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A periodic summation question
The periodic series can be solved by doing
I understand the solution but it is little lengthy process. Can't we apply any law to get the solution direct?
| Splitting the sum and performing an index shift gets you there pretty quick.
$$
\begin{align*}
\sum_{n=1}^{20} a_n
&= \sum_{n=1}^{20} \left(\frac{1}{n}-\frac{1}{n+2}\right)
= \sum_{n=1}^{20} \frac{1}{n} - \sum_{n=1}^{20} \frac{1}{n+2} \\
&= \sum_{n=1}^{20} \frac{1}{n} - \sum_{n=3}^{22} \frac{1}{n}
= \sum_{n=1}^{2} \frac{1}{n} - \sum_{n=21}^{22} \frac{1}{n} \\
&= \left(\frac 1 1+\frac 1 2\right) - \left(\frac 1 {21} + \frac 1 {22}\right)
\end{align*}
$$
This is a typical case of a telescoping sum as mentioned by Sharkos.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/472360",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Sum of largest two angles All the inner angles of a 7 sided polygon are obtuse, their sizes in degrees being distinct integers divisible by 9. What is the sum (in degree) of the largest two angles?
| Sum of all angles is $180^\circ \cdot (n-2) = 900^\circ$.
Denote angles as $a,b,c,d,e,f,g$ ($a<b<c<d<e<f<g$).
If all angles are obtuse and are integer numbers divisible by $9^\circ$, then they can be
$99^\circ, 108^\circ, 117^\circ, 126^\circ, 135^\circ, 144^\circ, 153^\circ, ...$
A).
If $a\ge 108^\circ$, then $b\ge 117^\circ, c\ge 126^\circ, d\ge 135^\circ, \ldots$, then
$a+b+c+d+e+f+g\ge (108+117+126+135+144+153+162)^\circ = 945^\circ>900^\circ.$
So, $a=99^\circ$.
B).
If $b\ge 117^\circ$, then $c\ge 126^\circ$, $\ldots$, then
$a+b+c+d+e+f+g\ge (99+117+126+135+144+153+162)^\circ= 936^\circ>900^\circ$.
So, $b=108^\circ$.
C).
Same way we can show, that $c=117^\circ$, ...
After that you can find possible values for $2$ largest angles.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/474333",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
$\lfloor \sqrt n+\sqrt {n+1}+\sqrt{n+2}+\sqrt{n+3}+\sqrt{n+4}\rfloor=\lfloor\sqrt {25n+49}\rfloor$ is true? I found the following relational expression by using computer:
For any natural number $n$,
$$\lfloor \sqrt n+\sqrt {n+1}+\sqrt{n+2}+\sqrt{n+3}+\sqrt{n+4}\rfloor=\lfloor\sqrt {25n+49}\rfloor.$$
Note that $\lfloor x\rfloor$ is the largest integer not greater than $x$.
I can neither prove this nor find any counterexample even by using computer.
Could you show me how to prove this? Or could you get an counterexample?
Update: I've just asked a related question.
Generalization of $\lfloor \sqrt n+\sqrt {n+1}+\sqrt{n+2}+\sqrt{n+3}+\sqrt{n+4}\rfloor=\lfloor\sqrt {25n+49}\rfloor$
| I want to proof this only by using basic arithmetic and therefore avoiding Jensen's Inequality and Taylor's Theorem.
It's simpler if one replaces $n$ by $n-2$. On then gets
$$ \lfloor \sqrt{n-2}+\sqrt{n-1}+\sqrt n+\sqrt {n+1}+\sqrt{n+2}\rfloor=\lfloor\sqrt {25n-1}\rfloor$$
It is necessary to show that
$$ \sqrt {25n-1} \lt \sqrt{n-2}+\sqrt{n-1}+\sqrt n+\sqrt {n+1}+\sqrt{n+2} \tag{1} $$
$$ \sqrt{n-2}+\sqrt{n-1}+\sqrt n+\sqrt {n+1}+\sqrt{n+2} \lt \sqrt {25n} \tag{2}$$
We first prove
$$ \sqrt{x-b}+\sqrt{x+b} \lt \sqrt{x-a}+\sqrt{x+a} , 0 \lt a \lt b \lt x \tag{3}$$
by squaring $(3)$ we get
$$ \sqrt{x-b}\sqrt{x+b} \lt \sqrt{x-a}\sqrt{x+a} $$
and by squaring this again we get
$$x^2-b^2 \lt x^2-a^2 $$
and therefore
$$ a^2 \lt b^2 $$
The arguments is also valid in the opposite direction , because all numbers we squared where positive.
$$ \sqrt{n-2}+\sqrt{n-1}+\sqrt n+\sqrt {n+1}+\sqrt{n+2} \lt 5 \sqrt{n}$$
follows immediately from
$$\sqrt{n-2}+\sqrt{n+2}=\sqrt{n}+\sqrt{n}$$
which can be deduced from $(3)$ and $b=2$ and $a=0$,
and from
$$\sqrt{n-1}+\sqrt{n+1}=\sqrt{n}+\sqrt{n}$$
which can be deduced from $(3)$ and $b=1$ and $a=0$,
So $(2)$ is proofen.
To proof $(1)$ we notice that
$$ \sqrt{n-2}+\sqrt{n-1}+\sqrt n+\sqrt {n+1}+\sqrt{n+2} \gt 2\sqrt{n-2}+\sqrt n+ 2 \sqrt{n+2} $$
because
$$\sqrt{n-1}+ \sqrt{n+1} \gt \sqrt{n-2}+ \sqrt{n+2}$$
So it is sufficiont to check if
$$ 2 \sqrt{n-2}+\sqrt n+ 2 \sqrt{n+2} \gt \sqrt{25n-1} $$
We square
$$ 2 \sqrt{n-2} + 2 \sqrt{n+2} \gt \sqrt{25n-1} - \sqrt{n} $$
and get
$$8\,\sqrt{n-2}\,\sqrt{n+2}+8\,n>-2\,\sqrt{n}\,\sqrt{25\,n-1}+26\,n-1$$
and bring the squareroot terms to the LHS and the non non-squareroot terms to the RHS of the inequation
$$ 2\,\sqrt{n}\,\sqrt{25\,n-1}+8\,\sqrt{n-2}\,\sqrt{n+2} \gt 18\,
n-1 $$
We repeate this process until we get
$$9216\,n^3-185664\,n^2+20544\,n-66049 \gt 0$$
The polynomial can be written as
$$\left(n-21\right)\,\left(9216\,n^2+7872\,n+185856\right)+3836927$$
so for $n \ge 21$ the polynomial is positive and $(1)$ is valid.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/477108",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "47",
"answer_count": 4,
"answer_id": 3
} |
Infinite Series $\sum\limits_{n=1}^{\infty}\frac{1}{\prod\limits_{k=1}^{m}(n+k)}$ How to prove the following equality?
$$\sum_{n=1}^{\infty}\frac{1}{\prod\limits_{k=1}^{m}(n+k)}=\frac{1}{(m-1)m!}.$$
| Addendum. The partial fraction identity at the top can be shown by induction on $m$. The case $m=1$ is trivial. The induction step gives
$$\frac{1}{(m-1)!} \sum_{k=0}^{m-1} (-1)^k \binom{m-1}{k} \frac{1}{n+k+1} \frac{1}{n+m+1}
\\ = \frac{1}{(m-1)!} \sum_{k=0}^{m-1} (-1)^k \binom{m-1}{k}
\frac{1}{m-k} \left( \frac{1}{n+k+1} - \frac{1}{n+m+1} \right).$$
The first part of the sum is
$$\frac{1}{(m-1)!} \sum_{k=0}^{m-1} (-1)^k \binom{m-1}{k}
\frac{1}{m-k} \frac{1}{n+k+1} =
\frac{1}{m!} \sum_{k=0}^{m-1} (-1)^k \binom{m}{k} \frac{1}{n+k+1}.$$
The second part is
$$- \frac{1}{n+m+1} \frac{1}{(m-1)!} \sum_{k=0}^{m-1} (-1)^k \binom{m-1}{k} \frac{1}{m-k}
= - \frac{1}{n+m+1} \frac{1}{m!} \sum_{k=0}^{m-1} (-1)^k \binom{m}{k} \\
= - \frac{1}{n+m+1} \frac{1}{m!} (0 - (-1)^m)
= \frac{1}{n+m+1} (-1)^m \frac{1}{m!}.$$
Putting these two together completes the induction.
This partial fraction identity was also used here at MSE.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/477174",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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Find the shortest distance between the point $(8,3,2)$ and the line through the points $(1,2,1)$ and $(0,4,0)$ "Find the shortest distance between the point $(8,3,2)$ and the line through the points $(1,2,1)$ and $(0,4,0)$"
$$P = (1,2,1), Q = (0,4,0), A = (8,3,2)$$
$OP$ = vector to $P$
$$PQ_ = (0,4,0) - (1,2,1)$$
I found that the equation of the line $L$ that passes through $(1,2,1)$ and $(0,4,0)$ is:
$$L = OP + PQ \, t;$$
$$L = (1,2,1) + (-1,2,-1) \, t .$$
However after this I'm not sure how to proceed. I can find PA_ then draw a line from $A$ to the line $L$... advice?
| The equation of the plane passing through the points $P$, $Q$, and $A$ is given by
$\begin{vmatrix}
x & y & z & 1 \\
1 & 2 & 1 & 1 \\
0 & 4 & 0 & 1 \\
8 & 3 & 2 & 1
\end{vmatrix}=0$.
After you compute the determinant, you get $3x-6y-15z+24=0$. Then the area $S$ of the triangle defined by the points is given by $S=\frac{1}{2}\sqrt{3^2+(-6)^2+(-15)^2}=\frac{1}{2}\sqrt{270}$. The area also satisfies $S=\tfrac{1}{2}hb$, where the height $h$ is measured from $A$ to the base $b$ connecting $P$ and $Q$. So $h=2S/b$ where $b=\sqrt{(1-0)^2+(2-4)^2+(1-0)^2}=\sqrt{6}$ is the distance from $P$ to $Q$. So, you get $h=\sqrt{270}/\sqrt{6}=3\sqrt{5}$, which is the shortest distance from $A$ to the line passing through $P$ and $Q$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/478765",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 1
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Closed-form expression for this definite integral Does this definite integral have a closed-form expression?
\begin{align*}
I &= \int_0^\infty \sqrt{ \frac{1}{2} \frac{1}{x} \left( \frac{1}{(1+x)^2} + \frac{z}{(1+xz)^2} \right) } \, dx \\
&= \frac{1}{\sqrt{2}} \int_0^\infty \frac{1}{x} \sqrt{ \frac{x}{1+x} \left( 1-\frac{x}{1+x} \right) + \frac{xz}{1+xz} \left( 1-\frac{xz}{1+xz} \right) } \, dx
\end{align*}
where $x>=0$, and $z>=0$ is constant.
I have tried the following substitution:
$$
u = \frac{xz}{1+xz}, \quad
x = \frac{1}{z} \frac{u}{(1-u)}, \quad
dx = \frac{1}{z} \frac{1}{(1-u)^2} du
$$
$$ I = \frac{1}{\sqrt{2}} \int_0^1 \frac{1}{\sqrt{u(1-u)}} \sqrt{ 1 + \frac{z}{((1-u)z+u)^2} } \, du $$
...but I can't seem to make any more progress. Using $u=\frac{x}{1+x}$ yields a similar expression which is also difficult to simplify.
Can you find a way to simplify this further?
Update (Nov 4, 2013): The integral can be rearranged to the following form, which might make it easier to match something from a table of known integrals:
$$ I = \frac{1}{\sqrt{2}} \sqrt{\frac{1{+}z}{z}} \int_0^1 u^{-\frac{1}{2}} (1{-}u)^{-\frac{1}{2}} \left( 1+\frac{1{-}z}{z} u \right)^{-1} \left( 1 + 2\frac{1{-}z}{1{+}z}u + \frac{(1{-}z)^2}{z(1{+}z)}u^2 \right)^{\frac{1}{2}} \, du $$
| $$
{\rm I}\left(z\right)
\equiv
\int_{0}^{\infty}\sqrt{\vphantom{\LARGE A^{A}}\frac{1}{2x}
\left\lbrack%
\frac{1}{\left(1 + x\right)^{2}} + \frac{z}{\left(1 + xz\right)^2}
\right\rbrack\,}\ {\rm d}x\,,
\qquad
\begin{array}{|rclcl}
\,\,{\rm I}\left(0\right)
& = &
{\rm I}\left(\infty\right)
& = &
{\sqrt{2\,} \over 2}\,\pi
\\[1mm]
\,\,{\rm I}\left(1\right) & = & \pi&&
\end{array}
$$
\begin{align}
{\rm I}\left(z\right)
&=
{\sqrt{2} \over 2}\int_{0}^{\infty}\!\!
{\sqrt{z\left(z + 1\right)x^{2} + 4zx + 1 + z\,}
\over
\left(x + 1\right)\left(1 + xz\right)}\,
{{\rm d}x \over \sqrt{x\,}}
\\[3mm]&=
{\sqrt{2} \over 2}\left(z + 1 \over z\right)^{1/2}\int_{0}^{\infty}
{\sqrt{x^{2} + 4\left(z + 1\right)^{-1}\,x + z^{-1}\,}
\over
\left(x + 1\right)\left(x + z^{-1}\right)}
{{\rm d}x \over \sqrt{x\,}}&
\end{align}
Equation $x^{2} + 4\left(z + 1\right)^{-1}\,x + z^{-1} = 0$ has the complex roots:
$$
x_{\pm}
=
-\,{2 \over z + 1}
\pm
{\rm i}\,{\left\vert z -1\right\vert \over \left(z + 1\right)\,\sqrt{z\,}}\,,
\qquad\mbox{Notice that}\quad
x_{+}x_{-} = \left\vert x_{\pm}\right\vert^{2} = z^{-1}
$$
Then,
${\rm I}\left(z\right)
=
\left(\sqrt{2}/2\right)\sqrt{a^{2} + 1\,}\ {\cal I}\left(a\right)$ where
$a = z^{-1/2}$.
\begin{eqnarray*}
{\cal I}\left(a\right)
& \equiv &
\int_{0}^{\infty}
{\sqrt{x^{2} + 4a^{2}\left(a^{2} + 1\right)^{-1}\,x + a^{2}\,}
\over
\left(x + 1\right)\left(x + a^{2}\right)}
{{\rm d}x \over \sqrt{x\,}}
\\
x_{\pm}
& = &
{-2 \pm {\rm i}\,\left\vert\,a^{2} - 1\right\vert
\over
a^{2} + 1}\,a^{2}
\end{eqnarray*}
$$
{\rm I}\left(z\right)
=
{\sqrt{2\,} \over 2}\,\left(z + 1 \over z\right)^{1/2}{\cal I}\left(1 \over \sqrt{z\,}\right)\,,
\qquad\qquad
{\cal I}\left(a\right)
=
{\sqrt{2\,} \over \sqrt{a^{2} + 1\,}}\,{\rm I}\left(1 \over a^{2}\right)
$$
$$
{\cal I}\left(a\right)
=
\int_{-\infty}^{\infty}
{\sqrt{x^{4} + 4a^{2}\left(a^{2} + 1\right)^{-1}\,x^{2} + a^{2}\,}
\over
\left(x^{2} + 1\right)\left(x^{2} + a^{2}\right)}
\,{\rm d}x
$$
Integration over the complex plane is possible. You have to take into account that
$$
x^{4} + 4a^{2}\left(a^{2} + 1\right)^{-1}\,x^{2} + a^{2}
=
\left(x - x_{\atop -}^{1/2}\right)\left(x + x_{\atop -}^{1/2}\right)
\left(x - x_{\atop +}^{1/2}\right)\left(x + x_{\atop +}^{1/2}\right)
$$
which introduces branch-cuts in the complex plane.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/479993",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Chinese remainder theorem :Algebraic solution I need help in a question:
It is required to find the smallest $4$-digit number that when divided by $12,15$, and $18$ leaves remainders $8,11$, and $14$ respectively. Here's how I've attempted:
Let the number be $a$, then $$a=12p+8 = 15q+11 = 18r+14$$
Hence, $p=(5q+1)/2$ and $r=(5q-1)/6$
So, $a=15q+11$
Now if I put $q=67$, $a=1016$ (wrong answer because $r$ is not an integer) .
So where did I go wrong in the algebraic method?
| Given:
$$x \equiv 8 \pmod{12} \Rightarrow x = 12a + 8 \\
x \equiv 11 \pmod{15} \Rightarrow x = 15b + 11\\
x \equiv 14 \pmod{18} \Rightarrow x = 18c + 14$$
We solve the equations pairwise:
$$1) \ 12a+8=15b+11 \Rightarrow 4a-5b=1 \Rightarrow \begin{cases}a=5n-1 \\ b=4n-1\end{cases}$$
$$2) \ 12a+8=18c+14 \Rightarrow 2a-3c=1 \Rightarrow \begin{cases}a=3m-1 \\ c=2m-1\end{cases}$$
Hence:
$$a=5n-1=3m-1 \Rightarrow a=15k-1.$$
Now we find possible values of $k$:
$$x=12a+8\ge 1000 \Rightarrow 12(15k-1)+8\ge 1000 \Rightarrow k\ge 5.6 \Rightarrow k=6.$$
Hence:
$$a=15\cdot 6-1=89 \Rightarrow x=12\cdot 89+8=1076.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/480046",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 5
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Calculate $ \lim_{x \to 4} \frac{3 -\sqrt{5 -x}}{1 -\sqrt{5 -x}} $ How evaluate the following limit?
$$ \lim_{x \to 4} \frac{3 -\sqrt{5 -x}}{1 -\sqrt{5 -x}} $$
I cannot apply L'Hopital because $ f(x) = 3 -\sqrt{5 -x} \neq 0 $ at $x = 5$
| Let $\sqrt{5+x}=3+h$
Then $5+x=9+6h+h^2$
Let $\sqrt{5-x}=1+k$
Then $5-x=1+2k+k^2$
Adding gives: $5+x+5-x=9+6h+h^2+1+2k+k^2$
$0=6h+h^2+2k+k^2$
When $x=4$, $\sqrt{5+x}=\sqrt 9 =3 \Rightarrow 3+h=3 \Rightarrow h=0$
Similarly $k=0$
Required expression is $\frac {3-\sqrt{5+x}}{1-\sqrt{5-x}}=\frac h k$
$6h=-2k-k^2-h^2$
$\frac h k= -\frac 2 6 -\frac k6-\frac {h^2} {6k}$
As $h$ and $k$ tend to 0, $\frac h k$ tends to $-\frac 1 3$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/480816",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 2
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limit of a recursive sequence:2 Let $$x_k = \frac{A}{1-C} x_{k-1} + \frac{B}{1-C}x_{k-2},$$ where $A, B, C$ are positive reals such that $A + B + C =1$. Let $$x_1 = 1$$ and $$x_2 = 1 + y,$$ with $y$ is positive. Which conditions should I impose on the parameters $A$, $B$, $C$, $y$ such that the sequence $x_k$ goes to infinite as $k$ goes to infinite (if this is possible) ?
| Let $a=\frac{A}{1-C}$ and $b=\frac{B}{1-C}$; then $a+b=1$. As was noted in the comments, this makes $x_k$ a convex combination of $x_{k-1}$ and $x_{k-2}$, and since $a,b>0$, $x_k$ lies strictly between $x_{k-1}$ and $x_{k-2}$. Thus, $1\le x_k\le 1+y$ for all $k\ge 1$, and the sequence is bounded.
The characteristic polynomial of the recurrence $x_k=ax_{k-1}+bx_{k-2}$ is $$t^2-at-b=t^2-at-(1-a)\;,$$ whose roots are $1$ and $a-1$. Thus, $x_k=\alpha(1^k)+\beta(a-1)^k=\alpha+\beta(a-1)^k=\alpha+\beta(-b)^k$ for some $\alpha$ and $\beta$. $1=x_1=\alpha-b\beta$, and $1+y=\alpha+b^2\beta$, so $(b^2+b)\beta=y$, $$\beta=\frac{y}{b^2+b}\;,$$ and $$\alpha=1+b\beta=1+\frac{y}{b+1}\;.$$ Thus,
$$x_k=1+\frac{y}{b+1}+\frac{(-1)^kb^ky}{b(b+1)}=1+\frac{y}{b+1}\left(1+(-1)^kb^{k-1}\right)$$
for all $k\ge 1$. Clearly $$\lim_{k\to\infty}x_k=1+\frac{y}{b+1}\;,$$ since $|b|<1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/483104",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to evaluate this limit: $\lim_{x\to 0}\frac{\sqrt{x+1}-1}{x} = \frac12$? How do I evaluate the limit of $$\lim_{x\to 0}\frac{\sqrt{x+1}-1}{x} = \frac{1}{2}$$? As $x$ approaches $0$, I know the answer is $\frac{1}{2}$, but I got this question wrong. I think you have to multiply by the conjugate of the numerator? Can someone explain the steps to this, I must be doing something wrong when I multiply.
| $\require{cancel}$
Multiply numerator and denominator by the conjugate of the numerator: $$\sqrt{x+1} + 1$$ then evaluate the limit.
When we multiply by the conjugate, recall how we factor the difference of squares: $$(\sqrt a - b) \cdot (\sqrt a + b) = (\sqrt a)^2 - b^2 = a - b^2$$
$$\dfrac {\sqrt{x+1} - 1}{x} \cdot \frac{\sqrt{x+1} + 1}{\sqrt{x+1} + 1} = \dfrac {(x + 1) - 1 }{x(\sqrt{x+1} + 1)}= \dfrac {\cancel{x}}{\cancel{x}(\sqrt{x+1} + 1)} = \dfrac 1{\sqrt{x + 1} + 1}$$
Now we need only to evaluate $$\lim_{x \to 0} \dfrac 1{\sqrt{x + 1} + 1}$$
I trust you can do that.
| {
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"url": "https://math.stackexchange.com/questions/489699",
"timestamp": "2023-03-29T00:00:00",
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What should be added to $x^4 + 2x^3 - 2x^2 + x - 1$ to make it exactly divisible by $x^2 + 2x - 3$? I'm a ninth grader so please try to explain the answer in simple terms .
I cant fully understand the explanation in my book .
It just assumes that the expression that should be added has a degree of 1.
I apologize if this question is too simple or just stupid but this is a genuine doubt.
| Imagine that we add $ax+b$ to the given polynomial, obtaining a new polynomial
$$P(x)= x^4 +2x^3-2x^2+x-1+ax+b.$$
Note that $x^2+2x-3=(x+3)(x-1)$. So if $x^2+2x-3$ divides our new polynomial $P(x)$, then $P(1)=0$ and $P(-3)=0$.
We have
$$P(1)=a+b+1, \qquad\text{and}\qquad P(-3)=-3a+b+5.$$
Solve the system of linear equations $a+b+1=0$, $-3a+b+5=0$ for $a$ and $b$.
Remark: We have skipped a logical step that in principle should not be skipped. If $ax+b$ is to work, our argument shows that $a$ and $b$ must satisfy the two equations. We have not shown that if $a$ and $b$ satisfy the two equations, then $ax+b$ automatically works. This is in fact true, but requires some theory.
| {
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"url": "https://math.stackexchange.com/questions/490744",
"timestamp": "2023-03-29T00:00:00",
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Use Laplace transform to solve the following initial–value problems. Use Laplace transform to solve the following initial–value problem.
$y′′′′ + 2y′′ + y = 0, y(0) = 1, y′(0) = −1, y′′(0) = 0, y′′′(0) = 2$
Answer
$s^4 L(s) - s^3y(0) -s^2 y'(0) - s y''(0) - y'''(0) +2[s^2L(s)-sy(0)-y'(0)] +L(s) \\\\$
I get the partial fraction part and got stuck, need help!
$L(s) =\frac{s^3 - s^2 +2s}{s^4 +2s^2 +1}= \frac{s-1}{s^2 +1}+\frac{s+1}{(s^2+1)^2} \:\:$Factorising the denominator I get: $(s^2+1)^2$
Please some let me know if Im heading in the wrong direction here.
| Great job getting to this point, now we just have to get the answer in forms we can work with or use the formal definitions for inverse Laplace transforms. I will use known forms.
We have (split up the numerators):
$$\dfrac{s-1}{s^2 +1}+\dfrac{s+1}{(s^2+1)^2} = \dfrac{s}{s^2+1} - \dfrac{1}{s^2+1} + \dfrac{s}{(s^2+1)^2} + \dfrac{1}{(s^2+1)^2}$$
The inverse LT of this (using Laplace tables) is given by:
$$y(x) = \cos x - \sin x + \dfrac{1}{2} x \sin x + \dfrac{1}{2} (\sin x -x \cos x)$$
So, some simple algebra yields:
$$y(x) = \cos x - \dfrac{1}{2}\sin x + \dfrac{1}{2} x \sin x - \dfrac{1}{2} x \cos x$$
You should verify this result satisfies the original ODE.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/492513",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Evaluating $\int_a^b \frac12 r^2\ \mathrm d\theta$ to find the area of an ellipse I'm finding the area of an ellipse given by $\frac{x^2}{a^2}+\frac{y^2}{b^2} = 1$. I know the answer should be $\pi ab$ (e.g. by Green's theorem). Since we can parameterize the ellipse as $\vec{r}(\theta) = (a\cos{\theta}, b\sin{\theta})$, we can write the polar equation of the ellipse as $r = \sqrt{a^2 \cos^2{\theta}+ b^2\sin^2{\theta}}$. And we can find the area enclosed by a curve $r(\theta)$ by integrating
$$\int_{\theta_1}^{\theta_2} \frac12 r^2 \ \mathrm d\theta.$$
So we should be able to find the area of the ellipse by
$$\frac12 \int_0^{2\pi} a^2 \cos^2{\theta} + b^2 \sin^2{\theta} \ \mathrm d\theta$$
$$= \frac{a^2}{2} \int_0^{2\pi} \cos^2{\theta}\ \mathrm d\theta + \frac{b^2}{2} \int_0^{2\pi} \sin^2{\theta} \ \mathrm d\theta$$
$$= \frac{a^2}{4} \int_0^{2\pi} 1 + \cos{2\theta}\ \mathrm d\theta + \frac{b^2}{4} \int_0^{2\pi} 1- \cos{2\theta}\ \mathrm d\theta$$
$$= \frac{a^2 + b^2}{4} (2\pi) + \frac{a^2-b^2}{4} \underbrace{\int_0^{2\pi} \cos{2\theta} \ \mathrm d\theta}_{\text{This is $0$}}$$
$$=\pi\frac{a^2+b^2}{2}.$$
First of all, this is not the area of an ellipse. Second of all, when I plug in $a=1$, $b=2$, this is not even the right value of the integral, as Wolfram Alpha tells me.
What am I doing wrong?
| Your question has been answered, so now we look at how to find the area, using your parametrization, which is a perfectly good one.
The area is the integral of $|y\,dx|$ (or alternately of $|x\,dy|$. over the appropriate interval.
We have $y=b\sin\theta$ and $dx=-a\sin\theta\,d\theta$. So the area is
$$\int_0^{2\pi} |-ab\sin^2\theta|\,d\theta.$$
Using $\sin^2\theta=\frac{1-\cos 2\theta}{2}$, we find that the area is
$$\int_0^{2\pi} ab\frac{1-\cos 2\theta}{2}\,d\theta.$$
This is indeed $\pi ab$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/493104",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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} |
lcm in $\mathbb{Z}[\sqrt{-5}]$ does not exists
I need to show that lcm of $2$ and $1+\sqrt{-5}$ does not exists in $\mathbb{Z}[\sqrt{-5}]$
Getting no idea about how to start, I was thinking when does lcm cannot exists!
| I suppose you know the squared modulus function $N(a+bi)=a^2+b^2$. In $\mathbb Z[\sqrt{-5}]$ this becomes
$$
N(a+b\sqrt{-5})=a^2+5b^2
$$
The values of $a^2+5b^2$ are sums of $a^2\in\{1,4,9,16,...\}$ and $5b^2\in\{5,20,...\}$ so we have $a^2+5b^2\in\{1,4,5,6,9,13,...\}$. In particular no element $x\in\mathbb Z[\sqrt{-5}]$ has $N(x)=2,3,7,etc.$.
Furthermore $N(xy)=N(x)N(y)$ can easily be shown. Now if $xy=1$ we have $N(x)N(y)=1$ showing that $N(x)=N(y)=1$. So $x$ is a unit iff $N(x)=1$. In fact $\pm 1$ are the units in $\mathbb Z[\sqrt{-5}]$.
With this we can see that any $z\in\mathbb Z[\sqrt{-5}]$ with $N(z)=4$ has to be irreducible. For if $z=xy$ we have $N(z)=N(x)N(y)=4$ and as stated earlier $N(x)=N(y)=2$ is impossible so assuming WLOG $N(x)<N(y)$ we get $N(x)=1$ for it to divide $4$, so $x$ is a unit showing that $z$ is irreducible. A similar argument shows that if $N(z)=6$ or $N(z)=9$, then $z$ is irreducible.
Now consider $N(6)=36=2^2\cdot 3^2=6\cdot 6=4\cdot 9$. The last two expressions are the only factorizations of $36$ into factors from $\{4,5,6,9,13,...\}$ (units have been left out). So if $6=xy$ we have $N(6)=N(x)N(y)=6\cdot 6=4\cdot 9$, so either $N(x)=N(y)=6$ or $N(x)=4$ and $N(y)=9$. If we find such factors they will be irreducible by the preceding paragraph. Now
$$
6=2\cdot 3=(1+\sqrt{-5})(1-\sqrt{-5})
$$
shows that we DO have two such irreducible factorizations of $6$ in $\mathbb Z[\sqrt{-5}]$. And up to multiplication by units these must be unique.
In particular the above equations show that $6$ is a common multiple of $2$ and $1+\sqrt{-5}$. So if $\text{lcm}(2,1+\sqrt{-5})$ existed it would divide 6. And it would divide $2(1+\sqrt{-5})$ as well. But because these factorizations are irreducible only $2$ and $1+\sqrt{-5}$ are candidates that divides both $6$ and $2(1+\sqrt{-5})$ at the same time. But none of them are common multiples of $2$ and $1+\sqrt{-5}$. So finally we may conclude that the least common multiple does not exist.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/497775",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Show that the equation $x^2+xy-y^2=3$ does not have integer solutions. Show that the equation $$x^2+xy-y^2=3$$ does not have integer solutions.
I solved the equation for $x$:
$x=\displaystyle \frac{-y\pm\sqrt{y^2+4(y^2+3)}}{2}$
$\displaystyle =\frac{-y\pm\sqrt{5y^2+12}}{2}$
I was then trying to show that $\sqrt{5y^2+12}$ can not be an integer using $r^2\equiv 12 \pmod{5y^2}$.
I got stuck here.
| Just note that $5y^2$ is a multiple of $5$ and hence it ends on $0$ or $5$. Thus, $5y^2+12$ ends in $2$ or $7$, but there are no perfect squares that have this endings.
| {
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"source": "stackexchange",
"question_score": "4",
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How to prove $\cos 36^{\circ} = (1+ \sqrt 5)/4$? Given $4 \cos^2 x -2\cos x -1 = 0$.
Use this to show that $\cos 36^{\circ} = (1+ \sqrt 5)/4$, $\cos 72^{\circ} = (-1+\sqrt 5)/4$
Your help is greatly appreciated! Thanks
| To derive this from fundamentals, note that
$$\sin{108^{\circ}} = \sin{72^{\circ}}$$
then use a double-angle and triple-angle forumla:
$$\sin{2 x} = 2 \sin{x} \cos{x}$$
$$\sin{3 x} = 3 \sin{x} - 4 \sin^3{x}$$
In this case, $x=36^{\circ}$. Setting the above two equations equal to each other results in the quadratic equation in question:
$$2 \cos{x} = 3 - 4 (1-\cos^2{x}) = 4 \cos^2{x}-1$$
The rest follows from the above discussion.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How can I show that $\begin{pmatrix} 1 & 1 \\ 0 & 1\end{pmatrix}^n = \begin{pmatrix} 1 & n \\ 0 & 1\end{pmatrix}$? Well, the original task was to figure out what the following expression evaluates to for any $n$.
$$\begin{pmatrix} 1 & 1 \\ 0 & 1\end{pmatrix}^{\large n}$$
By trying out different values of $n$, I found the pattern:
$$\begin{pmatrix} 1 & 1 \\ 0 & 1\end{pmatrix}^{\large n} = \begin{pmatrix} 1 & n \\ 0 & 1\end{pmatrix}$$
But I have yet to figure out how to prove it algebraically.
Suggestions?
| Use induction on $n$.
(1) Prove the base case (trivial), perhaps even establish the case for $n = 2$ (two base cases here are not necessary, but as you found, it helps reveal the pattern.)
(2) Then assume it holds for $n = k$.
(3) Finally, show that from this assumption, it holds for $n = k+1$.
You've established the base case(s). Now, (2) assume the inductive hypothesis (IH) $$\begin{pmatrix} 1 & 1 \\ 0 & 1\end{pmatrix}^k = \begin{pmatrix} 1 & k \\ 0 & 1\end{pmatrix}.$$
Then, $$\begin{pmatrix} 1 & 1\\ 0 & 1 \end{pmatrix}^{k + 1} = \begin{pmatrix} 1 & 1\\ 0 & 1\end{pmatrix} \begin{pmatrix} 1 & 1 \\ 0 & 1\end{pmatrix}^k \quad \overset{IH}{=} \quad \begin{pmatrix} 1 & 1\\ 0 & 1\end{pmatrix} \begin{pmatrix} 1 & k \\ 0 & 1\end{pmatrix}=\quad\cdots$$
I think you can take it from here!
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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sum of irrational numbers - are there nontrivial examples? I know that the sum of irrational numbers does not have to be irrational. For example $\sqrt2+\left(-\sqrt2\right)$ is equal to $0$. But what I am wondering is there any example where the sum of two irrational numbers isn't obviously rational like an integer and yet after, say 50 digits after the decimal point it turns out to be rational. Are there any such examples with something non-trivial going on behind the scenes?
| The examples below are not quite what you're asking for (because the results, while not obviously rational, wind up being nice integers), but they may nonetheless be amusing:
$$\sqrt{3 \; + \; 2\sqrt{2}} \; - \; \sqrt{3 \; - \; 2\sqrt{2}} \; = \; 2$$
$$\sqrt[3]{3\sqrt{21} \; + \; 8} \; - \; \sqrt[3]{3\sqrt{21} \; -\; 8} \; = \; 1$$
$$\sqrt[3]{2 \; + \; \sqrt{5}} \; - \; \sqrt[3]{-2 \; + \; \sqrt{5}} \; = \; 1$$
$$\sqrt[3]{10 \; + \; \sqrt{108}} \; - \; \sqrt[3]{-10 \; + \; \sqrt{108}} \; = \; 2$$
$$\sqrt[3]{9 \; + \; 4\sqrt{5}} \; + \; \sqrt[3]{9 \; - \; 4\sqrt{5}} \; = \; 3$$
How can these be verified? In the first example, squaring, then rearranging, then squaring works. In the second example, show that the numerical expression is a solution to $x^3 + 15x - 16 = 0$ and then show that $x=1$ is the only real solution to this cubic equation by observing that $x^3 + 15x - 16 = (x-1)(x^2+x+16).$ The third example winds up being the only real solution to $x^3 + 3x - 4 = (x-1)(x^2 + x + 4),$ the fourth example winds up being the only real solution to $x^3 + 6x - 20 = (x-2)(x^2 + 2x + 10),$ and the fifth example winds up being the only real solution to $x^3 - 3x - 18 = (x-3)(x^2 + 3x + 6).$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/499784",
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"source": "stackexchange",
"question_score": "7",
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} |
What is the smallest value of $x^2+y^2$ when $x+y=6$? If $ x+y=6 $ then what is the smallest possible value for $x^2+y^2$?
Please show me the working to show where I am going wrong!
Cheers
| When Usually found yourself stuck at such questions:
One can Go for Hit and Trial Method:
Since the value of
x+y = 6
Start with lowest possible combination you can think of
For x=1, y=5 : $x^2+y^2=26$
For x=2, y=4 : $x^2+y^2=20$
For x=3, y=3 : $x^2+y^2=18$
For x=4, y=2 : $x^2+y^2=20$
For x=5, y=1 : $x^2+y^2=26$
So the minimum is 18
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Show that $\gcd(a + b, a^2 + b^2) = 1$ or $2$ if $\gcd(a, b)=1$ Show that $\gcd(a + b, a^2 + b^2) = 1$ or $2$ if $\gcd(a, b)=1$.
I have absolutely no clue where to start and what to do, please provide complete proof and answer.
| We use repeatedly the identity $\gcd(x,y)=\gcd(x,y+kx)$ for any integer $k$.
For $k=-(a+b)$, we have $$\gcd(a+b,a^2+b^2)=\gcd(a+b,-2ab)$$
But $\gcd(a,b)=1$ so $\gcd(a,a+b)=1$ and similarly $\gcd(b,a+b)=1$. Combining, we get $\gcd(ab,a+b)=1$. Hence $\gcd(a+b,a^2+b^2)=\gcd(a+b,-2)$, which is $1$ or $2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/505106",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
How to prove $(1+x)^n\geq 1+nx+\frac{n(n-1)}{2}x^2$ for all $x\geq 0$ and $n\geq 1$? I've got most of the inductive work done but I'm stuck near the very end. I'm not so great with using induction when inequalities are involved, so I have no idea how to get what I need...
\begin{align}
(1+x)^{k+1}&\geq (1+x)\left[1+kx+\frac{k(k-1)}{2}x^2\right]\\
&=1+kx+\frac{k(k-1)}{2}x^2+x+kx^2+\frac{k(k-1)}{2}x^3\\
&=1+(k+1)x+kx^2+\frac{k(k-1)}{2}x^2+\frac{k(k-1)}{2}x^3
\end{align}
And here's where I have no clue how to continue. I thought of factoring out $kx^2$ from the remaining three terms, but I don't see how that can get me anywhere.
| We need to show $(1+x)^{k+1}\ge1+(k+1)x+\frac{(k+1)k}2x^2$
You already have
$$
\begin{align}
(1+x)^{k+1}&\geq1+(k+1)x+kx^2+\frac{k(k-1)}{2}x^2+\frac{k(k-1)}{2}x^3
\end{align}$$
$$=1+(k+1)x+x^2\frac{k(k+1)}2+\frac{k(k-1)}{2}x^3$$
which is $$\ge1+(k+1)x+x^2\frac{k(k+1)}2$$ if $\displaystyle\frac{k(k-1)}{2}x^3\ge0$ which is true for $x\ge0$ and $k\ge1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/507537",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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$\lim_{x\to0} \frac{x-\sin x}{x-\tan x}$ without using L'Hopital $$\lim_{x\to0} \frac{x-\sin x}{x-\tan x}=?$$
I tried using $\lim\limits_{x\to0} \frac{\sin x}x=1$.
But it doesn't work :/
| $$\frac{x - \sin(x)}{x - \tan(x)} = \frac{x - \sin(x)}{x^3} \cdot \frac{x^3}{x - \tan(x)}$$
Let $x = 3y$ and $x\to 0 \implies y\to 0$
$$\lim_{x\to0} \frac{x - \sin(x)}{x^3} = L $$
$$L = \lim_{y\to0}\frac{3y - \sin(3y)}{(3y)^3} = \lim_{y\to0} \frac 3 {27} \frac{y - \sin(y)}{y^3} + \lim_{y\to0} \frac{4}{27} \frac{\sin^3(y)}{y^3} = \frac{1}{9} L + \frac 4{27} $$
This gives
$$\lim_{x\to0}\frac{x - \sin(x)}{x^3} = \frac 1 6 $$
\begin{align*}
L &= \lim_{y\to0}\frac{ 3y - \tan(3y)}{27 y^3} \\
&= \lim_{y\to0} \frac{1}{(1 - 3\tan^2(y ))} \cdot \frac{3y(1 - 3\tan^2(y )) - 3 \tan(y) + \tan^3(y)}{27y^3}\\
&= \lim_{y\to0} \frac{1}{(1 - 3\tan^2(y ))} \cdot \left(
\frac 3 {27} \frac{y - \tan(y)}{y^3} + \frac 1 {27} \frac{\tan^3(y)}{y^3} - \frac 9 {27} \frac{y \tan^2(y)}{y^3 }
\right )\\
&= \frac {3L}{27} + \frac 1 {27} - \frac 1 3 \\
\end{align*}
This gives other limit to be $-1/3$, put it up and get your limit.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
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Matrix Algebra Question (Linear Algebra) Find all values of $a$ such that $A^3 = 2A$, where
$$A = \begin{bmatrix} -2 & 2 \\ -1 & a \end{bmatrix}.$$
The matrix I got for $A^3$ at the end didn't match up, but I probably made a multiplication mistake somewhere.
| Here's a different approach that avoids matrix multiplication in favor of determinants:
We want to find $a$ such that $A^3=2A$. This is equivalent to $A^3-2A=0$, and to $(A^2-2I)A=0$.
Thus either $A=0$, which is impossible, or $A^2-2I$ is a zero divisor.
Thus \begin{align}0&=\det (A^2-2I) \\ &= \det(A+\sqrt2I)\det(A-\sqrt2 I),\end{align} so one of these determinants must be $0$. That is, $(-2+\sqrt 2)(a+\sqrt 2)+2=0$ or $(-2-\sqrt 2)(a-\sqrt 2)+2=0$. These have the same sole solution: $a=2$.
Note that we get the value of $A^2$ too: plugging in $a=2$ gives $\det A = -2\ne 0$, so $A$ is not a zero divisor, so $A^2-2I=0$, so $A^2=2I$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/508896",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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How would I reduce my matrix even further? How would I reduce my matrix even further?
| I usually do also pivot reduction:
\begin{align}
\large
\begin{bmatrix}
2 & 2 & 1 & 2\\
-1 & 2 & -1 & -5\\
1 & -3 & 2 & 8
\end{bmatrix}
\xrightarrow{E_1(1/2)}
&\large
\begin{bmatrix}
1 & 1 & 1/2 & 1\\
-1 & 2 & -1 & -5\\
1 & -3 & 2 & 8
\end{bmatrix}
\\
\large
\xrightarrow{E_{21}(1)}
&\large
\begin{bmatrix}
1 & 1 & 1/2 & 1\\
0 & 3 & -1/2 & -4\\
1 & -3 & 2 & 8
\end{bmatrix}
\\
\large
\xrightarrow{E_{31}(-1)}
&\large
\begin{bmatrix}
1 & 1 & 1/2 & 1\\
0 & 3 & -1/2 & -4\\
0 & -4 & 3/2 & 7
\end{bmatrix}
\\
\large
\xrightarrow{E_{2}(1/3)}
&\large
\begin{bmatrix}
1 & 1 & 1/2 & 1\\
0 & 1 & -1/6 & -4/3\\
0 & -4 & 3/2 & 7
\end{bmatrix}
\\
\large
\xrightarrow{E_{32}(4)}
&\large
\begin{bmatrix}
1 & 1 & 1/2 & 1\\
0 & 1 & -1/6 & -4/3\\
0 & 0 & 5/6 & 5/3
\end{bmatrix}
\\
\large
\xrightarrow{E_{3}(6/5)}
&\large
\begin{bmatrix}
1 & 1 & 1/2 & 1\\
0 & 1 & -1/6 & -4/3\\
0 & 0 & 1 & 2
\end{bmatrix}
\\
\text{Backwards elimination starts}\\
\large
\xrightarrow{E_{23}(1/6)}
&\large
\begin{bmatrix}
1 & 1 & 1/2 & 1\\
0 & 1 & 0 & -1\\
0 & 0 & 1 & 2
\end{bmatrix}
\\
\large
\xrightarrow{E_{13}(-1/2)}
&\large
\begin{bmatrix}
1 & 1 & 0 & 0\\
0 & 1 & 0 & -1\\
0 & 0 & 1 & 2
\end{bmatrix}
\\
\large
\xrightarrow{E_{12}(-1)}
&\large
\begin{bmatrix}
1 & 0 & 0 & 1\\
0 & 1 & 0 & -1\\
0 & 0 & 1 & 2
\end{bmatrix}
\end{align}
In this way you can read directly the solution in the last column:
$$
x=1,\quad y=-1,\quad z=2.
$$
Notation.
*
*$E_{i}(c)$ (with $c\ne0$) is “multiply the $i$-th row by $c$”.
*$E_{ij}(d)$ (with $i\ne j$) is “sum to the $i$-th row the $j$-th row multiplied by $d$”.
| {
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"question_score": "2",
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If $\frac{\cos x}{\cos y}=\frac{a}{b}$ then $a\tan x +b\tan y$ equals If $$\frac{\cos x}{\cos y}=\frac{a}{b}$$ Then $$a \cdot\tan x +b \cdot\tan y$$ Equals to (options below):
(a) $(a+b) \cot\frac{x+y}{2}$
(b) $(a+b)\tan\frac{x+y}{2}$
(c) $(a+b)(\tan\frac{x}{2} +\tan\frac{y}{2})$
(d) $(a+b)(\cot\frac{x}{2}+\cot\frac{y}{2})$
My approach :
$$\frac{\cos x}{\cos y} = \frac{a}{b} $$
[ Using componendo and dividendo ]
$$\frac{\cos x +\cos y}{\cos x -\cos y} = \frac{a+b}{a-b}$$
$$=\frac{2\cos\frac{x+y}{2}\cos\frac{x-y}{2}}{2\sin\frac{x+y}{2}\sin\frac{y-x}{2}}$$
I'm stuck, I'd aprecciate any suggestions. Thanks.
| I am hoping that my calculations haven't gone wrong. Here is a method to proceed.
From your method you have $$\frac{2\cos\frac{x+y}{2}\cos\frac{x-y}{2}}{2\sin\frac{x+y}{2}\sin\frac{y-x}{2}} = -\cot\Bigl(\frac{x+y}{2}\Bigr)\cdot\Bigl(\frac{x-y}{2}\Bigr)=\frac{a+b}{a-b}$$
Now note that
\begin{align*}
\tan(x) &=\tan\left(\frac{x+y}{2} + \frac{x-y}{2}\right)
\\ &=\frac{\tan\left(\frac{x+y}{2}\right)+\tan\left(\frac{x-y}{2}\right)}{1-\tan\left(\frac{x+y}{2}\right)\cdot\tan\left(\frac{x-y}{2}\right)} \\ &=\frac{\tan\left(\frac{x+y}{2}\right)+\tan\left(\frac{x-y}{2}\right)}{1-\frac{a-b}{a+b}} \\ &=\frac{a+b}{2b} \times \tan\left(\frac{x+y}{2}\right)+\tan\left(\frac{x-y}{2}\right)
\end{align*}
Similary $$\tan(y) =\frac{a+b}{2a} \times \biggl\{\tan\left(\frac{x+y}{2}\right)-\tan\left(\frac{x-y}{2}\right)\biggr\}$$
Now just multiply the $\tan(x)$ quantity by $a$ and $\tan(y)$ quantity by $b$ and add both the sides and see if you get the answer.
| {
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Prove that $x^2 + xy + y^2 \ge 0$ by contradiction Using the fact that $x^2 + 2xy + y^2 = (x + y)^2 \ge 0$, show that the assumption $x^2 + xy + y^2 < 0$ leads to a contradiction...
So do I start off with...
"Assume that $x^2 + xy + y^2 <0$, then blah blah blah"?
It seems true...because then I go $(x^2 + 2xy + y^2) - (x^2 + xy + y^2) \ge 0$.
It becomes $2xy - xy \ge 0$, then $xy \ge 0$.
How is this a contradiction?
I think I'm missing some key point.
| Suppose that $x^2+xy+y^2<0$; then $x^2+2xy+y^2<xy$, so $(x+y)^2<xy$. Subtracting $3xy$ from both sides of the original inequality, we see that $x^2-2xy+y^2<-3xy$, so $(x-y)^2<-3xy$. Squares are non-negative, so on the one hand $xy>0$, and on the other hand $-3xy>0$ and therefore $xy<0$.
| {
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"timestamp": "2023-03-29T00:00:00",
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How to find the maximum of $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{a+b}+\frac{1}{a+c}$ given certain constraints. Let $a,b,c\ge 0,$ and such $a+b+c=1$. Find the maximum of:
$$\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}+\dfrac{1}{a+b}+\dfrac{1}{a+c}$$
My try:
$$\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}+\dfrac{1}{a+b}+\dfrac{1}{a+c}=\dfrac{1}{1-b-c}+\dfrac{1}{b}+\dfrac{1}{c}+\dfrac{1}{1-c}+\dfrac{1}{1-b}$$
Then I can't, thank you.
| Hint:
*
*Kuhn-Tucker Multipliers For inequality constraints
*Lagrange Multipliers For equality constraints
Another Hint:
The constraint $a+b+c=1$ makes a variable redundant. For example, $a$ may be substituted with $a = 1-b-c$ (which you have correctly identified), so only the inequality constraints need to be considered.
| {
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Series Convergence What does this series converge to?
$$ \sqrt{3\sqrt {5\sqrt {3\sqrt {5\sqrt \cdots}}}} $$
and also this?
$$ \sqrt{6+\sqrt {6+\sqrt {6+\sqrt {6+\sqrt \cdots}}}} $$
And, generally speaking, how should one approach these kind of questions?
| For $$\sqrt{6 + \sqrt{6 + \sqrt{6 +\dots}}}:$$
Let \begin{align*}
x &= \text{the given equation}\\
&= \sqrt{6 + \sqrt{6 + \sqrt{6 + \dots}}}
\end{align*}
Since the series is infinite, we can write
$x = \sqrt{6 + x}$,
or $$x^2 - x - 6 = 0.$$
Therefore, $x = 3$ or $x = -2$
Since the answer cannot be negative, reject $x = -2$. Therefore, $x = 3$.
Do the same thing for the first series.
| {
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Find an angle of an isosceles triangle $\triangle ABC$ is an isosceles triangle such that $AB=AC$ and $\angle BAC$=$20^\circ$. And a point D is on $\overline{AC}$ so that AD=BC, , How to find $\angle{DBC}$?
I could not get how to use the condition $AD=BC$ , How do I use the condition to find $\angle{DBC}$?
EDIT 1: With MvG's observation, we can prove the following fact.
If we set on a point $O$ in $\triangle{ABC}$ such that $\triangle{OBC}$ is a regular triangle, then $O$ is the circumcenter of $\triangle{BCD}$.
First, we will show if we set a point $E$ on the segment $AC$ such that $OE=OB=OC=BC$, then $D=E$.
Becuase $\triangle{ABC}$ is a isosceles triangle, the point $O$ is on the bisecting line of $\angle{BAC}$. $\angle{OAE}=20^\circ/2=10^\circ$.
And because $OE=OC$, $\angle{OCE}=\angle{OEC}=20^\circ$, $\angle{EOA}=20^\circ-10^\circ=10^\circ=\angle{EAO}$.
Therefore $\triangle{AOE}$ is an isosceles triangle such that $EA=EO$. so $AD=BC=AE$, $D=E$.
Now we can see the point $O$ is a circumcenter of the $\triangle{DBC}$ because $OB=OC=OD.$
By using this fact, we can find $\angle{DBC}=70^\circ$,
| One way to calculate this is to write sin laws for two triangles $ABD$ and $BDC$. Call the angle $\angle ABD=x$. Then we have:
$$
\frac{AD}{\sin x}=\frac{BD}{\sin 20},\frac{BC}{\sin (20+x)}=\frac{BD}{\sin 80}
$$
Using $AD=BC$ and $\sin 80=\cos 10$ we get the following:
$$
\frac{\sin x}{\sin 20}=\frac{\sin (20+x)}{\sin 80}\implies \sin x=2{\sin 10}\sin (20+x)\implies\\
\tan x=\frac{2\sin 10\sin 20}{1-2\sin 10\cos 20}
$$
Now consider the following identities:
$$
1-2\sin 10\cos 20=1-(\sin 30-\sin 10)=\frac{1}{2}+\sin 10=2\cos 10 \sin 20
$$
Replacing this result in previous equation we get:
$$
\tan x=\frac{2\sin 10\sin 20}{2\cos 10\sin 20}=\tan 10 \implies x=10
$$
and hence $\angle DBC=70$.
| {
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Derivation of the quadratic equation So everyone knows that when $ax^2+bx+c=0$,$$x = \dfrac{-b\pm\sqrt{b^2-4ac}}{2a}.$$
But why does it equal this? I learned this in maths not 2 weeks ago and it makes no sense to me
| Maybe we don't know that we should create a term that can be squared. But we do know that it would be easy to solve if we eliminate the $x$ term. We can use substitution of variables to eliminate it.
$ax^2+bx+c=0$
$x=y+d$
$a(y^2+2dy+d^2)+b(y+d)+c=0$
$ay^2+ad^2+(2ad+b)y+bd+c=0$
We want: $2ad+b=0$
$d=-\frac{b}{2a}$
$x=y-\frac{b}{2a}$
Substitute for $x$ in the original:
$a(y-\frac{b}{2a})^2+b(y-\frac{b}{2a})+c=0$
$ay^2-by+\frac{b^2}{4a}+by-\frac{b^2}{2a}+c=0$
$ay^2-\frac{b^2}{4a}+\frac{4ac}{4a}=0$
$y^2=\frac{b^2}{4a^2}-\frac{4ac}{4a^2}$
$y=\pm\frac{\sqrt{b^2-4ac}}{2a}$
Substitute for $y$ in $x=y-\frac{b}{2a}$:
$x=\pm\frac{\sqrt{b^2-4ac}}{2a}-\frac{b}{2a}$
$\therefore x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$
| {
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"timestamp": "2023-03-29T00:00:00",
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What the rest of the division $1^6+2^6+...+100^6$ by $7$?
What the rest of the division $1^6+2^6+...+100^6$ by $7$?
$1^6\equiv1\pmod7\\2^6\equiv64\equiv1\pmod7\\3^6\equiv729\equiv1\pmod7$
Apparently all the leftovers are $one$, I thought of using Fermat's Little Theorem, however the $(7,7 k) = 7$, so you can not generalize, I think. help please.
| Split the sum into $1^6$ to $7^6$, then $8^6$ to $14^6$, and so on up to $92^6$ to $98^6$, plus the little tail $99^6+100^6$.
Each of the full chunks of length $7$ has shape $(7k+1)^6+(7k+2)^6+\cdots +(7k+6)^6+(7k+7)^6$. So the sum of the remainders of such a chunk on division by $7$ is the same as the remainder of $1^6+2^6+\cdots+6^6+7^6$, which is $6$. (The $7^6$ term is congruent to $0$ modulo $7$, and each of the $6$ other terms is congruent to $1$ modulo $7$, by Fermat's Theorem, or by direct calculation.)
There are $14$ full chunks, whose remainder is the same as the remainder of $(6)(14)$, which is $0$.
That leaves $99^6$ and $100^6$, which each have remainder $1$. Thus the remainder of the full sum is $2$.
| {
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How prove this $p\equiv 1\pmod 8$ let $m,n$ are positive integer numbers,and such $m<n$,if
$$p=\dfrac{n^2+m^2}{\sqrt{n^2-m^2}}$$
is prime number.
show that
$$p\equiv 1 \pmod 8$$
My try:
let $$p=\dfrac{n^2+m^2}{\sqrt{n^2-m^2}}=\dfrac{n^2-m^2+2m^2}{\sqrt{n^2-m^2}}=\sqrt{n^2-m^2}+\dfrac{2m^2}{\sqrt{n^2-m^2}}$$
Then I can't
| If $p$ divided either of $n$ and $m$, then it must also divide the other, so dividing by $p$ yields (with $n = p\nu,\, m = p\mu$)
$$1 = \frac{\nu^2 + \mu^2}{\sqrt{\nu^2-\mu^2}},$$
from which it follows that $\nu = 1,\, \mu = 0$ contradicting the positivity of $n$ and $m$. So $p$ divides neither $n$ nor $m$.
Now, for
$$\frac{n^2+m^2}{\sqrt{n^2-m^2}}$$
to be an integer, we must have
$$n^2 - m^2 = k^2,$$
and hence
$$n^2 + m^2 = k^2 + 2m^2.$$
$p = 2$ is impossible, since $\dfrac{n^2+m^2}{\sqrt{n^2-m^2}} > \sqrt{n^2+m^2}$, which would force $n = m = 1$ to have $\sqrt{n^2+m^2} < 2$. So $p$ is an odd prime.
$p \mid n^2 + m^2$ implies that $-1$ is a quadratic residue modulo $p$, so $p \equiv 1 \pmod{4}$.
$p \mid k^2 + 2m^2$ implies that $-2$ is a quadratic residue modulo $p$, so $p \equiv 1 \pmod{8}$ or $p \equiv 3 \pmod{8}$. Since we have $p \equiv 1 \pmod{4}$ from above, it follows that $p \equiv 1 \pmod{8}$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Find a good candidate for a closed-form solution of this recurrence relation: $P(n-1)+n^2$. I want to find a candidate for this recurrence relation:
$$
P(n) = \left\{\begin{aligned}
&1 &&: n = 0\\
&P(n-1)+n^2 &&: n>0
\end{aligned}
\right.$$
Starting from 0 the first 8 values are 1,2,6,15,31,56,92,141.
I can't figure out a formula for this.
| Write $P(n) - P(n-1) = n^2 = 2{n \choose 2} + {n\choose 1}$. But ${n \choose 2} = {{n+1} \choose 3}-{n \choose 3}$, and ${n\choose 1} = {{n+1} \choose 2} - {n\choose 2}$, so we have $P(n) - P(n-1) = Q(n) - Q(n-1)$, where $Q(n) = 2{{n+1} \choose 3} + {{n+1} \choose 2} $. Since $P(0) = 1$ and $Q(0) = 0$, we can now conclude that $P(n) = Q(n) +1 = 2{{n+1} \choose 3} + {{n+1} \choose 2} + 1$.
Unfortunately, there is not a very nice closed form for this. But here's the best I can do (the first term here is the sum of the first $n$ squares, a fairly well-known quantity): $\frac{(n-\frac{1}{2})\cdot n\cdot (n+\frac{1}{2})}{3} + 1$
| {
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$3x\equiv7\pmod{11}, 5y\equiv9\pmod{11}$. Find the number which $x+y\pmod{11}$ is congruent to. Given that $3x\equiv7\pmod{11}, 5y\equiv9\pmod{11}$. Find the number which $x+y\pmod{11}$ is congruent to. I'm thinking $20\equiv9\pmod{11}$, But I am having trouble find a number $3x$ that is divisible by $3$? Is there a better way of solving this problem.
| From the first congruence, by multiplying by $4$, we conclude that $x\equiv 28\equiv 6\pmod{11}$.
From the second congruence, multiplying by $2$, conclude that $-y\equiv 18\equiv 7\pmod{11}$, so $y\equiv -7\equiv 4\pmod{11}$.
Add. We get $x+y\equiv 10\pmod{11}$.
Another way: Multiply the first congruence by $5$, the second by $3$, and add. We get
$15x+15y\equiv 62\pmod{11}$. This can be rewritten as $4(x+y)\equiv -4\pmod{11}$, which gives $x+y\equiv -1\pmod{11}$.
| {
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Integral $\int_0^\infty\frac{dx}{\frac{x^4-1}{x\cos(\pi\ln x)+1}+2x^2+2}$ I need your help with this integral:
$$\int_0^\infty\frac{dx}{\frac{x^4-1}{x\cos(\pi\ln x)+1}+2\,x^2+2}.$$
I wasn't able to evaluate it in a closed form, although an approximate numerical evaluation suggested its value could be $\frac{\pi}{4}$.
| Let us introduce the notation
$$\mathcal{I}=\int_0^{\infty}\frac{dx}{\frac{x^4-1}{x\cos(\pi\ln x)+1}+2x^2+2}.$$
*
*Now observe that
\begin{align}\frac{1}{\frac{x^4-1}{x\cos(\pi\ln x)+1}+2x^2+2}&=
\frac{1}{x^2+1}\cdot\frac{1}{\frac{x^2-1}{x\cos(\pi\ln x)+1}+2}=\\
&=\frac{1}{x^2+1}\cdot\frac{\cos(\pi\ln x)+\frac1x}{x+\frac1x+2\cos(\pi\ln x)}.
\end{align}
*Using this formula and making the change of variables $x\leftrightarrow \frac1x$, we can rewrite $\mathcal{I}$ as
$$\mathcal{I}=\int_0^{\infty}\frac{1}{x^2+1}\cdot\frac{\cos(\pi\ln x)+x}{x+\frac1x+2\cos(\pi\ln x)}dx.$$
*Summing the last representation with the initial one, we get
$$2\mathcal{I}=\int_0^{\infty}\frac{dx}{1+x^2}=\frac{\pi}{2}\quad \Longrightarrow\quad \mathcal{I}=\frac{\pi}{4}.$$
| {
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Logarithm Problem : Find the number of real solutions of the equation $2\log_2\log_2x+\log_{\frac{1}{2}}\log_2(2\sqrt{2}x)=1$
Find the number of real solutions of the equation $2\log_2\log_2x+\log_{\frac{1}{2}}\log_2(2\sqrt{2}x)=1$
My approach :
Solution : Here right hand side is constant term so convert it into log of same base as L.H.S. therefore, $1$ can be written as $\log_2\log_24$
$\implies 2\log_2\log_2x+\log_{\frac{1}{2}}\log_2(2\sqrt{2}x)= \log_2\log_24$
$\implies \log_2\log_2x^2 -\log_{2}\log_2(2\sqrt{2}x)= \log_2\log_24$
$\implies \log_2 \frac{\log_2x^2}{\log_2(2\sqrt2x)}= \log_2\log_24$
$\implies \frac{\log_2x^2}{\log_2(2\sqrt2x)}= \log_24$
Please suggest whether is it the right approach... thanks...
| I think you went off the rails in the second line. Here's what I get
$$2 \log_2{\log_2{x}} = \log_2{\log_2^2{x}}$$
so that
$$2 \log_2{\log_2{x}} + \log_{1/2}{\log_2{(2 \sqrt{2} x)}} = \log_2{\frac{\log_2^2{x}}{\log_2{(2 \sqrt{2} x)}}}$$
and the equation becomes
$$\log_2{\frac{\log_2^2{x}}{\log_2{(2 \sqrt{2} x)}}} = 1 \implies \frac{\log_2^2{x}}{\log_2{(2 \sqrt{2} x)}} = 2$$
or
$$\log_2^2{x} = 2 \log_2{2^{3/2}} + 2 \log_2{x} \implies \log_2^2{x} - 2 \log_2{x}-3=0$$
This implies that $\log_2{x}=3$ or $\log_2{x}=-1$. In the former case, we have that $x=8$; in the latter, we have $x=1/2$. However, in the latter case, we have a false solution, as $\log_2{\log_2{(1/2)}} = \log_2{(-1)}$ which is outside the realm of the reals. Thus, the only solution is at $x=8$.
| {
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Evaluate $\lfloor \frac{x}{m} \rfloor + \lfloor \frac{x+1}{m} \rfloor + \dots + \lfloor \frac{x+m-1}{m} \rfloor $ For any $x \in \mathbb{R}$ and $m \in \mathbb{N} $ evaluate $\lfloor \frac{x}{m} \rfloor + \lfloor \frac{x+1}{m} \rfloor + \dots + \lfloor \frac{x+m-1}{m} \rfloor $.
Well if $x=m$ then we obviously have $\lfloor \frac{x}{m} \rfloor + \lfloor \frac{x+1}{m} \rfloor + \dots + \lfloor \frac{x+m-1}{m} \rfloor = 1 + 1 + \dots + 1=m$.
If $x=-m$ then $\lfloor \frac{x}{m} \rfloor + \lfloor \frac{x+1}{m} \rfloor + \dots + \lfloor \frac{x+m-1}{m} \rfloor=-m $.
If $x=0$ then $\lfloor \frac{x}{m} \rfloor + \lfloor \frac{x+1}{m} \rfloor + \dots + \lfloor \frac{x+m-1}{m} \rfloor=0 $.
There are many such cases I can think of, but something tells me there might be an easier solution.
| This is known as Hermite's identity. The result will always be $\lfloor x \rfloor$.
| {
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A math contest problem $\int_0^1\ln\left(1+\frac{\ln^2x}{4\,\pi^2}\right)\frac{\ln(1-x)}x \ \mathrm dx$ A friend of mine sent me a math contest problem that I am not able to solve (he does not know a solution either). So, I thought I might ask you for help.
Prove:
$$\int_0^1\ln\left(1+\frac{\ln^2x}{4\,\pi^2}\right)\frac{\ln(1-x)}x dx=-\pi^2\left(4\,\zeta'(-1)+\frac23\right).$$
| Here is a solution: Let $I$ denote the integral. Then
\begin{align*}
I &= - \int_{0}^{1} \log \left( 1 + \frac{\log^{2}x}{4\pi^{2}}\right) \mathrm{Li}_{2}'(x) \, dx \\
&= \left[ -\log \left( 1 + \frac{\log^{2}x}{4\pi^{2}}\right) \mathrm{Li}_{2}(x) \right]_{0}^{1} + 2 \int_{0}^{1} \frac{\log x}{4\pi^{2} + \log^{2} x} \frac{\mathrm{Li}_{2}(x)}{x} \, dx \\
&= -2 \int_{0}^{\infty} \frac{t}{4\pi^{2} + t^{2}} \mathrm{Li}_{2}(e^{-t}) \, dt \qquad (x = e^{-t}) \\
&= -2 \sum_{n=1}^{\infty} \frac{1}{n^{2}} \int_{0}^{\infty} \frac{t}{4\pi^{2} + t^{2}} \, e^{-nt} \, dt \\
&= -2 \sum_{n=1}^{\infty} \frac{1}{n^{2}} \int_{0}^{\infty} \left( \int_{0}^{\infty} \cos (2\pi u) e^{-tu} \, du \right) e^{-nt} \, dt \\
&= -2 \sum_{n=1}^{\infty} \frac{1}{n^{2}} \int_{0}^{\infty} \frac{\cos (2\pi u)}{u + n} \, du \\
&= -2 \sum_{n=1}^{\infty} \frac{1}{n^{2}} \int_{0}^{\infty} \frac{\cos (2\pi n u)}{u + 1} \, du \qquad (u \mapsto nu) \\
&= -2 \sum_{n=1}^{\infty} \sum_{k=1}^{\infty} \frac{1}{n^{2}} \int_{0}^{1} \frac{\cos (2\pi n u)}{u + k} \, du \\
&= -2 \sum_{k=1}^{\infty} \int_{0}^{1} \frac{1}{u + k} \left( \sum_{n=1}^{\infty} \frac{\cos (2\pi n u)}{n^{2}} \right) \, du.
\end{align*}
Now we invoke the Fourier series of the Bernoulli polynomial $B_{2}(x)$:
$$ \sum_{n=1}^{\infty} \frac{\cos 2\pi nx}{n^{2}} = \pi^{2} B_{2}(x) = \pi^{2} \left( x^{2} - x + \frac{1}{6} \right), \quad 0 \leq x \leq 1. $$
Then it follows that
\begin{align*}
I &= -2\pi^{2} \sum_{k=1}^{\infty} \int_{0}^{1} \frac{u^{2} - u + \frac{1}{6}}{u + k} \, du \\
&= \pi^{2} \sum_{k=1}^{\infty} \left\{ 2k + 1 + 2\left( k^{2} + k + \frac{1}{6} \right) \log \left( \frac{k}{k+1} \right) \right\}.
\end{align*}
Now we consider the exponential of the partial sum:
\begin{align*}
&\exp \left[ \sum_{k=1}^{N} \left\{ 2k + 1 + 2\left( k^{2} + k + \frac{1}{6} \right) \log \left( \frac{k}{k+1} \right) \right\} \right] \\
&= e^{N^{2} + 2N} \prod_{k=1}^{N} \left( \frac{k}{k+1} \right)^{2k^{2} + 2k + \frac{1}{3}} \\
&= \frac{e^{N^{2} + 2N}}{(N + 1)^{2N^{2} + 2N + \frac{1}{3}}} \prod_{k=1}^{N} k^{4k} \\
&= \frac{e^{2N}}{\left(1 + \frac{1}{N}\right)^{2N^{2} + 2N + \frac{1}{3}}} \left\{ \frac{e^{N^{2}/4}}{N^{N^{2}/2 + N/2 + 1/12}} \prod_{k=1}^{N} k^{k} \right\}^{4}.
\end{align*}
In view of the definition of Glaisher-Kinkelin constant $A$, we have
$$ \lim_{N\to\infty} \exp \left[ \sum_{k=1}^{N} \left\{ 2k + 1 + 2\left( k^{2} + k + \frac{1}{6} \right) \log \left( \frac{k}{k+1} \right) \right\} \right] = \frac{A^{4}}{e}. $$
This, together with the identity $ \log A = \frac{1}{12} - \zeta'(-1)$, yields
$$ I = \pi^{2} ( 4 \log A - 1 ) = -\pi^{2} \left(4\zeta'(-1) + \frac{2}{3} \right) $$
as desired.
| {
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integrate $ \frac {(x^3 + 36)} {(x^2 + 36)}$ I know I have to use long division first, but I don't really know how to do it in this case
$$\int \frac{x^3 + 36}{x^2 + 36}dx$$
| HINT:
Using Partial fraction decomposition $$\frac{x^3+36}{x^2+36}=x+A+B\frac{\frac{d(x^2+36)}{dx}}{x^2+36}+C\frac1{x^2+36}$$
$$\implies \frac{x^3+36}{x^2+36}=x+A+\frac{2Bx}{x^2+36}+\frac C{x^2+36}$$
$$\implies x^3+36=(x+A)(x^2+36)+2Bx+C$$
$$\implies x^3+36=x^3+Ax^2+x(36+2B)+36A+C$$
Compare the coefficients of the different power of $x$ to find $A,B,C$
Now, $$\int\frac{d f(x)}{dx}dx=\int d f(x)=f(x)+C$$ and we know about $$\int\frac{dx}{x^2+6^2}$$
| {
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$a+b+c=0;a^2+b^2+c^2=1$ then $a^4+b^4+c^4$ is equal to what?
$a+b+c=0;a^2+b^2+c^2=1$ then $a^4+b^4+c^4$ is equal to what?
I tried to solve this problem, and I get $a^4+b^4+c^4 = 2(a^2b^2 + 1)$ but I'm not sure if it's correct
| Given $a+b+c = 0$ and $(a^2+b^2+c^2) = 1$, Now $(a^2+b^2+c^2)^2 = 1^2 = 1$
$(a^4+b^4+c^4)+2(a^2b^2+b^2c^2+c^2a^2) = 1.................(1)$
and from $(a+b+c)^2 = 0$,
we get $\displaystyle 1+2(ab+bc+ca) = 0\Rightarrow (ab+bc+ca) = -\frac{1}{2}$
again squaring both side , we get $\displaystyle (ab+bc+ca)^2 = \frac{1}{4}$
$\displaystyle (a^2b^2+b^2c^2+c^2a^2)+2abc(a+b+c) = \frac{1}{4}\Rightarrow (a^2b^2+b^2c^2+c^2a^2) = \frac{1}{4}$
So put in eqn.... $(1)$ , we get
$\displaystyle (a^4+b^4+c^4)+2\cdot \frac{1}{4} = 1\Rightarrow (a^4+b^4+c^4) = \frac{1}{2}$
| {
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"url": "https://math.stackexchange.com/questions/524335",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Standard inductive problem Question: Prove that $2^n \geq (n+1)^2$ for all $n \geq 6$.
I have tried to prove this below and I'm interested if my method was correct and if there is a simpler answer since my answer seems unnecessarily long for such a simple claim.
Inductive hypothesis $$2^n \geq (n+1)^2$$
We need to show that $2^{n+1} \geq (n+2)^2$ or alternatively
$$2^n2 \geq (n+1)^2 \frac{(n+2)^2}{(n+1)^2}$$
claim $$\frac{(n+2)^2}{(n+1)^2} <2, \forall n \geq 6$$
notice that it is true for $n=6$ and $$\frac{\frac{(n+2)^2}{(n+1)^2}}{\frac{((n+1)+2)^2}{((n+1)+1)^2}}<1$$ so $\frac{(n+2)^2}{(n+1)^2}$ is decreasing as $n$ gets larger so we have proven the above claim.
So we have that for all $n \geq 6$ we have $2^n \geq (n+1)^2$ and by our induction hypothesis and $\frac{(n+2)^2}{(n+1)^2} <2$ so we can conclude that $2^n2 \geq (n+1)^2 \frac{(n+2)^2}{(n+1)^2}$.
| You have $2^{n+1}=2\times 2^{n}\geq 2(n+1)^2$
Now, for $n\geq 6,$
$$2(n+1)^2-(n+2)^2$$
$$=2(n^2+2n+1)-(n^2+4+4n)$$
$$=2n^2+4n+2-n^2-4-4n$$
$$=n^2-2\geq0$$
$$\therefore 2(n+1)^2\geq(n+2)^2$$
$$\therefore 2^{n+1}\geq(n+2)^2$$
:)
| {
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"timestamp": "2023-03-29T00:00:00",
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Rabin's test for polynomial irreducibility over $\mathbb{F}_2$ I know that $f(x) = x^{169}+x^2+1$ is a reducible polynomial over $\mathbb{F}_2$. I want to show this using Rabin's irreduciblity test. First, I then have to check if $f$ is a divisor of $x^{2^{169}}+x$ and this is where I get stuck. That exponent is way too large to work with. How to handle such cases?
| This is probably not what you wanted to see, but it is easy to see that your polynomial is divisible by $x^2+x+1=(x^3-1)/(x-1)$. Here
$$
x^{169}+x^2+1=(x^{169}+x)+(x^2+x+1)=x(x^{3\cdot56}+1)+(x^2+x+1),
$$
and clearly $x^{3\cdot56}-1$ is divisible by $x^3-1$ and hence by $x^2+x+1$.
It is usually easy to check the divisibility of a `sparse' polynomial by the low degree polynomials: $x^2+x+1$, $x^3+x+1$, $x^3+x^2+1$, $x^4+x+1$, $x^4+x^3+1$, $x^4+x^3+x^2+x+1$, because they are respectively factors of $x^3-1$, $x^7-1$, $x^7-1$, $x^{15}-1$, $x^{15}-1$ and $x^5-1$.
| {
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"source": "stackexchange",
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showing algebraic inequality with arithmetic and harmonic means Let x, y, z be positive real numbers. Prove that $$\frac{x}{y+z}+\frac{y}{z+x}+\frac{z}{x+y} \ge \frac{3}{2}$$
This problem appears to be simple, but upon further work and lots of failed attempts, I am stuck. I have tried using arithmetic and harmonic means (which I am sure are the key) to show that there exists some number which is fits between these two, thus proving the inequality. I have also tried multiplying it out and simplifying and obtained:
$$1 + \frac{x^3 + y^3 + z^3 + xyz}{(x+y)(y+z)(z+x)} \ge \frac{3}{2} $$ This didn't really seem to help. Any guidance is greatly appreciated!
| $$A=\dfrac a{b+c}+\dfrac{b}{a+c}+\dfrac{c}{a+b}$$
$$B= \dfrac b{b+c}+ \dfrac{c}{c+a}+\dfrac{a}{a+b}$$
$$C=\dfrac c{b+c}+\dfrac{a}{a+c}+\dfrac{b}{b+a}$$
By A-G inequality
$$A+B=\dfrac{a+b}{b+c}+\dfrac{b+c}{a+c}+\dfrac{c+a}{b+a}\ge 3$$
$$A+C=\dfrac{a+c}{b+c}+\dfrac{a+b}{a+c}+\dfrac{b+c}{b+a}\ge3$$
so
+
$$(A+B)+(A+C)\ge6$$
$B+C=3$, we get $2A\ge3$, yeah
| {
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"timestamp": "2023-03-29T00:00:00",
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A $2 \times 2$ matrix $A$ such that $A^n$ is the identity matrix So basically determine a $2 \times 2$ matrix $A$ such that $A^n$ is an identity matrix, but none of $A^1, A^2,..., A^{n-1}$ are the identity matrix. (Hint: Think geometric mappings)
I don't understand this question at all, can someone help please?
| Here's a cute way of looking at it: let
$J = \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix}; \tag{1}$
then
$J^2 = -I. \tag{2}$
From (1) and (2) it follows that
$((\cos \theta)I + (\sin \theta)J)^n = ((\cos n \theta)I + (\sin n \theta)J); \tag{3}$
the proof of (3) is virtually identical to that of de Moivre's formula (see http://en.m.wikipedia.org/wiki/De_Moivre's_formula); indeed, the algebraic maneuvers are essentially the same in either case. Noting that
$(\cos \theta)I + (\sin \theta)J = \begin{bmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end {bmatrix}, \tag{4}$
it follows from (3) that
$\begin{bmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{bmatrix}^n = \begin{bmatrix} \cos n \theta & -\sin n \theta \\ \sin n \theta & \cos n \theta \end{bmatrix}. \tag{5}$
Now setting $\theta = \frac{2 \pi}{n}$, we let
$A = \begin{bmatrix} \cos \frac{2 \pi}{n} & -\sin \frac{2\pi}{n} \\ \sin \frac{2\pi}{n} & \cos \frac{2\pi}{n} \end{bmatrix}, \tag{6}$
and we see from (5) that
$A^n = \begin{bmatrix} \cos 2\pi & -\sin 2\pi \\ \sin 2\pi & \cos 2\pi \end{bmatrix} = I, \tag{7}$
whereas for $k$, $1 \le k \le n -1$,
$A^k = \begin{bmatrix} \cos \frac{2\pi k}{n} & -\sin \frac{2\pi k}{n} \\ \sin \frac{2\pi k}{n} & \cos \frac{2 \pi k}{n} \end{bmatrix} \ne I, \tag{8}$
establishing the desired result.
Hope this helps. Cheers,
and as always,
Fiat Lux!!!
| {
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Proof of $10^{n+1} -9n -10 \equiv 0 \pmod {81}$ I am trying to prove that $10^{n+1} -9n -10 \equiv 0 \pmod {81}$. I think that decomposing into 9 and then 9 again is the way to go, but I just cannot get there. Any help is greatly appreciated.
\emph{edit} I originally posted this a $9^n$ not $9n$. Apologies.
| $$
\begin{align}
&10^{n+1}-9n-10\\[9pt]
&=9\left(10\frac{10^n-1}{10-1}-n\right)\\
&=9\left(10\left(10^{n-1}+10^{n-2}+\dots+10+1\right)+9n-10n\right)\\
&=9\left(10\cdot9\left(\frac{10^{n-1}-1}{10-1}+\frac{10^{n-2}-1}{10-1}+\dots+\frac{10-1}{10-1}+\frac{1-1}{10-1}\right)+9n\right)\\
&=81\left(10\left(\frac{10^{n-1}-1}{10-1}+\frac{10^{n-2}-1}{10-1}+\dots+\frac{10-1}{10-1}+\frac{1-1}{10-1}\right)+n\right)
\end{align}
$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Inequality. $\frac{a}{\sqrt{b}}+\frac{b}{\sqrt{c}}+\frac{c}{\sqrt{a}}\geq3$ Let $a,b,c \in (0, \infty)$, with $a+b+c=3$. How can I prove that:
$$\frac{a}{\sqrt{b}}+\frac{b}{\sqrt{c}}+\frac{c}{\sqrt{a}}\geq3 ?$$.
I try to use Cauchy-Schwarz rewriting the inequality like :
$$\sum_{cyc}\frac{a\sqrt{b}}{b} \geq \frac{(\sum_{cyc}{\sqrt[4]{a^2b}})^2}{a+b+c}$$ but I don't obtain anything.
| We need to prove that
$$\sum_{cyc}\frac{a^2}{b}\geq3,$$ where $a$, $b$ and $c$ are positives such that $a^2+b^2+c^2=3.$
Indeed,
$$\sum_{cyc}\frac{a^2}{b}-3=\sum_{cyc}\left(\frac{a^2}{b}-2a+b\right)-\left(3-a-b-c\right)=$$
$$=\sum_{cyc}\frac{(a-b)^2}{b}-\frac{9-(a+b+c)^2}{3+a+b+c}=\sum_{cyc}\frac{(a-b)^2}{b}-\frac{3(a^2+b^2+c^2)-(a+b+c)^2}{3+a+b+c}=$$
$$\sum_{cyc}\frac{(a-b)^2}{b}-\frac{\sum\limits_{cyc}(a-b)^2}{3+a+b+c}=\sum_{cyc}\frac{(a-b)^2(3+a+c)}{b(3+a+b+c)}\geq0$$ and we are done!
| {
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Did I solve this limit problem correctly? $\lim_{x \to 3}\left(\frac{6-x}{3}\right)^{\tan \frac{\pi x}{6}}$ Need to solve this limit
$$\lim_{x \to 3}\left(\frac{6-x}{3}\right)^{\tan \frac{\pi x}{6}}$$
When I put $x$ in this expression I have indeterminate form $1^{\infty }.$
So I choose the way which was described on this page (2nd method)
http://www.vitutor.com/calculus/limits/one_infinity.html
$$e^{\lim_{x \to 3 }{\tan \frac{\pi x}{6}(\frac{6-x}{3}-1)}}=e^{\lim_{x \to 3 }{\tan \frac{\pi x}{6}(\frac{6-x}{3}-\frac{3}{3})}} = e^{\lim_{x \to 3 }{\tan \frac{\pi x}{6}(\frac{3-x}{3})}}$$
and now I just put $x$ in this expression and I got $$e^{\infty \cdot 0} = e^0 =1$$
P.S. If you don't see my formulas, then
Right click on the math, go to Math Settings > Scale All Math... and enter your favorite number bigger than 100.
| Your solution is correct halfway:
$$\lim_{x \to 3}\left(\frac{6-x}{3}\right)^{\tan \frac{\pi x}{6}}=\exp\left\{\lim_{x \to 3 }{\tan \frac{\pi x}{6}\left(\frac{6-x}{3}-1\right)}\right\}$$ Above you used two facts:
*
*$\lim(\exp)=\exp(\lim)$, which is the continuity of exponent
*$\frac{1+t}{t}\to1$ if $t\to0$: $$\lim_{x\to3} \tan\frac{\pi x}{6}\ln \frac{6-x}{3}=$$ $$\lim_{x\to3} \tan\frac{\pi x}{6}\ln\left(1+\frac{6-x}{3}-1\right)=\lim_{x\to3} \tan\frac{\pi x}{6}\ln\left(1+\frac{6-x}{3}-1\right)=$$ $$\lim_{x\to3} \tan\frac{\pi x}{6}\left(\frac{6-x}{3}-1\right)\frac{\ln\left(1+\frac{6-x}{3}-1\right)}{\frac{6-x}{3}-1}=$$ $$\lim_{x\to3} \tan\frac{\pi x}{6}\left(\frac{6-x}{3}-1\right)\cdot \underbrace{\lim_{t\to0}\frac{\ln\left(1+t\right)}{t}}_1$$
Here we set $t=\frac{6-x}{3}-1\to0$ as $x\to 3$
Now it remains to calculate $$\lim_{x\to3} \tan\frac{\pi x}{6}\left(\frac{6-x}{3}-1\right)=\lim_{x\to3} \tan\frac{\pi x}{6}\frac{3-x}{3}=\frac{1}{3}\lim_{x\to3} (3-x)\tan\frac{\pi x}{6}$$
if we try to substitute $x=3$ we will obtain $0\cdot\infty$, which is one of the indeterminate forms and we cannot immediately conclude that it is $0$. We must carefully proceed with addressing this indeterminate forms. Further computation can take may paths. For example, we can use the L'Hospital's Rule $$\lim_{x\to3} (3-x)\tan\frac{\pi x}{6}=\lim_{x\to3}\frac{3-x}{\cot\frac{\pi x}{6}}=\left[\frac{\infty}{\infty}\right]=\lim_{x\to3}\frac{(3-x)'}{\left(\cot\frac{\pi x}{6}\right)'}=$$ $$\lim_{x\to3}\frac{-1}{-\frac{\pi}{6}\csc^2\frac{\pi x}{6}}=\frac{6}{\pi}$$
We do not forget the $1/3$ and the $exp$ in front:
$$\lim_{x \to 3}\left(\frac{6-x}{3}\right)^{\tan \frac{\pi x}{6}}=\exp\frac{1}{3}\frac{6}{\pi}=e^\frac{2}{\pi}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Find the coefficient of a power of x in the product of polynomials - Link with Combinations? I came across a new set of problems while studying combinatorics which involves restrictions to several variables and use of multinomial theoram to evaluate the number of possible combinations of the variables subjected to those restrictions.For example
If we were to pick 6 balls out of a bag which contained 3 red , 4 green and 5 white identical balls then the method i've been taught is as follows :
Suppose
$a$ = red balls , $b$= green balls and $c$=white balls .
Now,
$a +b+c =6$ where $0<=a<=3,0<=b<=4,0<=c<=5.$
Now i am supposed to find the coefficient of $x^6$ in the following product :
$(1+x^1+x^2+x^3)(1+x^1+x^2+x^3+x^4)(1+x^1+x^2+x^3+x^4+x^5)$ and that is the solution.
I want to understand what's actually happening here since finding a coefficient seems to have no link with the possible combinations. THANKS.
| Consider for a moment the product
$$\left(x^0+x^1+x^2+x^3\right)\left(y^0+y^1+y^2+y^3+y^4\right)\left(z^0+z^1+z^2+z^3+z^4+z^5\right)\;.$$
When you multiply it out, you get a total of $4\cdot5\cdot6=120$ terms, each of the form $x^ay^bz^c$, where $0\le a\le 3$, $0\le b\le 4$, and $0\le c\le 5$.
Now change the $y$’s and $z$’s back to $x$’s to get the original product; when multiplied out, it’s still a sum of $120$ terms, which are now of the form $x^ax^bx^c=x^{a+b+c}$, where $0\le a\le 3$, $0\le b\le 4$, and $0\le c\le 5$. The lowest power of $x$ in the product is clearly $x^0$, and the highest is $x^{3+4+5}=x^{12}$.
Suppose that $0\le n\le 12$; what’s the coefficient of $x^n$ in this product when we collect like terms? It’s simply the number of products $x^ax^bx^c$ for which $a+b+c=n$. In other words, the coefficient of $x^n$ in this product is the number of solutions to the equation $a+b+c=n$ such that $a,b$, and $c$ are integers satisfying the inequalities $0\le a\le 3$, $0\le b\le 4$, and $0\le c\le 5$. In your ball problem that’s the number of distinguishable combinations of $n$ red, green, and white balls that can be drawn from that particular bag.
| {
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"timestamp": "2023-03-29T00:00:00",
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finding determinant as an function in given matrix Calculate the determinant of the following matrix as an explicit
function of $x$. (It is a polynomial in $x$. You are asked to find
all the coefficients.)
\begin{bmatrix}1 & x & x^{2} & x^{3} & x^{4}\\
x^{5} & x^{6} & x^{7} & x^{8} & x^{9}\\
0 & 0 & 0 & x^{10} & x^{11}\\
0 & 0 & 0 & x^{12} & x^{13}\\
0 & 0 & 0 & x^{14} & x^{15}
\end{bmatrix}
Can someone help me with this question?
| Hint:
$$\det\begin{pmatrix}A & 0\\ C & D\end{pmatrix} = \det\begin{pmatrix}A & B\\ 0 & D\end{pmatrix} = \det(A) \det(D)$$
Where $A,B,C,D$ are block matrices.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Finite Sum $\sum\limits_{k=1}^{m-1}\frac{1}{\sin^2\frac{k\pi}{m}}$
Question : Is the following true for any $m\in\mathbb N$?
$$\begin{align}\sum_{k=1}^{m-1}\frac{1}{\sin^2\frac{k\pi}{m}}=\frac{m^2-1}{3}\qquad(\star)\end{align}$$
Motivation : I reached $(\star)$ by using computer. It seems true, but I can't prove it. Can anyone help?
By the way, I've been able to prove $\sum_{n=1}^{\infty}\frac{1}{n^2}=\frac{{\pi}^2}{6}$ by using $(\star)$.
Proof : Let
$$f(x)=\frac{1}{\sin^2x}-\frac{1}{x^2}=\frac{(x-\sin x)(x+\sin x)}{x^2\sin^2 x}.$$
We know that $f(x)\gt0$ if $0\lt x\le {\pi}/{2}$, and that $\lim_{x\to 0}f(x)=1/3$. Hence, letting $f(0)=1/3$, we know that $f(x)$ is continuous and positive at $x=0$. Hence, since $f(x)\ (0\le x\le {\pi}/2)$ is bounded, there exists a constant $C$ such that $0\lt f(x)\lt C$. Hence, substituting $x={(k\pi)}/{(2n+1)}$ for this, we get
$$0\lt \frac{1}{\frac{2n+1}{{\pi}^2}\sin^2\frac{k\pi}{2n+1}}-\frac{1}{k^2}\lt\frac{{\pi}^2C}{(2n+1)^2}.$$
Then, the sum of these from $1$ to $n$ satisfies
$$0\lt\frac{{\pi}^2\cdot 2n(n+1)}{(2n+1)^2\cdot 3}-\sum_{k=1}^{n}\frac{1}{k^2}\lt\frac{{\pi}^2Cn}{(2n+1)^2}.$$
Here, we used $(\star)$. Then, considering $n\to\infty$ leads what we desired.
| Setting $\tan mx=0$ in Sum of tangent functions where arguments are in specific arithmetic series,
we get $$\binom m1\tan x-\binom m3\tan^3x+\cdots=0\ \ \ \ (1)$$
Multiplying both sides by $\cot^mx,$
$$\binom m1\cot^{m-1}x-\binom m3\cot^{m-3}x+\cdots=0\ \ \ \ (2)$$
$\tan mx=0\implies mx=n\pi$ where $n$ is any integer
$\implies$ the roots of $(2)$ are $\cot x$ where $x=\dfrac{n\pi}m$ where $1\le n\le m-1$
Using Vieta's formula, $$\sum_{r=1}^{m-1}\cot^2\dfrac{r\pi}m=\left(\sum_{r=1}^{m-1}\cot\dfrac{r\pi}m\right)^2-2\left(\sum_{r_1,r_2=1,r_1>r_2}^{m-1}\cot\dfrac{r_1\pi}m\cot\dfrac{r_2\pi}m\right)$$
$$=0-2\cdot\left(-\frac{\binom m3}{\binom m1}\right)=\cdots$$
$$\sum_{r=1}^{m-1}\frac1{\sin^2\dfrac{r\pi}m}=\sum_{r=1}^{m-1}\left(1+\cot^2\dfrac{r\pi}m\right)=m-1+\sum_{r=1}^{m-1}\cot^2\dfrac{r\pi}m=\cdots$$
Hope the steps are understandable
| {
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Prove that $ \int_{0}^{1}\frac{\sqrt{1-x^2}}{1-x^2\sin^2 \alpha}dx = \frac{\pi}{4\cos^2 \frac{\alpha}{2}}$. Prove that $\displaystyle \int_{0}^{1}\frac{\sqrt{1-x^2}}{1-x^2\sin^2 \alpha}dx = \frac{\pi}{4\cos^2 \frac{\alpha}{2}}$.
$\bf{My\; Try}::$ Let $x = \sin \theta$, Then $dx = \cos \theta d\theta$
$\displaystyle = \int _{0}^{1}\frac{\cos \theta }{1-\sin^2 \theta \cdot \sin^2 \alpha}\cdot \cos \theta d\theta = \int_{0}^{1}\frac{\cos ^2 \theta }{1-\sin^2 \theta \cdot \sin ^2 \alpha}d\theta$
$\displaystyle = \int_{0}^{1}\frac{\sec^2 \theta}{\sec^4 \theta -\tan^2 \theta \cdot \sec^2 \theta \cdot \sin^2 \alpha}d\theta = \int_{0}^{1}\frac{\sec^2 \theta }{\left(1+\tan ^2 \theta\right)^2-\tan^2 \theta \cdot \sec^2 \theta\cdot \sin^2 \alpha}d\theta$
Let $\tan \theta = t$ and $\sec^2 \theta d\theta = dt$
$\displaystyle \int_{0}^{1}\frac{1}{(1+t^2)^2-t^2 (1+t^2)\sin^2 \alpha}dt$
Now How can I solve after that
Help Required
Thanks
| Here is a much nicer way than my first attempt. Use $x=\sin(\theta)$ and $u=\tan(\theta)$
$$
\begin{align}
&\int_0^1\frac{\sqrt{1-x^2}}{1-x^2\sin^2(\alpha)}\,\mathrm{d}x\\
&=\int_0^{\pi/2}\frac{\cos^2(\theta)}{1-\sin^2(\theta)\sin^2(\alpha)}\,\mathrm{d}\theta\\
&=\int_0^{\pi/2}\frac{\mathrm{d}\theta}{\sec^2(\theta)-\tan^2(\theta)\sin^2(\alpha)}\\
&=\int_0^{\pi/2}\frac{\mathrm{d}\theta}{1+\tan^2(\theta)\cos^2(\alpha)}\\
&=\int_0^{\pi/2}\frac{\mathrm{d}\tan(\theta)}{(1+\tan^2(\theta)\cos^2(\alpha))(1+\tan^2(\theta))}\\
&=\frac1{\sin^2(\alpha)}\int_0^{\pi/2}\left(\frac1{1+\tan^2(\theta)}-\frac{\cos^2(\alpha)}{1+\tan^2(\theta)\cos^2(\alpha)}\right)\,\mathrm{d}\tan(\theta)\\
&=\frac1{\sin^2(\alpha)}\int_0^\infty\left(\frac1{1+u^2}-\frac{\cos^2(\alpha)}{1+u^2\cos^2(\alpha)}\right)\,\mathrm{d}u\\
&=\frac1{\sin^2(\alpha)}\left(\frac\pi2-\frac\pi2\cos(\alpha)\right)\\
&=\frac\pi2\frac1{1+\cos(\alpha)}\\
&=\frac\pi{4\cos^2(\alpha/2)}
\end{align}
$$
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Limit of $\frac{1}{a} + \frac{2}{a^2} + \cdots + \frac{n}{a^n}$ What is the limit of this sequence $\frac{1}{a} + \frac{2}{a^2} + \cdots + \frac{n}{a^n}$?
Where $a$ is a constant and $n \to \infty$.
If answered with proofs, it will be best.
| With $S_n = \frac{1}{a} + \frac{2}{a^2} + \frac{3}{a^3} + \cdots \frac{n}{a^n}$ and using the closed form of geometric sums,
$$
\begin{align}
\lim_{n \to \infty} S_n &= \frac{1}{a} + \frac{2}{a^2} + \frac{3}{a^3} + \cdots \\
&= \frac{1}{a} + \frac{1}{a^2} + \frac{1}{a^3} + \cdots \\
& \phantom{=\frac{1}{a}} + \frac{1}{a^2} + \frac{1}{a^3} + \cdots \\
& \phantom{=\frac{1}{a} + \frac{1}{a^2}} + \frac{1}{a^3} + \cdots \\
& \phantom{=\frac{1}{a} + \frac{1}{a^2} + \frac{1}{a^3}} \ddots \\
&= \frac{1}{a - 1} + \frac{1}{a} \cdot \frac{1}{a - 1} + \frac{1}{a^2} \cdot \frac{1}{a - 1} + \cdots \\
&= \frac{1}{a - 1} \left( 1 + \frac{1}{a} + \frac{1}{a^2} + \cdots \right) \\
&= \frac{1}{a - 1} \cdot \frac{a}{a - 1} \\
&= \frac{a}{(a - 1)^2}
\end{align}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/551234",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 0
} |
The number of integral solutions for the equation $x-y = x^2 + y^2 - xy$ Find the number of integral solutions for the equation $x-y = x^2 + y^2 - xy$ and the equation of type $x+y = x^2 + y^2 - xy$
| Rearranging we get $$x^2-x(y+1)+y^2+y=0$$ which is a Quadratic Equation in $x$
As $x$ must be real, the discriminant must be $\ge0$ i.e.,
$(y+1)^2-4(y^2+y)=-3y^2-2y+1\ge0$
$\iff 3y^2+2y-1\le0$
$\iff \{y-(-1)\}(y-\frac13)\le0$
$\iff -1\le y\le \frac13$
Now, use the fact that $y$ is integer
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/555235",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 2
} |
How prove this inequality $\frac{x^2y}{z}+\frac{y^2z}{x}+\frac{z^2x}{y}\ge x^2+y^2+z^2$ let $x\ge y\ge z\ge 0$,show that
$$\dfrac{x^2y}{z}+\dfrac{y^2z}{x}+\dfrac{z^2x}{y}\ge x^2+y^2+z^2$$
my try:
$$\Longleftrightarrow x^3y^2+y^3z^2+z^3x^2\ge xyz(x^2+y^2+z^2)$$
| Another way.
$$\sum_{cyc}\left(\frac{x^2y}{z}-x^2\right)=(y-z)^2\left(\frac{x}{z}+\frac{x}{y}-1\right)+(x-y)(x-z)\left(\frac{y}{z}+\frac{y}{x}-1\right)\geq0.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/555400",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 3
} |
How to find $x^4+y^4+z^4$ from equation? Please help me.
There are equations: $x+y+z=3, x^2+y^2+z^2=5$ and $x^3+y^3+z^3=7$. The question:
what is the result of $x^4+y^4+z^4$?
Ive tried to merge the equation and result in desperado. :(
Please explain with simple math as I'm only a junior high school student. Thx a lot
| We have in general if
$$
s_1:=x+y+z\textrm{, }s_2:=x^2+y^2+z^2\textrm{, }s_3:=x^3+y^3+z^3
$$
and
$$
s_4:=x^4+y^4+z^4
$$
and if
$$
\sigma_1=x+y+z\textrm{, }\sigma_2=xy+yz+zx\textrm{, }\sigma_3=xyz\tag 1
$$
Then
$$
s_1=\sigma_1\textrm{, }s_2=\sigma_1^2-2\sigma_2\textrm{, }s_3=\sigma_1^3-3\sigma_1\sigma_2+3\sigma_3\textrm{, }s_4=\sigma_1^4-4\sigma_1^2\sigma_2+2\sigma_2^2+4\sigma_1 \sigma_3\tag 2
$$
Hence if $s_1=3,s_2=5,s_3=7$, then
$$
\sigma_1=s_1=3\textrm{, }\sigma_2=2\textrm{, }\sigma_3=-2/3
$$
Hence
$$
s_4=x^4+y^4+z^4=9
$$
For to find $x^5+y^5+z^5$, we use
$$
(x+y+z)^5=x^5+y^5+z^5+20xyz(x^2+y^2+z^2)+30xyz(xy+yz+zx)+5 S,
$$
where
$$
S=\sum_{cyc}\left(x^4y+xy^4+2x^3y^2+2x^2y^3\right)
=\sum_{cyc}xy\left(x^3+y^3\right)+2x^2y^2z^2\sum_{cyc}\frac{x+y}{z^2}=
$$
$$
=\sum_{cyc}xy\left(x^3+y^3+z^3-z^3\right)+2\sigma_3^2\sum_{cyc}\frac{s_1-z}{z^2}=
$$
$$
=s_3\sum_{cyc}xy-\sigma_3s_2+2\sigma_3^2s_1\sum_{cyc}x^{-2}-2\sigma_3^2\sum_{cyc}x^{-1}=
$$
$$
=s_3\sigma_2-\sigma_3s_2+2\sigma_3^2s_1\frac{-\sigma_2^2+2\sigma_1\sigma_2}{\sigma_3^2}-2\sigma_3^2\frac{\sigma_2}{\sigma_3}.
$$
Since
$$
\sum_{cyc}\frac{1}{x^2}=\frac{2\sigma_1\sigma_2-2\sigma_2^2}{\sigma_3^2}\textrm{, }\sum_{cyc}\frac{1}{x}=\frac{\sigma_2}{\sigma_3}.
$$
Hence finaly in general
$$
s_5=x^5+y^5+z^5=\sigma_1^5-5\sigma_1^3\sigma_2+25\sigma_1\sigma_2^2-5\sigma_2\sigma_3-5\sigma_1^2(4\sigma_2+3\sigma_3).\tag 3
$$
and as application
$$
x^5+y^5+z^5=29/3.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/556726",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
What is the equation for a line tangent to a circle from a point outside the circle? I need to know the equation for a line tangent to a circle and through a point outside the circle. I have found a number of solutions which involve specific numbers for the circles equation and the point outside but I need a specific solution, i.e., I need an equation which gives me the $m$ and the $b$ in $f(x) = mx + b$ for this line.
| Given a circle $x^2 + y^2 = r^2$ and the point (a,b)
The line from the origin to (a,b) is $y = \frac{b}{a} * x$
The line and the circle intersect at $( \frac{a*r}{\sqrt{a^2+b^2}}, \frac{b*r}{\sqrt{a^2+b^2}} )$
The slope of the tangent is $\frac{-a}{b}$
The equation of the tangent is
$y - \frac{b*r}{\sqrt{a^2+b^2}} = \frac{-a}{b} * (x - \frac{a*r}{\sqrt{a^2+b^2}} )$
$y = \frac{-a}{b} x + \frac{\sqrt{a^2+b^2}}{b} r $
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/557036",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
} |
In a triangle $\angle A = 2\angle B$ iff $a^2 = b(b+c)$ Prove that in a triangle $ABC$, $\angle A = \angle 2B$, if and only if:
$$a^2 = b(b+c)$$
where $a, b, c$ are the sides opposite to $A, B, C$ respectively.
I attacked the problem using the Law of Sines, and tried to prove that if $\angle A$ was indeed equal to $2\angle B$ then the above equation would hold true. Then we can prove the converse of this to complete the proof.
From the Law of Sines,
$$a = 2R\sin A = 2R\sin (2B) = 4R\sin B\cos B$$
$$b = 2R\sin B$$
$$c = 2R\sin C = 2R\sin(180 - 3B) = 2R\sin(3B) = 2R(\sin B\cos(2B) + \sin(2B)\cos B)$$
$$=2R(\sin B(1 - 2\sin^2 B) +2\sin B\cos^2 B) = 2R(\sin B -2\sin^3 B + 2\sin B(1 - \sin^2B))$$
$$=\boxed{2R(3\sin B - 4\sin^3 B)}$$
Now,
$$\implies b(b+c) = 2R\sin B[2R\sin B + 2R(3\sin B - 4\sin^3 B)]$$
$$=4R^2\sin^2 B(1 + 3 - 4\sin^2 B)$$
$$=16R^2\sin^2 B\cos^2 B = a^2$$
Now, to prove the converse:
$$c = 2R\sin C = 2R\sin (180 - (A + B)) = 2R\sin(A+B) = 2R\sin A\cos B + 2R\sin B\cos A$$
$$a^2 = b(b+c)$$
$$\implies 4R^2\sin^2 A = 2R\sin B(2R\sin B + 2R\sin A\cos B + 2R\sin B\cos) $$
$$ = 4R^2\sin B(\sin B + \sin A\cos B + \sin B\cos A)$$
I have no idea how to proceed from here. I tried replacing $\sin A$ with $\sqrt{1 - \cos^2 B}$, but that doesn't yield any useful results.
| Double angle formula says
$$
\begin{align}
\sin(A)
&=2\sin(B)\cos(B)\\
&\implies\frac{\sin(A)}{\sin(B)}=2\cos(B)\tag{1}
\end{align}
$$
The formula for the sine of a sum yields
$$
\begin{align}
\sin(C)
&=\sin(A+B)\\
&=2\sin(B)\cos(B)\cos(B)+(2\cos^2(B)-1)\sin(B)\\
&=\sin(B)(4\cos^2(B)-1)\\
&\implies\frac{\sin(C)}{\sin(B)}=4\cos^2(B)-1\tag{2}
\end{align}
$$
Thus, the Law of Sines says
$$
\left(\frac ab\right)^2-1=\frac cb\implies a^2-b^2=bc\tag{3}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/557704",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
} |
Prove that $\sqrt{x}$ is continuous on its domain $[0, \infty).$ Prove that the function $\sqrt{x}$ is continuous on its domain $[0,\infty)$.
Proof.
Since $\sqrt{0} = 0, $ we consider the function $\sqrt{x}$ on $[a,\infty)$ where $a$ is real number and $s \neq 0.$ Let $\delta=2\sqrt{a}\epsilon.$ Then, $\forall x \in dom,$ and $\left | x-x_0\right | < \delta \Rightarrow \left| \sqrt{x}-\sqrt{x_0}\right| = \left| \frac{x-x_0}{ \sqrt{x}+\sqrt{x_0}} \right| < \left|\frac{\delta}{2\sqrt{a}}\right|=\epsilon.$
Can I do this?
| Formal Proof:
Let $\epsilon > 0$, $a = 0$. Choose $\delta = \epsilon^2$. Then $\left |x \right| < \delta$ implies $\left | \sqrt{x} \right | < \epsilon$, as desired.
Let $\epsilon > 0$, $a \in (0,\infty)$. Choose $\delta = \sqrt{a}\epsilon$. Then $\left |x - a \right | < \delta$ implies $\left | \sqrt{x} - \sqrt{a} \right | < \frac{\delta}{\left | \sqrt{x} + \sqrt{a} \right |} \leq \epsilon$, as desired.
Thus $f$ is continuous at $x = a$ for all $a \in [0, \infty)$.
Explanation:
The first part is more obvious in its design. We would like to arrive at the form $\left | \sqrt{x} \right | < \epsilon$, so we notice we may square both sides to reach $\left | x \right | < \epsilon^2$, meaning we may choose $\delta = \epsilon^2$, as these steps may be undone.
For the second part, we would like to arrive at the form $\left | \sqrt{x} - \sqrt{a} \right | < \epsilon$ from $\left |x - a\right| < \delta$. We note $x - a = (\sqrt{x} - \sqrt{a})(\sqrt{x}+\sqrt{a})$, so whatever $\delta$ we choose, we will arrive at $\left | \sqrt{x} - \sqrt{a} \right | < \frac{\delta}{\left | \sqrt{x} + \sqrt{a} \right |}$, and we wish this to be less than or equal to $\epsilon$. We cannot choose $\delta$ immediately because our expression still depends on $x$, but we may easily remedy this. Noting that $\sqrt{a} \neq 0$, $\left | \sqrt{x} + \sqrt{a} \right | \geq \left | \sqrt{a} \right |$, meaning $\frac{\delta}{\left | \sqrt{x} + \sqrt{a} \right |} \leq \frac{\delta}{\left | \sqrt{a} \right |}$. This means $\delta = \sqrt{a}\epsilon$ is a good choice, as it is only in terms of the given point $a$ and $\epsilon$.
The proof was split into two cases, $a = 0$ and $a > 0$, because $a = 0$ is rather easy to prove, and the proof of the second part would have been problematic if $a$ were allowed to be $0$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Find the maximum and minimum values of $A \cos t + B \sin t$ Let $A$ and $B$ be constants. Find the maximum and minimum values of $A \cos t + B \sin t$.
I differentiated the function and found the solution to it as follows:
$f'(x)= B \cos t - A \sin t$
$B \cos t - A \sin t = 0 $
$t = \cot^{-1}(\frac{A}{B})+\pi n$
However, I got stuck here on how to formulate the minimum and maximum points. Any explanation would be appreciated.
| If $A=B=0$, then
$$
\min_t(A\cos t+B\sin t)=\max_t(A\cos t+B\sin t)=0.
$$
If $(A,B) \ne (0,0)$, let $\theta \in [0,2\pi)$ such that
$$
\cos\theta=\frac{A}{\sqrt{A^2+B^2}},\quad \sin\theta=\frac{B}{\sqrt{A^2+B^2}}.
$$
Then
$$
A\cos t+B\sin t=\sqrt{A^2+B^2}(\cos t\cos\theta+\sin t\sin\theta)=\sqrt{A^2+B^2}\cos(t-\theta).
$$
Hence
$$
\min_t(A\cos t+B\sin t)=-\sqrt{A^2+B^2},\quad \max_t(A\cos t+B\sin t)=\sqrt{A^2+B^2}.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/560711",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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A positive integer is divisible by $3$ iff 3 divides the sum of its digits I am having trouble proving the two following questions:
*
*If $p|N$, $q|N$ and gcd(p,q)=1, then prove that $pq|N$
*If $x$ is non zero positive integer number, then prove that $3|x$ if and only if 3 divides the sum of all digits of $x$.
For both questions I tried to use theorems of discrete mathematics, but I could not find the way to solve them.
| (1) The hypotheses $p|N$ and $q|N$ give us two integers $m,k$ so that $N=mp$ and $N=kq$. This implies $mp=kq$ so that $q|mp$. Now, since $gcd(p,q)=1$ and $q|mp$ we know that $q|m$ (think about why this is true if it's not clear). Then $m=sq$ for some integer $s$. Putting this all together, we have $N=mp=sqp=s(pq)$ so $pq|N$.
(2) Here we have an if and only if statement so you'll have to prove two statements/directions(or both at once):
(i) if $3|x$ then $3$ divides the sum of the digits of $x$
(ii) if $3$ divides the sum of the digits of $x$ then $3$ divides $x$
The strategy is to write the number in base $10$, for example (not a proof):
$1356=1\cdot 10^3+3\cdot 10^2+5\cdot 10^1 + 6\cdot 10^0$
Now $10$ has remainder $1$ under division by $3$ so the remainder of $1356$ under division by $3$ is $1 \cdot 1^3 + 3\cdot 1^2 +5\cdot 1^1 +6\cdot 1^0=1+3+5+6$ which is exactly the sum of the digits. Then the remainder under division by $3$ of $1356$ and $1+3+5+6$ are the same, and $3|1356$ if and only if $3|1+3+5+6$.
Try to do this in general for some number $n$ with logical steps following the idea of the example above to write a formal proof.
| {
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What are the elements of the quotient ring $\mathbb{Z_3}[x]/(x^3 + x^2)$? $$R = \mathbb{Z_3}[x]/(x^3 + x^2).$$
As $\mathbb{Z_3}$ is a field we have that every polynomial in $\mathbb{Z_3}[x]/(x^3 + x^2)$ of degree less than ${x^3 + x^2}$ is a distinct element in $R$.
So I conclude the following are the elements of $R$.
$
\overline{0},
\overline{1},
\overline{2}$
$
\overline{x},
\overline{x + 1},
\overline{x + 2},
\overline{2x},
\overline{2x + 1},
\overline{2x + 2},
$
$
\overline{x^2},
\overline{x^2 + 1},
\overline{x^2 + 2},
\overline{x^2 +x},
\overline{x^2 + 2x},
\overline{x^2 + x + 1},
\overline{x^2 + x + 2},
\overline{x^2 + 2x + 1},
\overline{x^2 + 2x + 2},
$
$
\overline{2x^2},
\overline{2x^2 + 1},
\overline{2x^2 + 2},
\overline{2x^2 +x},
\overline{2x^2 + 2x},
\overline{2x^2 + x + 1},
\overline{2x^2 + x + 2},
\overline{2x^2 + 2x + 1},
\overline{2x^2 + 2x + 2},$
Also, noting that for each element in $R$ we have $3$ choices for the coefficient and $3$ choices for the exponent of $x \implies$ we have $3^3 = 27$ polynomials in $R$.
Have I got my facts right here?
| As pointed out in the comments, you are correct. The ring $\mathbb{Z}_3[x]/(x^3 + x^2)$ has $27$ elements, and they are precisely the ones you wrote down.
| {
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"timestamp": "2023-03-29T00:00:00",
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Sum involving binomial coefficient I want to analyse the following expression for $x \geq 0$:
$$
\sum_{k = 0}^n (-1)^{k+n} \binom{n+k}{2k} x^k
$$
I expect and want to prove that for $x \geq 4$, the expression tends to infinity as $n \to \infty$. I tried coming up with a closed expression, however found it difficult to cope with the upper $k$ in the binomial coefficient.
| First we try to find a closed form of the sum, preferably one whose asymptotics can be classified. Let $f_n$ be your sum. Then Sister Celine / Zeilberger's algorithm says that
$$f_n = (x-2) f_{n-1} - f_{n-2}.$$
The roots of the characteristic equation of this recurrence are
$$\rho_{1,2} = -1 + \frac{1}{2} x \pm \frac{\sqrt{x^2-4x}}{2}.$$
We are looking for a closed expression of the form
$$c_1\rho_1^n+c_2\rho_2^n$$
using $f_0 = 1$ and $f_1 = x-1$ and solving for $c_{1,2}$ we get that
$$c_{1,2} = \frac{1}{2} \pm \frac{1}{2} \frac{\sqrt{x^2-4x}}{x-4}$$
for an end result of
$$f_n =
\left(\frac{1}{2} + \frac{1}{2} \frac{\sqrt{x^2-4x}}{x-4}\right)
\left(-1 + \frac{1}{2} x + \frac{\sqrt{x^2-4x}}{2}\right)^n \\+
\left(\frac{1}{2} - \frac{1}{2} \frac{\sqrt{x^2-4x}}{x-4}\right)
\left(-1 + \frac{1}{2} x - \frac{\sqrt{x^2-4x}}{2}\right)^n.$$
We will now study the two exponential terms, showing that the second one vanishes and the first one produces exponential growth. Start by observing that
$$\frac{\sqrt{x^2-4x}}{2} = \frac{1}{2} x \sqrt{1-4/x}$$ which for $x\ge 4$ can be expanded into a convergent asymptotic series
$$\sqrt{1-4/x} = \sum_{q\ge 0} (-1)^q {1/2\choose q} \left(\frac{4}{x}\right)^q
\sim 1-\frac{2}{x} - \frac{2}{x^2}.$$
Therefore
$$-1 + \frac{1}{2} x - \frac{\sqrt{x^2-4x}}{2}
= -1 + \frac{1}{2} x - \frac{x}{2} \sqrt{1-4/x}
\\ \sim -1 + \frac{1}{2} x - \frac{x}{2} \left(1-\frac{2}{x} - \frac{2}{x^2}\right)
= \frac{1}{x}.$$
We see that the second term behaves like $1/x^n$ for $x\ge 4$ and vanishes as claimed.
Note that it is critical in this argument that the first terms of the series expansion of the root get canceled.
Finally we have for the first term that
$$-1 + \frac{1}{2} x + \frac{\sqrt{x^2-4x}}{2}
= -1 + \frac{1}{2} x + \frac{x}{2} \sqrt{1-4/x}
\\ \sim -1 + \frac{1}{2} x + \frac{x}{2} \left(1-\frac{2}{x} - \frac{2}{x^2}\right)
= x - 2 - \frac{1}{x}.$$
This is approximately $x-2$ and its $n$th powers go to infinity as claimed. The constants in front of the two exponential terms are independent of $n$ and do not influence the asymptotic behaviour.
Addendum. This matches the output from Wolfram Alpha.
| {
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$1 + \frac{1}{1+2}+ \frac{1}{1+2+3}+ ... + \frac{1}{1+2+3+...+n} = ?$ How do I simplify the following series
$$1 + \frac{1}{1+2}+ \frac{1}{1+2+3}+ \frac{1}{1+2+3+4} + \frac{1}{1+2+3+4+5} + \cdot\cdot\cdot + \frac{1}{1+2+3+\cdot\cdot\cdot+n}$$
From what I see, each term is the inverse of the sum of $n$ natural numbers.
Assuming there are $N$ terms in the given series,
$$a_N = \frac{1}{\sum\limits_{n = 1}^{N} n} = \frac{2}{n(n+1)}$$
$$\Rightarrow \sum\limits_{n=1}^{N} a_N = \sum\limits_{n=1}^{N} \frac{2}{n(n+1)}$$
... and I'm stuck.
I've never actually done this kind of problem before (am new to sequences & series).
So, a clear and detailed explanation of how to go about it would be most appreciated.
PS- I do not know how to do a telescoping series!!
| $$
\sum_{n = 1}^{N}{2 \over n\left(n + 1\right)}
=
2\sum_{n = 1}^{N}\left({1 \over n} - {1 \over n + 1}\right)
=
2\left[1 - {1 \over 2} + {1 \over 2} - {1 \over 3} + \cdots + {1 \over N}
-
{1 \over N + 1}\right]
$$
$$
\sum_{n = 1}^{N}{2 \over n\left(n + 1\right)}
=
2\left(1 - {1 \over N + 1}\right)\quad\to\quad \color{#0000ff}{\Large 2}\quad\mbox{when}\quad N \to \infty
$$
| {
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Recurrence sequence limit I would like to find the limit of
$$
\begin{cases}
a_1=\dfrac3{4} ,\, & \\
a_{n+1}=a_{n}\dfrac{n^2+2n}{n^2+2n+1}, & n \ge 1
\end{cases}
$$
I tried to use this - $\lim \limits_{n\to\infty}a_n=\lim \limits_{n\to\infty}a_{n+1}=L$, so $L=\lim \limits_{n\to\infty}a_n\dfrac{n^2+2n}{n^2+2n+1}=L\cdot1$ What does this result mean? Where did I make a mistake?
| $\begin{cases}
a_1=\dfrac3{4} ,\, & \\
a_{n+1}=a_{n}\dfrac{n^2+2n}{n^2+2n+1} & n \ge 1
\end{cases}
$
$a_{n+1}
=a_{n}\left(\dfrac{n^2+2n}{n^2+2n+1}\right)
=a_{n}\left(\dfrac{n(n+2)}{(n+1)^2}\right)
$
so
$\begin{align}
a_{n}
&=a_1\prod \limits_{k=1}^{n-1} \dfrac{a_{k+1}}{a_k}\\
&=a_1\prod \limits_{k=1}^{n-1} \dfrac{k(k+2)}{(k+1)^2}\\
&=a_1 \dfrac{\prod \limits_{k=1}^{n-1}k\prod \limits_{k=1}^{n-1}(k+2)}{\prod \limits_{k=1}^{n-1}(k+1)^2}\\
&=a_1 \dfrac{\prod \limits_{k=1}^{n-1}k\prod \limits_{k=3}^{n+1}k}{\prod \limits_{k=2}^{n}k^2}\\
&=a_1 \dfrac{\prod \limits_{k=1}^{n-1}k}{\prod \limits_{k=2}^{n}k}\dfrac{\prod \limits_{k=3}^{n+1}k}{\prod \limits_{k=2}^{n}k}\\
&=a_1 \dfrac{1}{n}\dfrac{n+1}{2}\\
&=a_1 \dfrac{n+1}{2n}\\
\end{align}
$
Since $a_1 = \dfrac34$,
$a_n
=\dfrac34 \dfrac{n+1}{2n}
= \dfrac{3(n+1)}{8n}
$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
Determinant of a n x n Matrix - Main Diagonal = 2, Sub- & Super-Diagonal = 1 I'm stuck with this one - Any tips?
The Problem:
Let $n \in \mathbb{N}.$ The following $n \times n$ matrix:
$$A = \left( \begin{array}{ccc}
2 & 1 & & & & ...\\
1 & 2 & 1 & & & ...\\
& 1 & 2 & 1 & & ...\\
& & 1 & 2 & 1 & ...\\
& & & 1 & ... & 1\\
... & ... & ... & ... & 1 &2
\end{array} \right) $$
e.g. for the main diagonal = 2, the sub- and superdiagonal = 1 .
Show with Induction that $\det(A) = n + 1$.
My solution approach: Laplace Expansion starting with the 2 in the bottom right corner $(a_{n+1,n+1})$. But how can I tell wether its positive or negative? After that I'm stuck with the 1 $(a_{n,n+1})$(the sub matrix matrix becomes ugly and I get a recursively solution). How can I formalize this in a proper manner?
| Why not develop directly wrt the first column? The subindex means the order of the square matrix:
$$\begin{vmatrix}2 & 1 & & & & ...\\
1 & 2 & 1 & & & ...\\
& 1 & 2 & 1 & & ...\\
& & 1 & 2 & 1 & ...\\
& & & 1 & ... & 1\\
... & ... & ... & ... & 1 &2\end{vmatrix}_n=2\begin{vmatrix}2 & 1 & & & & ...\\
1 & 2 & 1 & & & ...\\
& 1 & 2 & 1 & & ...\\
& & 1 & 2 & 1 & ...\\
& & & 1 & ... & 1\\
... & ... & ... & ... & 1 &2\end{vmatrix}_{n-1}-$$
$$-\begin{vmatrix}1 & 0 & & & & ...\\
1 & 2 & 1 & & & ...\\
& 1 & 2 & 1 & & ...\\
& & 1 & 2 & 1 & ...\\
& & & 1 & ... & 1\\
... & ... & ... & ... & 1 &2\end{vmatrix}_{n-1}\stackrel{\text{Ind. Hyp.}}=2\left[(n-1)+1\right]-\left[(n-2)+1\right]=$$
$$=2n-n+1=n+1$$
as the last determinant is easily developed wrt the first entry and we get a $\;(n-2)\times(n-2)\;$ copy of the original matrix!
| {
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"source": "stackexchange",
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"answer_count": 2,
"answer_id": 0
} |
Finding the limit of a matrix and showing if it exists For the matrix
\[
A=\frac{1}{10}\cdot \begin{pmatrix}
1 & 7 \\
7 & 1
\end{pmatrix}\]
I try to find the limit of $A^n$ when $n$ goes to infinity.
| Assuming you are good with guessing eigenvalues there is a a similar way which in this case is kind of more easy.
Because of
\[ \frac{1}{10}\begin{pmatrix} 1 & 7 \\ 7 & 1 \end{pmatrix} \cdot
\begin{pmatrix} 1 \\ 1 \end{pmatrix} = \frac{4}{5} \cdot \begin{pmatrix} 1 \\ 1 \end{pmatrix}\]
and
\[ \frac{1}{10}\begin{pmatrix} 1 & 7 \\ 7 & 1 \end{pmatrix} \cdot
\begin{pmatrix} 1 \\ -1 \end{pmatrix} = -\frac{3}{5} \cdot \begin{pmatrix} 1 \\ -1\end{pmatrix}\]
So instead of calculating the limit of $A^n$, I look at the limit of $A^n \cdot v$ for any $v$.
As the geometric sequence $q^n$ converges to zero iff $|q|<1$ we see that here $A^n\cdot v$ converges
to zero for any $v$. Hence $A^n$ must converge to zero.
| {
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"answer_count": 2,
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} |
Show that there is no natural number $n$ such that $3^7$ is the largest power of $3$ dividing $n!$
Show that there is no natural number $n$ such that $7$ is the largest power $a$ of $3$ for which $3^a$ divides $n!$
After doing some research, I could not understand how to start or what to do to demonstrate this.
We have $$E_3(n!)\neq7\;\;\forall n\in\mathbb{N}\\\left[\frac{n}{3} \right]+\left[\frac{n}{3^2} \right]+\left[\frac{n}{3^3} \right]+\dots\neq7$$I do not know where from, or what to do to solve it.
| For $n=15,\left[\frac{15}{3} \right]+\left[\frac{15}{3^2} \right]+\left[\frac{15}{3^3} \right]+\;...=6$
for $n=18$ (the next multiple of $3$) $\left[\frac{18}{3} \right]+\left[\frac{18}{3^2} \right]+\left[\frac{18}{3^3} \right]+\;...=8$
If $n\geq 18$ then $\left[\frac{n}{3} \right]+\left[\frac{n}{3^2} \right]+\left[\frac{n}{3^3} \right]+\;...\geq 8$
So there is no possibility for $7$
| {
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} |
Limit $\lim_{n\to \infty} n(\frac{1}{(n+1)^2}+\frac{1}{(n+2)^2}+\cdots+\frac{1}{(2n)^2})$ I need some help finding the limit of the following sequence:
$$\lim_{n\to \infty} a_n=n\left(\frac{1}{(n+1)^2}+\frac{1}{(n+2)^2}+\cdots+\frac{1}{(2n)^2}\right)$$
I can tell it is bounded by $\frac{1}{4}$ from below and decreasing from some point which tells me it is convergent.
I can't get anything else though.
So far we did not have integrals and we just started series, so this should be solved without using either.
Some hints would be very welcome :)
Thanks
| Notice
$$\frac{1}{(n+k-1)(n+k)} \ge \frac{1}{(n+k)^2}\ge \frac{1}{(n+k)(n+k+1)}$$
We have
$$\begin{align}a_n = n \sum_{k=1}^{n}\frac{1}{(n+k)^2}\le & n\sum_{k=1}^n\frac{1}{(n+k-1)(n+k)}= n \sum_{k=1}^n\left(\frac{1}{n+k-1}-\frac{1}{n+k}\right)\\ = & n \left(\frac{1}{n}- \frac{1}{2n}\right) = \frac12\\ \text{AND}\quad \ge & n\sum_{k=1}^n\frac{1}{(n+k)(n+k+1)} = n \sum_{k=1}^n\left(\frac{1}{n+k}-\frac{1}{n+k+1}\right)\\ = & n \left(\frac{1}{n+1} - \frac{1}{2n+1}\right) = \frac{n^2}{(n+1)(2n+1)}\end{align}$$
Since $\;\;\displaystyle \lim_{n\to\infty} \frac{n^2}{(n+1)(2n+1)} = \frac12,\;\;$ we get
$\;\;\displaystyle \lim_{n\to\infty} a_n = \frac12$.
| {
"language": "en",
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"answer_id": 1
} |
Proof of Heron's Formula for the area of a triangle
Let $a,b,c$ be the lengths of the sides of a triangle. The area is given by Heron's formula:
$$A = \sqrt{p(p-a)(p-b)(p-c)},$$
where $p$ is half the perimeter, or $p=\frac{a+b+c}{2}$.
Could you please provide the proof of this formula?
Thank you in advance.
| Nobody can provide the proof but many can provide a proof or perhaps many proofs.
Notice that the area must be a 2nd-degree homogeneous function of $a$, $b$, and $c$, for example, if you multiply $a$, $b$, and $c$ by $9$ then you multiply the area by $9^2=81$, etc.
Next, notice that the area must be $0$ if $a+b=c$: if the distance along one side plus the distance along another side is equal to the distance along the third side, then the three corners are on a straight line, so the area is $0$. For that reason $a+b-c$ should appear as a factor, i.e. as something you multiply by.
For the same reason, $b+c-a$ and $c+a-b$ should be factors.
Next, notice that if $a=b=c=0$, then the area must be $0$, so that's why $a+b+c$ should be a factor.
Now we have $(a+b+c)(a+b-c)(b+c-a)(c+a-b)$. That is homogeneous of degree $4$ rather than $2$, so take its square root and we have $\sqrt{(a+b+c)(a+b-c)(b+c-a)(c+a-b)}$, which is homogenous of degree $2$. That, then, should be proportional to the area.
Now let's see what the constant of proportionality is: The area of a right triangle with legs of length $1$ and hypotenuse of length $\sqrt2$ is $1/2$. Plugging in those three numbers for $a$, $b$, and $c$ we get
$$
\frac 1 2 =\text{constant}\times\sqrt{\left(1+1+\sqrt 2\right) \left(1+1-\sqrt 2\right)\left(1-1+\sqrt 2\right)\left(-1+1+\sqrt2\right)} = \text{constant}\times 2.
$$
So the "constant" is $\dfrac 1 4$ and finally we have
$$
\text{area} = \frac 1 4 \sqrt{(a+b+c)(a+b-c)(b+c-a)(c+a-b)}.
$$
That's Heron's formula.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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} |
Given that $z= \cos \theta + i\sin \theta$, prove that $\Re\left\{\dfrac{z-1}{z+1}\right\}=0$ Given that $z= \cos \theta + i\sin \theta$, prove that $\Re\left\{\dfrac{z-1}{z+1}\right\}=0$
How would I do this?
| Using $\displaystyle\cos2A=2\cos^2A-1=1-2\sin^2A $ and $\displaystyle\sin2A=2\cos A\sin A,$
$$\frac{\cos\theta+i\sin\theta-1}{\cos\theta+i\sin\theta+1}=\frac{-2\sin^2\frac{\theta}2+i2\sin\frac{\theta}2\cos\frac{\theta}2}{2\cos^2\frac{\theta}2+i2\sin\frac{\theta}2\cos\frac{\theta}2}$$
$$=\frac{2i\sin\frac{\theta}2\left(\cos\frac{\theta}2+i\sin\frac{\theta}2\right)}{2\cos\frac{\theta}2\left(\cos\frac{\theta}2+i\sin\frac{\theta}2\right)}=i\tan\frac{\theta}2$$ as $\displaystyle\cos\frac{\theta}2+i\sin\frac{\theta}2\ne0$
| {
"language": "en",
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"source": "stackexchange",
"question_score": "1",
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"answer_id": 3
} |
Distribution of function of two random variables Let $X$ be the number on a die roll, between 1 and 6. Let $Y$ be a random number which is uniformly distributed on $[0,1]$, independent of $X$. Let $Z = 10X + 10Y$.
What is the distribution of $Z$?
| Suppose $Y$ is another discrete random variables on $[0,1]$ with $\Pr(Y=0) = \Pr(Y=1) = \frac{1}{2}$. It helps to build the table of possible values of $Z = 10 X + 10 Y$:
$$
\begin{array}{c|cccccc} Z(X,Y) & X=1 & X=2 & X=3 & X=4 & X=5 & X+6 \cr \hline
Y=0 & Z=10 & Z=20 & Z=30 & Z=40 & Z=50 & Z=60 \cr
Y=1 & Z = 20 & Z=30 & Z=40 & Z=50 & Z=60 & Z=70
\end{array}
$$
Since there are 7 possible outcomes we compute their probabilities manually, e.g.:
$$ \begin{eqnarray}
\Pr(Z=10) &=& \Pr(X=1,Y=0) = \frac{1}{6} \cdot \frac{1}{2} \\
\Pr(Z=20) &=& \Pr(X=2,Y=0) + \Pr(X=1,Y=1) = \frac{1}{6} \cdot \frac{1}{2} + \frac{1}{6} \cdot \frac{1}{2}
\end{eqnarray}
$$
and so on...
| {
"language": "en",
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} |
$\varepsilon$-$\delta$ proof of $\sqrt{x+1}$ need to prove that $ \lim_{x\rightarrow 0 } \sqrt{1+x} = 1 $
proof of that is:
need to find a delta such that $ 0 < |x-1| < \delta \Rightarrow 1-\epsilon < \sqrt{x+1} < \epsilon + 1 $ if we choose $ \delta = (\epsilon + 1)^2 -2 $ and consider $ |x-1| < \delta = (\epsilon + 1)^2 - 2 $
$ 4 - (\epsilon + 1)^2 < x +1 < (\epsilon + 1)^2 $
$ \sqrt{4-(\epsilon + 1)^2} -1 < \sqrt{x+1} -1 < \epsilon$ but I need to show that $\sqrt{4-(\epsilon + 1)^2} -1 > -\epsilon $ before this proof is complete...
any help on how to finish the proof?
| You need to show that for all $\varepsilon > 0$ there exists a $\delta > 0$ such that $$0<|x|<\delta \implies |\sqrt{x+1}-1|<\varepsilon.$$
It's helpful in this case to multiply by the conjugate of $\sqrt{x+1}-1$, which is $\sqrt{x+1}+1$, so that the rightmost inequality becomes $$|\sqrt{x+1}-1|\cdot|\sqrt{x+1}+1|<\varepsilon(\sqrt{x+1} + 1)$$ $$|x|<\varepsilon(\sqrt{x+1}+1). \tag{1}$$
Now, if you require that $0 < |x| < 1$, then we know that $0<\sqrt{x+1}<\sqrt{2}$, so $1<\sqrt{x+1} + 1<\sqrt{2}+1.$
So we now know that if we require $0 < |x| < 1$, then $(1)$ tells us that $|x| < \varepsilon(\sqrt{2}+1)$, which we have also shown can be rearranged into the form $|\sqrt{x+1}-1| < \varepsilon.$
So, we take $\delta = \min\left(1,\ \varepsilon(\sqrt{2}+1)\ \right).$ Then, $$0<|x|<\delta \implies |\sqrt{x+1}-1|<\varepsilon$$
which is exactly what we wanted to show.
| {
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} |
De Moivre's formula to solve $\cos$ equation Use De Moivre’s formula to show that
$$
\cos\left(x\right) + \cos\left(3x\right) = 2\cos\left(x\right)\cos\left(2x\right)
$$
$$
\mbox{Show also that}\quad
\cos^{5}\left(x\right) = \frac{1}{16}\,\cos\left(5x\right) + \frac{5}{16}\,\cos\left(3x\right) + \frac{5}{8}\,\cos\left(x\right)
$$
Hence solve the equation $\cos\left(5x\right) = 16\cos^{5}\left(x\right)$ completely.
Express your answers as rational multiples of $\pi$.
| Let me first show you another problem using DeMoivre's formula, to give you the idea of how to do the first two parts. Namely, I will show that $$\cos(4x)=8\cos^4x-8\cos^2x+1,$$ using only DeMoivre's formula and the Pythagorean identity $\sin^2x=1-\cos^2x$.
First off, we know that $$\cos(4x)+i\sin(4x)=(\cos x+i\sin x)^4,$$ and by expanding the right-hand side, we have $$\cos(4x)+i\sin(4x)=\cos^4x+4i\cos^3x\sin x-6\cos^2x\sin^2x-4i\cos x\sin^3x+\sin^4x.$$ Gathering our real and imaginary terms gives us $$\cos(4x)+i\sin(4x)=\cos^4x-6\cos^2x\sin^2x+\sin^4x+i(4\cos^3x\sin x-4\cos x\sin^3x),$$ so, remembering that sine and cosine are real-valued functions on the reals, we have that $$\cos(4x)=\cos^4x-6\cos^2x\sin^2x+\sin^4x.$$ At that point, we can rewrite as $$\cos(4x)=\cos^4x-6\cos^2x\sin^2x+(\sin^2x)^2,$$ which means that $$\cos(4x)=\cos^4x-6\cos^2x(1-\cos^2x)+(1-\cos^2x)^2,$$ whence expanding and collecting like terms gives us $$\cos(4x)=8\cos^4x-8\cos^2x+1,$$ as desired.
Now, once we've proved that $$\cos(x)+\cos(3x)=2\cos(x)\cos(2x)\tag{$\heartsuit$}$$ and that $$\cos^5(x) = \frac1{16}\cos(5x) + \frac5{16}\cos(3x) + \frac58\cos(x),\tag{$\diamondsuit$}$$ let us assume that $$\cos(5x)=16\cos^5(x).\tag{$\spadesuit$}$$ By $(\spadesuit)$ and $(\diamondsuit),$ we can conclude that $$0=\cos(3x)+2\cos(x).$$ (Do you see how?) By $(\heartsuit),$ it then follows that $$0=2\cos(x)\cos(2x)+\cos(x)\\0=\cos(x)\bigl(2\cos(2x)+1\bigr).$$ (Do you see how?) Thus, we have $$\cos(x)=0$$ or $$\cos(2x)=-\frac12,$$ which I leave to you to solve.
| {
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How to prove $\frac{y^2-x^2}{x+y+1}=\pm1$ is a hyperbola? How to prove $\frac{y^2-x^2}{x+y+1}=\pm1$ is a hyperbola, knowing the canonical form is $\frac{y^2}{a^2}-\frac{x^2}{b^2}=\pm1$ where $a$ and $b$ are constants? Thanks !
| Taking the '+' sign, $$y^2-x^2=x+y+1\implies\left(y-\frac12\right)^2-\left(x+\frac12\right)^2=1^2$$
$$\implies\frac{\left(y-\frac12\right)^2}{1^2}-\frac{\left(x+\frac12\right)^2}{1^2}=1^2$$
Similarly for the '-' sign,
$$y^2-x^2=-(x+y+1)\implies\left(x-\frac12\right)^2-\left(y+\frac12\right)^2=1^2$$
Can you recognize $a,b$ here?
| {
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How find this $M(2)$ let matrix $$A(x)=\begin{bmatrix}
1&x\\
x&1
\end{bmatrix}$$
and consider the infinte matrix product
$$M(t)=\prod_{n=1}^{\infty}A(p^{-t}_{n})$$
where $p_{n}$ is the nth prime
Evaluate $M(2)$
My try: since
$$M(2)=\prod_{n=1}^{\infty}A(\dfrac{1}{p^2_{n}})=\begin{bmatrix}
1&\dfrac{1}{2^2}\\
\dfrac{1}{2^2}&1
\end{bmatrix}\cdot\begin{bmatrix}
1&\dfrac{1}{3^2}\\
\dfrac{1}{3^2}&1
\end{bmatrix}\cdot\begin{bmatrix}
1&\dfrac{1}{5^2}\\
\dfrac{1}{5^2}&1
\end{bmatrix}\cdots\begin{bmatrix}
1&\dfrac{1}{p_{n}^2}\\
\dfrac{1}{p_{n}^2}&1
\end{bmatrix}\cdots$$
then I can't,Thank you
| Your matrices form a commuting family and therefore can be simultaneously diagonalized. Let $P$ be the orthogonal matrix which simultaneously diagonalizes the family. We have
$$P = \begin{pmatrix}-\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}}\end{pmatrix}$$
Note that the eigenvalues of each matrix are $1\pm x$. Then we can write the product as
$$M(t) = P\begin{pmatrix}\prod_{i=1}^\infty\left(1 - \frac{1}{p_i^t}\right) & 0 \\ 0 & \prod_{i=1}^\infty\left(1 + \frac{1}{p_i^t}\right)\end{pmatrix}P^\mathrm{T}$$
The two infinite products are well known as
$$\prod_{i=1}^\infty\left(1 - \frac{1}{p_i^t}\right) = \frac{1}{\zeta(t)}$$
$$\prod_{i=1}^\infty\left(1 + \frac{1}{p_i^t}\right) = \frac{\zeta(t)}{\zeta(2t)}$$
where $\zeta$ is the Riemann zeta. We have $\zeta(2) = \frac{\pi^2}{6}$ and $\zeta(4) = \frac{\pi^4}{90}$ whence we have
$$M(2) = P\begin{pmatrix}\frac{6}{\pi^2} & 0 \\ 0 & \frac{15}{\pi^2}\end{pmatrix}P^\mathrm{T} = \frac{3}{2\pi^2}\begin{pmatrix}7 & 3\\3& 7\end{pmatrix}$$
| {
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2nd order differential equation I'm working on the following 2nd order ODE:
$$
x^2 y''+2(2x-1/b^2) y'+2(1-(a/b)^2)y=0,
$$
where $b\neq 0$. It's very similar to the equation for the generalized Bessel polynomials (see http://en.wikipedia.org/wiki/Bessel_polynomials ); There's a slight difference in the factor in front of $y$ though. Any advise on how to handle the beast? Thanks a lot.
| Mathematica tells me that after multiplying by a (horrible) power it's hypergeometric:
$
y(x)=c_1 \left(\frac{1}{x}\right)^{\frac{3 b^2-b \sqrt{8
a^2+b^2}}{2 b^2}} \, _1F_1\left(\frac{3}{2}-\frac{\sqrt{b^4+8 a^2 b^2}}{2
b^2};1-\frac{\sqrt{b^4+8 a^2 b^2}}{b^2};-\frac{2}{b^2 x}\right)$
$
\quad +\quad c_2
\left(\frac{1}{x}\right)^{\frac{b \sqrt{8 a^2+b^2}+3 b^2}{2
b^2}} \, _1F_1\left(\frac{\sqrt{b^4+8 a^2 b^2}}{2
b^2}+\frac{3}{2};\frac{\sqrt{b^4+8 a^2 b^2}}{b^2}+1;-\frac{2}{b^2
x}\right)
$
Enjoy!
| {
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does this series converge? $\sum_{n=1}^\infty{\left( \sqrt[3]{n+1} - \sqrt[3]{n-1} \right)^\alpha} $
show the the following series converge\diverge
$\sum_{n=1}^\infty{\left( \sqrt[3]{n+1} - \sqrt[3]{n-1} \right)^\alpha} $
all the test i tried failed (root test, ratio test,direct comparison)
please dont use integrals as this is out of the scope for me right now
| Hint rationalize:
$$\left( \sqrt[3]{n+1} - \sqrt[3]{n-1} \right)^\alpha = \left( \frac{2}{(\sqrt[3]{n+1})^2+ \sqrt[3]{n-1}\sqrt[3]{n+1}+ (\sqrt[3]{n-1})^2} \right)^\alpha$$
Compare with $$\sum\frac{1}{n^{\frac{2\alpha}{3}}}$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Checking on some convergent series I need some verification on the following 2 problems I attemped:
I have to show that the following series
is convergent: $$1-\frac{1}{3 \cdot 4}+\frac{1}{ 5 \cdot 4^2 }-\frac{1}{7 \cdot 4^3}+ \ldots$$ .
My Attempt: I notice that the general term is given by $$\,\,a_n=(-1)^{n}{1 \over {(2n+1)4^n}} \,\,\text{by ignoring the first term of the given series.}$$ Using the fact that An absolutely convergent series is convergent, $$\sum_{1}^{\infty}|a_n|=\sum_{1}^{\infty} {1 \over {(2n+1)4^n}}\le \sum_{1}^{\infty} {1 \over 4^n}=\sum_{1}^{\infty}{1 \over {2^{2n}}}$$ which is clearly convergent by comparing it with the p-series with $p >1$.
I have to show that the following series
is convergent:$$1-\frac{1}{2!}+\frac{1}{4!}-\frac{1}{6!}+ \ldots $$
My Attempt:$$1-\frac{1}{2!}+\frac{1}{4!}-\frac{1}{6!}+ \ldots \le 1+\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+ \frac{1}{4!}+\frac{1}{5!}+\frac{1}{6!}+\ldots $$. Now, using the fact that $n! >n^2$ for $n \ge 4$ and the fact that omitting first few terms of the series does not affect the characteristics of the series ,we see that $$\frac{1}{4!}-\frac{1}{6!}+\frac{1}{8!}+ \ldots \le \frac{1}{4!}+\frac{1}{5!}+\frac{1}{6!}+ \frac{1}{6!}+\frac{1}{7!}+\frac{1}{8!}+\ldots =\sum_{4}^{\infty}{1 \over n!} <\sum_{4}^{\infty}{1 \over n^2}$$ and it is clearly convergent by comparing it with the p-series with $p >1$.
Now,I am stuck on the third one.
I have to show that the following series
is convergent:$$\frac{\log 2}{2^2}-\frac{\log 3}{3^2}+\frac{\log 4}{4^2}- \ldots $$
I see that $$\frac{\log 2}{2^2}-\frac{\log 3}{3^2}+\frac{\log 4}{4^2}- \ldots \le \sum_{2}^{\infty} {{\log n} \over {n^2}}= ?? $$
Thanks and regards to all.
| Another approach for the last series is the alternating series test which proves the convergence since
i) $ a_{n+1} < a_n, \quad \forall n\geq 2, $
ii) $ a_n\longrightarrow_{n\to \infty} 0. $
Note: You can prove the $a_n$ is decreasing by using the first derivative test for the function
$$ f(x)=\frac{\ln x}{x^2}. $$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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} |
Find $\lim_{x \to - \infty} \left(\frac{4^{x+2}- 2\cdot3^{-x}}{4^{-x}+2\cdot3^{x+1}}\right)$ I am to find the limit of
$$\lim_{x \to - \infty} \left(\frac{4^{x+2}- 2\cdot3^{-x}}{4^{-x}+2\cdot3^{x+1}}\right)$$
so I used:
$$\lim_{x \to -\infty} = \lim_{x \to \infty}f(-x)$$
but I just can't solve it to the end.
| $\lim_{x \to -\infty} \left(\dfrac{4^{x+2}- 2\cdot3^{-x}}{4^{-x}+2\cdot3^{x+1}}\right)
=\lim_{x \to -\infty} \left(\dfrac{3^{-x}(4^{x+2}\cdot3^x- 2)}{3^{-x}(4^{-x}\cdot3^x+2\cdot3^{2x+1})}\right)
=\lim_{x \to -\infty} \left(\dfrac{4^{x+2}\cdot3^x- 2}{(3/4)^x+2\cdot3^{2x+1}}\right)$
The numerator converges to $-2$ while the denominator diverges to $\infty$, and so the limit is $0$.
| {
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Proof by Induction for a $f_3 + f_6 + · · · + f_{3n} = \frac{1}{2} (f_{3n+2} - 1)$ The Fibonacci numbers are defined as follows: $f_0 = 0,\ f_1 = 1$, and for $n ≥ 2,\ f_n = f_{n−1} +f_{n−2}$. Prove that for every positive integer $n$,
$$f_3 + f_6 + \ldots + f3_n = \frac{1}{2} (f_{3n+2} - 1) $$
Here are my steps:
Basis Step:
For $n ≥ 1$
$f_{3\cdot1} = \frac{1}{2} (f_{3\cdot 1+2} - 1) $
LHS $= f_3 = 2$
RHS $= \frac{1}{2} (f_5 - 1) = \frac{1}{2} (5 – 1) = 2$
Induction Step:
Assume that $f_3 + f_6 + \ldots + f_{3n} = \frac{1}{2} (f_{3n+2} - 1)$ for some $n ≥ 1$.
Show that $f_3 + f_6 + \ldots + f_{3n} + f_{3(n+1)} = \frac{1}{2} (f_{3(n+1)+2} – 1)$.
$$\frac{1}{2} (f_{3n+2} – 1) = \frac{1}{2} (f_{3(n+1)+2} – 1) + f_{3(n+1)}\ \text{(Basis)}\\ \text{LHS}=\frac{1}{2}(f_{3n+5}-1)+f_{3n+3}$$
I don't know how to proceed further. I understand what I have to do, but I don't know where to start reducing it using the fib definition.
| $$f_3 + f_6 + · · · + f_{3n} =\frac{1}{2}(f_{3n+2} - 1) $$
$$f_3+f_6 + ···+f_{3n}+f_{3(n+1)} =\frac{1}{2}(f_{3n+2} - 1)+f_{3n+3}= $$
$$=\frac{1}{2}((f_{3n+2}+f_{3n+3})+f_{3n+3}-1)=\frac{1}{2}((f_{3n+4}+f_{3n+3}) - 1)= $$
$$=\frac{1}{2}(f_{3n+5} - 1)=\frac{1}{2}(f_{3(n+1)+2} - 1) $$
| {
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"timestamp": "2023-03-29T00:00:00",
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Simplify $2^{(n-1)} + 2^{(n-2)} + .... + 2 + 1$ Simplify $2^{(n-1)} + 2^{(n-2)} + .... + 2 + 1$
I know the answer is $2^n - 1$, but how to simplify it?
| Forget about all the smart generic formulas. Just rewrite the last summand $1$ as $2-1$. You get
$$
2^{n-1} + 2^{n-2} + \ldots + 2^3 + 2^2 + 2 + 2 - 1.
$$
Now group the $2$'s together. $2 + 2 = 2^2$, so you get
$$
2^{n-1} + 2^{n-2} + \ldots + 2^3 + 2^2 + 2^2 - 1
$$
Now group $2^2$ and $2^2$ together to get $2^2 + 2^2 = 2^3$:
$$
2^{n-1} + 2^{n-2} + \ldots + 2^3 + 2^3 - 1
$$
If you go on in this fashion, you will eventually come to $2^n - 1$. The actual number of steps in this process depends on $n$, so technically this should be framed as a proof by induction.
| {
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"timestamp": "2023-03-29T00:00:00",
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} |
How to solve this reccurence relation? Let a,b,c be real numbers. Find the explicit formula for $f_n=af_{n-1}+b$ for $n \ge 1$ and $f_0 = c$
So I rewrote it as $f_n-af_{n-1}-b=0$ which gives the characteristic equation as $x^2-ax-b=0$. The quadratic formula gives roots $x= \frac{a+\sqrt{a^2+4b}}{-2}, \frac{a-\sqrt{a^2+4b}}{-2}$
Then $f_n=P_1(\frac{a+\sqrt{a^2+4b}}{-2})^n+P_2(\frac{a-\sqrt{a^2+4b}}{-2})^n$ and using the initial condition $t_0=c$ gives $C=P_1+P_2 \Rightarrow P_1=C-P_2$
So $(C-P_2)(\frac{a+\sqrt{a^2+4b}}{-2})^n+P_2(\frac{a-\sqrt{a^2+4b}}{-2})^n$ what next? I tried expanding but that didn't help. I know the answer is something like $cd^n-\frac{b}{a-1}+\frac{bd^n}{a-1}$
| If $a=1$, then $f_n=f_0+nb$, otherwise since $f_n=af_{n-1}+b$ we can subtract $\frac{b}{1-a}$ from both sides to get
$$
f_n-\frac{b}{1-a}=a\left(f_{n-1}-\frac{b}{1-a}\right)
$$
therefore,
$$
\begin{align}
f_n
&=\frac{b}{1-a}+a^n\left(f_0-\frac{b}{1-a}\right)\\
&=a^nf_0+b\frac{1-a^n}{1-a}
\end{align}
$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
Determine coordinates $x_c$ and $y_c$
Answer: $x_c=18.2$ and $y_c=9.5$.
Who can solve with explanation?
| Given the distributive property of the centroid, calling $A_i, i=1,2,3$ the areas of the triangle, rectangle and circle, respectively, and $P_i=(x_i,y_i)$ the respective centroids, we have
\begin{align}
x_c&=\frac{A_1x_1+A_2x_2-A_3x_3}{A_1+A_2-A_3}\\
&=\frac{a^2\cdot\frac{2}{3}a+4a^2\cdot2a-\pi r^2\cdot2a}{a^2+4a^2-\pi r^2}
\end{align}
and similarly for $y_c$
\begin{align}
y_c&=\frac{A_1y_1+A_2y_2-A_3y_3}{A_1+A_2-A_3}\\
&=\frac{a^2\cdot\frac{2}{3}a+4a^2\cdot a-\pi r^2\cdot a}{a^2+4a^2-\pi r^2}
\end{align}
given the centroid of the triangle $(2a/3, 2a/3)$ and the common centroid of the rectangle and of the circle $(2a,a)$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
finding the limit of a series with 3rd root this is not a homework question.
how do i reach the limit of:
i tried taking out $n^3$ and then i got $n-n$ which is $0$, but the true answer is $2/3$.
i can't understand why the answer is $2/3$ and what method to use here.
thank you very much in advance,
yaron.
| Perhaps easier to simplify first. We want to get rid of the cube roots, so remember that $a^3-b^3 = (a-b) \cdot \left(a^2 + ab + b^2\right)$. Note that
$$
\begin{split}
\sqrt[3]{n^3+2n^2+4} - \sqrt[3]{n^3+1}
&= \frac{\left(\sqrt[3]{n^3+2n^2+4} - \sqrt[3]{n^3+1}\right)
\left( {(n^3+2n^2+4)^{2/3} + (n^3+1)^{2/3} + \sqrt[3]{(n^3+2n^2+4)(n^3+1)}} \right)}
{(n^3+2n^2+4)^{2/3} + (n^3+1)^{2/3} + \sqrt[3]{(n^3+2n^2+4)(n^3+1)}} \\
&= \frac{n^3+2n^2+4 - n^3 - 1}
{(n^3+2n^2+4)^{2/3} + (n^3+1)^{2/3} + \sqrt[3]{(n^3+2n^2+4)(n^3+1)}} \\
&= \frac{2n^2+3}
{(n^3+2n^2+4)^{2/3} + (n^3+1)^{2/3} + \sqrt[3]{(n^3+2n^2+4)(n^3+1)}} \\
\end{split}
$$
Divide through by $n^2$ and with $n \to \infty$ this will look like $\frac{2}{1+1+1} = 2/3$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Complex Analysis - Cauchy Principal Value of Improper Integral I am having some difficulty with a P.V. Cauchy Integral problem, and after working on it for hours, I just cannot seem to find what I might be doing wrong.
Here is the problem:
Evaluate the Cauchy principal value of the given improper integral:
$$\int_{-\infty}^{\infty} \frac{1}{(x^2+4)^2} dx$$
My attempt at the problem is below:
Let $$f(z)=\frac{1}{(z^2+4)^2}=\frac{1}{(z+2i)^2(z-2i)^2}$$
Then:
$$\int_{-R}^{R}\frac{dx}{(x^2+4)^2} + \int_{C_R}\frac{dz}{(z^2+4)^2}=2 \pi i \left ( \operatorname*{Res}_{z=2i}\left ( \frac{1}{(z^2+4)^2} \right ) + \operatorname*{Res}_{z=-2i}\left ( \frac{1}{(z^2+4)^2} \right )\right )$$
$$=2 \pi i \left ( \operatorname*{Res}_{z=2i}\left ( \frac{1/(z+2i)^2}{(z-2i)^2} \right ) + \operatorname*{Res}_{z=-2i}\left ( \frac{1/(z-2i)^2}{(z+2i)^2} \right )\right )$$
Now, let $$\frac{p_1(z)}{q_1(z)}=\frac{1/(z+2i)^2}{(z-2i)^2} \text{and let} \frac{p_2(z)}{q_2(z)}=\frac{1/(z-2i)^2}{(z+2i)^2}$$
Then, we see that
$$p_1(2i)=\frac{1}{(4i)^2}\neq 0$$
$$q_1(2i)=(2i-2i)^2=0$$
$$q_1\prime(2i)=2(2i-2i)=0$$
Since $q_1\prime(2i)$ is not nonzero, we know that $\frac{p_1(z)}{q_1(z)}$ is not a simple pole.
So
$$\operatorname*{Res}_{z=2i}\left ( \frac{1/(z+2i)^2}{(z-2i)^2} \right )=\lim_{z\to2i}\left ( \frac{\mathrm{d} }{\mathrm{d} z} \left ( (z-2i)^2*\frac{1/(z+2i)^2}{(z-2i)^2} \right )\right )$$
$$=\lim_{z\to2i}\left ( \frac{\mathrm{d} }{\mathrm{d} z} \left ( \frac{1}{(z+2i)^2} \right )\right )$$
$$=\lim_{z\to 2i}\left ( -\frac{2}{(z+2i)^3} \right )$$
$$=-\frac{2}{(4i)^3}$$
Similarly,
$$\operatorname*{Res}_{z=-2i}\left ( \frac{p_2(z)}{q_2(z)} \right )=\operatorname*{Res}_{z=-2i}\left ( \frac{1/(z-2i)^2}{(z+2i)^2} \right )=\lim_{z\to-2i}\left ( \frac{\mathrm{d} }{\mathrm{d} z} \left ( (z+2i)^2*\frac{1/(z-2i)^2}{(z+2i)^2} \right )\right )$$
$$=\lim_{z\to-2i}\left ( \frac{\mathrm{d} }{\mathrm{d} z} \left ( \frac{1}{(z-2i)^2} \right )\right )$$
$$=\lim_{z\to -2i}\left ( -\frac{2}{(z-2i)^3} \right )$$
$$=-\frac{2}{(-4i)^3}$$
So finally,
$$\int_{-R}^{R}\frac{dx}{(x^2+4)^2} + \int_{C_R}\frac{dz}{(z^2+4)^2}=2 \pi i \left ( -\frac{2}{(4i)^3}-\frac{2}{(-4i)^3} \right )=0$$
So,
$$\int_{-R}^{R}\frac{dx}{(x^2+4)^2} =0- \int_{C_R}\frac{dz}{(z^2+4)^2}$$
$$=-\left ( \int_{C_1}\frac{1/(z+2i)^2}{(z-2i)^2}dz+\int_{C_2}\frac{1/(z-2i)^2}{(z+2i)^2}dz \right )$$
where $C_1$ contains the point $z=2i$ and $C_2$ contains the point $z=-2i$.
$$=-\left (2\pi i\left ( \frac{1}{(2i+2i)^2} \right )+2\pi i\left ( \frac{1}{(-2i-2i)^2} \right ) \right )$$
$$=-\frac{2\pi i}{(4i)^2}-\frac{2\pi i}{(-4i)^2}$$
$$=\frac{\pi i}{4}$$
The final answer should be $\frac{\pi}{16}$, however, which I cannot seem to get.
Any help is greatly appreciated!
| You should get $2\pi i \ Res(f(z)|_{z=2i} = 2\pi i \frac{d}{dz} \frac{1}{(z+2i)^2}|_{z=2i} = 2\pi i \frac{-2}{(4i)^3} = 2\pi i \frac{-2}{-64i} = \frac{\pi}{16}$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Spring Calculation - find mass
A spring with an $-kg$ mass and a damping constant $9$ can be held stretched $2.5 \text{ meters}$ beyond its natural length by a force of $7.5 \text{ Newtons}$. If the spring is stretched $5 \text{ meters}$ beyond its natural length and then released with zero velocity, find the mass that would produce critical damping.
My work:
The restoring force is $-kx$. Then
$$7.5 = -k(2.5) \\
-\frac{7.5}{2.5} = k \\
ma = -\frac{7.5x}{2.5} \\
my’’ + 9y’ + -3y = 0,\quad y(0) = 2.5, y(5) = 0 \\
\frac{-9 \pm \sqrt{81 + 4(m)(3)}}{2m} \\
-\frac{9}{2m} \pm \frac{\sqrt{81+12m}}{2m} \\
y = Ae^{-(9/2)x}\cos\left(\frac{\sqrt{81+12m}}{2m}x\right) + Be^{-(9/2)x}\sin\left(\frac{\sqrt{81+12m}}{2m}x\right) \\
2.5 = A + B\cdot 0 \\
0 = (2.5)e^{-45/2}\cos\left(\sqrt{81+12m}\frac{5}{2m}\right) + Be^{-45/2}\sin\left(\sqrt{81+12m}\frac{5}{2m}\right)$$
Any help would be appreciated
| A short, non insightful answer:
$$ \omega_0^2 = \frac{k}{m} $$
$$\alpha=\frac{c}{2m}$$
If the system is critically damped, $\omega_0^2 = \alpha^2$, so:
$$\frac{3}{m}=\frac{81}{4m^2} $$
$$m=6.75kg$$
If the system is overdamped damped, $\omega_0^2 \lt \alpha^2$
If the system is underdamped damped, $\omega_0^2 \gt \alpha^2$
| {
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"timestamp": "2023-03-29T00:00:00",
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What is $ \lim_{(x,y)\to(2,2)}\frac{x^4-y^4}{x^2 - y^2} $? I have limit:
$$
\lim_{(x,y)\to(2,2)}\frac{x^4-y^4}{x^2 - y^2}
$$
Why is the result $8$ ?
| The limit you are looking up to has a 'hole' in its value as $(x,y)\rightarrow(2,2)$. In order to eradicate that, we have to factor out the hole which is quite easy in this case. It can be done as follows:
$$\lim_{(x,y)\rightarrow(2,2)}\dfrac{x^4-y^4}{x^2-y^2}=\lim_{(x,y)\rightarrow(2,2)}\dfrac{(x^2+y^2)(x^2-y^2)}{x^2-y^2}=\lim_{(x,y)\rightarrow(2,2)}x^2+y^2$$
Now we have successfully removed the 'hole' in the value by getting rid of the $\dfrac00$ format of the expression.
Now, substituting the values of $x$ and $y$ in the expression gives us,
$$\lim_{(x,y)\rightarrow(2,2)}x^2+y^2=2^2+2^2=4+4=8$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove $\sin^2\theta + \cos^2\theta = 1$ How do you prove the following trigonometric identity: $$ \sin^2\theta+\cos^2\theta=1$$
I'm curious to know of the different ways of proving this depending on different characterizations of sine and cosine.
| Consider a right-angled triangle, $\Delta ABC$, where $\angle BAC = \theta$,
By the Pythagorean theorem,
$$ {AC}^2+{BC}^2 = {AB}^2 $$
Dividing by $AB^2$,
$$
\require{cancel}
\begin{align}
&\Rightarrow \frac{AC^2}{AB^2} + \frac{BC^2}{AB^2} = \frac{AB^2}{AB^2}\\
&\Rightarrow \Big(\frac{\text{opposite}}{\text{hypotenuse}}\Big)^2 + \Big(\frac{\text{adjacent}}{\text{hypotenuse}}\Big)^2 = \frac{\cancel{AB^2}}{\cancel{AB^2}} = 1\\
&\Rightarrow \boxed{\sin^2\theta + \cos^2\theta = 1}
\end{align}
$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "37",
"answer_count": 16,
"answer_id": 9
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Infinite Series $\sum\limits_{n=1}^\infty\frac{H_{2n+1}}{n^2}$ How can I prove that
$$\sum_{n=1}^\infty\frac{H_{2n+1}}{n^2}=\frac{11}{4}\zeta(3)+\zeta(2)+4\log(2)-4$$
I think this post can help me, but I'm not sure.
| Using the following nice rule: $$\sum_{n=1}^\infty a_{2n}=\sum_{n=1}^\infty a_{n}\left(\frac{1+(-1)^n}{2}\right)$$
We get
\begin{align}
S&=\sum_{n=1}^\infty\frac{H_{2n+1}}{n^2}=4\sum_{n=1}^\infty\frac{H_{2n+1}}{{(2n)}^2}=4\left(\frac12\sum_{n=1}^\infty\frac{H_{n+1}}{n^2}+\frac12\sum_{n=1}^\infty(-1)^n\frac{H_{n+1}}{n^2}\right)\\
&=2\sum_{n=1}^\infty\frac{H_n}{n^2}+2\sum_{n=1}^\infty(-1)^n\frac{H_n}{n^2}+2\sum_{n=1}^\infty\frac{1}{n^2(n+1)}+2\sum_{n=1}^\infty\frac{(-1)^n}{n^2(n+1)}\\
&=2\left(2\zeta(3)\right)+2\left(-\frac58\zeta(3)\right)+2\left(\zeta(2)-1\right)+2\left(2\ln2-\frac12\zeta(2)-1\right)\\
&=\frac{11}4\zeta(3)+\zeta(2)+4\ln2-4
\end{align}
Note that $\sum_{n=1}^\infty(-1)^n\frac{H_n}{n^2}$ was obtained from using the generating function where we set $x=-1$ :
$$\sum_{n=1}^\infty x^n\frac{H_n}{n^2}=\operatorname{Li}_3(x)-\operatorname{Li}_3(1-x)+\ln(1-x)\operatorname{Li}_2(1-x)+\frac12\ln x\ln^2(1-x)+\zeta(3)$$
| {
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solving an expression based on sin $\theta$ If $\sin^2 \theta = \frac{x^2 + y^2 + 1}{2x}$, then $x$ must be equal to what?
What does the following solution mean?
$0 \le \sin^2 \le 1$
This implies $0 \le \frac{x^2 + y^2 + 1 }{ 2x } \le 1$
This implies $\frac{(x - 1)^2 + y^2 }{2x} \le 0 $
This implies $x = 1$.
Can you please explain me the solution?
| Certainly, for any value of $\theta$, we have $-1 \leq \sin \theta \leq 1$, and therefore, $0 \leq \sin^2 \theta \leq 1$, which bounds the left-hand side of your equation
$$
\sin^2 \theta = \frac{x^2+y^2+1}{2x}.
$$
Since the left-hand side is bounded, so must be the right-hand side, since they are equal. Thus,
$$
0 \leq \frac{x^2+y^2+1}{2x} \leq 1.
$$
Assume $x>0$ and multiply both sides by $2x$, you get
$$
0 \leq x^2+y^2+1\leq 2x
$$
or if you consider the rght-hand side only:
$$
0 = 2x - 2x \geq x^2+y^2+1-2x = \left(x^2-2x+1\right)+y^2 = (x-1)^2+y^2.
$$
Thus, $(x-1)^2+y^2 \leq 0$. But how can a sum of two squares be zero -- both squares are non-negative?!! This only happens if both are zero by themselves. So $y^2 = 0$ (which implies $y=0$) and $(x-1)^2 = 0$ (which implies $x=1$).
Notice $x\neq 0$ otherwise you cannot divide by $2x$ and the problem is invalid. If $x < 0$ on the other hand, the inequality flips:
$$
0 \geq x^2 + y^2 + 1 \geq 2x
$$
but the left-hand side makes no sense, since for any real $x,y$, we must have $x^2+y^2+1 \geq 1 > 0$. So $x$ cannot be negative.
Thus the only acceptable solution is $x=1$ and $y=0$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Applying Fermat's Little Theorem: $6^{1987}$ divided by $37$ Find the remainder when $6^{1987}$ is divided by $37$.
Because 37 is prime we have: $6^{36}$ mod $37 = 1$. I tried to get a nice combination like: $1987 = 36 * 55 + 7$, so we would have $(6^{36})^{55}6^{7}=6^{1987}$.
Then, I've taken mod $37$, which is: $6^{1987}$ mod $37=1^{55}(6^7$ mod $37)$. I need to find $6^7$ mod $37$. What can I do from here?
Of course, any other method (solution) for finding the remainder would be great.
| $$6^2\equiv-1\pmod{37}$$
So $6^7\equiv (-1)^3\cdot 6\equiv 31\pmod{37}$
You could have utilized this from the start, since $6^2\equiv-1\implies 6^4\equiv1$. Thus, you could have written $1987=496\cdot 4+3$ and $$6^{1987}=(6^4)^{496}\cdot6^3\equiv 6^3\equiv 6^26\equiv-6\equiv31\pmod{37}$$
| {
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Prove: Exactly a quarter of 3-part partitions of numbers >2 equal to 0, 2, 10 mod 12 will make a triangle. Consider perimeters $>2$ equal to $0$, $2$, or $10 \mod(12)$. The sequence starts $10, 12, 14, 22, 24, 26, 34, 36, 38, 46, 48, ...$ and we can look at the three part partitions that make triangles.
$\{\{2, 4, 4\}, \{3, 3, 4\}\}$ out of $8$ partitions for $10$.
$\{\{2, 5, 5\}, \{3, 4, 5\}, \{4, 4, 4\}\}$ out of $12$ partitions for $12$.
$\{\{2, 6, 6\}, \{3, 5, 6\}, \{4, 4, 6\}, \{4, 5, 5\}\}$ out of $16$ partitions for $14$.
$\{\{2, 10, 10\}, \{3, 9, 10\}, \{4, 8, 10\}, \{5, 7, 10\}, \{6, 6, 10\}, \{4, 9, 9\}, \{5, 8, 9\}, \{6, 7, 9\}, \{6, 8, 8\}, \{7, 7, 8\}\}$ out of $40$ partitions for $22$.
It turns out that for any number $>2$ equal to $0$, $2$, or $10 \mod 12$, exactly one quarter of the $3$-part partitions can make a triangle. It's easy to verify by computer that this true for $n<1000$. Can someone find a proof?
Alternately, there is some $n$ not equal to $0$, $2$, or $10 \mod 12$ that also has a quarter of it's partitions making a triangle. What's the first exception?
| Let $P_\ell(n)$ denote the number of partitions of $n$ with exactly $\ell$ parts.
The case $n=0\mod 12$:
According to wikipedia, the number of partitions with exactly three parts is equal to
$$P_3(n)=\left[\frac{(n+3)^2}{12}\right]-\left\lfloor \frac n2+1\right\rfloor$$
where $[]$ denotes the nearest integer and $\lfloor\rfloor$ is the lower integer part.
Write $n=12 k$.
Then direct computations give $P_3(12 k)=12 k^2$.
Next notice that any $3$-partition that is not a triangle gives rise to a $2$-partition by summing the two smallest entries. Conversely, any $2$-partition $a+b=n$ with $a\leq b$ gives rise to $\lfloor a/2\rfloor$ $3$-partitions that are not triangles by decomposing $a$ in a $2$-partition. Note that there are precisely $6 k$ such $2$-partitions of $12 k$.
Consequently, the number of $3$-partitions that are not triangles is equal to
$\sum_{i=1}^{6 k} P_2(i)=\sum_{i=1}^{6 k} \lfloor i/2\rfloor=0+1+1+2+2+\dots+(3 k-1)+(3k -1) + 3k=9 k^2 = \frac 34 12 k^2.$
The case $n=-2\mod 12$: the proof works similarly:
$$P_3(12 k-2)=12 k^2-4k,$$
and the number of non-triangles is
$$\sum_{i=1}^{6k-1} \lfloor i/2\rfloor=9 k^2-3k.$$
The case $n=2\mod 12$:
$$P_3(12 k+2)=12 k^2+4k,$$
and the number of non-triangles is
$$\sum_{i=1}^{6k+1} \lfloor i/2\rfloor=9 k^2+3k.$$
| {
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Finding a vector close to vector $\vec{b}$ using $A^T$ and $A\vec{x}=\vec{b}$ I'm having a hard time understanding the rest of the steps after $A^TA\vec{x}=A^T\vec{b}$ to find $\vec{x}$
Problem:
Find the vector in $W= span\ \left(\right.\
\left[ \begin{array}{c} 1 \\ 0 \\ 1 \\ 1 \end{array} \right]$ ,
$\left[ \begin{array}{c} 1 \\ 0 \\ 1 \\ 1 \end{array} \right]$ ,
$\left[ \begin{array}{c} 1 \\ 0 \\ 1 \\ 1 \end{array} \right]\ \left.\right)$
which is closest to the vector $\vec{b}=\left[ \begin{array}{c} 3 \\ 4 \\ 5 \\ 6 \end{array} \right]$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ $
(Note that $\vec{v_1},\vec{v_2},\vec{v_3}$ in $W$ are linearly independent)
Solution:
$$W=Col(A)\ \ \ , \ \ \ A=\left[ \begin{array}{ccc}
\ 1 & 2 & -1 \\
\ 0 & 0 & 0 \\
\ 1 & -1 & 1 \\
\ 1 & 0 & 0 \end{array} \right]$$
$$------------$$
$$A\vec{x}=\vec{b}$$
$$A^TA\vec{x}=A^T\vec{b}$$
$$A^TA=\left[ \begin{array}{ccc}
\ 3 & 1 & 0 \\
\ 1 & 5 & -3 \\
\ 0 & -3 & 2 \\
\end{array} \right]
\ \ ,\ \
A^T\vec{b}=\left[ \begin{array}{c}
\ 14 \\
\ 1 \\
\ 2 \\
\end{array} \right]$$
$$(some\ term?)=\left[ \begin{array}{ccc}
\ 3 & 1 & 0 \\
\ 1 & 5 & -3 \\
\ 0 & -3 & 2 \\
\end{array}\ \ \right|\
\left. \begin{array}{c}
\ 14 \\
\ 1 \\
\ 2 \\
\end{array} \right]
=\left[ \begin{array}{ccc}
\ 1 & 0 & 0 \\
\ 0 & 1 & 0 \\
\ 0 & 0 & 1 \\
\end{array}\ \ \right|\
\left. \begin{array}{c}
\ 6 \\
\ -4 \\
\ -5 \\
\end{array} \right]\ \ ,\ \
\vec{x}=\left[ \begin{array}{c}
\ 6 \\
\ -4 \\
\ -5 \\
\end{array} \right]$$
$$------------$$
$$\vec{v}_{answer}=W\cdot \vec{x} =
\left[ \begin{array}{c}
\ 3 \\
\ 0 \\
\ 5 \\
\ 6 \end{array} \right]$$
My Confusion:
*
*Is there a name for this theorem or property : $A^TA\vec{x}=A^T\vec{b}$
*What is the process of simplification to take "$A^TA\vec{x}=A^T\vec{b}$" and get $\vec{x}$ by itself? The solution I'm given doesn't explain this well. ($\ \vec{x}=\{A^TA\ |\ A^T\vec{b}\}$ What?!)
*How does the $\vec{v}_{answer}=\{3,0,5,6\}$ relate to the vectors $\vec{v_1},\vec{v_2},\vec{v_3},$ and $A\vec{x}=\vec{b}$ ? For instance is $\vec{v}_{answer}$ interchangeable with $\vec{v_1},\vec{v_2},\vec{v_3},$ or is $\vec{v}_{answer}$ only used to find $\vec{b}$, ect.?
| First bullet point: This is called the normal equations.
| {
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$\cos^2\frac{1}{2}(\alpha-\beta)=\frac{3}{4}$ if........... Help please: If $\sin\alpha+\sin\beta= \sqrt{3} (\cos\beta-\cos\alpha)$ then show that $\cos^2\frac{1}{2}(\alpha-\beta)=\frac{3}{4}$ please tell me how can I approach
| Use the following identities:
$$\sin\alpha+\sin\beta=2\sin\left(\frac{\alpha+\beta}2\right)\cos\left(\frac{\alpha-\beta}2\right)\\
\cos\beta-\cos\alpha=-2\sin\left(\frac{\beta+\alpha}2\right)\sin\left(\frac{\beta-\alpha}2\right)$$
So the first condition is equivalent to
$$\cos\left(\frac{\alpha-\beta}2\right)=-\sqrt{3}\sin\left(\frac{\beta-\alpha}2\right)$$
I.e
$$\tan\left(\frac{\alpha-\beta}2\right)=\frac{1}{\sqrt{3}}$$
Then use that $1+\tan^2\theta=\frac1{\cos^2\theta}$
| {
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$0Let $a_n$ be a sequence of positive real numbers such that
$$a_n<\frac{a_{n-1}+a_{n-2}}{2}$$
Show that $a_n$ converges.
| It is clear that $(a_n)_{n\in\mathbb{N}}$ is bounded between $\ \max \{a_1,a_2\} $ and $0.$ Therefore by the Bolzano-Weierstrass theorem, $\ (a_n)_{n\in\mathbb{N}}\ $ has at least one limit (accumulation) point in the interval$\ [\ 0,\ \max \{a_1,a_2\}\ ].$
Now suppose, by way of contradiction, that $\ (a_n)\ $ has at least two limit points, $\ l_1\ $ and $\ l_2,\ $ with $\ l_1<l_2.\ $
Lemma: If $\ a_{n+1} > a_n,\ $ then $\ a_k < a_n + \frac{3}{4}\left( a_{n+1} -a_n\right)\quad \forall\ k\geq n+2.$
Next, since $\ l_1\ $ and $\ l_2\ $ are limit points, $\ \exists\ N_1\ $ such that, $\ a_{N_1} < \frac{l_1+l_2}{2},\ $ and $\ a_{N_1 + 1} > \frac{l_1+l_2}{2}.\ $ The Lemma implies then that:
$$\ a_k < a_{N_1} + \frac{3}{4}\left( a_{N_1+1} -a_{N_1}\right) \leq \frac{l_1+l_2}{2} + \frac{3}{4}\left( a_{N_1+1} - \frac{l_1+l_2}{2}\right)$$
$$\implies a_k - \frac{l_1+l_2}{2} < \frac{3}{4}\left( a_{N_1+1} - \frac{l_1+l_2}{2}\right) \quad \forall\ k\geq N_1+2.\quad (1) $$
Continuing, $\ \exists\ N_2>N_1+2\ $ such that, $\ a_{N_2} < \frac{l_1+l_2}{2},\ $ and $\ a_{N_2 + 1} > \frac{l_1+l_2}{2}.\ $ Since $\ N_2 > N_1 + 2,\ $ all $\ k\geq N_2\ $ satisfies $\ (1)\ $ also. Thus,
$$ a_k - \frac{l_1+l_2}{2} < \frac{3}{4}\left( a_{N_2+1} - \frac{l_1+l_2}{2}\right) < \frac{3}{4}\left( \frac{3}{4} \left( a_{N_1+1} - \frac{l_1+l_2}{2}\right) \right) = \left( \frac{3}{4} \right)^2 \left( a_{N_1+1} - \frac{l_1+l_2}{2}\right) $$
$$ \quad \forall\ k\geq N_2+2.\quad (2) $$
Continuing in this way, we can find and increasing sequence of integers $\ (n_j)_{j\in\mathbb{N}}\ $ such that for each $\ j,\ $ we have:
$$ a_k \leq \frac{l_1+l_2}{2} + \left(\frac{3}{4}\right)^j \left( a_{N_1 + 1} - \frac{l_1+l_2}{2} \right)\quad \forall\ k\geq N_j + 2. $$
But then, since $\ \left(\frac{3}{4}\right)^j \to 0,\ $ as $\ j\to\infty,\ $ it follows that $\ \displaystyle\limsup_k\ a_k\ \leq \frac{l_1+l_2}{2} $ contradicting the fact that $\ l_2\ $ is a limit point of $\ (a_n)_n.$
$$$$
I have left the proof of the lemma, and a couple of other details as an exercise to the reader, but these things should be relatively straightforward.
| {
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Doubts in Trigonometrical Inequalities I'm now studying Trigonometrical Inequalities, and I've just got struck when I have modified arguments to my trigonometrical functions, for example:
$\sqrt{2} - 2\sin\left(x - \dfrac{\pi}{3} \right) < 0$ when $-\pi < x < \pi$
With some work I've got: $\sin\left(x - \dfrac{\pi}{3} \right) > \dfrac{\sqrt{2}}{2}$
To find bounds: $\sin(x) = \dfrac{\sqrt{2}}{2},\ x = \dfrac{\pi}{4},\ \dfrac{3\pi}{4}$
Resolving to $x + \dfrac{\pi}{3} \implies \dfrac{\pi}{4} < x < \dfrac{3\pi}{4}$
But wolfram gives a way different result, where's my mistake ?
| First of all, you said that $\sin x = \frac{\sqrt{2}}{2}$ when $x = \frac{\pi}{4}, \frac{3\pi}{4}$ then concluded that $\sin x > \frac{\sqrt{2}}{2}$ when $\frac{\pi}{4} < x < \frac{3\pi}{4}$. While this is true, you should give some explanation here as it could be the case that $\sin x < \frac{\sqrt{2}}{2}$ for $\frac{\pi}{4} < x < \frac{3\pi}{4}$.
As $\sin x > \frac{\sqrt{2}}{2}$ for $\frac{\pi}{4} < x < \frac{3\pi}{4}$, $\sin(x - \frac{\pi}{3}) > \frac{\sqrt{2}}{2}$ for $\frac{\pi}{4} < x - \frac{\pi}{3} < \frac{3\pi}{4}$. By adding $\frac{\pi}{3}$ to each term in the inequality, we have $\frac{7\pi}{12} < x < \frac{13\pi}{12}$.
So, for every $k \in \mathbb{Z}$, we have $\sqrt{2} - 2\sin(x-\frac{\pi}{3}) > 0$ for $\frac{7\pi}{12}+2k\pi < x < \frac{13\pi}{12}+2k\pi$.
For $k = -1$ we have $-\frac{17\pi}{12} < x < -\frac{11\pi}{12}$, and for $k = 0$ we have $\frac{7\pi}{12} < x < \frac{13\pi}{12}$. As we are looking for $x$ which satisfy $-\pi < x < \pi$, these are the only $x$ we need to consider (for any other $k$, the corresponding inequalities do not allow for $x$ which also satisfy $-\pi < x < \pi$).
If $x$ satisfies $-\frac{17\pi}{12} < x < -\frac{11\pi}{12}$ and $-\pi < x < \pi$, then $-\pi < x < -\frac{11\pi}{12}$.
If $x$ satisfies $\frac{7\pi}{12} < x < \frac{13\pi}{12}$ and $-\pi < x < \pi$, then $\frac{7\pi}{12} < x < \pi$.
Therefore, for $-\pi < x < \pi$, $\sqrt{2} -2\sin(x-\frac{\pi}{3}) > 0$ for $-\pi < x < -\frac{11\pi}{12}$ and $\frac{7\pi}{12} < x < \pi$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Help finding the $\lim\limits_{x \to \infty} \frac{\sqrt[3]{x} - \sqrt[5]{x}}{\sqrt[3]{x} + \sqrt[5]{x}}$ I need help finding the $$\lim_{x \to \infty} \frac{\sqrt[3]{x} - \sqrt[5]{x}}{\sqrt[3]{x} + \sqrt[5]{x}}$$
I did the following:
$$\begin{align*}
\lim_{x \to \infty} \frac{\sqrt[3]{x} - \sqrt[5]{x}}{\sqrt[3]{x} + \sqrt[5]{x}}
=& \lim_{x \to \infty} \frac{(\sqrt[3]{x} - \sqrt[5]{x})(\sqrt[3]{x} + \sqrt[5]{x})}{(\sqrt[3]{x} + \sqrt[5]{x})(\sqrt[3]{x} + \sqrt[5]{x})}\\
\\
=& \lim_{x \to \infty} \frac{(\sqrt[3]{x})^2 - (\sqrt[5]{x})^2}{(\sqrt[3]{x})^2+2\sqrt[3]{x}\sqrt[5]{x}+(\sqrt[3]{x})^2}\\
\\
=& \lim_{x \to \infty} \frac{x^{2/3}-x^{2/5}}{x^{2/3}+2x^{1/15}+x^{2/5}}\\
\\
=& \lim_{x \to \infty} \frac{x^{4/15}}{2x^{17/15}}
\end{align*}$$
Somehow I get stuck. I am sure I did something wrong somewhere.. Can someone please help me out?
| Your major mistake : $x^a + x^b \ne x^{a+b}$. This is very wrong and a common mistake.
Did you know that $\sqrt[n]{x} = x^{1/n}$? Therefore the intial expression can be written as : $$\frac{x^{1/3} - x^{1/5}}{x^{1/3} + x^{1/5}} = \frac{x^{1/3}(1 - x^{1/5-1/3})}{x^{1/3}(1+x^{1/5-1/3})} = \frac{1-x^{-2/15}}{1+x^{-2/15}}$$
What do you know about $x^{-a}$ ($a>0$) as $x\longrightarrow \infty$?
| {
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Probability Of Rolling A Strictly Increasing Sequence On A Six-Sided Die
By rolling a six-sided die 6 times, a strictly increasing sequence of
numbers was obtained, what is the probability of such an event?
I have no ideas on how to attack this. It says, an increasing sequence for the $6$ times were obtained. So this means, that on the first roll, you get $1$, and on second you get $2$, on the third you get $3$, and so on. The total sample space is given by $6^6$ and the probability of obtaining the set $\{1,2,3,4,5,6\}$ is $\frac{1}{6^6}$? Please help.
| For each roll of the die, there are $6$ possible outcomes — ${1,2,3,4,5,6}$. You roll $6$ times, so there are $6$ possible outcomes for the first, for each of these there are $6$ possible outcomes for the second roll, and so on. This means there are $6 \cdot 6 \cdot 6 \cdot 6 \cdot 6 \cdot 6 = 6^6$ sequences possible. Only one of these is the sequence you want, so your chance of getting that sequence is $\frac{1}{6^6}$.
Alternatively, you can see it this way. There is a $\frac{1}{6}$ chance of getting a $1$ on your first roll, a $\frac{1}{6}$ chance of getting a $2$ on your second roll, and so on. This means the total probability of getting this particular sequence, is $\frac{1}{6} \cdot \frac{1}{6} \cdot \frac{1}{6} \cdot \frac{1}{6} \cdot \frac{1}{6} \cdot \frac{1}{6} = \frac{1}{6^6}$.
| {
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Prove that at least two of these inequalities are true: $|a-b|\le2$, $|b-c|\le2$, $|c-a|\le2$. It's given that: $$\begin{cases}a,b,c>0\\a+b+c\le4\\ab+bc+ac\ge4\end{cases}$$
Prove without using calculus that it's true that at least two of these are correct inequalities:$$\begin{cases}|a-b|\le2\\|b-c|\le2\\|c-a|\le2\end{cases}$$
If you think about it, we may as well prove that at least two of these are correct:$$\begin{cases}(a-b)^2\le4\\(b-c)^2\le4\\(c-a)^2\le4\end{cases}$$
We could square one of the inequalities (since both sides are positive):$$(a+b+c)^2\le16\Rightarrow a^2+b^2+c^2+2(ab+bc+ac)\le16$$
$$a^2+b^2+c^2+8\le a^2+b^2+c^2+2(ab+bc+ac)\le16\Rightarrow a^2+b^2+c^2\le 8$$
So now we know that: $$\begin{cases}a,b,c>0\\a+b+c\le4\\a^2+b^2+c^2\le8\\ab+bc+ac\ge4\end{cases}$$
Does anyone see how to solve this? Thanks.
| It is easy!
first, we know that
$a^2+b^2+c^2<=8$, therefore
$2a^2+2b^2+2c^2<=16$ ............. ine1
and we know $ab+ba+ca>=4$,so
$-2ab-2bc-2ca<=-8$ ......... ine2
so ine1+ine2, we get
$(a-b)^2+(b-c)^2+(a-c)^2<=8$
so ,it is easy to explain your inequalities:
$\begin{cases}(a-b)^2\le4\\(b-c)^2\le4\\(c-a)^2\le4\end{cases}$
at least two of there are right.
| {
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Find all solutions of ${\frac {1} {x} } + {\frac {1} {y} } +{\frac {1} {z}}=1$, where $x$, $y$ and $z$ are positive integers Find all solutions of ${\frac {1} {x} } + {\frac {1} {y} } +{\frac {1} {z}}=1$ , where $x,y,z$ are positive integers.
Found ten solutions $(x,y,z)$ as ${(3,3,3),(2,4,4),(4,2,4),(4,4,2),(2,3,6),(2,6,3),(3,6,2),(3,2,6),(6,2,3),(6,3,2)}$. Are these the only 10 solutions?
First, none of $x$, $y$ or $z$ can be $1$ ($x$, $y$ and $z$ are positive integers)
If I let $x=2$, then finding all solutions to $1/y+1/z = 1/2$ leads to $(4,4), (3,6)$ and $(6,3)$ which gives me $(x,y,z)$ as $(2,4,4), (2,3,6), (2,6,3)$ but this also means $(4,4,2), (4,2,4), (3,2,6), (3,6,2), (6,2,3), (6,3,2)$ are all valid triples for this equation.
If I let $x=3$, the only different values of $y$ and $z$ are $(3,3)$
How do I prove these are the only ten solutions? (without using any programming)
Known result: If we denote $d(n^2)$ as the number of divisors of $n^2$, then the number of solutions of ${\frac {1} {x} }+{\frac {1} {y} } = {\frac {1} {n} }$ = $d(n^2)$ (For positive $x$, $y$)
For ${\frac {1} {x} } + {\frac {1} {y} } +{\frac {1} {z}}=1$
$z = \frac{xy}{y(x-1)-x}$ where $xy \neq 0$
What happens after that?
Question is: how do we sho there are the only ten solutions? I'm not asking for a solution.
Assuming $x \le y \le z$
$1 \le y \le \frac{xy}{y(x-1)-x}$
$\Longrightarrow 1 \le x \le y \le \frac{2x}{x-1} $
Got the answer. I'll probably call @mathlove's answer. (Any additional answers I'll view later)
Liked @user44197 answer.
| They are the only possible solution. Proof is as follows:
Suppose that $d = gcd(x,y)$ and $x=d ~ x0$, $y=d ~y_0$ where $x_0$ and $y_0$ are co-prime.
Substituting in the original equation we get
$$ 1/x + 1/y + 1/z=1 \Rightarrow -d\,x_0\,y_0\,z+y_0\,z+x_0\,z+d\,x_0\,y_0 =0$$
Solving for $d$:
$$d=\frac{\left( y_0+x_0\right) \,z}{x_0\,y_0\,\left( z-1\right) }$$
Since $x_0$ and $y_0$ are co-prime, for $d$ to be an integer, $x_0\,y_0$ should divide $z$.
Hence we require
$$ z= k ~x_0 ~y_0$$
Substituting in the equation for $d$ and solving for $k$:
$$ k=\frac{d}{d\,x_0\,y_0-y_0-x_0}$$
This shows that $d$ is a multiple of $k$. Let
$$ d= \mu k$$. Then
$$k=\frac{k\,\mu}{k\,\mu\,x_0\,y_0-y_0-x_0}$$
Solving for $k$:
$$k=\frac{1}{x_0\,y_0}+\frac{1}{\mu\,y_0}+\frac{1}{\mu\,x_0}$$
This implies that $1 \le x_0 \le 3$, $1 \le y_0 \le 3$, $1 \le \mu\le 3$
It is possible to eliminate some of the 27 possible values since $k$ has to be an integers this will result in 12 possible values for $(x_0,y_0,\mu)$ and two of the solutions are repetitions giving the 10 solutions mentioned in the problem.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/616639",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 5,
"answer_id": 0
} |
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