Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Ellipse: Most direct algebraic demonstration that $\left|x\right|$ is maximal when $y=0$ Yes, I know this is the equation of an ellipse.
This is the starting point of the problem:
$$
r_1+r_2=2a,
$$
where
$$r_1\equiv\sqrt{(x+c)^2+y^2}; r_2\equiv\sqrt{(x-c)^2+y^2}.$$
I want to establish certain properties prior to deriving a parametric expression for the ellipse, or even the "standard form",
$$
\frac{x^2}{a^2}+\frac{y^2}{b^2}=1.
$$
I can do all of this with a couple tacks, a string and a pencil, but
I want to show things algebraically. Preferably in a way that is directly applicable to all second-degree plane curves.
The first property I wish to establish is the range of $x$ values.
I know from experience that the answer is going to be that the extreme
values of $x$ will occur when $y=0$. The question that I am asking
in this post is: what is the most direct (algebraic) way to show that $\left|x\right|$
is greatest when $y=0$?
I had hoped to establish that fact, and then use another (absolute value)
method of determining the actual values of $x$ when $y=0$. But, the best
I cam up with is this very long-winded demonstration.
First some definitions: Let $a,c,x,y\in\mathbb{R}$ where $a>c>0$
are constants, and $x$ and $y$ are variables such that, for $r_1\equiv\sqrt{(x+c)^2+y^2}$
and $r_2\equiv\sqrt{(x-c)^2+y^2}$ we have the equation
$$
r_1+r_2=2a.
$$
The degenerate case has been omitted.
I will presume the greatest value to be $\left|x\right|=a$, and take
the case of $x=a$. We square our equation
$$
(r_1+r_2)^2=4a^2
$$
$$
=r_1^2+2r_1 r_2+r_2^2
$$
$$
=(a+c)^2+y^2+2r_1 r_2+(a-c)^2+y^2
$$
$$
=2(a^2+c^2+y^2)+2r_1 r_2.
$$
So
$$
r_1(2a-r_1)=a^2-c^2-y^2
$$
$$
=2a\sqrt{(a+c)^2+y^2}-((a+c)^2+y^2)=a^2-c^2-y^2.
$$
Transpose, cancel and simplify
$$
2a\sqrt{(a+c)^2+y^2}=2a^2+2ac,
$$
$$
r_1=\sqrt{(a+c)^2+y^2}=a+c.
$$
And, therefore
$$
r_2=2a-r_1=a-c.
$$
Now, by definition we have
$$
\sqrt{\left(a+c\right)^2}\equiv\left|a+c\right|=r_1
$$
and
$$
\sqrt{(a-c)^2}\equiv\left|a-c\right|=r_2.
$$
If $y\ne0$ while $x=a$, then
$$
r_1=\sqrt{(a+c)^2+y^2}=\left|a+c\right|+\delta_1,
$$
and
$$
r_{2}=\sqrt{(a-c)^2+y^2}=\left|a-c\right|+\delta_2,
$$
where both $\delta_1>0$ and $\delta_2>0.$
Since $a>c>0,$ we have $r_1+r_2=2a+\delta_1+\delta_2$, in
contradiction to the equation we seek to satisfy. So $x=a\implies y=0.$
If $x>a$, let $\delta_3=x-a>0.$
So
$$
r_1+r_2=a+c+\delta_3+a-c+\delta_3>2a;
$$
another contradiction.
Thus far we have dertermined that $x=a\implies y=0,$ and for any
$y$ that $x\le a$.
If we set $x=-a,$ then
$$
(r_1+r_2)^2=4a^2
$$
$$
=r_1^2+2r_1 r_2+r_2^2
$$
$$
=(-a+c)^2+y^2+2r_1 r_2+(-a-c)^2+y^2
$$
$$
=2(a^2+c^2+y^2)+2r_1 r_2
$$
$$
r_{1}\left(2a-r_{1}\right)=a^{2}-c^{2}-y^{2}
$$
$$
2a\sqrt{\left(-a+c\right)^{2}+y^{2}}-\left(\left(-a+c\right)^{2}+y^{2}\right)=a^{2}-c^{2}-y^{2}
$$
$$
2a\sqrt{\left(-a+c\right)^{2}+y^{2}}=2a^{2}-2ac
$$
$$
r_{1}=\sqrt{\left(a-c\right)^{2}+y^{2}}=a-c>0.
$$
$$
r_{2}=2a-r_{1}=a+c.
$$
Since we now have
$$
r_{1}=\sqrt{\left(a-c\right)^{2}+y^{2}}=a-c,
$$
and
$$
r_{2}=\sqrt{\left(a+c\right)^{2}+y^{2}}=a+c,
$$
the argument showing that $x=-a\implies y=0,$ is essentially the
same. Likewise for the argument that $x\ge-a.$
And that's just the $x$ range.
NB: I don't "need" any of this to fully characterize the ellipse. My goal is to find multiple ways of examining the same construct.
| There is a very simple and short answer. Using the definition of monotonicity given in Fundamentals of Mathematics, Volume 1: Foundations of Mathematics: The Real Number System and Algebra Edited by H. Behnke, F. Bachmann, K. Fladt, W. Süss and H. Kunle, which is a monotonic function is such that
$$x<y\implies f\left(x\right)<f\left(y\right)$$
or such that
$$x<y\implies f\left(y\right)<f\left(x\right),$$
and using the results of the elementary theory of numbers developed in Chapter 1B-1 for exponents, the proof is straight forward. That was the part I was missing. The result is "intuitively obvious", but I did not have a theorem to apply. In particular, I had no proof of the the blatantly obvious fact
$$\sqrt{(x-c)^2+y^2}\ge\sqrt{(x-c)^2}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2825926",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
$x,y$ and $z$ are consecutive integers, $\frac {1}{x}+\frac {1}{y}+\frac {1}{z}$... $x,y$ and $z$ are consecutive integers, such that $\frac {1}{x}+ \frac {1}{y}+ \frac {1}{z} \gt \frac {1}{45} $, what is the biggest value of $x+y+z$ ?.
I assumed that $x$ was the smallest number so that I could express the other numbers as $x+1$ and $x+2$ and in the end I got to a cubic function but I didn't know how to find its roots. I probably didn't do anything important so I'd appreciate if you give me any hints or help. Thanks in advance.
I have thought about using the AM-HM.
| Write $x = y-1$ and $z = y+1$. Then we have $$\frac{1}{y-1} + \frac{1}{y} + \frac{1}{y+1} = \frac{2y}{y^2 - 1} + \frac{1}{y} = \frac{3y^2 - 1}{y^3 - y} > \frac{1}{45}$$ Since $3y^2 - 1< 3y^2$, this implies $$\frac{3y}{y^2 - 1} = \frac{3y^2}{y^3 - y} > \frac{1}{45} \iff 135y > y^2- 1 \iff y-\frac{1}{y} < 135$$ As $y$ is assumed to be a positive integer, it follows that $y\le 135$. From here, you can check that $y=135$ certainly works for the initial inequality, so the largest possible value of $x+y+z = 3y$ is $3\cdot 135 = 405$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2826850",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Find all possible Jordan forms Let $A \in M_3(\mathbb{C})$, and:
$\frac1{12}A=[A^2-7A+16I_3]^{-1}$
Find all possible Jordan Forms of A
(no need to show different block orders)
I was thinking of Algebraic manipulations such as:
$12A^{-1}=A^2-7A+16I_3 \space/*A$
$12I_3=A(A^2-7A+16I_3)\space/*A$
$A(A^2-7A+16I_3)-12I_3=0 \space/$
$A^3-7A^2+16A-12I_3=0$
$(A-3)(A-2)^2$
Meaning Eigenvalues are 2 and 3.
Notice that Geometric Multiplicity of $\lambda_2$ can be 1 or 2.
So biggest Joran block can be of the sizes 1 or 2
finally:
$$
\begin{pmatrix}
2 & 1 & 0 \\
0 & 2 & 0 \\
0 & 0 & 3 \\
\end{pmatrix}
$$
or
$$
\begin{pmatrix}
2 & 0 & 0 \\
0 & 2 & 0 \\
0 & 0 & 3 \\
\end{pmatrix}
$$
I'd be happy if you could confirm my solution.
| You obtained $A^3-7A^2+16A-12I_3=0$ therefore the minimal polynomial $m_A$ divides the polynomial $x^3-7x^2+16x-12=(x-2)^2(x-3)$.
The options for the minimal polynomial and respective Jordan forms are:
*
*$m_A(x) = x-2$
$$\pmatrix{2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2}$$
*$m_A(x) = (x-2)^2$
$$\pmatrix{2 & 1 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2}$$
*$m_A(x) = x-3$
$$\pmatrix{3 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 3}$$
*$m_A(x) = (x-2)(x-3)$
$$\pmatrix{2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 3} \quad\text{ or} \quad\pmatrix{2 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 3}$$
*$m_A(x) = (x-2)^2(x-3)$
$$\pmatrix{2 & 1 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 3}$$
Note that in all these cases the matrix $A^2 -7A+16I_3$ is invertible because the polynomial $x^2-7x+16$ has complex zeros, so $\frac1{12}A=[A^2-7A+16I_3]^{-1}$ is indeed equivalent to $A^3-7A^2+16A-12I_3=0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2828323",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Solve the recurrence $T(n) = T(n-1) + 2T(n-2) + 2$ $T(0) = 1,
T(1) = 0$
I ain't able to get answer from any of the methods.
Substitution:
$T(n) = x^n $
\begin{align}
& x^n = x^{n-1} + 2x^{n-2} + 2 \\
& x^2 = x + 2 + 2x^2\\
& x^2 + x + 2 =0
\end{align}
solving this I will get a complex root.
\begin{align}
x & = \frac{-1 \pm \sqrt{1-8}}{2}
x & = \frac{-1 \pm 7i}{2}
\end{align}
Now how to go further.
General:
\begin{align}
T(2) & = 4\\
T(3) & = 6\\
T(4) & = 16\\
T(5) & = 30\\
T(6) & = 64\\
T(7) & = 126\\
T(8) & = 256
\end{align}
From this i can deduce if $n$ is even then $2^n$.
I cant deduce for if $n$ is odd.
| Hint try to make it homogeneous, from
$$T(n) = T(n-1) + 2T(n-2) + 2$$
$$T(n+1) = T(n) + 2T(n-1) + 2$$
we have
$$T(n)-T(n+1)=T(n-1) + 2T(n-2) - T(n) - 2T(n-1) \iff \\
T(n+1)-2T(n)-T(n-1)+2T(n-2)=0$$
leading to characteristic polynomial
$$x^3-2x^2-x+2=0 \iff (x-2)(x-1)(x+1)=0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2830215",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Inequalities with absolute values My question is:
Show that for all $|x-1|+|x-2|+\dots+|x-10| > 23$
I have solved above problem as below,
If $x-1 > 0$ and $x-2 > 0$ and ......$x-10 > 0$ then
LHS $= x-1+x-2+x-3+\dots+x-10 = 10x-55 > 23$ (because $x>10$)
If $x-1<0$ and $x-2 < 0$ and ......$x-10 < 0$ then
LHS $= -x+1-x+2-x+3.....-x+10 = -10x+55 > 23 $ (because $x<1$)
If the above solution is wrong, please give me the correct method
| Just use triangle inequality:
$$|x-1|+|x-2|+\dots+|x-10| = |1-x|+|2-x|+\dots+|5-x|+ |x-6|+|x-7|\dots+|x-10| >$$ $$|1-x+2-x+...+x-10| = |15-6-7-8-9-10| = |-25|$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2830651",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 1
} |
Specific case of Sherman–Morrison formula I want to find an easy solution to a matrix inverse for a very special case.
My matrix is of the form: $$D+vv^{T}$$
I know that I can use Sherman–Morrison formula to get
$$(D+vv^{T})^{-1}=D^{-1}-{D^{-1} vv^{T} D^{-1} \over 1+v^{T}D^{-1}v}$$
My D is actually of a very simple diagonal form. It is a diagonal matrix with all elements $1-s^2$. Thus, it looks like this
\begin{pmatrix} 1-s^2 & 0 & 0 \\
0 & 1-s^2 & 0 \\
0 & 0 & 1-s^2 \end{pmatrix}
and can be represented as $$D=(1-s^2)I$$ so that I can derive $$D^{-1}=\frac{1}{1-s^2}I$$ with $s$ being a scalar (between 0 and 1).
I actually have an additional information that all elements of the vector $v$ would be $s$. So actually $v=s*u$ with $u=(1,\dots,n)^T$.
From here, I kind of struggle to simplify the Sherman–Morrison formula.
1) Is the following correct?
$$\frac{1}{1-s^2}I-{\frac{1}{1-s^2}I vv^{T} \frac{1}{1-s^2}I \over 1+v^{T} \frac{1}{1-s^2}I v}$$
$$\frac{1}{1-s^2}I-{\frac{1}{1-s^2}I s^2uu^{T} \frac{1}{1-s^2}I \over 1+su^{T} \frac{1}{1-s^2}I su}$$
$$\frac{1}{1-s^2}I-\frac{s^2}{1-2s^2+s^4} {I uu^{T} I \over 1+\frac{s^2}{1-s^2} u^{T}Iu}$$
$$\frac{1}{1-s^2}I-\frac{s^2}{1-2s^2+s^4} {uu^{T} \over 1+\frac{s^2}{1-s^2} n}$$
2) Do you see a chance to simplify it even further?
| \begin{align}
\frac1{1-s^2}\cdot I-\frac{s^2}{({1-s^2)(1+s^2(n-1))}}\cdot\mathbf{1}
,
\end{align}
that is, all diagonal elements are
\begin{align}
\frac{1+s^2(n-2)}{(1-s^2)(1+s^2(n-1))}
,
\end{align}
and all the other elements are
\begin{align}
-\frac{s^2}{(1-s^2)(1+s^2(n-1))}
.
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2831504",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
General solution of the ordinary differential equation $(D^4+D^2+1)y=0$ $(D^4+D^2+1)y=0$ where $D=\frac{d}{dx}$
$$D^4+D^2+1 =0 \Rightarrow D^4 + D^2 + \frac14 +\frac{{3}}{4} = 0 \Rightarrow (D^2 + \frac12) = \pm \frac{\sqrt{3\iota}}{2} \Rightarrow D^2 = \frac{-1}{2} \pm \frac{\sqrt{3\iota}}{2} \Rightarrow D= \pm \sqrt{\frac{-1}{2} \pm \frac{\sqrt{3\iota}}{2}} = \pm \sqrt{e^{\pm(2/3)\pi\iota}} = \pm e^{\pm(1/3)\pi\iota} $$
Is there a way to simplify this?
| The roots of $\lambda^4+\lambda^2+1$ are $$ \lambda_1=\frac{1}{2} + i\frac{\sqrt{3}}{2}\\ \lambda_2=\frac{1}{2} - i\frac{\sqrt{3}}{2}\\ \lambda_3=-\frac{1}{2} + i\frac{\sqrt{3}}{2}\\ \lambda_4=-\frac{1}{2} - i\frac{\sqrt{3}}{2} $$
Thus, the solution to the ODE is
$$ y(x) = \sum_{k=1}^4 a_k\exp(x\lambda_k) $$
If we write $\mu_{1,2} = \pm\Re(\lambda_1), \sigma = \Im(\lambda_1)$, then
$$ y(x) = \sum_{k=1}^2 [b_k\cos(x\sigma) + c_k\sin(x\sigma)]\exp(x\mu_k) $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2832619",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Proving the product of four consecutive integers, plus one, is a square I need some help with a Proof:
Let $m\in\mathbb{Z}$. Prove that if $m$ is the product of four consecutive integers, then $m+1$ is a perfect square.
I tried a direct proof where I said:
Assume $m$ is the product of four consecutive integers.
If $m$ is the product of four consecutive integers, then write $m=x(x+1)(x+2)(x+3)$ where $x$ is an integer.
Then $m=x(x+1)(x+2)(x+3)=x^4+6x^3+11x^2 +6x$.
Adding $1$ to both sides gives us:
$m+1=x^4+6x^3+11x^2+6x+1$.
I'm unsure how to proceed. I know I'm supposed to show $m$ is a perfect square, so I should somehow show that $m+1=a^2$ for some $a\in\mathbb{Z}$, but at this point, I can't alter the right hand side of the equation to get anything viable.
| This is another way of looking at @jwg's answer.
Let the four consecutive numbers be $a,b,c,d$ and let $t$ be the number half-way between $b$ and $c$. Then
\begin{align}
abcd + 1
&= \bigg(t-\frac 32\bigg)\bigg(t-\frac 12\bigg)
\bigg(t+\frac 12\bigg)\bigg(t+\frac 32\bigg) + 1\\
&= \bigg(t^2-\frac 94\bigg)\bigg(t^2-\frac 14\bigg) + 1\\
&= t^4 - \frac 52 t + \frac{25}{16}\\
&= \bigg(t^2 - \frac 54 \bigg)^2
\end{align}
Letting $t = a + \frac 32$, we get
\begin{align}
abcd + 1
&= \bigg(t^2 - \frac 54 \bigg)^2 \\
&= \bigg(a^2 +3a + \dfrac 94 - \frac 54 \bigg)^2 \\
&= (a^2 +3a + 1)^2
\end{align}
This suggests the following solution.
\begin{align}
abcd + 1
&= a(a+1)(a+2)(a+3) + 1 \\
&= a(a+3) \cdot (a+1)(a+2) + 1 \\
&= (a^2+3a) (a^2+3a+2) + 1 \\
&= (a^2+3a+1 \ - \ 1)(a^2+3a+1 \ + \ 1) + 1 \\
&= (a^2+3a+1)^2 - 1 + 1 \\
&= (a^2+3a+1)^2
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2832986",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "18",
"answer_count": 13,
"answer_id": 12
} |
Convergence of sum of reciprocals of binomial coefficients I am interested in evaluating the sums
$$
\sum_{k=1}^{\infty}{\binom {2k}{k}^{-n}},
$$
where $n$ is a positive integer. It is already known that for $n=1$ we have
$$
\sum_{k=1}^{\infty}\frac{1}{\binom {2k}{k}}=\frac{9+2\sqrt 3 \pi}{27}.
$$
Several papers analyze the properties of sums involving the above identity (such as this one), however I was not able to find any material relating to the cases $n>1$. I already know these sums converge for all positive integers $n>0$, however I would be interested in finding a nice closed form for them as in the case $n=1$. How would I go about this? Are there results available about such sums in literature?
| I think that, for $n>1$, you are entering in the world of hypergeometric functions.
Just have a look to the table and notice the patterns
$$\left(
\begin{array}{cc}
n & S_n \\
2 & \frac{1}{4} \, _3F_2\left(1,2,2;\frac{3}{2},\frac{3}{2};\frac{1}{16}\right) \\
3 & \frac{1}{8} \,
_4F_3\left(1,2,2,2;\frac{3}{2},\frac{3}{2},\frac{3}{2};\frac{1}{64}\right) \\
4 & \frac{1}{16} \,
_5F_4\left(1,2,2,2,2;\frac{3}{2},\frac{3}{2},\frac{3}{2},\frac{3}{2};\frac{1}{25
6}\right) \\
5 & \frac{1}{32} \,
_6F_5\left(1,2,2,2,2,2;\frac{3}{2},\frac{3}{2},\frac{3}{2},\frac{3}{2},\frac{3}{
2};\frac{1}{1024}\right)
\end{array}
\right)$$ What is interesting is that $\log(S_n)$ is almost a linear function of $n$ (almost $\log(S_n)=-n \log(2)$).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2833496",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Solve the equation $X^2+X=\text{a given matrix}$ I want to solve the quadratic matrix equation
$$X^2+X=\begin{pmatrix}1&1\\1&1\end{pmatrix}$$
If I put $X$ in the form
$$X=\begin{pmatrix}a&b\\c&d\end{pmatrix}$$
then I find complicated equations. Is there a simple way to tackle the problem without using diagonalization?
| There is a matrix $S$ such that
$$
J=S\begin{bmatrix} 2 & 0 \\ 0 & 0 \end{bmatrix}S^{-1}
$$
If you set $Y=S^{-1}XS$, then $X=SYS^{-1}$ and the equation becomes
$$
S(Y^2+Y)S^{-1}=S\begin{bmatrix} 2 & 0 \\ 0 & 0 \end{bmatrix}S^{-1}
$$
that simplifies to
$$
Y^2+Y=\begin{bmatrix} 2 & 0 \\ 0 & 0 \end{bmatrix}
$$
We can try and complete the square:
$$
4Y^2+4Y=\begin{bmatrix} 8 & 0 \\ 0 & 0 \end{bmatrix}
$$
so the equation becomes
$$
(2Y+I)^2=\begin{bmatrix} 9 & 0 \\ 0 & 1 \end{bmatrix}
$$
Now the problem is simpler; if $2Y+I=\left[\begin{smallmatrix}a&b\\c&d\end{smallmatrix}\right]$, then the equation becomes
$$
\begin{bmatrix}
a^2 + bc & ab + bd \\
ac + cd & bc + d^2
\end{bmatrix}=
\begin{bmatrix} 9 & 0 \\ 0 & 1 \end{bmatrix}
$$
Case 1: $b=0$; then $a=\pm3$, $d=\pm1$ and $c=0$
Case 2: $a+d=0$: then $a^2+bc=9$ and $a^2+bc=1$: a contradiction.
Thus we have four solutions and
$$
Y=\frac{1}{2}\left(\begin{bmatrix} \pm3 & 0 \\ 0 & \pm1 \end{bmatrix}-I\right)
$$
so
$$
X=\frac{1}{2}S\left(\begin{bmatrix} \pm3 & 0 \\ 0 & \pm1 \end{bmatrix}-I\right)S^{-1}
$$
How do you determine the matrix $S$? Just determine an eigenvector relative to $2$ and one relative to $0$:
$$
S=\begin{bmatrix} 1 & -1 \\ 1 & 1 \end{bmatrix}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2836028",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 7,
"answer_id": 5
} |
On the property $\lvert x + y \rvert \leq \lvert \lvert x \rvert + \lvert y \rvert \rvert$ for complex numbers Suppose we have complex numbers, $x$ and $y$. My question is: we know, from the triangle inequality, that $\lvert x + y \rvert \leq \lvert \lvert x \rvert + \lvert y \rvert \rvert$. It makes sense, geometrically, when the strict inequality would hold. Whenever, however, would we have equality? Would it be necessary that $\lvert x \rvert = \rvert y \rvert$, or is there some other necessary condition?
Thanks.
| Let $x=a+bi, y=c+di,$ where $a,b,c,d \in \mathbb{R}.$ Then $$|x+y|=|(a+c)+(b+d)i|=\sqrt{(a+c)^2+(b+d)^2},$$ and $$|x|+|y|=\sqrt{a^2+b^2}+\sqrt{c^2+d^2}.$$ Thus, the inequality is equivalent to $$\sqrt{(a+c)^2+(b+d)^2} \leq \sqrt{a^2+b^2}+\sqrt{c^2+d^2},$$by squaring both sides, namely,$$ac+bd \leq \sqrt{(a^2+b^2)(c^2+d^2)}.$$
But in fact, by Cauchy's inequality, we have $$(a^2+b^2)(c^2+d^2) \geq (ac+bd)^2.$$ Hence,$$\sqrt{(a^2+b^2)(c^2+d^2)} \geq \sqrt{(ac+bd)^2}=|ac+bd|\geq ac+bd.$$
The equality holds if and only if $ad=bc$ and $ac+bd \geq 0.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2841302",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Trig function bounded on interval (without calculus), prove that $x^{3/2}\sin x + \sqrt{9-x^3}\cos x \leq 3$.
If $0 \lt x \lt \dfrac{\pi}{2}$, prove that
$$x^{3/2}\sin x + \sqrt{9-x^3}\cos x \leq 3$$
This question must be done without calculus. First, I tried splitting it into the intervals $(0,\pi/4)$ and $(\pi/4, \pi/2)$, hoping that, $\sin x$ was bound tightly enough on the interval that it'd be less than 3 even if $\cos x = 1$ (which doesn't work -- letting $\sin x = \dfrac{1}{\sqrt{2}}$ and $\cos x = 1$ produces a result greater than 3).
The other thing I noticed was that inside the square root sign, we have $\sqrt{9-x^3} = \sqrt{(3-x^{3/2})(3+x^{3/2})}$, and an $x^{3/2}$ appears in the first term, but I'm not sure how useful the similarity there is.
Advice on how to proceed?
| By Cauchy Schwarz $$ 9 = \left(x^3+(9-x^3) \right) \left(\sin^2 x + \cos^2 x\right) \ge \left(x^{3/2}\sin x + \sqrt{9-x^3}\cos x\right)^2$$ and the result follows
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2843337",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 0
} |
Compute $\int_0^{2\pi} \frac 1{\sin^4x+\cos^4x}dx$
Evaluate $\int_0^{2\pi} \frac 1{\sin^4x+\cos^4x}dx$
My attempt:
$I=\int_0^{2\pi}\frac 1{\sin^4x+\cos^4x}dx=\int_0^{2\pi}\frac 1{(\sin^2x+\cos^2x)^2-2\sin^2(2x)}dx=\int_0^{2\pi}\frac {1}{1-2\sin^2(2x)}dx=\frac 12\int_0^{4\pi}\frac 1{1-2\sin^2(x)}dx=\frac 12 \int_0^{4\pi}\frac {1}{\cos(\frac{x}2)}dx=\int_0^{2\pi}\frac 1{\cos x}dx=0$
So it actually is:
$$I=2\int_0^{2\pi}\frac {1}{2-\sin^2(2x)}dx=\int_0^{4\pi}\frac{1}{2-\sin^2(x)}dx=\int_0^{4\pi}\frac 1{1+\cos^2x}dx$$
Now if I try to make the substituion $u=\tan(\frac x2)$ I get integral from $0$ to $0$...Why?
What I am doing wrong?
| You know
$$\sin^4x+\cos^4x=1-\dfrac12\sin^22x=1-\dfrac12\left(\dfrac{1-\cos4x}{2}\right)$$
is periodic with period $T=\dfrac{\pi}{2}$, so write
$$\int_0^{2\pi} \frac 1{\sin^4x+\cos^4x}dx=4\int_0^\frac{\pi}{2}\dfrac{(1+\tan^2x)^2}{1+\tan^4x}dx=4\int_0^\infty\dfrac{1+t^2}{1+t^4}dt=4\dfrac{\pi}{\sqrt{2}}=\color{blue}{2\sqrt{2}\pi}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2843759",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 8,
"answer_id": 2
} |
Prove by induction $\sum_{i=1}^n 3i(i+4) = \frac{(n)(n+1)(2n+13)}{2}$ In this problem I have to prove the following equation using mathematical induction:
$$\sum_{i=1}^n 3i(i+4) = \frac{(n)(n+1)(2n+13)}{2}$$
So far I've proved that $P_1$ is true, and written $P_k$ as $$15+36+63+...+3k(k+4)= \frac{(k)(k+1)(2k+13)}{2}$$
I then attempt to solve $P_{k+1}$. I start by using $P_k$ and substituting it into the equation to get $$\frac{(k)(k+1)(2k+13)}{2}+(3k+1)(k+5)=\frac{(k+1)(k+2)(2k+14)}{2}$$
I then multiply $(3k+1)(k+5)$ by $\frac{2}{2}$ to get $$\frac{(k)(k+1)(2k+13)(3k+1)(2k+10)}{2}=\frac{(k+1)(k+2)(2k+14)}{2}$$
Here its obvious that the LHS isnt going to be equal to the RHS, seeing as the LHS would have $k^5$ amd the RHS would have $k^3$. Any ideas?
| There are three silly mistakes here: First off, your expression on the RHS should be $\frac{(k+1)(k+2)(2k+15)}{2}$ since.
Secondly you somehow made the addition into multiplication here
$$\frac{(k)(k+1)(2k+13)}{2}+(3k+1)(k+5)=\frac{(k+1)(k+2)(2k+14)}{2}$$
And finally as Bernard said you should have a $3(k+1)=3k+3$ instead of $3k+1$ in your inductive step.
Try again starting from here
$$\frac{(k)(k+1)(2k+13)}{2}+(3k+3)(k+5)=\frac{(k+1)(k+2)(2k+15)}{2}$$
and comment if you need more help.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2845489",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Complex Expression is Real
If I have the a complex number $z \in \mathbb{C}$ with absolute value $|z| = 1$, how do I show that $-i \frac {z-1}{z+1}$ is real?
| A number $w$ is real iff $\operatorname{Im}(w) = \frac{1}{2i}(w - \bar w) = 0\,$. In this case:
$$\require{cancel}
\begin{align}
-i \frac {z-1}{z+1} - \overline{\left(-i \frac {z-1}{z+1}\right)} &= -i \frac {z-1}{z+1} - \left(+i \frac{\bar z - 1}{\bar z + 1}\right) \\
&= -i \cdot\frac{(z-1)(\bar z + 1)+ (\bar z -1)(z+1)}{(z+1)(\bar z + 1)} \\
&= -i \frac{|z|^2+\bcancel{z}-\cancel{\bar z}-1+|z|^2+\cancel{\bar z} - \bcancel{z} - 1}{|z+1|^2} = \\
&= -2i \frac{\cancel{|z|^2}-\cancel{1}}{|z+1|^2} \\
&= 0
\end{align}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2846928",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 3
} |
ellipsoid of greatest volume is a sphere The equation of an ellipsoid is $$f(x,y,x)=(\frac xa)^2+(\frac yb)^2+(\frac zc)^2=1$$.
Given that the volume of an ellipsoid is $$V=\frac43\pi abc$$ and the constraint $$L=a+b+c$$ L some positive constant. Show that the ellipsoid with greatest volume is a sphere.
I should use Lagrange multipliers for this question.
I tried doing $$\nabla f = \lambda\nabla V$$ which gave an anwser making no sense$$<\frac{2x}{a^2},\frac{2y}{b^2},\frac{2z}{c^2}> =\lambda <0,0,0>$$
So then I tried $$"\nabla"V=\lambda"\nabla"L$$
where $"\nabla"$ treats a as x, b as y, c as z, and got
$$4\pi/3<bc,ac,ab>=\lambda<1,1,1>$$
so $ab=ac=bc$ gives $a=b=c$ a sphere. But why am I allowed to use $"\nabla"$ as such
| Here the lagrangian reads
$$
L(a,b,c,x,y,z,\lambda,\mu) = \frac 43 \pi a b c +\lambda\left(\left(\frac{x}{a}\right)^2+\left(\frac{y}{b}\right)^2+\left(\frac{z}{c}\right)^2-1\right)+\mu(a+b+c-L_0)
$$
The stationary conditions give
$$
\nabla L = \left\{
\begin{array}{rcl}
-\frac{2 \lambda x^2}{a^3}+\mu +\frac{4 b c \pi }{3}=0 \\
-\frac{2 \lambda y^2}{b^3}+\mu +\frac{4 a c \pi }{3}=0 \\
-\frac{2 \lambda z^2}{c^3}+\mu +\frac{4 a b \pi }{3}=0 \\
\frac{2 \lambda x}{a^2}=0 \\
\frac{2 \lambda y}{b^2}=0 \\
\frac{2 \lambda z}{c^2}=0 \\
\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}-1=0 \\
a+b+c-L_0=0 \\
\end{array}
\right.
$$
and solving gives
$$
\left[
\begin{array}{ccccccccc}
a & b & c & x & y & z & \lambda & \mu & V\\
\frac{L_0}{3} & \frac{L_0}{3} & \frac{L_0}{3} & x & y & -\frac{1}{3} \sqrt{L_0^2-9 x^2-9 y^2} & 0 & -\frac{4 L_0^2
\pi }{27} & \frac{4 L_0^3 \pi }{81} \\
\frac{L_0}{3} & \frac{L_0}{3} & \frac{L_0}{3} & x & y & \frac{1}{3} \sqrt{L_0^2-9 x^2-9 y^2} & 0 & -\frac{4 L_0^2 \pi
}{27} & \frac{4 L_0^3 \pi }{81} \\
\end{array}
\right]
$$
As can be observed the volume is maximum for $a = b = c\;$ defining a sphere. Note also the found relationship
$$
9z^2 = L_0^2-9x^2-9y^2
$$
which is a sphere.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2847461",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Hints with this Integration Problem $$\int_{0}^{\infty}\frac{(x \cos x-\sin x)\cos (x/2)}{x^3}\mathrm d x $$
I tried solving it by substitution, and I did come close to the answer. But that was really long and tedious, I don't even know whether or not what I did was even correct.
Can anyone suggest me some other way? Hints, anything?
Cuz I can't think of anything other than substitution, thoughts?
| $\begin{align}J&=\int_{0}^{\infty}\frac{(x \cos x-\sin x)\cos (x/2)}{x^3}\mathrm d x\end{align}$
Perform integration by parts,
$\begin{align}J&=\left[-\frac{1}{2x^2}(x \cos x-\sin x)\cos (x/2)\right]_{0}^{\infty}+\\
&\int_0^{\infty}\frac{-\frac{1}{2}\sin\left( \frac{1}{2}x\right)(x \cos x-\sin x)-x\sin x\cos\left( \frac{1}{2}x\right)}{2x^2}dx\\
&=\int_0^{\infty}\frac{-\frac{1}{2}\sin\left( \frac{1}{2}x\right)(x \cos x-\sin x)-x\sin x\cos\left( \frac{1}{2}x\right)}{2x^2}\,dx\\
&=-\frac{1}{4}\int_0^{\infty}\frac{\sin\left( \frac{1}{2}x\right)\cos x}{x}\,dx+\frac{1}{4}\int_0^{\infty}\frac{\sin\left( \frac{1}{2}x\right)\sin x}{x^2}\,dx-\frac{1}{2}\int_0^{\infty}\frac{\cos\left( \frac{1}{2}x\right)\sin x}{x}\,dx\\
&=-\frac{1}{4}\int_0^{\infty}\frac{\sin\left( \frac{1}{2}x\right)\cos x}{x}\,dx-\frac{1}{2}\int_0^{\infty}\frac{\cos\left( \frac{1}{2}x\right)\sin x}{x}\,dx+\left[-\frac{\sin\left( \frac{1}{2}x\right)\sin x}{4x}\right]_0^{\infty}+\\
&\frac{1}{4}\int_0^{\infty}\frac{\sin\left( \frac{1}{2}x\right)\cos x}{x}\,dx+\frac{1}{8}\int_0^{\infty}\frac{\cos\left( \frac{1}{2}x\right)\sin x}{x}\,dx\\
&=-\frac{3}{8}\int_0^{\infty}\frac{\cos\left( \frac{1}{2}x\right)\sin x}{x}\,dx\\
&=-\frac{3}{16}\int_0^{\infty}\frac{\sin\left( \frac{3}{2}x\right)+\sin\left( \frac{1}{2}x\right)}{x}\,dx\\
&=-\frac{3}{16}\int_0^{\infty}\frac{\sin\left( \frac{3}{2}x\right)}{x}\,dx-\frac{3}{16}\int_0^{\infty}\frac{\sin\left( \frac{1}{2}x\right)}{x}\,dx
\end{align}$
In the first intégral perform the change of variable $y=\dfrac{3}{2}x$,
in the second intégral perform the change of variable $y=\dfrac{1}{2}x$,
$\begin{align}J&=-\frac{3}{16}\int_0^{\infty}\frac{\sin x }{x}\,dx-\frac{3}{16}\int_0^{\infty}\frac{\sin x }{x}\,dx\\
&=-\frac{3}{8}\int_0^{\infty}\frac{\sin x }{x}\,dx\\
\end{align}$
But,
It's well-known that,
$\begin{align}\int_0^{\infty}\frac{\sin x }{x}\,dx=\frac{\pi}{2}\end{align}$
Therefore,
$\begin{align}J&=-\frac{3}{8}\times \frac{\pi}{2}\\
&=\boxed{-\frac{3}{16}\pi}
\end{align}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2849961",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Diophantine equation $a^2+b^2+c^2=a^2b^2$ I am trying to find all non trivial integers for which $a^2+b^2+c^2=a^2b^2$.
As suggested I have tried working (mod 4).
This is what I've gotten so far:
Squares can have a remainder of 0 or 1 (mod 4).
So the rhs can be 0 or 1 (mod 4). It is 1 if and only if both $a^2$ and $b^2$ have remainder 1 (mod 4).
But if both $a^2$ and $b^2$ are 1 the lhs has remainder 2 or 3 (if $c^2$ also has remainder 1), a contradiction.
So $a^2,b^2$ and $c^2$ must be divisible by 4.
Is this reasoning correct? If I have made any mistakes please let me know.
| Just work from parity.
If $a,b$ are both odd then $a^2+b^2\equiv 2 \pmod 4$ and $a^2b^2\equiv 1 \pmod 4$ This would mean that $c^2\equiv -1\pmod 4$ which is impossible.
If $a=2A$ and $b$ is odd then we get $1+c^2\equiv 0\pmod 4$ which is, again, impossible.
If $a,b$ are both even then $c$ is also even. Writing them as $2A,2B,2C$ we deduce that $$A^2+B^2+C^2=4A^2B^2$$
As before: if $A,B$ are both odd then we get $C^2\equiv 2\pmod 4$, which is impossible. If $A$ is even and $B$ is odd we get $C^2\equiv -1\pmod 4$ which is impossible. Hence, again, $A,B$ are both even. This repeats (with higher powers of $2$ appearing on the right) so infinite descent finishes the job.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2850016",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Solutions of $f(x+y^{n})=f(x)+[f(y)]^{n}$. Consider the functional equation $f(x+y^{n})=f(x)+[f(y)]^{n}$ where $f:\mathbb R \to \mathbb R$ and $n$ is given integer $>1$. This equation was discussed yesterday and it was shown that $f$ is necessarily additive. Assuming continuity it was concluded that $f(x)\equiv cx$ for some $c$. [ Necessarily $c$ is an n-th root of unity]. If $n$ is even then the given functional equation gives $f(x+y^{n}) \geq f(x)$ which easily leads to the conclusion that $f$ is an increasing function. It follows that $f$ is Borel measurable; since any Borel measurable additive function if of the type $f(x)\equiv cx$ the assumption that $f$ is continuous is not necessary. My question is what can be said for $n$ odd? Can one use some trick to prove that $f$ is necessarily Borel measurable? Or is there a counter-example? Discontinuous additive functions are constructed using Hamel basis but I am unable to use this method to construct a counter-example. I would appreciate receiving any ideas about this question.
| This is to show that if $n=5$ (so that $f(x+y^5)=f(x)+f(y)^5$ for all $x,y\in\mathbb{R}$), then $f(x)=0$, $f(x)=x$ or $f(x)=-x$. I think you can generalize the argument, so I leave the general cases to you.
As you mentioned, $f(x+y)=f(x)+f(y)$ for all $x, y\in \mathbb{R}$. Additionally, we have $f(x^5)=f(x)^5$. Since $f$ is additive, $f(qx)=qf(x)$ for all $q\in\mathbb{Q}$, $x\in\mathbb{R}$.
Since $f( (x+y)^5)= f(x+y)^5 = ( f(x)+f(y))^5$, we have
$$
\begin{align}
f & (x^5+5x^4 y + 10x^3y^2 + 10x^2y^3+5xy^4+y^5)\\
&=f(x)^5+5f(x)^4 f(y) + 10 f(x)^3f(y)^2+10f(x)^2f(y)^3+5f(x)f(y)^4+f(y)^5.
\end{align}
$$
Then, the first and the last term cancel due to $f(x^5)=f(x)^5$, $f(y^5)=f(y)^5$.
Since $5x^4 y + 10x^3y^2 + 10x^2y^3+5xy^4=(x+y)(5x^3y+5x^2y^2+5xy^3)$, we have
$$
\begin{align}
f((x+y)(5x^3y+5x^2y^2+5xy^3))=(f(x)+f(y))(5f(x)^3f(y)+5f(x)^2f(y)^2+5f(x)f(y)^3)
\end{align}
$$
Now, impose $x+y=q\in\mathbb{Q}\backslash\{0\}$ and substitute $y=q-x$, then
$$
f(5x^3y+5x^2y^2+5xy^3)=f(1)\left(5f(x)^3f(y)+5f(x)^2f(y)^2+5f(x)f(y)^3\right),
$$
hence
$$
\begin{align}
f & (x^3(q-x)+x^2(q-x)^2+x(q-x)^3)\\
&=f(1)\left(f(x)^3(qf(1)-f(x))+f(x)^2 (qf(1)-f(x))^2+f(x)(qf(1)-f(x)^3)\right)
\end{align}
$$
This is a polynomial identity in $q$, and holds for infinitely many values of $q$. Thus, it should also hold for $q=0$. Thus, we obtain
$$
f(-x^4)=-f(1)f(x)^4
$$
So, $f(x^4)=f(1)f(x)^4$.
Since $f(1)=f(1)^5$, $f(1)=0, 1, \mathrm{or} \ -1$. If $f(1)=0$, then $f(x)=0$. If $f(1)=1$, then $f$ is increasing, so $f(x)=x$. If $f(1)=-1$, then $f$ is decreasing, so $f(x)=-x$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2850534",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 2,
"answer_id": 0
} |
How do you evaluate limit of $\frac{\sqrt{1+x^2} - \sqrt {1+x}}{\sqrt {1+x^3} - \sqrt {{1+x}}}$ when $x$ tends to $0$? I tried rationalization method and got as $\frac{x^2-x}{\sqrt {1+x^3} - \sqrt {1+x} ({\sqrt{1+x^2} + \sqrt {1+x}})}$. But i feel the denominator having power of 3 I may be doing it wrong.The answer should be 1. Please help.
| $\sqrt{1+x^n}=1+x^n/2+O(x^{2n})$ as $x\to0$.
In particular $\sqrt{1+x}=1+x/2+O(x^2)$, $\sqrt{1+x^2}=1+O(x^2)$
and $\sqrt{1+x^3}=1+O(x^2)$. Then your fraction is
$$\frac{1+O(x^2)-(1+x/2+O(x^2))}{1+O(x^2)-(1+x/2+O(x^2))}
=\frac{-x/2+O(x^2)}{-x/2+O(x^2)}$$
etc.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2852294",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 3
} |
Evaluating definite integral of $p(x)$
Let $p(x)$ be fifth degree polynomial such that $p(x)+1$ is divisible by $(x-1)^3$ and $p(x)-1 $is divisible by $(x+1)^3 $. Then find the value of the definite integral $$\int _{-10}^{10}p(x)dx$$
Attempt:
$p(x)-1 = (x+1)^3 Q(x)$
$p(x)+1 = (x-1)^3 H(x)$
Where $Q(x)$ and $H(x)$ are unknown quadratics.
But there's not sufficient information to find $Q$ and $H$ thats why I am unable to proceed.
Please provide only a guiding hint, I want to solve it myself.
| Note that if $(x-1)^3$ divides $p(x)+1$, then $(x-1)^2$ divides $p^{\prime}(x)$
Similarly,
if $(x+1)^3$ divides $p(x)-1$, then $(x-1)^2$ divides $p^{\prime}(x)$
Since given that the degree is $5$ then when we differentiate it becomes $4$.
Therefore, $p^{\prime}(x)=t(x+1)^2(x-1)^2=t(x^4-2x^2+1)$ where $t\in\mathbb{R}$
Then $$p(x)=\dfrac t5x^5-\dfrac{2t}{3}x^3+tx+b\mbox{ $\{$ for $b\in\mathbb{R}$\}$$}$$
Since $(x+1)^3$ divides, we have $$p(-1)=-\frac a5+\dfrac{2a}{3}-a+b=1......(1)$$
Since $(x-1)^3$ divides, we have $$p(1)=\dfrac a5-\dfrac{2a}{3}+a+b=-1.......(2)$$
Now add $(1)+(2)$ gives $b=1$ and $a=-\dfrac{15}{8}$
Therefore, $p(x)=-\dfrac38x^5+\dfrac54x^3-\dfrac{15}{8}x$
Now $$\int_{-10}^{10}p(x)dx=\int_{-10}^{10}\left(-\dfrac38x^5+\dfrac54x^3-\dfrac{15}{8}x\right)dx=0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2852904",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 2,
"answer_id": 0
} |
$\frac{x^2+y^2}{x-y}|1995$. Find positive integer $x,y$
$x,y$ are positive integer . Find all $x,y$ such that
$\frac{x^2+y^2}{x-y}|1995$.
My answer: ${x^2+y^2}|1995$.
Therefore, $x^2+y^2$ can be $1$, $3$, $5$, $7$, $19$, $15$, $21$, $35$, $57$, $95$, $133$, $105$, $285$, $399$, $665$ or $1\,995$.
I observed that only $5$ can be represented as the sum of two square ( I haven't check the last two number because they are very big).
So, $(x,y)=(1,2),(2,1)$.
Am I right??
I believe there is a easier method.I am looking for that. Please help me.
| Your first assertion that $x^2+y^2 \mid 1995$ doesn't follow from your hypotheses. Note that $(6,3)$ and $(38,19)$ are also solutions. Certainly $(x^2+y^2)/(x-y) >1995$ for $x>1995$, so there can be only finitely many solutions. I believe that $(399,1197)$ is the largest.
I observe that if $a=(x^2+y^2)/(x-y)$ and that $(x,y)$ is a solution, then so is $(a-x,y)$.
Edit: I also observe that every solution is of the form $(2y,y)$ or $(3y,y)$ and in all cases $a=5y$.
Another Edit. Let $k$ be a divisor of $1995$. Then you have $x^2+y^2=k(x-y).$ Put everything on one side and complete the square:
$$\left(x-\frac{k}{2}\right)^2+\left(y+\frac{k}{2}\right)^2 = \frac{k^2}{2}.$$
Multiply through by $4$:
$$(2x-k)^2 +(2y+k)^2 = 2k^2.$$
So find, in the usual way, all the ways of writing $2k^2$ as a sum of two squares. Since many of the prime divisors are congruent to $3$ mod $4$, there are not many solutions. Then solve for $x$ and $y$. Do this for each $k$ and you have all solutions. Here's one example. Take $k=35$. So we find all solutions to
$$u^2+v^2 = 2\cdot35^2$$
and these are $(7,49)$ and $(35,35)$. The second one leads to $x=0$, which is not positive. The first one gives $2x-35 = 7$ and $2y+35 = 49$ which leads to $(x,y) = (21,7)$. Then from my observation above $y=35-21 = 14$ gives a second solution. Lather, rinse, repeat for all $k$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2854476",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
Finding a holomorphic function with real part $u=\frac{x(1+x^2+y^2)}{1+2x^2-2y^2+(x^2+y^2)^2}$. I am asked to
find the holomorphic function $f(z)=f(x\pm iy)$ with real part
$$u=\frac{x(1+x^2+y^2)}{1+2x^2-2y^2+(x^2+y^2)^2}$$
and such that $f(0)=0$.
I have tried to apply Cauchy-Riemann equations directly but I don't seem to be getting anywhere. I've also tried writing $u$ in terms of $z$ and $\bar{z}$ and then applying $f'=2\partial_z{u}$, but the expression of the derivative seems unmanagable. Since $z$ and $\bar{z}$ are symmetric in the expression of $u$, I also have the relation $\partial_z{u}=\overline{\partial_{\overline{z}}u}$, which I've tried to plug into $\partial_{\overline{z}}f=0$. But I dont know how to proceed from here.
| Hint: In general, use these substitutions
\begin{align}
x=\frac{z+\bar{z}}{2}\\
y=\frac{z-\bar{z}}{2i}\\
|z|^2=x^2+y^2\\
=z\bar{z}
\end{align}
Edit: After substitution
\begin{align}
u
&= \dfrac{\frac{z+\bar{z}}{2}\left(1+z\overline{z}\right)}{1+z^2+\overline{z}^2+z^2\overline{z}^2}\\
&= \dfrac12\dfrac{z(1+\overline{z}^2)+\overline{z}(1+z^2)}{(1+z^2)(1+\overline{z}^2)}\\
&= \dfrac12\left(\dfrac{z}{1+z^2}+\dfrac{\overline{z}}{1+\overline{z}^2}\right)
\end{align}
then
$$v=\dfrac{1}{2i}\left(\dfrac{z}{1+z^2}-\dfrac{\overline{z}}{1+\overline{z}^2}\right)$$
and
$$\color{blue}{f(z)=\dfrac{z}{1+z^2}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2855183",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Find the probability that atleast one valve is defective. A factory A produces $10$% defective valves and another factory $B$ produces 20% defective valves.A bag contains $4$ valves of factory $A$ and $5$ valves of factory B.If two valves are drawn at random from the bag,find the probability that at least one valve is defective.
$P(\text{at least one valve is defective})=\\=1-P(\text{none of the two valves are defective})=\\=1-\left(\frac{\binom{4}{2}}{\binom{9}{2}}(0.9)^2+\frac{\binom{5}{2}}{\binom{9}{2}}(0.8)^2+\frac{\binom{4}{1}\binom{5}{1}}{\binom{9}{2}}(0.9)(0.8)\right)=\frac{517}{1800}$,
but the answer given is $\frac{303}{1800}$ I don't know where i am wrong.
| The answer is indeed $\frac{517}{1800}$, because:
$$P(\text{at least one of two defective})=1-P(\text{both normal})=\\
1-[P(ANAN)+P(ANBN)+P(BNAN)+P(BNBN)]=\\
1-\left[\frac{4}{9}\cdot \frac9{10}\cdot \frac38\cdot \frac9{10}+
\frac{4}{9}\cdot \frac9{10}\cdot \frac58\cdot \frac8{10}+
\frac{5}{9}\cdot \frac8{10}\cdot \frac48\cdot \frac9{10}+
\frac{5}{9}\cdot \frac8{10}\cdot \frac48\cdot \frac8{10}\right]=\\
1-\frac{1283}{1800}=\frac{517}{1800}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2857533",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Find $\frac{1}{x-y+z}$ from given fraction equations. $$\begin{align}
\frac{3}{x}\,-\,\frac{4}{y}\,+\,\frac{2}{z}\quad&=\quad3\\
\frac{2}{x}\,-\,\frac{8}{y}\,-\,\frac{1}{z}\quad&=\,-\,8\\
\frac{4}{x}\,-\,\frac{6}{y}\,-\,\frac{3}{z}\quad&=\quad1\\
\frac{1}{x-y+z}\quad&=\quad?
\end{align}$$
I tried to find solution by equaling $\frac{1}{x}=a, \frac{1}{y}=b\,$ and $\frac{1}{z}=c\,.$ However, $b=\frac{149}{86}\,$ and $c=\frac{14}{43}$ which led me to ugly solution or no intended correct solution at all. How to find solution to the above-shown fraction equation?
Thanks beforehand.
| Go on, go on…
$$x=\frac{43}{133}\qquad y=\frac{86}{149}\qquad z=\frac{43}{14}$$
$$\frac1{x-y+z}=\frac{5662}{15953}=0.3549\dots$$
Beauty is not a requirement of mathematical problems – especially in applied problems, and even in pure problems.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2860033",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Algebraic Inequality involving AM-GM-HM If $$a,b,c \;\epsilon \;R^+$$
Then show, $$\frac{bc}{b+c} + \frac{ab}{a+b} + \frac{ac}{a+c} \leq \frac{1}{2} \Bigl(a+b+c\Bigl)$$
I have solved a couple of problems using AM-GM but I have always struggled with selecting terms for the inequality.
I tried writing $a+b+c$ as $x$ and re-writing the inequality as $$\frac{bc}{x-a} + \frac{ab}{x-c} + \frac{ac}{x-b} \leq \frac{1}{2} \Bigl(x\Bigl)$$
This didn't help so instead I wrote it as $$2\Biggl(\frac {1}{a} + \frac{1}{b} + \frac{1}{c}\Biggl ) \leq \frac{1}{2} \Bigl(a+b+c\Bigl)$$
Not sure what to to do next.
Any help would be appreciated.
| Multiply both sides by $4$ and rearrange:
$$\frac{bc}{b+c} + \frac{ab}{a+b} + \frac{ac}{a+c} \leq \frac{1}{2} \Bigl(a+b+c\Bigl) \Rightarrow \\
\left(a+b-\frac{4ab}{a+b}\right)+\left(b+c-\frac{4bc}{b+c}\right)+\left(c+a-\frac{4ca}{c+a}\right)\ge 0 \Rightarrow \\
\frac{(a-b)^2}{a+b}+\frac{(b-c)^2}{b+c}+\frac{(c-a)^2}{c+a}\ge 0.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2861098",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Common roots of recursive defined polynomial I have a series of polynomials $P_j(x)$ given by the recursive formula
$$P_{j+1}=\frac{e_j}{c_j}xP_{j}-\frac{f_j}{c_j}P_{j-1}
$$
with $P_{-1} \equiv 0$, $P_0 \equiv 1$, where
$$c_j = (j+1)(j+2\kappa+1),\\
e_j = (2j+2\kappa+1)(j+\kappa+1),\\
f_j = (j+\kappa)(j+\kappa+1),\\
j=0, \dots, N-1.$$
For $\kappa=\dfrac{1}{2}$ numerical results indicate that the roots of the polynomial $P_i$ are a subset of the roots of $P_{2i+1}$. E.g. The most simple case:
$$P_1(x)=\frac{3}{2}x, \qquad x_1=0\\
P_2(x)=\frac{5}{2}x^2-\frac{5}{8}, \qquad x_{1,2}=\pm\frac{1}{2},\\
P_3(x)=\frac{35}{8}x^3-\frac{35}{16}x \qquad x_1=0, \ x_{2,3}=\pm-\frac{1}{\sqrt{2}}.$$
How can this be proved or disproved?
| Define $Q_n(x) = \dfrac{(2n)!!}{(2n - 1)!!} P_{n - 1}(x)$ for $n \geqslant 0$, where $(-1)!! = 0!! = 1$, then $Q_0(x) = 0$, $Q_1(x) = 2$, and for $n \geqslant 0$,$$
(n + 1)(n + 2) P_{n + 1}(x) = (2n + 2)\left( n + \frac{3}{2} \right) xP_n(x) - \left( n + \frac{1}{2} \right)\left( n + \frac{3}{2} \right) P_{n - 1}(x)\\ \Longrightarrow Q_{n + 2}(x) = 2xQ_{n + 1}(x) - Q_n(x).$$
To prove that $P_n(x) \mid P_{2n + 1}(x)$ for $n \geqslant -1$, it is equivalent to prove that $Q_n(x) \mid Q_{2n}(x)$ for $n \geqslant 0$. Now it suffices to restrict the domains of all $Q_n$'s to $\mathbb{R} \setminus (-1, 1)$. Solving the recurrence relation,$$
Q_n(x) = \frac{1}{\sqrt{x^2 - 1}} ((x + \sqrt{x^2 - 1})^n - (x - \sqrt{x^2 - 1})^n), \quad \forall n \geqslant 0
$$
then for any $n \geqslant 0$,$$
\frac{Q_{2n}(x)}{Q_n(x)} = (x + \sqrt{x^2 - 1})^n + (x - \sqrt{x^2 - 1})^n \in \mathbb{R}[x] \Longrightarrow Q_n(x) \mid Q_{2n}(x).
$$
Therefore, $P_n(x) \mid P_{2n + 1}(x)$ for any $n \geqslant -1$, which implies the set of all roots of $P_n$ (multiplicity counted) is a subset of those of $P_{2n + 1}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2862104",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
} |
exercise on double integration I am not sure if i am not understanding the concept of double integration right, or if i am just making a silly mistake, but here is the problem:
given the function $f(x,y)=24xy, y>0, x>0$ and $x+y<=1 $ fint the $E[X,Y]$
$E[X,Y] = \int \int 24x^2y^2dxdy$ =
$24 \int_0^xx^2 (\int_0^{1-x}y^2dy) dx $ =
$24 \int_0^xx^2 [\frac{y^3}{3}]_0^{1-x}dx $=
$24 \int_0^xx^2 \frac{(1-x)^3}{3}dx $=
$8 \int_0^xx^2 (1-x)^3dx $=
$8[\frac{x^3}{x} \frac{(1-x)^4}{4}\frac{-x^2}{2}]_0^1$
= $ 8-\frac {5}{12}$
However, my professor gave us different solution:
$E[X,Y] = \int \int 24x^2y^2dxdy$ =
$24 \int_0^xx^2 (\int_0^{1-x}y^2dy) dx $ =
$3 \int_0^1 x^2 (1-x)^3 dx = $
$ \frac{2}{15}$
What is the problem with my calculation?
Where does the 3 come from in $3 \int_0^1 x^2 (1-x)^3 dx $ ?
| I believe you either misread '8' as '3', or there is a typo on your prof's part.
\begin{align}
E &= \int_{x=0}^{x=1}\int_{y=0}^{y=1-x} 24x^2y^2 \, dy \, dx \\
&= 24\int_{x=0}^{x=1}x^2\int_{y=0}^{y=1-x} y^2 \, dy \,dx \\
&= 24\int_{x=0}^{x=1}x^2\left[\frac{y^3}{3}\right]_{y=0}^{y=1-x} \,dx \\
&= 8\int_{0}^{1}x^2(1-x)^3 \,dx \\
&= 8\int_{0}^{1}x^2(1-x)^3 \,dx \\
&= 8\int_{0}^{1}x^2(1-3x+3x^2-x^3) \,dx \\
&= 8\int_{0}^{1}x^2-3x^3+3x^4-x^5 \,dx \\
&= 8\left[\frac{x^3}{3}-\frac{3x^4}{4}+\frac{3x^5}{5}-\frac{x^6}{6}\right]_0^1\\
&= 8\left(\frac 13-\frac34+\frac35-\frac16\right)\\
&= 8\cdot\frac{1}{60} \\
&=\frac2{15}
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2865126",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Linear algebra inequality (1) Note that if $\| \mathbf{A} \mathbf{x} \|_{2} = 1$, then
$$\| \mathbf{A}^{T} \mathbf{A} \mathbf{x} \|_{2} \cdot \| \mathbf{x} \|_{2} \ge \langle \mathbf{A}^{T} \mathbf{A} \mathbf{x}, \mathbf{x} \rangle = \| \mathbf{A} \mathbf{x} \|_{2}^{2} = 1$$
and thus we have $\| \mathbf{A}^{T} \mathbf{A} \mathbf{x} \|_{2} \ge \frac{1}{\| \mathbf{x} \|_{2}}$.
(2) By (1), if $\| \mathbf{A} \mathbf{x}_{1} \|_{2} = \cdots = \| \mathbf{A} \mathbf{x}_{r} \|_{2} = 1$, we have
$$\| \mathbf{A}^{T} \mathbf{A} \mathbf{x}_{1} \|_{2}^{2} + \cdots + \| \mathbf{A}^{T} \mathbf{A} \mathbf{x}_{r} \|_{2}^{2} \ge \frac{1}{\| \mathbf{x}_{1} \|_{2}^{2}}+\cdots+\frac{1}{\| \mathbf{x}_{r} \|_{2}^{2}}.$$
Here is my question:
If $\{ \mathbf{A}\mathbf{x}_{1}, \cdots, \mathbf{A}\mathbf{x}_{r} \}$ is orthonormal, can we obtain a better inequality than (2)?
Thanks.
| Consider $A=I$ and $x_i=e_i$ the $i$-th vector of the canonical basis. Then the set $\{ \mathbf{A}\mathbf{x}_{1}, \cdots, \mathbf{A}\mathbf{x}_{r} \}$ is orthonormal, but :
$$\| \mathbf{A}^{T} \mathbf{A} \mathbf{x}_{1} \|_{2}^{2} + \cdots + \| \mathbf{A}^{T} \mathbf{A} \mathbf{x}_{r} \|_{2}^{2} =r= \frac{1}{\| \mathbf{x}_{1} \|_{2}^{2}}+\cdots+\frac{1}{\| \mathbf{x}_{r} \|_{2}^{2}}.$$
Hence the inequality cannot be "better" even if we take orthogonality into account.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2866653",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
derivative of an integral coming from an expectation I just wanted to check on the following result from this paper (a very rough sketch of the proof is given in the supplementary material, starting from eq 8).
$$\left.\frac{\partial}{\partial c} \left(\frac{1}{2 \pi q} \int \phi(\sqrt{q} z_1) \phi\left(\sqrt{q} u\right) e^\frac{-z_1^2}{2} e^\frac{-z_2^2}{2} \mathrm{d}z_1 \mathrm{d}z_2\right)\right|_{c=1} = \frac{1}{\sqrt{2 \pi}} \int \phi'(\sqrt{q} z)^2 e^{\frac{-z^2}{2}} \mathrm{d}z,$$
where $u = c z_1 + \sqrt{1 - c^2} z_2$.
So I started to write down the derivative
$$\begin{align}
& = \frac{1}{2 \pi q} \int \phi(\sqrt{q} z_1) \frac{\partial}{\partial c} \left(\phi\left(\sqrt{q} u\right) \right) e^\frac{-z_1^2}{2} e^\frac{-z_2^2}{2} \mathrm{d}z_1 \mathrm{d}z_2 \\
& = \frac{1}{2 \pi q} \int \phi(\sqrt{q} z_1) \phi'\left(\sqrt{q} u\right) \sqrt{q} \frac{\partial u}{\partial c} e^\frac{-z_1^2}{2} e^\frac{-z_2^2}{2} \mathrm{d}z_1 \mathrm{d}z_2 \\
& = \frac{1}{2 \pi q} \int \phi(\sqrt{q} z_1) \phi'\left(\sqrt{q} u\right) \sqrt{q} \left(z_1 - \frac{c z_2}{\sqrt{1 - c^2}}\right)e^\frac{-z_1^2}{2} e^\frac{-z_2^2}{2} \mathrm{d}z_1 \mathrm{d}z_2 \\
& = \frac{1}{2 \pi q} \int \phi(\sqrt{q} z_1) \phi'\left(\sqrt{q} u\right) \sqrt{q} z_1 e^\frac{-z_1^2}{2} e^\frac{-z_2^2}{2} \mathrm{d}z_1 \mathrm{d}z_2 \\
& \qquad - \frac{1}{2 \pi q} \int \phi(\sqrt{q} z_1) \phi'\left(\sqrt{q} u\right) \sqrt{q} \frac{c z_2}{\sqrt{1 - c^2}} e^\frac{-z_1^2}{2} e^\frac{-z_2^2}{2} \mathrm{d}z_1 \mathrm{d}z_2 \\
& = \frac{1}{2 \pi} \int \phi'(\sqrt{q} z_1) \phi'\left(\sqrt{q} u\right) e^\frac{-z_1^2}{2} e^\frac{-z_2^2}{2} \mathrm{d}z_1 \mathrm{d}z_2 \\
& \qquad - \frac{1}{2 \pi q} \int \phi(\sqrt{q} z_1) \phi'\left(\sqrt{q} u\right) \sqrt{q} \frac{c z_2}{\sqrt{1 - c^2}} e^\frac{-z_1^2}{2} e^\frac{-z_2^2}{2} \mathrm{d}z_1 \mathrm{d}z_2,
\end{align}$$
but then I got stuck with this second term. If we replace $c = 1$ in the expression, the expected result appears in the first term, but I have no idea how the second term should become zero. Definitely not with the factor $\frac{c}{\sqrt{1 - c^2}}$ in that second term.
Would anybody see where I made a mistake, missed something or how to proceed to get the desired result?
| I have found my own mistake: I was a little to fast in using the identity $\int f(x) x e^{\frac{-x^2}{2}} = \int f'(x) e^{\frac{- x^2}{2}}$ in the last line. Acutally, the last line should have been
$$\begin{align}
& = \frac{1}{2 \pi} \int \left[\phi'(\sqrt{q} z_1) \phi'\left(\sqrt{q} u\right) + c \phi(\sqrt{q} z_1) \phi''(\sqrt{q}u)\right] e^\frac{-z_1^2}{2} e^\frac{-z_2^2}{2} \mathrm{d}z_1 \mathrm{d}z_2 \\
& \qquad - \frac{1}{2 \pi q} \int \phi(\sqrt{q} z_1) \phi'\left(\sqrt{q} u\right) \sqrt{q} \frac{c z_2}{\sqrt{1 - c^2}} e^\frac{-z_1^2}{2} e^\frac{-z_2^2}{2} \mathrm{d}z_1 \mathrm{d}z_2,
\end{align}$$
because $u$ also depends on $z_1$, which I did not account for in the last line of my question. To get to the final result, we need to apply the identity again for the last term
$$\begin{align}
& = \frac{1}{2 \pi} \int \phi'(\sqrt{q} z_1) \phi'\left(\sqrt{q} u\right) e^\frac{-z_1^2}{2} e^\frac{-z_2^2}{2} \mathrm{d}z_1 \mathrm{d}z_2 \\
& \qquad + \frac{c}{2 \pi} \int \phi(\sqrt{q} z_1) \phi''(\sqrt{q}u) e^\frac{-z_1^2}{2} e^\frac{-z_2^2}{2} \mathrm{d}z_1 \mathrm{d}z_2 \\
& \qquad - \frac{c}{2 \pi} \int \phi(\sqrt{q} z_1) \phi''\left(\sqrt{q} u\right) e^\frac{-z_1^2}{2} e^\frac{-z_2^2}{2} \mathrm{d}z_1 \mathrm{d}z_2 \\
& = \frac{1}{2 \pi} \int \phi'(\sqrt{q} z_1) \phi'\left(\sqrt{q} u\right) e^\frac{-z_1^2}{2} e^\frac{-z_2^2}{2} \mathrm{d}z_1 \mathrm{d}z_2, \\
\end{align}$$
which leads to the desired result.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2867113",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Minimum value of $\frac{x^4+5x^2+7}{x^2+3}$ Minimum value of $$f(x)=\frac{x^4+5x^2+7}{x^2+3}$$
we have $f(x)$ as
$$f(x)=(x^2+3)+\frac{1}{x^2+3}-1$$
Now by $AM \gt GM$ we have
$$(x^2+3)+\frac{1}{x^2+3} \gt 2$$
But equality cannot occur since $$x^2+3 \ne \frac{1}{x^2+3}$$
But my question is without using calculus is there any way to find minimum using AM, GM?
| Let $a\geq 1$ be fixed. Then, $g:[a,\infty)\to\mathbb{R}$ defined by $$g(t):=t+\frac{1}{t}\text{ for all }t\in[a,\infty)$$ is minimized at $t=a$. To show this, write
$$g(t)-g(a)=(t-a)\left(1-\frac{1}{at}\right)\,.$$
With the Weighted AM-GM inequality, we note that
$$g(t)=a^2\,\left(\frac{t}{a^2}\right)+\frac{1}{t}\geq \left(a^2+1\right)\left(\left(\frac{t}{a^2}\right)^{a^2}\,\frac{1}{t}\right)^{\frac{1}{a^2+1}}=\left(a^2+1\right)\,\left(\frac{t^{a^2-1}}{a^{2a^2}}\right)^{\frac{1}{a^2+1}}\,\,.$$
Since $t\geq a\geq1$, we get
$$g(t)\geq \left(a^2+1\right)\,\left(\frac{a^{a^2-1}}{a^{2a^2}}\right)^{\frac{1}{a^2+1}}=\left(a^2+1\right)\,\frac{1}{a}=g(a)\,.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2867849",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 0
} |
Showing that $\left|\frac{z}{z^3+1}\right|\leq\frac{R}{R^3-1}$
I am trying to show that if $|z|=R>1$, then
$$\left|\frac{z}{z^3+1}\right|\leq\frac{R}{R^3-1}$$
I've been playing around with the above inequality and this is where I am at.
\begin{align}
\text{Consider} \ \left|z^3+1\right|&\leq\left|z^3\right|+1 \ \ \ \ \ \text{(by the triangle inequality)}\\
&=\left|z\right|^3+1 \\
&=R^3+1 \\ \\
\Rightarrow \left|z^3+1\right|&\leq R^3+1 \\
\frac{1}{\left|z^3+1\right|}&\geq\frac{1}{R^3+1} \\
\frac{|z|}{\left|z^3+1\right|}&\geq\frac{|z|}{R^3+1} \ \ \ \ \ \ \text{(inequality unchanged, |z|>1)} \\
\left|\frac{z}{z^3+1}\right|&\geq\frac{R}{R^3+1} \\
\end{align}
I'm unsure of how to yield the desired inequality. Have I made a mistake somewhere?
EDIT
If I used the inequality $$|z_1+z_2|\geq |z_1|-|z_2|$$ I believe this will work. My question is, does this inequality hold for real numbers as well?
| WLOG $z=R(\cos t+i\sin t)$ where $t$ is real
$$|z^3+1|^2=R^6+1+2R^3\cos3t\ge R^6+1-2R^3= (R^3-1)^2$$
as $\cos3t\ge-1$ and $R^3>1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2870628",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Show that $AD=2AP$ if and only if $BP \perp CP$. Let the incircle of $\triangle ABC$ contact $AB,AC,BC$ at $D,E,F$ respectively, and intersect $AF$ at a second point $P$. Show that $AD=2AP$ if and only if $BP \perp CP$.
| Here is a not-so-pretty partial solution. I shall prove that, if $AD=2\cdot AP$, then $BP\perp CP$. This is a trigonometric solution, so it is quite boring.
Without loss of generality, suppose that $\angle B\geq \angle C$. Write $a:=BC$, $b:=CA$, $c:=AB$, and $s:=\dfrac{a+b+c}{2}$. Note that $$AP\cdot AF=AD^2\,.$$
Thus,
$$PF=AF-AP=\frac{AF^2-AD^2}{AF}\,.$$
Let $M$ be the midpoint of $BC$. Then,
$$PM^2=PF^2+FM^2+2\cdot PF\cdot FM\cdot\cos(\phi)\,,\tag{*}$$
where $\phi:=\angle AFB$.
We now find
$$AF^2=AB^2+BF^2-2\cdot AB\cdot BF\cdot\cos(\angle ABC)=c^2+(s-b)^2-2c(s-b)\left(\frac{c^2+a^2-b^2}{2ca}\right)\,.$$
Thus,
$$AF^2=\frac{(s-a)}{2a}\left(a^2+(b+c)a-(b-c)^2\right)\,.\tag{#}$$
Therefore,
$$\cos(\phi)=\frac{AF^2+BF^2-AB^2}{2\cdot AF\cdot BF}=\frac{(b-c)(s-a)}{a\cdot AF}\,.$$
Note that $FM=\dfrac{b-c}{2}$.
From (*), we get
$$PM^2=\left(\frac{AF^2-AD^2}{AF}\right)^2+\left(\frac{b-c}{2}\right)^2+\left(\frac{AF^2-AD^2}{AF}\right)(b-c)\left(\frac{(b-c)(s-a)}{a\cdot AF}\right)\,.$$
Due to (#), we get
$$AF^2-AD^2=\frac{s-a}{2a}\left(2a^2-(b-c)^2\right)\,.$$
If $AD=2\cdot AP$, then $AF=2\cdot AD=2(s-a)=b+c-a$, since $AD=s-a$.
Using (#), we see that
$$(b+c-a)^2=AF^2=\frac{(b+c-a)}{4a}\left(a^2+(b+c)a-2(b-c)^2\right)\,,$$
so
$$0=a^2+(b+c)a-2(b-c)^2-4a(b+c-a)=5a^2-3(b+c)a-2(b-c)^2\,.$$ From (*), we have
$$PM^2=\left(\frac{3}{2}AD\right)^2+\left(\frac{b-c}{2}\right)^2+\frac{3}{4}(b-c)^2\frac{(s-a)}{a}\,.$$
Using $AD=s-a$, we conclude that
$$PM^2-\left(\frac{a}{2}\right)^2=\frac{\big(a-3(b+c)\big)\left(5a^2-3a(b+c)-2(b-c)^2\right)}{16a}=0\,.$$
Consequently, $PM=\dfrac{a}{2}=BM=CM$. That is, the circle with diameter $M$ passes through $P$, whence $\angle BPC=\dfrac{\pi}{2}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2871347",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
Prove inequality inside a triangle "Let $ABC$ be a triangle with centroid $G$. A line $PQ$ is drawed in the triangle such that it passes through $G$ and intersects the sides $AB$ and $AC$ in $P$ and $Q$ respectively."
Prove that:
$$\frac {PB \cdot QC}{PA\cdot QA} \le \frac {1}{4}$$
I made an aproximate drawing of this problem. Unfortunately i don't know how to start with this kind of problem and i need some hints. Thanks.
| Let $M\in AC$ and $N\in AB$ such that $MN||BC$, and $G\in MN.$
Also, let $\measuredangle MGQ=\theta$.
Thus, in the standard notation $$MG=GN=\frac{1}{3}a$$ and by law of sines we obtain:
$$\frac{PN}{\sin\theta}=\frac{\frac{1}{3}a}{\sin\beta}$$ or
$$PN=\frac{a\sin\theta}{3\sin\beta}.$$
Similarly,
$$MQ=\frac{a\sin\theta}{3\sin\gamma}.$$
Id est, we need to prove that
$$\frac{\left(\frac{1}{3}c+\frac{a\sin\theta}{3\sin\beta}\right)\left(\frac{1}{3}b-\frac{a\sin\theta}{3\sin\gamma}\right)}{\left(\frac{2}{3}c-\frac{a\sin\theta}{3\sin\beta}\right)\left(\frac{2}{3}b+\frac{a\sin\theta}{3\sin\gamma}\right)}\leq\frac{1}{4}$$ or
$$\frac{(\sin\gamma\sin\beta+\sin\alpha\sin\theta)(\sin\gamma\sin\beta-\sin\alpha\sin\theta)}{(2\sin\gamma\sin\beta-\sin\alpha\sin\theta)(2\sin\gamma\sin\beta+\sin\alpha\sin\theta)}\leq\frac{1}{4}$$ or
$$\sin^2\alpha\sin^2\theta\geq0.$$
Done!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2873086",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Prove that: $\frac{bc}{a^2+1}+\frac{ac}{b^2+1}+\frac{ab}{c^2+1}\leq \frac{3}{4}$ Given three positive numbers a,b,c satisfying $$a^2+b^2+c^2=1$$ Prove that: $$\frac{bc}{a^2+1}+\frac{ac}{b^2+1}+\frac{ab}{c^2+1}\leq \frac{3}{4}$$
The things I have done so far:
$$\sum \limits_{cyc}\frac{bc}{a^2+1}=\sum \limits_{cyc}\frac{bc}{2a^2+b^2+c^2}\leq \sum \limits_{cyc}\frac{bc}{2ab+2ac}$$
$$=\sum \limits_{cyc}\frac{bc}{2a(b+c)}\leq \frac{1}{4}\sum \limits_{cyc}\frac{(b+c)^2}{2a(b+c)}=\frac{1}{4}\sum \limits_{cyc}\frac{b+c}{2a}$$
$$=\frac{1}{8}.\frac{\sum \limits_{cyc}bc(b+c)}{abc}=\frac{1}{8}.\frac{\sum \limits_{cyc}bc(b+c)+3abc}{abc}-\frac{3}{8}$$
$$=\frac{1}{8}.\frac{(a+b+c)(ab+bc+ca)}{abc}-\frac{3}{8}$$
$$\leq \frac{1}{8}.\frac{\sqrt{3(a^2+b^2+c^2)}(a^2+b^2+c^2)}{abc}-\frac{3}{8}$$
$$=\frac{\sqrt{3}}{8abc}-\frac{3}{8}$$
I don't know what to do anymore.
| By AM-GM and C-S we obtain:
$$\sum_{cyc}\frac{bc}{a^2+1}=\sum_{cyc}\frac{bc}{2a^2+b^2+c^2}\leq\frac{1}{4}\sum_{cyc}\frac{(b+c)^2}{a^2+b^2+a^2+c^2}\leq$$
$$\leq\frac{1}{4}\sum_{cyc}\left(\frac{b^2}{a^2+b^2}+\frac{c^2}{a^2+c^2}\right)=\frac{1}{4}\sum_{cyc}\left(\frac{b^2}{a^2+b^2}+\frac{a^2}{b^2+a^2}\right)=\frac{3}{4}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2875162",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
First Order Separable differential Equation
Problem:
Solve the following differential equation:
\begin{eqnarray*}
6x^2y \, dx - (x^3 + 1) \, dy &=& 0 \\
\end{eqnarray*}
Answer:
This is a separable differential equation.
\begin{eqnarray*}
\frac{6x^2}{x^3+1} \, dx - \frac{dy}{y} &=& 0 \\
\int \frac{6x^2}{x^3+1} \, dx - \int \frac{dy}{y} &=& c_1 \\
2 \ln{|x^3+1|} - \ln{|y|} &=& c_1 \\
\ln{(x^3+1)^2} - \ln{|y|} &=& c_1 \\
\ln{ \Big( \frac{(x^3+1)^2}{|y|} \Big) } &=& c_1 \\
(x^3+1)^2 &=& c|y| \\
\end{eqnarray*}
However, the book gets:
\begin{eqnarray*}
(x^3+1)^2 &=& |cy| \\
\end{eqnarray*}
Is my answer different from the book's answer? I believe it is. What am I missing?Any idea of how to proceed?
Thanks,
Bob
| $$\ln (x^3+1)^2=c_1+ \ln |y|$$
For $y > 0$
$$\ln (x^3+1)^2=c_1+ \ln y$$
$$(x^3+1)^2=e^{c_1} y$$
For $y < 0$
$$\ln (x^3+1)^2=c_1+ \ln (-y)$$
$$(x^3+1)^2=-e^{c_1} y$$
so we have that
$$(x^3+1)^2=|e^{c_1} y|=e^{c_1}|y|$$
On the other hand you have that
$$|ab| = |a||b|$$
so
$$|cy| = |c||y|$$
When in your book they choose $c=-2$ you take $e^{c_1}=2$ thats all
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2875468",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
Prove that inequality $1+\frac{1}{2\sqrt{2}}+...+\frac{1}{n\sqrt{n}}<2\sqrt{2}$ Let $n$ is a natural number. Prove that inequality $$1+\frac{1}{2\sqrt{2}}+\frac{1}{3\sqrt{3}}+...+\frac{1}{n\sqrt{n}}<2\sqrt{2}$$
My try: $$\frac{1}{n\sqrt{n}}=\frac{\sqrt{n}}{n^2}<\frac{\sqrt{n}}{n\left(n-1\right)}=\sqrt{n}\left(\frac{1}{n}-\frac{1}{n-1}\right)=\left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n-1}}\right)\left(\frac{\sqrt{n}+\sqrt{n-1}}{\sqrt{n-1}}\right)<\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n-1}}$$
Please check my solution for me and give me some idea.
| The original inequality cannot be proved directly by induction but by the solution given by Jack D'Aurizio is interesting to note that we can also proceed by induction proving the stronger result
$$1+\frac{1}{2\sqrt{2}}+\frac{1}{3\sqrt{3}}+...+\frac{1}{n\sqrt{n}}<2\sqrt{2}-\frac{2}{\sqrt{n+\frac12}}<2\sqrt{2}$$
indeed it is true for $n=1$ since
$$1<2\sqrt 2-2\sqrt{\frac23}\approx 1.20$$
and by induction step we obtain
$$1+\frac{1}{2\sqrt{2}}+...+\frac{1}{(n+1)\sqrt{n+1}}\stackrel{Hyp.}<2\sqrt{2}-\frac{2}{\sqrt{n+\frac12}}+\frac{1}{(n+1)\sqrt{n+1}}\stackrel{?}<2\sqrt{2}-\frac{2}{\sqrt{n+\frac32}}$$
which is true indeed
$$\frac{1}{(n+1)\sqrt{n+1}}<\frac{2}{\sqrt{n+\frac12}}-\frac{2}{\sqrt{n+\frac32}} \iff \frac{1}{2n\sqrt{n}}<\frac{1}{\sqrt{n-\frac12}}-\frac{1}{\sqrt{n+\frac12}}$$
and by convexity of $f(x)=\frac1{\sqrt {n-x}}-\frac1{\sqrt {n+x}}$ we have
$$f'(x)=\frac1{\sqrt {(n-x)^3}}+\frac1{\sqrt {(n+x)^3}}$$
$$\frac1{2n\sqrt {n}}<f'(0)=\frac2{n\sqrt {n}}\le f(1/2)-f(0)=\frac{1}{\sqrt{n-\frac12}}-\frac{1}{\sqrt{n+\frac12}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2876221",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 4,
"answer_id": 3
} |
Area of a triangle with ratio of a side In the drawing, $AM=MN=NC$ and $\frac {BP}{PC}=\frac {5}{3}$, if the area of the gray region is 8, whats the area of $\triangle ABC$?
I saw that $\triangle ABN$ and $\triangle BNC$ have the same height but their bases are in ratio $2:1$, and the sum of the areas of those triangles gives the final answer, but i don't know how to continue and how to use the area of the gray triangle. I think i'm close.
Any hints?
| Add an auxiliary segment through P parallel to AC intersecting BN at Q and AB at S. Call the intersection of BN and MP R.
Let $h_1$ be the altitude of triangle QRP
Let $h_2$ be the altitude of triangle MRN
Let $h_3$ be the altitude of triangle PMC
Let $h_4$ be the altitude of triangle BSP
Let $h_5$ be the altitude of triangle BAC
$h_3$ = $h_1$ + $h_2$
$h_5$ = $h_4$ + $h_3$
Triangle BQP is similar to triangle BNC. $\frac {BP}{BC}=\frac {5}{8}$, so $\frac {QP}{NC}=\frac {QP}{MN}=\frac {5}{8}$. Triangle QRP is similar to triangle NRM. So $h_1$ = $\frac {5}{8}$$h_2$
$h_3$ = $h_1$ + $h_2$ = $\frac {5}{8}$$h_2$ + $h_2$ = $\frac {13}{8}$$h_2$
Triangle BSP is similar to triangle BAC, so $h_4$ = $\frac {5}{8}$$h_5$
$h_5$ = $h_4$ + $h_3$ = $\frac {5}{8}$$h_5$ + $h_3$
$\frac {3}{8}$$h_5$ = $h_3$, so $h_5$ = $\frac {8}{3}$$h_3$ = $\frac {13}{3}$$h_2$
Area of RMN = 8 = $\frac {1}{2}$(MN)$h_2$
Area of BAC = $\frac {1}{2}$(3MN)$h_5$ = $\frac {1}{2}$(3MN)$\frac {13}{3}$$h_2$ = 13($\frac {1}{2}$(MN)$h_2$) = (13)(8) = 104
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2878961",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Find the equation of the common tangents Find the equation of the common tangents to the parabola $y^2=4ax$ and $x^2=4by$.
My Attempt :
$y^2=4ax$ is the equation of parabola with focus at $(a,0)$
$x^2=4by$ is the equation of parabola with focus at $(0,b)$
The equation of the tangent to the parabola $y^2=4ax$ is
$$y=mx+\dfrac {a}{m}$$
$$m^2x-my+a=0$$
The equation of the tangent to the parabola $x^2=4by$ is:
$$x=m'y+\dfrac {b}{m'}$$
$$(m')^{2}y - m'x+b=0$$
| Let $A(at^2,2at)$ and $B(2bu,bu^2)$ be the points on $C_1:y^2=4ax$ and $C_2:x^2=4by$ respectively.
Equation of tangent of $C_1$ at $A$:
$$\frac{x}{-at^2}+\frac{y}{at}=1 \quad \cdots \cdots \: (1)$$
Equation of tangent of $C_2$ at $B$:
$$\frac{x}{bu}+\frac{y}{-bu^2}=1 \quad \cdots \cdots \: (2)$$
Comparing $(1)$ and $(2)$,
$$
\left \{
\begin{array}{rcl}
-at^2 &=& bu \\
at &=& -bu^2 \\
\end{array}
\right.$$
On solving,
$$t=\frac{1}{u}=-\sqrt[3]{\frac{b}{a}}$$
The common tangent is
$$-\frac{x}{\sqrt[3]{ab^2}}-\frac{y}{\sqrt[3]{a^2b}}=1$$
$$\fbox{$\sqrt[3]{a}\, x+\sqrt[3]{b} \, y+\sqrt[3]{a^2 b^2}=0$}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2879161",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
Are there functions satisfying $\int f(x)g(x)dx=\int f(x)dx \times \int g(x)dx$ Are there functions satisfying $\int f(x)g(x)dx=\int f(x)dx \times \int g(x)dx$
I came up with one which is
$f(x)=e^{x \sec ^2 \alpha}$ and $g(x)=e^{x \csc^2 \alpha}$
$$\int f(x)g(x)dx=\frac{e^{x(\sec^2 \alpha+\csc^2 \alpha)}}{\sec^2 \alpha +\csc ^2 \alpha}$$
Also
$$\int e^{x\sec^2 \alpha}dx \times \int e^{x \csc^2 \alpha} dx=\frac{e^{x(\sec^2 \alpha+\csc^2 \alpha)}}{\sec^2 \alpha \csc ^2 \alpha}=\frac{e^{x(\sec^2 \alpha+\csc^2 \alpha)}}{\sec^2 \alpha +\csc ^2 \alpha}$$
| Let
$F = \int f,
G = \int g
$.
I'll try to get $f$
in terms of $g$.
Differentiating
$\int fg
=\int f \int g
$,
$fg
=g\int f+f\int g
=g\int f+fG
$
or,
in terms of $F$,
$F'(g-G)
=gF$
so
$\dfrac{F'}{F}
=\dfrac{g}{g-G}
$
or
$(\ln(F))'
=\dfrac{g}{g-G}
$
or
$\ln F
=\int(\dfrac{g}{g-G})
$
so,
$ F
=e^{\int(\dfrac{g}{g-G})}
$
and,
differentiating,
$f
=\dfrac{g}{g-G}e^{\int(\dfrac{g}{g-G})}
$.
Example:
Let
$g = x^n$
so
$G = \dfrac{x^{n+1}}{n+1}
$.
$\dfrac{g}{g-G}
=\dfrac{x^n}{x^n-\dfrac{x^{n+1}}{n+1}}
=\dfrac{1}{1-\dfrac{x}{n+1}}
=\dfrac{n+1}{n+1-x}
$
so
$\int\dfrac{g}{g-G}
=\int\dfrac{n+1}{n+1-x}
= -(n + 1) \log(n - x + 1)
$
and
$f
=\dfrac{n+1}{n+1-x}e^{-(n + 1) \log(n - x + 1) }
=\dfrac{n+1}{n+1-x}\dfrac1{(n-x+1)^{n+1}}
=\dfrac{n+1}{(n-x+1)^{n+2}}
$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2879633",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Smallest cube ending in $2017$
What is the smallest cube ending in $2017$?
My try: I know that the only possible units digit is $3$,
$$(a+3)^3 = 2017 \mod 10^4\;$$
and
$$a^3 + 9a^2 + 27a = 1990\mod 10^4$$
I don't know how to proceed, I tried factoring and adding $10^5$ and $1990$ but when I saw the answer, this approach would take forever.
| $x^3\equiv 7 \pmod {10}$. So $x\equiv 3 \pmod {10}$.
Put $x=10y+3$, and compute $\pmod {100}$.
This yields $1000y^3+900y^2+270y+27 \equiv 17 \pmod {100}$. Thus $70y \equiv -10 \pmod {100}$, or equivalently, $7y \equiv -1 \pmod {10}$. Divide by $7$: $y \equiv 7 \pmod {10}$.
Put $y=10z+7$, hence $x=100z+73$. I think you see the pattern now: calculate modulo $1000$, it gives you the $\pmod{10}$ remainder of $z$, and one more iteration gives you the $\pmod {10000}$ remainder of $x$, which is also the smallest solution.
(Solution: $9073$)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2880243",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
"answer_count": 3,
"answer_id": 2
} |
Find $x$, given: $x^2 + \frac{9x^2}{(x+3)^2} = 16$ Here's an equation:
$$x^2 + \frac{9x^2}{(x+3)^2} = 16$$
First, I subtracted 16 from both sides and factored $x^2$ so I would get a quadratic equation, but with no success. Also, I can see that the equation can be rewritten as:
$$(x+4)(x-4) + \left(\frac{3x}{x+3}\right)^2 = 0$$
But I can't see how can I use that information.
What should I do?
| You can rewrite it as:
\begin{align*}
\left(\frac{x}{4}\right)^2 + \left(\frac{3x}{4(x+3)}\right)^2 & = 1.
\end{align*}
Let $\frac{x}{4}=\cos A$, then $\frac{3x}{4(x+3)}=\sin A$. Using $x=4 \cos A$ in the expression for $\sin A$, we get
\begin{align*}
12 (\cos A -\sin A) & = 8 \sin 2A\\
9(\cos A -\sin A)^2 & = 4 \sin ^22A\\
9(1-\sin 2A)&=4 \sin^2 2A.
\end{align*}
Now you have a quadratic in $\sin 2A$. Can you proceed from here?
Comment: (added)
This quadratic gives $\sin2A= -3$ or $\sin 2A=\frac{3}{4}$. The former cannot happen (for real values of $x$). The latter upon substituting back in terms of $x$ gives two real solutions $x=1\pm \sqrt{7}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2881725",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Find all pairs of intergers satisfying $x^2+11 = y^4 -xy$ and $y^2 + xy= 30 $
Find all pairs of intergers $(x,y)$ that satisfy the two following equations:
$x^2+11 = y^4 -xy$
$y^2 + xy= 30 $
Here's what I did:
$x^2+11 +(30) = y^4 -xy +(y^2 + xy)$
$x^2+41 = y^4 +y^2 $
$x^2 = y^4 +y^2 - 41$
$x^2 -49 = y^4 +y^2 - 41 - 49$
$(x+7)(x-7)= (y^2 +10) (y^2-9)$
And from here you get that two pairs can be: $(3,7)$ and $(-3,-7)$.
I think I've made some progress but don't know if those are the only two pairs that satisfy the equation.
Also I would like to see a different way of solving it since I think subtracting that $49$ was just luck.
| Continuing your calculation,
$(\dfrac{30-y^2}y)^2=y^4+y^2-41,y≠0 $
$⇒y^4-60y^2+900=y^2(y^4+y^2-41)$
$⇒y^6+19y^2-900=0$
$⇒y^2(y^4+19)=900$
This give integer solution $y^2=9$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2881848",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 0
} |
Finding the least sum of digits possible for an outcome of a function in prime numbers. Let $f(n)=p^4-5p^2+13$ simplified as $f(n)=(p^2-{5\over2})^2+{27\over4}$ where $p$ is an odd prime.
Find the least possible sum of digit of $f(n)$.
My findings:
After putting $p=(3n,3n+1,3n+2)$ in $p(n)$ and dividing $p(n)$ by $3$. I found that $p(n)$ is always divisible by $3$ except for $p=3$.
So the possible sum of digits is either divisible by $3$ or $13$. $\rightarrow$ (We get $p(3)=49$ $\Rightarrow$ $4+9=13$)
| Let $p \not = 5$, then by using Fermat's Little Theorem we get that $p^4 - 5p^2 + 13 \equiv 1 + 3 \equiv 4 \pmod 5$. So the last digit is either $4$ or $9$. However $f(p)$ is always odd number, so the last digit must be $9$. Thus the sum of digits of $f(p)$ for $p \not = 5$ is at least $9$. On the other side we have that $f(5) = 513$, whose digit sum is $9$.
In conclusion $9$ is the smallest sum of digits of $f(p)$ for an odd prime $p$.
REMARK: Note that the minimum occurs only for $p=5$. This is true as for $p \not = 5$ we get that the last digit is $9$ and as $f(p) \not = 9$, there must be another non-zero digit.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2883699",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How can I solve $x^\sqrt{y} +y^\sqrt{x} =\dfrac{49}{48}$ and $\sqrt{x}+\sqrt{y} =\dfrac72$? I have tried Wolfram Alpha and Mathematica to get the solution of the below system, but no result , I have used variable change $z=\sqrt{x}+\sqrt{y}$ for simplification but no result ,
$$
\left\{
\begin{array}{ll}
x^\sqrt{y} +y^\sqrt{x} &=\dfrac{49}{48} \\
\sqrt{x}+\sqrt{y} &=\dfrac72 \\
\end{array}
\right.
$$
How I can solve this?
| Doing the same as Jack D'Aurizio (eliminating $y$ from the second equation) and using a Taylor expansion around $x=0$, the first equation is
$$-\frac{1}{48}+\sqrt{x} \log \left(\frac{49}{4}\right)+x \left(\frac{1}{2} \log
^2\left(\frac{49}{4}\right)-\frac{4}{7}\right)+O\left(x^{3/2}\right)$$ the solution of which being
$$x=\frac{7}{96 \left(-2+343 \log ^2\left(\frac{7}{2}\right)+4 \log \left(\frac{7}{2}\right) \sqrt{7350 \log
^2\left(\frac{7}{2}\right)-84} \right)}\approx 0.0000679848$$ Using this as a starting guess, Newton iterates would be
$$\left(
\begin{array}{cc}
n & x_n \\
0 & 0.000067984814185381173107 \\
1 & 0.000067980800578179758426 \\
2 & 0.000067980800636422696813 \\
3 & 0.000067980800636422696826
\end{array}
\right)$$ which is the solution for twenty significant figures.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2884179",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Sum of the series $\frac{1}{2.4.6}+\frac{2}{3.5.7}+\frac{3}{4.6.8}+.....+\frac{n}{(n+1).(n+3).(n+5)}$. Sum to n terms and also to infinity of the following series:
$$\frac{1}{2.4.6}+\frac{2}{3.5.7}+\frac{3}{4.6.8}+.....+\frac{n}{(n+1).(n+3).(n+5)}$$
the solution provided by the book is
$$S_n=\frac{17}{96}-[\frac{1}{n+5}+\frac{1}{2(n+4)(n+5)}-\frac{1}{2(n+3)(n+4)}-\frac{3}{4(n+2)(n+3)(n+4)(n+5)}]$$
And $S_\infty=\dfrac{17}{96}$.Can anyone help me to explain how to get $S_n$.
Thanks in Advanced.
| $$S_n=\dfrac{n}{(n+1)(n+3)(n+5)}=\dfrac{A}{n+1}+\dfrac{B}{n+3}+\dfrac{C}{n+5}$$
then $A=-\dfrac18$, $B=\dfrac68$ and $C=-\dfrac58$ then
$$S_n=\dfrac{n}{(n+1)(n+3)(n+5)}=\dfrac18\left(\dfrac{1}{n+3}-\dfrac{1}{n+1}\right) - \dfrac58\left(\dfrac{1}{n+5}-\dfrac{1}{n+3}\right)$$
for using telescopic method we need add some terms:
$$S_n=\dfrac18\left(\dfrac{1}{n+3}\color{red}{-\dfrac{1}{n+2}}\right) + \dfrac18\left(\color{red}{\dfrac{1}{n+2}}-\dfrac{1}{n+1}\right) -\dfrac58 \left(\dfrac{1}{n+5}\color{red}{-\dfrac{1}{n+4}}\right) - \dfrac58\left(\color{red}{\dfrac{1}{n+4}}-\dfrac{1}{n+3}\right)$$
telescopic method says $\displaystyle\sum_{k=p}^q a_{k+1}-a_k=a_{q+1}-a_p$ therefore
$$S_n=\dfrac18\dfrac{1}{n+3} + \dfrac18\dfrac{1}{n+2} - \dfrac58 \dfrac{1}{n+5} - \dfrac58\dfrac{1}{n+4} + \dfrac{17}{96}$$
and $\color{blue}{S_\infty=\dfrac{17}{96}}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2886560",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
A finite alternate binomial sum with finite value (modulo 6) We use Iverson symbol ( [expr] is $1$ when expr is true else $0$), and $n\%6$ for $n$ modulo $6$. .
For any $n >= 2 $.
Let $ F(n) := \sum_{j>=1} (-1)^j {\binom {n-j-2} j } $
Prove that $F(n) = [n\%6 \in \{0,5\}] - [n\%6 \in \{2,3\}]$
In other words:
IF $n$ modulo $6$ is $0$ or $5$ then $F(n) = 1$
IF $n$ modulo $6$ is $2$ or $3$ then $F(n) =-1$
IF $n$ modulo $6$ is $1$ or $4$ then $F(n) =0$
It is true up to $n=15$ by hand, $40$ would be safe
Yet to what kind of identities does this belong? .And how to prove this?
Can we replace $-2$ by another constant and still have the same behaviour ( finite set of value with periodic recurrence.
Note : The motivation is to calculate
$ F(n) = \sum (-1)^k[a_1a_2a_3 \dots a_k=2^n] $. Where any $a_i$ is >= 3.
| Working with the following sum
$$\sum_{q=0}^n (-1)^q {n-q\choose q}$$
we have
$$\sum_{q=0}^n (-1)^q {n-q\choose n-2q}
= \sum_{q=0}^n (-1)^q [z^{n-2q}] (1+z)^{n-q}
\\ = [z^n] (1+z)^n \sum_{q=0}^n (-1)^q z^{2q} (1+z)^{-q}.$$
Now we get zero contribution when $2q\gt n$ hence we may write
$$[z^n] (1+z)^n \sum_{q\ge 0} (-1)^q z^{2q} (1+z)^{-q}
\\ = [z^n] (1+z)^n \frac{1}{1+z^2/(1+z)}
= [z^n] (1+z)^{n+1} \frac{1}{1+z+z^2}.$$
This is
$$\mathrm{Res}_{z=0}
\frac{1}{z^{n+1}} (1+z)^{n+1} \frac{1}{1+z+z^2}.$$
With the substitution $z/(1+z)=w$ or $z=w/(1-w)$ we have
$dz = 1/(1-w)^2 \; dw$ and we obtain
$$\mathrm{Res}_{w=0}
\frac{1}{w^{n+1}}
\frac{1}{1+w/(1-w)+w^2/(1-w)^2} \frac{1}{(1-w)^2}
\\ = \mathrm{Res}_{w=0}
\frac{1}{w^{n+1}} \frac{1}{(1-w)^2+w(1-w)+w^2}
\\ = \mathrm{Res}_{w=0}
\frac{1}{w^{n+1}} \frac{1}{1-w+w^2}
= [w^n] \frac{1}{1-w+w^2}.$$
Starting from
$$G(w) = \frac{1}{1-w+w^2}$$
we have $$(1-w+w^2) G(w) = 1$$
so that
$$[w^0] (1-w+w^2) G(w) = G_0 = [w^0] 1 = 1$$
and furthermore
$$[w^1] (1-w+w^2) G(w) = G_1-G_0 = [w^1] 1 = 0$$
This establishes the initial values $G_0=1$ and $G_1=1.$
Moreover for $n\ge 2$ we find
$$[w^n] (1-w+w^2) G(w) = G_n - G_{n-1} + G_{n-2}
= [w^n] 1 = 0$$
producing the recurrence
$$G_n = G_{n-1} - G_{n-2}.$$
This means $G_n$ only depends on the preceding two values. Starting
from $1,1$ we obtain by applying the recurrence
$$1,1,0,-1,-1,0,1,1,\ldots$$
We have reached the period (intial pair appears), thus proving the
claim.
We may also compute the generating function by snake oil method.
Start with
$$G(w) = \sum_{n\ge 0} w^n \sum_{q=0}^n (-1)^q {n-q\choose q}
= \sum_{q\ge 0} (-1)^q \sum_{n\ge q} w^n {n-q\choose q}
\\ = \sum_{q\ge 0} (-1)^q w^q \sum_{n\ge 0} w^n {n\choose q}
= \sum_{q\ge 0} (-1)^q w^q \sum_{n\ge q} w^n {n\choose q}
\\ = \sum_{q\ge 0} (-1)^q w^{2q} \sum_{n\ge 0} w^n {n+q\choose q}
= \sum_{q\ge 0} (-1)^q w^{2q} \frac{1}{(1-w)^{q+1}}
\\ = \frac{1}{1-w} \frac{1}{1+w^2/(1-w)}
= \frac{1}{1-w+w^2}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2887183",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Range of a rational function with radicals Find the range of the function
$$\frac{6}{5\sqrt{x^2-10x+29} - 2}$$
I tried using inverses, but the equation got super messy and I dont think its a good method for this problem.
$\frac{6}{5\sqrt{x^2-10x+29} - 2} = y$
getting the inverse,
$\frac{6}{5\sqrt{y^2-10y+29} - 2} = x$
$\frac{4x^2+24x+36}{25x^2}= y^2-10y +29$
Then it would be a quadratic function in y, but the discriminant becomes really big
$100- 116(\frac{4x^2+24x+36}{25x^2})$
| $\frac{6}{5\sqrt(x^2-10x+29)-2}=\frac{6}{5\sqrt((x-5)^2+4)-2}$
so you can see that there will be a maximum at $x=5$ and as $x$ tends to either positive or negative infinity the graph goes to zero.
So just plug in $x=5$ to get $\frac{3}{4}$ and you have a range $]0,\frac{3}{4}]$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2887536",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Find $\lim_{x \to 0 }\frac{(1+x)^{(1/2)} -1}{(1+x)^{(1/3)} -1}$ Find the limit without using L'hopital
$$\lim_{x \to 0 }\frac{(1+x)^{(1/2)} -1}{(1+x)^{(1/3)} -1}$$
I have tried multiply by the conjugate of the denominator and numerator but didn't work
Any hint would be appreciated
Thank you
| Using that $t^2-1=(t-1)(t+1)$ and $t^3-1=(t-1)(t^2+t+1)$ we have
$$\lim_{x \to 0}\dfrac{\sqrt{x+1}-1}{\sqrt[3]{x+1} -1}=\lim_{x \to 0}\left[\dfrac{\sqrt{x+1}-1}{\sqrt[3]{x+1} -1}\cdot \dfrac{\sqrt{x+1}+1}{\sqrt{x+1}+1}\cdot \dfrac{\sqrt[3]{(x+1)^2}+\sqrt[3]{x+1}+1}{\sqrt[3]{(x+1)^2}+\sqrt[3]{x+1}+1}\right]$$ and thus
$$\lim_{x \to 0}\dfrac{\sqrt{x+1}-1}{\sqrt[3]{x+1} -1}=\lim_{x \to 0}\left[\dfrac{\sqrt[3]{(x+1)^2}+\sqrt[3]{x+1}+1}{\sqrt{x+1}+1}\right]=\dfrac 32.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2888412",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 3
} |
Non linear first order ODE in in $\mathbb{R}^2$ 2 Let $a, b \in \mathbb{R}$ and consider the system of ODEs
$$\left\{ \begin{matrix} x' = ax+ by^2, & x(0) = x_0\\ y' = ay +bx^2, & y(0) = y_0 \end{matrix} \right. $$
where $x, y :\mathbb{R} \to \mathbb{C}.$
I wonder whether there is, in general, $F$ such that $F(x,y) =0.$
For example, if $a =0,$ and $b \neq 0,$ we can write
$$\frac{dx}{dy} = \frac{y^2}{x^2},$$ which implies that
$$x^3- y^3 + c = 0$$ for some constant $c.$
Thank you for any hint.
| $$\left\{ \begin{matrix} x' = ax+ by^2, & x(0) = x_0\\ y' = ay +bx^2, & y(0) = y_0 \end{matrix} \right.$$
Multiply first equation by $x^2$ and second equation by $y^2$
$$\left\{ \begin{matrix} x^2x' = ax^3+ by^2x^2 & . \\ y'y^2 = ay^3 +bx^2y^2 & . \end{matrix} \right.$$
$$x'x^2-ax^3=y'y^2-ay^3$$
$$\frac 13 (x^3)'-ax^3=\frac 13 (y^3)'-ay^3$$
$$\frac 13( x^3-y^3)'=a(x^3-y^3)$$
You can integrate, substitute $z=x^3-y^3$
$$z'=3az$$
$$ \implies x^3(t)=y^3(t)+Ke^{3at}$$
You can try to integrate the differential equation with exactness
$$\frac {dx}{dy}=\frac {ax+by^2}{ay+bx^2}$$
$$({ay+bx^2})dx-({ax+by^2})dy=0$$
$$Pdx+Qdy=0 \implies \partial_y P=a, \partial_x Q=-a $$ $$\implies \partial_y P -\partial_x Q=2a$$
Try an integrating factor $z=x^3-y^3$
And the formula for the integratinf factor $z=z(x,y)$ is
$$\boxed {\frac {d\mu}{\mu}=\frac {\partial_y P- \partial_x Q}{Q \partial_x z -P \partial_y z}}$$
$$\frac {d\mu}{\mu}=\frac {2a dz}{Q \partial_x z -P \partial_y z}$$
$$\frac {d\mu}{\mu}=-\frac {2 dz}{3z}$$
$$\mu =\frac 1 {\sqrt[3]{z^2}}$$
$$\mu (x,y) =\frac 1 {\sqrt[3]{(x^3-y^3)^2}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2891168",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Solving $|z-4i|=2|z+4|$ $$\begin{align}
|z-4i|&=2|z+4|\\[4pt]
|x+yi-4i|&=2|x+yi+4|\\[4pt]
|x+i(y-4)|&=2|(x+4)+iy|\\[4pt]
\sqrt{x^2+(y-4)^2}&=2\sqrt{(x+4)^2+y^2}\\[4pt]
(\sqrt{x^2+(y-4)^2})^2&=(2\sqrt{(x+4)^2+y^2})^2\\[4pt]
x^2+y^2-8y+16&= 4(x^2+8x+16+y^2)\\[4pt]
x^2+y^2-8y+16&= 4x^2+32x+64+4y^2\\[4pt]
0&= 3x^2+3y^2+32x+8y+48
\end{align}$$
Is it okay? Thank you
| So far so good.
You may continue to see the geometry of the solution as well.
$$3x^2+3y^2+32x+8y+48=0$$
$$x^2+y^2+(32/3)x+(8/3)y+16=0$$
Which is the equation of a circle.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2891300",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 2
} |
Can we factor multivariate polynomial $x+xy+y$? It seems to me we can only get $x(1+y)+y$ or $x+(x+1)y$, but maybe something better is possible with complex numbers?
It has zeros at $y=-\frac{x}{x+1}$, but how to use that fact?
| A possible factorization is :
$\begin{align}x+xy+y &=x+xy+y+1-1\\ &=x(y+1)+(y+1)-1 \\ &=(x+1)(y+1)-1 \\ &=\left(\sqrt{(x+1)(y+1)}\right)^2-1^2 ,\text{assuming $x,y \gt -1$} \\ &=\left(\sqrt{(x+1)(y+1)}+1\right)\cdot\left(\sqrt{(x+1)(y+1)}-1\right)\end{align}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2891748",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
How to find an affine map for given points I am trying to understand affine maps. In the book I am using there is the following example, which I don't understand. Given are the following points:
\begin{array}{lll}
p_0 = \begin{pmatrix} 1 \\ 1 \end{pmatrix},
&
p_1 = \begin{pmatrix} 1 \\ 2 \end{pmatrix},
&
p_2 = \begin{pmatrix} 0 \\ 1 \end{pmatrix},
\\
q_0 = \begin{pmatrix} -2 \\ -1 \end{pmatrix},
&
q_1 = \begin{pmatrix} \phantom{-}0 \\ -1 \end{pmatrix},
&
q_2 = \begin{pmatrix} -2 \\ -2 \end{pmatrix}.
\end{array}
Then the following vectors are computed
\begin{array}{ll}
\overrightarrow{p_0 p_1}
= \begin{pmatrix} 0 \\ 1 \end{pmatrix},
&
\overrightarrow{p_0 p_2}
= \begin{pmatrix} -1 \\ \phantom{-}0 \end{pmatrix},
\\
\overrightarrow{p_0 p_1}
= \begin{pmatrix} 2 \\ 0 \end{pmatrix},
&
\overrightarrow{p_0 p_1}
= \begin{pmatrix} \phantom{-}0 \\ -1 \end{pmatrix}.
\end{array}
I am asked to find an affine map such that $f(p_{i})=q_{i}$.
It now says that the matrix of the map determined by $F(\overrightarrow{p_0 p_i})=\overrightarrow{q_0 q_i}$, $i=1,2$ is given by
$$
A
= \begin{pmatrix}
0 & 2 \\
1 & 0
\end{pmatrix}.
$$
I don't understand how I get this matrix. I would say it is the following one, but I guess I am wrong. I did everything with repect to the bases $\{(0,1),(-1,0)\}$ and $\{(1,0),(0,1)\}$. I am not sure about the second basis, but since there is no other basis mentioned I thought it is the one the author has used too.
$$
\begin{pmatrix}
2 & \phantom{-}0 \\
0 & -1 \\
\end{pmatrix}
$$
Unfortunately I don't know where I making the mistake.
I would really appreciate some help. Thanks a lot in advance!
| Let $A=
\begin{pmatrix}
a & b \\
c & d \\
\end{pmatrix}$
Now, apply this matrix to your vectors $p_0p_1$ and $p_0p_2$
The first one gives:
$q_0q_1=A*p_0p_1$
So
$\begin{pmatrix}2 \\0 \end{pmatrix}=A* \begin{pmatrix}0 \\1 \end{pmatrix} = \begin{pmatrix} a*0 + b*1 \\ c*0 +d*1\\ \end{pmatrix}= \begin{pmatrix}b \\d \end{pmatrix}$
This gives you $b=2$ and $d=0$. You can use the other equation to get a and c.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2892086",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Find the maximum value of $A$ Let $a;b;c>0$ such that $a+b+c=6$. Find the maximum value of $A=a^2bc+a^2+2b^2+2c^2$
WLOG $b\ge c$. I see maximum value of $A=36$ at $(a;b;c)=(2;1;3)$
So i need to prove $A\le 36$. Or I will prove
$(a+b+c)^4\ge 36a^2bc+(a^2+2b^2+2c^2)(a+b+c)^2$
Or $(2a-b-c)(b^3+c^3+a^2b+a^2c+2ab^2+2ac^2-12abc-b^2c-bc^2)\ge 0$
Then Im stuck here, help me solve it.
| For $(a,b,c)\rightarrow(0,6,0)$ we have $A\rightarrow72$ and since
$$72-a^2bc-a^2-2b^2-2c^2\geq0$$ it's
$$2(a+b+c)^4-(a+b+c)^2(a^2+2b^2+2c^2)-36a^2bc\geq0$$ or
$$a^4+6a^3b+6a^3c+9a^2b^2+9a^2c^2-14a^2bc+4ab^3+4ac^3+20b^2ac+20c^2ab+4ab^3+4ac^3+8b^2c^2\geq0,$$
which is obvious, we see that the maximum does not exist, but the supremum is $72$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2894315",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
How to evaluate the integral $\int_{0}^{1} \frac{(\log {x})^2}{\sqrt{x^2(1-x)^2+x^2+(1-x)^2}}dx$ $$I=\int_{0}^{1} \frac{(\log {x})^2}{\sqrt{x^2(1-x)^2+x^2+(1-x)^2}}dx$$
My attempt:
$$I=\int_{0}^{1} \frac{(\log {x})^2}{\sqrt{x^2(1-x)^2-2x(1-x)+(x+1-x)^2}}dx$$
$$=\int_{0}^{1} \frac{(\log {x})^2}{\sqrt{(1-x(1-x))^2}}dx=\int_{0}^{1} \frac{(\log {x})^2}{(x^2-x+1)}dx$$
How do I proceed further next step?
| Here is another method:
\begin{eqnarray}
I&=&\int_{0}^{1} \frac{(\log {x})^2}{x^2-x+1}dx=\int_{0}^{1} \frac{(1+x)(\log {x})^2}{1+x^3}dx\\
&=&\int_{0}^{1}(1+x)\log^2x\sum_{k=0}^\infty(-1)^kx^{3k}dx\\
&=&\sum_{k=0}^\infty(-1)^k\int_0^1(x^{3k}+x^{3k+1})\log^2xdx\\
&=&2\sum_{k=0}^\infty(-1)^k\left[\frac{1}{(3k+1)^3}+\frac{1}{(3k+2)^3}\right]\\
&=&2\sum_{k=-\infty}^\infty(-1)^k\frac{1}{(3k+1)^3}\\
&=&-2\pi\text{Res}\left(\frac{1}{(3z+1)^3\sin(\pi z)},z=-\frac13\right)\\
&=&\frac{10\pi^3}{81\sqrt3}.
\end{eqnarray}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2896039",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Continued fraction of $ \sqrt{3} $ How can I solve the continued fraction of $\sqrt{3}$?
I know that a continued fraction is [1;1,2,1,2,1,2,....] -> this is the continued fraction of $\sqrt{3}$, but how can I solve? What are the steps?
I don't understand this proof.
Thank you so much.
| We know $\sqrt{3}$ lies between 1 and 2 (because $1^2 < 3 < 2^2$). So the first digit of the continued fraction expansion is 1. So
$\sqrt{3} = 1 + (\sqrt{3}-1)$
and $\sqrt{3}-1$ must be between $0$ and $1$. So next we want to find the reciprocal of $\sqrt{3}-1$:
$\frac{1}{\sqrt{3}-1} = \frac{\sqrt{3}+1}{(\sqrt{3}-1)(\sqrt{3}+1)} = \frac{\sqrt{3}+1}{2}$
$\frac{\sqrt{3}+1}{2}$ lies between $1$ and $2$ so the next term in the continued fraction is 1. So we have $\sqrt{3} = [1;1,\dots]$.
The remaining amount is $\frac{\sqrt{3}+1}{2}-1 = \frac{\sqrt{3}-1}{2}$ so next we want to find the reciprocal of $\frac{\sqrt{3}-1}{2}$:
$\frac{2}{\sqrt{3}-1} = \frac{2(\sqrt{3}+1)}{2} = \sqrt{3}+1$
$\sqrt{3}+1$ is between 2 and 3, so the next term in the continued fraction is 2. So we have $\sqrt{3} = [1;1,2,\dots]$.
The remaining amount is $(\sqrt{3}+1)-2=\sqrt{3}-1$. We could continue, but we have already calculated the reciprocal of $\sqrt{3}-1$ above, so now the continued fraction just repeats and we have:
$\sqrt{3} = [1;1,2,1,2,1,2, \dots]$
An alternative way of seeing why this continued fraction repeats is to note that
$(\sqrt{3}+1)^2 = 2\sqrt{3} + 4 = 2(\sqrt{3}+1) + 2$
so $\sqrt{3}+1$ is a solution to $x^2=2x+2$, which we can re-write as:
$x = 2 + \dfrac{2}{x} = 2 + \dfrac{1}{\dfrac{x}{2}} = 2 + \dfrac{1}{1 + \dfrac{1}{x}}$
so $\sqrt{3} + 1 = [2;1,2,1,2,1,\dots]$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2898083",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
The priority of limits How are the two expressions different?
$$\lim_{x\to0}\bigg\lfloor\frac{\sin{x}}{x}\bigg\rfloor$$
and $$\bigg\lfloor\lim_{x\to0}\frac{\sin{x}}{x}\bigg\rfloor$$
If limit is inside the floor function, Do I have to apply the limits first?
If this is the case, then, $$\lim_{x\to0}\bigg\lfloor\frac{\sin{x}}{x}\bigg\rfloor=0$$
$$\bigg\lfloor\lim_{x\to0}\frac{\sin{x}}{x}\bigg\rfloor=1$$
Am I solving this right?
Also how can I calculate,
$$\lim_{x\to0}\bigg\lfloor\frac{\sin{x}\cdot \tan{x}}{x^2}\bigg\rfloor$$
Thank you.
| Yes you are right indeed since
$$\frac{\sin x}{x}<1 \implies \bigg \lfloor\frac{\sin{x}}{x}\bigg\rfloor=0 \implies \lim_{x\to0} \bigg \lfloor\frac{\sin{x}}{x}\bigg\rfloor=0$$
and since
$$\lim_{x\to0} \frac{\sin{x}}{x}=1 \implies \bigg\lfloor\lim_{x\to0}\frac{\sin{x}}{x}\bigg\rfloor=1
$$
For the latter for $x$ sufficiently small we have
$$1<\frac{\sin{x}\cdot \tan{x}}{x^2}<2$$
indeed $\sin x>x-\frac{x^3} 6$ and $\tan x>x+\frac{x^3} 3$ and
$$\frac{\sin{x}\cdot \tan{x}}{x^2}>1+\frac{x^2}6-\frac{x^6}{18}$$
therefore
$$\lim_{x\to0}\bigg\lfloor\frac{\sin{x}\cdot \tan{x}}{x^2}\bigg\rfloor=1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2898467",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
Prove that $a+b+c\le 0$ $a,b,c$ are integer numbers different from zero and
$$\frac{a}{b+c^2}=\frac{a+c^2}{b}$$
Prove that $a+b+c\le 0$.
I know how to prove that $a+b<0$ but do not know what about $c$.
| We have $$ab=(a+c^2)(b+c^2)\\ab=ab+c^4+c^2(a+b)$$which means that $$a+b=-c^2\le -c$$or equivalently $$a+b+c\le 0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2901440",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Proof of onto under interval $(0,1)$ How do I prove that the function $$f:(0,1)\rightarrow \mathbb R$$ defined by:
$$f(x) = \frac{-2x+1}{(2x-1)^2-1}$$
is onto?
| Alternatively, set $t = 2x-1$ and for each $r \in \mathbb{R}$ we show we can find a number $t$ and then an $x \in (0,1)$ such that$f(x) = r$. We have: $\dfrac{t}{1-t^2}=r$. So this gives $t = r - rt^2\implies rt^2+t -r = 0$. If $r = 0 \implies t = 0\implies x = \dfrac{1}{2}$. If $r \neq 0$, $t = \dfrac{-1\pm \sqrt{1+4r^2}}{2r}$. If $r > 0$, take $t = \dfrac{-1+\sqrt{1+4r^2}}{2r}>0$, and $t = \dfrac{2r}{1+\sqrt{1+4r^2}}< \dfrac{2r}{\sqrt{1+4r^2}}<1\implies t \in (0,1)\implies x = \dfrac{t+1}{2} > \dfrac{1}{2} > 0$, and $x < \dfrac{1+1}{2} = 1\implies x \in (0,1)$. If $r < 0$, choose $t = \dfrac{-1+\sqrt{1+4r^2}}{2r}\implies x = \dfrac{-1+2r+\sqrt{1+4r^2}}{4r}$. Observe that with $r < 0\implies \sqrt{1+4r^2} < 1-2r$ ( clear ),and this yields $-1+2r+\sqrt{1+4r^2} < 0\implies x > 0$. To complete the proof, you show $x < 1$. But $x = \dfrac{1-2r-\sqrt{1+4r^2}}{-4r}< 1 \iff 1-2r-\sqrt{1+4r^2} < -4r\iff 1+2r < \sqrt{1+4r^2}\iff 4r < 0$, and this is true since $r < 0$ ( by assumption ) .
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2903110",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Evaluate $\lim_{x\rightarrow -\infty}{e^{\frac {1}{2-x}}\cdot\frac{x^2+2x-1}{x-2}}-x$ I want to find the following limit:
$$\lim_{x\rightarrow -\infty}{e^{\frac {1}{2-x}}\cdot\frac{x^2+2x-1}{x-2}}-x$$
This is what I do. I change the variable $t=-x$ and I have the following limit:
$$\lim_{t\rightarrow +\infty}{e^{\frac {1}{2+t}}\cdot{\frac{t^2-2t-1}{-t-2}}+t}=\lim_{t\rightarrow+\infty}{h(x)}$$
We have $e^{\frac{1}{2+t}}\rightarrow1$ for $t\rightarrow+\infty$
Therefore I think (this is the passage I'm less sure about)
$$h(x)\sim \frac{t^2-2t-1}{-t-2}+t=\frac{t^2-2t-1+t(-t-2)}{-t-2}=\frac{t^2-2t-1-t^2-2t}{-t-2}=\frac{-4t-1}{-t-2}\sim{\frac {-4t}{-t}}\rightarrow4$$
The solution should actually be $3$. Any hints on what I'm doing wrong?
| Let set $\frac {1}{2-x}=y \to 0$
$$\lim_{x\rightarrow -\infty}{e^{\frac {1}{2-x}}\cdot\frac{x^2+2x-1}{x-2}}-x=\lim_{y\rightarrow 0}{e^{y}\cdot\frac{\frac1{y^2}-\frac 6 y+7}{-\frac1y}}-2+\frac1y$$$$=\lim_{y\rightarrow 0}e^{y}\cdot\left(-\frac1{y}+6-7y\right)-2+\frac1y$$
then use $e^y=1+y+o(y)$ to obtain
$$(1+y+o(y))\cdot\left(-\frac1{y}+6-7y\right)-2+\frac1y$$
$$-\frac1{y}+6-7y-1+6y-7y^2+o(1)-2+\frac1y=3+o(1) \to 3$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2903718",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 0
} |
General formula for Differentiation Operator I was considering the Operator
$$
x\,\frac{\rm d}{{\rm d}x}
$$
and applying it $n$ times to an arbitrary function $f(x)$.
Is there a general formula for it?
I started with the first few
\begin{align}
\left(x\,\frac{\rm d}{{\rm d}x}\right)^{1} f(x) &= x f' \\
\left(x\,\frac{\rm d}{{\rm d}x}\right)^{2} f(x) &= x f' + x^2 f'' \\
\left(x\,\frac{\rm d}{{\rm d}x}\right)^{3} f(x) &= x f' + 3x^2 f'' + x^3 f''' \\
\left(x\,\frac{\rm d}{{\rm d}x}\right)^{4} f(x) &= x f' + 7 x^2 f'' + 6x^3 f''' + x^4 f'''' \\
\vdots
\end{align}
but I don't see any pattern yet.
Obviously the first and last coefficients are always $1$.
I figured if I start with any $n$ of the form
$$
\left(x\,\frac{\rm d}{{\rm d}x}\right)^{n} f(x) = \sum_{k=1}^n a_k \, x^k f^{(k)}(x) \tag{1}
$$
where $a_1=a_n=1$, then recursively
\begin{align}
\left(x\,\frac{\rm d}{{\rm d}x}\right)^{n+1} f(x) &= xf^{(1)}(x) + \sum_{k=2}^n \left(k \, a_k + a_{k-1}\right) x^k f^{(k)}(x) + x^{n+1} f^{(n+1)}(x) \\
&= \sum_{k=1}^{n+1} \left(k \, a_k + a_{k-1}\right) x^k f^{(k)}(x) \tag{2}
\end{align}
where $a_0=a_{n+1}=0$.
From (1) and (2) we get for fixed $k$ the coupled recurrence
$$
a_k(n+1) = k\, a_k(n) + a_{k-1}(n)
$$
which can be put in matrix form
\begin{align}
\begin{pmatrix} a_1(n+1) \\ a_2(n+1) \\ a_3(n+1) \\ \vdots \\ a_{n-1}(n+1) \\ a_n(n+1) \end{pmatrix} &= \begin{pmatrix} 1 & 0 & 0 & \dots & 0 & 0 \\ 1 & 2 & 0 & \dots & 0 & 0 \\ 0 & 1 & 3 & \dots & 0 & 0 \\ \vdots & \vdots & \vdots & \dots & \vdots & \vdots \\ 0 & 0 & 0 & \dots & n-1 & 0 \\ 0 & 0 & 0 & \dots & 1 & n \end{pmatrix} \begin{pmatrix} a_1(n) \\ a_2(n) \\ a_3(n) \\ \vdots \\ a_{n-1}(n) \\ a_n(n) \end{pmatrix} \\
&=\begin{pmatrix} 1 & 0 & 0 & \dots & 0 & 0 \\ 1 & 2 & 0 & \dots & 0 & 0 \\ 0 & 1 & 3 & \dots & 0 & 0 \\ \vdots & \vdots & \vdots & \dots & \vdots & \vdots \\ 0 & 0 & 0 & \dots & n-1 & 0 \\ 0 & 0 & 0 & \dots & 1 & n \end{pmatrix}^n \begin{pmatrix} 1 \\ 0 \\ 0 \\ \vdots \\ 0 \\ 0 \end{pmatrix} \, .
\end{align}
So either I'm too stupid or there is no obvious one.
| The coefficients are the Stirling numbers of the second kind.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2905016",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Given $\sin(t) + \cos(t) = a$, derive an expression in '$a$' for $(\cos(t))^4 + (\sin(t))^4$ I received this question from a student's trigonometry review assignment. After spending an embarrassing amount of time on it, I consulted others and learned that nobody has been able to solve this problem for multiple semesters. I wonder if there is a typo in the statement or if I just haven't been rigorous enough? Here is the question:
$$\text{Given } \sin{t} + \cos{t} = a, \text{ find an equivalent expression for } \sin^4{t} + \cos^4{t} \text{ in terms of } a.$$
Has anybody seen this one before? I tried (among other things) this (which is an approximation of an earlier attempt):
$(\sin{t} + \cos{t})^4$ and using whatever identities I could remember;
$(\sin{t} + \cos{t})^4 = \sin^4{t} + \cos^4(t) + 4\sin{t}\cos^3{t} + 6\sin^2{t}\cos^2{t} + 4\sin^3{t}\cos{t}$
so then
$\begin{align}
\sin^4{t} + \cos^4{t} &= (\sin{t} + \cos{t})^4 - 4\sin{t}\cos^3{t} - 6\sin^2{t}\cos^2{t} - 4\sin^3{t}\cos{t}\\
&= (\sin{t} + \cos{t})^4 - 2\sin{t}\cos{t}(2\cos^2{t} + 3\sin{t}\cos{t} + 2\sin^2{t})\\
&= (\sin{t} + \cos{t})^4 - 2\sin{t}\cos{t}(3\sin{t}\cos{t} + 4)\\
&= a^4 - 2\sin{t}\cos{t}(3\sin{t}\cos{t} + 4)
\end{align}$
Couldn't get further than this, felt like I was overthinking it.
| If $\sin x+\cos x=a,a^2=1+2\sin x\cos x\iff\sin x\cos x=\dfrac{a^2-1}2 $
$$\sin^{n+2}x+\cos^{n+2}x=\sin^nx(1-\cos^2x)+\cos^nx(1-\sin^2x)$$
If $I_m=\sin^mx+\cos^mx,I_2=1,I_0=2$
$$I_{n+2}=I_n-(\sin x\cos x)^2I_{n-2}$$
Here $n=2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2905869",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 2
} |
Find the smallest and highest value of the product $xyz$ Find the smallest and highest value of the product $xyz$ assuming that:
$x + y + z = 10$ and
$x^2 + y^2 + z^2 = 36$.
I calculated this:
$x+y+z=10 => (x+y+z)^2=10^2$
$x^2+y^2+z^2+2xy+2yz+2zx=100$
$(x^2+y^2+z^2+2xy+2yz+2zx)-(x^2+y^2+z^2)=100-36$
$2xy+2yz+2zx=64$
$xy+yz+zx=32$
I'm stuck. What is the next step to this exercise?
My idea is to show the equation using one variable and after computing the derivative reach global extremes.
| Consider the equation: $A^3-10A^2+32A-xyz=0$, whose roots are $x,y,z$. We need to find the maximum and minimum values of:
$$xyz=A^3-10A^2+32A=f(A).$$
FOC:
$$f'(A)=3A^2-20A+32=0 \Rightarrow A_1=\frac83; \ \ A_2=4.$$
SOC:
$$\begin{align}f''(A)&=6A-20; \\
f''\left(\frac83\right)&=-4<0 \Rightarrow f\left(\frac83\right)=\frac{896}{27} \ (\text{max});\\
f''(4)&=4>0 \Rightarrow f(4)=32 \ (\text{min}).\end{align}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2906604",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 0
} |
Solving $\left(\sqrt{3+2\sqrt{2}}\right)^x - \left(\sqrt{3-2\sqrt{2}}\right)^x = \frac32$
Given
$$\left(\sqrt{3+2\sqrt{2}}\right)^x - \left(\sqrt{3-2\sqrt{2}}\right)^x = \frac32$$
What is $x$?
I just can do this with that equation
$$\left(\sqrt{2+1+2\sqrt{2.1}}\right)^x - \left(\sqrt{2+1-2\sqrt{2.1}}\right)^x = \frac32$$
$$\left(\sqrt{({\sqrt{2}+\sqrt{1})^2}}\right)^x-\left(\sqrt{({\sqrt{2}-\sqrt{1})^2}}\right)^x = \frac32$$
$$\left(\sqrt{2}+1\right)^x - \left(\sqrt{2}-1\right)^x = \frac32$$
And i stuck there for a few hours and get nothing
Pliz help me
| Use the fact that $$\sqrt{3-2\sqrt{2}}=\frac{1}{\sqrt{3+2\sqrt{2}}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2908830",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Separate f(z) into real and imaginary parts I'm doing some practice problems about complex numbers. I want to break $f(z)=(1-z)/(1+z)$ into real and imaginary parts $u(x,y)$ and $v(x,y)$ such that $f(z)=u+iv$. I did a different problem like this correctly, so I think it's more the algebra than the complex number aspect that's hard for me here. I know the answer is
$$u=\frac{1-x^2-y^2}{(1+x)+y^2}, v=\frac{-2y}{(1+x)^2+y^2}$$
$z=x+iy$, so $$f(z)=\frac{1-(x+iy)}{1+(x+iy)}=\frac{1-x-iy}{1+x+iy}$$
I get that it needs to be split into two fractions so that the $i$ can be pulled out. I just have no idea how to make that work. Or the rest of the algebra for that matter.
| You just need to realise the denominator:
$$f(z) = \frac{(1-x) - iy}{(1+x) + iy} \times \frac{(1+x) - iy}{(1+x)-iy}$$ $$= \frac{(1-x)(1+x) -y^2}{(1+x)^2+y^2} + \frac{i(-y(1-x) - y(1+x))}{(1+x)^2 + y^2}$$ $$= \frac{1-x^2 - y^2}{(1+x)^2 + y^2} + i \frac{-y + yx - y - yx}{(1+x)^2 + y^2} $$ and so it is clear that $u = \frac{1-x^2-y^2}{(1+x)^2 + y^2}$ and $v = \frac{-2y}{(1+x)^2 + y^2}$ as required.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2909044",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 0
} |
Evaluate $ \lim_{x \to 0} \left( {\frac{1}{x^2}} - {\frac{1} {\sin^2 x} }\right) $ $$\lim_{x\to0}\left({\frac{1}{x^2}}-{\frac{1}{\sin^2x}}\right)$$
Using the L'Hospital Rule I obtained the value $-1/4$, but the answer is given to be $-1/3$. I can't find the mistake. Here's what I did; please point out the mistake.
\begin{align}
\lim_{x\to0}\left({\frac{1}{x^2}}-{\frac{1}{\sin^2x}}\right)&=\lim_{x\to0}\frac{(\sin x+x)(\sin x-x)}{(x\sin x)(x\sin x)} \\[1ex]
&=\lim_{x\to0}\left(\frac{\sin x+x}{x\sin x}\right)\lim_{x\to0}\left(\frac{\sin x-x}{x\sin x}\right) \\[1ex]
&=\lim_{x\to0}\left(\frac{\cos x+1}{\sin x+x\cos x}\right)\lim_{x\to0}\left(\frac{\cos x-1}{\sin x+x\cos x}\right) \\[1ex]
&=\lim_{x\to0}\:(\cos x+1)\,\lim_{x\to0}\left(\frac{\cos x-1}{(\sin x+x\cos x)^2}\right) \\[1ex]
&=\lim_{x\to0}\frac{-\sin x}{(\sin x+x\cos x)(2\cos x-x\sin x)} \\[1ex]
&=-\lim_{x\to0}\left[\frac{1}{1+\cos x\left(\frac{x}{\sin x}\right)}\right]\left(\frac{1}{2\cos x-x\sin x}\right) \\[1ex]
&=-\frac{1}{2}\left[\lim_{x\to0}\,\frac{1}{1+\cos x}\right] \\[1ex]
&=-\frac{1}{4}
\end{align}
| $$\lim_{x\to0}\frac{(\sin{x}+x)(\sin{x}-x)}{x\sin{x}\cdot x\sin{x}}$$
Here are some limits I remember that help me a lot, (easily derivable using L-Hopital)
$$\lim_{x\to0}\frac{\sin{x}-x}{x^3}=-\frac{1}{6}$$
$$\lim_{x\to0}\frac{x-\tan{x}}{x^3}=-\frac{1}{3}$$
$$\lim_{x\to 0}\frac{e^x-1-x}{x^2}=\frac{1}{2}$$
So using this,
$$\lim_{x\to0}\frac{x^2}{\sin^2x}\cdot \frac{(\sin{x}+x)}{x}\cdot \frac{(\sin{x}-x)}{x^3}$$
$$1\cdot2\cdot -\frac{1}{6}$$
$$-\frac{1}{3}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2910595",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 9,
"answer_id": 7
} |
If $\sin^8(x)+\cos^8(x)=48/128$, then find the value of $x$? If $$\sin^8(x)+\cos^8(x)=48/128,$$ then find the value of $x$? I tried this by De Moivre's theorem:
$$(\cos\theta+i\sin\theta)^n=\cos(n\theta)+i\sin(\theta)=e^{in\theta}$$
But could not proceed further please help.
| Let $c=cosx$ and $s=sinx$ :
$s^8+c^8= \frac{48}{2^7}*\frac{2}{2}$
$s^8+c^8= \frac{96}{2^8} $
$(\frac{G}{2^4})^2+(\frac{H}{2^4})^2=\frac{96}{2^8}\implies G^2+H^2=96 $
$\implies s=\pm\frac{\sqrt[4]{G}}{2} \wedge c=\pm\frac{\sqrt[4]{96-G^2}}{2}$
We gonna use: $s^2+c^2=1\rightarrow c^2=1-s^2$:
$\frac{\sqrt{96-G^2}}{4}=1-\frac{\sqrt{G}}{4}$
$\sqrt{96-G^2}=4-\sqrt{G}\rightarrow$ WolframAlpha finds the solution : $G\approx9.7587$
For $\sqrt{96-g^4}=4-g$ we get real solutions : $g_1\approx0.0406-e \vee g_2\approx\pi-0.0177;$ cause $g_1^2\neq G\implies g_2$ gonna be our result of $g^2=G$
.$(g\approx 3.12389354026129)$
So:
$s=\pm\frac{\sqrt{g}}{2}$
We consider for $sinx$ in the 1st quarter:
$\frac{\sqrt{3}}{2}<\frac{\sqrt{g}}{2}<\frac{\sqrt{\pi}}{2}$
$60^\circ<x<62.4^\circ$;cause of $arcsin(\frac{\sqrt{\pi}}{2})$
Direct:
$x=arcsin(\frac{\sqrt{g}}{2})\approx62.1^\circ$
History and altenative version (with different* solution):
All starts with:
$(c^4+is^4)(c^4-is^4)=\frac{6}{2^4}$
$(\frac{A}{2^2})^2+(\frac{B}{2^2})^2=\frac{6}{2^4}\implies A^2+B^2=6 $
$\implies c^4=\frac{A}{4} \wedge s^4=\frac{\sqrt{6-A^2}}{4}$
Altenative for:
$c=\pm\sqrt[4]{\frac{A}{4}} \wedge s=\pm\sqrt[4]{\frac{B}{4}}$ with $s^2+c^2=1$ :
$\sqrt{\frac{B}{4}}=1-\sqrt{\frac{A}{4}}$
$6-A^2=[2-\sqrt{A}]^2$
$6-A^2=4-2\sqrt{A}+A$
$0=A^2+A-2\sqrt{A}-2\rightarrow$ WolframAlpha finds the solution : $A\approx1.7049$
So:
$c=\pm\frac{\sqrt[4]{A}}{\sqrt{2}}$
We consider for $cosx$ in the 1st quarter:
$x=arccos(\frac{\sqrt[4]{A}}{\sqrt{2}})\approx36.1^\circ$
*Also: If I would stand by $s=\pm\frac{\sqrt[4]{A}}{\sqrt{2}}\rightarrow x\approx53.9^\circ$ (1st quarter)
Other possibilities via $E+F=(9)6$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2910845",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 7,
"answer_id": 4
} |
Show that $\frac{3^n}{n!}$ converges to $0$ I was wondering if this proof is correct.
$$\left|\frac{3^n}{n!} - 0\right| = \frac{3^n}{n!} \lt \frac{3^n}{2^n} \lt \varepsilon$$
So then
$$\frac{2^n}{3^n} \gt \frac{1}{\varepsilon}$$
$$\left(\frac{2}{3}\right)^n \gt \frac{1}{\varepsilon}$$
$$\log\left(\frac{2^n}{3^n}\right) \gt \log\left(\frac{1}{\varepsilon}\right)$$
$$n\log\left(\frac{2}{3}\right) \gt \log\left(\frac{1}{\varepsilon}\right)$$
$$n \gt \log\left(\frac{1}{\varepsilon}\right)/\log\left(\frac23\right)$$
So then any $N \gt \log(\frac1\varepsilon)\log(\frac23)$ yields the result we want, so that for all $n \gt N, \left|\frac{3^n}{n!} - 0\right| \lt ε$
Thanks in advance!
| You can modify your argument as follows:
You can assume that $n \geq 4$ because shifting the terms of a sequence does not affect its convergence or the value of convergence.
Then
$$\left|\frac{3^n}{n!} - 0\right| = \frac{3^n}{n!} =\frac{27}{3\times 2\times 1}
\frac{3^{n-3}}{n \times(n-1)\times\cdots\times 4}\lt 5 \times \frac{3^{n-3}}{4^{n-3}} \lt \varepsilon$$
Now rewrite your own argument to finish the proof:
$$(\frac{3}{4})^{n-3} < \frac{\varepsilon}{5}$$
$$\ln(\frac{3}{4})^{n-3} < \ln\frac{\varepsilon}{5}$$
$$(n-3)\ln(\frac{3}{4}) < \ln\frac{\varepsilon}{5}$$
$$n \geq \lceil \ln\left(\frac{\varepsilon}{5}\right)/\ln\left(\frac{3}{4}\right) \rceil + 3$$
where $\lceil \cdot \rceil$ denotes the ceiling function.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2911253",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 1
} |
Find the domain of x $\left \lfloor x \right \rfloor + \left \lfloor x+\frac{1}{2} \right \rfloor + \left \lfloor x-\frac{1}{3} \right \rfloor =8$ Find the domain of x
$$\left \lfloor x \right \rfloor + \left \lfloor x+\frac{1}{2} \right \rfloor + \left \lfloor x-\frac{1}{3} \right \rfloor =8$$
My approach
When x is an integer $x+x+x-1=8$ or x=3
But for case when x is not an integer i am not able to substitute, manually x is $\frac{10}{3}$ which i did by trial method.
My final answer is $3\le x <\frac{10}{3} $
| Let $x = k + y$, where $k$ is an integer part of $x$ ($k=\left \lfloor x \right \rfloor$) and $y$ is its fractional part. Consider $y$-s belonging to the following intervals
$\left[0, \frac{1}{3}\right)$, $\left[\frac{1}{3}, \frac{1}{2}\right)$ and $\left[\frac{1}{2}, 1\right)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2914774",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Find all non-negative integer $n$ that satisfy $f(x+1)+f(x-1)=\sqrt nf(x)$
Find all non-negative integer $n$ that there exists a non-periodic function $f:\Bbb R->\Bbb R\quad f(x+1)+f(x-1)=\sqrt nf(x)\forall x$.
My attempt:
$f(x+2)+f(x)=\sqrt n\cdot f(x+1)$
$\sqrt n \cdot (f(x+1)+f(x+3))=n\cdot f(x+2)$
$f(x+2)+f(x+4)=\sqrt n\cdot f(x+3)$
Therefore $$f(x)+f(x+4)=(n-2)\cdot f(x+2)\tag{*}$$
$n=0\implies f(x)=-f(x+2)\implies f(x)=f(x+4)$
$n=1\implies f(x+2)+f(x)=f(x+1)\,$ and $\,f(x+3)+f(x+1)=f(x+2)\implies f(x)=f(x+3)$
$n=2\quad(*)\implies f(x)+f(x+4)=0\,$ A similar argument hold for $n=0$.
$n=3\quad(*) \implies f(x)+f(x+4)=f(x+2).$ A similar argument hold for $n=1.$
$n=4\quad f(x)=x$ works.
$\forall n>4,f(x)$ can be defined recursively:
$i)\;$ $f(x)=f(\lfloor x\rfloor)\forall x$
$ii)\;$ $f(0)=0,\;f(1)=1$
$iii)\;$ $f(x)+f(x+2)=\sqrt n \cdot f(x+1)\;\forall x\in\Bbb Z$
It's easy to show that this satisfy the riginal functional equation, but we still need to prove it's non-periodic. It's sufficient to prove $iii)$ is monotonic.
$f(x+2)=\sqrt n\cdot f(x+1)-f(x)>2\cdot f(x+1)-f(x)\ge f(x+1)$
Is my proof correct?
Any help appreciated.
| This is not actually an answer, but more than a comment following the previous nice answers. If $a,b$ are the roots (real roots case where $ n\geq 4$ ) of the trionym $t^2-\sqrt{n}\,t+1$ then all the functions of the form : $f(x)=c_1a^x+c_2b^x$ have the desired attribute.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2917573",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
$\frac{1}{5}n^5+\frac{1}{3}n^3+\frac{7}{15}n$ is an integer for every integer n. Q: Prove that $\frac{1}{5}n^5+\frac{1}{3}n^3+\frac{7}{15}n$ is an integer for every integer n.
In fact, I already have a method, but it seems too long. So I would first (1)is there any problem with my logic? (2) Is there any better/faster method to do these kinds of questions?
My solution:
$$\frac{1}{5}n^5+\frac{1}{3}n^3+\frac{7}{15}n$$
$$=\frac{1}{15}(3n^5+5n^3+7n)$$
Then my next time is to use M.I to prove that $(3n^5+5n^3+7n)$ is divisible by $15$ for positive integers. And for negative integer, I switch it into -$(3n^5+5n^3+7n)$ and it also satisfies the divisibility of $15$.
But the above seems too clumsy. This course is "Intro to number theory". Is there any beter method to do it?
| Modulo $3$, your numerator is:
$$3n^5+5n^3+7n \equiv -n(n^2-1) $$
$$\equiv -n(n-1)(n+1)\pmod{3}.$$
The last expression is a product of three consecutive integers and hence divisible by $3$.
Modulo $5$, you have
$$3n^5+5n^3+7n \equiv -2n^5+2n $$ $$\equiv -2n(n^4-1) \pmod{5}.$$
If $n$ is divisible by $5$, then so is the last expression. If $n$ is not divisible by $5$, then $n^4 \equiv 1 \pmod{5}$ and still the last expression is divisible by $5$.
Since the numerator is divisible by both $3$ and $5$, it's divisible by $15.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2917871",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 0
} |
Greatest term of this expansion? If in the expansion of $(x+a)^{15}$ the $11^{\text{th}}$ term is G. M. i.e. geometric mean of $8^{\text{th}}$ and $12^{\text{th}}$ terms then which term is the greatest term among the terms in expansion of $(x+a)^{15}$?
| Applying the binomial theorem we have
\begin{align*}
(x+a)^{15}=\sum_{j=0}^{15}\binom{15}{j}x^ja^{15-j}
\end{align*}
Since the $11^{\text{th}}$ term is the geometric mean of the the $8^{\text{th}}$ and $12^{\text{th}}$ term, we have
\begin{align*}
\binom{15}{10}x^{10}a^5&=\sqrt{\binom{15}{7}x^7a^8\binom{15}{11}x^{11}a^4}\\
&=x^9a^6\sqrt{\binom{15}{7}\binom{15}{11}}\\
\end{align*}
Simplification with $a\ne 0$ gives
\begin{align*}
\frac{x}{a}&=\frac{\sqrt{\binom{15}{7}\binom{15}{11}}}{\binom{15}{10}}=\ldots=\frac{5}{77}\sqrt{231}\approx0.98\tag{1}
\end{align*}
In order to find the greatest term we compare successive terms we obtain
\begin{align*}
\binom{15}{j+1}x^{j+1}a^{14-j}&>\binom{15}{j}x^ja^{15-j}\tag{2}\\
\end{align*}
Simplification of (2) gives
\begin{align*}
\frac{x}{a}&>\binom{15}{j}\binom{15}{j+1}^{-1}=\ldots=\frac{j+1}{15-j}\\
\end{align*}
After rearrangement we get $j$ on the left-hand side and evaluation according (1) gives
\begin{align*}
j&<\frac{15\frac{x}{a}-1}{\frac{x}{a}+1}\bigg|_{\frac{x}{a}=\frac{5}{77}\sqrt{231}}\approx 6.8\tag{3}
\end{align*}
We conclude $j=6$ and the greatest term is according to (2) the term at the left-hand side of (2) which is (with $j+1=7$) the $8^{\text{th}}$ term
\begin{align*}
\color{blue}{\binom{15}{7}x^7a^8}
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2918651",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Solving $x^4 -10x^3 + 26x^2 -10x +1 = 0$. Find the root of the equation
$$x^4 -10x^3 + 26x^2 -10x +1 = 0$$
My attempt:
If $a,b,c$ and $d$ are a roots then $a+b+c+d = 10,$
$(a+b)(c+d) + ab +cd = 26$
$(a+b)cd +ab(c+d) = 10$
$abcd =1$
Now i'm not able procceed further.
| You can proceede also like this:
\begin{eqnarray}x^4 -10x^3 + 26x^2 -10x +1 &=& (x^4 -4x^3 + 6x^2 -4x +1)-6x(x^2-2x+1)+8x^2\\
&=& (x-1)^4-6x(x-1)^2+8x^2\\
&=& \Big((x-1)^2-4x\Big)\Big((x-1)^2-2x\Big)\\
&=& \Big(x^2-6x+1\Big)\Big(x^2-4x+1\Big)\\
\end{eqnarray}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2922660",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Solving a recurrence relation with floor function I'm having trouble solving this recurrence relation:
\begin{align}
T(n) &=
\begin{cases}
2\,T\big(\big\lfloor \frac{n}{\sqrt{2}} \big\rfloor - 5\big) + n^\frac{\pi}{2} &\text{if } n > 7 \\
1 &\text{otherwise}
\end{cases}
\end{align}
where $n \in \mathbb{N}$. I would prefer to find the explicit solution for $T(n)$, but just an asymptotic bound on the solution would be enough.
I guess this is going to be done via substitution method and through induction, but I have no idea how to set it up/solve it. I assume the master theorem cannot be used here, because of the floor function.
I found two similar questions, here and here, but I don't know how their solutions can be adapted to my question.
| Actually, after some manipulations, you can use the master theorem! Let us see how. First, let us prove the following lemma:
Lemma: The function $T$ is non-decreasing, i.e. $T(n) \leq T(n+1)$ for all $n \in \mathbb{N}$.
Proof. By strong induction on $n \in \mathbb{N}$.
Base cases: For all $0 \leq n \leq 6$, one has $T(n) = 1 = T(n+1)$. Moreover, $T(7) = 1 < 2\,T(0) + 8^{\pi/2} = T(8)$, as $\big\lfloor \frac{8}{\sqrt{2}} \big\rfloor =5$.
Inductive step: Let $n > 7$. The strong induction hypothesis is $T(k) \leq T(k+1)$ for all $0 \leq k < n$. The goal is to prove that $T(n) \leq T(n+1)$.
By definition,
\begin{align}
T(n) &= 2\,T\big(\big\lfloor \frac{n}{\sqrt{2}} \big\rfloor - 5\big) + n^\frac{\pi}{2}
&
T(n+1) &= 2\,T\big(\big\lfloor \frac{n+1}{\sqrt{2}} \big\rfloor - 5\big) + (n+1)^\frac{\pi}{2}\,.
\end{align}
According to the properties of the floor function, $\big\lfloor \frac{n+1}{\sqrt{2}} \big\rfloor \leq \big\lfloor \frac{n}{\sqrt{2}} \big\rfloor + \big\lfloor \frac{1}{\sqrt{2}} \big\rfloor + 1 = \big\lfloor \frac{n}{\sqrt{2}} \big\rfloor + 1$, and $\big\lfloor \frac{n}{\sqrt{2}} \big\rfloor \leq \big\lfloor \frac{n}{\sqrt{2}} \big\rfloor + \big\lfloor \frac{1}{\sqrt{2}} \big\rfloor \leq \big\lfloor \frac{n+1}{\sqrt{2}} \big\rfloor$, since $\sqrt{2} > 1$.
Therefore, there are only two cases:
*
*either $\big\lfloor \frac{n+1}{\sqrt{2}} \big\rfloor = \big\lfloor \frac{n}{\sqrt{2}} \big\rfloor$ and then $T(n) \leq 2\,T\big(\big\lfloor \frac{n}{\sqrt{2}} \big\rfloor - 5\big) + (n+1)^\frac{\pi}{2} = T(n+1)$, where the inequality holds because from $\frac{\pi}{2} > 0$ it follows that $n^\frac{\pi}{2} < (n+1)^\frac{\pi}{2}$ ;
*or $\big\lfloor \frac{n+1}{\sqrt{2}} \big\rfloor = \big\lfloor \frac{n}{\sqrt{2}} \big\rfloor + 1$; we can apply the strong induction hypothesis to $T \big(\big\lfloor \frac{n}{\sqrt{2}} \big\rfloor- 5 \big)$ because $\big\lfloor \frac{n}{\sqrt{2}} \big\rfloor - 5 < n$ (indeed, $n + 5 > n = \lfloor n \rfloor \geq \big\lfloor \frac{n}{\sqrt{2}} \big\rfloor $ since $\lfloor \cdot \rfloor$ is non-decreasing), so $T \big(\big\lfloor \frac{n}{\sqrt{2}} \big\rfloor- 5 \big)\leq T \big(\big\lfloor \frac{n}{\sqrt{2}} \big\rfloor- 5 + 1\big) = T \big(\big\lfloor \frac{n + 1}{\sqrt{2}} \big\rfloor- 5 \big)$ and hence $T(n) \leq 2\,T\big(\big\lfloor \frac{n}{\sqrt{2}} \big\rfloor - 5\big) + (n+1)^\frac{\pi}{2} = T(n+1)$. $\qquad\square$
As $T$ is non-decreasing by the lemma above (and $\big\lfloor \frac{n}{\sqrt{2}} \big\rfloor - 5 < \big\lfloor \frac{n}{\sqrt{2}} \big\rfloor \leq \frac{n}{\sqrt{2}}$), for $n > 7$ one has $T(n) = 2\,T\big(\big\lfloor \frac{n}{\sqrt{2}} \big\rfloor - 5\big) + n^\frac{\pi}{2} \leq 2\,T\big(\frac{n}{\sqrt{2}}\big) + n^\frac{\pi}{2}$. Therefore, if we set
\begin{align}
S(n) =
\begin{cases}
1 &\text{if } n = 0 \\
2\,S\big(\frac{n}{\sqrt{2}}\big) + n^\frac{\pi}{2} & \text{otherwise}
\end{cases}
\end{align}
then $T(n) \leq S(n)$ for all $n \in \mathbb{N}$ and so, for any function $g$, $S(n) \in O(g(n))$ implies $T(n) \in O(g(n))$, i.e. the fact that $S$ grows asymptotically no faster than $g$ implies that $T$ grows asymptotically no faster than $g$.
The point is that we can use the master theorem to find a $g$ such that $S(n) \in O(g(n))$. Using the same notations as in Wikipedia article:
\begin{align}
a &= 2 & b&= \sqrt{2} & c_\text{crit} &= \log_\sqrt{2} 2 = 2 & f(n) &= n^{\pi/2}
\end{align}
thus, $S(n) \in O(n^2)$ by the master theorem (since $\pi/2 < 2 = c_\text{crit}$), and hence $T(n) \in O(n^2)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2924180",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Resolution of first order differential equation I have difficulties to solve these two differential equations:
1) $ y'(x)=\frac{x-y(x)}{x+y(x)} $ with the initial condition $ y(1)=1 $ .I'm arrived to prove that $$ y=x(\sqrt{2-e^{-2(\ln x+c)}}-1) $$ but I don't know if it's correct. If it's right, how do I find the constant $ c $? Because WolframAlpha says that the solution is $ y(x)=\sqrt{2}\sqrt{x^2+1}-x $.
2) $ y'(x)=\frac{2y(x)-x}{2x-y(x)} $ . I'm arrived to prove that $ \frac{z-1}{(z+1)^3}=e^{2c}x^{2} $ but I don't know if it's correct. If it's right, how do I explain $ z $ to substitute it in $ y=xz $? Then, how do I find the constant $ c $ ?
Thanks for any help!
| HINT
First of all, notice that
\begin{align*}
y^{\prime} = \frac{x-y}{x+y} \Longleftrightarrow y^{\prime} = \frac{1 - \frac{y}{x}}{1 + \frac{y}{x}}
\end{align*}
Then, if we make $y = ux$, we get:
\begin{align*}
u + u^{\prime}x = \frac{1-u}{1+u} \Longleftrightarrow u^{\prime}x = \frac{1-u}{1+u} - u = \frac{1-2u-u^{2}}{1+u} \Longleftrightarrow \left[\frac{1+u}{2 - (1+u)^{2}}\right]u^{\prime} = \frac{1}{x}
\end{align*}
The same trick applies to the second case. Can you proceed from here?
EDIT
If we make $y = ux$, we get:
\begin{align*}
y^{\prime} = \frac{2y-x}{2x-y} & \Longleftrightarrow u + u^{\prime}x = \frac{2u-1}{2 - u}\\
& \Longleftrightarrow u^{\prime}x = \frac{2u-1}{2 - u} - u = \frac{u^{2}-1}{u-2}\\
& \Longleftrightarrow \left[\frac{u-2}{u^{2}-1}\right]u^{\prime} = \frac{1}{x}
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2925633",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 1
} |
Extend $f(x)=\frac{6x^3-11x^2-16x-4}{2x^2-5x-2}$ in a way it is continuous $\forall x \in \mathbb{R}$ Problem
If it is possible to extend $$f(x)=\frac{6x^3-11x^2-16x-4}{2x^2-5x-2}$$ in a way it is continuous $\forall x \in \mathbb{R}$. Extend all $x \not\in X$ where $X$ is domain of $f$. Also define domain $X$.
Attempt to solve
By solving $2x^2-5x-2=0$ we can get $\forall x \not\in X$.
$$ x= \frac{5\pm \sqrt{(-5)^2-4 \cdot 2\cdot(-2)}}{2 \cdot 2} $$
$$ x= \frac{5 \pm \sqrt{41}}{4} $$
Meaning :
$$ X=\mathbb{R}\setminus \{\frac{5-\sqrt{41}}{4},\frac{5+\sqrt{41}}{4}\} $$
In order to $f$ to be continuous $\forall x \
\not\in X$ have to have limits in these undefined points.
$$ \lim_{x \rightarrow \frac{5-\sqrt{41}}{4}} \frac{6x^3-11x^2-16x-4}{2x^2-5x-2}
=\frac{1}{4}(23-3\sqrt{41})$$
and
$$ \lim_{x \rightarrow \frac{5+\sqrt{41}}{4}}\frac{6x^3-11x^2-16x-4}{2x^2-5x-2} = \frac{1}{4}(23+3\sqrt{41})$$
Meaning if i set
$$
\begin{cases}
f(\frac{5-\sqrt{41}}{4})=\frac{1}{4}(23-3\sqrt{41}) \\
f(\frac{5+\sqrt{41}}{4}=\frac{1}{4}(23+3\sqrt{41})
\end{cases}
$$
is continous $\forall x \in \mathbb{R}$
Now only problem is that is this correct solution and if it is how do i solve it more simply than this ? I didn't compute these limits by hand, maybe there is more simpler solution ?
| One point here is that there is only a removable singularity in such a function if the numerator and denominator have a common root. Common factors of polynomials, which of course have identical roots, can be found by Euclid's extended division algorithm (here just dividing through gives what you need, but if there were a remainder the process could be iterated).
This process could be done entirely with polynomials having rational coefficients - no irrational numbers are required. At the end there may be a common factor which can cancel, and the remaining fraction will be in the equivalent of "lowest terms". At this stage, if the denominator still has a real root, each such root will be a singularity which cannot be removed.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2927664",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
How to find the value of $r$ such that $\frac{1}{\binom{9}{r}} - \frac{1}{\binom{10}{r}} = \frac{11}{6\binom{11}{r}}$? Find the value of $r$:
$$\frac{1}{\dbinom{9}{r}} - \frac{1}{\dbinom{10}{r}} = \frac{11}{6\dbinom{11}{r}}$$
I'm not sure where I should take this problem in order to isolate $r$. I seem to get many factorials that don't seem to combine.
| Making the problem more general
$$\frac{1}{\dbinom{n}{r}} - \frac{1}{\dbinom{n+1}{r}} = \frac{k}{\dbinom{n+2}{r}}$$ and using the same approach as N. F. Taussig in his/her answer, you should end with a quadratic equation in $r$
$$k r^2- \left(k(2 n+3 )+n+2\right)r+k\left( n^2+3 n+2\right)=0$$ If $k >0$, there are two positive roots $\left(\Delta=k^2+2 k (n+2) (2 n+3)+(n+2)^2,r_1+r_2>0,r_1r_2>0\right)$ and, as said in answers, you must discard the one which is > n.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2929538",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
How to find $a$ and $b$ in $\lim\limits_{x→0}\frac{\sqrt{ax+b}-2}{x}=1$? How to find $a$ and $b$ such that $$\lim_{x→0}\frac{\sqrt{ax+b}-2}{x}=1?$$
By setting the equation equal to one directly I was able to find out that $b = 4$. However, when I try to solve for $a$ I am stuck and left in circles. How would I go about solving this problem?
| You could also do it using binomial expansion or Taylor series (built at $x=0$)
$$\sqrt{a x+b}=\sqrt{b}+\frac{a x}{2 \sqrt{b}}-\frac{a^2 x^2}{8 b^{3/2}}+O\left(x^3\right)$$ making
$$\frac{\sqrt{ax+b}-2}{x}=\frac{-2+\sqrt{b}+\frac{a x}{2 \sqrt{b}}-\frac{a^2 x^2}{8 b^{3/2}}+O\left(x^3\right) }x=\frac{\sqrt{b}-2}{x}+\frac{a}{2 \sqrt{b}}-\frac{a^2 x}{8 b^{3/2}}+O\left(x^2\right)$$ So
$$\frac{a}{2 \sqrt{b}}=1 \qquad \text{and}\qquad \sqrt{b}-2=0$$ then $a,b$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2929813",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 2
} |
Find the nonzero vectors $u,v,w$ that are perpendicular to the vector $(1,1,1,1)$ and to each other Find the nonzero vectors $u,v,w$ that are perpendicular to the vector $(1,1,1,1)$ and to each other.
Answer:
If I follow algebra, then I get complicated results to solve it as follows:
Let $u=(u_1,u_2,u_3,u_4), \ v=(v_1,v_2,v_3,v_4) , \ w=(w_1,w_2,w_3,w_4)$
Then, $u \cdot (1,1,1,1)=v \cdot (1,1,1,1)=w \cdot (1,1,1,1)=0$
Also $u \cdot v=w \cdot u=v \cdot w=0$.
These gives us
$u_1+u_2+u_3+u_4=0, \\ v_1+v_2+v_3+v_4=0 , \\ w_1+w_2+w_3+w_4=0, \\ u_1v_1+u_2v_2+u_3v_3+u_4v_4=0, \\ u_1w_1+u_2w_2+u_3v_3+u_4v_4=0, \\ v_1w_1+v_2w_2+v_3w_3+v_4w_4=0. $
But how to solve for $u_i, v_i,w_i, \ i=1,2,3,4$ from here?
Does there exit any other easy method?
Help me out
| as columns
$$
\left(
\begin{array}{rrrr}
1&-1&-1&-1 \\
1& 1&-1&-1 \\
1&0 &2&-1 \\
1&0&0&3
\end{array}
\right)
$$
Pattern, done correctly, works in any dimension
$$
\left( \begin{array}{rrrrrrrrrr}
1 & -1 & -1 & -1 & -1 & -1 & -1 & -1 & -1 & -1 \\
1 & 1 & -1 & -1 & -1 & -1 & -1 & -1 & -1 & -1 \\
1 & 0 & 2 & -1 & -1 & -1 & -1 & -1 & -1 & -1 \\
1 & 0 & 0 & 3 & -1 & -1 & -1 & -1 & -1 & -1 \\
1 & 0 & 0 & 0 & 4 & -1 & -1 & -1 & -1 & -1 \\
1 & 0 & 0 & 0 & 0 & 5 & -1 & -1 & -1 & -1 \\
1 & 0 & 0 & 0 & 0 & 0 & 6 & -1 & -1 & -1 \\
1 & 0 & 0 & 0 & 0 & 0 & 0 & 7 & -1 & -1 \\
1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 8 & -1 \\
1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 9
\end{array}
\right).
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2932529",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Regarding solving the partial differential equation using the method of characteristics I would like to solve the p.d.e of $(a+b)f_a + (a-b)f_b -f =2$ subject to the initial condition $f(a,0)=0$ with $a > 0$.
So the characteristic equations will be $a_t(t,s)=a+b$, $b_t(t,s)=a-b$ and $f_t(t,s)=f+2$.
Also the parametric initial conditions will be $a(0,s)=s$, $b(0,s)=0$ and $f(0,s)=0$.
Since the Jacobian is $-s \neq 0$, we have a unique solution.
The solution of the equations using the o.d.e is therefore $a(t,s)=\frac{1}{4}se^{-\sqrt{2}t}((2+\sqrt{2})e^{2\sqrt{2}t}+2-\sqrt{2})$, $b(t,s) = \frac{1}{2\sqrt{2}}se^{-\sqrt{2}t}(e^{2\sqrt{2}t}-1)$ and $f(t,s) = 2e^t - 2$.
Now I would like to complete the computation by expressing $t$ and $s$ in terms of $a$ and $b$ (and thus expressing $f$ in terms of $a$ and $b$), but absolutely have no idea how to approach it. In fact I suspect the methodology I used is not correct. Any help will be greatly appreciated.
| I didn't check your calculus. Supposing that it is correct, the solution would be :
$$a(t,s)=\frac{1}{4}se^{-\sqrt{2}t}((2+\sqrt{2})e^{2\sqrt{2}t}+2-\sqrt{2})$$
$$b(t,s) = \frac{1}{2\sqrt{2}}se^{-\sqrt{2}t}(e^{2\sqrt{2}t}-1)$$
$$f(t,s) = 2e^t - 2$$
My answer is only about the method to find $f(a,b)$ explicitly.
Let $\quad X=\left(\frac{f-2}{2}\right)^{2\sqrt{2}}\quad;\quad e^{2\sqrt{2}t}=X$
$$a=\frac{1}{4}sX^{-1/2}((2+\sqrt{2})X+2-\sqrt{2})$$
$$b = \frac{1}{2\sqrt{2}}sX^{-1/2}(X-1)$$
We eliminate $s$ :
$$\frac{a}{b}=\frac{\frac{1}{4}((2+\sqrt{2})X+2-\sqrt{2})}{ \frac{1}{2\sqrt{2}}(X-1) }$$
$$\frac{a}{b}\frac{1}{2\sqrt{2}}(X-1)- \frac{1}{4}((2+\sqrt{2})X+2-\sqrt{2})=0$$
Solve this linear equation for $X$. This leads to $X$ as a function of $\frac{a}{b}$. Then :
$$f(a,b)=2+2X^{1/2\sqrt{2}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2933304",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
The lines that connect the intersections of the three circles around a triangle meet at the same point
For any $\Delta ABC$, construct three circles with radius $AB,CB,AC$ centered at $A,B,C$ respectively. Then, connect the intersections of the circles as shown in the picture. The lines are concurrent and meet at point $O$. I am curious which center this is among the thousands of the known triangle centers.
I know that I can find it by calculating the barycentrics and the trillinears, but could you teach me how?
| The lines through two points of intersection of two circles is called the radical axis; the axis is actually defined for any two non-concentric circles, and the radical axis theorem states that the axes for each pair of three given circles are concurrent. So, that property of the configuration in question is unsurprising.
As @darij notes in a comment, the point described is not a triangle center, as its description is not properly symmetric in the elements of the triangle. Nevertheless, I'll walk through finding the barycentric and trilinear coordinates, since OP wants to see how that's done.
Defining $a := |BC|$, $b := |CA|$, $c := |AB|$ as is typical, we can assign these (Cartesian) coordinates
$$A = (0,0) \qquad B=(c,0) \qquad C = (b \cos A, b \sin A)$$
Then we have
$$\begin{align}
\bigcirc A:\quad x^2 + y^2 &= c^2 \\
\bigcirc B:\quad x^2 + y^2 &= b^2 - 2 b c \cos A + 2 xc \quad\to\quad x^2+y^2= a^2 - c^2+ 2 x c \\
\bigcirc C:\quad x^2 + y^2 &= 2 b ( x\cos A + y\sin A)
\end{align}$$
The equations of the radical axes of two circles (intersecting or not) are dead-simple to calculate: simply combine the circle equations to eliminate $x^2$ and $y^2$ terms; here, simple subtraction works. Solving any two of the resulting equations, and manipulating appropriately, gives
$$P =
\left(\frac{-a^2 + 2 c^2}{2 c}, \frac{-a^4 + a^2 b^2 + 3 a^2 c^2 - 2 b^2 c^2}{
4 b c^2 \sin A}\right)$$
Writing $$P = \frac{\alpha A + \beta B + \gamma C}{\alpha+\beta+\gamma}$$ we can solve for $\alpha$, $\beta$, $\gamma$ to get
$$\begin{align}
\alpha:\beta:\gamma \quad=\quad &b^4 - 3 a^2 b^2 - \phantom{3}b^2 c^2 + 2 c^2 a^2 \\[4pt]
:\; &c^4 + 2 a^2 b^2 - 3 b^2 c^2 - \phantom{3}c^2 a^2 \\[4pt]
:\; &a^4 - \phantom{3}a^2 b^2 + 2 b^2 c^2 - 3 a^2 c^2
\end{align}$$
These are the barycentric coordinates of $P$. The trilinear coordinates of $P$ are $\alpha/a : \beta/b :\gamma/c$.
Given barycentrics $\alpha:\beta:\gamma$, we could search the Encyclopedia of Triangle Centers using a "normalized" value generated from specific edge-lengths $a=6$, $b=9$, $c=13$. In this case, we have
$$\frac{2|\triangle ABC|}{\alpha+\beta+\gamma}\;\frac{\alpha}{a} = 3.26441\ldots$$
This number doesn't not appear in the table, but we already knew not to expect it, since $P$ is not a symmetrically-defined triangle center. $\square$
For an example of a related point that is a Triangle Center, take $\bigcirc A$, $\bigcirc B$, $\bigcirc C$ to have respective radii $a$, $b$, $c$. (See the symmetry in the definition?) Using the techniques above, we find that the point of concurrency of the three circles is
$$P = \left( c - b \cos A, \frac{b - 2 c \cos A + b \cos A^2}{\sin A}\right)$$
so that it has barycentric coordinates
$$\begin{align}
\alpha:\beta:\gamma \quad=\quad &-3 a^4 + \phantom{3} b^4 + \phantom{3}c^4 + 2 a^2 b^2 + 2 a^2 c^2 - 2 b^2 c^2 \\[4pt]
:\;&\phantom{\,-3}a^4 - 3 b^4 + \phantom{3}c^4 + 2 a^2 b^2 - 2 a^2 c^2 + 2 b^2 c^2 \\[4pt]
:\;&\phantom{\,-3}a^4 + \phantom{3}b^4 - 3 c^4 - 2 a^2 b^2 + 2 a^2 c^2 + 2 b^2 c^2
\end{align}$$
The corresponding lookup number at the ETC is $19.2413\ldots$, which we find leads to the De Longchamps Point, $X(20)$. $\square$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2934958",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
How to transform the graph $\sin^{-1}(3x+1)$? So there is this question in which you have to transform the graph of $\sin^{-1}(3x+1)$ and so is the following:
$\sin^{-1}(3(x+\frac{1}{3}))$ with $D: -1 \leq x \leq 1$
However, if one applies the transformation of moving the graph $\frac{1}{3}$ to the left, it results in $\frac{2}{3}$. Now, you apply the transformation of horizontally compressing it by a factor of $\frac{1}{3}$, to get $\frac{2}{9}$. However, using graphing technology, the graph domain is $-\frac{2}{3} \leq x \leq 0$ which isn't known how to get.
Thanks.
| Here's why ...
$$\begin{align}
y &= \sin^{-1}\bigg(3(x-(-\frac{1}{3}))\bigg)\\
\sin y &= 3(x-(-\frac{1}{3}))\\
\frac{1}{3}\sin y &= x-(-\frac{1}{3})\\
\frac{1}{3}\sin y + (-\frac{1}{3}) &= x\\
\end{align}$$
so after algebraically flipping the graph $45^{\circ}$ we can easily see that we have a compression along the x-axis by a factor of three and a shift along the x-axis of magnitude $-\frac{1}{3}$. (I avoided the terms 'vertical' and 'horizontal' intentionally here to avoid confusion).
So, if $\sin y$ ranges over $[-1, 1]$, then $\frac{1}{3}\sin y$ ranges over $[-\frac{1}{3}, \frac{1}{3}]$, and $\frac{1}{3}\sin y + (-\frac{1}{3})$ ranges across $[-\frac{2}{3},0]$, but this is just the domain of $x$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2935831",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
How to know that $2n^3+9n^2+13n+6$ factors into $(n+1)(n+2)(2n+3)$? Apologies in advance, my math is very rusty.
I'm slowly working my way through Schaum's Outline of Discrete Math for some self-study, occasionally filling in large knowledge gaps in my grasp on algebra. In one of the supplementary questions I'm asked to prove (by induction) that:
$$
\sum^n_{i=1}i^2 = \frac{n(n+1)(2n+1)}{6}
$$
In the inductive step I add $(n+1)^2$ to either side and then try to work my way through, multiplying out the factors:
$$
\frac{n(n+1)(2n+1)}{6} + (n+1)^2 \\
= \frac{n(n+1)(2n+1)+6(n+1)^2}{6} \\
= \frac{(n^2+n)(2n+1)+6(n+1)^2}{6} \\
= \frac{2n^3+n^2+2n^2+n+6(n^2+2n+1)}{6} \\
= \frac{2n^3+n^2+2n^2+n+6n^2+12n+6}{6} \\
= \frac{2n^3+9n^2+13n+6}{6}
$$
I think everything up to this point is pretty trivial, but it's around this point that I get a bit lost. I know I need to factor the terms in the 3-degree polynomial in the numerator somehow, but I'm having a hard time with the actual mechanics given the lack of obvious common factors & the fact that 13 is prime.
I'm pretty sure the final result should look something like:
$$
\frac{(n+1)((n+1)+1)(2(n+1)+1)}{6} \\
= \frac{(n+1)(n+2)(2n+3)}{6}
$$
Multiplying the expected result out I can see the two statements are equal and I'm confident that
$$
\frac{2n^3+9n^2+13n+6}{6} = \frac{(n+1)(n+2)(2n+3)}{6}
$$
but I'm just not quite sure how to factor the polynomial myself to arrive at the final result.
Any hints on how to proceed from here, or what I need to be reading up on to get my head around this?
| Recall (in your question) the line ...
$$=\frac{(n^2+n)(2n+1)+6(n+1)^2}{6}$$
What followed, however was a missed opportunity, for $n^2+n$ factors as $n(n+1)$, so your expression becomes
$$\begin{align}
\frac{(n^2+n)(2n+1)+6(n+1)^2}{6} &= \frac{n(n+1)(2n+1)+6(n+1)^2}{6}\\
&= \frac{(n+1)\bigg(n(2n+1)+6(n+1)\bigg)}{6}\\
&= \frac{(n+1)(2n^2+n+6n+6)}{6}\\
&= \frac{(n+1)(2n^2+7n+6)}{6}\\
\end{align}
$$
and all that remains to do is to factor the quadratic $2n^2+7n+6$, and you're done. :-)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2935977",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 6,
"answer_id": 1
} |
How to solve this set of definite integrals? I'm looking to solve the integrals
$$
I_n=\int_0^\infty \frac{x^ne^{-2ax}}{\sqrt{x^4+1}}dx.
$$
for $a>0$.
This should be reduced to finding $I_0$ as numerically I find that
$$
I_n=\left(-2\right)^{-n}\frac{d^nI_0}{da^n}.
$$
Mathematica gives a result in terms of the MeijerG function $G^{51}_{15}\left(x|^{\frac{1}{4}}_{-\frac{1}{2},-\frac{1}{4},-\frac{1}{4},0,\frac{1}{4}}\right)$ but I cannot manage to prove it. I have looked for integral representations of this and hypergeometric functions in Gradshteyn but couldn't find anything useful.
Any help is much welcomed.
Thanks
| One way to get the result is to use the Mellin transform of the integral considered as a function of $a$:
\begin{align}
I_0&=\int_0^\infty \frac{e^{-2ax}}{\sqrt{x^4+1}}\,dx\\
\mathcal{M}\left[ I_0\right]&=\int_0^\infty a^{s-1}\,da\int_0^\infty \frac{e^{-2ax}}{\sqrt{x^4+1}}\,dx\\
&=\int_0^\infty \frac{1}{\sqrt{x^4+1}}\,dx \int_0^\infty a^{s-1}e^{-2ax}\,da\\
&=2^{-s}\Gamma(s)\int_0^\infty \frac{x^{-s}}{\sqrt{x^4+1}}\,dx
\end{align}
which is valid for $\Re(s)>0$. Now, changing $x=y^{1/4}$ in the integral, one obtains
\begin{align}
\mathcal{M}\left[ I_n\right]&=2^{-s-2}\Gamma\left(s\right)\int_0^\infty\frac{y^{\frac{-s-3}{4}}}{\sqrt{y+1}}\,dy\\
&=2^{-s-2}\Gamma(s)B\left(\frac{-s+1}{4},\frac{s+1}{4}\right)\\
&=\frac{2^{-s}}{4\sqrt{\pi}}\Gamma\left(s\right)\Gamma\left(\frac{-s+1}{4}\right)\Gamma\left(\frac{s+1}{4}\right)
\end{align}
(The integral is the Mellin transform of $1/\sqrt{y+1}$ taken at $(n-s+1)/4$). This result is valid for $0<\Re(s)<1$. Taking the inverse transform,
\begin{equation}
I_0=\frac{1}{4\sqrt{\pi}}\frac{1}{2i\pi}\int_{\sigma-i\infty}^{\sigma+i\infty}(2a)^{-s}\Gamma\left(s\right)\Gamma\left(\frac{-s+1}{4}\right)\Gamma\left(\frac{s+1}{4}\right)\,ds
\end{equation}
with $0<\sigma<1$. Changing $s=4t$, one can express
\begin{equation}
I_n=\frac{1}{\sqrt{\pi}}\frac{1}{2i\pi}\int_{\sigma-i\infty}^{\sigma+i\infty}(16a^4)^{-t}\Gamma\left(4t\right)\Gamma\left(t+\frac{1}{4}\right)\Gamma\left(\frac{1}{4}-t\right)\,dt
\end{equation}
Expanding $\Gamma(4t)$ using the duplication formula:
\begin{equation}
I_0=\frac{\sqrt{2}}{8\pi^2}\frac{1}{2i\pi}\int_{\sigma-i\infty}^{\sigma+i\infty}(\frac{a^4}{16})^{-t}\Gamma\left(t\right)\Gamma^2\left( t+\frac{1}{4} \right)\Gamma\left( t+\frac{1}{2} \right)\Gamma\left( t+\frac{3}{4} \right)\Gamma\left(\frac{1}{4}-t\right)\,dt
\end{equation}
With $b_1=\frac{1}{4},a_1=1,a_2=\frac{3}{4},a_3=\frac{3}{4},a_4=\frac{1}{2},a_5=\frac{1}{4}$ and $\sigma=1/8$, we can express the result using the integral representation of the Meijer function DLMF:
\begin{equation}
I_0=\frac{\sqrt{2}}{8\pi^2}{G^{1,5}_{5,1}}\left(\left.\frac{16}{a^4}\right|{1,\frac{3}{4},\frac{3}{4},\frac{1}{2},\frac{1}{4}\atop \frac{1}{4}}\right)
\end{equation}
which, using these identities, can be transformed into
\begin{align}
I_0&=\frac{\sqrt{2}}{8\pi^2}{G^{5,1}_{1,5}}\left(\left.\frac{a^4}{16}\right|{\frac{3}{4}\atop 0,\frac{1}{4},\frac{1}{4},\frac{1}{2},\frac{3}{4}}\right)\\
&=\frac{\sqrt{2}}{32\pi^2}a^2{G^{5,1}_{1,5}}\left(\left.\frac{a^4}{16}\right|{\frac{1}{4}\atop -\frac{1}{2},-\frac{1}{4},-\frac{1}{4},0,\frac{1}{4}}\right)
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2936069",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
Find a basis for the range of a transformation Supposed $T(v) = Av$ and say I have a matrix like this:
$$\begin{bmatrix} 1 & 2 & -2 \\ 2 & -1 & 0 \\ 1 & 7 & -6 \end{bmatrix}$$
and I want to find the basis for the range.
Is this right?
$$\begin{bmatrix} 1 & 2 & -2 \\ 2 & -1 & 0 \\ 1 & 7 & 6 \end{bmatrix}$$
$$\begin{bmatrix} 1 & 2 & -2 \\ 0 & -5 & 4 \\ 0 & 5 & -4 \end{bmatrix}$$
$$\begin{bmatrix} 1 & 2 & -2 \\ 0 & 1 & \frac{-4}{5} \\ 0 & 0 & 0 \end{bmatrix}$$
$$\begin{bmatrix} 1 & 0 & \frac{-2}{5} \\ 0 & 1 & \frac{-4}{5} \\ 0 & 0 & 0 \end{bmatrix}$$
question, how do you know the first two columns are linearly independent? In short, why does then taking the first two columns from the original matrix the basis for the range of T?
The basis I think is: $(\begin{bmatrix} 1 & 2 & 1 \end{bmatrix}, \begin{bmatrix} 2 & -1 & 7 \end{bmatrix})$. Is this right?
Most importantly, how do you know the first two rows in the reduced row form are linearly independent?
| Your reasoning is sound. Elementary row operations don't chance the linearly independentness of columns, so from the reduced row echelon form you arrived at you can see the two first columns are linearly independent, while the third can (allegedly) be written as a sum of the two first. In the original matrix then $\begin{bmatrix} 1 \\ 2 \\ 1\end{bmatrix}$ and $\begin{bmatrix} 2 \\ -1 \\ 7\end{bmatrix}$ must be linearly independent.
These vectors are in the range of $T$ since $$\begin{bmatrix} 1 & 2 & -1 \\ 2 & -1 & 0 \\ 1 & 7 & 6 \end{bmatrix} \begin{bmatrix} 1 \\ 0 \\ 0\end{bmatrix} = \begin{bmatrix} 1 \\ 2 \\ 1\end{bmatrix}$$
and
$$\begin{bmatrix} 1 & 2 & -1 \\ 2 & -1 & 0 \\ 1 & 7 & 6 \end{bmatrix} \begin{bmatrix} 0 \\ 1 \\ 0\end{bmatrix} = \begin{bmatrix} 2 \\ -1 \\ 7\end{bmatrix}$$
So since we know the range is 2-dimensional they must span the range.
That being said, there is a small mistake in your calculation as the original matrix is invertible.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2938641",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Proof: For all integers $x$ there exists an integer $y$ so that $3$ divides $x+y$ and $3$ divides $x-y$ We start of a proof by finding out if it is true or false, I am having trouble with this and I think it is true, because the negation (there exists and integer $x$ for all integers $y$ so that $3$ does not divides $x+y$ and $3$ does not divides $x-y$) says the same thing? So I tried to prove it for it being true.
First thing I tried was to use the definition of divides:
$3(k) = x+y$ and $3(p)=x - y$ for integers $k$ and $p$,
then I isolated for $x$ in one of the equations:
$3(k) - y = x$ so $3(p) = 3(k) - y - y$,
then this would become:
$3(p) = 3(k) - 2(y)$ and since $k$ and $y$ are integers $p$ is also an integer.
Would this reasoning be correct?
| Remember that $(a+b)\pmod n = [a\pmod n + b\pmod n] \pmod n$, for any $a$ and $b$ positive or negative.
So, if $x \pmod 3 = 0$, we are done, and $y=0$ for instance.
If $x \pmod 3 = 1$, we may set $y=2$ (or any other $y$ such that $y \pmod 3 = 2$) so that $(x+y) \pmod 3 = 3 \pmod 3 = 0$. But $(x-y) \pmod 3 = (1 + 1) \pmod 3 = 2 \neq 0$!
So the statment is not true and we cannot satisfy both $(x+y) \pmod 3 = 0$ and $(x-y) \pmod 3 = 0$.
Note: $-2 \pmod 3 = 1$, not $-2$!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2941102",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Show that $\frac{\sqrt{2}+\sqrt{2+\sqrt{3}}}{\sqrt{2}-\sqrt{2+\sqrt{3}}}=-3-2\sqrt{3}$
Show that $$\frac{\sqrt{2}+\sqrt{2+\sqrt{3}}}{\sqrt{2}-\sqrt{2+\sqrt{3}}}=-3-2\sqrt{3}.$$
My attempt: I could find a way to develop the LHS by un-nesting the double radical, figuring out that
$$\sqrt{2+\sqrt{3}}=\frac{\sqrt{6}+\sqrt{2}}{2}\ \ (1)$$
By substituting (1) in the LHS of the original expression,
it is straightforward to show that it is equal to the RHS.
But my problem is on how to show without the un-nesting, by another approach. I've used a standard approach of
multiplying both terms by $\sqrt{2}+\sqrt{2+\sqrt{3}}$, leading to
$$\frac{4+\sqrt{3}+2\sqrt{4+2\sqrt{3}}}{-\sqrt{3}}\Leftrightarrow \frac{(4+\sqrt{3}+2\sqrt{4+2\sqrt{3}})\sqrt{3}}{-3}.$$
But I was not able to show that this last expression is equal to RHS by this approach.
Hints and answers not using my first un-nesting approach (if possible) will be appreciated. Sorry if this is a duplicate.
| Since $$\sqrt{2+\sqrt{3}} = \sqrt{4+2\sqrt{3}\over 2} = {1+\sqrt{3}\over \sqrt{2}}$$
\begin{eqnarray}\frac{\sqrt{2}+\sqrt{2+\sqrt{3}}}{\sqrt{2}-\sqrt{2+\sqrt{3}}}&=& \frac{\sqrt{2}+{1+\sqrt{3}\over \sqrt{2}}}{\sqrt{2}-{1+\sqrt{3}\over \sqrt{2}}}\\
&=& \frac{2+1+\sqrt{3}}{2-1-\sqrt{3}}\\
&=& \frac{3+\sqrt{3}}{1-\sqrt{3}}=...\\
&=&-3-2\sqrt{3}
\end{eqnarray}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2941552",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
Show that Ratio Test gives no information for $3^{-1} + 5^{-1} + 3^{-2} + 5^{-2} + 3^{-3} + 5^{-3} + \cdots$
Q: Examine the series:
\begin{align*}
\frac{1}{3} + \frac{1}{5} + \frac{1}{3^2} + \frac{1}{5^2} + \frac{1}{3^3} + \frac{1}{5^3} + \frac{1}{3^4} + \frac{1}{5^4} + \cdots \\
\end{align*}
Prove that the Root Test shows that the series converges while the Ratio test gives no information.
I can use the Root Test to show convergence as the question requests. I will omit that. My issue is that when I apply the Ratio Test, it also seems to show convergence, when the problem asks me to show that the Ratio Test gives no information.
Applying the Ratio Test:
\begin{align*}
\lim\limits_{j \to \infty} \left| \frac{a_{j+1}}{a_j} \right| &= \lim\limits_{j \to \infty} \frac{\frac{1}{3^{j+1}} + \frac{1}{5^{j+1}}}{\frac{1}{3^j} + \frac{1}{5^j}} \\
&= \lim\limits_{j \to \infty} \frac{\frac{1}{3} + \frac{1}{5} \cdot \left( \frac{3}{5} \right)^j}{1 + \left( \frac{3}{5} \right)^j} \\
&= \frac{1}{3} < 1 \\
\end{align*}
And therefore the ratio test says that this series will converge which is not what the problem was asking for.
| Given the series is $$3^{-1} + 5^{-1} + 3^{-2} + 5^{-2} + 3^{-3} + 5^{-3} + \cdots=\dfrac13+\dfrac15+\dfrac{1}{3^2}+\dfrac1{5^2}+\dfrac1{3^3}+\dfrac1{5^3}+.......$$
Notice that from the above series we get
$$a_{2n-1}=\dfrac{1}{3^n}\mbox{ and }a_{2n}=\dfrac{1}{5^n}$$
Now consider taking $\dfrac{a_{2n+1}}{a_{2n}}$ and $\dfrac{a_{2n}}{a_{2n-1}}$
$$\dfrac{a_{2n+1}}{a_{2n}}=\dfrac{1}{3}\left(\dfrac{5}{3}\right)^n\mbox{ which is greater than }1$$
and now $$\dfrac{a_{2n}}{a_{2n-1}}=\left(\dfrac{3}{5}\right)^n\mbox{ which is less than }1$$
Therefore, from $\dfrac{a_{2n+1}}{a_{2n}}$ and $\dfrac{a_{2n}}{a_{2n-1}}$ we cannot conclude.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2947946",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Formalization/Verification of a beginner combinatorics problem I have the following task:
To build a computer chip, 4 non-distinct components are needed. Of 12 existing components, 2 are defective. Given that 4 components are chosen, what are the probabilities of the following combinations?
*
*No defective components
*Exactly one defective component
*Exactly two defective components
Progress so far:
The general approach I use is to consider the number of non-defective components over the total number of components for repeated decrements of each set.
*
*$\frac{10}{12} \cdot \frac{9}{11} \cdot \frac{8}{10} \cdot \frac{7}{9} \approx 0.4242 $
For the next question, I sum the probabilities of each case, depending on if the defective component is chosen last, second last, etc.
*$\frac{10}{12} \cdot \frac{9}{11} \cdot \frac{8}{10} \cdot \frac{2}{9} \ + \frac{10}{12} \cdot \frac{9}{11} \cdot \frac{2}{10} \cdot \frac{8}{9} \ + \frac{10}{12} \cdot \frac{2}{11} \cdot \frac{9}{10} \cdot \frac{8}{9} \ + \frac{2}{12} \cdot \frac{10}{11} \cdot \frac{9}{10} \cdot \frac{8}{9} $
I could employ a similar intuitive approach for 3. , however I feel like this is messy and not particularly rigorous. How do I answer these questions formally?
| What you have so far is good. To simplify case 2, you can notice that the probabilities are all the same: for each one you get $12\times11\times10\times 9$ on the bottom and $10\times9\times8\times 2$ on the top in some order. So the total probability is just the probability for a specific "pattern" (e.g. defective, good, good, good) times the number of patterns. The same approach works for 3: here there are $6=\binom 42$ patterns, each with probability $\frac{10}{12}\times\frac{9}{11}\times\frac{2}{10}\times\frac{1}{9}$. This works because the first "good" (wherever you put it) always has $10$ on top, the second "good" always has $9$, and so on.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2948470",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Show that $A^2 +B^2+C^2=D^2$ using the following diagram (tetrahedron) Slicing a corner off a square gives a right-angled triangle, as shown in the diagram below.
The lengths of the sides of this triangle are related by Pythagoras’s theorem: $a^
2 + b^
2 = c^
2$
. Show
that this two-dimensional setup generalises to three dimensions in the following way. Slice a corner
off a cube, as shown in the diagram below. This gives a tetrahedron in which three of the faces are
right-angled triangles, while the fourth is not. Let’s call the areas of the three right-angled faces $A,
B, C$ and the area of the fourth face $D$.
$A
^2 + B^
2 + C^
2 = D^2$
.
Can anyone explain to me what formula/how to go about doing this question? thank you.
| Let $XYZT$ be our tetrahedron, where $TX\perp TY$, $TX\perp TZ$, $TY\perp TZ$, $TX=x$, $TY=y$ and $TZ=z$.
Thus, $XY=\sqrt{x^2+y^2}$, $XZ=\sqrt{x^2+z^2},$ $YZ=\sqrt{y^2+z^2}$ and
$$S_{\Delta XYZ}=\frac{1}{4}\sqrt{\sum_{cyc}(2(x^2+y^2)(x^2+z^2)-(x^2+y^2)^2)}=$$
$$=\frac{1}{4}\sqrt{\sum_{cyc}(2x^4+6x^2y^2-2x^4-2x^2y^2)}=\frac{1}{2}\sqrt{x^2y^2+x^2z^2+y^2z^2}$$ and since
$$\left(\frac{1}{2}xy\right)^2+\left(\frac{1}{2}xz\right)^2+\left(\frac{1}{2}yz\right)^2=\left(\frac{1}{2}\sqrt{x^2y^2+x^2z^2+y^2z^2}\right)^2,$$ we are done!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2950559",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
} |
Generating Pythagorean triples I'm asked to generate Pythagorean triples from the polynomial identity:
$$(X^2-1)^2 + (2X)^2=(X^2+1)^2$$
By substituting rational numbers $\frac p q$ for $X$. However, Pythagorean triples are just as the name says, it, three numbers. If I would substitute this number I get:
$$(\left(\frac p q\right)^2-1)^2 + 4\left(\frac p q\right)^2=(\left(\frac p q\right)^2+1)^2$$
How would I get three integers from this? There are just two numbers involved, $p$ and $q$.
| If we take the equation asis, it is not true unless $x=0$ or$X=1$.
$$(x-1)^2+2x=(x+1)^2$$
$$(x^2-2X+1)+4X^2=(x^2+2X+1)$$
Subracting $x^2+2x+1$ from both sides we get
$$4X^2-4X=0\Rightarrow4x(X-1)=0$$
A variation of the identity will work for all $n$ in generating triples where $C-A)=2.$
$$\mathbf{A=4n^2-1}$$
$$\mathbf{B=4n}$$
$$\mathbf{C=4n^2+1}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2950734",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Limit of $\sin(\pi \sqrt{4n^2+n})$ We can notice that $\sqrt{4n^2 + n} = \sqrt{4n^2(1+ \frac{1}{4n})} = 2n\sqrt{1+\frac{1}{4n}}$. Therefore
$$\lim_{n \to \infty} \sin (2n\pi \sqrt{1+ \frac{1}{4n}}) \text{ will be an even number}$$
Because the square root becomes $1$ and we end up with an even number: $\sin(\text{even number})$
And sine of an even number is $0$. But apparently that is not the right answer.
| When $n\to\infty$, we have
$$
\sqrt{4n^2+n} = 2n\sqrt{1+\frac{1}{4n}}
= 2n\left(1+\frac{1}{8n} + o\!\left(\frac{1}{n}\right)\right)
= 2n + \frac{1}{4} + o(1) \tag{1}
$$
using the Taylor expansion of $\sqrt{1+x}$ for $x$ around $0$. Recalling that $$\forall a,b,\qquad \sin(a+b)=\sin a \cos b + \cos a \sin b \tag{2}$$ we get
$$\begin{align*}
\sin(\pi\sqrt{4n^2+n}) &=\sin\left(2n\pi + \frac{\pi}{4} + o(1)\right) \\
&= \sin(2n\pi) \cos\left(\frac{\pi}{4} + o(1)\right)
+ \cos(2n\pi) \sin\left(\frac{\pi}{4} + o(1)\right) \\
&= 0 + \sin\left(\frac{\pi}{4} + o(1)\right) \\
&\xrightarrow[n\to\infty]{} \sin\frac{\pi}{4} = \boxed{\frac{1}{\sqrt{2}}}
\end{align*}$$
using continuity of $\sin$ (as $\frac{\pi}{4} + o(1)\xrightarrow[n\to\infty]{}\frac{\pi}{4}$).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2953115",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Integration of $\int^{\pi/45}_0 \frac{x^2 \ln (1-x)}{1-x^3}\, dx$ How would you integrate:
$$I=\int^{\pi/45}_0 \frac{x^2 \ln (1-x)}{1-x^3} dx$$
My attempt:
$$du=\frac{x^2}{1-x^3}dx ⇒ u=\frac{-1}{3}\ln (1-x^3)$$
$$v= \ln (1-x) ⇒ dv=\frac{-1}{1-x}$$
$$I=\frac{-1}{3}\ln (1-x^3) \ln (1-x)-\int \frac{\ln (1-x^3)}{1-x}\,dx$$
$\ln (1-x^3)=\ln (1-x)+\ln (1+x+x^2)$
$$\int \frac{\ln (1-x^3)}{1-x}\,dx=\int \frac{\ln(1-x)}{1-x}\,dx+\int \frac{\ln (x^2+x+1)}{1-x}\,dx$$
$$I_1=\int \frac{\ln(1-x)}{1-x}\,dx=-\frac{1}{2} \ln^2 (1-x)$$
$$x^2+x+1=\left(x+\frac{1-i\sqrt3}{2}\right)\left(x+\frac{1+i\sqrt3}{2}\right)$$
$$\ln \left(x+\frac{1-i\sqrt3}{2}\right)= \ln (2x+1-i\sqrt 3)- \ln 2=\tan^{-1}-\frac{\sqrt3}{2x+1}-\ln 2$$
$$\ln \left(x+\frac{1+i\sqrt3}{2}\right)= \ln (2x-1+i\sqrt 3)- \ln 2=\tan^{-1}\frac{\sqrt3}{2x-1}-\ln 2$$
$$\int \frac{\ln (x^2+x+1)}{1-x} \,dx=\int \frac{\tan^{-1}-\frac{\sqrt3}{2x+1}}{1-x}\,dx+\int \frac{\tan^{-1}\frac{\sqrt3}{2x-1}}{1-x}\,dx+2 \ln2 \ln (1-x)$$
Can someone proceed?
| Hint: A better method
Using the partial fraction:
$$\frac{{{x}^{2}}}{1-{{x}^{3}}}=\frac{1}{3}\frac{1}{1-x}-\frac{1}{3}\frac{2x+1}{{{x}^{2}}+x+1}=\frac{1}{3}\frac{1}{1-x}-\frac{2}{3}\frac{1}{2x-i\sqrt{3}+1}-\frac{2}{3}\frac{1}{2x+i\sqrt{3}+1}$$
The integral becomes
$$\begin{align}
& I=\frac{1}{3}\int{\frac{\ln \left( 1-x \right)}{1-x}dx--\frac{2}{3}\int{\frac{\ln \left( 1-x \right)}{2x-i\sqrt{3}+1}dx}-\frac{2}{3}\int{\frac{\ln \left( 1-x \right)}{2x+i\sqrt{3}+1}dx}} \\
& \quad =-\frac{1}{6}{{\ln }^{2}}\left( 1-x \right)-\frac{2}{3}J-\frac{2}{3}K \\ \end{align}$$
Evaluating $J$ and $K$ are similar, for $J$ first set $u=1-x$, to get:
$$J=\int{\frac{\ln \left( u \right)}{2u-3+i\sqrt{3}}du}$$
Now use the formula:
$$\int{\frac{\ln \left( u \right)}{au+b}dx}=\frac{1}{a}\log \left( u \right)\log \left( \frac{au}{b}+1 \right)+\frac{1}{a}L{{i}_{2}}\left( -\frac{au}{b} \right)$$
where $L{{i}_{n}}$ is the poly-logarithm function.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2953564",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Prove $\ 1<\frac{1}{n+1}+\frac{1}{n+2}+ ... + \frac{1}{3n+1}<2 $ using Cauchy-Schwarz How do I prove this inequality
$$\ 1<\frac{1}{n+1}+\frac{1}{n+2}+ ... + \frac{1}{3n+1}<2 $$
I've tried to prove that $$\ \frac{1}{n+1}+\frac{1}{n+2}+ ... + \frac{1}{3n+1} \ $$ is less than $$\ \frac{1}{1}+\frac{1}{2}+ ... + \frac{1}{n} \ $$ and that this sum is either less than $2$ or equal to $2$ using C-S.
| One Approach
Simpler than Cauchy-Schwarz is to note that for $x\in(n,n+1)$,
$$
\frac1{n+1}\lt\frac1x\lt\frac1n\tag1
$$
which, when integrated in $x$ over $(n,n+1)$, gives
$$
\frac1{n+1}\lt\log\left(\frac{n+1}n\right)\lt\frac1n\tag2
$$
then
$$
\begin{align}
\sum_{k=n}^{3n}\frac1{k+1}
&\le\sum_{k=n}^{3n}\log\left(\frac{k+1}k\right)\\
&=\log\left(\frac{3n+1}n\right)\\[3pt]
&=\log\left(3+\frac1n\right)\tag3
\end{align}
$$
and
$$
\begin{align}
\sum_{k=n}^{3n}\frac1{k+1}
&\ge\sum_{k=n}^{3n}\log\left(\frac{k+2}{k+1}\right)\\
&=\log\left(\frac{3n+2}{n+1}\right)\\[3pt]
&=\log\left(3-\frac1{n+1}\right)\tag4
\end{align}
$$
Since $\log(4)\lt2$, inequality $(3)$ shows the upper bound for $n\ge1$.
Since $\log\left(\frac{11}4\right)\gt1$, inequality $(4)$ shows the lower bound for $n\ge3$. We need only verify the lower bound for $n=1$ and $n=2$.
A Second Approach, Using Cauchy-Schwarz
Simply multiplying the number of terms by the largest term gives
$$
\begin{align}
\sum_{k=n}^{3n}\frac1{k+1}
&\le\frac{2n+1}{n+1}\\
&=2-\frac1{n+1}\\[9pt]
&\lt2\tag5
\end{align}
$$
Applying Cauchy-Schwarz gives
$$
\begin{align}
\sum_{k=n}^{3n}\frac1{k+1}\sum_{k=n}^{3n}(k+1)
&\gt\left(\sum_{k=n}^{3n}1\right)^2\\
&=(2n+1)^2\\
&=\sum_{k=n}^{3n}(k+1)\tag6
\end{align}
$$
Equality does not hold in the first line of $(6)$ because the vectors are not parallel.
Therefore,
$$
\sum_{k=n}^{3n}\frac1{k+1}\gt1\tag7
$$
Inequalities $(5)$ and $(7)$ show that
$$
1\lt\sum_{k=n}^{3n}\frac1{k+1}\lt2\tag8
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2955021",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 1
} |
Find the remainder when $13^{13}$ is divided by $25$. Find the remainder when $13^{13}$ is divided by $25$.
Here is my attempt, which I think is too tedious:
Since $13^{2} \equiv 19 (\text{mod} \ 25),$ we have $13^{4} \equiv 19^{2} \equiv 11 (\text{mod} \ 25)$ and $13^{8} \equiv 121 \equiv 21 (\text{mod} \ 25).$
Finally, we have $13^{8+4} \equiv 13^{12} \equiv 21\times 11 \equiv 231 \equiv 6 (\text{mod} \ 25)$ and hence $13^{13} \equiv 3 (\text{mod} \ 25).$
Is there a less tedious way to find the remainder? Thank you.
| We have
*
*$2\cdot 13 = 26 \equiv 1 \mod 25$
*$\Rightarrow 13^{13} \cdot \color{blue}{2^{13}} \equiv 1 \mod 25$
*$\color{blue}{2^{13}} \equiv 2^{10}\cdot 2^3 \equiv -8 \color{blue}{\equiv 17 \mod 25}$
$$\Rightarrow \boxed{13^{13} \equiv 17^{-1} \equiv 3 \mod 25}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2956834",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Find the smallest positive integer $X$ such that $478^{870} \equiv X \ (\text{mod} \ 273)$ Appreciate if one could advise if my solution is correct. Here is my attempt of the problem:
Since $(273, 478) =1,$ by Euler's theorem, $478^{\phi(273)}=478^{144} \equiv 1 \ \ (\text{mod} \ 273) \implies 478^{864} \equiv 1 \ \ (\text{mod} \ 273).$
Next, $478^{2} \equiv 22 \ \ (\text{mod} \ 39) \implies 478^{6} \equiv 22^{3} \equiv 1 \ \ (\text{mod} \ 39)$ and $478^{\phi(7)} = 478^{6} \equiv 1 \ \ (\text{mod} \ 7) $
Hence, $478^{6} \equiv 1 \ \ (\text{mod} \ 273)$ and $478^{864+6}= 478^{870} \equiv 1 \ \ (\text{mod} \ 273)$
| Recall Carmichael's theorem: $a^{\lambda(m)} \equiv 1 \bmod m$ for all $a$ coprime with $m$.
Now $\lambda(273)=lcm(\phi(3),\phi(7),\phi(13))=lcm(2,6,12)=12$ and $870 \equiv 6 \bmod 12$.
Therefore
$
478^{870} \equiv 205^{6} \equiv (-68)^{6} \equiv 4^6 \cdot 17^6 \equiv 1 \bmod 273
$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2957622",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
$x=\sqrt[3]{\sqrt{5}+2}+\sqrt[3]{\sqrt{5}-2}$ is rational or irrational?
The number $x$ defined below is rational or irrational?
$$x=\sqrt[3]{\sqrt{5}+2}+\sqrt[3]{\sqrt{5}-2}$$
From: IMO 1973 - Longlist
My attempt (my real question is at the end):
the identity $a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^3-ab-ac-bc)$ when $a+b+c=0$, leads to $$a^3+b^3+c^3=3abc \tag{1}$$
Now considering
$$a=\sqrt[3]{\sqrt{5}+2},b=\sqrt[3]{\sqrt{5}-2},c=-x$$
from (1) it is true that
$$x^3-3x-2\sqrt{5}=0 \tag{2}$$
That is the number $x$ is a root from (2).
Note: By trial and error I've found that answer is $x=\sqrt{5}$ (the other 2 roots are complex), that is irrational. But my question is more subtle.
Question: Can I conclude just inspecting (2), judging by the coefficient $2\sqrt{5}$, that $x$ is irrational, without actually solving the equation? In a math contest that might be helpful, if possible, as it would avoid extra steps.
| Alternatively
$$x=\sqrt[3]{\sqrt{5}+2}+\sqrt[3]{\sqrt{5}-2}=\\
\sqrt[3]{\frac{8\sqrt{5}+16}{8}}+\sqrt[3]{\frac{8\sqrt{5}-16}{8}}=\\
\sqrt[3]{\frac{5\sqrt{5}+3\cdot5+3\sqrt{5}+1}{8}}+\sqrt[3]{\frac{5\sqrt{5}-3\cdot5+3\sqrt{5}-1}{8}}=\\
\sqrt[3]{\frac{\left(\sqrt{5}+1\right)^3}{2^3}}+\sqrt[3]{\frac{\left(\sqrt{5}-1\right)^3}{2^3}}=\\
\frac{\sqrt{5}+1}{2}+\frac{\sqrt{5}-1}{2}=\\
\sqrt{5}$$
is irrational.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2958501",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 4,
"answer_id": 0
} |
Average height of raindrops from height $h$
Raindrops drip from a spout at the edge of a roof and fall to the ground. Assume that the drops drip at a steady rate of $n$ drops per second (where $n$ is large) and that the height of the roof is $h$.
(c) What is the average height of these raindrops as $n$ grows larger when the first drop hits the ground?
First we need to find the time that the first raindrop hits the ground. We know from the projectile motion formula that
$$-\frac{1}{2}gt^2+h=0\Longrightarrow t=\sqrt{\frac{2h}{g}}$$
We then substitute it back into the first equation:
$$-\frac{1}{2}g{\left(\sqrt{\frac{2h}{g}}\right)}^2+h=0.$$
Then we solve for $h$:
$$h_{\text{avg}}=\frac{1}{2}g{\left(\sqrt{\frac{2h}{g}}\right)}^2$$
We add an additional factor since to find the limit of the average as n gets large. We get
$$\lim_{n\to\infty}\frac{1}{n}\sum_{i=1}^{n}\frac{1}{2}g{\left(\frac{i}{n}\sqrt{\frac{2h}{g}}\right)}^2=\lim_{n\to\infty}\frac{1}{n}\sum_{i=1}^{n}\frac{i^2}{n^2}{h}.$$
Ultimately we solve
$$h\lim_{n\to\infty}\sum_{i=1}^{n}\frac{i^2}{n^2}\frac{1}{n},$$
which can be converted into
$$h\int_{0}^{1}x^2\,\mathrm dx=\frac{1}{3}h.$$
(This is the distance from the roof down. The real height is $\dfrac{2}{3}h$.)
Am I correct? If not, how do I find the average height?
| $\def\peq{\mathrel{\phantom{=}}{}}$Assume instead that there are $n$ drops per $T$ seconds so that $n$ becomes dimensionless, thus the time interval between two drops is $Δt = \dfrac{T}{n}$.
It takes $t_1 = \sqrt{\dfrac{2h}{g}}$ for the first drop to hit the ground, and when it hits, there are $N = \left[ \dfrac{t_1}{Δt} \right] + 1$ drops in the air (including the first drop) and the $k$-th drop has traveled $t_k = t_1 - (k - 1)Δt$, which implies its current height is $h - \dfrac{1}{2} g t_k^2$. Therefore, the average height is\begin{align*}
&\peq \frac{1}{N} \sum_{k = 1}^N \left(h - \frac{1}{2} g t_k^2 \right) = h - \frac{g}{2N} \sum_{k = 1}^N (t_1 - (k - 1)Δt)^2\\
&= h - \frac{g}{2N} \left( N t_1^2 - 2t_1 Δt \sum_{k = 1}^N (k - 1) + (Δt)^2 \sum_{k = 1}^N (k - 1)^2 \right)\\
&= h - \frac{g}{2N} \left( N t_1^2 - N(N - 1) t_1 Δt + \frac{1}{6} (2N - 1)N(N - 1) (Δt)^2 \right)\\
&= h - \frac{1}{2} g t_1^2 + \frac{1}{2} (N - 1) g t_1 Δt - \frac{1}{12} (2N - 1)(N - 1) g(Δt)^2\\
&= \frac{1}{2} (N - 1) g t_1 Δt - \frac{1}{12} (2N - 1)(N - 1) g(Δt)^2.
\end{align*}
Note that $N Δt → t_1$ as $n → ∞$, thus\begin{align*}
&\peq \lim_{n → ∞} \left( \frac{1}{2} (N - 1) g t_1 Δt - \frac{1}{12} (2N - 1)(N - 1) g(Δt)^2 \right)\\
&= \frac{1}{2} g t_1^2 - \frac{1}{6} g t_1^2 = \frac{2}{3} h.
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2960018",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Find maximum value Given $0 \leq a,b,c \leq \dfrac{3}{2}$ satisfying $a+b+c=3$. Find the maximum value of $$N=a^3+b^3+c^3+4abc.$$
I think the equality does not occur when $a=b=c=1$ as usual. I get stuck in finding the strategy, especially I don't know what to do with $4abc$. Thank you a lot.
| For $a=b=\frac{3}{4}$ and $c=\frac{3}{2}$ we get a value $\frac{243}{32}.$
We'll prove that it's a maximal value.
Indeed, $a+b-c=3-2c\geq0,$ which says that there are non-negatives $x$, $y$ and $z$ for which
$a=y+z$, $b=x+z$ and $c=x+y$.
Id est, we need to prove that
$$\sum_{cyc}(x+y)^3+4\prod_{cyc}(x+y)\leq\frac{9(a+b+c)^3}{32}$$ or
$$\sum_{cyc}(x+y)^3+4\prod_{cyc}(x+y)\leq\frac{9(x+y+z)^3}{4}$$ or
$$\sum_{cyc}\left(x^3-x^2y-x^2z+\frac{22}{3}xyz\right)\geq0,$$ which is true by Schur.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2960306",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Evaluate $\int_{0}^{1} \log^2\left(\frac{\Gamma(x)}{\Gamma(1-x)}\right)\log^2\left(\frac{\Gamma(x)}{\Gamma(1+x)}\right) dx$
Evaluate :
$$\int_{0}^{1} \log^2\left(\frac{\Gamma(x)}{\Gamma(1-x)}\right)\log^2\left(\frac{\Gamma(x)}{\Gamma(1+x)}\right) dx$$
This is my attempt in below ... but what I want is simplify more to answer....thanks.
My own attempt:
| I really do not see how you could simplify the monster. I really wonder if a closed form expression could exist. I would try numerical integration.
We can get an approximate solution using the following series expansions at $x=0$.
$$\log (\Gamma (x))=-\log (x)-\gamma x+\frac{\pi ^2 x^2}{12}+\frac{x^3 \psi ^{(2)}(1)}{6}+\frac{\pi ^4
x^4}{360}+O\left(x^5\right)$$
$$\log (\Gamma (1-x))=\gamma x+\frac{\pi ^2 x^2}{12}-\frac{x^3 \psi ^{(2)}(1)}{6}+\frac{\pi ^4
x^4}{360}+O\left(x^5\right)$$
$$\log (\Gamma (1+x))=-\gamma x+\frac{\pi ^2 x^2}{12}+\frac{x^3 \psi ^{(2)}(1)}{6}+\frac{\pi ^4
x^4}{360}+O\left(x^5\right)$$ which would make the integrand to be
$$\log ^4(x)+4 \gamma x \log ^3(x)+4 \gamma ^2 x^2 \log ^2(x)-\frac{2}{3} x^3
\left(\psi ^{(2)}(1) \log ^3(x)\right)-\frac{4}{3} x^4 \left(\gamma \psi
^{(2)}(1) \log ^2(x)\right)+O\left(x^5\right)$$ and the antiderivative
$$x \left(\log ^4(x)-4 \log ^3(x)+12 \log ^2(x)-24 \log (x)+24\right)+\frac{1}{2}
\gamma x^2 \left(4 \log ^3(x)-6 \log ^2(x)+6 \log (x)-3\right)+\frac{4}{27}
\gamma ^2 x^3 \left(9 \log ^2(x)-6 \log (x)+2\right)+\frac{1}{192} x^4 \psi
^{(2)}(1) \left(-32 \log ^3(x)+24 \log ^2(x)-12 \log (x)+3\right)-\frac{4}{375}
x^5 \left(\gamma \psi ^{(2)}(1) \left(25 \log ^2(x)-10 \log
(x)+2\right)\right)+O\left(x^6\right)$$ Then, for the definite integral
$$\gamma \left(\frac{16 \zeta (3)}{375}-\frac{3}{2}\right)-\frac{\zeta
(3)}{32}+24+\frac{8 \gamma ^2}{27}\approx 23.2249$$ while the numerical integration leads to $23.2372$ that is to say in error by $0.05$%.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2961778",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
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.