Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Show that the number $\sqrt{2^1}+\sqrt{2^2}+\cdots+\sqrt{2^n}$ is not rational for any integer $n>0$ I've got the first part of the proof.
So we know that the odd exponents will give us an irrational answer, and the even ones will give us a rational answer. Therefore we have an alternating sum of rationals and irrationals. I can prove that an irrational plus a rational is irrational, so now we can combine all the alternating sum into a sum of irrationals. I can't prove that for this specific case that an irrational plus an irrational equals an irrational.
| Tricks, tricks, tricks.Here, it's the geometric summation.
$$
\sqrt{2} + \sqrt{2^2} + \ldots \sqrt{2^n} = 2^{\frac 12} + 2^\frac{2}{2} + \ldots + 2^\frac{n}{2}
$$
Use the geometric series summation formula:
$$
2^{\frac 12} + 2^\frac{2}{2} + \ldots + 2^\frac{n}{2} = (2^\frac 12)\frac{2^\frac{n+1}{2} - 1}{2^{\frac 12} - 1}
$$
Now, rationalize by multiplying top and bottom by $2^{\frac 12} + 1$:
$$
(2^\frac 12)\frac{2^\frac{n+1}{2} - 1}{2^{\frac 12} - 1} \times \frac{2^{\frac 12} + 1}{2^{\frac 12} + 1} = 2^{\frac 12} (2^{\frac 12} + 1)(2^{\frac{n+1}{2} - 1})
$$
expand this to get:
$$
{\frac 12} (2^{\frac 12} + 1)(2^{\frac{n+1}{2} - 1}) = 2^{n/2}(1 + \sqrt 2) = 2^{\frac n2} + 2^\frac{n+1}{2}
$$
Now, suppose that $n$ is even: Then $2^{n/2}$ is an integer, so the answer is of the form $a+b \sqrt 2$ where $a,b$ are integers. If this were rational, then $\sqrt 2$ would also be rational. Contradiction.
Now, suppose that $n$ is odd: Then $2^\frac{n+1}{2}$ is an integer, so again the answer is of the form $a+b \sqrt 2$ where $a,b$ are integers. If this were rational, then $\sqrt 2$ would also be rational. Contradiction yet again.
Hence, we can conclude that for all $n$, the quantity given is irrational.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2126732",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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there are infnitely many postive integer $n$ such $ \lfloor \sqrt{7}\cdot n \rfloor=k^2+1(k\in \Bbb{Z})$ show that: there are infnitely many postive integer $n$ such $$ \lfloor \sqrt{7}\cdot n \rfloor=k^2+1(k\in \Bbb{Z})$$
I think use pell equation to solve it. But I can't.
| I get that
there are an infinite number of $n$
such that
$\lfloor n\sqrt{d} \rfloor
=k^2-1
$,
not $k^2+1$.
However,
for $d$ such that
there are solutions to
$x^2-dy^2 = -3$,
such as $d=7$,
then there are $n$
such that
$\lfloor n \sqrt{d} \rfloor
= k^2+1
$.
This generalizes to $k^2 \pm j$
depending on the existence
of solutions to
$x^2-dy^2 = \pm m$
for different $m$.
Here we go.
As the OP stated,
the Pell equation comes into it.
We start with the fact that
there are an infinite
number of integer solutions to
$x^2-dy^2 = 1$,
where $d$ is square free.
For each of these,
$1
=x^2-dy^2
=(x-y\sqrt{d})(x+y\sqrt{d})
$
so
$(x-y\sqrt{d})
=\dfrac1{x+y\sqrt{d}}
$
or,
squaring,
$x^2-2xy\sqrt{d}+dy^2
=\dfrac1{(x+y\sqrt{d})^2}
$
or
$2xy\sqrt{d}
=x^2+dy^2-\dfrac1{(x+y\sqrt{d})^2}
=x^2+(x^2-1)-\dfrac1{(x+y\sqrt{d})^2}
=2x^2-1-\dfrac1{(x+y\sqrt{d})^2}
$
or
$xy\sqrt{d}
=x^2-\frac12(1+\dfrac1{(x+y\sqrt{d})^2})
$.
Since
$0 < \dfrac1{(x+y\sqrt{d})^2})
< \frac12$,
$\frac12
< \frac12(1+\dfrac1{(x+y\sqrt{d})^2})
< 1
$
so
$\lfloor xy\sqrt{d} \rfloor
=\lfloor x^2-\frac12(1+\dfrac1{(x+y\sqrt{d})^2}) \rfloor
= x^2-\lfloor\frac12(1+\dfrac1{(x+y\sqrt{d})^2}) \rfloor
= x^2-1
$.
This is not what is asked.
However,
if there is one solution to
$x^2-dy^2 = -1$,
then there are
an infinite number of solutions.
Modifying this calculation
we get
$x^2-2xy\sqrt{d}+dy^2
=\dfrac1{(x+y\sqrt{d})^2}
$
or
$2xy\sqrt{d}
=x^2+dy^2-\dfrac1{(x+y\sqrt{d})^2}
=x^2+(x^2+1)-\dfrac1{(x+y\sqrt{d})^2}
=2x^2+1-\dfrac1{(x+y\sqrt{d})^2}
$
or
$xy\sqrt{d}
=x^2+\frac12(1-\dfrac1{(x+y\sqrt{d})^2})
$
$\lfloor xy\sqrt{d} \rfloor
=\lfloor x^2+\frac12(1-\dfrac1{(x+y\sqrt{d})^2}) \rfloor
= x^2+\lfloor\frac12(1-\dfrac1{(x+y\sqrt{d})^2}) \rfloor
= x^2
$.
However,
there are no solutions to
$x^2-7y^2 = -1$,
so this does not hold.
However,
suppose there are
an infinite number of solutions to
$x^2-dy^2 = -m$.
Modifying this calculation
we get
$-m
=x^2-dy^2
=(x-y\sqrt{d})(x+y\sqrt{d})
$
or
$x-y\sqrt{d}
=\dfrac{-m}{x+y\sqrt{d}}
$.
Squaring,
$x^2-2xy\sqrt{d}+dy^2
=\dfrac{m^2}{(x+y\sqrt{d})^2}
$
or
$2xy\sqrt{d}
=x^2+dy^2-\dfrac{m^2}{(x+y\sqrt{d})^2}
=x^2+(x^2+m)-\dfrac{m^2}{(x+y\sqrt{d})^2}
=2x^2+m-\dfrac{m^2}{(x+y\sqrt{d})^2}
$
or
$xy\sqrt{d}
=x^2+\frac12(m-\dfrac{m^2}{(x+y\sqrt{d})^2})
$
so
$\lfloor xy\sqrt{d} \rfloor
=\lfloor x^2+\frac12(m-\dfrac{m^2}{(x+y\sqrt{d})^2}) \rfloor
= x^2+\lfloor\frac12(m-\dfrac{m^2}{(x+y\sqrt{d})^2}) \rfloor
$.
If $m$ is odd,
$m = 2j+1$,
then
$\lfloor xy\sqrt{d} \rfloor
= x^2+\lfloor\frac12(2j+1-\dfrac{m^2}{(x+y\sqrt{d})^2}) \rfloor
=x^2+j
$
once
$x+y\sqrt{d}
> m
$.
Since there are solutions to
$x^2-7y^2 = -3$
(e.g.,
$5^2-7\cdot 2^2 = -3$)
there are an infinite number of solutions,
so OP's statement is true.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2127039",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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If $xy+xz+yz=1+2xyz$ then $\sqrt{x}+\sqrt{y}+\sqrt{z}\geq2$. Let $x$, $y$ and $z$ be non-negative numbers such that $xy+xz+yz=1+2xyz$. Prove that:
$$\sqrt{x}+\sqrt{y}+\sqrt{z}\geq2$$
The equality occurs for $x=y=1$ and $z=0$.
I tried Lagrange Multipliers and more, but I don't see a proof.
| Let: $x=a^2 , y=b^2 , z=c^2$
So we must prove : $a+b+c \ge 2$
for nonnegative $a, b, c : \ a^2b^2+b^2c^2+c^2a^2=1+2a^2b^2c^2 $
$$p=a+b+c\ , \ q=ab+bc+ca \ , \ r=abc$$
$$q^2=1+2pr+2r^2 \Rightarrow q \ge 1$$
*
*$q \ge \dfrac{4}{3} \Rightarrow p^2 \ge 3q \ge 4$
*$1 \le q \le \dfrac{4}{3}\ , \ 2pr = q^2-1-2r^2 \le q^2-1$
By Shur we have : $p^3+9r \ge 4pq \Rightarrow p^4-4qp^2+\dfrac{9}{2}(q^2-1)\ge 0\Rightarrow$
$$p^2\ge 2q+\sqrt{\dfrac{9-q^2}{2}} \ge 4 \Leftrightarrow (q-1)(23-9q)\ge 0$$
Equality holdes for : $a=b=1 \ , \ c=0$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2127470",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
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Find all the differentiable functions $f$ with the property $f\left(x^4+y\right) = x^3f(x) + f(y)$ I'm having a hard time answering this question:
Find all the differentiable functions $f$ such that
$$\forall x,y\in\mathbb R : f\left(x^4+y\right) = x^3f(x) + f(y) \text.$$
I would love to get a hint about how to start since this is my first time answering this type of questions.
Thanks a lot.
| You can show that a function $ f : \mathbb R \to \mathbb R $ satisfies
$$ f \left( x ^ 4 + y \right) = x ^ 3 f ( x ) + f ( y ) \tag 0 \label 0 $$
for all $ x , y \in \mathbb R $ iff there is a constant $ a \in \mathbb R $ such that $ f ( x ) = a x $ for all $ x \in \mathbb R $, even without the assumption of differentiability. It's easy to check that functions of this form are solutions. We try to prove the converse.
Put $ y = 0 $ in \eqref{0} to get $ f \left( x ^ 4 \right) = x ^ 3 f ( x ) + f ( 0 ) $, which in particular for $ x = 1 $ shows that $ f ( 0 ) = 0 $ and hence
$$ f \left( x ^ 4 \right) = x ^ 3 f ( x ) \text . \tag 1 \label 1 $$
This lets you rewrite \eqref{0} as
$$ f \left( x ^ 4 + y \right) = f \left( x ^ 4 \right) + f ( y ) \text . $$
This means that for all $ x \in \mathbb R ^ { 0 + } $ and all $ y \in \mathbb R $ we have $ f ( x + y ) = f ( x ) + f ( y ) $. Noting that for any $ x \in \mathbb R $ we have $ | x | \ge 0 $ and $ x + | x | \ge 0 $, we can generalize this and get
$$ f ( x + y ) = f \big( ( | x | + x ) + ( y - | x | ) \big) = f ( | x | + x ) + f ( y - | x | ) \\
= f ( | x | ) + f ( x ) + f ( y - | x | ) = f ( x ) + f \big( | x | + ( y - | x | ) \big) \text , $$
and thus
$$ f ( x + y ) = f ( x ) + f ( y ) \tag 2 \label 2 $$
for all $ x , y \in \mathbb R $. Letting $ y = - x $ in \eqref{2} we have
$$ f ( - x ) = - f ( x ) \tag 3 \label 3 $$
and using induction and \eqref{2} we get
$$ f ( n x ) = n f ( x ) \tag 4 \label 4 $$
for every positive integer $ n $. Defining $ a = f ( 1 ) $, we can use \eqref{1} and \eqref{2} to get
$$ f \left( ( x + 1 ) ^ 4 \right) = ( x + 1 ) ^ 3 f ( x + 1 ) = ( x + 1 ) ^ 3 \big( f ( x ) + a \big) \text . \tag 5 \label 5 $$
Again, by \eqref{1}, \eqref{2} and \eqref{4} we have
$$ f \left( ( x + 1 ) ^ 4 \right) = f \left( x ^ 4 + 4 x ^ 3 + 6 x ^ 2 + 4 x + 1 \right) \\
= f \left( x ^ 4 \right) + f \left( 4 x ^ 3 \right) + f \left( 6 x ^ 2 \right) + f ( 4 x ) + a \\
= x ^ 3 f ( x ) + 4 f \left( x ^ 3 \right) + 6 f \left( x ^ 2 \right) + 4 f ( x ) + a \text . \tag 6 \label 6 $$
Combining \eqref{5} and \eqref{6} we get
$$ \left( x ^ 3 + 3 x ^ 2 + 3 x + 1 \right) \big( f ( x ) + a \big) = x ^ 3 f ( x ) + 4 f \left( x ^ 3 \right) + 6 f \left( x ^ 2 \right) + 4 f ( x ) + a \text . \tag 7 \label 7 $$
Similarly, calculating $ f ( x - 1 ) $ in two different ways, and this time also using \eqref{3}, we get
$$ \left( x ^ 3 - 3 x ^ 2 + 3 x - 1 \right) \big( f ( x ) - a \big) = x ^ 3 f ( x ) - 4 f \left( x ^ 3 \right) + 6 f \left( x ^ 2 \right) - 4 f ( x ) + a \text , $$
which together with \eqref{7} yields
$$ \left( x ^ 3 + 3 x \right) f ( x ) + a \left( 3 x ^ 2 + 1 \right) = x ^ 3 f ( x ) + 6 f \left( x ^ 2 \right) + a \text , $$
or simply
$$ f \left( x ^ 2 \right) = \frac 1 2 x f ( x ) + \frac a 2 x ^ 2 \text . $$
Similar to before, we calculate $ f \left( ( x + 1 ) ^ 2 \right) $ in two ways:
$$ f \left( ( x + 1 ) ^ 2 \right) = \frac 1 2 ( x + 1 ) \big( f ( x ) + a \big) + \frac a 2 ( x + 1 ) ^ 2 \text ; $$
$$ f \left( ( x + 1 ) ^ 2 \right) = f \left( x ^ 2 \right) + f ( 2 x ) + a = \frac 1 2 x f ( x ) + \frac a 2 x ^ 2 + 2 f ( x ) + a \text ; $$
and these together simplify to $ f ( x ) = a x $.
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "3",
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When does $2^n-1$ divide $3^n-1$?
Is it possible for some integer $n>1$ that $2^n-1\mid 3^n-1$ ?
I have tried many things, but nothing worked.
| I was looking for this as well, and eventually figured it out myself. So here's my solution for future reference. The short answer is, $2^n - 1$ never divides $3^n - 1$. Here's the proof, making use of the Jacobi symbol.
Assume $2^n - 1 \mid 3^n - 1$. If $n = 2k$ is even, then $2^n - 1 = 4^k - 1 \equiv 0 \bmod 3$. Consequently, $3$ must also divide $3^n - 1$, which is a contradiction. At the very least, we can already assume $n = 2k + 1$ is odd. Next, since $3^n \equiv 1 \bmod 2^n - 1$, from the properties of the Jacobi-symbol it follows that
\begin{equation}
1 = (\frac{1}{2^n - 1}) = (\frac{3^n}{2^n - 1}) = (\frac{3^{2k}}{2^n - 1}) \cdot (\frac{3}{2^n - 1}) = (\frac{3}{2^n - 1})
\end{equation}
However, using Jacobi's law of reciprocity we also know
\begin{equation}
(\frac{2^n - 1}{3}) = (\frac{3}{2^n - 1}) \cdot (\frac{2^n - 1}{3}) = (-1)^{\frac{3 - 1}{2}\frac{2^n - 2}{2}} = (-1)^{2^{n - 1} - 1} = -1
\end{equation}
The only quadratic non-residue $\bmod 3$ is $2$, therefore $2^n - 1 \equiv 2 \bmod 3$ or alternatively $2^n \equiv 0 \bmod 3$. Since this implies $3$ divides $2^n$, we again arrive at a contradiction.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Let $x,y$ be nonnegative integers satisfying $(xy-7)^2 = x^2+y^2$. Prove that $x+y = 7$
Let $x,y$ be nonnegative integers satisfying $(xy-7)^2 = x^2+y^2$. Prove that $x+y = 7$.
We can rewrite the given equation as $$(x+y)^2-(xy-6)^2 = (x+y-xy+6)(x+y+xy-6) = 13.$$ How do we continue?
| Note that we now have that either $$x+y-xy+6 = \pm 1 \tag{1}$$
$$x+y+xy-6=\pm 13 \tag{2}$$
or $$x+y-xy+6= \pm 13 \tag{3}$$ $$x+y+xy-6= \pm 1 \tag{4}$$
From your last equation. $\frac{(1)+(2)}{2}$ and $\frac{(3)+(4)}{2}$ both give that $$x+y=\pm 7$$From this, using the condition that $x$ and $y$ are non negative integers, we see that $x+y=7$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "2",
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How do i find sum $\sum_{r=1}^{n}\sin(3^{r-1}\theta)\sec(3^r\theta) $ How do i find sum
$$\sum_{r=1}^{n}\sin(3^{r-1}\theta)\sec(3^r\theta) $$
Needs hint to begin
thanks
| Note that this can be done through significant amounts of trigonemetric manipulation on $\frac{\sin x}{\cos 3x}$. So first,
$$\frac{\sin x}{\cos 3x}= \frac{1}{2} \cdot \frac{2 \cdot \sin x \cdot \cos x}{\cos x \cdot \cos 3x}= \frac{1}{2} \cdot \frac{\sin 2x}{\cos x \cdot \cos 3x}$$
Now, this becomes
$$ \frac{1}{2} \cdot \frac{ \sin(3x - x) }{ \cos x \cdot \cos3x }= \frac{1}{2} \cdot \frac{ \sin 3x \cdot \cos x - \cos 3x \cdot \sin x}{ \cos x \cdot \cos3x}$$
Which is $$ \frac{1}{2} \cdot \frac {\sin 3x \cdot \cos x}{\cos x \cdot \cos 3x} - \frac{1}{2} \cdot \frac{\cos 3x \cdot \sin x}{\cos x \cdot \cos 3x}= \frac{1}{2} \cdot \left (\frac{\sin 3x}{\cos 3x} - \frac{\sin x}{\cos x} \right )=\frac{\tan 3x-\tan x}{2}$$
So we have $$\frac{\sin x}{\cos 3x}=\frac{\tan 3x-\tan x}{2} \tag{1}$$
So $$\sum_{r=1}^{n}\sin(3^{r-1}\theta)\sec(3^r\theta) =\sum_{r=1}^{n}\frac{\sin(3^{r-1}\theta)}{\cos(3^r\theta)}=\frac{1}{2} \left( \sum_{r=1}^{n}\tan 3^r \theta-\tan 3^{r-1} \theta \right)$$From $(1)$. Note that this is a telescoping series. You can simplify this nicely.
| {
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"timestamp": "2023-03-29T00:00:00",
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Writing sins 3s in terms of sin s In the reduction below, I do not understand line 4 and 6. What identities were applied to line 3 and 5 to reach those conclusions? How were those identities introduced?
1 ) $\sin 3a $
2 ) $= \sin(2a +s)$
3 ) $= \sin2a ·\cos a + \cos 2a·\sin a$
4 ) $=(2\sin a·\cos a)\cos a + (\cos^2a - \sin^2a)\sin a$
5 ) $= 2\sin a·\cos²a + \cos²a·\sin a - \sin³a$
6 ) $ = 2\sin a(1-\sin²a) + (1 - \sin²a)\sin a - \sin³a$
7 ) $ = 2\sin a - 2\sin³a + \sin a - \sin³a - \sin³a$
8 ) $= 3\sin a - 4\sin³a$
This is important rewriting all forms of $\sin$ $n·s$ in terms of $\sin$ $s$. All help is greatly appreciated
| Following are the formula's used above -
1.) $\sin(a + b) = \sin a \cos b + \cos a \sin b$
2.) $\sin 2a = 2 \sin a \cos a$
3.) $\cos 2a = \cos^2 a - \sin^2 a$
4.) $\cos^2 a = 1 - \sin^2 a$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2135438",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Ptove that $3\prod\limits_{cyc}(x^2-x+1)\geq x^2y^2z^2-5xyz+\sum\limits_{cyc}(x^2y+x^2z)$ For all real numbers $x$, $y$ and $z$ prove that:
$$3(x^2-x+1)(y^2-y+1)(z^2-z+1)\geq x^2y^2z^2-5xyz+x^2y+x^2z+y^2x+y^2z+z^2x+z^2y$$
I tried $uvw$ and more. For example, this inequality is a quadratic inequality of $x$ and it's enough to prove it for $\Delta\ge0$, but it's nothing.
I found this problem here :artofproblemsolving.com/community/c6h140p779898
My motivation is the following.
There is the following easy similar inequality.
For all non-negative $x$, $y$ and $z$ prove that:
$$3(x^2-x+1)(y^2-y+1)(z^2-z+1)\geq1+xyz+x^2y^2z^2$$
| Let us prove that
$$3(x^2-x+1)(y^2-y+1)(z^2-z+1) \ge x^2y^2z^2 - 5xyz + x^2(y+z) + y^2(z+x) + z^2(x+y) + 1.$$
Proof: With the substitutions $y = 1 + u, \ z = 1+v$, we have
$$\mathrm{LHS} - \mathrm{RHS} = Ax^2 + Bx + C$$
where
\begin{align}
A &= (2 v^2+v+2) u^2+(v^2-v) u+2 v^2, \\
B &= -3 u^2 v^2-3 u^2 v-3 u v^2-4 u^2+2 u v-4 v^2, \\
C &= (3 v^2+2 v+2) u^2+(2 v^2-v) u+2 v^2.
\end{align}
$A \ge 0$ follows from $2 v^2+v+2 > 0$ and $4(2 v^2+v+2)\cdot 2v^2 - (v^2-v)^2 = 5v^2(3v^2+2v+3) \ge 0$.
$C \ge 0$ follows from $3 v^2+2 v+2 > 0$ and $4(3 v^2+2 v+2)\cdot 2v^2 - (2 v^2-v)^2 = 5v^2(4v^2+4v+3)\ge 0$.
We have
$$4AC - B^2 = 5u^2v^2\big[(3v^2+2v+3)u^2+(2v^2-2v)u+3v^2\big].$$
$4AC - B^2\ge 0$ follows from $3v^2+2v+3 > 0$ and
$4(3v^2+2v+3)\cdot 3v^2 - (2v^2-2v)^2 = 32v^2(v^2+v+1) \ge 0$.
We are done.
| {
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"timestamp": "2023-03-29T00:00:00",
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How to find limit of $\lim_{n\to\infty} (\sqrt[3]{n^2+5}-\sqrt[3]{n^2+3}$)? I am stuck on this limit.
$$\lim_{n\to\infty} \sqrt[3]{n^2+5}-\sqrt[3]{n^2+3}$$
I couldn't find the limit using the basic properties of limits, since that just yields: $$\infty-\infty$$ which is undefined. Could I get any hints for finding this limit?
| You can write $\sqrt[3]{1+x}=1+\frac{1}{3}x+O(x^2)$ for $x$ near zero.
So $$\sqrt[3]{n^2+3}=\sqrt[3]{n^2}\cdot \sqrt[3]{1+3/n^2} = n^{2/3}+\frac{1}{3}\frac{3}{n^{4/3}} + O(1/n^{10/3})$$
Similarly, $$\sqrt[3]{n^2+5}=\sqrt[3]{n^2}\cdot \sqrt[3]{1+5/n^2} = n^{2/3}+\frac{1}{3}\frac{5}{n^{4/3}} + O(1/n^{10/3})$$
So $$\sqrt[3]{n^2+5}-\sqrt[3]{n^2+3}=\frac{2}{3}n^{-4/3}+O(n^{-10/3})$$
Alternatively, you can use the mean value theorem. Let $f(x)=\sqrt[3]{x}$. Then for any $n$ there is a $c_n\in [n^2+3,n^2+5]$ such that:
$$f(n^2+5)-f(n^2+3)=((n^2+5)-(n^2+3))f'(c_n)=2f'(c_n)$$
Now, show that $0< f'(c_n)\leq f'(n^2+2)$ and $f'(n^2+2)\to 0.$
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove that: $\tan^6 20° - 33\tan^4 20° + 27\tan^2 20°=3$ Prove that: $\tan^6 20° - 33\tan^4 20° + 27\tan^2 20°=3$
My Attempt:
$$L.H.S=\tan^6 20 - 33\tan^4 20 + 27\tan^2 20°$$
$$=\tan^2 20°(\tan^4 20° - 33\tan^2 20°+27)$$
$$=(\sec^2 20° -1)(\tan^4 20° - 33\tan^2 20° + 27)$$
Please help me to continue from here..
| $$\tan60^{\circ}=\sqrt3$$ gives
$$\frac{3\tan20^{\circ}-\tan^320^{\circ}}{1-3\tan^220^{\circ}}=\sqrt3,$$
which gives $$\left(\frac{3\tan20^{\circ}-\tan^320^{\circ}}{1-3\tan^220^{\circ}}\right)^2=3,$$
which gives your identity.
| {
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Factoring irreducible fractions Is there a technique or a formula that factors irreducible fractions into products of polynomials, for example:
$$\frac1{x^8+1}=\frac{1}{(x^4+ \sqrt2x^2+1)(x^4- \sqrt2x^2+1)}$$
Also, is there a way to factor, in the same way as above, the following fraction:
$$\frac1{x^4-x^2+1}$$
I am not interested in the completition of the square. Thank you.
| To answer the "without completion of the square" part of the question, it is possible to actually determine the roots of the given polynomials in both cases. Then grouping each root with its complex conjugate gives a quadratic in $x$ which is a factor of the original polynomial.
In the case of $x^8+1$ for example, a pair of roots is $x = \cos(\frac{\pi}{8}) \pm i \sin(\frac{\pi}{8})\,$. The quadratic which has those roots is $x^2 - 2 \cos(\frac{\pi}{8})\,x +1\,$. Using the half-angle formula, $\cos(\frac{\pi}{8})= \frac{1}{2}\sqrt{2+\sqrt{2}}\,$. Therefore a factor of $x^8+1$ is:
$$x^2 - \sqrt{2+\sqrt{2}} \,x + 1$$
Since $x^8 +1$ is an even polynomial, another factor follows by substituting $x \mapsto -x\,$:
$$x^2 + \sqrt{2+\sqrt{2}} \,x + 1$$
Their product, which itself must be a factor of $x^8+1$ is:
$$\left(x^2 - \sqrt{2+\sqrt{2}} \,x + 1\right)\left(x^2 + \sqrt{2+\sqrt{2}} \,x + 1\right) \;=\; x^4 - \sqrt{2} \,x^2 + 1$$
Dividing $x^8+1$ by the above gives the other factor $\;x^4 + \sqrt{2} \,x^2 + 1\,$.
| {
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Which expressions is greater, $\frac{b}{c+d+b}$ or $\frac{1}{3}$? There are 3 kinds of animals: Cats, dogs, bears. # of cats is $c$, # of dogs is $d$ and # of bears is $b$. It's said that $$c>d>b.$$ The question is:
Which expression is greater $\dfrac{b}{c+d+b}$ or $\dfrac{1}{3}$?
The thing that I didn't understand is what $\dfrac{b}{c+d+b}$ actually means. And how can I solve for it?
| I prefer to keep things in fractional form, and to try to get the same numerator to compare.
$$\frac{b}{c+d+b} = \frac{b}{c+d+b}\cdot\frac{\frac{1}{b}}{\frac{1}{b}}= \frac{\frac{b}{b}}{\frac{c+d+b}{b}}=\frac{1}{\frac{c}{b}+\frac{d}{b}+1}\lt\frac{1}{3}$$
| {
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Number of consecutive integers such that the product is $1$ $\pmod n$ Define a function for $n$ $>$ $1$ which is $f(n)$ $=$ $k$, the maximum number of consecutive integers such that their product is $1$ $\pmod n$, and every integer $1 \leq x \leq k$, there exists $x$ consecutive integers such that their product is $1$ $\pmod n$.
For prime $n$, does $f(n)$ $=$ $n-2$?
Example (not a counterexample for the question above);
$n=11$, $f(11)$ $=$ $3$ because there are $1, 2, 3$ consecutive integers such that their product is $1$ $\pmod{11}$. But there are no $4$ consecutive integers such that their product is $1$ $\pmod {11}$. There are $9$ consecutive integers such that their product is $1$ $\pmod {11}$, but $f(11)$ $=$ $3$ due to the fact that the product of $4$ consecutive integers is never $1$ $\pmod {11}$.
$(1)$ $=$ $(1)$ $\pmod {11}$
$(3 \times 4)$, $(8 \times 7)$ $=$ $1$ $\pmod {11}$
$(5 \times 6 \times 7)$ $=$ $1$ $\pmod {11}$
Can anyone list some extraordinarily large values for $f(n)$, compared to most values of $f(n)$? The example is the largest I could come up with. Thanks for help, feedback, etc.
This question may be helpful.
| Another result along the lines of the one in the OP's link is the following:
Statement: If $p$ is prime and $f(p) > 3$, then $p \equiv \pm 1 \pmod{8}$.
Proof: Suppose $f(p) \geqslant 4$; then there exist 4 consecutive integers whose product is 1 mod $p$:
$$(k-1)k(k+1)(k+2) \equiv 1 \pmod{p}.$$
Simple algebraic manipulations show that the above is equivalent to
$$(k^2+k-1)^2 \equiv 2 \pmod{p},$$
hence 2 is a quadratic residue modulo $p$, which implies that $p \equiv \pm 1 \pmod{8}$.
Hence to get large values of $f(p)$, it suffices to check primes $p$ that are $\pm 1 \pmod{8}$ and $\pm 1 \pmod{5}$, hence $\pm 1$ or $\pm 9 \pmod{40}$. My computer tells me that of the first 50000 primes, the maximum value of $f(p)$ is $f(321721) = 15$, with $1, 2, 3, \ldots, 15$ runs starting respectively at at $$1, 46396, 240095, 51194, 37879, 124075, 94231, 137236, 142566, 31506, 193600, 140210, 234595, 121755, 12219$$
| {
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gcd of power plus one I know that $$\gcd(a^b-1,a^c-1)=a^{\gcd(b,c)}-1$$
which can be seen by expanding both terms into geometric series. Is there any such simplification for
$$\gcd(a^b+1,a^c+1)=?$$
EDIT: When working with polynomials instead of whole numbers, it is
$$gcd(x^n+1,x^m+1)=\begin{cases}a^{\gcd(n,m)}+1 & \nu_2(n) =\nu_2(m)\\1 &\text{else}\end{cases}$$
But this gives only a lower bound for the whole-number case.
| Let's first deal with the cases $a = 0$ and $a = 1$ separately.
For $a = 0$ we have $a^m + 1 = 1$ if $m > 0$ and $a^0 + 1 = 2$, so $\gcd(0^b+1, 0^c+1) = 1$ if $b > 0$ or $c > 0$, and $\gcd(0^0+1, 0^0+1) = 0^0+1$. So $\gcd(0^b+1, 0^c+1) = 0^{\gcd(b,c)} + 1$. For $a = 1$ we have $a^m + 1 = 2$ for all $m \in \mathbb{N}$ and hence $\gcd(1^b+1, 1^c+1) = \gcd(2,2) = 2 = 1^{\gcd(b,c)} + 1$.
Next we consider $a \geqslant 2$. We know
$$\gcd(a^{2b}-1, a^{2c}-1) = a^{\gcd(2b,2c)}-1 = a^{2\gcd(b,c)}-1 = (a^{\gcd(b,c)}-1)(a^{\gcd(b,c)}+1)$$
and $\gcd(a^b-1,a^c-1) = a^{gcd(b,c)}-1$, hence
$$\gcd(a^b+1,a^c+1) \mid \gcd\biggl((a^b+1)\frac{a^b-1}{a^{\gcd(b,c)}-1}, (a^c+1)\frac{a^c-1}{a^{\gcd(b,c)}-1}\biggr) = a^{\gcd(b,c)}+1.$$
If $\nu_2(b) = \nu_2(c)$, then
$$a^{\gcd(b,c)}+1 \mid \gcd(a^b+1,a^c+1),$$
so in this case we have $\gcd(a^b+1,a^c+1) = a^{\gcd(b,c)}+1$.
If $p$ is an odd prime dividing $a^m+1$, then $\operatorname{ord}_p(a) \mid 2m$ but $\operatorname{ord}_p(a) \nmid m$, hence it follows that $\nu_2\bigl(\operatorname{ord}_p(a)\bigr) = \nu_2(m) + 1$. Thus, if $\nu_2(b) \neq \nu_2(c)$, no odd prime divides $\gcd(a^b+1,a^c+1)$, i.e.
$$\gcd(a^b+1,a^c+1) = 2^k$$
for some $k$. If $a$ is even, then $k = 0$ since $a^{\max \{b,c\}}+1$ is odd. The square of an odd number is $\equiv 1 \pmod{4}$, so $\nu_2(a^m+1) = 1$ if $a$ is odd and $\nu_2(m) \geqslant 1$.
We thus have
$$\gcd(a^b+1,a^c+1) = \begin{cases} a^{\gcd(b,c)}+1 &\text{if } \nu_2(b) = \nu_2(c) \\ \qquad 2 &\text{if } \nu_2(b) \neq \nu_2(c)\text{ and } a \equiv 1 \pmod{2} \\ \qquad 1 &\text{if } \nu_2(b) \neq \nu_2(c) \text{ and } a \equiv 0 \pmod{2}\end{cases} \tag{$\ast$}$$
for $a \geqslant 2$. Though $\gcd(a^b+1,a^c+1) = a^{\gcd(b,c)}+1$ for $a\in \{0,1\}$ and arbitary $b,c \in \mathbb{N}$, this also matches $(\ast)$.
For $a < 0$, we have $\gcd(a^b+1,a^c+1) = \gcd(\lvert a\rvert^b - 1, \lvert a\rvert^c-1) = \lvert a\rvert^{\gcd(b,c)}-1$ if $b$ and $c$ are both odd and $\gcd(a^b+1,a^c+1) = \gcd(\lvert a\rvert^b+1, \lvert a\rvert^c+1)$ if both are even. These give nothing new. If one exponent is even and the other odd, say $b$ is even, then, using $d = -a$, we have $\gcd(a^b+1,a^c+1) = \gcd(d^b + 1, d^c - 1)$ and
$$\gcd(d^b+1,d^c-1) \mid \gcd(d^{2b}-1,d^c-1) = d^{\gcd(2b,c)}-1 = d^{\gcd(b,c)}-1 = \gcd(d^b-1,d^c-1).$$
Since $\gcd(d^b+1,d^b-1) = \gcd(d^b+1,2) \mid 2$, it follows that $\gcd(d^b+1,d^c-1) = 1$ if $d$ is even, and $\gcd(d^b+1,d^c-1) = 2$ if $d$ is odd. Noting that $\lvert a\rvert^m - 1 = -(a^m+1)$ for $a < 0$ and odd $m$, we see that $(\ast)$ holds for all $a \in \mathbb{Z}$ if we don't insist on a non-negative $\gcd$. If we do,
$$\gcd(a^b+1,a^c+1) = \begin{cases}\lvert a^{\gcd(b,c)}+1\rvert &\text{if } \nu_2(b) = \nu_2(c) \\ \qquad 2 &\text{if } \nu_2(b) \neq \nu_2(c)\text{ and } a \equiv 1 \pmod{2} \\ \qquad 1 &\text{if } \nu_2(b) \neq \nu_2(c) \text{ and } a \equiv 0 \pmod{2}\end{cases} \tag{$\ast\ast$}$$
holds for all $a\in \mathbb{Z}$ and $b,c \in \mathbb{N}$.
| {
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Maximize $P=a^2+b^2+c^2+ab+ac+bc$
For real numbers $a, b, c$ that satisfy $a + b + c = 6$ and $0\leq a,b,c \leq 4$,
maximize $P=a^2+b^2+c^2+ab+ac+bc$.
My try:
$$\begin{align} \begin{cases} a+b+c=6(1) \\ 0\leq a,b,c\leq4(2) \end{cases} \end{align}$$ $$(1)\Rightarrow \begin{align} \begin{cases} b+c=(6-a) \\b^2+c^2+bc=(6-a)^2-bc \end{cases} \end{align}$$ $$P=a^2+(b^2+c^2+bc)+a(b+c)=a^2+[(6-a)^2-bc]+a(6-a)$$ $$P=(a^2-12a+36)-bc=(a-6)^2-bc (2)\Rightarrow bc\leq 0 \Rightarrow P\geq (a-6)^2$$ When $bc=0 \Rightarrow [{\begin{matrix}b=0\\c=0\end{matrix}}(3)$. From $(1)$ and $(3)$, $\Rightarrow 2\leq a\leq 4(4)$ $P_{max} \implies |a-6|$ max satisfy $(4)$ $\implies a=2$ from $(1)$ and $(3)$ $\implies b=c=4$ $\implies P_{max}(a,b,c)=P(4;2;0)=28$
| The stationary points of a quadratic form over a triangle (or hexagon) are simple to locate through Lagrange's multipliers. Your quadratic form is positive definite with eigenvectors $(1,1,1)$, $(-1,0,1)$, $(-1,1,0)$ hence the maximum value is attained on the boundary of the given domain, by convexity. By analyzing the instances $a=0$ and $a=4$ we easily get that the maximum is $28$, attained at $\{a,b,c\}=\{0,2,4\}$.
| {
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Find all integers $k$ such that $\frac{k^4+k^3-5k^2+k}{k^2+k-6}$ is an integer. Find all integers $k$ such that $\frac{k^4+k^3-5k^2+k}{k^2+k-6}$ is an integer.
I tried factorization and ended up with $\frac{k(k^3+k^2-5k+1)}{(k-2)(k+3)}$. Please give me some hints. Thank you!
| HINT:
$$\frac{k^4+k^3-5k^2+k}{k^2+k-6}=\frac{(k^2+1)(k^2+k-6)+6}{k^2+k-6}$$
| {
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Prove that $\frac{x^2}{2} > x\cos x - \sin x$ for every $x\neq0$ I'm trying to prove that $\frac{x^2}{2} > x\cos x - \sin x$ for every $x\neq0$
$f(x) =\frac{x^2}{2} - x\cos x + \sin x $.
I need to prove that $f > 0 $ for every $x\neq 0$.
$f(0) = 0$.
$f'(x) = x(1+\sin x)$
for every $x > 0$,
$ f'(x) \geq 0$ and this means that $f$ is strictly growing in that interval meaning that for $x>0, f(x)>0$
However for negative values $f'(x)$ is not always positive. If so, how can I prove that $f >0 $ for those values as well?
| We have that $\frac{x\cos x-\sin x}{x^2}=\frac{d}{dx}\left(\frac{\sin x}{x}\right)$ is an odd function, hence in order to prove is is bounded by $\frac{1}{2}$ for any $x\in\mathbb{R}$ it is enough to show that its absolute value is bounded by $\frac{1}{2}$ for any $x\geq 0$.
If $x\geq 0$ we have $\sin(x)\leq\min(x,1)$, hence
$$ \forall t\geq 0,\qquad t\sin(t) \leq \min(t,t^2)\tag{1} $$
and by integrating both sides of $(1)$ over the interval $(0,x)$ we get:
$$ \forall x\geq 1,\qquad \sin(x)-x\cos(x)\leq \frac{x^2}{2}-\frac{1}{6} \tag{2}$$
$$ \forall x\in[0,1],\qquad \sin(x)-x\cos(x)\leq \frac{x^3}{6}\tag{3} $$
It follows that over the interval $\left[0,\sqrt{\frac{7}{3}+2\sqrt{\frac{7}{3}}}\right]$ the function $f(x)=\frac{\sin x-x\cos x}{x^2} $ is positive and bounded by $\frac{1}{5}\left(3-\sqrt{\frac{3}{7}}\right)$. On the other hand, by the Cauchy-Schwarz inequality we have $\left|x\cos(x)-\sin(x)\right|\leq \sqrt{x^2+1}$, hence
$$ \left|\frac{x\cos(x)-\sin(x)}{x^2}\right|\leq \frac{1}{5}\left(3-\sqrt{\frac{3}{7}}\right) \tag{4} $$
is trivial for any $x\geq \sqrt{\frac{7}{3}+2\sqrt{\frac{7}{3}}}$. This proves the slightly stronger inequality
$$ \forall x\in\mathbb{R},\qquad \frac{x\cos(x)-\sin(x)}{x^2}\leq \color{red}{\frac{1}{5}\left(3-\sqrt{\frac{3}{7}}\right)}=0.4690692658584\ldots\tag{5}$$
| {
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What is $\lim_{(x,y)\rightarrow(0,0)} \frac{x^3-y^3}{x^2-y^2}$? Given is the function $f(x,y)=\frac{x^3-y^3}{x^2-y^2}$ defined for $(x,y)\in\mathbb{R^2}, \, x^2\neq y^2$
I need to find the limit of $\lim_{(x,y)\rightarrow(0,0)} \frac{x^3-y^3}{x^2-y^2}$.
I have tried a few paths, $(x,mx),\, (x,mx^2), \, (x,sin(x)), \, (x,tan(x))$ but all of them yield the same result, $0$.
So I tried to prove the limit exists using the epsilon-delta definition. So:
$$\left|\frac{x^3-y^3}{x^2-y^2} - 0\right| = \left|\frac{(x-y)(x^2+y^2+xy)}{(x-y)(x+y)}\right|= \frac{x^2+y^2+xy}{\left|x+y\right|}<\epsilon$$
Here I am stuck and do not know how to proceed.
| The problem is that there is no limit, only pathwise limits (which of course depend on the path to zero).
To see this one could rewrite the expression:
$${x^3-y^3\over x^2-y^2} = {(x-y)(x^2+xy+y^2)\over(x-y)(x+y)} = {(x+y)x - y^2\over (x+y)} = x - {y^2\over x+y}$$
Now of course $x\to 0$, but the term $y^2/(x+y)$ can be selected arbitrarily by having $x = y^2/C - y$. So along the parabola $x = y^2/C - y$ aproaching $0$ we have the limit of the expression being $-C$.
But this can't happen if the limit actually exists.
| {
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Prove that for $x>0$ we have $\sin{x}<\frac{x}{\sqrt{1+\frac{x^2}{3}}}$ Prove the inequality:
$$\sin{x}<\frac{x}{\sqrt{1+\frac{x^2}{3}}}$$
for real $x>0$.
Any hints?
EDIT: Using calculus and derivatives is perfectly allowed.
| This is an incomplete answer.
Perhaps you can do it easily for $x>\sqrt{\frac{3}{2}}$
$1+\frac{x^2}{3}<x^2 \hspace{1cm} for \hspace{0.1cm}x>\sqrt{\frac{3}{2}}$
Then
$(sin^2x+cos^2x)(1+\frac{x^2}{3})<x^2 \hspace{1cm}$
Therefore
$(sin^2x)(1+\frac{x^2}{3})<(sin^2x+cos^2x)(1+\frac{x^2}{3})<x^2 \hspace{1cm} \implies $
$(sinx)\sqrt{(1+\frac{x^2}{3})}<x \implies \sin{x}<\frac{x}{\sqrt{1+\frac{x^2}{3}}}$
| {
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Bizarre Definite Integral Does the following equality hold?
$$\large \int_0^1 \frac{\tan^{-1}{\left(\frac{88\sqrt{21}}{215+36x^2}\right)}}{\sqrt{1-x^2}} \, \text{d}x = \frac{\pi^2}{6}$$
The supposed equality holds to 61 decimal places in Mathematica, which fails to numerically evaluate it after anything greater than 71 digits of working precision. I am unsure of it's correctness, and I struggle to prove it's correctness.
The only progress I have in solving this is the following identity, which holds for all real $x$:
$$\tan^{-1}{\left( \frac{11+6x}{4\sqrt{21}} \right )} + \tan^{-1}{\left( \frac{11-6x}{4\sqrt{21}} \right )} \equiv \tan^{-1}{\left(\frac{88\sqrt{21}}{215+36x^2}\right)}$$
I also tried the Euler Substitution $t^2 = \frac{1-x}{1+x}$ but it looks horrible.
Addition: Is there some kind of general form to this integral?
Side thoughts: Perhaps this is transformable into the Generalised Ahmed's Integral, or something similar.
| As pointed out in one of the comments, user @Start wearing purple demonstrated a very general approach for solving this kind of integral, see this. As an alternative approach, let me give a different argument that appeals to a specific property satisfied by OP's integral.
Step 1. (Reduction and the main claim) We begin by substituting $x = \cos(\theta/2)$. Then the integral equals
$$ \frac{1}{2} \int_{0}^{\pi} \arctan\left(\frac{88\sqrt{21}}{233+18\cos\theta}\right) \, d\theta = \frac{\pi}{4} - \frac{1}{2} \int_{0}^{\pi} \arctan\left(\frac{233+18\cos\theta}{88\sqrt{21}}\right) \, d\theta. $$
So it suffices to prove that
$$ \int_{0}^{\pi} \arctan\left(\frac{233+18\cos\theta}{88\sqrt{21}}\right) \, d\theta \stackrel{?}{=} \frac{\pi^2}{6}. \tag{1} $$
To evaluate this integral, let me give the punchline.
Claim. Let $0 < a <1$ and $b > 0$ satisfy $4a^2 - b^2 = \frac{4}{3}$. Then
$$ \int_{0}^{\pi} \arctan(a + b\cos\theta) \, d\theta = \frac{\pi^2}{6}. $$
Notice that $(a, b) = \left( \frac{233}{88\sqrt{21}}, \frac{18}{88\sqrt{21}} \right)$ satisfies the relation in the assertion of Claim. So we focus on proving this claim.
Step 2. (Definition and properties of $I$)
Now define $I(a, b)$ by
$$ I(a, b) = \int_{0}^{\pi} \arctan(a + b\cos\theta) \, d\theta. $$
From the substitution $\theta \mapsto \pi - \theta$, it is clear that $I(a,-b) = I(a, b)$. Then for $0 < a < 1$ and $0 < \theta < \pi$, we have
\begin{align*}
&\arctan(a + b\cos\theta) + \arctan(a - b\cos\theta) \\
&\hspace{1em}= \arctan\left( \frac{2a}{1-(a^2-b^2\cos^2\theta)} \right) \\
&\hspace{2em}= \arctan\left( \frac{4a}{2-2a^2+b^2+b^2\cos(2\theta)} \right) \\
&\hspace{3em}= \frac{\pi}{2} - \arctan\left( \frac{2-2a^2+b^2}{4a} + \frac{b^2}{4a}\cos(2\theta) \right).
\end{align*}
Plugging this back and exploiting the symmetry of cosine, we have
$$ I(a, b) = \frac{\pi^2}{4} - \frac{1}{2}I\left( \frac{2-2a^2+b^2}{4a}, \frac{b^2}{4a} \right). \tag{2} $$
Step 3. Now here comes the central observation. Let $(a, b)$ satisfy $0 < a < 1$ and $b > 0$, and define the sequence $(a_n, b_n)$ recursively by
$$ (a_0, b_0) = (a, b), \qquad (a_{n+1}, b_{n+1}) = \left( \frac{2-2a_n^2+b_n^2}{4a_n}, \frac{b_n^2}{4a_n} \right). $$
Observation. Assume that $4a^2 - b^2 = \frac{4}{3}$. Then for all $n \geq 0$ we have
$$ \frac{1}{\sqrt{3}} \leq a_{n+1} \leq a_n, \qquad 4a_n^2 - b_n^2 = \frac{4}{3}. $$
The proof is a tedious algebra, so we skip this. Now by this observation, we have $|a_n| < 1$ for all $n$. Then a recursive application of $\text{(2)}$ gives
$$ I(a, b) = \frac{\pi^2}{4}\sum_{k=0}^{n-1} \left(-\frac{1}{2}\right)^k + \left(-\frac{1}{2}\right)^n I(a_n, b_n). $$
Since $|I(a_n, b_n)| \leq \frac{\pi^2}{2}$ for all $n$, taking limist as $n\to\infty$ proves the claim.
Remark. (1) The condition $4a^2 - b^2 = \frac{4}{3}$ is crucial for our proof. For arbitrary starting point $(a, b)$, the sequence $(a_n, b_n)$ is dynamically unstable and hence the formula $\text{(2)}$ is not applicable.
(2) The claim is true for any $a > 0$ in view of the principle of analytic continuation.
(3) Again, @Start wearing purple's computation gives a more general result with a relatively economic computation: for all $a, b \in \Bbb{R}$,
$$ \int_{0}^{\pi} \arctan(a + b\cos\theta) \, d\theta = \pi \arg \left(1 + ia + \sqrt{b^2 + (1+ia)^2}\right). \tag{3} $$
This follows from the formula
$$ \int_{0}^{\pi} \log(1 + s \cos\theta) \, d\theta = \pi \log\left( \frac{1 + \sqrt{1-s^2}}{2} \right) $$
which is valid for any complex $s$ with $|s| < 1$. Our relation $4a^2 - b^2 = \frac{4}{3}$ ensures that the RHS of $\text{(3)}$ is always $\frac{\pi^2}{6}$, since $1 + ia + \sqrt{b^2 + (1+ia)^2} = (1+\sqrt{3}a)\left( 1 + \frac{i}{\sqrt{3}} \right)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2159282",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "27",
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$\lim_{ x \to 0 }\left( \frac{\sin 3x}{x^3}+\frac{a}{x^2}+b \right)=0$ if :
$$\lim_{ x \to 0 }\left( \frac{\sin 3x}{x^3}+\frac{a}{x^2}+b \right)=0$$
then $a+b=?$
Without the use of the L'Hôspital's Rule
My Try :
$$\lim_{ x \to 0 }\left( \frac{ax+bx^3+\sin 3x}{x^3} \right)=0$$
$$\lim_{ x \to 0 }\left( \frac{x(a+bx^2)+\sin 3x}{x^3} \right)=0$$
$$\lim_{ x \to 0 }x(a+bx^2)+\sin 3x=0 $$
now ?
| With Taylor expansion
$$
\begin{aligned}
\lim _{x\to 0}\left(\frac{\sin \left(3x\right)+ax+bx^3}{x^3}\right)
& = \lim _{x\to \:0}\left(\frac{\left(3x-\frac{9}{2}x^3+\frac{81}{40}x^5+o\left(x^5\right)\right)+ax+bx^3}{x^3}\right)
\\ & = \lim _{x\to \:0}\left(\frac{81x^2}{40}\right) = 0
\end{aligned}
$$
Then $a$ and $b$ must have the following values
$$a = -3, b = 9/2$$
So:
$$\lim _{x\to \:0}\left(\frac{\sin \left(3x\right)-3x+\frac{9}{2}x^3}{x^3}\right) = 0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2160584",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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The equation $a^x +a^{-x}=b$ Given $a$ is a fixed real number that is greater than one, how many real numbers $b$ are there such that the equation $^{x} + a^{-x} = $ has a unique real solution x?
To solve this problem, I tried to plug in $a = 3$.
So, I want to find how many reals $b$ there are such that $3^{x} + 3^{-x} = b$ has a unique real solution x.
Setting $3^x$ = $t$, I got that $t + \frac{1}{t} = b$. It follows that $t^2 - tb + 1 = 0$. From this, it can be obtained that $t = \frac{b \pm \sqrt{b^2-4}}{2}$.
I set the discriminant equal to zero: $b^{2}-4=0$, so $b = \pm 2$. This means that $t = \pm 1$. Since $t = 3^{x}$ (previously defined), $x$ is real only when $t = 1$.
From the work, doesn't it follow that there is only one real number $b$ (that is when $b$ = 2) such that there is one unique real solution $x$? But the answer I see says there are an infinite values of $b$ such that there is one real solution x.
What am I missing in my analysis?
| $a^0 + a^0 = 2$
For $b \ge 2$ then $a^{\log_a b} + a^{-\log_a b} > a^{\log_a b} = b$.
As $a^x + a^{-x}$ is continuous there must be at least one solution $a^x + a^{-x} = b; x > 0$.
If we note that $a^x + a^{-x} $ is strictly increasing (look at the derivative; $a^x\ln {a} - a^{-x}\ln{a} > 0$ as $a^x > a^{-x}$) the positive solution is unique. (Clearly if $x$ is a solution $-x$ is as well.)
Remains to show $a^{x}+ a^{-x} \ge 2$ with equality holding only if $x = 0$. It's irritating but straightforward to prove for positive $y$ that $y + 1/y \ge 2$ ($y + 1/y \ge 2 \iff y^2 + 1 \ge 2y \iff (y^2 - 2y + 1) = (y-1)^2 \ge 0$).
[actually as $a^{x} + a^{-x}$ is increasing if $x > 0$ and decreasing on $x < 0$, this result is automatic.)
So $b < 2$ has no solution. $b = 2$ has solution $x = 0$. and for $b > 2$ there will be two solutions, one positive, one negative, of equal magnitude.
| {
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"timestamp": "2023-03-29T00:00:00",
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To find area of quadrilateral $ABCD$ using the given co-ordinates. I am unable to get the answer to this question. The question is to find the area of a quadrilateral having its vertices as coordinates in order:
$A(3,-2)$; $B(4,0)$; $C(6,-3)$ and $D(5,-5)$.
I divided this into 2 triangles - $\Delta ABC$ and $\Delta BCD$. Now by the area of triangle formula, I am getting the total area to be $35$ square units but my tuition partner is getting a very different answer $7$ square units to be precise.
| Here's a graph so you can see what is going on visually.
By looking at the graph, or explicitly finding the equations of the lines $BC$, $AD$, $CD$, $AB$ we can see that $AB$ is parallel to $AD$ and $CD$. is parallel to $AB$.
Note that also $|BC| = |CD|$ and $|AB|=|CD|$.
Then we can deduce this quadrilateral is a parallelogram. The area is then equal to the product of the base and its height. That is the product of $AD$ and the perpendicular distance between $AD$ and $BC$.The equation of the line $AD$ is given by $y =-\frac{3}{2}x + \frac{5}{2}$
The gradient equation of the normal to this line is then given by $\frac{-1}{-\frac{3}{2}} = \frac{2}{3}$
This normal starts has a point on $AD$ and a corresponding point on $BC$
The equation of the normal is then given by $y = \frac{2}{3} x -4$
The equation of the line $BC$ is given by $y= -\frac{3}{2}x + 6$
Set these two equations equal to find the point of intersection:
$-\frac{3}{2}x + 6 = \frac{2}{3}x -4 \Rightarrow x =\frac{60}{13}, y = -\frac{12}{13}$
Let's call this point of intersection on $BC$ point $ E = (\frac{60}{13},-\frac{12}{13})$.
Then we have $|AE| = \sqrt{(\frac{60}{13} - 3)^{2} + (\frac{12}{13} - - 2)^{2}} = \sqrt{\frac{145}{13}}$
$|AD| = \sqrt{(3-5)^{2}+(-2+5)^{2}} = \sqrt{13}$
Then the area of $ABCD = \sqrt{\frac{145}{13}} \cdot \sqrt{13} = \sqrt{145}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2163818",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Convergence or divergence of the series $1+\frac{1}{2}-\frac{1}{3}+\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+...$ I'm taking undergrad real analysis. I'm trying to come up with a general formula for the series
\begin{equation}
1+\frac{1}{2}-\frac{1}{3}+\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+...
\end{equation} and cannot understand how to get the negative term to only show up with powers of $3$. Or is there another way to test the series. any help would be appreciated.
| $$ \begin{align}
&\quad\, \sum_{n=0}^{\infty}\left[+\frac{1}{3n+1}+\frac{1}{3n+2}\color{red}{-}\frac{1}{3n+3}\right] \\[2mm]
&= \sum_{n=0}^{\infty}\left[\frac{1}{3n+1}+\frac{1}{3n+2}\color{red}{-}\frac{2}{3n+3}+\frac{1}{3n+3}\right] \\[2mm]
&= \sum_{n=0}^{\infty}\left[\frac{1}{3n+1}+\frac{1}{3n+2}\color{red}{-}\frac{2}{3n+3}\right]+\sum_{n=0}^{\infty}\left[\frac{1}{3n+3}\right] \\[2mm]
&= \log{3}+\frac{1}{3}\sum_{n=1}^{\infty}\frac{1}{n} \color{red}{\,\,\rightarrow\,\,\infty} \\[2mm]
\end{align} $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2164942",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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Find the limit without L'hopital rule $\lim_{ x \to 1}\frac{1-\cot(\frac{π}{4}x)}{\sin πx}=$? Find the limit without L'hopital rule
$$\lim_{ x \to 1}\frac{1-\cot(\frac{π}{4}x)}{\sin πx}=?$$
My Try:
$$1-\cot( \frac{π}{4}x)=1-\frac{1}{\tan( \frac{π}{4}x)}=\frac{\tan( \frac{π}{4}x)-1}{\tan( \frac{π}{4}x)}\\\sin (πx)=\sin (\pi-\pi x)=-\sin \pi(x-1)\\\lim_{ x \to 1}\frac{1-\cot(\frac{π}{4}x)}{\sin πx}=\lim_{ x \to 1}\frac{\frac{\tan( \frac{π}{4}x)-1}{\tan( \frac{π}{4}x)}}{-\sin \pi(x-1)}\\u=x-1⇒x=u+1\\\lim_{ u \to 0}\frac{\frac{\tan( \frac{π}{4}(u+1))-1)}{\tan( \frac{π}{4}(u+1))}}{-\sin \pi u}$$
now ?
| Just another way using Taylor series.
Built around $t=a$, Taylor series are $$\cot(t)=\cot (a)- \left(1+\cot ^2(a)\right)(t-a)+O\left((t-a)^2\right)\tag1$$ $$\sin(t)=\sin (a)+ \cos (a)(t-a)+O\left((t-a)^2\right)\tag2$$ So, replacing by the proper values $$\cot \left(\frac{\pi x}{4}\right)=1-\frac{\pi}{2} (x-1)+O\left((x-1)^2\right)$$ $$\sin(\pi x)=-\pi (x-1)+O\left((x-1)^2\right)$$ from which it is easy to conclude.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2165038",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 7,
"answer_id": 5
} |
Finding maximum of $x+y$ Let x and y be real numbers satisfying $9x^{2} + 16y^{2} = 1$. Then $x + y$ is maximum when
a. $y = \frac{9x}{16}$
b. $y = −\frac{9x}{16}$
c. $y = \frac{4x}{3}$
d. $y = −\frac{4x}{3}$
| with Holder,s Inequality
$$(9x^2+16y^2)\cdot \left[\frac{1}{3^2}+\frac{1}{4^2}\right]\geq \bigg[\bigg(16y^2 \cdot \frac{1}{9}\bigg)^{\frac{1}{2}}+\bigg(9x^2 \cdot \frac{1}{16}\bigg)^{\frac{1}{2}}\bigg]^{2}$$
so $$(x+y)^2 \leq \frac{25}{144}\Leftrightarrow (x+y)\leq \frac{5}{12}$$
and equality hold when $$\frac{9x^2}{\frac{1}{3^2}} = \frac{16y^2}{\frac{1}{4^2}}\Leftrightarrow (9x^2)^2 = (16y^2)^2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2166855",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 8,
"answer_id": 7
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Find the closed form of a summation from $k=1$ to $n$ For my discrete mathematics class, I need to express this summation in closed form in terms of $n$,
$$\sum_{k=1}^n \left(6 + 2 \cdot \frac{k}{n}\right)^2. $$
I was in the hospital when they went over this in class, and they were not working out of the textbook. I've looked extensively online and found plenty of cases where people solved problems like this, but no one explained how they did it.
I know the first thing to do is to multiply out the expression, but I'm not sure how to proceed from there. Any help would be hugely appreciated.
Based on the responses you all have given me, I have an idea of how I should solve this. Can anyone verify that this is correct?
so:
$$\sum_{k=1}^n \left(6 + 2 \cdot \frac{k}{n}\right)^2 $$
$$\equiv \sum_{k=1}^n 36 + \sum_{k=1}^n \frac{24k}{n} + \sum_{k=1}^n \frac{4k^2}{n^2}$$
$$\equiv 36 + \frac{24}{n} \sum_{k=1}^n k + \frac{4}{n^2} \sum_{k=1}^n k^2$$
$$\equiv 36 + \frac{24}{n} \cdot \frac{n(n + 1)}{2} + \frac{4}{n^2} \cdot\frac{n(n + 1)(2n + 1)}{6}$$
..and then simplify? Is that correct? Thanks!
| Just do it.
$(6 + 2\frac kn)^2 = 36 + \frac 24n k + \frac 4{n^2}k^2$
So $\sum_{k=1}^n (6 + 2\frac kn)^2=\sum_{k=1}^n(36 + \frac 24n k + \frac 4{n^2}k^2)=$
$\sum_{k=1}^n 36 + \frac 24n\sum_{k=1}^n k + \frac4{n^2}\sum_{k=1}^n k^2=$
$36n + \frac 24n\frac {n(n+1)}2 + \frac 4{n^2}\frac{n(n+1)(2n+1)}6=$
$36n +12(n+1) +\frac 2{3n}(n+1)(2n+1)=$
$\frac {144n^2 + 12n +2(n^2+ 3n+1)}{3n}=$
$\frac {146n^2 + 18n + 2}{3n}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2168001",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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For $a,b,c<1$. Minimize $M=\frac{a^2\left(1-2b\right)}{b}+\frac{b^2\left(1-2c\right)}{c}+\frac{c^2\left(1-2a\right)}{a}$ For $0<a,b,c<1$ and $ab+bc+ca=1$, minimize $$M=\frac{a^2\left(1-2b\right)}{b}+\frac{b^2\left(1-2c\right)}{c}+\frac{c^2\left(1-2a\right)}{a}$$
My Try:
From $ab+bc+ca=1\Leftrightarrow \sqrt{3}\leq a+b+c<3$
We have: $M=\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}-2\left(a^2+b^2+c^2\right)$
$\ge\frac{\left(a+b+c\right)^2}{a+b+c}-2\left(a^2+b^2+c^2\right)-4\left(ab+bc+ca\right)+4\left(ab+bc+ca\right)$
$=a+b+c-2(a+b+c)^2+4 (1)$
Let $a+b+c=x (\sqrt {3}\leq x<3)$
We find Min of function $y=-2x^2+x+4$
And $y_{Min}=-2+\sqrt {3}$ when $x=\sqrt {3}$
Right or Wrong ?
| Let $a=b=c=\frac{1}{\sqrt3}$.
Hence, $M=\sqrt3-2$.
We'll prove that it's a minimal value.
Indeed, Let $c=\min\{a,b,c\}$. Hence, we need to prove that
$$\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}-\frac{2(a^2+b^2+c^2)}{\sqrt{ab+ac+bc}}\geq(\sqrt3-2)\sqrt{ab+ac+bc}$$ or
$$\sqrt{ab+ac+bc}\left(\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}-\sqrt{3(ab+ac+bc)}\right)\geq2(a^2+b^2+c^2-ab-ac-bc)$$ or
$$\frac{a^2}{b}+\frac{b^2}{a}-a-b+\frac{b^2}{c}+\frac{c^2}{a}-\frac{b^2}{a}-c+a+b+c-\sqrt{3(ab+ac+bc)}\geq$$
$$\geq2((a-b)^2+(c-a)(c-b))$$ or
$$\frac{(a-b)^2(a+b)}{ab}+\frac{(c-a)(c^2-b^2)}{ac}+a+b+c-\sqrt{3(ab+ac+bc)}\geq$$
$$\geq2((a-b)^2+(c-a)(c-b))$$ or
$$(a-b)^2\left(\frac{1}{a}+\frac{1}{b}-2\right)+(c-a)(c-b)\left(\frac{1}{a}+\frac{b}{ac}-2\right)+a+b+c-\sqrt{3(ab+ac+bc)}\geq0,$$
which is true because $\frac{1}{a}+\frac{1}{b}-2>0$, $\frac{1}{a}+\frac{b}{ac}-2\geq\frac{1}{a}+\frac{1}{a}-2>0$ and
$$a+b+c-\sqrt{3(ab+ac+bc)}=\frac{(a-b)^2+(c-a)(c-b)}{a+b+c+\sqrt{3(ab+ac+bc)}}\geq0.$$
Done!
The inequality
$$\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}-\frac{2(a^2+b^2+c^2)}{\sqrt{ab+ac+bc}}\geq(\sqrt3-2)\sqrt{ab+ac+bc}$$
is true for all positives $a$, $b$ and $c$, but my proof of the last statement is very ugly.
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "3",
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Why is this inequality true? $(a+b)^2\leq 2(a^2+b^2)$ Why is this inequality true? $a,b$ are real numbers.
$$
(a+b)^2=a^2+2ab+b^2\leq 2(a^2+b^2)
$$
I know $(a+b)^2=a^2+2ab+b^2 \geq 0$, but then?
| Because by C-S $$2(a^2+b^2)=(1^2+1^2)(a^2+b^2)\geq(1\cdot a+1\cdot b)^2=(a+b)^2$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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What does this series converge to, if anything? $$\arctan{1} + \arctan{\frac{1}{2}} + \arctan{\frac{1}{3}} + \arctan{\frac{1}{4}} ...= ?$$
The infinite series for arctan is
$$\arctan{x} = x - \frac{x^3}{3} + \frac{x^5}{5} - \frac{x^7}{7} ...$$
So I want to sum up the $\arctan{1\over n}$ where $n$ starts at $1$ and goes to infinity.
I originally thought the resulting series can be written this way:
$$(1 + \frac{1}{2} + \frac{1}{3} ...) - \frac{(1 + \frac{1}{2} + \frac{1}{3} ...)^3}{3} + \frac{(1 + \frac{1}{2} + \frac{1}{3} ...)^5}{5} - \frac{(1 + \frac{1}{2} + \frac{1}{3} ...)^7}{7} ...$$
But that is wrong. The right way to "insert" the series is:
$$1 + \frac{1}{2} + \frac{1}{3} ... - \frac{1}{3} - \frac{(\frac{1}{2})^3}{3} - \frac{(\frac{1}{3})^3}{3} ... + \frac{1}{5} + \frac{(\frac{1}{2})^5}{5} + \frac{(\frac{1}{3})^5}{5} ... ...$$
So it looks like a bunch of harmonic serieses manipulated. Harmonic series diverges, but I remember that doesn't necessarily mean a similar series diverges. I remember from Calculus 2 that, for example, $\lim_{n\to\infty}$ of $\sin(n)\over n$ converges to zero even though $\sin(n)$ does not converge.
So how do I figure out what this series converges to, if anything?
| It seems it has not yet been pointed out that your rearrangement does not work!
$\frac11 - \frac12 + \frac13 - \frac14 + \cdots = \ln(2) > \frac12$.
$\frac11 - \frac12 + \frac13 - \frac14 + \cdots \ne \color{red}{( \frac11 + \frac13 + \frac15 + \frac17 + \cdots ) - ( \frac12 + \frac14 + \frac16 + \frac18 \cdots )}$. [RHS is ill defined!]
$\frac11 - \frac12 + \frac13 - \frac14 + \cdots $
$\ \ne ( \frac11 - \frac12 - \frac14 + \frac15 + \frac17 - \frac18 - \cdots ) + \frac13 ( \frac11 - \frac12 - \frac14 + \frac15 + \frac17 - \frac18 - \cdots )$
$\ \quad + \frac1{3^2} ( \frac11 - \frac12 - \frac14 + \frac15 + \frac17 - \frac18 - \cdots ) + \frac1{3^3} ( \frac11 - \frac12 - \frac14 + \frac15 + \frac17 - \frac18 - \cdots ) + \cdots$
$\ = \frac12 \times ( \frac1{1 \times 2} - \frac1{4 \times 5} + \frac1{7 \times 8} - \cdots ) < \frac14$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2175526",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Expected number of numbers to be picked from $[0, 2n - 1]$ with repetition until two numbers sum to $2n$. Given we are picking numbers from $[0, 2n - 1]$ with repetition, how many numbers should we need to pick on the average, such that some two picked numbers sum up to $2n$.
Here is a simplified version of the question. Given we are picking numbers from the set $S = \{0, 1, 2, 3\}$ with repetition. How many numbers should we pick on the average, such that some two picked numbers sum up to 4. Monte-Carlo simulation shows that the answer is 4.222. I have no idea how to compute this analytically.
(Case n = 1) Consider the case when we are pick numbers with repetition from $S = \{0, 1\}$. For the expected number of numbers to be picked until some two picked numbers sum up to 2. We can compute the expectation like this:
$E = \frac{1}{2}(1 + E) + \frac{1}{2}(\frac{1}{2}(2) + \frac{1}{2}(1 + E_1))$, where E is the desired expectation and $E_1$ is the expected wait time until you draw a 1. So $E_1 = 2$, and $E = 4$.
| Let $E$ be the expected number of numbers drawn from $[0, 2n - 1]$ until two of the picked numbers
sum up to $2n$. Let $E_{x,y}$ be the expected number of numbers to be drawn until two of the
picked numbers sum up to $2n$, where $x$ is the number of distinct numbers seen so far belonging
to the set $S = \{1, 2, 3, ..., (n - 1), (n + 1), ..., (2n - 1)\}$. And, $y \in \{1, 2, 3, 4\}$ where 1 - no $0$ is seen, no $n$ is seen, 2 - $0$ is seen, no $n$ is seen, 3 - no $0$ is seen, $n$ is seen, 4 - $0$ is seen, $n$ is seen.
Then,
$$E = E_{0,1}$$
where,
$$E_{i,1} = \frac{1}{2n}(1 + E_{i,2}) + \frac{1}{2n}(1 + E_{i,3}) + \frac{i}{2n}(1 + E_{i,1}) + \frac{i}{2n}(1) + \frac{2n - 2i - 2}{2n}(1 + E_{i+1,1})$$
$$E_{i,2} = \frac{1}{2n}(1 + E_{i,2}) + \frac{1}{2n}(1 + E_{i,4}) + \frac{i}{2n}(1 + E_{i,2}) + \frac{i}{2n}(1) + \frac{2n - 2i - 2}{2n}(1 + E_{i+1,2})$$
$$E_{i,3} = \frac{1}{2n}(1 + E_{i,4}) + \frac{1}{2n}(1) + \frac{i}{2n}(1 + E_{i,3}) + \frac{i}{2n}(1) + \frac{2n - 2i - 2}{2n}(1 + E_{i+1,3})$$
$$E_{i,4} = \frac{1}{2n}(1 + E_{i,4}) + \frac{1}{2n}(1) + \frac{i}{2n}(1 + E_{i,4}) + \frac{i}{2n}(1) + \frac{2n - 2i - 2}{2n}(1 + E_{i+1,4})$$
The base cases are $E_{n,1} = 0, E_{n,2} = E_{n,3} = 0, E_{n,4} = 0$.
To solve this problem, all we need to do is to setup these equations and solve them. The equations can be solved simply by back-substitution. I have manually verified the answers for $n = 1$, $n = 2$ and $n = 3$.
| {
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"timestamp": "2023-03-29T00:00:00",
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integrate $\frac 1{x^2 + i}$ from $-\infty$ to $\infty$ I have a question regarding this complex integral. How do I calculate $$\int_{-\infty}^{\infty} \frac{1}{x^2 + i} \,{\rm d}x?$$ I keep getting the result $0$ and most probably doing it the wrong way.
| Hint. Alternatively, one may observe that
$$
\int_{-\infty}^\infty\frac{1}{1+x^4}\:dx=\int_{-\infty}^\infty\frac{1}{\left(x-\frac1x\right)^2+2}\:\frac{dx}{x^2}=\int_{-\infty}^\infty\frac{1}{x^2+2}\:dx=\frac{\pi }{\sqrt{2}}
$$similarly
$$
\int_{-\infty}^\infty\frac{x^2}{1+x^4}\:dx=\int_{-\infty}^\infty\frac{1}{\left(x-\frac1x\right)^2+2}\:dx=\int_{-\infty}^\infty\frac{1}{x^2+2}\:dx=\frac{\pi }{\sqrt{2}}
$$ then, by writing
$$
\int_{-\infty}^\infty\frac{dx}{x^2+i}=\int_{-\infty}^\infty\frac{x^2-i}{x^4-i^2}\:dx=\int_{-\infty}^\infty\frac{x^2\:dx}{1+x^4}-i\int_{-\infty}^\infty\frac{dx}{1+x^4}
$$ one may conclude with the preceding identities.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2182262",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Prove that $\int_0^1\frac{x\ln (1+x)}{1+x^2}dx=\frac{\pi^2}{96}+\frac{\ln^2 2}{8}$ We know that $$\int_0^1\frac{\ln (1+x)}{1+x^2}dx=\frac{\pi}8\ln 2,$$ but how about $$\int_0^1\frac{x\ln (1+x)}{1+x^2}dx?$$Prove that $$\int_0^1\frac{x\ln (1+x)}{1+x^2}dx=\frac{\pi^2}{96}+\frac{\ln^2 2}{8}.$$
| Put
\begin{equation*}
f(s) = \int_{0}^{1}\dfrac{x\ln(s+x)}{1+x^2}\, dx.
\end{equation*}
We want to determine $f(1)$. After differentiation we have
\begin{gather*}
f'(s) = \int_{0}^{1}\dfrac{x}{(s+x)(1+x^2)}\, dx = -\int_{0}^{1}\dfrac{s}{(s^2+1)(s+x)}\, dx +\int_{0}^{1}\dfrac{sx+1}{(s^2+1)(x^2+1)}\, dx =
\\[2ex]
-\dfrac{s\ln(1+s)}{1+s^2}+\dfrac{\ln s}{1+s^2}+\dfrac{s\ln 2 }{2(1+s^2)}+\dfrac{{\pi}}{4(1+s^2)}.
\end{gather*}
Now we integrate wrt $s$ between $0$ and $1$. That yields
\begin{equation*}
f(1)-f(0) = -f(1)+f(0) +\dfrac{\ln^2(2)}{4}+\dfrac{{\pi}^2}{16}.
\end{equation*}
Consequently
\begin{equation*}
f(1)= f(0) +\dfrac{\ln^2(2)}{8} + \dfrac{{\pi}^2}{32}. \tag{1}
\end{equation*}
But
\begin{equation*}
f(0) = \int_{0}^{1}\dfrac{x\ln(x)}{1+x^2}\, dx = \sum_{k=0}^{\infty}\int_{0}^{1}(-1)^kx^{2k+1}\ln x\, dx =\sum_{k=0}^{\infty}\dfrac{(-1)^{k+1}}{4(k+1)^{2}} = -\dfrac{\pi^2}{48}.
\end{equation*}
Finally we substitute that into (1).
\begin{equation*}
f(1) = \dfrac{\ln^2(2)}{8} + \dfrac{{\pi}^2}{96}.
\end{equation*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2182816",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
"answer_count": 5,
"answer_id": 3
} |
Using the definition of limit prove the following Using the definition of limit prove that
$$\lim_{x\to \infty}\frac{2x^2+x+1}{x^2-3x+1}=2$$
$$\lim_{x\to \infty}\frac{x-[x]}{x}=0$$ $[x]$ greatest integer function
Please help me to solve such type of problem using $\epsilon$-$K$ method.
| We need the following:
Definition We say that $\lim_{x\to\infty}f(x)=L$, where $L\in\Bbb R$, if for every $\epsilon>0$, we can find $K>0$ such that for all $x>K$, we have $|f(x)-L|<\epsilon$.
We have to prove that
$$\lim_{x\to \infty}\frac{2x^2+x+1}{x^2-3x+1}=2.$$
Proof: Let $\epsilon>0$. Choose $K\in\Bbb N$ such that $K>\frac{9}{2}$ and $\frac{1}{K}<\frac{\epsilon}{21}$. (Hope you get this.). Observe that
$$\begin{align}
x>\frac{9}{2}&\implies 2x>9\\
&\implies 3x-9>x\\
&\implies x-3>\frac{x}{3}\\
&\implies \frac{x-3}{7}>\frac{x}{21}\\
&\implies \frac{7}{x-3}<\frac{21}{x}.
\end{align}$$
Therefore, if $x>K>\frac{9}{2}$ then we get
$$\begin{align}
\bigg|\frac{2x^2+x+1}{x^2-3x+1}-2\bigg|&=\bigg|\frac{7x-1}{x^2-3x+1}\bigg|\\
&=\frac{7x-1}{x^2-3x+1}\\
&<\frac{7x}{x^2-3x+1}\\
&<\frac{7x}{x^2-3x}\\
&=\frac{7x}{x(x-3)}\\
&=\frac{7}{x-3}<\frac{21}{x}<\frac{21}{K}<\epsilon.
\end{align}$$
Apply the definition and we are done.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2182995",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Help with a Bernoulli DE Problem:
Solve the following differential equation.
\begin{eqnarray*}
x \frac{dy}{dx} + y &=& xy^3 \\
\end{eqnarray*}
Answer:
This is a Bernoulli equation so I use the substitution $z = y^{-2}$ to
reduce it to a linear equation.
\begin{eqnarray*}
\frac{dz}{dx} &=& -2y^{-3} \frac{dy}{dx} \\
-2xy^{-3} \frac{dy}{dx} - 2y^{-2} &=& -2x \\
\frac{dz}{dz} - 2z &=& -2x \\
\end{eqnarray*}
Now, I have reduced the equation to a linear equation, so I find $I$ an
integrating factor.
\begin{eqnarray*}
I &=& e^{\int {-2 dx} } = e^{-2x} \\
e^{-2x} \frac{dz}{dx} - 2e^{-2x}z &=& -2xe^{-2x} \\
D(ze^{-2x}) &=& -2xe^{-2x} dx \\
ze^{-2x} &=& -2 \int xe^{-2x} dx \\
\end{eqnarray*}
Now I need to evaluate the following integral:
\begin{eqnarray*}
\int xe^{-2x} dx \\
\end{eqnarray*}
This can be done using integration by parts with $u = x$ and
$dv = e^{-2x} dx$.
\begin{eqnarray*}
\int xe^{-2x} dx &=& \frac{xe^{-2x}}{-2} - \int { \frac{e^{-2x}}{-2} dx} \\
\int xe^{-2x} dx &=& \frac{-xe^{-2x}}{2} - \frac{e^{-2x}}{4} + C_1
\end{eqnarray*}
Now I substitute into the differential equation.
\begin{eqnarray*}
ze^{-2x} &=& -2 (\frac{-xe^{-2x}}{2} - \frac{e^{-2x}}{4} + C_1) \\
ze^{-2x} &=& \frac{2xe^{-2x}}{2} + \frac{2e^{-2x}}{4} + C \\
ze^{-2x} &=& xe^{-2x} + \frac{e^{-2x}}{2} + C \\
z &=& x + \frac{1}{2} + Ce^{2x} \\
y^{-2} &=& x + \frac{1}{2} + Ce^{2x} \\
y^{2} &=& \frac{1}{x + \frac{1}{2} + Ce^{2x}} \\
\end{eqnarray*}
However, the back of the book gets:
\begin{eqnarray*}
y^2 &=& \frac{1}{2x + Cx^2} \\
\end{eqnarray*}
I have good reason to believe that the back of the book is right. Could
somebody please tell me where I went wrong?
Thanks,
Bob
| Your problem is near the top: you have written \begin{align*}
-2xy^{-3} \frac{dy}{dx} -2y^{-2} &= -2x\\
\frac{dz}{dx} - 2z &= -2x.
\end{align*} Notice that there is an $x$ in the term on the far left which you dropped when you switched to $z$. This completely changes the problem. It should be $$x \frac{dz}{dx} - 2z = -2x \,\,\,\, \implies \,\,\,\, \frac{dz}{dx} - \frac{2}{x} z = -2.$$ Now the integrating factor is $x^{-2}$ so, we see $$x^{-2} \frac{dz}{dx} - 2x^{-3}z = -2x^{-2} \,\,\,\, \implies \,\,\,\, \frac{d}{dx}[x^{-2}z] = -2x^{-2}.$$ Integrating gives $$x^{-2}z = 2x^{-1} + C \,\,\,\, \implies \,\,\,\, z = 2x+Cx^2.$$ Since $z = y^{-2}$, this ends with $$y^2 = \frac{1}{2x+Cx^2}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2184805",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Help to solve: $\int 1/(x\sqrt{25-x^2})\ dx$ I'm a brand new student. Need some help to integrate this.
Perform this integration: $$\int \frac{1}{x\sqrt{25-x^2}}\ dx$$
I'm able to obtain in theta terms like this:
$$\frac 15 \ln|\cscθ-\cotθ|+C$$
But I have problems to convert in terms of "$x$"
Thanks a lot.
This is what is suggested by this image:
Using the substitution $x=5\sin{\theta}$ and $dx=5\cos{\theta}$, we obtain:
$$\int \frac{1}{5\sin{\theta}(5\cos{\theta})}\cdot 5\cos{\theta}~d\theta$$
Simplifying, we obtain:
$$\frac{1}{5}\int \frac{d\theta}{\sin{\theta}}$$
Using trigonometric identities:
$$\frac{1}{5}\int \csc{\theta}~d\theta$$
Integrating:
$$\frac{1}{5}\ln|\csc{\theta}-\cot{\theta}|+C$$
| HINT:
If trigonometric substitution is not mandatory, set $\sqrt{a^2-x^2}=y$
$$\int\dfrac{2x\ dx}{2x^2\sqrt{a^2-x^2}}=\int\dfrac{dy}{y^2-a^2}=?$$
Else setting $x=5\sin y$ where $-\dfrac\pi2\le y\le\dfrac\pi2\implies\cos y\ge0$
$\cos y=\sqrt{1-\left(\dfrac x5\right)^2}=\dfrac{\sqrt{25-x^2}}5$
$$\implies\int\dfrac{dx}{x\sqrt{25-x^2}}=\int\dfrac{5\cos y\ dy}{5\sin y(5\cos y)}=\dfrac{\ln|\csc y-\cot y|}5+K$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2191043",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
For what integer $n$ is $1+n$ divisible by $1+n^2$? For what integer $n$ is $1+n$ divisible by $1+n^2$?
I tried it in modulo $3$ and $8$ but to no avail.
Trial and error gave me answers $0$,$1$,$-2$ and $-3$.
| Supposing you mean $n^2+1$ is divisible by $n+1$, we have:
$$n^2+1=(n^2-1)+2= (n-1)(n+1)+2$$
We know that $(n-1)(n+1)$ is divisible by $n+1$, and the problem states that $n^2+1$ is also divisible by $n+1$, thus $n+1$ divides their difference $(n^2+1)-(n^2-1)=2$. So acceptable values for $n+1$ are $1$, $2$, $-1$, $-2$, thus $n=0, 1, -2, -3$ are all integer solutions for this problem.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2191170",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Maximize $P=\frac{1}{\sqrt{1+x^{2}}}+\frac{1}{\sqrt{1+y^{2}}}+\frac{1}{\sqrt{1+z^{2}}}$ For $x,y,z$ are positive real numbers that satisfy $xy+yz+xz=1$. Maximize $$P=\frac{1}{\sqrt{1+x^{2}}}+\frac{1}{\sqrt{1+y^{2}}}+\frac{1}{\sqrt{1+z^{2}}}.$$
I think if we let $x=\tan A;y=\tan B;z=\tan C$, then
$$P\Leftrightarrow \displaystyle \text {sin}A+\text {sin}B+ \text{sin}C\leq \frac{3}{2}. \hspace{2cm}(1)$$
But I can't prove $(1)$.
| Hint : You're making a mistake.
You have to prove this
$$
\cos A + \cos B + \cos C \le \frac{3}{2}
$$
because :
$$
1+\tan^2 \theta = \sec^2 \theta
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2193123",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Determine all real numbers x > 0 for which log equation is true.
Determine all real numbers x > 0 for which
$$\log_4 x - \log_x 16 = \frac{7}{6} - \log_x 8$$
Attemp at solution:
$$\frac{1}{\log_x 4} -\log_x 16 + \log_x 8 = \frac{7}{6}$$
$$\frac{1}{\log_x 4} + \log_x \frac{8}{16} = \frac{7}{6}$$
$$\frac{1+\log_x \frac{1}{2} \cdot \log_x 4}{\log_x 4} = \frac{7}{6}$$
$$\frac{1+(-2\log_x 2\cdot \log_x 2 )}{\log_x 4} = \frac{7}{6}$$
$$\frac{1+(-2\log_x 2\cdot \log_x 2 )}{2\log_x 2} = \frac{7}{6}$$
Not sure how to solve further
| Let
$$
u=\ln x,
$$
then the original equation becomes
$$
\dfrac{u}{2\ln2}-\dfrac{4\ln2}{u}=\dfrac76-\dfrac{3\ln2}{u},
$$
i.e.
$$
3u^2-(7\ln2)u-6(\ln2)^2=0.
$$
Then solution are
$$
u=\dfrac{7\ln2\pm\sqrt{49(\ln2)^2+72(\ln2)^2}}{6}=\dfrac{7\ln2\pm11\ln2}{6}=-\dfrac23\ln2,3\ln2=\ln2^{-2/3},\ln8.
$$
Hence
$$
x=2^{-2/3},8.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2194487",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Problem with inequality and number e Prove that, for every positive integer $n$, the following inequality holds:
$$n \{n! \times e\} \lt 1,$$ where $\{x\}$ denotes the fractional part function applied to number $x$.
I can;t think of any solution.
| Because
$$e=2+\frac{1}{2!}+\frac{1}{3!}+...+\frac{1}{n!}+...=$$
$$=2+\frac{1}{2!}+\frac{1}{3!}+...+\frac{1}{n!}+\frac{1}{(n+1)!}\left(1+\frac{1}{n+2}+\frac{1}{(n+2)(n+3)}+...\right)<$$
$$=2+\frac{1}{2!}+\frac{1}{3!}+...+\frac{1}{n!}+\frac{1}{(n+1)!}\left(1+\frac{1}{n+1}+\frac{1}{(n+1)^2}+...\right)=$$
$$=2+\frac{1}{2!}+\frac{1}{3!}+...+\frac{1}{n!}+\frac{1}{(n+1)!}\cdot\frac{1}{1-\frac{1}{n+1}}=$$
$$=2+\frac{1}{2!}+\frac{1}{3!}+...+\frac{1}{n!}+\frac{1}{n\cdot n!}.$$
Thus, $\{en!\}<\frac{1}{n}$ and we are done!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2195223",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Evaluating $\int _0^{\frac{\pi }{2}}\frac{1}{1+\cos \theta \cdot \cos x} \ \mathrm{d}x$ for $\theta \in (0, \pi)$ Within a problem with several steps, I am asked to show the following equality given that $\theta \in (0, \pi)$:
$$\pi\int _0^{\frac{\pi }{2}}\frac{1}{1+\cos \theta \cdot \cos x} \ \mathrm{d}x=\frac{\pi \theta}{\sin \theta}$$
I have no idea how to attack this. I don't see any clear trigonometric identity that I could apply. What would you suggest? Any hint/clue/help would be appreciated. Thanks.
You can check this post if you are interested in the rest of the problem.
| Letting $t=\tan \frac{x}{2} $ yields
$$
\begin{aligned}
& \int_0^{\frac{\pi}{2}} \frac{1}{1+\cos \theta \cos x} d x \\
= & \int_0^1 \frac{1}{1+\cos \theta \frac{1-t^2}{1+t^2}} \cdot \frac{2 d t}{1+t^2} \\
= & 2 \int_0^1 \frac{1}{(1+\cos \theta)+(1-\cos \theta) t^2} d t \\
= & \frac{2}{\sqrt{(1+\cos \theta)(1-\cos \theta)}}\left[\tan \left(\sqrt{\frac{1-\cos \theta}{1+\cos \theta}} \cdot t\right)\right]_0^1 \\
= & \frac{2}{\sin \theta} \cdot \frac{\theta}{2} \\
= & \frac{\theta}{\sin \theta}
\end{aligned}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2198902",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Prove this Generalizing AM-GM inequality
Let $n\ge 2$ and $a_{i} \ge 0,i=1,2,\cdots,n$, show that
$$(n-1)^{n-1}(a^n_{1}+a^n_{2}+\cdots+a^n_{n})+n^na_{1}a_{2}\cdots a_{n}\ge (a_{1}+a_{2}+\cdots+a_{n})^n$$
When $n=2$,
$$a^2_{2}+a^2_{2}+4a_{1}a_{2}=(a_{1}+a_{2})^2+2a_{1}a_{2}\ge (a_{1}+a_{2})^2$$
When $n=3$, it is
$$4(a^3_{1}+a^3_{2}+a^3_{3})+27a_{1}a_{2}a_{3}\ge (a_{1}+a_{2}+a_{3})^3$$
By
$$(a_{1}+a_{2}+a_{3})^3=a^3_{1}+a^3_{2}+a^3_{3}+3a_{1}a_{2}(a_{1}+a_{2})+3a_{1}a_{3}(a_{1}+a_{3})+3a_{2}a_{3}(a_{2}+a_{3})+6a_{1}a_{2}a_{3}$$
so it's enough to prove
$$a^3_{1}+a^3_{2}+a^3_{3}+7a_{1}a_{2}a_{3}\ge a_{1}a_{2}(a_{1}+a_{2})+a_{1}a_{3}(a_{1}+a_{3})+a_{2}a_{3}(a_{2}+a_{3})$$
which is clear by using Schur inequality:
$$a^3+b^3+c^3+3abc\ge ab(a+b)+bc(b+c)+ac(a+c)$$
| We can use the Vasc's EV-method. See here the corollary 1.7 (b):
https://www.emis.de/journals/JIPAM/images/059_06_JIPAM/059_06_www.pdf
Indeed, let $a_1\leq a_2\leq...\leq a_n$, $a_1+a_2+...+a_n=const$ and $a_1^n+a_2^n+...+a_n^n=const$.
Thus, by EV $a_1a_2...a_n$ gets a minimal value in the following cases.
*
*One of our variables is equal to $0$.
Let $a_n=0$.
In this case we need to prove that
$$(n-1)^{n-1}\sum_{k=1}^{n-1}a_k^n\geq\left(\sum_{k=1}^{n-1}a_k\right)^n$$ or
$$\frac{\sum\limits_{k=1}^{n-1}a_k^n}{n-1}\geq\left(\frac{\sum\limits_{k=1}^{n-1}a_k}{n-1}\right)^n,$$
which is Power Mean inequality;
*$a_1=x$ and $a_2=...=a_n=1$, where $0\leq x\leq1$.
We need to prove that $f(x)\geq0$, where
$$f(x)=(n-1)^{n-1}(x^n+n-1)+n^nx-(x+n-1)^n.$$
But $$f'(x)=n(n-1)^{n-1}x^{n-1}+n^n-n(x+n-1)^{n-1}\geq n^n-n\cdot (1+n-1)^{n-1}=0,$$
which says that $f$ is an increasing function.
Thus, $f(x)\geq f(0)=0$ and we are done!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2199804",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 2,
"answer_id": 0
} |
Finding general coefficients for $\frac{\partial^n}{\partial x^n} \left( \frac{1}{1 - e^{-x}} \right)$ I want to compute the following $n-th$ differentiation
\begin{align}
\frac{\partial^n}{\partial x^n} \left( \frac{1}{1 - e^{-x}} \right)
\end{align}
With the some help of computation tools i have
\begin{align}
\frac{\partial^n}{\partial x^n} \left( \frac{1}{1 - e^{-x}} \right)
= (-1)^n \sum_{k=1}^n a(k,n) \frac{e^-kx}{(1-e^{-x})^{k+1}}
\end{align}
Here what i want to do is find exact expression for $a(k,n)$
Here i state some examples
\begin{align}
n=1 \qquad : \qquad -\frac{e^{-x}}{\left(1-e^{-x}\right)^2} \\
n=2 \qquad : \qquad \frac{2 e^{-2 x}}{\left(1-e^{-x}\right)^3}+\frac{e^{-x}}{\left(1-e^{-x}\right)^2} \\
n=3 \qquad : \qquad -\frac{6 e^{-2 x}}{\left(1-e^{-x}\right)^3}-\frac{6 e^{-3 x}}{\left(1-e^{-x}\right)^4}-\frac{e^{-x}}{\left(1-e^{-x}\right)^2} \\
n=4 \qquad : \qquad \frac{14 e^{-2 x}}{\left(1-e^{-x}\right)^3}+\frac{36 e^{-3 x}}{\left(1-e^{-x}\right)^4}+\frac{24 e^{-4 x}}{\left(1-e^{-x}\right)^5}+\frac{e^{-x}}{\left(1-e^{-x}\right)^2} \\
n=5 \qquad : \qquad -\frac{30 e^{-2 x}}{\left(1-e^{-x}\right)^3}-\frac{150 e^{-3 x}}{\left(1-e^{-x}\right)^4}-\frac{240 e^{-4 x}}{\left(1-e^{-x}\right)^5}-\frac{120 e^{-5 x}}{\left(1-e^{-x}\right)^6}-\frac{e^{-x}}{\left(1-e^{-x}\right)^2}
\end{align}
and so on.
I am trying to construct the coefficients but got stuck.
can you have any ideas?
|
We can write
\begin{align*}
\frac{d^n}{dx^n}\left(\frac{1}{1-e^{-x}}\right)=(-1)^n\sum_{k=1}^n{n\brace k}k!\frac{e^{-kx}}{\left(1-e^{-kx}\right)^{k+1}}
\end{align*}
with the numbers ${n\brace k}$ denoting the Stirling numbers of the second kind.
The coefficients $a(k,n)={n\brace k}k!$ are archived as OEIS/A019538 and start with
\begin{array}{c|rrrrr}
{n\brace k}k!&1&2&3&4&5\\
\hline
1&1\\
2&1&2\\
3&1&6&6\\
4&1&14&36&24\\
5&1&30&150&240&120\\
\end{array}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2199942",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
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Computing the Jacobi symbol I have been asked to compute $\left(\frac{77}{257}\right)$ specifically using Jacobi symbols, showing all working. I have done the following:
$\left(\frac{77}{257}\right) =\left(\frac{257}{77}\right)$
=$\left(\frac{26}{77}\right)$ reducing $257 \bmod77$
=$\left(\frac{2}{77}\right)$$\left(\frac{13}{77}\right)$
Then here is where I'm not sure if what I have done is correct:
$(-1)\left(\frac{13}{77}\right)$ since $2$ is a quadratic non residue $\bmod77$? (Can I do this or is there a different way?
=$-\left(\frac{77}{13}\right)$ by flipping
=$-\left(\frac{12}{13}\right)$
=$-\left(\frac{2}{13}\right)^2$$\left(\frac{3}{13}\right)$
=$-\left(\frac{3}{13}\right)$
=$-\left(\frac{13}{3}\right)$
=$\left(\frac{1}{3}\right)$
=$-(1)$ since $1$ is a quadratic residue $\bmod3$
=$-1$
| We can use multiplicativity, and the Legendre symbol, since $257$ is prime.
Note that $257\equiv 1 \bmod 4$, and $7\equiv 11\equiv 3\bmod 4$. So we have
$$
\left(\frac{77}{257}\right)=\left(\frac{7}{257}\right)\cdot \left(\frac{11}{257}\right) = \left(\frac{257}{7}\right)\cdot \left(\frac{257}{11}\right)
= \left(\frac{5}{7}\right)\cdot \left(\frac{2^2}{11}\right)= \left(\frac{7}{5}\right)=\left(\frac{2}{5}\right)=-1.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2201040",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Prove that $\sqrt{\frac{2}{a}}+\sqrt{\frac{2}{b}}+\sqrt{\frac{2}{c}}\le\sqrt{\frac{a+b}{ab}}+\sqrt{\frac{b+c}{bc}}+\sqrt{\frac{c+a}{ca}}$ For $a,b,c$ are positive real number. Prove that
$$\sqrt{\frac{2}{a}}+\sqrt{\frac{2}{b}}+\sqrt{\frac{2}{c}}\le\sqrt{\frac{a+b}{ab}}+\sqrt{\frac{b+c}{bc}}+\sqrt{\frac{c+a}{ca}}$$
Let $\left(\dfrac{1}{a};\dfrac{1}{b};\dfrac{1}{c}\right)\rightarrow\left(x;y;z\right)$
We need prove $\sqrt{2x}+\sqrt{2y}+\sqrt{2z}\le\sqrt{x+y}+\sqrt{y+z}+\sqrt{z+x}$
We have: $\left(\sqrt{x+y}+\sqrt{y+z}+\sqrt{z+x}\right)^2=2\left(x+y+z\right)+2\left[\sqrt{\left(x+y\right)\left(y+z\right)}+\sqrt{\left(y+z\right)\left(z+x\right)}+\sqrt{\left(z+x\right)\left(x+y\right)}\right]$
By C-S we have: $\sqrt{\left(x+y\right)\left(y+z\right)}\ge\sqrt{xy}+\sqrt{yz}$
$\Rightarrow 2\sum\sqrt{\left(x+y\right)\left(y+z\right)}\ge4\left(\sqrt{xy}+\sqrt{yz}+\sqrt{xz}\right)$
$\Rightarrow LHS^2\ge 2(x+y+z+2\sqrt {xy}+2\sqrt {yz}+2\sqrt {xz})=2(\sqrt{x}+\sqrt{y}+\sqrt{z})=RHS^2$
Can do other way ?
| Hint:
Note $t \to \sqrt t$ is concave and $\left( 2x, 2y, 2z \right) \succ \left( x+y, y+z, z+x \right) $, so Karamata's inequality applies.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Linear Algebra II - Linear Transformation Assume that $T$ is a linear transformation. If $T: P_2 \to P_2$ where $P_2$ is the polynom vector space with polynomials of degree $2$, and $T(x+1) = x$, $T(x-1) =1$, $T(x^2)=0$ find $T(2+3x-x^2)$
I was just wondering if it is possible to do $$T(2+3x-x^2) $$ as ..
$$(a,b,c) = a(x+1) + b(x-1) + c(x^2)
= ax+a+bx-b+cx^2$$
Then let $a=2, b=-x, c= 2/x$
to get
$$T (2+3X-x^2) = 2T(x+1) -xT(x-1) +2/xT(x^2)$$
to eventually get $(-x^2+3x+1)$
| $a(x+1)+b(x-1)+cx^2=ax+a+bx-b+cx^2$ is a relevant quantity here, but remember that the scalars should be numbers, not things like $-x$ or $\frac2x$. Can you find numbers $a,b,c$ so that $a(x+1)+b(x-1)+cx^2=2+3x-x^2$? Once you find those numbers, then you can get your answer using the linearity of $T$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2204459",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find the minimum value of $P=\left(3+\frac{1}{a}+\frac{1}{b}\right)\left(3+\frac{1}{b}+\frac{1}{c}\right)\left(3+\frac{1}{c}+\frac{1}{a}\right)$ For $a,b,c$ are positive numbers satisfying $a+b+c\leq \frac{3}{2}$, find the minimum value of $$P=\left(3+\frac{1}{a}+\frac{1}{b}\right)\left(3+\frac{1}{b}+\frac{1}{c}\right)\left(3+\frac{1}{c}+\frac{1}{a}\right)$$
We have: $\frac{3}{2}\ge a+b+c \ge 3\sqrt[3]{abc}\Rightarrow abc\leq\frac{1}{8}$
We have $3+\frac{1}{a}+\frac{1}{b}\ge 7\sqrt[7]{\frac{1}{16a^2b^2}}$
$\Rightarrow P=\prod 7\sqrt[7]{\frac{1}{16a^2b^2}}=7^3$
When $a=b=c=\frac{1}{2}$
It looks wrong for $abc \le \frac{1}{8}$, and I need another way.
| By AM-GM $\frac{3}{2}\geq a+b+c\geq3\sqrt[3]{abc},$ which gives $\sqrt[3]{abc}\leq\frac{1}{2}$.
Thus, by Holder $$\left(3+\frac{1}{a}+\frac{1}{b}\right)\left(3+\frac{1}{b}+\frac{1}{c}\right)\left(3+\frac{1}{c}+\frac{1}{a}\right)\geq$$
$$\geq\left(3+\frac{1}{\sqrt[3]{abc}}+\frac{1}{\sqrt[3]{abc}}\right)^3\geq(3+2+2)^3=343.$$
The equality occurs for $a=b=c=\frac{1}{2}$.
Id est, the answer is $343$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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limit of an indetermined sequence I'm trying to find out the limits of a sequence of the type $\frac{\infty}{\infty}$, but I got stuck and am starting to get frustrated. The sequence in question is: $$\lim_{n \to \infty}\frac{\sqrt{n^2+3n+1} - \sqrt{n^2+3n-1}}{\ln(1+n) - \ln(2+n)}$$ I tried looking at its parts and figuring out their own limits to try and help me, but I didn't get anywhere useful. I tried rationalizing the square roots and this is what I have at the moment: $$\lim_{n \to \infty}\frac{\sqrt{n^2+3n+1} - \sqrt{n^2+3n-1}}{\ln(1+n) - \ln(2+n)}\times \frac{\sqrt{n^2+3n+1} + \sqrt{n^2+3n-1}}{\sqrt{n^2+3n+1} + \sqrt{n^2+3n-1}} = \lim_{n \to \infty}\frac{(n^2+3n+1)-(n^2+3n-1)}{(\ln(1+n) - \ln(2+n))(\sqrt{n^2+3n+1} + \sqrt{n^2+3n-1})}=\lim_{n \to \infty}\frac{2}{{(\ln(1+n) - \ln(2+n))(\sqrt{n^2+3n+1} + \sqrt{n^2+3n-1})}}$$
| Observe that
$$
\lim_{n\to\infty}(\sqrt{n^2+3n+1} - \sqrt{n^2+3n-1})=
\lim_{n\to\infty}\frac{2}{\sqrt{n^2+3n+1} + \sqrt{n^2+3n-1}}=0
$$
as well as
$$
\lim_{n\to\infty}(\ln(1+n)-\ln(n+2))=
\lim_{n\to\infty}\ln\frac{1+n}{n+2}=0
$$
so your limit is in the form $0/0$, rather than $\infty/\infty$.
On the other hand, you can observe that
$$
\lim_{n\to\infty}n(\sqrt{n^2+3n+1} - \sqrt{n^2+3n-1})=
\lim_{n\to\infty}\frac{2n}{\sqrt{n^2+3n+1} + \sqrt{n^2+3n-1}}=1
$$
and that
$$
\lim_{n\to\infty}n(\ln(1+n)-\ln(n+2))=
\lim_{n\to\infty}n\ln\frac{1+n}{n+2}
$$
is finite as well: with $t=1/n$, the limit becomes
$$
\lim_{t\to0^+}\frac{\ln(t+1)-\ln(1+2t)}{t}=-1
$$
So you can write your limit as
$$
\lim_{n \to \infty}\frac{n(\sqrt{n^2+3n+1} - \sqrt{n^2+3n-1})}{n(\ln(1+n) - \ln(2+n))}
$$
and finish up.
| {
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"timestamp": "2023-03-29T00:00:00",
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Does this integral have a closed form or asymptotic expansion? $\int_0^\infty \frac{\sin(\beta u)}{1+u^\alpha} du$ I'm interested in this following definite integral:
$$
\int_0^\infty du \frac{\sin(\beta u)}{1+u^\alpha},
$$
where $\beta>0$ and $\alpha\geq1$. Is there any closed form for this integral? I would be fine if the answer involves special functions, it would just be nice to have the answer in closed form.
I'm also interested in the behavior of this integral in the limit $\beta\gg1$. In particular, I'd like to know if it decays exponentially at large values of $\beta$.
| Assuming $\beta >0$, considering $$I_a=\int_0^\infty \frac{\sin(b u)}{1+u^a}\,du$$ it seems that $a=1$ corresponds to a very particular case $$I_1=\text{Ci}(b) \sin (b)-\text{Si}(b) \cos (b)+\frac{1}{2} \pi \cos (b)$$ where appear the sine and cosine integrals. $I_1$ seems to decrease as an hyperbola. Expanding $I_1$ for large values of $b$ up to $O\left(\frac{1}{b^9}\right)$ gives $$I_1\approx \frac{1}{b}-\frac{2}{b^3}+\frac{24}{b^5}-\frac{720}{b^7}+\cdots$$
Fo the other cases, using a CAS, it seems that $I_a$ systematically express in terms of the Meijer G function. For example
$$I_2=\frac{1}{4} \sqrt{\pi } b G_{1,3}^{2,1}\left(\frac{b^2}{4}|
\begin{array}{c}
0 \\
0,0,-\frac{1}{2}
\end{array}
\right)$$
$$I_3=\frac{G_{1,7}^{5,1}\left(\frac{b^6}{46656}|
\begin{array}{c}
\frac{5}{6} \\
\frac{1}{6},\frac{1}{3},\frac{1}{2},\frac{5}{6},\frac{5}{6},0,\frac{2}{3}
\end{array}
\right)}{2 \sqrt{3 \pi }}$$ $$I_4=\frac{1}{2} \sqrt{\frac{\pi }{2}} G_{1,5}^{3,1}\left(\frac{b^4}{256}|
\begin{array}{c}
\frac{3}{4} \\
\frac{1}{4},\frac{3}{4},\frac{3}{4},0,\frac{1}{2}
\end{array}
\right)$$ I have not been able to see any clear pattern.
For $b >1$, these functions start at $0$, go through a maximum value and seem to also decrease as hyperbolas at large values of $b$ (just as Taozi commented).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2207869",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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proving $\cos (A+B)>0$, if given angles $A$ and $B$ If $\displaystyle A=3\sin^{-1}\left(\frac{6}{11}\right)$ and $\displaystyle B = 3\cos^{-1}\left(\frac{4}{9}\right),$ then proving $\cos (A+B)>0$
Attempt: $$ 3\sin^{-1}\left(\frac{1}{2}\right)<3\sin^{-1}\left(\frac{6}{11}\right)<3\sin^{-1}\left(\frac{\sqrt{3}}{2}\right)\Rightarrow \frac{\pi}{2}<A<\pi$$
same way $$\displaystyle 3\cos^{-1}\left(0\right)<3\cos^{-1}\left(\frac{4}{9}\right)<3\cos^{-1}\left(\frac{1}{2}\right)\Rightarrow \frac{\pi}{2}<B<\frac{3\pi}{2}$$
so $$\pi<A+B<\frac{5\pi}{2}$$
could some help me how to prove $\cos (A+B)>0,$ thanks
| $A = 3\sin^{-1}(\frac{6}{11}), B = 3\cos^{-1}(\frac{4}{9})$
Let $C = \cos(\frac{A+B}{3})$
Note that $\cos (3x) = 4\cos^3x - 3\cos x$
Let $x=\frac{A+B}{3} \implies \cos x = \cos \frac{A}{3} \cos\frac{B}{3} -\sin\frac{A}{3}\sin\frac{B}{3}$
$\cos x = \cos (\sin^{-1}(\frac{6}{11}))\cos(\cos^{-1}(\frac{4}{9})) - \sin(\sin^{-1}(\frac{6}{11}))\sin(\cos^{-1}(\frac{4}{9}))$
Consider $y = \sin^{-1}(\frac{6}{11}) \implies \cos y = \frac{\sqrt{85}}{11}$
Consider $y = \cos^{-1}(\frac{4}{9}) \implies \sin y = \frac{\sqrt{65}}{9}$
Then $\cos x = \frac{\sqrt{85}}{11}\cdot \frac{4}{9} -\frac{6}{11}\cdot \frac{\sqrt{65}}{9} = \frac{4\sqrt{85}-6\sqrt{65}}{99}$
So then $\cos (3x) = 4\cdot (\frac{4\sqrt{85}-6\sqrt{65}}{99})^3-3\cdot (\frac{4\sqrt{85}-6\sqrt{65}}{99}) = 4\cdot (\frac{4\sqrt{85}-6\sqrt{65}}{99})^3+3\cdot (\frac{-4\sqrt{85}+6\sqrt{65}}{99})$
$\cos (3x) = \frac{134080\sqrt{85}-154080\sqrt{65})}{99^3} +\frac{-12\sqrt{85}+18\sqrt{65}}{99}$
$\cos (3x) = \cos(A+B) = \frac{(134080-117612)\sqrt{85} +(176418-154080)\sqrt{65}}{99^3} > 0$
and we're done
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 1
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Inverse function for $f(x)=\frac{x^2+1}{2x^2+1}$ Given that $f(x)=\frac{x^2+1}{2x^2+1}$ and $x\geq2$, find $f^{-1}(x)$. Prove that $f^{-1}f(x)=x$
My attempt,
Let $f^{-1}(x)=a$
$x=f(a)$
$x=\frac{a^2+1}{2a^2+1}$
$(2x-1)a^2+x-1=0$
$a=\pm\frac{\sqrt{-(2x-1)(x-1})}{2x-1}$
Since $x\geq2$ for $f(x)$
So, $f^{-1}(x)=\frac{\sqrt{-(2x-1)(x-1})}{2x-1}$
Am I correct for my $f^{-1}(x)?$ If I do, how do I proceed for proving part? It's tedious.
| HInt:$$f^{-1}(x)=\sqrt{\dfrac{-(x-1)}{2x-1}}\\\to\\f^{-1}f(x)=f^{-1}(\dfrac{x^2+1}{2x^2+1})=\\\sqrt{\dfrac{-(\dfrac{x^2+1}{2x^2+1}-1)}{2\dfrac{x^2+1}{2x^2+1}-1}}$$
can yoou go further ?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2208210",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Coefficient of $x^5$ in $(4x^2- \frac{1}{x^3})^8$ I took one $x$ out of bracket, and I got $\frac{1}{x^{24}} (4x^5-1)^8.$ Now, we get $x$ to the power of if we get $x$ to the power $29,$ then we divide it by $24,$ we get $x$ to the power $5,$ but i can't get $29,$ by multiplying $5.$
| The term in given expansion containing $x^5$ $=^8C_r(4x^2)^r(-1/x^3)^{8-r} = ^8C_r4^r(-1)^{8-r}x^{2r-3(8-r)} = ^8C_r4^r(-1)^{8-r}x^{5r-24}$
For $x^5$, $5r-24=5$, so $r=29/5$ which is not a whole number. So there is no term in the expansion of $(4x^2-(1/x^3))^8$ having term with $x^5$. So coefficient of $x^5$ in the given expansion is $\color{red}{0}$
| {
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"timestamp": "2023-03-29T00:00:00",
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Disjoint Cycle Question I'm working on this question:
List the permutations of $\{1,2,3\}$ in disjoint cycle form.
I already know what a disjoint cycle is. It's basically means that every cycle contains numbers that are not in any other cycle. So with that in mind, do I write all the possible permutations of
\begin{pmatrix}
1 & 2 & 3\\
? & ? & ?
\end{pmatrix}
such that I could write down all of it's permutations in disjoint cycle form? A push to get me started is all I'm asking for here.
EDIT
Please see finalized answer below.
| Here is my attempted:
\begin{pmatrix} 1 & 2 & 3 \\ 3 & 2 & 1 \end{pmatrix} is given by $(1 3)(2)$.
Then \begin{pmatrix} 1 & 2 & 3\\3 & 1 & 2 \end{pmatrix}is given by $(1 3 2)$.
Then \begin{pmatrix} 1 & 2 & 3\\ 2& 3 & 1\end{pmatrix} is given by $(1 2 3)$.
Then \begin{pmatrix} 1 & 2 & 3 \\2 & 1 & 3 \end{pmatrix} is given by $(1 2)(3)$.
Then \begin{pmatrix} 1 & 2 & 3 \\1 & 3 & 2 \end{pmatrix} is given by $(1)(2 3)$.
Lastly, \begin{pmatrix} 1 & 2 & 3\\1 & 2 & 3 \end{pmatrix} is given by $(1)(2)(3)$, but these are just the identity matrices, so we can eliminate $(2)$ and $(3)$ and are only left with $(1)$.
So therefore, the final list of permutations is given by $$(1),(12),(13),(23),(123),(132).$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Value of $\tan^2(5)+\tan^2(10)+\tan^2(15)+...+\tan^2(85)$ Calculate the exact value of $$\tan^2(5^\circ)+\tan^2(10^\circ)+\tan^2(15^\circ)+\cdots+\tan^2(85^\circ)$$
How to evaluate this sum of all these values? Is there a specific way? Thanks in advance.
| $$\tan^2(5^\circ)+\tan^2(10^\circ)+...+\tan^2(85^\circ) = \tan^2\bigg(\frac{5\pi}{180}\bigg) +\tan^2\bigg(\frac{10\pi}{180}\bigg)+...+\tan^2\bigg(\frac{85\pi}{180}\bigg)$$
$$= \sum_{r=1}^{17}\tan^2\bigg(\frac{r\pi}{2\cdot18}\bigg)$$
Now note this result: Prove that $\sum\limits_{k=1}^{n-1}\tan^{2}\frac{k \pi}{2n} = \frac{(n-1)(2n-1)}{3}$
We have $n = 18$ so then we get:
$$S = \frac{(18-1)(2\cdot 18 -1)}{3} =\frac{595}{3}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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If $a^3+a^2+a=9b^3+b^2+b$ and $a,b$ are integers then show $a-b$ is a perfect cube. If $a^3+a^2+a=9b^3+b^2+b$ and $a,b$ are integers then show $a-b$ is a perfect cube.
My attempt:I factorized it like below:
$(a-b)(a^2+b^2+ab+a+b+1)=8b^3=(2b)^3$
I take $gcd(a-b,a^2+b^2+ab+a+b+1)=d$
If $d=1$ then it is clear that $a-b$ is a perfect cube then consider $d>1$ then there is a $p$ that is prime and $p \mid d$.We have :
$p\mid a-b \Rightarrow p \mid (2b)^3 \Rightarrow p \mid 2b \Rightarrow p\mid 2$ or $p\mid b$
If $p\mid b$ then also $p\mid a$ (as $p\mid a-b$ holds). Then we will get to $p \mid 1$ because:
$ p \mid a^2+b^2+ab+a+b,p \mid a^2+b^2+ab+a+b+1 \Rightarrow p \mid 1$
which is clearly wrong then we have $p\mid 2$ so $p=2$ means $d=2^k$ where $k$ is a natural number including $0$.In the case $d=1$ we have the right result.So assemble $k \ge 1$.Because $2 \mid a-b$ we can conclude that $a,b$ have the same parity @Ghartal showed in his answer that if $a,b$ are both even we have a right result but if $a,b$ are both odd we don't.So maybe we have to prove $a,b$ can,t be both odd.
| It turns out that there are only two rational solutions of
$a^3+a^2+a = 9b^3+b^2+b$. One is the obvious integer solution
$(a,b)=(0,0)$, for which $a-b$ is the cube $0$.
The other is $(a,b) = (-7/5, -3/5)$, for which $a-b = -4/5$
is not a cube but $a,b$ are not integers. So $a-b$ is indeed
a cube for all the integer solutions, as desired.
The proof that there are no other rational solutions is elementary,
but requires Fermat's method of descent. The equation
$a^3+a^2+a = 9b^3+b^2+b$ defines a smooth cubic curve with
a rational point at $(a,b)=(0,0)$, so it's an elliptic curve
that we can bring to standard form $y^2=\text{cubic}(x)$ by
projecting from the origin.
Let $(a,b)$ be any rational solution other than $(0,0)$.
Then $b\neq 0$ because the nonzero solutions of $a^3+a^2+a$
are not rational (or even real). Write $a=mb$ and divide by $b$ to get
$$
(9-m^3) b^2 + (1-m^2) b + (1-m) = 0.
$$
This equation is quadratic in $b$, so it has rational solutions iff
the discriminant is a square. The discriminant is
$$
-3 m^4 + 4 m^3 - 2 m^2 + 36 m - 35 = (1-m)(3m-7)(m^2+2m+5).
$$
Taking $m=1$ yields $8b^2 = 0$, so we're back to the $b=0$ solution;
and $m=7/3$ yields $-4(5b+3)^2/27 = 0$, so $b=-3/5$ and $a=-7/5$.
We claim that there are no other solutions.
If $m\neq 1$ we may write $m = 1+(8/x)$ and calculate
$$
-3 m^4 + 4 m^3 - 2 m^2 + 36 m - 35 = \frac{2^8}{x^4} (x^3-2x^2-16x-48),
$$
so we seek rational points on the elliptic curve
$$
y^2 = x^3-2x^2-16x-48 = (x-6)(x^2+4x+8)
$$
other than the "point at infinity".
The point $(x,y)=(6,0)$ brings us back to $m = 1 + (8/6) = 7/3$,
and it turns out there are no others. Since the cubic on the right-hand side
has a linear factor, we can try to prove this using Fermat's descent,
which turns out to be enough. But these days there are other
resources we can try first: Cremona's mwrank or the
LMFDB, and either of these
tells us that there are no other solutions and that Fermat-style descent
suffices to prove it. In the case of mwrank we'd need a bit of
additional computation to exclude other torsion points, so let's go the
database route: enter the coefficients [0,-2,0,-16,-48] into the
"curve, label or isogeny class label" search window for
elliptic curves
over $\bf Q$, and find the entry for
this curve of
conductor 544 (which is small enough that it's also in Cremona's book).
We find that the curve has only the known two torsion points
and rank zero, and moreover that the 2-part of the Tate-Shafarevich group
is trivial, so a 2-descent suffices to establish rank zero.
This completes the proof.
| {
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Help with solving trig identity problem It's been 20 years since I did trig, and this one seems a little tricky. How would I solve
$$
\tan^2(x) -2\tan(x)=1
$$
with steps?
| $$\tan ^{ 2 }{ x-2\tan { x } -1=0 } \\ \tan ^{ 2 }{ x-2\tan { x } +1-2=0 } \\ { \left( \tan { x } -1 \right) }^{ 2 }-2=0\\ \tan { x-1=\pm \sqrt { 2 } } \\ \tan { x } =1\pm \sqrt { 2 } \\ x=\arctan { \left( 1\pm \sqrt { 2 } \right) +\pi n } \\ $$
| {
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"timestamp": "2023-03-29T00:00:00",
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Trig identities - stuck solving $\tan^2\theta = -\frac 32 \sec\theta$ Solve the equation on the interval $0\leq \theta < 2\pi$
$$\tan^2 \theta = -\frac{3}{2}\sec \theta $$
Here are the steps I have so far:
Identity: $\tan^2 \theta = \sec^2 \theta -1 $
Substitute:
$$\sec^2 \theta -1 = -\frac{3}{2}\sec \theta $$
$$2\sec^2 \theta -2 = {-3}\sec \theta $$
$$2\sec^2 \theta +3\sec \theta - 2 = 0 $$
Is this factoring correct?:
$$(2\sec\theta+4)(\sec\theta-1) = 0 $$
$$2\sec\theta+4 =0$$
$$2\sec\theta = -4 $$
$$\sec \theta = -2$$
$$(2\pi/3), (4\pi/3) $$
$$\sec\theta - 1 = 0$$
$$\sec\theta = 1 $$
$\sec\theta=1$ would evaluate to $0$, for some reason that is an invalid answer? (according to my assignment)
So is $\ (2\pi/3), (4\pi/3) $ the full answer?
| Set $\cos\theta=t$; then
$$
\tan^2\theta=\frac{1-t^2}{t^2}
\qquad
\sec\theta=\frac{1}{t}
$$
and the equation becomes
$$
\frac{1-t^2}{t^2}=-\frac{3}{2t}
$$
that is,
$$
2-2t^2=-3t
$$
and finally
$$
2t^2-3t-2=0
$$
The roots are $2$ and $-1/2$. So the equation reduces to
$$
\cos\theta=-\frac{1}{2}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2213674",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Prove that $\lim\limits_{x\rightarrow 4} \sqrt{2x+7} = \sqrt{15}$. Prove that $\displaystyle \lim_{x\rightarrow 4} \sqrt{2x+7} = \sqrt{15}$ using the epsilon-delta definition.
This is what I have, but I know my delta value is incorrect. My professor said that it was the right path but my delta is incorrect.
Proof: Let $\varepsilon>0$. Choose $\delta$ such that $0<\delta<\min(\varepsilon,1)$. This means that both $\delta<1$ and $\delta<\varepsilon$. Let $x\in\mathbb{R}$ such that $0<|2x-8|<\delta$. Since $\delta<1$, we have
$$\begin{array}{cccccc}
&-1 &< & 2x-8 & < & 1\\
\Rightarrow & 7 &<& 2x &<& 9 \\
\Rightarrow & 7/2 & < & x & < & 9/2
\end{array}$$
Since $7/2<x<9/2$,
$$\begin{array}{cccccc}
&7/2 & < & x & < & 9/2\\
\Rightarrow & 7 &<& 2x &<& 9 \\
\Rightarrow & 7+7 & < & 2x+7 & < & 9+7 \\
\Rightarrow & \sqrt{14} & < & \sqrt{2x+7} & < & \sqrt{16} \\
\Rightarrow & \sqrt{14} + \sqrt{15} & < & \sqrt{2x+7}+\sqrt{15} & < & \sqrt{16}+\sqrt{15}\\
\Rightarrow & \displaystyle \frac{1}{\sqrt{14} + \sqrt{15}} & > & \displaystyle \frac{1}{\sqrt{2x+7}+\sqrt{15}} & > & \displaystyle \frac{1}{\sqrt{16} + \sqrt{15}}\\
\end{array}$$
This implies $$\left|\frac{1}{\sqrt{2x+7}+\sqrt{15}}\right|< \frac{1}{\sqrt{14} + \sqrt{15}}<1.$$
Therefore,
$$\begin{align*}
\left|\sqrt{2x+7}-\sqrt{15}\right|
&= \left|\left(\sqrt{2x+7}-\sqrt{15}\right) \cdot \left(\frac{\sqrt{2x+7}+\sqrt{15}}{\sqrt{2x+7}+\sqrt{15}}\right)\right| \\
&= \left|2x+7-15\right| \cdot \left| \frac{1}{\sqrt{2x+7}+\sqrt{15}}\right|\\
&=\left|2x-8\right|\cdot \left|\frac{1}{\sqrt{2x+7}+\sqrt{15}}\right|\\
&< \delta \cdot 1 \\
&< \varepsilon \cdot 1\\
\end{align*}$$
Thus, $|\sqrt{2x+7}-\sqrt{15}|<\varepsilon$. So, $\displaystyle \lim_{x\rightarrow 4} \sqrt{2x+7} = \sqrt{15}$.
| Denote by $D$ the domain of $f$, where $f(x) = \sqrt{2x + 7}$.
Let $\varepsilon > 0$. Choose $\delta = (\sqrt{15}/2) \varepsilon$. Let $x \in D$. If $0 < |x-4| < \delta$, then $$\begin{aligned}[t] |f(x) - \sqrt{15}| = \biggl| \dfrac{(f(x) - \sqrt{15}\,)(f(x) + \sqrt{15}\,)}{f(x) + \sqrt{15}}\biggr| &= \dfrac{1}{f(x) + \sqrt{15}} \cdot |(2x+7)-15| \\ &= \dfrac{2}{f(x) + \sqrt{15}} \cdot |x-4| \\ &< \dfrac{2}{\sqrt{15}} \cdot \delta \\ &= \dfrac{2}{\sqrt{15}} \cdot \dfrac{\varepsilon \sqrt{15}}{2} = \varepsilon.\end{aligned} $$
Therefore, $\lim_{x \to 4} \sqrt{2x+7} = \sqrt{15}$.
Remember that if you are trying to prove $\lim_{x \to c} f(x) = L$ using the $\varepsilon$-$\delta$ definition, the structure of your proof should be like so:
"Let $\varepsilon > 0$."
[A choice for $\delta$ goes here.]
"Let $x \in \operatorname{dom}(f)$."
"Suppose $0 < |x-c| < \delta$."
[Proof that $|f(x) - L| < \varepsilon$ goes here.]
"Thus, $\lim_{x \to c} f(x) = L$."
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2215161",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 2,
"answer_id": 1
} |
Solve $\sum_{k=0}^{\infty}\frac{1}{1-16k^2}$ I could use some help on calculating this infinite sum: $\sum_{k=0}^{\infty}\frac{1}{1-16k^2}$. Included was that I had to start with a Fourier series for the function $f:\Re \to \Re: x \mapsto \sin(x)$ for $x\in[0, \frac{\pi}{2}[$, so let's start with that.
Let
\begin{eqnarray*}
g(x) = \frac{a_0}{2} + \sum_{k=1}^{\infty}a_k\cos(4kx) + \sum_{k=1}^{\infty}b_k\sin(4kx).
\end{eqnarray*}
This is the Fourier series for $f$.
With $a_k = \frac{4}{\pi}\int_{0}^{\frac{\pi}{2}}\sin(x)\cos(4kx)dx$ and $b_k = \frac{4}{\pi}\int_{0}^{\frac{\pi}{2}}\sin(x)\sin(4kx)dx$. Solving this leads to (or at least I found that): $a_k = \frac{4}{(1-16k^2)\pi}$, $b_k = \frac{16k}{(1-16k^2)\pi}$ for $k\geq1$ and $a_0 = \frac{4}{\pi}$. Bringing this to $g(x)$ gives:
\begin{eqnarray*}
g(x) = \frac{2}{\pi} + \sum_{k=1}^{\infty}\frac{4}{(1-16k^2)\pi}\cos(4kx) + \sum_{k=1}^{\infty}\frac{16k}{(1-16k^2)\pi}\sin(4kx).
\end{eqnarray*}
Since $f(x) \approx g(x)$, we can say that $f(0) = g(0)$. We get
\begin{eqnarray*}
\sin(0) = \frac{2}{\pi} + \sum_{k=1}^{\infty}\frac{4}{(1-16k^2)\pi}\cos(0) + \sum_{k=1}^{\infty}\frac{16k}{(1-16k^2)\pi}\sin(0),
\end{eqnarray*}
this becomes
\begin{eqnarray*}
0 = \frac{2}{\pi} + \sum_{k=1}^{\infty}\frac{4}{(1-16k^2)\pi}.
\end{eqnarray*}
We get
\begin{eqnarray*}
\frac{-2}{\pi} = \frac{4}{\pi}\sum_{k=1}^{\infty}\frac{1}{1-16k^2}
\end{eqnarray*} so
\begin{eqnarray*}
\sum_{k=1}^{\infty}\frac{1}{1-16k^2} = \frac{-1}{2}.
\end{eqnarray*} We need the sum from k = 0. The term $\frac{1}{1-16k^2}$ for k = 0 gives 1, so we add 1 to both sides. This leads to my solution
\begin{eqnarray*}
\sum_{k=0}^{\infty}\frac{1}{1-16k^2} = \frac{1}{2}.
\end{eqnarray*}
However, when approaching this sum numerically and using Wolfram, I find that the sum should be $\frac{4+\pi}{8}$. Could some help and point out where I went wrong with my approach? Thanks in advance
| First lets tidy the sum up a little
\begin{eqnarray*}
\sum_{k=0}^{\infty} \frac{1}{1-16k^2}=1-\sum_{k=1}^{\infty} \frac{1}{16k^2-1}
\end{eqnarray*}
Now partial fractions gives
\begin{eqnarray*}
\sum_{k=1}^{\infty} \frac{1}{16k^2-1}= \sum_{k=1}^{\infty} \left( \frac{1/2}{4k-1}-\frac{1/2}{4k+1} \right)
\end{eqnarray*}
Now use $\frac{1}{i}= \int_0^1 x^{i-1} dx $. We have
\begin{eqnarray*}
\sum_{k=1}^{\infty} \frac{1}{16k^2-1}= \frac{1}{2}\sum_{k=1}^{\infty}\int_0^1 \left( x^{4k-2}-x^{4k} \right) dx
\end{eqnarray*}
Now interchange the sum & the integral.... & perform the geometric sums
\begin{eqnarray*}
\sum_{k=1}^{\infty} \frac{1}{16k^2-1}=\frac{1}{2} \int_0^1 \left( \frac{x^2-x^4}{1-x^4} \right) dx = \frac{1}{2}\int_0^1 \left( \frac{x^2}{1-x^2} \right) dx =\frac{1}{2}(1-\frac{\pi}{4}).
\end{eqnarray*}
Now putting it all back together ... we have $\color{red}{\frac{1}{2}+\frac{\pi}{8}}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2216441",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 1
} |
$\lim_{k \to \infty} \frac{x_k}{k^2}$ Let $(x_k)$ be the sequence of real numbers defined as follows: $x_1=1$; $x_{k+1}=x_k+\sqrt {x_k}$ for $k>0$. Find $$\lim_{k \to \infty} \frac{x_k}{k^2}$$ My thought is to find some $y_k$ which is a function of $x_k$ and find the limit of this $y_k$, but I'm not sure if this will work, or what my $y_k$ should be.
| Here is an argument that avoids the Stolz-Cesaro theorem:
You can prove by induction that $\dfrac {k^2}{9} \le x_k \le \dfrac{(k+1)^2}4$ for all $k$. Clearly $\dfrac 19 \le x_1 \le 1$, $$x_k \ge \frac{k^2}{9} \implies x_{k+1} = x_k + \sqrt{x_k} \ge \frac{k^2}{9} + \frac{k}{3} > \frac{k^2 + 2k + 1}{9} = \frac{(k+1)^2}{9},$$ and $$ x_k \le \frac{(k+1)^2}{4} \implies x_{k+1} = x_k + \sqrt{x_k} \le \frac{(k+1)^2}{4} + \frac{k+1}2 = \frac{k^2 + 4k + 3}4 < \frac{(k+2)^2}4.$$
Let $I = \liminf \dfrac{x_k}{k^2}$ and $J = \limsup \dfrac{x_k}{k^2}$. The remarks above imply $\dfrac 19 \le I \le J \le \dfrac 14$. Fix $0 < \epsilon < I$. Then there exists an index $K$ so that $k \ge K$ implies $\dfrac{x_k}{k^2} > I - \epsilon$. Consequently $$k \ge K \implies \frac{x_{k+1} - x_k}{k} = \sqrt{ \frac{x_k}{k^2}} > \sqrt{I - \epsilon} \implies x_{k+1} - x_k \ge k \sqrt{I - \epsilon}.$$ Now let $N > K$. It follows that $$x_N - x_K = \sum_{k=K}^{N-1} (x_{k+1} - x_k) \ge \sqrt{I - \epsilon} \sum_{k=K}^{N-1} k = \sqrt{I - \epsilon} \left( \frac{N(N-1)}2 - \frac{K(K+1)}{2} \right).$$ Divide the last inequality by $N^2$ and let $N \to \infty$ to find that $$I = \liminf_{N \to \infty} \frac{x_N}{N^2} \ge \frac{\sqrt{I - \epsilon}}2.$$ Now take $\epsilon \to 0^+$ to find $I \ge \dfrac{\sqrt{I}}2$ which implies (since $I > 0$) that $I \ge \dfrac 14$.
It follows that $\dfrac 14 \le I \le J \le \dfrac 14$ so that $I = J = \dfrac 14$ and thus $\dfrac{x_k}{k^2} \to \dfrac 14.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2219385",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 2
} |
What will be the value of $\displaystyle\int_{0}^{1} \frac{1-\cos(x)}{x} dx$ for the first three decimal digit? $$\displaystyle\int_{0}^{1} \frac{1-\cos(x)}{x} dx$$
Can I somehow write the Taylor-series of $\frac{1-\cos(x)}{x}$?
| Observe we have
\begin{align}
1-\cos x = \sum^\infty_{n=1} (-1)^{n+1}\frac{x^{2n}}{(2n)!}
\end{align}
which means
\begin{align}
\int^1_0 \frac{1-\cos x}{x} \ dx = \sum^\infty_{n=1}(-1)^{n+1}\frac{1}{(2n)!}\int^1_0 x^{2n-1}\ dx = \sum^\infty_{n=1}(-1)^{n+1}\frac{1}{(2n)!}\frac{1}{2n}.
\end{align}
Hence
\begin{align}
\left|\sum^\infty_{n=N}(-1)^{n+1}\frac{1}{(2n)!2n}\right| \leq \sum^\infty_{n=N} \frac{1}{(2n)! 2n} \leq \sum^\infty_{n=N}\frac{1}{2^{2(n+1)}} = \sum^\infty_{n=N}\frac{1}{4^{n+1}} = \frac{3}{4^{N}}
\end{align}
when $N\geq 4$. In particular, if you want three digit accuracy, all you have to do is choose $N$ such that
\begin{align}
\frac{3}{4^N}<10^{-3}
\end{align}
which means you need $N\geq 6$. Hence sum up six terms to get at least 3 decimal places of accuracy.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2220293",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Let $n$ be an integer such that if $d | n$ then $d + 1 | n + 1.$ Show that $n$ is a prime number Let $n$ be an integer such that for any integer $d$, if $d | n$ then $d + 1 | n + 1$.
Show that $n$ is a
prime number or equal to $1$
My workout...
Suppose $n = xy$ for some postive integers $x$ and $y. x|n$, and hence we
assume without loss of generality $x + 1|n + 1.$
Then $x|xy + 1$.
However, $xy + x$ is divisible by $x$.
This implies $x − 1$ is divisible by $x$,
which is an impossibility if $x$ is not equal to $1$.
Therefore whenever $n$ is decomposed
into two factors, one must turn out to be $1$. Hence $n$ must be a prime.
I am looking for any other method which solves this quickly and more quickly...
| Suppose that $d\mid n$ and $d+1\mid n+1$.
Assume that $d\ne1$ and $d\ne n$. If $d^2\lt n$, we can use $\frac{n}{d}$ in place of $d$ to get that $d\lt n$ and $d^2\ge n$. Then
$$
\begin{align}
\frac{n}{d}-\frac{n+1}{d+1}
&=\frac{n-d}{d(d+1)}\\[3pt]
&\in(0,1)
\end{align}
$$
However, if both $\frac{n}{d}$ and $\frac{n+1}{d+1}$ are integers, their difference must be an integer, and we have a contradiction. Therefore, we have $d=1$ or $d=n$.
Since for each $d$ so that $d\mid n$ we have $d+1\mid n+1$, we must have that for each $d$ so that $d\mid n$, either $d=1$ or $d=n$. Therefore, $n$ is prime.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2220622",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Find locus of intersection of two lines with parameters $\require{cancel}$
I have a triangle
where $$DE || BC, A(a,0), B(0,b), b>0, a>0$$
I need to find the locus of the intersection between $BE$ and $CD$ (in terms of $a$ and $b$).
They didn't tell me more about this parallel line that cuts AB y and AC, therefore I thought I would need to create some variables such that
$$E (e,0) \quad \text{and} \quad D(e,d)$$
From here I get from Tales's theorem that
$$\frac{DE}{BC} = \frac{AE}{AC} \Rightarrow \frac db = \frac{a-e}{a} \Rightarrow d = \frac{b(a-e)}{a}$$
$$\sqrt{a^2-2ae+e^2+d^2} = AD \quad \text{Pythagoras theorem}$$
$$AB = \sqrt{a^2+b^2}$$
$$\tag{$()^2$}\frac{\sqrt{a^2-2ae+e^2+d^2}}{\sqrt{a^2+b^2}} = \frac db$$
$$\frac{a^2-2ae+e^2+d^2}{a^2+b^2} = \frac {d^2}{b^2}$$
$$\frac{a^2-2ae+e^2 + \frac{b^2\left(a^2-2ae+e^2\right)}{a^2}{}}{a^2+b^2} = \frac{\frac{\cancel{b^2}\left(a^2-2ae+e^2\right)}{a^2}}{\frac {\cancel{b^2}}{1}}$$
At the end I would get something like
$$(a^2-2ae+e^2)(a^2+b^2) = a^2(a^2-2ae+e^2) + b^2(a^2-2ae+e^2)$$
Which will not help. I'm doing something wrong. I wanted to find $e$ and $d$ in terms of $a$ and $b$ so I can find the locus I need.
| My idea is to try to express the coordinates of $F = (x,y)$, the intersection of $BE$ and $CD$, with $a$, $b$ and $E = (e,0)$. Let $G$ be the point on x-axis such that $FG\perp CE$. $$\frac{FG}{DE}+\frac{FG}{BC} = 1\implies y = FG = \frac{DE\cdot BC}{DE+BC}$$
With $\frac{DE}{BC} = \frac{a-e}{a}\implies DE = \frac{ab-be}{a}$, we have $y=\frac{ab-be}{2a-e}$, which is equivalent to:
$$e = \frac{ab-2ay}{b-y}$$
Further, we can derive a result for $x$:
$$\frac{EG}{CE} = \frac{FG}{BC}\implies \frac{e-x}{e} = \frac{a-e}{2a-e}\implies x=e-\frac{ae-e^2}{2a-e}$$
and we can put the above two equations together (by plugging $e$ expressed as $a$, $b$, $y$ into the above equation) to get the locus
$$x=a-\frac{2a}{b}\cdot y$$
or equivalently, $$y=\frac{b}{2}-\frac{b}{2a}\cdot x$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2220757",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Proving that a sequence of functions converge uniformly I have a sequence of functions:
$$S_n(x) = x\frac{1-x^n}{1-x}, \text{ where } x \in (-1, 1)$$
It is known that:
$$\lim_{n \to \infty}S_n(x) = \frac{x}{1-x}$$
thus $S_n(x)$ converges pointwise. I am to prove that is does not converge uniformly. I tried to estimate:
$$\left|S_n(x) - \frac{x}{1-x}\right|$$ using an $x \in (-1, 1)$ but it didn't work.
I would appreciate any help.
| First, we simplify the expression. Note that:
\begin{align}
\left|\frac{x(1-x^n)}{1-x} - \frac{x}{1-x}\right| &= \left|\frac{x(1-x^n) -x}{1-x}\right| \\
&=\left|\frac{-x^{n+1}}{1-x}\right| \\
&= \frac{x}{1-x}\cdot x^{n}
\end{align}
We want $\left|S_n(x) - \frac{x}{1-x} \right| < \epsilon$ for given $\epsilon$ and $n \ge N$ for some large enough $N$. For this to be the case, we would need an $N$ such that, for all $n\ge N$:
\begin{align*}
&\frac{x}{1-x}x^n < \epsilon \\
\iff & x^n < \epsilon\frac{1-x}{x} \\
\iff & n > \frac{\ln\left(\epsilon\frac{1-x}{x}\right)}{\ln x} = \frac{\ln \left(\epsilon(1-x)\right) - \ln x}{\ln x}
\end{align*}
But, as $x\to 1$ from the left, $\frac{\ln \left(\epsilon(1-x)\right) - \ln x}{\ln x} \to \infty$. So, no such $N$ exists.
Therefore, $S_n(x)$ does not converge uniformly.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2221152",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
"answer_id": 4
} |
How do I to find the limits of these? $$\lim _{x\to \frac{\pi }{4}}\left(\frac{1-\sin 2x}{\cos x -\sin x}\right)$$
$$\lim _{x\to \frac{\pi}{4}}\left(\frac{\cos^5x - \sin^5x}{\cos\:2x}\right)$$
I tried to solve but I get 0/0.
Help me to find the limit.
| $$
\begin{aligned}
\lim _{x\to \frac{\pi }{4}}\left(\frac{1-\sin2x}{\cos x\:-\sin x}\right)
& = \lim _{t\to 0}\left(\frac{1-sin2\left(t+\frac{\pi }{4}\right)}{\cos \left(t+\frac{\pi \:}{4}\right)\:-\sin \left(t+\frac{\pi \:}{4}\right)}\right)
\\& = \lim _{t\to 0}\left(-\frac{\sqrt{2}\left(1-\sin \left(2\right)\left(t+\frac{\pi \:}{4}\right)\right)}{2\sin \left(t\right)}\right)
\\& = \color{red}{\infty}
\end{aligned}
$$
$$
\begin{aligned}
\lim _{x\to \frac{\pi }{4}}\left(\frac{\cos ^5x\:-\:\sin ^5x}{\cos \:2x}\right)
& = \lim _{t\to 0}\left(\frac{\cos ^5\left(t+\frac{\pi }{4}\right)\:-\:\sin ^5\left(t+\frac{\pi \:}{4}\right)}{\cos \left(2\left(t+\frac{\pi \:}{4}\right)\right)}\right)
\\& = \lim _{t\to 0}\left(\frac{-\frac{\sqrt{2}}{4}\sin ^5\left(t\right)-\frac{\sqrt{2}\cdot 5}{2}\sin ^3\left(t\right)\cos ^2\left(t\right)-\frac{\sqrt{2}\cdot 5}{4}\cos ^4\left(t\right)\sin \left(t\right)}{\cos \left(2\left(t+\frac{\pi \:}{4}\right)\right)}\right)
\\& \approx \lim _{t\to 0}\left(\frac{-\frac{\sqrt{2}}{4}\left(t\right)^5-\frac{\sqrt{2}\cdot \:\:5}{2}\left(t\right)^3\left(1-\frac{t^2}{2}\right)^3-\frac{\sqrt{2}\cdot \:\:5}{4}\left(1-\frac{t^2}{2}\right)^4\left(t\right)}{\cos \:\left(2\left(t+\frac{\pi \:}{4}\right)\right)}\right)
\\& = \lim _{t\to 0}\left(-\frac{\sqrt{2}\left(15t^9-80t^7+104t^5-80t\right)}{64\sin \:\left(2t\right)}\right)
\\& \approx \lim _{t\to 0}\left(-\frac{\sqrt{2}\left(15t^9-80t^7+104t^5-80t\right)}{64\left(2t\right)}\right)
\\& = \color{red}{\frac{5}{4\sqrt{2}}}
\end{aligned}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2225797",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Limit of $S_n = \frac{1}{2n + 1} + \frac{1}{2n + 3} + ... + \frac{1}{4n - 1}$ when $n\to\infty$ $$S_n = \frac{1}{2n + 1} + \frac{1}{2n + 3} + ... + \frac{1}{4n - 1}$$
the task is to find $$\lim_{n \to \infty} S_n$$
I've tried different ways, but all I could do is to make an estimation that the limit is somewhere between 0.5 and 1, but that's not the exact answer.
Manually doing fist elements gives
$$\frac{1}{3}, \quad \frac{1}{5} + \frac{1}{7}, \quad \frac{1}{7} + \frac{1}{9} + \frac{1}{11}, \quad ...$$
I'm trying to restate it as
$$\lim_{n \to \infty} \bigg[ \sum_{i = 1}^{\infty} \frac{1}{2i - 1} - \sum_{i = 1}^{n} \frac{1}{2i - 1} - \sum_{i = 4n}^{\infty} \frac{1}{2i - 1} \bigg]$$
so that I cut off the beginning of the series and it's tail.
| I'll give a hint that follows user43208's reasoning (though NB that the formula he gives is missing a numerical factor from one term and so leads to an incorrect result).
Hint The quantity $S_n$ can be written as
$$\left(\frac{1}{2 n + 1} + \frac{1}{2 n + 2} + \cdots + \frac{1}{4 n - 1} + \frac{1}{4n}\right) \\ \quad- \left(\frac{1}{2n + 2} + \frac{1}{2 n + 4} + \cdots + \frac{1}{4 n - 2} + \frac{1}{4 n} \right),$$
and the second quantity in parentheses can be written as
$$\frac{1}{2}\left(\frac{1}{n + 1} + \frac{1}{n + 2} + \cdots + \frac{1}{2 n -
1} + \frac{1}{2 n}\right).$$
So we can write $$S_n = (H_{4n} - H_{2n}) - \tfrac{1}{2}(H_{2n} - H_n),$$ where $H_m$ is the $m$th harmonic number, $1 + \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{m}$. On the other hand, $H_n$ satisfies the asymptotic expansion $$H_m = \log m + \gamma + O\left(\frac{1}{m}\right) .$$ Here, $\gamma$ is the Euler-Mascheroni constant---in fact, it is usually essentially defined by this previous equation, but substituting the expansion in our above formula for $S_n$ shows that its value is irrelevant for purposes of evaluating our limit.
If the question had instead by tagged $\texttt{calculus}$, the intended method would presumably be to recognize $S_n$ as a Riemann sum, as Fedor Petrov's hint does. (NB the answer he states is also off by a numerical factor.)
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Help differentiating $\frac{(x-1)^2(x+2)^2}{(x+1)^2}$ I need help differentiating this expression below. I know you can use a mix of the product rule and quotient rule, but that is tedious and long. Is there a shorter method?
$$\dfrac{(x-1)^2(x+2)^2}{(x+1)^2}$$
| In fact we do not need a mix of quotient and product rule. Applying the product rule once is sufficient.
We obtain
\begin{align*}
\frac{d}{dx}&\frac{(x-1)^2(x+2)^2}{(x+1)^2}\\
&=\frac{d}{dx}\left[\color{blue}{(x^2+x-2)^2}\color{red}{(x+1)^{-2}}\right]\\
&=\color{blue}{2(x^2+x-2)(2x+1)}\color{red}{(x+1)^{-2}}+\color{blue}{(x^2+x-2)^2}\color{red}{(-2)(x+1)^{-3}}\\
&=\frac{2(x^2+x-2)}{(x+1)^3}\left[(2x+1)(x+1)-(x^2+x-2)\right]\\
&=\frac{2(x^2+x-2)(x^2+2x+3)}{(x+1)^3}
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2230982",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How do we integrate $x\sqrt{x^3 + 1}$? Tried to use integration by parts but just keep going round in circles. Any help appreciated, thanks.
| a possible answer can be found by Mathematica:
$$\frac{2 \left(x^5+x^2-3^{3/4} \sqrt{-\sqrt[6]{-1} \left(x+(-1)^{2/3}\right)}
\sqrt{(-1)^{2/3} x^2+\sqrt[3]{-1} x+1} \left((-1)^{5/6} F\left(\sin
^{-1}\left(\frac{\sqrt{-(-1)^{5/6}
(x+1)}}{\sqrt[4]{3}}\right)|\sqrt[3]{-1}\right)+\sqrt{3} E\left(\sin
^{-1}\left(\frac{\sqrt{-(-1)^{5/6}
(x+1)}}{\sqrt[4]{3}}\right)|\sqrt[3]{-1}\right)\right)\right)}{7
\sqrt{x^3+1}}$$
| {
"language": "en",
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"source": "stackexchange",
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Is the set $M = \{d(n) \mid n \in S\}$ unbounded?
Consider the set $S = \{c \in \mathbb{Z}^+ \mid a^2+b^2 = c^2, \text{for some} \text{ } \text{coprime positive integers} \text{ } a,b\}$. Let $d(n)$ denote the number of positive integer divisors of a positive integer $n$. Is the set $M = \{d(n) \mid n \in S\}$ unbounded?
In other words, I am wondering if the hypotenuse of a primitive right triangle can have an arbitrarily large amount of divisors. How can we relate the divisor function to a primitive right triangle?
| Notation: All variable-symbols are positive integers.
Theorem 1. If $p$ is prime and $p\equiv 1 \pmod 4$ there exist $a,b$ with $a^2+b^2=p.$ Note that we must have $\gcd(a,b)=1.$
Corollary. If $p$ is prime and $p\equiv 1 \pmod 4$, there exist $x,y$ with $p^2=x^2+y^2$ and $\gcd (x,y)=1.$ Proof: By Theorem 1, let $p=a^2+b^2$ . Let $x=|a^2-b^2|$ and $y=2ab.$ We have $a\ne b$ (else $p=2a^2$ is not prime), so $x\ne 0.$ We have $p^2=x^2+y^2.$
Now $p^2$ is odd so $|a^2-b^2|$ is odd so $\gcd (|a^2-b^2|,2ab)=\gcd(|a^2-b^2|,ab).$ And $\gcd (a,b)=1,$ so if $q$ is prime and $q|ab$ then $$[(q|a\land q\not | b)\lor (q\not |a\land q|b) ],$$ which implies $q\not |\;(a^2-b^2).$
Theorem 2. If $q$ is prime and $q\equiv 3 \pmod 4$ then $q$ does not divide any $a^2+1.$
Theorem 3. There is no largest prime congruent to $1 \pmod 4.$ Proof: If $S$ is a finite non-empty set of primes, each congruent to $1\pmod 4$ then any prime divisor of $(2\prod_{p\in S}p)^2+1$ does not belong to $S$, and is congruent to $1 \pmod 4,$ by Theorem 2.
Lemma. If $x_1^2=a_1^2+b_1^2$ and $x_2^2=a_2^2+b_2^2$ where $1=\gcd (x_1,x_2)=\gcd (a_1,b_1)=\gcd (a_2,b_2)$ then $\gcd (a_1a_2+b_1b_2, a_1b_2-a_2b_1)=1.$
Proof: By contradiction, suppose prime $q$ divides both $a_1a_2+b_1b_2$ and $a_1b_2-b_1a_2.$ Then modulo $q$ we have $$a_1a_2b_2\equiv -b_1b_2^2 \text { and }\;a_1b_2a_2\equiv b_1a_2^2$$ $$\text {implying }\; b_1(a+2^2+b_2^2)\equiv 0.$$ $$\text {Now }\; q|b_1\implies q|(a_1ca+2+b_1b_2)-b_1b_2=a_1a_2\implies q|a_2$$ (because if $q|b_1$ and $\gcd (a_1,b_1)=1$ and $q|a_1a_2$ then $q|a_2$). Similarly $$q|b_1\implies q|(a_1b_2-b_1a_2)+b_1a_2=a_1b_2\implies q|b_2.$$
So $q|b_1\implies \gcd (a_2,b_2)\geq q>1,$ which is false. (So far so good, but this is not the desired contradiction.) So from $b_1(a_2^2+b_2^2)\equiv 0\pmod q$ we conclude that $$q|(a_2^2+b_2^2)=x_2^2.$$
But by interchanging the subscripts 1,2 we also conclude that $$q|(a_1^2+b_1^2)=x_1^2.$$
This is a contradiction because $q$ is prime and $\gcd(x_1^2,x_2^2)=1. $ QED.
Putting this all together:
Let $p_1,..., p_{n+1}$ be $n+1$ distinct primes, each $\equiv 1 \pmod 4$.
Let $p_{n+1}^2=a_2^2+b_2^2$ with $\gcd (a_2,b_2)=1.$
If $\;\prod_{1=1}^np_i^2=a_1^2+b_1^2$ with $\gcd (a_1,b_1)=1,$ then $$\prod_{i=1}^{n+1}p_i^2=c^2+d^2$$ where $c=a_1a_2+b_1b_2$ and $d=|a_1b_2-a_2b_1|.$ And by the lemma we have $\gcd (c,d)=1.$
| {
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Proof of $\sum_{k=0}^{n}{(-1)^k{n \choose k}\frac{1}{(k+1)(k+2)}}=\frac{1}{n+2}$
Deduce that
$$
{n \choose 0}\frac{1}{1\cdot 2}-{n \choose 1}\frac{1}{2\cdot 3}+{n \choose 2}\frac{1}{3\cdot4}+\cdots+{n \choose n}\frac{1}{(n+1)\cdot(n+2)}=\frac{1}{n+2}
$$
Is induction a viable method to solve this problem? Or is there some other method which might be simpler?
| Observe that the sum is
$$\frac{1}{(n+1)(n+2)}
\sum_{k=0}^n {n+2\choose k+2} (-1)^{k+2}$$
which is
$$\frac{1}{(n+1)(n+2)}
\sum_{k=2}^{n+2} {n+2\choose k} (-1)^{k}
\\ = \frac{1}{(n+1)(n+2)}
\left(-{n+2\choose 0} (-1)^0 - {n+2\choose 1} (-1)^1\right)
\\ = \frac{1}{(n+1)(n+2)}
\left(-1 + n+2\right) = \frac{1}{n+2}.$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Prove this $(ab)^2+(bc)^2+(ca)^2\le ab+bc+ac$
Let$a$, $b$ and $c$ be non-negative numbers such that
$$a^2+b^2+c^2=a+b+c.$$
Show that
$$(ab)^2+(bc)^2+(ca)^2\le ab+bc+ac.$$
I tried the uvw's technique and BW and more but without some success.I think can use C-S solve it?
| I think $uvw$ helps here.
The homogenization gives:
$$(a^2b^2+a^2c^2+b^2c^2)(a+b+c)^2\leq(ab+ac+bc)(a^2+b^2+c^2)^2.$$
Hence, the condition it's $u=3u^2-2v^2$ and does not depend on $w^3$.
In another hand, we need to prove that $$3v^4-2uw^3\leq v^2,$$
which is a linear inequality of $w^3$, which says that
it remains to prove our inequality for an extremal value of $w^3$,
which happens in the following cases.
*
*$w^3=0$.
Since our new inequality is homogeneous, we can assume $c=0$ and $b=1$, which gives
$$a(a^2-a+1)(a-1)^2\geq0;$$
*$b=c=1$, which gives
$$(a-1)^2(2a^2+3a+4)\geq0.$$
Done!
But we can prove this by another way:
we need to prove that
$$\sum_{cyc}(a^4+2a^2b^2)\sum_{cyc}ab\geq\sum_{cyc}a^2b^2\sum_{cyc}(a^2+2bc)$$ or
$$\sum_{cyc}(a^5b+a^5c-a^4b^2-a^4c^2+a^4bc-a^2b^2c^2)\geq0$$ or
$$\sum_{cyc}ab(a^2-ab+b^2)(a-b)^2+\frac{1}{2}abc(a+b+c)\sum_{cyc}(a-b)^2\geq0$$
and we are done!
| {
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"timestamp": "2023-03-29T00:00:00",
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Analyzing quadratic forms $x^2-3xy+y^2$, $2xy+yz-3xz$ and $x^2+y^2+2xy-xt+2yt$. Positive, negative or indefinite?
By means of successive coordinate changes, write each one of the
quadratic forms below as a sum of terms of the type $\pm u^2$ and
decide wich ones are positive, negative or indefinite:
$$A(x,y) = x^2-3xy+y^2$$
$$B(x,y,z) = 2xy+yz-3xz$$
$$C(x,y,z,t) = x^2+y^2+2xy-xt+2yt$$
For $A$ I did $A(x,y) = x^2-3xy+y^2 = x^2-2xy+y^2-xy = (x-y)^2-xy$ which is indefinite.
For $B(x,y) = 2xy+yz-3xz$ I remembered that $(x+y+z)^2 = x^2 + 2 x y + y^2 + 2 x z + 2 y z + z^2$ so $2xy+yz-3xz = (x+y+z)^2 -x^2-y^2-z^2-yz+xz$ which won't help anything.
For $C(x,y,z,t)$ there's too much terms, how should I do it?
| All you need to do is to plug random numbers in. If an expression yields both positive and negative numbers, it is indefinite.
For example, for $B(x,y,z)$, when $x=1$, $y=3$, $z=2$, the expression becomes $2(1)(3) + (3)(2) - 3(1)(3) = 3$, which is positive. When $x=10$, $y=1$, $z=100$, the expression becomes $2(10)(1) + (1)(100) - 3(10)(100) = -2880$ which is negative. Therefore, $2xy + yz - 3xz$ is indefinite.
Try using this method to evaluate question c). Although this method is outside the rules, you can use this method to check any mistakes in $u^2$.
| {
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If $x^2 - 3x + 2$ is a factor of $x^4 - px^2 +q$, then find the value of $p$ and $q$ If $x^2 - 3x +2$ is a factor of $x^4-px^2+q$ then find the value of $p$ and $q$.
My attempt:
$$x^2-3x+2$$
$$x^2-2x-x+2$$
$$x(x-2)-1(x-2)$$
$$(x-1)(x-2)$$
How do I proceed further?
P.S: Edit after Deepak's comment!
| Using long division:
$$ \frac{x^4 - px^2 + q}{x^2-3x+2} = x^2 + 3x + \frac{(7-p)x^2 - 6x +q}{x^2-3x+2}.$$
Look at the last fraction. By inspection, the ratio will be equal to $2$ if
$$ 7-p = 2 \implies p = 5$$
$$q = 4$$
Note: Previously, I mistook $q$ for $9$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Better way of finding $\int {x^2 + 1 \over x^4 + x^2 + 1} dx$.
$$\int {x^2 + 1 \over x^4 + x^2 + 1} dx$$
By separating partial fractions,
$${Ax + B \over x^2 - x + 1} + {Cx + D \over x^2 + x + 1} = {x^2 + 1 \over x^4 + x^2 + 1} \\ \implies (Ax + B)(x^2 + x + 1) + (Cx + D)(x^2 - x + 1) = x^2 + 1$$
I get $$\begin{cases} A = -1/2 \\
B = 1/2\\
C = 1/2 \\
D = 1/2\end{cases}$$
For which the integrand becomes
$${1 \over 2}\int {1 - x\over x^2 - x + 1} dx + {1\over2} \int {x + 1\over x^2 + x + 1} dx$$
Now these two are easy enough to solve but still very tedious, not to mention that partial fractions was also very tedious.
Is there a less cumbersome way to solve this ? I tried to change the integrand of form $1 + 1/x^2$ so that I substitute $u = 1 - 1/x$ but was unsuccessful .
| Another way is to perform the awkward substitution $x=\frac{-\sqrt{3}+\sqrt{3+4t^2}}{2t}$ to simply get $\int\frac{dt}{\sqrt{3}(1+t^2)}$, but of course this approach just follows from reverse-engineering the straightforward solution through partial fraction decomposition.
| {
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If $\alpha$, $\beta$, $\gamma$, the angles made by a vector with $x, y, z$ axes, are in the ratio 1:2:3 then their values are? This is a question asked in a book:
If $\alpha$, $\beta$, $\gamma$ are angles made by a vector $x, y, z$ axes respectively, and $\alpha$:$\beta$:$\gamma$ = 1:2:3 then values of $\alpha$, $\beta$, $\gamma$ are?
I tried the obvious by taking $\alpha$ = k; $\beta$=2k; $\gamma$ = 3k; and substituting in $cos^2\alpha$+$cos^2\beta$+$cos^2\gamma$ = 1
but its complicating things. I'm must be missing something simple and obvious but I can't figure out what.
| If
$\cos^2 x +\cos^2(2x)+\cos^2(3x)=1
$,
then,
since
$\cos^2(z)
=(\cos(2z)+1)/2
$,
$1
=(\cos(2x)+\cos(4x)+\cos(6x)+3)/2
$
or
$\cos(2x)+\cos(4x)+\cos(6x)
= -1
$.
Letting
$y = 2x$,
$\begin{array}\\
-1
&=\cos(y)+\cos(2y)+\cos(3y)\\
&=\cos(y)+2\cos^2y-1+4\cos^3(y)-3\cos(y)\\
&=4\cos^3(y)+2\cos^2y-2\cos(y)-1\\
\end{array}
$
or
$4\cos^3(y)+2\cos^2y-2\cos(y)
=0
$.
Letting
$z = \cos y$,
$\begin{array}\\
0
&=2z^3+2z^2-z\\
&=z(2z^2+z-1)\\
&=z(z+1)(2z-1)
\qquad\text{just fixed an error here}\\
\end{array}
$
so
$z = 0, -1, \frac12
$.
Therefore
$y=\pi/2, \pi, \pi/3$
so
$x
= y/2
=\pi/4, \pi/2, \pi/6
$.
| {
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"timestamp": "2023-03-29T00:00:00",
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What is $1+x+x^2+x^3...$? What is the difference between these two series?
$$
\begin{align}
1+x+x^2+x^3+...+x^n+\mathcal O(x^{n+1})&=\frac{1}{1-x}\\
\\
1+r+r^2+r^3+...+r^{n-1}&=\frac{r^n-1}{r-1}\\
\end{align}
$$
I can't wrap my head around it; they both start with $1+x+x^2+x^3...$? Of course the first is the maclaurin series. But other than that, why don't they both yield the same result?
Thanks!
| Hint:
$$1+x+x^2+\dots+x^n=\frac{x^{n+1}-1}{x-1}$$
$$\frac{x^{n+1}-1}{x-1}=\frac{1-x^{n+1}}{1-x}=\frac1{1-x}-\frac{x^{n+1}}{1-x}=\frac1{1-x}+\mathcal O(x^{n+1})$$
which holds as $x\to0$, since $1-x\to1$.
Move $\mathcal O(x^{n+1})$ to the other side and you'll get
$$1+x+x^2+\dots+x^n+\mathcal O(x^{n+1})=\frac1{1-x}$$
| {
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If $a,b,c\in N$, such that $aaaaaa=b*c*bc*(b+b+c)*(c+c-b)(c-b)$, then find $a,b,c$.
If $a,b,c\in N$, such that $aaaaaa=b*c*bc*(b+b+c)*(c+c-b)(c-b)$, then find $a,b,c$. Where $aaaaaa$ is a six digit no. with each digit equal to $a$ and $bc$ is a 2-digit no. with digits $b$ and $c$.
I can think of expressing $aaaaaa$ as $a\frac{10^6-1}{10-1}$, but how to proceed?
| $aaaaaa=111111 \times a$. So lets factorise $111111=3 \times 7 \times 11 \times 13 \times 37$. Now check the factors on the right hand side ... $b=3$ , $ c=7$ ,$bc=37$ ,$(b+b+c)=13$,$(c+c-b)=11$, $(c-b)=4$. So $a= \color{red}{4}$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Evaluating $\int_C\frac{z+1}{z^2-2z}dz$, where $C$ is the circle $|z|=3$
Evaluate the contour integral $\int_C\frac{z+1}{z^2-2z}dz$ using Cauchy's residue theorem, where $C$ is the circle $|z|=3$.
I see that the function has 2 singularities, at 0 and 2, so I need to find the residue of each. By examining the Laurent series, I have the following:
$$f(z)=\left(\frac{z+1}{z}\right)\left(\frac{1}{z-2}\right)=\left(\frac{z+1}{z^2}\right)\left(\frac{1}{1-2/z}\right)=\left(\frac{1}{z}+\frac{1}{z^2}\right)\left(\frac{1}{1-2/z}\right)$$ and therefore $$f(z)=\left(\frac{1}{z}+\frac{1}{z^2}\right)\left(1-\frac{2}{z}+\frac{4}{z^2}-\frac{8}{z^3}+\cdots\right)=\frac{1}{z}-\frac{1}{z^2}+\frac{2}{z^3}-\frac{4}{z^4}+\cdots$$
so the residue at 0 is 1.
Similarly, $$f(z)=\left(\frac{z+1}{2(z-2)}\right)\left(\frac{1}{1+(z-2)/2}\right)=\left(\frac{1}{2}+\frac{3}{2(z-2)}\right)\left(\frac{1}{1+(z-2)/2}\right)$$ and so $$f(z)=\left(\frac{1}{2}+\frac{3}{2(z-2)}\right)\left(1-\frac{z-2}{2}+\frac{(z-2)^2}{4}-\frac{(z-2)^3}{8}+\cdots\right)$$
Thus $$f(z)=\frac{3}{2(z-2)}-\frac{1}{4}+\frac{1}{8}(z-2)-\frac{1}{16}(z-2)^2+\cdots$$
and so the residue at 2 is $\frac{3}{2}$.
So I think $\int_Cf(z)dz=2\pi i(1+\frac{3}{2})= 5\pi i$, but that's not what the book is telling me - the book says the answer should be $2\pi i$. What am I doing wrong?
| Everything about your approach is fine, and your method of finding the Laurent series by factoring out a geometric series is smart. You just made a mistake when you equated
$$\frac{1}{1-\frac{2}{z}} = 1-\frac{2}{z}+\frac{4}{z^2}-\frac{8}{z^3}+\cdots$$
The series on the right is actually an expansion of $\frac{1}{1+\frac{2}{z}}$, and it's the expansion at $z = \infty$ whereas you need the expansion at $z = 0$.
So what you should have is:
$$\frac{1}{1-\frac{2}{z}} = - \frac{z}{2} - \frac{z^2}{4} - \ldots - \frac{z^n}{2^n} + \ldots$$
Then you'll get the Laurent series expansion at $z = 0$,
$$\frac{z+1}{z^2 - 2z} = -\frac{1}{2z} - \frac{3}{4} - \frac{3 z}{8} - \frac{3 z^2}{16} - \ldots - \frac{3z^n}{2^{n+2}} - \ldots$$ and the residue
you find this way is correct.
| {
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If one root of the equation $ax^2+bx+c=0$ be the square of the other then which is true? If one root of the equation $ax^2+bx+c=0$ be the square of the other then which is true?
$1$. $a^3+b^3+c^3-3abc=0$
$2$. $a^3+b^3+bc^2=3abc$
$3$. $b^3+a^2c+ac^2=3abc$
$4$. none.
My Attempt:
Let one root be $\alpha $ then the other root will be $\alpha^2$. Then,
$$(x-\alpha)(x-\alpha^2)=0$$
$$x^2-x(\alpha^2+\alpha)+\alpha^3=0$$ Comparing with $ax^2+bx+c=0$ we get,
$$a=1$$
$$b=-(\alpha^2+\alpha)$$
$$c=\alpha^3$$
| Product of roots $= \alpha^3 = \dfrac{c}{a}$
Now $a\alpha^2+b\alpha+c = 0 \Rightarrow a^3 \alpha^6+b^3 \alpha^3+c^3 = 3abc \alpha^3$
Substitute for $\alpha^3$ in the above and after simplification we obtain $b^3+a^2c+ac^2=3abc$
In the above we assume that $\alpha \ne 0$, but its easy to see that the above relation holds even when $\alpha = 0$ and hence all cases are covered.
| {
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"url": "https://math.stackexchange.com/questions/2252741",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
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Prove $\int_{0}^{1}{{2x+\ln{1-x\over 1+x}\over x^3(1-x^2)^{1/2}}}dx=-(\pi/2)^2$ and $\int_{0}^{1}{2x+\ln{1-x\over 1+x}\over x^2(1-x^2)^{1/2}}dx=-2$ Motivated by this Question
Two similar integrals, but exhibit completely different closed forms
$$\int_{0}^{1}{{2x+\ln\left({1-x\over 1+x}\right)\over x^3\sqrt{1-x^2}}}\mathrm dx=-\color{blue}{\left({\pi\over 2}\right)^2}\tag1$$
and
$$\int_{0}^{1}{{2x+\ln\left({1-x\over 1+x}\right)\over x^2\sqrt{1-x^2}}}\mathrm dx=-\color{red}{2}\tag2$$
Here is my try:
I don't think substitution work here initially, but anyway I try
$u={1-x\over 1+x}\implies -{(1+x)^2\over 2}du=dx$, then $(1)$ becomes after a lengthy simplication to
$$\sqrt{2}\int_{0}^{1}{2\left({1-u\over 1+u}\right)+\ln u \over \sqrt{u}(1-u)^3}\cdot(1+u)^{11/2}\mathrm du\tag3$$
Or we split $(1)$ into
$$2\int_{0}^{1}{\mathrm dx\over x^2\sqrt{1-x^2}}+\int_{0}^{1}{\ln(1-x)\over x^3\sqrt{1-x^2}}\mathrm dx-\int_{0}^{1}{\ln(1+x)\over x^3\sqrt{1-x^2}}\mathrm dx=I_1+I_2-I_3\tag4$$
$I_1$ diverges.
How can we prove integrals $(1)$ and $(2)$?
| Here's a possible path to #1 with some arguably important details left out. So I'm making this community wiki in case someone else wants to come along and fill in those details or just completely fix this.
Recall:
$$ \ln(1+x) = \sum_{n=1}^{+\infty} \frac{(-1)^{n+1}}n x^n$$
So then:
$$ \ln(1-x) = \sum_{n=1}^{+\infty} \frac{(-1)^{n+1}}n (-x)^n = \sum_{n=1}^{+\infty} \frac{-1}n x^n$$
And:
\begin{align*}
2x + \ln\left(\frac{1-x}{1+x}\right) &= 2x + \ln(1-x) - \ln(1+x)\\[0.3cm]
&= 2x + \sum_{n=1}^{+\infty} \frac{-1}n x^n - \sum_{n=1}^{+\infty} \frac{(-1)^{n+1}}n x^n\\[0.3cm]
&= 2x + \left(-x -\frac12x^2 -\frac13x^3 -\cdots\right) - \left(x - \frac12 x^2 + \frac13 x^3 - \cdots\right)\\[0.3cm]
&= 2x -2 \left(x + \frac13x^3 + \frac15x^5 + \cdots\right)\\[0.3cm]
&= -2\sum_{n=1}^{+\infty} \frac{x^{2n+1}}{2n+1}
\end{align*}
Then (sweeping some convergence details under the rug here) we have:
\begin{align*}
\int_0^1 \frac{2x + \ln\left(\frac{1-x}{1+x}\right)}{x^3\sqrt{1-x^2}} \, dx &= \int_0^1 \frac{\displaystyle -2\sum_{n=1}^{+\infty} \frac{x^{2n+1}}{2n+1}}{x^3\sqrt{1-x^2}} \, dx\\[0.3cm]
&= -2\sum_{n=1}^{+\infty} \frac1{2n+1} \int_0^1 \frac{x^{2n-2}}{\sqrt{1-x^2}} \, dx
\end{align*}
Let $x = \sin \theta$. Then $$ \int_0^1 \frac{x^{2n-2}}{\sqrt{1-x^2}} \, dx = \int_0^{\pi/2} \sin^{2n-2} \theta \, d\theta. $$
Wolfram Alpha says: $$\int_0^{\pi/2} \sin^{2n-2} \theta \, d\theta = \frac{\sqrt\pi \ \Gamma\left(n-\frac12\right)}{2\Gamma(n)}$$
I suspect this can be proved by induction but I didn't quite get there. Anyway, now we have:
$$
\int_0^1 \frac{2x + \ln\left(\frac{1-x}{1+x}\right)}{x^3\sqrt{1-x^2}} \, dx = -2\sum_{n=1}^{+\infty} \frac{\sqrt\pi \ \Gamma\left(n-\frac12\right)}{2(2n+1) \Gamma(n)}$$
Not quite sure how to evaluate that summation or if it's even possible. But it does give the correct value.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2254767",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Solve real integrals $\int_0^{2\pi} \frac{d\theta}{1+a\cos(\theta)} = \frac{2\pi}{\sqrt{1-a^2}}$ using complex variables. I am trying to
verify that $\displaystyle\int_0^{2\pi} \frac{d\theta}{1+a\cos(\theta)} = \frac{2\pi}{\sqrt{1-a^2}}$, for $-1\lt a \lt 1$.
So far I replaced $\cos(\theta)$ with $\dfrac{z+\frac{1}{z}}{2}$ and $d\theta$ with $\dfrac{dz}{iz}$, and then simplified to get $$\frac 2 i \int_C \frac{dz}{az^2+2z+a}$$ (with $C$ being the unit circle). Then this function will have simple poles at $\frac{-1\pm\sqrt{1-a^2}}{a}$. Then I calculated $\operatorname{Res}\left(f: \dfrac{-1+\sqrt{1-a^2}} a\right) = \dfrac a {2\sqrt{1-a^2}}$ and $\operatorname{Res}\left(f: \dfrac{-1-\sqrt{1-a^2}} a \right) = \dfrac{-a}{2\sqrt{1-a^2}}$. But then when I apply the residue theorem, clearly the sum of the two residues is zero, and then the entire integral is zero. I'm sure there must be a mistake in my calculations somewhere, but I'm not sure where. I'm thinking maybe I need to use that a is between $-1$ and $1$. Any help/corrections are appreciated!
| By symmetry
$$ \int_{0}^{2\pi}\frac{d\theta}{1+a\cos\theta} = 2\int_{0}^{\pi}\frac{d\theta}{1-a^2\cos^2\theta} =4\int_{0}^{\pi/2}\frac{d\theta}{1-a^2\cos^2\theta}$$
and by setting $\theta=\arctan t$ the last integral turns into:
$$ 4\int_{0}^{+\infty}\frac{dt}{(1+t^2)-a^2} = 2\int_{\mathbb{R}}\frac{dt}{(1-a^2)+t^2} $$
that is:
$$ 4\pi i\,\text{Res}\left(\frac{1}{(1-a^2)+t^2},t=i\sqrt{1-a^2}\right)=4\pi i\lim_{t\to i\sqrt{1-a^2}}\frac{1}{t+i\sqrt{1-a^2}} $$
i.e. $\color{red}{\large\frac{2\pi}{\sqrt{1-a^2}}}$ as wanted.
It is always worth considering symmetry tricks, because they make our life easier both in the real and complex case.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2254949",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Solving Recurrence Relation with substitution How to solve T(n) = T(n-2) + n using iterative substitution
Base case:
T(0) = 1
T(1) = 1
Solve:
T(n) = T(n-2) + n
Currently I have:
T(n) = T(n-2) + n
= T(n-4) + n - 2 + n = T(n-4) + 2n - 2
= T(n-6) + n - 4 + n - 2 + n = T(n-6) + 3n - 6
= T(n-8) + n - 6 + n - 4 + n -2 + n = T(n-8) + 4n - 12
= T(n-10) + n - 8 + n - 6 + n - 4 + n - 2 + n = T(n-10) + 5n - 20
The pattern I see is:
$$\ T(n-2 \sum_{i=1}^k i) + n \sum_{i=0}^k i - \sum_{i=0}^{k-1} i(i+1) $$
but this may be wrong because I am completely stuck after this
| If $n$ is even,
$$\sum_{k=1}^{\frac{n}{2}}[T(2k)-T(2k-2)]=\sum_{k=1}^{\frac{n}{2}}2k$$
$$T(n)-T(0)=\frac{1}{2}\left(\frac{n}{2}\right)(2+n)$$
$$T(n)=\frac{1}{4}n^2+\frac{1}{2}n+1$$
If $n$ is odd,
$$\sum_{k=1}^{\frac{n-1}{2}}[T(2k+1)-T(2k-1)]=\sum_{k=1}^{\frac{n-1}{2}}(2k+1)$$
$$T(n)-T(1)=\frac{1}{2}\left(\frac{n-1}{2}\right)(3+n)$$
$$T(n)=\frac{1}{4}n^2+\frac{1}{2}n+\frac{1}{4}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2259243",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Congruence equation with polynomials So, I tried solving $x^{12} \equiv 37 \mod 41$. I tried to use indices technique (not sure if everyone is familiar with that) but ended up 3 solutions short (there are 4 solutions) and calculations were tedious, to say at least.
Could anyone provide a method for solving congruences like this one, possibly using same technique or something elementary?
Thank you very much!
| We would like to find a primitive gernerator for $\mathbb{Z}_{41}$ ... so lets try $2$ ...$2,4,8,16,32,23,5,10,20,40$. Now $40 \equiv -1 \mod 41$ so $2^{20}=1$ ... carrying on a bit ...
$39,\color{red}{37},33,25,9,18,36,31,21,1$. So it turns out that $x=2$ is a solution.
To find the other three solutions we shall use $2^{20}=1$
\begin{eqnarray*}
37 \equiv 2^{12} \equiv 2^{12+3 \times 20} \equiv 2^{72} \equiv (2^{6})^{12} \equiv 23^{12} \mod 41 \\
37 \equiv 2^{12} \equiv 2^{12+6 \times 20} \equiv 2^{132} \equiv (2^{11})^{12} \equiv 39^{12} \mod 41 \\
37 \equiv 2^{12} \equiv 2^{12+9 \times 20} \equiv 2^{192} \equiv (2^{16})^{12} \equiv 18^{12} \mod 41
\end{eqnarray*}
So the four solutions are $x=\color{red}{2,23,39,18}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2260676",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Proof that $1 \geq \sqrt{1-p} \cdot \sqrt{1-p+e}+\sqrt{p} \cdot \sqrt{p-e}$ with $0Is there a way to prove that
$$
1 \geq \sqrt{1-p} \cdot \sqrt{1-p+e}+\sqrt{p} \cdot \sqrt{p-e} \\
\leftrightarrow \sqrt{1-p} \cdot \sqrt{1-p}+\sqrt{p} \cdot \sqrt{p} \geq \sqrt{1-p} \cdot \sqrt{1-p+e}+\sqrt{p} \cdot \sqrt{p-e}
$$
with $0<p<1$ and $0<e<p$
I have been struggling on this all day, and I thought that one of you might have the answer? It may have to do with the concavity of the square-root function?
| Let $p=u^2$ and $p-e=v^2$, with $0\lt v\lt u\lt1$. The inequality becomes
$$1\ge\sqrt{1-u^2}\sqrt{1-v^2}+uv$$
or
$$1-uv\ge\sqrt{1-u^2}\sqrt{1-v^2}$$
Since $1-uv\gt0$, this inequality is equivalent to
$$(1-uv)^2\ge(1-u^2)(1-v^2)$$
which simplifies to $u^2-2uv+v^2\ge0$, which holds since $u^2-2uv+v^2=(u-v)^2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2261001",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
proving $ \lim_{x \rightarrow 1} \frac{1-\sqrt{x}}{1-x} = \frac{1}{2}$. I'm having trouble with proving $ \lim_{x \rightarrow 1} \frac{1-\sqrt{x}}{1-x} = \frac{1}{2}$.
So far I have:
$\frac{1-\sqrt{x}}{1-x} =\frac{1}{1+\sqrt{x}} $
If $x\in dom (\frac{1}{1+\sqrt{x}}$) and $|x-1| <\delta$ then $ |\frac{1}{1+\sqrt{x}}- \frac{1}{2} |< \epsilon$
So I started writing out $|\frac{1}{1+\sqrt{x}}- \frac{1}{2}|$, and got that it's equal to$ |\frac{1-x}{2(1+2\sqrt{x}+x)}|$. And since we have $|x-1| <\delta$ ,
$ |\frac{1-x}{2(1+2\sqrt{x}+x)}|< |\frac{\delta}{2(1+2\sqrt{x}+x)}|$.
I'm kind of stuck here. Any help on how to continue is much appreciated!
| So you arrived at:
$$\left|\frac{1-x}{2(1+2\sqrt{x}+x)}\right|
=\frac{\left|1-x\right|}{2\left|1+2\sqrt{x}+x\right|}$$
and $\delta$ gives you an upper bound for the numerator.
Hint: note that $x>0$ (domain) and look for an (obvious) lower bound for the denominator.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2263811",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
} |
Solving $x^{45} \equiv 7 \mod 113$ Pretty much as in the title, though a more general answer would also be nice. . I thought you could find in inverse of $45$ in mod $113$, then take the equation to that power. In this situation that gives:
$45^{-1} = 108 \mod 113$
$(x^{45})^{108} \equiv x^{45\times108} \equiv x^1 \equiv 7^{108}$
However this is wrong according to wolfram alpha, so I guess the above is complete nonsense. The correct answer is $83$
| $$\text{ Solve $x^{45} \equiv 7 \pmod{113}$}.$$
For all $\gcd(x,113)=1$: $\quad x^{112} \equiv 1 \pmod{113}$
$45 \times 5 \equiv 225 \equiv 2(112) + 1 \equiv 1 \pmod{112}$
$(x^{45})^{5} \equiv x^{225} \equiv x \pmod{113}$
$7^3 \equiv 343 \equiv 3(113) + 4 \equiv 4 \pmod{113}$
$7^5 \equiv 7^2 \times 7^3 \equiv 49\times 4 \equiv 196 \equiv 113 + 83 \equiv 83 \pmod{113}$
$x \equiv (x^{45})^{5} \equiv 7^5 \equiv 83$
Computing $45^{-1} \pmod {112}$
\begin{array}{r|r|rr|l}
& 112 & 1 & 0 & 112 = 1(112)+0(45)\\
-2 & 45 & 0 & 1 & 45 = 0(112)+1(45)\\
-2 & 22 & 1 & -2 & 22 = 1(112)-2(45)\\
& 1 & -2 & 5 & 1 = -2(112)+5(45)
\end{array}
Since $1 = -2(112)+5(45)$, then $5(45) = 1+2(112) \equiv 1 \pmod{112}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2264185",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How to show this set is a basis? I want to show that a basis for $\mathbb{R}^4$ is: $ \begin{pmatrix} 1 \\ 2 \\ 0 \\ 0 \end{pmatrix},\begin{pmatrix} 2 \\ 2 \\ 1 \\ -2 \end{pmatrix}, \begin{pmatrix} 1 \\ 4 \\ 0 \\ 0 \end{pmatrix}$ and $\begin{pmatrix} 0 \\ 0 \\ 0 \\ 1 \end{pmatrix}$.
I know that I need to show linear independence and that these span $\mathbb{R}^4$. However, this was a past exam question and I am not really sure that row reduction/solving systems (for only 4 marks) is what they want. Is there an easier/alternative way to show that they span the space and are LI?
Thank you :)
| While I suggest to show this using the determinant, here is one easy and quick way to do it without determinants:
As we have 4 vectors, we only need to show that they are linearly independent. Let $a,b,c,d\in \mathbb R$ with
$$a\cdot\begin{pmatrix} 1 \\ 2 \\ 0 \\ 0 \end{pmatrix}+b\cdot\begin{pmatrix} 2 \\ 2 \\ 1 \\ -2 \end{pmatrix} + c\cdot \begin{pmatrix} 1 \\ 4 \\ 0 \\ 0 \end{pmatrix}+d\cdot\begin{pmatrix} 0 \\ 0 \\ 0 \\ 1 \end{pmatrix}=\begin{pmatrix} 0 \\ 0 \\ 0 \\ 0 \end{pmatrix}.$$
It follows immediately that $b=0$ by looking at the third row of each vector, so this leaves us with
$$a\cdot\begin{pmatrix} 1 \\ 2 \\ 0 \\ 0 \end{pmatrix} + c\cdot \begin{pmatrix} 1 \\ 4 \\ 0 \\ 0 \end{pmatrix}+d\cdot\begin{pmatrix} 0 \\ 0 \\ 0 \\ 1 \end{pmatrix}=\begin{pmatrix} 0 \\ 0 \\ 0 \\ 0 \end{pmatrix}.$$
The same argument yields $d=0$ by looking at the fourth row and from
$$a\cdot\begin{pmatrix} 1 \\ 2 \\ 0 \\ 0 \end{pmatrix} + c\cdot \begin{pmatrix} 1 \\ 4 \\ 0 \\ 0 \end{pmatrix}=\begin{pmatrix} 0 \\ 0 \\ 0 \\ 0 \end{pmatrix}$$
we can easily conclude $a=c=0$ (otherwise we would get a contradiction with $a=-c$ and $a=-2c$).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2265484",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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If $\cos A+\cos B=p$ and $\sin A+\sin B=q$, then find $\cos\left( \frac {A+B}{2}\right)$ in terms of $p$ and $q$ If $\cos A+\cos B=p$ and $\sin A+\sin B=q$ then find $\cos \left( \dfrac {A+B}{2}\right)$ in terms of $p$ and $q$.
My Attempt:
$$\cos A+\cos B=p$$
$$2\cos \left( \dfrac {A+B}{2}\right)\cos \left( \dfrac {A-B}{2}\right)=p$$
And,
$$\sin A+ \sin B=q$$
$$2\sin \left( \dfrac {A+B}{2} \right)\cos \left( \dfrac {A-B}{2} \right)=q$$
Now,
$$\tan \left( \dfrac {A+B}{2}\right)=\dfrac {q}{p}$$.
How do I proceed further?
| Let, $\dfrac{A+B}{2}=x\implies\tan x=\dfrac{q}{p}=\dfrac{\text{height}}{\text{base}}$.
Now suppose for a right angle triangle the height is $aq$ units and base is $ap$ units $(a\neq0)$. So the length of hypotenuse is $=a\sqrt{q^2+p^2}$ units.
$\cos x=\dfrac{\text{base}}{\text{hypotenuse}}=\dfrac{ap}{a\sqrt{q^2+p^2}}=\dfrac{p}{\sqrt{q^2+p^2}}$.
$\implies\cos\left(\dfrac{A+B}{2}\right)=\dfrac{p}{\sqrt{q^2+p^2}}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2265706",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Parameterize Intersection of Surfaces I need to parameterize the intersection of $$4x^2 + y^2 + z^2 = 9\tag{1}$$ and $$z=x^2+y^2\tag{2}$$.
First, I'll solve (2) for $y^2$ and substitute the result into (1):
$$3x^2+z+z^2 = 9 \tag{3}$$
Next, I'll make the substitution $u=\sqrt{3}x$, such that we can complete the square in (3) by adding $1/4$ to each side and arrive at
$$\frac{u^2}{r^2} + \frac{(z+\frac{1}{2})^2}{r^2} = 1$$
where $r^2 = 9 + \frac{1}{4} = \frac{37}{4}$
Now I'll write a parameterization:
$$u = r\cos \phi \implies x(\phi) = \frac{1}{\sqrt{3}}r\cos\phi$$
$$z(\phi) = r\sin \phi -\frac{1}{2}$$
$$y(\phi) = \pm \sqrt{z-x^2} = \pm\left(\sqrt{r\sin \phi - \frac{1}{2} - \left(\frac{1}{\sqrt{3}}r\cos\phi\right)^2}\right)$$
such that we have two branches:
$$\mathbf{r}(\phi)_1 = \big<x(\phi), y(\phi), z(\phi)\big>$$
$$\mathbf{r}(\phi)_2 = \big<x(\phi), -y(\phi), z(\phi)\big>$$
Is this correct?
| I think that taking cylindrical coordinates is rather natural.
let us set $x=r \cos(t),y=r \sin(t)$, giving
$$\left\{\begin{array} .4 r^2 \cos^2 \theta +r^2 \sin^2 \theta+ z^2 & = & 9\\z & =& r^2 \end{array}\right. \ \iff \ \left\{\begin{array} . r^2 (1+ 3\cos^2 \theta) + r^4 & = & 9\\z & =& r^2 \end{array}\right.$$
Setting $R=r^2$, one has a quadratic equation:
$$R^2+R(1+ 3\cos^2 \theta)-9=0$$
whose solution is $R=\tfrac12(-1-3\cos^2 \theta + \sqrt{80 - 6 \cos^2\theta - 9 \cos^4 \theta})$
(we have not taken the other root, negative, which is impossible because $R=r^2 \geq 0$).
Thus the parametric representation of the intersection curve (of an ellipsoid and a paraboloid) in cylindrical coordinates is :
$$\left\{\begin{array}{rcl}r&=&\pm \sqrt{\tfrac12(-1-3\cos^2 \theta + \sqrt{80 - 6 \cos^2\theta - 9 \cos^4 \theta}})\\z & =& r^2 \end{array}\right.$$
The advantage of this representation is that it gives you a polar representation of the projection of the curve onto the horizontal plane.
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove by induction that $2^n + 4^n \leq 5^n$ I'm trying to prove by induction that $2^n + 4^n \leq 5^n$. Through some value plugging I've established that the induction must start from $n = 2$ because $2^2 + 4^2 \leq 5^2 \equiv 20 \leq 25$; for $n = 1$ it doesn't hold since $2 + 4 \geq 5$.
Now I assume that $2^k + 4^k \leq 5^k$ is true and I want to prove that implies $k+1$. Using the inductive hypothesis I multiply both sides by $4$ to get this:
$$4 \cdot 2^{k} + 4\cdot 4^{k} \leq 4 \cdot 5^{k}$$
$$2^{k+2} + 4^{k+1} \leq 4 \cdot 5^{k}$$
I will use again the induction hypothesis, this time I'll multiply both side by $5$ to get:
$$5 \cdot (2^{k} + 4^{k}) \leq 5^{k+1}$$
I can say that $2^{k+2} + 4^{k+1} \leq 5 \cdot (2^{k} + 4^{k})$ and $4 \cdot 5^{k} \leq 5^{k+1}$ so I concatenate them:
$$2^{k+2} + 4^{k+1} \leq 5 \cdot (2^{k} + 4^{k}) \leq 4 \cdot 5^{k} \leq 5^{k+1}$$
However this doesn't feel right. I'm assuming that $5 \cdot (2^{k} + 4^{k}) \leq 4 \cdot 5^{k}$ which there's no way I can be sure about. At this point I'm stuck since the whole reasoning seems wrong.
| You almost proved it in the beginning, then went somewhat off-course. You showed at first that $2^{k+2} + 4^{k+1} ≤ 4*5^k$ inductively. But now $4 < 5$, so $4*5^k < 5^{k+1}$ and hence $2^{k+2} + 4^{k+1} ≤ 5^{k+1}$. In fact, this shows that the inequality is strict for all $n ≥ 2$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find $b$ in $2\sin(\frac{x}{2}+b)$ given the graph Following is the graph of the function. $2\sin \left(\dfrac{x}{2}+b\right)$
I tried solving the equation using the point $\left (\dfrac{\pi}{2},2\right)$:
$$2 = 2\sin\left(\dfrac{x}{2}+b\right) \Leftrightarrow \\
1 = \sin\left(\dfrac{x}{2}+b\right)\Leftrightarrow \\
\frac{x}{2}+b = k\pi + (-1)^k\arcsin(1) \Leftrightarrow \\
b = 2k\pi$$
But my book $b = \dfrac{\pi}{4}$. How do I solve this?
| Note that $y=2 \sin \left( \frac{x}{2} \right)$ passes through the origin, and increases to its first peak at $x=\pi$.
The function $2 \sin \left( \frac{x}{2} + b \right)$ is a horizontal shift of $2 \sin \left( \frac{x}{2} \right)$. Since the picture shows a peak at $\frac{\pi}{2}$, we need to shift $y=2 \sin \left( \frac{x}{2} \right)$ to the left by $\frac{\pi}{2}$ to achieve this.
The function $2 \sin \left( \frac{x}{2} \right)$ shifted $\frac{\pi}{2}$ to the left is
$$
2 \sin \left( \frac{x+\frac{\pi}{2}}{2} \right) = 2 \sin \left( \frac{x}{2}+\frac{\pi}{4} \right)
$$
and so $b=\frac{\pi}{4}$.
| {
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"timestamp": "2023-03-29T00:00:00",
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"answer_id": 1
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$\frac{1 - e^{iz}}{z^2} = \frac{-iz}{z^2} + E(z)$ where $E(z)$ is bounded as $z \rightarrow 0?$ Was reading some notes and it states that $f(z) = \frac{1 - e^{iz}}{z^2}$ can be written as $f(x) = \frac{-iz}{z^2} + E(z)$ where $E(z)$ is bounded as $z \rightarrow 0.$ I don't exactly see why. Help is appreciated.
| Taylors series of $e^{iz}$ about $z=0$ is $$1+iz+\frac{(iz)^2}{2!}+\frac{(iz)^3}{3!}+\cdots$$ so $$\frac{1-e^{iz}}{z^2}=\frac{1}{z^2}\left(1-1-iz-\frac{(iz)^2}{2!}-\frac{(iz)^3}{3!}-\cdots\right)=\frac{-iz}{z^2}+E(z)$$ with $E(z)=-\frac{1}{z^2}\left(\frac{(iz)^2}{2!}+\frac{(iz)^3}{3!}+\cdots\right)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2268114",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
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System of Linear Congruences Find all $x$ such that
\begin{align}
x&\equiv 1 \pmod {12}\\
x&\equiv 4 \pmod {21}\\
x&\equiv 18 \pmod {35}
\end{align}
Im not quite sure if this system of linear congruence is solvable. Since
$\gcd(12,21) =3$, $\gcd (12,35)=1$ and $\gcd(21,35) = 7$, and the CRT states that "If(m1, m2) = 1, then the system has its complete solution a single resident class (mod m1.....mr).
| When the moduli are not coprime, you can proceed like this: $x\equiv 1 \pmod{12}$ so $x=1+12y$ for some $y$.
Then $x\equiv 1+12y \equiv 4 \pmod{21}$, which has solution $y\equiv 2 \pmod{7}$, so $y=2+7z$ for some $z$ and we have $x = 1+12(2+7z)=25+84z.$
Then $x \equiv 25+84z \equiv 18 \pmod{35}$ which has solution $z=2 \pmod{5}$.
So $x = 25+84(2+5w) = 193 + 420 w$, which is to say $x\equiv 193 \pmod{420}.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2272593",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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} |
How would you prove that $2^{50} < 3^{33}$ without directly calculating the values Could you generalise the question and get something along the lines of $n^{50} < (n+1)^{33}$ ?
| \begin{align}
3^2=2^3+1\quad&\Leftrightarrow\quad \underbrace{(3^2)^{17}}_{3^{34}}=(2^3+1)^{17}=[\text{binomial}]=2^{51}+17\cdot 2^{48}+\text{positive}\quad\Rightarrow\quad \\
&\Rightarrow\quad 3\cdot 3^{33}>2\cdot 2^{50}+\frac{17}{4}\cdot 2^{50}>3\cdot 2^{50}.
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2276227",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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} |
Solve $\frac{1}{x-1}\geq\frac{a}{x+1}$, provided $a>1$ Moving everything to the left,
$$\frac{1}{x-1}-\frac{a}{x+1}\geq0\Longleftrightarrow\frac{ax-x-a-1}{x^2-1}\leq 0.$$
Finding roots to nominator and denominator:
$$\begin{array}{lcl}
ax-x-a-1 & = & 0 \Leftrightarrow \frac{a+1}{a-1}\\
x^2-1 & = & 0 \Leftrightarrow x_1 =1 \ \text{och} \ x_2=-1.\\
\end{array}$$
And that's it. How to proceed?
| you have to solve $$\frac{x+1-a(x-1)}{x^2-1}\geq 0$$
the first case is given by
$$x+1-ax+a\geq 0$$ and $$x^2>1$$
this gives
$$x(1-a)\geq -(a+1)$$ and $$|x|>1$$
since $$a>1$$ we get
$$x\le \frac{a+1}{1-a}$$ and if $$x\geq 0$$ we get
the solution set
$$1<x\le \frac{a+1}{a-1}$$
can you proceed?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2276309",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Recursive sequence modulo 4 For all odd positive integers $k$, I define a recursive sequence by
$$
d_k=2+ {k\choose 1}d_{k-2} + {k\choose 2}d_{k-4} + \dots +{k\choose \frac{k-1}{2}}d_1\\
d_1=2
$$
I want to study this sequence modulo $4$. By induction, it is easy to see that $d_k$ is either $0$ or $2$. Computing this sequence I get
$$
2,0,2,2,0,2,2,0,2,2,0,2,2,0,2\dots (\mod 4)
$$
which made me think that
$$
d_k\equiv 0 (\mod 4)\text{ if and only if } k\equiv 0 (\mod 3)
$$
Do you have an idea how to prove that? I tried to prove but I don't find any nice behavior on the binomial coefficients that helps me.
| As in @A.P.'s answer, define
$$ \begin{eqnarray}
c_0 & = & 1 \\
c_n & = & 1 + \sum_{k = 1}^n {2n + 1 \choose k}c_{n - k}
\end{eqnarray} $$
so that $ d_{2n + 1} = 2c_n $. Let's prove by induction that $ c_n $ is even iff $ n \equiv 1 \mod 3 $.
$ \bullet $ Initialisation: $ c_0 = 1 $ is odd.
$ \bullet $ Induction: We have
$$ \begin{eqnarray}
c_n
& = & 1 + \sum_{k = 1}^n {2n + 1 \choose k}c_{n - k} \\
& \equiv & 1 + \sum_{k \in [\![1, n]\!], k \not\equiv n - 1\mod 3} {2n + 1 \choose k} \\
& = & 1 + \frac{\sum_{k = 1}^{2n} {2n + 1 \choose k}}2 - \sum_{k \in [\![1, n]\!], k \equiv n - 1\mod 3} {2n + 1 \choose k} \\
& = & 2^{2n} - \sum_{k \in [\![1, n]\!], k \equiv n - 1\mod 3} {2n + 1 \choose k} \\
& \equiv & \sum_{k \in [\![1, n]\!], k \equiv n - 1\mod 3} {2n + 1 \choose k} \mod 2 \\
& \equiv & \sum_{k \in [\![0, n]\!], k \equiv n - 1\mod 3} {2n + 1 \choose k} + \begin{cases}1 \text{ if $ 3 \mid n - 1 $} \\0 \text{ otherwise}\end{cases} \mod 2
\end{eqnarray} $$
Notice furthermore that $ k \equiv n - 1 \mod 3 \implies 2n + 1 - k \equiv n - 1 \mod 3 $. Hence
$$ \begin{eqnarray}
c_n \equiv \begin{cases}0 \text{ if $ 3 \mid n - 1 $} \\1 \text{ otherwise}\end{cases} \mod 2
& \iff & \sum_{k \in [\![0, n]\!], k \equiv n - 1\mod 3} {2n + 1 \choose k} \equiv 1 \mod 2 \\
& \iff & \sum_{k \in [\![0, 2n + 1]\!], k \equiv n - 1\mod 3} {2n + 1 \choose k} \equiv 2 \mod 4
\end{eqnarray} $$
Now denote
$$ r_{a, m} = \sum_{k \in [\![0, m]\!], k \equiv a\mod 3} {m \choose k} $$
$ r $ behaves like a Pascal triangle rolled around. That is, $ r_{a + 1, m + 1} = r_{a, m} + r_{a + 1, m} $ and $ r_{a + 3, m} = r_{a, m} $. We can thus easily calculate $ r_{0, m}, r_{1, m}, r_{2, m} $ modulo $ 4 $
$$ \begin{eqnarray}
m\quad & r_0 & r_1 & r_2 \\
0\quad & 1 & 0 & 0 \\
1\quad & 1 & 1 & 0 \\
2\quad & 1 & 2 & 1 \\
3\quad & 2 & 3 & 3 \\
4\quad & 1 & 1 & 2
\end{eqnarray} $$
We see that it enters a cycle of period $ 6 $ starting from $ m = 2 $ and that $ r_{n - 1, 2n + 1} $ is always $ 2 $, as wanted.
Note: The pattern highly suggests that proving $ c_n \equiv c_{n - 1} + c_{n - 2} \mod 2 $ is an option, but I couldn't find a way to make it work.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2276470",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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How do you derive the quadratic formula using calculus? The quadratic formula: $$f(x)=ax^2+bx+c=0$$
$$x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}$$
I remember a tutor once showing me a method for deriving the quadratic formula using calculus somehow. This was around 20 years ago and I can't even remember the tutor's name. I'd really like to learn this method. Just to clarify, I do know how to derive it using the "Completing the square" method.
I was linked to the solution here: https://www.google.com/amp/s/threesixty360.wordpress.com/2008/10/19/using-calculus-to-generate-the-quadratic-formula/amp/
But I am stuck at one step.
Start with: $$f(x)=ax^2+bx+c$$
We want: $$f(x)=0$$
The first derivative gives: $$f'(x)=2ax+b$$
Which leads to this: $$f(x)=c+\int_0^x (2at+b)dt$$
I can't see why the $t's$ were introduced here.
If anyone has any other methods I'd really like to see them also.
| Let
$$
y=ax^2+bx+c \tag{1}
$$
and $\tfrac{dy}{dx}=2ax+b$. Now let $g=\tfrac{dy}{dx}$ such that
$$
\begin{align*}
g &=2ax+b \tag{2} \\
x &= \dfrac{g-b}{2a} \tag{3}.
\end{align*}
$$
Substitute $(3)$ into $(1)$ and get
$$
\begin{align*}
y &=a {\left( \dfrac{g-b}{2a} \right)}^2+b {\left( \dfrac{g-b}{2a} \right)}+c \\
&= a \dfrac{g^2-gb+b^2}{4a^2} + \dfrac{bg-b^2}{2a}+c \\
&= \dfrac{g^2-gb+b^2}{4a} + \dfrac{2bg-2b^2}{4a}+ \frac{4ac}{4a} \\
y &= \dfrac{g^2-b^2 +4ac}{4a} \\
\end{align*}
$$
Set $y=0$ and solve for $g$.
$$
\begin{align*}
y &= \dfrac{g^2-b^2 +4ac}{4a} \\
0 &= \dfrac{g^2-b^2 +4ac}{4a} \\
0 &= g^2-b^2 +4ac \\
g &= \pm \sqrt{b^2-4ac} \tag{4} \\
\end{align*}
$$
Now substitute $(2)$ in for $g$ and solve fo $x$.
$$
\begin{align*}
g &= \pm \sqrt{b^2-4ac} \tag{4} \\
2ax+b &= \pm \sqrt{b^2-4ac} \\
x &= \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}.
\end{align*}
$$
Another way is: let $f(x)=ax^2+bx+c$ such that
$$
f(x)=\int_{0}^{x} \left( 2at+b \right) dt + c
$$
The $t$'s are introduced here because we cannot have $x$ as the variable of integration and also as a bound. Think of them as dummy variables; all the $t$'s will become $x$'s when you evaluate the integral.
Introduce the "u" substitution where $u=2at+b \implies dt= \tfrac{1}{2a}du$.
$$
f(x)=\int_{t=0}^{t=x} \dfrac{u}{2a} du + c
$$
Since we are now integrating with respect to $u$, so your new lower and upper bounds respectively become $u=b$ and $u=2ax+b$. Thus, you get
$$
\begin{align*}
f(x) &=\int_{b}^{2ax+b} \dfrac{u}{2a} du + c \\
& = \dfrac{1}{2a} \left( {\dfrac{ {\left( {2ax+b} \right)}^2}{2} - \dfrac{b^2}{2}} \right) + c\\
& = \dfrac{{\left( 2ax+b \right)}^2 }{4a} - \dfrac{b^2}{4a} + \dfrac{4ac}{4a}. \\
\end{align*}
$$
Set $f(x)=0$ and solve for $x$.
$$
\begin{align*}
0 & = \dfrac{{\left( 2ax+b \right)}^2 }{4a} - \dfrac{b^2}{4a} + \dfrac{4ac}{4a}. \\
0 & = \left( 2ax+b \right)^2 - b^2 + 4ac \\
\left( 2ax+b \right)^2 & = b^2 - 4ac \\
x &= \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a} \\
\end{align*}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2277358",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 4,
"answer_id": 2
} |
Solve for xy in terms of a and b $$x^2 + xy + y^2 = a$$
$$x + y = b$$
$xy = ?$
I tried this and did this:
$xy = a - x^2 - y^2$
$xy = a - (x^2 + y^2)$
$xy = a - (x + y)(x - y)$
$xy = a - b(x - y)$
At this point I can't think of anything to do to represent the $x - y$ part in terms of $a$ and $b$.
Any help would be appreciated.
| $b^2=(x+y)^2=x^2+2xy+y^2=a+xy$. Therefore $xy=b^2-a$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2278301",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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Three sides of a triangle- find general law (polya) The tree sides of a triangle are of lengths $l$, $m$, and $n$, respectively. The numbers $l, m,$ and $n$ are positive integers.
$l \le m \le n$
Find the number of different triangles of the described kind for a given 'n'. ( Take $n = 1, 2, 3, 4, 5, ..., n)$
Find a general law governing the dependence of the number of triangles on $n$.
from Mathematics and Plausible Reasoning .
what i find is :
if i write the three sequence
1- n . (n)
2-number of all possibilities of l and m for each n .(P)
3- number of possible triangle that we can construct according to triangle iniquity. (T)
is
$p = (1,2,6,10,15,21,28,26,45,55,...)$
$T= (1,2,4,6,9,12,16,20,25,30,36,42,...)$
$n= (1,2,3,4,5,6,7,8,9,10,11)$
some observation about these sequence
1- for $P$ :
* it's increasing by n
*$P_n$ - $T_n-1$ = $T_n$
such that $P_n$ = (number of possible values for l and m for such n )
$T_n$ = number of possible triangles for n
2- For $T$
$1,1+1,2+2,2+2+2,2+2+2+3,2+2+2+3+3,2+2+2+3+3+4,2+2+2+3+3+4+4,2+2+2+3+3+4+4+5,2+2+2+3+3+4+4+5+5,...$
How can i derive general law ??
| By the triangle inequality, for $l,m,n$ to make a triangle, $l+m>n$.
Fix $n$ and $m$, then we know that $n+1-m\leq l\leq m$. If $n+1-m\leq m$, then $\frac{n+1}{2}\leq m$ and these restrictions are not empty.
In such a case, there are $m-(n+1-m)+1=2m-n$ possible values for $l$. Now, we have two cases:
*
*If $n$ is odd, then $\frac{n+1}{2}$ is a natural number, so we need:
\begin{align*}
\sum_{m=\frac{n+1}{2}}^n(2m-n)&=2\left(\frac{n(n+1)}{2}\right)-2\left(\frac{\frac{n-1}{2}\cdot\frac{n+1}{2}}{2}\right)-n\left(n-\frac{n-1}{2}\right)\\
&=n(n+1)-\frac{1}{4}(n^2-1)-\frac{1}{2}n(n+1)\\
&=(n+1)\left(\frac{1}{2}n-\frac{1}{4}(n-1)\right)\\
&=\frac{1}{4}(n+1)^2
\end{align*}
*If $n$ is odd, then $\frac{n+1}{2}$ is not a natural number, so $\frac{n+1}{2}\leq m$ reduces to $\frac{n}{2}+1\leq m$. I'll leave the rest of the derivation as an exercise at this point.
| {
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"timestamp": "2023-03-29T00:00:00",
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