Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Showing that $x^x/(2x)!=0$ as $x$ approaches infinity How would I show
$$\lim_{x\to\infty} \frac{x^x}{(2x)!}=0$$
I know $x^x$ grows faster than $(2x)!$
So then would I do
$$\frac{x^x}{2x(2x-1)(2x-2)(2x-3)\cdots(2x-(2x-1))}$$
But how do I proceed.
| \begin{align}
\frac{x^x}{(2x)!} & = \frac{\overbrace{x\cdots x}^{x\text{ factors}}}{\underbrace{1\cdot2\cdot3\cdots x}_{x\text{ factors}} \cdot \underbrace{(x+1) \cdot (x+2) \cdots (2x)}_{x\text{ factors}}} \\[10pt]
& = \frac 1 {x!} \cdot \underbrace{\frac x {x+1} \cdot \frac x {x+2} \cdot \frac x {x+3} \cdots \frac x {x+x}}_\text{This is $<1$.} \\[10pt]
& < \frac 1 {x!} \to 0 \text{ as } x\to\infty.
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1445204",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 2
} |
Find the roots of the equation $x^2-[x]=4$ where $[x]$ is the greatest integer less than or equal to $x$. Show that there is no rational number $x$ satisfying the equation $x^2-[x]=4$. Find the roots of the equation $x^2-[x]=4$ where $[x]$ is the greatest integer less than or equal to $x$.
I tried to solve this equation but stuck.
For the first part i supposed $x=\frac{p}{q}$ be the solution of the equation.
$$\frac{p^2}{q^2}-4=\left[\frac{p}{q}\right]$$
Now LHS is rational and RHS is an integer. Therefore our assumption is wrong. There is contradiction. So its roots are not rational.
$x^2-4=[x]\implies (x-2)(x+2)=[x]$
How should i move ahead? Please guide me. Its roots are $-\sqrt2,\sqrt6$.
Is my method of solving the first part correct?
| Using $\displaystyle \bullet \lfloor x \rfloor \leq x$
So here Given $x^2-4 = \lfloor x \rfloor $
So we get $x^2-4\leq x\Rightarrow x^2-x-4\leq 0$
So we get $\displaystyle \frac{1-\sqrt{17}}{2}\leq x\leq \frac{1+\sqrt{17}}{2}$
So we get Approx $-1.6\leq x \leq 2.55$
So we get $\lfloor x \rfloor =-2,-1,0,1,2$
Now we will form Different cases.
$\; \bullet \; $ If $\lfloor x \rfloor = -2\Rightarrow -1.66\leq x<-1\;,$ Put into $x^2-4=\lfloor x \rfloor\;,$ we get $x^2-4=-2$
So we get $x^2=\left(\sqrt{2}\right)^2\Rightarrow x=\pm \sqrt{2}$
So we get $x=-\sqrt{2}\;,$ bcz above $-2\leq x<-1$.
$\; \bullet \; $ If $\lfloor x \rfloor = -1\Rightarrow -1\leq x<0\;,$ Put into $x^2-4=\lfloor x \rfloor\;,$ we get $x^2-4=-1$
So we get $x^2=\left(\sqrt{3}\right)^2\Rightarrow x=\pm \sqrt{3}$
So we get no real value of $x$ bcz above $-1\leq x<0$.
$\; \bullet \; $ If $\lfloor x \rfloor = 0\Rightarrow 0\leq x<1\;,$ Put into $x^2-4=\lfloor x \rfloor\;,$ we get $x^2-4=0$
So we get $x^2=\left(2\right)^2\Rightarrow x=\pm 2$
So we get no real value of $x$ bcz above $0 \leq x<1$.
$\; \bullet \; $ If $\lfloor x \rfloor = 1\Rightarrow 1\leq x<2\;,$ Put into $x^2-4=\lfloor x \rfloor\;,$ we get $x^2-4=1$
So we get $x^2=\left(\sqrt{5}\right)^2\Rightarrow x=\pm \sqrt{5}$
So we get no real value of $x$ bcz above $1\leq x<2$.
$\; \bullet \; $ If $\lfloor x \rfloor = 2\Rightarrow 2\leq x<2.55\;,$ Put into $x^2-4=\lfloor x \rfloor\;,$ we get $x^2-4=2$
So we get $x^2=\left(\sqrt{6}\right)^2\Rightarrow x=\pm \sqrt{6}$
So we get $x = \sqrt{6}$ bcz above $2\leq x<2.55$.
So we get $x=-\sqrt{2}$ and $x=\sqrt{6}$ are only two solution.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1446094",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 3,
"answer_id": 2
} |
Modulo polynomials, excluding selected solution combinations Given
$$
(x-a)(x-b)(x-c) \equiv 0\ \ \pmod {p}
$$
$$
(x-d)(x-e)(x-f)\ \equiv 0\ \ \pmod {q}
$$
x: unknown variable. p,q : known primes. a,b,c,d,e,f : known values.
Are there one or more modulo equations that will exclude the combination
$$
(x \equiv a \pmod {p})\ AND\ (x \equiv d \pmod {q})
$$
but still allow all other combinations of x solutions?
regards arthur
Edit
I tried $\left(mod\ pq\right)$ but couldn't get it to work. List all legal combinations and exclude the illegal one $\left(mod\ pq\right)$.
Let $$r_{jk} \equiv j \left(mod\ p\right)\ and\ \ r_{jk} \equiv k \left(mod\ q\right)$$
$$(x-r_{ae})(x-r_{af})(x-r_{bd})(x-r_{be})(x-r_{bf})(x-r_{cd})(x-r_{ce})(x-r_{cf}) \equiv 0 \left(mod\ pq\right)$$
But if $x = r_{ad}$ (illegal) then $x = a +k_1p$ and $x = d + k_2q$ then
$$(x-r_{ae})\dots(x-r_{bd})\dots \equiv (a+k_1p - (a + k_3p))\dots(d+k_2q-(d+k_4q))\dots\left(mod\ pq\right)$$
$$\equiv p(k_1-k_3)\dots q(k_2-k_4)\dots \equiv 0\left(mod\ pq\right)$$
The $x\equiv r_{ad}$ (illegal) produces a $p$ and a $q$ solving the equation $\left(mod\ pq\right)$.
Edit 2
My application will work with the simplified problem:
Let
$$(x-a)(x-b) \equiv 0\ \ \pmod {p}$$
$$(x-a)(x-b) \equiv 0\ \ \pmod {q}$$
Find equations that will allow
$$x \equiv a \ \pmod {p}\ \ and \ \ x \equiv a \ \pmod {q}$$
$$or$$
$$x \equiv b \ \pmod {p}\ \ and\ \ x \equiv b \ \pmod {q}$$
but block
$$x \equiv a \ \pmod {p}\ \ and \ \ x \equiv b \ \pmod {q}$$
$$or$$
$$x \equiv b \ \pmod {p}\ \ and\ \ x \equiv a \ \pmod {q}$$
where $p$ and $q$ are distinct.
| Let
$\left(\frac{a}{p}\right)=\left(\frac{a}{q}\right)=1$ ,
$\left(\frac{b}{p}\right)=\left(\frac{b}{q}\right)=-1$
$$x^{\frac{p-1}{2}} \equiv \pm1\ (mod\ p)\ \ , \ x^{\frac{q-1}{2}} \equiv \pm1\ (mod\ q)$$
$$x^{\frac{p-1}{2}} = \pm1\ + k_1 p\ \ ,\ x^{\frac{q-1}{2}} = \pm1\ + k_2 q$$
$$qx^{\frac{p-1}{2}} = \pm q\ + k_1 pq\ \ , \ px^{\frac{q-1}{2}} = \pm p\ + k_2p q$$
$$qx^{\frac{p-1}{2}} + px^{\frac{q-1}{2}} = \pm q\ \pm p\ + (k_1 + k_2)pq$$
$$qx^{\frac{p-1}{2}} + px^{\frac{q-1}{2}} = \pm q\ \pm p\ (mod\ pq)$$
To select $\left(\frac{x}{p}\right)\left(\frac{x}{q}\right)=1$ i.e. block
$\left(\frac{a}{p}\right)\left(\frac{b}{q}\right)=\left(\frac{b}{p}\right)\left(\frac{a}{q}\right)=-1$
$$qx^{\frac{p-1}{2}} + px^{\frac{q-1}{2}} \equiv \pm (q+p)\ (mod\ pq)$$
$$qx^{\frac{p-1}{2}} - px^{\frac{q-1}{2}} \equiv \pm (q-p)\ (mod\ pq)$$
if $d$ can be found where $d|\phi(pq)$ s.t. $a^d\equiv 1\ (mod\ pq)$ and $b^d\equiv -1\ (mod\ pq)$ then the $\pm$ sign can be removed s.t.
$$qx^{\frac{p-1}{2}} + px^{\frac{q-1}{2}} \equiv x^d (q+p)\ (mod\ pq)$$
I still need to determine if $d$ always exists.
Admittedly extra conditions were added. A more general solution would be welcome.
$$\left(x^{\frac{p-1}{2}}\right)^{odd} \equiv \pm1\ (mod\ p)\ \ , \ \left(x^{\frac{q-1}{2}}\right)^{odd} \equiv \pm1\ (mod\ q)$$
Multiply each side by itself an odd number of times preserves the sign of the $\pm1$. if $p \equiv q \equiv 3\ mod\ 4$ then $\frac{p-1}{2}$ and $\frac{q-1}{2}$ are odd.
$$x^{\frac{p-1}{2}\frac{q-1}{2}} \equiv \pm1\ (mod\ p)\ \ , \ x^{\frac{p-1}{2}\frac{q-1}{2}} \equiv \pm1\ (mod\ q)$$
$$x^{\frac{p-1}{2}\frac{q-1}{2}} \equiv \pm1\ (mod\ pq)$$
So if $p \equiv q \equiv 3\ mod\ 4$ then $d = \frac{\phi(pq)}{4}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1448164",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Solving $\lim_{x\to0}\frac{x}{\sqrt{3+x}-\sqrt{3-x}}$ I'm trying to resolve the $$\lim_{x\to0}\frac{x}{\sqrt{3+x}-\sqrt{3-x}}$$
First answer is $\frac{0}{0}$
Applying formula:
$$\lim_{x\to0}\frac{x}{\sqrt{3+x}-\sqrt{3-x}} = \lim_{x\to0}\frac{x(\sqrt{3+x}+\sqrt{3-x})}{(\sqrt{3+x}-\sqrt{3-x})(\sqrt{3+x}+\sqrt{3-x})}$$
And now:
$$\lim_{x\to0}\frac{x(\sqrt{3+x}+\sqrt{3-x})}{3+x-3+x} = \lim_{x\to0}\frac{x(\sqrt{3+x}+\sqrt{3-x})}{2x} = \lim_{x\to0}\frac{2\sqrt{3}}{x} = \infty$$
What I'm doing wrong? I know that answer is $\sqrt{3}$, but where is my mistake?
| Let $f(x) =\sqrt {3+x} - \sqrt {3-x}.$ Note $f(0)= 0.$ Our expression has the form
$$\frac{x-0}{f(x) - f(0)}.$$
By the definition of a derivative, the above $\to 1/f'(0)$ as $x\to 0.$ This is an easy computation.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1450330",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove that using induction that $\binom22+\dots+\binom n2 = \binom{n+1}2$ so I have this math problem where I have to prove this using induction. $$\begin{pmatrix}2\\2\end{pmatrix}+\begin{pmatrix}3\\2\end{pmatrix}+\begin{pmatrix}4\\2\end{pmatrix}+\cdots+\begin{pmatrix}n\\2\end{pmatrix}=\begin{pmatrix}n+1\\3\end{pmatrix}\text{ for } n\ge2$$
I start by checking the initial case $P(2)=\begin{pmatrix}2\\2\end{pmatrix}=1$ It's good.
I then assume P(k) is true. $P(k)=\begin{pmatrix}2\\2\end{pmatrix}+\begin{pmatrix}3\\2\end{pmatrix}+\begin{pmatrix}4\\2\end{pmatrix}+\cdots+\begin{pmatrix}k\\2\end{pmatrix}=\begin{pmatrix}k+1\\3\end{pmatrix}$
I then try to prove P(k+1) is also true. $P(k+1)=\begin{pmatrix}2\\2\end{pmatrix}+\begin{pmatrix}3\\2\end{pmatrix}+\begin{pmatrix}4\\2\end{pmatrix}+\cdots+\begin{pmatrix}k+1\\2\end{pmatrix}=\begin{pmatrix}k+2\\3\end{pmatrix}$
I know that $\begin{pmatrix}k+1\\3\end{pmatrix}=\begin{pmatrix}k\\2\end{pmatrix}+\begin{pmatrix}k\\3\end{pmatrix}$
Now I'm stuck here... I'm not sure how to finish this proof. Thanks.
| You said, $${k+1\choose3} = {k\choose2}+{k\choose3}$$
$$\implies {k+2\choose3} = {k+1\choose2}+{k+1\choose3}$$
Now use you induction hypothesis, i.e. $${2\choose2}+{3\choose2}+\dots+{k\choose2} = {k+1\choose3}$$
Substituting the value of ${k+1\choose3}$ in the above equation you get the desired result.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1450902",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Inequality involving an exponent I wish to prove the following inequality
$$x^{\frac{3}{x-1}} > 1 + \frac{1}{x} + \frac{1}{x^2} + \frac{1}{x^3}, \quad x > 1.$$
Graphically the above inequality appears to be true since if one plots
$$g(x) = x^{\frac{3}{x-1}} - \left (1 + \frac{1}{x} + \frac{1}{x^2} + \frac{1}{x^3} \right )$$
it appears as though $g(x) > 0$ for all $x > 1$. I have however been unable to prove analytically this is true.
I know
$$1 < 1 + \frac{1}{x} + \frac{1}{x^2} + \frac{1}{x^3} < 4 \quad \mbox{for all} \,\, x> 1$$
and
$$1 < x^{\frac{3}{x - 1}} < \mathrm{e}^3 \quad \mbox{for all} \,\, x > 1,$$
but neither of these bounds seem to help me very much.
Any pointers in the right direction would be greatly appreciated.
| We want to prove that
$$ \forall x\in(0,1),\qquad x^{\frac{3x}{x-1}}>1+x+x^2+x^3\tag{1}$$
but it is enough to prove that:
$$ \forall x\in(0,1),\qquad 3x\log(x)<(x-1)(x+x^2+x^3)=x^4-x\tag{2} $$
or:
$$ \forall x\in(0,1),\qquad 1+3\log(x) < x^3\tag{3} $$
or (by replacing $x$ with $z^{1/3}$):
$$ \forall z\in(0,1),\qquad 1+\log(z) < z \tag{4} $$
that is trivial by the concavity of the $\log$ function.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1452300",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find the Taylor series about $x = 1$ for $f(x) = \dfrac{1}{(x − 2)^2}$ . Find the Taylor series about $x = 1$ for $f(x) = \dfrac{1}{(x − 2)^2}$ . Express your answer in sigma notation, simplified as much as possible.
This is a practice question that I am having trouble with. I know how to get it about $x=0$, but it is the x=1 part that confuses me.
| $f(x)
= \dfrac{1}{(x − 2)^2}
$
Let $y = x-1$,
or $x= y+1$.
We want to find the expansion
about $y = 0$.
$\begin{array}\\
g(y)
&=f(y+1)\\
&= \dfrac{1}{(y-1)^2}\\
&= \dfrac{1}{(1-y)^2}\\
&=\sum_{n=0}^{\infty} (n+1)y^n\\
&=\sum_{n=0}^{\infty} (n+1)(x-1)^n\\
&\text{and we could stop here, but, for the fun of it,...}\\
&=\sum_{n=0}^{\infty} (n+1)\sum_{k=0}^n \binom{n}{k}x^k(-1)^{n-k}\\
&=\sum_{k=0}^{\infty}\sum_{n=k}^{\infty} (n+1) \binom{n}{k}x^k(-1)^{n-k}\\
&=\sum_{k=0}^{\infty}x^k(-1)^k\sum_{n=k}^{\infty} (n+1) \binom{n}{k}(-1)^{n}\\
&=\sum_{k=0}^{\infty}x^k(-1)^k\sum_{n=k}^{\infty} \frac{(n+1)!}{k!(n-k)!}(-1)^{n}\\
&=\sum_{k=0}^{\infty}(k+1)x^k(-1)^k\sum_{n=k}^{\infty} \frac{(n+1)!}{(k+1)!(n-k)!}(-1)^{n}\\
&=\sum_{k=0}^{\infty}(k+1)x^k(-1)^k\sum_{n=k}^{\infty} \binom{n+1}{k+1}(-1)^{n}\\
&=\sum_{k=0}^{\infty}(k+1)x^k(-1)^k\sum_{n=0}^{\infty} \binom{n+k+1}{k+1}(-1)^{n}\\
&\text{and this is clearly divergent, so please disregard the last 7 lines}\\
\end{array}
$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1452460",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Formula for $A^n$ where $n \in \{1, \ 2, \ \cdots \ \}$ for the matrix $A = \begin{bmatrix} 1 && b \\ 0 && 1 \end{bmatrix}$ Formula for $A^n$ where $n \in \{1, \ 2, \ \cdots \ \}$ for the matrix $A = \begin{bmatrix} 1 && b \\ 0 && 1 \end{bmatrix}.$
Please help with the question if you can, it is for my Linear Algebra class and I cannot find anything close to an answer.
EDIT: As advised in comments, I have tried this for the first few $n$'s:
$A=\begin{bmatrix}
1 & b \\
0 & 1
\end{bmatrix}$
$A^2=
\begin{bmatrix}
1 & b \\
0 & 1
\end{bmatrix}
\begin{bmatrix}
1 & b \\
0 & 1
\end{bmatrix}=
\begin{bmatrix}
1 & 2b \\
0 & 1
\end{bmatrix}$
$A^3=A^2\cdot A=
\begin{bmatrix}
1 & 2b \\
0 & 1
\end{bmatrix}
\begin{bmatrix}
1 & b \\
0 & 1
\end{bmatrix}=
\begin{bmatrix}
1 & b+2b \\
0 & 1
\end{bmatrix}=
\begin{bmatrix}
1 & 3b \\
0 & 1
\end{bmatrix}$
| $$A^n=\begin{pmatrix}1&nb\\0&1\end{pmatrix}$$
Suppose this is true until $n$. Then
$$A^{n+1}=A.A^n$$
Computing the right side almost completes the proof.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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how to show cubic function is one to one using assumption Okay I know the function $f(x)= x^3 +2x+3$ is one to one but how do I show that using the assumption that $f(a)=f(b)$ where $a=b$?
So if I do the process I have
$ a^3+2a+3 = b^3+2b+3$ my 3's cancel and i get
$a^3+2a= b^3+2b$
I know I can factor out an $a$ and $b$
$a(a^2+2)=b(b^2+2)$
but what do I do from here to show $a=b$?
| Notice that
$$a^3+2a= b^3+2b\iff a^3-b^3+2(a-b)=0\iff (a-b)(a^2+ab+b^2+2)=0...(1)$$
Since $$a^2+ab+b^2+2=\left(a+\frac{b}{2}\right)^2+\frac{b^2}{4}+2\ge 2$$
it follows that the only real solution of ($1$) is $a=b$. Then $f$ is one-to-one.
Another approach:
Since $f'(x)=3x^2+2>0$ for any $x\in\mathbb{R}$ it follows $f$ is increasing, then monotone, which implies it is one-to-one.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1454183",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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how can we show $\frac{\pi^2}{8} = 1 + \frac1{3^2} +\frac1{5^2} + \frac1{7^2} + …$? Let $f(x) = \frac4\pi \cdot (\sin x + \frac13 \sin (3x) + \frac15 \sin (5x) + \dots)$. If for $x=\frac\pi2$, we have
$$f(x) = \frac{4}{\pi} ( 1 - \frac13 +\frac15 - \frac17 + \dots) = 1$$
then obviously :
$$ 1 - \frac13 +\frac15 - \frac17 + \dots=\frac{\pi}{4}$$
Now how can we prove that:
$$\frac{\pi^2}{8} = 1 + \frac1{3^2} +\frac1{5^2} + \frac1{7^2} + \dots$$
| To prove the sum $\frac{\pi^2}{8}=1+\frac{1}{3^2}+\frac{1}{5^2}+\frac{1}{7^2}+\dotsc$ I don't see how it could be useful to refer to
$$1-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+\dotsc=\frac{\pi}{4}$$
It might, however, be convenient to recall the famous sum of Euler:
$$\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+\dotsc=\frac{\pi^2}{6}$$
and to think of possible variations of this sum in order to obtain your desired sum...
What terms are missing between these two sums?
How is this difference related to one of the original sums?
$\dotsc$
| {
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"url": "https://math.stackexchange.com/questions/1454960",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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To Substitute or Not ... $2\sin 2x = 3\tan x$, in interval $0\le x<360^\circ$ The Question
Solve $2\sin 2x = 3\tan x$, in interval $0\le x<360^\circ$
I thought this straightforward enough, used the identity $\sin^2 x + \cos^2 x \equiv 1$, shown below in the working, which clearly is not right. My question is what have I done wrong here - why is the substitution of $(1-\sin^2x)$ for $\cos^2x$ wrong?
My Solution
$$\Rightarrow 4\sin x \cos x = \frac{3\sin x}{\cos x}$$
$$\Rightarrow 4\sin x \cos^2 x = 3\sin x$$
$$\Rightarrow 4\sin x(1 - \sin^2 x) - 3\sin x = 0$$
$$\Rightarrow \sin x - 4\sin^3x = 0$$
$$\Rightarrow \sin x(1 - 4\sin^2 x) = 0$$
$$\Rightarrow \sin x = 0, \sin^2 x = \frac{1}{4}$$
The alarm bells started ringing at this point as $2\sin 90^\circ \neq 3\tan 45^\circ$
Correct Approach
So, looking at it, using the identity $\sin^2 x + \cos^2 x \equiv 1$ was the error.
So:
$$\Rightarrow 4\sin x \cos x = \frac{3\sin x}{\cos x}$$
$$\Rightarrow 4\sin x \cos^2 x = 3\sin x$$
$$\Rightarrow \sin x (4\cos^2 x - 3) = 0$$
$$\Rightarrow \sin x = 0, \cos^2 x = \frac{3}{4}$$
I'd be very grateful if anyone could explain why making the substitution I made was wrong. I thought they could be used as-and-when. Have I misunderstood something significant?
Or is is that the substitution can't be used because it gives the wrong answer? But this begs the question of how can you know when it's OK to use it.
Thanks in advance for your insights.
| Your original solution is correct as well. It's just that $x=45^\circ$ doesn't satisfy $\sin^2 x = \frac14$. If $\sin^2 x = \frac14$, then $\sin x = \pm\frac12$, while $\sin 45^\circ = \frac{1}{\sqrt{2}}$. The solutions for $\sin^2 x = \frac14$ with $0\leq x<360^\circ$ are $30^\circ, 150^\circ, 210^\circ$ and $330^\circ$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1456439",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Doing $\lim_{x\to0}\frac{\sin x - \tan x}{x^2\cdot\sin 2x}$ without L'Hopital Without L'Hopital,
$$\lim_{x\to0}\frac{\sin x - \tan x}{x^2\cdot\sin 2x}$$
This is
$$\frac{\sin x -\frac{\sin x}{\cos x}}{x^2\cdot\sin 2x} = \frac{\frac{\sin x \cdot \cos x - \sin x}{\cos x}}{x^2\cdot\sin 2x} = \frac{\sin x\cdot \cos x - \sin x}{x^2\cdot\sin 2x\cdot \cos x}$$
Split that:
$$\frac{\sin x\cdot \cos x}{x^2\cdot\sin 2x\cdot \cos x} - \frac{\sin x}{x^2\cdot\sin 2x\cdot \cos x}$$
In the left side, we can cancel the $\cos x$ and also apply $\frac{\sin x}{x} = 1$ once:
$$\frac{1}{x\cdot\sin 2x} - \frac{\sin x}{x^2\cdot\sin 2x\cdot \cos x}$$
That was probably a bad idea, since $x \cdot \sin2x$ will definitely be $0$... But anyway, let's keep going with the right side. There, we can apply the identity $\frac{\sin x}{x} = 1$ again:
$$\frac{1}{x\cdot\sin 2x} - \frac{1}{x\cdot\sin 2x\cdot \cos x}$$
Hey, I could get rid of the $\sin 2x$ on the left side if I multiply and divide by $2x$... the same on the right side:
$$\frac{1}{2x^2} - \frac{1}{2x^2\cdot \cos x}$$
Looking pretty, but sadly that's not going anywhere. What can I do?
| $$\frac{\sin x-\tan x}{x^{2}\sin 2x}=\frac{\sin x}{\sin 2x}\frac{\cos x -1}{x^{2} \cos x}=\frac{\sin x}{\sin 2x}\frac{-2\sin^{2}\frac{x}{2}}{x^{2}\cos x}$$
Now, $$\frac{\sin x}{\sin 2x}\to \frac{1}{2}$$
$$\frac{\sin^{2}\frac{x}{2}}{x^{2}}=1/4\frac{\sin^{2}\frac{x}{2}}{(x/2)^{2}}\to 1/4$$
And $\cos x \to 1$
So the limit is $-1/4$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1459308",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
Find the maximum of $\frac{(a+b+c)^3-27abc}{a^3+b^3+c^3-3abc}$ for nonnegative $a$, $b$, $c$.
Show that the maximum of
$$\frac{(a+b+c)^3-27abc}{a^3+b^3+c^3-3abc}$$
is $4$ for nonnegative $a$, $b$, $c$.
An elegant elementary solution is preferred.
Generally, is there an easy way to show that
$$
\frac{
\left( a_1 + \dots + a_n \right)^n - n^n \, a_1 \dots a_n
}
{
a_1^n + \dots + a_n^n - n \, a_1 \dots a_n
}
\le
\left( n-1 \right)^{n-1},
$$
for nonnegative $a_1, \dots, a_n$? In other words,
$$
r_n(a_1, \dots, a_n)
\equiv
\frac{
\dfrac{a_1^n+\dots+a_n^n}{n} - \,a_1 \dots a_n
}
{
\left(\dfrac{ a_1+\dots+a_n }{n} \right)^n - a_1 \dots a_n
}
\ge
\left( \frac{n}{n-1} \right)^{n-1}.
$$
This would imply that
$$
\lim_{n \rightarrow \infty} r_n(a_1, \dots, a_n)
\ge e.
$$
| Not a very elegant solution, I will admit.
We want to show that
$$ \frac{(a+b+c)^3-27abc}{a^3+b^3+c^3-3abc}\leq 4$$
which is equivalent to (since the denominator is non-negative by AM-GM)
$$(a+b+c)^3-27abc \leq 4(a^3+b^3+c^3) - 12abc$$
or
$$a^3 + b^3 + c^3 + 3a^2(b+c) + 3b^2(c+a)+3c^2(a+b) + 6abc - 27abc \leq 4(a^3+b^3+c^3)-12abc$$
which further simplifies to
$$ a^2(b+c) + b^2(c+a) + c^2(a+b) \leq a^3 + b^3 + c^3 + 3abc$$
which in turn is equivalent to
$$ a(a-b)(a-c) + b(b-c)(b-a) + c(c-a)(c-b) \geq 0$$
which is just Schur's inequality.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1460250",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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} |
Computing $\int (1 - \frac{3}{x^4})\exp(-\frac{x^{2}}{2}) dx$ How does one compute
$$\int \left(1 - \frac{3}{x^4}\right)\exp\left(-\frac{x^{2}}{2}\right) dx?$$
Mathematica gives $(x^{-3} - x^{-1})\exp(-x^2/2)$ which indeed the correct answer, but how does one get there?
Integration by parts gives $(y + y^{-3})e^{-y^2/2} + \int (y^2 + y^{-2})e^{-y^2/2} dy$, but I'm not sure what to do with the integral.
| i tired integration by parts to show to OP but i think it doesn't work
Let's compute this integral :
$$\int \left(1 - \frac{3}{x^4}\right)\exp\left(-\frac{x^{2}}{2}\right) dx?$$
indeed,
$$ u=e^{\dfrac{-x^{2}}{2}},dv = (1 - 3/x^4)\qquad du=-e^{-\dfrac{x^2}{2}}x,\, v=\frac{1}{x^3}+x $$
\begin{align}
\int \left(1 - \frac{3}{x^4}\right)\exp\left(-\frac{x^{2}}{2}\right)\,dx&=(e^{\dfrac{-x^{2}}{2}} )(\frac{1}{x^3}+x)\biggl|-\int (\frac{1}{x^3}+x)( -e^{-\dfrac{x^2}{2}}x)\,dx \\
&=\frac{x^4+1}{e^{\frac{x^2}{2}}x^3}\biggl|-\int (\frac{1}{x^3}+x)( -e^{-\dfrac{x^2}{2}}x)\,dx\\
&=\frac{x^4+1}{e^{\frac{x^2}{2}}x^3}\biggl|-\int (\frac{1}{x^3}+x)( -e^{-\dfrac{x^2}{2}}x)\,dx
\end{align}
let's calculate $$\int (\frac{1}{x^3}+x)( -e^{-\dfrac{x^2}{2}}x)\,dx$$
$$ u=e^{\dfrac{-x^{2}}{2}},dv = \frac{1}{x^3}+x\qquad du=-e^{-\dfrac{x^2}{2}}x,\, v=-\frac{1}{2x^2}+\frac{x^2}{2} $$
\begin{align}
\int (\frac{1}{x^3}+x)( -e^{-\dfrac{x^2}{2}}x)\,dx&=-(e^{\dfrac{-x^{2}}{2}})(-\frac{1}{2x^2}+\frac{x^2}{2})\biggl|-\int (-\frac{1}{2x^2}+\frac{x^2}{2})( -e^{-\dfrac{x^2}{2}}x)\,dx\\
&=
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1462529",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
} |
Evaluate Integral $\int\frac{1+x}{1+x^2}\ dx$ $\int\frac{1+x}{1+x^2}\ dx$
Let $u=1+x^2$
Then $du = 2x\ dx$
Here is my work.
Split integral $\int\frac{1}{1+x^2}\ dx$ + $\int\frac{x}{1+x^2}\ dx$
Integrate first integral term:
$\int\frac{1}{1+x^2}\ dx=tan^{-1}x$
$\int\frac{x}{1+x^2}\cdot\frac{du}{2x} $
I am stuck when it comes to the second integral term (I hope that is the right term). I have $\frac{du}{2x}$ It does get rid of the x. Should I carry the 2 over by placing outside the second integral.
| $\frac{x}{1+x^2}$ is something of the form $\frac{1}{2}\cdot\frac{f'}{f}$, hence:
$$ \int\frac{1+x}{1+x^2}\,dx = \arctan(x)+\frac{1}{2}\log(1+x^2)+C.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1464192",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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What is the expansion? I encounter the following formula in some textbook. However, I can not understand what the expansion in this formula is. Is there anyone giving some tips?
$$
\begin{align}
\alpha &= -\frac{iU}{2 \epsilon} \pm \sqrt{\frac{i\omega}{\epsilon} - \frac{U^{2}}{4 \epsilon^{2}}} \\
&= \frac{iU}{2 \epsilon} \pm \frac{iU}{2 \epsilon} \left [ 1 - \frac{2i\omega \epsilon}{U^{2}} + \frac{2 \omega^{2} \epsilon^{2}}{U^{4}} + O(\epsilon^{3}) \right ]
\end{align}
$$
Thank you!
| Hint:
They factored out $\;-\dfrac{U^2}{4\varepsilon^2}=\biggl(\dfrac{iU}{2\varepsilon}\biggr)^2$ under the radical:
$$\sqrt{\frac{i\omega}{\epsilon} - \frac{U^{2}}{4 \epsilon^{2}}}=\sqrt{\frac{-U^2}{4\varepsilon^2}\biggl(1-\frac{4i\omega\varepsilon}{U^2}\biggr)}=\dfrac{iU}{2\varepsilon}\sqrt{1-\frac{4i\omega\varepsilon\mathstrut}{U^2}},$$
and applied the development of $\sqrt{1-x}$ at order $2$:
$$\sqrt{1-x}=1-\frac12 x-\frac18x^2+O(x^3).$$
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "1",
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Show that if $a \neq b$ and a and b are positive then $\frac{a}{b}+\frac{b}{a}$ is never an integer
Some observations I made is for $\frac{a}{b}+\frac{b}{a}$, is either:
*
*the denominator has to be one,
*the numerator has to be a multiple of the denominator or
*the numerator and denominator have to be the same.
Obviously, with the given conditions case 3 is eliminated since $a \neq b$.
For case 1, if $b=1$, then $a=1$ which is a contradiction to the given condition that says $a \neq b$.
For case 2, I would think of examples. For example if $b=2$ then a multiple is $4$, so $a=4$. Then we have:
$$\frac{a}{b}+\frac{b}{a}=\frac{4}{2}+\frac{2}{4}=\frac{10}{4}$$
which is not an integer, but how would i proceed to show this case for all integers.
| (I assume that $a$ and $b$ are both integers...)
Let $x=\dfrac{a}{b}+\dfrac{b}{a}=\dfrac{a^2+b^2}{ab}$. We can assume that $a$ and $b$ are relatively prime, since substituting $a/g$ and $b/g$ instead of $a$ and $b$ doesn't change the value of $x$.
If $x$ is integer, than $ab$ should divide $a^2+b^2$, so
$$a|(a^2+b^2) \; \Rightarrow\; a|b^2$$
$$b|(a^2+b^2) \; \Rightarrow\; b|a^2$$
Since we assumed that $a$ and $b$ are relatively prime, these formulas hold only when $a=b=1$, which is contradicted to the condition $a\neq b$.
Therefore, $x\not\in \mathbb{Z}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1468189",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 1
} |
Maximum value of a variable from a system of 2 equations From the following $2$ equations, find the maximum value of $d$.
$a + b + c + d = 8$ and $ab + ac + ad + bc + bd + cd = 12$
How to go about with this problem?
Please help.
Thankyou.
| Without loss of generality we can assume
$a = p + q$, $b = p - q$, with $q > 0$.
The first equation is simplified as
(1)
$$
c + d = 8 - 2 \, p.
$$
Similarly, for the second equation, we have
\begin{align}
12 & = c d + (a + b)(c+ d) + a b \\
& = c d + 2\,p (c + d) + p^2 - q^2, \\
& = c d + 2\,p (8 - 2p) + p^2 - q^2 \\
& = c d + 16\,p - 3\,p^2 - q^2,
\end{align}
where we have used (1) on the third line.
Thus,
\begin{align}
c + d &= 8 - 2 \, p, \\
cd &= 12 -16 \, p + 3 \, p^2 + q^2
\end{align}
This means that $c$ and $d$ are the roots of the quadratic equation,
$$
x^2 - (8 - 2 \, p) \, x + 12 - 16 \, p + 3 \, p^2 +q^2 = 0.
$$
In other words,
\begin{align}
c &= 4 - p - \sqrt{4 + 16 \, p - 2 \, p^2 - q^2} \\
d &= 4 - p + \sqrt{4 + 16 \, p - 2 \, p^2 - q^2},
\end{align}
where we have assumed that $d$ gets the plus sign, because we want to maximize $d$.
Now the expression is $d$ obviously maximized with $q = 0$, so let us assume that,
\begin{align}
d &= 4 - p + \sqrt{4 + 16 \, p - 2 \, p^2} \\
&= (4 - p) + \sqrt{36 - 2 \, (4-p)^2}
\end{align}
Finally, let us ask for what value of $p$, $d$ is maximized?
For this, let us replace $4 - p = 3 \, \sqrt{2} \, \sin t$, then we have
\begin{align}
d &= 3 \sqrt{2} \, \left( \sin t + \sqrt{2} \cos t \right) \\
&= 3 \sqrt{2} \, \sqrt{3} \, \left(\sqrt{\frac 1 3} \sin t + \sqrt{\frac 2 3} \cos t \right) \\
&= 3 \sqrt{6} \, \sin\left(t + \tan^{-1}\sqrt2 \right) \\
&\le 3 \sqrt{6}.
\end{align}
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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} |
Evaluation of $ I = \int_{0}^{1}\frac{2x-(1+x^2)^2\cdot \cot^{-1}(x)}{(1+x^2)[1-(1+x^2)\cot^{-1}(x)]}dx$
If $\displaystyle I = \int_{0}^{1}\frac{2x-(1+x^2)^2\cdot \cot^{-1}(x)}{(1+x^2)[1-(1+x^2)\cot^{-1}(x)]}dx\;,$ Then value of $100(I-\ln 2) =$
$\bf{My\; Try::}$ Let $\cot^{-1}(x)=t\;,$ Then $\displaystyle \frac{1}{1+x^2}dx = -dt$ and changing limit, We get
$$\displaystyle I = -\int_{\frac{\pi}{2}}^{\frac{\pi}{4}}\frac{2\cot t-\csc^4 t\cdot t}{1-\csc^2 t\cdot t}dt = \int_{\frac{\pi}{4}}^{\frac{\pi}{2}}\left(\frac{2\cot t-\csc^4 t\cdot t}{1-\csc^2 t\cdot t}\right)dt$$
Now How can I solve after that, Help Required
Thanks
| You can start by rewriting your integrand as
$$
\begin{aligned}
1&+\frac{2x-(1+x^2)}{(1+x^2)\bigl[1-(1+x^2)\,\text{arccot}\,x\bigr]}\\
&=1+\frac{2x\bigl[1-(1+x^2)\,\text{arccot}\,x\bigr]+(1+x^2)\bigl[2x\,\text{arccot}\,x-1\bigr]}{(1+x^2)\bigl[1-(1+x^2)\,\text{arccot}\,x\bigr]}\\
&=1+\frac{2x}{1+x^2}+\frac{1-2x\,\text{arccot}\,x}{1-(1+x^2)\,\text{arccot}\,x}
\end{aligned}
$$
Now, we are really lucky! Noting that $D(1-(1+x^2)\,\text{arccot}\,x)=1-2x\,\text{arccot}\,x$, we find that
$$
\int \frac{2x-(1+x^2)^2 \,\text{arccot}\,x}{(1+x^2)[1-(1+x^2)\,\text{arccot}\,x]}\,dx=x+\log|1+x^2|+\log|1-(1+x^2)\,\text{arccot}\,x|+C.
$$
I leave the rest for you.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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If $\sqrt{1-x^2}+\sqrt{1-y^2}=a(x-y)$, then what is $\frac{dy}{dx}$? The question states:
If $\sqrt{1-x^2}+\sqrt{1-y^2}=a(x-y)$, then what is $\frac{dy}{dx}$?
The options are:
*
*$$\sqrt{\frac{1-x^2}{1-y^2}}$$
*$$\sqrt{\frac{1-y^2}{1-x^2}}$$
*$$\frac{1-x^2}{1-y^2}$$
*$$\frac{1-y^2}{1-x^2}$$
I tried differentiating the entire expression:
$$\begin{align}
\frac{-2x}{2\sqrt{1-x^2}}+\frac{-2y}{2\sqrt{1-y^2}}\frac{dy}{dx} &= a(1-\frac{dy}{dx}) \\
a + \frac{x}{\sqrt{1-x^2}} &= \frac{dy}{dx}(a-\frac{y}{\sqrt{1-y^2}}) \\
\frac{dy}{dx} &= (a+\frac{x}{\sqrt{1-x^2}})(a-\frac{y}{\sqrt{1-y^2}})^{-1}
\end{align}$$
I am not sure how to proceed now. Is there a better approach? Is it advisable to use my approach?
| Your approach is absolutely fine and the result that you have obtained is correct. If you handed me a homework like this I would be satisfied, since it is clear that you have understood the procedure of implicit differentiation.
Still, the problem is somewhat pedantic and asks you to do a final superfluous simplification: note that $a$ is absent from all of the available options, therefore express $a$ as $\dfrac {\sqrt{1-x^2} + \sqrt{1-y^2}} {x-y}$ and replace it in your own result. After a number of elementary algebraic simplifications you will obtain $\dfrac {\sqrt{1-y^2}} {\sqrt{1-x^2}}$, which is option 2.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1472086",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
$\binom{n}{0}-\frac{1}{(2+\sqrt3)^2}\binom{n}{2}+\frac{1}{(2+\sqrt3)^4}\binom{n}{4}-\frac{1}{(2+\sqrt3)^6}\binom{n}{6}+.....=$ If $n=12m$, where $m\in\Bbb{N}$, prove that
$$\binom{n}{0}-\frac{1}{(2+\sqrt3)^2}\binom{n}{2}+\frac{1}{(2+\sqrt3)^4}\binom{n}{4}-\frac{1}{(2+\sqrt3)^6}\binom{n}{6}+.....=(-1)^m\left(\frac{2\sqrt2}{1+\sqrt3}\right)^n$$
I tried to solve it but stuck after few steps:
$\binom{n}{0}-\frac{1}{(2+\sqrt3)^2}\binom{n}{2}+\frac{1}{(2+\sqrt3)^4}\binom{n}{4}-\frac{1}{(2+\sqrt3)^6}\binom{n}{6}+.....$
$=1-\frac{1}{(2+\sqrt3)^2}\frac{n(n-1)}{2}+\frac{1}{(2+\sqrt3)^4}\frac{n(n-1)(n-2)(n-3)(n-4)(n-5)}{6!}+....$
How should I prove this question?
| Hint: First try to prove that the sum can be simplified to
$$\binom{n}{0}-\frac{1}{(2+\sqrt3)^2}\binom{n}{2}+\frac{1}{(2+\sqrt3)^4}\binom{n}{4}-\frac{1}{(2+\sqrt3)^6}\binom{n}{6}+.....=\operatorname{Re}\left(\frac{1}{2+\sqrt3}+i\right)^n.$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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A problem on complex no. Q.If $z$ and $w$ be two complex no, such that $|z|=|w|=1$ and$|z+iw|=|z-iw|=2$.Then find z
I tried to break $z$ and $w$ into $a+ib$ form separately and carry on linear equation by substituting the values.But,It leads to me nowhere.Any hints?
| Let z = a + bi
Let w = c + di
iw = -d + ci
z + iw = (a-d) + i(b+c)
z - iw = (a+d) +i(b-c)
|z| = |w| = 1 => $a^2 + b^2 = c^2 + d^2 = 1$
|z + iw| =2 => $(a - d)^2 + (b + c)^2 = 4$ so $a^2 + b^2 + c^2 + d^2 - 2ad + 2bc =4$ so $2 + 2bc - 2ad =4$ so $bc - ad = 1$
Likewise |z - iw| =2 => $(a + d)^2 + (b - c)^2 = 4$ so $a^2 + b^2 + c^2 + d^2 + 2ad - 2bc =4$ so $2 + 2bc - 2ad =4$ so $ad - bc = 1$
So we have $bc - ad = 1$ and $ad - bc = 1$ which is impossible so, either it's late and I made a stupid error, or ... there's no solution.
| {
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"url": "https://math.stackexchange.com/questions/1474377",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Solve the recurrence relation $a_n = 4a_{n-1} - 3a_{n-2} + 2^n $
Solve the recurrence relation
$$a_n = 4a_{n-1} - 3a_{n-2} + 2^n $$
With initial conditions:
$a_1 = 1$
$a_2 = 11$
I have done similar recurrence relation problems to this, but none that were a non-homogeneous recurrence relation such as this one.
So far I have:
$$r^n = 4^{n-1} - 3^{n-2} $$
Divide both sides by $$\frac{1}{r^{n-2}}$$
Giving me this as my Auxiliary Equation:
$$ r^n - 4r + 3 = 0 $$
I then solved for the $r$ values and got $r = -4$ and $r = 1$
I am stumped from here as to where the non-homogeneous piece comes into play, any help is appreciated.
| Start by finding the general solution to the homogeneous recurrence relationship:
$$a_n = 4a_{n-1} - 3a_{n-2}$$
This has auxiliary equation $\lambda^2=4 \lambda-3$
$\lambda^2-4\lambda+3=0$
$\lambda_1=1, \lambda_2=3$
$$a_n = A(1)^n +B(3)^n$$
You want a particular solution to the non-homogeneous relationship.
Try $a_n=k(2)^n$
Then $a_{n-1}=\frac 12 k(2)^n$, $a_{n-2}=\frac 14 k(2)^n$
So $k(2)^n = 4\left (\frac 12 k(2)^n \right)-3 \left(\frac 14 k(2)^n \right)+(2)^n$
$k = 2k - \frac 34 k +1$
$k=-4$
Add this to the general solution to the homogeneous relationship to find the general solution to the non-homogeneous relationship.
$$a_n = A +B(3)^n-4(2)^n$$
Use the known values $a_1=1$ and $a_2=11$
$1=A+3B-8$
$11=A+9B-16$
gives $10=6B-8$
$6B=18$
$B=3$
$1=A+9-8$
$A=0$
$$a_n =3(3)^n-4 (2)^n$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1475469",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
} |
Proof of trigonometric relationship of angle bisector I have to prove the following equation:
$$CD=\frac{2ab\cos\frac{C}{2}}{a+b}$$ where $CD$ is the angle bisector of $C$ in $\Delta ABC $.
My attempt:
I've started by rearranging the equation in terms of $\cos \frac{C}{2}$ ,then by squaring and subtracting $\sin^2 \frac{C}{2}$ from both sides, yielding thus:
$$\cos^2 \frac{C}{2}-sin^2\frac{C}{2} =\frac{(a+b)^2 \cdot CD^2}{(2ab)^2}- \ \sin^2\frac{C}{2}\tag{1}$$
$$\cos C =\frac{(a+b)^2 \cdot CD^2}{(2ab)^2} - \frac{(s-a)(s-b)}{ab}\tag{2}$$
Now i express $ \cos C $ in terms of $a,b,c$ ,then I simplify by multiplying for $ab$ both sides and I multiply out $(s-a)(s-b)$ , getting now:
$$2(b^2+a^2-c^2)=\frac{(a+b)^2 \cdot CD^2}{4ab}-(ab+bc-b^2+c^2-a^2)\tag{3}$$
By simplifying this and rearranging for $CD^2$ I get:
$$CD^2=\frac{(a^2+b^2-c^2 +ab+ bc)(ab)}{(a+b)^2}\tag{4}$$
Finally by replacing $CD^2=ab-(abc^2)/(a+b)^2\tag{5}$
and doing all the algebraic manipulation i get the final result that
$a+c=2a$ which is clearly wrong...And here i am asking for humble help on math.stackexchange
| Drop perpendiculars $AF$ and $BE$ onto $CD$.
By the definition of cosine,
$$
\left|CF\right|=b\cos(C/2)\quad\text{and}\quad\left|CE\right|=a\cos(C/2)\tag{1}
$$
Since $\triangle AFC\sim\triangle BEC$ and $\triangle AFD\sim\triangle BED$, we get
$$
\frac{\left|CD\right|-\left|CE\right|}{\left|CF\right|-\left|CD\right|}=\frac{\left|DE\right|}{\left|DF\right|}=\frac{\left|BE\right|}{\left|AF\right|}=\frac ab\tag{2}
$$
Solving $(2)$ for $\left|CD\right|$ gives
$$
\left|CD\right|=\frac{a\left|CF\right|+b\left|CE\right|}{a+b}\tag{3}
$$
Applying $(1)$ to $(3)$ yields
$$
\left|CD\right|=\frac{2ab\cos(C/2)}{a+b}\tag{4}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1479557",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
$2=1$ Paradoxes repository I really like to use paradoxes in my math classes, in order to awaken the interest of my students. Concretely, these last weeks I am proposing paradoxes that achieve the conclusion that 2=1. After one week, I explain the solution in the blackboard and I propose a new one. For example, I posted the following one some months ago: What is wrong with the sum of these two series?
I would like to increase my repertoire of fake-proofs. I would be glad to read your proposals and discuss them! My students are 18 years old, so don't be too cruel :) Here is my own contribution:
\begin{equation}
y(x) = \tan x
\end{equation}
\begin{equation}
y^{\prime} = \frac{1}{\cos^{2} x}
\end{equation}
\begin{equation}
y^{\prime \prime} = \frac{2 \sin x}{\cos^{3} x}
\end{equation}
This can be rewritten as:
\begin{equation}
y^{\prime \prime} = \frac{2 \sin x}{\cos^{3} x} = \frac{2 \sin x}{\cos x \cdot \cos^{2} x} = 2 \tan x \cdot \frac{1}{\cos^{2} x} = 2yy^{\prime} = \left( y^{2} \right)^{\prime}
\end{equation}
Integrating both sides of the equation $y^{\prime \prime} = \left( y^{2} \right)^{\prime}$:
\begin{equation}
y^{\prime} = y^{2}
\end{equation}
And therefore
\begin{equation}
\frac{1}{\cos^{2} x} = \tan^{2} x
\end{equation}
Now, evalueting this equation at $x = \pi / 4$
\begin{equation}
\frac{1}{(\sqrt{2}/2)^{2}} = 1^{2}
\end{equation}
\begin{equation}
2 = 1
\end{equation}
| $(1 - x)(1 + x + x^2 + ..... )=$
$(1 + x + x^2 .... )(-x - x^2 - x^3 -......) = 1 + (x -x) + (x^2 - x^2)... = 1$ so
$1 + x + x^2 + .... = \frac{1}{1 - x}$
Let x = -1.
$ 1 - 1 + 1 - 1 + 1 - 1 .... = \frac{1}{1 -(-1)} = \frac 12$
but clearly $1 - 1 + 1 - 1 +... = (1-1) + (1-1) +... = 0$.
So $0 = \frac 12$ (and also 1, and -1).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1480488",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 15,
"answer_id": 8
} |
show this diophantine equation has at least is $3n+3\lfloor \frac{n+1}{3}\rfloor+1$ postive integer solution For any postive integer $n\ge 4$, let $s(n)$ denote the number of ordered pairs $(x,y,z)$ of positive integers for which
$$\color{red}{xy+yz+xz=n(x+y+z)}$$
show that
$$s(n)\ge 3n+3\lfloor \dfrac{n+1}{3}\rfloor+1$$
Thanks
| Here's an approach to construct the necessary quantity of solutions which uses the ideas suggested by individ:
So we fix some $x=k$ where $1\le k \le 10$ and obtain the equation
$$(y-n+k)(z-n+k)=n^2-kn+k^2$$
Now, we are searching for a factorization of the RHS.
There is one obvious factorization which works for all $k$:
$$n^2-kn+k^2=1 \cdot (n^2-kn+k^2)$$
This gives the solution $(x,y,z)=(k,n-k+1,k^2-nk+n^2+n-k)$.
Let's look at these solutions for $k$ running from $1$ to $\lceil \frac{n}{2} \rceil$. Here, it's easy to verify that for $n \ge 4$ all the 3 numbers are distinct and we have $x<y<z$.
Since we can have 6 permutations of each such solution, this gives $6 \cdot \lceil \frac{n}{2} \rceil \ge 6 \frac{n}{2}=3n$ solutions.
Also, we can find the solution $(x,y,z)=(n,n,n)$.
So we are just left to find $3\lfloor \frac{n+1}{3} \rfloor$ more solutions.
Therefore, we note that if we choose $k$ such that $n+k$ is divisible by 3 i.e. $k \equiv -n \mod 3$ then $n^2-kn+k^2 \equiv 3n^2 \equiv 0 \mod 3$ i.e. $n^2-kn+k^2$ is divisible by 3.
Hence, we can here construct solutions with $x=k, y=n+3-k$ and solve the remaining equation to $z=\frac{n(x+y)-xy}{3}=\frac{(n-k)(n+3)+k^2}{3}$.
Since we must choose $k \le n+2$ there are at least $\lfloor \frac{n+2}{3} \rfloor$ possible values of $k$.
This would give $6 \lfloor \frac{n+2}{3} \rfloor$ solutions for $(x,y,z)$ (again using the permutations) but it's easy to see that each solutions is counted twice so we obtain at least $3 \lfloor \frac{n+2}{3} \rfloor$ additional solutions.
So after checking that the three sets of constructed solutions are indeed disjoint (which is a bit tedious but essentially just casework and rough estimations) we see that we have indeed constructed at least
$$3n+3\lfloor \frac{n+2}{3} \rfloor +1$$
distinct solutions $(x,y,z)$ which is clearly sufficient.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1481351",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Given known and and unknown line equation and maximising area of triangle condition.
Q) If from a point $P\equiv(4,4)$ perpendiculars to the straight lines
$-3x+4y+5=0$ and $y=mx+7$ meet at $Q$ and $R$ and area of triangle
$PQR$ is maximum then $m=__ $ ?
First I drew the known line $-3x+4y+5=0$ and the unknown line $y=mx+7$ then I fixed the perpendicular distance from $P(4,4)$ as a fixed value $b$ .
I also know that area of a triangle $\dfrac{1}2 \times base \times height$ so to maximise this area we need $base=height$
I took $base$ as my assumed constant $b$
Then I found out $b=9/5$
Then I substituted in distance formula and I have no idea how to proceed further
$\frac{9}{5}=\frac{-4m+4+c}{\sqrt{m^{2}+1}}$
| First, I found the equation of lines which form the triangle.
*
*$y-4=\frac{-1}{m}(x-4)$ which is the line perpendicular to $y=mx+7$ and passing through $P(4,4)$
*$y-4=\frac{4}{3}(x-4)$ represents the line perpendicular to $3x+4y+5=0$ and passing through $P(4,4)$
Next, I need information about vertices of the triangle. One of the vertex can be obtained by solving $y-4=\frac{-1}{m}(x-4)$ and $y=mx+7$ which is
$$h=\frac{4-3m}{1+m^2}$$
and $$k=\frac{4m-3m^2}{1+m^2}+7$$
Now, area of triangle is given by $\frac{1}{2}ab\sin\theta$. It easy to observe that one of the side is fixed. (The side formed by given point and fixed line). Let that side be $a$. Now, all that remains is to maximize $b\sin\theta$ which represents the altitude of the triangle.
To get altitude of the triangle, I need to find the perpendicular distance from $(h,k)$ to side $a$.
Side $a$ is given by the line $y-4=\frac{4}{3}(x-4)$ which can be written as $$3y-4x+4=0$$
the perpendicular distance formula, $$d(m)=\frac{3k-4h+4}{\sqrt{4^2+3^2}}$$
The graph of $d(m)$ looks like this:
So, when the lines are perpendicular, you get maximum area of triangle.
That is, $m=\frac{4}{3}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1483352",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Find solutions of $2^m\cdot p^2+1=q^5$ $2^m\cdot p^2+1=q^5$
$p$ and $q$ are prime numbers
find $p$ and $q$
I think it will be useful to transfer $1$ to the other side of the equation
$2^m\cdot p^2=(q-1)(q^4+q^3+q^2+q+1)$
and we know $gcd(q-1,q^4+q^3+q^2+q+1)=1$ or $5$
we know if
I)$q-1|2^m \implies p^2|q^4+q^3+q^2+q+1$ or
II)$2^m|q-1 \implies q^4+q^3+q^2+q+1|p^2$
if $gcd(q-1,q^4+q^3+q^2+q+1)=5 \implies$
I)$5|2^m \implies$
Inconsistency
II)$5|p^2 \implies p=5$
$\implies gcd(q-1,q^4+q^3+q^2+q+1)=1$
But I went to this part of the problem and more of this I could not continue
| You can write the equation as:
$$2^m\cdot p^2=q^5-1$$
From here you can note that the prime number $q>3$.
Noe you obtain:
$$2^m\cdot p^2=(q-1)(q^4+q^3+q^2+q+1)$$
therefore you can note that $q^4+q^3+q^2+q+1$ is odd; this yields $$2^m=q-1$$
and $$p^2=q^4+q^3+q^2+q+1$$
but $p^2=q^4+q^3+q^2+q+1$ has solutions only for $p=11$ and $q=3$.
Indeed we can write $$\left(q^2+\frac{q}{2}\right)^2={q^4+q^3}+\frac{q^2}{4}<{q^4+q^3}+q^2+q+1 \\ \frac{q^2}{4}<q^2+q+1 $$ and on the other hand $$ \left(q^2+\frac{q+2}{2}\right)^2=q^4+q^3+2q^2+\frac{q^2+4q+4}{4}>q^4+q^3+q^2+q+1 \\ {q^4+q^3}+\frac{9}{4}q^2{+q+1}>{q^4+q^3}+q^2{+q+1} \\ \frac{9}{4}q^2>q^2.$$ From here, $q$ cannot be even, and for some odd $q$ we must have $$\left(q^2+\frac{q+1}{2}\right)^2={q^4+q^3+q^2}+\frac{q^2+2q+1}{4}={q^4+q^3+q^2}+q+1 \\ q^2+2q+1=4q+4 \\ q^2-2q-3=(q-3)(q+1)=0,$$ from here $q=3$. In particular, $$3^4+3^3+3^2+3+1=11^2$$
therefore the only solutions are $p=11$, $q=3$ and $m=1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1483444",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Does $c\cdot\sum\limits_{n=k}^{\infty}a_{n}= \sum\limits_{n=k}^{\infty}c\cdot a_{n}$ provided that the series converge? I am struggling to find what is wrong about this reasoning when calculating a series that does not start at $n=0$.
For instance, let $S = \sum\limits_{n=2}^{\infty}\left(\frac{1}{2}\right)^n$.
Then $\left(\frac{1}{2}\right)^2S=\left(\frac{1}{2}\right)^2\sum\limits_{n=2}^{\infty}\left(\frac{1}{2}\right)^n$ (I)
And:
$$\left(\frac{1}{2}\right)^2S=\left(\frac{1}{2}\right)^2\sum\limits_{n=2}^{\infty}\left(\frac{1}{2}\right)^n=\sum\limits_{n=2}^{\infty}\left(\left(\frac{1}{2}\right)^n\cdot\left(\frac{1}{2}\right)^2\right)=\sum\limits_{k=0}^{\infty}\left(\frac{1}{2}\right)^k=2$$
But that means (from I):
$$\left(\frac{1}{2}\right)^2S = 2 \Rightarrow \dfrac{\frac{1}{4}}{\frac{1}{4}}S=\dfrac{2}{\frac{1}{4}}\Rightarrow S = 8$$
Which is clearly wrong since:
$$\sum\limits_{n=2}^{\infty}\left(\frac{1}{2}\right)^n=\sum\limits_{n=0}^{\infty}\left(\frac{1}{2}\right)^n-\sum\limits_{n=0}^{1}\left(\frac{1}{2}\right)^n=\sum\limits_{n=0}^{\infty}\left(\frac{1}{2}\right)^n - 1 -\frac{1}{2}=\frac{1}{2}$$
I suspect that there is something behind this step $\left(\frac{1}{2}\right)^2\sum\limits_{n=2}^{\infty}\left(\frac{1}{2}\right)^n=\sum\limits_{n=2}^{\infty}\left(\left(\frac{1}{2}\right)^n\cdot\left(\frac{1}{2}\right)^2\right)$ that I am missing.
So my question is:
Considering that $\sum\limits_{n=k}^{\infty}a_{n}$ is well defined (converges?), does:
$$c\cdot\sum\limits_{n=k}^{\infty}a_{n}= \sum\limits_{n=k}^{\infty}c\cdot a_{n}$$
And if it does, why does the reasoning presented above is wrong?
| If the others' help was not sufficient, let me explain it to you like this:
You tried to calculate $S = \sum\limits_{n=2}^{\infty}\left(\frac{1}{2}\right)^n.$ The first terms of this sum are $$S=\frac14 + \frac18 + \frac1{16}+\dots.$$You tried to do this by premultiplying by $1/4,$ probably because you wanted the first term to be $1$ to use the well-known expression $1 + x + x^2+\dots = 1/(1 - x)$.
However: if you multiply both sides by $1/4$ the first term is instead $1/16,$ which is not $1$ and therefore not what you were trying for at all! In fact we should expect (since you meant to multiply by $4$ and instead multiplied by $1/4$) that you are off by a factor of $16$, and in fact you are: you got $8$ when you expected $1/2.$
We can use the corrected form of your reasoning to discover that $$\sum_{k=n}^\infty x^k = x^n \sum_{k=n}^\infty x^{k-n} = x^n \sum_{m=0}^\infty x^m = \frac{x^n}{1 - x},$$ where the "proper" substitution $m = k - n$ correctly maps our first index $k=n$ to the value $m = 0.$
In a little more detail: your error was to premultiply by $x^{-n}$ and substitute instead $m = k + n,$ both perfectly valid, but then you said that this resulting series starts from $m = 0$ when it does not: it starts from $m = 2n$ when we substitute $k=n$ into that relation for $m$ there.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1483616",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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(-8)^(4/3) is equals with 16 or (-16)*(-1)^(1/3)? 1. $(-8)^{4/3}=\bigl((-8){^4\bigr)^{1/3}}=4096^{1/3}=16$.
2.
$$
\begin{align*}
(-8)^{4/3} &= (-8)^{1+1/3} \\
&= -8\times(-8)^{1/3} \\
&= -8\times (-1)^{1/3}\times 8^{1/3} \\
&= -2\times 8\times (-1)^{1/3} \\
&= -16\times (-1)^{1/3}.
\end{align*}
$$
So, which is the correct?
| Let $a$ be any real number. We can show that the equation
$$
x^3 = a
$$
has one and only one real root. It is called the cube root of $a$ and
is denoted $a^{1/3}$ or $\sqrt[3]{a}$.
From $(-1)^3 = -1$, we conclude that $(-1)^{1/3} = -1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1485666",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
The limit of $(1-\cos^{1/3} x)/(1-\cos x^{1/3})$ as $x\to 0$ I have a problem to find answer for this limit.
$$\lim_{x\to 0}\frac{1-{\cos^{1/3} x}}{1-\cos(x^{1/3})}$$
I did it like this $$\dfrac{\dfrac{\sin x}{3×\sqrt[3]{\cos x²}}}{\sin \sqrt[3]x×\dfrac 1{3×\sqrt[3]{x²}}}$$
| We need two standard limits $$\lim_{x \to 0}\frac{1 - \cos x}{x^{2}} = \frac{1}{2},\,\,\lim_{x \to a}\frac{x^{n} - a^{n}}{x - a} = na^{n - 1}$$ We can proceed as follows
\begin{align}
L &= \lim_{x \to 0}\frac{1 - \cos^{1/3}x}{1 - \cos x^{1/3}}\notag\\
&= \lim_{x \to 0}\frac{1 - \cos^{1/3}x}{1 - \cos x}\cdot\frac{1 - \cos x}{1 - \cos x^{1/3}}\notag\\
&= \lim_{t \to 1}\frac{1 - t^{1/3}}{1 - t}\cdot\lim_{x \to 0}\frac{1 - \cos x}{1 - \cos x^{1/3}}\text{ (putting }t = \cos x)\notag\\
&= \frac{1}{3}\cdot 1^{-2/3}\cdot\lim_{x \to 0}\frac{1 - \cos x}{x^{2}}\cdot\frac{x^{2}}{1 - \cos x^{1/3}}\notag\\
&= \frac{1}{6}\cdot\lim_{x \to 0}x^{4/3}\cdot\frac{x^{2/3}}{1 - \cos x^{1/3}}\notag\\
&= \frac{1}{6}\cdot 0\cdot 2 = 0\notag
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1485968",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find all five solutions of the equation $z^5+z^4+z^3+z^2+z+1 = 0$ $z^5+z^4+z^3+z^2+z+1 = 0$
I can't figure this out can someone offer any suggestions?
Factoring it into $(z+1)(z^4+z^2+1)$ didn't do anything but show -1 is one solution.
I solved for all roots of $z^4 = -4$ but the structure for this example was more simple.
| Note that $(z-1)(1+z+z^2+z^3+z^4+z^5) = z^6-1$. Solve $z^6-1=0$ and discard the $z=1$ solution (which comes from the $z-1$ factor).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1486464",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "17",
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What is the best way of resolving an expression with square roots in denominator? I was resolving the following question from my textbook:
Write each of the following expressions as a single fraction, simplifying your answer where possible:
$4 - \frac{1}{\sqrt{12}} + \frac{10}{\sqrt{3}}$
I have resolved it this way:
$ \frac{4\sqrt{12}\sqrt{3}}{\sqrt{12}\sqrt{3}} -
\frac{\sqrt{3}}{\sqrt{12}\sqrt{3}} +
\frac{10\sqrt{12}}{\sqrt{12}\sqrt{3}} = \frac{4\sqrt{36} -
\sqrt{3}+10\sqrt{4}\sqrt{3}}{\sqrt{36}}=\frac{24 - \sqrt{3} +
20\sqrt{3}}{6}$
while in the textbook it is resolved like this:
$4 - \frac{1}{\sqrt{4 \times 3}} + \frac{10}{\sqrt{3}} = 4 -
\frac{1}{2\sqrt{3}} + \frac{10}{\sqrt{3}} = \frac{8\sqrt{3} - 1 +
20}{2\sqrt{3}}= \frac{8\sqrt{3} + 19}{2\sqrt{3}}$
I am wondering did I do it correctly and if so, which way is better?
| Your answer of $$\frac{24 - \sqrt{3} +
20\sqrt{3}}{6}=\frac{24 +
19\sqrt{3}}{6}$$
matches the answer given by the book within a factor of $\sqrt 3\over \sqrt 3$:
$$\frac{24 +
19\sqrt{3}}{6}=\frac{8\sqrt3+19}{2\sqrt 3}$$
Your method was effective in solving the problem, although putting $\sqrt{12}\sqrt 3$ as the denominator does not follow the "usual" method of coming up with a "least common multiple" for the final denominator; in this case, we have $(12,3)=3$, and therefore we would use $\sqrt {12}=2\sqrt 3$ as the final denominator and avoid a couple extra multiplications in the process. But another intent in solving problems like this is often to "rationalize the denominator", which your method achieves quite effectively. In the end, you'll want to first make sure that you understand the problem and the solution, and then follow the guidelines of the instructor you are working under: if you have a professor or teacher, match what their expectations are in the final answer form; if it's just a book or self-learning, try to match "common" notation and approaches that you see.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Evaluating a logarithmic integral with square roots We need to evaluate $$I = \int x\ln\left(x^2 + a^2 + \sqrt{x^2 - a^2}\right)\, \mathrm{d}x$$ so we using $u = \ln f(x)$ and $\mathrm{d}v = x$ so that $\mathrm{d}u = \frac{f'(x)}{f(x)}$ and $v = \frac{x^2}{2}$ (where $f(x) =$ argument of the logarithm yielding:
$$I = \frac{x^2}{2}\ln f(x) - \frac{1}{2} \int \frac{x^2\left(2x + x(x^2 - a^2)^{-1/2}\right)}{x^2 + a^2 + \sqrt{x^2 - a^2}} \, \mathrm{d}x$$
I can't seem to progress from there, I was wondering if you could provide me with any pushes/solutions in the right direction or an altogether new way of approaching the problem cleverly. Thank you!
| Using substitution will be more successful that 'by parts'.
Let $t=x^2-a^2$ so $dt=2xdx$
Then the integral becomes:
$$ I=\frac{1}{2}\int \ln(t + \sqrt{t}+2a^2)dt$$
Next using 'by parts':
Let $u=\ln(t+\sqrt{t}+2a^2)$ and let $dv=1$
We get: $v=t$ and
$$du = \frac{1+\frac{1}{2}t^{-\frac{1}{2}}}{t+\sqrt{t}+2a^2}dt$$
$$I=t \ln(t+\sqrt{t}+2a^2)-\int\frac{1+\frac{1}{2}t^{-\frac{1}{2}}}{t+\sqrt{t}+2a^2} tdt$$
$$I=t \ln(t+\sqrt{t}+2a^2)-\int\frac{t+\frac{1}{2}\sqrt{t}}{t+\sqrt{t}+2a^2} dt$$
Next let: $t=w^2$ so $dt=2wdw$
$$I=t \ln(t+\sqrt{t}+2a^2)-\int\frac{w^2+\frac{1}{2}w}{w^2+w+2a^2} 2wdw$$
$$I=t \ln(t+\sqrt{t}+2a^2)-\int\frac{2w^3+w^2}{w^2+w+2a^2} dw$$
$$I=t \ln(t+\sqrt{t}+2a^2)-\int (2w-1) dw-\int\frac{w-2a^2w+a^2}{w^2+w+2a^2} dw$$
$$I=t \ln(t+\sqrt{t}+2a^2)-w^2+w-\int\frac{\frac{1}{2}(2w+1)(1-2a^2)-\frac{1}{2}+2a^2}{w^2+w+2a^2} dw$$
$$I=t \ln(t+\sqrt{t}+2a^2)-w^2+w-\int\frac{\frac{1}{2}(2w+1)(1-2a^2)}{w^2+w+2a^2} dw-\int\frac{-\frac{1}{2}+2a^2}{w^2+w+2a^2} dw$$
For the first integral (on the last line) let: $y=w^2+w+a^2$ so $dy=(2w+1)dw$
$$I=t \ln(t+\sqrt{t}+2a^2)-w^2+w-\int\frac{\frac{1}{2}(1-2a^2)}{y} dy-\int\frac{-\frac{1}{2}+2a^2}{(w+\frac{1}{2})^2+2a^2-\frac{1}{4}} dw$$
$$I=t \ln(t+\sqrt{t}+2a^2)-w^2+w-\frac{1}{2}(1-2a^2)\ln|y|-\frac{-\frac{1}{2}+2a^2}{\sqrt{2a^2-\frac{1}{4}}}\arctan\left(\frac{w+\frac{1}{2}}{\sqrt{2a^2-\frac{1}{4}}}\right)+c,c\in\mathbb{R}$$
Then substitute back in the various expressions of $x$.
**Comparing this to wolframalpha's result I can tell I've messed up my constants along the way but the main points - which integration techniques to use are there.
P.S. I'll see if I can find time to improve this answer but I've already spend ages on it.
P.P.S. This would be significantly less messy looking and easier to do if $x^2+a^2$ and $x^2-a^2$ were the same. Did you copy the question down correctly??
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to Compute $\frac{d}{dx}\left(\left(1+x^2\right)^x\right)$? This is what I worked out:
Let $y = (1 + x^2)^x$ and let $a = 1 + x^2$
Then, by the chain rule of differentiation:
$\frac{dy}{dx} = \frac{dy}{da}\cdot\frac{da}{dx} = a^x\cdot ln(a) \cdot 2x$
$\frac{dy}{dx} = (1 + x^2)^x \cdot ln(1 + x^2) \cdot 2x $
But when I try to verify the result on WolframAlpha, I get this.
What have I done wrong?
| In order to calculate $\frac{d}{dx}\left(\left(1+x^2\right)^x\right)$ then this expression can be placed into the form
\begin{align}
\frac{d}{dx} \left(\left(1+x^2\right)^x\right) &= \frac{d}{dx}\left[ e^{x \, \ln(1+x^2)} \right] \\
&= e^{x \, \ln(1+x^2)} \, \left[ x \, \frac{d}{dx} \ln(1+x^2) + \ln(1+x^2) \right] \\
&= (1+x^2)^{x} \, \left[ \frac{2 \, x^2}{1+x^2} + \ln(1+x^2) \right]
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1492320",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Find $\lim_{x\to0}\frac1{x^3}\left(\left(\frac{2+\cos x}{3}\right)^x-1\right)$ without L'Hopital's rule
Find the following limit
$$\lim_{x\to0}\frac1{x^3}\left(\left(\frac{2+\cos x}{3}\right)^x-1\right)$$
without using L'Hopital's rule.
I tried to solve this using fundamental limits such as $\lim_{x\to0}\left(1+x\right)^\frac1x=e,\lim_{x\to0}x^x=1$ and equivalent infinitesimals at $x\to0$ such as $x\sim\sin x,a^x\sim1+x\ln a,(1+x)^a\sim1+ax$. This is what I did so far:
$$\begin{align}\lim_{x\to0}\frac1{x^3}\left(\left(\frac{2+\cos x}{3}\right)^x-1\right)&=\lim_{x\to0}\frac{\left(\frac{3-(1-\cos x)}3\right)^x-1}{x^3}\\&=\lim_{x\to0}\frac{\left(\frac{3-\frac12x^2}3\right)^x-1}{x^3}\\&=\lim_{x\to0}\frac{\left(1-\frac16x^2\right)^x-1}{x^3}\end{align}$$
I used fact that $x\sim\sin x$ at $x\to0$. After that, I tried to simplify $\left(1-\frac16x^2\right)^x$. Problem is because this is not indeterminate, so I cannot use infinitesimals here. Can this be solved algebraically using fundamental limits, or I need different approach?
| Consider first $$A=\left(\frac{2+\cos (x)}{3}\right)^x$$ $$\log(A)=x\log\left(\frac{2+\cos (x)}{3}\right)$$ Now, Taylor series for $\cos(x)$ and then for $\log(1+y)$$$\log(A)=x\log\left(\frac{2+\cos (x)}{3}\right)=x\log\left(\frac{2+1-\frac{x^2}{2}+\cdots}{3}\right)$$ $$\log(A)=x\log\left(\frac{3-\frac{x^2}{2}+\cdots}{3}\right)=x\log\left(1-\frac{x^2}{6}+\cdots\right)$$ $$\log(A)=x\left(-\frac{x^2}{6}+\cdots \right)=-\frac{x^3}{6}+\cdots$$ So $$A\approx e^{-x^3/6+\cdots}$$ Now using Taylor for $e^y$ $$B=\frac1{x^3}\left(\left(\frac{2+\cos x}{3}\right)^x-1\right)\approx\frac1{x^3}\left( e^{-x^3/6}-1\right)=\frac1{x^3}\left(1- \frac {x^3} {6}-1\right)= -\frac {1} {6}$$
If you use one more term in each expansion, you would find $$B=-\frac{1}{6}+\frac{x^3}{72}+\cdots$$ Plotting the curves for $-1<x<1$, you could be amazed to see to see how close is the curve defined by the original expression and the last approximation; they fiffer by less than $0.0002$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Two similar integration about continued fractions Prove that \begin{align*}
\int_0^{+\infty} \cfrac{\sin nx}{x + \cfrac{1}{x + \cfrac{2}{x + \cfrac{3}{x + \cdots}}}} \, dx &= \cfrac{\sqrt {\cfrac{\pi }{2}} }{n + \cfrac{1}{n + \cfrac{2}{n + \cfrac{3}{n + \cdots}}}}\\
\int_0^{+\infty} \cfrac{\sin \cfrac{n\pi x}{2}}{x + \cfrac{1^2}{x + \cfrac{2^2}{x + \cfrac{3^2}{x + \cdots}}}} \, dx &= \cfrac{1}{n + \cfrac{1^2}{n + \cfrac{2^2}{n + \cfrac{3^2}{n + \cdots}}}}.\end{align*}
We can follow this:
For the first one, we use
$$\cfrac{1}{x + \cfrac{1}{x + \cfrac{2}{x + \cdots}}} = e^{x^2/2} \int_x^\infty e^{-t^2/2} \, dt .$$
Then we must prove
$$\int_0^{+\infty} \sin nx \cdot e^{x^2/2} \, dx \int_x^\infty e^{-t^2/2 \, dt} = \sqrt {\cfrac{\pi }{2}} \cdot e^{n^2/2} \int_n^\infty e^{-t^2/2} \, dt .$$
For the second one, we use
$$\cfrac{1}{x + \cfrac{1^2}{x + \cfrac{2^2}{x + \cfrac{3^2}{x + \cdots }}}} = 2\sum_{n = 1}^\infty \cfrac{(-1)^{n + 1}}{x + 2n - 1} = 2 \int_0^1 \cfrac{t^x}{1 + t^2} \, dt.$$
We need to show
$$\int_0^1 \cfrac{2n\pi}{(1 + x^2)(n^2 \pi ^2 + 4\ln^2 x)}dx = \int_0^1 \frac{x^n}{1 + x^2} \, dx .$$
But how can we continue?
| Hint
Note
$$4\int_0^{+\infty}\cos(4t\ln x)\cdot e^{-2n\pi t} \, dt = \frac{2n\pi}{n^2\pi^2 + 4\ln^2 x}$$
then you only find
$$\int_0^{+\infty}\left(\int_0^1 \frac{\cos(4t\ln x)}{1+x^2} \, dx\right) e^{-2n\pi t} \, dt$$
you only let
$x=e^{-u}$, then
$$\int_0^1 \frac{\cos(4t\ln x)}{1+x^2} \, dx = \int_0^{+\infty} \frac{\cos(4tu)}{2\cosh u}$$
then it is well know,and you can do it(if you can't,you can see this simaler problem:Solve $\int \limits_{0}^{\infty} \frac{\cos(x)}{\cosh(x)} dx$ without complex integration.
| {
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"url": "https://math.stackexchange.com/questions/1500112",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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prove trig equivalence $$\sin x + \sin y = 2\sin\left(\frac{x+y}{2}\right)\cos\left(\frac{x-y}{2}\right)$$
I want to use the the $e^{ix}$ identities but I'm not sure if it can be done that way, let alone how to do it.
Any tips would be appreciated.
| \begin{align}
\sin u \cos v + \cos u \sin v & = \sin(u+v) \\
\sin u \cos v - \cos u \sin v & = \sin(u-v)
\end{align}
Apply the above in the case where $u = \dfrac{x+y} 2$ and $v=\dfrac{x-y} 2$:
\begin{align}
\sin\left(\frac{x+y}{2}\right) \cos\left(\frac{x-y}{2}\right) + \cos\left(\frac{x+y}{2}\right)\sin\left(\frac{x-y}{2}\right) & = \sin \left( \frac{x+y}{2} + \frac{x-y}{2} \right) \\[10pt]
\sin\left(\frac{x+y}{2}\right) \cos\left(\frac{x-y}{2}\right) - \cos\left(\frac{x+y}{2}\right)\sin\left(\frac{x-y}{2}\right) & = \sin \left( \frac{x+y}{2} - \frac{x-y}{2} \right)
\end{align}
Or in other words:
\begin{align}
\sin\left(\frac{x+y}{2}\right) \cos\left(\frac{x-y}{2}\right) + \cos\left(\frac{x+y}{2}\right)\sin\left(\frac{x-y}{2}\right) & = \sin x \\[10pt]
\sin\left(\frac{x+y}{2}\right) \cos\left(\frac{x-y}{2}\right) - \cos\left(\frac{x+y}{2}\right)\sin\left(\frac{x-y}{2}\right) & = \sin y
\end{align}
Add left sides and add right sides:
$$
2 \sin\left(\frac{x+y}{2}\right) \cos\left(\frac{x-y}{2}\right) = \sin x + \sin y.
$$
| {
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"url": "https://math.stackexchange.com/questions/1500472",
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"source": "stackexchange",
"question_score": "5",
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We have $12$ balls numbered with $1, 2, 3, 4, 5, 6, 7, 8, 9, 9, 9, 9$ in a bin.
We have $12$ balls in a bin, numbered from $1$ to $8$ and the other $4$ numbered with the number $9$. We take one of them, write down the number and then put back into the bin again. We do this $3$ times. Which is the probability of obtaining $9$ when we multiply those $3$ numbers obtained?
My try:
There is $\frac{12!}{4!}$ possible cases of extracting $3$ balls from $12$ and obtain a different number on each extraction.
To calculate favorable cases, we see that we need to extract $3$, $1$
and $3$ or $9$, $1$, $1$, up to order of the extractions. We only have
that two options.
How many ways we have to obtain two threes and a one in our extractions? $$\binom{12}{1}\cdot\binom{12}{1}\cdot\binom{12}{1}=36.$$
How many ways we have to obtain one nine and two one in our extractions? $$\binom{12}{4}\cdot\binom{12}{1}\cdot\binom{12}{1}=71316$$
So the probability is:$$P\{\text{obtain a 9 multiplying the results}\}=\frac{36+71316}{12\cdot11\cdot10\cdot\dots\cdot5}$$
I don't know if I did it correctly. Thanks.
| I think your methodology is off. For instance, $\binom{12}{1}\cdot\binom{12}{1}\cdot\binom{12}{1} = 12^3$ (not $36$), is the number of ways to choose any three balls, not to choose two $3$'s and a $1$.
Think about them as $12$ balls: if we pick three of them with replacement, we have $12^3$ possible sequences of balls (where the $9$'s have the same numerical value but are thought of as different balls). You're correct that the only sequences that work are when we have two $3$'s and a $1$ or when we have two $1$'s and a $9$.
The total number of ways to draw two $3$'s and a $1$ is $\binom{3}{1} =3$ (we have to choose which of the three balls will be a $1$).
The total number of ways to draw one $9$'s and two $1$'s is $\binom{3}{1} \cdot 4 = 12$ (we have to choose which of the three balls will be a $9$ and then choose which $9$-ball to draw).
Thus, we have a probability of $$\mathbb{P}(\text{multiplying to }9) =\frac{3 + 12}{12^3} = \frac{5}{576}.$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Find limit of the expression $$\lim_{x \to 0}\frac{\cos(x) - 8x\sin(x/2) - \cos(3x)}{x^4}$$
I think I should replace by equivalent, such as $\sin(x)$ ~ $x$, but got nothing.
Thank you for answers, but what about solving without using L'Hôpital's rule?
| $\cos(x)-\cos(3x) = 2\sin(x)\sin(2x) = 4\sin\left(\frac{x}{2}\right)\cos\left(\frac{x}{2}\right)\sin(2x)$, hence we want to compute:
$$ \lim_{x\to 0}\frac{4\sin\left(\frac{x}{2}\right)}{x}\cdot\frac{\sin(2x)-2x}{x^3}=2\cdot\lim_{x\to 0}\frac{\sin(2x)-2x}{x^3}=16\cdot\lim_{z\to 0}\frac{\sin(z)-z}{z^3} $$
that equals $-\frac{8}{3}$, by applying twice De l'Hopital theorem, then subtract:
$$ \lim_{x\to 0}\frac{4\sin\left(\frac{x}{2}\right)}{x}\cdot\frac{\sin(2x)(1-\cos\frac{x}{2})}{x^3}=4\cdot\lim_{x\to 0}\frac{\sin(2x)\sin^2\left(\frac{x}{4}\right)}{x^3}=\frac{1}{2}. $$
The given limit is so $-\frac{8}{3}-\frac{1}{2}=\color{red}{\Large -\frac{19}{6}}$.
To prove the crucial part, i.e. $\lim_{x\to 0}\frac{x-\sin(x)}{x^3}=\frac{1}{6}$, without derivatives, you may assume that the limit just exists and equals $L$. Then:
$$ L = \lim_{x\to 0}\frac{2x-\sin(2x)}{8x^3} = \lim_{x\to 0}\frac{x-\sin(x)\cos(x)}{4x^3}=\frac{L}{4}+\lim_{x\to 0}\frac{\sin(x)\sin^2\left(\frac{x}{2}\right)}{2x^3} $$
and that leads to $L=\frac{L}{4}+\frac{1}{8}$, from which $L=\frac{1}{6}$.
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "1",
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Find probability of having a female in every group 10 male students and 5 female students are split into 5 groups (every grup consists of 3 students). What is the probability that there is a female student in every group?
I should solve this with combinatorics. Any idea on how to appraoch the problem?
| total possible: ${15\choose 3}{12\choose 3}{9\choose 3}{6\choose 3}{3\choose 3}$
one female in every group: ${10\choose 2}{5\choose 1}{8\choose 2}{4\choose 1}{6\choose 2}{3\choose 1}{4\choose 2}{2\choose 1}{2\choose 2}{1\choose 1}$
solution: ${{10\choose 2}{5\choose 1}{8\choose 2}{4\choose 1}{6\choose 2}{3\choose 1}{4\choose 2}{2\choose 1}{2\choose 2}{1\choose 1}}\over{{15\choose 3}{12\choose 3}{9\choose 3}{6\choose 3}{3\choose 3}}$
| {
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"source": "stackexchange",
"question_score": "1",
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Find : $\sqrt[6]{\frac{\sqrt{2}+(-\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}i)^7}{(\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}i)^{11}}}$ in its algebraic form. Find : $$\sqrt[6]{\frac{\sqrt{2}+(-\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}i)^7}{(\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}i)^{11}}}$$ in its algebraic form.
Now, I kinda think it would not be wise to try to expand this, but rather apply de Moivre formula on the complex number in the numerator and denominator, then simplify that complex number within the root, and once again apply moivres formula.
I have tried but then I get the expression:$\sqrt[6]{\frac{\sqrt{2}+\cos{\frac{21 \pi}{4}+i\sin{\frac{21 \pi }{4}}}}{\cos \frac{11\pi}{4}+i\sin\frac{11 \pi}{4}}}$ and don't know what to do with it.
| The binomial quantities being raised to powers are easy to simplify: $$\sqrt[6]{\frac{\sqrt{2}-\frac{\sqrt{2}}{2}-\frac{\sqrt{2}}{2}i}{-\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}i}}$$
Now it's quite easy to continue. $$\sqrt[6]{\frac{\frac{\sqrt{2}}{2}-\frac{\sqrt{2}}{2}i}{-\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}i}}=\sqrt[6]{-1}$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Show that $\left[\alpha\right]+\left[\alpha+\frac{1}{m}\right]+\cdots + \left[\alpha+\frac{m-1}{m}\right] = \left[m\alpha\right]$ Let $m$ be a positive integer and let $\alpha$ be a real number. Show that
$$\left[\alpha\right]+\left[\alpha+\frac{1}{m}\right]+\cdots + \left[\alpha+\frac{m-1}{m}\right] = \left[m\alpha\right]$$ where $\left[\ \ \right]$ is the integral part function.
Remark: Attempt to prove this result by induction but I failed in the inductive step, I have been unable use the induction hypothesis.
| This is known as Hermite's identity. A nice proof is due to Matsuoka "On a Proof of Hermite's Identity", AMM 71:10 (1964), pp. 1115, DOI: 10.2307/2311413.
Let:
$\begin{align}
f(x)
= \lfloor m x \rfloor
- \lfloor x \rfloor
- \left\lfloor x + \frac{1}{m} \right\rfloor
- \left\lfloor x + \frac{2}{m} \right\rfloor
- \dotsb
- \left\lfloor x + \frac{m - 1}{m} \right\rfloor
\end{align}$
For real $\alpha, \beta$ it is $\lfloor \alpha + 1 \rfloor - \lfloor \beta + 1 \rfloor
= \lfloor \alpha \rfloor - \lfloor \beta \rfloor$, so that:
$\begin{align}
f\left( x + \frac{1}{m} \right)
&= \lfloor m x + 1 \rfloor
- \left\lfloor x + \frac{1}{m}\right\rfloor
- \left\lfloor x + \frac{2}{m} \right\rfloor
- \dotsb
- \left\lfloor x + \frac{m - 1}{m} \right\rfloor
- \lfloor x + 1 \rfloor \\
&= \lfloor m x \rfloor
- \lfloor x \rfloor
- \left\lfloor x + \frac{1}{m} \right\rfloor
- \left\lfloor x + \frac{2}{m} \right\rfloor
- \dotsb
- \left\lfloor x + \frac{m - 1}{m} \right\rfloor \\
&= f(x)
\end{align}$
On the other hand, for $0 \le x < 1 /m$ it is $f(x) = 0$, and so $f(x) = 0$ for all $x$, as was to be proved.
| {
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"timestamp": "2023-03-29T00:00:00",
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Write out the Form of the Partial Fraction Decomposition: $\int\frac{x^6}{x^2-4}$ Write out the Form of the Partial Fraction Decomposition: $\int\frac{x^6}{x^2-4}$
The book says use long division my answer was $x^3+\frac{4x^3}{x^2-4}$
The answer manual is $\frac{x^6}{x^2-4}=x^4+4x^2+16+\frac{64}{(x+2)(x-2)}$
| Notice, the formula $a^3-b^3=(a-b)(a^2+b^2+ab)$,
Now, re-arrange the numerator as follows $$\frac{x^6}{x^2-4}=\frac{x^6-4^3+4^3}{x^2-4}$$
$$=\frac{((x^2)^3-4^3)+4^3}{x^2-4}$$
$$=\frac{(x^2)^3-4^3}{x^2-4}+\frac{64}{x^2-4}$$ $$=\frac{(x^2-4)((x^2)^2+4^2+4x^2)}{x^2-4}+\frac{64}{x^2-4}$$
$$=\frac{(x^2-4)(x^4+4x^2+16)}{x^2-4}+\frac{64}{x^2-4}$$
$$=x^4+4x^2+16+\frac{64}{x^2-4}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1510978",
"timestamp": "2023-03-29T00:00:00",
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Find all possible remainders.
Suppose $b\in\mathbb Z$. Find all possible remainders of $b^3$ divided by $7$.
I know $b^6=1\bmod7$ which means $(b^2)^3=1\bmod7$ so all square numbers^3 leave $1$ as a remainder, but how to continue?
| Notice that this depends only on the value of $b$ mod $7$, since $(b+7k)^3 = b^3 + (\text{multiple of 7})$. So we can work wlog with $b = 0, 1, \dots, 6$.
Your method is also faulty: just because $b^6 \equiv 1 \pmod{7}$, that doesn't mean $b^2 \equiv 1 \pmod{7}$. For example, $3^2 \equiv 2 \pmod{7}$.
There's a really brute-force way to do it, modulo 7. I'll use $=$ instead of $\equiv$ to save a bit of typing:
$0^3 = 0, 1^3 = 1, 2^3 = 8 = 1, 3^3 = 27 = -1, 4^3 = 64 = 1, 5^3 = (-2)^3 = -1, 6^3 = (-1)^3 = -1$
yielding the answers $0, 1, -1$.
Really, you only need to check $0, 1, 2, 3$ (and then negate those answers), because $(7-3)^3 \equiv (-3)^3 = - (3^3)$.
| {
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Am I correctly finding the standard matrix?
Question:
Let $F: \Bbb R^3 \to \Bbb R^3$ be the linear transformation satisfying
$F(1,0,1)=(-3,-3,1)$,$F(0,1,0)=(0,1,1)$, and $F(0,1,1)=(2,-2,1)$. Find
the standard matrix $A$ of $F$.
My Approach:
I used a method that I haven't been taught, so I'm not sure if I am correct and allowed to do this. Can you guys see if this is acceptable.
Since:
$$F\begin{pmatrix}1\\0\\1\end{pmatrix}=\begin{pmatrix}3\\-3\\1\end{pmatrix}$$
$$F\begin{pmatrix}0\\1\\0\end{pmatrix}=\begin{pmatrix}0\\1\\1\end{pmatrix}$$
$$F\begin{pmatrix}0\\1\\1\end{pmatrix}=\begin{pmatrix}2\\-2\\1\end{pmatrix}$$
We want to find:
$$\text{standard matrix:}F\begin{pmatrix}1&0&0\\0&1&0\\0&0&1\end{pmatrix}=?$$
So what I did was:
$$\therefore \text{standard matrix:}F\begin{pmatrix}1&0&0\\0&1&0\\0&0&1\end{pmatrix}=\begin{pmatrix}3&0&2\\-1&0&-1\\-2&1&-1\end{pmatrix}$$
Is this method correct?
| Let's check your solution,
So you got $T(1,0,0)=(3,-1,-2)$, $~T(0,1,0)=(0,0,1)$ and $~T(0,0,1)=(2,-1,-1)$
Now $T(1,0,1)=(3,-3,1)$
also $~T(1,0,1)=T(e_1+0.e_2+e_3)=Te_1+Te_3=(3,-1,-2)+(2,-1,-1)=(5,-2,-3) \not=(3,-3,1).$
| {
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Help with $\lim \frac {{1}^{p} + 3^p + ... + (2n+1)^p}{n^{p+1}}$. I'm using Stolz–Cesàro theorem to show that this limit = $\frac {2^p}{P+1}$. I took $(X_{n})$ = $\sum_{i=0}^{n}{2i+1}^p$ and $(Y_n)$ = $n^{p+1}$. Then, by the theorem, $\lim = \frac {X_{n+1}-X_n}{Y_{n+1}-Y_n}$.
In the numerator I get $(2n+3)^{p}$. I've tried to expand this and I have leftover terms $$2 + \frac{6}{p} + \frac{9}{2p^{2}}+...$$ after canceling out. I don't know what to do with these, or if I did the expansion correctly.
| Try to link the summation to a Riemann sum:
\begin{align}
& \frac{1^p + 3^p + \cdots + (2n + 1)^p}{(n + 1)^{p + 1}} \\
= & \frac{1}{2}\sum_{i = 1}^{n + 1}\left(\frac{2i - 1}{n + 1}\right)^p\times\frac{2}{n + 1} \\
\to & \frac{1}{2} \int_0^2 x^p dx = \frac{2^p}{p + 1},
\end{align}
during which we partitioned interval $[0, 2]$ by an equally-spaced partition
$$\left\{0, \frac{1}{n + 1}, \frac{3}{n + 1}, \ldots, \frac{2n + 1}{n + 1}, 2\right\}.$$
Since $\lim_{n \to \infty} \frac{(n + 1)^p}{n^p} = 1$, the result follows.
| {
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How to find the Minimum of the Function. Minimize $xy$ on the ellipse $b^2x^2+a^2y^2=a^2b^2$
So what I did first was take the gradient of $f$ and $g$
$∇ f = (y,x)$ $\qquad$ $∇ g= (2x^2b^2,a^22y)$
Then we do the lagrange multiplier
$y= 2x^2b^2λ$
$x=2ya^2λ$
Then we equate the functions. By multiplying.
$(y= 2x^2b^2λ)y^2a^2$
$(x=2ya^2λ )x^2b^2$
$y^3=2x^2y^2a^2b^2λ$
$x^3=2x^2ya^2b^2λ$
But this is where I get stuck. Setting them equal to each other gets me this.
$2x^2y^2a^2b^2= 2x^2ya^2b^2$
$y=1$
I think there is another method to solve this problem than the one I specifed does anyone know what that method is?
| $$L(x,y,z)=xy+\lambda(b^2x^2+a^2y^2-a^2b^2)$$
$$\nabla_{x,y,\lambda} L(x,y,z)=\langle y+2\lambda b^2 x, x+2\lambda a^2 y , b^2x^2+a^2y^2-a^2b^2 \rangle=\langle 0,0,0 \rangle$$
$$y=-2 \lambda b^2 x$$
$$x=-2 \lambda a^2 y$$
assuming $\lambda \ne0$:
$$\frac{y}{x}=-2 \lambda b^2$$
$$\frac{y}{x}=\frac{1}{-2 \lambda a^2}$$
$$-2 \lambda b^2=\frac{1}{-2 \lambda a^2}$$
or
$$\lambda=\frac{\pm1}{2 a b}$$
$$y=\mp \frac{ b}{a}x$$
$$b^2x^2+a^2(\frac{b^2}{a^2})x^2=a^2b^2$$
$$x=\pm\frac{a}{\sqrt2}$$
$$y=\mp\frac{b}{\sqrt2}$$
$$xy=\frac{-ab}{2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1518265",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Solving an equation by telling the value of $x^2+y^2$. I have a problem solving an equation. The equation is:
$xy+x+y=44$ and $x^2y+xy^2=448$
and we have to tell the value of $x^2+y^2$
First I tried solving this by doing the following:
$xy+x+y=44~\to~x+y=44-xy~\to~x^2+2xy+y^2=44^2-88xy+x^2y^2~\Rightarrow$
$\Rightarrow~x^2+y^2=44^2-90xy+x^2y^2$
But from here I didn't know what to do. Could you help me in solving this equation?
| $$(x+y)=44-xy..............1$$
$$xy(x+y)=448........................2$$
substitute 1 in 2
$$xy(44-xy)=448$$
$$x^2y^2-44xy+448=0$$
use the quadratic formula
$$xy=22\pm6$$
or
$$x^2y^2=(22\pm6)^2$$
now square the equ.1
$$x^2+2xy+y^2=44^2-88xy+x^2y^2$$
$$x^2+y^2=44^2-90xy+x^2y^2$$
hence
$$x^2+y^2=44^2-90(22\pm6)+(22\pm6)^2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1518594",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 4
} |
Which natural numbers satisfy $2^n > n^2$?
Which natural numbers satisfy $2^n > n^2$ ?
My work. Step 1: $n = 1 $, $2^1 > 1^2$. True.
For $n = k$, $2^k > k^2$. For $n = k+1$,
$$ 2^{(k+1)} > (k+1)^2 \\
2\cdot 2^k>k^2+2k+1 \\
2^k+2^k > k^2+2k+1$$
*
*$2^k > k^2 \text{ - from step 1}$
*$2^k > k^2+2k+1$
How I can find the numbers now?
| (A rather informal approach, just for fun!)
As $n$ increases by one, the expression $2^n$ doubles.
As $n$ increases by one, the expression $n^2$ increases by the $(n+1)$st odd.
This is because $n^2$ becomes $(n+1)^2 = n^2 + (2n+1)$ where $2n+1$ is the $(n+1)$st odd.
Note that the expressions are equal when $n = 4$: $2^4 = 16 = 4^2$.
As $n$ increases by one, the left hand side doubles, which means adding $16$. But, the right hand side increases by the $(4+1)$th odd, i.e., the fifth odd, $9$, going from $4^2 = 16$ to $5^2 = 25$.
So, after being equal when $n=4$, the exponential expression increases by $16$ whereas the quadratic expression increases by $9$, which means the former is now greater than the latter. Moreover, each subsequent increase by one in $n$ is leading the left hand side to grow by twice as much as before, whereas the right hand side grows by two more than the previous iteration. This means that the exponential expression will continue to be greater than the quadratic expression for $n \geq 5$.
We already saw the inequality does not hold for $n = 4$, since both sides are equal, and so all that remains now is to check the earlier natural numbers:
When $n=1$, we have $2^1 = 2 > 1 = 1^2$; when $n=2$, we have equality; when $n = 3$, we have that $2^3 = 8 < 9 = 3^2$. Whether to include $0$ in $\mathbb{N}$ is a function of your course, but let us at least note for completeness that $2^0 = 1 > 0 = 0^2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1518774",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
Find the determinent of a $4 \times 4$ matrix with the letter $a$ in it Any idea how to compute the determinant of $4 \times 4$ matrix $A$ when
\begin{equation}
A = \begin{bmatrix} 1 & 4 & 8 & 1\\ 0 & 30 & 1 & 0 \\ 0 & 2 & 0 & 0 \\ 1 & 2 & 9 & a \end{bmatrix}.
\end{equation}
The $a$ in the $A_{44}$ location is really confusing me!
| Just expand across the third row, as @Surb suggested:
\begin{equation}
|A| = -2 \left|
\begin{pmatrix}
1 & 8 & 1\\
0 & 1 & 0 \\
1 & 9 & a
\end{pmatrix} \right|
=-2 \left|
\begin{pmatrix}
1 & 1 \\ 1 & a
\end{pmatrix} \right| = -2(a - 1) = 2 - 2a.
\end{equation}
Another option is to expand down the fourth column:
\begin{align}
| A| &=
-1 \left|\begin{pmatrix} 0 & 30 & 1 \\ 0 & 2 & 0 \\ 1 & 2 & 9 \end{pmatrix} \right| + a \left |
\begin{pmatrix} 1 & 4 & 8 \\ 0 & 30 & 1 \\ 0 & 2 & 0 \end{pmatrix} \right |.
\end{align}
The two determinants on the right simplify greatly by expanding down the first column:
\begin{equation}
\left|\begin{pmatrix} 0 & 30 & 1 \\ 0 & 2 & 0 \\ 1 & 2 & 9 \end{pmatrix} \right|
= 1\left| \begin{pmatrix} 30 & 1 \\ 2 & 0 \end{pmatrix} \right| = -2
\end{equation}
and
\begin{equation}
\left |
\begin{pmatrix} 1 & 4 & 8 \\ 0 & 30 & 1 \\ 0 & 2 & 0 \end{pmatrix} \right |
= 1 \left| \begin{pmatrix} 30 & 1 \\ 2 & 0 \end{pmatrix} \right| = -2.
\end{equation}
So we see that $|A| = 2 - 2a$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1520819",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Prove the identity $\sum^{n}_{k=0}\binom{m+k}{k} = \binom{n+m+1}{n}$
Let $n,m \in \mathbb{N}$. Prove the identity $$\sum^{n}_{k=0}\binom{m+k}{k} = \binom{n+m+1}{n}$$
This seems very similar to Vandermonde identity, which states that for nonnegative integers we have $\sum^{m}_{k=0}\binom{m}{k}\binom{n}{r-k} = \binom{m+n}{r}$. But, clearly this identity is somehow different from it. Any ideas?
| We can write $\displaystyle \sum^{n}_{k=0}\binom{m+k}{k} = \sum^{n}_{k=0}\binom{m+k}{m} = \binom{m+0}{m}+\binom{m+1}{m}+........+\binom{m+n}{m}.$
Now Using Coefficient of $x^r$ in $(1+x)^{t} $ is $\displaystyle = \binom{t}{r}.$
So we can write above series as...
Coefficient of $x^m$ in $$\displaystyle \left[(1+x)^m+(1+x)^{m+1}+..........+(1+x)^{m+n}\right] = \frac{(1+x)^{m+n+1}-(1+x)^{m}}{(1+x)-1} = \frac{(1+x)^{m+n+1}-(1+x)^{m}}{x}$$
above we have used Sum of Geometric Progression.
So we get Coefficient of $x^{m+1}$ in $\displaystyle \left[(1+x)^{m+n+1}-(1+x)^{m}\right] = \binom{m+n+1}{m+1} = \binom{m+n+1}{n}.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1521147",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 1
} |
Finding the vertex and focus of a rotated parabola So I begun with the following equation : $x^2+2xy+y^2+2\sqrt{2}x-2\sqrt{2}y+4=0$
I transformed it in the following : $y'=\frac{x'^2}{2}+1$
I had to do a rotation of $\frac{\pi}{4}$ of the xy axis. (counter clock wise) My question is how do you find the focus and vertex of the rotated parabola ? I think that the vertex is going to be (0,1) but I'm not sure if this is the case when we rotate it.
Thank you!
| I'm going to assume the OP wanted the vertex and focus of the original tilted parabola since they already had rotated it to a standard form where those values are easier to find.
Rather than rotating, its convenient to do everything in place since we're looking for the focus and by definition: for any point on the parabola the distance to the focus is equal to the distance to the directrix.
First let the focus F = (a,b) and the directrix is a linear equation y = mx + c
We need to find the point P (p, mp + c) on the directrix such that the segment between it
and an arbitrary point on the parabola is perpendicular to the directrix i.e. $-\dfrac{1} {m} = \dfrac{(y - mp - c)}{x - p}$
Solving for p we get $p = \dfrac{1}{1 + m^2}(my + x- mc)$ and $P = \left(\dfrac{1}{1 + m^2}(my + x- mc), \dfrac{m}{1 + m^2}(my + x - mc) + c \right)$
Then our general equation for the parabola is:
$(x-a)^2 + (y-b)^2 = (x - \dfrac{1}{1 + m^2}(my + x- mc))^2 + (y - \dfrac{m}{1 + m^2}(my + x - mc) - c)^2$
This simplifies nicely to $(x-a)^2 + (y-b)^2 = \dfrac{1}{1 + m^2}(mx - y + c)^2$
If we move everything to the lhs we get this form:
$x^2 + m^2y - 2mxy + (-2(m^2 + 1)a -2c)x + (-2(m^2 + 1)b + 2c)y + ((m^2 + 1)(a^2 + b^2
) - c^2) = 0$
So if the coefficient of $x^2$ in the original equation is 1 we can easily find m.
This is already the case in our parabola so here m = 1.
Now we're left with $(x-a)^2 + (y - b^2) = \dfrac{1}{2}(x - y + c)^2$
Corresponding with the original equation gives us a system of equations:
*
*$2a^2 + 2b^2 - c^2 = 4$
*$-4a -2c = 2 \sqrt{2}$
*$-4b +2c = -2 \sqrt{2}$
Solving for c and then substituting back in to find a and b we get:
$(x + \frac{3}{4}\sqrt{2})^2 + (y - \frac{3}{4}\sqrt{2})^2 = \frac{1}{2}(x - y + \frac{\sqrt{2}}{2})^2$
The first part is done. The focus $F = \left( -\dfrac{3}{4}\sqrt{2}, \dfrac{3}{4}\sqrt{2} \right)$
The vertex is halfway between F and the directrix. So first we look at the axis of symmetry y = -x + d which goes through F. Solving d = 0. This intersects the directrix where $-x = x + \frac{\sqrt{2}}{2}$ at I = $\left(-\dfrac{1}{4}\sqrt{2}, \dfrac{1}{4}\sqrt{2}\right)$
And the vertex V is at the midpoint of the segment between F and I. $V = \left(-\dfrac{\sqrt{2}}{2},\dfrac{\sqrt{2}}{2} \right)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1521728",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Convergence of $\sum _{n=2}^{\infty }\:\frac{1}{n\ln\left(n\right)}$ Is this convergent or not?
$$\sum _{n=2}^{\infty }\:\frac{1}{n\ln\left(n\right)}$$
I tried using the ratio test but the limit is giving me 1, which doesn't help me. I don't think I'm supposed to use the integral test, since we haven't studied it.
| If you are familiar with Cauchy Condensation Test, the condens series becomes
$$\sum_{n} 2^n \frac{1}{2^n \ln (2^n)}$$
If not, you can use the idea of the test:
$$\frac{1}{2 \ln2 } \geq \frac{1}{2 \ln 2 } \\
\frac{1}{3 \ln 3 } \geq \frac{1}{4 \ln 2^2 } \\
\frac{1}{4 \ln 4 } \geq \frac{1}{4 \ln 2^2 } \\
\frac{1}{5 \ln 5 } \geq \frac{1}{8 \ln 2^3 } \\
.....\\
\frac{1}{2^n \ln 2^n } \geq \frac{1}{2^n \ln 2^n } $$
By adding you get
$$s_{2^n} \geq \frac{1}{2 \ln 2 } +\frac{1}{2 \ln 2^2 } +...+\frac{1}{2 \ln 2^n }=\frac{1}{2 \ln 2} (1+\frac12+\frac13+...+\frac1n)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1522832",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
} |
Centre of gravity of eighth of a sphere Could I ask for help to show where I'm going wrong?
Question
Find the position of the center of gravity of that part of a thin spherical shell x^2 + y^2 + z^2 = a^2 which exists in the first octant.
My Answer
Working in spherical coordinates:
The $2_{nd}$ moment of area of a small area element $dA$ about the z-axis is:
$dA=(a\sin\theta)^2\,\delta{A}$
(This is because in spherical coordinates $(r,\theta,\phi)$ the perpendicular distance of an area element of the shell from the z-axis is $a\sin\theta$ and the product of area and this distance squared is the 2nd moment of area)
Now, in this problem, in spherical coordinates the elemental area is given by:
$\delta{A}=a^2\,\sin\theta\,\delta\theta\,\delta\phi$
(this comes from the general result that in spherical coordinates $(r,\theta,\phi)$ the elemental area is given by $\delta{A}=r^2\,\sin\theta\,\delta\theta\,\delta\phi\,$ but in this problem r = a is constant)
So we have that the elemental $2_{nd}$ moment of area is given by:
$a^2\sin\theta\,(a\sin\theta)^2\,\delta\theta\,\delta\phi$ =
$a^4sin^{3}\theta\,\delta\theta\,\delta\phi\,\therefore$
Total $2_{nd}$ moment of area = $a^4\int_{\phi=0}^{\pi/2}\,\int_{\theta=0}^{\pi/2}sin^3\theta\,d\theta\,d\phi\,=$
$a^4\int_{\theta=0}^{\pi/2}\frac{2}{3}\,d\phi\,=$
$a^4\left[\frac{2}{3}\phi\right]_0^{\pi/2}\,=$
$a^4(\frac{2}{3}\frac{\pi}{2})\,=\,\frac{a^4\pi}{3}$
Now, we can equate this with the radius of gyration (k) of the eighth-of-a-sphere and its total area (A) as follows:
$\frac{a^4\pi}{3}=Ak^2=\frac{4\pi a^2}{8}k^2$
(Dividing by eight as we only have one eighth of a sphere)
And so, this leads to:
$k=\frac{\sqrt{2}a}{\sqrt{3}}$
And as the radius of gyration k is the distance at which a point mass of the same mass as the eighth-of-a-sphere will produce an equivalent moment as the whole eighth-of-a-sphere we have that the x-compnent of the centre of mass of the whole object is $\frac{\sqrt{2}a}{\sqrt{3}}$ and so by symmetry the answer to the whole problem is $(\frac{\sqrt{2}a}{\sqrt{3}},\frac{\sqrt{2}a}{\sqrt{3}},\frac{\sqrt{2}a}{\sqrt{3}})$
However, I believe the answer should be:
$(a/2,a/2,a/2)$
Where have I gone wrong?
| edit: Never mind this answer. I mistakenly though you were talking about a solid eight of a sphere.
Why use spherical coordinates at all?
Let $A$ be the one-eight sphere located in the positive octant. The total volume of $A$ is just $\frac{a^3\pi}{6}$. So to calculate the $z$ component of the center of mass, and by symmetry the others, we need to calculate $$\frac{6}{a^3\pi}\int_AzdV$$
But for a fixed $z$ the horizontal slice through A at height $z$ is just a quarter circle of radius $\sqrt{a^2-z^2}$ which has a very convenient area of simply $\frac{\pi}{4}(a^2-z^2)$. So we can compute as follows
$$\begin{align}
\frac{6}{a^3\pi}\int_AzdV &= \frac{6}{a^3\pi} \int_0^a z\cdot \frac{\pi}{4}(a^2-z^2)dz\\
&=\frac{3}{2a^3}\int_0^a a^2z-z^3dz \\
&=\frac{3}{2a^3}\left[\frac{a^2z^2}{2}-\frac{z^4}{4}\right]_0^a \\
&=\frac{3}{2a^3}(\frac{a^4}{2}-\frac{a^4}{4}) \\
&=\frac{3}{2a^3}\cdot\frac{a^4}{4} = \frac{3}{8}a
\end{align}$$
So the center of mass is at $(\frac{3}{8}a, \frac{3}{8}a, \frac{3}{8}a)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1525625",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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How to compute eigenvalues of big $5×5$ matrix (symmetric matrix) . How to compute eigenvalues of big $5×5$ matrix (symmetric matrix) .
$\left(\begin{matrix}
0 & 0 & 0 & 0 & 1\\
0 & 0 & 1 & 1 & 0\\
0 & 1 & 0 & 1 & 0\\
0 & 1 & 1 & 0 & 0\\
1 & 0 & 0 & 0 & 0\\
\end{matrix}\right)
$
By characteristics equation will be lengthy process . If you are going to suggest "eigenvalues by inspection", please explain in steps .
| There is a lot of $0$... you can compute it as usual...
$$\left|\begin{matrix}-t&0&0&0&1\\0&-t&1&1&0\\ 0&1&-t&1&0\\0&1&1&-t&0\\1&0&0&0&-t\end{matrix}\right|=
-t\left|\begin{matrix}
-t&1&1&0\\ 1&-t&1&0\\1&1&-t&0\\0&0&0&-t
\end{matrix}\right|+
\left|\begin{matrix}
0&0&0&1\\-t&1&1&0\\ 1&-t&1&0\\1&1&-t&0
\end{matrix}\right|$$
$$=-t^2\left|\begin{matrix}
-t&1&1\\1&-t&1\\1&1&-t
\end{matrix}\right|-
\left|\begin{matrix}
-t&1&1\\1&-t&1\\1&1&-t
\end{matrix}\right|
=-(t^2+1)\left|\begin{matrix}
-t&1&1\\1&-t&1\\1&1&-t
\end{matrix}\right|=...$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1526988",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove that $\lfloor{(2+\sqrt{3})^n}\rfloor$ is an odd number? How can we prove that $\lfloor{(2+\sqrt{3})^n}\rfloor$ is an odd number? My teacher did something like this:
$$\lfloor{(2+\sqrt{3})^n}\rfloor=(2+\sqrt{3})^n+(2-\sqrt{3})^n-1$$
And she said a few more words, but I understood nothing. Could you please tell me what she meant by that?
| $$(2+\sqrt{3})^n=\binom{n}{0}2^{n}\sqrt{3}^{0}+ \binom{n}{1}2^{n-1}\sqrt{3}^{1}+\binom{n}{2}2^{n-2}\sqrt{3}^{2}+...+\binom{n}{n}2^{0}\sqrt{3}^{n}\\
(2-\sqrt{3})^n=\binom{n}{0}2^{n}\sqrt{3}^{0}- \binom{n}{1}2^{n-1}\sqrt{3}^{1}+\binom{n}{2}2^{n-2}\sqrt{3}^{2}+...+(-1)^n\binom{n}{n}2^{0}\sqrt{3}^{n}$$ note that for sum of them
$$ (2+\sqrt{3})^n+(2-\sqrt{3})^n=2\left ( \binom{n}{0}2^{n}\sqrt{3}^{0}+ \binom{n}{2}2^{n-2}\sqrt{3}^{2}+...+\binom{n}{n}2^{0}\sqrt{3}^{n} \right )=2Q\\if \space n=even$$
$$ (2+\sqrt{3})^n+(2-\sqrt{3})^n=2\left ( \binom{n}{0}2^{n}\sqrt{3}^{0}+ \binom{n}{2}2^{n-2}\sqrt{3}^{2}+...+\binom{n-1}{n}2^{1}\sqrt{3}^{n-1} \right )=2Q\\if \space n=odd$$
both sum is integer ,and now
$$ \left \lfloor (2+\sqrt{3})^n \right \rfloor=\left \lfloor 2Q-(2-\sqrt{3})^n \right \rfloor$$ note that
$$0<(2-\sqrt{3})^n<1\\2Q-1<2Q-(2-\sqrt{3})^n<2Q\\$$ so
$$ \left \lfloor (2+\sqrt{3})^n \right \rfloor=\left \lfloor 2Q-(2-\sqrt{3})^n \right \rfloor=2Q-1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1527827",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Find the limit of the sequence $\left( \sqrt {2n^{2}+n}-\sqrt {2n^{2}+2n}\right) _{n\in N}$ My answer is as follows, but I'm not sure with this:
$\lim _{n\rightarrow \infty }\dfrac {\sqrt {2n^{2}+n}}{\sqrt {2n^{2}+2n}}=\lim _{n\rightarrow \infty }\left( \dfrac {2n^{2}+n}{2n^{2}+2n}\right) ^{\dfrac {1}{2}}$
$\lim _{n\rightarrow \infty }\dfrac {2n^{2}+n}{2n^{2}+2n}=\lim _{n\rightarrow \infty }\dfrac {2+\dfrac {1}{n}}{2+\dfrac {2}{n}}$
since $\lim _{n\rightarrow \infty }\dfrac {1}{n}=0$, $\lim _{n\rightarrow \infty }\dfrac {2n^{2}+n}{2n^{2}+2n}=1$
hence $\lim _{n\rightarrow \infty }\left( \dfrac {2n^{2}+n}{2n^{2}+2n}\right) ^{\dfrac {1}{2}}=\left( 1\right) ^{\dfrac {1}{2}}=1$ (by composite rule)
hence $\sqrt {2n^{2}+n}=\sqrt {2n^{2}+2n}$ as $n\rightarrow \infty $
so $\lim _{n\rightarrow \infty }\left( \sqrt {2n^{2}+n}-\sqrt {2n^{2}+2n}\right) =0$
| Multiplying it by
$$\frac{\sqrt{2n^2+n}+\sqrt{2n^2+2n}}{\sqrt{2n^2+n}+\sqrt{2n^2+2n}}$$
gives
$$\begin{align}&\lim_{n\to\infty}(\sqrt{2n^2+n}-\sqrt{2n^2+2n})\\\\&=\lim_{n\to\infty}(\sqrt{2n^2+n}-\sqrt{2n^2+2n})\cdot\frac{\sqrt{2n^2+n}+\sqrt{2n^2+2n}}{\sqrt{2n^2+n}+\sqrt{2n^2+2n}}\\\\&=\lim_{n\to\infty}\frac{(2n^2+n)-(2n^2+2n)}{\sqrt{2n^2+n}+\sqrt{2n^2+2n}}\\\\&=\lim_{n\to\infty}\frac{-n}{\sqrt{2n^2+n}+\sqrt{2n^2+2n}}\\\\&=\lim_{n\to\infty}\frac{-1}{\sqrt{2+\frac 1n}+\sqrt{2+\frac{2}{n}}}\\\\&=-\frac{1}{2\sqrt 2}\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1529154",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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Find the coefficient of partial expansion $\frac{2x+3}{(1-x)(1+0.5x+0.5x^2)}$ I want to decompose the equation:
$$\frac{2x+3}{(1-x)(1+0.5x+0.5x^2)}=\frac{A}{1-x}+\frac{B}{1+0.5x+0.5x^2}$$
I found $A$ by multiple both side with $1-x$ and plug $x=1$. However, it is so difficult to find $B$. Could you help me to find $B$
| \begin{equation*}
f(x)=\frac{x+\frac{3}{2}}{(x-1)(2+x+x^{2})}=\frac{A}{x-1}+\frac{Bx+C}{%
x^{2}+x+2}
\end{equation*}
cover-up method leads to
\begin{equation*}
A=\left. f(x)\left( x-1\right) \right\vert _{x=1}=\left. \frac{x+\frac{3}{2}%
}{(2+x+x^{2})}\right\vert _{x=1}=\frac{\frac{5}{2}}{4}=\frac{5}{8}.
\end{equation*}
generalized cover-up method leads to
\begin{equation*}
Bx+C=\left. f(x)\left( x^{2}+x+2\right) \right\vert _{x^{2}+x=-2}=\left.
\frac{x+\frac{3}{2}}{(x-1)}\right\vert _{x^{2}+x=-2}
\end{equation*}
To find $B$ and $C$, multiply bottom and top of the fraction $\frac{x+\frac{3%
}{2}}{(x-1)}$ by $(x-s)$ for some appropriate $s$ such that the denominator
(bottom) will be as follows
\begin{equation*}
(x-1)(x-s)=x^{2}-(s+1)x+s=(x^{2}+x)+s
\end{equation*}
so we choose
\begin{equation*}
-(s+1)=+1,\ then\ s=-2
\end{equation*}
so
\begin{eqnarray*}
Bx+C &=&\left. f(x)\left( x^{2}+x+2\right) \right\vert _{x^{2}+x=-2}=\left.
\frac{x+\frac{3}{2}}{(x-1)}\right\vert _{x^{2}+x=-2} \\
&=&\left. \frac{\left( x+\frac{3}{2}\right) (x+2)}{(x-1)(x+2)}\right\vert
_{x^{2}+x=-2} \\
&=&\left. \frac{x^{2}+2x+\frac{3}{2}x+3}{\left( x^{2}+x\right) -2}%
\right\vert _{x^{2}+x=-2} \\
&=&\left. \frac{(x^{2}+x)+\frac{5}{2}x+3}{\left( x^{2}+x\right) -2}%
\right\vert _{x^{2}+x=-2},\ replace\ (x^{2}+x)\ by\ -2 \\
&=&\left. \frac{-2+\frac{5}{2}x+3}{-2-2}\right\vert _{x^{2}+x=-2} \\
&=&\frac{\frac{5}{2}x+1}{-4} \\
&=&-\frac{5}{8}x-\frac{1}{4},\ then\ B=-\frac{5}{8},\ and\ C=-\frac{1}{4},
\end{eqnarray*}
therefore
\begin{equation*}
\frac{x+\frac{3}{2}}{(x-1)(2+x+x^{2})}=\frac{5/8}{x-1}+\frac{(-5/8)x+(-1/4)}{%
x^{2}+x+2}.\ \ \ \ \blacksquare
\end{equation*}
EDIT: Actualy, when we replace $x^2+x$ by $-2$ we do not need to know what is the value of $x$ itself!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1529509",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Equations of the two lines through the origin which intersect the line $\frac{x-3}{2}=\frac{y-3}{1}=\frac{z}{1}$ at an angle of $\frac{\pi}{3}$ Find the equations of the two lines through the origin which intersect the line $\frac{x-3}{2}=\frac{y-3}{1}=\frac{z}{1}$ at an angle of $\frac{\pi}{3}$.
Let the direction ratios of the two required lines be $(a_1,b_1,c_1)$ and $(a_2,b_2,c_2)$.
Therefore the two equations are
$\frac{x-0}{a_1}=\frac{y-0}{b_1}=\frac{z-0}{c_1}$ and $\frac{x-0}{a_2}=\frac{y-0}{b_2}=\frac{z-0}{c_2}$
As these lines are making an angle of $\frac{\pi}{3}$ with $\frac{x-3}{2}=\frac{y-3}{1}=\frac{z}{1}$.
So,$\cos\frac{\pi}{3}=\frac{2a_1+b_1+c_1}{\sqrt6\sqrt{a_1^2+b_1^2+c_1^2}}$
And $\cos\frac{\pi}{3}=\frac{2a_2+b_2+c_2}{\sqrt6\sqrt{a_2^2+b_2^2+c_2^2}}$
But i am stuck here.How i can solve three variables with one equation.The book gives answer as $\frac{x}{1}=\frac{y}{2}=\frac{z}{-1}$ and $\frac{x}{-1}=\frac{y}{1}=\frac{z}{-2}$
Please help me.
| Hint:
Write the equation of the line in parametric form, so that an arbitrary point is expressed in terms of just one letter. Then you can use the dot product to form a quadratic equation in the parameter giving you two solutions.
So we have$$\frac{x-3}{2}=\frac{y-3}{1}=\frac z1=\lambda$$
$$\Rightarrow \underline{r}=\left(\begin{matrix}x\\y\\z\end{matrix}\right)=\left(\begin{matrix}2\lambda+3\\ \lambda+3\\ \lambda\end{matrix}\right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1530416",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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What is $ \lim_{(x,y) \rightarrow (0,0)} \frac{x^3-y^3}{x^2-y^2} $ Find the following limit
$$ \lim_{(x,y) \rightarrow (0,0)} \frac{x^3-y^3}{x^2-y^2}$$
This question has been bugging me for some time. Couldn't find it anywhere on the Internet
| You can cancel a common factor:
$$\frac{x^3-y^3}{x^2-y^2}=\frac{(x-y)(x^2+xy+y^2)}{(x-y)(x+y)}=\frac{x^2+xy+y^2}{x+y} $$
(This also continuously extends the function to most of the line $x=y$).
Along any straight line $(x,y)=(at,bt)$, $t\to 0$, this is $\frac{a^2 +ab+b^2}{a+b}\cdot t$ and tends to $0$. But this of course does not work for the case $b=-a$ where the expression is undefined.
However, convergence along straight lines is not convergence!
If we consider $(x,y)=(t+t^2,-t)$, $t\to 0$, we arrive at
$$\frac{(t+t^2)^2+(t+t^2)(-t)+(-t)^2}{t+t^2-t} =\frac{t^2+t^3+t^4}{t^2}=1+t+t^2\to 1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1531919",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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} |
Prove $\lim_{n\rightarrow \infty} 2^n \sqrt{2-x_n}=\pi$ using the half angle identities. Given is the sequence $x_1=0,\; x_{n+1}=\sqrt{2+x_n}$. Prove: $$\lim_{n\rightarrow \infty} 2^n \sqrt{2-x_n}=\pi$$
Hint:
Use the following formulas:
$$\cos\left(\frac{x}{2}\right)=\sqrt{\frac{1+\cos x}{2}}$$
$$\sin\left(\frac{x}{2}\right)=\sqrt{\frac{1-\cos x}{2}}$$
Any idea how to solve this problem?
| Note that as $x_1 \le 2$, $x_n \le 2 \forall n$. This can be shown inductively.
Thus, we may write $x_n = 2 \cos \theta_n$ for some $\theta_n \in [0, \pi/2]$, with $\theta_1 = \pi/2$
Now, $2 \cos \theta_{n+1} = \sqrt{2(1+ \cos \theta_n)} = 2 \cos \frac{\theta_n}2 \implies \theta_{n+1} = \frac{\theta_n}{2} = \frac{\theta_1}{2^n} = \frac{\pi}{2^{n+1}}$.
This implies that $2^n\sqrt{2 - x_n} = 2^{n+1} \sin \frac{\pi}{2^{n+1}}$. Finish off by using $\frac{\sin x}{x} \to 1$ as $x \to 0$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Calculate $\lim_{x \to \infty} x^{3/2}(\sqrt[3]{x^3 + 1} - \sqrt[3]{x^3 - 1})$ Calculate $$\lim_{x \to \infty} x^{3/2}(\sqrt[3]{x^3 + 1} - \sqrt[3]{x^3 - 1})$$
I found out that we can simplify it by multiplying:
$$\lim_{x \to \infty} x^{3/2}\frac {{x^3 + 1} - (x^3 - 1)} {\sqrt[3]{(x^3 + 1)^2} + \sqrt[3]{(x^3 + 1)(x^3 - 1)} + \sqrt[3]{(x^3 - 1)^2}} = \lim_{x \to \infty} x^{3/2}\frac {2} {\sqrt[3]{(x^3 + 1)^2} + \sqrt[3]{(x^3 + 1)(x^3 - 1)} + \sqrt[3]{(x^3 - 1)^2}}$$
So what am I supposed to do next?
| Dividing both the numerator and the denominator by $x^{3/2}$ gives
$$\frac{2}{(x^{3/2}+2x^{-3/2}+x^{-9/2})^{1/3}+(x^{3/2}-x^{-9/2})^{1/3}+(x^{3/2}-2x^{-3/2}+x^{-9/2})^{1/3}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1537313",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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String in an kleene star alphabet Let Σ = {a, b}.
How many strings of length 10 are in the language (bb + aab)*?
If this a matter of writing them out or is there a formula to it?
| We could also associate a geometric series with $$(bb+aab)^\star=\varepsilon+(bb+abb)+(bb+abb)^2+\ldots$$ Since each word in this language is uniquely identified by the occurrences of $bb$ and $aab$ parts and since we are only interested in the length of the substrings $bb$ and $aab$, we consider the series
\begin{align*}
\frac{1}{1-(x^2+x^3)}=1+(x^2+x^3)+(x^2+x^3)^2+\ldots
\end{align*}
and calculate the coefficient of $x^{10}$ of this series:
\begin{align*}
[x^{10}]\frac{1}{1-x^2(1+x^3)}&=[x^{10}]\sum_{k=0}^{\infty}x^{2k}(1+x)^k\\
&=[x^{10}]\sum_{k=4}^5x^{2k}(1+x)^k\\
&=\sum_{k=4}^{5}[x^{10-2k}](1+x)^k\\
&=\binom{5}{2}+\binom{5}{0}\\
\end{align*}
In short: We have $\binom{5}{0}=1$ possibilities to generate a word of length $10$ with the string $bb$
\begin{align*}
(bb)(bb)(bb)(bb)(bb)
\end{align*}
Next we can replace two strings $abb$ with three of $bb$ giving $\binom{5}{2}$ and there are no further possibilities.
| {
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Show $p$ prime s.t. $p \not\equiv 1 \mod 3$ is represented by the binary quadratic equation. I am working on the following question:
Let $p>3$ be a prime such that $p \not\equiv 1 \mod 3$. Show that $p$ is not represented by the binary quadratic equation $f(x, y) = x^2 + xy + y^2$.
I would appreciate any help.
| Remember that you can just add up mods.
If $x \equiv 1 \bmod 3$ and $y \equiv 1 \bmod 3$, then $x^2 + xy + y^2 \equiv 0 \bmod 3$.
If $x \equiv 1 \bmod 3$ and $y \equiv 2 \bmod 3$, then $x^2 + xy + y^2 \equiv 1 \bmod 3$.
If $x \equiv 1 \bmod 3$ and $y \equiv 0 \bmod 3$, then $x^2 + xy + y^2 \equiv 1 \bmod 3$.
If $x \equiv 2 \bmod 3$ and $y \equiv 2 \bmod 3$, then $x^2 + xy + y^2 \equiv 0 \bmod 3$.
Obviously we needn't worry about both $x$ and $y$ satisfying $0 \mod 3$, and as for the other possibilities with $x$ and $y$ switched, we can ignore them "without any loss of generality."
Since $x^2 + xy + y^2 \equiv 2 \bmod 3$ is impossible, no $p \equiv 2 \bmod 3$ can be represented by it.
| {
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"source": "stackexchange",
"question_score": "6",
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A new approach to find value of $x^2+\frac{1}{x^2}$ When I was teaching in a college class, I write this question on board.
If we now $\bf{x+\dfrac{1}{x}=4}$ show the value of $\bf{x^2+\dfrac{1}{x^2}=14}$
Some student asks me for a multi idea to show or prove that.
I tried these : $$ $$1 :$$\left(x+\frac{1}{x}\right)^2=4^2\\x^2+\frac{1}{x^2}+2x\times \frac{1}{x}=16\\x^2+\frac{1}{x^2}=16-2 $$
2:solving quadratic equation ,and putting one of roots$$x+\frac{1}{x}=4\\\frac{x^2+1}{x}=4\\x^2+1=4x\\x^2-4x+1=0\\x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}=\frac{-(-4)\pm \sqrt{(-4)^2-4(1)(+1)}}{2}=\\x=\frac{4\pm \sqrt{12}}{2}=2\pm \sqrt{3}\\x=2+\sqrt3 \to x^2=4+4\sqrt3+3=7+4\sqrt3\\x^2+\frac{1}{x^2}=7+4\sqrt3+\frac{1}{7+4\sqrt3}=\\7+4\sqrt3+\frac{1}{7+4\sqrt3}\cdot\frac{7-4\sqrt3}{7-4\sqrt3}=\\7+4\sqrt3+\frac{7-4\sqrt3}{49-16\cdot 3}=\\7+4\sqrt3+\frac{7-4\sqrt3}{1}=14$$
3: visual approach .assume side length of a square is $x+\frac{1}{x}=4$
now I am looking for new idea to proof.Any hint will be appreciated.(more visual proof - geometrical - trigonometrical - using complex numbers ...)
| Here's a trigonometric approach:
Let $x=\tan\theta=\sin\theta/\cos\theta$. Then
$$4=x+{1\over x}={\sin\theta\over\cos\theta}+{\cos\theta\over\sin\theta}={1\over\sin\theta\cos\theta}={2\over\sin2\theta}$$
so $\sin2\theta=1/2$. Now
$$x^2+{1\over x^2}={\sin^2\theta\over\cos^2\theta}+{\cos^2\theta\over\sin^2\theta}={1\over\cos^2\theta}-1+{1\over\sin^2\theta}-1={1\over\sin^2\theta\cos^2\theta}-2\\
={4\over\sin^22\theta}-2=16-2=14$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Find the limit $\lim\limits_{x\to\infty} f(x)= \lim\limits_{x \to \infty} \left(\frac{x}{x+1}\right)^x$ I need to find the following:
$$\lim_{x\to\infty} f(x)= \lim_{x \to \infty}\left (\frac{x}{x+1} \right )^x$$
I know that this limit = $\frac{1}{e}$ from plugging it into a calculator, but I have to prove it without using the fact that:
$$\lim_{x\to\infty} f(x)=\lim_{x \to \infty}\left (\frac{x}{x+k} \right )^x=\frac{1}{e^k}$$
I started by exponentiating:
$$\lim_{x\to\infty} f(x)=\lim_{x \to \infty}e^{\ln \left (\frac{x}{x+1} \right )^x}$$
and from here I've dropped the exponent $x$ in front of the $\ln$, and from here I'm getting stuck. Should I separate the $\ln$ like this?
$$\lim_{x\to\infty} f(x)=\lim_{x \to \infty}e^{x(\ln(x)-\ln(x+1))}$$
This doesn't seem to be leading me down the right path, but I'm not sure how else to do it. Is there a way to apply L'Hopital? If so, how?
| First you set the entire equation equal to y:
$$
(\frac {x}{(x+1)})^x = y
$$
We can then insert both sides into $ln(x)$:
$$
ln((\frac x{(x+1)})^x) = ln(y)
$$
Then pull out the x power:
$$
xln(\frac x{(x+1)}) = ln(y)
$$
Split the natural log:
$$
xln(x) - xln(x+1) = ln(y)
$$
Shift the left term into the bottom of a fraction and add it to the other term:
$$
\frac {1}{\frac 1{xln(x)}} - xln(x+1) = ln(y)
$$
$$
\frac {1}{\frac 1{xln(x)}} - \frac {\frac {xln(x+1)}{xln(x)}}{\frac 1{xln(x)}} = ln(y)
$$
$$
\frac {1 - \frac {ln(x+1)}{ln(x)}}{\frac 1{xln(x)}} = ln(y)
$$
Taking the limits of the top and bottom yields the form $\frac 00$. Therefore, L'hopetal's Rule can now be applied:
$$
\frac {\frac {(x+1)ln(x)-xln(x+1)}{ln^2(x)}}{\frac {ln(x) + \frac xx}{(xln(x))^2}} = ln(y)
$$
A little simplification and canceling...
$$
\frac {x^3ln(x)+x^2ln(x)-x^3ln(x+1)}{ln(x) + 1} = ln(y)
$$
L'Hopetals rule still applies.
$$
\frac {2x^2ln(x) + \frac {x^3}x + 2xln(x) + \frac {x^2}x - 3x^2ln(x+1) - \frac {x^3}{x+1}}{\frac 1x} = ln(y)
$$
Simplify:
$$
2x^3ln(x) + x^3 + 2x^2ln(x) + x^2 - 3x^3ln(x+1) - \frac {x^4}{x+1} = ln(y)
$$
Put everything back under $x+1$
$$
\frac {(2x^3ln(x) + x^3 + 2x^2ln(x) + x^2 - 3x^3ln(x+1))(x+1)}{x+1} - \frac {x^4}{x+1} = ln(y)
$$
$$
\frac {2x^4ln(x) + 2x^3ln(x) + x^4 + x^3 + 2x^2ln(x) + 2x^3ln(x) + x^2 + x^3 - 3x^3ln(x+1) - 3x^4ln(x+1) - x^4}{x+1} = ln(y)
$$
$$
\frac {2x^4ln(x) + 4x^3ln(x) + 2x^3 + 2x^2ln(x) + x^2 - 3x^3ln(x+1) - 3x^4ln(x+1)}{x+1} = ln(y)
$$
$$
\frac {ln(x)(2x^4 + 4x^3 + 2x^2) + 2x^3 + x^2 - ln((x+1)(3x^3 + 3x^4)}{x+1} = ln(y)
$$
$$
\frac {ln((x)^{2x^4 + 4x^3 + 2x^2}) + 2x^3 + x^2 - ln((x+1)^{3x^3 + 3x^4}}{x+1} = ln(y)
$$
$$
\frac {ln(\frac{(x)^{2x^4 + 4x^3 + 2x^2}}{(x+1)^{3x^3 + 3x^4}}) + 2x^3 + x^2 }{x+1} = ln(y)
$$
$$
2x^2 + \frac {ln(\frac{(x)^{2x^4 + 4x^3 + 2x^2}}{(x+1)^{3x^3 + 3x^4}}) - x^2 }{x+1} = ln(y)
$$
This will continue on and on until you get $\frac 1e$
At this point I am stuck, but I believe I have well illustrated the way you would do this WITHOUT substituting values in other known limits. Unfortunately it is too late at night for me and I just can't see how to reduce it so that L'hopetals rule isn't infinite. (At least, it appears like it'd be applied infinitely from my perspective.)
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Power series summation Trying to find the sum of the following infinite series:
$$ \displaystyle\sum_{n=1}^{\infty}\frac{{(-1)}^{n-1}}{(2n-1)3^{n-1}}$$
Any ideas on how to find this sum?
| Note that for suitable $t$,
$$\frac{1}{1+t^2}=1-t^2+t^4-t^6+\cdots.$$
Integrate term by term from $0$ to $x$. We get
$$\arctan x=x-\frac{x^3}{3}+\frac{x^5}{5}-\frac{x^7}{7}+\cdots.$$
Divide by $x$. We get
$$\frac{\arctan x}{x}=1-\frac{x^2}{3}+\frac{x^4}{5}-\frac{x^6}{7}+\cdots.$$
Finally, let $x=\frac{1}{\sqrt{3}}$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Is $a\sqrt[3]{2} + b\sqrt[3]{4}$ irrational? I need to prove that $$ a\sqrt[3]{2} + b\sqrt[3]{4}$$ is irrational, while $a$,$b $ are non zero rationals.
I know that $\sqrt[3]{2} + \sqrt[3]{4}$ is irrational and I also know how to prove it, but I can't think of any reasonable implication that would state: $a\sqrt[3]{2} + b\sqrt[3]{4}$ is rational $\implies$ $\sqrt[3]{2} + \sqrt[3]{4}$ is rational, so I could show it is a contradiction.
| If $b\ne0$, then $a\sqrt[3]{2} + b\sqrt[3]{4}$ is rational iff $\sqrt[3]{2}$ is a root of a quadratic polynomial over $\mathbb Q$.
$\sqrt[3]{2}$ is a root of $X^3-2$, which is irreducible over $\mathbb Q$.
So, every polynomial having $\sqrt[3]{2}$ as root must be a multiple of $X^3-2$. In particular, $\sqrt[3]{2}$ is not a root of a quadratic polynomial over $\mathbb Q$.
Therefore,
if $a\sqrt[3]{2} + b\sqrt[3]{4}$ is rational, then $b=0$.
But $\sqrt[3]{2}$ is irrational, and so $a=0$.
All this can be summarized as:
$1, \sqrt[3]{2}, \sqrt[3]{4}$ are linearly independent over $\mathbb Q$ and so the only way that $a\sqrt[3]{2} + b\sqrt[3]{4}$ is rational is when $a=b=0$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Double radical proof I'm trying to prove that
$$
\sqrt{A+\sqrt{B}}=\sqrt{\frac{A+C}{2}}+\sqrt{\frac{A-C}{2}}
$$
With
$$
C=\sqrt{A^2 - B}
$$
How can I handle this?
Edit: obviously is easy that this holds when you know the r.h.s., but my question is: how to get the r.h.s. when you only know the l.h.s.
| Well assume that $\sqrt{a+\sqrt{b}}$ can be written as sum of 2 square roots
$$\sqrt{a+\sqrt{b}}=\sqrt{x}+\sqrt{y}\\a+\sqrt{b}=x+y+\sqrt{4xy}\\a=x+y\\b=4xy\\x=a-y\\b=4(a-y)y\\b=4ay-4y^2\\4y^2-4ay+b=0\\y_{1,2}=\frac{4a\pm\sqrt{16a^2-16b}}{8}\\y_{1,2}=\frac{a\pm\sqrt{a^2-b}}{2}\\x_{1,2}=\frac{a\mp\sqrt{a^2-b}}{2}$$
Now it $x_1=y_2$ and $x_2=y_1$ so that doesn't matter at all,now set $C=\sqrt{a^2-b}$ and you get your formula
| {
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"timestamp": "2023-03-29T00:00:00",
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Show this function of two variables $f(x,y) = \frac{2x^2y^3}{x^4+y^6}$ continuous at (0,0) Is the function $f(x,y) = \frac{2x^2y^3}{x^4+y^6}$ continuous at $(0, 0)$?
Trying to use the epsilon-delta definition of continuity to prove this but can't figure it out.
$\sqrt{x^2+y^2} < \delta$ implies $\left|\frac{2x^2y^3}{x^4+y^6}\right| < \epsilon$.
Using polar coordinates $r < \delta$ implies
$$\left|\frac{2r\cos^2(\theta)\sin^2(\theta)}{\cos^4(\theta)+r^2\sin^6(\theta)}\right| < \epsilon^4$$
$$\left|\frac{2r\cos^2(\theta)\sin^2(\theta)}{\cos^4(\theta)+r^2\sin^6(\theta)}\right| \leq \left|\frac{2r}{\cos^4(\theta)+r^2\sin^6(\theta)}\right|$$
But from here I cant simplify the equation further.
| Note $f(x,y)$ is continuous at $(0,0)$ if $\lim_{(x,y)\to (0,0)}f(x,y)=f(0,0)$, which is equivalent to for all $(x_n,y_n)$ such that $x_n\to 0,y_n\to 0$, we have $\lim f(x_n,y_n)$ exists and equal to $f(0,0)$.
In your example, the function is not continuous at $(0,0)$. To show this, we show $\lim_{(x,y)\to (0,0)}f(x,y)$ does not exist. In particular, we find $x^1_n\to 0,y^1_n\to 0$ and $x^2_n\to 0,y^2_n\to 0$, but $f(x_n^1,y_n^1)$ and $f(x_n^2,y_n^2)$ converges to different valueS.
Let $x^1_n=(\frac{1}{n})^3\to 0 ,y^1_n=(\frac{1}{n})^2\to 0$, you can check $f(x_n^1,y_n^1)\to 2$.
Let $x^2_n=(\frac{1}{n})^3\to 0 ,y^2_n=2(\frac{1}{n})^2\to 0$, you can check $f(x_n^2,y_n^2)\to 16$
| {
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What's the expression $( \cos 6x + 6 \cos 4x + 15 \cos 2x + 10 ) / ( \cos 5x + 5 \cos 3x + 10 \cos x ) $ equal to? $$\frac{ \cos 6x + 6 \cos 4x + 15 \cos 2x + 10 }{ \cos 5x + 5 \cos 3x + 10 \cos x }$$
My approach so far : Tried to represent the denominator as a factor of numerator by manipulating numerator's $\cos 6x = \cos (5x+x)$ , $\cos 4x = \cos (3x+x)$ , so on .. but then $\sin x$ come up which make it more complex to solve .
The options for the answer are:
A) $\cos 2x$.
B) $2 \cos x$.
C) $\cos^2 x$.
D) $1 + \cos x$
| $$
2^6 \cos^6 x = (e^{ix}+e^{-ix})^6=e^{6ix}+6e^{4ix}+15e^{2ix} +20+15e^{-2ix}+6e^{-4x}+e^{-6ix} \\
= 2(\cos 6x + 6 \cos 4x +15 \cos 2x +10)
$$
and so forth
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1545398",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 0
} |
Find $A^x$ where A is a $2\times 2$ matrix So, I need to calculate $A^x$ where
$A=\pmatrix{4&-3\\3&-2}$
First of all, $A^x = e^{x\ln A}$, so I get that
$\ln A = \pmatrix{2&0\\-1&1}$
Then I use that
$f(A)= T^{-1}f(J)T$
Here $J$ is the Jordan Canonical form. I got that
$J=\pmatrix{1&1\\0&1}$
and
$T=\pmatrix{3&1\\2&0}$, $T^{-1}=\pmatrix{0&\frac{1}{2}\\1&\frac{-3}{2}}$
What should I do next? Any help would be greatly appreciated.
| The result is this one
\begin{equation}
A^x=\left(\begin{array}{c}
\begin{array}{ccccc}
3x+1 & -3x\\
3x & -3x+1\\
\end{array}\end{array}\right)
\end{equation}
Now I have to exit, but as soon as I'm back I will explain it if needed...
Edit>>:
There are many ways to reach the same answer. The more plain one is that:
\begin{equation}
A^x=\left(\begin{array}{c}
\begin{array}{ccccc}
ax+b & cx+d\\
ex+f & gx+h\\
\end{array}\end{array}\right)
\end{equation}
since $A^0=I$ we have $d,f=0$ and $b,h=1$, and since $A^1=A$ we have $c=-3$ and $e=3$.
Then imposing
$$
a+b=4, g+h=-2,
$$
we obtain $a=3$ and $g=-3$.
So for example
\begin{equation}
A^{10}=\left(\begin{array}{c}
\begin{array}{ccccc}
31 & -30\\
30 & -29\\
\end{array}\end{array}\right)
\end{equation}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1546154",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Improper integral $\int_{-\infty}^{+\infty}\frac{1}{a x^2 + bx + c} dx$, $a > 0,a x^2 + bx + c>0$ I have a question about improper integral. If you can help me , I appreciate that.
If a > 0 and the graph of $y=a x^2 + bx + c$ lies entirely above the x-axis, show that
$$
\int_{-\infty}^{+\infty} \frac{dx}{a x^2 + bx + c}=\frac{2 \pi}{\sqrt{4ac - b^2}}.
$$
| Note that if $ax^2+bx+c$ has real roots then the integral doesn't exists, since we will end up with an integral of the form $$\dfrac1{a}\int_{-\infty}^{\infty} \dfrac{dx}{(x-\alpha)(x-\beta)}$$where $\alpha, \beta \in \mathbb{R}$, which clearly diverges. Hence, we can assume that the discriminant is strictly negative, i.e., $4ac -b^2 > 0$.
Given that $4ac-b^2>0$, we have $ax^2+bx+c = a\left(x+\dfrac{b}{2a}\right)^2 + c - \dfrac{b^2}{4a}$. Hence, letting $y=x+\dfrac{b}{2a}$, we have
\begin{align}
\int_{-\infty}^{\infty} \dfrac{dx}{ax^2+bx+c} & = \int_{-\infty}^{\infty} \dfrac{dy}{ay^2 + c-\dfrac{b^2}{4a}} = \dfrac1{a} \dfrac1{\sqrt{c/a-b^2/(4a^2)}}\left. \arctan\left(\dfrac{y}{\sqrt{c/a-b^2/(4a^2)}}\right)\right \vert_{-\infty}^{\infty}\\
& = \dfrac{2\pi}{\sqrt{4ac-b^2}}
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1547421",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Let $a$, $b$, and $c$ be positive real numbers with $aSource: AoPS
My attempt:
$$(a+b+c)^2=a^2+b^2+c^2+2(ab+bc+ca)\implies144=50+2(ab+bc+ca) $$
$$\implies ab+bc+ca=47$$
and
$$a^3+b^3+c^3-3abc=(a^2+b^2+c^2-(ab+bc+ca))(a+b+c)\implies216-3abc=12(50-47)$$$$\implies abc=(216-36)/3=60$$
So, $ab=\frac{60}{c}$
Now, $$a^3+b^3+c^3=216\implies a^3+b^3=216-c^3$$$$\implies(a+b)(a^2+b^2-ab)=216-c^3$$
In this equation, I substituted $a+b = 12 - c$, $a^2+b^2=50-c^2$, $ab=60/c$, and got a fourth degree polynomial in $c$ with complex roots which is terribly wrong.
Question: Where did I go wrong and how should I proceed?
| Note that you have found the three elementary symmetric polynomials in the variables $a,b,c$, which determine a cubic polynomial whose roots are $a,b,c$. So $a,b,c$ are the solutions to
\begin{equation*}
\begin{aligned}
&\mathrel{\phantom{=}} x^3 - (a+b+c)x^2 + (ab+bc+ac)x - abc \\
&= x^3 - 12x^2 + 47x - 60 \\
&= (x - 3)(x - 4)(x - 5).
\end{aligned}
\end{equation*}
Given the condition $a < b < c$, we have $a = 3, b = 4, c = 5$, and so $a + 2b + 3c = 26$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1549587",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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Easy way of Compute this limit Easy way to compute: $$\lim_{x\to \:0\:}\left(\left(\frac{a^x-x\cdot \ln\left(a\right)}{b^x-x\cdot \ln\left(b\right)}\right)^{\frac{1}{x^2}}\right)$$
| Let the desired limit be $L$. Then we have
\begin{align}
\log L &= \log\left(\lim_{x \to 0}\left(\frac{a^{x} - x\log a}{b^{x} - x\log b}\right)^{1/x^{2}}\right)\notag\\
&= \lim_{x \to 0}\log\left(\frac{a^{x} - x\log a}{b^{x} - x\log b}\right)^{1/x^{2}}\text{ (via continuity of log)}\notag\\
&= \lim_{x \to 0}\frac{1}{x^{2}}\log\left(\frac{a^{x} - x\log a}{b^{x} - x\log b}\right)\notag\\
&= \lim_{x \to 0}\frac{1}{x^{2}}\log\left(1 + \frac{a^{x} - x\log a}{b^{x} - x\log b} - 1\right)\notag\\
&= \lim_{x \to 0}\frac{1}{x^{2}}\log\left(1 + \frac{a^{x} - b^{x} - x(\log a - \log b)}{b^{x} - x\log b}\right)\notag\\
&= \lim_{x \to 0}\frac{1}{x^{2}}\cdot\dfrac{a^{x} - b^{x} - x(\log a - \log b)}{b^{x} - x\log b}\cdot\dfrac{\log\left(1 + \dfrac{a^{x} - b^{x} - x(\log a - \log b)}{b^{x} - x\log b}\right)}{\dfrac{a^{x} - b^{x} - x(\log a - \log b)}{b^{x} - x\log b}}\notag\\
&= \lim_{x \to 0}\frac{1}{x^{2}}\cdot\dfrac{a^{x} - b^{x} - x(\log a - \log b)}{b^{x} - x\log b}\cdot 1\notag\\
&= \lim_{x \to 0}\frac{a^{x} - b^{x} - x(\log a - \log b)}{x^{2}}\notag\\
&= \lim_{x \to 0}\frac{a^{x} - 1 - x\log a}{x^{2}} - \frac{b^{x} - 1 - x\log b}{x^{2}}\notag\\
&= \lim_{x \to 0}(\log a)^{2}\cdot\frac{a^{x} - 1 - x\log a}{(x\log a)^{2}} - (\log b)^{2}\cdot\frac{b^{x} - 1 - x\log b}{(x\log b)^{2}}\notag\\
&= (\log a)^{2}\lim_{u \to 0}\frac{e^{u} - 1 - u}{u^{2}} - (\log b)^{2}\lim_{v \to 0}\frac{e^{v} - 1 - v}{v^{2}}\text{ (putting }u = x\log a, v = x\log b)\notag\\
&= \{(\log a)^{2} - (\log b)^{2}\}\lim_{u \to 0}\frac{e^{u} - 1 - u}{u^{2}}\notag
\end{align}
The last limit is easily seen to be $1/2$ either via one application of L'Hospital's Rule or through Taylor's series. Hence $$\log L = \frac{(\log a)^{2} - (\log b)^{2}}{2}$$ and therefore $$L = \exp\left(\frac{(\log a)^{2} - (\log b)^{2}}{2}\right)$$
| {
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"url": "https://math.stackexchange.com/questions/1552536",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Integral of $\frac{1}{x^2+4}$ Different approach underneath is a brief method of partial fractions integration on the problem given in the title
Using a standard trigonometric result it is known that:
$$ \int \frac{1}{x^2+4}dx=\frac{1}{2}\tan^{-1}(\frac{x}{2})+C$$
But also:
$$\frac{1}{x^2+4}=\frac{A}{x+2i}+\frac{B}{x-2i}$$
Hence using partial fractions,
$$1=A(x-2i)+B(x+2i)$$
Let $x=2i$
$$\therefore 1=4Bi$$
$$\text{Hence } B=\frac{-i}{4}$$
Let $x=-2i$
$$\therefore 1=-4Ai$$
$$\text{Hence } A=\frac{i}{4}$$
Hence,
$$\frac{1}{x^2+4}=\frac{i}{4(x+2i)}-\frac{i}{4(x-2i)}$$
Or,
$$\frac{1}{x^2+4}=\frac{i}{4}(\frac{1}{x+2i}-\frac{i}{x-2i})$$
Hence, it is quite easy to see that:
\begin{align*}
\int \frac{1}{x^2+4}dx &=\int \frac{i}{4}(\frac{1}{x+2i}-\frac{1}{x-2i}) dx\\
&=\frac{i}{4}\int \frac{1}{x+2i}-\frac{1}{x-2i} dx\\
&=\frac{i}{4}(\log(x+2i)-\log(x-2i))+C\\
\end{align*}
We know that the principal value of log of a complex number can be calculated by the following formula:
$$\log(x+yi)=\log(x^2+y^2)+arg(x+yi)$$
Hence,
\begin{align*}
\int \frac{1}{x^2+4}dx &=\frac{i}{4}(\log(x+2i)-\log(x-2i))+C\\
&=\frac{i}{4}(\log(x^2+4)+\arg(x+2i)-\log(x^2+4)-\arg(x-2i)+C\\
&=\frac{i}{4}(\arg(x+2i)-\arg(x-2i))+C\\
\end{align*}
Now, consider cases. If x is bigger than 0, the argument of a complex number is always defined as $\arg(x+yi)=\tan^{-1}(\frac{y}{x})$
So our integral becomes
$$\int \frac{1}{x^2+4}dx=\frac{i}{4}(\tan^{-1}(\frac{2}{x})-\tan^{-1}(\frac{-2}{x}))+C$$
And since arctan is an odd function
\begin{align*}
\int \frac{1}{x^2+4}dx&=\frac{i}{4}(2\tan^{-1}(\frac{2}{x}))+C\\&=\frac{i}{2}\tan^{-1}(\frac{2}{x})+C\\
\end{align*}
Now, if $x$ is less than $0$ and $y$ is less than $0$ then argument of a complex number becomes $\tan^{-1}(\frac{y}{x})-\pi$ and if $x$ is less than $0$ and $y$ is more than $0$ the argument becomes $\tan^{-1}(\frac{y}{x})+\pi$ Also as in a previous case becasue arctan is an odd function, the integral becomes
\begin{align*}
\int \frac{1}{x^2+4}dx&=\frac{i}{4}(\tan^{-1}(\frac{2}{x})-\tan^{-1}(\frac{-2}{x})+2\pi)+C\\
&=\frac{i}{4}(2\tan^{-1}(\frac{2}{x})+2\pi)+C\\
&=\frac{i}{2}\tan^{-1}(\frac{2}{x})+D
\end{align*}
Hence
$$\int \frac{1}{x^2+4}dx=\frac{i}{2}(\tan^{-1}\frac{2}{x}), x\in\mathbb R\ \land x\neq0$$
Now, obviously, it is not the same, as the trig identity, it has a pole at x=0 while the original integral doesn't and most importantly it is not a real number. This was done by me purely for recreational purposes but now it frustrates me.
Can it be done this way? Did I basically did a mistake or maybe I missed something crucial?. thanks in advance.
EDIT
After using the correct definition of principal value of log and nice property about arctan (both given to me by you guys)
we know that
$$\log(x+yi)=\frac{1}{2}\log(x^2+y^2)+i\arg(x+yi)$$
and hence
*** becomes
$$\frac{i}{4}(i(\arg(x+2i)-\arg(x-2i)))$$
=$$\frac{-1}{4}((\arg(x+2i)-\arg(x-2i)))$$
So the answe becomes
$$\int \frac{1}{x^2+4}dx=\frac{-1}{2}\tan^{-1}(\frac{2}{x})+D$$
And since $$\tan^{-1}(\frac{2}{x})=\frac{\pi}{2}-\tan^{-1}(\frac{x}{2})$$
the result is $$\int \frac{1}{x^2+4}dx=\frac{-1}{2}(\frac{\pi}{2}-\tan^{-1}(\frac{x}{2}))+D$$
OR$$\int \frac{1}{x^2+4}dx=\frac{1}{2}\tan^{-1}(\frac{x}{2})+E$$ as required
Thanks :)
| Also
$$ tan^{-1}(\frac 2x )= cot^{-1}(\frac x2 ) = \frac \pi 2 -tan^{-1}(\frac x2 ) $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1555671",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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Find $\min \left((x-2)^2+y^2\right)$ $s.t. \>x^2\leq ky^2+1$, $x\geq 0$ Consider the problem $$\min \left((x-2)^2+y^2\right)$$ $$s.t. \>x^2\leq ky^2+1$$
$$x\geq 0$$
where $k \in \mathbb{R}$ is a parameter of the problem.
Determine the status of the point $(1,0)$ for each value of $k$. For what values of $k$ is the point $(1,0)$ is a KKT point, local min,global min.
Solution so far
$L(x,\lambda)=(x-2)^2+y^2+\lambda_1(x^2-ky^2-1)-\lambda_2 x$
*
*$2(x-2)+2 \lambda_1 x-\lambda_2=0$
*$2y-2k \lambda_1 y=0$
*$\lambda_1\geq 0, x^2\leq ky^2+1,\lambda_1 (x^2-ky^2-1)=0$
*$\lambda_2\geq 0, x\geq 0, -\lambda_2 x= 0$
case $\lambda_1 >0, \lambda_2=0$
The third condition implies $x^2-ky^2-1=0$ But since $\lambda_2=0$ we have $$2y-2\lambda_1 ky=2y(1-k\lambda_1)=0$$
So $y=0, 1=k\lambda_1$ and $(x^2-1)=0 \implies x=1,-1$
Observe $2(1-2)+2\lambda_1=0 \implies \lambda_1=1 \implies k=1$
$@(1,0),k=1,\lambda=(1,0)$
| You are looking for the closest point to the point $(2,0)$ in an area limited by the hyperbole $x^2-ky^2=1$ on the right side and the y axis on the left side.
The solutions of the minimisation problem are clearly on the hyperbole.
So $x^2-ky^2=1$, $\lambda_2=0$
If $y \neq 0$ then $k=\frac{1}{\lambda_1}$ and $x(1+\frac{1}{k}) = 2$
$$x=\frac{2k}{k+1}$$
$$y=\pm \sqrt{\frac{(\frac{2k}{k+1})^2-1}{k}}$$
$$y=\pm \sqrt{\frac{3k^2-2k-1}{k(k+1)^2}}$$
So $(1,0)$ is solution when $k$ is so that $1\le (\frac{2k}{k+1}-2)^2+\frac{3k^2-2k-1}{k(k+1)^2}=\frac{3k^2+2k-1}{k(k+1)^2} = \frac{k-1/3}{k(k+1)}$
$$\iff k^2+k\le k-\frac{1}{3}$$
So $(1,0)$ is never a solution
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1556781",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove this binomial identity using the following equality: Use the equation $\frac{(1-x^2)^n}{(1-x)^n} = (1+x)^n$ to prove the following identity:
$\displaystyle \sum_{k=0}^\frac{m}{2}(-1)^k{n\choose k}{n+m-2k-1\choose n-1}={m\choose n}$, $m\leqslant n$ and $m$ even
I am really at a loss for how to do this. Where did the $m$ and condition $m$ even come from?
| Let me just briefly present a different generating function.
Suppose we seek to evaluate
$$\sum_{k=0}^{\lfloor m/2\rfloor}
{n\choose k} (-1)^k {m-2k+n-1\choose n-1}$$
where $m\le n$ and introduce
$${m-2k+n-1\choose n-1} = {m-2k+n-1\choose m-2k}
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^{m-2k+1}} (1+z)^{m-2k+n-1} \; dz$$
which has the property that it is zero when $2k\gt m$ so we may set
the upper limit in the sum to $n,$ getting
$$\frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^{m+1}} (1+z)^{m+n-1}
\sum_{k=0}^n {n\choose k} (-1)^k \frac{z^{2k}}{(1+z)^{2k}} \; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^{m+1}} (1+z)^{m+n-1}
\left(1-\frac{z^2}{(1+z)^2}\right)^n \; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^{m+1}} (1+z)^{m-n-1}
(1+2z)^n \; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{(1+z)^m}{z^{m}}
\frac{1}{z(1+z)}
\frac{(1+2z)^n}{(1+z)^n} \; dz.$$
Now put $$\frac{1+2z}{1+z} = u
\quad\text{so that}\quad
z = -\frac{u-1}{u-2},\; 1+z = -\frac{1}{u-2},\;
\frac{1+z}{z} = \frac{1}{u-1},\;
\\ \quad \frac{1}{z(1+z)} = \frac{(u-2)^2}{u-1}
\quad\text{and}\quad
dz = \frac{1}{(u-2)^2} \; du$$
to get for the integral
$$\frac{1}{2\pi i}
\int_{|u-1|=\gamma}
\frac{1}{(u-1)^m} \frac{(u-2)^2}{u-1} u^n
\frac{1}{(u-2)^2} \; du
\\ = \frac{1}{2\pi i}
\int_{|u-1|=\gamma}
\frac{1}{(u-1)^{m+1}} u^n \; du.$$
This is
$$[(u-1)^m] u^n = [(u-1)^m] \sum_{q=0}^n {n\choose q} (u-1)^q
= {n\choose m}.$$
This solution is more complicated than the obvious one
but it serves to illustrate some aspects of the method.
Concerning the choice of $\epsilon$ and $\gamma$ the closest that the
image of $|z|=\epsilon$ which is $1+\frac{z}{1+z}$, gets to one, is
$\epsilon/(1+\epsilon)$ so that must be the upper bound for $\gamma.$
Taking $\epsilon=1/3$ and $\gamma = 1/5$ will work. Note also that $u=
1+z+\cdots$ makes one turn around one.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1558659",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Inverse of an exponential function I am having difficulties forming the inverse of this
$f(x) = 3 \cdot2^{3x+1} \cdot 5^{3x-1}$.
What I have done so far:
$3 \cdot 2^{3y} \cdot 2^1 \cdot 5^{3y}\cdot5^{-1} \Leftrightarrow 3\cdot 2\cdot \frac{1}{5}\cdot (5\cdot 2)^{3y} \Leftrightarrow \frac{6}{5}\cdot 10^{3y} \Leftrightarrow \ln \frac{6}{5}\cdot \ln10\cdot 3y$
| $x = \frac{\log \left(\frac{y}{6}\right)+\log (5)}{3 (\log (2)+\log (5))}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1558941",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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find all integers $a,b,c$ such that $a^2=bc+1,$ $b^2=ca+1.$ This was one of the problems in a math contest in India. This is how I tried it:
Subtracting the second equation from the first one gives $a^2-b^2=bc-ca$ or $(a-b)(a+b)=c(b-a)$. $a-b=-(b-a)$. Therefore a+b has to be -c. Thus all integers satisfying the condition $a+b=-c$ are such that $a^2=bc+1$ and $b^2=ca+1$. Also if $a-b=b-a,$ it means $a=b.$ Then both $b-a$ and $a-b$ are $0.$ Hence $c$ can thus be any integer. Hence, there are two groups of integers satisfying the required condition $-a=b$ with any $c$ and $a+b=-c.$
Please let me know whether my solution is correct or not.
| $c=-(a+b)$ is not a general solution: for example with $a=b=2$ and $c=-4$ would give $4=-8+1$, which is not correct.
Instead, in cases of $c=-(a+b)$, you need to solve $a^2=-ab-b^2+1$, i.e. $a =\frac{-b \pm\sqrt{4-3b^2}}{2}$ and that only has integer solutions when $b=-1,0,1$ in which case $a=0\text{ or }1, -1\text{ or }1, -1\text{ or }0$ and $c=1\text{ or }0, 1\text{ or }-1, 0\text{ or }-1$ respectively.
$a=b$ with any $c$ is not a general solution: for example with $a=b=2$ and $c=-4$ would give $4=-8+1$, which is not correct.
Instead, in cases of $a=b$, you need to solve $a^2=ac+1$, i.e. $a=b=\frac{c \pm\sqrt{c^2+4}}{2}$ and that only has a solution in integers when $c=0$ and so $a=b=\pm 1$.
So your solutions are:
$$ a=0, \, b=-1,\, c=1$$
$$ a=1, \, b=-1,\, c=0$$
$$ a=-1, \, b=0,\, c=1$$
$$ a=1, \, b=0,\, c=-1$$
$$ a=-1, \, b=1,\, c=0$$
$$ a=0, \, b=1,\, c=-1$$
$$ a=-1, \, b=-1,\, c=0$$
$$ a=1, \,b=-1,\, c=0$$
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "1",
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Prove the identity $\sum\limits_{k=0}^n\left(x-\frac{k}{n}\right)^2 \binom{n}{k}x^k(1-x)^{n-k}=\frac{1}{n}x(1-x)$ for $0 \leq x \leq 1$ I'd like to prove this identity:
$$\sum_{k=0}^n\left(x-\frac{k}{n}\right)^2 \binom{n}{k}x^k(1-x)^{n-k}=\frac{1}{n}x(1-x)$$
for $x\in[0,1]$ and $n\in\mathbb{N}$.
I've worked on this problem for the last few hours without success. I have tried induction and I have tried re-writing the expression as many ways as I could think of. Any help would be greatly appreciated. Thanks.
| Brute force approach. There might be a more elegant approach.
Multiply by $n^2$, and you want:
$$\sum_{k=0}^n\left(nx-k\right)^2 \binom{n}{k}x^k(1-x)^{n-k}=nx(1-x)$$
First part:
$$\sum_{k=0}^n n^2x^2\binom{n}{k}x^k(1-x)^{n-k} = n^2x^2(x+(1-x))^n = n^2x^2.\tag{1}$$
Second part:
$$\begin{align}
\sum_{k=0}^n 2nkx \binom{n}{k}x^ky^{n-k} &= 2nx^2\sum_{k=0}^n k\binom{n}{k}x^{k-1}y^{n-k}\\
&=2nx^2\frac{d}{dx}(x+y)^n \\
&= 2nx^2n(x+y)^{n-1}\end{align}$$
Letting $y=1-x$, then we get:
$$\sum_{k=0}^n 2nkx \binom{n}{k}x^k(1-x)^{n-k} = 2n^2x^2\tag{2}$$
Part 3:
$$\begin{align}
\sum k^2\binom{n}{k}x^ky^{n-k} &= x\frac{d}{dx}\sum_{k=0}^nk\binom{n}{k}x^ky^{n-k}\\
&=\left(x\frac{d}{dx}\right)^2(x+y)^n\\
&=x\frac{d}{dx}\left(xn(x+y)^{n-1}\right)\\
&=x\left(n(x+y)^{n-1} + xn(n-1)(x+y)^{n-2}\right)\\
\end{align}$$
Letting $y=1-x$ we have:
$$\sum k^2\binom{n}{k}x^ky^{n-k}=xn + n(n-1)x^2\tag{3}$$
Computing $(1)-(2)+(3)$ we get:
$$\sum_{k=0}^n\left(nx-k\right)^2 \binom{n}{k}x^k(1-x)^k=n^2x^2-2n^2x^2+n(n-1)x^2 + nx=nx-nx^2=nx(1-x)$$
A more algebraic way might amount to noting that:
$$\sum_{k=0}^n ((n-k)x -ky)^2\binom{n}{k}x^ky^{n-k}$$
is a symmetric polynomial in $x,y$, and thus can be written in terms of $x+y,xy$. Not sure where to go from there, however.
| {
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"timestamp": "2023-03-29T00:00:00",
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find $ \cfrac{1}{r^2} +\cfrac{1}{s^2} +\cfrac{1}{t^2}$ given that $r,s,t$ are the roots of $x^3-6x^2+5x-7=0$
I am asked to find $$ \cfrac{1}{r^2} +\cfrac{1}{s^2} +\cfrac{1}{t^2}$$
given that $r,s,t$ are the roots of $x^3-6x^2+5x-7=0$ .
So what I did was to get the polynomial whose roots are the reciprocals of $r,s,t$ ,namely $$-7x^3+5x^2-6x+1=0$$
From that, I've then used the $2^{nd}$ Newton'sum $a_ns_2+a_{n-1}s_1+2a_{n-2}=0$.
Now given that $s_1=- \cfrac{a_{2}}{a_3}$ I have in my case $s_1=\cfrac{5}{7}$ (I've turned my polynomial into a monic polynomial) .
Therefore my Newton's sum is $$s_2-\cfrac{5}{7}s_1+2\cdot\cfrac{6}{7}=0$$
$$s_2-\cfrac{5}{7}\cdot\cfrac{5}{7} +\cfrac{12}{7}=0$$
Solving for $s_2$ I get $s_2=-\cfrac{59}{49}\approx -1,204 $
Now ,I've just started studying this stuff so I am not sure I've applied it all right(the hardest part for me is to get the signs right),so I would be really grateful if someone could verify my work .
Thanks in advance.
| Still another approach. We have that
$$ M=\begin{pmatrix}0 & 0 & 7 \\ 1 & 0 & -5 \\ 0 & 1 & 6\end{pmatrix}$$
is the companion matrix of your polynomial, and:
$$ \frac{1}{r^2}+\frac{1}{s^2}+\frac{1}{t^2}=\text{Tr}(M^{-2})=\text{Tr}\,\begin{pmatrix}0 & 7 & 42 \\ 0 & -5 & -23 \\ 1 & 6 & 31\end{pmatrix}^{-1} $$
hence:
$$ \frac{1}{r^2}+\frac{1}{s^2}+\frac{1}{t^2}=\text{Tr}\,\begin{pmatrix}-\frac{17}{49} & \frac{5}{7} & 1 \\ -\frac{23}{49} & -\frac{6}{7} & 0 \\ \frac{5}{49} & \frac{1}{7} & 0\end{pmatrix} = \color{red}{-\frac{59}{49}}.$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove that $\cos\left(\frac{2\pi}{n}\right)+\cos\left(\frac{4\pi}{n}\right)+\ldots+\cos\left(\frac{2(n-1)\pi}{n}\right)=-1$ May you help on how to start, or where to look for the following question?
By using the $n$-th roots of the unity, show that:
*
*$\cos\left(\frac{2\pi}{n}\right)+\cos\left(\frac{4\pi}{n}\right)+\ldots+\cos\left(\frac{2(n-1)\pi}{n}\right)=-1$
*$\sin\left(\frac{2\pi}{n}\right)+\sin\left(\frac{4\pi}{n}\right)+\ldots+\sin\left(\frac{2(n-1)\pi}{n}\right)=0$
Can we prove the above by induction? Thank you very much in advance.
| Consider $n$-th roots of the unity $z$, we have $1+z+z^2+\cdots+z^{n-1}=0$, i.e. $$1+(\cos\frac{2\pi}{n}+i\sin\frac{2\pi}{n})+(\cos\frac{2\pi}{n}+i\sin\frac{2\pi}{n})^2+\cdots+(\cos\frac{2\pi}{n}+i\sin\frac{2\pi}{n})^{n-1}=0,$$ by De Moivre's formula, we have $$1+\cos\left(\frac{2\pi}{n}\right)+\cos\left(\frac{4\pi}{n}\right)+\ldots+\cos\left(\frac{2(n-1)\pi}{n}\right)+i\left(\sin\left(\frac{2\pi}{n}\right)+\sin\left(\frac{4\pi}{n}\right)+\ldots+\sin\left(\frac{2(n-1)\pi}{n}\right)\right)=0,$$ which implies that $$1+\cos\left(\frac{2\pi}{n}\right)+\cos\left(\frac{4\pi}{n}\right)+\ldots+\cos\left(\frac{2(n-1)\pi}{n}\right)=0,$$ and $$\sin\left(\frac{2\pi}{n}\right)+\sin\left(\frac{4\pi}{n}\right)+\ldots+\sin\left(\frac{2(n-1)\pi}{n}\right)=0$$
Update: we couldn't get any relation formula on $\cos(\frac{2\pi}{n})$ and $\cos(\frac{2\pi}{n+1})$ if consider induction on $n$, even if we separate the question into even/odd parts. But the even case is easy since $$\cos\frac{2j\pi}{n}+\cos\frac{(n+2j)\pi}{n}=0,$$ hence the summation only left $\cos\pi=-1$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Generating function, finding coefficient (decomposing) I just started learning about generating functions, and there is a problem that I have the solution to, but I'm wondering if there is a better general method to solve problems of that kind.
If I want to find the coefficient of $x^8$ in
$$\left( 1+x+x^2+x^3+x^4 \right) \sum_{k=0}^{\infty} (-1)^k {-2 \choose k} x^k, $$
is there a better method than to decompose it as
$${-2\choose 8}-{-2\choose 7}+{-2\choose 6}-{-2\choose5}+{-2\choose4}$$
$$={9\choose8}+{8\choose7}+{7\choose6}+{6\choose5}+{5\choose4}$$
$=35$?
It is the suggested method in the literature, but it seems to me there should be a better way, since this method would take a lot of time if the coefficient was, for example, the coefficient of $x^{20}$ or larger.
Edit
I found a somewhat similar thread here, and tried to apply that method, but that gives me the wrong answer. That method dealt with partitions, so that in my case the term $x^8$ can arise from $x^4 \cdot p(4,2)x^4, x^3 \cdot p(5,2)x^5, x^2 \cdot p(6,2)x^6, x \cdot p(7,2)x^7$ and $1\cdot p(8,2)x^8.$
Counting all such partitions gives me the answer 25, where it should be 35...
| You could try:
$\begin{align*}
(1 + z + z^2 + z^3 + z^4) \sum_{k \ge 0} (-1)^k \binom{-2}{k} z^k
&= \frac{1 - z^5}{1 - z} \cdot (1 - z)^{-2} \\
&= (1 - z^5) \cdot (1 - z)^{-3} \\
&= (1 - z^5) \cdot \sum_{k \ge 0} (-1)^k \binom{-3}{k} z^k \\
&= (1 - z^5) \cdot \sum_{k \ge 0} \binom{k + 2}{2} z^k
\end{align*}$
so that:
$\begin{align*}
[z^8] (1 + z + z^2 + z^3 + z^4) \sum_{k \ge 0} (-1)^k \binom{-2}{k} z^k
&= [z^8] (1 - z^5) \cdot \sum_{k \ge 0} \binom{k + 2}{2} z^k \\
&= \binom{8 + 2}{2} - \binom{3 + 2}{2} \\
&= 35
\end{align*}$
| {
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"timestamp": "2023-03-29T00:00:00",
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Evaluating limit by sandwich theorem: $\lim_{n\to\ \infty}\frac{1}{1+n^2} +\frac{2}{2+n^2}+\cdots+\frac{n}{n+n^2}$ $$\lim_{n\to\ \infty}\frac{1}{1+n^2} +\frac{2}{2+n^2}+\cdots+\frac{n}{n+n^2}$$
For using Sandwich theorem I need two functions such that $g(x)<f(x)<h(x)$
$$\frac{1}{n+n^2} +\frac{2}{n+n^2}+\cdots+\frac{n}{n+n^2} \leq \frac{1}{1+n^2} +\frac{2}{2+n^2}+\cdots+\frac{n}{n+n^2} $$
But I can't find function greater than the given which will help me evaluate the limit.
| HINT:
If you are familiar with the harmonic numbers, you can also proof:
$$\lim_{n\to\infty}\frac{1}{1+n^2}+\frac{2}{2+n^2}+\cdots+\frac{n}{n+n^2}=\lim_{n\to\infty}\sum_{m=1}^{n}\frac{m}{m+n^2}=$$
$$\lim_{n\to\infty}n\left(n\text{H}_{n^2}-n\text{H}_{n^2+n}+1\right)=\frac{1}{2}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Go from A to D in three equal steps Given two parallel lines $r$ and $s$, line $p$, perpendicular to both, and points $A$ and $D$ on different sides of $p$ with respect to the parallel lines, how can I prove the existence of two points, $B$ and $C$, respectively on $r$ and $s$, such that segments $AB$, $BC$ and $CD$ are congruent?
I'd prefer a constructive proof, for it to be a more general solution useful to solve this question too. (also, geometric proofs are fun)
Here are two examples I constructed in reverse (I fitted the lines to the points and not viceversa):
| Let $a,x,d,y$ the distances shown in the picture and $t=\overline{AB}=\overline{BC}=\overline{CD}$
Then we have
$$a^2+x^2=t^2\tag{1}$$
$$d^2+y^2=t^2\tag{2}$$
$$(x-y)^2+e^2=t^2\tag{3}$$
If we add $a^2d^2$ to $(3)$ and reorder the terms we get
$$x^2+a^2-2xy+y^2+b^2+e^2=t^2+a^2+d^2$$
an further
$$2xy=t^2+e^2-a^2-d^2\tag{4}$$
if we substitute by $(1)$ and $(2)$.
Squaring $(5)$ and substituting again by $(1)$ and $(2)$ gives
$$4(t^2-a^2)(t^2-d^2)=(t^2+e^2-a^2-d^2)^2\tag{5}$$
$(5)$ is an equation of degree $4$ in $t$ but quadratic in $t^2$:
$$3t^4-2t^2(a^2+d^2+e^2)+(2a^2d^2+2a^2e^2+2d^2e^2)-(a^4+d^4+e^4)=0$$
So it can be easily solved.
The discriminant $\Delta$ of the quadratic equation satisfies
$$\frac{1}{8}\Delta=(a^2-d^2)^2+(a^2-e^2)^2+(d^2-e^2)^2\ge0$$
For the second picture we could start with the equations
$$a^2+x^2=t^2$$
$$d^2+y^2=t^2$$
$$(x+y)^2+e^2=t^2\tag{6}$$
but we would get the same solutions because the equations differ only in the sign of $y$ in $(6)$
| {
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Find $ \int \frac{1}{2\sin(x)-3\cos(x)}dx$.
Find $\displaystyle \int \dfrac{1}{2\sin(x)-3\cos(x)}dx$.
My book said to solve this by saying $u = \tan \left(\dfrac{x}{2} \right)$ since $\cos(x) = \dfrac{1-u^2}{1+u^2}$ and $\sin(x) = \dfrac{2u}{1+u^2}$. I don't see how this will help since $du = \dfrac{1}{\cos(x)+1}dx $. How will we get that in the integrand?
| Hint:
the suggested substitution works well because:
$$
\cos x+1=\frac{1-u^2}{1+u^2}+1=\frac{2}{1+u^2}
$$
so:
$$
du=\frac{1}{2\cos^2(x/2)}dx= \frac{dx}{\cos x+1} \Rightarrow dx=\frac{2\,du}{1+u^2}
$$
the integral becomes:
$$
2\int\frac{du}{3u^2+4u-3}
$$
that, completing the square, becomes:
$$
2\int\frac{du}{(\sqrt{3}u+2\sqrt{3}/3)^2-13/3}
$$
that can be solved with the substitution $t=\sqrt{3}u+2\sqrt{3}/3$ and factorizing $13/3$.
| {
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"timestamp": "2023-03-29T00:00:00",
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I want to calculate the limit of: $\lim_{x \to 0} \left(\frac{2^x+8^x}{2} \right)^\frac{1}{x} $ I want to calculate the limit of: $$\lim_{x \to 0} \left(\frac{2^x+8^x}{2} \right)^\frac{1}{x} $$
or prove that it does not exist. Now I know the result is $4$, but I am having trouble getting to it. Any ideas would be greatly appreciated.
| Notice, $$\lim_{x\to 0}\left(\frac{2^x+8^x}{2}\right)^{1/x}$$
$$=\lim_{x\to 0}\left(\frac{2^x+2^{3x}}{2}\right)^{1/x}$$
$$=\lim_{x\to 0}\exp\frac{1}{x}\ln\left(\frac{2^x+2^{3x}}{2}\right)$$
Applying L'hospital's rule for $\frac 00$ form $$=\lim_{x\to 0}\exp\frac{\frac{d}{dx}\ln\left(\frac{2^x+ 2^{3x}}{2}\right)}{\frac{d}{dx}(x)}$$
$$=\lim_{x\to 0}\exp\frac{\frac{1}{\left(\frac{2^x+2^{3x}}{2}\right)}\left(\frac{2^x\ln 2+3\cdot 2^{3x}\ln 2}{2}\right)}{1}$$
$$=\exp\frac{\left(\frac{2^0\ln 2+3\cdot 2^{0}\ln 2}{2}\right)}{\left(\frac{2^0+2^{0}}{2}\right)}=e^{2\ln 2}=2^2=\color{red}{4}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Find $n$ if the area between the curve of $y=x^n$ and the $y$ axis is $3$ times the area between the curve and the $x$-axis
For this question i tried to find the area for red and blue sections, and equate them by red= 3 blue.
However, it didnt work out and I got $b-a = 3(b^n - a^n)$ for the outcome.
In the book, it says we dont have to deal with the red area....
Thank you.
| The blue area is $$\int_a^b x^n dx = \frac{b^{n+1}-a^{n+1}}{n+1}.$$
The red (orange?) area is the area of the big rectangle with sides, $b$, $b^n$, minus the small (white) rectangle with sides $a$, $a^n$, minus the above integral. That is $$b^{n+1}-a^{n+1}-\frac{b^{n+1}-a^{n+1}}{n+1}.$$ Hence, we need $$ b^{n+1}-a^{n+1}-\frac{b^{n+1}-a^{n+1}}{n+1} = 3 \cdot \frac{b^{n+1}-a^{n+1}}{n+1}. $$ Simplifying a bit gives $$ (n+1)(b^{n+1} - a^{n+1}) = 4 (b^{n+1} - a^{n+1}). $$ So we have $$(n - 3)(b^{n+1}-a^{n+1})=0.$$ Since $a$ and $b$ are arbitrary, we require $n=3$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Involutory matrix $2 \times 2$ I want to find out how many $2 \times 2$ involutory matrices are there over $\mathbb Z_{26}$.
$ $
Is there any formula to calculate this?
$ $
Thanks for your help.
| Let $p$ be an odd prime, and $A=\left(\begin{matrix}a&b\\c&d\end{matrix}\right)\in M_2(\mathbb Z/p\mathbb Z)$.
Then $A^2=I$ is equivalent to $a^2+bc=1,$ $d^2+bc=1,$ $b(a+d)=0,$ $c(a+d)=0.$
If $a+d\ne0$, then $b=c=0$, so $a^2=d^2=1$, and there are $2$ solutions.
If $a+d=0$, then consider the following cases:
(i) $a=0$. Then $bc=1$, that is, $c=b^{-1}$ and find $p-1$ solutions.
(ii) $a=1$. Then $bc=0$, and we have $2p-1$ solutions.
(iii) $a=-1$. Similarly, we get $2p-1$ solutions.
(iv) $a\ne0,\pm1$. Then $c=(1-a^2)b^{-1}$, so there are $(p-1)(p-3)$ solutions.
Summing up we get $p^2+p+2$ matrices $2\times 2$ over $\mathbb Z/p\mathbb Z$.
| {
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Evaluate $\prod \frac {2k - 1} {2k}$ I came across to this: evaluate $$\prod_{k = 1}^{n} \frac {2k - 1} {2k}.$$
I brought it to a closed-form expression in asymptotics but I suspect my asymptotics is bad and I also want to see different solutions, hence the question.
My original solution: Write
$$\begin {eqnarray}
\prod_{k = 1}^{n} \frac {2k - 1} {2k}
& = & \exp \left (\sum_{k = 1}^{n} \log \left (1 - \frac {1} {2k}\right) \right) \nonumber \\
& = & \exp \left (- \sum_{k = 1}^{n} \sum_{m = 1}^{\infty} \frac {1} {(2k)^m m}\right).
\end {eqnarray}$$
Let $$S = \sum_{m = 1}^{\infty} \frac {1} {(2k)^m m} \tag {1}.$$ Multiply both sides of $(1)$ by $2k$ to get $2k S < S + 1$, and by $4k$ to get $4kS > S + 2$. Hence, $$\frac {2} {4k - 1} < S < \frac {1} {2k - 1}.$$ Then, $$\frac {1} {2} \left( \log (4n - 1) - \log 3 \right) - 2 + O \left(\frac {1} {n^2}\right) < \sum_{k = 1}^{n} \sum_{m = 1}^{\infty} \frac {1} {(2k)^m m} < \frac {1} {2} \log (2n - 1) - 1 + O \left(\frac {1} {n^2}\right).$$ Hence, $$\sqrt {\frac {c} {2n - 1}} < \prod_{k = 1}^{n} \frac {2k - 1} {2k} < \sqrt {\frac {3 c^2} {4n - 1}},$$ where $c$ is a constant. Any better approach?
| $$\prod_{k=1}^{n}\frac{2k-1}{2k}=\frac{2-1}{2}\cdot\frac{4-1}{4}\cdot\frac{6-1}{6}\cdot\dots\cdot\frac{2n-1}{2n}=$$
$$\frac{1}{2}\cdot\frac{3}{4}\cdot\frac{5}{6}\cdot\dots\cdot\frac{2n-1}{2n}=\frac{1\cdot3\cdot5\dots\cdot(2n-1)}{2\cdot4\cdot6\dots\cdot2n}=\frac{\frac{2^n\Gamma\left(n+\frac{1}{2}\right)}{\sqrt{\pi}}}{2^n\Gamma\left(n+1\right)}=\frac{\Gamma\left(n+\frac{1}{2}\right)}{n!\sqrt{\pi}}$$
| {
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Derive quadratic formula I cannot understand how quadratic formula to solve for $x$ was derived.
On this website, it explains the steps
Following I understand
but I cannot understand how they got
$b/2a$
and why they are squaring it
$b^2/4a^2$
Really, I am baffled!
| $x^2+\frac{b}{a}x+K^2=(x+K)^2$
$x^2+\frac{b}{a}x+K^2=x^2+2xK+K^2$
$\frac{b}{a}x=2xK$
$k=\frac{b}{2a}$
Then we know that
$x^2+\frac{b}{a}x+(\frac{b}{2a})^2=(x+\frac{b}{2a})^2$
$x^2+\frac{b}{a}x+\frac{b^2}{4a^2}=(x+\frac{b}{2a})^2$
This process is called completing the square.
Now we can factor a perfect square. Is this answer sufficient?
| {
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Prove that $0 \leq ab^2-ba^2 \leq \frac{1}{4}$ with $0 \leq a \leq b \leq 1$.
Let $a$ and $b$ be real numbers such that $0 \leq a \leq b \leq 1$. Prove that $0 \leq ab^2-ba^2 \leq \dfrac{1}{4}$.
Attempt
We can see that $ab^2-ba^2 = ab(b-a)$, so it is obvious that it is greater than or equal to $0$. But how do I show it is also less than or equal to $\dfrac{1}{4}$?
| There are more generic ways of solving such problems.
Since $f(a,b) = ab^2 - ba^2$ is continuous and differentiable, and the domain $0 \le a \le b \le 1$ is a compact simplex, it attains its max and min at one of the following points:
*
*A point where $0 < a < b < 1$ and $\nabla f (a,b) = 0$, i.e. $f_x(a,b) = f_y(a,b) = 0$
*A point where $a = 0$, $0 < b < 1$, and $f_y(a,b) = 0$
*A point where $0 < a < 1$, $b = 1$, and $f_x(a,b) = 0$
*A point where $0 < a = b < 1$, and $f_x(a,b) + f_y(a,b) = 0$
*A point where $a = b = 0$ or $a = b = 1$.
Calculate that $f_x(a,b) = b^2 - 2ab$, $f_y(a,b) = 2ab - a^2$.
There are no points of type (1).
Type (2) points are anything on the line, and here $a = 0$ so $f(a,b) = 0$.
Type (3) points satisfy $b = 1$, so $1 - 2a = 0$, so $a = \frac12$, and here $f(a,b) = \frac12 - \frac14 = \frac14$. Type (4) points all have $f(a,b) = 0$.
Finally, the two type (5) points are $f(0,0) = 0$ and $f(1,1) = 0$.
Thus the minimum of $f$ is $0$ and the maximum is $\frac14$, on this domain.
| {
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Arithmetic: Prove that is multiple of 30 Prove that $n^{19}-n^7$ is multiple of $30$
I've seen $6$ can divide it because
$$n^{19}-n^7=n^7(n^{12}-1) = n^7(n^6+1)(n^6-1)=n^4(n^6+1)(n^3-1)n^3(n^3+1)$$
And there are three consecutive numbers, so, at least one is multiple of $3$ and up to two even numbers.
But, how to prove that is multiple of $5$?
| Notice, one should re-write & factorize as follows $$n^{19}-n^7=n^7(n^{12}-1)$$
$$=n\cdot \underbrace{n^6\color{blue}{(n^{6}-1)}\color{red}{ (n^{6}+1)}}_{\text{divisible by 5}}$$
$$=n^7\color{blue}{(n^{3}-1)(n^3+1)}\color{red}{(n^{2}+1)(n^4-n^2+1)}$$
$$=n^7\color{blue}{(n-1)(n^2+n+1) (n+1)(n^2-n+1)}\color{red}{(n^{2}+1)(n^4-n^2+1)}$$
$$=n^4\underbrace{(n-1)n(n+1)}_{\text{divisible by 3!}}\ \underbrace{n^2(n^2+1)}_{\text{divisible by 2! }}(n^8+n^4+1)$$
thus it clear that $\color{blue}{n^{19}-n^7}$ is divisible by $5\times 3!\times 2!=60$ i.e. it is divisible by $\color{red}{30}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1588009",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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} |
Subadditivity of square root function Any hint to prove
$$ \sqrt{x+y} \le \sqrt{x} + \sqrt{y}, \qquad \forall x,y \ge 0 $$
| Another:
$x,y \geq 0$ then the product $\sqrt{x} \sqrt{y}$ have sence and now
$$\sqrt{x + y} \leq \sqrt{x + 2\sqrt{x}\sqrt{y} + y} = \sqrt{(\sqrt{x} + \sqrt{y})^2} = \sqrt{x} + \sqrt{y}$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Problem related to non-coplanar unit vectors,equally inclined to one another at an angle $\theta$ I've lately been facing lot of trouble in solving vector equations.Like the one below :
Let $a,b,c$ be non-coplanar unit vectors,equally inclined to one
another at an angle $\theta$.If $a×b+b×c=pa+qb+rc$,find the scalars
p,q and r in terms of $\theta$.
What would be the shortest method to solve this problem?
In my book they took nearly 2 pages!But I guess there might be a shorter method to solve such type of problems.What say ?
| First of all, observe that we can safely assume $0 < \theta < \frac{2}{3} \pi$. Indeed, it makes sense to consider $\theta$ to be the smallest (positive) angle between, say, $a$ and $b$. Furthermore, clearly for $\theta = 0$ we have $a = b = c$, while for $\theta = \frac{2}{3} \pi$ we have $a,b,c$ coplanar.
Now, recall that
$$
a \cdot (b \times c) = c \cdot (a\times b)
$$
coincides with the volume of the parallelepiped with sides defined by $a,b,c$, i.e.
$$
V = \sqrt{1 + 2 \cos^3\theta - 3 \cos^2\theta}.
$$
Since we're after a representation of a vector in the $\Bbb{R}^3$-basis $\{a,b,c\}$, let's consider the system of linear equations obtained by taking the dot product of $a \times b + b \times c = pa + qb + rc$ by $a,b$, and $c$, respectively:
$$
\begin{cases}
a \cdot (a \times b) + a \cdot (b \times c) = p(a \cdot a) + q(a \cdot b) + r(a \cdot c) \\
b \cdot (a \times b) + b \cdot (b \times c) = p(b \cdot a) + q(b \cdot b) + r(b \cdot c) \\
c \cdot (a \times b) + c \cdot (b \times c) = p(c \cdot a) + q(c \cdot b) + r(c \cdot c).
\end{cases}
$$
Recalling that $u \cdot (u \times v) = 0$ for every $u,v \in \Bbb{R}^3$, that $u \cdot u = 1$ for every unit vector $u$, and that $a \cdot b = b \cdot c = a \cdot c = \cos\theta$, this becomes
$$
\begin{cases}
p + q \cos\theta + r \cos\theta = V \\
p \cos\theta + q + r \cos\theta = 0 \\
p \cos\theta + q \cos\theta + r = V
\end{cases}
\tag{1} \label{eq:1}
$$
or, in matrix form
$$
A \mathbf{p} =
\begin{pmatrix}
1 & \cos\theta & \cos\theta \\
\cos\theta & 1 & \cos\theta \\
\cos\theta & \cos\theta & 1
\end{pmatrix}
\begin{pmatrix}
p \\
q \\
r
\end{pmatrix}
=
\begin{pmatrix}
V \\
0 \\
V
\end{pmatrix}.
$$
Interestingly, for $0 < \theta < \frac{2}{3} \pi$ we have that $\det A = V^2 > 0$ (try to see why), so $\eqref{eq:1}$ has exactly one solution, which is†
$$
\begin{pmatrix}
\dfrac{1-\cos\theta}{V},\;
\dfrac{2 (\cos\theta-1) \cos\theta}{V},\;
\dfrac{1-\cos\theta}{V}
\end{pmatrix}^T.
$$
† Since doing this by hand is a bit tedious, here's how to compute it in Mathematica
m = {{1, Cos[t], Cos[t]}, {Cos[t], 1, Cos[t]}, {Cos[t], Cos[t], 1}};
v = Sqrt[Det[m]];
p = Assuming[0 <= t <= 2/3 Pi, LinearSolve[m, {v, 0, v}]] // Simplify
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Finding a combinatorial recurrence relation with three variables This question is from the generating functionology textbook,
Let $f(n, m,k)$ be the number of strings of n $0$’s and $1$’s that contain
exactly $m$ $1$’s, no $k$ of which are consecutive.
Find a recurrence formula for $f$. It should have $f(n, m,k)$
on the left side, and exactly three terms on the right.
There were previous examples involving binomial coefficients and partitions, so I am sure the method of finding the recurrence is similar (splitting the total into piles with n and without n etc)
This just seems on another level of complexity, any help would be much appreciated.
| We can break down $f(n,m,k)$ into how many consecutive ones it ends with
$$
f(n,m,k)=\left\{\begin{array}{}
0&\text{if }m<0\text{ or }n<m&\text{impossible}\\[9pt]
[n<k]&\text{if }n=m&\text{only ones}\\
\displaystyle\sum_{j=0}^{k-1}f(n-j-1,m-j,k)&\text{if }n\gt m&\begin{array}{l}\text{ends with a zero}\\[-4pt]\text{and $j$ ones}\end{array}
\end{array}\right.
$$
where $[\cdots]$ are Iverson Brackets.
Here is a table for $k=3$:
$$
\begin{array}{r|rr}
&0&1&2&3&4&5&6&7&8&9&n\\\hline
0&1&1&1&1&1&1&1&1&1&1\\
1&0&1&2&3&4&5&6&7&8&9\\
2&0&0&1&3&6&10&15&21&28&36\\
3&0&0&0&0&2&7&16&30&50&77\\
4&0&0&0&0&0&1&6&19&45&90\\
5&0&0&0&0&0&0&0&3&16&51\\
6&0&0&0&0&0&0&0&0&1&10\\
7&0&0&0&0&0&0&0&0&0&0\\
m
\end{array}
$$
| {
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"url": "https://math.stackexchange.com/questions/1589355",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove that $\frac{1}{a}+\frac{1}{b}+\frac{1}{c} \geq 2 \left (\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a} \right) \geq \frac{9}{a+b+c}$
Let $a,b,$ and $c$ be positive real numbers, prove that $\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c} \geq 2 \left (\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a} \right) \geq \dfrac{9}{a+b+c}$.
Should I use AM-GM for the expression in the middle of the inequality? We have $a+b \geq 2\sqrt{ab}$ etc.?
| Yes, you can! By AM-GM we obtain:
$$\sum_{cyc}\frac{2}{a+b}\leq\sum_{cyc}\frac{1}{\sqrt{ab}}\leq\sum_{cyc}\frac{\frac{1}{a}+\frac{1}{b}}{2}=\sum_{cyc}\frac{1}{a}.$$
The right inequality we can prove also by AM-GM:
$$2\sum_{cyc}\frac{1}{a+b}=\frac{1}{a+b+c}\sum_{cyc}(a+b)\sum_{cyc}\frac{1}{a+b}\geq$$
$$\geq\frac{1}{a+b+c}\cdot3\sqrt[3]{\prod_{cyc}(a+b)}\cdot\frac{3}{\sqrt[3]{\prod\limits_{cyc}(a+b)}}=\frac{9}{a+b+c}.$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove that ${x^7-1 \over x-1}=y^5-1$ has no integer solutions I want to show that $${x^7-1 \over x-1}=y^5-1$$
cannot have any integer solutions. The only observation I have made so far is that the left hand side is the $7$th cyclotomic polynomial
$$\Phi_7(x)= {x^7-1 \over x-1}=x^6+x^5+x^4+x^3+x^2+1$$
If I remember correctly cyclotomic polynomials are irreducible. Now can I use this property to arrive at the conclusion or should I try to approach by contradiction and assume $\Phi_7(a)=b^5-1$ for some integers $a$ and $b$? The only problem is that I don't see where I would look for an easy contradiction. Any hints?
Edit
I also see that the right hand side can be factored as
$$(y-1)(y^4+y^3+y^2+y+1)=(y-1)\Phi_5(y)\implies \frac{\Phi_7(x)}{\Phi_5(y)}=y-1$$
which seems like it could give the result if we prove that the two cyclotomic polynomials have no common factors. How could this be done?
Edit: This is IMO2006 Shortlised Problem N5 (RUS).
| A start to solving this equation is to first prove that if $x$ is an integer and $p$ is a prime divisor of the left hand side $\frac{x^{7}-1}{x-1}$ then either $p=7$ or $p \equiv 1(\mod{7})$.
Proof: First notice by Fermat's little theorem that $x^{p-1}-1$ is divisible by $p$. Also by our hypothesis $x^{7}-1$ is divisible by $p$. Now suppose that $7$ does not divide $p-1$. Then $\gcd(p-1,7) =1$, so there exists integers $a$ and $b$ such that $7a + (p-1)b = 1$. We subsequently see
$$ x \equiv x^{7a+(p-1)b} \equiv (x^{7})^{a} \cdot (x^{p-1})^{b} \equiv 1 (\mod p)$$
and so then
$$\frac{x^{7}-1}{x-1}= 1 + x + \text{ ... } + x^{6} \equiv 7 (\mod p)$$
So we have that $p$ divides $7$, therefore $p=7$ must hold if we have that $p \equiv 1(\mod 7)$ does not, as we stated. $\square$
We have now shown that every positive divisor of $\frac{x^{7}-1}{x-1}$ meets either $d \equiv 1(\mod 7)$ or $d \equiv 0$.
Now assuming $(x,y)$ is some integer solution to our problem, we notice that $y-1>0$ since $\frac{x^{7}-1}{x-1} > 1 \ \forall \ x \neq 1$. Then since $y-1$ divides $\frac{x^{7}-1}{x-1} = y^{5}-1$ we have $y \equiv 1(\mod 7)$ or $y \equiv 2(\mod 7)$. Evaluating $1+y+y^{2}+y^{3}+y^{4}$ in both possible cases will contradict the fact that our positive divisor $1+y+y^{2}+y^{3}+y^{4}$ of $\frac{x^{7}-1}{x-1}$ is congruent to either $0$ or $1 (\mod 7)$. Your result then follows.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
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Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.