Q
stringlengths 70
13.7k
| A
stringlengths 28
13.2k
| meta
dict |
|---|---|---|
Can a triangle ABC be made if $\frac{\cos A}{2}=\frac{\cos B}{3}=\frac{\cos C}{7}$
I would like to know the simplest approach to find out whether a triangle ABC will be made if $$\frac{\cos A}{2}=\frac{\cos B}{3}=\frac{\cos C}{7}$$
The counterpart questions for sine and tangent can be handled as follows:
*
*If $\dfrac{\sin A}{2}=\dfrac{\sin B}{3}=\dfrac{\sin C}{7}$, we can rule out triangle because by the Sine Rule $a=2k$, $b=3k$, $c=7k \implies a+b <c.$
*If $\dfrac{\tan A}{2}=\dfrac{\tan B}{3}=\dfrac{\tan C}{7}$, we can see that a triangle will be made as $\tan A=2k, \tan B =3k,\tan C=7k$, when inserted in the identity $\tan A+ \tan B+ \tan C= \tan A \tan B \tan C \implies k=\sqrt{2/7}$.
|
Hint use may use the identity
$$\cos^2A+\cos^2B+\cos^2C+2\cos A\cos B\cos C=1$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3834491",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 2
}
|
Understanding convergence of $a_n := \frac{2n^3 +n^2 +3}{n^3-4}$ We proved by definition that the sequence $a_n := \frac{2n^3 +n^2 +3}{n^3-4}$ converges.
Let $\epsilon > 0$. Choose $N := \lceil \frac{24}{\epsilon} \rceil + 2 $.
Then for all $n \geq N$ it holds, that
$$|a_n - 2| = \big |\frac{2n^3+n^2+3}{n^3-4} - \frac{2n^3-8}{n^3-4} \big | = \big | \frac{n^2+11}{n^3-4} \big | = \frac{n^2+11}{n^3-4} \leq \frac{n^2+11}{n^3 - \frac{1}{2}n^3} \leq \frac{12n^2}{\frac{1}{2} n^3} = \frac{24}{n} < \epsilon $$
What I don't understand is how we get to $$- \frac{2n^3-8}{n^3-4} $$
Could someone explain? Thanks
|
I like to do these by dividing out the highest power of the variable. You get $\dfrac{2n^3+n^2+3}{n^3-4}=\dfrac{2+1/n+3/n^3}{1-4/n^3}\to2/1=2$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3834940",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
}
|
Induction proof of a known harmonic sum I want to prove that $$\frac{1}{2} + \frac{1}{4} + \dots + \frac{1}{2^n} \leq 1$$ only by induction!
I check for the first one, $\frac12 \leq 1 $ correct.
Then I assume for $n=k$ : $$\frac12 + \frac14 + \dots + \frac{1}{2^k} \leq 1$$
And Try and prove for $n=k+1$
$$\frac12 + \frac14 + \dots + \frac{1}{2^k} + \frac{1}{2^{k+1}} \leq 1$$
But I know that $$\frac12 + \frac14 + \dots + \frac{1}{2^k} \leq \frac12 + \frac14 + \dots + \frac{1}{2^k} + \frac{1}{2^{k+1}} \leq 1$$ and so:
But I am stuck, this tells me that the sum for $n=k+1$ is always $1$ , not $S \leq 1$ I am so confused, because I can't use the geometric series sum formula.. any help would be appreciated!
|
Hint : Prove the following stronger hypothesis induction :
$$\frac{1}{2} + \frac{1}{4} + \dots + \frac{1}{2^n} = 1 - \frac{1}{2^n}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3837387",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
}
|
Maximizing $\frac{y+1}{x+2}$ when $(x-3)^2 + (y-3)^2 = 6$ Suppose $x$ and $y$ are real numbers such that $(x-3)^2 + (y-3)^2 = 6.$ Than, maximize $\frac{y+1}{x+2}.$
I do in fact realize that this is a double post, but it's a 5 year old question and I don't feel as if it is appropriate to bump it. I did as the original post hinted towards, setting $k = \frac{y+1}{x+2}$ and than writing the given equation as $$(x-3)^2 + (k(x+2) - 4)^2 = 6.$$ I than proceeded to expand and simplify, which gave me $$x^2(k^2 + 1) + x(4k^2 - 8k - 6) + (4k^2 - 16k + 19).$$ However, I am unsure where to go from here. Should I use the discriminant now?
|
Going from what you have (and avoiding calculus):
$(k^2 + 1)x^2 + (4k^2 -4k-3)x + (k^2 - 16k + 19)= 0$
If we plug this into the quadratic formula:
$x = \frac {-b \pm \sqrt {b^2 - 4ac}}{2a}$
When $b^2 - 4ac < 0$ the formula breaks.
Setting $b^2 - 4ac = 0$ will give us the extreme values of $k.$
$(4k^2 - 8k - 6)^2 - 4(k^2+1)(4k^2 - 16k + 19) \ge 0\\
(16k^4 - 64k^3 + (-48+64)k^2 + 96k + 36) - (16k^4-64k^3 + 4(19+4)k^2 -64k + 76)\ge 0\\
-76k^2 + 160k - 40 \ge 0\\
19k^2 - 40k + 10 \le 0\\
$
And now use the quadratic formula again to find the range of $k.$
$\frac {20 - \sqrt {210}}{19}\le k \le \frac {20 + \sqrt {210}}{19}
$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3837653",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
}
|
Calculate the value of the following limit So I have given the limit:
$$\lim_{x\to0} \frac{2\sin\left ( e^{-\frac{x^2}{2}} -\cos x \right)}{(\arctan(\sinh(x^2))^2}$$
I have been struggling for hours with it. Since i got the undefined form when i put $x=0$ i tried out with L'Hopital method and I come to this point:
$$\lim_{x\to 0} \frac{2\cos(e^{-\frac{x^2}{2}}-\cos x)(e^{-\frac{x^2}{2}}(-x)+\sin x)}{\frac{4x \arctan(\sinh(x^2))\cosh(x^2)}{\sinh(x^2)+1}}$$
Still here when i substitute x with $0$ i still get $0$. I tried factorising the x, i also tried using the identities $cos(A-B)$ and so on, but nothing.
I think the answer that should come out is $\frac{1}{6}$
I would be very thankful for your help,
Annalisa
|
We have that
$$\frac{2\sin\left ( e^{\frac{-x^2}2}-\cos x \right )}{(\arctan(\sinh(x^2))^2}
=$$
$$=\left(\frac{\sinh(x^2)}{\arctan(\sinh(x^2)}\right)^2\cdot\left(\frac{x^2}{\sinh(x^2)}\right)^2\cdot \frac{\sin\left ( e^{\frac{-x^2}2}-\cos x \right )}{e^{\frac{-x^2}2}-\cos x }\cdot2\cdot\frac{e^{\frac{-x^2}2}-\cos x }{x^4} $$
and since by standard limits
$$\left(\frac{\sinh(x^2)}{\arctan(\sinh(x^2)}\right)^2\to 1, \quad\left(\frac{\sinh(x^2)}{x^2}\right)^2\to 1, \quad \frac{\sin\left ( e^{\frac{-x^2}2}-\cos x \right )}{e^{\frac{-x^2}2}-\cos x } \to 1$$
we reduce to
$$\lim_{x\to 0} \frac{2\sin\left ( e^{\frac{-x^2}2}-\cos x \right )}{(\arctan(\sinh(x^2))^2}
=2\cdot \lim_{x\to 0}\frac{e^{\frac{-x^2}2}-\cos x }{x^4} = 2 \cdot \frac1{12}= \frac16$$
which can be shown by Taylor's expansion or by l'Hospital as follows
$$\lim_{x\to 0}\frac{e^{\frac{-x^2}2}-\cos x }{x^4}\stackrel{H.R.}=\lim_{x\to 0}\frac{-xe^{\frac{-x^2}2}+\sin x }{4x^3}\stackrel{H.R.}=\lim_{x\to 0}\frac{x^2e^{\frac{-x^2}2}-e^{\frac{-x^2}2}+\cos x }{12x^2}=$$
$$=\lim_{x\to 0}\frac{e^{\frac{-x^2}2}}{12}+\lim_{x\to 0}\frac{1-e^{\frac{-x^2}2}+\cos x-1 }{12x^2}=\frac1{12}+0=\frac1{12}$$
indeed by standard limits
$$\frac{1-e^{\frac{-x^2}2}+\cos x-1 }{12x^2}=\frac{1}{24}\frac{e^{\frac{-x^2}2}-1}{-\frac{x^2}2}-\frac{1}{12}\frac{1-\cos x }{x^2}=0$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3838421",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
}
|
If $|ax^2 + bx + c| \le 1$ for all $x$ in $[0,1]$ then evaluate $|a|$, $|b|$ and $|c|$
Solve the following:
If $|ax^2 + bx + c| \le 1$ for all $x$ in $[0,1]$, then
i) $|a| \le 8$
ii) $|b| \ge 8$
iii) $|c| \le 1$
iv) $|a| + |b| + |c| \le 17$
Solution from my textbook:
Put $x = 0, 1 \text{ and } \frac{1} {2}$ to get:
$$|c| \le 1$$
$$|a + b + c| \le 1$$
$$|a + 2b + 4c| \le 4$$
From the above three equations, we get,
$|b| \le 8 \text{ and } |a| \le 8$.
Therefore, $|a| + |b| + |c| \le 17$
However, I don't understand: how, from the three equations, can you find the values of $|a|$ and $|b|$?
|
Using triangle inequality to simplify the other answer, we have for example:
$$|b+3c|=|a+2b+4c-(a+b+c)|\le|a+2b+4c|+|a+b+c|\le4+1=5$$
$$|b|=|b+3c-3c|\le |b+3c|+|3c|\le5+3=8$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3839626",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
}
|
Determining whether a subset is a subspace The problems are as follows:
Determine whether the following subsets of $\mathbb{R}^3$ are subspaces of $\mathbb{R}^3$.
*
*$A = \{(u^2, v^2, w^2) \,|\, u, v, w \in \mathbb{R} \}$,
*$B = \left\{(a, b, c) \,|\,
\begin{pmatrix}
a & b & c\\
1 & 2 & 0\\
0 & 1 & 2
\end{pmatrix} \text{is not invertible}\right\}$,
*$C = \left\{(x, y, z) \,|\, \begin{pmatrix}
1 & 2 & 3\\
4 & 5 & 6\\
7 & 8 & 9
\end{pmatrix}
\begin{pmatrix}
x\\
y\\
z
\end{pmatrix} =
\begin{pmatrix}
2x\\
2y\\
2z
\end{pmatrix}\right\}$.
Here's what I've tried so far:
*
*$A$ is a subspace of $\mathbb{R}^3$ as it contains the $0$ vector (?).
*The matrix is not invertible, meaning that the determinant is equal to $0$. With this in mind, computing the determinant of the matrix yields $4a - 2b + c = 0$. The original subset can thus be represented as $B = \left\{\left(\frac{2s - t}{4}, s, t\right) \,|\, s, t \in \mathbb{R}\right\}$; i.e. $B = \text{span}\left\{(\frac{1}{2}, 1, 0), (-\frac{1}{4}, 0, 1)\right\}$, a plane in $\mathbb{R}^3$.
*Solving for the linear system,
$$
\begin{aligned}
\begin{pmatrix}
1 & 2 & 3\\
4 & 5 & 6\\
7 & 8 & 9
\end{pmatrix}
\begin{pmatrix}
x\\
y\\
z
\end{pmatrix} &=
\begin{pmatrix}
2x\\
2y\\
2z
\end{pmatrix}\\
x
\begin{pmatrix}
1\\
4\\
7
\end{pmatrix}
+ y
\begin{pmatrix}
2\\
5\\
8
\end{pmatrix}
+ z
\begin{pmatrix}
3\\
6\\
9
\end{pmatrix}
&=
\begin{pmatrix}
2x\\
2y\\
2z
\end{pmatrix}\\
x
\begin{pmatrix}
-1\\
4\\
7
\end{pmatrix}
+ y
\begin{pmatrix}
2\\
3\\
8
\end{pmatrix}
+ z
\begin{pmatrix}
3\\
6\\
7
\end{pmatrix}
&=
\begin{pmatrix}
0\\
0\\
0
\end{pmatrix}
\end{aligned}$$
Converting to row echelon form gives the trivial solution $x = 0, y = 0, \text{ and } z = 0$; $C$ only contains the $0$ vector.
Is my reasoning correct? Also, quite a bit of the computations have been omitted in this post for brevity. Regardless of this, are my answers well justified?
Thank you.
|
You proof seems fine to me but for case $A$ we should observe that in general
$$(u_1^2, v_1^2, w_1^2) + (u_2^2, v_2^2, w_2^2)\neq ((u_1+u_2)^2, (v_1+v_2)^2, (w_1+w_2)^2)$$
and you can easily find by yourself an example where equality doesn't hold which is a necessary step to complete the proof.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3839873",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
}
|
Find the area bounded by the curve $x^4+y^4=x^2+y^2$ I am stuck with this problem which deals with evaluating an Area
The problem reads :
Find the area bounded by the curve $x^4+y^4=x^2+y^2$.
I tried factorizing the expression and expressing $y$ in terms of $x$, not able to proceed with that idea. Someone please help me out.
|
It is possible to do this without using polar coordinates- but not easy. I would start by noting that only even powers of x and y are involved so the graph of this is symmetric about the x and y axes. That means that we can calculate the area for x and y positive and multiply by 4.
There is a little problem in that y is not a function of x even in the first quadrant. Differentiating with respect to y gives $4x^3x'+ 4y^3= 2xx'+ 2y$, $(4x^3- 2x)x'= 2y- 4y^3$. Setting x'= 0, for a maximum or minimum, $2y- 4y^3= 2y(1- 2y^2)= 0$. That has three roots, y= 0, $y= \frac{\sqrt{2}}{2}$, and $y= -\frac{\sqrt{2}}{2}$. y= 0 gives $x^4= x^2$ in the original equation which gives $x^2= 1$, $x= \pm 1$.
$y= \frac{\sqrt{2}}{2}$ gives $x^4- x^2= \frac{1}{4}- \frac{1}{2}= -\frac{1}{4}$. Letting $z= x^2$, $z^2- z+ \frac{1}{4}= (z- \frac{1}{2})^2= 0$ so $z^2= x^2= \frac{1}{2}$. Since we are in the first quadrant, $x= \frac{\sqrt{2}}{2}$.
For x between 0 and $\frac{\sqrt{2}}{2}$ we can solve y^4- y^2+ (x^4- x^2)= 0 using the quadratic formula: $y= \frac{1+ \sqrt{1- 4(x^4- x^2)}}{2}$.
$\frac{1}{2}\int_0^\frac{\sqrt{2}}{2} 1+ \sqrt{1- 4(x^4-x^2)}dx$. But then we have to integrate from x= $\frac{\sqrt{2}}{2}$ to 1: $\int_\frac{\sqrt{2}}{2}^1 \sqrt{1- 4(x^4- x^2)}dx$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3840121",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
}
|
Prove that the diophantine equation $(xz+1)(yz+1)=az^{k}+1$ has infinitely many solutions in positive integers. Given two positive integers $a$ and $k>3$ : From experimental data, it appears the diophantine equation
$(xz+1)(yz+1)=az^{k}+1$
has infinitely many solutions in positive integers $x,y, z$.
To motivate the question, it can easily be shown that if $k <3$, the given diophantine equation has no solutions in positive integers $x, y ,z$ with $z>a$.
Proof: $(xz+1)(yz+1)=az^{k}+1$ may be simplified to $xyz^{2}+(x+y)z=az^{k}$. If $k=1$, this reduces to $xyz+x+y=a$. Its clear that $a>z$ therefore there are no positive integral solutions in $x$ and $y$ when $z>a$. if $k=2$, we have the reduced equation $xyz+x+y=az$. We have $z$ | $x+y$, $z \le(x+y) \le xy$. Therefore $LHS=xyz+x+y>z^{2}$. Because $RHS=az$, we must have $a>z$ thus there are no solutions in positive integers $x ,y$ when $z>a$.
I would like to prove that given two positive integers $a$ and $k>3 $, the diophantine equation $(xz+1)(yz+1)=az^{k}+1$ has infinitely many positive integer solutions $x, y, z$. I do not know how to start the proof.
|
Getting there. Here is $k=4.$ a family of solutions to
$$ a z^4 + 1 = (xz+1)(yz+1) $$
is parametrized by integer $t$ with
$$ y=at $$
$$ x = a^4 t^5 - at $$
$$ z = a^2 t^3 $$
Both sides of the equation are
$$ a^9 t^{12} + 1 $$
=======================================
For that matter, we can take care of all $k \neq 0 \pmod 3$ this way.
When $k > 3$ and $k \equiv 1 \pmod 3,$ we may take
$$ y = a^{\frac{2k-5}{3}} \; t^{k-3} $$
$$ z = a^2 t^3 $$
followed by
$$ x = y \left( y^2 z^2 - 1 \right) $$
When $k > 3$ and $k \equiv 2 \pmod 3,$ we may take
$$ y = a^{\frac{k+1}{3}} \; t^{k-3} $$
$$ z = a t^3 $$
followed by
$$ x = y \left( y^2 z^2 - 1 \right) $$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3840780",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 2,
"answer_id": 0
}
|
Closed Form Solution for a Recurrence Relation We start out with: $a_n = 3a_{n-1} - 7, a_0 = 2$. Is the following valid?
$$a_{n-1} = 3(3a_{n-2}-7)-7 \\ a_{n-1} = 3^2a_{n-2}-(3\cdot-7) - 7 \\ \vdots \\ a_k = 3^ka_{n-k} -(3^{k-1}\cdot-7) \cdots -7\\ \text{Let $k = n$} \\ a_n = 3^ka_0 - (3^{n-1}\cdot-7)-\cdots-7$$
I don't know where to go from here. I would I make a closed form out of this?
|
As Greg Martin has already pointed out, your closed form cannot be right, because it doesn’t satisfy the recurrence. You could also see whether it generates the right values for $a_1,a_2$, and $a_3$, say, and find that it doesn’t.
There is a slightly better way to organize this sort of ‘unwinding’ of a simple recurrence, but make sure that you do the algebra correctly:
$$\begin{align*}
a_n&=3a_{n-1}-7\\
&=3(3a_{n-2}-7)-7\\
&=3^2a_{n-2}-3\cdot7-7\\
&=3^2(3a_{n-3}-7)-3\cdot7-7\\
&=3^3a_{n-3}-3^2\cdot7-3\cdot7-7\\
&\;\;\vdots\\
&=3^ka_{n-k}-7\sum_{i=0}^{k-1}3^i\\
&\;\;\vdots\\
&=3^na_0-7\sum_{i=0}^{n-1}3^i\\
&=2\cdot3^n-7\cdot\frac{3^n-1}{3-1}\\
&=2\cdot3^n-\frac{7\cdot3^n-7}2\\
&=\frac12(7-3\cdot3^n)\\
&=\frac12(7-3^{n+1})
\end{align*}$$
Note that it’s generally best not to do too much simplification at each stage: too much simplification tends to obscure the pattern. And when you’re done, you should always check to be sure that your closed form satisfies the recurrence:
$$3\left(\frac12(7-3^n)\right)-7=\frac32\cdot7-\frac12\cdot3^{n+1}-7=\frac12(3^{n+1}-7)\,.$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3841476",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
}
|
$\frac{a^3}{b+c}+\frac{b^3}{a+c}+\frac{c^3}{a+b}\geq \frac{3}{2}$ for $a,b,c\in\mathbb{R}^+$ with $abc=1$ Suppose that $a,b,c$ are positive reals such that $abc=1$. Prove that $$\frac{a^3}{b+c}+\frac{b^3}{a+c}+\frac{c^3}{a+b}\geq \frac{3}{2}.$$
Hint: Use Titu's lemma.
My approach: I am trying to use Titu's lemma directly but it is not working. I meant that I wrote each term in the following way: $$\frac{a^3}{b+c}=\frac{a^3bc}{bc(b+c)}=\frac{a^2}{bc(b+c)}.$$ Then I applied Titu's lemma to the sum $$\frac{a^2}{bc(b+c)}+\frac{b^2}{ac(a+c)}+\frac{c^2}{ab(a+b)}\geq \frac{(a+b+c)^2}{bc(b+c)+ac(a+c)+ab(a+b)}=\frac{\sigma_1^2}{\sigma_1\sigma_2-3},$$ where $\sigma_1=a+b+c$ and $\sigma_2=ab+ac+bc$ are elementary symmetric functions.
This is what I got so far.
Would thankful if someone shows correct solution.
|
Also, by Holder and AM-GM $$\sum_{cyc}\frac{a^3}{b+c}\geq\frac{(a+b+c)^3}{3\sum\limits_{cyc}(b+c)}=\frac{1}{6}(a+b+c)^2\geq\frac{9}{6}=\frac{3}{2}.$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3842654",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
}
|
Find coefficient of $x^3y^4z^5$ in polynomial $(x + y + z)^8(x + y + z + 1)^8$
Find coefficient of $x^3y^4z^5$ in polynomial $(x + y + z)^8(x + y + z + 1)^8$
It is pretty easy to see that our goal is to choose from each multiplier in this polynomial $x,y,z,1$ in certain amounts. Since sum of powers equals $12$ and we have $16$ multipliers, we have to choose four $1$'s. There are $\displaystyle {8 \choose 4}$ ways to choose $1$ as part of our product (actually, four $1$'s). But how do we handle what has left? It seems like in such solution we have to consider many cases.
|
Objective: To find coefficient of $\ x^3y^4z^5\ $ in $\ (x + y + z)^8(x + y + z + 1)^8$
First expand $$(1+(x+y+z))^8=\sum_{r=0}^{8} {8 \choose r}(x+y+z)^r$$
Now multiplying $(1+(x+y+z))^8$ with $(x+y+z)^8$ gives us,
$$(x+y+z)^8(1+(x+y+z))^8=\sum_{r=0}^{8} {8 \choose r}(x+y+z)^{r+8} \label{1} \tag{1}$$
Now using multinomial theorem, coefficient of $x^3y^4z^5$ in $(x+y+z)^n$ is $\Large{}{n \choose {3,\ 4,\ 5}}$ only when $n=12$.
Applying the same logic on equation $(1)$, we get the coefficient as
$$\sum_{r=0}^{8} {8 \choose r}* {8+r \choose {3,\ 4,\ 5}}$$ so we can fix $r=4$ and the answer we get is $${8\choose4}*{12\choose{3,4,5}}$$
For computationally rigorous answer see Wolfram Alpha coefficient finder
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3844565",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
}
|
How to construct a solution to a homogeneous system of linear equations? given the system of linear equations
$\begin{array}{rrrrr}x_{1} & + & x_{2} & + & 2 x_{3} & = & 2 \\ 2 x_{1} & + & 5 x_{2} & - & x_{3} & = & 2 \\ 3 x_{1} & + & 6 x_{2} & + & x_{3} & = & 4\end{array}$
I found the solutions
$x=\left(\begin{array}{c}\frac{8}{3}-\frac{11}{3} t \\ -\frac{2}{3}+\frac{5}{3} t \\ t\end{array}\right)$ ,
$y=\left(\begin{array}{ccc}\frac{6}{5} -\frac{11s}{5} \\ s\\ \frac{2}{5}+\frac{3s}{5}\end{array}\right)$
Now I have to construct a solution to the following homogenous system out of x and y:
$\begin{array}{rrrrr}x_{1} & + & x_{2} & + & 2 x_{3} & = & 0 \\ 2 x_{1} & + & 5 x_{2} & - & x_{3} & = & 0 \\ 3 x_{1} & + & 6 x_{2} & + & x_{3} & = & 0\end{array}$
How can this be done?
|
Since both $x$ and $y$ are solutions of the original system, $x-y$ is a solution of the homogenous system.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3847342",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
How to solve $\sqrt{2} + \sqrt{3} - \frac{m}{n} > \frac{1}{cn^4}, \forall n \ge 1$? If $\frac{m}{n}$ is an irreducible fraction such that $1 > \sqrt{2} + \sqrt{3} - \frac{m}{n} > 0$, then there exists a constant $c > 0$ so that $$\sqrt{2} + \sqrt{3} - \frac{m}{n} > \frac{1}{cn^4}, \forall n \ge 1$$
From the first inequation we can show that $4n > m \ $ and $\ m > 2n$.
I was also intuitively thinking that as $n$ gets bigger and bigger $m$ can approximate $(\sqrt{2} +\sqrt{3})n$ better and better so then $\sqrt{2} + \sqrt{3} - \frac{m}{n}$ tends to $0$.
|
It is an old result of Liouville :https://en.m.wikipedia.org/wiki/Diophantine_approximation
It can be applied because your number $\sqrt{2}+\sqrt{3}$ is algebraic of degree $4$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3847915",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
Remainder of dividing polynomial of the $n$th degree
What is the remainder when dividing the polynomial
$$P(x)=x^n+x^{n-1}+\cdots+x+1$$ with the polynomial
$$x^3-x$$ if $n$ is a natural odd number?
So, what I know so far is:
$$P(x)=Q(x)D(x)+R(x)$$
In this case I'll call $Q(x) = x^3-x$
$$Q(x) = 0 \iff x=\pm1$$
So from here:
$\begin{array} {l}P(1):\qquad&1^n+1^{n-1}+\cdots+1^1+1=R(1)\\P(-1):&(-1)^n+(-1)^{n-1}+\cdots+(-1)^1+1=R(-1)\end{array}$
Here is where I assumed that $R(x)=ax^2+b$ since $Q(x)=x^3-x$
And from the equations (assuming $n = \{2k+1 \mid k\in\mathbb{N}\}$):
$\begin{array}{l}P(1):\\&R(1)=n\\P(-1):\\&R(-1)=0\end{array}$
And here I get to:
$$\left\{
\begin{array}{ll}
ax^2+by &=n \\
ax^2+by &=0 \\
\end{array}
\right.$$
I would like to know where did I go wrong? Is my assumption for $R(x)$ incorrect, my calculation of the $P(1)$ and $P(-1)$ wrong or is there something else I didn't think about?
|
Your approach is fine but you've missed the root $x=0$ of $x^3-x$.
We get $R(0)=P(0)=1$, $R(1)=P(1)=n+1$, $R(-1)=P(-1)=0$.
The quadratic polynomial that interpolates this data is $\frac{n-1}{2} x^2+\frac{n+1}{2}x+1$.
The same approach works when $n$ is even (the interpolation data is different because $P(-1)=1$) and we get $\frac{n}{2} x^2+\frac{n}{2}x+1$.
In both cases, we get the nice recurrence: $R_{n+2}=R_n+x^2+x$, with $R_0=1$ and $R_1=x+1$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3849311",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
}
|
Triples satisfying three equations with six solutions What are all triples $(x, y, z)$ such that all three of the following equations are satisfied:$$x(y + z - 5) = 7,$$$$y(x + z - 5) = 7,$$$$x^2 + y^2 = 50.$$According to Wolfram Alpha we know that $(-7,-1,5)$, $(-5,-5,43/5)$, $(-1,-7,5)$, $(1,7,5)$, $(5,5,7/5)$, $(7,1,5)$ are the only solutions. But how do we know that there are no more solutions?
|
By subtracting the first two equations, we have $$(z-5)(x-y)=0$$
Let's consider two cases:
*
*If $z=5$, then we have $xy=7$ and $x^2+y^2=50$.
$$x^2+\frac{49}{x^2}=50$$
$$x^4-50x^2+49=0$$
$$(x^2-49)(x^2-1)=0$$
Solving this would give us $4$ solutions, corresponding to when $x=\pm 7, \pm 1$. Given $z$ and $x$, $y$ is determined.
*
*If $x=y$, then $2x^2=50$, $x=\pm 5$, this gives us another $2$ solutions. Knowing $x$ and $y$, $z$ is determined.
Hence in total, we have $6$ solutions.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3850732",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
a recursion in roots of the polynomial let $P(x)$ is a polynomial which satisfies property $\psi$ where property $\psi$ is given by
whenever r is a root of $P(x) = 0$ then $r^2 - 4$ is also a root of the given equation.
i) if $P(x)$ is a quadratic polynomial of the form $x^2 + ax + b$ then find all the possible equations which satisfy $\psi$ and has distinct real roots.
ii) if $P(x)$ is a cubic polynomial of the form $x^3 + ax^2+bx+c$ then find all the cubic equations which do not share any root with the equations calculated in part (i)
My attempt
for part (i)
let $r$ is a root of $P(x)$ then $r^2 - 4$ should also be a root of the equation
and as $r^2 -4$ is a root then is $(r^2-4)^2-4$ so atleast two of them should be equal otherwise the $P(x)$ would be the zero polynomial which is a contradiction so
either $$ r = r^2 - 4 \ or \ r = r = (r^2-4)^2-4 $$ as the third equation leads to complex roots
and again the first is false as the roots should be distinct so $r$ satisfies $(r^2-4)^2-4 = r$
but I am not able to solve further and also not sure that what i have done is correct
I just need the idea to solve the cubic part
by calculator i found four values of r for first part but in the question it was mentioned that there are only 2 polynomials
|
Part i):
Let $r, s$ be the two roots of $P$.
By assumption, $r^2 - 4$ and $s^2 - 4$ all belong to the set $\{r, s\}$. Thus there are several possibilities:
*
*$r^2 - 4 = r$, $s^2 - 4 = s$.
In this case, $r$ and $s$ are the two roots of $x^2 - x - 4$, and we have $P(x) = x^2 - x - 4$.
*$r^2 - 4 = r$, $s^2 - 4 = r$.
In this case, $r$ is one of the two roots of $x^2 - x - 4$, which are $\frac {1 \pm \sqrt{17}}2$, and since $s^2 = 4 + r = r^2$, we have $s = - r$.
Therefore $P(x) = (x - r)(x + r) = x^2 - r^2 = x^2 - \frac{9 \pm \sqrt{17}}2$.
*$r^2 - 4 = s$, $s^2 - 4 = r$.
In this case, we have $(r^2 - 4)^2 - 4 = r$, which gives $(r^2 - r - 4)(r^2 + r - 3) = 0$.
Since $r\neq s$, we have $r^2 - r - 4 \neq 0$, so $r$ is a root of the polynomial $x^2 + x - 3$.
The same argument shows that $s$ is also a root of $x^2 + x - 3$. Therefore $P(x) = x^2 + x - 3$.
Part ii):
We now have three roots $r, s, t$ (which are a priori not necessarily real numbers).
The number $r^2 - 4$ cannot be equal to $r$, so it is one of $s$ and $t$, let's say $s$.
And $s^2 - 4$ cannot be equal to neither $s$ nor $r$ (otherwise $s$ satisfies $(s^2 - 4)^2 - 4 = s$ and coincides with one of the previous roots), so it must be $t$.
Finally, $t^2 - 4$ cannot be equal to neither $t$ nor $s$, for the same reason as above, so it must be $r$.
Therefore we have $((r^2 - 4)^2 - 4)^2 - 4 = r$. After factorization, we get: $$(r^2 - r - 4) (r^3 - r^2 - 6r + 7) (r^3 + 2r^2 - 3r - 5) = 0.$$ The first factor contains only previous roots, so we see that $r^3 - r^2 - 6r + 7 = 0$ or $r^3 + 2r^2 - 3r - 5 = 0$.
If $r^3 - r^2 - 6r + 7 = 0$, then $s = r^2 - 4$ also satisfies $s^3 - s^2 - 6s + 7 = 0$, as can be verifies as follows: $$(r^2 - 4)^3 - (r^2 - 4)^2 - 6(r^2 - 4) + 7 = (r^3 - r^2 - 6r + 7) (r^3 + r^2 - 6r - 7) = 0.$$
For the same reason, $t$ also satisfies $t^3 - t^2 - 6t + 7 = 0$. Therefore $r, s, t$ are exactly the three different roots of the polynomial $x^3 - x^2 - 6x + 7$, so that $P(x) = x^3 - x^2 - 6x + 7$.
Similarly, if $r^3 + 2r^2 - 3r - 5 = 0$, then we conclude that $P(x) = x^3 + 2x^2 - 3x - 5$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3851144",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
prove $\sum_{cyc}\frac{a^2}{a+2b^2}\ge 1$
prove $$\sum_{cyc}\frac{a^2}{a+2b^2}\ge 1$$ holds for all positives $a,b,c$ when $\sqrt{a}+\sqrt{b}+\sqrt{c}=3$ or $ab+bc+ca=3$
Background Taking $\sqrt{a}+\sqrt{b}+\sqrt{c}=3$ This was left as an exercise to the reader in the book 'Secrets in inequalities'.This comes under the section Cauchy Reverse technique'.i.e the sum is rewritten as :$$\sum_{cyc} a- \frac{ab^2}{a+2b^2}\ge a-\frac{2}{3}\sum_{cyc}{(ab)}^{2/3}$$ which is true by AM-GM.($a+b^2+b^2\ge 3{(a. b^4)}^{1/3}$)
By QM-AM inequality $$\sum_{cyc}a\ge \frac{{ \left(\sum \sqrt{a} \right)}^2}{3}=3$$.
we are left to prove that $$\sum_{cyc}{(ab)}^{2/3}\le 3$$ .But i am not able to prove this .Even the case when $ab+bc+ca=3$ seems difficult to me.
Please note I am looking for a solution using this cuchy reverse technique and AM-GM only.
|
For $ab+bc+ca=3.$ We need to prove
$$ a+b+c \geqslant 1 + \frac{2}{3}\sum (ab)^{2/3}.$$
From condition we get
$$a+b+c \geqslant \sqrt{3(ab+bc+ca)}=3.$$
Now, using the AM-GM inequality we have
$$\sum (ab)^{2/3} = \sum \sqrt[3]{1 \cdot ab \cdot ab} \leqslant \sum \frac{1+2ab}{3} = 1+\frac{2}{3}(ab+bc+ca)=3.$$
So
$$1 + \frac{2}{3}\sum (ab)^{2/3} \leqslant 1 + \frac{2}{3} \cdot 3=3 \leqslant a+b+c.$$
Done.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3853509",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
}
|
Multiple proofs of $\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\geq\frac{3}2$ Here is my question:
Let $a,b,c\in\mathbb{R^+}$, prove that $$\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\geq\frac{3}2$$
Here is my solution:
From C-S inequality, we get $$\sum_{cyc}(a+b)\sum_{cyc}\frac{1}{a+b}\geq3^2,$$
which is equivalent to $$2(a+b+c) \sum_{cyc}\frac{1}{a+b}\geq9$$
or $$\frac{a+b+c}{a+b}+ \frac{a+b+c}{b+c}+ \frac{a+b+c}{c+a} \geq\frac{9}2$$
or $$\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\geq\frac{9}2-3=\frac{3}2$$
QED.
Here is my problem:
My teacher said that this question could be done in a lot of ways, one of which is by AM-GM inequality. However, after observing the original inequality, I still couldn’t get a clue.
I am stuck since I completely don’t know where to apply AM-GM. I am thinking about substitute something in the original inequality with some new positive numbers and continue. Is my thought in the right direction?
In addition, other approaches without AM-GM will be appreciated as well. Thanks for help.
P.S.
I am new to Mathematics SE, so if my post can be improved, make sure to let me know through the comment section. Thanks a lot.
|
One answer using AM-GM.
From AM-GM, we obtain two inequalities:
*
*$a^{\frac{3}2}+b^{\frac{3}2}+b^{\frac{3}2}\geq3a^\frac{1}2b$
*$a^{\frac{3}2}+c^{\frac{3}2}+c^{\frac{3}2}\geq3a^\frac{1}2c$
Add them up and we get:
$$2(a^{\frac{3}2}+b^{\frac{3}2}+c^{\frac{3}2})\geq3a^\frac{1}2(b+c)\iff\frac{a}{b+c}\geq\frac{a^\frac{3}2}{a^{\frac{3}2}+b^{\frac{3}2}+c^{\frac{3}2}}$$
Therefore
$$\sum_{cyc}\frac{a}{b+c}\geq\frac{3}2\sum_{cyc}\frac{a^\frac{3}2}{a^{\frac{3}2}+b^{\frac{3}2}+c^{\frac{3}2}}=\frac{3}2$$QED.
Note: This inequality is called Nesbitt's inequality, and in this link you can see a variety of approaches (as already commented by Dr.Mathva). This approach isn't included in it, so I think you'll probably like this as well.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3860625",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
}
|
Why is -8 $\equiv$ 6 mod 7? I read in a book that $-8 \equiv 6 \bmod 7$ which means that $-8$ and $6$ leave the same remainder when divided by $7.$
The remainder when $-8$ is divided by $7$ is $-1.$ But when $6$ is divided by $7,$ isn't the remainder $6$?
I recognise that we can write $7\cdot1 - 1=6$, so from here it seems that the remainder is $-1.$ Why is the above reasoning (remainder${}= 6$) incorrect?
|
The definition is $a\equiv b\mod n$ iff $n$ divides $a-b$.
Here $-8\equiv 6\mod 7$ since $7$ divides $-8-6 = -14$.
Another definition is that $a\equiv b\mod n$ iff both leave the same remainder modulo $n$. But note that the remainder is between $0$ and $n-1$.
Here $-8$ and $6$ have the same remainder $6$ modulo $7$, since $-8 = (-2)\cdot 7 + 6$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3864264",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 0
}
|
Remainder Theorem: $f(x) = x^3+1$ divided by $\operatorname {d}(x)= x$ Take for example $\operatorname {f}(x) = x^3+1$ divided by $\operatorname {d}(x)= x-1$ $$\frac {x^3+1}{x-1}$$
The Remainder Theorem tells us that the remainder will be $$\operatorname {f}(zero-of-\operatorname {d}(x))$$
In this case the remainder should be $$\operatorname {f}(1) = 1^3+1 = 2$$
so the remainder should be 2 for any value of $x$
But for $x=3$
$$f(3) = 3^3+1$$
when this is divided by $x-1$ which is $3-1$ which equals to $2$, $$\frac {3^3+1}{3-1}$$ $$= \frac {27+1}{2}$$ $$= \frac {28}{2}$$ $$= 14, remainder = 0 ≠ 2$$
Please explain me why this happened
|
$$\frac{x^3+1}{x-1}=x^2+x+1+\frac{2}{x-1}$$
Here $f(x)=x^3+1$, so the remainder when $f(x)$ is divided by the linear factor $x-1$ is a constant 2 as seen above and which is nothing but $f(1)=2$.
If the divisor is quadratic, the remainder is linear function of $x$ or a constant.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3864990",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
}
|
$n$ is odd if and only if there exists an $a \in \mathbb{Z}$ such that $n^4=16a+1$ Write a formal proof.
Suppose that $n \in ℕ$. Prove $n$ is odd if and only if there exists $a \in \mathbb{Z}$ such that $n^4 = 16a + 1$.
There exists $k\in \mathbb{Z}$ such that $n=2k+1$. So I've used this formula for odd numbers
$$n^4=(2k+1)^4 = 16k^4+32k^3+24k^2+8k+1$$
since our goal is to match this to the above $n^4$,
$$n^4=8(2k^4+4k^3+3k^2+k)+1$$
unfortunately this form does not match $16a+1$
This is where I got stuck and could use help on the proof.
A random solution I thought of is below, however I'm looking for a better answer.
Let $a=2k$ ; I randomly though of a substitution.
$$n^4=8(2*16a^4+16*8a^3+3*4a^2+2a)+1$$
$$n^4=16(16a^4+64a^3+6a^2+a)+1$$
this has the form $16a+1$ with '$(16a^4+64a^3+6a^2+a)$ = an integer'.
Note
Thanks everyone for the help the proof totally makes sense now!
Special shoutout to @lulu and @fleablood
The part that solidified it for me was the breakdown of the 2 cases
*(3+1)2, k being even or odd!
|
If $n$ is odd then there exists an integer $k$ so that $n$ can be written as $n=2k +1$. And with respect to that $k, n^4 = (2k+1)^4 = 16k^4 + 4*8k^3 + 6*4k^2 + 4k + 1=16k^4 + 32k^2 + 24k^2 + 4k +1$.
And $n^4= 16(k^4 + 2k^3 + \frac {3k^2 + k}2) + 1= 16(k^4 + 2k^3 + \frac {k(3k+1)}2) + 1$.
So if we can prove $\frac {k(3k+1)}2$ is always an integer we are done. If $k$ is even than $\frac k2$ is an integer. and $\frac k2(3k+1)$ is an integer. If $k$ is odd then $3k+1$ is even so $\frac {3k+1}2$ is an integer.
........
In hindsight we could have realized that if $n$ is odd then $n$ can be written as $4k \pm 1$ for some $k$ and some choose of $+/-$.
Then $n^4 = (4k \pm 1)^4 = 4^4k^4 \pm 4*4^3k^3 + 6*4^2k^2 \pm 4*4k + 1= 16(16k^4 \pm 16k^3 +6k^2 \pm k) + 1$.
And if we want to be pedantic masochists $n^4 = 16a+1$ when $a = (\frac {n-1}2)^4 + 2(\frac {n-1}2)^3 + \frac {(n-1)(3n-1)}4$, which, as $n$ is odd is an integer.
======
That proves if $n$ is odd then there is an integer $a$ so that $n^4 = 16a + 1$.
If $n^4 = 16a+1$ then $n^4$ is odd and $n$ is odd. (So the other direction is facile.)
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3865992",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
}
|
How can you calculate the rank of an nxn matrix with the given conditions? Let $A=(a_{i,j})$ a square matrix whose elements are:
*
*$0$ if $i=j$.
*$1$ if $j>i$.
*$-1$ if $j<i$.
Is there a simple way to find its rank?
|
that was neat. Same rank as the matrix with entry $1$ when $j=i+1,$ then $-1$ when $j=i-1,$ otherwise $0$
Congruence:
$$ \tiny
\left(
\begin{array}{rrrrr}
1&0&0&0&0 \\
0&1&0&0&0 \\
0&-1&1&0&0 \\
0&0&-1&1&0 \\
0&0&0&-1&1 \\
\end{array}
\right)
\left(
\begin{array}{rrrrr}
0&1&1&1&1\\
-1&0&1&1&1 \\
-1&-1&0&1&1 \\
-1&-1&-1&0&1 \\
-1&-1&-1&-1&0 \\
\end{array}
\right)
\left(
\begin{array}{rrrrr}
1&0&0&0&0 \\
0&1&-1&0&0 \\
0&0&1&-1&0 \\
0&0&0&1&-1 \\
0&0&0&0&1 \\
\end{array}
\right) =
\left(
\begin{array}{rrrrr}
0&1&0&0&0 \\
-1&0&1&0&0 \\
0&-1&0&1&0 \\
0&0&-1&0&1 \\
0&0&0&-1&0 \\
\end{array}
\right)
$$
The final matrix is the Skew Normal Form in Newman, especially pages 56-60.
The process is now repeated with $E$ until the canonical form
described by the theorem is obtained, and the procedure makes it clear
that the rank of $A$ must be even.
Once the matrix is in this normal form, the new null (zero eigenvector)vector(s) becomes quickly apparent: none when $n$ even, exactly one when $n$ is odd, namely $(1,0,1,0,1)^T$ and multiplied by any constant.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3867140",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
}
|
Let $x,y,z$ are the lengths of sides of a triangle such that $x+y+z=2$. Find the range of $xy+yz+xz-xyz$ .
Let $x,y,z$ are the lengths of sides of a triangle such that $x+y+z=2$. Find the range of $xy+yz+xz-xyz$ .
What I Tried: I have tried by doing $(x+y+z)(x+y+z)=2*2 = 4.$
Also I got $2(xy+yz+xz)+(x^2+y^2+z^2)=4$.
But I think it is a wrong method to proceed...
|
Given any three positive numbers $x,y,z$, it can form the sides of a non-degenerate triangle if and only if we can find three positive numbers $u, v, w$ such that
$$x = v +w,\quad y = u + w\quad\text{ and }\quad z = u +v$$
(This is known as the Ravi's substituion).
In terms of them, the condition $x + y + z = 2$ is equivalent to $u + v + w = 1$.
Under this condition, we have
$$\begin{cases}
u &= 1 - (v+w) = 1 - x\\
v &= 1 - (u+w) = 1 - y\\
w &= 1 - (u+v) = 1 - z
\end{cases}$$
Notice
$$(1-x)(1-y)(1-z) = 1 - (x+y+z) + (xy+yz+zx) - xyz$$
The expression at hand equals to
$$\begin{align}xy+yz+zx - xyz
&= (1-x)(1-y)(1-z) - 1 + (x+y+z)\\
&= uvw - 1 + 2 = 1+uvw
\end{align}$$
By AM $\ge$ GM, we know $uvw \le \left(\frac{u+v+w}{3}\right)^3 = \frac1{27}$. Since $u,v,w > 0$, we have
$$1 < xy+yz+zx - xyz \le \frac{28}{27}$$
for any valid choice of $x,y,z$.
In fact, if we set $(u,v,w)$ to $(t,t,1-2t)$ and let $t$ increases from $0$ to $\frac13$, the expression $1+uvw$ increases from $1$ continuously to $\frac{28}{27}$. From this, we can deduce the range of the expression $xy + yz + xz - xyz$ is $(1,\frac{28}{27}]$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3869454",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
}
|
Find $\lim_{x \to 3} \frac{x^{15} - 3^{15} - 15 \cdot 3^{14}(x-3)}{(x-3)^2}$ Please help me find:
$$
\lim_{x \to 3} \frac{x^{15} - 3^{15} - 15 \cdot 3^{14}(x-3)}{(x-3)^2}
$$
I cannot use L'Hospital's rule.
I tried to eliminate $x-3$, but I have no idea what to do next.
$$
\lim_{x \to 3} \frac{x^{15} - 3^{15} - 15 \cdot 3^{14}(x-3)}{(x-3)^2} =
\lim_{x \to 3} \frac{x^{14} + 3x^{13} + ... + 3^{14} - 15 \cdot 3^{14}}{x-3}
$$
|
Now, $$\frac{x^{14}+3x^{13}+...+3^{13}x-14\cdot3^{14}}{x-3}=\frac{x^{14}-3^{14}}{x-3}+3\cdot\frac{x^{13}-3^{13}}{x-3}+...+3^{13}\rightarrow$$
$$=14\cdot3^{13}+3\cdot13\cdot3^{12}+...+3^{13}=(14+13+...+1)3^{13}=105\cdot3^{13}=35\cdot3^{14}.$$
I used the following:
$$a^n-b^n=(a-b)\left(a^{n-1}+a^{n-2}b+...+ab^{n-2}+b^{n-1}\right)$$ for natural $n\geq2$.
We see that in the big brackets we have $n$ terms $a^n$ for $a=b$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3869567",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
}
|
Evaluating the challenging sum $\sum _{k=1}^{\infty }\frac{H_{2k}}{k^3\:4^k}\binom{2k}{k}$. I managed to evaluate the sum, my approach can be found $\underline{\operatorname{below as an answer}}$, I'd truly appreciate if any of you could share new methods to evaluate this series, thank you.
The following are the short proofs of the other series I encountered during my approach.
$$\sum _{k=1}^{\infty }\frac{x^k}{4^k}\binom{2k}{k}=\frac{1}{\sqrt{1-x}}-1\tag{$\ast$}$$
$$\sum _{k=1}^{\infty }\frac{x^k}{k\:4^k}\binom{2k}{k}=\:-2\ln \left(1+\sqrt{1-x}\right)+2\ln \left(2\right)$$
$$\sum _{k=1}^{\infty }\frac{1}{k\:4^k}\binom{2k}{k}=\:2\ln \left(2\right)$$
On $\left(\ast\right)$ perform the following manipulations.
$$-\sum _{k=1}^{\infty }\frac{1}{4^k}\binom{2k}{k}\int _0^1x^{k-1}\ln \left(x\right)\:dx=-\int _0^1\frac{\ln \left(x\right)\left(1-\sqrt{1-x}\right)}{x\sqrt{1-x}}\:dx$$
$$\sum _{k=1}^{\infty }\frac{1}{k^2\:4^k}\binom{2k}{k}=-2\int _0^1\frac{\ln \left(1-x^2\right)}{1+x}\:dx$$
$$=-2\int _0^1\frac{\ln \left(1-x\right)}{1+x}\:dx-2\int _0^1\frac{\ln \left(1+x\right)}{1+x}\:dx$$
$$=2\operatorname{Li}_2\left(\frac{1}{2}\right)-\ln ^2\left(2\right)$$
$$\sum _{k=1}^{\infty }\frac{1}{k^2\:4^k}\binom{2k}{k}=\zeta \left(2\right)-2\ln ^2\left(2\right)$$
|
First part.
Evaluating $$\int _0^1\frac{\operatorname{Li}_3\left(x\right)}{x\sqrt{1-x^2}}\:dx$$
$$\int _0^1\frac{\operatorname{Li}_3\left(x\right)}{x\sqrt{1-x^2}}\:dx=\frac{1}{2}\int _0^1\ln ^2\left(t\right)\left(\int _0^1\frac{1}{\left(1-tx\right)\sqrt{1-x^2}}\:dx\right)dt$$
$$=\frac{\pi }{4}\int _0^1\frac{\ln ^2\left(t\right)}{\sqrt{1-t^2}}\:dt+\frac{1}{2}\int _0^1\frac{\ln ^2\left(t\right)\arcsin \left(t\right)}{\sqrt{1-t^2}}\:dt$$
$$=\frac{\pi }{4}\int _0^{\frac{\pi }{2}}\ln ^2\left(\sin \left(x\right)\right)\:dx+\frac{1}{2}\int _0^{\frac{\pi }{2}}x\ln ^2\left(\sin \left(x\right)\right)\:dx$$
Using the Beta function for the first integral and using this result for the $2$nd one proved there we conclude that:
$$\int _0^1\frac{\operatorname{Li}_3\left(x\right)}{x\sqrt{1-x^2}}\:dx=\frac{41}{64}\zeta \left(4\right)+\frac{1}{2}\operatorname{Li}_4\left(\frac{1}{2}\right)+\ln ^2\left(2\right)\zeta \left(2\right)+\frac{1}{48}\ln ^4\left(2\right)$$
Second part.
Finding $$\sum _{k=1}^{\infty }\frac{H_{2k}}{k^3\:4^k}\binom{2k}{k}$$
By employing $1+\sum _{k=1}^{\infty }\frac{x^{2k}}{4^k}\binom{2k}{k}=\frac{1}{\sqrt{1-x^2}}$ on the previous integral we have:
$$\int _0^1\frac{\operatorname{Li}_3\left(x\right)}{x\sqrt{1-x^2}}\:dx=\int _0^1\frac{\operatorname{Li}_3\left(x\right)}{x}\:dx+\sum _{k=1}^{\infty }\frac{1}{4^k}\binom{2k}{k}\int _0^1x^{2k-1}\operatorname{Li}_3\left(x\right)\:dx$$
$$=\zeta \left(4\right)+\sum _{k=1}^{\infty }\frac{1}{4^k}\binom{2k}{k}\left(\frac{1}{2k}\zeta \left(3\right)-\frac{1}{2k}\int _0^1x^{2k-1}\operatorname{Li}_2\left(x\right)\:dx\right)$$
$$\int _0^1\frac{\operatorname{Li}_3\left(x\right)}{x\sqrt{1-x^2}}\:dx=\zeta \left(4\right)+\frac{1}{2}\zeta \left(3\right)\sum _{k=1}^{\infty }\frac{1}{k\:4^k}\binom{2k}{k}-\frac{1}{4}\zeta \left(2\right)\sum _{k=1}^{\infty }\frac{1}{k^2\:4^k}\binom{2k}{k}$$
$$+\frac{1}{8}\sum _{k=1}^{\infty }\frac{H_{2k}}{k^3\:4^k}\binom{2k}{k}$$
$$\sum _{k=1}^{\infty }\frac{H_{2k}}{k^3\:4^k}\binom{2k}{k}=8\int _0^1\frac{\operatorname{Li}_3\left(x\right)}{x\sqrt{1-x^2}}\:dx-3\zeta \left(4\right)-8\ln \left(2\right)\zeta \left(3\right)-4\ln ^2\left(2\right)\zeta \left(2\right)$$
Using the result found on the first part we finalize.
$$\sum _{k=1}^{\infty }\frac{H_{2k}}{k^3\:4^k}\binom{2k}{k}=\frac{17}{8}\zeta \left(4\right)+4\operatorname{Li}_4\left(\frac{1}{2}\right)-8\ln \left(2\right)\zeta \left(3\right)+4\ln ^2\left(2\right)\zeta \left(2\right)+\frac{1}{6}\ln ^4\left(2\right)$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3870288",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
"answer_count": 1,
"answer_id": 0
}
|
$\sqrt{1-x^2}$ is not differentiable at $x = 1$. Please give me an easier proof if exists.
Prove that $\sqrt{1-x^2}$ is not differentiable at $x = 1$.
My proof is here:
Let $0 < h \leq 2$.
$\frac{\sqrt{1-(1-h)^2}-\sqrt{1-1^2}}{-h} = \frac{\sqrt{2h-h^2}}{-h} = \frac{2h-h^2}{-h\sqrt{2h-h^2}} = \frac{h-2}{\sqrt{2h-h^2}}$.
So, $\lim_{h\to 0+} \frac{\sqrt{1-(1-h)^2}-\sqrt{1-1^2}}{-h} = -\infty$.
So, $\sqrt{1-x^2}$ is not differentiable at $x = 1$.
I want an easier proof.
Is there any general theorem or general proposition to prove the above fact?
By the way, the following proposition is not true. $f(x) = x^{\frac{1}{3}}$ is not differentiable at $x = 0$ and $g(x) = x^3$ is differentiable at $x = 0$ but $f(g(x)) = x$ is differentiable at $x=0$.
If $f(x)$ is not differentiable at $x = a$ and $g(x)$ is differentiable at $x = b$ and $g(b) = a$, then $f(g(x))$ is not differentiable at $x = b$.
So, we cannot prove as follows:
$\sqrt{x}$ is not differentiable at $x = 0$. And $1-x^2$ is differentiable at $x = 1$. So, $\sqrt{1-x^2}$ is not differentiable at $x = 1$.
|
The derivative of $x\mapsto\sqrt{1-x^2}$ exists only for $-1<x<1$:
$$f'(x)=-\frac{x}{\sqrt{1-x^2}}$$
Since the function is continuous on $[-1,1]$, we can compute the limits of the derivative. Then
\begin{align}
\lim_{x\to1^+}f'(x)&=i\infty\\
\lim_{x\to 1^-}f'(x)&=-\infty
\end{align}
so the function is not differentiable at $x=1$ as we have a two-sided limit (see this also).
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3870737",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 0
}
|
Find the internal angle of the triangle between $\overrightarrow a \& \overrightarrow b $. If $\overrightarrow a = \hat j + \sqrt 3 \hat k;\overrightarrow b = - \hat j + \sqrt 3 \hat k;\overrightarrow c = 2\sqrt 3 \hat k$ form a triangle then find the internal angle of the triangle between $\overrightarrow a \& \overrightarrow b $.
My approach is as follow
Let $\overrightarrow a + \overrightarrow b = \overrightarrow c $
$\overrightarrow a = \hat j + \sqrt 3 \hat k;\overrightarrow b = - \hat j + \sqrt 3 \hat k;\overrightarrow c = 2\sqrt 3 \hat k$
$ \Rightarrow {\left( {\overrightarrow a + \overrightarrow b } \right)^2} = {\left( {\overrightarrow c } \right)^2} \Rightarrow {\overrightarrow a ^2} + {\overrightarrow b ^2} + 2\overrightarrow a .\overrightarrow b = {\overrightarrow c ^2}$
$ \Rightarrow \overrightarrow a .\overrightarrow b = 2 \Rightarrow $
$\cos \theta = \frac{{\overrightarrow a .\overrightarrow b }}{{\left| {\overrightarrow a } \right|.\left| {\overrightarrow b } \right|}} = \frac{1}{2} = \frac{\pi }{3}$
But official answer is $\frac{2\pi }{3}$ where I am making mistake
|
By your work $$\cos\theta=-\frac{2}{2\cdot2}-\frac{1}{2}.$$
Let $\vec{a}=\vec{AB},$ $\vec{b}=\vec{BC}$ and $\vec{c}=\vec{AC}.$
Thus, $$\cos\theta=\cos\measuredangle ABC=\cos\left(180^{\circ}-\measuredangle(\vec{a},\vec{b})\right)=-\frac{1}{2}.$$$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3873356",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
}
|
Is this proof regarding odd perfect numbers valid? (Edited in response to a comment.)
Here are my:
QUESTIONS
(1) Is this proof regarding odd perfect numbers valid, particularly the part where it says
$$\dfrac{2n^2}{D(n^2)} \neq (q + 1)?$$
(2) If the proof is incorrect, how can the argument be mended to produce a valid proof?
Let $N = q^k n^2$ be an odd perfect number with special/Euler prime $q$. (That is, $\gcd(q,n)=1$ and $q \equiv k \equiv 1 \pmod 4$.) Let $\sigma(x)$ denote the sum of the divisors of $x \in \mathbb{N}$.
Define
$$D(n^2) := 2n^2 - \sigma(n^2)$$
to be the deficiency of the non-Euler part $n^2$.
Define
$$I(n^2) := \dfrac{\sigma(n^2)}{n^2}$$
to be the abundancy index of $n^2$.
Since $N$ is perfect, then we have
$$\sigma(q^k)\sigma(n^2)=\sigma(q^k n^2)=\sigma(N)=2N=2q^k n^2$$
from which it follows that
$$I(N)=2 \iff I(q^k)I(n^2)=2.$$
But since $q$ is prime, $I(q^k)$ can be rewritten as
$$I(q^k) = \dfrac{q^{k+1} - 1}{q^k (q - 1)}$$
which can be bounded as follows (since $k \equiv 1 \pmod 4$ implies that $k \geq 1$)
$$\frac{q+1}{q} = I(q) \leq I(q^k) < \dfrac{q^{k+1}}{q^k (q - 1)} = \frac{q}{q - 1}.$$
This implies that
$$\dfrac{2(q - 1)}{q} < I(n^2) = \dfrac{2}{I(q^k)} \leq \dfrac{2q}{q + 1}$$
from which it follows that
$$\dfrac{2}{q+1} \leq 2 - I(n^2) = \dfrac{D(n^2)}{n^2} < \dfrac{2}{q},$$
which means that
$$q < \dfrac{2n^2}{D(n^2)} \leq (q + 1).$$
Since $2n^2$ is even and $D(n^2) = 2n^2 - \sigma(n^2)$ is always odd (since $n^2$ is a square), then $$\dfrac{2n^2}{D(n^2)}$$
cannot be an integer. This implies that
$$\dfrac{2n^2}{D(n^2)} \neq (q + 1)$$
which implies that
$$\dfrac{n^2}{D(n^2)} \neq \dfrac{q+1}{2}$$
and
$$2 - I(n^2) = \dfrac{D(n^2)}{n^2} \neq \dfrac{2}{q+1}.$$
Finally, we obtain
$$I(n^2) \neq \bigg(2 - \dfrac{2}{q+1}\bigg) = \dfrac{2q}{q+1}$$
which implies that
$$I(q^k) = \dfrac{2}{I(n^2)} \neq \dfrac{q+1}{q},$$
thus resulting in
$$k \neq 1.$$
|
(1)
Your proof looks invalid to me.
You are saying "Since $2n^2$ is even and $D(n^2) = 2n^2 - \sigma(n^2)$ is always odd (since $n^2$ is a square), then $\dfrac{2n^2}{D(n^2)}$ cannot be an integer".
This is not true since $\frac{\text{even}}{\text{odd}}$ can be an integer. For example, $\frac{6}{3}=2$.
Since $D(n^2)$ is odd, we have
$$\frac{2n^2}{D(n^2)}\in\mathbb Z\iff D(n^2)\mid n^2$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3873500",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
Prove that $\alpha^n<\sum_{k=1}^n {2n \choose n+k}k^2<\beta^n$, when $0<\alpha<4<\beta$ are constants
Let $\alpha,\beta$ constants such that $0<\alpha<4<\beta$. Prove that $\exists n_0.\forall n>n_0$ $$\alpha^n<\sum_{k=1}^n {2n \choose n+k}k^2<\beta^n$$
My attempt:
$$\sum_{k=1}^n {2n \choose n+k}k^2=\sum_{k=1}^n \frac{(2n)!}{(n+k)!(n-k)!}k^2=\sum_{k=1}^n \frac{(n-k+1)\cdot\ldots \cdot (n-1)n}{(n+1)\cdot\ldots \cdot(n+k)}\cdot \frac{(2n)!}{(n!)^2}k^2=\sum_{k=1}^n \frac{(n-k+1)\cdot\ldots \cdot (n-1)n}{(n+1)\cdot\ldots \cdot(n+k)}\cdot{2n \choose n}k^2$$
Using the Stirling Approximation we get
$$\sum_{k=1}^n k^2\frac{(n-k+1)\cdot\ldots \cdot (n-1)n}{(n+1)\cdot\ldots \cdot(n+k)}\cdot(1+o(1))4^n\sqrt{\frac{1}{\pi n}}=\sum_{k=1}^n k^2\frac{(n-k+1)\cdot\ldots \cdot (n-1)n}{(n+1)\cdot\ldots \cdot(n+k)}\cdot\Theta(\frac{4^n}{\sqrt{n}})$$
Will this get me anywhere? Or is there another solution? Thanks!
|
\begin{align*}
\sum_{k=1}^n\binom{2n}{n+k}k^2
&=\frac12\sum_{k=-n}^{n}\binom{2n}{n+k}k^2
\\&=\frac12\sum_{k=0}^{2n}\binom{2n}{k}(k-n)^2
\\&=\frac18\left.\frac{d^2}{dx^2}\sum_{k=0}^{2n}\binom{2n}{k}e^{2(k-n)x}\right|_{x=0}
\\&=\frac18\left.\frac{d^2}{dx^2}(e^x+e^{-x})^{2n}\right|_{x=0}
\\&=2^{2n-3}\left.\frac{d^2}{dx^2}\cosh^{2n}x\right|_{x=0}
\\&=2^{2n-2}n\big((2n-1)\cosh^{2n-2}x\sinh^2 x+\cosh^{2n}x\big)_{x=0}
\\&=4^{n-1}n.
\end{align*}
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3875034",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
}
|
How to define $\delta$ to prove $\lim\limits_{x \to 9} \sqrt{x-5} = 2$ When tackling this problem, parting from the assumption that $\lvert \sqrt{x-5} - 2\rvert \lt \epsilon$, I arrived through arithmetical manipulation at $$\frac{\lvert x-9\rvert}{\lvert \sqrt{x-5} + 2\rvert} \lt \epsilon,$$ which has the $\lvert x-9\rvert$ needed to prove that for a certain $\delta$, $0<|x-9|<\delta \Rightarrow |\sqrt{x-5} + 2|<\epsilon$.
However, I don't know how to eliminate the $x$ in ${\lvert \sqrt{x-5} + 2\rvert}$, such that I can define $\delta$ solely based on $\epsilon$. I would appreciate any help.
|
Assuming $|x-9|<\delta$ we have
$$
\left| \sqrt{x-5} - 2 \right|
= \left| \frac{(\sqrt{x-5} - 2)(\sqrt{x-5} + 2)}{\sqrt{x-5} + 2} \right|
= \left| \frac{x-9}{\sqrt{x-5} + 2} \right|
< \frac{\delta}{\sqrt{x-5} + 2}
.
$$
Now note that $\frac{1}{\sqrt{x-5} + 2} < \frac{1}{2}$ as long as $x-5>0.$ The latter can be achieved by making sure that $|x-9|<4.$ Therefore we will require $\delta<4.$ Then we can just take $\delta\leq 2\epsilon$ and we can continue:
$$
\cdots < \frac{\delta}{2} \leq \epsilon.
$$
Thus, given $\epsilon>0$, take $\delta=\min(2\epsilon,4).$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3876334",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
}
|
Let $n$ be an integer. If the tens digit of $n^2$ is 7, what is the units digit of $n^2$?
Let $n$ be an integer. If the tens digit of $n^2$ is 7, what is the units digit of $n^2$?
So $n^2 \equiv 7 \pmod{100}$? If this is the case then this can be written as $n^2 = 100k +7$, where $k \in \Bbb Z.$
Here one can see that no matter what the choice of $k$, the units digit will be $7$. Thus $n^2 \equiv 7 \pmod{10}.$ However this was wrong. The correct answer is $\textbf{6}.$
What am I doing wrong here? It seems that $n^2 \equiv 7 \pmod{100}$ doesn't hold. If the tens digit is $7$ should I have that $n^2 \equiv 7k \pmod{100}$, where $k$ represents the unit digit of $70$ and not a multiplication?
|
Hint You are looking for adigit $k$ such that
$$n^2 \equiv 70+ k \pmod{100}$$
By the Chinese Remainder Theorem this is equivalent to
$$n^2 \equiv 2+ k \pmod{4}\\
n^2 \equiv k-5\pmod{25}
$$
The quadratic residues modulo $4$ are $0,1$, therefore $k \in \{ 2,3, 6,7 \}$. You have now to figure for which of those $k-5$ is a quadratic residue modulo $25$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3880913",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 1
}
|
Hard limit $\lim_{x\to\infty}\Bigg((x(x+1)(x+2))^{\frac{1}{3}}-\Big(2(x+1)-\Big(x^x(x+1)^{x+1}(x+2)^{x+2}\Big)^{\frac{1}{3x+3}}\Big)\Bigg)=0$ Prove that :
$$\lim_{x\to\infty}\Bigg((x(x+1)(x+2))^{\frac{1}{3}}-\Big(2(x+1)-\Big(x^x(x+1)^{x+1}(x+2)^{x+2}\Big)^{\frac{1}{3x+3}}\Big)\Bigg)=0$$
I can prove that :
$$\lim_{x\to\infty}\frac{(x(x+1)(x+2))^{\frac{1}{3}}-2(x+1)}{-\Big(x^x(x+1)^{x+1}(x+2)^{x+2}\Big)^{\frac{1}{3x+3}}}=1$$
Using the Hospital rule but it doesn't help here .Moreover I have tried power series without success .
Any helps is welcome .
Thanks in advance
|
It suffices to note that as $x\to +\infty$,
$$\left(1+\frac{a}{x}\right)^{m\frac{x+b}{x+c}}=
\exp\left(m\frac{1+b/x}{1+c/x}\log\left(1+\frac{a}{x}\right)\right)=1+\frac{ma}{x}+o(1/x).$$
Then,
\begin{align*}(x(x+1)(x+2))^{\frac{1}{3}}&=x\left(1+\frac{1}{x}\right)^{\frac{1}{3}}\left(1+\frac{2}{x}\right)^{\frac{1}{3}}\\
&=x\left(1+\frac{1}{3x}+\frac{2}{3x}+o(1/x)\right)=x+1+o(1)
\end{align*}
and
\begin{align*}\Big(x^x(x+1)^{x+1}(x+2)^{x+2}\Big)^{\frac{1}{3x+3}}&=
x\left(1+\frac{1}{x}\right)^{\frac{x+1}{3x+3}}\left(1+\frac{2}{x}\right)^{\frac{x+2}{3x+3}}\\
&=x\left(1+\frac{1}{3x}+\frac{2}{3x}+o(1/x)\right)=x+1+o(1).
\end{align*}
Then
$$(x(x+1)(x+2))^{\frac{1}{3}}-\Big(2(x+1)-\Big(x^x(x+1)^{x+1}(x+2)^{x+2}\Big)^{\frac{1}{3x+3}}\Big)\\=
x+1+o(1)-\Big(2(x+1)-\Big(x+1+o(1)\Big)=o(1)
$$
and the given limit easily follows.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3881843",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
}
|
Prove that $\frac{1}{\sqrt{1}} + \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{3}} + ... + \frac{1}{\sqrt{n}} \leq 3\sqrt{n+1} - 3$ Prove that $\frac{1}{\sqrt{1}} + \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{3}} + ... + \frac{1}{\sqrt{n}} \leq 3\sqrt{n+1} - 3$ for every natural $n$.
I've already tried to write it like this:
$$\frac{\sqrt{1}}{1} + \frac{\sqrt{2}}{2} + \frac{\sqrt{3}}{3} + ... + \frac{\sqrt{n}}{n} \leq 3\sqrt{n+1} - 3$$
$$\frac{\sqrt{2}}{2} + \frac{\sqrt{3}}{3} + ... + \frac{\sqrt{n}}{n} \leq 3\sqrt{n+1} - 4$$
but I don't know that to do next or if it's the right way prove this.
|
By induction we have
*
*base case: $n=1 \implies 1 \le 3\sqrt 2-3$
*inductive step: we assume true
$$\frac{1}{\sqrt{1}} + \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{3}} + … + \frac{1}{\sqrt{n}} \leq 3\sqrt{n+1} - 3 \tag 1$$
and we need to prove that
$$\frac{1}{\sqrt{1}} + \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{3}} + … + \frac{1}{\sqrt{n}}+ \frac{1}{\sqrt{n+1}} \stackrel{(1)}\leq 3\sqrt{n+1} - 3+\frac{1}{\sqrt{n+1}} \stackrel{?}\le 3\sqrt{n+2} - 3$$
and last inequality is true indeed
$$ 3\sqrt{n+1} +\frac{1}{\sqrt{n+1}}\le 3\sqrt{n+2} $$
$$ 3(n+1) +1\le 3\sqrt{(n+2)(n+1)} $$
$$ 3n+4\le 3\sqrt{(n+2)(n+1)} $$
$$ 9n^2+24n+16\le 9n^2+27n+18 $$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3886334",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
}
|
Why I cannot get the same answer if I do substitution with $x=a \cos\theta$ for $\int \sqrt{a^2-x^2}dx$ compared with substitution $x=a \sin\theta$? Why I cannot get the same answer if I do substitution with $x=a \cos\theta$ for $\int \sqrt{a^2-x^2}dx$ compared with substitution $x=a \sin\theta$?
$\int \sqrt{a^2-x^2}dx = \int \sqrt{a^2 - a^2 \cos^2 \theta} d\theta = \int a \sin\theta d\theta = \int a \sin\theta(-a \sin\theta)d\theta = -a^2 \int sin^2 \theta = -a^2[\frac{\theta}{2} - \frac{\sin(2\theta)}{4}]+C$
However, the correct answer seems to be $a^2(\frac{\theta}{2} + \frac{\sin(2\theta)}{4}) + C$
|
Method$\#1:$
If $\theta=\arccos\dfrac xa=\dfrac\pi2-\arcsin\dfrac xa,$
$\cos\theta=\dfrac xa,0\le\theta\le\pi$
$$dx=-a\sin\theta\ d\theta\text{ and }\sqrt{a^2-x^2}=a\sin\theta$$
$$\int\sqrt{a^2-x^2}\ dx=-\int a^2\sin^2\theta\ d\theta=\dfrac{a^2}2\int(\cos2\theta-1)\ d\theta=\dfrac{a^2\sin2\theta}4-\dfrac{a^2\theta}2$$
Now $\sin2\theta=2\sin\theta\cos\theta=?$
Method$\#2:$
If $\theta=\arcsin\dfrac xa,$
$\sin\theta=\dfrac xa,-\dfrac\pi2\le\theta\le\dfrac\pi2$
$$dx=a\cos\theta\ d\theta\text{ and }\sqrt{a^2-x^2}=a\cos\theta$$
$$\int\sqrt{a^2-x^2}\ dx=\int a^2\cos^2\theta\ d\theta=\dfrac{a^2}2\int(\cos2\theta+1)\ d\theta=\dfrac{a^2\sin2\theta}4+\dfrac{a^2\theta}2$$
Now $\sin2\theta=2\sin\theta\cos\theta=?$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3892104",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
}
|
How can I calculate Slope and Y-intercept in Multiple Regression? What is the formula for Slope and Y-intercept in Multiple Linear Regression? We can easily find Slope and Y-intercept of Linear Regression meaning the data having only one Independent Variable?
Is there any general way to calculate it?
|
Suppose you have the following regression function: $ y_i = \beta_{0} + \beta_{1} x_{i1} + \cdots + \beta_{p} x_{ip} + \varepsilon_i$, where $\varepsilon_i$ is the random part (white noise). Here you have $p+1$ parameters. To estimate the the parameters $b_0,b_1,\ldots, b_p$ we need the following matrix and vectors.
$\mathbf{y} = \begin{pmatrix} y_1 \\ y_2 \\ \vdots \\ y_n \end{pmatrix},X = \begin{pmatrix} 1 & x_{11} & \cdots & x_{1p} \\
1 & x_{21} & \cdots & x_{2p} \\
\vdots & \vdots & \ddots & \vdots \\
1 & x_{n1} & \cdots & x_{np}
\end{pmatrix}, \textbf β = \begin{pmatrix} \beta_0 \\ \beta_1 \\ \beta_2 \\ \vdots \\ \beta_p \end{pmatrix}, $
Suppose you have the following data
Here is $p=2$ and $n=17$. Then the corresponding vector and matrix is
$\mathbf{y} = \begin{pmatrix} 251.3 \\ 251.3 \\ \vdots \\ 349.0 \end{pmatrix},X = \begin{pmatrix} 1 & 41.9 & 29.1 \\
1 & 34.4 & 29.3 \\
\vdots & \vdots & \vdots \\
1 & 77.8 & 32.9
\end{pmatrix}$
The estimated parameters are
$ \begin{pmatrix} \hat \beta_0 \\ \hat \beta_1 \\ \hat \beta_2 \\ \end{pmatrix}=\hat \beta=(\mathbf{X^{'}X)^{-1}}\cdot \mathbf{X^{'}}\cdot \mathbf{y}$
Here $X^{'}$ denotes the transpose of $X$ and $(X^{'}X)^{-1}$ the inverse of $X^{'}X$
The result is $\begin{pmatrix} \hat \beta_0 \\ \hat \beta_1 \\ \hat \beta_2 \\ \end{pmatrix}=\begin{pmatrix} −153.5 \\ 1.24 \\ 12.08 \\ \end{pmatrix}$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3894085",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
Prove by induction $\left(1-\frac{1}{2}\right)\ldots\left(1-\frac{1}{2^{n}}\right)\geq\frac{1}{4}+\frac{1}{2^{n+1}}$ Prove by induction $\left(1-\frac{1}{2}\right)\left(1-\frac{1}{4}\right)\left(1-\frac{1}{8}\right)\ldots\left(1-\frac{1}{2^{n}}\right)\geq\frac{1}{4}+\frac{1}{2^{n+1}}$
What would be the best way to solve this by the induction method?
|
To prove a statement by induction, we must first prove the statement works for $n=1$, then extend that statement by supposing that the given statement is true for $n$, then proving it maintains truth for $n+1$. I am also assuming $n$ is an positive integer.
The logic behind that is if the statement works for $n=1$, then it will work for $n+1=2$, and if it works for $n=2$... well you get the point.
In this case for $n=1$, we have $(1- \frac{1}{2})\geq \frac{1}{4} + \frac{1}{2^{(1)+1}}$
or $\frac{1}{2} \geq \frac{1}{4} + \frac{1}{4}$ which is true.
Now since that statement is true for $n=1$, its time to do the induction step.
We can now assume $(1- \frac{1}{2^n})\geq \frac{1}{4} + \frac{1}{2^{n+1}}$ holds true for n.
Consider $n+1$:
$(1- \frac{1}{2^{n+1}})\geq \frac{1}{4} + \frac{1}{2^{(n+1)+1}}$
Simplifying gives you $(1- \frac{1}{2^{n+1}})\geq \frac{1}{4} + \frac{1}{2^{n+2}}$
Taking it one step further: $1- \frac{1}{2}\cdot\frac{1}{2^{n}}\geq \frac{1}{4} + \frac{1}{4}\cdot\frac{1}{2^{n}}$
$1\geq \frac{1}{2^n}$ or $2^n \geq 1$, which is true because n is a positive integer.
So if $(1- \frac{1}{2^n})\geq \frac{1}{4} + \frac{1}{2^{n+1}}$ is true for $n$, it will work for $n+1$
And that's all you need to prove an induction.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3894217",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
}
|
The diagram shows an equilateral triangle $ADE$ inside a square $ABCD$ . What is the value of $\frac{[\Delta ADE]}{[\Delta DEC]}$ .
The diagram shows an equilateral triangle $ADE$ inside a square $ABCD$ . What is the value of $\frac{[\Delta ADE]}{[\Delta DEC]}$ .
What I Tried: Here is the diagram :-
You can see I marked the angles as required. Now let $AB = x$ . We then have :-
$$[\Delta ADE] = \frac{\sqrt{3}}{4}x^2$$
Now from here :- https://www.quora.com/What-is-the-ratio-of-sides-of-a-30-75-75-angle-triangle-without-sine-rule , I could understand and show that :- $$ED : DC : CE = \bigg(\frac{\sqrt{3} + 1}{2} : \frac{\sqrt{3} + 1}{2} : 1\bigg)$$
So let $EC = k$ , $CD = DE = \frac{(\sqrt{3} + 1)k}{2}$ .
From here :- $$x = \frac{(\sqrt{3} + 1)k}{2}$$
$$\rightarrow k = EC = \frac{2x}{(\sqrt{3} + 1)}$$
Now, we can find area by Heron's Formula. We have :- $$s = x + \frac{x}{(\sqrt{3} + 1)}$$
$$\rightarrow s = \frac{x\sqrt{3} + 2x}{(\sqrt{3} + 1)}$$
So :- $[\Delta DEC] = \sqrt{s(s-a)(s-b)(s-c)}$
$$\rightarrow \sqrt{\Bigg(\frac{(x\sqrt{3} + 2x)}{(\sqrt{3} + 1)}\Bigg)\Bigg(\frac{x}{(\sqrt{3} + 1)}\Bigg)\Bigg(\frac{x}{(\sqrt{3} + 1)}\Bigg)\Bigg(\frac{(x\sqrt{3})}{(\sqrt{3} + 1)}\Bigg)}$$
This looks like really a complicated expression, and I really am not going to attempt to simplify this. So can anyone give me a different solution?
Thank You.
|
By Area of triangle using trigonometry
Area (DEC) $ = \frac12 (CD)(DE) \sin (30^\circ) = \frac 14 x^2$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3896680",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
}
|
Prove the identity $\frac{1}{1+x^{1}}+\cdots+\frac{2^{n}}{1+x^{2^{n}}}=\frac{1}{x-1}+\frac{2^{n+1}}{1+x^{2^{n+1}}}$ Show that for all non-negatives integers $n$, it is true that
$$\frac{1}{1+x^{1}}+\frac{2}{1+x^{2}}+\frac{4}{1+x^{4}}+\frac{8}{1+x^{8}}+\cdots+\frac{2^{n}}{1+x^{2^{n}}}=\frac{1}{x-1}+\frac{2^{n+1}}{1+x^{2^{n+1}}}$$
For $x \ne 1$, $x \ne -1$
I tried to solved it by appliying geometric series but I got the following
after doing the $2f\left(x\right)-f\left(x\right)$ step, where $f\left(x\right)=\frac{1}{1+x^{1}}+\frac{2}{1+x^{2}}+\frac{4}{1+x^{4}}+\frac{8}{1+x^{8}}+\cdots+\frac{2^{n}}{1+x^{2^{n}}}$
$$f\left(x\right)=\frac{1}{x-1}+\frac{2^{n+1}}{1+x^{2^{n}}}$$
Not sure if Fermat's theorem is needed in order to solve it
|
A proof by induction is elementary and not difficult. Let $$S_n(x) = \sum_{k=0}^n \frac{2^k}{1 + x^{2^k}}, \quad P_n(x) = \frac{1}{x-1} + \frac{2^{n+1}}{\color{red}{1 - x^{2^{n+1}}}}.$$ Note that you have a typographical error for $P_n(x)$; the correct expression is shown in red text. The claim is that for all nonnegative integers $n$, we have $S_n = P_n$. The proof of the inductive step is simply
$$\begin{align}
S_{n+1}(x) &= S_n(x) + \frac{2^{n+1}}{1 + x^{2^{n+1}}} \\
&= P_n(x) + \frac{2^{n+1}}{1 + x^{2^{n+1}}} \\
&= \frac{1}{x-1} + \frac{2^{n+1}}{1 - x^{2^{n+1}}} + \frac{2^{n+1}}{1 + x^{2^{n+1}}} \\
&= \frac{1}{x-1} + 2^{n+1} \left( \frac{(1 + x^{2^{n+1}}) + (1 - x^{2^{n+1}})}{(1 - x^{2^{n+1}})(1 + x^{2^{n+1}})} \right) \\
&= \frac{1}{x-1} + 2^{n+1} \left( \frac{2}{1 - x^{2^{n+2}}} \right) \\
&= P_{n+1}(x).
\end{align}$$
The way that this simplifies also suggests that a direct proof is possible via algebraic simplification of a telescoping sum, e.g., $$\frac{2^{k+1}}{1 - x^{2^{k+1}}} = \frac{2^k}{1 - x^{2^k}} + \frac{2^k}{1 + x^{2^k}}$$ suggests defining $$a_k (x) = \frac{2^k}{1 + x^{2^k}}, \quad b_k(x) = \frac{2^k}{1 - x^{2^k}} $$ hence $$a_k(x) = b_{k+1}(x) - b_k(x)$$ and $$S_n(x) = \sum_{k=0}^n a_k(x) = \sum_{k=0}^n b_{k+1}(x) - b_k(x) = b_{n+1}(x) - b_0(x) = P_n(x).$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3897133",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
}
|
By using the properties of definite integrals, evaluate $\int_0^{\pi}\frac{x}{1+\sin x}dx$
By using the properties of definite integrals, evaluate $\int_0^{\pi}\frac{x}{1+\sin x}dx$.
My attempt:
(Using the property $\int_0^{2a}f(x)dx=\int_0^a(f(x)+f(2a-x))dx$)
$$\int_0^{2\frac{\pi}{2}}\frac{x}{1+\sin x}dx=\int_0^{\frac{\pi}{2}}(\frac{x}{1+\sin x}+\frac{\pi-x}{1+\sin x})dx$$$$=\pi\int_0^{\frac{\pi}{2}}\frac1{1+\sin x}dx=\pi\int_0^{\frac{\pi}{2}}\frac{1-\sin x}{\cos^2x}dx$$$$=\pi\int_0^{\frac{\pi}{2}}(\sec^2x-\sec x\tan x)dx=\pi[\tan x-\sec x]_0^{\frac{\pi}{2}}$$
Now I am stuck. I understand there might be other ways of solving it, but what's wrong in my method? Why am I not getting the answer?
|
Alternatively, I like to try to change sine to cosine and then tackle it by the properties of odd and even functions.
First of all, I use the substitution $y=\dfrac{\pi}{2}-x,$ then
$$
\begin{aligned}
I &=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{\frac{\pi}{2}-y}{1+\cos y} d y \\
&=\frac{\pi}{2} \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{d y}{1+\cos y}-\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{y}{1+\cos y} d y \\
&=\frac{\pi}{2} \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{d y}{1+\cos y} \quad \text { (By the fact that } \frac{y}{1+\cos y} \textrm{ is odd.}) \\
&=\frac{\pi}{2} \int_{0}^{\frac{\pi}{2}} \sec ^{2} \frac{y}{2} d y \\
&=\pi\left[\tan \frac{y}{2}\right]_{0}^{\frac{\pi}{2}} \\
&=\pi
\end{aligned}
$$
:|D Wish you enjoy the proof!
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3899307",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
}
|
Recurring Sequence with Exponent Working with recurring sequences and generating functions, I'm generally lost on solving a general expression of $a_n$ for any $n$ when the next part of the sequence, that is $a_{n+1}$, is in the form of an exponent, such that $a_n = a_{n-1} +k^{n-1}$, where k is some constant. I have no clue on how to approach this problem.
I've solved the Fibonacci sequence by subtracting the two pervious terms and shifting the sequence, but it does not seem to work here.
I'm particularly working with $a_n = 2a_{n-1} + 5^{n-1}$, but the sequence expands extremely fast. The base case, $a_{0} = 1$.
Any help would be appreciated!
|
We use ordinary generating functions. Let $A(x) = \sum_{i=0}^n a_n x^n$. Then we have (summing from $n=1$)
\begin{align}
a_n &= 2a_{n-1} + 5^{n-1},\\
\sum_{n=1}^\infty a_nx^n &= \sum_{n=1}^\infty 2a_{n-1} x^n + \sum_{n=1}^\infty 5^{n-1} x^n,\\
A(x) - a_0 &= 2x\sum_{n=1}^\infty a_{n-1} x^{n-1} + x\sum_{n=1}^\infty 5^{n-1} x^{n-1},\\
A(x) - 1 &= 2x\sum_{n=0}^\infty a_{n} x^{n} + x\sum_{n=0}^\infty 5^{n} x^{n},\\
A(x) - 1 &= 2xA(x) + \frac{x}{1-5x},\\
A(x) - 2xA(x) &= \frac{x}{1-5x} + 1,\\
A(x) &= \frac{x}{(1-2x)(1-5x)} + \frac{1}{1-2x}.\\
\end{align}
Now we use partial fraction decomposition and a bit of algebra to obtain
\begin{align}
A(x) &= \frac{1}{3(1-5x)} - \frac{1}{3(1-2x)} + \frac{1}{1-2x}\\
&= \frac{1}{3} \left(\frac{1}{(1-5x)} + \frac{2}{(1-2x)}\right)\\
&= \frac{1}{3} \left(\sum_{n=0}^{\infty} 5^nx^n + 2\sum_{n=0}^{\infty}2^nx^n \right).
\end{align}
From here we see $$a_n = \frac{5^n + 2^{n+1}}{3}.$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3901080",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
}
|
$16$ people, $8$ men, $8$ women, divide the group to $8$ couples, what is the chance for exactly $3$ male couples? We have a group of $16$ people, $8$ men, $8$ women.
We divide them to $8$ couples.
Let $X$ Be the number of couples make of $2$ men.
Calculate: $P(X = 3)$
I am not sure how to approach this.
I thought that this has an hypergeometric distribution.
Therefore i chose:
$$
N = 16, n = 16, K = 8, k = 6
$$
For:
$$
\frac{\binom{K}{k} \binom{N-K}{n-k}}{\binom{N}{n}}
$$
But its wont gonna work.
Another way i thought is to choose the couples like this:
$$
\frac{\binom{8}{2} \binom{6}{2} \binom{4}{2} \binom{8}{1} \binom{7}{1} \binom{6}{2} \binom{4}{2}}{\frac{16!}{2^8}}
$$
The logic wass:
But again its not working.
I would like some help
|
The number of ways to make $n$ couples from $2n$ people is
$$\frac{(2n)!}{n!\cdot 2^n}$$
This is derived by starting like you did, multiplying
$$\begin{align}
&\quad\binom{2n}{2} \cdot \binom{2n-2}{2} \cdot \binom{2n-4}{2} \cdots \binom{4}{2} \cdot \binom{2}{2} \\
\\
=& \quad\frac{2n(2n-1)}{2} \cdot \frac{(2n-2)(2n-3)}{2} \cdot \frac{(2n-4)(2n-5)}{2} \cdots \frac{(2)(1)}{2} \\
\\
=&\quad \frac{(2n)!}{2^n}
\end{align}$$
However, the above includes an order for the couples chosen which we don't want, so the above has to be divided by $n!$
Now to count the number of ways of choosing exactly three male couples, we start by choosing the males to form the three couples, then apply the above formula once, then choose the two females for the two lone males, then apply the above formula a second time to pair up the remaining six females.
$$
\quad \binom{8}{6} \cdot \frac{6!}{3!\cdot 2^3} \cdot 8 \cdot 7 \cdot \frac{6!}{3!\cdot 2^3} = 352,800
$$
The number of ways to make eight couples with no restriction is
$$\frac{16!}{8!\cdot 2^8} = 2,027,025$$
Thus the required probability is
$$\frac{352,800}{2,027,025} = \boxed{\frac{224}{1287}}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3901380",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
}
|
A convergence test for series I need to prove the following result: let $\left( a_{n}\right) \subset
\left( 0,\infty \right) $ such that
\begin{equation*}
\frac{a_{n+1}}{a_{n}}=1-\frac{1}{n}-\frac{x_{n}}{n\ln n},
\quad
(X)
\end{equation*}
where $x_{n}\geq x>1$. Prove that
\begin{equation*}
\sum_{n\geq 1}a_{n}
\end{equation*}
is convergent. In addition, if
\begin{equation*}
\frac{a_{n+1}}{a_{n}}=1-\frac{1}{n}-\frac{x_{n}}{n\ln n},
\end{equation*}
where $x_{n}\leq x<1$, then
\begin{equation*}
\sum_{n\geq 1}a_{n}
\end{equation*}
is divergent.
For the first asertion it is clear. More precisely, the condition (X) can be
rewritten as follows
\begin{equation*}
a_{n}\left[ \left( n-1\right) \ln n-1\right] -a_{n+1}n\ln n=\left(
x_{n}-1\right) a_{n}.
\end{equation*}
Since $x_{n}\geq x>1$, then $x_{n}-1\geq x-1>0$
\begin{equation*}
a_{n}\left[ \left( n-1\right) \ln n-1\right] -a_{n+1}n\ln n\geq \left(
x-1\right) a_{n},
\end{equation*}
and, using the inequality
\begin{equation*}
\left( n-1\right) \ln \left( n-1\right) >\left( n-1\right) \ln n-1,
\end{equation*}
we obtain that
\begin{equation*}
a_{n}\left( n-1\right) \ln \left( n-1\right) -a_{n+1}n\ln n\geq \left(
x-1\right) a_{n}>0.
\end{equation*}
Hence $u_{n}=a_{n}\left( n-1\right) \ln \left( n-1\right) ,n\geq 2$, is
strictly decreasing, so it converges. Whence the telescoping sum associated
with $\left( u_{n}\right) $, that is
\begin{equation*}
\sum_{n\geq 1}\left( u_{n}-u_{n+1}\right) ,
\end{equation*}
is convergent. But
\begin{equation*}
0<\left( x-1\right) a_{n}\leq u_{n}-u_{n+1},
\end{equation*}
for all $n\geq 2$. By comparison test, it follows that the series
\begin{equation*}
\sum_{n\geq 1}a_{n}
\end{equation*}
is convergent.
Now the problem is the second assertion. Since
\begin{equation*}
a_{n}\left[ \left( n-1\right) \ln n-1\right] -a_{n+1}n\ln n=\left(
x_{n}-1\right) a_{n}.
\end{equation*}
and since $x_{n}\leq x<1$, then $x_{n}-1\leq x-1<0$,
\begin{equation*}
a_{n}\left[ \left( n-1\right) \ln n-1\right] -a_{n+1}n\ln n\leq \left(
x-1\right) a_{n}<0,
\end{equation*}
If one can guarantee that
\begin{equation*}
a_{n}\left( n-1\right) \ln \left( n-1\right) -a_{n+1}n\ln n\leq \left(
x-1\right) a_{n}<0,
\quad(XX)
\end{equation*}
for all $n\geq 2$, then it follows that $\left( u_{n}\right) $ is
increasing. So
\begin{equation*}
u_{n}=a_{n}\left( n-1\right) \ln \left( n-1\right) >\alpha ,
\end{equation*}
for all $n\geq 2$ and some $\alpha \in \left( 0,\infty \right) $. Hence, for
all $n\geq 2$, we have
\begin{equation*}
\frac{\alpha }{\left( n-1\right) \ln \left( n-1\right) }<a_{n}
\end{equation*}
and, since $\sum \frac{1}{\left( n-1\right) \ln \left( n-1\right) }$ is
divergent, it follows that the series $\sum a_{n}$ is divergent. How can I
prove that (XX), if it is correct of course. If not, how can we tackle the
proof of the second assertion?
|
Here is a method which can be applied to both cases : let's do the first one, the other one will be similar. We have $a_n > 0$ such that
$$\frac{a_{n+1}}{a_n} = 1 - \frac{1}{n} - \frac{x_n}{n \ln (n)}$$
with $x_n \geq x > 1$ for a real number $x$. Let $y=(1+x)/2$. Consider $$u_n = \frac{1}{n \ln^{y}(n)}$$
One has
$$\frac{u_{n+1}}{u_n} = \frac{n \ln^y(n)}{(n+1)\ln^y(n+1)} = \left(\left(1+ \frac{1}{n} \right)\left( \frac{\ln(n+1)}{\ln(n)}\right)^y \right)^{-1}$$ $$ =\left(\left(1+ \frac{1}{n} \right)\left( 1+\frac{\ln(1+\frac{1}{n})}{\ln(n)}\right)^y \right)^{-1}=\left(\left(1+ \frac{1}{n} \right)\left( 1+\frac{1}{n\ln(n)}+o\left(\frac{1}{n\ln(n)} \right)\right)^y \right)^{-1}$$
$$=\left(\left(1+ \frac{1}{n} \right)\left( 1+\frac{y}{n\ln(n)}+o\left(\frac{1}{n\ln(n)} \right)\right) \right)^{-1} =\left(1+ \frac{1}{n} +\frac{y}{n\ln(n)}+o\left(\frac{1}{n\ln(n)} \right) \right)^{-1} $$ $$=1- \frac{1}{n} -\frac{y}{n\ln(n)}+o\left(\frac{1}{n\ln(n)} \right) $$
Now, because $$x_n - y \geq \frac{x-1}{2} > 0$$
you see that from a certain rank, you will have
$$\frac{u_{n+1}}{u_n} \geq \frac{a_{n+1}}{a_n}$$
Because the series $\sum u_n$ is convergent, you deduce that $\sum a_n$ is also convergent by comparison.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3902066",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
Contour integration with pole I wanted to integrate
$$\oint \frac{\sqrt{z+5}}{z^5}$$
around a circular contour radius $1$ center $0$.
So the function has pole at 0. How can I proceed from here?
|
Cauchy's integral formula is
$$f^{(n)} (z_0) = \frac{n!}{2\pi i} \oint \frac{f(z)}{(z-z_0)^{n+1}} dz$$
so
$$ \oint \frac{\sqrt{z+5}}{z^5} dz = \frac{2\pi i}{4!} \,\left.\frac{d^4}{dz^4}(z+5)^{1/2}\right|_{z=0}$$
$$ \oint \frac{\sqrt{z+5}}{z^5} dz = -\frac{2\pi i}{4!} \, \left. \frac{15}{16(5+z)^{7/2}}\right|_{z=0}$$
$$ \oint \frac{\sqrt{z+5}}{z^5} dz = -\frac{\pi i}{64\cdot 5^{5/2}} $$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3902611",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
Find the flux integral of the vector field $F=(0,0,-1)$ through the cone $z=\sqrt{x^2+y^2}$, $x^2+y^2 \leq 1$
A uniform fluid flowing vertically downward (heavy rain) is described by the vector field $F=(0,0,-1)$. Find the total flow through the cone $z=\sqrt{x^2+y^2}$, $x^2+y^2 \leq 1$.
b)Now consider $F=(-\frac{\sqrt{2}}{2},0,\frac{-\sqrt{2}}{2})$ and Find the total flow through the cone $z=\sqrt{x^2+y^2}$, $x^2+y^2 \leq 1$
Attempt a)
We shall use the expresion
$$\int {F dS}=\int \int_{D}F \cdot ( T_{u} \times T_{v})=\int \int_D (-Pg_x-Qg_y+R)dA$$
Notice that the region $S$ alredy was given $g(x,y)=z(x,y)=\sqrt{x^2+y^2}$
$$g_x=\frac{x}{\sqrt{x^2+y^2}}$$
$$g_y=\frac{y}{\sqrt{x^2+y^2}}$$
doing the substitution on the integral
$$\iint_{D}(-0g_x-0g_y+(-1))dA$$
Since $x^2+y^2 \leq 1$ we shall use polar cordinates
$$x=r \cos \theta $$
$$y=r \sin \theta$$
where $r\in [0,1]$ and $\theta \in [0,2\pi]$
Finally doing the integration
$$ \int_{0}^{2 \pi} \int_{0}^{1}-rdrd \theta=-\pi $$
Now for $b)
We shall use the expresion
$$\int {F dS}=\int \int_{D}F \cdot ( T_{u} \times T_{v})=\int \int_D (-Pg_x-Qg_y+R)dA$$
Notice that the region $S$ alredy was given $g(x,y)=z(x,y)=\sqrt{x^2+y^2}$
$$g_x=\frac{x}{\sqrt{x^2+y^2}}$$
$$g_y=\frac{y}{\sqrt{x^2+y^2}}$$
doing the substitution on the integral
$$\iint_{D}(\frac{\sqrt{2}}{2}g_x-0g_y+(\frac{-\sqrt{2}}{2}))dA$$
$$\frac{\sqrt{2}}{2} \iint_{D}(\frac{x}{\sqrt{x^2+y^2}}-1)dA$$
Usinng polar coordinates we get
$$\int_{0}^{2 \pi} \int_{0}^{1} (cos \theta-1)r dr d\theta=-\frac{\pi}{\sqrt{2}}$$
Are my answers right or probably i do a mistake applying the formula.
|
Please note your normal vector is pointing upward. As per question, You need normal vector pointing downward through the cone surface. So the signs of your final answer will change otherwise it looks fine.
From cross product, you get $(-\frac{x}{\sqrt{x^2+y^2}}, -\frac{y}{\sqrt{x^2+y^2}}, 1)$.
Please see the third coordinate is positive that gives us upward direction but given we want downward direction, you should take the normal vector $(\frac{x}{\sqrt{x^2+y^2}}, \frac{y}{\sqrt{x^2+y^2}}, -1)$.
Now please do the same calculations and you will get opposite sign in both your answers.
By the way, instead of converting to polar coordinates at the time of integration, you could also parametrize the cone as below -
$T(r, \theta) = (r \cos \theta, r \sin \theta, r) \,$ $(0 \leq r \leq 1, 0 \leq \theta \leq 2\pi)$
So you get $T_{\theta} \times T_{r} = \, (r \cos \theta, r \sin \theta, -r)$
Integral becomes $\iint_D F.(T_{\theta} \times T_{r}) \, dr \, d\theta$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3905372",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
Prove $\sin x\ge x-\frac{x^3}{3!}$
Prove $$\sin x\ge x-\frac{x^3}{3!}$$ for $x\ge 0$.
I know that a calculus/continuity proof exists, but I am curious if this can be proved without that. Here is a sketch I have made.;
we will use for $x\ge 0$;
$$\sin x\le x\le \tan x$$
proof of this see here
I got a weaker bound :
$$\tan\left(\frac{x}{2}\right)\ge \frac{x}{2}$$
$$\sin\left(\frac{x}2\right)\cos\left(\frac{x}2\right)\ge \frac{x}{2}\cos^2\left(\frac{x}{2}\right)$$
$$\sin x= x\left[1-\sin^2\left(\frac{x}{2}\right)\right]\ge x\left(1-\frac{x^2}{4}\right)= x-\frac{x^3}{4}$$
Thus $$\sin x \ge x-\frac{x^3}{4}$$
As you can see is there any way I can take this to strengthen the inequality, to get $$\sin x\ge x-\frac{x^3}{6}.$$
Also could this possibly be extended by strengthening inequality further to get the Taylor series. I think I am going too far!
|
Well, from the celebrated triplication formula, we get $$\sin3x=3\,\sin x-4\,\sin^3x.$$ This, of course, means that $$3^{k}\sin x/3^k-3^{k+1}\sin x/3^{k+1}=-4\cdot3^{k}\,\sin^3 x/3^{k+1}.$$ Summing the telescopic series and observing $3^{k}\sin x/3^k\to x$ as $k\to\infty$, we get
$$\sin x-x= -4\,\sum^\infty_{k=0}3^{k}\,\sin^3 x/3^{k+1},$$ i.e.
$$x-\sin x= 4\,\sum^\infty_{k=0}3^{k}\,\sin^3 x/3^{k+1}.$$
Since $\sin x\le x$ for non-negative $x$, this means
$$x-\sin x\le4\,\sum^\infty_{k=0}3^k\,\left(x/3^{k+1}\right)^3=4\cdot\frac1{27}\cdot\frac1{1-\frac19}\,x^3=\frac16\,x^3.$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3905903",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
}
|
A circle of radius 1 is inscribed inside a regular octagon (a polygon with eight sides of length b). Calculate the octagon’s perimeter and its area.
A circle of radius 1 is inscribed inside a regular octagon (a polygon with eight sides of length b). Calculate the octagon’s perimeter and its area.
Hint: Split the octagon into eight isosceles triangles.
What I did is divide the entire circle into 16 parts like the following:
Side A has length 1.
Side C has a length of 1 + x (x being the difference between the circle and the vertex of the octagon).
Side B has length b/2.
Angle AC is $\frac{360}{12}=22.5$.
Angle AB is 90 degrees.
Angle BC is 67.5 degrees.
I know that
$\tan{(22.5)}=\frac{b}{2}\times\frac{1}{1}$
The perimeter then is 16 times $\frac{b}{2}:
$16\times2\tan{(22.5)}=P_b$
$32\tan{(22.5)}=P_b$
However, the answer in the textbook is:
What am I doing wrong?
|
To add onto @JoshuaWang, tan($22.5^o$)=tan$\frac{\frac π4}{2}$
Then using the half angle formula for tangent which can be found from the sine and cosine half angle formulas:
$tan^2\frac π8$= $\frac{1-cos(π/4)}{1+cos(π/4)}$=$\frac{1-\frac{1}{\sqrt 2}}{1+\frac{1}{\sqrt 2}}$= $\frac{\frac{\sqrt2}{\sqrt2}-\frac{1}{\sqrt 2}}{\frac{\sqrt2}{\sqrt2}+\frac{1}{\sqrt 2}}$=$\frac{\sqrt2-1}{\sqrt2+1}$*$\frac{\sqrt2-1}{\sqrt2-1}$
Which simplifies to the the original numerator squared divided by original denominator multiplied by its conjugate:
$\frac{3-2\sqrt2}{\sqrt2 ^2-1^2}$=$3-2\sqrt2$
Taking the positive square root gets us tan$\frac π8$=$ \sqrt {3-2\sqrt2}$.
Then let tan$\frac π8$=$\sqrt c$=$\sqrt a\pm\sqrt b$→$\sqrt c ^2$=$(\sqrt a\pm\sqrt b)^2$=$a+b\pm2\sqrt {ab}$= $3-2\sqrt2$.
Then, in order to simplify the nested square roots, the system of equations has to be solved from the last part of the previous step. It would be easier to use the negative branch:
$a+b-\sqrt {ab}$= $3-2\sqrt2$⇔
a+b=3
$\sqrt {ab}=\sqrt2$
Solving this gets us that a=2 and b=1:
tan $\frac π8$=$tan22.5^o$=$\sqrt{3-2\sqrt2}$=$\sqrt2-1$.
In conclusion:
P=8($\sqrt2-1$)=8 $\sqrt2-8$,A=16 ($\sqrt2-1$)=16 $\sqrt2-16$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3907436",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
}
|
Problem of minimum in physics.
It is necessary to go from a point $A(0,0)$ to a point $B(a,b)$ walking from A to $P(x,0)$ with a speed $v_1$ and then until the point B with a speed $v_2$.
Find where is the point in which it is necessary to abandon x axis in order to have the minimum time to complete the path.
I've called $x$ the distance of A from P.
The function is $$t(x)=\frac{x}{v_1}+\frac{\sqrt{(a-x)^2+b^2}}{v_2}$$
$$\frac{dt}{dx}=\frac{1}{v_1}+\frac{(a-x)*(-1)}{v_2* \sqrt{(a-x)^2+b^2}}=0 \Rightarrow v_2*\sqrt{(a-x)^2+b^2}=v_1*(a-x) \Rightarrow x^2(v_2^2-v_1^2)+x(-2av_2^2+2av_1^2)+(a^2v_2^2+b^2v_2^2-v_1^2a^2)=0$$
The solutions of this equation are $x_{1/2} =a \pm \frac{v_2 b}{\sqrt{v_1^2-v_2^2}} $
Putting $\frac{dt}{dx}>0$ it is verified for $0<x<a - \frac{v_2 b}{\sqrt{v_1^2-v_2^2}}$ and for $x>a + \frac{v_2 b}{\sqrt{v_1^2-v_2^2}}$.
So the minimum point is for $a + \frac{v_2 b}{\sqrt{v_1^2-v_2^2}}$
and the minimum time is
$$t(a + \frac{v_2 b}{\sqrt{v_1^2-v_2^2}})= \frac{b (v_1^2+v_2^2)}{v_1v_2 \sqrt{v_1^2-v_2^2}}+\frac{a}{v_1}$$
I'm not sure that I haven't done mistakes, because in the solution on the book it indicates
$a - \frac{v_2 b}{\sqrt{v_1^2-v_2^2}} $ as the solution of minimum.
and the minimum time
$$t(a - \frac{v_2 b}{\sqrt{v_1^2-v_2^2}})= \frac{b\sqrt{v_1^2-v_2^2}}{v_1v_2}+\frac{a}{v_1}$$
but, in particular , it distinguishes the case of $\frac{v_2^2}{v_1^2}\ge \frac{a^2}{a^2+b^2}$
in which the minimum time is possible walking from A to B directly.
|
You are correct that
$$ \frac{dt}{dx} = \frac{1}{v_1} - \frac{a-x}{v_2 \sqrt{(a-x)^2+b^2}} \tag1$$
and that $\frac{dt}{dx} = 0$ implies
$$ v_2 \sqrt{(a-x)^2+b^2} = v_1 (a-x), \tag2$$
which (I think--I did not check your arithmetic here) implies
$$ x^2(v_2^2-v_1^2)+x(-2av_2^2+2av_1^2)+(a^2v_2^2+b^2v_2^2-v_1^2a^2) = 0. \tag3$$
But you seem to be assuming (incorrectly) that Equation $(3)$ implies Equation $(2)$.
It does not. Equation $(3)$ does imply
$$ v_2 \sqrt{(a-x)^2+b^2} = \pm v_1 (a-x), $$
but this is not the same thing.
Also, the region where the left side of Equation $(3)$ is positive does not correspond to the region where $\frac{dt}{dx}$ is positive.
What Equation $(3)$ gives you is two possible points at which $\frac{dt}{dx}$ might be zero. To determine which points are actual zeros, compare each of them against Equation $(2)$.
Assuming that $v_1$ and $v_2$ both are positive,
in Equation $(2)$ we see that the left side is positive and that
therefore $a - x$ must also be positive.
That is, $x < a.$ Hence of the two roots of Equation $(3)$, the only one that can be a zero of $\frac{dt}{dx}$ is the one that is less than $a,$ namely
$$ x_0 = a - \frac{v_2 b}{\sqrt{v_1^2-v_2^2}}. $$
To determine whether this is a local minimum or a local maximum,
you must examine the sign of $\frac{dt}{dx}$ for values of $x$
less than $x_0$ or greater than $x_0$,
not the values of the left-hand side of Equation $(3).$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3908495",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
}
|
How many 5-digit numbers such that the sum of their digits is a multiple of 4? My textbook has the same problem, except for 5-digit numbers the sum of whose digits is a multiple of 5. The general approach is:
Select the first 4 digits. Number of possible ways of doing this = $9\times 10 \times 10 \times 10$ since the ten thousands-place digit cannot be zero. The sum of these 4 digits is $5k$, $5k+1$, $5k+2$, $5k+3$, or $5k+4$. If it's $5k$, the last digit may only be $0$ or $5$. Sum is $5k+1 \implies$ last digit is $4$ or $9$, and so on. In each case, there are only $2$ possible last digits.
Therefore the total number of such 5-digit numbers is $9\times 10 \times 10 \times 10 \times 2 = 18000$.
When we apply this approach to the "sum of digits is a multiple of 4" problem, we get,
Sum of first 4 digits is $4k \implies$ last digit is $0$, $4$, or $8$.
Sum of first 4 digits is $4k+1 \implies$ last digit is $3$ or $7$.
Sum of first 4 digits is $4k+2 \implies$ last digit is $2$ or $6$.
Sum of first 4 digits is $4k+3 \implies$ last digit is $1$, $5$ or $9$.
In two of these cases, we get three possible last digits and in the other two, we get two possible last digits. Our general approach clearly breaks down here. One way to go around this would be to ask ourselves, "how many ways can we select four digits such their sum is $4k$ or $4k+3$", and multiply that number by 3; repeat for $4k+2$ and $4k+1$ but multiply by 2. Add both these numbers to get the answer.
This seems too long. Is there a simpler way to do this?
|
Denote by $a_i(n)$ $\>(0\leq i\leq3)$ the number of $n$-digit decimal strings having a digit sum leaving remainder $i$ modulo $4$, and collect the $a_i(n)$ into $a(n):=\bigl(a_0(n), a_1(n), a_2(n), a_3(n)\bigr)$, the latter considered as a column vector. We then have $$a(1)=(2,3,2,2)\ .$$ For the latter digits the multiplicities of the possible remainders are $(3,3,2,2)$, since the digit ${\tt 0}$ with remainder $0$ is allowed.
What we have written already allows to compute $$a_0(2)=3\cdot2+2\cdot3+2\cdot2+3\cdot2=22\ .$$
Indeed, when the first digit has remainder $0$ the second also has to have remainder $0$, and when the first digit has remainder $1$ the second has to have remainder $3$, etcetera. This idea, used for all possible remainders, leads to the recursion
$$ a(n+1)=A\ a(n)\qquad(n\geq1),\qquad A:=\left[\matrix{3&2&2&3\cr3&3&2&2\cr2&3&3&2\cr2&2&3&3\cr}\right]\ .$$
In this way we obtain
$$a(5)=A^4\ a(1)=\bigl(22500, 22498, 22500, 22502\bigr)\ ,$$
so that the number we are looking for is $22\,500$. In order to compute, e.g., $a(23)$ we better find a scheme for computing the higher powers of the "cyclic" matrix $A$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3909479",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
}
|
Show that $(\sqrt{3}+1)^{2n}+(\sqrt{3}-1)^{2n}$ and $(\sqrt{3}+1)^{2n+1}+(1-\sqrt{3})^{2n+1}$.........? Show that $(\sqrt{3}+1)^{2n}+(\sqrt{3}-1)^{2n}$ and $(\sqrt{3}+1)^{2n+1}+(1-\sqrt{3})^{2n+1}$ are both divisible by $2^{n+1}$. Is this the highest power of $2$ dividing either of the numbers?
I am not well versed with binomial theorem hence I am not able to proceed. I am also interested in the different ways this problem can be solved.
Any help would be appreciated.
|
From binomial expansion it follows that
$$[(2+\sqrt{3})^n+(2-\sqrt{3})^n]$$ $$=[2^n+{n \choose 2} 2^{n-2} 3+{n \choose 4}2^{n-4} 3^2+...]=2K, K \in N$$
And $$[(2+\sqrt{3})^n-(2-\sqrt{3})^n]$$ $$=2\sqrt{3}[2^{n-1}+{n \choose 1} 2^{n-1}+{n \choose 3}2^{n-3} 3+...]=2\sqrt{3}L, L \in N$$
$$S_1=(1+\sqrt{3})^{2n}+(1-\sqrt{3})^{2n}=2^n[(2+\sqrt{3})^n+(2-\sqrt{3})^n]=2^{n+1} K.$$
$$S_2=(1+\sqrt{3})^{2n+1}+(1-\sqrt{3})^{2n+1}=S_1+2^n \sqrt{3}[(2+\sqrt{3})^{n}-(2-\sqrt{3})^n]=3.2^{n+1}L$$
Hence, both $S_1$ and $S_2$ are divisible by $2^{n+1}$.
By binomial expansio
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3910795",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
}
|
Proving that, if $f(k)=\prod_{i=1}^ka_i+\sum_{b=1}^{k-1}(1-a_{k-b})\prod_{i=1}^ba_{k-b+i}$, then $f(k+1)=f(k)\cdot a_{k+1}+(1 - a_k)a_{k+1}$
Given that $$f(k) = \prod_{i=1}^k a_i + \sum_{b=1}^{k-1} (1-a_{k-b}) \prod_{i=1}^b a_{k-b+i}$$ for all $k$ where $(a_1, a_2, a_3, \ldots )$ are random constants, prove that: $$f(k+1) = f(k) \cdot a_{k+1} + (1 - a_k)a_{k+1}$$
So far, my current work is:
$$f(k+1) = \prod_{i=1}^{k+1} a_i + \sum_{b=1}^{k} (1-a_{k-b+1}) \prod_{i=1}^b a_{k-b+i+1} = a_{k+1}\prod_{i=1}^{k+1} a_i + \sum_{b=1}^{k} (1-a_{k-b+1}) \prod_{i=1}^b a_{k-b+i+1}.$$
So, we have to show that: $$a_{k+1}\prod_{i=1}^{k+1} a_i + \sum_{b=1}^{k} (1-a_{k-b+1}) \prod_{i=1}^b a_{k-b+i+1} = a_{k+1}\prod_{i=1}^{k+1} a_i + a_{k+1}\sum_{b=1}^{k-1} (1-a_{k-b}) \prod_{i=1}^b a_{k-b+i} + (1-a_k)(a_{k+1}).$$
The $a_{k+1}\prod_{i=1}^{k+1} a_i$ cancel out on both sides, leaving us to prove: $$\sum_{b=1}^{k} (1-a_{k-b+1}) \prod_{i=1}^b a_{k-b+i+1} = a_{k+1}\sum_{b=1}^{k-1} (1-a_{k-b}) \prod_{i=1}^b a_{k-b+i} + (1-a_k)(a_{k+1}).$$
Turning the max value of $i$ on the first summation to $k-1$ and changing the second series of products a bit to match the first one gives:
\begin{align}
\sum_{b=1}^{k-1} (1-a_{k-b+1}) \prod_{i=1}^b a_{k-b+i+1} + (1-a_1)\prod_{i=1}^k a_{k-b+i+1} = &\\
a_{k+1} \sum_{b=1}^{k-1} \left( (1-a_{k-b}) \prod_{i=1}^b a_{k-b+i+1} \cdot \frac{a_{k-b+1}}{a_{k+1}}\right) + (1-a_k)(a_{k+1}).
\end{align}
We can cancel out the $a_{k+1}$ terms on the RHS, leaving us to prove: $$\sum_{b=1}^{k-1} (1-a_{k-b+1}) \prod_{i=1}^b a_{k-b+i+1} + (1-a_1)\prod_{i=1}^k a_{k-b+i+1} = \sum_{b=1}^{k-1} \left( (1-a_{k-b}) a_{k-b+1} \prod_{i=1}^b a_{k-b+i+1}\right) + (1-a_k)(a_{k+1}).$$
However, I don't know how to continue on from here. Any help would be greatly appreciated.
|
We can considerably simplify $f(k)$ which makes the proof easy.
We obtain
\begin{align*}
\color{blue}{f(k)}&=\prod_{i=1}^k a_i+\sum_{b=1}^{k-1}\left(1-a_{k-b}\right)\prod_{i=1}^b a_{k-b+i}\\
&=\prod_{i=1}^k a_i+\sum_{b=1}^{k-1}\left(1-a_{b}\right)\prod_{i=1}^{k-b} a_{b+i}\tag{1}\\
&=\prod_{i=1}^k a_i+\sum_{b=1}^{k-1}\prod_{i=1}^{k-b} a_{b+i}-\sum_{b=1}^{k-1}a_b\prod_{i=1}^{k-b} a_{b+i}\\
&=\prod_{i=1}^k a_i+\sum_{b=2}^{k}\prod_{i=1}^{k-b+1} a_{b-1+i}-\sum_{b=1}^{k-1}\prod_{i=0}^{k-b} a_{b+i}\tag{2}\\
&=\prod_{i=1}^k a_i+\sum_{b=2}^{k}\prod_{i=0}^{k-b} a_{b+i}-\sum_{b=1}^{k-1}\prod_{i=0}^{k-b} a_{b+i}\tag{3}\\
&=\prod_{i=1}^k a_i+a_k-\prod_{i=0}^{k-1} a_{i+1}\tag{4}\\
&=\prod_{i=1}^k a_i+a_k-\prod_{i=1}^{k} a_{i}\tag{5}\\
&\,\,\color{blue}{=a_k}\tag{6}
\end{align*}
Comment:
*
*In (1) we change the order of summation of the sum $b\to k-b$.
*In (2) we shift the index of $b$ by one in the left sum and merge the factor $a_b$ into the product of the right sum.
*In (3) we shift the index of the product in the left sum by one and observe the sums are telescoping.
*In (4) we do the telescoping.
*In (5) we shift the index of the right-hand product and simplify in the next line.
With $f(k)=a_k$ the proof is simple, since we have
\begin{align*}
\color{blue}{f(k) a_{k+1} + (1 - a_k)a_{k+1}}&= a_ka_{k+1}+(1-a_k)a_{k+1}=a_{k+1}\\
&\color{blue}{=f(k+1)}
\end{align*}
and the claim follows.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3915158",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
}
|
Equation of Circle touching a straight line and passing through the centroid of a triangle and a particular point If the equation $3{x^2}y + 2x{y^2} - 18xy = 0$ represent the sides of a triangle whose centroid is G, then the equation of the circle which touches the line $x-y+1=0$ and passing through G and $(1,1)$ is
|
${x^2} + {y^2} - x - y = 0$
B: ${x^2} + {y^2} - 9x - y + 8 = 0$
C: ${x^2} + {y^2} - 4x + 2y = 0$
D: ${x^2} + {y^2} + 9x + y + 8 = 0$
I got the answer but had the choice not been provided how would I find the solution.
My steps are elaborated below
$3{x^2}y + 2x{y^2} - 18xy = 0 \Rightarrow xy\left( {3x + 2y - 18} \right) = 0$
The triangle is formed by line x=0, y=0 and 3x+2y-18=0
Centroid $G = \left( {\frac{{6 + 0 + 0}}{3},\frac{{9 + 0 + 0}}{3}} \right) = \left( {2,3} \right)$
Let the centre of the circle is (h,k)
Circle passes through (2,3) and (1,1), hence the radius is
${r^2} = {\left( {h - 1} \right)^2} + {\left( {k - 1} \right)^2} = {\left( {h - 2} \right)^2} + {\left( {k - 3} \right)^2} \Rightarrow 2h + 4k = 11$
${x^2} + {y^2} + 2gx + 2fy + c + \lambda \left( {x - y + 1} \right) = 0$, represent the equation of the circle touching the line x-y+1=0 and passing through (2,3) and (1,1)
${x^2} + {y^2} + 2x\left( {g + \frac{\lambda }{2}} \right) + 2y\left( {f - \frac{\lambda }{2}} \right) + c + \lambda = 0$
$Centre:\left( { - \left( {g - \frac{\lambda }{2}} \right), - \left( {f - \frac{\lambda }{2}} \right)} \right)$
Putting it in 2h+4k=1
$ - 2\left( {g - \frac{\lambda }{2}} \right) - 4\left( {f - \frac{\lambda }{2}} \right) = 11 \Rightarrow - 2g - 4f + \lambda + 2\lambda = 11 \Rightarrow 2g + 4f = 3\lambda - 11$
By Hit and Trial I am getting choice (B) as it satisfies $2h+4k=11$ where $h=\frac{9}{2}$ and $k=\frac{1}{2}$.
Had the option not been given how do I proceed
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3917805",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 0
}
|
Prove inverse trigonometric equation $2\tan^{-1}2=\pi-\cos^{-1}\frac{3}{5}$ The question is:
Prove that: $2tan^{-1}2=\pi-cos^{-1}\frac{3}{5}$ Hint: use the fact that $tan(\pi-x)=-tanx$
I did the question without using the Hint, but I don't know how to do it using the hint.
Quick working out of what I've done:
\begin{aligned}
\text { If } \theta &=\tan ^{-1} 2 \\
\tan \theta &=2 \\
0 & < \theta < \frac{\pi}{2}
\end{aligned}
\begin{aligned}
\cos 2 \theta &=2 \cos ^{2} \theta-1 \\
&=2 \times\left(\frac{2}{\sqrt{5}}\right)^{2}-1 \\
&=\frac{3}{5} \\
2 \theta =& \cos ^{-1} \frac{3}{5}, \quad \text { since } 0 < 2\theta < \pi
\end{aligned}
\begin{array}{l}
2 \tan ^{-1} 2=\cos ^{-1} \frac{3}{5} \text { . } \\
\text { Note: } \cos ^{-1} x \text { has point symmetry } \\
\text { in }\left(0, \frac{\pi}{2}\right) \text { . }
\end{array}
$$
\begin{array}{l}
\cos ^{-1} x+\cos ^{-1}(-x)=\pi \\
\cos ^{-1} \frac{3}{5}=\pi-\cos ^{-1}\left(-\frac{3}{5}\right) \\
\therefore \quad 2 \tan ^{-1} 2=\pi-\cos ^{-1}\left(-\frac{3}{5}\right)
\end{array}
$$
But I didn't use the Hint given in the question for this working out. How do I use the hint? Thank you !
|
Hint:
Using Inverse trigonometric function identity doubt: $\tan^{-1}x+\tan^{-1}y =-\pi+\tan^{-1}\left(\frac{x+y}{1-xy}\right)$, when $x<0$, $y<0$, and $xy>1$
$$2\tan^{-1}2=\pi+\tan^{-1}\dfrac{2\cdot2}{1-2^2}=\pi+\tan^{-1}\left(?\right)$$
Now using Principal values $\tan^{-1}(-y)=-\tan^{-1}y$
Again, if $\tan^{-1}\dfrac43=u,0<u<\dfrac\pi2$ as $\dfrac43>0$
$\cos u=\dfrac1{+\sqrt{1+\tan^2y}}=?$
Can you take it home from here?
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3918968",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 4
}
|
What is the relationship between $\frac{1}{3}x_1+\frac{1}{3}x_2+\frac{1}{3}x_3$ and $x_1^{\frac{1}{3}}x_2^{\frac{1}{3}}x_3^{\frac{1}{3}}$? Is $\frac{1}{3}x_1+\frac{1}{3}x_2+\frac{1}{3}x_3$ > or < or = $x_1^{\frac{1}{3}}x_2^{\frac{1}{3}}x_3^{\frac{1}{3}}$, given $x_1$, $x_2$ and $x_3$ are all positive?
I know there exists a real number $a\in(1,2)$ such that $\frac{1}{3}x=x^{\frac {1}{3}}$, but there are three variables even though they are all positive.
Also I know how to simplify $\frac{1}{3}x_1+\frac{1}{3}x_2+\frac{1}{3}x_3-x_1^{\frac{1}{3}}x_2^{\frac{1}{3}}x_3^{\frac{1}{3}}$, but then I get stuck too.
I also tried to relate this to the roots of polynomial, but I don't quite know how.
Could anyone help please?
|
As you said, $x_1$,$x_2$,$x_3$ are positive reals. Furthermore, by AM-GM inequality $\frac{x_1 + x_2 + x_3}{3} \geq (x_1)^\frac{1}{3}(x_2)^\frac{1}{3}(x_3)^\frac{1}{3}$
so the proof is using just AM-GM inequality.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3921922",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
}
|
Find scale, rotation, and translation that maps three points to three points in 3D Given three source points $\mathbf{P}_1 = [x_1\ y_1\ z_1]^T,$
$\mathbf{P}_2 = [x_2\ y_2\ z_2]^T,$ $\mathbf{P}_3 = [x_3\ y_3\ z_3]^T,$ and
three destination points $\mathbf{P}'_1 = [x'_1\ y'_1\ z'_1]^T,$
$\mathbf{P}'_2 = [x'_2\ y'_2\ z'_2]^T,$ $\mathbf{P'}_3 = [x'_3\ y'_3\ z'_3]^T,$
how do we find the scale $S,$ rotation $R,$ and translation $T$ that transforms
the source points into the destination points as in the following equation?
$$
\left[\begin{array}{ccc}
s_x & & \\
& s_y & \\
& & s_z \\
\end{array}\right]
\left[\begin{array}{cccc}
r_{11} & r_{12} & r_{13} & t_x \\
r_{21} & r_{22} & r_{23} & t_y \\
r_{31} & r_{32} & r_{33} & t_z
\end{array}\right]
\left[\begin{array}{ccc}
x_1 & x_2 & x_3 \\
y_1 & y_2 & y_3 \\
z_1 & z_2 & z_3 \\
1 & 1 & 1
\end{array}\right]
=
\left[\begin{array}{ccc}
x'_1 & x'_2 & x'_3 \\
y'_1 & y'_2 & y'_3 \\
z'_1 & z'_2 & z'_3
\end{array}\right]
$$
Since the $3 \times 3$ rotation $R$ is orthonormal and can be parameterized by three rotation
angles (e.g. yaw, pitch, roll) there are 9 equations and 9 unknowns. Ideally the scale would be uniform $(s_x = s_y = s_z)$ resulting in a similarity transformation, but then the system would be over constrained. As long as the source and destination points are not co-linear, this should have a unique solution.
One could solve for a general affine transformation $A$ and decompose this, but this would require 4 non-coplanar source and destination points and may include shears.
EDIT: A possible approach is to solve for the general affine case
$X A = B$
$$
\left[\begin{array}{cccc}
x_1 & y_1 & z_1 & 1\\
x_2 & y_2 & z_2 & 1\\
x_3 & y_3 & z_3 & 1
\end{array}\right]
\left[\begin{array}{ccc}
a_{11} & a_{21} & a_{31} \\
a_{12} & a_{22} & a_{32} \\
a_{13} & a_{23} & a_{33} \\
a_{14} & a_{24} & a_{34} \\
\end{array}\right]
=
\left[\begin{array}{ccc}
x'_1 & y'_1 & z'_1 \\
x'_2 & y'_2 & z'_2 \\
x'_3 & y'_3 & z'_3
\end{array}\right]
$$
which is under-constrained since there are 9 equations and 12 unknowns which means there are infinite solutions.
Therefore, following the recipe here, we will find the solution $A'$ that has the minimum norm $\|A\|.$
The normal equation for an underdetermined system is
$$
A^* = X^T (X X^T)^{-1} B
$$
from which we can get the optimal value via SVD.
The SVD for $X$ is
$$
\begin{eqnarray}
X &=& U \Sigma V^T \\
&=&
\left[\begin{array}{ccc}
\cdot & \cdot & \cdot \\
\cdot & \cdot & \cdot \\
\cdot & \cdot & \cdot
\end{array}\right]
\left[\begin{array}{cccc}
\sigma_1 & & & \cdot \\
& \sigma_2 & & \cdot \\
& & \sigma_3 & \cdot
\end{array}\right]
\left[\begin{array}{cccc}
\cdot & \cdot & \cdot & \cdot \\
\cdot & \cdot & \cdot & \cdot\\
\cdot & \cdot & \cdot & \cdot\\
\cdot & \cdot & \cdot & \cdot
\end{array}\right]^T
\end{eqnarray}
$$
where $\sigma_1 \geq \sigma_2 \geq \sigma_3 \geq 0$ are the
singular values (if $\sigma_3 = 0$ then the source points are
co-linear, i.e., $X$'s rank is less than 3) and $U$ and $V$ are
orthonormal matrices.
Plugging this into our normal equation we have
$$
\begin{eqnarray}
A^* &=& V \Sigma^T (\Sigma \Sigma^T )^{-1} U^T B \\
&=&
V
\left[\begin{array}{ccc}
\frac{1}{\sigma_1} & & \\
& \frac{1}{\sigma_2} & & \\
& & \frac{1}{\sigma_3} \\
\cdot & \cdot & \cdot
\end{array}\right]
U^T B
\end{eqnarray}
$$
where the matrix above containing the reciprocals of the singular values
is the pseudo-inverse of $\Sigma.$
The hope is the solution $A^*$ which the minimal norm from the set of all solutions will be a similar transformation. Yet to be verified.
|
The assumption is that this can be accomplished with a similarity transformation.
We use the following sequence of transformations to compose the desired transformation:
*
*Translate $\mathbf{P}_0$ to the origin;
*Orient $\triangle \mathbf{P}_0 \mathbf{P}_1 \mathbf{P}_2$
to align with $\triangle \mathbf{P}'_0 \mathbf{P}'_1 \mathbf{P}'_2$ via rotation $R;$
*Uniform scale by
$s = \frac{\|\mathbf{P}'_1 - \mathbf{P}'_0\|}{\|\mathbf{P}_1 - \mathbf{P}_0\|}$;
*Translate to $\mathbf{P'}_0$
$$
\begin{eqnarray}
S &=&
\left[\begin{array}{cccc}
1 & 0 & 0 & x_0' \\
0 & 1 & 0 & y_0' \\
0 & 0 & 1 & z_0' \\
0 & 0 & 0 & 1
\end{array}\right]
\left[\begin{array}{cccc}
s & 0 & 0 & 0 \\
0 & s & 0 & 0 \\
0 & 0 & s & 0 \\
0 & 0 & 0 & 1
\end{array}\right]
\left[\begin{array}{cccc}
r_{11} & r_{12} & r_{13} & 0 \\
r_{21} & r_{32} & r_{23} & 0 \\
r_{31} & r_{32} & r_{33} & 0 \\
0 & 0 & 0 & 1
\end{array}\right]
\left[\begin{array}{cccc}
1 & 0 & 0 & -x_0 \\
0 & 1 & 0 & -y_0 \\
0 & 0 & 1 & -z_0 \\
0 & 0 & 0 & 1
\end{array}\right] \\
&=&
%\left[sI\ | \mathbf{P}'_0 \right] \left[R\ | -R \mathbf{P}_0 \right]
\left[\begin{array}{cccc}
s & 0 & 0 & x_0' \\
0 & s & 0 & y_0' \\
0 & 0 & s & z_0' \\
0 & 0 & 0 & 1
\end{array}\right]
\left[\begin{array}{cccc}
r_{11} & r_{12} & r_{13} & -x''_0 \\
r_{21} & r_{32} & r_{23} & -y''_0 \\
r_{31} & r_{32} & r_{33} & -z''_0 \\
0 & 0 & 0 & 1
\end{array}\right]\ \ \ \ \mbox{(let $\mathbf{P}''_0 = R \mathbf{P}_0)$} \\
&=&
\left[\begin{array}{cccc}
s r_{11} & s r_{12} & s r_{13} & x'_0 - s x''_0 \\
s r_{21} & s r_{32} & s r_{23} & y'_0 - s y''_0 \\
s r_{31} & s r_{32} & s r_{33} & z'_0 - s z''_0 \\
0 & 0 & 0 & 1
\end{array}\right]
\end{eqnarray}
$$
To find the rotation $R$ we align the edge $\overline{\mathbf{P}_0 \mathbf{P}_1}$ with the edge $\overline{\mathbf{P}'_0 \mathbf{P}'_1}$ and align the plane of
$\triangle \mathbf{P}_0 \mathbf{P}_1 \mathbf{P}_2$ with the plane of
$\triangle \mathbf{P}'_0 \mathbf{P}'_1 \mathbf{P}'_2.$
We define the source orthonormal coordinate axes as
$$
\begin{eqnarray}
\mathbf{X} &=& \frac{\mathbf{P}_1 - \mathbf{P}_0}{\|\mathbf{P}_1 - \mathbf{P}_0 \|}\\
\mathbf{Z} &=& \frac{(\mathbf{P}_2 - \mathbf{P}_0) \times \mathbf{X}}{\| (\mathbf{P}_2 - \mathbf{P}_0) \times \mathbf{X}\|} \\
\mathbf{Y} &=& \mathbf{X} \times \mathbf{Z}
\end{eqnarray}
$$
and the target coordinate axes as
$$
\begin{eqnarray}
\mathbf{X}' &=& \frac{\mathbf{P}'_1 - \mathbf{P}'_0}{\|\mathbf{P}'_1 - \mathbf{P}'_0 \|}\\
\mathbf{Z}' &=& \frac{(\mathbf{P}'_2 - \mathbf{P}'_0) \times \mathbf{X}'}{\| (\mathbf{P}'_2 - \mathbf{P}'_0) \times \mathbf{X}'\|} \\
\mathbf{Y}' &=& \mathbf{X}' \times \mathbf{Z}'
\end{eqnarray}
$$
We are looking for the rotation $R$ that maps the source coordinate axes
into the target axes:
$$
\begin{eqnarray}
R \left[ \mathbf{X}\ \mathbf{Y}\ \mathbf{Z} \right]
&=&
\left[ \mathbf{X}'\ \mathbf{Y}'\ \mathbf{Z}' \right] \\
R &=& \left[ \mathbf{X}'\ \mathbf{Y}'\ \mathbf{Z}' \right]
\left[ \mathbf{X}\ \mathbf{Y}\ \mathbf{Z} \right]^T %\\
% &=& \left[ \mathbf{X}'\ \mathbf{Y}'\ \mathbf{Z}' \right]
% \left[ \begin{array}{c}
% \mathbf{X}^T \\
% \mathbf{Y}^T \\
% \mathbf{Z}^T
% \end{array}
% \right]
\end{eqnarray}
$$
So we have a uniform scale $s,$ rotation $R,$ and translation
$\mathbf{P}'_0 - s R \mathbf{P}_0.$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3922403",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
Is this a correct way to show that $\sum_{n \geq 0} \frac{n^3}{n!}=5e$ Is this a correct way to show that $\sum_{n \geq 0} \frac{n^3}{n!}=5e$ ?
$$S_3 = \sum_{n \geq 0} \frac{n^3}{n!}=\sum_{n \geq 1} \frac{n^2}{(n-1)!} \implies$$
$$S'_3=S_3-e=\sum_{n \geq 1} \frac{n^2-1^2}{(n-1)!}=\sum_{n \geq 2} \frac{n+1}{(n-2)!} \implies$$
$$S'_3-3e=\sum_{n \geq 2} \frac{(n+1)-3}{(n-2)!}=\sum_{n \geq 3} \frac{1}{(n-3)!}=e\implies$$
$$S'_3-3e=e\iff S_3=5e$$
Exploring $ \sum_{n=0}^\infty \frac{n^p}{n!} = B_pe$, particularly $p = 2$. : One of the answers shows how $\sum_{n \geq 0} \frac{n^2}{n!}=2e$ and asserts that $\sum_{n \geq 0} \frac{n^3}{n!}=5e$ can be showed in the same manner of reasoning. Is my "proof" correct?
|
Seems good to me. An alternate way - which is slightly neater - would be to first write $n^3$ in a convenient way:
\begin{align} n^3 &= n^2(n-1) + n^2 \\
&= n(n-1)(n-2) + 2n(n-1) + n^2 \\
&= n(n-1)(n-2) + 2n(n-1) + n(n-1) + n.\end{align}
Then, since all the sums involved converge, \begin{align} \sum_{n \ge 0} \frac{n^3}{n!} &= \sum_{n \ge 0} \frac{n(n-1)(n-2)}{n!} + \frac{3n(n-1)}{n!} + \frac{n}{n!} \\ &= \sum_{n\ge 3} \frac{1}{(n-3)!} + \sum_{n \ge 2} \frac{3}{(n-2)!} + \sum_{n \ge 1} \frac{1}{(n-1)!} = 5e.\end{align}
Expressions such as $n(n-1)(n-2)$ are called falling factorials:
$(n)_k = n (n-1) \dots(n-k+1)$. These are sometimes also denoted $n^{\underline{k}}$. There is a standard way to write powers as falling factorials: $$ n^k = \sum_{i = 0}^k \begin{Bmatrix} n \\ i\end{Bmatrix} (n)_i, $$ where $\begin{Bmatrix} n \\ i\end{Bmatrix}$ are the Stirling numbers of the second kind. Using this, and the observation that $\frac{(n)_i}{n!} = \frac{1}{(n-i)!},$ you can generalise the above decomposition to any $k$ instead of $3$, and derive a similar result, that $$ \sum_{n \ge 0} \frac{n^k}{n!} = \left( \sum_{i = 0}^k \begin{Bmatrix} n \\ i\end{Bmatrix}\right) e = B_k e,$$ where $B_k$ are the Bell numbers, that are simply the sum in the parenthesis above.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3925523",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
}
|
On the convergence of improper integrals Question
Determine the value of $r$ for which the integral $$\int^{\infty}_1 (\frac r {x + 1} - \frac {3x} {2x^2 + r})\ \mathrm {d}x$$ is convergent and evaluate the integral for that value of $r$.
My working
It is trivial to see that the integral is divergent for $r = 0\ $, so let $r \neq 0\ $.
If $r \neq 0\ $, then
\begin{align}
\int^{\infty}_1 (\frac r {x + 1} - \frac {3x} {2x^2 + r})\ \mathrm {d}x & =
\lim\limits_{a\to\infty}[r\ln(x + 1) - \frac 3 4 \ln|2x^2 + r|]^{x = a}_{x = 1}
\\[5 mm] & =
\lim\limits_{a\to\infty}[r\ln(a + 1) - \frac 3 4 \ln|2a^2 + r| - r\ln 2 + \frac 3 4 \ln|2 + r|]
\\[5 mm] & =
\lim\limits_{a\to\infty}\left(r\ln\frac {a + 1} 2 + \frac 3 4 \ln\left|\frac {2 + r} {2a^2 + r}\right|\right)
\end{align}
This question just appeared in my final calculus examination this afternoon and I got stuck here ):. We have practiced such questions before but I cannot seem to get this one. I would usually proceed by trying L'Hopital's Rule since we want the limit to exist. I know I probably will not be touching calculus for quite a while from now, but I still want to figure this out. Any intuitions as to how I should proceed will be greatly appreciated :)
|
I have finally managed to answer the question :)
It is trivial to see that the integral is divergent for $r = 0\ $, so let $r \neq 0$.
If $r \neq 0$, note that we can always choose $a$ large enough such that $$\int^{\infty}_1 (\frac r {x + 1} - \frac {3x} {2x^2 + r})\ \mathrm {d}x = \int^a_1 (\frac r {x + 1} - \frac {3x} {2x^2 + r})\ \mathrm {d}x + \int^{\infty}_a (\frac r {x + 1} - \frac {3x} {2x^2 + r})\ \mathrm {d}x,$$ where $$\int^a_1 (\frac r {x + 1} - \frac {3x} {2x^2 + r})\ \mathrm {d}x$$ is always convergent $\forall\ r \in \mathbb {R}$.
Then, the problem reduces to checking the convergence of $$\int^{\infty}_a (\frac r {x + 1} - \frac {3x} {2x^2 + r})\ \mathrm {d}x,$$ where
\begin{align}
\int^{\infty}_a (\frac r {x + 1} - \frac {3x} {2x^2 + r})\ \mathrm {d}x & =
\lim\limits_{b\to\infty}[r\ln(x + 1) - \frac 3 4 \ln(2x^2 + r)]^{x = b}_{x = a}
\\[5 mm] & =
\lim\limits_{b\to\infty}[r\ln(b + 1) - \frac 3 4 \ln(2b^2 + r) - r\ln (a + 1) + \frac 3 4 \ln(2a^2 + r)]
\\[5 mm] & =
\ln \left[\lim\limits_{b\to\infty}\frac {(b + 1)^r} {(2b^2 + r)^{\frac 3 4}}\right] - r\ln (a + 1) + \frac 3 4 \ln(2a^2 + r)
\\[5 mm] & =
\ln \left[\lim\limits_{b\to\infty}\frac {b^{r - \frac 3 2}(1 + \frac 1 b)^r} {(2 + \frac r {b^2})^{\frac 3 4}}\right] - r\ln (a + 1) + \frac 3 4 \ln(2a^2 + r).
\end{align}
Observe that, due to the condition imposed on $a$, the last two terms, namely, $- r\ln (a + 1)$ and $\frac 3 4 \ln (2a^2 + r)$ always exist. Moreover, as $b \rightarrow \infty$, $(1 + \frac 1 b)^r$ and $(2 + \frac r {b^2})^{\frac 3 4}$ are always convergent, so it suffices to check the convergence of $\lim\limits_{b\to\infty}b^{r - \frac 3 2}$ and it is easy to see that $r$ can only take on one value, that is $\frac 3 2$.
Thus, when $r = \frac 3 2$,
\begin{align}
\int^{\infty}_1 (\frac r {x + 1} - \frac {3x} {2x^2 + r})\ \mathrm {d}x & =
\lim\limits_{c\to\infty}[\frac 3 2\ln(x + 1) - \frac 3 4 \ln(2x^2 + \frac 3 2)]^{x = c}_{x = 1}
\\[5 mm] & =
\lim\limits_{c\to\infty}[\frac 3 2\ln(c + 1) - \frac 3 4 \ln\frac {4c^2 + 3} 2 - \frac 3 2\ln 2 + \frac 3 4 \ln\frac 7 2]
\\[5 mm] & =
\frac 3 4\lim\limits_{c\to\infty}[\ln(c + 1)^2 - \ln(4c^2 + 3)] - \frac 3 2\ln 2 + \frac 3 4 \ln 7
\\[5 mm] & =
\frac 3 4\lim\limits_{c\to\infty}\ln\frac {(c + 1)^2} {4c^2 + 3} - \frac 3 2\ln 2 + \frac 3 4 \ln 7
\\[5 mm] & =
\frac 3 4\ln\frac 1 4 - \frac 3 2\ln 2 + \frac 3 4 \ln 7
\\[5 mm] & =
\frac 3 4 \ln 7 - 3\ln 2
\end{align}
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3925912",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
}
|
To derive quantity $k=\frac{2-p^2}{p+2}$ for positive rational $p$ such that $p^2>2$ and $(p+k)^2>2$ I am reading Baby Rudin and stuck at one small point.
$A$ is set of all positive rationals $p$ such that $p^2<2$ and $B$ is set of all positive rationals $p$ such that $p^2>2$. We wanted to show $A$ contains no largest element and $B$ contains no smallest element. So for any such $p$ in either set, author associated number $q$ as,
$q=p-\frac{p^2-2}{p+2}$, which works fine.
I was trying to figure out how to obtain the quantity $\frac{p^2-2}{p+2}$.
I did it for set $A$ as follows.
So, we have positive rational $p$ such that $p^2<2$. We want some $k>0$ such that $(p+k)^2 <2$
i.e. $p^2+2pk+k^2<2$
As all three quantities are positive, we can say
$p^2+2pk<2$, which gives
$k<\frac{2-p^2}{2p}$.
But as $p^2<2$, we have $p<2 \Rightarrow p+p=2p<p+2$
Therefore, we can choose $k=\frac{2-p^2}{p+2}$.
But I am stuck while deriving it for set $B$.
I did following.
So, we have positive rational $p$ such that $p^2>2$. We want some $k<0$ such that $(p+k)^2 >2$
i.e. $p^2+2pk+k^2>2$.
From this point, I can say $p^2+k^2>2$, but not sure how this will take me to the desired result.
Any hint or help. Thank you.
|
We get $$(p+k)^2 >2\implies p+k\gt\sqrt 2\implies k\gt \sqrt 2-p$$
Multiplying $\sqrt 2-p$ by $\dfrac{\sqrt 2+p}{\sqrt 2+p}\ (=1)$ gives $\dfrac{2-p^2}{\sqrt 2+p}$, and so we have
$$k\gt \dfrac{2-p^2}{\sqrt 2+p}$$
It follows from $2\gt \sqrt 2$ and $2-p^2\lt 0$ that $\dfrac{2-p^2}{2+p}\gt\dfrac{2-p^2}{\sqrt 2+p}$.
Therefore, we can choose $k=\dfrac{2-p^2}{2+p}$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3926373",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
}
|
How to prove the inequality $abc(a+b+c)^2≤(a^3+b^3+c^3)(ab+bc+ca)$? I need to prove something like that:
For $a,b,c>0$ prove: $abc(a+b+c)^2≤(a^3+b^3+c^3)(ab+bc+ca)$.
I know that $3abc≤(a^3+b^3+c^3)$, but then I derived $3(ab+bc+ca) ≤ (a+b+c)^2$, I can't move on.
Can anyone help me?
|
Hint: Using CS inequality,
$$(a^3+b^3+c^3)\left(\frac1a+\frac1b+\frac1c\right)\geqslant (a+b+c)^2$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3926463",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
}
|
Why $\int{\frac{dr}{r\sqrt{c^{2}r^4 -1}}} = -\frac{1}{2} \arcsin\left(\frac{1}{cr^2}\right)$? I must solve the following integral, where $c$ is a constant $$\int{\frac{dr}{r\sqrt{c^{2}r^4 -1}}}$$ When I compute and do the calculations, I obtain that $$\int{\frac{dr}{r\sqrt{c^{2}r^4 -1}}} = \frac{1}{2} \arctan \left(\sqrt{c^{2}r^4 -1} \right)$$ I have rectified my calculations many times and I always get the same result, but in the book I'm studying I have that $$\int{\frac{dr}{r\sqrt{c^{2}r^4 -1}}} = -\frac{1}{2} \arcsin\left(\frac{1}{cr^2}\right)$$ Why? Is it some mistake in the book or is it my own mistake? The book I mention is: The calculus of variations by Van Brunt page 47, there the equation arises through an example on the invariance of the Eule-Lagrange equation.
Attached images:
|
In fact
$$ \tan\bigg(\cos^{-1}\bigg(\frac{1}{cr^2}\bigg)\bigg)=\frac{\sin\bigg(\cos^{-1}\bigg(\frac{1}{cr^2}\bigg)\bigg)}{\cos\bigg(\cos^{-1}\bigg(\frac{1}{cr^2}\bigg)\bigg)}=\frac{\sqrt{1-\frac1{c^2r^4}}}{\frac{1}{cr^2}}=\sqrt{c^2r^4-1}. $$
Since $\frac{1}{cr^2}>0$, $\cos^{-1}\bigg(\frac{1}{cr^2}\bigg)\in(0,\pi/2)$ and hence
$$ \cos^{-1}\bigg(\frac{1}{cr^2}\bigg)\bigg)=\tan^{-1}\sqrt{c^2r^4-1}. $$
Using
$$ \sin^{-1} x+\cos^{-1} x=\frac{\pi}{2} $$
and hence
$$ -\sin^{-1}x=\cos^{-1}x-\frac{\pi}{2}. $$
one has
$$ -\sin^{-1}\bigg(\frac{1}{cr^2}\bigg)=\cos^{-1}\bigg(\frac{1}{cr^2}\bigg)-\frac{\pi}{2}=\tan^{-1}\bigg(\sqrt{c^2r^4-1}\bigg)-\frac{\pi}{2}. $$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3928172",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
What is a simple formula that can create the patterns $+1 +2 +3 +4$, or $+1 +3 +5 +7$, or $+1 +4 + 7 +10$, etc. Originally I was looking for some simpler more intuitive way to appreciate the value of squaring something. I looked for how much it was increasing by every time I add $+1$ to the size of what was being squared. so in other words $3$ squared is $+5$ more than $2$ squared, and $4$ squared is $+7$ more than $3$ squared, and $5$ squared is $9$ more than $4$ squared. You can see that whenever you increase the number being squared by $1$, the amount increased itself increases by $+2$ more than the previous increase. ie, the increase itself increases by a rate of $+ 2$.
So
$$- 1^2 - 0^2 = +1$$
$$- 2^2 - 1^2 = +3$$
$$- 3^2 - 2^2 = +5$$
$$- 4^2 - 3^2 = +7$$
$$- 5^2 - 4^2 = +9$$
and
*
*the difference between the $+3$ and $+1$ is $2$
*the difference between the $+5$ and $+3$ is $2$
*the difference between the $+7$ and $+5$ is $2$
*the difference between the $+9$ and $+7$ is $2$ etc...
So I say, it's "increase bonus" is at a rate of +2. That's definitely not a correct mathematical term, but in my ignorance that's what i'm calling it.
This struck me as very interesting, because I used to make little wc3 mods as a kid and they gave tools that made same pattern. What I became interested in, was what if I wanted it to increase by a pattern of say +1, +2, +3, +4, +5. (for an increase bonus of +1). Or what if I wanted it to be +1, +4, +7, +10, +13 (an increase bonus of +3)?
My main question is, is there a formula where I can easily change the "increase bonus" to whatever rate I want? While I can do this manually, I don't know how to represent it as a simple, calculable formula.
My side question is... what is this formula / branch of mathematics actually officially called?
|
A polynomial of degree $n$ will have the $n$th finite difference equal to $n!$ times the leading coefficient of the polynomial. In your example, you have
\begin{array}{|c|c|c|c|}
\hline
x & x^2 & \text{1st difference} & \text{2nd difference} \\
\hline
1 & 1 & & \\
2 & 4 & 3 & \\
3 & 9 & 5 & 2 \\
4 & 16 & 7 & 2 \\
5 & 25 & 9 & 2 \\
\hline
\end{array}
If you want the bonus to be $+3$ instead, you can use $\frac32x^2$ instead of $x^2$:
\begin{array}{|c|c|c|c|}
\hline
x & \frac32x^2 & \text{1st difference} & \text{2nd difference} \\
\hline
1 & \frac32 & & \\
2 & 6 & \frac92 & \\
3 & \frac{27}2 & \frac{15}2 & 3 \\
4 & 24 & \frac{21}2 & 3 \\
5 & \frac{75}2 & \frac{27}2 & 3 \\
\hline
\end{array}
Or, if you want instead the third finite difference to be $+3$, you can use $\frac12x^3:$
\begin{array}{|c|c|c|c|c|}
\hline
x & \frac12x^3 & \text{1st difference} & \text{2nd difference} & \text{3rd difference}\\
\hline
1 & \frac12 & & & \\
2 & 4 & \frac72 & & \\
3 & \frac{27}2 & \frac{19}2 & 6 & \\
4 & 32 & \frac{37}2 & 9 & 3\\
5 & \frac{125}2 & \frac{61}2 & 12 & 3\\
\hline
\end{array}
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3934864",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
Number Theory : Solve the system of congruence $28x+17y\equiv 18 \pmod{41}$ and $31x+11y\equiv 35\pmod{41}$ Number Theory : Solve the system of congruence
(1) $28x+17y\equiv 18 \pmod{41}$
(2) $31x+11y\equiv 35\pmod{41}$
Attempt :
we know that equation (1) and (2) are in the same$\pmod{41}$. so we can use Modular arithmetic
lets multiply the first equation (1) by $31$ and the second equation(2) by $28$.
(1) $31\cdot(28x+17y)\equiv 31\cdot18 \pmod{41}$
(1) $868x+527y\equiv 558\pmod{41}$
(2) $28\cdot(31x+11y)\equiv 28\cdot35\pmod{41}$
(2) $868x+308y\equiv 980\pmod{41}$
So finally we can subtract equation (1) from (2) we get:
$219y\equiv -422\pmod{41}$
lets check the $\gcd(219,41)$ by Euclidian algorithm :
$219 = 41\cdot 5 + 14$
$41= 14\cdot 2 + 13$
$14= 13\cdot 1 + 1$
$\gcd(219,41)=1$
Hence, because the $\gcd$ is equal to $1$ we can find the Inverse and multiply the equation by the Inverse to find $y$.
$219y\equiv 1\pmod{41}$
$219a = 1+41k$ , the $41k$ must end with the digit of $8$ for $1+$digit $8$ will be $9$ so $k$ must be multiply of number with end digit of $8$.
I don't know how to continue from here .
|
Alternatively, use linear algebra. The system is
$$
\begin{pmatrix} 28 & 17 \\ 31 & 11 \end{pmatrix}
\begin{pmatrix} x \\ y\end{pmatrix}
=
\begin{pmatrix} 18 \\ 35 \end{pmatrix}
$$
The inverse of the matrix is
$$
-\frac{1}{219}\begin{pmatrix} \hphantom-11 & -17 \\ -31 & \hphantom-28 \end{pmatrix}
$$
It remains to find the inverse of $219$ mod $41$, which is $3$ by the extended Euclidean algorithm.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3935020",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 5,
"answer_id": 1
}
|
Simplifying $\log_x 25 - \left(\frac{\log_5 x}{4} - 2\log_5 x\right)$ My teacher gave me the following expression to change into a single expression:
$$\log_x 25 - \left(\frac{\log_5 x}{4} - 2\log_5 x\right)$$
This is what I got after working it out:
$$\log_x 25 - \log_5 \left(\frac{x}{4x^2}\right)$$
|
$$\begin{align}
& \log_{x}25 - \left(\dfrac{\log_{5}x}{4} - 2\log_{5}x\right) \\
= & \log_{x}5^2 + \dfrac{7}{4}\log_{5}x \\
= & 2\log_{x}5 + \dfrac{7}{4}\log_{5}x \\
= & \dfrac{2}{\log_{5}x} + \dfrac{7}{4}\log_{5}x \\
= & \dfrac{7\log_{5}x + 8}{4\log_{5}x} \\
= & \dfrac{\log_{5}x^7 + \log_{5}5^8}{\log_{5}x^4} \\
= & \dfrac{\log_{5}(5^8 \cdot x^7)}{\log_{5}x^4} \\
= & \log_{x^4}(5^8 \cdot x^7) \\
\end{align}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3936543",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
}
|
Determine the number of three-digit numbers that are divisible by 3 Determine the number of three-digit numbers that are divisible by 3 and that you can form with 3 different numbers among 1, 2, 3, 5, 6, 8
I found 5 numbers or 5 different ways with a sum that is divisible by 3. And each of them has 6 combinations because 3* 2* 1=6 so the total number should be 5*3=30 three-digit numbers that are divisible by 3. But I'm not sure if it's right or if there's a better solution.
|
$$\begin{align}2,5,8&\equiv -1\pmod 3\\3,6&\equiv 0\pmod 3\\1&\equiv 1\pmod 3\end{align}$$
There are $3\times 2\times 1=6$ ways to select one number from each group and $1$ way to select three numbers from a single group, for a total of $7$ combinations.
There are $3!=6$ permutations of each combination, giving a total of $7\times 6=42$ three-digit numbers.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3938180",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
show that $\frac{a}{b} + \frac{b}{c} + \frac{c}{a} \geq \frac{1}{a} + \frac{1}{b} + \frac{1}{c}$ if $abc \geq 1 $ We want to show that $\frac{a}{b} + \frac{b}{c} + \frac{c}{a} \geq \frac{1}{a} + \frac{1}{b} + \frac{1}{c}$ if $abc \geq 1 $
I've tried to use AM-GM directly on the LHS but that obviously failed.
Simplifying the question gives $a^2b + b^2c + c^2a \geq ab + bc + ac$
The fact that this is $a(ab) + b(bc) + c(ca) \geq ab + bc + ac$ makes me think that there's a way to proceed from here, but I'm not quite sure how.
Note : a,b,c are real positive numbers
|
$$\begin{align}\frac{a}{b} + \frac{b}{c} + \frac{c}{a}
&\geq \frac{(abc)^{1/3}}{a} + \frac{(abc)^{1/3}}{b} + \frac{(abc)^{1/3}}{c}\\
&\geq \frac{1}{a} + \frac{1}{b} + \frac{1}{c}
\end{align}$$
The second inequality is obvious while the first one comes from Karamata's Majorization Inequality with the convexity of $f(x)=e^x$ and
$$\left(\ln\dfrac{(abc)^{1/3}}{c}, \ln\dfrac{(abc)^{1/3}}{b}, \ln\dfrac{(abc)^{1/3}}{a}\right) \prec \left(\ln\dfrac{a}{b}, \ln\dfrac{b}{c}, \ln\dfrac{c}{a}\right)
$$
Let (WLOG) $\;\ln\dfrac{a}{b}\ge\ln\dfrac{b}{c}\ge\ln\dfrac{c}{a}$
We have $\;\dfrac{a}{b}\ge\dfrac{b}{c}\ge\dfrac{c}{a}$ thus, $\;ac\ge b^2\;$, $a^2\ge bc\;$ and $\;ab\ge c^2\;$ hence, we get $a\ge b$ and $a\ge c$ thus,
$$\dfrac{a}{b}\ge \dfrac{(abc)^{1/3}}{c} \implies \ln\dfrac{a}{b}\ge \ln\dfrac{(abc)^{1/3}}{c}$$
$$\dfrac{c}{a}\le \dfrac{(abc)^{1/3}}{a} \implies \ln\dfrac{c}{a}\le \ln\dfrac{(abc)^{1/3}}{a}$$
$$0=\ln\dfrac{(abc)^{1/3}}{c}+ \ln\dfrac{(abc)^{1/3}}{b}+ \ln\dfrac{(abc)^{1/3}}{a}=
\ln\dfrac{a}{b}+ \ln\dfrac{b}{c}+ \ln\dfrac{c}{a}
$$
so that our majorization rule holds, hence first inequality holds.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3939742",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
}
|
Integrating $\int{\frac{x\,dx}{x^2 + 3x -4}}$ I saw this "Beat the Integral" problem and wanted to be sure I was approaching it correctly.
The integral is $$\int{\frac{x}{x^2 + 3x -4}dx}$$
So I decide I want to decompose this, because it's a degree-2 polynomial in the denominator and a degree-1 in the numerator, so that'd be a go-to. Also, since the degree of th enumerator is one less than the denominator we can maybe treat this as $du/u$ which would imply a natural log, though we don't quite know that yet.
So we decompose: I know that $x^2+3x-4$ can be factored as $(x-1)(x+4)$.
That gets me
$$\int{\frac{x}{(x-1)(x+4)}dx}$$
so I can do this:
$$\frac{Ax}{x-1}+\frac{B}{x+4} = \frac{x}{(x-1)(x+4)}$$
Which implies
$$A(x+4)+B(x-1) = x$$
and since the roots are at $x=1$ and $x=-4$, I can set it up like this:
$$5A = 1;-5B=1 \text{ and } A = \frac{1}{5}, B=-\frac{1}{5}$$
Leading to:
$$\frac{x}{5(x-1)}-\frac{1}{5(x+4)}$$
Which I can set up the integral like so:
$$\int{\frac{x}{5(x-1)}-\frac{1}{5(x+4)}dx}=\int{\frac{x}{5(x-1)}dx-\int{\frac{1}{5(x+4)}dx}}$$
I can now integrate by addition here $$\int{\frac{x}{x-1}}dx=\int{\frac{x-1+1}{x-1}}dx=x+\int{\frac{1}{x-1}}dx=x-\ln(x-1)$$
And doing the same thing for the second term and bringing back my $\frac{1}{5}$
$$\frac{1}{5}(x-\ln(x-1)+\ln(x+4))$$
I suspect there is a further simplification I could do. On a problem like this I also saw it integrated as an arctangent, but that seemed needlessly complex? In any case I was curious if I did this correctly.
|
$$I=\int\frac{x}{x^2+3x+4}dx=\frac12\int\frac{2x+3}{x^2+3x+4}-\frac{3}{x^2+3x+4}$$
$$I=\frac12\ln|x^2+3x+4|-\frac32\int\frac{1}{(x+3/2)^2+7/4}dx$$
then you can let $u=x+3/2,\,du=dx$ so you have $u^2+(\sqrt{7}/2)^2=\frac74\left((2u/\sqrt{7})^2+1\right)$ now let $v=2u/\sqrt{7},\,dv=2/\sqrt{7}du$ now you have an integral of the form:
$$\int\frac{1}{v^2+1}dv$$ which can be easily solved using the substitution $v=\tan(t)$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3940216",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 3
}
|
Evaluating $\int_{-\infty}^{\infty} \int_{-\infty}^{\infty}{ {{1}\over{2\pi}} e^{{{-1}\over{2}}(4x^2-2xy+3y^2)} ~\mathrm dx \mathrm dy}$ How to calculate that integral? I have no idea what substitution to use. I think polar coordinates will be needed here.
$$\displaystyle \int_{-\infty}^{\infty} \int_{-\infty}^{\infty}{ {{1}\over{2\pi}} e^{{{-1}\over{2}}(4x^2-2xy+3y^2)} ~\mathrm dx \mathrm dy}$$
|
You can even do it with $(x,y)$. Complete the square first
$$\int{ \frac{1}{2\pi}} e^{{{-1}\over{2}}(4x^2-2xy+3y^2)} dx=\frac{e^{-\frac{11 y^2}{8}} \text{erf}\left(\frac{4 x-y}{2 \sqrt{2}}\right)}{4
\sqrt{2 \pi }}$$
$$\int_{-\infty}^\infty{ \frac{1}{2\pi}} e^{{{-1}\over{2}}(4x^2-2xy+3y^2)} dx=\frac{e^{-\frac{11 y^2}{8}}}{2 \sqrt{2 \pi }}$$
$$\int \frac{e^{-\frac{11 y^2}{8}}}{2 \sqrt{2 \pi }}\,dy=\frac{\text{erf}\left(\frac{1}{2} \sqrt{\frac{11}{2}} y\right)}{2 \sqrt{11}}$$
$$\int_{-\infty}^\infty \frac{e^{-\frac{11 y^2}{8}}}{2 \sqrt{2 \pi }}\,dy=\frac{1}{\sqrt{11}}$$
In fact, for the most general case
$$I=\int_{-\infty}^{\infty} \int_{-\infty}^{\infty}\frac 1{2\pi} e^{-(ax^2+bxy+by^2)}\,dx\,dy=\frac{1}{\sqrt{4 a c-b^2}}$$ if
$$\Re(a)>0\land \Re\left(\frac{b^2}{a}\right)<4 \Re(c)$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3946976",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
}
|
Prove $\frac{\ln(1+x)}{1+x}=x-\left(1+\frac{1}{2}\right)x^2+\left(1+\frac{1}{2}+\frac{1}{3}\right)x^3-\ldots$ I have to prove that :
$$\frac{\ln(1+x)}{1+x}=x-\left(1+\frac{1}{2}\right)x^2+\left(1+\frac{1}{2}+\frac{1}{3}\right)x^3-\ldots$$
I already know that the Maclaurin series of $\ln(1+x)$ is $x-\frac{1}{2}x^2+\frac{1}{3}x^3-\frac{1}{4}x^4+\frac{1}{5}x^5+\ldots$ But how can I use it to prove the first claim? Should I just multiply $\frac{1}{1+x}$ to the Maclaurin series of $\ln(1+x)$? Still, I don't know how to do that.
I tried to use the derivatives of $\frac{\ln(1+x)}{1+x}$ then substitute that into the Maclaurin formula. But I don't think that's the right way to do this efficiently.
|
You must multiply the series
$$\log(1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+\frac{x^5}{5}+O\left(x^6\right)$$
$$\frac 1{1+x}=1-x+x^2-x^3+x^4-x^5+O\left(x^6\right)$$
$$\frac{\log(1+x)}{1+x}=\left(x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+\frac{x^5}{5}+O\left(x^6\right) \right) \times$$ $$\left(1-x+x^2-x^3+x^4-x^5+O\left(x^6\right) \right) $$
$$\frac{\log(1+x)}{1+x}=x-\frac{3 x^2}{2}+\frac{11 x^3}{6}-\frac{25 x^4}{12}+\frac{137
x^5}{60}+O\left(x^6\right)$$
In fact, almost as you wrote in the post
$$\frac{\log(1+x)}{1+x}=\sum_{n=1}^\infty (-1)^{n+1} H_n \,x^n$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3947763",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
}
|
(Multiplicative) inverse of $\alpha = (\sqrt[3]{7})^2 + 3\sqrt[3]{7} + 1 \in\mathbb Q[\sqrt[3]{7}]$ Set $\mathbb Q[\sqrt[3]{7}] = \{F(\sqrt[3]{7}) \mid F ∈ Q[x]\}$ is a field (with the usual addition and the usual multiplication).
Calculate the (multiplicative) inverse of
$$\alpha = (\sqrt[3]{7})^2 + 3\sqrt[3]{7} + 1 \in \mathbb Q[\sqrt[3]{7}]$$
Note: Application of the Euclidean algorithm to the polynomials $x^2 + 3x + 1$ and $x^3 - 7$ could help.
Attempt:
I know there is a multiplcative inverse $\beta$ with $\alpha\beta=1$, and that one should exist in $\mathbb Q[\sqrt[3]{7}]$, but do not know how I would go about expressing it in any form simpler than $\frac1{(\sqrt[3]{7})^2 + 3\sqrt[3]{7} + 1}$.
How can I determine a simpler way to express the value of $\frac1{(\sqrt[3]{7})^2 + 3\sqrt[3]{7} + 1}$?
|
You actually have two possibilities:
*
*Either, denoting $x=\sqrt[3]7$, you try to find a linearcombination $ax^2+bx+c$ such that
$\;(ax^2+bx+)(x^2+3x+1)=1$, which leads to solving the linear system
$$\begin{cases}a+3b+c=0\\7a+b+3c=0\\21a+7b+c=1\end{cases},$$
which can be solved finding the reduced row echelon form of the augmented matrix
$$\left[\begin{array}{rrr|l}1&3&1&0\\ 7&1&3&0\\ 21&7&1&1\end{array}\right].$$
*Or, extending a bit the hint, you apply the extended Euclidean algorithm to the polynomials $X^3-7$ and $X^2+3X+1$, which are coprime, to obtain a Bézout's relation
$$u(X)(X^2+3X+1)+v(X)(X^3-1)=1,$$
which, substituting $x$ to $X$, shows the inverse of $x^2+3x+1$ is $u(x)$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3948181",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
}
|
Prove $\frac{1}{2}\cdot\frac{3}{4}\cdot...\cdot\frac{2n-1}{2n}<\frac{1}{\sqrt{3n}}$ for all $n$. Prove for all $n$: $\frac{1}{2}\cdot\frac{3}{4}\cdot...\cdot\frac{2n-1}{2n}<\frac{1}{\sqrt{3n}}$.
Using induction, I tried the brain-dead method and went straight for $$\frac{2n+1}{2n+2}\cdot\frac{1}{\sqrt{3n}}<\frac{1}{\sqrt{3n+3}}$$ $$...$$ $$1<0.$$ After embarrassing myself, I looked around and I found this thread. Using induction, we can then easily prove $$\frac{1}{2}\cdot\frac{3}{4}\cdot...\cdot\frac{2n-1}{2n}\leq\frac{1}{\sqrt{3n+1}}$$ $$\frac{1}{2}\cdot\frac{3}{4}\cdot...\cdot\frac{2n-1}{2n}\leq\frac{1}{\sqrt{3n+1}}<\frac{1}{\sqrt{3n}}.$$ This gets me to the original problem. But in a problem solving standpoint, how do you think to use $\frac{1}{\sqrt{3n+1}}$? Is there some point in the first induction that leads to this idea? Or is there a better method than the above?
|
A First Approach
$$
\begin{align}
n\prod_{k=1}^n\left(\frac{2k-1}{2k}\right)^2
&=\frac14\prod_{k=2}^n\left(\frac{2k-1}{2k}\right)^2\frac{k}{k-1}\tag{1a}\\
&=\frac14\prod_{k=2}^n\frac{2k-1}{2k}\frac{2k-1}{2k-2}\tag{1b}\\
&=\frac14\prod_{k=2}^n\frac{\color{#C00}{k-1/2}}{\color{#090}{k}}\frac{\color{#75F}{k-1/2}}{\color{#C90}{k-1}}\tag{1c}\\
&=\frac14\color{#C00}{\frac{\Gamma(n+1/2)}{\Gamma(3/2)}}\color{#090}{\frac{\Gamma(2)}{\Gamma(n+1)}}\color{#75F}{\frac{\Gamma(n+1/2)}{\Gamma(3/2)}}\color{#C90}{\frac{\Gamma(1)}{\Gamma(n)}}\tag{1d}\\[3pt]
&=\frac1\pi\frac{\Gamma(n+1/2)^2}{\Gamma(n+1)\,\Gamma(n)}\tag{1e}\\[3pt]
&\le\frac1\pi\tag{1f}
\end{align}
$$
Explanation:
$\text{(1a)}$: pull the $k=1$ term out front and bring $n$ inside as a telescoping product
$\text{(1b)}$: rearrange terms
$\text{(1c)}$: divide numerator and denominator by $2$
$\text{(1d)}$: write the products as ratios of the Gamma function, using $\Gamma(x+1)=x\,\Gamma(x)$
$\text{(1e)}$: collect terms using $\Gamma(1)=\Gamma(2)=1$ and $\Gamma(3/2)=\sqrt\pi/2$
$\text{(1f)}$: $\Gamma(x)$ is log-convex
Thus, we get the stronger
$$
\prod_{k=1}^n\frac{2k-1}{2k}\le\frac1{\sqrt{\pi n}}\tag2
$$
A Slightly Simpler Approach with a Better Bound
$$
\begin{align}
\prod_{k=1}^n\frac{2k-1}{2k}
&=\prod_{k=1}^n\frac{(2k-1)2k}{4k^2}\tag{3a}\\
&=\frac1{4^n}\binom{2n}{n}\tag{3b}\\
&\le\frac1{\sqrt{\pi\!\left(n+\frac14\right)}}\tag{3c}
\end{align}
$$
Explanation:
$\text{(3a)}$: multiply numerator and denominator by $2k$
$\text{(3b)}$: $\prod\limits_{k=1}^n(2k-1)2k=(2n)!$ and $\prod\limits_{k=1}^n2k=2^nn!$
$\text{(3c)}$: inequality $(9)$ from this answer
In fact, using inequality $(9)$ from this answer, we get
$$
\frac1{\sqrt{\pi\!\left(n+\frac13\right)}}\le\prod_{k=1}^n\frac{2k-1}{2k}\le\frac1{\sqrt{\pi\!\left(n+\frac14\right)}}\tag4
$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3953464",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
}
|
Assuming the expression converges, determine the largest integer $n\le 9,000,000$ for which $\sqrt {n+\sqrt {n+\sqrt {n+\cdots}}}$ is rational. What I have done is that I supposed $y=\sqrt {n+\sqrt {n+\sqrt {n+\cdots}}}$.
So, $y^2-n=y$, which is $y^2-y+\frac{1}{4}=n+\frac{1}{4}$. So, $(y-\frac{1}{2})^2=n+{1\over 4}=\frac{4n+1}{4}$.
Hence, $y={1\over 2}\pm\frac{\sqrt {4n+1}}{2}$ is a rational, which means $4n+1$ must be a perfect square.
Then I let $m^2=4n+1$. So, $(m-1)(m+1)=4n$, and that is where I got stuck.
Obviously $m$ must be an odd integer and $n$ CANNOT be a perfect square. But I have no idea what to do next because $n$ can be very big... Could anyone share some ideas?
|
Let $m=2k+1$. Varying $k$ gives all possible values of $n$ for $y$ to be rational:
$$m^2=4k^2+4k+1=4(k^2+k)+1\implies n=k^2+k=k(k+1)\le9\cdot10^6$$
It is easily seen that the solution corresponds to $k=2999$, i.e. $n=8997000$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3953592",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
}
|
partial fractions when the fraction cannot be decomposed I am trying to find partial fractions of $\frac {1}{(x^2+1)^2}$. All the coefficients I get are zeros except the coefficient for the constant term which is 1, leaving me with the fraction I started with, so it seems like the fraction cannot be decomposed. How can I then go about writing this fraction as the sum $\frac {1}{2}\left [\frac {1}{x^2+1}-\frac{x^2-1}{(x^2+1)^2}\right]$? Is it just manipulation and trial and error?
|
In the real-valued partial fraction method, your term
$$
\frac{1}{(x^2+1)^2}
$$
is the best you can do. Evaluation of integrals
$$
\int\frac{dx}{(ax^2+bx+c)^n}
$$
is done as follows: For $n=1$, complete the square and it is an arctangent. For $n>1$ there is a reduction formula from integration by parts:
\begin{align}
\int\frac{dx}{(ax^2+bx+c)^n} &= \frac{2ax+b}{(n-1)(4ac-b^2)(ax^2+bx+c)^{n-1}}
\\ &\qquad+\frac{(2n-3)2a}{(n-1)(4ac-b^2)}
\int\frac{dx}{(ax^2+bx+c)^{n-1}}
\end{align}
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3957126",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 2
}
|
Finding probability of all outcomes occurring in n repetitions Suppose you have numbers 1, 2, 3, 4. You can pick each of these numbers with probability 1/4. After n iid selections, what is the probability that you have seen all 4 numbers? Express this as a function of n.
For example, if n=5, and you draw the numbers 1,1,3,2,1 then this is unsuccessful, as 4 was never picked. If you draw 3,4,2,3,4, this is also unsuccessful as 1 was never picked. However, 3,2,2,1,4 is successful.
I am aware you can use a discrete Markov chain and find the n-step transition probability matrix on Wolfram, but is there a way to find this without a calculator and with reasoning alone?
|
You may just count systematically as follows:
Number of sequences of length $n$ containing only $1$ digit: $$\color{blue}{\Rightarrow 4}$$
Number of sequences of length $n$ containing exactly $2$ digits:
*
*Choose two of the digits: $\color{blue}{\binom 42}$
*Number of sequences of length $n$ containing at least one of these digits: $\color{blue}{2^n}$
*Number of sequences of length $n$ containing only one of the $2$ selected digits: $\color{blue}{2}$
$$\color{blue}{\Rightarrow \binom 42\left(2^n-2\right)}$$
Number of sequences of length $n$ containing exactly $3$ digits:
*
*Choose three of the digits: $\color{blue}{\binom 43}$
*Number of sequences of length $n$ containing at least one of these digits: $\color{blue}{3^n}$
*Number of sequences of length $n$ containing at least one of only two of the $3$ selected digits: $\color{blue}{\binom{3}{2}2^n}$
*Number of sequences of length $n$ containing only one of the $3$ selected digits: $\color{blue}{3}$
$$\color{blue}{\Rightarrow \binom 43\left(3^n- \binom{3}{2}2^n+3\right)}$$
All together, number of sequences where at least one digit is missing:
$$\color{blue}{4 + \binom 42\left(2^n-2\right) + \binom 43\left(3^n- \binom{3}{2}2^n+3\right)=4-6\cdot 2^n+4\cdot 3^n}$$
Now, the probability you are looking for is
$$\frac{4^n - \left(4-6\cdot 2^n+4\cdot 3^n\right)}{4^n}=\boxed{1-\frac 1{4^{n-1}}+6\frac 1{2^n}-4\left(\frac 34\right)^n}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3960242",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
}
|
Solve the equation $\sqrt{45x^2-30x+1}=7+6x-9x^2$ Solve the equation $$\sqrt{45x^2-30x+1}=7+6x-9x^2.$$
So we have $\sqrt{45x^2-30x+1}=7+6x-9x^2\iff \begin{cases}7+6x-9x^2\ge0\\45x^2-30x+1=(7+6x-9x^2)^2\end{cases}.$ The inequality gives $x\in\left[\dfrac{1-2\sqrt{2}}{3};\dfrac{1+2\sqrt{2}}{3}\right].$ I am not sure how to deal with the equation. Thank you in advance!
|
Let $u=9x^2-6x$ and rewrite the equation as
$$\sqrt{5u+1}=7-u$$
Squaring both sides gives $5u+1=49-14u+u^2$, or
$$u^2-19u+48=(u-16)(u-3)=0$$
We see that $u=16$ is not a solution, since $\sqrt{81}\not=-9$. This leaves us with $u=3$, which is a valid solution, since $\sqrt{16}=4$, and from this we have
$$9x^2-6x=3\implies3x^2-2x-1=(x-1)(3x+1)=0$$
which gives $x=1$ and $x=-1/3$ as the complete solution set.
Remark: What makes this work so nicely is that $45:30=9:6$. If the coefficients of $x^2$ and $x$ on the two sides hadn't been in proportion, the solution would have been much more involved.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3962818",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
}
|
Prove that $a ( n ) = b( n + 2)$
let $a(n)$ denotes the number of ways of expressing the positive integer $n$ as an ordered sum of 1's and 2's. Let $b(n)$ denote the number of ways of expressing n as an ordered sum of integers greater than 1. prove that $a(n) = b(n+2)$. for $n=1,2,3...$
My approach:
$a(1) = 1$ (only) , $b(3) = 3$
$a(2) = 2$, because $2=1+1=2$ ,and $b(4)=2$, because $4=2+2=4$
$a(3) = 3$, because $3=1+1+1=1+2=2+1$ and, $b(5) = 3$ , because $5=2+3=3+2=5$
$a(4)= 5$, because $4=1+1+1+1=2+2=1+1+2=1+2+1=2+1+1$ and, $b(6)=5$, because $6=3+3=2+4=4+2=2+2+2=6$
$a(5)=8$, because $5=1+1+1+1+1=2+1+1+1=1+2+1+1=1+1+2+1=1+1+1+2=2+2+1=2+1+2=1+2+2$ and, $b(7)=8$, because
$7=3+2+2=2+3+2=2+2+3=3+4=4+3=2+5=5+2=7$
By this way i am able to show that $a(n)=b(n+2)$. but is there any general method for this problem. I mean any recursion relation which i can understand.
Background:-This problem is from pathfinder for Olympiad mathematics.
|
Hint: Observe that the sequence $1, 2, 3, 5, 8, \ldots $ looks like the Fibonacci sequence, offset by 1
Hint: Show that $a_n$ is indeed the Fibonacci sequence (offset by 1), because $ a_{n+2} = a_{n+1 } + a_{n} $, with initial conditions $ a_1 = 1, a_2 = 2$.
Hint: Show that $b_n, n \geq 2$ is indeed the Fibonacci sequence, because $b_{n} = b_{n-2} + b_{n-3} + b_{n-4} + \ldots + b_3 + b_2 + 1$, and $ b_2 = 1, b_3 = 1$.
Hence, $ a_n, n \geq 1 $ and $ b_n, n \geq 3$ are the Fibonacci sequence offset by 1.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3963094",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
}
|
Find $\lim_{x \to \infty}{\frac{(x+1)^{\frac{2}{3}}-(x-1)^{\frac{2}{3}}}{(x+2)^{\frac{2}{3}}-(x-2)^{\frac{2}{3}}}}$ As stated in the title.
My attempt. Dividing through $(x-2)^{\frac{2}{3}}$.
$$L=\lim_{x \to \infty}{\frac{(x+1)^{\frac{2}{3}}-(x-1)^{\frac{2}{3}}}{(x+2)^{\frac{2}{3}}-(x-2)^{\frac{2}{3}}}}=\lim_{x \to \infty}\frac{(\frac{x+1}{x-2})^{\frac{2}{3}}-(\frac{x-1}{x-2})^{\frac{2}{3}}}{(\frac{x+2}{x-2})^{\frac{2}{3}}-1}$$
L'Hopital
$$L=\lim_{x \to \infty}\frac{\frac{2}{3}(\frac{x+1}{x_2})^{-\frac{1}{3}}(\frac{x-2-(x+1)}{(x-2)^2})-\frac{2}{3}(\frac{x-1}{x-2})^{-\frac{1}{3}}(\frac{x-2-(x-1)}{(x-2)^2})}{\frac{2}{3}(\frac{x+2}{x-2})^{-\frac{1}{3}}(\frac{x-2-(x+2)}{(x-2)^2})}=\lim_{x\ \to \infty}{\frac{3(x+1)^{-\frac{1}{3}}-(-1)(x-1)^{-\frac{1}{3}}}{(x+2)^{-\frac{1}{3}}(-4)}}$$
$$=\lim_{x \to \infty}{\frac{3(1+\frac{1}{x})^{-\frac{1}{3}}-(1-\frac{1}{x})^{-\frac{1}{3}}}{4(1+\frac{2}{x})^{-\frac{1}{3}}}}=\frac{3-1}{(4)(1)}=\frac{1}{2}.$$
Is this correct and is there a more elegant way of doing it?
EDIT: strictly speaking L'Hopital is not applicable with $x \to \infty$ so just got lucky here...
|
Here is how I would do it without using L'hopital, or series expansion :
Using the identity $ a^{3}-b^{3}=\left(a-b\right)\left(a^{2}+ab+b^{2}\right) $, we have : $$ \left(x+1\right)^{2}-\left(x-1\right)^{2}=\left(\left(x+1\right)^{\frac{2}{3}}-\left(x-1\right)^{\frac{2}{3}}\right)\left(\left(x+1\right)^{\frac{4}{3}}+\left(x^{2}-1\right)^{\frac{2}{3}}+\left(x-1\right)^{\frac{4}{3}}\right) $$
And : $$ \left(x+2\right)^{2}-\left(x-2\right)^{2}=\left(\left(x+2\right)^{\frac{2}{3}}-\left(x-2\right)^{\frac{2}{3}}\right)\left(\left(x+2\right)^{\frac{4}{3}}+\left(x^{2}-4\right)^{\frac{2}{3}}+\left(x-2\right)^{\frac{4}{3}}\right) $$
Thus : \begin{aligned}\lim_{x\to +\infty}{\frac{\left(x+1\right)^{\frac{2}{3}}-\left(x-1\right)^{\frac{2}{3}}}{\left(x+2\right)^{\frac{2}{3}}-\left(x-2\right)^{\frac{2}{3}}}}&=\lim_{x\to +\infty}{\frac{\left(\left(x+1\right)^{2}-\left(x-1\right)^{2}\right)\left(\left(x+2\right)^{\frac{4}{3}}+\left(x^{2}-4\right)^{\frac{2}{3}}+\left(x-2\right)^{\frac{4}{3}}\right)}{\left(\left(x+2\right)^{2}-\left(x-2\right)^{2}\right)\left(\left(x+1\right)^{\frac{4}{3}}+\left(x^{2}-1\right)^{\frac{2}{3}}+\left(x-1\right)^{\frac{4}{3}}\right)}}\\ &=\lim_{x\to +\infty}{\frac{\left(1+\frac{2}{x}\right)^{\frac{4}{3}}+\left(1-\frac{4}{x^{2}}\right)^{\frac{2}{3}}+\left(1-\frac{2}{x}\right)^{\frac{4}{3}}}{2\left(\left(1+\frac{1}{x}\right)^{\frac{4}{3}}+\left(1-\frac{1}{x^{2}}\right)^{\frac{2}{3}}+\left(1-\frac{1}{x}\right)^{\frac{4}{3}}\right)}}\\ &=\frac{1}{2}\end{aligned}
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3963376",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 2
}
|
The $2$nd, the $1$st and the $3$rd term of an arithmetic progression form a geometric progression An arithmetic progression is given with a common difference $\ne0.$ The $2nd$, the $1st$ and the $3rd$ term of the ap form a geometric progression. Find the common ratio.
So we have the ap: $a_1,a_1+d,a_1+2d$ and the gp: $a_1+d,a_1,a_1+2d.$ How can we find $q?$ Thank you in advance!
I have tried to solve the problem using a system, but I am not sure which equations we can use. For example, $a_1^2=(a_1+d)(a_1+2d)$ and $\dfrac{a_1+d}{2}=a_1+(a_1+2d)$ hold, but I don't see how to use them.
Why is it important for the common difference of the AP to be $\ne0?$ Then we will have a constant sequence in which all terms are equal, right?
|
The three terms of a geometric series $a_0, a_1= a_0 + d, a_2 = a_0+2d$ or if it is easier $a_0= a_1-d, a_1, a_2= a_1 + d$.
The three terms of a geometric series is $b_0, b_1=b_0\cdot r, b_2=b_0\cdot r^2$ or if it is easier $b_0=\frac {b_1}r, b_1, b_2= b_1\cdot r$.
We are told that $a_1, a_1 -d, a_1 +d$ form a geometric series. so
$\frac {b_1}r = a_1$. $b_1 = a_1-d$, $b_1r = a_1 + d$.
So Solve for $r$.
========
substitute $b_1$ with $a_1 -d$ and
$\frac {a_1 -d}r = a_1$ and $(a_1 -d)r = a_1 + d$.
So $r = \frac {a_1 -d}{a_1} = 1 - \frac d{a_1}$ and $r = \frac {a_1 + d}{a_1-d}=1 + \frac {2d}{a_1-d}$
So $-\frac d{a_1} = \frac {2d}{a_1-d}$ so $-\frac 1{a_1} = \frac 2{a_1-d}$ so
$-(a_1 -d) = 2a_1$ so $d = 3a_1$.
Substituting $3a_1$ for $d$ we get
$r = \frac {a_1 -d}{a_1} = \frac {-2a_1}{a_1} = -2$ and $r = \frac {a_1 + d}{a_1-d}= \frac {4a_1}{-2a_1} = -2$.
So $r = -2$.
......
Furthermore the three terms as an arithmetic series $a_0 = a_1-d =-2a_1; a_1 = a_1; a_2= a_1 + d = 4a_1$.
And as a geometric series they are $b_0 = a_1; b_1 =-2a_1; b_2 = 4a_1$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3964832",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 3
}
|
How to do integral $\int\frac1{x^3+x+1}dx$? I've been stuck by this integral. I only knew that roots of the cubic equation are complex.
Please help me with this. $$\int\frac{dx}{x^3+x+1}$$
|
Since $x^3+x+1=(x-a)((x-b)^2+c^2)$ with$$\begin{align}a&:=\sqrt[3]{\frac{\sqrt{93}-9}{18}}-\sqrt[3]{\frac{2/3}{\sqrt{93}-9}},\\b&:=\frac{1}{\sqrt[3]{12(\sqrt{93}-9)}}-\sqrt[3]{\frac{\sqrt{93}-9}{144}},\\c&:=\sqrt{3}\left(\frac{1}{\sqrt[3]{12(\sqrt{93}-9)}}+\sqrt[3]{\frac{\sqrt{93}-9}{144}}\right),\end{align}$$the integrand is as follows (double-check my partial fractions):$$\frac{1}{(x-a)((x-b)^2+c^2)}=\frac{1}{(b-a)^2+C^2}\left(\frac{1}{x-a}+\frac{(b-a)-(x-b)}{(x-b)^2+c^2}\right).$$So the most general antiderivative is$$\frac{1}{(b-a)^2+C^2}\left(\ln\frac{|x-a|}{\sqrt{(x-b)^2+c^2}}+\frac{b-a}{c}\arctan\frac{x-b}{c}\right)+C,$$where the locally constant function $C$ can have different values either side of $x=a$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3967108",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
}
|
Given $f(x) = x^2 + \ln (1 + \frac{1}{x})$, prove that $\forall x > 0, x \in \mathbb{R}: f(x) \geq \frac{1 + 2\ln(2)}{4}$ Given $f(x) = x^2 + \ln (1 + \frac{1}{x})$, prove that $\forall x\in(0,+\infty): f(x) \geq \frac{1 + 2\ln(2)}4$.
That would be same as proving that $x^2 + \ln (\frac{x + 1}{x}) - \frac{1 + \ln(4)}{4} \geq 0$, isn't?
I have also found the derivative $f'(x) = \frac{2x^3 + 2x^2 - 1}{x^2 + x}$ and that there is a solution of $f(x)$ in the interval of $\left[\frac12, \frac1{\sqrt 2}\right]$.
But from there I am stuck, I am not able to see how to link all the concepts.
Could someone give me a hint?
Thank you and Happy New Year 2021 :)
|
You are almost there.
Since $f(0) = f(+\infty) = +\infty$ and there is a unique solution $x_0 > 0$ such that $f'(x_0)=0$ (because $2x^2(1+x)$ is an increasing function of $x$) we only need to show that $f(x_0) > \frac{1+2 \ln 2}{4}$.
Since $x_0 \in \left(\frac 12, 1\right)$, then
$$f(x_0) = x_0^2 + \ln \left( 1+\frac{1}{x_0}\right)> \left(\frac 12\right)^2 + \ln \left( 1 + \frac{1}{1} \right) = \frac{1+4\ln 2}{4}> \frac{1+2\ln2}{4}.\blacksquare$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3968778",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
}
|
Given that $a = \sqrt[3]4 + \sqrt[3]2 + 1$, find $\frac{3}{a} + \frac{3}{a^2} + \frac{1}{a^3}$.
Given that $a = \sqrt[3]4 + \sqrt[3]2 + 1$, find $\frac{3}{a} + \frac{3}{a^2} + \frac{1}{a^3}$.
What I Tried: I only figured out that:- $$\rightarrow a = 2^\frac{2}{3} + 2^\frac{1}{3} + 2^\frac{0}{3}$$ Yet this does not help me anywhere. Perhaps I have to multiply something with $a$ only so that the expression becomes usable, what what to multiply?
Next, the expression we need to find is $\frac{3a^2 + 3a + 1}{a^3}$ , which I did not find any cool factorizations, and do not think anything will help here.
Can anyone help me?
|
Hint: As $$a=\frac{(\sqrt[3]{4}+\sqrt[3]{2}+1)(\sqrt[3]{2}-1)}{\sqrt[3]{2}-1}=\frac{1}{\sqrt[3]{2}-1} $$ $${\left(\frac{1}{a}+1\right)}^3=2$$ can you proceed ....
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3969900",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
}
|
Simplify, $\sqrt[3]\frac{(\sqrt{a-1} - \sqrt{a})^5}{(\sqrt{a-1} + \sqrt{a})} + \sqrt[3]\frac{(\sqrt{a-1} + \sqrt{a})^5}{(\sqrt{a} - \sqrt{a-1})}$.
Simplify, $\sqrt[3]\frac{(\sqrt{a-1} - \sqrt{a})^5}{(\sqrt{a-1} + \sqrt{a})} + \sqrt[3]\frac{(\sqrt{a-1}
+ \sqrt{a})^5}{(\sqrt{a} - \sqrt{a-1})}$.
What I Tried: I thought of substituting $\sqrt{a - 1} = x$ , $\sqrt{a} = y$ .
This gives :- $$\rightarrow \sqrt[3]\frac{(x - y)^5}{(x + y)} + \sqrt[3]\frac{(x + y)^5}{(y - x)}$$
But I was not able to find any good factorisation for this. I even took some help from Wolfram Alpha and it gives me this :-
Another thing I thought of was to substitute only $\sqrt{a} = x$. This would give :-
$$\rightarrow \sqrt[3]\frac{(\sqrt{(x + 1)(x - 1)} - x)^5}{(2x^2 - 1)} + \sqrt[3]\frac{(\sqrt{(x + 1)(x - 1)} + x)^5}{1}$$
This looks more or less simpler to work with, but unfortunately I could not get any ideas.
Can anyone help me?
|
\begin{eqnarray*}
\sqrt[3]\frac{(\sqrt{a-1} - \sqrt{a})^5}{(\sqrt{a-1} + \sqrt{a})} + \sqrt[3]\frac{(\sqrt{a-1}
+ \sqrt{a})^5}{(\sqrt{a} - \sqrt{a-1})} = \frac{-(\sqrt{a-1} -\sqrt{a})^2 + (\sqrt{a-1} -\sqrt{a})^2}{\sqrt[3]{(\sqrt{a} - \sqrt{a-1})(\sqrt{a} + \sqrt{a-1})}} \\
= \frac{-(a-1+a-2 \sqrt{a(a-1)})+ a-1+a+2 \sqrt{a(a-1)}}{\sqrt[3]{a-(a-1)}}=4\sqrt{a(a-1)}
\end{eqnarray*}
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3970199",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
}
|
Consider the function $f(x) = \frac{1}{3^x + \sqrt{3}}.$ Find :- $\sqrt{3}[f(-5) + f(-4) + ... + f(3) + f(4) + f(5) + f(6)]$.
Consider the function $f(x) = \frac{1}{3^x + \sqrt{3}}.$ Find :-
$\sqrt{3}[f(-5) + f(-4) + ... + f(3) + f(4) + f(5) + f(6)]$.
What I Tried: I checked similar questions and answers in the Art of Problem Solving here and here and tried to get some ideas.
First thing which I did is thinking of pairing the values, I took for example, $f(-1)$ and $f(1)$.
We have :-
$$\rightarrow f(-1) = \frac{1}{\frac{1}{3} + \sqrt{3}} = \frac{3\sqrt{3} + 1}{3}$$
$$\rightarrow f(1) = \frac{1}{3 + \sqrt{3}}$$
Adding both gives $\frac{7 + 6\sqrt{3}}{12 + 10\sqrt{3}}$, which more or less looks like a random sum.
So my idea of pairing did not work, or at least I couldn't pair them nicely or missed a pattern. So how would I start solving it?
Can anyone help?
|
Let $\sqrt3=a$
$$\dfrac{f(x)}a=\dfrac1{a^{2x-1}+1}$$
If $\dfrac{a^{2x-1}}{1+a^{2x-1}}=f(px+q)=\dfrac1{1+a^{2(px+q)-1}}$
$\implies 1-2x=2(px+q)-1$
$\implies p=-1,q=1$
$$\implies f(x)+f(1-x)=a$$
Can you take it from here?
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3970350",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 2
}
|
Explanation for $\int_{\frac{\pi}{2}+(j-1)\pi}^{\frac{\pi}{2}+j\pi} \frac{|\cos(t)|}{\frac{\pi}{2}+j\pi} \,dt = \frac{2}{\frac{\pi}{2}+j\pi} $ I don't see how / why one can rewrite the integral as following:
$$\int_{\frac{\pi}{2}+(j-1)\pi}^{\frac{\pi}{2}+j\pi} \frac{|\cos(t)|}{\frac{\pi}{2}+j\pi} \,dt = \frac{2}{\frac{\pi}{2}+j\pi} $$
I think this should be rather easy, but I don't see what I'm missing.
|
We can write
\begin{align*}
\int_{\frac{\pi}{2}+(j-1)\pi}^{\frac{\pi}{2}+j\pi} \frac{|\cos(t)|}{\frac{\pi}{2}+j\pi} \ dt & = \frac{1}{\frac{\pi}{2}+j\pi} \int_{\frac{-\pi}{2}+j\pi}^{\frac{\pi}{2}+j\pi}|\cos(t)| \ dt\\
&\overset{\color{blue}{u=t-j\pi} }{=} \frac{1}{\frac{\pi}{2}+j\pi} \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}|\cos(u+j\pi)| \ du\\
\end{align*}
and if $j$ is an integer, we know that
$$|\cos(u + j\pi)| = |(-1)^j \cos(u)| = |\cos(u)|
$$
so we get\begin{align*}
\frac{1}{\frac{\pi}{2}+j\pi} \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}|\cos(u+j\pi)| \ du & = \frac{1}{\frac{\pi}{2}+j\pi} \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}|\cos(u)| \ du
\end{align*}
and on the interval $\left( -\frac{\pi}{2}, \frac{\pi}{2} \right)$ we know that cosine is positive, so on this interval (which is the one we're integrating on) we get $|\cos(u)| = \cos(u)$. And lastly we see
\begin{align*}
\frac{1}{\frac{\pi}{2}+j\pi} \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}|\cos(u)| \ du & = \frac{1}{\frac{\pi}{2}+j\pi} \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\cos(u) \ du\\
& = \frac{1}{\frac{\pi}{2}+j\pi} \sin(u)\Biggr|_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\\
& = \frac{1}{\frac{\pi}{2}+j\pi} \left(1 - (-1)\right)\\
& = \frac{2}{\frac{\pi}{2}+j\pi}
\end{align*}
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3972672",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
}
|
Evaluate $\int_{2}^{7} \frac{x}{1-\sqrt{2+x}} d x$ We have the following integral:
$$
\int_{2}^{7} \frac{x}{1-\sqrt{2+x}}\, dx
$$
And this is my solution, which seems to be wrong, and I am failing to see where exactly I failed at:
We have $u=1-\sqrt{2+x}, x=u^2-2u-1, dx=-2\sqrt{2+x}\, du$, and we know that $x\geq -2$ and thus $u\leq 1$:
\begin{align}
\int_{2}^{7} \frac{x}{1-\sqrt{2+x}}\, dx
&=-2\int_{-1}^{-2} \frac{(u^2+2u-1)(u-1)}{u}\, du\\
&= -2 \left( \int_{-1}^{-2} u^2 d u + \int_{-1}^{-2} u\, du +\int_{-1}^{-2} -3\, du + \int_{-1}^{-2} \frac{1}{u}\, du \right) \\
&= -2\left[\frac{u^3}{3}+\frac{u^2}{2}-3u+\ln{|u|}\right]_{-1}^{-2}\approx -18
\end{align}
Can someone please help me pinpoint the issue?
|
The first thing I would like to point out is that since you want to do a substitution, it is really extra (unnecessary) work writing $dx$ in terms of $x$ (where the aim of substitution is to "get rid" of the initial variable and simplify the integral in the process).
With that out of the way, let $u = 1 - \sqrt {2 + x}$, then $x = u^2 - 2u - 1$ and $\mathrm{d}x = 2u - 2$.
Now, observe that, $\forall\ n \in \mathbb {R}$, $\sqrt n \geq 0$, so we note that $u \leq 1$.
\begin{align}
\int_{2}^{7} \frac x {1 - \sqrt {2 + x}}\ \mathrm {d}x & =
2\int_{-1}^{-2} \frac {(u^2 - 2u - 1)(u - 1)} u\ \mathrm {d}u
\\[5 mm] & =
2\int_{-1}^{-2} \frac {u^3 - 3u^2 + u + 1} u\ \mathrm {d}u
\\[5 mm] & =
2\left[\frac {u^3} 3 - \frac {3u^2} 2 + u + \ln|u|\right]^{-2}_{-1}
\\[5 mm] & =
2\ln 2 -\frac {47} 3
\end{align}
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3975385",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
}
|
Let $a,b,c\in\mathbb{Z}$, $1Let $a,b,c\in\mathbb{Z}$, $1<a<10$, $c$ is a prime number and $f(x)=ax^2+bx+c$. If $f(f(1))=f(f(2))=f(f(3))$, find $f'(f(1))+f(f'(2))+f'(f(3))$
My attempt:
\begin{align*}
f'(x)&=2ax+b\\
(f(f(x)))'&=f'(f(x))f'(x)\\
f'(f(x))&=\frac{(f(f(x)))'}{f'(x)}\\
\end{align*}
|
Okay here's the outline of how you solve this.
first, you write down the system of equations
$$
0=f(f(2))-f(f(1))=(3a+b)*(b+5a^2+3ab+2ac) \\
0=f(f(3))-f(f(2))=(5a+b)*(b+13a^2+5ab+2ac)
$$
and this leaves you with four possibilities:
(a) $(3a+b=0)$ and $(5a+b=0)$ which is impossible since $a \ne 0$
(b) $(3a+b=0)$ and $(b+13a^2+5ab+2ac=0)$
in which case we obtain $b=-3a$, then our second condition turns into $2ac-2a^2-3a=0$, which further yields $c=a+\frac{3}{2}$, and we see no acceptable solutions here since both $a$ and $c$ are asked to be integers.
(c) $(b+13a^2+5ab+2ac=0)$ and $(b+5a^2+3ab+2ac=0)$ which, if we subtract one of the equations from the other, gives us $b=-4a$ and, shortly after, $p=\frac{7a+4}{2}$, and we can just check the values of $a$ from $2$ to $9$ to find the only solution at $a=6, b=-24, c=23$.
(d) $(5a+b=0)$ and $(b+5a^2+3ab+2ac=0)$, in which case we get $b=-5a$, the second condition simplifies to $10a-2c+5$, which gives us $c=5a+\frac{5}{2}$, and we also see that $c$ will never be integer if $a$ is integer, and $a$ is always integer.
This leaves us with the only acceptable solution found in (c). As I already mentioned above, putting $(a,b,c)=(6,-24,23)$ into $f'(f(1))+f(f'(2))+f'(f(3))$ gives you $95$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3975895",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
}
|
How to solve this equation $(x-1)(x+2) + 4(x-1)\sqrt{\dfrac{x+2}{x-1}} = 12$? $(x-1)(x+2) + 4(x-1)\sqrt{\dfrac{x+2}{x-1}} = 12$
$Domain: x\in (-\infty;-2]\cup(1;+\infty)$
$x = 2\longrightarrow$ Done
How to prove $x = \dfrac{-1-3\sqrt{17}}{2}$ is the last solution? With raise both side by power of two?
Any better way such as factoring, grouping,...?
Please help!!
|
Assuming $x\neq 1$ you can rewrite your equation as
$$(x-1)(x+2) + 4\sqrt{(x-1)(x+2)} = 12$$
Then call $z = \sqrt{(x+1)(x-2)}$ and solve the quadratic associated equation, and then you pullback to $x$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3978740",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
}
|
Find n where $(2n+1)2^{4n+5} = 3 \pmod{7}$ For $n$ normal number, the book solved it like this:
If $n$ can be divided by $3$ (which is $n = 3k$) then $n = 21L + 9$.
If $n$ can't be divided by 3(Which is either $n = 3k +1$ or $n = 3k + 2$) then $n = 21L + 1$ or $n = 21L + 2$ .
But I didn't solve it like this.
My logic is that since 3 is a prime number then $(2n+1)2^{4n+5} = 3 \pmod{7}$ means either $2^{4n+5} = 3\pmod{7}$ and $ 2n + 1 = 1\pmod{7} $ or the other way around.
But since there is no $n$ value that can make $2^{4n+5} = 3\pmod{7}$ then it means:
$ 2n + 1 = 3 \pmod{7}$ and by calculating we find in the end $n = 7k' + 1$
And $ 2^{4n+5} = 1\pmod{7}$ and by calculating we find that $n = 3k''$
So it means that $n = 7k' + 1$ AND $n = 3k''$
So, I know that my solution is faulty but can anyone explains to me the right solution or point out where I went wrong?
|
This is somewhere between a comment and an answer.
The simplest way to solve this is that the value of $(2n+1)2^{4n+5}$ is precisely a function of $n \mod 42 = 6 \times 7$. More precisely, $2n+1 \mod 7$ is precisely a function of $n \mod 7$ while $2^{4n+5} \mod 7$ is precisely a function of $n \mod 6$. For each $i \in \{0,1,2,3,4,5\}$, you can check directly what $n \mod 7$ must be--in particular, what $(2n+1) \mod 7$ must be, given $n \mod 6$--in particular, what is $2^{4n+5} \mod 7$, which is a function of $n \mod 6$.
For example, for the case where $n = 0 \mod 6$ , we see that $2^{4n+5}= 1 \times 2^5 = 32 = 4 \mod 7$. So $2^{4n+5}$ is $4 \mod 7$ for this case where $n = 0 \mod 6$. So for the case where $n = 0 \mod 6$, the integer $n$ must satisfy the equation $(2n+1) \times 4 = 3 \mod 7$, which gives $n = 6 \mod 7$. So for the case where $n = 0 \mod 6$, the integer $n$ must also be $6 \mod 7$ which gives $n = 6 \mod 42$. What about the case where $n = 1 \mod 6$. What about each of the cases where $n \mod 6 = i=2,3,4,5$.
This in fact would work if $2n+1, 4n+5$ were each replaced by any two polynomials in $n$. Can you see why this is.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3984075",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
}
|
Maximum value of modulus in exponential form I am trying to find the maximum value of
$\left|e^{i\theta}-2\right|+\left|e^{i\theta}+2\right|\mbox{ for }0\le\theta\le2\pi$.
I can replace $e^{i\theta}$ with $a+ib$ and then proceed to get a function of $a$;
Namely;
$$\sqrt{5+4a}+\sqrt{5-4a}$$
I can then use calculus to find the maximum value of this function of $a$.
My question is: Are there any more efficient ways to solve this problem?
|
Let $f(\theta) = \left|e^{i\theta}-2\right|+\left|e^{i\theta}+2\right|$. Note that $f(\theta)\ge0$. We have $$f^2(\theta) = |e^{i\theta}-2|^2 + |e^{i\theta}+2|^2 + 2|e^{2i\theta}-4| = (e^{i\theta}- 2)(e^{-i\theta}- 2) + (e^{i\theta}+2)(e^{-i\theta}+2) + 2|e^{2i\theta}-4| = 1 -4\cos(\theta) +4 +1+4\cos(\theta)+4 +2|e^{2i\theta}-4| = 10+ 2|e^{2i\theta}-4|$$Also we know that $$|e^{2i\theta}-4|\le |e^{2i\theta}| +|4| = 5$$So we have $$f^2(\theta)\le10+2\times 5 = 20$$We can conclude that $$0\le f(\theta)\le \sqrt{20} = 2\sqrt{5}$$If we can find $\theta_0$ such that $f(\theta_0) = 2\sqrt{5}$, we are done. Indeed, we have $f(\frac{\pi}{2}) = 2\sqrt{5}$. Also, this can be seen by $$f^2(\theta) = 20 \implies |e^{2i\theta} - 4| = 5 \implies |e^{2i\theta} - 4|^2 = 25 \implies 1 - 8\cos(2\theta) + 16 = 25 \implies 8\cos(2\theta) = -8 \implies \cos(2\theta) = -1$$Which has solutions $\theta = \frac{\pi}{2},\frac{3\pi}{2}$ in $0\le \theta \le 2\pi$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3985577",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
}
|
Number of integers at most $x$ with exactly two distinct prime factors I wish to find an asymptotic for the number of integers not exceeding $x$ with exactly two distinct prime factors. Here is a starting point:
Throughout $p$ and $q$ are primes. We are interested in $\frac{1}{2}\#\{(p,q,\alpha,\beta): p^{\alpha}q^{\beta} < x\}$. (The factor $\frac{1}{2}$ arises since e.g. for $x=10^4$ we get $2^{3}\cdot 5^4$ once from $p=2$, $q=5$, $\alpha = 3$, $\beta = 4$ and once from $p=5$, $q=2$, $\alpha = 4$, $\beta = 3$.) Another way to write it is $\frac{1}{2}\#\{(p,q,\alpha,\beta): q<x, p<(x/q^{\beta})^{1/\alpha} \} = \frac{1}{2}\sum_{q<x}\sum_{\alpha, \beta \geq 1}\sum_{p<(x/q^{\beta})^{1/\alpha}}1$. Taking care of the inner sum with the Prime Number Theorem, the count becomes
$$\frac{1}{2}\sum_{q<x}\sum_{\substack{\alpha, \beta \geq 1}} \bigg(\frac{x^{1/\alpha}}{q^{\beta/\alpha}\log \frac{x^{1/\alpha}}{q^{\beta/\alpha}}} + O\bigg(\frac{x^{1/\alpha}}{q^{\beta/\alpha}\log^2 \frac{x^{1/\alpha}}{q^{\beta/\alpha}}}\bigg)\bigg)$$
Any idea how to conviently proceed with the latter nasty sums (or perhaps another approach)? Any help appreciated!
|
The asymptotic is
$$\frac{x\log \log x}{\log x}\,.$$
More generally, a result of Landau (1900) is that
$$\pi_k(x) \sim \rho_k(x) \sim \sigma_k(x) \sim \frac{x(\log \log x)^{k-1}}{(k-1)!\log x}\,,$$
where
\begin{align}
\pi_k(x) &= \# \{ n \leqslant x : \omega(n) = \Omega(n) = k\}\,,\\
\rho_k(x) &= \#\{ n \leqslant x : \omega(n) = k\}\,,\\
\sigma_k(x) &= \#\{ n \leqslant x : \Omega(n) = k\}\,,
\end{align}
and $\omega(n)$ is the number of distinct prime factors of $n$, while $\Omega(n)$ is the number of prime factors of $n$ counted with multiplicity,.
You are specifically asking for the asymptotics of $\rho_2$.
If we only want to show that, not Landau's result in full generality, the IMO easiest way is to obtain the asymptotics for $\pi_2(x)$, and then show that the numbers $p^{\alpha} q^{\beta}$ with $\alpha > 1$ or $\beta > 1$ are asymptotically negligible.
By symmetry, we can assume $\alpha > 1$, and we can ignore the condition $q \neq p$ (that can only make us overestimate the count, so if it still remains negligible, a fortiori the true count is negligible).
Given a prime $p$ and an exponent $\alpha > 1$, the count of numbers $p^{\alpha} q^{\beta} \leqslant x$, where $q$ is a prime (possibly equal to $p$) is the number of prime powers (including primes) not exceeding $x/p^{\alpha}$. Splitting into two parts according as $p^{\alpha} > \sqrt{x}$, for the part with $p^{\alpha} > \sqrt{x}$ we obtain the upper bound
$$\sum_{\substack{\sqrt{x} < p^{\alpha} \leqslant x/2 \\ \alpha \geqslant 2}} \sqrt{x} \leqslant \sqrt{x}\cdot \pi(\sqrt{x}) \leqslant C_1\cdot \frac{x}{\log x}$$
and for the part with $p^{\alpha} \leqslant \sqrt{x}$, since there are only $O(\sqrt{y})$ perfect powers not exceeding $y$ we obtain an upper bound of
$$C_2\cdot\sum_{\substack{p^{\alpha} \leqslant \sqrt{x} \\ \alpha \geqslant 2}} \pi\biggl(\frac{x}{p^{\alpha}}\biggr) \leqslant C_3\cdot \sum_{\substack{p^{\alpha} \leqslant \sqrt{x} \\ \alpha \geqslant 2}} \frac{x}{p^{\alpha}\log \frac{x}{p^{\alpha}}} \leqslant 2C_3\frac{x}{\log x} \sum_{\substack{p^{\alpha} \leqslant \sqrt{x} \\ \alpha \geqslant 2}} \frac{1}{p^{\alpha}} \leqslant C_4\frac{x}{\log x}\,.$$
Altogether,
$$\rho_2(x) - \pi_2(x) \in O\biggl(\frac{x}{\log x}\biggr)\,,$$
which is of smaller order than $\frac{x\log \log x}{\log x}$.
It remains to prove the assertion about $\pi_2$. It is easy to see
$$\pi_2(x) = \sum_{p \leqslant \sqrt{x}} \Biggl(\pi\biggl(\frac{x}{p}\biggr) - \pi(p)\Biggr)\,,$$
and
$$\sum_{p \leqslant \sqrt{x}} \pi(p) = \frac{\pi(\sqrt{x})\bigl(\pi(\sqrt{x}) + 1\bigr)}{2} \in O\biggl(\frac{x}{(\log x)^2}\biggr)\,,$$
whence we obtain
\begin{align}
\pi_2(x) &\sim \sum_{p \leqslant \sqrt{x}} \pi\biggl(\frac{x}{p}\biggr) \\
&\sim \int_2^{\sqrt{x}} \frac{\pi(x/u)}{\log u}\,du \\
&\sim \int_2^{\sqrt{x}} \frac{x}{u\log u(\log x - \log u)}\,du \\
&= x\int_{\log 2}^{\log \sqrt{x}} \frac{dv}{v(\log x - v)} \\
&= \frac{x}{\log x}\int_{\log 2}^{\log \sqrt{x}}\frac{1}{v} + \frac{1}{\log x - v}\,dv \\
&= \frac{x}{\log x} \int_{\log 2}^{\log \frac{x}{2}} \frac{dw}{w} \\
&= \frac{x}{\log x}\biggl(\log \log \frac{x}{2} - \log \log 2\biggr)\\
&\sim \frac{x\log \log x}{\log x}\,.
\end{align}
In that, the second asymptotic equality is admittedly not obvious, it is using a lemma by Landau. We can avoid using that, at the cost of a messier calculation, for example
\begin{align}
\sum_{p \leqslant \sqrt{x}} \pi\biggl(\frac{x}{p}\biggr) &\sim \sum_{p \leqslant \sqrt{x}} \frac{x}{p(\log x - \log p)} \\
&= \int_{2-\varepsilon}^{\sqrt{x}} \frac{x}{u(\log x - \log u)}\,d\pi(u) \\
&= \pi(\sqrt{x})\frac{\sqrt{x}}{\log \sqrt{x}} - \int_{2}^{\sqrt{x}} \pi(u) \frac{d}{du}\biggl(\frac{x}{u(\log x - \log u)}\biggr)\,du \\
&\sim -\int_{2}^{\sqrt{x}} \operatorname{Li}(u)\frac{d}{du}\biggl(\frac{x}{u(\log x - \log u)}\biggr)\,du \\
&= -\operatorname{Li}(\sqrt{x})\frac{\sqrt{x}}{\log \sqrt{x}}
+ \int_2^{\sqrt{x}} \frac{x}{u\log u(\log x - \log u)}\,du
\end{align}
using the asymptotics of $\pi(x)$ and Riemann-Stieltjes integrals. The asymptotic equalities here are fairly straightforward to establish.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3987005",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
}
|
Discussion on the prolongation of the solution I have the following
$$(PC)\begin{cases}y'=\dfrac{1}{x+1}y-\dfrac{4x}{x+1}y^2\\y(3)=\dfrac{2}{b} \end{cases}\text{ with }b\ne 0$$
Determine for which values of $b$, it exists at least one solution defined in $I=[0,4]$.
$\textbf{My attempt:}$ I noticed that the function $f(x,y):=\dfrac{1}{x+1}y-\dfrac{4x}{x+1}y^2$ is defined in $(-\infty,-1)\cup (-1,+\infty)$ and, since $3\in(-1,+\infty)$, I studied the Cauchy problem for $x>-1$.
The differential equation admits a unique local solution and a unique prolongation in its maximal domain.
We have a Bernoulli equation so I substituted $z(x)=\dfrac{1}{y(x)}$ and this substitution defines the new Cauchy problem:
$$(PC')\begin{cases}z'+\dfrac{1}{x+1}z=\dfrac{4x}{x+1}\\z(3)=\dfrac{b}{2} \end{cases}$$
and I got the solution $y(x)=\dfrac{x+1}{2x^2+2b-18}$, so I imposed $2x^2+2b-18\ne 0$ and $y(x)\ne 0$ for the substitution that I did$\implies|x|\ne\sqrt{9-b}$ and $x\ne -1$. Then I considered the solution in
$$(-\infty,-\sqrt{9-b})\cup (-\sqrt{9-b},\sqrt{9-b})\cup (\sqrt{9-b},+\infty)\bigcap (-1,+\infty).$$
If we want $I\subseteq I_{\text{SOL}}$ I think that it's correct to impose that $(-\sqrt{9-b},\sqrt{9-b})$ must contain $[0,4]\implies$ $$\begin{cases} -\sqrt{9-b}<0\text{ or}\\\sqrt{9-b}>4\end{cases}\implies\begin{cases}b<9\text{ or}\\b<-7\end{cases}.$$ For the other case I imposed that $[0,4]\subseteq (\sqrt{9-b},+\infty)$, so
$\sqrt{9-b}> 0$ which defines only the condition $b< 9$.
I always have difficulties in the study of the prolongation of the solution on a certain interval... :-(
|
You found correctly that
$$
(1+x)z(x)=2(x^2+C)\implies 4·\frac{b}{2}=2·(9+C)
$$
The solution $y$ becomes singular where $z=\frac1y$ has a root. As $(x+1)$ has no root on $[0,4]$, it remains to consider the other side. If $C>0$ then there is no problem. For negative $C$ to avoid roots on the interval one needs $C<-16$. So either
*
*$C=b-9>0\iff b>9$ or
*$C=b-9<-16\iff b<-7$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3987698",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
How to determine h(x) in a polynomial partial fraction decompostion Im susposed to do a partial fractional division;
$$ \frac{-2x^2 + 8x - 9} {(x-1) (x-3)^2}$$ Now I used the formula and this is what I got;
$$\frac {A}{(x-1)} + \frac{B}{x-3} + \frac{C}{(x-3)^2}$$
Now I've put the 3 fractions together,and got this
$$ \frac{A(x-3)^2+B(x-3)(x-1) +C(x-1)} {(x-1) (x-3)} $$
Now I looked at the starting function looked what my x^2 ,x terms were on the right on the left, solved it and got this
$$f(x) = \frac{-2} {4(x+1)} + \frac{-5} {4(x-3)} + \frac {3} {2(x-1)^2} $$
Now after looking at the formula I'd realise that I'm missing my h(x) the whole part if you will ( not sure how it is said in english) Now the h(x) doesnt exist when m < n but in my case m = n so I should also be having that? But how do I get that part? the formula says simply h(x) + the A B and C parts.How do I calculate h(x)?
EDIT: To clarify what h(x) means and what I was exactly trying to figure out.
$$f(x) = h(x) + \frac{r(x)} {q(x)}$$
Where m = deg r and n = deg q
|
You made a mistake when you put three fractions together:
$$
\frac{A(x-3)^2+B(x-1)(x-3)+C(x-1)}{(x-1)(x-3)^\color{red}{2}}.
$$
Now you set
$$
A(x-3)^2+B(x-1)(x-3)+C(x-1)=-2x^2+8x-9
$$
to find $A,B,C$.
*
*Set $x=1$: $4A=-2+8-9$.
*Set $x=3$: $2C=-18+24-9$.
*Set $x=2$: $A-B+C=-8+16-9$.
Now it should be very straightforward to go on.
Notes.
If you write
$$
f(x)=\frac{-2x^2+8x-9}{(x-1)(x-3)^2}
$$
then you can find polynomials $h,p,q$ so that
$$
f(x)=h(x)+\frac{p(x)}{q(x)}
$$
where the $\deg p<\deg q$.
This is the preliminary step for partial fraction decomposition.
But note that
$$
\deg (-2x^2+8x-9)<\deg (x-1)(x-3)^2
$$
you have $h(x)=0$, $p(x)=-2x^2+8x-9$ and $q(x)=(x-1)(x-3)^2$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3990210",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
}
|
Euler's proof of Fermat's Last Theorem for cubes I'm working through Richard Freidburg's An Adventurer's Guide to Number Theory, and on p. 149, chapter 7, he presents Euler's proof of Fermat's Last Theorem for cubes.
He starts with $z^3 = x^3 + y^3$, where $x, y, z$ have no factors in common. This implies that $z$ is even and that $x$ and $y$ are odd.
He doesn't state explicitly, but I think his proof assumes $x > y$.
Because $x$ and $y$ are both odd, $x + y$ is even and $x - y$ is also even.
He defines $p = (x + y)/2$, $q = (x - y)/2$, therefore $x = p + q$ and $y = p - q$.
By substitution, binomial expansion and factoring, he converts $z^3 = x^3 + y^3$ to $z^3 = 2p(p^2+3q^2)$
He then presents 4 'remarks', that the reader should prove. I think I've got 1, 2, and 4 figured out & need help with 3:
Remark 1: $p$ and $q$ have no common factor
Proof: Suppose $p$ and $q$ have a common factor $a$, then $x = a(p\prime + q\prime)$ and $y = a(p\prime - q\prime)$, therefore $a$ is a common factor of both $x$ and $y$, but $x$ and $y$ have no common factors
Remark 2: $p^2 + 3q^2$ is odd
Proof: $x = p + q$, $x$ is odd, therefore either $p$ is odd or $q$ is odd. Both cannot be odd, nor can both be even. If $p$ is odd then $q$ is even and $p^2$ is odd, $q^2$ is even and thus $3q^2$ is also even, and $p^2 + 3q^2$ is an odd number plus an even number, hence an odd number. Conversely, if $p$ is even, then $p^2$ is also even, and $q$ is odd, as is $q^2$, thus $3q^2$ is also odd. In this case $p^2 + 3q^2$ is an even number plus an odd number, hence an odd number.
Remark 3: $q$ is not divisible by 3
(Hints provided by the author: For this you must use the fact proven earlier in this chapter, that every cube is congruent to 0, 1 or -1 modulo 9. If $q$ is divisible by 3, then $x \equiv y$ modulo 3, and hence $x^3 \equiv y^3$ modulo 9.)
Remark 4: $p$ and $p^2 + 3q^2$ have no prime factor in common, except possibly 3.
Proof: Follows from remark 1. $p$ and $q$ have no common factors, therefore $p^2$ and $q^2$ have no common factors. However if 3 is a factor of $p$, then $p$ and $3q^2$ share 3 as a common factor.
|
Well, if $x^3$ and $y^3$ are congruent to $0$ mod $9$, then so is $z^3$, which contradicts "no common factors". If they are both $1$ (resp $-1$) then $z^3$ is $2$ (resp $-2$) which contradicts that "cube should be congruent to $0,1,-1$".
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3993277",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
}
|
Using quadratic residue show that $2b^2+ 3$ cannot be a divisor of $a^2-2$
I have tried working with residue of $2$ modulo $p$ where $p$ is a divisor of $2b^2+3$ but couldn't stablish the result.
|
You have the right idea. Consider there are integers $a$ and $b$ with $2b^2 + 3 \mid a^2 - 2$. With $2b^2 + 3$, it's always odd and $\gt 1$, so it has at least one prime factor and all of these prime factors are odd. Let $p$ be any of these prime factors. Since $p \mid 2b^2 + 3$, then $p \mid a^2 - 2$. Thus, as you stated, $2$ must be a quadratic residue modulo $p$. The table in the Law of quadratic reciprocity section shows $2$ is a quadratic residue mod $p$ if and only if
$$p \equiv 1, 7 \pmod{8} \tag{1}\label{eq1A}$$
Note since $7 \equiv -1 \pmod{8}$, the product of any number of these factors will only be congruent to either $1$ or $7$ mod $8$.
However, if $b$ is even, then $2b^2$ has a factor of $2 \times 2^2 = 8$, so $2b^2 + 3 \equiv 3 \pmod{8}$. Alternatively, if $b$ is odd, then $b^2 \equiv 1 \pmod{8}$, so $2b^2 + 3 \equiv 2 + 3 \equiv 5 \pmod{8}$. In both cases, it's not congruent to $1$ or $7$ mod $8$, meaning $2b^2 + 3$ must have at least one odd prime factor for which $2$ is not a quadratic residue. Thus, this means $p \not\mid a^2 - 2$, so
$$2b^2 + 3 \not\mid a^2 - 2 \tag{2}\label{eq2A}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3995118",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
Finding the intervals of increase and decrease and concavity of $y=\frac{\sqrt{1+|x-2|}}{1+|x|}$ I have the function $f(x)=\frac{\sqrt{1+|x-2|}}{1+|x|}$, which I need to analyze. I have already proven that $\lim\limits_{x \to +-\infty}{f(x)}=0$, therefore the function has only a horizontal asymptote at $y=0$. Then I looked at three cases for the derivative
*
*$x>2 \Longrightarrow f'(x)=\frac{1-3x}{2\sqrt{x-1}(1+x)^2}$
*$x \in[0,2]\Longrightarrow f'(x)=\frac{x-7}{x(1+x)^2\sqrt{3-x}}$
*$x<0 \Longrightarrow f'(x)=\frac{5-x}{2(1-x)^2\sqrt{3-x}}$
Now I need to look at the intervals of concavity of the function. I suppose I have to differentiate all derivatives for a second time and look at the results in each of the corresponding intervals. But this would require too much calculations. Isn't there a simpler way to find the intervals of increase/decrease and of concavity?
|
$$f(x)=\left\{
\begin{array}{ll}
\frac{\sqrt{3-x}}{1-x} & x<0 \\
\frac{\sqrt{3-x}}{x+1} & 0 \le x<2 \\
\frac{\sqrt{x-1}}{x+1} & x\geq 2 \\
\end{array}\right.
$$
$$
f'(x)=
\left \{
\begin{array}{ll}
\frac{5-x}{2 \sqrt{3-x} (x-1)^2} & x<0 \\
\frac{x-7}{2 \sqrt{3-x} (x+1)^2} & 0<x<2 \\
\frac{3-x}{2 \sqrt{x-1} (x+1)^2} & x\ge2 \\
\end{array}\right.
$$
$$
f''(x)=
\left\{
\begin{array}{ll}
\frac{-3 x^2+30 x-59}{4 (3-x)^{3/2} (x-1)^3} & x<0 \\
\frac{3 x^2-42 x+83}{4 (3-x)^{3/2} (x+1)^3} & 0<x<2 \\
\frac{3 x^2-18 x+11}{4 (x-1)^{3/2} (x+1)^3} & x>2 \\
\end{array}\right.
$$
Plot below, for reference
$$...$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3995882",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
}
|
Show that there is only one real root of the equation $1 + a^2 + ax - x^3$ I have to show that for every $a>0$ the equation $1 + a^2 + ax - x^3 = 0$, has exactly one solution. I have made the graph of the function $f(x) = 1 + a^2 + ax - x^3$ on desmos and I can clearly see that. But, how can I prove it explicity?
Thanks.
|
Oh... for shucks and giggles.
All cubic have at least one real solution. So let $x=r$ be one real root.
Then $f(x)=(1+a^2) + ax -x^3 = (x-r)(Ax^2 + Bx + C) = Ax^3 + (B-Ar)x^2 + (C-Br)x -Cr$ for some
$A=-1; B+r=0; C-Br= a; -Cr=1+a^2$
So $B = -r; C = -\frac {1+a^2}r$ (assuming $r \ne 0$ but as $f(0) = 1+a^2 \ne 0$ we know $r \ne 0)$.
And $-\frac {1+a^2}r + r^2 = a$.
So $f(x) = (x-r)(-x^2 -rx -\frac {1+a^2}r)$
Any other root of $f(x)$ must be $x = \frac {-r \pm \sqrt{r^2 -4\frac {1+a^2}r}}2=\frac {-r\pm \sqrt{r^2 - 4(r^2 -a)}}2$
But $r^2 \ge 0$ and $a > 0$ so $r^2 - 4(r^2 -a)=-3r^2 -a=-(r^2 + a) < 0$. so those solutions can never exist.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3996005",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 2
}
|
How can I prove $\sqrt{3}+\sqrt{4}+...+\sqrt{n}<\frac{1}{4}n^2$ for all whole numbers $n \geq 3$? $\sqrt{3}+\sqrt{4}+...+\sqrt{n}<\frac{1}{4}n^2 ; n \geq 3$
Induction first step:
$n=3$
$\sqrt{3}<\frac{9}{4}$ which is true
Induction second step:
$n=k$
$\sqrt{3}+\sqrt{4}+...+\sqrt{k}<\frac{1}{4}k^2$
Induction third step:
$n=k+1$
$\sqrt{3}+\sqrt{4}+...+\sqrt{k}+\sqrt{k+1}<\frac{1}{4}(k+1)^2$
From the assumption in the second step we know that: $\sqrt{3}+\sqrt{4}+...+\sqrt{k}<\frac{1}{4}k^2$
From this we can assume that: $\sqrt{3}+\sqrt{4}+...+\sqrt{k}+\sqrt{k+1}<\frac{1}{4}k^2+\sqrt{k+1}$
So we know that $\frac{1}{4}k^2+\sqrt{k+1}$ is bigger than $\sqrt{3}+\sqrt{4}+...+\sqrt{k}+\sqrt{k+1}$. We can then replace $\sqrt{3}+\sqrt{4}+...+\sqrt{k}+\sqrt{k+1}$ in the statement that we want to prove with $\frac{1}{4}k^2+\sqrt{k+1}$
So:
$$\frac{1}{4}k^2+\sqrt{k+1}<\frac{1}{4}(k+1)^2$$
$$16k+16<4k^2+4k+1$$
$$0<4k^2-12k-15$$
When I solve this I get $\left( \frac{3+2\sqrt{6}}{2};+\infty \right)$ for positive numbers
--Edit--
$3\not\in\left( \frac{3+2\sqrt{6}}{2};+\infty \right)$, but we manually proved above that the statement is true for $3$.
In this case, we proved by induction that this statement is true for $n \geq 4$. We don't need to rewrite our proof but only include a new first induction step.
*Induction first step:
$n=4$
$\sqrt{3}+\sqrt{4}<\frac{16}{4}$ which is true
And because we manually proved for $3$ we can say that this statement is true for all whole numbers $n \geq 3$
Is this correct?
--
Thanks
|
You have a statement $S(n)$. You attempt to show $S(3)$, and that $S(n)$ implies $S(n+1)$ for all $n \geq 3$. Unfortunately this works only for $n \geq 4$.
You solve the conundrum by showing both $S(3)$ and $S(4)$, and since the induction statement is true for $n \geq 4$, $S(n)$ is also true for all $n \geq 5$.
It happens quite often that the induction statement is only true for $n \geq N$. In that case you prove $S(N)$ by hand and this proves $S(n)$ for all $n \geq N$. Then you try to prove $S(n-1)$, $S(N-2)$ and so on.
But $S(N)$ might also be false. In that case, if you find an $m \geq N$ where $S(m)$ is true, you know $S(n)$ is true for all $n \geq m$, and you try to prove $S(m-1)$ and so on as well. That would happen if you tried to prove $\text{LHS}<\text{RHS}-1000000$. You might not find any such $m$ because the statement is false for all $n$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4001045",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
A triangle $ABC$ with $c=60,l_c=10$ and $P=125$ A triangle $ABC$ is given with $c=60,l_c=10$ and $P=125$, where $AB=c$, $l_c$ is the angle bisector of $\measuredangle C$ and $P$ is the perimeter. Find $a$ and $b$.
Here are my thoughts:
We have $\begin{cases}\dfrac{m}{n}=\dfrac{b}{a}\\l_c^2=ab-mn\end{cases}.$ From the first equation we can get $$\dfrac{m+n}{n}=\dfrac{a+b}{a}$$ or $$\dfrac{12}{n}=\dfrac{13}{a}.$$ Then we can substitute $l_c=10$ into the second equation to get $$100=ab-mn.$$ I am stuck here. Thank you in advance!
Katherine
|
Length of angle bisector is given by
$\displaystyle l_c^2 = \frac{ab}{(a+b)^2} ((a+b)^2 - c^2)$
$a+b = 125-60 = 65, c = 60$
Leads to $\displaystyle 10^2 = \frac{ab}{65^2}(65^2-60^2)$
i.e., $ab = 676 \implies a(65-a) = 676$
We get a quadratic $a^2 - 65a + 676 = 0 \implies a = 13, 52$
So $a = 13, b = 52$ or $a = 52, b = 13$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4003400",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
}
|
Calculate $\displaystyle \lim_{x \to 3} \frac{\sqrt{19-x} - 2\sqrt[4]{13+x}}{\sqrt[3]{11-x} - x + 1}$ Calculate:
$$\displaystyle \lim_{x \to 3} \frac{\sqrt{19-x} - 2\sqrt[4]{13+x}}{\sqrt[3]{11-x} - x + 1}$$
The problem with that case is that the roots are in different powers so multiplication in nominator and denominator by conjugate is not an option (at least I think it's not).
|
Here is a simple strategy. Find the limits of those radicals and subtract that limit from them.
Thus for numerator the term $\sqrt{19-x}\to 4$ and hence subtract and add $4$ to get $$\sqrt{16-x}-4+4$$ The other term in numerator is $2\sqrt[4]{13+x}$ which tends to $4$ and replace it with $$2\sqrt[4]{13+x}-4+4$$ and see that that those extra $4$'s added in both terms cancel out (well such limit problems are designed in such manner that numerator and denominator tend to $0$ and thus we have the desired cancelation here). Thus numerator equals $$\sqrt{19-x}-4-(2\sqrt[4]{13+x}-4)$$ and similarly denominator equals $$\sqrt[3]{11-x}-2-(x-3)$$ Now put $x-3=h$ so that $h\to 0$ and divide each term in numerator and denominator by $h$ to get $$\lim_{h\to 0}\dfrac{\dfrac{\sqrt{16-h}-4}{h}-2\cdot\dfrac{\sqrt[4]{16+h}-2}{h}}{\dfrac{\sqrt[3]{8-h}-2}{h}-1}$$ You can now evaluate the limit of three fractions easily and get the final answer as $$\frac{(-1/8)-2(1/32)} {(-1/12)-1} =\frac{9}{52}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4004897",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 4
}
|
Prove that $\left \lfloor \frac{1+\lfloor na+1/a\rfloor}{a} \right \rfloor=n$ If $a \geq \frac{1+\sqrt{5}}{2}$ and $n \in \Bbb W$, prove that
$$\left \lfloor \frac{1+\left\lfloor \frac{1+na^2}{a}\right\rfloor}{a} \right \rfloor=n.$$
I could prove only when $a$ is an integer, that is $a \geq 2$. If $a \in \Bbb Z$ we have: $$\left \lfloor \frac{1+na^2}{a} \right \rfloor=na+\left\lfloor \frac{1}{a} \right\rfloor=na$$ so we get: $$\left \lfloor \frac{1+\left\lfloor \frac{1+na^2}{a}\right\rfloor}{a} \right \rfloor=\left \lfloor \frac{1+na}{a} \right \rfloor=n+\left\lfloor \frac{1}{a} \right\rfloor=n.$$ But what if $a \notin \Bbb Z$?
|
We need to prove
*
*$$ \frac{1+\lfloor \frac{1+na^2}{a}\rfloor}{a} \geqslant n $$
*$$ \frac{1+\lfloor \frac{1+na^2}{a}\rfloor}{a} < n+1 $$
*
*$$ 1+\left\lfloor \frac{1+na^2}{a}\right\rfloor > \frac{1+na^2}{a} > an$$
so dividing by $a$ we obtain what we wanted to.
*$$ \frac{1+\lfloor \frac{1+na^2}{a}\rfloor}{a} \leqslant \frac{1+\frac{1+na^2}{a}}{a}$$
and
$$ 1+na^2 \leqslant na^2+a^2-a $$
since $0 \leqslant a^2 - a-1 = \left(a - \frac{1+\sqrt{5}}{2}\right)\left(a - \frac{1-\sqrt{5}}{2}\right)$, so after some transformations we arrive at
$$ \frac{1+\lfloor \frac{1+na^2}{a}\rfloor}{a} \leqslant n+1 $$
Now, the equality may hold only if both inequalities we used are actually equalities, so $a = \frac{1+\sqrt{5}}{2}$ and $\frac{1+na^2}{a}$ is an integer. But since $a^2=a+1$,
$$ \frac{1+na^2}{a} = \frac{(n+1) + na}{a} = \frac{n+1}{a} + n, $$
which is irrational, in particular it can't be integer.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4006683",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
}
|
Prove that the algebraic expression is greater than zero Prove that $x^8 - x^5 + x^2 -x + 1 >0$ for $x \in \mathbb{R}$.
It's from the (junior) high-school competition and the idea is that not everyone there knows calculus, so that I'm looking for more "basic" justification.
My idea was to use AM-GM: $x^8 + x^2 \geq 2\sqrt{x^{10}} = 2x^5$. Thus,
$$x^8 - x^5 + x^2 -x + 1 \geq x^5-x + 1 $$
So that we can now focus on proving that $x^5 - x + 1 > 0 $ but I don't really see how to do so without calculus...
|
Let's complete the alternate sign polynomial with $\begin{cases}a=x^{8}-x^{5}+x^{2}-x+1\\b=-x^{7}+x^{6}+x^{4}-x^{3}\end{cases}$
When $x\le 0$ then $(x^5+x)\le 0$ and since $(x^8+x^2+1)\ge 1$ we conclude easily that $a>0$.
So let's focus on the case $x>0$.
Notice that $a+b=\dfrac{x^9+1}{x+1}>0$
But since $b=(1-x)(x^6-x^3)=-x^3(1-x)^2(x^2+x+1)\le 0$
We can conclude that $a=(a+b)+(-b)>0$ in this interval.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4007910",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 7,
"answer_id": 5
}
|
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.