & a_{21} \\
a_{33} & a_{32} & a_{31}
\end{pmatrix}
$$
Thus you get
\begin{cases}
a_{31}=a_{13} \\
a_{32}=a_{12} \\
a_{33}=a_{11} \\
a_{21}=a_{23}
\end{cases}
and so the linear system has rank $4$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3155767",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
Recursion Question using Generating Functions Here is my question:
Consider the recurrence,
$$a_{n+1}=2a_n+(-1)^n$$
with initial condition,
$$a_0=0$$
Find and prove a formula for $a_n$.
I don't really know how to prove this formula
I tried going with a generating function method, but that kind of led nowhere.
|
The sequence generated by the recurrence relation is;
$$0, 1, 1, 3, 5, 11, 21, \dots$$
Write the recurrence relation as;
$$a_n-2a_{n-1}=(-1)(-1)^n$$
Get the generating function, $GF$ in the standard way;
$$GF=0+x+x^2+3x^3+5x^4+11x^5+21x^6+\dots$$
$$-2xGF=0+0x-2x^2-2x^3-6x^4-10x^5-22x^6+\dots$$
$$(1-2x)GF=0+x-x^2+x^3-x^4+x^5-x^6+x^7-x^8+ \dots$$
$$(1-2x)GF=-\big(\frac{1}{1+x}\big)+1$$
$$GF=\frac{x}{(1-2x)(1+x)}$$
Use partial fractions to get;
$$GF=\frac{1}{3}\times \frac{1}{1-2x}-\frac{1}{3}\times \frac{1}{1+x}$$
These are standard bits that translate directly into the formula;
$$a_n=\frac{1}{3}2^n-\frac{1}{3}(-1)^n$$
or
$$a_n=\frac{2^n-(-1)^n}{3}$$
Check this gives the sequence expected, which it does!
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3156168",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
}
|
Sum of all values that satisfy $\frac{x^{x-3}}{x}=\frac{x}{x^\frac{4}{x}}$.
What is the sum of all values of $x$ that satisfy the equation
$\frac{x^{x-3}}{x}=\frac{x}{x^\frac{4}{x}}$?
I start off by cross multiplying.
$$x^2=x^{x-3+\frac{4}{x}}$$
Taking the square root of both sides gives me:
$$x=\pm x^{\frac{1}{2}x-\frac{3}{2}+\frac{4}{2x}}$$
I start with the positive side first:
$$1=\frac{1}{2}x-\frac{3}{2}+\frac{4}{2x}$$$$x^2-5x+4=0$$$$x=4, 1$$
Now, I start with the negative side:
$$x=-x^{\frac{1}{2}x-\frac{3}{2}+\frac{4}{2x}}$$$$x^\frac{x^2-5x+4}{2x}=-1$$
I can't log both sides because of the $-1$, so I raise both sides by a power of $2x$ to get rid of the fraction. Because $2x$ is even, the RHS becomes a $1$.
$$x^{x^2-5x+4}=1$$
Now, taking the log of both sides, I have
$$(x^2-5x+4)\cdot\operatorname{log}(x)=0$$
Dividing both sides by $\operatorname{log}(x)$, I get $x^2-5x+4=0$ again, which should give me $x=4, 1$ again. So thus, the answer should be $4+1=5$.
However, this is wrong, as the answer key says that the answer is $-1+1+4=\boxed{4}$. I have checked $-1$ as a solution on all of my equations, and all of them work. How can I derive the $-1$ out from my equations? Or, if my approach or any of my equations are wrong, how do I get the answer $x=-1$?
|
I start off by cross multiplying.
$$x^2=x^{x-3+\frac{4}{x}}$$
Taking the square root of both sides gives me:
$$x=\pm x^{\frac{1}{2}x-\frac{3}{2}+\frac{4}{2x}}$$
I start with the positive side first:
$$1=\frac{1}{2}x-\frac{3}{2}+\frac{4}{2x}$$$$x^2-5x+4=0$$$$x=\boxed{4}, \boxed{1}$$
Now, I start with the negative side:
$$x=-x^{\frac{1}{2}x-\frac{3}{2}+\frac{4}{2x}}$$$$x^\frac{x^2-5x+4}{2x}=-1$$
Absolute value-ing both sides gives $$\lvert x^\frac{x^2-5x+4}{2x}\rvert=\lvert -1\rvert$$
Now, we use the property that $\lvert x^y\rvert=\lvert x\rvert^y$ to get $$\lvert x\rvert^\frac{x^2-5x+4}{2x}=1$$
Logging both sides gives:
$$\operatorname{log}\lvert x\rvert^\frac{x^2-5x+4}{2x}=\log1$$$$\frac{x^2-5x+4}{2x}\cdot\operatorname{log}\lvert x\rvert=0$$$$\frac{\operatorname{log}\lvert x\rvert\cdot (x^2-5x+4)}{2x}=0$$
Solving for $\operatorname{log}\lvert x\rvert$(multiplying by $2x$ and dividing by $x^2-5x+4$) gives:$$\operatorname{log}\lvert x\rvert\cdot (x^2-5x+4)=0$$$$\operatorname{log}\lvert x\rvert=0$$
Now, solving for $x$ gives:
$$10^{\operatorname{log}\lvert x\rvert}=10^0$$$$\lvert x\rvert=1$$$$x=\boxed{\pm1}$$
So the three solutions are $x=-1, 1, 4$ which gives $-1+1+4=\boxed{4}$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3160152",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
}
|
Compute $I = \int_{0}^{\infty} \frac{8}{\gamma^{4}} x^{3} e^{-\frac{2}{\gamma^{2}}x^{2}} e^{jxt} dx$ I want to evaluate the following integral:
$$I = \int_{0}^{\infty} \frac{8}{\gamma^{4}} x^{3} e^{-\frac{2}{\gamma^{2}}x^{2}} e^{jxt} dx $$
Where $\gamma \in \mathbb{R}$ and $j = \sqrt{-1}$.
The first thing I do is let:
$$ u = \frac{2}{\gamma} x \quad \Rightarrow \quad du = \frac{2}{\gamma} dx$$
$$ x = \frac{\gamma}{2} u \quad \Rightarrow \quad dx = \frac{\gamma}{2} du$$
After substituting and cancellation, we get:
$$ I = \frac{1}{2} \int_{0}^{\infty} u^{3} e^{-\frac{1}{2} u^{2} + j \frac{1}{2} \gamma tu} du$$
We complete the square, giving:
$$ I = \frac{1}{2} \int_{0}^{\infty} u^{3} e^{-\frac{1}{2} (u - j\frac{1}{2} \gamma t)^{2} - \frac{1}{8} \gamma^{2} t^{2}} du$$
$$ I = \frac{1}{2} e^{-\frac{1}{8} \gamma^{2} t^{2}} \int_{0}^{\infty} u^{3} e^{-\frac{1}{2} (u - j\frac{1}{2} \gamma t)^{2}} du$$
This is where I am unsure of what to do. My approach is as follows:
We want to solve:
$$ I = \frac{1}{2}e^{-\frac{1}{8} \gamma^{2} t^{2}} \underset{R \rightarrow \infty}{lim} \int_{0}^{R} u^{3} e^{-\frac{1}{2} (u - j\frac{1}{2} \gamma t)^{2}} du$$
We let:
$$ z = u - j \frac{1}{2} \gamma t \quad \Rightarrow \quad dy = du$$
$$ u = y + j\frac{1}{2} \gamma t \quad \Rightarrow \quad du = dy$$
$$ b = R - j\frac{1}{2} \gamma t \quad a = -j \frac{1}{2} \gamma t $$
Therefore:
$$ I = \frac{1}{2}e^{-\frac{1}{8} \gamma^{2} t^{2}} \underset{R \rightarrow \infty}{lim} \int_{ -j \frac{1}{2} \gamma t}^{R - j \frac{1}{2} \gamma t} (y- j \frac{1}{2} \gamma t)^{3} e^{-\frac{1}{2} y^{2}} dy$$
We create a contour $\Gamma = C_{1} + C_{2} + C_{3} + C_{4} = [-j \frac{1}{2} \gamma t, R-j \frac{1}{2} \gamma t] + [R-j \frac{1}{2} \gamma t, R] + [R, 0] + [0, -j \frac{1}{2} \gamma t]$
Since $ \underset{y \rightarrow \infty}{lim}\frac{(y - j \frac{1}{2} \gamma t)^{3}}{e^{\frac{1}{2}y^{2}}} = 0 $, the integral over the vertical contour $C_{3}$ is zero.
However I still need to evaluate the following integrals:
$$ I_{1} = \underset{R \rightarrow \infty}{lim} \int_{0}^{R} (y- j \frac{1}{2} \gamma t)^{3} e^{-\frac{1}{2} y^{2}} dy $$
$$ I_{2} = \int_{-j \frac{1}{2} \gamma t}^{0} (y- j \frac{1}{2} \gamma t)^{3} e^{-\frac{1}{2} y^{2}} dy $$
This is where I am getting stuck.
Is this the right approach? Am I on the right track? How can I solve those two integrals? How can I solve my original problem?
Thanks!
|
Well, in general we have:
$$\mathcal{I}_\text{n}:=\int_0^{\infty}x^\text{n}\cdot\exp\left(\alpha \cdot x^{\text{n}-1}\right)\cdot\exp\left(\beta\cdot x\right)\space\text{d}x\tag1$$
Using the 'evaluating integrals over the positive real axis' property of the Laplace transform, we can write:
$$\mathcal{I}_\text{n}=\int_0^\infty\mathcal{L}_x\left[\exp\left(\alpha \cdot x^{\text{n}-1}\right)\cdot\exp\left(\beta\cdot x\right)\right]_{\left(\text{s}\right)}\cdot\mathcal{L}_x^{-1}\left[x^\text{n}\right]_{\left(\text{s}\right)}\space\text{ds}\tag2$$
Using the properties of the Laplace transform:
*
*$$\mathcal{L}_x\left[\exp\left(\alpha \cdot x^{\text{n}-1}\right)\cdot\exp\left(\beta\cdot x\right)\right]_{\left(\text{s}\right)}=\mathcal{L}_x\left[\exp\left(\alpha \cdot x^{\text{n}-1}\right)\right]_{\left(\text{s}-\beta\right)}\tag3$$
*$$\mathcal{L}_x\left[\exp\left(\alpha \cdot x^{\text{n}-1}\right)\right]_{\left(\text{s}-\beta\right)}=\sum_{\text{k}=0}^\infty\frac{\alpha^\text{k}}{\text{k}!}\cdot\mathcal{L}_x\left[x^{\text{k}\left(\text{n}-1\right)}\right]_{\left(\text{s}-\beta\right)}=\sum_{\text{k}=0}^\infty\frac{\alpha^\text{k}}{\text{k}!}\cdot\frac{\Gamma\left(1+\text{k}\left(\text{n}-1\right)\right)}{\left(\text{s}-\beta\right)^{1+\text{k}\left(\text{n}-1\right)}}\tag4$$
*$$\mathcal{L}_x^{-1}\left[x^\text{n}\right]_{\left(\text{s}\right)}=\frac{1}{\Gamma\left(-\text{n}\right)}\cdot\frac{1}{\text{s}^{1+\text{n}}}\tag5$$
So, we get:
$$\mathcal{I}_\text{n}=\int_0^\infty\left\{\sum_{\text{k}=0}^\infty\frac{\alpha^\text{k}}{\text{k}!}\cdot\frac{\Gamma\left(1+\text{k}\left(\text{n}-1\right)\right)}{\left(\text{s}-\beta\right)^{1+\text{k}\left(\text{n}-1\right)}}\right\}\cdot\frac{1}{\Gamma\left(-\text{n}\right)}\cdot\frac{1}{\text{s}^{1+\text{n}}}\space\text{ds}=$$
$$\sum_{\text{k}=0}^\infty\frac{\alpha^\text{k}}{\text{k}!}\cdot\frac{\Gamma\left(1+\text{k}\left(\text{n}-1\right)\right)}{\Gamma\left(-\text{n}\right)}\cdot\int_0^\infty\frac{\text{s}^{-1-\text{n}}}{\left(\text{s}-\beta\right)^{1+\text{k}\left(\text{n}-1\right)}}\space\text{ds}\tag6$$
Using the property:
$$\int_0^\infty\frac{x^\text{m}}{\left(x-t\right)^\text{p}}\space\text{d}x=\frac{1}{\left(-t\right)^{\text{p}-1-\text{m}}}\cdot\frac{\Gamma\left(1+\text{m}\right)\cdot\Gamma\left(\text{p}-\text{m}-1\right)}{\Gamma\left(\text{p}\right)}\tag7$$
Gives:
$$\mathcal{I}_\text{n}=\sum_{\text{k}=0}^\infty\frac{\alpha^\text{k}}{\text{k}!}\cdot\left(-1\right)^{\text{k}-\text{n}\left(1+\text{k}\right)-1}\cdot\beta^{\text{k}-\text{n}\left(1+\text{k}\right)-1}\cdot\Gamma\left(1+\text{k}\left(\text{n}-1\right)+\text{n}\right)\tag8$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3161867",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
}
|
Need to compute $\sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)(2n+2)3^{n+1}}$. Is my solution correct? $$\sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)(2n+2)3^{n+1}}$$
Since $$\tan^{-1}x=\int \frac{1}{1+x^{2}} dx=\int (1-x^{2}+x^{4}+...)dx=x-\frac{x
^3}{3}+\frac{x^5}{5}+...$$
so$$\int \tan^{-1}x dx=\int (x-\frac{x^3}{3}+\frac{x^5}{5}+...)dx=\frac {x^2}{2}-\frac{x^4}{3\cdot4}+\frac {x^6}{5\cdot 6}+...=\sum_{n=0}^{\infty} \frac{(-1)^nx^{2n+2}}{(2n+1)(2n+2)}$$
Therefore, $$\int_{0}^{1/{\sqrt 3}}\tan^{-1}xdx=\sum_{n=0}^{\infty}\frac{(-1)^n}{(2n+1)(2n+2)3^{n+1}}$$
$$\Rightarrow \frac{\tan^{-1}(\frac{1}{\sqrt 3})}{\sqrt 3}-\frac{\ln(1+\frac{1}{3})}{2}=\frac{\pi}{6\sqrt3}-\frac{\ln\frac{4}{3}}{2}$$
Is my method correct?
|
You series equals
$$ \frac{1}{3}\sum_{n\geq 0}\frac{(-1)^n}{(2n+1)3^n}-\sum_{n\geq 0}\frac{(-1)^n}{(2n+2)3^{n+1}} $$
or
$$ \frac{1}{\sqrt{3}}\arctan\frac{1}{\sqrt{3}}+\frac{1}{2}\sum_{n\geq 1}\frac{(-1)^n}{n 3^n}=\frac{\pi}{6\sqrt{3}}-\frac{1}{2}\log\left(1+\frac{1}{3}\right). $$
No integrals, just partial fraction decomposition, reindexing and the Maclaurin series of $\arctan(x)$ and $\log(1+x)$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3162027",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
}
|
induction proof: number of compositions of a positive number $n$ is $2^{n-1}$ So my problem was to find the formula for the number of composition for a positive number $n$. Then prove its validity. I found the formula which was $2^{n-1}$. Having trouble with the induction.
$$Claim:2^0+2^1+2^2+...+(2^{n-2}+1)=2^{n-1}$$
When $n=5$ $$ 1+2+4+(2^{5-2}+1)=2^{5-1}$$
$$ 1+2+4+(2^{3}+1)=2^{4}$$
$$ 1+2+4+(8+1)=16$$
$$ 16=16$$
Assume that $n=k$
$$2^0+2^1+2^2+...+(2^{k-2}+1)=2^{k-1}$$
Now show for $k+1$
$$2^0+2^1+2^2+...+(2^{k-1}+1)=2^{k}$$
Don't know what to do after. Help is appreciated.
$$2^0+2^1+2^2+...+(2^k)+(2^{k-1}+1)=2^{k}$$
|
Let's return to the original problem. A composition of a positive integer $n$ is a way of writing $n$ as the sum of a sequence of strictly positive integers.
For $n = 5$, there are indeed $2^{5 - 1} = 2^4 = 16$ compositions since each composition corresponds to placing or omitting an addition sign in the four spaces between successive ones in a row of five ones.
\begin{array}{c c}
\text{composition} & \text{representation}\\ \hline
5 & 1 1 1 1 1\\
4 + 1 & 1 1 1 1 + 1\\
3 + 2 & 1 1 1 + 1 1\\
3 + 1 + 1 & 1 1 1 + 1 + 1\\
2 + 3 & 1 1 + 1 1 1\\
2 + 2 + 1 & 1 1 + 1 1 + 1\\
2 + 1 + 2 & 1 1 + 1 + 1 1\\
2 + 1 + 1 + 1 & 1 1 + 1 + 1 + 1\\
1 + 4 & 1 + 1 1 1 1\\
1 + 3 + 1 & 1 + 1 1 1 + 1\\
1 + 2 + 2 & 1 + 1 1 + 1 1\\
1 + 2 + 1 + 1 & 1 + 1 1 + 1 + 1\\
1 + 1 + 3 & 1 + 1 + 1 1 1\\
1 + 1 + 2 + 1 & 1 + 1 + 1 1 + 1\\
1 + 1 + 1 + 2 & 1 + 1 + 1 + 1 1\\
1 + 1 + 1 + 1 + 1 & 1 + 1 + 1 + 1 + 1
\end{array}
Using this observation, we can write a combinatorial proof. A positive integer $n$ has $2^{n - 1}$ compositions since each composition is uniquely determined by choosing to include or omit an addition sign in each of the $n - 1$ spaces between successive ones in a row of $n$ ones.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3163924",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 1
}
|
Equation of plane containing a point and a line
Find the equation of the plane containing the point $A(0,1,-1)$ and the line
$(d) : \begin{cases} 2x - y + z + 1 = 0 \\ x + y + 1 = 0 \end{cases}$
Where should I start? I was thinking about writing the normal vectors for the line and make their cross product or something like that, but I don't really understand what's going on here. I would really appreciate some help.
|
A plane in $\mathbb{R}^{3}$ satisfies an equation of the type
\begin{equation}
ax+by+cz+d=0. \tag{1}
\end{equation}
You can reduce this problem to that of finding a plane defined by 3 points. So just find 2 points on the line $ d $, using the given equations. For $x=0$ and $x=1$ you get, respectively, the points $B\left(
0,-1,-2\right) $ and $C\left( 1,-2,-5\right) $. To find 3 out of all the constants $a,b,c,d$ you need to substitute the coordinates of $A,B,C$ into $(1)$:
\begin{eqnarray*}
\left\{
\begin{array}{c}
b-c+d=0 \\
-b-2c+d=0 \\
a-2b-5c+d=0
\end{array}
\right. &\Leftrightarrow &\left\{
\begin{array}{c}
b-c+d=0 \\
-3c+2d=0 \\
a-2b-5c+d=0
\end{array}
\right. \Leftrightarrow \ldots
&\Leftrightarrow &\left\{
\begin{array}{c}
b+\frac{1}{2}c=0 \\
d=\frac{3}{2}c \\
a-2b-5c+\frac{3}{2}c=0,
\end{array}
\right.
\end{eqnarray*}
whose solution in terms of $c$ is $d=\frac{3}{2}c,a=\frac{5}{2}c,b=-\frac{1}{2}c$. So
$$
\frac{5}{2}cx-\frac{1}{2}cy+cz+\frac{3}{2}c=0
$$
and, finally, dividing by $c/2$ we obtain the equation of the plane
$$
5x-y+2z+3=0.\tag{2}
$$
Plot of plane $(2)$, line $d$ and points $A, B, C$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3164984",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
}
|
Find minimum perimeter of the triangle circumscribing semicircle The following diagram shows triangle circumscribing a semi circle of unit radius. Find minimum perimeter of triangle
My try:
Letting $$AP=AQ=x$$
By power of a point we have:
$$BP^2=OB^2-1$$ where $O$ is center of the circle.
Also $$CQ^2=OC^2-1$$
Let $$OB=y$$
$$OC=z$$
Then $$BP=\sqrt{y^2-1}$$
$$CQ=\sqrt{z^2-1}$$
So the perimeter of triangle is:
$$P=2x+y+z+\sqrt{y^2-1}+\sqrt{z^2-1}$$
Any help from here?
|
Let us calculate the perimeter from the figure.
$OB = \dfrac{1}{\sin\alpha}, PB=\dfrac{1}{\tan\alpha},AP= \dfrac{1}{\tan(\pi/2 -\dfrac{\alpha+\theta}{2})},AQ= \dfrac{1}{\tan(\pi/2 -\dfrac{\alpha+\theta}{2})},QC=\dfrac{1}{\tan\theta}, OC =\dfrac{1}{\sin\theta} $
So total perimeter is $ p = \dfrac{1}{\sin\alpha}+\dfrac{1}{\tan\alpha}+2\dfrac{1}{\tan(\pi/2 -\dfrac{\alpha+\theta}{2})}+\dfrac{1}{\tan\theta}+\dfrac{1}{\sin\theta}$
differentiating p wrt alpha and theta and setting them to zero we get.
$-{\csc}^2\alpha +\sec^2\dfrac{\alpha+\theta}{2}- \csc\alpha\cot\alpha = 0$
$-{\csc}^2\theta +\sec^2\dfrac{\alpha+\theta}{2}- \csc\theta\cot\theta = 0$
from above we get
${\csc}^2\alpha + \csc\alpha\cot\alpha ={\csc}^2\theta + \csc\theta\cot\theta$
After squaring both side and simplification/cancelling terms, we get
${(\cos\alpha - \cos\theta)}^2=0$ that gives $\alpha= \theta$, using this back to
$-{\csc}^2\alpha +\sec^2\dfrac{\alpha+\theta}{2}- \csc\alpha\cot\alpha = 0$
we get
$\tan^2\alpha = 1 + \cos\alpha$ from which we get
$\cos^3\alpha+2\cos^2\alpha-1=0$ or $\cos^3\alpha+\cos^2\alpha+\cos^2\alpha+\cos^1\alpha-\cos^1\alpha-1=0$
or $(\cos\alpha+1)(\cos^2\alpha+\cos^1\alpha-1) = 0$
the only possible solution from above is $\cos\alpha = \dfrac{(\sqrt 5 -1)}{2}$ or $\alpha=\theta = 51.827$ (approx)
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3166836",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
}
|
Proving a sequence has limit using an epsilon - N argument I have two separate sequences that I using this approach on.
$a_n=\frac{n^3-2n^2+3}{2n^3+7n}$ & $a_n=\frac{n^3}{2^n}$
Proof 1
$a_n=\frac{n^3-2n^2+3}{2n^3+7n}\to \frac{1}{2}$
$$\text{Let } \epsilon \gt 0 \,\, \exists N \;\;\forall n\geq N \; \left\lvert {a_n-\frac{1}{2}}\right\rvert \lt \epsilon$$
$$\left\lvert {\frac{n^3-2n^2+3}{2n^3+7n}-\frac{1}{2}}\right\rvert=\left\lvert {\frac{(2n^3-4n^2+6)-(2n^3+7n)}{4n^3+14n}}\right\rvert=\left\lvert {\frac{-4n^2-7n+6}{4n^3+14n}}\right\rvert \lt \frac{1}{2n} \lt \epsilon$$
$$\frac{1}{2\epsilon} \lt n$$
I am having difficulty in showing exactly why, $\left\lvert {\frac{-4n^2-7n+6}{4n^3+14n}}\right\rvert \lt \frac{1}{2n}$
Proof 2
$a_n=\frac{n^3}{2^n} \to 0$
$$\text{Let } \epsilon \gt 0 \,\, \exists N \;\;\forall n\geq N \; \left\lvert {a_n-0}\right\rvert \lt \epsilon$$
$$\left\lvert {\frac{n^3}{2^n}}\right\rvert=\frac{n^3}{2^n} \lt \frac{?}{?} \lt \epsilon$$
I am lost as to where to proceed on this one. Any input would be greatly appreciated . Thanks
|
On the first proof:
It would be simplest to show that $\left|\dfrac{4n^2+1}{4n^3+14n}\right|<\dfrac{1}{n}$ since, clearly $4n^3+n<4n^3+14n$.
On the second proof:
For $n\ge16$ we have that $n^4\le2^n$ from which it follows that $\dfrac{n^3}{2^n}\le\dfrac{1}{n}$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3167746",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
How to prove the inequality $a/(b+c)+b/(a+c)+c/(a+b) \ge 3/2$ Suppose $a>0, b>0, c>0$.
Prove that:
$$a+b+c \ge \frac{3}{2}\cdot [(a+b)(a+c)(b+c)]^{\frac{1}{3}}$$
Hence or otherwise prove:
$$\color{blue}{\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}\ge \frac{3}{2}}$$
|
Hint: Substitute $$b+c=x,a+c=y,a+b=z$$ so $$a=\frac{-x+y+z}{2}$$
$$b=\frac{x-y+z}{2}$$
$$c=\frac{x+y-z}{2}$$
And we get
$$\frac{-x+y+z}{2x}+\frac{x-y+z}{2y}+\frac{x+y-z}{2z}\geq \frac{3}{2}$$
Can you finish?
And we get $$\frac{x}{y}+\frac{y}{x}+\frac{y}{z}+\frac{z}{y}+\frac{x}{z}+\frac{z}{x}\geq 6$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3168215",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
}
|
Prove that $a+b+2\sqrt{ab+c^2}$ cannot be a prime number Prove that the number
$$a+b+2\sqrt{ab+c^2}$$
cannot be a prime number for any positive integer numbers $a,b,c$.
My attempt:
Suppose that $p=a+b+2\sqrt{ab+c^2}$ is a prime. WLOG assume that $a \geq b$. From the equality we have
$$\left(a+b\right)^2+p^2-2p\left(a+b\right)=4ab+4c^2 \Leftrightarrow \left(a-b-2c\right)\left(a-b+2c\right)=p\left(2a+2b-p\right)$$
Since $0<a-b+2c<a+b+2c<p$, we must have $p |\left(a-b-2c\right)$. But then I don't know how to continue the way. Please help me. Thank you.
|
Your work has almost finished the problem. $p|(a-b-2c)$ is the key property.
If $a-b-2c>0$, then $a-b-2c\ge p$. Hence $a\ge b+2c+p$, a contradiction since $a<p$.
If $a-b-2c<0$, then $a-b-2c\le -p$. Hence $a+p\le b+2c$, so $2a+b+2\sqrt{ab+c^2}\le b+2c$, so $a+\sqrt{ab+c^2}\le c$, again a contradiction.
Finally, if $a-b-2c=0$, then $a=b+2c$. But then $p=(b+2c)+b+2\sqrt{ab+c^2}$ is even.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3171455",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
}
|
show this inequality to $\sum_{cyc} \frac {a^3b}{(3a+2b)^3} \ge \sum_{cyc} \frac {a^2bc}{(2a+2b+c)^3} $ Let $a,b$ and $c$ be positive real numbers. Prove that
$$\sum_{cyc} \frac {a^3b}{(3a+2b)^3} \ge \sum_{cyc} \frac {a^2bc}{(2a+2b+c)^3} $$
This problem is from Iran 3rd round-2017-Algebra final exam-P3,Now I can't find this inequality have solve it,maybe it seem can use integral to solve it?
my attempts:
I took $p=3a+2b,$ $2a+2b+c$. $k=3$ and I wanted to use this integral:
$$\dfrac{1}{p^k}=\dfrac{1}{\Gamma(k)}\int_{0}^{+\infty}t^{k-1}e^{-pt}dt$$
but I don't see how it helps.
|
Using binomial inequality for
$$|c-a|<2a+2b+c,$$
one can get
$$\dfrac1{(3a+2b)^3}=\dfrac1{(2a+2b+c)^3}\left(1-\dfrac{c-a}{2a+2b+c}\right)^{-3}
\ge \dfrac1{(2a+2b+c)^3}\left(1+3\dfrac{c-a}{2a+2b+c}\right),$$
$$\dfrac{a^3b}{(3a+2b)^3}-\dfrac{a^2bc}{(2a+2b+c)^3} \ge \dfrac{a^2b}{(2a+2b+c)^3}\left(a-c-3a\dfrac{c-a}{2a+2b+c}\right),\tag1$$
The issue inequality can be presented in the form of
$$S\ge 0\tag2,$$
where
$$S = \sum\limits_\bigcirc \left(\dfrac{a^3b}{(3a+2b)^3}-\dfrac{a^2bc}{(2a+2b+c)^3}\right) \ge \sum\limits_\bigcirc\dfrac{a^2b(a-c)}{(2a+2b+c)^3}
+3\sum\limits_\bigcirc\dfrac{a^3b(a-c)}{(2a+2b+c)^4}.$$
In according with the rearrangement inequality for the productions of
$$a\cdot a\cdot a\cdot \dfrac b{2a+2b+c}
+ b\cdot b\cdot b\cdot \dfrac c{2b+2c+a}
+ c\cdot c\cdot c\cdot \dfrac a{2c+2a+b}$$
and
$$a\cdot a\cdot c\cdot \dfrac b{2a+2b+c}
+ b\cdot b\cdot a\cdot \dfrac c{2b+2c+a}
+ c\cdot c\cdot b\cdot \dfrac a{2c+2a+b}$$
between $a$ and $c$ (where the ratio can be placed arbitrary), one can write
$$\sum\limits_\bigcirc\dfrac{a^3b}{(2a+2b+c)^3} \ge
\sum\limits_\bigcirc\dfrac{a^2bc}{(2a+2b+c)^3}$$
and similarly
$$\sum\limits_\bigcirc\dfrac{a^4b}{(2a+2b+c)^4} \ge
\sum\limits_\bigcirc\dfrac{a^3bc}{(2a+2b+c)^4},$$
so $(1)$ is valid.
$\textbf{Proved.}$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3172778",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
}
|
Interpreting images representing geometric series I understand the formula for infinite geometric series as
$$S = \frac{a_{1}}{1-r}$$ if $0<r<1$
However I'm having trouble applying it to these images
It seems to me that in the first image, the first square represents 1/4 of the entire square
For the second and third images, the respective rectangle and triangle make up 1/2 of the entire square.
Not sure what to do with this. Does it mean that for the first image for example, the image is $$\sum_{n=0}^\infty \left(\frac{1}{4}\right)^n$$
|
1) The area of the first shaded square is a fourth part of the original square: $\frac{1}{4}$. The area of the second shaded square is a fourth of a fourth of the original area: $\frac{1}{4}\cdot\frac{1}{4}$. The area of the third square would be a fourth of that: $\frac{1}{4}\cdot\frac{1}{4}\cdot\frac{1}{4}$. Do you see the pattern?
$$
\frac{1}{4}+\left(\frac{1}{4}\cdot\frac{1}{4}\right)+\left(\frac{1}{4}\cdot\frac{1}{4}\cdot\frac{1}{4}\right)+...=\\
\left(\frac{1}{4}\right)^1+\left(\frac{1}{4}\right)^2+\left(\frac{1}{4}\right)^3+...=\\
\sum_{n=1}^{\infty}\left(\frac{1}{4}\right)^n=
\sum_{n=0}^{\infty}\left(\frac{1}{4}\right)^n-1=
\frac{1}{1-\frac{1}{4}}-1=\frac{4}{3}-1=\frac{1}{3}.
$$
2) The first rectangle is area $\frac{1}{2}$. The second is a half of the original area divided by four $\frac{1}{2}\cdot\frac{1}{4}$. The third part is one fourt of that $\frac{1}{2}\cdot\frac{1}{4}\cdot\frac{1}{4}$:
$$
\frac{1}{2}+\left(\frac{1}{2}\cdot\frac{1}{4}\right)+\left(\frac{1}{2}\cdot\frac{1}{4}\cdot\frac{1}{4}\right)+...=\\
\frac{1}{2}\left(\frac{1}{4}\right)^0+\frac{1}{2}\left(\frac{1}{4}\right)^1+
\frac{1}{2}\left(\frac{1}{4}\right)^2+...=\\
\sum_{n=0}^{\infty}\frac{1}{2}\left(\frac{1}{4}\right)^n=\frac{1}{2}\cdot\frac{1}{1-\frac{1}{4}}=\frac{2}{3}.
$$
3) The first triangle is area $\frac{1}{2}$. The second triangle is area $\frac{1}{2}\cdot\frac{1}{4}$ (a fourth part of a half of the original triangle). The third triangle is going to have area $\frac{1}{2}\cdot\frac{1}{4}\cdot\frac{1}{4}$. I think you see that the pattern is the same as in the previous case:
$$
\frac{1}{2}+\left(\frac{1}{2}\cdot\frac{1}{4}\right)+\left(\frac{1}{2}\cdot\frac{1}{4}\cdot\frac{1}{4}\right)+...=\\
\frac{1}{2}\left(\frac{1}{4}\right)^0+\frac{1}{2}\left(\frac{1}{4}\right)^1+
\frac{1}{2}\left(\frac{1}{4}\right)^2+...=\\
\sum_{n=0}^{\infty}\frac{1}{2}\left(\frac{1}{4}\right)^n=\frac{1}{2}\cdot\frac{1}{1-\frac{1}{4}}=\frac{2}{3}.
$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3177728",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 4
}
|
Are there any other methods to apply to solving simultaneous equations? We are asked to solve for $x$ and $y$ in the following pair of simultaneous equations:
$$\begin{align}3x+2y&=36 \tag1\\ 5x+4y&=64\tag2\end{align}$$
I can multiply $(1)$ by $2$, yielding $6x + 4y = 72$, and subtracting $(2)$ from this new equation eliminates $4y$ to solve strictly for $x$; i.e. $6x - 5x = 72 - 64 \Rightarrow x = 8$. Substituting $x=8$ into $(2)$ reveals that $y=6$.
I could also subtract $(1)$ from $(2)$ and divide by $2$, yielding $x+y=14$. Let $$\begin{align}3x+3y - y &= 36 \tag{1a}\\ 5x + 5y - y &= 64\tag{1b}\end{align}$$ then expand brackets, and it follows that $42 - y = 36$ and $70 - y = 64$, thus revealing $y=6$ and so $x = 14 - 6 = 8$.
I can even use matrices!
$(1)$ and $(2)$ could be written in matrix form:
$$\begin{align}\begin{bmatrix} 3 &2 \\ 5 &4\end{bmatrix}\begin{bmatrix} x \\ y\end{bmatrix}&=\begin{bmatrix}36 \\ 64\end{bmatrix}\tag3 \\ \begin{bmatrix} x \\ y\end{bmatrix} &= {\begin{bmatrix} 3 &2 \\ 5 &4\end{bmatrix}}^{-1}\begin{bmatrix}36 \\ 64\end{bmatrix} \\ &= \frac{1}{2}\begin{bmatrix}4 &-2 \\ -5 &3\end{bmatrix}\begin{bmatrix}36 \\ 64\end{bmatrix} \\ &=\frac12\begin{bmatrix} 16 \\ 12\end{bmatrix} \\ &= \begin{bmatrix} 8 \\ 6\end{bmatrix} \\ \\ \therefore x&=8 \\ \therefore y&= 6\end{align}$$
Question
Are there any other methods to solve for both $x$ and $y$?
|
If you prefer using parametric form or your equations are already in parametric form, this is how you can proceed:
We know that $(0,18)$ is a solution to $3x+2y=36$ and $(0,16)$ is a solution to $5x + 4y = 64$. Therefore the equations in parametric form become:
$${ \begin{pmatrix} 0 \\ 18 \\ \end{pmatrix} } + t_1 { \begin{pmatrix} 2 \\ -3 \\ \end{pmatrix} } \tag{3}$$
$$\pmatrix { \begin{matrix} 0 \\ 16 \\ \end{matrix} } + t_2 { \begin{pmatrix} 4 \\ -5 \\ \end{pmatrix} } \tag{4} $$
Equate the $x$ and $y$ coordinates:
$$0+2t_1 = 0+4t_2 \Rightarrow t_1 = 2t_2 \tag{5}$$
$$18-3t_1 = 16-5t_2 \Rightarrow 18-6t_2 = 16-5t_2 \text{ (using (5) }) \Rightarrow2 = t_2 \tag{6}$$
and now substitute back into $(4)$:
$$\pmatrix { \begin{matrix} 0 \\ 16 \\ \end{matrix} } + 2 { \begin{pmatrix} 4 \\ -5 \\ \end{pmatrix} } =\pmatrix { \begin{matrix} 8 \\ 6 \\ \end{matrix} } $$
i.e $x=8, y=6$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3180580",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "32",
"answer_count": 14,
"answer_id": 13
}
|
Show that a determinant equals the product of another determinant and a polynomial function without calculating Show without calculating the determinant, that
$$
\det\left(\begin{bmatrix}
a_{1}+b_{1}x & a_{1}-b_{1}x & c_{1}\\
a_{2}+b_{2}x & a_{2}-b_{2}x & c_{2}\\
a_{3}+b_{3}x & a_{3}-b_{3}x& c_{3}\\
\end{bmatrix}\right)
=
-2x
\det\left(\begin{bmatrix}
a_{1} & b_{1}x & c_{1}\\
a_{2} & b_{2}x & c_{2}\\
a_{3} & b_{3}x& c_{3}\\
\end{bmatrix}\right)
$$
I've tried using the multilinearity of the determinant, but i didn't get very far and now im stuck.
|
It is indeed multilinearity you want to use. Write
$$
a =
\begin{bmatrix}
a_{1}\\
a_{2}\\
a_{3}
\end{bmatrix},
\quad
b =
\begin{bmatrix}
b_{1}\\
b_{2}\\
b_{3}
\end{bmatrix},
\quad
c =
\begin{bmatrix}
c_{1}\\
c_{2}\\
c_{3}
\end{bmatrix},
$$
then
\begin{align*}
\det(\begin{bmatrix}a + b x & a - b x & c\end{bmatrix})
&=
\det(\begin{bmatrix}a & a & c\end{bmatrix})
- x^{2}
\det(\begin{bmatrix}b & b & c\end{bmatrix})
\\&- x
\det(\begin{bmatrix}a & b & c\end{bmatrix})
+ x
\det(\begin{bmatrix}b & a & c\end{bmatrix}).
\end{align*}
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3181006",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
}
|
Use generating functions to solve the recurrence relation Use generating functions to solve $a_n = 3a_{n-1} - 2a_{n-2} + 2^n + (n+1)3^n$.
What I have so far, not sure if I forgot to do something or am missing out on something obvious:
Define $$G(x) = \sum_{n=0}^{\infty} a_nx^n$$
Then, $G(x) = a_0 + a_1x + a_2x^2 + \sum_{n=3}^{\infty} a_nx^n$
= $a_0 + a_1x + a_2x^2 + \sum_{n=3}^{\infty} (3a_{n-1} - 2a_{n-2} + 2^n + (n+1)3^n)x^n$
= $a_0 + a_1x + a_2x^2 + \sum_{n=3}^{\infty} 3a_{n-1}x^n - \sum_{n=3}^{\infty} 2a_{n-2}x^n + \sum_{n=3}^{\infty} 2^nx^n + \sum_{n=3}^{\infty} (n+1)3^nx^n$
= $a_0 + a_1x + a_2x^2 + 3\sum_{n=3}^{\infty} a_{n-1}x^n - 2\sum_{n=3}^{\infty} a_{n-2}x^n + \sum_{n=3}^{\infty} (2x)^n + \sum_{n=3}^{\infty} (n+1)(3x)^n$
=$a_0 + a_1x + a_2x^2 + 3x\sum_{n=3}^{\infty} a_{n-1}x^{n-1} - 2x^2\sum_{n=3}^{\infty} a_{n-2}x^{n-2} + \sum_{n=3}^{\infty} (2x)^n +
\sum_{n=3}^{\infty} (n+1)(3x)^n$
= $a_0 + a_1x + a_2x^2 + 3x\sum_{n=2}^{\infty} a_nx^n - 2x^2\sum_{n=1}^{\infty} a_nx^n + \sum_{n=3}^{\infty} (2x)^n +
\sum_{n=3}^{\infty} (n+1)(3x)^n$
|
After making these replacements, you will have $G(x)$ on both sides, and a couple of other summations you can (hopefully?) find closed forms for. Then you can solve for $G(x)$.
$$
\sum_{n=1}^\infty a_nx^n=G(x)-a_0\hspace{1.3cm}
$$
$$
\sum_{n=2}^\infty a_nx^n=G(x)-a_0-a_1x
$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3182852",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
}
|
Evaluating $\sum_{n=1}^\infty\frac{(H_n)^2}{n}\frac{\binom{2n}n}{4^n}$
Question: How can we evaluate $$\sum_{n=1}^\infty\frac{(H_n)^2}{n}\frac{\binom{2n}n}{4^n},$$where $H_n=\frac11+\frac12+\cdots+\frac1n$?
Quick Results
This series converges because $$\frac{(H_n)^2}{n}\frac{\binom{2n}n}{4^n}=O\left(\frac{\ln^2n}{n^{3/2}}\right).$$
My Attempt
Recall the integral representation of harmonic number $$H_n=\int_0^1\frac{1-x^n}{1-x}d x$$
we have $$
S=\sum_{n=1}^\infty\frac1n\frac{\binom{2n}n}{4^n}\iint_{[0,1]^2}\frac{(1-x^n)(1-y^n)}{(1-x)(1-y)}d xd y\\
=\tiny\iint_{[0,1]^2}\frac{x y \log (4)-2 x y \log \left(\sqrt{1-x}+1\right)-2 x y \log \left(\sqrt{1-y}+1\right)+2 x y \log \left(\frac{1}{2} \left(\sqrt{1-x y}+1\right)\right)}{\left(\sqrt{1-x y}-1\right) \left(\sqrt{1-x y}+1\right)}dxdy\\
$$
This integral is too hard for me and Mathematica to compute. Numerical integration returns $12.6178$, it agrees with the numerical summation of the original series. I tried to integrate with respect to $x$, but failed.
|
From this paper, Eq $(13)$ page 4, we have
$$\sum_{n=1}^\infty H_n\binom{2n}n x^n=\frac{2}{\sqrt{1-4x}}\ln\left(\frac{1+\sqrt{1-4x}}{2\sqrt{1-4x}}\right).$$
Replace $x$ by $\frac{x}{4}$, multiply both sides by $-\frac{\ln(1-x)}{x}$ then integrate using $-\int_0^1 x^{n-1} \ln(1-x)dx=\frac{H_n}{n}$, we have
$$\sum_{n=1}^\infty \frac{H_n^2}{n4^n}\binom{2n}n=-\int_0^1\frac{\ln(1-x)}{x\sqrt{1-x}}\ln\left(\frac{1+\sqrt{1-x}}{2\sqrt{1-x}}\right)dx$$
$$\overset{\sqrt{1-x}=y}{=}-8\int_0^1\frac{\ln y}{1-y^2}\ln\left(\frac{1+y}{2y}\right)dy$$
$$\overset{y=\frac{1-x}{1+x}}{=}4\int_0^1\frac{\ln^2(1-x)}{x}dx-4\int_0^1\frac{\ln(1-x)\ln(1+x)}{x}dx$$
$$=4[2\zeta(3)]-4\left[-\frac58\zeta(3)\right]=\frac{21}{2}\zeta(3).$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3184675",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 3,
"answer_id": 1
}
|
$f$ dividing by $x + 1$ have remainder 4, when dividing with $x^2 + 1$ have remainder 2x+3. Find remainder dividing polynomial with($x+1$)($x^2+1$) Problem: The polynomial $f$ dividing by ($x + 1$) gives the remainder 4, and when dividing with ($x^2 + 1$) gives the remainder (2x+3). Determine the remainder when dividing the polynomial with ($x + 1$)($x^2 + 1$)?
My attempt:
By Polynomial remainder theorem we know that
$$f(x)=q_1(x)(x+1)+4$$
$$f(x)=q_2(x)(x^2+1)+(2x+3)$$
By putting $x=1$ we know that $f(-1)=4, f(i)=2i+3, f(-i)=-2i+3$
We want to find $r(x)$ such that: $$f(x)=(x+1)(x^2+1)q_3(x) + r(x) .$$
By applying the previous idea we know that $f(-1)=r(-1)=4$, but the same idea $r(i)=2i+3$ and $r(-i)=-2i+3$ but this is only three point and we need to determinant polynomial $r(x)$ of degree 3...
Please solve without modular arithmetic.
|
Upon applying the law: $\,ab\bmod ac = a(b\bmod c) = $ Mod Distributive Law we obtain
$(f\!-\!(2x\!+\!3))\bmod{(x^2\!+\!1)(x\!+\!1)}\,=\, (x^2\!+\!1){\Huge[}\dfrac{\overbrace{f\!-\!(2x\!+\!3)}^{\large 4\,-\,(2(\color{#c00}{\bf -1})+3)}}{\underbrace{x^2\!+\!1}_{\large (\color{#c00}{\bf -1})^2+1\ \ \ \ }}\underbrace{\bmod{x\!+\!1}}_{\Large x\ \equiv\ \color{#c00}{\bf {-}1}}{\Huge]}=\, (x^2\!+\!1)\left[\dfrac{3}{2}\right]$
So $\, f\equiv 2x\!+\!3 + \dfrac{3}2(x^2\!+\!1).\ $ Note that we did not need to solve any equations - just evaluate.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3185530",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
}
|
Multiplying two generating functions I am trying to complete exercise 10 from here. It says to find $a_7$ of the sequence with generating function $\frac{2}{(1−x)^2} \cdot \frac{x}{1−x−x^2}$. I wrote down the first $7$ numbers of both sequences and got $2, 4, 6, 8, 10, 12, 14$ and $1, 1, 2, 3, 5, 8, 13, 21$. I then tried to multiply it out as using distribution as described on the page
$$AB = a_0b_0 + (a_0b_1 + a_1b_0)x + (a_0b_2 + a_1b_1 + a_2b_0)x^2 + \dots$$ where $A = a_0 + a_1x + a_2x^2 + \ldots$ and $B = b_0 + b_1x + b_2x^2 + \ldots$.
However, this is incorrect in addition to coming up with closed forms for each of the individual generating functions and adding them.
|
The two sequences are,
$$2, 4, 6, 8, 10, 12, 14, 16, \dots$$
and,
$$0, 1, 1, 2, 3, 5, 8, 13, 21, \dots$$
Note that with generating functions the initial term is $a_0$, "term zero".
So, $a_7$ is actually the $8^{th}$ term.
The convolution of these two sequences, also called the Cauchy product, is found by multiplying the generating series associated with these sequences together.
In other words we're going to multiply,
$$2+4x+6x^2+8x^3+10x^4+12x^5+14x^6+16x^7+\dots$$
by,
$$0+x+x^2+2x^3+3x^4+5x^5+8x^6+13x^7+...$$
I'll do all eight terms up to $a_7$ as it'll make the pattern more obvious,
$$(0\times 2)$$
$$+(1\times 2+0\times 4)x$$
$$+(1\times 2+1\times 4+0\times 2)x^2$$
$$+(2\times 2+1\times 4+1\times 6+0\times 8)x^3$$
$$+(3\times 2+2\times 4+1\times 6+1\times 8+0\times 10)x^4$$
$$+(5\times 2+3\times 4+2\times 6+1\times 8+1\times 10+0\times 12)x^5$$
$$+(8\times 2+5\times 4+3\times 6+2\times 8+1\times 10+1\times 12+0\times 14)x^6$$
$$+(13\times 2+8\times 4+5\times 6+3\times 8+2\times 10+1\times 12+1\times
14+0\times 16)x^7$$
$$+\dots$$
It's the coefficient of $x^7$ we are after,
$$(26+32+30+24+20+12+14)x^7$$
$$=(158)x^7$$
which is, indeed, 158, as you were expecting.
Not the easiest of examples on account of the 0 and pair of 1s at the start of the Fibonacci sequence.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3187623",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
}
|
Number of real roots of $p_n(x)=1+2x+3x^2+....+(n+1)x^n$ if $n$ is an odd integer
If $n$ be an odd integer. Then find the number of real roots of the polynomial equation $p_n(x)=1+2x+3x^2+....+(n+1)x^n$
$$
p_n(x)=1+2x+3x^2+....+(n+1)x^n\\
x.p_n(x)=x+2x^2+....nx^n+(n+1)x^{n+1}\\
p(x)[1-x]=1+x+x^2+....+x^n-(n+1)x^{n+1}\\
p(x)=\frac{1+x+x^2+....+x^{n}}{1-x}-\frac{(n+1)x^{n+1}}{1-x}=\frac{x^{n+1}-1}{x-1}\frac{1}{1-x}-\frac{(n+1)x^{n+1}}{1-x}\\
=\frac{x^{n+1}-1-(n+1)(x-1)x^{n+1}}{-(x-1)^2}=\frac{x^{n+1}-1+(n+1)x^{n+1}-(n+1)x^{n+2}}{-(x-1)^2}\\
=\frac{(n+2)x^{n+1}-(n+1)x^{n+2}-1}{-(x-1)^2}=\frac{1-(n+2)x^{n+1}+(n+1)x^{n+2}}{(x-1)^2}=0\\
\implies \boxed{(n+2)x^{n+1}-(n+1)x^{n+2}=1}
$$
I think I am stuck with my attempt, how can I find the real solutions ?
|
$$
p(x)=1+2x+3x^2+\ldots+(n+1)x^n;\\
xp(x)=x+2x^2+3x^3+\ldots+(n+1)x^{n+1};\\
(1-x)p(x)=1+x+x^2+\ldots+x^n-(n+1)x^{n+1}=\frac{1-x^{n+1}}{1-x}-(n+1)x^{n+1}\\
=\frac{1-x^{n+1}-(1-x)(n+1)x^{n+1}}{1-x}=0.\\
\Rightarrow1-x^{n+1}-(1-x)(n+1)x^{n+1}=0
$$
Note that $x=1$ is clearly not a solution of the original equation. As a result, the above equation is actually equivalant to your original equation, apart from having an extra root $x=1$.
Here is the equation we are trying to solve:
$$
1-x^{n+1}-(1-x)(n+1)x^{n+1}\equiv (n+1)x^{n+2}-(n+2)x^{n+1}+1=0\\
\Leftrightarrow(n+1)x^{n+2}=(n+2)x^{n+1}-1
$$
Since the original equation has no roots for $x\geq 0$, we now focus on the case $x<0$. When $x<0$, LHS is decreasing toward $-\infty$; RHS is increasing towards $\infty$. LHS>RHS for $x=0$. This tells us that the original equation has only one root.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3188364",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
}
|
Definite Integral of $\int_0^1\frac{dx}{\sqrt {x(1-x)}}$ We have to calculate value of the following integral :
$$\int_0^1\cfrac{dx}{\sqrt {x(1-x)}} \qquad \qquad (2)$$
What i've done for (2) :
\begin{align}
& = \int_0^1\cfrac{dx}{\sqrt {x(1-x)}} \\
& = \int_0^1\cfrac{dx}{\sqrt {x-x^2}} \\
& = \int_0^1\cfrac{dx}{\sqrt {(x^2-x+\frac 14)-\frac 14 }} \\
& = \int_0^1\cfrac{dx}{\sqrt {(x-\frac 12)^2-(\frac 12)^2 }} \\
& = \cfrac {1}{2}\int_0^1\cfrac{\sec \theta \tan \theta \ d\theta}{\sqrt {(\frac 12\sec \theta)^2-(\frac 12)^2 }} I\ used\ trigonometric\ substitution \ u=a\sec \theta, by \ it's \ form \ u^2-a^2 \\
& = \cfrac {1}{2}\int_0^1\cfrac{\sec \theta \tan \theta \ d\theta}{\sqrt {(\frac 14\sec^2 \theta)-\frac 14 }} \\
& = \cfrac {1}{2}\int_0^1\cfrac{\sec \theta \tan \theta \ d\theta}{\sqrt {\frac 14(\sec ^2\theta-1)}} \ using \\tan^2\theta=\sec^2\theta-1 \\
& = \cfrac {1}{2}\int_0^1\cfrac{\sec \theta \tan \theta \ d\theta}{\sqrt {\frac 12(\sqrt{\tan^2\theta) }}} \\
& = \int_0^1\sec\theta d\theta = \sec\theta \tan \theta |_0^1 \\
\end{align}
But i got problems calculating $\theta$ value, using trigonometric substitution, any help?
|
Letting $x=\sin^2 \theta$ yields
$$
\begin{aligned}
\int_0^1 \frac{d x}{\sqrt{x(1-x)}} &=\int_0^{\frac{\pi}{2}} \frac{2 \sin \theta \cos \theta d \theta}{\sin \theta \cos \theta} =[2 \theta]_0^{\frac{\pi}{2}} =\pi
\end{aligned}
$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3191401",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
}
|
Solution to $\sqrt{\sqrt{x + 5} + 5} = x$ There are natural numbers $a$, $b$, and $c$ such that the solution to the equation
\begin{equation*}
\sqrt{\sqrt{x + 5} + 5} = x
\end{equation*}
is $\displaystyle{\frac{a + \sqrt{b}}{c}}$. Evaluate $a + b + c$.
I am not sure where I saw this problem. My guess is that it was from a high school math competition. The solution to the equation is $\frac{1 + \sqrt{21}}{2}$. This suggests use of the quadratic formula.
The solution set to the given equation is a subset of the solution set to
\begin{equation*}
x^{2} - 5 = \sqrt{x + 5} ,
\end{equation*}
\begin{equation*}
x^{4} - 10x^{2} + 25 = x + 5
\end{equation*}
\begin{equation*}
x^{4} - 10x^{2} - x + 20 = 0 .
\end{equation*}
Using the quartic equation (or Wolfram), the solutions to this equation are computed to be
\begin{equation*}
\frac{1 \pm \sqrt{21}}{2} , \qquad \frac{-1 \pm \sqrt{17}}{2} .
\end{equation*}
|
Note that if $x = \sqrt{x+5}$ then $x = \sqrt{\sqrt{x+5}+5}$.
So, try solving $x = \sqrt{x+5}$. This is a quadratic.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3191768",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
}
|
How does one prove such an equation? The problem occurred to me while I was trying to solve a problem in planimetry using analytic geometry.
for $b$ between $-\frac{1}2$ and $1$ :
$\sqrt{2+\sqrt{3-3b^2}+b} = \sqrt{2-2b}+ \sqrt{2-\sqrt{3-3b^2}+b}$
|
Hint:
set $b=\cos t$:
\begin{align}
\sqrt{2+\cos t+\sqrt3\sqrt{1-\cos^2t}}
&\overset?=
\sqrt2\sqrt{\vphantom{\sqrt3}1-\cos t}+ \sqrt{2+\cos t-\sqrt3\sqrt{1-\cos^2t}}
,\\
\sqrt{2+\cos t+\sqrt3\sin t}
&\overset?=
\sqrt2\sqrt{\vphantom{\sqrt3}1-\cos t}+ \sqrt{2+\cos t-\sqrt3\sin t}
,\\
\sqrt{1+\tfrac12\cos t+\tfrac{\sqrt3}2\sin t}
&\overset?=
\sqrt{\vphantom{\tfrac{\sqrt3}2}1-\cos t}+ \sqrt{1+\tfrac12\cos t-\tfrac{\sqrt3}2\sin t}
\\
\dots
\\
\text{(use trigonometry to combine and get rid of radicals)}
\\
\dots,
\end{align}
and it turns out that
on the given interval
both sides are equivalent to the same expression
\begin{align}
\sin(\tfrac12t+\tfrac\pi3)
.
\end{align}
Edit:
\begin{align}
\sqrt{1+\cos\tfrac\pi3\cos t+\sin\tfrac\pi3\sin t}
&\overset?=
\sqrt{\vphantom{\tfrac{\sqrt3}2}1-(1-2\sin^2\tfrac t2)}+
\sqrt{1+\cos\tfrac\pi3\cos t-\sin\tfrac\pi3\sin t}
,\\
\sqrt{1+\cos(t-\tfrac\pi3)}
&\overset?=
\sqrt{\vphantom{\tfrac{\sqrt3}2}1-(1-2\sin^2\tfrac t2)}+
\sqrt{1+\cos(t+\tfrac\pi3)}
,\\
\sqrt{1+2\cos^2\frac{t-\tfrac\pi3}2-1}
&\overset?=
\sqrt2\sin\tfrac t2+
\sqrt{1+2\cos^2\frac{t+\tfrac\pi3}2-1}
,\\
\cos(\tfrac t2-\tfrac\pi6)
&\overset?=
\sin\tfrac t2+
\cos(\tfrac t2+\tfrac\pi6)
,\\
\cos\tfrac t2\cos\tfrac\pi6+
\sin\tfrac t2\sin\tfrac\pi6
&\overset?=
\sin\tfrac t2+
\cos\tfrac t2\cos\tfrac\pi6-
\sin\tfrac t2\sin\tfrac\pi6
,\\
\cos\tfrac t2\cos\tfrac\pi6+
\sin\tfrac t2\sin\tfrac\pi6
&\overset?=
\cos\tfrac t2\cos\tfrac\pi6+
\sin\tfrac t2\sin\tfrac\pi6
.
\end{align}
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3193543",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
}
|
Complex number cannot arive at $\frac{9}{2}-\frac{9}{2}i$ with problem $\frac{4+i}{i}+\frac{3-4i}{1-i}$ I am asked to evaluate: $\frac{4+i}{i}+\frac{3-4i}{1-i}$
The provided solution is: $\frac{9}{2}-\frac{9}{2}i$
I arrived at a divide by zero error which must be incorrect. My working:
$\frac{4+i}{i}$, complex conjugate is $-i$ so:
$\frac{-i(4+i)}{-i*i}$
= $\frac{-4i+i^2}{i^2}$
= $\frac{-4i--1}{-1}$
= $-4i+1$
Then the next part:
$\frac{3-4i}{1-i}$ complex conjugate is $1+i$ so:
$\frac{(1+i)(3-4i)}{(1+i)(1-i)}$
= $\frac{3-4i+3i-4i^2}{1-i^2}$
= $\frac{7-i}{0}$ # 1 + -1 = 0
How can I arrive at $\frac{9}{2}-\frac{9}{2}i$?
|
You've missed a minus sign there:
$\frac{4+i}{i}=\frac{4i-1}{-1}=1-4i$
The second one is
$\frac{3-4i}{1-i}=\frac{(3-4i)(1+i)}{(1-i)(1+i)}=\frac{7-i}{2}$
Hence the sum is $1-4i+\frac{7}{2}-\frac{i}{2}=\frac{9}{2}-\frac{9i}{2}$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3193823",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
}
|
to prove $x^2 + y^2+1\ge xy + y + x$ $$x^2 + y^2+1\ge xy + y + x$$
$x$ and $ y$ belong to all real numbers
my attempt
$(u-2)^2\ge0\Rightarrow \frac{u^2}{4}+1\ge u $
let $u=x+y\Rightarrow \frac{(x+y)^2}{4}+1\ge x+y$
$\Rightarrow (x+y)^2+1\ge \frac{3}{4}(x+y)^2+(x+y)$
$but \frac{(x+y)^2}{4} \ge xy $ by AM-GM inequality
$\Rightarrow (x+y)^2+1\ge \frac{3}{4}(x+y)^2+(x+y)\ge3xy+(x+y)$
hence $\Rightarrow x^2 + y^2+2xy+1\ge 3xy+x+y$
are the steps correct and is there any other better way??
|
Prove
$$f(x,y)=x^2+y^2+1-x-xy-y\geq 0.$$
$$f_x=2x-1-y$$
$$f_y=2y-1-x$$
$$f_x=0\implies y=2x-1$$
$$f_y=0\implies x=2y-1$$
$$y=2(2y-1)-1=4y-3\implies y=1$$
$$x=2(2x-1)-1=4x-3\implies x=1$$
There's a stationary point at $(1,1)$.
$$f_{xx}=2,f_{xy}=f_{yx}=-1,f_{yy}=2.$$
Since $$f_{xx}f_{yy}-f_{xy}^2=4-(-1)^2=3>0\text{ and }f_{xx}=2>0$$
then that stationary point is a minimum. Furthermore $f(1,1)=0\geq 0$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3197998",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
}
|
Given three non-negatve numbers $a,b,c$. Prove that $1+a^{2}+b^{2}+c^{2}+4abc\geqq a+b+c+ab+bc+ca$ .
Given three non-negatve numbers $a, b, c$. Prove that:
$$1+ a^{2}+ b^{2}+ c^{2}+ 4abc\geqq a+ b+ c+ ab+ bc+ ca$$
Let $t= a+ b+ c$, we have to prove
$$\left(\!\frac{1}{t^{3}}- \frac{1}{t^{2}}+ \frac{1}{t}\!\right)\sum a^{3}+ \left(\!\frac{3}{t^{3}}- \frac{3}{t^{2}}\!\right)\left (\!\sum a^{2}b+ \sum a^{2}c\!\right )+ \left(\!\frac{6}{t^{3}}- \frac{6}{t^{2}}- \frac{3}{t}+ 4\!\right)abc\geqq 0$$
If $0< t< 1$ so
$${\rm LHS}\geqq \left(\frac{3}{t}+ 1\right)\left(\frac{3}{t}- 2\right)^{2}abc\geqq 0$$
If $1< t$ so
$${\rm LHS}= \left(\!\frac{3}{t^{2}}- \frac{3}{t^{3}}\!\right)(\!{\rm Schur.3}\!)+ \frac{1}{t}\left(\!\frac{2}{t}- 1\!\right)^{2}(\!{\rm a.m.}- {\rm g.m.}\!)+ \left(\!\frac{3}{t}+ 1\!\right)+ \left(\!\frac{3}{t}- 2\!\right)^{2}abc\geqq 0$$
(Can you find the way without deviding two cases as above?)
|
Another way.
Since $$\prod_{cyc}(2a-1)^2=\prod_{cyc}((2a-1)(2b-1))\geq0$$ and our inequality is symmetric, we can assume that $$(2a-1)(2b-1)\geq0,$$ which gives
$$c(2a-1)(2b-1)\geq0$$ or
$$4abc\geq 2ac+2bc-c.$$
Thus, $$1+a^2+b^2+c^2+4abc-a-b-c-ab-ac-bc\ge$$
$$\geq1+c^2+a^2+b^2-ab+ac+bc-a-b-2c\geq$$
$$\geq1+c^2+\frac{1}{4}(a+b)^2+(a+b-2)c-a-b=$$
$$=c^2+(a+b-2)c+\left(\frac{a+b}{2}-1\right)^2=\left(c+\frac{a+b}{2}-1\right)^2\geq0.$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3198229",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
}
|
Jordan normal as transformation with respect to the basis of eigenvectors I have the following matrix
$$A = \begin{pmatrix}
2 & 0 & 1 & -3 \\
0 & 2 & 10 & 4 \\
0 & 0 & 2 & 0 \\
0 & 0 & 0 & 3 \\
\end{pmatrix}$$
and its Jordan normal form is
$$J = \begin{pmatrix}
2 & 0 & 0 & 0 \\
0 & 2 & 1 & 0 \\
0 & 0 & 2 & 0 \\
0 & 0 & 0 & 3 \\
\end{pmatrix}$$
with the linearly independent set of eigenvectors:
$$P = \begin{pmatrix}
0 & 1 & 0 & -3 \\
1 & 10 & 0 & 4 \\
0 & 0 & 1 & 0 \\
0 & 0 & 0 & 1 \\
\end{pmatrix}$$
where $J = P^{-1} A P$. I am told that the following relations hold:
$$A\mathbf{v_{1}} = 2\mathbf{v_{1}}, \qquad A\mathbf{v_{2}} = 2\mathbf{v_{2}}, \qquad A\mathbf{v_{3}} = \mathbf{v_{2}} + 2\mathbf{v_{3}}, \qquad A\mathbf{v_{4}} = 3\mathbf{v_{4}}$$
In the case of $\mathbf{v_{1}}, \mathbf{v_{2}}, \mathbf{v_{4}}$, it makes total sense since this is due to the fundamental relation $A\mathbf{v} = \lambda \mathbf{v}$. Also, I see how $A\mathbf{v_{3}} = A_{2} +2A_{3}$, where $A_{2}$ and $A_{3}$ are the second and third column vectors of thre matrix $A$. This is simply multiplying $\mathbf{A}$ by $\mathbf{v_{3}}$. However, my intuition is failing me in seeing how
$$A\mathbf{v_{3}} = \mathbf{v_{2}} + 2\mathbf{v_{3}}$$
I am told that this is simply because $J$ represents the transformation corresponding to $A$ with respect to the basis $\big\{\mathbf{v_{1}}, \mathbf{v_{2}}, \mathbf{v_{3}}, \mathbf{v_{4}}\big\}$. Any pointers that can help me understand this?
|
B= $\big\{ v_{1},v_{2}, v_{3}, v_{4} \big\} $
$P^{-1}_{B} [A \mathbf{v_{3}}] = [A \mathbf{v_{3}}]_{B} = \begin{bmatrix}
0 \\
1 \\
2 \\
0\\
\end{bmatrix}$
... from the change of basis formula: $x = P_{B}[x]_{B}$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3202610",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
}
|
Minimum value of $\frac{2-\cos(x)}{\sin(x)}$ without differentiation I have to find the minimum value of the expression
$$\frac{2 - \cos x}{ \sin x}$$
Also $x$ lies between $0$ to $\pi$. One way is to find the minima using differentiation. But it is not taught in my grade so my teacher asked me to do it without differentiation.
Here's what I did
Let$$\frac{2 - \cos x }{ \sin x} = y~,$$
so that $$ (2-y \sin x)^2 = 1 - \sin^2(x) \\
\implies \sin^2(x) \cdot (y^2 +1) - 4 y \sin x +3=0$$
Now I am struck. I tried using Discriminant $\ge{0}$ but no use as our variable $\sin x$ lies between $0$and $1$. Please help.
|
You can use AM-GM as follows for $x \in (0,\pi)$:
$$\frac{2-\cos(x)}{\sin(x)}= \frac{2-(\cos^2(x/2) - \sin^2(x/2))}{2\sin(x/2)\cos(x/2)}$$
$$= \frac{\cos^2(x/2) +3\sin^2(x/2)}{2\sin(x/2)\cos(x/2)}\stackrel{AM-GM}{\geq} \frac{\sqrt{\cos^2(x/2) \cdot 3\sin^2(x/2)}}{\sin(x/2)\cos(x/2)} = \sqrt{3}$$
Additional note after comment:
Note that according to AM-GM you have equality if and only if
$$\cos^2(x/2) = 3\sin^2(x/2) \Leftrightarrow \tan^2(x/2) = \frac{1}{3} \stackrel{x \in (0,\pi)}{\Leftrightarrow}x = \frac{\pi}{3}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3202732",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
}
|
Prime numbers of the form $p=m^2+n^2$ such that $p \mid m^3+n^3-4$
Find all prime numbers $p$, for which there are positive integers $m$ and $n$ such that $p=m^2+n^2$ and $p \mid m^3+n^3-4$.
I simplified this a little bit.
$$m^2+n^2 \mid m^3+n^3-4 =(m+n)(m^2+n^2-mn)-4 \\ \Longrightarrow m^2+n^2 \mid (m+n)mn+4 $$
The only case that $m$ and $n$ can both be odd is $m=n=1$, which leads to $p=2$. If one of $m$ and $n$ is $1$ (For example $n=1$), then $p=m^2+1$ and
$$m^2+1 \mid m^2+m+4 \Longrightarrow m^2+1 \mid m+3 \Longrightarrow m=2$$
Which gives $p=5$.
For $m,n>1$ with different parity, I could not find a strong argument. Can you guys give it a try?
|
There are no primes with this property other than $p=2$ and $p=5$; here is a proof.
Suppose that $m,n>1$ and $p=m^2+n^2$ is a prime dividing $m^3+n^3-4$. As you have observed, we have then
$$ mn(m+n) \equiv -4 \pmod p. $$
We apply two operations to this congruence: squaring and multiplying by $m+n$; keeping in mind that $(m+n)^2\equiv 2mn\pmod p$, this gives
\begin{align*}
2m^3n^3 &\equiv 16\pmod p, \\
2m^2n^2 &\equiv -4(m+n)\pmod p;
\end{align*}
that is,
\begin{align*}
m^3n^3 &\equiv 8\pmod p, \tag{1} \\
m^2n^2 &\equiv -2(m+n)\pmod p. \tag{2}
\end{align*}
In view of $m^3n^3-8=(mn-2)(m^2n^2+2mn+4)$, from (1) we conclude that either $mn\equiv 2\pmod p$, or $m^2n^2+2mn+4\equiv 0\pmod p$. In the former case (2) leads to $p\mid m+n+2$, and as a result to $p=m^2+n^2\le m+n+2$, which is easily seen to be possible only if $m=2$ and $n=1$, or vice versa, contradicting the assumption $m,n>1$. Suppose thus that
$$ m^2n^2+2mn+4\equiv 0\pmod p. $$
Combined with (2), this gives
$$ m+n \equiv mn+2\pmod p, $$
whence
$$ p = m^2+n^2 \le mn-m-n+2. $$
But, recalling that $m,n>1$,
$$ m^2+n^2 \ge 2mn = mn - m - n + 2 + ((m+1)(n+1) - 3) > mn -m -n + 2, $$
a contradiction.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3203678",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
}
|
Can this determinant expression ever equal $0$? My issue is that when calculating the eigenvalues of a matrix, there are cases where the eigenvalue is $0$, and so I was trying to confirm this by calculating the determinant and checking when it is equal to $0$.
The setup is: $a^2 - 4bc \neq 0$, $bc \neq 0$, $(a,b,c) \in \mathbb R^3$, and $n \geq 1$ is an integer
and I have this expression:
\begin{equation}
\dfrac1{\sqrt{a^2-4bc}} \left( \left( \dfrac{a + \sqrt{a^2-4bc}}2\right)^{n+1} - \left( \dfrac{a - \sqrt{a^2-4bc}}2\right)^{n+1}\right) = 0
\end{equation}
Which I simplified to
\begin{equation}
\left( \dfrac{a + \sqrt{a^2-4bc}}2\right)^{n+1} = \left( \dfrac{a - \sqrt{a^2-4bc}}2\right)^{n+1} \implies \sqrt{a^2-4bc} = -\sqrt{a^2-4bc}
\end{equation}
From here, I am unsure how to get a solution, since $a^2 - 4bc \neq 0$, and for real numbers, $x$, I think the only case for $+x = -x$ is when $x = 0$, which is not true here. So since the square root can be complex, I am unsure if there are cases where this can be true.
The eigenvalue expression is
\begin{equation}
\lambda_k = a + 2b\sqrt{\frac{c}{b}}\text{cos}(\frac{k\pi}{n+1}), \qquad k \in \{1, \dots, n\}
\end{equation}
I am not experienced with complex numbers, but using the eigenvalue equations, it seems like the case of $b = 1 \text{, } c = 1, a= -2\cdot\text{cos}(\frac{1\pi}{n+1})$ (for any $n$, but I used $n = 5$), should lead to a $0$ determinant/singular matrix, but this results (for $n=5$) in $a^2 - 4bc = -1$, and so it'd require $\sqrt{-1} = -\sqrt{-1}$, but I can't see how that'd be true.
Thanks in advance for the help! (Also if you want to see the matrix I am working with, this question has the determinant expression: How to compute the determinant of a tridiagonal matrix with constant diagonals?)
|
From $$\left(\frac{a+\sqrt{a^2-4bc}}2\right)^{n+1}=\left(\frac{a-\sqrt{a^2-4bc}}2\right)^{n+1} $$
we get, by dividing by the non-zero(!) rught hand side
$$ \left(\frac{a+\sqrt{a^2-4bc}}{a-\sqrt{a^2-4bc}}\right)^{n+1}=1,$$
or after making the denominator rational,
$$ \left(\frac{(a+\sqrt{a^2-4bc})^2}{4bc}\right)^{n+1}=1,$$
so the condition is rather
$$ \tag1\frac{(a+\sqrt{a^2-4bc})^2}{4bc}=e^{\frac{2k\pi i}{n+1}},\quad 0\le k\le n.$$
Now distinguish cases depending on the value of $bc$.
If $bc<0$, the left hand side in $(1)$ is real and negative, and we arrive at
$$ \left|a+\sqrt{a^2-4bc}\right|=2\sqrt{-bc},\quad n\text{ odd}, k=\frac{n+1}2.$$
If $0<bc<\frac14a^2$, the left hand side in $(1)$ is real and positive, and we arrive at
$$ \left|a+\sqrt{a^2-4bc}\right|=2\sqrt{bc},\quad n\text{ arbitrary}, k=0.$$
Finally, if $bc>\frac14a^2$, we really encounter complex solutions, namely for
$$ a=\sqrt{4bc}\cos \phi, \quad(n+1)\phi\equiv 0\pmod{2\pi}.$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3207065",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
How to prove an inequality using mathematical induction? I have to prove the following:
$$ \sqrt{x_1} + \sqrt{x_2} +...+\sqrt{x_n} \ge \sqrt{x_1 + x_2 + ... + x_n}$$
For every $n \ge 2$ and $x_1, x_2, ..., x_n \in \Bbb N$
Here's my attempt:
Consider $P(n): \sqrt{x_1} + \sqrt{x_2} +...+\sqrt{x_n} \ge \sqrt{x_1 + x_2 + ... + x_n}$
$$P(2): \sqrt{x_1} + \sqrt{x_2} \ge \sqrt{x_1 + x_2}$$
$$ x_1 + x_2 + 2\sqrt{x_1x_2} \ge x_1 + x_2$$
Which is true because $2\sqrt{x_1x_2} > 0$.
$$P(n + 1): \sqrt{x_1} + \sqrt{x_2} + ... + \sqrt{x_n} + \sqrt{x_{n+1}} \ge \sqrt{x_1 + x_2 +...+ x_n + x_{n+1}}$$
From the hypothesis we have:
$$\sqrt{x_1} + \sqrt{x_2} + ... + \sqrt{x_n} + \sqrt{x_{n+1}} \ge \sqrt{x_1 + x_2 + ... + x_n} + \sqrt{x_{n + 1}} \ge \sqrt{x_1 + x_2 +...+ x_n + x_{n+1}}$$
Squaring both sides of the right part:
$$ x_1 + x_2 + ... + x_n + x_{n + 1} + 2\sqrt{x_{n+1}(x_1 + x_2 +... + x_n)} \ge x_1 + x_2 +...+ x_n + x_{n+1} $$
Which is true, hence $P(n + 1)$ is true as well.
I'm not sure if I did it correctly?
|
Without induction. For $any$ non-negative reals $x_1,...,x_n$ let $x_j=(y_j)^2$ for each $j,$ with each $y_j\ge 0.$ The inequality is then $$y_1+...+y_n\ge \sqrt {(y_1)^2+...+(y_n)^2}.$$ Since both sides are non-negative reals, this is equivalent to $$(y_1+...+y_n)^2\ge (y_1)^2+...+(y_n)^2.$$ Expand the LHS of this and you see that each term $(y_j)^2$ appears on the LHS, and (if $n>1$) all the other terms are non-negative, while $(y_1)^2,...,(y_n)^2$ are the $only$ terms on the RHS, so the LHS is $\ge$ the RHS.
E.g. For $a,b,c\ge 0$ we have $(a+b)^2=a^2+b^2+2ab\ge a^2+b^2,$ and
$(a+b+c)^2=a^2+b^2+c^2+2ab+2bc+2ca\ge a^2+b^2+c^2.$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3208412",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
}
|
Compute Moore-Penrose Pseudoinverse of a 3x3 matrix
Given a 3x3 matrix A= \begin{pmatrix}
1 & 1 & 1\\
1 & 1 &1 \\
1 & 1 & 1
\end{pmatrix} Find Psuedoinverse of the above matrix
For the above matrix I am getting it as singular.
Also the rank of the above matrix is 1 so tried to calculate left and right inverse of A but am getting stuck at calculating them as they are coming out to be singular.
Does the psuedoinverse of the matrix exists or the procedure followed to find it is incorrect.
|
Note that, if $A$ is the given matrix, then $A^2 = 3A$, and $A^3 = 9A$. Also, $A$ is symmetric. From this information, I claim that $B = \frac{1}{9}A$ is the pseudoinverse of $A$. Note that
\begin{align*}
ABA &= \frac{1}{9}A^3 = A \\
BAB &= \frac{1}{9^2}A^3 = \frac{1}{9}A = B \\
AB &= \frac{1}{3}A \\
BA &= \frac{1}{3}A,
\end{align*}
where $\frac{1}{3} A$ is symmetric. Hence, $B$ is the pseudoinverse of $A$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3209171",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
}
|
Solving limit without L'Hôpital's rule: $\lim\limits_{x \to 0} \frac{\sqrt{1+\tan x}-\sqrt{1+\sin x}}{x^3}$ How can I solve this limit without L'Hôpital's rule?
$$\begin{align}\lim\limits_{x \to 0} \frac{\sqrt{1+\tan x}-\sqrt{1+\sin x}}{x^3}&=\lim\limits_{x \to 0} \frac{\tan x-\sin x}{x^3(\sqrt{1+\tan x}+\sqrt{1+\sin x})}\\&=\lim\limits_{x \to 0} \frac{\frac{\tan x-\sin x}{x^3}}{\sqrt{1+\tan x}+\sqrt{1+\sin x}}\end{align}$$
I can't proceed anymore from here.
|
We can work asymptotically also:
We denote the limit by $L$
As $x \to 0$
$$L=\frac{(\sqrt{1+\tan x}-1) -(\sqrt{1+\sin x}-1)}{x^3} \sim \frac{1}{2} \,\,\left[ \frac{\tan x (1-\cos x)}{x^3} \right] \sim \frac{1}{4}\,\,\frac{\tan x}{x} \sim \frac{1}{4}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3210646",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
}
|
How do I determine the limit of this sequence? I need a refresher.
I would assume that the answer would be $\infty$, but I am not quite sure.
The problem reads:
Determine the limit of this sequence? $c_n$ = $\sqrt {n^2+n} - \sqrt {n^2 - n}$ .
I would need a refresher on this. That's all thank you.
|
Multiply $\sqrt{n^2+n} - \sqrt {n^2 - n}$ by its conjugate (it looks exactly the same way except that it has a plus sign instead of a minus sign between the square roots):
$$\require{cancel}
\frac{\sqrt {n^2+n} - \sqrt {n^2 - n}}{1}\cdot\frac{\sqrt {n^2+n} + \sqrt {n^2 - n}}{\sqrt {n^2+n} + \sqrt {n^2 - n}}=\\
\frac{\left(\sqrt {n^2+n}\right)^2 - \left(\sqrt {n^2 - n}\right)^2}{\sqrt {n^2+n} + \sqrt {n^2 - n}}=\\
\frac{|n^2+n| - |n^2 -n|}{\sqrt {n^2+n} + \sqrt {n^2 - n}}=
\frac{\cancel{n^2}+n \cancel{- n^2} +n}{\sqrt {n^2+n} + \sqrt {n^2 - n}}=\\
\frac{2n}{|n|\sqrt {1+\frac{1}{n}} + |n|\sqrt {1 - \frac{1}{n}}}=
\frac{2\cancel{n}}{\cancel{n}\left(\sqrt {1+\frac{1}{n}} + \sqrt {1 - \frac{1}{n}}\right)}=\\
\frac{2}{\sqrt {1+\frac{1}{n}} + \sqrt {1 - \frac{1}{n}}}
\xrightarrow{n\rightarrow+\infty}
\frac{2}{\sqrt{1+0}+\sqrt{1-0}}=\frac{2}{1+1}=\frac{2}{2}=1.
$$
So, the limit is going to be $1$. Notice that when you pull any expression out of a square root sign, you actually need to surround it with absolute value bars like this: $|n^2+n|$, $|n|$, etc. And then you need to make sure whether you're allowed to just drop those absolute value bars. In our case here, we're totally safe. That's because the square roots of all the expressions containing the variable $n$ are always positive quantities.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3211606",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
}
|
Solve $4+\frac{1}{x}-\frac{1}{x^2}$ using quadratic formula I am to solve for x using the quadratic formula:
$$4+\frac{1}{x}-\frac{1}{x^2}=0$$
The solution provided in the answers section is: $\dfrac{-1\pm\sqrt{17}}{8}$ whereas I arrived at something entirely different: $$\dfrac{\frac{1}{x}\pm\sqrt{\frac{1}{x^2}+\frac{16}{x}}}{\frac{2}{x}}$$
Here's my working:
Start with $$4+\frac{1}{x}-\frac{1}{x^2}=0$$
Rearranging into standard form:
$$-\frac{1}{x^2}+\frac{1}{x}+4=0$$
Multiply by $-1$ to get a positive leading coefficient $a$:
$$\frac{1}{x^2}-\frac{1}{x}-4=0$$
I'm not sure how to determine my inputs $a,b$ and $c$ with these fractions but I guess $a=\dfrac{1}{x^2}$, $b=\dfrac{1}{x}$ and $c=-4$.
Plugging into quadratic function:
$$x = \frac{-\frac{1}{x}\pm\sqrt{\frac{1}{x^2}+\frac{16}{x}}}{\frac{2}{x}}$$
I find this challenging due to the coefficients $a$ and $b$ being fractions.
How can I apply the quadratic formula to $4+\dfrac{1}{x}-\dfrac{1}{x^2}=0$ to arrive at $\dfrac{-1\pm\sqrt{17}}{8}$?
|
Multiplying both sides by $x^2$ will result in
$$4x^2+x-1=0$$
Now, with $a=4, b=1, c=-1$ use the quadratic formula and let us know what you get.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3212108",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 6,
"answer_id": 1
}
|
An algebra problem: Let $a+b+c=0$. Prove that $a^2+b^2+c^2 =6/5$ Let $a,b,c$ be nonzero real numbers such that $a+b+c=0$ and $a^3+b^3+c^3 = a^5 +b^5 +c^5$. Prove that $a^2+b^2+c^2 =6/5$
I tried to expand $(a+b+c)^5$ but I can't get term of $a^2+b^2+c^2$.
|
Hint: With $$c=-a-b$$ we get in $$a^3+b^3+c^3-a^5-b^5-c^5=0$$ the equation
$$ab(a+b)(5a^2+5ab+5b^2-3)=0$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3212646",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
}
|
Remove the discontinuity of $f(x) = \frac{x^2-11x+28}{x-7}$ at $f(7)$? I need to remove the discontinuity of $f(x) = \frac{x^2-11x+28}{x-7}$ at $f(7)$, and find out what $f(7)$ equals. I am not sure what I've done wrong, but I'm getting 33, which the website I'm using tells me is wrong.
Please help me figure out what I'm doing wrong here.
My Steps:
$$\frac{x^2-11x+28}{x-7} \cdot \frac{x+7}{x+7}$$
$$=\frac{x^3+7x^2-11x^2-77x+28x+196}{x^2-49}$$
$$=\frac{x^3-4x^2-49x+196}{x^2-49}$$
$$=\frac{x^2(x-4)-49(x+4)}{x^2-49}$$
$$=(x-4)(x+4)$$
plug in $7$ for $x$.
$$(7-4)(7+4)=7^2-16=33$$
|
x=7 is a root of $x^2-11x+28$. So write $x^2-11x+28=(x-7)p(x)$. Then $f(7)=p(7)$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3213721",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
}
|
Prove that $\sum ^{n} _{k=1} \sin(kx) \le \frac{1}{|\sin(\frac{x}{2})|}$
Prove that $\sum ^{n} _{k=1} \sin(kx) \le \frac{1}{|\sin(\frac{x}{2})|}$
I am doing some task and I wanted to see solution which is in my book. However there is inequality $\sum ^{n} _{k=1} \sin(kx) \le \frac{1}{|\sin(\frac{x}{2})|}$ and I don't understand why it is true.Author of this inequality wanted to show that a sequence of partial sums series $ \sin(nx)$ is uniformly limited.Can you help me?
|
$\sum\limits_{k=1}^n\sin{kx}\sin{\frac{x}{2}}=\frac{1}{2}\sum\limits_{k=1}^n(\cos(kx-\frac{x}{2})-\cos(kx+\frac{x}{2}))=\frac{1}{2}(\cos(\frac{x}{2})-\cos(\frac{3x}{2})+\cos(\frac{3x}{2})-\cos(\frac{5x}{2})+...)=\frac{1}{2}(\cos{\frac{x}{2}}-\cos\frac{(2n+1)x}{2})$
So $|\sum\limits_{k=1}^n\sin{kx}|\leq \frac{\frac{1+1}{2}}{|\sin{\frac{x}{2}}|}= \frac{1}{|\sin{\frac{x}{2}}|}$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3216908",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
}
|
Proof that $ p^2 \equiv 1 \mod{30} \vee p^2 \equiv 19 \mod{30} $ for $p>5$ Proof that $$ p^2 \equiv 1 \mod{30} \vee p^2 \equiv 19 \mod{30} $$
for $p>5$ and $p$ is prime.
$\newcommand{\Mod}[1]{\ (\mathrm{mod}\ #1)}$
My try
Let show that
$$p^2 - 1 \equiv 0 \Mod{30} \vee p^2 - 1\equiv 18 \Mod{30}$$
Let check $$p^2 -1 = (p-1)(p+1) $$
We know that (for example from here) that this is dividable by $2$ and by $3$ so by $6$
Let consider $5$ cases:
$$\exists_k p=5k \rightarrow \mbox{false because p is prime}$$
$$\exists_k p=5k+1 \rightarrow p^2 - 1 = 5k(5k+2) \rightarrow \mbox{ dividable by 6 and by 5 so we have rest 0}$$
$$\exists_k p=5k+2 \rightarrow p^2 - 1 = (5k+1)(5k+3) \rightarrow \mbox{ ??? }$$
$$\exists_k p=5k+3 \rightarrow p^2 - 1 = (5k+2)(5k+4) \rightarrow \mbox{ ??? }$$
$$\exists_k p=5k+4 \rightarrow p^2 - 1 = (5k+3)(5k+5) \rightarrow \mbox{ dividable by 6 and by 5 so we have rest 0}$$
I have stucked with $???$ cases...
|
To extend your approach:
Let $p=5k+r$ with $r=1,2,-1,-2.$
As you note, when $r=1,-1$ you get $p^2\equiv 1\pmod{5}$ and hence $p^2\equiv 1\pmod{30},$ since you've already shown $p^2\equiv 1\pmod 6.$
In the other cases, you need to deduce additional properties about $k,$ because just $p=5k\pm 2$ doesn't let us deduce it alone.
If $r=\pm 2$ then $0\equiv p^2-1\equiv (r-1-k)(r+1-k)\pmod{6}.$
So when $r=2,$ we need $(1-k)(3-k)$ divisible by $6$ and hence $k$ is odd and $k\equiv 0,1\pmod{3},$ which means $k\equiv 1,3\pmod{6}.$
When $r=-2,$ we need $(1+k)(3+k)$ divisible by $6$, so you need $k$ odd and $k\equiv 0,2\pmod{3}$ which means $k\equiv 3,5\pmod{6}.$
These two cases can be written $k\equiv r\pm 1\pmod{6}$ for $r\in \{-2,2\}.$ Writing $k=6m+r\pm 1$ you get:
$$p=5k+r=30m+5r\pm 5 +r=30m+6r\pm 5,$$ and so:
$$p^2-1 \equiv (6r\pm 5)^2 =36r^2\pm 60r+25=6r^2+ 25=49\equiv 19\pmod{30}$$ sihce $r^2=4.$
An easier proof:
Start with $p^2\equiv 1\pmod 6$ if $\gcd(p,6)=1$ and $p^2\equiv \pm 1\pmod{5}$ if $\gcd(p,5)=1.$
Now,
*
*Show if $m\equiv 1\pmod{6}$ and $m\equiv 1\pmod{5}$ then $m\equiv 1\pmod{30}.$
*Show if $m\equiv 1\pmod{6}$ and $m\equiv -1\pmod{5}$ then $m\equiv 19\pmod{30}.$
The first follows since $6\mid m-1$ and $5\mid m-1$ and $\gcd(6,5)=1$ means $30=6\cdot 5\mid m-1.$
The second is is an application of Chinese remainder theorem.
You actually can get the stronger result, that $p^2\equiv 1\pmod{24},$ when $\gcd(p,6)=1,$ and hence:
If $p$ is prime and $p>5$ then $$p^2\equiv 1,49\pmod{120}.$$
This is not just true for primes $p>5,$ but for any integer $p$ with $\gcd(p,30)=1.$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3219882",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 3
}
|
Convergence of powers of matrices in Jordan Canonical Form (Jordan Normal Form) I've actually been stuck on this for a bit while studying for an exam, so would appreciate any help.
The problem involves testing whether $\lim\limits_{m \to \infty}$ $A^m$ exists.
From lecture notes, I know that the limit above exists for a matrix A when A has no eigenvalue $\lambda > \lvert 1\rvert$. I also know this limit exists if and only if the limit exists for the Jordan canonical form for the matrix. I was given the following theorem in class with respect to matrices in JCF, however I'm unable to apply it (perhaps poor understanding of Jordan canonical matrices):
"Let A be an nxn matrix. Then $\lim\limits_{m \to \infty}$ $A^m$ only exists if:
Matrix A has an eigenvalue $ \lvert \lambda\rvert = 1$, then $\lambda = 1$ and all the blocks in the Jordan normal form $J(A)$ of the form $J_k(1)$ must have $k = 1$.
I am unable to understand how to apply this to the following matrices to determine whether $\lim\limits_{m \to \infty}$ $A^m$ exists.
1) $J(A_1)$ = $$
\begin{pmatrix}
1/2 & 1 & 0 & 0 & 0 \\
0 & 1/2 & 1 & 0 & 0 \\
0 & 0 & 1/2 & 0 & 0 \\
0 & 0 & 0 & 1 & 0 \\
0 & 0 & 0 & 0 & 1 \\
\end{pmatrix}
$$
2) $J(A_2)$ = $$
\begin{pmatrix}
1/2 & 1 & 0 & 0 & 0 \\
0 & 1/2 & 1 & 0 & 0 \\
0 & 0 & 1/2 & 0 & 0 \\
0 & 0 & 0 & 1 & 1 \\
0 & 0 & 0 & 0 & 1
\end{pmatrix}
$$
3) $J(A_3)$: $$
\begin{pmatrix}
1 & 0 & 0 \\
0 & -1 & 0 \\
0 & 0 & 1 \\
\end{pmatrix}
$$
|
For the first matrix, the Jordan blocks are:
$$J(A_1)=
\begin{pmatrix}
\color{red}{1/2} & \color{red}1 & \color{red}0 & 0 & 0 \\
\color{red}0 & \color{red}{1/2} & \color{red}1 & 0 & 0 \\
\color{red}0 & \color{red}0 & \color{red}{1/2} & 0 & 0 \\
0 & 0 & 0 & \color{blue}1 & 0 \\
0 & 0 & 0 & 0 & \color{brown}1 \\
\end{pmatrix}.
$$
In particular, there is a $3 \times 3$ block with eigenvalue $1/2$ (convergent, since $|1/2| < 1$) and two $1 \times 1$ blocks with eigenvalue $1$ (convergent even though $|1| = 1$, since the blocks are only $1 \times 1$). So, the powers of $A_1$ are convergent.
For the second matrix, the Jordan blocks are:
$$J(A_2) =
\begin{pmatrix}
\color{red}{1/2} & \color{red}1 & \color{red}0 & 0 & 0 \\
\color{red}0 & \color{red}{1/2} & \color{red}1 & 0 & 0 \\
\color{red}0 & \color{red}0 & \color{red}{1/2} & 0 & 0 \\
0 & 0 & 0 & \color{blue}1 & \color{blue}1 \\
0 & 0 & 0 & \color{blue}0 & \color{blue}1
\end{pmatrix}.
$$
The $3 \times 3$ Jordan block is still convergent, but now the two $1 \times 1$ Jordan blocks have been replaced by a $2 \times 2$ Jordan block with eigenvalue $1$. Since there is a Jordan block corresponding to $1$ that is larger than $1 \times 1$, the powers of $A_2$ are divergent.
For the third matrix, the Jordan blocks are:
$$J(A_3) =
\begin{pmatrix}
\color{red}1 & 0 & 0 \\
0 & \color{blue}{-1} & 0 \\
0 & 0 & \color{brown}1 \\
\end{pmatrix}.
$$
In this case, there are three $1 \times 1$ Jordan blocks. The two blocks with eigenvalue $1$ are not a problem, but the one with eigenvalue $-1$ is an a problem (while $|-1| = 1$, we don't have $-1 = 1$). So, the powers of $A_3$ are not convergent (it's not hard to compute $J(A_3)^n$ and verify that it alternates, rather than converging).
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3220367",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
How do I evaluate the following integral using residue theory? This is how I've approached solving the integral so far$$\int_{0}^{2\pi}\frac{d\theta}{4-2\sqrt{3}\cos(\theta)}=\frac{1}{2}\int_{0}^{2\pi}\frac{d\theta}{2-\sqrt{3}\cos(\theta)}$$
First, I substituted $$\cos(\theta)=\bigg(z+\frac{1}{z}\bigg)$$ into the given integral to get
$$\frac{1}{2}\int_{0}^{2\pi}\frac{dz}{\bigg[2-\sqrt{3}\bigg(z + \dfrac{1}{z}\bigg)\bigg]iz}=\frac{1}{2}\int_{0}^{2\pi}\frac{dz}{2iz-\sqrt{3}iz^{2}-\sqrt{3}i}$$
Then I factor out $\sqrt{3}i$ in the denominator to get
$$\frac{1}{2\sqrt{3}i}\int_{0}^{2\pi}\frac{dz}{\dfrac{2z}{\sqrt{3}}-z^{2}-1}=\frac{1}{2\sqrt{3}i}\int_{0}^{2\pi}\frac{dz}{(z + i\sqrt{2/3} - 1/\sqrt{3})(z - i\sqrt{2/3} - 1/\sqrt{3})}$$
Now, using $$f(z)=\frac{dz}{(z + i\sqrt{2/3} - 1/\sqrt{3})(z - i\sqrt{2/3} - 1/\sqrt{3})}$$
I get the residues
Res$\bigg(f; - i\sqrt{2/3} + 1/\sqrt{3}\bigg)=\dfrac{i}{2}\sqrt{3/2}$
Res$\bigg(f; i\sqrt{2/3} + 1/\sqrt{3}\bigg)=-\dfrac{i}{2}\sqrt{3/2}$
This means that the integral is $0$. But in Mathematica, I get $\pi$ as the answer. Also, I tried substituting $\bigg(\dfrac{2\sqrt{3}}{z} + \dfrac{z}{2\sqrt{3}}\bigg) = \cos(\theta)$, but that didn't seem to work ( I got a negative number). So I don't know where I'm going wrong.
|
You need to make the substitution
$$\cos\theta=\frac12\left(z+\frac1z\right).$$
You then get the contour integral
$$\text{constant}\int_C\frac{dz}{z^2-(4/\sqrt3)z+1}$$
where $C$ is the unit circle. Now the function has one pole inside
and one outside $C$ and you'll get a non-zero answer.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3220532",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
}
|
Evaluate $\lim\limits_{x \to 0}\frac{e^{(1+x)^{\frac{1}{x}}}-(1+x)^{\frac{e}{x}}}{x^2}$ Solution
Expanding $(1+x)^{\frac{1}{x}}$ at $x=0$ by Taylor's Formula，we obtain
\begin{align*}
(1+x)^{\frac{1}{x}}&=\exp\left[\frac{\ln(1+x)}{x}\right]=\exp \left(\frac{x-\frac{x^2}{2}+\frac{x^3}{3}+\cdots }{x}\right)\\
&=\exp \left(1-\frac{x}{2}+\frac{x^2}{3}+\cdots \right)=e\cdot\exp \left(-\frac{x}{2}+\frac{x^2}{3}+\cdots \right) \\
&=e\left[1+\left(-\dfrac{x}{2}+\dfrac{x^2}{3}+\cdots \right)+\frac{1}{2!}\left(-\dfrac{x}{2}+\dfrac{x^2}{3}+\cdots \right)^2+\cdots\right]\\
&=e\left(1-\frac{x}{2}+\frac{11}{24}x^2+\cdots\right)\\
&=e-\frac{ex}{2}+\frac{11}{24}ex^2+\cdots
\end{align*}
Likewise, expanding $e^{(1+x)^{\frac{1}{x}}}$ at $x=0$, we obtain
\begin{align*}
e^{(1+x)^{\frac{1}{x}}}&=(e^e)^{1-\frac{x}{2}+\frac{11}{24}x^2-\cdots}=e^e\cdot (e^e)^{-\frac{x}{2}+\frac{11}{24}x^2+\cdots}\\
&=e^e\cdot\left[1+\left(-\frac{x}{2}+\frac{11}{24}x^2+\cdots\right)\ln e^e+\frac{1}{2!}\left(-\frac{x}{2}+\frac{11}{24}x^2+\cdots\right)^2\ln^2 e^e+\cdots\right]\\
&=e^e \cdot\left[1-\frac{ex}{2}+\frac{1}{24}(11e+3e^2)x^2+\cdots\right]
\end{align*}
Expanding $(1+x)^{\frac{e}{x}}$ at $x=0$, it follows that
\begin{align*}
(1+x)^{\frac{e}{x}}&=\exp\left[\frac{e\ln(1+x)}{x}\right]=\exp \left(e\cdot\frac{x-\frac{x^2}{2}+\frac{x^3}{3}+\cdots }{x}\right)\\
&=\exp \left(e-\frac{ex}{2}+\frac{ex^2}{3}+\cdots \right)=e^e\cdot\exp \left(-\frac{ex}{2}+\frac{ex^2}{3}+\cdots \right) \\
&=e^e\left[1+\left(-\frac{ex}{2}+\frac{ex^2}{3}+\cdots \right)+\frac{1}{2!}\left(-\frac{ex}{2}+\frac{ex^2}{3}+\cdots \right)^2+\cdots\right]\\
&=e^e\left[1-\frac{ex}{2}+\frac{1}{24}e(8+3e)x^2+\cdots\right]
\end{align*}
Therefore
\begin{align*}
&\lim_{x \to 0}\frac{e^{(1+x)^{\frac{1}{x}}}-(1+x)^{\frac{e}{x}}}{x^2}\\
=&\lim_{x \to 0}\frac{e^e\cdot\left[1-\dfrac{ex}{2}+\dfrac{1}{24}e(11+3e)x^2+\cdots\right]-e^e\cdot\left[1-\dfrac{ex}{2}+\dfrac{1}{24}e(8+3e)x^2+\cdots\right]}{x^2}\\
=&e^e\cdot\frac{1}{8}e\\
=&\frac{1}{8}e^{e+1}
\end{align*}
Please check. Is there any more simpler solution?
|
Another Solution
\begin{align*}
&\lim_{x \to 0+}\frac{e^{(1+x)^{\frac{1}{x}}}-(1+x)^{\frac{e}{x}}}{x^2}\\
=&\lim_{x \to 0+}\frac{e^{(1+x)^{\frac{1}{x}}}-e^{\frac{e}{x}\ln(1+x)}}{x^2}\\
=&\lim_{x \to 0+}\left[e^{\frac{e}{x}\ln(1+x)}\cdot\frac{e^{(1+x)^{\frac{1}{x}}-\frac{e}{x}\ln(1+x)}-1}{x^2}\right]\\
=&e^e\cdot \lim_{x \to 0+}\frac{e^{(1+x)^{\frac{1}{x}}-\frac{e}{x}\ln(1+x)}-1}{x^2}\\
=&e^e\cdot \lim_{x \to 0+}\frac{(1+x)^{\frac{1}{x}}-\frac{e}{x}\ln(1+x)}{x^2}\\
=&e^e\cdot \lim_{x \to 0+}\frac{(e-\frac{1}{2}ex+\frac{11}{24}ex^2+\cdots)-\frac{e}{x}(x-\frac{x^2}{2}+\frac{x^3}{3}+\cdots)}{x^2}\\
=&e^e\cdot \frac{1}{8}e\\
=&\frac{e^{e+1}}{8}.
\end{align*}
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3220990",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 5,
"answer_id": 2
}
|
The integral solution of $x^{2}-y^{3}=1 (x>1,y>1) $? I know it's a special case of catalan's conjecture.Wiki says its only solution is $(3,2)$.But I cannot work it out.Any help will be appreciated.
|
You can re-write the expression as $y^3=x^2-1=(x+1)(x-1)$. Now we can do some analysis. The only common factor that $x+1$ and $x-1$ can have is $2$.
*
*Case 1- They have no common factor. In that case, both $x+1$ and $x-1$ are cubes. This is impossible, as the smallest difference between cubes is $2^3-1^3=7$.
*Case 2- They have one common factor- namely $2$. Then for $(x-1)(x+1)$ to be a cube, $x-1=2a^3$ and $x+1=2^2b^3$, or $x-1=2^2a^3$ and $x+1=2b^3$. Here $a,b$ are positive numbers different from $2$. Hence, $|2a^3-2^2b^3|=2$ or $|2^2a^3-2b^3|=2$. We can now see that the only way that this is possible is that $a=b=1$.
Hence, $x=3, y=2$ is the only solution
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3223241",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
}
|
Compute polynomial $p(x)$ if $x^5=1,\, x\neq 1$ [reducing mod $\textit{simpler}$ multiples] The following question was asked on a high school test, where the students were given a few minutes per question, at most:
Given that,
$$P(x)=x^{104}+x^{93}+x^{82}+x^{71}+1$$
and,
$$Q(x)=x^4+x^3+x^2+x+1$$
what is the remainder of $P(x)$ divided by $Q(x)$?
The given answer was:
Let $Q(x)=0$. Multiplying both sides by $x-1$:
$$(x-1)(x^4+x^3+x^2+x+1)=0 \implies x^5 - 1=0 \implies x^5 = 1$$
Substituting $x^5=1$ in $P(x)$ gives $x^4+x^3+x^2+x+1$. Thus,
$$P(x)\equiv\mathbf0\pmod{Q(x)}$$
Obviously, a student is required to come up with a “trick” rather than doing brute force polynomial division. How is the student supposed to think of the suggested method? Is it obvious? How else could one approach the problem?
|
Let $a$ be zero of $x^4+x^3+x^2+x+1=0$. Obviously $a\ne 1$. Then $$a^4+a^3+a^2+a+1=0$$
so multiply this with $a-1$ we get $$a^5=1$$ (You can get this also from geometric series $$a^n+a^{n-1}+...+a^2+a+1 = {a^{n+1}-1\over a-1}$$ by putting $n=4$).
But then \begin{eqnarray} Q(a) &=& a^{100}\cdot a^4+a^{90}\cdot a^3+a^{80}\cdot a^2+a^{70}\cdot a+1\\ &=& a^4+a^3+a^2+a+1\\&=&0\end{eqnarray}
So each zero of $Q(x)$ is also a zero of $P(x)$ and since all 4 zeroes of $Q(x)$ are different we have $Q(x)\mid P(x)$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3224765",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "63",
"answer_count": 6,
"answer_id": 0
}
|
Pattern in Squared Numbers and their Digit Sum So this has been boggling my mind for some time now. On spring break I was really bored and started messing around with numbers when I noticed something. I was squaring each number (1-9) when I realized I could get the products down to a single digit using the digit sum.
1^2=1 | 1^3=1 | 1^4=1...
2^2=4 | 2^3=8 | 2^4=16=1+6=7...
3^2=9 | 3^3=27=2+7=9...
4^2=16=1+6=7...
5^2=25=2+5=7...
6^2=36=3+6=9...
7^2=49=4+9=13=1+3=4...
8^2=64=6+4=10=1...
9^2=81=8+1=9...
So on and so forth. I realized there was many patterns were emerging.
X |x^2|x^3|x^4|x^5|x^6|x^7
1 = 1 | 1 | 1 | 1 | 1 | 1 | 1
2 = 2 | 4 | 8 | 7 | 5 | 1 | 2 | 4 | 8 | 7 | 5
3 = 3 | 9 | 9 | 9 | 9 | 9 | 9 | 9 | 9 | 9 | 9
4 = 4 | 7 | 1 | 4 | 7 | 1 | 4 | 7 | 1 | 4 | 7
5 = 5 | 7 | 8 | 4 | 2 | 1 | 5 | 7 | 8 | 4 | 2
6 = 6 | 9 | 9 | 9 | 9 | 9 | 9 | 9 | 9 | 9 | 9
7 = 7 | 4 | 1 | 7 | 4 | 1 | 7 | 4 | 1 | 7 | 4
8 = 8 | 1 | 8 | 1 | 8 | 1 | 8 | 1 | 8 | 1 | 8
9 = 9 | 9 | 9 | 9 | 9 | 9 | 9 | 9 | 9 | 9 | 9
(The numbers follow these patterns)
without 3, 6, or 9
ebb and flow of numbers squared
|
When you take the repeated digit sum you are finding the remainder on division by $9$ except you are getting $9$ instead of $0$. This is the classic divisibility test for $9$. When you take the powers, you are effectively doing that $\bmod 9$. That shows why you get stuck with $3,6,9.$ Once you square it you have two factors of $3$ and every successive power is also divisible by $9$. $\bmod 9$ we have $8 \equiv -1$, so when you square $8$ you get $1$, then another multiply gets $8$ and so on.
The numbers coprime to $9$, of which there are $\phi(9)=6,$ form a multiplcative group. Each one will cycle with a period that divides into $6$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3228462",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
How can I order these numbers without a calculator?
*
*Classify the following numbers as rational or irrational. Then place them in order on a number line:
$$\pi^2, -\pi^3, 10, 31/13, \sqrt{13}, 2018/2019, -17, 41000$$
I know $\pi$ is irrational so $\pi^2$ and $-\pi^3$ are irrational. 10, 31/30, 2018/2019, -17, 41000 are all rational because then can be written in the form $\frac{a}{b}, a,b \in \mathbb{Z}, b\neq 0$
I'm not sure how I can look at $\sqrt{13}$ and determine if it's rational or irrational.
For ordering them I am approximating $\pi$ as 3, so
$\pi^2 \approx 9$ and
$-\pi^3 \approx -27$.
$31/13 \approx 30/10 = 3$
$\sqrt{9}<\sqrt{13}<\sqrt{16}$ so $3<\sqrt{13}<4$
$2018/2019 \approx 1$
If I were to order these, I would say:
$-\pi^3, -17, 2018/2019, 31/13, \sqrt{13}, \pi^2, 41000$
*Put these numbers in order from least to greatest (no calculator):
$$10^8, 5^{12}, 2^{24}$$
All I can think of is that $2^{10} \approx 10^3$ so
$2^{24} = (2^{10})^{2}* 2^{4} \approx (10^3)^2 *2^4 = 10^6 *16 <10^{8}$
So $2^{24} < 10^8$
Not sure how $5^{12}$ fits in.
|
For the second list we can see that $5 \gt 2^2$ as $5 \gt 4$ so if we take both sides to the power of $12$ we get $5^{12} \gt 2^{24}$. Then we can compare $10^8$ and $2^{24}$, we can split $10$ into $2*5$, so $10^8=2^8*5^8$. We claim $10^8 \gt 2^{24}$, which leads to $2^8*5^8 \gt 2^{24}$, then taking $2^8$ over, $5^8 \gt 2^{16}$. Then by our logic before $5 \gt 2^2$ we can prove the claim!
So the order is $5^{12} \gt 10^8 \gt 2^{24}$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3230359",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
}
|
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.