Q
stringlengths 70
13.7k
| A
stringlengths 28
13.2k
| meta
dict |
|---|---|---|
Taylor series of $\frac{x-2}{\sqrt{x^2-4x+8}}$ at $x = 2$. Where is the mistake? $$\frac{x-2}{\sqrt{x^2-4x+8}} \quad \text{at}\ x_0=2$$
Let $t = x-2$. then as $x\to 2$, $t\to 0$ and we can find the Maclaurin Series in terms of $t$.
\begin{align*}
\frac{x-2}{\sqrt{x^2-4x+8}} &=\frac{x-2}{\sqrt{(x-2)^2+4}}\\
&=\frac{t}{\sqrt{t^2+4}}\\
&=\frac{1}{2}\frac{t}{\sqrt{1+\left(\frac{t}{2}\right)^2}}
\end{align*}
Since the Maclaurin Series
$$
\left(1+\left(\frac{t}{2}\right)^2\right)^{-1/2} = \sum_{k=0}^n{-1/2 \choose k}\left(\frac{t}{2}\right)^{2k} + o(t^{2n})$$
Then
$$\frac{t}{2}\left(1+\left(\frac{t}{2}\right)^2\right)^{-1/2} = \sum_{k=0}^n{-1/2 \choose k}\left(\frac{t}{2}\right)^{2k+1} + o(t^{2n})$$
Here we have
\begin{align*}
{-1/2 \choose k} &= \frac{-\frac{1}{2}\left(-\frac{1}{2}-1\right)\left(-\frac{1}{2}-2\right)...\left(-\frac{1}{2}-(k-1)\right)}{k!}\\
&=\frac{-\frac{1}{2}\left(-\frac{3}{2}\right)\left(-\frac{5}{2}\right)...\left(-\frac{2k-1}{2}\right)}{k!}
\end{align*}
If we denote $1\cdot 3\cdot 5\cdot ... \cdot (2k-1) = (2k-1)!!$ we get
$${-1/2 \choose k} = \frac{(-1)^k(2k-1)!!}{2^k k!}$$
And finally
\begin{align*}\frac{t}{2}\left(1+\left(\frac{t}{2}\right)^2\right)^{-1/2} &= \sum_{k=0}^n{-1/2 \choose k}\left(\frac{t}{2}\right)^{2k+1} + o(t^{2n})\\
&= \frac{t}{2} + \sum_{k=1}^n\frac{(-1)^k(2k-1)!!}{2^kk!}\left(\frac{t}{2}\right)^{2k+1} + o(t^{2n})
\end{align*}
And in terms of $x$:
$$
\frac{x-2}{\sqrt{x^2-4x+8}}\Bigg\rvert_{x_0=2} = \frac{x-2}{2} + \sum_{k=1}^n\frac{(-1)^k(2k-1)!!}{2^kk!}\left(\frac{x-2}{2}\right)^{2k+1} + o((x-2)^{2n})$$
But the textbook's answer is
$$
\frac{x-2}{2} + \sum_{k=1}^{n-1}\frac{(-1)^k(2k-1)!!}{2^{2k+1}(2k)!!}(x-2)^{2k+1} + o((x-2)^{2n})$$
I don't get why the sum goes just to $n-1$ instead of $n$ and instead of $2^{3k+1}k!$ in the denominator, as I got, it is $2^{2k+1}(2k)!!$
What am I missing?
| There's no error in what you did, you only "wasted" two orders of magnitude in your remainder term.
The terms of the textbook's answer look different from yours, but are in fact the same, since
$$(2k)!! = 2^k\cdot k!$$
for one definition of the double factorial of even natural numbers (there's another definition that differs from the used one by a factor of $\sqrt{\frac{2}{\pi}}$), thus
$$2^{3k+1}\cdot k! = 2^{2k+1}(2k)!!$$
and consequently we see that the terms are identical.
You wasted one order of magnitude in the remainder term by writing
$$\biggl(1 + \Bigl(\frac{t}{2}\Bigr)^2\biggr)^{-1/2} = \sum_{k = 0}^n \binom{-\frac{1}{2}}{k} \biggl(\frac{t}{2}\biggr)^{2k} + o(t^{2n})\tag{$\ast$}$$
and interpreting the polynomial as the Taylor polynomial of order $2n$. It is also the Taylor polynomial of order $2n+1$, and therefore one can write the remainder term as $o(t^{2n+1})$. The next order of magnitude you wasted in the remainder term was when you multiplied $(\ast)$ with $\frac{t}{2}$, since $\frac{t}{2}o(t^m) = o(t^{m+1})$, and you didn't increase the exponent.
So with these orders taken into account, we find
$$\frac{x-2}{\sqrt{x^2-4x+8}} = \frac{x-2}{2} + \sum_{k = 1}^m \frac{(-1)^k (2k-1)!!}{2^{2k+1}(2k)!!}(x-2)^{2k+1} + o\bigl((x-2)^{2m+2}\bigr)$$
for any fixed $m\in \mathbb{N}$. So if we want to have a remainder term in $o(t^{2n})$, it suffices to take $m = n-1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2021155",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Can Every Irrational, but Non-Transcendental be Expressed as a Repeating Continued Fraction? So, as far as I understand any rational number can be expressed as a continued fraction of finite length
for example:
$$\frac {7}{17} = \cfrac {1}{2 + \cfrac{1}{2 + \cfrac{1}{3}}}$$
which for convenience I will express as $[0,2,2,3]$
Now I also understand that some irrationals can be written as infinite continued fractions
$$\sqrt{2} =1 + \cfrac{1}{2 +\cfrac{1}{2 + \cfrac{1}{\cdots}}}$$
expressible as $[1,\overline{2}]$
Now my question, can all irrational numbers that can be expressed as the root of a polynomial (algebraic irrational) be expressed as a repeating continued fraction? How would we go about proving this if it is true?
| If a simple continued fraction repeats but does not terminate then it converges to a second-degree algebraic number. ("Simple" means all of the numerators are $1$.)
First, look at the case where the repeating part begins immediately:
$$
x = a_1 + \cfrac 1 {a_2 + \cfrac 1 {a_3 + \cfrac 1 {\ddots \cfrac{}{a_n + \cfrac 1 {a_1 + _{\large\ddots}}}}}}
$$
Then you have
$$
x = a_1 + \cfrac 1 {a_2 + \cfrac 1 {a_3 + \cfrac 1 {\ddots \cfrac{}{a_n + \cfrac 1 x}}}}
$$
This reduces to a quadratic equation in $x$ with integer coefficients.
To see that reduction to a quadratic equation, start with
$$
a_{n-1} + \cfrac 1 {a_n + \cfrac 1 x} = a_{n-1} + \frac{x}{a_n x + 1} = \frac{ (a_{n-1} a_n + 1) x + a_n }{a_{n-1} x + 1}
$$
So you have
$$
a_{n-2} + \cfrac 1 {a_{n-1} + \cfrac{x}{a_n x + 1}} = a_{n-2} + \frac{a_n x + 1}{a_{n-1}(a_n x + 1) + x } = \text{etc.}
$$
Keep going until you have
$$
x = \frac{\text{a first-degree polynomial in } x}{\text{another first-degree polyonomial in } x},
$$
then multiply both sides by the denominator to get a quadratic equation.
If the repeating part doesn't start at the beginning, you can work on the repeating part, getting it to be the solution of a quadratic equation, then repeatedly rationalize denominators until you're done.
Here's a concrete example:
$$
x = 3 + \cfrac 1 {2 + \cfrac 1 {5 + \cfrac 1 {2 + \cfrac 1 {5 + \cdots}}}}
$$
with $2,5$ repeating.
One can write
$$
y = 3 + \cfrac 1 x = 3 + \cfrac 1 {2 + \cfrac 1 {5 + \cfrac 1 x}}
$$
So
$$
x = 2 + \cfrac 1 {5 + \cfrac 1 x} = 2 + \frac x {5x+1} = \frac{11x+2}{5x+1}
$$
$$
x = \frac{11x+2}{5x+1}
$$
$$
5x^2 + x = 11x+2
$$
$$
5x^2 - 10 x - 2 = 0.
$$
$$
x = \frac{10 \pm \sqrt{140}}{10} = \frac{5 \pm \sqrt{35}} 5.
$$
Since the value we seek is positive, we have
$$
x= \frac{5 + \sqrt{35}} 5.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2022951",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
How can we show $\cos^6x+\sin^6x=1-3\sin^2x \cos^2x$? How can we simplify $\cos^6x+\sin^6x$ to $1−3\sin^2x\cos^2x$?
One reasonable approach seems to be using $\left(\cos^2x+\sin^2x\right)^3=1$, since it contains the terms $\cos^6x$ and $\sin^6x$. Another possibility would be replacing all occurrences of $\sin^2x$ by $1-\cos^2x$ on both sides and then comparing the results.
Are there other solutions, simpler approaches?
I found some other questions about the same expression, but they simplify this to another form: Finding $\sin^6 x+\cos^6 x$, what am I doing wrong here?, Alternative proof of $\cos^6{\theta}+\sin^6{\theta}=\frac{1}{8}(5+3\cos{4\theta})$ and Simplify $2(\sin^6x + \cos^6x) - 3(\sin^4 x + \cos^4 x) + 1$. Perhaps also Prove that $\sin^{6}{\frac{\theta}{2}}+\cos^{6}{\frac{\theta}{2}}=\frac{1}{4}(1+3\cos^2\theta)$ can be considered similar. The expression $\cos^6x+\sin^6x$ also appears in this integral Find $ \int \frac {\tan 2x} {\sqrt {\cos^6x +\sin^6x}} dx $ but again it is transformed to a different from then required here.
Note: The main reason for posting this is that this question was deleted, but I still think that the answers there might be useful for people learning trigonometry. Hopefully this new question would not be closed for lack of context. And the answers from the deleted question could be moved here.
| $$(\cos^2 x+\sin^2x)^3=1^3=\cos^6x+3\cos^4x\sin^2x+3\cos^2x\sin^4x+\sin^6x$$
$$\cos^6x+\sin^6x=1-3\cos^2x\sin^2x(\cos^2x+\sin^2x)$$
so
$$\cos^6x+\sin^6x=1-3\cos^2x\sin^2x$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2023931",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 11,
"answer_id": 7
} |
Show that $1^3 + 2^3 + ... + n^3 = (n(n+1)/2)^2$ by induction I have some with proving by induction. I cannot find a solution for the inductive step: $1^3 + 2^3 + ... + n^3 = (n(n+1)/2)^2$
I already did the induction steps:
Basis: P(1) = $1^3 = (1(1+1)/2)^2$ (This is true)
Inductive step: Assume $P(k) = ((k)(k+1)/2)^2$
To be proven: $((k)(k+1)/2)^2 + (k+1)^3 = ((k+1)(k+2)/2)^2$
My problem is that I do not know how I can put the $ + (k+1)^3$ inside $((k)(k+1)/2)^2$.
Simplifying the left and right part of the statement does not help:
Simplifying the left side: $((k)(k+1)/2)^2 + (k+1)^3 = ((k^2+k)/2)^2 + (k+1)^3 $
Simplifying the right side: $((k+1)(k+2)/2)^2 = ((k^2+3k+2)/2)^2$
So i am left with: $((k^2+k)/2)^2 + (k+1)^3 = ((k^2+3k+2)/2)^2$
That is the same as: $1/4 (k^2+k)^2 + (k+1)^3 = 1/4((k^2+3k+2))^2$
Going further with the left side:$1/4 (k^2+k)^2 + (k+1)^3 = (1/4)(k^4 + 2k^3 + k^2) + k^3+3 k^2+3 k+1$
Going further with the right side: $1/4((k^2+3k+2))^2 = 1/4 (k^4+6 k^3+13 k^2+12 k+4)$
Now I am stuck with: $(1/4)(k^4 + 2k^3 + k^2) + k^3+3 k^2+3 k+1 = 1/4 (k^4+6 k^3+13 k^2+12 k+4)$
Now I am kind of left with garbage. Am I missing something? What do I do wrong? Where can I find a good resources to learn how to solve this issue?
| Take your "to be proven" equation and divide through by $(k+1)^2$ and multiply through by $4$. This gives $k^2+4(k+1)=(k+2)^2$ as the equation to be proven, and this is trivial.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2024028",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 2
} |
System of four equations of four variables including second powers. I've been tasked with solving the following system of equations and it seems like I am stuck:
$$a-x^2=y$$$$a-y^2=z$$$$a-z^2=t$$$$a-t^2=x,$$where $a$ is a real number, for which $0\leq a\leq 1$. I thought the best way would be to subtract some equations from each other and the exploit $x^2-y^2=(x+y)(x-y)$. Even some estimates could be useful, since we have an estimate for $a$. However, I put this system of equations to WolframAlfa and the solutions (depending on $a$) looked very uneasy. In general, I have got very little experience in solving quadratic systems of equations like this one, could somebody please point me in the right direction? Thanks a lot!
| from the first equation we get
$$a-(a-x^2)^2=z$$ then we obtain
$$a-(a-(a-x^2)^2)^2=t$$
$$a-(a^2+(a-x^2)^4-2a(a-x^2)^2)=t$$ and finally we obtain
$$a-(a^2+(a-x^2)^4-2a(a-x^2)^2)^2=x$$
and now you can try to get $x$
and this is what i'm get:
$$a^8+a^7 \left(-8 x^2-4\right)+a^6 \left(28 x^4+24 x^2+6\right)+a^5 \left(-56 x^6-60 x^4-24
x^2-6\right)+a^4 \left(70 x^8+80 x^6+36 x^4+16 x^2+5\right)+a^3 \left(-56 x^{10}-60
x^8-24 x^6-16 x^4-8 x^2-2\right)+a^2 \left(28 x^{12}+24 x^{10}+6 x^8+8 x^6+4
x^4+1\right)+a \left(-8 x^{14}-4 x^{12}-2 x^8-1\right)=x \left(-x^{15}-1\right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2027492",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
$\forall n\in\mathbb N$ prove that at least one of the number $3^{3n}-2^{3n}$,$3^{3n}+2^{3n}$ is divisible by $35$. $3^{3n}-2^{3n}=27^n-8^n=(27-8)(27^{n-1}+27^{n-2}\cdot 8+...+27^1\cdot8^{n-2}+8^{n-1})$
If $n$ is even,
$3^{3n}+2^{3n}=27^n+8^n=(27+8)(27^{n-1}-27^{n-2}\cdot 8+...-27^1\cdot8^{n-2}+8^{n-1})$
If $n$ is even and a power of $2$, $3^{3n}+2^{3n}$ can't be factorized.
If $n$ is even, $n=m\cdot 2^k,m>1,k>0$ and $m$ is odd $\Rightarrow$
$$27^n+8^n=(27^{2^k}+8^{2^k})\sum_{i=1}^m 27^{(m-i)2^k}(-b^{2^k})^{i-1}$$
How to check divisibility using these cases?
Reference
| if n is odd
$3^{3n} + 2^{3n} = (27+8)(27^{n-1} - 27^{n-2}8 +\cdots + 8^{n-1})$
and $35|(3^{3n} + 2^{3n})$
if $n$ is even, $n = 2k:$
$(3^{3n}-2^{3n}) = (3^{n}-2^{n})(3^{2n} + 3^n2^n + 2^{2n})$
$(3^{n}-2^{n}) = (3^{2k} - 2^{2k}) = (3^2 - 2^2)(3^{2k-1} +\cdots + 2^{2k-1})\\
5|(3^{2k}-2^{2k})$
What is left?
showing that $7|(3^{6k} - 2^{6k})$
Lets use Fermat's little theorem on this one
when $p$ is prime and $p$ does not divide $a,$
$a^{p-1} \equiv 1\pmod p$
$(3^{6k} - 2^{6k}) \equiv 1-1 \equiv 0\pmod 7$
We probably should have brought in the modular arithmetic earlier, but you had already started by factoring.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2027993",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Find possible number of triangles with integer sides for a given inradius inradius = $5$
possible triangle sides
$(25, 20, 15)$
$(37, 35, 12)
(39, 28, 17)$
...
Find formula to find other possible sides of triangles.
| Let $a$, $b$, $c$ be the integer lengths of the triangle sides, and $r$ the inradius, which I suppose to be integer too. It is well known that $r(a+b+c)$ is twice the area of triangle, so by squaring Heron's formula we have:
$$
r^2(a+b+c)={1\over4}(a+b-c)(a-b+c)(-a+b+c).
$$
The three factors in parentheses on the right hand side of that equation must be all even or all odd, because the sum of any two of them is twice the length of a side. But their product must be divisible by $4$, so they are all even numbers and we can set:
$$
2x=(a+b-c),\quad 2y=(a-b+c),\quad 2z=(-a+b+c).
$$
After substitution, the equation becomes
$$
r^2(x+y+z)=xyz,
$$
where $x$, $y$, $z$ are positive integers and $a=x+y$, $b=x+z$, $c=y+z$. We can solve for $z$ to get:
$$
z={r^2(x+y)\over xy-r^2}.
$$
All solutions can be obtained from that formula, by choosing $x$ and $y$ such that $xy>r^2$ and $z$ is integer. If we suppose, without loss of generality, that $z\ge x$ and $z\ge y$, then $2z\ge x+y$, which entails $xy\le 3r^2$. So we must check only a limited number of cases.
A complete search in the case $r=5$ can be readily carried out and the only solutions for $(a,b,c)$ are the following:
(11, 60, 61), (12, 35, 37), (12, 153, 159), (13, 68, 75), (15, 20, 25),
(15, 377, 388), (17, 28, 39), (17, 87, 100), (27, 29, 52), (27, 676, 701),
(28, 351, 377), (31, 156, 185), (36, 91, 125), (39, 76, 113), (51, 52, 101).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2028918",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Help with $\int \cos^6{(x)} \,dx$ Problem:
\begin{eqnarray*}
\int \cos^6{(x)} dx \\
\end{eqnarray*}
Answer:
\begin{eqnarray*}
\int \cos^4{(x)} \,\, dx &=& \int { \cos^2{(x)}(\cos^2{(x)}) } \,\, dx \\
\int \cos^4{(x)} \,\, dx &=& \int { \frac{(1+\cos(2x))^2}{4} } \,\, dx \\
\int \cos^4{(x)} \,\, dx &=&
\int { \frac{\cos^2(2x)^2 + 2\cos(2x)+1}{4} } \,\, dx \\
\int \cos^4{(x)} \,\, dx &=&
\int { \frac{(\frac{1+\cos(4x)}{2} + 2\cos(2x)+1}{4} } \,\, dx \\
\int \cos^4{(x)} \,\, dx &=&
\int { \frac{1+\cos(4x) + 4\cos(2x)+2}{8} } \,\, dx \\
\int \cos^4{(x)} \,\, dx &=&
\int { \frac{\cos(4x) + 4\cos(2x)+3}{8} } \,\, dx \\
\int \cos^4{(x)} \,\, dx &=& \frac{\sin(4x)+ 8 \sin(2x)+12x}{32} \\
\text{Let }I_6 &=& \int \cos^6{(x)} \,\, dx \\
\end{eqnarray*}
To perform this integration, I use integration by parts with
$u = \cos^5(x)$ and $dv = \cos(x) dx$.
\begin{eqnarray*}
I_6 &=& \sin(x)\cos^5(x) - \int \sin(x) 5\cos^4(x)(-\sin(x)) \,\, dx \\
I_6 &=& \sin(x)\cos^5(x) + \int 5\cos^4(x)(\sin(x))^2 \,\, dx \\
I_6 &=& \sin(x)\cos^5(x) + \int 5\cos^4(x)(1 - \cos(x))^2 \,\, dx \\
I_6 &=& \sin(x)\cos^5(x) + \int 5\cos^4(x) \,\, dx - 5I_6 \\
6I_6 &=& \sin(x)\cos^5(x) + \int 5\cos^4(x) \,\, dx \\
6I_6 &=& \sin(x)\cos^5(x) + \frac{5\sin(4x)+ 40 \sin(2x)+60x}{32} + C_1 \\
6I_6 &=& \frac{32\sin(x)\cos^5(x) + 5\sin(4x)+ 40 \sin(2x)+60x}{32} + C_1 \\
I_6 &=& \frac{32\sin(x)\cos^5(x) + 5\sin(4x)+ 40 \sin(2x)+60x}{192} + C \\
\end{eqnarray*}
I believe that the above result is wrong. Using an online integral
calculator, I get:
\begin{eqnarray*}
I_6 &=& \frac{\sin(6x) + 9\sin(4x) + 45 \sin(2x) + 60x}{192} + C \\
\end{eqnarray*}
I am hoping that somebody can tell me where I went wrong.
Bob
| You could also employ the binomial theorem
\begin{align}
\cos^6x&=\frac1{64}(e^{ix}+e^{-ix})^6\\
&=\frac1{64}(e^{i6x}+6e^{i4x}+15e^{i2x}+20+15e^{-i2x}+6e^{-i4x}+e^{-i6x})
\\
&=\frac1{32}(\cos(6x)+6\cos(4x)+15\cos(2x)+10).
\end{align}
Which now is easy to integrate.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2032178",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
summation of a infinite series i want to know if this infinite sum $\sum_{n=1}^{\infty}$ $\frac{a(a+1)(a+2)......(a+n-1)}{b(b+1)(b+2)......(b+n-1)}$ converges or diverges ? where a>0 and b>a+1.
if the sum converges what is the sum i.e where it will converge? i need a concrete explanation
i found some inequalities $\frac{a(a+1)}{b(b+1)}$ <$\frac{a}{b+1}$ and $\frac{a(a+1)(a+2)}{b(b+1)(b+2)}$< $\frac{a}{b+2}$ .......and continuing $\frac{a(a+1)....(a+n-1)}{b(b+1)....(b+n-1)}$ < $\frac{a}{b+n-1}$
this inequalities can be useful for this problem
| Let $u_n=\dfrac{a(a+1)\ldots(a+n-1)}{b(b+1)\ldots(b+n-1)}$
We have $\dfrac{u_{n+1}}{u_n}=1-\dfrac{b-a}{n}+o(\dfrac{1}{n})$
Hence by Raabe Duhamel :
If $b-a>1$ the infinite sum converge
If $b-a<1$ the infinite sum diverge.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2034148",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
How to use quaternions express number How to use quaternions express number like 49 as sums of four squares
based on their factorizations?
I just start to learn quaternion in my math class,
I tired to find my class note :
Quaternions look like a + b i + c j + d k .
Hmmmm ... i guess a,b,c,d are integers.Real numbers commutate i,j,k?
and..sum of four squares will look like this?
$8 = 4+4+0+0 $
$9 = 4+4+1+0 $
$10 = 4+4+1+1$
$49 =？ $
| Quaternions $q=a+b\mathbf{i}+c\mathbf{j}+d\mathbf{k}$ have a "norm" defined by
$$ |a+b\mathbf{i}+c\mathbf{j}+d\mathbf{k}|^2 = a^2+b^2+c^2+d^2. $$
This is a generalization of $|a+bi|=a^2+b^2$ defined for complex numbers. Indeed, it is still multiplicative: we have $|xy|=|x||y|$ for quaternions $x$ and $y$.
How can we use this idea to write whole numbers as sums of four squares? Let's do a more illustrative example first: say $42$. To write $42=a^2+b^2+c^2+d^2$ we want to find a quaternion $a+b\mathbf{i}+c\mathbf{j}+d\mathbf{k}$ with square norm $42$. We can factor $42$ with smaller numbers in various ways, for example as $6\cdot 7$ or $3\cdot 14$. So to write $42$ as $|x|^2$, it is sufficient to find $x$s for which $|x|^2$ is $6$ and $7$, or $3$ and $14$. Indeed, we have
$$ 6=0^2+1^2+1^2+2^2 \\ 7=1^2+1^2+1^2+2^2 $$
(one may permute the summands any way one likes) and then
$$ \begin{array}{l} 6=|\mathbf{i}+\mathbf{j}+2\mathbf{k}|^2 \\ 7 =|1+\mathbf{i}+\mathbf{j}+2\mathbf{k}|^2 \end{array}$$
which implies
$$ \begin{array}{ll} 42 & =|(\mathbf{i}+\mathbf{j}+2\mathbf{k})(1+\mathbf{i}+\mathbf{j}+2\mathbf{k})|^2 \\ & =|-6+\mathbf{i}+\mathbf{j}+2\mathbf{k}|^2 \\ & =6^2+1^2+1^2+2^2. \end{array} $$
Alternatively,
$$ \begin{array}{r} 3 = 1^2+1^2+0^2+1^2 \\ 14 = 0^2+1^2+2^2+3^2 \end{array} $$
which tells us
$$ \begin{array}{l} \phantom{1}3 = |1+\mathbf{i}+\mathbf{k}|^2 \\ 14 = |\mathbf{i}+2\mathbf{j}+3\mathbf{k}|^2 \end{array} $$
which implies
$$ \begin{array}{ll} 42 & = |(1+\mathbf{i}+\mathbf{k})(\mathbf{i}+2\mathbf{j}+3\mathbf{k})|^2 \\ & = |-4-\mathbf{i}+5\mathbf{k}|^2 \\ & = 4^2+1^2+0^2+5^2. \end{array} $$
I'll let you play around with $49=7\cdot7$. We can write $7=1^2+1^2+2^2+1^2$, but there are different ways to permute the summands. See what they tell you.
Note that you will need to know how to multiply two quaternions to do this. This involves using the distributive property and the multiplication table for $\mathbf{i},\mathbf{j},\mathbf{k}$ (or a helpful mnemonic).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2034842",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Prove $4a^2+b^2+1\ge2ab+2a+b$ Prove $4a^2+b^2+1\ge2ab+2a+b$
$4a^2+b^2+1-2ab-2a-b\ge0$
$(2)^2(a)^2+(b)^2+1-2ab-2a-b\ge0$
Any help from here? I am not seeing how this can be factored
| High school solution:
$$
4a^2+b^2+1-2ab-2a-b=\frac{(2a-1)^2+(2a-b)^2+(b-1)^2}{2}\ge 0.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2035680",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 3
} |
Show that $(1+\frac{x}{n})^n \rightarrow e^x$ uniformly on any bounded interval of the real line.
Show that $(1+\frac{x}{n})^n \rightarrow e^x$ uniformly on any bounded interval of the real line.
I am trying to argue from the definition of uniform convergence for a sequence of real-valued functions, but am struggling quite a lot. My efforts so far have concentrated on trying to find a sequences,
${a_n}$ which tends to zero, such that
$$|(1+\frac{x}{n})^n -e^x |\leq a_n$$ for all $n$. But I have been unsuccessful thus far. All help is greatly appreciated.
| To find your sequence $(a_n)$ where $|(1 + x/n)^n - e^x| \leqslant a_n \to 0$ -- proving uniform convergence on any bounded interval -- use the inequality $\ln(1+y) \leqslant y$.
We have for $0 \leqslant y < 1$,
$$1+y \leqslant e^y = \sum_{k=0}^{\infty} \frac{y^k}{k!} \leqslant \sum_{k=0}^{\infty} y^k = \frac1{1-y},$$
Take $y = x/n$. It follows that for $n$ sufficiently large
$$1 + \frac{x}{n} \leqslant e^{x/n} \leqslant \left(1 - \frac{x}{n}\right)^{-1},$$
and
$$\left(1 + \frac{x}{n}\right)^n \leqslant e^x \leqslant \left(1 - \frac{x}{n}\right)^{-n}.$$
The second inequality implies that
$$e^{-x} \geqslant \left(1 - \frac{x}{n}\right)^{n}.$$
Using Bernoulli's inequality $(1 - x^2/n^2)^n \geqslant 1 - x^2/n.$
Hence,
$$0 \leqslant e^{x} - \left(1+ \frac{x}{n}\right)^n = e^{x}\left[1 - e^{-x}\left(1+ \frac{x}{n}\right)^{n}\right]\\ \leqslant e^{x}\left[1 - \left(1+ \frac{x}{n}\right)^{n}\left(1- \frac{x}{n}\right)^{n}\right]\\= e^{x}\left[1 - \left(1- \frac{x^2}{n^2}\right)^{n}\right]\leqslant e^{x}\frac{x^2}{n}.$$
Therefore, for all $x \in [0,K]$, we have as $n \to \infty$
$$0 \leqslant \left|e^{x} - \left(1+ \frac{x}{n}\right)^n\right| \leqslant e^K\frac{K^2}{n} \rightarrow 0.$$
An almost identical argument for $y \geqslant 0$ shows that
$$0 \leqslant e^{-y} - \left(1- \frac{y}{n}\right)^n \leqslant e^{-y}\frac{y^2}{n}.$$
Thus if $-y = x \in [-L,0]$ we have
$$0 \leqslant \left|e^{x} - \left(1+ \frac{x}{n}\right)^n\right| = \left|e^{-y} - \left(1- \frac{y}{n}\right)^n\right| \leqslant \frac{L^2}{n} \rightarrow 0.$$
proving uniform convergence on any bounded interval.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2036694",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 3,
"answer_id": 0
} |
how to find mod 100 in the following case what is the remainder when $3^{999}$ divided by 100?
Is there is any short method to find remainder ?
i tried
$3^{1}$ divided by 100=3
$3^{2}$ divided by 100=9
$3^{3}$ divided by 100=27
$3^{4}$ divided by 100=81
but unable to proceed further
| There are some good more-direct methods in the other answers, but let me just help you "proceed further".
Note that "$x \bmod 100$" effectively means the remainder of $x$ when divided by $100$, and $x \equiv y \bmod 100$ means that $x$ and $y$ have the same remainder when divided by $100$.
So you already have:
\begin{align}
3^1 &\equiv 3 \bmod 100\\
3^2 &\equiv 9 \bmod 100\\
3^3 &\equiv 27 \bmod 100\\
3^4 &\equiv 81 \bmod 100\\
\end{align}
but at this point we're not really finding remainders because the values are less than $100$. So the next step $3^5=243$ means we actually start doing that.
$243 \equiv 43 \bmod 100$. Now the step after that gets interesting, because we can ignore that $200$ we just threw away. Even when we multiply it by $3$ (or $9$, etc), it's still a multiple of $100$ and makes no difference to the remainder, so we can just forget it in subsequent steps:
\begin{align}
3^5 &\equiv 243 \equiv 43 \bmod 100\\
3^6 &\equiv 43\cdot 3 \equiv 129 \equiv 29 \bmod 100\\
3^7 &\equiv 29\cdot 3 \equiv 87 \bmod 100\\
3^8 &\equiv 87\cdot 3 \equiv 261 \equiv 61 \bmod 100\\
3^9 &\equiv 61\cdot 3 \equiv 183 \equiv 83 \bmod 100\\
3^{10} &\equiv 83\cdot 3 \equiv 249 \equiv 49 \bmod 100\\
\end{align}
And so on. What you will find on continuing this process is that $3^{20}\equiv 1 \bmod 100$, and this means that the remainder values cycle from there onwards, allowing look up your result from where the $999$ exponent falls in that cycle.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2039110",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
Solving difference equation by z transform I have the following difference equation:
$$\ \ y(k+2) - 2y(k+1) +2y(k) = x(k) \,$$
where x(k) is an input of the form $\ x(k) = cos(\pi k)\ $, we also have the initial value conditions
$$\ y(0) = 1\,$$
$$\ y(1) = 1\,$$
I got to the following equation by applying the properties of time shifting:
$$\ Z^2Y[Z] - Z^2y(0) - Zy(1) - 2ZY[Z] + 2Zy(0) + 2Y[Z] = X[Z]\,$$
My goal was to get to a transfer function in order to analyze how the system behaves, but it doesn't seem like it's possible since the initial conditions are not zero. Are there other approaches to this problem? (P.S. It needs to be solved by Z transform)
| For the difference equation:
$$y(k+2) - 2y(k+1) +2y(k) = x(k)$$
We have:
$$
\begin{align}
x(k)&= \cos(\pi k) & y(0)&=1 & y(1)&=1\
\end{align}
$$
Z-transforming both sides of the equation:
$$
\begin{align}
\mathcal{Z}\left[y(k+2) - 2y(k+1) +2y(k)\right]&=\mathcal{Z}\left[x(k)\right]\\
z^2Y(z) - z^2y(0) - zy(1) - 2zY(z) + 2zy(0) + 2Y(z) &= X(z)\\
Y(z)(z^2-2z+2)-z^2+z&=X(z)\\
\frac{Y(z)}{X(z)}(z^2-2z+2)-\frac{(z^2-z)}{X(z)}&=1\\
\frac{Y(z)}{X(z)}&=\frac{1}{(z^2-2z+2)}+\frac{(z^2-z)}{X(z)(z^2-2z+2)}
\end{align}
$$
Using a Z-transform table, we know that:
$$\mathcal{Z}[\cos(ak)]=\frac{z(z-\cos(a))}{z^2-2z\cos(a)+1}$$
Then:
$$X(z)=\mathcal{Z}[\cos(\pi k)]=\frac{(z^2+z)}{z^2+2z+1}$$
Substituting it back on the equation, we have:
$$
\begin{align}
\frac{Y(z)}{X(z)}&=\frac{1}{(z^2-2z+2)}+\frac{z(z-1)(z^2+2z+1)}{z(z+1)(z^2-2z+2)}\\
&=\frac{(z+1)(z^2-2z+2)+(z-1)(z^2+2z+1)(z^2-2z+2)}{(z+1)(z^2-2z+2)^2}\\
&=\frac{(z+1)+(z-1)(z^2+2z+1)}{(z+1)(z^2-2z+2)}\\
&=\frac{z+1+z^3+2z^2+z-z^2-2z-1}{(z+1)(z^2-2z+2)}\\
&=\frac{z^2(z+1)}{(z+1)(z^2-2z+2)}\\
&=\frac{z^2}{(z^2-2z+2)}
\end{align}
$$
If what I did is correct, this is an unstable system, once we have Zeros:$\{0,0\}$
Poles:$\{1\pm j\}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2039954",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Use generating functions to solve nonhomogeneous recurrence relation The recurrence relation is
$$a_n = a_{n-1} + a_{n-2} + n,\quad n\ge 2$$
with initial conditions $a_0 = 0$ and $a_1 = 1$.
I know I need to convert the recurrence into series and I have broken it down, but am struggling with getting it into a proper form to do partial fractions.
I have:
$$f(x) = a_0 + a_1x + x\sum_{n\ge 2} a_{n-1}x^{n-1} + x^2\sum_{n\ge 2} a_{n-2}x^{n-2} + \frac{1}{(1-x)^2}$$
Any insight/help is much appreciated!
| Just manipulate the generating function;
\begin{align}
f(x)=x+\sum_{n=2}^\infty a_nx^n=&x+x\sum_{n=2}^\infty a_{n-1}x^{n-1}+x^2\sum_{n=2}^\infty a_{n-2}x^{n-2}+x\sum_{n=2}^\infty nx^{n-1}\\
=&x+xf(x)+x^2f(x)+x\,\frac{d}{dx}\sum_{n=2}^\infty x^{n}\\
=&x+xf(x)+x^2f(x)+x\,\frac{d}{dx}\underbrace{\left(\frac{1}{1-x}-x-1\right)}_{x^2/(1-x)}\\
=&x+xf(x)+x^2f(x)+x\,\frac{2x-x^2}{(1-x)^2},
\end{align}
and now solve for $f(x)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2042797",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Find the sum of $\displaystyle\sum_{n=3}^\infty \frac{2^n-1}{3^n}$ My work:$$\sum_{n=3}^\infty \frac{2^n-1}{3^n}$$
$$a_1 = \frac{2^3-1}{3^3}$$
$$a_1 = \frac{7}{27}, r=\frac{2}{3}$$
$$S_N=\frac{\frac{7}{27}}{1-\frac{2}{3}} = \frac{7}{9}$$
The correct answer is $\frac{5}{6}$
| This is not a geometric series on its own, it's the difference of two of them
$$\sum_{n\ge 3} \left({2\over 3}\right)^n-\sum_{n\ge 3}\left({1\over 3}\right)^n = {8/27\over 1-2/3}-{1/27\over 1-1/3}={8\over 9}-{1\over 18}={5\over 6}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2045321",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
$\lim_{N\rightarrow\infty}\left< \psi{\left(N_k+1\right)} - \psi{\left(N_j+1\right)} \right >$ Question
What is
\begin{align}
\lim_{N\rightarrow\infty}\left< \psi{\left(N_k+1\right)} - \psi{\left(N_j+1\right)} \right >?
\end{align}
Here, $N_k$ and $N_j$ are multinomially distributed random variable with expected values $\left<N_k\right> = N\,p_k$ and $\left<N_j\right> = N\,p_j$, respectively; $0<p_k, p_j <1$; and $\psi$ is the digamma function.
Context
This question implicitly solves the questions in [1] and [2]. The solution is based on a related question [3]; and specifically makes use of of Srivatsan's second method therein. At one point below, a binomial identity in [4] is used.
Approach
To solve this problem, we generalize Srivatsan's second method that is derived for a related problem (see [3]).
\begin{align}
\left< \psi{\left(N_k+1\right)} \right >
&
= - \gamma + \sum\limits_{N_k=0}^{N}{\left[\binom{N}{N_k}\, p_k^{N_k}\,\left( 1 - p_k\right)^{N-N_k} \, \sum\limits_{t=1}^{N_k}{\dfrac{1}{t}}\right]}
\\
&
= - \gamma + \sum\limits_{N_1=0}^{N}{\left[\binom{N}{N_k}\, p_k^{N_k}\,\left( 1 - p_k\right)^{N-N_k} \, \sum\limits_{t=1}^{N_k}{ \int_{0}^{1}{x^{t-1}} }\right]}\,dx
\\
&
= - \gamma + \int_{0}^{1} \sum\limits_{N_1=0}^{N}{\left[\binom{N}{N_k} \, p_k^{N_k}\,\left( 1 - p_k\right)^{N-N_k} \, { \dfrac{ x^{N_1} -1 }{x - 1} }\right]}\,dx
\\
&
= - \gamma + \int_{0}^{1} { { \dfrac{\sum\limits_{N_1=0}^{N}\binom{N}{N_k} \, p_k^{N_k}\,\left( 1 - p_k\right)^{N-N_k} \, x^{N_1} - \sum\limits_{N_1=0}^{N}\binom{N}{N_k} \, p_k^{N_k}\,\left( 1 - p_k\right)^{N-N_k} \, 1 }{x - 1} } }\,dx
\\
&
= - \gamma + \int_{0}^{1} { { \dfrac{ \left(1 - p_k + x\,p_k\right)^N - 1 }{x - 1} } }\,dx
\end{align}
In the step above, we use an identity found in [4]. Furhter, $\gamma$ is the Euler-Mascheroni constant.
Next, make the substitution $y = 1 - p_k + x\,p_k$ so $\dfrac{dy}{p_k} = dx$
\begin{align}
\left< \psi{\left(N_k+1\right)} \right >
&
= - \gamma + \int_{1-p_k}^{1} { { \dfrac{ y^N - 1 }{ \dfrac{y - 1 + p_k}{p_k} - 1} } }\, \dfrac{dy}{p_k}
\\
&
= - \gamma + \int_{1-p_k}^{1} { { \dfrac{ y^N - 1 }{ y - 1 + p_k - p_k} } }\, dy
\\
&
= - \gamma + \int_{1-p_k}^{1} { { \dfrac{ y^N - 1 }{y - 1} } }\, dy
\\
&
= - \gamma + \int_{1-p_k}^{1} { \sum\limits_{k=0}^{N-1}{y^k} }\, dy
\\
&
= - \gamma + \left. \sum\limits_{k=1}^{N}{\dfrac{y^k}{k}} \right|_{1-p_k}^{1}
\\
&
= - \gamma + H_N - \sum\limits_{k=1}^{N}{\dfrac{\left(1 - p_k\right)^k}{k}}
\end{align}
where $H_{N}$ is the $N^{\textrm{th}}$ harmonic number.
If one notes that
\begin{equation}
- \ln (1 - z) = z + \frac{z^2}{2} + \frac{z^3}{3}\cdots = \sum\limits_{k=1}^{\infty}{\dfrac{ z ^k}{k}}
\end{equation}
is valid for $\left|z\right| \leq 1$, then the summation $\sum\limits_{k=1}^{N}{\dfrac{\left(1 - p_k\right)^k}{k}}$ is understood to be the $n$th partial sum of the Taylor series expansion of $ -\ln{\left(1-y\right)}$ evaluated at $y= 1- p_k$. Therfore, with respect to the question at hand, we write
\begin{align}
\lim_{N\rightarrow \infty }\left< \psi{\left(N_k+1\right)} - \psi{\left(N_j+1\right)} \right >
&
= \left[- \gamma + \lim_{N\rightarrow \infty }\left\{H_N\right\} - \left(- \log{\left(1- \left[1 - p_k\right]\right)}\right) \right] -\left[- \gamma + \lim_{N\rightarrow \infty }\left\{H_N\right\} - \left(- \log{\left(1- \left[1 - p_j\right]\right)}\right) \right]
\\
&
= \log{\left( p_k \right)} - \log{\left( p_j \right)}
\end{align}
Solution
Finally, we find
\begin{align}
\lim_{N\rightarrow \infty }\left< \psi{\left(N_k+1\right)} - \psi{\left(N_j+1\right)} \right >
= \log{\left( \dfrac{p_k}{p_j} \right)}.
\end{align}
Citations
[1] Bounding $\sum\limits_{k=1}^n{ \binom{n}{k} \, \frac{\left(-1\right)^k}{k} \left[ p_1^k - p_2^k \right] } $
[2] Identity and bounding of ${ \sum\limits_{k=1}^N{ \binom{N}{k}\,\dfrac{p^k \, \left( - 1 \right)^{k} }{k} } } $ when $0<p<1$?
[3] Proofs of $\lim\limits_{n \to \infty} \left(H_n - 2^{-n} \sum\limits_{k=1}^n \binom{n}{k} H_k\right) = \log 2$
[4] Covariance of product of two functions of two binomial distributions
| My answer is correct. Problem solved.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2047100",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
How many solutions does $x^2 + 3x +1 \equiv 0\, \pmod{101}$ have? $x^2 + 3x +1 \equiv 0 \pmod{101}$. To solve this I found the determinant $D = 5 \pmod{101}$). Using the Legendre symbol,
$$\left(\frac{5}{101}\right) = \left(\frac{101}{5}\right) \equiv \left(\frac{1}{5}\right) \equiv 1,$$
$\therefore$ The equations have a solution.
My question is how I can find out how many solutions it has?
| An idea to tackle this and similar questions using as small numbers and multiples of $\;101\;$ as possible (when possible, of course...).
Observe that $\;5=-96\pmod {101}\;$ , and $\;96=2^5\cdot3\;$ , so we can try to deal with these apparently easier numbers. Since we have $\;96=16\cdot6\;$, we have:
$$4^2=16=2^4\;,\;\;101\cdot2=202-6=196=14^2$$
and thus we have:
$$\;\sqrt{16}=4\;,\;\;\sqrt{-6}=14\implies \sqrt{16\cdot(-6)=-96}=4\cdot14=56$$
and thus also $\;\sqrt{-96}=-56=45\;$ , so $\;x^2=5\pmod{101}\implies x=\pm56=56,\,45\;$
Finally, your quadratic's solutions are ($\pmod{101}$ ):
$$x_{1,2}=\frac{-3\pm56}2=\begin{cases}\frac{98+56}2=77\\{}\\\frac{98-56}2=21\end{cases}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2047575",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
Solutions of $x^3 + (x+4)^2 = y^2$ I want to solve the above equation in integers. I'm pretty sure the only solutions are $(x,y) = (0, \pm 4)$ but I'm not sure how to prove it.
| When $x=0$, $y^2 = 4^2$ and hence $y=\pm 4$. Thus $(x,y) = (0,\pm 4)$ are solutions. We will show there are no other solutions.
Suppose that $x \neq 0$. We have \begin{align}
x^3 = y^2 - (x+4)^2 = (y+x+4)(y-x-4)
\end{align} Let $d$ be the greatest common divisor of $y+x+4$ and $y-x-4$. We claim that $d$ is a power of 2. If $d$ has an odd prime factor $p > 1$, then clearly, $p$ divides $x^3$ and hence $x$. Also since $p$ divides both $y+x+4$ and $y-x-4$, it divides their sum, $2y$ and since $p$ is odd, $p$ divides $y$. But then $p$ must divide 4, a contradiction.
Thus $d = 2^k$ for some $k \geq 0$.Suppose that $k=0$. That is, $y+x+4$ and $y-x-4$ are coprime. Then, there exists integers $m,n$ such that $y+x+4 = m^3$ and $y-x-4=n^3$. We have $x=mn$ and $2x+8 = m^3-n^3$. Clearly, $m,n$ are both odd and hence $|m-n| \geq 2$.
Case 1 $x=mn > 0$, $m > n$
$$2mn+8 = m^3-n^3 = (m-n)((m-n)^2+3mn) \geq 6mn$$ and hence $mn \leq 2$. Since $m,n$ are both odd integers, it follows that $mn \geq 3$. Thus no solutions in this case.
Case 2 $x = mn >0$, $m < n$
We have $$2mn+8 = (m-n)^3 + 3(m-n)mn \leq -8-6mn $$
This gives $mn \leq -2$, a contradiction to the assumption $mn > 0$.
Case 3 $x = mn <0$, $m > n$
We have $2mn + 8 = m^3 - n^3 > 0$ and hence $mn > -4$. Since $m,n$ are odd integers, the possibilities are $mn = -1, -3$. Thus $m = 1, n=-1$ or $m=3, n=-3$. Putting $m=1, n=-1$, we get $-1 + 27 = y^2$, and $y $ is not an integer. For $m=3, n=-1$, we have $-27+1 = y^2$ again an impossibility.
Case 4 $x = mn < 0$, $m < n$
Here, $m < 0$ and $n > 0$.
$$2mn + 8 = m^3 - n^3 < -m^2 - n^2$$ giving $m^2+n^2+2mn = (m+n)^2 < -8$, an impossibility.
Suppose that $d = 2^k$ and $k >0$. Since $2^{2k}$ divides $x^3$, it follows that $k$ must be a multiple of $3$. We need to discuss two cases: $k > 3$ and $k=3$. Suppose that $k > 3$. Then, $2^6 | x$. But $2^6 | (y+x+4) - (y-x-4) = 2x+8$. This implies $2^6$ divides 8, a contradiction.
Suppose that $k=3$. Then $y+x+4 = 8a, y-x-4 = 8b$ where $a,b$ are coprime.
We have $x^3 = 64ab$ and hence both $a,b$ must be cubes of integers. Let $a =m^3, b=n^3$. Note that both $m,n$ are odd integers. Also $x=4mn$ and $2x+8 = 8(m^3-n^3)$. Thus $mn+1 = m^3-n^3$ and since $m, n$ are both odd integers, $|m-n| \geq 2$. If $mn > 0$, then $m >n$ and $mn+1 = (m-n)((m-n)^2 + 3mn) \geq 6mn$ and we get $5mn \leq 1$, a contradiction. If $mn < 0$, two cases arise:
$m < 0, \, n > 0$: $2mn + 1 < mn+1 = m^3 - n^3 \leq -m^2 - n^2$ giving $m^2+n^2+ 2mn \leq -1$, an impossibility.
$m > 0,\, n < 0$: $mn+1 = m^3 - n^3 > 0$ and $mn > -1$. Since $mn < 0$ and $m, n$ are integers, this is impossible.
Thus the only solutions of the given equation are $x=0, y=\pm 4$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2047812",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
Finding all solutions to $3x + 4y \equiv 1 \pmod 7$ Find all solutions $\pmod 7$: $3x + 4y$ is congruent to $1 \pmod 7$.
I have tried writing out the various equations such as $3x + 4y = 1$, $3x + 4y = 8$, $3x + 4y = 15$, etc., but I do not know how to find the finite solution.
| Consider the equation
$$ 3x + 4y = 1 (\mod 7).$$
$ (3,4,1) = 1 | 7$, so there are $7\times 1 = 7$ solutions mod $7$. I’ll solve the equation using a reduction trick.
The given equation is equivalent to
$3x + 4y + 7z = 1$ for some $z$.
Set $$ w=\frac {3x+4y}{(3,4)} $$.
Then
$(3,4)w + 7z = 1, w + 7z = 1$.
$w_0 = -6, z_0 = 1$ is a particular solution. The general solution is
$w = −6 + 7s, z = 1 − s$.
Substitute for $w$:
$$ \frac {3x+4y}{(3,4)} =-6+7s$$, $$\Rightarrow 3x + 4y = −6 + 7s$$
$x_0 = -2, y_0 = \frac{7s}{4}$, is a particular solution. The general solution is
$x = -2 + 4t, y = \frac {7s}{4}-3t $.
$t = 0, 1, . . ., 6$ will produce distinct values of y mod 7. Note, however, that $s$ and $s + r$ produce $7s$ and $7s + 7r$, which are congruent mod $7$. That is, adding a multiple of $1$ to a given value of $s$ makes the
$7s$ term in $x$ repeat itself mod $7$. So I can get all possibilities for $x$ mod $7$ by letting $s = 0$.
All together, the distinct solutions mod $7$ are:
$$ x = -2 + 4t, y = \frac{7s}{4}-3t$$
where $s = 0$ and $t = 0,1...6$. Hope it helps you.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2049003",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Limit of a sequence including infinite product. $\lim\limits_{n \to\infty}\prod_{k=1}^n \left(1+\frac{k}{n^2}\right)$ I need to find the limit of the following sequence:
$$\lim\limits_{n \to\infty}\prod_{k=1}^n \left(1+\frac{k}{n^2}\right)$$
| $$\left(1+\frac{k}{n^2} \right)\left(1+\frac{n-k}{n^2} \right)=\left(1+\frac{1}{n} +\frac{k(n-k)}{n^4} \right)$$
Then we have
$$\left(1+\frac{1}{n} \right) \leq \left(1+\frac{1}{n} +\frac{k(n-k)}{n^4} \right) \leq \left(1+\frac{1}{n} +\frac{1}{n^2} \right)=\left(1 +\frac{n+1}{n^2} \right)$$
Therefore
$$\left(1+\frac{1}{n} \right) ^\frac{n}{2} \leq \prod_{k=1}^n\left(1+\frac{k}{n^2}\right) \leq \left(1 +\frac{n+1}{n^2} \right)^\frac{n}{2}$$
and squeeze it.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2052624",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 4,
"answer_id": 2
} |
Limit of recursive sequence defined by $a_0=0$, $a_{n+1}=\frac12\left(a_n+\sqrt{a_n^2+\frac{1}{4^n}}\right)$ Given the following sequence: $a_0=0$, $$a_{n+1}=\frac12\left(a_n+\sqrt{a_n^2+\frac{1}{4^n}}\right),\ \forall n\ge 0.$$
Find $\lim\limits_{n\to\infty}a_n$.
| By the given recurrence, $a_{n+1}$ is a root of $x^2-a_n x-\frac{1}{4^{n+1}}$, hence:
$$ \sum_{n\geq 0} a_{n+1}(a_{n+1}-a_n) = \sum_{n\geq 0}\frac{1}{4^{n+1}} = \frac{1}{3}. \tag{1}$$
By assuming $\lim_{n\to +\infty}a_n = L$ and applying summation by parts, we get:
$$ \frac{1}{3} = L^2 - \sum_{n\geq 0}(a_{n+2}-a_{n+1})a_{n+1} \tag{2}$$
but:
$$ \sum_{n\geq 0}(a_{n+2}-a_{n+1})a_{n+1} = -\frac{1}{4}+\sum_{n\geq 0} a_{n}(a_{n+1}-a_n)= \frac{1}{12}-\sum_{n\geq 0}(a_{n+1}-a_n)^2\tag{3} $$
implies
$$ L^2 = \frac{5}{12}-\sum_{n\geq 0}(a_{n+1}-a_n)^2 \tag{4}$$
so $L\leq\sqrt{\frac{5}{12}}$ but $L^2\geq \frac{5}{12}-\sum_{n\geq 0}\frac{1}{16^{n+1}a_n^2}$. Numerically,
$$ L \approx 0.63661977236758\ldots \tag{5}$$
On the other hand, by setting $a_n = \frac{b_n}{2^n}$ we get $b_{n+1}=b_n+\sqrt{b_n^2+1}$, that is associated with the halving formula for the cotangent function. That leads to:
$$ a_n = \frac{1}{2^n}\cot\left(\frac{\pi}{2^{n+1}}\right),\qquad L=\lim_{n\to +\infty}a_n = \color{blue}{\frac{2}{\pi}}. \tag{6} $$
It is very interesting to notice that by combining $(6)$ and $(4)$ we get a fast-converging algorithm for the numerical computation of $\frac{1}{\pi^2}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2054026",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
} |
Let $a,b,c$ be the length of sides of a triangle then prove that $a^2b(a-b)+b^2c(b-c)+c^2a(c-a)\ge0$
Let $a,b,c$ be the length of sides of a triangle then prove that:
$a^2b(a-b)+b^2c(b-c)+c^2a(c-a)\ge0$
Please help me!!!
| $$a^2b(a-b)+b^2c(b-c)+c^2a(c-a)
=\dfrac{1}{2}[(a+b-c)(b+c-a)(a-b)^2+(b+c-a)(a+c-b)(b-c)^2+(a+c-b)(a+b-c)(c-a)^2]
≥0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2055559",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 1
} |
Curious limits with tanh and sin These two limits can be easily solved by using De l'Hopital Rule multiple times (I think), but I suspect that there could be an easier way... Is there?
\begin{gather}
\lim_{x\to 0} \frac{\tanh^2 x - \sin^2 x}{x^4} \\
\lim_{x\to 0} \frac{\sinh^2 x - \tan^2 x}{x^4}
\end{gather}
Thanks for your attention!
| There is: Taylor's formula at order $4$:
*
*$\tanh x=x-\dfrac{x^3}3+o(x^4)$, hence
$$\tanh^2 x=\Bigl(x-\dfrac{x^3}3+o(x^4)\Bigl)^2=x^2-\dfrac{2x^4}3+o(x^4).$$
*$\sin^2x=\frac12(1-\cos 2x)=\frac12\Bigl(1-1+\dfrac{4x^2}2-\dfrac{16x^4}{24}+o(x^4)\Bigr)=x^2-\dfrac{x^4}3+o(x^4)$.
Thus
$$\frac{\tanh^2 x-\sin^2x}{x^4}=\frac{-\dfrac{x^4}3+o(x^4)}{x^4}=-\frac13+o(1)\to-\frac13.$$
Similarly:
*
*$\tan x=x+\dfrac{x^3}3+o(x^4)$, hence
$\tan^2 x=x^2+\dfrac{2x^4}3+o(x^4).$
*$\sinh^2x=\frac12(\cosh 2x-1)=\frac12\Bigl(1+\dfrac{4x^2}2+\dfrac{16x^4}{24}+o(x^4)-1\Bigr)=x^2+\dfrac{x^4}3+o(x^4)$.
Thus
$$\frac{\sinh^2 x-\tan^2x}{x^4}=\frac{-\dfrac{x^4}3+o(x^4)}{x^4}\to-\frac13.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2056180",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 1
} |
First coin is tossed 3 times. $X$=heads after 3 tosses.Second coin is tossed $X$ times and$Y$represents total heads.Find $P(X>=2|Y =1) $ There are 2 fair coins. First coin is tossed 3 times. $X$ represents number of heads in the 3 tosses.After this second coin is tossed $X$ number of times, where $Y$ represents number of heads.Find $P(X>=2|Y =1) $
Here, by statistical reasoning it can be concluded that the answer is $(5/9)$, but it appears to be $(11/18)$.
My attempt:
P(X greater than or equal to 2 given Y=1)=1- P(X is 0 or 1 given Y=1)
Which is,
= 1- P(X is one given Y is one)
= $1- (4/9)$
Which is indeed (5/9).
| Can you tell me how does it appear to be $\frac{11}{18}$? Is it by simulation or something else? I'm giving the mathematical argument below. Your conclusion based on statistical reasoning is absolutely fine! There seems to be some holes in whatever procedure that gives you $\frac{11}{18}$ as the answer.
First you need the joint distribution of $(X,Y)$. Note that
\begin{align}
P(X=x, Y=y)&=P(Y=y|X=x)P(X=x)\\
&=\binom{x}{y}\bigg(\frac{1}{2}\bigg)^x\cdot\binom{3}{x}\bigg(\frac{1}{2}\bigg)^3\\
&=\binom{x}{y}\binom{3}{x}\bigg(\frac{1}{2}\bigg)^{x+3}
\end{align}
Armed with the joint distribution, now we can calculate
\begin{align}
&\frac{P(X \geq 2, Y=1)}{P(Y=1)}\\
&=\frac{P(X=2,Y=1)+P(X=3,Y=1)}{P(X=0,Y=1)+P(X=1,Y=1)+P(X=2,Y=1)+P(X=3,Y=1)}\\
&=\frac{\binom{2}{1}\binom{3}{2}(\frac{1}{2})^5+\binom{3}{1}\binom{3}{3}(\frac{1}{2})^6}{0+\binom{1}{1}\binom{3}{1}(\frac{1}{2})^4+\binom{2}{1}\binom{3}{2}(\frac{1}{2})^5+\binom{3}{1}\binom{3}{3}(\frac{1}{2})^6}\\
&=\frac{5}{9}
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2059497",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
If $\frac{\sqrt{31+\sqrt{31+\sqrt{31+ \cdots}}}}{\sqrt{1+\sqrt{1+ \sqrt{1+ \cdots}}}}=a-\sqrt b$, find the value of $a+b$
$\frac{\sqrt{31+\sqrt{31+\sqrt{31+ \cdots}}}}{\sqrt{1+\sqrt{1+ \sqrt{1+ \cdots}}}}=a-\sqrt b$ where $a,b$ are natural numbers. Find the value of $a+b$.
I am not able to proceed with solving this question as I have no idea as to how I can calculate $\frac{\sqrt{31+\sqrt{31+\sqrt{31+ \cdots}}}}{\sqrt{1+\sqrt{1+ \sqrt{1+ \cdots}}}}$. A small hint would do.
| Hint: To find
$$\sqrt{a+\sqrt{a+\cdots}} $$
solve the equation
$$x = \sqrt{a+x}$$
As for the answer, I get $11$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2059779",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
All fractions which can be written simultaneously in the forms $\frac{7k-5}{5k-3}$ and $\frac{6l-1}{4l-3}$ Find all fractions which can be written simultaneously in the forms $\frac{7k-5}{5k-3}$ and $\frac{6l-1}{4l-3}$
for some integers $k,l$.
Please check my answer and tell me is correct or not....
$$\frac{43}{31},\frac{31}{27},1,\frac{55}{39},\frac{5}{3},\frac{61}{43},\frac{19}{13},\frac{13}{9}$$
| We have that $$\frac{7k-5}{5k-3}=\frac{6l-1}{4l-3}\iff kl+8k+l=6.$$ That is, if $k\ne -1,$
$$l=2\frac{3-4k}{k+1}=-2\left(4-\frac{7}{k+1}\right)=-8+\frac{14}{k+1}.$$ Since $l$ has to be an integer $k+1$ must divide $14.$ So, we have that $k\in\{-15,-8,-3,-2,0,1,6,13\}.$
Note that $k\ne -1$ since if $k=-1$ the equation $kl+8k+l=6$ gives $-8=6$ which doesn't hold.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2060154",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
} |
Simplifying $\frac { \sqrt2 + \sqrt 6}{\sqrt2 + \sqrt3}$? Is there anything else you can do to reduce it to something "nicer" other than multiplying it by $\dfrac {\sqrt3 - \sqrt2}{\sqrt3 - \sqrt2}$ and get $\sqrt 6 -2 + \sqrt {18} - \sqrt {12}$? The reason I think there's a nicer form is because the previous problem in the book was to simplify $\sqrt{3+ 2 \sqrt 2} - \sqrt{3 - 2 \sqrt 2}$, which simplifies nicely to $\sqrt{( \sqrt 2+1)^2} - \sqrt{ ( \sqrt 2 - 1)^2} = 2.$
| You are correct. Multiply by the conjugate which is $\sqrt{2}-\sqrt{3}$
After multiplying by the conjugate we have $\frac{2-\sqrt6+\sqrt{12}-\sqrt{18}}{-1}$ which gives $-2+\sqrt6-\sqrt{12}+\sqrt{18}$. The only further simplification is as follows: Simplify $\sqrt{12}$ and $\sqrt{18}$ and we have $-2+\sqrt6-2\sqrt3+3\sqrt2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2060719",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Polynomials and the IMO I have been stuck on this problem, as I am unable to proceed properly. The problem is as follows:
If $f(x) = (x+2x^2+\cdots nx^n)^2 = a_2x^2 + a_3x^3 +\cdots a_{2n}x^{2n},$ prove that $$a_{n+1} + a_{n+2} +\cdots +a_{2n} = \binom{n+1}{2}\frac{5n^2+5n+2}{12}$$
I tried expanding the LHS but only ended up with a bunch of expressions. Any help is appreciated.
| Hint: Induction works. Let $S_n$ be the sum on the LHS of the desired inequality. Then
$$
S_1=1=\binom{1+1}{2}\frac{5\cdot 1^2+5\cdot 1+2}{12}.
$$
From
\begin{aligned}
&\quad(x+2x^2+\cdots+nx^n+(n+1)x^{n+1})^2\\
&=(x+2x^2+\cdots+nx^n)^2+2(n+1)(x+2x^2+\cdots+nx^n)x^{n+1}+(n+1)^2x^{2(n+1)}\\
\end{aligned}
we have
$$
\boxed{S_{n+1}=S_n+2(n+1)\frac{n(n+1)}{2}+(n+1)^2-\varphi_n}\tag{$*$}
$$
where $\varphi_n$ is the cofficient of $x^{n+1}$ in the expansion of $(x+2x^2+\cdots+nx^n)^2$:
\begin{aligned}
\varphi_n&=1\cdot n+2(n-1)+3(n-2)+\cdots+n\cdot 1\\
&=\sum_{i=1}^ni(n+1-i)\\
&=(n+1)\sum_{i=1}^ni-\sum_{i=1}^ni^2\\
&=\frac{1}{6}n(n+1)(n+2).
\end{aligned}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2062208",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
show $\frac{y}{x^2+y^2} $ is harmonic except at $y=0,x=0$ Let $f(z)=u(x,y)+iv(x,y) $
where $$ f(z)=u(x,y)=\frac{y}{x^2+y^2}$$
show $u(x,y)$ is harmonic except at $z=0$
Attempt
$$ u=\frac{y}{x^2+y^2}=y(x^2+y^2)^{-1} $$
Partial derivatives with x
$$\begin{aligned}
u_x&= y *(x^2+y^2)^{-2}*-1*2x
\\ &= -y*2x(x^2+y^2)^{-2}=-2xy(x^2+y^2)^{-2}
\\ u_{xx} &= -2xy*(x^2+y^2)^{-3}*-2*2x+-2y*(x^2+y^2)^{-2}
\\&= \frac{-2xy}{(x^2+y^2)^3}*-4x +\frac{-2y}{(x^2+y^2)^2}
\\&=\frac{8x^2y}{(x^2+y^2)^3}+\frac{-2y}{(x^2+y^2)^2}
\\
\end{aligned} $$
Partial Derivatives with y
$$\begin{aligned}
u_y&=1(x^2+y^2)^{-1}+y*(x^2+y^2)^{-2}*-1*2y
\\ &=(x^2+y^2)^{-1}-2y^2(x^2+y^2)^{-2}
\\ u_{yy}&=-1(x^2y^2)^{-2}*2y -2*2y(x^2+y^2)^{-2}-2y^2*-2(x^2+y^2)^{-3}*2y
\\ &=-2y(x^2+y^2)^{-2}-4y(x^2+y^2)^{-2}+8y^3(x^2+y^2)^{-3}
\\ &=\frac{-6y}{(x^2+y^2)^2} + 8y^3(x^2+y^2)^{-3}
\end{aligned} $$
From here need to show that $u_{xx}+u_{yy}=0$ and technically say why the other partials are continous right?? This was a test question whith 3 lines of paper by the way
| $$\begin{align*}&u_x=-\frac{2xy}{(x^2+y^2)^2}\;&,\;\;&u_{xx}=-\frac{2y(x^2+y^2)-8x^2y}{(x^2+y^2)^3}=\frac{6x^2y-2y^3}{(x^2+y^2)^3}\\{}\\
&u_y=\frac{x^2+y^2-2y^2}{(x^2+y^2)^2}=\frac{x^2-y^2}{(x^2+y^2)^2}\;&,\;\;&u_{yy}=\frac{-2y(x^2+y^2)-4y(x^2-y^2)}{(x^2+y^2)^3}=\frac{-6x^2y+2y^3}{(x^2+y^2)^3}\end{align*}$$
and now just check that we certainly get $\;\nabla f=u_{xx}+u_{yy}=0\;$ . Observe the function $\;f\;$ is defined everywhere except at the origin, and the partial derivatives above, all of them (first and second order) are continuous everywhere where $\;f\;$ is defined.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2062523",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Does it converge? $\sum_{n=1}^{\infty} \arcsin^n\frac{1}{n}$ Does it converge? $\sum_\limits{n=1}^{\infty} \arcsin^n\frac{1}{n}$
I'm trying to solve it with the ratio test and using L'hopital, but it does not seem to go well. Can anyone help please?
| It converges drastically.
Since,
for $0 \le x \le \pi/2$
we have
$1 \ge \dfrac{\sin x}{x}
\ge \dfrac{2}{\pi}
$,
so that
$1 \ge \dfrac{ x}{\arcsin x}
\ge \dfrac{2}{\pi}
$.
Therefore
$1
\le \dfrac{\arcsin x}{ x}
\le \dfrac{\pi}{2}
$,
so
$\arcsin x
\le \dfrac{\pi x}{2}
$.
In particular,
$\arcsin \dfrac1{n}
\le \dfrac{\pi }{2n}
$.
Therefore
$\sum_\limits{n=1}^{\infty} \arcsin^{1+c}\dfrac{1}{n}
$
converges for any
$c > 1$
by comparison with
$\sum_\limits{n=1}^{\infty} \dfrac{1}{n^{1+c}}
$.
Note that
$\sum_\limits{n=1}^{\infty} \arcsin\dfrac{1}{n}
$
diverges since
$\arcsin\dfrac{1}{n}
\ge \dfrac1{n}
$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2062656",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
How do I solve quadratic equations when the coefficients are complex and real? I needed to solve this: $$x^2 + (2i-3)x + 2-4i = 0 $$
I tried the quadratic formula but it didn't work. So how do I solve this without "guessing" roots? If I guess $x=2$ it works; then I can divide the polynomial and find the other root; but I can't "guess" a root.
$b^2-4ac=4i-3$, now I have to work with $\sqrt{4-3i}$ which I don't know how. Apparently $4i-3$ is equal to $(1+2i)^2$, but I don't know how to get to this answer, so I am stuck.
| The quadratic equation works for real and imaginary coefficients (but maybe without as much geometric intuitiveness when there are imaginary coefficients). You can "check" this by trying to use the quadratic equation and then substituting back in your answers to see if we get $0$. We have $a=1$, $b=2i-3$, $c=2-4i$: $$x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}=\dfrac{-2i+3\pm\sqrt{(2i-3)^2-4(2-4i)}}{2} $$ $$= \dfrac{-2i+3\pm\sqrt{-4+9-12i-8+16i}}{2}=\dfrac{-2i+3\pm\sqrt{4i-3}}{2}$$
Not pretty, but a quick check in your equation of $x^2+(2i-3)x+2-4i$ with $x=\dfrac{-2i+3+\sqrt{4i-3}}{2}$ gives: $$\left(\dfrac{-2i+3+\sqrt{4i-3}}{2}\right)^2+(2i-3)\cdot\left(\dfrac{-2i+3+\sqrt{4i-3}}{2}\right)+2-4i$$ $$= \dfrac{(-2i+3)^2+4i-3+2(-2i+3)\sqrt{4i-3}}{4}+\dfrac{-(-2i+3)^2+(2i-3)\sqrt{4i-3}}{2}+2-4i$$ $$=\frac{-(-4+9-12i)+4i-3-2(2i-3)\sqrt{10i-3}+2(2i-3)\sqrt{10i-3}}{4}+2-4i$$ $$=\dfrac{-8+16i}{4}+2-4i=0\qquad\checkmark$$
and the other root should (hopefully) follow the same arithmetic, but still work just the same.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2067993",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 8,
"answer_id": 1
} |
Finding the position vector of the centre of a circle of intersection. If there are two spheres with vector equations:
$(r-a)^2=A^2$ and $(r-b)^2=B^2$
where, intuitively, "a" and "b" are vectors representing the centre of the spheres and "A" and "B" are scalars representing the radius of the spheres,
Assuming that they intersect each others, how to show that the position vector of the centre of the circle of intersection is given by:
$$\frac{a+b}{2} + \frac{(A^2-B^2)(b-a)}{2(b-a)^2}$$
| Consider any plane containing the sphere's centers $\mathbf a$ and $\mathbf b$. If there is an intersecting circle, it will also contain two points from this circle. Let us consider one and call it $\mathbf c$.
This triangle is illustrated here as follows:
As you can see, the sides of the triangles coincide with the radii of the spheres ($A$ and $B$) and the connection of the sphere's center ($\left\|\mathbf b - \mathbf a\right\|$). Moreover, the altitude of the triangle coincides with a radius of the interseting circle (denoted by $R$) and the altitude's foot is the center of the circle. I denote it by $\mathbf x$. Since $\mathbf x$ is on the line through $\mathbf a$ and $\mathbf b$, we can express it as $\mathbf x = \mathbf a + t\cdot(\mathbf b - \mathbf a)$, where $t \in (0,1)$. We therefore have $\left\|\mathbf x - \mathbf a\right\| = t \cdot \left\|\mathbf b - \mathbf a\right\|$ and $\left\|\mathbf b - \mathbf x\right\| = (1-t) \cdot \left\|\mathbf b - \mathbf a\right\|$
To compute $t$ consider the two right triangles that have the altitude in common. From Pythagoras we have $$R^2 = A^2 - t^2\left\|\mathbf b - \mathbf a\right\|^2 = B^2-(1-t)^2\left\|\mathbf b - \mathbf a\right\|^2.$$
Solving for $t$ gives $$A^2-B^2 = (t^2-1+2t-t^2)\left\|\mathbf b - \mathbf a\right\|^2 \quad \Rightarrow t = \frac{A^2-B^2}{2\left\|\mathbf b - \mathbf a\right\|^2} + \frac{1}{2}.$$
Finally, since $\mathbf x = \mathbf a + t\cdot(\mathbf b - \mathbf a)$, the result is $$\mathbf x = \mathbf a + \frac{A^2-B^2}{2\left\|\mathbf b - \mathbf a\right\|^2} (\mathbf b - \mathbf a) + \frac{1}{2}(\mathbf b - \mathbf a) = \frac{\mathbf a + \mathbf b}{2} + \frac{A^2-B^2}{2\left\|\mathbf b - \mathbf a\right\|^2} (\mathbf b - \mathbf a).$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2068409",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Given a positive number n, how many tuples $(a_1,...,a_k)$ are there such that $a_1+..+a_k=n$ with two extra constraints The problem was: Given a positive integer $n$, how many tuples $(a_1,...,a_k)$ of positive integers are there such that $a_1+a_2+...+a_k=n$. And $0< a_1 \le a_2 \le a_3 \le...\le a_k$. Also, $a_k-a_1$ is either $0$ or $1$.
Here is what I did:
For $n=1$, there is one way $1=1$.
For $n=2$, there are $2$ ways, $2=1+1,2=2$
For $n=3$, there are $3$ ways, $3=1+1+1,3=1+2,3=3$
For $n=4$, there are $4$ ways, $4=1+1+1+1,4=1+1+2,4=2+2,4=4$
So it seems that there are $n$ tuples that satisfies the three conditions for each $n$. But I'm not sure how to prove it.
| Every solution is a tuple in order of increasing integers. Also, $a_k=a_1$ or $a_k=a_1+1$. Therefore, we have all of the $a_1$s at the beginning and $a_1+1$s at the end. We can say there are $l$ instances of $a_1$ and thus $k-l$ instances of $a_1+1$. Since the sum of the tuples is $n$, we find:
$$l*a_1+(k-l)*(a_1+1)=n$$
Simplify the left side:
$$a_1k+k-l=n$$
Now, $l$ is the number of instances of $a_1$ in the sequence. Thus, $l$ is at least $1$ and at most $k$. Therefore, $k-l$ is at least $0$ and at most $k-1$, so $k-l$ is basically the remainder of $n$ when divided by $k$ and $a_1$ is the quotient. Thus, by the Division Theorem, we know there are unique integer solutions of $a_1$ and $l$ for this equation for any $k$ and $n$.
However, the problem says that $a_1 > 0$, so we need to exclude the solutions where $a_1=0$. When $a_1=0$, we get:
$$0k+k-l=n \implies k-l=n$$
Since $k-l$ is at most $k-1$, this means $n$ is at most $k-1$ and thus $n < k$. Therefore, exclude all solutions where $k > n$.
Also, obviously, $k \geq 1$. This leaves us with the solutions $k \geq 1$ and $k \leq n$, so there are $n$ solutions.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2068875",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
Finding the value of $f'(0)$ If $f$ is a quadratic function such that $f(0)=1$ and
$$\int\frac{f(x)}{x^2(x+1)^3}dx$$
is a rational function, how can we find the value of $f'(0)$? I am totally clueless to this. Any tip on how to start? If you wish to give details, then many thanks to you.
| Let $f(x) = ax^2 + bx + c.$ As $f(0) = 1,$ we have that $f(x) = ax^2 + bx + 1.$ Then $$I=\int \frac{f(x) dx}{x^2(x+1)^3} = \int \frac{a}{(x+1)^3}+\frac{b}{x(x+1)^3}+\frac{1}{x^2(x+1)^3}\,dx \\\stackrel{Partial Fractions}{=} -\frac 1x + \frac{b-a-1}{2 (1 + x)^2} + \frac{-2 + b}{1 + x} + (b-3)\log(x) - (b-3)\log(1 + x) + C.$$
As the function must be rational, we must have $b=3,$ which leads $f'(0) = \left.2ax+b\right|_{x=0} = \boxed{3}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2070180",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
} |
Choose $a, b$ so that $\cos(x) - \frac{1+ax^2}{1+bx^2}$ would be as infinitely small as possible on ${x \to 0}$ using Taylor polynomial $$\cos(x) - \frac{1+ax^2}{1+bx^2} \text{ on } x \to 0$$
If $\displaystyle \cos(x) = 1 - \frac{x^2}{2} + \frac{x^4}{4!} - \cdots $
Then we should choose $a, b$ in a such way that it's Taylor series is close to this.
However, I'm not sure how to approach this. I tried to take several derivates of second term to see its value on $x_0 = 0$, but it becomes complicated and I don't see general formula for $n$-th derivative at point zero to find $a$ and $b$.
| The quick-and-dirty method:
\begin{align*}
f(x) = \frac{1 + a x^2}{1 + bx^2} &= (1 + a x^2) \left( 1 - b x^2 + b^2 x^4 - b^3 x^6 + \cdots \right) \\&= 1 - (b - a) x^2 + (b^2 - ab) x^4 - (b^3 - a b^2) x^6 + \cdots
\end{align*}
We want $b - a = \frac{1}{2}$ and $b (b-a) = \frac{1}{24}$, so that (at least) the first three terms in the Taylor series of $f(x)$ and $\cos x$ agree. This implies that $b = \frac{1}{12}$ and $a = -\frac{5}{12}$; with this choice, we have
$$
f(x) = \frac{1 - \frac{5}{12} x^2}{1 + \frac{1}{12} x^2} = 1 - \frac{x^2}{2} + \frac{x^4}{24} - \frac{x^6}{288} + \cdots
$$
which agrees with $\cos x$ up to the $\mathcal{O}(x^6)$ term.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2070748",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 0
} |
How many ways can 8 teachers be distributed among $4 $ schools? There are several ways that the teachers can be divided amongst $4$ schools, namely here are the possible choices I came up with:
$1) 1 1 1 5$
$2) 1 1 2 4$
$3) 1 1 3 3$
$4) 1 2 2 3$
$5) 2 2 2 2$
now given the fact that say $2213$ is the same as $1 2 2 3$ it was omitted. With out repeats I believe these 5 are the only possibilities.
1)
${8 \choose 5} \times {3 \choose 1} \times {2 \choose 1} \times {1 \choose 1}$:
$\frac{8!}{5!3!} \times \frac{3!}{1!2!} \times \frac{2!}{1!1!} \times 1$
Which comes out to
$56 \times 3 \times 2 \times 1 = 336$
2)
${8 \choose 4} \times {4 \choose 2} \times {2 \choose 1} \times {1 \choose 1}$:
$\frac{8!}{4!4!} \times \frac{4!}{2!2!} \times \frac{2!}{1!1!} \times 1$
Which comes out to
$70 \times 6 \times 2 \times 1= 840$
3)
${8 \choose 3} \times {5 \choose 3} \times {2 \choose 1} \times {1 \choose 1}$
$\frac{8!}{3!5!} \times \frac{5!}{3!2!} \times \frac{2!}{1!1!} \times 1$
Which comes out to
$56 \times 10 \times \times 2 = 1,120$
4)
${8 \choose 3} \times {5 \choose 2} \times {3 \choose 2} \times {1 \choose 1}$
$\frac{8!}{3!5!} \times \frac{5!}{2!3!} \times \frac{3!}{2!1!} \times \frac{1!}{1!0!}$
Which comes out to:
If 8 new teachers are to be divided amongst 4 new schools how many divisions are possible?
$56 \times 10 \times 3 \times 1= 1,680$
5)
${8 \choose 2} \times {6 \choose 2} \times {4 \choose 2} \times {2 \choose 2}$
$\frac{8!}{2!6!} \times \frac{6!}{2!4!} \times \frac{4!}{2!2!} \times \frac{2!}{2!0!}$
Which comes out to:
$28 \times 15 \times 6 \times 1 = 2,520$
What am I missing?
| With teachers and schools distinct and every school must have at least one teacher assigned to it, I would approach via inclusion-exclusion.
Let the schools be labeled $a,b,c,d$. Let the teachers be labeled $1,2,3,\dots,8$.
Let event $A$ be the event that school $a$ has no teachers assigned to it. Similarly $B,C,D$ are the events that school $b,c,d$ have no teachers assigned to them respectively.
Then $|A\cup B\cup C\cup D|$ is the set of ways to distribute the teachers which is "bad", yielding a school without a teacher in it.
$|A|$ (and similarly $|B|,|C|,|D|$) can be described as the functions from $\{1,2,3,\dots,8\}$ to $\{b,c,d\}$. From earlier example, we know that there are $3^8$ such functions.
$|A\cap B|$ (and similarly $|A\cap C|,|A\cap D|,\dots$) can be described as the functions from $\{1,2,\dots,8\}$ to $\{c,d\}$. From earlier example, we know that there are $2^8$ such functions.
Similarly, we calculate $|A\cap B\cap C|=1^8$ and $|A\cap B\cap C\cap D|=0^8$
Applying inclusion-exclusion then, we have $|A\cup B\cup C\cup D|=\binom{4}{1}3^8-\binom{4}{2}2^8+\binom{4}{3}1^8-\binom{4}{4}0^8=4\cdot 3^8-6\cdot 2^8+4=24712$
As those were the "bad" outcomes out of the $4^8$ total number of ways to distribute the teachers regardless, there are then $4^8-24712=40824$ good ways to distribute the teachers.
A shorter solution using more complicated formulae which take some time to learn appears in the chart I linked to in the comments above. We are able to describe this as trying to count the number of surjective functions from $\{1,2,\dots,8\}$ to $\{a,b,c,d\}$.
We can accomplish this by first counting how many ways there are to partition $\{1,2,\dots,8\}$ into $4$ nonempty subsets, and then choose how to label the subsets.
We can accomplish the first step in $\left\{\begin{array}{}8\\4\end{array}\right\}$ number of ways. (here $\left\{\begin{smallmatrix}n\\r\end{smallmatrix}\right\}$ represents the stirling number of the second kind $S(n,r)$). We can accomplish the second step in $4!$ number of ways, yielding a final total of $4!\cdot \left\{\begin{array}{}8\\4\end{array}\right\}=40824$ ways. (wolfram)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2074092",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 0
} |
Find the maximum power of $24$ in $(48!)^2$? Find the maximum power of $24$ in $(48!)^2$ ?
How to approach for such questions ?
| Divide $48$ by $2$ repeatedly and add the quotient omitting any remainder,
$48/2=24, 24/2=12, 12/2=6, 6/2=3, 3/2=1$(with remainder as $1$). Therefore adding the quotients,i.e, $24+12+6+3+1=46$ $\Longrightarrow$ there are $46$ $2$s in $48!$ and $92$ $2$s in $(48!)^2$.
Similarly, divide 48 by 3 repeatedly and add the quotients omitting any remainder,
$48/3=16, 16/3=5$ (with remainder as $2$), $5/3=1$ (with remainder as $2$). Adding these quotients gives , $16+5+1=22$ $\Longrightarrow$ There are $22$ $3$s in $48!$ and hence $44$ $3$s in $(48!)^2$.
Now, the number of $24$s ($2^3\times3$)that can be made from $92$ $2$s and $44$ $3$s is $30$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2075577",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 3,
"answer_id": 2
} |
Natural numbers $a$ for which $ \int\frac{x^2-12}{(x^2-6x+a)^2}dx$ is a rational function possible values of $a$ for which $\displaystyle f(x) = \int\frac{x^2-12}{(x^2-6x+a)^2}dx$ is a rational function for $a\in N$
$\displaystyle \frac{d}{dx}f(x)=\frac{x^2-12}{(x^2-6x+a)}$ assuming $\displaystyle f(x) = \frac{px+q}{x^2-6x+a}$
$\displaystyle \frac{(x^2-6x+a)p-(px+q)(2x-6)}{(x^2-6x+a)^2} = \frac{x^2-12}{(x^2-6x+a)^2}$
after camparing coff. $p=-1,q=0,a=12$
but answer is $a=12$ as well as $a=9$ is also given
could some help me
| The polynomial $x^2-6x+a$ has two roots in the complex plane, given by
$$ \zeta_a^\pm = 3\pm\sqrt{9-a}$$
so that $x^2-6x+a = (x-\zeta_a^+)(x-\zeta_a^-)$. Let we perform the partial fraction decomposition of the integrand function given such information:
$$\begin{eqnarray*} \frac{x^2-12}{(x^2-6x+a)^2}&=&\frac{x^2-12}{(\zeta_a^+-\zeta_a^-)^2}\left(\frac{1}{x-\zeta_a^+}-\frac{1}{x-\zeta_a^-}\right)^2 \\&=&\frac{x^2-12}{4(9-a)}\left(\frac{1}{(x-\zeta_a^+)^2}+\frac{1}{(x-\zeta_a^-)^2}-\frac{2}{x^2-6x+a}\right)\end{eqnarray*}$$
to deduce, by termwise integration, that:
$$ \int\frac{(x^2-12)\,dx}{(x^2-6x+a)^2}=\frac{1}{2}\left(\frac{(6-a)x+3(12-a)}{(a-9)(x^2-6x+a)}+\frac{\color{blue}{(a-12)}}{(a-9)^{3/2}}\,\text{arctanh}\left(\frac{x-3}{\color{green}{\sqrt{9-a}}}\right)\right) $$
so the only values for which the $\arctan$/$\text{arctanh}$ part disappears are $\color{blue}{a=12}$ and $\color{green}{a=9}$.
In such cases, we have:
$$ \int\frac{(x^2-12)\,dx}{(x^2-6x+\color{blue}{12})^2}=-\frac{x}{x^2-6x+12},\qquad \int\frac{(x^2-12)\,dx}{(x^2-6x+\color{green}{9})^2}=-\frac{x^2-3x-1}{(x-3)^3}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2077361",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How to simplify an expression that does not have a common factor I am trying to simplify this expression :
$$9a^4 + 12a^2b^2 + 4b^4$$
So I ended up having this :
$$(3a^2)^2 + 2(3a^2)(2b^2) + (2b^2)^2$$
However, after that I don't know how to keep on simplifying the equation, it is explained that the answer is $(3a^2 + 2b^2)^2$ because the expression is equivalent to $(x + y)^2$ but I don't understand how they get to that ?
| Notice that
$$(x+y)^2=x^2+2xy+y^2$$
If $x=3a^2$ and $y=2b^2$, we end up with
$$(3a^2+2b^2)^2=(3a^2)^2+2(3a^2)(2b^2)+(2b^2)^2\\=9a^4+12a^2b^2+4b^4$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2077624",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 3
} |
$\sum\limits_{k=1}^{n-1}\frac{1}{\sqrt{k(n-k)}}\geq 1$ proof
$\sum_\limits{k=1}^{n-1}\frac{1}{\sqrt{k(n-k)}}\geq 1$ for all $n\geq 2$
Basecase n=2
$\sum_\limits{k=1}^{2-1}\frac{1}{\sqrt{k(2-k)}}=1\geq 1$
Assumption
$\sum_\limits{k=1}^{n-1}\frac{1}{\sqrt{k(n-k)}}\geq 1$ holds for some $n$
Claim
$\sum_\limits{k=1}^{n}\frac{1}{\sqrt{k(n+1-k)}}\geq 1$ holds too
Step
Assume $n$ is odd
$\sum_\limits{k=1}^{n}\frac{1}{\sqrt{k(n+1-k)}}=\frac{1}{\sqrt{n}}+\frac{1}{\sqrt{2(n-1)}}+...+\frac{1}{\sqrt{n}}$
Then, to determine which of those terms is the smallest, we want to find the maximum of $k(n+1-k)$.
The deriviative would be $n+1-2k$. So $k=\frac{n+1}{2}$, that's why we assume $n$ is odd in this step.
So our smallest term looks like: $\frac{1}{\sqrt{(\frac{n+1}{2})^2}}=\frac{2}{n+1}$
And since we add this term $n$ times, the sum is bounded below by $\frac{2}{n+1}n=\frac{2n}{n+1}$.
Via induction it is very easy to see, that $\frac{2n}{n+1}>1$ for all n>1, which is all we care about.
Now, how do I proceed for even $n$?
| Using $\sqrt{ab}\leq \frac{a+b}{2}$ it follows that $\frac{1}{\sqrt{k(n-k)}}\geq \frac{2}{n}$, therefore lhs $\geq \frac{2(n-1)}{n}\geq1$ for $n\geq 2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2077990",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
If $ f(x) = \frac{\sin^2 x+\sin x-1}{\sin^2 x-\sin x+2},$ then value of $f(x)$ lies in the interval If $\displaystyle f(x) = \frac{\sin^2 x+\sin x-1}{\sin^2 x-\sin x+2},$ then value of $f(x)$ lies in the interval
assume $\sin x= t$ where $|\sin x|\leq 1$
let $\displaystyle y = \frac{t^2+t-1}{t^2-t+1}$
$\displaystyle yt^2-yt+y=t^2+t-1$
$(y-1)t^2-(y+1)t+(y-1)=0$
for real roots $(y+1)^2-4(y-1)^2\geq 0$ or $(y+1)^2-(2y-2)^2\geq0$
$(3y-3)(3y-1)\leq 0$ or $\displaystyle \frac{1}{3}\leq y\leq 1$
but walframalpha shows different answer
https://www.wolframalpha.com/input/?i=range+of+f(x)+%3D+%5Cfrac%7B%5Csin%5E2+x%2B%5Csin+x-1%7D%7B%5Csin%5E2+x-%5Csin+x%2B2%7D
could some help me with this, thanks
| Where DeepSea has left off, let $2t-3=-u\implies1\le u\le5$
As $u>0,$
$$y=1-\dfrac{4u}{u^2-4u+11}=1-\dfrac4{u+\dfrac{11}u-4}$$
Now $u+\dfrac{11}u=\left(\sqrt u-\sqrt{\dfrac{11}u}\right)^2+2\sqrt{11}\ge2\sqrt{11}$
$\implies\dfrac1{u+\dfrac{11}u-4}\le\dfrac1{2\sqrt{11}-4}=\dfrac{2\sqrt{11}+4}{44-4^2}=\dfrac{\sqrt{11}+2}{14}$
$\implies y=1-\dfrac4{u+\dfrac{11}u-4}\ge1-\dfrac{4(\sqrt{11}+2)}{14}$
Again with $1\le u\le5,$ clearly, $\dfrac u{u^2-4u+11}=\dfrac u{7+(u-2)^2}>0$
Trying to prove $\dfrac u{u^2-4u+11}\ge\dfrac18$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2081524",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Prove that $(2+ \sqrt5)^{\frac13} + (2- \sqrt5)^{\frac13}$ is an integer When checked in calculator it is 1. But how to prove it?
Also it is not a normal addition like $x+ \frac1x$ which needs direct rationalization. So I just need something to proceed.
| Let $x=(2+\sqrt 5)^{1/3}$ and $y=(2-\sqrt 5)^{1/3} .$
We have $5>x^3>4$ and $0>y^3>-1$ so $2>x>1$ and $0>y>-1,$ so $2>x+y>0.$ Therefore $$x+y\in \mathbb Z \iff x+y=1.$$ We have $xy= (x^3y^3)^{1/3}=(-1)^{1/3}=-1$ and $x^3+y^3=4.$ So we have $$(x+y)^3=x^3+y^3+3xy(x+y)=4-3(x+y).$$ So $x+y$ is a positive solution to the equation $z^3+3z=4.$ The function $f(z)=z^3+3z$ is strictly increasing for $z>0$ (As it is the sum of the strictly increasing functions $z^3$ and $3z$.)
So $f(z)=4$ has at most one positive solution, which, if it exists, is equal to $x+y.$ Therefore $$x+y\in \mathbb Z \iff 1^3+3\cdot 1=4.$$
For insight on why this kind of thing happens, see Cardano's Method for solving cubic equations.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2082836",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 8,
"answer_id": 6
} |
Calculate the antiderivative of a given function Consider the function $f : \left[ 0, \frac{\pi}{4} \right)$, $f(x) = \frac{(\cos x + \sin x)^n}{(\cos x - \sin x)^{n + 2}}$, where $n \in \mathbb{N}^*$. Find the antiderivative $F$ of $f$ such that $F(0) = \frac{1}{2(n + 1)}$.
I've noticed that $(\cos x - \sin x)' = -(\cos x + \sin x)$, but I can't figure out how to use this in finding $F$.
Thank you!
| We're trying to integrate
$$ \int f(x) dx = \int \frac{(\cos x + \sin x)^n}{(\cos x - \sin x)^{n + 2}} dx,$$
Divide numerator and denominator by $\cos^{n+2}(x)$:
$$ \int \frac{1}{\cos^2(x)}\frac{(1 + \tan x)^n}{(1 - \tan x)^{n + 2}} dx,$$
Now substitute $t = \tan(x)$; $dt = \frac{1}{\cos^2(x)}dx$:
\begin{align} \int \frac{(1 + t)^n}{(1 - t)^{n + 2}} dt &= \frac{1}{2}\int \frac{(1 + t)^n(1-t+1+t)}{(1 - t)^{n + 2}}\\
& = \frac{1}{2}\int \frac{(1 + t)^n(1-t)^n((1-t)+(1+t))}{(1 - t)^{2n + 2}}dt\\
& = \frac{1}{2}\int \frac{(1-t)^{n+1}(1 + t)^n+(1+t)^{n+1}(1 - t)^n}{(1 - t)^{2n + 2}}dt\\
& = \frac{1}{2(n+1)}\int \frac{d}{dt}\left(\frac{1+t}{1-t}\right)^{n+1}dt\\
& = \frac{1}{2(n+1)}\left(\frac{1+t}{1-t}\right)^{n+1} + c\\
& = \frac{1}{2(n+1)}\left(\frac{1+\tan(x)}{1-\tan(x)}\right)^{n+1} + c
\end{align}
Since $\tan(0) = 0$, you'll want $c = 0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2083394",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Doing definite integration $\int_0^{\pi/4}\frac{x\,dx}{\cos x(\cos x + \sin x)}$ We have to solve the following integration
$$
\int_0^{\pi/4}\frac{x\,dx}{\cos x(\cos x + \sin x)}
$$
I divided both in Nr and Dr by $\cos^2 x$.
But after that I stuck.
| Let $$I(x)= \int^{\frac{\pi}{4}}_0 \frac{x}{\cos(x)(\cos(x)+\sin(x))} dx$$
$$ I(x)= \int^{\frac{\pi}{4}}_0 \frac{x}{\cos(x)(\sqrt2\sin(x+\frac{\pi}{4}))} dx$$
$$ I(x)= \int^{\frac{\pi}{4}}_0 \frac{x}{\sqrt2\cos(x)\sin(x+\frac{\pi}{4})} dx$$
Since $$ I(x)=I(\frac{\pi}{4}-x)$$
$$ I(\frac{\pi}{4}-x)=\int^{\frac{\pi}{4}}_0 \frac{\frac{\pi}{4}-x}{\cos(\frac{\pi}{4}-x)(\sqrt2\sin(\frac{\pi}{4}-x+\frac{\pi}{4}))} dx$$
$$ I(\frac{\pi}{4}-x) =\int^{\frac{\pi}{4}}_0 \frac{\frac{\pi}{4}-x}{\sqrt2 \cos(x)\sin(x+\frac{\pi}{4})} dx$$
But $$ I(x)=I(\frac{\pi}{4}-x)$$
so $$ I+I =\int^{\frac{\pi}{4}}_0 \frac{\frac{\pi}{4}-x}{\sqrt2 \cos(x)\sin(x+\frac{\pi}{4})}+\int^{\frac{\pi}{4}}_0 \frac{x}{\sqrt2\cos(x)\sin(x+\frac{\pi}{4})} dx$$
$$ 2I = \frac{\pi}{4\sqrt2}\int^{\frac{\pi}{4}}_0 \frac{1}{\cos(x)\sin(x+\frac{\pi}{4})} dx$$
Can you continue?
$$ I = \frac{\sqrt2\pi}{16}\int^{\frac{\pi}{4}}_0 \frac{1}{\cos(x)(\frac{1}{\sqrt{2}})(\cos(x)+\sin(x))} dx$$
$$ I = \frac{\pi}{8}\int^{\frac{\pi}{4}}_0 \frac{\sec(x)}{\sin(x)+\cos(x)}\cdot\frac{\frac{1}{\cos(x)}}{\frac{1}{\cos(x)}} dx$$
$$ I = \frac{\pi}{8}\int^{\frac{\pi}{4}}_0 \frac{\sec^2(x)}{\tan(x)+1} dx$$
$$ I = \frac{\pi}{8} \left[\ln|\tan(x)+1 \right]^{\frac{\pi}{4}}_0$$
$$ I = \frac{\pi}{8} \cdot \ln(2) $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2084474",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 6,
"answer_id": 1
} |
Finding the numbers $n$ such that $2n+5$ is composite. Let $n$ be a positive integer greater than zero. I write
$$a_n =
\begin{cases}
1 , &\text{ if } n=0 \\
1 , &\text{ if } n=1 \\
n(n-1), & \text{ if $2n-1$ is prime} \\
3-n, & \text{ otherwise}
\end{cases}$$
The sequence goes like this $$1,1,2,6,12,-2,30,42,-5,72,90,-8,132,-10,-11,\ldots$$ I would like to prove the following two claims.
claim 1 : If $a_n>0$ and
${a_n \above 1.5 pt 3} \notin \mathbb{Q}$ then $\sqrt{4a_n+1}$ is prime.
The table below illustrates what I am seeing:
\begin{array}{| l | l | l | l }
\hline
n & a_n & {a_n \above 1.5 pt 3} & \sqrt{4a_n+1}\\ \hline
0 & 1 & .333333.. & 2.2360679.. \\
1 & 1 & .333333.. & 3 \\
2 & 2 & .666666.. & 3 \\
3 & 6 & 2 & 5 \\
4 & 12 & 4 & 7 \\
6 & 30 & 10 & 11 \\
7 & 42 & 14 & 13 \\
9 & 72 & 24 & 17 & \\
10 & 90 & 30 & 19 \\
12 & 132 & 44 & 23 \\
15 & 210 & 70 & 29 \\
16 & 240 & 80 & 31 \\
19 & 342 & 114 & 37 \\
21 & 420 & 140 & 41 \\
22 & 462 & 154 & 43 \\
\hline
\end{array}
claim 2: If $a_n<0$ then $2a_n+5$ is composite
| $n=13$ is a counter example for claim 2.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2085307",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
Proving $a\cos x+b\sin x=\sqrt{a^2+b^2}\cos(x-\alpha)$
Show that $a\cos x+b\sin x=\sqrt{a^2+b^2}\cos(x-\alpha)$ and find the correct phase angle $\alpha$.
This is my proof.
Let $x$ and $\alpha$ be the the angles in a right triangle with sides $a$, $b$ and $c$, as shown in the figure. Then, $c=\sqrt{a^2+b^2}$. The left-hand side is $a\cos x+b\sin x=\frac{ab}{c}+\frac{ab}{c}=2\frac{ab}{c}$. The right-hand side is $\sqrt{a^2+b^2}\cos(x-\alpha)=c\left(\cos{x}\cos{\alpha}+\sin{x}\sin{\alpha}\right)=c\left(\frac{ab}{c^2}+\frac{ab}{c^2}\right)=2\frac{ab}{c}$.
Is my proof valid? Is there a more general way to prove it?
For the second part of the question, I think it should be $\alpha=\arccos\frac{a}{\sqrt{a^2+b^2}}=\arcsin\frac{b}{\sqrt{a^2+b^2}}$. Is this correct?
| Let $a = r \cos \alpha$ and $b = r \sin \alpha$. Then,
$
a\cos x + b\sin x\\
= r\cos \alpha \cos x + r\sin \alpha \sin x\\
= r \cos (x - \alpha)
$
since we know that $\cos \theta \cos \phi + \sin \theta \sin \phi = \cos (\theta - \phi)$. Now, we already have $a = r \cos \alpha$ and $b = r \sin \alpha$. Squaring and adding both of these,
$
a^2 + b^2 = r^2(\cos^2 \alpha + \sin^2 \alpha) = r^2 \\
\implies r = \sqrt{a^2 + b^2}
$
which gives us our desired result.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2085687",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 2
} |
Maximum value of $|x-y|$ Given $x,y\in\mathbb R$ such that
$$5x^2+5y^2-6xy\ =\ 8$$
find the maximum value of $|x-y|$.
My attempt
$5x^2 - 6yx + (5y^2-8)\ =\ 0$
$x\ =\ \dfrac{6y\pm\sqrt{(6y)^2-4(5)(5y^2-8)}}{10} = \dfrac{6y\pm\sqrt{160-64y^2}}{10} = \dfrac{3y\pm2\sqrt{10-4y^2}}{10}$
$5y^2 - 6xy + (5x^2-8)\ =\ 0$
$y\ =\ \dfrac{3x\pm2\sqrt{10-4x^2}}{10}$
$|x-y|\ =\ \dfrac{\left|3(y-x)\pm2\sqrt{10-4y^2}\mp2\sqrt{10-4x^2}\right|}{10}$
What do I do from here?
| Hint: We can write $$P= 5x^2 +5y^2-6xy-8 =5 (x-y)^2+4xy-8 =5 (x-y)^2 +(x+y)^2-(x-y)^2-8=4 (x-y)^2 +(x+y)^2-8$$ Can you take it from here?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2087496",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Find the $\gcd[x+y+z; x^2+xy+z^2; y^2+yz+z^2; z^2+zx+x^2]$ What I have done:
There exists a non-zero integer $t$ such:
$$x+y+z=kt$$
$$x^2+xy+y^2=ut$$
$$y^2+yz+z^2=vt$$
$$z^2+zx+x^2=wt$$
$\implies$
$$(x-y)(x+y+z)=(u-v)t$$
$$(y-z)(x+y+z)=(v-w)t$$
$$(z-x)(x+y+z)=(w-u)t$$
$\implies$
$$\dfrac{x+y+z}{t}= \dfrac{u-v}{x-y}=\dfrac{v-w}{y-z}=\dfrac{w-u}{z-x}=k$$
$\implies$
$$x+y+z=kt $$
$$k(x-y)=u-v$$
$$k(y-z)=v-w $$
$$k(z-x)=w-u $$
$\implies$
$$u=w+k(x-z) $$
$$v=w+k(y-z)$$
$$w=w $$
$\implies$
$$x+y+z=kt$$
$$x^2+xy+y^2=[w+k(x-z)]t$$
$$y^2+yz+z^2=[w+k(y-z)t]t$$
$$z^2+zx+x^2=wt$$
$\implies$
| $gcd(x+y+z,x^2+xy+z^2,y^2+yz+z^2,z^2+zx+x^2)=1$
If $(x+y+z,x^2+xy+z^2,y^2+yz+z^2,z^2+zx+x^2)$ shared a common factor, the existence of an algebraic factor would soon be apparent after trying a few numerical examples.
So I used $(x,y,z)=(17,23,31),(2,3,5),(3,7,11)$ and found $gcd=1$ in each case.
I realise this method might seem unscientific, but I’ve yet to find an example where it fails.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2088438",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
If $f(x+1)+f(x-1)= \sqrt{2}\cdot f(x)$, find the period of $f(x)$. If $f(x+1)+f(x-1)= \sqrt{2}\cdot f(x)$, then the period of $f(x)$ is?
I tried replacing $x$ with $x-1$ and stuff, but didn't lead me to anything...
| Make $x=x-1$
$$f(x)+f(x-2)=\sqrt{2}f(x-1) \tag {1}$$
Make $x=x+1$
$$f(x+2)+f(x)=\sqrt{2}f(x+1) \tag {2}$$
Now sum $(1)$ and $(2)$:
$$2f(x)+f(x+2)+f(x-2)=\sqrt{2}(f(x+1)+f(x-1))=2f(x) $$
$$f(x+2)+f(x-2)=0$$
Now make $x=x+2$ then $f(x+4)=-f(x)$.
Make $x+4$ and get $f(x+8)=-f(x+4)=f(x)$, then the period is $8$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2088809",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 0
} |
Find the sum of first $n$ terms of the series: $\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+\cdots +\frac{1}{n\times(n+1)}$
I have the series $$\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+\cdots +\frac{1}{n\times(n+1)}$$
I know the following formulas: $$1+2+3+\cdots +n=\frac {n (n+1)}{2}\tag1$$ $$1^2+2^2+3^2+\cdots +n^2=\frac{n(n+1)(2n+1)}{6}\tag2$$ $$1^3+2^3+3^3+\cdots +n^3=\left(\frac{n(n+1)}{2}\right)^2\tag3$$ But none of $(1)(2)$ and $(3)$ worked.
Please help___.
| Hint: We can write $$\frac {1}{n(n+1)} =\frac {1}{n} -\frac {1}{n+1} $$ Can you take it from here?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2090435",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
$4x^2-5xy+4y^2=19$ with $Z=x^2+y^2$, find $\dfrac1{Z_{\max}}+\dfrac1{Z_{\min}}$ I have no idea how to approach this question at all. I've tried to find the maximum and minimum of the quadratic but i am too confused on what to do afterwards.
| *
*From the given equation $\,xy = \cfrac{4(x^2+y^2) - 19}{5} = \cfrac{4 Z - 19}{5}\,$. But for $\forall x,y \in \mathbb{R}$ the inequality holds $\,xy \le \cfrac{1}{2}(x^2+y^2)\,$ so $\,\cfrac{4 Z - 19}{5} \le \cfrac{Z}{2} \iff \boxed{Z \le \cfrac{38}{3}}\,$
*$0 \le (x+y)^2 = x^2+y^2+2xy = Z + 2 \cfrac{4 Z - 19}{5}=\cfrac{13 Z - 38}{5} \implies \boxed{Z \ge \cfrac{38}{13}}$
$Z_{min}=\cfrac{38}{13}$ is attained for $x=-y=\sqrt{\cfrac{19}{13}}$ and $Z_{max}=\cfrac{38}{3}$ is attained for $x=y=\sqrt{\cfrac{19}{3}}\,$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2091264",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
Maximum of $f(x) =(1+\cos x)\cdot \sin(\frac{x}{2})$ on $x \in (0, \pi)$ I've attempted to solve this question by using $$f(x) = 2\sin\frac{x}{2}\cos^2\frac{x}{2} \leq \frac{(2\sin\frac{x}{2}+\cos^2\frac{x}{2})^2}{2}$$ but it results in the wrong answer every time. Is there another way to solve this question and is there anything I can do to change my method to make it work.
| When you have $2 \sin \frac{x}{2} (1-\sin^2 \frac x2) = 2\sin \frac x2 - 2\sin^3 \frac x2$, where $x$ varies between $0$ and $\pi$, hence we can let $y= \sin \frac x2$, then $y$ varies between $0$ and $1$, and we are trying to find the maximum value of $2y-2y^3$. Now, we can just use ordinary differentiation, $2-6y^2=0\implies y^2 = \frac 13$, and in this case, the value of $2y-2y^3$ is $2y(1-y^2)=\frac{2}{3} \times \frac{2}{\sqrt 3}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2091339",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
What values can $2^j-3^k$ have? What values can $2^j-3^k$ have?
E.g.,
$$
2^2-3^1=1\\
2^2-3^0=3\\
2^3-3^1=5\\
2^4-3^2=7
$$
Can every number not divisible by $2$ or $3$ be written as $2^j-3^k$? If not, why?
| we try to find an $n$ coprime to $6$ such that $2^j-3^k$ doesn't cover all options $\bmod n$.
Since we want $2^j$ to cover a small number of cases we are going to try with $n=2^m-1$.
We find that the order of $3\bmod 511$ is $12$ and the order of $2\bmod 511$ is clearly $9$.
Thus only a small fraction of residues $\bmod 511$ are covered by $2^j-3^k$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2092541",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
Solving irrational inequalities: $\sqrt{x^2-7x+10}+9\log_4{x/8}\geq 2x+\sqrt{14x-20-2x^2}-13$
Solving the following inequalities: $$\sqrt{x^2-7x+10}+9\log_4{x/8}\geq 2x+\sqrt{14x-20-2x^2}-13$$
I have been solving some questions on inequalities lately and I have come across this question, but I can't figure out a way to solve this...it contains both log and square root. Here's how I approached the question:
$\sqrt{x^2-7x+10}+9\log_4{\frac{x}{8}}\geq 2x+\sqrt{14x-20-2x^2}-13$
$\sqrt{x^2-7x+10}+9\log_4{x}-3/2\geq 2x+\sqrt{(-2)(x^2-7x+10)}-13$
$(1-2i)\sqrt{x^2-7x+10}+\geq 2x-9\log_4{x}+3/2-13$
$(1-2i)\sqrt{(x-5)(x-2)}+\geq 2x-9\log_4{x}-23/2$
I don't know how to proceed further. It would be great if I could get a hint.
| Hint:
Let $x \in \mathbb R$
If $$\sqrt{x^2-7x+10}=\sqrt{(x-2)(x-5)}$$ then $x\le 2$ or$x \ge5$
If $$\sqrt{-2(x^2-7x+10)}=\sqrt{-2(x-2)(x-5)}$$ then $2\le x \le 5$
Hence, $x=2$ or $x=5$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2093159",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Graph the inequality $1<|2z-6|<2$. How to graph the inequality $1<|2z-6|<2$ in the complex plane?
My work so far is below:
I assume $z$ is complex so it can be represented as $z=a+bi$
\begin{align}
1<|2z-6|<2 \\
1<2|(a+bi)-3|<2 \\
\frac{1}{2}<|a-3+bi|<1\\
\frac{1}{2}<|a-3+bi|<1 \\
\frac{1}{2}<\sqrt{a^2-6a+9+b^2}<1 \\
\frac{-35}{4}<a^2-6a+b^2<-8 \\
\end{align}
From there I am confused.
(Note that I am assuming that $z$ is complex as the this part of a question for a complex analysis class and we are just starting to cover the basics of complex numbers).
| hint
$$1<|2z-6|<2 \Leftrightarrow\frac{1}{2}<|z-3|<1 \Leftrightarrow \frac{1}{4}<|z-3|^2<1$$
$$\frac{1}{4}<(a-3)^2+b^2<1$$
What is:
$$(a-3)^2+b^2<1?$$
and
$$(a-3)^2+b^2>1/4?$$
what about the intersection?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2096580",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Find $xyz$ given that $x + z + y = 5$, $x^2 + z^2 + y^2 = 21$, $x^3 + z^3 + y^3 = 80$ I was looking back in my junk, then I found this:
$$x + z + y = 5$$
$$x^2 + z^2 + y^2 = 21$$
$$x^3 + z^3 + y^3 = 80$$
What is the value of $xyz$?
A) $5$
B) $4$
C) $1$
D) $-4$
E) $-5$
It's pretty easy, any chances of solving this question? I already have the
answer for this, but I didn't fully understand.
Thanks for the attention.
| x + y + z = 5
On squaring both sides,
$(x + y + z)^2 = 25$
$x^2 + y^2 + z^2 + 2xy + 2yz + 2zx = 25$
$21 + 2xy + 2yz + 2zx = 25$
$2xy + 2yz + 2zx = 25 - 21$
$2xy + 2yz + 2zx = 4$
$xy + yz + zx = 2$
Also,
$x^3 + y^3 + z^3 - 3xyz = (x + y + z)(x^2 + y^2 + z^2 - xy - yz - zx)$
Putting values,
$80 - 3xyz = (5)\left[21 - (xy + yz + zx)\right]$
80 - 3xyz = (5)(21 - 2)
80 - 3xyz = 95
-3xyz = 15
xyz = -5
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2097444",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 0
} |
Find $y'$ given $y\,\sin\,x^3=x\,\sin\,y^3$? The problem is
$$y\,\sin\,x^3=x\,\sin\,y^3$$
Find the $y'$
The answer is
Can some explain how to do this, please help.
| $y \sin x^3=x\sin y^3→$ Differentiating both sides with respect to $x .$
$⇒y(\cos x^3×3x^2)+y′\sin x^3 =\sin y^3+x \cos y^3×3y^2y′ $
$⇒3x^2y \cos x^3+y′\sin x^3=\sin y^3+3y^2x \cos y^3y′$
$⇒y′[\sin x^3−3y^2x\cos y^3]=\sin y^3−3x^2y \cos x^3$
$⇒y′=\dfrac{\sin y^3−3x^2y \cos x^3}{\sin x^3−3y^2x \cos y^3}$
I was searching it and I got the complete solution from here,
Implicit Differentiation
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2104203",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
Compute $\lim \limits_{n \to \infty} \frac{\sqrt[4]{n^4+4n}\,-\,\sqrt[3]{n^3+3n}}{\sqrt[5]{n^5+1}\,-\, \sqrt[5]{n^5+n}}\cdot \frac{1}{n^2}$ I need to compute:
$\lim \limits_{n \to \infty} \frac{\sqrt[4]{n^4+4n}\,-\,\sqrt[3]{n^3+3n}}{\sqrt[5]{n^5+1}\,-\, \sqrt[5]{n^5+n}}\cdot\frac{1}{n^2}$
I tried this
$\lim \limits_{n \to \infty} \frac{n}{n}\frac{\sqrt[4]{n^4+4n}\,-\,\sqrt[3]{n^3+3n}}{\sqrt[5]{n^5+1}\,-\, \sqrt[5]{n^5+n}}\cdot\frac{1}{n^2}=\lim \limits_{n \to \infty} \frac{\sqrt[4]{1+\frac{4}{n^3}}\,-\,\sqrt[3]{1+\frac{3}{n^2}}}{\sqrt[5]{1+\frac{1}{n^5}}\,-\, \sqrt[5]{1+\frac{1}{n^4}}}\cdot\frac{1}{n^2}$
I get stuck and I will have to probably use different method, would someone give me an advice, how to approach to this problem?
I also thought about multiplying it by $\frac{(a+b)}{(a+b)}$, but I am not really sure how.
| The answer is 5.
I expect you're supposed to use the generalized binomial theorem here: namely, $(1+x)^n = \sum_{k=0}^{\infty}\left(\frac{n^{\underline{k}}}{k!}x^k\right)$, where $n^{\underline{k}}$ means $1$ for $k=0$, or $n(n-1)(n-2)\cdots(n-k+1)$ otherwise.
$$\lim \limits_{n \to \infty} \frac{\sqrt[4]{1+\frac{4}{n^3}}\,-\,\sqrt[3]{1+\frac{3}{n^2}}}{\sqrt[5]{1+\frac{1}{n^5}}\,-\, \sqrt[5]{1+\frac{1}{n^4}}}\cdot\frac{1}{n^2} = $$
$$\lim \limits_{n \to \infty} \frac{\left(1 +\frac{1}{n^3} + \mathrm{O}(n^{-6}) \right)\,-\,\left(1+\frac{1}{n^2} + \mathrm{O}(n^{-4})\right)}{\left(1+\frac{1}{5n^5}+\mathrm{O}(n^{-10})\right)\,-\, \left({1+\frac{1}{5n^4} +\mathrm{O}(n^{-8})}\right)}\cdot\frac{1}{n^2} =$$
$$\lim \limits_{n \to \infty} \frac{5 + \mathrm{O}(n^{-3}) \,-\,5n + \mathrm{O}(n^{-1})}{1+\mathrm{O}(n^{-5})\,-\, {n +\mathrm{O}(n^{-3})}} =$$
$$\lim_{n\to\infty}\frac{5n}{n} = 5$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2104344",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Proving an alternate quadratic formula It is well known that the quadratic formula for $ax^2+bx+c=0$ is given by$$x=\dfrac {-b\pm\sqrt{b^2-4ac}}{2a}\tag1$$
Where $a\ne0$. However, I read somewhere else that given $ax^2+bx+c=0$, we have another solution for $x$ as$$x=\dfrac {-2c}{b\pm\sqrt{b^2-4ac}}\tag2$$
Where $c\ne0$. In fact, $(2)$ gives solutions for $0x^2+bx+c=0$!
Question:
*
*How would you prove $(2)$?
*Why is $(2)$ somewhat similar to $\dfrac 1{(1)}$ but with $2a$ replaced with $-2c$?
| There are a few errors in the question as posed.
First, the quadratic formula is
$$x = \frac{-b \pm \sqrt {b^2-4ac} }{2a}$$
Note the factor of $a$ in the denominator, missing from the OP. [EDIT -- I see this has been fixed now.]
Second, you write:
In fact, (2) gives solutions for $0x^2+bx+c=0$!
This is not true, and does not even make sense. The equation $0x^2+bx+c=0$ is equivalent to $bx = -c$, whose only solution is $x=-c/b$ (assuming $b\ne 0$).
In any case, to turn (1) into (2), just multiply by $-b \mp \sqrt{b^2-4ac}$ in both the numerator and the denominator:
$$x = \frac{-b \pm \sqrt {b^2-4ac} }{2a} \cdot \frac{-b \mp \sqrt{b^2-4ac}}{-b \mp \sqrt{b^2-4ac}}$$
The numerator now has the form $(X \pm Y)(Y \pm X)$, which simplifies to just $X^2-Y^2$. So we have
$$x = \frac{1}{2a} \frac{ \left( -b \right)^2 - \left(b^2-4ac\right)}{-b \mp \sqrt{b^2-4ac}}$$
Cleaning up, the $(-b)^2-b^2$ in the numerator cancels out; the $2a$ in the denominator reduces against the $4ac$ in the numerator, leaving just $2c$; and you can move a negative sign out of the denominator and into the numerator, giving the alternative form of the quadratic formula you wanted.
Finally, you ask
Why is (2) somewhat similar to $\frac{1}{(1)}$ but with $2a$ replaced with $−2c$?
As Ross Millikan says in his answer, if in the original equation $ax^2 + bx + c =0$ we assume that $x=0$ is not a solution (which is equivalent to assuming that $c\ne 0$), then we can divide the entire equation through by $x^2$, obtaining
$$c \left(\frac{1}{x}\right)^2 + b\left(\frac1x\right) + a = 0$$
If we set $u=\frac1x$, then this is $cu^2 + bu + a = 0$, and the quadratic equation tells us
$$u = \frac{ - b \pm \sqrt{b^2 - 4ac} }{2c}$$
Finally we get
$$x = \frac{2c}{-b \pm \sqrt{b^2-4ac}}$$
and you can pull the negative sign out of the denominator into the numerator. So the reason the equations are so similar is because of a duality in the equation: interchanging $a$ with $c$ and simultaneously replacing $x$ with $1/x$ changes one quadratic equation into an equivalent one.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2105240",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 4,
"answer_id": 1
} |
Find linear transformation given its matrix representation
If $$A = \left[ \begin{matrix} 1 & -1 & 2 \\
-2 & 1 & -1 \\
1 & 2 & 3
\end{matrix} \right]$$ is the matrix representation of a linear transformation $T: P_2(x) \to P_2(x)$ with respect to the bases
$\{1-x, x(1-x), x(1+x)\}$ and $\{1, 1+x, 1+x^2\}$, then find $T$.
I have tried it but could not able to get the solution.
| The relation between $A$ and $T$ is
$$A\left[a + bx + cx^2\right]_{\mathcal B} = \left[T\left(a + bx + cx^2\right)\right]_{\mathcal B'}$$
where $\mathcal B$ is the first basis, $\mathcal B'$ is the second basis, and $[ \cdot ]_{\mathcal B}$ denotes the coordinate matrix in basis $\mathcal B.$
To find $T,$ we compute as follows:
$$\begin{align}
A\left[a + bx + cx^2\right]_{\mathcal B} & = A\left[a(1 - x) + \frac{a + b - c}2 x(1 - x) + \frac{a + b + c}2 x(1 + x)\right]_{\mathcal B}\\
& = \begin{bmatrix} 1 & -1 & 2 \\ -2 & 1 & -1 \\ 1 & 2 & 3\end{bmatrix} \frac12 \begin{bmatrix} 2a\\ a + b - c\\ a + b + c\end{bmatrix}\\
& = \frac12 \begin{bmatrix} 3a + b + 3c\\ -4a - 2c\\ 7a + 5b + c\end{bmatrix}.\end{align}$$
Comparing that last expression to the right side of the first displayed equation, we see that
$$\begin{align}
T\left(a + bx + cx^2\right) & = \tfrac12 \left((3a + b + 3c) \cdot 1 + (-4a - 2c)(1 + x) + (7a + 5b + c)\left(1 + x^2\right)\right)\\
& = \tfrac12 \left((6a + 6b + 2c) - (4a + 2c)x + (7a + 5b + c)x^2\right).\end{align}$$
Thus, $T$ is given by
$$T\left(a + bx + cx^2\right) = 3a + 3b + c - (2a + c)x + \tfrac12 (7a + 5b + c)x^2.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2108758",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Probability with balls and a box - complementary event of "exactly" There are $12$ balls in a box: $b$ blue, $y$ yellow and $3$ red balls. Three balls are randomly chosen. If the probability of choosing one blue, one yellow and one red is $3/11$, find the number of yellow balls in a box.
Attempt:
If $t$ is the total number of balls in a box, then:
$t=3+b+y$
$A$: "We choose exactly one blue, yellow and red ball."
$$P(A)=\frac{b}{t}\cdot\frac{y}{t-1}\cdot\frac{3}{t-2}=\frac{3}{11}$$
Substituting $t$ gives
$$P(A)=\frac{b}{3+b+y}\cdot\frac{y}{2+b+y}\cdot\frac{3}{1+b+y}=\frac{3}{11}$$
Now we have one equation with two unknowns, so we need to define another event.
Because we already know $P(A)$, that new event should be the complementary event of $P(A)$.
How to define that event and how to evaluate it?
| We have,
b + y + 3 = 12
b + y = 9
b = 9 - y
Probability = $\frac{\binom{9 - y}1 \times \binom{y}1 \times \binom{3}1}{\binom{12}3}$
$\frac 3{11} = \frac{\frac{(9 - y)!}{(8 - y)!.1!} \times \frac{(y)!}{(y - 1)!.1!} \times \frac{3!}{2!.1!}}{\frac{12!}{3!.9!}}$
$\frac 3{11} = \frac{\frac{(9 - y)(8 - y)!}{(8 - y)!} \times \frac{(y)(y - 1)!}{(y - 1)!} \times 3}{2.11.10}$
$\frac 3{11} = \frac{(9 - y) \times (y) \times 3}{2.11.10}$
$20 = (9 - y)(y)$
$ y^2 - 9y + 20 = 0$
On solving yellow y = 4 or y = 5.
Then blue b = 5 or b = 4.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2108979",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Solve $x=\pm\frac{y'}{\sqrt{(y')^2+1}}$
$$x=\pm\frac{y'}{\sqrt{(y')^2+1}}$$
$$x=\pm\frac{y'}{\sqrt{(y')^2+1}}$$
$$x^2=\frac{(y')^2}{{(y')^2+1}}$$
$$x^2(y')^2+x^2=(y')^2$$
$$(y')^2[x^2-1]=-x^2$$
$$(y')^2=\frac{-x^2}{x^2-1}$$
$$y'=\pm \sqrt{\frac{-x^2}{x^2-1}}$$
Have I got it wrong? as there is no ODE
| Here is a solution
$$y'=\pm \sqrt {\frac{-x^2}{x^2-1}} \\ \frac{dy}{dx} =\pm \frac x {\sqrt {1-x^2}} \\ \int dy = \pm \int \frac x {\sqrt {1-x^2}} \, dx =\pm \frac 1 2 \int \frac {d(1-x^2)}{\sqrt{ 1-x^2}}=\pm \sqrt {1-x^2} \\ y=\pm \sqrt{1-x^2} +C$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2110777",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Solve the equation: $\sin 3x=2\cos^3x$ Solve the equation :
$$\sin 3x=2\cos^3x$$
my try :
$\sin 3x=3\sin x-4\sin^3x$
$\cos^2x=1-\sin^2x$
so:
$$3\sin x-4\sin^3x=2((1-\sin^2x)(\cos x))$$
then ?
| $$2\cos^3x=3\sin x-4\sin^3x$$
Divide both sides by $\cos^3x$
$$2=3\tan x(1+\tan^2x)-4\tan^3x$$
$$\tan^3x-3\tan x+2=0$$ which is a cubic equation in $\tan x$
Clearly, one of the roots is $\tan x=1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2111959",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
How to solve a system of 3 trigonometric equations How to solve system of three trigonometric equations:
$(\sin x)^2 (\cos y)^2 = 4 \cos x \sin y\tag1$
$(\sin y)^2 (\cos z)^2 = 4 \cos y \sin z \tag2$
$1- \sqrt{\sin z}(1+\sqrt{\cos x})=\sqrt{\frac{1-\sin y}{1+\sin y}}\tag3$
and to verify that
$\sin x=\sqrt{2}(\sqrt{2}+1)\sqrt{\sqrt{10}-3}(\sqrt{5}-2)\\
\sin y=(\sqrt{2}-1)^2(\sqrt{10}-3)\\
\sin z=(2\sqrt{2}+\sqrt{5}-\sqrt{12+4\sqrt{10}})^2(\sqrt{5}+\sqrt{2}-\sqrt{6+2\sqrt{10}})^2$
not is the only solution, but that many others exist?
The simplest solution is:
$\sin x=2^{5/4}\big(\sqrt{2}-1\big)$
$\sin y=\frac{1}{\sqrt{2}}$
$\sin z=3-2\sqrt{2}$.
Three other solutions are:
1)
$\sin x=\sqrt{2\sqrt{2}-2}$
$\sin y=\sqrt{2}-1$
$\sin z=5+4\sqrt{2}-\sqrt{56+40\sqrt{2}}$,
2)
$\sin x=\frac{\sqrt{2}}{2}$
$\sin y=3-2\sqrt{2}$
$\sin z=\big(\sqrt{2}+1\big)^{2}\big(2^{1/4}-1\big)^{4}$
3)
$\sin x=\frac{2^{5/4}a^{2}c. d^{6}\varphi^{3}}{b^{2}}$
$\sin y=\frac{d^4}{\sqrt{2}\varphi^{4}}$
$\sin z=a^{4}b^{4}c^{2}\varphi^{6}$
where
$$a=\sqrt{2}-1$$
$$b=5^{1/4}-\sqrt{2}$$
$$c=\sqrt{10}-3$$
$$d=\frac{5^{1/4}-1}{\sqrt{2}}$$
$\varphi$ is the golden ratio.
| This is based on modular equations of degree 2. If $k, l$ are elliptic moduli with $l$ of degree 2 over $k$ then we have $$l=\frac{1-k'}{1+k'}\tag{1}$$ where $k'=\sqrt{1-k^2}$.
One should observe that the relation between $\sin x, \sin y$ is such that $\sin y$ is of degree 2 over $\sin x$. This is easily verified by writing $k, l$ in place of $\sin x, \sin y$ in the equation given in question. Doing so we get $$k^2(1-l^2)=4k'l$$ This gives us a quadratic equation in $l$ as $$k^2l^2+4k'l-k^2=0$$ or $$l=\frac{-2k'\pm\sqrt{4k'^2+k^4}}{k^2}$$ Choosing $+$ sign as $l>0$ we get $$l=\frac{2-k^2-2k'}{k^2}=\frac{(1-k')^2}{1-k'^2}=\frac{1-k'}{1+k'}$$ as in equation $(1)$.
Similarly $\sin z$ of degree 2 over $\sin y$ and hence $\sin z$ is of degree 4 over $\sin x$. Let $l_1=\sin z$ so that $$l_1=\frac{1-l'}{1+l'}$$ Next we have $$1+\sqrt{\cos x} =1+\sqrt{k'}=1+\sqrt{\frac{1-l}{1+l}}$$ The equation to be proved can now be rewritten as $$\sqrt{\sin z} =\dfrac{1-\sqrt{\dfrac{1-l}{1+l}}} {1+\sqrt{\dfrac{1-l} {1+l} } } $$ or $$l_1=\sin z=\frac{2-2\sqrt{1-l^2} } {2+2\sqrt{1-l^2}}=\frac{1-l'}{1+l'} $$ as expected.
The given numerical values for the moduli are taken from a table of singular moduli. In particular the value $$\sin y=(\sqrt{10}-3)(\sqrt{2}-1)^2$$ is $k_{10}$ where $$k_n=\frac{\vartheta_2^2(e^{-\pi\sqrt{n}})}{\vartheta_3^2(e^{-\pi\sqrt{n}})} $$ and $$\vartheta_2(q)=\sum_{r\in\mathbb{Z}}q^{(r+(1/2))^2},\,\vartheta_3(q)=\sum_{r\in\mathbb {Z}} q^{r^2} $$ And therefore the values of $\sin x, \sin z$ are $k_{5/2},k_{40}$ respectively.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2112277",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Find two numbers whose $AM+...$ Find two numbers whose $AM + GM =25$ and $AM:GM=5:3$.
My Attempt;
Given,
$\frac {AM}{GM}=\frac {5}{3} = k (let) $
$\AM=5k,
GM=3k$.
Also,
$AM+GM=25$
$5k+3k=25$
$8k=25$
$k=\frac {25}{8}$.
Am I going right? Or, is there any other simple alternative.?
| Let the numbers be $a^2$ and $b^2$.
Then we have :
$$
\frac{a^2+b^2}{2} +ab=25
\implies (a+b)^2=50
\implies a+b=5\sqrt{2}
$$
Let $AM$ be $5x$ and $GM$ be $3x$. Then, $x=\frac{25}{8}$ and the $AM$ is $\frac{125}{8}$ and the $GM$ is $\frac{75}{8}$.
Hence
$$
a^2+b^2=2AM=\frac{125}{4} \\
2ab=2GM=\frac{75}{4} \\
a^2+b^2-2ab=\frac{125-75}{4}\\
a-b=\sqrt{\frac{50}{4}}=\frac{5\sqrt{2}}{2}
$$
Find $a$ and $b$ from the two equations and $a^2$ and $b^2$ are your answers.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2113371",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Help Determinant Binary Matrix I was messing around with some matrices and found the following result.
Let $A_n$ be the $(2n) \times (2n)$ matrix consisting of elements $$a_{ij} = \begin{cases} 1 & \text{if } (i,j) \leq (n,n) \text{ and } i \neq j \\ 1 & \text{if } (i,j) > (n,n) \text{ and } i \neq j \\ 0 & \text{otherwise}. \end{cases} $$
Then, the determinant of $A_n$ is given by $$\text{det}(A_n) = (n-1)^2.$$
Example: $$A_2 = \begin{pmatrix} 0 & 1 & 0 & 0 \\ 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 0 \end{pmatrix}, A_3 = \begin{pmatrix} 0 & 1 & 1 & 0 & 0 & 0 \\ 1 & 0 & 1 & 0 & 0 & 0 \\ 1 & 1 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 1 & 1 \\ 0 & 0 & 0 & 1 & 0 & 1 \\ 0 & 0 & 0 & 1 & 1 & 0 \\ \end{pmatrix},$$ with det$(A_2)$ and det$(A_3)$ being $1$ and $4$ respectively. I was wondering if anybody could prove this statement for me.
| The matrix is diagonal by blocks. so the determinant is equal to the product of the determinants of each block.
So we just have to prove that:
$\det\begin{pmatrix}0 & 1 & \dots & 1 \\ 1 & 0 & \dots & 1 \\ & &\vdots \\ 1 & 1 & \dots & 0\\ \end{pmatrix}=\pm(n-1)$
This is done in various ways here, in fact it is equal to $(-1)^n(n-1)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2113878",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
$\tan {\frac{A}{2}} + \tan {\frac{B}{2}} +\tan{\frac{C}{2}} \geq 4 - \sqrt {3} $
In a triangle $ABC$ with one angle exceeding $\frac {2}{3} \pi$, prove that
$\tan {\frac{A}{2}} + \tan {\frac{B}{2}} + \tan{\frac{C}{2}} \geq 4 - \sqrt {3} $
I tried expanding that half angle, applying AM-GM on various sets, using Sine rule and Napier's Analogy, but without success.
Can anyone provide a hint ?
Also, how does the left hand side of the inequality behave when the condition of one angle exceeding $\frac {2} {3 }\pi$ is removed?
Thanks in advance :) .
| Let $ \tan\dfrac A2 =x, \tan\dfrac B2=y, \tan\dfrac C2=z,\tan \dfrac{A}{2} + \tan \dfrac{B}{2} + \tan\dfrac{C}{2} = w$. WLOG $x \ge \sqrt{3}$.
Note that we have the identity $\tan\left(\dfrac{A+B}2\right)=\tan\left(\dfrac{\pi- C}2\right)$
$$\implies\dfrac{x+y}{1-xy}=\dfrac1{z}$$
$$\implies xy+yz+zx=1 \le \frac{(y+z)^2}{4}+x(w-x)=\frac{(w-x)^2}{4}+x(w-x) $$
This gives us $$w \ge \sqrt{4+4x^2}-x \ge 4- \sqrt{3}$$
From the fact that $\sqrt{4+4x^2}-x $ is an increasing function.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2114781",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Proving a well-known inequality using S.O.S Using $AM-GM$ inequality, it is easy to show for $a,b,c>0$, $$\frac{a}{b} + \frac{b}{c} + \frac{c}{a} \ge 3.$$
However, I can't seem to find an S.O.S form for $a,b,c$
$$f(a,b,c) = \frac{a}{b} + \frac{b}{c} + \frac{c}{a} - 3 = \sum_{cyc}S_A(b-c)^2 \ge 0.$$
Update:
Please note that I'm looking for an S.O.S form for $a, b, c$, or a proof that there is no S.O.S form for $a, b, c$. Substituting other variables may help to solve the problem using the S.O.S method, but those are S.O.S forms for some other variables, not $a, b, c$.
| Let $c=\min\{a,b,c\}$. Hence,
$$\frac{a}{b}+\frac{b}{c}+\frac{c}{a}-3=\frac{a}{b}+\frac{b}{a}-2+\frac{b}{c}+\frac{c}{a}-\frac{b}{a}-1=\frac{(a-b)^2}{ab}+\frac{(c-a)(c-b)}{ac}\geq0$$
From here we can get a SOS form:
$$\frac{a}{b}+\frac{b}{c}+\frac{c}{a}-3=\frac{1}{6abc}\sum_{cyc}(a-b)^2(3c+a-b)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2116233",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
"answer_id": 0
} |
Find all prime solutions of equation $5x^2-7x+1=y^2.$ Find all prime solutions of the equation $5x^2-7x+1=y^2.$
It is easy to see that
$y^2+2x^2=1 \mod 7.$ Since $\mod 7$-residues are $1,2,4$ it follows that $y^2=4 \mod 7$, $x^2=2 \mod 7$ or $y=2,5 \mod 7$ and $x=3,4 \mod 7.$
In the same way from $y^2+2x=1 \mod 5$ we have that $y^2=1 \mod 5$ and $x=0 \mod 5$ or $y^2=-1 \mod 5$ and $x=4 \mod 5.$
How put together the two cases?
Computer find two prime solutions $(3,5)$ and $(11,23).$
| Since $(0,1)$ is a solution to $5x^2-7x+1=y^2$, it can be used to parametrize all rational solutions to $5x^2-7x+1=y^2$. That will give us:
$$x:=\frac{-2ab - 7b^2}{a^2 - 5b^2}$$
and
$$y:=\frac{a}{b}x+1$$
where $a,b\in \mathbb{Z}$, $b\neq 0$.
Since $x$ is prime, it follows that either $b=1$ or $x=b$.
*
*case $b=1$
This gives us $x:=(-2a - 7)/(a^2 - 5)$. Since $x\in \mathbb{Z}$,
we get $a=\pm 2$, and then $x=11$ or $x=3$.
Those will give $y=23$ or $y=5$, repectively.
*case $x=b$.
We have that $\frac{-2ab - 7b^2}{a^2 - 5b^2}=b$ gives
$$(*) \hspace{2cm} a(a + 2) =5b^2 - 7b.$$
Since $y=a+1$ and $x=b$ are prime, and $x=2$ or $y=2$ do not give solutions to $5x^2-7x+1=y^2$, we conclude that $y=a+1$ and $x=b$ are ODD primes. In particular, $a(a + 2)\equiv 0 \mod 4$.
Now reducing (*) $\mod 4$, contradicts the fact that $b$ is odd. Thus, the case $x=b$ does not occur.
Therefore $(x,y)= (11,23)$ and $(x,y)=(3,5)$ are the only solutions
where both coordinates are prime numbers.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2117054",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 5,
"answer_id": 0
} |
Quadratic equation, find $1/x_1^3+1/x_2^3$ In an exam there is given the general equation for quadratic: $ax^2+bx+c=0$.
It is asking: what does $\dfrac{1}{{x_1}^3}+\dfrac{1}{{x_2}^3}$ equal?
| Let $\dfrac1{x^3}=y\iff x^3=?$
$$(-c)^3=(ax^2+bx)^3\iff -c^3=a^3(x^3)^2+b^3(x^3)+3ab(x^3)(-c)$$
$$\iff-c^3=\dfrac{a^3}{y^2}+\dfrac{b^3}y-\dfrac{3abc}y$$
$$\iff c^3y^2+(b^3-3abc)y+a^3=0$$ whose roots are $\dfrac1{x_1^3},\dfrac1{x_2^3}$
Can you apply Vieta's formula now?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2119399",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
How to use finite differences to determine an equation of a polynomial given consecutive integer $x$ and corresponding $y$ coordinates of the graph? This chart is given:
for $x=-3$, $y=-9$
for $x=-2$, $y=3$
for $x=-1$, $y=3$
for $x=0$, $y=-3$
for $x=1$, $y=-9$
for $x=2$, $y=-9$
for $x=3$, $y=3$
I found the finite differences to be 6 and degree of the polynomial with the given points to be 3 ($n=3$). But how do I find the polynomial function rule out of this information? Given no factors? Please help with a short solution. Thanks a lot in advance. I also did $y=kx^3$, $-9=k(-3)^3$, $k=1/3$. But of course the rule isn't $y=1/3x^3$.
| Once you discover that the third difference is constant you know that the polynomial is third degree. Denote it by $P(x)$.
We can see that the horizontal lines $y=3$ and $y=-9$ cross the graph of $y=P(x)$ three times each. Therefore the graph of the polynomial $Q(x)=P(x)-3$ has $x$-intercepts $(-2,0),\,(-1,0)$ and $(3,0)$. Therefore for some leading coefficient $a$
\begin{eqnarray}
P(x)-3&=&a(x+2)(x+1)(x-3)\\
P(x)&=&a(x+2)(x+1)(x-3)+3\\
&=&ax^3-7ax-6a+3
\end{eqnarray}
We know that when $x=0$, $y=-3$ so $a=1$. Therefore
$$P(x)=x^3-7x-3$$
The same result can be had using the three values of $-9$ since the graph of $y=P(x)+9$ has $x$-intercepts $(-3,0),\,(1,0)$ and $(2,0)$. Therefore
\begin{eqnarray}
P(x)+9&=&a(x+3)(x-1)(x-2)\\
P(x)&=&a(x+3)(x-1)(x-2)-9\\
&=&ax^3-7ax+6a-9
\end{eqnarray}
In this case, $6a-9=-3$ so $a=1$. And we obtain the same result.
Note: This approach works only when an $n$th degree polynomial sequence $\{y_k\}$ has $n$ values of $x_k$ for which the $y_k$ values are equal. In general you need the method used by Ahmed S. Attaalla in his answer.
Suppose $P$ is a degree $n$ polynomial with $P(x_k)=y_k$ and that for $n$ different values of $k$ we have $y_k=c$. Then for those same $n$ values of $k$ it will be true that the polynomial $Q(x_k)=P(x_k)-c$ will have $y_k=0$ since for those values of $k$, $Q(x_k)=P(x_k)-c=c-c=0$. Knowing the $n$ $x$-intercepts of $Q(x)$ allows us to factor $Q(x)$. Adding $c$ to $Q(x)$ gives us $P(x)$. We can then use one of the given values of $(x,P(x))$ to find the value of $a$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2120280",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Fourier Series/ fourier transform What is the Fourier series of the following piece-wise function?
$$
f(x) = \begin{cases}
0 & -1 \leq x < -0.5 \\
\cos (3 \pi x) & -0.5 < x < 0.5 \\
0 & 0.5 \leq x < 1
\end{cases}
$$
| Define problem
Piecewise function: Resolve $f(x)$ into a left, center, and right piece
$$
\begin{align}
%
l(x) &= 0, \qquad \qquad \, -1 \le x < -\frac{1}{2} \\
%
c(x) &= \cos \left( 3\pi x \right) \quad\ -\frac{1}{2} \le x \le \frac{1}{2} \\
%
r(x) &= 0, \qquad \qquad \ \ \ \frac{1}{2} \le x \le 1
%
\end{align}
$$
The length of the domain is $2$.
Find the Fourier expansion
$$
f(x) = \frac{1}{2}a_{0} +
\sum_{k=1}^{\infty} \left(
a_{k} \cos \left( k \pi x \right) +
\color{gray}{b_{k} \sin \left( k \pi x \right)}
\right)
$$
where the amplitudes are given by
$$
\begin{align}
%
a_{0} &= \int_{-1}^{1} f(x) dx \\
%
a_{k} &= \int_{-1}^{1} f(x) \cos \left( k \pi x \right) dx \\
%
\color{gray}{b_{k}} &= \color{gray}{\int_{-1}^{1} f(x) \sin \left( k \pi x \right) dx} \\
%
\end{align}
$$
Basic integrals
Center piece
$$
\begin{align}
%
\int_{-\frac{1}{2}}^{\frac{1}{2}} c(x) dx &= -\frac{2}{3 \pi } \\
%
\int_{-\frac{1}{2}}^{\frac{1}{2}} c(x) \sin \left( k \pi x \right) dx &= 0 \\
\end{align}
$$
For $m\ne n$,
$$
\int \cos (m \pi x) \cos (n \pi x) \, dx =
\left( 2\pi \right)^{-1}
\left(
\frac{\sin (\pi (m-n) x )}{m-n}+\frac{\sin (\pi (m+n) x )}{m+n}
\right)
$$
$$
\int \cos (3 \pi x) \cos (3 \pi x) \, dx = \frac{x}{2}+\frac{\sin (6 \pi x)}{12 \pi }
$$
Amplitudes
$$
\begin{align}
a_{0} &= -\frac{2}{3 \pi }, \\
a_{3} &= \frac{1}{2}.
\end{align}
$$
For $k=1,2,\dots..$
$$
a_{2k} = \frac{6 \cos \left( \pi k \right)}{\pi \left((2k)^2-9\right)}
$$
Convergence sequence
The parameter $n$ represents the highest frequency term in the series,
$$
f(x) = \frac{1}{2}a_{0} +
\sum_{k=1}^{n}
a_{k} \cos \left( \frac{k \pi x}{2} \right)
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2120851",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Find the value of $x$ that satisfy the equation: $3^{11}+3^{11}+3^{11} = 3^x$ I have this question:
$$3^{11}+3^{11}+3^{11} = 3^x$$
Find the value of $x$
| Answer:
$$3^{11}+3^{11}+3^{11} = 3^x\implies 3^{10}(3^{1} + 3^{1} + 3^{1}) =3^x$$
$$(3^1 + 3^1+ 3^1) = \frac {3^x} {3^{10}}$$
$$9 = \frac{3^x}{3^{10}}$$
$$3^2 = \frac {3^x}{3^{10}}$$
Hence:
$$2 = x -10$$
$$x = 12$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2122390",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Common roots of a quadratic equation If the quadratic equations, $ x^2
+ bx + c = 0$ and $ bx^2
+ cx + 1 = 0 $ have a common root then we have to prove that either $b + c + 1 = 0$ or $ b^2
+ c^2 + 1 = bc + b + c$.
I tried it a lot , but not able to proceed further .
Can anybody provide me a hint ?
| We have that both these equations have a common root. Solving these equations for $x^2$ and $x $, we get, $$x^2=\frac {b-c^2}{c-b^2} \text { and } x =\frac {bc-1}{c-b^2} $$ Now eliminating $x $, we get, $$b^3+c^3+1-3bc=0$$ $$\Rightarrow (b+c)^3-3bc (b+c) +1 -3bc =0$$ $$\Rightarrow (b+c+1)(b^2-(c+1)b +(c^2-c+1))=0$$ and the result follows. Hope it helps.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2122484",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Find the determinant of order $100$ Find the determinant of order $100$:
$$D=\begin{vmatrix}
5 &5 &5 &\ldots &5 &5 &-1\\
5 &5 &5 &\ldots &5 &-1 &5\\
5 &5 &5 &\ldots &-1 &5 &5\\
\vdots &\vdots &\vdots &\ddots &\vdots &\vdots &\vdots\\
5 &5 &-1 &\ldots &5 &5 &5\\
5 &-1 &5 &\ldots &5 &5 &5\\
-1 &5 &5 &\ldots &5 &5 &5
\end{vmatrix}$$
I think I should be using recurrence relations here but I'm not entirely sure how that method works. I tried this:
Multiplying the first row by $(-1)$ and adding it to all rows:
$$D=\begin{vmatrix}
5 &5 &5 &\ldots &5 &5 &-1\\
5 &5 &5 &\ldots &5 &-1 &5\\
5 &5 &5 &\ldots &-1 &5 &5\\
\vdots &\vdots &\vdots &\ddots &\vdots &\vdots &\vdots\\
5 &5 &-1 &\ldots &5 &5 &5\\
5 &-1 &5 &\ldots &5 &5 &5\\
-1 &5 &5 &\ldots &5 &5 &5
\end{vmatrix}=\begin{vmatrix}
5 &5 &5 &\ldots &5 &5 &-1\\
0 &0 &0 &\ldots &0 &-6 &6\\
0 &0 &0 &\ldots &-6 &0 &6\\
\vdots &\vdots &\vdots &\ddots &\vdots &\vdots &\vdots\\
0 &0 &-6 &\ldots &0 &0 &6\\\
0 &-6 &0 &\ldots &0 &0 &6\\
-6 &0 &0 &\ldots &0 &0 &6
\end{vmatrix}$$
Applying Laplace's method to the first column
$$D=5\begin{vmatrix}
0 &0 &\ldots &0 &-6 &6\\
0 &0 &\ldots &-6 &0 &6\\
\vdots &\vdots &\ddots &\vdots &\vdots &\vdots\\
0 &-6 &\ldots &0 &0 &6\\
-6 &0 &\ldots &0 &0 &6
\end{vmatrix}+6\begin{vmatrix}
5 &5 &\ldots &5 &5 &-1\\
0 &0 &\ldots &0 &-6 &6\\
0 &0 &\ldots &-6 &0 &6\\
\vdots &\vdots &\ddots &\vdots &\vdots &\vdots\\
0 &-6 &\ldots &0 &0 &6\\
-6 &0 &\ldots &0 &0 &6
\end{vmatrix}$$
I can see that this one is $D$ but of order $99$...Is this leading anywhere? How would you solve this?
| Starting from
$$D=\begin{vmatrix}
5 &5 &5 &\ldots &5 &5 &-1\\
5 &5 &5 &\ldots &5 &-1 &5\\
5 &5 &5 &\ldots &-1 &5 &5\\
\vdots &\vdots &\vdots &\ddots &\vdots &\vdots &\vdots\\
5 &5 &-1 &\ldots &5 &5 &5\\
5 &-1 &5 &\ldots &5 &5 &5\\
-1 &5 &5 &\ldots &5 &5 &5
\end{vmatrix}=\begin{vmatrix}
5 &5 &5 &\ldots &5 &5 &-1\\
0 &0 &0 &\ldots &0 &-6 &6\\
0 &0 &0 &\ldots &-6 &0 &6\\
\vdots &\vdots &\vdots &\ddots &\vdots &\vdots &\vdots\\
0 &0 &-6 &\ldots &0 &0 &6\\\
0 &-6 &0 &\ldots &0 &0 &6\\
-6 &0 &0 &\ldots &0 &0 &6
\end{vmatrix}$$
you can add each column to the last column to obtain
$$ D=\begin{vmatrix}
5 &5 &5 &\ldots &5 &5 &-1+99\times5\\
0 &0 &0 &\ldots &0 &-6 &0\\
0 &0 &0 &\ldots &-6 &0 &0\\
\vdots &\vdots &\vdots &\ddots &\vdots &\vdots &\vdots\\
0 &0 &-6 &\ldots &0 &0 &0\\\
0 &-6 &0 &\ldots &0 &0 &0\\
-6 &0 &0 &\ldots &0 &0 &0 \end{vmatrix}.
$$
Now the matrix is upper-tridiagonal (along the not so conventional diagonal I have to admit). By Leibniz formula (or Laplace along the last column), it can be seen that the determinant is the product of the diagonal elements. This yields
$$D= (-1+99\times5)(-6)^{99} =-494\times 6^{99}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2122803",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 3
} |
finding limit with $\cos$ function occur $n$ times Finding $\displaystyle \lim_{x\rightarrow 0}\frac{1-\cos(1-\cos(1-\cos(1-\cdots \cdots (1-\cos x))))}{x^{2^n}}$
where number of $\cos$ is $n$ times
when $x\rightarrow 0$ then $\displaystyle 1-\cos x = 2\sin^2 \frac{x}{2} \rightarrow 2\frac{x}{2} = x$
so $1-\cos (1-\cos x) = 1-\cos x$
some help me., thanks
| Take the simple case
$$\frac{1-\cos(1-\cos x)}{x^4}$$
$$=\frac{1-\cos(1-\cos x)}{(1-\cos x)^2}\left(\frac{1-\cos x}{x^2}\right)^2$$
$$\rightarrow \frac{1}{2}\left(\frac{1}{2}\right)^2$$
Can you see how to do the induction ?
So you have in general if $f(n)=1-\cos(1-\cos(1-\cos(1-\cdots \cdots (1-\cos x))))$
$$\frac{f(n)}{x^{2^n}}=\frac{f(n)}{f(n-1)^2}\left(\frac{f(n-1)}{x^{2^{n-1}}}\right)^2\rightarrow \frac{1}{2}L_{n-1}^2$$
And if we write $L_n=\left(\frac{1}{2}\right)^{e_n}$
we get the recurrence relation
$$e_n=2e_{n-1}+1$$ which means
$$e_n=2^n-1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2124023",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
Let $a,b,c$ be positive real numbers such that $abc =1$ Let $a,b,c$ be positive real numbers such that $abc=1$. Prove that $$a^2+b^2+c^2\geq a+b+c$$.
Also, state the condition for equality.
My Attempt,
$a,b,c$ are real and positive numbers, then
$$(a-1)^2+(b-1)^2+(c-1)^2\ge 0$$
$$a^2-2a+1+ b^2-2b+1+c^2-2c+1\ge 0$$
$$a^2+b^2+ c^2-2(a+b+c)+3\ge 0$$.
I have made a start in this way, but I am not sure if this works. Please help me, with any simple and beautiful method.
| Note that from $$(a-1)^2+(b-1)^2+(c-1)^2 \ge 0$$we have $$a^2+b^2+ c^2 \ge 2(a+b+c)-3$$
By AM-GM $$a+b+c \ge 3 \sqrt[3]{abc}=3 \implies a+b+c-3 \ge 0$$
So $$a^2+b^2+ c^2 \ge 2(a+b+c)-3=a+b+c+(a+b+c-3) \ge a+b+c$$
EDIT
Here is a simple proof of AM-GM when $n=3$. We will prove $$\frac{a+b+c}{3} \ge \sqrt[3]{abc}$$
Let $x=\sqrt[3]{a}, y=\sqrt[3]{b}, z=\sqrt[3]{c}$. The problem is equivalent to proving $$x^3+y^3+z^3-3xyz \ge 0$$
From here, we know $$x^3+y^3+z^3-3xyz=(x+y+z)(x^2+y^2+z^2-xy-yz-zx)$$
This can simplify to $$x^3+y^3+z^3-3xyz=\frac{1}{2}(x+y+z)((x-y)^2+(y-z)^2+(z-x)^2) \ge 0$$
So we are done.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2125952",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Prove the polynomial $f(x) = 3x^3+x^2+2x+1155$ has no root in $\mathbb{Z}$ Prove the polynomial $f(x) = 3x^3+x^2+2x+1155$ has no root in $\mathbb{Z}$
The hint says If $f$ has a root in $\mathbb{Z}$, then $f$ has also a root in $\mathbb{Z}/2\mathbb{Z}$. I'm still confused
| If $f(x) = 0$ then $f(x) \equiv 0 \mod n$ for all $n$.
So $f(x) \equiv 0 \mod 2$.
So $3x^3+x^2+2x+1155 \equiv 0 \mod 2$
$x^3 + x^2 + 1 \equiv 0 \mod 2$
$x^3 + x^2 \equiv 1 \mod 2$
Notice $0^k \equiv 0 \mod n$ and $1^k \equiv 1 \mod n$ and and $x$ is either $x \equiv 0 \mod 2$ or $x \equiv 1 \mod2$ we have $x^k \equiv x \mod 2$.
So
$x^3 + x^2 \equiv x+x = 2x \equiv 0 \mod 2$
This contradicts $x^3 + x^2 \equiv 1 \mod2$.
....
Actually if we had started with the observation $x^k \equiv x \mod 2$ we could have done simply
$3x^3 + x^2 +2x + 1155 \equiv$
$3x + x +2x +1155 \equiv$
$6x + 1155 \equiv$
$1155 \equiv$
$1 \not \equiv 0 \mod 2$.
....
Alternatively if we never made the observation $x^k \equiv x \mod 2$ we could have broken
$x^3 + x^2 \equiv 1 \mod 2$ into two cases: $x \equiv 1 \mod 2$ and $x\equiv 0 \mod 2$ and not Case 1 yields $1 + 1 \equiv 0 \mod 2$ and the second $0+0 \equiv 0 \mod 2$.
Or we could have noted:
$x^3 + x^2 = x^2(x+1)$ and note them most be of different parity.
.....
Or if on this $\mod 2$ notation doesn't same natural, we could have simply said from the start.
If $x \in \mathbb Z$ either $x$ is even or odd.
If $x$ is even $mx^k$ is even so
$3x^2 + x^2 + 2x + 1115 = even + even + even + odd = odd \ne 0$.
If $x$ is odd then $mx^k$ is odd if $m$ is odd and $mx^k$ is even if $m$ is even so
$3x^2 + x^2 + 2x + 1115 = odd + odd + even + odd = odd \ne 0$.
=====
And if you want to get really annoying...
Let $x = 2y + i$ where $i$ is either $1$ or $0$.
$f(x) = 3(2y+i)^3 + (2y + i)^2 + 2(2y+i) + (2k+1)$ where $k = \frac {1155 -1}2$.
$= EVEN + 3i^3 + EVEN + i^2 + EVEN + 1$
$= EVEN + (3i^3 + i^2) + 1$
$= EVEN + (\{0 + 0|3 + 1\}) + 1$
$= EVEN + 1$
$= ODD \ne 0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2126843",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
If:$(\sqrt{x + 9})^{\frac{1}{3}} - (\sqrt{x-9})^{\frac{1}{3}} = 3$, Find $x^2$
If:$$\sqrt[3]{(x + 9)} - \sqrt[3]{(x-9)} = 3$$
Find $x^2$
I can't seem to solve this question. Any hints or solutions is welcomed.
| Put $a=(x+9)^{1/3},b=(x-9)^{1/3}$
$$a-b=3\\(a-b)^3=3^3\\a^3-3ab(a-b)+b^3=27$$
Now you already know that $a-b=3$ so
$$a^3-9ab-b^3=27$$
Now plugging back $a,b$
$$((x+9)^{1/3})^3-9(x+9)^{1/3}(x-9)^{1/3}-((x-9)^{1/3})^3=27\\x+9-9(x^2-81)^{1/3}-(x-9)=27\\18-9(x^2-81)^{1/3}=27\\(x^2-81)^{1/3}=-1\\x^2-81=-1\\x^2=80$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2127400",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
Solving a univariable identity that satisfies the following relationship:
Question: Consider the following equations:$$\begin{align*}1^2+2^2+2^3 & =3^2\\2^2+3^2+6^2 & =7^2\\3^2+4^2+12^2 & =13^2\\4^2+5^2+20^2 & =21^2\end{align*}$$
State a one variable identity that is suggested by these examples.
Since the question asked for a uni-variable identity, I assumed the form$$(ax+b)^2+(cx+d)^2+(ex+f)^2=(gx+h)^2\tag1$$
And equated coefficients to get an undermined system. Namely,$$\begin{align*} & a^3+c^3+e^3=g^3\\ & ab+cd+ef=gh\\ & b^2+d^2+f^2=h^2\end{align*}$$
And solving for integer solutions (I set $(a,c,e,g)=(1,2,2,3)$ and solved the remaining system) to get$$(x+2)^2+(2x+4)^2+(2x+4)^2=(3x+6)^2$$
Which works for $x\in\mathbb{Z}$. However, the question asked for an identity that gave the examples listed above. Something my formula clearly isn't capable of.
So my actual question is simple: How would you go about solving this problem?
| Notice that the $2$nd and $3$rd term are respectively $2,2$ then $3,6$ then $4,12$ then $5,20$.Now dividing the $3$rd with the second you get $1,2,3,4$ respectively you can guess that $3$rd term is $n$ times bigger then $2$nd and you can see that the $2$nd is $(n+1)$.Also the $4$th term is $3$rd minus one.
Putting that together on has
$$n^2+(n+1)^2+(n(n+1))^2=(n(n+1)+1)^2$$
As given by S.C.B
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2128295",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
help prove inequality when $3\left( {{a^4} + {b^4} + {c^4}} \right) + 10 = 7\left( {{a^2} + {b^2} + {c^2}} \right)$ Given the positive real numbers $a,b,c$ satisfy $3\left( {{a^4} + {b^4} + {c^4}} \right) + 10 = 7\left( {{a^2} + {b^2} + {c^2}} \right)$. Prove that $$\frac{{{a^2}}}{{b + 2c}} + \frac{{{b^2}}}{{c + 2a}} + \frac{{{c^2}}}{{a + 2b}} \ge \frac{{16}}{{27}}$$
AM-GM we have $\frac{{{a^2}}}{{b + 2c}} +\dfrac{a^2(b+2c)}{9} \geq \dfrac{2}{3}a^2$
$=>P \geq \dfrac{2}{3}(a^2+b^2+c^2)-\dfrac{1}{9}(a^2b+b^2c+c^2a+2(ca^2+ab^2+bc^2))$
Cauchy-Schwarz $a^2b+b^2c+c^2a \leq \sqrt{(a^2+b^2+c^2)(a^2b^2+b^2c^2+c^2a^2)} \leq \dfrac{1}{\sqrt{3}}(a^2+b^2+c^2)\sqrt{a^2+b^2+c^2}$
And $ab^2+bc^2+ca^2 \leq \dfrac{1}{\sqrt{3}}(a^2+b^2+c^2)\sqrt{a^2+b^2+c^2}$
$=>P \geq \dfrac{2}{3}t^2-\dfrac{\sqrt3}{9}t^3$
$3\left( {{a^4} + {b^4} + {c^4}} \right) + 10 = 7\left( {{a^2} + {b^2} + {c^2}} \right)$=>$3\left( {{a^4} + {b^4} + {c^4}} \right) + 10 = 7\left( {{a^2} + {b^2} + {c^2}} \right) \geq (a^2+b^2+c^2)^2+10$ $=>2 \leq t^2 \leq 5$
We have
$f(t)=\dfrac{2}{3}t^2-\dfrac{\sqrt3}{9}t^3$
$f'(t)=\dfrac{4t}{3}-\dfrac{\sqrt3t^2}{3}$
We have $Minimize=\dfrac{\sqrt{6}}{3}$ but "=" ... (But I can't continue, help me)
| By C-S $$\sum_{cyc}\frac{a^2}{b+2c}\geq\frac{(a+b+c)^2}{\sum\limits_{cyc}(b+2c)}=\frac{a+b+c}{3}.$$
Thus, it remains to prove that $$a+b+c\geq\frac{16}{9}$$ or
$$\sum_{cyc}\left(a-\frac{16}{27}+\frac{1}{2}\left(3a^4-7a^2+\frac{10}{3}\right)\right)\geq0$$ or
$$\sum_{cyc}(81a^4-189a^2+54a+58)\geq0,$$
which is true because
$$81a^4-189a^2+54a+58=27(3a^4-7a^2+2a+2)+4=27(a-1)^2(3a^2+6a+2)+4>0.$$
Done!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2128974",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
How to integrate $\int_{0}^{1} \frac{1-x}{1+x} \frac{dx}{\sqrt{x^4 + ax^2 + 1}}$? The question is how to show the identity
$$ \int_{0}^{1} \frac{1-x}{1+x} \cdot \frac{dx}{\sqrt{x^4 + ax^2 + 1}} = \frac{1}{\sqrt{a+2}} \log\left( 1 + \frac{\sqrt{a+2}}{2} \right), \tag{$a>-2$} $$
I checked this numerically for several cases, but even Mathematica 11 could not manage this symbolically for general $a$, except for some special cases like $a = 0, 1, 2$.
Addendum. Here are some backgrounds and my ideas:
*
*This integral came from my personal attempt to find the pattern for the integral
$$ J(a, b) := \int_{0}^{1} \frac{1-x}{1+x} \cdot \frac{dx}{\sqrt{1 + ax^2 + bx^4}}. $$
This drew my attention as we have the following identity
$$ \int_{0}^{\infty} \frac{x}{x+1} \cdot \frac{dx}{\sqrt{4x^4 + 8x^3 + 12x^2 + 8x + 1}} = J(6,-3), $$
where the LHS is the integral from this question. So establishing the claim in this question amounts to showing that $J(6,-3) = \frac{1}{2}\log 3 - \frac{1}{3}\log 2$, though I am skeptical that $J(a, b)$ has a nice closed form for every pair of parameters $(a, b)$.
*A possible idea is to write
\begin{align*}
&\int_{0}^{1} \frac{1-x}{1+x} \cdot \frac{dx}{\sqrt{x^4 + ax^2 + 1}} \\
&\hspace{5em}= \int_{0}^{1} \frac{(x^{-2} + 1) - 2x^{-1}}{x^{-1} - x} \cdot \frac{dx}{\sqrt{(x^{-1} - x)^2 + a + 2}}
\end{align*}
This follows from a simple algebraic manipulation. This suggests that we might be able to apply Glasser's master theorem, though in a less trivial way.
I do not believe that this is particularly hard, but I literally have not enough time to think about this now. So I guess it is a good time to seek help.
| $$\int_0^1 \frac{1-x}{1+x}\frac{dx}{\sqrt{x^4+ax^2+1}}\overset{\large \frac{1-x}{1+x}=t}=\int_0^1 \frac{2t}{\sqrt{(a+2)t^4-2(a-6)t^2+(a+2)}}dt$$
$$\overset{t^2=x}=\frac{1}{\sqrt{a+2}}\int_0^1 \frac{dx}{\sqrt{x^2-2\left(\frac{a-6}{a+2}\right)x+1}}=\frac{1}{\sqrt{a+2}}\ln \left(1+\frac{\sqrt{a+2}}{2}\right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2129537",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "20",
"answer_count": 2,
"answer_id": 1
} |
Is there a systematic way to find irreducible polynomials? I am preparing for an exam and I'm looking for a systematic way to find all irreducible polynomials. I am given the following problem > Find all irreducible polynomials of degree 3 with coefficients over Z2.
I know that a polynomial is reducible if it can be represented as a product of two or more polynomials, but I can't find an efficient algorithm which can help me to list all the irreducible polynomials.
Any suggestions?
| If $f(x)\in \mathbb Z_2[x]$ and has degree $3$,then $f(x)=x^3+ax^2+bx+c$ and we have the possible states:
$a=0,b=0,c=0 \Rightarrow f(x)=x^3$
$a=1,b=0,c=0 \Rightarrow f(x)=x^3+x^2$
$a=1,b=1,c=0 \Rightarrow f(x)=x^3+x^2+x$
$a=1,b=1,c=1 \Rightarrow f(x)=x^3+x^2+x+1$
$a=1,b=0,c=1 \Rightarrow f(x)=x^3+x^2+1$
$a=0,b=1,c=0 \Rightarrow f(x)=x^3+x$
$a=0,b=1,c=1 \Rightarrow f(x)=x^3+x+1$
$a=0,b=0,c=1 \Rightarrow f(x)=x^3+1$
$f(x)$ is irreducible polynomial in $\mathbb Z_2[x]$ if $0$ or $1$ isnot roots, then $f(x)=x^3+x^2+1$ and $f(x)=x^3+x+1$ are irreducible polynomials in $\mathbb Z_2[x]$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2131638",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Evaluate the integral $\int \frac{x^2(x-2)}{(x-1)^2}dx$
Find $$\int \frac{x^2(x-2)}{(x-1)^2} dx .$$
My attempt:
$$\int \frac{x^2(x-2)}{(x-1)^2}dx = \int \frac{x^3-2x^2}{x^2-2x+1}dx $$
By applying polynomial division, it follows that
$$\frac{x^3-2x^2}{x^2-2x+1} = x + \frac{-x}{x^2-2x+1}$$
Hence $$\int \frac{x^3-2x^2}{x^2-2x+1}dx = \int \left(x + \frac{-x}{x^2-2x+1}\right) dx =\int x \,dx + \int \frac{-x}{x^2-2x+1} dx \\ = \frac{x^2}{2} + C + \int \frac{-x}{x^2-2x+1} dx $$ Now using substitution $u:= x^2-2x+1$ and $du = (2x-2)\,dx $ we get $dx= \frac{du}{2x+2}$.
Substituting dx in the integral:
$$\frac{x^2}{2} + C + \int \frac{-x}{u} \frac{1}{2x-2} du =\frac{x^2}{2} + C + \int \frac{-x}{u(2x-2)} du $$
I am stuck here. I do not see how using substitution has, or could have helped solve the problem. I am aware that there are other techniques for solving an integral, but I have been only taught substitution and would like to solve the problem accordingly.
Thanks
| With a shift, $t=x-1$,
$$\int \frac{(t+1)^2(t-1)}{t^2} dt=\int\left(t+1-\frac1t-\frac1{t^2}\right)dt\\
=\frac{(x-1)^2}2+(x-1)-\ln|x-1|+\frac1{x-1}+C.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2135003",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 7,
"answer_id": 0
} |
A quadratic polynomial $f(x)$ satisfy $ f(x) = \left[\frac{f(x+1)-f((x-1)^2)}{2}\right]^2$ for all real $x$. then $f(x)$ is A quadratic polynomial $f(x)$ satisfy $\displaystyle f(x) = \left[\frac{f(x+1)-f((x-1)^2)}{2}\right]^2$ for all real $x$. then $f(x)$ is
Let $f(x) = ax^2+bx+c$, then substitute $x=0$ in functional equation
$\displaystyle f(0) = \left(\frac{f(1)-f(1)}{2}\right) = 0$ So $c=0$
now $\displaystyle f'(x) = 2\left[\frac{f(x+1)-f((x-1)^2)}{2}\right]\cdot \left(\frac{f'(x+1)-f((x-1)^2)2(x-1)}{2}\right)$,
then substitute $x=0$
So we have $b=0$
could some help me to find value of $a$, thanks
| With $f(x)=ax^2$ and $f(x) = \left[\frac{f(x+1)-f((x-1)^2)}{2}\right]^2$ we get
$ax^2=\left[\frac{a(x+1)^2-a(x-1)^4}{2}\right]^2$.
Now take $x=1$ and look what happens....
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2136629",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Find the value of $\binom{2000}{2} + \binom{2000}{5} + \binom{2000}{8} + \cdots \binom{2000}{2000}$
Find the value of $\binom{2000}{2} + \binom{2000}{5} + \binom{2000}{8} + \cdots \binom{2000}{2000}$
I've seen many complex proofs. I am looking for an elementary proof. I know the fact that $\binom{2000}{0} + \binom{2000}{1} + \binom{2000}{2} + \cdots \binom{2000}{2000} = 2^{2000}$. This may help here.
| Let $w=\exp(2\pi i/3)$. Then $w^3=1$ and $1+w+w^2=0$, and hence
$$
(1+1)^{2000}+w(1+w)^{2000}+w^2(1+w^2)^{2000}\\=\sum_{k=0}^{2000}\Bigg(\binom{2000}{k}+\binom{2000}{k}w^{k+1}+\binom{2000}{k}w^{2k+2}\Bigg)
\\=3 \Bigg(\binom{2000}{2}+\binom{2000}{5}+\cdots+\binom{2000 }{2000}\Bigg),
$$
since
$$
1+w^{k+1}+w^{2k+2}=\left\{\begin{array}{ccc}
3 & \text{if} & k=2\mod 3,\\
0 & \text{if} & k\ne2\mod 3.
\end{array}
\right.
$$
Next
$$
1+w=-w^2\quad\Rightarrow\quad(1+w)^{2000}=w^{4000}=w\\
1+w^2=-w\quad\Rightarrow\quad(1+w^2)^{2000}=w^{2000}=w^2
$$
Finally
$$
(1+1)^{2000}+w(1+w)^{2000}+w^2(1+w^2)^{2000}=2^{2000}+w+w^2=2^{2000}-1.
$$
Hence
$$
\binom{2000}{2}+\binom{2000}{5}+\cdots+\binom{2000 }{2000}=\frac{1}{3}\big(2^{2000}-1\big)
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2136738",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 1
} |
In right triangle $ABC$ ($\angle A=90$), $E$ is a point on $AC$.Find $AE$ given that...
In right triangle $ABC$ ($\angle A=90$), $AD$ is a height and $E$ is a point on $AC$ so that $BE=EC$ and $CD=CE=EB=1$. $\color {red} {Without}$ using trigonometric relations find $AE$.
I do have a solution USING trigonometric relations ( $AE=\sqrt[3]{2}-1$ ),but it seems other solutions are troublesome.My attempt led to a complicated polynomial equation...
| Let $AE=x$.
Hence, $AD=\sqrt{(x+1)^2-1}=\sqrt{x^2+2x}$ and since $AD^2=BD\cdot DC$,
we obtain $BD=x^2+2x$, which says that $AB=\sqrt{(x^2+2x)^2+x^2+2x}$ and
$$BE=\sqrt{(x^2+2x)^2+x^2+2x+x^2}.$$
Thus,$$(x^2+2x)^2+x^2+2x+x^2=1$$ or
$$x^4+4x^3+6x^2+2x-1=0$$ or
$$(x+1)(x^3+3x^2+3x-1)=0$$ or
$$(x+1)^3=2$$ or
$$x=\sqrt[3]2-1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2138416",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Prove, by contradiction, that, if $cx^2 + bx + a$ has no rational root, then $ax^2 + bx + c$ has no rational root. Proposition: Suppose that $a$, $b$, and $c$ are real numbers with $c \not = 0$. Prove, by contradiction, that, if $cx^2 + bx + a$ has no rational root, then $ax^2 + bx + c$ has no rational root.
Hypothesis: $cx^2 + bx + a$ has no rational root where $a$, $b$, and $c$ are real numbers with $c \not = 0$.
Conclusion: $ax^2 + bx + c$ has no rational root
To form a proof by contradiction, we take the negation of the conclusion:
$\neg B$: $ax^2 + bx + c$ has a rational root.
We now have a suitable hypothesis and conclusion for proof by contradiction:
A (Hypothesis): $cx^2 + bx + a$ has no rational root where $a$, $b$, and $c$ are real numbers with $c \not = 0$.
A1: $ax^2 + bx + c$ has a rational root.
Given that this is a proof by contradiction, we can work forward from both the hypothesis and conclusion, as shown above.
My Workings
A2: Let $x = \dfrac{p}{q}$ where $p$ and $q \not = 0$ are integers. This is the definition of a rational number (in this case, $x$): A rational number is any number that can be expressed as the quotient/fraction of two integers.
A3: $a\left(\dfrac{p}{q}\right)^2 + b\left(\dfrac{p}{q}\right) + c = 0$
$\implies \dfrac{ap^2}{q^2} + \dfrac{bp}{q} + c = 0$ where $q \not = 0$.
$\implies ap^2 + bpq + cq^2 = 0$
A4: $ap^2 + bpq + cq^2 = 0$ where $c \not = 0$
$\implies ap^2 + bpq = -cq^2$ where $-cq \not = 0$ since $c \not = 0$ and $q \not = 0$.
A5: $ap^2 + bpq + cq^2 = 0$ where $ap^2 + bpq \not = 0$ and $cq^2 \not = 0$.
But $ap^2 + bpq + cq^2 = 0$? Contradiction. $Q.E.D.$
I would greatly appreciate it if people could please take the time to review my proof and provide feedback on its correctness.
| To continue what you started
$ap^2 + bpq + cq^2 = 0$ divide both sides by $p^2$.
$a + b\frac qp + c(\frac {q^2}{p^2}) = 0$
So $\frac pc$ is a rational solution to $cx^2 + bx + a = 0$. A contradiction.
Worth noting: if $w$ is a solution to $ax^2 + bx + c = 0$ ($a \ne 0; c\ne 0$) then $\frac 1w$ is a solution to $cx^2 +bx +a=0$. And if $w \ne 0$ then $w$ is rational if and only $\frac 1w $ is rational.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2139056",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Solve an integral $\int\frac{\cos^3 x}{\sin^3 x+\cos^3 x}dx$ Solve an integral $$\int\frac{\cos^3 x}{\sin^3 x+\cos^3 x}dx$$
I tried to divide the numerator and denominator by $\cos^4 x$ to get $\sec x$ function but the term ${\sin^3 x}/{\cos^4 x}$ gives $\tan^2 x\sec^2 x\sin x$. How to get rid of $\sin x$ term?
| I wasn't really able to come up with a better (elegant) method other than the following:
$$\int \frac{\cos^3 x}{\sin^3 x + \cos^3 x} \mathrm{d}x = \int \frac{1}{1 + \tan^3 x} \mathrm{d}x$$
Now, using the substitution, $t = \tan x \implies \frac{\mathrm{d}t}{1+t^2} = \mathrm{d}x$, we get
$$= \int \frac{1}{(1 + t^2)(1+t^3)} \mathrm{d}t$$
Decomposing it into partial fraction (copying from W|A):
$$= \int \frac{1}{6(t+1)} + \frac{t+1}{2(t^2+1)} - \frac{2t-1}{3(t^2-t+1)} \mathrm{d}t \\ = \frac 16\ln t + \frac 14\ln (t^2+1) + \frac 12\arctan t - \frac 13 \ln (t^2-t+1) + C$$
Substituting back $t = \tan x$
$$\frac 16 \ln \tan x + \frac 12 \ln \sec x -\frac 13 \ln (\sec^2x - \tan x) + \frac x2 + C$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2139198",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
If a,b and c are sides of a triangle, then prove that the following polynomial has no real roots This is the polynomial: $$a^2x^2+(b^2+a^2-c^2)x+b^2=0$$
Now this is my progress:
Assuming l,m, and n are sides of a triangle, then $$|m-n|\lt l\lt m+n$$
Also, if a second degree polynomial in the form $kx^2+px+q$ has real roots, then $$p^2-4kq\ge 0$$
In this case, if there are no real roots, $$(b^2+a^2-c^2)^2-4a^2b^2\lt 0$$
$$b^2+a^2-c^2\lt 2ab$$
But by the first equation, $$a+b\gt c$$
$$a^2+b^2-c^2\gt 2ab$$
Wich actually proves by contradiction there ARE real roots. What did I do wrong?
| The following answers the What did I do wrong? part of the question
In this case, if there are no real roots, $$(b^2+a^2-c^2)^2-4a^2b^2\lt 0$$
$$b^2+a^2-c^2\lt 2ab$$
The two inequalities are not equivalent, since $b^2+a^2-c^2$ can be either positive or negative (remember that it is $0$ for right triangles by Pythagora's). What the first inequality is equivalent with, instead, is:
$$\left|b^2+a^2-c^2\right|\lt 2ab \quad \iff \quad -2ab \lt b^2+a^2-c^2\lt 2ab \tag{1}$$
But by the first equation, $$a+b\gt c$$
$$a^2+b^2-c^2\gt 2ab$$
The second inequality does not follow, in particular $2ab$ has the wrong sign. Instead, if you square the first inequality you get: $$a^2+b^2 - c^2 \gt -2ab \tag{2}$$
If you square $|a-b| \lt c$ though, then you get:
$$a^2+b^2 - c^2 \lt 2ab \tag{3}$$
Note that the last two inequalities $(2),(3)$ put together give $(1)$ and therefore complete the proof.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2139802",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Minimize $\big(3+2a^2\big)\big(3+2b^2\big)\big(3+2c^2\big)$ if $a+b+c=3$ and $(a,b,c) > 0$
Minimize $\big(3+2a^2\big)\big(3+2b^2\big)\big(3+2c^2\big)$ if $a+b+c=3$ and $(a,b,c) > 0$.
I expanded the brackets and applied AM-GM on all of the eight terms to get :
$$\big(3+2a^2\big)\big(3+2b^2\big)\big(3+2c^2\big) \geq 3\sqrt{3}abc$$
, which is horribly weak !
I can not use equality constraint whichever method I use. Thanks to Wolfram|Alpha, I know the answer is $125$ when $(a,b,c) \equiv (1,1,1).$
Any help would be appreciated. :)
| Set $g(a,b,c)=a+b+c-3$ and $f(a,b,c)=(3+2a^2)(3+2b^2)(3+2c^2)$. We shall use the Lagrange multipliers method.
We have to minimize $h(a,b,c)=f(a,b,c)-\lambda\cdot g(a,b,c)$.
$\frac{d\ h(a,b,c)}{d\ a}=0$
$\frac{d\ h(a,b,c)}{d\ b}=0$
$\frac{d\ h(a,b,c)}{d\ c}=0$
$\frac{d\ h(a,b,c)}{d\ \lambda}=0$
You obtain
$4a(3+2b^2)(3+2c^2)-\lambda=0$
$4b(3+2a^2)(3+2c^2)-\lambda=0$
$4c(3+2a^2)(3+2b^2)-\lambda=0$
$a+b+c=3$
Solve the above system and you have the solution. Indeed, you'll find that the only real solutions are given when $a=b=c$ and hence from last equation you have $a=b=c=1$. Hence the min is $125$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2141009",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 2
} |
What is the Conjugate of $\frac{1}{4\sqrt{3}-7}$ I'm having problems finding the conjugate of $\frac{1}{4\sqrt{3}-7}$
The answer I get is as follows: ${-4\sqrt{3}-7}$
However the answer given is ${-4\sqrt{3}+7}$
Here are my workings out...
$\frac{1}{4\sqrt{3}-7} * \frac{4\sqrt{3}+7}{4\sqrt{3}+7} = \frac{4\sqrt{3}+7}{-1} = {-4\sqrt{3}-7}$
I'd appreciate any guidance on where I've gone wrong.
| Well, after all the changes:
$$(4\sqrt3-7)(4\sqrt3+7)=48-49=-1$$
and then indeed:
$$\frac1{4\sqrt3-7}=-\left(4\sqrt3+7\right)=-4\sqrt3-7$$
and yes: the conjugate is $\;\frac1{4\sqrt3+7}\;$ , and what you say is the answer given is neither the conjugate nor the product of the original expression by its conjugate.
Added following an idea by projectilemotion: Perhaps the idea is first to rationalize the expression and then to find its conjugate:
$$\text{Rationalizing:}\;\;\frac1{4\sqrt3-7}\cdot\frac{4\sqrt3+7}{4\sqrt3+7}=-4\sqrt3-7$$
and now the rightmost expression's conjugate indeed is: $\;-4\sqrt3\color{Red}+7\;$ .....tadaaah!
It's hard to tell what they meant without knowing a priori their definitions... BTW, in this case and for me, the conjugate could as well be $\;4\sqrt3-7=-7+4\sqrt3\;$ ...Funny.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2141589",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Compute $\lim\limits_{x \to 0} \frac{ \sqrt[3]{1+\sin x} - \sqrt[3]{1- \sin x }}{x}$
Compute $\lim\limits_{x \to 0} \dfrac{ \sqrt[3]{1+\sin x} - \sqrt[3]{1- \sin x }}{x}$
Original question was to solve $\lim\limits_{x \to 0} \dfrac{ \sqrt[3]{1+ x } - \sqrt[3]{1-x }}{x}$ and it was solved by adding and subtracting 1 in denominator. Making it in the form $\lim\limits_{x \to a}\dfrac {x^n - a^n}{x-a} = ax^{n-1}$
How to solve for above limit without using lhopitals rule?
| If you already know how to find the limit $\lim\limits_{x \to 0} \dfrac{ \sqrt[3]{1+ x } - \sqrt[3]{1-x }}{x}$ and also that $\dfrac{\sin x}x$ tends to 1, then you can simply use
$$
\frac {\sqrt[3]{1+\sin x} - \sqrt[3]{1- \sin x}}{x} =
\frac {\sqrt[3]{1+\sin x} - \sqrt[3]{1- \sin x}}{\sin x} \cdot \frac{\sin x}x.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2143650",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
Let $\alpha = \omega + \omega^2 + \omega^4$ and $\beta = \omega^3 + \omega^5 + \omega^6$. Let $\omega$ be a complex number such that $\omega^7 = 1$ and $\omega \neq 1$. Let $\alpha = \omega + \omega^2 + \omega^4$ and $\beta = \omega^3 + \omega^5 + \omega^6$. Then $\alpha$ and $\beta$ are roots of the quadratic
[x^2 + px + q = 0]
for some integers $p$ and $q$. Find the ordered pair $(p,q)$.
I got that $p=-1$ but does not know how to go on to finding $q$. All help is appreciated!
| Hint. Observe that
$$
\alpha\beta = \omega^4+\omega^5+\omega^6+3\omega^7+\omega^8+\omega^9+\omega^{10}
= 2+\omega^4(1+\omega+\omega+\cdots+\omega^6)
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2144834",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
How does one show that $\sum_{k=1}^{n}\sin\left({\pi\over 2}\cdot{4k-1\over 2n+1}\right)={1\over 2\sin\left({\pi\over 2}\cdot{1\over 2n+1}\right)}?$ Consider
$$\sum_{k=1}^{n}\sin\left({\pi\over 2}\cdot{4k-1\over 2n+1}\right)=S\tag1$$
How does one show that $$S={1\over 2\sin\left({\pi\over 2}\cdot{1\over 2n+1}\right)}?$$
An attempt:
Let $$A={\pi\over 2(2n+1)}$$
$$\sin(4Ak-A)=\sin(4Ak)\cos(A)-\sin(A)\cos(4Ak)\tag2$$
$$\cos(A)\sum_{k=1}^{n}\sin(4Ak)-\sin(A)\sum_{k=1}^{n}\cos(4Ak)\tag3$$
$$\sin(4Ak)=4\sin(Ak)\cos(Ak)-8\sin^3(Ak)\cos(Ak)\tag4$$
$$\cos(4Ak)=8\cos^4(Ak)-8\cos^2(Ak)+1\tag5$$
Substituting $(4)$ and $(5)$ into $(3)$ is going to be quite messy. So how else can we prove $(1)$?
| Using Lagrange Identities:
$$ \sum_{n=1}^{N}\sin\left(\theta n\right)=+\frac{\cot\left(\theta/2\right)}{2}-\frac{\cos\left(\theta N+\theta/2\right)}{2\sin\left(\theta/2\right)} \quad\&\quad \sum_{n=1}^{N}\cos\left(\theta n\right)=-\frac{1}{2}+\frac{\sin\left(\theta N+\theta/2\right)}{2\sin\left(\theta/2\right)}$$
One could simplify $(3)$ through a different path:
$$
\begin{align}
\color{red}{S} &= \sum_{k=1}^{n}\sin\left(\frac{\pi}{2}\,\frac{4k-1}{2n+1}\right) = \sum_{k=1}^{n}\sin\left(4A k-A\right) \\[3mm]
&= \cos\left(A\right)\sum_{k=1}^{n}\sin\left(4A k\right)-\sin\left(A\right)\sum_{k=1}^{n}\cos\left(4A k\right) \\[3mm]
&= \cos\left(A\right)\left[\frac{\cot\left(2A\right)}{2}-\frac{\cos\left(4A n+2A\right)}{2\sin\left(2A\right)}\right]+\sin\left(A\right)\left[\frac{1}{2}-\frac{\sin\left(4A n+2A\right)}{2\sin\left(2A\right)}\right] \\[3mm]
&= \frac{\cos\left(A\right)\cos\left(2A\right)+\sin\left(A\right)\sin\left(2A\right)}{2\sin\left(2A\right)}-\frac{\cos\left(A\right)\cos\left(4A n+2A\right)+\sin\left(A\right)\sin\left(4A n+2A\right)}{2\sin\left(2A\right)} \\[3mm]
&= \frac{\cos\left(2A-A\right)}{2\sin\left(2A\right)}-\frac{\cos\left(4A n+2A-A\right)}{2\sin\left(2A\right)} \qquad\qquad\qquad\left\{\color{blue}{\small 4A n+2A = 2(2n+1)A = \pi}\right\} \\[3mm]
&= \frac{\cos\left(A\right)}{2\sin\left(2A\right)}-\frac{\cos\left(\pi-A\right)}{2\sin\left(2A\right)} = \frac{\cos\left(A\right)}{2\sin\left(2A\right)}+\frac{\cos\left(A\right)}{2\sin\left(2A\right)} \qquad\left\{\color{blue}{\small \cos(\pi-A)=-\cos(A)}\right\} \\[3mm]
&= \frac{\cos\left(A\right)}{\sin\left(2A\right)} = \frac{\cos\left(A\right)}{2\sin\left(A\right)\cos\left(A\right)} = \frac{1}{2\sin\left(A\right)} = \color{red}{\frac{1}{2\sin\left(\frac{\pi/2}{2n+1}\right)}}
\end{align}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2149149",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Other Idea to show an inequality $\dfrac{1}{\sqrt 1}+\dfrac{1}{\sqrt 2}+\dfrac{1}{\sqrt 3}+\cdots+\dfrac{1}{\sqrt n}\geq \sqrt n$ $$\dfrac{1}{\sqrt 1}+\dfrac{1}{\sqrt 2}+\dfrac{1}{\sqrt 3}+\cdots+\dfrac{1}{\sqrt n}\geq \sqrt n$$
I want to prove this by Induction
$$n=1 \checkmark\\
n=k \to \dfrac{1}{\sqrt 1}+\dfrac{1}{\sqrt 2}+\dfrac{1}{\sqrt 3}+\cdots+\dfrac{1}{\sqrt k}\geq \sqrt k\\
n=k+1 \to \dfrac{1}{\sqrt 1}+\dfrac{1}{\sqrt 2}+\dfrac{1}{\sqrt 3}+\cdots+\dfrac{1}{\sqrt {k+1}}\geq \sqrt {k+1}$$ so $$\dfrac{1}{\sqrt 1}+\dfrac{1}{\sqrt 2}+\dfrac{1}{\sqrt 3}+\cdots+\dfrac{1}{\sqrt k}+\dfrac{1}{\sqrt {k+1}}\geq \sqrt k+\dfrac{1}{\sqrt {k+1}}$$now we prove that $$\sqrt k+\dfrac{1}{\sqrt {k+1}} >\sqrt{k+1} \\\sqrt{k(k+1)}+1 \geq k+1 \\ k(k+1) \geq k^2 \\k+1 \geq k \checkmark$$ and the second method like below ,
and I want to know is there more Idia to show this proof ? forexample combinatorics proofs , or using integrals ,or fourier series ,....
Is there a close form for this summation ?
any help will be appreciated .
| $$\begin{cases}\dfrac{1}{\sqrt 1}\geq \dfrac{1}{\sqrt n}\\+\dfrac{1}{\sqrt 2}\geq \dfrac{1}{\sqrt n}\\+\dfrac{1}{\sqrt 3}\geq \dfrac{1}{\sqrt n}\\ \vdots\\+\dfrac{1}{\sqrt n}\geq \dfrac{1}{\sqrt n}\end{cases} \\\\$$
sum of left hands is $\underbrace{\dfrac{1}{\sqrt 1}+\dfrac{1}{\sqrt 2}+\dfrac{1}{\sqrt 3}+\cdots+\dfrac{1}{\sqrt n}}$ sum of the right hands is $n\times \dfrac{1}{\sqrt n}$ so
$$\dfrac{1}{\sqrt 1}+\dfrac{1}{\sqrt 2}+\dfrac{1}{\sqrt 3}+\cdots+\dfrac{1}{\sqrt n} \geq n\dfrac{1}{\sqrt n}=\dfrac{\sqrt{n^2}}{\sqrt{n}}=\sqrt{n} \checkmark$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2149448",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 6,
"answer_id": 3
} |
Decomposition of self-adjoint operator Suppose we have some self adjoint operator, given by either a matrix $$\begin{pmatrix}
1 & 2 \\
2 & 1
\end{pmatrix}$$ or a function $f \longmapsto xf$ on $L^2[-1,1]$. Is there a quick way of decomposing these self adjoint operators into the difference of positive operators?
| Compose the operator with the projections on the positive and negative part of the spectrum and take the difference. For your matrix $A$, you need to find the eigenvalues which are the roots of
$$ \lambda^2 - \operatorname{tr}(A)\lambda + \det(A) = \lambda^2 - 2\lambda -3 = (\lambda + 1)(\lambda - 3).$$
An eigenvector associated to $\lambda = 3$ is $v_1 = (1,1)^T$ and an eigenvector associated to $\lambda = -1$ is $v_2 = (-1,1)^T$. The orthogonal projection onto $\operatorname{span} \{ v_1 \}$ is given by
$$ P_1 = \begin{pmatrix} \frac{1}{2} & \frac{1}{2} \\ \frac{1}{2} & \frac{1}{2} \end{pmatrix} $$
and the orthogonal projection onto $\operatorname{span} \{ v_2 \} = \operatorname{span} \{ v_1 \}^{\perp}$ is given by
$$ P_2 = I - P = \begin{pmatrix} \frac{1}{2} & -\frac{1}{2} \\ -\frac{1}{2} & \frac{1}{2} \end{pmatrix}.$$
Finally, we have
$$ A = A(P_1 + P_2) = AP_1 - (-AP_2) = \begin{pmatrix} \frac{3}{2} & \frac{3}{2} \\ \frac{3}{2} & \frac{3}{2} \end{pmatrix} - \begin{pmatrix} \frac{1}{2} & -\frac{1}{2} \\ -\frac{1}{2} & \frac{1}{2}\end{pmatrix}$$
where $AP_1$ is positive (with spectrum $\{ 0, 3 \}$) and $-AP_2$ is positive (with spectrum $\{ 0, 1 \}$).
For the operator $(M(f))(x) = x f(x)$ on $L^2([-1,1])$ the situation is easier because it is already "diagonal". Just define
$$ g_1(x) = \begin{cases} x & x \geq 0, \\ 0 & x < 0 \end{cases} $$
and let $P_1(f)(x) := g_1(x) f(x)$. Then $P_1$ is positive (with spectrum $[0,1])$ and $M = P_1 - (P_1 - M)$ where $P_1 - M$ is also positive.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2150602",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Use infinite series to prove Use infinite series to prove that
$$\arcsin{x}\lt \frac{x}{1-x^2},$$ for $0\lt x\lt1$.
| Let $f(x)=\frac{x}{1-x^2}-\arcsin{x}$.
Hence, $$f'(x)=\frac{1-x^2-x(-2x)}{(1-x^2)^2}-\frac{1}{\sqrt{1-x^2}}=\frac{1+x^2-\sqrt{(1-x^2)^3}}{(1-x^2)^2}=$$
$$=\frac{ x^2(x^4-2x^2+5)}{(1+x^2-\sqrt{(1-x^2)^3})(1-x^2)^2}>0.$$
Thus, $f(x)>f(0)=0$ and we are done!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2152374",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Showing that $\sum_{n=1}^{\infty}\sum_{m=1}^{x-1} \frac{m}{n^2 x^2 - m^2}=\frac{1}{2}H_{x-1}$ Consider $(1)$, $H_n$ is the nth-harmonic number
$$\sum_{n=1}^{\infty}\left({1\over (nx)^2-1}+{2\over (nx)^2-2^2}+{3\over (nx)^2-3^2}+\cdots+{x-1\over (nx)^2-(x-1)^{2}}\right)=S\tag1$$
$x\ge2$
How does one show that $$\color{blue}{S={H_{x-1}\over 2}}?$$
An attempt:[Edited]
Because $(1)$ is the difference of two squares form, so it becomes
$$\sum_{n=1}^{\infty}\left[\left({1\over nx-1}+{1\over nx-2}+{1\over nx-3}+\cdots+{x-1\over nx-(x-1)}\right)-\left({1\over nx+1}+{1\over nx+2}+{1\over nx+3}+\cdots+{x-1\over nx+(x-1)}\right)\right]=2S\tag2$$
Not sure how to proceed next
| Well, you have made a small calculation mistake. We have, $$\frac {1}{n^2x^2-1} = \frac {1}{(nx-1)(nx+1)} = \frac {1}{2}[\frac {1}{nx-1}-\frac {1}{nx+1}] $$ $$\frac {2}{n^2x^2-4} = \frac {2}{(nx-2)(nx+2)} = \frac {1}{2}[\frac {1}{nx-2}-\frac {1}{nx+2}] $$ $$\frac {3}{n^2x^2-9} = \frac {3}{(nx-3)(nx+3)} = \frac {1}{2}[\frac {1}{nx-3}-\frac {1}{nx+3}] $$ $$\vdots $$ $$\frac {x-1}{n^2x^2-(x-1)^2} = \frac {1}{2}[\frac {1}{nx-(x-1)} - \frac {1}{nx+(x-1)}] $$
Now with the numerator as $1$ in all cases, it will easily telescope for $n=1$ to $\infty $ and for $x\geq 2$, giving us, $$S = \frac{1}{2}[\frac {1}{1} + \frac {1}{2} + \frac{1}{3} + \cdots +\frac {1}{x-1}] = \frac {H_{x-1}}{2} $$
Hope it helps.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2152635",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Subsets and Splits